Precalculus, 2nd Edition

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second edition

John w. Coburn St. Louis Community College at Florissant Valley

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PRECALCULUS, SECOND EDITION Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright © 2010 by The McGraw-Hill Companies, Inc. All rights reserved. Previous edition © 2007. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 DOW/DOW 0 9 ISBN 978–0–07–351942–5 MHID 0–07–351942–1 ISBN 978–0–07–336086–7 (Annotated Instructor’s Edition) MHID 0–07–336086–4 Editorial Director: Stewart K. Mattson Sponsoring Editor: Dawn R. Bercier Senior Developmental Editor: Michelle L. Flomenhoft Developmental Editor: Katie White Marketing Manager: John Osgood Senior Project Manager: Vicki Krug Senior Production Supervisor: Sherry L. Kane Senior Media Project Manager: Sandra M. Schnee

Designer: Laurie B. Janssen Cover Designer: Christopher Reese (USE) Cover Image: © Georgette Douwma/Gettyimages Senior Photo Research Coordinator: John C. Leland Supplement Producer: Mary Jane Lampe Compositor: Aptara®, Inc. Typeface: 10.5/12 Times Roman Printer: R. R. Donnelley Willard, OH

Chapter 1 Opener: © Royalty-Free/CORBIS; pg. 12: NASA/RF; pg. 30: PhotoLinK/Getty Images/RF; pg. 68 top: © Brand X Pictures/ PunchStock/RF; pg. 68 bottom: Photodisc Collection/Getty Images/RF. Chapter 2 Opener: © Royalty-Free/CORBIS; pg. 139: Siede Preis/Getty Images/RF; pg. 140: The McGraw-Hill Companies, Inc./Ken Cavanagh Photographer; pg. 155: Steve Cole/Getty Images/RF; pg. 172: Alan and Sandy Carey/Getty Images/RF; pg. 183: Courtesy John Coburn; pg. 201 top: Patrick Clark/Getty Images/RF; pg. 201 bottom: © Digital Vision/PunchStock/RF. Chapter 3 Opener: © 1997 IMS Communications Ltd./Capstone Design. All Rights Reserved/ RF; pg. 238: © Adalberto Rios/Sexto Sol/Getty Images/RF; pg. 240: © Royalty-Free/CORBIS; pg. 255: © Royalty-Free/CORBIS; pg. 288: © Royalty-Free/CORBIS; pg. 314: © Royalty-Free/CORBIS; pg. 320: © Royalty-Free/CORBIS. Chapter 4 Opener: © Andrew Ward/Life File/Getty Images/RF; pg. 364 left: © Geostock/Getty Images/RF: pg. 364 right: © Lawrence M. Sawyer/Getty Images/RF; pg. 373: Photography by G.K. Gilbert, courtesy U.S. Geological Survery; pg. 374: © Lars Niki/RF; pg. 378: © Medioimages/Superstock/RF; pg. 395: StockTrek/Getty Images/RF; pg. 415: Courtesy Simon Thomas. Chapter 5 Opener: Digital Vision/RF; pg. 434: © Jules Frazier/ Getty Images/RF; pg. 438: © Karl Weatherly/Getty Images/RF; pg. 468: © Royalty-Free/CORBIS; pg. 510: © Royalty-Free/CORBIS; pg. 527: Royalty-Free/CORBIS. Chapter 6 Opener: © Digital Vision/Getty Images/RF; pg. 617 © John Wong/Getty Images/RF. Chapter 7 Opener: © Royalty-Free/CORBIS/RF Chapter 8 Opener: © Royalty-Free/CORBIS; pg. 730: © The McGraw-Hill Companies, Inc./Jill Braaten, photographer; pg. 731: © Royalty-Free/CORBIS; pg: 742: © Creatas/PunchStock/RF; pg. 751: Royalty-Free/CORBIS. Chapter 9 Opener: © Mark Downey/Getty Images/RF; pg. 859: © Brand X Pictures/PunchStock/RF; pg. 860: © Digital Vision/Getty Images/ RF; pg. 864: © H. Wiesenhofer/PhotoLink/Getty Images/RF; pg. 873 top: © Jim Wehtje/Getty Images/RF; pg. 873 middle: © Creatas/ PunchStock/RF; pg. 873 bottom: © Edmond Van Hoorick/Getty Images/RF; pg. 882: © The McGraw-Hill Companies, Inc./Jill Braaten, photographer/RF; pg. 922: © PhotoLink/Getty Images/RF. Chapter 10 Opener: © Doug Menuez/Getty Images/RF; pg. 955: RoyaltyFree/CORBIS; pg. 964: © Andersen Ross/Getty Images/RF. Chapter 11 Opener: © Royalty-Free/CORBIS. Appendices Pg. A-12: © Photodisc/Getty Images/RF; pg. A-50: © Glen Allison/Getty Images/RF. Library of Congress Cataloging-in-Publication Data Coburn, John W. Precalculus / John W. Coburn. —2nd ed. p. cm. Includes index. ISBN 978–0–07–351942–5—ISBN 0–07–351942–1 (hard copy : alk. paper) QA331.3.C63 2010 510--dc22 2008050984

www.mhhe.com

1. Functions.

2. Trigonometry.

I. Title.

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Brief Contents Preface vi Index of Applications

1 CHAPTER 2 CHAPTER 3 CHAPTER 4 CHAPTER 5 CHAPTER 6 CHAPTER 7 CHAPTER 8 CHAPTER 9 C H A P T E R 10 C H A P T E R 11 CHAPTER

xxxv

Equations and Inequalities

1

Relations, Functions, and Graphs 83 Polynomial and Rational Functions 219 Exponential and Logarithmic Functions 341 An Introduction to Trigonometric Functions 425 Trigonometric Identities, Inverses, and Equations 543 Applications of Trigonometry

633

Systems of Equations and Inequalities

719

Analytical Geometry and the Conic Sections 831 Additional Topics in Algebra 939 Bridges to Calculus: An Introduction to Limits 1023

Appendix I

A Review of Basic Concepts and Skills

Appendix II

More on Synthetic Division

Appendix III

More on Matrices

Appendix IV

Deriving the Equation of a Conic

Appendix V

Selected Proofs A-60

Appendix VI

Families of Polar Curves A-63

A-54

A-56 A-58

Student Answer Appendix (SE only)

SA-1

Instructor Answer Appendix (AIE only) Index

A-1

IA-1

I-1

Additional Topics Online (Visit www.mhhe.com/coburn) R.7 Geometry Review with Unit Conversions R.8 Expressions, Tables and Graphing Calculators 5.0 An Introduction to Cycles and Periodic Functions 7.7 Complex Numbers in Exponential Form 7.8 Trigonometry, Complex Numbers and Cubic Equations 10.8 Conditional Probability and Expected Value 10.9 Probability and the Normal Curve with Applications 11.2B Properties of Limits with an Introduction to the Precise Definition iii

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About the Author Background

John Coburn grew up in the Hawaiian Islands, the seventh of sixteen children. John’s mother and father were both teachers. John’s mother taught English and his father, as fate would have it, held advanced degrees in physics, chemistry, and mathematics. Whereas John’s father was well known, well respected, and a talented mathematician, John had to work very hard to see the connections so necessary for success in mathematics. In many ways, his writing is born of this experience.

Education

In 1979 John received a bachelor’s degree in education from the University of Hawaii. After working in the business world for a number of years, John returned to his first love by accepting a teaching position in high school mathematics and in 1987 was recognized as Teacher of the Year. Soon afterward John decided to seek a master’s degree, which he received two years later from the University of Oklahoma.

Teaching Experience

John is now a full professor at the Florissant Valley campus of St. Louis Community College where he has taught mathematics for the last eighteen years. During h time there he has received numerous nominations as an his o outstanding teacher by the local chapter of Phi Theta Kappa, a was recognized as Post-Secondary Teacher of the Year and i 2004 by Mathematics Educators of Greater St. Louis in ( (MEGSL). John is a member of the following organizations: N National Council of Teachers of Mathematics (NCTM), M Missouri Council of Teachers of Mathematics (MCTM), M Mathematics Educators of Greater Saint Louis (MEGSL), A American Mathematical Association of two Year Colleges ( (AMATYC), Missouri Mathematical Association of two Y Colleges (MoMATYC), Missouri Community College Year A Association (MCCA), and Mathematics Association of A America (MAA).

Personal Interests

Some of John’s other interests include body surfing, snorkeling, and beach combing whenever he gets the chance. In addition, John’s loves include his family, music, athletics, games, and all things beautiful. John hopes that this love of life comes through in the writing, and serves to make the learning experience an interesting and engaging one for all students.

Dedication To my wife and best friend Helen, whose love, support, and willingness to sacrifice never faltered.

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About the Cover Coral reefs support an extraordinary biodiversity as they C aare home to over 4000 species of tropical or reef fish. In addition, coral reefs are iimmensely beneficial to humans; buffering coastal regions from strong waves and sstorms, providing millions of people with food and jobs, and prompting advances iin modern medicine. Similar to a reef, a precalculus course is unique because of its diverse population oof students. Nearly every major is represented in this course, featuring students w with a wide range of backgrounds and skill sets. Just like the variety of the fish in tthe sea rely on the coral reefs to survive, the assortment of students in precalculus rrely on succeeding in this course in order to further pursue their degree, as well as ttheir career goals.

From the Author

directio ns. This is flue nce of needs, idea s, desires, and con hty mig a of ult res the is t ion. This tex the most dive rse in all of edu cat of one is e ienc aud d nde inte the easily und ers tan dable, as pre par atio n, bac kgr oun ds, var ying deg ree s of of ge ran e wid a h wit us to e Ou r stu den ts com ses incl ude those to exc item ent. In add itio n, our clas thy apa m fro y var t tha ls leve t and inte res fut ure eng inee rs and uiremen t, as wel l as our cou ntr y’s req ion cat edu l era gen a y onl wou ld nee ding nee ds of so dive rse a pop ulat ion the ting mee is ge llen cha st ate scie ntis ts. To say our gre cam e to min d, rsit y, the ima ge of a cor al ree f dive this on ing lect ref In t. men be an und ers tate pop ulat ion, wit h ana logy. We hav e a hug ely dive rse the of th eng str the by uck str and I was on the ree f for the ir wit h all the inh abitant s dep end ing ce, pla ting mee n mo com a as f the ree n. experiences pur pose, nou rishmen t, and directio of the most daunting and cha llenging one n bee has rse cou this for Wr iting a text that most text s on g exp erience left a nagging sense chin tea my an, beg I ore bef g Lon in my life. addition, they app ear ed t wit h so diverse an audience. In nec con to ity abil the ed lack t rke nections, the ma terse a dev elopmen t to ma ke con too ts, cep con d buil to ork mew to off er too sca nt a fra foster a love of to dev elop long-ter m retention or s set e rcis exe ir the in t por sup and insu fficient , cur ious interest, s seemed to lack a sense of rea lism tion lica app the lar, ticu par In . mathematics everyday exp erience. task of and/or connections to a studen t’s te a bet ter text, I set about the wri to ire des ng stro a and d min Wit h all of this in re sup por tive aging tool for studen ts, and a mo eng re mo a e om bec ld wou ed hop creating what I erie nce, and an ear ly rsit y of my own edu cat iona l exp dive the on g win Dra s. tor ruc too l for inst con trib ute d to the tex t’s s, and per spectiv es, I beli eve has view s, ure cult nt ere diff to re diverse exp osu re and bet ter connections wit h our mo to , end the in e hop I and le, unique and engaging sty ers, foc us peo ple, incl uding ma nuscrip t rev iew 400 n tha re mo m fro ck dba fee aud ienc e. Hav ing con nec tion s in the inva luable to help ing me hon e the was rs, uto trib con and ts, pan tici res t gro up par it the re was also a desire to inte adm I , nce erie exp this of wth love wit h boo k. As a coll ate ral outgro aga in and aga in, why we fell in us ind rem to s— tor ruc inst the and eng age our selv es, —John Cob urn ma thema tics in the firs t pla ce. v

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Making Connections . . . Precalculus tends to be a challenging course for many students. They don’t see the connections that precalculus has to their life or why it is so critical that they take and pass this course for both technical and nontechnical careers alike. Others may enter into this course underprepared or improperly placed and with very little motivation. Instructors are faced with several challenges as well. They are given the task of improving pass rates and student retention while energizing a classroom full of students comprised of nearly every major. Furthermore, it can be difficult to distinguish between students who are likely to succeed and students who may struggle until after the first test is given. The goal of the Coburn series is to provide both students and instructors with tools to address these challenges, as well as the diversity of the students taking this course, so that you can experience greater success in precalculus. For instance, the comprehensive exercise sets have a range of difficulty that provides very strong support for weaker students, while advanced students are challenged to reach even further. The rest of this preface further explains the tools that John Coburn and McGraw-Hill have developed and how they can be used to connect students to precalculus and connect instructors to their students.

The Coburn Precalculus Series provides you with strong tools to achieve better outcomes in your Precalculus course as follows:

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▶

Better Student Preparedness

▶

Increased Student Engagement

▶

Solid Skill Development

▶

Strong Connections

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▶

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Better Student Preparedness

No two students have the same strengths and weaknesses in mathematics. Typically students will enter any math course with different preparedness levels. For most students who have trouble retaining or recalling concepts learned in past courses, basic review is simply not enough to sustain them successfully throughout the course. Moreover, instructors whose main focus is to prepare students for the next course do not have adequate time in or out of class to individually help each student with review material. ALEKS Prep uniquely assesses each student to determine their individual strengths and weaknesses and informs the student of their capabilities using a personalized pie chart. From there, students begin learning through ALEKS via a personalized learning path uniquely designed for each student. ALEKS Prep interacts with students like a private tutor and provides a safe learning environment to remediate their individual knowledge gaps of the course pre-requisite material outside of class. ALEKS Prep is the only learning tool that empowers students by giving them an opportunity to remediate individual knowledge gaps and improve their chances for success. ALEKS Prep is especially effective when used in conjunction with ALEKS Placement and ALEKS 3.0 course-based software. ▶

Increased Student Engagement

What makes John Coburn’s applications unique is that he is constantly thinking mathematically. John’s applications are spawned during a trip to Chicago, a phone call with his brother or sister, or even while watching the evening news for the latest headlines. John literally takes notes on things that he sees in everyday life and connects these situations to math. This truly makes for relevant applications that are born from real-life experiences as opposed to applications that can seem fictitious or contrived. ▶

Solid Skill Development

The Coburn series intentionally relates the examples to the exercise sets so there is a strong connection between what students are learning while working through the examples in each section and the homework exercises that they complete. In turn, students who attempt to work the exercises first can surely rely on the examples to offer support as needed. Because of how well the examples and exercises are connected, key concepts are easily understood and students have plenty of help when using the book outside of class. There are also an abundance of exercise types to choose from to ensure that homework challenges a wide variety of skills. Furthermore, John reconnects students to earlier chapter material with Mid-Chapter Checks; students have praised these exercises for helping them understand what key concepts require additional practice. ▶

Strong Connections

John Coburn’s experience in the classroom and his strong connections to how students comprehend the material are evident in his writing style. This is demonstrated by the way he provides a tight weave from topic to topic and fosters an environment that doesn’t just focus on procedures but illustrates the big picture, which is something that so often is sacrificed in this course. Moreover, he deploys a clear and supportive writing style, providing the students with a tool they can depend on when the teacher is not available, when they miss a day of class, or simply when working on their own.

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Better Student Preparedness . . . Experience Student Success! ALEKS ALEKS ALEK S is a unique uni niqu q e online math tool that uses adaptive questioning and artificial intelligence to correctly place, prepare, and remediate students . . . all in one product! Institutional case studies have shown that ALEKS has improved pass rates by over 20% versus traditional online homework and by over 30% compared to using a text alone. By offering each student an individualized learning path, ALEKS directs students to work on the math topics that they are ready to learn. Also, to help students keep pace in their course, instructors can correlate ALEKS to their textbook or syllabus in seconds. To learn more about how ALEKS can be used to boost student performance, please visit www.aleks.com/highered/math or contact your McGraw-Hill representative.

Easy Graphing Utility! ALEKS Pie

Students can answer graphing problems with ease!

Each student is given their own individualized learning path.

Course Calendar Instructors can schedule assignments and reminders for students.

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. . . Through New ALEKS Instructor Module Enhanced Functionality and Streamlined Interface Help to Save Instructor Time The new ALEKS Instructor Module features enhanced functionality and streamlined interface based on research with ALEKS instructors and homework management instructors. Paired with powerful assignment driven features, textbook integration, and extensive content flexibility, the new ALEKS Instructor Module simplifies administrative tasks and makes ALEKS more powerful than ever.

New Gradebook! Instructors can seamlessly track student scores on automatically graded assignments. They can also easily adjust the weighting and grading scale of each assignment.

Gradebook view for all students Gradebook view for an individual student

Track Student Progress Through Detailed Reporting Instructors can track student progress through automated reports and robust reporting features.

Automatically Graded Assignments Instructors can easily assign homework, quizzes, tests, and assessments to all or select students. Deadline extensions can also be created for select students.

Learn more about ALEKS by visiting

www.aleks.com/highered/math / / or contact your McGraw-Hill representative. Select topics for each assignment

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Increased Student Engagement . . . Through Meaningful Applications enn mat hematics een wee bettwe on betw tion ecti the con nec uires that student s exp erie nce th req ful ning mea s atic hem mat mit men t to Ma king also the result of a pow erf ul com is text This in. live y the ld wor and its impact on the and with car efully monitor ed hav ing close ties to the exa mples, lity, qua est high the of s tion lica provide app levels of diff iculty. ers came from a cur ious, my own diverse life experiences, oth of n bor e wer es mpl exa e thes of , and see the Ma ny e upon the every day events of life seiz to one ws allo t tha y foll ry lucid, and even visiona not ebo ok was use d a the bac kgr oun d. My eve r-p res ent in s atic hem mat ful ning mea or t is the genesis sign ific ant that sudden bur st of inspiration tha or n, atio erv obs ual cas t tha ture thousa nd times to cap libr ary of ref erence and sup por ted at home by a substan tial e wer se The s. tion lica app ding mod ern ma rve l of for out stan cur ren t eve nts, and of cou rse our and ory hist h bot ard tow eye res ear ch boo ks, an t, ref lect ion, and resear ch, (som etim es long) per iod of tho ugh a er Aft et. ern Int he l—t too ch student s while a resear e so that it wou ld resona te wit h rcis exe the of ing ord rew a and followed by a wor ding —JC mea ning ful app lication was bor n. filling the need, a sign ificant and

2

▶ Chapter Openers highlight Chapter Connections, an interesting application

exercise from the chapter chapter, and provide a list of other real real-world world connections to give context for students who wonder how math relates to them.

“I especially like the depth and variety of applications in this textbook.

Other Precalculus texts the department considered did not share this strength. In particular, there is a clear effort on the part of the author to include realistic examples showing how such math can be utilized in the real world. —George Alexander, Madison Area Technical College

”

▶ Examples throughout the text feature word problems,

providing students with a starting point for how to solve these types of problems in their exercise sets.

“ One of this text’s strongest features is the wide range of applications exercises. As an instructor, I can choose which exercises fit my teaching style as well as the student interest level.

CHAPTER CONNECTIONS

Relations, Functions, and Graphs

Viewing a function in terms of an equation, a table of values, and the related graph, often brings a clearer understanding of the relationships involved. For example, the power generated by a wind turbine is often modeled 8v 3 by the function P1v2 , where P is 125

2.1

Rectangular Coordinates; Graphing Circles and Other Relations 152

2.2

Graphs of Linear Equations 165

the power in watts and v is the wind velocity in miles per hour. While the formula enables us to predict the power generated for a given wind speed, the graph offers a visual representation of this relationship, where we note a rapid growth in power output as the wind speed increases. This application appears as Exercise 107 in Section 2.6.

2.3

Linear Graphs and Rates of Change 178

Check out these other real-world connections:

2.4

Functions, Function Notation, and the Graph of a Function 190

2.5

Analyzing the Graph of a Function 206

2.6

The Toolbox Functions and Transformations 225

2.7

Piecewise-Defined Functions 240

2.8

The Algebra and Composition of Functions 254

CHAPTER OUTLINE

Earthquake Area (Section 2.1, Exercise 84) Height of an Arrow (Section 2.5, Exercise 61) Garbage Collected per Number of Garbage Trucks (Section 2.2, Exercise 42) Number of People Connected to the Internet (Section 2.3, Exercise 109)

”

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—Stephen Toner, Victor Valley College

▶ Application Exercises at the end of each section are the hallmark of

the Coburn series. Never contrived, always creative, and born out of the author’s life and experiences, each application tells a story and appeals to a variety of teaching styles, disciplines, backgrounds, and interests.

“ [The application problems] answered the question, ‘When are we ever going to use this?’ ”

—Student class tester at Metropolitan Community College–Longview

▶ Math M th iin A Action ti A Applets, l t llocated d online, li enable bl students d to work k

collaboratively as they manipulate applets that apply mathematical concepts in real-world contexts. x

EXAMPLE 10

Determining the Domain and Range from the Context Paul’s 1993 Voyager has a 20-gal tank and gets 18 mpg. The number of miles he can drive (his range) depends on how much gas is in the tank. As a function we have M1g2 18g, where M(g) represents the total distance in miles and g represents the gallons of gas in the tank. Find the domain and range.

Solution C. You’ve just learned how to use function notation and evaluate functions

Begin evaluating at x 0, since the tank cannot hold less than zero gallons. On a full tank the maximum range of the van is 20 # 18 360 miles or M1g2 3 0, 360 4 . Because of the tank’s size, the domain is g 3 0, 20 4 . Now try Exercises 94 through 101

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Through Timely Examples set the sta ge the imp ortance of exa mp les that te rsta ove to t icul diff be ld wou it In mathematics, exa mp le that was too eriences hav e falt ered due to an exp l iona cat edu few a t No . ning for lear this ser ies, a car efu l and ce, or had a dist rac ting result. In diff icul t, a poor fit, out of seq uen direct focus on the that wer e timely and clear, wit h a les mp exa ct sele to de ma was deliberate eff ort d to link pre vious concep ts possible, they wer e fur ther designe e her ryw Eve d. han at l skil or t or kno ws, concep cep ts to com e. As a tra ined edu cat con for k wor und gro the lay to to cur ren t idea s, and sequence of car efully bef ore it’s ever asked, and a timely en oft is n stio que a wer ans to e nex t logical, the best tim ma king each new idea simply the , ard reg this in way long a go constructed exa mples can ws in unnoticed matica l matur ity of a student gro the ma the l, sfu ces suc en Wh . step even anticipated supposed to be that way. —JC incr ements, as though it was just

“ The author does a great job in describing the

▶ Titles have been added to Examples in this edition to

examples and how they are to be written. In the examples, the author shows step by step ways to do just one problem . . . this makes for a better understanding of what is being done.

highlight relevant learning objectives and reinforce the importance of speaking mathematically using vocabulary.

”

▶ Annotations located to the right of the solution sequence

—Michael Gordon, student class tester at Navarro College

help the student recognize which property or procedure is being applied. ▶ “Now Try” boxes immediately following EXAMPLE 3

Examples guide students to specific matched exercises at the end of the section, helping them identify exactly which homework problems coincide with each discussed concept.

Solution

Solving a System Using Substitution 4x y 4 Solve using substitution: e . y x 2 Since y x 2, we can replace y with x 4x 1x 5x

2 in the first rst equation. e

4 first equation 4x 4 substitute x 2 for y simplify 4 2 x result 5 The x-coordinate is 25. To find the y-coordinate, substitute 25 for x into either of the original equations. Substituting in the second equation gives y x 2 second equation 2 2 2 substitute for x 5 5 12 2 10 10 2 12 , 1 5 5 5 5 5 2 12 The solution to the system is 1 5, 5 2. Verify by substituting 25 for x and 12 5 for y into both equations.

▶ Graphical Support Boxes, located after

selected examples, visually reinforce algebraicc concepts with a corresponding graphing g calculator example.

“ I thought the author did a good job of explaining

y 22 2

Now try Exercises 23 through 32

the content by using examples, because there was an example of every kind of problem.

”

—Brittney Pruitt, student class tester at Metropolitan Community College–Longview

each group of examples. I have not seen this in other texts and it is a really nice addition. I usually tell my students which examples correspond to which exercises, so this will save time and effort on my part.

”

GRAPHICAL SUPPORT Graphing the lines from Example 8 as Y1 and Y2 on a graphing calculator, we note the lines do appear to be parallel (they actually must be since they have identical slopes). Using the ZOOM 8:ZInteger feature of the TI-84 Plus we can quickly verify that Y2 indeed contains the point ( 6, 1).

“ I particularly like the ‘Now Try exercises . . .’ after

31

—Scott Berthiaume, Edison State College 47

47

“ The incorporation of technology and graphing calculator 31

usage . . . is excellent. For the faculty that do not use the technology it is easily skipped. It is very detailed for the students or faculty that [do] use technology.

”

—Rita Marie O’Brien, Navarro College

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Solid Skill Development . . . Through Exercises idea s. I con str uct ed in sup por t of eac h sec tion’s ma in es rcis exe of lth wea a d ude incl I hav e wea ker stu den ts, whi le ort to pro vide str ong sup por t for eff an in e, car at gre h wit set h rcis es to eac r. I also design ed the var ious exe the fur n eve ch rea to ts den stu cha llenging adv anc ed exercises allow —t he qua ntity and qua lity of the ors eav end g chin tea ir the in s tor sup por t inst ruc ions, and to illus tra te stu den ts thr oug h diff icul t calc ulat de gui to ies unit ort opp us ero for num ues.—JC imp ortant pro blem-so lving tech niq

MID-CHAPTER CHECK

Mid-Chapter Checks Mid-Chapter Checks provide students with a good stopping place to assess their knowledge before moving on to the second half of the chapter.

1. Compute 1x3 8x2 7x 142 1x 22 using long division and write the result in two ways: (a) dividend 1quotient21divisor2 remainder and remainder dividend (b) . 1quotient2 divisor divisor 2. Given that x 2 is a factor of f 1x2 2x4 x3 8x2 x 6, use the rational zeroes theorem to write f(x) in completely factored form.

End-of-Section Exercise Sets

9. Use the Guidelines for Graphing to draw the graph of q1x2 x3 5x2 2x 8. 10. When fighter pilots train for dogfighting, a “harddeck” is usually established below which no competitive activity can take place. The polynomial graph given shows Maverick’s altitude above and below this hard-deck during a 5-sec interval. a. What is the minimum ibl d

Altitude A (100s of feet)

1.3 EXERCISES

▶ Concepts and Vocabulary exercises to help students

recall and retain important terms.

CONCEPTS AND VOCABULARY Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. When multiplying or dividing by a negative quantity, we the inequality to maintain a true statement. 2. To write an absolute value equation or inequality in the absolute value simplified form, we

▶ Developing Your Skills exercises to provide

practice of relevant concepts just learned with increasing levels of difficulty.

“ Some of our instructors would mainly assign the developing your skills and working with formula problems, however, I would focus on the writing, research and decision making [in] extending the concept. The flexibility is one of the things I like about the Coburn text.

”

—Sherry Meier, Illinois State University

contextual applications of well-known formulas. ▶ Extending the Concept exercises that require

communication of topics, synthesis of related concepts, and the use of higher-order thinking skills. ▶ Maintaining Your Skills exercises that address

skills from previous sections to help students retain previously learning knowledge.

Describe each solution set (assume k answer.

b 6

5. ax

0). Justify your

k

DEVELOPING YOUR SKILLS Solve each absolute value equation. Write the solution in set notation.

7. 2 m

1

8. 3 n

5

9.

3x

10.

2y

7

3

14 5 3

Solve each absolute value inequality. Write solutions in interval notation.

25. x 27.

2

6

15

29.

2 3 m

5v

1 4

4

14

11. 2 4v

5

6.5

10.3

31. 3 p

12. 7 2w

5

6.3

11.2

33. 3b

13.

7p

3

6

5

35. 4

14.

3q

4

3

5

37. `

4

7

26. y

2 7 4

28.

8 6 9

30.

5 6 8

32. 5 q

6

11 3z

4x

5 3

1

9

12 6 7 1 ` 2

7 6

3 3 7 7

2n 3w

2 2

6 6 8

2

7

34. 2c

3

5 6 1

36. 2

7u

38. `

2y

3 4

7

8

4

15 3 ` 6 8 16

WORKING WITH FORMULAS 55. Spring Oscillation | d

▶ W Working ki with ith Formulas F l exercises i tto ddemonstrate t t

4. The absolute value inequality 3x 6 6 12 is true when 3x 6 7 and 3x 6 6 .

x|

L

A weight attached to a spring hangs at rest a distance of x in. off the ground. If the weight is pulled down (stretched) a distance of L inches and released, the weight begins to bounce and its distance d off the ground must satisfy the indicated formula. If x equals 4 ft and the spring is stretched 3 in. and released, solve the inequality to find what distances

56. A “Fair” Coin `

h

50 5

`

1.645

If we flipped a coin 100 times, we expect “heads” to come up about 50 times if the coin is “fair.” In a study of probability, it can be shown that the number of heads h that appears in such an experiment must satisfy the given inequality to be considered “fair.” (a) Solve this inequality for h.

EXTENDING THE CONCEPT 67. Determine the value or values (if any) that will make the equation or inequality true. x x 8 a. x b. x 2 2 x x x 6x c. x d. x 3 x 3 e. 2x 1

3 2x has only one 68. The equation 5 2x solution. Find it and explain why there is only one.

MAINTAINING YOUR SKILLS 69. (R.4) Factor the expression completely: 18x3 21x2 60x. 70. (1.1) Solve V2

2W for C A

(physics).

72. (1.2) Solve the inequality, then write the solution set in interval notation: 312x 2

52 7 21x

12

7.

1

“

He not only has exercises for skill development, but also problems for ‘extending the concept’ and ‘maintaining your skills,’ which our current text does not have. I also like the mid-chapter checks provided. All these give Coburn an advantage in my view.

”

—Randy Ross, Morehead State University

xii

“ The strongest feature seems to be the wide variety of

exercises included at the end of each section. There are plenty of drill problems along with good applications.

”

—Jason Pallett, Metropolitan Community College–Longview

cob19421_fm_i-xl.indd Page xiii 1/20/09 1:00:07 AM user

End-of-Chapter Review Material Exercises located at the end of the chapter provide students with the tools they need to prepare for a quiz or test. Each chapter features the following: ▶

Chapter Summary and Concept Reviews that present key concepts with corresponding exercises by section in a format easily used by students.

▶

Mixed Reviews that offer more practice on topics from the entire chapter, arranged in random order requiring students to identify problem types and solution strategies on their own.

▶

Practice Tests that give students the opportunity to check their knowledge and prepare for classroom quizzes, tests, and other assessments.

“ We always did reviews and a quiz before the actual test; it helped a lot.”

—Melissa Cowan, student class tester Metropolitan Community College–Longview

▶

Cumulative Reviews that are presented at the end of each chapter help students retain previously learned skills and concepts by revisiting important ideas from earlier chapters (starting with Chapter 2).

“ The cumulative review is very good and is considerably better than some of the books I have reviewed/used. I have found these to be wonderful practice for the final exam.

”

—Sarah Clifton, Southeastern Louisiana University

/Users/user/Desktop/Temp Work/2008/DECEMBER/01-12-08/VGP_01_12_08/BACKUP

“ The summary and concept review was very helpful

because it breaks down each section. That is what helps me the most.

”

—Brittany Pratt, student class tester at Baton Rouge Community College

S U M M A RY A N D C O N C E P T R E V I EW SECTION SECTI ION 1.1

Linear Equations, Formulas, and Problem Solving

KEY CONCEPTS • An equation is a statement that two expressions are equal. • Replacement values that make an equation true are called solutions or roots. • Equivalent equations are those that have the same solution set. • To solve an equation we use the distributive property and the properties of equality to write a sequence of simpler, equivalent equations until the solution is obvious. A guide for solving linear equations appears on page 75. • If an equation contains fractions, multiply both sides by the LCD of all denominators, then solve. • Solutions to an equation can be checked using back-substitution, by replacing the variable with the proposed solution and verifying the left-hand expression is equal to the right. • An equation can be: 1. an identity, one that is always true, with a solution set of all real numbers. 2. a contradiction, one that is never true, with the empty set as the solution set. 3. conditional, or one that is true/false depending on the value(s) input. • To solve formulas for a specified variable, focus on the object variable and apply properties of equality to write this variable in terms of all others. • The basic elements of good problem solving include: 1. Gathering and organizing information 2. Making the problem visual 3. Developing an equation model 4. Using the model to solve the application For a complete review, see the problem-solving guide on page 78.

C U M U L AT I V E R E V I E W C H A P T E R S 1 – 2 1. Perform the division by factoring the numerator: 1x3 5x2 2x 102 1x 52. x 6 5 and

2. Find the solution set for: 2 3x 2 6 8.

3. The area of a circle is 69 cm2. Find the circumference of the same circle. 4. The surface area of a cylinder is A 2 r2 Write r in terms of A and h (solve for r). 5. Solve for x:

213

x2

5x

41x

6. Evaluate without using a calculator: a

12 27 b 8

2 rh. 7.

2 3

.

18. Determine if the following relation is a function. If not, how is the definition of a function violated?

7. Find the slope of each line: a. through the points: 1 4, 72 and (2, 5). b. a line with equation 3x 5y 20.

Michelangelo

8. Graph using transformations of a parent function. 1x 2 3. a. f 1x2 x 2 3. b. f 1x2 9. Graph the line passing through 1 3, 22 with a slope of m 12, then state its equation.

Graphing Calculator icons appear next to exercises where important concepts can be supported by the use of graphing technology.

Homework Selection Guide

111. Given f 1x2 1 f # g21x2, 1 f

3x2 6x and g1x2 x g2 1x2, and 1g f 2 1 22.

Parnassus

Titian

La Giocanda

Raphael

The School of Athens

Giorgione

Jupiter and Io

da Vinci

Venus of Urbino

Correggio

The Tempest

19. Solve by completing the square. Answer in both exact and approximate form: 2x2 49 20x

110. Show that x 1 5i is a solution to x2 2x 26 0.

▶

16. Simplify the radical expressions: 10 172 1 a. b. 4 12 17. Determine which of the following statements are false, and state why. a. ( ( ( ( ( ( ( ( b. c. ( ( ( ( d. ( ( ( (

2 find:

12. Graph by plotting the y-intercept, then counting ¢y m to find additional points: y 13x 2 ¢x 13. Graph the piecewise defined function x2 4 x 6 2 f 1x2 e and determine 2 x 8 x 1 the following:

20. Solve using the quadratic formula. If solutions are complex, write them in a bi form. 51 2x2 20x 21. The National Geographic Atlas of the World is a very large, rectangular book with an almost inexhaustible panoply of information about the world we live in. The length of the front cover is 16 cm more than its width, and the area of the cover is 1457 cm2. Use this information to write an equation model, then use the quadratic formula to determine the length and width

A list of suggested homework exercises has been provided for each section of the text (Annotated Instructor’s Edition only). This feature may prove especially useful for departments that encourage consistency among many sections, or those having a large adjunct population. The feature was also designed as a convenience to instructors, enabling them to develop an inventory of exercises that is more in tune with the course as they like to teach it. The Guide provides prescreened and preselected p assignments at four different levels: Core, Standard, Extended, and In Depth. and . 1. After a vertical , points on the graph are • Core: These assignments go right to the heart of the 3. The vertex of h1x2 31x 52 9 is at farther from the x-axis. After a vertical , and the graph opens . points on the graph are closer to the x-axis. material, offering a minimal selection of exercises that cover the primary concepts and solution strategies of the section, along with a small selection of the best applications. • Standard: The assignments at this level include the Core exercises, while providing for additional practice without included as well as a greater variety of excessive drill. A wider assortment of the possible variations on a theme are included, applications. • Extended: Assignments from the Extended category expand on the Standard exercises to include more applications, as well as some conceptual or theory-based questions. Exercises may include selected items from the Concepts and Vocabulary, Working with Formulas, and the Extending the Thought categories of the exercise sets. • In Depth: The In Depth assignments represent a more comprehensive look at the material from each section, while attempting to keep the assignment manageable for students. These include a selection of the most popular and highest-quality exercises from each category of the exercise set, with an additional emphasis on Maintaining Your Skills. reflections

stretch

2

compression

( 5,

9)

upward

HOMEWORK SELECTION GUIDE

Core: 7–59 every other odd, 61–73 odd, 75–91 every other odd, 93–101 odd, 105, 107 (33 Exercises) Standard: 1–4 all, 7–59 every other odd, 61, 63–73 odd, 75–91 every other odd, 93–101 odd, 105, 107, 109 (38 Exercises)

Extended: 1–4 all, 7–59 every other odd, 61, 63–73 odd, 75–91 every other odd, 93–101 odd, 103, 105, 106, 107, 109, 111, 112, 114, 117 (44 Exercises) In Depth: 1–6 all, 7–59 every other odd, 61, 63–73 odd, 75–91 every other odd, 93–101 odd, 103, 104, 105–110 all, 111–117 all (54 Exercises)

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Strong Connections . . . Through a Conversational Writing Style ext, text, atics te matics mostt promin ent featur es of a mathem he m ly tthe ably g ab gu argu arguab aree ar onss ar on ttions ti cati ca pplililica pp aapplica nd app es aand es pllles ampl am exam examp ililee ex While Whil Wh ts udents s st studen er. It may be true that some t ogether th em togeth ndds them bbinds hatt bi ha t tthat iliity bility d bi t l andd readab itii style t h writing ’ the it’s for an example similar to the don’t read the text, and that others open the text only when looking studen ts who do (read the text), exercise they’re curren tly working. But when they do and for those ts in a form and at a level it’s important they have a text that “speak s to them,” relating concep ts in and keep their interest, they understand and can relate to. Ideally this text will draw studen third time, until it becomes becoming a positiv e experience and bringing them back a second and of their text (as more that just habitual. At this point, studen ts might begin to see the true value learning on equal footing with a source of problems—p un intended), and it becomes a resour ce for —JC any other form of supplementa l instruction.

Conversational Writing Style John Coburn’s experience in the classroom and his strong connections to how students comprehend the material are evident in his writing style. He uses a conversational and supportive writing style, providing the students with a tool they can depend on when the teacher is not available, when they miss a day of class, or simply when working on their own. The effort John has put into the writing is representative of his unofficial mantra: “If you want more students to reach the top, you gotta put a few more rungs on the ladder.”

“ The author does a fine job with his narrative.

His explanations are very clear and concise. I really like his explanations better than in my current text.

”

—Tammy Potter, Gadsden State College

“ The author does an excellent job of

engagement and it is easily seen that he is conscious of student learning styles.

”

—Conrad Krueger, San Antonio College

Through Student Involvement nt How do you design a student-friendly textbook? We decided to get students involved by hosting two separate focus groups. During these sessions we asked students to advise us on how they use their books, what pedagogical elements are useful, which elements are distracting and not useful, as well as general feedback on page layout. During this process there were times when we thought, “Now why hasn’t anyone ever thought of that before?” Clearly these student focus groups were invaluable. Taking direct student feedback and incorporating what is feasible and doesn’t detract from instructor use of the text is the best way to design a truly student-friendly text. The next two pages will highlight what we learned from students so you can see for yourself how their feedback played an important role in the development of the Coburn series.

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1.1 Linear Equations, Formulas, and Problem Solving

Students said that Learning Objectives should clearly define the goals of each section.

Learning Objectives In Section 1.1 you will learn how to:

A. Solve linear equations using properties of equality

In a study of algebra, you will encounter many families of equations, or groups of equations that share common characteristics. Of interest to us here is the family of linear equations in one variable, a study that lays the foundation for understanding more advanced families. In addition to solving linear equations, we’ll use the skills we develop to solve for a specified variable in a formula, a practice widely used in science, business, industry, and research.

B. Recognize equations that are identities or contradictions

C. Solve for a specified variable in a formula or literal equation

A. Solving Linear Equations Using Properties of Equality An equation is a statement that two expressions are equal. From the expressions 31x 12 x and x 7, we can form the equation

D. Use the problem-solving guide to solve various problem types

31x

12

x

Solution

A. You’ve just learned how to solve linear equations using properties of equality

31x

x

12

x

2

11

9

1

7

8

0

3

7

1

1

6

2

5

5

7

Solving a Linear Equation with Fractional Coefficients Solve for n: 14 1n

D Described ib d by b students t d t as one off th the most useful features in a math text, Caution Boxes signal a student to stop and take note in order to avoid mistakes in problem solving.

7.

x

which is a linear equation in one variable. To solve an equation, we attempt to find a specific input or xvalue that will make the equation true, meaning the left-hand expression will be equal to the right. Using

EXAMPLE 2

Students asked for Check Points throughout each section to alert them when a specific learning objective has been covered and to reinforce the use of correct mathematical terms.

Table 1.1 x

1 4 1n 1 4n

82 2

2 2 1 4n 1 41 4 n2 n n n

1 2 1n

82

2

1 2 1n 1 2n 1 2n 41 12 n

62 3 3 32 12

2n 12 12

62.

original equation distributive property combine like terms multiply both sides by LCD

4

distributive property subtract 2n multiply by

Verify the solution is n

1

12 using back-substitution. Now try Exercises 13 through 30

1 22 4 2 6 3 The slope of this line is

8 2 3 1 6 2 The slope of this line is 12.

4

2 3 .

Now try Exercises 33 through 40

CAUTION

When using the slope formula, try to avoid these common errors. 1. The order that the x- and y-coordinates are subtracted must be consistent, since

y 2 x2

y

y 1 x1

y

2

1

x2 .

x1

2. The vertical change (involving the y-values) always occurs in the numerator: y 2 x2

y 1 x1

x

2

y2

x

1

y1 .

3. When x1 or y1 is negative, use parentheses when substituting into the formula to prevent confusing the negative sign with the subtraction operation.

Students told us that the color red should only be used for things that are really important. Also, anything significant should be included in the body of the text; marginal readings imply optional.

Actually, the slope value does much more than quantify the slope of a line, it expresses a rate of change between the quantities measured along each axis. In appli¢y change in y cations of slope, the ratio change in x is symbolized as ¢x. The symbol ¢ is the Greek letter delta and has come to represent a change in some quantity, and the notation m ¢y is read, “slope is equal to the change in y over the change in x.” Interpreting ¢x slope as a rate of change has many significant applications in college algebra and beyond.

EXAMPLE 8

Examples are called out in the margins so they are easy for students to spot. Solution

Determining the Domain of an Expression 6 Determine the domain of the expression . State the result in set notation, x 2 graphically, and using interval notation. Set the denominator equal to zero and solve: x 2 is outside the domain and must be excluded.

• Set notation: 5x|x

,x

• Graph: 1 0 1 • Interval notation: x

Examples are “boxed” so students can clearly see where they begin and end

)) 2

0 yields x

2. This means

3

4

5

1 q, 22 ´ 12, q2 Now try Exercises 61 through 68

A second area where allowable values are a concern involves the square root operation. Recall that 149 7 since 7 # 7 49. However, 1 49 cannot be written as the product of two real numbers since 1 72 # 1 72 49 and 7 # 7 49. In other words, 1X represents a real number only if the radicand is positive or zero. If X represents an algebraic expression, the domain of 1X is 5X|X 06 . EXAMPLE 9

Determining the Domain of an Expression Determine the domain of 1x and in interval notation.

Solution

3. State the domain in set notation, graphically,

The radicand must represent a nonnegative number. Solving x x 3.

• Set notation: 5x|x • Graph:

Students told us they liked when the examples were linked to the exercises.

2

26

4

[

3

2

• Interval notation: x

3

0 gives

36 1

0

1

2

3, q2 Now try Exercises 69 through 76

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Solve using the zero product property. Be sure each equation is in standard form and factor out any common factors before attempting to solve. Check all answers in the original equation.

x3

7. 22x 9. 3x3

Students told us that directions should be in bold so they are easily distinguishable from the problems.

9x2

7x2

6x

8 . x3

13x2

42x

10. 7x2

15x

2x3

11. 2x4

3x3

9x2

12.

13. 2x4

16x

0

14. x4

4x

5x2

20 16. x3

12

2

15. x3 17. 4x 19. 2x3

3x

12x2

3

x

81

27x

7x3

4x2

22. x4

3x3

23. x4

81

2x4

41.

9x3 42.

64x

0

18

9x

2x2

2

3

7

7x

40.

43.

x

60

2

21. x4

4

18. x

10x

20. 9x

7x2

3x

44.

3

28x

9x2

10 6 3

n

n

7 p

5

2a

2a

2

1

2a

18 n

2

2 3

p

6 5

5a

3 3

a

3

3n 2n 1

1

2

n

6

2

a

6n

5

x

1 5p

2

2

p

2x

20 n

2

7

x

1

x

5

x

2x

1

7

x

4n 3n 1

Solve for the variable indicated.

27x 45.

0 0

1 f

1 f1

1 x

1 ; for f f2

46.

; for r

48. q

1 y

1 ; for z z

24. x

1

25. x4

256

0

R

26. x4

625

0

49. V

50. s

27. x6

2x4

x2

1 2 r h; for h 3

1 2 gt ; for g 2

6

28. x

3x

4

51. V

52. V

29. x5

x3

4 3 r ; for r3 3

1 2 r h; for r2 3

30. x5

9x3

31. x6

1

32. x6

64

47. I

2

16x

2

0 48

0

8x2

8

0

x2

9

0

33. 34. 35. 36.

2 x

0

1 x

x

3

m2

2

a

3m

4 3y

pf p

f

; for p

313x

5

9

54. a.

214x

1

10 b.

13x

b. x 5

1

3 b. 21 7

2

3 3 d. 1 2x

b. 3 1 3 3 1 5p 2 3 1 6x 7 5 6 4 3 3 d. 31 x 3 21 2x 17 3

3

1

3

15x

1

3x

3

9 4x

x 7

3 1 3x

7

7

4

c.

1

57. a. 1x 9 1x 9 b. x 3 223 x c. 1x 2 12x 2

7 3

53. a.

56. a.

1 m

3

21 a 2y

x

5

3 m

r

3 55. a. 2 1 3m 3 1 2m 3 c. 5

5 2

1

E

Solve each equation and check your solutions by substitution. Identify any extraneous roots.

0

Solve each equation. Identify any extraneous roots.

Because students spend a lot of time in the exercise section of a text, they said that a white background is hard on their eyes . . . so we used a soft, off-white color for the background

14

39. x

5

velocity of 160 ft/sec and a height of 240 ft, it runs out of fuel and becomes a projectile. a. How high is the rocket three seconds later? Four seconds later? b. How long will it take the rocket to attain a height of 640 ft? c. How many times is a height of 384 ft attained? When do these occur? d. How many seconds until the rocket returns to the ground?

88. Composite figures—gelatin capsules: The gelatin capsules manufactured for cold and flu medications are shaped like a cylinder with a hemisphere on each end. The interior volume V of each capsule r2h, where h is can be modeled by V 43 r3 the height of the cylindrical portion and r is its radius. If the cylindrical portion of the capsule is 8 mm long 1h 8 mm2, what radius would give the capsule a volume that is numerically equal to 15 times this radius?

93. Printing newspapers: The editor of the school newspaper notes the college’s new copier can complete the required print run in 20 min, while the back-up copier took 30 min to do the same amount of work. How long would it take if both copiers are used? 94. Filling a sink: The cold water faucet can fill a sink in 2 min. The drain can empty a full sink in 3 min. If the faucet were left on and the drain was left open, how long would it take to fill the sink?

89. Running shoes: When a popular running shoe is priced at $70, The Shoe House will sell 15 pairs each week. Using a survey, they have determined that for each decrease of $2 in price, 3 additional pairs will be sold each week. What selling price will give a weekly revenue of $2250?

95. Triathalon competition: As one part of a Mountain-Man triathalon, participants must row a canoe 5 mi down river (with the current), circle a buoy and row 5 mi back up river (against the current) to the starting point. If the current is flowing at a steady rate of 4 mph and Tom Chaney made the round-trip in 3 hr, how fast can he row in still water? (Hint: The time rowing down river and the time rowing up river must add up to 3 hr.)

90. Cell phone charges: A cell phone service sells 48 subscriptions each month if their monthly fee is $30. Using a survey, they find that for each decrease of $1, 6 additional subscribers will join. What charge(s) will result in a monthly revenue of $2160?

96. Flight time: The flight distance from Cincinnati, Ohio, to Chicago, Illinois, is approximately 300 mi. On a recent round-trip between these cities in my private plane, I encountered a steady 25 mph headwind on the way to Chicago, with a 25 mph tailwind on the return trip. If my total flying time

Projectile height: In the absence of resistance, the height of an object that is projected upward can be modeled by the equation h 16t2 vt k, where h represents the height of the object (in feet) t sec after it has been thrown, v represents the initial velocity (in feet per second), and k represents the height of the object when t 0 (before it has

WORKING WITH FORMULAS

Students said having a lot of icons was confusing. The graphing calculator is the only icon used in the exercise sets; no unnecessary icons are used

xvi

79. Lateral surface area of a cone: S The lateral surface area (surface area excluding the base) S of a cone is given by the formula shown, where r is the radius of the base and h is the height of the cone. (a) Solve the equation for h. (b) Find the surface area of a cone that has a radius of 6 m and a height of 10 m. Answer in simplest form.

r2r 2

h2

h

r

80. Painted area on a canvas: A

4x2

60x x

104

A rectangular canvas is to contain a small painting with an area of 52 in2, and requires 2-in. margins on the left and right, with 1-in. margins on the top and bottom for framing. The total area of such a canvas is given by the formula shown, where x is the height of the painted area. a. What is the area A of the canvas if the height of the painting is x 10 in.? b. If the area of the canvas is A 120 in2, what are the dimensions of the painted area?

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Connections to Calculus Co with a strong sense that more Like many of you who have taught calculus, I’ve often been left To this end I’ve included a could be done in Preca lculus, to prepar e our studen ts for calculus. , which is design ed to more closely tie their experience in featur e called a timely choice of topics PreCa lculus to their future studies. This end-of-chapter featur e makes t, providing studen ts with a from within each chapter to illustrate their use in a calculus contex —JC more solid footing as they step up to the calculus level.

5

The Connections to Calculus feature is included at the end of Chapters 1–10 to highlight the connections between algebraic concepts presented in this course, and the calculus concepts to be learned in a later course. This feature includes exposition, examples, and exercises to reinforce the material. Each chapter opener provides a preview of the Connection to Calculus coming at the end of the chapter.

An Introduction to Trigonometric Functions

CONNECTIONS TO CALCULUS While right triangles have a number of meaningful applications as a problem-solving tool, they can also help to rewrite certain expressions in preparation for the tools of calculus, and introduce us to an alternative method for graphing relations and functions using polar coordinates. As things stand, some functions and relations are much easier to graph in polar coordinates, and converting between the two systems is closely connected to a study of right triangles.

Drawing a diagram to visualize relationships and develop information is an important element of good problem solving. This is no less true in calculus, where it is often a fundamental part of understanding the question being asked. As a precursor to applications involving trig substitutions, we’ll illustrate how right triangle diagrams are used to rewrite trigonometric functions of ␪ as algebraic functions of x. 䊳

䊳

x Using x ⫽ 5 sin ␪, we obtain sin ␪ ⫽ . From our 5 work in Chapter 5, we know the right triangle opp definition of sin ␪ is , and we draw a triangle hyp with side x oriented opposite an angle ␪, and label a hypotenuse of 5 (see figure). To find an expression for the adjacent side, we use the Pythagorean theorem: 1adj2 2 ⫹ x2 ⫽ 52 1adj2 2 ⫽ 25 ⫺ x2 adj ⫽ 225 ⫺ x2

5

5–115

䊳

䊳

This application appears as Exercise 85 in Section 5.6

5.5 Transformations and Applications of Trigonometric Graphs 486 5.6

The Trigonometry of Right Triangles 499

5.7 Trigonometry and the Coordinate Plane 512

The trigonometry of right triangles also plays an important role in a study of calculus, as we seek to simplify certain expressions, or convert from one system of graphing to another. The Connections to Calculus feature following Chapter 5 offers additional practice and insight in preparation for a study of calculus. 425

x

Adjacent side

Pythagorean theorem isolate term result (length must be positive); ⫺5 6 x 6 5

䊳

Using Right Triangle Diagrams to Rewrite Trig Expressions Find expressions for tan ␪ and csc ␪, given sec ␪ ⫽

Solution

While rainbows have been admired for centuries for their beauty and color, understanding the physics of rainbows is of fairly recent origin. Answers to questions regarding their seven color constitution, the order the colors appear, the circular shape of the bow, and their relationship to moisture all have answers deeply rooted in mathematics. The relationship between light and color can be understood in terms of trigonometr y, with questions regarding the apparent height of the rainbow, as well as the height of other natural and man-made phenomena, found using the trigonometr y of right triangles.

␪

Now try Exercises 1 through 4

䊳

5.3 Graphs of the Sine and Cosine Functions; Cosecant and Secant Functions 454

Connections to Calculus

Using this triangle and the standard definition of the remaining trig functions, we 5 225 ⫺ x2 5 x find cos ␪ ⫽ , sec ␪ ⫽ , and csc ␪ ⫽ . , tan ␪ ⫽ x 5 225 ⫺ x2 225 ⫺ x2

EXAMPLE 2

5.1 Angle Measure, Special Triangles, and Special Angles 426

Using Right Triangle Diagrams to Rewrite Trig Expressions Use the equation x ⫽ 5 sin ␪ and a right triangle diagram to write cos ␪, tan ␪, sec ␪, and csc ␪ as functions of x.

Solution

CHAPTER OUTLINE 5.2 Unit Circles and the Trigonometry of Real Numbers 440

5.4 Graphs of Tangent and Cotangent Functions 474

Right Triangle Relationships

EXAMPLE 1

CHAPTER CONNECTIONS

hyp , we draw a right triangle adj diagram as in Example 1, with a hypotenuse of 2 2u ⫹ 144 and a side u adjacent to angle ␪. For hyp opp tan ␪ ⫽ and csc ␪ ⫽ , we use the opp hyp

2u2 ⫹ 144 . u

is to come in calculus and to get them familiar with some notation is the correct one and hopefully it will motivate them to study the algebra in the chapters a little more. —Terry Hobbs, Metropolitan Community

”

With sec ␪ ⫽

冪u2 ⫹ 144 ␪

“ I think the approach of giving them a taste of what

opp

College, Maple Woods

u

539

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Connections to Calculus Co with a strong sense that more Like many of you who have taught calculus, I’ve often been left To this end I’ve included a could be done in Preca lculus, to prepar e our studen ts for calculus. , which is design ed to more closely tie their experience in featur e called es a timely choice of topics PreCa lculus to their future studies. This end-of-chapter featur e provid t, giving studen ts a more solid from within each chapter to illustrate their use in a calculus contex —JC footing as they step up to the calculus level.

5

The Connections to Calculus feature is included at the end of Chapters 1–10 to highlight the connections between algebraic concepts presented in this course, and the calculus concepts to be learned in a later course. This feature includes exposition, examples, and exercises to reinforce the material. Each chapter opener provides a preview of the Connection to Calculus coming at the end of the chapter.

An Introduction to Trigonometric Functions

CONNECTIONS TO CALCULUS While right triangles have a number of meaningful applications as a problem-solving tool, they can also help to rewrite certain expressions in preparation for the tools of calculus, and introduce us to an alternative method for graphing relations and functions using polar coordinates. As things stand, some functions and relations are much easier to graph in polar coordinates, and converting between the two systems is closely connected to a study of right triangles.

Drawing a diagram to visualize relationships and develop information is an important element of good problem solving. This is no less true in calculus, where it is often a fundamental part of understanding the question being asked. As a precursor to applications involving trig substitutions, we’ll illustrate how right triangle diagrams are used to rewrite trigonometric functions of ␪ as algebraic functions of x. 䊳

䊳

x Using x ⫽ 5 sin ␪, we obtain sin ␪ ⫽ . From our 5 work in Chapter 5, we know the right triangle opp definition of sin ␪ is , and we draw a triangle hyp with side x oriented opposite an angle ␪, and label a hypotenuse of 5 (see figure). To find an expression for the adjacent side, we use the Pythagorean theorem: 1adj2 2 ⫹ x2 ⫽ 52 1adj2 2 ⫽ 25 ⫺ x2 adj ⫽ 225 ⫺ x2

5

5–115

䊳

䊳

This application appears as Exercise 85 in Section 5.6

5.5 Transformations and Applications of Trigonometric Graphs 486 5.6

The Trigonometry of Right Triangles 499

5.7 Trigonometry and the Coordinate Plane 512

The trigonometry of right triangles also plays an important role in a study of calculus, as we seek to simplify certain expressions, or convert from one system of graphing to another. The Connections to Calculus feature following Chapter 5 offers additional practice and insight in preparation for a study of calculus. 425

x

Adjacent side

Pythagorean theorem isolate term result (length must be positive); ⫺5 6 x 6 5

䊳

Using Right Triangle Diagrams to Rewrite Trig Expressions Find expressions for tan ␪ and csc ␪, given sec ␪ ⫽

Solution

While rainbows have been admired for centuries for their beauty and color, understanding the physics of rainbows is of fairly recent origin. Answers to questions regarding their seven color constitution, the order the colors appear, the circular shape of the bow, and their relationship to moisture all have answers deeply rooted in mathematics. The relationship between light and color can be understood in terms of trigonometr y, with questions regarding the apparent height of the rainbow, as well as the height of other natural and man-made phenomena, found using the trigonometr y of right triangles.

␪

Now try Exercises 1 through 4

䊳

5.3 Graphs of the Sine and Cosine Functions; Cosecant and Secant Functions 454

Connections to Calculus

Using this triangle and the standard definition of the remaining trig functions, we 5 225 ⫺ x2 5 x find cos ␪ ⫽ , sec ␪ ⫽ , and csc ␪ ⫽ . , tan ␪ ⫽ x 5 225 ⫺ x2 225 ⫺ x2

EXAMPLE 2

5.1 Angle Measure, Special Triangles, and Special Angles 426

Using Right Triangle Diagrams to Rewrite Trig Expressions Use the equation x ⫽ 5 sin ␪ and a right triangle diagram to write cos ␪, tan ␪, sec ␪, and csc ␪ as functions of x.

Solution

CHAPTER OUTLINE 5.2 Unit Circles and the Trigonometry of Real Numbers 440

5.4 Graphs of Tangent and Cotangent Functions 474

Right Triangle Relationships

EXAMPLE 1

CHAPTER CONNECTIONS

hyp , we draw a right triangle adj diagram as in Example 1, with a hypotenuse of 2 2u ⫹ 144 and a side u adjacent to angle ␪. For hyp opp tan ␪ ⫽ and csc ␪ ⫽ , we use the opp hyp

2u2 ⫹ 144 . u

is to come in calculus and to get them familiar with some notation is the correct one and hopefully it will motivate them to study the algebra in the chapters a little more. —Terry Hobbs, Metropolitan Community

”

With sec ␪ ⫽

冪u2 ⫹ 144 ␪

“ I think the approach of giving them a taste of what

opp

College, Maple Woods

u

539

xvii

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• Coverage of base e as an alternative to base 10 or b is addressed in one section (4.2) as opposed to two sections as in the first edition. • Likewise, coverage of properties of logs and log equations is found in the same section (4.4). • A clear introduction to fundamental logarithmic properties has also been added to Section 4.4. • Applications have been added and improved throughout the chapter.

CHAPTER

5

Introduction to Trigonometric Functions

• Section 5.1 includes improved DMS to decimal degrees conversion coverage, improved introduction to standard 45-45-90 and 30-60-90 triangles, better illustrations of longitude and latitude applications, and streamlined/clarified coverage of angular and linear velocity. • Section 5.2 includes a table showing summary of trig functions of special angles. • Section 5.3 has improved coverage of secant and cosecant graphs. • Section 5.4 has a strengthened connection between y ⫽ tan x and y ⫽ (sin x)/cos(x). • Section 5.5 has an improved introduction to transformations, and a clearer distinction between phase angle and phase shift. • Section 5.6 has improved coverage of co-functions, and better illustrations for angles of elevation and depression. • Section 5.7 has improved applications, and the connection between f and f-1 is introduced.

CHAPTER

6

Trigonometric Identities, Inverses, and Equations

• Section 6.1 has an increased emphasis on what an identity is (the definition of an identity), as well as an additional example of quadrant and sign analysis. • Section 6.2 has a better introduction to clarify goals, as well as an improved format for verifying identities. • Section 6.3 has improved coverage of the co-function identities, as well as extended coverage of the sum and difference identities. • Section 6.5 has a strengthened connection between inverse functions and drawn diagrams, improved coverage on evaluating the inverse trig functions, and more real-world applications of inverse trig functions.

CHAPTER

7

Applications of Trigonometry

• Section 7.1 has consolidated coverage of the ambiguous case. • Section 7.2 has expanded coverage of computing areas using trig, as well as six new contextual applications of triangular area using trig. • Section 7.3 has improved discussion, coverage, and illustrations of vector subtraction, and stronger connections between solutions using components, and solutions using the law of cosines. • Section 7.5 has additional real-world applications of complex numbers (AC circuits).

CHAPTER

8

Systems of Equations and Inequalities

• Section 8.1 includes improved coverage of equivalent systems in addition to more examples and exercises having to do with distance and navigation. • Section 8.2 features improved coverage of dependent and inconsistent systems. • Section 8.3 has improved coverage of partial fractions. • New applications of linear programming are found in Section 8.4. • Section 8.5 features an added example of Gauss-Jordan Elimination. • Section 8.6 includes better sequencing of examples and improved coverage of matrix properties. • Coverage of determinants has been streamlined with more development given to determinants in Section 8.8.

CHAPTER

9

Analytical Geometry

• Section 9.1 presents a brief introduction to analytical geometry to provide a better bridge to the conic sections and show why cone/conic connection is important. • Greater emphasis on the connection between ellipses and circles is featured in section 9.2. • Exercises requiring the movement from graph to equation have been added throughout the chapter. • Section 9.5 now covers non-linear systems to include parabolas.

CHAPTER

10

Additional Topics in Algebra

• The exposition has been revised throughout Chapter 10 for increased clarity and improved flow of topics.

CHAPTER • • • • • •

11

Bridge to Calculus: An Introduction to Limits

New applications-based chapter on limits. Addresses the question of why limits are important. Close connections made between current (new) ideas and previous material. Information and concepts come in small, understandable increments. Includes extensive exercise sets, with numerous applications. Includes optional (online) section with an introduction to the precise definition of a limit.

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Making Connections . . . Through 360° Development McGraw-Hill’s 360° Development Process is an ongoing, never-ending, market-oriented approach to building accurate and innovative print and digital products. It is dedicated to continual large-scale and incremental improvement driven by multiple customer feedback loops and checkpoints. This process is initiated during the early planning stages of our new products, intensifies during the development and production stages, and then begins again on publication, in anticipation of the next edition. A key principle in the development of any mathematics text is its ability to adapt to teaching specifications in a universal way. The only way to do so is by contacting those

universal voices—and learning from their suggestions. We are confident that our book has the most current content the industry has to offer, thus pushing our desire for accuracy to the highest standard possible. In order to accomplish this, we have moved through an arduous road to production. Extensive and open-minded advice is critical in the production of a superior text. We engaged over 400 instructors and students to provide us guidance in the development of the second edition. By investing in this extensive endeavor, McGraw-Hill delivers to you a product suite that has been created, refined, tested, and validated to be a successful tool in your course.

Board of Advisors A hand-picked group of trusted teachers active in the College Algebra and Precalculus course areas served as the chief advisors and consultants to the author and editorial team with regards to manuscript development. The Board of Advisors reviewed the manuscript in two drafts; served as a sounding board for pedagogical, media, and design concerns; approved organizational changes; and attended a symposium to confirm the manuscript’s readiness for publication. Bill Forrest, Baton Rouge Community College Marc Grether, University of North Texas Sharon Hamsa, Metropolitan Community College –Longview Max Hibbs, Blinn College Terry Hobbs, Metropolitan Community College– Maple Woods Klay Kruczek, Western Oregon University Rita Marie O’Brien’s , Navarro College

Nancy Matthews, University of Oklahoma Rebecca Muller, Southeastern Louisiana University Jason Pallett, Metropolitan Community College Kevin Ratliff, Blue Ridge Community College Stephen Toner, Victor Valley College

Accuracy Panel A selected trio of key instructors served as the chief advisors for the accuracy and clarity of the text and solutions manual. These individuals reviewed the final manuscript, the page proofs in first and revised rounds, as well as the writing and accuracy check of the instructor’s solutions manuals. This trio, in addition to several other accuracy professionals, gives you the assurance of accuracy. J.D. Herdlick, St. Louis Community College–Meramac Richard A. Pescarino, St. Louis Community College–Florissant Valley Nathan G. Wilson, St. Louis Community College–Meramac

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Student Focus Groups Two student focus groups were held at Illinois State University and Southeastern Louisiana University to engage students in the development process and provide feedback as to how the design of a textbook impacts homework and study habits in the College Algebra and Precalculus course areas. Francisco Arceo, Illinois State University Dave Cepko, Illinois State University Andrea Connell, Illinois State University Brian Lau, Illinois State University Daniel Nathan Mielneczek, Illinois State University Mingaile Orakauskaite, Illinois State University Todd Michael Rapnikas, Illinois State University Bethany Rollet, Illinois State University Teddy Schrishuhn, Illinois State University Josh Schultz, Illinois State University Andy Thurman, Illinois State University Candace Banos, Southeastern Louisiana University Nicholas Curtis, Southeastern Louisiana University

M. D. “Boots” Feltenberger, Southeastern Louisiana University Regina Foreman, Southeastern Louisiana University Ashley Lae, Southeastern Louisiana University Jessica Smith, Southeastern Louisiana University Ashley Youngblood, Southeastern Louisiana University

Special Thanks Sherry Meier, Illinois State University Rebecca Muller, Southeastern Louisiana University Anne Schmidt, Illinois State University

Instructor Focus Groups Focus groups held at Baton Rouge Community College and ORMATYC provided feedback on the new Connections to Calculus feature in Precalculus, and shed light on the coverage of review material in this course. User focus groups at Southeastern Louisiana University and Madison Area Technical College confirmed the organizational changes planned for the second edition, provided feedback on the interior design, and helped us enhance and refine the strengths of the first edition. Virginia Adelmann, Southeastern Louisiana University George Alexander, Madison Area Technical College Kenneth R. Anderson, Chemeketa Community College Wayne G.Barber, Chemeketa Community College Thomas Dick, Oregon State University Vickie Flanders, Baton Rouge Community College Bill Forrest, Baton Rouge Community College Susan B. Guidroz, Southeastern Louisiana University Christopher Guillory, Baton Rouge Community College Cynthia Harrison, Baton Rouge Community College Judy Jones, Madison Area Technical College Lucyna Kabza, Southeastern Louisiana University Ann Kirkpatrick, Southeastern Louisiana University Sunmi Ku, Bellevue Community College

Pamela Larson, Madison Area Technical College Jennifer Laveglia, Bellevue Community College DeShea Miller, Southeastern Louisiana University Elizabeth Miller, Southeastern Louisiana University Rebecca Muller, Southeastern Louisiana University Donna W. Newman, Baton Rouge Community College Scott L. Peterson, Oregon State University Ronald Posey, Baton Rouge Community College Ronni Settoon, Southeastern Louisiana University Jeganathan Sriskandarajah, Madison Area Technical College Martha Stevens, Bellevue Community College Mark J. Stigge, Baton Rouge Community College Nataliya Svyeshnikova, Southeastern Louisiana University

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JJohn Jo ohn hn N. N C. C. S Szeto, zeeto to, Southeastern SSo out utheas hheeas aste terrn nL Louisiana ouis ou isiana isi is n University Universsit ityy Christina Ch hrriist stin ina C. ina C. T Terranova, erra er r no ra nova va, Sout va, va SSoutheastern So out uthe h astern Louisiana he na University Uni U n versity Amy S. VanWey, Clackamas Clackamaas Community Comm mun unit ityy College Coll Co lleg ege Andria Andr An dria ia V Villines, Vil illi line nes, s, Be Bell Bellevue llev evue ue Community Com C ommu muni nity ty College College l

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Jeff Weaver, Baton Rouge Community College Ana Wills, Southeastern Louisiana University Randall G. Wills, Southeastern Louisiana University Xuezheng Wu, Madison Area Technical College

Developmental Symposia McGraw-Hill conducted two symposia directly related to the development of Coburn’s second edition. These events were an opportunity for editors from McGraw-Hill to gather information about the needs and challenges of instructors teaching these courses and confirm the direction of the second edition. Rohan Dalpatadu, University of Nevada–Las Vegas Franco Fedele, University of West Florida Bill Forrest, Baton Rouge Community College Marc Grether, University of North Texas Sharon Hamsa, Metropolitan Community College–Longview Derek Hein, Southern Utah University Rebecca Heiskell, Mountain View College Terry Hobbs, Metropolitan Community College– Maple Woods Klay Kruczek, Western Oregon University Nancy Matthews, University of Oklahoma Sherry Meier, Illinois State University Mary Ann (Molly) Misko, Gadsden State Community College

Rita Marie O’Brien, Navarro College Jason Pallett, Metropolitan Community College– Longview Christopher Parks, Indiana University–Bloomington Vicki Partin, Bluegrass Community College Philip Pina, Florida Atlantic University–Boca Nancy Ressler, Oakton Community College, Des Plaines Campus Vicki Schell, Pensacola Junior College Kenan Shahla, Antelope Valley College Linda Tansil, Southeast Missouri State University Stephen Toner, Victor Valley College Christine Walker, Utah Valley State College

Diary Reviews and Class Tests Users of the first edition, Said Ngobi and Stephen Toner of Victor Valley College, provided chapter-by chapter feedback in diary form based on their experience using the text. Board of Advisors members facilitated class tests of the manuscript for a given topic. Both instructors and students returned questionnaires detailing their thoughts on the effectiveness of the text’s features.

Class Tests Instructors Bill Forrest, Baton Rouge Community College Marc Grether, University of North Texas Sharon Hamsa, Metropolitan Community College–Longview Rita Marie O’Brien’s , Navarro College

Students Cynthia Aguilar, Navarro College Michalann Amoroso, Baton Rouge Community College Chelsea Asbill, Navarro College Sandra Atkins, University of North Texas Robert Basom, University of North Texas Cynthia Beasley, Navarro College Michael Bermingham, University of North Texas Jennifer Bickham, Metropolitan Community College–Longview Rachel Brokmeier, Baton Rouge Community College Amy Brugg, University of North Texas

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Zach Burke, University of North Texas Shaina Canlas, University of North Texas Kristin Chambers, University of North Texas Brad Chatelain, Baton Rouge Community College Yu Yi Chen, Baton Rouge Community College Jasmyn Clark, Baton Rouge Community College Belinda Copsey, Navarro College Melissa Cowan, Metropolitan Community College–Longview Katlin Crooks, Baton Rouge Community College Rachele Dudley, University of North Texas Kevin Ekstrom, University of North Texas Jade Fernberg, University of North Texas Joseph Louis Fino, Jr., Baton Rouge Community College Shannon M. Fleming, University of North Texas Travis Flowers, University of North Texas Teresa Foxx, University of North Texas Michael Giulietti, University of North Texas Michael Gordon, Navarro College

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Hayley Hentzen, University of North Texas Courtney Hodge, University of North Texas Janice Hollaway, Navarro College Weslon Hull, Baton Rouge Community College Sarah James, Baton Rouge Community College Georlin Johnson, Baton Rouge Community College Michael Jones, Navarro College Robert Koon, Metropolitan Community College–Longview Ben Lenfant, Baton Rouge Community College Colin Luke, Baton Rouge Community College Lester Maloney, Baton Rouge Community College Ana Mariscal, Navarro College Tracy Ann Nguyen, Baton Rouge Community College Alexandra Ortiz, University of North Texas Robert T. R. Paine, Baton Rouge Community College Kade Parent, Baton Rouge Community College Brittany Louise Pratt, Baton Rouge Community College

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Brittney Pruitt, Metropolitan Community College–Longview Paul Rachal, Baton Rouge Community College Matt Rawls, Baton Rouge Community College Adam Reichert, Metropolitan Community College–Longview Ryan Rodney, Baton Rouge Community College Cody Scallan, Baton Rouge Community College Laura Shafer, University of North Texas Natina Simpson, Navarro College Stephanie Sims, Metropolitan Community College–Longview Cassie Snow, University of North Texas Justin Stewart, Metropolitan Community College–Longview Marjorie Tulana, Navarro College Ashleigh Variest, Baton Rouge Community College James A. Wann, Navarro College Amber Wendleton, Metropolitan Community College–Longview Eric Williams, Metropolitan Community College–Longview Katy Wood, Metropolitan Community College–Longview

Developmental Editing The manuscript has been impacted by numerous developmental editors who edited for clarity and consistency. Efforts resulted in cutting length from the manuscript, while retaining a conversational and casual narrative style. Editorial work also ensured the positive visual impact of art and photo placement. First Edition Chapter Reviews and Manuscript Reviews Over 200 instructors participated in postpublication single chapter reviews of the first edition and helped the team build the revision plan for the second edition. Over 100 teachers and academics from across the country reviewed the current edition text, the proposed second edition table of contents, and first-draft second edition manuscript to give feedback on reworked narrative, design changes, pedagogical enhancements, and organizational changes. This feedback was summarized by the book team and used to guide the direction of the second-draft manuscript. Scott Adamson, Chandler-Gilbert Community College Teresa Adsit, University of Wisconsin–Green Bay Ebrahim Ahmadizadeh, Northampton Community College George M. Alexander, Madison Area Technical College Frances Alvarado, University of Texas–Pan American Deb Anderson, Antelope Valley College Philip Anderson, South Plains College Michael Anderson, West Virginia State University Jeff Anderson, Winona State University Raul Aparicio, Blinn College Judith Barclay, Cuesta College Laurie Battle, Georgia College and State University Annette Benbow, Tarrant County College–Northwest Amy Benvie, Florida Gulf Coast University

Scott Berthiaume, Edison State College Wes Black, Illinois Valley Community College Arlene Blasius, SUNY College of Old Westbury Caroline Maher Boulis, Lee University Amin Boumenir, University of West Georgia Terence Brenner, Hostos Community College Gail Brooks, McLennan Community College G. Robert Carlson, Victor Valley College Hope Carr, East Mississippi Community College Denise Chellsen, Cuesta College Kim Christensen, Metropolitan Community College– Maple Woods Lisa Christman, University of Central Arkansas John Church, Metropolitan Community College–Longview Sarah Clifton, Southeastern Louisiana University David Collins, Southwestern Illinois College Sarah V. Cook, Washburn University Rhonda Creech, Southeast Kentucky Community and Technical College Raymond L. Crownover, Gateway College of Evangelism Marc Cullison, Connors State College Steven Cunningham, San Antonio College Callie Daniels, St. Charles Community College John Denney, Northeast Texas Community College Donna Densmore, Bossier Parish Community College Alok Dhital, University of New Mexico–Gallup

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James JJa aam mees Michael m mes Miichae M cchhae ael Du ael D Dubrowsky ubr brow owsky owsk skk y Wa W Wayne ayn yne Community Com Co mm mmunity mm College Colleg egee Brad Dyer, Community College Bra Br ad D Dyye yer, r, Hazzard Hazzzar Ha a d Co Comm mm mun unit ityy & Technical Co it Coll lleg egee Sally Edwards, Johnson County Coun u ty Community Commu muni nity ty College Coolle C lege g John Jo hn Elliott, E Ell llio iott tt,, St St.. Louis Loui Lo uiss Community Comm Co mmun unit ityy College–Meramec Coll Co lleg ege–M Meramec Gay Ellis, Missouri State University Barbara Elzey, Bluegrass Community College Dennis Evans, Concordia University Wisconsin Samantha Fay, University of Central Arkansas Victoria Fischer, California State University–Monterey Bay Dorothy French, Community College of Philadelphia Eric Garcia, South Texas College Laurice Garrett, Edison College Ramona Gartman, Gadsden State Community College– Ayers Campus Scott Gaulke, University of Wisconsin–Eau Claire Scott Gordon, University of West Georgia Teri Graville, Southern Illinois University Edwardsville Marc Grether, University of North Texas Shane Griffith, Lee University Gary Grohs, Elgin Community College Peter Haberman, Portland Community College Joseph Harris, Gulf Coast Community College Margret Hathaway, Kansas City Community College Tom Hayes, Montana State University Bill Heider, Hibbling Community College Max Hibbs, Blinn College Terry Hobbs, Metropolitan Community College–Maple Woods Sharon Holmes, Tarrant County College–Southeast Jamie Holtin, Freed-Hardeman University Brian Hons, San Antonio College Kevin Hopkins, Southwest Baptist University Teresa Houston, East Mississippi Community College Keith Hubbard, Stephen F. Austin State University Jeffrey Hughes, Hinds Community College–Raymond Matthew Isom, Arizona State University Dwayne Jennings, Union University Judy Jones, Madison Area Technical College Lucyna Kabza, Southeastern Louisiana University Aida Kadic-Galeb, University of Tampa Cheryl Kane, University of Nebraska Rahim Karimpour, Southern Illinois University Edwardsville Ryan Kasha, Valencia Community College David Kay, Moorpark College Jong Kim, Long Beach City College Lynette King, Gadsden State Community College Carolyn Kistner, St. Petersburg College Barbara Kniepkamp, Southern Illinois University Edwardsville Susan Knights, Boise State University Stephanie Kolitsch, University of Tennessee at Martin

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Delphy Shaulis, University of Colorado–Boulder Jennifer Simonton, Southwestern Illinois College David Slay, McNeese State University David Snyder, Texas State University at San Marcos Larry L. Southard, Florida Gulf Coast University Lee Ann Spahr, Durham Technical Community College Jeganathan Sriskandarajah, Madison Area Technical College Adam Stinchcombe, Eastern Arizona College Pam Stogsdill, Bossier Parish Community College Eleanor Storey, Front Range Community College Kathy Stover, College of Southern Idaho Mary Teel, University of North Texas Carlie Thompson, Southeast Kentucky Community & Technical College Bob Tilidetzke, Charleston Southern University Stephen Toner, Victor Valley College

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Thomas Tunnell, Illinois Valley Community College Carol Ulsafer, University of Montana John Van Eps, California Polytechnic State University–San Luis Obispo Andrea Vorwark, Metropolitan Community College–Maple Woods Jim Voss, Front Range Community College Jennifer Walsh, Daytona State College Jiantian Wang, Kean University Sheryl Webb, Tennessee Technological University Bill Weber, Fort Hays State University John Weglarz, Kirkwood Community College Tressa White, Arkansas State University–Newport Cheryl Winter, Metropolitan Community College–Blue River Kenneth Word, Central Texas College Laurie Yourk, Dickinson State University

Acknowledgments I first want to express a deep appreciation for the guidance, comments and suggestions offered by all reviewers of the manuscript. I have once again found their collegial exchange of ideas and experience very refreshing and instructive, and always helping to create a better learning tool for our students. I would especially like to thank Vicki Krug for her uncanny ability to bring innumerable pieces from all directions into a unified whole; Patricia Steele for her eagle-eyed attention to detail; Katie White and Michelle Flomenhoft for their helpful suggestions, infinite patience, tireless efforts, and steady hand in bringing the manuscript to completion; John Osgood for his ready wit and creative energies, Laurie Janssen and our magnificent design team, and Dawn Bercier, the master of this large ship, whose indefatigable spirit kept the ship on course through trial and tempest, and brought us all safely to port. In truth, my hat is off to all the fine people at McGraw-Hill

for their continuing support and belief in this series. A final word of thanks must go to Rick Armstrong, whose depth of knowledge, experience, and mathematical connections seems endless; J. D. Herdlick for his friendship and his ability to fill an instant and sudden need, Anne Marie Mosher for her contributions to various features of the text, Jennifer McNeilly for her review of the Limits Chapter, Mitch Levy for his consultation on the exercise sets, Stephen Toner for his work on the videos, Rosemary Karr for her meticulous work on the solutions manuals, Donna Gerker for her work on the preformatted tests, Jay Miller and Carrie Green for their invaluable ability to catch what everyone else misses; and to Rick Pescarino, Nate Wilson, and all of my colleagues at St. Louis Community College, whose friendship, encouragement and love of mathematics makes going to work each day a joy.

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Making Connections . . . Through Supplements *All online supplements are available through the book’s website: www.mhhe.com/coburn.

Instructor Supplements ⦁

⦁ ⦁

⦁

Computerized Test Bank Online: Utilizing Brownstone Diploma® algorithm-based testing software enables users to create customized exams quickly. Instructor’s Solutions Manual: Provides comprehensive, worked-out solutions to all exercises in the text. Annotated Instructor’s Edition: Contains all answers to exercises in the text, which are printed in a second color, adjacent to corresponding exercises, for ease of use by the instructor. PowerPoint Slides: Fully editable slides that follow the textbook.

Student Supplements ⦁ ⦁

Student Solutions Manual provides comprehensive, worked-out solutions to all of the odd-numbered exercises. Videos • Interactive video lectures are provided for each section in the text, which explain to the students how to do key problem types, as well as highlighting common mistakes to avoid. • Exercise videos provide step-by-step instruction for the key exercises which students will most wish to see worked out. • Graphing calculator videos help students master the most essential calculator skills used in the college algebra course. • The videos are closed-captioned for the hearing impaired, subtitled in Spanish, and meet the Americans with Disabilities Act Standards for Accessible Design.

www.mhhe.com/coburn McGraw-Hill’s MathZone is a complete online homework system for mathematics and statistics. Instructors can assign textbook-specific content from over 40 McGraw-Hill titles as well as customize the level of feedback students receive, including the ability to have students show their work for any given exercise. Assignable content includes an array of videos and other multimedia along with algorithmic exercises, providing study tools for students with many different learning styles. Within MathZone, a diagnostic assessment tool powered by ALEKS® is available to measure student preparedness and provide detailed reporting and personalized remediation. MathZone also helps ensure consistent assignment delivery across several sections through a course administration function and makes sharing courses with other instructors easy. For additional study help students have access to NetTutor™, a robust online live tutoring service that incorporates whiteboard technology to communicate mathematics. The tutoring schedules are built around peak homework times to best accommodate student schedules. Instructors can also take advantage of this whiteboard by setting up a Live Classroom for online office hours or a review session with students. For more information, visit the book’s website (www.mhhe.com/ coburn) or contact your local McGraw-Hill sales representative (www.mhhe.com/rep).

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www.aleks.com ALEKS (Assessment and LEarning in Knowledge Spaces) is a dynamic online learning system for mathematics education, available over the Web 24/7. ALEKS assesses students, accurately determines their knowledge, and then guides them to the material that they are most ready to learn. With a variety of reports, Textbook Integration Plus, quizzes, and homework assignment capabilities, ALEKS offers flexibility and ease of use for instructors. • ALEKS uses artificial intelligence to determine exactly what each student knows and is ready to learn. ALEKS remediates student gaps and provides highly efficient learning and improved learning outcomes • ALEKS is a comprehensive curriculum that aligns with syllabi or specified textbooks. Used in conjunction with McGraw-Hill texts, students also receive links to text-specific videos, multimedia tutorials, and textbook pages. • Textbook Integration Plus allows ALEKS to be automatically aligned with syllabi or specified McGraw-Hill textbooks with instructor chosen dates, chapter goals, homework, and quizzes. • ALEKS with AI-2 gives instructors increased control over the scope and sequence of student learning. Students using ALEKS demonstrate a steadily increasing mastery of the content of the course. • ALEKS offers a dynamic classroom management system that enables instructors to monitor and direct student progress towards mastery of course objectives.

ALEKS Prep/Remediation: • Helps instructors meet the challenge of remediating unequally prepared or improperly placed students. • Assesses students on their pre-requisite knowledge needed for the course they are entering (i.e. Calculus students are tested on Precalculus knowledge). • Based on the assessment, students are prescribed a unique and efficient learning path specific to address their strengths and weaknesses. • Students can address pre-requisite knowledge gaps outside of class freeing the instructor to use class time pursuing course outcomes.

Electronic Textbook: CourseSmart is a new way for faculty to find and review eTextbooks. It’s also a great option for students who are interested in accessing their course materials digitally and saving money. CourseSmart offers thousands of the most commonly adopted textbooks across hundreds of courses from a wide variety of higher education publishers. It is the only place for faculty to review and compare the full text of a textbook online, providing immediate access without the environmental impact of requesting a print exam copy. At CourseSmart, students can save up to 50% off the cost of a print book, reduce their impact on the environment, and gain access to powerful web tools for learning including full text search, notes and highlighting, and email tools for sharing notes between classmates. www.CourseSmart.com Primis: You can customize this text with McGraw-Hill/Primis Online. A digital database offers you the flexibility to customize your course including material from the largest online collection of textbooks, readings, and cases. Primis leads the way in customized eBooks with hundreds of titles available at prices that save your students over 20% off bookstore prices. Additional information is available at 800-228-0634.

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Contents Preface vi Index of Applications

CHAPTER

1

xxxv

Equations and Inequalities

1

1.1 Linear Equations, Formulas, and Problem Solving 2 Technology Highlight: Using a Graphing Calculator as an Investigative Tool 9

1.2 Linear Inequalities in One Variable 14 1.3 Absolute Value Equations and Inequalities 24 Technology Highlight: Absolute Value Equations and Inequalities 28 Mid-Chapter Check 31 Reinforcing Basic Concepts: Using Distance to Understand Absolute Value Equations and Inequalities 32

1.4 Complex Numbers 33 1.5 Solving Quadratic Equations 42 Technology Highlight: The Discriminant 51

1.6 Solving Other Types of Equations 56 Summary and Concept Review 70 Mixed Review 75 Practice Test 75 Calculator Exploration and Discovery: Evaluating Expressions and Looking for Patterns 76 Strengthening Core Skills: An Alternative Method for Checking Solutions to Quadratic Equations 77 Connections to Calculus: Solving Various Types of Equations; Absolute Value Inequalities and Delta/Epsilon form 79

CHAPTER

2

Relations, Functions, and Graphs 83 2.1 Rectangular Coordinates; Graphing Circles and Other Relations 84 Technology Highlight: The Graph of a Circle

92

2.2 Graphs of Linear Equations 97 Technology Highlight: Linear Equations, Window Size, and Friendly Windows 105

2.3 Linear Graphs and Rates of Change 110 2.4 Functions, Function Notation, and the Graph of a Function 122 Mid-Chapter Check 137 Reinforcing Basic Concepts: The Various Forms of a Linear Equation

138

2.5 Analyzing the Graph of a Function 138 Technology Highlight: Locating Zeroes, Maximums, and Minimums 149

2.6 The Toolbox Functions and Transformations 157 Technology Highlight: Function Families

166

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2.7 2 .7 Piecewise-Defined Piecewisse-Defined Functions 172 Technology Te ech c no nolo l gy lo g Highlight: Piecewise-Defined Functions

179

2.8 2. .8 TThe he A Alg Algebra lgebra and Composition of Functions 186 Teec nology Highlight: Composite Functions 196 TTechnology Tech Summary and Concept Review 202 Mixed Review 209 Practice Test 211 Calculator Exploration and Discovery: Using a Simple Program to Explore Transformations 212 Strengthening Core Skills: Transformations via Composition 213 Cumulative Review: Chapters 1–2 214 Connections to Calculus: Rates of Change and the Difference Quotient; Transformations and the Area Under a Curve 215

CHAPTER

3

Polynomial and Rational Functions 219 3.1 Quadratic Functions and Applications 220 Technology Highlight: Estimating Irrational Zeroes 225

3.2 Synthetic Division; The Remainder and Factor Theorems 230 3.3 The Zeroes of Polynomial Functions 242 Technology Highlight: The Intermediate Value Theorem and Split Screen Viewing 252

3.4 Graphing Polynomial Functions 257 Mid-Chapter Check 271 Reinforcing Basic Concepts: Approximating Real Zeroes

271

3.5 Graphing Rational Functions 272 Technology Highlight: Rational Functions and Appropriate Domains

282

3.6 Additional Insights into Rational Functions 289 Technology Highlight: Removable Discontinuities

297

3.7 Polynomial and Rational Inequalities 303 Technology Highlight: Polynomial and Rational Inequalities

310

3.8 Variation: Function Models in Action 316 Summary and Concept Review 326 Mixed Review 331 Practice Test 332 Calculator Exploration and Discovery: Complex Zeroes, Repeated Zeroes, and Inequalities 334 Strengthening Core Skills: Solving Inequalities Using the Push Principle 334 Cumulative Review: Chapters 1–3 335 Connections to Calculus: Graphing Techniques 337

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Exponential and Logarithmic Functions 341 4.1 One-to-One and Inverse Functions 342 Technology Highlight: Investigating Inverse Functions

349

4.2 Exponential Functions 354 Technology Highlight: Solving Exponential Equations Graphically

361

4.3 Logarithms and Logarithmic Functions 366 Mid-Chapter Check 379 Reinforcing Basic Concepts: Linear and Logarithm Scales 380

4.4 Properties of Logarithms; Solving Exponential/Logarithmic Equations 381 4.5 Applications from Business, Finance, and Science 397 Technology Highlight: Exploring Compound Interest 404 Summary and Concept Review 410 Mixed Review 414 Practice Test 415 Calculator Exploration and Discovery: Investigating Logistic Equations 416 Strengthening Core Skills: Understanding Properties of Logarithms 418 Cumulative Review: Chapters 1–4 418 Connections to Calculus: Properties of Logarithms; Area Functions; Expressions Involving ex 421

CHAPTER

5

An Introduction to Trigonometric Functions 425 5.1 Angle Measure, Special Triangles, and Special Angles 426 5.2 Unit Circles and the Trigonometry of Real Numbers 440 5.3 Graphs of Sine and Cosine Functions; Cosecant and Secant

rain drop

5.4

5.5 5.6 5.7

Functions 454 Technology Highlight: Exploring Amplitudes and Periods 465 Graphs of Tangent and Cotangent Functions 471 Technology Highlight: Zeroes, Asymptotes, and the Tangent/Cotangent Functions 478 Mid-Chapter Check 484 Reinforcing Basic Concepts: Trigonometry of the Real Numbers and the Wrapping Function 485 Transformations and Applications of Trigonometric Graphs 486 Technology Highlight: Locating Zeroes, Roots, and x-Intercepts 494 The Trigonometry of Right Triangles 499 Trigonometry and the Coordinate Plane 512 Summary and Concept Review 522 Mixed Review 530 Practice Test 532 Calculator Exploration and Discovery: Variable Amplitudes and Modeling the Tides 534 Strengthening Core Skills: Standard Angles, Reference Angles, and the Trig Functions 535 Cumulative Review: Chapters 1–5 536 Connections to Calculus: Right Triangle Relationships; Algebraic Form and Polar Form 539 Contents

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Trigonometric Identities, Inverses, and Equations 543 6.1 6.2 6.3 6.4

Fundamental Identities and Families of Identities 544 Constructing and Verifying Identities 552 The Sum and Difference Identities 558 The Double-Angle, Half-Angle, and Product-to-Sum Identities 568 Mid-Chapter Check 580 Reviewing Basic Concepts: Identities—Connections and Relationships 581 6.5 The Inverse Trig Functions and Their Applications 582 Technology Highlight: More on Inverse Functions 592 6.6 Solving Basic Trig Equations 599 Technology Highlight: Solving Equations Graphically 605 6.7 General Trig Equations and Applications 610 Summary and Concept Review 619 Mixed Review 623 Practice Test 625 Calculator Exploration and Discovery: Seeing the Beats as the Beats Go On 626 Strengthening Core Skills: Trigonometric Equations and Inequalities 627 Cumulative Review: Chapters 1–6 628 Connections to Calculus: Simplifying Expressions Using a Trigonometric Substitution; Trigonometric Identities and Equations 630

CHAPTER

7

Applications of Trigonometry

633

7.1 Oblique Triangles and the Law of Sines 634 7.2 The Law of Cosines; the Area of a Triangle 646 7.3 Vectors and Vector Diagrams 658 Technology Highlight: Vector Components Given the Magnitude and the Angle ␪ 668 Mid-Chapter Check 673 Reinforcing Basic Concepts: Scaled Drawings and the Laws of Sine and Cosine 673 7.4 Vector Applications and the Dot Product 674 7.5 Complex Numbers in Trigonometric Form 687 7.6 De Moivre’s Theorem and the Theorem on nth Roots 698 Summary and Concept Review 705 Mixed Review 709 Practice Test 710 Calculator Exploration and Discovery: Investigating Projectile Motion 712 Strengthening Core Skills: Vectors and Static Equilibrium 713 Cumulative Review: Chapters 1–7 713 Connections to Calculus: Trigonometry and Problem Solving; Vectors in Three Dimensions 715

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Systems of Equations and Inequalities

719

8.1 Linear Systems in Two Variables with Applications 720 Technology Highlight: Solving Systems Graphically

727

8.2 Linear Systems in Three Variables with Applications 732 Technology Highlight: More on Parameterized Solutions

739

8.3 Partial Fraction Decomposition 743 8.4 Systems of Inequalities and Linear Programming 755 Technology Highlight: Systems of Linear Inequalities 763 Mid-Chapter Check 768 Reinforcing Basic Concepts: Window Size and Graphing Technology 768

8.5 Solving Linear Systems Using Matrices and Row Operations 769 Technology Highlight: Solving Systems Using Matrices and Calculating Technology 775

8.6 The Algebra of Matrices 780 8.7 Solving Linear Systems Using Matrix Equations 791 8.8 Applications of Matrices and Determinants: Cramer’s Rule, Geometry, and More 806 Summary and Concept Review 814 Mixed Review 819 Practice Test 820 Calculator Exploration and Discovery: Optimal Solutions and Linear Programming 822 Strengthening Core Skills: Augmented Matrices and Matrix Inverses 823 Cumulative Review: Chapters 1–8 825 Connections to Calculus: More on Partial Fraction Decomposition; The Geometry of Vector and Determinants 827

CHAPTER

9

Analytical Geometry and the Conic Sections 831 9.1 A Brief Introduction to Analytic Geometry 832 9.2 The Circle and the Ellipse 839 9.3 The Hyperbola 852 Technology Highlight: Studying Hyperbolas

861

9.4 The Analytic Parabola 866 Mid-Chapter Check 874 Reinforcing Basic Concepts: Ellipses and Hyperbolas with Rational/Irrational Values of a and b 875

9.5 Nonlinear Systems of Equations and Inequalities 875 9.6 Polar Coordinates, Equations, and Graphs 883 9.7 More on Conic Sections: Rotation of Axes and Polar Form 896 Technology Highlight: Investigating the Eccentricity e

907

9.8 Parametric Equations and Graphs 913 Technology Highlight: Exploring Parametric Graphs 919 Summary and Concept Review 924 Mixed Review 929

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Practice Test 930 Calculator Exploration and Discovery: Conic Rotations in Polar Form 931 Strengthening Core Skills: Simplifying and Streamlining Computations for the Rotation of Axes 932 Cumulative Review: Chapters 1–9 934 Connections to Calculus: Polar Graphs and Instantaneous Rates of Change; Systems of Polar Equations 935

CHAPTER

10

Additional Topics in Algebra 939 10.1 Sequences and Series 940 Technology Highlight: Studying Sequences and Series

945

10.2 Arithmetic Sequences 949 10.3 Geometric Sequences 956 10.4 Mathematical Induction 966 Mid-Chapter Check 973 Reinforcing Basic Concepts: Applications of Summation

974

10.5 Counting Techniques 975 Technology Highlight: Calculating Permutations and Combinations

981

10.6 Introduction to Probability 987 Technology Highlight: Principles of Quick-Counting, Combinations, and Probability 992

10.7 The Binomial Theorem 999 Summary and Concept Review 1007 Mixed Review 1011 Practice Test 1013 Calculator Exploration and Discovery: Infinite Series, Finite Results Strengthening Core Skills: Probability, Quick Counting, and Card Games 1015 Cumulative Review: Chapters 1–10 1016 Connections to Calculus: Applications of Summation 1019

CHAPTER

11

1014

Bridges to Calculus–An Introduction to Limits 1023 11.1 An Introduction to Limits Using Tables and Graphs 1024 11.2 The Properties of Limits 1034 Mid-Chapter Check 1043

11.3 Continuity and More on Limits 1044 11.4 Applications of Limits: Instantaneous Rates of Change and the Area Under a Curve 1056 Summary and Concept Review 1066 Mixed Review 1069 Practice Test 1070 Cumulative Review: Chapters 1–11 1071

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Appendix I

A Review of Basic Concepts and Skills

A-1

Appendix II

The Language, Notation, Numbers of Mathematics A-1 Algebraic Expressions and the Properties of Real Numbers A-8 Exponents, Scientific Notation, and a Review of Polynomials A-14 Factoring Polynomials A-24 Rational Expressions A-31 Radicals and Rational Exponents A-40 More on Synthetic Division A-54

Appendix III

More on Matrices

Appendix IV

Deriving the Equation of a Conic

Appendix V

Selected Proofs A-60

Appendix VI

Families of Polar Curves A-63

A-56 A-58

Student Answer Appendix SA-1 Instructor Answer Appendix (AIE only) Index I-1

IA-1

Additional Topics Online (Visit www.mhhe.com/coburn) R.7 Geometry Review with Unit Conversions R.8 Expressions, Tables and Graphing Calculators 5.0 An Introduction to Cycles and Periodic Functions 7.7 Complex Numbers in Exponential Form 7.8 Trigonometry, Complex Numbers and Cubic Equations 10.8 Conditional Probability and Expected Value 10.9 Probability and the Normal Curve with Applications 11.2B Properties of Limits with an Introduction to the Precise Definition

Contents

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Precalculus

Coburn’s Precalculus Series College Algebra, Second Edition Review 䉬 Equations and Inequalities 䉬 Relations, Functions, and Graphs 䉬 Polynomial and Rational Functions 䉬 Exponential and Logarithmic Functions 䉬 Systems of Equations and Inequalities 䉬 Matrices 䉬 Geometry and Conic Sections 䉬 Additional Topics in Algebra ISBN 0-07-351941-3, ISBN 978-0-07351941-8

College Algebra Essentials, Second Edition Review 䉬 Equations and Inequalities 䉬 Relations, Functions, and Graphs 䉬 Polynomial and Rational Functions 䉬 Exponential and Logarithmic Functions 䉬 Systems of Equations and Inequalities ISBN 0-07-351968-5, ISBN 978-0-07351968-5

Algebra and Trigonometry, Second Edition Review 䉬 Equations and Inequalities 䉬 Relations, Functions, and Graphs 䉬 Polynomial and Rational Functions 䉬 Exponential and Logarithmic Functions 䉬 Trigonometric Functions 䉬 Trigonometric Identities, Inverses and Equations 䉬 Applications of Trigonometry 䉬 Systems of Equations and Inequalities 䉬 Matrices 䉬 Geometry and Conic Sections 䉬 Additional Topics in Algebra ISBN 0-07-351952-9, ISBN 978-0-07-351952-4

Precalculus, Second Edition Equations and Inequalities 䉬 Relations, Functions, and Graphs 䉬 Polynomial and Rational Functions 䉬 Exponential and Logarithmic Functions 䉬 Trigonometric Functions 䉬 Trigonometric Identities, Inverses and Equations 䉬 Applications of Trigonometry 䉬 Systems of Equations and Inequalities, and Matrices 䉬 Geometry and Conic Sections 䉬 Additional Topics in Algebra 䉬 Limits ISBN 0-07-351942-1, ISBN 978-0-07351942-5

Trigonometry, Second Edition—Coming in 2010! Introduction to Trigonometry 䉬 Right Triangles & Static Trigonometry 䉬 Radian Measure & Dynamic Trigonometry 䉬 Trigonometric Graphs and Models 䉬 Trigonometric Identities 䉬 Inverse Functions and Trigonometric Equations 䉬 Applications of Trigonometry 䉬 Trigonometric Connections to Algebra ISBN 0-07-351948-0, ISBN 978-0-07351948-7

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Index of Applications ARCHITECTURE/DESIGN Capitol Building, 531 Chrysler Building, 924 CN Tower, 509, 731 corporate headquarters design, 930 decorative fireplaces, 850 Eiffel Tower, 509, 731, 924 elliptical arches, 850 flagpole height, 709 fortress height, 644 Great Pyramid of Giza, 706 indoor waterfall height, 537 John Hancock building height, 814 lawn dimensions, A–13 Petronas Tower, 509 pitch of a roof, 107 rafter length of roof, 644 roof pitch, A–13 Sears Tower, 618, 814 seating arrangements, 977–978, 983, 985 service door width, 530 St. Louis Arch, 1071 Stratosphere Tower, 551 suspension bridges, 12 Taipei 101, 510 Washington Monument, 522, 673 water tower height, 883 window area, 71 window height, 505

ART/FINE ARTS/THEATER art and mathematics, 566 art show viewing angles, 597 candle-making, 768 cartoon characters, 803 company logo design, 609 cornucopia composition, 1013 decorative gardens, 850 Mozart arias, 804 origami, 579 original value of collector’s items, 774–775 rare books, 778, 1013 rare coins, 1013 Rolling Stones, 803–804 seating capacity, 953, 955 showroom design, 850 soft drinks sold, 800 ticket sales, 56

BIOLOGY/ZOOLOGY allometry, 74, 76 animal birth weight, 416, 948

animal gestation periods, 742 animal lifespan, 1014 animal seasonal weight of a bear, 1012 animal sinusoidal movement, 469 animal territories, 895 bacteria growth, 364, 408, 964 endangered species list, 172 fruit fly population, 403 insect movement, 218 insect population, 269–270, 338–339, 496 nocturnal animals, 91–92 scavenger bird flight path, 892 species preservation, 948 temperature and cricket chirps, 109 wildlife population fluctuations, 488, 495, 496 wildlife population growth, 69, 408, 414, 955 wingspan of birds, 742

BUSINESS/ECONOMICS account balance/service fees, 22 advertising and sales, 182, 374, 378, 394, 415 annual profit, 974 annual sales, 598 annuities, 401, 407–408, 414, 416 auction purchases, 742 balance of payments, 270 billboard design, 657 billboard viewing angle, 597, 715–716 break-even analysis, 726, 730, 879, 881–882 business loans, 738–739 canned goods cost, 821 car/moving van rental cost, 23, 135, 208 cell phone charges, 68, 184–185 cereal package weight, 31 commission sales, 104–105 copper tubing cost, 324, 332 cost/revenue/profit, 54, 73, 194, 199, 200, 242, 336 customer service, 996, 1013–1014 depreciation, 12, 108, 116–117, 121, 361, 364, 395, 411, 416, 883, 948, 964, 1013, 1055 DVD rentals, 411, 1006 employee productivity, 1016, 1065 envelope size, 67 equation models revenue, 64, 365 equipment aging, 964 food ordering cost, 790 fruit cost, 813

gasoline cost, 769 gifts to grandchildren, 766 gold coins, 742 guns versus butter revenue, 766 home appreciation, 379, 395, 1012 hourly wage, 948 housecleaning business, 64 households holding stock, 183 inflation, 365, 948, 965 inimizing cost, 762–763, 767 internet commerce, 325 land sales, 714 manufacturing cost, 287, 294–295, 299–300, 329–331, 333, 790, 911 market demand/consumer interest, 286 market share, 390–391 maximizing profit/revenue, 74–76, 200, 223–224, 227–229, 610, 628, 761–762, 766, 767, 816, 821, 825, 865, A–23, A–53 merchandise revenue, 64, 813 mileage rate, 28, 109, 130, 135, 316 minimum wage, 101, A–13 mixture exercises, 8–9, 13, 71, 110, 725, 729, 742, 768, 815 natural gas prices, 183 overtime wage, 184 packaging material cost, 300 paper size, 62, 67, 821 personnel decisions, 983, 984 phone service charges, 184 plant production, 789 pollution removal cost, 1060–1061 postage cost, 179, 184, A–13 pricing for undeveloped lots, 657 pricing strategies, 228 printing and publishing, 68, 300–301 profit/loss, 137, 153 quality control stress test, 30 recycling cost, 285–286 repair cost, A–13 research and development, 199 resource allocation, 804 running shoes cost, 68 salary calculations, 20, 322, 965, 1008 sales goals, 104, 789, 955 seasonal revenue, 76, 617, 623, 624, 626, 934 service call cost, 135 sizes of merchandise revenue, 790–791, 800, 803, 813 stock prices, A–39 stock purchase, 945

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stock value, 30 student loans, 964 supply and demand, 323, 324, 730, 882, 934 textbook shipments, A–30 typing speed, A–39 USPS package size regulations, 12 vehicle sales, 198 video game cost, 924 working in pairs, 64–65 work per unit time, 107

CHEMISTRY chemical mixtures, 13, 742 concentration and dilution, 287 pH levels, 376, 378 photochromatic sunglasses, 365

COMMUNICATION cable installations, 779 cable length, A–49 cell phone subscriptions, 55 e-mail addresses, 986 Internet connections, 121 Internet rebates, 783–784 parabolic dish, 870, 873 phone call volume, 324–325 phone numbers, 985 phone service charges, 184 radio broadcast range, 96, 874 radio tower cables, 566 spending on media, 95 television programming, 985 touch-tone phones, 575, 579 walkie-talkie range, 711

COMPUTERS animations, 955 consultant salaries, 30 elastic rebound, 965 ownership, 996–997

CONSTRUCTION bathroom tile pattern, 530 building codes, 296–297 building height, 511 ceiling fan angles, 521, 522 condominiums in New York, 653 deck dimensions, 882 fence an area, 229 flooded basement, 74 home cost per square foot, 106 home improvement, 789–790

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home ventilation, 378 lift capacity, 22 manufacturing cylindrical vents, 881 pitch of a roof, 107 ramp support height, 529 safe load, 323, 325, 333, 396 sewer line slope, 107 tool rental, 23, 1018

CRIMINAL JUSTICE/LEGAL STUDIES prison population, 121 speeding fines, 352 stopping distance, 171, 324

DEMOGRAPHICS age reporting, 184 committee composition, 981, 984, 985 customers at the mall, 327 eating out, 121 fighter pilot training, 271 Goldsboro, 354, 365, 409 households holding stock, 183 Internet connections, 121 lottery numbers, 980, 981 military expenditures, 184 military volunteer service, 786–787 minimizing response time, 79–80 multiple births, 183, 418 newspapers published, 182 opinion polls, 1006 paperback book thickness, 322 pencil use, 1071 per capita spending on police protection, 177–178 population density, 285 population doubling time, 354, 964 population growth, 394, 409, 754, 964, 1065; A–39 raffle tickets, 1012 spending on media, 95 surveillance camera, 657 tablecloth border, 851 tourist population, 240–241 visitor attendance, 617

EDUCATION/TEACHING average grades, 19 college costs, 109 course scheduling, 982, 983 credit hours taught, 1009 grade point averages, 22, 288

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learning curves, 395 memory retention, 286, 378, A–39 scholarships, 985 Stooge IQ, 778 true/false quizzes, 997

ENGINEERING Civil traffic and travel time, 336 traffic volume, 30, 269 Electrical AC circuits, 41, 470, 566, 693–697, 704, 709, 710 impedance calculations, 41, 510, 694, 697 relay stations, 967–968 resistance, 314–315, 324, 767, 852 switches, 997–998 voltage calculations, 41 Mechanical kinetic energy, 323 machine gears, 579 parabolic reflectors, 872, 873 solar furnace, 873 trolley cable length, 410 velocity, 438–439, 524, 531, 532, 965 wind-powered energy, 69, 171, 324, 353, A–49 – A–50

ENVIRONMENTAL STUDIES chemical waste removal, 282 cleanup time, 324 endangered species list, 172 energy rationing, 183 forest fires, 201 fuel consumption, 320–321 ice thickness, 617 landfill volume, 107 oil spills, 195 pollution removal, 65, 69, 285, 288, 333, 1060–1061 pollution testing, 1006 recycling cost, 285–286 stocking a lake, 394, 714, 948 storage barrels, 974, 1017 sulphur dioxide emissions, 628 waste production, 629 water rationing, 183 water usage, 533 wildlife population growth, 69 wind-powered energy, 171, 324, 353

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FINANCE annuities, 401, 407–408, 414, 416 charitable giving, 790, 955 compound interest, 200, 210, 397–399, 404–408, 410, 414–416, 440, 714, 1006, 1043–1044 debt load, 241 diversifying investments, 742, 758–759 heating and cooling subsidies, 23 inheritance tax, 409 interest compounded continuously, 400, 407, 537, 629, 825, 1017, 1043–1044 interest earnings, 108, 324, 378 investment growth, 376, 407, 821 investment in coins, 121 investment returns, 731, 812 investment strategies, 399, 779 investment value, 266 mortgage interest, 154, 408 mortgage payments, 408 simple interest, 397, 406, 414 sinking funds, 402, 408

GEOGRAPHY/GEOLOGY avalanche conditions, 596 Bermuda Triangle, 711 canyon height, 509 canyon width, 645 cliff height, 429, 504–505 continents range of altitude, 23 contour maps, 510, 531 Coral Triangle, 712 cradle of civilization, 12 distance between major cities, 314, 433, 438, 511, 643, 655, 673, 778 distance between states, 532 distance to fire tower, 645, 710 earthquake epicenter, 95 earthquake magnitude, 372, 373, 377, 380, 412, 415 home location, 911 land area of island nations, 731, 821 land area of various states, 7 land tract dimensions, 882 map distance, 644, 645 mountain height, 628, 645 Mount Rushmore height, 527 Mount Tortolas height, 714 natural gas prices, 183 Nile River Delta, 657 oceanography, 251 rock formation width, 934 tall structure height, 482

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terrain model, 339–340 tidal waves, 468 tide prediction, 136, 469, 533, 534–535 topographical maps, 439 tree height, 509, 705 Triangle Peak height, 711 volcano width, 649 Yukon Territory, 657

HISTORY American Flag dimensions, 882 child prodigies, 768 Civil Rights Act, 742 Declaration of Independence, 742 famous composers, 804 important dates in U.S. history, 731 major wars, 742 postage cost, 179, 184 Statue of Liberty, 322 Zeno’s Paradox, 1014–1015

INTERNATIONAL STUDIES currency conversion, 200–201 shoe sizing, 23, 200 telephone traffic, 51

MATHEMATICS angle of intersection, 520, 551 angle of rotation, 452–453 angle of slice of bread, 529 arc length, 872, 955 area of circle, 122, 353 circular segment, 609, 623 ellipse, 849 equilateral triangle, 322, 438, 696 first quadrant triangle, 299 frustum, A–50 inscribed circle, 96 inscribed polygon, 1024–1025, 1031 inscribed square, 95 inscribed triangle, 96 a lune, 484 Norman window, 812 parabolic segment, 872, 874, 882 parallelogram, 520, 812 polygon, 483, 550, 551, 1031 rectangle, 23, 67, 74, 214, 717, 766, 789 a sector, 438 triangle, 23, 521, 557, 625, 687, 711, 766, 778, 812

average rate of change, 131, 145, 155–156, 211, 242 circle properties, 838 circumscribed triangles, 645 clock angles, 522, 579, 624 collinear points, 812 combinations and permutations book selection, 984, 1013 coins, 979, 982, 984 colored balls, 984, 1013 committee formation, 981, 984, 985 cornucopia composition, 1013 course schedules, 982, 983 finishing a race, 978, 984 free-throws, 975 horse racing, 983, 1010 key rings, 1012 letters, 983, 1010, 1012 license plates, 982, 984, 1013 lock, 976, 982, 1010 lottery numbers, 980, 981 menu items, 981, 982, 984 numbers, 982–983 outfits, 982, 1010 personnel, 983, 984, 1010 photo arrangements, 979, 983, 1010 report cards, 982 Scrabble, 977, 979, 983 seating arrangements, 977–978, 983, 985 security code PIN, 976, 982, 1072 song selection, 984 spinner, 975, 982 team members, 984, 986 tournament finalists, 983 complex numbers, 41 absolute value, 41, 182, 185, 254 binomial cubes, 41, 696 Girolamo Cardano, 41 square roots, 42, 254 complex polynomials, 55 composite figures, 68 conic sections degenerate cases, 865 hyperbola, 152 rotation of, 911 volume, A–30 consecutive integers, 12, 23, 67, 69 curve fitting, 804 cylindrical tank dimensions, 882 difference identity, 567

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difference quotient, 215–216, 567, 1056–1057 discriminant of cubic equation, 704 quadratic, 51–52, 55 reduced cubic, 313 distance from point to line, 837 distance from point to plane, 741 equipoise cylinder, 348–349 factorial formulas, 984 focal chords ellipse, 847, 906, 911, 928 hyperbola, 864 parabola, 873 folium of Descartes, 313, 922 friction, 483 geometric formulas, 510, 778, 779, 812 geometry, 62–63 Heron’s Formula, 653, 657, 712, 1034 instantaneous rate of change, 1058–1061, 1065, 1068, 1069 inverse of matrix, 803 involute of circle, 521 joint angles, 672, 711 law of cosines, 655 linear equation, 120–121 line segment distance, 453, 483 mathematics and art, 415 maximum and minimum values, 255 number puzzles, 62, 67 perimeter of elllipse, 849 hexagon, 657 pentagon, 657 polygon, 551 rectangle, 789 trapezoid, 609, 655 triangle, 656, 657, 687 Pick’s theorem, 135 polar coordinates, 894 polar curves and cosine, 895 polar form of an ellipse, 910 equation of a line, 910 equation of a rose, 930 polar graphs, 894–895 powers and roots, 704 powers of the imaginary unit, 972 probability appointment scheduling, 1011, 1013–1014 balls in a bag, 997 binomial, 1005

xxxviii

coin toss, 365, 998 dice, 987–990, 993–995, 1010, 1012 dominos, 994 drawing a card, 987–991, 993–995, 1010, 1012 emergency response routes, 994 fair coin, 30, 993, 996 free-throws, 1004 full production status, 994 minutes on hold, 996 money, 1017–1018 points in quadrants, 997 selection family member, 988 group, 990–991, 993, 995, 998 letter, 989, 998 location, 994 number, 994, 996, 998 person with certain characteristics, 992–995, 998, 1012, 1014 pool balls, 995, 996 spinning a spinner, 364, 988, 993, 994, 996 winning the lottery, 365, 998 Pythagorean Theorem, 468, 536, 557, 567 Pythagorean triples, 452 quadratic solutions, 77 quartic polynomials, 269 radius of a ripple, 201, 209 radius of a sphere, 352 radius of circumscribed circle, 643 range of projectiles, 318, 574, 579, 598, 608, 624, 685, 712 rational equation, 65 rational function, 742 rectangle method, 1019–1021, 1068–1070, 1072 rectangular solid dimensions, 741 regressions and parameters, 923 right triangle, 438, 523 sand dune function, 182 semi-hyperbola equation, 864 similar triangles, 436 slant height, 616–617 special triangles, 436, 437, 439, 511, 523 Spiral of Archimedes, 521 Stirling’s Formula, 984, 986 sum of consecutive cubes, 315 consecutive squares, 315 cubes of the first n natural numbers, 964 first n cubes, 972

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first n natural numbers, 954 first n terms of the sequence, 947 natural numbers, 972 perfect numbers, 998 sum-to-product identities, 575, 624 surface area of cone, 67, 882, A–50 cube, 322 cylinder, 11, 54–55, 156, 200, 299, 301, 882 open cylinder, 302 pyramid, 706 sphere, 322–323, 882 spherical cap, 301–302 tangent functions, 482 telescoping sums, 754 trigonometric form of linear equation, 616 trigonometric graphs, 153 triple angle formula for sine, 642 USPS package size regulations, 12 vertex/intercept formula, 227 volume of an egg, 325 box, A–30 cone, 353, 616, 766, 882 cube, 135, 255 cylinder/cylindrical shells, 135, 156, 301, 616, 766, 882 hemispherical wash basin, 63 open box, 240, 332 prism, 812 pyramid, 812 rectangular box, 227, 255 sphere, 170, 882 sphere circumscribed by cylinder, 13 spherical cap, 301–302 spherical shell, A–30 Witch of Agnesi, 922 working with identities, 579, 581–582

MEDICINE/NURSING/ NUTRITION/DIETETICS/HEALTH animal diets, 804 appointment scheduling, 1011 bacteria response to antibiotics, 1065 board of directors, 998 body mass index, 22 deaths due to heart disease, 210 eardrum pressure, 566 female physicians, 109 fertility rates, 136 fetus weight, 136

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heart rate, 618 height vs. weight, 107 hodophobia, 998 human life expectancy, 108 ideal weight, 135 lithotripsy, 850 medication in the bloodstream, 286, 333, 395 multiple births, 183, 418 prescription drugs, 121 saline mixtures, 8–9 smokers, 109 time of death, 394 training diets, 804–805

METEOROLOGY altitude and atmospheric pressure, 373, 377, 378 atmospheric pressure, 391–392, 413 jet stream altitude, 30 monthly rainfall, 528, 530 rainbow height, 510 reservoir water levels, 270, 1009, 1066 river discharge rate, 238, 613–614, 617 storm location, 860 temperature and altitude, 116, 352, 394 and atmospheric pressure, 373, 377 conversions, 135 drop, 155, A–7 fluctuation, 487, 495, 537, 955, A–7 mountain stream, 617 record low, 626 scale agreement, 729 water level during drought, 108

MUSIC arias, 804 composers, 804 Rolling Stones, 803–804

PHYSICS/ASTRONOMY/ PLANETARY STUDIES acceleration, 121, 608 airline altitude change, 100 boiling temperature of water, 108 Boyle’s Law, 319–320 Brewster’s law, 566 charged particles, 324, 864, A–23 climb rate, aircraft, 107 comet orbit, 912, 930 comet path, 859–860 creating a vacuum, 965

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daylight hours, 495, 497, 498 deflection of a beam, 314 density of objects, 256 depth of a dive, 228, 332 distance between planets, 12, 453, 639–640, 643 distance to equator, 470, 484 drag resistance on a boat, 255 elastic rebound, 965 electron motion, 923 fluid mechanics, 566 fluid motion, 170, A–31 force and work measurement, 671, 677– 678, 685, 686, 707, 708, 710, 711 force on inclined plane, 595 gravity effects of, 323 free-fall, 147–148, 170–171, 211, 322, 353, 999 heat flow on cylindrical pipe, 596 hydrostatics, 468 illumination of a surface, 508, 550, 717 instantaneous velocity, 1058, 1064, 1065, 1069, 1070 Kepler’s third law of planetary motion, A–49 light beam distance, 477–478 light beam velocity, 484 light intensity, 325, 557, A–23 magnets attractive force, 536 Malus’ law, 578 mixture exercises, 13 motion dectection, 643 Newton’s Law of Cooling, 360, 364, 395 Newton’s law of universal gravitation, 325 nuclear power, 864 particle motion, 80, 923, 924 pendulums, 324, 331, 497, 961, 964, 1013 planet orbits, 69, 439, 453, 469, 850–851, 905–906, 910–911, 919, 928–932, 1016 position of reflected image, 482 projected image, 322, 352 projectile height, 50, 54–55, 68, 73, 74, 155, 215, 224, 228, 229, 326, 598, 821, 924, 1064 range, 318, 579, 598, 608, 624, 685, 712, 918 velocity, 145–146, 153, 155, 574, 598, 624, 671, 683, 686, 708, 711 radar detection, 95, 640, 643

radioactive Carbon-14 dating, 409 radioactive decay, 365, 413, 419, 484 radioactive half-life, 396, 404, 409, 419 radio broadcast range, 96, 874 satellite orbit, 597 Snell’s law, 608–609 sound intensity, 377, 379, 380 sound waves, 492, 493, 498 spaceship velocity, 395 space-time relationships, A–30 – A–31 speed of sound, 120, 578 spring oscillation, 30, 497 star intensity, 377 static equilibrium, 684, 685, 710, 713, 714 supernova expansion, 201 temperature scales, 135, 468–469 tethered plane, 706 tethered weight, 710 thermal conductivity, 804 traveling waves, 566 uniform motion, 8, 12, 68, 71, 110, 726, 729, 730–731 vectors, 659, 666, 667, 669, 671, 677–678, 683, 684, 708, 713, 716, 717, 865 velocity of a particle, 309–310 visible light, 469 volume and pressure, 319–320 wave motion, 497 weight on other planets, 317, 324

POLITICS conservatives and liberals in senate, 201 dependency on foreign oil, 183 federal deficit (historical data), 154 government deficits, 255, 333 government investment, 198 military expenditures, 184 Supreme Court Justices, 107 tax reform, 731 U.S. International trade balance, 55

SOCIAL SCIENCE/HUMAN SERVICES memory retention, 286, 378

SPORTS/LEISURE admission price, 184 amusement park attendance, 241 arcades, 184 archery, 922, 997, 1014 athletic performance, 325 average bowling score, 76, 825

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baseball, 922 card value, 108 exponential decay of pitcher’s mound, 364 basketball final scores, 778 free-throw percentage, 975, 1004 height of players, 75 batting averages, 1006 bicycle speed, 434, 440 butterfly stroke, 22 carnival games, 521 circus clowns, 567, 598, 820 club memberships, 790 Clue, 985 darts, 997 distance to a target, 643–644 distance to movie screen, 557, 591–592, 718 Ellipse Park, 847 ferris wheel, 610 fishing, 30 fish tank dimensions, 882 fitness club membership, 820 football, 922–923 football player weight, 22 go-cart track, 929 golf ball to hole distance, 470, 531, 596 golf swing, 608 gymnastics, 521 high dives, 521 high-wire walking, 510, 710 horse racing, 983 hot-air balloon volume, 75 jogging distance, 217 kiteboarding speed, 32 kite height, A–49 lacrosse pass, 711 marathon training, 1062–1063 motocross miles per hour, 32 mowing the park, 537 official ball size, 30 Olympic anxiety of skater, 339–340 orienteering, 242, 706 pay-per-view subscriptions, 378

xl

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playing cards Trumps, 731 poker probabilities, 1015–1016 public park usage, 617, 1018 race track area, 851 for recruits, 30 regimen training, 618, 974 river race, 707 roller coaster design, 609 roping a steer, 711 roping a wild stallion, 684 sail dimensions, 882 scooter speed, 865 Scrabble, 977, 979, 983 sculpture, 415 shot-put, 533 ski ramp, 596, 598 snowcone dimensions, 596 soccer all-star competition, 711 ball height and velocity, 1056–1057 shooting angles, 597, 930 softball toss, A–13 spelunking, 12 Star Trek—The Next Generation spacecraft, 322 stunt pilots, 864 swimming pool volume of water, 241 tandem bicycle trip, 1013 team rosters, 984, 986 tennis court dimensions, 55 tic-tac-toe, 986 treasure hunt, 710 tug-of-war, 684 Twister, 985 watering the lawn, 537 Yahtzee, 985 yoga, 533

bridge length, 558 distance between automated carts, 717 distance between flying airplanes, 439 distance between ships, 439, 529 distance from shore, 533, 698 distress call from stranded boat, 883–884 driver sight angle of sidewalk, 718 emissions testing, 522 figure eight formation, 895 flight headings, 655–656, 672, 687, 710, 714, 1017, 1071 flying clubs, 864 flying speed, 509, 646 gasoline cost, 769 height of blimp, 645 highway sign, 673 horsepower, 419 hydrofoil service, 989 nautical distance, 656 parallel/nonparallel roads, 109 perpendicular/nonperpendicular course headings, 109 propeller manufacturing, 895 radar detection, 518, 857, 858, 864–865 runway length, 378, 656 screw jack use, 566 search area at sea, 714 searchlight angle, 717 ship course and speed, 672, 912 submarine depth, 30 suspension bridges, 12 tow forces, 671, 707, 710 trailer dimensions, 882 train speed, 509 trip planning, 655–656 tunnel clearance, 881 tunnel length, 656 windshield wiper armature, 511

TRANSPORTATION aerial distance, 656 aircraft carrier distance from home port, 107 aircraft N-Numbers, 985 airplane altitude, 100 airplane course and speed, 672, 687, 710, 711, 714, 852, 1071, A–13

WOMEN’S ISSUES female physicians, 109 fertility rates, 136 fetus weight, 136 multiple births, 183, 418

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Precalculus—

1 CHAPTER CONNECTIONS

Equations and Inequalities CHAPTER OUTLINE 1.1 Linear Equations, Formulas, and Problem Solving 2 1.2 Linear Inequalities in One Variable 14 1.3 Absolute Value Equations and Inequalities 24 1.4 Complex Numbers 33

In the world of professional sports, very strict specifications are put in place to ensure that competition is fair, with no one person or team having an unfair advantage. In professional hockey, the goal must be 48 in. high and 72 in. wide, as measured from inside the posts. In baseball, the bats used can have a maximum diameter of 69.8 mm, and a maximum length of 1006.8 mm, while in professional golf, the diameter d of a golf ball must be within 0.1 mm of the established standard of 42.7 mm. In the latter case, we can express this requirement as d  42.7 6 0.1. This application appears as Exercise 65 in Section 1.3.

1.5 Solving Quadratic Equations 42 1.6 Solving Other Types of Equations 56

The foundation and study of calculus involves analyzing very small differences similar to the one above. The Connections to Calculus for Chapter 1 expands on the notation and language used in this analysis, and Connections introduces how the absolute value concept contributes to this foundation. The need to solve a broad range to Calculus of equation types is also explored. 1

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Precalculus—

1.1 Linear Equations, Formulas, and Problem Solving Learning Objectives

In a study of algebra, you will encounter many families of equations, or groups of equations that share common characteristics. Of interest to us here is the family of linear equations in one variable, a study that lays the foundation for understanding more advanced families. In addition to solving linear equations, we’ll use the skills we develop to solve for a specified variable in a formula, a practice widely used in science, business, industry, and research.

In Section 1.1 you will learn how to:

A. Solve linear equations using properties of equality

B. Recognize equations that are identities or contradictions

A. Solving Linear Equations Using Properties of Equality

C. Solve for a specified

An equation is a statement that two expressions are equal. From the expressions 31x  12  x and x  7, we can form the equation

variable in a formula or literal equation

D. Use the problem-solving

31x  12  x  x  7,

guide to solve various problem types

CAUTION

Table 1.1 x

31x  12  x

x  7

2

11

9

1

7

8

which is a linear equation in one variable. To solve 0 an equation, we attempt to find a specific input or x1 value that will make the equation true, meaning the 2 left-hand expression will be equal to the right. Using 3 Table 1.1, we find that 31x  12  x  x  7 is a 4 true equation when x is replaced by 2, and is a false equation otherwise. Replacement values that make the equation true are called solutions or roots of the equation. 

3

7

1

6

5

5

9

4

13

3

From Appendix I.F, an algebraic expression is a sum or difference of algebraic terms. Algebraic expressions can be simplified, evaluated or written in an equivalent form, but cannot be “solved,” since we’re not seeking a specific value of the unknown.

Solving equations using a table is too time consuming to be practical. Instead we attempt to write a sequence of equivalent equations, each one simpler than the one before, until we reach a point where the solution is obvious. Equivalent equations are those that have the same solution set, and are obtained by using the distributive property to simplify the expressions on each side of the equation, and the additive and multiplicative properties of equality to obtain an equation of the form x  constant. The Additive Property of Equality

The Multiplicative Property of Equality

If A, B, and C represent algebraic expressions and A  B,

If A, B, and C represent algebraic expressions and A  B , B A then AC  BC and  , 1C  02 C C

then A  C  B  C

In words, the additive property says that like quantities, numbers or terms can be added to both sides of an equation. A similar statement can be made for the multiplicative property. These properties are combined into a general guide for solving linear equations, which you’ve likely encountered in your previous studies. Note that not all steps in the guide are required to solve every equation.

2

1-2

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Precalculus—

1-3

3

Section 1.1 Linear Equations, Formulas, and Problem Solving

Guide to Solving Linear Equations in One Variable

• Eliminate parentheses using the distributive property, then combine any like terms. • Use the additive property of equality to write the equation with all variable terms on one side, and all constants on the other. Simplify each side.

• Use the multiplicative property of equality to obtain an equation of the form x  constant.

• For applications, answer in a complete sentence and include any units of measure indicated. For our first example, we’ll use the equation 31x  12  x  x  7 from our initial discussion. EXAMPLE 1



Solving a Linear Equation Using Properties of Equality Solve for x: 31x  12  x  x  7.

Solution



31x  12  x  x  7 3x  3  x  x  7 4x  3  x  7 5x  3  7 5x  10 x2

original equation distributive property combine like terms add x to both sides (additive property of equality) add 3 to both sides (additive property of equality) multiply both sides by 15 or divide both sides by 5 (multiplicative property of equality)

As we noted in Table 1.1, the solution is x  2. Now try Exercises 7 through 12



To check a solution by substitution means we substitute the solution back into the original equation (this is sometimes called back-substitution), and verify the left-hand side is equal to the right. For Example 1 we have: 31x  12  x  x  7 312  12  2  2  7 3112  2  5 5  5✓

original equation substitute 2 for x simplify solution checks

If any coefficients in an equation are fractional, multiply both sides by the least common denominator (LCD) to clear the fractions. Since any decimal number can be written in fraction form, the same idea can be applied to decimal coefficients. EXAMPLE 2



Solving a Linear Equation with Fractional Coefficients Solve for n: 14 1n  82  2  12 1n  62.

Solution



A. You’ve just learned how to solve linear equations using properties of equality

1 4 1n  82 1 4n  2

 2  12 1n  62  2  12 n  3 1 1 4n  2n  3 41 14 n2  41 12 n  32 n  2n  12 n  12 n  12

original equation distributive property combine like terms multiply both sides by LCD  4 distributive property subtract 2n multiply by 1

Verify the solution is n  12 using back-substitution. Now try Exercises 13 through 30



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Precalculus—

4

1-4

CHAPTER 1 Equations and Inequalities

B. Identities and Contradictions Example 1 illustrates what is called a conditional equation, since the equation is true for x  2, but false for all other values of x. The equation in Example 2 is also conditional. An identity is an equation that is always true, no matter what value is substituted for the variable. For instance, 21x  32  2x  6 is an identity with a solution set of all real numbers, written as 5x 0 x  6, or x 1q, q 2 in interval notation. Contradictions are equations that are never true, no matter what real number is substituted for the variable. The equations x  3  x  1 and 3  1 are contradictions. To state the solution set for a contradiction, we use the symbol “” (the null set) or “{ }” (the empty set). Recognizing these special equations will prevent some surprise and indecision in later chapters. EXAMPLE 3



Solving an Equation That Is a Contradiction Solve for x: 21x  42  10x  8  413x  12 , and state the solution set.

Solution



21x  42  10x  8  413x  12 2x  8  10x  8  12x  4 12x  8  12x  12 8  12

original equation distributive property combine like terms subtract 12x

Since 8 is never equal to 12, the original equation is a contradiction. The solution is the empty set { }. Now try Exercises 31 through 36

B. You’ve just learned how to recognize equations that are identities or contradictions



In Example 3, our attempt to solve for x ended with all variables being eliminated, leaving an equation that is always false—a contradiction 18 is never equal to 12). There is nothing wrong with the solution process, the result is simply telling us the original equation has no solution. In other equations, the variables may once again be eliminated, but leave a result that is always true—an identity.

C. Solving for a Specified Variable in Literal Equations A formula is an equation that models a known relationship between two or more quantities. A literal equation is simply one that has two or more variables. Formulas are a type of literal equation, but not every literal equation is a formula. For example, the formula A  P  PRT models the growth of money in an account earning simple interest, where A represents the total amount accumulated, P is the initial deposit, R is the annual interest rate, and T is the number of years the money is left on deposit. To describe A  P  PRT , we might say the formula has been “solved for A” or that “A is written in terms of P, R, and T.” In some cases, before using a formula it may be convenient to solve for one of the other variables, say P. In this case, P is called the object variable. EXAMPLE 4



Solving for Specified Variable Given A  P  PRT , write P in terms of A, R, and T (solve for P).

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Precalculus—

1-5

5

Section 1.1 Linear Equations, Formulas, and Problem Solving

Solution



Since the object variable occurs in more than one term, we first apply the distributive property. A  P  PRT A  P11  RT2 P11  RT2 A  1  RT 11  RT2 A P 1  RT

focus on P — the object variable factor out P solve for P [divide by (1RT )]

result

Now try Exercises 37 through 48



We solve literal equations for a specified variable using the same methods we used for other equations and formulas. Remember that it’s good practice to focus on the object variable to help guide you through the solution process, as again shown in Example 5. EXAMPLE 5



Solving for a Specified Variable Given 2x  3y  15, write y in terms of x (solve for y).

Solution



WORTHY OF NOTE

2x  3y  15 3y  2x  15 1 3 13y2

 y

In Example 5, notice that in the second step we wrote the subtraction of 2x as 2x  15 instead of 15  2x. For reasons that become clear later in this chapter, we generally write variable terms before constant terms.

1 3 12x 2 3 x 

 152 5

focus on the object variable subtract 2x (isolate term with y) multiply by 13 (solve for y) distribute and simplify

Now try Exercises 49 through 54



Literal Equations and General Solutions Solving literal equations for a specified variable can help us develop the general solution for an entire family of equations. This is demonstrated here for the family of linear equations written in the form ax  b  c. A side-by-side comparison with a specific linear equation demonstrates that identical ideas are used. Specific Equation 2x  3  15 2x  15  3 x

15  3 2

Literal Equation focus on object variable subtract constant divide by coefficient

ax  b  c ax  c  b x

cb a

Of course the solution on the left would be written as x  6 and checked in the original equation. On the right we now have a general formula for all equations of the form ax  b  c. EXAMPLE 6



Solving Equations of the Form ax  b  c Using the General Formula Solve 6x  1  25 using the formula just developed, and check your solution in the original equation.

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Precalculus—

6

1-6

CHAPTER 1 Equations and Inequalities

Solution



WORTHY OF NOTE Developing a general solution for the linear equation ax  b  c seems to have little practical use. But in Section 1.5 we’ll use this idea to develop a general solution for quadratic equations, a result with much greater significance. C. You’ve just learned how to solve for a specified variable in a formula or literal equation

For this equation, a  6, b  1, and c  25, this gives cb → Check: x 6x  1  25 a 25  112  6142  1  25 6 24  24  1  25 6  4 25  25 ✓ Now try Exercises 55 through 60



D. Using the Problem-Solving Guide Becoming a good problem solver is an evolutionary process. Over time and with continued effort, your problem-solving skills grow, as will your ability to solve a wider range of applications. Most good problem solvers develop the following characteristics:

• A positive attitude • A mastery of basic facts • Strong mental arithmetic skills

• Good mental-visual skills • Good estimation skills • A willingness to persevere

These characteristics form a solid basis for applying what we call the ProblemSolving Guide, which simply organizes the basic elements of good problem solving. Using this guide will help save you from two common stumbling blocks—indecision and not knowing where to start. Problem-Solving Guide • Gather and organize information. Read the problem several times, forming a mental picture as you read. Highlight key phrases. List given information, including any related formulas. Clearly identify what you are asked to find. • Make the problem visual. Draw and label a diagram or create a table of values, as appropriate. This will help you see how different parts of the problem fit together. • Develop an equation model. Assign a variable to represent what you are asked to find and build any related expressions referred to in the exercise. Write an equation model from the information given in the exercise. Carefully reread the exercise to double-check your equation model. • Use the model and given information to solve the problem. Substitute given values, then simplify and solve. State the answer in sentence form, and check that the answer is reasonable. Include any units of measure indicated.

General Modeling Exercises In Appendix I.B, we learned to translate word phrases into symbols. This skill is used to build equations from information given in paragraph form. Sometimes the variable occurs more than once in the equation, because two different items in the same exercise are related. If the relationship involves a comparison of size, we often use line segments or bar graphs to model the relative sizes.

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Precalculus—

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7

Section 1.1 Linear Equations, Formulas, and Problem Solving

EXAMPLE 7



Solving an Application Using the Problem-Solving Guide The largest state in the United States is Alaska (AK), which covers an area that is 230 square miles (mi2) more than 500 times that of the smallest state, Rhode Island (RI). If they have a combined area of 616,460 mi2, how many square miles does each cover?

Solution



Combined area is 616,460 mi2, AK covers 230 more than 500 times the area of RI.

gather and organize information highlight any key phrases

616,460

…

230

make the problem visual

500 times

Rhode Island’s area R

Alaska

Let R represent the area of Rhode Island. Then 500R  230 represents Alaska’s area.

assign a variable build related expressions

Rhode Island’s area  Alaska’s area  Total R  1500R  2302  616,460 501R  616,230 R  1230

write the equation model combine like terms, subtract 230 divide by 501

2

Rhode Island covers an area of 1230 mi , while Alaska covers an area of 500112302  230  615,230 mi2. Now try Exercises 63 through 68



Consecutive Integer Exercises Exercises involving consecutive integers offer excellent practice in assigning variables to unknown quantities, building related expressions, and the problem-solving process in general. We sometimes work with consecutive odd integers or consecutive even integers as well. EXAMPLE 8



Solving a Problem Involving Consecutive Odd Integers The sum of three consecutive odd integers is 69. What are the integers?

Solution



The sum of three consecutive odd integers . . . 2

2

2

2

gather/organize information highlight any key phrases

2

WORTHY OF NOTE The number line illustration in Example 8 shows that consecutive odd integers are two units apart and the related expressions were built accordingly: n, n  2, n  4, and so on. In particular, we cannot use n, n  1, n  3, . . . because n and n  1 are not two units apart. If we know the exercise involves even integers instead, the same model is used, since even integers are also two units apart. For consecutive integers, the labels are n, n  1, n  2, and so on.

4 3 2 1

odd

odd

0

1

odd

2

3

odd

4

n n1 n2 n3 n4

odd

odd

odd

Let n represent the smallest consecutive odd integer, then n  2 represents the second odd integer and 1n  22  2  n  4 represents the third.

In words: first  second  third odd integer  69 n  1n  22  1n  42  69 3n  6  69 3n  63 n  21

make the problem visual

assign a variable build related expressions

write the equation model equation model combine like terms subtract 6 divide by 3

The odd integers are n  21, n  2  23, and n  4  25. 21  23  25  69 ✓ Now try Exercises 69 through 72



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1-8

CHAPTER 1 Equations and Inequalities

Uniform Motion (Distance, Rate, Time) Exercises Uniform motion problems have many variations, and it’s important to draw a good diagram when you get started. Recall that if speed is constant, the distance traveled is equal to the rate of speed multiplied by the time in motion: D  RT. EXAMPLE 9



Solving a Problem Involving Uniform Motion I live 260 mi from a popular mountain retreat. On my way there to do some mountain biking, my car had engine trouble—forcing me to bike the rest of the way. If I drove 2 hr longer than I biked and averaged 60 miles per hour driving and 10 miles per hour biking, how many hours did I spend pedaling to the resort?

Solution



The sum of the two distances must be 260 mi. The rates are given, and the driving time is 2 hr more than biking time.

Home

gather/organize information highlight any key phrases make the problem visual

Driving

Biking

D1  RT

D2  rt

Resort

D1  D2  Total distance 260 miles

Let t represent the biking time, then T  t  2 represents time spent driving. D1  D2  260 RT  rt  260 601t  22  10t  260 70t  120  260 70t  140 t2

assign a variable build related expressions write the equation model RT  D1, rt  D2 substitute t  2 for T, 60 for R, 10 for r distribute and combine like terms subtract 120 divide by 70

I rode my bike for t  2 hr, after driving t  2  4 hr. Now try Exercises 73 through 76



Exercises Involving Mixtures Mixture problems offer another opportunity to refine our problem-solving skills while using many elements from the problem-solving guide. They also lend themselves to a very useful mental-visual image and have many practical applications. EXAMPLE 10



Solving an Application Involving Mixtures As a nasal decongestant, doctors sometimes prescribe saline solutions with a concentration between 6% and 20%. In “the old days,” pharmacists had to create different mixtures, but only needed to stock these concentrations, since any percentage in between could be obtained using a mixture. An order comes in for a 15% solution. How many milliliters (mL) of the 20% solution must be mixed with 10 mL of the 6% solution to obtain the desired 15% solution?

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Section 1.1 Linear Equations, Formulas, and Problem Solving

Solution



WORTHY OF NOTE For mixture exercises, an estimate assuming equal amounts of each liquid can be helpful. For example, assume we use 10 mL of the 6% solution and 10 mL of the 20% solution. The final concentration would be halfway in between, 6 2 20  13%. This is too low a concentration (we need a 15% solution), so we know that more than 10 mL of the stronger (20%) solution must be used.

D. You’ve just learned how to use the problem-solving guide to solve various problem types

Only 6% and 20% concentrations are available; mix a 20% solution with 10 mL of a 6% solution 20% solution

gather/organize information highlight any key phrases

6% solution ? mL

10 mL make the problem visual

(10  ?) mL 15% solution

Let x represent the amount of 20% solution, then 10  x represents the total amount of 15% solution. 1st quantity times its concentration

1010.062 0.6

2nd quantity times its concentration

 

x10.22 0.2x

assign a variable build related expressions

1st2nd quantity times desired concentration

 110  x2 10.152  1.5  0.15x 0.2x  0.9  0.15x 0.05x  0.9 x  18

write equation model distribute/simplify subtract 0.6 subtract 0.15x divide by 0.05

To obtain a 15% solution, 18 mL of the 20% solution must be mixed with 10 mL of the 6% solution. Now try Exercises 77 through 84



TECHNOLOGY HIGHLIGHT

Using a Graphing Calculator as an Investigative Tool The mixture concept can be applied in a wide variety of ways, including mixing zinc and copper to get bronze, different kinds of nuts for the holidays, diversifying investments, or mixing two acid solutions in order to get a desired concentration. Whether the value of each part in the mix is monetary or a percent of concentration, the general mixture equation has this form: Quantity 1 # Value I  Quantity 2 # Value II  Total quantity # Desired value Graphing calculators are a great tool for exploring this relationship, because the TABLE feature enables us to test the result of various mixtures in an instant. Suppose 10 oz of an 80% glycerin solution are to be mixed with an unknown amount of a 40% solution. How much of the 40% solution is used if a 56% solution is needed? To begin, we might consider that using equal amounts of the 40% and 80% solutions would result in a 60% concentration (halfway between 40% and 80%). To illustrate, let C represent the final concentration of the mix. 1010.82  1010.42  110  102C 8  4  20C 12  20C 0.6  C

equal amounts simplify add divide by 20

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Precalculus—

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CHAPTER 1 Equations and Inequalities

Figure 1.1

Figure 1.2

Figure 1.3

Since this is too high a concentration (a 56%  0.56 solution is desired), we know more of the weaker solution should be used. To explore the relationship further, assume x oz of the 40% solution are used and enter the resulting equation on the Y = screen as Y1  .81102  .4X. Enter the result of the mix as Y2  .56110  X2 (see Figure 1.1). Next, set up a TABLE using 2nd WINDOW (TBLSET) with TblStart  10, ¢Tbl  1, and the calculator set in Indpnt: AUTO mode (see Figure 1.2). Finally, access the TABLE results using 2nd GRAPH (TABLE). The resulting screen is shown in Figure 1.3, where we note that 15 oz of the 40% solution should be used (the equation is true when X is 15: Y1  Y2 2. Exercise 1:

Use this idea to solve Exercises 81 and 82 from the Exercises.

1.1 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

1. A(n) is an equation that is always true, regardless of the value. 2. A(n) is an equation that is always false, regardless of the value. 3. A(n)



equation is an equation having or more unknowns.

4. For the equation S  2r2  2rh, we can say that S is written in terms of and . 5. Discuss/Explain the three tests used to identify a linear equation. Give examples and counterexamples in your discussion. 6. Discuss/Explain each of the four basic parts of the problem-solving guide. Include a solved example in your discussion.

DEVELOPING YOUR SKILLS

Solve each equation. Check your answer by substitution.

7. 4x  31x  22  18  x 8. 15  2x  41x  12  9

9. 21  12v  172  7  3v

10. 12  5w  9  16w  72

11. 8  13b  52  5  21b  12 12. 2a  41a  12  3  12a  12

Solve each equation.

13. 15 1b  102  7  13 1b  92

14. 61 1n  122  14 1n  82  2 15. 32 1m  62  1 2 16. 45 1n  102 

8 9

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Section 1.1 Linear Equations, Formulas, and Problem Solving

17. 12x  5  13x  7

18. 4  23y  12y  5

x3 x 19.  7 5 3

z z4 20. 2 6 2

21. 15  6 

3p 8

22. 15 

2q  21 9

23. 0.2124  7.5a2  6.1  4.1 24. 0.4117  4.25b2  3.15  4.16

25. 6.2v  12.1v  52  1.1  3.7v

26. 7.9  2.6w  1.5w  19.1  2.1w2

39. C  2r for r (geometry) 40. V  LWH for W (geometry) 41.

P1V1 P2V2 for T2 (science)  T1 T2

42.

P1 C  2 for P2 (communication) P2 d

43. V  43r2h for h (geometry) 44. V  13r2h for h (geometry) 45. Sn  na

a1  an b for n (sequences) 2

h1b1  b2 2 for h (geometry) 2

27.

n 2 n   2 5 3

46. A 

28.

2 m m   3 5 4

47. S  B  12PS for P (geometry)

p p 29. 3p   5   2p  6 4 6 30.

q q  1  3q  2  4q  6 8

Identify the following equations as an identity, a contradiction, or a conditional equation, then state the solution.

31. 314z  52  15z  20  3z 32. 5x  9  2  512  x2  1 33. 8  813n  52  5  611  n2 34. 2a  41a  12  1  312a  12 35. 414x  52  6  218x  72 36. 15x  32  2x  11  41x  22 Solve for the specified variable in each formula or literal equation.

48. s  12gt2  vt for g (physics) 49. Ax  By  C for y 50. 2x  3y  6 for y 51. 56x  38y  2 for y 52. 23x  79y  12 for y

53. y  3  4 5 1x  102 for y

54. y  4  2 15 1x  102 for y

The following equations are given in ax  b  c form. Solve by identifying the value of a, b, and c, then using cb the formula x  . a

55. 3x  2  19 56. 7x  5  47 57. 6x  1  33 58. 4x  9  43

37. P  C  CM for C (retail)

59. 7x  13  27

38. S  P  PD for P (retail)

60. 3x  4  25



11

WORKING WITH FORMULAS

61. Surface area of a cylinder: SA  2r2  2rh The surface area of a cylinder is given by the formula shown, where h is the height of the cylinder and r is the radius of the base. Find the height of a cylinder that has a radius of 8 cm and a surface area of 1256 cm2. Use   3.14.

62. Using the equation-solving process for Exercise 61 as a model, solve the formula SA  2r2  2rh for h.

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Precalculus—

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1-12

CHAPTER 1 Equations and Inequalities

APPLICATIONS

Solve by building an equation model and using the problem-solving guidelines as needed. General Modeling Exercises

63. Two spelunkers (cave explorers) were exploring different branches of an underground cavern. The first was able to descend 198 ft farther than twice the second. If the first spelunker descended a 1218 ft, how far was the second spelunker able to descend? 64. The area near the joining of the Tigris and Euphrates Rivers (in modern Iraq) has often been called the Cradle of Civilization, since the area has evidence of many ancient cultures. The length of the Euphrates River exceeds that of the Tigris by 620 mi. If they have a combined length of 2880 mi, how long is each river? 65. U.S. postal regulations require that a package Girth can have a maximum combined length and girth (distance around) L of 108 in. A shipping H carton is constructed so that it has a width of W 14 in., a height of 12 in., and can be cut or folded to various lengths. What is the maximum length that can be used? Source: www.USPS.com

66. Hi-Tech Home Improvements buys a fleet of identical trucks that cost $32,750 each. The company is allowed to depreciate the value of their trucks for tax purposes by $5250 per year. If company policies dictate that older trucks must be sold once their value declines to $6500, approximately how many years will they keep these trucks? 67. The longest suspension bridge in the world is the Akashi Kaikyo (Japan) with a length of 6532 feet. Japan is also home to the Shimotsui Straight bridge. The Akashi Kaikyo bridge is 364 ft more than twice the length of the Shimotsui bridge. How long is the Shimotsui bridge? Source: www.guinnessworldrecords.com

68. The Mars rover Spirit landed on January 3, 2004. Just over 1 yr later, on January 14, 2005, the Huygens probe landed on Titan (one of Saturn’s moons). At their closest approach, the distance from the Earth to Saturn is 29 million mi more than 21 times the distance from the Earth to Mars. If the distance to Saturn is 743 million mi, what is the distance to Mars?

Consecutive Integer Exercises

69. Find two consecutive even integers such that the sum of twice the smaller integer plus the larger integer is one hundred forty-six. 70. When the smaller of two consecutive integers is added to three times the larger, the result is fiftyone. Find the smaller integer. 71. Seven times the first of two consecutive odd integers is equal to five times the second. Find each integer. 72. Find three consecutive even integers where the sum of triple the first and twice the second is eight more than four times the third. Uniform Motion Exercises

73. At 9:00 A.M., Linda leaves work on a business trip, gets on the interstate, and sets her cruise control at 60 mph. At 9:30 A.M., Bruce notices she’s left her briefcase and cell phone, and immediately starts after her driving 75 mph. At what time will Bruce catch up with Linda? 74. A plane flying at 300 mph has a 3-hr head start on a “chase plane,” which has a speed of 800 mph. How far from the airport will the chase plane overtake the first plane? 75. Jeff had a job interview in a nearby city 72 mi away. On the first leg of the trip he drove an average of 30 mph through a long construction zone, but was able to drive 60 mph after passing through this zone. If driving time for the trip was 112 hr, how long was he driving in the construction zone? 76. At a high-school cross-country meet, Jared jogged 8 mph for the first part of the race, then increased his speed to 12 mph for the second part. If the race was 21 mi long and Jared finished in 2 hr, how far did he jog at the faster pace?

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Section 1.1 Linear Equations, Formulas, and Problem Solving

Mixture Exercises Give the total amount of the mix that results and the percent concentration or worth of the mix.

77. Two quarts of 100% orange juice are mixed with 2 quarts of water (0% juice). 78. Ten pints of a 40% acid are combined with 10 pints of an 80% acid. 79. Eight pounds of premium coffee beans worth $2.50 per pound are mixed with 8 lb of standard beans worth $1.10 per pound. 80. A rancher mixes 50 lb of a custom feed blend costing $1.80 per pound, with 50 lb of cheap cottonseed worth $0.60 per pound. Solve each application of the mixture concept.

81. To help sell more of a lower grade meat, a butcher mixes some premium ground beef worth $3.10/lb,



with 8 lb of lower grade ground beef worth $2.05/lb. If the result was an intermediate grade of ground beef worth $2.68/lb, how much premium ground beef was used? 82. Knowing that the camping/hiking season has arrived, a nutrition outlet is mixing GORP (Good Old Raisins and Peanuts) for the anticipated customers. How many pounds of peanuts worth $1.29/lb, should be mixed with 20 lb of deluxe raisins worth $1.89/lb, to obtain a mix that will sell for $1.49/lb? 83. How many pounds of walnuts at 84¢/lb should be mixed with 20 lb of pecans at $1.20/lb to give a mixture worth $1.04/lb? 84. How many pounds of cheese worth 81¢/lb must be mixed with 10 lb cheese worth $1.29/lb to make a mixture worth $1.11/lb?

EXTENDING THE THOUGHT

85. Look up and read the following article. Then turn in a one page summary. “Don’t Give Up!,” William H. Kraus, Mathematics Teacher, Volume 86, Number 2, February 1993: pages 110–112. 86. A chemist has four solutions of a very rare and expensive chemical that are 15% acid (cost $120 per ounce), 20% acid (cost $180 per ounce), 35% acid (cost $280 per ounce) and 45% acid (cost $359 per ounce). She requires 200 oz of a 29% acid solution. Find the combination of any two of these concentrations that will minimize the total cost of the mix. 87. P, Q, R, S, T, and U represent numbers. The arrows in the figure show the sum of the two or three numbers added in the indicated direction



13

(Example: Q  T  23). Find P  Q  R  S  T  U. P

Q

26

S

30 40

R

T 19

U 23

34

88. Given a sphere circumscribed by a cylinder, verify the volume of the sphere is 23 that of the cylinder.

MAINTAINING YOUR SKILLS

89. (R.1) Simplify the expression using the order of operations. 2  62  4  8 90. (R.3) Name the coefficient of each term in the expression: 3v3  v2  3v  7

91. (R.4) Factor each expression: a. 4x2  9 b. x3  27 92. (R.2) Identify the property illustrated: 6 7

# 5 # 21  67 # 21 # 5

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Precalculus—

1.2 Linear Inequalities in One Variable There are many real-world situations where the mathematical model leads to a statement of inequality rather than equality. Here are a few examples:

Learning Objectives In Section 1.2 you will learn how to:

Clarice wants to buy a house costing $85,000 or less. To earn a “B,” Shantë must score more than 90% on the final exam. To escape the Earth’s gravity, a rocket must travel 25,000 mph or more.

A. Solve inequalities and state solution sets

B. Solve linear inequalities C. Solve compound

While conditional linear equations in one variable have a single solution, linear inequalities often have an infinite number of solutions—which means we must develop additional methods for writing a solution set.

inequalities

D. Solve applications of inequalities

A. Inequalities and Solution Sets The set of numbers that satisfy an inequality is called the solution set. Instead of using a simple inequality to write solution sets, we will often use (1) a form of set notation, (2) a number line graph, or (3) interval notation. Interval notation is a symbolic way of indicating a selected interval of the real number line. When a number acts as the boundary point for an interval (also called an endpoint), we use a left bracket “[” or a right bracket “]” to indicate inclusion of the endpoint. If the boundary point is not included, we use a left parenthesis “(” or right parenthesis “).”

WORTHY OF NOTE Some texts will use an open dot “º” to mark the location of an endpoint that is not included, and a closed dot “•” for an included endpoint.

EXAMPLE 1



Using Inequalities in Context Model the given phrase using the correct inequality symbol. Then state the result in set notation, graphically, and in interval notation: “If the ball had traveled at least one more foot in the air, it would have been a home run.”

Solution



WORTHY OF NOTE Since infinity is really a concept and not a number, it is never included (using a bracket) as an endpoint for an interval.

Let d represent additional distance: d  1.

• Set notation: 5d| d  16 • Graph 2 1 0 1[ 2 3 4 • Interval notation: d  3 1, q2

5

Now try Exercises 7 through 18



The “” symbol says the number d is an element of the set or interval given. The “ q ” symbol represents positive infinity and indicates the interval continues forever to the right. Note that the endpoints of an interval must occur in the same order as on the number line (smaller value on the left; larger value on the right). A short summary of other possibilities is given here. Many variations are possible.

Conditions ( a  b) x is greater than k x is less than or equal to k

A. You’ve just learned how to solve inequalities and state solution sets

14

Set Notation 5x| x 7 k6 5x| x  k6

Number Line

x  1k, q2

) k

5x | a 6 x 6 b6

)

)

a

b

x is less than b and greater than or equal to a

5x |a  x 6 b6

[

)

a

b

5x |x 6 a or x 7 b6

x  1q, k4

[ k

x is less than b and greater than a

x is less than a or x is greater than b

Interval Notation

x  1a, b2 x  3a, b2

)

)

a

b

x  1q, a2 ´ 1b, q 2

1-14

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Precalculus—

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Section 1.2 Linear Inequalities in One Variable

15

B. Solving Linear Inequalities A linear inequality resembles a linear equality in many respects: Linear Inequality

Related Linear Equation

(1)

x 6 3

x3

(2)

3 p  2   12 8

3 p  2  12 8

A linear inequality in one variable is one that can be written in the form ax  b 6 c, where a, b, and c   and a  0. This definition and the following properties also apply when other inequality symbols are used. Solutions to simple inequalities are easy to spot. For instance, x  2 is a solution to x 6 3 since 2 6 3. For more involved inequalities we use the additive property of inequality and the multiplicative property of inequality. Similar to solving equations, we solve inequalities by isolating the variable on one side to obtain a solution form such as variable 6 number. The Additive Property of Inequality If A, B, and C represent algebraic expressions and A 6 B, then A  C 6 B  C Like quantities (numbers or terms) can be added to both sides of an inequality. While there is little difference between the additive property of equality and the additive property of inequality, there is an important difference between the multiplicative property of equality and the multiplicative property of inequality. To illustrate, we begin with 2 6 5. Multiplying both sides by positive three yields 6 6 15, a true inequality. But notice what happens when we multiply both sides by negative three: 2 6 5

original inequality

2132 6 5132 6 6 15

multiply by negative three false

This is a false inequality, because 6 is to the right of 15 on the number line. Multiplying (or dividing) an inequality by a negative quantity reverses the order relationship between two quantities (we say it changes the sense of the inequality). We must compensate for this by reversing the inequality symbol. 6 7 15

change direction of symbol to maintain a true statement

For this reason, the multiplicative property of inequality is stated in two parts. The Multiplicative Property of Inequality If A, B, and C represent algebraic expressions and A 6 B, then AC 6 BC

If A, B, and C represent algebraic expressions and A 6 B, then AC 7 BC

if C is a positive quantity (inequality symbol remains the same).

if C is a negative quantity (inequality symbol must be reversed).

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CHAPTER 1 Equations and Inequalities

EXAMPLE 2



Solving an Inequality Solve the inequality, then graph the solution set and write it in interval notation: 2 1 5 3 x  2  6.

Solution



1 5 2 x  3 2 6 1 5 2 6a x  b  162 3 2 6 4x  3  5 4x  2 1 x  2

WORTHY OF NOTE

EXAMPLE 3

clear fractions (multiply by LCD) simplify subtract 3 divide by 4, reverse inequality sign

1 2

• Graph:

As an alternative to multiplying or dividing by a negative value, the additive property of inequality can be used to ensure the variable term will be positive. From Example 2, the inequality 4x  2 can be written as 2  4x by adding 4x to both sides and subtracting 2 from both sides. This gives the solution 12  x, which is equivalent to x  12.

original inequality

3 2 1

[

0

• Interval notation: x 

1

2 3 1 3 2, q 2

4

Now try Exercises 19 through 28



To check a linear inequality, you often have an infinite number of choices—any number from the solution set/interval. If a test value from the solution interval results in a true inequality, all numbers in the interval are solutions. For Example 2, using x  0 results in the true statement 12  56 ✓. Some inequalities have all real numbers as the solution set: 5x | x  6, while other inequalities have no solutions, with the answer given as the empty set: { }. 

Solving Inequalities Solve the inequality and write the solution in set notation: a. 7  13x  52  21x  42  5x b. 31x  42  5 6 21x  32  x

Solution



a. 7  13x  52  21x  42  5x 7  3x  5  2x  8  5x 3x  2  3x  8 2  8

original inequality distributive property combine like terms add 3x

Since the resulting statement is always true, the original inequality is true for all real numbers. The solution is 5x |x  6 . b. 31x  42  5 6 21x  32  x original inequality 3x  12  5 6 2x  6  x distribute 3x  7 6 3x  6 combine like terms 7 6 6 subtract 3x B. You’ve just learned how to solve linear inequalities

Since the resulting statement is always false, the original inequality is false for all real numbers. The solution is { }. Now try Exercises 29 through 34



C. Solving Compound Inequalities In some applications of inequalities, we must consider more than one solution interval. These are called compound inequalities, and require us to take a close look at the

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Precalculus—

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Section 1.2 Linear Inequalities in One Variable

17

operations of union “ ´ ” and intersection “ ¨”. The intersection of two sets A and B, written A ¨ B, is the set of all elements common to both sets. The union of two sets A and B, written A ´ B, is the set of all elements that are in either set. When stating the union of two sets, repetitions are unnecessary. EXAMPLE 4



Finding the Union and Intersection of Two Sets

Solution



A ¨ B is the set of all elements in both A and B: A ¨ B  51, 2, 36. A ´ B is the set of all elements in either A or B: A ´ B  52, 1, 0, 1, 2, 3, 4, 56.

WORTHY OF NOTE For the long term, it may help to rephrase the distinction as follows. The intersection is a selection of elements that are common to two sets, while the union is a collection of the elements from two sets (with no repetitions).

EXAMPLE 5

For set A  52, 1, 0, 1, 2, 36 and set B  51, 2, 3, 4, 56, determine A ¨ B and A ´ B.

Now try Exercises 35 through 40



Notice the intersection of two sets is described using the word “and,” while the union of two sets is described using the word “or.” When compound inequalities are formed using these words, the solution is modeled after the ideas from Example 4. If “and” is used, the solutions must satisfy both inequalities. If “or” is used, the solutions can satisfy either inequality. 

Solving a Compound Inequality Solve the compound inequality, then write the solution in interval notation: 3x  1 6 4 or 4x  3 6 6.

Solution WORTHY OF NOTE



Begin with the statement as given: 3x  1 6 4 3x 6 3

4x  3 6 6 original statement 4x 6 9 isolate variable term 9 x 7 1 or x 6  solve for x, reverse first inequality symbol 4 The solution x 7 1 or x 6 94 is better understood by graphing each interval separately, then selecting both intervals (the union).

The graphs from Example 5 clearly show the solution consists of two disjoint (disconnected) intervals. This is reflected in the “or” statement: x 6 94 or x 7 1, and in the interval notation. Also, note the solution x 6 94 or x 7 1 is not equivalent to 94 7 x 7 1, as there is no single number that is both greater than 1 and less than 94 at the same time.

x  1:

x 9 : 4 x  9 or x  1: 4

or or

8 7 6 5 4 3 2 1

)

0

1

2

3

4

5

6

0

1

2

3

4

5

6

0

1

2

3

4

5

6

9 4

)

8 7 6 5 4 3 2 1

9 4

)

8 7 6 5 4 3 2 1

)

9 Interval notation: x  aq,  b ´ 11, q 2. 4 Now try Exercises 41 and 42 EXAMPLE 6



Solving a Compound Inequality Solve the compound inequality, then write the solution in interval notation: 3x  5 7 13 and 3x  5 6 1.

Solution



Begin with the statement as given: and 3x  5 7 13 3x  5 6 1 and 3x 7 18 3x 6 6 x 7 6 and x 6 2

original statement subtract five divide by 3



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The solution x 7 6 and x 6 2 can best be understood by graphing each interval separately, then noting where they intersect.

WORTHY OF NOTE The inequality a 6 b (a is less than b) can equivalently be written as b 7 a (b is greater than a). In Example 6, the solution is read, “ x 7 6 and x 6 2,” but if we rewrite the first inequality as 6 6 x (with the “arrowhead” still pointing at 62, we have 6 6 x and x 6 2 and can clearly see that x must be in the single interval between 6 and 2.

EXAMPLE 7

Solution

x  6: x 2: x  6 and x 2:

)

8 7 6 5 4 3 2 1

)

8 7 6 5 4 3 2 1

0

1

2

3

4

5

6

0

1

2

3

4

5

6

0

1

2

3

4

5

6

)

)

18

8 7 6 5 4 3 2 1

Interval notation: x  16, 22.

Now try Exercises 43 through 54



The solution from Example 6 consists of the single interval 16, 22, indicating the original inequality could actually be joined and written as 6 6 x 6 2, called a joint or compound inequality (see Worthy of Note). We solve joint inequalities in much the same way as linear inequalities, but must remember they have three parts (left, middle, and right). This means operations must be applied to all three parts in each step of the solution process, to obtain a solution form such as smaller number 6 x 6 larger number. The same ideas apply when other inequality symbols are used. 



C. You’ve just learned how to solve compound inequalities

Solving a Compound Inequality Solve the compound inequality, then graph the solution set and write it in interval 2x  5 notation: 1 7  6. 3 2x  5  6 original inequality 1 7 3 3 6 2x  5  18 multiply all parts by 3; reverse the inequality symbols 8 6 2x  13 subtract 5 from all parts 13 4 6 x  divide all parts by 2 2

• Graph:

)

5 4 3 2 1

13 2 0

• Interval notation: x  14,

1 13 2 4

2

3

4

5

6

[

7

8

Now try Exercises 55 through 60



D. Applications of Inequalities Domain and Allowable Values

Figure 1.4

Table 1.2 One application of inequalities involves the concept of allowable . values. Consider the expression 24 As Table 1.2 suggests, we can 24 x x x evaluate this expression using any real number other than zero, since 24 6 4 the expression 0 is undefined. Using set notation the allowable values 12 2 are written 5x | x  , x  06 . To graph the solution we must be 1 careful to exclude zero, as shown in Figure 1.4. 48 2 The graph gives us a snapshot of the solution using interval 0 error notation, which is written as a union of two disjoint (disconnected) intervals so as to exclude zero: x  1q, 02 ´ 10, q 2 . The set of allowable values is referred to as the domain of the expression. Allowable values are said to be “in the domain” of the expression; values that are not allowed are said to be “outside the domain.” When the denominator of a fraction contains a variable expression, values that cause a denominator )) of zero are outside the domain. 3 2 1 0 1 2 3

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19

Section 1.2 Linear Inequalities in One Variable

EXAMPLE 8



Determining the Domain of an Expression 6 Determine the domain of the expression . State the result in set notation, x2 graphically, and using interval notation.

Solution



Set the denominator equal to zero and solve: x  2  0 yields x  2. This means 2 is outside the domain and must be excluded.

• Set notation: 5x | x  , x  26

• Graph: 1 0 1 )2) 3 4 5 • Interval notation: x  1q, 22 ´ 12, q 2 Now try Exercises 61 through 68



A second area where allowable values are a concern involves the square root operation. Recall that 149  7 since 7 # 7  49. However, 149 cannot be written as the product of two real numbers since 172 # 172  49 and 7 # 7  49. In other words, 1X represents a real number only if the radicand is positive or zero. If X represents an algebraic expression, the domain of 1X is 5X |X  06 . EXAMPLE 9



Determining the Domain of an Expression Determine the domain of 1x  3. State the domain in set notation, graphically, and in interval notation.

Solution



The radicand must represent a nonnegative number. Solving x  3  0 gives x  3.

• Set notation: 5x | x  36 • Graph:

[

4 3 2 1

0

1

• Interval notation: x  3, q 2

2

Now try Exercises 69 through 76



Inequalities are widely used to help gather information, and to make comparisons that will lead to informed decisions. Here, the problem-solving guide is once again a valuable tool. EXAMPLE 10



Using an Inequality to Compute Desired Test Scores Justin earned scores of 78, 72, and 86 on the first three out of four exams. What score must he earn on the fourth exam to have an average of at least 80?

Solution



Gather and organize information; highlight any key phrases. First the scores: 78, 72, 86. An average of at least 80 means A  80. Make the problem visual. Test 1

Test 2

Test 3

Test 4

Computed Average

Minimum

78

72

86

x

78  72  86  x 4

80

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CHAPTER 1 Equations and Inequalities

Assign a variable; build related expressions. Let x represent Justin’s score on the fourth exam, then represents his average score. 78  72  86  x  80 4

78  72  86  x 4

average must be greater than or equal to 80

Write the equation model and solve. 78  72  86  x  320 236  x  320 x  84

multiply by 4 simplify solve for x (subtract 236)

Justin must score at least an 84 on the last exam to earn an 80 average. Now try Exercises 79 through 86



As your problem-solving skills improve, the process outlined in the problemsolving guide naturally becomes less formal, as we work more directly toward the equation model. See Example 11. EXAMPLE 11



Using an Inequality to Make a Financial Decision As Margaret starts her new job, her employer offers two salary options. Plan 1 is base pay of $1475/mo plus 3% of sales. Plan 2 is base pay of $500/mo plus 15% of sales. What level of monthly sales is needed for her to earn more under Plan 2?

Solution



D. You’ve just learned how to solve applications of inequalities

Let x represent her monthly sales in dollars. The equation model for Plan 1 would be 0.03x  1475; for Plan 2 we have 0.15x  500. To find the sales volume needed for her to earn more under Plan 2, we solve the inequality 0.15x  500 0.12x  500 0.12x x

7 7 7 7

0.03x  1475 1475 975 8125

Plan 2 7 Plan 1 subtract 0.03x subtract 500 divide by 0.12

If Margaret can generate more than $8125 in monthly sales, she will earn more under Plan 2. Now try Exercises 87 and 88



1.2 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

1. For inequalities, the three ways of writing a solution set are notation, a number line graph, and notation.

2. The mathematical sentence 3x  5 6 7 is a(n) inequality, while 2 6 3x  5 6 7 is a(n) inequality.

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Section 1.2 Linear Inequalities in One Variable

of sets A and B is written A  B. of sets A and B is written A ´ B.

3. The The

5. Discuss/Explain how the concept of domain and allowable values relates to rational and radical expressions. Include a few examples.



21

4. The intersection of set A with set B is the set of elements in A B. The union of set A with set B is the set of elements in A B. 6. Discuss/Explain why the inequality symbol must be reversed when multiplying or dividing by a negative quantity. Include a few examples.

DEVELOPING YOUR SKILLS

Use an inequality to write a mathematical model for each statement.

Solve each inequality and write the solution in set notation.

7. To qualify for a secretarial position, a person must type at least 45 words per minute.

29. 7  21x  32  4x  61x  32

8. The balance in a checking account must remain above $1000 or a fee is charged.

31. 413x  52  18 6 215x  12  2x

9. To bake properly, a turkey must be kept between the temperatures of 250° and 450°.

33. 61p  12  2p  212p  32

10. To fly effectively, the airliner must cruise at or between altitudes of 30,000 and 35,000 ft. Graph each inequality on a number line.

11. y 6 3

12. x 7 2

13. m  5

14. n  4

15. x  1

16. x  3

17. 5 7 x 7 2

18. 3 6 y  4

Write the solution set illustrated on each graph in set notation and interval notation.

19. 20. 21. 22.

[

3 2 1

0

3 2 1

[

3 2 1

[

3 2 1

1

)

0

0

0

2

1

[

1

1

3

2

2

2

2

34. 91w  12  3w  215  3w2  1 Determine the intersection and union of sets A, B, C, and D as indicated, given A  53, 2, 1, 0, 1, 2, 36, B  52, 4, 6, 86, C  54, 2, 0, 2, 46, and D  54, 5, 6, 76.

35. A  B and A ´ B

36. A  C and A ´ C

37. A  D and A ´ D

38. B  C and B ´ C

39. B  D and B ´ D

40. C  D and C ´ D

Express the compound inequalities graphically and in interval notation.

41. x 6 2 or x 7 1

42. x 6 5 or x 7 5

43. x 6 5 and x  2

44. x  4 and x 6 3

45. x  3 and x  1

46. x  5 and x  7

Solve the compound inequalities and graph the solution set.

3

)

3

47. 41x  12  20 or x  6 7 9 4

23. 5a  11  2a  5

49. 2x  7  3 and 2x  0 50. 3x  5  17 and 5x  0 51. 35x  12 7 2 3x

3 10

and 4x 7 1

  0 and 3x 6 2 5 6

3x x  6 3 or x  1 7 5 8 4 2x x 54.  6 2 or x  3 7 2 5 10 55. 3  2x  5 6 7 56. 2 6 3x  4  19 53.

25. 21n  32  4  5n  1 26. 51x  22  3 6 3x  11 28.

48. 31x  22 7 15 or x  3  1

52.

24. 8n  5 7 2n  12

3x x  6 4 8 4

32. 8  16  5m2 7 9m  13  4m2

3

Solve the inequality and write the solution in set notation. Then graph the solution and write it in interval notation.

27.

30. 3  61x  52  217  3x2  1

2y y  6 2 5 10

57. 0.5  0.3  x  1.7

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CHAPTER 1 Equations and Inequalities

58. 8.2 6 1.4  x 6 0.9 59. 7 6

34

x  1  11

60. 21  23x  9 6 7 Determine the domain of each expression. Write your answer in interval notation.

61.

12 m

62.

6 n

63.

5 y7

64.

4 x3

65.

a5 6a  3

66.

m5 8m  4

67.

15 3x  12

68.

7 2x  6

Determine the domain for each expression. Write your answer in interval notation.

69. 1x  2

70. 1y  7

71. 13n  12

72. 12m  5

73. 2b 

74. 2a 

4 3

75. 18  4y 

76. 112  2x

WORKING WITH FORMULAS

77. Body mass index: B 

704W H2

The U.S. government publishes a body mass index formula to help people consider the risk of heart disease. An index “B” of 27 or more means that a person is at risk. Here W represents weight in pounds and H represents height in inches. (a) Solve the formula for W. (b) If your height is 5¿8– what range of weights will help ensure you remain safe from the risk of heart disease? Source: www.surgeongeneral.gov/topics.



3 4

78. Lift capacity: 75S  125B  750 The capacity in pounds of the lift used by a roofing company to place roofing shingles and buckets of roofing nails on rooftops is modeled by the formula shown, where S represents packs of shingles and B represents buckets of nails. Use the formula to find (a) the largest number of shingle packs that can be lifted, (b) the largest number of nail buckets that can be lifted, and (c) the largest number of shingle packs that can be lifted along with three nail buckets.

APPLICATIONS

Write an inequality to model the given information and solve.

79. Exam scores: Jacques is going to college on an academic scholarship that requires him to maintain at least a 75% average in all of his classes. So far he has scored 82%, 76%, 65%, and 71% on four exams. What scores are possible on his last exam that will enable him to keep his scholarship? 80. Timed trials: In the first three trials of the 100-m butterfly, Johann had times of 50.2, 49.8, and 50.9 sec. How fast must he swim the final timed trial to have an average time of 50 sec? 81. Checking account balance: If the average daily balance in a certain checking account drops below $1000, the bank charges the customer a $7.50 service fee. The table gives the daily balance for

one customer. What must the daily balance be for Friday to avoid a service charge?

Weekday

Balance

Monday

$1125

Tuesday

$850

Wednesday

$625

Thursday

$400

82. Average weight: In the Lineman Weight National Football League, Left tackle 318 lb many consider an Left guard 322 lb offensive line to be “small” if the average Center 326 lb weight of the five down Right guard 315 lb linemen is less than Right tackle ? 325 lb. Using the table, what must the weight of the right tackle be so that the line will not be considered too small?

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83. Area of a rectangle: Given the rectangle shown, what is the range of values for the width, in order to keep the area less than 150 m2?

w

84. Area of a triangle: Using the triangle shown, find the height that will guarantee an area equal to or greater than 48 in2.

h

12 in.

85. Heating and cooling subsidies: As long as the outside temperature is over 45°F and less than 85°F 145 6 F 6 852, the city does not issue heating or

87. Power tool rentals: Sunshine Equipment Co. rents its power tools for a $20 fee, plus $4.50/hr. Kealoha’s Rentals offers the same tools for an $11 fee plus $6.00/hr. How many hours h must a tool be rented to make the cost at Sunshine a better deal? 88. Moving van rentals: Davis Truck Rentals will rent a moving van for $15.75/day plus $0.35 per mile. Bertz Van Rentals will rent the same van for $25/day plus $0.30 per mile. How many miles m must the van be driven to make the cost at Bertz a better deal?

EXTENDING THE CONCEPT

89. Use your local library, the Internet, or another resource to find the highest and lowest point on each of the seven continents. Express the range of altitudes for each continent as a joint inequality. Which continent has the greatest range? 90. The sum of two consecutive even integers is greater than or equal to 12 and less than or equal to 22. List all possible values for the two integers. Place the correct inequality symbol in the blank to make the statement true.

91. If m 7 0 and n 6 0, then mn 

cooling subsidies for low-income families. What is the corresponding range of Celsius temperatures C? Recall that F  95 C  32. 86. U.S. and European shoe sizes: To convert a European male shoe size “E” to an American male shoe size “A,” the formula A  0.76E  23 can be used. Lillian has five sons in the U.S. military, with shoe sizes ranging from size 9 to size 14 19  A  142. What is the corresponding range of European sizes? Round to the nearest half-size.

20 m



23

Section 1.2 Linear Inequalities in One Variable

92. If m 7 n and p 7 0, then mp

np.

93. If m 6 n and p 7 0, then mp

np.

94. If m  n and p 6 0, then mp

np. n.

95. If m 7 n, then m 96. If m 6 n, then

1 m

1 n.

97. If m 7 0 and n 6 0, then m2 98. If m 0, then m

3

n. 0.

0.

MAINTAINING YOUR SKILLS

99. (R.2) Translate into an algebraic expression: eight subtracted from twice a number. 100. (1.1) Solve: 41x  72  3  2x  1

101. (R.3) Simplify the algebraic expression: 21 59x  12  1 16x  32.

102. (1.1) Solve: 45m  23  12

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Precalculus—

1.3 Absolute Value Equations and Inequalities While the equations x  1  5 and x  1  5 are similar in many respects, note the first has only the solution x  4, while either x  4 or x  6 will satisfy the second. The fact there are two solutions shouldn’t surprise us, as it’s a natural result of how absolute value is defined.

Learning Objectives In Section 1.3 you will learn how to:

A. Solve absolute value equations

B. Solve “less than”

A. Solving Absolute Value Equations

absolute value inequalities

The absolute value of a number x can be thought of as its distance from zero on the number line, regardless of direction. This means x  4 will have two solutions, since there are two numbers that are four units from zero: x  4 and x  4 (see Figure 1.5).

C. Solve “greater than” absolute value inequalities

D. Solve applications involving absolute value

Exactly 4 units from zero

Figure 1.5

5 4

Exactly 4 units from zero 3 2 1

0

1

2

3

4

5

This basic idea can be extended to include situations where the quantity within absolute value bars is an algebraic expression, and suggests the following property.

WORTHY OF NOTE Note if k 6 0, the equation X  k has no solutions since the absolute value of any quantity is always positive or zero. On a related note, we can verify that if k  0, the equation X  0 has only the solution X  0.

Property of Absolute Value Equations If X represents an algebraic expression and k is a positive real number, then X  k implies X  k or X  k As the statement of this property suggests, it can only be applied after the absolute value expression has been isolated on one side.

EXAMPLE 1



Solving an Absolute Value Equation Solve: 5x  7  2  13.

Solution



Begin by isolating the absolute value expression. 5x  7  2  13 5x  7  15 x  7  3

original equation subtract 2 divide by 5 (simplified form)

Now consider x  7 as the variable expression “X” in the property of absolute value equations, giving x  7  3 x4

or or

x73 x  10

apply the property of absolute value equations add 7

Substituting into the original equation verifies the solution set is {4, 10}. Now try Exercises 7 through 18

CAUTION

24





For equations like those in Example 1, be careful not to treat the absolute value bars as simple grouping symbols. The equation 51x  72  2  13 has only the solution x  10, and “misses” the second solution since it yields x  7  3 in simplified form. The equation 5x  7  2  13 simplifies to x  7  3 and there are actually two solutions.

1-24

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Section 1.3 Absolute Value Equations and Inequalities

25

Absolute value equations come in many different forms. Always begin by isolating the absolute value expression, then apply the property of absolute value equations to solve. EXAMPLE 2



Solving an Absolute Value Equation Solve:

Solution



Check



WORTHY OF NOTE As illustrated in both Examples 1 and 2, the property we use to solve absolute value equations can only be applied after the absolute value term has been isolated. As you will see, the same is true for the properties used to solve absolute value inequalities.

2 `5  x `  9  8 3

2 `5  x `  9  8 original equation 3 2 ` 5  x `  17 add 9 3 apply the property of absolute 2 2 value equations 5  x  17 5  x  17 or 3 3 2 2  x  22 or subtract 5  x  12 3 3 x  33 or x  18 multiply by 32 2 2 For x  33: ` 5  1332 `  9  8 For x  18: ` 5  1182 `  9  8 3 3 05  21112 0  9  8 0 5  2162 0  9  8 0 5  22 0  9  8 0 5  12 0  9  8 0 17 0  9  8 0 17 0  9  8 17  9  8 17  9  8 8  8✓ 8  8✓

Both solutions check. The solution set is 518, 336.

Now try Exercises 19 through 22



For some equations, it’s helpful to apply the multiplicative property of absolute value: Multiplicative Property of Absolute Value If A and B represent algebraic expressions, then AB  AB. Note that if A  1 the property says B  1 B  B. More generally the property is applied where A is any constant.

EXAMPLE 3



Solution



A. You’ve just learned how to solve absolute value equations

Solving Equations Using the Multiplicative Property of Absolute Value Solve: 2x  5  13. 2x  5  13 2x  8 2x  8 2x  8 x  4 or x  4

original equation subtract 5 apply multiplicative property of absolute value simplify divide by 2

x4

apply property of absolute value equations

Both solutions check. The solution set is 54, 46. Now try Exercises 23 and 24



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CHAPTER 1 Equations and Inequalities

B. Solving “Less Than” Absolute Value Inequalities Absolute value inequalities can be solved using the basic concept underlying the property of absolute value equalities. Whereas the equation x  4 asks for all numbers x whose distance from zero is equal to 4, the inequality x 6 4 asks for all numbers x whose distance from zero is less than 4. Distance from zero is less than 4

Figure 1.6

Property I can also be applied when the “” symbol is used. Also notice that if k 6 0, the solution is the empty set since the absolute value of any quantity is always positive or zero.

0

1

2

3

)

4

5

Property I: Absolute Value Inequalities If X represents an algebraic expression and k is a positive real number, then X 6 k implies k 6 X 6 k



Solving “Less Than” Absolute Value Inequalities Solve the inequalities: 3x  2 a. 1 4

Solution

3 2 1

As Figure 1.6 illustrates, the solutions are x 7 4 and x 6 4, which can be written as the joint inequality 4 6 x 6 4. This idea can likewise be extended to include the absolute value of an algebraic expression X as follows.

WORTHY OF NOTE

EXAMPLE 4

)

5 4



WORTHY OF NOTE As with the inequalities from Section 1.2, solutions to absolute value inequalities can be checked using a test value. For Example 4(a), substituting x  0 from the solution interval yields: 1  1✓ 2 B. You’ve just learned how to solve less than absolute value inequalities

a.

3x  2 1 4 3x  2  4 4  3x  2  4 6  3x  2 2 2  x  3

b. 2x  7 6 5 original inequality multiply by 4 apply Property I subtract 2 from all three parts divide all three parts by 3

The solution interval is 3 2, 23 4. b. 2x  7 6 5

original inequality

Since the absolute value of any quantity is always positive or zero, the solution for this inequality is the empty set: { }. Now try Exercises 25 through 38



C. Solving “Greater Than” Absolute Value Inequalities For “greater than” inequalities, consider x 7 4. Now we’re asked to find all numbers x whose distance from zero is greater than 4. As Figure 1.7 shows, solutions are found in the interval to the left of 4, or to the right of 4. The fact the intervals are disjoint

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27

Section 1.3 Absolute Value Equations and Inequalities

(disconnected) is reflected in this graph, in the inequalities x 6 4 or x 7 4, as well as the interval notation x  1q, 42 ´ 14, q 2. Distance from zero is greater than 4

)

7 6 5 4 3 2 1

Figure 1.7

0

1

2

3

)

4

Distance from zero is greater than 4 5

6

7

As before, we can extend this idea to include algebraic expressions, as follows: Property II: Absolute Value Inequalities If X represents an algebraic expression and k is a positive real number, then X 7 k implies X 6 k or X 7 k

EXAMPLE 5



Solving “Greater Than” Absolute Value Inequalities Solve the inequalities: 1 x a.  ` 3  ` 6 2 3 2

Solution



b. 5x  2  

3 2

a. Note the exercise is given as a less than inequality, but as we multiply both sides by 3, we must reverse the inequality symbol. 

3

x 1 ` 3  ` 6 2 3 2 x `3  ` 7 6 2

original inequality multiply by 3, reverse the symbol

x x 6 6 or 3  7 6 2 2 x x 6 9 or 7 3 2 2 x 6 18 or x 7 6

apply Property II

subtract 3 multiply by 2

Property II yields the disjoint intervals x  1q, 182 ´ 16, q2 as the solution. )

30 24 18 12 6

b. 5x  2   C. You’ve just learned how to solve greater than absolute value inequalities

3 2

0

)

6

12

18

24

30

original inequality

Since the absolute value of any quantity is always positive or zero, the solution for this inequality is all real numbers: x  . Now try Exercises 39 through 54

CAUTION





Be sure you note the difference between the individual solutions of an absolute value equation, and the solution intervals that often result from solving absolute value inequalities. The solution 52, 56 indicates that both x  2 and x  5 are solutions, while the solution 32, 52 indicates that all numbers between 2 and 5, including 2, are solutions.

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CHAPTER 1 Equations and Inequalities

D. Applications Involving Absolute Value Applications of absolute value often involve finding a range of values for which a given statement is true. Many times, the equation or inequality used must be modeled after a given description or from given information, as in Example 6. EXAMPLE 6



Solving Applications Involving Absolute Value Inequalities For new cars, the number of miles per gallon (mpg) a car will get is heavily dependent on whether it is used mainly for short trips and city driving, or primarily on the highway for longer trips. For a certain car, the number of miles per gallon that a driver can expect varies by no more than 6.5 mpg above or below its field tested average of 28.4 mpg. What range of mileage values can a driver expect for this car?

Solution



Field tested average: 28.4 mpg mileage varies by no more than 6.5 mpg 6.5

gather information highlight key phrases

6.5

28.4

make the problem visual

Let m represent the miles per gallon a driver can expect. Then the difference between m and 28.4 can be no more than 6.5, or m  28.4  6.5. m  28.4  6.5 6.5  m  28.4  6.5 21.9  m  34.9

D. You’ve just learned how to solve applications involving absolute value

assign a variable write an equation model equation model apply Property I add 28.4 to all three parts

The mileage that a driver can expect ranges from a low of 21.9 mpg to a high of 34.9 mpg. Now try Exercises 57 through 64



TECHNOLOGY HIGHLIGHT

Absolute Value Equations and Inequalities Graphing calculators can explore and solve inequalities in many different Figure 1.8 ways. Here we’ll use a table of values and a relational test. To begin we’ll consider the equation 2 | x  3|  1  5 by entering the left-hand side as Y1 on the Y = screen. The calculator does not use absolute value bars the way they’re written, and the equation is actually entered as Y1  2 abs 1X  32  1 (see Figure 1.8). The “abs(” notation is accessed by pressing (NUM) 1 (option 1 gives only the left parenthesis, you MATH , must supply the right). Preset the TABLE as in the previous Highlight (page 10). By scrolling through the table (use the up and down Figure 1.9 arrows), we find Y1  5 when x  1 or x  5 (see Figure 1.9). Although we could also solve the inequality 2|x  3 |  1  5 using the table (the solution interval is x  3 1, 5 4 2, a relational test can help. Relational tests have the calculator return a “1” if a given statement is true, and a “0” otherwise. Enter Y2  Y1  5, by accessing Y1 using VARS (Y-VARS) 1:Function ENTER , and the “” symbol using 2nd MATH (TEST) [the “less than or equal to” symbol is option 6]. Returning to the table shows Y1  5 is true for 1  x  5 (see Figure 1.9). Use a table and a relational test to help solve the following inequalities. Verify the result algebraically. Exercise 1:

3|x  1|  2  7

Exercise 2:

2|x  2|  5  1

Exercise 3:

1  4 x  3  1

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Section 1.3 Absolute Value Equations and Inequalities

29

1.3 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. When multiplying or dividing by a negative quantity, we the inequality to maintain a true statement. 2. To write an absolute value equation or inequality in simplified form, we the absolute value expression on one side. 3. The absolute value equation 2x  3  7 is true when 2x  3  or when 2x  3  . 

4. The absolute value inequality 3x  6 6 12 is true when 3x  6 7 and 3x  6 6 . Describe each solution set (assume k  0). Justify your answer.

5. ax  b 6 k 6. ax  b 7 k

DEVELOPING YOUR SKILLS

Solve each absolute value equation. Write the solution in set notation.

Solve each absolute value inequality. Write solutions in interval notation.

7. 2m  1  7  3

25. x  2  7

26. y  1  3

8. 3n  5  14  2

27. 3 m  2 7 4

28. 2 n  3 7 7

9. 3x  5  6  15 10. 2y  3  4  14

29.

5v  1 8 6 9 4

30.

3w  2 6 6 8 2

11. 24v  5  6.5  10.3

31. 3 p  4  5 6 8

32. 5q  2  7  8

12. 72w  5  6.3  11.2

33. 3b  11  6  9

34. 2c  3  5 6 1

13. 7p  3  6  5

35. 4  3z  12 6 7

36. 2  7u  7  4

14. 3q  4  3  5

37. `

15. 2b  3  4 16. 3c  5  6 17. 23x  17  5 18. 52y  14  6 19. 3 `

w  4 `  1  4 2

20. 2 ` 3 

v `  1  5 3

4x  5 1 7  `  3 2 6

38. `

2y  3 3 15  ` 6 4 8 16

39. n  3 7 7

40. m  1 7 5

41. 2w  5  11 q 5 1 43.   2 6 3

42. 5v  3  23 p 3 9 44.   5 2 4

45. 35  7d  9  15

46. 52c  7  1  11

47. 4z  9  6  4

48. 5u  3  8 7 6

49. 45  2h  9 7 11 50. 37  2k  11 7 10 51. 3.94q  5  8.7  22.5

21. 8.7p  7.5  26.6  8.2

52. 0.92p  7  16.11  10.89

22. 5.3q  9.2  6.7  43.8

53. 2 6 ` 3m 

23. 8.72.5x  26.6  8.2 24. 5.31.25n  6.7  43.8

1 4 `  5 5 5 3 54. 4  `  2n `  4 4

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30 

WORKING WITH FORMULAS

55. Spring Oscillation | d  x |  L A weight attached to a spring hangs at rest a distance of x in. off the ground. If the weight is pulled down (stretched) a distance of L inches and released, the weight begins to bounce and its distance d off the ground must satisfy the indicated formula. If x equals 4 ft and the spring is stretched 3 in. and released, solve the inequality to find what distances from the ground the weight will oscillate between.



1-30

CHAPTER 1 Equations and Inequalities

56. A “Fair” Coin `

h  50 `  1.645 5

If we flipped a coin 100 times, we expect “heads” to come up about 50 times if the coin is “fair.” In a study of probability, it can be shown that the number of heads h that appears in such an experiment must satisfy the given inequality to be considered “fair.” (a) Solve this inequality for h. (b) If you flipped a coin 100 times and obtained 40 heads, is the coin “fair”?

APPLICATIONS

Solve each application of absolute value.

57. Altitude of jet stream: To take advantage of the jet stream, an airplane must fly at a height h (in feet) that satisfies the inequality h  35,050  2550. Solve the inequality and determine if an altitude of 34,000 ft will place the plane in the jet stream. 58. Quality control tests: In order to satisfy quality control, the marble columns a company produces must earn a stress test score S that satisfies the inequality S  17,750  275. Solve the inequality and determine if a score of 17,500 is in the passing range. 59. Submarine depth: The sonar operator on a submarine detects an old World War II submarine net and must decide to detour over or under the net. The computer gives him a depth model d  394  20 7 164, where d is the depth in feet that represents safe passage. At what depth should the submarine travel to go under or over the net? Answer using simple inequalities. 60. Optimal fishing depth: When deep-sea fishing, the optimal depths d (in feet) for catching a certain type of fish satisfy the inequality 28d  350  1400 6 0. Find the range of depths that offer the best fishing. Answer using simple inequalities. For Exercises 61 through 64, (a) develop a model that uses an absolute value inequality, and (b) solve.

61. Stock value: My stock in MMM Corporation fluctuated a great deal in 2009, but never by more than $3.35 from its current value. If the stock is worth $37.58 today, what was its range in 2009?

62. Traffic studies: On a given day, the volume of traffic at a busy intersection averages 726 cars per hour (cph). During rush hour the volume is much higher, during “off hours” much lighter. Find the range of this volume if it never varies by more than 235 cph from the average. 63. Physical training for recruits: For all recruits in the 3rd Armored Battalion, the average number of sit-ups is 125. For an individual recruit, the amount varies by no more than 23 sit-ups from the battalion average. Find the range of sit-ups for this battalion. 64. Computer consultant salaries: The national average salary for a computer consultant is $53,336. For a large computer firm, the salaries offered to their employees varies by no more than $11,994 from this national average. Find the range of salaries offered by this company. 65. According to the official rules for golf, baseball, pool, and bowling, (a) golf balls must be within 0.03 mm of d  42.7 mm, (b) baseballs must be within 1.01 mm of d  73.78 mm, (c) billiard balls must be within 0.127 mm of d  57.150 mm, and (d) bowling balls must be within 12.05 mm of d  2171.05 mm. Write each statement using an absolute value inequality, then (e) determine which sport gives the least tolerance t width of interval b for the diameter of the ball. at  average value

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Mid-Chapter Check

66. The machines that fill boxes of breakfast cereal are programmed to fill each box within a certain tolerance. If the box is overfilled, the company loses money. If it is underfilled, it is considered unsuitable for sale. Suppose that boxes marked “14 ounces” of cereal must be filled to within 

0.1 oz. Write this relationship as an absolute value inequality, then solve the inequality and explain what your answer means. Let W represent weight.

EXTENDING THE CONCEPT

67. Determine the value or values (if any) that will make the equation or inequality true. x a. x  x  8 b. x  2  2 c. x  x  x  x d. x  3  6x e. 2x  1  x  3 

31

68. The equation 5  2x  3  2x has only one solution. Find it and explain why there is only one.

MAINTAINING YOUR SKILLS

69. (R.4) Factor the expression completely: 18x3  21x2  60x. 70. (1.1) Solve V2  71. (R.6) Simplify

2W for  (physics). CA

72. (1.2) Solve the inequality, then write the solution set in interval notation: 312x  52 7 21x  12  7.

1

by rationalizing the 3  23 denominator. State the result in exact form and approximate form (to hundredths):

MID-CHAPTER CHECK 1. Solve each equation. If the equation is an identity or contradiction, so state and name the solution set. r a.  5  2 3 b. 512x  12  4  9x  7 c. m  21m  32  1  1m  72 3 1 d. y  3  y  2 5 2 3 1 e. 15j  22  1 j  42  j 2 2 f. 0.61x  32  0.3  1.8 Solve for the variable specified. 2. H  16t2  v0t; for v0

3. S  2x2  x2y; for x 4. Solve each inequality and graph the solution set. a. 5x  16  11 or 3x  2  4 1 5 3 1 6 x  b. 2 12 6 4 5. Determine the domain of each expression. Write your answer in interval notation. a.

3x  1 2x  5

b. 217  6x

6. Solve the following absolute value equations. Write the solution in set notation. 2 11 a. d  5  1  7 b. 5  s  3  3 2

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1-32

CHAPTER 1 Equations and Inequalities

7. Solve the following absolute value inequalities. Write solutions in interval notation. a. 3q  4  2 6 10 x b. `  2 `  5  5 3 8. Solve the following absolute value inequalities. Write solutions in interval notation. a. 3.1d  2  1.1  7.3 1  y 11 2 7 b. 3 2 c. 5k  2  3 6 4

9. Motocross: An enduro motocross motorcyclist averages 30 mph through the first part of a 115-mi course, and 50 mph though the second part. If the rider took 2 hr and 50 min to complete the course, how long was she on the first part? 10. Kiteboarding: With the correct sized kite, a person can kiteboard when the wind is blowing at a speed w (in mph) that satisfies the inequality w  17  9. Solve the inequality and determine if a person can kiteboard with a windspeed of 9 mph.

REINFORCING BASIC CONCEPTS x  3  4 can be read, “the distance between 3 and an unknown number is equal to 4.” The advantage of reading it in this way (instead of the absolute value of x minus 3 is 4), is that a much clearer visualization is formed, giving a constant reminder there are two solutions. In diagram form we have Figure 1.10.

Using Distance to Understand Absolute Value Equations and Inequalities In Appendix I.A we noted that for any two numbers a and b on the number line, the distance between a and b is written a  b or b  a. In exactly the same way, the equation Distance between 3 and x is 4.

Figure 1.10

5 4 3 2

4 units 1

0

1

4 units 2

3

From this we note the solution is x  1 or x  7. In the case of an inequality such as x  2  3, we rewrite the inequality as x  122   3 and read it, “the distance between 2 and an unknown number is less than Distance between ⫺2 and x is less than or equal to 3.

Figure 1.11

⫺8 ⫺7 ⫺6

Figure 1.12

⫺6 ⫺5 ⫺4 ⫺3

3 units

3 units 0

6

7

8

9

Distance between ⫺2 and x is less than or equal to 3.

3 units

⫺5 ⫺4 ⫺3 ⫺2 ⫺1

ⴚ2 ⫺1

5

or equal to 3.” With some practice, visualizing this relationship mentally enables a quick statement of the solution: x  3 5, 14. In diagram form we have Figure 1.11.

Equations and inequalities where the coefficient of x is not 1 still lend themselves to this form of conceptual understanding. For 2x  1  3 we read, “the distance between 1 Distance between 1 and 2x is greater than or equal to 3.

4

Distance between 3 and x is 4.

0

1

2

3

4

5

and twice an unknown number is greater than or equal to 3.” On the number line (Figure 1.12), the number 3 units to the right of 1 is 4, and the number 3 units to the left of 1 is 2. 3 units

1

For 2x  2, x  1, and for 2x  4, x  2, and the solution is x  1q, 14 ´ 3 2, q 2. Attempt to solve the following equations and inequalities by visualizing a number line. Check all results algebraically.

6

2

3

Distance between 1 and 2x is greater than or equal to 3. 4

5

6

7

8

Exercise 1: x  2  5 Exercise 2: x  1  4 Exercise 3: 2x  3  5

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Precalculus—

1.4 Complex Numbers Learning Objectives In Section 1.4 you will learn how to:

A. Identify and simplify imaginary and complex numbers

B. Add and subtract complex numbers

C. Multiply complex numbers and find powers of i

For centuries, even the most prominent mathematicians refused to work with equations like x2  1  0. Using the principal of square roots gave the “solutions” x  11 and x   11, which they found baffling and mysterious, since there is no real number whose square is 1. In this section, we’ll see how this “mystery” was finally resolved.

A. Identifying and Simplifying Imaginary and Complex Numbers The equation x2  1 has no real solutions, since the square of any real number is positive. But if we apply the principle of square roots we get x  11 and x   11, which seem to check when substituted into the original equation:

D. Divide complex numbers

x2  1  0

(1)

1 112  1  0 2

1  1  0✓

(2)

1112  1  0 2

1  1  0✓

original equation substitute 11 for x answer “checks” substitute  11 for x answer “checks”

This observation likely played a part in prompting Renaissance mathematicians to study such numbers in greater depth, as they reasoned that while these were not real number solutions, they must be solutions of a new and different kind. Their study eventually resulted in the introduction of the set of imaginary numbers and the imaginary unit i, as follows. Imaginary Numbers and the Imaginary Unit

• Imaginary numbers are those of the form 1k, where k is a positive real number. • The imaginary unit i represents the number whose square is 1: i2  1 and i  11 WORTHY OF NOTE It was René Descartes (in 1637) who first used the term imaginary to describe these numbers; Leonhard Euler (in 1777) who introduced the letter i to represent 11; and Carl F. Gauss (in 1831) who first used the phrase complex number to describe solutions that had both a real number part and an imaginary part. For more on complex numbers and their story, see www.mhhe.com/coburn

As a convenience to understanding and working with imaginary numbers, we rewrite them in terms of i, allowing that the product property of radicals 1 1AB  1A1B2 still applies if only one of the radicands is negative. For 13, we have 11 # 3  1113  i 13. In general, we simply state the following property. Rewriting Imaginary Numbers

• For any positive real number k, 1k  i 1k. For 120 we have: 120  i 120  i 14 # 5  2i 15, and we say the expression has been simplified and written in terms of i. Note that we’ve written the result with the unit “i” in front of the radical to prevent it being interpreted as being under the radical. In symbols, 2i 15  2 15i  215i. The solutions to x2  1 also serve to illustrate that for k 7 0, there are two solutions to x2  k, namely, i 1k and i1k. In other words, every negative number has two square roots, one positive and one negative. The first of these, i 1k, is called the principal square root of k.

1-33

33

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1-34

CHAPTER 1 Equations and Inequalities

EXAMPLE 1



Simplifying Imaginary Numbers Rewrite the imaginary numbers in terms of i and simplify if possible. a. 17 b. 181 c. 124 d. 3116

Solution



a. 17  i 17

b. 181  i 181  9i d. 3116  3i 116  3i142  12i

c. 124  i 124  i 14 # 6  2i 16

Now try Exercises 7 through 12

EXAMPLE 2





Writing an Expression in Terms of i 6  116 6  116 and x  are not real, but are known 2 2 6  116 to be solutions of x2  6x  13  0. Simplify . 2 Using the i notation, we have The numbers x 

Solution



6  i 116 6  116  2 2 6  4i  2 213  2i2  2  3  2i

WORTHY OF NOTE 6  4i from 2 the solution of Example 2 can also be simplified by rewriting it as two separate terms, then simplifying each term: 6  4i 6 4i   2 2 2  3  2i. The expression

write in i notation

simplify

factor numerator reduce

Now try Exercises 13 through 16



The result in Example 2 contains both a real number part 132 and an imaginary part 12i2. Numbers of this type are called complex numbers. Complex Numbers Complex numbers are numbers that can be written in the form a  bi, where a and b are real numbers and i  11. The expression a  bi is called the standard form of a complex number. From this definition we note that all real numbers are also complex numbers, since a  0i is complex with b  0. In addition, all imaginary numbers are complex numbers, since 0  bi is a complex number with a  0.

EXAMPLE 3



Writing Complex Numbers in Standard Form Write each complex number in the form a  bi, and identify the values of a and b. 4  3 125 a. 2  149 b. 112 c. 7 d. 20

Solution



a. 2  149  2  i 149  2  7i a  2, b  7

b. 112  0  i 112  0  2i13 a  0, b  213

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35

Section 1.4 Complex Numbers

c. 7  7  0i a  7, b  0

d.

4  3 125 4  3i125  20 20 4  15i  20 3 1   i 5 4 1 3 a ,b 5 4 Now try Exercises 17 through 24

A. You’ve just learned how to identify and simplify imaginary and complex numbers



Complex numbers complete the development of our “numerical landscape.” Sets of numbers and their relationships are represented in Figure 1.13, which shows how some sets of numbers are nested within larger sets and highlights the fact that complex numbers consist of a real number part (any number within the orange rectangle), and an imaginary number part (any number within the yellow rectangle).

C (complex): Numbers of the form a  bi, where a, b  R and i  1.

Q (rational): {qp, where p, q  z and q  0}

H (irrational): Numbers that cannot be written as the ratio of two integers; a real number that is not rational. 2, 7, 10, 0.070070007... and so on.

Z (integer): {... , 2, 1, 0, 1, 2, ...} W (whole): {0, 1, 2, 3, ...} N (natural): {1, 2, 3, ...}

i (imaginary): Numbers of the form k, where k > 0 7 9 0.25 a  bi, where a  0 i3

5i

3 i 4

R (real): All rational and irrational numbers: a  bi, where a  R and b  0.

Figure 1.13

B. Adding and Subtracting Complex Numbers The sum and difference of two polynomials is computed by identifying and combining like terms. The sum or difference of two complex numbers is computed in a similar way, by adding the real number parts from each, and the imaginary parts from each. Notice in Example 4 that the commutative, associative, and distributive properties also apply to complex numbers. EXAMPLE 4



Adding and Subtracting Complex Numbers Perform the indicated operation and write the result in a  bi form. a. 12  3i2  15  2i2 b. 15  4i2  12  12i2

Solution



a. 12  3i2  15  2i2  2  3i  152  2i  2  152  3i  2i  32  152 4  13i  2i2  3  5i

original sum distribute commute terms group like terms result

b. 15  4i2  12  12 i2  5  4i  2  12 i  5  2  14i2  12 i  15  22  3 14i2  12 i4  3  14  122i

original difference distribute commute terms group like terms result

Now try Exercises 25 through 30



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1-36

CHAPTER 1 Equations and Inequalities

C. Multiplying Complex Numbers; Powers of i

B. You’ve just learned how to add and subtract complex numbers

EXAMPLE 5



The product of two complex numbers is computed using the distributive property and the F-O-I-L process in the same way we apply these to binomials. If any result gives a factor of i 2, remember that i2  1.

Multiplying Complex Numbers Find the indicated product and write the answer in a  bi form. a. 1419 b. 16 12  132 c. 16  5i214  i2 d. 12  3i212  3i2

Solution



a. 14 19  i 14 # i 19  2i # 3i  6i2  6  0i

rewrite in terms of i simplify multiply result 1i 2  12

b. 16 12  132  i 1612  i 132 terms of i  2i 16  i2 118 distribute  2i 16  112 1912 i 2  1  2i 16  312 simplify  3 12  2i 16 standard

rewrite in

form

c. 16  5i214  i2  162142  6i  15i2 142  15i21i2  24  6i  120i2  152i2  24  6i  120i2  152112  29  14i

d. 12  3i212  3i2  122 2  13i2 2 i # i  i2  4  9i2 i 2  1  4  9112 result  13  0i F-O-I-L

1A  B21A  B2  A2  B 2 13i2 2  9i 2

i 2  1 result

Now try Exercises 31 through 48

CAUTION

WORTHY OF NOTE Notice that the product of a complex number and its conjugate also gives us a method for factoring the sum of two squares using complex numbers! For the expression x2  4, the factored form would be 1x  2i 21x  2i 2. For more on this idea, see Exercise 79.





When computing with imaginary and complex numbers, always write the square root of a negative number in terms of i before you begin, as shown in Examples 5(a) and 5(b). Otherwise we get conflicting results, since 14 19  136  6 if we multiply the radicands first, which is an incorrect result because the original factors were imaginary. See Exercise 80.

Recall that expressions 2x  5 and 2x  5 are called binomial conjugates. In the same way, a  bi and a  bi are called complex conjugates. Note from Example 5(d) that the product of the complex number a  bi with its complex conjugate a  bi is a real number. This relationship is useful when rationalizing expressions with a complex number in the denominator, and we generalize the result as follows: Product of Complex Conjugates For a complex number a  bi and its conjugate a  bi, their product 1a  bi2 1a  bi2 is the real number a2  b2; 1a  bi21a  bi2  a2  b2

Showing that 1a  bi21a  bi2  a2  b2 is left as an exercise (see Exercise 79), but from here on, when asked to compute the product of complex conjugates, simply refer to the formula as illustrated here: 13  5i213  5i2  132 2  52 or 34.

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Section 1.4 Complex Numbers

37

These operations on complex numbers enable us to verify complex solutions by substitution, in the same way we verify solutions for real numbers. In Example 2 we stated that x  3  2i was one solution to x2  6x  13  0. This is verified here. EXAMPLE 6



Checking a Complex Root by Substitution Verify that x  3  2i is a solution to x2  6x  13  0.

Solution



x2  6x  13  0 original equation 13  2i2  613  2i2  13  0 substitute 3  2i for x 132 2  2132 12i2  12i2 2  18  12i  13  0 square and distribute 9  12i  4i2  12i  5  0 simplify 2 9  142  5  0 combine terms 112i  12i  0; i  12 0  0✓ 2

Now try Exercises 49 through 56

EXAMPLE 7





Checking a Complex Root by Substitution Show that x  2  i13 is a solution of x2  4x  7.

Solution



x2  4x  7 12  i 132 2  412  i 132  7 4  4i 13  1i 132 2  8  4i 13  7 4  4i 13  3  8  4i 13  7 7  7✓

original equation substitute 2  i 13 for x square and distribute 1i132 2  3 solution checks

Now try Exercises 57 through 60



The imaginary unit i has another interesting and useful property. Since i  11 and i2  1, we know that i3  i2 # i  112i  i and i4  1i2 2 2  1. We can now simplify any higher power of i by rewriting the expression in terms of i4. i5  i4 # i  i i6  i4 # i2  1 i7  i4 # i3  i i8  1i4 2 2  1

Notice the powers of i “cycle through” the four values i, 1, i and 1. In more advanced classes, powers of complex numbers play an important role, and next we learn to reduce higher powers using the power property of exponents and i4  1. Essentially, we divide the exponent on i by 4, then use the remainder to compute the value of the expression. For i35, 35  4  8 remainder 3, showing i35  1i4 2 8 # i3  i. EXAMPLE 8



Simplifying Higher Powers of i Simplify: a. i22

Solution



b. i28

a. i22  1i4 2 5 # 1i2 2  112 5 112  1

c. i57

d. i75

b. i28  1i4 2 7  112 7 1

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1-38

CHAPTER 1 Equations and Inequalities

C. You’ve just learned how to multiply complex numbers and find powers of i

c. i57  1i4 2 14 # i  112 14i i

d. i75  1i4 2 18 # 1i3 2  112 18 1i2  i Now try Exercises 61 and 62



D. Division of Complex Numbers 3i actually have a radical in the denominator. To 2i divide complex numbers, we simply apply our earlier method of rationalizing denominators (Appendix I.F), but this time using a complex conjugate. Since i  11, expressions like

EXAMPLE 9



Dividing Complex Numbers Divide and write each result in a  bi form. 2 3i 6  136 a. b. c. 5i 2i 3  19

Solution



2 2 #5i  5i 5i 5i 215  i2  2 5  12 10  2i  26 10 2   i 26 26 1 5  i  13 13 6  136 6  i 136  c. 3  19 3  i 19 6  6i  3  3i

a.

b.

3i 3i 2i #  2i 2i 2i 6  3i  2i  i2  22  1 2 6  5i  112  5 5 5i 5  5i    5 5 5 1i

convert to i notation

simplify

The expression can be further simplified by reducing common factors. 

611  i2 2 311  i2

factor and reduce

Now try Exercises 63 through 68



Operations on complex numbers can be checked using inverse operations, just as we do for real numbers. To check the answer 1  i from Example 9(b), we multiply it by the divisor: 11  i2 12  i2  2  i  2i  i2  2  i  112

D. You’ve just learned how to divide complex numbers

2i1  3  i✓

Several checks are asked for in the exercises.

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1-39

Section 1.4 Complex Numbers

39

1.4 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

1. Given the complex number 3  2i, its complex conjugate is . 2. The product 13  2i213  2i2 gives the real number .

4  6i12 is written in the standard 2 form a  bi, then a  and b  .

3. If the expression



4. For i  11, i2  , i4  , i6  , and 8 3 5 7 i  ,i  ,i  ,i  , and i9 

.

5. Discuss/Explain which is correct: a. 14 # 19  1142 192  136  6 b. 14 # 19  2i # 3i  6i2  6

6. Compare/Contrast the product 11  12211  132 with the product 11  i 12211  i132. What is the same? What is different?

DEVELOPING YOUR SKILLS

Simplify each radical (if possible). If imaginary, rewrite in terms of i and simplify.

7. a. 116 c. 127

b. 149 d. 172

8. a. 181 c. 164

b. 1169 d. 198

9. a.  118 c. 3 125

b.  150 d. 2 19

10. a.  132 c. 3 1144

b.  175 d. 2 181

11. a. 119 12 c. A 25

b. 131 9 d. A 32

12. a. 117 45 c. A 36

b. 153 49 d. A 75

Write each complex number in the standard form a  bi and clearly identify the values of a and b.

2  14 13. a. 2

6  127 b. 3

14. a.

16  18 2

b.

4  3120 2

15. a.

8  116 2

b.

10  150 5

6  172 16. a. 4

12  1200 b. 8

17. a. 5

b. 3i

18. a. 2

b. 4i

19. a. 2 181

b.

132 8

20. a. 3136

b.

175 15

21. a. 4  150

b. 5  127

22. a. 2  148

b. 7  175

23. a.

14  198 8

b.

5  1250 10

24. a.

21  163 12

b.

8  127 6

Perform the addition or subtraction. Write the result in a  bi form.

25. a. 112  142  17  192 b. 13  1252  11  1812 c. 111  11082  12  1482

26. a. 17  1722  18  1502 b. 1 13  122  1 112  182 c. 1 120  132  1 15  1122 27. a. 12  3i2  15  i2 b. 15  2i2  13  2i2 c. 16  5i2  14  3i2

28. a. 12  5i2  13  i2 b. 17  4i2  12  3i2 c. 12.5  3.1i2  14.3  2.4i2

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CHAPTER 1 Equations and Inequalities

29. a. 13.7  6.1i2  11  5.9i2 2 3 b. a8  ib  a7  ib 4 3 1 5 c. a6  ib  a4  ib 8 2 30. a. 19.4  8.7i2  16.5  4.1i2 7 3 b. a3  ib  a11  ib 5 15 5 3 c. a4  ib  a13  ib 6 8 Multiply and write your answer in a  bi form.

Use substitution to determine if the value shown is a solution to the given equation.

49. x2  36  0; x  6 50. x2  16  0; x  4 51. x2  49  0; x  7i 52. x2  25  0; x  5i

53. 1x  32 2  9; x  3  3i

54. 1x  12 2  4; x  1  2i

55. x2  2x  5  0; x  1  2i 56. x2  6x  13  0; x  3  2i

31. a. 5i # 13i2

b. 14i214i2

57. x2  4x  9  0; x  2  i 15

b. 713  5i2

58. x2  2x  4  0; x  1  13 i

33. a. 7i15  3i2

b. 6i13  7i2

34. a. 14  2i213  2i2 b. 12  3i215  i2

59. Show that x  1  4i is a solution to x2  2x  17  0. Then show its complex conjugate 1  4i is also a solution.

36. a. 15  2i217  3i2 b. 14  i217  2i2

60. Show that x  2  3 12 i is a solution to x2  4x  22  0. Then show its complex conjugate 2  3 12 i is also a solution.

32. a. 312  3i2

35. a. 13  2i212  3i2 b. 13  2i211  i2

For each complex number, name the complex conjugate. Then find the product.

Simplify using powers of i.

37. a. 4  5i

b. 3  i12

61. a. i48

b. i26

c. i39

d. i53

38. a. 2  i

b. 1  i 15

62. a. i36

b. i50

c. i19

d. i65

39. a. 7i

b.

1 2

 23i

40. a. 5i

b.

3 4

 15i

Compute the special products and write your answer in a  bi form.

41. a. 14  5i214  5i2 b. 17  5i217  5i2

42. a. 12  7i212  7i2 b. 12  i212  i2

43. a. 13  i 122 13  i 122 b. 1 16  23i21 16  23i2 44. a. 15  i 132 15  i 132 b. 1 12  34i21 12  34i2 45. a. 12  3i2 2

b. 13  4i2 2

47. a. 12  5i2 2

b. 13  i122 2

46. a. 12  i2 2

48. a. 12  5i2 2

b. 13  i2 2

b. 12  i132 2

Divide and write your answer in a  bi form. Check your answer using multiplication.

63. a.

2 149

b.

4 125

64. a.

2 1  14

b.

3 2  19

65. a.

7 3  2i

b.

5 2  3i

66. a.

6 1  3i

b.

7 7  2i

67. a.

3  4i 4i

b.

2  3i 3i

68. a.

4  8i 2  4i

b.

3  2i 6  4i

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Section 1.4 Complex Numbers

WORKING WITH FORMULAS

69. Absolute value of a complex number: a  bi  2a2  b2 The absolute value of any complex number a  bi (sometimes called the modulus of the number) is computed by taking the square root of the sum of the squares of a and b. Find the absolute value of the given complex numbers. a. |2  3i| b. |4  3i | c. | 3  12 i| 

70. Binomial cubes: 1A  B2 3  A3  3A2B  3AB2  B3 The cube of any binomial can be found using the formula shown, where A and B are the terms of the binomial. Use the formula to compute 11  2i2 3 (note A  1 and B  2i2.

APPLICATIONS

71. Dawn of imaginary numbers: In a day when imaginary numbers were imperfectly understood, Girolamo Cardano (1501–1576) once posed the problem, “Find two numbers that have a sum of 10 and whose product is 40.” In other words, A  B  10 and AB  40. Although the solution is routine today, at the time the problem posed an enormous challenge. Verify that A  5  115i and B  5  115i satisfy these conditions. 72. Verifying calculations using i: Suppose Cardano had said, “Find two numbers that have a sum of 4 and a product of 7” (see Exercise 71). Verify that A  2  13i and B  2  13i satisfy these conditions. Although it may seem odd, imaginary numbers have several applications in the real world. Many of these involve a study of electrical circuits, in particular alternating current or AC circuits. Briefly, the components of an AC circuit are current I (in amperes), voltage V (in volts), and the impedance Z (in ohms). The impedance of an electrical circuit is a measure of the total opposition to the flow of current through the circuit and is calculated as Z  R  iXL  iXC where R represents a pure resistance, XC represents the capacitance, and XL represents the inductance. Each of these is also measured in ohms (symbolized by ). 

41

73. Find the impedance Z if R  7 , XL  6 , and XC  11 . 74. Find the impedance Z if R  9.2 , XL  5.6 , and XC  8.3 . The voltage V (in volts) across any element in an AC circuit is calculated as a product of the current I and the impedance Z: V  IZ.

75. Find the voltage in a circuit with a current I  3  2i amperes and an impedance of Z  5  5i . 76. Find the voltage in a circuit with a current I  2  3i amperes and an impedance of Z  4  2i . In an AC circuit, the total impedance (in ohms) is given Z1Z2 by Z  , where Z represents the total impedance Z1  Z2 of a circuit that has Z1 and Z2 wired in parallel.

77. Find the total impedance Z if Z1  1  2i and Z2  3  2i. 78. Find the total impedance Z if Z1  3  i and Z2  2  i.

EXTENDING THE CONCEPT

79. Up to this point, we’ve said that expressions like x2  9 and p2  7 are factorable: x2  9  1x  32 1x  32

and

p  7  1p  172 1p  172, 2

while x  9 and p  7 are prime. More correctly, we should state that x2  9 and p2  7 2

2

are nonfactorable using real numbers, since they actually can be factored if complex numbers are used. From 1a  bi2 1a  bi2  a2  b2 we note a2  b2  1a  bi2 1a  bi2, showing x2  9  1x  3i2 1x  3i2 and

p2  7  1p  i 1721p  i 172.

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Use this idea to factor the following.

a. x  36

b. m  3

c. n2  12

d. 4x2  49

2

2

80. In this section, we noted that the product property of radicals 1AB  1A1B, can still be applied when at most one of the factors is negative. So what happens if both are negative? First consider the expression 14 # 25. What happens if you first multiply in the radicand, then compute the square root? Next consider the product 14 # 125. Rewrite each factor using the i notation, then compute the product. Do you get the same result as before? What can you say about 14 # 25 and 14 # 125? 

1-42

CHAPTER 1 Equations and Inequalities

81. Simplify the expression i17 13  4i2  3i3 11  2i2 2. 82. While it is a simple concept for real numbers, the square root of a complex number is much more involved due to the interplay between its real and imaginary parts. For z  a  bi the square root of z can be found using the formula: 12 1 1z  a  i 1z  a2, where the sign 1z  2 is chosen to match the sign of b (see Exercise 69). Use the formula to find the square root of each complex number, then check by squaring. a. z  7  24i c. z  4  3i

b. z  5  12i

MAINTAINING YOUR SKILLS

83. (R.7) State the perimeter and area formulas for: (a) squares, (b) rectangles, (c) triangles, and (d) circles.

85. (1.1) John can run 10 m/sec, while Rick can only run 9 m/sec. If Rick gets a 2-sec head start, who will hit the 200-m finish line first?

84. (R.1) Write the symbols in words and state True/False. a. 6   b.  (  c. 103  53, 4, 5, p6 d.   

86. (R.4) Factor the following expressions completely. a. x4  16 b. n3  27 c. x3  x2  x  1 d. 4n2m  12nm2  9m3

1.5 Solving Quadratic Equations Learning Objectives In Section 1.5 you will learn how to:

A. Solve quadratic equations using the zero product property

B. Solve quadratic equations using the square root property of equality

C. Solve quadratic equations by completing the square

D. Solve quadratic equations using the quadratic formula

E. Use the discriminant to identify solutions

F. Solve applications of quadratic equations

In Section 1.1 we solved the equation ax  b  c for x to establish a general solution for all linear equations of this form. In this section, we’ll establish a general solution for the quadratic equation ax2  bx  c  0, 1a  02 using a process known as completing the square. Other applications of completing the square include the graphing of parabolas, circles, and other relations from the family of conic sections.

A. Quadratic Equations and the Zero Product Property A quadratic equation is one that can be written in the form ax2  bx  c  0, where a, b, and c are real numbers and a  0. As shown, the equation is written in standard form, meaning the terms are in decreasing order of degree and the equation is set equal to zero. Quadratic Equations A quadratic equation is one that can be written in the form ax2  bx  c  0, with a, b, c  , and a  0.

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Section 1.5 Solving Quadratic Equations

Notice that a is the leading coefficient, b is the coefficient of the linear (first degree) term, and c is a constant. All quadratic equations have degree two, but can have one, two, or three terms. The equation n2  81  0 is a quadratic equation with two terms, where a  1, b  0, and c  81. EXAMPLE 1



Determining Whether an Equation Is Quadratic State whether the given equation is quadratic. If yes, identify coefficients a, b, and c. 3 a. 2x2  18  0 b. z  12  3z2  0 c. x50 4 d. z3  2z2  7z  8 e. 0.8x2  0

Solution



Standard Form

Quadratic

Coefficients

WORTHY OF NOTE

a.

2x  18  0

yes, deg 2

a2

The word quadratic comes from the Latin word quadratum, meaning square. The word historically refers to the “four sidedness” of a square, but mathematically to the area of a square. Hence its application to polynomials of the form ax2  bx  c— the variable of the leading term is squared.

b.

3z2  z  12  0

yes, deg 2

a  3

c.

3 x50 4

no, deg 1

(linear equation)

d.

z3  2z2  7z  8  0

no, deg 3

(cubic equation)

e.

0.8x  0

yes, deg 2

2

2

b0

a  0.8

c  18

b1

b0

c  12

c0

Now try Exercises 7 through 18



With quadratic and other polynomial equations, we generally cannot isolate the variable on one side using only properties of equality, because the variable is raised to different powers. Instead we attempt to solve the equation by factoring and applying the zero product property. Zero Product Property If A and B represent real numbers or real valued expressions and A # B  0, then A  0 or B  0. In words, the property says, If the product of any two (or more) factors is equal to zero, then at least one of the factors must be equal to zero. We can use this property to solve higher degree equations after rewriting them in terms of equations with lesser degree. As with linear equations, values that make the original equation true are called solutions or roots of the equation.

EXAMPLE 2



Solving Equations Using the Zero Product Property Solve by writing the equations in factored form and applying the zero product property. a. 3x2  5x b. 5x  2x2  3 c. 4x2  12x  9

Solution



a.

3x2  5x given equation b. 5x  2x2  3 given equation 3x2  5x  0 standard form 2x2  5x  3  0 standard form factor x 13x  52  0 12x  12 1x  32  0 factor x  0 or 3x  5  0 set factors equal to zero 2x  1  0 or x  3  0 set factors equal (zero product property) to zero (zero product property) 5 1 x  0 or x  result x or x  3 result 3 2

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c.

4x2  12x  9 4x  12x  9  0 12x  32 12x  32  0 2x  3  0 or 2x  3  0 3 3 x x or 2 2 2

given equation standard form factor set factors equal to zero (zero product property) result

3 This equation has only the solution x  , which we call a repeated root. 2 Now try Exercises 19 through 42



CAUTION

A. You’ve just learned how to solve quadratic equations using the zero product property



Consider the equation x2  2x  3  12. While the left-hand side is factorable, the result

is 1x  321x  12  12 and finding a solution becomes a “guessing game” because the equation is not set equal to zero. If you misapply the zero factor property and say that x  3  12 or x  1  12, the “solutions” are x  15 or x  11, which are both incorrect! After subtracting 12 from both sides x2  2x  3  12 becomes x2  2x  15  0, giving 1x  521x  32  0 with solutions x  5 or x  3.

B. Solving Quadratic Equations Using the Square Root Property of Equality The equation x2  9 can be solved by factoring. In standard form we have x2  9  0 (note b  02, then 1x  32 1x  32  0. The solutions are x  3 or x  3, which are simply the positive and negative square roots of 9. This result suggests an alternative method for solving equations of the form X2  k, known as the square root property of equality.

WORTHY OF NOTE In Section R.6 we noted that for any real number a, 2a2  a. From Example 3(a), solving the equation by taking the square root of both sides produces 2x2  294. This is equivalent to x  294, again showing this equation must have two solutions, x   294 and x  294.

EXAMPLE 3

Square Root Property of Equality If X represents an algebraic expression and X2  k, then X  1k or X   1k; also written as X  1k



Solving an Equation Using the Square Root Property of Equality Use the square root property of equality to solve each equation. a. 4x2  3  6 b. x2  12  0 c. 1x  52 2  24

Solution



a. 4x2  3  6 9 x2  4 9 9 x or x   A4 A4 3 3 x or x   2 2

original equation subtract 3, divide by 4

square root property of equality

simplify radicals

This equation has two rational solutions.

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b. x2  12  0 x2  12 x  112 or x  2i 13 or

original equation subtract 12

x   112 x  2i 13

square root property of equality simplify radicals

This equation has two complex solutions. c. 1x  52 2  24 original equation x  5  124 or x  5   124 square root property of equality x  5  2 16 x  5  2 16 solve for x and simplify radicals This equation has two irrational solutions.

B. You’ve just learned how to solve quadratic equations using the square root property of equality

Now try Exercises 43 through 58 CAUTION

45





For equations of the form 1x  d 2 2  k [see Example 3(c)], you should resist the temptation to expand the binomial square in an attempt to simplify the equation and solve by factoring—many times the result is nonfactorable. Any equation of the form 1x  d 2 2  k can quickly be solved using the square root property of equality.

Answers written using radicals are called exact or closed form solutions. Actually checking the exact solutions is a nice application of fundamental skills. Let’s check x  5  2 16 from Example 3(c). check:

1x  52 2  24 15  2 16  52 2  24 12 162 2  24 4162  24 24  24✓

original equation substitute 5  2 16 for x simplify 12 162 2  4162

result checks 1x  5  2 16 also checks)

C. Solving Quadratic Equations by Completing the Square

Again consider 1x  52 2  24 from Example 3(c). If we had first expanded the binomial square, we would have obtained x2  10x  25  24, then x2  10x  1  0 in standard form. Note that this equation cannot be solved by factoring. Reversing this process leads us to a strategy for solving nonfactorable quadratic equations, by creating a perfect square trinomial from the quadratic and linear terms. This process is known as completing the square. To transform x2  10x  1  0 back into x2  10x  25  24 [which we would then rewrite as 1x  52 2  24 and solve], we subtract 1 from both sides, then add 25: x2  10x  1  0 x2  10x  1 x  10x  25  1  25 2

1x  52  24 2

subtract 1 add 25 factor, simplify

In general, after subtracting the constant term, the number that “completes the square” is found by squaring 12 the coefficient of the linear term: 12 11022  25. See Exercises 59 through 64 for additional practice. EXAMPLE 4



Solving a Quadratic Equation by Completing the Square Solve by completing the square: x2  13  6x.

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Solution



x2  13  6x x  6x  13  0 x2  6x  __  13  ___ 3 1 12 2 162 4 2  9 2 x  6x  9  13  9 1x  32 2  4 x  3  14 or x  3   14 x  3  2i or x  3  2i 2

original equation standard form subtract 13 to make room for new constant compute 3 A 12 B 1linear coefficient 2 4 2

add 9 to both sides (completing the square) factor and simplify square root property of equality simplify radicals and solve for x

Now try Exercises 65 through 74



The process of completing the square can be applied to any quadratic equation with a leading coefficient of 1. If the leading coefficient is not 1, we simply divide through by a before beginning, which brings us to this summary of the process. WORTHY OF NOTE

Completing the Square to Solve a Quadratic Equation

It’s helpful to note that the number you’re squaring in 1 b b step three, c # d  , 2 a 2a turns out to be the constant term in the factored form. From Example 4, the number we squared was A 12 B 162  3, and the binomial square was 1x  32 2.

EXAMPLE 5

To solve ax2  bx  c  0 by completing the square: 1. Subtract the constant c from both sides. 2. Divide both sides by the leading coefficient a. 1 b 2 3. Compute c # d and add the result to both sides. 2 a 4. Factor left-hand side as a binomial square; simplify right-hand side. 5. Solve using the square root property of equality.



Solving a Quadratic Equation by Completing the Square Solve by completing the square: 3x2  1  4x.

Solution



3x2  1  4x 3x  4x  1  0 3x2  4x  1 4 1 x2  x   3 3 4 4 1 4 x2  x    3 9 3 9 2 2 7 ax  b  3 9 2 7 x  or x  3 A9 2 17 x  or 3 3 x  0.22 or

original equation

2

C. You’ve just learned how to solve quadratic equations by completing the square

standard form (nonfactorable) subtract 1 divide by 3 c

1 4 2 4 4 1 b 2 d  c a ba b d  ; add 2 a 2 3 9 9

1 3 factor and simplify a  b 3 9

7 2  3 A9 17 2 x  3 3 x  1.55

square root property of equality solve for x and simplify (exact form) approximate form (to hundredths)

Now try Exercises 75 through 82

CAUTION





For many of the skills/processes needed in a study of algebra, it’s actually easier to work with the fractional form of a number, rather than the decimal form. For example, com9 puting A 23 B 2 is easier than computing 10.62 2, and finding 216 is much easier than finding 10.5625.

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47

D. Solving Quadratic Equations Using the Quadratic Formula In Section 1.1 we found a general solution for the linear equation ax  b  c by comparing it to 2x  3  15. Here we’ll use a similar idea to find a general solution for quadratic equations. In a side-by-side format, we’ll solve the equations 2x2  5x  3  0 and ax2  bx  c  0 by completing the square. Note the similarities. 2x2  5x  3  0 2x2  5x 

 3

2 1 c 1linear coefficient2 d 2

5 25 25 3 x2  x    2 16 16 2 5 2 25 3 ax  b   4 16 2

left side factors as a binomial square

determine LCDs

5 2 1 ax  b  4 16

simplify right side

x

5 1  4 B 16

x

5 1  4 4

square root property of equality

simplify radicals

5 1 x  4 4

5  1 4

or

1 b 2 b2 c a bd  2 2 a 4a

add to both sides

5 2 25 24 ax  b   4 16 16

x

b2 b b2 c x2  x  2  2  a a 4a 4a b 2 b2 c ax  b  2  a 2a 4a 2 2 b b 4ac ax  b  2  2 2a 4a 4a 2 2 b b  4ac ax  b  2a 4a2 b b2  4ac x  2a B 4a2 b 2b2  4ac x  2a 2a x

solve for x

5  1 4

x

combine terms

5  1 4

solutions

x

 c

c b x2  x  ____   a a

divide by lead coefficient

1 5 2 25 c a bd  2 2 16

x

ax2  bx 

subtract constant term

5 3 x2  x  ___   2 2

x

ax2  bx  c  0

given equations

b  2b2  4ac 2a

or

b 2b2  4ac  2a 2a

b  2b2  4ac 2a

x

b  2b2  4ac 2a

On the left, our final solutions are x  1 or x  32. The general solution is called the quadratic formula, which can be used to solve any equation belonging to the quadratic family. Quadratic Formula If ax2  bx  c  0, with a, b, and c   and a  0, then x

b  2b2  4ac 2a also written x 

CAUTION



or

x

b  2b2  4ac ; 2a

b  2b2  4ac . 2a

It’s very important to note the values of a, b, and c come from an equation written in standard form. For 3x2  5x  7, a  3 and b  5, but c Z 7! In standard form we have 3x2  5x  7  0, and note the value for use in the formula is actually c  7.

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CHAPTER 1 Equations and Inequalities

EXAMPLE 6



Solving Quadratic Equations Using the Quadratic Formula Solve 4x2  1  8x using the quadratic formula. State the solution(s) in both exact and approximate form. Check one of the exact solutions in the original equation.

Solution



Begin by writing the equation in standard form and identifying the values of a, b, and c. 4x2  1  8x 4x  8x  1  0 a  4, b  8, c  1 182  2182 2  4142 112 x 2142 8  148 8  164  16  x 8 8 8 4 13 8  4 13   x 8 8 8 13 13 x1 or x  1  2 2 or x  0.13 x  1.87

original equation

2

Check



D. You’ve just learned how to solve quadratic equations using the quadratic formula

standard form

substitute 4 for a, 8 for b, and 1 for c

simplify

rationalize the radical (see following Caution)

exact solutions approximate solutions

4x2  1  8x 13 2 13 4a1  b  1  8a1  b 2 2 13 3 4 c 1  2a b  d  1  8  4 13 2 4 4  4 13  3  1  8  4 13 8  4 13  8  4 13 ✓

original equation substitute 1 

13 2

for x

square binomial; distribute distribute result checks

Now try Exercises 83 through 112



1

CAUTION



For

8  4 13 8  413 , be careful not to incorrectly “cancel the eights” as in  1  413. 8 8 1

No! Use a calculator to verify that the results are not equivalent. Both terms in the numerator are divided by 8 and we must either rewrite the expression as separate terms (as above) or factor the numerator to see if the expression simplifies further: 1 4 12  132 8  4 13 2  13 13   , which is equivalent to 1  . 8 8 2 2 2

E. The Discriminant of the Quadratic Formula Recall that 1X represents a real number only for X 0. Since the quadratic formula contains the radical 2b2  4ac, the expression b2  4ac, called the discriminant, will determine the nature (real or complex) and the number of solutions to a given quadratic equation. The Discriminant of the Quadratic Formula For ax2  bx  c  0, a  0, 1. If b2  4ac  0, the equation has one real root. 2. If b2  4ac 7 0, the equation has two real roots. 3. If b2  4ac 6 0, the equation has two complex roots.

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Section 1.5 Solving Quadratic Equations

Further analysis of the discriminant reveals even more concerning the nature of quadratic solutions. If a, b, and c are rational and the discriminant is a perfect square, there will be two rational roots, which means the original equation can be solved by factoring. If the discriminant is not a perfect square, there will be two irrational roots that are conjugates. If the discriminant is zero there is one rational root, and the original equation is a perfect square trinomial. EXAMPLE 7



Using the Discriminant to Analyze Solutions Use the discriminant to determine if the equation given has any real root(s). If so, state whether the roots are rational or irrational, and whether the quadratic expression is factorable. a. 2x2  5x  2  0 b. x2  4x  7  0 c. 4x2  20x  25  0

Solution



a. a  2, b  5, c  2 b. a  1, b  4, c  7 c. a  4, b  20, c  25 b2  4ac  152 2  4122 122 b2  4ac  142 2  4112 172 b2  4ac  1202 2  41421252 9  12 0 Since 9 7 0, S two rational roots, factorable

Since 12 6 0, S two complex roots, nonfactorable

Since b2  4ac  0, S one rational root, factorable

Now try Exercises 113 through 124



In Example 7(b), b2  4ac  12 and the quadratic formula shows 4  112 x . After simplifying, we find the solutions are the complex conjugates 2 x  2  i 13 or x  2  i 13. In general, when b2  4ac 6 0, the solutions will be complex conjugates. Complex Solutions The complex solutions of a quadratic equation with real coefficients occur in conjugate pairs. EXAMPLE 8



Solving Quadratic Equations Using the Quadratic Formula Solve: 2x2  6x  5  0.

Solution



With a  2, b  6, and c  5, the discriminant becomes 162 2  4122152  4, showing there will be two complex roots. The quadratic formula then yields b  2b2  4ac 2a 162  14 x 2122 x

6  2i 4 3 1 x  i 2 2 x

E. You’ve just learned how to use the discriminant to identify solutions

quadratic formula

b 2  4ac  4, substitute 2 for a, and 6 for b simplify, write in i form

solutions are complex conjugates

Now try Exercises 125 through 130



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Summary of Solution Methods for ax2  bx  c  0

WORTHY OF NOTE While it’s possible to solve by b completing the square if is a a fraction or an odd number (see Example 5), the process is usually most efficient when b is an even number. This is a one observation you could use when selecting a solution method.

1. If b  0, isolate x and use the square root property of equality. 2. If c  0, factor out the GCF and solve using the zero product property. 3. If no coefficient is zero, you can attempt to solve by a. factoring the trinomial b. completing the square c. using the quadratic formula

F. Applications of the Quadratic Formula A projectile is any object that is thrown, shot, or projected upward with no sustaining source of propulsion. The height of the projectile at time t is modeled by the equation h  16t2  vt  k, where h is the height of the object in feet, t is the elapsed time in seconds, and v is the initial velocity in feet per second. The constant k represents the initial height of the object above ground level, as when a person releases an object 5 ft above the ground in a throwing motion. If the person were on a cliff 60 ft high, k would be 65 ft.

EXAMPLE 9



Solving an Application of Quadratic Equations A person standing on a cliff 60 ft high, throws a ball upward with an initial velocity of 102 ft/sec (assume the ball is released 5 ft above where the person is standing). Find (a) the height of the object after 3 sec and (b) how many seconds until the ball hits the ground at the base of the cliff.

Solution



5 ft

Using the given information, we have h  16t2  102t  65. To find the height after 3 sec, substitute t  3. a. h  16t2  102t  65 original equation 2  16132  102132  65 substitute 3 for t result  227

60 ft

50

After 3 sec, the ball is 227 ft above the ground. b. When the ball hits the ground at the base of the cliff, it has a height of zero. Substitute h  0 and solve using the quadratic formula. 0  16t2  102t  65 b  2b2  4ac t 2a 11022  211022 2  411621652 t 21162 102  114,564 t 32

a  16, b  102, c  65 quadratic formula

substitute 16 for a, 102 for b, 65 for c

simplify

Since we’re trying to find the time in seconds, we go directly to the approximate form of the answer. t  0.58

or

t  6.96

approximate solutions

The ball will strike the base of the cliff about 7 sec later. Since t represents time, the solution t  0.58 does not apply. Now try Exercises 133 through 140



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EXAMPLE 10



Solving Applications Using the Quadratic Formula For the years 1995 to 2002, the amount A of annual international telephone traffic (in billions of minutes) can be modeled by A  0.3x2  8.9x  61.8, where x  0 represents the year 1995 [Source: Data from the 2005 Statistical Abstract of the United States, Table 1372, page 870]. If this trend continues, in what year will the annual number of minutes reach or surpass 275 billion minutes?

Solution



We are essentially asked to solve A  0.3x2  8.9x  61.8, when A  275. 275  0.3x2  8.9x  61.8 0  0.3x2  8.9x  213.2

given equation subtract 275

For a  0.3, b  8.9, and c  213.2, the quadratic formula gives b  2b2  4ac 2a 8.9  218.92 2  410.321213.22 x 210.32 8.9  1335.05 x 0.6 x  15.7 or x  45.3

x

F. You’ve just learned how to solve applications of quadratic equations

quadratic formula

substitute known values

simplify result

We disregard the negative solution (since x represents time), and find the annual number of international telephone minutes will reach or surpass 275 billion 15.7 years after 1995, or in the year 2010. Now try Exercises 141 and 142



TECHNOLOGY HIGHLIGHT

The Discriminant Quadratic equations play an important role in a study of College Algebra, forming a bridge between our previous and current studies, and the more advanced equations to come. As seen in this section, the discriminant of the quadratic formula 1b2  4ac2 reveals the type and number of solutions, and whether the original equation can be solved by factoring (the discriminant is a perfect square). It will often be helpful to have this information in advance of trying to solve or graph the equation. Since this will be done for each new equation, the discriminant is a prime candidate for a short program. To begin a new program press PRGM (NEW) ENTER . The calculator will prompt you to name the program using the green ALPHA letters (eight letters max), then allow you to start entering program lines. In PRGM mode, pressing PRGM once again will bring up menus that contain all needed commands. For very basic programs, these commands will be in the I/O (Input/Output) submenu, with the most common options being 2:Prompt, 3:Disp, and 8:CLRHOME. As you can see, we have named our program DISCRMNT. PROGRAM:DISCRMNT :CLRHOME

Clears the home screen, places cursor in upper left corner

:DISP "DISCRIMINANT "

Displays the word DISCRIMINANT as user information

:DISP "B24AC"

Displays B2  4AC as user information

:DISP ""

Displays a blank line (for formatting)

:Prompt A, B, C

Prompts the user to enter the values of A, B, and C

:B24AC → D

Computes B2  4AC using given values and stores result in memory location D —continued

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:CLRHOME

Clears the home screen, places cursor in upper left corner

:DISP "DISCRIMINANT IS:" Displays the words DISCRIMINANT IS as user information :DISP D

Displays the computed value of D

Exercise 1:

Run the program for x2  3x  10  0 and x2  5x  14  0 to verify that both can be solved by factoring. What do you notice?

Exercise 2:

Run the program for 25x2  90x  81  0 and 4x2  20x  25  0, then check to see if each is a perfect square trinomial. What do you notice?

Exercise 3:

Run the program for y  x2  2x  10 and y  x2  2x  5. Do these equations have real number solutions? Why or why not?

Exercise 4:

b  2D and 2a solutions can quickly be found. Solve the equations in Exercises 1–3 above. Once the discriminant D is known, the quadratic formula becomes x 

1.5 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

1. A polynomial equation is in standard form when written in order of degree and set equal to .

4. The quantity b2  4ac is called the of the quadratic equation. If b2  4ac 7 0, there are real roots.

2. The solution x  2  13 is called an form of the solution. Using a calculator, we find the form is x  3.732.

5. According to the summary on page 50, what method should be used to solve 4x2  5x  0? What are the solutions?

3. To solve a quadratic equation by completing the square, the coefficient of the term must be a .

6. Discuss/Explain why this version of the quadratic formula is incorrect: x  b 



2b2  4ac 2a

DEVELOPING YOUR SKILLS

Determine whether each equation is quadratic. If so, identify the coefficients a, b, and c. If not, discuss why.

7. 2x  15  x  0 2

9. 11.

8. 21  x  4x  0 2

2 x70 3

10. 12  4x  9

1 2 x  6x 4

12. 0.5x  0.25x2

13. 2x2  7  0

14. 5  4x2

15. 3x2  9x  5  2x3  0

16. z2  6z  9  z3  0

17. 1x  12 2  1x  12  4  9

18. 1x  52 2  1x  52  4  17 Solve using the zero factor property. Be sure each equation is in standard form and factor out any common factors before attempting to solve. Check all answers in the original equation.

19. x2  15  2x

20. z2  10z  21

21. m2  8m  16

22. 10n  n2  25

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23. 5p2  10p  0

24. 6q2  18q  0

67. p2  6p  3  0

68. n2  4n  10

25. 14h2  7h

26. 9w  6w2

69. p2  6p  4

70. x2  8x  1  0

27. a2  17  8

28. b2  8  12

71. m2  3m  1

72. n2  5n  2  0

29. g2  18g  70  11

73. n2  5n  5

74. w2  7w  3  0

30. h2  14h  2  51

75. 2x2  7x  4

76. 3w2  8w  4  0

31. m3  5m2  9m  45  0

77. 2n2  3n  9  0

78. 2p2  5p  1

32. n3  3n2  4n  12  0

79. 4p2  3p  2  0

80. 3x2  5x  6  0

81. m2  7m  4

82. a2  15  4a

33. 1c  122c  15  30

34. 1d  102d  10  6

Solve each equation using the most efficient method: factoring, square root property of equality, or the quadratic formula. Write your answer in both exact and approximate form (rounded to hundredths). Check one of the exact solutions in the original equation.

35. 9  1r  52r  33 36. 7  1s  42s  28

37. 1t  421t  72  54

38. 1g  1721g  22  20

83. x2  3x  18

84. w2  6w  1  0

39. 2x2  4x  30  0

85. 4m2  25  0

86. 4a2  4a  1

40. 3z2  12z  36  0

87. 4n2  8n  1  0

88. 2x2  4x  5  0

41. 2w2  5w  3

89. 6w2  w  2

90. 3a2  5a  6  0

42. 3v2  v  2

91. 4m2  12m  15

92. 3p2  p  0

93. 4n2  9  0

94. 4x2  x  3

95. 5w2  6w  8

96. 3m2  7m  6  0

97. 3a2  a  2  0

98. 3n2  2n  3  0

Solve the following equations using the square root property of equality. Write answers in exact form and approximate form rounded to hundredths. If there are no real solutions, so state.

43. m2  16

44. p2  49

45. y2  28  0

46. m2  20  0

47. p2  36  0

48. n2  5  0

49. x2 

50. y2  13 9

21 16

51. 1n  32 2  36 53. 1w  52 2  3

55. 1x  32  7  2 2

57. 1m  22  2

18 49

100. 2x2  x  3  0 102. 3m2  2  5m

103. 2a2  5  3a

104. n2  4n  8  0

105. 2p2  4p  11  0

106. 8x2  5x  1  0

52. 1p  52 2  49

2 1 107. w2  w  3 9

108.

56. 1m  112  5  3

109. 0.2a2  1.2a  0.9  0

54. 1m  42 2  5 2

58. 1x  52 2  12 25

59. x2  6x 

60. y2  10y 

61. n2  3n 

62. x2  5x 

2 63. p2  p  3

3 64. x2  x  2

Solve by completing the square. Write your answers in both exact form and approximate form rounded to the hundredths place. If there are no real solutions, so state.

65. x  6x  5

99. 5p2  6p  3 101. 5w2  w  1

Fill in the blank so the result is a perfect square trinomial, then factor into a binomial square.

2

53

66. m  8m  12 2

1 5 2 8 m  m 0 4 3 6

110. 5.4n2  8.1n  9  0 111.

8 2 2 p 3 p 7 21

112.

5 2 16 3 x  x 9 15 2

Use the discriminant to determine whether the given equation has irrational, rational, repeated, or complex roots. Also state whether the original equation is factorable using integers, but do not solve for x.

113. 3x2  2x  1  0 114. 2x2  5x  3  0 115. 4x  x2  13  0 116. 10x  x2  41  0

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117. 15x2  x  6  0

118. 10x2  11x  35  0

Solve the quadratic equations given. Simplify each result.

119. 4x2  6x  5  0 120. 5x2  3  2x

125. 6x  2x2  5  0 126. 17  2x2  10x

121. 2x2  8  9x

122. x2  4  7x

127. 5x2  5  5x

128. x2  2x  19

123. 4x2  12x  9

124. 9x2  4  12x

129. 2x2  5x  11

130. 4x  3  5x2



WORKING WITH FORMULAS

131. Height of a projectile: h  16t2  vt If an object is projected vertically upward from ground level with no continuing source of propulsion, the height of the object (in feet) is modeled by the equation shown, where v is the initial velocity, and t is the time in seconds. Use the quadratic formula to solve for t in terms of v and h. (Hint: Set the equation equal to zero and identify the coefficients as before.) 

132. Surface area of a cylinder: A  2r2  2rh The surface area of a cylinder is given by the formula shown, where h is the height and r is the radius of the base. The equation can be considered a quadratic in the variable r. Use the quadratic formula to solve for r in terms of h and A. (Hint: Rewrite the equation in standard form and identify the coefficients as before.)

APPLICATIONS

133. Height of a projectile: The height of an object thrown upward from the roof of a building 408 ft tall, with an initial velocity of 96 ft/sec, is given by the equation h  16t2  96t  408, where h represents the height of the object after t seconds. How long will it take the object to hit the ground? Answer in exact form and decimal form rounded to the nearest hundredth. 134. Height of a projectile: The height of an object thrown upward from the floor of a canyon 106 ft deep, with an initial velocity of 120 ft/sec, is given by the equation h  16t2  120t  106, where h represents the height of the object after t seconds. How long will it take the object to rise to the height of the canyon wall? Answer in exact form and decimal form rounded to hundredths. 135. Cost, revenue, and profit: The revenue for a manufacturer of microwave ovens is given by the equation R  x140  13x2, where revenue is in thousands of dollars and x thousand ovens are manufactured and sold. What is the minimum number of microwave ovens that must be sold to bring in a revenue of $900,000? 136. Cost, revenue, and profit: The revenue for a manufacturer of computer printers is given by the equation R  x130  0.4x2 , where revenue is in thousands of dollars and x thousand printers are manufactured and sold. What is the minimum

number of printers that must be sold to bring in a revenue of $440,000? 137. Cost, revenue, and profit: The cost of raw materials to produce plastic toys is given by the cost equation C  2x  35, where x is the number of toys in hundreds. The total income (revenue) from the sale of these toys is given by R  x2  122x  1965. (a) Determine the profit equation 1profit  revenue  cost2. During the Christmas season, the owners of the company decide to manufacture and donate as many toys as they can, without taking a loss (i.e., they break even: profit or P  02. (b) How many toys will they produce for charity? 138. Cost, revenue, and profit: The cost to produce bottled spring water is given by the cost equation C  16x  63, where x is the number of bottles in thousands. The total revenue from the sale of these bottles is given by the equation R  x2  326x  18,463. (a) Determine the profit equation 1profit  revenue  cost2. (b) After a bad flood contaminates the drinking water of a nearby community, the owners decide to bottle and donate as many bottles of water as they can, without taking a loss (i.e., they break even: profit or P  0). How many bottles will they produce for the flood victims?

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139. Height of an arrow: If an object is projected vertically upward from ground level with no continuing source of propulsion, its height (in feet) is modeled by the equation h  16t2  vt, where v is the initial velocity and t is the time in seconds. Use the quadratic formula to solve for t, given an arrow is shot into the air with v  144 ft/sec and h  260 ft. See Exercise 131. 140. Surface area of a cylinder: The surface area of a cylinder is given by A  2r2  2rh, where h is the height and r is the radius of the base. The equation can be considered a quadratic in the variable r. Use the quadratic formula to solve for r, given A  4710 cm2 and h  35 cm. See Exercise 132. 141. Cell phone subscribers: For the years 1995 to 2002, the number N of cellular phone subscribers (in millions) can be modeled by the equation N  17.4x2  36.1x  83.3, where x  0 represents the year 1995 [Source: Data from the 2005 Statistical Abstract of the United States, Table 1372, page 870]. If this trend continued, in what year did the number of subscribers reach or surpass 3750 million? 

55

Section 1.5 Solving Quadratic Equations

142. U.S. international trade balance: For the years 1995 to 2003, the international trade balance B (in millions of dollars) can be approximated by the equation B  3.1x2  4.5x  19.9, where x  0 represents the year 1995 [Source: Data from the 2005 Statistical Abstract of the United States, Table 1278, page 799]. If this trend continues, in what year will the trade balance reach a deficit of $750 million dollars or more? 143. Tennis court dimensions: A regulation tennis court for a doubles match is laid out so that its length is 6 ft more than two Exercises 143 times its width. The area of the and 144 doubles court is 2808 ft2. What is the length and width of the doubles court? 144. Tennis court dimensions: A regulation tennis court for a singles match is laid out so that its length is 3 ft less than three times its width. The area of the singles court is 2106 ft2. What is the length and width of the singles court?

Singles Doubles

EXTENDING THE CONCEPT

145. Using the discriminant: Each of the following equations can easily be solved by factoring, since a  1. Using the discriminant, we can create factorable equations with identical values for b and c, but where a  1. For instance, x2  3x  10  0 and 4x2  3x  10  0 can both be solved by factoring. Find similar equations 1a  12 for the quadratics given here. (Hint: The discriminant b2  4ac must be a perfect square.) a. x2  6x  16  0 b. x2  5x  14  0 c. x2  x  6  0 146. Using the discriminant: For what values of c will the equation 9x2  12x  c  0 have a. no real roots b. one rational root c. two real roots d. two integer roots

Complex polynomials: Many techniques applied to solve polynomial equations with real coefficients can be applied to solve polynomial equations with complex coefficients. Here we apply the idea to carefully chosen quadratic equations, as a more general application must wait until a future course, when the square root of a complex number is fully developed. Solve each equation 1 using the quadratic formula, noting that  i. i

147. z2  3iz  10 148. z2  9iz  22 149. 4iz2  5z  6i  0 150. 2iz2  9z  26i  0

151. 0.5z2  17  i2z  16  7i2  0

152. 0.5z2  14  3i2z  19  12i2  0

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MAINTAINING YOUR SKILLS

153. (R.7) State the formula for the perimeter and area of each figure illustrated. a. b. L

154. (1.3) Factor and solve the following equations: a. x2  5x  36  0 b. 4x2  25  0 c. x3  6x2  4x  24  0

r

155. (1.1) A total of 900 tickets were sold for a recent concert and $25,000 was collected. If good seats were $30 and cheap seats were $20, how many of each type were sold?

W

c.

d.

b1

156. (1.1) Solve for C: P  C  Ct. c

h

a b2

h

c

b

1.6 Solving Other Types of Equations Learning Objectives

The ability to solve linear and quadratic equations is the foundation on which a large percentage of our future studies are built. Both are closely linked to the solution of other equation types, as well as to the graphs of these equations. In this section, we get our first glimpse of these connections, as we learn to solve certain polynomial, rational, radical, and other equations.

In Section 1.6 you will learn how to:

A. Solve polynomial equations of higher degree

B. Solve rational equations C. Solve radical equations and equations with rational exponents

D. Solve equations in quadratic form

E. Solve applications of various equation types

A. Polynomial Equations of Higher Degree In standard form, linear and quadratic equations have a known number of terms, so we commonly represent their coefficients using the early letters of the alphabet, as in ax2  bx  c  0. However, these equations belong to the larger family of polynomial equations. To write a general polynomial, where the number of terms is unknown, we often represent the coefficients using subscripts on a single variable, such as a1, a2, a3, and so on. A polynomial equation of degree n has the form anxn  an1xn1  p  a1x1  a0  0 where an, an1, p , a1, a0 are real numbers and an  0. Factorable polynomials of degree 3 and higher can also be solved using the zero product property and fundamental algebra skills. As with linear equations, values that make an equation true are called solutions or roots to the equation.

EXAMPLE 1



Solving Polynomials by Factoring Solve by factoring: 2x3  20x  3x2.

Solution



2x3  20x  3x2 given equation standard form 2x  3x2  20x  0 common factor is x x 12x2  3x  202  0 factored form x 12x  52 1x  42  0 x  0 or 2x  5  0 or x  4  0 zero product property 5 result x  0 or x  2 or x  4 Substituting these values into the original equation verifies they are solutions. 3

Now try Exercises 7 through 14



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Section 1.6 Solving Other Types of Equations

EXAMPLE 2



57

Solving Higher Degree Equations Solve each equation by factoring: a. x3  7x  21  3x2 b. x4  16  0

Solution



x3  7x  21  3x2

a.

x  3x  7x  21  0 x2 1x  32  71x  32  0 1x  32 1x2  72  0 x  3  0 or x2  7  0 x  3 or x2  7 x  17 3

b.

2

given equation standard form; factor by grouping remove common factors from each group factored form zero product property isolate variables square root property of equality

The solutions are x  3, x  17, and x   17. given equation x4  16  0 2 2 factor as a difference of squares 1x  421x  42  0 1x2  421x  22 1x  22  0 factor x 2  4 x2  4  0 or x  2  0 or x  2  0 zero product property x2  4 or x  2 or x  2 isolate variables square root property of equality x  14 Since  14  2i, the solutions are x  2i, x  2i, x  2, and x  2. Now try Exercises 15 through 32

A. You’ve just learned how to solve polynomial equations of higher degree



In Examples 1 and 2, we were able to solve higher degree polynomials by “breaking them down” into linear and quadratic forms. This basic idea can be applied to other kinds of equations as well, by rewriting them as equivalent linear and/or quadratic equations. For future use, it will be helpful to note that for a third-degree equation in the standard form ax3  bx2  cx  d  0, a solution using factoring by grouping is always possible when ad  bc.

B. Rational Equations In Section 1.1 we solved linear equations using basic properties of equality. If any equation contained fractional terms, we “cleared the fractions” using the least common denominator (LCD). We can also use this idea to solve rational equations, or equations that contain rational expressions. Solving Rational Equations 1. Identify and exclude any values that cause a zero denominator. 2. Multiply both sides by the LCD and simplify (this will eliminate all denominators). 3. Solve the resulting equation. 4. Check all solutions in the original equation.

EXAMPLE 3



Solving a Rational Equation Solve for m:

1 4 2  .  2 m m1 m m

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Solution



Since m2  m  m1m  12, the LCD is m1m  12, where m  0 and m  1. 4 2 1 b  m1m  12 c d  m m1 m1m  12 21m  12  m  4 2m  2  m  4 m6

m1m  12 a

multiply by LCD simplify—denominators are eliminated distribute solve for m

Checking by substitution we have: 2 1  m m1 1 2  162 162  1 1 1  3 5 3 5  15 15 2 15

    

4 m m 4 2 162  162 4 30 2 15 2 ✓ 15 2

original equation substitute 6 for m

simplify

common denominator

result

Now try Exercises 33 through 38



Multiplying both sides of an equation by a variable sometimes introduces a solution that satisfies the resulting equation, but not the original equation—the one we’re trying to solve. Such “solutions” are called extraneous roots and illustrate the need to check all apparent solutions in the original equation. In the case of rational equations, we are particularly aware that any value that causes a zero denominator is outside the domain and cannot be a solution. EXAMPLE 4



Solving a Rational Equation Solve: x 

Solution



4x 12 1 . x3 x3

The LCD is x  3, where x  3. 4x 12 b  1x  32a1  b x3 x3 x2  3x  12  x  3  4x x2  8x  15  0 1x  321x  52  0 x  3 or x  5

1x  32ax 

multiply both sides by LCD simplify—denominators are eliminated set equation equal to zero factor zero factor property

Checking shows x  3 is an extraneous root, and x  5 is the only valid solution. Now try Exercises 39 through 44



In many fields of study, formulas involving rational expressions are used as equation models. Frequently, we need to solve these equations for one variable in terms of others, a skill closely related to our work in Section 1.1.

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Section 1.6 Solving Other Types of Equations

EXAMPLE 5



Solving for a Specified Variable in a Formula Solve for the indicated variable: S 

Solution

S



a 1r

11  r2S  11  r2a

WORTHY OF NOTE

a for r. 1r

LCD is 1  r

a b 1r

S  Sr  a Sr  a  S aS r S Sa r ;S0 S

Generally, we should try to write rational answers with the fewest number of negative signs possible. Multiplying the numerator and denominator in Example a 5 by 1 gave r  S  S , a more acceptable answer.

multiply both sides by 11  r2 simplify—denominator is eliminated isolate term with r solve for r (divide both sides by S ) multiply numerator/denominator by 1

Now try Exercises 45 through 52 B. You’ve just learned how to solve rational equations

59



C. Radical Equations and Equations with Rational Exponents A radical equation is any equation that contains terms with a variable in the radicand. To solve a radical equation, we attempt to isolate a radical term on one side, then apply the appropriate nth power to free up the radicand and solve for the unknown. This is an application of the power property of equality. The Power Property of Equality n

n

If 1 u and v are real-valued expressions and 1 u  v, n then 1 1 u2 n  vn u  vn for n an integer, n  2. Raising both sides of an equation to an even power can also introduce a false solution (extraneous root). Note that by inspection, the equation x  2  1x has only the solution x  4. But the equation 1x  22 2  x (obtained by squaring both sides) has both x  4 and x  1 as solutions, yet x  1 does not satisfy the original equation. This means we should check all solutions of an equation where an even power is applied. EXAMPLE 6



Solving Radical Equations Solve each radical equation: a. 13x  2  12  x  10

Solution



a. 13x  2  12  x  10 13x  2  x  2 1 13x  22 2  1x  22 2 3x  2  x2  4x  4 0  x2  7x  6 0  1x  62 1x  12 x  6  0 or x  1  0 x  6 or x  1

3 b. 2 1 x540

original equation isolate radical term (subtract 12) apply power property, power is even simplify; square binomial set equal to zero factor apply zero product property result, check for extraneous roots

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Check



Check



13162  2  12  162  10 116  12  16 16  16 ✓

x  6:

x  1:

13112  2  12  112  10 11  12  11 13  11 x

The only solution is x  6; x  1 is extraneous. 3 b. 2 1 x540 3 1 x  5  2 3 1 1x  52 3  122 3 x  5  8 x  3

original equation isolate radical term (subtract 4, divide by 2) apply power property, power is odd 3 simplify: 1 1 x  52 3  x  5

solve

Substituting 3 for x in the original equation verifies it is a solution. Now try Exercises 53 through 56



Sometimes squaring both sides of an equation still results in an equation with a radical term, but often there is one fewer than before. In this case, we simply repeat the process, as indicated by the flowchart in Figure 1.14.

Figure 1.14 Radical Equations

EXAMPLE 7



Solve the equation: 1x  15  1x  3  2.

Isolate radical term

Solution



Apply power property

Does the result contain a radical?

NO

Solve using properties of equality

Solving Radical Equations

YES

Check



1x  15  1x  3  2 1x  15  1x  3  2 1 1x  152 2  1 1x  3  22 2 x  15  1x  32  4 1x  3  4 x  15  x  41x  3  7 8  4 1x  3 2  1x  3 4x3 1x 1x  15  1x  3  2 1112  15  1112  3  2 116  14  2 422 2  2✓

original equation isolate one radical power property 1A  B2 2; A  1x 3, B  2 simplify isolate radical divide by four power property possible solution

original equation substitute 1 for x simplify solution checks

Now try Exercises 57 and 58 Check results in original equation



Since rational exponents are so closely related to radicals, the solution process for each is very similar. The goal is still to “undo” the radical (rational exponent) and solve for the unknown.

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61

Power Property of Equality For real-valued expression u and v, with positive integers m, n, and If m is odd

m

and u  v, m n n then A u n B m  vm

and u n  v1v 7 02, m n n then A u n B m  vm n u  vm

n

u  vm



Solving Equations with Rational Exponents Solve each equation: 3 a. 31x  12 4  9  15

Solution

Check





3

a. 31x  12 4  9  15 3 1x  12 4  8 3 4 4 3 1x  12 443  83 x  1  16 x  15 3 4

3115  12  9  15 1 3 A 164 B 3  9  15 3122 3  9  15 3182  9  15 15  15 ✓ b.

C. You’ve just learned how to solve radical equations and equations with rational exponents

in lowest terms:

If m is even

m n

EXAMPLE 8

m n

1x  32  4 2 3 3 3 1x  32 342  42 x  3  8 x38 2 3

b. 1x  32 3  4 2

original equation; mn  34 isolate variable term (add 9, divide by 3) apply power property, note m is odd simplify 383  A 83 B 4  164 4

1

result

substitute 15 for x in the original equation simplify, rewrite exponent 4 1 16  2

23  8 solution checks original equation; mn  23 apply power property, note m is even simplify 342  A 42 B 3  84 3

1

result

The solutions are 3  8  11 and 3  8  5. Verify by checking both in the original equation. Now try Exercises 59 through 64

CAUTION





As you continue solving equations with radicals and rational exponents, be careful not to arbitrarily place the “” sign in front of terms given in radical form. The expression 118 indicates the positive square root of 18, where 118  312. The equation x2  18 becomes x  118 after applying the power property, with solutions x  312 1x  312, x  3122, since the square of either number produces 18.

D. Equations in Quadratic Form In Appendix I.D we used a technique called u-substitution to factor expressions in quadratic form. The following equations are in quadratic form since the degree of the 2 1 leading term is twice the degree of the middle term: x3  3x3  10  0, 1x2  x2 2  81x2  x2  12  0 and x  13 1x  4  4  0 [Note: The last equation can be rewritten as 1x  42  31x  42 2  0]. A u-substitution will help to solve these equations by factoring. The first equation appears in Example 9, the other two are in Exercises 70 and 74, respectively.

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EXAMPLE 9



Solving Equations in Quadratic Form Solve using a u-substitution: 2 1 a. x3  3x3  10  0

Solution



b. x4  36  5x2

a. This equation is in quadratic form since it can be rewritten as: 1 1 the degree2 of leading term is twice that of A x3 B 2  3 A x3 B 1  10  0, where 1 second term. If we let u  x3, then u2  x3 and the equation becomes u2  3u1  10  0 which is factorable. 1u  521u  22  0 or u5 u  2 1 1 x3  5 or x3  2 1 1 A x3 B 3  53 or A x3 B 3  122 3 x  125 or x  8

factor solution in terms of u 1

resubstitute x 3 for u

cube both sides: 13 132  1 solve for x

Both solutions check. b. In the standard form x4  5x2  36  0, we note the equation is also in quadratic form, since it can be written as 1x2 2 2  51x2 2 1  36  0. If we let u  x2, then u2  x4 and the equation becomes u2  5u  36  0, which is factorable. 1u  921u  u9 x2  9 x   19 x  3

D. You’ve just learned how to solve equations in quadratic form

42  0 or u  4 or x2  4 or x   14 or x   2i

factor solution in terms of u resubstitute x 2 for u square root property simplify

The solutions are x  3, x  3, x  2i, and x  2i. Verify that all solutions check. Now try Exercises 65 through 78



E. Applications Applications of the skills from this section come in many forms. Number puzzles and consecutive integer exercises help develop the ability to translate written information into algebraic forms (see Exercises 81 through 84). Applications involving geometry or a stated relationship between two quantities often depend on these skills, and in many scientific fields, equation models involving radicals and rational exponents are commonplace (see Exercises 99 and 100). EXAMPLE 10



Solving a Geometry Application A legal size sheet of typing paper has a length equal to 3 in. less than twice its width. If the area of the paper is 119 in2, find the length and width.

Solution



Let W represent the width of the paper. Then 2W represents twice the width, and 2W  3 represents three less than twice the width: L  2W  3: 1length2 1width2  area 12W  32 1W2  119

verbal model substitute 2W  3 for length

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Since the equation is not set equal to zero, multiply and write the equation in standard form. 2W2  3W  119 2W  3W  119  0 12W  1721W  72  0 W  17 2 or W  7 2

distribute subtract 119 factor solve

We ignore W  7, since the width cannot be negative. The width of the paper is 17 1 17 2  82 in. and the length is L  2 A 2 B  3 or 14 in. Now try Exercises 85 and 86

EXAMPLE 11



Solving a Geometry Application A hemispherical wash basin has a radius of 6 in. The volume of water in the basin can be modeled by V  6h2  3 h3, where h is the height of the water (see diagram). At what height h is the volume of water numerically equal to 15 times the height h?

Solution





h

We are essentially asked to solve V  6h2  3 h3 when V  15h. The equation becomes 15h  6h2 

 3 h 3

 3 h  6h2  15h  0 3 h3  18h2  45h  0 h1h2  18h  452  0 h1h  32 1h  152  0 h  0 or h  3 or h  15

original equation, substitute15h for V

standard form multiply by 3 factor out h factored form result

The “solution” h  0 can be discounted since there would be no water in the basin, and h  15 is too large for this context (the radius is only 6 in.). The only solution that fits this context is h  3. Check



 3 h 3  15132  6132 2  132 3 3  45  6192  1272 3 45  54  9 45  45 ✓ 15h  6h2 

resulting equation substitute 3 for h

apply exponents simplify result checks

Now try Exercises 87 and 88



In this section, we noted that extraneous roots can occur when (1) both sides of an equation are multiplied by a variable term (as when solving rational equations) and (2) when both sides of an equation are raised to an even power (as when solving certain radical equations or equations with rational exponents). Example 11 illustrates a third

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way that extraneous roots can occur, as when a solution checks out fine algebraically, but does not fit the context or physical constraints of the situation.

Revenue Models In a free-market economy, we know that if the price of an item is decreased, more people will buy it. This is why stores have sales and bargain days. But if the item is sold too cheaply, revenue starts to decline because less money is coming in—even though more sales are made. This phenomenon is analyzed in Example 12, where we use the revenue formula revenue  price # number of sales or R  P # S. EXAMPLE 12



Solving a Revenue Application When a popular printer is priced at $300, Compu-Store will sell 15 printers per week. Using a survey, they find that for each decrease of $8, two additional sales will be made. What price will result in weekly revenue of $6500?

Solution



Let x represent the number of times the price is decreased by $8. Then 300  8x represents the new price. Since sales increase by 2 each time the price is decreased, 15  2x represents the total sales. RP#S 6500  1300  8x2 115  2x2 6500  4500  600x  120x  16x2 0  16x2  480x  2000 0  x2  30x  125 0  1x  52 1x  252 x  5 or x  25

revenue model R  6500, P  300  8x, S  15  2x multiply binomials simplify and write in standard form divide by 16 factor result

Surprisingly, the store’s weekly revenue will be $6500 after 5 decreases of $8 each ($40 total), or 25 price decreases of $8 each ($200 total). The related selling prices are 300  5182  $260 and 300  25182  $100. To maximize profit, the manager of Compu-Store decides to go with the $260 selling price. Now try Exercises 89 and 90



Applications of rational equations can also take many forms. Work and uniform motion exercises help us develop important skills that can be used with more complex equation models. A work example follows here. For more on uniform motion, see Exercises 95 and 96. EXAMPLE 13



Solving a Work Application Lyf can clean a client’s house in 5 hr, while it takes his partner Angie 4 hr to clean the same house. Both of them want to go to the Cubs’ game today, which starts in 212 hr. If they work together, will they see the first pitch?

Solution



After 1 hr, Lyf has cleaned 15 and Angie has cleaned 14 of the house, so together 1 1 9 1 1 5  4  20 or 45% of the house has been cleaned. After 2 hr, 2 A 5 B  2 A 4 B 2 1 9 or 5  2  10 or 90% of the house is clean. We can use these two illustrations to form an equation model where H represents hours worked: 1 1 Ha b  Ha b  1 clean house 11  100% 2. 5 4

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1 1 Ha b  Ha b  1 5 4 1 1 20Ha b  20Ha b  11202 5 4 4H  5H  20 9H  20 20 H 9

equation model

multiply by LCD of 20 simplify, denominators are eliminated combine like terms solve for H

It will take Lyf and Angie 229 hr (about 2 hr and 13 min) to clean the house. Yes! They will make the first pitch, since Wrigley Field is only 10 min away. Now try Exercises 93 and 94

EXAMPLE 14





Solving an Application Involving a Rational Equation In Verano City, the cost C to remove industrial waste from drinking water is given 80P , where P is the percent of total pollutants removed by the equation C  100  P and C is the cost in thousands of dollars. If the City Council budgets $1,520,000 for the removal of these pollutants, what percentage of the waste will be removed?

Solution



E. You’ve just learned how to solve applications of various equation types

80P 100  P 80P 1520  100  P 15201100  P2  80P 152,000  1600P 95  P C

equation model substitute 1520 for C multiply by LCD of 1100  P 2 distribute and simplify result

On a budget of $1,520,000, 95% of the pollutants will be removed. Now try Exercises 97 and 98



1.6 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

1. For rational equations, values that cause a zero denominator must be . 2. The equation or formula for revenue models is revenue  . 3. “False solutions” to a rational or radical equation are also called roots.

4. Factorable polynomial equations can be solved using the property. 5. Discuss/Explain the power property of equality as it relates to rational exponents and properties of 2 reciprocals. Use the equation 1x  22 3  9 for your discussion. 6. One factored form of an equation is shown. Discuss/Explain why x  8 and x  1 are not solutions to the equation, and what must be done to find the actual solutions: 21x  821x  12  16.

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DEVELOPING YOUR SKILLS

Solve using the zero product property. Be sure each equation is in standard form and factor out any common factors before attempting to solve. Check all answers in the original equation.

7. 22x  x3  9x2

39. x  40.

2x 10 x1 x5 x5

41.

20 6 5  2  n3 n2 n n6

42.

2 1 7   2 p2 p  3 p  5p  6

43.

2a2  5 a 3  2  2a  1 a3 2a  5a  3

44.

3n 4n 18   2n  1 3n  1 6n  n  1

8. x3  13x2  42x

9. 3x3  7x2  6x

10. 7x2  15x  2x3

11. 2x4  3x3  9x2

12. 7x2  2x4  9x3

13. 2x4  16x  0

14. x4  64x  0

15. x3  4x  5x2  20 16. x3  18  9x  2x2 17. 4x  12  3x  x 2

18. x  7  7x  x

3

2

3

19. 2x3  12x2  10x  60 20. 9x  81  27x  3x 2

3

21. x4  7x3  4x2  28x 23. x  81  0 24. x4  1  0 25. x4  256  0 26. x4  625  0 27. x6  2x4  x2  2  0 28. x6  3x4  16x2  48  0 29. x5  x3  8x2  8  0 30. x5  9x3  x2  9  0 31. x6  1  0 32. x6  64  0 Solve each equation. Identify any extraneous roots.

33.

1 5 2   2 x x1 x x

5 3 1 34.  2  m m3 m  3m 35.

3 21  a2 a1

36.

4 7  2y  3 3y  5

37.

1 1 1   2 3y 4y y

3 1 1 38.   2 5x 2x x

2

Solve for the variable indicated.

22. x4  3x3  9x2  27x 4

2x 14 1 x7 x7

45.

1 1 1   ; for f f f1 f2

47. I 

E ; for r Rr

46.

1 1 1   ; for z z x y

48. q 

pf ; for p pf

1 49. V  r2h; for h 3

1 50. s  gt2; for g 2

4 51. V  r3; for r3 3

1 52. V  r2h; for r2 3

Solve each equation and check your solutions by substitution. Identify any extraneous roots.

53. a. 313x  5  9

b. x  13x  1  3

54. a. 214x  1  10 b. 5  15x  1  x 3 3 55. a. 2  1 b. 2 1 3m  1 7  3x  3  7 3 1 2m  3 3 3 c.  2  3 d. 1 2x  9  1 3x  7 5 3 3 56. a. 3  1 b. 3 1 5p  2 3  4x  7  4 3 1 6x  7 c.  5  6 4 3 3 d. 31 x  3  21 2x  17

57. a. b. c. d.

1x  9  1x  9 x  3  223  x 1x  2  12x  2 112x  9  124x  3

58. a. b. c. d.

1x  7  1x  1 12x  31  x  2 13x  1x  3  3 13x  4  17x  2

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Write the equation in simplified form, then solve. Check all answers by substitution. 3 5

59. x  17  9 5 2

61. 0.3x  39  42

3 4

60. 2x  47  7

4

Use a u-substitution to solve each radical equation.

64. 31x  22 5  29  19 Solve each equation using a u-substitution. Check all answers.

65. x  2x  15  0

66. x3  9x  8  0

69. 1x2  32 2  1x2  32  2  0

75. x  4  7 1x  4 77. 2 1x  10  8  31x  102 78. 41x  3  31x  32  4

WORKING WITH FORMULAS

79. Lateral surface area of a cone: S  r 2r 2  h2 The lateral surface area (surface area excluding the base) S of a cone is given by the formula shown, where r is the radius of the base and h is the height of the cone. (a) Solve the equation for h. (b) Find the surface area of a cone that has a radius of 6 m and a height of 10 m. Answer in simplest form.



74. x  3 1x  4  4  0 76. 21x  12  5 1x  1  2

3 2

67. x4  24x2  25  0 68. x4  37x2  36  0



71. x2  3x1  4  0

73. x4  13x2  36  0

63. 21x  52  11  7

1 3

70. 1x2  x2 2  81x2  x2  12  0 72. x2  2x1  35  0

5

62. 0.5x3  92  43

2 3

2 3

67

h

r

80. Painted area on a canvas: A 

4x2  60x  104 x

A rectangular canvas is to contain a small painting with an area of 52 in2, and requires 2-in. margins on the left and right, with 1-in. margins on the top and bottom for framing. The total area of such a canvas is given by the formula shown, where x is the height of the painted area. a. What is the area A of the canvas if the height of the painting is x  10 in.? b. If the area of the canvas is A  120 in2, what are the dimensions of the painted area?

APPLICATIONS

Find all real numbers that satisfy the following descriptions.

81. When the cube of a number is added to twice its square, the result is equal to 18 more than 9 times the number. 82. Four times a number decreased by 20 is equal to the cube of the number decreased by 5 times its square. 83. Find three consecutive even integers such that 4 times the largest plus the fourth power of the smallest is equal to the square of the remaining even integer increased by 24. 84. Find three consecutive integers such that the sum of twice the largest and the fourth power of the smallest is equal to the square of the remaining integer increased by 75. 85. Envelope sizes: Large mailing envelopes often come in standard sizes, with 5- by 7-in. and 9- by

12-in. envelopes being the most common. The next larger size envelope has an area of 143 in2, with a length that is 2 in. longer than the width. What are the dimensions of the larger envelope? 86. Paper sizes: Letter size paper is 8.5 in. by 11 in. Legal size paper is 812 in. by 14 in. The next larger (common) size of paper has an area of 187 in2, with a length that is 6 in. longer than the width. What are the dimensions of the Ledger size paper?

Letter

Legal Ledger

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87. Composite figures— grain silos: Grain silos can be described as a hemisphere sitting atop a cylinder. The interior volume V of the silo can be modeled by V  23r3  r2h, where h is the height of a cylinder with radius r. For a cylinder 6 m tall, what radius would give the silo a volume that is numerically equal to 24 times this radius? 88. Composite figures—gelatin capsules: The gelatin capsules manufactured for cold and flu medications are shaped like a cylinder with a hemisphere on each end. The interior volume V of each capsule can be modeled by V  43r3  r2h, where h is the height of the cylindrical portion and r is its radius. If the cylindrical portion of the capsule is 8 mm long 1h  8 mm2, what radius would give the capsule a volume that is numerically equal to 15 times this radius?

been thrown). Use this information to complete the following problems.

91. From the base of a canyon that is 480 feet deep (below ground level S 4802, a slingshot is used to shoot a pebble upward toward the canyon’s rim. If the initial velocity is 176 ft per second: a. How far is the pebble below the rim after 4 sec? b. How long until the pebble returns to the bottom of the canyon? c. What happens at t  5 and t  6 sec? Discuss and explain. 92. A model rocket blasts off. A short time later, at a velocity of 160 ft/sec and a height of 240 ft, it runs out of fuel and becomes a projectile. a. How high is the rocket three seconds later? Four seconds later? b. How long will it take the rocket to attain a height of 640 ft? c. How many times is a height of 384 ft attained? When do these occur? d. How many seconds until the rocket returns to the ground? 93. Printing newspapers: The editor of the school newspaper notes the college’s new copier can complete the required print run in 20 min, while the back-up copier took 30 min to do the same amount of work. How long would it take if both copiers are used?

89. Running shoes: When a popular running shoe is priced at $70, The Shoe House will sell 15 pairs each week. Using a survey, they have determined that for each decrease of $2 in price, 3 additional pairs will be sold each week. What selling price will give a weekly revenue of $2250? 90. Cell phone charges: A cell phone service sells 48 subscriptions each month if their monthly fee is $30. Using a survey, they find that for each decrease of $1, 6 additional subscribers will join. What charge(s) will result in a monthly revenue of $2160? Projectile height: In the absence of resistance, the height of an object that is projected upward can be modeled by the equation h  16t2  vt  k, where h represents the height of the object (in feet) t sec after it has been thrown, v represents the initial velocity (in feet per second), and k represents the height of the object when t  0 (before it has

94. Filling a sink: The cold water faucet can fill a sink in 2 min. The drain can empty a full sink in 3 min. If the faucet were left on and the drain was left open, how long would it take to fill the sink? 95. Triathalon competition: As one part of a Mountain-Man triathalon, participants must row a canoe 5 mi down river (with the current), circle a buoy and row 5 mi back up river (against the current) to the starting point. If the current is flowing at a steady rate of 4 mph and Tom Chaney made the round-trip in 3 hr, how fast can he row in still water? (Hint: The time rowing down river and the time rowing up river must add up to 3 hr.) 96. Flight time: The flight distance from Cincinnati, Ohio, to Chicago, Illinois, is approximately 300 mi. On a recent round-trip between these cities in my private plane, I encountered a steady 25 mph headwind on the way to Chicago, with a 25 mph tailwind on the return trip. If my total flying time

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came to exactly 5 hr, what was my flying time to Chicago? What was my flying time back to Cincinnati? (Hint: The flight time between the two cities must add up to 5 hr.)

modeled by T  0.407R2, where R is the maximum radius of the planet’s orbit in millions of miles (Kepler’s third law of planetary motion). Use the equation to approximate the maximum radius of each orbit, given the number of days it takes for one revolution. (See Appendix I.F, Exercises 45 and 46.) a. Mercury: 88 days b. Venus: 225 days c. Earth: 365 days d. Mars: 687 days e. Jupiter: 4,333 days f. Saturn: 10,759 days

97. Pollution removal: For a steel mill, the cost C (in millions of dollars) to remove toxins from the 92P , where resulting sludge is given by C  100  P P is the percent of the toxins removed. What percent can be removed if the mill spends $100,000,000 on the cleanup? Round to tenths of a percent. 98. Wildlife populations: The Department of Wildlife introduces 60 elk into a new game reserve. It is projected that the size of the herd will grow 1016  3t2 , where according to the equation N  1  0.05t N is the number of elk and t is the time in years. If recent counts find 225 elk, approximately how many years have passed? (See Appendix I.E, Exercise 66.) 99. Planetary motion: The time T (in days) for a planet to make one revolution around the sun is



100. Wind-powered energy: If a wind-powered generator is delivering P units of power, the velocity V of the wind (in miles per hour) can be 3 P , where k is a constant determined using V  Ak that depends on the size and efficiency of the generator. Given k  0.004, approximately how many units of power are being delivered if the wind is blowing at 27 miles per hour? (See Appendix I.F, Exercise 48.)

EXTENDING THE CONCEPT 8 1  , a student x x3 multiplied by the LCD x1x  32, simplified, and got this result: 3  8x  1x  32. Identify and fix the mistake, then find the correct solution(s).

101. To solve the equation 3 

102. The expression x2  7 is not factorable using integer values. But the expression can be written in the form x2  1 172 2, enabling us to factor it as a binomial and its conjugate: 1x  172 1x  172. Use this idea to solve the following equations: a. x2  5  0 b. n2  19  0 c. 4v2  11  0 d. 9w2  11  0

69

Determine the values of x for which each expression represents a real number.

103.

1x  1 x2  4

104.

x2  4 1x  1

105. As an extension of working with absolute values, try the following exercises. Recall that for X  k, X  k or X  k. a. x2  2x  25  10 b. x2  5x  10  4 c. x2  4  x  2 d. x2  9  x  3 e. x2  7x  x  7 f. x2  5x  2  x  5

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MAINTAINING YOUR SKILLS

106. (1.1) Two jets take off on parallel runways going in opposite directions. The first travels at a rate of 250 mph and the second at 325 mph. How long until they are 980 miles apart?

108. (R.3) Simplify using properties of exponents: 21  12x2 0  2x0

109. (1.2) Graph the relation given: 2x  3 6 7 and x  2 7 1

107. (R.6) Find the missing side. 12 cm

10 cm

S U M M A RY A N D C O N C E P T R E V I E W SECTION 1.1

Linear Equations, Formulas, and Problem Solving

KEY CONCEPTS • An equation is a statement that two expressions are equal. • Replacement values that make an equation true are called solutions or roots. • Equivalent equations are those that have the same solution set. • To solve an equation we use the distributive property and the properties of equality to write a sequence of simpler, equivalent equations until the solution is obvious. A guide for solving linear equations appears on page 3. • If an equation contains fractions, multiply both sides by the LCD of all denominators, then solve. • Solutions to an equation can be checked using back-substitution, by replacing the variable with the proposed solution and verifying the left-hand expression is equal to the right. • An equation can be: 1. an identity, one that is always true, with a solution set of all real numbers. 2. a contradiction, one that is never true, with the empty set as the solution set. 3. conditional, or one that is true/false depending on the value(s) input. • To solve formulas for a specified variable, focus on the object variable and apply properties of equality to write this variable in terms of all others. • The basic elements of good problem solving include: 1. Gathering and organizing information 2. Making the problem visual 3. Developing an equation model 4. Using the model to solve the application For a complete review, see the problem-solving guide on page 6. EXERCISES 1. Use substitution to determine if the indicated value is a solution to the equation given. 1 3 5 3 a. 6x  12  x2  41x  52, x  6 b. b  2  b  16, b  8 c. 4d  2    3d, d  4 2 2 2 Solve each equation. 2. 2b  7  5 5.

1 2 3 x  2 3 4

3. 312n  62  1  7 6. 6p  13p  52  9  31p  32

4. 4m  5  11m  2 5g g 1 7.   3   6 2 12

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71

Solve for the specified variable in each formula or literal equation. 8. V  r2h for h 9. P  2L  2W for L 10. ax  b  c for x

11. 2x  3y  6 for y

Use the problem-solving guidelines (page 6) to solve the following applications. 12. At a large family reunion, two kegs of lemonade are available. One is 2% sugar (too sour) and the second is 7% sugar (too sweet). How many gallons of the 2% keg, must be mixed with 12 gallons of the 7% keg to get a 5% mix? 13. A rectangular window with a width of 3 ft and a height of 4 ft is topped by a semi-circular window. Find the total area of the window. 14. Two cyclists start from the same location and ride in opposite directions, one riding at 15 mph and the other at 18 mph. If their radio phones have a range of 22 mi, how many minutes will they be able to communicate?

SECTION 1.2

Linear Inequalities in One Variable

KEY CONCEPTS • Inequalities are solved using properties similar to those for solving equalities (see page 15). The one exception is the multiplicative property of inequality, since the truth of the resulting statement depends on whether a positive or negative quantity is used. • Solutions to an inequality can be graphed on a number line, stated using a simple inequality, or expressed using set or interval notation. • For two sets A and B: A intersect B 1A  B2 is the set of elements in both A and B (i.e., elements common to both sets). A union B 1A ´ B2 is the set of elements in either A or B (i.e., all elements from either set). • Compound inequalities are formed using the conjunctions “and”/“or.” These can be either a joint inequality as in 3 6 x  5, or a disjoint inequality, as in x 6 2 or x 7 7. EXERCISES Use inequality symbols to write a mathematical model for each statement. 15. You must be 35 yr old or older to run for president of the United States. 16. A child must be under 2 yr of age to be admitted free. 17. The speed limit on many interstate highways is 65 mph. 18. Our caloric intake should not be less than 1200 calories per day. Solve the inequality and write the solution using interval notation. 19. 7x 7 35

3 20.  m 6 6 5

21. 213m  22  8

22. 1 6

23. 4 6 2b  8 and 3b  5 7 32

24. 51x  32 7 7 or x  5.2 7 2.9

1 x25 3

25. Find the allowable values for each of the following. Write your answer in interval notation. a.

7 n3

b.

5 2x  3

c. 1x  5

d. 13n  18

26. Latoya has earned grades of 72%, 95%, 83%, and 79% on her first four exams. What grade must she make on her fifth and last exam so that her average is 85% or more?

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SECTION 1.3

Absolute Value Equations and Inequalities

KEY CONCEPTS • To solve absolute value equations and inequalities, begin by writing the equation in simplified form, with the absolute value isolated on one side. • If X represents an algebraic expression and k is a nonnegative constant: • Absolute value equations: X  k is equivalent to X  k or X  k X 6 k is equivalent to k 6 X 6 k • “Less than” inequalities: • “Greater than” inequalities: X 7 k is equivalent to X 6 k or X 7 k • These properties also apply when the symbols “” or “”are used. • If the absolute value quantity has been isolated on the left, the solution to a less-than inequality will be a single interval, while the solution to a greater-than inequality will consist of two disjoint intervals. • The multiplicative property states that for algebraic expressions A and B, AB  AB. EXERCISES Solve each equation or inequality. Write solutions to inequalities in interval notation. 27. 7  0 x  3 0 28. 2x  2  10 29. 2x  3  13 2x  5 x 30. 31. 3x  2  2 6 14 32. `  9 `  7 89 3 2 33. 3x  5  4 34. 3x  1 6 9 35. 2x  1 7 4 3x  2 36. 5m  2  12  8 37.  6  10 2 38. Monthly rainfall received in Omaha, Nebraska, rarely varies by more than 1.7 in. from an average of 2.5 in. per month. (a) Use this information to write an absolute value inequality model, then (b) solve the inequality to find the highest and lowest amounts of monthly rainfall for this city.

SECTION 1.4

Complex Numbers

KEY CONCEPTS • The italicized i represents the number whose square is 1. This means i2  1 and i  11. • Larger powers of i can be simplified using i4  1. • For k 7 0, 1k  i1k and we say the expression has been written in terms of i. • The standard form of a complex number is a  bi, • The commutative, associative, and distributive where a is the real number part and bi is the properties also apply to complex numbers and are imaginary number part. used to perform basic operations. • To add or subtract complex numbers, combine the • To multiply complex numbers, use the F-O-I-L like terms. method and simplify. For any complex number its complex a  bi, • • To find a quotient of complex numbers, multiply the conjugate is a  bi. numerator and denominator by the conjugate of the denominator. • The product of a complex number and its conjugate is a real number. EXERCISES Simplify each expression and write the result in standard form. 39. 172

40. 6 148

42. 1316

43. i57

41.

10  150 5

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73

Perform the operation indicated and write the result in standard form. 5i 44. 15  2i2 2 45. 46. 13  5i2  12  2i2 1  2i 47. 12  3i212  3i2

48. 4i13  5i2

Use substitution to show the given complex number and its conjugate are solutions to the equation shown. 49. x2  9  34; x  5i 50. x2  4x  9  0; x  2  i 25

SECTION 1.5

Solving Quadratic Equations

KEY CONCEPTS • The standard form of a quadratic equation is ax2  bx  c  0, where a, b, and c are real numbers and a  0. In words, we say the equation is written in decreasing order of degree and set equal to zero. • The coefficient of the squared term a is called the leading coefficient, b is called the linear coefficient, and c is called the constant term. The square root property of equality states that if X 2  k, where k  0, then X  1k or X   1k. • • Factorable quadratics can be solved using the zero product property, which states that if the product of two factors is zero, then one, the other, or both must be equal to zero. Symbolically, if A # B  0, then A  0 or B  0. • Quadratic equations can also be solved by completing the square, or using the quadratic formula. • If the discriminant b2  4ac  0, the equation has one real (repeated) root. If b2  4ac 7 0, the equation has two real roots; and if b2  4ac 6 0, the equation has two complex roots. EXERCISES 51. Determine whether the given equation is quadratic. If so, write the equation in standard form and identify the values of a, b, and c. a. 3  2x2 b. 7  2x  11 c. 99  x2  8x d. 20  4  x2 52. Solve by factoring. a. x2  3x  10  0 b. 2x2  50  0 c. 3x2  15  4x d. x3  3x2  4x  12 53. Solve using the square root property of equality. a. x2  9  0 b. 21x  22 2  1  11 c. 3x2  15  0 d. 2x2  4  46 54. Solve by completing the square. Give real number solutions in exact and approximate form. a. x2  2x  15 b. x2  6x  16 c. 4x  2x2  3 d. 3x2  7x  2 55. Solve using the quadratic formula. Give solutions in both exact and approximate form. a. x2  4x  9 b. 4x2  7  12x c. 2x2  6x  5  0 Solve the following quadratic applications. For 56 and 57, recall the height of a projectile is modeled by h  16t2  v0t  k. 56. A projectile is fired upward from ground level with an initial velocity of 96 ft/sec. (a) To the nearest tenth of a second, how long until the object first reaches a height of 100 ft? (b) How long until the object is again at 100 ft? (c) How many seconds until it returns to the ground? 57. A person throws a rock upward from the top of an 80-ft cliff with an initial velocity of 64 ft/sec. (a) To the nearest tenth of a second, how long until the object is 120 ft high? (b) How long until the object is again at 120 ft? (c) How many seconds until the object hits the ground at the base of the cliff? 58. The manager of a large, 14-screen movie theater finds that if he charges $2.50 per person for the matinee, the average daily attendance is 4000 people. With every increase of 25 cents the attendance drops an average of 200 people. (a) What admission price will bring in a revenue of $11,250? (b) How many people will purchase tickets at this price?

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59. After a storm, the Johnson’s basement flooded and the water needed to be pumped out. A cleanup crew is sent out with two powerful pumps to do the job. Working alone (if one of the pumps were needed at another job), the larger pump would be able to clear the basement in 3 hr less time than the smaller pump alone. Working together, the two pumps can clear the basement in 2 hr. How long would it take the smaller pump alone?

SECTION 1.6

Solving Other Types of Equations

KEY CONCEPTS • Certain equations of higher degree can be solved using factoring skills and the zero product property. • To solve rational equations, clear denominators using the LCD, noting values that must be excluded. • Multiplying an equation by a variable quantity sometimes introduces extraneous solutions. Check all results in the original equation. • To solve radical equations, isolate the radical on one side, then apply the appropriate “nth power” to free up the radicand. Repeat the process if needed. See flowchart on page 60. • For equations with a rational exponent mn, isolate the variable term and raise both sides to the mn power. If m is even, there will be two real solutions. • Any equation that can be written in the form u2  bu  c  0, where u represents an algebraic expression, is said to be in quadratic form and can be solved using u-substitution and standard approaches. EXERCISES Solve by factoring. 60. x3  7x2  3x  21

61. 3x3  5x2  2x

62. x4  8x  0

63. x4 

Solve each equation. 3 7 1 64.   5x 10 4x 2n 3 n2  20   2 n2 n4 n  2n  8 68. 31x  4  x  4 1 70. 3ax  b 4



3h 1 7   2 h3 h h  3h 2 2x  7 67. 35 2 65.

66.

32

1 0 16

8 9

72. 1x2  3x2 2  141x2  3x2  40  0

69. 13x  4  2  1x  2 2

71. 215x  22 3  17  1 73. x4  7x2  18

74. The science of allometry studies the growth of one aspect of an organism relative to the entire organism or to a set standard. Allometry tells us that the amount of food F (in kilocalories per day) an herbivore must eat to 3 survive is related to its weight W (in grams) and can be approximated by the equation F  1.5W4. a. How many kilocalories per day are required by a 160-kg gorilla 1160 kg  160,000 g2? b. If an herbivore requires 40,500 kilocalories per day, how much does it weigh?

75. The area of a common stenographer’s tablet, commonly called a steno book, is 54 in2. The length of the tablet is 3 in. more than the width. Model the situation with a quadratic equation and find the dimensions of the tablet. 76. A batter has just flied out to the catcher, who catches the ball while standing on home plate. If the batter made contact with the ball at a height of 4 ft and the ball left the bat with an initial velocity of 128 ft/sec, how long will it take the ball to reach a height of 116 ft? How high is the ball 5 sec after contact? If the catcher catches the ball at a height of 4 ft, how long was it airborne? 77. Using a survey, a firewood distributor finds that if they charge $50 per load, they will sell 40 loads each winter month. For each decrease of $2, five additional loads will be sold. What selling price(s) will result in new monthly revenue of $2520?

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Practice Test

75

MIXED REVIEW 1. Find the allowable values for each expression. Write your response in interval notation. 10 5 a. b. 3x 4 1x  8 2. Perform the operations indicated. a. 118  150 b. 11  2i2 2 3i c. d. 12  i132 12  i 132 1i 3. Solve each equation or inequality. a. 2x3  4x2  50x  100 b. 3x4  375x  0 c. 23x  1  12 4 x d. 3 `  5 `  12 e. v3  81 3 1 f. 21x  12 4  6 Solve for the variable indicated. 1 2 4. V  r2h  r3; for h 5. 3x  4y  12; for y 3 3 Solve as indicated, using the method of your choice. 6. a. 20  4x  8 6 56 b. 2x  7  12 and 3  4x 7 5

17. a. 12v  3  3  v 3 2 3 b. 2 x 9 1 x  11  0 c. 1x  7  12x  1 18. The local Lion’s Club rents out two banquet halls for large meetings and other events. The records show that when they charge $250 per day for use of the halls, there are an average of 156 bookings per year. For every increase of $20 per day, there will be three less bookings. (a) What price per day will bring in $61,950 for the year? (b) How many bookings will there be at the price from part (a)? 19. The Jefferson College basketball team has two guards who are 6¿3– tall and two forwards who are 6¿7– tall. How tall must their center be to ensure the “starting five” will have an average height of at least 6¿6–? 20. The volume of an inflatable hot-air balloon can be approximated using the formulas for a hemisphere and a cone: V  23r3  13r2h. Assume the conical portion has height h  24 ft. During inflation, what is the radius of the balloon at the moment the volume of air is numerically equal to 126 times this radius?

7. a. 5x  12x  32  3x  415  x2  3 n 5 4 b.  2  2   n 5 3 15 8. 5x1x  102 1x  12  0 9. x2  18x  77  0

10. 3x2  10  5  x  x2

11. 4x2  5  19

12. 31x  52 2  3  30

13. 25x2  16  40x

14. 3x2  7x  3  0

15. 2x4  50  0 x 2 2 1 1 b.  0  2  x 5x  12 n1 2 n 1 2x 36 x c.  2  x3 x3 x 9

16. a.

PRACTICE TEST 1. Solve each equation. 2 a.  x  5  7  1x  32 3 b. 5.7  3.1x  14.5  41x  1.52

c. P  C  kC; for C d. 22x  5  17  11 2. How much water that is 102°F must be mixed with 25 gal of water at 91°F, so that the resulting temperature of the water will be 97°F?

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3. Solve each equation or inequality. 2 a.  x  7 6 19 5 b. 1 6 3  x  8 2 1 x13 c. x  3 6 9 or 2 3 5 7 1 d. x  3   2 4 4 2 e.  x  1  5 6 7 3

revenue of $405? (b) How many tins will be sold at the price from part (a)? 16. Due to the seasonal nature of the business, the revenue of Wet Willey’s Water World can be modeled by the equation r  3t2  42t  135, where t is the time in months 1t  1 corresponds to January) and r is the dollar revenue in thousands. (a) What month does Wet Willey’s open? (b) What month does Wet Willey’s close? (c) Does Wet Willey’s bring in more revenue in July or August? How much more?

4. To make the bowling team, Jacques needs a threegame average of 160. If he bowled 141 and 162 for the first two games, what score S must be obtained in the third game so that his average is at least 160? 5. z2  7z  30  0

6. x2  25  0

7. 1x  12  3  0

8. x  16  17x

2

4

8  120 6

a. x  y

9. 3x  20x  12 10. 4x3  8x2  9x  18  0 2x x  16 2   2 x3 x2 x x6 4 5x 2 2 12. x3 x 9 13. 1x  1  12x  7 2

11.

1  4

15. The Spanish Club at Rock Hill Community College has decided to sell tins of gourmet popcorn as a fundraiser. The suggested selling price is $3.00 per tin, but Maria, who also belongs to the Math Club, decides to take a survey to see if they can increase “the fruits of their labor.” The survey shows it’s likely that 120 tins will be sold on campus at the $3.00 price, and for each price increase of $0.10, 2 fewer tins will be sold. (a) What price per tin will bring in a

18. i39

1 13 13 1  i and y   i find 2 2 2 2 b. x  y c. xy

2

2

14. 1x  32

17.

19. Given x 

Solve each equation.

2 3

Simplify each expression.

20. Compute the quotient:

3i . 1i

21. Find the product: 13i  5215  3i2. 22. Show x  2  3i is a solution of x2  4x  13  0. 23. Solve by completing the square. a. 2x2  20x  49  0 b. 2x2  5x  4 24. Solve using the quadratic formula. a. 3x2  2  6x b. x2  2x  10 25. Allometric studies tell us that the necessary food intake F (in grams per day) of nonpasserine birds (birds other than song birds and other small3 birds) can be modeled by the equation F  0.3W4, where W is the bird’s weight in grams. (a) If my Greenwinged macaw weighs 1296 g, what is her anticipated daily food intake? (b) If my blue-headed pionus consumes 19.2 g per day, what is his estimated weight?

C A L C U L AT O R E X P L O R AT I O N A N D D I S C O V E RY Evaluating Expressions and Looking for Patterns These “explorations” are designed to explore the full potential of a graphing calculator, as well as to use this potential to investigate patterns and discover connections that might otherwise be overlooked. In this Exploration and Discovery, we point out the various ways an expression can be evaluated on a graphing calculator. Some ways seem easier, faster, and/or better than others, but each has

advantages and disadvantages depending on the task at hand, and it will help to be aware of them all for future use. One way to evaluate an expression is to use the TABLE feature of a graphing calculator, with the expression entered as Y1 on the Y = screen. If you want the calculator to generate inputs, use the 2nd WINDOW (TBLSET) screen to indicate a starting value

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Strengthening Core Skills

1TblStart2 and an increment value 1 ¢Tbl2 , and set the calculator in Indpnt: AUTO ASK mode (to input specific values, the calculator should be in Indpnt: AUTO ASK mode). After pressing 2nd GRAPH (TABLE), the calculator shows the corresponding input and output values. For help with the basic TABLE feature of the TI-84 Plus, you can visit Section R.7 at www.mhhe.com/coburn. Expressions can also be evaluated on the home screen for a single value or a series of values. Enter the expres3 sion 4x  5 on the Y = screen (see Figure 1.16) and use 2nd MODE (QUIT) to get back to the home screen. To evaluate this expression, access Y1 using VARS (Y-VARS), and use the first option 1:Function ENTER . This brings us to a submenu where any of the equations Y1 through Y0 (actually Y10) can be accessed. Since the default setting is the one we need 1:Y1, simply press ENTER and Y1 appears on the home screen. To evaluate a single input, simply enclose it in parentheses. To evaluate more than one input, enter the numbers as a set of values with the set enclosed in parentheses. In Figure 1.17, Y1 has been evaluated for x  4, then simultaneously for x  4, 2, 0, and 2. A third way to evaluate expressions is using a list, with the desired inputs entered in List 1 (L1), and List 2 (L2) defined in terms of L1. For example, L2  34L1  5 will return the same values for inputs of 4, 2, 0, and 2 seen previously on the home screen (remember to clear the lists first). Lists are accessed by pressing STAT 1:Edit. Enter the numbers 4, 2, 0 and 2 in L1, then use the right arrow to move to L2. It is important to note that you next press the up arrow key so that the cursor overlies L2. The bottom of the screen now reads L2= (see Figure 1.18) and the calculator is waiting for us to define L2. After entering L2  34L1  5 and pressing ENTER we obtain the same outputs as before (see Figure 1.19).

The advantage of using the “list” method is that we can further explore or experiment with the output values in a search for patterns. Exercise 1: Evaluate the expression 0.2L1  3 on the list screen, using consecutive integer inputs from 6 to 6 inclusive. What do you notice about the outputs? Exercise 2: Evaluate the expression 12 L1  19.1 on the list screen, using consecutive integer inputs from 6 to 6 inclusive. We suspect there is a pattern to the output values, but this time the pattern is very difficult to see. Compute the difference between a few successive outputs from L2 [for Example L2112  L2122 4 . What do you notice?

77

Figure 1.16

Figure 1.17

Figure 1.18

Figure 1.19

STRENGTHENING CORE SKILLS An Alternative Method for Checking Solutions to Quadratic Equations To solve x2  2x  15  0 by factoring, students will often begin by looking for two numbers whose product is 15 (the constant term) and whose sum is 2 (the linear coefficient). The two numbers are 5 and 3 since 152132  15 and 5  3  2. In factored form, we have 1x  521x  32  0 with solutions x1  5 and x2  3. When these solutions are compared to the original coefficients, we can still see the sum/product relationship, but note that while 152132  15 still gives the constant term, 5  132  2 gives the linear coefficient with opposite sign. Although more difficult to accomplish,

this method can be applied to any factorable quadratic equation ax2  bx  c  0 if we divide through by a, c b giving x2  x   0. For 2x2  x  3  0, we a a 1 3 divide both sides by 2 and obtain x2  x   0, 2 2 3 then look for two numbers whose product is  and 2 1 3 whose sum is  . The numbers are  and 1 2 2

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3 3 1 3 since a b112   and   1   , showing the 2 2 2 2 3 solutions are x1  and x2  1. We again note the 2 3 c product of the solutions is the constant   , and the a 2 sum of the solutions is the linear coefficient with opposite 1 b sign:   . No one actually promotes this method for a 2 solving trinomials where a  1, but it does illustrate an important and useful concept: c b If x1 and x2 are the two roots of x2  x   0, a a c b then x1x2  and x1  x2   . a a Justification for this can be found by taking the product 2b2  4ac b and sum of the general solutions x1   2a 2a 2 b 2b  4ac  . Although the computation and x2  2a 2a looks impressive, the product can be computed as a binomial times its conjugate, and the radical parts add to zero for the sum, each yielding the results as already stated.

This observation provides a useful technique for checking solutions to a quadratic equation, even those having irrational or complex roots! Check the solutions shown in these exercises. Exercise 1: 2x2  5x  7  0 7 x1  2 x2  1 Exercise 2: 2x2  4x  7  0 2  312 x1  2 2  312 x2  2 Exercise 3: x2  10x  37  0 x1  5  2 13 i x2  5  2 13 i Exercise 4: Verify this sum/product check by computing the sum and product of the general solutions.

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CONNECTIONS TO CALCULUS Chapter 1 actually highlights numerous concepts and skills that transfer directly into a study of calculus. In the Chapter 1 opener, we noted that analyzing very small differences is one such skill, with this task carried out using the absolute value concept. The ability to solve a wide variety of equation types will also be a factor of your success in calculus. Here we’ll explore how these concepts and skills are “connected.”

Solving Various Types of Equations The need to solve equations of various types occurs frequently in both differential and integral calculus, and the required skills will span a broad range of your algebraic experience. Here we’ll solve a type of radical equation that occurs frequently in a study of optimization [finding the maximum or minimum value(s) of a function]. EXAMPLE 1



Minimizing Response Time A boater is 70 yd away from a straight shoreline when she gets an emergency call from her home, 400 yd down shore. Knowing she can row at 200 yd/min and run at 300 yd/min, how far down shore should she land the boat to make it home in the shortest time possible?

Solution



As with other forms of problem solving, drawing an accurate sketch is an important first step.

x

400  x Run

Home

70 yd Row

Boat

From the diagram, we note the rowing distance will be 2x2  4900 (using the Pythagorean theorem), and the running distance will be 400  x (total minus distance downshore). distance From the relationship time  , we find the total time required to reach rate 2x2  4900 400  x  home is t1x2  . Using the tools of calculus it can be 200 300 shown that the distance x down shore that results in the shortest possible time, 1 x  is a zero of T1x2  . Find the zero(es) of T(x) and state the 2 300 200 2x  4900 result in both exact and approximate form.

C2C1–1

79

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Connections to Calculus

Solution



Begin by isolating the radical on one side. x



1 300

200 2x  4900 300x  200 2x2  4900 1.5x  2x2  4900 2.25x2  x2  4900 1.25x2  4900 x2  3920 x  13920 x  28 15  62.6 2

add

1 300

clear denominators divide by 200 square both sides subtract x 2 divide by 1.25 solve for x; x 7 0 (distance) simplify radical (exact form) approximate form

The boater should row to a spot about 63 yd down shore, then run the remaining 337 yd. Now try Exercises 1 and 2



In addition to radical equations, equations involving rational exponents are often seen in a study of calculus. Many times, solving these equations involves combining the basic properties of exponents with other familiar skills such as factoring, or in this case, factoring least powers. EXAMPLE 2



Modeling the Motion of a Particle Suppose the motion of an object floating in turbulent water is modeled by the function d1t2  1t 1t2  9t  222, where d(t) represents the displacement (in meters) at t sec. Using the tools of calculus, it can be shown that the velocity v of 1 5 3 27 1 the particle is given by v1t2  t2  t2  11t2. Find any time(s) t when the 2 2 particle is motionless 1v  02.

Solution



Set the equation equal to zero and factor out the fraction and least power.

4 1 1 5t2a bt2 2

1 5 32 27 1 t  t2  11t2  0 2 2 2 1 1 1 1  27t2a bt2  22a b t2  0 2 2 1 12 2 t 15t  27t  222  0 2 1 12 t 15t  222 1t  12  0 2 1 22 or t  1 t2  0; t  5

original equation

rewrite to help factor

1 12 t (least power) 2

common factor

factor the trinomial

result

The particle is temporarily motionless at t  4.4 sec and t  1 sec. Now try Exercises 3 and 4



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Connections to Calculus

81

Absolute Value Inequalities and Delta/Epsilon Form While the terms may mean little to you now, the concept of absolute value plays an important role in the precise definition of a limit, intervals of convergence, and derivatives involving logarithmic functions. In the case of limits, the study of calculus concerns itself with very small differences, as in the difference between the number 3 itself, and a number very close to 3.

x2  9 . From the implicit domain and the figures x3 shown, we see that f 1x2 (shown as Y1) is not defined at 3, but is defined for any number near 3. The figures also suggest that when x is a number very close to 3, f (x) is a number very close to 6. Alternatively, we might say, “if the difference between x and 3 is very small, the difference between f (x) and 6 is very small.” The most convenient way to express this idea and make it practical is through the use of absolute value (which allows that the difference can be either positive or negative). Using the symbols  (delta) and  (epsilon) to represent very small (and possibly unequal) numbers, we can write this phrase in delta/epsilon form as Consider the function f 1x2 

if x  3 6 , then  f 1x2  6 6  For now, we’ll simply practice translating similar relationships from words into symbols, leaving any definitive conclusions for our study of limits in Chapter 11, or a future study of calculus. EXAMPLE 3



Using Delta/Epsilon Form Use a graphing calculator to explore the value of g1x2 

x2  3x  10 when x is x2

near 2, then write the relationship in delta/epsilon form. Solution



Using a graphing calculator and the approach outlined above produces the tables in the figures.

From these, it appears that, “if the difference between x and 2 is very small, the difference between f(x) and 7 is very small.” In delta/epsilon form: if x  2 6 , then  f 1x2  7 6 . Now try Exercises 5 through 8



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At first, modeling this relationship may seem like a minor accomplishment. But historically and in a practical sense, it is actually a major achievement as it enables us to “tame the infinite,” since we can now verify that no matter how small  is, there is a corresponding  that guarantees  f 1x2  7 6 

whenever

f(x) is infinitely close to 7

x  2 6  x is infinitely close to 2

This observation leads directly to the precise definition of a limit, the type of “limit” referred to in our Introduction to Calculus, found in the Preface (page 000). As noted there, such limits will enable us to find a precise formula for the instantaneous speed of the cue ball as it falls, and a precise formula for the volume of an irregular solid.

CONNECTIONS TO CALCULUS EXERCISES Solve the following equations.

1. To find the length of a rectangle with maximum area that can be circumscribed by a circle of radius 3 in. x2  0, requires that we solve 29  x2  29  x2 where the length of the rectangle is 2x. To the nearest hundredth, what is the length of the rectangle? 2. To find the height of an isosceles triangle with maximum area that can be inscribed in a circle of radius r  5 in. requires that we solve 5x x2 2  225  x   0, 225  x2 225  x2 where the height is 5  x. What is the height of the triangle? 3. If the motion of a particle in turbulent air is modeled by d  1t 12t2  9t  182, the velocity 3 1 27 1 of the particle is given by v  5t2  t2  9t2 2 (d in meters, t in seconds). Find any time(s) t when velocity v  0. 4. In order for a light source to provide maximum (circular) illumination to a workroom, the light must be hung at a certain height. While the

complete development requires trigonometry, we find that maximum illumination is obtained at the solutions of the equation shown, where h is the height of the light, k is a constant, and the radius of illumination is 12 ft. Solve the equation for h by factoring the least power and simplifying the 3 1 1h2  122 2 2  3h2 1h2  122 2 2  0. result: k 1h2  122 2 3 Use a graphing calculator to explore the value of the function given for values of x near the one indicated. Then write the relationship in words and in delta/epsilon form.

5. h1x2 

4x2  9 3 ;x 2x  3 2

6. v1x2 

x3  27 ; x  3 x3

7. w1x2 

7x3  28x ;x2 x2  4

8. F1x2 

x2  7x ;x0 x

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Precalculus—

2 CHAPTER CONNECTIONS

Relations, Functions, and Graphs CHAPTER OUTLINE 2.1 Rectangular Coordinates; Graphing Circles and Other Relations 84 2.2 Graphs of Linear Equations 97 2.3 Linear Graphs and Rates of Change 110 2.4 Functions, Function Notation, and the Graph of a Function 123 2.5 Analyzing the Graph of a Function 138 2.6 The Toolbox Functions and Transformations 157 2.7 Piecewise-Defined Functions 172

From the rate at which your computer can download a large file, to the rate a bacteria culture grows in the production of penicillin, science, medicine, sports, and industry all have a great interest in the rate at which change takes place. Where some measures of change have a tremendous impact on civilization (faster drying cement, stronger metal alloys, better communication), other measures of change quantify and track improvements in various areas of human endeavor. For instance, by making slight modifications in the tip or tail of an arrow, an olympic archer can alter the velocity of the arrow. Using the average rate of change formula, we can find the average velocity of the arrow for any time interval. This application appears as Exercise 61 in Section 2.5.

2.8 The Algebra and Composition of Functions 186

In this chapter, we use the average rate of change formula to develop the difference quotient. In a calculus course, the difference quotient is combined with the concept of a limit to find the instantaneous velocity of Connections the arrow at any time t. The Connections to Calculus for Chapter 2 highlights the algebraic skills necessary to Calculus to make the transition from the discrete view to the instantaneous view a smooth one. 83

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Precalculus—

2.1 Rectangular Coordinates; Graphing Circles and Other Relations Learning Objectives

In everyday life, we encounter a large variety of relationships. For instance, the time it takes us to get to work is related to our average speed; the monthly cost of heating a home is related to the average outdoor temperature; and in many cases, the amount of our charitable giving is related to changes in the cost of living. In each case we say that a relation exists between the two quantities.

In Section 2.1 you will learn how to:

A. Express a relation in mapping notation and ordered pair form

B. Graph a relation C. Develop the equation of a circle using the distance and midpoint formulas

D. Graph circles

WORTHY OF NOTE

EXAMPLE 1

Figure 2.1 In the most general sense, a relation is simply a P B correspondence between two sets. Relations can be represented in many different ways and may even Missy April 12 Jeff be very “unmathematical,” like the one shown in Nov 11 Angie Figure 2.1 between a set of people and the set of their Sept 10 Megan corresponding birthdays. If P represents the set of Nov 28 people and B represents the set of birthdays, we say Mackenzie May 7 Michael that elements of P correspond to elements of B, or the April 14 Mitchell birthday relation maps elements of P to elements of B. Using what is called mapping notation, we might simply write P S B. Figure 2.2 The bar graph in Figure 2.2 is also 155 ($145) 145 an example of a relation. In the graph, 135 each year is related to average annual ($123) 125 consumer spending on Internet media 115 (music downloads, Internet radio, Web105 based news articles, etc.). As an alterna($98) 95 tive to mapping or a bar graph, the ($85) 85 relation could also be represented using 75 ($69) ordered pairs. For example, the 65 ordered pair (3, 98) would indicate that in 2003, spending per person on Internet 2 3 5 1 7 media averaged $98 in the United Year (1 → 2001) States. Over a long period of time, we Source: 2006 Statistical Abstract of the United States could collect many ordered pairs of the form (t, s), where consumer spending s depends on the time t. For this reason we often call the second coordinate of an ordered pair (in this case s) the dependent variable, with the first coordinate designated as the independent variable. In this form, the set of all first coordinates is called the domain of the relation. The set of all second coordinates is called the range. Consumer spending (dollars per year)

From a purely practical standpoint, we note that while it is possible for two different people to share the same birthday, it is quite impossible for the same person to have two different birthdays. Later, this observation will help us mark the difference between a relation and a function.

A. Relations, Mapping Notation, and Ordered Pairs



Expressing a Relation as a Mapping and in Ordered Pair Form Represent the relation from Figure 2.2 in mapping notation and ordered pair form, then state its domain and range.

Solution



A. You’ve just learned how to express a relation in mapping notation and ordered pair form

84

Let t represent the year and s represent consumer spending. The mapping t S s gives the diagram shown. In ordered pair form we have (1, 69), (2, 85), (3, 98), (5, 123), and (7, 145). The domain is {1, 2, 3, 5, 7}, the range is {69, 85, 98, 123, 145}.

t

s

1 2 3 5 7

69 85 98 123 145

Now try Exercises 7 through 12



For more on this relation, see Exercise 81. 2-2

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Table 2.1 y  x  1 x

y

4

5

2

3

0

1

2

1

4

3

B. The Graph of a Relation Relations can also be stated in equation form. The equation y  x  1 expresses a relation where each y-value is one less than the corresponding x-value (see Table 2.1). The equation x  y expresses a relation where each x-value corresponds to the absolute value of y (see Table 2.2). In each case, the relation is the set of all ordered pairs (x, y) that create a true statement when substituted, and a few ordered pair solutions are shown in the tables for each equation. Relations can be expressed graphically using a rectangular coordinate system. It consists of a horizontal number line (the x-axis) and a vertical number line (the y-axis) intersecting at their zero marks. The Figure 2.3 point of intersection is called the origin. The x- and y y-axes create a flat, two-dimensional surface called 5 the xy-plane and divide the plane into four regions 4 called quadrants. These are labeled using a capital 3 QII QI 2 “Q” (for quadrant) and the Roman numerals I through 1 IV, beginning in the upper right and moving counterclockwise (Figure 2.3). The grid lines shown denote 5 4 3 2 11 1 2 3 4 5 x the integer values on each axis and further divide the 2 QIII QIV plane into a coordinate grid, where every point in 3 4 the plane corresponds to an ordered pair. Since a 5 point at the origin has not moved along either axis, it has coordinates (0, 0). To plot a point (x, y) means we place a dot at its location in the xy-plane. A few of the Figure 2.4 ordered pairs from y  x  1 are plotted in Figure y 5 2.4, where a noticeable pattern emerges—the points seem to lie along a straight line. (4, 3) If a relation is defined by a set of ordered pairs, the graph of the relation is simply the plotted points. The (2, 1) graph of a relation in equation form, such as y  x  1, 5 x is the set of all ordered pairs (x, y) that make the equa- 5 (0, 1) tion true. We generally use only a few select points to (2, 3) determine the shape of a graph, then draw a straight line (4, 5) or smooth curve through these points, as indicated by 5 any patterns formed.

Table 2.2 x  y x

y

2

2

1

1

0

0

1

1

2

2

EXAMPLE 2



Graphing Relations Graph the relations y  x  1 and x  y using the ordered pairs given earlier.

Solution



For y  x  1, we plot the points then connect them with a straight line (Figure 2.5). For x  y, the plotted points form a V-shaped graph made up of two half lines (Figure 2.6). Figure 2.5 5

Figure 2.6

y yx1

y 5

x  y (3, 3)

(4, 3)

(2, 2)

(2, 1) (0, 0) 5

5

x

5

5

(0, 1)

(2, 3)

x

(2, 2) (3, 3)

5

5

Now try Exercises 13 through 16



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While we used only a few points to graph the relations in Example 2, they are actually made up of an infinite number of ordered pairs that satisfy each equation, including those that might be rational or irrational. All of these points together make these graphs continuous, which for our purposes means you can draw the entire graph without lifting your pencil from the paper. Actually, a majority of graphs cannot be drawn using only a straight line or directed line segments. In these cases, we rely on a “sufficient number” of points to outline the basic shape of the graph, then connect the points with a smooth curve. As your experience with graphing increases, this “sufficient number of points” tends to get smaller as you learn to anticipate what the graph of a given relation should look like.

WORTHY OF NOTE As the graphs in Example 2 indicate, arrowheads are used where appropriate to indicate the infinite extension of a graph.

EXAMPLE 3



Graphing Relations Graph the following relations by completing the tables given. a. y  x2  2x b. y  29  x2 c. x  y2

Solution



For each relation, we use each x-input in turn to determine the related y-output(s), if they exist. Results can be entered in a table and the ordered pairs used to draw the graph. a. y  x2  2x Figure 2.7 y

x

y

(x, y) Ordered Pairs

4

24

(4, 24)

3

15

(3, 15)

2

8

(2, 8)

1

3

(1, 3)

0 1

0 1

2

0

(2, 0)

3

3

(3, 3)

4

8

(4, 8)

(0, 0)

(4, 8)

(2, 8) y  x2  2x

5

(1, 3)

(3, 3) (2, 0)

(0, 0) 5

5

(1, 1)

x

(1, 1)

2

The result is a fairly common graph (Figure 2.7), called a vertical parabola. Although (4, 24) and 13, 152 cannot be plotted here, the arrowheads indicate an infinite extension of the graph, which will include these points. y  29  x2

b. x

y

Figure 2.8

(x, y) Ordered Pairs

4

not real

—

3

0

(3, 0)

2

15

(2, 15)

1

212

(1, 212)

0

3

(0, 3)

1

212

(1, 212)

2

15

(2, 15)

3

0

(3, 0)

4

not real

—

y  9  x2

y 5

(1, 22) (2, 5) (3, 0)

(0, 3) (1, 22) (2, 5) (3, 0)

5

5

5

x

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The result is the graph of a semicircle (Figure 2.8). The points with irrational coordinates were graphed by estimating their location. Note that when x 6 3 or x 7 3, the relation y  29  x2 does not represent a real number and no points can be graphed. Also note that no arrowheads are used since the graph terminates at (3, 0) and (3, 0). c. Similar to x  y, the relation x  y2 is defined only for x  0 since y2 is always nonnegative (1  y2 has no real solutions). In addition, we reason that each positive x-value will correspond to two y-values. For example, given x  4, (4, 2) and (4, 2) are both solutions. x  y2

B. You’ve just learned how to graph a relation

Figure 2.9

x

y

(x, y) Ordered Pairs

2

not real

—

1

y 5

x  y2 (4, 2)

(2, 2)

not real

—

0

0

(0, 0)

1

1, 1

(1, 1) and (1, 1)

2

12, 12

(2, 12) and (2, 12)

3

13, 13

(3, 13) and (3, 13)

4

2, 2

(4, 2) and (4, 2)

(0, 0) 5

5

5

x

(2, 2) (4, 2)

This is the graph of a horizontal parabola (Figure 2.9). Now try Exercises 17 through 24



C. The Equation of a Circle Using the midpoint and distance formulas, we can develop the equation of another very important relation, that of a circle. As the name suggests, the midpoint of a line segment is located halfway between the endpoints. On a standard number line, the midpoint of the line segment with endpoints 1 and 5 is 3, but more important, note that 6 15   3. This 3 is the average distance (from zero) of 1 unit and 5 units: 2 2 observation can be extended to find the midpoint between any two points (x1, y1) and (x2, y2). We simply find the average distance between the x-coordinates and the average distance between the y-coordinates. The Midpoint Formula Given any line segment with endpoints P1  1x1, y1 2 and P2  1x2, y2 2 , the midpoint M is given by M: a

x1  x2 y1  y2 , b 2 2

The midpoint formula can be used in many different ways. Here we’ll use it to find the coordinates of the center of a circle.

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CHAPTER 2 Relations, Functions, and Graphs



EXAMPLE 4

Using the Midpoint Formula The diameter of a circle has endpoints at P1  13, 22 and P2  15, 42 . Use the midpoint formula to find the coordinates of the center, then plot this point.



Solution

x1  x2 y1  y2 , b 2 2 3  5 2  4 , b M: a 2 2 2 2 M: a , b  11, 12 2 2

Midpoint: a

y 5

P2

(1, 1) 5

5

x

P1 5

The center is at (1, 1), which we graph directly on the diameter as shown. Now try Exercises 25 through 34



Figure 2.10 y

c

The Distance Formula

(x2, y2)

In addition to a line segment’s midpoint, we are often interested in the length of the segment. For any two points (x1, y1) and (x2, y2) not lying on a horizontal or vertical line, a right triangle can be formed as in Figure 2.10. Regardless of the triangle’s orientation, the length of side a (the horizontal segment or base of the triangle) will have length x2  x1 units, with side b (the vertical segment or height) having length y2  y1 units. From the Pythagorean theorem (Appendix I.F), we see that c2  a2  b2 corresponds to c2  1 x2  x1 2 2  1 y2  y1 2 2. By taking the square root of both sides we obtain the length of the hypotenuse, which is identical to the distance between these two points: c  21x2  x1 2 2  1y2  y1 2 2. The result is called the distance formula, although it’s most often written using d for distance, rather than c. Note the absolute value bars are dropped from the formula, since the square of any quantity is always nonnegative. This also means that either point can be used as the initial point in the computation.

b

x

a

(x1, y1)

(x2, y1)

P2

The Distance Formula

P1

Given any two points P1  1x1, y1 2 and P2  1x2, y2 2, the straight line distance between them is

b   y2 y1

d

d  21x2  x1 2 2  1y2  y1 2 2

a   x2 x1

EXAMPLE 5



Using the Distance Formula Use the distance formula to find the diameter of the circle from Example 4.

Solution



For 1x1, y1 2  13, 22 and 1x2, y2 2  15, 42, the distance formula gives d  21x2  x1 2 2  1y2  y1 2 2

 2 3 5  132 4 2  34  122 4 2

 282  62  1100  10

The diameter of the circle is 10 units long. Now try Exercises 35 through 38



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EXAMPLE 6



Determining if Three Points Form a Right Triangle Use the distance formula to determine if the following points are the vertices of a right triangle: (8, 1), (2, 9), and (10, 0)

Solution



We begin by finding the distance between each pair of points, then attempt to apply the Pythagorean theorem. For 1x1, y1 2  18, 12, 1x2, y2 2  12, 92 : For 1x2, y2 2  12, 92, 1x3, y3 2  110, 02 : d  21x2  x1 2 2  1y2  y1 2 2

 2 3 2  182 4 2  19  12 2  262  82  1100  10

For 1x1, y1 2  18, 12, 1x3, y3 2  110, 02 : d  21x3  x1 2 2  1y3  y1 2 2

 2 3 10  182 4 2  10  12 2  2182  112 2  1325  5113

d  21x3  x2 2 2  1y3  y2 2 2

 23 10  122 4 2  10  92 2

 2122  192 2  1225  15

Using the unsimplified form, we clearly see that a 2  b 2  c 2 corresponds to 1 11002 2  1 12252 2  1 13252 2, a true statement. Yes, the triangle is a right triangle. Now try Exercises 39 through 44



A circle can be defined as the set of all points in a plane that are a fixed distance called the radius, from a fixed point called the center. Since the definition involves distance, we can construct the general equation of a circle using the distance formula. Assume the center has coordinates (h, k), and let (x, y) represent any point on the graph. Since the distance between these points is equal to the radius r, the distance formula yields: 21x  h2 2  1y  k2 2  r. Squaring both sides gives the equation of a circle in standard form: 1x  h2 2  1y  k2 2  r2. The Equation of a Circle A circle of radius r with center at (h, k) has the equation 1x  h2 2  1y  k2 2  r2 If h  0 and k  0, the circle is centered at (0, 0) and the graph is a central circle with equation x2  y2  r2. At other values for h or k, the center is at (h, k) with no change in the radius. Note that an open dot is used for the center, as it’s actually a point of reference and not a part of the actual graph.

y

Circle with center at (h, k) r

k

(x, y)

(h, k)

Central circle

(x  h)2  (y  k)2  r2 r

(x, y)

(0, 0)

h

x

x2  y2  r2

EXAMPLE 7



Finding the Equation of a Circle

Solution



Since the center is at (0, 1) we have h  0, k  1, and r  4. Using the standard form 1x  h2 2  1y  k2 2  r2 we obtain

Find the equation of a circle with center 10, 1) and radius 4. 1x  02 2  3y  112 4 2  42 x2  1y  12 2  16

substitute 0 for h, 1 for k, and 4 for r simplify

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The graph of x2  1y  12 2  16 is shown in the figure. y (0, 3) Circle

r4 (4, 1)

Center: (0, 1) x Radius: r  4 (4, 1) Diameter: 2r  8

(0, 1)

C. You’ve just learned how to develop the equation of a circle using the distance and midpoint formulas

(0, 5)

Now try Exercises 45 through 62



D. The Graph of a Circle The graph of a circle can be obtained by first identifying the coordinates of the center and the length of the radius from the equation in standard form. After plotting the center point, we count a distance of r units left and right of center in the horizontal direction, and up and down from center in the vertical direction, obtaining four points on the circle. Neatly graph a circle containing these four points. EXAMPLE 8



Graphing a Circle

Solution



Comparing the given equation with the standard form, we find the center is at 12, 32 and the radius is r  213  3.5.

Graph the circle represented by 1x  22 2  1y  32 2  12. Clearly label the center and radius.

1x  h2 2  1y  k2 2  r2 ↓ ↓ ↓ 1x  22 2  1y  32 2  12 h  2 k  3 h2 k  3

standard form given equation

r2  12 r  112  2 13  3.5

radius must be positive

Plot the center (2, 3) and count approximately 3.5 units in the horizontal and vertical directions. Complete the circle by freehand drawing or using a compass. The graph shown is obtained. y Some coordinates are approximate

Circle (2, 0.5) x

r ~ 3.5 (1.5, 3)

(2, 3)

Center: (2, 3) Radius: r  2兹3

Endpoints of horizontal diameter (5.5, 3) (2  2兹3, 3) and (2  2兹3, 3) Endpoints of vertical diameter (2, 3  2兹3) and (2, 3  2兹3)

(2, 6.5)

Now try Exercises 63 through 68



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In Example 8, note the equation is composed of binomial squares in both x and y. By expanding the binomials and collecting like terms, we can write the equation of the circle in the general form:

WORTHY OF NOTE After writing the equation in standard form, it is possible to end up with a constant that is zero or negative. In the first case, the graph is a single point. In the second case, no graph is possible since roots of the equation will be complex numbers. These are called degenerate cases. See Exercise 91.

EXAMPLE 9

1x  22 2  1y  32 2  12 x  4x  4  y2  6y  9  12 x2  y2  4x  6y  1  0 2

standard form expand binomials combine like terms—general form

For future reference, observe the general form contains a sum of second-degree terms in x and y, and that both terms have the same coefficient (in this case, “1”). Since this form of the equation was derived by squaring binomials, it seems reasonable to assume we can go back to the standard form by creating binomial squares in x and y. This is accomplished by completing the square. 

Finding the Center and Radius of a Circle Find the center and radius of the circle with equation x2  y2  2x  4y  4  0. Then sketch its graph and label the center and radius.

Solution



To find the center and radius, we complete the square in both x and y. x2  y2  2x  4y  4  0 1x2  2x  __ 2  1y2  4y  __ 2  4 1x2  2x  12  1y2  4y  42  4  1  4

given equation group x-terms and y-terms; add 4 complete each binomial square

1x  12 2  1y  22 2  9

factor and simplify

The center is at 11, 22 and the radius is r  19  3. (1, 5)

(4, 2)

y

r3 (1, 2)

(2, 2)

(1, 1)

Circle x Center: (1, 2) Radius: r  3

Now try Exercises 69 through 80

EXAMPLE 10





Applying the Equation of a Circle To aid in a study of nocturnal animals, some naturalists install a motion detector near a popular watering hole. The device has a range of 10 m in any direction. Assume the water hole has coordinates (0, 0) and the device is placed at (2, 1). a. Write the equation of the circle that models the maximum effective range of the device. b. Use the distance formula to determine if the device will detect a badger that is approaching the water and is now at coordinates (11, 5).

y 5

10

5

x

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Solution



a. Since the device is at (2, 1) and the radius (or reach) of detection is 10 m, any movement in the interior of the circle defined by 1x  22 2  1y  12 2  102 will be detected. b. Using the points (2, 1) and (11, 5) in the distance formula yields: d  21x2  x1 2 2  1y2  y1 2 2

 2111  22  35  112 4 2

 29  142  181  16  197  9.85 2

D. You’ve just learned how to graph circles

2

distance formula 2

substitute given values simplify compute squares result

Since 9.85 6 10, the badger is within range of the device and will be detected. Now try Exercises 83 through 88



TECHNOLOGY HIGHLIGHT

The Graph of a Circle When using a graphing calculator to study circles, it is important to keep two things in mind. First, we must modify the equation of the circle before it can be graphed using this technology. Second, most standard viewing windows have the x- and y-values preset at 3 10, 104 even though the calculator screen is not square. This tends to compress the y-values and give a skewed image of the graph. Consider the relation x2  y2  25, which we know is the equation of a circle centered at (0, 0) with radius r  5. To enable the calculator to graph this relation, we must define it in two pieces by solving for y: x2  y 2  25 y 2  25  x2 y   225  x2

original equation isolate y 2

Figure 2.11 10

solve for y

Note that we can separate this result into two parts, 10 10 enabling the calculator to draw the circle: Y1  225  x2 gives the “upper half” of the circle, and Y2  225  x2 gives the “lower half.” Enter these on the Y = screen (note that Y2  Y1 can be used instead of reentering the entire 10 expression: VARS ENTER ). But if we graph Y1 and Y2 Figure 2.12 on the standard screen, the result appears more oval than 10 circular (Figure 2.11). One way to fix this is to use the ZOOM 5:ZSquare option, which places the tick marks equally spaced on both axes, instead of trying to force both to display points 15.2 15.2 from 10 to 10 (see Figure 2.12). Although it is a much improved graph, the circle does not appear “closed” as the calculator lacks sufficient pixels to show the proper curvature. A second alternative is to manually set a “friendly” window. 10 Using Xmin  9.4, Xmax  9.4, Ymin  6.2, and Ymax  6.2 will generate a better graph, which we can use to study the relation more closely. Note that we can jump between the upper and lower halves of the circle using the up or down arrows. Exercise 1: Graph the circle defined by x2  y2  36 using a friendly window, then use the TRACE feature to find the value of y when x  3.6. Now find the value of y when x  4.8. Explain why the values seem “interchangeable.” Exercise 2: Graph the circle defined by 1x  32 2  y2  16 using a friendly window, then use the feature to find the value of the y-intercepts. Show you get the same intercepts by computation.

TRACE

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93

2.1 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.



1. If a relation is defined by a set of ordered pairs, the domain is the set of all components, the range is the set of all components.

4. For x2  y2  25, the center of the circle is at and the length of the radius is units. The graph is called a circle.

2. For the equation y  x  5 and the ordered pair (x, y), x is referred to as the input or variable, while y is called the or dependent variable.

5. Discuss/Explain how to find the center and radius of the circle defined by the equation x2  y2  6x  7. How would this circle differ from the one defined by x2  y2  6y  7?

3. A circle is defined as the set of all points that are an equal distance, called the , from a given point, called the .

6. In Example 3b we graphed the semicircle defined by y  29  x2. Discuss how you would obtain the equation of the full circle from this equation, and how the two equations are related.

DEVELOPING YOUR SKILLS

Represent each relation in mapping notation, then state the domain and range.

GPA

7.

4.00 3.75 3.50 3.25 3.00 2.75 2.50 2.25 2.00 0

2 13. y   x  1 3 x

1

2

3

4

5

Year in college

Efficiency rating

8.

95 90 85 80 75 70 65 60 55 0

Complete each table using the given equation. For Exercises 15 and 16, each input may correspond to two outputs (be sure to find both if they exist). Use these points to graph the relation.

2

3

4

5

6

Month

State the domain and range of each relation.

9. {(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)} 10. {(2, 4), (3, 5), (1, 3), (4, 5), (2, 3)} 11. {(4, 0), (1, 5), (2, 4), (4, 2), (3, 3)} 12. {(1, 1), (0, 4), (2, 5), (3, 4), (2, 3)}

x

6

8

3

4

0

0

3

4

y

6

8

8

10

15. x  2  y

16. y  1  x

x

1

y

5 14. y   x  3 4

y

x

2

0

0

1

1

3

3

5

6

6

7

7

y

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17. y  x2  1 x

18. y  x2  3 x

y

2

2

1

0

0

2

1

3

2

4

3

19. y  225  x2 x

x

y

4

12

3

5

0

0

2

3

3

5

4

12

21. x  1  y x

y

2

5

3

4

4

2

5

1.25

6

1

11

23. y  2x  1 3

x

y

54321 1 2 3 4 5

1 2 3 4 5 x

1 2 3 4 5 x

33.

y 5 4 3 2 1

34.

1 2 3 4 5 x

y 5 4 3 2 1 54321 1 2 3 4 5

1 2 3 4 5 x

36. Use the distance formula to find the length of the line segment in Exercise 32.

y

37. Use the distance formula to find the length of the diameter for the circle in Exercise 33. 38. Use the distance formula to find the length of the diameter for the circle in Exercise 34.

24. y  1x  12 x

5 4 3 2 1

35. Use the distance formula to find the length of the line segment in Exercise 31.

22. y  2  x x

y

32.

Find the center of each circle with the diameter shown.

54321 1 2 3 4 5

2

10

y 5 4 3 2 1 54321 1 2 3 4 5

20. y  2169  x2

y

2

31.

y

3

Find the midpoint of each segment.

3

y

In Exercises 39 to 44, three points that form the vertices of a triangle are given. Use the distance formula to determine if any of the triangles are right triangles.

39. (5, 2), (0, 3), (4, 4)

9

2

2

1

1

0

41. (4, 3), (7, 1), (3, 2)

0

1

4

2

42. (3, 7), (2, 2), (5, 5)

7

3

40. (7, 0), (1, 0), (7, 4)

43. (3, 2), (1, 5), (6, 4) 44. (0, 0), (5, 2), (2, 5)

Find the midpoint of each segment with the given endpoints.

25. (1, 8), (5, 6)

26. (5, 6), (6, 8)

27. (4.5, 9.2), (3.1, 9.8) 28. (5.2, 7.1), (6.3, 7.1) 3 1 2 1 3 1 3 5 29. a ,  b, a , b 30. a ,  b, a , b 5 3 10 4 4 3 8 6

Find the equation of a circle satisfying the conditions given, then sketch its graph.

45. center (0, 0), radius 3 46. center (0, 0), radius 6 47. center (5, 0), radius 13 48. center (0, 4), radius 15 49. center (4, 3), radius 2 50. center (3, 8), radius 9 51. center (7, 4), radius 17

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52. center (2, 5), radius 16 53. center (1, 2), diameter 6 54. center (2, 3), diameter 10 55. center (4, 5), diameter 4 13

67. 1x  42 2  y2  81

68. x2  1y  32 2  49 Write each equation in standard form to find the center and radius of the circle. Then sketch the graph.

56. center (5, 1), diameter 4 15

69. x2  y2  10x  12y  4  0

57. center at (7, 1), graph contains the point (1, 7)

70. x2  y2  6x  8y  6  0

58. center at (8, 3), graph contains the point (3, 15)

71. x2  y2  10x  4y  4  0

59. center at (3, 4), graph contains the point (7, 9)

72. x2  y2  6x  4y  12  0

60. center at (5, 2), graph contains the point (1, 3)

73. x2  y2  6y  5  0

61. diameter has endpoints (5, 1) and (5, 7)

74. x2  y2  8x  12  0

62. diameter has endpoints (2, 3) and (8, 3)

75. x2  y2  4x  10y  18  0

Identify the center and radius of each circle, then graph. Also state the domain and range of the relation.

63. 1x  22 2  1y  32 2  4 64. 1x  52 2  1y  12 2  9

65. 1x  12 2  1y  22 2  12 66. 1x  72 2  1y  42 2  20 

76. x2  y2  8x  14y  47  0 77. x2  y2  14x  12  0 78. x2  y2  22y  5  0 79. 2x2  2y2  12x  20y  4  0 80. 3x2  3y2  24x  18y  3  0

WORKING WITH FORMULAS

81. Spending on Internet media: s  12.5t  59 The data from Example 1 is closely modeled by the formula shown, where t represents the year (t  0 corresponds to the year 2000) and s represents the average amount spent per person, per year in the United States. (a) List five ordered pairs for this relation using t  1, 2, 3, 5, 7. Does the model give a good approximation of the actual data? (b) According to the model, what will be the average amount spent on Internet media in the year 2008? (c) According to the model, in what year will annual spending surpass $196? (d) Use the table to graph this relation. 

95

Section 2.1 Rectangular Coordinates; Graphing Circles and Other Relations

82. Area of an inscribed square: A  2r2 The area of a square inscribed in a circle is found by using the formula given where r is the radius of the circle. Find the area of the inscribed square shown.

y

(5, 0) x

APPLICATIONS

83. Radar detection: A luxury liner is located at map coordinates (5, 12) and has a radar system with a range of 25 nautical miles in any direction. (a) Write the equation of the circle that models the range of the ship’s radar, and (b) Use the distance formula to determine if the radar can pick up the liner’s sister ship located at coordinates (15, 36).

84. Earthquake range: The epicenter (point of origin) of a large earthquake was located at map coordinates (3, 7), with the quake being felt up to 12 mi away. (a) Write the equation of the circle that models the range of the earthquake’s effect. (b) Use the distance formula to determine if a person living at coordinates (13, 1) would have felt the quake.

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85. Inscribed circle: Find the equation for both the red and blue circles, then find the area of the region shaded in blue.

y

(2, 0) x

87. Radio broadcast range: Two radio stations may not use the same frequency if their broadcast areas overlap. Suppose station KXRQ has a broadcast area bounded by x2  y2  8x  6y  0 and WLRT has a broadcast area bounded by x2  y2  10x  4y  0. Graph the circle representing each broadcast area on the same grid to determine if both stations may broadcast on the same frequency.



y (3, 4)

x

88. Radio broadcast range: The emergency radio broadcast system is designed to alert the population by relaying an emergency signal to all points of the country. A signal is sent from a station whose broadcast area is bounded by x2  y2  2500 (x and y in miles) and the signal is picked up and relayed by a transmitter with range 1x  202 2  1y  302 2  900. Graph the circle representing each broadcast area on the same grid to determine the greatest distance from the original station that this signal can be received. Be sure to scale the axes appropriately.

EXTENDING THE THOUGHT

89. Although we use the word “domain” extensively in mathematics, it is also commonly seen in literature and heard in everyday conversation. Using a collegelevel dictionary, look up and write out the various meanings of the word, noting how closely the definitions given are related to its mathematical use. 90. Consider the following statement, then determine whether it is true or false and discuss why. A graph will exhibit some form of symmetry if, given a point that is h units from the x-axis, k units from the y-axis, and d units from the origin, there is a second point



86. Inscribed triangle: The area of an equilateral triangle inscribed in a circle is given 3 13 2 r, by the formula A  4 where r is the radius of the circle. Find the area of the equilateral triangle shown.

on the graph that is a like distance from the origin and each axis. 91. When completing the square to find the center and radius of a circle, we sometimes encounter a value for r2 that is negative or zero. These are called degenerate cases. If r2 6 0, no circle is possible, while if r2  0, the “graph” of the circle is simply the point (h, k). Find the center and radius of the following circles (if possible). a. x2  y2  12x  4y  40  0 b. x2  y2  2x  8y  8  0 c. x2  y2  6x  10y  35  0

MAINTAINING YOUR SKILLS

92. (1.3) Solve the absolute value inequality and write the solution in interval notation. w  2 1 5   3 4 6 93. (R.1) Give an example of each of the following: a. a whole number that is not a natural number b. a natural number that is not a whole number c. a rational number that is not an integer

d. an integer that is not a rational number e. a rational number that is not a real number f. a real number that is not a rational number. 94. (1.5) Solve x2  13  6x using the quadratic equation. Simplify the result. 95. (1.6) Solve 1  1n  3  n and check solutions by substitution. If a solution is extraneous, so state.

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Precalculus—

2.2 Graphs of Linear Equations In preparation for sketching graphs of other relations, we’ll first consider the characteristics of linear graphs. While linear graphs are fairly simple models, they have many substantive and meaningful applications. Figure 2.13 For instance, most of us are aware that 155 ($145) music and video downloads have been 145 increasing in popularity since they were 135 first introduced. A close look at Example ($123) 125 1 of Section 2.1 reveals that spending on 115 music downloads and Internet radio 105 ($98) increased from $69 per person per year in 95 ($85) 2001 to $145 in 2007 (Figure 2.13). 85 From an investor’s or a producer’s point 75 ($69) of view, there is a very high interest in the 65 questions, How fast are sales increasing? 3 5 1 2 7 Can this relationship be modeled matheYear (1 → 2001) matically to help predict sales in future years? Answers to these and other ques- Source: 2006 SAUS tions are precisely what our study in this section is all about.

Learning Objectives In Section 2.2 you will learn how to:

A. Graph linear equations using the intercept method

Consumer spending (dollars per year)

B. Find the slope of a line C. Graph horizontal and vertical lines

D. Identify parallel and perpendicular lines

E. Apply linear equations in context

A. The Graph of a Linear Equation A linear equation can be identified using these three tests: (1) the exponent on any variable is one, (2) no variable occurs in a denominator, and (3) no two variables are multiplied together. The equation 3y  9 is a linear equation in one variable, while 2x  3y  12 and y  32 x  4 are linear equations in two variables. In general, we have the following definition: Linear Equations A linear equation is one that can be written in the form ax  by  c where a and b are not simultaneously zero. The most basic method for graphing a line is to simply plot a few points, then draw a straight line through the points. EXAMPLE 1



Graphing a Linear Equation in Two Variables Graph the equation 3x  2y  4 by plotting points.

Solution

WORTHY OF NOTE If you cannot draw a straight line through the plotted points, a computational error has been made. All points satisfying a linear equation lie on a straight line.



y

Selecting x  2, x  0, x  1, and x  4 as inputs, we compute the related outputs and enter the ordered pairs in a table. The result is x input 2

y output

(2, 5)

0

2

(0, 2)

1

0.5

(1, 12 )

4

5

(0, 2) (1, q)

(x, y) ordered pairs

5

4

(2, 5)

(4, 4)

5

5

(4, 4) 5

Now try Exercises 7 through 12 2-15

x



97

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Note the line in Example 1 crosses the y-axis at (0, 2), and this point is called the y-intercept of the line. In general, y-intercepts have the form (0, y). Although difficult to see graphically, substituting 0 for y and solving for x shows the line crosses the x-axis at (43 , 0) and this point is called the x-intercept. In general, x-intercepts have the form (x, 0). The x- and y-intercepts are usually easier to calculate than other points (since y  0 or x  0, respectively) and we often graph linear equations using only these two points. This is called the intercept method for graphing linear equations. The Intercept Method 1. Substitute 0 for x and solve for y. This will give the y-intercept (0, y). 2. Substitute 0 for y and solve for x. This will give the x-intercept (x, 0). 3. Plot the intercepts and use them to graph a straight line. EXAMPLE 2



Graphing Lines Using the Intercept Method Graph 3x  2y  9 using the intercept method.

Solution



Substitute 0 for x (y-intercept) 3102  2y  9 2y  9 9 y 2 9 a0, b 2

Substitute 0 for y (x-intercept) 3x  2102  9 3x  9 x3 13, 02

5

y 3x  2y  9

冢0, t 冣

(3, 0) 5

A. You’ve just learned how to graph linear equations using the intercept method

5

x

5

Now try Exercises 13 through 32



B. The Slope of a Line After the x- and y-intercepts, we next consider the slope of a line. We see applications of the concept in many diverse occupations, including the grade of a highway (trucking), the pitch of a roof (carpentry), the climb of an airplane Figure 2.14 (flying), the drainage of a field (landscaping), and y the slope of a mountain (parks and recreation). y (x2, y2) 2 While the general concept is an intuitive one, we seek to quantify the concept (assign it a numeric y2  y1 value) for purposes of comparison and decision rise making. In each of the preceding examples, slope is a measure of “steepness,” as defined by the ratio (x1, y1) vertical change horizontal change . Using a line segment through y1 arbitrary points P1  1x1, y1 2 and P2  1x2, y2 2 , we x2  x1 run can create the right triangle shown in Figure 2.14. x The figure illustrates that the vertical change or the x2 x1

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Section 2.2 Graphs of Linear Equations

change in y (also called the rise) is simply the difference in y-coordinates: y2  y1. The horizontal change or change in x (also called the run) is the difference in x-coordinates: x2  x1. In algebra, we typically use the letter “m” to represent slope, y y change in y 1 giving m  x22   x1 as the change in x . The result is called the slope formula.

WORTHY OF NOTE While the original reason that “m” was chosen for slope is uncertain, some have speculated that it was because in French, the verb for “to climb” is monter. Others say it could be due to the “modulus of slope,” the word modulus meaning a numeric measure of a given property, in this case the inclination of a line.

EXAMPLE 3

The Slope Formula Given two points P1  1x1, y1 2 and P2  1x2, y2 2 , the slope of any nonvertical line through P1 and P2 is y2  y1 m x2  x1 where x2  x1. 

Using the Slope Formula Find the slope of the line through the given points. a. (2, 1) and (8, 4) b. (2, 6) and (4, 2)

Solution



a. For P1  12, 12 and P2  18, 42 , y2  y1 m x2  x1 41  82 3 1   6 2 The slope of this line is 12.

b. For P1  12, 62 and P2  14, 22, y2  y1 m x2  x1 26  4  122 4 2   6 3 The slope of this line is 2 3 . Now try Exercises 33 through 40

CAUTION





When using the slope formula, try to avoid these common errors. 1. The order that the x- and y-coordinates are subtracted must be consistent, since

y  y 2 1 x2  x1

y  y

 x21 

1

x2 .

2. The vertical change (involving the y-values) always occurs in the numerator: y  y 2 1 x2  x1

x  x

 y22 

1

y1 .

3. When x1 or y1 is negative, use parentheses when substituting into the formula to prevent confusing the negative sign with the subtraction operation.

Actually, the slope value does much more than quantify the slope of a line, it expresses a rate of change between the quantities measured along each axis. In appli¢y change in y cations of slope, the ratio change in x is symbolized as ¢x . The symbol ¢ is the Greek letter delta and has come to represent a change in some quantity, and the notation ¢y m  ¢x is read, “slope is equal to the change in y over the change in x.” Interpreting slope as a rate of change has many significant applications in college algebra and beyond. EXAMPLE 4



Interpreting the Slope Formula as a Rate of Change Jimmy works on the assembly line for an auto parts remanufacturing company. By 9:00 A.M. his group has assembled 29 carburetors. By 12:00 noon, they have completed 87 carburetors. Assuming the relationship is linear, find the slope of the line and discuss its meaning in this context.

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Solution



First write the information as ordered pairs using c to represent the carburetors assembled and t to represent time. This gives 1t1, c1 2  19, 292 and 1t2, c2 2  112, 872. The slope formula then gives: c2  c1 ¢c 87  29   ¢t t2  t1 12  9 58 or 19.3  3

WORTHY OF NOTE Actually, the assignment of (t1, c1) to (9, 29) and (t2, c2) to (12, 87) was arbitrary. The slope ratio will be the same as long as the order of subtraction is the same. In other words, if we reverse this assignment and use 1t1, c1 2  112, 872 and 1t2, c2 2  19, 292 , we have  87 58 58 m  29 9  12  3  3 .

assembled Here the slope ratio measures carburetors , and we see that Jimmy’s group can hours assemble 58 carburetors every 3 hr, or about 1913 carburetors per hour.

Now try Exercises 41 through 44

Positive and Negative Slope If you’ve ever traveled by air, you’ve likely heard the announcement, “Ladies and gentlemen, please return to your seats and fasten your seat belts as we begin our descent.” For a time, the descent of the airplane follows a linear path, but now the slope of the line is negative since the altitude of the plane is decreasing. Positive and negative slopes, as well as the rate of change they represent, are important characteristics of linear graphs. In Example 3a, the slope was a positive number (m 7 0) and the line will slope upward from left to right since the y-values are increasing. If m 6 0, the slope of the line is negative and the line slopes downward as you move left to right since y-values are decreasing.

m  0, positive slope y-values increase from left to right

EXAMPLE 5





m  0, negative slope y-values decrease from left to right

Applying Slope to Changes in Altitude At a horizontal distance of 10 mi after take-off, an airline pilot receives instructions to decrease altitude from their current level of 20,000 ft. A short time later, they are 17.5 mi from the airport at an altitude of 10,000 ft. Find the slope ratio for the descent of the plane and discuss its meaning in this context. Recall that 1 mi  5280 ft.

Solution



Let a represent the altitude of the plane and d its horizontal distance from the airport. Converting all measures to feet, we have 1d1, a1 2  152,800, 20,0002 and 1d2, a2 2  192,400, 10,0002 , giving a 2  a1 10,000  20,000 ¢a   ¢d d2  d1 92,400  52,800 10,000 25   39,600 99

B. You’ve just learned how to find the slope of a line

¢altitude Since this slope ratio measures ¢distance , we note the plane decreased 25 ft in altitude for every 99 ft it traveled horizontally.

Now try Exercises 45 through 48



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Section 2.2 Graphs of Linear Equations

C. Horizontal Lines and Vertical Lines Horizontal and vertical lines have a number of important applications, from finding the boundaries of a given graph, to performing certain tests on nonlinear graphs. To better understand them, consider that in one dimension, the graph of x  2 is a single point (Figure 2.15), indicating a location on Figure 2.15 the number line 2 units from zero in the posx2 itive direction. In two dimensions, the equation x  2 represents all points with an 5 4 3 2 1 0 1 2 3 4 5 x-coordinate of 2. A few of these are graphed in Figure 2.16, but since there are an infinite number, we end up with a solid vertical line whose equation is x  2 (Figure 2.17). Figure 2.16

Figure 2.17

y 5

y (2, 5)

5

x2

(2, 3) (2, 1) 5

(2, 1)

5

x

5

5

x

(2, 3) 5

The same idea can be applied to horizontal lines. In two dimensions, the equation y  4 represents all points with a y-coordinate of positive 4, and there are an infinite number of these as well. The result is a solid horizontal line whose equation is y  4. See Exercises 49–54.

WORTHY OF NOTE If we write the equation x  2 in the form ax  by  c, the equation becomes x  0y  2, since the original equation has no y-variable. Notice that regardless of the value chosen for y, x will always be 2 and we end up with the set of ordered pairs (2, y), which gives us a vertical line.

EXAMPLE 6

5

Vertical Lines

Horizontal Lines

The equation of a vertical line is

The equation of a horizontal line is

xh

yk

where (h, 0) is the x-intercept.

where (0, k) is the y-intercept.

So far, the slope formula has only been applied to lines that were nonhorizontal or nonvertical. So what is the slope of a horizontal line? On an intuitive level, we expect that a perfectly level highway would have an incline or slope of zero. In general, for any two points on a horizontal line, y2  y1 and y2  y1  0, giving a slope of m  x2 0 x1  0. For any two points on a vertical line, x2  x1 and x2  x1  0, making y  y the slope ratio undefined: m  2 0 1.



The Slope of a Vertical Line

The Slope of a Horizontal Line

The slope of any vertical line is undefined.

The slope of any horizontal line is zero.

Calculating Slopes The federal minimum wage remained constant from 1997 through 2006. However, the buying power (in 1996 dollars) of these wage earners fell each year due to inflation (see Table 2.3). This decrease in buying power is approximated by the red line shown.

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a. Using the data or graph, find the slope of the line segment representing the minimum wage. b. Select two points on the line representing buying power to approximate the slope of the line segment, and explain what it means in this context. Table 2.3

Solution



WORTHY OF NOTE In the context of lines, try to avoid saying that a horizontal line has “no slope,” since it’s unclear whether a slope of zero or an undefined slope is intended.

C. You’ve just learned how to graph horizontal and vertical lines

5.15

Minimum wage w

Buying power p

1997

5.15

5.03

1998

5.15

4.96

1999

5.15

4.85

2000

5.15

4.69

2001

5.15

4.56

2002

5.15

4.49

4.15

2003

5.15

4.39

4.05

2004

5.15

4.28

2005

5.15

4.14

2006

5.15

4.04

5.05 4.95 4.85 4.75 4.65 4.55 4.45 4.35 4.25

19

97 19 98 19 9 20 9 00 20 01 20 02 20 03 20 04 20 0 20 5 06

Wages/Buying power

Time t (years)

Time in years

a. Since the minimum wage did not increase or decrease from 1997 to 2006, the line segment has slope m  0. b. The points (1997, 5.03) and (2006, 4.04) from the table appear to be on or close to the line drawn. For buying power p and time t, the slope formula yields: p2  p 1 ¢p  ¢t t2  t1 4.04  5.03  2006  1997 0.99 0.11   9 1 The buying power of a minimum wage worker decreased by 11¢ per year during this time period. Now try Exercises 55 and 56



D. Parallel and Perpendicular Lines Two lines in the same plane that never intersect are called parallel lines. When we place these lines on the coordinate grid, we find that “never intersect” is equivalent to saying “the lines have equal slopes but different y-intercepts.” In Figure 2.18, notice the rise ¢y and run of each line is identical, and that by counting ¢x both lines have slope m  34. y

Figure 2.18

5

Generic plane L 1

Run L2

L1 Run

Rise

L2

Rise

5

5

5

Coordinate plane

x

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Section 2.2 Graphs of Linear Equations

Parallel Lines Given L1 and L2 are distinct, nonvertical lines with slopes of m1 and m2, respectively. 1. If m1  m2, then L1 is parallel to L2. 2. If L1 is parallel to L2, then m1  m2. In symbols we write L1 7 L2. Any two vertical lines (undefined slope) are parallel. EXAMPLE 7A



Determining Whether Two Lines Are Parallel Teladango Park has been mapped out on a rectangular coordinate system, with a ranger station at (0, 0). BJ and Kapi are at coordinates 124, 182 and have set a direct course for the pond at (11, 10). Dave and Becky are at (27, 1) and are heading straight to the lookout tower at (2, 21). Are they hiking on parallel or nonparallel courses?

Solution



To respond, we compute the slope of each trek across the park. For BJ and Kapi: For Dave and Becky: y2  y1 x2  x1 10  1182  11  1242 28 4   35 5

m

y2  y1 x2  x1 21  1  2  1272 20 4   25 5

m

Since the slopes are equal, the couples are hiking on parallel courses.

Two lines in the same plane that intersect at right angles are called perpendicular lines. Using the coordinate grid, we note that intersect at right angles suggests that their rise 4 slopes are negative reciprocals. From Figure 2.19, the ratio rise run for L1 is 3 , the ratio run 3 for L2 is 4 . Alternatively, we can say their slopes have a product of 1, since m1 # m2  1 implies m1  m12. Figure 2.19

Generic plane

y

L1

5

L1

Run Rise

Rise Run 5

5

L2

WORTHY OF NOTE Since m1 # m2  1 implies m1  m12, we can easily find the slope of a line perpendicular to a second line whose slope is given—just find the reciprocal and make it 3 negative. For m1  7 7 m2  3, and for m1  5, m2  15.

x

L2 5

Coordinate plane

Perpendicular Lines Given L1 and L2 are distinct, nonvertical lines with slopes of m1 and m2, respectively. 1. If m1 # m2  1, then L1 is perpendicular to L2. 2. If L1 is perpendicular to L2, then m1 # m2  1. In symbols we write L1  L2. Any vertical line (undefined slope) is perpendicular to any horizontal line (slope m  0).

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EXAMPLE 7B



Determining Whether Two Lines Are Perpendicular

Solution



For a right triangle to be formed, two of the lines through these points must be perpendicular (forming a right angle). From Figure 2.20, it appears a right triangle is formed, but we must verify that two of the sides are perpendicular. Using the slope formula, we have:

The three points P1  15, 12, P2  13, 22 , and P3  13, 22 form the vertices of a triangle. Use these points to draw the triangle, then use the slope formula to determine if they form a right triangle.

For P1 and P2 2  1 35 3 3   2 2

m1 

Figure 2.20 y 5

P1

P3

For P1 and P3

5

x

5

P2

21 3  5 1  8

m2 

5

For P2 and P3

2  122 3  3 2 4   6 3

m3  D. You’ve just learned how to identify parallel and perpendicular lines

Since m1 # m3  1, the triangle has a right angle and must be a right triangle.

Now try Exercises 57 through 68



E. Applications of Linear Equations The graph of a linear equation can be used to help solve many applied problems. If the numbers you’re working with are either very small or very large, scale the axes appropriately. This can be done by letting each tic mark represent a smaller or larger unit so the data points given will fit on the grid. Also, many applications use only nonnegative values and although points with negative coordinates may be used to graph a line, only ordered pairs in QI can be meaningfully interpreted. EXAMPLE 8



Applying a Linear Equation Model-Commission Sales Use the information given to create a linear equation model in two variables, then graph the line and use the graph to answer the question: A salesperson gets a daily $20 meal allowance plus $7.50 for every item she sells. How many sales are needed for a daily income of $125?



Let x represent sales and y represent income. This gives verbal model: Daily income (y) equals $7.5 per sale 1x2  $20 for meals equation model: y  7.5x  20 Using x  0 and x  10, we find (0, 20) and (10, 95) are points on this graph. From the graph, we estimate that 14 sales are needed to generate a daily income of $125.00. Substituting x  14 into the equation verifies that (14, 125) is indeed on the graph:

y y  7.5x  20

150

Income

Solution

(10, 95)

100 50

(0, 20) 0

2

4

6

8 10 12 14 16

Sales

x

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105

y  7.5x  20  7.51142  20  105  20  125 ✓

E. You’ve just learned how to apply linear equations in context

Now try Exercises 71 through 74



TECHNOLOGY HIGHLIGHT

Linear Equations, Window Size, and Friendly Windows To graph linear equations on the TI-84 Plus, we (1) solve the equation for the variable y, (2) enter the equation on the Y = screen, and (3) GRAPH the equation and adjust the WINDOW if necessary. 1. Solve the equation for y. For the equation 2x  3y  3, we have 2x  3y  3

Figure 2.21

given equation

3y  2x  3 2 y x1 3

subtract 2x from each side divide both sides by 3

2. Enter the equation on the Y = screen. On the Y = screen, enter 23 x  1. Note that for some calculators parentheses are needed to group 12  32x, to prevent the Figure 2.22 calculator from interpreting this term as 2  13x2. 10 3. GRAPH the equation, adjust the WINDOW . Since much of our work is centered at (0, 0) on the coordinate grid, the calculator’s default settings have a domain of x  310, 104 and a range of y  310, 104, as shown in 10 10 Figure 2.21. This is referred to as the WINDOW size. To graph the line in this window, it is easiest to use the ZOOM key and select 6:ZStandard, which resets the window to these default 10 settings. The graph is shown in Figure 2.22. The Xscl and Yscl entries give the scale used on each axis, indicating that each “tic mark” represents 1 unit. Graphing calculators have many features that enable us to find ordered pairs on a line. One is the ( 2nd GRAPH ) (TABLE) feature we have seen previously. We can also use the calculator’s TRACE feature. As the name implies, this feature enables us to trace along the line by moving a blinking cursor using the left and right arrow keys. The calculator simultaneously displays the coordinates of the current location of the cursor. After pressing the TRACE button, the cursor appears automatically— usually at the y-intercept. Moving the cursor left and right, note the coordinates changing at the bottom of the screen. The point (3.4042553, 3.2695035) is on the line and satisfies the equation of the line. The calculator is displaying decimal values because the screen is exactly 95 pixels wide, 47 pixels to the left of the y-axis, and 47 pixels to the right. This means that each time you press the left or right arrow, the x-value changes by 1/47—which is not a nice round number. To TRACE through “friendlier” values, we can use the

ZOOM

4:ZDecimal feature, which sets Xmin  4.7

and Xmax  4.7, or 8:Zinteger, which sets Xmin  47 and Xmax  47. Press ZOOM 4:ZDecimal and the calculator will automatically regraph the line. Now when you TRACE the line, “friendly” decimal values are displayed. Exercise 1: Use the Y1  23 x  1.

ZOOM

4:ZDecimal and TRACE features to identify the x- and y-intercepts for

Exercise 2: Use the ZOOM 8:Zinteger and TRACE features to graph the line 79x  55y  869, then identify the x- and y-intercepts.

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2.2 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.



1. To find the x-intercept of a line, substitute for y and solve for x. To find the y-intercept, substitute for x and solve for y.

4. The slope of a horizontal line is , the slope of a vertical line is , and the slopes of two parallel lines are .

2. The slope formula is m   , and indicates a rate of change between the x- and y-variables.

5. Discuss/Explain If m1  2.1 and m2  2.01, will the lines intersect? If m1  23 and m2   23 , are the lines perpendicular?

3. If m 6 0, the slope of the line is line slopes from left to right.

6. Discuss/Explain the relationship between the slope formula, the Pythagorean theorem, and the distance formula. Include several illustrations.

and the

DEVELOPING YOUR SKILLS

Create a table of values for each equation and sketch the graph.

7. 2x  3y  6 x

9. y  x

8. 3x  5y  10 x

y

3 x4 2 y

10. y 

y

5 x3 3 x

y

11. If you completed Exercise 9, verify that (3, 0.5) and (12, 19 4 ) also satisfy the equation given. Do these points appear to be on the graph you sketched? 12. If you completed Exercise 10, verify that 37 (1.5, 5.5) and 1 11 2 , 6 2 also satisfy the equation given. Do these points appear to be on the graph you sketched?

Graph the following equations using the intercept method. Plot a third point as a check.

13. 3x  y  6

14. 2x  y  12

15. 5y  x  5

16. 4y  x  8

17. 5x  2y  6

18. 3y  4x  9

19. 2x  5y  4

20. 6x  4y  8

21. 2x  3y  12 1 23. y   x 2 25. y  25  50x 2 27. y   x  2 5 29. 2y  3x  0

22. 3x  2y  6 2 24. y  x 3 26. y  30  60x 3 28. y  x  2 4 30. y  3x  0

31. 3y  4x  12

32. 2x  5y  8

Compute the slope of the line through the given points, ¢y then graph the line and use m  ¢x to find two additional points on the line. Answers may vary.

33. (3, 5), (4, 6)

34. (2, 3), (5, 8)

35. (10, 3), (4, 5)

36. (3, 1), (0, 7)

37. (1, 8), (3, 7)

38. (5, 5), (0, 5)

39. (3, 6), (4, 2)

40. (2, 4), (3, 1)

41. The graph shown models the relationship between the cost of a new home and the size of the home in square feet. (a) Determine the slope of the line and

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interpret what the slope ratio means in this context and (b) estimate the cost of a 3000 ft2 home. Exercise 41

Exercise 42 1200 960

Volume (m3)

Cost ($1000s)

500

250

720 480 240

0

1

2

3

4

5

0

ft2 (1000s)

50

100

Trucks

42. The graph shown models the relationship between the volume of garbage that is dumped in a landfill and the number of commercial garbage trucks that enter the site. (a) Determine the slope of the line and interpret what the slope ratio means in this context and (b) estimate the number of trucks entering the site daily if 1000 m3 of garbage is dumped per day. 43. The graph shown models the relationship between the distance of an aircraft carrier from its home port and the number of hours since departure. (a) Determine the slope of the line and interpret what the slope ratio means in this context and (b) estimate the distance from port after 8.25 hours. Exercise 43

Exercise 44

150

0

10

Hours

20

250

0

47. Sewer line slope: Fascinated at how quickly the plumber was working, Ryan watched with great interest as the new sewer line was laid from the house to the main line, a distance of 48 ft. At the edge of the house, the sewer line was six in. under ground. If the plumber tied in to the main line at a depth of 18 in., what is the slope of the (sewer) line? What does this slope indicate? 48. Slope (pitch) of a roof: A contractor goes to a lumber yard to purchase some trusses (the triangular frames) for the roof of a house. Many sizes are available, so the contractor takes some measurements to ensure the roof will have the desired slope. In one case, the height of the truss (base to ridge) was 4 ft, with a width of 24 ft (eave to eave). Find the slope of the roof if these trusses are used. What does this slope indicate? Graph each line using two or three ordered pairs that satisfy the equation.

500

Circuit boards

Distance (mi)

300

46. Rate of climb: Shortly after takeoff, a plane increases altitude at a constant (linear) rate. In 5 min the altitude is 10,000 feet. Fifteen minutes after takeoff, the plane has reached its cruising altitude of 32,000 ft. (a) Find the slope of the line and discuss its meaning in this context and (b) determine how long it takes the plane to climb from 12,200 feet to 25,400 feet.

5

10

Hours

44. The graph shown models the relationship between the number of circuit boards that have been assembled at a factory and the number of hours since starting time. (a) Determine the slope of the line and interpret what the slope ratio means in this context and (b) estimate how many hours the factory has been running if 225 circuit boards have been assembled. 45. Height and weight: While there are many exceptions, numerous studies have shown a close relationship between an average height and average weight. Suppose a person 70 in. tall weighs 165 lb, while a person 64 in. tall weighs 142 lb. Assuming the relationship is linear, (a) find the slope of the line and discuss its meaning in this context and (b) determine how many pounds are added for each inch of height.

49. x  3

50. y  4

51. x  2

52. y  2

Write the equation for each line L1 and L2 shown. Specifically state their point of intersection. y

53.

L1

54.

L1

L2

4 2 4

2

2 2 4

4

x

4

2

y 5 4 3 2 1 1 2 3 4 5

L2 2

4

x

55. The table given shows the total number of justices j sitting on the Supreme Court of the United States for selected time periods t (in decades), along with the number of nonmale, nonwhite justices n for the same years. (a) Use the data to graph the linear relationship between t and j, then determine the slope of the line and discuss its meaning in this context. (b) Use the data to graph the linear relationship between t and n, then determine the slope of the line and discuss its meaning.

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Exercise 55 Time t (1960 S 0)

Justices j

Nonwhite, nonmale n

0

9

0

10

9

1

20

9

2

30

9

3

40

9

4

50

9

5 (est)

56. The table shown gives the boiling temperature t of water as related to the altitude h. Use the data to graph the linear relationship between h and t, then determine the slope of the line and discuss its meaning in this context. Exercise 56 Altitude h (ft)



Boiling Temperature t (F)

Two points on L1 and two points on L2 are given. Use the slope formula to determine if lines L1 and L2 are parallel, perpendicular, or neither.

57. L1: (2, 0) and (0, 6) L2: (1, 8) and (0, 5)

58. L1: (1, 10) and (1, 7) L2: (0, 3) and (1, 5)

59. L1: (3, 4) and (0, 1) 60. L1: (6, 2) and (8, 2) L2: (5, 1) and (3, 0) L2: (0, 0) and (4, 4) 61. L1: (6, 3) and (8, 7) L2: (7, 2) and (6, 0)

62. L1: (5, 1) and (4, 4) L2: (4, 7) and (8, 10)

In Exercises 63 to 68, three points that form the vertices of a triangle are given. Use the points to draw the triangle, then use the slope formula to determine if any of the triangles are right triangles. Also see Exercises 39–44 in Section 2.1.

63. (5, 2), (0, 3), (4, 4) 64. (7, 0), (1, 0), (7, 4)

0

212.0

65. (4, 3), (7, 1), (3, 2)

1000

210.2

2000

208.4

66. (3, 7), (2, 2), (5, 5)

3000

206.6

67. (3, 2), (1, 5), (6, 4)

4000

204.8

68. (0, 0), (5, 2), (2, 5)

5000

203.0

6000

201.2

WORKING WITH FORMULAS

69. Human life expectancy: L  0.11T  74.2 The average number of years that human beings live has been steadily increasing over the years due to better living conditions and improved medical care. This relationship is modeled by the formula shown, where L is the average life expectancy and T is number of years since 1980. (a) What was the life expectancy in the year 2000? (b) In what year will average life expectancy reach 77.5 yr?



2-26

CHAPTER 2 Relations, Functions, and Graphs

7 b(5000)T 100 If $5000 dollars is invested in an account paying 7% simple interest, the amount of interest earned is given by the formula shown, where I is the interest and T is the time in years. (a) How much interest is earned in 5 yr? (b) How much is earned in 10 yr? (c) Use the two points (5 yr, interest) and (10 yr, interest) to calculate the slope of this line. What do you notice?

70. Interest earnings: I  a

APPLICATIONS

For exercises 71 to 74, use the information given to build a linear equation model, then use the equation to respond.

71. Business depreciation: A business purchases a copier for $8500 and anticipates it will depreciate in value $1250 per year. a. What is the copier’s value after 4 yr of use? b. How many years will it take for this copier’s value to decrease to $2250?

72. Baseball card value: After purchasing an autographed baseball card for $85, its value increases by $1.50 per year. a. What is the card’s value 7 yr after purchase? b. How many years will it take for this card’s value to reach $100? 73. Water level: During a long drought, the water level in a local lake decreased at a rate of 3 in. per month. The water level before the drought was 300 in.

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a. What was the water level after 9 months of drought? b. How many months will it take for the water level to decrease to 20 ft? 74. Gas mileage: When empty, a large dump-truck gets about 15 mi per gallon. It is estimated that for each 3 tons of cargo it hauls, gas mileage decreases by 34 mi per gallon. a. If 10 tons of cargo is being carried, what is the truck’s mileage? b. If the truck’s mileage is down to 10 mi per gallon, how much weight is it carrying? 75. Parallel/nonparallel roads: Aberville is 38 mi north and 12 mi west of Boschertown, with a straight road “farm and machinery road” (FM 1960) connecting the two cities. In the next county, Crownsburg is 30 mi north and 9.5 mi west of Dower, and these cities are likewise connected by a straight road (FM 830). If the two roads continued indefinitely in both directions, would they intersect at some point? 76. Perpendicular/nonperpendicular course headings: Two shrimp trawlers depart Charleston Harbor at the same time. One heads for the shrimping grounds located 12 mi north and 3 mi east of the harbor. The other heads for a point 2 mi south and 8 mi east of the harbor. Assuming the harbor is at (0, 0), are the routes of the trawlers perpendicular? If so, how far apart are the boats when they reach their destinations (to the nearest one-tenth mi)? 77. Cost of college: For the years 1980 to 2000, the cost of tuition and fees per semester (in constant dollars) at a public 4-yr college can be approximated by the equation y  144x  621, where y represents the cost in dollars and x  0 

109

represents the year 1980. Use the equation to find: (a) the cost of tuition and fees in 2002 and (b) the year this cost will exceed $5250. Source: 2001 New York Times Almanac, p. 356

78. Female physicians: In 1960 only about 7% of physicians were female. Soon after, this percentage began to grow dramatically. For the years 1980 to 2002, the percentage of physicians that were female can be approximated by the equation y  0.72x  11, where y represents the percentage (as a whole number) and x  0 represents the year 1980. Use the equation to find: (a) the percentage of physicians that were female in 1992 and (b) the projected year this percentage will exceed 30%. Source: Data from the 2004 Statistical Abstract of the United States, Table 149

79. Decrease in smokers: For the years 1980 to 2002, the percentage of the U.S. adult population who were smokers can be approximated by the equation 7 x  32, where y represents the percentage y  15 of smokers (as a whole number) and x  0 represents 1980. Use the equation to find: (a) the percentage of adults who smoked in the year 2000 and (b) the year the percentage of smokers is projected to fall below 20%. Source: Statistical Abstract of the United States, various years

80. Temperature and cricket chirps: Biologists have found a strong relationship between temperature and the number of times a cricket chirps. This is modeled by the equation T  N4  40, where N is the number of times the cricket chirps per minute and T is the temperature in Fahrenheit. Use the equation to find: (a) the outdoor temperature if the cricket is chirping 48 times per minute and (b) the number of times a cricket chirps if the temperature is 70°.

EXTENDING THE CONCEPT

81. If the lines 4y  2x  5 and 3y  ax  2 are perpendicular, what is the value of a? 82. Let m1, m2, m3, and m4 be the slopes of lines L1, L2, L3, and L4, respectively. Which of the following statements is true? a. m4 6 m1 6 m3 6 m2 y L2 L1 b. m3 6 m2 6 m4 6 m1 L3 c. m3 6 m4 6 m2 6 m1 L4 x d. m1 6 m3 6 m4 6 m2 e. m1 6 m4 6 m3 6 m2 83. An arithmetic sequence is a sequence of numbers where each successive term is found by adding a

fixed constant, called the common difference d, to the preceding term. For instance 3, 7, 11, 15, . . . is an arithmetic sequence with d  4. The formula for the “nth term” tn of an arithmetic sequence is a linear equation of the form tn  t1  1n  12d , where d is the common difference and t1 is the first term of the sequence. Use the equation to find the term specified for each sequence. a. 2, 9, 16, 23, 30, . . . ; 21st term b. 7, 4, 1, 2, 5, . . . ; 31st term c. 5.10, 5.25, 5.40, 5.55, . . . ; 27th term 9 d. 32, 94, 3, 15 4 , 2 , . . . ; 17th term

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MAINTAINING YOUR SKILLS

84. (1.1) Simplify the equation, then solve. Check your answer by substitution: 3x2  3  4x  6  4x2  31x  52

86. (1.1) How many gallons of a 35% brine solution must be mixed with 12 gal of a 55% brine solution in order to get a 45% solution?

85. (R.7) Identify the following formulas:

87. (1.1) Two boats leave the harbor at Lahaina, Maui, going in opposite directions. One travels at 15 mph and the other at 20 mph. How long until they are 70 mi apart?

P  2L  2W V  r2h

V  LWH C  2r

2.3 Linear Graphs and Rates of Change Learning Objectives

The concept of slope is an important part of mathematics, because it gives us a way to measure and compare change. The value of an automobile changes with time, the circumference of a circle increases as the radius increases, and the tension in a spring grows the more it is stretched. The real world is filled with examples of how one change affects another, and slope helps us understand how these changes are related.

In Section 2.3 you will learn how to:

A. Write a linear equation in slope-intercept form

B. Use slope-intercept form to graph linear equations

A. Linear Equations and Slope-Intercept Form

C. Write a linear equation in point-slope form

D. Apply the slope-intercept form and point-slope form in context

EXAMPLE 1



In Section 1.1, formulas and literal equations were written in an alternate form by solving for an object variable. The new form made using the formula more efficient. Solving for y in equations of the form ax  by  c offers similar advantages to linear graphs and their applications. Solving for y in ax  by  c Solve 2y  6x  4 for y, then evaluate at x  4, x  0, and x  13.

Solution



2y  6x  4 2y  6x  4 y  3x  2

given equation add 6x divide by 2

Since the coefficients are integers, evaluate the function mentally. Inputs are multiplied by 3, then increased by 2, yielding the ordered pairs (4, 14), (0, 2), and 113, 12 . Now try Exercises 7 through 12



This form of the equation (where y has been written in terms of x) enables us to quickly identify what operations are performed on x in order to obtain y. For y  3x  2, multiply inputs by 3, then add 2. EXAMPLE 2



Solving for y in ax  by  c Solve the linear equation 3y  2x  6 for y, then identify the new coefficient of x and the constant term.

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Section 2.3 Linear Graphs and Rates of Change

Solution



given equation add 2x divide by 3

The new coefficient of x is 23 and the constant term is 2.

WORTHY OF NOTE In Example 2, the final form can be written y  23x  2 as shown (inputs are multiplied by two-thirds, then increased by 2), or written 2x as y   2 (inputs are 3 multiplied by two, the result divided by 3 and this amount increased by 2). The two forms are equivalent.

EXAMPLE 3

3y  2x  6 3y  2x  6 2 y x2 3



Now try Exercises 13 through 18

When the coefficient of x is rational, it’s helpful to select inputs that are multiples of the denominator if the context or application requires us to evaluate the equation. This enables us to perform most operations mentally. For y  23x  2, possible inputs might be x  9, 6, 0, 3, 6, and so on. See Exercises 19 through 24. In Section 2.2, linear equations were graphed using the intercept method. When a linear equation is written with y in terms of x, we notice a powerful connection between the graph and its equation, and one that highlights the primary characteristics of a linear graph. 

Noting Relationships between an Equation and Its Graph Find the intercepts of 4x  5y  20 and use them to graph the line. Then, a. Use the intercepts to calculate the slope of the line, then b. Write the equation with y in terms of x and compare the calculated slope and y-intercept to the equation in this form. Comment on what you notice.

Solution



A. You’ve just learned how to write a linear equation in slope-intercept form

Substituting 0 for x in 4x  5y  20, we find the y-intercept is 10, 42. Substituting 0 for y gives an x-intercept of 15, 02 . The graph is displayed here. ¢y , the slope is a. By calculation or counting ¢x 4 m  5. b. Solving for y: 4x  5y  20 5y  4x  20 4 y x4 5

y 5 4 3 2

(5, 0)

1

5 4 3 2 1 1

4

subtract 4x

2

3

4

5

x

2 3

given equation

1

(0, 4)

5

divide by 5

The slope value seems to be the coefficient of x, while the y-intercept is the constant term. Now try Exercises 25 through 30



B. Slope-Intercept Form and the Graph of a Line After solving a linear equation for y, an input of x  0 causes the “x-term” to become zero, so the y-intercept is automatically the constant term. As Example 3 illustrates, we can also identify the slope of the line—it is the coefficient of x. In general, a linear equation of the form y  mx  b is said to be in slope-intercept form, since the slope of the line is m and the y-intercept is (0, b). Slope-Intercept Form For a nonvertical line whose equation is y  mx  b, the slope of the line is m and the y-intercept is (0, b).

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EXAMPLE 4



Finding the Slope-Intercept Form Write each equation in slope-intercept form and identify the slope and y-intercept of each line. a. 3x  2y  9 b. y  x  5 c. 2y  x

Solution



a. 3x  2y  9

b. y  x  5

2y  3x  9 3 9 y x 2 2 3 9 m ,b 2 2 9 y-intercept a0,  b 2

y  x  5 y  1x  5 m  1, b  5

c. 2y  x x y 2 1 y x 2 1 m ,b0 2

y-intercept (0, 5)

y-intercept (0, 0)

Now try Exercises 31 through 38



If the slope and y-intercept of a linear equation are known or can be found, we can construct its equation by substituting these values directly into the slope-intercept form y  mx  b. EXAMPLE 5



y

Finding the Equation of a Line from Its Graph

5

Find the slope-intercept form of the line shown.

Solution



Using 13, 22 and 11, 22 in the slope formula, ¢y or by simply counting , the slope is m  42 or 21. ¢x By inspection we see the y-intercept is (0, 4). Substituting 21 for m and 4 for b in the slopeintercept form we obtain the equation y  2x  4.

5

5

x

5

Now try Exercises 39 through 44



Actually, if the slope is known and we have any point (x, y) on the line, we can still construct the equation since the given point must satisfy the equation of the line. In this case, we’re treating y  mx  b as a simple formula, solving for b after substituting known values for m, x, and y. EXAMPLE 6



Using y  mx  b as a Formula

Solution



Using y  mx  b as a “formula,” we have m  45, x  5, and y  2.

Find the equation of a line that has slope m  45 and contains 15, 22. y  mx  b 2  45 152  b 2  4  b 6b

slope-intercept form substitute 45 for m, 5 for x, and 2 for y simplify solve for b

The equation of the line is y  45x  6. Now try Exercises 45 through 50



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Writing a linear equation in slope-intercept form enables us to draw its graph with a minimum of effort, since we can easily locate the y-intercept and a second point using ¢y ¢y 2 m . For instance,  means count down 2 and right 3 from a known point. ¢x ¢x 3 EXAMPLE 7



Graphing a Line Using Slope-Intercept Form Write 3y  5x  9 in slope-intercept form, then graph the line using the y-intercept and slope.

Solution



3y  5x  9 3y  5x  9 y  53x  3

y  fx  3 y

given equation

y f x

Rise 5

divide by 3

The slope is m  and the y-intercept is (0, 3). ¢y 5  (up 5 and Plot the y-intercept, then use ¢x 3 right 3—shown in blue) to find another point on the line (shown in red). Finish by drawing a line through these points.

5 3

Noting the fraction is equal to 5 3 , we could also begin at ¢y 5 (0, 3) and count  ¢x 3 (down 5 and left 3) to find an additional point on the line: (3, 2). Also, for any ¢y a negative slope  , ¢x b a a a note    . b b b

(3, 8)

isolate y term

5 3

WORTHY OF NOTE

Run 3

(0, 3)

5

5

x

2

Now try Exercises 51 through 62



For a discussion of what graphing method might be most efficient for a given linear equation, see Exercises 103 and 115.

Parallel and Perpendicular Lines From Section 2.2 we know parallel lines have equal slopes: m1  m2, and perpendicular 1 lines have slopes with a product of 1: m1 # m2  1 or m1   . In some applim2 cations, we need to find the equation of a second line parallel or perpendicular to a given line, through a given point. Using the slope-intercept form makes this a simple four-step process. Finding the Equation of a Line Parallel or Perpendicular to a Given Line 1. Identify the slope m1 of the given line. 2. Find the slope m2 of the new line using the parallel or perpendicular relationship. 3. Use m2 with the point (x, y) in the “formula” y  mx  b and solve for b. 4. The desired equation will be y  m2x  b.

EXAMPLE 8



Finding the Equation of a Parallel Line

Solution



Begin by writing the equation in slope-intercept form to identify the slope.

Find the equation of a line that goes through 16, 12 and is parallel to 2x  3y  6. 2x  3y  6 3y  2x  6 y  2 3 x  2

given line isolate y term result

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The original line has slope m1  2 3 and this will also be the slope of any line parallel to it. Using m2  2 with 1x, y2 S 16, 12 we have 3 y  mx  b 2 1  162  b 3 1  4  b 5  b

The equation of the new line is y 

slope-intercept form substitute 2 3 for m, 6 for x, and 1 for y simplify solve for b 2 3 x

 5. Now try Exercises 63 through 76



GRAPHICAL SUPPORT Graphing the lines from Example 8 as Y1 and Y2 on a graphing calculator, we note the lines do appear to be parallel (they actually must be since they have identical slopes). Using the ZOOM 8:ZInteger feature of the TI-84 Plus we can quickly verify that Y2 indeed contains the point (6, 1).

31

47

47

31

For any nonlinear graph, a straight line drawn through two points on the graph is called a secant line. The slope of the secant line, and lines parallel and perpendicular to this line, play fundamental roles in the further development of the rate-of-change concept. EXAMPLE 9



Finding Equations for Parallel and Perpendicular Lines A secant line is drawn using the points (4, 0) and (2, 2) on the graph of the function shown. Find the equation of a line that is: a. parallel to the secant line through (1, 4) b. perpendicular to the secant line through (1, 4).

Solution



Either by using the slope formula or counting m

WORTHY OF NOTE The word “secant” comes from the Latin word secare, meaning “to cut.” Hence a secant line is one that cuts through a graph, as opposed to a tangent line, which touches the graph at only one point.

¢y , we find the secant line has slope ¢x

1 2  . 6 3

a. For the parallel line through (1, 4), m2  y  mx  b 1 4  112  b 3 1 12   b 3 3 13  b 3

1 . 3

y 5

slope-intercept form substitute 1 3 for m, 1 for x, and 4 for y

5

5

simplify

result

The equation of the parallel line (in blue) is y 

(1, 4)

1 13 x . 3 3

5

x

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b. For the line perpendicular through (1, 4), m2  3. y  mx  b 4  3112  b 4  3  b 1  b B. You’ve just learned how to use the slope-intercept form to graph linear equations

y 5

slope-intercept form substitute 3 for m, 1 for x, and 4 for y simplify

5

5

x

result

The equation of the perpendicular line (in yellow) is y  3x  1.

(1, 4)

5

Now try Exercises 77 through 82



C. Linear Equations in Point-Slope Form As an alternative to using y  mx  b, we can find the equation of the line using the y2  y1  m, and the fact that the slope of a line is constant. For a given slope formula x2  x1 slope m, we can let (x1, y1) represent a given point on the line and (x, y) represent any y  y1  m. Isolating the “y” terms other point on the line, and the formula becomes x  x1 on one side gives a new form for the equation of a line, called the point-slope form: y  y1 m x  x1 1x  x1 2 y  y1 a b  m1x  x1 2 x  x1 1 y  y1  m1x  x1 2

slope formula multiply both sides by 1x  x1 2 simplify S point-slope form

The Point-Slope Form of a Linear Equation For a nonvertical line whose equation is y  y1  m1x  x1 2 , the slope of the line is m and (x1, y1) is a point on the line. While using y  mx  b as in Example 6 may appear to be easier, both the y-intercept form and point-slope form have their own advantages and it will help to be familiar with both. EXAMPLE 10



Using y  y1  m1x  x1 2 as a Formula

Find the equation of a line in point-slope form, if m  23 and (3, 3) is on the line. Then graph the line. Solution



C. You’ve just learned how to write a linear equation in point-slope form

y  y1  m1x  x1 2 2 y  132  3 x  132 4 3 2 y  3  1x  32 3

y y  3  s (x  3)

point-slope form

5

substitute 23 for m; (3, 3) for (x1, y1) simplify, point-slope form

¢y 2  to To graph the line, plot (3, 3) and use ¢x 3 find additional points on the line.

x3

5

5

x

y2 (3, 3) 5

Now try Exercises 83 through 94



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D. Applications of Linear Equations As a mathematical tool, linear equations rank among the most common, powerful, and versatile. In all cases, it’s important to remember that slope represents a rate of change. ¢y The notation m  literally means the quantity measured along the y-axis, is chang¢x ing with respect to changes in the quantity measured along the x-axis. EXAMPLE 11



Relating Temperature to Altitude In meteorological studies, atmospheric temperature depends on the altitude according to the formula T  3.5h  58.6, where T represents the approximate Fahrenheit temperature at height h (in thousands of feet). a. Interpret the meaning of the slope in this context. b. Determine the temperature at an altitude of 12,000 ft. c. If the temperature is 10°F what is the approximate altitude?

Solution



3.5 ¢T  , ¢h 1 meaning the temperature drops 3.5°F for every 1000-ft increase in altitude. b. Since height is in thousands, use h  12.

a. Notice that h is the input variable and T is the output. This shows

T  3.5h  58.6  3.51122  58.6  16.6

original function substitute 12 for h result

At a height of 12,000 ft, the temperature is about 17°F. c. Replacing T with 10 and solving gives 10  3.5h  58.6 68.6  3.5h 19.6  h

substitute 10 for T simplify result

The temperature is 10°F at a height of 19.6  1000  19,600 ft. Now try Exercises 105 and 106



In some applications, the relationship is known to be linear but only a few points on the line are given. In this case, we can use two of the known data points to calculate the slope, then the point-slope form to find an equation model. One such application is linear depreciation, as when a government allows businesses to depreciate vehicles and equipment over time (the less a piece of equipment is worth, the less you pay in taxes). EXAMPLE 12A



Using Point-Slope Form to Find an Equation Model Five years after purchase, the auditor of a newspaper company estimates the value of their printing press is $60,000. Eight years after its purchase, the value of the press had depreciated to $42,000. Find a linear equation that models this depreciation and discuss the slope and y-intercept in context.

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Solution



Since the value of the press depends on time, the ordered pairs have the form (time, value) or (t, v) where time is the input, and value is the output. This means the ordered pairs are (5, 60,000) and (8, 42,000). v2  v1 t2  t1 42,000  60,000  85 6000 18,000   3 1

m

WORTHY OF NOTE Actually, it doesn’t matter which of the two points are used in Example 12A. Once the point (5, 60,000) is plotted, a constant slope of m  6000 will “drive” the line through (8, 42,000). If we first graph (8, 42,000), the same slope would “drive” the line through (5, 60,000). Convince yourself by reworking the problem using the other point.

117

slope formula 1t1, v1 2  15, 60,0002; 1t2, v2 2  18, 42,0002 simplify and reduce

6000 ¢value  , indicating the printing press loses ¢time 1 $6000 in value with each passing year. The slope of the line is

v  v1  m1t  t1 2 v  60,000  60001t  52 v  60,000  6000t  30,000 v  6000t  90,000

point-slope form substitute 6000 for m; (5, 60,000) for (t1, v1) simplify solve for v

The depreciation equation is v  6000t  90,000. The v-intercept (0, 90,000) indicates the original value (cost) of the equipment was $90,000.

Once the depreciation equation is found, it represents the (time, value) relationship for all future (and intermediate) ages of the press. In other words, we can now predict the value of the press for any given year. However, note that some equation models are valid for only a set period of time, and each model should be used with care. EXAMPLE 12B



Using an Equation Model to Gather Information From Example 12A, a. How much will the press be worth after 11 yr? b. How many years until the value of the equipment is less than $9,000? c. Is this equation model valid for t  18 yr (why or why not)?

Solution



a. Find the value v when t  11: v  6000t  90,000 v  60001112  90,000  24,000

equation model substitute 11 for t result (11, 24,000)

After 11 yr, the printing press will only be worth $24,000. b. “. . . value is less than $9000” means v 6 9000: v 6000t  90,000 6000t t D. You’ve just learned how to apply the slope-intercept form and point-slope form in context

6 6 6 7

9000 9000 81,000 13.5

value at time t substitute 6000t  90,000 for v subtract 90,000 divide by 6000, reverse inequality symbol

After 13.5 yr, the printing press will be worth less than $9000. c. Since substituting 18 for t gives a negative quantity, the equation model is not valid for t  18. In the current context, the model is only valid while v  0 and we note the domain of the function is t  30, 15 4 . Now try Exercises 107 through 112



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2.3 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

4. The equation y  y1  m1x  x1 2 is called the form of a line.

7 1. For the equation y   x  3, the slope 4 is and the y-intercept is . ¢cost 2. The notation indicates the ¢time changing in response to changes in

is .

3. Line 1 has a slope of 0.4. The slope of any line perpendicular to line 1 is . 

5. Discuss/Explain how to graph a line using only the slope and a point on the line (no equations).

6. Given m  35 and 15, 62 is on the line. Compare and contrast finding the equation of the line using y  mx  b versus y  y1  m1x  x1 2.

DEVELOPING YOUR SKILLS

Solve each equation for y and evaluate the result using x  5, x  2, x  0, x  1, and x  3.

7. 4x  5y  10

8. 3y  2x  9

9. 0.4x  0.2y  1.4 10. 0.2x  0.7y  2.1 11.

1 3x



1 5y

 1

12.

1 7y



1 3x

2

For each equation, solve for y and identify the new coefficient of x and new constant term.

13. 6x  3y  9

14. 9y  4x  18

15. 0.5x  0.3y  2.1 16. 0.7x  0.6y  2.4 17.

5 6x



1 7y



47

18.

7 12 y



4 15 x



Write each equation in slope-intercept form (solve for y), then identify the slope and y-intercept.

31. 2x  3y  6

32. 4y  3x  12

33. 5x  4y  20

34. y  2x  4

35. x  3y

36. 2x  5y

37. 3x  4y  12  0

38. 5y  3x  20  0

For Exercises 39 to 50, use the slope-intercept form to state the equation of each line.

39.

Evaluate each equation by selecting three inputs that will result in integer values. Then graph each line.

19. y  43x  5

20. y  54x  1

21. y  32x  2

22. y  25x  3

23. y  16x  4

24. y  13x  3

Find the x- and y-intercepts for each line, then (a) use these two points to calculate the slope of the line, (b) write the equation with y in terms of x (solve for y) and compare the calculated slope and y-intercept to the equation from part (b). Comment on what you notice.

25. 3x  4y  12

26. 3y  2x  6

27. 2x  5y  10

28. 2x  3y  9

29. 4x  5y  15

30. 5y  6x  25

40.

y 5 4 3 2 1

7 6

54321 1 2 (3, 1) 3 4 5

41.

(3, 3) (0, 1) 1 2 3 4 5 x

(5, 5)

y 5 4 (0, 3) 3 2 1

54321 1 2 3 4 5

(5, 1)

1 2 3 4 5 x

y

(1, 0)

5 4 3 (0, 3) 2 1

54321 1 2 (2, 3) 3 4 5

1 2 3 4 5 x

42. m  2; y-intercept 43. m  3; y-intercept 10, 32 10, 22 44. m  32; y-intercept 10, 42

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45.

46.

y 10,000

1600

6000

1200

4000

800

2000

400 14

16

18

Write the lines in slope-intercept form and state whether they are parallel, perpendicular, or neither.

y 2000

8000

12

47.

119

Section 2.3 Linear Graphs and Rates of Change

20 x

8

10

12

14

16 x

y 1500

71. 4y  5x  8 5y  4x  15

72. 3y  2x  6 2x  3y  3

73. 2x  5y  20 4x  3y  18

74. 5y  11x  135 11y  5x  77

75. 4x  6y  12 2x  3y  6

76. 3x  4y  12 6x  8y  2

1200

A secant line is one that intersects a graph at two or more points. For each graph given, find the equation of the line (a) parallel and (b) perpendicular to the secant line, through the point indicated.

900 600 300 26

28

30

32

34 x

48. m  4; 13, 22 is on the line

77.

78.

y 5

y 5

49. m  2; 15, 32 is on the line

(1, 3)

50. m  32; 14, 72 is on the line

5

Write each equation in slope-intercept form, then use the slope and intercept to graph the line.

51. 3x  5y  20 53. 2x  3y  15

52. 2y  x  4

79.

57. y 

1 3 x

2

58. y 

60. y  3x  4

61. y  12x  3

62. y  3 2 x  2

Find the equation of the line using the information given. Write answers in slope-intercept form.

63. parallel to 2x  5y  10, through the point 15, 22

64. parallel to 6x  9y  27, through the point 13, 52

65. perpendicular to 5y  3x  9, through the point 16, 32 66. perpendicular to x  4y  7, through the point 15, 32

67. parallel to 12x  5y  65, through the point 12, 12

68. parallel to 15y  8x  50, through the point 13, 42 69. parallel to y  3, through the point (2, 5)

70. perpendicular to y  3 through the point (2, 5)

y 5

(1, 3)

5

5

5 x

5

2

59. y  2x  5

80.

y

56. y  52x  1 4 5 x

5 x

5

5

54. 3x  2y  4

5

(2, 4)

5

Graph each linear equation using the y-intercept and slope determined from each equation.

55. y  23x  3

5 x

81.

5 x

5

82.

y 5

(1, 2.5)

y 5

(1, 3)

5

5 x

5

5 x

(0, 2) 5

5

Find the equation of the line in point-slope form, then graph the line.

83. m  2; P1  12, 52

84. m  1; P1  12, 32

85. P1  13, 42, P2  111, 12 86. P1  11, 62, P2  15, 12

87. m  0.5; P1  11.8, 3.12

88. m  1.5; P1  10.75, 0.1252

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Find the equation of the line in point-slope form, and state the meaning of the slope in context—what information is the slope giving us?

89.

90.

0

x

1 2 3 4 5 6 7 8 9

10 9 8 7 6 5 4 3 2 1

y E

0

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 x Hours of television per day

60 40 20

1

2

3

4

5

x

0 1 2 3 4 5 6 7 8 9 10 x Independent investors (1000s)

y F

x

x y G

x

x y H

x

x

97. At first I ran at a steady pace, then I got tired and walked the rest of the way. 98. While on my daily walk, I had to run for a while when I was chased by a stray dog.

8 6 4

99. I climbed up a tree, then I jumped out.

2 0

60

65

70

75

80

x

Temperature in °F

Using the concept of slope, match each description with the graph that best illustrates it. Assume time is scaled on the horizontal axes, and height, speed, or distance

100. I steadily swam laps at the pool yesterday. 101. I walked toward the candy machine, stared at it for a while then changed my mind and walked back. 102. For practice, the girls’ track team did a series of 25-m sprints, with a brief rest in between.

WORKING WITH FORMULAS

103. General linear equation: ax  by  c The general equation of a line is shown here, where a, b, and c are real numbers, with a and b not simultaneously zero. Solve the equation for y and note the slope (coefficient of x) and y-intercept (constant term). Use these to find the slope and y-intercept of the following lines, without solving for y or computing points. a. 3x  4y  8 b. 2x  5y  15 c. 5x  6y  12 d. 3y  5x  9



x

10

Rainfall per month (in inches)



y D

96. After hitting the ball, I began trotting around the bases shouting, “Ooh, ooh, ooh!” When I saw it wasn’t a home run, I began sprinting.

y Eggs per hen per week

Cattle raised per acre

80

0

10 9 8 7 6 5 4 3 2 1

94.

y 100

y C

95. While driving today, I got stopped by a state trooper. After she warned me to slow down, I continued on my way.

y Online brokerage houses

Student’s final grade (%) (includes extra credit)

100 90 80 70 60 50 40 30 20 10

93.

x

1 2 3 4 5 6 7 8 9

Year (1990 → 0)

92.

y

y B

x

Sales (in thousands)

91.

y A

y Typewriters in service (in ten thousands)

Income (in thousands)

y 10 9 8 7 6 5 4 3 2 1

from the origin (as the case may be) is scaled on the vertical axis.

104. Intercept/Intercept form of a linear y x equation:   1 h k The x- and y-intercepts of a line can also be found by writing the equation in the form shown (with the equation set equal to 1). The x-intercept will be (h, 0) and the y-intercept will be (0, k). Find the x- and y-intercepts of the following lines using this method: (a) 2x  5y  10, (b) 3x  4y  12, and (c) 5x  4y  8. How is the slope of each line related to the values of h and k?

APPLICATIONS

105. Speed of sound: The speed of sound as it travels through the air depends on the temperature of the air according to the function V  35C  331, where V represents the velocity of the sound waves in meters per second (m/s), at a temperature of C° Celsius.

a. Interpret the meaning of the slope and y-intercept in this context. b. Determine the speed of sound at a temperature of 20°C. c. If the speed of sound is measured at 361 m/s, what is the temperature of the air?

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106. Acceleration: A driver going down a straight highway is traveling 60 ft/sec (about 41 mph) on cruise control, when he begins accelerating at a rate of 5.2 ft/sec2. The final velocity of the car is given by V  26 5 t  60, where V is the velocity at time t. (a) Interpret the meaning of the slope and y-intercept in this context. (b) Determine the velocity of the car after 9.4 seconds. (c) If the car is traveling at 100 ft/sec, for how long did it accelerate? 107. Investing in coins: The purchase of a “collector’s item” is often made in hopes the item will increase in value. In 1998, Mark purchased a 1909-S VDB Lincoln Cent (in fair condition) for $150. By the year 2004, its value had grown to $190. (a) Use the relation (time since purchase, value) with t  0 corresponding to 1998 to find a linear equation modeling the value of the coin. (b) Discuss what the slope and y-intercept indicate in this context. (c) How much will the penny be worth in 2009? (d) How many years after purchase will the penny’s value exceed $250? (e) If the penny is now worth $170, how many years has Mark owned the penny? 108. Depreciation: Once a piece of equipment is put into service, its value begins to depreciate. A business purchases some computer equipment for $18,500. At the end of a 2-yr period, the value of the equipment has decreased to $11,500. (a) Use the relation (time since purchase, value) to find a linear equation modeling the value of the equipment. (b) Discuss what the slope and y-intercept indicate in this context. (c) What is the equipment’s value after 4 yr? (d) How many years after purchase will the value decrease to $6000? (e) Generally, companies will sell used equipment while it still has value and use the funds to purchase new equipment. According to the function, how many years will it take this equipment to depreciate in value to $1000? 109. Internet connections: The number of households that are hooked up to the Internet (homes that are online) has been increasing steadily in recent years. In 1995, approximately 9 million homes were online. By 2001 this figure had climbed to about 51 million. (a) Use the relation (year, homes online) with t  0 corresponding to 1995 to find an 

Section 2.3 Linear Graphs and Rates of Change

121

equation model for the number of homes online. (b) Discuss what the slope indicates in this context. (c) According to this model, in what year did the first homes begin to come online? (d) If the rate of change stays constant, how many households will be on the Internet in 2006? (e) How many years after 1995 will there be over 100 million households connected? (f) If there are 115 million households connected, what year is it? Source: 2004 Statistical Abstract of the United States, Table 965

110. Prescription drugs: Retail sales of prescription drugs have been increasing steadily in recent years. In 1995, retail sales hit $72 billion. By the year 2000, sales had grown to about $146 billion. (a) Use the relation (year, retail sales of prescription drugs) with t  0 corresponding to 1995 to find a linear equation modeling the growth of retail sales. (b) Discuss what the slope indicates in this context. (c) According to this model, in what year will sales reach $250 billion? (d) According to the model, what was the value of retail prescription drug sales in 2005? (e) How many years after 1995 will retail sales exceed $279 billion? (f) If yearly sales totaled $294 billion, what year is it? Source: 2004 Statistical Abstract of the United States, Table 122

111. Prison population: In 1990, the number of persons sentenced and serving time in state and federal institutions was approximately 740,000. By the year 2000, this figure had grown to nearly 1,320,000. (a) Find a linear equation with t  0 corresponding to 1990 that models this data, (b) discuss the slope ratio in context, and (c) use the equation to estimate the prison population in 2007 if this trend continues. Source: Bureau of Justice Statistics at www.ojp.usdoj.gov/bjs

112. Eating out: In 1990, Americans bought an average of 143 meals per year at restaurants. This phenomenon continued to grow in popularity and in the year 2000, the average reached 170 meals per year. (a) Find a linear equation with t  0 corresponding to 1990 that models this growth, (b) discuss the slope ratio in context, and (c) use the equation to estimate the average number of times an American will eat at a restaurant in 2006 if the trend continues. Source: The NPD Group, Inc., National Eating Trends, 2002

EXTENDING THE CONCEPT

113. Locate and read the following article. Then turn in a one-page summary. “Linear Function Saves Carpenter’s Time,” Richard Crouse, Mathematics Teacher, Volume 83, Number 5, May 1990: pp. 400–401.

114. The general form of a linear equation is ax  by  c, where a and b are not simultaneously zero. (a) Find the x- and y-intercepts using the general form (substitute 0 for x, then 0 for y). Based on what you see, when does the intercept method work most efficiently? (b) Find the slope

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and y-intercept using the general form (solve for y). Based on what you see, when does the intercept method work most efficiently?. 115. Match the correct graph to the conditions stated for m and b. There are more choices than graphs. a. m 6 0, b 6 0 b. m 7 0, b 6 0 c. m 6 0, b 7 0 d. m 7 0, b 7 0 e. m  0, b 7 0 f. m 6 0, b  0 g. m 7 0, b  0 h. m  0, b 6 0 

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CHAPTER 2 Relations, Functions, and Graphs

(1)

y

(2)

y

(3)

x

(4)

y

x

y

(5)

x

x

y

(6)

y

x

x

MAINTAINING YOUR SKILLS

116. (2.2) Determine the domain: a. y  12x  5 5 b. y  2 2x  3x  2

119. (R.7) Compute the area of the circular sidewalk shown here. Use your calculator’s value of  and round the answer (only) to hundredths. 10 yd

117. (1.5) Solve using the quadratic formula. Answer in exact and approximate form: 3x2  10x  9. 118. (1.1) Three equations follow. One is an identity, another is a contradiction, and a third has a solution. State which is which.

8 yd

21x  52  13  1  9  7  2x

21x  42  13  1  9  7  2x 21x  52  13  1  9  7  2x

2.4 Functions, Function Notation, and the Graph of a Function Learning Objectives In Section 2.4 you will learn how to:

A. Distinguish the graph of a function from that of a relation

B. Determine the domain and range of a function

C. Use function notation and evaluate functions

D. Apply the rate-of-change concept to nonlinear functions

In this section we introduce one of the most central ideas in mathematics—the concept of a function. Functions can model the cause-and-effect relationship that is so important to using mathematics as a decision-making tool. In addition, the study will help to unify and expand on many ideas that are already familiar.

A. Functions and Relations There is a special type of relation that merits further attention. A function is a relation where each element of the domain corresponds to exactly one element of the range. In other words, for each first coordinate or input value, there is only one possible second coordinate or output. Functions A function is a relation that pairs each element from the domain with exactly one element from the range.

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Section 2.4 Functions, Function Notation, and the Graph of a Function

If the relation is defined by a mapping, we need only check that each element of the domain is mapped to exactly one element of the range. This is indeed the case for the mapping P S B from Figure 2.1 (page 152), where we saw that each person corresponded to only one birthday, and that it was impossible for one person to be born on two different days. For the relation x  y  shown in Figure 2.6 (page 153), each element of the domain except zero is paired with more than one element of the range. The relation x  y  is not a function. EXAMPLE 1



Determining Whether a Relation Is a Function Three different relations are given in mapping notation below. Determine whether each relation is a function. a. b. c.

Solution



Person

Room

Pet

Weight (lbs)

War

Year

Marie Pesky Bo Johnny Rick Annie Reece

270 268 274 276 272 282

Fido

450 550 2 40 8 3

Civil War

1963

Bossy Silver Frisky Polly

World War I

1950

World War II

1939

Korean War

1917

Vietnam War

1861

Relation (a) is a function, since each person corresponds to exactly one room. This relation pairs math professors with their respective office numbers. Notice that while two people can be in one office, it is impossible for one person to physically be in two different offices. Relation (b) is not a function, since we cannot tell whether Polly the Parrot weighs 2 lb or 3 lb (one element of the domain is mapped to two elements of the range). Relation (c) is a function, where each major war is paired with the year it began. Now try Exercises 7 through 10



If the relation is defined by a set of ordered pairs or a set of individual and distinct plotted points, we need only check that no two points have the same first coordinate with a different second coordinate. EXAMPLE 2



Identifying Functions Two relations named f and g are given; f is stated as a set of ordered pairs, while g is given as a set of plotted points. Determine whether each is a function. f: 13, 02, 11, 42, 12, 52, 14, 22, 13, 22, 13, 62, 10, 12, (4, 5), and (6, 1)

Solution

WORTHY OF NOTE The definition of a function can also be stated in ordered pair form: A function is a set of ordered pairs (x, y), in which each first component is paired with only one second component.



The relation f is not a function, since 3 is paired with two different outputs: (3, 02 and (3, 22 . The relation g shown in the figure is a function. Each input corresponds to exactly one output, otherwise one point would be directly above the other and have the same first coordinate.

g

5

y (0, 5)

(4, 2) (3, 1)

(2, 1) 5

5

x

(4, 1) (1, 3) 5

Now try Exercises 11 through 18



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The graphs of y  x  1 and x  y from Section 2.1 offer additional insight into the definition of a function. Figure 2.23 shows the line y  x  1 with emphasis on the plotted points (4, 3) and 13, 42. The vertical movement shown from the x-axis to a point on the graph illustrates the pairing of a given x-value with one related y-value. Note the vertical line shows only one related y-value ( x  4 is paired with only y  3). Figure 2.24 gives the graph of x  y, highlighting the points (4, 4) and (4, 4). The vertical movement shown here branches in two directions, associating one x-value with more than one y-value. This shows the relation y  x  1 is also a function, while the relation x  y is not. Figure 2.24

Figure 2.23 y yx1

5

y

x  y (4, 4)

5

(4, 3) (2, 2) (0, 0) 5

5

5

x

5

x

(2, 2) (3, 4)

(4, 4)

5

5

This “vertical connection” of a location on the x-axis to a point on the graph can be generalized into a vertical line test for functions. Vertical Line Test A given graph is the graph of a function, if and only if every vertical line intersects the graph in at most one point. Applying the test to the graph in Figure 2.23 helps to illustrate that the graph of any nonvertical line is a function. EXAMPLE 3



Using the Vertical Line Test Use the vertical line test to determine if any of the relations shown (from Section 2.1) are functions.

Solution



Visualize a vertical line on each coordinate grid (shown in solid blue), then mentally shift the line to the left and right as shown in Figures 2.25, 2.26, and 2.27 (dashed lines). In Figures 2.25 and 2.26, every vertical line intersects the graph only once, indicating both y  x2  2x and y  29  x2 are functions. In Figure 2.27, a vertical line intersects the graph twice for any x 7 0. The relation x  y2 is not a function. Figure 2.25

Figure 2.26

y (4, 8)

(2, 8) y

x2

 2x

5

Figure 2.27 y

y y  9  x2 (0, 3)

5

(4, 2) (2, 2)

5

(1, 3)

(3, 0)

(3, 3)

(0, 0)

(0, 0)

(3, 0)

5

5

x

5

5

(2, 0)

5

5 2

y2  x

(1, 1)

x 5

5

x

(2, 2) (4, 2)

Now try Exercises 19 through 30



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Section 2.4 Functions, Function Notation, and the Graph of a Function

EXAMPLE 4



Using the Vertical Line Test Use a table of values to graph the relations defined by a. y  x  b. y  1x, then use the vertical line test to determine whether each relation is a function.

Solution



WORTHY OF NOTE For relations and functions, a good way to view the distinction is to consider a mail carrier. It is possible for the carrier to put more than one letter into the same mailbox (more than one x going to the same y), but quite impossible for the carrier to place the same letter in two different boxes (one x going to two y’s).

a. For y  x , using input values from x  4 to x  4 produces the following table and graph (Figure 2.28). Note the result is a V-shaped graph that “opens upward.” The point (0, 0) of this absolute value graph is called the vertex. Since any vertical line will intersect the graph in at most one point, this is the graph of a function. y  x Figure 2.28 x

y  x

4

4

3

3

2

2

1

1

0

0

1

1

2

2

3

3

4

4

y 5

5

x

5

5

b. For y  1x, values less than zero do not produce a real number, so our graph actually begins at (0, 0) (see Figure 2.29). Completing the table for nonnegative values produces the graph shown, which appears to rise to the right and remains in the first quadrant. Since any vertical line will intersect this graph in at most one place, y  1x is also a function. Figure 2.29

y  1x x

y  1x

0

0

1

1

2

12  1.4

3

13  1.7

4

y 5

5

5

x

2

A. You’ve just learned how to distinguish the graph of a function from that of a relation

5

Now try Exercises 31 through 34



B. The Domain and Range of a Function Vertical Boundary Lines and the Domain In addition to its use as a graphical test for functions, a vertical line can help determine the domain of a function from its graph. For the graph of y  1x (Figure 2.29), a vertical line will not intersect the graph until x  0, and then will intersect the graph for all values x  0 (showing the function is defined for these values). These vertical boundary lines indicate the domain is x  3 0, q 2 . For the graph of y  x (Figure 2.28), a vertical line will intersect the graph (or its infinite extension) for all values of x, and the

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domain is x  1q, q 2 . Using vertical lines in this way also affirms the domain of y  x  1 (Section 2.1, Figure 2.5) is x  1q, q 2 while the domain of the relation x  y (Section 2.1, Figure 2.6) is x  30, q 2 .

Range and Horizontal Boundary Lines The range of a relation can be found using a horizontal “boundary line,” since it will associate a value on the y-axis with a point on the graph (if it exists). Simply visualize a horizontal line and move the line up or down until you determine the graph will always intersect the line, or will no longer intersect the line. This will give you the boundaries of the range. Mentally applying this idea to the graph of y  1x (Figure 2.29) shows the range is y  30, q 2. Although shaped very differently, a horizontal boundary line shows the range of y  x (Figure 2.28) is also y  30, q 2. EXAMPLE 5



Determining the Domain and Range of a Function Use a table of values to graph the functions defined by 3 a. y  x2 b. y  1 x Then use boundary lines to determine the domain and range of each.

Solution



a. For y  x2, it seems convenient to use inputs from x  3 to x  3, producing the following table and graph. Note the result is a basic parabola that “opens upward” (both ends point in the positive y direction), with a vertex at (0, 0). Figure 2.30 shows a vertical line will intersect the graph or its extension anywhere it is placed. The domain is x  1  q, q 2 . Figure 2.31 shows a horizontal line will intersect the graph only for values of y that are greater than or equal to 0. The range is y  30, q 2 . Figure 2.30

Squaring Function x

yx

2

3

9

2

4

1

1

0

0

1

1

2

4

3

9

5

5

Figure 2.31

y y  x2

5

5

5

x

5

y y  x2

5

x

5

3 b. For y  1x, we select points that are perfect cubes where possible, then a few others to round out the graph. The resulting table and graph are shown, and we notice there is a “pivot point” at (0, 0) called a point of inflection, and the ends of the graph point in opposite directions. Figure 2.32 shows a vertical line will intersect the graph or its extension anywhere it is placed. Figure 2.33 shows a horizontal line will likewise always intersect the graph. The domain is x  1q, q 2 , and the range is y  1q, q 2 .

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Cube Root Function x

y  2x

8

2

4

 1.6

1

1

0

0

1

1

4

 1.6

8

2

Figure 2.33

Figure 2.32

3

5

3 y y  x

10

127

5

10

x

3 y y  x

10

5

10

x

5

Now try Exercises 35 through 46



Implied Domains When stated in equation form, the domain of a function is implicitly given by the expression used to define it, since the expression will dictate the allowable values (Section 1.2). The implied domain is the set of all real numbers for which the function represents a real number. If the function involves a rational expression, the domain will exclude any input that causes a denominator of zero. If the function involves a square root expression, the domain will exclude inputs that create a negative radicand. EXAMPLE 6



Determining Implied Domains State the domain of each function using interval notation. 3 a. y  b. y  12x  3 x2 x5 c. y  2 d. y  x2  5x  7 x 9

Solution



a. By inspection, we note an x-value of 2 gives a zero denominator and must be excluded. The domain is x  1q, 22 ´ 12, q 2. b. Since the radicand must be nonnegative, we solve the inequality 2x  3  0, 3 giving x  3 2 . The domain is x  3 2 , q 2. c. To prevent division by zero, inputs of 3 and 3 must be excluded (set x2  9  0 and solve by factoring). The domain is x  1q, 32 ´ 13, 32 ´ 13, q 2 . Note that x  5 is in the domain 0  0 is defined. since 16 d. Since squaring a number and multiplying a number by a constant are defined for all reals, the domain is x  1q, q 2. Now try Exercises 47 through 64

EXAMPLE 7



Determining Implied Domains Determine the domain of each function: 2x 7 a. y  b. y  Ax  3 14x  5



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Solution

B. You’ve just learned how to determine the domain and range of a function

7 7  0 (for the radicand) and x  3  0 , we must have Ax  3 x3 (for the denominator). Since the numerator is always positive, we need x  3 7 0, which gives x 7 3. The domain is x  13, q 2 . 2x b. For y  , we must have 4x  5  0 and 14x  5  0. This indicates 14x  5 4x  5 7 0 or x 7 54. The domain is x  154, q 2 .

a. For y 

Now try Exercises 65 through 68



C. Function Notation Figure 2.34 x

In our study of functions, you’ve likely noticed that the relationship between input and output values is an important one. To highlight this fact, think of a function as a simple machine, which can process inputs using a stated sequence of operations, then deliver a single output. The inputs are x-values, a program we’ll name f performs the operations on x, and y is the resulting output (see Figure 2.34). Once again we see that “the value of y depends on the value of x,” or simply “y is a function of x.” Notationally, we write “y is a function of x” as y  f 1x2 using function notation. You are already familiar with letting a variable represent a number. Here we do something quite different, as the letter f is used to represent a sequence of operations to be performed on x. Consider the function y  2x  1, which we’ll now write as f 1x2  2x  1 [since y  f 1x2 ]. In words the function says, “divide inputs by 2, then add 1.” To evaluate the function at x  4 (Figure 2.35) we have:

Input f Sequence of operations on x as defined by f(x)

Output

y

input 4

↓

x ↓ f 1x2   1 2

input 4

Figure 2.35 4

Input

4 f 142   1 2 21

f(x) Divide inputs by 2 then add 1 4 +1 2

3 Output

3

Instead of saying, “. . . when x  4, the value of the function is 3,” we simply say “f of 4 is 3,” or write f 142  3. Note that the ordered pair (4, 3) is equivalent to (4, f(4)). CAUTION

EXAMPLE 8





Although f(x) is the favored notation for a “function of x,” other letters can also be used. For example, g(x) and h(x) also denote functions of x, where g and h represent a different sequence of operations on the x-inputs. It is also important to remember that these represent function values and not the product of two variables: f1x2  f # 1x2.

Evaluating a Function

Given f 1x2  2x2  4x, find 3 a. f 122 b. f a b 2

c. f 12a2  214a2 2  8a

d. f 1a  12

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Solution



a.

c.

f 1x2  2x2  4x f 122  2122 2  4122  8  182  16

b.

f 1x2  2x2  4x f 12a2  212a2 2  412a2  8a2  8a

d.

129

f 1x2  2x2  4x 3 2 3 3 f a b  2a b  4a b 2 2 2 3 9  6 2 2

f 1x2  2x2  4x f 1a  12  21a  12 2  41a  12  21a2  2a  12  4a  4  2a2  4a  2  4a  4  2a2  2 Now try Exercises 69 through 84



Graphs are an important part of studying functions, and learning to read and interpret them correctly is a high priority. A graph highlights and emphasizes the allimportant input/output relationship that defines a function. In this study, we hope to firmly establish that the following statements are synonymous: 1. 2. 3. 4. EXAMPLE 9A



f 122  5 12, f 122 2  12, 52 12, 52 is on the graph of f, and When x  2, f 1x2  5

Reading a Graph For the functions f (x) and g(x) whose graphs are shown in Figures 2.36 and 2.37 a. State the domain of the function. b. Evaluate the function at x  2. c. Determine the value(s) of x for which y  3. d. State the range of the function. Figure 2.36 y 5



y 4

3

3

2

2

1

1 1

2

3

g(x)

5

4

5 4 3 2 1 1

Solution

Figure 2.37

f(x)

4

5

x

5 4 3 2 1 1

2

2

3

3

1

2

3

4

5

x

For f(x), a. The graph is a continuous line segment with endpoints at (4, 3) and (5, 3), so we state the domain in interval notation. Using a vertical boundary line we note the smallest input is 4 and the largest is 5. The domain is x  34, 5 4. b. The graph shows an input of x  2 corresponds to y  1: f 122  1 since (2, 1) is a point on the graph. c. For f 1x2  3 (or y  3) the input value must be x  5 since (5, 3) is the point on the graph. d. Using a horizontal boundary line, the smallest output value is 3 and the largest is 3. The range is y  3 3, 34 .

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For g(x), a. Since the graph is pointwise defined, we state the domain as the set of first coordinates: D  54, 2, 0, 2, 46. b. An input of x  2 corresponds to y  2: g122  2 since (2, 2) is on the graph. c. For g1x2  3 (or y  32 the input value must be x  4, since (4, 3) is a point on the graph. d. The range is the set of all second coordinates: R  51, 0, 1, 2, 36. EXAMPLE 9B

Solution





Reading a Graph

Use the graph of f 1x2 given to answer the following questions: a. What is the value of f 122 ? (2, 4) b. What value(s) of x satisfy f 1x2  1?

y 5

f (x) a. The notation f 122 says to find the value of the (0, 1) function f when x  2. Expressed graphically, (3, 1) we go to x  2, locate the corresponding point 5 on the graph of f (blue arrows), and find that f 122  4. b. For f 1x2  1, we’re looking for x-inputs that result in an output of y  1 3since y  f 1x2 4 . 5 From the graph, we note there are two points with a y-coordinate of 1, namely, (3, 1) and (0, 1). This shows f 132  1, f 102  1, and the required x-values are x  3 and x  0.

5

Now try Exercises 85 through 90

x



In many applications involving functions, the domain and range can be determined by the context or situation given. EXAMPLE 10



Determining the Domain and Range from the Context Paul’s 1993 Voyager has a 20-gal tank and gets 18 mpg. The number of miles he can drive (his range) depends on how much gas is in the tank. As a function we have M1g2  18g, where M(g) represents the total distance in miles and g represents the gallons of gas in the tank. Find the domain and range.

Solution



C. You’ve just learned how to use function notation and evaluate functions

Begin evaluating at x  0, since the tank cannot hold less than zero gallons. On a full tank the maximum range of the van is 20 # 18  360 miles or M1g2  30, 360 4 . Because of the tank’s size, the domain is g  30, 20 4. Now try Exercises 94 through 101



D. Average Rates of Change As noted in Section 2.3, one of the defining characteristics of a linear function is that ¢y the rate of change m  is constant. For nonlinear functions the rate of change is ¢x not constant, but we can use a related concept called the average rate of change to study these functions.

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Section 2.4 Functions, Function Notation, and the Graph of a Function

Average Rate of Change For a function that is smooth and continuous on the interval containing x1 and x2, the average rate of change between x1 and x2 is given by ¢y y2  y1  x2  x1 ¢x which is the slope of the secant line through (x1, y1) and (x2, y2) EXAMPLE 11



Calculating Average Rates of Change The graph shown displays the number of units shipped of vinyl records, cassette tapes, and CDs for the period 1980 to 2005. Units shipped in millions

1000

CDs

900

Units shipped (millions)

800 700 600 500 400 300

Cassettes

200

Vinyl

100

80

82

84

86

88

90

92

94

96

98

100

102

104

Year

Vinyl

Cassette

CDs

1980

323

110

0

1982

244

182

0

1984

205

332

6

1986

125

345

53

1988

72

450

150

1990

12

442

287

1992

2

366

408

1994

2

345

662

1996

3

225

779

1998

3

159

847

2000

2

76

942

2004

1

5

767

2005

1

3

705

106

Year (80 → 1980) Source: Swivel.com

a. Find the average rate of change in CDs shipped and in cassettes shipped from 1994 to 1998. What do you notice? b. Does it appear that the rate of increase in CDs shipped was greater from 1986 to 1992, or from 1992 to 1996? Compute the average rate of change for each period and comment on what you find. Solution



Using 1980 as year zero (1980 S 0), we have the following: a. CDs Cassettes 1994: 114, 6622, 1998: 118, 8472 1994: 114, 3452, 1998: 118, 1592 ¢y ¢y 847  662 159  345   ¢x 18  14 ¢x 18  14 185 186   4 4  46.25  46.5 The decrease in the number of cassettes shipped was roughly equal to the increase in the number of CDs shipped (about 46,000,000 per year).

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b. From the graph, the secant line for 1992 to 1996 appears to have a greater slope. 1986–1992 CDs 1986: 16, 532, 1992: 112, 4082 ¢y 408  53  ¢x 12  6 355  6  59.16

D. You’ve just learned how to apply the rate-of-change concept to nonlinear functions

1992–1996 CDs 1992: 112, 4082, 1996: 116, 7792 ¢y 779  408  ¢x 16  12 371  4  92.75

For 1986 to 1992: m  59.2; for 1992 to 1996: m  92.75, a growth rate much higher than the earlier period. Now try Exercises 102 and 103



2.4 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. If a relation is given in ordered pair form, we state the domain by listing all of the coordinates in a set. 2. A relation is a function if each element of the is paired with element of the range. 3. The set of output values for a function is called the of the function.



4. Write using function notation: The function f evaluated at 3 is negative 5: 5. Discuss/Explain why the relation y  x2 is a function, while the relation x  y2 is not. Justify your response using graphs, ordered pairs, and so on. 6. Discuss/Explain the process of finding the domain and range of a function given its graph, using vertical and horizontal boundary lines. Include a few illustrative examples.

DEVELOPING YOUR SKILLS

Determine whether the mappings shown represent functions or nonfunctions. If a nonfunction, explain how the definition of a function is violated.

7.

Woman

Country

Indira Gandhi Clara Barton Margaret Thatcher Maria Montessori Susan B. Anthony

Britain U.S. Italy India

8.

Book

Author

Hawaii Roots Shogun 20,000 Leagues Under the Sea Where the Red Fern Grows

Rawls Verne Haley Clavell Michener

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Section 2.4 Functions, Function Notation, and the Graph of a Function

Basketball star

Reported height

MJ The Mailman The Doctor The Iceman The Shaq

7'1" 6'6" 6'7" 6'9" 7'2"

10.

Country

Language

Canada Japan Brazil Tahiti Ecuador

Japanese Spanish French Portuguese English

y

21.

5

5

y

5

5

5

5

5 x

5 x

5

y

30.

5

5

5 x

(1, 4)

(0, 2) (5, 3) 5

y

y

28.

y

29.

5

5

5 x

5 x

y

18.

5

5 x

5

5

(4, 2)

(4, 2)

17.

5

5

5 x

y

(3, 4) (1, 3)

(5, 0)

5

y

26.

5

5

(3, 5)

5 x

5

5

27.

5 x

y

16. (2, 4)

5

5

5 x

5

14. (1, 81), (2, 64), (3, 49), (5, 36), (8, 25), (13, 16), (21, 9), (34, 4), and (55, 1)

(1, 1)

5

5

13. (9, 10), (7, 6), (6, 10), (4, 1), (2, 2), (1, 8), (0, 2), (2, 7), and (6, 4)

(3, 4)

y

24.

5

25.

5 x

5

5

12. (7, 5), (5, 3), (4, 0), (3, 5), (1, 6), (0, 9), (2, 8), (3, 2), and (5, 7)

y

5

5 x

y

23.

11. (3, 0), (1, 4), (2, 5), (4, 2), (5, 6), (3, 6), (0, 1), (4, 5), and (6, 1)

5

5

5

Determine whether the relations indicated represent functions or nonfunctions. If the relation is a nonfunction, explain how the definition of a function is violated.

15.

y

22.

5

5

(3, 4)

5

(3, 4)

(2, 3)

5

(3, 3) (1, 2)

(5, 1)

(1, 1)

5

5

5 x

5 x

(5, 2) (2, 4)

(1, 4)

(4, 5)

5

(3, 2)

31. y  x

5

Determine whether or not the relations given represent a function. If not, explain how the definition of a function is violated. y

19.

y

20.

5

Graph each relation using a table, then use the vertical line test to determine if the relation is a function.

33. y  1x  22 2

5 x

5

y

5

y

36.

5

5

5 x 5

5

34. x  y  2

Determine whether or not the relations indicated represent a function, then determine the domain and range of each.

35. 5

3 32. y  2 x

5 x

5

5 x

5 5

5

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37.

5

5

5

5 x

y

5

y

40.

5

5

5 x

5 x

5

5

63. y 

1x  2 2x  5

64. y 

1x  1 3x  2

5 x

5

y

5

y

44.

5

5

5 x

5 x

5

y

66. g1x2 

4 A3  x

67. h1x2 

2 14  x

68. p1x2 

7 15  x

y

46.

2 70. f 1x2  x  5 3

71. f 1x2  3x2  4x

72.

73. h1x2 

3 x

74. h1x2 

2 x2

75. h1x2 

5x x

76. h1x2 

4x x

5

77. g1r2  2r

78. g1r2  2rh

79. g1r2  r

80. g1r2  r2h

2

5

5 x

5

5 x

5

f 1x2  2x2  3x

Determine the value of g(4), g(32 ), g(2c), and g(c  3), then simplify as much as possible.

5

5

5 Ax  2

Determine the value of h(3), h(23), h(3a), and h(a  2), then simplify as much as possible.

5

5

65. f 1x2 

1 69. f 1x2  x  3 2

5

5

5 x

45.

x4 x  2x  15 2

y

42.

5

43.

62. y2 

2

Determine the value of f(6), f(32 ), f(2c), and f(c  1), then simplify as much as possible.

5

y

41.

x x  3x  10

5

5

60. y  x  2  3

61. y1  5 x

5

39.

59. y  2x  1

y

38.

5

5

Determine the value of p(5), p(32 ), p(3a), and p(a  1), then simplify as much as possible.

81. p1x2  12x  3

82. p1x2  14x  1

3x  5 x2 2

Determine the domain of the following functions.

47. f 1x2 

3 x5

49. h1a2  13a  5 51. v1x2 

x2 x2  25

v5 53. u  2 v  18 55. y 

17 x  123 25

57. m  n2  3n  10

48. g1x2 

2 3x

50. p1a2  15a  2 52. w1x2  54. p  56. y 

x4 x2  49

83. p1x2 

84. p1x2 

2x2  3 x2

Use the graph of each function given to (a) state the domain, (b) state the range, (c) evaluate f(2), and (d) find the value(s) x for which f 1x2  k (k a constant). Assume all results are integer-valued.

85. k  4

86. k  3 y

q7

y

5

5

q2  12 11 x  89 19

58. s  t2  3t  10

5

5 x

5

5

5 x

5

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87. k  1

88. k  3

89. k  2

y

5

5 x

5

y

5

5

5

5 x

5

5

5

5 x

5

5

5 x

5

WORKING WITH FORMULAS

91. Ideal weight for males: W(H)  92H  151 The ideal weight for an adult male can be modeled by the function shown, where W is his weight in pounds and H is his height in inches. (a) Find the ideal weight for a male who is 75 in. tall. (b) If I am 72 in. tall and weigh 210 lb, how much weight should I lose? 92. Celsius to Fahrenheit conversions: C  59(F  32) The relationship between Fahrenheit degrees and degrees Celsius is modeled by the function shown. (a) What is the Celsius temperature if °F  41? (b) Use the formula to solve for F in terms of C, then substitute the result from part (a). What do you notice? 

90. k  1 y

y

5



135

Section 2.4 Functions, Function Notation, and the Graph of a Function

1 93. Pick’s theorem: A  B  I  1 2 Picks theorem is an interesting yet little known formula for computing the area of a polygon drawn in the Cartesian coordinate system. The formula can be applied as long as the vertices of the polygon are lattice points (both x and y are integers). If B represents the number of lattice points lying directly on the boundary of the polygon (including the vertices), and I represents the number of points in the interior, the area of the polygon is given by the formula shown. Use some graph paper to carefully draw a triangle with vertices at (3, 1), (3, 9), and (7, 6), then use Pick’s theorem to compute the triangle’s area.

APPLICATIONS

94. Gas mileage: John’s old ’87 LeBaron has a 15-gal gas tank and gets 23 mpg. The number of miles he can drive is a function of how much gas is in the tank. (a) Write this relationship in equation form and (b) determine the domain and range of the function in this context. 95. Gas mileage: Jackie has a gas-powered model boat with a 5-oz gas tank. The boat will run for 2.5 min on each ounce. The number of minutes she can operate the boat is a function of how much gas is in the tank. (a) Write this relationship in equation form and (b) determine the domain and range of the function in this context. 96. Volume of a cube: The volume of a cube depends on the length of the sides. In other words, volume is a function of the sides: V1s2  s3. (a) In practical terms, what is the domain of this function? (b) Evaluate V(6.25) and (c) evaluate the function for s  2x2. 97. Volume of a cylinder: For a fixed radius of 10 cm, the volume of a cylinder depends on its height. In other words, volume is a function of height:

V1h2  100h. (a) In practical terms, what is the domain of this function? (b) Evaluate V(7.5) and 8 (c) evaluate the function for h  .  98. Rental charges: Temporary Transportation Inc. rents cars (local rentals only) for a flat fee of $19.50 and an hourly charge of $12.50. This means that cost is a function of the hours the car is rented plus the flat fee. (a) Write this relationship in equation form; (b) find the cost if the car is rented for 3.5 hr; (c) determine how long the car was rented if the bill came to $119.75; and (d) determine the domain and range of the function in this context, if your budget limits you to paying a maximum of $150 for the rental. 99. Cost of a service call: Paul’s Plumbing charges a flat fee of $50 per service call plus an hourly rate of $42.50. This means that cost is a function of the hours the job takes to complete plus the flat fee. (a) Write this relationship in equation form; (b) find the cost of a service call that takes 212 hr; (c) find the number of hours the job took if the

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charge came to $262.50; and (d) determine the domain and range of the function in this context, if your insurance company has agreed to pay for all charges over $500 for the service call. 100. Predicting tides: The graph shown approximates the height of the tides at Fair Haven, New Brunswick, for a 12-hr period. (a) Is this the graph of a function? Why? (b) Approximately what time did high tide occur? (c) How high is the tide at 6 P.M.? (d) What time(s) will the tide be 2.5 m? 5

Meters

4 3 2 1

5

7

9

11

1 A.M.

4.0

3

Time

101. Predicting tides: The graph shown approximates the height of the tides at Apia, Western Samoa, for a 12-hr period. (a) Is this the graph of a function? Why? (b) Approximately what time did low tide occur? (c) How high is the tide at 2 A.M.? (d) What time(s) will the tide be 0.7 m? Meters

1.0



6

8

10

12 2 A.M.

4

Time

3600 3400

Full term

3200

(40, 3200)

2800

(36, 2600) 2400 2000

(32, 1600)

1600 1200

20

30

40

50

60

70

80

90

100

110

Source: Statistical History of the United States from Colonial Times to Present

(29, 1100)

800

2.0

10

3800

Weight (g)

102. Weight of a fetus: The growth rate of a fetus in the mother’s womb (by weight in grams) is modeled by the graph shown here, beginning with the 25th week of

3.0

1.0

0.5

4 P.M.

103. Fertility rates: Over the years, fertility rates for (60, 3.6) (10, 3.4) women in the (20, 3.2) (50, 3.0) United States (average number (70, 2.4) of children per (40, 2.2) (90, 2.0) woman) have (80, 1.8) varied a great deal, though in the twenty-first Year (10 → 1910) century they’ve begun to level out. The graph shown models this fertility rate for most of the twentieth century. (a) Calculate the average rate of change from the years 1920 to 1940. Is the slope of the secant line positive or negative? Discuss what the slope means in this context. (b) Calculate the average rate of change from the year 1940 to 1950. Is the slope of the secant line positive or negative? Discuss what the slope means in this context. (c) Was the fertility rate increasing faster from 1940 to 1950, or from 1980 to 1990? Compare the slope of both secant lines and comment. Rate (children per woman)

3 P.M.

gestation. (a) Calculate the average rate of change (slope of the secant line) between the 25th week and the 29th week. Is the slope of the secant line positive or negative? Discuss what the slope means in this context. (b) Is the fetus gaining weight faster between the 25th and 29th week, or between the 32nd and 36th week? Compare the slopes of both secant lines and discuss.

(25, 900) 24

26

28

30

32

34

36

38

40

42

Age (weeks)

EXTENDING THE CONCEPT

Distance in meters

104. A father challenges his son to a 400-m race, depicted in the graph shown here.

b. Approximately how many meters behind was the second place finisher? c. Estimate the number of seconds the father was in the lead in this race. d. How many times during the race were the father and son tied?

400 300 200 100 0

10

20

30

40

50

60

70

Time in seconds Father:

Son:

a. Who won and what was the approximate winning time?

80

105. Sketch the graph of f 1x2  x, then discuss how you could use this graph to obtain the graph of F1x2  x without computing additional points. x What would the graph of g1x2  look like? x

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106. Sketch the graph of f 1x2  x2  4, then discuss how you could use this graph to obtain the graph of F1x2  x2  4 without computing additional x2  4 points. Determine what the graph of g1x2  2 x 4 would look like. 107. If the equation of a function is given, the domain is implicitly defined by input values that generate real

137

Section 2.4 Functions, Function Notation, and the Graph of a Function

valued outputs. But unless the graph is given or can be easily sketched, we must attempt to find the range analytically by solving for x in terms of y. We should note that sometimes this is an easy task, while at other times it is virtually impossible and we must rely on other methods. For the following functions, determine the implicit domain and find the range by 3 2 solving for x in terms of y. a. y  xx   2 b. y  x  3

MAINTAINING YOUR SKILLS

108. (2.2) Which line has a steeper slope, the line through (5, 3) and (2, 6), or the line through (0, 4) and (9, 4)?

110. (1.5) Solve the equation using the quadratic formula, then check the result(s) using substitution: x2  4x  1  0

109. (R.6) Compute the sum and product indicated: a. 124  6 154  16 b. 12  132 12  132

111. (R.4) Factor the following polynomials completely: a. x3  3x2  25x  75 b. 2x2  13x  24 c. 8x3  125

MID-CHAPTER CHECK

2. Find the slope of the line passing through the given points: 13, 82 and 14, 102 . 3. In 2002, Data.com lost $2 million. In 2003, they lost $0.5 million. Will the slope of the line through these points be positive or negative? Why? Calculate the slope. Were you correct? Write the slope as a unit rate and explain what it means in this context.

Exercises 5 and 6 L1

y 5 L 2

5

5 x

5

Exercises 7 and 8 y 5

h(x)

5

5 x

8. Judging from the appearance of the graph alone, compare the average rate of change from x  1 to x  2 to the rate of change from x  4 to x  5. Which rate of change is larger? How is that demonstrated graphically? Exercise 9 F 9. Find a linear function that models the graph of F(p) given. F(p) Explain the slope of the line in this context, then use your model to predict the fox population when the pheasant P population is 20,000. Pheasant population (1000s) Fox population (in 100s)

1. Sketch the graph of the line 4x  3y  12. Plot and label at least three points.

10

9 8 7 6 5 4 3 2 1

0

1

2

3

4

5

6

7

5

5

4. Sketch the line passing through (1, 4) with slope m  2 3 (plot and label at least two points). Then find the equation of the line perpendicular to this line through (1, 4).

5

5 x

y

c.

5

5

5 x

5

5

5 x

5

5

6. Write the equation for line L2 shown. Is this the graph of a function? Discuss why or why not. 7. For the graph of function h(x) shown, (a) determine the value of h(2); (b) state the domain; (c) determine the value of x for which h1x2  3; and (d) state the range.

9 10

10. State the domain and range for each function below. y y a. b. 5

5. Write the equation for line L1 shown. Is this the graph of a function? Discuss why or why not.

8

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REINFORCING BASIC CONCEPTS The Various Forms of a Linear Equation In a study of mathematics, getting a glimpse of the “big picture” can be an enormous help. Learning mathematics is like building a skyscraper: The final height of the skyscraper ultimately depends on the strength of the foundation and quality of the frame supporting each new floor as it is built. Our work with linear functions and their graphs, while having a number of useful applications, is actually the foundation on which much of your future work will be built. The study of quadratic and polynomial functions and their applications all have their roots in linear equations. For this reason, it’s important that you gain a certain fluency with linear functions—even to a point where things come to you effortlessly and automatically. This level of performance requires a strong desire and a sustained effort. We begin by reviewing the basic facts a student MUST know to reach this level. MUST is an acronym for memorize, understand, synthesize, and teach others. Don’t be satisfied until you’ve done all four. Given points (x1, y1) and (x2, y2): Forms and Formulas slope formula point-slope form slope-intercept form standard form y2  y1 m y  y1  m1x  x1 2 y  mx  b Ax  By  C x2  x1 given any two points given slope m and given slope m and also used in linear y-intercept (0, b) systems (Chapter 6) on the line any point (x1, y1) Characteristics of Lines y-intercept x-intercept increasing decreasing (0, y) (x, 0) m 7 0 m 6 0 let x  0, let y  0, line slants upward line slants downward solve for y solve for x from left to right from left to right Practice for Speed and Accuracy For the two points given, (a) compute the slope of the line and state whether the line is increasing or decreasing; (b) find the equation of the line using point-slope form; (c) write the equation in slope-intercept form; (d) write the equation in standard form; and (e) find the x- and y-intercepts and graph the line. 1. P1(0, 5); P2(6, 7) 4. P1 15, 42; P2 13, 22

2. P1(3, 2); P2(0, 9) 5. P1 12, 52; P2 16, 12

3. P1(3, 2); P2(9, 5) 6. P1 12, 72; P2 18, 22

2.5 Analyzing the Graph of a Function Learning Objectives In Section 2.5 you will learn how to:

A. Determine whether a function is even, odd, or neither

B. Determine intervals where a function is positive or negative

C. Determine where a function is increasing or decreasing

D. Identify the maximum and minimum values of a function

E. Develop a formula to calculate rates of change for any function

In this section, we’ll consolidate and refine many of the ideas we’ve encountered related to functions. When functions and graphs are applied as real-world models, we create a numeric and visual representation that enables an informed response to questions involving maximum efficiency, positive returns, increasing costs, and other relationships that can have a great impact on our lives.

A. Graphs and Symmetry While the domain and range of a function will remain dominant themes in our study, for the moment we turn our attention to other characteristics of a function’s graph. We begin with the concept of symmetry.

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Symmetry with Respect to the y-Axis

Consider the graph of f 1x2  x4  4x2 shown in Figure 2.38, where the portion of the graph to the left of the y-axis appears to be a mirror image of the portion to the right. A function is symmetric to the y-axis if, given any point (x, y) on the graph, the point 1x, y2 is also on the graph. We note that 11, 32 is on the graph, as is 11, 32, and that 12, 02 is an x-intercept of the graph, as is (2, 0). Functions that are symmetric to the y-axis are also known as even functions and in general we have:

Figure 2.38 5

y f(x)  x4  4x2 (2.2, ~4)

(2.2, ~4)

(2, 0)

(2, 0)

5

5

x

(1, 3) 5 (1, 3)

Even Functions: y-Axis Symmetry A function f is an even function if and only if, for each point (x, y) on the graph of f, the point (x, y) is also on the graph. In function notation f 1x2  f 1x2

Symmetry can be a great help in graphing new functions, enabling us to plot fewer points, and to complete the graph using properties of symmetry.

EXAMPLE 1



Graphing an Even Function Using Symmetry a. The function g(x) in Figure 2.39 is known to be even. Draw the complete graph (only the left half is shown). Figure 2.39 2 y b. Show that h1x2  x3 is an even function using 5 the arbitrary value x  k [show h1k2  h1k2 ], g(x) then sketch the complete graph using h(0), (1, 2) h(1), h(8), and y-axis symmetry. (1, 2)

Solution



a. To complete the graph of g (see Figure 2.39) use the points (4, 1), (2, 3), (1, 2), and y-axis symmetry to find additional points. The corresponding ordered pairs are (4, 1), (2, 3), and (1, 2), which we use to help draw a “mirror image” of the partial graph given. 2 b. To prove that h1x2  x3 is an even function, we must show h1k2  h1k2 for any 2 1 constant k. After writing x3 as 3x2 4 3, we have: h1k2  h1k2

3 1k2 4  3 1k2 4 2

The proof can also be 2 demonstrated by writing x3 1 as A x3 B 2, and you are asked to complete this proof in Exercise 82.

2

2 1k2  2 1k2 3

WORTHY OF NOTE

1 3

2

3

(4, 1)

(2, 3)

2

3 2 3 2 2 k 2 k✓

(2, 3) 5

Figure 2.40 y 5

(8, 4)

first step of proof 1 3

(4, 1) 5 x

5

evaluate h 1k2 and h (k )

(1, 1)

radical form

10

result: 1k2 2  k 2

Using h102  0, h112  1, and h182  4 with y-axis symmetry produces the graph shown in Figure 2.40.

h(x)

(8, 4)

(1, 1) (0, 0)

10

x

5

Now try Exercises 7 through 12



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Symmetry with Respect to the Origin Another common form of symmetry is known as symmetry to the origin. As the name implies, the graph is somehow “centered” at (0, 0). This form of symmetry is easy to see for closed figures with their center at (0, 0), like certain polygons, circles, and ellipses (these will exhibit both y-axis symmetry and symmetry to the origin). Note the relation graphed in Figure 2.41 contains the points (3, 3) and (3, 3), along with (1, 4) and (1, 4). But the function f(x) in Figure 2.42 also contains these points and is, in the same sense, symmetric to the origin (the paired points are on opposite sides of the x- and y-axes, and a like distance from the origin). Figure 2.41

Figure 2.42

y

y

5

5

(1, 4)

(1, 4)

(3, 3)

(3, 3)

5

5

x

f(x)

5

5

(3, 3) (1, 4)

x

(3, 3) (1, 4)

5

5

Functions symmetric to the origin are known as odd functions and in general we have: Odd Functions: Symmetry about the Origin A function f is an odd function if and only if, for each point (x, y) on the graph of f, the point (x, y) is also on the graph. In function notation f 1x2  f 1x2

EXAMPLE 2



Graphing an Odd Function Using Symmetry a. In Figure 2.43, the function g(x) given is known to be odd. Draw the complete graph (only the left half is shown). b. Show that h1x2  x3  4x is an odd function using the arbitrary value x  k 3show h1x2  h1x2 4 , then sketch the graph using h122 , h112 , h(0), and odd symmetry.

Solution



a. To complete the graph of g, use the points (6, 3), (4, 0), and (2, 2) and odd symmetry to find additional points. The corresponding ordered pairs are (6, 3), (4, 0), and (2, 2), which we use to help draw a “mirror image” of the partial graph given (see Figure 2.43). Figure 2.43

Figure 2.44

y

y

10

5

(1, 3)

g(x)

WORTHY OF NOTE While the graph of an even function may or may not include the point (0, 0), the graph of an odd function will always contain this point.

(6, 3)

(2, 2) (4, 0)

10

h(x)

(4, 0)

(2, 0) x (6, 3) 10

(2, 2)

5

(2, 0) (0, 0)

5

(1, 3) 10

5

x

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b. To prove that h1x2  x3  4x is an odd function, we must show that h1k2  h1k2. h1k2  h1k2

1k2  41k2  3 k3  4k 4 k3  4k  k3  4k ✓ 3

A. You’ve just learned how to determine whether a function is even, odd, or neither

Using h122  0, h112  3, and h102  0 with symmetry about the origin produces the graph shown in Figure 2.44. Now try Exercises 13 through 24



B. Intervals Where a Function Is Positive or Negative

Consider the graph of f 1x2  x2  4 shown in Figure 2.45, which has x-intercepts at (2, 0) and (2, 0). Since x-intercepts have the form (x, 0) they are also called the zeroes of the function (the x-input causes an output of 0). Just as zero on the number line separates negative numbers from positive numbers, the zeroes of a function that crosses the x-axis separate x-intervals where a function is negative from x-intervals where the function is positive. Noting that outputs ( y-values) are positive in Quadrants I and II, f 1x2 7 0 in intervals where its graph is above the x-axis. Conversely, f 1x2 6 0 in x-intervals where its graph is below the x-axis. To illustrate, compare the graph of f in Figure 2.45, with that of g in Figure 2.46. Figure 2.45 5

(2, 0)

Figure 2.46

y f(x)  x2  4

5

y g(x)  (x  4)2

(2, 0)

5

5

x

3

(4, 0)

5

x

(0, 4) 5

WORTHY OF NOTE These observations form the basis for studying polynomials of higher degree, where we extend the idea to factors of the form 1x  r2 n in a study of roots of multiplicity (also see the Calculator Exploration and Discovery feature in this chapter).

EXAMPLE 3

5

The graph of f is a parabola, with x-intercepts of (2, 0) and (2, 0). Using our previous observations, we note f 1x2  0 for x  1q, 2 4 ´ 32, q 2 and f 1x2 6 0 for x  12, 22 . The graph of g is also a parabola, but is entirely above or on the x-axis, showing g1x2  0 for x  . The difference is that zeroes coming from factors of the form ( x  r) (with degree 1) allow the graph to cross the x-axis. The zeroes of f came from 1x  221x  22  0. Zeroes that come from factors of the form 1x  r2 2 (with degree 2) cause the graph to “bounce” off the x-axis since all outputs must be nonnegative. The zero of g came from 1x  42 2  0. 

Solving an Inequality Using a Graph Use the graph of g1x2  x3  2x2  4x  8 given to solve the inequalities a. g1x2  0 b. g1x2 6 0

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Solution



From the graph, the zeroes of g (x-intercepts) occur at (2, 0) and (2, 0). a. For g1x2  0, the graph must be on or above the x-axis, meaning the solution is x  32, q 2 . b. For g1x2 6 0, the graph must be below the x-axis, and the solution is x  1q, 22 . As we might have anticipated from the graph, factoring by grouping gives g1x2  1x  221x  22 2, with the graph crossing the x-axis at 2, and bouncing off the x-axis (intersects without crossing) at x  2.

y (0, 8) g(x) 5

5

x

5 2

Now try Exercises 25 through 28



Even if the function is not a polynomial, the zeroes can still be used to find x-intervals where the function is positive or negative.

EXAMPLE 4

Solution





B. You’ve just learned how to determine intervals where a function is positive or negative

y

Solving an Inequality Using a Graph

For the graph of r 1x2  1x  1  2 shown, solve a. r 1x2  0 b. r 1x2 7 0 a. The only zero of r is at (3, 0). The graph is on or below the x-axis for x  3 1, 34 , so r 1x2  0 in this interval. b. The graph is above the x-axis for x  13, q 2 , and r 1x2 7 0 in this interval.

10

r(x) 10

10

x

10

Now try Exercises 29 through 32



C. Intervals Where a Function Is Increasing or Decreasing In our study of linear graphs, we said a graph was increasing if it “rose” when viewed from left to right. More generally, we say the graph of a function is increasing on a given interval if larger and larger x-values produce larger and larger y-values. This suggests the following tests for intervals where a function is increasing or decreasing. Increasing and Decreasing Functions Given an interval I that is a subset of the domain, with x1 and x2 in I and x2 7 x1, 1. A function is increasing on I if f 1x2 2 7 f 1x1 2 for all x1 and x2 in I (larger inputs produce larger outputs). 2. A function is decreasing on I if f 1x2 2 6 f 1x1 2 for all x1 and x2 in I (larger inputs produce smaller outputs). 3. A function is constant on I if f 1x2 2  f 1x1 2 for all x1 and x2 in I (larger inputs produce identical outputs).

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f (x)

f (x) is increasing on I

f (x)

f(x2)

f (x) is decreasing on I

f(x) is constant on I

f (x)

f (x1)

f(x1)

f (x2)

f (x2)

f (x1)

f (x1)

f (x1) x1

x2

x

x1

Interval I

Interval I

x2  x1 and f (x2)  f (x1) for all x  I graph rises when viewed from left to right

x

Questions about the behavior of a function are asked with respect to the y outputs: where is the function positive, where is the function increasing, etc. Due to the input/ output, cause/effect nature of functions, the response is given in terms of x, that is, what is causing outputs to be negative, or to be decreasing.

x2  x1 and f(x2)  f(x1) for all x  I graph is level when viewed from left to right

x2  x1 and f (x2)  f (x1) for all x  I graph falls when viewed from left to right

1 7 2

x2 7 x1

and

and

f 112 7 f 122 8 7 7



Figure 2.47 10



y f(x)  x2  4x  5 (2, 9) (0, 5)

(1, 0)

(5, 0)

5

5

x

10

x  (3, 2)

f 1x2 2 7 f 1x1 2

Finding Intervals Where a Function Is Increasing or Decreasing

y 5

Use the graph of v(x) given to name the interval(s) where v is increasing, decreasing, or constant. Solution

x

x2

x1 Interval I

Consider the graph of f 1x2  x2  4x  5 in Figure 2.47. Since the graph opens downward with the vertex at (2, 9), the function must increase until it reaches this maximum value at x  2, and decrease thereafter. Notationally we’ll write this as f 1x2c for x  1q, 22 and f 1x2T for x  12, q 2. Using the interval 13, 22 shown, we see that any larger input value from the interval will indeed produce a larger output value, and f 1x2c on the interval. For instance,

WORTHY OF NOTE

EXAMPLE 5

x2

f(x2)

f(x1)

f (x2)

From left to right, the graph of v increases until leveling off at (2, 2), then it remains constant until reaching (1, 2). The graph then increases once again until reaching a peak at (3, 5) and decreases thereafter. The result is v 1x2c for x  1q, 22 ´ 11, 32, v1x2T for x  13, q2, and v(x) is constant for x  12, 12 .

v(x)

5

5

x

5

Now try Exercises 33 through 36



Notice the graph of f in Figure 2.47 and the graph of v in Example 5 have something in common. It appears that both the far left and far right branches of each graph point downward (in the negative y-direction). We say that the end behavior of both graphs is identical, which is the term used to describe what happens to a graph as x becomes very large. For x 7 0, we say a graph is, “up on the right” or “down on the right,” depending on the direction the “end” is pointing. For x 6 0, we say the graph is “up on the left” or “down on the left,” as the case may be.

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EXAMPLE 6



Describing the End Behavior of a Graph

y

The graph of f 1x2  x  3x is shown. Use the graph to name intervals where f is increasing or decreasing, and comment on the end-behavior of the graph.

5

3

Solution



C. You’ve just learned how to determine where a function is increasing or decreasing

From the graph we observe that: f 1x2c for x  1q, 12 ´ 11, q 2 , and f 1x2T for x  11, 12 . The end behavior of the graph is down on the left, up on the right (down/up).

f(x)  x2  3x

5

5

x

5

Now try Exercises 37 through 40



D. More on Maximum and Minimum Values The y-coordinate of the vertex of a parabola where a 6 0, and the y-coordinate of “peaks” from other graphs are called maximum values. A global maximum (also called an absolute maximum) names the largest range value over the entire domain. A local maximum (also called a relative maximum) gives the largest range value in a specified interval; and an endpoint maximum can occur at an endpoint of the domain. The same can be said for the corresponding minimum values. We will soon develop the ability to locate maximum and minimum values for quadratic and other functions. In future courses, methods are developed to help locate maximum and minimum values for almost any function. For now, our work will rely chiefly on a function’s graph.

EXAMPLE 7



Analyzing Characteristics of a Graph Analyze the graph of function f shown in Figure 2.48. Include specific mention of a. domain and range, b. intervals where f is increasing or decreasing, c. maximum (max) and minimum (min) values, d. intervals where f 1x2  0 and f 1x2 6 0, e. whether the function is even, odd, or neither.

Solution



D. You’ve just learned how to identify the maximum and minimum values of a function

a. Using vertical and horizontal boundary lines show the domain is x  , with range: y  1q, 7 4 . b. f 1x2c for x  1q, 32 ´ 11, 52 shown in blue in Figure 2.49, and f 1x2T for x  13, 12 ´ 15, q 2 as shown in red. c. From Part (b) we find that y  5 at (3, 5) and y  7 at (5, 7) are local maximums, with a local minimum of y  1 at (1, 1). The point (5, 7) is also a global maximum (there is no global minimum). d. f 1x2  0 for x  3 6, 84 ; f 1x2 6 0 for x  1q, 62 ´ 18, q 2 e. The function is neither even nor odd.

Figure 2.48 y 10

(5, 7) f(x)

(3, 5)

(1, 1) 10

10

x

10

Figure 2.49 y 10

(5, 7) (3, 5) (6, 0)

(1, 1)

10

(8, 0) 10 x

10

Now try Exercises 41 through 48



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The ideas presented here can be applied to functions of all kinds, including rational functions, piecewise-defined functions, step functions, and so on. There is a wide variety of applications in Exercises 51 through 58.

E. Rates of Change and the Difference Quotient We complete our study of graphs by revisiting the concept of average rates of change. In many business, scientific, and economic applications, it is this attribute of a function that draws the most attention. In Section 2.4 we computed average rates of change by selecting two points from a graph, and computing the slope of the secant line: ¢y y2  y 1 m  . With a simple change of notation, we can use the function’s equax2  x1 ¢x tion rather than relying on a graph. Note that y2 corresponds to the function evaluated at x2: y2  f 1x2 2 . Likewise, y1  f 1x1 2 . Substituting these into the slope formula yields f 1x2 2  f 1x1 2 ¢y , giving the average rate of change between x1 and x2 for any func x2  x1 ¢x tion f (assuming the function is smooth and continuous between x1 and x2). Average Rate of Change For a function f and [x1, x2] a subset of the domain, the average rate of change between x1 and x2 is f 1x2 2  f 1x1 2 ¢y  , x1  x2 x2  x1 ¢x

Average Rates of Change Applied to Projectile Velocity A projectile is any object that is thrown, shot, or cast upward, with no continuing source of propulsion. The object’s height (in feet) after t sec is modeled by the function h1t2  16t2  vt  k, where v is the initial velocity of the projectile, and k is the height of the object at contact. For instance, if a soccer ball is kicked upward from ground level (k  0) with an initial speed of 64 ft/sec, the height of the ball t sec later is h1t2  16t2  64t. From Section 2.5, we recognize the graph will be a parabola and evaluating the function for t  0 to 4 produces Table 2.4 and the graph shown in Figure 2.50. Experience tells us the ball is traveling at a faster rate immediately after being kicked, as compared to when it nears its maximum height where it ¢height momentarily stops, then begins its descent. In other words, the rate of change ¢time has a larger value at any time prior to reaching its maximum height. To quantify this we’ll compute the average rate of change between t  0.5 and t  1, and compare it to the average rate of change between t  1 and t  1.5. Table 2.4 WORTHY OF NOTE Keep in mind the graph of h represents the relationship between the soccer ball’s height in feet and the elapsed time t. It does not model the actual path of the ball.

Time in seconds

Figure 2.50

Height in feet

0

0

1

48

2

64

3

48

4

0

h(t) 80

(2, 64) 60

(3, 48)

(1, 48) 40 20

0

1

2

3

4

5

t

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EXAMPLE 8



Calculating Average Rates of Change For the projectile function h1t2  16t2  64t, find a. the average rate of change for t  30.5, 1 4 b. the average rate of change for t  31, 1.5 4 . Then graph the secant lines representing these average rates of change and comment.

Solution



Using the given intervals in the formula a.

h112  h10.52 ¢h  ¢t 1  10.52 48  28  0.5  40

b.

h1t2 2  h1t1 2 ¢h  yields ¢t t2  t1

h11.52  h112 ¢h  ¢t 1.5  1 60  48  0.5  24

For t  30.5, 1 4 , the average rate of change is meaning the height of the ball is increasing at an average rate of 40 ft/sec. For t  31, 1.5 4 , the average rate of change has slowed to 24 1 , and the soccer ball’s height is increasing at only 24 ft/sec. The secant lines representing these rates of change are shown in the figure, where we note the line from the first interval (in red), has a steeper slope than the line from the second interval (in blue). 40 1,

h(t)

80 60

(1.5, 60) (1, 48)

40

(0.5, 28) 20

(4, 0)

(0, 0) 0

1

2

3

4

Now try Exercises 59 through 64

5

t 

¢y for ¢x each new interval. Using a slightly different approach, we can develop a general formula for the average rate of change. This is done by selecting a point x1  x from the domain, then a point x2  x  h that is very close to x. Here, h  0 is assumed to be a small, arbitrary constant, meaning the interval [x, x  h] is very small as well. Substituting x  h for x2 and x for x1 in the rate of change formula gives f 1x  h2  f 1x2 f 1x  h2  f 1x2 ¢y . The result is called the difference quotient   ¢x 1x  h2  x h and represents the average rate of change between x and x  h, or equivalently, the slope of the secant line for this interval. The approach in Example 8 works very well, but requires us to recalculate

The Difference Quotient For a function f (x) and constant h  0, f 1x  h2  f 1x2 h is the difference quotient for f.

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Note the formula has three parts: (1) the function f evaluated at x  h S f 1x  h2 , (2) the function f itself, and (3) the constant h. For convenience, the expression f 1x  h2 can be evaluated and simplified prior to its use in the difference quotient. (1)

(2)

f 1x  h2  f 1x2 h (3)

EXAMPLE 9



Computing a Difference Quotient and Average Rates of Change For a. b. c.

Solution



f 1x2  x2  4x, Compute the difference quotient. Find the average rate of change in the intervals [1.9, 2.0] and [3.6, 3.7]. Sketch the graph of f along with the secant lines and comment on what you notice.

a. For f 1x2  x2  4x, f 1x  h2  1x  h2 2  41x  h2  x2  2xh  h2  4x  4h Using this result in the difference quotient yields,

f 1x  h2  f 1x2 1x2  2xh  h2  4x  4h2  1x2  4x2  h h 2 2 x  2xh  h  4x  4h  x2  4x  h 2 2xh  h  4h  h h12x  h  42  h  2x  4  h b. For the interval [1.9, 2.0], x  1.9 and h  0.1. The slope of the secant line is ¢y  211.92  4  0.1  0.1. For the ¢x 5 interval [3.6, 3.7], x  3.6 and h  0.1. The slope of this secant line is ¢y  213.62  4  0.1  3.3. ¢x 4 c. After sketching the graph of f and the secant lines from each interval (see the figure), we note the slope of the first line (in red) is negative and very near zero, while the slope of 5 the second (in blue) is positive and very steep.

substitute into the difference quotient

eliminate parentheses

combine like terms

factor out h result

y

6

Now try Exercises 65 through 76

x



You might be familiar with Galileo Galilei and his studies of gravity. According to popular history, he demonstrated that unequal weights will fall equal distances in equal time periods, by dropping cannonballs from the upper floors of the Leaning Tower of Pisa. Neglecting air resistance, this distance an object falls is modeled by the function d1t2  16t2, where d(t) represents the distance fallen after t sec. Due to the effects of gravity, the velocity of the object increases as it falls. In other words, the ¢distance velocity or the average rate of change is a nonconstant (increasing) rate of ¢time change. We can analyze this rate of change using the difference quotient.

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EXAMPLE 10



Applying the Difference Quotient in Context A construction worker drops a heavy wrench from atop the girder of new skyscraper. Use the function d1t2  16t2 to a. Compute the distance the wrench has fallen after 2 sec and after 7 sec. b. Find a formula for the velocity of the wrench (average rate of change in distance per unit time). c. Use the formula to find the rate of change in the intervals [2, 2.01] and [7, 7.01]. d. Graph the function and the secant lines representing the average rate of change. Comment on what you notice.

Solution



a. Substituting t  2 and t  7 in the given function yields d122  16122 2  16142  64

d172  16172 2  161492  784

evaluate d 1t 2  16t 2 square input multiply

After 2 sec, the wrench has fallen 64 ft; after 7 sec, the wrench has fallen 784 ft. b. For d1t2  16t2, d1t  h2  161t  h2 2, which we compute separately. d1t  h2  161t  h2 2  161t2  2th  h2 2  16t2  32th  16h2

substitute t  h for t square binomial distribute 16

Using this result in the difference quotient yields

116t2  32th  16h2 2  16t2 d1t  h2  d1t2  h h 2 16t  32th  16h2  16t2  h 2 32th  16h  h h132t  16h2  h  32t  16h

substitute into the difference quotient

eliminate parentheses

combine like terms

factor out h and simplify result

For any number of seconds t and h a small increment of time thereafter, the 32t  16h distance  velocity of the wrench is modeled by . time 1 c. For the interval 3 t, t  h 4  32, 2.01 4, t  2 and h  0.01: 32122  1610.012 ¢distance  ¢time 1  64  0.16  64.16

substitute 2 for t and 0.01 for h

Two seconds after being dropped, the velocity of the wrench is approximately 64.16 ft/sec. For the interval 3 t, t  h4  37, 7.01 4 , t  7 and h  0.01: 32172  1610.012 ¢distance  ¢time 1  224  0.16  224.16

substitute 7 for t and 0.01 for h

Seven seconds after being dropped, the velocity of the wrench is approximately 224.16 ft/sec (about 153 mph).

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d.

149

y 1000

Distance fallen (ft)

800

600

400

200

E. You’ve learned how to develop a formula to calculate rates of change for any function

0 1

2

3

4

5

6

7

8

9

10

x

Time in seconds

The velocity increases with time, as indicated by the steepness of each secant line. Now try Exercises 77 and 78



TECHNOLOGY HIGHLIGHT

Locating Zeroes, Maximums, and Minimums Figure 2.51

Figure 2.52 10

▲

Graphically, the zeroes of a function appear as x-intercepts with coordinates (x, 0). An estimate for these zeroes can easily be found using a graphing calculator. To illustrate, enter the function y  x2  8x  9 on the Y = screen and graph it using the standard window ( ZOOM 6). We access the option for finding zeroes by pressing 2nd TRACE (CALC), which displays the screen shown in Figure 2.51. Pressing the number “2” selects 2:zero and returns you to the graph, where you’re asked to enter a “Left Bound.” The calculator is asking you to narrow the area it has to search. Select any number conveniently to the left of the x-intercept you’re interested in. For this graph, we entered a left bound of “0” (press ENTER ). The calculator marks this choice with a “ ” marker (pointing to the right), then asks you to enter a “Right Bound.” Select any value to the right of the x-intercept, but be sure the value you enter bounds 10 only one intercept (see Figure 2.52). For this graph, a choice of 10 would include both x-intercepts, while a choice of 3 would bound only the intercept on the left. After entering 3, the calculator asks for a “Guess.” This option is used when there is more than one zero in the interval, and most of the time we’ll bypass this option by pressing ENTER again. The calculator then finds the zero in the selected interval (if it exists), with the coordinates displayed at the bottom of the screen (Figure 2.53). The maximum and minimum values of a function are located in the same way. Enter y  x3  3x  2 on the Y = screen and 10 graph the function. As seen in Figure 2.54, it appears a local maximum occurs near x  1. To check, we access the CALC 4:maximum option, which returns you to the graph and asks you for a Left Bound, a Right Bound, and a Guess as before. After entering a left bound of “3” and a right bound of “0,” and

10

10

Figure 2.53 10

10

10

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Figure 2.54

Figure 2.55

5

5

4

4

4

4

5

▲

▲

5

bypassing the Guess option (note the “ ” and “ ” markers), the calculator locates the maximum you selected, and again displays the coordinates. Due to the algorithm used by the calculator to find these values, a decimal number is sometimes displayed, even if the actual value is an integer (see Figure 2.55). Use a calculator to find all zeroes and to locate the local maximum and minimum values. Round to the nearest hundredth as needed. Exercise 1: y  2x2  4x  5

Exercise 2: y  w3  3w  1

Exercise 3: y  x2  8x  9

Exercise 4: y  x3  2x2  4x  8

Exercise 5: y  x4  5x2  2x

Exercise 6: y  x1x  4

2.5 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. The graph of a polynomial will cross through the x-axis at zeroes of factors of degree 1, and off the x-axis at the zeroes from linear factors of degree 2.

2. If f 1x2  f 1x2 for all x in the domain, we say that f is an function and symmetric to the axis. If f 1x2  f 1x2 , the function is and symmetric to the .

3. If f 1x2 2 7 f 1x1 2 for x1 6 x2 for all x in a given interval, the function is in the interval. 

4. If f 1c2  f 1x2 for all x in a specified interval, we say that f (c) is a local for this interval. 5. Discuss/Explain the following statement and give an example of the conclusion it makes. “If a function f is decreasing to the left of (c, f (c)) and increasing to the right of (c, f (c)), then f (c) is either a local or a global minimum.” 6. Without referring to notes or textbook, list as many features/attributes as you can that are related to analyzing the graph of a function. Include details on how to locate or determine each attribute.

DEVELOPING YOUR SKILLS

The following functions are known to be even. Complete each graph using symmetry.

7.

8.

y 5

5

5 x

5

y 10

10 x

10

10

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Determine whether the following functions are even: f 1k2  f 1k2 .

27. f 1x2  x4  2x2  1; f 1x2 7 0 y

9. f 1x2  7x  3x  5 10. p1x2  2x  6x  1 2

5

4

1 1 11. g1x2  x4  5x2  1 12. q1x2  2  x 3 x

5

The following functions are known to be odd. Complete each graph using symmetry.

13.

14.

y 10

5 x

5

28. f 1x2  x3  2x2  4x  8; f 1x2  0

y 10

y

1 5 10

10 x

10

5 x

10 x 5

10

10

Determine whether the following functions are odd: f 1k2  f 1k2 . 3 15. f 1x2  41 xx

1 16. g1x2  x3  6x 2

17. p1x2  3x3  5x2  1

18. q1x2 

1 x x

3 29. p1x2  1 x  1  1; p1x2  0 y 5

5

Determine whether the following functions are even, odd, or neither.

19. w1x2  x3  x2

3 20. q1x2  x2  3x 4

1 3 21. p1x2  2 1x  x3 4

22. g1x2  x3  7x

23. v1x2  x3  3x

24. f 1x2  x4  7x2  30

Use the graphs given to solve the inequalities indicated. Write all answers in interval notation.

25. f 1x2  x  3x  x  3; f 1x2  0 3

2

5 x

p(x)

5

30. q1x2  1x  1  2; q1x2 7 0 y 5

q(x) 5

5 x

5

31. f 1x2  1x  12 3  1; f 1x2  0 y

5

y

5

5 5

f(x)

5 x

5 x

5 5

26. f 1x2  x3  2x2  4x  8; f 1x2 7 0

32. g1x2  1x  12 3  1; g1x2 6 0 y 5

y

5

5

5 x

g(x) 5

5 1

5 x

151

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Name the interval(s) where the following functions are increasing, decreasing, or constant. Write answers using interval notation. Assume all endpoints have integer values.

33. y  V1x2

34. y  H1x2 y

y 5

10

10

10 x

5

5 x

H(x)

For Exercises 41 through 48, determine the following (answer in interval notation as appropriate): (a) domain and range of the function; (b) zeroes of the function; (c) interval(s) where the function is greater than or equal to zero, or less than or equal to zero; (d) interval(s) where the function is increasing, decreasing, or constant; and (e) location of any local max or min value(s).

5

y (2, 5)

y

5

5

10

(1, 0)

35. y  f 1x2

42. y  f 1x2

41. y  H1x2

36. y  g1x2

(3.5, 0)

(3, 0)

5

5

5 x

5 x

y

y

10 5 (0, 5)

10

f(x)

8

43. y  g1x2

g(x)

6

10

5

10 x

44. y  h1x2

y

4

y 5

2

10

2

4

6

8

5

x

10

5

For Exercises 37 through 40, determine (a) interval(s) where the function is increasing, decreasing or constant, and (b) comment on the end behavior.

37. p1x2  0.51x  22 3

3 38. q1x2   2 x1

y

2

5

5

45. y  Y1

46. y  Y2 y

5

y

5

(0, 4)

(2, 0)

x

2

y

5

5 x

g(x)

5

(1, 0)

5

5 x

5

5

39. y  f 1x2

5 x

(0, 1)

5

5

5 x

5

5

47. p1x2  1x  32 3  1

40. y  g1x2 y

y

5

5 x

48. q1x2  x  5  3 y

y

10

10

5

10 8

10

5 3



10 x

5 x

6

10

10 x

4 2

10

10

2

4

6

8

10

x

WORKING WITH FORMULAS

49. Conic sections—hyperbola: y  13 24x2  36 While the conic sections are not covered in detail until later in the course, we’ve already developed a number of tools that will help us understand these relations and their graphs. The equation here gives the “upper branches” of a hyperbola, as shown in the figure. Find the following by analyzing the

y

equation: (a) the domain and range; (b) the zeroes of the relation; (c) interval(s) where y is increasing or decreasing; and (d) whether the relation is even, odd, or neither.

5

f(x) 5

5 x

5

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Section 2.5 Analyzing the Graph of a Function

50. Trigonometric graphs: y  sin1x2 and y  cos1x2 The trigonometric functions are also studied at some future time, but we can apply the same tools to analyze the graphs of these functions as well. The graphs of y  sin x and y  cos x are given, graphed over the interval x  3 180, 3604 degrees. Use them to find (a) the range of the functions; (b) the zeroes of the functions; (c) interval(s) where

y is increasing/decreasing; (d) location of minimum/maximum values; and (e) whether each relation is even, odd, or neither. y

y

(90, 1)

1

1

y  cos x

y  sin x

(90, 0) 90

90

180

270

90

360 x

1



90

180

270

360 x

1

APPLICATIONS

Height (feet)

51. Catapults and projectiles: Catapults have a long and interesting history that dates back to ancient times, when they were used to launch javelins, rocks, and other projectiles. The diagram given illustrates the path of the projectile after release, which follows a parabolic arc. Use the graph to determine the following: 80 70 60 50 40 30

20

60

100

140

180

220

260

Distance (feet)

a. State the domain and range of the projectile. b. What is the maximum height of the projectile? c. How far from the catapult did the projectile reach its maximum height? d. Did the projectile clear the castle wall, which was 40 ft high and 210 ft away? e. On what interval was the height of the projectile increasing? f. On what interval was the height of the projectile decreasing? P (millions of dollars)

52. Profit and loss: The profit of DeBartolo Construction Inc. is illustrated by the graph shown. Use the graph to t (years since 1990) estimate the point(s) or the interval(s) for which the profit P was: a. increasing b. decreasing c. constant d. a maximum 16 12 8 4 0 4 8

1 2 3 4 5 6 7 8 9 10

e. f. g. h.

a minimum positive negative zero

53. Functions and rational exponents: The graph of 2 f 1x2  x3  1 is shown. Use the graph to find: a. domain and range of the function b. zeroes of the function c. interval(s) where f 1x2  0 or f 1x2 6 0 d. interval(s) where f (x) is increasing, decreasing, or constant e. location of any max or min value(s) Exercise 53

Exercise 54

y

y

5

5

(1, 0) (1, 0) 5

(0, 1)

5

(3, 0) 5 x

(3, 0) (0, 1)

5

5 x

5

54. Analyzing a graph: Given h1x2  x2  4  5, whose graph is shown, use the graph to find: a. domain and range of the function b. zeroes of the function c. interval(s) where h1x2  0 or h1x2  0 d. interval(s) where f(x) is increasing, decreasing, or constant e. location of any max or min value(s)

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c. location of the maximum and minimum values d. the one-year period with the greatest rate of increase and the one-year period with the greatest rate of decrease

I(t)  rate of interest (%) for years 1972 to 1996

55. Analyzing interest rates: The graph shown approximates the average annual interest rates on 30-yr fixed mortgages, rounded to the nearest 14 % . Use the graph to estimate the following (write all answers in interval notation). a. domain and range b. interval(s) where I(t) is increasing, decreasing, or constant

Source: 1998 Wall Street Journal Almanac, p. 446; 2004 Statistical Abstract of the United States, Table 1178

16 15 14 13 12 11 10 9 8 7

t

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

Year (1972 → 72)

D(t): Federal Deficit (in billions)

56. Analyzing the deficit: The following graph approximates the federal deficit of the United States. Use the graph to estimate the following (write answers in interval notation). a. the domain and range b. interval(s) where D(t) is increasing, decreasing, or constant

c. the location of the maximum and minimum values d. the one-year period with the greatest rate of increase, and the one-year period with the greatest rate of decrease Source: 2005 Statistical Abstract of the United States, Table 461

240 160 80 0 80 160 240 320 400

t

75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102

Year (1975 → 75)

57. Constructing a graph: Draw the function f that has the following characteristics, then state the zeroes and the location of all maximum and minimum values. [Hint: Write them as (c, f(c)).] a. Domain: x  110, q 2 b. Range: y  16, q2 c. f 102  0; f 142  0 d. f 1x2c for x  110, 62 ´ 12, 22 ´ 14, q 2 e. f 1x2T for x  16, 22 ´ 12, 42 f. f 1x2  0 for x  3 8, 4 4 ´ 3 0, q 2 g. f 1x2 6 0 for x  1q, 82 ´ 14, 02

58. Constructing a graph: Draw the function g that has the following characteristics, then state the zeroes and the location of all maximum and minimum values. [Hint: Write them as (c, g(c)).] a. Domain: x  1q, 82 b. Range: y  36, q 2 c. g102  4.5; g162  0 d. g1x2c for x  16, 32 ´ 16, 82 e. g1x2T for x  1q, 62 ´ 13, 62 f. g1x2  0 for x  1q, 9 4 ´ 33, 82 g. g1x2 6 0 for x  19, 32

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Section 2.5 Analyzing the Graph of a Function

For Exercises 59 to 64, use the formula for the average f 1x2 2  f 1x1 2 rate of change . x2  x1

height of the rocket after t sec (assume the rocket was shot from ground level). a. Find the rocket’s height at t  1 and t  2 sec. b. Find the rocket’s height at t  3 sec. c. Would you expect the average rate of change to be greater between t  1 and t  2, or between t  2 and t  3? Why? d. Calculate each rate of change and discuss your answer.

59. Average rate of change: For f 1x2  x3, (a) calculate the average rate of change for the interval x  2 and x  1 and (b) calculate the average rate of change for the interval x  1 and x  2. (c) What do you notice about the answers from parts (a) and (b)? (d) Sketch the graph of this function along with the lines representing these average rates of change and comment on what you notice. 60. Average rate of change: Knowing the general 3 shape of the graph for f 1x2  1x, (a) is the average rate of change greater between x  0 and x  1 or between x  7 and x  8? Why? (b) Calculate the rate of change for these intervals and verify your response. (c) Approximately how many times greater is the rate of change? 61. Height of an arrow: If an arrow is shot vertically from a bow with an initial speed of 192 ft/sec, the height of the arrow can be modeled by the function h1t2  16t2  192t, where h(t) represents the height of the arrow after t sec (assume the arrow was shot from ground level).

a. What is the arrow’s height at t  1 sec? b. What is the arrow’s height at t  2 sec? c. What is the average rate of change from t  1 to t  2? d. What is the rate of change from t  10 to t  11? Why is it the same as (c) except for the sign? 62. Height of a water rocket: Although they have been around for decades, water rockets continue to be a popular toy. A plastic rocket is filled with water and then pressurized using a handheld pump. The rocket is then released and off it goes! If the rocket has an initial velocity of 96 ft/sec, the height of the rocket can be modeled by the function h1t2  16t2  96t, where h(t) represents the

155

63. Velocity of a falling object: The impact velocity of an object dropped from a height is modeled by v  12gs, where v is the velocity in feet per second (ignoring air resistance), g is the acceleration due to gravity (32 ft/sec2 near the Earth’s surface), and s is the height from which the object is dropped. a. Find the velocity at s  5 ft and s  10 ft. b. Find the velocity at s  15 ft and s  20 ft. c. Would you expect the average rate of change to be greater between s  5 and s  10, or between s  15 and s  20? d. Calculate each rate of change and discuss your answer. 64. Temperature drop: One day in November, the town of Coldwater was hit by a sudden winter storm that caused temperatures to plummet. During the storm, the temperature T (in degrees Fahrenheit) could be modeled by the function T1h2  0.8h2  16h  60, where h is the number of hours since the storm began. Graph the function and use this information to answer the following questions. a. What was the temperature as the storm began? b. How many hours until the temperature dropped below zero degrees? c. How many hours did the temperature remain below zero? d. What was the coldest temperature recorded during this storm? Compute and simplify the difference quotient f 1x  h2  f 1x2 for each function given. h

65. f 1x2  2x  3

66. g1x2  4x  1

67. h1x2  x  3

68. p1x2  x2  2

69. q1x2  x2  2x  3

70. r1x2  x2  5x  2

71. f 1x2 

72. g1x2 

2

2 x

3 x

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day, the distance is approximated by the function d1h2  1.5 1h, where d(h) represents the viewing distance (in miles) at height h (in feet). Find the average rate of change in the intervals (a) [9, 9.01] and (b) [225, 225.01]. Then (c) graph the function along with the lines representing the average rates of change and comment on what you notice.

Use the difference quotient to find: (a) a rate of change formula for the functions given and (b)/(c) calculate the rate of change in the intervals shown. Then (d) sketch the graph of each function along with the secant lines and comment on what you notice.

73. g1x2  x2  2x 74. h1x2  x2  6x [3.0, 2.9], [0.50, 0.51] [1.9, 2.0], [5.0, 5.01]

78. A special magnifying lens is crafted and installed in an overhead projector. When the projector is x ft from the screen, the size P(x) of the projected image is x2. Find the average rate of change for P1x2  x2 in the intervals (a) [1, 1.01] and (b) [4, 4.01]. Then (c) graph the function along with the lines representing the average rates of change and comment on what you notice.

75. g1x2  x3  1 [2.1, 2], [0.40, 0.41] 76. r1x2  1x (Hint: Rationalize the numerator.) [1, 1.1], [4, 4.1] 77. The distance that a person can see depends on how high they’re standing above level ground. On a clear 

EXTENDING THE THOUGHT

79. Does the function shown have a maximum value? Does it have a minimum value? Discuss/explain/justify why or why not.

c. By approximately how many seconds? d. Who was leading at t  40 seconds? e. During the race, how many seconds was the daughter in the lead? f. During the race, how many seconds was the mother in the lead?

y 5

5

5 x

5

81. Draw a general function f (x) that has a local maximum at (a, f (a)) and a local minimum at (b, f (b)) but with f 1a2 6 f 1b2 .

80. The graph drawn here depicts a 400-m race between a mother and her daughter. Analyze the graph to answer questions (a) through (f). a. Who wins the race, the mother or daughter? b. By approximately how many meters? Mother

2

82. Verify that h1x2  x3 is an even function, by first 1 rewriting h as h1x2  1x3 2 2.

Daughter

Distance (meters)

400 300 200 100

10

20

30

40

50

60

70

80

Time (seconds) 

MAINTAINING YOUR SKILLS 86. (R.7) Find the surface area and volume of the cylinder shown.

83. (1.5) Solve the given quadratic equation three different ways: (a) factoring, (b) completing the square, and (c) using the quadratic formula: x2  8x  20  0 y

36 cm 12 cm

5

84. (R.5) Find the (a) sum and (b) product of the rational 3 3 expressions and . x2 2x 85. (2.3) Write the equation of the line shown, in the form y  mx  b.

5

5 x

5

Exercise 85

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Precalculus—

2.6 The Toolbox Functions and Transformations Learning Objectives In Section 2.6 you will learn how to:

A. Identify basic characteristics of the toolbox functions

B. Perform vertical/ horizontal shifts of a basic graph

C. Perform vertical/ horizontal reflections of a basic graph

D. Perform vertical stretches and compressions of a basic graph

E. Perform transformations on a general function f(x)

Many applications of mathematics require that we select a function known to fit the context, or build a function model from the information supplied. So far we’ve looked extensively at linear functions, and have introduced the absolute value, squaring, square root, cubing, and cube root functions. These are the six toolbox functions, so called because they give us a variety of “tools” to model the real world. In the same way a study of arithmetic depends heavily on the multiplication table, a study of algebra and mathematical modeling depends (in large part) on a solid working knowledge of these functions.

A. The Toolbox Functions While we can accurately graph a line using only two points, most toolbox functions require more points to show all of the graph’s important features. However, our work is greatly simplified in that each function belongs to a function family, in which all graphs from a given family share the characteristics of one basic graph, called the parent function. This means the number of points required for graphing will quickly decrease as we start anticipating what the graph of a given function should look like. The parent functions and their identifying characteristics are summarized here.

The Toolbox Functions Identity function

Absolute value function y

y 5

5

x

f(x)  x

3

3

2

2

1

1

f(x)  x 5

5

x

x

f(x)  |x|

3

3

2

2

1

1

0

0

0

0

1

1

1

1

2

2

3

3

2

2

3

3

5

Domain: x 僆 (q, q), Range: y 僆 (q, q) Symmetry: odd Increasing: x 僆 (q, q) End behavior: down on the left/up on the right

Squaring function

5

Domain: x 僆 (q, q), Range: y 僆 [0, q) Symmetry: even Decreasing: x 僆 (q, 0); Increasing: x 僆 (0, q ) End behavior: up on the left/up on the right Vertex at (0, 0)

Square root function y

y

5

5

x

f(x)  x2

x

f(x)  1x

3

9

2

–

2

4

1

1

0

0

1

1

1

1

2

 1.41

2

4

3

 1.73

3

9

4

2

2-75

x

5

x

Domain: x 僆 (q, q), Range: y 僆 [0, q) Symmetry: even Decreasing: x 僆 (q, 0); Increasing: x 僆 (0, q) End behavior: up on the left/up on the right Vertex at (0, 0)

1

–

0

0

5

x

Domain: x 僆 [0, q), Range: y 僆 [0, q) Symmetry: neither even nor odd Increasing: x 僆 (0, q) End behavior: up on the right Initial point at (0, 0)

157

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Cubing function x

f(x)  x3

3

27

2

8

1

1

0

0

1

Cube root function y

y

x

3 f(x)  2 x

27

3

8

2

1

1

0

0

1

1

1

2

8

8

2

3

27

27

3

10

5

x

Domain: x 僆 (q, q), Range: y 僆 (q, q) Symmetry: odd Increasing: x 僆 (q, q) End behavior: down on the left/up on the right Point of inflection at (0, 0)

5

f(x)  3 x

10

10

x

5

Domain: x 僆 (q, q), Range: y 僆 (q, q) Symmetry: odd Increasing: x 僆 (q, q) End behavior: down on the left/up on the right Point of inflection at (0, 0)

In applications of the toolbox functions, the parent graph may be altered and/or shifted from its original position, yet the graph will still retain its basic shape and features. The result is called a transformation of the parent graph. Analyzing the new graph (as in Section 2.5) will often provide the answers needed. EXAMPLE 1

Solution





Identifying the Characteristics of a Transformed Graph The graph of f 1x2  x2  2x  3 is given. Use the graph to identify each of the features or characteristics indicated. a. function family b. domain and range c. vertex d. max or min value(s) e. end behavior f. x- and y-intercept(s)

y 5

5

5

x

5

a. The graph is a parabola, from the squaring function family. b. domain: x  1q, q 2 ; range: y  34, q 2 c. vertex: (1, 4) d. minimum value y  4 at (1, 4) e. end-behavior: up/up f. y-intercept: (0, 3); x-intercepts: (1, 0) and (3, 0) Now try Exercises 7 through 34

A. You’ve just learned how to identify basic characteristics of the toolbox functions



Note that we can algebraically verify the x-intercepts by substituting 0 for f(x) and solving the equation by factoring. This gives 0  1x  121x  32 , with solutions x  1 and x  3. It’s also worth noting that while the parabola is no longer symmetric to the y-axis, it is symmetric to the vertical line x  1. This line is called the axis of symmetry for the parabola, and will always be a vertical line that goes through the vertex.

B. Vertical and Horizontal Shifts As we study specific transformations of a graph, try to develop a global view as the transformations can be applied to any function. When these are applied to the toolbox

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Section 2.6 The Toolbox Functions and Transformations

functions, we rely on characteristic features of the parent function to assist in completing the transformed graph.

Vertical Translations We’ll first investigate vertical translations or vertical shifts of the toolbox functions, using the absolute value function to illustrate. EXAMPLE 2



Solution



Graphing Vertical Translations

Construct a table of values for f 1x2  x, g1x2  x  1, and h1x2  x  3 and graph the functions on the same coordinate grid. Then discuss what you observe. A table of values for all three functions is given, with the corresponding graphs shown in the figure. x

f (x)  |x|

g(x)  |x|  1

h(x)  |x|  3

3

3

4

0

2

2

3

1

1

1

2

2

0

0

1

3

1

1

2

2

2

2

3

1

3

3

4

0

(3, 4)5

y g(x)  x  1

(3, 3) (3, 0)

1

f(x)  x

5

5

x

h(x)  x  3 5

Note that outputs of g(x) are one more than the outputs for f (x), and that each point on the graph of f has been shifted upward 1 unit to form the graph of g. Similarly, each point on the graph of f has been shifted downward 3 units to form the graph of h. Since h1x2  f 1x2  3. Now try Exercises 35 through 42



We describe the transformations in Example 2 as a vertical shift or vertical translation of a basic graph. The graph of g is the graph of f shifted up 1 unit, and the graph of h is the graph of f shifted down 3 units. In general, we have the following: Vertical Translations of a Basic Graph Given k 7 0 and any function whose graph is determined by y  f 1x2 , 1. The graph of y  f 1x2  k is the graph of f(x) shifted upward k units. 2. The graph of y  f 1x2  k is the graph of f(x) shifted downward k units.

Horizontal Translations The graph of a parent function can also be shifted left or right. This happens when we alter the inputs to the basic function, as opposed to adding or subtracting something to the basic function itself. For Y1  x2  2 note that we first square inputs, then add 2, which results in a vertical shift. For Y2  1x  22 2, we add 2 to x prior to squaring and since the input values are affected, we might anticipate the graph will shift along the x-axis—horizontally.

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EXAMPLE 3



Graphing Horizontal Translations

Solution



Both f and g belong to the quadratic family and their graphs are parabolas. A table of values is shown along with the corresponding graphs.

Construct a table of values for f 1x2  x2 and g1x2  1x  22 2, then graph the functions on the same grid and discuss what you observe.

f (x )  x2

x

y

g(x)  (x  2)2

3

9

1

2

4

0

1

1

1

0

0

4

1

1

9

2

4

16

3

9

25

9 8

(3, 9)

(1, 9)

7

f(x)  x2

6 5

(0, 4)

4

(2, 4)

3

g(x)  (x  2)2

2 1

5 4 3 2 1 1

1

2

3

4

5

x

It is apparent the graphs of g and f are identical, but the graph of g has been shifted horizontally 2 units left. Now try Exercises 43 through 46



We describe the transformation in Example 3 as a horizontal shift or horizontal translation of a basic graph. The graph of g is the graph of f, shifted 2 units to the left. Once again it seems reasonable that since input values were altered, the shift must be horizontal rather than vertical. From this example, we also learn the direction of the shift is opposite the sign: y  1x  22 2 is 2 units to the left of y  x2. Although it may seem counterintuitive, the shift opposite the sign can be “seen” by locating the new x-intercept, which in this case is also the vertex. Substituting 0 for y gives 0  1x  22 2 with x  2, as shown in the graph. In general, we have Horizontal Translations of a Basic Graph Given h 7 0 and any function whose graph is determined by y  f 1x2 , 1. The graph of y  f 1x  h2 is the graph of f(x) shifted to the left h units. 2. The graph of y  f 1x  h2 is the graph of f(x) shifted to the right h units. EXAMPLE 4



Graphing Horizontal Translations Sketch the graphs of g1x2  x  2 and h1x2  1x  3 using a horizontal shift of the parent function and a few characteristic points (not a table of values).

Solution



The graph of g1x2  x  2 (Figure 2.56) is the absolute value function shifted 2 units to the right (shift the vertex and two other points from y  x 2 . The graph of h1x2  1x  3 (Figure 2.57) is a square root function, shifted 3 units to the left (shift the initial point and one or two points from y  1x).

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Figure 2.56 5

Figure 2.57

y g(x)  x  2

y h(x)  x  3

(1, 3)

5

(6, 3)

(5, 3) 5

Vertex

(2, 0)

5

(1, 2)

x 4

5

(3, 0)

x

B. You’ve just learned how to perform vertical/horizontal shifts of a basic graph

Now try Exercises 47 through 50



C. Vertical and Horizontal Reflections The next transformation we investigate is called a vertical reflection, in which we compare the function Y1  f 1x2 with the negative of the function: Y2  f 1x2 .

Vertical Reflections EXAMPLE 5



Graphing Vertical Reflections Construct a table of values for Y1  x2 and Y2  x2, then graph the functions on the same grid and discuss what you observe.

Solution



A table of values is given for both functions, along with the corresponding graphs. y 5

x

Y1  x2

Y2  x2

2

4

4

1

1

1

0

0

0

1

1

1

2

4

4

Y1  x2

(2, 4)

5 4 3 2 1

Y2  x2

1

2

3

4

5

x

(2, 4) 5

As you might have anticipated, the outputs for f and g differ only in sign. Each output is a reflection of the other, being an equal distance from the x-axis but on opposite sides. Now try Exercises 51 and 52



The vertical reflection in Example 5 is called a reflection across the x-axis. In general, Vertical Reflections of a Basic Graph For any function y  f 1x2 , the graph of y  f 1x2 is the graph of f(x) reflected across the x-axis.

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Horizontal Reflections It’s also possible for a graph to be reflected horizontally across the y-axis. Just as we noted that f (x) versus f 1x2 resulted in a vertical reflection, f (x) versus f 1x2 results in a horizontal reflection. EXAMPLE 6



Graphing a Horizontal Reflection

Solution



A table of values is given here, along with the corresponding graphs.

Construct a table of values for f 1x2  1x and g1x2  1x, then graph the functions on the same coordinate grid and discuss what you observe.

x

f 1x2  1x

g1x2  1x

4

not real

2

2

not real

12  1.41

1

not real

1

0

0

0

1

1

not real

2

12  1.41

not real

4

2

not real

y (4, 2)

(4, 2)

2

g(x)  兹x

f(x)  兹x

1

5 4 3 2 1

1

2

3

4

5

x

1 2

The graph of g is the same as the graph of f, but it has been reflected across the y-axis. A study of the domain shows why— f represents a real number only for nonnegative inputs, so its graph occurs to the right of the y-axis, while g represents a real number for nonpositive inputs, so its graph occurs to the left. Now try Exercises 53 and 54



The transformation in Example 6 is called a horizontal reflection of a basic graph. In general, Horizontal Reflections of a Basic Graph C. You’ve just learned how to perform vertical/horizontal reflections of a basic graph

For any function y  f 1x2 , the graph of y  f 1x2 is the graph of f (x) reflected across the y-axis.

D. Vertically Stretching/Compressing a Basic Graph As the words “stretching” and “compressing” imply, the graph of a basic function can also become elongated or flattened after certain transformations are applied. However, even these transformations preserve the key characteristics of the graph. EXAMPLE 7



Stretching and Compressing a Basic Graph

Construct a table of values for f 1x2  x2, g1x2  3x2, and h1x2  13x2, then graph the functions on the same grid and discuss what you observe.

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Solution



A table of values is given for all three functions, along with the corresponding graphs.

x

f (x)  x 2

g(x)  3x 2

h(x)  13 x 2

3

9

27

3

2

4

12

4 3

1

1

3

1 3

0

0

0

0

1

1

3

1 3

2

4

12

4 3

3

9

27

3

y g(x)  3x2

(2, 12)

(2, 4)

f(x)  x2

10

h(x)  ax2 (2, d) 5 4 3 2 1

1

2

3

4

5

x

4

The outputs of g are triple those of f, making these outputs farther from the x-axis and stretching g upward (making the graph more narrow). The outputs of h are one-third those of f, and the graph of h is compressed downward, with its outputs closer to the x-axis (making the graph wider).

WORTHY OF NOTE In a study of trigonometry, you’ll find that a basic graph can also be stretched or compressed horizontally, a phenomenon known as frequency variations.

Now try Exercises 55 through 62



The transformations in Example 7 are called vertical stretches or compressions of a basic graph. In general, Stretches and Compressions of a Basic Graph

D. You’ve just learned how to perform vertical stretches and compressions of a basic graph

For any function y  f 1x2 , the graph of y  af 1x2 is 1. the graph of f (x) stretched vertically if a 7 1, 2. the graph of f (x) compressed vertically if 0 6 a 6 1.

E. Transformations of a General Function If more than one transformation is applied to a basic graph, it’s helpful to use the following sequence for graphing the new function. General Transformations of a Basic Graph Given a function y  f 1x2 , the graph of y  af 1x  h2  k can be obtained by applying the following sequence of transformations: 1. horizontal shifts 2. reflections 3. stretches or compressions 4. vertical shifts We generally use a few characteristic points to track the transformations involved, then draw the transformed graph through the new location of these points. EXAMPLE 8



Graphing Functions Using Transformations Use transformations of a parent function to sketch the graphs of 3 a. g1x2  1x  22 2  3 b. h1x2  2 1 x21

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Solution



a. The graph of g is a parabola, shifted left 2 units, reflected across the x-axis, and shifted up 3 units. This sequence of transformations in shown in Figures 2.58 through 2.60.

Figure 2.58 y  (x 

Figure 2.59

y

2)2

y  x2

5

(4, 4)

5

Figure 2.60

y y  (x  2)2

5

y g(x)  (x  2)2  3

(2, 3)

(0, 4)

(2, 0) 5

(0, 2) Vertex

5

x

5

5

x

5

(4, 4)

5

5

(0, 4)

x

5

Reflected across the x-axis

Shifted left 2 units

5

(0, 1)

(4, 1)

Shifted up 3

b. The graph of h is a cube root function, shifted right 2, stretched by a factor of 2, then shifted down 1. This sequence is shown in Figures 2.61 through 2.63. Figure 2.61 y 5

Figure 2.63

Figure 2.62

3

y y  2兹x  2 3

y  兹x  2

5

5

3 y h(x)  2兹x 21

(3, 2) (3, 1) (2, 0) Inflection (1, 1)

4

(2, 0) 6

x

4

x

4

(2, 1)

(1, 2)

6

x

(1, 3)

5

5

Shifted right 2

(3, 1) 6

5

Stretched by a factor of 2

Shifted down 1

Now try Exercises 63 through 92

Parent Function quadratic: absolute value: cube root: general:

Transformation of Parent Function y  21x  32 2  1 y  2x  3  1 3 y  21 x31 y  2f 1x  32  1

yx y  x 3 y 1 x y  f 1x2 2

In each case, the transformation involves a horizontal shift right 3, a vertical reflection, a vertical stretch, and a vertical shift up 1. Since the shifts are the same regardless of the initial function, we can generalize the results to any function f(x). General Function

y  af 1x  h2  k vertical reflections vertical stretches and compressions

S

y  f 1x2

Transformed Function S

Since the shape of the initial graph does not change when translations or reflections are applied, these are called rigid transformations. Stretches and compressions of a basic graph are called nonrigid transformations, as the graph is distended in some way.

It’s important to note that the transformations can actually be applied to any function, even those that are new and unfamiliar. Consider the following pattern:

S

WORTHY OF NOTE



horizontal shift h units, opposite direction of sign

vertical shift k units, same direction as sign

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Also bear in mind that the graph will be reflected across the y-axis (horizontally) if x is replaced with x. Use this illustration to complete Exercise 9. Remember—if the graph of a function is shifted, the individual points on the graph are likewise shifted. EXAMPLE 9



Graphing Transformations of a General Function

Solution



For g, the graph of f is (1) shifted horizontally 1 unit left, (2) reflected across the x-axis, and (3) shifted vertically 2 units down. The final result is shown in Figure 2.65.

Given the graph of f (x) shown in Figure 2.64, graph g1x2  f 1x  12  2.

Figure 2.65

Figure 2.64 y

y 5

5

(2, 3)

f (x)

g (x) (1, 1) (0, 0) 5

5

x

5

5

(3, 2) (1, 2)

(5, 2) (2, 3) 5

(3, 5)

x

5

Now try Exercises 93 through 96



Using the general equation y  af 1x  h2  k, we can identify the vertex, initial point, or inflection point of any toolbox function and sketch its graph. Given the graph of a toolbox function, we can likewise identify these points and reconstruct its equation. We first identify the function family and the location (h, k) of the characteristic point. By selecting one other point (x, y) on the graph, we then use the general equation as a formula (substituting h, k, and the x- and y-values of the second point) to solve for a and complete the equation. EXAMPLE 10



Writing the Equation of a Function Given Its Graph Find the equation of the toolbox function f (x) shown in Figure 2.66.

Solution



y 5

The function f belongs to the absolute value family. The vertex (h, k) is at (1, 2). For an additional point, choose the x-intercept (3, 0) and work as follows: y  ax  h  k 0  a 132  1  2

E. You’ve just learned how to perform transformations on a general function f(x)

Figure 2.66

0  4a  2 2  4a 1  a 2

general equation

f(x) 5

x

Now try Exercises 97 through 102



substitute 1 for h and 2 for k, substitute 3 for x and 0 for y

5

simplify subtract 2

5

solve for a

The equation for f is y  12x  1  2.

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TECHNOLOGY HIGHLIGHT

Function Families Graphing calculators are able to display a number of graphs Figure 2.67 simultaneously, making them a wonderful tool for studying families of functions. Let’s begin by entering the function y  |x| [actually y  abs( x) MATH ] as Y1 on the Y = screen. Next, we enter different variations of the function, but always in terms of its variable name “Y1.” This enables us to simply change the basic function, and observe how the changes affect the graph. Recall that to access the function name Y1 press VARS (to access the Y-VARS menu) ENTER (to access the function variables Figure 2.68 menu) and ENTER (to select Y1). Enter the functions Y2  Y1  3 10 and Y3  Y1  6 (see Figure 2.67). Graph all three functions in the ZOOM 6:ZStandard window. The calculator draws each graph in the order they were entered and you can always 10 10 identify the functions by pressing the TRACE key and then the up arrow or down arrow keys. In the upper left corner of the window shown in Figure 2.68, the calculator identifies which function the cursor is currently on. Most 10 importantly, note that all functions in this family maintain the same “V” shape. Next, change Y1 to Y1  abs1x  32 , leaving Y2 and Y3 as is. What do you notice when these are graphed again? Exercise 1: Change Y1 to Y1  1x and graph, then enter Y1  1x  3 and graph once again. What do you observe? What comparisons can be made with the translations of Y1  abs1x2 ?

Exercise 2: Change Y1 to Y1  x2 and graph, then enter Y1  1x  32 2 and graph once again. What do you observe? What comparisons can be made with the translations of Y1  abs1x2 and Y1  1x?

2.6 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. After a vertical , points on the graph are farther from the x-axis. After a vertical , points on the graph are closer to the x-axis.

2. Transformations that change only the location of a graph and not its shape or form, include and .

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3. The vertex of h1x2  31x  52 2  9 is at and the graph opens . 5. Given the graph of a general function f(x), discuss/ explain how the graph of F1x2  2f 1x  12  3 can be obtained. If (0, 5), (6, 7), and 19, 42 are on the graph of f, where do they end up on the graph of F?



167

Section 2.6 The Toolbox Functions and Transformations

4. The inflection point of f 1x2  21x  42 3  11 is at and the end behavior is , . 6. Discuss/Explain why the shift of f 1x2  x2  3 is a vertical shift of 3 units in the positive direction, while the shift of g1x2  1x  32 2 is a horizontal shift 3 units in the negative direction. Include several examples linked to a table of values.

DEVELOPING YOUR SKILLS

By carefully inspecting each graph given, (a) indentify the function family; (b) describe or identify the end behavior, vertex, axis of symmetry, and x- and y-intercepts; and (c) determine the domain and range. Assume required features have integer values.

7. f 1x2  x2  4x

8. g1x2  x2  2x

y

For each graph given, (a) indentify the function family; (b) describe or identify the end behavior, initial point, and x- and y-intercepts; and (c) determine the domain and range. Assume required features have integer values.

13. p1x2  21x  4  2 14. q1x2  2 1x  4  2 y

y

5

y

5

5

5

p(x)

5

5 x

5

5

5 x

5

10. q1x2  x2  2x  8

y

5

10

11. f 1x2  x2  4x  5

10 x

5

5 x

5 x

f(x)

5

17. g1x2  2 14  x

y

18. h1x2  2 1x  1  4

y

10

10

5

r(x)

5

12. g1x2  x2  6x  5

y

y 5

5

10

5

5

y

y

5 x

5 x

q(x)

15. r 1x2  314  x  3 16. f 1x2  21x  1  4

10

5

5

5

5

9. p1x2  x2  2x  3

5 x

y 5

5

g(x) h(x) 10

10 x

10

10

10 x

10

5

5 x

5

5

5 x

5

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For each graph given, (a) indentify the function family; (b) describe or identify the end behavior, vertex, axis of symmetry, and x- and y-intercepts; and (c) determine the domain and range. Assume required features have integer values.

19. p1x2  2x  1  4

27. h1x2  x3  1 y

p(x) h(x) 5

5

5 x

5 x

y

5

5

p(x)

5 x

5

5

q(x)

5

y 5

5

20. q1x2  3x  2  3

y

3 28. p1x2   2 x1

5

3 29. q1x2  2 x11

3 30. r1x2  2 x  11

5 x

y

y 5

5 5

5

21. r1x2  2x  1  6

22. f1x2  3x  2  6

y

5

5

5 x

q(x)

5 x

r(x)

y 4

6

5

5

r(x) 5 5

5 x

5 x

f(x) 6

4

23. g1x2  3x  6

31.

24. h1x2  2x  1

y

For Exercises 31–34, identify and state the characteristic features of each graph, including (as applicable) the function family, domain, range, intercepts, vertex, point of inflection, and end behavior. y

g(x)

y 5

g(x)

5

5

5 x

5 x

h(x)

5 x

5

33.

For each graph given, (a) indentify the function family; (b) describe or identify the end behavior, inflection point, and x- and y-intercepts; and (c) determine the domain and range. Assume required features have integer values. Be sure to note the scaling of each axis.

25. f 1x2  1x  12 3

26. g1x2  1x  12 3

y

5

5

5 x

5 x

g(x)

5

5

5 x

5

g1x2  1x  2,

h1x2  1x  3

37. p1x2  x, q1x2  x  5, r1x2  x  2 38. p1x2  x2,

5

y 5

3 3 3 36. f 1x2  2 x, g1x2  2 x  3, h1x2  2 x1

g(x)

5 x

34.

f(x)

35. f 1x2  1x,

5

f(x)

y 5

Use a table of values to graph the functions given on the same grid. Comment on what you observe.

y

5

5

5

5 x

4

4

5

y 5

6

6

5

32.

f(x)

5

q1x2  x2  4, r1x2  x2  1

Sketch each graph using transformations of a parent function (without a table of values).

39. f 1x2  x3  2 41. h1x2  x2  3

40. g1x2  1x  4 42. Y1  x  3

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c.

Use a table of values to graph the functions given on the same grid. Comment on what you observe.

43. p1x2  x2,

44. f 1x2  1x,

d.

y

q1x2  1x  32 2

y

x

g1x2  1x  4

x

45. Y1  x, Y2  x  1 46. h1x2  x3,

H1x2  1x  22 3

e.

f.

y

y

Sketch each graph using transformations of a parent function (without a table of values).

47. p1x2  1x  32 2

48. Y1  1x  1

51. g1x2  x

52. Y2   1x

x

3 50. f 1x2  1 x2

49. h1x2  x  3 53. f 1x2  2x

54. g1x2  1x2

3

g.

y

x

x

q1x2  2x2, r1x2  12x2

56. f 1x2  1x, g1x2  4 1x, 57. Y1  x, Y2  3x, Y3  58. u1x2  x3,

h.

y

3

Use a table of values to graph the functions given on the same grid. Comment on what you observe.

55. p1x2  x2,

x

v1x2  2x3,

h1x2  14 1x

i.

j.

y

y

1 3 x

w1x2  15x3

x

x

Sketch each graph using transformations of a parent function (without a table of values). 3 59. f 1x2  4 2 x

61. p1x2 

60. g1x2  2x 62. q1x2 

1 3 3x

k.

Use the characteristics of each function family to match a given function to its corresponding graph. The graphs are not scaled—make your selection based on a careful comparison.

63. f 1x2  12x3

x

64. f 1x2  2 3 x  2

65. f 1x2  1x  32  2

66. f 1x2  1x  1  1

67. f 1x2  x  4  1

68. f 1x2   1x  6

69. f 1x2   1x  6  1 70. f 1x2  x  1 71. f 1x2  1x  42  3

72. f 1x2  x  2  5

2

73. f 1x2  1x  3  1 a. y

74. f 1x2  1x  32  5 y b. 2

x

y

x

Graph each function using shifts of a parent function and a few characteristic points. Clearly state and indicate the transformations used and identify the location of all vertices, initial points, and/or inflection points.

3

2

l.

y

3 4 1x

x

75. f 1x2  1x  2  1

76. g1x2  1x  3  2

79. p1x2  1x  32 3  1

80. q1x2  1x  22 3  1

77. h1x2  1x  32 2  2

78. H1x2  1x  22 2  5

3 81. Y1  1 x12

3 82. Y2  1 x31

83. f 1x2  x  3  2

84. g1x2  x  4  2

85. h1x2  21x  12 2  3 86. H1x2  12x  2  3 3 87. p1x2  13 1x  22 3  1 88. q1x2  51 x12

89. Y1  2 1x  1  3 90. Y2  3 1x  2  1 91. h1x2  15 1x  32 2  1

92. H1x2  2 x  3  4

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Apply the transformations indicated for the graph of the general functions given.

93.

y 5

94.

f(x)

y 5

g(x)

97.

(1, 4) (4, 4)

Use the graph given and the points indicated to determine the equation of the function shown using the general form y  af (x  h)  k.

98.

y 5

y (5, 6)

(3, 2)

5

(1, 2) 5

5

5 x

5 x

(4, 2) 5

y 5

(2, 2) 5

a. b. c. d.

g1x2  2 g1x2  3 2g1x  12 1 2 g1x  12  2

96.

h(x)

y 5

a. b. c. d. 

99.

5 x

4

100.

y

(0, 4)

y (4, 5) 5

(6, 4.5)

5

p(x)

r(x) 4

H(x)

3(3, 0)

5

x

(5, 1)

x

5

3

5

5 x

5

101.

102.

y 5

(1, 3)

(2, 4)

h1x2  3 h1x  22 h1x  22  1 1 4 h1x2  5

5 x

(2, 0)

y (3, 7)

7

(1, 4) 5

a. b. c. d.

H1x  32 H1x2  1 2H1x  32 1 3 H1x  22  1

f(x) 8

h(x) 2 x

(4, 0)

3 5

7 x 3

(0, 2)

WORKING WITH FORMULAS

103. Volume of a sphere: V1r2  43r3 The volume of a sphere is given by the function shown, where V(r) is the volume in cubic units and r is the radius. Note this function belongs to the cubic family of functions. Approximate the value of 4 3  to one decimal place, then graph the function on the interval [0, 3]. From your graph, estimate the volume of a sphere with radius 2.5 in. Then compute the actual volume. Are the results close?



5

(2, 0)

(1, 0) 5

5

f(x) (0, 4)

(1, 3)

(4, 4)

5 x

5

a. f 1x  22 b. f 1x2  3 c. 12 f 1x  12 d. f 1x2  1 95.

g(x)

(2, 0) 5

104. Fluid motion: V1h2  41h  20 Suppose the velocity of a fluid flowing from an open tank (no top) through an opening in its side is given by the function shown, where V(h) is the velocity of the fluid (in feet per second) at water height h (in feet). Note this function belongs to the square root family of functions. An open tank is 25 ft deep and filled to the brim with fluid. Use a table of values to graph the function 25 ft on the interval [0, 25]. From your graph, estimate the velocity of the fluid when the water level is 7 ft, then find the actual velocity. Are the answers close? If the fluid velocity is 5 ft/sec, how high is the water in the tank?

APPLICATIONS

105. Gravity, distance, time: After being released, the time it takes an object to fall x ft is given by the function T1x2  14 1x, where T(x) is in seconds. Describe the transformation applied to obtain the

graph of T from the graph of y  1x, then sketch the graph of T for x  30, 100 4 . How long would it take an object to hit the ground if it were dropped from a height of 81 ft?

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106. Stopping distance: In certain weather conditions, accident investigators will use the function v1x2  4.9 1x to estimate the speed of a car (in miles per hour) that has been involved in an accident, based on the length of the skid marks x (in feet). Describe the transformation applied to obtain the graph of v from the graph of y  1x, then sketch the graph of v for x  3 0, 400 4. If the skid marks were 225 ft long, how fast was the car traveling? Is this point on your graph? 107. Wind power: The power P generated by a certain 8 3 v wind turbine is given by the function P1v2  125 where P(v) is the power in watts at wind velocity v (in miles per hour). (a) Describe the transformation applied to obtain the graph of P from the graph of y  v3, then sketch the graph of P for v  30, 25 4 (scale the axes appropriately). (b) How much power is being generated when the wind is blowing at 15 mph? (c) Calculate the rate of change ¢P ¢v in the intervals [8, 10] and [28, 30]. What do you notice? 108. Wind power: If the power P (in watts) being generated by a wind turbine is known, the velocity of the wind can be determined using the function 

3 v1P2  1 52 2 2 P. Describe the transformation applied to obtain the graph of v from the graph of 3 y 2 P, then sketch the graph of v for P  3 0, 512 4 (scale the axes appropriately). How fast is the wind blowing if 343W of power is being generated?

109. Acceleration due to gravity: The distance a ball rolls down an inclined plane is given by the function d1t2  2t2, where d(t) represents the distance in feet after t sec. (a) Describe the transformation applied to obtain the graph of d from the graph of y  t2, then sketch the graph of d for t  30, 3 4. (b) How far has the ball rolled after 2.5 sec? (c) Calculate the rate of change ¢d ¢t in the intervals [1, 1.5] and [3, 3.5]. What do you notice? 110. Acceleration due to gravity: The velocity of a steel ball bearing as it rolls down an inclined plane is given by the function v1t2  4t, where v(t) represents the velocity in feet per second after t sec. Describe the transformation applied to obtain the graph of v from the graph of y  t, then sketch the graph of v for t  3 0, 3 4. What is the velocity of the ball bearing after 2.5 sec?

EXTENDING THE CONCEPT

111. Carefully graph the functions f 1x2  x and g1x2  21x on the same coordinate grid. From the graph, in what interval is the graph of g(x) above the graph of f (x)? Pick a number (call it h) from this interval and substitute it in both functions. Is g1h2 7 f 1h2? In what interval is the graph of g(x) below the graph of f (x)? Pick a number from this interval (call it k) and substitute it in both functions. Is g1k2 6 f 1k2? 

171

112. Sketch the graph of f 1x2  2x  3  8 using transformations of the parent function, then determine the area of the region in quadrant I that is beneath the graph and bounded by the vertical lines x  0 and x  6.

113. Sketch the graph of f 1x2  x2  4, then sketch the graph of F1x2  x2  4 using your intuition and the meaning of absolute value (not a table of values). What happens to the graph?

MAINTAINING YOUR SKILLS

114. (2.1) Find the distance between the points 113, 92 and 17, 122, and the slope of the line containing these points. 32 in. 32 in.

115. (R.7) Find the perimeter and area of the figure shown (note the units).

2 ft 38 in.

2 1 1 7 116. (1.1) Solve for x: x   x  . 3 4 2 12 117. (2.5) Without graphing, state intervals where f 1x2c and f 1x2T for f 1x2  1x  42 2  3.

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2.7 Piecewise-Defined Functions Learning Objectives

Most of the functions we’ve studied thus far have been smooth and continuous. Although “smooth” and “continuous” are defined more formally in advanced courses, for our purposes smooth simply means the graph has no sharp turns or jagged edges, and continuous means you can draw the entire graph without lifting your pencil. In this section, we study a special class of functions, called piecewise-defined functions, whose graphs may be various combinations of smooth/not smooth and continuous/not continuous. The absolute value function is one example (see Exercise 31). Such functions have a tremendous number of applications in the real world.

In Section 2.7 you will learn how to:

A. State the equation and domain of a piecewisedefined function

B. Graph functions that are piecewise-defined

C. Solve applications involving piecewisedefined functions

A. The Domain of a Piecewise-Defined Function For the years 1990 to 2000, the American bald eagle remained on the nation’s endangered species list, although the number of breeding pairs was growing slowly. After 2000, the population of eagles grew at a much faster rate, and they were removed from the list soon afterward. From Table 2.5 and plotted points modeling this growth (see Figure 2.69), we observe that a linear model would fit the period from 1992 to 2000 very well, but a line with greater slope would be needed for the years 2000 to 2006 and (perhaps) beyond.

Table 2.5

Figure 2.69

Bald Eagle Breeding Pairs

Year

Bald Eagle Breeding Pairs

2

3700

10

6500

4

4400

12

7600

6

5100

14

8700

8

5700

16

9800

Source: www.fws.gov/midwest/eagle/population 1990 corresponds to year 0.

WORTHY OF NOTE For the years 1992 to 2000, we can estimate the growth in breeding pairs ¢pairs ¢time using the points (2, 3700) and (10, 6500) in the slope formula. The result is 350 1 , or 350 pairs per year. For 2000 to 2006, using (10, 6500) and (16, 9800) shows the rate of growth is significantly larger: ¢pairs 550 ¢years  1 or 550 pairs per year.

172

10,000 9,000

Bald eagle breeding pairs

Year

8,000 7,000 6,000 5,000 4,000 3,000

0

2

4

6

8

10

12

14

16

18

t (years since 1990)

The combination of these two lines would be a single function that modeled the population of breeding pairs from 1990 to 2006, but it would be defined in two pieces. This is an example of a piecewise-defined function. The notation for these functions is a large “left brace” indicating the equations it groups are part of a single function. Using selected data points and techniques from Section 2.3, we find equations that could represent each piece are p1t2  350t  3000 2-90

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for 0  t  10 and p1t2  550t  1000 for t 7 10, where p(t) is the number of breeding pairs in year t. The complete function is then written:

WORTHY OF NOTE In Figure 2.69, note that we indicated the exclusion of t  10 from the second piece of the function using an open half-circle.

EXAMPLE 1

function name

function pieces

domain of each piece

350t  3000 p1t2  e 550t  1000



2  t  10 t 7 10

Writing the Equation and Domain of a Piecewise-Defined Function y

The linear piece of the function shown has an equation of y  2x  10. The equation of the quadratic piece is y  x 2  9x  14. Write the related piecewise-defined function, and state the domain of each piece by inspecting the graph. Solution



A. You’ve just learned how to state the equation and domain of a piecewisedefined function

10 8

f(x) 6

From the graph we note the linear portion is defined between 0 and 3, with these endpoints included as indicated by the closed dots. The domain here is 0  x  3. The quadratic portion begins at x  3 but does not include 3, as indicated by the half-circle notation. The equation is function name

function pieces

2x  10 f 1x2  e 2 x  9x  14

4

(3, 4)

2

0

2

4

6

8

10

x

domain

0x3 3 6 x7 Now try Exercises 7 and 8



Piecewise-defined functions can be composed of more than two pieces, and can involve functions of many kinds.

B. Graphing Piecewise-Defined Functions As with other functions, piecewise-defined functions can be graphed by simply plotting points. Careful attention must be paid to the domain of each piece, both to evaluate the function correctly and to consider the inclusion/exclusion of endpoints. In addition, try to keep the transformations of a basic function in mind, as this will often help graph the function more efficiently. EXAMPLE 2



Graphing a Piecewise-Defined Function Graph the function by plotting points, then state its domain and range: h1x2  e

Solution



x  2 2 1x  1  1

5  x 6 1 x  1

The first piece of h is a line with negative slope, while the second is a transformed square root function. Using the endpoints of each domain specified and a few additional points, we obtain the following: For h1x2  x  2, 5  x 6 1, x

x

h(x)

3

1

1

3

1

0

1

1

1

3

3

5

h(x)

For h1x2  2 1x  1  1, x  1,

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After plotting the points from the first piece, we connect them with a line segment noting the left endpoint is included, while the right endpoint is not (indicated using a semicircle around the point). Then we plot the points from the second piece and draw a square root graph, noting the left endpoint here is included, and the graph rises to the right. From the graph we note the complete domain of h is x  35, q 2 , and the range is y  31, q 2 .

h(x) 5

h(x)  x  2 h(x)  2 x  1 1 5

5

x

5

Now try Exercises 9 through 14



As an alternative to plotting points, we can graph each piece of the function using transformations of a basic graph, then erase those parts that are outside of the corresponding domain. Repeat this procedure for each piece of the function. One interesting and highly instructive aspect of these functions is the opportunity to investigate restrictions on their domain and the ranges that result.

Piecewise and Continuous Functions

EXAMPLE 3



Graphing a Piecewise-Defined Function Graph the function and state its domain and range: f 1x2  e

Solution



1x  32 2  12 3

0 6 x6 x 7 6

The first piece of f is a basic parabola, shifted three units right, reflected across the x-axis (opening downward), and shifted 12 units up. The vertex is at (3, 12) and the axis of symmetry is x  3, producing the following graphs. 1. Graph first piece of f (Figure 2.70).

2. Erase portion outside domain of 0 6 x  6 (Figure 2.71).

Figure 2.70

Figure 2.71 y

y 12

y  1(x  3)2  12

12

10

10

8

8

6

6

4

4

2

2

1

1 2 3 4 5 6 7 8 9 10

x

1

y  1(x  3)2  12

1 2 3 4 5 6 7 8 9 10

The second function is simply a horizontal line through (0, 3).

x

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3. Graph second piece of f (Figure 2.72).

4. Erase portion outside domain of x 7 6 (Figure 2.73).

Figure 2.72

Figure 2.73

y

y

12

12

y  1(x  3)2  12

10

10

8

8

6

6

y3

4

f (x)

4

2

2

1

1 2 3 4 5 6 7 8 9 10

1

x

1 2 3 4 5 6 7 8 9 10

x

The domain of f is x  10, q 2, and the corresponding range is y  33, 124. Now try Exercises 15 through 18



Piecewise and Discontinuous Functions Notice that although the function in Example 3 was piecewise-defined, the graph was actually continuous—we could draw the entire graph without lifting our pencil. Piecewise graphs also come in the discontinuous variety, which makes the domain and range issues all the more important. EXAMPLE 4



Graphing a Discontinuous Piecewise-Defined Function Graph g(x) and state the domain and range: g1x2  e

Solution



12x  6 x  6  10

0x4 4 6 x9

The first piece of g is a line, with y-intercept (0, 6) and slope 1. Graph first piece of g (Figure 2.74).

¢y ¢x

 12.

2. Erase portion outside domain of 0  x  4 (Figure 2.75).

Figure 2.74

Figure 2.75

y

y

10

10

8

8

6

6

y  qx  6

4

4

2

2

1

2

3

4

5

6

7

8

9 10

x

y  qx  6

1

2

3

4

5

6

7

8

9 10

x

The second is an absolute value function, shifted right 6 units, reflected across the x-axis, then shifted up 10 units.

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3. Graph second piece of g (Figure 2.76).

WORTHY OF NOTE As you graph piecewisedefined functions, keep in mind that they are functions and the end result must pass the vertical line test. This is especially important when we are drawing each piece as a complete graph, then erasing portions outside the effective domain.

4. Erase portion outside domain of 4 6 x  9 (Figure 2.77).

Figure 2.76

Figure 2.77 y

y  x  6  10

y 10

10

8

8

6

6

4

4

2

2

1

2

3

4

5

6

7

8

9 10

x

g(x)

1

2

3

4

5

6

7

8

9 10

x

Note that the left endpoint of the absolute value portion is not included (this piece is not defined at x  4), signified by the open dot. The result is a discontinuous graph, as there is no way to draw the graph other than by jumping the pencil from where one piece ends to where the next begins. Using a vertical boundary line, we note the domain of g includes all values between 0 and 9 inclusive: x  30, 94. Using a horizontal boundary line shows the smallest y-value is 4 and the largest is 10, but no range values exist between 6 and 7. The range is y  34, 6 4 ´ 3 7, 104. Now try Exercises 19 through 22 EXAMPLE 5





Graphing a Discontinuous Function The given piecewise-defined function is not continuous. Graph h(x) to see why, then comment on what could be done to make it continuous. x2  4 h1x2  • x  2 1

Solution



x2 x2

The first piece of h is unfamiliar to us, so we elect to graph it by plotting points, noting x  2 is outside the domain. This produces the table shown in Figure 2.78. After connecting the points, the graph of h turns out to be a straight line, but with no corresponding y-value for x  2. This leaves a “hole” in the graph at (2, 4), as designated by the open dot. Figure 2.78

WORTHY OF NOTE The discontinuity illustrated here is called a removable discontinuity, as the discontinuity can be removed by redefining a piece of the function. Note that after factoring the first piece, the denominator is a factor of the numerator, and writing the result in lowest terms 1x  22 1x  22 gives h1x2  x  2  x  2, x  2. This is precisely the equation of the line in Figure 2.78 3 h1x2  x  2 4 .

x

h(x)

4

2

2

0

0

2

2

—

4

6

Figure 2.79

y

y

5

5

5

5

5

x

5

5

x

5

The second piece is point-wise defined, and its graph is simply the point (2, 1) shown in Figure 2.79. It’s interesting to note that while the domain of h is all real numbers (h is defined at all points), the range is y  1q, 42 ´ 14, q2 as the function never takes on the value y  4. In order for h to be continuous, we would need to redefine the second piece as y  4 when x  2. Now try Exercises 23 through 26



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To develop these concepts more fully, it will help to practice finding the equation of a piecewise-defined function given its graph, a process similar to that of Example 10 in Section 2.6. EXAMPLE 6



Determining the Equation of a Piecewise-Defined Function Determine the equation of the piecewise-defined function shown, including the domain for each piece.

Solution



y 5

y

¢ By counting ¢x from (2, 5) to (1, 1), we find the linear portion has slope m  2, and the y-intercept must be (0, 1). The equation of the line is y  2x  1. The second piece appears to be a parabola with vertex (h, k) at (3, 5). Using this vertex with the point (1, 1) in the general form y  a1x  h2 2  k gives

y  a1x  h2 2  k 1  a11  32 2  5 4  a122 2 4  4a 1  a

4

6

x

5

general form substitute 1 for x, 1 for y, 3 for h, 5 for k simplify; subtract 5 122 2  4 divide by 4

The equation of the parabola is y  1x  32 2  5. Considering the domains shown in the figure, the equation of this piecewise-defined function must be B. You’ve just learned how to graph functions that are piecewise-defined

p1x2  e

2x  1 1x  32 2  5

2  x  1 x 7 1

Now try Exercises 27 through 30



C. Applications of Piecewise-Defined Functions The number of applications for piecewise-defined functions is practically limitless. It is actually fairly rare for a single function to accurately model a situation over a long period of time. Laws change, spending habits change, and technology can bring abrupt alterations in many areas of our lives. To accurately model these changes often requires a piecewise-defined function. EXAMPLE 7



Modeling with a Piecewise-Defined Function For the first half of the twentieth century, per capita spending on police protection can be modeled by S1t2  0.54t  12, where S(t) represents per capita spending on police protection in year t (1900 corresponds to year 0). After 1950, perhaps due to the growth of American cities, this spending greatly increased: S1t2  3.65t  144. Write these as a piecewise-defined function S(t), state the domain for each piece,

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then graph the function. According to this model, how much was spent (per capita) on police protection in 2000? How much will be spent in 2010? Source: Data taken from the Statistical Abstract of the United States for various years.

Solution



function name

S1t2  e

function pieces

effective domain

0.54t  12 3.65t  144

0  t  50 t 7 50

Since both pieces are linear, we can graph each part using two points. For the first function, S102  12 and S1502  39. For the second function S1502  39 and S1802  148. The graph for each piece is shown in the figure. Evaluating S at t  100: S1t2  3.65t  144 S11002  3.6511002  144  365  144  221

240

S(t)

200

(80, 148)

160 120 80 40 0

(50, 39) 10 20 30 40 50 60 70 80 90 100 110

t

About $221 per capita was spent on police protection in the year 2000. For 2010, the model indicates that $257.50 per capita will be spent: S11102  257.5. Now try Exercises 33 through 44



Step Functions The last group of piecewise-defined functions we’ll explore are the step functions, so called because the pieces of the function form a series of horizontal steps. These functions find frequent application in the way consumers are charged for services, and have a number of applications in number theory. Perhaps the most common is called the greatest integer function, though recently its alternative name, floor function, has gained popularity (see Figure 2.80). This is in large part due to an improvement in notation and as a better contrast to ceiling functions. The floor function of a real number x, denoted f 1x2  : x ; or Œ x œ (we will use the first), is the largest integer less than or equal to x. For instance, : 5.9;  5, : 7;  7, and :3.4 ;  4. In contrast, the ceiling function C1x2  0 (only y is positive)

QI x > 0, y > 0 (both x and y are positive)

sin  is positive

All functions are positive

tan  is positive

cos  is positive

QIII x < 0, y < 0 (both x and y are negative)

EXAMPLE 5





QIV x > 0, y < 0 (only x is positive)

Evaluating Trig Functions for a Rotation  Evaluate the six trig functions for  

Solution

x

y q

5 . 4

5 terminates in QIII, so 4 5  r     . The associated point is 4 4 12 12 , b since x 6 0 and y 6 0 in QIII. a 2 2

  5 4

A rotation of



2`

r  d

√22 , √22  3 2

x

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This yields cosa

5 12 b 4 2

sina

5 12 b 4 2

tana

5 b1 4

12 is 12 after rationalizing, we have 2 5 5 5 seca b   12 csca b   12 cota b  1 4 4 4

Noting the reciprocal of 

C. You’ve just learned how to define the six trig functions in terms of a point on the unit circle

Now try Exercises 37 through 40



D. The Trigonometry of Real Numbers Defining the trig functions in terms of a point on the Figure 5.29 unit circle is precisely what we needed to work with y 3 s 4 √2, √2 them as functions of real numbers. This is because  2 2  when r  1 and  is in radians, the length of the subtended arc is numerically the same as the sr  d   3 measure of the angle: s  112 1 s  ! This means 4  we can view any function of  as a like function of arc 1x length s, where s   (see the Reinforcing Basic Concepts feature following this section.). As a compromise the variable t is commonly used, with t representing either the amount of rotation or the length of the arc. As such we will assume t is a unitless quantity, although there are other reasons 3 for this assumption. In Figure 5.29, a rotation of   is subtended by an arc length 4 3  of s  (about 2.356 units). The reference angle for  is , which we will now 4 4 refer to as a reference arc. As you work through the remaining examples and the exercises that follow, it will often help to draw a quick sketch similar to that in Figure 5.29 to determine the quadrant of the terminal side, the reference arc, and the sign of each function. 

EXAMPLE 6

Evaluating Trig Functions for a Real Number t Evaluate the six trig functions for the given value of t. 3 11 a. t  b. t  6 2



Solution y q

  11 6

x



r  k 2`

√32 , 12  3 2

11 , the arc terminates in QIV where x 7 0 and y 6 0. The 6  reference arc is and from our previous work we know the corresponding 6 13 1 ,  b. This gives point (x, y) is a 2 2

a. For t 

11 13 b 6 2 2 13 11 b seca 6 3 cosa

11 1 b 6 2 11 csca b  2 6

sina

11 13 b 6 3 11 cota b   13 6

tana

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3 is a quadrantal angle and the associated point is 10, 12. 2 This yields

y q

b. t 

  3 2

3 b0 2 3 seca b  undefined 2 cosa



2`

(0, 1)

x

3 b  1 2 3 csca b  1 2

sina

3 b  undefined 2 3 cot a b  0 2

tana

Now try Exercises 41 through 44

3 2



As Example 6(b) indicates, as functions of a real number the concept of domain comes into play. From their definition it is apparent there are no restrictions on the domain of cosine and sine, but the domains of the other functions must be restricted to exclude division by zero. For functions with x in the denominator, we cast out the  odd multiples of , since the x-coordinate of the related quadrantal points is zero: 2  3 S 10, 12, S 10, 12, and so on. The excluded values can be stated as 2 2  t   k for all integers k. For functions with y in the denominator, we cast out all 2 multiples of  1t  k for all integers k) since the y-coordinate of these points is zero: 0 S 11, 02,  S 11, 02, 2 S 11, 02, and so on. The Domains of the Trig Functions as Functions of a Real Number For t   and k  , the domains of the trig functions are: cos t  x

sin t  y

t

t

1 sec t  ; x  0 x  t   k 2

1 csc t  ; y  0 y

y ;x0 x  t   k 2 x cot t  ; y  0 y

t  k

t  k

tan t 

For a given point (x, y) on the unit circle associated with the real number t, the value of each function at t can still be determined even if t is unknown. EXAMPLE 7



Finding Function Values Given a Point on the Unit Circle

Solution



24 Using the definitions from the previous box we have cos t  7 25 , sin t  25 , and sin t 24 25 tan t  cos t  7. The values of the reciprocal functions are then sec t  7 , 25 7 csc t  24, and cot t  24 .

D. You’ve just learned how to define the six trig functions in terms of a real number t

24 Given 1 7 25 , 25 2 is a point on the unit circle corresponding to a real number t, find the value of all six trig functions of t.

Now try Exercises 45 through 70



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E. Finding a Real Number t Whose Function Value Is Known In Example 7, we were able to determine the values of the trig functions even though t was unknown. In many cases, however, we need to find the value of t. For instance, what is the value of t given 13 cos t   with t in QII? Exercises of 2 this type fall into two broad categories: (1) you recognize the given number as one of the special values: 1 12 13 13 , , , 13, 1 f ;  e 0, , or 2 2 2 3 (2) you don’t. If you recognize a special value, you can often name the real number t after a careful consideration of the related quadrant and required sign.

Figure 5.30 (0, 1)

y

 12 , √32  √22 , √22  √32 , 12  k

d

u

(1, 0) x

 but 2 remember—all other special values can be found using reference arcs and the symmetry of the circle. The diagram in Figure 5.30 reviews these special values for 0  t 

EXAMPLE 8



Finding t for Given Values and Conditions Find the value of t that corresponds to the given function values. 12 a. cos t   b. tan t  13; t in QIII ; t in QII 2

Solution



a. The cosine function is negative in QII and QIII, where x 6 0. We recognize 12  as a standard value for sine and cosine, related to certain multiples of 2 3  t  . In QII, we have t  . 4 4 b. The tangent function is positive in QI and QIII, where x and y have like signs. We recognize 13 as a standard value for tangent and cotangent, related to  4 certain multiples of t  . For tangent in QIII, we have t  . 3 3 Now try Exercises 71 through 94



If the given function value is not one of the special values, properties of the inverse trigonometric functions must be used to find the associated value of t. The inverse functions are developed in Section 6.5. Using radian measure and the unit circle is much more than a simple convenience to trigonometry and its applications. Whether the unit is 1 cm, 1 m, 1 km, or even 1 light-year, using 1 unit designations serves to simplify a great many practical applications, including those involving the arc length formula, s  r. See Exercises 97 through 104. The following table summarizes the relationship between a special arc t (t in QI) and the value of each trig function at t. Due to the frequent use of these relationships, students are encouraged to commit them to memory.

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E. You’ve just learned how to find the real number t corresponding to given values of sin t, cos t, and tan t

t

sin t

cos t

tan t

csc t

sec t

cot t

0

0

1

0

undefined

1

undefined

 6

1 2

13 2

1 13  3 13

2

2 213  3 13

13

 4

12 2

12 2

1

12

12

1

 3

13 2

1 2

13

2 2 13  3 13

2

1 13  3 13

 2

1

0

undefined

1

undefined

0

5.2 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. A central circle is symmetric to the axis and to the .

axis, the

5 2. Since 1 13 , 12 13 2 is on the unit circle, the point in QII is also on the circle.

3. On a unit circle, cos t 

, sin t 

1 and tan t  ; while  x 1 x , and  .  y y



,

,

4. On a unit circle with  in radians, the length of a(n) is numerically the same as the measure of the , since for s  r, s   when r  1. 5. Discuss/Explain how knowing only one point on the unit circle, actually gives the location of four points. Why is this helpful to a study of the circular functions? 6. A student is asked to find t using a calculator, given sin t  0.5592 with t in QII. The answer submitted is t  sin1 0.5592  34°. Discuss/Explain why this answer is not correct. What is the correct response?

DEVELOPING YOUR SKILLS

Given the point is on a unit circle, complete the ordered pair (x, y) for the quadrant indicated. For Exercises 7 to 14, answer in radical form as needed. For Exercises 15 to 18, round results to four decimal places.

7. 1x, 0.82; QIII

9. a

5 , yb; QIV 13

111 11. a , yb; QI 6 13. a

111 , yb; QII 4

8. 10.6, y2; QII

10. ax, 

8 b; QIV 17

113 12. ax,  b; QIII 7 14. ax,

16 b; QI 5

15. 1x, 0.21372 ; QIII

16. (0.9909, y); QIV

17. (x, 0.1198); QII

18. (0.5449, y); QI

Verify the point given is on a unit circle, then use symmetry to find three more points on the circle. Results for Exercises 19 to 22 are exact, results for Exercises 23 to 26 are approximate.

19. a 21. a

13 1 , b 2 2

111 5 , b 6 6

20. a

17 3 , b 4 4

22. a

16 13 , b 3 3

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23. (0.3325, 0.9431)

25. 10.9937, 0.11212

24. 10.7707, 0.63722

26. 10.2029, 0.97922

   : : triangle with a hypotenuse of length 6 3 2 1 13 1 to verify that a , b is a point on the unit circle. 2 2

40. a. sin   c. sina b 2

b. sin 0 3 d. sina b 2

27. Use a

28. Use the results from Exercise 27 to find three additional points on the circle and name the quadrant of each point. Find the reference angle associated with each rotation, then find the associated point (x, y) on the unit circle.

29.  

5 4

31.   

5 6

11 33.   4 35.  

25 6

30.  

5 3

32.   

7 4

11 34.   3 36.  

39 4

Without the use of a calculator, state the exact value of the trig functions for the given angle. A diagram may help.

 37. a. sina b 4 5 c. sina b 4 9 e. sina b 4 5 g. sina b 4  38. a. tana b 3 4 c. tana b 3 7 e. tana b 3 4 g. tana b 3 39. a. cos   c. cosa b 2

3 b 4 7 d. sina b 4  f. sina b 4 11 b h. sina 4

b. sina

2 b 3 5 d. tana b 3  f. tana b 3 10 b h. tana 3

b. tana

b. cos 0 3 d. cosa b 2

Use the symmetry of the circle and reference arcs as needed to state the exact value of the trig functions for the given real number, without the use of a calculator. A diagram may help.

 41. a. cosa b 6 7 c. cosa b 6 13 b e. cosa 6 5 g. cosa b 6

5 b 6 11 b d. cosa 6  f. cosa b 6 23 b h. cosa 6

b. cosa

 42. a. csca b 6 7 c. csca b 6 13 b e. csca 6 11 b g. csca 6

5 b 6 11 b d. csca 6  f. csca b 6 17 b h. csca 6

b. csca

43. a. tan   c. tana b 2

b. tan 0 3 d. tana b 2

44. a. cot   c. cota b 2

b. cot 0 3 d. cota b 2

Given (x, y) is a point on a unit circle corresponding to t, find the value of all six circular functions of t.

45.

y (0.8, 0.6) t

46.

(1, 0) x

y

t

(1, 0) x





15 , 8  17 17

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47.

451

figure to estimate function values to one decimal place (use a straightedge). Check results using a calculator.

y

Exercises 59 to 70 t

(1, 0) x

y

q

1.5

2.0

1.0

2.5

5 , 12   13 13

0.5

48.

3.0 

y

1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1

0 x 6.0

3.5

t (1, 0)



24 7  , 25 25

x



5.5 4.0 4.5 3 2

5.0

y

49.

 t

5 , √11 6 6



(1, 0) x

60. cos 2.75

61. cos 5.5

62. sin 4.0

63. tan 0.8

64. sec 3.75

65. csc 2.0

66. cot 0.5

69. tana t

(1, 0) x



68. sina

5 b 8

8 b 5

70. seca

8 b 5

Without using a calculator, find the value of t in [0, 2 ) that corresponds to the following functions.



√5 , 2 3 3

2 121 b 51. a , 5 5 1 212 53. a ,  b 3 3 1 13 55. a , b 2 2 12 12 57. a , b 2 2

5 b 8

67. cosa

y

50.

59. sin 0.75

17 3 , b 4 4 2 16 1 54. a , b 5 5 13 1 56. a , b 2 2 12 17 58. a , b 3 3 52. a

On a unit circle, the real number t can represent either the amount of rotation or the length of the arc when we associate t with a point (x, y) on the circle. In the circle diagram shown, the real number t in radians is marked off along the circumference. For Exercises 59 through 70, name the quadrant in which t terminates and use the

71. sin t 

13 ; t in QII 2

1 72. cos t  ; t in QIV 2 73. cos t  

23 ; t in QIII 2

1 74. sin t   ; t in QIV 2 75. tan t   13; t in QII 76. sec t  2; t in QIII 77. sin t  1; t is quadrantal 78. cos t  1; t is quadrantal

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Without using a calculator, find the two values of t (where possible) in [0, 2 ) that make each equation true.

79. sec t   12 81. tan t undefined 83. cos t  

12 2

85. sin t  0

2 13 82. csc t undefined

80. csc t  

84. sin t 

12 2

86. cos t  1

87. Given 1 34, 45 2 is a point on the unit circle that corresponds to t. Find the coordinates of the point corresponding to (a) t and (b) t  .

7 24 88. Given 125 , 25 2 is a point on the unit circle that corresponds to t. Find the coordinates of the point corresponding to (a) t   and (b) t  .

Find an additional value of t in [0, 2 ) that makes the equation true.

89. sin 0.8  0.7174 90. cos 2.12  0.5220 91. cos 4.5  0.2108 92. sin 5.23  0.8690 93. tan 0.4  0.4228 94. sec 5.7  1.1980



WORKING WITH FORMULAS

95. From Pythagorean triples to points on the x y unit circle: 1x, y, r2 S a , , 1b r r While not strictly a “formula,” dividing a Pythagorean triple by r is a simple algorithm for rewriting any Pythagorean triple as a triple with hypotenuse 1. This enables us to identify certain points on a unit circle, and to evaluate the six trig functions of the related acute angle. Rewrite each x y triple as a triple with hypotenuse 1, verify a , b is r r a point on the unit circle, and evaluate the six trig functions using this point. a. (5, 12, 13) b. (7, 24, 25) c. (12, 35, 37) d. (9, 40, 41)



 96. The sine and cosine of 12k  12 ; k   4 In the solution to Example 8(a), we mentioned 12  were standard values for sine and cosine, 2  “related to certain multiples of .” Actually, we 4  meant “odd multiples of .” The odd multiples of 4  are given by the “formula” shown, where k is 4  any integer. (a) What multiples of are generated 4 by k  3, 2, 1, 0, 1, 2, 3? (b) Find similar formulas for Example 8(b), where 13 is a standard value for tangent and cotangent, “related to certain  multiples of .” 6

APPLICATIONS

97. Laying new sod: When new sod is laid, a heavy roller is used to press the sod down to ensure good contact with the ground 1 ft beneath. The radius of the roller is 1 ft. (a) Through what angle (in radians) has the roller turned after being pulled across 5 ft of yard? (b) What angle must the roller turn through to press a length of 30 ft?

98. Cable winch: A large winch with a radius of 1 ft winds in 3 ft of cable. (a) Through what angle (in radians) has it turned? (b) What angle must it turn through in order to winch in 12.5 ft of cable?

Exercise 98

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Section 5.2 Unit Circles and the Trigonometry of Real Numbers

103. Compact disk circumference: A standard compact disk has a radius of 6 cm. Call this length “1 unit.” Mark a starting point on any large surface, then carefully roll the compact disk along this line without slippage, through one full revolution (2 rad) and mark this spot. Take an accurate measurement of the resulting line segment. Is the result close to 2 “units” (2  6 cm)? Exercise 104 104. Verifying s  r: On a protractor, carefully measure the distance from the middle of the protractor’s eye to the edge of the eye 1 unit protractor along the 0° mark, to the nearest half-millimeter. Call this length “1 unit.” Then use a ruler to draw a straight line on a blank sheet of paper, and with the protractor on edge, start the zero degree mark at one end of the line, carefully roll the protractor until it reaches 1 radian 157.3°2 , and mark this spot. Now measure the length of the line segment created. Is it very close to 1 “unit” long? 70 110

80 100

90 90

100 80

110 70

12 60 0

13 50 0

4 14 0 0

60 0 12 50 0 13

20 160

10 170

0 180

180 0



170 10

102. If you include the dwarf planet Pluto, Jupiter is the middle (fifth of nine) planet from the Sun. Suppose

160 20

101. If the Earth travels through an angle of 2.5 rad about the Sun, (a) what distance in astronomical units (AU) has it traveled? (b) How many AU does it take for one complete orbit around the Sun?

1500 3

Interplanetary measurement: In the year 1905, astronomers began using astronomical units or AU to study the distances between the celestial bodies of our solar system. One AU represents the average distance between the Earth and the Sun, which is about 93 million miles. Pluto is roughly 39.24 AU from the Sun.

0 14 0 4

100. Barrel races: In the barrel races popular at some family reunions, contestants stand on a hard rubber barrel with a radius of 1 cubit (1 cubit  18 in.), and try to “walk the barrel” from the start line to the finish line without falling. (a) What distance (in cubits) is traveled as the barrel is walked through an angle of 4.5 rad? (b) If the race is 25 cubits long, through what angle will the winning barrel walker walk the barrel?

astronomers had decided to use its average distance from the Sun as 1 AU. In this case, 1 AU would be 480 million miles. If Jupiter travels through an angle of 4 rad about the Sun, (a) what distance in the “new” astronomical units (AU) has it traveled? (b) How many of the new AU does it take to complete one-half an orbit about the Sun? (c) What distance in the new AU is the dwarf planet Pluto from the Sun?

3 1500

99. Wiring an apartment: In the wiring of an apartment complex, electrical wire is being pulled from a spool with radius 1 decimeter (1 dm  10 cm). (a) What length (in decimeters) is removed as the spool turns through 5 rad? (b) How many decimeters are removed in one complete turn 1t  22 of the spool?

EXTENDING THE CONCEPT

105. In this section, we discussed the domain of the circular functions, but said very little about their range. Review the concepts presented here and determine the range of y  cos t and y  sin t. In other words, what are the smallest and largest output values we can expect? sin t , what can you say about the cos t range of the tangent function?

106. Since tan t 

Use the radian grid given with Exercises 59–70 to answer Exercises 107 and 108.

107. Given cos12t2  0.6 with the terminal side of the arc in QII, (a) what is the value of 2t? (b) What quadrant is t in? (c) What is the value of cos t? (d) Does cos12t2  2cos t? 108. Given sin12t2  0.8 with the terminal side of the arc in QIII, (a) what is the value of 2t? (b) What quadrant is t in? (c) What is the value of sin t? (d) Does sin12t2  2sin t?

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MAINTAINING YOUR SKILLS

109. (2.1) Given the points (3, 4) and (5, 2) find a. the distance between them b. the midpoint between them c. the slope of the line through them.

111. (1.3) Solve each equation: a. 2x  1  3  7 b. 2 1x  1  3  7 112. (3.2) Use the rational zeroes theorem to solve the equation completely, given x  3 is one root.

110. (4.3) Use a calculator to find the value of each expression, then explain the results. a. log 2  log 5  ______ b. log 20  log 2  ______

x4  x3  3x2  3x  18  0

5.3 Graphs of the Sine and Cosine Functions; Cosecant and Secant Functions Learning Objectives In Section 5.3 you will learn how to:

A. Graph f1t2  sin t using special values and symmetry

B. Graph f1t2  cos t using special values and symmetry

As with the graphs of other functions, trigonometric graphs contribute a great deal toward the understanding of each trig function and its applications. For now, our primary interest is the general shape of each basic graph and some of the transformations that can be applied. We will also learn to analyze each graph, and to capitalize on the features that enable us to apply the functions as real-world models.

A. Graphing f(t)  sin t Consider the following table of values (Table 5.1) for sin t and the special angles in QI. Table 5.1

C. Graph sine and cosine functions with various amplitudes and periods

t

0

 6

 4

 3

 2

sin t

0

1 2

12 2

13 2

1

D. Investigate graphs of the reciprocal functions f1t2  csc 1Bt2 and f1t2  sec 1Bt2

E. Write the equation for a given graph

 to  2 (QII), special values taken from the unit circle show sine values are decreasing from 1 to 0, but through the same output values as in QI. See Figures 5.31 through 5.33. Observe that in this interval, sine values are increasing from 0 to 1. From

Figure 5.31

Figure 5.32

y (0, 1)

y (0, 1)

 12 , √32 

Figure 5.33 y (0, 1)

√22 , √22  2 3

3 4

(1, 0) x

(1, 0)

2 23 sin a b  3 2

√32 , 12  5 6

(1, 0) x

(1, 0)

3 22 sin a b  4 2

(1, 0) x

(1, 0)

5 1 sin a b  6 2

With this information we can extend our table of values through , noting that sin   0 (see Table 5.2).

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Section 5.3 Graphs of the Sine and Cosine Functions; Cosecant and Secant Functions

Table 5.2 t

0

 6

 4

 3

 2

2 3

3 4

5 6



sin t

0

1 2

12 2

13 2

1

13 2

12 2

1 2

0

Using the symmetry of the circle and the fact that y is negative in QIII and QIV, we can complete the table for values between  and 2. EXAMPLE 1



Finding Function Values Using Symmetry Use the symmetry of the unit circle and reference arcs of special values to complete Table 5.3. Recall that y is negative in QIII and QIV. Table 5.3 t

7 6



5 4

4 3

3 2

5 3

7 4

11 6

2

sin t

 12 , sin t   depending on 4 2  1 the quadrant of the terminal side. Similarly, for any reference arc of , sin t   , 6 2 13  while any reference arc of will give sin t   . The completed table is 3 2 shown in Table 5.4. Table 5.4 Symmetry shows that for any odd multiple of t 

t



7 6

sin t

0



5 4

1 2



4 3

12 2



3 2

13 2

1

5 3 

13 2

7 4 

11 6

12 2



1 2

2 0

Now try Exercises 7 and 8



1 12 13  0.5,  0.71, and  0.87, we plot these points and 2 2 2 connect them with a smooth curve to graph y  sin t in the interval 30, 24 . The first Noting that

five plotted points are labeled in Figure 5.34. Figure 5.34 

␲, 6

0.5

 ␲4 , 0.71

 ␲3 , 0.87

sin t

 ␲2 , 1

1

ng asi cre De

si

ng

0.5

rea



In c

Solution

(0, 0) 0.5 1

␲ 2

␲

3␲ 2

2␲

t

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Expanding the table from 2 to 4 using reference arcs and the unit circle 13  b  sin a b since shows that function values begin to repeat. For example, sin a 6 6  9   r  ; sin a b  sin a b since r  , and so on. Functions that cycle through a 6 4 4 4 set pattern of values are said to be periodic functions. Periodic Functions A function f is said to be periodic if there is a positive number P such that f 1t  P2  f 1t2 for all t in the domain. The smallest number P for which this occurs is called the period of f. For the sine function we have sin t  sin1t  22, as in sin a

13 b 6

 9   2b and sin a b  sin a  2b, with the idea extending to all other real 6 4 4 numbers t: sin t  sin1t  2k2 for all integers k. The sine function is periodic with period P  2. Although we initially focused on positive values of t in 30, 2 4 , t 6 0 and k 6 0 are certainly possibilities and we note the graph of y  sin t extends infinitely in both directions (see Figure 5.35).

sin a

Figure 5.35 ␲

 2 , 1

y 1

y  sin t

0.5

4␲  3

␲

␲ 3

2␲ 3





␲

 2 , 1

0.5

␲ 3

2␲ 3

␲

t

4␲ 3

1

Finally, both the graph and the unit circle confirm that the range of y  sin t is 31, 14 , and that y  sin t is an odd function. In particular, the graph   shows sina b  sina b, and the unit circle 2 2 shows (Figure 5.36) sin t  y, and sin1t2  y, from which we obtain sin1t2  sin t by substitution. As a handy reference, the following box summarizes the main characteristics of y  sin t.

Figure 5.36 y (0, 1)

y  sin t ( x, y) t

(1, 0)

(1, 0) t

(0, 1)

(x, y)

x

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Section 5.3 Graphs of the Sine and Cosine Functions; Cosecant and Secant Functions

Characteristics of f(t)  sin t For all real numbers t and integers k, Domain 1q, q 2

Range 3 1, 14

Period

Symmetry

Maximum value

Minimum value

odd

sin t  1  at t   2k 2

sin t  1 3  2k at t  2

Decreasing

Zeroes

 3 b a , 2 2

t  k

sin1t2  sin t Increasing

a0,

EXAMPLE 2



Solution



 3 b ´ a , 2b 2 2

2

Using the Period of sin t to Find Function Values

Use the characteristics of f 1t2  sin t to match the given value of t to the correct value of sin t.   17 11 a. t  a  8b b. t   c. t  d. t  21 e. t  4 6 2 2 1 12 I. sin t  1 II. sin t   III. sin t  1 IV. sin t  V. sin t  0 2 2    8b  sin , the correct match is (IV). 4 4   Since sin a b  sin , the correct match is (II). 6 6 17   Since sin a b  sina  8b  sin , the correct match is (I). 2 2 2 Since sin 1212  sin1  202  sin , the correct match is (V). 3 3 11 b  sin a  4b  sin a b, the correct match is (III). Since sin a 2 2 2

a. Since sin a b. c. d. e.



Now try Exercises 9 and 10

Many of the transformations applied to algebraic graphs can also be applied to trigonometric graphs. These transformations may stretch, reflect, or translate the graph, but it will still retain its basic shape. In numerous applications it will help if you’re able to draw a quick, accurate sketch of the transformations involving f 1t2  sin t. To assist this effort, we’ll begin with the interval 3 0, 2 4, combine the characteristics just listed with some simple geometry, and offer the following four-step process. Steps I through IV are illustrated in Figures 5.37 through 5.40. Figure 5.38

Figure 5.37 y

y 1

1

0 2␲

1

t

0 ␲ 2

1

␲

3␲ 2

2␲

t

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Figure 5.39

Figure 5.40 y

y 1

1

Increasing 0 ␲ 2

1

␲

3␲ 2

2␲

t

Decreasing

0 ␲ 2

␲

3␲ 2

2␲

t

1

Draw the y-axis, mark zero halfway up, with 1 and 1 an equal distance from this zero. Then draw an extended t-axis and tick mark 2 to the extreme right (Figure 5.37). Step II: On the t-axis, mark halfway between 0 and 2 and label it “,” mark 3  . Halfway halfway between  on either side and label the marks and 2 2 between these you can draw additional tick marks to represent the remain ing multiples of (Figure 5.38). 4 Step III: Next, lightly draw a rectangular frame, which we’ll call the reference rectangle, P  2 units wide and 2 units tall, centered on the t-axis and with the y-axis along one side (Figure 5.39). Step IV: Knowing y  sin t is positive and increasing in QI, that the range is 31, 1 4, that the zeroes are 0, , and 2, and that maximum and minimum values occur halfway between the zeroes (since there is no horizontal shift), we can draw a reliable graph of y  sin t by partitioning the rectangle into four equal parts to locate these values (note bold tick-marks). We will call this partitioning of the reference rectangle the rule of fourths, since we are then P scaling the t-axis in increments of (Figure 5.40). 4 Step I:

EXAMPLE 3



Graphing y  sin t Using a Reference Rectangle  3 d. Use steps I through IV to draw a sketch of y  sin t for the interval c , 2 2

Solution



A. You’ve just learned how to graph f1t2  sin t using special values and symmetry

Start by completing steps I and II, then y  1 extend the t-axis to include  . Beginning Increasing Decreasing 2  at  , draw a reference rectangle 2 units ␲ ␲ ␲ 3␲ t 2 2 2 2 wide and 2 units tall, centered on the x-axis 1 3 aending at b. After applying the rule of 2 fourths, we note the zeroes occur at t  0 and t  , with the max/min values spaced equally between and on either side. Plot these points and connect them with a smooth curve (see the figure). Now try Exercises 11 and 12



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Section 5.3 Graphs of the Sine and Cosine Functions; Cosecant and Secant Functions

B. Graphing f(t)  cos t

With the graph of f1t2  sin t established, sketching the graph of f 1t2  cos t is a very natural next step. First, note that when t  0, cos t  1 so the graph of y  cos t 1 13 b, will begin at (0, 1) in the interval 30, 2 4 . Second, we’ve seen a ,  2 2 12 13 1 12 a ,  b and a , b are all points on the unit circle since they satisfy 2 2 2 2 x2  y2  1. Since cos t  x and sin t  y, the equation cos2t  sin2t  1 can be 1 13 obtained by direct substitution. This means if sin t   , then cos t   and 2 2 vice versa, with the signs taken from the appropriate quadrant. The table of values for cosine then becomes a simple variation of the table for sine, as shown in Table 5.5 for t  30,  4. Table 5.5 t

0

 6

 4

 3

 2

2 3

3 4

5 6



sin t

0

1  0.5 2

12  0.71 2

13  0.87 2

1

13  0.87 2

12  0.71 2

1  0.5 2

0

cos t

1

13  0.87 2

12  0.71 2

1  0.5 2

0

1   0.5 2



12  0.71 2



13  0.87 2

1

The same values can be taken from the unit circle, but this view requires much less effort and easily extends to values of t in 3, 2 4. Using the points from Table 5.5 and its extension through 3 , 2 4 , we can draw the graph of y  cos t in 3 0, 2 4 and identify where the function is increasing and decreasing in this interval. See Figure 5.41. Figure 5.41 0.87  ␲ , 0.71 4

1

D

␲

2

0

g sin rea ec

0.5

 ␲3 , 0.5  ␲2 , 0

␲ 2

␲

g

cos t

Inc rea sin



␲, 6

3␲ 2

2␲

t

0.5

1

The function is decreasing for t in 10, 2, and increasing for t in 1, 22. The end  result appears to be the graph of y  sin t shifted to the left units, a fact more easily 2  seen if we extend the graph to  as shown. This is in fact the case, and 2 is a relationship we will later prove in Chapter 6. Like y  sin t, the function y  cos t is periodic with period P  2, with the graph extending infinitely in both directions. Finally, we note that cosine is an even function, meaning cos1t2  cos t for all   t in the domain. For instance, cos a b  cos a b  0 (see Figure 5.41). Here is a 2 2 summary of important characteristics of the cosine function.

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Characteristics of f(t)  cos t For all real numbers t and integers k, Domain 1q, q2

Range 31, 1 4

Period

Symmetry

Maximum value

Minimum value

even cos1t2  cos t

cos t  1 at t  2k

cos t  1 at t    2k

Increasing

Decreasing

Zeroes

1, 22

EXAMPLE 4



10, 2



B. You’ve just learned how to graph f1t2  cos t using special values and symmetry

t

  k 2

Graphing y  cos t Using a Reference Rectangle Draw a sketch of y  cos t for t in c,

Solution

2

3 d. 2

y After completing steps I and II, extend the negative x-axis to include 1 y  cos t . Beginning at , draw a Decreasing reference rectangle 2 units wide and 2 units tall, centered on the ␲ ␲ 3␲ ␲ 0 ␲ t 2 2 2 x-axis. After applying the rule of Increasing fourths, we note the zeroes will 1 occur at t  /2 and t  /2, with the max/min values spaced equally between these zeroes and on either side 1at t  , t  0, and t  2. Finally, we extend the graph to include 3/2.

Now try Exercises 13 and 14



WORTHY OF NOTE

C. Graphing y  A sin(Bt) and y  A cos(Bt)

Note that the equations y  A sin t and y  A cos t both indicate y is a function of t, with no reference to the unit circle definitions cos t  x and sin t  y.

In many applications, trig functions have maximum and minimum values other than 1 and 1, and periods other than 2. For instance, in tropical regions the maximum and minimum temperatures may vary by no more than 20°, while for desert regions this difference may be 40° or more. This variation is modeled by the amplitude of sine and cosine functions.

Amplitude and the Coefficient A (assume B  1) For functions of the form y  A sin t and y  A cos t, let M represent the Maximum Mm value and m the minimum value of the functions. Then the quantity gives the 2 Mm average value of the function, while gives the amplitude of the function. 2 Amplitude is the maximum displacement from the average value in the positive or negative direction. It is represented by A, with A playing a role similar to that seen for algebraic graphs 3 Af 1t2 vertically stretches or compresses the graph of f, and reflects it across the t-axis if A 6 0 4. Graphs of the form y  sin t (and y  cos t) can quickly be sketched with any amplitude by noting (1) the zeroes of the function remain fixed since sin t  0 implies A sin t  0, and (2) the maximum and minimum values are A and A, respectively, since sin t  1 or 1 implies A sin t  A or A. Note this implies the reference rectangle will be 2A units tall and P units wide. Connecting the points that result with a smooth curve will complete the graph.

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EXAMPLE 5



Graphing y  A sin t Where A  1

Solution



With an amplitude of  A  4, the reference rectangle will be 2142  8 units tall, by 2 units wide. Using the rule of fourths, the zeroes are still t  0, t  , and t  2, with the max/min values spaced equally between. The maximum value is  3 4 sin a b  4112  4, with a minimum value of 4 sin a b  4112  4. 2 2 Connecting these points with a “sine curve” gives the graph shown 1y  sin t is also shown for comparison).

Draw a sketch of y  4 sin t in the interval 3 0, 2 4.

y  4 sin t

4

Zeroes remain fixed ␲

␲ 2

3␲ 2

y  sin t

2␲

t

4

Now try Exercises 15 through 20



Period and the Coefficient B While basic sine and cosine functions have a period of 2, in many applications the period may be very long (tsunami’s) or very short (electromagnetic waves). For the equations y  A sin1Bt2 and y  A cos1Bt2, the period depends on the value of B. To see why, consider the function y  cos12t2 and Table 5.6. Multiplying input values by 2 means each cycle will be completed twice as fast. The table shows that y  cos12t2 completes a full cycle in 30,  4 , giving a period of P   (Figure 5.42, red graph). Table 5.6 t

0

 4

 2

3 4



2t

0

 2



3 2

2

cos(2t)

1

0

1

0

1

Dividing input values by 2 (or multiplying by 12 2 will cause the function to complete a cycle only half as fast, doubling the time required to complete a full cycle. Table 5.7 shows y  cos A 12t B completes only one-half cycle in 2 (Figure 5.42, blue graph). Table 5.7 (values in blue are approximate) t

0

 4

 2

3 4



5 4

3 2

7 4

2

1 t 2

0

 8

 4

3 8

 2

5 8

3 4

7 8



1 cos a tb 2

1

0.92

12 2

0.38

0

0.38

0.92

1



12 2

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Figure 5.42

The graphs of y  cos t, y  cos12t2,

and y  cos A 12t B shown in Figure 5.42 y  cos(2t) 1 clearly illustrate this relationship and how the value of B affects the period of a graph. To find the period for arbitrary values ␲ 2 of B, the formula P  is used. Note for B 1 2 y  cos12t2, B  2 and P   , as 2 2 1 1 shown. For y  cos a tb, B  , and P   4. 2 2 1/2 y

y  cos 12 t

y  cos t

2␲

3␲

t

4␲

Period Formula for Sine and Cosine For B a real number and functions y  A sin1Bt2 and y  A cos1Bt2, 2 . P B To sketch these functions for periods other than 2, we still use a reference rectangle of height 2A and length P, then break the enclosed t-axis in four equal parts to help draw the graph. In general, if the period is “very large” one full cycle is appropriate for the graph. If the period is very small, graph at least two cycles. Note the value of B in Example 6 includes a factor of . This actually happens quite frequently in applications of the trig functions. EXAMPLE 6



Solution



Graphing y  A cos(Bt), Where A, B  1

Draw a sketch of y  2 cos10.4t2 for t in 3, 2 4 .

The amplitude is A  2, so the reference rectangle will be 2122  4 units high. Since A 6 0 the graph will be vertically reflected across the t-axis. The period is 2 P  5 (note the factors of  reduce to 1), so the reference rectangle will 0.4 be 5 units in length. Breaking the t-axis into four parts within the frame (rule of fourths) gives A 14 B 5  54 units, indicating that we should scale the t-axis in multiples 1 15 10 of 4. Note the zeroes occur at 5 4 and 4 , with a maximum value at 4 . In cases where the  factor reduces, we scale the t-axis as a “standard” number line, and estimate the location of multiples of . For practical reasons, we first draw the unreflected graph (shown in blue) for guidance in drawing the reflected graph, which is then extended to fit the given interval. y y  2cos(0.4␲t)

2 ␲

C. You’ve just learned how to graph sine and cosine functions with various amplitudes and periods

3

␲

1

2

1

1

2

3

2␲

4

5

6

t

1 2

y  2 cos(0.4␲t)

Now try Exercises 21 through 32



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D. Graphs of y  csc(Bt) and y  sec(Bt) The graphs of these reciprocal functions follow quite naturally from the graphs of y  A sin1Bt2 and y  A cos1Bt2, by using these observations: (1) you cannot divide by zero, (2) the reciprocal of a very small number is a very large number (and vice versa), and (3) the reciprocal of 1 is 1. Just as with rational functions, division 1 by zero creates a vertical asymptote, so the graph of y  csc t  will have a sin t vertical asymptote at every point where sin t  0. This occurs at t  k, where k is an integer 1p2, , 0, , 2, p2. Further, when csc1Bt2  1, sin1Bt2  1 since the reciprocal of 1 and 1 are still 1 and 1, respectively. Finally, due to observation 2, the graph of the cosecant function will be increasing when the sine function is decreasing, and decreasing when the sine function is increasing. In most cases, we graph y  csc1Bt2 by drawing a sketch of y  sin1Bt2, then using these observations as demonstrated in Example 7. In doing so, we discover that the period of the cosecant function is also 2 and that y  csc1Bt2 is an odd function. EXAMPLE 7



Graphing a Cosecant Function

Solution



The related sine function is y  sin t, which means we’ll draw a rectangular frame 2 2A  2 units high. The period is P   2, so the reference frame will be 2 1 units in length. Breaking the t-axis into four parts within the frame means each tick 1 2  mark will be a b a b  units apart, with the asymptotes occurring at 0, , 4 1 2 and 2. A partial table and the resulting graph are shown.

Graph the function y  csc t for t  3 0, 4 4 .

y

1

t

t

sin t

0

0

 6

1  0.5 2

 4

12  0.71 2

 3  2

13  0.87 2

2  1.41 12 2  1.15 13

1

1

csc t 1 S undefined 0 2 2 1

Now try Exercises 33 and 34

D. You’ve just learned how to investigate graphs of the reciprocal functions f(t)  csc(Bt) and f(t)  sec(Bt)



Similar observations can be made regarding y  sec1Bt2 and its relationship to y  cos1Bt2 (see Exercises 8, 35, and 36). The most important characteristics of the cosecant and secant functions are summarized in the following box. For these functions, there is no discussion of amplitude, and no mention is made of their zeroes since neither graph intersects the t-axis.

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Characteristics of f(t)  csc t and f(t)  sec t For all real numbers t and integers k, y  sec t

y  csc t Domain

t  k

Range

Asymptotes

Domain

Range

t  k

 t   k 2

1q, 14 ´ 31, q 2

1q, 1 4 ´ 3 1, q 2

Asymptotes

t

Period

Symmetry

Period

Symmetry

2

odd csc1t2  csc t

2

even sec1t2  sec t

  k 2

E. Writing Equations from Graphs Mathematical concepts are best reinforced by working with them in both “forward and reverse.” Where graphs are concerned, this means we should attempt to find the equation of a given graph, rather than only using an equation to sketch the graph. Exercises of this type require that you become very familiar with the graph’s basic characteristics and how each is expressed as part of the equation. EXAMPLE 8



Determining the Equation of a Given Graph The graph shown here is of the form y  A sin1Bt2. Find the value of A and B. y 2

y  A sin(Bt)



 2

 2



3 2

2

t

2

Solution



By inspection, the graph has an amplitude of A  2 and a period of P  To find B we used the period formula P  2 B 3 2  2 B 3B  4 4 B 3 P

E. You’ve just learned how to write the equation for a given graph

3 . 2

2 3 , substituting for P and solving. B 2

period formula

substitute

3 for P; B 7 0 2

multiply by 2B solve for B

The result is B  43, which gives us the equation y  2 sin A 43t B . Now try Exercises 37 through 58



There are a number of interesting applications of this “graph to equation” process in the exercise set. See Exercises 61 to 72.

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TECHNOLOGY HIGHLIGHT

Exploring Amplitudes and Periods In practice, trig applications offer an immense range of coefficients, creating amplitudes that are sometimes very large and sometimes extremely small, as well as periods ranging from nanoseconds, to many years. This Technology Highlight is designed to help you use the calculator more effectively in the study of these functions. To begin, we note that many calculators offer a preset ZOOM option that automatically sets a window size convenient to many trig graphs. The resulting WINDOW after pressing ZOOM 7:ZTrig on a TI-84 Plus is shown in Figure 5.43 for a calculator set in Radian MODE . In Section 5.3 we noted that a change in amplitude will not change the location of the zeroes or max/min values. On 1 the Y = screen, enter Y1  sin x, Y2  sin x, Y3  2 sin x, 6.2 2 and Y4  4 sin x , then use ZOOM 7:ZTrig to graph the functions. As you see in Figure 5.44, each graph rises to the expected amplitude at the expected location, while “holding on” to the zeroes. To explore concepts related to the coefficient B and the 1 period of a trig function, enter Y1  sina xb and Y2  sin12x2 2 on the Y = screen and graph using ZOOM 7:ZTrig. While the result is “acceptable,” the graphs are difficult to read and 0 compare, so we manually change the window size to obtain a better view (Figure 5.45). A true test of effective calculator use comes when the amplitude or period is a very large or very small number. For instance, the tone you hear while pressing “5” on your telephone is actually a combination of the tones modeled by Y1  sin 3 217702t 4 and Y2  sin32113362t4 . Graphing these functions requires a careful analysis of the period, otherwise the graph can appear garbled, misleading, or difficult to 10 read —try graphing Y1 on the ZOOM 7:ZTrig or ZOOM 6:ZStandard screens (see Figure 5.46). First note A  1, 2 1 and P  or . With a period this short, even 2770 770 graphing the function from Xmin  1 to Xmax  1 gives a distorted graph. Setting Xmin to 1/770, Xmax to 1/770, and Xscl to (1/770)/10 gives the graph in Figure 5.47, which can be used to investigate characteristics of the function. Exercise 1: Graph the second tone Y2  sin 32113362t4 and find its value at t  0.00025 sec.

Figure 5.43

Figure 5.44 4

6.2

4

Figure 5.45 1.4

2␲

1.4

Figure 5.46 10

10

10

Figure 5.47 1.4

1  770

1 770

Exercise 2: Graph the function Y1  950 sin10.005t2 on a “friendly” window and find the value at x  550. 1.4

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5.3 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. For the sine function, output values are  the interval c 0, d . 2

in

2. For the cosine function, output values are  in the interval c 0 , d . 2 3. For the sine and cosine functions, the domain is and the range is . 

4. The amplitude of sine and cosine is defined to be the maximum from the value in the positive and negative directions. 5. Discuss/Describe the four-step process outlined in this section for the graphing of basic trig functions. Include a worked-out example and a detailed explanation. 6. Discuss/Explain how you would determine the domain and range of y  sec x. Where is this function undefined? Why? Graph y  2 sec12t2 using y  2 cos12t2. What do you notice?

DEVELOPING YOUR SKILLS 7. Use the symmetry of the unit circle and reference arcs of standard values to complete a table of values for y  cos t in the interval t  3 , 2 4 .

8. Use the standard values for y  cos t for t  3 , 24 to create a table of values for y  sec t on the same interval. Use the characteristics of f1t2  sin t to match the given value of t to the correct value of sin t.

  10b 6 15 t 4 21 t 2 1 sin t  2 12 sin t  2

9. a. t  a c. e. II. IV.

  12b 4 23 c. t  2 25 e. t   4

10. a. t  a

b. t  

 4

12 2 12 IV. sin t  2 II. sin t  

III. sin t  0 V. sin t  1

Use steps I through IV given in this section to draw a sketch of each graph.

11. y  sin t for t  c

3  , d 2 2

12. y  sin t for t  3,  4

d. t  13

 13. y  cos t for t  c , 2 d 2

I. sin t  0

 5 14. y  cos t for t  c , d 2 2

III. sin t  1 V. sin t   b. t 

12 2

11 6

d. t  19 I. sin t  

1 2

Use a reference rectangle and the rule of fourths to draw an accurate sketch of the following functions through two complete cycles—one where t  0, and one where t  0. Clearly state the amplitude and period as you begin.

15. y  3 sin t

16. y  4 sin t

17. y  2 cos t

18. y  3 cos t

19. y 

1 sin t 2

20. y 

3 sin t 4

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21. y  sin12t2

22. y  cos12t2

23. y  0.8 cos12t2

24. y  1.7 sin14t2

1 25. f 1t2  4 cos a tb 2

3 26. y  3 cosa tb 4

27. f1t2  3 sin14t2

28. g1t2  5 cos18t2

5 29. y  4 sin a tb 3

2 30. y  2.5 cos a tb 5

31. f 1t2  2 sin1256t2

g.

2

i.

0

t

4

4

j.

1 12

0 2

1 6

1 4

1 3

5 12

t

40. y  3 cos12t2

1 41. y  2 csc a tb 2

1 42. y  2 sec a tb 4

3 43. f 1t2  cos10.4t2 4

7 44. g1t2  cos10.8t2 4

45. y  sec18t2

46. y  csc112t2

47. y  4 sin1144t2 a. 2 y

48. y  4 cos172t2 y b.

8␲

1 12

1 6

1 4

1 3

5 12

␲ 4

␲ 2

3␲ 4

␲

t

t

y

0

l.

y 2

t

y 2 1

␲ 4

0

39. y  3 sin12t2

6␲

4

1

38. y  2 sin14t2

4␲

4

2

4

k.

2␲

2

1

37. y  2 cos14t2

␲ 2

3␲ 4

␲

0

t

1

2

2

The graphs shown are of the form y  A cos(Bt) or y  A csc(Bt). Use the characteristics illustrated for each graph to determine its equation.

49.

50.

y 1 0.5

4 ␲ 4

␲ 8

0 0.5

␲ 2

3␲ 8

5␲ t 8

2

2 5

1 5

0 4

1

51.

y 8

3 5

4 5

t

1

8

52.

y 0.8 0.4

y 0.4 0.2

1 2␲

0 ␲

0

2␲

3␲

4␲

5␲ t

1

1

2

2

d.

y 4

1 144

2

1 72

1 48

1 36

5 t 144

4␲

5␲

6␲ t

53.

y

1 144

2

1 72

1 48

1 36

5 t 144

4␲

6␲

t

0

0.4

0.2

0.8

0.4

54.

y 6

f.

y

2

y 4 2

␲ 2

␲

3␲ 2

2␲

t

0 2 4

0

6

4

4

4

3␲

4

0

4

0

2␲

␲ 4

␲ 2

3␲ 4

␲

t

␲

2␲

3␲

4␲

t

y 1.2

2

0

2

␲

0

2

e.

8␲

y

Clearly state the amplitude and period of each function, then match it with the corresponding graph.

c.

6␲

4

36. f 1t2  3 sec12t2

1

4␲

2

34. g1t2  2 csc14t2

35. y  2 sec t

2␲

2

2

Draw the graph of each function by first sketching the related sine and cosine graphs, and applying the observations made in this section.

y 4 2

0

32. g1t2  3 cos1184t2

33. y  3 csc t

h.

y 4

␲ 2

␲

3␲ 2

2␲

t

1

2

3

4

5 t

0

1.2

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Match each graph to its equation, then graphically estimate the points of intersection. Confirm or contradict your estimate(s) by substituting the values into the given equations using a calculator.

55. y  cos x;1 y  sin x

56. y  cos x; y  sin12x2

y

y

1

1

0.5

0.5

0 0.5

␲ 2

␲

3␲ 2

2␲ x

1



0 0.5

57. y  2 cos x; y  2 sin13x2

58. y  2 cos12x2; y  2 sin1x2

y

y

2

2

1

1

0 1

␲ 2

␲

3␲ 2

0

2␲ x

␲ 2

␲

3␲ 2

1 2

1

2

2 x

3 2

1

2

2␲ x

1

WORKING WITH FORMULAS

59. The Pythagorean theorem in trigonometric form: sin2  cos2  1 The formula shown is commonly known as a Pythagorean identity and is introduced more formally in Chapter 6. It is derived by noting that on a unit circle, cos t  x and sin t  y, while 15 x2  y2  1. Given that sin t  113 , use the formula to find the value of cos t in Quadrant I. What is the Pythagorean triple associated with these values of x and y?



5-44

CHAPTER 5 An Introduction to Trigonometric Functions

60. Hydrostatics, surface tension, and contact 2 cos  angles: y  kr ␪

Capillary

y The height that a liquid will Tube rise in a capillary tube is given by the formula shown, where Liquid r is the radius of the tube,  is the contact angle of the liquid (the meniscus),  is the surface tension of the liquid-vapor film, and k is a constant that depends on the weight-density of the liquid. How high will the liquid rise given that the surface tension   0.2706, the tube has radius r  0.2 cm, the contact angle   22.5°, and k  1.25?

APPLICATIONS

Tidal waves: Tsunamis, also known as tidal waves, are ocean waves produced by earthquakes or other upheavals in the Earth’s crust and can move through the water undetected for hundreds of miles at great speed. While traveling in the open ocean, these waves can be represented by a sine graph with a very long wavelength (period) and a very small amplitude. Tsunami waves only attain a monstrous size as they approach the shore, and represent a very different phenomenon than the ocean swells created by heavy winds over an extended period of time. Height 61. A graph modeling a in feet 2 tsunami wave is given in 1 the figure. (a) What is 20 40 60 80 100 Miles 1 the height of the tsunami 2 wave (from crest to trough)? Note that h  0 is considered the level of a calm ocean. (b) What is the tsunami’s wavelength? (c) Find the equation for this wave.

62. A heavy wind is kicking up ocean swells approximately 10 ft high (from crest to trough), with wavelengths of 250 ft. (a) Find the equation that models these swells. (b) Graph the equation. (c) Determine the height of a wave measured 200 ft from the trough of the previous wave. Sinusoidal models: The sine and cosine functions are of great importance to meteorological studies, as when modeling the temperature based on the time of day, the illumination of the Moon as it goes through its phases, or even the prediction of tidal motion.

63. The graph given shows the deviation from the average daily temperature for the hours of a given day, with t  0

4

Temperature deviation

2 0 2 4

t 4

8

12

16

20

24

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 64. The equation y  7 sin a tb models the height of 6 the tide along a certain coastal area, as compared to average sea level. Assuming t  0 is midnight, (a) graph this function over a 12-hr period. (b) What will the height of the tide be at 5 A.M.? (c) Is the tide rising or falling at this time? Sinusoidal movements: Many animals exhibit a wavelike motion in their movements, as in the tail of a shark as it swims in a straight line or the wingtips of a large bird in flight. Such movements can be modeled by a sine or cosine function and will vary depending on the animal’s size, speed, and other factors.

65. The graph shown models Distance in inches 20 the position of a shark’s 10 t sec tail at time t, as measured 2 3 4 5 1 10 to the left (negative) and 20 right (positive) of a straight line along its length. (a) Use the graph to determine the related equation. (b) Is the tail to the right, left, or at center when t  6.5 sec? How far? (c) Would you say the shark is “swimming leisurely,” or “chasing its prey”? Justify your answer. 66. The State Fish of Hawaii is the humuhumunukunukuapua’a, a small colorful fish found abundantly in coastal waters. Suppose the tail motion of an adult fish is modeled by the equation d1t2  sin115t2 with d(t) representing the position of the fish’s tail at time t, as measured in inches to the left (negative) or right (positive) of a straight line along its length. (a) Graph the equation over two periods. (b) Is the tail to the left or right of center at t  2.7 sec? How far? (c) Would you say this fish is “swimming leisurely,” or “running for cover”? Justify your answer. Kinetic energy: The kinetic energy a planet possesses as it orbits the Sun can be modeled by a cosine function. When the planet is at its apogee (greatest distance from the Sun), its kinetic energy is at its lowest point as it slows down and “turns around” to head back toward the Sun. The kinetic energy is at its highest when the planet “whips around the Sun” to begin a new orbit.

67. Two graphs are given here. (a) Which of the graphs could represent the kinetic energy of a planet

orbiting the Sun if the planet is at its perigee (closest distance to the Sun) when t  0? (b) For what value(s) of t does this planet possess 62.5% of its maximum kinetic energy with the kinetic energy increasing? (c) What is the orbital period of this planet? a. 100 b. 100 75

Percent of KE

corresponding to 6 A.M. (a) Use the graph to determine the related equation. (b) Use the equation to find the deviation at t  11 (5 P.M.) and confirm that this point is on the graph. (c) If the average temperature for this day was 72°, what was the temperature at midnight?

Percent of KE

5-45

50 25 0

75 50 25 0

12 24 36 48 60 72 84 96

12 24 36 48 60 72 84 96

t days

t days

68. The potential energy of the planet is the antipode of its kinetic energy, meaning when kinetic energy is at 100%, the potential energy is 0%, and when kinetic energy is at 0% the potential energy is at 100%. (a) How is the graph of the kinetic energy related to the graph of the potential energy? In other words, what transformation could be applied to the kinetic energy graph to obtain the potential energy graph? (b) If the kinetic energy is at 62.5% and increasing [as in Graph 67(b)], what can be said about the potential energy in the planet’s orbit at this time? Visible light: One of the narrowest bands in the electromagnetic spectrum is the region involving visible light. The wavelengths (periods) of visible light vary from 400 nanometers (purple/violet colors) to 700 nanometers (bright red). The approximate wavelengths of the other colors are shown in the diagram. Violet

Blue

400

Green

Yellow Orange

500

600

Red

700

69. The equations for the colors in this spectrum have 2 the form y  sin1t2, where gives the length  of the sine wave. (a) What color is represented by  tb? (b) What color is the equation y  sina 240 represented by the equation y  sin a

 tb? 310

70. Name the color represented by each of the graphs (a) and (b) here and write the related equation. a. 1 y t (nanometers) 0

1

300

600

900

1200

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470

b.

y 1

t (nanometers) 0

300

600

900

1200

1

Alternating current: Surprisingly, even characteristics of the electric current supplied to your home can be modeled by sine or cosine functions. For alternating current (AC), the amount of current I (in amps) at time t can be modeled by I  A sin1t2, where A represents the maximum current that is produced, and  is related to the frequency at which the generators turn to produce the current. 

71. Find the equation of the household current modeled by the graph, then use the equation to determine I when t  0.045 sec. Verify that the resulting ordered pair is on the graph.

Exercise 71 Current I 30 15

t sec 15

1 50

1 25

3 50

2 25

1 10

30

72. If the voltage produced by an AC circuit is modeled by the equation E  155 sin1120t2, (a) what is the period and amplitude of the related graph? (b) What voltage is produced when t  0.2?

EXTENDING THE CONCEPT

73. For y  A sin1Bx2 and y  A cos1Bx2, the Mm expression gives the average value of the 2 function, where M and m represent the maximum and minimum values, respectively. What was the average value of every function graphed in this section? Compute a table of values for y  2 sin t  3, and note its maximum and minimum values. What is the average value of this function? What transformation has been applied to change the average value of the function? Can you name the average value of y  2 cos t  1 by inspection?



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CHAPTER 5 An Introduction to Trigonometric Functions

2 B came from, consider that if B  1, the graph of y  sin1Bt2  sin11t2 completes one cycle from 1t  0 to 1t  2. If B  1, y  sin1Bt2 completes one cycle from Bt  0 to Bt  2. Discuss how this observation validates the period formula.

74. To understand where the period formula P 

75. The tone you hear when pressing the digit “9” on your telephone is actually a combination of two separate tones, which can be modeled by the functions f 1t2  sin 3 218522t 4 and g1t2  sin 32 114772t4. Which of the two functions has the shortest period? By carefully scaling the axes, graph the function having the shorter period using the steps I through IV discussed in this section.

MAINTAINING YOUR SKILLS

76. (5.2) Given sin 1.12  0.9, find an additional value of t in 30, 22 that makes the equation sin t  0.9 true. Exercise 77 77. (5.1) Use a special triangle to calculate the distance from the ball to the pin on the seventh hole, given the ball is in a straight line with the 100-yd plate, as shown in the 100 yd figure. 60 100 yd

78. (5.1) Invercargill, New Zealand, is at 46°14¿24– south latitude. If the Earth has a radius of 3960 mi, how far is Invercargill from the equator? 79. (1.4) Given z1  1  i and z2  2  5i, compute the following: a. z1  z2 b. z1  z2 c. z1z2 z2 d. z1

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5.4 Graphs of Tangent and Cotangent Functions Learning Objectives

Unlike the other four trig functions, tangent and cotangent have no maximum or minimum value on any open interval of their domain. However, it is precisely this unique feature that adds to their value as mathematical models. Collectively, the six functions give scientists the tools they need to study, explore, and investigate a wide range of phenomena, extending our understanding of the world around us.

In Section 5.4 you will learn how to:

A. Graph y  tan t using asymptotes, zeroes, sin t and the ratio cos t

A. The Graph of y  tan t

B. Graph y  cot t using asymptotes, zeroes, cos t and the ratio sin t

C. Identify and discuss important characteristics of y  tan t and y  cot t

D. Graph y  A tan1Bt2 and y  A cot1Bt2 with various values of A and B

Like the secant and cosecant functions, tangent is defined in terms of a ratio, creating asymptotic behavior at the zeroes of the denominator. In terms of the unit circle, y   tan t  , which means in 3 , 2 4, vertical asymptotes occur at t   , t  , and x 2 2 3 , since the x-coordinate on the unit circle is zero (see Figure 5.48). We further note 2 tan t  0 when the y-coordinate is zero, so the function will have t-intercepts at t  , 0, , and 2 in the same interval. This produces the framework for graphing the tangent function shown in Figure 5.49.

E. Solve applications of

Figure 5.49

y  tan t and y  cot t

tan t

Figure 5.48

Asymptotes at odd multiples of

y (0, 1)



(x, y)

4

 2 

2

t (1, 0)

(0, 0)

(0, 1) y tan t  x

t-intercepts at integer multiples of 

2 

 2

2

2

3 2

t

4

(1, 0)

x

Knowing the graph must go through these zeroes and approach the asymptotes, we are left with determining the direction of the approach. This can be discovered by noting that in QI, the y-coordinates of points on the unit circle start at 0 and increase, y while the x-values start at 1 and decrease. This means the ratio defining tan t is x  increasing, and in fact becomes infinitely large as t gets very close to . A similar 2 observation can be made for a negative rotation of t in QIV. Using the additional points   provided by tan a b  1 and tan a b  1, we find the graph of tan t is increasing 4 4   throughout the interval a , b and that the function has a period of . We also note 2 2 y  tan t is an odd function (symmetric about the origin), since tan1t2  tan t as evidenced by the two points just computed. The completed graph is shown in Figure 5.50 with the primary interval in red. Figure 5.50 tan t 4

 4 , 1

2  

 4 , 1



2

 2

2



3 2

2

t

4 

5-47



471

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CHAPTER 5 An Introduction to Trigonometric Functions

y The graph can also be developed by noting sin t  y, cos t  x, and tan t  . x sin t This gives tan t  by direct substitution and we can quickly complete a table of cos t values for tan t, as shown in Example 1. These and other relationships between the trig functions will be fully explored in Chapter 6. EXAMPLE 1



Constructing a Table of Values for f1t2  tan t y Complete Table 5.8 shown for tan t  using the values given for sin t and cos t, x then graph the function by plotting points. Table 5.8 t

0

 6

 4

 3

 2

2 3

3 4

5 6



sin t  y

0

1 2

12 2

13 2

1

13 2

12 2

1 2

0

cos t  x

1

13 2

12 2

1 2

0



tan t 

Solution



1 2



12 2



13 2

1

y x

For the noninteger values of x and y, the “twos will cancel” each time we compute y . This means we can simply list the ratio of numerators. The resulting points are x shown in Table 5.9, along with the plotted points. The graph shown in Figure 5.51 was completed using symmetry and the previous observations. Table 5.9

t

0

 6

 4

 3

 2

2 3

3 4

5 6



sin t  y

0

1 2

12 2

13 2

1

13 2

12 2

1 2

0

cos t  x

1

13 2

12 2

1 2

0



y x

0

1

23  1.7

undefined

tan t 

1 23

 0.58

1 2



23

12 2



1



13 2 1 23

1 0

Figure 5.51 

 6 , 0.58

f (t)



 4 , 1

4

tan t

2 

2



 3 , 1.7

 2

2



3

2

t 4

3

4

3

2



6

5

2

, 1

, 0.58

, 1.7



Now try Exercises 7 and 8



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473

Section 5.4 Graphs of Tangent and Cotangent Functions

Additional values can be found using a calculator as needed. For future use and reference, it will help to recognize the approximate decimal equivalent of all special 1  0.58. See values and radian angles. In particular, note that 13  1.73 and 13 Exercises 9 through 14.

A. You’ve just learned how to graph y  tan t using asymptotes, zeroes, and the sin t ratio cos t

B. The Graph of y  cot t Since the cotangent function is also defined in terms of a ratio, it too displays asymptotic behavior at the zeroes of the denominator, with t-intercepts at the zeroes of the x numerator. Like the tangent function, cot t  can be written in terms of cos t  x y cos t and sin t  y: cot t  , and the graph obtained by plotting points. sin t EXAMPLE 2



Constructing a Table of Values for f1t2  cot t x for t in 30,  4 using its ratio relationship y with cos t and sin t. Use the results to graph the function for t in 1, 22. Complete a table of values for cot t 

Solution



The completed table is shown here. In this interval, the cotangent function has  asymptotes at 0 and  since y  0 at these points, and has a t-intercept at since 2 x  0. The graph shown in Figure 5.52 was completed using the period P  .

t

0

 6

 4

 3

 2

2 3

3 4

5 6



sin t  y

0

1 2

12 2

13 2

1

13 2

12 2

1 2

0

cos t  x

1

13 2

12 2

1 2

0



undefined

23

1

cot t 

x y

1



0

23

1 2



1

12 2

1

23



13 2

 23

1 undefined

Figure 5.52 cot t 4 2



 2

2

 2



3 2

2

t

4 





Now try Exercises 15 and 16 B. You’ve just learned how to graph y  cot t using asymptotes, zeroes, and the cos t ratio sin t



C. Characteristics of y  tan t and y  cot t The most important characteristics of the tangent and cotangent functions are summarized in the following box. There is no discussion of amplitude, maximum, or minimum values, since maximum or minimum values do not exist. For future use and

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reference, perhaps the most significant characteristic distinguishing tan t from cot t is that tan t increases, while cot t decreases over their respective domains. Also note that due to symmetry, the zeroes of each function are always located halfway between the asymptotes. Characteristics of f(t)  tan t and f(t)  cot t For all real numbers t and integers k, y  tan t Domain

 t   k 2 Period 

y  cot t Range

Asymptotes

R Behavior increasing

EXAMPLE 3



 t   k 2 Symmetry odd tan1t2  tan t

Domain

Range

Asymptotes

t  k

R

t  k

Period

Behavior decreasing

Symmetry



odd cot1t2  cot t

Using the Period of f1t2  tan t to Find Additional Points  7 13 1 , what can you say about tan a b, tan a b, and Given tan a b  6 6 6 13 5 tan a b? 6

Solution



7   by a multiple of : tana b  tana  b, 6 6 6 5 13   tana b  tana  2b and tana b  tana  b. Since the period of 6 6 6 6 1 the tangent function is P  , all of these expressions have a value of . 13 Each value of t differs from

Now try Exercises 17 through 22

C. You’ve just learned how to identify and discuss important characteristics of y  tan t and y  cot t



Since the tangent function is more common than the cotangent, many needed calculations will first be done using the tangent function and its properties, then  reciprocated. For instance, to evaluate cota b we reason that cot t is an odd 6   function, so cota b  cota b. Since cotangent is the reciprocal of tangent and 6 6  1  tana b  , cota b   13. See Exercises 23 and 24. 6 6 13

D. Graphing y  A tan1Bt2 and y  A cot1Bt2 The Coefficient A: Vertical Stretches and Compressions For the tangent and cotangent functions, the role of coefficient A is best seen through an analogy from basic algebra (the concept of amplitude is foreign to these functions). Consider the graph of y  x3 (Figure 5.53). Comparing the parent function y  x3 with

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475

Section 5.4 Graphs of Tangent and Cotangent Functions

functions y  Ax3, the graph is stretched vertically if  A  7 1 (see Figure 5.54) and compressed if 0 6  A  6 1. In the latter case the graph becomes very “flat” near the zeroes, as shown in Figure 5.55. Figure 5.53

Figure 5.54

Figure 5.55

y  x3 y

y  4x3; A  4 y

y  14 x3; A  y

1 4

x

x

x

While cubic functions are not asymptotic, they are a good illustration of A’s effect on the tangent and cotangent functions. Fractional values of A 1  A  6 12 compress the graph, flattening it out near its zeroes. Numerically, this is because a fractional part of  a small quantity is an even smaller quantity. For instance, compare tana b with 6 1    1 tana b. To two decimal places, tana b  0.57, while tana b  0.14, so the 4 6 6 4 6 graph must be “nearer the t-axis” at this value.

EXAMPLE 4



Comparing the Graph of f1t2  tan t and g1t2  A tan t Draw a “comparative sketch” of y  tan t and y  14 tan t on the same axis and discuss similarities and differences. Use the interval 3, 2 4 .

Solution



Both graphs will maintain their essential features (zeroes, asymptotes, period, increasing, and so on). However, the graph of y  14 tan t is vertically compressed, causing it to flatten out near its zeroes and changing how the graph approaches its asymptotes in each interval. y y  tan t y  14 tan t

4 2





 2

2

 2



3 2

2

t

4

Now try Exercises 25 through 28



The Coefficient B: The Period of Tangent and Cotangent WORTHY OF NOTE It may be easier to interpret the phrase “twice as fast” as 2P   and “one-half as fast” as 12P  . In each case, solving for P gives the correct interval for the period of the new function.

Like the other trig functions, the value of B has a material impact on the period of the function, and with the same effect. The graph of y  cot12t2 completes a cycle twice 1  versus P  b, while y  cota tb completes a cycle 2 2 one-half as fast 1P  2 versus P  2. This reasoning leads us to a period formula for tangent and cotangent, namely,  P  , where B is the coefficient of the input variable. B as fast as y  cot t aP 

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Similar to the four-step process used to graph sine and cosine functions, we can  graph tangent and cotangent functions using a rectangle P  units in length and 2A B units high, centered on the primary interval. After dividing the length of the rectangle into fourths, the t-intercept will always be the halfway point, with y-values of  A  occuring at the 41 and 34 marks. See Example 5.

EXAMPLE 5



Graphing y  A cot1Bt2 for A, B,  1

Solution



For y  3 cot12t2,  A   3 which results in a vertical stretch, and B  2 which

Sketch the graph of y  3 cot12t2 over the interval 3 ,  4.

 . The function is still undefined at t  0 and is asymptotic there, 2  then at all integer multiples of P  . We also know the graph is decreasing, with 2  3 zeroes of the function halfway between the asymptotes. The inputs t  and t  8 8 3 1 3   a the and marks between 0 and b yield the points a , 3b and a , 3b, which 4 4 2 8 8 we’ll use along with the period and symmetry of the function to complete the graph: gives a period of

y y  3 cot(2t) 6

 8 , 3

3 



 2



 2

t

6

, 3 3 8

Now try Exercises 29 through 40



As with the trig functions from Section 5.3, it is possible to determine the equation of a tangent or cotangent function from a given graph. Where previously we used the amplitude, period, and max/min values to obtain our equation, here we first determine the period of the function by calculating the “distance” between asymptotes, then choose any convenient point on the graph (other than a t-intercept) and substitute in the equation to solve for A. EXAMPLE 6



Constructing the Equation for a Given Graph Find the equation of the graph, given it’s of the form y  A tan1Bt2. y  A tan(Bt)

y 3 2 1 

2  3

  3

1 2 3

 3

2 3



, 2



2

t

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Section 5.4 Graphs of Tangent and Cotangent Functions

Solution



D. You’ve just learned how to graph y  A tan1Bt2 and y  A cot1Bt2 with various values of A and B

477

  and t  , we find the 3 3 2   2 . To find the value of B we substitute period is P   a b  for P in 3 3 3 3 3  3 P  and find B  (verify). This gives the equation y  A tan a tb. B 2 2   To find A, we take the point a , 2b shown, and use t  with y  2 to 2 2 solve for A: Using the primary interval and the asymptotes at t  

3 y  A tana tb 2 3  2  A tan c a ba b d 2 2 3 2  A tana b 4 2 A 3 tana b 4 2 The equation of the graph is y  2

substitute

3 for B 2

substitute 2 for y and

 for t 2

multiply solve for A

result

tan1 32t2. Now try Exercises 41 through 46



E. Applications of Tangent and Cotangent Functions We end this section with one example of how tangent and cotangent functions can be applied. Numerous others can be found in the exercise set. EXAMPLE 7



Applications of y  A tan1Bt2 : Modeling the Movement of a Light Beam One evening, in port during a Semester at Sea, Richard is debating a project choice for his Precalculus class. Looking out his porthole, he notices a revolving light turning at a constant speed near the corner of a long warehouse. The light throws its beam along the length of the warehouse, then disappears into the air, and then returns time and time again. Suddenly—Richard has his project. He notes the time it takes the beam to traverse the warehouse wall is very close to 4 sec, and in the morning he measures the wall’s length at 127.26 m. His project? Modeling the distance of the beam from the corner of the warehouse as a function of time using a tangent function. Can you help?

Solution



The equation model will have the form D1t2  A tan1Bt2, where D(t) is the distance (in meters) of the beam from the corner after t sec. The distance along the wall is measured in positive values so we’re using only 12 the period of the function, giving 12P  4 (the beam “disappears” at t  4) so P  8. Substitution in the period   formula gives B  and the equation D  A tana tb. 8 8

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Knowing the beam travels 127.26 m in about 4 sec (when it disappears into infinity), we’ll use t  3.9 and D  127.26 in order to solve for A and complete our equation model (see note following this example).  A tana tb  D 8  A tan c 13.92 d  127.26 8 127.26 A  tan c 13.92 d 8 5

equation model

substitute 127.26 for D and 3.9 for t solve for A

result

One equation approximating the distance of the beam from the corner of the  warehouse is D1t2  5 tana tb. 8 Now try Exercises 49 through 52

E. You’ve just learned how to solve applications of y  tan t and y  cot t



For Example 7, we should note the choice of 3.9 for t was arbitrary, and while we obtained an “acceptable” model, different values of A would be generated for other choices. For instance, t  3.95 gives A  2.5, while t  3.99 gives A  0.5. The true value of A depends on the distance of the light from the corner of the warehouse wall. In any case, it’s interesting to note that at t  2 sec (one-half the time it takes the beam to disappear), the beam has traveled only 5m from the corner of the building:  D122  5 tana b  5 m. Although the light is rotating at a constant angular speed, 4 the speed of the beam along the wall increases dramatically as t gets close to 4 sec.

TECHNOLOGY HIGHLIGHT

Zeroes, Asymptotes, and the Tangent/Cotangent Functions In this Technology Highlight we’ll explore the tangent and cotangent functions from the perspective of their ratio definition. While we could easily use Y1  tan x to generate and explore the graph, we would miss an opportunity to note the many important connections that emerge from a ratio definition perspective. To begin, enter Y1 Y1  sin x, Y2  cos x, and Y3  , as shown in Figure 5.56 [recall Y2 that function variables are accessed using VARS (Y-VARS) ENTER (1:Function)]. Note that Y2 has been disabled by overlaying the cursor on the equal sign and pressing ENTER . In addition, note the slash next to Y1 is more bold than the other slashes. The TI-84 Plus offers options that help distinguish between graphs when more than one is being displayed, and we selected a bold line for Y1 by moving the cursor to the far left position and repeatedly pressing ENTER until the desired option appeared. Pressing ZOOM 7:ZTrig at this point produces the screen shown in Figure 5.57, where we note that tan x is zero everywhere that sin x

Figure 5.56

Figure 5.57 4

6.2

6.2

4

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Section 5.4 Graphs of Tangent and Cotangent Functions

Figure 5.58

sin x , but is a cos x point that is often overlooked. Going back to the Y = screen and disabling Y1 while enabling Y2 will produce the graph shown in Figure 5.58. is zero. This is hardly surprising since tan x 

4

6.2

6.2

Exercise 1: What do you notice about the zeroes of cos x as they relate to the graph of Y3  tan x? Y1 Exercise 2: Go to the Y = screen and change Y3 from Y2 Y2 (tangent) to (cotangent), then repeat the Y1 previous investigation regarding y  sin x and y  cos x.

4

5.4 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. The period of y  tan t and y  cot t is ________. To find the period of y  tan1Bt2 and y  cot1Bt2, the formula _________ is used. 2. The function y  tan t is ___________ everywhere it is defined. The function y  cot t is ___________ everywhere it is defined. 3. Tan t and cot t are _______ functions, so f 1t2  11 b,  0.268, then ___________. If tana 12 11 b  _________. tana 12 

4. The asymptotes of y  _________ are located at  odd multiples of . The asymptotes of y  2 _________ are located at integer multiples of . 5. Discuss/Explain how you can obtain a table of values for y  cot t (a) given the values for y  sin t and y  cos t, and (b) given the values for y  tan t. 6. Explain/Discuss how the zeroes of y  sin t and y  cos t are related to the graphs of y  tan t and y  cot t. How can these relationships help graph functions of the form y  A tan1Bt2 and y  A cot1Bt2 ?

DEVELOPING YOUR SKILLS

Use the values given for sin t and cos t to complete the tables.

7.

8. t



7 6

sin t  y

0



cos t  x

1

tan t 

y x



5 4

1 2



12 2

13 2



12 2

4 3 

3 2

3 2

13 2

1

sin t  y

1

1 2

0

cos t  x

0



tan t 

y x

5 3 

13 2 1 2

7 4 

12 2

12 2

11 6

2

1 2

0

13 2

1



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9. Without reference to a text or calculator, attempt to name the decimal equivalent of the following values to one decimal place.  2

 4

 6

12

12 2



3 2

13

13 2

15.

2 13

10. Without reference to a text or calculator, attempt to name the decimal equivalent of the following values to one decimal place.  3

Use the values given for sin t and cos t to complete the tables.

1 13

11. State the value of each expression without the use of a calculator.   a. tana b b. cota b 4 6  3 c. cota b d. tana b 4 3

t



7 6

sin t  y

0



cos t  x

1

cot t 

1 2



12 2

13 2



12 2

4 3 

3 2

13 2

1

1 2

0



x y

16. 3 2 sin t  y

1

cos t  x

0

12. State the value of t without the use of a calculator.  a. cota b b. tan  2 5 5 c. tana b d. cota b 4 6

cot t 

13. State the value of t without the use of a calculator, given t  3 0, 22 terminates in the quadrant indicated. a. tan t  1, t in QIV b. cot t  13, t in QIII 1 , t in QIV c. cot t   13 d. tan t  1, t in QII

18. Given t 

14. State the value of each expression without the use of a calculator, given t  3 0, 22 terminates in the quadrant indicated. a. cot t  1, t in QI b. tan t   13, t in QII 1 , t in QI c. tan t  13 d. cot t  1, t in QIII



5 4

5 3 

13 2

7 4 

1 2

12 2

12 2

11 6

2

1 2

0

13 2

1



x y

11 is a solution to tan t  7.6, use the 24 period of the function to name three additional solutions. Check your answer using a calculator.

17. Given t 

7 is a solution to cot t  0.77, use the 24 period of the function to name three additional solutions. Check your answer using a calculator.

19. Given t  1.5 is a solution to cot t  0.07, use the period of the function to name three additional solutions. Check your answers using a calculator. 20. Given t  1.25 is a solution to tan t  3, use the period of the function to name three additional solutions. Check your answers using a calculator. Verify the value shown for t is a solution to the equation given, then use the period of the function to name all real roots. Check two of these roots on a calculator.

21. t 

 ; tan t  0.3249 10

22. t  

 ; tan t  0.1989 16

23. t 

 ; cot t  2  13 12

24. t 

5 ; cot t  2  13 12

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5-57 Graph each function over the interval indicated, noting the period, asymptotes, zeroes, and value of A. Include a comparative sketch of y  tan t or y  cot t as indicated.

Find the equation of each graph, given it is of the form y  A tan1Bt2.

41.

y

25. f 1t2  2 tan t; 3 2, 24 26. g1t2 

9 

 2 , 3

1 tan t; 3 2, 2 4 2

2





3 2

 2

27. h1t2  3 cot t; 3 2, 24 28. r1t2 

1 cot t; 3 2, 2 4 4

42.

  31. y  cot14t2; c  , d 4 4

 3

 6



 6

1

39. f 1t2  2 cot1t2; 3 1, 14

1  cota tb; 3 4, 44 2 4

t

Find the equation of each graph, given it is of the form y  A cot1Bt2 .

43.

 14 , 2√3 

y 3

3

2

1

1

2

3

t

1 2

t

3

44.

y

 9 , √3  1

1 35. y  5 cota tb; 33, 3 4 3

 38. y  4 tana tb; 32, 2 4 2

 2

2

1

12 ,  2 

3

1 1 37. y  3 tan12t2; c  , d 2 2

 3

y

 2

  33. y  2 tan14t2; c  , d 4 4

  1 36. y  cot 12t2; c  , d 2 2 2

t

1

1 32. y  cota tb; 32, 2 4 2

1 34. y  4 tana tb; 32, 2 4 2



2

  29. y  tan12t2; c  , d 2 2 1 30. y  tana tb; 34, 4 4 4

 2

9

Graph each function over the interval indicated, noting the period, asymptotes, zeroes, and value of A and B.

40. p1t2 

481

Section 5.4 Graphs of Tangent and Cotangent Functions



1 2



1 3



1 6

1 6

1 3

3

3  and t   are solutions to 8 8 cot13t2  tan t, use a graphing calculator to find two additional solutions in 30, 24 .

45. Given that t  

46. Given t  16 is a solution to tan12t2  cot1t2, use a graphing calculator to find two additional solutions in 31, 14 .

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WORKING WITH FORMULAS 48. Position of an image reflected from a spherical h lens: tan   sk

47. The height of an object calculated from a d distance: h  cot u  cot v The height h of a tall structure can be computed using two h angles of elevation measured some u v distance apart along a d xd straight line with the object. This height is given by the formula shown, where d is the distance between the two points from which angles u and v were measured. Find the height h of a building if u  40°, v  65°, and d  100 ft.



The equation Lens shown is used to help locate the h position of an ␪ k P P image reflected Object Reflected by a spherical image mirror, where s s is the distance of the object from the lens along a horizontal axis,  is the angle of elevation from this axis, h is the altitude of the right triangle indicated, and k is distance from the lens to the foot of altitude h. Find the distance k  given h  3 mm,   , and that the object is 24 24 mm from the lens.

APPLICATIONS

Tangent function data models: Model the data in Exercises 49 and 50 using the function y  A tan(Bx). State the period of the function, the location of the asymptotes, the value of A, and name the point (x, y) used to calculate A (answers may vary). Use your equation model to evaluate the function at x  2 and x  2. What observations can you make? Also see Exercise 58.

49.

50.

Input

Output

Input

Output

6

q

1

1.4

5

20

2

3

4

9.7

3

5.2

3

5.2

4

9.7

2

3

5

20

1

1.4

6

q

0

0

Input

Output

Input

Output

3

q

0.5

6.4

2.5

91.3

1

13.7

2

44.3

1.5

23.7

1.5

23.7

2

44.3

1

13.7

2.5

91.3

0.5

6.4

3

q

0

0

Exercise 51 51. As part of a lab setup, a laser pen is made to swivel on a large protractor as illustrated in the figure. For their lab project, students are asked to take the Distance  instrument to one end of (degrees) (cm) a long hallway and 0 0 measure the distance of 10 2.1 the projected beam relative to the angle the 20 4.4 pen is being held, and 30 6.9 collect the data in a 40 10.1 table. Use the data to 50 14.3 find a function of the 60 20.8 form y  A tan1B2. 70 33.0 State the period of the function, the location of 80 68.1 the asymptotes, the value 89 687.5 of A, and name the point (, y) you used to calculate A (answers may vary). Based on the result, can you approximate the length of the laser pen? Note that in degrees, the 180° . period formula for tangent is P  B Laser Light



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CHAPTER 5 An Introduction to Trigonometric Functions

52. Use the equation model obtained in Exercise 51 to compare the values given by the equation with the actual data. As a percentage, what was the largest deviation between the two?

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5-59 Exercise 53 53. Circumscribed polygons: The perimeter of a regular polygon circumscribed about a circle of radius r is  r given by P  2nr tana b, n where n is the number of sides 1n  32 and r is the radius of the circle. Given r  10 cm, (a) What is the circumference of the circle? (b) What is the perimeter of the polygon when n  4? Why? (c) Calculate the perimeter of the polygon for n  10, 20, 30, and 100. What do you notice?

54. Circumscribed polygons: The area of a regular polygon circumscribed about a circle of radius r is  given by A  nr2tana b, where n is the number n of sides 1n  32 and r is the radius of the circle. Given r  10 cm, a. What is the area of the circle? b. What is the area of the polygon when n  4? Why? c. Calculate the area of the polygon for n  10, 20, 30, and 100. What do you notice? Coefficients of friction: Material Coefficient Pulling someone on a steel on steel 0.74 sled is much easier copper on glass 0.53 during the winter than in the summer, due to glass on glass 0.94 a phenomenon known copper on steel 0.68 as the coefficient of wood on wood 0.5 friction. The friction between the sled’s skids and the snow is much lower than the friction between the skids and the dry ground or pavement. Basically, the coefficient of friction is defined by the relationship m  tan , where  is the angle at which a block composed of one material will slide down an inclined plane made of another material, with a constant velocity. Coefficients of friction have been established experimentally for many materials and a short list is shown here.

55. Graph the function   tan , with  in degrees over the interval 3 0°, 60° 4 and use the graph to estimate solutions to the following. Confirm or contradict your estimates using a calculator. a. A block of copper is placed on a sheet of steel, which is slowly inclined. Is the block of copper moving when the angle of inclination is 30°? At what angle of inclination will the copper block be moving with a constant velocity down the incline?

Section 5.4 Graphs of Tangent and Cotangent Functions

483

b. A block of copper is placed on a sheet of castiron. As the cast-iron sheet is slowly inclined, the copper block begins sliding at a constant velocity when the angle of inclination is approximately 46.5°. What is the coefficient of friction for copper on cast-iron? c. Why do you suppose coefficients of friction greater than   2.5 are extremely rare? Give an example of two materials that likely have a high m-value. 56. Graph the function   tan  with  in radians 5 d and use the graph to over the interval c 0, 12 estimate solutions to the following. Confirm or contradict your estimates using a calculator. a. A block of glass is placed on a sheet of glass, which is slowly inclined. Is the block of glass  moving when the angle of inclination is ? 4 What is the smallest angle of inclination for which the glass block will be moving with a constant velocity down the incline (rounded to four decimal places)? b. A block of Teflon is placed on a sheet of steel. As the steel sheet is slowly inclined, the Teflon block begins sliding at a constant velocity when the angle of inclination is approximately 0.04. What is the coefficient of friction for Teflon on steel? c. Why do you suppose coefficients of friction less than   0.04 are extremely rare for two solid materials? Give an example of two materials that likely have a very low m value. 57. Tangent lines: The actual definition of the word tangent comes from the tan  Latin tangere, meaning “to touch.” In mathematics, a tangent line touches the  1 graph of a circle at only one point and function values for tan  are obtained from the length of the line segment tangent to a unit circle. a. What is the length of the line segment when   80°? b. If the line segment is 16.35 units long, what is the value of ? c. Can the line segment ever be greater than 100 units long? Why or why not? d. How does your answer to (c) relate to the asymptotic behavior of the graph?

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EXTENDING THE CONCEPT

58. Rework Exercises 49 and 50, obtaining a new equation for the data using a different ordered pair to compute the value of A. What do you notice? Try yet another ordered pair and calculate A once again for another equation Y2. Complete a table of values using the given inputs, with the outputs of the three equations generated (original, Y1, and Y2). Does any one equation seem to model the data better than the others? Are all of the equation models “acceptable”? Please comment. 59. Regarding Example 7, we can use the standard distance/rate/time formula D  RT to compute the 

5-60

CHAPTER 5 An Introduction to Trigonometric Functions

average velocity of the beam of light along the wall D in any interval of time: R  . For example, using T  D1t2  5 tana tb, the average velocity in the 8 D122  D102  2.5 m/sec. interval [0, 2] is 20 Calculate the average velocity of the beam in the time intervals [2, 3], [3, 3.5], and [3.5, 3.8] sec. What do you notice? How would the average velocity of the beam in the interval [3.9, 3.99] sec compare?

MAINTAINING YOUR SKILLS

60. (5.1) A lune is a section of surface area on a sphere, which is subtended by an  angle  at the r circumference. For  in radians, the surface area of a lune is A  2r2, where r is the radius of the sphere. Find the area of a lune on the surface of the Earth which is subtended by an angle of 15°. Assume the radius of the Earth is 6373 km. 61. (3.4/3.5) Find the y-intercept, x-intercept(s), and all asymptotes of each function, but do not graph. 3x2  9x x1 a. h1x2  b. t1x2  2 2 2x  8 x  4x

c. p1x2 

x2  1 x2

62. (5.2) State the points on the unit circle that   3 3 correspond to t  0, , , , , , and 2. 4 2 4 2  What is the value of tana b? Why? 2 63. (4.1) The radioactive element potassium-42 is sometimes used as a tracer in certain biological experiments, and its decay can be modeled by the formula Q1t2  Q0e0.055t, where Q(t) is the amount that remains after t hours. If 15 grams (g) of potassium-42 are initially present, how many hours until only 10 g remain?

MID-CHAPTER CHECK 1. The city of Las Vegas, Nevada, is located at 36°06¿36– north latitude, 115°04¿48– west longitude. (a) Convert both measures to decimal Exercise 2 degrees. (b) If the radius of y the Earth is 3960 mi, how far north of the equator is Las 86 cm Vegas? 2. Find the angle subtended by the arc shown in the figure, then determine the area of the sector.

3. Evaluate without using a 7 calculator: (a) cot 60° and (b) sin a b. 4  4. Evaluate using a calculator: (a) sec a b and 12 (b) tan 83.6°.

 20 cm

x

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Reinforcing Basic Concepts

Exercise 5

5. Complete the ordered pair indicated on the unit circle in the figure and find the value of all six trigonometric functions at this point.

9. On a unit circle, if arc t has length 5.94, (a) in what quadrant does it terminate? (b) What is its reference arc? (c) Of sin t, cos t, and tan t, which are negative for this value of t?

y

 1

6. For the point on the unit circle in Exercise 5, find the related angle t in both degrees (to tenths) and radians (to ten-thousandths).

x

√53 , y

7. Name the location of the asymptotes and graph  y  3 tana tb for t  3 2, 2 4. 2

10. For the graph given here, (a) clearly state the amplitude and period; (b) find the equation of the graph; (c) graphically find f 12 and then confirm/contradict your estimation using a calculator.

Exercise 10 y 8

f(t)

4

0

 4

 2

3 4



5 4

3 2

t

4 8

8. Clearly state the amplitude and period, then sketch  the graph: y  3 cosa tb. 2

REINFORCING BASIC CONCEPTS Trigonometry of the Real Numbers and the Wrapping Function The circular functions are sometimes discussed in terms of what is called a wrapping function, in which the real number line is literally wrapped around the unit circle. This approach can help illustrate how the trig functions can be seen as functions of the real numbers, and apart from any reference to a right triangle. Figure 5.59 shows (1) a unit circle with the location of certain points on the circumference clearly marked and (2) a number line that has been marked in multiples of  to coincide with the length of the special arcs (integers are shown in the background). Figure 5.60 shows this same 12 number line wrapped counterclockwise around the unit circle in the positive direction. Note how the resulting diagram   12 12 12 , b on the unit circle: cos  confirms that an arc of length t  is associated with the point a and 4 2 2 4 2  12 5 13 1 5 13 5 1 sin  , b: cos   . ; while an arc of length of t  is associated with the point a and sin 4 2 6 2 2 6 2 6 2 Use this information to complete the exercises given. Figure 5.60

Figure 5.59 12, √32  √22 , √22 



√3 1  2, 2

(1, 0)



(0, 1) y

 12 , √32  √22 , √22  √32 , 12 

 45 4

7 2 12 3 3 2 4

1

x 0

 12

 6

 4

 3

2 5  12 2

3

7 2 3 5 11  12 3 4 6 12

t

5 6 11 12 3 

y

 2

5 12  3 1

45

 4

 4

 6

 12

0 x

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CHAPTER 5 An Introduction to Trigonometric Functions

1. What is the ordered pair associated with an arc length of t  2. What arc length t is associated with the ordered pair a

2 ? What is the value of cos t? sin t? 3

13 1 , b? Is cos t positive or negative? Why? 2 2

3. If we continued to wrap this number line all the way around the circle, in what quadrant would an arc length of 11 t terminate? Would sin t be positive or negative? 6 4. Suppose we wrapped a number line with negative values clockwise around the unit circle. In what quadrant would 5 an arc length of t   terminate? What is cos t? sin t? What positive rotation terminates at the same point? 3

5.5 Transformations and Applications of Trigonometric Graphs Learning Objectives In Section 5.5 you will learn how to:

A. Apply vertical translations in context

B. Apply horizontal translations in context

From your algebra experience, you may remember beginning with a study of linear graphs, then moving on to quadratic graphs and their characteristics. By combining and extending the knowledge you gained, you were able to investigate and understand a variety of polynomial graphs—along with some powerful applications. A study of trigonometry follows a similar pattern, and by “combining and extending” our understanding of the basic trig graphs, we’ll look at some powerful applications in this section.

C. Solve applications

A. Vertical Translations: y  A sin1Bt2  D

involving harmonic motion

Figure 5.61 C 15

6

15

12

18

24

t

On any given day, outdoor temperatures tend to follow a sinusoidal pattern, or a pattern that can be modeled by a sine function. As the sun rises, the morning temperature begins to warm and rise until reaching its high in the late afternoon, then begins to cool during the early evening and nighttime hours until falling to its nighttime low just prior to sunrise. Next morning, the cycle begins again. In the northern latitudes where the winters are very cold, it’s not unreasonable to assume an average daily temperature of 0°C 132°F2, and a temperature graph in degrees Celsius that looks like the one in Figure 5.61. For the moment, we’ll assume that 2 t  0 corresponds to 12:00 noon. Note that A  15 and P  24, yielding 24  B  or B  . 12 If you live in a more temperate area, the daily temperatures still follow a sinusoidal pattern, but the average temperature could be much higher. This is an example of a vertical shift, and is the role D plays in the equation y  A sin1Bt2  D. All other aspects of a graph remain the same; it is simply shifted D units up if D 7 0 and D units down if D 6 0. As in Section 5.3, for maximum value M and minimum value m, Mm Mm gives the amplitude A of a sine curve, while gives the average value D. 2 2

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Section 5.5 Transformations and Applications of Trigonometric Graphs

EXAMPLE 1

Solution





Modeling Temperature Using a Sine Function On a fine day in Galveston, Texas, the high temperature might be about 85°F with an overnight low of 61°F. a. Find a sinusoidal equation model for the daily temperature. b. Sketch the graph. c. Approximate what time(s) of day the temperature is 65°F. Assume t  0 corresponds to 12:00 noon.  a. We first note the period is still P  24, so B  , and the equation model 12  85  61 Mm will have the form y  A sina tb  D. Using  , we find 12 2 2 85  61 the average value D  73, with amplitude A   12. The resulting 2  equation is y  12 sina tb  73. 12 b. To sketch the graph, use a reference rectangle 2A  24 units tall and P  24 units wide, along with the rule of fourths to locate zeroes and max/min values (see Figure 5.62). Then lightly sketch a sine curve through these points and  within the rectangle as shown. This is the graph of y  12 sina tb  0. 12 Using an appropriate scale, shift the rectangle and plotted points vertically upward 73 units and carefully draw the finished graph through the points and within the rectangle (see Figure 5.63). Figure 5.62

Figure 5.63

F

90

12

␲

y  12 sin 12 t

6

85

t (hours)

WORTHY OF NOTE Recall from Section 5.5 that transformations of any function y  f1x2 remain consistent regardless of the function f used. For the sine function, the transformation y  af1x  h2  k is more commonly written y  A sin1t  C2  D, and A gives a vertical stretch or compression, C is a horizontal shift opposite the sign, and D is a vertical shift, as seen in Example 1.

487

0

F

␲

y  12 sin 12 t  73

80 75

6

12

18

24

6 12

Average value

70 65 60

(c) t (hours)

0 6

12

18

24

 tb  73. Note the brokenline notation 12 “ ” in Figure 5.63 indicates that certain values along an axis are unused (in this case, we skipped 0° to 60°2, and we began scaling the axis with the values needed. This gives the graph of y  12 sina

c. As indicated in Figure 5.63, the temperature hits 65° twice, at about 15 and 21 hr after 12:00 noon, or at 3:00 A.M. and 9:00 A.M. Verify by computing f(15) and f(21). Now try Exercises 7 through 18



Sinusoidal graphs actually include both sine and cosine graphs, the difference being that sine graphs begin at the average value, while cosine graphs begin at the maximum value. Sometimes it’s more advantageous to use one over the other, but equivalent forms can easily be found. In Example 2, a cosine function is used to model an animal population that fluctuates sinusoidally due to changes in food supplies.

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CHAPTER 5 An Introduction to Trigonometric Functions

EXAMPLE 2



Modeling Population Fluctuations Using a Cosine Function The population of a certain animal species can be modeled by the function  P1t2  1200 cos a tb  9000, where P1t2 represents the population in year t. 5 Use the model to a. b. c. d.

Solution



Find the period of the function. Graph the function over one period. Find the maximum and minimum values. Estimate the number of years the population is less than 8000.  2 , the period is P   10, meaning the population of this 5 /5 species fluctuates over a 10-yr cycle.

a. Since B 

b. Use a reference rectangle (2A  2400 by P  10 units) and the rule of fourths to locate zeroes and max/min values, then sketch the unshifted graph  y  1200 cos a tb. With P  10, these occur at t  0, 2.5, 5, 7.5, and 10 5 (see Figure 5.64). Shift this graph upward 9000 units (using an appropriate scale) to obtain the graph of P(t) shown in Figure 5.65. Figure 5.65

Figure 5.64 P 1500

P

␲

10,500

y  1200 cos  5 t

␲

P(t)  1200 cos  5 t  9000

10,000

1000

9500

500

t (years)

9000

10

8500

Average value

0 500 1000 1500

2

4

6

8

8000

(d)

7500

t (years)

0 2

A. You’ve just learned how to apply vertical translations in context

4

6

8

10

c. The maximum value is 9000  1200  10,200 and the minimum value is 9000  1200  7800. d. As determined from the graph, the population drops below 8000 animals for approximately 2 yr. Verify by computing P(4) and P(6). Now try Exercises 19 and 20



B. Horizontal Translations: y  A sin1Bt  C2  D In some cases, scientists would rather “benchmark” their study of sinusoidal phenomena by placing the average value at t  0 instead of a maximum value (as in Example 2), or by placing the maximum or minimum value at t  0 instead of the average value (as in Example 1). Rather than make additional studies or recompute using available data, we can simply shift these graphs using a horizontal translation. To help understand how, consider the graph of y  x2. The graph is a parabola, concave up, with a vertex at the origin. Comparing this function with y1  1x  32 2 and

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Section 5.5 Transformations and Applications of Trigonometric Graphs

y2  1x  32 2, we note y1 is simply the parent graph shifted 3 units right, and y2 is the parent graph shifted 3 units left (“opposite the sign”). See Figures 5.66 through 5.68. While quadratic functions have no maximum value if A  0, these graphs are a good reminder of how a basic graph can be horizontally shifted. We simply replace the independent variable x with 1x  h2 or t with 1t  h2, where h is the desired shift and the sign is chosen depending on the direction of the shift. Figure 5.66 y  x2

x



Figure 5.68

y1  (x  3)2

y2  (x  3)2

y

y

EXAMPLE 3

Figure 5.67

y

x

3

3

x

Investigating Horizontal Shifts of Trigonometric Graphs Use a horizontal translation to shift the graph from Example 2 so that the average population begins at t  0. Verify the result on a graphing calculator, then find a sine function that gives the same graph as the shifted cosine function.

11,000

0

10

7000

Solution



 For P1t2  1200 cosa tb  9000 from Example 2, the average value first occurs 5 at t  2.5. For the average value to occur at t  0, we must shift the graph to the right  2.5 units. Replacing t with 1t  2.52 gives P1t2  1200 cos c 1t  2.52 d  9000. 5 A graphing calculator shows the desired result is obtained (see figure). The new graph appears to be a sine function with the same amplitude and period, and the  equation is y  1200 sina tb  9000. 5 Now try Exercises 21 and 22

WORTHY OF NOTE When the function  P1t2  1200 cos c 1t  2.52 d 5  9000 is written in standard form as P1t2  1200   cos c t  d  9000, we 5 2 can easily see why they are equivalent to P1t2  1200  sina tb  9000. Using the 5 cofunction relationship,    cos c t  d  sina tb. 5 2 5



 Equations like P1t2  1200 cos c 1t  2.52 d  9000 from Example 3 are said 5 to be written in shifted form, since we can easily tell the magnitude and direction of the shift. To obtain the standard form we distribute the value of B:   P1t2  1200 cosa t  b  9000. In general, the standard form of a sinusoidal 5 2 equation (using either a cosine or sine function) is written y  A sin1Bt  C2  D, with the shifted form found by factoring out B from Bt  C : y  A sin1Bt  C2  D S y  A sin c B at 

C bd  D B

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CHAPTER 5 An Introduction to Trigonometric Functions

In either case, C gives what is known as the phase angle of the function, and is used in a study of AC circuits and other areas, to discuss how far a given function is C “out of phase” with a reference function. In the latter case, is simply the horizontal B shift (or phase shift) of the function and gives the magnitude and direction of this shift (opposite the sign). Characteristics of Sinusoidal Models Transformations of the graph of y  sin t are written as y  A sin1Bt2, where 1.  A  gives the amplitude of the graph, or the maximum displacement from the average value. 2 2. B is related to the period P of the graph according to the ratio P  B (the interval required for one complete cycle). Translations of y  A sin1Bt2 can be written as follows:

WORTHY OF NOTE It’s important that you don’t confuse the standard form with the shifted form. Each has a place and purpose, but the horizontal shift can be identified only by focusing on the change in an independent variable. Even though the equations y  41x  32 2 and y  12x  62 2 are equivalent, only the first explicitly shows that y  4x2 has been shifted three units left. Likewise y  sin 321t  32 4 and y  sin12t  62 are equivalent, but only the first explicitly gives the horizontal shift (three units left). Applications involving a horizontal shift come in an infinite variety, and the shifts are generally not uniform or standard.

EXAMPLE 4

Standard form

Shifted form

C bd  D B C 3. In either case, C is called the phase angle of the graph, while  gives the B magnitude and direction of the horizontal shift (opposite the given sign). y  A sin1Bt  C2  D

y  A sin c Bat 

4. D gives the vertical shift of the graph, and the location of the average value. The shift will be in the same direction as the given sign.

Knowing where each cycle begins and ends is a helpful part of sketching a graph of the equation model. The primary interval for a sinusoidal graph can be found by solving the inequality 0  Bt  C 6 2, with the reference rectangle and rule of fourths giving the zeroes, max/min values, and a sketch of the graph in this interval. The graph can then be extended in either direction, and shifted vertically as needed. 

Analyzing the Transformation of a Trig Function Identify the amplitude, period, horizontal shift, vertical shift (average value), and endpoints of the primary interval.  3 y  2.5 sina t  b6 4 4

Solution



The equation gives an amplitude of  A   2.5, with an average value of D  6. The maximum value will be y  2.5112  6  8.5, with a minimum of 2  , the period is P   8. To find the 4 /4   3 b horizontal shift, we factor out to write the equation in shifted form: a t  4 4 4  1t  32. The horizontal shift is 3 units left. For the endpoints of the primary interval 4  we solve 0  1t  32 6 2, which gives 3  t 6 5. 4 y  2.5112  6  3.5. With B 

Now try Exercises 23 through 34



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Section 5.5 Transformations and Applications of Trigonometric Graphs

GRAPHICAL SUPPORT  The analysis of y  2.5 sin c 1t  32 d  6 from 4 Example 4 can be verified on a graphing calculator. Enter the function as Y1 on the Y = screen and set an appropriate window size using the information gathered. Press the TRACE key and 3 ENTER and the calculator gives the average value y  6 as output. Repeating this for x  5 shows one complete cycle has been completed.

10

3

7

0

To help gain a better understanding of sinusoidal functions, their graphs, and the role the coefficients A, B, C, and D play, it’s often helpful to reconstruct the equation of a given graph. EXAMPLE 5



Determining the Equation of a Trig Function from Its Graph Determine the equation of the given graph using a sine function.

Solution

B. You’ve just learned how to apply horizontal translations in context



From the graph it is apparent the maximum value is 300, with a minimum of 50.This gives a value 300  50 300  50 of  175 for D and  125 2 2 for A. The graph completes one cycle from t  2  to t  18, showing P  18  2  16 and B  . 8 The average value first occurs at t  2, so the basic graph has been shifted to the right 2 units.  The equation is y  125 sin c 1t  22 d  175. 8

y 350 300 250 200 150 100 50 0

4

8

12

16

20

24

Now try Exercises 35 through 44

t



C. Simple Harmonic Motion: y  A sin1Bt2 or y  A cos1Bt2 The periodic motion of springs, tides, sound, and other phenomena all exhibit what is known as harmonic motion, which can be modeled using sinusoidal functions.

Harmonic Models—Springs Consider a spring hanging from a beam with a weight attached to one end. When the weight is at rest, we say it is in equilibrium, or has zero displacement from center. Stretching the spring and then releasing it causes the weight to “bounce up and down,” with its displacement from center neatly modeled over time by a sine wave (see Figure 5.69).

Figure 5.69 At rest

Stretched

Released

4

4

4

2

2

2

0

0

0

2

2

2

4

4

4

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For objects in harmonic motion (there are other harmonic models), the input variable t is always a time unit (seconds, minutes, days, etc.), so in addition to the period of the sinusoid, we are very interested in its frequency—the number of cycles it completes per unit time (see Figure 5.70). Since the period gives the time required to complete one 1 B . cycle, the frequency f is given by f   P 2 EXAMPLE 6



Figure 5.70 Harmonic motion Displacement (cm) 4

t (seconds) 0 0.5

1.0

1.5

2.0

2.5

4

Applications of Sine and Cosine: Harmonic Motion For the harmonic motion modeled by the sinusoid in Figure 5.70, a. Find an equation of the form y  A cos1Bt2 . b. Determine the frequency. c. Use the equation to find the position of the weight at t  1.8 sec.

Solution



a. By inspection the graph has an amplitude  A   3 and a period P  2. After 2 , we obtain B   and the equation y  3 cos1t2. substitution into P  B 1 b. Frequency is the reciprocal of the period so f  , showing one-half a cycle is 2 completed each second (as the graph indicates). c. Evaluating the model at t  1.8 gives y  3 cos 311.82 4  2.43, meaning the weight is 2.43 cm below the equilibrium point at this time. Now try Exercises 47 through 50



Harmonic Models—Sound Waves A second example of harmonic motion is the production of sound. For the purposes of this study, we’ll look at musical notes. The vibration of matter produces a pressure wave or sound energy, which in turn vibrates the eardrum. Through the intricate structure of the middle ear, this sound energy is converted into mechanical energy and sent to the inner ear where it is converted to nerve impulses and transmitted to the brain. If the sound wave has a high frequency, the eardrum vibrates with greater frequency, which the brain interprets as a “high-pitched” sound. The intensity of the sound wave can also be transmitted to the brain via these mechanisms, and if the arriving sound wave has a high amplitude, the eardrum vibrates more forcefully and the sound is interpreted as “loud” by the brain. These characteristics are neatly modeled using y  A sin1Bt2 . For the moment we will focus on the frequency, keeping the amplitude constant at A  1. The musical note known as A4 or “the A above middle C” is produced with a frequency of 440 vibrations per second, or 440 hertz (Hz) (this is the note most often used in the tuning of pianos and other musical instruments). For any given note, the same note one octave higher will have double the frequency, and the same note one octave 1 lower will have one-half the frequency. In addition, with f  the value of P 1 B  2a b can always be expressed as B  2f , so A4 has the equation P y  sin 344012t2 4 (after rearranging the factors). The same note one octave lower is A3

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and has the equation y  sin 322012t2 4 , with one-half the frequency. To draw the representative graphs, we must scale the t-axis in very small increments (seconds  103) 1  0.0023 for A4, and since P  440 1  0.0045 for A3. Both are graphed P 220 in Figure 5.71, where we see that the higher note completes two cycles in the same interval that the lower note completes one. EXAMPLE 7



Figure 5.71 A4 A3

y

y  sin[440(2␲t)] y  sin[220(2␲t)]

1

t (sec  103) 0 1

2

3

4

5

1

Applications of Sine and Cosine: Sound Frequencies The table here gives the frequencies for three octaves of the 12 “chromatic” notes with frequencies between 110 Hz and 840 Hz. Two of the 36 notes are graphed in the figure. Which two? y 1

0

1

Solution



C. You’ve just learned how to solve applications involving harmonic motion

y1  sin[ f (2␲t)]

Frequency by Octave

y2  sin[ f (2␲t)] t (sec  1.0 2.0 3.0 4.0 5.0 6.0 7.0

103)

Note

Octave 3

Octave 4

Octave 5

A

110.00

220.00

440.00

A#

116.54

233.08

466.16

B

123.48

246.96

493.92

C

130.82

261.64

523.28

C#

138.60

277.20

554.40

D

146.84

293.68

587.36

D#

155.56

311.12

622.24

E

164.82

329.24

659.28

F

174.62

349.24

698.48

F#

185.00

370.00

740.00

G

196.00

392.00

784.00

G#

207.66

415.32

830.64

Since amplitudes are equal, the only difference is the frequency and period of the notes. It appears that y1 has a period of about 0.004 sec, giving a frequency of 1  250 Hz—very likely a B4 (in bold). The graph of y2 has a period of about 0.004 1 0.006, for a frequency of  167 Hz—probably an E3 (also in bold). 0.006 Now try Exercises 51 through 54



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TECHNOLOGY HIGHLIGHT

Locating Zeroes, Roots, and x-Intercepts As you know, the zeroes of a function are input values that cause an output of zero. Graphically, these show up as x-intercepts and once a function is graphed they can be located (if they exist) using the 2nd CALC 2:zero feature. This feature is similar to the 3:minimum and 4:maximum features, in that we have the calculator search a specified interval by giving a left bound and a right bound. To illustrate,  enter Y1  3 sina xb  1 on the Y = screen and graph it 2

Figure 5.72 4

6.2

6.2

4 using the ZOOM 7:ZTrig option. The resulting graph shows there are six zeroes in this interval and we’ll locate the first negative root. Knowing the 7:Trig option

uses tick marks that are spaced every CALC

  units, this root is in the interval a,  b. After pressing 2 2

2nd

2:zero the calculator returns you to the graph, and requests a “Left Bound,” (see Figure 5.72).

We enter  (press

ENTER

) and the calculator marks this choice with a “ N ” marker (pointing to the right),  then asks for a “Right Bound.” After entering  , the calculator marks this with a “ > ” marker and asks 2 for a “Guess.” Bypass this option by pressing ENTER once again (see Figure 5.73). The calculator searches the interval until it locates a zero (Figure 5.74) or displays an error message indicating it was unable to comply (no zeroes in the interval). Use these ideas to locate the zeroes of the following functions in [0, ].

Figure 5.73

Figure 5.74

4

6.2

6.2

4

Exercise 1: y  2 cos1t2  1 Exercise 3: y 

4

3 tan12x2  1 2

6.2

6.2

4

Exercise 2: y  0.5 sin 3 1t  22 4 Exercise 4: y  x3  cos x

5.5 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. A sinusoidal wave is one that can be modeled by functions of the form ______________ or _______________.

2. The graph of y  sin x  k is the graph of y  sin x shifted __________ k units. The graph of y  sin1x  h2 is the graph of y  sin x shifted __________ h units.

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3. To find the primary interval of a sinusoidal graph, solve the inequality ____________. 5. Explain/Discuss the difference between the standard form of a sinusoidal equation, and the shifted form. How do you obtain one from the other? For what benefit?



495

4. Given the period P, the frequency is __________, and given the frequency f, the value of B is __________. 6. Write out a step-by-step procedure for sketching  1 the graph of y  30 sina t  b  10. Include 2 2 use of the reference rectangle, primary interval, zeroes, max/mins, and so on. Be complete and thorough.

DEVELOPING YOUR SKILLS

Use the graphs given to (a) state the amplitude A and period P of the function; (b) estimate the value at x  14; and (c) estimate the interval in [0, P] where f (x)  20.

7.

50

8.

f (x)

6

12

18

24

30

50

f (x)

x

5

50

10 15 20 25

30

x

50

Use the graphs given to (a) state the amplitude A and period P of the function; (b) estimate the value at x  2; and (c) estimate the interval in [0, P], where f (x)  100.

9.

10.

f (x) 250

3 3 9 3 15 9 21 6 27 4 2 4 4 2 4 4

125

x

250

f (x)

1

2

3

4

5

6

7

8 x

125

Use the information given to write a sinusoidal equation 2 and sketch its graph. Recall B  . P

11. Max: 100, min: 20, P  30 12. Max: 95, min: 40, P  24 13. Max: 20, min: 4, P  360 14. Max: 12,000, min: 6500, P  10 Use the information given to write a sinusoidal equation, sketch its graph, and answer the question posed.

15. In Geneva, Switzerland, the daily temperature in January ranges from an average high of 39°F to an average low of 29°F. (a) Find a sinusoidal equation model for the daily temperature; (b) sketch the graph; and (c) approximate the time(s) each January day the temperature reaches the freezing point (32°F). Assume t  0 corresponds to noon. Source: 2004 Statistical Abstract of the United States, Table 1331.

16. In Nairobi, Kenya, the daily temperature in January ranges from an average high of 77°F to an average low of 58°F. (a) Find a sinusoidal equation model for the daily temperature; (b) sketch the graph; and (c) approximate the time(s) each January day the temperature reaches a comfortable 72°F. Assume t  0 corresponds to noon. Source: 2004 Statistical Abstract of the United States, Table 1331.

17. In Oslo, Norway, the number of hours of daylight reaches a low of 6 hr in January, and a high of nearly 18.8 hr in July. (a) Find a sinusoidal equation model for the number of daylight hours each month; (b) sketch the graph; and (c) approximate the number of days each year there are more than 15 hr of daylight. Use 1 month  30.5 days. Assume t  0 corresponds to January 1. Source: www.visitnorway.com/templates.

18. In Vancouver, British Columbia, the number of hours of daylight reaches a low of 8.3 hr in January, and a high of nearly 16.2 hr in July. (a) Find a sinusoidal equation model for the number of daylight hours each month; (b) sketch the graph; and (c) approximate the number of days each year there are more than 15 hr of daylight. Use 1 month  30.5 days. Assume t  0 corresponds to January 1. Source: www.bcpassport.com/vital/temp.

19. Recent studies seem to indicate the population of North American porcupine (Erethizon dorsatum) varies sinusoidally with the solar (sunspot) cycle due to its effects on Earth’s ecosystems. Suppose the population of this species in a certain locality is modeled by the 2 function P1t2  250 cosa tb  950, where P(t) 11 represents the population of porcupines in year t. Use the model to (a) find the period of the function; (b) graph the function over one period; (c) find the maximum and minimum values; and

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(d) estimate the number of years the population is less than 740 animals. Source: Ilya Klvana, McGill University (Montreal), Master of Science thesis paper, November 2002.

20. The population of mosquitoes in a given area is primarily influenced by precipitation, humidity, and temperature. In tropical regions, these tend to fluctuate sinusoidally in the course of a year. Using trap counts and statistical projections, fairly accurate estimates of a mosquito population can be obtained. Suppose the population in a certain region was modeled by the function  P1t2  50 cosa tb  950, where P(t) was the 26 mosquito population (in thousands) in week t of the year. Use the model to (a) find the period of the function; (b) graph the function over one period; (c) find the maximum and minimum population values; and (d) estimate the number of weeks the population is less than 915,000. 21. Use a horizontal translation to shift the graph from Exercise 19 so that the average population of the North American porcupine begins at t  0. Verify results on a graphing calculator, then find a sine function that gives the same graph as the shifted cosine function.

29. f 1t2  24.5 sin c

 30. g1t2  40.6 sin c 1t  42 d  13.4 6 5  b  92 31. g1t2  28 sina t  6 12 32. f 1t2  90 sina

34. y  1450 sina

 24. y  560 sin c 1t  42 d 4   25. h1t2  sina t  b 6 3 26. r1t2  sina

2  t b 10 5

  27. y  sina t  b 4 6  5 28. y  sina t  b 3 12

3  t  b  2050 4 8

Find the equation of the graph given. Write answers in the form y  A sin1Bt  C2  D.

35.

37.

36.

y 700 600 500 400 300 200 100 0

6

12

18

t 24

0

38.

y

25

50

75

100

t 125

0

40.

t 90

180

270

t 8

24

12

18

24

t 6

12

 1t  2.52 d  15.5 10

43. h1t2  3 sin14t  2 44. p1t2  2 cosa3t 

 b 2

36

y

 1t  22 d  55 4

42. g1t2  24.5 sin c

30

6000 5000 4000 3000 2000 1000 18

Sketch one complete period of each function.

41. f 1t2  25 sin c

32

t 6

0

360

16

y 140 120 100 80 60 40 20

y 12 10 8 6 4 2 0

y 140 120 100 80 60 40 20

20 18 16 14 12 10 8 0

39.

 23. y  120 sin c 1t  62 d 12

  t  b  120 10 5

  33. y  2500 sina t  b  3150 4 12

22. Use a horizontal translation to shift the graph from Exercise 20 so that the average population of mosquitoes begins at t  0. Verify results on a graphing calculator, then find a sine function that gives the same graph as the shifted cosine function. Identify the amplitude (A), period (P), horizontal shift (HS), vertical shift (VS), and endpoints of the primary interval (PI) for each function given.

 1t  2.52 d  15.5 10

24

30

36

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WORKING WITH FORMULAS

45. The relationship between the coefficient B, the frequency f, and the period P In many applications of trigonometric functions, the equation y  A sin1Bt2 is written as y  A sin 3 12f 2t4 , where B  2f . Justify the new 1 2 equation using f  and P  . In other words, P B explain how A sin(Bt) becomes A sin 3 12f2t 4 , as though you were trying to help another student with the ideas involved.



497

46. Number of daylight hours: 2 K 1t  792 d  12 D1t2  sin c 2 365 The number of daylight hours for a particular day of the year is modeled by the formula given, where D(t) is the number of daylight hours on day t of the year and K is a constant related to the total variation of daylight hours, latitude of the location, and other factors. For the city of Reykjavik, Iceland, K  17, while for Detroit, Michigan, K  6. How many hours of daylight will each city receive on June 30 (the 182nd day of the year)?

APPLICATIONS

47. Harmonic motion: A weight on the end of a spring is oscillating in harmonic motion. The equation model for the oscillations is  d1t2  6 sina tb, where d is the 2 distance (in centimeters) from the equilibrium point in t sec. a. What is the period of the motion? What is the frequency of the motion? b. What is the displacement from equilibrium at t  2.5? Is the weight moving toward the equilibrium point or away from equilibrium at this time? c. What is the displacement from equilibrium at t  3.5? Is the weight moving toward the equilibrium point or away from equilibrium at this time? d. How far does the weight move between t  1 and t  1.5 sec? What is the average velocity for this interval? Do you expect a greater or lesser velocity for t  1.75 to t  2? Explain why. 48. Harmonic motion: The bob on the end of a 24-in. pendulum is oscillating in harmonic motion. The equation model for the oscillations is d1t2  20 cos14t2 , where d is the distance (in inches) from the equilibrium point, t sec after being released from one side.

d

d

a. What is the period of the motion? What is the frequency of the motion? b. What is the displacement from equilibrium at t  0.25 sec? Is the weight moving toward the equilibrium point or away from equilibrium at this time? c. What is the displacement from equilibrium at t  1.3 sec? Is the weight moving toward the equilibrium point or away from equilibrium at this time? d. How far does the bob move between t  0.25 and t  0.35 sec? What is its average velocity for this interval? Do you expect a greater velocity for the interval t  0.55 to t  0.6? Explain why. 49. Harmonic motion: A simple pendulum 36 in. in length is oscillating in harmonic motion. The bob at the end of the pendulum swings through an arc of 30 in. (from the far left to the far right, or one-half cycle) in about 0.8 sec. What is the equation model for this harmonic motion? 50. Harmonic motion: As part of a study of wave motion, the motion of a floater is observed as a series of uniform ripples of water move beneath it. By careful observation, it is noted that the floater bobs up and down through a distance of 2.5 cm 1 every sec. What is the equation model for this 3 harmonic motion?

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51. Sound waves: Two of the musical notes from the chart on page 493 are graphed in the figure. Use the graphs given to determine which two. y

y2  sin[ f (2␲t)]

1

t (sec  103)

0 2 1

4

6

8

10

y1  sin[ f (2␲t)]

52. Sound waves: Two chromatic notes not on the chart from page 493 are graphed in the figure. Use the graphs and the discussion regarding octaves to determine which two. Note the scale of the t-axis has been changed to hundredths of a second. y 1

y2  sin[ f (2␲t)] t (sec  102)

0 0.4

0.8

1.2

1.6

2.0

y1  sin[ f (2␲t)]

5-74 Daylight hours model: Solve using a graphing calculator and the formula given in Exercise 46.

55. For the city of Caracas, Venezuela, K  1.3, while for Tokyo, Japan, K  4.8. a. How many hours of daylight will each city receive on January 15th (the 15th day of the year)? b. Graph the equations modeling the hours of daylight on the same screen. Then determine (i) what days of the year these two cities will have the same number of hours of daylight, and (ii) the number of days each year that each city receives 11.5 hr or less of daylight. 56. For the city of Houston, Texas, K  3.8, while for Pocatello, Idaho, K  6.2. a. How many hours of daylight will each city receive on December 15 (the 349th day of the year)? b. Graph the equations modeling the hours of daylight on the same screen. Then determine (i) how many days each year Pocatello receives more daylight than Houston, and (ii) the number of days each year that each city receives 13.5 hr or more of daylight.

1

Sound waves: Use the chart on page 493 to write the equation for each note in the form y  sin 3f(2t) 4 and clearly state the period of each note.

53. notes D3 and G4 

54. the notes A5 and C#3

EXTENDING THE CONCEPT

57. The formulas we use in mathematics can sometimes seem very mysterious. We know they “work,” and we can graph and evaluate them—but where did they come from? Consider the formula for the number of daylight hours from Exercise 46: K 2 D1t2  sin c 1t  792 d  12. 2 365 a. We know that the addition of 12 represents a vertical shift, but what does a vertical shift of 12 mean in this context? b. We also know the factor 1t  792 represents a phase shift of 79 to the right. But what does a horizontal (phase) shift of 79 mean in this context? K c. Finally, the coefficient represents a change 2 in amplitude, but what does a change of amplitude mean in this context? Why is the coefficient bigger for the northern latitudes?

58. Use a graphing calculator to graph the equation 3x  2 sin12x2  1.5. f 1x2  2 a. Determine the interval between each peak of the graph. What do you notice? 3x b. Graph g1x2   1.5 on the same screen 2 and comment on what you observe. c. What would the graph of 3x f 1x2    2 sin12x2  1.5 look like? 2 What is the x-intercept?

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MAINTAINING YOUR SKILLS

59. (5.1) In what quadrant does the arc t  3.7 terminate? What is the reference arc?

61. (1.4) Compute the sum, difference, product and quotient of 1  i 15 and 1  i 15.

60. (3.1) Given f 1x2  31x  12 2  4, name the vertex and solve the inequality f 1x2 7 0.

62. (5.3/5.4) Sketch the graph of (a) y  cos t in the interval [0, 2) and (b) y  tan t in the interval  3 b. a , 2 2

5.6 The Trigonometry of Right Triangles Learning Objectives In Section 5.6 you will learn how to:

A. Find values of the six trigonometric functions from their ratio definitions

Over a long period of time, what began as a study of chord lengths by Hipparchus, Ptolemy, Aryabhata, and others became a systematic application of the ratios of the sides of a right triangle. In this section, we develop the sine, cosine, and tangent functions from a right triangle perspective, and explore certain relationships that exist between them. This view of the trig functions also leads to a number of significant applications.

B. Solve a right triangle given one angle and one side

C. Solve a right triangle given two sides

D. Use cofunctions and complements to write equivalent expressions

E. Solve applications involving angles of elevation and depression

A. Trigonometric Ratios and Their Values In Section 5.1, we looked at applications involving 45-45-90 and 30-60-90 triangles, using the fixed ratios that exist between their sides. To apply this concept more generally using other right triangles, each side is given a specific name using its location relative to a specified angle. For the 30-60-90 triangle in Figure 5.75(a), the side opposite (opp) and the side adjacent (adj) are named with respect to the 30° angle, with the hypotenuse (hyp) always across from the right angle. Likewise for the 45-45-90 triangle in Figure 5.75(b). Figure 5.75

F. Solve general applications of right triangles 60

hyp 2x

opp x

30 adj √3x

for 30 opp 1 2 hyp

hyp √2x

adj √3  2 hyp

45

opp x

45 adj x

opp  1  √3 3 adj √3 (a)

for 45 opp 1   √2 2 hyp √2 adj  1  √2 2 hyp √2 opp  adj

1 1

(b)

Using these designations to define the various trig ratios, we can now develop a systematic method for applying them. Note that the x’s “cancel” in each ratio, reminding us the ratios are independent of the triangle’s size (if two triangles are similar, the ratio of corresponding sides is constant). Ancient mathematicians were able to find values for the ratios corresponding to any acute angle in a right triangle, and realized that naming each ratio would be opp adj opp S sine, S cosine, and S tangent. Since each helpful. These names are hyp hyp adj ratio depends on the measure of an acute angle , they are often referred to as functions of an acute angle and written in function form. sine  

opp hyp

cosine  

adj hyp

tangent  

opp adj

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hyp opp instead of , also play a sigopp hyp nificant role in this view of trigonometry, and are likewise given names: The reciprocal of these ratios, for example,

cosecant  

hyp opp

secant  

hyp adj

cotangent  

adj opp

The definitions hold regardless of the triangle’s orientation or which of the acute angles is used. In actual use, each function name is written in abbreviated form as sin , cos , tan , csc , sec , and cot  respectively. Note that based on these designations, we have the following reciprocal relationships:

WORTHY OF NOTE Over the years, a number of memory tools have been invented to help students recall these ratios correctly. One such tool is the acronym SOH CAH TOA, from the first letter of the function and the corresponding ratio. It is often recited as, “Sit On a Horse, Canter Away Hurriedly, To Other Adventures.” Try making up a memory tool of your own.

sin  

1 csc 

cos  

1 sec 

tan  

1 cot 

csc  

1 sin 

sec  

1 cos 

cot  

1 tan 

In general: Trigonometric Functions of an Acute Angle sin  

a c

cos  

b c

a tan   b

 c

a

sin  

b c

cos  

a c

tan  

b a

 b

Now that these ratios have been formally named, we can state values of all six functions given sufficient information about a right triangle. EXAMPLE 1



Finding Function Values Using a Right Triangle Given sin   47, find the values of the remaining trig functions.

Solution



opp 4  , we draw a triangle with a side of 4 units opposite a designated 7 hyp angle , and label a hypotenuse of 7 (see the figure). Using the Pythagorean theorem we find the length of the adjacent side: adj  272  42  133. The ratios are For sin  

4 7 7 csc   4

sin  

133 7 7 sec   133

cos  

4 133 133 cot   4

tan  

7

4

 adj

Now try Exercises 7 through 12 A. You’ve just learned how to find values of the six trigonometric functions from their ratio definitions



Note that due to the properties of similar triangles, identical results would be 2 8  47  14  16 obtained using any ratio of sides that is equal to 74. In other words, 3.5 28 and so on, will all give the same value for sin .

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Section 5.6 The Trigonometry of Right Triangles

B. Solving Right Triangles Given One Angle and One Side Example 1 gave values of the trig functions for an unknown angle . Using the special triangles, we can state the value of each trig function for 30°, 45°, and 60° based on the related ratio (see Table 5.10). These values are used extensively in a study of trigonometry and must be committed to memory.

60

hyp 2x

opp x

Table 5.10

30 adj √3x hyp √2x

45

opp x



sin 

cos 

tan 

csc 

sec 

cot 

30°

1 2

13 2

1 13  3 13

2

2 213  3 13

13

45°

12 2

12 2

1

12

12

1

60°

13 2

1 2

13

2 2 13  3 13

2

1 13  3 13

45 adj x

To solve a right triangle means to find the measure of all three angles and all three sides. This is accomplished using combinations of the Pythagorean theorem, the properties of triangles, and the trigonometric ratios. We will adopt the convention of naming each angle with a capital letter at the vertex or using a Greek letter on the interior. Each side is labeled using the related lowercase letter from the angle opposite. The complete solution should be organized in table form as in Example 2. Note the quantities shown in bold were given, and the remaining values were found using the techniques mentioned. EXAMPLE 2



Solving a Right Triangle Solve the triangle given.

Solution



Applying the sine ratio (since the side opposite 30° is given), we have: sin 30°  For side c:

17.9 c c sin 30°  17.9 17.9 c sin 30°  35.8 sin 30° 

sin 30° 

B

opposite hypotenuse

multiply by c

c

17.9

divide by sin 30° 

opp . hyp

1 2

30

C

Using the Pythagorean theorem shows b  31, and since A and B are complements, B  60°. Note the results would have been identical if the special ratios from the 30-60-90 triangle were applied. The hypotenuse is twice the shorter side: c  2117.92  35.8, and the longer side is 13 times the shorter: b  17.91 132  31.

A

b

result

Angles

Sides

A  30

a  17.9

B  60°

b  31

C  90

c  35.8

Now try Exercises 13 through 16



Prior to the widespread availability of handheld calculators, a table of values was used to find sin , cos , and tan  for nonstandard angles. Table 5.11 shows the sine of 49° 30¿ is approximately 0.7604.

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Table 5.11 sin  

10

20

30

45

0.7071

0.7092

0.7112

0.7133

46

0.7193

0.7214

0.7234

0.7254

47

0.7314

0.7333

0.7353

0.7373

48

0.7431

0.7451

0.7470

0.7490

49 "

0.7547

0.7566

0.7585

0.7604

"

0

Today these trig values are programmed into your calculator and we can retrieve them with the push of a button (or two). To find the sine of 48°, make sure your calculator is in degree MODE , then press the key, 48, and ENTER . The result should be very close to 0.7431 as the table indicates. EXAMPLE 3



Solving a Right Triangle Solve the triangle shown in the figure.

Solution



We know B  58° since A  B  90°. We can find length b using the tangent function: 24 b b tan 32°  24 24 b tan 32°  38.41 mm tan 32° 

tan 32° 

opp adj

B

multiply by b

c

24 mm

divide by tan 32°

A result

32

C

b

We can find the length c by simply applying the Pythagorean theorem, or by using another trig ratio and a known angle. For side c:

24 c c sin 32°  24 24 c sin 32°  45.29 mm sin 32° 

sin 32° 

opp hyp

multiply by c divide by sin 32°

Angles

Sides

A  32

a  24

B  58°

b  38.41

C  90

c  45.29

result

The complete solution is shown in the table. Now try Exercises 17 through 22

B. You’ve just learned how to solve a right triangle given one angle and one side

When solving a right triangle, any of the triangle relationships can be employed: (1) angles must sum to 180°, (2) Pythagorean theorem, (3) special triangles, and (4) the trigonometric functions of an acute angle. However, the resulting equation must have only one unknown or it cannot be used. For the triangle shown in Figure 5.76, we cannot begin with the Pythagorean theorem since sides a and b are unknown, and tan 51° is unusable for the same reason. Since the b hypotenuse is given, we could begin with cos 51°  and solve 152 a for b, or with sin 51°  and solve for a, then work out a com152 plete solution. Verify that a  118.13 ft and b  95.66 ft.



Figure 5.76 B

a

152 ft

A 51

b

C

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Section 5.6 The Trigonometry of Right Triangles

C. Solving Right Triangles Given Two Sides The partial table for sin  given earlier was also used in times past to find an angle whose sine was known, meaning if sin   0.7604, then  must be 49.5° (see the last line of Table 5.11). The modern notation for “an angle whose sine is known” is   sin1x or   arcsin x, where x is the known value for sin . The values for the acute angles   sin1x,   cos1x, and   tan1x are also programmed into your calculator and are generally accessed using the or 2nd keys with the related , , or key. With these we are completely equip to find all six measures of a right triangle, given at least one side and any two other measures. EXAMPLE 4



Solving a Right Triangle Solve the triangle given in the figure.

Solution



C. You’ve just learned how to solve a right triangle given two sides



Since the hypotenuse is unknown, we cannot begin with the sine or cosine ratios. The opposite and adjacent 17 sides for  are known, so we use tan . For tan   25 17 we find   tan1a b  34.2° [verify that 25 17 tan134.2°2  0.6795992982  4 . Since  and  are 25 complements,   90  34.2  55.8°. The Pythagorean theorem shows the hypotenuse is about 30.23 m.

c

17 m

 25 m

Angles   34.2°

Sides a  17

  55.8°

b  25

  90

c  30.23

Now try Exercises 23 through 54



D. Using Cofunctions and Complements to Write Equivalent Expressions WORTHY OF NOTE The word cosine is actually a shortened form of the words “complement of sine,” a designation suggested by Edmund Gunter around 1620 since the sine of an angle is equal to the cosine of its complement 3 sine12  cosine190°  2 4 .

In Figure 5.77,  and  must be complements since we have a right triangle, and the sum of the three angles must be 180°. The complementary angles in a right triangle have a unique relationship that is often used. Specifically     90° means a a   90°  . Note that sin   and cos   . This means Figure 5.77 c c sin   cos  or sin   cos190°  2 by substitution.  In words, “The sine of an angle is equal to the cosine of its c a complement.” For this reason sine and cosine are called cofunctions (hence the name cosine), as are secant/cosecant,  and tangent/cotangent. As a test, we use a calculator to check b the statement sin 52.3°  cos190  52.32° sin 52.3°  cos 37.7° 0.791223533  0.791223533 ✓ To verify the cofunction relationship for sec  and csc , recall their reciprocal relationship to cosine and sine, respectively. sec 52.3°  csc  37.7° 1 1  cos 52.3° sin 37.7° 1.635250666  1.635250666 ✓ The cofunction relationship for tan  and cot  can similarly be verified.

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Summary of Cofunctions sine and cosine sin   cos190  2 cos   sin190  2

tangent and cotangent

secant and cosecant sec   csc190  2 csc   sec190  2

tan   cot190  2 cot   tan190  2

For use in Example 5 and elsewhere in the text, note the expression tan215° is simply a more convenient way of writing 1tan 15°2 2. EXAMPLE 5



Applying the Cofunction Relationship Given cot 75°  2  13 in exact form, find the exact value of tan215° using a cofunction. Check the result using a calculator.

Solution



Using cot 75°  tan190°  75°2  tan 15° gives cot275°  tan215°  12  132 2  4  413  3  7  4 13

D. You’ve just learned how to use cofunctions and complements to write equivalent expressions

cofunctions substitute known value square as indicated result

Using a calculator, we verify tan215°  0.0717967697  7  413. Now try Exercises 55 through 68



E. Applications Using Angles of Elevation/Depression While the name seems self-descriptive, in more formal terms an angle of elevation is defined to be the acute angle formed by a horizontal line of orientation (parallel to level ground) and the line of sight (see Figure 5.78). An angle of depression is likewise defined but involves a line of sight that is below the horizontal line of orientation (Figure 5.79). Figure 5.79

Figure 5.78 t

igh

s of

e Lin  S angle of elevation  Line of orientation

Line of orientation 

 S angle of depression

Lin

eo

fs

igh

t

Angles of elevation/depression make distance and length computations of all sizes a relatively easy matter and are extensively used by surveyors, engineers, astronomers, and even the casual observer who is familiar with the basics of trigonometry. EXAMPLE 6



Applying Angles of Elevation In Example 4 from Section 5.1, a group of campers used a 45-45-90 triangle to estimate the height of a cliff. It was a time consuming process as they had to wait until mid-morning for the shadow of the cliff to make the needed 45° angle. If the campsite was 250 yd from the base of the cliff and the angle of elevation was 40° at that point, how tall is the cliff?

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Section 5.6 The Trigonometry of Right Triangles

Solution



As described we want to know the height of the opposite side, given the adjacent side, so we use the tangent function. For height h: h 250 250 tan 40°  h 209.8  h tan 40° 

E. You’ve just learned how to solve applications involving angles of elevation and depression

tan 40° 

opp adj

multiply by 250 result 1tan 40°  0.83912

h

The cliff is approximately 209.8 yd high (about 629 ft).

40 Angle of elevation 250 yd

Now try Exercises 71 through 76



F. Additional Applications of Right Triangles In their widest and most beneficial use, the trig functions of acute angles are used with other problem-solving skills, such as drawing a diagram, labeling unknowns, working the solution out in stages, and so on. Example 7 serves to illustrate some of these combinations. EXAMPLE 7



Applying Angles of Elevation and Depression From his hotel room window on the sixth floor, Singh notices some window washers high above him on the hotel across the street. Curious as to their height above ground, he quickly estimates the buildings are 50 ft apart, the angle of elevation to the workers is about 80°, and the angle of depression to the base of the hotel is about 50°. a. How high above ground is the window of Singh’s hotel room? b. How high above ground are the workers?

Solution



a. Begin by drawing a diagram of the situation (see figure). To find the height of the window we’ll use the tangent ratio, since the adjacent side of the angle is known, and the opposite side is the height we desire. For the height h1:

(not to scale) h2

h1 50 50 tan 50°  h1 59.6  h1 tan 50° 

tan 50° 

opp adj

solve for h1

result 1tan 50°  1.19182

The window is approximately 59.6 ft above ground. h2 opp b. For the height h2: tan 80°  tan 80°  adj 50 50 tan 80°  h2 solve for h2 283.6  h2 result 1tan 80°  5.67132

80° 50° h1

The workers are approximately 283.6  59.6  343.2 ft above ground.

x 50 ft F. You’ve just learned how to solve general applications of right triangles

Now try Exercises 77 through 80 There are a number of additional, interesting applications in the exercise set.



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5.6 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. The phrase, “an angle whose tangent is known,” is written notationally as . 7 2. Given sin   24 , csc   are .

because they

3. The sine of an angle is the ratio of the side to the . 

4. The cosine of an angle is the ratio of the side to the . 5. Discuss/Explain exactly what is meant when you are asked to “solve a triangle.” Include an illustrative example. 6. Given an acute angle and the length of the adjacent leg, which four (of the six) trig functions could be used to begin solving the triangle?

DEVELOPING YOUR SKILLS

Use the function value given to determine the value of the other five trig functions of the acute angle . Answer in exact form (a diagram will help).

7. cos  

5 13

8. sin  

20 29

9. tan  

84 13

10. sec  

53 45

2 12. cos   3

2 11. cot   11

Solve each triangle using trig functions of an acute angle . Give a complete answer (in table form) using exact values.

13.

15.

9.9 mm

A

a

c

30

14.

b

17.

B c 22

18.

B

c

89 in.

A 49

B

c

A

b

C

b b

a

81.9 m

C

Solve the triangles shown and write answers in table form. Round sides to the nearest 100th of a unit. Verify that angles sum to 180 and that the three sides satisfy (approximately) the Pythagorean theorem.

A

420 ft

C

A

B

C

C

c

45

14 m

196 cm

B 45

B

a

16.

C

19.

B

60 c

A A

58 5.6 mi C

b

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20.

21.

B

625 mm

C

B

45.

238 ft

a C

51 A

b

65 19.5 cm

A

46.

B

22.

28

5 mi

c

A

45.8 m

A

C

47.

23. sin 27°

24. cos 72°

25. tan 40°

26. cot 57.3°

27. sec 40.9°

28. csc 39°

29. sin 65°

30. tan 84.1°

Use a calculator to find the acute angle whose corresponding ratio is given. Round to the nearest 10th of a degree. For Exercises 31 through 38, use Exercises 23 through 30 to answer.

31. sin A  0.4540

32. cos B  0.3090

33. tan   0.8391

34. cot A  0.6420

35. sec B  1.3230

36. csc   1.5890

37. sin A  0.9063

38. tan B  9.6768

39. tan   0.9896

40. cos   0.7408

41. sin   0.3453

42. tan   3.1336

Select an appropriate function to find the angle indicated (round to 10ths of a degree).

43. 6m  18 m 

6.2 mi

a

Use a calculator to find the value of each expression, rounded to four decimal places.

44.

 18.7 cm

c

b

507

15 in. 14 in.

20 mm

42 mm

B

48.

A

207 yd

221 yd

Draw a right triangle ABC as shown, using the information given. Then select an appropriate ratio to find the side indicated. Round to the nearest 100th. Exercises 49 to 54 B

c

a

A b

49. A  25°

C

50. B  55°

c  52 mm

b  31 ft

find side a

find side c

51. A  32°

52. B  29.6°

a  1.9 mi

c  9.5 yd

find side b

find side a

53. A  62.3°

54. B  12.5°

b  82.5 furlongs

a  32.8 km

find side c

find side b

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Use a calculator to evaluate each pair of functions and comment on what you notice.

55. sin 25°, cos 65°

64. 

sin 

cos 

tan 

sin190  2

tan190  2

csc 

sec 

cot 

45°

56. sin 57°, cos 33°

cos190  2

57. tan 5°, cot 85° 58. sec 40°, csc 50° Based on your observations in Exercises 55 to 58, fill in the blank so that the functions given are equal.

59. sin 47°, cos ___

Evaluate the following expressions without a calculator, using the cofunction relationship and the following exact forms: sec 75  16  12; tan 75  2  13.

65. 16 csc 15° 66. csc215°

60. cos ___, sin 12°

67. cot215°

61. cot 69°, tan ___

68. 13 cot 15°

62. csc 17°, sec ___ Complete the following tables without referring to the text or using a calculator.

63. 

sin 

cos 

tan 

sin190  2

tan190  2

csc 

sec 

cot 

30° cos190  2



WORKING WITH FORMULAS

69. The sine of an angle between two sides of a 2A triangle: sin   ab If the area A and two sides a and b of a triangle are known, the sine of the angle between the two sides is given by the formula shown. Find the angle  for the triangle below given A  38.9, and use it to solve the triangle. (Hint: Apply the same concept to angle  or .) ␥

17 ␪

8 ␤ 24

70. Illumination of a surface: E 

I cos  d2

The illumination E of a surface by a light source is a measure of the luminous flux per unit area that reaches the surface. The value of E [in lumens (lm)

per square foot] is given by the 90 cd (about 75 W) formula shown, where d is the distance from the light source (in feet), I is the intensity of the light [in candelas (cd)], and  is the angle the light source makes with the vertical. For reading a book, an illumination E of at least 18 lm/ft2 is recommended. Assuming the open book is lying on a horizontal surface, 65° how far away should a light source be placed if it has an intensity of 90 cd (about 75 W) and the light flux makes an angle of 65° with the book’s surface (i.e.,   25°)?

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509

APPLICATIONS

71. Angle of elevation: For a person standing 100 m from the center of the base of the Eiffel Tower, the angle of elevation to the top of the tower is 71.6°. How tall is the Eiffel Tower? 72. Angle of depression: A person standing near the top of the Eiffel Tower notices a car wreck some distance from the tower. If the angle of depression from the person’s eyes to the wreck is 32°, how far away is the accident from the base of the tower? See Exercise 71. 73. Angle of elevation: In 2001, the tallest building in the world was the Petronas Tower I in Kuala Lumpur, Malaysia. For a person standing 25.9 ft from the base of the tower, the angle of elevation to the top of the tower is 89°. How tall is the Petronas tower? 74. Angle of depression: A person standing on the top of the Petronas Tower I looks out across the city and pinpoints her residence. If the angle of depression from the person’s eyes to her home is 5°, how far away (in feet and in miles) is the residence from the base of the tower? See Exercise 73. 75. Crop duster’s speed: While standing near the edge of a farmer’s field, Johnny watches a crop duster dust the farmer’s field for insect control. Curious as to the plane’s speed during each drop, Johnny attempts an estimate 50 ft using the angle of  rotation from one end of the field to the other, while standing 50 ft from one corner. Using a stopwatch he finds the plane makes each pass in 2.35 sec. If the angle of rotation was 83°, how fast (in miles per hour) is the plane flying as it applies the insecticide? 76. Train speed: While driving to their next gig, Josh and the boys get stuck in a line of cars at a railroad crossing as the gates go down. As the sleek, speedy express train approaches, Josh decides to pass the time estimating its speed. He spots a large oak tree beside the track some distance away, and figures the angle of rotation from the crossing to the tree is about 80°. If their car is 60 ft from the crossing and it takes the train 3 sec to reach the tree, how fast is the train moving in miles per hour?

77. Height of a climber: A local Outdoors Club has just hiked to the south rim of a large canyon, when they spot a climber attempting to scale the taller northern face. Knowing the distance between the sheer walls of the northern and southern faces of the canyon is approximately 175 yd, they attempt to compute the distance remaining for the climbers to reach the top of the northern rim. Using a homemade transit, they sight an angle of depression of 55° to the bottom of the north face, and angles of elevation of 24° and 30° to the climbers and top of the northern rim respectively. (a) How high is the southern rim of the canyon? (b) How high is the northern rim? (c) How much farther until the climber reaches the top?

24

30

55 175 yd

78. Observing wildlife: From her elevated observation post 300 ft away, a naturalist spots a troop of baboons high up in a tree. Using the small transit attached to her telescope, she finds the angle of depression to the bottom of this tree is 14°, while the angle of elevation to the top of the tree is 25°. The angle of elevation to the troop of baboons is 21°. Use this information to find (a) the height of the observation post, (b) the height of the baboons’ tree, and (c) the height of the baboons above ground. 79. Angle of elevation: The tallest free-standing tower in the world is the CNN Tower in Toronto, Canada. The tower includes a rotating restaurant high above the ground. From a distance of 500 ft the angle 66.5 of elevation to the pinnacle 74.6 of the tower is 74.6°. The angle of elevation to the 500 ft restaurant from the same vantage point is 66.5°. How tall is the CNN Tower? How far below the pinnacle of the tower is the restaurant located?

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80. Angle of elevation: In August 2004, Taipei 101 captured the record as the world’s tallest building, according to the Council on Tall Buildings and Urban Habitat [Source: www.ctbuh.org]. Measured at a point 108 m from its base, the angle of elevation to the top of the spire is 78°. From a distance of about 95 m, the angle of elevation to the top of the roof is also 78°. How tall is Taipei 101 from street level to the top of the spire? How tall is the spire itself? Alternating current: In AC (alternating current) applications, the Z relationship between measures known as the impedance (Z), resistance (R),  and the phase angle 12 can be R demonstrated using a right triangle. Both the resistance and the impedance are measured in ohms 12 .

81. Find the impedance Z if the phase angle  is 34°, and the resistance R is 320 . 82. Find the phase angle  if the impedance Z is 420 , and the resistance R is 290 .

83. Contour maps: In the A figure shown, the contour interval is 175 m (each concentric line represents an increase of 175 m in B elevation), and the scale of horizontal distances is 1 cm  500 m. (a) Find the vertical change from A to B (the increase in elevation); (b) use a proportion to find the horizontal change between points A and B if the measured distance on the map is 2.4 cm; and (c) draw the corresponding right triangle and use it to estimate the length of the trail up the mountain side that connects A and B, then use trig to compute the approximate angle of incline as the hiker climbs from point A to point B. 84. Contour maps: In the figure shown, the contour interval is 150 m (each concentric line represents an increase of 150 m in elevation), and B the scale of horizontal distances is 1 cm  250 m. A (a) Find the vertical change from A to B (the increase in elevation); (b) use a proportion to find the

horizontal change between points A and B if the measured distance on the map is 4.5 cm; and (c) draw the corresponding right triangle and use it to estimate the length of the trail up the mountain side that connects A and B, then use trig to compute the approximate angle of incline as the hiker climbs from point A to point B. 85. Height of a rainbow: While visiting the Lapahoehoe Memorial on the island of Hawaii, Bruce and Carma see a spectacularly vivid rainbow arching over the bay. Bruce speculates the rainbow is 500 ft away, while Carma estimates the angle of elevation to the highest point of the rainbow is about 42°. What was the approximate height of the rainbow? 86. High-wire walking: As part of a circus act, a high-wire walker not only “walks the wire,” she walks a wire that is set at an incline of 10° to the horizontal! If the length of the (inclined) wire is 25.39 m, (a) how much higher is the wire set at the destination pole than at the departure pole? (b) How far apart are the poles? 87. Diagonal of a cube: A cubical box has a diagonal measure of 35 x cm. (a) Find the dimensions of the box and (b) the angle  that the diagonal makes at the lower corner of the box.

d  35 cm 

x

x

88. Diagonal of a rectangular h parallelepiped: A  50 rectangular box has a 70 cm width of 50 cm and a length of 70 cm. (a) Find the height h that ensures the diagonal across the middle of the box will be 90 cm and (b) the angle  that the diagonal makes at the lower corner of the box.

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EXTENDING THE CONCEPT

d 89. The formula h  cot u  cot v can be used to calculate the height h of a building when distance x is unknown but u v distance d is known (see the d xd diagram). Use the ratios for x cot u and cot v to derive the formula (note x is “absent” from the formula).

h

90. Use the diagram given to derive a formula for the height h of the taller u building in terms of the height x of the x shorter building v and the ratios d for tan u and tan v. Then use the formula to find h given the shorter building is 75 m tall with u  40° and v  50°.



511

Section 5.6 The Trigonometry of Right Triangles

91. The radius of the Earth North Pole at the equator (0° N h latitude) is approx␪N r imately 3960 mi. Rh R Beijing, China, is ␪ located at 39.5° N R latitude, 116° E longitude. Philadelphia, Pennsylvania, is located at the same latitude, but at 75° W South Pole longitude. (a) Use the diagram given and a cofunction relationship to find the radius r of the Earth (parallel to the equator) at this latitude; (b) use the arc length formula to compute the shortest distance between these two cities along this latitude; and (c) if the supersonic Concorde flew a direct flight between Beijing and Philadelphia along this latitude, approximate the flight time assuming a cruising speed of 1250 mph. Note: The shortest distance is actually traversed by heading northward, using the arc of a “great circle” that goes through these two cities.

MAINTAINING YOUR SKILLS

92. (1.5) Solve by factoring: a. g2  9g  0 b. g2  9  0 c. g2  9g  10  0 d. g2  9g  10  0 e. g3  9g2  10g  90  0 y 93. (2.5) For the graph of T(x) 3 T(x) given, (a) name the local maximums and minimums, 5 5 (b) the zeroes of T, 3 (c) intervals where T1x2T and T1x2c, and (d) intervals where T1x2 7 0 and T1x2 6 0.

94. (5.1) The armature for the rear windshield wiper has a length of 24 in., with a rubber wiper blade that is 20 in. long. What area of my rear windshield is cleaned as the armature swings back-and-forth through an angle of 110°?

x

95. (5.1) The boxes used to ship some washing machines are perfect cubes with edges measuring 38 in. Use a special triangle to find the length of the diagonal d of one side, and the length of the interior diagonal D (through the middle of the box).

D

d

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Precalculus—

5.7 Trigonometry and the Coordinate Plane Learning Objectives In Section 5.7 you will learn how to:

A. Define the trigonometric functions using the coordinates of a point in QI

A. Trigonometric Ratios and the Point P(x, y)

B. Use reference angles to evaluate the trig functions for any angle

C. Solve applications using the trig functions of any angle

EXAMPLE 1

This section tends to bridge the study of static trigonometry and the angles of a right triangle, with the study of dynamic trigonometry and the unit circle. This is accomplished by noting that the domain of the trig functions (unlike a triangle point of view) need not be restricted to acute angles. We’ll soon see that the domain can be extended to include trig functions of any angle, a view that greatly facilitates our work in Chapter 7, where many applications involve angles greater than 90°.



Figure 5.80 Regardless of where a right triangle is situated or how it y is oriented, each trig function can be defined as a given ratio of sides with respect to a given angle. In this light, consider a 30-60-90 triangle placed in the first quadrant with the 30° angle at the origin and the longer side along (5√3, 5) the x-axis. From our previous review of similar triangles, 5 60 the trig ratios will have the same value regardless of the 10 5 triangle’s size so for convenience, we’ll use a hypotenuse of 10. This gives sides of 5, 513, and 10, and from the 30 diagram in Figure 5.80 we note the point (x, y) marking 10 x 5√3 the vertex of the 60° angle has coordinates (5 13, 5). Further, the diagram shows that sin 30°, cos 30°, and tan 30° can all be expressed in adj opp y 5 5 13 x   1sine2,   1cosine2, and terms of these coordinates since r r hyp 10 hyp 10 opp y 5   1tangent2, where r is the length of the hypotenuse. Each result x adj 5 13 1 13 , and reduces to the more familiar values seen earlier: sin 30°  , cos 30°  2 2 1 13 tan 30°   . This suggests we can define the six trig functions in terms 3 13 of x, y, and r, where r  2x2  y2. Consider that the slope of the line coincident with the hypotenuse is 5 13 rise   , and since the line goes through the origin its equation must be run 3 5 13 13 x. Any point (x, y) on this line will be at the 60° vertex of a right triangle y 3 formed by drawing a perpendicular line from the point (x, y) to the x-axis. As Example 1 shows, we obtain the special values for sin 30°, cos 30°, and tan 30° regardless of the point chosen.

Evaluating Trig Functions Using x, y, and r Pick an arbitrary point in QI that satisfies y 

13 x, 3

y 10

y  √3 x 3

then draw the corresponding right triangle and

(6, 2√3)

evaluate sin 30°, cos 30°, and tan 30°. Solution

512



The coefficient of x has a denominator of 3, so we choose a multiple of 3 for convenience. For x  6 13 162  2 13. As seen in the figure, we have y  3

4√3 10

30

30 6

2√3 10

x

10

5-88

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Section 5.7 Trigonometry and the Coordinate Plane

the point (6, 2 13) is on the line and at the vertex of the 60° angle. Evaluating the trig functions at 30°, we obtain: y 2 13  r 4 13 1  2

6 x  r 4 13 6 13 13   2 4 13 13

sin 30° 

tan 30° 

cos 30° 



y 213  x 6

13 3

Now try Exercises 7 and 8



In general, consider any two points (x, y) and (X, Y) on an arbitrary line y  kx, at corresponding distances r and R from the origin (Figure 5.81). Because the triangles y Y x X formed are similar, we have  ,  , and so on, and we conclude that the value x X r R of the trig functions are indeed independent of the point chosen. Figure 5.81

Figure 5.82

y

y 10

(X, Y) y



(x, y) r  x

y  kx

y

Y x

2√3

X

6 30

210

√3 x 3

(6, 2√3) 10

x

4√3

(6, 2√3)

Viewing the trig functions in terms of x, y, and r produces significant results. In 13 x from Example 1 also extends into QIII, and Figure 5.82, we note the line y  3 creates another 30° angle whose vertex is at the origin (since vertical angles are equal). The sine, cosine, and tangent functions can still be evaluated for this angle, but in QIII both x and y are negative. If we consider the angle in QIII to be a positive rotation of 210° 1180°  30°2, we can evaluate the trig functions using the values of x, y, and r from any point on the terminal side, since these are fixed by the 30° angle created and are the same as those in QI except for their sign: sin 210° 

y 2 13  r 4 13 1  2

x 6  r 4 13 13  2

cos 210° 

y 213  x 6 13  3

tan 210° 

For any rotation  and a point (x, y) on the terminal side, the distance r can be found using r  2x2  y2 and the six trig functions likewise evaluated. Note that evaluating them correctly depends on the quadrant of the terminal side, since this will dictate the signs for x and y. Students are strongly encouraged to make these quadrant and sign observations the first step in any solution process. In summary, we have

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CHAPTER 5 An Introduction to Trigonometric Functions

Trigonometric Functions of Any Angle Given P(x, y) is any point on the terminal side of angle  in standard position, with r  2x2  y2 1r 7 02 the distance from the origin to (x, y). The six trigonometric functions of  are sin  

y r

cos  

x r

y x

tan   x0

csc  

r y

sec  

y0

EXAMPLE 2



r x

cot  

x0

x y

y0

Evaluating Trig Functions Given the Terminal Side is on y  mx Given that P(x, y) is a point on the terminal side of angle  in standard position, find the value of sin  and cos , if a. The terminal side is in QII and coincident with the line y  12 5 x, b. The terminal side is in QIV and coincident with the line y  12 5 x.

Solution



a. Select any convenient point in QII that satisfies this equation. We select x  5 since x is negative in QII, which gives y  12 and the point (5, 12). Solving for r gives r  2152 2  1122 2  13. The ratios are sin  

y 12  r 13

cos  

x 5  r 13

b. In QIV we select x  10 since x is positive in QIV, giving y  24 and the point (10, 24). Solving for r gives r  21102 2  1242 2  26. The ratios are y 24  r 26 12  13

10 x  r 26 5  13

sin  

cos  

Now try Exercises 9 through 12



In Example 2, note the ratios are the same in QII and QIV except for their sign. We will soon use this observation to great advantage. EXAMPLE 3



Evaluating Trig Functions Given a Point P Find the value of the six trigonometric functions given P15, 52 is on the terminal side of angle  in standard position.

Solution



For P15, 52 we have x 6 0 and y 7 0 so the terminal side is in QII. Solving for r yields r  2152 2  152 2  150  5 12. For x  5, y  5, and r  5 12, we obtain y 5  r 5 12 12  2

sin  

5 x  r 5 12 12  2

cos  

tan  

y 5  x 5

 1

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The remaining functions can be evaluated using reciprocals. csc  

2  12 12

sec   

2   12 12

cot   1

Note the connection between these results and the special values for   45°.

Figure 5.83 y

Now try Exercises 13 through 28



(0, b)

Now that we’ve defined the trig functions in terms of ratios involving x, y, and r, the question arises as to their value at the quadrantal angles. For 90° and 270°, any point on the terminal side of the angle has an x-value of zero, meaning tan 90°, sec 90°, tan 270°, and sec 270° are all undefined since x  0 is in the denominator. Similarly, at 180° and 360°, the y-value of any point on the terminal side is zero, so cot 180°, csc 180°, cot 360°, and csc 360° are likewise undefined (see Figure 5.83).

180 (a, 0)

(a, 0) 90

x

(0, b)

EXAMPLE 4



Evaluating the Trig Functions for   90k, k an Integer Evaluate the six trig functions for   270°.

Solution



Here,  is the quadrantal angle whose terminal side separates QIII and QIV. Since the evaluation is independent of the point chosen on this side, we choose (0, 1) for convenience, giving r  1. For x  0, y  1, and r  1 we obtain sin  

1  1 1

cos  

0 0 1

tan  

1 1undefined2 0

The remaining ratios can be evaluated using reciprocals. csc   1

sec  

1 1undefined2 0

cot  

0 0 1

Now try Exercises 29 and 30



Results for the quadrantal angles are summarized in Table 5.12. Table 5.12 

A. You’ve just learned how to define the trigonometric functions using the coordinates of a point in QI

0° S 11, 02

90° S 10, 12

180° S 11, 02

270° S 10, 12

sin   0

y r

cos  

x r

tan  

1

0

1

0

0

1

1

0

y x

csc  

r y

sec  

r x

cot  

x y

undefined

1

undefined

undefined

1

undefined

0

0

undefined

1

undefined

undefined

1

undefined

0

B. Reference Angles and the Trig Functions of Any Angle Recall that for any angle  in standard position, the acute angle r formed by the terminal side and the x-axis is called the reference angle. Several examples of this definition are illustrated in Figures 5.84 through 5.87 for  7 0 in degrees.

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Figure 5.84

Figure 5.85

Figure 5.86

y

5

5

5

Figure 5.87

y

y

y

5

(x, y) (x, y) r



5

5

x

r

5

 5

x



5

r



r

5

x

5

5

x

(x, y) (x, y) 5

5

5

180    270 r    180

90    180 r  180  

EXAMPLE 5



5

270    360 r  360  

Finding Reference Angles Determine the reference angle for a. 315° b. 150° c. 121°

Solution



360    450 r    360

d. 425°

Begin by mentally visualizing each angle and the quadrant where it terminates. a. 315° is a QIV angle: c. 121° is a QIII angle: r  360°  315°  45° r  180°  121°  59° b. 150° is a QII angle: d. 425° is a QI angle: r  180°  150°  30° r  425°  360°  65° Now try Exercises 31 through 42



The reference angles from Examples 5(a) and 5(b) were special angles, which means we automatically know the absolute value of the trig ratios using r . The best way to remember the signs of the trig Figure 5.88 functions is to keep in mind that sine is associated with y, cosine with x, and tangent with both x and y Quadrant II Quadrant I (r is always positive). In addition, there are several Sine All mnemonic devices (memory tools) to assist you. One is positive are positive is to use the first letter of the function that is positive in each quadrant and create a catchy acronym. For Cosine Tangent instance ASTC S All Students Take Classes (see is positive is positive Figure 5.88). Note that a trig function and its recipQuadrant IV Quadrant III rocal function will always have the same sign. EXAMPLE 6



Evaluating Trig Functions Using r

y

Use a reference angle to evaluate sin , cos , and tan  for   315°. Solution



The terminal side is in QIV where x is positive and y is negative. With r  45°, we have: 12 12 cos 315°  sin 315°   2 2 tan 315°  1

8

315 

r  45 8

Now try Exercises 43 through 54

x



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Section 5.7 Trigonometry and the Coordinate Plane



EXAMPLE 7

Finding Function Values Using a Quadrant and Sign Analysis Given sin  



Solution

517

5 and cos  6 0, find the value of the other ratios. 13

Always begin with a quadrant and sign analysis: sin  is positive in QI and QII, while cos  is negative in QII and QIII. Both conditions are satisfied in QII only. For r  13 and y  5, the Pythagorean theorem shows x  2132  52  1144  12. 5 12 With  in QII this gives cos   and tan   . The reciprocal values are 13 12 12 13 13 csc   , sec   , and cot   . 5 12 5 Now try Exercises 55 through 62



In our everyday experience, there are many Figure 5.89 y actions and activities where angles greater than or equal to 360° are applied. Some common instances are a professional basketball player who “does a   495 three-sixty” (360°) while going to the hoop, a diver 45   135 who completes a “two-and-a-half” (900°) off the high x board, and a skater who executes a perfect triple axel (312 turns or 1260°). As these examples suggest, angles   225 greater than 360° must still terminate on a quadrantal axis, or in one of the four quadrants, allowing a reference angle to be found and the functions to be evaluated for any angle regardless of size. Figure 5.89 illustrates that   135°,   225°, and   495° are all coterminal, with each having a reference angle of 45°. EXAMPLE 8



Evaluating Trig Functions of Any Angle

Solution



The angles are coterminal and terminate in QII, where x 6 0 and y 7 0. With 12 12 r  45° we have sin 135°  , cos 1225°2   , and tan 495°  1. 2 2

B. You’ve just learned how to use reference angles to evaluate the trig functions for any angle

Now try Exercises 63 through 74



8

Since 360° is one full rotation, all angles   360°k will be coterminal for any integer k. For angles with a very large magnitude, we can find the quadrant of the terminal side by subtracting as many integer multiples of 360° as needed from the angle. 1908  5.3 and 1908  360152  108°. This angle is in QII with For   1908°, 360 r  72°. See Exercises 75 through 90.

5

C. Applications of the Trig Functions of Any Angle

Figure 5.90 y

  150 r  30

Evaluate sin 135°, cos 1225°2 , and tan 495°.

  30 x

One of the most basic uses of coterminal angles is determining all values of  that 1 satisfy a stated relationship. For example, by now you are aware that if sin   2 (positive one-half), then   30° or   150° (see Figure 5.90). But this is also true for all angles coterminal with these two, and we would write the solutions as   30°  360°k and   150°  360°k for all integers k.

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EXAMPLE 9



Finding All Angles that Satisfy a Given Equation Find all angles satisfying the relationship given. Answer in degrees. 12 a. cos    b. tan   1.3764 2

Solution



Figure 5.92 y 8

tan   1.3764   126 r  54

Figure 5.91 y 8

√2

cos    2

r  45 r  45

  135 x

b. Tangent is negative in QII and QIV. For 1.3764 we find r using a calculator: 2nd (tan 1) 1.3764 ENTER shows 1 tan 11.37642  54, so r  54°. Two solutions are   180°  54°  126° from QII, and in QIV   360°  54°  306°. The result is   126°  360°k and   306°  360°k. Note these can be combined into the single statement   126°  180°k. See Figure 5.92.

5

r  54

a. Cosine is negative in QII and QIII. Recognizing 12 cos 45°  , we reason r  45° and two 2 solutions are   135° from QII and   225° from QIII. For all values of  satisfying the relationship, we have   135°  360°k and   225°  360°k. See Figure 5.91.

x

Now try Exercises 93 through 100



We close this section with an additional application of the concepts related to trigonometric functions of any angle. EXAMPLE 10



Applications of Coterminal Angles: Location on Radar A radar operator calls the captain over to her screen saying, “Sir, we have an unidentified aircraft heading 20° (20° east of due north or a standard 70° rotation). I think it’s a UFO.” The captain asks, “What makes you think so?” To which the sailor replies, “Because it’s at 5000 ft and not moving!” Name all angles for which the UFO causes a “blip” to occur on the radar screen.

Solution



y

Blip!

70

Since radar typically sweeps out a 360° angle, a blip will occur on the screen for all angles   70°  360°k, where k is an integer. Now try Exercises 101 through 106

C. You’ve just learned how to solve applications using the trig functions of any angle

x



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519

5.7 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. An angle is in standard position if its vertex is at the and the initial side is along the . 2. A(n) angle is one where the side is coincident with one of the coordinate axes. 3. Angles formed by a counterclockwise rotation are angles. Angles formed by a rotation are negative angles. 

4. For any angle , its reference angle r is the positive angle formed by the side and the nearest x-axis. 5. Discuss the similarities and differences between the trigonometry of right triangles and the trigonometry of any angle. 6. Let T(x) represent any one of the six basic trig functions. Explain why the equation T1x2  k will always have exactly two solutions in [0, 2) if x is not a quadrantal angle.

DEVELOPING YOUR SKILLS 7. Draw a 30-60-90 triangle with the 60° angle at the origin and the short side along the positive x-axis. Determine the slope and equation of the line coincident with the hypotenuse, then pick any point on this line and evaluate sin 60°, cos 60°, and tan 60°. Comment on what you notice. 8. Draw a 45-45-90 triangle with a 45° angle at the origin and one side along the positive x-axis. Determine the slope and equation of the line coincident with the hypotenuse, then pick any point on this line and evaluate sin 45°, cos 45°, and tan 45. Comment on what you notice.

Graph each linear equation and state the quadrants it traverses. Then pick one point on the line from each quadrant and evaluate the functions sin , cos  and tan  using these points.

3 9. y  x 4 11. y  

13 x 3

10. y 

5 x 12

12. y  

13 x 2

Find the value of the six trigonometric functions given P(x, y) is on the terminal side of angle , with  in standard position.

13. (8, 15)

14. (7, 24)

15. (20, 21)

16. (3, 1)

17. (7.5, 7.5)

18. (9, 9)

19. (4 13, 4)

20. (6, 6 13)

21. (2, 8)

22. (6, 15)

23. (3.75, 2.5)

24. (6.75, 9)

5 2 25. a , b 9 3

7 3 26. a ,  b 4 16

1 15 27. a ,  b 4 2

28. a

13 22 , b 5 25

29. Evaluate the six trig functions in terms of x, y, and r for   90°. 30. Evaluate the six trig functions in terms of x, y, and r for   180°. Name the reference angle r for the angle  given.

31.   120°

32.   210°

33.   135°

34.   315°

35.   45°

36.   240°

37.   112°

38.   179°

39.   500°

40.   750°

41.   168.4°

42.   328.2°

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State the quadrant of the terminal side of , using the information given.

Find two positive and two negative angles that are coterminal with the angle given. Answers will vary.

43. sin  7 0, cos  6 0

63. 52°

64. 12°

44. cos  6 0, tan  6 0

65. 87.5°

66. 22.8°

45. tan  6 0, sin  7 0

67. 225°

68. 175°

46. sec  7 0, tan  7 0

69. 107°

70. 215°

Find the exact value of sin , cos , and tan  using reference angles.

47.   330°

48.   390°

49.   45°

50.   120°

51.   240°

52.   315°

53.   150°

54.   210°

For the information given, find the values of x, y, and r. Clearly indicate the quadrant of the terminal side of , then state the values of the six trig functions of .

55. cos  

4 and sin  6 0 5

56. tan   

12 and cos  7 0 5

37 57. csc    and tan  7 0 35 58. sin   

20 and cot  6 0 29

Evaluate in exact form as indicated.

71. sin 120°, cos 240°, tan 480° 72. sin 225°, cos 585°, tan 495° 73. sin 30°, cos 390°, tan 690° 74. sin 210°, cos 570°, tan 150° Find the exact value of sin , cos , and tan  using reference angles.

75.   600°

76.   480°

77.   840°

78.   930°

79.   570°

80.   495°

81.   1230°

82.   3270°

For each exercise, state the quadrant of the terminal side and the sign of the function in that quadrant. Then evaluate the expression using a calculator. Round to four decimal places.

83. sin 719°

84. cos 528°

59. csc   3 and cos  7 0

85. tan 419°

86. sec 621°

60. csc   2 and cos  7 0

87. csc 681°

88. tan 995°

7 61. sin    and sec  6 0 8

89. cos 805°

90. sin 772°

62. cos   

5 and sin  6 0 12

WORKING WITH FORMULAS

91. The area of a parallelogram: A  ab sin  The area of a parallelogram is given by the formula shown, where a and b are the lengths of the sides and  is the angle between them. Use the formula to complete the following: (a) find the area of a parallelogram with sides a  9 and b  21 given   50°. (b) What is the smallest integer value of  where the area is greater than 150 units2? (c) State what happens when   90°. (d) How can you find the area of a triangle using this formula?

92. The angle between two intersecting lines: m2  m1 tan   1  m2m1 Given line 1 and line 2 with slopes m1 and m2, respectively, the angle between the two lines is given by the formula shown. Find the angle  if the equation of line 1 is y1  34x  2 and line 2 has equation y2  23x  5.

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APPLICATIONS

Find all angles satisfying the stated relationship. For standard angles, express your answer in exact form. For nonstandard values, use a calculator and round function values to tenths.

Exercise 103

12 94. sin   2

1 93. cos   2 95. sin   

he enter the water feet first or head first? Through what angle did he turn from takeoff until the moment he entered the water?

13 2

96. tan   

97. sin   0.8754

13 1

98. cos   0.2378

99. tan   2.3512

100. cos   0.0562

101. Nonacute angles: At a Exercise 101 recent carnival, one of the games on the midway was played using a large spinner that turns clockwise. On Jorge’s spin the number 25 began at the 12 o’clock (top/center) position, returned to this position five times during the spin and stopped at the 3 o’clock position. What angle  did the spinner spin through? Name all angles that are coterminal with . 102. Nonacute angles: One of the four blades on a ceiling fan has a decal on it and begins at a designated “12 o’clock” position. Turning the switch on and then immediately off, causes the blade to make over three complete, counterclockwise rotations, with the blade stopping at the 8 o’clock position. What angle  did the blade turn through? Name all angles that are coterminal with . 103. High dives: As part of a diving competition, David executes a perfect reverse two-and-a-half flip. Does

104. Gymnastics: While working out on a trampoline, Charlene does three complete, forward flips and then belly-flops on the trampoline before returning to the upright position. What angle did she turn through from the start of Exercise 105 this maneuver to the y 10 moment she belly-flops? 105. Spiral of Archimedes: The graph shown is called the spiral of Archimedes. Through what angle  has the spiral turned, given the spiral terminates at 16, 22 as indicated? 106. Involute of a circle: The graph shown is called the involute of a circle. Through what angle  has the involute turned, given the graph terminates at 14, 3.52 as indicated?

10

10 x

0

(6, 2) 10

Exercise 106 y 10

10

0

(4, 3.5) 10

Area bounded by chord and circumference: Find the area of the shaded region, rounded to the nearest 100th. Note the area of a triangle is one-half the area of a parallelogram (see Exercise 91).

107.

108.

150 18 in.

252 20 ft

10 x

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109. In an elementary study of trigonometry, the hands of a clock are often studied because of the angle relationship that exists between the hands. For example, at 3 o’clock, the angle between the two hands is a right angle and measures 90°. a. What is the angle between the two hands at 1 o’clock? 2 o’clock? Explain why. b. What is the angle between the two hands at 6:30? 7:00? 7:30? Explain why. c. Name four times at which the hands will form a 45° angle. 110. In the diagram shown, the indicated ray is of arbitrary length. (a) Through what additional angle  would the ray have to be rotated to create triangle ABC? (b) What will be the length of side AC once the triangle is complete? 

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CHAPTER 5 An Introduction to Trigonometric Functions

111. Referring to Exercise 102, suppose the fan blade had a radius of 20 in. and is turning at a rate of 12 revolutions per second. (a) Find the angle the blade turns through in 3 sec. (b) Find the circumference of the circle traced out by the tip of the blade. (c) Find the total distance traveled by the blade tip in 10 sec. (d) Find the speed, in miles per hour, that the tip of the blade is traveling.

y 5

C



B 5

A

(3, 2)  5 x

5

MAINTAINING YOUR SKILLS

112. (5.1) For emissions testing, automobiles are held stationary while a heavy roller installed in the floor allows the wheels to turn freely. If the large wheels of a customized pickup have a radius of 18 in. and are turning at 300 revolutions per minute, what speed is the odometer of the truck reading in miles per hour? 113. (5.2) Jazon is standing 117 ft from the base of the Washington Monument in Washington, D.C. If his eyes are 5 ft above level ground and he must hold

his head at a 78° angle from horizontal to see the top of the monument (the angle of elevation of 78°2, estimate the height of the monument. Answer to the nearest tenth of a foot. 114. (4.4) Solve for t. Answer in both exact and approximate form: 250  150e0.05t  202. 115. (2.3) Find the equation of the line perpendicular to 4x  5y  15 that contains the point 14, 32.

S U M M A RY A N D C O N C E P T R E V I E W SECTION 5.1

Angle Measure, Special Triangles, and Special Angles

KEY CONCEPTS • An angle is defined as the joining of two rays at a common endpoint called the vertex. • An angle in standard position has its vertex at the origin and its initial side on the positive x-axis. • Two angles in standard position are coterminal if they have the same terminal side. • A counterclockwise rotation gives a positive angle, a clockwise rotation gives a negative angle.

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Summary and Concept Review

• One 11°2 degree is defined to be • • • • • • • • • • • • •

523

1 of a full revolution. One (1) radian is the measure of a central angle 360 subtended by an arc equal in length to the radius. Degrees can be divided into a smaller unit called minutes: 1°  60¿; minutes can be divided into a smaller unit called seconds: 1¿  60–. This implies 1°  3600–. Two angles are complementary if they sum to 90° and supplementary if they sum to 180°. Properties of triangles: (I) the sum of the angles is 180°; (II) the combined length of any two sides must exceed that of the third side and; (III) larger angles are opposite larger sides. Given two triangles, if all three corresponding angles are equal, the triangles are said to be similar. If two triangles are similar, then corresponding sides are in proportion. In a 45-45-90 triangle, the sides are in the proportion 1x: 1x: 12x. In a 30-60-90 triangle, the sides are in the proportion 1x: 13x: 2x. The formula for arc length: s  r,  in radians. 1 The formula for the area of a circular sector: A  r2,  in radians. 2  180° To convert degree measure to radians, multiply by ; for radians to degrees, multiply by .  180°     Special angle conversions: 30°  , 45°  , 60°  , 90°  . 6 4 3 2 A location north or south of the equator is given in degrees latitude; a location east or west of the Greenwich Meridian is given in degrees longitude.  Angular velocity is a rate of rotation per unit time:   . t r Linear velocity is a change in position per unit time: V  or V  r. t

EXERCISES 1. Convert 147°36¿48– to decimal degrees. 2. Convert 32.87° to degrees, minutes, and seconds. 3. All of the right triangles given are similar. Find the dimensions of the largest triangle. Exercise 3

Exercise 4

16.875 3 6

4

60



d

40

0y

d

4. Use special angles/special triangles to find the length of the bridge needed to cross the lake shown in the figure. 5. Convert to degrees:

2 . 3

7. Find the arc length if r  5 and   57°.

6. Convert to radians: 210°. 8. Evaluate without using a calculator: 7 sina b. 6

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Find the angle, radius, arc length, and/or area as needed, until all values are known. y y 9. 10. 11. 96 in.

y 152 m2

s 1.7



2.3 15 cm

r

x

8

x

x

12. With great effort, 5-year-old Mackenzie has just rolled her bowling ball down the lane, and it is traveling painfully slow. So slow, in fact, that you can count the number of revolutions the ball makes using the finger holes as a reference. (a) If the ball is rolling at 1.5 revolutions per second, what is the angular velocity? (b) If the ball’s radius is 5 in., what is its linear velocity in feet per second? (c) If the distance to the first pin is 60 feet and the ball is true, how many seconds until it hits?

SECTION 5.2

Unit Circles and the Trigonometry of Real Numbers

KEY CONCEPTS • A central unit circle is a circle with radius 1 unit having its center at the origin. • A central circle is symmetric to both axes and the origin. This means that if (a, b) is a point on the circle, then 1a, b2, 1a, b2 , and 1a, b2 are also on the circle and satisfy the equation of the circle. On a unit circle with  in radians, the length of a subtended arc is numerically the same as the subtended angle, • making the arc a “circular number line” and associating any given rotation with a unique real number. • A reference angle is defined to be the acute angle formed by the terminal side of a given angle and the x-axis. For functions of a real number we refer to a reference arc rather than a reference angle. • For any real number t and a point on the unit circle associated with t, we have: cos t  x

sin t  y

y x x0

tan t 

1 x x0

sec t 

1 y y0

csc t 

x y y0

cot t 

• Given the specific value of any function, the related real number t or angle  can be found using a reference arc/angle, or the sin1, cos1, or tan1 features of a calculator.

EXERCISES 113 , yb is on a unit circle, find y if the point is in QIV, then use the symmetry of the circle to locate 7 three other points.

13. Given a

3 17 b is on the unit circle, find the value of all six trig functions of t without the use of a 14. Given a ,  4 4 calculator. 2 . 15. Without using a calculator, find two values in [0, 2) that make the equation true: csc t  13 16. Use a calculator to find the value of t that corresponds to the situation described: cos t  0.7641 with t in QII. 17. A crane used for lifting heavy equipment has a winch-drum with a 1-yd radius. (a) If 59 ft of cable has been wound in while lifting some equipment to the roof-top of a building, what radian angle has the drum turned through? (b) What angle must the drum turn through to wind in 75 ft of cable?

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525

Graphs of the Sine and Cosine Functions; Cosecant and Secant Functions

SECTION 5.3

KEY CONCEPTS • Graphing sine and cosine functions using the special values from the unit circle results in a periodic, wavelike graph with domain 1q, q 2.  6 , 0.87

 4 , 0.71

cos t

 6 , 0.5

 3 , 0.5

1

 2 ,

0.5

0

sin t

 3 , 0.87

 2 , 1

1

0.5

Decreasing (0, 0)

 4 , 0.71

 2

Increasing 

3 2

2

t

(0, 0)

0.5

0.5

1

1

 2



3 2

2

t

• The characteristics of each graph play a vital role in their contextual application, and these are summarized on • • • • •

• •

pages 457 and 460. The amplitude of a sine or cosine graph is the maximum displacement from the average value. For y  A sin1Bt2 and y  A cos1Bt2, the amplitude is A. The period of a periodic function is the smallest interval required to complete one cycle. 2 For y  A sin1Bt2 and y  A cos1Bt2, P  gives the period. B If A 7 1, the graph is vertically stretched, if 0 6 A 6 1 the graph is vertically compressed, and if A 6 0 the graph is reflected across the x-axis. If B 7 1, the graph is horizontally compressed (the period is smaller/shorter); if B 6 1 the graph is horizontally stretched (the period is larger/longer). 2 To graph y  A sin1Bt2 or A cos (Bt), draw a reference rectangle 2A units high and P  units wide, B centered on the x-axis, then use the rule of fourths to locate zeroes and max/min values. Connect these points with a smooth curve. 1 The graph of y  sec t  will be asymptotic everywhere cos t  0, increasing where cos t is decreasing, cos t and decreasing where cos t is increasing. 1 The graph of y  csc t  will be asymptotic everywhere sin t  0, increasing where sin t is decreasing, sin t and decreasing where sin t is increasing.

EXERCISES Use a reference rectangle and the rule of fourths to draw an accurate sketch of the following functions through at least one full period. Clearly state the amplitude (as applicable) and period as you begin. 18. y  3 sin t 19. y  3 sec t 20. y  cos12t2 21. y  1.7 sin14t2 22. f1t2  2 cos14t2 23. g1t2  3 sin1398t2

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The given graphs are of the form y  A sin1Bt2 and y  A csc1Bt2 . Determine the equation of each graph. y 24. 25. 8 y 1 6 0.5

4 2

(0, 0)

 6

 3

0.5

 2

2 3

5 t 6

(0, 0) 2 4

1 3

2 3

1

4 3

5 3

t

6 1

8

26. Referring to the chart of colors visible in the electromagnetic spectrum (page 469), what color is represented by   tb? By y  sina tb? the equation y  sina 270 320

SECTION 5.4

Graphs of Tangent and Cotangent Functions

KEY CONCEPTS y  2 (0, 1) • Since tan t is defined in terms of the ratio x , the graph will be asymptotic everywhere x  0 (x, y)  on the unit circle, meaning all odd multiples of . 0 2  (1, 0) (1, 0) x x • Since cot t is defined in terms of the ratio y , the graph will be asymptotic everywhere y  0 3 (0, 1) on the unit circle, meaning all integer multiples of . 2 The graph of is increasing everywhere it is defined; the graph of is y  tan t y  cot t • decreasing everywhere it is defined. • The characteristics of each graph play a vital role in their contextual application, and these are summarized on page 474. • For the more general tangent and cotangent graphs y  A tan1Bt2 and y  A cot1Bt2, if A 7 1, the graph is vertically stretched, if 0 6 A 6 1 the graph is vertically compressed, and if A 6 0 the graph is reflected across the x-axis. • If B 7 1, the graph is horizontally compressed (the period is smaller/shorter); if B 6 1 the graph is horizontally stretched (the period is larger/longer).  • To graph y  A tan1Bt2, note A tan(Bt) is zero at t  0. Compute the period P  and draw asymptotes a B P distance of on either side of the y-axis. Plot zeroes halfway between the asymptotes and use symmetry to 2 complete the graph.  • To graph y  A cot1Bt2, note it is asymptotic at t  0. Compute the period P  and draw asymptotes a distance B P on either side of the y-axis. Plot zeroes halfway between the asymptotes and use symmetry to complete the graph. EXERCISES 27. State the value of each expression without the aid of a calculator: 7  tana b cota b 4 3 28. State the value of each expression without the aid of a calculator, given that t terminates in QII. 1 tan1 1 132 cot1a b 13

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1 29. Graph y  6 tana tb in the interval [2, 2]. 2

30. Graph y 

527

1 cot12t2 in the interval [1, 1]. 2

31. Use the period of y  cot t to name three additional solutions to cot t  0.0208, given t  1.55 is a solution. Many solutions are possible.

32. Given t  0.4444 is a solution to cot1 1t2  2.1, use an analysis of signs and quadrants to name an additional solution in [0, 2). d 33. Find the approximate height of Mount Rushmore, using h  and the values shown. cot u  cot v

h (not to scale) v  40

u  25 144 m

34. Model the data in the table using a tangent function. Clearly state the period, the value of A, and the location of the asymptotes.

SECTION 5.5

Input

Output

Input

Output

6

q

1

1.4

5

19.4

2

3

4

9

3

5.2

3

5.2

4

9

2

3

5

19.4

1

1.4

6

q

0

0

Transformations and Applications of Trigonometric Graphs

KEY CONCEPTS • Many everyday phenomena follow a sinusoidal pattern, or a pattern that can be modeled by a sine or cosine function (e.g., daily temperatures, hours of daylight, and more). • To obtain accurate equation models of sinusoidal phenomena, vertical and horizontal shifts of a basic function are used. • The equation y  A sin1Bt C2 D is called the standard form of a general sinusoid. The equation C y  A sin c B at b d D is called the shifted form of a general sinusoid. B • In either form, D represents the average value of the function and a vertical shift D units upward if D 7 0, Mm M m  D,  A. D units downward if D 6 0. For a maximum value M and minimum value m, 2 2 C • The shifted form y  A sin c Bat b d D enables us to quickly identify the horizontal shift of the function: B C units in a direction opposite the given sign. B • To graph a shifted sinusoid, locate the primary interval by solving 0  Bt C 6 2, then use a reference rectangle along with the rule of fourths to sketch the graph in this interval. The graph can then be extended as needed, then shifted vertically D units.

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• One basic application of sinusoidal graphs involves phenomena in harmonic motion, or motion that can be

modeled by functions of the form y  A sin1Bt2 or y  A cos1Bt2 (with no horizontal or vertical shift). • If the period P and critical points (X, M) and (x, m) of a sinusoidal function are known, a model of the form y  A sin1Bx C2 D can be obtained: B

2 P

A

Mm 2

D

M m 2

C

3  Bx 2

EXERCISES For each equation given, (a) identify/clearly state the amplitude, period, horizontal shift, and vertical shift; then (b) graph the equation using the primary interval, a reference rectangle, and rule of fourths.   3 35. y  240 sin c 1t  32 d 520 36. y  3.2 cosa t b 6.4 6 4 2 For each graph given, identify the amplitude, period, horizontal shift, and vertical shift, and give the equation of the graph. 37. 350 y 38. 210 y 300

180

250

150

200

120

150

90

100

60

50

30

0 6

12

18

0

24 t

 4

 2

3 4



t

39. Monthly precipitation in Cheyenne, Wyoming, can be modeled by a sine function, by using the average precipitation for July (2.26 in.) as a maximum (actually slightly higher in May), and the average precipitation for February (0.44 in.) as a minimum. Assume t  0 corresponds to March. (a) Use the information to construct a sinusoidal model, and (b) use the model to estimate the inches of precipitation Cheyenne receives in August 1t  52 and December 1t  92. Source: 2004 Statistical Abstract of the United States, Table 380.

SECTION 5.6

The Trigonometry of Right Triangles

KEY CONCEPTS • The sides of a right triangle can be named relative to their location with respect to a given angle. B 

B Hypotenuse

A

 Side adjacent 

Hypotenuse

Side opposite  A

C

Side adjacent 

Side opposite 

C

• The ratios of two sides with respect to a given angle are named as follows: sin  

opp hyp

cos  

adj hyp

tan  

opp adj

• The reciprocal of the ratios above play a vital role and are likewise given special names: hyp opp 1 csc   sin 

csc  

hyp adj 1 sec   cos 

sec  

adj opp 1 cot   tan  cot  

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Summary and Concept Review

• Each function of  is equal to the cofunction of its complement. For instance, the complement of sine is cosine

and sin   cos190°  2. • To solve a right triangle means to apply any combination of the trig functions, along with the triangle properties, until all sides and all angles are known. • An angle of elevation is the angle formed by a horizontal line of sight (parallel to level ground) and the true line of sight. An angle of depression is likewise formed, but with the line of sight below the line of orientation.

EXERCISES 40. Use a calculator to solve for A: a. cos 37°  A b. cos A  0.4340

41. Rewrite each expression in terms of a cofunction. a. tan 57.4° b. sin119°30¿15– 2

Solve each triangle. Round angles to the nearest tenth and sides to the nearest hundredth. B B 42. 43. c

20 m c

A

89 in.

49

C

21 m

A

C

b

44. Josephine is to weld a vertical support to a 20-m ramp so that the incline is exactly 15°. What is the height h of the support that must be used? 45. From the observation deck of a seaside building 480 m high, Armando sees two fishing boats in the distance. The angle of depression to the nearer boat is 63.5°, while for the boat farther away the angle is 45°. (a) How far out to sea is the nearer boat? (b) How far apart are the two boats?

20 m

h

15 45 63.5 480 m

46. A slice of bread is roughly 14 cm by 10 cm. If the slice is cut diagonally in half, what acute angles are formed?

SECTION 5.7

Trigonometry and the Coordinate Plane

KEY CONCEPTS • In standard position, the terminal sides of 0°, 90°, 180°, 270°, and 360° angles coincide with one of the axes and are called quadrantal angles. • By placing a right triangle in the coordinate plane with one acute angle at the origin and one side along the x-axis, we note the trig functions can be defined in terms of a point P(x, y) on the hypotenuse. • Given P(x, y) is any point on the terminal side of an angle  in standard position. Then r  2x2 y2 is the distance from the origin to this point. The six trigonometric functions of  are defined as y r r x csc   sec   cot   x x y y x0 y0 x0 y0 • A reference angle r is defined to be the acute angle formed by the terminal side of a given angle  and the x-axis. • Reference angles can be used to evaluate the trig functions of any nonquadrantal angle, since the values are fixed by the ratio of sides and the signs are dictated by the quadrant of the terminal side. sin  

y r

cos  

x r

tan  

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• If the value of a trig function and the quadrant of the terminal side are known, the related angle  can be found using a reference arc/angle, or the sin1, cos1, or tan1 features of a calculator. • If  is a solution to sin   k, then   360°k is also a solution for any integer k.

EXERCISES 47. Find two positive angles and two negative angles that are coterminal with   207°. 48. Name the reference angle for the angles given:   152°   521°   210° 49. Find the value of the six trigonometric functions, given P(x, y) is on the terminal side of angle  in standard position. a. P112, 352 b. 112, 182 50. Find the values of x, y, and r using the information given, and state the quadrant of the terminal side of . Then state the values of the six trig functions of . 4 12 a. cos   ; sin  6 0 b. tan    ; cos  7 0 5 5 51. Find all angles satisfying the stated relationship. For standard angles, express your answer in exact form. For nonstandard angles, use a calculator and round to the nearest tenth. a. tan   1

b. cos  

13 2

c. tan   4.0108

d. sin   0.4540

MIXED REVIEW 1. For the graph of periodic function f given, state the (a) amplitude, (b) average value, (c) period, and (d) value of f(4).

2. Name two values in 3 0, 22 where tan t  1.

Exercise 1 y 30

y  f(x)

25 20 15 10 5 4

8

12

16

1 3. Name two values in 3 0, 22 where cos t   . 2 8 4. Given sin   with  in QII, state the value 1185 of the other five trig functions. 5. Convert to DMS form: 220.8138°. 6. Find two negative angles and two positive angles that are coterminal with (a) 57° and (b) 135°. Exercise 7 7. To finish the top row of the tile pattern on our bathroom wall, 12– by 12– tiles must be cut diagonally. Use a standard triangle to find the length of each cut and the width of the wall covered by tiles.

x

8. The service door into the foyer of a large office building is 36– wide by 78– tall. The building manager has ordered a large wall painting 85– by 85– to add some atmosphere to the foyer area. (a) Can the painting be brought in the service door? (b) If so, at what two integer-valued angles (with respect to level ground) could the painting be tilted? Exercise 9 9. Find the arc length and area of the shaded sector. 10. Monthly precipitation in Minneapolis, Minnesota, can ␪ be modeled by a sine function, by using the average precipitation for August (4.05 in.) as a (4√3, 4) maximum (actually slightly higher in June), and the average precipitation for February (0.79 in.) as a minimum. Assume t  0 corresponds to April. (a) Use the information to construct a sinusoidal model, and (b) Use the model to approximate the inches of precipitation Minneapolis receives in July (t  3) and December (t  8). Source: 2004 Statistical Abstract of the United States, Table 380

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11. Convert from DMS to decimal degrees: 86° 54¿ 54–. 12. Name the reference angle r for the angle  given. 5 5 a. 735° b. 135° c. d.  6 3 13. Find the value of all six trig functions of , given the point 115, 82 is on the terminal side. 12 12 , b is a point on the unit circle 14. Verify that a 2 2 and find the value of all six trig functions at this point. 15. On your approach shot to the ninth green, the Global Positioning System (GPS) your cart is equipped with 115.47 yd tells you the pin is 115.47 yd away. The distance plate states the straight line  100 yd distance to the hole is 100 yd (see the diagram). Relative to a straight line between the plate and the hole, at what acute angle  should you hit the shot? 16. The electricity supply lines to the top of Lone Eagle Plateau must be replaced, and the B new lines will be run A in conduit buried 1:150 slightly beneath the 1 in.  200 ft surface. The scale of elevation is 1:150 (each closed figure indicates an increase in 150 ft of elevation), and the scale of horizontal distance is 1 in.  200 ft. (a) Find the increase in elevation from point A to point B, (b) use a proportion to find the horizontal distance from A to B if the measured distance on the map is 214 in., (c) draw the corresponding right triangle and use it to estimate the length of conduit needed from A to B and the angle of incline the installers will experience while installing the conduit.

Mixed Review

531

17. A salad spinner consists of a colander basket inside a large bowl, and is used to wash and dry lettuce and other salad ingredients. The spinner is turned at about 3 revolutions per second. (a) Find the angular velocity and (b) find the linear velocity of a point of the circumference if the basket has a 20 cm radius. 18. Solve each equation in [0, 2) without the use of a calculator. If the expression is undefined, so state.  a. x  sina b b. sec x  12 4  c. cota b  x d. cos   x 2  2 13 e. csc x  f. tana b  x 3 2 19. State the amplitude, period, horizontal shift, vertical shift, and endpoints of the primary interval (as applicable), then sketch the graph using a reference rectangle and the rule of fourths.  7 a. y  5 cos12t2  8 b. y  sin c 1x  12 d 2 2  1 c. y  2 tana tb d. y  3 secax  b 4 2 20. Virtually everyone is familiar with the Statue of Liberty in New York Bay, but fewer know that America is home to a second “Statue of Liberty” standing proudly atop the iron dome of the Capitol Building. From a distance of 600 ft, the angle of elevation from ground level to the top of the statue (from the east side) is 25.60°. The angle of elevation to the base of the statue is 24.07°. How tall is the statue Freedom (the name sculptor Thomas Crawford gave this statue)?

H

25.6 600 ft

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PRACTICE TEST 3. Find two negative angles and two positive angles that are coterminal with   30°. Many solutions are possible.

1. State the complement and supplement of 35°. 2. Name the reference angle of each angle given. a. 225° b. 510° 25 7 c. d. 6 3

4. Convert from DMS to decimal degrees or decimal degrees to DMS as indicated. a. 100°45¿18– to decimal degrees b. 48.2125° to DMS

5. Four Corners USA is the point at which Utah, Colorado, Arizona, and New Mexico meet. The southern border of Colorado, the western border of Kansas, and the point P where Colorado, Nebraska, and Kansas meet, very nearly approximates a 30-60-90 triangle. If the western border of Kansas is 215 mi long, (a) what is the distance from Four Corners USA to point P? (b) How long is Colorado’s southern border?

Exercise 5

Colorado

Nebraska P Kansas

Utah Arizona

New Mexico

6. Complete the table from memory using exact values. If a function is undefined, so state. t

sin t

cos t

tan t

csc t

sec t

cot t

0 2 3 7 6 5 4 5 3 13 6

2 and tan  6 0, find the value of the 5 other five trig functions of .

7. Given cos  

1 212 b is a point on the unit circle, 8. Verify that a ,  3 3 then find the value of all six trig functions associated with this point. 9. In order to take pictures of a dance troupe as it performs, a camera crew rides in a cart on tracks that trace a circular arc. The radius of the arc is 75 ft, and

from end to end the cart sweeps out an angle of 172.5° in 20 seconds. Use this information to find (a) the length of the track in feet and inches, (b) the angular velocity of the cart, and (c) the linear velocity of the cart in both ft/sec and mph. 10. Solve the triangle shown. Answer in table form. Exercise 10 A

C 15.0 cm

57 B

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Practice Test

11. The “plow” is a yoga position in which a person lying on their back brings their feet up, over, and behind their head and touches them to the floor. If distance from hip to shoulder (at the right angle) is 57 cm and from hip to toes is 88 cm, find the distance from shoulders to toes and the angle formed at the hips. Exercise 11 Hips Legs

Torso

Arms

Head

Toes

12. While doing some night fishing, you round a peninsula and a tall light house comes into view. Taking a sighting, you find the angle of elevation to the top of the lighthouse is 25°. If the lighthouse is known to be 27 m tall, how far from the lighthouse are you?

the day, with a minimum usage of 157,000 gallons in the cool of the night. Assume t  0 corresponds to 6:00 A.M. (a) Use the information to construct a sinusoidal model, and (b) Use the model to approximate water usage at 4:00 P.M. and 4:00 A.M. 15. State the domain, range, period, and amplitude (if it exists), then graph the function over 1 period.  a. y  2 sin a tb b. y  sec t 5 c. y  2 tan13t2 16. State the amplitude, period, horizontal shift, vertical shift, and endpoints of the primary interval. Then sketch the graph using a reference rectangle and the rule of fourths: y  12 sina3t 

 b  19. 4

17. An athlete throwing the shot-put begins his first attempt facing due east, completes three and onehalf turns and launches the shot facing due west. What angle did his body turn through? 18. State the domain, range, and period, then sketch the graph in 30, 22.

Exercise 12

1 b. y  cota tb 2

a. y  tan12t2 27 m 25

13. Find the value of t 3 0, 2 4 satisfying the conditions given. 1 a. sin t   , t in QIII 2 b. sec t 

533

213 , t in QIV 3

c. tan t  1, t in QII 14. In arid communities, daily water usage can often be approximated using a sinusoidal model. Suppose water consumption in the city of Caliente del Sol reaches a maximum of 525,000 gallons in the heat of

19. Due to tidal motions, the depth of water in Brentwood Bay varies sinusoidally as shown in the diagram, where time is in hours and depth is in feet. Find an equation that models the depth of water at time t. Exercise 19 Depth (ft) 20 16 12 8 4

Time (hours) 0

4

8

12

16

20

24

20. Find the value of t satisfying the given conditions. a. sin t  0.7568; t in QIII b. sec t  1.5; t in QII

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C A L C U L AT O R E X P L O R AT I O N A N D D I S C O V E RY Variable Amplitudes and Modeling the Tides Tidal motion is often too complex to be modeled by a single sine function. In this Exploration and Discovery, we’ll look at a method that combines two sine functions to help model a tidal motion with variable amplitude. In the process, we’ll use much of what we know about the amplitude, horizontal shifts and vertical shifts of a sine function, helping to reinforce these important concepts and broaden our understanding about how they can be applied. The graph in Figure 5.93 shows three days of tidal motion for Davis Inlet, Canada. Figure 5.93 Height (m) 2.4

2.3 1.9

2

2.4 2.0

1.9

1

0.9 0 Midnight

0.8

0.7 Noon

0.7

0.7

Midnight

Noon

Midnight

0.6 Noon

t Midnight

As you can see, the amplitude of the graph varies, and there is no single sine function that can serve as a model. However, notice that the amplitude varies predictably, and that the high tides and low tides can independently be modeled by a sine function. To simplify our exploration, we will use the assumption that tides have an exact 24-hr period (close, but no), that variations between high and low tides takes place every 12 hr (again close but not exactly true), and the variation between the “low-high” (1.9 m) and the “high-high” (2.4 m) is uniform. A similar assumption is made for the low tides. The result is the graph in Figure 5.94. Figure 5.94 Height (m)

2.4

2.4 1.9

2

2.4 1.9

1.9

1

0.9 0 Midnight

0.7

0.9

0.7

0.9

0.7 t

Noon

Midnight

Noon

Midnight

Noon

Midnight

First consider the high tides, which vary from a maximum of 2.4 to a minimum of 1.9. Using the ideas from Section 5.7 to construct an equation model gives 2.4  1.9 2.4 1.9 A  0.25 and D   2.15. With 2 2 a period of P  24 hr we obtain the equation  Y1  0.25 sina xb 2.15. Using 0.9 and 0.7 as the 12

maximum and minimum low tides, similar calculations  yield the equation Y2  0.1 sina xb 0.8 (verify this). 12 Graphing these two functions over a 24-hr period yields the graph in Figure 5.95, where we note the high and low values are correct, but the two functions are in phase with each other. As can be determined from Figure 5.94, we want the high tide model to start at the Figure 5.95 average value and 3 decrease, and the low tide equation model to start at high-low and decrease. Replac- 0 24 ing x with x  12 in Y1 and x with x 6 in Y2 accomplishes this result (see Figure 0 5.96). Now comes the Figure 5.96 fun part! Since Y1 rep3 resents the low/high maximum values for high tide, and Y2 represents the 24 low/high minimum 0 values for low tide, the amplitude and average value for the tidal 0 motion at Davis Inlet Y1  Y2 Figure 5.97 are A  2 3 Y1 Y2 ! and D  2 By entering 0 24 Y1  Y2 Y3  and 2 Y1 Y2 Y4  , the 2 0 equation for the tidal motion (with its variable amplitude) will have the form Y5  Y3 sin1Bx C2 Y4, where the value of B and C must be determined. The key here is to note there is only a 12-hr difference between the changes in amplitude,  so P  12 (instead of 24) and B  for this function. 6 function. Also, from the graph (Figure 5.94) we note the tidal motion begins at a minimum and increases, indicating a shift of 3 units to the right is required. Replacing x with x  3 gives the equation modeling these tides, and the final  equation is Y5  Y3 sin c 1x  32 d Y4. Figure 5.97 6

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Strengthening Core Skills

gives a screen shot of Y1, Y2, and Y5 in the interval [0, 24]. The tidal graph from Figure 5.94 is shown in Figure 5.98 with Y3 and Y4 superimposed on it. Figure 5.98 Height (m)

2.4

2.4 1.9

2

2.4 Y1

1.9

1.9

Y5

535

Exercise 1: The website www.tides.com/tcpred.htm offers both tide and current predictions for various locations around the world, in both numeric and graphical form. In addition, data for the “two” high tides and “two” low tides are clearly highlighted. Select a coastal area where tidal motion is similar to that of Davis Inlet, and repeat this exercise. Compare your model to the actual data given on the website. How good was the fit?

1

0.9 0 Midnight

0.7

0.9

0.7

0.9

0.7

Y2 t

Noon

Midnight

Noon

Midnight

Noon

Midnight

STRENGTHENING CORE SKILLS Standard Angles, Reference Angles, and the Trig Functions A review of the main ideas discussed in this chapter indicates there are four of what might be called “core skills.” These are skills that (a) play a fundamental part in the acquisition of concepts, (b) hold the overall structure together as we move from concept to concept, and (c) are ones we return to again and again throughout our study. The first of these is (1) knowing the standard angles and standard values. These values are “standard” because no estimation, interpolation, or special methods are required to name their value, and each can be expressed as a single factor. This gives them a great advantage in that further conceptual development can take place without the main points being obscured by large expressions or decimal approximations. Knowing the value of the trig functions for each standard angle will serve you very well throughout this study. Know the chart on page ••• and the ideas that led to it. The standard angles/values brought us to the trigonometry of any angle, forming a strong bridge to the second core skill: Figure 5.99 (2) using reference y angles to determine sin r  12 2 the value of the trig functions in each (√3, 1) r  30 r  30 (√3, 1) quadrant. For review, 2 2 a 30-60-90 triangle 2 2 x will always have sides 2 2 that are in the propor1x: 13x: 2x, (√3, 1) r  30 r  30 (√3, 1) tion regardless of its size. 2 This means for any , angle where r  30°, sin   12 or sin   12 since the ratio is fixed but the sign depends on the quadrant of : sin 30°  12

[QI], sin 150°  12 [QII], sin 210°  12 [QIII], sin 330°  12 [QII], and so on (see Figure 5.99). In turn, the reference Figure 5.100 angles led us to a third core y 2 cos r  √3 skill, helping us realize that 2 if  was not a quadrantal (√3, 1) angle, (3) equations like 210 2 13 r  30 150 cos12   must have 2 2 2 x r  30 two solutions in 30, 360°2. 2 From the standard angles (√3, 1) and standard values we 2 learn to recognize that for 13 cos    , r  30°, 2 which will occur as a reference angle in the two quadrants where cosine is negative, QII and QIII. The solutions in 30, 360°2 are   150° and   210° (see Figure 5.100). Of necessity, this brings us to the fourth core skill, (4) effective use of a calculator. The standard angles are a wonderful vehicle for introducing the basic ideas of trigonometry, and actually occur quite frequently in realworld applications. But by far, most of the values we encounter will be nonstandard values where r must be found using a calculator. However, once r is found, the reason and reckoning inherent in these ideas can be directly applied. The Summary and Concept Review Exercises, as well as the Practice Test offer ample opportunities to refine these skills, so that they will serve you well in future chapters as we continue our attempts to explain and understand the world around us in mathematical terms.

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CHAPTER 5 An Introduction to Trigonometric Functions

Exercise 1: Fill in the table from memory. t

 6

0

 4

 2

 3

a. 2 sin t 13  0 b. 312 cos t 4  1

sin t  y

c. 13 tan t 2  1

cos t  x tan t 

Exercise 2: Solve each equation in 3 0, 22 without the use of a calculator.

d. 12 sec t 1  3

y x

2 3

3 4

5 6



7 6

5 4

Exercise 3: Solve each equation in 30, 22 using a calculator and rounding answers to four decimal places. a. 16 sin t  2  1 b. 3 12 cos t 12  0 1 1 c. 3 tan t   2 4 d. 2 sec t  5

C U M U L AT I V E R E V I E W C H A P T E R S 1 – 5 1. Solve the inequality given: 2x 1  3 6 5 2. Find the domain of the function: y  2x  2x  15 80 3. Given that tan   , draw a right triangle that 39 corresponds to this ratio, then use the Pythagorean theorem to find the length of the missing side. Finally, find the two acute angles. 2

4. Without a calculator, what values in 3 0, 22 make 13 ? the equation true: sin t   2 17 3 b is a point on the unit circle 5. Given a ,  4 4 corresponding to t, find all six trig functions of t. State the domain and range of each function shown: 6. y  f1x2

7. a. f1x2  12x  3 y

2x b. g1x2  2 x  49

5

5

5 x

f(x) 5

8. y  T1x2 x

T(x)

0

7

1

5

2

3

3

1

4

1

5

3

6

5

9. Analyze the graph of the function in Exercise 6, including: (a) maximum and minimum values; (b) intervals where f1x2  0 and f1x2 6 0; (c) intervals where f is increasing or decreasing; and (d) any symmetry noted. Assume the features you are to describe have integer values. 10. The attractive force that exists between two magnets varies inversely as the square of the distance between them. If the attractive force is 1.5 newtons (N) at a distance of 10 cm, how close are the magnets when the attractive force reaches 5 N?

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Cumulative Review Chapters 1–5

11. The world’s tallest indoor waterfall is in Detroit, Michigan, in the lobby of the International Center Building. Standing 66 ft from the base of the falls, the angle of elevation is 60°. How tall is the waterfall? 12. It’s a warm, lazy Saturday and Hank is watching a county maintenance crew mow the park across the street. He notices the mower takes 29 sec to pass through 77° of rotation from one end of the park to the other. If the corner of the park is 60 ft directly across the street from his house, (a) how wide is the park? (b) How fast (in mph) does the mower travel as it cuts the grass?

537

19. Determine the equation of the graph shown given it is of the form y  A sin1Bt C2 D. y 2 1  4

0 1

 2

3 4

 x

2

13. Graph using transformations of a parent function: 1 f1x2   2. x 1

20. In London, the average temperatures on a summer day range from a high of 72°F to a low of 56°F (Source: 2004 Statistical Abstract of the United States, Table 1331). Use this information to write a sinusoidal equation model, assuming the low temperature occurs at 6:00 A.M. Clearly state the amplitude, average value, period, and horizontal shift.

14. Graph using transformations of a parent function: g1x2  ex1  2.

21. The graph of a function f(x) is given. Sketch the graph of f1 1x2.

15. Find f12 for all six trig functions, given the point P19, 402 is a point on the terminal side of the angle. Then find the angle  in degrees, rounded to tenths. 16. Given t  5.37, (a) in what quadrant does the arc terminate? (b) What is the reference arc? (c) Find the value of sin t rounded to four decimal places. 17. A jet-stream water sprinkler shoots water a distance of 15 m and turns back-and-forth through an angle of t  1.2 rad. (a) What is the length of the arc that the sprinkler reaches? (b) What is the area in m2 of the yard that is watered? 18. Determine the equation of graph shown given it is of the form y  A tan1Bt2.

y 5

5

5 x

5

22. The volume of a spherical cap is given by h2 13r  h2. Solve for r in terms of V and h. V 3 h r

y 4

23. Find the slope and y-intercept: 3x  4y  8.

2 3

4



2



 4 2

 8 , 1

4

 4

 2

3 4

t

24. Solve by factoring: 4x3  8x2  9x 18  0. 25. At what interest rate will $1000 grow to $2275 in 12 yr if compounded continuously?

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CONNECTIONS TO CALCULUS While right triangles have a number of meaningful applications as a problem-solving tool, they can also help to rewrite certain expressions in preparation for the tools of calculus, and introduce us to an alternative method for graphing relations and functions using polar coordinates. As things stand, some functions and relations are much easier to graph in polar coordinates, and converting between the two systems is closely connected to a study of right triangles.

Right Triangle Relationships Drawing a diagram to visualize relationships and develop information is an important element of good problem solving. This is no less true in calculus, where it is often a fundamental part of understanding the question being asked. As a precursor to applications involving trig substitutions, we’ll illustrate how right triangle diagrams are used to rewrite trigonometric functions of  as algebraic functions of x. EXAMPLE 1



Using Right Triangle Diagrams to Rewrite Trig Expressions Use the equation x  5 sin  and a right triangle diagram to write cos , tan , sec , and csc  as functions of x.

Solution



x Using x  5 sin , we obtain sin   . From our 5 work in Chapter 5, we know the right triangle opp definition of sin  is , and we draw a triangle hyp with side x oriented opposite an angle , and label a hypotenuse of 5 (see figure). To find an expression for the adjacent side, we use the Pythagorean theorem: 1adj2 2  x2  52 1adj2 2  25  x2 adj  225  x2

5

x

␪ Adjacent side

Pythagorean theorem isolate term result (length must be positive); 5 6 x 6 5

Using this triangle and the standard definition of the remaining trig functions, we 225  x2 5 5 x find cos   , tan   , sec   , and csc   . 2 2 x 5 225  x 225  x Now try Exercises 1 through 4

EXAMPLE 2



Using Right Triangle Diagrams to Rewrite Trig Expressions Find expressions for tan  and csc , given sec  

Solution

5–115





hyp , we draw a right triangle adj diagram as in Example 1, with a hypotenuse of 2u2  144 and a side u adjacent to angle . For hyp opp and csc   , we use the tan   opp hyp

2u2  144 . u

With sec  

u2  144 ␪

opp

u

539

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Connections to Calculus

Pythagorean theorem to find an expression for the opposite side: opp2  u2  1 2u2  1442 2 opp2  u2  u2  144 opp2  144 opp  12

Pythagorean theorem square radical subtract u 2 result (length must be positive)

With an opposite side of 12 units, the figure shows tan   csc  

2u2  144 . 12

12 and u

Now try Exercises 5 and 6



Converting from Rectangular Coordinates to Trigonometric (Polar) Form Using the equations x  r cos , y  r sin , and x2  y2  r2 from Section 5.2, we can rewrite functions of x given in rectangular form as equations of  in trigonometric (polar) form. In Chapter 9 we’ll see how this offers us certain advantages. Note that while algebraic equations are often written with y in terms of x, polar equations are written with r in terms of . EXAMPLE 3



Converting from Rectangular to Polar Form Rewrite the equation 2x  3y  6 in trigonometric form using the substitutions indicated and solving for r. Note the given equation is that of a line with x-intercept (3, 0) and y-intercept (0, 2).

Solution



Using x  r cos  and y  r sin  we proceed as follows: 2x  3y  6 21r cos 2  31r sin 2  6

given substitute r cos  for x, r sin  for y factor out r

r 32 cos   3 sin  4  6 r

6 2 cos   3 sin 

solve for r

This is the equation of the same line, but in trigonometric form. Now try Exercises 7 through 10



While it is somewhat simplistic (there are other subtleties involved), we can verify the equation obtained in Example 3 produces the same line as 2x  3y  6 by evaluating the equation at   0° to find a point on the positive x-axis,   90° to find a point of the positive y-axis, and   135° to find a point in QII. For   0°:

r

6 2 cos 0  3 sin 0

6 6  2112  3102 2 3



For   90°:

r

6 2 cos 90  3 sin 90

6 6  2102  3112 3 2



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Precalculus—

5-117

541

Connections to Calculus

For   135°:

r

y

6 2 cos 135  3 sin 135

10 8

6   12 12 12 2a b  3a b 2 2 2 6

6 4

6冪2

⫺10 ⫺8 ⫺6 ⫺4 ⫺2 ⫺2

12   6 12  8.5 12

⫺4



2

4

6

8 10

135⬚

⫺6 ⫺8

Using a distance r from the origin at each  given, we note that all three points are on the line 2x  3y  6, as shown in the figure. EXAMPLE 4

2

⫺10

Converting from Polar to Rectangular Form Rewrite the equation r  4 cos  in rectangular form using the relationships x  r cos , y  r sin , and/or x2  y2  r2. Identify the resulting equation.

Solution

x From x  r cos  we have cos   . Substituting into the given equation we have r



r  4 cos  x r  4a b r r2  4x x2  y2  4x 2 x  4x  y2  0 1x2  4x  42  y2  4 1x  22 2  y2  4

given x substitute for cos  r multiply by r substitute x 2  y 2 for r 2 set equal to 0 complete the square in x standard form

The result shows r  4 cos  is a trigonometric form for the equation of a circle with center at (2, 0) and radius r  2. Now try Exercises 11 through 14



Connections to Calculus Exercises Use the diagram given to find the remaining side of the triangle. Then write the six trig functions as functions of x.

1.

2. hyp

4

x⫺2

冪2x2 ⫹ 8

␪

␪ 冪x2 ⫹ 8x

adj

Use the equation given and a sketch of the corresponding right triangle diagram to write the remaining five trig functions as functions of x.

3. x  4 tan 

4. x  5 sec 

5. Find expressions for cot  and sec , given 2u2  169 csc   . u

6. Find expressions for sin  and cos , given 2 15 . cot   x

x

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Connections to Calculus

Use x  r cos , y  r sin , and x2  y2  r2 to rewrite the expressions in trigonometric form.

7. y  2 9. y  2x  3

1 8. y  x2 4

10. 1x  22 2  y2  4

Use x  r cos , y  r sin , and x2  y2  r2 to rewrite the expressions in rectangular form, then identify the equation as that of a line, circle, vertical parabola, or horizontal parabola.

11. r  5 sin 

13. r 13 cos   2 sin 2  6

12. r 

4 1  sin 

14. r 

4 2  2 cos 

y aHint: sin   .b r x aHint: cos   .b r

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Precalculus—

6 CHAPTER CONNECTIONS

Trigonometric Identities, Inverses, and Equations CHAPTER OUTLINE 6.1 Fundamental Identities and Families of Identities 544 6.2 Constructing and Verifying Identities 552 6.3 The Sum and Difference Identities 558

Have you ever noticed that people who arrive early at a movie tend to choose seats about halfway up the theater’s incline and in the middle of a row? More than likely, this is due to a phenomenon called the optimal viewing angle, or the angle formed by the viewer’s eyes and the top and bottom of the screen. Seats located in this area maximize the viewing angle, with the measure of the angle depending on factors such as the distance from the floor to the bottom of the screen, the height of the screen, the location of a seat, and the incline of the auditorium. Here, trigonometric functions and identities play an important role. This application appears as Exercise 59 of Section 6.2.

6.4 The Double-Angle, Half-Angle, and Product-to-Sum Identities 568 6.5 The Inverse Trig Functions and Their Applications 582 6.6 Solving Basic Trig Equations 599 6.7 General Trig Equations and Applications 610 Trigonometric equations, identities, and substitutions also play a vital role in a study of calculus, helping Connections to simplify complex expressions, or rewrite an expression in a form more suitable for the tools of calculus. 543 to Calculus These connections are explored in the Connections to Calculus feature following Chapter 6.

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6.1 Fundamental Identities and Families of Identities Learning Objectives In Section 6.1 you will learn how to:

A. Use fundamental identities to help understand and recognize identity “families”

In this section we begin laying the foundation necessary to work with identities successfully. The cornerstone of this effort is a healthy respect for the fundamental identities and vital role they play. Students are strongly encouraged to do more than memorize them—they should be internalized, meaning they must become a natural and instinctive part of your core mathematical knowledge.

A. Fundamental Identities and Identity Families

B. Verify other identities using the fundamental identities and basic algebra skills

C. Use fundamental identities to express a given trig function in terms of the other five

D. Use counterexamples and contradictions to show an equation is not an identity

WORTHY OF NOTE

An identity is an equation that is true for all elements in the domain. In trigonometry, some identities result directly from the way the trig functions are defined. For instance, y y 1 r 1  , and the identity sin   since sin   and csc   , immediately r r y csc  csc  follows. We call identities of this type fundamental identities. Successfully working with other identities will depend a great deal on your mastery of these fundamental types. For convenience, the definition of the trig functions are reviewed here, followed by the fundamental identities that result. Given point P(x, y) on the unit circle, and the central angle  associated with P, we have 2x2  y2  1 and y tan   ; x  0 cos   x sin   y x 1 1 x sec   ; x  0 csc   ; y  0 cot   ; y  0 x y y Fundamental Trigonometric Identities

The word fundamental itself means, “a basis or foundation supporting an essential structure or function” (Merriam Webster).

Reciprocal identities

1 csc  1 cos   sec  1 tan   cot  sin  

Ratio identities

sin  cos  sec  tan   csc  cos  cot   sin  tan  

Pythagorean identities

Identities due to symmetry

sin2  cos2  1

sin12  sin 

tan2  1  sec2

cos12  cos 

1  cot2  csc2

tan12  tan 

These identities seem to naturally separate themselves into the four groups or families listed, with each group having additional relationships that can be inferred from the definitions. For instance, since sin  is the reciprocal of csc , csc  must be the reciprocal of sin . Similar statements can be made regarding cos  and sec  as well as tan  and cot . Recognizing these additional “family members” enlarges the number of identities you can work with, and will help you use them more effectively. In particular, since they are reciprocals: sin  csc   1, cos  sec   1, tan  cot   1. See Exercises 7 and 8. EXAMPLE 1



Identifying Families of Identities Use algebra to write four additional identities that belong to the Pythagorean family.

544

6-2

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Section 6.1 Fundamental Identities and Families of Identities

Solution



545

Starting with sin2  cos2  1, sin2  cos2  1 • sin2  1  cos2 • sin   21  cos2 sin2  cos2  1 • cos2  1  sin2 • cos   21  sin2

original identity subtract cos2 take square root original identity subtract sin2 take square root

For the identities involving a radical, the choice of sign will depend on the quadrant of the terminal side. Now try Exercises 9 and 10

A. You’ve just learned how to use fundamental identities to help understand and recognize identity “families”



The four additional Pythagorean identities are marked with a “•” in Example 1. The fact that each of them represents an equality gives us more options when attempting to verify or prove more complex identities. For instance, since cos2  1  sin2, we can replace cos2 with 1  sin2, or replace 1  sin2 with cos2, any time they occur in an expression. Note there are many other members of this family, since similar steps can be performed on the other Pythagorean identities. In fact, each of the fundamental identities can be similarly rewritten and there are a variety of exercises at the end of this section for practice.

B. Verifying an Identity Using Algebra Note that we cannot prove an equation is an identity by repeatedly substituting input values and obtaining a true equation. This would be an infinite exercise and we might easily miss a value or even a range of values for which the equation is false. Instead we attempt to rewrite one side of the equation until we obtain a match with the other side, so there can be no doubt. As hinted at earlier, this is done using basic algebra skills combined with the fundamental identities and the substitution principle. For now we’ll focus on verifying identities by using algebra. In Section 6.2 we’ll introduce some guidelines and ideas that will help you verify a wider range of identities. EXAMPLE 2



Using Algebra to Help Verify an Identity Use the distributive property to verify that sin 1csc   sin 2  cos2 is an identity.

Solution



Use the distributive property to simplify the left-hand side. sin 1csc   sin 2  sin  csc   sin2  1  sin2  cos2

distribute substitute 1 for sin  csc  1  sin2  cos2

Since we were able to transform the left-hand side into a duplicate of the right, there can be no doubt the original equation is an identity. Now try Exercises 11 through 20



Often we must factor an expression, rather than multiply, to begin the verification process. EXAMPLE 3



Using Algebra to Help Verify an Identity Verify that 1  cot2x sec2x  cot2x is an identity.

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Solution



The left side is as simple as it gets. The terms on the right side have a common factor and we begin there. cot2x sec2x  cot2x  cot2x 1sec2x  12  cot2x tan2x  1cot x tan x2 2  12  1

factor out cot2x substitute tan2x for sec2x  1 power property of exponents cot x tan x  1

Now try Exercises 21 through 28



Examples 2 and 3 show you can begin the verification process on either the left or right side of the equation, whichever seems more convenient. Example 4 shows how the special products 1A  B21A  B2  A2  B2 and/or 1A  B2 2  A2  2AB  B2 can be used in the verification process. EXAMPLE 4



Using a Special Product to Help Verify an Identity Use a special product and fundamental identities to verify that 1sin x  cos x2 2  1  2 sin(x) cos x is an identity.

Solution



Begin by squaring the left-hand side, in hopes of using a Pythagorean identity. 1sin x  cos x2 2  sin2x  2 sin x cos x  cos2x  sin2x  cos2x  2 sin x cos x  1  2 sin x cos x

binomial square rewrite terms substitute 1 for sin2x  cos2x

At this point we appear to be off by a sign, but quickly recall that sine is on odd function and sin x  sin1x2. By writing 1  2 sin x cos x as 1  21sin x21cos x2, we can complete the verification:  1  21sin x21cos x2  1  2 sin( x) cos x ✓

rewrite expression to obtain sin x substitute sin1x2 for sin x

Now try Exercises 29 through 34 B. You’ve just learned how to verify other identities using the fundamental identities and basic algebra skills



Another common method used to verify identities is simplification by combining sin2u A C AD  BC  cos u, the right. For sec u  terms, using the model   cos u B D BD 1 sin2u  cos2u hand side immediately becomes , which gives  sec u. See cos u cos u Exercises 35 through 40.

C. Writing One Function in Terms of Another Any one of the six trigonometric functions can be written in terms of any of the other functions using fundamental identities. The process involved offers practice in working with identities, highlights how each function is related to the other, and has practical applications in verifying more complex identities. EXAMPLE 5



Writing One Trig Function in Terms of Another Write the function cos x in terms of the tangent function.

Solution



Begin by noting these functions share “common ground” via sec x, since 1 sec2x  1  tan2x and cos x  . Starting with sec2x, sec x sec2x  1  tan2x sec x  21  tan2x

Pythagorean identity square roots

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Section 6.1 Fundamental Identities and Families of Identities

We can now substitute 21  tan2x for sec x in cos x  cos x 

1 21  tan x 2

WORTHY OF NOTE

EXAMPLE 6

substitute  21  tan2x for sec x 

Example 5 also reminds us of a very important point — the sign we choose for the final answer is dependent on the terminal side of the angle. If the terminal side is in QI or QIV we chose the positive sign since cos x 7 0 in those quadrants. If the angle terminates in QII or QIII, the final answer is negative since cos x 6 0 in those quadrants. Similar to our work in Chapter 5, given the value of cot t and the quadrant of t, the fundamental identities enable us to find the value of the other five functions at t. In fact, this is generally true for any given trig function and real number or angle t. 

Using a Known Value and Quadrant Analysis to Find Other Function Values Given cot t 

Solution

1 . sec x

Now try Exercises 41 through 46

It is important to note the stipulation “valid where both are defined” does not preclude a difference in the domains of each function. The result of Example 5 is indeed an identity, even though the expressions have unequal domains.

547



9 with t in QIV, find the value of the other five functions at t. 40

40 follows immediately, since cotangent and tangent 9 are reciprocals. The value of sec t can be found using sec2t  1  tan2t. The function value tan t  

sec2t  1  tan2t 40 2  1  a b 9 81 1600   81 81 1681  81 41 sec t   9

Pythagorean identity substitute  square 

40 for tan t 9

40 81 , substitute for 1 9 81

combine terms

take square roots

41 . This automatically gives 9 9 40 (reciprocal identities), and we find sin t   using sin2t  1  cos2t cos t  41 41 sin t or the ratio identity tan t  (verify). cos t Since sec t is positive in QIV, we have sec t 

C. You’ve just learned how to use fundamental identities to express a given trig function in terms of the other five

Now try Exercises 47 through 55



D. Showing an Equation Is Not an Identity To show an equation is not an identity, we need only find a single value for which the functions involved are defined but the equation is false. This can often be done by trial and error, or even by inspection. To illustrate the process, we’ll use two common misconceptions that arise in working with identities. EXAMPLE 7



Showing an Equation is Not an Identity Show the equations given are not identities. a. sin12x2  2 sin x b. cos1  2  cos   cos 

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Solution



a. The assumption here seems to be that we can factor out the coefficient from the argument. By inspection we note the amplitude of sin(2x) is A  1, while the amplitude of 2 sin x is A  2. This means they cannot possibly be equal for all values of x, although they are equal for integer multiples of . Verify they are  not equivalent using x  or other standard values. 6 GRAPHICAL SUPPORT While not a definitive method of proof, a graphing calculator can be used to investigate whether an equation is an identity. Since the left and right members of the equation must 3␲  be equal for all values (where they are 2 defined), their graphs must be identical. Graphing the functions from Example 7(a) as Y1 and Y2 shows the equation s in12x2  2 s in x is definitely not an identity.

3

3␲ 2

3

b. The assumption here is that we can distribute function values. This is similar to saying 1x  4  1x  2, a statement obviously false for all values except x  0. Here we’ll substitute convenient values to prove the equation false, 3  namely,   and   . 4 4 cosa D. You’ve just learned how to use counterexamples and contradictions to show an equation is not an identity

3  3   b  cosa b  cosa b 4 4 4 4 12 12  cos    2 2 1  0

substitute

  for  and for  3 4

simplify result is false

Now try Exercises 56 through 62



6.1 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. Three fundamental ratio identities are ? ? ? , and cot   tan   , tan   . cos  csc  sin  2. When applying identities due to symmetry, and cos 1x2 cot x  sin 1x2 tan x  .

4. Using a calculator we find sec245°  and 3 tan 45°  1  . We also find sec2225°  and 3 tan 225°  1  . Is the equation sec2  3 tan   1 an identity? A C AD  BC   to add the B D BD following terms, and comment on this process versus “finding a common denominator:” cos x sin x .  sec x sin x

5. Use the pattern .

3. To show an equation is not an identity, we must find at least value(s) where both sides of the equation are defined, but which makes the equation .

6. Name at least four algebraic skills that are used with the fundamental identities in order to rewrite a trigonometric expression. Use algebra to quickly rewrite 1sin x  cos x2 2.

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Section 6.1 Fundamental Identities and Families of Identities

549

DEVELOPING YOUR SKILLS

Starting with the ratio identity given, use substitution and fundamental identities to write four new identities belonging to the ratio family. Answers may vary.

7. tan x 

sin x cos x

8. cot x 

cos x sin x

Starting with the Pythagorean identity given, use algebra to write four additional identities belonging to the Pythagorean family. Answers may vary.

9. 1  tan2x  sec2x

10. 1  cot2x  csc2x

Verify the equation is an identity using multiplication and fundamental identities.

11. sin x cot x  cos x

12. cos x tan x  sin x

13. sec2x cot2x  csc2x

14. csc2x tan2x  sec2x

30.

11  tan x2 2  sec x  2 sin x sec x

31. 11  sin x2 3 1  sin1x2 4  cos2x

32. 1sec x  12 3 sec1x2  14  tan2x 33. 34.

1csc x  cot x21csc x  cot x2  cot x tan x

1sec x  tan x21sec x  tan x2  sin x csc x

Verify the equation is an identity using fundamental A C AD  BC identities and   to combine terms. B D BD

35.

cos2x sin x   csc x sin x 1

16. tan x 1cot x  tan x2  sec2x

36.

tan2 sec   cos   sec  1

18. cot x 1tan x  cot x2  csc2x

37.

tan x sin x sin x  1   csc x cos x cot x

38.

cot x cos x cos x  1   sec x sin x tan x

39.

sec x sec x csc x csc x  tan x 40.   cot x  sec x cos x csc x sin x

15. cos x 1sec x  cos x2  sin2x 17. sin x 1csc x  sin x2  cos2x

19. tan x 1csc x  cot x2  sec x  1 20. cot x 1sec x  tan x2  csc x  1

Verify the equation is an identity using factoring and fundamental identities.

21. tan2x csc2x  tan2x  1 22. sin2x cot2x  sin2x  1

Write the given function entirely in terms of the second function indicated.

sin x cos x  sin x 23.  tan x cos x  cos2x

41. tan x in terms of sin x

42. tan x in terms of sec x

43. sec x in terms of cot x

44. sec x in terms of sin x

sin x cos x  cos x  cot x sin x  sin2x

45. cot x in terms of sin x

46. cot x in terms of csc x

24. 25.

1  sin x  sec x cos x  cos x sin x

For the function f 12 and the quadrant in which  terminates, state the value of the other five trig functions.

20 with  in QII 29

1  cos x  csc x 26. sin x  cos x sin x

47. cos  

sin x tan x  sin x  cos x 27. tan x  tan2x

48. sin  

12 with  in QII 37

cos x cot x  cos x  sin x 28. cot x  cot2x

49. tan  

15 with  in QIII 8

50. sec  

45 with  in QIV 27

51. cot  

x with  in QI 5

Verify the equation is an identity using special products and fundamental identities.

29.

1sin x  cos x2 2  sec x  2 sin x cos x

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52. csc  

7 with  in QII x

53. sin   

7 with  in QIII 13

  57. cosa b  cos   cosa  b 4 4 58. cos122  2 cos  59. tan122  2 tan 

23 54. cos   with  in QIV 25

tan   60. tana b  4 tan 4

9 55. sec    with  in QII 7

61. cos2  sin2  1 62. 2sin2x  9  sin x  3

Show that the following equations are not identities.

56. sina  

  b  sin   sina b 3 3

WORKING WITH FORMULAS

63. The illuminance of a point on a surface by a I cos  source of light: E  r2 The illuminance E (in lumens/m2) of a point on a horizontal surface is given by the formula shown, where I is the intensity of the light source in lumens, r is the distance in meters from the light source to the point, and  is the complement of the angle  (in degrees) made by the light source and the horizontal surface. Calculate the illuminance if I  800 lumens, and the flashlight is held so that the distance r is 2 m while the angle  is 40°.

64. The area of regular polygon: A  a

nx2 cos1 n 2 b 4 sin1 n 2

The area of a regular polygon is given by the formula shown, where n represents the number of sides and x is the length of each side. a. Rewrite the formula in terms of a single trig function. b. Verify the formula for a square with sides of 8 m. c. Find the area of a dodecagon (12 sides) with 10-in. sides.

Exercise 63 Intensity I

r

Illuminance E ␣ 

APPLICATIONS

Writing a given expression in an alternative form is an idea used at all levels of mathematics. In future classes, it is often helpful to decompose a power into smaller powers (as in writing A3 as A # A2) or to rewrite an expression using known identities so that it can be factored.

65. Show that cos3x can be written as cos x 11  sin2x2 .

68. Show that cot3x can be written as cot x1csc2x  12 . 69. Show tan2x sec x  4 tan2 x can be factored into 1sec x  421sec x  12 1sec x  12 . 70. Show 2 sin2x cos x  13 sin2x can be factored into 11  cos x2 11  cos x2 12 cos x  132 .

66. Show that tan3x can be written as tan x1sec2x  12 .

71. Show cos2x sin x  cos2x can be factored into 111  sin x211  sin x2 2.

67. Show that tan x  tan3x can be written as tan x(sec2x).

72. Show 2 cot2x csc x  212 cot2x can be factored into 21csc x  122 1csc x  12 1csc x  12 .

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Section 6.1 Fundamental Identities and Families of Identities

Many applications of fundamental identities involve geometric figures, as in Exercises 73 and 74.

73. Area of a polygon: The area of a regular polygon that has been circumscribed about a circle of radius sin1 n 2 r (see figure) is given by the formula A  nr2 , cos1 n 2 where n represents the number of sides. (a) Rewrite the formula r in terms of a single trig function; (b) verify the formula for a square circumscribed about a circle with radius 4 m; and (c) find the area of a dodecagon (12 sides) circumscribed about the same circle. 74. Perimeter of a polygon: The perimeter of a regular polygon circumscribed about a circle of sin1 n 2 radius r is given by the formula P  2nr , cos1 n 2 

where n represents the number of sides. (a) Rewrite the formula in terms of a single trig function; (b) verify the formula for a square circumscribed about a circle with radius 4 m; and (c) Find the perimeter of a dodecagon (12 sides) circumscribed about the same circle. 75. Angle of intersection: At their point of intersection, the angle  between any two nonparallel lines satisfies the relationship 1m2  m1 2cos   sin   m1m2sin , where m1 and m2 represent the slopes of the two lines. Rewrite the equation in terms of a single trig function. 76. Angle of intersection: Use the result of Exercise 2 75 to find the angle between the lines Y1  x  3 5 7 and Y2  x  1. 3 77. Angle of intersection: Use the result of Exercise 75 to find the angle between the lines Y1  3x  1 and Y2  2x  7.

EXTENDING THE CONCEPT

78. The word tangent literally means “to touch,” which in mathematics we take to mean touches in only and exactly one point. In the figure, the circle has a radius of 1 and the vertical line is 

551

y sin ␪ 1

tan ␪

␪ cos ␪

x

“tangent” to the circle at the x-axis. The figure can be used to verify the Pythagorean identity for sine and cosine, as well as the ratio identity for tangent. Discuss/Explain how. 79. Use factoring and fundamental identities to help find the x-intercepts of f in 3 0, 22 .

f 12  2 sin4  23 sin3  2 sin2  23 sin .

MAINTAINING YOUR SKILLS

80. (4.6) Solve for x: 2351 

2500 1  e1.015x

81. (5.6) Standing 265 ft from the base of the Strastosphere Tower in Las Vegas, Nevada, the angle of elevation to the top of the tower is about 77°. Approximate the height of the tower to the nearest foot.

82. (3.3) Use the rational zeroes theorem and other “tools” to find all zeroes of the function f 1x2  2x4  9x3  4x2  36x  16. 83. (5.3) Use a reference rectangle and the rule of fourths to sketch the graph of y  2 sin12t2 for t in [0, 2).

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Precalculus—

6.2 Constructing and Verifying Identities Learning Objectives

In Section 6.1, our primary goal was to illustrate how basic algebra skills are used to help rewrite trigonometric expressions. In this section, we’ll sharpen and refine these skills so they can be applied more generally, as we develop the ability to verify a much wider range of identities.

In Section 6.2 you will learn how to:

A. Create and verify a new identity

B. Verify general identities

A. Creating and Verifying Identities

In Example 2 of Section 6.1, we showed sin  1csc   sin 2  cos2 was an identity by transforming the left-hand side into cos2. There, the instructions were very specific: “Use the distributive property to . . .” When verifying identities, one of the biggest issues students face is that the directions are deliberately vague — because there is no single, fail-proof approach for verifying an identity. This sometimes leaves students feeling they don’t know where to start, or what to do first. To help overcome this discomfort, we’ll first create an identity by substituting fundamental identities into a given expression, then reverse these steps to get back the original expression. This return to the original illustrates the essence of verifying identities, namely, if two things are equal, one can be substituted for the other at any time. The process may seem arbitrary (actually—it is), and the steps could vary. But try to keep the underlying message in mind, rather than any specific steps. When working with identities, there is actually no right place to start, and the process begins by using the substitution principle to create an equivalent expression as you work toward the expression you’re trying to match. EXAMPLE 1



Creating and Verifying an Identity Starting with the expression csc x  cot x, use fundamental identities to rewrite the expression and create a new identity. Then verify the identity by reversing the steps.

Solution



csc x  cot x cos x 1   sin x sin x 1  cos x  sin x

original expression substitute reciprocal and ratio identities

write as a single term

1  cos x sin x

Resulting identity



csc x  cot x 

Verify identity



Working with the right-hand side, we reverse each step with a view toward the original expression. 1 cos x 1  cos x   sin x sin x sin x  csc x  cot x

rewrite as individual terms substitute reciprocal and ratio identities

Now try Exercises 7 through 9



In actual practice, all you’ll see is this instruction, “Verify the following is an identity: 1  cos x csc x  cot x  ,” and it will be up to you to employ the algebra and fundasin x mental identities needed.

552

6-10

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Section 6.2 Constructing and Verifying Identities

EXAMPLE 2



553

Creating and Verifying an Identity Starting with the expression 2 tan x sec x, use fundamental identities to rewrite the expression and create a new identity. Then verify the identity by reversing the steps.

Solution



2 tan x sec x sin x # 1 2# cos x cos x 2 sin x  cos2x 2 sin x  1  sin2x 2 sin x 1  sin2x

original expression substitute ratio and reciprocal identities

multiply substitute 1  sin2x for cos2x

Resulting identity



2 tan x sec x 

Verify identity



Working with the right-hand side, we reverse each step with a view toward the original expression.

A. You’ve just learned how to create and verify a new identity

identity

2 sin x 2 sin x  1  sin2x cos2x sin x # 1 2# cos x cos x  2 tan x sec x

substitute cos2x for 1  sin2x substitute cos x # cos x for cos2x substitute ratio and reciprocal identities

Now try Exercises 10 through 12



B. Verifying Identities We’re now ready to put these ideas, and the ideas from Section 6.1, to work for us. When verifying identities we attempt to mold, change, or rewrite one side of the equality until we obtain a match with the other side. What follows is a collection of the ideas and methods we’ve observed so far, which we’ll call the Guidelines for Verifying Identities. But remember, there really is no right place to start. Think things over for a moment, then attempt a substitution, simplification, or operation and see where it leads. If you hit a dead end, that’s okay! Just back up and try something else. Guidelines for Verifying Identities WORTHY OF NOTE When verifying identities, it is actually permissible to work on each side of the equality independently, in the effort to create a “match.” But properties of equality can never be used, since we cannot assume an equality exists.

1. As a general rule, work on only one side of the identity. • We cannot assume the equation is true, so properties of equality cannot be applied. • We verify the identity by changing the form of one side until we get a match with the other. 2. Work with the more complex side, as it is easier to reduce/simplify than to “build.” 3. If an expression contains more than one term, it is often helpful to combine A C AD  BC terms using   . B D BD 4. Converting all functions to sines and cosines can be helpful. 5. Apply other algebra skills as appropriate: distribute, factor, multiply by a conjugate, and so on. 6. Know the fundamental identities inside out, upside down, and backward — they are the key!

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Note how these ideas are employed in Examples 3 through 5, particularly the frequent use of fundamental identities. EXAMPLE 3



Verifying an Identity Verify the identity: sin2 tan2  tan2  sin2.

Solution



As a general rule, the side with the greater number of terms or the side with rational terms is considered “more complex,” so we begin with the right-hand side. tan2  sin2     

sin2  sin2 cos2 sin2 # 1 2 2  sin  1 cos  sin2 sec2  sin2 sin2 1sec2  12 sin2 tan2

substitute

sin2 cos2

for tan2

decompose rational term substitute sec2 for factor out sin2

1 cos2

substitute tan2 for sec2  1

Now try Exercises 13 through 18



Example 3 involved factoring out a common expression. Just as often, we’ll need to multiply numerators and denominators by a common expression, as in Example 4. EXAMPLE 4



Verifying an Identity by Multiplying Conjugates Verify the identity:

Solution



1  cos t cos t  . 1  sec t tan2t

Both sides of the identity have a single term and one is really no more complex than the other. As a matter of choice we begin with the left side. Noting the denominator on the left has the term sec t, with a corresponding term of tan2t to the right, we reason that multiplication by a conjugate might be productive. cos t 1  sec t cos t a ba b 1  sec t 1  sec t 1  sec t cos t  1  1  sec2t cos t  1  tan2t 1  cos t  tan2t

multiply above and below by the conjugate distribute: cos t sec t  1, 1A  B 2 1A  B2  A 2  B 2 substitute tan2t for 1  sec2t 11  tan2t  sec2t 1 1  sec2t  tan2t 2 multiply above and below by 1

Now try Exercises 19 through 22



Example 4 highlights the need to be very familiar with families of identities. To replace 1  sec2t, we had to use tan2t, not simply tan2t, since the related Pythagorean identity is 1  tan2t  sec2t. As noted in the Guidelines, combining rational terms is often helpful. At this point, A C AD  BC students are encouraged to work with the pattern   as a means of B D BD combing rational terms quickly and efficiently.

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Section 6.2 Constructing and Verifying Identities

EXAMPLE 5



Verifying an Identity by Combining Terms Verify the identity:

Solution



B. You’ve just learned how to verify general identities

sec x tan2x  cos2x sin x   . sec x sin x tan x

We begin with the left-hand side. sec2x  sin2x sin x sec x   sec x sin x sin x sec x 11  tan2x2  11  cos2x2  sin x 1 b a ba cos x 1 tan2x  cos2x  tan x

combine terms:

A C AD  BC   B D BD

substitute 1  tan2x for sec2x, 1 1  cos2x for sin2x, for sec x, cos x simplify numerator, substitute tan x for

sin x cos x

Now try Exercises 23 through 28



Identities come in an infinite variety and it would be impossible to illustrate all variations. Using the general ideas and skills presented should prepare you to verify any of those given in the exercise set, as well as those you encounter in your future studies. See Exercises 29 through 58.

6.2 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. If two expressions are equal, then one may be for the other at any time and the result will be equivalent.



4. Converting all terms to functions of may help verify an identity.

and

2. We verify an identity by changing the form of one side, working until we the other side.

5. Discuss/Explain why you must not add, subtract, multiply, or divide both sides of the equation when verifying identities.

3. To verify an identity, always begin with the more expression, since it is easier to than to .

6. Discuss/Explain the difference between operating on both sides of an equation (see Exercise 5) and working on each side independently.

DEVELOPING YOUR SKILLS

Using algebra and the fundamental identities, rewrite each given expression to create a new identity relationship. Then verify your identity by reversing the steps. Answers will vary.

7. sec x  tan x

8. 1cos x  sin x2 2

9. 11  sin2x2sec x 11.

sin x  sin x cos x sin2x

10. 2 cot x csc x

12. 1cos x  sin x2 1cos x  sin x2

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38.

cos x  sec x  sin2x sec x

39.

1  tan x sec x csc x  sin x

16. cot x cos x  csc x  sin x

40.

1  cot x csc x sec x  cos x

17.

cos x  csc x  sin x tan x

41.

1  sin x  1tan x  sec x2 2 1  sin x

18.

sin x  sec x  cos x cot x

42.

19.

cos   sec   tan  1  sin 

1  cos x  1csc x  cot x2 2 1  cos x

43.

20.

sin   csc   cot  1  cos 

cos x  sin x cos x  sin x  1  tan x 1  tan x

44.

21.

cos x 1  sin x  cos x 1  sin x

sin x  cos x 1  cot x  1  cot x sin x  cos x

45.

22.

sin x 1  cos x  sin x 1  cos x

tan2x  cot2x  csc x sec x tan x  cot x

46.

23.

cos x cot2x  sin2x csc x   cos x csc x cot x

cot x  tan x  sin x cos x cot2x  tan2x

47.

cot x  1  sin2x cot x  tan x

24.

1 1  csc2x sec2x 2  cos x sin2x

48.

tan x  1  cos2x cot x  tan x

25.

sin x sin x   2 tan2x 1  sin x 1  sin x

49.

sec4x  tan4x 1 sec2x  tan2x

26.

cos x cos x   2 cot2x 1  cos x 1  cos x

50.

27.

cot x cot x   2 sec x 1  csc x 1  csc x

csc4x  cot4x 1 csc2x  cot2x

51.

cos4x  sin4x  2  sec2x cos2x

28.

tan x tan x   2 csc x 1  sec x 1  sec x

52.

29.

sec2x  tan2x 1  cot2x

sin4x  cos4x  2  csc2x sin2x

Verify that the following equations are identities.

13. cos2x tan2x  1  cos2x 14. sin2x cot2x  1  sin2x 15. tan x  cot x  sec x csc x

30.

csc2x  cot2x 1  tan2x

31. sin x 1cot x  csc x2  sin x 2

2

2

2

53. 1sec x  tan x2 2 

1sin x  12 2 cos2x

1cos x  12 2

32. cos2x 1tan2x  sec2x2  cos2x

54. 1csc x  cot x2 2 

34. sin x tan x  cos x  sec x

55.

cos x csc x sec x  cos x sin x    cos x sec x sin x sin x

35.

sec x  sin x cot x  tan x

56.

cos x sec x csc x  sin x sin x    cos x csc x cos x sin x

36.

csc x  cos x cot x  tan x

57.

sin4x  cos4x sin x  cos x 3 3  1  sin x cos x sin x  cos x

37.

sin x  csc x  cos2x csc x

58.

sin4x  cos4x sin x  cos x 3 3  1  sin x cos x sin x  cos x

33. cos x cot x  sin x  csc x

sin2x

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Section 6.2 Constructing and Verifying Identities

557

WORKING WITH FORMULAS

59. Distance to top of movie screen: d 2  120  x cos 2 2  120  x sin 2 2 At a theater, the optimum viewing angle depends on a number of factors, like the height of the screen, the incline of the auditorium, the location of a seat, the height of your eyes while seated, and so on. One of the measures needed to find the “best” seat is the distance from your eyes to

60. The area of triangle ABC: A 

(not to scale)

d 20 ft

the top of the screen. For a theater with the dimensions shown, this distance is given by the formula here (x is the diagonal distance from the horizontal floor to your seat). (a) Show the formula is equivalent to 800  40x 1cos   sin 2  x2. (b) Find the distance d if   18° and you are sitting in the eighth row with the rows spaced 3 ft apart.



3 ft

c2 sin A sin B 2 sin C

If one side and three angles of a triangle are known, its area can be computed using this formula, where side c is opposite angle C. Find the area of the triangle shown in the diagram. C

D

75

x 3 ft



20 ft



A

45 20 cm

60 B

APPLICATIONS

61. Pythagorean theorem: For the triangle shown, (a) find an expression for the length of the hypotenuse in terms of h tan x and cot x, then √cot x determine the length of √tan x the hypotenuse when x  1.5 rad; (b) show the expression you found in part (a) is equivalent to h  1csc x sec x and recompute the length of the hypotenuse using this expression. Did the answers match? 62. Pythagorean theorem: For the triangle shown, (a) find an expression for the area of the triangle in terms of cot x and cos Exercise 62 x, then determine its  cos x area given x  ; 6 (b) show the expression cot x you found in part (a) is equivalent to 1 A  1csc x  sin x2 and recompute the area 2 using this expression. Did the answers match?

63. Viewing distance: Referring to Exercise 59, find a formula for D—the distance from this patron’s eyes to the bottom of the movie screen. Simplify the result using a Pythagorean identity, then find the value of D. 64. Viewing angle: Referring to Exercises 59 and 63, once d and D are known, the viewing angle  (the angle subtended by the movie screen and the viewer’s eyes) can be found using the formula d2  D2  202 cos   . Find the value of cos  2dD for this particular theater, person, and seat. 65. Intensity of light: In a study of the luminous intensity of light, the expression I1cos  sin   can occur. 21I1cos 2 2  1I2sin 2 2 Simplify the equation for the moment I1  I2. 66. Intensity of light: Referring to Exercise 65, find the angle  given I1  I2 and   60°.

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EXTENDING THE CONCEPT

67. Just as the points P(x, y) on the unit circle x2  y2  1 are used to name the circular trigonometric functions, the points P(x, y) on the unit hyperbola x2  y2  1 are used to name what are called the hyperbolic trigonometric functions. The hyperbolic functions are used extensively in many of the applied sciences. The identities for these functions have many similarities to those for the circular functions, but also have some significant



6-16

CHAPTER 6 Trigonometric Identities, Inverses, and Equations

differences. Using the Internet or the resources of a library, do some research on the functions sinh t, cosh t, and tanh t, where t is any real number. In particular, see how the Pythagorean identities compare/contrast between the two forms of trigonometry. 68. Verify the identity

sin6x  cos6x  1  sin2x cos2x. sin4x  cos4x

69. Use factoring to show the equation is an identity: sin4x  2 sin2x cos2x  cos4x  1.

MAINTAINING YOUR SKILLS

70. (3.5) Graph the rational function given. x1 h1x2  2 x 4 27 3 , b is a point on the unit 4 4 circle, then state the values of sin t, cos t, and tan t associated with this point.

71. (5.2) Verify that a

72. (5.7) Use an appropriate trig ratio to find the length of the bridge needed to cross the lake shown in the figure.

Exercise 72 400 yd 62

d

73. (2.5) Graph using transformations of a basic function: f 1x2  2x  3  6

6.3 The Sum and Difference Identities Learning Objectives In Section 6.3 you will learn how to:

A. Develop and use sum and difference identities for cosine

B. Use the cofunction identities to develop the sum and difference identities for sine and tangent

C. Use the sum and difference identities to verify other identities

Figure 6.1

The sum and difference formulas for sine and cosine have a long and ancient history. Originally developed to help study the motion of celestial bodies, they were used centuries later to develop more complex concepts, such as the derivatives of the trig functions, complex number theory, and the study wave motion in different mediums. These identities are also used to find exact results (in radical form) for many nonstandard angles, a result of great importance to the ancient astronomers and still of notable mathematical significance today.

A. The Sum and Difference Identities for Cosine On a unit circle with center C, consider the point A on the terminal side of angle , and point B on the terminal side of angle , as shown in Figure 6.1. Since r  1, the coordinates of A and B are 1cos , sin 2 and 1cos , sin 2, respectively. Using the distance formula, we find that AB is equal to AB  21cos   cos 2 2  1sin   sin 2 2

 2cos   2 cos  cos   cos   sin   2 sin  sin   sin  2

A (cos , sin ) 1

 C





(cos , sin ) B

2

2

2

 21cos2  sin22  1cos2  sin22  2 cos  cos   2 sin  sin   22  2 cos  cos   2 sin  sin 

binomial squares regroup

cos2u  sin2u  1

With no loss of generality, we can rotate sector ACB clockwise, until side CB coincides with the x-axis. This creates new coordinates of (1, 0) for B, and new coordinates of 1cos1  2, sin1  2 2 for A, but the distance AB remains unchanged! (see Figure 6.2). Recomputing the distance gives

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Section 6.3 The Sum and Difference Identities

559

AB  23cos1  2  1 4 2  3sin1  2  0 4 2

Figure 6.2

 2cos2 1  2  2 cos1  2  1  sin2 1  2

 23 cos2 1  2  sin2 1  2 4  2 cos1  2  1 (cos(  ), sin(  )) A 1 C



(1, 0)

 12  2 cos1  2

Since both expressions represent the same distance, we can set them equal to each other and solve for cos1  2. 12  2 cos1  2  12  2 cos  cos   2 sin  sin 

B

2  2 cos1  2  2  2 cos  cos   2 sin  sin  2 cos1  2  2 cos  cos   2 sin  sin  cos1  2  cos  cos   sin  sin 

AB  AB property of radicals subtract 2 divide both sides by 2

The result is called the difference identity for cosine. The sum identity for cosine follows immediately, by substituting  for . cos1  2  cos  cos   sin  sin 

cos1  3 4 2  cos  cos12  sin  sin12 cos1  2  cos  cos   sin  sin 

difference identity substitute  for  cos 12  cos ; sin 12  sin 

The sum and difference identities can be used to find exact values for the trig functions of certain angles (values written in nondecimal form using radicals), simplify expressions, and to establish additional identities. EXAMPLE 1



Finding Exact Values for Non-Standard Angles Use the sum and difference identities for cosine to find exact values for a. cos 15°  cos145°  30°2 b. cos 75°  cos145°  30°2 Check results on a calculator.

Solution WORTHY OF NOTE Be aware that cos160°  30°2  cos 60°  cos 30° 1 13 a0   b and in general 2 2 f 1a  b2  f 1a2  f 1b2.



Each involves a direct application of the related identity, and uses special values. a. difference identity cos1  2  cos  cos   sin  sin  cos145°  30°2  cos 45° cos 30°  sin 45° sin 30°   45°,   30° 12 13 12 1 a ba ba ba b standard values 2 2 2 2 16  12 combine terms cos 15°  4 b.

To 10 decimal places, cos 15°  0.9659258263. cos1  2  cos  cos   sin  sin  cos145°  30°2  cos 45° cos 30°  sin 45° sin 30° 13 12 1 12 ba ba ba b a 2 2 2 2 16  12 cos 75°  4

sum identity   45°,   30° standard values

combine terms

To 10 decimal places, cos 75°  0.2588190451. Now try Exercises 7 through 12



These identities are listed here using the “” and “ ” notation to avoid needless repetition. In their application, use both upper symbols or both lower symbols depending on whether you’re evaluating the cosine of a sum or difference of two angles. As with the other identities, these can be rewritten to form other members of the identity family, as when they are used to consolidate a larger expression. This is shown in Example 2.

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The Sum and Difference Identities for Cosine cosine family: cos1  2  cos  cos   sin  sin 

functions repeat, signs alternate

cos  cos   sin  sin   cos1  2

EXAMPLE 2



can be used to expand or contract

Using a Sum/Difference Identity to Simplify an Expression Write as a single expression in cosine and evaluate: cos 57° cos 78°  sin 57° sin 78°

Solution



Since the functions repeat and are expressed as a difference, we use the sum identity for cosine to rewrite the difference as a single expression. cos  cos   sin  sin   cos1  2 cos 57° cos 78°  sin 57° sin 78°  cos157°  78°2 The expression is equal to cos 135°  

sum identity for cosine   57°,   78°

12 . 2 Now try Exercises 13 through 16



The sum and difference identities can be used to evaluate the cosine of the sum of two angles, even when they are not adjacent, or even expressed in terms of cosine.

EXAMPLE 3



Computing the Cosine of a Sum 5 Given sin   13 with the terminal side in QI, and tan   24 7 with the terminal side in QII. Compute the value of cos1  2.

Solution



Figure 6.3

To use the sum formula we need the value of cos , sin , cos , and sin . Using the given information about the quadrants along with the Pythagorean theorem, we draw the triangles shown in Figures 6.3 and 6.4, yielding the values that follow.

y

cos  

13 ␣ 12 5  2

5

Using cos1  2  cos  cos   sin  sin  gives this result:

x

122



132

cos1  2  a

12 7 5 24 b a b  a b a b 13 25 13 25 84 120   325 325 204  325

Figure 6.4 y

25 24 ␤ 27

Now try Exercises 17 and 18

x

(27)2



242

5 7 24 12 1QI2, sin   1QI2, cos    1QII2, and sin   1QII2 13 13 25 25





252

B. The Sum and Difference Identities for Sine and Tangent A. You’ve just learned how to develop and use sum and difference identities for cosine

The cofunction identities were actually introduced in Section 5.1, using the comple mentary angles in a right triangle. In this section we’ll verify that cosa  b  sin  2

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Section 6.3 The Sum and Difference Identities

WORTHY OF NOTE

and sina

It is worth pointing out that in Example 3, if we approximate the values of  and  using tables or a calculator, we find   22.62° and   106.26°. Sure enough, cos122.62°  106.26°2 204 325 !

obtain

561

  b  cos . For the first, we use the difference identity for cosine to 2

cosa

    b  cos cos   sin sin  2 2 2  102cos   112sin   sin 

For the second, we use cosa

   b  sin , and replace  with the real number  t. 2 2

This gives cosa cosa

  b  sin  2

cofunction identity for cosine

    c  t d b  sina  tb 2 2 2 cos t  sina

  tb 2

replace  with result, note c

 t 2

   a  tb d  t 2 2

  tb  cos t for any 2 real number t. Both identities can be written in terms of the real number t. See Exercises 19 through 24. This establishes the cofunction relationship for sine: sina

The Cofunction Identities cosa

  tb  sin t 2

sina

  tb  cos t 2

The sum and difference identities for sine can easily be developed using cofunction  identities. Since sin t  cosa  tb, we need only rename t as the sum 1  2 or 2 the difference 1  2 and work from there.   tb 2

cofunction identity

  1  2 d 2

substitute 1  2 for t

sin t  cosa sin1  2  cos c

 cos c a  cos a

  b   d 2

   b cos   sin a  b sin  2 2

sin 1  2  sin  cos   cos  sin 

regroup argument apply difference identity for cosine result

The difference identity for sine is likewise developed. The sum and difference identities for tangent can be derived using ratio identities and their derivation is left as an exercise (see Exercise 78).

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CHAPTER 6 Trigonometric Identities, Inverses, and Equations

The Sum and Difference Identities for Sine and Tangent sine family: sin1  2  sin  cos   cos  sin  sin  cos   cos  sin   sin1  2 tan   tan  1 tan  tan  tan   tan   tan1  2 1  tan  tan 

tangent family: tan1  2 

functions alternate, signs repeat can be used to expand or contract

signs match original in numerator signs alternate in denominator can be used to expand or contract

EXAMPLE 4A



Simplifying Expressions Using Sum/Difference Identities

Solution



Since the functions in each term alternate and the expression is written as a sum, we use the sum identity for sine:

Write as a single expression in sine: sin 12t2 cos t  cos12t2 sin t. sin  cos   cos  sin   sin 1  2

sin12t2cos t  cos 12t2 sin t  sin 12t  t2

sum identity for sine substitute 2t for  and t for 

The expression is equal to sin(3t).

EXAMPLE 4B



Simplifying Expressions Using Sum/Difference Identities Use the sum or difference identity for tangent to find the exact value of tan

Solution



11 must be the sum or difference of two 12 2  11 standard angles. A casual inspection reveals   . This gives 12 3 4 Since an exact value is requested,

tan   tan  1  tan  tan   2 tan a b  tana b 2 3 4  tana  b 3 4 2  1  tana b tana b 3 4  13  1  1  1 132112 1  13  1  13 tan1  2 

B. You’ve just learned how to use the cofunction identities to develop the sum and difference identities for sine and tangent

11 . 12

sum identity for tangent



2  , 3 4

tan a

2  b   13, tan a b  1 3 4

simplify expression

Now try Exercises 25 through 54



C. Verifying Other Identities Once the sum and difference identities are established, we can simply add these to the tools we use to verify other identities.

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Section 6.3 The Sum and Difference Identities

EXAMPLE 5



Verifying an Identity Verify that tana 

Solution



 tan   1 is an identity. b 4 tan   1

Using a direct application of the difference formula for tangent we obtain   4 tana  b   4 1  tan  tan 4 tan   1 tan   1   1  tan  tan   1 tan   tan

  ,  

 4

 tan a b  1 4

Now try Exercises 55 through 60

EXAMPLE 6



563



Verifying an Identity Verify that sin1  2sin1  2  sin2  sin2 is an identity.

Solution



Using the sum and difference formulas for sine we obtain

sin1  2sin1  2  1sin  cos   cos  sin 2 1sin  cos   cos  sin 2  sin2 cos2  cos2 sin2

C. You’ve just learned how to use the sum and difference identities to verify other identities

1A  B21A  B2  A2  B2

 sin  11  sin 2  11  sin 2 sin 

use cos2x  1  sin2x to write the expression solely in terms of sine

 sin2  sin2 sin2  sin2  sin2 sin2

distribute

 sin   sin 

simplify

2

2

2

2

2

2

Now try Exercises 61 through 68



6.3 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. Since tan 45°  tan 60° 7 1, we know tan 45°  tan 60°  tan 105° is since tan  6 0 in . 2. To find an exact value for tan 105°, use the sum identity for tangent with a  and b  . 3. For the cosine sum/difference identities, the functions in each term, with the sign between them. 4. For the sine sum/difference identities, the functions in each term, with the sign between them.

5. Discuss/Explain how we know the exact value for 11 2  cos  cosa  b will be negative, prior 12 3 4 to applying any identity. sin1  2 cos1  2 is an identity, even though the arguments of cosine have been reversed. Then verify the identity.

6. Discuss/Explain why tan1  2 

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DEVELOPING YOUR SKILLS

Find the exact value of the expression given using a sum or difference identity. Some simplifications may involve using symmetry and the formulas for negatives.

7. cos 105°

8. cos 135°

9. cos a

10. cos a

7 b 12

5 b 12

12. a. cosa

b. cos a

   b 4 3

Rewrite as a single expression in cosine.

13. cos172 cos122  sin172 sin122

15. cos 183° cos 153°  sin 183° sin 153°

cot  

4 17. For sin    with terminal side in QIV and 5 5 tan    with terminal side in QII, find 12 cos1  2. 112 with terminal side in QII and 113 89 sec    with terminal side in QII, find 39 cos1  2.

18. For sin  

Use a cofunction identity to write an equivalent expression.

 b 10

23. sin a

  b 6

5 21. tan a b 12 24. cos a

Rewrite as a single expression.

25. sin13x2 cos15x2  cos13x2 sin15x2 x x x x 26. sina b cosa b  cosa b sina b 2 3 2 3

4 11 b  tan a b 21 21 31. 11 4 1  tan a b tan a b 21 21 tan a

33. For cos   

5 7 5 7 16. cos a b cosa b  sina b sina b 36 36 36 36

22. sec a

5 11 5 11 b cosa b  cos a b sina b 24 24 24 24

tan a

Find the exact value of the given expressions.

20. sin 18°

30. sin a

 3 b  tan a b 20 10 32. 3  1  tan a b tana b 20 10

    14. cosa b cosa b  sina b sina b 3 6 3 6

19. cos 57°

x x tana b  tana b 2 8 28. x x 1  tana b tana b 2 8

29. sin 137° cos 47°  cos 137° sin 47°

b. cos1120°  45°2

   b 6 4

tan152  tan122 1  tan152 tan122

Find the exact value of the given expressions.

Use sum/difference identities to verify that both expressions give the same result.

11. a. cos145°  30°2

27.

  b 3

7 with terminal side in QII and 25

15 with terminal side in QIII, find 8

a. sin1  2

b. tan1  2

29 with terminal side in QI and 20 12 cos    with terminal side in QII, find 37 a. sin1  2 b. tan1  2

34. For csc  

Find the exact value of the expression given using a sum or difference identity. Some simplifications may involve using symmetry and the formulas for negatives.

35. sin 105°

36. sin 175°2

37. sin a

38. sin a

5 b 12

11 b 12

39. tan 150°

40. tan 75°

41. tan a

42. tan a

2 b 3

 b 12

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Section 6.3 The Sum and Difference Identities

53. For the figure indicated, show that      and compute the following:

Use sum/difference identities to verify that both expressions give the same result.

43. a. sin 145°30°2

   b 3 4   b. sin a  b 4 6

44. a. sin a

b. sin 1135°  120°2

a. sin 

19 19 5 b given 2   . See 12 12 12 Exercises 10 and 37.

a. sin 

c. tan1  2

51. Use the diagram indicated to compute the following: b. cos A

c. tan A Exercise 52

Exercise 51

12 



55. sin1  2  sin 

c. tan1  2

b. cos1  2

5

Verify each identity.

60 50. Given  and  are obtuse angles with tan    11 35 and sin   , find 37

a. sin A

6

c. tan1  2

b. cos1  2

c. tan 

Exercise 54



28 49. Given  and  are obtuse angles with sin   53 13 and cos    , find 85

a. sin1  2

b. cos 

8

8 48. Given  and  are acute angles with cos   17 25 and sec   , find 7

a. sin1  2



54. For the figure indicated, show that      and compute the following:

c. tan1  2

b. cos1  2

28 



12 47. Given  and  are acute angles with sin   13 35 and tan   , find 12

a. sin1  2

45

24

32

46. Find cos a

b. cos1  2

c. tan 

Exercise 53

45. Find sin 255° given 150°  105°  255°. See Exercises 7 and 35.

a. sin1  2

b. cos 

56. cos1  2  cos  57. cos ax 

22  b 1cos x  sin x2 4 2

58. sin ax 

12  b 1sin x  cos x2 4 2

59. tanax 

1  tan x  b 4 1  tan x

60. tanax 

tan x  1  b 4 tan x  1

61. cos1  2  cos1  2  2 cos  cos  62. sin1  2  sin1  2  2 sin  sin  15

30 

A

5

 

64. sin12t2  2 sin t cos t

45

8

15

12

52. Use the diagram indicated to compute the following: a. sin 

b. cos 

63. cos12t2  cos2t  sin2t

c. tan 

65. sin13t2  4 sin3t  3 sin t 66. cos13t2  4 cos3t  3 cos t 67. Use a difference identity to show  12 cos ax  b  1cos x  sin x2. 4 2 68. Use sum/difference identities to show   sinax  b  sinax  b  12 sin x. 4 4

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WORKING WITH FORMULAS

69. Force required to maintain equilibrium using a Wk tan1p  2 screw jack: F  c The force required to maintain equilibrium when a screw jack is used can be modeled by the formula shown, where p is the pitch angle of the screw, W is the weight of the load,  is the angle of friction, with k and c being constants related to a particular jack. Simplify the formula using the



6-24

CHAPTER 6 Trigonometric Identities, Inverses, and Equations

 difference formula for tangent given p  6  and   . 4 70. Brewster’s law: reflection and refraction of n2 unpolarized light: tan p  n1 Brewster’s law of optics states that when unpolarized light strikes a dielectric surface, the transmitted light rays and the reflected light rays are perpendicular to each other. The proof of Brewster’s law involves the expression  n1sin p  n2 sin a  p b. Use the 2 difference identity for sine to verify that this expression leads to Brewster’s law.

APPLICATIONS

71. AC circuits: In a study of AC circuits, the equation cos s cos t sometimes arises. Use a sum R C sin1s  t2 identity and algebra to show this equation is 1 equivalent to R  . C1tan s  tan t2 72. Fluid mechanics: In studies of fluid mechanics, the equation 1V1sin   2V2sin1  2 sometimes arises. Use a difference identity to show that if 1V1  2V2, the equation is equivalent to cos   cot  sin   1. 73. Art and mathematics: When working in twopoint geometric perspective, artists must scale their work to fit on the paper or canvas they are using. In A tan  doing so, the equation  arises. B tan190°  2 Rewrite the expression on the right in terms of sine and cosine, then use the difference identities to A show the equation can be rewritten as  tan2. B

75. Pressure on the eardrum: If a frequency generator is placed a certain distance from the ear, the pressure on the eardrum can be modeled by the function P1 1t2  A sin12ft2, where f is the frequency and t is the time in seconds. If a second frequency generator with identical settings is placed slightly closer to the ear, its pressure on the eardrum could be represented by P2 1t2  A sin12ft  C2 , where  C is a constant. Show that if C  , the total 2 pressure on the eardrum 3P1 1t2  P2 1t2 4 is P1t2  A3 sin12ft2  cos12ft2 4 . 76. Angle between two cables: Two cables used to steady a radio tower are attached to the tower at heights of 5 ft and 35 ft, with both secured to a stake 12 ft from the tower (see figure). Find the value of cos , where  is the angle between the upper and lower cables. Exercise 76

74. Traveling waves: If two waves of the same frequency, velocity, and amplitude are traveling along a string in opposite directions, they can be represented by the equations Y1  A sin1kx  t2 and Y2  A sin1kx  t2 . Use the sum and difference formulas for sine to show the result YR  Y1  Y2 of these waves can be expressed as YR  2A sin1kx2cos1t2.

35 ft

5 ft

 12 ft

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77. Difference quotient: Given f1x2  sin x, show that the difference quotient results in the expression sin h cos h  1  cos x a b. sin x h h

78. Difference identity: Derive the difference identity sin1  2 for tangent using tan1  2  . cos1  2 (Hint: After applying the difference identities, divide the numerator and denominator by cos  cos .)

EXTENDING THE CONCEPT

A family of identities called the angle reduction formulas, will be of use in our study of complex numbers and other areas. These formulas use the period of a function to reduce large angles to an angle in [0, 360°) or [0, 2) having an equivalent function value: (1) cos1t  2k2  cos t; (2) sin1t  2k2  sin t . Use the reduction formulas to find values for the following functions (note the formulas can also be expressed in degrees).

(cos(  ), sin(  )) (cos , sin ) D  

d 

(1, 0)

81. sin a

82. sin 2385° Exercise 84

83. An alternative method of proving the difference formula for cosine uses a unit circle and the fact that equal arcs are subtended by equal chords (D  d in the diagram). Using a combination of algebra, the distance formula, and a Pythagorean identity, show that cos1  2  cos  cos   sin  sin  (start by computing D2 and d2). Then discuss/explain how the sum identity can be found using the fact that   12. 84. A proof without words: Verify the Pythagorean theorem for each right triangle in the diagram, then discuss/explain how the diagram offers a proof of the sum identities for sine and cosine. Be detailed and thorough.

A

1

sin B

sin B sin A cos B B

A

cos B sin A

80. cos a



(cos , sin )

91 b 6

79. cos 1665° 41 b 6

Exercise 83

sin B cos A



567

Section 6.3 The Sum and Difference Identities

cos B cos A

MAINTAINING YOUR SKILLS

85. (5.3/5.4) State the period of the functions given:   a. y  3 sin a x  b 8 3  b. y  4 tan a2x  b 4

87. (5.2) Clarence the Clown is about to be shot from a circus cannon to a safety net on the other side of the main tent. If the cannon is 30 ft long and must be aimed at 40° for Clarence to hit the net, the end of the cannon must be how high from ground level?

86. (2.7) Graph the piecewise-defined function given:

88. (2.3) Find the equation of the line parallel to 2x  5y  10, containing the point (5, 2). Write your answer in standard form.

3 f 1x2  • x2 x

x 6 1 1  x  1 x 7 1

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Precalculus—

6.4 The Double-Angle, Half-Angle, and Product-to-Sum Identities Learning Objectives In Section 6.4 you will learn how to:

A. Derive and use the double-angle identities for cosine, tangent, and sine

B. Develop and use the power reduction and half-angle identities

C. Derive and use the product-to-sum and sum-to-product identities

D. Solve applications using these identities

The derivation of the sum and difference identities in Section 6.3 was a “watershed event” in the study of identities. By making various substitutions, they lead us very naturally to many new identity families, giving us a heightened ability to simplify expressions, solve equations, find exact values, and model real-world phenomena. In fact, many of the identities are applied in very practical ways, as in a study of projectile motion and the conic sections (Chapter 10). In addition, one of the most profound principles discovered in the eighteenth and nineteenth centuries was that electricity, light, and sound could all be studied using sinusoidal waves. These waves often interact with each other, creating the phenomena known as reflection, diffraction, superposition, interference, standing waves, and others. The product-to-sum and sum-to-product identities play a fundamental role in the investigation and study of these phenomena.

A. The Double-Angle Identities The double-angle identities for sine, cosine, and tangent can all be derived using the related sum identities with two equal angles 1  2. We’ll illustrate the process here for the cosine of twice an angle. cos1  2  cos  cos   sin  sin 

sum identity for cosine

cos1  2  cos  cos   sin  sin 

assume    and substitute  for 

cos122  cos   sin  2

2

simplify—double-angle identity for cosine

Using the Pythagorean identity cos   sin2  1, we can easily find two additional members of this family, which are often quite useful. For cos2  1  sin2 we have 2

cos122  cos2  sin2

 11  sin 2  sin  2

2

cos122  1  2 sin  2

double-angle identity for cosine substitute 1  sin2 for cos2 double-angle in terms of sine

Using sin   1  cos  we obtain an additional form: 2

2

cos122  cos2  sin2

 cos   11  cos 2 2

2

cos122  2 cos2  1

double-angle identity for cosine substitute 1  cos2 for sin2 double-angle in terms of cosine

The derivations of sin122 and tan122 are likewise developed and are asked for in Exercise 103. The double-angle identities are collected here for your convenience. The Double-Angle Identities cosine:

cos122  cos2  sin2

sine: sin122  2 sin  cos 

 1  2 sin  2

 2 cos2  1 tangent:

568

tan122 

2 tan  1  tan2

6-26

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Section 6.4 The Double-Angle, Half-Angle, and Product-to-Sum Identities

EXAMPLE 1



Using a Double-Angle Identity to Find Function Values 5 Given sin   , find the value of cos122. 8

Solution



Using the double-angle identity for cosine interms of sine, we find cos122  1  2 sin2 5 2  1  2a b 8 25 1 32 7  32

double-angle in terms of sine substitute

5 for sin  8

5 2 25 2a b  8 32 result

5 7 If sin   , then cos122  . 8 32 Now try Exercises 7 through 20



Like the fundamental identities, the double-angle identities can be used to verify or develop others. In Example 2, we explore one of many multiple-angle identities, verifying that cos132 can be rewritten as 4 cos3  3 cos  (in terms of powers of cos ).

EXAMPLE 2



Verifying a Multiple Angle Identity Verify that cos132  4 cos3  3 cos  is an identity.

Solution



Use the sum identity for cosine, with   2 and   . Note that our goal is an expression using cosines only, with no multiple angles. cos1  2  cos  cos   sin  sin  cos12  2  cos122 cos   sin122sin  cos132  12 cos2  12 cos   12 sin  cos 2 sin   2 cos3  cos   2 cos  sin2  2 cos3  cos   2 cos 11  cos22  2 cos3  cos   2 cos   2 cos3  4 cos3  3 cos 

sum identity for cosine substitute 2 for  and  for  substitute for cos122 and sin122 multiply substitute 1  cos2 for sin2 multiply combine terms

Now try Exercises 21 and 22

EXAMPLE 3





Using a Double-Angle Formula to Find Exact Values Find the exact value of sin 22.5° cos 22.5°.

Solution



A product of sines and cosines having the same argument hints at the double-angle identity for sine. Using sin122  2 sin  cos  and dividing by 2 gives

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CHAPTER 6 Trigonometric Identities, Inverses, and Equations

sin122 2 sin12 322.5° 4 2 sin 22.5° cos 22.5°  2 sin 45°  2 12 2 12   2 4 sin  cos  

A. You’ve just learned how to derive and use the doubleangle identities for cosine, tangent, and sine

double-angle identity for sine

replace  with 22.5°

multiply

sin 45° 

12 2

Now try Exercises 23 through 30



B. The Power Reduction and Half-Angle Identities Expressions having a trigonometric function raised to a power occur quite frequently in various applications. We can rewrite even powers of these trig functions in terms of an expression containing only cosine to the power 1, using what are called the power reduction identities. This makes the expression easier to use and evaluate. It can legitimately be argued that the power reduction identities are actually members of the double-angle family, as all three are a direct consequence. To find identities for cos2x and sin2x, we solve the related double-angle identity involving cos (2x). 1  2 sin2  cos122

cos122 in terms of sine

2 sin   cos122  1 2

sin2 

1  cos122 2

subtract 1, then divide by 2 power reduction identity for sine

Using the same approach for cos2 gives cos2 

1  cos122 . The identity for 2

2 tan  (see Exercise 104), but in this case 1  tan2 1  cos122 sin2u it’s easier to use the identity tan2u  . 2 . The result is 1  cos122 cos u tan2 can be derived from tan122 

The Power Reduction Identities cos2 

EXAMPLE 4



1  cos122 2

sin2 

1  cos122 2

tan2 

1  cos122 1  cos122

Using a Power Reduction Formula Write 8 sin4x in terms of an expression containing only cosines to the power 1.

Solution



8 sin4x  81sin2x2 2 1  cos12x2 2  8c d 2  2 31  2 cos12x2  cos2 12x2 4 1  cos14x2  2 c 1  2 cos12x2  d 2  2  4 cos12x2  1  cos14x2  3  4 cos12x2  cos14x2

original expression substitute

1  cos12x2 2

for sin2x

multiply substitute

1  cos14x2 2

for cos2(2x)

multiply result

Now try Exercises 31 through 36



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Section 6.4 The Double-Angle, Half-Angle, and Product-to-Sum Identities

571

The half-angle identities follow directly from those above, using algebra and a 1  cos122 , we first take square roots and simple change of variable. For cos2  2 u 1  cos122 . Using the substitution u  2 gives   , obtain cos    A 2 2 and making these substitutions results in the half-angle identity for cosine: u 1  cos u cosa b   , where the radical’s sign depends on the quadrant in which 2 A 2 u u 1  cos u , terminates. Using the same substitution for sine gives sin a b   2 2 A 2 u u 1  cos u . In the case of tan a b, we can and for the tangent identity, tan a b   2 A 1  cos u 2 actually develop identities that are free of radicals by rationalizing the denominator or numerator. We’ll illustrate the former, leaving the latter as an exercise (see Exercise 102). u 11  cos u2 11  cos u2 tan a b   2 A 11  cos u2 11  cos u2 11  cos u2 2  B 1  cos2u

 B  `

multiply by the conjugate

rewrite

11  cos u2 2

Pythagorean identity

sin2u

1  cos u ` sin u

2x2  x

u Since 1  cos u 7 0 and sin u has the same sign as tan a b for all u in its domain, 2 u 1  cos u the relationship can simply be written tan a b  . 2 sin u The Half-Angle Identities u 1  cos u cosa b   2 A 2

u 1  cos u sin a b   2 A 2

u 1  cos u tan a b  2 sin u

EXAMPLE 5



u 1  cos u tan a b   2 A 1  cos u

sin u u tan a b  2 1  cos u

Using Half-Angle Formulas to Find Exact Values Use the half-angle identities to find exact values for (a) sin 15° and (b) tan 15°.

Solution



Noting that 15° is one-half the standard angle 30°, we can find each value by applying the respective half-angle identity with u  30° in Quadrant I.

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30 1  cos 30 b 2 A 2 13 1 2  Q 2 12  13 sin 15°  2

a. sin a

1  cos 30 30 b 2 sin 30 13 1 2  2  13 tan 15°  1 2

b. tan a

Now try Exercises 37 through 48

EXAMPLE 6



Using Half-Angle Formulas to Find Exact Values For cos   

Solution



B. You’ve just learned how to develop and use the power reduction and half-angle identities



7   and  in QIII, find exact values of sin a b and cos a b. 25 2 2

 3   3 , we know must be in QII S 6 6 and 2 2 2 2 4   we choose our signs accordingly: sin a b 7 0 and cos a b 6 0. 2 2

With  in QIII S  6  6

 1  cos  sin a b  2 A 2 7 1  a b 25  2 Q 16 4   A 25 5

 1  cos  cos a b   2 A 2 7 1  a b 25  2 Q 9 3   A 25 5 Now try Exercises 49 through 64



C. The Product-to-Sum Identities As mentioned in the introduction, the product-to-sum and sum-to-product identities are of immense importance to the study of any phenomenon that travels in waves, like light and sound. In fact, the tones you hear as you dial a telephone are actually the sum of two sound waves interacting with each other. Each derivation of a product-to-sum identity is very similar (see Exercise 105), and we illustrate by deriving the identity for cos  cos . Beginning with the sum and difference identities for cosine, we have cos  cos   sin  sin   cos1  2  cos  cos   sin  sin   cos1  2 2 cos  cos   cos1  2  cos1  2 1 cos  cos   3 cos1  2  cos1  2 4 2 The identities from this family are listed here.

cosine of a difference cosine of a sum combine equations divide by 2

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573

The Product-to-Sum Identities 1 cos  cos   3 cos1  2  cos1  2 4 2

sin  sin  

1 sin  cos   3 sin1  2  sin1  2 4 2

1 cos  sin   3 sin1  2  sin1  2 4 2

EXAMPLE 7



1 3cos1  2  cos1  2 4 2

Rewriting a Product as an Equivalent Sum Using Identities Write the product 2 cos(27t) cos(15t) as the sum of two cosine functions.

Solution



This is a direct application of the product-to-sum identity, with   27t and   15t. 1 cos  cos   3 cos1  2  cos1  2 4 2 1 2 cos127t2cos115t2  2a b 3cos127t  15t2  cos127t  15t2 4 2  cos112t2  cos142t2

product-to-sum identity

substitute result

Now try Exercises 65 through 73



There are times we find it necessary to “work in the other direction,” writing a sum of two trig functions as a product. This family of identities can be derived from the product-to-sum identities using a change of variable. We’ll illustrate the process for sin u  sin v. You are asked for the derivation of cos u  cos v in Exercise 106. To begin, we use 2  u  v and 2  u  v. This creates the sum 2  2  2u and the difference 2  2  2v, yielding     u and     v, respectively. uv uv , which all Dividing the original expressions by 2 gives   and   2 2 together make the derivation a matter of direct substitution. Using these values in any product-to-sum identity gives the related sum-to-product, as shown here. 1 sin  cos   3 sin1  2  sin1  2 4 2 sina

product-to-sum identity (sum of sines)

uv 1 uv b cosa b  1sin u  sin v2 2 2 2

substitute

uv uv for , for , 2 2

substitute u for    and v for    multiply by 2

uv uv b cosa b  sin u  sin v 2 sina 2 2 The sum-to-product identities follow. The Sum-to-Product Identities

uv uv b cosa b 2 2

cos u  cos v  2 cosa

uv uv b cosa b 2 2

sin u  sin v  2 sina

sin u  sin v  2 cosa

uv uv b sina b 2 2

cos u  cos v  2 sina

uv uv b sina b 2 2

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EXAMPLE 8



Rewriting a Sum as an Equivalent Product Using Identities Given y1  sin112t2 and y2  sin110t2, express y1  y2 as a product of trigonometric functions.

Solution



This is a direct application of the sum-to-product identity sin u  sin v, with u  12t and v  10t. uv uv b cosa b 2 2 12t  10t 12t  10t b cosa b sin112t2  sin110t2  2 sina 2 2  2 sin111t2 cos1t2 sin u  sin v  2 sina

C. You’ve just learned how to derive and use the product-to-sum and sumto-product identities

sum-to-product identity

substitute 12t for u and 10t for v substitute

Now try Exercises 74 through 82



For a mixed variety of identities, see Exercises 83–100.

D. Applications of Identities In more advanced mathematics courses, rewriting an expression using identities enables the extension or completion of a task that would otherwise be very difficult (or even impossible). In addition, there are a number of practical applications in the physical sciences.

Projectile Motion A projectile is any object that is thrown, shot, kicked, dropped, or otherwise given an initial velocity, but lacking a continuing source of propulsion. If air resistance is ignored, the range of the projectile depends only on its initial velocity v and the angle  at which it is propelled. This phenomenon is modeled by the function 1 2 r 12  v sin  cos . 16 EXAMPLE 9



Using Identities to Solve an Application a. Use an identity to show r12 

1 2 v sin  cos  is equivalent to 16

r 12 

1 2 v sin122. 32 b. If the projectile is thrown with an initial velocity of v  96 ft/sec, how far will it travel if   15°? c. From the result of part (a), determine what angle  will give the maximum range for the projectile. Solution



a. Note that we can use a double-angle identity if we rewrite the coefficient. 1 1 Writing as 2a b and commuting the factors gives 16 32 1 1 2 r 12  a bv 12 sin  cos 2  a b v2sin122. 32 32 1 b. With v  96 ft/sec and   15°, the formula gives r 115°2  a b1962 2sin 30°. 32 Evaluating the result shows the projectile travels a horizontal distance of 144 ft.

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Section 6.4 The Double-Angle, Half-Angle, and Product-to-Sum Identities

c. For any initial velocity v, r 12 will be maximized when sin122 is a maximum. This occurs when sin122  1, meaning 2  90° and   45°. The maximum range is achieved when the projectile is released at an angle of 45°. 

Now try Exercises 109 and 110 GRAPHICAL SUPPORT The result in Example 9(c) can be verified graphically by assuming an initial velocity of 96 ft/sec and entering the function 1 r12  1962 2sin122  288 sin122 as Y1 on a 32 0 graphing calculator. With an amplitude of 288 and results confined to the first quadrant, we set an appropriate window, graph the function, and use the 2nd TRACE (CALC 4:maximum) feature. As shown in the figure, the max occurs at   45°.

300

90

0

Sound Waves Each tone you hear on a touch-tone phone is actually the combination of precisely two B sound waves with different frequencies (frequency f is defined as f  ). This is why 2 the tones you hear sound identical, regardless of what phone you use. The sumto-product and product-to-sum formulas help us to understand, study, and use sound in very powerful and practical ways, like sending faxes and using other electronic media.

EXAMPLE 10



Using an Identity to Solve an Application On a touch-tone phone, the sound created by pressing 5 is produced by combining a sound wave with frequency 1336 cycles/sec, with another wave having frequency 770 cycles/sec. Their respective equations are y1  cos12 1336t2 and y2  cos12 770t2, with the resultant wave being y  y1  y2 or y  cos12672t2  cos11540t2. Rewrite this sum as a product.

Solution



1

2

3

697 cps

4

5

6

770 cps

7

8

9

852 cps

*

0

#

941 cps

1209 cps

1477 cps

1336 cps

This is a direct application of the sum-to-product identity, with u  2672t and v  1540t. Computing one-half the sum/difference of u and v gives 2672t  1540t 2672t  1540t  2106t and  566t. 2 2 uv uv b cos a b 2 2 cos12672t2  cos11540t2  2 cos12106t2cos1566t2 cos u  cos v  2 cos a

sum-to-product identity substitute 2672t for u and 1540t for v

Now try Exercises 111 and 112

D. You’ve just learned how to solve applications using identities



Note we can identify the button pressed when the wave is written as a sum. If we have only the resulting wave (written as a product), the product-to-sum formula must be used to identify which button was pressed. Additional applications requiring the use of identities can be found in Exercises 113 through 117.

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6.4 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. The double-angle identities can be derived using the identities with   . For cos122 we expand cos1  2 using . 2. If  is in QIII then 180° 6  6 270° and in

since

6

 6 2

 must be 2 .

3. Multiple-angle identities can be derived using the sum and difference identities. For sin (3x) use sin  ( ). 

4. For the half-angle identities the sign preceding the u radical depends on the in which . 2 5. Explain/Discuss how the three different identities u for tana b are related. Verify that 2 sin x 1  cos x  . sin x 1  cos x 7 6. In Example 6, we were given cos    and  25 in QIII. Discuss how the result would differ if we stipulate that  is in QII instead.

DEVELOPING YOUR SKILLS

Find exact values for sin(2), cos(2), and tan(2) using the information given.

7. sin  

5 ;  in QII 13

9. cos   

21 ;  in QII 29

9 63 ;  in QII 10. sin    ;  in QIII 41 65

11. tan  

13 ;  in QIII 84

13. sin  

48 ; cos  6 0 73

14. cos   

8. cos   

12. sec  

53 ;  in QI 28

8 ; tan  7 0 17

5 15. csc   ; sec  6 0 3 16. cot   

80 ; cos  7 0 39

Find exact values for sin , cos , and tan  using the information given.

24 17. sin122  ; 2 in QII 25 18. sin122  

240 ; 2 in QIII 289

19. cos122   20. cos122 

41 ; 2 in QII 841

120 ; 2 in QIV 169

21. Verify the following identity: sin132  3 sin   4 sin3 22. Verify the following identity: cos142  8 cos4  8 cos2  1 Use a double-angle identity to find exact values for the following expressions.

23. cos 75° sin 75°

24. cos215°  sin215°

 25. 1  2 sin2 a b 8

26. 2 cos2 a

27.

2 tan 22.5° 1  tan222.5°

28.

 b1 12

 2 2 tan 1 12

 1  tan2 1 12 2

29. Use a double-angle identity to rewrite 9 sin (3x) cos (3x) as a single function. [Hint: 9  92 122 .] 30. Use a double-angle identity to rewrite 2.5  5 sin2x as a single term. [Hint: Factor out a constant.]

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Rewrite in terms of an expression containing only cosines to the power 1.

31. sin2x cos2x

32. sin4x cos2x

33. 3 cos4x

34. cos4x sin4x

35. 2 sin6x

36. 4 cos6x

Use a half-angle identity to find exact values for sin , cos , and tan  for the given value of .

37.   22.5°

38.   75°

 39.   12

5 40.   12

41.   67.5°

42.   112.5°

43.  

3 8

44.  

11 12

Use the results of Exercises 37–40 and a half-angle identity to find the exact value.

45. sin 11.25° 47. sin a

46. tan 37.5°

 b 24

48. cos a

5 b 24

Use a half-angle identity to rewrite each expression as a single, nonradical function.

1  cos 30° 49. A 2

1  cos 45° 50. A 2

1  cos142 51. A 1  cos142

1  cos16x2 52. sin16x2

sin12x2 53. 1  cos12x2

1211  cos x2 54. 1  cos x

   Find exact values for sin a b, cos a b, and tan a b 2 2 2 using the information given.

55. sin  

12 ;  is obtuse 13

56. cos   

8 ;  is obtuse 17

4 57. cos    ;  in QII 5 58. sin   

7 ;  in QIII 25

35 59. tan    ;  in QII 12

60. sec   

577

65 ;  in QIII 33

61. sin  

15 ;  is acute 113

62. cos  

48 ;  is acute 73

63. cot  

3 21 ; 6  6 20 2

64. csc  

41  ; 6  6  9 2

Write each product as a sum using the product-to-sum identities.

65. sin142 sin182

66. cos1152 sin132

7t 3t 67. 2 cosa b cosa b 2 2

9t 5t 68. 2 sina b sina b 2 2

69. 2 cos11979t2 cos1439t2 70. 2 cos12150t2 cos1268t2 Find the exact value using product-to-sum identities.

71. 2 cos 15° sin 135° 72. sina

7  b cosa b 8 8

73. sina

 7 b sina b 12 12

Write each sum as a product using the sum-to-product identities.

74. cos19h2  cos14h2 76. sina

75. sin114k2  sin141k2

11x 5x 5x 7x b  sina b 77. cosa b  cosa b 8 8 6 6

78. cos1697t2  cos 11447t2

79. cos1852t2  cos11209t2 Find the exact value using sum-to-product identities.

80. cos 75°  cos15° 81. sina

13 17 b  sina b 12 12

82. sina

7 11 b  sina b 12 12

Verify the following identities.

83.

2 sin x cos x  tan12x2 cos2x  sin2x

84.

1  2 sin2 x  cot12x2 2 sin x cos x

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85. 1sin x  cos x2 2  1  sin 12x2

86. 1sin x  12  sin x  cos 12x2 2

2

4

87. cos182  cos2 142  sin2 142

88. sin 14x2  4 sin x cos x 11  2 sin2x2 89.

cos 122 sin2

 cot   1 2

90. csc2  2  91. tan 122 

cos 122 sin2

2 cot   tan 

92. cot   tan  

2 cos 122 sin 122

93. tan x  cot x  2 csc 12x2 94. csc 12x2 

1 csc x sec x 2

x x 95. cos2 a b  sin2 a b  cos x 2 2 x x 96. 1  2 sin a b  cos a b 4 2 2

97. 1  sin2 122  1  4 sin2  4 sin4



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CHAPTER 6 Trigonometric Identities, Inverses, and Equations

x 98. 2 cos2 a b  1  cos x 2 99. 100.

sin1120t2  sin180t2  cot120t2 cos1120t2  cos180t2 sin m  sin n mn  tan a b cos m  cos n 2

 2 101. Show sin2  11  cos 2 2  c 2 sin a b d . 2 1  cos u u 102. Show that tan a b   is equivalent 2 A 1  cos u sin u to by rationalizing the numerator. 1  cos u

103. Derive the identity for sin 122 and tan 122 using sin 1  2 and tan 1  2, where   .

104. Derive the identity for tan2 12 using 2 tan12 tan122  . Hint: Solve for tan2 and 1  tan2 12 work in terms of sines and cosines. 105. Derive the product-to-sum identity for sin  sin . 106. Derive the sum-to-product identity for cos u  cos v.

WORKING WITH FORMULAS

107. Supersonic speeds, the sound barrier, and Mach  numbers: M  csc a b 2 The speed of sound varies with temperature and altitude. At 32°F, sound travels about 742 mi/hr at sea level. A jet-plane flying faster than the speed of sound (called supersonic speed) has “broken the sound barrier.” The plane projects threedimensional sound waves about the nose of the craft that form the shape of a cone. The cone intersects the Earth along a hyperbolic path, with a sonic boom being heard by anyone along this path. The ratio of the plane’s speed to the speed of sound is

called its Mach number M, meaning a plane flying at M  3.2 is traveling 3.2 times the speed of sound. This Mach number can be determined using the formula given here, where  is the vertex angle of the cone described. For the following exercises, use the formula to find M or  as required. For parts (a) and (b), answer in exact form (using a half-angle identity) and approximate form. a.   30° b.   45° c. M  2 108. Malus’s law: I  I0 cos2 When a beam of plane-polarized light with intensity I0 hits an analyzer, the intensity I of the transmitted beam of light can be found using the formula shown, where  is the angle formed between the transmission axes of the polarizer and the analyzer. Find the intensity of the beam when   15° and I0  300 candelas (cd). Answer in exact form (using a power reduction identity) and approximate form.

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579

APPLICATIONS

Range of a projectile: Exercises 109 and 110 refer to Example 9. In Example 9, we noted that the range of a projectile was maximized at   45°. If  7 45° or  6 45°, the projectile falls short of its maximum potential distance. In Exercises 109 and 110 assume that the projectile has an initial velocity of 96 ft/sec.

109. Compute how many feet short of maximum the projectile falls if (a)   22.5° and (b)   67.5°. Answer in both exact and approximate form. 110. Use a calculator to compute how many feet short of maximum the projectile falls if (a)   40° and   50° and (b)   37.5° and   52.5°. Do you see a pattern? Discuss/explain what you notice and experiment with other values to confirm your observations. Touch-tone phones: The diagram given in Example 10 shows the various frequencies used to create the tones for a touch-tone phone. One button is randomly pressed and the resultant wave is modeled by y(t) shown. Use a product-tosum identity to write the expression as a sum and determine the button pressed.

111. y 1t2  2 cos 12150t2cos 1268t2 112. y 1t2  2 cos 11906t2cos 1512t2

113. Clock angles: Kirkland City has a large clock atop city hall, with a minute hand that is 3 ft d long. Claire and Monica independently attempt to devise a function that will track the distance between the tip of the minute hand at t minutes between the hours, and the tip of the minute hand when it is in the vertical position as shown. Claire finds the t function d1t2  ` 6 sina b ` , while Monica devises 60 t d1t2  18 c 1  cos a b d . Use the identities A 30 from this section to show the functions are equivalent. ␪ L

21.6 cm

114. Origami: The Japanese art of origami involves the repeated folding of a single piece of paper to create various art forms. When the upper right corner of

28 cm

a rectangular 21.6-cm by 28-cm piece of paper is folded down until the corner is flush with the other side, the length L of the fold is 10.8 related to the angle  by L  . (a) Show sin  cos2 21.6 sec  this is equivalent to L  , (b) find the sin122 length of the fold if   30°, and (c) find the angle  if L  28.8 cm. 115. Machine gears: A machine part involves two gears. The first has a radius of 2 cm and the second a radius of 1 cm, so the smaller gear ␪ 2 cm turns twice as fast as the larger gear. Let  represent the angle of rotation in the larger gear, measured from a vertical and downward 2␪ 1 m h starting position. Let P be a P point on the circumference of the smaller gear, starting at the vertical and downward position. Four engineers working on an improved design for this component devise functions that track the height of point P above the horizontal plane shown, for a rotation of ° by the larger gear. The functions they develop are: Engineer A: f 12  sin 12  90°2  1; Engineer B: g12  2 sin2; Engineer C: k12  1  sin2  cos2; and Engineer D: h12  1  cos 122. Use any of the identities you’ve learned so far to show these four functions are equivalent. 116. Working with identities: Compute the value of sin 15° two ways, first using the half-angle identity for sine, and second using the difference identity for sine. (a) Find a decimal approximation for each to show the results are equivalent and (b) verify algebraically that they are equivalent. (Hint: Square both sides.) 117. Working with identities: Compute the value of cos 15° two ways, first using the half-angle identity for cosine, and second using the difference identity for cosine. (a) Find a decimal approximation for each to show the results are equivalent and (b) verify algebraically that they are equivalent. (Hint: Square both sides.)

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EXTENDING THE CONCEPT

118. Can you find three distinct, real numbers whose sum is equal to their product? A little known fact from trigonometry stipulates that for any triangle, the sum of the tangents of the angles is equal to the products of their tangents. Use a calculator to test this statement, recalling the three angles must sum to 180°. Our website at www.mhhe.com/coburn shows a method that enables you to verify the statement using tangents that are all rational values. Exercise 119

1 ␪ 2

1

sin ␪

119. A proof without words: From elementary geometry we have the following: (a) an angle inscribed in a semicircle is a right angle; and (b) the measure of an 

6-38

CHAPTER 6 Trigonometric Identities, Inverses, and Equations

s

␪ cos ␪

inscribed angle (vertex on the circumference) is one-half the measure of its intercepted arc. Discuss/explain how the unit-circle diagram offers sin x x a proof that tan a b  . Be detailed and 2 1  cos x thorough. 120. Using   30° and repeatedly applying the halfangle identity for cosine, show that cos 3.75° is 12  12  12  13 equal to . Verify the result 2 using a calculator, then use the patterns noted to write the value of cos 1.875° in closed form (also verify this result). As  becomes very small, what appears to be happening to the value of cos ?

MAINTAINING YOUR SKILLS

121. (3.3) Use the rational roots theorem to find all zeroes of x4  x3  8x2  6x  12  0. 122. (5.1) The hypotenuse of a certain right triangle is twice the shortest side. Solve the triangle. 63 123. (5.3) Verify that 1 16 65 , 65 2 is on the unit circle, then find tan  and sec  to verify 1  tan2  sec2.

124. (5.5) Write the equation of the function graphed in terms of a sine function of the form y  A sin 1Bx  C2  D.

y 3 2 1 1 2 3

␲

MID-CHAPTER CHECK 1. Verify the identity using a multiplication: sin x 1csc x  sin x2  cos2x 2. Verify the identity by factoring: cos2x  cot2x  cos2x cot2x 3. Verify the identity by combining terms: 2 sin x cos x   cos x sin x sec x csc x 4. Show the equation given is not an identity. 1  sec2x  tan2x 5. Verify each identity. sin3x  cos3x  1  sin x cos x a. sin x  cos x b.

1  cos x 1  sec x  0 csc x cot x

6. Verify each identity. sec2x  tan2x a.  cos2x sec2x cot x  tan x b.  cos2x  sin2x csc x sec x 56 7. Given  and  are obtuse angles with sin   65 80 and tan    , find 39 a. sin 1  2 b. cos 1  2 c. tan 1  2

2␲ x

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Reinforcing Basic Concepts

8. Use the diagram shown to compute sin A, cos A, and tan A.

Exercise 8 

7 with  in QIII, find the value of 25 sin 122, cos 122 , and tan 122 .

10. Given sin   

60⬚ A

48

581

15 17 14 and  in QII, find exact values of   sina b and cosa b. 2 2

9. Given cos   

REINFORCING BASIC CONCEPTS Identities—Connections and Relationships It is a well-known fact that information is retained longer and used more effectively when it is organized, sequential, and connected. In this Strengthening Core Skills (SCS), we attempt to do just that with our study of identities. In flowchart form we’ll show that the entire range of identities has only two tiers, and that the fundamental identities and the sum and difference identities are really the keys to the entire range of identities. Beginning with the right triangle definition of sine, cosine, and tangent, the reciprocal identities and ratio identities are more semantic (word related) than mathematical, and the Pythagorean identities follow naturally from the properties of right triangles. These form the first tier. Basic Definitions sin  

opp hyp

cos  

adj hyp

tan  

opp adj

Fundamental Identities defined

Reciprocal Identities 1  csc  sin  1  sec  cos  1  cot  tan 

defined

Ratio Identities sin   tan  cos  cos   cot  sin  sec   tan  csc 

derived

Pythagorean Identities sin2  cos2  1

tan2  1  sec2

1  cot2  csc2

(divide by cos2)

(divide by sin2)

The reciprocal and ratio identities are actually defined, while the Pythagorean identities are derived from these two families. In addition, the identity sin2  cos2  1 is the only Pythagorean identity we actually need to memorize; the other two follow by division of cos2 and sin2 as indicated. In virtually the same way, the sum and difference identities for sine and cosine are the only identities that need to be memorized, as all other identities in the second tier flow from these.

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CHAPTER 6 Trigonometric Identities, Inverses, and Equations

Sum/Difference Identities

cos 1  2  cos  cos   sin  sin  sin 1  2  sin  cos   cos  sin 

Double-Angle Identities Power Reduction Identities use    solve for cos2, sin2 in in sum identities related cos122 identity

Half-Angle Identities solve for cos , sin  and use   u/2 in the power reduction identities

Product-to-Sum Identities combine various sum/difference identities

sin 122  2 sin  cos 

cos2 

1  cos 122 2

u 1  cos u cos a b   2 A 2

see Section 6.4

cos 122  cos2  sin2

sin2 

1  cos 122 2

u 1  cos u sin a b   2 A 2

see Section 6.4

cos 122  2 cos2  1

(use sin2  1  cos2)

cos 122  1  2 sin2 (use cos2  1  sin2)

2 2 Exercise 1: Starting with the identity sin   cos   1, derive the other two Pythagorean identities.

Exercise 2: Starting with the identity cos 1  2  cos  cos   sin  sin , derive the double-angle identities for cosine.

6.5 The Inverse Trig Functions and Their Applications Learning Objectives In Section 6.5 you will learn how to:

A. Find and graph the inverse sine function and evaluate related expressions

B. Find and graph the inverse cosine and tangent functions and evaluate related expressions

C. Apply the definition and notation of inverse trig functions to simplify compositions

D. Find and graph inverse functions for sec x, csc x, and cot x

E. Solve applications involving inverse functions

While we usually associate the number  with the features of a circle, it also occurs in some “interesting” places, such as the study of normal (bell) curves, Bessel functions, Stirling’s formula, Fourier series, Laplace transforms, and infinite series. In much the same way, the trigonometric functions are surprisingly versatile, finding their way into a study of complex numbers and vectors, the simplification of algebraic expressions, and finding the area under certain curves—applications that are hugely important in a continuing study of mathematics. As you’ll see, a study of the inverse trig functions helps support these fascinating applications.

A. The Inverse Sine Function In Section 4.1 we established that only one-to-one functions have an inverse. All six trig functions fail the horizontal line test and are not one-to-one as given. However, by suitably restricting the domain, a one-to-one function can be defined that makes finding an inverse possible. For the sine function, it seems natural to choose   the interval c  , d since it is centrally located and the sine function attains all 2 2 possible range values in this interval. A graph of y  sin x is shown in Figure 6.5, with the portion corresponding to this interval colored in red. Note the range is still 31, 1 4 (Figure 6.6).

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6-41 Figure 6.5  2

Figure 6.6

y  2





 2

2

3 2

2

x



2

1



 2 , 1



 2

1

1 2

Figure 6.7

y

y y  sin1 x

y  sin x

y  sin x

1



583

Section 6.5 The Inverse Trig Functions and Their Applications

1

y  sin x

1

 2,

1

 2

x



 2 , 1 

2   2 ,

1

 2

1

x

1 1

1 

2



2

yx

We can obtain an implicit equation for the inverse of y  sin x by interchanging x- and y-values, obtaining x  sin y. By accepted convention, the explicit form of the inverse sine function is written y  sin1x or y  arcsin x. Since domain and range values have been interchanged, the domain of y  sin1x is 31, 1 4 and the range is   c  , d . The graph of y  sin1x can be found by reflecting the portion in red across 2 2 the line y  x and using the endpoints of the domain and range (see Figure 6.7).

WORTHY OF NOTE In Example 4 of Section 4.1, we noted that by suitably restricting the domain of y  x 2, a one-to-one function could be defined that made finding an inverse function possible. Specifically, for f1x2  x 2; x  0, f 1 1x2  1x.

The Inverse Sine Function For y  sin x with domain   c  , d and range 31, 1 4, 2 2 the inverse sine function is

␲ 2

y y  sin1 x ␲

1, 2 

1

y  sin1x or y  arcsin x,

1 ␲

2

  with domain 31, 1 4 and range c  , d . 2 2

1

␲ 2

x

1

y  sin1x if and only if sin y  x

1,

␲2 

␲

2

From the implicit form x  sin y, we learn to interpret the inverse function as, “y is the number or angle whose sine is x.” Learning to read and interpret the explicit form in this way will be helpful. That is, y  sin1x means “y is the number or angle whose sine is x.” y  sin 1x 3 x  sin y EXAMPLE 1



x  sin y 3 y  sin 1x

Evaluating y  sin 1x Using Special Values Evaluate the inverse sine function for the values given: 13 1 a. y  sin1a b. y  arcsin a b c. y  sin12 b 2 2

Solution



  For x in 31, 1 4 and y in c  , d , 2 2 13 13 a. y  sin1a b: y is the number or angle whose sine is 2 2 13 13  1 sin y  , so sin1a b . 2 2 3

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1 1 b. y  arcsin a b: y is the arc or angle whose sine is  2 2 1  1 1 sin y   , so arcsin a b   . 2 2 6 1 c. y  sin 122: y is the number or angle whose sine is 2 1 sin y  2. Since 2 is not in 31, 14 , sin1 122 is undefined. Now try Exercises 7 through 12

Table 6.1 x

sin x

2

1

3

 13 2

4

 12 2

6

12

0

0

 6

1 2

 4

12 2

 3

13 2

 2

1

EXAMPLE 2



1 13 and sin y   2 2   each have an infinite number of solutions, but only one solution in c  , d . 2 2 1 13 , 1, and so onb, y  sin1x can be When x is one of the standard values a0, , 2 2 evaluated by reading a standard table “in reverse.” For y  arcsin 112, we locate the number 1 in the right-hand column of Table 6.1, and note the “number or angle  whose sine is 1,” is  . If x is between 1 and 1 but is not a standard value, we can 2 use the sin1 function on a calculator, which is most often the 2nd or function for . In Examples 1a and 1b, note that the equations sin y 



Evaluating y  sin1x Using a Calculator Evaluate each inverse sine function twice. First in radians rounded to four decimal places, then in degrees to the nearest tenth. a. y  sin10.8492 b. y  arcsin 10.23172

Solution WORTHY OF NOTE The sin1x notation for the inverse sine function is a carryover from the f 1 1x2 notation for a general inverse function, and likewise has nothing to do with the reciprocal of the function. The arcsin x notation derives from our work in radians on the unit circle, where y  arcsin x can be interpreted as “y is an arc whose sine is x.”



For x in 31, 14 , we evaluate y  sin1x. a. y  sin10.8492: With the calculator in radian MODE , use the keystrokes 2nd 1 0.8492 ) 10.84922  1.0145 radians. In degree ENTER . We find sin 1 MODE , the same sequence of keystrokes gives sin 10.84922  58.1° (note that 1.0145 rad  58.1°2 . b. y  arcsin 10.23172: In radian MODE , we find sin1 10.23172  0.2338 rad. In degree MODE , sin1 10.23172  13.4°.

Now try Exercises 13 through 16



From our work in Section 4.1, we know that if f and g are inverses, 1 f  g21x2  x and 1g  f 21x2  x. This suggests the following properties. Inverse Function Properties for Sine For f 1x2  sin x and g1x2  sin1x:

I. 1f  g21x2  sin 1sin1x2  x for x in 31, 14 and   II. 1g  f 21x2  sin1 1sin x2  x for x in c  , d 2 2

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Section 6.5 The Inverse Trig Functions and Their Applications



EXAMPLE 3

Evaluating Expressions Using Inverse Function Properties Evaluate each expression and verify the result on a calculator. 1  5 a. sin c sin1a b d b. arcsin c sin a b d c. sin1 c sin a b d 2 4 6



Solution

1 1 1 a. sin c sin1a b d  , since is in 31, 1 4 Property I 2 2 2      b. arcsin c sin a b d  , since is in c  , d Property II 4 4 4 2 2 5 5 5   c. sin1 c sin a b d  , since is not in c  , d . 6 6 6 2 2 This doesn’t mean the expression cannot be evaluated, only that we cannot use   5 5  Property II. Since sin a b  sina b, sin1 c asin b d  sin1 c sin a b d  . 6 6 6 6 6 The calculator verification for each is shown in Figures 6.8 and 6.9. Note    0.5236 and  0.7854. 6 4 Figure 6.8

Figure 6.9

Parts (a) and (b)

Part (c)

A. You’ve just learned how to find and graph the inverse sine function and evaluate related expressions

Now try Exercises 17 through 24



B. The Inverse Cosine and Inverse Tangent Functions Like the sine function, the cosine function is not one-to-one and its domain must also be restricted to develop an inverse function. For convenience we choose the interval x  3 0,  4 since it is again somewhat central and takes on all of its range values in this interval. A graph of the cosine function, with the interval corresponding to this interval shown in red, is given in Figure 6.10. Note the range is still 3 1, 14 (Figure 6.11). Figure 6.10

y

3

y  cos1 (x)

2

y  cos(x)

1

(1, ␲)

3

2

␲ 2

Figure 6.12

Figure 6.11

y

␲ 2

␲

3␲ 2

2␲

␲

2

1

3 2

(0, 1)

(0, 1) x

y

␲ 2

y  cos(x)

␲

x (␲, 1)

␲

2

1

(1, 0) y  cos (x)

␲

x (␲, 1)

For the implicit equation of inverse cosine, y  cos x becomes x  cos y, with the corresponding explicit forms being y  cos1x or y  arccos x. By reflecting the graph of y  cos x across the line y  x, we obtain the graph of y  cos1x shown in Figure 6.12.

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The Inverse Cosine Function For y  cos x with domain 3 0,  4 and range 3 1, 1 4 , the inverse cosine function is

(1, ␲)

y  cos1x or y  arccos x

y  cos1 (x)

with domain 31, 1 4 and range 30,  4.



␲ 2



(1, 0)

1

␲

x

Evaluating y  cos 1x Using Special Values Evaluate the inverse cosine for the values given: 13 b a. y  cos10 b. y  arccos a 2

Solution

2 1

y  cos1x if and only if cos y  x

EXAMPLE 4

y 3

c. y  cos1

For x in 3 1, 14 and y in 30,  4 , a. y  cos10: y is the number or angle whose cosine is 0 1 cos y  0.  This shows cos10  . 2 13 b. y  arccos a b: y is the arc or angle whose cosine is 2 13 13 5 13  1 cos y   . This shows arccosa b . 2 2 2 6 c. y  cos1: y is the number or angle whose cosine is  1 cos y  . Since   3 1, 14, cos1 is undefined. Now try Exercises 25 through 34



Knowing that y  cos x and y  cos1x are inverse functions enables us to state inverse function properties similar to those for sine. Inverse Function Properties for Cosine For f 1x2  cos x and g1x2  cos1x: I. 1 f  g21x2  cos1cos1x2  x for x in 3 1, 14 and II. 1g  f 21x2  cos1 1cos x2  x for x in 3 0,  4

EXAMPLE 5



Evaluating Expressions Using Inverse Function Properties Evaluate each expression. a. cos 3 cos1 10.732 4

Solution



b. arccos c cos a

 bd 12

c. cos1 c cos a

a. cos 3cos1 10.732 4  0.73, since 0.73 is in 3 1, 14    b. arccos c cos a b d  , since is in 30,  4 12 12 12

Property I Property II

4 bd 3

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587

4 4 4 bd  , since is not in 30,  4 . 3 3 3 This expression cannot be evaluated using Property II. Since 4 2 4 2 2 cos a b  cos a b, cos1 c cos a b d  cos1 c cos a b d  . 3 3 3 3 3 The results can also be verified using a calculator. c. cos1 c cosa

Now try Exercises 35 through 42

For the tangent function, we likewise restrict the domain to obtain a one-to-one   function, with the most common choice being a , b. The corresponding range 2 2 Figure 6.14 is . The implicit equation for the inverse tangent y function is x  tan y with the explicit forms 3 y  tan1x or y  arctan x. With the domain and 1 2 y  tan x ␲ range interchanged, the domain of y  tan1x is 2 1   ␲ 1, 4  , and the range is a , b. The graph of y  tan x 2 2 ␲ ␲ x   1 for x in a , b is shown in red (Figure 6.13), with  ␲2 2 2 1,  ␲4  2 the inverse function y  tan1x shown in blue (Figure 6.14). 3

Figure 6.13 y  tan x

y

3 2 1 ␲

␲

 4 , 1 ␲

 ␲4 , 1

x

1 2 3

The Inverse Tangent Function

Inverse Function Properties for Tangent

  For y  tan x with domain a , b and 2 2 range , the inverse tangent function is y  tan 1x or y  arctan x,   with domain  and range a , b. 2 2 y  tan1x if and only if tan y  x

EXAMPLE 6





For f 1x2  tan x and g1x2  tan1x:

I. 1 f  g2 1x2  tan1tan1x2  x for x in  and

  II. 1g  f 21x2  tan1 1tan x2  x for x in a , b. 2 2

Evaluating Expressions Involving Inverse Tangent Evaluate each expression. a. tan1 1 132 b. arctan 3tan 10.892 4

Solution



B. You’ve just learned how to find and graph the inverse cosine and tangent functions and evaluate related expressions

  For x in  and y in a , b, 2 2   a. tan1 1 132   , since tan a b   13 3 3   b. arctan 3 tan10.892 4  0.89, since 0.89 is in a , b 2 2

Property II

Now try Exercises 43 through 52



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C. Using the Inverse Trig Functions to Evaluate Compositions 1 In the context of angle measure, the expression y  sin1a b represents an angle— 2 1 the angle y whose sine is  . It seems natural to ask, “What happens if we take the 2 1 tangent of this angle?” In other words, what does the expression tan c sin1a b d 2  mean? Similarly, if y  cos a b represents a real number between 1 and 1, how do 3  we compute sin1 c cos a b d ? Expressions like these occur in many fields of study. 3 EXAMPLE 7



Simplifying Expressions Involving Inverse Trig Functions Simplify each expression: 1  a. tan c arcsin a b d b. sin1 c cos a b d 2 3

Solution

WORTHY OF NOTE To verify the result of Example 8, we can actually 8 find the value of sin1a b 17 on a calculator, then take the tangent of the result. See the figure.



1   1 a. In Example 1 we found arcsin a b   . Substituting  for arcsin a b 2 6 6 2  13 13 1 gives tan a b   , showing tan c arcsina b d   . 6 3 2 3  1  b. For sin1 c cosa b d , we begin with the inner function cos a b  . 3 3 2 1 1  Substituting for cos a b gives sin1a b. With the appropriate checks 2 3 2 1    satisfied we have sin1a b  , showing sin1 c cos a b d  . 2 6 3 6 Now try Exercises 53 through 64



If the argument is not a special value and we need the Figure 6.15 answer in exact form, we can draw the triangle described by the inner expression using the definition of the trigonometric functions as ratios. In other words, for either 17 8 8 1 y or   sin a b, we draw a triangle with hypotenuse 17 ␪ 17 and side 8 opposite  to model the statement, “an angle adj opp 8  whose sine is ” (see Figure 6.15). Using the Pythagorean theorem, we find 17 hyp the adjacent side is 15 and can now name any of the other trig functions.

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EXAMPLE 8



Solution



Using a Diagram to Evaluate an Expression Involving Inverse Trig Functions 8 Evaluate the expression tan c sin1a b d . 17 The expression tan c sin1a where   sin1a

8 b d is equivalent to tan , 17

8 b with  in 17

15

␪

  c  , d (QIV or QI). For 2 2 8 sin    1sin  6 02,  must be in 17 QIII or QIV. To satisfy both,  must be in QIV. From the figure 8 8 8 we note tan    , showing tan c sin1a b d   . 15 17 15

17

8 (15, 8)

Now try Exercises 65 through 72



These ideas apply even when one side of the triangle is unknown. In other words, x b, since “ is an angle whose we can still draw a triangle for   cos1a 2 2x  16 adj x cosine is  .” 2 hyp 2x  16 EXAMPLE 9



Using a Diagram to Evaluate an Expression Involving Inverse Trig Functions Evaluate the expression tan c cos1a

x

2x  16 function is defined for the expression given. Solution



Rewrite tan c cos1a   cos1a

C. You’ve just learned how to apply the definition and notation of inverse trig functions to simplify compositions

x

x 2x2  16

2

b d . Assume x 7 0 and the inverse

b d as tan , where

b. Draw a triangle with

2x2  16 side x adjacent to  and a hypotenuse of 2x2  16. The Pythagorean theorem gives

√x2  16

opp

␪ x

x2  opp2  1 2x2  162 2, which leads to opp2  1x2  162  x2 giving opp  116  4. This shows x 4 b d  (see the figure). tan   tan c cos1a 2 x 2x  16 Now try Exercises 73 through 76



D. The Inverse Functions for Secant, Cosecant, and Cotangent As with the other functions, we restrict the domain of the secant, cosecant, and cotangent functions to obtain a one-to-one function that is invertible (an inverse can be found). Once again the choice is arbitrary, and some domains are easier to work with than others in more advanced mathematics. For y  sec x, we’ve chosen the “most

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Figure 6.16

Figure 6.17 y  sec 1 x

y  sec x

y

y 3

y  sec x y 僆 (, 1] 傼 [1, )

2

x 僆 (, 1] 傼 [1, ) y 僆 [0, ␲2 ) 傼 ( ␲2 , ␲]

1

␲

␲

1

x ␲

x

␲

1

1

2

2

3

3

intuitive” restriction, one that seems more centrally located (nearer the origin). The graph of y  sec x is reproduced here, along with its inverse function (see Figures 6.16 and 6.17). The domain, range, and graphs of the functions y  csc1x and y  cot1x are asked for in the Exercises (see Exercise 100). The functions y  sec1x, y  csc1x, and y  cot1x can be evaluated by noting their relationship to y  cos1x, y  sin1x, and y  tan1x, respectively. For y  sec1x, we have

WORTHY OF NOTE While the domains of y  cot1x and y  tan1x both include all real numbers, evaluating cot1x 1 using tan1a b involves the x restriction x  0. To maintain consistency, the equation  cot1x   tan1x is often 2 used. The graph of  y   tan1x is that of 2 y  tan1x reflected across  the x-axis and shifted 2 units up, with the result identical to the graph of y  cot1x.

EXAMPLE 10

y  sec1x

2

x 僆 [0, ␲2 ) 傼 ( ␲2 , ␲]

3

sec y  x 1 1  x sec y 1 cos y  x 1 y  cos1a b x 1 sec1x  cos1a b x

definition of inverse function property of reciprocals

reciprocal ratio

rewrite using inverse function notation substitute sec1x for y

1 In other words, to find the value of y  sec1x, evaluate y  cos1a b, x  1. x 1 Similarly, the expression csc1x can be evaluated using sin1a b, x  1. The x 1 expression cot x can likewise be evaluated using an inverse tangent function: 1 cot1x  tan1a b. x 

Evaluating an Inverse Trig Function Evaluate using a calculator only if necessary: 2  a. sec1a b. cot1a b b 12 13

Solution



D. You’ve just learned how to find and graph inverse functions for sec x, csc x, and cot x

2 13 b, we evaluate cos1a b. 2 13 Since this is a standard value, no calculator is needed and the result is 30°.  12 b. For cot1a b, find tan1a b on a calculator:  12  12 cot1a b  tan1a b  1.3147.  12 a. From our previous discussion, for sec1a

Now try Exercises 77 through 86



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591

A summary of the highlights from this section follows. Summary of Inverse Function Properties and Compositions 1. For sin x and sin1x,

2. For cos x and cos1x,

  sin1 1sin x2  x, for any x in the interval c  , d 2 2 3. For tan x and tan1x,

cos1 1cos x2  x, for any x in the interval 3 0, 4 1 4. To evaluate sec1x, use cos1a b, x  1, x

sin1sin1x2  x, for any x in the interval 3 1, 14

tan1tan1x2  x, for any real number x   tan1 1tan x2  x, for any x in the interval a , b 2 2

cos1cos1x2  x, for any x in the interval 31, 1 4

1 csc1x, use sin1a b, x  1 x cot1x, use

  tan1x, for all real numbers x 2

E. Applications of Inverse Trig Functions We close this section with one example of the many ways that inverse functions can be applied. EXAMPLE 11



Using Inverse Trig Functions to Find Viewing Angles Believe it or not, the drive-in Figure 6.18 movie theaters that were so popular in the 1950s are making a comeback! If you arrive early, you can park in 30 ft one of the coveted “center spots,” but if you arrive late, you might have to park very ␣ close and strain your neck to ␪ 10 ft watch the movie. Surprisingly, ␤ the maximum viewing angle x (not the most comfortable viewing angle in this case) is actually very close to the front. Assume the base of a 30-ft screen is 10 ft above eye level (see Figure 6.18). a. Use the inverse function concept to find expressions for angle  and angle . b. Use the result of Part (a) to find an expression for the viewing angle . c. Use a calculator to find the viewing angle  (to tenths of a degree) for distances of 15, 25, 35, and 45 ft, then to determine the distance x (to tenths of a foot) that maximizes the viewing angle.

Solution



a. The side opposite  is 10 ft, and we want to know x — the adjacent side. This 10 10 suggests we use tan   , giving   tan1a b. In the same way, we find x x 40 that   tan1a b. x b. From the diagram we note that     , and substituting for  and  40 10 directly gives   tan1a b  tan1a b. x x

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40 10 b  tan1a b, a graphing calculator gives x x Figure 6.19 approximate viewing angles of 50 35.8 °, 36.2°, 32.9°, and 29.1°, for x  15, 25, 35, and 45 ft, respectively. From these data, we note the distance x that makes  a maximum must be between 15 and 35 ft, 0 and using 2nd TRACE (CALC) 4:maximum shows  is a maximum of 36.9° at a distance of 20 ft from the screen (see Figure 6.19).

c. After we enter Y1  tan1a

E. You’ve just learned how to solve applications involving inverse functions

100

0

Now try Exercises 89 through 95



TECHNOLOGY HIGHLIGHT

More on Inverse Functions The domain and range of the inverse functions for sine, cosine, and tangent are preprogrammed into most graphing calculators, making them an ideal tool for reinforcing the concepts involved. In particular, sin x  y implies that sin1y  x only if 90°  y  90° and 1  x  1. For a stark reminder of this fact we’ll use the TABLE feature of the grapher. Begin by using the TBLSET screen ( 2nd WINDOW

Figure 6.20

) to set TblStart  90 with ¢Tbl  30. After placing the

calculator in degree MODE , go to the Y = screen and input Y1  sin x, Y2  sin1x, and Y3  Y2 1Y1 2 (the composition Y2  Y1). Figure 6.21 Then disable Y2 [turn it off—Y3 will read it anyway) so that both Y1 and Y3 will be displayed simultaneously on the TABLE screen. Pressing GRAPH brings up the TABLE shown in Figure 6.20, where we 2nd note the inputs are standard angles, the outputs in Y1 are the (expected) standard values, and the outputs in Y3 return the original standard values. Now scroll upward until 180° is at the top of the X column (Figure 6.21), and note that Y3 continues to return standard angles from the interval 390°, 90°4 —a stark reminder that while the expression sin 150°  0.5, sin1 1sin 150°2  150°. Once again we note that while sin1 1sin 150°2 can be evaluated, it cannot be evaluated directly using the inverse function properties. Use these ideas to complete the following exercises. Exercise 1: Go through an exercise similar to the one here using Y1  cos x and Y2  cos1x. Remember to modify the TBLSET to accommodate the restricted domain for cosine. Exercise 2: Complete parts (a) and (b) using the TABLE from Exercise 1. Complete parts (c) and (d) without a calculator. 1 a. cos 1cos 150°2 1 c. cos 1cos 120°2

1 b. cos 1cos 210°2 1 d. cos 1cos 240°2

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6.5 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. All six trigonometric functions fail the test and therefore are not to .

5. Most calculators do not have a key for evaluating an expression like sec15. Explain how it is done using the key.

2. The two most common ways of writing the inverse function for y  sin x are and .

6. Discuss/Explain what is meant by the implicit form of an inverse function and the explicit form. Give algebraic and trigonometric examples.

3. The domain for the inverse sine function is and the range is . 

4. The domain for the inverse cosine function is and the range is .

DEVELOPING YOUR SKILLS

The tables here show values of sin , cos , and tan  for   [180 to 210]. The restricted domain used to develop the inverse functions is shaded. Use the

information from these tables to complete the exercises that follow.

y  cos 

y  sin 

y  tan 



sin 



sin 



cos 



cos 



tan 



tan 

180°

0

30°

1 2

180°

1

30°

13 2

180°

0

30°

13 3

150°



1 2

60°

13 2

150°

13 2

60°

1 2

150°

13 3

60°

13

13 2

90°

1

120°



1 2

90°

0

120°

13

90°

—

120°

13 2

90°

0

120°



1 2

90°

—

120°

 13

13 2

150°

1 2

60°

1 2

150°

13 2

60°

13

150°



1 2

180°

0

30°

13 2

180°

30°



13 3

180°

0

210°



0

1

210°

0

210°

13

120°



90° 60°

1 

30°



0

0

1 2





1 

13 2

0

Use the preceding tables to fill in each blank (principal values only). 7.

sin 0  0  sin a b  6

sin10 

1  arcsin a b  2 6

8.

sin   0 sin 120° 

13 2

5 1 b 6 2

1 sin1a b  2

sin 160°2  

 sin a b  1 2

sin1 112 

sin 180° 

sin a

sin10  sin1a

13 b 2

13 13 b arcsina 2 2 arcsin 0  0

13 3

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6-52

CHAPTER 6 Trigonometric Identities, Inverses, and Equations

Evaluate without the aid of calculators or tables, keeping the domain and range of each function in mind. Answer in radians.

12 b 9. sin1a 2

13 b 10. arcsina 2 1 12. arcsina b 2

11. sin11

Evaluate using a calculator, keeping the domain and range of each function in mind. Answer in radians to the nearest ten-thousandth and in degrees to the nearest tenth.

7 14. arcsina b 8

13. arcsin 0.8892 15. sin

1

1

Evaluate each expression.

17. sin c sin1a

12 bd 2

18. sin c arcsina

 19. arcsin c sina b d 3 1

21. sin

23. sin 1sin

1

13 bd 2

20. sin1 1sin 30°2 2 22. arcsin c sina bd 3

1sin 135°2

3 24. sin c arcsina b d 5

0.82052

1 27. cos1a b 2

29. cos1 112

 cos a b  6 cos 120°  

arccos a

33. cos1a

1

cos   1

26.

cos160°2 

cos 1 2

 13 cos a b  6 2 cos1120°2  cos 122  1

112 

1 cos1a b  2 13 cos1a b 2 1 arccos a b  120° 2 1

cos 1 

4 32. arccosa b 7

15 b 3

34. cos1a

16  1 b 5

 35. arccos c cosa b d 4

36. cos1 1cos 60°2

37. cos 1cos1 0.55602

38. cos c arccos a

39. cos c cos1a 41. cos1 c cosa

12 bd 2

5 bd 4

40. cos c arccos a

8 bd 17

13 bd 2

42. arccos 1cos 44.2°2

Use the tables presented before Exercise 7 to fill in each blank. Convert from radians to degrees as needed.

43.

tan10 

tan 0  0  tan a b  3 tan 30° 

arctan1132   13 3

 tan a b  3

13  b 2 6

1 arccos a b  2

1 2

30. arccos (0)

31. arccos 0.1352

cos11 

cos 0  1

13 b 2

Evaluate using a calculator. Answer in radians to the nearest ten-thousandth, degrees to the nearest tenth.

Use the tables given prior to Exercise 7 to fill in each blank (principal values only).

25.

28. arccosa

Evaluate each expression.

1  15 b 16. sin a 2

1 a b 17

Evaluate without the aid of calculators or tables. Answer in radians.

44.

tan 1150°2 

13 b 3

tan1 1 132  13 3

tan   0 tan 120°   13  tan a b  4

arctan a

 3

tan1a

13 b 3

tan10 

arctan 1 132  arctan 1 

 4

Evaluate without the aid of calculators or tables.

45. tan1a

13 b 3

47. arctan1 132

46. arctan112 48. tan10

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Evaluate using a calculator, keeping the domain and range of each function in mind. Answer in radians to the nearest ten-thousandth and in degrees to the nearest tenth.

49. tan1 12.052

50. tan1 10.32672

29 51. arctana b 21

52. arctan1162

53. sin1 c cosa

69. sin c cos1a

5 bd 2x

1 56. sec c arcsina b d 2

76. tan c sec1a

29  x2 bd x

12 bd 2

1 58. cot c cos1a b d 2 60. arcsin1cos 135°2

1

64. cos

77.

arcsec 2 

sec130°2 

2 13

78.

sec160°2  2 seca

10√3

7 2 b 6 13

2 b 13

arcsec 2  arcseca

2 b 13

arcsec 1  sec1 2  60°

sec160°2 

20

arcseca

 3

sec1 112  

sec12 

sec1360°2  1

0.4

bd

sec11 

sec 0  1  seca b  3

2 c seca b d 3

10

x

Use the tables given prior to Exercise 7 to help fill in each blank.

62. cot(arccos 1)

66.

␪

␪

0.3

Evaluate using a calculator only as necessary. √x 2

6

74. tan c arcseca

13 bd 2

Use the diagrams below to write the value of: (a) sin , (b) cos , and (c) tan .

67.

123 bd 12

212  x2

 c csca b d 4

0.5

3x bd 5

11 bd 61

72. tan c cos1a

75. cos c sin1a

61. tan1sin112

65.

15 bd 2

73. cot c arcsina

Explain why the following expressions are not defined.

63. sin

70. cos c sin1a

 54. cos1 c sina b d 3

59. arccos3 sin130°2 4

1

7 bd 25

2 bd 3

55. tan c arccosa 57. csc c sin1a

Evaluate each expression by drawing a right triangle and labeling the sides.

71. sin c tan1a

Simplify each expression without using a calculator.

⫺ 36

x

68. √100 ⫹ 9x2

10

␪ 3x



595

Section 6.5 The Inverse Trig Functions and Their Applications

␪

79. arccsc 2

80. csc1a

2 b 13

81. cot1 13

82. arccot112

83. arcsec 5.789

84. cot1a

85. sec1 17

86. arccsc 2.9875

17 b 2

WORKING WITH FORMULAS

87. The force normal to an object on an inclined plane: FN  mg cos  When an object is on an inclined plane, the normal force is the force acting perpendicular to the plane and away from the force of gravity, and is measured in a unit called newtons (N). The magnitude of this force depends on the angle of incline of the plane according to the formula

above, where m is the mass of the object in kilograms and g is the force of gravity (9.8 m/sec2). Given m  225 g, find (a) FN for   15° and   45° and (b)  for FN  1 N and FN  2 N. FN

FN 225 ␪

g

g

5k

kg

22

␪

g

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88. Heat flow on a cylindrical pipe: T  1T0  TR 2 sina Fan When a circular pipe is exposed to a fan-driven source of heat, the Heat source temperature of the air reaching the pipe is y greatest at the point nearest to the source (see diagram). As you move around the r circumference of the pipe away from the Pipe source, the temperature of the air reaching the x2  y 2  r 2 pipe gradually decreases. One



6-54

CHAPTER 6 Trigonometric Identities, Inverses, and Equations

x

y 2x  y2 2

b  TR; y  0

possible model of this phenomenon is given by the formula shown, where T is the temperature of the air at a point (x, y) on the circumference of a pipe with outer radius r  2x2  y2, T0 is the temperature of the air at the source, and TR is the surrounding room temperature. Assuming T0  220°F, TR  72° and r  5 cm: (a) Find the temperature of the air at the points (0, 5), (3, 4), (4, 3), (4.58, 2), and (4.9, 1). (b) Why is the temperature decreasing for this sequence of points? (c) Simplify the formula using r  5 and use it to find two points on the pipe’s circumference where the temperature of the air is 113°.

APPLICATIONS

89. Snowcone dimensions: Made in the Shade Snowcones sells a colossal size cone that uses a conical cup holding 20 oz of ice and liquid. The cup is 20 cm tall and has a radius of 5.35 cm. Find the angle  formed by a cross-section of the cup.

Exercise 89 5.35 cm

20 cm

␪

90. Avalanche conditions: Winter avalanches occur for many reasons, one being the slope of the mountain. Avalanches Exercise 90 seem to occur most often for slopes between 35° and 60° (snow gradually slides off steeper slopes). The slopes at a local ski resort have an 2000 ft average rise of 2000 ft for each horizontal run of 2559 ft. Is this resort ␪ prone to avalanches? 2559 ft Find the angle  and respond.

Exercise 91 91. Distance to hole: A popular story on the PGA Tour has Gerry Yang, Tiger H Woods’ teammate at Stanford and occasional caddie, using the Pythagorean theorem to find the distance Tiger needed to reach a 150 yd particular hole. Suppose you notice a marker in the ground stating that M ␪ the straight line B Marker 48 yd distance from the marker to the hole (H ) is 150 yd. If your ball B is 48 yd from the marker (M ) and angle BMH is a right angle, determine the angle  and your straight line distance from the hole.

92. Ski jumps: At a waterskiing contest on a large lake, skiers use a ramp rising out of the water that is 30 ft long and 10 ft high at the high end. What angle  does the ramp make with the lake?

Exercise 92

10 ft

␪ 30 ft

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Section 6.5 The Inverse Trig Functions and Their Applications

93. Viewing angles for advertising: A 25-ft-wide billboard is erected perpendicular to a straight highway, with the closer edge 50 ft away (see figure). Assume the advertisement on the billboard is most easily read when the viewing angle is 10.5° or more. (a) Use inverse functions to find an expression for the viewing angle . (b) Use a calculator to help determine the distance d (to tenths of a foot) for which the viewing angle is greater than 10.5°. (c) What distance d maximizes this viewing angle?

Exercise 94

2.5 ft

␣

1.5 ft

␪ ␤

x

Exercise 93

calculator to help determine the distance x (to tenths of a foot) that maximizes this viewing angle.

50 ft 25 ft

95. Shooting angles and shots on goal: A soccer player is on a breakaway and is dribbling just inside the right sideline toward the opposing goal (see figure). As the defense closes in, she has just a few seconds to decide when to shoot. (a) Use inverse functions to find an expression for the shooting angle . (b) Use a calculator to help determine the distance d (to tenths of a foot) that will maximize the shooting angle for the dimensions shown.

d ␪

Exercise 95 24 ft

70 ft

(goal area)

94. Viewing angles at an art show: At an art show, a painting 2.5 ft in height is hung on a wall so that its base is 1.5 ft above the eye level of an average viewer (see figure). (a) Use inverse functions to find expressions for angles  and . (b) Use the result to find an expression for the viewing angle . (c) Use a

(penalty area)

␪

d

(sideline) 

EXTENDING THE CONCEPT

Consider a satellite orbiting at an altitude of x mi above the Earth. The distance d from the satellite to the horizon and the length s of the corresponding arc of the Earth are shown in the diagram. Earth

96. To find the distance d we use the formula d  22rx  x2. (a) Show how this formula was developed using the

Pythagorean theorem. (b) Find a formula for the angle  in terms of r and x, then a formula for the arc length s. d

x s r

␪ r Center of the Earth

97. If the Earth has a radius of 3960 mi and the satellite is orbiting at an altitude of 150 mi, (a) what is the measure of angle ? (b) how much longer is d than s?

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A projectile is any object that is shot, thrown, slung, or otherwise projected and has no continuing source of propulsion. The horizontal and vertical position of the projectile depends on its initial velocity, angle of projection, and height of release (air resistance is neglected). The horizontal position of the projectile is given by x  v0cos  t, while its vertical position is modeled by y  y0  v0sin  t  16t2, where y0 is the height it is projected from,  is the projection angle, and t is the elapsed time in seconds.

98. A circus clown is shot out of a specially made cannon at an angle of 55°, 55° 10 ft with an initial velocity of 85 ft/sec, and the end of the cannon is 10 ft high. a. Find the position of the safety net (distance from the cannon and height from the ground) if the clown hits the net after 4.3 sec. b. Find the angle at which the clown was shot if the initial velocity was 75 ft/sec and the clown hits a net that is placed 175.5 ft away after 3.5 sec.



6-56 Exercise 99 99. A winter ski jumper leaves the ski-jump with an initial velocity of 70 ft/sec at an 10 70 ft/sec (0, 0). angle of 10°. Assume the jump-off point has coordinates (0, 0). a. What is the horizontal position of the skier after 6 sec? b. What is the vertical position of the skier after 6 sec? c. What diagonal distance (down the mountain side) was traveled if the skier touched down after being airborne for 6 sec?

100. Suppose the domain of y  csc x was restricted to   x  c  , 0b ´ a0, d , and the domain of 2 2 y  cot x to x  10, 2. (a) Would these functions then be one-to-one? (b) What are the corresponding ranges? (c) State the domain and range of y  csc1x and y  cot1x. (d) Graph each function.

MAINTAINING YOUR SKILLS 103. (3.7) Solve the inequality f 1x2  0 using zeroes and end behavior given f 1x2  x3  9x.

101. (6.4) Use the triangle given with a double-angle identity to find the exact value of sin122. ␣

␪ √85

89

6

39

␤ 7

80

102. (6.3) Use the triangle given with a sum identity to find the exact value of sin1  2.

104. (2.3) In 2000, Space Tourists Inc. sold 28 low-orbit travel packages. By 2005, yearly sales of the loworbit package had grown to 105. Assuming the growth is linear, (a) find the equation that models this growth 12000 S t  02 , (b) discuss the meaning of the slope in this context, and (c) use the equation to project the number of packages that will be sold in 2010.

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Precalculus—

6.6 Solving Basic Trig Equations Learning Objectives In Section 6.6 you will learn how to:

A. Use a graph to gain information about principal roots, roots in [0, 2), and roots in 

B. Use inverse functions to solve trig equations for the principal root

In this section, we’ll take the elements of basic equation solving and use them to help solve trig equations, or equations containing trigonometric functions. All of the algebraic techniques previously used can be applied to these equations, including the properties of equality and all forms of factoring (common terms, difference of squares, etc.). As with polynomial equations, we continue to be concerned with the number of solutions as well as with the solutions themselves, but there is one major difference. There is no “algebra” that can transform a function like sin x  12 into x  solution. For that we rely on the inverse trig functions from Section 6.5.

A. The Principal Root, Roots in [0, 2 ), and Real Roots

C. Solve trig equations for roots in [0, 2 ) or [0, 360°)

In a study of polynomial equations, making a connection between the degree of an equation, its graph, and its possible roots, helped give insights as to the number, location, and nature of the roots. Similarly, keeping graphs of basic trig functions constantly in mind helps you gain information regarding the solutions to trig equations. When solving trig equations, we refer to the solution found using sin1, cos1, and tan1 as the principal root. You will alternatively be asked to find (1) the principal root, (2) solutions in 30, 22 or 30°, 360°2, or (3) solutions from the set of real numbers . For convenience, graphs of the basic sine, cosine, and tangent functions are repeated in Figures 6.22 through 6.24. Take a mental snapshot of them and keep them close at hand.

D. Solve trig equations for roots in 

E. Solve trig equations using fundamental identities

F. Solve trig equations using graphing technology

Figure 6.22 1

Figure 6.24

Figure 6.23

y

y

y y  sin x

1

y  cos x

3

y  tan x

2 2





2

2

x

1



 1

2

1

x 2



1



2

x

2 3

EXAMPLE 1



Visualizing Solutions Graphically Consider the equation sin x  23. Using a graph of y  sin x and y  23, a. State the quadrant of the principal root. b. State the number of roots in 3 0, 22 and their quadrants. c. Comment on the number of real roots.

Solution WORTHY OF NOTE  b 2 as Quadrant I or QI, regardless of whether we’re discussing the unit circle or the graph of the function. In Example 1b, the solutions correspond to those found in QI and QII on the unit circle, where sin x is also positive. Note that we refer to a0,



y We begin by drawing a quick sketch of y  sin x 2 y  sin x 1 and y  3, noting that solutions will occur where ys the graphs intersect. a. The sketch shows the principal root occurs 2   2 x  between 0 and in QI. 1 2 b. For 3 0, 22 we note the graphs intersect twice and there will be two solutions in this interval. c. Since the graphs of y  sin x and y  23 extend infinitely in both directions, they will intersect an infinite number of times — but at regular intervals! Once a root is found, adding integer multiples of 2 (the period of sine) to this root will give the location of additional roots.

Now try Exercises 7 through 10 6-57



599

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CHAPTER 6 Trigonometric Identities, Inverses, and Equations

A. You’ve just learned how to use a graph to gain information about principal roots, roots in [0, 2 ), and roots in 

When this process is applied to the equation tan x  2, the graph shows the principal root occurs  between  and 0 in QIV (see Figure 6.25). In the 2 interval 30, 22 the graphs intersect twice, in QII and QIV where tan x is negative (graphically—below the x-axis). As in Example 1, the graphs continue infinitely and will intersect an infinite number of times— but again at regular intervals! Once a root is found, adding integer multiples of  (the period of tangent) to this root will give the location of other roots.

Figure 6.25 y 3

y  tan x

2 1 2␲

␲

y  2

1

␲

2␲

x

2 3

B. Inverse Functions and Principal Roots To solve equations having a single variable term, the basic goal is to isolate the variable term and apply the inverse function or operation. This is true for algebraic equations like 2x  1  0, 2 1x  1  0, or 2x2  1  0, and for trig equations like 2 sin x  1  0. In each case we would add 1 to both sides, divide by 2, then apply the appropriate inverse. When the inverse trig functions are applied, the result is only the principal root and other solutions may exist depending on the interval under consideration. EXAMPLE 2



Finding Principal Roots Find the principal root of 13 tan x  1  0.

Solution



We begin by isolating the variable term, then apply the inverse function. 13 tan x  1  0

B. You’ve just learned how to use inverse functions to solve trig equations for the principal root

tan x 

Table 6.2 

sin 

cos 

0

0

1

 6  4

1 2 12 2

13 2 12 2

 3

13 2

1 2

1

0

 2 2 3

13 2



3 4

12 2

12  2

5 6

1 2



0



1 2

13 2

1

1 13

tan1 1tan x2  tan1a x

given equation

 6

add 1 and divide by 13

1 b 13

apply inverse tangent to both sides

result (exact form)

Now try Exercises 11 through 28



Equations like the one in Example 2 demonstrate the need to be very familiar with the functions of a special angle. They are frequently used in equations and applications to ensure results don’t get so messy they obscure the main ideas. For convenience, the values of sin  and cos  are repeated in Table 6.2 for x  3 0,  4 . Using symmetry and the appropriate sign, the table can easily be extended to all values in 30, 22 . Using the reciprocal and ratio relationships, values for the other trig functions can also be found.

C. Solving Trig Equations for Roots in [0, 2 ) or [0, 360) To find multiple solutions to a trig equation, we simply take the reference angle of the principal root, and use this angle to find all solutions within a specified range. A mental image of the graph still guides us, and the standard table of values (also held in memory) allows for a quick solution to many equations.

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Section 6.6 Solving Basic Trig Equations

EXAMPLE 3



Finding Solutions in [0, 2)

Solution



Isolate the variable term, then apply the inverse function.

For 2 cos   12  0, find all solutions in 3 0, 22. 2 cos   12  0

given equation

12 cos    2

WORTHY OF NOTE Note how the graph of a trig function displays the information regarding quadrants. From the graph of y  cos x we “read” that cosine is negative in QII and QIII [the lower “hump” of the graph is below the x-axis in 1/2, 3/22 ] and positive in QI and QIV [the graph is above the x-axis in the intervals 10, /22 and 13/2, 22 ].

subtract 12 and divide by 2

cos1 1cos 2  cos1a 

12 b 2

3 4

apply inverse cosine to both sides

result

3  as the principal root, we know r  . 4 4 Since cos x is negative in QII and QIII, the second 5 . The second solution could also have solution is 4 been found from memory, recognition, or symmetry on the unit circle. Our (mental) graph verifies these are the only solutions in 30, 22 . With

1

2␲

y y  cos x

␲

␲

2␲

x

√2

1

y  2

Now try Exercises 29 through 34 EXAMPLE 4



Finding Solutions in [0, 2)

Solution



As with the other equations having a single variable term, we try to isolate this term or attempt a solution by factoring.



For tan2x  1  0, find all solutions in 30, 22 .

tan2x  1  0 2tan2x   11 tan x  1

given equation add 1 to both sides and take square roots result

The algebra gives tan x  1 or tan x  1 and we solve each equation independently. tan x  1 tan1 1tan x2  tan1 112  x 4

tan x  1 tan1 1tan x2  tan1 112  x 4

 is in the 4 specified interval. With tan x positive in QI and  5 . While x   is QIII, a second solution is 4 4 not in the interval, we still use it as a reference angle in QII and QIV (for tan x  12 and find 3 7 . The four solutions the solutions x  and 4 4  3 5 7 , , which is supported are x  , , and 4 4 4 4 by the graph shown. Of the principal roots, only x 

apply inverse tangent principal roots

y 3

y  tan x

2

y1

y 1

1

␲

2␲

x

1 2 3

Now try Exercises 35 through 42



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CHAPTER 6 Trigonometric Identities, Inverses, and Equations

For any trig function that is not equal to a standard value, we can use a calculator to approximate the principal root or leave the result in exact form, and apply the same ideas to this root to find all solutions in the interval.

EXAMPLE 5



Finding Solutions in [0, 360°)

Solution



Use a u-substitution to simplify the equation and help select an appropriate strategy. For u  cos , the equation becomes 3u2  u  2  0 and factoring seems the best approach. The factored form is 1u  12 13u  22  0, with solutions u  1 and u  23. Re-substituting cos  for u gives

Find all solutions in 30°, 360°2 for 3 cos2  cos   2  0.

cos   1

cos  

equations from factored form

2 cos1 1cos 2  cos1a b 3   48.2°

cos1 1cos 2  cos1 112   180°

C. You’ve just learned how to solve trig equations for roots in [0, 2 ) or [0, 360°)

2 3

Both principal roots are in the specified interval. The first is quadrantal, the second was found using a calculator and is approximately 48.2°. With cos x positive in QI and QIV, a second solution is 1360  48.22°  311.8°. The three solutions are 48.2°, 180°, and 311.8° although only   180° is exact.

apply inverse cosine principal roots y y  cos x

1

360

ys

180

180 1

360

x

y 1

Now try Exercises 43 through 50



D. Solving Trig Equations for All Real Roots () As we noted, the intersections of a trig function with a horizontal line occur at regular, predictable intervals. This makes finding solutions from the set of real numbers a simple matter of extending the solutions we found in 3 0, 22 or 3 0°, 360°2. To illustrate, consider 3 the solutions to Example 3. For 2 cos   12  0, we found the solutions   4 5 . For solutions in , we note the “predictable interval” between roots is and   4 identical to the period of the function. This means all real solutions will be represented by 3 5   2k and    2k, k   (k is an integer). Both are illustrated in 4 4 Figures 6.26 and 6.27 with the primary solution indicated with a “*.”

Figure 6.26 y

  f  2k

y  cos 

1

43 2 

Figure 6.27



2 3 4

etc.

etc. 2

s 4

 x

2 h

43 2 



2 3 4

etc.

 x

etc. 2

z 4

y  cos x

1

2 f

y

  h  2k

z 4

2 f

2 h

m 4

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Section 6.6 Solving Basic Trig Equations

EXAMPLE 6



Finding Solutions in  Find all real solutions to 13 tan x  1  0.

Solution



y

y  tan x z

etc. 

In Example 2 we found the  principal root was x  . Since the 3 6 tangent function has a period of , adding integer multiples of  to this root will identify all solutions:  x   k, k  , as illustrated here. 6

3

l



2



k

'





m

 

 etc. 2

3

y  0.58

x

3

Now try Exercises 51 through 56



These fundamental ideas can be extended to many different situations. When asked to find all real solutions, be sure you find all roots in a stipulated interval before naming solutions by applying the period of the function. For instance, cos x  0 has two  3 d , which we can quickly extend to find all real solutions in 3 0, 22 c x  and x  2 2 roots. But using x  cos10 or a calculator limits us to the single (principal) root 3  x  , and we’d miss all solutions stemming from . Note that solutions involving 2 2 multiples of an angle (or fractional parts of an angle) should likewise be “handled with care,” as in Example 7. EXAMPLE 7



Finding Solutions in 

Solution



Since we have a common factor of cos x, we begin by rewriting the equation as cos x 3 2 sin12x2  1 4  0 and solve using the zero factor property. The resulting equations are cos x  0 and 2 sin12x2  1  0 S sin 12x2  12.

WORTHY OF NOTE When solving trig equations that involve arguments other than a single variable, a u-substitution is sometimes used. For Example 7, substituting u for 2x gives the 1 equation sin u  , making 2  it “easier to see” that u  6 1 (since is a special value), 2  and therefore 2x  and 6  x . 12

D. You’ve just learned how to solve trig equations for roots in 

Find all real solutions to 2 sin 12x2 cos x  cos x  0.

cos x  0

sin 12x2 

In 30, 22, cos x  0 has solutions x 

1 2

equations from factored form

 3  and x  , giving x   2k 2 2 2

3  2k as solutions in . Note these can actually be combined and 2  1 written as x   k, k  . The solution process for sin 12x2  yields 2 2  5 2x  and 2x  . Since we seek all real roots, we first extend each solution 6 6 by 2k before dividing by 2, otherwise multiple solutions would be overlooked. and x 

  2k 6  x  k 12

2x 

5  2k 6 5 x  k 12

2x 

solutions from sin 12x2  12 ; k   divide by 2

Now try Exercises 57 through 66



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CHAPTER 6 Trigonometric Identities, Inverses, and Equations

E. Trig Equations and Trig Identities In the process of solving trig equations, we sometimes employ fundamental identities to help simplify an equation, or to make factoring or some other method possible. EXAMPLE 8



Solving Trig Equations Using an Identity

Solution



With a mixture of functions, exponents, and arguments, the equation is almost impossible to solve as it stands. But we can eliminate the sine function using the identity cos 122  cos2  sin2, leaving a quadratic equation in cos x.

Find all solutions in 30°, 360°2 for cos 122  sin2  3 cos   1.

cos122  sin2  3 cos   1 cos2  sin2  sin2  3 cos   1 cos2  3 cos   1 cos2  3 cos   1  0

given equation substitute cos2  sin2 for cos122 combine like terms subtract 1

Let’s substitute u for cos  to give us a simpler view of the equation. This gives u2  3u  1  0, which is clearly not factorable over the integers. Using the quadratic formula with a  1, b  3, and c  1 gives

E. You’ve just learned how to solve trig equations using fundamental identities

3  2132 2  4112 112 u quadratic formula in u 2112 3  113  simplified 2 To four decimal places we have u  3.3028 and u  0.3028. To answer in terms of the original variable we re-substitute cos  for u, realizing that cos   3.3028 has no solution, so solutions in 3 0°, 360°2 must be provided by cos   0.3028 and occur in QII and QIII. The solutions are   cos1 10.30282  107.6° and 360°  107.6°  252.4° to the nearest tenth of a degree. Now try Exercises 67 through 82



F. Trig Equations and Graphing Technology A majority of the trig equations you’ll encounter in your studies can be solved using the ideas and methods presented here. But there are some equations that cannot be solved using standard methods because they mix polynomial functions (linear, quadratic, and so on) that can be solved using algebraic methods, with what are called transcendental functions (trigonometric, logarithmic, and so on). By definition, transcendental functions are those that transcend the reach of standard algebraic methods. These kinds of equations serve to highlight the value of graphing and calculating technology to today’s problem solvers. EXAMPLE 9



Solving Trig Equations Using Technology Use a graphing calculator in radian mode to find all real roots of 3x 2 sin x   2  0. Round solutions to four decimal places. 5

Solution



When using graphing technology our initial concern is the size of the viewing window. After carefully entering the equation on the Y = screen, we note the term 2 sin x will never be larger than 2 or less than 2 for any real number x. On the 3x other hand, the term becomes larger for larger values of x, which would seem 5 3x to cause 2 sin x  to “grow” as x gets larger. We conclude the standard 5 window is a good place to start, and the resulting graph is shown in Figure 6.28.

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Figure 6.28

Figure 6.29 10

10

10

0

10

10

10

F. You’ve just learned how to solve trig equations using graphing technology

20

From this screen it appears there are three real roots, but to be sure none are hidden to the right, we extend the Xmax value to 20 (Figure 6.29). Using 2nd TRACE CALC 2:zero, we follow the prompts and enter a left bound of 0 (a number to the left of the zero) and a right bound of 2 (a number to the right of the zero—see Figure 6.29). If you can visually approximate the root, the calculator prompts you for a GUESS, otherwise just bypass the request by pressing ENTER . The smallest root is approximately x  0.8435. Repeating this sequence we find the other roots are x  3.0593 and x  5.5541. Now try Exercises 83 through 88



TECHNOLOGY HIGHLIGHT

Solving Equations Graphically Some equations are very difficult to solve analytically, and even with the use of a graphing calculator, a strong combination of analytical skills with technical skills is required to state the solution set. Consider 1 1 the equation 5 sin a xb  5  cot a xb and solutions in 32, 22 . There appears to be no quick analytical 2 2 solution, and the first attempt at a graphical solution holds Figure 6.30 1 4 some hidden surprises. Enter Y1  5 sin a xb  5 and 2 1 Y2  on the Y = screen. Pressing ZOOM 7:ZTrig gives tan1 12x2 the screen in Figure 6.30, where we note there are at least two and possibly three solutions, depending on how the sine graph intersects the cotangent graph. We are also uncertain as to   whether the graphs intersect again between  and . 2 2 Increasing the maximum Y-value to Ymax  8 shows they do indeed. But once again, are there now three or four solutions? In situations like this it may be helpful to use the zeroes method for solving graphically. On the Y = screen, disable Y1 and Y2 and enter Y3 as Y1  Y2. Pressing ZOOM 7:ZTrig at this point clearly shows that there are four solutions in this interval CALC (Figure 6.31), which can easily be found using 2nd 2:zero: x  5.7543, 4.0094, 3.1416, and 0.3390. Use these ideas to find solutions in 32, 22 for the exercises that follow. Exercise 1:

11  sin x2 2  cos 12x2  4 cos x11  sin x2

Exercise 2:

x 4 sin x  2 cos2 a b 2

2

2

4

Figure 6.31 4

2

2

4

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6.6 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if necessary.

4. For tan x  1, the principal root is solutions in 30, 22 are and and an expression for all real roots is

1. For simple equations, a mental graph will tell us the quadrant of the root, the number of roots in , and show a pattern for all roots.

12 the principal root is 2 solutions in 3 0, 22 are and and an expression for all real roots is ; k  .



, .

3 and 4 y  cos x have two solutions in 30, 22, even though the period of y  tan x is , while the period of y  cos x is 2.

5. Discuss/Explain/Illustrate why tan x 

2. Solving trig equations is similar to solving algebraic equations, in that we first the variable term, then apply the appropriate function. 3. For sin x 

,

1 has four solutions in 2 30, 22 . Explain how these solutions can be viewed as the vertices of a square inscribed in the unit circle.

6. The equation sin2x 

, , and

DEVELOPING YOUR SKILLS 3 7. For the equation sin x   and the graphs of 4 3 y  sin x and y   given, state (a) the quadrant 4 of the principal root and (b) the number of roots in 3 0, 22. Exercise 7 1

2␲

␲

Exercise 8

y y  sin x

␲

1

2␲

2␲ x

y 1

  34

␲

y y  cos x 3

yE

␲

2␲ x

1

8. For the equation cos x  34 and the graphs of y  cos x and y  34 given, state (a) the quadrant of the principal root and (b) the number of roots in 3 0, 22.

9. Given the graph y  tan x shown here, draw the horizontal line y  1.5 and then for tan x  1.5, state (a) the quadrant of the principal root and (b) the number of roots in 3 0, 22.

Exercise 9 3

2␲

Exercise 10

y y  tan x

3

2

2

1

1

␲

␲

2␲ x

1

2␲ 3␲␲ ␲ 21 2

2

2

3

3

y

y  sec x

␲ 2

␲ 3␲ 2␲ x 2

10. Given the graph of y  sec x shown, draw the horizontal line y  54 and then for sec x  54 , state (a) the quadrant of the principal root and (b) the number of roots in 3 0, 22.  11. The table that follows shows  in multiples of 6 4 between 0 and , with the values for sin  given. 3 Complete the table without a calculator or references using your knowledge of the unit circle, the signs of f 12 in each quadrant, memory/recognition, sin  tan   , and so on. cos 

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Exercise 11 

sin 

0

0

0

1

 6

1 2

 4

12 2

 3

13 2

 2

0

 2

1

3 4

2 3

13 2



5 6

1 2

5 4

 7 6 4 3

cos 

Exercise 12 tan 



sin 

cos  tan 



12 2

1

38. 3 sec2x  6

39. 4 csc2x  8

40. 6 13 cos2x  3 13

41. 4 12 sin2x  4 12

42.

607

5 4 2 cos2x   3 6 3

Solve the following equations by factoring. State all real solutions in radians using the exact form where possible and rounded to four decimal places if the result is not a standard value.

43. 3 cos2  14 cos   5  0

12 2

44. 6 tan2  2 13 tan   0

0

3 2

0

46. 2 sin2x  7 sin x  4 47. sec2x  6 sec x  16

1  2

7 4

12 2

2

1



13 2



 between 0 and 4 2, with the values for cos  given. Complete the table without a calculator or references using your knowledge of the unit circle, the signs of f 12 in sin  each quadrant, memory/recognition, tan   , cos  and so on.

12. The table shows  in multiples of

Find the principal root of each equation.

45. 2 cos x sin x  cos x  0 48. 2 cos3x  cos2x  0

49. 4 sin2x  1  0

50. 4 cos2x  3  0 Find all real solutions. Note that identities are not required to solve these exercises.

51. 2 sin x  12

52. 2 cos x  1

53. 4 cos x  212

54. 4 sin x  213

55. 13 tan x   13

56. 213 tan x  2

57. 6 cos12x2  3

58. 2 sin13x2   12

59. 13 tan12x2   13

60. 2 13 tan13x2  6

1 61. 213 cosa xb  2 13 3

13. 2 cos x  12

14. 2 sin x  1

15. 4 sin x  2 12

16. 4 cos x  2 13

1 62. 8 sina xb  4 13 2

17. 13 tan x  1

18. 213 tan x  2

19. 2 13 sin x  3

20. 3 12 csc x  6

63. 12 cos x sin12x2  3 cos x  0

21. 6 cos x  6

22. 4 sec x  8

7 7 23. cos x  8 16

5 5 24.  sin x  3 6

25. 2  4 sin 

26.  tan x  0

27. 513  10 cos 

28. 4 13  4 tan 

Find all solutions in [0, 2 ).

29. 9 sin x  3.5  1

30. 6.2 cos x  4  7.1

31. 8 tan x  713   13 1 3 7 32. sec x    2 4 4

2 5 3 33. cot x    3 6 2

34. 110 sin x  5513 35. 4 cos2x  3 36. 4 sin2x  1

37. 7 tan2x  21

64. 13 sin x tan12x2  sin x  0 65. cos13x2 csc12x2  2 cos13x2  0

66. 13 sin12x2 sec 12x2  2 sin12x2  0 Solve each equation using calculator and inverse trig functions to determine the principal root (not by graphing). Clearly state (a) the principal root and (b) all real roots.

67. 3 cos x  1

68. 5 sin x  2

69. 12 sec x  3  7

70. 13 csc x  2  11

71.

1 1 sin122  2 3

73. 5 cos122  1  0

72.

2 1 cos122  5 4

74. 6 sin122  3  2

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Solve the following equations using an identity. State all real solutions in radians using the exact form where possible and rounded to four decimal places if the result is not a standard value.

75. cos2x  sin2x 

1 2

1 1 77. 2 cosa xbcos x  2 sina xbsin x  1 2 2 78. 12 sin 12x2 cos 13x2  12 sin13x2cos12x2  1

80. 1cos   sin 2 2  2

81. cos122  2 sin2  3 sin   0 82. 3 sin122  cos2 122  1  0

83. 5 cos x  x  3

85. cos 12x2  x  3 2

87. x2  sin12x2  1

84. 3 sin x  x  4

86. sin2 12x2  2x  1

88. cos12x2  x2  5

WORKING WITH FORMULAS

89. Range of a projectile: R 

5 2 v sin122 49

The distance a projectile travels is called its range and is modeled by the formula shown, where R is the range in meters, v is the initial velocity in meters per second, and  is the angle of release. Two friends are standing 16 m apart playing catch. If the first throw has an initial velocity of 15 m/sec, what two angles will insure the ball travels the 16 m between the friends?



79. 1cos   sin 2 2  1

Find all roots in [0, 2 ) using a graphing calculator. State answers in radians rounded to four decimal places.

76. 4 sin2x  4 cos2x  213



6-66

CHAPTER 6 Trigonometric Identities, Inverses, and Equations

90. Fine-tuning a golf swing: 1club head to shoulder2 2  1club length2 2  1arm length2 2  2 1club length21arm length2cos  A golf pro is taking 37 in. specific measurements on a client’s swing to help 39 in.  improve her game. If the angle  is too small, the ball is hit late and “too 27 in. thin” (you top the ball). If  is too large, the ball is hit early and “too fat” (you scoop the ball). Approximate the angle  formed by the club and the extended (left) arm using the given measurements and formula shown.

APPLICATIONS

Acceleration due to gravity: When a steel ball is released down an inclined plane, the rate of the ball’s acceleration depends on the angle of incline. The acceleration can be approximated by the formula A12  9.8 sin , where  is in degrees and the acceleration is measured in meters per second/per second. To the nearest tenth of a degree,

91. What angle produces an acceleration of 0 m/sec2 when the ball is released? Explain why this is reasonable. 92. What angle produces an acceleration of 9.8 m/sec2? What does this tell you about the acceleration due to gravity? 93. What angle produces an acceleration of 5 m/sec2? Will the angle be larger or smaller for an acceleration of 4.5 m/sec2? 94. Will an angle producing an acceleration of 2.5 m/sec2 be one-half the angle required for an acceleration of 5 m/sec2? Explore and discuss.

Exercises 95 to 98 Snell’s law states that when a ray of light passes light incidence reflection from one medium into another, the sine of the   angle of incidence a   varies directly with the refraction sine of the angle of new medium  refraction b (see the figure). This phenomenon sin()  k sin() is modeled by the formula sin   k sin , where k is called the index of refraction. Note the angle  is the angle at which the light strikes the surface, so that   90°  . Use this information to work Exercises 95 to 98.

95. A ray of light passes from air into water, striking the water at an angle of 55°. Find the angle of incidence  and the angle of refraction , if the index of refraction for water is k  1.33.

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96. A ray of light passes from air into a diamond, striking the surface at an angle of 75°. Find the angle of incidence  and the angle of refraction , if the index of refraction for a diamond is k  2.42. 97. Find the index of refraction for ethyl alcohol if a beam of light strikes the surface of this medium at an angle of 40° and produces an angle of refraction   34.3°. Use this index to find the angle of incidence if a second beam of light created an angle of refraction measuring 15°. 98. Find the index of refraction for rutile (a type of mineral) if a beam of light strikes the surface of this medium at an angle of 30° and produces an angle of refraction   18.7°. Use this index to find the angle of incidence if a second beam of light created an angle of refraction measuring 10°. 99. Roller coaster design: As part of a science fair project, Hadra Loading platform builds a scale model of a roller coaster using the equation 1 y  5 sina xb  7, where y is the height of the 2 model in inches and x is the distance from the “loading platform” in inches. (a) How high is the platform? (b) What distances from the platform does the model attain a height of 9.5 in.? 100. Company logo: Part of the logo for an engineering firm was modeled by a cosine function. The logo was then manufactured in steel and installed on the entrance marquee of the home office. The position and size of the logo is modeled by the function y  9 cos x  15, where y is the height of the graph above the base of the marquee in inches and 

609

Section 6.6 Solving Basic Trig Equations

x represents the distance from the edge of the marquee. Assume the graph begins flush with the edge. (a) How far above the base is the beginning of the cosine graph? (b) What distances from the edge does the graph attain a height of 19.5 in.?

Entrance marquee

International Engineering

Geometry applications: Solve Exercises 101 and 102 graphically using a calculator. For Exercise 101, give  in radians rounded to four decimal places. For Exercise 102, answer in degrees to the nearest tenth of a degree.

101. The area of a circular segment (the shaded portion shown) is given by the 1 formula A  r2 1  sin 2, 2 where  is in radians. If the circle has a radius of 10 cm, find the angle  that gives an area of 12 cm2.

r

r



Exercise 102 b

102. The perimeter of a trapezoid with parallel sides B and b, h altitude h, and base angles    and  is given by the formula B P  B  b  h1csc   csc 2. If b  30 m, B  40 m, h  10 m, and   45°, find the angle  that gives a perimeter of 105 m.

EXTENDING THE CONCEPT

103. Find all real solutions to 5 cos x  x  x in two ways. First use a calculator with Y1  5 cos x  x and Y2  x to determine the regular intervals between points of intersection. Second, simplify by adding x to both sides, and draw a quick sketch of the result to locate x-intercepts. Explain why both methods give the same result, even though the first presents you with a very different graph. 104. Once the fundamental ideas of solving a given family of equations is understood and practiced, a

student usually begins to generalize them—making the numbers or symbols used in the equation irrelevant. (a) Use the inverse sine function to find the principal root of y  A sin1Bx  C2  D, by solving for x in terms of y, A, B, C, and D. (b) Solve the following equation using the techniques addressed in this section, and then using the  1 “formula” from Part (a) 5  2 sin a x  b  3. 2 4 Do the results agree?

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MAINTAINING YOUR SKILLS

105. (1.4) Use a substitution to show that x  2  i is a zero of f 1x2  x2  4x  5. 106. (3.1) Currently, tickets to productions of the Shakespeare Community Theater cost $10.00, with an average attendance of 250 people. Due to market research, the theater director believes that for each $0.50 reduction in price, 25 more people will attend. What ticket price will maximize the theater’s revenue? What will the average attendance projected to become at that price?

107. (6.5) Evaluate without using a calculator: 1 a. tan c sin1a b d b. sin 3tan1 112 4 2 108. (5.1) The largest Ferris wheel in the world, located in Yokohama, Japan, has a radius of 50 m. To the nearest hundredth of a meter, how far does a seat on the rim travel as the wheel turns through   292.5°?

6.7 General Trig Equations and Applications Learning Objectives In Section 6.7 you will learn how to:

A. Use additional algebraic techniques to solve trig equations

B. Solve trig equations using multiple angle, sum and difference, and sumto-product identities

At this point you’re likely beginning to understand the true value of trigonometry to the scientific world. Essentially, any phenomenon that is cyclic or periodic is beyond the reach of polynomial (and other) functions, and may require trig for an accurate understanding. And while there is an abundance of trig applications in oceanography, astronomy, meteorology, geology, zoology, and engineering, their value is not limited to the hard sciences. There are also rich applications in business and economics, and a growing number of modern artists are creating works based on attributes of the trig functions. In this section, we try to place some of these applications within your reach, with the exercise set offering an appealing variety from many of these fields.

C. Solve trig equations of the form A sin1Bx  C2  D  k

A. Trig Equations and Algebraic Methods

D. Use a combination of skills to model and solve a variety of applications

We begin this section with a follow-up to Section 6.6, by introducing trig equations that require slightly more sophisticated methods to work out a solution.

EXAMPLE 1



Solving a Trig Equation by Squaring Both Sides

Solution



Our first instinct might be to rewrite the equation in terms of sine and cosine, but that simply leads to a similar equation that still has two different functions 3 13 cos x  sin x  14 . Instead, we square both sides and see if the Pythagorean identity 1  tan2x  sec2x will be of use. Prior to squaring, we separate the functions on opposite sides to avoid the mixed term 2 tan x sec x.

Find all solutions in 30, 22: sec x  tan x  13.

sec x  tan x  13 1sec x2 2  1 13  tan x2 2 sec2x  3  2 13 tan x  tan2x

given equation subtract tan x and square result

Since sec2x  1  tan2x, we substitute directly and obtain an equation in tangent alone. 1  tan2x  3  2 13 tan x  tan2x 2  213 tan x 1  tan x 13 tan x 7 0 in QI and QIII

substitute 1  tan2x for sec2x simplify solve for tan x

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611

7  [QI] and [QIII]. Since squaring an 6 6 equation sometimes introduces extraneous roots, both should be checked in the  original equation. The check shows only x  is a solution. 6 The proposed solutions are x 

Now try Exercises 7 through 12



Here is one additional example that uses a factoring strategy commonly employed when an equation has more than three terms. EXAMPLE 2



Solving a Trig Equation by Factoring

Solution



The four terms in the equation share no common factors, so we attempt to factor by grouping. We could factor 2 cos  from the first two terms but instead elect to group the sin2 terms and begin there.

Find all solutions in 3 0°, 360°2: 8 sin2 cos   2 cos   4 sin2  1  0.

8 sin2 cos   2 cos   4 sin2  1 18 sin2 cos   4 sin22  12 cos   12 4 sin212 cos   12  112 cos   12 12 cos   12 14 sin2  12

0 0 0 0

given equation rearrange and group terms remove common factors remove common binomial factors

Using the zero factor property, we write two equations and solve each independently. 2 cos   1  0 2 cos   1 cos  

1 2

cos  7 0 in QI and QIV   60°, 300° A. You’ve just learned how to use additional algebraic techniques to solve trig equations

4 sin2  1  0 1 sin2  4 sin   

resulting equations isolate variable term

1 2

solve

sin  7 0 in QI and QII sin  6 0 in QIII and QIV   30°, 150°, 210°, 330°

solutions

Initially factoring 2 cos  from the first two terms and proceeding from there would have produced the same result. Now try Exercises 13 through 16



B. Solving Trig Equations Using Various Identities To solve equations effectively, a student should strive to develop all of the necessary “tools.” Certainly the underlying concepts and graphical connections are of primary importance, as are the related algebraic skills. But to solve trig equations effectively we must also have a ready command of commonly used identities. Observe how Example 3 combines a double-angle identity with factoring by grouping. EXAMPLE 3



Using Identities and Algebra to Solve a Trig Equation

Solution



Noting that one of the terms involves a double angle, we attempt to replace that term to make factoring a possibility. Using the double identity for sine, we have

Find all solutions in 30, 22: 3 sin12x2  2 sin x  3 cos x  1. Round solutions to four decimal places as necessary.

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3 12 sin x cos x2  2 sin x  3 cos x  1 16 sin x cos x  2 sin x2  13 cos x  12  0 2 sin x 13 cos x  12  113 cos x  12  0 13 cos x  12 12 sin x  12  0

substitute 2 sin x cos x for sin (2x) set equal zero and group terms factor using 3 cos x  1 common binomial factor

Use the zero factor property to solve each equation independently. 3 cos x  1  0 1 cos x   3 cos x 6 0 in QII and QIII x  1.9106, 4.3726 B. You’ve just learned how to solve trig equations using multiple angle, sum and difference, and sum-to-product identities

resulting equations 2 sin x  1  0 1 isolate variable term sin x  2 sin x 7 0 in QI and QII  5 x , solutions 6 6

Should you prefer the exact form, the solutions from the cosine equation could be 1 1 written as x  cos1a b and x  2  cos1a b. 3 3 Now try Exercises 17 through 26



C. Solving Equations of the Form A sin (Bx  C)  D  k You may remember equations of this form from Section 5.7. They actually occur quite frequently in the investigation of many natural phenomena and in the modeling of data from a periodic or seasonal context. Solving these equations requires a good combination of algebra skills with the fundamentals of trig. EXAMPLE 4



Solution



Solving Equations That Involve Transformations   Given f 1x2  160 sina x  b  320 and x  30, 22 , for what real numbers x 3 3 is f (x) less than 240?

We reason that to find values where f 1x2 6 240, we should begin by finding values where f 1x2  240. The result is   160 sin a x  b  320  240 equation 3 3   sin a x  b  0.5 subtract 320 and divide by 160; isolate variable term 3 3    At this point we elect to use a u-substitution for a x  b  1x  12 to obtain 3 3 3 a “clearer view.” sin u  0.5

substitute u for

 1x  12 3

sin u 6 0 in QIII and QIV u

7 6

u

11 6

solutions in u

 1x  12 for u and solve. 3  11  1x  12  re-substitute 1x  12 for u 3 3 6 11 3 x1 multiply both sides by  2 x  4.5 solutions

To complete the solution we re-substitute  7 1x  12  3 6 7 x1 2 x  2.5

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Section 6.7 General Trig Equations and Applications

We now know f 1x2  240 when x  2.5 and x  4.5 but when will f (x) be less

C. You’ve just learned how to solve trig equations of the form A sin 1Bx  C2  D  k

than 240? By analyzing the equation, we find the function has period of  2 P    6 and is shifted to the left units. This would indicate the graph peaks 3 3 early in the interval 3 0, 22 with a “valley” in the interior. We conclude f 1x2 6 240 in the interval (2.5, 4.5).

Now try Exercises 27 through 30



GRAPHICAL SUPPORT Support for the result in Example 4 can be

500

obtained by graphing the equation over the specified interval. Enter   Y1  160 sin a x  b  320 on the Y = 3 3 screen, then Y2  240. After locating points of

0

intersection, we note the graphs indeed verify

2

0

that in the interval 30, 22, f 1x2 6 240 for x  12.5, 4.52.

There is a mixed variety of equation types in Exercises 31 through 40.

D. Applications Using Trigonometric Equations Figure 6.32

Using characteristics of the trig functions, we can often generalize and extend many of the formulas that are familiar to you. For example, the formulas for the volume of a right circular cylinder and a right circular cone are well known, but what about the volume of a nonright figure (see Figure 6.32)? Here, trigonometry provides the answer, as the most general volume formula is V  V0 sin , where V0 is a “standard” volume formula and  is the complement of angle of deflection (see Exercises 43 and 44). As for other applications, consider the following from the environmental sciences. Natural scientists are very interested in the discharge rate of major rivers, as this gives an indication of rainfall over the inland area served by the river. In addition, the discharge rate has a large impact on the freshwater and saltwater zones found at the river’s estuary (where it empties into the sea).

␣ h

␪

EXAMPLE 5



Solving an Equation Modeling the Discharge Rate of a River For May through December, the discharge rate of the Ganges River (Bangladesh) 2  b  17,760 where t  1 can be modeled by D 1t2  16,580 sina t  3 3 represents May 1, and D(t) is the discharge rate in m3/sec. Source: Global River Discharge Database Project; www.rivdis.sr.unh.edu.

a. What is the discharge rate in mid-October? b. For what months (within this interval) is the discharge rate over 26,050 m3/sec?

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Solution



a. To find the discharge rate in mid-October we simply evaluate the function at t  6.5: 2  b  17,760 D 1t2  16,580 sin a t  3 3 2  d  17,760 D 16.52  16,580 sin c 16.52  3 3  1180

given function substitute 6.5 for t compute result on a calculator

In mid-October the discharge rate is 1180 m3/sec. b. We first find when the rate is equal to 26,050 m3/sec: D1t2  26,050. 2  b  17,760 26,050  16,580 sin a t  3 3 2  b 0.5  sin a t  3 3

substitute 26,050 for D(t )

subtract 17,760; divide by 16,580

2  b we obtain the equation Using a u-substitution for a t  3 3 0.5  sin u sin u 7 0 in QI and QII u

 6

u

5 6

solutions in u

 2  b  1t  22 for u and solve. To complete the solution we re-substitute a t  3 3 3   1t  22  3 6 t  2  0.5 t  2.5

 5 1t  22  3 6 t  2  2.5 t  4.5

 1t  22 for u 3 3 multiply both sides by  solutions re-substitute

The Ganges River will have a flow rate of over 26,050 m3/sec between mid-June (2.5) and mid-August (4.5). Now try Exercises 45 through 48



GRAPHICAL SUPPORT To obtain a graphical view of the solution to Example 5, enter  2 Y1  16,580 sin a t  b  17,760 on the Y = screen, then Y2  26,050. To set 3 3 an appropriate window, note the amplitude is 40,000

16,580 and that the graph has been vertically shifted by 17,760. Also note the x-axis 0

r

represents months 5 through 12. After locating points of intersection, we note the graphs verify that in the interval [1, 9] D 1t2 7 26,050 for t  (2.5, 4.5).

0

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D. You’ve just learned how to use a combination of skills to model and solve a variety of applications

615

There is a variety of additional exercises in the Exercise Set. See Exercises 49 through 54.

6.7 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. The three Pythagorean identities are , and .

,

2. When an equation contains two functions from a Pythagorean identity, sometimes both sides will lead to a solution. 3. One strategy to solve equations with four terms and no common factors is by . 

4. To combine two sine or cosine terms with different arguments, we can use the to formulas. 5. Regarding Example 5, discuss/explain how to determine the months of the year the discharge rate is under 26,050 m3/sec, using the solution set given. 6. Regarding Example 6, discuss/explain how to determine the months of the year the revenue projection is under $1250 using the solution set given.

DEVELOPING YOUR SKILLS

Solve each equation in [0, 2) using the method indicated. Round nonstandard values to four decimal places. • Squaring both sides

16 7. sin x  cos x  2 9. tan x  sec x  1 4 11. cos x  sin x  3 • Factor by grouping

8. cot x  csc x  13

24. sin17x2 cos 14x2  sin 15x2  cos17x2 sin14x2  cos x  0

12. sec x  tan x  2

25. sec4x  2 sec2x tan2x  tan4x  tan2x 26. tan4x  2 sec2x tan2x  sec4x  cot2x

14. 4 sin x cos x  213 sin x  2 cos x  13  0 15. 3 tan2x cos x  3 cos x  2  2 tan2x 16. 413 sin x sec x  13 sec x  2  8 sin x 2

2

• Using identities

1  cot2x 2 cot2x

23. cos 13x2  cos 15x2 cos 12x2  sin 15x2 sin 12x2  1  0

10. sin x  cos x  12

13. cot x csc x  2 cot x  csc x  2  0

17.

x x 22. 2 cos2 a b  3 sin a b  3  0 3 3

18.

1  tan2x 4  2 3 tan x

19. 3 cos12x2  7 sin x  5  0 20. 3 cos12x2  cos x  1  0 x x 21. 2 sin2a b  3 cos a b  0 2 2

State the period P of each function and find all solutions in [0, P). Round to four decimal places as needed.

  27. 250 sin a x  b  125  0 6 3   28. 7512 sec a x  b  150  0 4 6 29. 1235 cos a

  x  b  772  1750 12 4

  30. 0.075 sin a x  b  0.023  0.068 2 3

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• Using any appropriate method to solve.

31. cos x  sin x 

37. sec x cosa

12 2

38. sin a

32. 5 sec x  2 tan x  8  0 2

33.

13 1  cos2x  2 2 tan x

  xbcsc x  13 2

39. sec2x tana

34. 5 csc2x  5 cot x  5  0 35. csc x  cot x  1 

36.

12 1  sin2x  2 2 cot x

40. 2 tana

  xb  4 2

13   xbsin2x  2 2

WORKING WITH FORMULAS

41. The equation of a line in trigonometric form: y

D  x cos  sin 

Exercise 41 y

The trigonometric form of a linear equation is given by the formula shown, where D D is the perpendicular distance ␪ from the origin to the line x and  is the angle between the perpendicular segment and the x-axis. For each pair of perpendicular lines given, (a) find the point (a, b) of their intersection; (b) compute the distance b D  2a2  b2 and the angle   tan1a b, and a give the equation of the line in trigonometric form; and (c) use the GRAPH or the 2nd GRAPH TABLE feature of a graphing calculator to verify that both equations name the same line. 1 I. L1: y  x  5 II. L1: y   x  5 2 L2: y  x L2: y  2x 

  xb  1 2

13 4 13 x 3 3 L2: y  13x

III. L1: y  

42. Rewriting y  a cos x  b sin x as a single function: y  k sin(x  ) Linear terms of sine and cosine can be rewritten as a single function using the formula shown, where a k  2a2  b2 and   sin1a b. Rewrite the k equations given using these relationships and verify they are equivalent using the GRAPH or the 2nd GRAPH TABLE feature of a graphing calculator: a. y  2 cos x  2 13 sin x b. y  4 cos x  3 sin x The ability to rewrite a trigonometric equation in simpler form has a tremendous number of applications in graphing, equation solving, working with identities, and solving applications.

APPLICATIONS

43. Volume of a cylinder: The volume Exercise 43 of a cylinder is given by the formula V  r2h sin , where r is ␣ the radius and h is the height of the cylinder, and  is the indicated h complement of the angle of deflection . Note that when  ␪   , the formula becomes that 2  of a right circular cylinder (if   , then h is 2 called the slant height or lateral height of the cylinder). An old farm silo is built in the form of a right circular cylinder with a radius of 10 ft and a

height of 25 ft. After an earthquake, the silo became tilted with an angle of deflection   5°. (a) Find the volume of the silo before the earthquake. (b) Find the volume of the silo after the earthquake. (c) What angle  is required to bring the original volume of the silo down 2%? 44. Volume of a cone: The volume of a cone is given 1 by the formula V  r2h sin , where r is the 3 radius and h is the height of the cone, and  is the indicated complement of the angle of deflection .  Note that when   , the formula becomes that 2  of a right circular cone (if   , then h is called 2

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the slant height or lateral height of the cone). As part of a sculpture exhibit, an artist is constructing three such structures each with a radius of 2 m and a slant height of 3 m. (a) Find the volume of the sculptures if the angle of deflection is   15°. (b) What angle  was used if the volume of each sculpture is 12 m3?

Exercise 44

h

␣

␪

45. River discharge rate: For June through February, the discharge rate of the La Corcovada River (Venezuela) can be modeled by the function  9 D1t2  36 sina t  b  44, where t represents 4 4 the months of the year with t  1 corresponding to June, and D(t) is the discharge rate in cubic meters per second. (a) What is the discharge rate in midSeptember? (b) For what months of the year is the discharge rate over 50 m3/sec? Source: Global River Discharge Database Project; www.rivdis.sr.unh.edu.

46. River discharge rate: For February through June, the average monthly discharge of the Point Wolfe River (Canada) can be modeled by the function  D1t2  4.6 sina t  3b  7.4, where t represents 2 the months of the year with t  1 corresponding to February, and D(t) is the discharge rate in cubic meters/second. (a) What is the discharge rate in mid-March 1t  2.52 ? (b) For what months of the year is the discharge rate less than 7.5 m3/sec? Source: Global River Discharge Database Project; www.rivdis.sr.unh.edu.

47. Seasonal sales: Hank’s Heating Oil is a very seasonal enterprise, with sales in the winter far exceeding sales in the summer. Monthly sales for the company can be modeled by   S1x2  1600 cosa x  b  5100, where S1x2 6 12 is the average sales in month x 1x  1 S January2. (a) What is the average sales amount for July? (b) For what months of the year are sales less than $4000? 48. Seasonal income: As a roofing company employee, Mark’s income fluctuates with the seasons and the availability of work. For the past several years his average monthly income could be approximated by   the function I1m2  2100 sina m  b  3520, 6 2 where I(m) represents income in month m 1m  1 S January2. (a) What is Mark’s average

617

monthly income in October? (b) For what months of the year is his average monthly income over $4500? 49. Seasonal ice thickness: The average thickness of the ice covering an arctic lake can be modeled by  the function T1x2  9 cosa xb  15, where T(x) 6 is the average thickness in month x 1x  1 S January2. (a) How thick is the ice in mid-March? (b) For what months of the year is the ice at most 10.5 in. thick? 50. Seasonal temperatures: The function   T1x2  19 sina x  b  53 models the average 6 2 monthly temperature of the water in a mountain stream, where T(x) is the temperature 1°F2 of the water in month x 1x  1 S January2 . (a) What is the temperature of the water in October? (b) What two months are most likely to give a temperature reading of 62°F? (c) For what months of the year is the temperature below 50°F? 51. Coffee sales: Coffee sales fluctuate with the weather, with a great deal more coffee sold in the winter than in the summer. For Joe’s Diner, assume 2  x  b  29 the function G1x2  21 cosa 365 2 models daily coffee sales (for non-leap years), where G(x) is the number of gallons sold and x represents the days of the year 1x  1 S January 12. (a) How many gallons are projected to be sold on March 21? (b) For what days of the year are more than 40 gal of coffee sold? 52. Park attendance: Attendance at a popular state park varies with the weather, with a great deal more visitors coming in during the summer months. Assume daily attendance at the park can be modeled 2 x  b  545 by the function V1x2  437 cosa 365 (for non-leap years), where V(x) gives the number of visitors on day x 1x  1 S January 12 . (a) Approximately how many people visited the park on November 1 111  30.5  335.52? (b) For what days of the year are there more than 900 visitors?

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53. Exercise routine: As part of his yearly physical, Manu Tuiosamoa’s heart rate is closely monitored during a 12-min, cardiovascular exercise routine. His heart rate in beats per minute (bpm) is modeled  by the function B1x2  58 cosa x  b  126 6 where x represents the duration of the workout in minutes. (a) What was his resting heart rate? (b) What was his heart rate 5 min into the workout? (c) At what times during the workout was his heart rate over 170 bpm?

begun. (a) What is the initial grade for her workout? (b) What is the grade at x  4 min? (c) At G1x2  4.9%, how long has she been working out? (d) What is the duration of the treadmill workout?

54. Exercise routine: As part of her workout routine, Sara Lee programs her treadmill to begin at a slight initial grade (angle of incline), gradually increase to a maximum grade, then gradually decrease back to the original grade. For the duration of her workout, the grade is modeled by the function  G1x2  3 cosa x  b  4, where G(x) is the 5 percent grade x minutes after the workout has 

EXTENDING THE CONCEPT

55. As we saw in Chapter 6, cosine is the cofunction of sine and each can be expressed in terms of the other:   cosa  b  sin  and sina  b  cos . 2 2 This implies that either function can be used to model the phenomenon described in this section by adjusting the phase shift. By experimentation, (a) find a model using cosine that will produce results identical to the sine function in Exercise 50 and (b) find a model using sine that will produce results identical to the cosine function in Exercise 51.



MAINTAINING YOUR SKILLS

58. (5.7) Find f 12 for all six trig functions, given P151, 682 is on the terminal side. 59. (3.4) Sketch the graph of f by locating its zeroes and using end behavior: f 1x2  x4  3x3  4x. 60. (4.3) Use a calculator and the change-of-base formula to find the value of log5279. 61. (5.6) The Sears Tower in Chicago, Illinois, remains one of the tallest structures in the world. The top of the roof reaches 1450 ft above the street below and the antenna extends an additional 280 ft

56. Use multiple identities to find all real solutions for the equation given: sin15x2  sin12x2cos x  cos12x2sin x  0. Exercise 57 57. A rectangular parallelepiped with square ends has 12 edges and six surfaces. If the sum of all edges is 176 cm and the total surface area is 1288 cm2, find (a) the length of the ␪ diagonal of the parallelepiped (shown in bold) and (b) the angle the diagonal makes with the base (two answers are possible). Exercise 61

into the air. Find the viewing angle  for the antenna from a distance of 1000 ft (the angle formed from the base of the antenna to its top).

280 ft

␪

1450 ft

1000 ft

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S U M M A RY A N D C O N C E P T R E V I E W SECTION 6.1

Fundamental Identities and Families of Identities

KEY CONCEPTS • The fundamental identities include the reciprocal, ratio, and Pythagorean identities. • A given identity can algebraically be rewritten to obtain other identities in an identity “family.” • Standard algebraic skills like distribution, factoring, combining terms, and special products play an important role in working with identities. A C AD  BC gives an efficient method for combining rational terms. • The pattern   B D BD • Using fundamental identities, a given trig function can be expressed in terms of any other trig function. • Once the value of a given trig function is known, the value of the other five can be uniquely determined using fundamental identities, if the quadrant of the terminal side is known. • To show an equation is not an identity, find any one value where the expressions are defined but the equation is false, or graph both functions on a calculator to see if the graphs are identical. EXERCISES Verify using the method specified and fundamental identities. 1. multiplication 2. factoring sin x 1csc x  sin x2  cos2x

3. special products 1sec x  tan x21sec x  tan x2  sin x csc x

tan2x csc x  csc x  csc x sec2x

4. combine terms using A C AD  BC   B D BD 2 tan2x sec x sin x  csc x csc x

Find the value of all six trigonometric functions using the information given. 25 12 5. cos    ;  in QIII 6. sec   ;  in QIV 37 23

SECTION 6.2

Constructing and Verifying Identities

KEY CONCEPTS • The steps used to verify an identity must be reversible. • If two expressions are equal, one may be substituted for the other and the result will be equivalent. • To verify an identity we mold, change, substitute, and rewrite one side until we “match” the other side. • Verifying identities often involves a combination of algebraic skills with the fundamental trig identities. A collection and summary of the Guidelines for Verifying Identities can be found on page 553. EXERCISES Rewrite each expression to create a new identity, then verify the identity by reversing the steps. cos x  sin x cos x 7. csc x  cot x 8. cos2x Verify that each equation is an identity. csc2x 11  cos2x2 9.  cot2x tan2x

10.

cot x csc x  cot x 1cos x  csc x2  sec x tan x

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CHAPTER 6 Trigonometric Identities, Inverses, and Equations

sin4x  cos4x  tan x  cot x sin x cos x

SECTION 6.3

12.

1sin x  cos x2 2  csc x sec x  2 sin x cos x

The Sum and Difference Identities

KEY CONCEPTS The sum and difference identities can be used to • Find exact values for nonstandard angles that are a sum or difference of two standard angles. • Verify the cofunction identities and to rewrite a given function in terms of its cofunction. • Find coterminal angles in 30, 360°2 for very large angles (the angle reduction formulas). • Evaluate the difference quotient for sin x, cos x, and tan x. • Rewrite a sum as a single expression: cos  cos   sin  sin   cos1  2 . The sum and difference identities for sine and cosine can be remembered by noting • For cos1  2 , the function repeats and the signs alternate: cos1  2  cos  cos   sin  sin  • For sin1  2 the signs repeat and the functions alternate: sin1  2  sin  cos   cos  sin 

EXERCISES Find exact values for the following expressions using sum and difference formulas.  13. a. cos 75° b. tana b 14. a. tan 15° 12 Evaluate exactly using sum and difference formulas. 15. a. cos 109° cos 71°  sin 109° sin 71°

b. sina

 b 12

b. sin 139° cos 19°  cos 139° sin 19°

Rewrite as a single expression using sum and difference formulas. 16. a. cos13x2cos12x2  sin13x2sin12x2

x 3x x 3x b. sina bcosa b  cosa bsina b 4 8 4 8

Evaluate exactly using sum and difference formulas, by reducing the angle to an angle in 3 0, 360°2 or 3 0, 22. 57 b 17. a. cos 1170° b. sina 4 Use a cofunction identity to write an equivalent expression for the one given.  x 18. a. cosa b b. sinax  b 8 12 19. Verify that both expressions yield the same result using sum and difference formulas. tan 15°  tan145°  30°2 and tan 15°  tan1135°  120°2 . 20. Use sum and difference formulas to verify the following identity.   cosax  b  cosax  b  13 cos x 6 6

SECTION 6.4

The Double-Angle, Half-Angle, and Product-to-Sum Identities

KEY CONCEPTS • When multiple angle identities (identities involving n) are used to find exact values, the terminal side of  must be determined so the appropriate sign can be used. • The power reduction identities for cos2x and sin2x are closely related to the double-angle identities, and can be derived directly from cos12x2  2 cos2x  1 and cos12x2  1  2 sin2x.

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• The half-angle identities can be developed from the power reduction identities by using a change of variable and taking square roots. The sign is then chosen based on the quadrant of the half angle. • The product-to-sum and sum-to-product identities can be derived using the sum and difference formulas, and have important applications in many areas of science.

EXERCISES Find exact values for sin122, cos122 , and tan122 using the information given. 13 29 21. a. cos   ;  in QIV b. csc    ;  in QIII 85 20 Find exact values for sin , cos , and tan  using the information given. 336 41 22. a. cos122   b. sin122   ;  in QII ;  in QII 841 625 Find exact values using the appropriate double-angle identity. 23. a. cos222.5°  sin222.5°

b. 1  2 sin2 a

 b 12

Find exact values for sin  and cos  using the appropriate half-angle identity. 5 24. a.   67.5° b.   8   Find exact values for sina b and cosa b using the given information. 2 2 65 24 25. a. cos   ; 0° 6  6 360°;  in QIV b. csc    ; 90° 6  6 0;  in QIV 25 33 26. Verify the equation is an identity. cos132  cos  2 tan2  cos132  cos  sec2  2

27. Solve using a sum-to-product formula. cos13x2  cos x  0

  28. The area of an isosceles triangle (two equal sides) is given by the formula A  x2sina b cosa b, where the 2 2 equal sides have length x and the vertex angle measures °. (a) Use this formula and the half-angle identities to find the area of an isosceles triangle with vertex angle   30° and equal sides of 12 cm. (b) Use substitution and   1 a double-angle identity to verify that x2sina b cosa b  x2sin , then recompute the triangle’s area. Do the 2 2 2 results match?

SECTION 6.5

The Inverse Trig Functions and Their Applications

KEY CONCEPTS • In order to create one-to-one functions, the domains of y  sin t, y  cos t, and y  tan t are restricted as follows:     (a) y  sin t, t  c  , d ; (b) y  cos t, t 3 0,  4 ; and (c) y  tan t; t  a , b. 2 2 2 2 • For y  sin x, the inverse function is given implicitly as x  sin y and explicitly as y  sin1x or y  arcsin x. • The expression y  sin1x is read, “y is the angle or real number whose sine is x.” The other inverse functions are similarly read/understood. • For y  cos x, the inverse function is given implicitly as x  cos y and explicitly as y  cos1x or y  arccos x. • For y  tan x, the inverse function is given implicitly as x  tan y and explicitly as y  tan1x or y  arctan x.

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• The domains of y  sec x, y  csc x, and y  cot x are likewise restricted to create one-to-one functions:

    b ´ a ,  d ; (b) y  csc t, t  c  , 0b ´ a0, d ; and (c) y  cot t, t  10, 2. 2 2 2 2 • In some applications, inverse functions occur in a composition with other trig functions, with the expression best evaluated by drawing a diagram using the ratio definition of the trig functions. 1 1 • To evaluate y  sec1t, we use y  cos1a b; for y  cot1t, use tan1a b; and so on. t t • Trigonometric substitutions can be used to simplify certain algebraic expressions. (a) y  sec t; t  c 0,

EXERCISES Evaluate without the aid of calculators or tables. State answers in both radians and degrees in exact form. 12 13 29. y  sin1a 30. y  csc12 31. y  arccosa b b 2 2 Evaluate the following using a calculator, keeping the domain and range of each function in mind. Answer in radians to the nearest ten-thousandth and in degrees to the nearest tenth. Some may be undefined. 7 32. y  tan14.3165 33. y  sin10.8892 34. f 1x2  arccosa b 8 Evaluate the following without the aid of a calculator. Some may be undefined. 1  35. sin c sin1a b d 36. arcsec c seca b d 37. cos 1cos122 2 4 Evaluate the following using a calculator. Some may be undefined. 38. sin1 1sin 1.02452

39. arccos3 cos 160°2 4

Evaluate each expression by drawing a right triangle and labeling the sides. 12 7 41. sin c cos1a b d 42. tan c arcsec a b d 37 3x

40. cot1 c cot a 43. cot c sin1a

11 bd 4

x 281  x2

bd

Use an inverse function to solve the following equations for  in terms of x. 44. x  5 cos 

SECTION 6.6

45. 7 13 sec   x

46. x  4 sin a 

 b 6

Solving Basic Trig Equations

KEY CONCEPTS • When solving trig equations, we often consider either the principal root, roots in 30, 22, or all real roots. • Keeping the graph of each function in mind helps to determine the desired solution set. • After isolating the trigonometric term containing the variable, we solve by applying the appropriate inverse function, realizing the result is only the principal root. • Once the principal root is found, roots in 30, 22 or all real roots can be found using reference angles and the period of the function under consideration. • Trig identities can be used to obtain an equation that can be solved by factoring or other solution methods. EXERCISES Solve each equation without the aid of a calculator (all solutions are standard values). Clearly state (a) the principal root; (b) all solutions in the interval 3 0, 22 ; and (c) all real roots. 47. 2 sin x  12 48. 3 sec x  6 49. 8 tan x  7 13   13

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Mixed Review

Solve using a calculator and the inverse trig functions (not by graphing). Clearly state (a) the principal root; (b) solutions in 3 0, 22; and (c) all real roots. Answer in radians to the nearest ten-thousandth as needed. 2 1 50. 9 cos x  4 51. sin122  52. 12 csc x  3  7 5 4 53. The area of a circular segment (the shaded portion shown in the diagram) is given by the 1 formula A  r2 1  sin 2 , where  is in radians. If the circle has a radius of 10 cm, find 2 the angle  that gives an area of 12 cm2.

SECTION 6.7

r

␪

r

General Trig Equations and Applications

KEY CONCEPTS • In addition to the basic solution methods from Section 6.6, additional strategies include squaring both sides, factoring by grouping, and using the full range of identities to simplify an equation. • Many applications result in equations of the form Asin1Bx  C2  D  k. To solve, isolate the factor sin1Bx  C2 (subtract D and divide by A), then apply the inverse function. • Once the principal root is found, roots in 3 0, 22 or all real roots can be found using reference angles and the period of the function under consideration. EXERCISES Find solutions in 3 0, 22 using the method indicated. Round nonstandard values to four decimal places. 54. squaring both sides 55. using identities sin x  cos x 

16 2

3 cos12x2  7 sin x  5  0

56. factor by grouping 4 sin x cos x  213 sin x  2 cos x  13  0

57. using any appropriate method csc x  cot x  1

State the period P of each function and find all solutions in [0, P). Round to four decimal places as needed.   58. 750 sina x  b  120  0 6 2

  59. 80 cos a x  b  40 12  0 3 4

60. The revenue earned by Waipahu Joe’s Tanning Lotions fluctuates with the seasons, with a great deal more lotion   sold in the summer than in the winter. The function R1x2  15 sin a x  b  30 models the monthly sales of 6 2 lotion nationwide, where R(x) is the revenue in thousands of dollars and x represents the months of the year (x  1 S Jan). (a) How much revenue is projected for July? (b) For what months of the year does revenue exceed $37,000?

MIXED REVIEW Find the value of all six trig functions using the information given. 1. csc  

1117 ;  in QII 6

4 2. tan1a b   3

Find the exact value of each expression using a sum or difference identity. 3. tan 255°

4. cos a

19 b 12

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Evaluate each expression by drawing a right triangle and labeling the sides appropriately. 5. tan c arccsca

10 bd x

6. sin c sec1a

264  x2 bd x

7. Solve for x in the interval [0, 2). Round to four decimal places as needed:   100 sin a x  b  80  100 4 6 8. Without the aid of a calculator, find: (a) the principal roots, (b) all solutions in [0, 2) and (c) all real solutions: 1cos x  12 3 2 cos2 1x2  1 4  0 9. The horizontal distance R that an object will travel when it is projected at angle  with initial velocity v 1 is given by the equation R  v2 sin  cos . 16 a. Use an identity to show this equation can be 1 written as R  v2 sin122. 32 b. Use this equation to show why the horizontal distance traveled by the object is the same for any two complementary angles.

Evaluate without the aid of calculator or tables. Answer in both radians and degrees. 15. y  arcsec1122

16. y  sin10

17. y  tan1 13 18. Verify the following identities using a sum formula. a. sin12x2  2 sin x cos x b. cos12x2  cos2x  sin2x Use an inverse function to solve each equation for  in terms of x. 19.

20. 2 12 csc a 

x  tan  10

 bx 4

21. On a large clock, the distance from the tip of the hour hand to the base of the “12” can be approxit mated by the function D1t2  ` 8 sina b `  2, 12 where D(t) is this distance in feet at time t in hours. Use this function to approximate (a) the time of day when the hand is 6 ft from the 12 and 10 ft from the 12 and (b) the distance between the tip and the 12 at 4:00. Check your answer graphically. 22. The figure shows a smaller pentagon inscribed within a larger pentagon. Find the measure of angle  using the diagram and equation given: 3.22  112  9.42  2111219.42cos 

␪ R

␪

10. The profits of Red-Bud Nursery can be modeled by a sinusoid, with profit peaking twice each year. Given profits reach a yearly low of $4000 in midJanuary (month 1.5), and a yearly high of $14,000 in mid-April (month 4.5). (a) Construct an equation for their yearly profits. (b) Use the model to find their profits for August. (c) Name the other month at which profit peaks. 11. Find the exact value of 2 cos2a

 b  1 using an 12

appropriate identity. Verify the following identity. 1  cos2  sin2  1  cos122 tan2 1cos t  sin t2 2  cot t  2cos2t 13. tan t 12.

x x 14. Find exact values for sina b and cosa b using the 2 2 information given. 6 ; 540° 6 x 6 630° a. sin x  7.5 11.7  ;0 6 x 6 b. sec x  4.5 2

9.4 mm 3.2 mm

11 mm

23. Find the value of each expression using sum-toproduct and half-angle identities (without using a calculator). a. sin 172.5°  sin 52.5° b. cos 172.5°  cos 52.5° 24. Given 100 sin t  70, use a calculator to find (a) the principal root, (b) all solutions in [0, 2], and (c) all real solutions. Round to the nearest ten-thousandth. 25. Use the product-to-sum formulas to find the exact value of 13 7 7 13 bcosa b b. sina bsina b a. sina 24 24 24 24

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Practice Test

625

PRACTICE TEST Verify each identity using fundamental identities and the method specified. 1. special products 1csc x  cot x2 1csc x  cot x2  cos x sec x 2. factoring

13. Evaluate without the aid of calculators or tables. 1 1 b a. y  tan1a b. y  sin c sin1a b d 2 13 c. y  arccos1cos 30°2

sin3x  cos3x  sin x  cos x 1  cos x sin x

14. Evaluate the following. Use a calculator for part (a), give exact answers for part (b), and find the value of the expression in part (c) without using a calculator. Some may be undefined. a. y  sin10.7528 b. y  arctan1tan 78.5°2

3. Find the value of all six trigonometric functions 48 given cos   ;  in QIV 73 4. Find the exact value of tan 15° using a sum or difference formula.

c. y  sec1 c seca

5. Rewrite as a single expression and evaluate: cos 81° cos 36°  sin 81° sin 36° 6. Evaluate cos 1935° exactly using an angle reduction formula. 7. Use sum and difference formulas to verify   sinax  b  sinax  b  12 cos x. 4 4

16. cot c cos1a

  10. Find exact values for sina b and cosa b given 2 2 12 tan   ;  in QI 35

␣

56 bd 33

bd 125  x2 17. Solve without the aid of a calculator (all solutions are standard values). Clearly state (a) the principal root, (b) all solutions in the interval 30, 22, and (c) all real roots. I. 8 cos x  412 II. 13 sec x  2  4

9. Use a double-angle identity to evaluate 2 cos275°  1.

b

7 bd 24

Evaluate the expressions by drawing a right triangle and labeling the sides. 15. cos c tan1a

8. Find exact values for sin , cos , and tan  given 161 cos122   ;  in QI 289

11. The area of a triangle is given geometrically as 1 A  base # height. The 2 trigonometric formula for

find values for sin  and cos , which are used in a conversion formula. Find sin  and cos  for 17x2  5 13xy  2y2  0, assuming 2 in QI.

a

c

1 the triangle’s area is A  bc sin , where  is the 2 angle formed by the sides b and c. In a certain triangle, b  8, c  10, and   22.5°. Use the formula for A given here and a half-angle identity to find the area of the triangle in exact form. 12. The equation Ax2  Bxy  Cy2  0 can be written in an alternative form that makes it easier to graph. This is done by eliminating the mixed xy-term using B the relation tan122  to find . We can then AC

x

18. Solve each equation using a calculator and inverse trig functions to find the principal root (not by graphing). Then state (a) the principal root, (b) all solutions in the interval 30, 22 , and (c) all real roots. 1 2 I. sin12x2  II. 3 cos12x2  0.8  0 3 4 19. Solve the equations graphically in the indicated interval using a graphing calculator. State answers in radians rounded to the nearest ten-thousandth. a. 3 cos12x  12  sin x; x  3, 4 b. 2 1x  1  3 cos2x; x  30, 22 20. Solve the following equations for x  3 0, 22 using a combination of identities and/or factoring. State solutions in radians using the exact form where possible. a. 2 sin x sin12x2  sin12x2  0 1 b. 1cos x  sin x2 2  2

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CHAPTER 6 Trigonometric Identities, Inverses, and Equations

Solve each equation in 3 0, 22 by squaring both sides, factoring, using identities or by using any appropriate method. Round nonstandard values to four decimal places. 21. 3 sin12x2  cos x  0  3 5 2 22. sina2x  b   3 6 2 6 23. The revenue for Otake’s Mower Repair is very seasonal, with business in the summer months far exceeding business in the winter months. Monthly revenue for the company can be modeled by the 4  b  12.5, where function R1x2  7.5 cosa x  6 3 R(x) is the average revenue (in thousands of dollars) for month x 1x  1 S Jan2. (a) What is the average

revenue for September? (b) For what months of the year is revenue at least $12,500? 24. The lowest temperature on record for the even months of the year are given in the table for the city of Denver, Colorado. The equation y  35.223 sin10.576x  2.5892  6 is a fairly accurate model for this data. Use the equation to estimate the record low temperature for the odd numbered months.

Month (Jan S 1)

Low Temp. (F )

2

30

4

2

6

30

8

41

10

3

12

25

Source: 2004 Statistical Abstract of the United States, Table 379.

25. Write the product as a sum using a product-to-sum identity: 2 cos11979t2cos1439t2.

C A L C U L AT O R E X P L O R AT I O N A N D D I S C O V E RY Seeing the Beats as the Beats Go On When two sound waves of slightly different frequencies are combined, the resultant wave varies periodically in amplitude over time. These Figure 6.33 amplitude pulsations are called beats. In this Exploration and Discovery, we’ll look at ways to “see” the beats more clearly on a graphing calculator, by representing sound waves very simplistically as Y1  cos1mt2 and Figure 6.34 Y2  cos1nt2 and 3 noting a relationship between m, n, and the number of beats in 3 0, 24. Using a 0 2␲ sum-to-product formula, we can represent the resultant wave as a 3 single term. For Y1  cos112t2 and Y2  cos18t2 the result is 12t  8t 12t  8t b cosa b 2 2  2 cos110t2cos12t2

cos112t2  cos18t2  2 cosa

The window used and resulting graph are shown in Figures 6.33 and 6.34, and it appears that “silence” occurs

four times in this interval— where the graph of the combined waves is tangent to (bounces off of) the x-axis. This indicates a total of four beats. Note the number of beats is equal to the difference m  n: 12  8  4. Further experimentation will show this is not a coincidence, and this enables us to construct two additional functions that will 0 frame these pulsations and make them easier to see. Since the maximum amplitude of the resulting wave

Figure 6.35

Figure 6.36 3

2␲

3

k is 2, we use functions of the form  2 cosa xb to con2 struct the frame, where k is the number of beats in the interval 1m  n  k2. For Y1  cos112t2 and Y2  cos18t2,

12  8  2 and the functions we use will 2 be Y2  2 cos12x2 and Y3  2 cos12x2 as shown in Figure 6.35. The result is shown in Figure 6.36, where the we have k 

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627

Strengthening Core Skills

frame clearly shows the four beats or more precisely, the four moments of silence. For each exercise, (a) express the sum Y1  Y2 as a product, (b) graph YR on a graphing calculator for x  3 0, 24 and identify the number of beats in this interk val, and (c) determine what value of k in 2 cosa xb 2

would be used to frame the resultant YR, then enter these as Y2 and Y3 to check the result. Exercise 1: Exercise 2: Exercise 3: Exercise 4:

Y1 Y1 Y1 Y1

   

cos114t2; cos112t2; cos114t2; cos111t2;

Y2  Y2  Y2  Y2 

cos18t2 cos19t2 cos16t2 cos110t2

STRENGTHENING CORE SKILLS Trigonometric Equations and Inequalities The ability to draw a quick graph of the trigonometric functions is a tremendous help in understanding equations and inequalities. A basic sketch can help reveal the number of solutions in 3 0, 22 and the quadrant of each solution. For nonstandard angles, the value given by the inverse function can then be used as a basis for stating the solution set for all real numbers. We’ll illustrate the process using a few simple examples, then generalize our observations to solve more realistic applications. Consider the function f 1x2  2 sinx  1, a sine wave with amplitude 2, and a vertical translation of  1. To find intervals in 3 0, 22 where f 1x2 7 2.5, we reason that f has a maximum of 3  2112  1 and a minimum of 1  2112  1, since 1  sin x  1. With no phase shift and a standard period of 2, we can easily draw a quick sketch of f by vertically translating x-intercepts and max/min points 1 unit up. After Figure 6.37 drawing the line y  2.5 (see y y  2.5 Figure 6.37), it appears there are 3 f (x) two intersections in the interval, one in QI and one in QII. More ␲ 2␲ x importantly, it is clear that f 1x2 7 2.5 between these two 3 solutions. Substituting 2.5 for f (x) in f 1x2  2 sin x  1, we solve for sin x to obtain sin x  0.75, which we use to state the solution in exact form: f 1x2 7 2.5 for x  1sin10.75,   sin10.752. In approximate form the solution interval is x  10.85, 2.292. If the function involves a horizontal shift, the graphical analysis will reveal which intervals should be chosen to satisfy the given inequality. Illustration 1  Given g1x2  3 sinax  g1x2  1.2 for x  30, 22.

 b  1, solve 4

Solution  Plot the x-intercepts and maximum/minimum values for a standard sine wave with amplitude 3, then shift  these points units to the right. Then shift each point one 4

unit down and draw a sine wave through the points (see Figure 6.38). This sketch along with the graph of y  1.2 is sufficient to reveal that solutions to g1x2  1.2 occur in QI and QIII, with solutions to g1x2  1.2 outside this interval. Substituting 1.2 for g(x) and isolat-

Figure 6.38 y 3

f (x) ␲

2␲ x

4

ing the sine function we obtain sin ax 

 1 b , 4 15

1  after taking the inverse b 15 4 sine of both sides. This is the QI solution, with 1  x  c   sin1a b d  being the solution in 15 4 QIII. In approximate form the solution interval is x  30, 0.72 4 ´ 33.99, 2 4. then x  sin1a

The basic ideas remain the same regardless of the complexity of the equation. Remember—our current goal is not a supremely accurate graph, just a sketch that will guide us to the solution using the inverse functions and the correct quadrants. Perhaps that greatest challenge is recallC ing that when B  1, the horizontal shift is  , but other B than this a fairly accurate sketch can quickly be obtained. Practice with these ideas by solving the following inequalities within the intervals specified. f 1x2  3 sin x  2; f 1x2 7 3.7; x  30, 22  Exercise 2: g1x2  4 sin ax  b  1; 3 g1x2  2; x  30, 22   Exercise 3: h1x2  125 sin a x  b  175; 6 2 h1x2  150; x  30, 122 2  Exercise 4: f 1x2  15,750 sin a x  b  19,250; 360 4 f 1x2 7 25,250; x  30, 3602

Exercise 1:

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CHAPTER 6 Trigonometric Identities, Inverses, and Equations

C U M U L AT I V E R E V I E W C H A P T E R S 1 – 6 1. Find f 12 for all six trig

10. Solve for x:

Exercise 2

functions, given P113, 842 is on the terminal side with  in QII.

60

3

21x  32 4  1  55.

11. Solve for x: 3x 

1   5 10 2

12. The Earth has a radius of 3960 mi. Tokyo, Japan, is

2. Find the lengths of the

missing sides.

5√3

3. Verify that x  2  13

is a zero of g1x2  x2  4x  1.

4. Determine the domain of r1x2  29  x2. Answer

in interval notation. 5. Standing 5 mi (26,400 ft) from the base of Mount

Logan (Yukon) the angle of elevation to the summit is 36° 56¿ . How much taller is Mount McKinley (Alaska) which stands at 20,320 ft high? 6. Use the Guidelines for Graphing Polynomial

Functions to sketch the graph of f 1x2  x3  3x2  4. 7. Use the Guidelines for Graphing Rational Functions

x1 x2  4 8. The Petronas Towers in Malaysia are two of the tallest structures in the world. The top of the roof reaches 1483 ft above the street below and the stainless steel pinnacles extend an additional 241 ft into the air (see figure). Find the viewing angle  for the pinnacles from a distance of 1000 ft (the angle formed from the base of the antennae to its top). to sketch the graph of h1x2 

241 ft

1483 ft ␪

located at 35.4° N latitude, very near the 139° E latitude line. Adelaide, Australia, is at 34.6° S latitude, and also very near 139° E latitude. How many miles separate the two cities? 13. Since 1970, sulphur dioxide emissions in the United

States have been decreasing at a nearly linear rate. In 1970, about 31 million tons were emitted into the atmosphere. In 2000, the amount had decreased to approximately 16 million tons. (a) Find a linear equation that models sulphur dioxide emissions. (b) Discuss the meaning of the slope ratio in this context. (c) Use the equation model to estimate the emissions in 1985, and project the emission for 2010. Source: 2004 Statistical Abstract of the United States, Table 360.

14. List the three Pythagorean identities and three

identities equivalent to cos122.   b  168, what values 6 2 of x in 3 0, 22 satisfy f 1x2 7 330.5?

15. For f1x2  325 cosa x 

16. Write as a single logarithmic expression in simplest

form: log1x2  92  log1x  12  log1x2  2x  32.

17. After doing some market research, the manager of a

sporting goods store finds that when a four-pack of premium tennis balls are priced at $9 per pack, 20 packs per day are sold. For each decrease of $0.25, 1 additional pack per day will be sold. Find the price at which four-packs of tennis balls should be sold in order to maximize the store’s revenue on this item. Exercise 18 18. Write the equation of the y 6 function whose graph is given, in terms of a sine function. 19. Verify that the following is an

identity:

cos x  1 cos x  sec x  1 tan2x

8␲ 4␲

4␲

6

1000 ft

9. A wheel with radius 45 cm is turning at 5

revolutions per second. Find the linear velocity of a point on the rim in kilometers per hour, rounded to the nearest 10th of a kilometer.

8␲ x

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Cumulative Review Chapters 1–6

20. The graph of a function f(x) is shown. Given the

zeroes are x   4 and x   12, estimate the following: a. the domain and range of the function b. intervals where f 1x2 7 0 and f 1x2  0 c. intervals where f 1x2T and f 1x2c d. name the location of all local maximums and minimums

21. Use the triangle shown to find the exact value of

sin122.

629

23. The amount of waste product released by a

manufacturing company varies according to its production schedule, which is much heavier during the summer months and lighter in the winter. Waste product amount reaches a maximum of 32.5 tons in the month of July, and falls to a minimum of 21.7 tons in January 1t  12 . (a) Use this information to build a sinusoidal equation that models the amount of waste produced each month. (b) During what months of the year does output exceed 30 tons? 24. At what interest rate will $2500 grow to $3500 if it’s

left on deposit for 6 yr and interest is compounded continuously?

␪ √202

9

11

22. Use the triangle shown to find the exact value of

sin1  2.

␣ 51 ␤ 68

25. Identify each geometric formula:

a. y  r2h

b. y  LWH

c. y  2r

1 d. y  bh 2

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Precalculus—

CONNECTIONS TO CALCULUS In calculus, as in college algebra, we often encounter expressions that are difficult to use in their given form, and so attempt to write the expression in an alternative form more suitable to the task at hand. Often, we see algebra and trigonometry working together to achieve this goal.

Simplifying Expressions Using a Trigonometric Substitution For instance, it is difficult to apply certain concepts from calculus to the equation x y , and we attempt to rewrite the expression using a trig substitution and 29  x2 a Pythagorean Identity. When doing so, we’re careful to ensure the substitution used represents a one-to-one function, and that the substitution maintains the integrity of the domain. EXAMPLE 1



Simplifying Algebraic Expressions Using Trigonometry Simplify y 

x

using the substitution x  3 sin  for 

29  x then verify that the result is equivalent to the original function. Solution



y   

3 sin 

29  13 sin 2 2 3 sin  29  9 sin2 3 sin  2911  sin22 3 sin 

29 cos  3 sin   3 cos   tan  2

2

  6  6 , 2 2

substitute 3 sin  for x 13 sin 2 2  9 sin2 factor

substitute cos2 for 1  sin2

29 cos2  3 cos  since 

   2 2

result

Now try Exercises 1 through 6



Using the notation for inverse functions, we can rewrite y  tan  as a function of x and use a calculator to compare it with the original function. For x  3 sin  x x x we obtain  sin  or   sin1a b. Substituting sin1a b for  in y  tan  3 3 3 x gives y  tan c sin1a b d . With the calculator in radian radian MODE , enter 3 x x Y1  and Y2  tan c sin1a b d on the 2 3 29  x Y= screen. Using TblStart  3 (due to the domain), the resulting table seems to indicate that the functions are indeed equivalent (see the figure).

630

6–88

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Connections to Calculus

631

Trigonometric Identities and Equations While the tools of calculus are very powerful, it is the application of rudimentary concepts that makes them work. Earlier, we saw how basic algebra skills were needed to simplify expressions that resulted from applications of calculus. Here we illustrate the use of basic trigonometric skills combined with basic algebra skills to achieve the same end. EXAMPLE 2



Finding Maximum and Minimum Values Using the tools of calculus, it can be shown that the maximum and/or minimum 1  cot  values of f 12  will occur at the zero(s) of the function csc  f 12 

csc 1csc22  11  cot2 1csc  cot 2 csc2

.

Simplify the right-hand side and use the result to find the location of any maximum or minimum values that occur in the interval 3 0, 22. Solution



Begin by simplifying the numerator. f 12 

csc 1csc22  1csc  cot   csc  cot22

csc2 3 csc   csc  cot   csc  cot2  csc2 2 csc 1cot   cot   csc22  csc2 1csc2  12  cot   csc2  csc  cot   1  csc 

distribute

simplify

factor, commute terms

simplify, substitute csc2  1 for cot2

result

This shows that f 12  0 when cot   1, or when  

  k, k  . In the 4

5  and . The function f has a maximum value of 4 4  5 . 12 at , with a minimum value of 12 at 4 4

interval 30, 22, this gives  

Now try Exercises 7 through 10



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Connections to Calculus

Connections to Calculus Exercises For the functions given, (a) use the substitution indicated to find an equivalent function of , (b) rewrite the resulting function in terms of x using an inverse trig function, and (c) use the TABLE feature of a graphing calculator to verify the two functions are equivalent for x  0. 2 1. y  2169  x ; x

2 2. y  2144  x ; x

  x  12 sin ,   a , b 2 2

  x  13 tan ,   a , b 2 2 Rewrite the following expressions using the substitution indicated.

3.

4.

x2

; x  4 sin 

5.

281  x2 ; x  9 sin  x

6.

216  x

2

x 29  x2 8

1x  42 2 2

3

; x  3 tan 

; x  2 sec 

Using the tools of calculus, it can be shown that for each function f(x) given, the zeroes of f (x) give the location of any maximum and/or minimum values. Find the location of these values in the interval [0, 2), using trig identities as needed to solve f (x)  0. Verify solutions using a graphing calculator.

7. f 1x2  f 1x2 

1  cos x ; sec x sec x1sin x2  11  cos x2sec x tan x

8. f 1x2  sin x tan x;

sec2x

f 1x2  sin x sec2x  tan x cos x

9. f 1x2  2 sin x cos x; f 1x2  2 sin x1sin x2  2 cos x cos x

10. f 1x2 

f 1x2 

cos x ; 2  sin x 12  sin x2 1sin x2  cos x cos x 12  sin x2 2

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PreCalculus—

7 CHAPTER CONNECTIONS

Applications of Trigonometry CHAPTER OUTLINE 7.1 Oblique Triangles and the Law of Sines 634 7.2 The Law of Cosines; the Area of a Triangle 646

When an airline pilot charts a course, it’s not as simple as pointing the airplane in the right direction. Wind currents must be taken into consideration, and compensated for by additional thrust or a change of heading that will help equalize the force of the wind and keep the plane flying in the desired direction. The effect of these forces working together can be modeled using a carefully drawn vector diagram, and with the aid of trigonometry, a pilot can easily determine any adjustments in navigation needed. This application appears as Exercise 85 in Section 7.3.

7.3 Vectors and Vector Diagrams 658 7.4 Vector Applications and the Dot Product 674 7.5 Complex Numbers in Trigonometric Form 687 7.6 De Moivre’s Theorem and the Theorem on nth Roots 698 As with other forms of problem solving, drawing an accurate sketch or diagram of the relationships involved has a large impact on our ability to understand vector applications and other applications of trigonometry. This is certainly no less true in a calculus course and is the subject of the Chapter 7 Connections to Calculus Connections to Calculus. 633

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7.1 Oblique Triangles and the Law of Sines Learning Objectives In Section 7.1 you will learn how to:

A. Develop the law of sines, and use it to solve ASA and AAS triangles

B. Solve SSA triangles (the ambiguous case) using the law of sines

Figure 7.1

Many applications of trigonometry involve oblique triangles, or triangles that do not have a 90° angle. For example, suppose a trolley carries passengers from ground level up to a mountain chateau, as shown in Figure 7.1. Assuming the cable could be held taut, what is its approximate length? Can we also determine the slant height of the mountain? To answer questions like these, we’ll develop techniques that enable us to solve acute and obtuse triangles using fundamental trigonometric relationships.

23 2000 m

67

C. Use the law of sines to solve applications

WORTHY OF NOTE As with right triangles, solving an oblique triangle involves determining the lengths of all three sides and the measures of all three angles.

A. The Law of Sines and Unique Solutions Consider the oblique triangle ABC pictured in Figure 7.2 Figure 7.2. Since it is not a right triangle, it seems B the trigonometric ratios studied earlier cannot be applied. But if we draw the altitude h (from vertex B), two right triangles are formed that share c a h a common side. By applying the sine ratio to angles A and C, we can develop a relationship that will help us solve the triangle. h A C For A we have sin A  or h  c sin A. For b c h C we have sin C  or h  a sin C. Since both products are equal to h, the transitive a property gives c sin A  a sin C, which leads to c sin A  a sin C

since h  h

c sin A a sin C  ac ac

divide by ac

sin C sin A  a c

simplify

Using the same triangle and the altitude drawn from C (Figure 7.3), we note a h similar relationship involving angles A and B: sin A  or h  b sin A, and b sin B h sin A  sin B  or h  a sin B. As before, we can then write . If A is obtuse, a a b the altitude h actually falls outside the triangle, as shown in Figure 7.4. In this case, consider that sin1180°  2  sin  from the difference formula for sines (Exercise 55, h Section 6.3). In the figure we note sin1180°  2   sin , yielding h  c sin  c Figure 7.4

Figure 7.3 B

B

a

c

h

h

A

634

b

a c

180  ␣ C

␣ A

b

C

7-2

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635

Section 7.1 Oblique Triangles and the Law of Sines

and the preceding relationship can now be stated using any pair of angles and corresponding sides. The result is called the law of sines, which is usually stated by combing the three possible proportions. The Law of Sines For any triangle ABC, the ratio of the sine of an angle to the side opposite that angle is constant: sin A sin B sin C   a c b As a proportional relationship, the law requires that we have three parts in order to solve for the fourth. This suggests the following possibilities: 1. 2. 3. 4. 5.

two angles and an included side (ASA) two angles and a side opposite one of these angles (AAS) two sides and a angle opposite one of these sides (SSA) two sides and an included angle (SAS) three sides (SSS)

Each of these possibilities is diagrammed in Figures 7.5 through 7.9. Figure 7.5 A

Figure 7.6

ASA

Figure 7.7

AAS

A Angle

A SSA

WORTHY OF NOTE When working with triangles, keeping these basic properties in mind will prevent errors and assist in their solution:

Side Angle B

1. The angles must sum to 180°. 2. The combined length of any two sides must exceed the length of the third side. 3. Longer sides will be opposite larger angles. 4. This sine of an angle cannot be greater than 1. 5. For y  10, 12 , the equation y  sin  has two solutions in (0°, 180°) that are supplements.

EXAMPLE 1

Side

Angle C

Angle B

Side

C

Figure 7.8 Angle

B

Side

Angle C

Figure 7.9

SAS

SSS

Side

Side

Side

Side Side

Since applying the law of sines requires we have a given side opposite a known angle, it cannot be used in the case of SAS or SSS triangles. These require the law of cosines, which we will develop in Section 7.2. In the case of ASA and AAS triangles, a unique triangle is formed since the measure of the third angle is fixed by the two angles given (they must sum to 180°) and the remaining sides must be of fixed length. 

Solving a Triangle Using the Law of Sines Solve the triangle shown, and state your answer using a table.

Solution



This is not a right triangle, so the standard ratios cannot be used. Since B and C are given, we know A  180°  1110°  32°2  38°. With A and side a, we have

B 110

39.0 cm

c A

32 b

C

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7-4

CHAPTER 7 Applications of Trigonometry

sin B sin A  a b sin 110° sin 38°  39 b b sin 38°  39 sin 110° 39 sin 110° b sin 38° b  59.5

WORTHY OF NOTE Although not a definitive check, always review the solution table to ensure the smallest side is opposite the smallest angle, the largest side is opposite the largest angle, and so on. If this is not the case, you should go back and check your work.

Repeating this procedure using

law of sines applied to A and B

substitute given values multiply by 39b divide by sin 38° result

sin C sin A  shows side c  33.6 cm. In table a c

form we have A. You’ve just learned to develop the law of sines and use it to solve ASA and AAS triangles

Angles

Sides (cm)

A  38°

a  39.0

B  110

b  59.5

C  32

c  33.6

Now try Exercises 7 through 24

B. Solving SSA Triangles—The Ambiguous Case

Figure 7.10 15

12

25

15

12 25

15



To understand the concept of unique and nonunique solutions regarding the law of sines, consider an instructor who asks a large group of students to draw a triangle with sides of 15 and 12 units, and a nonincluded 25° angle. Unavoidably, three different solutions will be offered (see Figure 7.10). For the SSA case, there is some doubt as to the number of solutions possible, or whether a solution even exists. To further understand why, consider a triangle with side c  30 cm, A  30°, and side a opposite the 30° angle (Figure 7.11— note the length of side b is yet to be determined). From our work with 30-60-90 triangles, we know if a  15 cm, it is exactly the length needed to form a right triangle (Figure 7.12). Figure 7.11

12

Figure 7.12 B

B

25 c  30 cm A

60

c  30 cm

a

a  15 cm

One solution

30 b  15√3

30 C

b

A

C

By varying the length of side a, we note three other possibilities. If side a 6 15 cm, no triangle is possible since a is too short to contact side b (Figure 7.13), while if 15 cm 6 side a 6 30 cm, two triangles are possible since side a will then intersect side b at two points, C1 and C2 (Figure 7.14). For future use, note that when two triangles are possible, angles C1 and C2 must be supplements since an-isosceles triangle is formed. Finally, if side a 7 30 cm, it will Figure 7.14

Figure 7.13

B

B c  30 cm

c  30 cm

a  10 cm

No 30 solution

A

b

C

A

Two a  20 cm solutions

30 C2 b

C1

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Figure 7.15

WORTHY OF NOTE

B

The case where three angles are known (AAA) is not considered since we then have a family of similar triangles, with infinitely many solutions.

c  30 cm

a  35 cm

30 A

C

b

intersect side b only once, forming the obtuse triangle shown in Figure 7.15, where we’ve assumed a  35 cm. Since the final solution is in doubt until we do further work, the SSA case is called the ambiguous case of the law of sines. EXAMPLE 2



Analyzing the Ambiguous Case of the Law of Sines Given triangle ABC with A  45° and side c  10012 mm, a. What length for side a will produce a right triangle where C  90°? b. How many triangles can be formed if side c  100√2 a  90 mm? c. If side a  120 mm, how many triangles can be formed? 45 A d. If side a  145 mm, how many triangles can be formed?

Solution



B

a  ?? mm

C

a. Recognizing the sides of a 45-45-90 triangle are in proportion according to 1x:1x: 12x, side a must be 100 mm for a right triangle to be formed. b. If a  90 mm, it will be too short to contact side b and no triangle is possible. c. As shown in Figure 7.16, if a  120 mm, it will contact side b in two distinct places and two triangles are possible. d. If a  145 mm, it will contact side b only once, since it is longer than side c and will “miss” side b as it pivots around B (see Figure 7.17). One triangle is possible. Figure 7.16 Figure 7.17 B

B

a  120 mm

c  100√2 45 A

C2

C1

a  145 mm

c  100√2 Misses side b A

C1

45 b

Now try Exercises 25 and 26



For a better understanding of the SSA (ambiguous) case, scaled drawings can initially be used along with a metric ruler and protractor. Begin with a horizontal line segment of undetermined length to represent the third (unknown) side, and use the protractor to draw the given angle on either the left or right side of this segment (we chose the left). Then use the metric ruler to draw an adjacent side of appropriate length, choosing a scale that enables a complete diagram. For instance, if the given sides are 3 ft and 5 ft, use 3 cm and 5 cm instead 11 cm  1 ft2. If the sides are 80 mi and 120 mi, use 8 cm and 12 cm 11 cm  10 mi2, and so on. Once the adjacent side is drawn, start at the free endpoint and draw a vertical segment to represent the remaining side. A careful sketch will often indicate whether none, one, or two triangles are possible (see the Reinforcing Basic Concepts feature on page 673).

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EXAMPLE 3



Solving the Ambiguous Case of the Law of Sines Solve the triangle with side b  100 ft, side c  60 ft, and C  28.0°.

Solution



Two sides and an angle opposite are given (SSA), A and we draw a diagram to help determine the possibilities. Draw the horizontal segment of b  100 ft c  60 ft some length and use a protractor to mark C  28°. Then draw a segment 10 cm long 28 (to represent b  100 ft) as the adjacent side of B1 the angle, with a vertical segment 6 cm long from C the free end of b (to represent c  60 ft). It seems apparent that side c will intersect the horizontal side in two places (see figure), and two triangles are possible. We apply the law of sines to solve the first triangle, whose features we’ll note with a subscript of 1. sin B1 sin C  c b sin B1 sin 28°  100 60 5 sin B1  sin 28° 3 B1  51.5°

law of sines

substitute

solve for sin B1 apply arcsine

Since B1  B2  180°, we know B2  128.5°. These values give 100.5° and 23.5° as the measures of A1 and A2, respectively. By once again applying the law of sines to each triangle, we find side a1  125.7 ft and a2  51.0 ft. See Figure 7.18.

WORTHY OF NOTE In Example 3, we found B2 using the property that states the angles in a triangle must sum to 180. We could also view B1 as a QI reference angle, which also gives a QII solution of 1180  51.52  128.5.

Angles

Sides (ft)

A1  100.5°

a1  125.7°

Angles

Sides (ft)

A2  23.5°

a2  51.0

B1  51.5°

b  100

B2  128.5°

b  100

C  28

c  60

C  28

c  60

Figure 7.18 First solution

A b  100 ft C

28.0

B2

b  100 ft

c  60 ft B1

C

28.0

Second solution

A1

100.5

a1 ≈ 125.7 ft

b  100 ft

c  60 ft 51.5

B1

C

A2

23.5

c  60 ft

28.0 128.5

a2 ≈ 51.0 ft B2

Now try Exercises 27 through 32



Admittedly, the scaled drawing approach has some drawbacks — it takes time to draw the diagrams and is of little use if the situation is a close call. It does, however, offer a deeper understanding of the subtleties involved in solving the SSA case. Instead of a scaled drawing, we can use a simple sketch as a guide, while keeping in mind the properties mentioned in the Worthy of Note on page 635.

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EXAMPLE 4



Solving the Ambiguous Case of the Law of Sines Solve the triangle with side a  220 ft, side b  200 ft, and A  40°.

Solution



The information given is again SSA, and we apply the law of sines with this in mind. sin B sin A  a b sin 40° sin B  220 200 200 sin 40° sin B  220 B1  35.7°

C

law of sines

b  200 ft

a  220 ft

substitute

solve for sin B

40

A

a

B1

apply arcsine

This is the solution from Quadrant I. The QII solution is about 1180  35.72°  144.3°. At this point our solution tables have this form: Angles

B. You’ve just learned how to solve SSA triangles (the ambiguous case) using the law of sines

Sides (ft)

Angles

Sides (ft)

A  40

a  220

A  40

a  220

B1  35.7°

b  200

B2  144.3°

C1 

c1 

C2 

b  200 c2 

It seems reasonable to once again find the remaining angles and finish by reapplying the law of sines, but observe that the sum of the two angles from the second solution already exceeds 180°: 40°  144.3°  188.3°! This means no second solution is possible (side a is too long). We find that C1  104.3°, and applying the law of sines gives a value of c1  331.7 ft. Now try Exercises 33 through 44



C. Applications of the Law of Sines As “ambiguous” as it is, the ambiguous case has a number of applications in engineering, astronomy, physics, and other areas. Here is an example from astronomy. EXAMPLE 5



Solving an Application of the Ambiguous Case — Planetary Distance The planet Venus can be seen from Earth with the naked eye, but as the diagram indicates, the position of Venus is uncertain (we are unable to tell if Venus is in the near position or the far position). Given the Earth is 93 million miles from the Sun and Venus is 67 million miles from the Sun, determine the closest and farthest possible distances that separate the planets in this alignment. Assume a viewing angle of   18° and that the orbits of both planets are roughly circular.

Solution



A close look at the information and diagram shows a SSA case. Begin by applying the law of sines where E S Earth, V S Venus, and S S Sun. sin V sin E  e v sin V sin 18°  67 93 93 sin 18° sin V  67 V  25.4°

Venus

law of sines

67 substitute given values

Venus ␪

solve for sin V apply arcsine

Earth

67 93

Sun

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This is the angle V1 formed when Venus is farthest away. The angle V2 at the closer distance is 180°  25.4°  154.6°. At this point, our solution tables have this form: Angles

Sides (106 mi)

E  18

e  67

V1  25.4°

v  93

S1 

s1 

Angles E  18

Sides (106 mi) e  67

V2  154.6°

v  93

S2 

s2 

For S1 and S2 we have S1  180  118  25.4°2  136.6° (larger angle) and S2  180  118  154.6°2  7.4° (smaller angle). Re-applying the law of sines for s1 shows the farther distance between the planets is about 149 million miles. Solving for s2 shows that the closer distance is approximately 28 million miles. Now try Exercises 47 and 48 EXAMPLE 6





Solving an Application of the Ambiguous Case — Radar Detection As shown in Figure 7.19, a radar ship is 30.0 mi off shore when a large fleet of ships leaves port at an angle of 43.0°. Figure 7.19 a. If the maximum range of the ship’s radar is 20.0 mi, Fleet will the departing fleet be detected? b. If the maximum range of the ship’s radar is 25.0 mi, Radar how far from port is the fleet when it is first detected? 20 mi

Solution



a. This is again the SSA (ambiguous) case. Applying the law of sines gives sin 43° sin   20 30 30 sin 43° sin   20 sin   1.02299754

Radar ship

43.0 30 mi Port

law of sines solve for sin  result

No triangle is possible and the departing fleet will not be detected. b. If the radar has a range of 25.0 mi, the radar beam will intersect the projected course of the fleet in two places. sin  sin 43°  law of sines 25 30 30 sin 43° sin   solve for sin  25   54.9° apply arcsine Figure 7.20

Radar 25 mi Radar ship

␪

125.1

␣ 43.0 30 mi Port

C. You’ve just learned how to use the law of sines to solve applications

This is the acute angle  related to the farthest point from port at which the fleet could be detected (see Figure 7.20). For the second triangle, we have 180°  54.9°  125.1° (the obtuse angle) giving a measure of 180°  1125.1°  43°2  11.9° for angle . For d as the side opposite  we have sin 43° sin 11.9°  25 d 25 sin 11.9° d sin 43°  7.6

law of sines solve for d simplify

This shows the fleet is first detected about 7.6 mi from port. Now try Exercises 49 and 50



There are a number of additional, interesting applications in the exercise set (see Exercises 51 through 70).

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7.1 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. For the law of sines, if two sides and an angle opposite one side are given, this is referred to as the case, since the solution is in doubt until further work. 2. Two inviolate properties of a triangle that can be used to help solve the ambiguous case are: (a) the angles must sum to and (b) no sine ratio can exceed . 3. For positive k, the equation sin   k has two solutions, one in Quadrant and the other in Quadrant . 

4. After a triangle is solved, you should always check to ensure that the side is opposite the angle. 5. In your own words, explain why the AAS case results in a unique solution while the SSA case does not. Give supporting diagrams. 6. Explain why no triangle is possible in each case: a. A  34°, B  73°, C  52°, a  14¿, b  22¿, c  18¿ b. A  42°, B  57°, C  81°, a  7–, b  9–, c  22–

DEVELOPING YOUR SKILLS

Solve each of the following equations for the unknown part (if possible). Round sides to the nearest hundredth and degrees to the nearest tenth.

7.

sin 32° sin 18.5°  a 15

sin C sin 19°  48.5 43.2

12.

15. side b  10 13 in. A  30° B  60° 16.

121 29

19.

A  45° B  45° side c  1512 mi

20.

21.

B  103.4° side a  42.7 km C  19.6°

22.

sin 38° sin B  125 190

Solve each triangle using the law of sines. If the law of sines cannot be used, state why. Draw and label a triangle or label the triangle given before you begin.

13. side a  75 cm A  38° B  64°

89 yd

sin 30° sin 52°  b 12

sin B sin 105° 10.  3.14 6.28

sin 63° sin C 9.  21.9 18.6 11.

8.

18.

13

14. side b  385 m B  47° A  108° 23.

112 0.8 cm

98 56

24.

17.

47

33

19 in. 102

126.2 mi

22

7.2 m 27

A  20.4° side c  12.9 mi B  63.4°

37

27.5 cm

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Answer each question and justify your response using a diagram, but do not solve.

25. Given ^ABC with A  30° and side c  20 cm, (a) what length for side a will produce a right triangle? (b) How many triangles can be formed if side a  8 cm? (c) If side a  12 cm, how many triangles can be formed? (d) If side a  25 cm, how many triangles can be formed?

35.

36. c

a

33. a

58 ft c 67 ft b

59

C

34.

465 mm

59

C

b

37.

25

10 2.6  c

B

mi

a

mi

Use the law of sines to determine if no triangle, one triangle, or two triangles can be formed from the diagrams given (diagrams may not be to scale), then solve. If two solutions exist, solve both completely. Note the arrowhead marks the side of undetermined length.

62 b

38.

C

B 6 .8  c

5.9



10 13

km

10 13 km a

A

51 b

For Exercises 39 to 44, assume the law of sines is being applied to solve a triangle. Solve for the unknown angle (if possible), then determine if a second angle (0    180) exists that also satisfies the proportion.

39.

sin A sin 48°  12 27

40.

sin B sin 60°  32 9

41.

sin 57° sin C  35.6 40.2

42.

sin 65° sin B  5.2 4.9

43.

sin A sin 15°  280 52

44.

sin B sin 29°  121 321

a 432 cm

c



398 mm

25

32. side b  24.9 km B  45° side a  32.8 km

38 b 382 cm

B

10

31. side c  58 mi C  59° side b  67 mi

A

b



30. side c  10 13 in. A  60° side a  15 in.

A

38

C

28. side a  36.5 yd B  67° side b  12.9 yd

29. side c  25.8 mi A  30° side a  12.9 mi

6.7 km

2.9

27. side b  385 m B  67° side a  490 m

c

10.9 km a

26. Given ^ABC with A  60° and side c  6 13 m, (a) what length for side a will produce a right triangle? (b) How many triangles can be formed if side a  8 m? (c) If side a  10 m, how many triangles can be formed? (d) If side a  15 m, how many triangles can be formed? Solve using the law of sines and a scaled drawing. If two triangles exist, solve both completely.

B

C

WORKING WITH FORMULAS

45. Triple angle formula for sine: sin132  3 sin   4 sin3 Most students are familiar with the double angle formula for sine: sin122  2 sin  cos . The

triple angle formula for sine is given here. Use the formula to find an exact value for sin 135°, then verify the result using a reference angle.

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46. Radius of a circumscribed circle: R 

b 2 sin B

Given ^ABC is circumscribed by a circle of radius R, the radius of the circle can be found using the formula shown, where side b is opposite angle B. Find the radius of the circle shown. 

643

Section 7.1 Oblique Triangles and the Law of Sines

B 51 c A

R a

b  15 cm

C

APPLICATIONS

47. Planetary distances: In a solar system that parallels our own, the planet Sorus can be seen from a Class M planet Exercise 47 with the naked eye, but Sorus as the diagram indicates, the position 51 of Sorus is uncertain. Sorus 51 Assume the orbits of Sun both planets are ␪ 82 roughly circular and that the viewing angle  Class M is about 20°. If the Class M planet is 82 million miles from its sun and Sorus is 51 million miles from this sun, determine the closest and farthest possible distances that separate the planets in this alignment. 48. Planetary distances: In a solar system that parallels our own, the planet Cirrus can be seen from a Class M planet Exercise 48 with the naked eye, but Cirrus as the diagram indicates, the position 70 of Cirrus is uncertain. 70 Assume the orbits of Sun Cirrus ␪ both planets are 105 roughly circular and Class M that the viewing angle  is about 15°. If the Class M planet is 105 million miles from its sun and Cirrus is 70 million miles from this sun, determine the closest and farthest possible distances that separate the planets in this alignment. 49. Radar detection: A radar ship is 15.0 mi off shore from a major port when a large fleet of ships leaves the port at the 35.0° angle shown. (a) If the maximum range of the Radar ship’s radar is 8.0 mi, will 8 mi the departing fleet be 35.0 detected? (b) If the 15 mi Radar Port maximum range of the ship ship’s radar is 12 mi, how far from port is the fleet when it is first detected?

50. Motion detection: To notify environmentalists of the presence of big game, motion detectors are installed 200 yd from a Exercise 50 watering hole. A pride of lions has just visited the hole and is leaving the Range 90 yd area at the 29.0° angle 29.0 shown. (a) If the Motion 200 yd Water maximum range of the detector motion detector is 90 yd, will the pride be detected? (b) If the maximum range of the motion detector is 120 yd, how far from the watering hole is the pride when first detected? Exercise 51 51. Distance between R cities: The cities of 80 km Van Gogh, 55 km Rembrandt, Pissarro, 40 and Seurat are S P situated as shown in V the diagram. Assume that triangle RSP is isosceles and use the law of sines to find the distance between Van Gogh and Seurat, and between Van Gogh and Pissarro. Exercise 52 52. Distance between M V O cities: The cities of 33 Mozart, Rossini, Offenbach, and Verdi 100 km 75 km are situated as shown in the diagram. Assume that triangle ROV is R isosceles and use the law of sines to find the distance between Mozart and Verdi, and between Mozart and Offenbach.

53. Distance to target: To practice for a competition, an archer stands as shown in the diagram and attempts to hit a moving target. (a) If the archer has a maximum effective 246 ft range of about 180 ft, can the target be 55 hit?

Exercise 53

Archer

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(b) What is the shortest range the archer can have and still hit the target? (c) If the archer’s range is 215 ft and the target is moving at 10 ft/sec, how many seconds is the target within range? 54. Distance to target: As part of an All-Star competition, a quarterback stands as shown in the diagram and attempts to hit a moving target Quarterback with a football. (a) If the quarterback has a maximum effective 50 yd range of about 35 yd, can the target be hit? (b) What is the shortest range the 53 quarterback can have and still hit the target? (c) If the quarterback’s range is 45 yd and the target is moving at 5 yd/sec, how many seconds is the target within range? In Exercises 55 and 56, three rods are attached via pivot joints so the rods can be manipulated to form a triangle. How many triangles can be formed if angle B must measure 26? If one triangle, solve it. If two, solve both. Diagrams are not drawn to scale.

55.

C

B

A

8

cm

cm

cm

12

??

56.

C

B

4

A

. in

11 . in

?? .

in

In the diagrams given, the measure of angle C and the length of sides a and c are fixed. Side c can be rotated at pivot point B. Solve any triangles that can be formed. (Hint: Begin by using the grid to find lengths a and c, then find angle C.) 57.

7-12

CHAPTER 7 Applications of Trigonometry

Length of a rafter: Determine the length of both roof rafters in the diagrams given.

59.

B Rafter c

Rafter a

53

32 C

A

42 ft

60.

B Rafter a

Rafter c

29

45 A

C

50 ft

61. Map distance: A cartographer is using aerial photographs to prepare a map for publication. The distance from Sexton to Rhymes is known to be 27.2 km. Using a protractor, the map T maker measures an angle of 96° from Sexton to Tarryson (a newly developed area) S 96 27 .2 58 and an angle of km 58° from Rhymes R to Tarryson. Compute each unknown distance. 62. Height of a fortress: An ancient fortress is built on a steep hillside, with the base of the fortress walls making a 102° angle with the hill. At the moment the fortress casts a 112-ft shadow, the angle of elevation from the tip of the shadow to the top of the wall is 32°. What is the distance from the base of the fortress to the top of the tower? Exercise 62

y a

B

C

c A x

58.

y

B

a

102 C

32 112 ft

c

x

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63. Distance to a fire: In Yellowstone Park, a fire is spotted by park rangers stationed in two towers that are known to be 5 mi apart. Using the line between them as a baseline, tower A reports the fire is at an angle of 39°, while tower B reports an angle of 58°. How far is the fire from the closer tower? 64. Width of a Exercise 64 canyon: To find the distance across Waimea Canyon (on the island of Kauai), a surveyor A 1000 m B marks a 1000-m 28 110 baseline along the southern rim. Using a transit, she sights on a large C rock formation on the north rim, and finds the angles indicated. How wide is the canyon from point B to point C? Exercise 65 65. Height of a blimp: When the GoodYear Blimp is viewed from the field-level bleachers near the southern end-zone of a 70 62 football stadium, the 145 yd N S angle of elevation is 62°. From the fieldlevel bleachers near the northern endzone, the angle of elevation is 70°. Find the height of the blimp if the distance from the southern bleachers to the northern bleachers is 145 yd. Eat at Ron

' s . . . ..

..

66. Height of a blimp: The rock-n-roll group Pink Floyd just finished their most recent tour and has moored their touring blimp at a hangar near the airport in Indianapolis, Indiana. From an unknown distance away, the angle of elevation is measured at 26.5°. After moving 110 yd closer, the angle of 

645

elevation has become 48.3°. At what height is the blimp moored? Exercises 67 and 68 67. Circumscribed triangles: A triangle is circumscribed within 5 cm the upper semicircle 27 63 drawn in the figure. Use 52 38 the law of sines to solve the triangle given the measures shown. What is the diameter of the circle? What do you notice about the triangle? 68. Circumscribed triangles: A triangle is circumscribed within the lower semicircle shown. Use the law of sines to solve the triangle given the measures shown. How long is the longer chord? What do you notice about the triangle? 69. Height of a mountain: Exercise 69 Approaching from the west, a group of hikers notes the angle of elevation to the summit of a steep 48 35 mountain is 35° at a 1250 m distance of 1250 meters. Arriving at the base of the mountain, they estimate this side of the mountain has an average slope of 48°. (a) Find the slant height of the mountain’s west side. (b) Find the slant height of the east side of the mountain, if the east side has an average slope of 65°. (c) How tall is the mountain? 70. Distance on a map: Coffeyville and Liberal, Kansas, lie along the state’s southern border and are roughly 298 miles apart. Olathe, Kansas, is Exercise 70 very near the state’s eastern border at an KANSAS Olathe angle of 23° with Liberal and 72° with Liberal Coffeyville 72 Coffeyville 23 298 mi (using the southern border as one side of the angle). (a) Compute the distance between these cities. (b) What is the shortest (straight line) distance from Olathe to the southern border of Kansas?

EXTENDING THE CONCEPT

71. Solve the triangle shown in three ways—first by using the law of sines, second using right triangle trigonometry, and third using the standard 30-60-90 triangle. Was one method “easier” than the others? Use these connections to express the irrational number 13 as a quotient of two trigonometric

functions of an angle. Can you find a similar expression for 12?

10.2 cm 30

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72. Use the law of sines and any needed identities to solve the triangles shown. 20 m

Exercise 74 74. Lines L1 and L2 shown are parallel. The three triangles between these lines all share the same base (in bold). Explain why all three triangles must have the same area.

B 2x

50 m

x 73. Similar to the law of A C sines, there is a law of tangents. The law says for any triangle 1 tan c 1A  B2 d 2 ab .  ABC, ab 1 tan c 1A  B2 d 2 Use the law of tangents to solve the triangle shown. B 45 cm 19 

L2

75. A UFO is sighted on a direct line between the towns of Batesville and Cave City, sitting stationary in the sky. The towns are 13 mi apart as the crow flies. A student in Batesville calls a friend in Cave City and both take measurements of the angle of elevation: 35° from Batesville and 42° from Cave City. Suddenly the UFO zips across the sky at a level altitude heading directly for Cave City, then stops and hovers long enough for an additional measurement from Batesville: 24°. If the UFO was in motion for 1.2 sec, at what average speed (in mph) did it travel?

31

A

L1

C

MAINTAINING YOUR SKILLS

76. (6.7) Find all solutions to the equation 2 sin x  cos12x2

78. (3.3) Write an equation for the real polynomial with smallest degree, possible, having the solutions x  2, x  1, and x  1  2i.

77. (6.2) Prove the given identity: tan2x  sin2x  tan2x sin2x

79. (2.3) Given the points 15, 32 and (4, 2), find (a) the equation of the line containing these points and (b) the distance between these points.

7.2 The Law of Cosines; the Area of a Triangle Learning Objectives In Section 7.2 you will learn how to:

A. Apply the law of cosines when two sides and an included angle are known (SAS)

B. Apply the law of cosines when three sides are known (SSS)

C. Solve applications using the law of cosines

D. Use trigonometry to find the area of a triangle

The distance formula d  21x2  x1 2 2  1y2  y1 2 2 is traditionally developed by placing two arbitrary points on a rectangular coordinate system and using the Pythagorean theorem. The relationship known as the law of cosines is developed in much the same way, but this time by using three arbitrary points (the vertices of a triangle). After giving the location of one vertex in trigonometric form, we obtain a formula that enables us to solve SSS and SAS triangles, which cannot be solved using the law of sines alone.

A. The Law of Cosines and SAS Triangles In situations where all three sides are known (but no angles), the law of sines cannot be applied. The same is true when two sides and the angle between them are known, since we must have an angle opposite one of the sides. In these two cases (Figure 7.21), side-side-side (SSS) and side-angle-side (SAS), we use the law of cosines. Figure 7.21 Law of Sines cannot be applied. B c  7 ft A

B

SSS

a  16 ft

b  18 ft

C

c  7 ft A

95 SAS b

a  16 ft C

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Figure 7.22 To solve these cases, it’s evident we need additional insight on the unknown angles. Cony C (b cos ␪, b sin ␪) sider a general triangle ABC on the rectangular coordinate system conveniently placed with b vertex A at the origin, side c along the x-axis, and y a the vertex C at some point (x, y) in QI (Figure A ␪ x c 7.22). Note cos   giving x  b cos , and B xc x (0, 0) b x y sin   or y  b sin . This means we can b write the point (x, y) as (b cos , b sin ) as shown, and use the Pythagorean theorem with side x  c to find the length of side a of the exterior, right triangle. It follows that

WORTHY OF NOTE Keep in mind that the sum of any two sides of a triangle must be greater than the remaining side. For example, if a  7, B  20, and C  12, no triangle is possible (see the figure). 12 cm

7 cm 20 cm

a2  1x  c2 2  y2  1b cos   c2 2  1b sin 2 2  b2cos2  2bc cos   c2  b2sin2  b2cos2  b2sin2  c2  2bc cos   b2 1cos2  sin22  c2  2bc cos   b2  c2  2bc cos 

Pythagorean theorem substitute b cos  for x and b sin  for y square binomial, square term rearrange terms factor out b2 substitute 1 for cos2  sin2

We now have a formula relating all three sides and an included angle. Since the naming of the angles is purely arbitrary, the formula can be used in any of the three forms shown. For the derivation of the formula where B is acute, see Exercise 61. The Law of Cosines For any triangle ABC and corresponding sides a, b, and c,

"

"

a2  b2  c2  2bc cos A b2  a2  c2  2ac cos B c2  a2  b2  2ab cos C

Note the relationship between the indicated angle and the squared term. In words, the law of cosines says that the square of any side is equal to the sums of the squares of the other two sides, minus twice their product times the cosine of the included angle. It is interesting to note that if the included angle is 90°, the formula reduces to the Pythagorean theorem since cos 90°  0. EXAMPLE 1



Verifying the Law of Cosines B

For the triangle shown, verify: a. c2  a2  b2  2ab cos C b. b2  a2  c2  2ac cos B Solution



c  20

Note the included angle C is a right angle. A a. c2  a2  b2  2ab cos C 2 2 2 20  10  110 132  2110 132 1102cos 90° 400  100  300  0  400 ✓ b. b2  a2  c2  2ac cos B 110 132 2  102  202  21102 1202cos 60° 1 300  100  400  400a b 2  500  200  300 ✓

30 30 b  10 √3

60 a  10 C

Now try Exercises 7 through 14



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CAUTION

When evaluating the law of cosines, a common error is to combine the coefficient of cos  with the squared terms (the terms shown in blue): a2  b2  c2  2bc cos A. Be sure to use the correct order of operations when simplifying the expression.

Once additional information about the triangle is known, the law of sines can be used to complete the solution. EXAMPLE 2



Solving a Triangle Using the Law of Cosines—SAS Solve the triangle shown. Write the solution in table form.

Solution



WORTHY OF NOTE After using the law of cosines, we often use the law of sines to complete a solution. With a little foresight, we can avoid the ambiguous case—since the ambiguous case occurs only if  could be obtuse (the largest angle of the triangle). After calculating the third side of a SAS triangle using the law of cosines, use the law of sines to find the smallest angle, since it cannot be obtuse. For SSS triangles, using the law of cosines to find the largest angle will ensure that when the second angle is found using the law of sines, it cannot be obtuse.

B 95

c  7.0 ft

The given information is SAS. Apply the law of A cosines with respect to side b and B: 2 2 2 law of cosines with respect to b b  a  c  2ac cos B 2 2 2 b  1162  172  21162 172cos 95° substitute known values simplify  324.522886 b  18.0 1324.522886  18.0

a  16.0 ft C

b

We now have side b opposite B, and complete the solution using the law of sines, selecting the smaller angle to avoid the ambiguous case (we could apply the law of cosines again, if we chose). sin B sin C  c b sin 95° sin C  7 18 sin 95° sin C  7 # 18 7 sin 95° C  sin1a b 18  22.8°

A. You’ve just learned how to apply the law of cosines when two sides and an included angle are known (SAS)

law of sines applied to C and B

substitute given values solve for sin C apply sin1

Angles result

Sides (ft)

A  62.2° a  16.0

For the remaining angle, C: 180°  195°  22.8°2  62.2°. The finished solution is shown in the table (given information is in bold).

B  95.0 b  18.0 C  22.8° c  7.0

Now try Exercises 15 through 26



B. The Law of Cosines and SSS Triangles When three sides of a triangle are given, we use the law of cosines to find any one of the three angles. As a good practice, we first find the largest angle, or the angle opposite the largest side. This will ensure that the remaining two angles are acute, avoiding the ambiguous case if the law of sines is used to complete the solution.

EXAMPLE 3



Solving a Triangle Using the Law of Cosines—SSS Solve the triangle shown. Write the solution in table form, with angles rounded to tenths of a degree.

C b  25 m A

c  28 m

a  15 m B

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Solution



The information is given as SSS. Since side c is the longest side, we apply the law of cosines with respect to side c and C: c2  a2  b2  2ab cos C 282  1152 2  1252 2  211521252cos C 784  850  750 cos C 66  750 cos C 0.088  cos C cos10.088  C 85.0  C

law of cosines with respect to c substitute known values simplify isolate variable term divide solve for C result

We now have side c opposite C and finish up using the law of sines. sin C sin A  a c sin 85° sin A  15 28 sin 85° sin A  15 # 28  0.5336757311 A  sin10.5336757311  32.3° B. You’ve just learned how to apply the law of cosines when three sides are known (SSS)

law of sines applied to A and C

substitute given values solve for sin A simplify solve for A result

Since the remaining angle must be acute, we compute it directly. B: 180°  185°  32.3°2  62.7°. The finished solution is shown in the table, with the information originally given shown in bold.

Angles

Sides (m)

A  32.3°

a  15

B  62.7°

b  25

C  85°

c  28

Now try Exercises 27 through 34



C. Applications Using the Law of Cosines As with the law of sines, the law of cosines has a large number of applications from very diverse fields including geometry, navigation, surveying, and astronomy, as well as being put to use in solving recreational exercises (see Exercises 37 through 40). EXAMPLE 4



Solving an Application of the Law of Cosines—Geological Surveys A volcanologist needs to measure the distance across the base of an active volcano. Distance AB is measured at 1.5 km, while distance AC is 3.2 km. Using a theodolite (a sighting instrument used by surveyors), angle BAC is found to be 95.7°. What is the distance across the base?

Solution



C. You’ve just learned how to solve applications using the law of cosines

The information is given as SAS. To find the distance BC across the base of the volcano, we apply the law of cosines with respect to A.

B

C

1.5 km

a2  b2  c2  2bc cos A  11.52 2  13.22 2  211.52 13.22cos 95.7°  13.44347 a  3.7

3.2 km 95.7 A

law of cosines with respect to a substitute known values simplify solve for a

The volcano is approximately 3.7 km wide at its base. Now try Exercises 41 through 52



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A variety of additional applications can be found in the exercise set (see Exercises 45 through 52).

D. Trigonometry and the Area of a Triangle While you’re likely familiar with the most common formula for a triangle’s area, A  12 bh, there are actually over 20 formulas for computing this area. Many involve basic trigonometric ideas, and we’ll use some of these ideas to develop three additional formulas here. Figure 7.23 For A  12bh, recall that b represents the length of a designated base, and h represents the B length of the altitude drawn to that base (see Figure 7.23). If the height h is unknown, but a c sides a and b with angle C between them are h h known, h can be found using sin C  , giving a h  a sin C. Figure 7.24 indicates the same C b A result is obtained if C is obtuse, since Figure 7.24 sin1180°  C2  sin C. Substituting B for in the formula A  12 b gives A  12 b , or A  12 ab sin C in more common form. Since naming the angles in a c triangle is arbitrary, the formulas A  12 bc sin A h a 1 and A  2 ac sin B can likewise be obtained. 180  C

C

b

A

Area Given Two Sides and an Included Angle (SAS) 1. A 

1 ab sin C 2

2. A 

1 bc sin A 2

3. A 

1 ac sin B 2

In words, the formulas say the area of a triangle is equal to one-half the product of two sides times the sine of the angle between them.

EXAMPLE 5



Finding the Area of a Nonright Triangle Find the area of ^ ABC, if a  16.2 cm, b  25.6 cm, and C  28.3°.

Solution



A

Since sides a and b and angle C are given, we apply the first formula. 1 ab sin C 2 1  116.22 125.62 sin 28.3° 2  98.3 cm2

A

area formula

25.6 cm

C

28.3 16.2 cm

B

substitute 15.2 for a, 25.6 for b, and 28.3° for C result

The area of this triangle is approximately 98.3 cm2. Now try Exercises 53 and 54



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Using these formulas, a second formula type requiring two angles and one side 2A 1 (AAS or ASA) can be developed. Solving for b in A  bc sin A gives b  . 2 c sin A 2A 1 Likewise, solving for a in A  ac sin B yields a  . Substituting these for b 2 c sin B 1 and a in A  ab sin C gives 2 A 2A 

1# # # a b sin C 2

given formula

2A # 2A # sin C c sin B c sin A

substitute

c2 sin A # sin B  2A # sin C

2A 2A for a, for b ; multiply by 2 c sin B c sin A

multiply by c sin A # c sin B ; divide by 2A

2

c sin A sin B A 2 sin C

solve for A

As with the previous formula, versions relying on side a or side b can also be found. Area Given Two Angles and Any Side (AAS/ASA) 1. A 

EXAMPLE 6



c2 # sin A # sin B 2 sin C

2. A 

a2 # sin B # sin C 2 sin A

3. A 

b2 # sin A # sin C 2 sin B

Finding the Area of a Nonright Triangle Find the area of ^ABC if a  34.5 ft, B  87.9°, and C  29.3°.

Solution



A

Since side a is given, we apply the second version of the formula. First we find the measure of angle A, then make the appropriate substitutions: A  180°  187.9  29.32°  62.8° a2sin B sin C A 2 sin A 134.52 2sin 87.9° sin 29.3°  2 sin 62.8°  327.2 ft2

C

29.3 34.5 ft

87.9 B

area formula—side a substitute 34.5 for a, 87.9° for B, 29.3° for C, and 62.8° for A simplify

The area of this triangle is approximately 327.2 ft2. Now try Exercises 55 and 56



Our final formula for a triangle’s area is a useful addition to the other two, as it requires only the lengths of the three sides. The development of the formula requires only a Pythagorean identity and solving for the angle C in the law of cosines, as follows. a2  b2  2ab cos C  c2

law of cosines

a  b  c  2ab cos C

add 2ab cos C, subtract c2

a2  b2  c2  cos C 2ab

divide by 2ab

2

2

2

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Beginning with our first area formula, we then have 1 ab sin C 2

previous area formula



1 ab 21  cos2C 2

sin2C  cos2C  1 S sin C  21  cos2C



a2  b2  c2 2 1 ab 1  a b 2 B 2ab

substitute

A

a2  b2  c2 for cos C 2ab

and can find the area of any triangle given its three sides. While the formula certainly serves this purpose, it is not so easy to use. By working algebraically and using the perimeter of the triangle, we can derive a more elegant version. 1 2 2 a2  b2  c2 2 a b c1  a b d 4 2ab

square both sides



1 2 2 a2  b2  c2 a2  b2  c2 a b c1  a bd c1  a bd 4 2ab 2ab

factor as a difference of squares



1 2 2 2ab  a2  b2  c2 2ab  a2  b2  c2 ab c dc d 4 2ab 2ab

1

2 2 2 1a2  2ab  b2 2  c2 1 1a  2ab  b 2  c c dc d 4 2 2

rewrite/regroup numerator; cancel a2b2

A2 



1 3 1a  b2 2  c2 4 3 c2  1a  b2 2 4 16 1 1a  b  c21a  b  c2 1c  a  b2 1c  a  b2  16 

2ab ; combine terms 2ab

factor (binomial squares)

factor (difference of squares)

For the perimeter p  a  b  c, we note the following relationships: a  b  c  p  2c

c  a  b  p  2b

c  a  b  p  2a

and making the appropriate substitutions gives 

1 p1p  2c21p  2b21p  2a2 16

substitute

While this would provide a usable formula for the area in terms of the perimeter, we p abc  . Since can refine it further using the semiperimeter s  2 2 1 4 1  a b , we can write the expression as 16 2 

p p  2c p  2b p  2a a ba ba b 2 2 2 2

rewrite expression



p p p p a  cb a  bb a  ab 2 2 2 2

simplify

 s1s  c21s  b2 1s  a2

substitute s for

p 2

Taking the square root of each side produces what is known as Heron’s formula. A  1s1s  a21s  b2 1s  c2

Heron’s formula

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Heron’s Formula Given ^ ABC with sides a, b, and c and semiperimeter s 

abc , 2

the area of the triangle is A  1s1s  a21s  b2 1s  c2 EXAMPLE 7



Solving an Application of Heron’s Formula—Construction Planning A New York City developer wants to build condominiums on the triangular lot formed by Greenwich, Watts, and Canal Streets. How many square meters does the developer have to work with if the frontage along each street is approximately 34.1 m, 43.5 m, and 62.4 m, respectively?

Solution



The perimeter of the lot is p  34.1  43.5  62.4  140 m, so s  70 m. By direct substitution we obtain A  1s1s  a21s  b21s  c2

D. You’ve just learned how to use trigonometry to find the area of a triangle

Heron’s formula

 170170  34.12170  43.52170  62.42

substitute known values

 170135.92126.5217.62

simplify

 1506,118.2

multiply

 711.4

result

The developer has about 711.4 m2 of land to work with. Now try Exercises 57 and 58



For a derivation of Heron’s formula that does not depend on trigonometry, see Appendix V.

7.2 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. When the information given is SSS or SAS, the law of is used to solve the triangle. 2. Fill in the blank so that the law of cosines is complete: c2  a2  b2  cos C 3. If the law of cosines is applied to a right triangle, the result is the same as the theorem, since cos 90°  0.

4. Write out which version of the law of cosines you would use to begin solving the A triangle shown:

B

52 m 17 37 m

C

5. Solve the triangle in Exercise 4 using only the law of cosines, then by using the law of cosines followed by the law of sines. Which method was more efficient? 6. Begin with a2  b2  c2  2bc cos A and write cos A in terms of a, b, and c (solve for cos A). Why must b2  c2  a2 6 2bc hold in order for a solution to exist?

653

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DEVELOPING YOUR SKILLS

Determine whether the law of cosines can be used to begin the solution process for each triangle.

7.

8. A

B 28 km

9.

15 mi 12 mi

11.

B

10.

A

15 in.

85

A

30 km

C

12 in.

15 km

C

C

6.8 AU

C

49

23. side c  25.8 mi B  30° side a  12.9 mi Solve using the law of cosines (if possible). Label each triangle appropriately before you begin.

24.

92

88

B

B

12.

B

12.5 yd

29 49 cm

C

C

52.4 km

80

70

15 yd

25. B

A 538 mm

32.5 ft

577 mi C

Solve each of the following equations for the unknown part.

27. side c  10 13 in. side b  6 13 in. side a  15 13 in.

30. B

432 cm 382 cm

16. 12.9  15.2  9.8  2115.2219.82cos C 2

2

17. a2  92  72  2192 172cos 52°

19. 102  122  152  21122 1152cos A

A

31.

A

2.3 

21. side a  75 cm C  38° side b  32 cm

10 25

2

Solve each triangle using the law of cosines.

2.9  10 25 m

i

20. 202  182  98  2118221982cos B 2

C

208 cm

mi

18. b2  3.92  9.52  213.9219.52cos 30°

28. side a  282 ft side b  129 ft side c  300 ft

29. side a  32.8 km side b  24.9 km side c  12.4 km

15. 42  52  62  2152 162cos B

2

C

A

105

2

27.5 ft

141

26.6 km

30 45 1114.7 mi

A

C

B

A

B 816 mi

29 465 mm

A

B

14.

6.7 km C

26.

50 km 30

C

98

A 8 yd

For each triangle, verify all three forms of the law of cosines.

13.

B 10.9 km

B

50 cm A

A

22. side b  385 m C  67° side a  490 m

C

25

4.1

0 1

mi

B

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32.

33. side a  12 13 yd side b  12.9 yd side c  9.2 yd

3

10 1
8 6.

4.9

2.9
1013 km

km

C

B


34. side a  36.5 AU side b  12.9 AU side c  22 AU

A 13 km
10

WORKING WITH FORMULAS

35. Alternative form for the law of cosines: B

b2  c2  a2 2bc 52 m 39 m By solving the law of cosines for the cosine of the angle, the formula A C 37 m can be written as shown. Derive this formula (solve for cos
), beginning from a2  b2  c2  2bc cos A, then use this form to begin the solution of the triangle given. cos A 




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36. The Perimeter of a Trapezoid: P  a  b  h1csc   csc ) a The perimeter of a trapezoid can be h found using the formula shown,   where a and b b represent the lengths of the parallel sides, h is the height of the trapezoid, and  and  are the base angles. Find the perimeter of Trapezoid Park (to the nearest foot) if a  5000 ft, b  7500 ft, and h  2000 ft, with base angles   42° and   78°.

APPLICATIONS

37. Distance between cities: The satellite Mercury II measures its distance from Portland and from Green Bay using radio waves as shown. Using an on-board sighting device, the satellite determines that
M is 99°. How many miles is it from Portland to Green Bay?

38. Distance between cities: Voyager VII measures its distance from Los Angeles and from San Francisco using radio waves as shown. Using an on-board sighting device, the satellite determines
V is 95°. How many kilometers separate Los Angeles and San Francisco? Voyager VII

Mercury II

V

M

95

99

488 km

311 km

1435 mi

692 mi S

P

L

G San Francisco

Portland

m

Green Bay

WORTHY OF NOTE In navigation, there are two basic methods for defining a course. Headings are understood to be the amount of rotation from due north in the clockwise direction 10 
6 360°2 . Bearings give the number of degrees East or West from a due North or due South orientation, hence the angle indicated is always less than 90°. For instance, the bearing N 25° W and a heading of 335° would indicate the same direction.

Los Angeles

v

39. Trip planning: A business executive is going to fly the corporate jet from Providence to College Cove. Exercise 39 North West

College Cove C

East South

p  198 mi

M Mannerly Main

m  354 mi

c  423 mi

P Providence

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She calculates the distances shown using a map, with Mannerly Main for reference since it is due east of Providence. What is the measure of angle P? What heading should she set for this trip? 40. Trip planning: A troop of Scouts is planning a hike from Montgomery to Pattonville. They calculate the distances shown using a map, using Bradleyton for reference since it is due east of Montgomery. What is the measure of angle M? What heading should they set for this trip? Exercise 40 p  21 mi

M

B

b  18 mi

m  10 mi P

41. Runway length: Surveyors are measuring a large, marshy area outside of the city as part of a feasibility study for the construction of a new airport. Using a theodolite and the markers shown gives the information indicated. If the main runway must be at least 11,000 ft long, and environmental concerns are satisfied, can the airport be constructed at this site (recall that 1 mi  5280 ft)?

43. Aerial distance: Two planes leave Los Angeles International Airport at the same time. One travels due west (at heading 270°) with a cruising speed of 450 mph, going to Tokyo, Japan, with a group that seeks tranquility at the foot of Mount Fuji. The other travels at heading 225° with a cruising speed of 425 mph, going to Brisbane, Australia, with a group seeking adventure in the Great Outback. Approximate the distance between the planes after 5 hr of flight. 44. Nautical distance: Two ships leave Honolulu Harbor at the same time. One travels 15 knots (nautical miles per hour) at heading 150°, and is going to the Marquesas Islands (Crosby, Stills, and Nash). The other travels 12 knots at heading 200°, and is going to the Samoan Islands (Samoa, le galu a tu). How far apart are the two ships after 10 hr? 45. Geoboard geometry: A rubber band is placed on a geoboard (a board with all pegs 1 cm apart) as shown. Approximate the perimeter of the triangle formed by the rubber band and the angle formed at each vertex. (Hint: Use a standard triangle to find A and length AB.) Exercise 45 B

C A 1.8 mi

2.6 mi 51

42. Tunnel length: An engineering firm decides to bid on a proposed tunnel through Harvest Mountain. In order to find the tunnel’s length, the measurements shown are taken. (a) How long will the tunnel be? (b) Due to previous tunneling experience, the firm estimates a cost of $5000 per foot for boring through this type of rock and constructing the tunnel according to required specifications. If management insists on a 25% profit, what will be their minimum bid to the nearest hundred?

685 yd

610 yd 79

46. Geoboard geometry: A rubber band is placed on a geoboard as shown. Approximate the perimeter of the triangle formed by the rubber band and the angle formed at each vertex. (Hint: Use a Pythagorean triple, then find angle A.) Exercise 46

B

C A

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In Exercises 47 and 48, three rods are attached via pivot joints so the rods can be manipulated to form a triangle. Find the three angles of the triangle formed.

47.

C B A

9 cm

12 cm 20 cm

48. A

B C 6 in.

11 in. 16 in.

Exercise 49

49. Pentagon perimeter: Find the perimeter of a regular pentagon that is circumscribed by a circle with radius r  10 cm.

m

10 c

50. Hexagon perimeter: Find the perimeter of a regular hexagon that is circumscribed by a circle with radius r  15 cm.

Exercise 55 55. Pricing for undeveloped lots: Undeveloped land in a popular resort area is selling for $3,000,000/acre. Given the dimensions of the lot 42° 65° shown, (a) find what 299 ft percent of a full acre is being purchased (to the nearest whole percent), and (b) compute the cost of the lot. Recall that 1 acre  43,560 ft2. 56. Area of the Nile River Delta: The Nile River Delta is one of the world’s largest. The delta begins slightly up river from the Egyptian capitol (Cairo) and stretches along the Mediterranean from Alexandria in the west to Port Said in the east (over 240 km). Approximate the area of this rich agricultural region using the two triangles shown.

Alexandria

20° 42°

Solve the following triangles. Round sides and angles to the nearest tenth. (Hint: Use Pythagorean triples.)

51.

52.

y A

C

218 km Nile Delta

20° 45°

Port Said

Cairo

y A

B

B

x

C

x

Exercise 53 53. Billboard design: Creative Designs iNc. has designed a flashy, new billboard for "Spice one of its clients. Using a rectangular highway up" billboard measuring 20 ft ␣ by 30 ft, the primary advertising area is a triangle formed using the diagonal of the billboard as one side, and one-half the base as another (see figure). Use the dimensions given to find the angle  formed at the corner, then compute the area of the triangle using two sides and this included angle.

54. Area caught by Exercise 54 surveillance camera: A stationary 110 ft surveillance camera is 225 ft set up to monitor 38° activity in the parking lot of a shopping mall. If the camera has a 38° field of vision, how many square feet of the parking lot can it tape using the dimensions given?

57. Area of the Yukon Territory: The Yukon Territory in northwest Canada is roughly triangular in shape with sides of 1289 km, 1063 km, and 922 km. What is the approximate area covered by this territory?

YUKON

58. Alternate method for computing area: Referring to Exercise 53, since the dimensions of the billboard are known, all three sides of the triangle can actually be determined. Find the length of the sides rounded to the nearest whole, then use Heron’s formula to find the area of the triangle. How close was your answer to that in Exercise 53?

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EXTENDING THE CONCEPT

59. No matter how hard I try, I cannot solve the triangle shown. Why? Exercise 59 Sputnik 10 B

502 mi

387 mi

A

902 mi

C

60. In Figure 7.22 (page 647, note that if the x-coordinate of vertex B is greater than the x-coordinate of vertex C, B becomes acute, and C obtuse. How does this change the relationship 

between x and c? Verify the law of cosines remains unchanged. Exercise 61 61. For the triangle shown, B verify that 117 c  b cos A  a cos B, 53.9 mi 37 mi then use two different 25 38 forms of the law of A C 78 mi cosines to show this relationship holds for any triangle ABC. 62. Most students are familiar with this double-angle formula for cosine: cos122  cos2  sin2. The triple angle formula for cosine is cos132  4 cos3  3 cos . Use the formula to find an exact value for cos 135°. Show that you get the same result as when using a reference angle.

MAINTAINING YOUR SKILLS

63. (4.4) Write the expression as a single term in simplest form: 2 log24  2 log23  2 log26 64. (5.4) State exact forms for each of the following:   7 sina b, cosa b, and tana b. 6 6 3

65. (5.7) Use fundamental identities to find the values of all six trig functions that satisfy the conditions. 5 sin x   and cos x 7 0. 13

66. (3.2) Use synthetic division to show f 122 7 0 for f 1x2  x4  x3  7x2  x  6.

7.3 Vectors and Vector Diagrams Learning Objectives In Section 7.3 you will learn how to:

A. Represent a vector quantity geometrically

B. Represent a vector

The study of vectors is closely connected to the study of force, motion, velocity, and other related phenomena. Vectors enable us to quantify certain characteristics of these phenomena and to physically represent their magnitude and direction with a simple model. To quantify something means we assign it a relative numeric value for purposes of study and comparison. While very uncomplicated, this model turns out to be a powerful mathematical tool.

quantity graphically

C. Perform defined operations on vectors

D. Represent a vector quantity algebraically and find unit vectors

E. Use vector diagrams to solve applications

A. The Notation and Geometry of Vectors Measurements involving time, area, volume, energy, and temperature are called scalar measurements or scalar quantities because each can be adequately described by their magnitude alone and the appropriate unit or “scale.” The related real number is simply called a scalar. Concepts that require more than a single quantity to describe their attributes are called vector quantities. Examples might include Figure 7.25 force, velocity, and displacement, which require knowing a A B magnitude and direction to describe them completely. To begin our study, consider two identical airplanes flying Line AB at 300 mph, on a parallel course and in the same direction. A B Although we don’t know how far apart they are, what direction they’re flying, or if one is “ahead” of the other, we can still Segment AB model, “300 mph on a parallel course,” using directed line segA B ments (Figure 7.25). Drawing these segments parallel with the Directed segment AB arrowheads pointing the same way models the direction of

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flight, while drawing segments the same length indicates the velocities are equal. The directed segment used to represent a vector quantity is simply called a vector. In this case the length of the vector models the magnitude of the velocity, while the arrowhead indicates the direction of travel. The origin of the segment is called the initial point, with the arrowhead pointing to the terminal point. Both are labeled using capital letters as shown in Figure 7.26 and we call this a geometric representation of the vectors. Vectors can be !named using the initial and terminal points that define them (initial ! point first) as in AB and CD , or using a bold, small case letter with the favorites being v (first letter of the word vector) and u. Other small Figure 7.27 case, bold letters can be used and subscripted vector names (v1, v2, v3, . . .) are also common. Two vectors are equal if they have ! the same magnitude ! and direction. For! u  AB ! and v  CD , we can say u  v or AB  CD since both airplanes are v 300 flying at the same speed and in the same direction mph (Figure 7.26). Based on these conventions, it seems reasonx able to represent an airplane flying at 600 mph 150 600 with a vector that is twice as long as u and v, and w mph mph one flying at 150 mph with a vector that is half as long. If all planes are flying in the same direction on a parallel course, we can represent them geometrically as shown in Figure 7.27, and state that w  2v, x  12v, and w  4x. The multiplication of a vector by a constant is called scalar multiplication, since the product changes only the scale or size of the vector and not its direction. Finally, consider the airplane represented by vector v2, flying at 200 mph on a parallel course but in the opposite direction (see Figure 7.28). In this case, the directed 2 segment will be 200 300  3 as long as v and point in the opposite or “negative” direction. In perspective we can now state: v2  23v, v2  13w, v2  43x, or any equivalent form of these equations.

Figure 7.26 B D

300 mph A

659

Section 7.3 Vectors and Vector Diagrams

300 mph C

Figure 7.28

v 300 mph v2 200 mph

EXAMPLE 1



Using Geometric Vectors to Model Forces Acting on a Point Two tugboats are attempting to free a barge that is stuck on a sand bar. One is pulling with a force of 2000 newtons (N) in a certain direction, the other is pulling with a force of 1500 N in a direction that is perpendicular to the first. Represent the situation geometrically using vectors.

Solution



A. You’ve just learned how to represent a vector quantity geometrically

We could once again draw a vector of arbitrary length and let it represent the 2000-N force applied by the first tugboat. For better perspective, we can actually use a ruler and choose a convenient length, say 6 cm. We then represent the pulling force of the second tug 3 with a vector that is 1500 2000  4 as long (4.5 cm), drawn at a 90° angle with relation to the first. Note that many correct solutions are possible, depending on the direction of the first vector drawn.

3  35

N

Now try Exercises 7 through 12



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B. Vectors and the Rectangular Coordinate System Representing vectors geometrically (with a directed Figure 7.29 line segment) is fine for simple comparisons, but y many applications involve numerous vectors acting on (3, 4) a single point or changes in a vector quantity over time. For these situations, a graphical representation v u in the coordinate plane helps to analyze this interac3 tion. The only question is where to place the vector x on the grid, and the answer is—it really doesn’t matw 5 4 ter. Consider the three vectors shown in Figure 7.29. From the initial point of each, counting four units in the vertical direction, then three units in the horizontal direction, puts us at the terminal point. This shows the vectors are all 5 units long (since a 3-4-5 triangle is formed) and are all parallel ¢y 4  ). In other words, they are equivalent vectors. (since slopes are equal: ¢x 3 Since a vector’s location is unimportant, we can replace any given vector with a unique and equivalent vector whose initial point is (0, 0), called the position vector. WORTHY OF NOTE

Position Vectors

For vector u, the initial and terminal points are (
5,
1) and (
2, 3), respectively, yielding the position vector H
2
1
52, 3
1
12I  H3, 4I as before.

For a vector v with initial point (x1, y1) and terminal point (x2, y2), the position vector for v is v  Hx2
x1, y2
y1I,

an equivalent vector with initial point (0, 0) and terminal point 1x2
x1, y2,
y1 2.

For instance, the initial and terminal points of vector w in Figure 7.29 are 12,
42 and (5, 0), respectively, with 15
2, 0
1
42 2  13, 42. Since (3, 4) is also the terminal point of v (whose initial point is at the origin), v is the position vector for u and w. This observation also indicates that every geometric vector in the xy-plane corresponds to a unique ordered pair of real numbers (a, b), with a as the horizontal component and b as the vertical component of the vector. As indicated, we denote the vector in component form as Ha, bI, using the new notation to prevent confusing vector Ha, bI with the ordered pair (a, b). Finally, while each of the vectors in Figure 7.29 has a component form of H3, 4I, the horizontal and vertical components can be read directly only from v  H3, 4I, giving it a distinct advantage.

EXAMPLE 2




Solution




Verifying the Components of a Position Vector

Vector v  H12,
5I has initial point (
4, 3). a. Find the coordinates of the terminal point. b. Verify the position vector for v is also H12,
5I and find its length. a. Since v has a horizontal component of 12 and a 12 ␪r vertical component of
5, we add 12 to the xcoordinate and
5 to the y-coordinate of the initial v point. This gives a terminal point of 112  1
42,
5  32  18,
22. H12,
5I b. To verify we use the initial and terminal points to compute Hx2
x1, y2,
y1I, giving a position vector y of H8
1
42,
2
3I  H12,
5I. To find its length we can use either the Pythagorean theorem or simply note that a 5-12-13 Pythagorean triple is formed. Vector v has a length of 13 units.

x
5

Now try Exercises 13 through 20




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Figure 7.30 For the remainder of this section, vector v  Ha, bI will y refer to the unique position vector for all those equivalent to v. Upon considering the graph of Ha, bI (shown Ha, bI in QI for convenience in Figure 7.30), several things are |v| immediately evident. The length or magnitude of the vecb tor, which is denoted v, can be determined using the ␪r Pythagorean theorem: v  2a2  b2. In addition, basic a x trigonometry shows the horizontal component can be a found using cos   or a  vcos , with the vertical component being v b sin   or b  vsin . Finally, we note the angle  can be determined using v b b tan   a b, or r  tan1a b and the quadrant of v. a a

Vector Components in Trig Form For a position vector v  Ha, bI and angle , we have horizontal component: a  vcos  vertical component: b  vsin , where

b r  tan a b and a

y

1

Ha, bI

v  2a  b 2

2

v

␪

b ␪r

x

a

The ability to model characteristics of a vector using these equations is a huge benefit to solving applications, since we must often work out solutions using only the partial information given. EXAMPLE 3

Solution





Finding the Magnitude and Direction Angle of a Vector For v1  H2.5, 6I and v2  H313, 3I, a. Graph each vector and name the quadrant where it is located. b. Find their magnitudes. c. Find the angle  for each vector (round to tenths of a degree as needed).

y

H3√3, 3I v2

␪1

␪2 x

v1

a. The graphs of v1 and v2 are shown in the H2.5, 6I figure. Using the signs of each coordinate, we note that v1 is in QIII, and v2 is in QI. v2  213 132 2  132 2 b. v1  212.52 2  162 2  16.25  36  127  9  136  142.25  6.5 6 6 3 c. For v1: r  tan1a b For v2: r  tan1a b 2.5 3 13 13  tan1a  tan1 12.42  67.4° b  30° 3 In QIII,   247.4°. In QI,   30°. Now try Exercises 21 through 24



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CHAPTER 7 Applications of Trigonometry



EXAMPLE 4



Solution

Finding the Horizontal and Vertical Components of a Vector The vector v  Ha, bI is in QIII, has a magnitude of v  21, and forms an angle of 25° with the negative x-axis (Figure 7.31). Find the horizontal and vertical components of the vector, rounded to tenths.

Figure 7.31 205 x 25 |v|  21

Begin by graphing the vector and setting up the equations for its components. For r  25°,   205°.

Ha, bI y

Figure 7.32

For the horizontal component: For the vertical component: a  vcos  b  vsin   21 cos 205°  21 sin 205°  19  8.9 With v in QIII, its component form is approximately H19, 8.9I. As a check, we apply the Pythagorean theorem: 21192 2  18.92 2  21 ✓. See Figure 7.32.

19

x

25

8.9

|v|  21 Ha, bI y B. You’ve just learned how to represent a vector quantity graphically



C. Operations on Vectors and Vector Properties The operations defined for vectors have a close knit graphical representation. Consider a local park having a large pond with pathways around both sides, so that a park visitor can enjoy the view from either side. Suppose v  H8, 2I is the position vector representing a person who decides to turn to the right at the pond, while u  H2, 6I represents a person who decides to first turn left. At (8, 2) the first person changes direction and walks to (10, 8) on the other side of the pond, while the second person arrives at (2, 6) and turns to head for (10, 8) as well. This is shown graphically in Figure 7.33 and demonstrates that (1) a parallelogram is formed (opposite sides equal and parallel), (2) the path taken is unimportant relative to the destination, and (3) the coordinates of the destination represent the sum of corresponding coordinates from the terminal points of u and v: 12, 62  18, 22  12  8, 6  22  110, 82. In other words, the result of adding u and v gives the new position vector u  v  w, called the resultant or the resultant vector. Note the resultant vector is a diagonal of the parallelogram formed. Geometrically or graphically, the addition of vectors can be viewed as a “tail-to-tip” combination of one with another, by shifting one vector (without changing its direction) so that its tail (initial point) is at the tip (terminal point) of the other vector. This is illustrated in Figures 7.34 through 7.36.

Figure 7.33 y (100 8) (2 2

Now try Exercises 25 through 30

v

) x

Figure 7.34

Figure 7.35

Given vectors u and v

Figure 7.36

x

u

v

Shift vector u

Shift vector v x

v

u

u uv

uv

v Tail of v to the tip of u

y

u

v H10, 9I

H10, 9I y

Tail of u to the tip of v

y

x

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Figure 7.37

The subtraction of vectors can be understood as either as u  v or u  1v2. Since the location of a vector is unimportant relative to the information it carries, vector subtraction can be interpreted as the tipto-tip diagonal of the parallelogram from vector addition. In Figures 7.34 to 7.36, assume u  H1, 5I and v  H9, 4I. Then u  v  H1, 5I  H9, 4I  H1  9, 5  4I giving the position vector H8, 1I. By repositioning this vector with its tail at the tip of v, we note the new vector points directly at u, forming the diagonal (see Figure 7.37). Scalar multiplication of vectors also has a graphical representation that corresponds to the geometric description given earlier.

WORTHY OF NOTE The geometry of vector subtraction is a key part of resolving a vector into orthogonal components that are nonquadrantal. Applications of this concept are wide ranging, and include thrust and drag forces, tension and stress limits in a cable, and others.

x

v u uv (repositioned) u v

y

Operations on Vectors Given vectors u  Ha, bI, v  Hc, dI, and a scalar k,

1. u  v  Ha  c, b  dI 2. u  v  Ha  c, b  dI 3. ku  Hka, kbI for k  

If k 7 0, the new vector points in the same direction as u. If k 6 0, the new vector points in the opposite direction as u.

EXAMPLE 5

Solution





Representing Operations on Vectors Graphically

Given u  H3, 2I and v  H4, 6I compute each of the following and represent the result graphically: 1 1 a. 2u b. v c. 2u  v 2 2 Note the relationship between part (c) and parts (a) and (b). a. 2u  2H3, 2I

b.

 H6, 4I y

1 1 v  H4, 6I 2 2  H2, 3I

c. 2u 

y

1 v  H6, 4I  H2, 3I 2  H8, 1I y

H6, 4I

H6, 4I 2u

2u

u H3, 2I

x

x

qv

H2, 3I

H8, 1I

qv

x

H2, 3I

v  H4, 6I

Now try Exercises 31 through 48



The properties that guide operations on vectors closely resemble the familiar properties of real numbers. Note we define the zero vector 0  H0, 0I as one having no magnitude or direction.

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Properties of Vectors For vector quantities u, v, and w and real numbers c and k, 1. 3. 5. 7. 9.

1u  u uvvu 1u  v2  w  u  1v  w2 u0u k1u  v2  ku  kv

2. 4. 6. 8. 10.

0u  0  k0 u  v  u  1v2 1ck2u  c1ku2  k1cu2 u  1u2  0 1c  k2u  cu  ku

Proof of Property 3

For u  Ha, bI and v  Hc, dI, we have

u  v  Ha, bI  Hc, d I

 Ha  c, b  d I

 Hc  a, d  bI

C. You’ve just learned how to perform defined operations on vectors

 Hc, dI  Ha, bI vu

sum of u and v vector addition commutative property vector addition result

Proofs of the other properties are similarly derived (see Exercises 89 through 97).

D. Algebraic Vectors, Unit Vectors, and i, j Form

While the bold, small case v and the Ha, bI notation for vectors has served us well, we now introduce an alternative form that is somewhat better suited to the algebra of vectors, and is used extensively in some of the physical sciences. Consider the vector H1, 0I, a vector 1 unit in length extending along the x-axis. It is called the horizontal unit vector and given the special designation i (not to be confused with the imaginary unit i  11). Likewise, the vector H0, 1I is called the vertical unit vector and given the designation j (see Figure 7.38). Using scalar multiplication, the unit vector along the negative x-axis is i and along the negative y-axis is Figure 7.38 j. Similarly, the vector 4i represents a position vector 4 units long y (0, 1) along the x-axis, and 5j represents a position vector 5 units long along the negative y-axis. Using these conventions, any nonquadrantal vector Ha, bI can be written as a linear combination of i and j, with j (1, 0) a and b expressed as multiples of i and j, respectively: ai  bj. These x i ideas can easily be generalized and applied to any vector. WORTHY OF NOTE

Algebraic Vectors and i, j Form

Earlier we stated, “Two vectors were equal if they have the same magnitude and direction.” Note that this means two vectors are equal if their components are equal.

EXAMPLE 6

For the unit vectors i  H1, 0I and j  H0, 1I, any arbitrary vector v  Ha, bI can be written as a linear combination of i and j: v  ai  bj Graphically, v is being expressed as the resultant of a vector sum. 

Finding the Horizontal and Vertical Components of Algebraic Vectors Vector u is in QII, has a magnitude of 15, and makes an angle of 20° with the negative x-axis. a. Graph the vector. b. Find the horizontal and vertical components (round to one decimal place) then write u in component form. c. Write u in terms of i and j.

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Solution



b. Horizontal Component Figure 7.40 Ha, bI x

a  vcos   15 cos 160°  14.1

y

|u|  15

y

Vertical Component

b  vsin   15 sin 160 °  5.1

Ha, bI |u|  15   20

160 x

With the vector in QII, u  H14.1, 5.1I in component form. c. In terms of i and j we have u  14.1i  5.1j. See Figure 7.40

5.1

20

Figure 7.39

a. The vector is graphed in Figure 7.39.

14.1

Now try Exercises 49 through 62

Some applications require that we find a nonhorizontal, nonvertical vector one unit in length, having the same direction as a given vector v. To understand how this is done, consider vector v  H6, 8I. Using the Pythagorean theorem we find v  10, and can form a 6-8-10 triangle using the horizontal and vertical components (Figure 7.41). Knowing that similar triangles have sides that are proportional, we can find a unit vector in the same direction as v by dividing all three sides by 10, giving a triangle with sides 35, 45, and 1. The new vector “u” (along the hypotenuse) indeed points in the same direction since we have merely shortened v, and is a unit vector 3 2 16 4 2 9   1. In retrospect, we have divided the components since a b  a b S 5 5 25 25 of vector v by its magnitude v (or multiplied components by the reciprocal of v) to H6, 8I v 6 8 3 4  h , i  h , i . In general we obtain the desired unit vector:  v 10 10 10 5 5 have the following:

Figure 7.41 y

H6, 8I v

8 10 6



x

Unit Vectors For any nonzero vector v  Ha, bI  ai  bj, the vector v a b u  i j 2 2 2 v 2a  b 2a  b2 is a unit vector in the same direction as v. You are asked to verify this relationship in Exercise 100. In summary, for vector v  6i  8j, we find v  262  82  10, so the unit vector pointing in the same v 3 4 direction is  i  j. See Exercises 63 through 74. v 5 5 EXAMPLE 7



Using Unit Vectors to Find Coincident Vectors Vectors u and v form the 37° angle illustrated in the figure. Find the vector w (in red), which points in the same direction as v (is coincident with v) and forms the base of the right triangle shown. y

Solution



WORTHY OF NOTE In this context w is called the projection of u on v, an idea applied more extensively in Section 7.4 D. You’ve just learned how to represent a vector quantity algebraically and find unit vectors

Using the Pythagorean theorem we find u  7.3 and v  10. Using the cosine of 37° the magnitude of w is then w  7.3 cos 37° or about 5.8. To ensure that w will point in the same direction as v, we simply multiply the 5.8 magnitude by the unit H8, 6I v  15.82H0.8, 0.6I, vector for v: w  15.82 v 10 and we find that w  H4.6, 3.5I. As a check we use the Pythagorean theorem: 24.62  3.52  133.41  5.8.

H2, 7I

u

H8, 6I v

37 w

Now try Exercises 75 through 78

x



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E. Vector Diagrams and Vector Applications Applications of vectors are virtually unlimited, with many of these in the applied sciences. Here we’ll look at two applications that are an extension of our work in this section. In Section 7.4 we’ll see how vectors can be applied in a number of other creative and useful ways. In Example 1, two tugboats were pulling on a barge to dislodge it from a sand bar, with the pulling force of each represented by a vector. Using our knowledge of vector components, vector addition, and resultant forces (a force exerted along the resultant), we can now determine the direction and magnitude of the resultant force if we know the angle formed by one of the vector forces and the barge. EXAMPLE 8



Solving an Application of Vectors—Force Vectors Acting on a Barge Two tugboats are attempting to free a barge that is stuck on a sand bar, and are exerting the forces shown in Figure 7.42. Find the magnitude and direction of the resultant force.

Solution



Begin by orienting the diagram on a coordinate grid (see Figure 7.43). Since the angle between the vectors is 90°, we know the acute angle formed by the first tugboat and the x-axis is 55°. With this information, we can write each vector in “i, j” form and add the vectors to find the resultant.

Figure 7.42

2000 N 1500 N

35°

For vector v1 (in QII): Horizontal Component

Vertical Component

a  v1cos   2000 cos 125°  1147

b  v1sin   2000 sin 125°  1638

v1  1147i  1638j.

Figure 7.43

For vector v2 (in QI):

y

Horizontal Component

Vertical Component

a  v2cos   1500 cos 35°  1229

b  v2sin   1500 sin 35°  860

v2  1229i  860j.

This gives a resultant of v1  v2  11147i  1638j2  11229i  860j2  82i  2498j, with magnitude v1  v2  2822  24982  2499 N. To find the direction 2498 of the force, we have r  tan1a b, or about 88°. 82

v1 2000 N

125v

2

55

35

1500 N x

Now try Exercises 81 and 82



It’s worth noting that a single tugboat pulling at 88° with a force of 2499 N would have the same effect as the two tugs in the original diagram. In other words, the resultant vector 82i  2498j truly represents the “result” of the two forces. Knowing that the location of a vector is unimportant enables us to model and solve a great number of seemingly unrelated applications. Although the final example concerns aviation, headings, and crosswinds, the solution process has a striking similarity to the “tugboat” example just discussed. In navigation, headings involve a single

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667

angle, which is understood to be the amount of rotation from due north in the clockwise direction. Several headings are illustrated in Figures 7.44 through 7.47. Figure 7.44

Figure 7.45

Figure 7.46

Figure 7.47

North West

East

Heading 30

Heading 330

30 115

South

210

330

Heading 210 Heading 115

In order to keep an airplane on course, the captain must consider the direction and speed of any wind currents, since the plane’s true course (relative to the ground) will be affected. Both the plane and the wind can be represented by vectors, with the plane’s true course being the resultant vector. 

EXAMPLE 9

Solving an Application of Vectors—Airplane Navigation An airplane is flying at 240 mph, heading 75°, when it suddenly encounters a strong, 60 mph wind blowing from the southwest, heading 10°. What is the actual course and speed of the plane (relative to the ground) as it flies through this wind?



Solution

Figure 7.48 y

Begin by drawing a vector p to represent the speed and direction of the airplane (Figure 7.48). Since the heading is 75°, the angle between the vector and the x-axis must be 15°. For convenience (and because location is unimportant) we draw it as a 60  14 as long, and position vector. Note the vector w representing the wind will be 240 can also be drawn as a position vector—with an acute 80° angle. To find the resultant, we first find the components of each vector, then add. For vector w (in QI): Horizontal Component

a  wcos   60 cos 80°  10.4 w

15 x

For vector p (in QI): Horizontal Component

y

wp p

80

Vertical Component

a  pcos  b  psin   240 cos 15°  240 sin 15°  231.8  62.1 p  231.8i  62.1j.

Figure 7.49

w

b  wsin   60 sin 80°  59.1

w  10.4i  59.1j.

p

80

Vertical Component

15 x

The resultant is w  p  110.4i  59.1j2  1231.8i  62.1j2  242.2i  121.2j, with magnitude w  p  21242.22 2  1121.22 2  270.8 mph (see Figure 7.49). 121.2 b, To find the heading of the plane relative to the ground we use r  tan1a 242.2 which shows r  26.6°. The plane is flying on a course heading of 90°  26.6°  63.4° at a speed of about 270.8 mph relative to the ground. Note the airplane has actually “increased speed” due to the wind. Now try Exercises 83 through 86



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Applications like those in Examples 8 and 9 can also be solved using what is called the parallelogram method, which takes its name from the tailto-tip vector addition noted earlier (See Figure 7.50). The resultant will be a diagonal of the parallelogram, whose magnitude can be found using the law of cosines. For Example 9, we note the parallelogram has two acute angles of 180  152°  65°, and since the adjacent angles must sum to 180°, the obtuse angles must be 115°. Using the law of cosines,

WORTHY OF NOTE Be aware that using the rounded values of intermediate calculations may cause slight variations in the final result. In Example 9, if we calculate w  p  160 cos 80°  240 cos 15°2i  160 sin 80°  240 sin 15°2j, then find w  p, the result is actually closer to 270.9 mph.

w  p2  p2  w2  2pw cos 115°

y

w

p

80 15

x

law of cosines substitute 240 for p, 60 for w

 73371.40594

compute result

2

w  p  270.9

115

wp

 240  60  2124021602 cos 115° 2

E. You’ve just learned how to use vector diagrams to solve applications

Figure 7.50

take square roots

Note this answer is slightly more accurate, since there was no rounding until the final stage.

TECHNOLOGY HIGHLIGHT

Vector Components Given the Magnitude and the Angle  The TABLE feature of a graphing calculator can help us find the horizontal and vertical components of any vector with ease. Consider the vector v shown in Figure 7.51, which has a magnitude of 9.5 with   15°. Knowing this magnitude is used in both computations, first store 9.5 in storage location A: 9.5 ALPHA STO MATH . Next, enter the expressions for the horizontal and vertical components as Y1 and Y2 on the Y = screen (see Figure 7.52). Note that storing the magnitude 9.5 in memory will prevent our having to alter Y1 and Y2 as we apply these ideas to other values of . As an additional check, note that Y3 recomputes the magnitude of the vector using the components generated in Y1 and Y2. To access ENTER and select the desired function. Although our the function variables we press: VARS primary interest is the components for   15°, we use the TBLSET screen to begin at TblStart  0°, ¢Tbl  5, and have it count AUTOmatically, so we can make additional observations. Pressing (TABLE) brings up the screen shown in Figure 7.53. As expected, at   0° the horizontal component is the same as the magnitude and the vertical component is zero. At   15° we have the components of the vector pictured in Figure 7.51, approximately H9.18, 2.46I. If the angle were increased to   30°, a 30-60-90 triangle could be formed and one component should be 13 times the other. Sure enough, 1314.752  8.2272. 2nd

GRAPH

Figure 7.52

Figure 7.51 y v 15 x

Figure 7.53

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Exercise 1: If   45°, what would you know about the lengths of the horizontal and vertical components? Scroll down to   45° to verify. Exercise 2: If   60°, what would you know about the lengths of the horizontal and vertical components? Scroll down to   60° to verify. Exercise 3: We used column Y3 as a double check on the magnitude of v for any given . What would this value be for   45° and   60°? Press the right arrow to verify. What do you notice?

7.3 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. Measurements that can be described using a single number are called quantities. 2.

quantities require more than a single number to describe their attributes. Examples are force, velocity, and displacement.

3. To represent a vector quantity geometrically we use a segment. 

4. Two vectors are equal if they have the same and . 5. Discuss/Explain the geometric interpretation of vector addition. Give several examples and illustrations. 6. Describe the process of finding a resultant vector given the magnitude and direction of two arbitrary vectors u and v. Follow-up with an example.

DEVELOPING YOUR SKILLS

Draw the comparative geometric vectors indicated.

7. Three oceanic research vessels are traveling on a parallel course in the same direction, mapping the ocean floor. One ship is traveling at 12 knots (nautical miles per hour), one at 9 knots, and the third at 6 knots. 8. As part of family reunion activities, the Williams Clan is at a bowling alley and using three lanes. Being amateurs they all roll the ball straight on, aiming for the 1 pin. Grand Dad in Lane 1 rolls his ball at 50 ft/sec. Papa in Lane 2 lets it rip at 60 ft/sec, while Junior in Lane 3 can muster only 30 ft/sec. 9. Vector v1 is a geometric vector representing a boat traveling at 20 knots. Vectors v2, v3, and v4 are geometric vectors representing boats traveling at 10 knots, 15 knots, and 25 knots, respectively. Draw these vectors given that v2 and v3 are traveling the same direction and parallel to v1, while v4 is traveling in the opposite direction and parallel to v1.

10. Vector F1 is a geometric vector representing a force of 50 N. Vectors F2, F3, and F4 are geometric vectors representing forces of 25 N, 35 N, and 65 N, respectively. Draw these vectors given that F2 and F3 are applied in the same direction and parallel to F1, while F4 is applied in the opposite direction and parallel to F1. Represent each situation described using geometric vectors.

11. Two tractors are pulling at a stump in an effort to clear land for more crops. The Massey-Ferguson is pulling with a force of 250 N, while the John Deere is pulling with a force of 210 N. The chains attached to the stump and each tractor form a 25° angle. 12. In an effort to get their mule up and plowing again, Jackson and Rupert are pulling on ropes attached to the mule’s harness. Jackson pulls with 200 lb of force, while Rupert, who is really upset, pulls with 220 lb of force. The angle between their ropes is 16°.

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Draw the vector v indicated, then graph the equivalent position vector.

Use the graphs of vectors a, b, c, d, e, f, g, and h given to determine if the following statements are true or false.

13. initial point 13, 22; terminal point (4, 5)

y

14. initial point 14, 42; terminal point (2, 3) 15. initial point 15, 32; terminal point 11, 22

b

16. initial point (1, 4); terminal point 12, 22

For each vector v  Ha, bI and initial point (x, y) given, find the coordinates of the terminal point and the magnitude v of the vector.

17. v  H7, 2I; initial point 12, 32 18. v  H6, 1I; initial point 15, 22

19. v  H3, 5I; initial point (2, 6)

20. v  H8, 2I; initial point 13, 52

a

23. H2, 5I

39. c  f  h

40. b  h  c

41. d  e  h

42. d  f  0

For the vectors u and v shown, compute u  v and u  v and represent each result graphically.

43.

44.

y

27. w  140.5;   41°; QIV

y x

H4, 1I

v u

H1, 4I H3, 6I

H7, 2I

v

u

x

45.

46.

y v

H8, 3I

u

y

x

H1, 3I

H4, 4I H5, 2I u v

x

47.

26. u  25;   32°; QIII

f

38. f  e  g

For Exercises 25 through 30, the magnitude of a vector is given, along with the quadrant of the terminal point and the angle it makes with the nearest x-axis. Find the horizontal and vertical components of each vector and write the result in component form.

25. v  12;   25°; QII

e

37. a  c  b

22. H7, 6I 24. H8, 6I

c

x

For each position vector given, (a) graph the vector and name the quadrant, (b) compute its magnitude, and (c) find the acute angle  formed by the vector and the nearest x-axis.

21. H8, 3I

d g h

48.

y v

u

H5, 3I

y

x

H4, 3I

H2, 3I

v u H5, 1I

28. p  15;   65°; QI

x

29. q  10;   15°; QIII 30. r  4.75;   62°; QII For each pair of vectors u and v given, compute (a) through (d) and illustrate the indicated operations graphically.

a. u  v c. 2u  1.5v

b. u  v d. u  2v

31. u  H2, 3I; v  H3, 6I

32. u  H3, 4I; v  H0, 5I 33. u  H7, 2I; v  H1, 6I

34. u  H5, 3I; v  H6, 4I 35. u  H4, 2I; v  H1, 4I 36. u  H7, 3I; v  H7, 3I

Graph each vector and write it as a linear combination of i and j. Then compute its magnitude.

49. u  H8, 15I

51. p  H3.2, 5.7I

50. v  H5, 12I

52. q  H7.5, 3.4I

For each vector here, r represents the acute angle formed by the vector and the x-axis. (a) Graph each vector, (b) find the horizontal and vertical components and write the vector in component form, and (c) write the vector in i, j form. Round to the nearest tenth.

53. v in QIII, v  12, r  16° 54. u in QII, u  10.5, r  25° 55. w in QI, w  9.5, r  74.5° 56. v in QIV, v  20, r  32.6°

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a. v1  v2  p

b. v1  v2  q

c. 2v1  1.5v2  r

d. v1  2v2  s

57. v1  2i  3j; v2  4i  5j 58. v1  7.8i  4.2j; v2  5j

69. 3.5i  12j

70. 9.6i  18j

73. 6i  11j

74. 2.5i  7.2j

71. v1  H13, 3I

75.



H10, 4I

p

H7, 4I p

52

62. v1  213i  6j; v2  4 13i  2j

67. 20i  21j

y

H2, 7I

61. v1  12i  4j; v2  4i

65. p  H20, 21I

76.

y

60. v1  6.8i  9j; v2  4i  9j

Find a unit vector pointing in the same direction as the vector given. Verify that a unit vector was found.

72. v2  H4, 7I

Vectors p and q form the angle indicated in each diagram. Find the vector r that points in the same direction as q and forms the base of the right triangle shown.

59. v1  512i  7j; v2  3 12i  5j

63. u  H7, 24I

671

Section 7.3 Vectors and Vector Diagrams

q

23 r

r

H9, 1I q

x

77.

78.

y r

36

64. v  H15, 36I

q

y

x H2, 7I

H8, 3I

p

66. q  H12, 35I

x

p

q

H10, 5I

H4, 6I

48 r

68. 4i  7.5j

x

WORKING WITH FORMULAS

The magnitude of a vector in three dimensions: v  2a2  b2  c2

79. The magnitude of a vector in three dimensional space is given by the formula shown, where the components of the position vector v are Ha, b, cI. Find the magnitude of v if v  H5, 9, 10I. 

80. Find a cardboard box of any size and carefully measure its length, width, and height. Then use the given formula to find the magnitude of the box’s diagonal. Verify your calculation by direct measurement.

APPLICATIONS

81. Tow forces: A large van has careened off of the road into a ditch, and two tow trucks are attempting to winch it out. The cable W1 from the first winch exerts a force of 900 lb, while the cable  32 from the second exerts a force of 700 lb. W2 Determine the angle  for the first tow truck that will bring the van directly out of the ditch and along the line indicated.

82. Tow forces: Two tugboats are pulling a large ship into dry dock. The first is pulling with a force of 1250 N and the second with a force of 1750 N. Determine the angle  for the second tugboat that will keep the ship moving straight forward and into the dock.

83. Projectile components: An arrow is shot into the air at an angle of 37° with an initial velocity of 100 ft/sec. Compute the horizontal and vertical components of the representative vector.

84. Projectile components: A football is punted (kicked) into the air at an angle of 42° with an initial velocity of 20 m/sec. Compute the horizontal and vertical components of the representative vector.

T1

40  T2

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85. Headings and cross-winds: An airplane is flying at 250 mph on a heading of 75°. There is a strong, 35 mph wind blowing from the southwest on a heading of 10°. What is the true course and speed of the plane (relative to the ground)?

86. Headings and currents: A cruise ship is traveling at 16 knots on a heading of 300°. There is a strong water current flowing at 6 knots from the northwest on a heading of 120°. What is the true course and speed of the cruise ship?

The lights used in a dentist’s office are multijointed so they can be configured in multiple ways to accommodate various needs. As a simple model, consider such a light that has the three joints, as illustrated. The first segment has a length of 45 cm, the second is 40 cm in length, and the third is 35 cm.

87. If the joints of the light are positioned so a straight line is formed and the angle made with the horizontal is 15°, determine the approximate coordinates of the joint nearest the light.

88. If the first segment is rotated 75° above horizontal, the second segment 30° (below the horizontal), and the third segment is parallel to the horizontal, determine the approximate coordinates of the joint nearest the light.

30 15



75

EXTENDING THE CONCEPT

For the arbitrary vectors u  Ha, bI, v  Hc, dI, and w  He, f I and the scalars c and k, prove the following vector properties using the properties of real numbers.

89. 1u  u

91. u  v  u  1v2

90. 0u  0  k0

99. Show that the sum of the vectors given, which form the sides of a closed polygon, is the zero vector. Assume all vectors have integer coordinates and each tick mark is 1 unit. y

92. 1u  v2  w  u  1v  w2 93. 1ck2u  c1ku2  k1cu2 94. u  0  u

95. u  1u2  0

96. k1u  v2  ku  kv 97. 1c  k2u  cu  ku 98. Consider an airplane flying at 200 mph at a heading of 45°. Compute the groundspeed of the plane under the following conditions. A strong, 40-mph wind is blowing (a) in the same direction; (b) in the direction of due north (0°); (c) in the direction heading 315°; (d) in the direction heading 270°; and (e) in the direction heading 225°. What did you notice about the groundspeed for (a) and (b)? Explain why the plane’s speed is greater than 200 mph for (a) and (b), but less than 200 mph for the others.

672

s r t

p

v

u

x

100. Verify that for v  ai  bj and v v  2a2  b2,  1. v v (Hint: Create the vector u  and find its magnitude.) v 101. Referring to Exercises 87 and 88, suppose the dentist needed the pivot joint at the light (the furthest joint from the wall) to be at (80, 20) for a certain patient or procedure. Find at least one set of “joint angles” that will make this possible.

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Reinforcing Basic Concepts

MAINTAINING YOUR SKILLS

102. (6.1) Derive the other two common versions of the Pythagorean identities, given sin2x  cos2x  1. 103. (2.5) Evaluate each expression for x  3 (if possible): 5 a. y  ln12x  72 b. y  x3 1 c. y  x5 A3

104. (6.5) Evaluate the expression csc c tan1a

55 b d by 48

drawing a representative triangle. 105. (3.4) Graph the function g1x2  x3  7x and find its zeroes.

MID-CHAPTER CHECK sin B sin A  , solve for sin B. a b 2. Given b2  a2  c2  2ac cos B, solve for cos B.

1. Beginning with

Solve the triangles shown below using any appropriate method. 3.

B 207 m C 250 m

31

A

B

4.

A

8. Modeled after an Egyptian obelisk, the Washington Monument (Washington, D.C.) is one of the tallest masonry buildings in the world. Find the height of the monument given the measurements shown (see the figure). 58 70

17 cm 21 cm

the sign casts a 75 ft shadow. Find the height of the sign if the angle of elevation (measured from a horizontal line) from the tip of the shadow to the top of the sign is 65°.

C 25 cm

9. The circles shown here have 44 m radii of 4 cm, 9 cm, and 12 cm, and are tangent to each other. Find the angles formed by the line segments joining their centers. 

Solve the triangles described below using the law of sines. If more than one triangle exists, solve both.



5. A  44°, a  2.1 km, c  2.8 km



6. C  27°, a  70 yd, c  100 yd 7. A large highway sign is erected on a steep hillside that is inclined 45° from the horizontal. At 9:00 A.M.

75 ft 45

10. On her delivery route, Judy drives 23 miles to Columbus, then 17 mi to 17 mi Drake, then back home to Balboa. Use the diagram 21 given to find the distance C from Drake to Balboa.

D B 23 mi

REINFORCING BASIC CONCEPTS Scaled Drawings and the Laws of Sine and Cosine In mathematics, there are few things as satisfying as the tactile verification of a concept or computation. In this Reinforcing Basic Concepts, we’ll use scaled drawings to

verify the relationships stated by the law of sines and the law of cosines. First, gather a blank sheet of paper, a ruler marked in centimeters/millimeters, and a protractor. When working with scale models, always measure and mark as carefully as possible. The greater the care, the better the 673

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Figure 7.54 results. For the first illustration (see Figure 7.54), we’ll 18 cm draw a 20-cm horizontal line ? segment near the bottom of 35 the paper, then use the left endpoint to mark off a 35° 20 cm angle. Draw the second side a length of 18 cm. Our first goal is to compute the length of the side needed to complete the triangle, then verify our computation by measurement. Since the current “triangle” is SAS, we use the law of cosines. Label the 35° as A, the top vertex as B, and the right endpoint as C.

measure these angles from the diagram using your protractor. How close was the computed measure to the actual measure?

90

180

a2  b2  c2  2bc cos A

0

law of cosines with respect to a

 1202  1182  21202 1182cos 35 2

2

substitute known values

 724  589.8

simplify (round to 10)

 134.2

combine terms

a  11.6

solve for a

The computed length of side a is 11.6 cm, and if you took great care in drawing your diagram, you’ll find the missing side is indeed very close to this length. Exercise 1: Finish solving the triangle above using the law of sines. Once you’ve computed B and C,

For the second illustraFigure 7.55 tion (see Figure 7.55), draw any arbitrary triangle on a separate blank sheet, noting 13.3 cm 15.3 cm that the larger the triangle, the easier it is to measure the 21.2 cm angles. After you’ve drawn it, measure the length of each side to the nearest millimeter (our triangle turned out to be 21.2 cm  13.3 cm  15.3 cm). Now use the law of cosines to find one angle, then the law of sines to solve the triangle. The computations for our triangle gave angles of 95.4°, 45.9°, and 38.7°. What angles did your computations give? Finally, use your protractor to measure the angles of the triangle you drew. With careful drawings, the measured results are often remarkably accurate! Exercise 2: Using sides of 18 cm and 15 cm, draw a 35° angle, a 50° angle, and a 70° angle, then complete each triangle by connecting the endpoints. Use the law of cosines to compute the length of this third side, then actually measure each one. Was the actual length close to the computed length?

7.4 Vector Applications and the Dot Product Learning Objectives In Section 7.4 you will learn how to:

A. Use vectors to investigate forces in equilibrium

In Section 7.3 we introduced the concept of a vector, with its geometric, graphical, and algebraic representations. We also looked at operations on vectors and employed vector diagrams to solve basic applications. In this section we introduce additional ideas that enable us to solve a variety of new applications, while laying a strong foundation for future studies.

B. Find the components of one vector along another

C. Solve applications involving work

D. Compute dot products and the angle between two vectors

E. Find the projection of one vector along another and resolve a vector into orthogonal components

F. Use vectors to develop an equation for nonvertical projectile motion, and solve related applications

A. Vectors and Equilibrium Much like the intuitive meaning of the word, vector forces are in equilibrium when they “counterbalance” each other. The simplest example is two vector forces of equal magnitude acting on the same point but in opposite directions. Similar to a tug-of-war with both sides equally matched, no one wins. If vector F1 has a magnitude of 500 lb in the positive direction, F1  H500, 0I would need vector F2  H500, 0I to counter it. If the forces are nonquadrantal, we intuitively sense the components must still sum to zero, and that F3  H600, 200I would need F4  H600, 200I for equilibrium to occur (see Figure 7.56). In other words, two vectors are in equilibrium when their sum is

Figure 7.56 y

H600, 200I F4 x

F3

H600, 200I

F3  F4  H600, 200I  H600, 200I  H0, 0I  0

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the zero vector 0. If the forces have unequal magnitudes or do not pull in opposite directions, recall a resultant vector F  Fa  Fb can be found that represents the combined force. Equilibrium will then occur by adding the vector 1 (F) and this vector is sometimes called the equilibriant. These ideas can be extended to include any number of vector forces acting on the same point. In general, we have the following: Vectors and Equilibrium Given vectors F1, F2, . . . , Fn acting on a point P, 1. The resultant vector is F  F1  F2  # # #  Fn. 2. Equilibrium for these forces requires the vector 1F, where F  112F  0

EXAMPLE 1



Finding the Equilibriant for Vector Forces

y

Two force vectors F1 and F2 act on the point P as shown. Find a force F3 so equilibrium will occur, and sketch it on the grid. Solution



A. You’ve just learned how to use vectors to investigate forces in equilibrium

5

F1 4.5

Begin by finding the horizontal and vertical components of each vector. For F1 we have 5 H4.5 cos 64°, 4.5 sin 64°I  H2.0, 4.0I, and for F2 we have H6.3 cos 18°, 6.3 sin 18°I  H6.0, 1.9I. The resultant vector is F  F1  F2  H4.0, 5.9I, meaning H4.0, 5.9I equilibrium will occur by applying the force 1F  H4.0, 5.9I (see figure).

F2

6.3 18

64 P

5

x

5

Now try Exercises 7 through 20



B. The Component of u along v: compvu As in Example 1, many simple applications involve position vectors where the angle and horizontal/vertical components are known or can easily be found. In these situations, the components are often quadrantal, that is, they lie along the x- and y-axes and meet at a right angle. Many other applications require us to find components of a vector that are nonquadrantal, with one of the components parallel to, or lying along a second vector. Given vectors u and v, as shown in Figure 7.57, we symbolize the component of u that lies along v as compvu, noting its value is simply ucos  since adj compvu . As the diagrams further indicate, compvu  ucos  regardless cos    u hyp Figure 7.57 u

v

u ␪

p vu ␪

com

u u

v ␪

v

u

p vu

u comp v 0␪q

com

q␪␲

u

of how the vectors are oriented. Note that even when the components of a vector do not lie along the x- or y-axes, they are still orthogonal (meet at a 90° angle). It is important to note that compvu is a scalar quantity (not a vector), giving only the magnitude of this component (the vector projection of u along v is studied later in this section). From these developments we make the following observations regarding the angle  at which vectors u and v meet:

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Vectors and the Component of u Along v Given vectors u and v, which meet at an angle , 1. compvu  ucos . 2. If 0 6  6 90°, compvu 7 0; if 90° 6  6 180°, compvu 6 0. 3. If   0, u and v have the same direction and compvu  u. 4. If   90°, u and v are orthogonal and compvu  0. 5. If   180°, u and v have opposite directions and compvu  u.

EXAMPLE 2



Finding the Component of Vector G Along Vector v Given the vectors G and v with G  850 lb as shown in the figure, find compvG.

Solution



Using Gcos   compvG we have 850 cos 65°  359 lb. The component of G along v is about 359 pounds.

p vG com 65

v

850 lb

G

Now try Exercises 21 through 26



One interesting application of equilibrium and compvu involves the force of gravity acting on an object placed on a ramp or an inclined plane. The greater the incline, the greater the tendency of the object to slide down the plane (for this study, we assume there is no friction between the object and the plane). While the force of gravity continues to pull straight downward (represented by the vector G in Figure 7.58), G is now the resultant of a force acting parallel to the plane along vector v (causing the object to slide) and a force acting perpendicular to the plane along vector p (causing the object to press against the plane). If we knew the component of G along v (indicated by the shorter, bold segment), we would know the force required to keep the object stationary as the two forces must be opposites. Note that G forms a right angle with the base of the inclined plane (see Figure 7.59), meaning that  and  must be complementary angles. Also note that since the location of a vector is unimportant, vector p has been repositioned for clarity. Figure 7.58

Figure 7.59 850 ␤

v

lb

850 ␤ v

␣

␣

p

p G

G

EXAMPLE 3A



Finding Components of Force for an Object on a Ramp A 850-lb object is sitting on a ramp that is inclined at 25°. Find the force needed to hold the object stationary (in equilibrium).

Solution



lb

Given   25°, we know   65°. This means the component of G along the inclined plane is compvG  850 cos 65° or about 359 lb. A force v of 359 lb is required to keep the object from sliding down the incline (compare to Example 2).

compvG

850 ␤

lb

850 lb 25 G

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EXAMPLE 3B

Solution





B. You’ve just learned how to find the components of one vector along another

A winch is being used to haul a 2000-lb block of granite up a ramp that is inclined at 15°. If the winch has a maximum tow rating of 500 lb, will it be successful? We again need the component of G along the inclined plane: compvG  2000 cos 75  518 lb. Since the capacity of the winch is exceeded, the attempt will likely not be successful.

0 lb

200 ␤ v

15 2000 lb

G

Now try Exercises 27 through 30



C. Vector Applications Involving Work Figure 7.60

In common, everyday usage, work is understood to involve the exertion of energy or force to move an object a certain distance. For example, digging a ditch is hard work and involves moving dirt (exerting a force) from the trench to the bankside (over a certain distance). In an office, moving a filing cabinet likewise involves work. If the filing cabinet is heavier, or the distance it needs to be moved is greater, more work is required to move it (Figures 7.60 and 7.61). To determine how much work was done by each person, we need to quantify the concept. Consider a constant force F, applied to move an object a distance D in the same direction as the force. In this case, work is defined as the product of the force applied and the distance the object is moved: Work  Force  Distance or W  FD. If the force is given in pounds and the distance in feet, the amount of work is measured in a unit called foot-pounds (ft-lb). If the force is in newtons and the distance in meters, the amount of work is measured in newton-meters (N-m).

D

Figure 7.61

D

EXAMPLE 4



Solving Applications of Vectors — Work and Force Parallel to the Direction of Movement While rearranging the office, Carrie must apply a force of 55.8 N to relocate a filing cabinet 4.5 m, while Bernard applies a 77.5 N force to move a second cabinet 3.2 m. Who did the most work?

Solution



For Carrie: W   F D  155.8214.52  251.1 N-m

For Bernard: W   F D  177.5213.22  248 N-m

Carrie did 251.1  248  3.1 N-m more work than Bernard. Now try Exercises 31 and 32



In many applications of work, the force F is not applied parallel to the direction of movement, as illustrated in Figures 7.62 and 7.63. In calculating the amount of work done, the general concept of force  distance is preserved, but only the component of force in the direction of movement is used. In Figure 7.62

Figure 7.63

F 30°

F D

25°

D

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terms of the component forces discussed earlier, if F is a constant force applied at angle  to the direction of movement, the amount of work done is the component of force along D times the distance the object is moved.

WORTHY OF NOTE In the formula W   F  cos   D, observe that if   0, we have the old formula for work when the force is applied in the direction of movement W  FD. If   0, cos   1 and the “effective force” on the object becomes  F cos .

EXAMPLE 5

Force Vectors and Work W Given a force F applied in the direction of movement at the acute angle  to an object, and D the distance it is moved, W  Fcos   D

F ␪

D Fcos ␪



Solving an Application of Vectors — Work and Force Applied at Angle  to the Direction of Movement To help move heavy pieces of furniture across the floor, movers sometime employ a body harness similar to that used for a plow horse. A mover applies a constant 200-lb force to drag a piano 100 ft down a long hallway and into another room. If the straps make a 40° angle with the direction of movement, find the amount of work performed.

Solution



40° D

The component of force in the direction of movement is 200 cos 40° or about 153 lb. The amount of work done is W  15311002  15,300 ft-lb. Now try Exercises 35 through 40



These ideas can be generalized to include work problems where the component of force in the direction of motion is along a nonhorizontal vector v. Consider Example 6. EXAMPLE 6

Solution





C. You’ve just learned how to solve applications involving work

Solving an Application of Vectors —Forces Along a Nonhorizontal Vector

The force vector F  H5, 12I moves an object along the vector v  H15.44, 2I as shown. Find the amount of work required to move the object along the entire length of v. Assume force is in pounds and distance in feet.

y

H5, 12I F

To begin, we first determine the angle between the vectors. H15.44, 2I 2 1 12 1 v 60 In this case we have   tan a b  tan a b  60°. x 5 15.44 For F  13 (5-12-13 triangle), the component of force in the direction of motion is compvF  13 cos 60°  6.5. With v  2115.442 2  122 2  15.57, the work required is W  compvF  v or 16.52 115.572  101.2 ft-lb. Now try Exercises 41 through 44



D. Dot Products and the Angle Between Two Vectors When the component of force in the direction of motion lies along a nonhorizontal vector (as in Example 6), the work performed can actually be computed more efficiently using an operation called the dot product. For any two vectors u and v, the dot product u # v is equivalent to compvu  v, yet is much easier to compute (for the proof of u # v  compvu  v, see Appendix V). The operation is defined as follows:

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679

The Dot Product u # v Given vectors u  Ha, bI and v  Hc, dI, u # v  Ha, bI # Hc, dI  ac  bd. In words, it is the real number found by taking the sum of corresponding component products. EXAMPLE 7



Using the Dot Product to Determine Force Along a Nonhorizontal Vector Verify the answer to Example 6 using the dot product u # v.

Solution



For u  H5, 12I and v  H15.44, 2I, we have u # v  H5, 12I # H15.44, 2I giving 5115.442  12122  101.2. The result is 101.2, as in Example 6. Now try Exercises 45 through 48



Note that dot products can also be used in the simpler case where the direction of motion is along a horizontal distance (Examples 4 and 5). While the dot product offers a powerful and efficient way to compute the work performed, it has many other applications; for example, to find the angle between two vectors. Consider that for any two u v # (solve for vectors u and v, u # v  ucos   v, leading directly to cos   u v cos ). In summary, The Angle  Between Two Vectors Given the nonzero vectors u and v: u # v cos   and u v

Figure 7.64 y

  cos1a

u # v b u v

(x, y)

u  1 ␪

In the special case where u and v are unit vectors, this simplifies to cos   u # v since u  v  1. This relationship is shown in Figure 7.64. The dot product u # v gives compvu  v, but v  1 and the component of u along v is simply the adjacent side of a right triangle whose hypotenuse is 1. Hence u # v  cos .

x

v compvu  x

v  1

EXAMPLE 8



Determining the Angle Between Two Vectors Find the angle between the vectors given. a. u  H3, 4I; v  H5, 12I b. v1  2i  3j; v2  6i  4j

Solution



u # v u v 3 4 # 5 12  h , i h , i 5 5 13 13 15 48   65 65 33  65 33   cos1a b 65  59.5°

a. cos  

v1 v2 # v1 v2 2 3 6 4  h , i # h , i 113 113 152 152 12 12   1676 1676 0  0 26

b. cos  

  cos10  90° Now try Exercises 49 through 66



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Note we have implicitly shown that if u # v  0, then u is orthogonal to v. As with other vector operations, recognizing certain properties of the dot product will enable us to work with them more efficiently. Properties of the Dot Product Given vectors u, v, and w and a constant k, 2. u # u  u2 4. k1u # v2  ku # v  u # kv u#v u # v 6.  u v uv

1. u # v  v # u 3. w # 1u  v2  w # u  w # v 5. 0 # u  u # 0  0

Property 6 offers an alternative to unit vectors when finding cos  — the dot product of the vectors can be computed first, and the result divided by the product of u#v their magnitudes: cos   . Proofs of the first two properties are given here. uv Proofs of the others have a similar development (see Exercises 79 through 82). For any two nonzero vectors u  Ha, bI and v  Hc, dI: Property 1: u # v  Ha, bI # Hc, dI  ac  bd  ca  db  Hc, dI # Ha, bI v#u D. You’ve just learned how to compute dot products and the angle between two vectors

Property 2: u # u  Ha, bI # Ha, bI  a2  b2  u2 (since u  2a2  b2)

Using compvu  ucos  and u # v  compvu  v, we can also state the following relationships, which give us some flexibility on how we approach applications of the dot product. For any two vectors u  Ha, bI and v  Hc, dI: (1) u # v  ac  bd (2) u # v  ucos   v (3) u # v  compvu  v u#v  cos  (4) uv u#v  compvu (5) v

standard computation of the dot product alternative computation of the dot product replace ucos  in (2) with compvu divide (2) by scalars u and v divide (3) by v

E. Vector Projections and Orthogonal Components In work problems and other simple applications, it is enough to find and apply compvu (Figure 7.65). However, applications involving thrust and drag forces, tension and stress limits in a cable, electronic circuits, and cartoon animations often require that we also find the vector form of compvu. This is called the projection of u along v or projvu, and is a vector in the same direction of v with magnitude compvu (Figures 7.66 and 7.67). Figure 7.65

Figure 7.66

u

u

v

v ␪

u comp v r) a l a c s (

Figure 7.67 u

␪

u proj v r) o t c e v (

v ␪

u proj v r) o t c (ve

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v has a length of one and points in the same direction as v v (see Example 7, Section 7.3). Using v, so projvu can be computed as compvu  v equation (5) above and the properties shown earlier, an alternative formula for projvu can be found that is usually easier to simplify: By its design, the unit vector

projvu  compvu  u#v v  v v u#v  2 v v 

v v

definition of a projection

substitute

u#v for compvu v

rewrite factors

Vector Projections Given vectors u and v, the projection of u along v is the vector projvu  a

EXAMPLE 9A Solution





Finding the Projection of One Vector Along Another Given u  H7, 1I and v  H6, 6I, find projvu. To begin, find u # v and v. u # v  H7, 1I # H6, 6I  42  6  36 u#v bv v2 36 a b H6, 6I 72  H3, 3I

projvu  a

v  262  62  172  612

y (6, 6) v

(7, 1)

u

projvu

x

projection of u along v substitute 36 for u # v, 172 for v, and H6, 6I for v result

A useful consequence of computing projvu is we can then resolve the vector u into orthogonal components that need not be quadrantal. One component will be parallel to v and the other perpendicular to v (the dashed line in the diagram in Example 9A). In general terms, this means we can write u as the vector sum u1  u2, where u1  projvu and u2  u  u1 (note u1 7 v).

WORTHY OF NOTE Note that u2  u  u1 is the shorter diagonal of the parallelogram formed by the vectors u and u1  projvu. This can also be seen in the graph supplied for Example 9B.

EXAMPLE 9B

u#v bv v2

Resolving a Vector into Orthogonal Components Given vectors u, v, and projvu, u can be resolved into the orthogonal components u1 and u2, where u  u1  u2, u1  projvu, and u2  u  u1. 

Resolving a Vector into Orthogonal Components

Given u  H2, 8I and v  H8, 6I, resolve u into orthogonal components u1 and u2, where u1 7 v and u2v. Also verify u1u2.

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Solution



Once again, begin by finding u # v and v.

projvu  a

u#v

bv v2 64 a bH8, 6I 100  H5.12, 3.84I

E. You’ve just learned how to find the projection of one vector along another and resolve a vector into orthogonal components

y

v  282  62  2100  10

u # v  H2, 8I # H8, 6I  16  48  64

(2, 8) u u2

(8, 6) v

projection of u along v u1  projvu

␪

substitute 64 for u # v, 10 for v, and H8, 6I for v

x

result

For projvu  u1  H5.12, 3.84I, we have u2  u  u1  H2, 8I  H5.12, 3.84I  H3.12, 4.16I . To verify u1u2, we need only show u1 # u2  0: u1 # u2  H5.12, 3.84I # H3.12, 4.16I  15.12213.122  13.842 14.162  0✓ Now try Exercises 67 through 72



F. Vectors and the Height of a Projectile Our final application of vectors involves projectile motion. A projectile is any object that is thrown or projected upward, with no source of propulsion to sustain its motion. In this case, the only force acting on the projectile is gravity (air resistance is neglected), so the maximum height and the range of the projectile depend solely on its initial velocity and the angle  at which it is projected. In a college algebra course, the equation y  v0t  16t2 is developed to model the height in feet (at time t) of a projectile thrown vertically upward with initial velocity of v0 feet per second. Here, we’ll modify the equation slightly to take into account that the object is now moving horizontally as well as vertically. As you can see in Figure 7.68, the vector v representing the initial velocity, as well as the velocity vector at other times, can easily be decomposed into horizontal and vertical components. This will enable us to find a more general relationship for the position of the projectile. For now, we’ll let vy represent the component of velocity in the vertical (y) direction, and vx represent the component of velocity in the horizontal (x) direction. Since gravity acts only in the vertical (and negative) direction, the horizontal component of the velocity remains constant at vx  vcos . Using D  RT, the x-coordinate of the projectile at time t is x  1 vcos 2t. For the vertical component vy we use the projectile equation developed earlier, substituting vsin  for v0, since the angle of projection is no longer 90°. This gives the y-coordinate at time t as y  v0t  16t2  1 vsin 2t  16t2. Figure 7.68 y

vy

vy

␪3  0 vy  0

v ␪2

vx

␪4

vx v ␪1

vx

vy vx

x

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683

Projectile Motion Given an object is projected upward from the origin with initial velocity v at angle °. The x-coordinate of its position at time t is x  1 vcos 2t. The y-coordinate of its position at time t is y  1 vsin 2t  16t2.

EXAMPLE 10



Solving an Application of Vectors —Projectile Motion An arrow is shot upward with an initial velocity of 150 ft/sec at an angle of 50°. a. Find the position of the arrow after 2 sec. b. How many seconds does it take to reach a height of 190 ft?

Solution



a. Using the preceding equations yields these coordinates for its position at t  2: x  1 vcos 2t  1150 cos 50°2 122  193

y  1 vsin 2t  16t2  1150 sin 50°2 122  16122 2  166

The arrow has traveled a horizontal distance of about 193 ft and is 166 ft high. b. To find the time required to reach 190 ft in height, set the equation for the y coordinate equal to 190, which yields a quadratic equation in t:

F. You’ve just learned how to use vectors to develop an equation for nonvertical, projectile motion and solve related applications

y  1 vsin 2t  16t2 190  1150 sin 50°2t  16t2 0  161t2 2  115t  190

equation for y substitute 150 for v and 50° for  150 sin 50  115

Using the quadratic formula we find that t  2.6 sec and t  4.6 sec are solutions. This makes sense, since the arrow reaches a given height once on the way up and again on the way down, as long as it hasn’t reached its maximum height. Now try Exercises 73 through 78



For more on projectile motion, see the Calculator Exploration and Discovery feature at the end of this chapter.

7.4 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. Vector forces are in when they counterbalance each other. Such vectors have a sum of . 2. The component of a vector u along another vector v is written notationally as , and is computed as . 3. Two vectors that meet at a right angle are said to be .

4. The component of u along v is a quantity. The projection of u along v is a . 5. Explain/Discuss exactly what information the dot product of two vectors gives us. Illustrate with a few examples. 6. Compare and contrast the projectile equations y  v0t  16t2 and y  1v0sin 2t  16t2. Discuss similarities/differences using illustrative examples.

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DEVELOPING YOUR SKILLS

The force vectors given are acting on a common point P. Find an additional force vector so that equilibrium takes place.

7. F1  H8, 3I; F2  H2, 5I

Find the component of u along v (compute compvu) for the vectors u and v given.

21.

u

8. F1  H2, 7I; F2  H5, 3I

v

50 kg

9. F1  H2, 7I; F2  H2, 7I; F3  H5, 4I 42

10. F1  H3, 10I; F2  H10, 3I; F3  H9, 2I

22. u

11. F1  5i  2j; F2  i  10j

3.5 tons

12. F1  7i  6j; F2  8i  3j 13. F1  2.5i  4.7j; F2  0.3i  6.9j; F3  12j 14. F1  3 12i  2 13j; F2  2i  7j; F3  5i  2 13j 15.

F1

16.

y

v

128

23. 65

F1

v

y

1525 lb 20

F2

10 104

6 25 20 9

F2 x

F3

17 19

115 35 18

u x

24.

F3

17. The force vectors F1 and F2 are simultaneously acting on a point P. Find a third vector F3 so that equilibrium takes place if F1  H19, 10I and F2  H5, 17I.

221 lb v

18. The force vectors F1, F2, and F3 are simultaneously acting on a point P. Find a fourth vector F4 so that equilibrium takes place if F1  H12, 2I, F2  H6, 17I, and F3  H3, 15I. 19. A new “Survivor” game y F2 involves a three-team tugF1 of-war. Teams 1 and 2 are 2210 2500 pulling with the magnitude 50 40 and at the angles indicated x ?F3 in the diagram. If the teams are currently in a stalemate, find the magnitude and angle of the rope held by team 3. 20. Three cowhands have roped y F1 a wild stallion and are attempting to hold him steady. The first and second F2 200 170 cowhands are pulling with 75 19 the magnitude and at the ?F3 angles indicated in the diagram. If the stallion is held fast by the three cowhands, find the magnitude and angle of the rope from the third cowhand.

u

x

70

25. 30 3010 kg

v

u

26.

u v

2 tons 115

27. Static equilibrium: A 500-lb crate is sitting on a ramp that is inclined at 35°. Find the force needed to hold the object stationary.

500

lb

500 lb 35 G

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28. Static equilibrium: A 1200-lb skiff is 1200 lb being pulled from a lake, using a boat 1200 lb ramp inclined at 20°. Find the G minimum force needed to dock the skiff. 29. Static equilibrium: A 325-kg carton is sitting on a ramp, held stationary by ␪ 225 kg of tension in a restraining rope. Find the ramps’s angle of incline.



20

325

kg

225 kg

325 kg

G

30. Static 1.75 tons ? ton equilibrium: s A heavy dump truck is being 18 winched up a ramp with an G 18° incline. Approximate the weight of the truck if the winch is working at its maximum capacity of 1.75 tons and the truck is barely moving. 31. While rearranging the patio furniture, Rick has to push the weighted base of the umbrella stand 15 m. If he uses a constant force of 75 N, how much work did he do? 32. Vinny’s car just broke down in the middle of the road. Luckily, a buddy is with him and offers to steer if Vinny will get out and push. If he pushes with a constant force of 185 N to move the car 30 m, how much work did he do?

WORKING WITH FORMULAS

The range of a projectile: R 

v2sin  cos  16

33. The range of a projected object (total horizontal distance traveled) is given by the formula shown, where v is the initial velocity and  is the angle at which it is projected. If an arrow leaves the bow traveling 175 ft/sec at an angle of 45°, what horizontal distance will it travel? 

685

34. A collegiate javelin thrower releases the javelin at a 40° angle, with an initial velocity of about 95 ft/sec. If the NCAA record is 280 ft, will this throw break the record? What is the smallest angle of release that will break this record? If the javelin were released at the optimum 45°, by how many feet would the record be broken?

APPLICATIONS

35. Plowing a field: An old-time farmer is plowing his field with a mule. How much work does the mule do in plowing one length of a field 300 ft long, if it pulls the plow with a constant force of 250 lb and the straps make a 30° angle with the horizontal.

37. Tough-man contest: As part of a “tough-man” contest, participants are required to pull a bus along a level street for 100 ft. If one contestant did 45,000 ft-lb of work to accomplish the task and the straps used made an angle of 5° with the street, find the tension in the strap during the pull.

36. Pulling a sled: To enjoy a beautiful snowy day, a mother is pulling her three 32° children on a sled along a level street. How much work (play) is done if the street is 100 ft long and she pulls with a constant force of 55 lb with the tow-rope making an angle of 32° with the street?

38. Moving supplies: An arctic explorer is hauling supplies from the supply hut to her tent, a distance of 150 ft, in a sled she is dragging behind her. If 9000 ft-lb of work was done and the straps used made an angle of 25° with the snow-covered ground, find the tension in the strap during the task. 39. Wheelbarrow rides: To break up the monotony of a long, hot, boring Saturday, a father decides to (carefully) give his kids a ride in a wheelbarrow. He applies a force of 30 N to move the “load” 100 m, then stops to rest. Find the amount of work done if the wheelbarrow makes an angle of 20° with level ground while in motion.

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40. Mowing the lawn: A home owner applies a force of 40 N to push her lawn mower back and forth across the back yard. Find the amount of work done if the yard is 50 m long, requires 24 passes to get the lawn mowed, and the mower arm makes an angle of 39° with the level ground. Force vectors: For the force vector F and vector v given, find the amount of work required to move an object along the entire length of v. Assume force is in pounds and distance in feet.

41. F  H15, 10I; v  H50, 5I 42. F  H5, 12I; v  H25, 10I

58. u  H5, 4I; v  H9, 11I 59. u  2i  6j; v  9i  3j 60. u  3 12i  2j; v  2 12i  6j Find compvu for the vectors u and v given.

61. u  H3, 5I; v  H7, 1I 62. u  H3, 5I; v  H7, 1I 63. u  7i  4j; v  10j 64. u  8i; v  10i  3j 65. u  7 12i  3j; v  6i  5 13j 66. u  3 12i  6j; v  2i  5 15j For each pair of vectors given, (a) find the projection of u along v (compute projvu) and (b) resolve u into vectors u1 and u2, where u1 7v and u2v.

43. F  H8, 2I; v  H15, 1I

67. u  H2, 6I; v  H8, 3I

44. F  H15, 3I; v  H24, 20I

68. u  H3, 8I; v  H12, 3I

45. Use the dot product to verify the solution to Exercise 41.

69. u  H2, 8I; v  H6, 1I

46. Use the dot product to verify the solution to Exercise 42.

71. u  10i  5j; v  12i  2j

47. Use the dot product to verify the solution to Exercise 43. 48. Use the dot product to verify the solution to Exercise 44. For each pair of vectors given, (a) compute the dot product p # q and (b) find the angle between the vectors to the nearest tenth of a degree.

70. u  H4.2, 3I; v  H5, 8.3I 72. u  3i  9j; v  5i  3j Projectile motion: A projectile is launched from a catapult with the initial velocity v0 and angle  indicated. Find (a) the position of the object after 3 sec and (b) the time required to reach a height of 250 ft.

73. v0  250 ft/sec;   60° 74. v0  300 ft/sec;   55°

49. p  H5, 2I; q  H3, 7I

75. v0  200 ft/sec;   45°

50. p  H3, 6I; q  H2,5I

76. v0  500 ft/sec;   70°

51. p  2i  3j; q  6i  4j

77. At the circus, a “human cannon ball” is shot from a large cannon with an initial velocity of 90 ft/sec at an angle of 65° from the horizontal. How high is the acrobat after 1.2 sec? How long until the acrobat is again at this same height?

52. p  4i  3j; q  6i  8j 53. p  712i  3j; q  212i  9j 54. p  12i  3j; q  312i  5j Determine if the pair of vectors given are orthogonal.

55. u  H7, 2I; v  H4, 14I 56. u  H3.5, 2.1I; v  H6, 10I 57. u  H6, 3I; v  H8, 15I

78. A center fielder runs down a long hit by an opposing batter and whirls to throw the ball to the infield to keep the hitter to a double. If the initial velocity of the throw is 130 ft/sec and the ball is released at an angle of 30° with level ground, how high is the ball after 1.5 sec? How long until the ball again reaches this same height?

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EXTENDING THE CONCEPT

For the arbitrary vectors u  H a, bI, v  Hc, d I, and w  He, f I and the scalar k, prove the following vector properties using the properties of real numbers.

represent the slopes of the vectors. Find the angle between the vectors 1i  5j and 5i  2j using each equation and comment on which you found more efficient. Then see if you can find a geometric connection between the two equations.

79. w # 1u  v2  w # u  w # v 80. k1u # v2  ku # v  u # kv 81. 0 # u  u # 0  0

82.

84. Use the equations for the horizontal and vertical components of the projected object’s position to obtain the equation of trajectory 16 y  1tan 2x  2 2 x2. This is a quadratic v cos  equation in x. What can you say about its graph? Include comments about the concavity, x-intercepts, maximum height, and so on.

u#v

u # v  u v uv

u#v for finding the uv angle between two vectors, the equation m2  m1 can be used, where m1 and m2 tan   1  m2m1

83. As alternative to cos  



687

Section 7.4 Vector Applications and the Dot Product

MAINTAINING YOUR SKILLS

85. (4.4) Solve for t: 2.9e0.25t  7.6  438

88. (7.3) A plane is flying 200 mph at heading 30°, with a 40 mph wind blowing from due west. Find the true course and speed of the plane.

86. (5.5) Graph the function using a reference rectangle and the rule of fourths:  y  3 cosa2  b 4 87. (7.2) Solve the triangle shown, then A compute its perimeter and area.

Plane

200 mph 30

C 250 m

Wind 40 mph

32 172 m B

7.5 Complex Numbers in Trigonometric Form Learning Objectives In Section 7.5 you will learn how to:

A. Graph a complex number B. Write a complex number in trigonometric form

C. Convert from trigonometric form to rectangular form

D. Interpret products and quotients geometrically

E. Compute products and quotients in trigonometric form

F. Solve applications involving complex numbers (optional)

Once the set of complex numbers became recognized and defined, the related basic operations matured very quickly. With little modification—sums, differences, products, quotients, and powers all lent themselves fairly well to the algebraic techniques used for real numbers. But roots of complex numbers did not yield so easily and additional tools and techniques were needed. Writing complex numbers in trigonometric form enables us to find complex roots (Section 7.6) and in some cases, makes computing products, quotients, and powers more efficient.

A. Graphing Complex Numbers In previous sections we defined a vector quantity as one that required more than a single component to describe its attributes. The complex number z  a  bi certainly fits this description, since both a real number “component” and an imaginary “component” are needed to define it. In many respects, we can treat complex numbers in the same way we treated vectors and in fact, there is much we can learn from this connection. Since both axes in the xy-plane have real number values, it’s not possible to graph a complex number in  (the real plane). However, in the same way we used

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the x-axis for the horizontal component of a vector and the y-axis for the vertical, we can let the x-axis represent the real valued part of a complex number and the y-axis the imaginary part. The result is called the complex plane . Every point (a, b) in  can be associated with a complex number a  bi, and any complex number a  bi can be associated with a point (a, b) in  (Figure 7.69). The point (a, b) can also be regarded as the terminal point of a position vector representing the complex number, generally named using the letter z.

WORTHY OF NOTE Surprisingly, the study of complex numbers matured much earlier than the study of vectors, and representing complex numbers as directed line segments actually preceded their application to a vector quantity.

EXAMPLE 1



Figure 7.69 Imaginary yi axis b

(a + bi) z x a Real axis

Graphing Complex Numbers

yi

Graph the complex numbers below on the same complex plane. a. z1  2  6i b. z2  5  4i c. z3  5 d. z4  4i Solution



(5, 4)

(0, 4) z4

z2 z3

The graph of each complex number is shown in the figure.

(5, 0) x

z1 (2, 6)

Now try Exercises 7 through 10

A. You’ve just learned how to graph a complex number

Figure 7.70 z2 (2, 3)

z

(5, 2) z2

In Example 1, you likely noticed that from a vector perspective, z2 is the “resultant vector” for the sum z3  z4. To investigate further, consider z1  12  3i2, z2  15  2i2, and the sum z1  z2  z shown in Figure 7.70. The figure helps to confirm that the sum of complex numbers can be illustrated geometrically using the parallelogram (tail-to-tip) method employed for vectors in Section 7.4.

B. Complex Numbers in Trigonometric Form

yi (3, 5)



z1 x

The complex number z  a  bi is said to be in rectangular form since it can be graphed using the rectangular coordinates of the complex plane. Complex numbers can also be written in trigonometric form. Similar to how  x  represents the distance between the real number x and zero, z represents the distance between Figure 7.71 (a, b) and the origin in the complex yi plane, and is computed as z  2a2  b2. With any nonzero z, we can also associate z  a  bi an angle , which is the angle in standard r  b position whose terminal side coincides x with the graph of z. If we let r represent r a a z, Figure 7.71 shows cos   and r b sin   , yielding r cos   a and r sin   b. The appropriate substitutions into r a  bi give the trigonometric form: z  a  bi  r cos   r sin  # i

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Factoring out r and writing the imaginary unit as the lead factor of sin  gives the b relationship in its more common form, z  r 1cos   i sin 2, where tan   . a WORTHY OF NOTE

The Trigonometric Form of a Complex Number

While it is true the trigonometric form can more generally be written as z  r 3cos 1  2k2  i sin 1  2k2 4 for k  , the result is identical for any integer k and we will select  so that 0   6 2 or 0°   6 360°, depending on whether we are working in radians or degrees.

For the complex number z  a  bi and angle  shown, z  r1cos   i sin 2 b is the trigonometric form of z, where r  2a2  b2, and tan   ; a  0. a yi • r  z represents the magnitude of z (also called the modulus). z  a  bi •  is often referred to as the argument of z. r  b

r

x

a

b b Be sure to note that for tan   , tan1a b is equal to r (the reference angle a a for ) and the value of  will ultimately depend on the quadrant of z. EXAMPLE 2



Converting a Complex Number from Rectangular to Trigonometric Form State the quadrant of the complex number, then write each in trigonometric form. a. z1  2  2i b. z2  6  2i

Solution

yi 2  r

3

2 x

r

2  2i

v

3 yi 5 6  2i r   r 6 x

4



Knowing that modulus r and angle  are needed for the trigonometric form, we first determine these values. Once again, to find the correct value of , it’s important to note the quadrant of the complex number. a. z1  2  2i; QIII b. z  6  2i; QI 2 2 r  2122  122 r  2162 2  122 2  18  2 12  140  2 110 2 2 r  tan1a b r  tan1a b 2 6 1  tan1a b  tan1 112 3  1  z is in QI, so   tan1a b 4 3 5 1 with z1 in QIII,   . z  2 110 acos c tan1a b d  4 3 1 5 5 i sin c tan1a b d b z1  2 12 c cos a b  i sin a b d 3 4 4 See the figure.

5

Now try Exercises 11 through 26



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B. You’ve just learned how to write a complex number in trigonometric form

WORTHY OF NOTE Using the triangle diagrams from Section 6.5, 1 cos ctan1a bd and 3 1 sin ctan1a b d can easily be 3 evaluated and used to verify 1 2110 cis ctan1a bd  6  2i. 3

EXAMPLE 3

Since the angle  is repeated for both cosine and sine, we often use an abbreviated notation for the trigonometric form, called “cis” (sis) notation: z  r 1cos   i sin 2  r cis . The results of Example 2(a) and 2(b) would then be 5 1 written z  2 12 cis a b and z  2 110 cis c tan1a b d , respectively. 4 3 b As in Example 2b, when r  tan1a b is not a standard angle we either answer a in exact form as shown, or use a four-decimal-place approximation: 2110 cis10.32182.

C. Converting from Trigonometric Form to Rectangular Form Converting from trigonometric form back to rectangular form is simply a matter of evaluating r cis . This can be done regardless of whether  is a standard angle or in b the form tan1a b, since in the latter case we can construct a right triangle with side a b opposite  and side a adjacent , and find the needed values as in Section 6.5. 

Converting a Complex Number from Trigonometric to Rectangular Form Graph the following complex numbers, then write them in rectangular form.  5 a. z  12 cis a b b. z  13 cis c tan1a b d 6 12

Solution



 , which yields the 6 graph in Figure 7.72. In the nonabbreviated   form we have z  12 c cos a b  i sin a b d . 6 6 Evaluating within the brackets gives 1 13  i d  6 13  6i. z  12 c 2 2 5 b. For r  13 and   tan1a b, we have 12 the graph shown in Figure 7.73. Here we obtain the rectangular form directly from the diagram with z  12  5i. Verify by 5 noting that for   tan1a b, 12 5 12 cos   , and sin   , meaning 5 13 13 z  131cos   i sin 2  13 c

C. You’ve just learned how to convert from trigonometric form to rectangular form

Figure 7.72

a. We have r  12 and  

12 5  i d  12  5i. 13 13

yi z (6√3  6i) 12 k x

Figure 7.73 yi z (12  5i) 13 5   tan112 

12 x

Now try Exercises 27 through 34



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D. Interpreting Products and Quotients Geometrically The multiplication and division of complex numbers has some geometric connections that can help us understand their computation in trigonometric form. Note the relationship between the modulus and argument of the following product, with the moduli (plural of modulus) and arguments from each factor. EXAMPLE 4



Noting Graphical Connections for the Product of Two Complex Numbers For z1  3  3i and z2  0  2i, a. Graph the complex numbers and compute their moduli and arguments. b. Compute and graph the product z1z2 and find its modulus and argument. Discuss any connections you see between the factors and the resulting product.

Solution



a. The graphs of z1 and z2 are shown in the figure. For the modulus and argument we have: z1  3  3i; QI r  2132 2  132 2  118  3 12 1

  tan 1 1   45°

D. You’ve just learned how to interpret products and quotients geometrically

yi

z3

z2  0  2i; 1quadrantal2 r  2 directly

z1

135 z2 45

x

  90° directly

b. The product z1z2 is 13  3i2 12i2  6  6i, which is in QII. The modulus is 2162 2  162 2  172  6 12, with an argument of r  tan1 112 or 135° (QII). Note the product of the two moduli is equal to the modulus of the final product: 2 # 3 12  6 12. Also note that the sum of the arguments for z1 and z2 is equal to the argument of the product: 45°  90°  135°! Now try Exercises 35 and 36



A similar geometric connection exists for the division of complex numbers. This connection is explored in Exercises 37 and 38 of the exercise set.

E. Products and Quotients in Trigonometric Form The connections in Example 4 are not a coincidence, and can be proven to hold for all complex numbers. Consider any two nonzero complex numbers z1  r1 1cos   i sin 2 and z2  r2 1cos   i sin 2. For the product z1z2 we have product in trig form z1z2  r1 1cos   i sin 2 r2 1cos   i sin 2  r1r2 3 1cos   i sin 21cos   i sin 2 4 rearrange factors  r1r2 3cos  cos   i sin  cos   i sin  cos   i2sin  sin  4  r1r2 3 1cos  cos   sin  sin 2  i1sin  cos   sin  cos 2 4

 r1r2 3cos1  2  i sin1  2 4

F-O-I-L commute terms; i 2  1

use sum/difference identities for sine/cosine

In words, the proof says that to multiply complex numbers in trigonometric form, we multiply the moduli and add the arguments. For division, we divide the moduli and subtract the arguments. The proof for division resembles that for multiplication and is asked for in Exercise 71.

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Products and Quotients of Complex Numbers in Trigonometric Form For the complex numbers z1  r1 1cos   i sin 2 and z2  r2 1cos   i sin 2,

z1z2  r1r2 3cos1  2  i sin1  2 4 and

z1 r1  3cos1  2  i sin1  2 4 , z2  0. z2 r2 EXAMPLE 5



Multiplying Complex Numbers in Trigonometric Form For z1  3  13i and z2  13  1i, a. Write z1 and z2 in trigonometric form and compute z1z2. z1 b. Compute the quotient in trigonometric form. z2 c. Verify the product using the rectangular form.

Solution



a. For z1 in QII we find r  2 13 and   150°, for z2 in QI, r  2 and   30°. In trigonometric form, z1  2 131cos 150°  i sin 150°2 and z2  21cos 30°  i sin 30°2: z1z2  2 131cos 150°  i sin 150°2 # 21cos 30°  i sin 30°2  2 13 # 2 3cos1150°  30°2  i sin1150°  30°2 4  4 131cos 180°  i sin 180°2  4 13 11  0i2  4 13 2 131cos 150°  i sin 150°2 z1  z2 21cos 30°  i sin 30°2  133cos1150°  30°2  i sin1150°  30°2 4  131cos 120°  i sin 120°2 1 13 ib  13a  2 2 3 13  i  2 2 c. z1z2  13  13i21 13  1i2  3 13  3i  3i  13i2  413

multiply moduli, add arguments

b.

divide moduli, subtract arguments

Now try Exercises 39 through 46

E. You’ve just learned how to compute products and quotients in trigonometric form



Converting to trigonometric form for multiplication and division seems too clumsy for practical use, as we can often compute these results more efficiently in rectangular form. However, this approach leads to powers and roots of complex numbers, an indispensable part of advanced equation solving, and these are not easily found in rectangular form. In any case, note that the power and simplicity of computing products/quotients in trigonometric form is highly magnified when the complex numbers are given in trig form: 112 cis 50°2 13 cis 20°2  36 cis 70° See Exercises 47 through 50.

12 cis 50°  4 cis 30°. 3 cis 20°

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F. (Optional) Applications of Complex Numbers Somewhat surprisingly, complex numbers Figure 7.74 have several applications in the real Magnetic flux world. Many of these involve a study of electricity, and in particular AC (alternating current) circuits. In simplistic terms, when an armature (molded wire) is rotated in a S N uniform magnetic field, a voltage V is generated that depends on the strength of the field. As the armature is rotated, the voltage varies between a maximum and a minimum value, with the amount of voltage modeled by V12  Vmaxsin1B2, with  in degrees. Here, Vmax represents the maximum voltage attained, and the input variable  represents the angle the armature makes with the magnetic flux, indicated in Figure 7.74 by the dashed arrows between the magnets. When the armature is perpendicular to the flux, we say   0°. At   0° and   180°, no voltage is produced, while at   90° and   270°, the voltage reaches its maximum and minimum values respectively (hence the name alternating current). Many electric dryers and other large appliances are labeled as 220 volt (V) appliances, but use an alternating current that varies from 311 V to 311 V (see Worthy of Note). This means when   52°, V152°2  311 sin152°2  245 V is being generated. In practical applications, we use time t as the independent variable, rather the angle of the armature. These large appliances usually operate with a frequency of 60 cycles per 1 1 2 , we obtain second, or 1 cycle every of a second aP  b. Using B  60 60 P B  120 and our equation model becomes V1t2  311 sin1120t2 with t in radians. This variation in voltage is an excellent example of a simple harmonic model.

WORTHY OF NOTE You may have wondered why we’re using an amplitude of 311 for a 220-V appliance. Due to the nature of the sine wave, the average value of an alternating current is always zero and gives no useful information about the voltage generated. Instead, the root-mean-square (rms) of the voltage is given on most appliances. While the maximum voltage is 311 V, the rms voltage is 311  220 V. See 12 Exercise 72.

EXAMPLE 6



Analyzing Alternating Current Using Trigonometry Use the equation V1t2  311 sin1120t2 to: a. Create a table of values illustrating the voltage produced every thousandth of a 1 second for the first half-cycle at   0.008b. 120 b. Use a graphing calculator to find the times t in this half-cycle when 160 V is being produced.

Solution



a. Starting at t  0 and using increments of 0.001 sec produces the table shown. 400

0

0.0165

400

Time t

Voltage

0

0

0.001

114.5

0.002

212.9

0.003

281.4

0.004

310.4

0.005

295.8

0.006

239.6

0.007

149.8

0.008

39.9

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b. From the table we note V1t2  160 when t  10.001, 0.0022 and t  10.006, 0.0072 . Using the intersection of graphs method places these values at t  0.0014 and t  0.0069 (see graph). Now try Exercises 53 and 54 Figure 7.77 Voltage leads current by 90 (phase angle   90)

sin 

90

180



270

Figure 7.78 Voltage lags current by 90 (phase angle   90)

sin 

90

180



270

Figure 7.79 yi XL R x

XC

WORTHY OF NOTE While mathematicians generally use the symbol i to represent 11, the “i” is used in other fields to represent an electric current so the symbol j  11 is used instead. In conformance with this convention, we will temporarily use j for 11 as well.

EXAMPLE 7





The chief components of AC circuits are voltage (V ) and current (I ). Due to the nature of how the current is generated, V and I can be modeled by sine functions. Other characteristics of electricity include pure resistance (R), Figure 7.75 inductive reactance (XL), and capacitive reactance (XC) (see Figure 7.75). Each of these is measured in a unit called XL XC R ohms (2, while current I is measured in amperes (A), and B C D voltages are measured in volts (V). These components of A electricity are related by fixed and inherent traits, which include the following: (1) voltage across a Figure 7.76 resistor is always in phase with the current, Voltage and current sin  meaning the phase shift or phase angle between are in phase (phase angle   0) them is 0° (Figure 7.76);(2) voltage across an inductor leads the current by 90° (Figure 7.77); (3) voltage across a capacitor lags the current by 90 180 270  90° (Figure 7.78); and (4) voltage is equal to the product of the current times the resistance or reactance: V  IR, V  IXL, and V  IXC. Different combinations of R, XL, and XC in a combined (series) circuit alter the phase angle and the resulting voltage. Since voltage across a resistance is always in phase with the current (trait 1), we can model the resistance as a vector along the positive real axis (since the phase angle is 0°). For traits (2) and (3), XL is modeled on the positive imaginary axis since voltage leads current by 90°, and XC on the negative imaginary axis since voltage lags current by 90° (see Figure 7.79). These natural characteristics make the complex plane a perfect fit for describing the characteristics of the circuit. Consider a series circuit (Figure 7.75), where R  12 , XL  9 , and XC  4 . For a current of I  2 amps through this circuit, the voltage across each individual element would be VR  122 1122  24 V (A to B), VL  122192  18 V (B to C), and VC  122142  8 V (C to D). However, the resulting voltage across this circuit cannot be an arithmetic sum, since R is real while XL and XC are represented by imaginary numbers. The joint effect of resistance (R) and reactance (XL, XC) in a circuit is called the impedance, denoted by the letter Z, and is a measure of the total resistance to the flow of electrons. It is computed Z  R  XL j  XC j (see Worthy of Note), due to the phase angle relationship of the voltage in each element (XL and XC point in opposite directions, hence the subtraction). The expression for Z is more commonly written R  1XL  XC 2 j, where we more clearly note Z is a complex number whose magnitude and angle with the x-axis can be found as before: XL  XC b. The angle  represents the  Z   2R2  1XL  XC 2 2 and r  tan1a R phase angle between the voltage and current brought about by this combination of elements. The resulting voltage of the circuit is then calculated as the product of the current with the magnitude of the impedance, or VRLC  I  Z  (Z is also measured in ohms, ). Finding the Impedence and Phase Angle of the Current in a Circuit For the circuit diagrammed in the figure, (a) find the magnitude of Z, the phase angle between current and voltage, and write the result in trigonometric form; and (b) find the total voltage across this circuit.

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Solution XL

R A

12 Ω

B

9Ω




XC C

4Ω

D

F. You have just learned how to solve applications involving complex numbers

695

a. Using the values given, we find Z
R  1XL  XC 2 j
12  19  42 j
12  5j (QI). This gives a magnitude of


Z

21122 2  152 2
1169
13 , with a phase angle of 5

tan1a b  22.6° (voltage leads the current by about 22.6°). 12 In trigonometric form Z  13 cis 22.6°. b. With I
2 amps, the total voltage across this circuit is VRLC
I
Z

21132
26 V. Now try Exercises 55 through 68




7.5 EXERCISES


CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. For a complex number written in the form z
r 1cos
 i sin
2, r is called the and
is called the .   2. The complex number z
2 c cosa b  i sina b d 4 4 can be written as the abbreviated “cis” notation as . 3. To multiply complex numbers in trigonometric form, we the moduli and the arguments.


4. To divide complex numbers in trigonometric form, we the moduli and the arguments. 5. Write z
1  13i in trigonometric form and explain why the argument is

240° instead of 60° as indicated by your calculator. 6. Discuss the similarities between finding the components of a vector and writing a complex number in trigonometric form.

DEVELOPING YOUR SKILLS

Graph the complex numbers z1, z2, and z3 given, then express one as the sum of the other two.

7. z1
7  2i z2
8  6i z3
1  4i 9. z1
2  5i z2
1  7i z3
3  2i

8. z1
2  7i z2
3  4i z3
1  3i 10. z1
2  6i z2
7  2i z3
5  4i

State the quadrant of each complex number, then write it in trigonometric form. For Exercises 11 through 14, answer in degrees. For 15 through 18, answer in radians.

11. 2  2i

12. 7  7i

13. 513  5i

14. 2  213i

15. 3 12  3 12i

16. 517  5 17i

17. 4 13  4i

18. 6  6 13i

Write each complex number in trigonometric form. For Exercises 19 through 22, answer in degrees using both an exact form and an approximate form, rounding to tenths. For 23 through 26, answer in radians using both an exact form and an approximate form, rounding to four decimal places.

19. 8  6i

20. 9  12i

21. 5  12i

22. 8  15i

23. 6  17.5i

24. 30  5.5i

25. 6  10i

26. 12  4i

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Graph each complex number using its trigonometric form, then convert each to rectangular form.

 27. 2 cis a b 4

 28. 12 cis a b 6

 29. 4 13 cis a b 3

30. 5 13 cis a

31. 17 cis c tan1a

15 bd 8

7 b 6

3 32. 10 cis c tan1a b d 4

z1 using the z2 trigonometric form. Answer in exact rectangular form where possible, otherwise round all values to two decimal places. Compute the product z1z2 and quotient

40. z1 

41. z1  2 13  0i

42. z1  0  6i 12

33. 6 cis c   tan1a

5 bd 111

z2  

34. 4 cis c   tan1a

17 bd 3

43. z1  9 c cos a

For the complex numbers z1 and z2 given, find their moduli r1 and r2 and arguments 1 and 2. Then compute their product in rectangular form. For modulus r and argument  of the product, verify that r1r2  r and 1  2  .

35. z1  2  2i; z2  3  3i 36. z1  1  13i;

z2  3  13i

For the complex numbers z1 and z2 given, find their moduli r1 and r2 and arguments 1 and 2. Then compute their quotient in rectangular form. For modulus r and argument  of the quotient, verify that r1  r and 1  2  . r2

37. z1  13  i; z2  1  13i 38. z1   13  i; z2  3  0i 

7i 13 21  2 2

z2 

3i 16 3 12  2 2

  b  i sin a b d 15 15 2 2 z2  1.8 c cos a b  i sin a b d 3 3 3 3 b  i sin a b d 5 5   z2  8.4 c cos a b  i sin a b d 5 5

44. z1  2 c cos a

45. z1  101cos 60°  i sin 60°2 z2  41cos 30°  i sin 30°2 46. z1  71cos 120°  i sin 120°2 z2  21cos 300°  i sin 300°2 47. z1  5 12 cis 210° z2  2 12 cis 30°

48. z1  5 13 cis 240° z2  13 cis 90°

49. z1  6 cis 82° z2  1.5 cis 27°

50. z1  1.6 cis 59° z2  8 cis 275°

WORKING WITH FORMULAS

51. Equilateral triangles in the complex plane: u2  v2  w2  uv  uw  vw If the line segments connecting the complex numbers u, v, and w form the vertices of an equilateral triangle, the formula shown above holds true. Verify that u  2  13i, v  10  13i, and w  6  5 13i form the vertices of an equilateral triangle using the distance formula, then verify the formula given. 

5 13 5  i 2 2 z2  0  6i

39. z1  4 13  4i 3 13 3  i z2  2 2

52. The cube of a complex number: 1A  B2 3  A3  3A2B  3AB2  B3 The cube of any binomial can be found using the formula here, where A and B are the terms of the binomial. Use the formula to compute the cube of 1  2i (note A  1 and B  2i2.

APPLICATIONS

53. Electric current: In the United States, electric power is supplied to homes and offices via a “120 V circuit,” using an alternating current that varies from 170 V to 170 V, at a frequency of 60 cycles/sec. (a) Write the voltage equation for U.S.

households, (b) create a table of values illustrating the voltage produced every thousandth of a second for the first half-cycle, and (c) find the first time t in this half-cycle when exactly 140 V is being produced.

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54. Electric current: While the electricity supplied in Europe is still not quite uniform, most countries employ 230-V circuits, using an alternating current that varies from 325 V to 325 V. However, the frequency is only 50 cycles per second. (a) Write the voltage equation for these European countries, (b) create a table of values illustrating the voltage produced every thousandth of a second for the first half-cycle, and (c) find the first time t in this halfcycle when exactly 215 V is being produced.

trigonometric form and find the voltage in each circuit. Recall V  IZ.

61. I  13  1j A and Z  5  5j  62. I  13  1j A and Z  2  2j  63. I  3  2j A and Z  2  3.75j  64. I  4  3j A and Z  2  4j 

AC circuits: For the circuits indicated in Exercises 55 through 60, (a) find the magnitude of Z, the phase angle between current and voltage, and write the result in trigonometric form; and (b) find the total voltage across this circuit. Recall Z  R  (XL  XC) j and |Z |  1R2  (XL  XC)2. Exercises 55 through 58

55. R  15 , XL  12 , and XC  4 , with I  3 A.

A

56. R  24 , XL  12 , and XC  5 , with I  2.5 A. 57. R  7 , XL  6 , and XC  11 , with I  1.8 A. 58. R  9.2 , XL  5.6 , and XC  8.3 , with I  2.0 A.

XL

XC

R B

D

Exercises 59 and 60 A

XL B

C

59. R  12  and XL  5 , with I  1.7 A. 60. R  35  and XL  12 , with I  4 A. AC circuits — voltage: The current I and the impedance Z for certain AC circuits are given. Write I and Z in



AC circuits — current: If the voltage and impedance are known, the current I in the circuit is calculated as the V quotient I  . Write V and Z in trigonometric form to Z find the current in each circuit.

65. V  2  2 13j and Z  4  4j  66. V  413  4j and Z  1  1j  67. V  3  4j and Z  4  7.5j 

C

R

697

68. V  2.8  9.6j and Z  1.4  4.8j  Parallel circuits: For AC circuits wired in parallel, the Z1Z2 total impedance is given by Z  , where Z1 and Z1  Z2 Z2 represent the impedance in each branch. Find the total impedance for the values given. Compute the product in the numerator using trigonometric form, and the sum in the denominator in rectangular form.

69. Z1  1  2j and Z2  3  2j 70. Z1  3  j and Z2  2  j

EXTENDING THE CONCEPT

71. Verify/prove that for the complex numbers z1  r1 1cos   i sin 2 and z2  r2 1cos   i sin 2, z1 r1  3cos1  2  i sin1  2 4. z2 r2 72. Using the Internet, a trade manual, or some other resource, find the voltage and frequency at which electricity is supplied to most of Japan (oddly enough — two different frequencies are in common use). As in Example 6, the voltage given will likely be the root-mean-square (rms) voltage. Use the information to find the true voltage and the equation model for voltage in most of Japan.

73. Recall that two lines are perpendicular if their slopes have a product of 1. For the directed line segment representing the complex number z1  7  24i, find complex numbers z2 and z3 whose directed line segments are perpendicular to z1 and have a magnitude one-fifth as large. 74. The magnitude of the impedance is  Z   2R2  1XL  XC 2 2. If R, XL, and XC are all nonzero, what conditions would make the magnitude of Z as small as possible?

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MAINTAINING YOUR SKILLS

75. (6.7) Solve for x  3 0, 22:  350  750 sina2x  b  25 4 76. (3.6) Name all asymptotes of the function 1  x3 h1x2  x2

78. (7.2) A ship is spotted by two observation posts that are 4 mi apart. Using the line between them for reference, the first post reports the ship is at an angle of 41°, while the second reports an angle of 63°, as shown. How far is the ship from the closest post? Exercise 78

77. (2.7) Graph the piecewise-defined function given: 2 x 6 2 f 1x2  • x2 2  x 6 1 x x 1

41

63 4 mi

7.6 De Moivre’s Theorem and the Theorem on nth Roots Learning Objectives In Section 7.6 you will learn how to:

A. Use De Moivre’s theorem to raise complex numbers to any power

B. Use De Moivre’s theorem to check solutions to polynomial equations

C. Use the nth roots theorem to find the nth roots of a complex number

The material in this section represents some of the most significant developments in the history of mathematics. After hundreds of years of struggle, mathematical scientists had not only come to recognize the existence of complex numbers, but were able to make operations on them commonplace and routine. This allowed for the unification of many ideas related to the study of polynomial equations, and answered questions that had puzzled scientists from many different fields for centuries. In this section, we will look at two fairly simple theorems that actually represent over 1000 years in the evolution of mathematical thought.

A. De Moivre’s Theorem Having found acceptable means for applying the four basic operations to complex numbers, our attention naturally shifts to the computation of powers and roots. Without them, we’d remain wholly unable to offer complete solutions to polynomial equations and find solutions for many applications. The computation of powers, squares, and cubes offer little challenge, as they can be computed easily using the formula for binomial squares 3 1A  B2 2  A2  2AB  B2 4 or by applying the binomial theorem. For larger powers, the binomial theorem becomes too time consuming and a more efficient method is desired. The key here is to use the trigonometric form of the complex number. In Section 7.5, we noted the product of two complex numbers involved multiplying their moduli and adding their arguments: For z1  r1 1cos 1  i sin 1 2 and z2  r2 1cos 2  i sin 2 2 we have z1z2  r1r 3cos11  2 2  i sin11  2 2 4

For the square of a complex number, r1  r2 and 1  2. Using  itself yields z2  r2 3cos1  2  i sin1  2 4  r2 3cos122  i sin122 4

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699

Multiplying this result by z  r 1cos   i sin 2 to compute z3 gives

r2 3cos122  i sin122 4 r1cos   i sin 2  r3 3cos12  2  i sin12  2 4

WORTHY OF NOTE

 r3 3cos132  i sin132 4 .

Sometimes the argument of cosine and sine becomes very large after applying De Moivre’s theorem. In these cases, we use the fact that     360°k and     2 k represent coterminal angles for integers k, and use the coterminal angle  where 0   6 360° or 0   6 2.

EXAMPLE 1

The result can be extended further and generalized into De Moivre’s theorem. De Moivre’s Theorem For any positive integer n, and z  r 1cos   i sin 2,

zn  rn 3cos1n2  i sin1n2 4

For a proof of the theorem where n is an integer and n  1, see Appendix V. 

Using De Moivre’s Theorem to Compute the Power of a Complex Number Use De Moivre’s theorem to compute z9, given z  12  12i.

Solution



1 2 12 1 2 . With z in QIII, tan   1 yields a b  a b  2 2 2 B 5 5 5 12  . The trigonometric form is z  c cosa b  i sina b d and applying 4 2 4 4 the theorem with n  9 gives

Here we have r 

5 5 12 9 b c cosa9 # b  i sina9 # bd 2 4 4 45 45 12 c cosa b  i sina bd  32 4 4 5 5 12 c cosa b  i sina b d  32 4 4 12 12 12  ib a  32 2 2 1 1   i 32 32

z9  a

A. You’ve just learned how to use De Moivre’s theorem to raise complex numbers to any power

De Moivre’s theorem

simplify

coterminal angles

evaluate functions

result

Now try Exercises 7 through 14



As with products and quotients, if the complex number is given in trigonometric form, computing any power of the number is both elegant and efficient. For instance, if z  2 cis 40°, then z4  16 cis 160°. See Exercises 15 through 18. For cases where  is not a standard angle, De Moivre’s theorem requires an intriguing application of the skills developed in Chapter 6, including the use of multiple angle b identities and working from a right triangle drawn relative to r  tan1a b. a See Exercises 57 and 58.

B. Checking Solutions to Polynomial Equations One application of De Moivre’s theorem is checking the complex roots of a polynomial, as in Example 2.

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EXAMPLE 2



Using De Moivre’s Theorem to Check Solutions to a Polynomial Equation Use De Moivre’s theorem to show that z  2  2i is a solution to z4  3z3  38z2  128z  144  0.

Solution



We will apply the theorem to the third and fourth degree terms, and compute the square directly. Since z is in QIII, the trigonometric form is z  2 12 cis 225°. In the following illustration, note that 900° and 180° are coterminal, as are 675° and 315°. 12  2i2 4  12 122 4cis14 # 225°2  12122 4cis 900°  64 cis 180°  6411  0i2  64

12  2i2 3  12 122 3cis13 # 225°2  12 122 3cis 675°  12122 3cis 315° 12 12  ib  1612a 2 2  16  16i

12  2i2 2  4  8i  12i2 2  4  8i  4i2  4  8i  4  0  8i  8i

Substituting back into the original equation gives B. You’ve just learned to use De Moivre’s theorem to check solutions to polynomial equations

1z4  3z3  38z2  128z  144  0 11642  3116  16i2  3818i2  12812  2i2  144  0 64  48  48i  304i  256  256i  144  0 164  48  256  1442  148  304  2562i  0 0  0✓ Now try Exercises 19 through 26



Regarding Example 2, we know from a study of algebra that complex roots must occur in conjugate pairs, meaning 2  2i is also a root. This equation actually has two real and two complex roots, with z  9 and z  2 being the two real roots.

C. The nth Roots Theorem Having looked at De Moivre’s theorem, which raises a complex number to any power, we now consider the nth roots theorem, which will compute the nth roots of a complex number. If we allow that De Moivre’s theorem also holds for rational values 1 , instead of only the integers n illustrated previously, the formula for computing an n nth root would be a direct result: 1 1 1 1 zn  r n c cosa b  i sina b d n n   n  1 r c cosa b  i sina b d n n

De Moivre’s theorem

simplify

However, this formula would find only the principal nth root! In other words, periodic solutions would be ignored. As in Section 7.5, it’s worth noting the most general form of a complex number is z  r 3 cos1  360°k2  i sin1  360°k2 4 , for k  . When De Moivre’s theorem is applied to this form for integers n, we obtain z n  r n 3cos1n  360°kn2  i sin1n  360°kn2 4 , which returns a result identical to 1 r n 3 cos 1n2  i sin 1n2 4 . However, for the rational exponent , the general form takes n additional solutions into account and will return all n, nth roots. 1 1 1 1 z n  r n ecos c 1  360°k2 d  i sin c 1  360°k2 d f n n  360°k  360°k n  1r c cosa  b  i sina  bd n n n n

De Moivre’s theorem for rational exponents simplify

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The nth Roots Theorem For z  r 1cos   i sin 2, a positive integer n, and r  , z has exactly n distinct nth roots determined by 360°k  360°k  n n 1 z  1 r c cosa  b  i sina  bd n n n n where k  0, 1, 2, # # # , n  1. For ease of computation, it helps to note that once the argument for the principal  360°k 2 360 root is found using k  0,  simply adds aor b to the previous n n n n argument for k  1, 2, 3, # # # , n  1. In Example 3 you’re asked to find the three cube roots of 1, also called the cube roots of unity, and graph the results. The nth roots of unity play a significant role in the solution of many polynomial equations. For an in-depth study of this connection, visit www.mhhe.com/coburn and go to Section 7.8: Trigonometry, Complex Numbers and Cubic Equations. EXAMPLE 3



Finding nth Roots Use the nth roots theorem to solve the equation x3  1  0. Write the results in rectangular form and graph.

Solution



From x3  1  0, we have x3  1 and must find the three cube roots of unity. As before, we begin in trigonometric form: 1  0i  11cos 0°  i sin 0°2. 3 3 With n  3, r  1, and   0°, we have 1 r  1 1  1, 0° 360°k   0°  120°k. The principal and √3 yi 3 3 q, ] root (k  0) is z0  11cos 0°  i sin 0°2  1. Adding 120° to each previous argument, we find z2 the other roots are z1  11cos 120°  i sin 120°2 z2  11cos 240°  i sin 240°2 .

120

13 1 i, and In rectangular form these are   2 2 13 1 i, as shown in the figure.   2 2

120 z1 120

(1, 0) x

z3 √3

q, ]

Now try Exercises 27 through 40

EXAMPLE 4





Finding nth Roots Use the nth roots theorem to find the five fifth roots of z  16 13  16i.

Solution



In trigonometric form, 1613  16i  321cos 30°  i sin 30°2. With 5 5 n  5, r  32, and   30°, we have 1 r 1 32  2, and 30° 360°k   6°  72°k. The principal root is z0  21cos 6°  i sin 6°2. 5 5 Adding 72° to each previous argument, we find the other four roots are z1  21cos 78°  i sin 78°2 z3  21cos 222°  i sin 222°2

z2  21cos 150°  i sin 150°2 z4  21cos 294°  i sin 294°2 Now try Exercises 41 through 44



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Figure 7.80 Of the five roots in Example 4, only z2  21cos 150°  i sin 150°2 uses a standard yi z1  2 cis 78 angle. Applying De Moivre’s theorem with n  5 5 gives 12 cis 150°2  32 cis 750°  32 cis 30° z2  2 cis 150 or 1613  16i.✓ See Exercise 54. As a consequence of the arguments in a z0  2 cis 6 2 360° x , the solution being uniformly separated by n graphs of complex roots are equally spaced z3  2 cis 222 about a circle of radius r. The five fifth roots z4  2 cis 294 from Example 3 are shown in Figure 7.80 (note each argument differs by 72°). For additional insight into roots of complex numbers, we reason that the nth roots of a complex number must also be complex. To find the four fourth roots of z  8  8 13i  161cos 60°  i sin 60°2, we seek a number of the form r 1cos   i sin 2 such that 3 r 1cos   i sin 2 4 4  161cos 60°  i sin 60°2. Applying De Moivre’s theorem to the left-hand side and equating equivalent parts we obtain

r4 3cos142  i sin142 4  161cos 60°  i sin 60°2, which leads to r4  16 and

4  60° From this it is obvious that r  2, but as with similar equations solved in Chapter 6, the equation 4  60° has multiple solutions. To find them, we first add 360°k to 60°, then solve for . 4  60°  360°k 60°  360°k 4  15°  90°k



add 360°k divide by 4 result

For convenience, we start with k  0, 1, 2, and so on, which leads to For k  0:

  15°  90°102

For k  1:

 15° For k  2:

  15°  90°122  195°

  15°  90°112  105°

For k  3:

  15°  90°132  285°

At this point it should strike us that we have four roots—exactly the number required. Indeed, using k  4 gives   15°  90°142  375°, which is coterminal with the 15° obtained when k  0. Hence, the four fourth roots are

C. You’ve just learned how to use the nth roots theorem to find the nth roots of a complex number

z0  21cos 15°  i sin 15°2

z1  21cos 105°  i sin 105°2

z2  21cos 195°  i sin 195°2

z3  21cos 285°  i sin 285°2 .

The check for these solutions is asked for in Exercise 53. As a final note, it must have struck the mathematicians who pioneered these discoveries with some amazement that complex numbers and the trigonometric functions should be so closely related. The amazement must have been all the more profound upon discovering an additional connection between complex numbers and exponential functions. For more on these connections, visit www.mhhe.com/coburn and review Section 7.7: Complex Numbers in Exponential Form.

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7.6 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. For z  r 1cos   i sin 2, z5 is computed as according to theorem. 2. If z  6i, then z raised to an power will be real and z raised to an power will be since   . .

3. One application of De Moivre’s theorem is to check solutions to a polynomial equation. 

4. The nth roots of a complex number are equally spaced on a circle of radius r, since their arguments all differ by degrees or radians. 5. From Example 4, go ahead and compute the value of z5, z6, and z7. What do you notice? Discuss how this reaffirms that there are exactly n, nth roots.

6. Use a calculator to find 11  3i2 4. Then use it again to find the fourth root of the result. What do you notice? Explain the discrepancy and then resolve it using the nth roots theorem to find all four roots.

DEVELOPING YOUR SKILLS

Use De Moivre’s theorem to compute the following. Clearly state the value of r, n, and  before you begin.

7. 13  3i2 4

8. 12  2i2 6

9. 11  13i2 3

1 13 11. a  ib 2 2 13. a

12 12  ib 2 2

6

15. 14 cis 330°2 3 17. a

27. five fifth roots of unity

10. 1 13  i2 3

5

5 12 cis 135°b 2

28. six sixth roots of unity

12. a

13 1  ib 2 2

14. a

12 12  ib 2 2

6

29. five fifth roots of 243 30. three cube roots of 8 5

16. 14 cis 300°2 3 18. a

Find the nth roots indicated by writing and solving the related equation.

8 12 cis 135°b 2

Use De Moivre’s theorem to verify the solution given for each polynomial equation.

19. z4  3z3  6z2  12z  40  0; z  2i 20. z4  z3  7z2  9z  18  0; z  3i 21. z4  6z3  19z2  6z  18  0; z  3  3i 22. 2z4  3z3  4z2  2z  12  0; z  1  i 23. z5  z4  4z3  4z2  16z  16  0; z  13  i 24. z5  z4  16z3  16z2  256z  256  0; z  2 13  2i 25. z4  4z3  7z2  6z  10  0; z  1  2i 26. z4  2z3  7z2  28z  52  0; z  3  2i

31. three cube roots of 27i 32. five fifth roots of 32i Solve each equation using the nth roots theorem.

33. x5  32  0

34. x5  243  0

35. x3  27i  0

36. x3  64i  0

37. x5  12  12i  0

38. x5  1  13i  0

39. Solve the equation x3  1  0 by factoring it as the difference of cubes and applying the quadratic formula. Compare results to those obtained in Example 3. 40. Use the nth roots theorem to find the four fourth roots of unity, then find all solutions to x4  1  0 by factoring it as a difference of squares. What do you notice?

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Use the nth roots theorem to find the nth roots. Clearly state r, n, and
(from the trigonometric form of z) as you begin. Answer in exact form when possible, otherwise use a four decimal place approximation.

42. five fifth roots of 16
1613i 43. four fourth roots of
7
7i 44. three cube roots of 9  9i

41. four fourth roots of
8  813i


WORKING WITH FORMULAS

The discriminant of a cubic equation: D


4p3  27q2 108

For cubic equations of the form z3  pz  q  0, where p and q are real numbers, one solution has the form q q z  3
 1D  3

1D, where D is called the A 2 A 2


discriminant. Compute the value of D for the cubic equations given, then use the nth roots theorem to find q q the three cube roots of
 1D and

1D in 2 2 trigonometric form (also see Exercises 61 and 62).

45. z3
6z  4  0

46. z3
12z
8  0

APPLICATIONS

47. Powers and roots: Just after Example 4, the four fourth roots of z  8  813i were given as

48. Powers and roots: In Example 4 we found the five fifth roots of z  1613  16i were

z0  21cos 15°  i sin 15°2

z0  21cos 6°  i sin 6°2

z1  21cos 105°  i sin 105°2

z1  21cos 78°  i sin 78°2

z2  21cos 195°  i sin 195°2

z2  21cos 150°  i sin 150°2

z3  21cos 285°  i sin 285°2.

z3  21cos 222°  i sin 222°2

Verify these are the four fourth roots of z  8  813i using a calculator and De Moivre’s theorem.

z4  21cos 294°  i sin 294°2 Verify these are the five fifth roots of 1613  16i using a calculator and De Moivre’s theorem.

Electrical circuits: For an AC circuit with three branches wired in parallel, the total impedance is given by Z1Z2Z3 ZT  , where Z1, Z2, and Z3 represent the impedance in each branch of the circuit. If the impedance Z1Z2  Z1Z3  Z2Z3 in each branch is identical, Z1  Z2  Z3  Z, and the numerator becomes Z 3 and the denominator becomes 3Z 2, (a) use De Moivre’s theorem to calculate the numerator and denominator for each value of Z given, (b) find the total impedance by Z3 Z computing the quotient 2 , and (c) verify your result is identical to . 3 3Z

49. Z  3  4j in all three branches


50. Z  5 13  5j in all three branches

EXTENDING THE CONCEPT

In Chapter 6, you were asked to verify that sin13
2  3 sin

4 sin3
and cos14
2  8 cos2

8 cos2
 1 were identities (Section 6.4, Exercises 21 and 22). For 17 z  3  17i, verify
z
 4 and
 tan
1a b, then 3 draw a right triangle with 17 opposite
and 3 adjacent to
. Discuss how this right triangle and the identities given can be used in conjunction with De Moivre’s theorem to find the exact value of the powers given (also see Exercises 53 and 54).

51. 13  17i2 3

52. 13  17i2 4

For cases where
is not a standard angle, working toward an exact answer using De Moivre’s theorem requires the use of multiple angle identities and drawing the right triangle b related to
 tan
1a b. For Exercises 53 and 54, use De a Moivre’s theorem to compute the complex powers by (a) constructing the related right triangle for
, (b) evaluating sin14
2 using two applications of double-angle identities, and (c) evaluating cos14
2 using a Pythagorean identity and the computed value of sin14
2.

53. z  11  2i2 4

54. 12  15i2 4

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Summary and Concept Review

The solutions to the cubic equations in Exercises 45 and 46 (repeated in Exercises 55 and 56) can be found by adding the q q cube roots of   1D and   1D that have arguments summing to 360°. 2 2

55. Find the roots of z3  6z  4  0 

56. Find the roots of z3  12z  8  0

MAINTAINING YOUR SKILLS

57. (6.2) Prove the following is a identity: tan2x 1  cos x  cos x sec x  1

59. (2.3) Find the equation of the line whose graph is given.

y 5

5 x

1

58. (3.3) Given f 1x2  2x2  3x, determine: 1 f 112, f a b, f(a) and f 1a  h2. 3

60. (5.2) Solve the triangle given. Round lengths to hundredths of a meter.

B 52

213 m

C

A

S U M M A RY A N D C O N C E P T R E V I E W SECTION 7.1

Oblique Triangles and the Law of Sines

KEY CONCEPTS sin A sin C sin B  . • In any triangle, the ratio of the sine of an angle to its opposite side is constant: a  c b • The law of sines requires a known angle, a side opposite this angle and an additional side or angle, hence cannot be applied for SSS and SAS triangles. • For AAS and ASA triangles, the law of sines yields a unique solution. • When given two sides of a triangle and an angle opposite one of these sides (SSA), the number of solutions is in doubt, giving rise to the designation, “the ambiguous case.” • SSA triangles may have no solution, one solution, or two solutions, depending on the length of the side opposite the given angle. • When solving triangles, always remember: • The sum of all angles must be 180°: A  B  C  180°. • The sum of any two sides must exceed the length of the remaining side. • Longer sides are opposite larger angles. • k  sin1 has no solution for k 7 1. • k  sin1 has two solutions in 30, 360°2 for 0 6 k 6 1. EXERCISES Solve the following triangles. 1.

A

2. B

293 cm B

21

123 C

A 28 142 52 yd C

3. A tree is growing vertically on a hillside. Find the height of the tree if it makes an angle of 110° with the hillside and the angle of elevation from the base of the hill to the top of the tree is 25° at a distance of 70 ft.

25

110

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4. Find two values of  that will make the equation true:

sin 50 sin   . 14 31

5. Solve using the law of sines. If two solutions exist, find both (figure not drawn to scale). C 105 cm

67 cm

35 A

6. Jasmine is flying her tethered, gas-powered airplane at a local park, where a group of bystanders is watching from a distance of 60 ft, as shown. If the tether has a radius of 35 ft and one of the bystanders walks away at an angle of 40°, will he get hit by the plane? What is the smallest angle of exit he could take (to the nearest whole) without being struck by Jasmine’s plane?

SECTION 7.2

Radius 35 ft Jasmine

40 60 ft Bystanders

The Law of Cosines; the Area of a Triangle

KEY CONCEPTS • The law of cosines is used to solve SSS and SAS triangles. • The law of cosines states that in any triangle, the square of any side is equal to the sums of the squares of the other two sides, minus twice their product times the cosine of the included angle: a2  b2  c2  2bc cos A • When using the law of cosines to solve a SSS triangle, always begin with the largest angle or the angle opposite the largest side. • The area of a nonright triangle can be found using the following formulas. The choice of formula depends on the information given. • two sides a and b • two angles A and B • three sides a, b, and c abc with included angle c with included side c with S  1 c2 sin A sin B 2 A  ab sin C A 2 2 sin C A  2s1s  a2 1s  b21s  c2

EXERCISES 98 167 m 325 m 7. Solve for B: 92  122  152  211221152 cos B 8. Use the law of cosines to find the missing side. 1250 yd 9. While preparing for the day’s orienteering meet, Rick finds that the distances between the first three markers he wants to pick up are 1250 yd, 1820 yd, and 720 yd. Find the 720 yd measure of each angle in the triangle formed so that Rick is sure to find all three markers. 1820 yd 10. The Great Pyramid of Giza, also known Khufu’s pyramid, is the sole remaining member of the Seven Wonders of the Ancient World. It was built as a tomb for the Egyptian pharaoh Khufu from the fourth dynasty. This square pyramid is made up of four isosceles triangles, each with a base of 230.0 m and a slant height of about 218.7 m. Approximate the total surface area of Khufu’s pyramid (excluding the base).

SECTION 7.3

Vectors and Vector Diagrams

KEY CONCEPTS • Quantities/concepts that can be described using a single number are called scalar quantities. Examples are time, perimeter, area, volume, energy, temperature, weight, and so on. • Quantities/concepts that require more than a single number to describe their attributes are called vector quantities. Examples are force, velocity, displacement, pressure, and so on.

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• Vectors can be represented using directed line segments to indicate magnitude and direction. The origin of the segment is called the initial point, with the arrowhead pointing to the terminal point. When used solely for comparative analysis, they are called geometric vectors. • Two vectors are equal if they have the same magnitude and direction. • Vectors can be represented graphically in the xy-plane by naming the initial and terminal points of the vector or by giving the magnitude and angle of the related position vector [initial point at (0, 0)]. • For a vector with initial point (x1, y1) and terminal point (x2, y2), the related position vector can be written in the component form Ha, bI, where a  x2  x1 and b  y2  y1. • For a vector written in the component form Ha, bI, a is called the horizontal component and b is called the vertical component of the vector. • For vector v  H a, bI, the magnitude of v is v  2a2  b2. • Vector components can also be written in trigonometric form. See page 661. • For u  Ha, bI, v  Hc, d I, and any scalar k, we have the following operations defined: u  v  Ha  c, b  d I

u  v  Ha  c, b  d I

ku  Hka, kbI for k  R

If k 7 0, the new vector has the same direction as u; k 6 0, the opposite direction. • Vectors can be written in algebraic form using i, j notation, where i is an x-axis unit vector and j is a y-axis unit vector. The vector Ha, bI is written as a linear combination of i and j: Ha, bI  ai  bj. v • For any nonzero vector v, vector u  is a unit vector in the same direction as v. v • In aviation and shipping, the heading of a ship or plane is understood to be the amount of rotation from due north in the clockwise direction.

EXERCISES 11. Graph the vector v  H9, 5I, then compute its magnitude and direction angle. 12. Write the vector u  H8, 3I in i, j form and compute its magnitude and direction angle. 13. Approximate the horizontal and vertical components of the vector u, where u  18 and   52°. 14. Compute 2u  v, then find the magnitude and direction of the resultant: u  H3, 5I and v  H2, 8I. 15. Find a unit vector that points in the same direction as u  7i  12j. 16. Without computing, if u  H9, 2I and v  H2, 8I, will the resultant sum lie in Quadrant I or II? Why? 17. It’s once again time for the Great River Race, a 12 -mi swim across the Panache River. If Karl fails to take the river’s 1-mph current into account and he swims the race at 3 mph, how far from the finish marker does he end up when he makes it to the other side? 18 18. Two Coast Guard vessels are towing a large yacht into port. The first is pulling with a ␪ force of 928 N and the second with a force of 850 N. Determine the angle  for the second Coast Guard vessel that will keep the ship moving safely in a straight line.

SECTION 7.4

Vector Applications and the Dot Product

KEY CONCEPTS • Vector forces are in equilibrium when the sum of their components is the zero vector. • When the components of vector u are nonquadrantal, with one of its components lying along vector v, we call this component the “component of u along v” or compvu. • For vectors u and v, compvu  ucos , where  is the angle between u and v. • Work done is computed as the product of the constant force F applied, times the distance D the force is applied: W  F # D. • If force is not applied parallel to the direction of movement, only the component of the force in the direction of movement is used in the computation of work. If u is a force vector not parallel to the direction of vector v, the equation becomes W  compvu # v.

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• For vectors u  Ha, bI and v  Hc, d I, the dot product u # v is defined as the scalar ac  bd. • The dot product u # v is equivalent to compuv # v and to uvcos . u#v u # v • The angle between two vectors can be computed using cos    . u v uv u#v b v. v2 • Given vectors u and v and projvu, u can be resolved into the orthogonal components u1 and u2 where u  u1  u2, u1  projvu, and u2  u  u1. • The horizontal distance x a projectile travels in t seconds is x  1 vcos 2t. • The vertical height y of a projectile after t seconds is y  1 vsin 2t  16t2, where v is the magnitude of the initial velocity, and  is the angle of projection. • Given vectors u and v, the projection of u along v is the vector projvu defined by projvu  a

EXERCISES 19. For the force vectors F1 and F2 given, find the resultant and an additional force vector so that equilibrium takes place: F1  H20, 70I; F2  H45, 53I. 20. Find compvu for u  12i  16j and v  19i  13j. 21. Find the component d that ensures vectors u and v are orthogonal: u  H2, 9I and v  H18, d I. 22. Compute p # q and find the angle between them: p  H5, 2I; q  H4, 7I. 23. Given force vector F  H50, 15I and v  H85, 6I, find the work required to move an object along the entire length of v. Assume force is in pounds and distance in feet. 24. A 650-lb crate is sitting on a ramp that is inclined at 40°. Find the force needed to hold the crate stationary. 25. An arctic explorer is hauling supplies from the supply hut to her tent, a distance of lb 120 ft, in a sled she is dragging behind her. If the straps used make an angle of 25° 650 with the snow-covered ground and she pulls with a constant force of 75 lb, find the amount of work done. 650 lb 40 26. A projectile is launched from a sling-shot with an initial velocity of v0  280 ft/sec at an angle of   50°. Find (a) the position of the object after 1.5 sec and (b) the G time required to reach a height of 150 ft.

SECTION 7.5

Complex Numbers in Trigonometric Form

KEY CONCEPTS • A complex number a  bi  1a, b2 can be written in trigonometric form by Imaginary yi (a, b) axis noting (from its graph) that a  r cos  and b  r sin : a  bi  (r cos ␪, r sin ␪) b r1cos   i sin 2. • The angle  is called the argument of z and r is called the modulus of z. r • The argument of a complex number z is not unique, since any rotation of x ␪   2k (k an integer) will yield a coterminal angle. a Real • To convert from trigonometric to rectangular form, evaluate cos  and sin  axis and multiply by the modulus. • To multiply complex numbers in trig form, multiply the moduli and add the arguments. To divide complex numbers in trig form, divide the moduli and subtract the arguments. • Complex numbers have numerous real-world applications, particularly in a study of AC electrical circuits. • The impedance of an AC circuit is given as Z  R  j1XL  XC 2, where R is a pure resistance, XC is the capacitive reactance, XL is the inductive reactance, and j  11. XL  XC • Z is a complex number with magnitude Z  2R2  1XL  XC 2 2 and phase angle   tan1a b R ( represents the angle between the voltage and current). V • In an AC circuit, voltage V  IZ ; current I  . Z

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Mixed Review

EXERCISES 27. Write in trigonometric

28. Write in rectangular form:  form: z  1  13i z  3 12 c cisa b d 4 z1   30. For z1  8 cisa b and z2  2 cisa b, compute z1z2 and . z2 4 6

29. Graph in the complex plane: z  51cos 30°  i sin 30°2

31. Find the current I in a circuit where V  4 13  4j and Z  1  13j . 32. In the VRLC series circuit shown, R  10 , XL  8 , and XC  5 . Find the magnitude of Z and the phase angle between current and voltage. Express the result in trigonometric form.

SECTION 7.6

XL

R A

10 Ω

B

8Ω

XC C

5Ω

D

De Moivre’s Theorem and the Theorem on nth Roots

KEY CONCEPTS • For complex number z  r1cos   i sin 2, zn  rn 3cos1n2  i sin1n2 4 (De Moivre’s theorem). • De Moivre’s theorem can be used to check complex solutions of polynomial equations. 2k  2k  n n b  i sina  b d , for • For complex number z  r1cos   i sin 2 , 2z  2r c cosa  n n n n k  1, 2, 3,p, n  1 (nth roots theorem). • The nth roots of a complex number are equally spaced around a circle of radius r in the complex plane.

EXERCISES 33. Use De Moivre’s theorem to compute the value of 11  i 132 5. 34. Use De Moivre’s theorem to verify that z  1  i is a solution of z4  z3  2z2  2z  4  0. 35. Use the nth roots theorem to find the three cube roots of 125i. 36. Solve the equation using the nth roots theorem: x3  216  0. 37. Given that z  2  2i is a fourth root of 64, state the other three roots. 38. Solve using the quadratic formula and the nth roots theorem: z4  6z2  25  0. 39. Use De Moivre’s theorem to verify the three roots of 125i found in Exercise 35.

MIXED REVIEW Solve each triangle using either the law of sines or law of cosines (whichever is appropriate) then find the area of each. 1. B 27

19 in. 112

A

C

2.

C 31 cm 37 A

B 52 cm

3. Find the horizontal and vertical components of the vector u, where u  21 and   40°. 4. Compute 2u  v, then find the magnitude and direction of the resultant: u  H6, 3I, v  H2, 8I. 5. Find the height of a flagpole that sits atop a hill, if it makes an angle 122 of 122° with the 35 hillside, and the angle of elevation between the side of the hill to the top of the flagpole is 35° at a distance of 120 ft.

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6. A 900-lb crate is sitting on a ramp that is inclined at 28°. Find the force needed to hold the object stationary. Exercise 6 900

13. a. Graph the complex number using the rectangular form, then convert to trigonometric form: z  4  4i. b. Graph the complex number using the trigonometric form, then convert to rectangular form: z  61cos 120°  i sin 120°2.

lb

900 lb 28 G

7. A jet plane is flying at 750 mph on a heading of 30°. There is a strong, 50-mph wind blowing from due south (heading 0°). What is the true course and speed of the plane (relative to the ground)? 8. Graph the vector v  H8, 5I, then compute its magnitude and direction.

12. Given the vectors p  H5, 2I and q  H4, 7I, use the dot product p # q to find the angle between them.

C 36 m

9. Solve using the law of sines. If two solutions exist, find both.

24 m

31

A

Exercise 10 10. A local Outdoors Club C sponsors a treasure hunt 0.9 mi activity for its members, 0.7 mi and has placed surprise B packages at the corners of the triangular park 1.2 mi A shown. Find the measure of each angle to help club members find their way to the treasure.

11. As part of a lab demonstrating centrifugal Teacher Students and centripetal forces, a 20 ft 35 physics teacher is whirling 10 ft a tethered weight above Radius her head while a group of students looks on from a distance of 20 ft as shown. If the tether has a radius of 10 ft and a student departs at the 35° angle shown, will the student be struck by the weight? What is the smallest angle of exit the student could take (to the nearest whole) without being struck by the whirling weight?

14 a. Verify that z  4  5i and its conjugate are solutions to z2  8z  41  0. b. Solve using the quadratic formula: z2  6iz  7  0 15. Two tractors are dragging a large, fallen tree into the brush pile 10 that’s being ␪ prepared for a large Fourth of July bonfire. The first is pulling with a force of 418 N and the second with a force of 320 N. Determine the angle  for the second tractor that will keep the tree headed straight for the brush pile. 16. Given z1  81cos 45°  i sin 45°2 and z2  41cos 15°  i sin 15°2 compute: a. the product z1z2

b. the quotient

z1 z2

17. Given the vectors u  12i  16j and v  19i  13j, find compvu and projvu. 18. Find the result using De Moivre’s theorem: 12 13  2i2 6. 19. Use the nth roots theorem to find the four fourth roots of 2  2i 13. 20. The impedance of an AC circuit is Z  R  j1XL  XC 2. The voltage across the circuit is VRLC  IZ. Given R  12 , XL  15.2 , and XC  9.4 , write Z in trigonometric form and find the voltage in the circuit if the current is I  6.5 A.

PRACTICE TEST 1. Within the Kilimanjaro Game Reserve, a fire is spotted by park rangers stationed in two towers 39 68 known to be 10 mi apart. 10 mi Using the line between them as a baseline, tower A A B reports the fire is at an angle of 39°, while tower B reports an angle of 68°. How far is the fire from the closer tower?

Exercise 2 2. At the circus, Mac and Joe are watching a high-wire act from first-row seats on opposite Acrobat sides of the center ring. Find the height of the performing acrobat at the instant Mac measures an angle of elevation 72 of 68° while Joe measures an Joe 68 angle of 72°. Assume Mac and Mac Joe are sitting 100 ft apart.

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3. Three rods are C B attached via two A joints and shaped 6 in. 15 in. into a triangle. How ?? in. many triangles can be formed if the angle at the joint B must measure 20°? If two triangles can be formed, solve both. 4. Jackie and Sam are rounding up cattle in the brush country, and are Range communicating via walkie3 mi 32 talkie. Jackie is at the water 6 mi Dead hole and Sam is at Dead Water hole Oak Oak, which are 6 mi apart. Sam finds some strays and heads them home at the 32° indicated. (a) If the maximum range of Jackie’s unit is 3 mi, will she be able to communicate with Sam as he heads home? (b) If the maximum range were 4 mi, how far from Dead Oak is Sam when he is first contacted by Jackie? 5. As part of an All-Star Exercise 5 competition, a group of soccer players (forwards) stand where shown in the diagram and attempt to hit a moving target with a twohanded overhead pass. If a Overhead 35 yd player has a maximum pass effective range of approximately (a) 25 yd, can 53 the target be hit? (b) about 28 yd, how many “effective” throws can be made? (c) 35 yd and the target is moving at 5 yd/sec, how many seconds is the target within range? 6. The summit of Triangle 3.5 mi Peak can only be reached from one side, using a trail 24 straight up the side that is 5 mi approximately 3.5 mi long. If the mountain is 5 mi wide at its base and the trail makes a 24° angle with the horizontal, (a) what is the approximate length of the opposing side? (b) How tall is the peak (in feet)?

Practice Test

711

helicopter relative to the ground? Draw a diagram as part of your solution. 9. Two mules walking along a river bank are pulling a heavy barge up river. The first is pulling with a force of 250 N and the second with a force of 210 N. Determine 30 the angle  for the second ␪ mule that will ensure the barge stays midriver and does not collide with the shore. Exercise 10 10. Along a production line, 22 various tools are attached to the ceiling with a multijointed arm 42 so that workers can draw one down, position it for use, then 58 Joint move it up out of the way for the next tool (see the diagram). If the first segment is 100 cm, the second is 75 cm, and the third is 50 cm, determine the approximate coordinates of the last joint. 11. Three ranch hands have roped a run-away steer and are attempting to hold him steady. F1 F2 The first and second ranch hands 150 are pulling with the magnitude 110 and at the angles indicated in the 67 42 diagram. If the steer is held fast ?F by the efforts of all three, find 3 the magnitude of the tension and angle of the rope from the third cowhand. 12. For u  H9, 5I and v  H2, 6I, (a) compute the angle between u and v; (b) find the projection of u along v (find projvu; and (c) resolve u into vectors u1 and u2, where u1 7 v and u2v.

13. A lacrosse player flips a long pass to a teammate way down field who is near the opponent’s goal. If the initial velocity of the pass is 110 ft/sec and the ball is released at an angle of 50° with level ground, how high is the ball after 2 sec? How long until the ball again reaches this same height? 14. Compute the quotient

z1 , given z2

7. The Bermuda Triangle is 1025 mi B generally thought to be the M triangle formed by Miami, 977 mi Florida, San Juan, Puerto 1020 mi Rico, and Bermuda itself. P If the distances between these locations are the 1025 mi, 1020 mi, and 977 mi indicated, find the measure of each angle and the area of the Bermuda Triangle.

16. Use De Moivre’s theorem to compute the value of 1 13  i2 4.

8. A helicopter is flying at 90 mph on a heading of 40°. A 20-mph wind is blowing from the NE on a heading of 190°. What is the true course and speed of the

18. Use the nth roots theorem to solve x3  125i  0.

  z1  6 15 cisa b and z2  3 15 cisa b. 8 12 15. Compute the product z  z1z2 in trigonometric form, then verify z1z2  z and 1  2  : z1  6  6i; z2  4  4 13i

17. Use De Moivre’s theorem to verify 2  213i is a solution to z5  3z3  64z2  192  0.

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19. Solve using u-substitution, the quadratic formula, and the nth roots theorem: z4  6z2  58  0. 20. Due to its huge biodiversity, preserving Southeast Asia’s Coral Triangle has become a top priority for conservationists. Stretching from the northern Philippines (P), south along the coast of Borneo (B) to the Lesser Sunda Islands (L), then eastward to the Solomon Islands (S), this area is home to over 75% of all coral species known. Use Heron’s formula to help find the total area of this natural wonderland, given the dimensions shown.

1275 mi B 2010 mi

L

Halmahera 2390 mi 23⬚

1690 mi

1590 mi

S

C A L C U L AT O R E X P L O R AT I O N A N D D I S C O V E RY Investigating Projectile Motion Figure 7.81 There are two important aspects of projectile motion that were not discussed earlier, the range of the projectile and the optimum angle  that will maximize this range. Both can be explored using the equations for the horizontal and Figure 7.82 vertical components of the projectile’s position: horizontal S 1 vcos 2t and vertical S 1 vsin 2 t  16t2. In Example 10 of Section 7.4, an arrow was shot from a bow with initial velocity v  150 ft/sec at an angle of   50°. Enter Figure 7.83 the equations above on the Y= screen as Y1 and Y2, using these values (Figure 7.81). Then set up the TABLE using TblStart  0, ¢Tbl  0.5 and the AUTO mode. The resulting table is shown in Figure 7.82, where Y1 represents the horizontal distance the arrow has traveled, and Y2 represents the height of the arrow. To find the range of the arrow, scroll downward until the height (Y2) shows a value that is less than or equal to zero (the arrow has hit the ground). As Figure 7.83 shows, this happens somewhere between t  7 and t  7.5 sec. We could now change the TBLSET settings to TblStart  0 and ¢Tbl  0.1 to get a better approximation of the time the arrow is in flight (it’s just less than 7.2 sec) and the horizontal range of the arrow (about 692.4 ft), but our main interest is how to compute these values exactly. We begin with the equation for the arrow’s vertical position y  1 vsin 2t  16t2. Since the object returns to Earth when y  0, we substitute 0 for y and factor out

t: 0  t1 vsin   16t2. Solving for t gives t  0 or vsin  t . Since the component of velocity in the hori16 zontal direction is vcos , the basic distance relationship D  r # t gives the horizontal range of R  vcos  # v2sin  cos  vsin  or . Checking the values given for 16 16 the arrow (v  150 ft/sec and   50°) verifies the range is R  692.4. But what about the maximum possible range for the arrow? Using v  150 for R results in an equation 1502sin  cos  , which we can 16 enter as Y3 and investigate for various . After carefully entering R12 as Y3 and resetting TBLSET to TblStart  30 and ¢Tbl  5, the TABLE in Figure 7.84 shows a maximum range of about 703 ft at 45°. Resetting TBLSET to TblStart  40 and ¢Tbl  1 verifies this fact. For each of the following exercises, find (a) the height of the projectile after 1.75 sec, (b) the maximum height of the projectile, (c) the range of the projectile, and (d) the number of seconds the proFigure 7.84 jectile is airborne. in theta only: R12 

Exercise 1: A javelin is thrown with an initial velocity of 85 ft/sec at an angle of 42°. Exercise 2: A cannon ball is shot with an initial velocity of 1120 ft/sec at an angle of 30°. Exercise 3: A baseball is hit with an initial velocity of 120 ft/sec at an angle of 50°. Will it clear the center field fence, 10 ft high and 375 ft away? Exercise 4: A field goal (American football) is kicked with an initial velocity of 65 ft/sec at an angle of 35°. Will it clear the crossbar, 10 ft high and 40 yd away?

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STRENGTHENING CORE SKILLS Vectors and Static Equilibrium In Sections 7.3 and 7.4, the concepts of vector forces, resultant forces, and equilibrium were studied extensively. A nice extension of these concepts involves what is called static equilibrium. Assuming that only coplanar forces are acting on an object, the object is said to be in static equilibrium if the sum of all vector forces acting on it is 0. This implies that the object is stationary, since the forces all counterbalance each other. The methods involved are simple and direct, with a wonderful connection to the systems of equations you’ve likely seen previously. Consider the following example. Illustration 1  As part of their training, prospective FBI agents must move hand-over-hand across a rope strung between two towers. An agent-in-training weighing 180 lb is two-thirds of the way u v across, causing the rope to deflect from the horizontal at the angles 9 14 shown. What is the w tension in each part of 180 lb the rope at this point? Solution  We have three concurrent forces acting on the point where the agent grasps the rope. Begin by drawing a vector diagram u and computing the components of 9 each force, using the i, j notation. 180 Note that w  180j. u  ucos19°2i  usin19°2j  0.9877ui  0.1564uj v  vcos114°2i  vsin114°2j  0.9703vi  0.2419vj

y

v 14

x

w

For equilibrium, all vector forces must sum to the zero vector: u  v  w  0, which results in the following

equation: 0.9877|u |i  0.1564|u |j  0.9703| v|i  0.2419| v | j  180j  0i  0j. Factoring out i and j from the left-hand side yields 10.9877|u |  0.9703 |v | 2i  10.1564 |u|  0.2419 | v |  1802j  0i  0j. Since any two vectors are equal only when corresponding components are equal, we obtain a system in the two variables u and v: e

0.9877u  0.9703v 2  0 . 0.1564u  0.2419v  180  0

Solving the system using matrix equations and a calculator (or any desired method), gives u  447 lb and v  455 lb. At first it may seem surprising that the vector forces (tension) in each part of the rope are so much greater than the 180-lb the agent weighs. But with a 180-lb object hanging from the middle of the rope, the tension required to keep the rope taut (with small angles of deflection) must be very great. This should become more obvious to you after you work Exercise 2. Exercise 1: A 500-lb crate is suspended by two 25 20 ropes attached 500 lb to the ceiling rafters. Find the tension in each rope. Exercise 2: Two people 45 45 team up to carry a 150-lb 150 lb weight by passing a rope through an eyelet in the object. Find the tension in each rope. Exercise 3: Referring to Illustration 1, if the rope has a tension limit of 600-lb (before it snaps), can a 200-lb agent make it across?

C U M U L AT I V E R E V I E W C H A P T E R S 1 – 7 1. Solve using a standard triangle. a  20, b  ____, c  ____   30°,   ____,   ____ A 2. Solve using trigonometric ratios. a  ____, b  ____, c  82  ____,   63°,   ____

B ␤ 20 m 30

C B

82 ft A

␣

3. A torus is a donut-shaped solid figure. Its surface area is given by the formula

63 C

A  2 1R2  r2 2 , where R is the outer radius of the donut, and r is the inner radius. Solve the formula for R in terms of r and A. 4. For a complex number a  bi, (a) verify the sum of a complex number and its conjugate is a real number, and (b) verify the product of a complex number and its conjugate is a real number. 5. State the value of all six trig functions given 3 tan    with cos  7 0. 4

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6. Sketch the graph of y  4 cosa x  b using 6 3 transformations of y  cos x. 7. Solve using the quadratic formula: 5x2  8x  2  0. 8. Solve by completing the square: 3x2  72x  427  0. 9. Given cos 53°
0.6 and cos 72°
0.3, approximate the value of cos 19° and cos 125° without using a calculator. 10. Find all real values of x that satisfy the equation 13  2 sin12x2  2 13. State the answer in degrees. 11. a. Given that 2008 ft 1 acre  43,560 ft2, find the cost of a lot with the 25.9
dimensions shown (to the 1475 ft nearest dollar) if land in this area is going for $4500 per acre. North b. After an accident at (5, 12) 12 sea, a search and 10 rescue team decides 8 to focus their efforts 6 on the area shown due (12, 5) 4 to prevailing winds 2 and currents. Find the 0 2 4 6 8 10 12 14 East (0, 0) distances between each vertex (use Pythagorean triples and a special triangle) and the number of square miles in the search area.

15.

C 31 cm

B

37


52 cm

A

16. A commercial fishery stocks a lake with 250 fish. Based on previous experience, the population of fish is expected to grow according to the model 12,000 P1t2  , where t is the time in months. 1  25e0.2t From on this model, (a) how many months are required for the population to grow to 7500 fish? (b) If the fishery expects to harvest three-fourths of the fish population in 2 yr, approximately how many fish will be taken? 17. A 900-lb crate is sitting on a ramp which is inclined at 28°. Find the force needed to hold the object stationary.

lb

900

900 lb 28


18. A jet plane is flying at G 750 mph on a heading of 30°. There is a strong, 50-mph wind blowing from due south (heading 0°). What is the true course and speed of the plane (relative to the ground)? 19. Use the Guidelines for Graphing to sketch the graph of function f given, then use it to solve f1x2 6 0. f 1x2  x3  4x2  x  6

20. Use the Guidelines for Graphing to sketch the graph of function g given, then use it to name the intervals where g1x2T and g1x2c. x2  4 x2  1

12. State the domain of each function: a. f1x2  12x  3 b. g1x2  logb 1x  32 x3 c. h1x2  2 x 5 d. v1x2  2x2  x  6

21. Find 11  13i2 8 using De Moivre’s theorem.

13. Write the following formulas from memory: a. slope formula b. midpoint formula c. quadratic formula d. distance formula e. interest formula (compounded continuously)

24. Mount Tortolas lies on the Argentine-Chilean border. When viewed from a distance of 5 mi, the angle of elevation to the top of the peak is 38°. How tall is Mount Tortolas? State the answer in feet.

Solve each triangle using the law of sines or the law of cosines, whichever is appropriate.

g1x2 

22. Solve ln1x  22  ln1x  32  ln14x2 .

23. If I saved $200 each month in an annuity program that paid 8% annual interest compounded monthly, how long would it take to save $10,000?

25. The graph given is of the form y  A sin1Bx  C2. Find the values of A, B, and C. y 2

14. A

1

27


19 in.




2

C

112


B

1 2


2




3
2

2


x

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CONNECTIONS TO CALCULUS Chapter 7 unites our study of trigonometry with many of the skills you likely developed in previous coursework. Certainly algebraic skills continue to play a significant role, but here we’ve chosen to review and highlight certain problem-solving skills as well. From the vector diagrams seen in Chapter 7 to the triangle diagrams used in numerous applications, a careful sketch can highlight relationships and suggest a solution process, and this will certainly serve us well throughout the calculus sequence.

Trigonometry and Problem Solving Perhaps to a higher degree in a calculus course, the quality of the diagrams we draw for problem solving have a greater impact on our ability to understand and solve applications. Draw the diagrams very carefully and to scale, if possible. Label the various parts of the diagram, make a list of all given information, assign variable names to unknown quantities, and build any needed relationships—in short, use the total accumulation of your problem-solving skills. EXAMPLE 1



Determining Sight Angles A 20-ft-tall road sign is to be placed along a busy highway by attaching it to a support pole, with the bottom edge of the sign 13 ft above the ground. a. Find a simplified expression for the “sight angle” (the angle at a person’s eyes subtended by the sign) in terms of x, when the driver is x ft away from the base of the pole (assume that eye level for an average driver is 3 ft above the ground). b. Find the sight angle at the moment a driver is 50 ft away.

Solution



a. Draw a car that is x ft away from a road sign with the dimensions described. Note that the triangle from eye level 20 ft to the top of the sign, and the triangle from eye level to the bottom of the sign are right triangles, while the sight angle is not. Naming the acute angle of the larger right triangle (at the eyes of the 10 ft   driver) , and the acute angle of the smaller right triangle  (see the  figure), we see that the sight angle x 3 ft is then   . Since we need to Car 3 ft involve the distance x in our equation model, it 30 appears the tangent function should be used, giving tan   and x 10 tan   . Knowing the identity for tan1  2 is expressed in terms of x tan   tan  tan  and tan  individually, we have tan 1  2  . Our 1  tan  tan  algebraic expression is then 10 30  x x tan1  2  30 10 1  a ba b x x 20 x  300 1 2 x

7–83

make substitutions

simplify

715

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20 x  2 x  300 x2 20x  2 x  300

combine terms

result

The expression for the sight angle is     tan1a

20x b x2  300

b. When x  50 ft,     tan1a

1000 b, or about 19.65°. For comparison, the 2800 sight angle at 25 ft is about 28.39°, and at 100 ft the angle is about 10.99°. Now try Exercises 1 through 4



Vectors in Three Dimensions z Many elements of a study of vectors in two dimensions transfer directly to the study of vectors in three dimensions. To help visualize H3, 4, 2I a three-dimensional vector, consider a large 3 ft  4 ft  2 ft cardboard box placed snugly in the corner of a room. The floor of y the room forms the positive xy-plane (the axes are where the walls meet the floor, the x-axis to the left), with the third dimension (the x z-axis) formed where the two walls meet (see the figure). The origin is then 10, 0, 02 , where the three axes meet, and the position vector H3, 4, 2I forms an interior diagonal of the box. For three-dimensional vectors u  Ha, b, cI and v  Hd, e, f I, the formulas for the magnitude of v and the dot product u # v are an extension of the two-dimensional cases, and are offered here without proof:

u  2a2  b2  c2, and u # v  ad  be  cf. EXAMPLE 2



Solution



Finding the Angle between Two Vectors

For vector u  H3, 4, 2I forming the interior diagonal of the box and v  H3, 4, 0I forming the diagonal of the base, a. Find u, then verify the result using the Pythagorean theorem and v. u#v b. Use the formula cos   to find the angle  between u and v, then verify uv the result using right-triangle trig. a. First find u  232  42  22  129. To verify, we use v  232  42  02  5 and the height of the box in the Pythagorean theorem (since the triangle formed by (0, 0, 0), (3, 4, 0), and (3, 4, 2) is a right triangle): v2  22  25  4  29  u2 ✓. u#v 9  16  0 5 5 , so   cos1a , cos    b  21.8°. uv 5 129 129 129 2 To verify, we use v  5 and the height of the box, giving   tan1a b  21.8° ✓. 5

b. For cos  

Now try Exercises 5 through 10



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717

Connections to Calculus Exercises For the following applications, draw a diagram and develop an appropriate equation model. Note that the emphasis on these exercises is shared equally between the quality of diagram and developing the correct equation model.

1. A jet plane is flying at a constant altitude of 9000 ft, straight toward a searchlight on the ground that is tracking it. (a) Find an expression for the angle the searchlight makes with the ground in terms of the horizontal distance d from the light to a point on the ground directly below the plane. (b) Find the angle  at the moment d  12,850 ft. 2. A light is hung from the rafters at a height h m above the floor, providing a circle of illumination. The illuminance E (in lumens/square meter or lm/m2) at any point P on the floor varies directly as the cosine of the angle  at which the rays leave the light source toward P, and inversely as the square of the height h of the light source. Using k as the constant of variation, (a) write an expression for the illuminance in terms of k, h, and . (b) If the distance from the light source to point P is 13 m, write an expression for the illuminance in terms of k and h alone. (c) For k  15,600, what is the height of the light source if E  100 lm/m2? (d) For k and h as in part (c), at what angle  is the illuminance 83 lm/m2.

3. In a large factory, two automated carts run on concentric circular tracks. The inner track has a radius of 24 m, while the outer track has a radius of 32 m. (a) Write an expression for the straightline distance x between the two carts in terms of the angle  formed by the line segments from the center to each cart. (b) Determine the straightline distance between the carts at the moment   150°. (c) At what angle  are the carts 45 m apart? 4. A rectangle is inscribed in a semicircle of radius r. A straight line from the center of the circle to the point where a vertex of the rectangle meets the circle, forms an angle . (a) Find an expression for the area of the rectangle in terms of r and a sine function. (Hint: Write the length and width of the rectangle in terms of sine and cosine.) (b) Find the area of the circle if r  5 cm and   26°. (c) At what angle  does the rectangle have an area of 22 cm2?

Solve each application of three-dimensional vectors using a careful sketch drawn from the information given. The emphasis on these exercises continues to be shared equally between the quality of diagram and developing the correct equation model.

5. A rectangular playground is 64 ft long and 48 ft wide. There is a light pole for evening play in one corner of the playground, with the light placed 18 ft high. (a) Find v the length of the vector formed from the top of the light pole to the opposite corner of the playground. (b) If u is the vector formed by the opposite end of the playground and the base of the pole, find the angle between these two vectors u#v using cos   and verify the result using right uv triangle trig. 6. A gallery is about to open an exhibit in a room that is 22 ft long, 24 ft wide, and 16 ft high. The edge of the door to the exhibit is on the 22-ft side, 4 ft from one wall. For special effects lighting as a guest enters the door, let u be the vector representing the distance from the edge of the door (at the floor) to the farthest bottom corner of the room, and let v be the vector from the same point to the farthest upper corner. (a) Find v, then (b) find the angle between u u#v and v using cos   and verify the result uv using right triangle trig.

7. At a small jungle airport, bush pilots are practicing their approach to drop zones for food aid. There is an observation post 525 ft from the center of one end of an 1800-ft straight dirt runway. At the moment the pilot of the first drop plane is over the opposite end of the runway, her altitude is 1000 ft. Let d be the vector representing the distance from the observation post to the opposite end of the runway, and let D represent the vector for the straight-line distance from the post to the plane at this moment. (a) Find D, then (b) find the angle between d and D using d#D cos   and verify the result using right dD triangle trig. 8. An ornithologist is in the field studying what ornithologists study (in this case, a formation of geese), and is 180 m south and 385 m west of where she parked her car. Let u be the vector representing the distance between the ornithologist and her car, and let v be the vector representing the straight-line distance from the ornithologist to the formation. If the lead goose flies directly over her car at an altitude of 319 m, (a) find v then, (b) find u#v the angle between u and v using cos   and uv verify the result using right triangle trig.

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9. A person is sitting in the center seat, in the center row, on the inclined seating of a new movie theater. The movie screen is 25 ft tall with the bottom of the screen 8 ft above a level floor. The person’s eye level is 15 ft above the floor and the person’s seat is a horizontal distance of x ft from the movie screen. (a) Write an expression for the person’s sight angle to the screen in terms of x. (b) Find the sight angle if the person is 46 ft from the screen.

7–86

10. The main road in front of a school runs north/south along the east side. A sidewalk runs east/west along the south side of the school, leading to a crosswalk. A secondary road intersects the west side of the main road near the crosswalk, forming a 30 angle with the sidewalk. For the safety of the school children using the sidewalk, engineers are interested in the sight angle that drivers have as they drive on the secondary road approaching this intersection. (a) Find an expression for the sight angle of a driver in terms of x, the distance between their car and a point on the sidewalk 100 ft from the intersection. (b) Find the sight angle at the moment x  125 ft.

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Precalculus—

8 CHAPTER CONNECTIONS

Systems of Equations and Inequalities CHAPTER OUTLINE 8.1 Linear Systems in Two Variables with Applications 720 8.2 Linear Systems in Three Variables with Applications 732 8.3 Partial Fraction Decomposition 743 8.4 Systems of Inequalities and Linear Programming 755 8.5 Solving Linear Systems Using Matrices and Row Operations 769 8.6 The Algebra of Matrices 780 8.7 Solving Linear Systems Using Matrix Equations 791 8.8 Applications of Matrices and Determinants: Cramer’s Rule, Geometry, and More 806

The applications of systems of equations are virtually limitless, ranging from the extremely large systems used by governments to make economic forecasts, to the systems used in operations research to help companies minimize cost and risk. The idea of using a system to solve problems has been around for centuries, with mathematicians through the years often posing puzzle-like questions to their peers for fun and challenge. For instance, if you add the gestation periods for an elephant, camel, and rhino, the result is 1520 days. The gestation period for a rhino exceeds that of a camel by 58 days, while twice the camel’s decreased by 162 gives that of an elephant. How long from conception to birth for each animal? This application appears as Exercise 57 of Section 8.2

Recall that for exponential functions, the rate of growth is proportional to the current population as shown in (1) below. For logistics functions, the rate of growth is proportional to the current population times the difference between the carrying capacity c and the current population (2). For further analysis, this relationship Connections can be rewritten as in (3), with a technique known as partial fraction decomposition then applied. This to Calculus technique is presented in Section 8.3, and further explored in the Connections to Calculus feature for Chapter 8. (1)

¢P  kP ¢t

(2)

¢P  kP1c
P2 ¢t

(3)

¢P  k¢t P1c
P2 719

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Precalculus—

8.1 Linear Systems in Two Variables with Applications Learning Objectives

In earlier chapters, we used linear equations in two variables to model a number of realworld situations. Graphing these equations gave us a visual image of how the variables were related, and helped us better understand this relationship. In many applications, two different measures of the independent variable must be considered simultaneously, leading to a system of two linear equations in two unknowns. Here, a graphical presentation once again supports a better understanding, as we explore systems and their many applications.

In Section 8.1 you will learn how to:

A. Verify ordered pair solutions

B. Solve linear systems by graphing

C. Solve linear systems by substitution

A. Solutions to a System of Equations

D. Solve linear systems by elimination

E. Recognize inconsistent systems and dependent systems

F. Use a system of equations to model and solve applications

Children

700

9a  5c  3100

e

500

(150, 350)

300

0

100

300

500

700

Adults

EXAMPLE 1

a  c  500 9a  5c  3100



amount of revenue

Verifying Solutions to a System Verify that (150, 350) is a solution to e

Solution

number of tickets

We note that both equations are linear and will have different slope values, so their graphs must intersect at some point. Since every point on a line satisfies the equation of that line, this point of intersection must satisfy both equations simultaneously and is the solution to the system. The figure that accompanies Example 1 shows the point of intersecion for this system is (150, 350).

a  c  500

100

A system of equations is a set of two or more equations for which a common solution is sought. Systems are widely used to model and solve applications when the information given enables the relationship between variables to be stated in different ways. For example, consider an amusement park that brought in $3100 in revenue by charging $9.00 for adults and $5.00 for children, while selling 500 tickets. Using a for adult and c for children, we could write one equation modeling the number of tickets sold: a  c  500, and a second modeling the amount of revenue brought in: 9a  5c  3100. To show that we’re considering both equations simultaneously, a large “left brace” is used and the result is called a system of two equations in two variables:



a  c  500 . 9a  5c  3100

Substitute the 150 for a and 350 for c in each equation. a  c  500 first equation 11502  13502  500 500  500 ✓

A. You’ve just learned how to verify ordered pair solutions

9a  5c  3100 second equation 911502  513502  3100 3100  3100 ✓

Since (150, 350) satisfies both equations, it is the solution to the system and we find the park sold 150 adult tickets and 350 tickets for children. Now try Exercises 7 through 18



B. Solving Systems Graphically To solve a system of equations means we apply various methods in an attempt to find ordered pair solutions. As Example 1 suggests, one method for finding solutions is to graph the system. Any method for graphing the lines can be employed, but to keep important concepts fresh, the slope-intercept method is used here.

720

8-2

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Section 8.1 Linear Systems in Two Variables with Applications

EXAMPLE 2



Solution



Solving a System Graphically

4x  3y  9 . 2x  y  5 First write each equation in slope-intercept form (solve for y): 4 y x3 4x  3y  9 3 • S• 2x  y  5 y  2x  5 Solve the system by graphing: e

4 For the first line, ¢y ¢x  3 with y-intercept 10, 32. 2 The second equation yields ¢y ¢x  1 with 10, 52 as the y-intercept. Both are then graphed on the grid as shown. The point of intersection appears to be (3, 1), and checking this point in both equations gives

B. You’ve just learned how to solve linear systems by graphing

4x  3y  9 4132  3112  9 99✓

y 5

y  2x  5

(3, 1) 5

5

x

y  43 x  3

(0, 3)

2x  y  5 substitute 3 2132  112  5 for x and 1 for y 5  5 ✓

5

(0, 5)

This verifies that (3, 1) is the solution to the system. Now try Exercises 19 through 22



C. Solving Systems by Substitution While a graphical approach best illustrates why the solution must be an ordered pair, it does have one obvious drawback — noninteger solutions are difficult to spot. The 4x  y  4 ordered pair 1 25, 12 , but this would be difficult to 5 2 is the solution to e yx2 “pinpoint” as a precise location on a hand-drawn graph. To overcome this limitation, we next consider a method known as substitution. The method involves converting a system of two equations in two variables into a single equation in one variable by using 4x  y  4 an appropriate substitution. For e , the second equation says “y is two more yx2 than x.” We reason that all points on this line are related this way, including the point where this line intersects the other. For this reason, we can substitute x  2 for y in the first equation, obtaining a single equation in x. EXAMPLE 3



Solution



Solving a System Using Substitution 4x  y  4 Solve using substitution: e . yx2 Since y  x  2, we can replace y with x  2 in the first equation. first equation 4x  y  4 4x  1x  22  4 substitute x  2 for y 5x  2  4 simplify 2 x result 5 The x-coordinate is 52. To find the y-coordinate, substitute 25 for x into either of the original equations. Substituting in the second equation gives yx2 second equation 2 2  2 substitute for x 5 5 12 2 10 10 2 12   ,   1 5 5 5 5 5 2 12 The solution to the system is 1 25, 12 5 2. Verify by substituting 5 for x and 5 for y into both equations.

Now try Exercises 23 through 32



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CHAPTER 8 Systems of Equations and Inequalities

If neither equation allows an immediate substitution, we first solve for one of the variables, either x or y, and then substitute. The method is summarized here, and can actually be used with either like variables or like variable expressions. See Exercises 57 to 60. Solving Systems Using Substitution

C. You’ve just learned how to solve linear systems by substitution

1. Solve one of the equations for x in terms of y or y in terms of x. 2. Substitute for the appropriate variable in the other equation and solve for the variable that remains. 3. Substitute the value from step 2 into either of the original equations and solve for the other unknown. 4. Write the answer as an ordered pair and check the solution in both original equations.

D. Solving Systems Using Elimination 2x  5y  13 , where solving for any one of the variables 2x  3y  7 will result in fractional values. The substitution method can still be used, but often the elimination method is more efficient. The method takes its name from what happens when you add certain equations in a system (by adding the like terms from each). If the coefficients of either x or y are additive inverses — they sum to zero and are eliminated. For the system shown, adding the equations produces 2y  6, giving y  3, then x  1 using back-substitution (verify). When neither variable term meets this condition, we can multiply one or both equations by a nonzero constant to “match up” the coefficients, so an elimination will take place. In doing so, we create an equivalent system of equations, meaning one 7x  4y  16 that has the same solution as the original system. For e , multiplying 3x  2y  6 7x  4y  16 the second equation by 2 produces e , giving x  4 after “adding 6x  4y  12 the equations.” Note the three systems produced are equivalent, and have the solution (4, 3) ( y  3 was found using back-substitution). Now consider the system e

1. e

7x  4y  16 3x  2y  6

2. e

7x  4y  16 6x  4y  12

3. e

7x  4y  16 x 4

In summary, Operations that Produce an Equivalent System 1. Changing the order of the equations. 2. Replacing an equation by a nonzero constant multiple of that equation. 3. Replacing an equation with the sum of two equations from the system. Before beginning a solution using elimination, check to make sure the equations are written in the standard form Ax  By  C, so that like terms will appear above/below each other. Throughout this chapter, we will use R1 to represent the equation in row 1 of the system, R2 to represent the equation in row 2, and so on. These designations are used to help describe and document the steps being used to solve a system, as in Example 4 where 2R1  R2 indicates the first equation has been multiplied by two, with the result added to the second equation.

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Section 8.1 Linear Systems in Two Variables with Applications

EXAMPLE 4



Solution



723

Solving a System by Elimination 2x  3y  7 Solve using elimination: e 6y  5x  4 The second equation is not in standard form, so we re-write the system as 2x  3y  7 e . If we “add the equations” now, we would get 7x  3y  11, with 5x  6y  4 neither variable eliminated. However, if we multiply both sides of the first equation by 2, the y-coefficients will be additive inverses. The sum then results in an equation with x as the only unknown. 2R1 4x  6y  14  R2 5x  6y  4 add sum 9x  0y  18 9x  18 x  2 solve for x Substituting 2 for x back into either of the original equations yields y  1. The ordered pair solution is (2, 1). Verify using the original equations.

{

WORTHY OF NOTE As the elimination method involves adding two equations, it is sometimes referred to as the addition method for solving systems.

Now try Exercises 33 through 38



The elimination method is summarized here. If either equation has fraction or decimal coefficients, we can “clear” them using an appropriate constant multiplier. Solving Systems Using Elimination 1. Write each equation in standard form: Ax  By  C. 2. Multiply one or both equations by a constant that will create coefficients of x (or y) that are additive inverses. 3. Combine the two equations using vertical addition and solve for the variable that remains. 4. Substitute the value from step 3 into either of the original equations and solve for the other unknown. 5. Write the answer as an ordered pair and check the solution in both original equations.

EXAMPLE 5



Solving a System Using Elimination 5 x  34y  14 . Solve using elimination: e 81 2 2x  3y  1

Solution



Multiplying the first equation by 8(8R1) and the second equation by 6(6R2) will clear the fractions from each. 5x  6y  2 8R1 81 1 58 2x  81 1 34 2y  81 1 14 2 e6 1 Se 6 2 3x  4y  6 6R2 1 1 2 2x  1 1 3 2y  6112

The x-terms can now be eliminated if we use 3R1  15R22 . 3R1 15x  18y  6  5R2 15x  20y  30 0x  2y  24 sum y  12

{

D. You’ve just learned how to solve linear systems by elimination

add

solve for y

Substituting y  12 in either of the original equations yields x  14, and the solution is 114, 122. Verify by substituting in both equations. Now try Exercises 39 through 44



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CAUTION

Be sure to multiply all terms (on both sides) of the equation when using a constant multiplier. Also, note that for Example 5, we could have eliminated the y-terms using 2R1 with 3R2.

E. Inconsistent and Dependent Systems A system having at least one solution is called a consistent system. As seen in Example 2, if the lines have different slopes, they intersect at a single point and the system has exactly one solution. Here, the lines are independent of each other and the system is called an independent system. If the lines have equal slopes and the same y-intercept, they are identical or coincident lines. Since one is right atop the other, they intersect at all points, and the system has an infinite number of solutions. Here, one line depends on the other and the system is called a dependent system. Using substitution or elimination on a dependent system results in the elimination of all variable terms and leaves a statement that is always true, such as 0  0 or some other simple identity.

EXAMPLE 6



Solving a Dependent System Solve using elimination: e

Solution



WORTHY OF NOTE When writing the solution to a dependent system using a parameter, the solution can be written in many different ways. For instance, if we let p  4b for the first coordinate of the solution to Example 6, we have 314b2  3  3b  3 as 4 the second coordinate, and the solution becomes (4b, 3b  3) for any constant b.

3x  4y  12 . 6x  24  8y

y 5

Writing the system in standard form gives 3x  4y  12 . By applying 2R1, we can e 6x  8y  24 eliminate the variable x: 2R1 6x  8y  24  e R2 6x  8y  24 add sum 0x  0y  0 variables are eliminated 0  0 true statement

3x  4y  12

(0, 3) (4, 0)

5

6x  24  8y

5

x

5

Although we didn’t expect it, both variables were eliminated and the final statement is true 10  02. This indicates the system is dependent, which the graph verifies (the lines are coincident). Writing both equations in slope-intercept form shows they represent the same line. e

3x  4y  12 6x  8y  24

e

4y  3x  12 8y  6x  24

3 y x3 4 μ 3 y x3 4

The solutions of a dependent system are often written in set notation as the set of ordered pairs (x, y), where y is a specified function of x. For Example 6 the solution would be 51x, y2 0 y  34x  36. Using an ordered pair with an arbitrary 3p variable, called a parameter, is also common: ap,  3b. 4 Now try Exercises 45 through 56

E. You’ve just learned how to recognize inconsistent and dependent systems



Finally, if the lines have equal slopes and different y-intercepts, they are parallel and the system will have no solution. A system with no solutions is called an inconsistent system. An “inconsistent system” produces an “inconsistent answer,” such as 12  0 or some other false statement when substitution or elimination is applied. In other words, all variable terms are once again eliminated, but the remaining statement is false. A summary of the three possibilities is shown here for arbitrary slope m and y-intercept (0, b).

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Section 8.1 Linear Systems in Two Variables with Applications

Independent m1  m2

Dependent m1  m2, b1  b2

y

Inconsistent m1  m2, b1  b2

y

y

x

One point in common

x

All points in common

x

No points in common

F. Systems and Modeling In previous chapters, we solved numerous real-world applications by writing all given relationships in terms of a single variable. Many situations are easier to model using a system of equations with each relationship modeled independently using two variables. We begin here with a mixture application. Although they appear in many different forms (coin problems, metal alloys, investments, merchandising, and so on), mixture problems all have a similar theme. Generally one equation is related to quantity (how much of each item is being combined) and one equation is related to value (what is the value of each item being combined). EXAMPLE 7



Solving a Mixture Application A jeweler is commissioned to create a piece of artwork that will weigh 14 oz and consist of 75% gold. She has on hand two alloys that are 60% and 80% gold, respectively. How much of each should she use?

Solution

WORTHY OF NOTE As an estimation tool, note that if equal amounts of the 60% and 80% alloys were used (7 oz each), the result would be a 70% alloy (halfway in between). Since a 75% alloy is needed, more of the 80% gold will be used.



Let x represent ounces of the 60% alloy and y represent ounces of the 80% alloy. The first equation must be x  y  14, since the piece of art must weigh exactly 14 oz (this is the quantity equation). The x ounces are 60% gold, the y ounces are 80% gold, and the 14 oz will be 75% gold. This gives the value equation: x  y  14 0.6x  0.8y  0.751142. The system is e (after clearing decimals). 6x  8y  105 Solving for y in the first equation gives y  14  x. Substituting 14  x for y in the second equation gives 6x  8y  105 6x  8114  x2  105 2x  112  105 7 x 2

second equation substitute 14  x for y simplify solve for x

Substituting 72 for x in the first equation gives y  21 2 . She should use 3.5 oz of the 60% alloy and 10.5 oz of the 80% alloy. Now try Exercises 63 through 70



Systems of equations also play a significant role in cost-based pricing in the business world. The costs involved in running a business can broadly be understood as either a fixed cost k or a variable cost v. Fixed costs might include the monthly rent paid for facilities, which remains the same regardless of how many items are produced and sold. Variable costs would include the cost of materials needed to produce the item, which depends on the number of items made. The total cost can then be modeled by

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C1x2  vx  k for x number of items. Once a selling price p has been determined, the revenue equation is simply R1x2  px (price times number of items sold). We can now set up and solve a system of equations that will determine how many items must be sold to break even, performing what is called a break-even analysis. EXAMPLE 8



Solving an Application of Systems: Break-Even Analysis In home businesses that produce items to sell on Ebay®, fixed costs are easily determined by rent and utilities, and variable costs by the price of materials needed to produce the item. Karen’s home business makes large, decorative candles for all occasions. The cost of materials is $3.50 per candle, and her rent and utilities average $900 per month. If her candles sell for $9.50, how many candles must be sold each month to break even?

Solution



WORTHY OF NOTE This break-even concept can also be applied in studies of supply and demand, as well as in the decision to buy a new car or appliance that will enable you to break even over time due to energy and efficiency savings.

EXAMPLE 9

Let x represent the number of candles sold. Her total cost is C1x2  3.5x  900 (variable cost plus fixed cost), and projected revenue is R1x2  9.5x. This gives the C 1x2  3.5x  900 system e . To break even, Cost  Revenue which gives R1x2  9.5x 9.5x  3.5x  900 6x  900 x  150 The analysis shows that Karen must sell 150 candles each month to break even. Now try Exercises 71 through 74



Our final example involves an application of uniform motion (distance  rate # time ), and explores concepts of great importance to the navigation of ships and airplanes. As a simple illustration, if you’ve ever walked at your normal rate r on the “moving walkways” at an airport, you likely noticed an increase in your total speed. This is because the resulting speed combines your walking rate r with the speed w of the walkway: total speed  r  w. If you walk in the opposite direction of the walkway, your total speed is much slower, as now total speed  r  w. This same phenomenon is observed when an airplane is flying with or against the wind, or a ship is sailing with or against the current. 

Solving an Application of Systems — Uniform Motion An airplane flying due south from St. Louis, Missouri, to Baton Rouge, Louisiana, uses a strong, steady tailwind to complete the trip in only 2.5 hr. On the return trip, the same wind slows the flight and it takes 3 hr to get back. If the flight distance between these cities is 912 km, what is the cruising speed of the airplane (speed with no wind)? How fast is the wind blowing?

Solution



F. You’ve just learned how to use a system of equations to model and solve applications

Let r represent the rate of the plane and w the rate of the wind. Since D  RT , the flight to Baton Rouge can be modeled by 912  1r  w2 12.52 , and the return flight 912  2.5r  2.5w R1 by 912  1r  w2132 . This produces the system e . Using and 912  3r  3w 2.5 364.8  r  w R2 gives the equivalent system e , which is easily solved using 304  r  w 3 elimination with R1  R2. 668.8  2r 334.4  r

R1  R2 divide by 2

The cruising speed of the plane (with no wind) is 334.4 kph. Using r  w  304 shows the wind is blowing at 30.4 kph. Now try Exercises 75 through 78



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TECHNOLOGY HIGHLIGHT

Solving Systems Graphically When used with care, graphing calculators offer an accurate way to solve linear systems and to check 4x  y  4 solution(s) obtained by hand. We’ll illustrate using the system from Example 3: e , where we yx2 2 12 found the solution was 1 5, 5 2 .

Figure 8.1

1. Solve for y in both equations: y  4x  4 e y x2 3. Graph using

ZOOM

6

2. Enter the equations as

10

Y1  4x  4 Y2  x  2 4. Press 2nd TRACE (CALC) 5

Y1  4x  4 Y2  x  2

ENTER

ENTER

10

10

ENTER

to have the calculator compute the point of intersection.

10

The coordinates of the intersection appear as decimal fractions at the bottom of the screen (Figure 8.1). In step 4, The first ENTER selects Y1, the second ENTER selects Y2 and the third ENTER bypasses the GUESS option (this option is most often used if the graphs intersect at more than one point). The calculator automatically registers the x-coordinate as its most recent entry, and from the home screen, converting it to a standard fraction (using MATH 1: Frac ENTER ) shows x  25 . You can also get an approximate solution by tracing along either line towards the point of intersection using the TRACE key and the left or right arrows. Solve each system graphically, using a graphing calculator.

Exercise 1:

e

3x  y  7 y  5x  1

Exercise 2:

e

2x  3y  3 6  8x  3y

8.1 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. Systems that have no solution are called systems. 2. Systems having at least one solution are called systems. 3. If the lines in a system intersect at a single point, the system is said to be and . 4. If the lines in a system are coincident, the system is referred to as and .

5. The given systems are equivalent. How do we obtain the second system from the first? 1 5 2 x y 4x  3y  10 3 2 3 e • 2x  4y  10 0.2x  0.4y  1 2x  5y  8 , which solution method would 3x  4y  5 be more efficient, substitution or elimination? Discuss/Explain why.

6. For e

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DEVELOPING YOUR SKILLS

Show the lines in each system would intersect in a single point by writing the equations in slope-intercept form.

7. e

7x  4y  24 4x  3y  15

8. e

0.3x  0.4y  2 0.5x  0.2y  4

An ordered pair is a solution to an equation if it makes the equation true. Given the graph shown here, determine which equation(s) have the indicated point as a solution. If the point satisfies more than one equation, write the system for which it is a solution.

y 5

3x  2y  6

yx2 A

F 5

5 x

C D

5

9. A

10. B

11. C

12. D

13. E

14. F

Substitute the x- and y-values indicated by the ordered pair to determine if it solves the system.

15. e

3x  y  11 13, 22 5x  y  13;

16. e

3x  7y  4 16, 22 7x  8y  21;

17. e

8x  24y  17 7 5 a , b 8 12 12x  30y  2;

18. e

4x  15y  7 1 1 a , b 8x  21y  11; 2 3

Solve each system by graphing. If the coordinates do not appear to be integers, estimate the solution to the nearest tenth (indicate that your solution is an estimate).

19. e

3x  2y  12 x y9

21. e

5x  2y  4 3x  y  2 22. e x  3y  15 5x  3y  12

20. e

5x  2y  2 3x  y  10

Solve each system using substitution. Write solutions as an ordered pair.

23. e

x  5y  9 x  2y  6

24. e

4x  5y  7 2x  5  y

25. e

y  23x  7 3x  2y  19

26. e

2x  y  6 y  34x  1

27. e

3x  4y  24 5x  y  17

29. e

0.7x  2y  5 0.8x  y  7.4 30. e x  1.4y  11.4 0.6x  1.5y  9.3

31. e

5x  6y  2 x  2y  6

28. e

32. e

3x  2y  19 x  4y  3

2x  5y  5 8x  y  6

B

E

x  3y  3

Identify the equation and variable that makes the substitution method easiest to use. Then solve the system.

Solve using elimination. In some cases, the system must first be written in standard form.

33. e

2x  4y  10 3x  4y  5

34. e

x  5y  8 x  2y  6

35. e

4x  3y  1 3y  5x  19

36. e

5y  3x  5 3x  2y  19

37. e

2x  3y  17 4x  5y  12

38. e

2y  5x  2 4x  17  6y

39. e

0.5x  0.4y  0.2 0.2x  0.3y  0.8 40. e 0.3y  1.3  0.2x 0.3x  0.4y  1.3

41. e

0.32m  0.12n  1.44 0.24m  0.08n  1.04

42. e

0.06g  0.35h  0.67 0.12g  0.25h  0.44

43. e

16u  14v  4 1 2 2 u  3 v  11

x  13y  2 1 2x  5y  3 3

44. e 43

Solve using any method and identify the system as consistent, inconsistent, or dependent.

45. e

4x  34y  14 9x  58y  13

2 xy2 46. e 3 2y  56x  9

47. e

0.2y  0.3x  4 1.2x  0.4y  5 48. e 0.6x  0.4y  1 0.5y  1.5x  2

49. e

6x  22  y 3x  12y  11

51. e

10x  35y  5 2x  3y  4 52. e y  0.25x x  2.5y

53. e

7a  b  25 2a  5b  14

50. e

54. e

15  5y  9x 3x  53y  5

2m  3n  1 5m  6n  4

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55. e

4a  2  3b 6b  2a  7

56. e

3p  2q  4 9p  4q  3

The substitution method can be used for like variables or for like expressions. Solve the following systems, using the expression common to both equations (do not solve for x or y alone). 

2x  4y  6 x  12  4y

59. e

5x  11y  21 11y  5  8x

58. e

8x  3y  24 8x  5y  36

60. e

6x  5y  16 5y  6x  4

WORKING WITH FORMULAS

61. Uniform motion with current: e

1R  C2T1  D1 1R  C2T2  D2

The formula shown can be used to solve uniform motion problems involving a current, where D represents distance traveled, R is the rate of the object with no current, C is the speed of the current, and T is the time. Chan-Li rows 9 mi up river (against the current) in 3 hr. It only took him 1 hr to row 5 mi downstream (with the current). How fast was the current? How fast can he row in still water? 

57. e

729

62. Fahrenheit and Celsius temperatures: F y  95x  32 e 5 y  9 1x  322 C Many people are familiar with temperature measurement in degrees Celsius and degrees Fahrenheit, but few realize that the equations are linear and there is one temperature at which the two scales agree. Solve the system using the method of your choice and find this temperature.

APPLICATIONS

Solve each application by modeling the situation with a linear system. Be sure to clearly indicate what each variable represents. Mixture

63. Theater productions: At a recent production of A Comedy of Errors, the Community Theater brought in a total of $30,495 in revenue. If adult tickets were $9 and children’s tickets were $6.50, how many tickets of each type were sold if 3800 tickets in all were sold? 64. Milk-fat requirements: A dietician needs to mix 10 gal of milk that is 212 % milk fat for the day’s rounds. He has some milk that is 4% milk fat and some that is 112 % milk fat. How much of each should be used? 65. Filling the family cars: Cherokee just filled both of the family vehicles at a service station. The total cost for 20 gal of regular unleaded and 17 gal of premium unleaded was $144.89. The premium gas was $0.10 more per gallon than the regular gas. Find the price per gallon for each type of gasoline. 66. Household cleaners: As a cleaning agent, a solution that is 24% vinegar is often used. How much pure (100%) vinegar and 5% vinegar must be mixed to obtain 50 oz of a 24% solution?

67. Alumni contributions: A wealthy alumnus donated $10,000 to his alma mater. The college used the funds to make a loan to a science major at 7% interest and a loan to a nursing student at 6% interest. That year the college earned $635 in interest. How much was loaned to each student? 68. Investing in bonds: A total of $12,000 is invested in two municipal bonds, one paying 10.5% and the other 12% simple interest. Last year the annual interest earned on the two investments was $1335. How much was invested at each rate? 69. Saving money: Bryan has been doing odd jobs around the house, trying to earn enough money to buy a new Dirt-Surfer©. He saves all quarters and dimes in his piggy bank, while he places all nickels and pennies in a drawer to spend. So far, he has 225 coins in the piggy bank, worth a total of $45.00. How many of the coins are quarters? How many are dimes? 70. Coin investments: In 1990, Molly attended a coin auction and purchased some rare “Flowing Hair” fifty-cent pieces, and a number of very rare twocent pieces from the Civil War Era. If she bought 47 coins with a face value of $10.06, how many of each denomination did she buy?

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71. Lawn service: Dave and his sons run a lawn service, which includes mowing, edging, trimming, and aerating a lawn. His fixed cost includes insurance, his salary, and monthly payments on equipment, and amounts to $4000/mo. The variable costs include gas, oil, hourly wages for his employees, and miscellaneous expenses, which run about $75 per lawn. The average charge for fullservice lawn care is $115 per visit. Do a breakeven analysis to (a) determine how many lawns Dave must service each month to break even and (b) the revenue required to break even. 72. Production of mini-microwave ovens: Due to high market demand, a manufacturer decides to introduce a new line of mini-microwave ovens for personal and office use. By using existing factory space and retraining some employees, fixed costs are estimated at $8400/mo. The components to assemble and test each microwave are expected to run $45 per unit. If market research shows consumers are willing to pay at least $69 for this product, find (a) how many units must be made and sold each month to break even and (b) the revenue required to break even.

74. Digital music: Market research has indicated that by 2010, sales of MP3 portables will mushroom into a $70 billion dollar market. With a market this large, competition is often fierce — with suppliers fighting to earn and hold market shares. For x million MP3 players sold, supply is modeled by y  10.5x  25, where y is the current market price (in dollars). The related demand equation might be y  5.20x  140. (a) How many million MP3 players will be supplied at a market price of $88? What will the demand be at this price? Is supply less than demand? (b) How many million MP3 players will be supplied at a market price of $114? What will the demand be at this price? Is demand less than supply? (c) To the nearest cent, at what price does the market reach equilibrium? How many units are being supplied/demanded?

In a market economy, the availability of goods is closely related to the market price. Suppliers are willing to produce more of the item at a higher price (the supply), with consumers willing to buy more of the item at a lower price (the demand). This is called the law of supply and demand. When supply and demand are equal, both the buyer and seller are satisfied with the current price and we have market equilibrium.

73. Farm commodities: One area where the law of supply and demand is clearly at work is farm commodities. Both growers and consumers watch this relationship closely, and use data collected by government agencies to track the relationship and make adjustments, as when a farmer decides to convert a large portion of her farmland from corn to soybeans to improve profits. Suppose that for x billion bushels of soybeans, supply is modeled by y  1.5x  3, where y is the current market price (in dollars per bushel). The related demand equation might be y  2.20x  12. (a) How many billion bushels will be supplied at a market price of $5.40? What will the demand be at this price? Is supply less than demand? (b) How many billion bushels will be supplied at a market price of $7.05? What will the demand be at this price? Is demand less than supply? (c) To the nearest cent, at what price does the market reach equilibrium? How many bushels are being supplied/demanded?

Uniform Motion

75. Canoeing on a stream: On a recent camping trip, it took Molly and Sharon 2 hr to row 4 mi upstream from the drop in point to the campsite. After a leisurely weekend of camping, fishing, and relaxation, they rowed back downstream to the drop in point in just 30 min. Use this information to find (a) the speed of the current and (b) the speed Sharon and Molly would be rowing in still water. 76. Taking a luxury cruise: A luxury ship is taking a Caribbean cruise from Caracas, Venezuela, to just off the coast of Belize City on the Yucatan Peninsula, a distance of 1435 mi. En route they encounter the Caribbean Current, which flows to the northwest, parallel to the coastline. From Caracas to the Belize coast, the trip took 70 hr. After a few days of fun in the sun, the ship leaves for Caracas, with the return trip taking 82 hr. Use

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this information to find (a) the speed of the Caribbean Current and (b) the cruising speed of the ship. 77. Airport walkways: As part of an algebra field trip, Jason takes his class to the airport to use their moving walkways for a demonstration. The class measures the longest walkway, which turns out to be 256 ft long. Using a stop watch, Jason shows it takes him just 32 sec to complete the walk going in the same direction as the walkway. Walking in a direction opposite the walkway, it takes him 320 sec — 10 times as long! The next day in class, Jason hands out a two-question quiz: (1) What was the speed of the walkway in feet per second? (2) What is my (Jason’s) normal walking speed? Create the answer key for this quiz. 78. Racing pigeons: The American Racing Pigeon Union often sponsors opportunities for owners to fly their birds in friendly competitions. During a recent competition, Steve’s birds were liberated in Topeka, Kansas, and headed almost due north to their loft in Sioux Falls, South Dakota, a distance of 308 mi. During the flight, they encountered a steady wind from the north and the trip took 4.4 hr. The next month, Steve took his birds to a competition in Grand Forks, North Dakota, with the birds heading almost due south to home, also a distance of 308 mi. This time the birds were aided by the same wind from the north, and the trip took only 3.5 hr. Use this information to (a) find the racing speed of Steve’s birds and (b) find the speed of the wind.



Descriptive Translation

79. Important dates in U.S. history: If you sum the year that the Declaration of Independence was signed and the year that the Civil War ended, you get 3641. There are 89 yr that separate the two events. What year was the Declaration signed? What year did the Civil War end? 80. Architectual wonders: When it was first constructed in 1889, the Eiffel Tower in Paris, France, was the tallest structure in the world. In 1975, the CN Tower in Toronto, Canada, became the world’s tallest structure. The CN Tower is 153 ft less than twice the height of the Eiffel Tower, and the sum of their heights is 2799 ft. How tall is each tower? 81. Pacific islands land area: In the South Pacific, the island nations of Tahiti and Tonga have a combined land area of 692 mi2. Tahiti’s land area is 112 mi2 more than Tonga’s. What is the land area of each island group? 82. Card games: On a cold winter night, in the lobby of a beautiful hotel in Sante Fe, New Mexico, Marc and Klay just barely beat John and Steve in a close game of Trumps. If the sum of the team scores was 990 points, and there was a 12-point margin of victory, what was the final score?

EXTENDING THE CONCEPT

83. Answer using observations only — no calculations. Is the given system consistent/independent, consistent/dependent, or inconsistent? y  5x  2 Explain/Discuss your answer. e y  5.01x  1.9 84. Federal income tax reform has been a hot political topic for many years. Suppose tax plan A calls for a flat tax of 20% tax on all income (no deductions or loopholes). Tax plan B requires taxpayers to pay 

731

$5000 plus 10% of all income. For what income level do both plans require the same tax? 85. Suppose a certain amount of money was invested at 6% per year, and another amount at 8.5% per year, with a total return of $1250. If the amounts invested at each rate were switched, the yearly income would have been $1375. To the nearest whole dollar, how much was invested at each rate?

MAINTAINING YOUR SKILLS

86. (2.6) Given the parent function f 1x2  x, sketch the graph of F1x2  x  3 2. 87. (5.1) Find two positive and two negative angles that are coterminal with   112°.

88. (4.4) Solve for x (rounded to the nearest thousandth): 33  77.5e0.0052x  8.37 89. (6.2) Verify that identity.

sin x  csc x  cos2x is an csc x

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The transition to systems of three equations in three variables requires a fair amount of “visual gymnastics” along with good organizational skills. Although the techniques used are identical and similar results are obtained, the third equation and variable give us more to track, and we must work more carefully toward the solution.

In Section 8.2 you will learn how to:

A. Visualize a solution in three dimensions

B. Check ordered triple

A. Visualizing Solutions in Three Dimensions

solutions

C. Solve linear systems in three variables

D. Recognize inconsistent and dependent systems

E. Use a system of three equations in three variables to solve applications

WORTHY OF NOTE We can visualize the location of a point in space by considering a large rectangular box 2 ft long  3 ft wide  1 ft tall, placed snugly in the corner of a room. The floor is the xy-plane, one wall is the xz-plane, and the other wall is the yz-plane. The z-axis is formed where the two walls meet and the corner of the room is the origin (0, 0, 0). To find the corner of the box located at (2, 3, 1), first locate the point (2, 3) in the xy-plane (the floor), then move up 1 ft.

Figure 8.2

Figure 8.3

z

z

(0, 0, 6)

(0, 0, 6)

(2, 3, 1) y

2 units along x

nit z el

all par



x

ly lle ara

x

sp

nit

(6, 0, 0)

(6, 0, 0)

y

(0, 6, 0)

1u

(0, 6, 0)

3u

EXAMPLE 1

The solution to an equation in one variable is the single number that satisfies the equation. For x  1  3, the solution is x  2 and its graph is a single point on the number line, a one-dimensional graph. The solution to an equation in two variables, such as x  y  3, is an ordered pair (x, y) that satisfies the equation. When we graph this solution set, the result is a line on the xy-coordinate grid, a two-dimensional graph. The solutions to an equation in three variables, such as x  y  z  6, are the ordered triples (x, y, z) that satisfy the equation. When we graph this solution set, the result is a plane in space, a graph in three dimensions. Recall a plane is a flat surface having infinite length and width, but no depth. We can graph this plane using the intercept method and the result is shown in Figure 8.2. For graphs in three dimensions, the xy-plane is parallel to the ground (the y-axis points to the right) and z is the vertical axis. To find an additional point on this plane, we use any three numbers whose sum is 6, such as (2, 3, 1). Move 2 units along the x-axis, 3 units parallel to the y-axis, and 1 unit parallel to the z-axis, as shown in Figure 8.3.

Finding Solutions to an Equation in Three Variables Use a guess-and-check method to find four additional points on the plane determined by x  y  z  6.

Solution



A. You’ve just learned how to visualize a solution in three dimensions

We can begin by letting x  0, then use any combination of y and z that sum to 6. Two examples are (0, 2, 4) and (0, 5, 1). We could also select any two values for x and y, then determine a value for z that results in a sum of 6. Two examples are 12, 9, 12 and 18, 3, 12. Now try Exercises 7 through 10



B. Solutions to a System of Three Equations in Three Variables When solving a system of three equations in three variables, remember each equation represents a plane in space. These planes can intersect in various ways, creating 732

8-14

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different possibilities for a solution set (see Figures 8.4 to 8.7). The system could have a unique solution (a, b, c), if the planes intersect at a single point (Figure 8.4) (the point satisfies all three equations simultaneously). If the planes intersect in a line (Figure 8.5), the system is linearly dependent and there are an infinite number of solutions. Unlike the two-dimensional case, the equation of a line in three dimensions is somewhat complex, and the coordinates of all points on this line are usually represented by a specialized ordered triple, which we use to state the solution set. If the planes intersect at all points, the system has coincident dependence (see Figure 8.6). This indicates the equations of the system differ by only a constant multiple—they are all “disguised forms” of the same equation. The solution set is any ordered triple (a, b, c) satisfying this equation. Finally, the system may have no solutions. This can happen a number of different ways, most notably if the planes intersect as shown in Figure 8.7 (other possibilities are discussed in the exercises). In the case of “no solutions,” an ordered triple may satisfy none of the equations, only one of the equations, only two of the equations, but not all three equations. Figure 8.4

Unique solution

Figure 8.5

Figure 8.6

Figure 8.7

Linear dependence

Coincident dependence

No solutions

EXAMPLE 2



Determining If an Ordered Triple Is a Solution

Solution



Substitute 1 for x, 2 for y, and 3 for z in the first system. x  4y  z  10 112  4122  132  10 10  10 true a. • 2x  5y  8z  4 S • 2112  5122  8132  4 S • 16  4 false x  2y  3z  4 112  2122  3132  4 4  4 true

B. You’ve just learned how to check ordered triple solutions

Determine if the ordered triple 11, 2, 32 is a solution to the systems shown. x  4y  z  10 3x  2y  z  4 a. • 2x  5y  8z  4 b. • 2x  3y  2z  2 x  2y  3z  4 x  y  2z  9

No, the ordered triple 11, 2, 32 is not a solution to the first system. Now use the same substitutions in the second system. 3x  2y  z  4 3112  2122  132  4 4  4 true b. • 2x  3y  2z  2 S • 2112  3122  2132  2 S • 2  2 true x  y  2z  9 112  122  2132  9 9  9 true The ordered triple 11, 2, 32 is a solution to the second system only.

Now try Exercises 11 and 12



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C. Solving Systems of Three Equations in Three Variables Using Elimination From Section 8.1, we know that two systems of equations are equivalent if they have the same solution set. The systems 2x  y  2z  7 2x  y  2z  7 • x  y  z  1 and • y  4z  5 2y  z  3 z1

are equivalent, as both have the unique solution 13, 1, 12. In addition, it is evident that the second system can be solved more easily, since R2 and R3 have fewer variables than the first system. In the simpler system, mentally substituting 1 for z into R2 immediately gives y  1, and these values can be back-substituted into the first equation to find that x  3. This observation guides us to a general approach for solving larger systems—we would like to eliminate variables in the second and third equations, until we obtain an equivalent system that can easily be solved by back-substitution. To begin, let’s review the three operations that “transform” a given system, and produce an equivalent system. Operations That Produce an Equivalent System 1. Changing the order of the equations. 2. Replacing an equation by a nonzero constant multiple of that equation. 3. Replacing an equation with the sum of two equations from the system. Building on the ideas from Section 8.1, we develop the following approach for solving a system of three equations in three variables. Solving a System of Three Equations in Three Variables 1. Write each equation in standard form: Ax  By  Cz  D. 2. If the “x” term in any equation has a coefficient of 1, interchange equations (if necessary) so this equation becomes R1. 3. Use the x-term in R1 to eliminate the x-terms from R2 and R3. The original R1, with the new R2 and R3, form an equivalent system that contains a smaller “subsystem” of two equations in two variables. 4. Solve the subsystem and keep the result as the new R3. The result is an equivalent system that can be solved using back-substitution. 2x  y  2z  7 We’ll begin by solving the system • x  y  z  1 using the elimination 2y  z  3 method and the procedure outlined. In Example 3, the notation 2R1  R2 S R2 indicates the equation in row 1 has been multiplied by 2 and added to the equation in row 2, with the result placed in the system as the new row 2. EXAMPLE 3



Solving a System of Three Equations in Three Variables 2x  y  2z  7 Solve using elimination: • x  y  z  1. 2y  z  3

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Solution



1. The system is in standard form. 2. If the x-term in any equation has a coefficient of 1, interchange equations so this equation becomes R1. 2x  y  2z  7 • x  y  z  1 2y  z  3

R2 4 R1 S

x  y  z  1 • 2x  y  2z  7 2y  z  3

3. Use R1 to eliminate the x-term in R2 and R3. Since R3 has no x-term, the only elimination needed is the x-term from R2. Using 2R1  R2 will eliminate this term: 2R1 2x  2y  2z  2  2x  y  2z  7 R2 0x  1y  4z  5 y  4z  5

sum simplify

The new R2 is y  4z  5. The original R1 and R3, along with the new R2 form an equivalent system that contains a smaller subsystem x  y  z  1 x  y  z  1 2R1  R2 S R2 S • y  4z  5 • 2x  y  2z  7 R3 S R3 2y  z  3 2y  z  3

new equivalent system

4. Solve the subsystem for either y or z, and keep the result as a new R3. We choose to eliminate y using 2R2  R3: 2R2 2y  8z  10  R3 2y  z  3 0y  7z  7 z 1

sum simplify

The new R3 is z  1. x  y  z  1 x  y  z  1 2R2  R3 S R3 S • y  4z  5 • y  4z  5  2y  z  3 z1

new equivalent system

The new R3, along with the original R1 and R2 from step 3, form an equivalent system that can be solved using back-substitution. Substituting 1 for z in R2 yields y  1. Substituting 1 for z and 1 for y in R1 yields x  3. The solution is 13, 1, 12. Now try Exercises 13 through 18



While not absolutely needed for the elimination process, there are two reasons for wanting the coefficient of x to be “1” in R1. First, it makes the elimination method more efficient since we can more easily see what to use as a multiplier. Second, it lays the foundation for developing other methods of solving larger systems. If no equation has an x-coefficient of 1, we simply use the y- or z-variable instead (see Example 7). Since solutions to larger systems generally are worked out in stages, we will sometimes track the transformations used by writing them between the original system and the equivalent system, rather than to the left as we did in Section 8.1. Here is an additional example illustrating the elimination process, but in abbreviated form. Verify the calculations indicated using a separate sheet.

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EXAMPLE 4



Solution



Solving a System of Three Equations in Three Variables 5y  2x  z  8 Solve using elimination: • x  3z  2y  13 . z  3y  x  5 2x  5y  z  8 1. Write the equations in standard form: • x  2y  3z  13 x  3y  z  5 2x  5y  z  8 x  3y  z  5 • x  2y  3z  13 2. • x  2y  3z  13 R3 4 R1 S x  3y  z  5 2x  5y  z  8

equivalent system

3. Using R1  R2 will eliminate the x-term from R2, yielding 5y  2z  18. Using 2R1  R3 eliminates the x-term from R3, yielding 11y  z  18. x  3y  z  5 R1  R2 S R2 S • x  2y  3z  13 2R1  R3 S R3 2x  5y  z  8

x  3y  z  5 equivalent • 5y  2z  18 system 11y  z  18

4. Using 2R3  R2 will eliminate z from the subsystem, leaving 27y  54. x  3y  z  5 x  3y  z  5 2R3  R2 S R3 equivalent S • • 5y  2z  18 5y  2z  18 system 11y  z  18 27y  54 C. You’ve learned just how to solve linear systems in three variables

Solving for y in R3 shows y  2. Substituting 2 for y in R2 yields z  4, and substituting 2 for y and 4 for z in R1 shows x  3. The solution is (3, 2, 4). Now try Exercises 19 through 24



D. Inconsistent and Dependent Systems As mentioned, it is possible for larger systems to have no solutions or an infinite number of solutions. As with our work in Section 8.1, an inconsistent system (no solutions) will produce inconsistent results, ending with a statement such as 0  3 or some other contradiction. EXAMPLE 5



Solution



Attempting to Solve an Inconsistent System 2x  y  3z  3 Solve using elimination: • 3x  2y  4z  2 . 4x  2y  6z  7 1. This system has no equation where the coefficient of x is 1. 2. We can still use R1 to begin the solution process, but this time we’ll use the variable y since it does have coefficient 1. Using 2R1  R2 eliminates the y-term from R2, leaving 7x  2z  4. But using 2R1  R3 to eliminate the y-term from R3 results in a contradiction: 2R1 4x  2y  6z  6 2R1 4x  2y  6z  6   R2 3x  2y  4z  2 4x  2y  6z  7 R3 7x  2z  4 0x  0y  0z  1 0  1 contradiction We conclude the system is inconsistent. The answer is the empty set , and we need work no further. Now try Exercises 25 and 26



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Unlike our work with systems having only two variables, systems in three variables can have two forms of dependence — linear dependence or coincident dependence. To help understand linear dependence, consider a system of two equations in 2x  3y  z  5 . Each of these equations represents a plane, three variables: e x  3y  2z  1 and unless the planes are parallel, their intersection will be a line (see Figure 8.5). As in Section 8.1, we can state solutions to a dependent system using set notation with two of the variables written in terms of the third, or as an ordered triple using a parameter. The relationships named can then be used to generate specific solutions to the system. Systems with two equations and two variables or three equations and three variables are called square systems, meaning there are exactly as many equations as there are variables. A system of linear equations cannot have a unique solution unless there are at least as many equations as there are variables in the system. EXAMPLE 6



Solving a Dependent System Solve using elimination: e

Solution



2x  3y  z  5 . x  3y  2z  1

Using R1  R2 eliminates the y-term from R2, yielding x  z  4. This means (x, y, z) will satisfy both equations only when x  z  4 (the x-coordinate must be 4 less than the z-coordinate). Since x is written in terms of z, we substitute z  4 for x in either equation to find how y is related to z. Using R2 we have: 1z  42  3y  2z  1, which yields y  z  1 (verify). This means the y-coordinate of the solution must be 1 less than z. In set notation the solution is 5 1x, y, z,2 | x  z  4, y  z  1, z  6. For z  2, 0, and 3, the solutions would be 16, 3, 22, 14, 1, 02, and 11, 2, 32 , respectively. Verify that these satisfy both equations. Using p as our parameter, the solution could be written 1p  4, p  1, p2 in parameterized form. Now try Exercises 27 through 30



The system in Example 6 was nonsquare, and we knew ahead of time the system would be dependent. The system in Example 7 is square, but only by applying the elimination process can we determine the nature of its solution(s). EXAMPLE 7



Solving a Dependent System 3x  2y  z  1 Solve using elimination: • 2x  y  z  5 . 10x  2y 8

Solution



This system has no equation where the coefficient of x is 1. We will still use R1, but we’ll try to eliminate z in R2 (there is no z-term in R3). Using R1  R2 eliminates the z-term from R2, yielding 5x  y  4. 3x  2y  z  1 3x  2y  z  1 R1  R2 S R2 • 5x  y 4 • 2x  y  z  5 R3 S R3 S 10x  2y 8 10x  2y 8 We next solve the subsystem. Using 2R2  R3 eliminates the y-term in R3, but also all other terms: 2R2 10x  2y  8  10x  2y  8 R3 0x  0y  0 sum 0  0 result

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Since R3 is the same as 2R2, the system is linearly dependent and equivalent to 3x  2y  z  1 e . We can solve for y in R2 to write y in terms of x: y  5x  4. 5x  y 4 Substituting 5x  4 for y in R1 enables us to also write z in terms of x: R1 3x  2y  z  1 3x  215x  42  z  1 substitute 5x  4 distribute 3x  10x  8  z  1 simplify 7x  z  9 z  7x  9 solve for z

D. You’ve just learned how to recognize inconsistent and dependent systems

for y

The solution set is 51x, y, z2 | x  , y  5x  4, z  7x  926. Three of the infinite number of solutions are 10, 4, 92 for x  0, 12, 6, 52 for x  2, and 11, 9, 162 for x  1. Verify these triples satisfy all three equations. Again using the parameter p, the solution could be written as 1p, 5p  4, 7p  92 in parameterized form. Now try Exercises 31 through 34



Solutions to linearly dependent systems can actually be written in terms of either x, y, or z, depending on which variable is eliminated in the first step and the variable we elect to solve for afterward. For coincident dependence the equations in a system differ by only a constant multiple. After applying the elimination process — all variables are eliminated from the other equations, leaving statements that are always true (such as 2  2 or some other). See Exercises 35 and 36. For additional practice solving various kinds of systems, see Exercises 37 to 51.

E. Applications Applications of larger systems are simply an extension of our work with systems of two equations in two variables. Once again, the applications come in a variety of forms and from many fields. In the world of business and finance, systems can be used to diversify investments or spread out liabilities, a financial strategy hinted at in Example 8. EXAMPLE 8



Modeling the Finances of a Business A small business borrowed $225,000 from three different lenders to expand their product line. The interest rates were 5%, 6%, and 7%. Find how much was borrowed at each rate if the annual interest came to $13,000 and twice as much was borrowed at the 5% rate than was borrowed at the 7% rate.

Solution



Let x, y, and z represent the amount borrowed at 5%, 6%, and 7%, respectively. This means our first equation is x  y  z  225 (in thousands). The second equation is determined by the total interest paid, which was $13,000: 0.05x  0.06y  0.07z  13. The third is found by carefully reading the problem. “twice as much was borrowed at the 5% rate than was borrowed at the 7% rate”, or x  2z. x  y  z  225 These equations form the system: • 0.05x  0.06y  0.07z  13. The x-term of x  2z the first equation has a coefficient of 1. Written in standard form we have:

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x  y  z  225 R1 • 5x  6y  7z  1300 R2 (multiplied by 100) x  2z  0 R3 Using 5R1  R2 will eliminate the x term in R2, while R1  R3 will eliminate the x-term in R3. 5R1 5x  5y  5z  1125  R2 5x  6y  7z  1300 y  2z  175

R1  R3

x  y  z  225 x

 2z  0 y  3z  225

The new R2 is y  2z  175, and the new R3 (after multiplying by 1) is x  y  z  225 y  3z  225, yielding the equivalent system • y  2z  175. y  3z  225 E. You’ve just learned how to use a system of three equations in three variables to solve applications

Solving the 2  2 subsystem using R2  R3 yields z  50. Back-substitution shows y  75 and x  100, yielding the solution (100, 75, 50). This means $50,000 was borrowed at the 7% rate, $75,000 was borrowed at 6%, and $100,000 at 5%. Now try Exercises 54 through 63



TECHNOLOGY HIGHLIGHT

More on Parameterized Solutions For linearly dependent systems, a graphing calculator can be used to both find and check possible solutions using the parameters Y1, Y2, and Y3. This is done by assigning the chosen parameter to Y1, then using Y2 and Y3 to form the other coordinates of the solution. We can then build the equations in the system using Y1, Y2, and Y3 in place of x, y, and z. The system from Example 7 is 3x  2y  z  1 • 2x  y  z  5 , which we found had solutions of the form 1x, 5x  4, 7x  92. We first form the 10x  2y 8 solution using Y1  X, Y2  5Y1  4 (for y), and Y3  7Y1  9 (for z). Then we form the equations in the system using Y4  3Y1  2Y2  Y3, Y5  2Y1  Y2  Y3 , and Y6  10Y1  2Y2 (see Figure 8.8). After setting up the table (set on AUTO), solutions can be found by enabling only Y1, Y2, and Y3, which gives values of x, y, and z, respectively (see Figure 8.9—use the right arrow to view Y3). By enabling Y4, Y5, and Y6 you can verify that for any value of the parameter, the first equation is equal to 1, the second is equal to 5, and the third is equal to 8 (see Figure 8.10—use the right arrow to view Y6).

Figure 8.8

Figure 8.9

Figure 8.10

Exercise 1: Use the ideas from this Technology Highlight to (a) find four specific solutions to Example 6, (b) check multiple variations of the solution given, and (c) determine if 19, 6, 52, 12, 1, 22, and (6, 2, 4) are solutions.

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8.2 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. The solution to an equation in three variables is an ordered . 2. The graph of the solutions to an equation in three variables is a(n) . 3. Systems that have the same solution set are called . 

4. If a 3  3 system is linearly dependent, the ordered triple solutions can be written in terms of a single variable called a(n) . 5. Find a value of z that makes the ordered triple 12, 5, z2 a solution to 2x  y  z  4. Discuss/Explain how this is accomplished. 6. Explain the difference between linear dependence and coincident dependence, and describe how the equations are related.

DEVELOPING YOUR SKILLS

Find any four ordered triples that satisfy the equation given.

7. x  2y  z  9 9. x  y  2z  6

8. 3x  y  z  8 10. 2x  y  3z  12

Determine if the given ordered triples are solutions to the system.

x  y  2z  1 11. • 4x  y  3z  3 ; 3x  2y  z  4

(0, 3, 2) (3, 4, 1)

2x  3y  z  9 12. • 5x  2y  z  32; 14, 5, 22 x  y  2z  13 15, 4, 112 Solve each system using elimination and back-substitution.

x  y  2z  10 13. • x  z1 z4

x  y  2z  1 14. • 4x  y 3 3x 6

x  3y  2z  16 x  y  5z  1 15. • 2y  3z  1 16. • 4x  y 1 8y  13z  7 3x  2y 8 2x  y  4z  7 2x  3y  4z  18 17. • x  2y  5z  13 18. • x  2y  z  4 y  4z  9 4x  z  19 x  y  2z  10 x  y  2z 1 19. • x  y  z  7 20. • 4x  y  3z  3 2x  y  z  5 3x  2y  z  4

3x  y  2z  3 2x  3y  2z  0 21. • x  2y  3z  10 22. • 3x  4y  z  20 4x  8y  5z  5 x  2y  z  16 3x  y  z  6 23. • 2x  2y  z  5 2x  y  z  5

2x  3y  2z  7 24. • x  y  2z  5 2x  2y  3z  7

Solve using the elimination method. If a system is inconsistent or dependent, so state. For systems with linear dependence, write solutions in set notation and as an ordered triple in terms of a parameter.

3x  y  2z  3 2x  y  3z  8 25. • x  2y  3z  1 26. • 3x  4y  z  4 4x  8y  12z  7 4x  2y  6z  5 27. e

4x  y  3z  8 4x  y  2z  9 28. e x  2y  3z  2 3x  y  5z  5

29. e

6x  3y  7z  2 3x  4y  z  6

30. e

2x  4y  5z  2 3x  2y  3z  7

Solve using elimination. If the system is linearly dependent, state the general solution in terms of a parameter. Different forms of the solution are possible.

3x  4y  5z  5 31. •x  2y  3z  3 3x  2y  z  1 5x  3y  2z  4 32. • 9x  5y  4z  12 3x  y  2z  12

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x  2y  3z  1 33. • 3x  5y  8z  7 x  y  2z  5

x  5y  4z  3 42. • 2x  9y  7z  2 3x  14y  11z  5

2x  3y  5z  3 34. • 5x  7y  12z  8 x  y  2z  2

2x  3y  5z  4 43. • x  y  2z  3 x  3y  4z  1

Solve using elimination. If the system has coincident dependence, state the solution in set notation.

0.2x  1.2y  2.4z  1 35. • 0.5x  3y  6z  2.5 x  6y  12z  5

44.

6x  3y  9z  21 36. • 4x  2y  6z  14 2x  y  3z  7

x  2y  z  1 37. • x  z3 2x  y  z  3

u

1 1 1 x y z2 6 3 2 1 1 3 x y z9 4 3 2 1 1 x y z2 2 2

45.

u

y x   2 3 2x y  3 x  2y  6

z 2 2 z8 3z 6 2

Some applications of systems lead to systems similar to those that follow. Solve using elimination.

Solve using the elimination method. If a system is inconsistent or dependent, so state. For systems with linear dependence, write the answer in terms of a parameter. For coincident dependence, state the solution in set notation.

3x  5y  z  11 38. • 2x  y  3z  12 y  2z  4

2x  5y  4z  6 39. • x  2.5y  2z  3 3x  7.5y  6z  9

46. •

2A  B  3C  21 B C1 A B  4

A  3B  2C  11 47. • 2B  C  9 B  2C  8 A 2C  7 48. • 2A  3B 8 3A  6B  8C  33 A  2B 5 49. • B  3C  7 2A  B  C  1

x  2y  2z  6 40. • 2x  6y  3z  13 3x  4y  z  11

C  2 50. • 5A  2C  5 4B  9C  16

4x  5y  6z  5 41. • 2x  3y  3z  0 x  2y  3z  5



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51. • 2A 

C3 3C  10 3B  4C  11

WORKING WITH FORMULAS

52. Dimensions of a rectangular solid: 2w  2h  P1 P2  16 cm (top) • 2l  2w  P2 P1 14 cm P3 18 cm (small side) 2l  2h  P3 h (large side)

w

l

Using the formula shown, the dimensions of a rectangular solid can be found if the perimeters of the three distinct faces are known. Find the dimensions of the solid shown.

53. Distance from a point (x, y, z) to the plane Ax  By  Cz  D Ax  By  Cz  D: ` ` 2A2  B2  C2 The perpendicular distance from a given point (x, y, z) to the plane defined by Ax  By  Cz  D is given by the formula shown. Consider the plane given in Figure 8.2 1x  y  z  62. What is the distance from this plane to the point (3, 4, 5)?

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APPLICATIONS

Solve the following applications by setting up and solving a system of three equations in three variables. Note that some equations may have only two of the three variables used to create the system. Investment/Finance and Simple Interest Problems

54. Investing the winnings: After winning $280,000 in the lottery, Maurika decided to place the money in three different investments: a certificate of deposit paying 4%, a money market certificate paying 5%, and some Aa bonds paying 7%. After 1 yr she earned $15,400 in interest. Find how much was invested at each rate if $20,000 more was invested at 7% than at 5%. 55. Purchase at auction: At an auction, a wealthy collector paid $7,000,000 for three paintings: a Monet, a Picasso, and a van Gogh. The Monet cost $800,000 more than the Picasso. The price of the van Gogh was $200,000 more than twice the price of the Monet. What was the price of each painting? Descriptive Translation

56. Major wars: The United States has fought three major wars in modern times: World War II, the Korean War, and the Vietnam War. If you sum the years that each conflict ended, the result is 5871. The Vietnam War ended 20 years after the Korean War and 28 years after World War II. In what year did each end? 57. Animal gestation periods: The average gestation period (in days) of an elephant, rhinoceros, and camel sum to 1520 days. The gestation period of a rhino is 58 days longer than that of a camel. Twice the camel’s gestation period decreased by 162 gives the gestation period of an elephant. What is the gestation period of each? 58. Moments in U.S. history: If you sum the year the Declaration of Independence was signed, the year the 13th Amendment to the Constitution abolished slavery, and the year the Civil Rights Act was signed, the total would be 5605. Ninety-nine years separate the 13th Amendment and the Civil Rights Act. The Civil Rights Act was signed 188 years after the Declaration of Independence. What year was each signed? 59. Aviary wingspan: If you combine the wingspan of the California Condor, the Wandering Albatross (see photo), and the prehistoric Quetzalcoatlus, you get an astonishing 18.6 m (over 60 ft). If the wingspan of the Quetzalcoatlus is equal to five times that of the Wandering Albatross minus twice

that of the California Condor, and six times the wingspan of the Condor is equal to five times the wingspan of the Albatross, what is the wingspan of each? Mixtures

60. Chemical mixtures: A chemist mixes three different solutions with concentrations of 20%, 30%, and 45% glucose to obtain 10 L of a 38% glucose solution. If the amount of 30% solution used is 1 L more than twice the amount of 20% solution used, find the amount of each solution used. 61. Value of gold coins: As part of a promotion, a local bank invites its customers to view a large sack full of $5, $10, and $20 gold pieces, promising to give the sack to the first person able to state the number of coins for each denomination. Customers are told there are exactly 250 coins, with a total face value of $1875. If there are also seven times as many $5 gold pieces as $20 gold pieces, how many of each denomination are there? 62. Rewriting a rational function: It can be shown that the rational function V1x2  3x  11 can be written as a sum of the 3 x  3x2  x  3 Bx  C A  2 terms , where the coefficients x3 x 1 A B 0 A, B, and C are solutions to • 3B  C  3 . A  3C  11 Find the missing coefficients and verify your answer by adding the terms. 63. Rewriting a rational function: It can be shown that the rational function V1x2  x9 can be written as a sum of the 3 x  6x2  9x B C A  terms  , where the x x3 1x  32 2 coefficients A, B, and C are solutions to A B 0 • 6A  3B  C  1 . Find the missing 9A  9 coefficients and verify your answer by adding the terms.

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EXTENDING THE CONCEPT

x  2y  z  2 64. The system • x  2y  kz  5 is inconsistent if 2x  4y  4z  10 , and dependent if k  . k

a. 9 cm d. 12 cm

b. 10 cm e. 13 cm

65. One form of the equation of a circle is x2  y2  Dx  Ey  F  0. Use a system to find the equation of the circle through the points 12, 12, 14, 32, and 12, 52. 66. The lengths of each side of the squares A, B, C, D, E, F, G, H, and I (the smallest square) shown are whole numbers. Square B has sides of 15 cm and square G has sides of 7 cm. What are the dimensions of square D? 

743

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c. 11 cm

B

A

G

H

F

C D

E

MAINTAINING YOUR SKILLS

67. (7.3) Given u  H1, 7I and v  H3, 12I, compute u  4v and 3u  v. 68. (5.2) Given cot A  1.6831, use a calculator to find the acute angle A to the nearest tenth of a degree. 69. (4.4) Solve the logarithmic equation: log1x  22  logx  log3 70. (2.5) Analyze the graph of g shown. Clearly state the domain and range, the zeroes of g, intervals

where g1x2 7 0, intervals where g1x2 6 0, local maximums or minimums, and intervals where the function is increasing or decreasing. Assume each tick mark is one unit and estimate endpoints to the nearest tenths.

y

x

8.3 Partial Fraction Decomposition Learning Objectives In Section 8.3 you will learn how to:

A. Set up a decomposition template to help rewrite a rational expression as the sum of its partial fractions

B. Decompose a rational expression using convenient values

C. Decompose a rational expression by equating coefficients and using a system of equations

D. Apply partial fraction decomposition to a telescoping sum

One application of linear systems often seen in higher mathematics involves rewriting a rational expression as a sum of its partial fractions. While most often used as a prelude to the application of other mathematical techniques, practical applications of the process range from a study of chemical reactions to the analysis of thermodynamic experiments.

A. Setting Up a Decomposition Template P1x2 , where P and Q are polynomials Q1x2 and Q1x2  0. The addition of rational expressions is widely taught in courses prior to college algebra, and involves combining two rational expressions into a single term using a common denominator. For the decomposition of rational expressions, we seek to reverse this process. To begin, we make the following observations:

Recall that a rational expression is one of the form

5 7  , noting both terms are proper fractions (the x2 x3 degree of the numerator is less than the degree of the denominator) and have distinct linear denominators.

1. Consider the sum

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71x  32 51x  22 7 5    x2 x3 1x  22 1x  32 1x  321x  22 71x  32  51x  22  1x  221x  32 12x  11  1x  22 1x  32

common denominators

combine numerators

result

Assuming we didn’t have the original sum to look at, reversing the process would require us to begin with the template A B 12x  11   1x  221x  32 x2 x3 and solve for the constants A and B. We know the numerators must be constant, otherwise the fraction(s) would be improper while the original expression is not. 5 3 2. Consider the sum  2 , again noting both terms are proper x1 x  2x  1 fractions. 3 5 5 3  2   x1 x1 1x  12 1x  12 x  2x  1 31x  12 5   1x  12 1x  12 1x  121x  12 13x  32  5  1x  12 1x  12 3x  2  1x  12 2

factor denominators

common denominators

combine numerators

result

Note that while the new denominator is the repeated factor 1x  12 2, both 1x  12 and 1x  12 2 were denominators in the original sum. Assuming we didn’t know the original sum, reversing the process would require us to begin with the template 3x  2 A B   x1 1x  12 2 1x  12 2 and solve for the constants A and B. As before, the numerator of the first term must be constant. While the second term would still be a proper fraction if the numerator were linear (degree 1), the denominator is a repeated linear factor and using a single constant in the numerator of all such fractions will ensure we obtain unique values for A and B (see Exercise 58). Note that for any repeated linear factor 1x  a2 n in the original denominator, terms of the form A3 An1 An A2 A1 ...  must appear in 2  3  n1  xa 1x  a2 n 1x  a2 1x  a2 1x  a2 the decomposition template, although some of these numerators may turn out to be zero. EXAMPLE 1



Setting Up the Decomposition Template for Linear Factors Write the decomposition template for x1 x8 a. 2 b. 2 2x  5x  3 x  6x  9

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Solution



745

x8 . Since the denominator 12x  32 1x  12 consists of two distinct linear factors, the decomposition template is

a. Factoring the denominator gives

x8 A B   decomposition template 12x  321x  12 2x  3 x1 x1 b. After factoring the denominator we have , where the denominator 1x  32 2 consists of a repeated linear factor. Using our previous observations, the decomposition template would be x1 A B  2  x3 1x  32 1x  32 2

decomposition template

Now try Exercises 7 through 16



When both distinct and repeated linear factors are present in the denominator, the decomposition template maintains the elements illustrated above. Each distinct linear factor appearing in the denominator, and all powers of a repeated linear factor, will have a constant numerator. EXAMPLE 2



Setting Up the Decomposition Template for Repeated Linear Factors Write the decomposition template for

Solution



x2  4x  15 . x3  2x2  x

x2  4x  15 x2  4x  15 or after factoring 2 x1x  2x  12 x1x  12 2 completely. With a distinct linear factor of x, and the repeated linear factor 1x  12 2, the decomposition template becomes Factoring the denominator gives

B A C x2  4x  15    2 x x1 x1x  12 1x  12 2

decomposition template

Now try Exercises 17 through 20



Returning to our observations: 4 2x  3 , noting the denominator of the first term is linear, 3. Consider the sum  2 x x 1 while the denominator of the second is an irreducible quadratic. 41x2  12 12x  32x 4 2x  3  2   2 2 x x 1 x1x  12 1x  12x 2 14x  42  12x2  3x2  x1x2  12 2 6x  3x  4  x1x2  12

find common denominator

combine numerators

result

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Here, reversing the process would require us to begin with the template 6x2  3x  4 Bx  C A   2 2 x x1x  12 x 1 allowing that the numerator of the second term might be linear since it is quadratic but not a repeated linear factor, and noting the fraction would still be proper in cases where B  0. 1 x2 , where we note the denominator of 4. Finally, consider the sum 2  2 x 3 1x  32 2 the first term is an irreducible quadratic, with the denominator of the second term being the same factor with multiplicity two. 11x2  32 1 x2 x2    2 2 2 2 2 2 x 3 1x  32 1x  32 1x  32 1x  321x2  32 1x2  32  1x  22  1x2  321x2  32 2 x x1  1x2  32 2

WORTHY OF NOTE Note that the second term in the decomposition template would still be a proper fraction if the numerator were quadratic or cubic, but since the denominator is a repeated quadratic factor, using only a linear form ensures we obtain unique values for all coefficients.

EXAMPLE 3

common denominators

combine numerators

result after simplifying

Reversing the process would require us to begin with the template x2  x  1 Ax  B Cx  D  2 2 2  2 1x  32 x 3 1x  32 2 allowing that the numerator of either term might be nonconstant for the reasons in observation 3. Similar to our reasoning in observation 2, all powers of a repeated quadratic factor must be present in the decomposition template. 

Setting Up the Decomposition Template for Quadratic Factors Write the decomposition template for x2  10x  1 x2 a. b. 1x  121x2  3x  42 1x2  22 3

Solution



a. With the denominator having one distinct linear factor and one irreducible quadratic factor, the decomposition template would be x2  10x  1 Bx  C A  2  2 x1 1x  121x  3x  42 x  3x  4

decomposition template

b. The denominator consists of a repeated quadratic factor. Using our previous observations, the decomposition template would be Ax  B Ex  F x2 Cx  D  2 2 3  2 2  1x  22 x 2 1x  22 1x2  22 3

decomposition template

Now try Exercises 21 through 24



When both distinct and repeated factors are present in the denominator, the decomposition template maintains the essential elements determined by observations 1 through 4. Using these observations, we can formulate a general approach to the decomposition template.

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The Decomposition Template For the rational expression

A. You’ve just learned how to set up a decomposition template

P1x2 and constants A, B, C, D, . . . , Q1x2

1. If the degree of P is greater than or equal to the degree of Q, find the quotient and remainder using polynomial division. Only the remainder portion need be decomposed into partial fractions. 2. Factor Q completely into linear factors and irreducible quadratic factors. 3. For the linear factors, each distinct linear factor and each power of a repeated linear factor must appear in the decomposition template and have a constant numerator. 4. For the irreducible quadratic factors, each distinct quadratic factor and each power of a repeated quadratic factor must appear in the decomposition template and have a linear numerator.

B. Decomposition Using Convenient Values Once the decomposition template is obtained, we multiply both sides of the equation by the factored form of the original denominator and simplify. The resulting equation is an identity (a true statement for all real numbers x), and in many cases, all that’s required is a choice of convenient values for x to identify the constants A, B, C, and so on. EXAMPLE 4



Decomposing a Rational Expression Using Convenient Values Decompose the expression

Solution



4x  11 into partial fractions. x  7x  10 2

4x  11 , and we note there are two distinct 1x  521x  22 linear factors in the denominator. The decomposition template will be Factoring the denominator gives

A B 4x  11   1x  52 1x  22 x5 x2

decomposition template

Multiplying both sides by 1x  52 1x  22 clears all denominators and yields 4x  11  A1x  22  B1x  52

clear denominators

Since the equation must be true for all x, using x  5 will conveniently eliminate the term with B, and enable us to solve for A directly: 4152  11  A15  22  B15  52 20  11  3A  B102 9  3A 3A

substitute 5 for x simplify term with B is eliminated solve for A

To find B, we repeat this procedure, using a value that conveniently eliminates the term with A, namely x  2. 4x  11  A1x  22  B1x  52 4122  11  A12  22  B12  52 8  11  A102  3B 3  3B 1B

original equation substitute 2 for x simplify term with A is eliminated solve for B

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With A  3 and B  1, the complete decomposition is 4x  11 3 1   1x  521x  22 x5 x2 which can be checked by adding the fractions on the right. Now try Exercises 25 through 28

EXAMPLE 5



Decomposing a Rational Expression Using Convenient Values Decompose the expression

Solution





9 into partial fractions. 1x  52 1x  7x  102 2

9 9 or in 1x  521x  22 1x  52 1x  221x  52 2 simplified form. With one distinct linear factor and a repeated linear factor, the A B C 9   decomposition template will be . 2  x2 x5 1x  221x  52 1x  52 2 Multiplying both sides by 1x  22 1x  52 2 clears all denominators and yields Factoring the denominator gives

9  A1x  52 2  B1x  22 1x  52  C1x  22

Using x  5 will conveniently eliminate the terms with A and B, enabling us to solve for C directly: 9  A15  52 2  B15  22 15  52  C15  22 9  A102  B132 102  3C 9  3C 3  C

substitute 5 for x simplify terms with A and B are eliminated solve for C

Using x  2 will conveniently eliminate the terms with B and C, enabling us to solve for A: 9  A1x  52 2  B1x  221x  52  C1x  22 9  A12  52 2  B12  22 12  52  C12  22 9  A132 2  B102132  C102 9  9A 1A

original equation substitute 2 for x simplify terms with B and C are eliminated solve for A

To find the value of B, we can substitute A  1 and C  3 into the previous equation, and any value of x that does not eliminate B. For efficiency’s sake, we often elect to use x  0 or x  1 for this purpose (if possible). 9  A1x  52 2  B1x  221x  52  C1x  22 9  110  52 2  B10  22 10  52  310  22 9  25  10B  6 10  10B 1  B

original equation substitute 1 for A, 3 for C, 0 for x simplify

solve for B

With A  1, B  1, and C  3 , the complete decomposition is 9 1 3 1   , which can also be written as 2  x  2 x  5 1x  22 1x  52 1x  52 2 1 1 3   . x2 x5 1x  52 2 Now try Exercises 29 and 30



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Section 8.3 Partial Fraction Decomposition

EXAMPLE 6



Decomposing a Rational Expression Using Convenient Values Decompose the expression

Solution



749

3x  11 into partial fractions. x  3x2  x  3 3

After inspection, we note the denominator can be factored by grouping, resulting 3x  11 in the expression . With one distinct linear factor and one irreducible 1x  321x2  12 quadratic factor, the decomposition template will be Bx  C 3x  11 A  2  . x3 1x  321x2  12 x 1

Multiplying both sides by 1x  32 1x2  12 yields

3x  11  A1x2  12  1Bx  C2 1x  32

clear denominators

Using x  3 will conveniently eliminate the term with B and C, giving 3132  11  A132  12  1B 33 4  C213  32 20  10A  1B 33 4  C2 102 20  10A 2A

substitute 3 for x simplify term with B and C is eliminated solve for A

Noting x  0 will conveniently eliminate the term with B; we substitute 0 for x and 2 for A in order to solve for C: 3x  11  A1x2  12  1Bx  C2 1x  32 3102  11  2102  12  1B 30 4  C2 10  32 11  2  C132 9  3C 3  C

original equation substitute 2 for A and 0 for x simplify solve for C result

To find the value of B, we substitute 2 for A, 3 for C, and any value of x that does not eliminate B. Here we chose x  1. 3x  11  A1x2  12  1Bx  C2 1x  32 3112  11  2112  12  1B 31 4  3 34 2 11  32 14  2122  1B  32122 14  4  2B  6 4  2B 2  B

original equation substitute 2 for A, 3 for C, and 1 for x simplify distribute solve for B result

With A  2, B  2, and C  3, the complete decomposition is

B. You’ve just learned how to decompose a rational expression using convenient values

2x  3 2 3x  11 .  2  2 x3 1x  32 1x  12 x 1 The result can be actually be written with fewer negative signs, as 2 2x  3  2 . Check the result by combining these fractions. x3 x 1 Now try Exercises 31 through 38



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C. Decomposition Using a System of Equations As an alternative to using convenient values, a system of equations can be set up by multiplying out the right-hand side (after clearing fractions) and equating coefficients of the terms with like degrees. Here we’ll re-solve Example 6 using this method. EXAMPLE 7



Decomposing a Rational Expression Using a System of Equations 3x  11 into partial fractions by equating x3  3x2  x  3 coefficients of like degree terms and using a system. Decompose the expression

Solution



From Example 6 we obtain A Bx  C 3x  11   2 2 x3 1x  321x  12 x 1

decomposition template

Multiplying both sides by 1x  32 1x2  12 yields

3x  11  A1x2  12  1Bx  C21x  32  Ax2  A  Bx2  3Bx  Cx  3C  1A  B2x2  13B  C2x  1A  3C2

clear denominators distribute/FOIL collect like terms

By comparing the like terms on the left and right, we find A  B  0, AB 0 3B  C  3, and A  3C  11, resulting in the system • 3B  C  3 . A  3C  11 3B  C  3 Using A  B (from R1) in R3 results in the 2  2 subsystem e , B  3C  11 and 3R1  R2 of this system yields 10B  20, giving B  2 as before. Backsubstitution again verifies C  3 and A  2. Now try Exercises 39 and 40



In some cases, there are no “convenient values” to use and a system of equations is our best and only approach. More often than not, this happens when one or more of the denominators are irreducible quadratic factors. EXAMPLE 8



Decomposing a Rational Expression Using a System of Equations 5x2  2x  12 into partial fractions by equating x4  6x2  9 coefficients of like degree terms and using a system. Decompose the expression

Solution



By inspection or using a u-substitution, we find the denominator factors into a perfect square: x4  6x2  9  1x2  32 2. The decomposition template is then Cx  D Ax  B 5x2  2x  12  2  2 2 2 1x  32 x 3 1x  32 2

After clearing fractions we obtain 5x2  2x  12  1Ax  B21x2  32  Cx  D. The only “candidate” for a convenient value is x  0, but this still leaves the unknowns B and D. Instead, we multiply out the right-hand side and equate coefficients of like terms as before. 5x2  2x  12  1Ax  B2 1x2  32  Cx  D  Ax3  3Ax  Bx2  3B  Cx  D  Ax3  Bx2  13A  C2x  13B  D2

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By equating coefficients we find A
0, since the left-hand side has no cubic term, and B
5 by direct comparison. This yields the system A
0 B
5 µ 3A  C
2 3B  D
12 C. You’ve just learned how to decompose a rational expression by equating coefficients and using a system of equations

With A
0 the third equation shows C
2, and by substituting 5 for B in the fourth equation we find that D
3. The final form of the decomposition is 5 5x2  2x  12 2x  3
2  2 2 2 1x  32 x 3 1x  32 2 Now try Exercises 41 through 46




D. Partial Fractions and Telescoping Sums From movies or the popular media (Pirates of the Caribbean, Dances With Wolves, etc), you might be aware that the telescopes of old were retractable. Since they were constructed as a series of nested tubes, you could compress the length with the lengths of the interior tubes being “negated” (see figure). In a similar fashion, some very extensive sums, called telescoping sums, can be rewritten in a more “compressed” form, where the interior terms are likewise negated and the resulting sum easily computed. EXAMPLE 9




Using Decomposition to Evaluate a Telescoping Sum 3 x x Evaluate the function for x
1, 2, 3, and 4, then find the sum of these four terms. Decompose the expression into partial fractions. Evaluate the decomposed form for x
1, 2, 3, and 4, leaving each result in unsimplified form. Rewrite the sum from part (a) using the decomposed form of each term, and see if you can detect a pattern that enables you to compute the original sum more efficiently. Use the pattern noted in part (d) to find the following sum: 3 3 3 3 .  #  # p 1#2 2 3 3 4 24 # 25 3 3 3 3 3 1 1 3   
   11 42 93 16  4 2 2 4 20

For t1x2
a. b. c. d.

e.

Solution




a.

2

30 10 5 3    20 20 20 20 48 12

20 5




b. The denominator has two distinct linear factors and the decomposition template will be B 3 A
 x x1x  12 x1

decomposition template

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Multiplying both sides by x1x  12 clears all denominators and gives 3  A1x  12  Bx

clear denominators

Since the equation must be true for all x, using x 
1 will eliminate the term with A, and enable us to solve for B directly: 3  A1
1  12  B1
12 3 
B
3  B

substitute
1 for x simplify, term with A is eliminated solve for B

Repeat this procedure using x  1 to solve for A. 3  A1x  12  Bx 3  A11  12
3112 6  2A 3A

from template substitute
3 for B, 1 for x add 3 to both sides result

With A  3 and B 
1, the decomposition is 3 3 3 
x x1x  12 x1

decomposed form

3 3 , we have
x x1 3 3 3 3 3 3 3 3 t112 
t122 
t132 
t142 
1 11 2 21 3 31 4 41 3 3 3 3 3 3 3 3



2 2 3 3 4 4 5 d. Replacing each term in the given sum by the equivalent term in decomposed form yields c. For t1x2 

3 3 3 3 3 3 3 3 3 3 3 3    a
ba
ba
ba
b 11 42 93 16  4 1 2 2 3 3 4 4 5 and we note that all interior terms will cancel (add to zero) leaving only the first 3 15 3 12 3
 . and last terms of the sum. The result is
 1 5 5 5 5 e. From the pattern in part (d) we note that regardless of the number of terms in the sum, all interior terms will cancel (add to zero) leaving only the first and last terms. 3 3 3 3 3 3 3 3 3 3 3 3 a
ba
ba
b...a
b  #  # ... 24 # 25 1#2 2 3 3 4 1 2 2 3 3 4 24 25 3 75 3 3

1 25 25 25 72 The result can be checked using a graphing calculator.  25



Now try Exercises 51 through 54

D. You’ve just learned how to apply partial fraction decomposition to a telescoping sum




As a final note, if the degree of the numerator is greater than the degree of the denominator in the original expression, divide using long division and apply the methods above to the remainder polynomial. For instance, you can check that 3x3  6x2  5x
7 2x
7 , and decomposing the remainder polynomial  3x  2 x  2x  1 1x  12 2 2 9 gives a final result of 3x 
. x1 1x  12 2

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8.3 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. In order to rewrite a rational expression as the sum of its partial fractions, we must set up a decomposition . 2. Before beginning the process of partial fraction decomposition, the rational expression must be . If not, use polynomial division. 3. If the denominator of a rational expression contains a factor, each power of the factor must appear in the decomposition template and have a constant numerator. 

4. If the denominator of a rational expression contains a distinct irreducible quadratic factor, it must appear in the decomposition template with a numerator. 8x  3 5. Discuss/Explain the process of writing 2 as a x x sum of partial fractions. 6. Discuss/Explain the first steps of the process of x2 writing 2 as a sum of partial fractions. x  7x  12

DEVELOPING YOUR SKILLS

The exercises below are designed solely to reinforce the various possibilities for decomposing a rational expression. All are proper fractions whose denominators are completely factored. Set up the partial fraction decomposition using appropriate numerators, but do not solve.

7.

3x  2 1x  32 1x  22

8.

4x  1 1x  22 1x  52

22. 23.

7x2  3x  1 1x  12 1x2  2x  52 x3  3x  2 1x  121x2  22 2

24.

2x3  3x2  4x  1 x1x2  32 2

Decompose each rational expression into partial fractions using convenient values.

25.

2x  27 2x  x  15

26.

11x  6 5x  4x  12

27.

8x2  3x  7 x3  x

28.

x2  24x  12 x3  4x

2x2  3x  4 12. 1x  32 1x  121x  22

29.

3x2  7x  1 x3  2x2  x

30.

2x2  7x  28 x3  4x2  4x

x2  5 13. x1x  321x  12

31.

2x3  2x2  6x 3x3  3x2  3x  5 32. x4  3x2  2 x4  4x2  3

33.

6x2  x  13 x3  2x2  3x  6

34.

3x2  4x  1 x3  1

35.

x4  3x2  2x  1 x5  2x3  x

36.

3x4  11x2  x  12 x5  4x3  4x

37.

3x3  2x2  7x  3 2x3  x  6 38. x4  x3  3x2 x4  2x3  3x2

39.

3x2  10x  4 8  x3

9.

2x  5 1x  12 2

10.

3x2  2x  5 11. 1x  12 1x  221x  32

x2  2x  4 15. 1x  52 3 x2  x  1 17. 2 x 1x  22

x3  2x  5 19. 2 x 1x  52 2

2x2  3 21. 1x  32 1x2  5x  72

x7 1x  32 2

x2  7 14. 1x  421x  22x x2  2x  3 16. 1x  42 3

x2  3x  5 18. 1x  321x  22 2 3x2  5 20. 2 x 12x  12 2

2

40.

2

2x2  14x  7 x  2x2  5x  10 3

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When the denominator of the rational expression contains repeated factors, using a system of equations sometimes offers a more efficient way to find the needed coefficients. Decompose each rational expression into partial fractions by equating coefficients and using a system of equations.

41.

5x  13 2 x  6x  9

42.

14  3x 2 x  8x  16

43.

2x3  x2  5x  1 x4  2x2  1

44.

x3  5x2  6x  37 x4  14x2  49

45.

2x2  4x  5 x3  3x2  3x  1

46.

5x2  20x  21 x3  6x2  12x  8



WORKING WITH FORMULAS

Logistics equations and population size: Logistics equations are often used to model the growth 1 of various populations. Initially growth is very near to exponential, but then due to certain limiting factors (food, limited resources, space, etc.), growth slows and reaches a limit called the carrying PaP102  P102 Pb c capacity c. In the process of solving the logistic equation for P(t), we often need to decompose the expression shown into partial fractions, where P(0) represents the initial population. Decompose the expression for the following values given. 47. P102  100, c  10



48. P102  80, c  20

49. P102  10, c  100

50. P102  20, c  80

APPLICATIONS

Telescoping sums: For each function given, use the method illustrated in Example 9 to find a pattern that enables you to compute the sum shown quickly. 1 ; x x 1 1 1 1  #  # p # 1 2 2 3 3 4 49 # 50

53. t1x2 

2 ; x x 2 2 2 2  #  # p # 1 2 2 3 3 4 19 # 20

54. t1x2 

51. t1x2 

52. t1x2 



2

2

1 ; 12x  1212x  12 1 1 1 1   p # # # 1 3 3 5 5 7 123 # 125 2x  1 ; 1x  12 1x2  2x  22 3 5 21 1  #  # p # 1 2 2 5 5 10 101 # 122 2

EXTENDING THE CONCEPT

Decompose the following expressions into partial fractions.

55.

ln x  2 1ln x  22 1ln x  12 2

56.

1  ex 1ex  32 1e2x  2ex  12

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Section 8.4 Systems of Inequalities and Linear Programming

x2 57. As written, the rational expression given cannot be 1x  12 11  x2 decomposed using the standard template (two distinct linear factors). Try to apply this template and see what happens. What can be done to decompose the expression?



755

58. Try to decompose the rational expression 3x  1 using the decomposition template x2 Bx  C A  (note the second rational expression x x2 in this template is a proper fraction). (a) How does this affect the decomposition process? Is the decomposition unique? (b) Next, use the standard template as outlined in this section. What do you notice?

MAINTAINING YOUR SKILLS

59. (3.2) Use polynomial division to rewrite r1x2 2x3  3x2  13x  5 in the form q1x2  . 2 d1x2 x x6 60. (3.2) Use the remainder theorem to find f(3) for f 1x2  2x3  x2  10. 61. (6.2) Verify that

62. (6.5) Evaluate the following expressions: 13 bd a. cos c cos1a 2  b. cos1 c cosa b d 6

cos3  cot   cos  sin  is an sin 

identity.

8.4 Systems of Inequalities and Linear Programming Learning Objectives In Section 8.4 you will learn how to:

A. Solve a linear inequality in two variables

B. Solve a system of linear

In this section, we’ll build on many of the ideas from Section 8.3, with a more direct focus on systems of linear inequalities. While systems of linear equations have an unlimited number of applications, there are many situations that can only be modeled using linear inequalities. For example, many decisions in business and industry are based on a large number of limitations or constraints, with many different ways these constraints can be satisfied.

inequalities

C. Solve applications using a system of linear inequalities

D. Solve applications using linear programming

A. Linear Inequalities in Two Variables A linear equation in two variables is any equation that can be written in the form Ax  By  C, where A and B are real numbers, not simultaneously equal to zero. A linear inequality in two variables is similarly defined, with the “  ” sign replaced by the “ 6 ,” “ 7 ,” “  ,” or “  ” symbol: Ax  By 6 C Ax  By  C

Ax  By 7 C Ax  By  C

Solving a linear inequality in two variables has many similarities with the one variable case. For one variable, we graph the boundary point on a number line, decide whether the endpoint is included or excluded, and shade the appropriate half line. For x  1  3, we have the solution x  2 with the endpoint included and the line shaded to the left (Figure 8.11):

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Figure 8.11

 3

Figure 8.12

 2

1

[ 0

2

1

3

Interval notation: x  (, 2]

y 5

For linear inequalities in two variables, we graph a boundary line, decide whether the boundary line is included or excluded, and shade the appropriate half plane. For x  y  3, the boundary line x  y  3 is graphed in Figure 8.12. Note it divides the coordinate plane into two regions called half planes, and it forms the boundary between the two regions. If the boundary is included in the solution set, we graph it using a solid line. If the boundary is excluded, a dashed line is used. Recall that solutions to a linear equation are ordered pairs that make the equation true. We use a similar idea to find or verify solutions to linear inequalities. If any one point in a half plane makes the inequality true, all points in that half plane will satisfy the inequality.

(0, 3)

xy3

Upper half plane

5

(4, 1)

Lower half plane

5

x

5

EXAMPLE 1



Checking Solutions to an Inequality in Two Variables Determine whether the given ordered pairs are solutions to x  2y 6 2: a. 14, 32 b. 12, 12

Solution



a. Substitute 4 for x and 3 for y: 142  2132 10 14, 32 is a solution. b. Substitute 2 for x and 1 for y: 122  2112 4 12, 12 is not a solution.

6 2 6 2

substitute 4 for x, 3 for y

6 2 6 2

substitute 2 for x, 1 for y

true

false

Now try Exercises 7 through 10



WORTHY OF NOTE

Earlier we graphed linear equations by plotting a small number of ordered pairs or by solving for y and using the slope-intercept method. The line represented all ordered pairs that made the equation true, meaning the left-hand expression was equal to the right-hand expression. To graph linear inequalities, we reason that if the line represents all ordered pairs that make the expressions equal, then any point not on that line must make the expressions unequal—either greater than or less than. These ordered pair solutions must lie in one of the half planes formed by the line, which we shade to indicate the solution region. Note this implies the boundary line for any inequality is determined by the related equation, temporarily replacing the inequality symbol with an “” sign.

This relationship is often called the trichotomy axiom or the “three-part truth.” Given any two quantities, they are either equal to each other, or the first is less than the second, or the first is greater than the second.

EXAMPLE 2



Solving an Inequality in Two Variables Solve the inequality x  2y  2.

Solution



The related equation and boundary line is x  2y  2. Since the inequality is inclusive (less than or equal to), we graph a solid line. Using the intercepts, we graph the line through (0, 1) and 12, 02 shown in Figure 8.13. To determine the solution region and which side to shade, we select (0, 0) as a test point, which results in a true statement: 102  2102  2✓. Since (0, 0) is in the “lower” half plane, we shade this side of the boundary (see Figure 8.14).

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Figure 8.13

Figure 8.14

y

y

5

Upper half plane

5

(4, 3)

(4, 3)

(0, 1)

(0, 1) (2, 0)

(2, 0) 5

x  2y  2

(0, 0) Test point

5

Lower half plane

x

5

5

x  2y  2

5

x

(0, 0) Test point

5

Now try Exercises 11 through 14

A. You’ve just learned how to solve a linear inequality in two variables



The same solution would be obtained if we first solve for y and graph the boundary line using the slope-intercept method. However, using the slope-intercept method offers a distinct advantage — test points are no longer necessary since solutions to “less than” inequalities will always appear below the boundary line and solutions to “greater than” inequalities appear above the line. Written in slope-intercept form, the inequality from Example 2 is y  12x  1. Note that (0, 0) still results in a true statement, but the “less than or equal to” symbol now indicates directly that solutions will be found in the lower half plane. This observation leads to our general approach for solving linear inequalities: Solving a Linear Inequality 1. Graph the boundary line by solving for y and using the slope-intercept form. • Use a solid line if the boundary is included in the solution set. • Use a dashed line if the boundary is excluded from the solution set. 2. For “greater than” inequalities shade the upper half plane. For “less than” inequalities shade the lower half plane.

B. Solving Systems of Linear Inequalities To solve a system of inequalities, we apply the procedure outlined above to all inequalities in the system, and note the ordered pairs that satisfy all inequalities simultaneously. In other words, we find the intersection of all solution regions (where they overlap), which then represents the solution for the system. In the case of vertical boundary lines, the designations “above” or “below” the line cannot be applied, and instead we simply note that for any vertical line x  k, points with x-coordinates larger than k will occur to the right. EXAMPLE 3

Solution



Solving a System of Linear Inequalities 2x  y  4 Solve the system of inequalities: e . xy 6 2



Solving for y, we obtain y  2x  4 and y 7 x  2. The line y  2x  4 will be a solid boundary line (included), while y  x  2 will be dashed (not included). Both inequalities are “greater than” and so we shade the upper half plane for each. The regions overlap and form the solution region (the lavender region shown). This sequence of events is illustrated here:

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CHAPTER 8 Systems of Equations and Inequalities Shade above y  x  2 (in pink) y

Shade above y  2x  4 (in blue) y 5

Overlapping region y

5

2x  y  4

5

2x  y  4

Solution region

2x  y  4 xy2

5

5

x

5

5

x

xy2 5

5

x

Corner point 5

5

5

The solutions are all ordered pairs found in this region and its included boundaries. To verify the result, test the point (2, 3) from inside the region, 15, 22 from outside the region (the point (2, 0) is not a solution since x  y 6 2). Now try Exercises 15 through 42 B. You’ve just learned how to solve a system of linear inequalities



For further reference, the point of intersection (2, 0) is called a corner point or vertex of the solution region. If the point of intersection is not easily found from the graph, we can find it by solving a linear system using the two lines. For Example 3, the system is e

2x  y  4 xy2

and solving by elimination gives 3x  6, x  2, and (2, 0) as the point of intersection.

C. Applications of Systems of Linear Inequalities Systems of inequalities give us a way to model the decision-making process when certain constraints must be satisfied. A constraint is a fact or consideration that somehow limits or governs possible solutions, like the number of acres a farmer plants — which may be limited by time, size of land, government regulation, and so on. EXAMPLE 4



Solving Applications of Linear Inequalities As part of their retirement planning, James and Lily decide to invest up to $30,000 in two separate investment vehicles. The first is a bond issue paying 9% and the second is a money market certificate paying 5%. A financial adviser suggests they invest at least $10,000 in the certificate and not more than $15,000 in bonds. What various amounts can be invested in each?

Solution



Consider the ordered pairs (B, C) where B represents the money invested in bonds and C the money invested in the certificate. Since they plan to invest no more than $30,000, the investment constraint would be B  C  30 (in thousands). Following the adviser’s recommendations, the constraints on C each investment would be B  15 and C  10. B  15 40 Since they cannot invest less than zero dollars, the Solution (0, 6) last two constraints are B  0 and C  0. region QI 30 B  C  30 B  15 20 (15, 15) C  10 C  10 10 B0 (0, 10) C0 (15, 10) The resulting system is shown in the figure, and 10 20 30 40 B indicates solutions will be in the first quadrant.

u

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C. You’ve just learned how to solve applications using a system of linear inequalities

759

There is a vertical boundary line at B  15 with shading to the left (less than) and a horizontal boundary line at C  10 with shading above (greater than). After graphing C  30  B, we see the solution region is a quadrilateral with vertices at (0, 10), (0, 30), (15, 10), and (15, 15), as shown. Now try Exercises 53 and 54



D. Linear Programming To become as profitable as possible, corporations look for ways to maximize their revenue and minimize their costs, while keeping up with delivery schedules and product demand. To operate at peak efficiency, plant managers must find ways to maximize productivity, while minimizing related costs and considering employee welfare, union agreements, and other factors. Problems where the goal is to maximize or minimize the value of a given quantity under certain constraints or restrictions are called programming problems. The quantity we seek to maximize or minimize is called the objective function. For situations where linear programming is used, the objective function is given as a linear function in two variables and is denoted f(x, y). A function in two variables is evaluated in much the same way as a single variable function. To evaluate f 1x, y2  2x  3y at the point (4, 5), we substitute 4 for x and 5 for y: f 14, 52  2142  3152  23. EXAMPLE 5



Determining Maximum Values Determine which of the following ordered pairs maximizes the value of f 1x, y2  5x  4y: (0, 6), (5, 0), (0, 0), or (4, 2).

Solution



Organizing our work in table form gives Given Point (0, 6) (5, 0) (0, 0) (4, 2)

Evaluate f 1x, y2  5x  4y

f 10, 62  5102  4162  24

f 15, 02  5152  4102  25

f 10, 02  5102  4102  0

f 14, 22  5142  4122  28

The function f 1x, y2  5x  4y is maximized at (4, 2). Now try Exercises 43 through 46

Figure 8.15

Convex

Not convex



When the objective is stated as a linear function in two variables and the constraints are expressed as a system of linear inequalities, we have what is called a linear programming problem. The systems of inequalities solved earlier produced a solution region that was either bounded (as in Example 4) or unbounded (as in Example 3). We interpret the word bounded to mean we can enclose the solution region within a circle of appropriate size. If we cannot draw a circle around the region because it extends indefinitely in some direction, the region is said to be unbounded. In this study, we will consider only situations that produce a bounded solution region, meaning the region will have three or more vertices. The regions we study will also be convex, meaning that for any two points in the feasible region, the line segment between them is also in the region (Figure 8.15). Under these conditions, it can be shown that the optimal solution(s) must occur at one of the corner points of the solution region, also called the feasible region.

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EXAMPLE 6



Finding the Maximum of an Objective Function

Solution



4 Begin by noting that the solutions must be in QI, (1, 3) 3 since x  0 and y  0. Graph the boundary lines Feasible 2 region y  x  4 and y  3x  6, shading the lower 1 half plane in each case since they are “less than” 5 4 3 2 1 1 2 3 4 5 x 1 inequalities. This produces the feasible region 2 shown in lavender. There are four corner points to this region: (0, 0), (0, 4), (2, 0), and (1, 3). Three of these points are intercepts and can be found quickly. The point (1, 3) was found by xy4 solving the system e . Knowing that the objective function will be 3x  y  6 maximized at one of the corner points, we test them in the objective function, using a table to organize our work.

Find the maximum value of the objective function f 1x, y2  2x  y given the xy4 y 3x  y  6 8 constraints shown: μ . x0 7 (0, 6) 6 y0 5

Corner Point (0, 0) (0, 4) (2, 0) (1, 3)

Objective Function f 1x, y2  2x  y

f 10, 02  2102  102  0

f 10, 42  2102  142  4

f 12, 02  2122  102  4

f 11, 32  2112  132  5

The objective function f 1x, y2  2x  y is maximized at (1, 3). Now try Exercises 47 through 50 Figure 8.16 y 8 7 6 5 4

(1, 3)

3 2 1 5 4 3 2 1 1

1

K1

2

3

4

5

K5 K3

x



To help understand why solutions must occur at a vertex, note the objective function f(x, y) is maximized using only (x, y) ordered pairs from the feasible region. If we let K represent this maximum value, the function from Example 6 becomes K  2x  y or y  2x  K, which is a line with slope 2 and y-intercept K. The table in Example 6 suggests that K should range from 0 to 5 and graphing y  2x  K for K  1, K  3, and K  5 produces the family of parallel lines shown in Figure 8.16. Note that values of K larger than 5 will cause the line to miss the solution region, and the maximum value of 5 occurs where the line intersects the feasible region at the vertex (1, 3). These observations lead to the following principles, which we offer without a formal proof. Linear Programming Solutions 1. If the feasible region is convex and bounded, a maximum and a minimum value exist. 2. If a unique solution exists, it will occur at a vertex of the feasible region. 3. If more than one solution exists, at least one of them occurs at a vertex of the feasible region with others on a boundary line. 4. If the feasible region is unbounded, a linear programming problem may have no solutions. Solving linear programming problems depends in large part on two things: (1) identifying the objective and the decision variables (what each variable represents

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761

in context), and (2) using the decision variables to write the objective function and constraint inequalities. This brings us to our five-step approach for solving linear programming applications. Solving Linear Programming Applications 1. Identify the main objective and the decision variables (descriptive variables may help) and write the objective function in terms of these variables. 2. Organize all information in a table, with the decision variables and constraints heading up the columns, and their components leading each row. 3. Complete the table using the information given, and write the constraint inequalities using the decision variables, constraints, and the domain. 4. Graph the constraint inequalities, determine the feasible region, and identify all corner points. 5. Test these points in the objective function to determine the optimal solution(s).

EXAMPLE 7



Solving an Application of Linear Programming The owner of a snack food business wants to create two nut mixes for the holiday season. The regular mix will have 14 oz of peanuts and 4 oz of cashews, while the deluxe mix will have 12 oz of peanuts and 6 oz of cashews. The owner estimates he will make a profit of $3 on the regular mixes and $4 on the deluxe mixes. How many of each should be made in order to maximize profit, if only 840 oz of peanuts and 348 oz of cashews are available?

Solution



Our objective is to maximize profit, and the decision variables could be r to represent the regular mixes sold, and d for the number of deluxe mixes. This gives P1r, d2  $3r  $4d as our objective function. The information is organized in Table 8.1, using the variables r, d, and the constraints to head each column. Since the mixes are composed of peanuts and cashews, these lead the rows in the table. Table 8.1  P1r, d2  $3r T Regular r

$4 d T Deluxe d

Constraints: Total Ounces Available

Peanuts

14

12

840

Cashews

4

6

348

After filling in the appropriate values, reading the table from left to right along the “peanut” row and the “cashew” row, gives the constraint inequalities 14r  12d  840 and 4r  6d  348. Realizing we won’t be making a negative number of mixes, the remaining constraints are r  0 and d  0. The complete system is 14r  12d  840 4r  6d  348 μ r0 d0 Note once again that the solutions must be in QI, since r  0 and d  0. Graphing the first two inequalities using slope-intercept form gives d  76r  70 and d  23r  58 producing the feasible region shown in lavender. The four corner

d 100 90 80 70 60 50 40 30 20

Feasible region

10 10 20 30 40 50 60 70 80 90 100

r

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points are (0, 0), (60, 0), (0, 58), and (24, 42). Three of these points are intercepts and can be read from a table of values or the graph itself. The point (24, 42) 14r  12d  840 was found by solving the system e . 4r  6d  348 Knowing the objective function will be maximized at one of these points, we test them in the objective function (Table 8.2). Table 8.2 Corner Point (0, 0) (60, 0)

Objective Function P1r, d2  $3r  $4d P10, 02  $3102  $4102  0 P160, 02  $31602  $4102  $180

(0, 58)

P10, 582  $3102  $41582  $232

(24, 42)

P124, 422  $31242  $41422  $240

Profit will be maximized if 24 boxes of the regular mix and 42 boxes of the deluxe mix are made and sold. Now try Exercises 55 through 60



Linear programming can also be used to minimize an objective function, as in Example 8. EXAMPLE 8



Minimizing Costs Using Linear Programming A beverage producer needs to minimize shipping costs from its two primary plants in Kansas City (KC) and St. Louis (STL). All wholesale orders within the state are shipped from one of these plants. An outlet in Macon orders 200 cases of soft drinks on the same day an order for 240 cases comes from Springfield. The plant in KC has 300 cases ready to ship and the plant in STL has 200 cases. The cost of shipping each case to Macon is $0.50 from KC, and $0.70 from STL. The cost of shipping each case to Springfield is $0.60 from KC, and $0.65 from STL. How many cases should be shipped from each warehouse to minimize costs?

Solution



Our objective is to minimize costs, which depends on the number of cases shipped from each plant. To begin we use the following assignments: A S cases shipped from KC to Macon B S cases shipped from KC to Springfield C S cases shipped from STL to Macon D S cases shipped from STL to Springfield From this information, the equation for total cost T is T  0.5A  0.6B  0.7C  0.65D, an equation in four variables. To make the cost equation more manageable, note since Macon ordered 200 cases, A  C  200. Similarly, Springfield ordered 240 cases, so B  D  240. After solving for C and D, respectively, these equations enable us to substitute for C and D, resulting in an equation with just two variables. For C  200  A and D  240  B we have T1A, B2  0.5A  0.6B  0.71200  A2  0.651240  B2  0.5A  0.6B  140  0.7A  156  0.65B  296  0.2A  0.05B The constraints involving the KC plant are A  B  300 with A  0, B  0. The constraints for the STL plant are C  D  200 with C  0, D  0. Since we want

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a system in terms of A and B only, we again substitute C  200  A and D  240  B in all the STL inequalities: C  D  200 1200  A2  1240  B2  200

STL inequalities substitute 200  A for C, 240  B for D

440  A  B  200 240  A  B

C0 200  A  0 200  A

D0 240  B  0 240  B

simplify result

Combining the new STL constraints with those from KC produces the following system and solution. All points of intersection were read from the graph or located using the related system of equations. 400

u

A  B  300 A  B  240 A  200 B  240 A0 B0

B

A  200

300

B  240

(60, 240) 200

100

Feasible region

(200, 100)

(200, 40) 100

200

A 300

400

To find the minimum cost, we check each vertex in the objective function. Objective Function T1A, B2  296  0.2 A  0.05B

Vertices (0, 240)

P10, 2402  296  0.2102  0.0512402  $284

(60, 240)

P160, 2402  296  0.21602  0.0512402  $272

(200, 100)

P1200, 1002  296  0.212002  0.0511002  $251

(200, 40)

P1200, 402  296  0.212002  0.051402  $254

The minimum cost occurs when A  200 and B  100, meaning the producer should ship the following quantities:

D. You’ve just learned how to solve applications using linear programming

A S cases shipped from KC to Macon  200 B S cases shipped from KC to Springfield  100 C S cases shipped from STL to Macon  0 D S cases shipped from STL to Springfield  140 Now try Exercises 61 and 62



TECHNOLOGY HIGHLIGHT

Systems of Linear Inequalities Solving systems of linear inequalities on the TI-84 Plus involves three steps, which are performed on both equations: (1) enter the related equations in Y1 and Y2 (solve for y in each equation) to create the boundary lines, (2) graph both lines and test the resulting half planes, and (3) shade the appropriate half plane. Since many real-world applications of linear inequalities preclude the use of negative numbers, we set Xmin  0 and Ymin  0 for the WINDOW size. Xmax and Ymax will depend on the equations given. We illustrate 3x  2y 6 14 by solving the system e . x  2y 6 8 —continued

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1. Enter the related equations. For 3x  2y  14, we have y 
1.5  7. For x  2y  8, we have y 
0.5x  4. Enter these as Y1 and Y2 on the Y = screen.

Figure 8.17

2. Graph the boundary lines. Note the x- and y-intercepts of both lines are less than 10, so we can graph them using a friendly window where x
[0, 9.4] and y
[0, 6.2]. After setting the window, press GRAPH to graph the lines. 3. Shade the appropriate half plane. Since both equations are in slope-intercept form, we shade below both lines for the less than inequalities, using the “ ” feature located to the far left of Y1 and Y2. Simply overlay the diagonal line and press ENTER repeatedly until the symbol appears (Figure 8.17). After pressing the GRAPH key, the calculator draws both lines and shades the appropriate regions (Figure 8.18). Note the calculator uses two different kinds of shading. This makes it easy to identify the solution region—it will be the “checker0 board area” where the horizontal and vertical lines cross. As a final check, you could navigate the position marker into the solution region and test a few points in both equations.

Figure 8.18 6.2

9.4

0

Use these ideas to solve the following systems of linear inequalities. Assume all solutions lie in Quadrant I. Exercise 1: e

y  2x 6 8 y x 6 6

Exercise 2: e

3x  y 6 8 xy 6 4

Exercise 3: e


4x
y 7
9
3x
y 7
7

8.4 EXERCISES


CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. Any line y  mx  b drawn in the coordinate plane divides the plane into two regions called . 2. For the line y  mx  b drawn in the coordinate plane, solutions to y 7 mx  b are found in the region the line. 3. The overlapping region of two or more linear inequalities in a system is called the region.


4. If a linear programming problem has a unique solution (x, y), it must be a of the feasible region. 5. Suppose two boundary lines in a system of linear inequalities intersect, but the point of intersection is not a vertex of the feasible region. Describe how this is possible. 6. Describe the conditions necessary for a linear programming problem to have multiple solutions. (Hint: Consider the diagram in Figure 8.16, and the slope of the line from the objective function.)

DEVELOPING YOUR SKILLS

Determine whether the ordered pairs given are solutions.

7. 2x  y 7 3; (0, 0), 13,
52, 1
3,
42, 1
3, 92

8. 3x
y 7 5; (0, 0), 14,
12 , 1
1,
52, 11,
22

9. 4x
2y 
8; (0, 0), 1
3, 52 , 1
3,
22 , 1
1, 12

10. 3x  5y  15; (0, 0), 13, 52, 1
1, 62, 17,
32

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Solve the linear inequalities by shading the appropriate half plane.

11. x  2y 6 8

12. x  3y 7 6

13. 2x  3y  9

14. 4x  5y  15

Use the equations given to write the system of linear inequalities represented by each graph.

39.

Determine whether the ordered pairs given are solutions to the accompanying system.

5y  x  10 ; 15. e 5y  2x  5 12, 12, 15, 42, 16, 22, 18, 2.22 8y  7x  56 16. • 3y  4x  12 y  4; 11, 52, 14, 62, 18, 52, 15, 32

Solve each system of inequalities by graphing the solution region. Verify the solution using a test point.

17. e

x  2y  1 2x  y  2

18. e

x  5y 6 5 x  2y  1

19. e

3x  y 7 4 x 7 2y

20. e

3x  2y y  4x  3

21. e

2x  y 6 4 2y 7 3x  6

22. e

x  2y 6 7 2x  y 7 5

x 7 3y  2 23. e x  3y  6

2x  5y 6 15 24. e 3x  2y 7 6

25. e

5x  4y  20 x1y

26. e

10x  4y  20 5x  2y 7 1

27. e

0.2x 7 0.3y  1 0.3x  0.5y  0.6

28. e

x 7 0.4y  2.2 x  0.9y  1.2

29. •

3 x 2 4y  6x  12

30. •

3x  4y 7 12 2 y 6 x 3

y

2 3 x y1 3 4 31. μ 1 x  2y  3 2

2 1 x y5 2 5 32. μ 5 x  2y  5 6

x  y  4 33. • 2x  y  4 x  1, y  0

2x  y  5 34. • x  3y  6 x1

yx3 35. • x  2y  4 y0

4y 6 3x  12 36. • x  0 yx1

2x  3y  18 37. • x  0 y0

8x  5y  40 38. • x  0 y0

54321 1 2 3 4 5

41.

40.

y 5 4 3 2 1

yx1

54321 1 2 3 4 5

54321 1 2 3 4 5

1 2 3 4 5 x

xy3

42.

y 5 4 3 2 1

yx1 1 2 3 4 5 x

xy3

y 5 4 3 2 1

yx1 1 2 3 4 5 x

xy3

y 5 4 3 2 1 54321 1 2 3 4 5

yx1 1 2 3 4 5 x

xy3

Determine which of the ordered pairs given produces the maximum value of f (x, y).

43. f 1x, y2  12x  10y; (0, 0), (0, 8.5), (7, 0), (5, 3) 44. f 1x, y2  50x  45y; (0, 0), (0, 21), (15, 0), (7.5, 12.5)

Determine which of the ordered pairs given produces the minimum value of f (x, y).

45. f 1x, y2  8x  15y; (0, 20), (35, 0), (5, 15), (12, 11)

46. f 1x, y2  75x  80y; (0, 9), (10, 0), (4, 5), (5, 4) For Exercises 47 and 48, find the maximum value of the objective function f 1x, y2  8x  5y given the constraints shown.

x  2y  6 3x  y  8 47. μ x0 y0

2x  y  7 x  2y  5 48. μ x0 y0

For Exercises 49 and 50, find the minimum value of the objective function f1x, y2  36x  40y given the constraints shown.

3x  2y  18 3x  4y  24 49. μ x0 y0

2x  y  10 x  4y  3 50. μ x2 y0

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Area Formulas

51. The area of a triangle is usually given as A  12 BH, where B and H represent the base and height, respectively. The area of a rectangle can be stated as A  BH. If the base of both the triangle and rectangle is equal to 20 in., what are the possible values for H if the triangle must have an area greater than 50 in2 and the rectangle must have an area less than 200 in2? 

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Volume Formulas

52. The volume of a cone is V  13r 2h, where r is the radius of the base and h is the height. The volume of a cylinder is V  r 2h. If the radius of both the cone and cylinder is equal to 10 cm, what are the possible values for h if the cone must have a volume greater than 200 cm3 and the volume of the cylinder must be less than 850 cm3?

APPLICATIONS

Write a system of linear inequalities that models the information given, then solve.

53. Gifts to grandchildren: Grandpa Augustus is considering how to divide a $50,000 gift between his two grandchildren, Julius and Anthony. After weighing their respective positions in life and family responsibilities, he decides he must bequeath at least $20,000 to Julius, but no more than $25,000 to Anthony. Determine the possible ways that Grandpa can divide the $50,000. 54. Guns versus butter: Every year, governments around the world have to make the decision as to how much of their revenue must be spent on national defense and domestic improvements (guns versus butter). Suppose total revenue for these two needs was $120 billion, and a government decides they need to spend at least $42 billion on butter and no more than $80 billion on defense. Determine the possible amounts that can go toward each need. Solve the following linear programming problems.

55. Land/crop allocation: A farmer has 500 acres of land to plant corn and soybeans. During the last few years, market prices have been stable and the farmer anticipates a profit of $900 per acre on the corn harvest and $800 per acre on the soybeans. The farmer must take into account the time it takes to plant and harvest each crop, which is 3 hr/acre for corn and 2 hr/acre for soybeans. If the farmer has at most 1300 hr to plant, care for, and harvest each crop, how many acres of each crop should be planted in order to maximize profits? 56. Coffee blends: The owner of a coffee shop has decided to introduce two new blends of coffee in order to attract new customers—a Deluxe Blend and a Savory Blend. Each pound of the deluxe blend contains 30% Colombian and 20% Arabian coffee, while each pound of the savory blend

contains 35% Colombian and 15% Arabian coffee (the remainder of each is made up of cheap and plentiful domestic varieties). The profit on the deluxe blend will be $1.25 per pound, while the profit on the savory blend will be $1.40 per pound. How many pounds of each should the owner make in order to maximize profit, if only 455 lb of Colombian coffee and 250 lb of Arabian coffee are currently available? 57. Manufacturing screws: A machine shop manufactures two types of screws—sheet metal screws and wood screws, using three different machines. Machine Moe can make a sheet metal screw in 20 sec and a wood screw in 5 sec. Machine Larry can make a sheet metal screw in 5 sec and a wood screw in 20 sec. Machine Curly, the newest machine (nyuk, nyuk) can make a sheet metal screw in 15 sec and a wood screw in 15 sec. (Shemp couldn’t get a job because he failed the math portion of the employment exam.) Each machine can operate for only 3 hr each day before shutting down for maintenance. If sheet metal screws sell for 10 cents and wood screws sell for 12 cents, how many of each type should the machines be programmed to make in order to maximize revenue? (Hint: Standardize time units.) 58. Hauling hazardous waste: A waste disposal company is contracted to haul away some hazardous waste material. A full container of liquid waste weighs 800 lb and has a volume of 20 ft3. A full container of solid waste weighs 600 lb and has a volume of 30 ft3. The trucks used can carry at most 10 tons (20,000 lb) and have a carrying volume of 800 ft3. If the trucking company makes $300 for disposing of liquid waste and $400 for disposing of solid waste, what is the maximum revenue per truck that can be generated?

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59. Maximizing profit—food service: P. Barrett & Justin, Inc., is starting up a fast-food restaurant specializing in peanut butter and jelly sandwiches. Some of the peanut butter varieties are smooth, crunchy, reduced fat, and reduced sugar. The jellies will include those expected and common, as well as some exotic varieties such as kiwi and mango. Independent research has determined the two most popular sandwiches will be the traditional P&J (smooth peanut butter and grape jelly), and the Double-T (three slices of bread). A traditional P&J uses 2 oz of peanut butter and 3 oz of jelly. The Double-T uses 4 oz of peanut butter and 5 oz of jelly. The traditional sandwich will be priced at $2.00, and a Double-T at $3.50. If the restaurant has 250 oz of smooth peanut butter and 345 oz of grape jelly on hand for opening day, how many of each should they make and sell to maximize revenue? 60. Maximizing profit—construction materials: Mooney and Sons produces and sells two varieties of concrete mixes. The mixes are packaged in 50-lb bags. Type A is appropriate for finish work, and contains 20 lb of cement and 30 lb of sand. Type B is appropriate for foundation and footing work, and contains 10 lb of cement and 20 lb of sand. The remaining weight comes from gravel aggregate. The profit on type A is $1.20/bag, while the profit on type B is $0.90/bag. How many bags of each should the company make to maximize profit, if 

2750 lb of cement and 4500 lb of sand are currently available? 61. Minimizing shipping costs: An oil company is trying to minimize shipping costs from its two primary refineries in Tulsa, Oklahoma, and Houston, Texas. All orders within the region are shipped from one of these two refineries. An order for 220,000 gal comes in from a location in Colorado, and another for 250,000 gal from a location in Mississippi. The Tulsa refinery has 320,000 gal ready to ship, while the Houston refinery has 240,000 gal. The cost of transporting each gallon to Colorado is $0.05 from Tulsa and $0.075 from Houston. The cost of transporting each gallon to Mississippi is $0.06 from Tulsa and $0.065 from Houston. How many gallons should be distributed from each refinery to minimize the cost of filling both orders? 62. Minimizing transportation costs: Robert’s Las Vegas Tours needs to drive 375 people and 19,450 lb of luggage from Salt Lake City, Utah, to Las Vegas, Nevada, and can charter buses from two companies. The buses from company X carry 45 passengers and 2750 lb of luggage at a cost of $1250 per trip. Company Y offers buses that carry 60 passengers and 2800 lb of luggage at a cost of $1350 per trip. How many buses should be chartered from each company in order for Robert to minimize the cost?

EXTENDING THE CONCEPT

63. Graph the feasible region formed by the system x0 y0 μ . How would you describe this region? y3 x3 Select random points within the region or on any boundary line and evaluate the objective function f1x, y2  4.5x  7.2y. At what point (x, y) will this 

767

function be maximized? How does this relate to optimal solutions to a linear programing problem?

64. Find the maximum value of the objective function f 1x, y2  22x  15y given the constraints 2x  5y  24 3x  4y  29 x  6y  26 . x0 y0

u

MAINTAINING YOUR SKILLS

65. (5.3) Given the point 13, 42 is on the terminal side of angle  with  in standard position, find a. cos  b. csc  c. cot 

67. (3.8) The resistance to current flow in copper wire varies directly as its length and inversely as the square of its diameter. A wire 8 m long with a 0.004-m diameter has a resistance of 1500 . Find the resistance in a wire of like material that is 2.7 m long with a 0.005-m diameter.

66. (3.7) Solve the rational inequality. Write your x2 7 0 answer in interval notation. 2 x 9

68. (6.4) Use a half-angle identity to find an exact 7 value for cosa b. 12

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MID-CHAPTER CHECK 1. Solve using the substitution method. State whether the system is consistent, inconsistent, or dependent. x  3y  2 e 2x  y  3 2. Solve the system using elimination. State whether the system is consistent, inconsistent, or dependent: e

x  3y  4 2x  y  13

4. Determine whether the ordered triple is a solution to the system. 12, 0, 32

5. The system given is a dependent system. Without solving, state why. x  2y  3z  3 • 2x  4y  6z  6 x  2y  5z  1 6. Solve the system of equations: •

2x  3y  4z  4 • x  2y  z  0 3  2y  2z  1 8. Decompose the expression partial fractions.

3. Solve using a system of linear equations and any method you choose. How many ounces of a 40% acid should be mixed with 10 oz of a 64% acid, to obtain a 48% acid solution?

5x  2y  4z  22 • 2x  3y  z  1 3x  6y  z  2

7. Solve using elimination:

x  2y  3z  4 2y  z  7 5y  2z  4

15  x2  4x into 1x  121x  22 2

9. Child prodigies: If you add Mozart’s age when he wrote his first symphony, with the age of American chess player Paul Morphy when he began dominating the international chess scene, and the age of Blaise Pascal when he formulated his wellknown Essai pour les coniques (Essay on Conics), the sum is 37. At the time of each event, Paul Morphy’s age was 3 yr less than twice Mozart’s, and Pascal was 3 yr older than Morphy. Set up a system of equations and find the age of each. 10. Candle manufacturing: David and Karen make table candles and holiday candles and sell them out of their home. Dave works 10 min and Karen works 30 min on each table candle, while Karen works 40 min and David works 20 min on each holiday candle. Dave can work at most 3 hr and 20 min (200 min) per day on the home business, while Karen can work at most 7 hr. If table candles sell for $4 and holiday candles sell for $6, how many of each should made to maximize their revenue?

REINFORCING BASIC CONCEPTS Window Size and Graphing Technology Since most substantial applications involve noninteger values, technology can play an important role in applying mathematical models. However, with its use comes a heavy responsibility to use it carefully. A very real effort must be made to determine the best approach and to secure a reasonable estimate. This is the only way to guard against (the inevitable) keystroke errors, or ensure a window size that properly displays the results.

Rationale On October 1, 1999, the newspaper USA TODAY ran an article titled, “Bad Math added up to Doomed Mars Craft.”

The article told of how a $125,000,000.00 spacecraft was lost, apparently because the team of scientists that plotted the course for the craft used U.S. units of measurement, while the team of scientists guiding the craft were using metric units. NASA’s space chief was later quoted, “The problem here was not the error, it was the failure of . . . the checks and balances in our process to detect the error.” No matter how powerful the technology, always try to begin your problem-solving efforts with an estimate. Begin by exploring the context of the problem, asking questions about the range of possibilities: How fast can a human run? How much does a new car cost? What is a reasonable price for a ticket? What is the total available to invest? There is no calculating involved in these estimates,

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they simply rely on “horse sense” and human experience. In many applied problems, the input and output values must be positive — which means the solution will appear in the first quadrant, narrowing the possibilities considerably. This information will be used to set the viewing window of your graphing calculator, in preparation for solving the problem using a system and graphing technology. Illustration 1  Erin just filled both her boat and Blazer with gas, at a total cost of $211.14. She purchased 35.7 gallons of premium for her boat and 15.3 gal of regular for her Blazer. Premium gasoline cost $0.10 per gallon more than regular. What was the cost per gallon of each grade of gasoline? Solution  Asking how much you paid for gas the last time you filled up should serve as a fair estimate. Certainly (in 2008) a cost of $6.00 or more per gallon in the United

769

States is too high, and a cost of $1.50 per gallon or less would be too low. Also, we can estimate a solution by assuming that both kinds of gasoline cost the same. This would mean 51 gal were purchased for about $211, and a quick division would place the estimate at near 211 51  $4.14 per gallon. A good viewing window would be restricted to the first quadrant (since cost 7 0) with maximum values of Xmax  6 and Ymax  6. Exercise 1: Solve Illustration 1 using graphing technology. Exercise 2: Re-solve Exercises 63 and 64 from Section 8.1 using graphing technology. Verify results are identical.

8.5 Solving Linear Systems Using Matrices and Row Operations Learning Objectives

Just as synthetic division streamlines the process of polynomial division, matrices and row operations streamline the process of solving systems using elimination. With the equations of the system in standard form, the location of the variable terms and constant terms are set, and we simply apply the elimination process on the coefficients and constants.

In Section 8.5 you will learn how to:

A. State the size of a matrix and identify its entries

B. Form the augmented

A. Introduction to Matrices

matrix of a system of equations

C. Solve a system of equations using row operations

D. Recognize inconsistent and dependent systems

E. Solve applications using linear systems

EXAMPLE 1A



In general terms, a matrix is simply a rectangular arrangement of numbers, called the entries of the matrix. Matrices (plural of matrix) are denoted by enclosing the entries between a left and right bracket, and named using a capital letter, such as 2 1 3 1 3 2 A c d and B  £ 4 6 2 § . They occur in many different sizes 5 1 1 1 0 1 as defined by the number of rows and columns each has, with the number of rows always given first. Matrix A is said to be a 2  3 (two by three) matrix, since it has two rows and three columns. Matrix B is a 3  3 (three by three) matrix.

Identifying the Size and Entries of a Matrix Determine the size of each matrix and identify the entry located in the second row and first column. 3 2 0.2 0.5 0.7 3.3 5 § 0.3 1 2 § a. C  £ 1 b. D  £ 0.4 4 3 2.1 0.1 0.6 4.1

Solution



a. Matrix C is 3  2. The row 2, column 1 entry is 1. b. Matrix D is 3  4. The row 2, column 1 entry is 0.4.

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If a matrix has the same number of rows and columns, it’s called a square matrix. Matrix B above is a square matrix, while matrix A is not. For square matrices, the values on a diagonal line from the upper left to the lower right are called the diagonal entries and are said to be on the diagonal of the matrix. When solving systems using matrices, much of our focus is on these diagonal entries.

EXAMPLE 1B



Identifying the Diagonal Entries of a Square Matrix Name the diagonal entries of each matrix. 0.2 1 4 d a. E  c b. F  £ 0.4 2 3 2.1

Solution



0.5 0.3 0.1

0.7 1 § 0.6

a. The diagonal entries of matrix E are 1 and 3. b. For matrix F, the diagonal entries are 0.2, 0.3, and 0.6.

A. You’ve just learned how to state the size of a matrix and identify its entries

Now try Exercises 7 through 9



B. The Augmented Matrix of a System of Equations A system of equations can be written in matrix form by augmenting or joining the coefficient matrix, formed by the variable coefficients, with the matrix of constants. 2x  3y  z  1 2 3 1 0 1 § with z  2 is £ 1 The coefficient matrix for the system • x  1 3 4 x  3y  4z  5 column 1 for the coefficients of x, column 2 for the coefficients of y, and so on. The 1 matrix of constants is £ 2 § . These two are joined to form the augmented matrix, 5 2 3 1 1 0 1 2§. with a dotted line often used to separate the two as shown here: £ 1 1 3 4 5 It’s important to note the use of a zero placeholder for the y-variable in the second row of the matrix, signifying there is no y-variable in the corresponding equation. EXAMPLE 2

Solution





Forming Augmented Matrices Form the augmented matrix for each system, and name the diagonal entries of each coefficient matrix. 1 x  4y  z  10 y  7 2x  2x  y  11 a. e b. • 2x  5y  8z  4 c. • x  23y  56z  11 12 x  3y  2 x  2y  3z  7 2y  z  3 1 11 2x  y  11 ¡ 2 d c a. e 1 3 2 x  3y  2 Diagonal entries: 2 and 3. x  4y  z  10 1 ¡ £2 b. • 2x  5y  8z  4 x  2y  3z  7 1 Diagonal entries: 1, 5, and 3.

4 5 2

1 8 3

10 4 § 7

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 c. • x  1 2x

1 y  7 2 2 5 11 ¡ £1 3 y  6 z  12 2y  z  3 0

1

0

7

2 3

5 6

11 12

2

1

3

771

§

Diagonal entries: 12, 23, and 1. Now try Exercises 10 through 12



This process can easily be reversed to write a system of equations from a given augmented matrix. EXAMPLE 3



Writing the System Corresponding to an Augmented Matrix Write the system of equations corresponding to each matrix. 1 4 1 10 3 5 14 a. c d b. £ 0 3 10 7 § 0 1 4 0 0 1 1

Solution



B. You’ve just learned how to form the augmented matrix of a system of equations

a. c

3 0 1 b. £ 0 0

5 1 4 3 0

14 ¡ 3x  5y  14 e d 0x  1y  4 4 1 10 1x  4y  1z  10 ¡ 10 7 § • 0x  3y  10z  7 1 1 0x  0y  1z  1 Now try Exercises 13 through 18



C. Solving a System Using Matrices When a system of equations is written in augmented matrix form, we can solve the system by applying the same operations to each row of the matrix, that would be applied to the equations in the system. In this context, the operations are referred to as elementary row operations. Elementary Row Operations 1. Any two rows in a matrix can be interchanged. 2. The elements of any row can be multiplied by a nonzero constant. 3. Any two rows can be added together, and the sum used to replace one of the rows. In this section, we’ll use these operations to triangularize the augmented matrix, employing a solution method known as Gaussian elimination. A matrix is said to be in triangular form when all of the entries below the diagonal are zero. For example, 1 4 1 10 1 4 1 10 the matrix £ 0 3 10 7 § is in triangular form: £ 0 3 10 7 §. 0 0 1 1 0 0 1 1 x  4y  z  10 In system form we have •  3y  10z  7 , meaning a matrix written in trianz 1 gular form can be used to solve the system using back-substitution. We’ll illustrate by 1x  4y  1z  4 solving • 2x  5y  8z  15, using elimination to the left, and row operations on the 1x  3y  3z  1 augmented matrix to the right. As before, R1 represents the first equation in the system and the first row of the matrix, R2 represents equation 2 and row 2, and so on.

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The calculations involved are shown for the first stage only and are designed to offer a careful comparison. In actual practice, the format shown in Example 4 is used. Elimination (System of Equations)

Row Operations (Augmented Matrix)

1x  4y  1z  4 • 2x  5y  8z  15 1x  3y  3z  1

1 £2 1

1 8 3

4 5 3

4 15 § 1

To eliminate the x-term in R2, we use 2R1  R2 S R2. For R3 the operations would be 1R1  R3 S R3. Identical operations are performed on the matrix, which begins the process of triangularizing the matrix. System Form

Matrix Form

2R1  R2 New R2

2x  8y  2z  8

1R1  R3 New R3

1x  4y  1z  4

2x  5y  8z  15  3y  10z  7

1x  3y  3z  1  1y  2z  3

2R1  R2 New R2

2

1R1  R3 New R3

1

8

2

8

2 5 8 0 3 10

15 7

4

1

4

1 3 3 1 0 1 2 3

As always, we should look for opportunities to simplify any equation in the system (and any row in the matrix). Note that 1R3 will make the coefficients and related matrix entries positive. Here is the new system and matrix. New System •

1x 

4y  1z  4 3y  10z  7 1y  2z  3

New Matrix 1 £0 0

4 3 1

1 10 2

4 7§ 3

On the left, we would finish by solving the 2  2 subsystem using R2  3R3 S R3. In matrix form, we eliminate the corresponding entry (third row, second column) to triangularize the matrix. 1 £0 0 WORTHY OF NOTE The procedure outlined for solving systems using matrices is virtually identical to that for solving systems by elimination. Using a 3  3 system for illustration, the “zeroes below the first diagonal entry” indicates we’ve eliminated the x-term from R2 and R3, the “zeroes below the second entry” indicates we’ve eliminated the y-term from the subsystem, and the division “to obtain a ‘1’ in the final entry” indicates we have just solved for z.

4 3 1

1 10 2

4 1 R2  3R3 S R3 7 § ¬¬¬¬¡ £0 3 0

4 3 0

1 10 16

4 7§ 16

Dividing R3 by 16 gives z  1 in the system, and entries of 0 0 1 1 in the augmented matrix. Completing the solution by back-substitution in the system gives the ordered triple (1, 1, 1). See Exercises 19 through 27. The general solution process is summarized here. Solving Systems by Triangularizing the Augmented Matrix 1. 2. 3. 4. 5.

Write the system as an augmented matrix. Use row operations to obtain zeroes below the first diagonal entry. Use row operations to obtain zeroes below the second diagonal entry. Continue until the matrix is triangularized (entries below diagonal are zero). Divide to obtain a “1” in the last diagonal entry (if it is nonzero), then convert to equation form and solve using back-substitution.

Note: At each stage, look for opportunities to simplify row entries using multiplication or division. Also, to begin the process any equation with an x-coefficient of 1 can be made R1 by interchanging the equations.

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EXAMPLE 4




Solving Systems Using the Augmented Matrix 2x  y
2z 
7 Solve by triangularizing the augmented matrix: • x  y  z 
1
2y
z 
3

Solution




2x  y
2z 
7 • x  y  z 
1
2y
z 
3 1 £2 0

1 1
2

1
2
1


1
7 §
3

1 £0 0

1 1
2

1 4
1


1 5 §
3

matrix form S

2 £1 0

1 1
2


2 1
1


7
1 §
3

R1 4 R2

1 £2 0

1 1
2

1
2
1


1
7 §
3


2R1  R2 S R2

1 £0 0

1
1
2

1
4
1


1
5 §
3


1R2 S R2

1 £0 0

1 1
2

1 4
1


1 5 §
3

2R2  R3 S R3

1 £0 0

1 1 0

1 4 7


1 5 § 7

R3 S R3 7

1 £0 0

1 1 0

1 4 1


1 5 § 1

x  y  z 
1 y  4z  5 . Converting the augmented matrix back into equation form yields • z1 Back-substitution shows the solution is 1
3, 1, 12. Now try Exercises 28 through 32

C. You’ve just learned how to solve a system of equations using row operations

EXAMPLE 5







The process used in Example 4 is also called Gaussian elimination (Carl Friedrich Gauss, 1777–1855), with the last matrix written in row-echelon form. It’s possible to solve a system entirely using only the augmented matrix, by continuing to use row operations until the diagonal entries are 1’s, with 0’s for all other entries: 1 0 0 a £ 0 1 0 b § . The process is then called Gauss-Jordan elimination (Wilhelm 0 0 1 c Jordan, 1842–1899), with the final matrix written in reduced row-echelon form (see Appendix III). Note that with Gauss-Jordan elimination, our initial focus is less on getting 1’s along the diagonal, and more on obtaining zeroes for all entries other than the diagonal entries. This will enable us to work with integer values in the solution process.

Solving a System Using Gauss-Jordan Elimination 2x  5z
15  2y Solve using Gauss-Jordan elimination • 2x  3y 
1  z. 4y  z 
7 standard form

Solution




2x
2y  5z  15 • 2x  3y
1z 
1 0x  4y  1z 
7 2
2 5 £ 0 5
6 0 4 1

matrix form S

2
2 5 £ 2 3
1 0 4 1

15 2R2  5R1 S R1 10
16 § £0
7
4R2  5R3 S R3 0

0 5 0

13
6 29

15
1 §
7


R1  R2 S R2

43
16 § 29

R3 S R3 29

2
2 5 £ 0 5
6 0 4 1

15
16 §
7

10 £0 0

43
16 § 1

0 5 0

13
6 1

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10 £0 0

0 5 0

13 43 6 16 § 1 1

13R3  R1 S R1 6R3  R2 S R2

10 £0 0

0 5 0

0 0 1

30 10 § 1

R1 S R1 10

1 £0 0

R2 S R2 5

0 1 0

0 0 1

3 2 § 1

The final matrix shows the solution is (3, 2, 1). Now try Exercises 33 through 36



D. Inconsistent and Dependent Systems Due to the strong link between a linear system and its augmented matrix, inconsistent and dependent systems can be recognized just as in Sections 5.1 and 5.2. An inconsistent system will yield an inconsistent or contradictory statement such as 0  12, meaning all entries in a row of the matrix of coefficients are zero, but the constant is not. A linearly dependent system will yield an identity statement such as 0  0, meaning all entries in one row of the matrix are zero. If the system has coincident dependence, there will be only one nonzero row of coefficients. EXAMPLE 6

Solution



x  y  5z  3 2z  1 Solve the system using Gauss-Jordan elimination: • x  2x  y  z  0 Solving a Dependent System



x  y  5z  3 • x  2z  1 2x  y  z  0 1 1 £ 1 0 2 1

5 2 1

3 1 § 0

standard form S

R1  R2 S R2 2R1  R3 S R3

x  y  5z  3 • x  0y  2z  1 2x  y  z  0 1 £0 0

1 5 1 3 3 9

3 2 § 6

matrix form S

1R2  R1 S R1 3R2  R3 S R3

1 1 £ 1 0 2 1 1 £0 0

0 2 1 3 0 0

5 2 1

3 1 § 0

1 2§ 0

Since all entries in the last row are zeroes and it’s the only row of zeroes, we conclude the system is x  2z  1 linearly dependent and equivalent to e . As in Chapter 5, we demonstrate this dependence y  3z  2 by writing the (x, y, z) solution in terms of a parameter. Solving for y in R2 gives y in terms of z: y  3z  2. Solving for x in R1 gives x in terms of z: x  2z  1. As written, the solutions all depend on z: x  2z  1, y  3z  2, and z  z. Selecting p as the parameter (or some other “neutral” variable), we write the solution as 12p  1, 3p  2, p2. Two of the infinite number of solutions would be (1, 2, 0) for p  0, and 11, 1, 12 for p  1. Test these triples in the original equations. Now try Exercises 37 through 45

D. You’ve just learned how to recognize inconsistent and dependent systems



E. Solving Applications Using Matrices As in other areas, solving applications using systems relies heavily on the ability to mathematically model information given verbally or in context. As you work through the exercises, read each problem carefully. Look for relationships that yield a system of two equations in two variables, three equations in three variables and so on. EXAMPLE 7



Determining the Original Value of Collector’s Items A museum purchases a famous painting, a ruby tiara, and a rare coin for its collection, spending a total of $30,000. One year later, the painting has tripled in value, while the tiara and the coin have doubled in value. The items now have a total value of $75,000. Find the purchase price of each if the original price of the painting was $1000 more than twice the coin.

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Section 8.5 Solving Linear Systems Using Matrices and Row Operations

Solution



Let P represent the price of the painting, T the tiara, and C the coin. Total spent was $30,000: S P  T  C  30,000 One year later: S 3P  2T  2C  75,000 Value of painting versus coin: S P  2C  1000

P  T  C  30000 1P  1T  1C  30000 • 3P  2T  2C  75000 standard form S • 3P  2T  2C  75000 P  2C  1000 1P  0T  2C  1000 1 £3 1 1 £0 0

1 2 0 1 1 1

1 2 2

30000 75000 § 1000

1 1 3

30000 15000 § 29000

3R1  R2 S R2 1R1  R3 S R3

1R2  R3 S R3

1 £0 0

1 1 1

1 £0 0

1 1 0

1 1 3 1 1 2

30000 15000 § 29000

30000 15000 § 14000

matrix form S

1R2 S R2 1R3 S R3 R3 S R3 2

1 £3 1

1 2 0

1 2 2

30000 75000 § 1000

1 £0 0

1 1 1

1 1 3

30000 15000 § 29000

1 £0 0

1 1 0

1 1 1

30000 15000 § 7000

From R3 of the triangularized form, C  $7000 directly. Since R2 represents T  C  15,000, we find the tiara was purchased for T  $8000. Substituting these values into the first equation shows the painting was purchased for $15,000. The solution is (15,000, 8000, 7000). Now try Exercises 48 through 55

E. You’ve just learned how to solve applications using linear systems



TECHNOLOGY HIGHLIGHT

Solving Systems Using Matrices and Calculating Technology Graphing calculators offer a very efficient way to solve systems using matrices. Once the system has been written in matrix form, it can easily be entered and solved by asking the calculator to instantly perform the row operations needed. Pressing

2nd

Xⴚ1

(MATRIX) gives a screen similar to the one

shown in Figure 8.19, where we begin by selecting the EDIT option (push the right arrow twice). Pressing ENTER places you on a screen where you can EDIT matrix A, changing the size as needed. Using the 3  4 matrix from Example 4, we press 3 and ENTER , then 4 and ENTER , giving the screen shown in Figure 8.20. The dash marks to the right indicate that there is a fourth column that cannot be seen, but that comes into view as you enter the elements of the matrix. Begin entering the first row of the matrix resulting from R1 4 R2, which has entries {1, 1, 1, 1}. Press ENTER after each entry and the cursor automatically goes to the next position in the matrix (note that the TI-84 Plus automatically shifts left and right to allow all four columns to be entered). After entering the second row {2, 1, 2, 7} and the third row {0, 2, 1, 3}, the completed matrix should look like the one shown in Figure 8.21 (the matrix is currently shifted to the right, showing the fourth column). To write this matrix in reduced row-echelon form (rref) we return to the home screen by pressing CLEAR

2nd

key for a clean home screen. To access the rref function, press

Figure 8.19

Figure 8.20

(QUIT). Press the

MODE 2nd

X

ⴚ1

(MATRIX) and select

Figure 8.21

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the MATH option, then scroll upward (or downward) until you get to B:rref. Pressing ENTER places this function on the home screen, where we must tell it to perform the rref operation on matrix [A]. Press 2nd Xⴚ1 (MATRIX) to select a matrix (notice that matrix NAMES is automatically highlighted. Press ENTER to select matrix [A] as the object of the rref function. After pressing ENTER the calculator quickly computes the reduced row-echelon form and displays it on the screen as in Figure 8.22. The solution is easily read as x  3, y  1, and z  1, as we found in Example 4. Use these ideas to complete the following. Exercise 1:

Figure 8.22

Use this method to solve the 2  2 system from Exercise 30.

Exercise 2: Use this method to solve the 3  3 system from Exercise 32.

8.5 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. A matrix with the same number of rows and columns is called a(n) matrix. 2. When the coefficient matrix is used with the matrix of constants, the result is a(n) matrix. 2 4 3 by d is a 1 2 1 matrix. The entry in the second row and third column is .

3. Matrix A  c



4. Given matrix B shown here, the diagonal entries are , , and . 1 4 3 B  £ 1 5 2§ 3 2 1 5. The notation 2R1  R2 S R2 indicates that an equivalent matrix is formed by performing what operations/replacements? 6. Describe how to tell an inconsistent system apart from a dependent system when solving using matrix methods (row reduction).

DEVELOPING YOUR SKILLS

Determine the size (order) of each matrix and identify the third row and second column entry. If the matrix given is a square matrix, identify the diagonal entries.

1 7. £ 2.1 3 1 1 9. ≥ 5 2

0 1 § 5.8 0 3 1 3

1 8. £ 1 5 4 7 ¥ 2 9

0 3 1

4 7 § 2

Form the augmented matrix, then name the diagonal entries of the coefficient matrix.

2x  3y  2z  7 10. • x  y  2z  5 3x  2y  z  11 x  2y  z  1 11. • x  z3 2x  y  z  3 2x  3y  z  5 12. • 2y  z  7 x  y  2z  5

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Write the system of equations for each matrix. Then use back-substitution to find its solution.

13. c

1 0

4 1

d

5 1 2

1 14. c 0

5 1

15 d 2

1 15. £ 0 0

2 1 0

1 2 1

0 2§ 3

1 16. £ 0 0

0 1 0

7 5 1

5 15 § 26

1 17. £ 0 0

3 1 0

4 32 1

29

1 18. £ 0 0

2 1 0

1

3

21 2

3

§

2 3 § 22 7

1 6

1

Perform the indicated row operation(s) and write the new matrix.

19. c

3 2

1 2

5

7 20. c 4

1 2R1 S R1, d 5R1  R2 S R2 4 3 14R2 S R2, d 12 R1 4 R2

4 8

2 21. £ 5 1

1 8 3

3 22. £ 1 4

2 1 1

0 2 3

0 6 § R1 4 R2, 2 4R1  R3 S R3

3 23. £ 6 4

1 1 2

1 1 3

8 10 § 2R1  R2 S R2, 22 4R1  3R3 S R3

2 24. £ 3 4

1 1 3

0 3 3

1 1 2

4 5 § R1 4 R3, 5R1  R2 S R2 2

3 0 § 3R1  2R2 S R2, 2R1  R3 S R3 3

What row operations would produce zeroes beneath the first entry in the diagonal?

1 25. £ 2 3

3 4 1

0 1 2

2 1§ 9

1 26. £ 3 5

1 0 3

4 1 2

3 5 § 3

1 27. £ 5 4

2 1 3

0 2 3

10 6 § 2

777

Solve each system by triangularizing the augmented matrix and using back-substitution. Simplify by clearing fractions or decimals before beginning.

28. e

2y  5x  4 5x  2  4y

29. e

0.15g  0.35h  0.5 0.12g  0.25h  0.1

30. e

15u  14v  1 1 1 10 u  2 v  7

x  2y  2z  7 31. • 2x  2y  z  5 3x  y  z  6

2x  3y  2z  7 32. • x  y  2z  5 3x  2y  z  11

x  2y  z  1 33. • x  z3 2x  y  z  3

2x  3y  z  5 34. • 2y  z  7 x  y  2z  5

x  y  2z  2 35. • x  y  z  1 2x  y  z  4

x  y  2z  1 36. • 4x  y  3z  3 3x  2y  z  4

Solve each system by triangularizing the augmented matrix and using back-substitution. If the system is linearly dependent, give the solution in terms of a parameter. If the system has coincident dependence, answer in set notation as in Chapter 5.

4x  8y  8z  24 37. • 2x  6y  3z  13 3x  4y  z  11 3x  y  z  2 38. • x  2y  3z  1 2x  3y  5z  3 x  3y  5z  20 39. • 2x  3y  4z  16 x  2y  3z  12 x  2y  3z  6 40. • x  y  2z  4 3x  6y  9z  18

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3x  4y  2z  2 41. • 23x  2y  z  1 6x  8y  4z  4

x  2y  z  4 44. • 3x  4y  z  4 6x  8y  2z  8

2x  y  3z  1 42. • 4x  2y  6z  2 10x  5y  15z  5

2x  4y  3z  4 45. • 5x  6y  7z  12 x  2y  z  4

43. •



2x  y  3z  1 2y  6z  2 x  12y  32z  5

WORKING WITH FORMULAS

Area of a triangle in the plane: 1 A   1x1 y2  x2 y1  x2 y3  x3 y2  x3 y1  x1 y3 2 2 The area of a triangle in the plane is given by the formula shown, where the vertices of the triangle are located at the points (x1, y1), (x2, y2), and (x3, y3), and the sign is chosen to ensure a positive value.



46. Find the area of a triangle whose vertices are 11, 32, (5, 2), and (1, 8). 47. Find the area of a triangle whose vertices are 16, 22, 15, 42 , and 11, 72.

APPLICATIONS

Model each problem using a system of linear equations. Then solve using the augmented matrix. Descriptive Translation

48. The distance (via air travel) from Los Angeles (LA), California, to Saint Louis (STL), Missouri, to Cincinnati (CIN), Ohio, to New York City (NYC), New York, is approximately 2480 mi. Find the distances between each city if the distance from LA to STL is 50 mi more than five times the distance between STL and CIN and 110 mi less than three times the distance from CIN to NYC.

New York City St. Louis

Cincinnati

Los Angeles

title game and the Heat won by 3 points, what was the final score? 50. Moe is lecturing Larry and Curly once again (Moe, Larry, and Curly of The Three Stooges fame) claiming he is twice as smart as Larry and three times as smart as Curly. If he is correct and the sum of their IQs is 165, what is the IQ of each stooge? 51. A collector of rare books buys a handwritten, autographed copy of Edgar Allan Poe’s Annabel Lee, an original advance copy of L. Frank Baum’s The Wonderful Wizard of Oz, and a first print copy of The Caine Mutiny by Herman Wouk, paying a total of $100,000. Find the cost of each one, given that the cost of Annabel Lee and twice the cost of The Caine Mutiny sum to the price paid for The Wonderful Wizard of Oz, and The Caine Mutiny cost twice as much as Annabel Lee. Geometry

49. In the 2006 NBA Championship Series, Dwayne Wade of the Miami Heat carried his team to the title after the first two games were lost to the Dallas Mavericks. If 187 points were scored in the

52. A right triangle has a hypotenuse of 39 m. If the perimeter is 90 m, and the longer leg is 6 m longer than twice the shorter leg, find the dimensions of the triangle.

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53. In triangle ABC, the sum of angles A and C is equal to three times angle B. Angle C is 10 degrees more than twice angle B. Find the measure of each angle. Investment and Finance

54. Suppose $10,000 is invested in three different investment vehicles paying 5%, 7%, and 9% annual interest. Find the amount invested at each rate if the interest earned after 1 yr is $760 and the amount invested at 9% is equal to the sum of the amounts invested at 5% and 7%.



55. The trustee of a union’s pension fund has invested the funds in three ways: a savings fund paying 4% annual interest, a money market fund paying 7%, and government bonds paying 8%. Find the amount invested in each if the interest earned after one year is $0.178 million and the amount in government bonds is $0.3 million more than twice the amount in money market funds. The total amount invested is $2.5 million dollars.

EXTENDING THE CONCEPT

56. In previous sections, we noted that one condition for a 3  3 system to be dependent was for the third equation to be a linear combination of the other two. To test this, write any two (different) equations using the same three variables, then form a third equation by performing some combination of elementary row operations. Solve the resulting 3  3 system. What do you notice? 57. Given the drawing shown, use a system of equations and the matrix method to find the measure of the angles labeled as x and y. Recall that vertical angles



779

Section 8.5 Solving Linear Systems Using Matrices and Row Operations

are equal and that the sum of the angles in a triangle is 180°. 71 y

x

(x  59)

58. The system given here has a solution of 11, 2, 32. Find the value of a and b. 1 £ 2b 2a

a 2a 7

b 5 3b

1 13 § 8

MAINTAINING YOUR SKILLS

59. (7.5) a. Convert z1  1  3i to trigonometric form. 2 b. Convert z2  5 cisa b to rectangular 3 form. 60. (5.4) State the exact value of the following trig functions: 5  a. sina b b. cosa b 6 4 2 3 c. tana b d. csca b 3 2 61. (4.4) Since 2005, cable installations for an Internet company have been modeled by the function C1t2  15 ln 1t  12 , where C(t) represents cable installations in thousands, t yr after 2005. In what

year will the number of installations be greater than 30,000? 62. (4.5) If a set amount of money p is deposited regularly (daily, weekly, monthly, etc.) n times per year at a fixed interest rate r, the amount of money A accumulated in t years is given by the formula shown. If a parent deposits $250 per month for 18 yr at 4.6% beginning when her first child was born, how much has been accumulated to help pay for college expenses? r nt p c a1  b  1 d n A r n

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Precalculus—

8.6 The Algebra of Matrices Learning Objectives

Matrices serve a much wider purpose than just a convenient method for solving systems. To understand their broader application, we need to know more about matrix theory, the various ways matrices can be combined, and some of their more practical uses. The common operations of addition, subtraction, multiplication, and division are all defined for matrices, as are other operations. Practical applications of matrix theory can be found in the social sciences, inventory management, genetics, operations research, engineering, and many other fields.

In Section 8.6 you will learn how to:

A. Determine if two matrices are equal

B. Add and subtract matrices

C. Compute the product of two matrices

A. Equality of Matrices To effectively study matrix algebra, we first give matrices a more general definition. For the general matrix A, all entries will be denoted using the lowercase letter “a,” with their position in the matrix designated by the dual subscript aij. The letter “i ” gives the row and the letter “j ” gives the column of the entry’s location. The general m  n matrix A is written col 1 row 1 S a11 row 2 S l a21 a31 row 3 S row i S row m S

p

col 2 a12 a22 a32

col 3 a13 a23 a33

p p p

col j a1j a2j a3j

p p p

o ai1

o ai2

o ai3

p

o aij

p

am1

o am2

o am3

p

o amj

p

z o

col n a1n a2n } a3n o p ain

aij is a general matrix element

o { amn

The size of a matrix is also referred to as its order, and we say the order of general matrix A is m  n. Note that diagonal entries have the same row and column number, aij, where i  j. Also, where the general entry of matrix A is aij, the general entry of matrix B is bij, of matrix C is cij, and so on. EXAMPLE 1



Identifying the Order and Entries of a Matrix State the order of each matrix and name the entries corresponding to a22, a31; b22, b31; and c22, c31. a. A  c

Solution



1 2

4 d 3

3 b. B  £ 1 4

2 5 § 3

0.2 c. C  £ 1 2.1

0.5 0.3 0.1

0.7 1 § 0.6

a. matrix A: order 2  2. Entry a22  3 (the row 2, column 2 entry is 3). There is no a31 entry (A is only 2  2). b. matrix B: order 3  2. Entry b22  5, entry b31  4. c. matrix C: order 3  3. Entry c22  0.3, entry c31  2.1. Now try Exercises 7 through 12



Equality of Matrices Two matrices are equal if they have the same order and their corresponding entries are equal. In symbols, this means that A  B if aij  bij for all i and j.

780

8-62

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Section 8.6 The Algebra of Matrices

EXAMPLE 2

Solution





A. You’ve just learned how to determine if two matrices are equal

781

Determining If Two Matrices Are Equal Determine whether the following statements are true, false, or conditional. If false, explain why. If conditional, find values that will make the statement true. 3 2 3 2 1 4 3 2 1 a. c b. £ 1 d d  c 5 §  c d 2 3 4 1 5 4 3 4 3 1 4 a  2 2b d  c d c. c 2 3 c 3 1 4 3 2 a. c d  c d is false. The matrices have the same order 2 3 4 1 and entries, but corresponding entries are not equal. 3 2 3 2 1 b. £ 1 5 §  c d is false. Their orders are not equal. 5 4 3 4 3 1 4 a  2 2b d  c d is conditional. The statement is true when c. c 2 3 c 3 a  2  1 1a  32, 2b  4 1b  22, c  2, and is false otherwise. Now try Exercises 13 through 16



B. Addition and Subtraction of Matrices A sum or difference of matrices is found by combining the corresponding entries. This limits the operations to matrices of like orders, so that every entry in one matrix has a “corresponding entry” in the other. This also means the result is a new matrix of like order, whose entries are the corresponding sums or differences. Note that since aij represents a general entry of matrix A, [aij] represents the entire matrix. Addition and Subtraction of Matrices Given matrices A, B, and C having like orders. The sum A  B  C, where 3aij  bij 4  3cij 4. EXAMPLE 3



The difference A  B  D, where 3aij  bij 4  3dij 4.

Adding and Subtracting Matrices Compute the sum or difference of the matrices indicated. 2 6 A  £1 0 § 1 3 a. A  C

Solution



2 a. A  C  £ 1 1

3 B c 5 b. A  B

6 0 §  3 23  £ 11 1  142

2 4

1 d 3

3 C £ 1 4

2 5 § 3

c. C  A

3 2 £ 1 5 § sum of A and C 4 3 6  122 5 4 05 § £ 2 5§ 3  3 3 0

add corresponding entries

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B. You’ve just learned how to add and subtract matrices

2 6 Addition and subtraction are not defined 3 2 1 b. A  B  £ 1 0 § c d for matrices of unlike order. 5 4 3 1 3 3 2 2 6 c. C  A  £ 1 5 §  £1 0 § difference of C and A 4 3 1 3 32 2  6 1 8  £ 11 50 §  £ 0 5 § subtract corresponding entries 4  1 3  132 5 6 Now try Exercises 17 through 20



Since the addition of two matrices is defined as the sum of corresponding entries, we find the properties of matrix addition closely resemble those of real number addition. Properties of Matrix Addition Given matrices A, B, C, and Z are m  I. A  B  B  A II. 1A  B2  C  A  1B  C2 III. A  Z  Z  A  A IV. A  1A2  1A2  A  Z

n matrices, with Z the zero matrix. Then, matrix addition is commutative matrix addition is associative Z is the additive identity A is the additive inverse of A

C. Matrices and Multiplication The algebraic terms 2a and ab have counterparts in matrix algebra. The product 2A represents a constant times a matrix and is called scalar multiplication. The product AB represents the product of two matrices.

Scalar Multiplication Scalar multiplication is defined by taking the product of the constant with each entry in the matrix, forming a new matrix of like size. In symbols, for any real number k and matrix A, kA  3 kaij 4. Similar to standard algebraic properties, A represents the scalar product 1 # A and any subtraction can be rewritten as an algebraic sum: A  B  A  1B2. As noted in the properties box, for any matrix A, the sum A  1A2 will yield the zero matrix Z, a matrix of like size whose entries are all zeroes. Also note that matrix A is the additive inverse for A, while Z is the additive identity. EXAMPLE 4



Computing Operations on Matrices 4 Given A  £ 12 0

3 3 1 § and B  £ 0 3 4

a. 12 B Solution



3 1 1 a. B  a b £ 0 2 2 4 1 12 2 132  £ 1 12 2 102 1 12 2 142

2 6 § , compute the following: 0.4

b. 4A  12 B 2 6 § 0.4 3 1 12 2122 2 1 12 2162 §  £ 0 2 1 12 2 10.42

1 3 § 0.2

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1 1 b. 4A  B  4A  a b B rewrite using algebraic addition 2 2 142 142 142132 112 2122 112 2 132  £ 142 1 12 2 142112 §  £ 112 2 102 112 2162 § 1 142 102 142132 12 2 142 112 210.42 32 16 12 1  £ 2 4 §  £ 0 3 § simplify 0 12 2 0.2 16  132 2 12  1 11 35 2  £ 2  0 4  132 §  £ 2 7 § result 02 12  10.22 2 11.8 Now try Exercises 21 through 24



Matrix Multiplication Consider a cable company offering three different levels of Internet service: Bronze — fast, Silver—very fast, and Gold—lightning fast. Table 8.3 shows the number and types of programs sold to households and businesses for the week. Each program has an incentive package consisting of a rebate and a certain number of free weeks, as shown in Table 8.4. Table 8.3 Matrix A Bronze

Silver

Table 8.4 Matrix B Gold

Rebate

Free Weeks

Homes

40

20

25

Bronze

$15

2

Businesses

10

15

45

Silver

$25

4

Gold

$35

6

To compute the amount of rebate money the cable company paid to households for the week, we would take the first row (R1) in Table 8.3 and multiply by the corresponding entries (bronze with bronze, silver with silver, and so on) in the first column (C1) of Table 8.4 and add these products. In matrix form, we have 15 3 40 20 254 # £ 25 §  40 # 15  20 # 25  25 # 35  $1975. Using R1 of Table 8.3 35 with C2 from Table 8.4 gives the number of free weeks awarded to homes: 2 340 20 25 4 # £ 4 §  40 # 2  20 # 4  25 # 6  310. Using the second row (R2) of 6 Table 8.3 with the two columns from Table 8.4 will give the amount of rebate money and the number of free weeks, respectively, awarded to business customers. When all computations are complete, the result is a product matrix P with order 2  2. This is because the product of R1 from matrix A, with C1 from matrix B, gives the entry in 15 2 40 20 25 1975 310 # d £ 25 4 §  c d. position P11 of the product matrix: c 10 15 45 2100 350 35 6 Likewise, the product R1 # C2 will give entry P12 (310), the product of R2 with C1

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will give P21 (2100), and so on. This “row  column” multiplication can be generalized, and leads to the following. Given m  n matrix A and s  t matrix B, A 1m  n2

c

B 1s  t 2

A 1m  n2

c

B 1s  t 2

c

matrix multiplication is possible only when ns

c

result will be an m  t matrix

In more formal terms, we have the following definition of matrix multiplication. Matrix Multiplication Given the m  n matrix A  3aij 4 and the s  t matrix B  3 bij 4. If n  s, then matrix multiplication is possible and the product AB is an m  t matrix P  3pij 4, where pij is product of the ith row of A with the jth column of B. In less formal terms, matrix multiplication involves multiplying the row entries of the first matrix with the corresponding column entries of the second, and adding them together. In Example 5, two of the matrix products [parts (a) and (b)] are shown in full detail, with the first entry of the product matrix color-coded.

EXAMPLE 5



Multiplying Matrices Given the matrices A through E shown here, compute the following products: a. AB

A c

Solution



2 3

b. CD 1 d 4

B c

4 6

3 d 1

c. DC 2 1 C £ 1 0 4 1

1 4 3 2 d  c dc 4 6 1 36 122142  112162 Computation: c 132 142  142 162

a. AB  c

2 3

d. AE 3 2 § 1

2 D  £4 0

5 d 13 122132  112 112 d 132132  142112

0 2 7 2 1 3 2 5 1 b. CD  £ 1 0 2 § £ 4 1 1 §  £ 2 11 3 § 12 16 7 4 1 1 0 3 2 122122  112142  (3)(0) Computation: £ 112 122  102 142  122(0) 142 122  112 142  112(0)

e. EA 5 1 2 1 1 § E £ 3 3 2 1 A B (2  2) (2  2)

c

A B (2  2) (2  2)

c

multiplication is possible since 2  2 C D (3  3) (3  3)

c

c

c result will be a 2  2 matrix

C D (3  3) (3  3)

c

multiplication is possible since 3  3

122 152  112112  (3)132 112 152  102 112  122 132 142 152  112 112  112 132

2 5 1 2 1 3 5 3 15 c. DC  £ 4 1 1 § £ 1 0 2 §  £ 5 5 9 § 0 3 2 4 1 1 5 2 8

c

1 0 § 2

c result will be a 3  3 matrix

122 112  112 112  (3)122 112112  102112  122122 § 142112  112112  112122

D (3  3)

c

C (3  3)

D C (3  3) (3  3)

c

c

multiplication is possible since 3  3

c

result will be a 3  3 matrix

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Section 8.6 The Algebra of Matrices

d. AE  c

2 3

2 EA  £ 3 1

e.

2 1 d£ 3 4 1 1 2 0 §c 3 2

A (2  2)

1 0 § 2

c

E (3  2)

c

multiplication is not possible since 2  3 E (3  2)

6 3 § 9

1 1 d  £ 6 4 4

c

A (2  2)

E A (3  2) (2  2)

c

c

multiplication is possible since 2  2

c

result will be a 3  2 matrix

Now try Exercises 25 through 36



Example 5 shows that in general, matrix multiplication is not commutative. Parts (b) and (c) show CD  DC since we get different results, and parts (d) and (e) show AE  EA, since AE is not defined while EA is. Operations on matrices can be a laborious process for larger matrices and for matrices with noninteger or large entries. For these, we can turn to available technology for assistance.This shifts our focus from a meticulous computation of entries, to carefully entering each matrix into the calculator, double-checking each entry, and appraising results to see if they’re reasonable. EXAMPLE 6



Using Technology for Matrix Operations Use a calculator to compute the difference A  B for the matrices given. 2 11

A  £ 0.9 0 Solution



0.5 6

4 § 5 12

B £

1 6 11 25

4

7 10

0 5 9

0.75 5 § 5 12

The entries for matrix A are shown in Figure 8.23. After entering matrix B, exit to MODE (QUIT)], call up matrix A, press the ⴚ the home screen [ 2nd (subtract) key, then call up matrix B and press ENTER . The calculator quickly finds the difference and displays the results shown in Figure 8.24. The last line on the screen shows the result can be stored for future use in a new matrix C by pressing the STO key, calling up matrix C, and pressing ENTER . Figure 8.23

Figure 8.25

3 4

6 5

Figure 8.24

Now try Exercises 37 through 40



In Figure 8.24 the dots to the right on the calculator screen indicate there are additional digits or matrix columns that can’t fit on the display, as often happens with larger matrices or decimal numbers. Sometimes, converting entries to fraction form will provide a display that’s easier to read. Here, this is done by calling up the matrix C, and using the MATH 1:  Frac option. After pressing ENTER , all entries are converted to fractions in simplest form (where possible), as in Figure 8.25. The third column can be viewed by pressing the right arrow.

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EXAMPLE 7



Using Technology for Matrix Operations Use a calculator to compute the product AB.

Solution



2 1 A ≥ 6 3

Carefully enter matrices A and B into the MODE (QUIT) calculator, then press 2nd to get to the home screen. Use [A][B] ENTER , and the calculator finds the product shown in the figure.

2 1 AB  ≥ 6 3

3 5 0 2

0 1 2 4 ¥ £ 0.5 2 2 1

0.7 3.2 3 4

3 5 0 2

0 1 2 4 ¥ B  £ 0.5 2 2 1

A (4  3)

c

0.7 3.2 3 4

B (3  3)

A (4  3)

c

c

multiplication is possible since 3  3

1 3 § 4 B (3  3)

c

result will be a 4  3 matrix

1 3 § 4

Now try Exercises 41 through 52



Properties of Matrix Multiplication Earlier, Example 5 demonstrated that matrix multiplication is not commutative. Here is a group of properties that do hold for matrices. You are asked to check these properties in the exercise set using various matrices. See Exercises 53 through 56. Properties of Matrix Multiplication Given matrices A, B, and C for which the products are defined: I. A1BC2  1AB2C II. A1B  C2  AB  AC

III. 1B  C2A  BA  CA IV. k1A  B2  kA  kB

matrix multiplication is associative matrix multiplication is distributive from the left matrix multiplication is distributive from the right a constant k can be distributed over addition

We close this section with an application of matrix multiplication. There are many other interesting applications in the exercise set.

EXAMPLE 8



Using Matrix Multiplication to Track Volunteer Enlistments In a certain country, the number of males and females that will join the military depends on their age. This information is stored in matrix A (Table 8.5). The likelihood a volunteer will join a particular branch of the military also depends on their age, with this information stored in matrix B (Table 8.6). (a) Compute the product P  AB and discuss/interpret what is indicated by the entries P11, P13, and P24 of the product matrix. (b) How many males are expected to join the Navy this year?

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Table 8.6 Matrix B

Table 8.5 Matrix A A

Solution



B

Age Groups

Likelihood of Joining

Sex

18–19

20–21

22–23

Age Group

Army

Navy

Air Force

Marines

Female

1000

1500

500

18–19

0.42

0.28

0.17

0.13

Male

2500

3000

2000

20–21

0.38

0.26

0.27

0.09

22–23

0.33

0.25

0.35

0.07

a. Matrix A has order 2  3 and matrix B has order 3  4. The product matrix P can be found and is a 2  4 matrix. Carefully enter the matrices in your calculator. Figure 8.26 shows the entries of matrix B. Using 3A 4 3 B4 ENTER , the calculator finds the product matrix shown in Figure 8.27. Pressing the right arrow shows the complete product matrix is P c

1155 2850

795 1980

750 1935

300 d. 735

The entry P11 is the product of R1 from A and C1 from B, and indicates that for the year, 1155 females are projected to join the Army. In like manner, entry P13 shows that 750 females are projected to join the Air Force. Entry P24 indicates that 735 males are projected to join the Marines. Figure 8.26

Figure 8.27

C. You’ve just learned how to compute the product of two matrices

b. The product R2 (males) # C2 (Navy) gives P22  1980, meaning 1980 males are expected to join the Navy. Now try Exercise 59 through 66



8.6 EXERCISES CONCEPTS AND VOCABULARY Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. Two matrices are equal if they are like size and the corresponding entries are equal. In symbols, A  B  if . 2. The sum of two matrices (of like size) is found by adding the corresponding entries. In symbols, AB . 3. The product of a constant times a matrix is called multiplication.

4. The size of a matrix is also referred to as its 1 2 3 d is The order of A  c . 4 5 6

.

5. Give two reasons why matrix multiplication is generally not commutative. Include several examples using matrices of various sizes. 6. Discuss the conditions under which matrix multiplication is defined. Include several examples using matrices of various sizes.

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DEVELOPING YOUR SKILLS

State the order of each matrix and name the entries in positions a12 and a23 if they exist. Then name the position aij of the 5 in each.

7. c

19 8. £ 11 § 5

3 d 7

1 5

3 5

2 9. c 0 2 11. £ 0 5

2 10. £ 0.1 0.3

0.5 d 6 7 1§ 4

1 8 1

89 12. £ 13 2

0.4 5§ 3 55 8 1

34 5 1

21 3§ 0

11 116

14 132 7 5 2 5

3 2 14. ≥ 1 2 2 15. £ 2b 0

3 5 9

18 1 d  c 164 4 13 10 1.5 ¥  c 1 0.5 3

a c 4 §  £6 3c 0

2p  1 16. £ 1 q5

5 12 9

2 12 d 8

2 4 12 1.4 0.4

1.3 d 0.3

4 a § 6

3 5 3b

5 3r 3p

9 7 0 §  £ 1 2r 2

2q 0 § 8

For matrices A through H as given, perform the indicated operation(s), if possible. Do not use a calculator. If an operation cannot be completed, state why.

A c

2 5

2 C  £ 0.2 1 1 E  £0 4

2 B £ 1 § 3

3 d 8 0.5 5 § 3 2 1 3

0 2 § 6

1 D  £0 0 6 F c 12

0 1 0 3 0

0 0§ 1 9 d 6

2 1 3

0 2 § 6

H c

3 d 2

8 5

17. A  H

18. E  G

19. F  H

20. G  D

21. 3H  2A

22. 2E  3G

23.

Determine if the following statements are true, false, or conditional. If false, explain why. If conditional, find values of a, b, c, p, q, and r that will make the statement true.

13. c

1 G £ 0 4

1 E  3D 2

2 24. F  F 3

25. ED

26. DE

27. AH

28. HA

29. FD

30. FH

31. HF

32. EB

33. H 2

34. F 2

35. FE

36. EF

For matrices A through H as given, use a calculator to perform the indicated operation(s), if possible. If an operation cannot be completed, state why.

A c

5 3

C c

13 2

13 3

13

2 13

1 E  £0 4 0

G

£ 1 2 1 4

4 d 9

2 1 3 3 4 3 8 11 16

B c d

1 0

1 D  £0 0

0 d 1 0 1 0

0 0.52 2 § F c 1.021 6 1 4 1 8 § 1 16

3

H  c 19 1 19

0 0§ 1 0.002 1.27

4 57 5 d 57

37. C  H

38. A  H

39. E  G

40. G  E

41. AH

42. HA

43. EG

44. GE

45. HB

46. BH

47. DG

48. GD

49. C

2

51. FG

50. E 2 52. AF

1.032 d 0.019

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For Exercises 53 through 56, use a calculator and matrices A, B, and C to verify each statement.

53. Matrix multiplication is not generally commutative: (a) AB  BA, (b) AC  CA, and (c) BC  CB.

1 A £ 2 4

3 7 0

54. Matrix multiplication is distributive from the left: A1B  C2  AB  AC.

45 C  £ 6 21

1 10 28



c

5 1 § 6

0.3 B  £ 2.5 1

0.4 2 0.5

1.2 0.9 § 0.2

3 15 § 36

55. Matrix multiplication is distributive from the right: 1B  C2A  BA  CA. 56. Matrix multiplication is associative: 1AB2C  A1BC2.

WORKING WITH FORMULAS

2 W

2 d 0

#

c

L Perimeter d  c d W Area

rectangles shown, then check the results using P  2L  2W and A  LW. 57.

The perimeter and area of a rectangle can be simultaneously calculated using the matrix formula shown, where L represents the length and W represents the width of the rectangle. Use the matrix formula and your calculator to find the perimeter and area of the 

6.374 cm

4.35 cm

58.

5.02 cm

3.75 cm

APPLICATIONS

59. Custom T’s designs and sells specialty T-shirts and sweatshirts, with plants in Verdi and Minsk. The company offers this apparel in three quality levels: standard, deluxe, and premium. Last fall the Verdi office produced 3820 standard, 2460 deluxe, and 1540 premium T-shirts, along with 1960 standard, 1240 deluxe, and 920 premium sweatshirts. The Minsk office produced 4220 standard, 2960 deluxe, and 1640 premium T-shirts, along with 2960 standard, 3240 deluxe, and 820 premium sweatshirts in the same time period. a. Write a 3  2 “production matrix” for each plant 3V S Verdi, M S Minsk], with a T-shirt column, a sweatshirt column, and three rows showing how many of the different types of apparel were manufactured. b. Use the matrices from Part (a) to determine how many more or less articles of clothing were produced by Minsk than Verdi. c. Use scalar multiplication to find how many shirts of each type will be made at Verdi and Minsk next fall, if each is expecting a 4% increase in business. d. What will be Custom T’s total production next fall (from both plants), for each type of apparel? 60. Terry’s Tire Store sells automobile and truck tires through three retail outlets. Sales at the Cahokia store for the months of January, February, and March

broke down as follows: 350, 420, and 530 auto tires and 220, 180, and 140 truck tires. The Shady Oak branch sold 430, 560, and 690 auto tires and 280, 320, and 220 truck tires during the same 3 months. Sales figures for the downtown store were 864, 980, and 1236 auto tires and 535, 542, and 332 truck tires. a. Write a 2  3 “sales matrix” for each store 3C S Cahokia, S S Shady Oak, D S Downtown], with January, February, and March columns, and two rows showing the sales of auto and truck tires respectively. b. Use the matrices from Part (a) to determine how many more or fewer tires of each type the downtown store sold (each month) over the other two stores combined. c. Market trends indicate that for the same three months in the following year, the Cahokia store will likely experience a 10% increase in sales, the Shady Oak store a 3% decrease, with sales at the downtown store remaining level (no change). What will be the combined monthly sales from all three stores next year, for each type of tire? 61. Home improvements: Dream-Makers Home Improvements specializes in replacement windows, replacement doors, and new siding. During the peak season, the number of contracts that came from various parts of the city (North, South, East, and West) are shown in matrix C. The average

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profit per contract is shown in matrix P. Compute the product PC and discuss what each entry of the product matrix represents. N Windows 9 Doors £ 7 Siding 2

S 6 5 3

E W 5 4 7 6§  C 5 2

Windows Doors 3 1500 500

Siding 2500 4  P

62. Classical music: Station 90.7—The Home of Classical Music —is having their annual fund drive. Being a loyal listener, Mitchell decides that for the next 3 days he will donate money according to his favorite composers, by the number of times their music comes on the air: $3 for every piece by Mozart (M ), $2.50 for every piece by Beethoven (B ), and $2 for every piece by Vivaldi (V ). This information is displayed in matrix D. The number of pieces he heard from each composer is displayed in matrix C. Compute the product DC and discuss what each entry of the product matrix represents. Mon. Tue. Wed. 3 5 2 4§  C 3 3 M B V 33 2.5 2 4  D

M 4 B £3 V 2

63. Pizza and salad: The science department and math department of a local college are at a pre-semester retreat, and decide to have pizza, salads, and soft drinks for lunch. The quantity of food ordered by each department is shown in matrix Q. The cost of the food item at each restaurant is shown in matrix C using the published prices from three popular restaurants: Pizza Home (PH), Papa Jeff’s (PJ), and Dynamos (D). a. What is the total cost to the math department if the food is ordered from Pizza Home? b. What is the total cost to the science department if the food is ordered from Papa Jeff’s? c. Compute the product QC and discuss the meaning of each entry in the product matrix. Pizza Science 8 c Math 10 PH Pizza 8 Salad £ 1.5 Drink 0.90

Salad 12 8 PJ 7.5 1.75 1

Drink 20 d Q 18 D 10 2 § C 0.75

64. Manufacturing pool tables: Cue Ball Incorporated makes three types of pool tables, for homes, commercial use, and professional use. The amount of time required to pack, load, and install each is summarized in matrix T, with all times in hours. The cost of these components in dollars per hour, is summarized in matrix C for two of its warehouses, one on the west coast and the other in the midwest. a. What is the cost to package, load, and install a commercial pool table from the coastal warehouse? b. What is the cost to package, load, and install a commercial pool table from the warehouse in the midwest? c. Compute the product TC and discuss the meaning of each entry in the product matrix. Pack Load Install Home 1 0.2 1.5 Comm £ 1.5 0.5 2.2 §  T Prof 1.75 0.75 2.5 Coast Midwest Pack 10 8 Load £ 12 10.5 §  C Install 13.5 12.5 65. Joining a club: Each school year, among the students planning to join a club, the likelihood a student joins a particular club depends on their class standing. This information is stored in matrix C. The number of males and females from each class that are projected to join a club each year is stored in matrix J. Compute the product JC and use the result to answer the following: a. Approximately how many females joined the chess club? b. Approximately how many males joined the writing club? c. What does the entry P13 of the product matrix tells us? Fresh Female 25 c Male 22 Spanish Fresh 0.6 Soph £ 0.5 Junior 0.4

Soph Junior 18 21 d J 19 18 Chess 0.1 0.2 0.2

Writing 0.3 0.3 §  C 0.4

66. Designer shirts: The SweatShirt Shoppe sells three types of designs on its products: stenciled (S), embossed (E), and applique (A). The quantity of each size sold is shown in matrix Q. The retail price

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of each sweatshirt depends on its size and whether it was finished by hand or machine. Retail prices are shown in matrix C. Assuming all stock is sold, a. How much revenue was generated by the large sweatshirts? b. How much revenue was generated by the extralarge sweatshirts? c. What does the entry P11 of the product matrix QC tell us? 

S med 30 large £ 60 x-large 50 Hand S 40 E £ 60 A 90

E 30 50 40

791

A 15 20 §  Q 30

Machine 25 40 §  C 60

EXTENDING THE CONCEPT

67. For the matrix A shown, use your calculator to compute A2, A3, A4, and A5. Do you notice a pattern? Try to write a “matrix formula” for An, where n is a positive integer, then use your formula to find A6. Check results using a calculator. 1 0 1 A  £1 1 1§ 1 0 1 68. The matrix M  c

2 3

1 d has some very 2

M 3, M 4, and M 5, then discuss what you find. Try to find/create another 2  2 matrix that has similar properties. 69. For the “matrix equation” 1 0 2 1 # a b d  c d , use matrix c d c c d 0 1 3 2 multiplication and two systems of equations to find the entries a, b, c, and d that make the equation true.

interesting properties. Compute the powers M 2, 

MAINTAINING YOUR SKILLS

70. (5.2) Solve the system using elimination. x  2y  z  3 • 2x  y  3z  5 5x  3y  2z  2

72. (7.6) Solve z4  81i  0 using the nth roots theorem. Leave your answer in trigonometric form. 73. (3.2) Find the quotient using synthetic division, then check using multiplication.

71. (6.5) Evaluate cos1cos10.32112 .

x3  9x  10 x2

8.7 Solving Linear Systems Using Matrix Equations Learning Objectives In Section 8.7 you will learn how to:

A. Recognize the identity matrix for multiplication

While using matrices and row operations offers a degree of efficiency in solving systems, we are still required to solve for each variable individually. Using matrix multiplication we can actually rewrite a given system as a single matrix equation, in which the solutions are computed simultaneously. As with other kinds of equations, the use of identities and inverses are involved, which we now develop in the context of matrices.

B. Find the inverse of a square matrix

C. Solve systems using matrix equations

D. Use determinants to find whether a matrix is invertible

A. Multiplication and Identity Matrices From the properties of real numbers, 1 is the identity for multiplication since n # 1  1 # n  n. A similar identity exists for matrix multiplication. Consider the 2  2 1 4 matrix A  c d . While matrix multiplication is not generally commutative, 2 3

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if we can find a matrix B where AB  BA  A, then B is a prime candidate for the identity matrix, which is denoted I. For the products AB and BA to be possible and have the same order as A, we note B must also be a 2  2 matrix. Using the arbitrary matrix a b B c d , we have the following. c d EXAMPLE 1A



Solving AB  A to Find the Identity Matrix For c

1 4 a b 1 4 d c d  c d , use matrix multiplication, the equality of 2 3 c d 2 3 matrices, and systems of equations to find the value of a, b, c, and d. Solution



a  4c b  4d 1 4 d  c d. 2a  3c 2b  3d 2 3 Since corresponding entries must be equal (shown by matching colors), we can find a  4c  1 b  4d  4 a, b, c, and d by solving the systems e and e . For 2a  3c  2 2b  3d  3 the first system, 2R1  R2 shows a  1 and c  0. Using 2R1  R2 for the second 1 0 shows b  0 and d  1. It appears c d is a candidate for the identity matrix. 0 1 The product on the left gives c



Before we name B as the identity matrix, we must show that AB  BA  A. EXAMPLE 1B



Verifying AB  BA  A Given A  c

Solution



1 2

4 1 d and B  c 3 0

0 d , determine if AB  A and BA  A. 1

1 4 1 0 d d c 2 3 0 1 1112  4102 1102  4112 d  c 2112  3102 2102  3112 1 4  c d A✓ 2 3

AB  c

BA  c

1 0 1 4 d c d 0 1 2 3 1112  0122 1142  0132  c d 0112  1122 0142  1132 1 4  c d A✓ 2 3

Since AB  A  BA, B is the identity matrix I. Now try Exercises 7 through 10 By replacing the entries of A  c



1 4 d with those of the general matrix 2 3 a11 a12 1 0 c d , we can show that I  c d is the identity for all 2  2 matrices. In a21 a22 0 1 considering the identity for larger matrices, we find that only square matrices have inverses, since AI  IA is the primary requirement (the multiplication must be possible in both directions). This is commonly referred to as multiplication from the right and multiplication from the left. Using the same procedure as before we can show 1 0 0 £ 0 1 0 § is the identity for 3  3 matrices (denoted I3). The n  n identity 0 0 1 matrix In consists of 1’s down the main diagonal and 0’s for all other entries. Also, the identity In for a square matrix is unique.

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Figure 8.28

A. You’ve just learned how to recognize the identity matrix for multiplication

793

As in Section 9.2, a graphing calculator can be used to investigate operations on 2 5 1 matrices and matrix properties. For the 3  3 matrix A  £ 4 1 1 § and 0 3 2 1 0 0 I3  £ 0 1 0 § , a calculator will confirm that AI3  A  I3 A. Carefully enter A 0 0 1 into your calculator as matrix A, and I3 as matrix B. Figure 8.28 shows AB  A and after pressing ENTER , the calculator will verify BA  A, although the screen cannot display the result without scrolling. See Exercises 11 through 14.

B. The Inverse of a Matrix Again from the properties of real numbers, we know the multiplicative inverse for a is 1 a1  1a  02, since the products a # a1 and a1 # a yield the identity 1. To show a 6 5 d and that a similar inverse exists for matrices, consider the square matrix A  c 2 2 a b d . If we can find a matrix B, where AB  BA  I, then an arbitrary matrix B  c c d B is a prime candidate for the inverse matrix of A, which is denoted A1. Proceeding as in Examples 1A and 1B gives the result shown in Example 2. EXAMPLE 2A



Solving AB  I to find A1 For c

6 5 a b 1 0 d c d  c d , use matrix multiplication, the equality of matrices, 2 2 c d 0 1 and systems of equations to find the entries of B. Solution



6a  5c 6b  5d 1 0 d  c d . Since corresponding 2a  2c 2b  2d 0 1 entries must be equal (shown by matching colors), we find the values of a, b, c, and 6b  5d  0 6a  5c  1 . Using 3R2  R1 d by solving the systems e and e 2b  2d  1 2a  2c  0 for the first system shows a  1 and c  1, while 3R2  R1 for the second a b 1 2.5 d  c d is the system shows b  2.5 and d  3. Matrix B  c c d 1 3 prime candidate for A1. The product on the left gives c

To determine if A1 has truly been found, we check to see if multiplication from the right and multiplication from the left yields the matrix I: AB  BA  I. EXAMPLE 2B



Verifying B  A1 For the matrices A  c if AB  BA  I.

6 2

5 1 2.5 d and B  c d from Example 2A, determine 2 1 3

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Solution



6 5 1 2.5 d c d 2 2 1 3 6112  5112 612.52  5132  c d 2112  2112 212.52  2132 1 0  c d✓ 0 1

AB  c

1 2.5 6 5 d c d 1 3 2 2 1162  12.52 122 1152  12.52122  c d 1162  3122 1152  3122 1 0  c d✓ 0 1

BA  c

Since AB  BA  I, we conclude B  A 1. Now try Exercises 15 through 22



These observations guide us to the following definition of an inverse matrix. The Inverse of a Matrix Given an n  n matrix A, if there exists an n  n matrix A1 such that AA1  A1A  In, then A1 is the inverse of matrix A. We will soon discover that while only square matrices have inverses, not every square matrix has an inverse. If an inverse exists, the matrix is said to be invertible. For 2  2 matrices that are invertible, a simple formula exists for computing the inverse. The formula is derived in the Strengthening Core Skills feature at the end of Chapter 8. The Inverse of a 2  2 Matrix If A  c

1 a b d d , then A1  c c d ad  bc c

b d provided ad  bc  0 a

To “test” the formula, again consider the matrix A  c c  2, and d  2: B. You’ve just learned how to find the inverse of a square matrix

A1  

6 2

5 d , where a  6, b  5, 2

1 2 5 c d 162122  152122 2 6

1 2 5 1 c d  c 2 2 6 1

2.5 d✓ 3

See Exercises 63 through 66 for more practice with this formula. Almost without exception, real-world applications involve much larger matrices, with entries that are not integer-valued. Although the equality of matrices method from Example 2 can be extended to find the inverse of larger matrices, the process becomes very tedious and too time consuming to be useful. As an alternative, the augmented matrix method can be used. This process is discussed in the Strengthening Core Skills feature at the end Chapter 8 (see page 823). For practical reasons, we will rely on a calculator to produce these larger inverse matrices. This is done by (1) carefully entering a square matrix A into the calculator, (2) returning to the home screen and (3) calling up matrix A and pressing the Xⴚ1 key and ENTER to find A1. In the context of matrices, calculators are programmed to compute an inverse matrix, rather than to somehow find a reciprocal. See Exercises 23 through 26.

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795

C. Solving Systems Using Matrix Equations One reason matrix multiplication has its row  column definition is to assist in writing a linear system of equations as a single matrix equation. The equation consists of the matrix of constants B on the right, and a product of the coefficient matrix A with x  4y  z  10 the matrix of variables X on the left: AX  B. For • 2x  5y  3z  7, the matrix 8x  y  2z  11 1 4 1 x 10 equation is £ 2 5 3 § £ y §  £ 7 § . Note that computing the product on the left 8 1 2 z 11 will yield the original system. Once written as a matrix equation, the system can be solved using an inverse matrix and the following sequence. If A represents the matrix of coefficients, X the matrix of variables, B the matrix of constants, and I the appropriate identity, the sequence is 112 122 132 142 152

AX  B A1 1AX2  A1B 1A1A2X  A1B IX  A1B X  A1B

matrix equation multiply from the left by the inverse of A associative property A1 A  I IX  X

Lines 1 through 5 illustrate the steps that make the method work. In actual practice, after carefully entering the matrices, only step 5 is used when solving matrix equations using technology. Once matrix A is entered, the calculator will automatically find and use A1 as we enter A1B.

EXAMPLE 3



Using Technology to Solve a Matrix Equation Use a calculator and a matrix equation to solve the system x  4y  z  10 • 2x  5y  3z  7. 8x  y  2z  11

Solution



1 4 1 x 10 As before, the matrix equation is £ 2 5 3 § £ y §  £ 7 § . 8 1 2 z 11 Carefully enter (and double-check) the matrix of coefficients as matrix A in your calculator, and the matrix of constants as matrix B. The product A1B shows the solution is x  2, y  3, z  4. Verify by substitution.

Now try Exercises 27 through 44



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The matrix equation method does have a few shortcomings. Consider the system 4 10 x 8 d c d  c d . After entering whose corresponding matrix equation is c 2 5 y 13 the matrix of coefficients A and matrix of constants B, attempting to compute A1B results in the error message shown in Figure 8.29. The calculator is unable to return a solution due to something called a “singular matrix.” To investigate further, we 4 10 d using the formula for a 2  2 matrix. With attempt to find A1 for c 2 5 a  4, b  10, c  2, and d  5, we have

Figure 8.29

1 d c ad  bc c 1 5 10  c d 0 2 4

A1 

C. You’ve just learned how to solve systems using matrix equations

b 1 5 d  c a 142 152  1102 122 2

10 d 4

Since division by zero is undefined, we conclude that matrix A has no inverse. A matrix having no inverse is said to be singular or noninvertible. Solving systems using matrix equations is only possible when the matrix of coefficients is nonsingular.

D. Determinants and Singular Matrices As a practical matter, it becomes important to know ahead of time whether a particular matrix has an inverse. To help with this, we introduce one additional operation on a square matrix, that of calculating its determinant. For a 1  1 matrix the determinant a11 a12 d , the determinant of A, written is the entry itself. For a 2  2 matrix A  c a21 a22 as det(A) or denoted with vertical bars as A, is computed as a difference of diagonal products beginning with the upper-left entry:

det1A2  `

a11 a21

2nd diagonal product a12 `  a11a22  a21a12 a22 1st diagonal product

The Determinant of a 2  2 Matrix Given any 2  2 matrix A  c

a12 d, a22 det1A2  A  a11a22  a21a12

EXAMPLE 4



a11 a21

Calculating Determinants Compute the determinant of each matrix given. 3 2 a. B  c d 1 6 5 2 1 b. C  c d 1 3 4 4 10 c. D  c d 2 5

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Solution




3 2 `  132 1
62
112 122 
20 1
6 b. Determinants are only defined for square matrices. 4
10 `  142 152
1
221
102  20
20  0 c. det1D2  `
2 5 a. det1B2  `

Now try Exercises 45 through 48




4
10 d was zero, and this
2 5 is the same matrix we earlier found had no inverse. This observation can be extended to larger matrices and offers the connection we seek between a given matrix, its inverse, and matrix equations. Notice from Example 4c, the determinant of c

Singular Matrices If A is a square matrix and det1A2  0, the inverse matrix does not exist and A is said to be singular or noninvertible.

WORTHY OF NOTE For the determinant of a general n  n matrix using cofactors, see Appendix III.

In summary, inverses exist only for square matrices, but not every square matrix has an inverse. If the determinant of a square matrix is zero, an inverse does not exist and the method of matrix equations cannot be used to solve the system. To use the determinant test for a 3  3 system, we need to compute a 3  3 determinant. At first glance, our experience with 2  2 determinants appears to be of little help. However, every entry in a 3  3 matrix is associated with a smaller 2  2 matrix, formed by deleting the row and column of that entry and using the entries that remain. These 2  2’s are called the associated minor matrices or simply the minors. Using a general matrix of coefficients, we’ll identify the minors associated with the entries in the first row. a11 a12 £ a21 a22 a31 a32

a13 a23 § a33

Entry: a11 associated minor a22 a23 c d a32 a33

a11 £ a21 a31

a12 a13 a22 a23 § a32 a33

Entry: a12 associated minor a21 a23 c d a31 a33

a11 £ a21 a31

a12 a13 a22 a23 § a32 a33

Entry: a13 associated minor a21 a22 d c a31 a32

To illustrate, consider the system shown, and (1) form the matrix of coefficients, (2) identify the minor matrices associated with the entries in the first row, and (3) compute the determinant of each minor. 2x  3y
z  1 • x
4y  2z 
3 3x  y 
1 2 122 £ 1 3

3
4 1


1 2§ 0

Entry a11: 2 associated minor
4 2 c d 1 0

2 (1) Matrix of coefficients £ 1 3 2 £1 3

3
4 1


1 2§ 0

Entry a12: 3 associated minor 1 2 c d 3 0

2 £1 3

3
1
4 2 § 1 0 3
4 1


1 2§ 0

Entry a13:
1 associated minor 1
4 c d 3 1

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(3) Determinant of minor

142 102  112122  2

Determinant of minor

Determinant of minor

112 102  132122  6 112 112  132142  13

For computing a 3  3 determinant, we illustrate a technique called expansion by minors. The Determinant of a 3  3 Matrix — Expansion by Minors matrix M For the matrix M shown, det(M) is the unique number computed as follows: a11 a12 a13 1. Select any row or column and form the product of each £ a21 a22 a23 § entry with its minor matrix. The illustration here uses the a31 a32 a33 entries in row 1: det1M2  a11 `

a22 a23 a21 `  a12 ` a32 a33 a31

a23 a21 a22 `  a13 ` ` a33 a31 a32

2. The signs used between terms of the expansion depends on the row or column chosen, according to the sign chart shown.

Sign Chart  £  

  

  § 

The determinant of a matrix is unique and any row or column can be used. For this reason, it’s helpful to select the row or column having the most zero, positive, and/or smaller entries. EXAMPLE 5



Calculating a 3  3 Determinant 2 1 Compute the determinant of M  £ 1 1 2 1

Solution



3 0 §. 4

Since the second row has the “smallest” entries as well as a zero entry, we compute the determinant using this row. According to the sign chart, the signs of the terms will be negative–positive–negative, giving 1 3 2 3 2 1 `  112 ` `  102 ` ` 1 4 2 4 2 1  114  32  11218  62  10212  22  7  (2)  0  9 S The value of det1M2 is 9.

det1M2  112 `

Now try Exercises 49 through 54



Try computing the determinant of M two more times, using a different row or column each time. Since the determinant is unique, you should obtain the same result. There are actually other alternatives for computing a 3  3 determinant. The first is called determinants by column rotation, and takes advantage of patterns generated from the expansion of minors. This method is applied to the matrix shown, which uses alphabetical entries for simplicity. a det £ d g

b e h

c f §  a1ei  fh2  b1di  fg2  c1dh  eg2  aei  afh  bdi  bfg  cdh  ceg i  aei  bfg  cdh  afh  bdi  ceg

expansion using R1 distribute rewrite result

Although history is unsure of who should be credited, notice that if you repeat the first two columns to the right of the given matrix (“rotation of columns”), identical

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products are obtained using the six diagonals formed—three in the downward direction using addition, three in the upward direction using subtraction. a £d g

b e h

gec c f§ i aei

hfa idb a b d e g h bfg cdh

Adding the products in blue (regardless of sign) and subtracting the products in red (regardless of sign) gives the determinant. This method is more efficient than expansion by minors, but can only be used for 3  3 matrices! EXAMPLE 6



Calculating det(A) Using Column Rotation 1 Use the column rotation method to find the determinant of A  £ 2 3

Solution



5 3 8 0 § . 11 1

Rotate columns 1 and 2 to the right as above, and compute the diagonal products. 1 £ 2 3

72 0 10 1 5 5 3 8 0 § 2 8 3 11 11 1 8 0 66

Adding the products in blue (regardless of sign) and subtracting the products in red (regardless of sign) shows det1A2  4: 8  0  66  72  0  1102  4.

Now try Exercises 55 through 58



The final method is presented in the Extending the Concept feature of the Exercise Set, and shows that if certain conditions are met, the determinant of a matrix can be found using its triangularized form. EXAMPLE 7



Solving a System after Verifying A is Invertible Given the system shown here, (1) form the matrix equation AX  B; (2) compute the determinant of the coefficient matrix and determine if you can proceed; and (3) if so, solve the system using a matrix equation. 2x  1y  3z  11  1 • 1x  1y 2x  1y  4z  8

Solution



1. Form the matrix equation AX  B: 2 £ 1 2

1 3 x 11 1 0 § £ y §  £ 1 § 1 4 z 8

2. Since det(A) is nonzero (from Example 5), we can proceed.

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3. For X  A1B, input A1B on your calculator and press X

4 9 £ 49 1 9

7 9 29 4 9

1 3 1 3§ 1 3

11 3 £ 1 §  £ 2 § 8 1

ENTER

.

calculator computes and uses A1 in one step

The solution is the ordered triple 13, 2, 12

Now try Exercises 59 through 62



We close this section with an application involving a 4  4 system. EXAMPLE 8



Solving an Application Using Technology and Matrix Equations A local theater sells four sizes of soft drinks: 32 oz @ $2.25; 24 oz @ $1.90; 16 oz @ $1.50; and 12 oz @ $1.20/each. As part of a “free guest pass” promotion, the manager asks employees to try and determine the number of each size sold, given the following information: (1) the total revenue from soft drinks was $719.80; (2) there were 9096 oz of soft drink sold; (3) there was a total of 394 soft drinks sold; and (4) the number of 24-oz and 12-oz drinks sold was 12 more than the number of 32-oz and 16-oz drinks sold. Write a system of equations that models this information, then solve the system using a matrix equation.

Solution



D. You’ve just learned how to use determinants to find whether a matrix is invertible

If we let x, l, m, and s represent the number of 32-oz, 24-oz, 16-oz, and 12-oz soft drinks sold, the following system is produced: revenue: 2.25x  1.90l  1.50m  1.20s  719.8 ounces sold: 32x  24l  16m  12s  9096 μ quantity sold: x  l  m  s  394 relationship between l  s  x  m  12 amounts sold: When written as a matrix equation the system becomes: 2.25 32 ≥ 1 1

1.9 24 1 1

1.5 1.2 x 719.8 16 12 l 9096 ¥ ≥ ¥  ≥ ¥ 1 1 m 394 1 1 s 12

To solve, carefully enter the matrix of coefficients as matrix A, and the matrix of constants as matrix B, then compute A1B  X [verify det1A2  0 4 . This gives a solution of 1x, l, m, s2  1112, 151, 79, 522.

Now try Exercises 67 to 78



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Section 8.7 Solving Linear Systems Using Matrix Equations

8.7 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. The n  n identity matrix In consists of 1’s down the and for all other entries. 2. Given square matrices A and B of like size, B is the inverse of A if   . Notationally we write B  . 3. The product of a square matrix A and its inverse A1 yields the matrix.

5. Explain why inverses exist only for square matrices, then discuss why some square matrices do not have an inverse. Illustrate each point with an example. 6. What is the connection between the determinant of a 2  2 matrix and the formula for finding its inverse? Use the connection to create a 2  2 matrix that is invertible, and another that is not.

4. If the determinant of a matrix is zero, the matrix is said to be or , meaning no inverse exists.



DEVELOPING YOUR SKILLS

Use matrix multiplication, equality of matrices, and the a b 1 0 arbitrary matrix given to show that c d  c d. c d 0 1

7. A  c

2 3

5 a b 2 dc d  c 7 c d 3

5 d 7

8. A  c

9 5

7 a b 9 dc d  c 4 c d 5

7 d 4

9. A  c

0.4 0.3

0.6 a b 0.4 dc d  c 0.3 0.2 c d

0.6 d 0.2

1

10. A  c 21 3

1 4 1d 8

c

1 a b d  c 21 c d 3

1 4 1d 8

1 0 0 0 d , I3  £ 0 1 0 § , and 1 0 0 1 1 0 0 0 0 1 0 0 I4  ≥ ¥ , show AI  IA  A for the 0 0 1 0 0 0 0 1 matrices of like size. Use a calculator for Exercise 14.

1 For I2  c 0

11. c

3 4

8 d 10

12. c

0.5 0.2 d 0.7 0.3

4 1 13. £ 9 5 0 2

6 3§ 1

9 2 14. ≥ 4 0

1 3 1 0 5 3 ¥ 6 1 0 2 4 1

Find the inverse of each 2  2 matrix using matrix multiplication, equality of matrices, and a system of equations.

15. c

5 2

4 d 2

16. c

1 0

5 d 4

17. c

1 4

3 d 10

18. c

2 0.4 d 1 0.8

Demonstrate that B  A1, by showing AB  BA  I. Do not use a calculator.

19. A  c

1 5 d 2 9

20. A  c

2 4

6 d 11

B c

9 5 d 2 1

B c

5.5 2

3 d 1

21. A  c

4 0

5 d 2

1

5 8 1d 2

B c4 0

22. A  c

2 5 d 3 4 4

B  c 73 7

5 7 2d 7

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Use a calculator to find A1  B, then confirm the inverse by showing AB  BA  I.

2 3 1 23. A  £ 5 2 4 § 2 0 1 0.5 24. A  £ 0 1

0.2 0.3 0.4

0.1 0.6 § 0.3

7 5 9 25. A  £ 1 2 2 12 1 0 ≥ 26. A  12 12 0

3 0 § 5

6 4 12 12

3 0 8 12 ¥ 0 0 0 12

Write each system in the form of a matrix equation. Do not solve.

27. e

2x  3y  9 5x  7y  8

28. e

0.5x  0.6y  0.6 0.7x  0.4y  0.375

x  2y  z  1 29. • x  z  3 2x  y  z  3 2x  3y  2z  4 30. • 14x  25y  34z  1 3 2x  1.3y  3z  5 2w  x  4y  5z  3 2w  5x  y  3z  4 31. μ 3w  x  6y  z  1 w  4x  5y  z  9 1.5w  2.1x  0.4y  z  1 0.2w  2.6x  y  5.8 32. μ 3.2x  z  2.7 1.6w  4x  5y  2.6z  1.8 Write each system as a matrix equation and solve (if possible) using inverse matrices and your calculator. If the coefficient matrix is singular, write no solution.

33. e

0.05x  3.2y  15.8 0.02x  2.4y  12.08

34. e

0.3x  1.1y  3.5 0.5x  2.9y  10.1

1 u  14v  1 35. e 16 2 2 u  3 v  2

36. e

12a  13b  15 16a  3b  17

37. e

1 3 5 8 a  5b  6 5 3 4 16 a  2 b  5

38. e

3 12a  2 13b  12 5 12a  3 13b  1

0.2x  1.6y  2z  1.9 39. • 0.4x  y  0.6z  1 0.8x  3.2y  0.4z  0.2 1.7x  2.3y  2z  41.5 40. • 1.4x  0.9y  1.6z  10 0.8x  1.8y  0.5z  16.5 x  2y  2z  6 41. • 2x  1.5y  1.8z  2.8 2 1 3 11 3 x  2y  5 z  30 42.

4x  5y  6z  5  35y  54z  2 3 0.5x  2.4y  5z  5

• 18x

2w  3x  4y  5z  3 0.2w  2.6x  y  0.4z  2.4 43. μ 3w  3.2x  2.8y  z  6.1 1.6w  4x  5y  2.6z  9.8 2w  5x  3y  4z  7 1.6w  4.2y  1.8z  5.4 44. μ 3w  6.7x  9y  4z  8.5 0.7x  0.9z  0.9 Compute the determinant of each matrix and state whether an inverse matrix exists. Do not use a calculator.

45. c

4 3

47. c

1.2 0.3

7 d 5 0.8 d 0.2

46. c

0.6 0.4

0.3 d 0.5

48. c

2 3

6 d 9

Compute the determinant of each matrix without using a calculator. If the determinant is zero, write singular matrix.

1 49. A  £ 0 2

0 2 1 1 § 1 4

2 2 50. B  £ 0 1 4 4

1 2§ 0

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2 3 51. C  £ 0 6 1 1.5 1 52. D  £ 2.5 3

Compute the determinant of each matrix using the column rotation method.

4 2 § 2

2 0.8 5 2 § 0 2.5

Use a calculator to compute the determinant of each matrix. If the determinant is zero, write singular matrix. If the determinant is nonzero, find A 1 and store the 2nd Xⴚ1 2: [B] ENTER ). result as matrix B ( STO Then verify each inverse by showing AB  BA  I.

1 2 53. A  ≥ 8 0

0 5 15 8

3 4 0 1 ¥ 6 5 4 1

1 2 0 1 54. M  ≥ 1 0 2 1



1 3 2 1

1 2 ¥ 3 4

2 55. £ 4 1

3 1 1 5 § 0 2

56.

3 £ 1 3

1 57. £ 3 4

1 2 2 4 § 3 1

58.

5 6 2 £ 2 1 2 § 3 4 1

2 4 2 0 § 1 5

For each system shown, form the matrix equation AX  B; compute the determinant of the coefficient matrix and determine if you can proceed; and if possible, solve the system using the matrix equation.

x  2y  2z  7 59. • 2x  2y  z  5 60. 3x  y  z  6

2x  3y  2z  7 • x  y  2z  5 3x  2y  z  11

x  3y  4z  1 61. • 4x  y  5z  7 62. 3x  2y  z  3

5x  2y  z  1 • 3x  4y  9z  2 4x  3y  5z  6

WORKING WITH FORMULAS

The inverse of a 2  2 matrix: a b 1 #c d d S A1  A c c d ad  bc c

b d a

The inverse of a 2  2 matrix can be found using the formula shown, as long as ad  bc  0. Use the formula to find inverses for the matrices here (if possible), then verify by showing A # A 1  A # A 1  I.



803

Section 8.7 Solving Linear Systems Using Matrix Equations

5 d 1

63. A  c

3 2

65. C  c

0.3 0.6

0.4 d 0.8

64. B  c 66. c

2 5

0.2 0.4

3 d 4

0.3 d 0.6

APPLICATIONS

Solve each application using a matrix equation. Descriptive Translation

67. Convenience store sales: The local Moto-Mart sells four different sizes of Slushies—behemoth, 60 oz @ $2.59; gargantuan, 48 oz @ $2.29; mammoth, 36 oz @ $1.99; and jumbo, 24 oz @ $1.59. As part of a promotion, the owner offers free gas to any customer who can tell how many of each size were sold last week, given the following information: (1) The total revenue for the Slushies was $402.29; (2) 7884 ounces were sold; (3) a total of 191 Slushies were sold; and (4) the number of behemoth Slushies sold was one more than the number of jumbo. How many of each size were sold?

68. Cartoon characters: In America, four of the most beloved cartoon characters are Foghorn Leghorn, Elmer Fudd, Bugs Bunny, and Tweety Bird. Suppose that Bugs Bunny is four times as tall as Tweety Bird. Elmer Fudd is as tall as the combined height of Bugs Bunny and Tweety Bird. Foghorn Leghorn is 20 cm taller than the combined height of Elmer Fudd and Tweety Bird. The combined height of all four characters is 500 cm. How tall is each one? 69. Rolling Stones music: One of the most prolific and popular rock-and-roll bands of all time is the Rolling Stones. Four of their many great hits include: Jumpin’ Jack Flash, Tumbling Dice, You Can’t Always Get What You Want, and Wild Horses. The total playing time of all four songs is 20.75 min.

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The combined playing time of Jumpin’ Jack Flash and Tumbling Dice equals that of You Can’t Always Get What You Want. Wild Horses is 2 min longer than Jumpin’ Jack Flash, and You Can’t Always Get What You Want is twice as long as Tumbling Dice. Find the playing time of each song. 70. Mozart’s arias: Mozart wrote some of vocal music’s most memorable arias in his operas, including Tamino’s Aria, Papageno’s Aria, the Champagne Aria, and the Catalogue Aria. The total playing time of all four arias is 14.3 min. Papageno’s Aria is 3 min shorter than the Catalogue Aria. The Champagne Aria is 2.7 min shorter than Tamino’s Aria. The combined time of Tamino’s Aria and Papageno’s Aria is five times that of the Champagne Aria. Find the playing time of all four arias. Manufacturing

71. Resource allocation: Time Pieces Inc. manufactures four different types of grandfather clocks. Each clock requires these four stages: (1) assembly, (2) installing the clockworks, (3) inspection and testing, and (4) packaging for delivery. The time required for each stage is shown in the table, for each of the four clock types. At the end of a busy week, the owner determines that personnel on the assembly line worked for 262 hours, the installation crews for 160 hours, the testing department for 29 hours, and the packaging department for 68 hours. How many clocks of each type were made? Dept. Assemble

Clock A

Clock B

Clock C

Clock D

2.2

2.5

2.75

3

Install

1.2

1.4

1.8

2

Test

0.2

0.25

0.3

0.5

Pack

0.5

0.55

0.75

1.0

72. Resource allocation: Figurines Inc. makes and sells four sizes of metal figurines, mostly historical figures and celebrities. Each figurine goes through four stages of development: (1) casting, (2) trimming, (3) polishing, and (4) painting. The time required for each stage is shown in the table, for each of the four sizes. At the end of a busy week, the manager finds that the casting department put in 62 hr, and the trimming department worked for 93.5 hr, with the polishing and painting departments logging 138 hr and 358 hr respectively. How many figurines of each type were made? Dept. Casting

Small

Medium

Large

X-Large

0.5

0.6

0.75

1

Trimming

0.8

0.9

1.1

1.5

Polishing

1.2

1.4

1.7

2

Painting

2.5

3.5

4.5

6

73. Thermal conductivity: In lab experiments designed to measure the heat conductivity of a square metal plate of 64°C uniform density, the edges are held at four different (constant) p2 p1 temperatures. The 70°C 80°C mean-value principle p3 p4 from physics tells us that the temperature 96°C at a given point pi on the plate, is equal to the average temperature of nearby points. Use this information to form a system of four equations in four variables, and determine the temperature at interior points p1, p2, p3, and p4 on the plate shown. (Hint: Use the temperature of the four points closest to each.) 74. Thermal conductivity: Repeat Exercise 73 if (a) the temperatures at the top and bottom of the plate were increased by 10°, with the temperatures at the left and right edges decreased by 10° (what do you notice?); (b) the temperature at the top and the temperature to the left were decreased by 10°, with the temperatures at the bottom and right held at their original temperature. Curve Fitting

75. Cubic fit: Find a cubic function of the form y  ax3  bx2  cx  d such that 14, 62, 11, 02, 11, 162, and (3, 8) are on the graph of the function. 76. Cubic fit: Find a cubic function of the form y  ax3  bx2  cx  d such that 12, 52, 10, 12, 12, 32, and (3, 25) are on the graph of the function. Nutrition

77. Animal diets: A zoo dietician needs to create a specialized diet that regulates an animal’s intake of fat, carbohydrates, and protein during a meal. The table given shows three different foods and the amount of these nutrients (in grams) that each ounce of food provides. How many ounces of each should the dietician recommend to supply 20 g of fat, 30 g of carbohydrates, and 44 g of protein? Nutrient

Food I

Food II

Food III

Fat

2

4

3

Carb.

4

2

5

Protein

5

6

7

78. Training diet: A physical trainer is designing a workout diet for one of her clients, and wants to supply him with 24 g of fat, 244 g of

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carbohydrates, and 40 g of protein for the noontime meal. The table given shows three different foods and the amount of these nutrients (in grams) that each ounce of food provides. How many ounces of each should the trainer recommend? 

Food I

Food II

Food III

Fat

2

5

0

Carb.

10

15

18

Protein

2

10

0.75

EXTENDING THE CONCEPT

79. Some matrix applications require that you solve a matrix equation of the form AX  B  C, where A, B, and C are matrices with the appropriate number of rows and columns and A1 exists. Investigate the solution process for such equations using A  c

2 5

3 4 12 d, B  c d, C  c d , and 4 9 4

x X  c d , then solve AX  B  C for X y symbolically (using A1, I, and so on). 80. It is possible for the matrix of coefficients to be singular, yet for solutions to exist. If the system is dependent instead of inconsistent, there may be infinitely many solutions and the solution set must be written using a parameter or the set notation seen previously. Try solving the exercise given here using matrix equations. If this is not possible, discuss why, then solve using elimination. If the system is dependent, find at least two sets of three fractions that fit the criteria.The sum of the two smaller fractions equals the larger, the larger less the smaller equals the “middle” fraction, and four times the smaller fraction equals the sum of the other two. 

Nutrient

805

81. Another alternative for finding determinants uses the triangularized form of a matrix and is offered without proof: If nonsingular matrix A is written in triangularized form without exchanging any rows and without using the operations kRi to replace any row (k a constant), then det(A) is equal to the product of resulting diagonal entries. Compute the determinant of each matrix using this method. Be careful not to interchange rows and do not replace any row by a multiple of that row in the process. 1 2 3 2 5 1 a. £ 4 5 6 § b. £ 2 3 4 § 2 5 3 4 6 5 2 4 1 3 1 4 c. £ 5 7 2 § d. £ 0 2 6 § 3 8 1 2 1 3 82. Find 2  2 nonzero matrices A and B whose product gives the zero matrix c

0 0

0 d. 0

MAINTAINING YOUR SKILLS

83. (5.5) Find the amplitude and period of y  125 cos13t2.

85. (1.3) Solve the absolute value inequality: 32x  5  7  19.

84. (2.5/4.3) Match each equation to its related graph. Justify your answers.

86. (6.6) Find all solutions of 7 tan2x  21 in 30, 22.

y  log2 1x  22

a.

y 5 4 3 2 1 1 1 2 3 4 5 6

x2

1 2 3 4 5 6 7 8 9 10 x

y  log2x  2 b.

y 5 4 3 2 1 1 1 2 3 4 5 6

1 2 3 4 5 6 7 8 9 10 x

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8.8

Applications of Matrices and Determinants: Cramer’s Rule, Geometry, and More

Learning Objectives

Introduction

In Section 8.8 you will learn how to:

In addition to their use in solving systems, matrices can be used to accomplish such diverse things as finding the volume of a three-dimensional solid or establishing certain geometrical relationships in the coordinate plane. Numerous uses are also found in higher mathematics, such as checking whether solutions to a differential equation are linearly independent.

A. Solve a system using determinants and Cramer’s Rule

B. Use determinants in applications involving geometry in the coordinate plane

C. Decompose a rational expression into partial fractions using matrices and technology

A. Solving Systems Using Determinants and Cramer’s Rule In addition to identifying singular matrices, determinants can actually be used to develop a formula approach for the solution of a system. Consider the following illustration, in which we solve a general 2  2 system by modeling the process after a specific 2  2 system. With a view toward a solution involving determinants, the coefficients of x are written as a11 and a21 in the general system, and the coefficients of y are a12 and a22. Specific System 2x  5y  9 e 3x  4y  10

General System a11x  a12y  c1 e a21x  a22y  c2

eliminate the x-term in R2

eliminate the x-term in R2 a21R1  a11R2

3R1  2R2 sums to zero

e

sums to zero

3 # 2x  3 # 5y  3 # 9 2 # 3x  2 # 4y  2 # 10 2 # 4y  3 # 5y  2 # 10  3 # 9

e

a21a11x  a21a12y  a21c1 a11a21x  a11a22y  a11c2 a11a22y  a21a12y  a11c2  a21c1

Notice the x-terms sum to zero in both systems. We are deliberately leaving the solution on the left unsimplified to show the pattern developing on the right. Next we solve for y. Factor Out y

12 # 4  3 # 52y  2 # 10  3 # 9 2 # 10  3 # 9 y 2#43#5

1a11a22

Factor Out y  a21a12 2y  a11c2  a21c1 a11c2  a21c1 y a11a22  a21a12

On the left we find y  7 7  1 and back-substitution shows x  2. But more importantly, on the right we obtain a formula for the y-value of any 2  2 system: a11c2  a21c1 y . If we had chosen to solve for x, the “formula” solution would be a11a22  a21a12 a22c1  a12c2 x . Note these formulas are defined only if a11a22  a21a12  0. You a11a22  a21a12 may have already noticed, but this denominator is the determinant of the matrix of a11 a12 d from the previous section! Since the numerator is also a coefficients c a21 a22 difference of two products, we investigate the possibility that it too can be expressed as a determinant. Working backward, we’re able to reconstruct the numerator for x in c1 a12 d , where it is apparent this matrix was formed by replacdeterminant form as c c2 a22 ing the coefficients of the x-variables with the constant terms.

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807

Forming the Numerator of the Solution for x `

a11 a21

a12 ` a22

`

a12 ` a22

`

c1 c2

a12 ` a22

(removed) replace with constants

remove coefficients of x

In a similar fashion, the numerator for y can be written in determinant form as a11 c1 c d , or the determinant formed by replacing the coefficients of the y-variables a21 c2 with the constant terms. If we use the notation Dy for this determinant, Dx for the determinant where x coefficients were replaced by the constants, and D as the determinant for the matrix of coefficients, the solutions can be written as shown next, with the result known as Cramer’s rule. Cramer’s Rule applied to 2  2 Systems Given a 2  2 system of linear equations a11  a12  c1 e a21  a22  c2 the solution is the ordered pair 1x, y2, where c1 a12 ` ` Dy c2 a22 Dx x  a  and y  a 11 12 D D ` `a a22 21

a11 c1 ` a21 c2 a11 a12 ` ` a21 a22 `

provided D  0. EXAMPLE 1



Solving a 2  2 System Using Cramer’s Rule Use Cramer’s rule to solve the system e

Solution



For x 

2x  5y  9 . 3x  4y  10

Dy Dx and y  , begin by finding the value of D, Dx, and Dy. D D

2 5 ` 3 4 122142  132152  7

D `

9 5 ` 10 4 192 142  1102 152  14

Dx  `

Dy  `

2 9 ` 3 10 122 1102  132192 7

Dy Dx 14 7   2 and y    1. The solution is 12, 12. D 7 D 7 Check by substituting these values into the original equations.

This gives x 

Now try Exercises 7 through 14



Regardless of the method used to solve a system, always be aware that a consistent, y  2x  3 inconsistent, or dependent system is possible. The system e yields 4x  6  2y 2x  y  3 2 1 e in standard form, with D  ` `  122122  142112  0. 4x  2y  6 4 2

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We stop here since Cramer’s rule cannot be applied, knowing the system is either inconsistent or dependent. To find out which, we write the equations in function form y  2x  3 (solve for y). The result is e , showing the system consists of two parallel y  2x  3 lines and has no solutions.

Cramer’s Rule for 3  3 Systems Cramer’s rule can be extended to a 3  3 system of linear equations, using the same pattern as for 2  2 systems. Given the general 3  3 system a11x  a12y  a13z  c1 • a21x  a22y  a23z  c2, a31x  a32y  a33z  c3 Dy Dz Dx , y  , and z  , where Dx, Dy, and Dz are again formed D D D by replacing the coefficients of the indicated variable with the constants, and D is the determinant of the matrix of coefficients.

the solutions are x 

Cramer’s Rule Applied to 3  3 Systems Given a 3  3 system of linear equations a11x  a12y  a13z  c1 • a21x  a22y  a23z  c2 a31x  a32y  a33z  c3

The solution is an ordered triple 1x, y, z2, where c1 † c2 c3 Dx x  D a11 † a21 a31 a11 † a21 Dz a31 z  D a11 † a21 a31

EXAMPLE 2



a12 a22 a32 a12 a22 a32 a12 a22 a32 a12 a22 a32

Solving a 3  3 System Using Cramer’s Rule x  2y  3z  1 Solve using Cramer’s rule: • 2x  y  5z  1 3x  3y  4z  2

a11 a13 † a21 a23 † Dy a33 a31 y  D a13 a11 a23 † † a21 a33 a31 c1 c2 † c3 , provided D  0. a13 a23 † a33

c1 c2 c3 a12 a22 a32

a13 a23 † a33 a13 a23 † a33

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Solution



Begin by computing the determinant of the matrix of coefficients, to ensure that Cramer’s rule can be applied. Using the third row, we have 1 D  † 2 3

2 1 3

3 2 3 1 3 1 2 5 †  3 ` ` 3` ` 4` ` 1 5 2 5 2 1 4  3172  3112  4132  6 Since D  0 we continue, electing to compute the remaining determinants using a calculator. 1 2 3 1 1 3 1 2 1 Dx  † 1 1 5 †  12 Dy  † 2 1 5 †  0 1 1 †  6 Dz  † 2 2 3 4 3 2 4 3 3 2 Dy Dz Dx 12 0 6 The solution is x    2, y    0, and z    1, or 12, 0, 12 D 6 D 6 D 6 in triple form. Check this solution in the original equations. 

Now try Exercises 15 through 22 A. You’ve just learned how to solve a system using determinants and Cramer’s rule

B. Determinants, Geometry, and the Coordinate Plane As mentioned in the introduction, the use of determinants extends far beyond solving systems of equations. Here, we’ll demonstrate how determinants can be used to find the area of a triangle whose vertices are given as three points in the coordinate plane. The Area of a Triangle in the xy-Plane Given a triangle with vertices at (x1, y1), (x2, y2), and (x3, y3). The area is given by the absolute value of one-half the determinant of T, x1 y1 1 det1T2 where T  £ x2 y2 1 § : Area  ` ` 2 x3 y3 1

EXAMPLE 3 Solution





Finding the Area of a Triangle in the xy-Plane

Find the area of a triangle with vertices at (3, 1), 12, 32, and (1, 7).

Begin by forming matrix T and computing det(T):

y

1 3 1 1 1 †  † 2 3 1 † 1 1 7 1  313  72  112  12  1114  32  12  3  1172  26 det1T2 26 Compute the Area: A  ` `  ` ` 2 2  13 The area of this triangle is 13 units2.

x1 det1T2  † x2 x3

7

y1 y2 y3

(1, 7)

6 5 4

(2, 3)

3 2

(3, 1)

1 5 4 3 2 1 1

1

2

3

4

5

x

2 3

Now try Exercises 25 through 30



As an extension of this determinant formula, what if the three points were collinear? After a moment, it may occur to you that the formula would give an area of 0 units2, since no triangle could be formed. This gives rise to a test for collinear points.

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B. You’ve just learned how to use determinants in applications involving geometry in the coordinate plane

Test for Collinear Points Three points (x1, y1), (x2, y2), and (x3, y3) are collinear if the determinant of A is zero, x1 y1 1 where A  † x2 y2 1 † . x3 y3 1 See Exercises 31 through 36.

C. More on Partial Fraction Decomposition Occasionally, the decomposition template for a rational expression becomes rather extensive. Instead of attempting a solution using convenient values or row operations as in Section 8.3, solutions can more readily be found using a matrix equation and technology. EXAMPLE 4



Decomposing a Rational Expression Using Matrices and Technology Use matrix equations and a graphing calculator to decompose the given expression 12x3  62x2  102x  56 into partial fractions: . x4  6x3  12x2  8x

Solution



The degree of the numerator is less than that of the denominator, so we begin by factoring the denominator. After removing the common factor of x and applying the rational roots theorem with synthetic division 1c  22, the completely 12x3  62x2  102x  56 factored form is . The decomposed form will be x1x  22 3 3 2 B A C D 12x  62x  102x  56     . x x1x  22 3 1x  22 1 1x  22 2 1x  22 3 Clearing denominators and simplifying yields 12x3  62x2  102  56  A1x  22 3  Bx1x  22 2  Cx1x  22  Dx clear denominators After expanding the powers on the right, grouping like terms, and factoring, we obtain 12x3  62x2  102x  56  1A  B2x3  16A  4B  C2x2  112A  4B  2C  D2x  8A

By equating the coefficients of like terms, the following system and matrix equation are obtained: A  B  12 1 6A  4B  C  62 6 μ S≥ 12A  4B  2C  D  102 12 8A  56 8

1 4 4 0

0 1 2 0

0 A 12 0 B 62 ¥≥ ¥  ≥ ¥. 1 C 102 0 D 56

After carefully entering the matrices F (coefficients) and G (constants), we obtain the solution A  7, B  5, C  0, and D  2 as shown in the figure. The decomposed form is C. You’ve just learned how to decompose a rational expression into partial fractions using matrices and technology

12x3  62x2  102x  56 7 2 5    . 3 1 x x1x  22 1x  22 1x  22 3 Now try Exercises 37 through 42

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8.8 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. The determinant `

a11 a21

a12 ` is evaluated as: a22

4. The three points (x1, y1), (x2, y2), and (x3, y3) are x1 y1 1 collinear if T   † x2 y2 1 † has a value of x3 y3 1

. 2.

rule uses a ratio of determinants to solve for the unknowns in a system.

3. Given the matrix of coefficients D, the matrix Dx is formed by replacing the coefficients of x with the terms. 

5. Discuss/Explain the process of writing

.

8x  3 as a x2  x

sum of partial fractions. 6. Discuss/Explain why Cramer’s rule cannot be applied if D  0. Use an example to illustrate.

DEVELOPING YOUR SKILLS

Write the determinants D, Dx, and Dy for the systems given. Do not solve.

7. e

2x  5y  7 3x  4y  1

8. e

x  5y  12 3x  2y  8

Solve each system of equations using Cramer’s rule, if possible. Do not use a calculator.

9. e

4x  y  11 3x  5y  60

y x  1 8 4 11. μ y x  6 5 2 13. e

0.6x  0.3y  8 0.8x  0.4y  3

10. e

x  2y  11 y  2x  13

2 3 7 x y 3 8 5 12. μ 5 3 11 x y 6 4 10 14. e

2.5x  6y  1.5 0.5x  1.2y  3.6

Write the determinants D, Dx, Dy, and Dz for the systems given, then determine if a solution using Cramer’s rule is possible by computing the value of D without the use of a calculator (do not solve the system). Try to determine how the system from Part (b) is related to the system in Part (a).

4x  y  2z  5 15. a. • 3x  2y  z  8 x  5y  3z  3 4x  y  2z  5 b. • 3x  2y  z  8 x  y  z  3

2x  3z  2 16. a. • x  5y  z  12 3x  2y  z  8 2x  3z  2 b. • x  5y  z  12 3x  5y  2z  8 Use Cramer’s rule to solve each system of equations.

x  2y  5z  10 17. • 3x  4y  z  10 x  y  z  2

x  3y  5z  6 18. • 2x  4y  6z  14 9x  6y  3z  3

y  2z  1 19. • 4x  5y  8z  8 8x  9z  9 x  2y  5z  10 20. • 3x  z  8 y  z  3 w  2x  3y  8 x  3y  5z  22 21. μ 4w  5x  5 y  3z  11 w  2x  3y  z  11 3w  2y  6z  13 22. μ 2x  4y  5z  16 3x  4z  5

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WORKING WITH FORMULAS

L r2 Area of a Norman window: A  †
† . The determinant shown can be used to find the area of a Norman  W 2 W window (rectangle
half-circle) with length L, width W, and radius r  . Find the area of the following windows. 2

23.

24. 16 in. 20 in. 58 cm

32 cm


APPLICATIONS

Area of a triangle: Find the area of the triangle with the vertices given. Assume units are in cm.

25. (2, 1), (3, 7), and (5, 3)

26. 12, 32, 13, 42, and 16, 12 Area of a parallelogram: Find the area of the parallelogram with vertices given (Hint: Use two triangles.) Assume units are in ft.

27. 14, 22, 16, 12, 13, 12, and (5, 2) 28. 15, 62, 15, 02, 15, 42 , and 15, 22

Volume of a pyramid: The volume of a triangular pyramid is given by the formula V  13 Bh, where B represents the area of the triangular base and h is the height of the pyramid. Find the volume of a triangular pyramid whose height is given and whose base has the coordinates shown. Assume units are in m.

29. h  6 m; vertices (3, 5), 14, 22, and 11, 62

30. h  7.5 m; vertices 12, 32, 13, 42, and 16, 12 Volume of a prism: The volume of a right prism is given by the formula V  Bh, where B represents the area of the base and h is the height of the prism. Find the volume of a triangular prism whose height is given and whose base has the coordinates shown. Assume units are in inches.

31. (3, 0), (9, 0), (6, 4), h  8 32. (2, 3), (9, 5), (0, 6), h  5

Test for collinear points: Determine if the following sets of points are collinear.

33. (1, 5), 12, 12 , and (4, 11) 34. (1, 1), 13, 52, and 12, 92

35. 12.5, 5.22, 11.2, 5.62 , and 12.2, 8.52 36. 10.5, 1.252, 12.8, 3.752, and (3, 6.25)

Verifying points are collinear: For each linear equation given, substitute the first two points to verify they are solutions. Then use the test for collinear points to determine if the third point is also a solution. Verify by direct substitution.

37. 2x  3y  7; 12, 12, 11.3, 3.22, 13.1, 4.42 38. 5x
2y  4; 12, 32, 13.5, 6.752, 12.7, 8.752

Decompose each rational expressing using a matrix equation and a graphing calculator.

39.

x4  x2  2x
1 x5  2x3
x

40.

3x4
13x2
x  12 x5  4x3
4x

41.

x3  17x2
76x  98 1x2  6x
921x2  2x  32

42.

16x3  66x2
98x  54 12x2  3x214x2  12x
92

Write a linear system that models each application. Then solve using Cramer’s rule.

43. Return on investments: If $15,000 is invested at a certain interest rate and $25,000 is invested at another interest rate, the total return was $2900. If the investments were reversed the return would be $2700. What was the interest rate paid on each investment?

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44. CD and DVD clearance sale: To generate interest in a music store clearance sale, the manger sets out a large box full of $2 CDs and $7 DVDs, with an advertised price of $800 for the lot. When asked how many of each are in the box, the manager will only say the box holds a total of 200 disks. How many CDs and DVDs are in the box? 45. Cost of fruit: Many years ago, two pounds of apples, 2 lb of kiwi, and 10 lb of pears cost $3.26. Three pounds of apples, 2 lb of kiwi, and 7 lb of pears cost $2.98. Two pounds of apples, 3 lb of kiwi, and 6 lb of pears cost $2.89. Find the cost of a pound of each fruit. 46. Vending machine receipts: The vending machines at an amusement park are stocked with various candies that sell for 5¢, 10¢, and 25¢. At week’s end, the park collected $54.30 from one of the machines. How many of each type of candy were sold, if 484 total sales were made and the machine vended twice as many 5¢ candies as 25¢ candies?



813

47. Coffee blends: To make its morning coffees, a coffee shop uses three kinds of beans costing $1.90/lb, $2.25/lb, and $3.50/lb, respectively. By the end of the week, the shop went through 24 lb of coffee beans, having a total value of $58. Find how many pounds of each type of bean were used, given that the number of pounds used of the cheapest beans was four more than the most expensive beans. 48. Manufacturing ball bearings: A ball bearing producer makes three sizes of bearings, half-inchdiameter size weighing 30 grams (g), a threequarter-inch-diameter size weighing 105 g, and a 1-in.-diameter size weighing 250 g. A large storage container holds ball bearings that have been rejected due to small defects. If the net weight of the container’s contents is 95.6 kilograms (95,600 g), and automated tallies show 920 bearings have been rejected, how many of each size is in the reject bin, given that statistical studies show there are twice as many rejects of the smallest bearing, as compared to the largest?

EXTENDING THE CONCEPT

49. Solve the given system four different ways: (1) elimination, (2) row reduction, (3) Cramer’s rule, and (4) using a matrix equation. Which method seems to be the least error-prone? Which method seems most efficient (takes the least time)? Discuss the advantages and drawbacks of each method. x  3y  5z  6 • 2x  4y  6z  14 9x  6y  3z  3 50. Find the area of the pentagon whose vertices are: 15, 52, 15, 52, (8, 6), 18, 62, and (0, 12.5). 51. The polynomial form for the equation of a circle is x2  y2  Dx  Ey  F  0. Find the equation of the circle that contains the points 11, 72, (2, 8), and 15, 12. 52. For square matrix A, calculating A2 is a simple matter of matrix multiplication. But what of those applications that require higher powers? a b For any matrix A  c d , we define the c d characteristic polynomial of A to be a b p12  ` `  1  a2 1  d2  cb. c d

The Cayley-Hamilton theorem states that if the original matrix A is substituted for , the result must be zero, or in other words, every square matrix satisfies its own characteristic equation. This fact is often used to compute higher powers of a square matrix, since all higher powers can then be expressed in terms of A. Specifically, for 1 2 1 2 A c d , p12  ` `  3 4 3 4 1  121  42  132122  2  5  2, with the theorem stating that A2  5A  2I2  0 (note I2 is used to ensure we remain in the context of 2  2 matrices). For A2  5A  2I2, we have the follow sequence for any power of A: 3 A  A2A A4  A3A  15A  2I2 2A  127A  10I2 2A 2  5A  2A  27A2  10A  515A  2I2 2  2A  2715A  2I2 2  10A  27A  10I2  145A  54I2 2 1 d , use the Cayley-Hamilton 4 1 3 theorem to find A , A4, and A5. Verify A3 using scalar multiplication and matrix addition. For A  c

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MAINTAINING YOUR SKILLS

53. (3.4) Graph the polynomial using information about end behavior, y-intercept, x-intercept(s), and midinterval points: f 1x2  x3  2x2  7x  6. 54. (3.3) Which is the graph (left or right) of g1x2  x  1  3? Justify your answer. y

y

5 4 3 2 1 54321 1 2 3 4 5

5 4 3 2 1 1 2 3 4 5 x

54321 1 2 3 4 5

55. (5.6) Solve the triangle: side a  8.7 in. side b  11.2 in. A  49.0° 56. (5.4) Graph y  3 tan12t2 over the interval 312, 12 4. Note the period, asymptotes, zeroes and value of A.

1 2 3 4 5 x

S U M M A RY A N D C O N C E P T R E V I E W SECTION 8.1

Linear Systems in Two Variables with Applications

KEY CONCEPTS • A solution to a linear system in two variables is any ordered pair (x, y) that makes all equations in the system true. • Since every point on the graph of a line satisfies the equation of that line, a point where two lines intersect must satisfy both equations and is a solution of the system. • A system with at least one solution is called a consistent system. • If the lines have different slopes, there is a unique solution to the system (they intersect at a single point). The system is called a consistent and independent system. • If the lines have equal slopes and the same y-intercept, they form identical or coincident lines. Since one line is right atop the other, they intersect at all points with an infinite number of solutions. The system is called a consistent and dependent system. • If the lines have equal slopes but different y-intercepts, they will never intersect. The system has no solution and is called an inconsistent system. EXERCISES Solve each system by graphing. If the solution is not a lattice point on the graph (x- and y-values both integers), indicate your solution is an estimate. If the system is inconsistent or dependent, so state. 3x  2y  4 0.2x  0.5y  1.4 2x  y  2 1. e 2. e 3. e x  3y  8 x  0.3y  1.4 x  2y  4 Solve using substitution. Indicate whether each system is consistent, inconsistent, or dependent. Write unique solutions as an ordered pair. y5x xy4 x  2y  3 4. e 5. e 6. e 0.4x  0.3y  1.7 x  4y  1 2x  2y  13 Solve using elimination. Indicate whether each system is consistent, inconsistent, or dependent. Write unique solutions as an ordered pair. 2x  4y  10 x  5y  8 2x  3y  6 7. e 8. e 9. e 3x  4y  5 x  2y  6 2.4x  3.6y  6 10. When it was first constructed in 1968, the John Hancock building in Chicago, Ilinois, was the tallest structure in the world. In 1985, the Sears Tower in Chicago became the world’s tallest structure. The Sears Tower is 323 ft taller than the John Hancock Building, and the sum of their heights is 2577 ft. How tall is each structure?

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Summary and Concept Review

815

Linear Systems in Three Variables with Applications

KEY CONCEPTS • The graph of a linear equation in three variables is a plane. • A linear system in three variables has the following possible solution sets: • If the planes intersect at a point, the system could have one unique solution (x, y, z). • If the planes intersect at a line, the system has linear dependence and the solution (x, y, z) can be written as linear combinations of a single variable (a parameter). • If the planes are coincident, the equations in the system differ by a constant multiple, meaning they are all “disguised forms” of the same equation. The solutions have coincident dependence, and the solution set can be represented by any one of the equations. • In all other cases, the system has no solutions and is an inconsistent system. EXERCISES Solve using elimination. If a system is inconsistent or dependent, so state. For systems with linear dependence, give the answer as an ordered triple using a parameter. x  y  2z  1 x  y  2z  2 3x  y  2z  3 11. • 4x  y  3z  3 12. • x  y  z  1 13. • x  2y  3z  1 3x  2y  z  4 2x  y  z  4 4x  8y  12z  7 Solve using a system of three equations in three variables. 14. A large coin jar is full of nickels, dimes, and quarters. There are 217 coins in all. The number of nickels is 12 more than the number of quarters. The value of the dimes is $4.90 more than the value of the nickels. How many coins of each type are in the bank? 15. If the point (x, y) is on the graph of a parabola, it must satisfy the equation y  ax2  bx  c. Use a system of three equations in three variables to find the equation of the parabola containing the points 10, 92, (2, 7), and (6, 15).

SECTION 8.3

Partial Fraction Decomposition

KEY CONCEPTS • Partial fraction decomposition is an attempt to reverse the process of adding rational expressions. • The process begins by factoring the denominator and setting up a “decomposition template.” A1 , A1 a constant. • For each linear factor 1x  c2 of the denominator, the template will have a term of the form xc • For each irreducible quadratic factor 1ax2  bx  c2 of the denominator, the template will have a term of the form A1x  B1 for constants A1 and B1. ax2  bx  c Ai for i  1 to n. • If the factor 1x  c2 is repeated n times the template will have n terms of the form 1x  c2 i • If the irreducible quadratic factor is repeated n times, the template will have n terms of the form Ai x  Bi , for i  1 to n. 1ax2  bx  c2 i • After the template has been set 1original expression  decomposition template2, multiply both sides by the LCD and (1) use convenient values or (2) simplify the right-hand side and equate coefficients to find the constants Ai and Bi needed. Note that some of the coefficients found may be zero. • If the degree of the numerator is greater than the degree of the denominator, divide using long division and decompose the remainder portion.

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CHAPTER 8 Systems of Equations and Inequalities

EXERCISES Decompose each of the following expressions into partial fractions. 6x  14 16x  1 16. 2 17. x  2x  15 2x2  5x  3 2x2  3x  19 2x2  13x  24 19. 3 20. x  5x2  3x  15 x3  27 2x2  8x  33 x2  15x  22 22. 3 23. x  x2  8x  12 x3  3x2  9x  5

SECTION 8.4

x2  2x  9 x3  4x2  x  4 6x2  2x  7 21. x3  1 2x2  x  1 24. 4 x  6x2  9 18.

Systems of Inequalities and Linear Programming

KEY CONCEPTS • To solve a system of inequalities, we find the intersecting or overlapping regions of the solution regions from the individual inequalities. The common area is called the feasible region. • The process known as linear programming seeks to maximize or minimize the value of a given quantity under certain constraints or restrictions. • The quantity we attempt to maximize or minimize is called the objective function. • The solution(s) to a linear programming problem occur at one of the corner points of the feasible region. • The process of solving a linear programming application contains these six steps: • Identify the main objective and the decision variables. • Write the objective function in terms of these variables. • Organize all information in a table, using the decision variables and constraints. • Fill in the table with the information given and write the constraint inequalities. • Graph the constraint inequalities and determine the feasible region. • Identify all corner points of the feasible region and test these points in the objective function. EXERCISES Graph the solution region for each system of linear inequalities and verify the solution using a test point. x  y 7 2 x  4y 5 x  2y 1 25. e 26. e 27. e x  y 6 4 x  2y 0 2x  y 2 28. Carefully graph the feasible region for the system of inequalities shown, then maximize the xy 7 2x  y 10 objective function: f 1x, y2  30x  45y μ 2x  3y 18 x 0 y 0 29. After retiring, Oliver and Lisa Douglas buy and work a small farm (near Hooterville) that consists mostly of milk cows and egg-laying chickens. Although the price of a commodity is rarely stable, suppose that milk sales bring in an average of $85 per cow and egg sales an average of $50 per chicken over a period of time. During this time period, the new ranchers estimate that care and feeding of the animals took about 3 hr per cow and 2 hr per chicken, while maintaining the related equipment took 2 hr per cow and 1 hr per chicken. How many animals of each type should be maintained in order to maximize profits, if at most 1000 hr can be spent on care and feeding, and at most 525 hr on equipment maintenance?

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SECTION 8.5

Summary and Concept Review

817

Solving Linear Systems Using Matrices and Row Operations

KEY CONCEPTS • A matrix is a rectangular arrangement of numbers. An m  n matrix has m rows and n columns. • The matrix derived from a system of linear equations is called the augmented matrix. It is created by augmenting the coefficient matrix (formed by the variable coefficients) with the matrix of constants. • One matrix method for solving systems involves the augmented matrix and row-reduction. • If possible, interchange equations so that the coefficient of x is a “1” in R1. • Write the system in augmented matrix form (coefficient matrix with matrix of constants). • Use row operations to obtain zeroes below the first entry of the diagonal. • Use row operations to obtain zeroes below the second entry of the diagonal. • Continue until the matrix is triangularized (all entries below the diagonal are zero). • Convert the augmented matrix back into equation form and solve for z. EXERCISES Solve by triangularizing the matrix. Use a calculator for Exercise 32. x  2y  6 30. e 4x  3y  4

SECTION 8.6

x  2y  2z  7 31. • 2x  2y  z  5 3x  y  z  6

2w  x  2y  3z  19 w  2x  y  4z  15 32. μ x  2y  z  1 3w  2x  5z  60

The Algebra of Matrices

KEY CONCEPTS • The entries of a matrix are denoted aij, where i gives the row and j gives the column of its location. • The m  n size of a matrix is also referred to as its order. • Two matrices A and B are equal if corresponding entries are equal: A  B if aij  bij. • The sum or difference of two matrices is found by combining corresponding entries: A  B  3aij  bij 4. • The identity matrix for addition is an m  n matrix whose entries are all zeroes. • The product of a constant times a matrix is called scalar multiplication, and is found by taking the product of the scalar with each entry, forming a new matrix of like size. For matrix A: kA  kaij. • Matrix multiplication is performed as row entry  column entry according to the following procedure: For an m  n matrix A  3 aij 4 and a p  q matrix B  3 bij 4, matrix multiplication is possible if n  p, and the result will be an m  q matrix P. In symbols A # B  3pij 4, where pij is product of the ith row of A with the jth column of B. • When technology is used to perform operations on matrices, the focus shifts from a meticulous computation of new entries, to carefully entering each matrix into the calculator, double checking that each entry is correct, and appraising the results to see if they are reasonable. EXERCISES Compute the operations indicated below (if possible), using the following matrices. 1 3 4 1 3 7 6 4 4 A  c 1 7 d B c d C  £ 5 2 0 § 1 2 8 8 6 3 2 33. A  B 34. B  A 35. C  B 36. 8A 38. C  D 39. D  C 40. BC 41. 4D

2 3 D  £ 0.5 1 4 0.1 37. BA 42. CD

0 1 § 5

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CHAPTER 8 Systems of Equations and Inequalities

SECTION 8.7

Solving Linear Systems Using Matrix Equations

KEY CONCEPTS • The identity matrix I for multiplication has 1s on the main diagonal and 0s for all other entries. For any n  n matrix A, the identity matrix is also n  n, where AI  IA  A. • For an n  n (square) matrix A, the inverse matrix for multiplication is a matrix B such that AB  BA  I. Only square matrices have an inverse. For matrix A the inverse is denoted A1. • Any n  n system of equations can be written as a matrix equation and solved (if solutions exist) using an inverse 2 3 1 2x  3y  z  5 x 5 # 2 4 § £ y §  £ 13 § . matrix. For • 3x  2y  4z  13, the matrix equation is £ 3 1 5 2 x  5y  2z  3 z 3 • Every square matrix has a real number associated with it called its determinant. For a 2  2 matrix a11 a12 A ` ` , the determinant A is the difference of diagonal products: a11a22  a21a12. a21 a22 • If the determinant of a matrix is zero, the matrix is said to be singular or non-invertible. • For matrix equations, if the coefficient matrix is non-invertible, the system has no unique solution. EXERCISES Complete Exercises 43 through 45 using the following matrices: 1 0 0.2 0.2 2 1 10 A c d B c d C c d D c 0 1 0.6 0.4 3 1 15 43. Exactly one of the matrices given is singular. Compute each determinant to identify it. 44. Show that AB  BA  B. What can you conclude about the matrix A? 45. Show that BC  CB  I. What can you conclude about the matrix C? Complete Exercises 46 through 49 using these matrices: 1 2 3 1 1 1 E  £ 2 1 5 § F £ 0 1 0 § 1 1 2 2 1 1

1 G  £0 0

0 1 0

0 0§ 1

46. Exactly one of the matrices above is singular. Use a calculator to determine which one. 47. Show that GF  FG  F. What can you conclude about the matrix G? 48. Show that FH  HF  I. What can you conclude about the matrix H? 49. Verify that EH  HE and EF  FE and comment. Solve manually using a matrix equation. 2x  5y  14 50. e 3y  4x  14 Solve using a matrix equation and your calculator. 0.5x  2.2y  3z  8 51. • 0.6x  y  2z  7.2 x  1.5y  0.2z  2.6

6 d 9

1 H £ 0 2

0 1 1

1 0 § 1

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Mixed Review

SECTION 8.8

819

Applications of Matrices and Determinants, Cramer’s Rule, Geometry, and More

KEY CONCEPTS • Cramer’s rule uses a ratio of determinants to solve systems of equations (if they exist). a11 a12 ` is a11a22  a21a12. • The determinant of the 2  2 matrix ` a21 a22 • To compute the value of 3  3 and larger determinants, a calculator is generally used. • Determinants can be used to find the area of a triangle in the plane if the vertices are known, and as a test to see if three points are collinear. • A system of equations can be used to write a rational expression as a sum of its partial fractions. EXERCISES Solve using Cramer’s rule. 52. e

2x  y  2 53. • x  y  5z  12 3x  2y  z  8

5x  6y  8 10x  2y  9

2x  y  z  1 54. • x  2y  z  5 3x  y  2z  8

55. Find the area of a triangle whose vertices have the coordinates (6, 1), 11, 62 , and 16, 22. 56. Use a matrix equation and a graphing calculator to find the partial fraction decomposition for

7x2  5x  17 . x3  2x2  3x  6

MIXED REVIEW 1. Write the equations in each system in slope-intercept form, then state whether the system is consistent/independent, consistent/dependent, or inconsistent. Do not solve. 3x  5y  10 a. e 6x  20  10y 4x  3y  9 b. e 2x  5y  10 x  3y  9 c. e 6y  2x  10 2. Solve by graphing. e

x  2y  6 2x  y  9

2x  3y  5 x  5y  17 7x  4y  5 4. Solve using elimination. e 3x  2y  9 3. Solve using a substitution. e

Solve using elimination. x  2y  3z  4 5. • 3x  4y  z  1 2x  6y  z  1

0.1x  0.2y  z  1.7 6. • 0.3x  y  0.1z  3.6 0.2x  0.1y  0.2z  1.7 Solve using row operations to triangularize the matrix. 1 2 x y3 2 3 7. μ 2 1 x y1 5 4 2x  y  4z  11 8. • x  3y  z  4 3x  2y  z  7 Decompose the rational expressions into partial fractions. 13  x x x6 7x2  3x  4 10. 3 x  3x2  x  3 9.

2

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CHAPTER 8 Systems of Equations and Inequalities

Compute as indicated for A c

2 0

C c

1 2

2 0 § 4 3 0 1 D  £ 1 2 0 § 1 1 4

1 B  £3 2

1 d 3 4 0

2 d 1

11. a. 2AC

b. CD

12. a. BA

b. CB  4A

13. Solve using a matrix equation: x  2z  5 • 2y  z  4 x  2y  3 14. Use a matrix equation and a calculator to solve: 2 1 w  x  y  z  3 2 3 3 5 41 xy z 4 8 8 2 3 27 wx z 5 10 10 w  2x  3y  4z  16

∂

15. Solve using Cramer’s rule: x  5y  2z  1 • 2x  3y  z  3 3x  y  3z  2 16. A triangle in the coordinate plane has vertices 11, 72, (6, 2), and (1, 5). Use the geometry of matrices to find the area of the triangle. 17. Graph the solution region for the system of 4x  2y 14 2x  3y 15 inequalities. μ y 0 x 0

18. Maximize P1x, y2  2.5x  3.75y, given xy 8 x  2y 14 μ 4x  3y 30 x, y 0 19. It’s the end of another big day at the circus, and the clowns are putting away their riding equipment—a motley collection of unicycles, bicycles, and tricycles. As she loads them into the storage shed, Trixie counts 21 cycles in all with a total of 40 wheels. In addition, she notes the number of bicycles is one fewer than twice the number of tricycles. How many cycles of each type do the clowns use? 20. A local fitness center is offering incentives in an effort to boost membership. If you buy a year’s membership (Y), you receive a $50 rebate and six tickets to a St. Louis Cardinals home game. For a half-year membership (H), you receive a $30 rebate and four tickets to a Cardinals home game. For a monthly trial membership (M), you receive a $10 rebate and two tickets to a Cardinals home game. During the last month, male clients purchased 40 one-year, 52 half-year, and 70 monthly memberships, while female clients purchased 50 one-year, 44 half-year, and 60 monthly memberships. Write the number of sales of each type to males and females as a 2  3 matrix, and the amount of the rebates and number of Cards tickets awarded per type of membership as a 3  2 matrix. Use these matrices to determine (a) the total amount of rebate money paid to males and (b) the number of Cardinals tickets awarded to females.

PRACTICE TEST Solve each system and state whether the system is consistent, inconsistent, or dependent. 1. Solve graphically: e

3x  2y  12 x  4y  10

2. Solve using substitution: e

3x  y  2 7x  4y  6

3. Solve using elimination: e

5x  8y  1 3x  7y  5

4. Solve using elimination: x  2y  z  4 • 2x  3y  5z  27 5x  y  4z  27

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Practice Test

Three cans of C, 2 of B, and 2 of P cost $1.73. One can of C, 4 of B, and 3 of P cost $1.92. What is the price of a single can of C, B, and P?

5. Given matrices A and B, compute: 2 a. A  B b. B c. AB 5 d. A1 e. A 3 2 3 3 d B c d 5 4 3 5 6. Given matrices C and D, use a calculator to find: a. C  D b. 0.6D c. DC 1 d. D e. D A c

0.5 C  £ 0.4 0.1

0 0.5 0.4

0.2 0 § 0.1

0.5 D  £ 0.1 0.3

0.1 0.1 0.4

0.2 0 § 0.8

7. Solve using matrices and row reduction: 4x  5y  6z  5 • 2x  3y  3z  0 x  2y  3z  5 8. Solve using Cramer’s rule: 2x  3y  z  3 • x  2y  z  4 x  y  2z  1 9. Solve using a matrix equation and your calculator: e

821

2x  5y  11 4x  7y  4

10. Solve using a matrix equation and your calculator: x  2y  2z  7 • 2x  2y  z  5 3x  y  z  6 Create a system of equations to model each exercise, then solve using the method of your choice. 11. The perimeter of a “legal-size” paper is 114.3 cm. The length of the paper is 7.62 cm less than twice the width. Find the dimensions of a legal-size sheet of paper. 12. The island nations of Tahiti and Tonga have a combined land area of 692 mi2. Tahiti’s land area is 112 mi2 more than Tonga’s. What is the land area of each island group? 13. Many years ago, two cans of corn (C), 3 cans of green beans (B), and 1 can of peas (P) cost $1.39.

14. After inheriting $30,000 from a rich aunt, David decides to place the money in three different investments: a savings account paying 5%, a bond account paying 7%, and a stock account paying 9%. After 1 yr he earned $2080 in interest. Find how much was invested at each rate if $8000 less was invested at 9% than at 7%. Using a radar gun placed in a protective cage beneath a falling baseball, the following data points are taken: (1, 128), (2, 80), (2.5, 44). Assume the data are of the form (time in seconds, distance from radar gun). 15. Use the data to set up a system of three equations in three variables to find the quadratic equation that models the ball’s height at time t. 16. From what height is the baseball being released? In how many seconds will it strike the cage protecting the radar gun? 17. Solve the system of inequalities by graphing. e

xy 2 x  2y 8

18. Maximize the objective function: P  50x  12y x  2y 8 • 8x  5y 40 x, y 0 Solve the linear programming problem. 19. A company manufactures two types of T-shirts, a plain T-shirt and a deluxe monogrammed T-shirt. To produce a plain shirt requires 1 hr of working time on machine A and 2 hr on machine B. To produce a deluxe shirt requires 1 hr on machine A and 3 hr on machine B. Machine A is available for at most 50 hr/week, while machine B is available for at most 120 hr/week. If a plain shirt can be sold at a profit of $4.25 each and a deluxe shirt can be sold at a profit of $5.00 each, how many of each should be manufactured to maximize the profit? 20. Decompose the expression into partial fractions: 4x2  4x  3 . x3  27

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CHAPTER 8 Systems of Equations and Inequalities

C A L C U L AT O R E X P L O R AT I O N A N D D I S C O V E RY Optimal Solutions and Linear Programming In Section 8.4 we learned, “If a linear programming problem has a unique solution, it must occur at a vertex.” Although we explored the reason why using the feasible region and a family of lines from the objective function, it sometimes helps to have a good “old fashioned,” point-bypoint verification to support the facts. In this exercise, we’ll use a graphing calculator to explore various areas of the feasible region, repeatedly evaluating the objective function to see where the maximal values (optimal solutions) seem to “congregate.” If all goes as expected, ordered pairs nearest to a vertex should give relatively larger values. To demonstrate, we’ll use Example 6 from Section 8.4, stated below. Example 6  Find the maximum value of the objective function f 1x, y2  2x  y given the constraints shown in xy 4 3x  y 6 the system μ . x 0 y 0 y Solution  Begin by noting the 7 (0, 6) solutions must be in QI, since x 0 and y 0. Graph the bound(1, 3) Feasible ary lines y  x  4 and region y  3x  6, shading the lower 5 x half plane in each case since they 5 are “less than” inequalities. This 3 produces the feasible region shown in lavender. There are four corner points to this region: (0, 0), (0, 4), (2, 0), and (1, 3), the latter found by solving xy4 the system: e . 3x  y  6

To explore this feasible region in terms of the objective function f 1x, y2  2x  y, enter the boundary lines Y1  x  4 and Y2  3x  6 on the Y = screen. However, instead of shading below the lines to show the feasible region (using the feature to the extreme left), we shade above both lines (using the feature) so that the feasible region remains clear. Setting the window size at x  30, 3 4 and Figure 8.30 proy  3 1.5, 4 4 4 duces Figure 8.30. 1.5 Using Ymin  1.5 will leave a blank area just below QI that allows you to cleanly explore the feasible 0 3 region as the x- and yvalues are displayed. 1.5 Next we place the

calculator in “split-screen” mode so that we can view the graph and the home screen simultaneously. Press the MODE key and notice the second-to-last line reads Full Horiz G-T. The Full (screen) mode is the default operating mode. The Horiz mode splits the screen horizontally, placing the graph directly above a shorter home screen. The G-T mode splits the screen vertically, with the graph to the left and a table to the right. Highlight Horiz, then press ENTER and GRAPH to have the calculator reset the screen in this mode. The TI-84 Plus has a free-moving cursor (a cursor you can move around without Figure 6.31 actually tracing a curve). Pressing the left or right arrow brings it into view (Figure 8.31). A useful feature of the free-moving cursor is that it automatically stores the current X value as the variable X ( X,T,␪,n or ALPHA STO ) and the current Y value as the variable Y ( ALPHA 1), which enables us to evaluate the objective function f 1x, y2  2x  y right on the home screen as we explore various corners of the feasible region. To access the graph and free-moving cursor you must press GRAPH each time, and to access the home screen you must press 2nd MODE (QUIT) each time. Begin by moving the cursor to the upper-left corner of the region, near the yintercept [we stopped at (~0.0957, 3.262 4 . Once you have the cursor “tucked up into the corner,” press 2nd MODE (QUIT) to get to the home screen, then enter the objective function: 2X  Y. Pressing ENTER automatically evaluates the function for the values indicated by the cursor’s location (Figure 8.32). It appears Figure 8.32 the value of the objective function for points (x, y) in this corner are close to 4, and it’s no accident that at the corner point (0, 4) the maximum value is in fact 4. Now let’s explore the area in the lower-right of the feasible region. Figure 8.33 Press GRAPH and move the cursor using the arrow keys until it is “tucked” over into the lower-right corner [we stopped at (~1.72, 0.32 4 . Pressing ENTER 2nd recalls 2X  Y and pressing ENTER reevaluates the objective function at these new X and Y values (Figure 8.33). It appears that values of the

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Strengthening Core Skills

Figure 8.34 objective function are also close to 4 in this corner, and it’s again no accident that at the corner point (2, 0) the maximum value is 4. Finally, press GRAPH to explore the region in the upper-right corner, where the lines intersect. Move the cursor to this vicinity, locate it very near the point of intersection [we stopped at (~0.957, 2.7162 4 and return to the home screen and evaluate (Figure 8.34). The value of the objective function is near 5 in this corner of the region, and at the corner point (1, 3) the maximum value is 5. This investigation can be repeated for any feasible region and for any number of points within the region, and serves to support the statement that, “If a linear programming problem has a unique solution, it must occur at a vertex.”

823

Exercise 1: Use the ideas discussed here to explore the solution to Example 7 from Section 8.4. Exercise 2: The feasible region for the system given to the right has five corner points. Use the ideas here to explore the area near each corner point of the feasible region to determine which point is the likely 8x  3y 30 candidate to produce the maximum value of the objective function 5x  4y 23 f 1x, y2  3.5x  y. Then solve the μ x  2y 10 linear programming problem to verify x, y 0 your guess. Exercise 3: The feasible region for the system given to the right has four corner points. Use the ideas here to explore the area near each corner point of the feasible region to determine which point is the likely candidate to 2x  2y 15 produce the minimum value of the objecxy 6 tive function f 1x, y2  2x  4y. Then μ x  4y 9 solve the linear programming problem to x, y 0 verify your guess.

STRENGTHENING CORE SKILLS Augmented Matrices and Matrix Inverses The formula for finding the inverse of a 2  2 matrix has its roots in the more general method of computing the inverse of an n  n matrix. This involves augmenting a square matrix M with its corresponding identity In on the right (forming an n  2n matrix), and using row operations to transform M into the identity. In some sense, as the original matrix is transformed, the “identity part” keeps track of the operations we used to convert M and we can use the results to “get back home,” so to speak. We’ll illustrate with the 2  2 matrix from Section 8.7, Example 3, where we found that 1 2.5 6 5 6 5 c d was the inverse matrix for c d . We begin by augmenting c d with the 2  2 identity. 1 3 2 2 2 2 c c

6 0

6 2 5 1

5 1 0 6 5 1 0 d  2R1  6R2 ¡ R2 c d 2 0 1 0 2 2 6 1 0 6 0 6 15 d  5R2  R1 ¡ R1 c d 1 3 0 1 1 3

R2 6 ¡ R2 c 2 0 1 R1 ¡ R1 c 6 0

5 1 0 1

1 1 1 1

0 d 3 2.5 d 3

As you can see, the identity is automatically transformed into the inverse matrix when this method is applied. a b d results in the formula given earlier. Performing similar row operations on the general matrix c c d c

£

a

b

0

1

a c

b d

1 c ad  bc

0 a b 1 0 d cR1  aR2 ¡ R2 c d 1 0 ad  bc c a a b 1 0 R2 § ¡ R2 £ a c ad  bc 0 1 ad  bc ad  bc bc ba a 0 1 0 ad  bc ad  bc §  bR2  R1 ¡ R1 ≥ ¥ a a c 0 1 ad  bc ad  bc ad  bc 1 0

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CHAPTER 8 Systems of Equations and Inequalities

Finding a common denominator for

≥

a

0

0

1

bc bc ad  bc ad  1 and combining like terms gives   . ad  bc ad  bc ad  bc ad  bc

d ba 1 0 ad  bc R1 ad  bc ¡ R1 ≥ ¥ a a c 0 1 ad  bc ad  bc d b ad  bc ad  bc This shows A1  ≥ ¥ a c ad  bc ad  bc

b ad  bc ¥. a ad  bc

ad ad  bc c ad  bc

1 produces the familiar formula. As you might imagine, attempting this on a general 3  3 ad  bc matrix is problematic at best, and instead we simply apply the procedure to any given 3  3 matrix. Here we’ll use the 2 1 0 3 2 § . augmented matrix method to find A1, given A  £ 1 3 1 2 and factoring out

2 £ 1 3

1 3 1

0 2 2

1 0 0

0 1 0

0 R1  2R2 ¡ R2 2 £0 0§ 1 3R1  2R3 ¡ R3 0 14 £ 0 5R2  7R3 ¡ R3 0 R2  7R1 ¡ R1

14 R3 ¡ R3 £ 0 8 0 4R3  R2 ¡ R2 14 £ 0 4R3  R1 ¡ R1 0

1 7 5 4 4 8

0 7 0 0 7 0

0 4 4

6 1 16

1 3 1

0 1 2 § £ 1 2 2

0.5 1 1.25

2 2 10

0 0§ 2 0 0§ 14

6 1 2

2 2 1.25

0 0 § 1.75

0 0 1

14 7 2

7 7 1.25

7 7 § 1.75

0 7 0

0.5 1 1 §  £0 1.75 0

0 1 0

Exercise 1: Use the preceding inverse and a matrix equation to solve the system 2x  y  2 • x  3y  2z  15. 3x  y  2z  9

0 2 0

4 4 1

1 R2 R1 1 ¡ R1 and ¡ R2 produces the inverse matrix A  £ 1 Using 14 7 2 To verify, we show AA1  I: 2 £ 1 3

1 1 3

0.5 1 1.25 0 0 § ✓. 1

0.5 1 §. 1.75

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Cumulative Review Chapters 1–8

825

C U M U L AT I V E R E V I E W C H A P T E R S 1– 8 1. Solve each equation by factoring. a. 9x2  12x  4 b. x2  7x  0 c. 3x3  15x2
6x  30 d. x3  4x
3x2 2. Solve for x. 3 3x 1
2  x3 x
2 x x6 3. A torus is a donut-shaped solid figure. Its surface area is given by the formula A  2 1R2  r2 2, where R is the outer radius of the donut, and r is the inner radius. Solve the formula for R in terms of r and A. 4. State the value of all six trig functions given (21, 28) is a point on the terminal side of
.   5. Sketch the graph of y  4 cosa x  b using 6 3 transformations of y  cos x. 6. A jai alai player in an open-air court becomes frustrated and flings the pelota out and beyond the court. If the initial velocity of the ball is 136 mph and it is released at height of 5 ft, (a) how high is the ball after 3 sec? (b) What is the maximum height of the ball? (c) How long until the ball hits the ground (hopefully in an unpopulated area)? Recall the projectile equation is h1t2  16t2
v0t
k. 7. For a complex number a
bi, (a) verify the sum of a complex number and its conjugate is a real number. (b) Verify the product of a complex number and its conjuate is a real number. 8. Solve using the quadratic formula: 5x2
8x
2  0. 9. Solve by completing the square: 3x2  72x
427  0. 10. Given cos 53°
0.6 and cos 72°
0.3, approximate the value of cos 19° and cos 125° without using a calculator.

11. Given A 13 4 , y B is a point on the unit circle, find the value of sin
, cos
, and tan
if y 7 0. 12. State the domain of each function: a. f 1x2  12x  3 b. g 1x2  logb 1x
32 x
3 c. h 1x2  2 x 5

13. Write the following formulas from memory: a. slope formula b. midpoint formula c. quadratic formula d. distance formula e. interest formula (compounded continuously) 14. Find the equation of the line perpendicular to the line 4x
5y  20, that contains the point (0, 1). 15. For the force vectors F1 and F2 given, find a third force vector F3 that will bring these vectors into equilibrium: F1 H5, 12I F2 H8, 6I 16. A commercial fishery stocks a lake with 250 fish. Based on previous experience, the population of fish is expected to grow according to the model 12,000 P 1t2  , where t is the time in months. 1
25e0.2t From on this model, (a) how many months are required for the population to grow to 7500 fish? (b) If the fishery expects to harvest three-fourths of the fish population in 2 yr, approximately how many fish will be taken? 17. Evaluate each expression by drawing a right triangle and labeling the sides. x bd a. sec c sin1a 2121
x2 b. sin c csc1a

29
x2 bd x

18. An luxury ship is traveling at 15 mph on a heading of 10°. There is a strong, 12 mph ocean current flowing from the southeast, at a heading of 330°. What is the true course and speed of the ship? 19. Use the Guidelines for Graphing to sketch the graph of function f given, then use it to solve f 1x2 6 0: f 1x2  x3  4x2
x
6 20. Use the Guidelines for Graphing to sketch the graph of function g given, then use it to name the intervals x2  4 where g 1x2T and g 1x2c: g1x2  2 x 1 8 21. Find 11  23i2 using De Moivre’s theorem. 22. Solve ln1x
22
ln1x  32  ln14x2.

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CHAPTER 8 Systems of Equations and Inequalities

23. If I saved $200 each month in an annuity program that paid 8% annual interest compounded monthly, how long would it take to save $10,000?

25. The graph given is of the form y
A sin1Bx  C2. Find the values of A, B, and C. y

24. Mount Tortolas lies on the Argentine-Chilean border. When viewed from a distance of 5 mi, the angle of elevation to the top of the peak is 38°. How tall is Mount Tortolas? State the answer in feet.

2 1




2

1 2


2




3
2

2


x

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CONNECTIONS TO CALCULUS As mentioned in the chapter opener, in calculus as well as algebra we often look for expressions that fit a special form, then apply a general formula to all expressions of that form. Here this general idea is applied to partial fraction decomposition, a technique we studied in Section 8.3 and an important part of integral calculus.

More on Partial Fraction Decomposition 10 as a sum of its partial fractions, we would use the x  25 A B 10   template and proceed as in Section 8.3. Using the 1x  52 1x  52 x5 x5 ideas developed there, we’ll attempt to develop a decomposition formula that works 2k for all rational expressions of the form 2 , a form that certainly applies to the x  k2 expression at hand. To rewrite the expression

EXAMPLE 1



Solution



2

Decomposing Special Forms

2k Use the techniques from Section 8.3 to rewrite the expression 2 as the sum x  k2 of its partial fractions. After factoring the denominator, we obtain the following template: A B 2k   1x  k21x  k2 xk xk 2k  A1x  k2  B1x  k2  Ax  Ak  Bx  Bk  1A  B2x  1B  A2k

decomposition template clear denominators distribute collect like terms

AB0 , yielding the solutions B  1 and A  1. As it BA2 2k turns out, any rational expression of the form 2 can be written in decomposed x  k2 1 1 form as  . xk xk This gives the system e

Now try Exercises 1 and 2

EXAMPLE 2



Solution



Decomposing Special Forms

10 Use the “formula” from Example 1 to decompose the expression 2 as the x  25 sum of its partial fractions, then verify the result. 10 2k 10 , we note 2k  10 and k  5, so 2 must 2 with 2 x k x  25 x  25 1 1 be equal to . To check, we combine the fractions using the LCD  x5 x5 11x  52  11x  52 x  5  x  5 10 1x  52 1x  52, giving  or 2 ✓. 1x  521x  52 1x  52 1x  52 x  25 By comparing

2

Now try Exercises 3 through 8

8–109





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Connections to Calculus

The Geometry of Vectors and Determinants Although vectors (Section 7.3) and determinants (Section 8.8) are studied in two separate chapters in Precalculus, in a calculus course you’ll see there is an intriguing connection between the two. In Section 7.3 we noted the tip-to-tail addition of vectors u and v formed a parallelogram, and used the parallelogram to illustrate a vector sum. In a calculus course, we say the vectors span the parallelogram, and in certain applications we need to know the area of this

Figure C2C 8.1 Ha, bI Hc, dI

u#v for the uv angle
between them, it can shown that the area of the parallelogram is the absolute a b ` (see Figure C2C 8.1). For a complete proof, see value of the determinant ` c d Appendix V. figure. Using the vectors u  Ha, bI and v  Hc, dI and relation cos


EXAMPLE 3




Solution




Using Determinants to Find Area

For the vectors u  H12, 5I and v  H8, 15I, a. Compute the vector sum w  u  v using the parallelogram (tail-to-tip) method. b. Compute w and find the angle
between u and w. 1 c. Use the formula A  uwsin
from Section 7.2 to find the area of the 2 triangle formed. a b 12 5 `  ` ` to find the area of the d. Use the determinant ` c d 8 15 parallelogram spanned by these vectors, then verify the area is twice as large as the triangle found in part (c). a. The vector sum is H12, 5I  H8, 15I  H20, 20I  w.

b. w  2202  202  1800  20 12. For u # w  121202  51202  340, 340 17 and u  13 (verify this), we have cos
 and
 cos
1a b. 1132 120 122 1312 This indicates

22.38°. 1 area formula c. A  uwsin
2 1
1132120 122sin 22.38° substitute 2 result
70 units2 20 H20, 20I d. For u  H12, 5I and v  H8, 15I, 18 16 a b 12 5 H8, 15I the determinant ` `  ` `  14 c d 8 15 12 11221152
182152  140 units2✓, and the 10 area of the parallelogram is indeed twice 8 6 the area of the triangle. H12, 5I

4 2 2

4

6

8 10 12 14 16 18 20

Now try Exercises 9 through 14




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Connections to Calculus

829

Connections to Calculus Exercises Use the techniques from Section 8.3 to complete Exercises 1 and 2.

1. A common form of partial fraction decomposition involves a constant over a factorable quadratic expression. (a) Rewrite the expression k as the sum of partial fractions to 1x  a2 1x  b2 AB0 . (b) Use show the system obtained is e Ab  Ba  k the fact that A  B  0 to show the system can be AB0 rewritten as e . (c) Solve for A in the A1b  a2  k second equation and use the result to quickly write 16 as the sum of its partial fractions. 1x  32 1x  52

2. Another common form involves a linear expression x  k over a factorable quadratic expression. xk (a) Rewrite the expression as the 1x  a21x  b2 sum of partial fractions to show the system obtained AB1 is e . (b) Use the fact that A  B  1 Ab  Ba  k to show that system can be written as AB1 e . (c) Solve for A in the A1b  a2  a  k second equation, and use the result to write x  10 as the sum of its partial fractions. 1x  321x  42

Use the “formulas” developed in Example 1 and Exercises 1 and 2 to decompose each expression into a sum of partial fractions.

3.

6 x 9

4.

18 x  81

5.

22 x  3x  28

6.

15 x  3x  4

7.

x  11 x  13x  40

8.

x  17 x  9x  14

2

2

2

2

2

2

Find the area of the parallelogram spanned by the vectors given, then verify your answer using the method shown in Example 3.

9. u  H2, 8I and v  H15, 3I

10. u  H4, 1I and v  H4, 4I

11. 6i  3j and 6i  3j

Similar to how the absolute value of `

12. 9i and 7j a b ` gives the area of a parallelogram spanned by Ha, bI and Hc, dI, the absolute c d

a b c value of the determinant † d e f † gives the volume of a parallelepiped spanned by Ha, b, c,I, Hd, e, fI, and Hg, h, iI in g h i three dimensions.

13. Find the volume of the parallelepiped spanned by H5, 0, 0I, H0, 6, 0I, and H0, 0, 8I using the determinant formula, then verify your answer using a more familiar formula.

14. Find the volume of the parallelepiped spanned by u  H2, 5, 8I, H3, 0, 4I, and H0, 10, 7I.

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9 CHAPTER CONNECTIONS

Analytical Geometry and the Conic Sections CHAPTER OUTLINE 9.1 A Brief Introduction to Analytical Geometry 832 9.2 The Circle and the Ellipse 839 9.3 The Hyperbola 852 9.4 The Analytic Parabola 866

Cathedral windows are just one place where we can appreciate the beauty and intricacy of polar graphs. The characteristics and features of these graphs are explored in this chapter, and introduce a historic alternative to graphing relations by simply computing points (x, y). While graphing relations using a rectangular coordinate system is much more widely known, both polar graphs and rectangular graphs seemed to have matured simultaneously, in the early 1600s. The graphs shown here appear as Exercises 57 and 65 in Section 9.6.

9.5 Nonlinear Systems of Equations and Inequalities 875 9.6 Polar Coordinates, Equations, and Graphs 883 9.7 More on the Conic Sections: Rotation of Axes and Polar Form 896 9.8 Parametric Equations and Graphs 913 When using polar curves in the design of windows or other forms of architechture, finding where two graphs intersect aids in the calculation of the area between curves and in determining the arclength of certain portions of the graph. These are important considerations in a study of calculus, and techniques for Connections to Calculus working with systems of polar equations are explored in the Connections to Calculus for Chapter 9.

831

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9.1 A Brief Introduction to Analytical Geometry Learning Objectives

Generally speaking, analytical geometry is a study of geometry using the tools of algebra and a coordinate system. These tools include the midpoint and distance formulas; the algebra of parallel, perpendicular, and intersecting lines; and other tools that help establish geometric concepts. In this section, we’ll use these tools to verify certain relationships, then use these relationships to introduce a family of curves known as the conic sections.

In Section 9.1 you will learn how to:

A. Verify theorems from basic geometry involving the distance between two points

A. Verifying Relationships from Plane Geometry

B. Verify that points (x, y) are an equal distance from a given point and a given line

C. Use the defining

For the most part, the algebraic tools used in this study were introduced in previous chapters. As the midpoint and distance formulas play a central role, they are restated here for convenience. Algebraic Tools Used in Analytical Geometry

characteristics of a conic section to find its equation

Given two points P1  1x1, y1 2 and P2  1x2, y2 2 in the xy-plane. Midpoint Formula The midpoint of line segment P1P2 is x1  x2 y1  y2 1x, y2  a , b 2 2

Distance Formula The distance from P1 to P2 is d  21x2  x1 2 2  1y2  y1 2 2

These formulas can be used to verify the conclusion of many theorems from Euclidean geometry, while providing important links to an understanding of the conic sections. EXAMPLE 1



Verifying a Theorem from Basic Geometry A theorem from basic geometry states: The midpoint of the hypotenuse of a right triangle is an equal distance from all three vertices. Verify this statement for the right triangle formed by 14, 22, 14, 22 , and (4, 4).

Solution



y 5

(0, 1)

5 After the plotting points and drawing a P triangle, we note the hypotenuse has endpoints (4, 2) 14, 22 and (4, 4), with midpoint 4  142 4  122 a , b  10, 12. Using the 2 2 distance formula to find the distance from (0, 1) to (4, 4) gives

R (4, 4) M 5

x

Q (4, 2) 5

d  210  42 2  11  42 2  2142 2  132 2  125

5 From the definition of midpoint, (0, 1) is also 5 units from 14, 22 . Checking the distance from (0, 1) to the vertex 14, 22 gives

d  214  02 2  12  12 2  242  132 2  125 5 The midpoint of the hypotenuse is an equal distance from all three vertices (see the figure). Now try Exercises 7 through 12

832



9-2

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833

Section 9.1 A Brief Introduction to Analytical Geometry

A. You’ve just learned how to verify theorems from basic geometry involving the distance between two points

Recall from Section 2.1 that a circle is the set of all points that are an equal distance (called the radius) from a given point (called the center). If all three vertices of a triangle lie on the circumference of a circle, we say the circle circumscribes the triangle. Based on our earlier work, it appears we could also state the theorem in Example 1 as For any circle in the xy-plane whose center (h, k) is the midpoint of the hypotenuse of a right triangle, all vertices lie on the circle defined by 1x  h2 2  1y  k2 2  1 d2 2 2 , where d is the length of the hypotenuse. See Figure 9.1 and Exercises 13 through 20.

Figure 9.1 y 5

(4, 4)

(0, 1) 5

x

5

(4, 2)

(4, 2) 5

B. The Distance between a Point and a Line Figure 9.2

In a study of analytical geometry, we are also interested in the distance d between a point and a line. This is always defined as the perpendicular distance, or the length of a line segment perpendicular to the given line, with the given point and the point of intersection as endpoints (see Figure 9.2).

EXAMPLE 2



d

Locating Points That Are an Equal Distance from a Given Point and Line In Figure 9.3, the origin (0, 0) is seen to be an equal distance from the point (0, 2) and the line y  2. Show that the following points are also an equal distance from (0, 2) and y  2: a. 12, 12 2

Solution



b. (4, 2)

c. (8, 8)

Since the given line is horizontal, the perpendicular distance from the line to each point can be found by vertically counting the units. It remains to show that this is also the distance from the given point to (0, 2) (see Figure 9.4). a. The distance from 12, 12 2 to y  2 Figure 9.3 is 2.5 units. The distance from y 12, 12 2 to (0, 2) is d  210  22 2  12  0.52 2  2122 2  11.52 2  16.25  2.5 ✓

b. The distance from (4, 2) to y  2 is 4 units. The distance from (4, 2) to (0, 2) is d  210  42  12  22 2

 2142 2  102 2  116 4 ✓

5

(0, 2) (0, 0)

5

2 3

5

x

y  2

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CHAPTER 9 Analytical Geometry and the Conic Sections

c. The distance from (8, 8) to y  2 is 10 units. The distance from (8, 8) to (0, 2) is

Figure 9.4 y

d  210  82 2  12  82 2

d1  d2

 2182 2  162 2  1100  10 ✓

(8, 8)

d1

5

(0, 2)

(4, 2)

d2

(2, 0.5) (0, 0)

5

B. You’ve just learned how to verify that points (x, y) are an equal distance from a given point and a given line

5

x

y  2

3

Now try Exercises 23 through 26



C. Characteristics of the Conic Sections Figure 9.5 Axis

Nappe Generator Vertex

Examples 1 and 2 bring us one step closer to the wider application of these ideas in a study of the conic sections. But before the connection is clearly made, we’ll introduce some background on this family of curves. In common use, a cone might bring to mind the conical paper cups found at a water cooler. The point of the cone is called the vertex and the sheet of paper forming the sides is called a nappe. In mathematical terms, a cone has two nappes, formed by rotating a nonvertical line (called the generator), about a vertical line (called the axis), at their point of intersection—the vertex (see Figure 9.5). The conic sections are so named because all curves in the family can be formed by a section of the cone, or more precisely the intersection of a plane and a cone. Figure 9.6 shows that if the plane does not go through the vertex, the intersection will produce a circle, ellipse, parabola, or hyperbola.

Nappe

Figure 9.6

WORTHY OF NOTE If the plane does go through the vertex, the result is a single point, a single line (if the plane contains the generator), or a pair of intersecting lines (if the plane contains the axis).

Circle

Parabola

Ellipse

Hyperbola

Circle

Ellipse

Parabola

Hyperbola

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Section 9.1 A Brief Introduction to Analytical Geometry

835

Figure 9.7 The connection we seek to make is that y each conic section can be defined in terms of the distance between points in the plane, as in Example 1, or the distance between a given point and a line, as in Example 2. In Example (8, 8) 1, we noted the points 14, 22, 14, 22, and (4, 4) were all on a circle of radius 5 with d1 5 center (0, 1), in line with the analytic definiFocus tion of a circle: A circle is the set of all points d2 that are an equal distance (called the radius) (4, 2) (0, 2) from a given point (called the center). In Example 2, you may have noticed 5 5 (0, 0) x that the points seemed to form the right y  2 branch of a parabola (see Figure 9.7), and directrix d  d in fact, this example illustrates the analytic 1 2 definition of a parabola: A parabola is the set of all points that are an equal distance from a given point (called the focus), and a given line (called the directrix). The focus and directrix are not actually part of the graph, they are simply used to locate points on the graph. For this reason all foci (plural of focus) will be represented by a “*” symbol rather than a point.

EXAMPLE 3



Finding an Equation for All Points That Form a Certain Parabola With Example 2 as a pattern, use the analytic definition to find a formula (equation) for the set of all points that form the parabola.

Solution



Use the ordered pair (x, y) to represent an arbitrary point on the parabola. Since any point on the line y  2 has coordinates 1x, 22 , we set the distance from 1x, 22 to (x, y) equal to the distance from (0, 2) to (x, y). The result is 21x  x2 2  3y  122 4 2  21x  02 2  1y  22 2 21y  22  2x  1y  22 1y  22 2  x2  1y  22 2 y2  4y  4  x2  y2  4y  4 8y  x2 1 y  x2 8 2

2

2

distances are equal simplify power property expand binomials simplify result

All points satisfying these conditions are on the parabola defined by y  18x2. Now try Exercises 27 and 28



At this point, it seems reasonable to ask what happens when the distance from the focus to (x, y) is less than the distance from the directrix to (x, y). For example, what if the distance is only two-thirds as long? As you might guess, the result is one of the other conic sections, in this case an ellipse. If the distance from the focus to a point (x, y) is greater than the distance from the directrix to (x, y), one branch of a hyperbola is formed. While we will defer a development of their general equations until later in the chapter, the following diagrams serve to illustrate this relationship for the ellipse, and show why we refer to the conic sections as a family of curves. In Figure 9.8, the line segment from the focus to each point on the graph (shown in blue), is exactly two-thirds the length of the line segment from the directrix to the same point (shown in red). Note the graph of these points forms the right half of an ellipse. In Figure 9.9, the lines and points forming the first half are moved to the background to more clearly show the remaining points that form the complete graph.

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CHAPTER 9 Analytical Geometry and the Conic Sections

Figure 9.8

EXAMPLE 4



Figure 9.9

Finding an Equation for All Points That Form a Certain Ellipse Suppose we arbitrarily select the point (1, 0) as a focus and the (vertical) line x  4 as the directrix. Use these to find an equation for the set of all points where the distance from the focus to a point (x, y) is 12 the distance from the directrix to (x, y).

Solution



Since any point on the line x  4 has coordinates (4, y), we have Distance from 11, 02 to 1x, y2 

1 3 distance from 14, y2 to 1x, y2 4 2

1 21x  12 2  3y  102 4 2  21x  42 2  1y  y2 2 2 1 21x  12 2  y2  21x  42 2 2 1 1x  12 2  y2  1x  42 2 4 1 x2  2x  1  y2  1x2  8x  162 4 1 x2  2x  1  y2  x2  2x  4 4 3 2 x  y2  3 4 3x2  4y2  12

in words

resulting equation

simplify

power property

expand binomials

distribute

simplify: 1x 2 

1 2 3 2 x  x 4 4

polynomial form

All points satisfying these conditions are on the ellipse defined by 3x2  4y2  12. Now try Exercises 29 and 30

C. You’ve just learned how to use the defining characteristics of a conic section to find its equation

Actually, any given ellipse has two foci (see Figure 9.10) and the equation from Example 4 could also have been developed using the left focus (with the directrix also on the left). This symmetrical relationship leads us to an alternative definition for the ellipse, which we will explore further in Section 9.2. For foci f1 and f2, an ellipse is the set of all points (x, y) where the sum of the distances from f1 to (x, y) and f2 to (x, y) is constant. See Figure 9.11 and Exercises 31 and 32. Both the focus/directrix definition and the two foci definition have merit, and simply tend to call out different characteristics and applications of the ellipse. The hyperbola also has a focus/directrix definition and a two foci definition. See Exercises 33 and 34.



Figure 9.10 f1

f2

Figure 9.11 d1

d2

f1 d3 (x, y)

d4 d1  d2  d3  d4

f2

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837

Section 9.1 A Brief Introduction to Analytical Geometry

9.1 EXERCISES 

CONCEPTS AND VOCABULARY 4. The conic sections are formed by the intersection of a and a .

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. Analytical geometry is a study of using the tools of . 2. The distance formula is d  the midpoint formula is M 

; .

3. The distance between a point and a line always refers to the distance. 

6. A circle is defined relative to an equal distance between two . A parabola is defined relative to an equal distance between a and a .

DEVELOPING YOUR SKILLS

The three points given form a right triangle. Find the midpoint of the hypotenuse and verify that the midpoint is an equal distance from all three vertices.

7. P1  15, 22 P2  11, 22 P3  15, 62 9. P1  12, 12 P2  16, 52 P3  12, 72

11. P1  110, 212 P2  16, 92 P3  13, 32

8. P1  13, 22 P2  13, 142 P3  18, 22

10. P1  10, 52 P2  16, 42 P3  16, 12

12. P1  16, 62 P2  112, 182 P3  120, 422

13. Find the equation of the circle that circumscribes the triangle in Exercise 7. 14. Find the equation of the circle that circumscribes the triangle in Exercise 8. 15. Find the equation of the circle that circumscribes the triangle in Exercise 9. 

5. If a plane intersects a cone at its vertex, the result is a , a line, or a pair of lines.

16. Find the equation of the circle that circumscribes the triangle in Exercise 10. 17. Find the equation of the circle that circumscribes the triangle in Exercise 11. 18. Find the equation of the circle that circumscribes the triangle in Exercise 12. 19. Of the following six points, four are an equal distance from the point A(2, 3) and two are not. (a) Identify which four, and (b) find any two additional points that are this same (nonvertical, nonhorizontal) distance from (2, 3): C110, 82 E13, 92 B(7, 15) D(9, 14) G12  2 130, 102 F15, 4  3 1102 20. Of the following six points, four are an equal distance from the point P11, 42 and two are not. (a) Identify which four, and (b) Find any two additional points that are the same (nonvertical, nonhorizontal) distance from (1, 4). Q19, 102 R(5, 12) S17, 112 T14, 4  5 132 U11  4 16, 62 V17, 4  1512

WORKING WITH FORMULAS

The Perpendicular Distance from a Point to a Line: Ax1  By1  C d ` ` 2A2  B2 The perpendicular distance from a point (x1, y1) to a given line can be found using the formula shown, where Ax  By  C  0 is the equation of the line in standard form (A, B, and C are integers).

21. Use the formula to verify that P16, 22 and Q(6, 4) are an equal distance from the line y  12x  3. 22. Find the value(s) for y that ensure (1, y) is this same distance from y  12x  3.

Exercise 21 (x1, y1) d

Ax  By  C  0

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CHAPTER 9 Analytical Geometry and the Conic Sections

APPLICATIONS

23. Of the following four points, three are an equal distance from the point A(0, 1) and the line y  1. (a) Identify which three, and (b) find any two additional points that satisfy these conditions. B16, 92 C14, 42 D1212, 62 E14 12, 82 24. Of the following four points, three are an equal distance from the point P(2, 4) and the line y  4. (a) Identify which three, and (b) find any two additional points that satisfy these conditions. R12  4 12, 32 S110, 42 Q110, 92 T12  415, 52 25. Consider a fixed point 10, 42 and a fixed line y  4. Verify that the distance from each point to 10, 42 , is equal to the distance from the point to the line y  4. 25 C1412, 22 A14, 12 Ba10,  b 4 D1815, 202 26. Consider a fixed point 10, 22 and a fixed line y  2. Verify that the distance from each point to 10, 22 , is equal to the distance from the point to the line y  2. 9 R1415, 102 P112, 182 Q a6,  b 2 S1416, 122 27. The points from Exercise 25 are on the graph of a parabola. Find the equation of the parabola. 28. The points from Exercise 26 are on the graph of a parabola. Find the equation of the parabola.

29. Using 10, 22 as the focus and the horizontal line y  8 as the directrix, find an equation for the set of all points (x, y) where the distance from the focus to (x, y) is one-half the distance from the directrix to (x, y).

points (x, y) where the distance from the focus to (x, y) is two-thirds the distance from the directrix to (x, y). Exercise 31

31. From Exercise 29, verify the points 13, 22 and 1 112, 02 are on the ellipse defined by 4x2  3y2  48. Then verify that d1  d2  d3  d4.

y 5

f2 (0, 2)

(3, 2) d1 5

32. From Exercise 30, verify the points 14, 10 3 2 and 13, 1152 are on the ellipse defined by 5x2  9y2  180. Then verify that d1  d2  d3  d4.

( 12, 0)

d3

5 x

d2 (0, 2)

f1

d4

5

Exercise 32 y 5

4, j d1 (4, 0)

f

d2

2 33. From the focus/directrix 6 f1 (4, 0) 6 x definition of a hyperbola: d3 d4 If the distance from the (3,  15) 5 focus to a point (x, y) is greater than the distance from the directrix to (x, y), one branch of a hyperbola is formed. Using (2, 0) as the focus and the vertical line x  12 as the directrix, find an equation for the set of all points (x, y) where the distance from the focus to (x, y), is twice the distance from the directrix to (x, y).

34. From the two foci definition 5 of a hyperbola: For foci f1 and f2, a hyperbola is the (2, 0) d1 f2 set of all points (x, y) where 5 the difference of the d4 distances from f1 to (x, y) d3 and f2 to (x, y) is constant. (3, 2 6) 5 Verify the points (2, 3) and 13, 2162 are on the graph of the hyperbola from Exercise 33. Then verify d1  d2  d3  d4.

y (2, 3) d2 (2, 0) f1 5 x

30. Using (4, 0) as the focus and the vertical line x  9 as the directrix, find an equation for the set of all 

EXTENDING THE CONCEPT

35. Do some reading or research on the orthocenter of a triangle, and the centroid of a triangle. How are they found? What are their properties? Use the ideas and skills from this section to find the (a) orthocenter and (b) centroid of the triangle formed by the points A18, 22, B12, 62 , and C(4, 0).

36. Properties of a circle: A theorem from elementary geometry states: If a radius is perpendicular to a chord, it bisects the chord. Verify this is true for the circle, radii, and chords shown.

y 5

(3, 4) Q (4, 2) T

C

P 5

5 x

S (2, 4) U 5

R

(4, 3)

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37. Verify that points C 12, 32 and D12 12, 162 are points on the ellipse with foci at A 12, 02 and B 12, 02 , by verifying d1AC2  d1BC2  d1AD2  d1BD2. The expression that results has the form 

839

2A  B  2A  B, which prior to the common use of technology had to be simplified using the formula 1A  B  1A  B  2a  1b, where a  2A and b  41A2  B 2 2 . Use this relationship to verify the equation above.

MAINTAINING YOUR SKILLS

38. (6.4) Verify the following is an identity: cos12x2  sin2x  cot2 1x2 1  cos2x 39. (6.7) Find all solutions in 3 0, 22  225  600  825 sinax  b 6

40. (4.4) Solve for x in both exact and approximate form: 10 a. 5  1  9e0.5x b. 345  5e0.4x  75 x2  9 . x2  4 Clearly label all intercepts and asymptotes.

41. (3.5) Sketch a complete graph of h1x2 

9.2 The Circle and the Ellipse Learning Objectives

In Section 9.1, we introduced the equation of an ellipse using analytical geometry and the focus-directrix definition. Here we’ll take a different approach, and use the equation of a circle to demonstrate that a circle is simply a special ellipse. In doing so, we’ll establish a relationship between the foci and vertices of the ellipse, that enables us to apply these characteristics in context.

In Section 9.2 you will learn how to:

A. Use the characteristics of a circle and its graph to understand the equation of an ellipse B. Use the equation of an ellipse to graph central and noncentral ellipses C. Locate the foci of an ellipse and use the foci and other features to write the equation D. Solve applications involving the foci

EXAMPLE 1

A. The Equation and Graph of a Circle Recall that the equation of a circle with radius r and center at (h, k) is 1x  h2 2  1y  k2 2  r2.

As in Section 2.1, the standard form can be used to construct the equation of the circle given the center and radius as in Example 1, or to graph the circle as in Example 2. 

Determining the Equation of a Circle Given Its Center and Radius Find the equation of a circle with radius 5 and center at (2, 1).

Solution



With a center of (2, 1), we have h  2, k  1, and r  5. Making the corresponding substitutions into the standard form we obtain 1x  h2 2  1y  k2 2  r 2 1x  22 2  3y  112 4 2  52 1x  22 2  1y  12 2  25

standard form substitute 2 for h, 1 for k, and 5 for r simplify

The equation of this circle is 1x  22  1y  12 2  25. 2

Now try Exercises 7 through 12



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If the equation is given in polynomial form, recall that we first complete the square in x and y to identify the center and radius. EXAMPLE 2



Completing the Square to Graph a Circle Find the center and radius of the circle whose equation is given, then sketch its graph: x2  y2  6x  4y  3  0.

Solution



Begin by completing the square in both x and y.

1x2  6x  __ 2  1y2  4y  __ 2  3 1x2  6x  92  1y2  4y  42  3  9  4 adds 9 to left side

adds 4 to left side

1x  32 2  1y  22 2 

group x- and y-terms; add 3

complete the square add 9  4 to right side 16 factor and simplify

The center is at (3, 2), with radius is r  116  4. y 3

(3, 2)

Circle Center at (3, 2)

2

8

r4 (3, 2)

(1, 2)

7

Radius: r  4

x

Diameter: 2r  8

(7, 2)

Endpoints of horizontal diameter (1, 2) and (7, 2) Endpoints of vertical diameter (3, 2) and (3, 6)

(3, 6)

Now try Exercises 13 through 18



The equation of a circle in standard form provides a useful link to some of the other conic sections, and is obtained by setting the equation equal to 1. In the case of a circle, this means we simply divide by r2. 1x  h2 2  1y  k2 2  r2

1x  h2 2

r

A. You’ve just learned how to use the characteristics of a circle and its graph to understand the equation of an ellipse

2



1y  k2 r2

standard form

2

1

divide by r 2

In this form, the value of r in each denominator gives the horizontal and vertical distances, respectively, from the center to the graph. This is not so important in the case of a circle, since this distance is the same in any direction. But for other conics, these horizontal and vertical distances are not the same, making the new form a valuable tool for graphing. To distinguish the horizontal from the vertical distance, r2 is replaced by a2 in the “x-term” (horizontal distance), and by b2 in the “y-term” (vertical distance).

B. The Equation of an Ellipse It then seems reasonable to ask, “What happens to the graph when a  b?” To answer, 1x  32 2 1y  22 2   1 (after consider the equation from Example 2. We have 42 42 1x  32 2 1y  22 2 dividing by 16), which we now compare to   1, where a  4 42 32

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WORTHY OF NOTE If you point a flashlight at the floor keeping it perpendicular to the ground, a circle is formed with the bulb pointing directly at the center and every point along the outer edge of the beam an equal distance from this center. If you hold the flashlight at an angle, the circle is elongated and becomes an ellipse, with the bulb pointing directly at one focus.

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Section 9.2 The Circle and the Ellipse

and b  3. The center of the graph is still at (3, 2), since h  3 and k  2 remain unchanged. Substituting y  2 to find additional points, eliminates the y-term and gives two values for x: 1x  32 2 42



12  22 2

32 1x  32 2

1

01 42 1x  32 2  16 x3 4 x34 x  7 and x  1

substitute 2 for y

simplify multiply by 42  16 property of square roots add 3

This shows the horizontal distance from the center to the graph is still a  4, and the points (1, 22 and (7, 2) are on the graph (see Figure 9.12). Similarly, for x  3 we have 1y  22 2  9, giving y  5 and y  1, and showing the vertical distance from the center to the graph is b  3, with points (3, 1) and (3, 5) also on the graph. Using this information to sketch the curve reveals the “circle” is elongated and has become an ellipse. For an ellipse, the longest distance across the graph is called the major axis, with the endpoints of the major axis called vertices. The segment perpendicular to and bisecting the major axis (with its endpoints on the ellipse) is called the minor axis, as shown in see Figure 9.13. Figure 9.12 y 3

(3, 1)

Figure 9.13 b3

2

(1, 2)

8

a4

(3, 2) 5

x

Major axis

(7, 2)

a

b Vertex

Vertex Ellipse

(3, 5) The case where a>b

Minor axis

• If a 7 b, the major axis is horizontal (parallel to the x-axis) with length 2a, and the minor axis is vertical with length 2b (see Example 3). • If b 7 a the major axis is vertical (parallel to the y-axis) with length 2b, and the minor axis is horizontal with length 2a (see Example 4). Generalizing this observation we obtain the equation of an ellipse in standard form. The Equation of an Ellipse in Standard Form 1x  h2 2



1y  k2 2

 1. a2 b2 If a  b the equation represents the graph of an ellipse with center at (h, k). • a gives the horizontal distance from center to graph. • b gives the vertical distance from center to graph. Given

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EXAMPLE 3



Graphing a Horizontal Ellipse Sketch the graph of 1x  22 2 1y  12 2   1. 25 9

Solution



y Ellipse (2, 2)

Noting a  b, we have an ellipse with center 1h, k2  12, 12. The horizontal distance from the center to the graph is a  5, and the vertical distance from the center to the graph is b  3. After plotting the corresponding points and connecting them with a smooth curve, we obtain the graph shown.

(3, 1)

b3 a5 (2, 1)

x

(7, 1)

(2, 4)

Now try Exercises 19 through 24



WORTHY OF NOTE In general, for the equation Ax2  By2  F ( A, B, F 7 0), the equation represents a circle if A  B, and an ellipse if A  B.

EXAMPLE 4

As with the circle, the equation of an ellipse can be given in polynomial form, and here our knowledge of circles is helpful. For the equation 25x2  4y2  100, we know the graph cannot be a circle since the coefficients are unequal, and the center of the graph must be at the origin since h  k  0. To actually draw the graph, we convert the equation to standard form. 

Graphing a Vertical Ellipse For 25x2  4y2  100, (a) write the equation in standard form and identify the center and the values of a and b, (b) identify the major and minor axes and name the vertices, and (c) sketch the graph.

Solution



The coefficients of x2 and y2 are unequal, and 25, 4, and 100 have like signs. The equation represents an ellipse with center at (0, 0). To obtain standard form: a. 25x2  4y2  100 given equation 4y2 25x2  1 divide by 100 100 100 y2 x2  1 standard form 4 25 y2 x2  1 write denominators in squared form; a  2, b  5 22 52 b. The result shows a  2 and b  5, indicating the major axis will be vertical and the minor axis will be horizontal. With the center at the origin, the x-intercepts will be (2, 0) and 12, 02, with the vertices (and y-intercepts) at (0, 5) and 10, 52. c. Plotting these intercepts and sketching the ellipse results in the graph shown. y (0, 5)

Vertical ellipse Center at (0, 0)

b5 (2, 0)

Endpoints of major axis (vertices) (0, 5) and (0, 5)

(2, 0) a2

x

Endpoints of minor axis (2, 0) and (2, 0) Length of major axis 2b: 2(5)  10 Length of minor axis 2a: 2(2)  4

(0, 5)

Now try Exercises 25 through 36



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WORTHY OF NOTE After writing the equation in standard form, it is possible to end up with a constant that is zero or negative. In the first case, the graph is a single point. In the second case, no graph is possible since roots of the equation will be complex numbers. These are called degenerate cases. See Exercise 78.

EXAMPLE 5



If the center of the ellipse is not at the origin, the polynomial form has additional linear terms and we must first complete the square in x and y, then write the equation in standard form to sketch the graph (see the Reinforcing Basic Concepts feature for more on completing the square). Figure 9.14 illustrates how the central ellipse and the shifted ellipse are related.

843

Figure 9.14 y Ellipse with center at (h, k) k

(h, k)

(x  h)2 (y  k)2 1  b2 a2 ab

Central ellipse (0, b) (a, 0) (a, 0) (0, 0) (0, b)

All points shift h units horizontally, k units vertically, opposite the sign

x h x2 y2  2 1 a2 b ab

Completing the Square to Graph an Ellipse Sketch the graph of 25x2  4y2  150x  16y  141  0.

Solution



The coefficients of x2 and y2 are unequal and have like signs, and we assume the equation represents an ellipse but wait until we have the factored form to be certain. 25x2  4y2  150x  16y  141  0 25x2  150x  4y2  16y  141 251x2  6x  __ 2  41y2  4y  __ 2  141 251x2  6x  92  41y2  4y  42  141  225  16 c

c

adds 25192  225

c

add 225  16 to right

c

given equation (polynomial form) group like terms; subtract 141 factor out leading coefficient from each group complete the square

adds 4142  16

251x  32 2  41y  22 2 41y  22 2 251x  32 2  100 100 1y  22 2 1x  32 2  4 25 2 1y  22 2 1x  32  22 52 The result is a vertical ellipse with center at 13, 22, with a  2 and b  5. The vertices are a vertical distance of 5 units from center, and the endpoints of the minor axis are a horizontal distance of 2 units from center. Note this is the same ellipse as in Example 4, but shifted 3 units left and 2 up.

 100 100  100

factor divide both sides by 100

1

simplify (standard form)

1

write denominators in squared form

(3, 7)

y

Vertical ellipse Center at (3, 2)

(5, 2)

(3, 2)

Endpoints of major axis (vertices) (3, 3) and (3, 7) (1, 2)

Endpoints of minor axis (5, 2) and (1, 2) x Length of major axis 2b: 2(5)  10 Length of minor axis 2a: 2(2)  4

(3, 3)

Now try Exercises 37 through 44 B. You’ve just learned how to use the equation of an ellipse to graph central and noncentral ellipses



C. The Foci of an Ellipse In Section 9.1, we noted that an ellipse could also be defined in terms of two special points called the foci. The Museum of Science and Industry in Chicago, Illinois (http://www.msichicago.org), has a permanent exhibit called the Whispering Gallery. The construction of the room is based on some of the reflective properties of an ellipse. If two people stand at designated points in the room and one of them

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whispers very softly, the other person can hear the whisper quite clearly—even though they are over 40 ft apart! The point where each person stands is a focus of the ellipse. This reflective property also applies to light and radiation, giving the ellipse some powerful applications in science, medicine, acoustics, and other areas. To understand and appreciate these applications, we introduce the analytic definition of an ellipse. WORTHY OF NOTE

Definition of an Ellipse

You can easily draw an ellipse that satisfies the definition. Press two pushpins (these form the foci of the ellipse) halfway down into a piece of heavy cardboard about 6 in. apart. Take an 8-in. piece of string and loop each end around the pins. Use a pencil to draw the string taut and keep it taut as you move the pencil in a circular motion—and the result is an ellipse! A different length of string or a different distance between the foci will produce a different ellipse.

Given two fixed points f1 and f2 in a plane, an ellipse is the set of all points (x, y) where the distance from f1 to (x, y) added to the distance from f2 to (x, y) remains constant.

y P(x, y) d1

d1  d2  k f1

The fixed points f1 and f2 are called the foci of the ellipse, and the points P(x, y) are on the graph of the ellipse.

To find the equation of an ellipse in terms of a and b we combine the definition just given with the distance formula. Consider the ellipse shown in Figure 9.15 (for calculating ease we use a central ellipse). Note the vertices have coordinates 1a, 02 and (a, 0), and the endpoints of the minor axis have coordinates 10, b2 and (0, b) as before. It is customary to assign foci the coordinates f1 S 1c, 02 and f2 S 1c, 02. We can calculate the distance between (c, 0) and any point P(x, y) on the ellipse using the distance formula:

6 in. 8 in.

d2

d1  d 2  k

Figure 9.15 y (0, b) P(x, y) (a, 0) (c, 0)

(c, 0)

(0, b)

21x  c2 2  1y  02 2

Likewise the distance between 1c, 02 and any point (x, y) is 21x  c2 2  1y  02 2

According to the definition, the sum must be constant: 21x  c2 2  y2  21x  c2 2  y2  k EXAMPLE 6



Finding the Value of k from the Definition of an Ellipse Use the definition of an ellipse and the diagram given to determine the constant k used for this ellipse following (also see the following Worthy of Note). Note that a  5, b  3, and c  4. y (0, 3)

P(3, 2.4)

(5, 0) (4, 0) (0, 3)

x

f2

(4, 0)

(5, 0) x

(a, 0) x

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Solution




21x  c2 2  1y  02 2  21x  c2 2  1y  02 2
k

213  42  12.4  02  213  42  12.4  02
k 2

WORTHY OF NOTE Note that if the foci are coincident (both at the origin) the “ellipse” will actually be a circle with radius k ; 2x2  y2  2x2  y2
k 2 k2 leads to x2  y2
. In 4 Example 1 we found k
10, 10 giving
5, and if we used 2 the “string” to draw the circle, the pencil would be 5 units from the center, creating a circle of radius 5.

2

2

2

2112  2.4  27  2.4
k 16.76  154.76
k 2.6  7.4
k 10
k The constant used for this ellipse is 10 units. 2

2

2

2

given substitute add simplify radicals compute square roots result

Now try Exercises 45 through 48 In Example 6, the sum of the distances could also be found by moving the point (x, y) to the location of a vertex (a, 0), then using the symmetry of the ellipse. The sum is identical to the length of the major axis, since the overlapping part of the string from (c, 0) to (a, 0) is the same length as from (a, 0) to (c, 0) (see Figure 9.16). This shows the constant k is equal to 2a regardless of the distance between foci. As we noted, the result is




Figure 9.16 y

d1

d2

(a, 0) (c, 0)

21x  c2 2  y2  21x  c2 2  y2
2a

(c, 0)

(a, 0) x

These two segments are equal: d1  d2
2a substitute 2a for k

The details for simplifying this expression are given in Appendix IV, and the result is very close to the standard form seen previously: y2 x2 
1 a2 a2  c2 y2 y2 x2 x2 with 
1 
1, we might a2 b2 a2 a2  c2 suspect that b2
a2  c2, and this is indeed the case. Note from Example 6 the relationship yields By comparing the standard form

b2
a2  c2 3 2
52  42 9
25  16 Additionally, when we consider that (0, b) is Figure 9.17 a point on the ellipse, the distance from (0, b) to y (c, 0) must be equal to a due to symmetry (the (0, b) “constant distance” used to form the ellipse is always 2a). We then see in Figure 9.17, that a a b2  c2
a2 (Pythagorean Theorem), yielding b (a, 0) (a, 0) 2 2 2 b
a  c as above. x (c, 0) (c, 0) With this development, we now have the ability to locate the foci of any ellipse—an important step toward using the ellipse in prac(0, b) tical applications. Because we’re often asked to find the location of the foci, it’s best to rewrite the relationship in terms of c2, using absolute value bars to allow for a major axis that is vertical: c2

a2  b2
.

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EXAMPLE 7



Completing the Square to Graph an Ellipse and Locate the Foci For the ellipse defined by 25x2  9y2  100x  54y  44  0, find the coordinates of the center, vertices, foci, and endpoints of the minor axis. Then sketch the graph.

Solution



25x2  9y2  100x  54y  44  0 25x2  100x  9y2  54y  44 2 251x  4x  __ 2  91y2  6y  __ 2  44 251x2  4x  42  91y2  6y  92  44  100  81 c

c adds 25142  100

c

given group terms; add 44 factor out lead coefficients

add 100  81 to right-hand side

c adds 9192  81

251x  22 2  91y  32 3 91y  32 2 251x  22 2  225 225 1y  32 2 1x  22 2  9 25 2 1x  22 1y  32 2  32 52

 225 225  225

factored form divide by 225

1

simplify (standard form)

1

write denominators in squared form

The result shows a vertical ellipse with a  3 and b  5. The center of the ellipse is at (2, 3). The vertices are a vertical distance of b  5 units from center at (2, 8) and (2, 2). The endpoints of the minor axis are a horizontal distance of a  3 units from center at (1, 3) and (5, 3). To locate the foci, we use the foci formula for an ellipse: c2  a2  b2, giving c2  32  52  16. This shows the foci “✹” are located a vertical distance of 4 units from center at (2, 7) and (2, 1). y (2, 8)

Vertical ellipse Center at (2, 3)

(2, 7)

(1, 3)

(2, 3)

(2, 1) (2, 2)

Endpoints of major axis (vertices) (2, 8) and (2, 2) (5, 3)

x

Endpoints of minor axis (1, 3) and (5, 3) Location of foci (2, 7) and (2, 1) Length of major axis: 2b  2(5)  10 Length of minor axis: 2a  2(3)  6

Now try Exercises 49 through 54



For future reference, remember the foci of an ellipse always occur on the major axis, with a 7 c and a2 7 c2 for a horizontal ellipse. This makes it easier to remember the foci formula for ellipses: c2  a2  b2. Since a2 is larger, it must be decreased by b2 to equal c2. If any two of the values for a, b, and c are known, the relationship between them can be used to construct the equation of the ellipse. EXAMPLE 8



Finding the Equation of an Ellipse Find the equation of the ellipse (in standard form) that has foci at (0, 2) and (0, 2), with a minor axis 6 units in length.

Solution



Since the foci must be on the major axis, we know this is a vertical and central ellipse with c  2 and c2  4. The minor axis has a length of 2a  6 units, meaning a  3 and a2  9. To find b2, use the foci equation and solve.

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Section 9.2 The Circle and the Ellipse

c2  a2  b2 4  9  b2 4  9  b2 4  9  b2 2 b2  5 b  13

LOOKING AHEAD For the hyperbola, we’ll find that c 7 a, and the formula for the foci of a hyperbola will be c2  a2  b2.

foci equation (ellipse) substitute

(0, √13)

solve result

2

C. You’ve just learned how to locate the foci of an ellipse and use the foci and other features to write the equation

y

(0, 2) 2

Since we know b must be greater than a (the major axis is always longer), b2  5 can be y2 x2  1. discarded. The standard form is 2  3 1 2132 2

(3, 0)

(3, 0) x

(0, 2) (0, √13)

Now try Exercises 55 through 64



D. Applications Involving Foci Applications involving the foci of a conic section can take various forms. In many cases, only partial information about the conic section is available and the ideas from Example 8 must be used to “fill in the gaps.” In other applications, we must rewrite a known or given equation to find information related to the values of a, b, and c. EXAMPLE 9



Solving Applications Using the Characteristics of an Ellipse In Washington, D.C., there is a park called the Ellipse located between the White House and the Washington Monument. The park is surrounded by a path that forms an ellipse with the length of the major axis being about 1502 ft and the minor axis having a length of 1280 ft. Suppose the park manager wants to install water fountains at the location of the foci. Find the distance between the fountains rounded to the nearest foot.

Solution



Since the major axis has length 2a  1502, we know a  751 and a2  564,001. The minor axis has length 2b  1280, meaning b  640 and b2  409,600. To find c, use the foci equation:

D. You’ve just learned how to solve applications involving the foci

c2  a2  b2  564,001  409,600  154,401 c  393 and c  393 The distance between the water fountains would be 213932  786 ft. Now try Exercises 65 through 76



9.2 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. For an ellipse, the relationship between a, b, and c is given by the foci equation , since c 6 a or c 6 b.

2. The greatest distance across an ellipse is called the and the endpoints are called .

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3. For a vertical ellipse, the length of the minor axis is and the length of the major axis is .

4. To write the equation 2x2  y2  6x  7 in standard form, the in x.

5. Explain/Discuss how the relations a 7 b, a  b and a 6 b affect the graph of a conic section with 1y  k2 2 1x  h2 2   1. equation a2 b2

6. Suppose foci are located at (3, 2) and (5, 2). Discuss/Explain the conditions necessary for the graph to be an ellipse.

DEVELOPING YOUR SKILLS

Find the equation of a circle satisfying the conditions given.

7. center (0, 0), radius 7 8. center (0, 0), radius 9 9. center (5, 0), radius 13

For each exercise, (a) write the equation in standard form, then identify the center and the values of a and b, (b) state the coordinates of the vertices and the coordinates of the endpoints of the minor axis, and (c) sketch the graph.

10. center (0, 4), radius 15

25. x2  4y2  16

26. 9x2  y2  36

11. diameter has endpoints (4, 9) and (2, 1)

27. 16x2  9y2  144

28. 25x2  9y2  225

12. diameter has endpoints (2, 32 , and (3, 9)

29. 2x2  5y2  10

30. 3x2  7y2  21

Identify the center and radius of each circle, then sketch its graph.

13. x2  y2  12x  10y  52  0

Identify each equation as that of an ellipse or circle, then sketch its graph.

31. 1x  12 2  41y  22 2  16

14. x2  y2  8x  6y  11  0

32. 91x  22 2  1y  32 2  36

15. x2  y2  4x  10y  4  0

33. 21x  22 2  21y  42 2  18

16. x2  y2  4x  6y  3  0 17. x2  y2  6x  5  0 18. x2  y2  8y  5  0 Sketch the graph of each ellipse.

1x  12 2 1y  22 2 19.  1 9 16 1x  32 2 1y  12 2 20.  1 4 25 21.

1x  22 2 1y  32 2  1 25 4

1x  52 2 1y  22 2 22.  1 1 16 23.

1x  12 2 1y  22 2  1 16 9 1x  12 1y  32  1 36 9 2

24.

2

34. 1x  62 2  y2  49

35. 41x  12 2  91y  42 2  36 36. 251x  32 2  41y  22 2  100 Complete the square in both x and y to write each equation in standard form. Then draw a complete graph of the relation and identify all important features.

37. 4x2  y2  6y  5  0 38. x2  3y2  8x  7  0 39. x2  4y2  8y  4x  8  0 40. 3x2  y2  8y  12x  8  0 41. 5x2  2y2  20y  30x  75  0 42. 4x2  9y2  16x  18y  11  0 43. 2x2  5y2  12x  20y  12  0 44. 6x2  3y2  24x  18y  3  0

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Use the definition of an ellipse to find the constant k used for each ellipse (figures are not drawn to scale).

45.

53. 6x2  24x  9y2  36y  6  0 54. 5x2  50x  2y2  12y  93  0

y (0, 8) (6, 6.4) (a, 0) (6, 0)

(6, 0)

(a, 0) x

55. vertices at (6, 0) and (6, 0); foci at (4, 0) and (4, 0) 56. vertices at (8, 0) and (8, 0); foci at (5, 0) and (5, 0)

(0, 8)

46.

Find the equation of an ellipse (in standard form) that satisfies the following conditions:

y (0, 12)

57. foci at (3, 6) and (3, 2); length of minor axis: 6 units

(9, 9.6) (a, 0) (9, 0)

(9, 0)

(a, 0) x

58. foci at (4, 3) and (8, 3); length of minor axis: 8 units

(0, 12)

47.

(0, b) y (0, 8)

(4.8, 6)

Use the characteristics of an ellipse and the graph given to write the related equation and find the location of the foci.

59.

60.

y

y

(6, 0) x

(6, 0) (0, 8)

x

(0, b)

48.

x

(0, b) y

(0, 28) (96, 0) x

(96, 0) (0, 28)

(76.8, 60)

61.

62.

y

y

(0, b)

Find the coordinates of the (a) center, (b) vertices, (c) foci, and (d) endpoints of the minor axis. Then (e) sketch the graph.

x

x

49. 4x2  25y2  16x  50y  59  0 50. 9x2  16y2  54x  64y  1  0 51. 25x2  16y2  200x  96y  144  0 52. 49x2  4y2  196x  40y  100  0 

WORKING WITH FORMULAS

63. Area of an Ellipse: A  ab The area of an ellipse is given by the formula shown, where a is the distance from the center to the graph in the horizontal direction and b is the distance from center to graph in the vertical direction. Find the area of the ellipse defined by 16x2  9y2  144.

a2  b2 B 2 The perimeter of an ellipse can be approximated by the formula shown, where a represents the length of the semimajor axis and b represents the length of the semiminor axis. Find the perimeter of the y2 x2 ellipse defined by the equation   1. 49 4

64. The Perimeter of an Ellipse: P  2

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APPLICATIONS

65. Decorative fireplaces: A bricklayer intends to build an elliptical fireplace 3 ft high and 8 ft wide, with two glass doors that open at the middle. The hinges to these doors are to be screwed onto a spine that is perpendicular to the hearth and goes through the foci of the ellipse. How far from center will the spines be located? How tall will each spine be?

for the best result? Round to the nearest hundredth. Exercise 68 Vertex Focus Lithotripter

8 ft

3 ft

Spines

66. Decorative gardens: A retired math teacher decides to present her husband with a beautiful elliptical garden to help celebrate their 50th anniversary. The ellipse is to be 8 m long and 5 m across, with decorative fountains located at the foci. How far from the center of the ellipse should the fountains be located (round to the nearest 100th of a meter)? How far apart are the fountains? 67. Attracting attention to art: As part of an art show, a gallery owner asks a student from the local university to design a unique exhibit that will highlight one of the more significant pieces in the collection, an ancient sculpture. The student decides to create an elliptical showroom with reflective walls, with a rotating laser light on a stand at one focus, and the sculpture placed at the other focus on a stand of equal height. The laser light then points continually at the sculpture as it rotates. If the elliptical room is 24 ft long and 16 ft wide, how far from the center of the ellipse should the stands be located (round to the nearest 10th of a foot)? How far apart are the stands? 68. Medical procedures: The medical procedure called lithotripsy is a noninvasive medical procedure that is used to break up kidney and bladder stones in the body. A machine called a lithotripter uses its three-dimensional semielliptical shape and the foci properties of an ellipse to concentrate shock waves generated at one focus, on a kidney stone located at the other focus (see diagram — not drawn to scale). If the lithotripter has a length (semimajor axis) of 16 cm and a radius (semiminor axis) of 10 cm, how far from the vertex should a kidney stone be located

69. Elliptical arches: In some Exercise 69 situations, bridges are built using uniform 8 ft elliptical archways as 60 ft shown in the figure given. Find the equation of the ellipse forming each arch if it has a total width of 30 ft and a maximum center height (above level ground) of 8 ft. What is the height of a point 9 ft to the right of the center of each arch? 70. Elliptical arches: An elliptical arch bridge is built across a one lane highway. The arch is 20 ft across and has a maximum center height of 12 ft. Will a farm truck hauling a load 10 ft wide with a clearance height of 11 ft be able to go through the bridge without damage? (Hint: See Exercise 69.)

As a planet orbits around the Sun, it traces out an ellipse. If the center of the ellipse were placed at (0, 0) on a coordinate grid, the Sun would be actually off-centered (located at the focus of the ellipse). Use this information and the graphs provided to complete Exercises 71 through 74.

71. Orbit of Mercury: The approximate orbit of the planet Mercury is shown in the figure given. Find an equation that models this orbit.

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Exercise 71

(1.296 million miles per day), how many days does it take Mars to orbit the Sun? (Hint: Use the formula from Exercise 64).

y

Sun x

70.5 million miles

Mercury

72 million miles

72. Orbit of Pluto: The approximate orbit of the dwarf planet Pluto is shown in the figure given. Find an equation that models this orbit. Exercise 72 y

Sun x

3540 million miles

Pluto

3650 million miles

73. Planetary orbits: Except for small variations, a planet’s orbit around the Sun is elliptical with the Sun at one focus. The aphelion (maximum distance from the Sun) of the planet Mars is approximately 156 million miles, while the perihelion (minimum distance from the Sun) of Mars is about 128 million miles. Use this information to find the lengths of the semimajor and semiminor axes, rounded to the nearest million. If Mars has an orbital velocity of 54,000 miles per hour



851

74. Planetary orbits: The aphelion (maximum distance from the Sun) of the planet Saturn is approximately 940 million miles, while the perihelion (minimum distance from the Sun) of Saturn is about 840 million miles. Use this information to find the lengths of the semimajor and semiminor axes, rounded to the nearest million. If Saturn has an orbital velocity of 21,650 miles per hour (about 0.52 million miles per day), how many days does it take Saturn to orbit the Sun? How many years? 75. Area of a race track: Suppose the Toronado 500 is a car race that is run on an elliptical track. The track is bounded by two ellipses with equations of 4x2  9y2  900 and 9x2  25y2  900, where x any y are in hundreds of yards. Use the formula given in Exercise 63 to find the area of the race track. Exercise 76 76. Area of a border: The table cloth for a large oval table is elliptical in shape. It is designed with two concentric ellipses (one within the other) as shown in the figure. The equation of the outer ellipse is 9x2  25y2  225, and the equation of the inner ellipse is 4x2  16y2  64 with x and y in feet. Use the formula given in Exercise 63 to find the area of the border of the tablecloth.

EXTENDING THE THOUGHT

Exercise 77 77. When graphing the conic sections, it is often helpful y Focal to use what is called a focal chords chord, as it gives additional points on the graph with very x little effort. A focal chord is a line segment through a focus (perpendicular to the major or transverse axis), with the endpoints on the graph. For an ellipse, 2 the length of the focal chord is given by L  2m n , where m is the length of the semiminor axis, and n is the length of the semimajor axis. The focus will always be the midpoint of this line segment. Find the length of the focal chord for the ellipse

x2 81

y2

 36  1 and the coordinates of the endpoints. Verify (by substituting into the equation) that these endpoints are indeed points on the graph, then use them to help complete the graph. 78. For the equation 6x2  36x  3y2  24y  74  28, does the equation appear to be that of a circle, ellipse, or parabola? Write the equation in factored form. What do you notice? What can you say about the graph of this equation? 79. Verify that for the ellipse the focal chord is

2b2 . a

y2 x2 2  2  1, the length of a b

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MAINTAINING YOUR SKILLS

80. (4.4) Evaluate the expression using the change-ofbase formula: log320. 81. (3.8) The resistance R to current flow in an electrical wire varies directly as the length L of the wire and inversely as the square of its diameter d. (a) Write the equation of variation; (b) find the constant of variation if a wire 2 m long with diameter d  0.005 m has a resistance of 240 ohms (); and (c) find the resistance in a similar wire 3 m long and 0.006 m in diameter.

82. (7.6) Use De Moivre’s theorem to compute the value of z  11  13i2 6. 83. (8.4) Find the true direction and groundspeed of the airplane shown, given the direction and speed of the wind (indicated in blue).

Exercise 83 250 mph heading 20

N

30 mph heading 90

9.3 The Hyperbola Learning Objectives In Section 9.3 you will learn how to:

A. Use the equation of a hyperbola to graph central and noncentral hyperbolas

B. Distinguish between the

As seen in Section 9.1 (see Figure 9.18), a hyperbola is a conic section formed by a plane that cuts both nappes of a right circular cone. A hyperbola has two symmetric parts called branches, which open in opposite directions. Although the branches appear to resemble parabolas, we will soon discover they are actually a very different curve.

Figure 9.18 Axis

Hyperbola

equations of a circle, ellipse, and hyperbola

A. The Equation of a Hyperbola

C. Locate the foci of a hyperbola and use the foci and other features to write its equation

In Section 9.2, we noted that for the equation Ax2  By2  F, if A  B, the equation is that of a circle, if A  B, the equation represents an ellipse. Both cases contain a sum of second-degree terms. Perhaps driven by curiosity, we might wonder what happens if the equation has a difference of second-degree terms. Consider the equation 9x2  16y2  144. It appears the graph will be centered at (0, 0) since no shifts are applied (h and k are both zero). Using the intercept method to graph this equation reveals an entirely new curve, called a hyperbola.

D. Solve applications involving foci

EXAMPLE 1



Graphing a Central Hyperbola Graph the equation 9x2  16y2  144 using intercepts and additional points as needed.

Solution



9x2  16y2 9102 2  16y2 16y2 y2

 144  144  144  9

given substitute 0 for x simplify divide by 16

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Since y2 can never be negative, we conclude that the graph has no y-intercepts. Substituting y  0 to find the x-intercepts gives 9x2  16y2  144 9x2  16102 2  144 9x2  144 x2  16 x  116 and x   116 x  4 and x  4 (4, 0) and 14, 02

given substitute 0 for y simplify divide by 9 square root property simplify x-intercepts

Knowing the graph has no y-intercepts, we select inputs greater than 4 and less than 4 to help sketch the graph. Using x  5 and x  5 yields 9x2  16y2 9152 2  16y2 91252  16y2 225  16y2 16y2

 144  144  144  144  81 81 y2  16 9 9 y y 4 4 y  2.25 y  2.25 15, 2.252 15, 2.252

9x2  16y2 9152 2  16y2 91252  16y2 225  16y2 16y2

given substitute for x

52  152 2  25 simplify subtract 225 divide by 16

9 4 y  2.25 15, 2.252 y

square root property decimal form ordered pairs

Plotting these points and connecting them with a smooth curve, while knowing there are no y-intercepts, produces the graph in the figure. The point at the origin (in blue) is not a part of the graph, and is given only to indicate the “center” of the hyperbola.

 144  144  144  144  81 81 y2  16 9 y 4 y  2.25 15, 2.252 y Hyperbola

(5, 2.25) (5, 2.25) (4, 0)

(4, 0) (0, 0)

x

(5, 2.25) (5, 2.25)

Now try Exercises 7 through 22



Since the hyperbola crosses a horizontal line of symmetry, it is referred to as a horizontal hyperbola. The points 14, 02 and (4, 0) are called vertices, and the center of the hyperbola is always the point halfway between them. If the center is at the origin, we have a central hyperbola. The line passing through the center and both vertices is called the transverse axis (vertices are always on the transverse axis), and the line passing through the center and perpendicular to this axis is called the conjugate axis (see Figure 9.19). In Example 1, the coefficient of x2 was positive and we were subtracting 2 16y : 9x2  16y2  144. The result was a horizontal hyperbola. If the y2-term is

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positive and we subtract the term containing x2, the result is a vertical hyperbola (Figure 9.20). Figure 9.19

Figure 9.20

y Conjugate axis

Center Vertex

Transverse axis

Transverse axis Vertex



x

Vertex Center



x

Vertical hyperbola

Identifying the Axes, Vertices, and Center of a Hyperbola from Its Graph For the hyperbola shown, state the location of the vertices and the equation of the transverse axis. Then identify the location of the center and the equation of the conjugate axis.

Solution

Conjugate axis

Vertex

Horizontal hyperbola

EXAMPLE 2

y

By inspection we locate the vertices at (0, 0) and (0, 4). The equation of the transverse axis is x  0. The center is halfway between the vertices at (0, 2), meaning the equation of the conjugate axis is y  2.

y

5

5

5

x

5

Now try Exercises 23 through 26



Standard Form As with the ellipse, the polynomial form of the equation is helpful for identifying hyperbolas, but not very helpful when it comes to graphing a hyperbola (since we still must go through the laborious process of finding additional points). For graphing, standard form is once again preferred. Consider the hyperbola 9x2  16y2  144 from Example 1. To write the equation in standard form, we y2 x2 divide by 144 and obtain 2  2  1. By comparing the standard form to the graph, 4 3 we note a  4 represents the distance from center to vertices, similar to the way we used a previously. But since the graph has no y-intercepts, what could b  3 represent? The answer lies in the fact that branches of a hyperbola are asymptotic, meaning they will approach and become very close to imaginary lines that can be used to sketch the graph. b For a central hyperbola, the slopes of the asymptotic lines are given by the ratios and a b b b  , with the related equations being y  x and y   x. The graph from Example 1 a a a is repeated in Figure 9.21, with the asymptotes drawn. For a clearer understanding of how the equations for the asymptotes were determined, see Exercise 88.

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A second method of drawing the asymptotes involves drawing a central rectangle with dimensions 2a by 2b, as shown in Figure 9.22. The asymptotes will be the extended diagonals of this rectangle. This brings us to the equation of a hyperbola in standard form. Figure 9.21

Figure 9.22 y

y Slope m 

 34

Slope m 

3 4

run a4

rise b3 (4, 0)

(4, 0)

(4, 0)

2b

x

(0, 0)

x

2a

Slope method

Central rectangle method

The Equation of a Hyperbola in Standard Form The equation 1x  h2 a2

2



1y  k2 b2

The equation 2

1

represents a horizontal hyperbola with center (h, k) • transverse axis y  k • conjugate axis x  h • a gives the distance from center to vertices.

1y  k2 2 b2



1x  h2 2 a2

1

represents a vertical hyperbola with center (h, k) • transverse axis x  h • conjugate axis y  k • b gives the distance from center to vertices.

b • Asymptotes can be drawn by starting at (h, k) and using slopes m   . a

EXAMPLE 3



Graphing a Hyperbola Using Its Equation in Standard Form Sketch the graph of 161x  22 2  91y  12 2  144. Label the center, vertices, and asymptotes.

Solution



Begin by noting a difference of the second-degree terms, with the x2-term occurring first. This means we’ll be graphing a horizontal hyperbola whose center is at (2, 1). Continue by writing the equation in standard form. 161x  22 2  91y  12 2  144 161x  22 2 91y  12 2 144   144 144 144 1x  22 2 1y  12 2  1 9 16 1x  22 2 1y  12 2  1 32 42

given equation divide by 144

simplify

write denominators in squared form

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Since a  3 the vertices are a horizontal distance of 3 units from the center (2, 1), giving 12  3, 12 S 15, 12 and 12  3, 12 S 11, 12 . After plotting the center and b 4 vertices, we can begin at the center and count off slopes of m     , or a 3 draw a rectangle centered at (2, 1) with dimensions 2132  6 (horizontal dimension) by 2142  8 (vertical dimension) to sketch the asymptotes. The complete graph is shown here. Horizontal hyperbola

y

Center at (2, 1)

md

Vertices at (1, 1) and (5, 1) Transverse axis: y  1 Conjugate axis: x  2

(2, 1) (1, 1)

(5, 1) x

2(3)  6 m  d



Width of rectangle horizontal dimension and distance between vertices 2a  2(3)  6



Length of rectangle (vertical dimension) 2b  2(4)  8

Now try Exercises 27 through 38



Polynomial Form If the equation is given as a polynomial in expanded form, complete the square in x and y, then write the equation in standard form.

EXAMPLE 4



Graphing a Hyperbola by Completing the Square Graph the equation 9y2  x2  54y  4x  68  0.

Solution



Since the y2-term occurs first, we assume the equation represents a vertical hyperbola, but wait for the factored form to be sure (see Exercise 87). 9y2  x2  54y  4x  68  0 9y2  54y  x2  4x  68 91y2  6y  ____ 2  11x2  4x  ____ 2  68 91y2  6y  92  11x2  4x  42  68  81  142 c

c

adds 9 19 2  81

c

c

adds 1 14 2  4

add 81  14 2 to right

91y  32 2  11x  22 2  9 1y  32 2 1x  22 2  1 1 9 1y  32 2 1x  22 2  1 12 32

given collect like-variable terms; subtract 68 factor out 9 from y-terms and 1 from x-terms complete the square

factor S vertical hyperbola divide by 9 (standard form)

write denominators in squared form

The center of the hyperbola is 12, 32 with a  3, b  1, and a transverse axis of x  2. The vertices are at 12, 3  12 and 12, 3 12 S 12, 22 and 12, 42 . After plotting the center and vertices, we draw a rectangle centered at 12, 32 with a horizontal “width” of 2132  6 and a vertical “length” of 2112  2 to sketch the asymptotes. The completed graph is given in the figure.

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857

Vertical hyperbola

y

Center at (2, 3) Vertices at (2, 2) and (2, 4)

m  3 1

A. You’ve just learned how to use the equation of a hyperbola to graph central and noncentral hyperbolas

(2, 2)

Transverse axis: x  2 Conjugate axis: y  3

x 1 m3 (2, 3) center

Width of rectangle (horizontal dimension) 2a  2(3)  6

(2, 4)



Length of rectangle vertical dimension and distance between vertices 2b  2(1)  2



Now try Exercises 39 through 48



B. Distinguishing between the Equations of a Circle, Ellipse, and Hyperbola So far we’ve explored numerous graphs of circles, ellipses, and hyperbolas. In Example 5 we’ll attempt to identify a given conic section from its equation alone (without graphing the equation). As you’ve seen, the corresponding equations have unique characteristics that can help distinguish one from the other. EXAMPLE 5



Identifying a Conic Section from Its Equation Identify each equation as that of a circle, ellipse, or hyperbola. Justify your choice and name the center, but do not draw the graphs. a. y2  36  9x2 b. 4x2  16  4y2 c. x2  225  25y2 d. 25x2  100  4y2 2 2 e. 31x  22  41y  32  12 f. 41x  52 2  36  91y  42 2

Solution



B. You’ve just learned how to distinguish between the equations of a circle, ellipse, and hyperbola

a. Writing the equation in factored form gives y2  9x2  36 1h  0, k  02. Since the equation contains a difference of second-degree terms, it is the equation of a (vertical) hyperbola. The center is at (0, 0). b. Rewriting the equation as 4x2  4y2  16 and dividing by 4 gives x2  y2  4. The equation represents a circle of radius 2, with the center at (0, 0). c. Writing the equation as x2  25y2  225 we note a sum of second-degree terms with unequal coefficients. The equation is that of an ellipse, with the center at (0, 0). d. Rewriting the equation as 25x2  4y2  100 we note the equation contains a difference of second-degree terms. The equation represents a central (horizontal) hyperbola, whose center is at (0, 0). e. The equation is in factored form and contains a sum of second-degree terms with unequal coefficients. This is the equation of an ellipse with the center at 12, 32 . f. Rewriting the equation as 41x  52 2  91y  42 2  36 we note a difference of second-degree terms. The equation represents a horizontal hyperbola with center 15, 42. Now try Exercises 49 through 60



C. The Foci of a Hyperbola Like the ellipse, the foci of a hyperbola play an important part in their application. A long distance radio navigation system (called LORAN for short), can be used to determine the location of ships and airplanes and is based on the characteristics of a hyperbola

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(see Exercises 85 and 86). Hyperbolic mirrors are also used in some telescopes, and have the property that a beam of light directed at one focus will be reflected to the second focus. To understand and appreciate these applications, we use the analytic definition of a hyperbola: Definition of a Hyperbola Given two fixed points f1 and f2 in a plane, a hyperbola is the set of all points (x, y) such that the distance from f2 to (x, y) subtracted from the distance from f1 to (x, y) is a positive constant. In symbols,

y

d1

d1  d2
k, k 7 0

(x, y) d2

The fixed points f1 and f2 are called the foci of the hyperbola, and the points (x, y) are on the graph of the hyperbola.

f1

f2

x

d1  d2
k k>0

Figure 9.23

As with the analytic definition of the ellipse, it can be shown that the constant k is again equal to 2a (for horizontal hyperbolas). To find the equation of a hyperbola in terms of a and b, we use an approach similar to that of the ellipse (see Appendix IV), y2 x2 and the result is identical to that seen earlier: 2  2
1 where b2
c2  a2 (see a b Figure 9.23). We now have the ability to find the foci of any hyperbola—and can use this information in many significant applications. Since the location of the foci play such an important role, it is best to remember the relationship as c2
a2  b2 (called the foci formula for hyperbolas), noting that for a hyperbola, c 7 a and c2 7 a2 (also c 7 b and c2 7 b2).

y

(x, y)

(c, 0) x

(c, 0) (a, 0)

(a, 0)

EXAMPLE 6




Graphing a Hyperbola and Identifying Its Foci by Completing the Square. For the hyperbola defined by 7x2  9y2  14x  72y  200
0, find the coordinates of the center, vertices, foci, and the dimensions of the central rectangle. Then sketch the graph.

Solution




7x2  9y2  14x  72y  200
0 given 2 2 7x  14x  9y  72y
200 group terms; add 200 71x2  2x  ____ 2  91y2  8y  ____ 2
200 factor out leading coefficients 2 2 71x  2x  12  91y  8y  162
200  7  11442 complete the square c

c

adds 7112
7

c

c

adds 9 116 2
144

71x  12 2  91y  42 2
63 1x  12 2 1y  42 2 
1 9 7 1x  12 2 1y  42 2 
1 32 1 172 2

S add 7  11442 to right-hand side factored form

divide by 63 and simplify

write denominators in squared form

This is a horizontal hyperbola with a
3 1a2
92 and b
17 1b2
72. The center is at (1, 4), with vertices 12, 42 and (4, 4). Using the foci formula c2
a2  b2 yields c2
9  7
16, showing the foci are 13, 42 and (5, 4) (4 units from center). The central rectangle is 2 17
5.29 by 2132
6.

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859

Drawing the rectangle and sketching the asymptotes to complete the graph, results in the graph shown. Horizontal hyperbola

y

Center at (1, 4) Vertices at (2, 4) and (4, 4) (3, 4)

(2, 4)

(1, 4)

Transverse axis: y  4 Conjugate axis: x  1 Location of foci: (3, 4) and (5, 4)

(5, 4) x

(4, 4)



Width of rectangle horizontal dimension and distance between vertices 2a  2(3)  6



Length of rectangle (vertical dimension) 2b  2(√7) ≈ 5.29

C. You’ve just learned how to locate the foci of a hyperbola and use the foci and other features to write its equation

Now try Exercises 61 through 70



As with the ellipse, if any two of the values for a, b, and c are known, the relationship between them can be used to construct the equation of the hyperbola. See Exercises 71 through 78.

D. Applications Involving Foci Applications involving the foci of a conic section can take many forms. As before, only partial information about the hyperbola may be available, and we’ll determine a solution by manipulating a given equation, or constructing an equation from given facts. EXAMPLE 7



Applying the Properties of a Hyperbola—The Path of a Comet Comets with a high velocity cannot be captured by the Sun’s gravity, and are slung around the Sun in a hyperbolic path with the Sun at one focus. If the path illustrated by the graph shown is modeled by the equation 2116x2  400y2  846,400, how close did the comet get to the Sun? Assume units are in millions of miles and round to the nearest million.

Solution



We are essentially asked to find the distance between a vertex and focus. Begin by writing the equation in standard form: 2116x2  400y2  846,400 y2 x2  1 400 2116 y2 x2  1 202 462

given divide by 846,400 y write denominators in squared form

This is a horizontal hyperbola with a  20 1a2  4002 and b  46 1b2  21162. Use the foci formula to find c2 and c.

(0, 0) x

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c2  a2  b2 c2  400  2116 c2  2516 c  50 and c  50 Since a  20 and c  50, the comet came within 50  20  30 million miles of the Sun. Now try Exercises 81 through 84

EXAMPLE 8





Applying the Properties of a Hyperbola—The Location of a Storm Two amateur meteorologists, living 4 km apart (4000 m), see a storm approaching. The one farthest from the storm hears a loud clap of thunder 9 sec after the one nearest. Assuming the speed of sound is 340 m/sec, determine an equation that models possible locations for the storm at this time.

Solution



D. You’ve just learned how to solve applications involving foci

Let M1 represent the meteorologist nearest the storm and M2 the farthest. Since M2 heard the thunder 9 sec after M1, M2 must be 9 # 340  3060 m farther away from the storm S. In other words, M2S  M1S  3060. The set of all points that satisfy this description fit the definition of a hyperbola, and we’ll use this fact to develop an equation model for possible locations of the storm. Let’s place the information on a coordinate grid. For convenience, we’ll use y the straight line distance S between M1 and M2 as the 2 x-axis, with the origin an equal distance from each. 1 With the constant difference M2 M1 equal to 3060, we have 3 2 1 1 2 3 x in 1000s 2a  3060, a  1530 from 1 the definition of a hyperbola, 2 y2 x2  giving 2 2  1. With 1530 b c  2000 m (the distance from the origin to M1 or M2), we find the value of b using the equation c2  a2  b2: 20002  15302  b2 or b2  120002 2  115302 2  1,659,100  12882. The equation that models possible locations of the storm is y2 x2   1. 15302 12882 Now try Exercises 85 and 86



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Section 9.3 The Hyperbola

TECHNOLOGY HIGHLIGHT

Studying Hyperbolas As with the circle and ellipse, the hyperbola must also be defined in two pieces in order to use a graphing calculator to study its graph. Consider the relation 4x2  9y2  36. From our work in this section, we know this is the equation of a horizontal hyperbola centered at (0, 0). Solving for y gives 4x2  9y2  36

original equation

9y  36  4x 2

y2 

2

isolate y2-term

36  4x2 9

y

divide by 9

36  4x2 B 9

solve for y

Figure 9.24

36  4x2 B 9 36  4x2 gives the “upper half” of the hyperbola, and Y2   gives B 9 the “lower half.” In Figure 9.24, note the use of parentheses on the Y = screen to ensure we’re taking the square root of the entire expression. Entering these on the Y = screen, graphing them with the window shown, and pressing the TRACE key gives the graph shown in Figure 9.25. Note the location of the cursor at x  0, but no y-value is displayed. This is because the hyperbola is not defined at x  0. Press the right arrow key and walk the cursor to the right until the y-values begin appearing. In fact, they begin to appear at (3, 0), which is one of the vertices of the hyperbola. We could also graph the asymptotes 1 y   23 x2 by entering the lines as Y3 and Y4 on the Y = screen. The resulting graph is shown in Figure 9.26 using the standard window (the TRACE key has been pushed and the down arrow used to highlight Y2). Use these ideas to complete the following exercises. We can again separate this result into two parts: Y1 

Figure 9.26

Figure 9.25

10

6

10

10

6

10

10

10

Exercise 1: Graph the hyperbola 25y2  4x2  100 using a friendly window. What are the coordinates of the vertices? Use the TRACE feature to find the value(s) of y when x  4. Determine (from the graph) the value(s) of y when x  4, then verify your response using the TABLE feature. Exercise 2: Graph the hyperbola 9x2  16y2  144 using the standard window. Then determine the equations of the asymptotes and graph these as well. Why do the asymptotes intersect at the origin? When will the asymptotes not intersect at the origin?

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9.3 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

4. The center of the hyperbola defined by 1x  22 2

1. The line that passes through the vertices of a hyperbola is called the axis. 2. The conjugate axis is axis and contains the

to the of the hyperbola.

3. The center of a hyperbola is located between the vertices.



42



1y  32 2 52

 1 is at

.

5. Compare/Contrast the two methods used to find the asymptotes of a hyperbola. Include an example illustrating both methods. 6. Explore/Explain why A1x  h2 2  B1y  k2 2  F results in a hyperbola regardless of whether A  B or A  B. Illustrate with an example.

DEVELOPING YOUR SKILLS

Graph each hyperbola. Label the center, vertices, and any additional points used.

7.

y2 x2  1 9 4 2

8.

2

y2 x2  1 16 9 2

y x  1 4 9

10.

y x  1 25 16

11.

y2 x2  1 49 16

12.

y2 x2  1 25 9

13.

y2 x2  1 36 16

14.

y2 x2  1 81 16

15.

2

y x  1 9 1 2

16.

2

24.

y 10

10

y 10

10

10 x

10 x

2

9.

2

23.

2

2

2

2

10

25.

17.

y x  1 12 4

18.

y x  1 9 18

19.

y2 x2  1 9 9

20.

y2 x2  1 4 4

21.

y2 x2  1 36 25

22.

y2 x2  1 16 4

For the graphs given, state the location of the vertices and the equation of the transverse axis. Then identify the location of the center and the equation of the conjugate axis. Note the scale used on each axis.

26.

y 10

10

y x  1 1 4

10

10 x

10

y 10

10

10 x

10

Sketch a complete graph of each equation, including the asymptotes. Be sure to identify the center and vertices.

27.

1y  12 2 x2  1 4 25

1x  22 2 y2 28.  1 4 9

1x  32 2 1y  22 2 29.  1 36 49 30.

1x  22 2 1y  12 2  1 9 4

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31. 32.

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Section 9.3 The Hyperbola

1y  12 2 1x  52 2  1 7 9 1y  32 2 1x  22 2  1 16 5

33. 1x  22  41y  12  16 2

Use the definition of a hyperbola to find the distance between the vertices and the dimensions of the rectangle centered at (h, k). Figures are not drawn to scale. Note that Exercises 63 and 64 are vertical hyperbolas.

61.

62.

y

y

2

34. 91x  12 2  1y  32 2  81

(15, 6.75)

(5, 2.25)

35. 21y  32 2  51x  12 2  50

(a, 0) (5, 0)

(5, 0)

x

(15, 0)

(a, 0)

36. 91y  42 2  51x  32 2  45

(a, 0) (15, 0) x

(a, 0)

37. 121x  42 2  51y  32 2  60 38. 81x  42 2  31y  32 2  24

63.

39. 16x2  9y2  144 40. 16x2  25y2  400 41. 9y  4x  36 2

2

64.

y

y (0, 13)

(0, 10) (0, b) (0, b)

(6, 7.5)

(9, 6.25)

x

(0, b) (0, b)

x

(0, 10) (0, 13)

42. 25y2  4x2  100 43. 12x2  9y2  72 44. 36x2  20y2  180 45. 4x  y  40x  4y  60  0 2

2

46. x2  4y2  12x  16y  16  0

Find and list the coordinates of the (a) center, (b) vertices, (c) foci, and (d) dimensions of the central rectangle. Then (e) sketch the graph, including the asymptotes.

47. x2  4y2  24y  4x  36  0

65. 4x2  9y2  24x  72y  144  0

48. 9x  4y  18x  24y  9  0

66. 4x2  36y2  40x  144y  188  0

2

2

Classify each equation as that of a circle, ellipse, or hyperbola. Justify your response.

67. 16x2  4y2  24y  100  0 68. 81x2  162x  4y2  243  0

49. 4x2  4y2  24

69. 9x2  3y2  54x  12y  33  0

50. 9y  4x  36

70. 10x2  60x  5y2  20y  20  0

2

2

51. x2  y2  2x  4y  4 52. x2  y2  6y  7 53. 2x2  4y2  8 54. 36x2  25y2  900 55. x2  5  2y2 56. x  y2  3x2  9 57. 2x2  2y2  x  20 58. 2y2  3  6x2  8 59. 16x  5y  3x  4y  538 2

2

60. 9x2  9y2  9x  12y  4  0

Find the equation of the hyperbola (in standard form) that satisfies the following conditions:

71. vertices at (6, 0) and (6, 0); foci at (8, 0) and (8, 0) 72. vertices at (4, 0) and (4, 0); foci at (6, 0) and (6, 0) 73. foci at 12, 3122 and 12, 3122; length of conjugate axis: 6 units

74. foci at (5, 2) and (7, 2); length of conjugate axis: 8 units

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Use the characteristics of a hyperbola and the graph given to write the related equation and state the location of the foci (75 and 76) or the dimensions of the central rectangle (77 and 78).

75.

y 5 4 3 2 1 54321 1 2 3 4 5



1 2 3 4 5 x

76.

y 10 8 6 4 2 108642 2 4 6 8 10

77.

y 5 4 3 2 1 54321 1 2 3 4 5

1 2 3 4 5 x

78.

y 10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

2 4 6 8 10 x

WORKING WITH FORMULAS 36  4x2 B 9 The “upper half” of a certain hyperbola is given by the equation shown. (a) Simplify the radicand, (b) state the domain of the expression, and (c) enter the expression as Y1 on a graphing calculator and graph. What is the equation for the “lower half” of this hyperbola?

79. Equation of a semi-hyperbola: y 

2b2 a The focal chords of a hyperbola are line segments parallel to the conjugate axis with endpoints on the

80. Focal chord of a hyperbola: L 



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CHAPTER 9 Analytical Geometry and the Conic Sections

y hyperbola, and containing 10 8 certain points f1 and f2 called 6 the foci (see grid). The length 4 f f 2 of the chord is given by the formula shown. Use it to find 1086422 2 4 6 4 the length of the focal chord 6 8 for the hyperbola indicated, 10 then compare the calculated value with the length estimated from the given graph: 1y  12 2 1x  22 2   1. 4 5 1

2

8 10 x

APPLICATIONS

81. Stunt pilots: At an air show, a stunt plane dives along a hyperbolic path whose vertex is directly over the grandstands. If the plane’s flight path can be modeled by the hyperbola 25y2  1600x2  40,000, what is the minimum altitude of the plane as it passes over the stands? Assume x and y are in yards. 82. Flying clubs: To test their skill as pilots, the members of a flight club attempt to drop sandbags on a target placed in an open field, by diving along a hyperbolic path whose vertex is directly over the target area. If the flight path of the plane flown by the club’s president is modeled by 9y2  16x2  14,400, what is the minimum altitude of her plane as it passes over the target? Assume x and y are in feet. 83. Nuclear cooling towers: The natural draft cooling towers for nuclear power stations are called hyperboloids of one sheet. The

perpendicular cross sections of these hyperboloids form two branches of a hyperbola. Suppose the central cross section of one such tower is modeled by the hyperbola 1600x2  4001y  502 2  640,000. What is the minimum distance between the sides of the tower? Assume x and y are in feet. 84. Charged particles: It has been shown that when like particles with a common charge are hurled at each other, they deflect and travel along paths that are hyperbolic. Suppose the path of two such particles is modeled by the hyperbola x2  9y2  36. What is the minimum distance between the particles as they approach each other? Assume x and y are in microns. 85. Locating a ship using radar: Under certain conditions, the properties of a hyperbola can be used to help locate the position of a ship. Suppose two radio stations are located 100 km apart along a straight shoreline. A ship is sailing parallel to the shore and is 60 km out to sea. The ship sends out a distress call that is picked up by the closer station

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Section 9.3 The Hyperbola

in 0.4 milliseconds (msec—one-thousandth of a second), while it takes 0.5 msec to reach the station that is farther away. Radio waves travel at a speed of approximately 300 km/msec. Use this information to find the equation of a hyperbola that will help you find the location of the ship, then find the coordinates of the ship. (Hint: Draw the hyperbola on a coordinate system with the radio stations on the x-axis at the foci, then use the definition of a hyperbola.) 86. Locating a plane using radar: Two radio stations are located 80 km apart along a straight shoreline, when a “mayday” call (a plea for immediate help) is received from a plane that is about to ditch in the ocean (attempt a water landing). The plane was flying at low altitude, parallel to the shoreline, and 20 km out when it ran into trouble. The plane’s distress call is picked up by the closer station in 0.1 msec, while it takes 0.3 msec to reach the other. Use this information to construct the equation of a hyperbola that will help you find the location of the ditched plane, then find the coordinates of the plane. Also see Exercise 85. 87. It is possible for the plane to intersect only the vertex of the cone or to be tangent to the sides. These are called degenerate cases of a conic section. Many times we’re unable to tell if the equation represents a degenerate case until it’s written in standard form. Write the following equations in standard form and comment. a. 4x2  32x  y2  4y  60  0 b. x2  4x  5y2  40y  84  0

865

88. For a greater understanding as to why the branches y2 x2 of a hyperbola are asymptotic, solve 2  2  1 a b for y, then consider what happens as x S q (note that x2  k  x2 for large x). 89. Which has a greater area: (a) The central rectangle of the hyperbola given by 1x  52 2  1y  42 2  57, (b) the circle given by 1x  52 2  1y  42 2  57, or (c) the ellipse given by 91x  52 2  101y  42 2  570? 90. Find the equation of the circle shown, given the equation of the hyperbola: 91x  22 2  251y  32 2  225 y

Focus

Focus x

91. Find the equation of the ellipse shown, given the equation of the hyperbola and (2, 0) is on the graph of the ellipse. The hyperbola and ellipse share the same foci: 91x  22 2  251y  32 2  225 y

Focus

Focus x

92. Verify that for a central hyperbola, a circle that circumscribes the central rectangle must also go through both foci. 

MAINTAINING YOUR SKILLS

93. (7.3) In weight-lifting competitions, Ursula Unger has shown she can lift up to 350 lb. Use a vector analysis to determine whether she will be able to pull the crate up the frictionless ramp shown.

700

lb

25

94. (5.1) The wheels on a motorized scooter are rotating at 403 rpm. If the wheels have a 2.5 in. radius, how fast is the scooter traveling in miles per hour?

95. (1.4) The number z  1  i 12 is a solution to two out of the three equations given. Which two? a. x4  4  0 b. x3  6x2  11x  12  0 c. x2  2x  3  0 96. (5.4) A government-approved company is licensed to haul toxic waste. Each container of solid waste weighs 800 lb and has a volume of 100 ft3. Each container of liquid waste weighs 1000 lb and is 60 ft3 in volume. The revenue from hauling solid waste is $300 per container, while the revenue from liquid waste is $375 per container. The truck used by this company has a weight capacity of 39.8 tons and a volume capacity of 6960 ft3. What combination of solid and liquid weight containers will produce the maximum revenue?

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Precalculus—

9.4 The Analytic Parabola Learning Objectives In Section 9.4 you will learn how to:

A. Graph parabolas with a horizontal axis of symmetry

B. Identify and use the focus-directrix form of the equation of a parabola

C. Solve applications of

Figure 9.27 In previous coursework, you likely learned that the graph of a quadratic function was a parabola. Parabo- Parabola las are actually the fourth and final member of the Axis family of conic sections, and as we saw in Section 9.1, the graph can be obtained by observing the interGenerator section of a plane and a cone. If the plane is parallel to the generator of the cone (shown as a dark line in Figure 9.27), the intersection of the plane with one nappe forms a parabola. In this section we develop the general equation of a parabola from its analytic definition, opening a new realm of applications that extends far beyond those involving only zeroes and extreme values.

the analytic parabola

A. Parabolas with a Horizontal Axis An introductory study of parabolas generally involves those with a vertical axis, defined by the equation y  ax2  bx  c. Unlike the previous conic sections, this equation has only one seconddegree (squared) term in x and defines a function. As a review of our work in Section 3.1, the primary characteristics are listed here and illustrated in Figure 9.28.

Figure 9.28 1. Opens upward

y

4. Axis of symmetry

2. y-intercept

3. x-intercepts

x

5. Vertex

Vertical Parabolas For a second-degree equation of the form y  ax2  bx  c, the graph is a vertical parabola with these characteristics: 1. opens upward if a 7 0; 2. y-intercept: (0, c) downward if a 6 0. (substitute 0 for x) b 3. x-intercept(s): substitute 4. axis of symmetry: x  ; 2a 0 for y and solve. 5. vertex: a

b , yb 2a

See Exercises 7 through 12 for additional review and practice.

Horizontal Parabolas Similar to our study of horizontal and vertical hyperbolas, the graph of a parabola can open to the right or left, as well as up or down. After interchanging the variables x and y in the standard equation, we obtain the parabola x  ay2  by  c, noting the resulting graph will be a reflection about the line y  x. Here, the axis of symmetry is a horizontal line and factoring or the quadratic formula is used to find the y-intercepts (if they exist). Note that although the graph is still a parabola—it is not the graph of a function.

866

9-36

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Section 9.4 The Analytic Parabola

Horizontal Parabolas For a second-degree equation of the form x  ay2  by  c, the graph is a horizontal parabola with these characteristics: 1. opens right if a 7 0, 2. x-intercept: (c, 0) left if a 6 0. (substitute 0 for y) b 3. y-intercept(s): substitute 4. axis of symmetry: y  , 2a 0 for x and solve. 5. vertex: ax, EXAMPLE 1



b b 2a

Graphing a Horizontal Parabola Graph the relation whose equation is x  y2  3y  4, then state the domain and range of the relation.

Solution



Since the equation has a single squared term in y, the graph will be a horizontal parabola. With a 7 0 1a  12, the parabola opens to the right. The x-intercept is 14, 02. Factoring shows the y-intercepts are y  4 and y  1. The axis of symmetry is y  3 2  1.5, and substituting this value into the original equation gives x  6.25. The coordinates of the vertex are 16.25, 1.52. Using horizontal and vertical boundary lines we find the domain for this relation is x  3 6.25, q 2 and the range is y  1q, q 2. The graph is shown.

y 10

(0, 1)

(4, 0) 10

10

(6.25, 1.5)

x

y  1.5 (0, 4) 10

Now try Exercises 13 through 18



As with the vertical parabola, the equation of a horizontal parabola can be written as a transformation: x  a1y  k2 2  h by completing the square. Note that in this case, the vertical shift is k units opposite the sign, with a horizontal shift of h units in the same direction as the sign. EXAMPLE 2



Graphing a Horizontal Parabola by Completing the Square Graph by completing the square: x  2y2  8y  9.

Solution



Using the original equation, we note the graph will be a horizontal parabola opening to the left 1a  22 and have an x-intercept of 19, 02. Completing the square gives x  21y2  4y  42  9  8, so x  21y  22 2  1. The vertex is at 11, 22 and y  2 is the axis of symmetry. This means there are no y-intercepts, a fact that comes to light when we attempt to solve the equation after substituting 0 for x: 21y  22 2  1  0

y 10

(9, 0)

(0, 1)

10

y  2

x

10

(1, 2)

(9, 4)

substitute 0 for x

1y  22 2  

A. You’ve just learned how to graph parabolas with a horizontal axis of symmetry

1 isolate squared term 2 The equation has no real roots. Using symmetry, the point 19, 42 is also on the graph. After plotting these points we obtain the graph shown. Now try Exercises 19 through 36



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B. The Focus-Directrix Form of the Equation of a Parabola As with the ellipse and hyperbola, many significant applications of the parabola rely on its analytical definition rather than its algebraic form. From the construction of radio telescopes to the manufacture of flashlights, the location of the focus of a parabola is critical. To understand these and other applications, we use the analytic definition of a parabola first introduced in Section 9.1. Definition of a Parabola Given a fixed point f and fixed line D in the plane, a parabola is the set of all points (x, y) such that the distance from f to (x, y) is equal to the distance from line D to (x, y). The fixed point f is the focus of the parabola, and the fixed line is the directrix.

d1

(x, y)

f d2 Vertex d1  d2

The general equation of a parabola can be obtained by combining this definition with the distance formula. With no loss of generality, we can assume the parabola shown in the definition box is oriented in the plane with the vertex at (0, 0) and the focus at (0, p). As the diagram in Figure 9.29 indicates, this gives the directrix an equation of y  p with all points on D having coordinates of 1x, p2. Using d1  d2, the distance formula yields

WORTHY OF NOTE For the analytic parabola, we use p to designate the focus, since c is so commonly used as the constant term in y  ax2  bx  c.

Figure 9.29 y

d1

(0, p) F

P(x, y) d2

y  p (0, 0) D

(0, p)

D

21x  02 2  1y  p2 2 1x  02 2  1y  p2 2 x2  y2  2py  p2 x2  2py x2

 21x  x2 2  1y  p2 2  1x  x2 2  1y  p2 2  0  y2  2py  p2  2py  4py

from definition square both sides simplify; expand binomials subtract p 2 and y 2 isolate x 2

The resulting equation is called the focus-directrix form of a vertical parabola with vertex at (0, 0). If we had begun by orienting the parabola so it opened to the right, we would have obtained the equation of a horizontal parabola with vertex (0, 0): y2  4px.

x (x, p)

The Equation of a Parabola in Focus-Directrix Form Vertical Parabola x2  4py focus (0, p), directrix: y  p If p 7 0, opens upward. If p 6 0, opens downward.

Horizontal Parabola y2  4px focus at (p, 0), directrix: x  p If p 7 0, opens to the right. If p 6 0, opens to the left.

For a parabola, note there is only one second-degree term. EXAMPLE 3



Locating the Focus and Directrix of a Parabola Find the vertex, focus, and directrix for the parabola defined by x2  12y. Then sketch the graph, including the focus and directrix.

Solution



Since the x-term is squared and no shifts have been applied, the graph will be a vertical parabola with a vertex of (0, 0). Use a direct comparison between the given equation and the focus-directrix form to determine the value of p: x2  12y given equation T x2  4py focus-directrix form

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This shows:

y

4p  12 p  3

10

Since p  3 1p 6 02, the parabola opens downward, with the focus at 10, 32 and directrix y  3. To complete the graph we need a few additional points. Since 36 (62) is divisible by 12, we can use inputs of x  6 and x  6, giving the points 16, 32 and 16, 32. Note the axis of symmetry is x  0. The graph is shown.

x0 y3

D (0, 0) 10

(6, 3)

f

(0, 3)

10 x (6, 3)

10

Now try Exercises 37 through 48

As an alternative to calculating additional points to sketch the graph, we can use what is called the focal chord of the parabola. Similar to the ellipse and hyperbola, the focal chord is a line segment that contains the focus, is parallel to the directrix, and has its endpoints on the graph. Using the definition of a parabola and the diagram in Figure 9.30, we see the horizontal distance from f to (x, y) is 2p. Since d1  d2, a line segment parallel to the directrix from the focus to the graph will also have a length of 2p, and the focal chord of any parabola has a total length of 4p. Note that in Example 3, the points we happened to choose were actually the endpoints of the focal chord. Finally, if the vertex of a vertical parabola is shifted to (h, k), the equation will have the form 1x  h2 2  4p1y  k2. As with the other conic sections, both the horizontal and vertical shifts are “opposite the sign.”

Figure 9.30 y 2p f d1 Vertex y  p

P

(x, y) d2 p x p D

EXAMPLE 4





Locating the Focus and Directrix of a Parabola Find the vertex, focus, and directrix for the parabola whose equation is given, then sketch the graph, including the focus and directrix: x2  6x  12y  15  0.

Solution



Since only the x-term is squared, the graph will be a vertical parabola. To find the concavity, vertex, focus, and directrix, we complete the square in x and use a direct comparison between the shifted form and the focus-directrix form: x2  6x  12y  15  0 x2  6x  ____  12y  15 x2  6x  9  12y  24 1x  32 2  121y  22

given equation complete the square in x add 9 to both sides factor

Notice the parabola has been shifted 3 right and 2 up, so all features of the parabola will likewise be shifted. Since we have 4p  12 (the coefficient of the linear term), we know p  3 1p 6 02 and the parabola opens downward. If the parabola were in standard position, the vertex would be at (0, 0), the focus at 10, 32 and the y directrix a horizontal line at y  3. But since the 10 x3 parabola is shifted 3 right and 2 up, we add 3 to all y5 x-values and 2 to all y-values to locate the features (3, 2) of the shifted parabola. The vertex is at (9, 1) (3, 1) 10  3, 0  22  13, 22. The focus is 10 10 x 10  3, 3  22  13, 12 and the directrix is (3, 1) y  3  2  5. Finally, the horizontal distance from the focus to the graph is 2p  6 units (since 4p  12), giving us the additional points 13, 12 10 and 19, 12. The graph is shown. Now try Exercises 49 through 60



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In many cases, we need to construct the equation of the parabola when only partial information in known, as illustrated in Example 5. EXAMPLE 5



Constructing the Equation of a Parabola Find the equation of the parabola with vertex (4, 4) and focus (4, 1). Then graph the parabola using the equation and focal chord.

Solution



B. You’ve just learned how to identify and use the focus directrix form of the equation of a parabola

As the vertex and focus are on a vertical line, we have a vertical parabola with general equation 1x  h2 2  4p1x  k2 . The distance p from vertex to focus is 3 units, and with the focus below the y vertex, the parabola opens downward y7 so p  3. Using the focal chord, the horizontal distance from (4, 1) to the 5 (4, 4) graph is 2p  2132   6, giving points 12, 12 and (10, 1). The vertex is shifted 4 units right and 4 units up (4, 1) (10, 1) (2, 1) from (0, 0), showing h  4 and k  4, and the equation of the parabola must 5 10 x be 1x  42 2  121y  42 , with 2 x4 directrix y  7. The graph is shown. Now try Exercises 61 through 76



C. Applications of the Analytic Parabola Here is just one of the many ways the analytic definition of a parabola can be applied. There are several others in the exercise set. EXAMPLE 6



Locating the Focus of a Parabolic Receiver The diagram shows the cross section of a radio antenna dish. Engineers have located a point on the cross section that is 0.75 m above and 6 m to the right of the vertex. At what coordinates should the engineers build the focus of the antenna?

Solution



(6, 0.75) (0, 0)

By inspection we see this is a vertical parabola with center at (0, 0). This means its equation must be of the form x2  4py. Because we know (6, 0.75) is a point on this graph, we can substitute (6, 0.75) in this equation and solve for p: x2  4py 162 2  4p10.752 36  3p p  12

C. You’ve just learned how to solve an application of the analytic parabola

Focus

equation for vertical parabola, vertex at (0, 0) substitute 6 for x and 0.75 for y simplify result

With p  12, we see that the focus must be located at (0, 12), or 12 m directly above the vertex. Now try Exercises 79 through 86



Note that in many cases, the focus of a parabolic dish may be taller than the rim of the dish.

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Section 9.4 The Analytic Parabola

9.4 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. The equation x  ay2  by  c is that of a(n) parabola, opening to the if a 7 0 and to the left if . 2. If point P is on the graph of a parabola with directrix D, the distance from P to line D is equal to the distance between P and the of the parabola. 3. Given y2  4px, the focus is at equation of the directrix is 

and the .

4. Given x2  16y, the value of p is the coordinates of the focus are

and .

5. Discuss/explain how to find the vertex, directrix, and focus from the equation 1x  h2 2  4p1y  k2. 6. If a horizontal parabola has a vertex of 12, 32 with a 7 0, what can you say about the y-intercepts? Will the graph always have an x-intercept? Explain.

DEVELOPING YOUR SKILLS

Find the x- and y-intercepts (if they exist) and the vertex of the parabola. Then sketch the graph by using symmetry and a few additional points or completing the square and shifting a parent function. Scale the axes as needed to comfortably fit the graph and state the domain and range.

7. y  x2  2x  3 9. y  2x2  8x  10 11. y  2x  5x  7 2

8. y  x2  6x  5 10. y  3x2  12x  15 12. y  2x  7x  3 2

Find the x- and y-intercepts (if they exist) and the vertex of the graph. Then sketch the graph using symmetry and a few additional points (scale the axes as needed). Finally, state the domain and range of the relation.

13. x  y2  2y  3

14. x  y2  4y  12

15. x  y  6y  7

16. x  y  8y  12

17. x  y  8y  16

18. x  y2  6y  9

2 2

2

Sketch using symmetry and shifts of a basic function. Be sure to find the x- and y-intercepts (if they exist) and the vertex of the graph, then state the domain and range of the relation.

27. x  y2  10y  4

28. x  y2  12y  5

29. x  3  8y  2y2

30. x  2  12y  3y2

33. x  1y  32 2  2

34. x  1y  12 2  4

31. y  1x  22 2  3

35. x  21y  32 2  1

32. y  1x  22 2  4

36. x  21y  32 2  5

Find the vertex, focus, and directrix for the parabolas defined by the equations given, then use this information to sketch a complete graph (illustrate and name these features). For Exercises 43 to 60, also include the focal chord.

37. x2  8y

38. x2  16y

39. x2  24y

40. x2  20y

41. x2  6y

42. x2  18y

43. y2  4x

44. y2  12x

45. y2  18x

46. y2  20x

47. y2  10x

48. y2  14x

49. x2  8x  8y  16  0 50. x2  10x  12y  25  0

19. x  y2  6y

20. x  y2  8y

51. x2  14x  24y  1  0

21. x  y2  4

22. x  y2  9

52. x2  10x  12y  1  0

23. x  y2  2y  1

24. x  y2  4y  4

53. 3x2  24x  12y  12  0

25. x  y2  y  6

26. x  y2  4y  5

54. 2x2  8x  16y  24  0

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55. y2  12y  20x  36  0 56. y  6y  16x  9  0 2

57. y2  6y  4x  1  0 58. y  2y  8x  9  0 2

59. 2y2  20y  8x  2  0 60. 3y2  18y  12x  3  0

For the graphs in Exercises 73–76, only two of the following four features are displayed: vertex, focus, directrix, and endpoints of the focal chord. Find the remaining two features and the equation of the parabola.

73.

For Exercises 61–72, find the equation of the parabola in standard form that satisfies the conditions given:

62. focus: (0, 3) directrix: y  3

63. focus: (4, 0) directrix: x  4

64. focus: (3, 0) directrix: x  3

65. focus: (0, 5) directrix: y  5

66. focus: (5, 0) directrix: x  5

67. vertex: (2, 2) focus: (1, 2)

68. vertex: (4, 1) focus: (1, 1)

69. vertex: (4, 7) focus: (4, 4)

70. vertex: (3, 4) focus: (3, 1)



y (1, 4)

2 2

6 2

4

6

2 4

y5

4

x

(2, 2)

2

(1, 4) 4

x  3

2

4

6

10 x

8

2

75.

76.

y

y

6 2

(4, 0)

4

(4, 2)

4

2

4

2

2

4

6

8

10 x

2

(2, 2) 6

2

2

x

4

2

y  6

WORKING WITH FORMULAS

77. The area of a right parabolic segment: A  23 ab A right parabolic segment is that part of a parabola formed by a line perpendicular to its axis, which cuts the parabola. The area of this segment is given by the formula shown, where y 10 b is the length of the chord 8 cutting the parabola and a is (3, 4) 6 4 the perpendicular distance 2 from the vertex to this chord. 1086422 2 4 6 8 10 x 4 What is the area of the 6 8 parabolic segment shown in 10 the figure?



74.

y 4

61. focus: (0, 2) directrix: y  2

72. focus: (1, 2) directrix: x  5

71. focus: (3, 4) directrix: y  0

78. The arc length of a right parabolic segment: b2 4a  2b2  16a2 1 2b2  16a2  lna b 2 8a b Although a fairly simple B concept, finding the length of the parabolic a arc traversed by a projectile requires a good A C deal of computation. To b find the length of the arc ABC shown, we use the formula given where a is the maximum height attained by the projectile, b is the horizontal distance it traveled, and “ln” represents the natural log function. Suppose a baseball thrown from centerfield reaches a maximum height of 20 ft and traverses an arc length of 340 ft. Will the ball reach the catcher 310 ft away without bouncing?

APPLICATIONS

79. Parabolic car headlights: The cross section of a typical car headlight can be modeled by an equation similar to 25x  16y2, where x and y are in inches and x  3 0, 44 . Use this information to graph the relation for the indicated domain.

80. Parabolic flashlights: The cross section of a typical flashlight reflector can be modeled by an equation similar to 4x  y2, where x and y are in centimeters and x  3 0, 2.25 4. Use this information to graph the relation for the indicated domain.

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Exercise 80

81. Parabolic sound receivers: Sound technicians at professional sports events often use parabolic receivers as they move along the sidelines. If a two-dimensional cross section of the receiver is modeled by the equation y2  54x, and is 36 in. in diameter, how deep is the parabolic receiver? What is the location of the focus? [Hint: Graph the parabola on the coordinate grid (scale the axes).]

telescopes to look for y radio signals from possible intelligent species in outer x space. The radio telescopes are actually parabolic dishes that vary in size from a few feet to hundreds of feet in diameter. If a particular radio telescope is 100 ft in diameter and has a cross section modeled by the equation x2  167y, how deep is the parabolic dish? What is the location of the focus? [Hint: Graph the parabola on the coordinate grid (scale the axes).]

y

82. Parabolic sound receivers: Private investigators will often use a smaller and less expensive parabolic receiver (see Exercise 81) to gather information for their clients. If a two-dimensional cross section of the receiver is modeled by the equation y2  24x, and the receiver is 12 in. in diameter, how deep is the parabolic dish? What is the location of the focus? 83. Parabolic radio wave receivers: The program known as S.E.T.I. (Search for Extra-Terrestrial Intelligence) identifies a group of scientists using radio 

873

x

84. Solar furnace: Another form of technology that uses a parabolic dish is called a solar furnace. In general, the rays of the Sun are reflected by the dish and concentrated at the focus, producing extremely high temperatures. Suppose the dish of one of these parabolic reflectors had a 30-ft diameter and a cross section modeled by the equation x2  50y. How deep is the parabolic dish? What is the location of the focus? 85. The reflector of a large, commercial flashlight has the shape of a parabolic dish, with a diameter of 10 cm and a depth of 5 cm. What equation will the engineers and technicians use for the manufacture of the dish? How far from the vertex (the lowest point of the dish) will the bulb be placed? (Hint: Analyze the information using a coordinate system.) 86. The reflector of an industrial spot light has the shape of a parabolic dish with a diameter of 120 cm. What is the depth of the dish if the correct placement of the bulb is 11.25 cm above the vertex (the lowest point of the dish)? What equation will the engineers and technicians use for the manufacture of the dish? (Hint: Analyze the information using a coordinate system.)

EXTENDING THE CONCEPT

87. In a study of quadratic graphs from the equation y  ax2  bx  c, no mention is made of a parabola’s focus and directrix. Generally, when a  1, the focus of a parabola is very near its vertex. Complete the square of the function y  2x2  8x and write the result in the form 1x  h2 2  4p1y  k2. What is the value of p? What are the coordinates of the vertex? 88. Like the ellipse and hyperbola, the focal chord of a parabola (also called the latus rectum)

Exercise 88 can be used to help y sketch its graph. From our 7 6 y5 earlier work, we know the 5 endpoints of the focal 4 3 (3, 2) chord are 2p units from the 2 1 focus. Write the equation x 642 1 2 4 6 8 10 12 14 12y  15  x2  6x (3, 1) 2 (9, 1) in the form 4p1y  k2  1x  h2 2, and use the endpoints of the focal chord to help graph the parabola.

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y 89. In Exercise 77, a formula was given for the area of a (3, 5) right parabolic segment. The (0, 4) area of an oblique parabolic segment (the line segment cutting the parabola is not (6, 3) perpendicular to the axis) is more complex, as it involves locating the point where a line parallel to this

x

segment is tangent (touches at only one point) to the parabola. The formula is A  43T, where T represents the area of the triangle formed by the endpoints of the segment and this point of tangency. What is the area of the parabolic segment shown (assuming the lines are parallel)? See Section 8.5, Exercises 46 and 47 and Section 8.8, Example 3.

MAINTAINING YOUR SKILLS

90. (3.3/3.4) Use the function f 1x2  x5  2x4  17x3  34x2  18x  36 to comment and give illustrations of the tools available for working with polynomials: (a) synthetic division, (b) rational roots theorem, (c) the remainder and factor theorems, (d) the test for x  1 and x  1, (e) the upper/lower bounds property, (f) Descartes’ rule of signs, and (g) roots of multiplicity (bounces, cuts, alternating intervals). 91. (1.6) Find all roots (real and complex) to the equation x6  64  0. (Hint: Begin by factoring the expression as the difference of two perfect squares.)

92. (6.6) Find all real solutions to sec   1.1547 (round to the nearest degree). 93. (6.5) The graph shown displays the variation in daylight from an average of 12 hours per day (i.e., the maximum is 3 15 hours and the minimum is 9). Use the graph to approximate 0 the number of days in a 60 120 180 240 300 year there are 10.5 or less hours of daylight. 3 Day of year Answers may vary. Hours



9-44

CHAPTER 9 Analytical Geometry and the Conic Sections

360

MID-CHAPTER CHECK c.

Sketch the graph of each conic section. 1. 1x  42  1y  32  9 2

2

2. x2  y2  10x  4y  4  0 3.

1x  22 2 1y  32 2  1 16 1

108642 2 4 6 8 10

4. 9x2  4y2  18x  24y  9  0 5.

1x  32 2 1y  42 2  1 9 4

6. 9x2  4y2  18x  24y  63  0 7. Find the equation of each relation and state its domain and range. a. b. (3, 5)

(5, 1)

y

5 4 3 2 (1, 1) 1

54321 1 2 3 (3, 3)4 5

y 10 8 6 4 (1, 2) 2

1 2 3 4 5 x

108642 2 4 6 8 10

y 10 8 6 4 2

(3, 6) (7, 2) 2 4 6 8 10 x

(3, 2)

2 4 6 8 10 x

(3, 4)

8. Solve the following system of inequalities by graphing. y2 x2  1 25 • 100 x2  1y  42 2  36 9. Find the equation of the ellipse (in standard form) if the vertices are (4, 0) and (4, 0) and the distance between the foci is 4 13 units. 10. The radio signal emanating from a tall radio tower spreads evenly in all directions with a range of 50 mi. If the tower is located at coordinates (20, 30) and my home is at coordinates (10, 78), will I be able to pick up this station on my home radio? Assume coordinates are in miles.

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875

REINFORCING BASIC CONCEPTS Ellipses and Hyperbolas with Rational/Irrational Values of a and b Using the process known as completing the square, we were able to convert from the polynomial form of a conic section to the standard form. However, for some equations, values of a and b are somewhat difficult to identify, since the coefficients are not factors. Consider the equation 20x2  120x  27y2  54y  192  0 the equation of an ellipse. 20x2  120x  27y2  54y  192  0 201x2  6x  ____ 2  271y2  2y  ____ 2  192 201x2  6x  92  271y2  2y  12  192  27  180 201x  32 2  271y  12 2  15 41x  32 2 91y  12 2  1 3 5

original equation subtract 192, begin process complete the square in x and y factor and simplify standard form

Unfortunately, we cannot easily identify the values of a and b, since the coefficients of each binomial square were not “1.” In these cases, we can write the equation in standard form by using a simple property of fractions— the numerator and denominator of any fraction can be divided by the same quantity to obtain an equivalent fraction. Although the result may look odd, it can nevertheless be applied here, giving a result of 1y  12 2 1x  32 2   1. We can now identify a and b by writing these denominators in squared form, 3/4 5/9 1x  32 2 1y  12 2   1. The values of a and b are now easily seen as which gives the following expression: 13 2 15 2 b b a a 2 3 a  0.866 and b  0.745. Use this idea to complete the following exercises. Exercise 1: Identify the values of a and b by writing the equation 100x2  400x  18y2  108y  230  0 in standard form. Exercise 2: Identify the values of a and b by writing the equation 28x2  56x  48y2  192y  195  0 in standard form. Exercise 3: Write the equation in standard form, then identify the values of a and b and use them to graph the ellipse. 251y  12 2 41x  32 2  1 49 36

Exercise 4: Write the equation in standard form, then identify the values of a and b and use them to graph the hyperbola. 91x  32 2 41y  12 2  1 80 81

9.5 Nonlinear Systems of Equations and Inequalities Learning Objectives In Section 9.5 you will learn how to:

A. Visualize possible

Equations where the variables have exponents other than 1 or that are transcendental (like logarithmic and exponential equations) are all nonlinear equations. A nonlinear system of equations has at least one nonlinear equation, and these occur in a great variety.

solutions

B. Solve nonlinear systems using substitution

C. Solve nonlinear systems using elimination

D. Solve nonlinear systems of inequalities

E. Solve applications of nonlinear systems

A. Possible Solutions for a Nonlinear System When solving nonlinear systems, it is often helpful to visualize the graphs of each equation in the system. This can help determine the number of possible intersections and further assist the solution process.

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EXAMPLE 1



Visualizing the Number of Possible Intersections for the Graphs in a System Identify each equation in the system as the equation of a line, parabola, circle, ellipse, or hyperbola. Then determine the number of solutions possible by 4x2  9y2  36 considering the different ways the graphs might intersect: e . 2x  3y  6 Finally, solve the system by graphing.

Solution



The first equation contains a sum of second-degree terms with unequal coefficients, and we recognize this as a central ellipse. The second equation is obviously linear. This means the system may have no solution, one solution, or two solutions, as shown in Figure 9.31. The graph of the system is shown in Figure 9.32 and the two points of intersection appear to be (3, 0) and (0, 2). After checking these in the original equations we find that both are solutions to the system. Figure 9.31

Figure 9.32 y

y No solutions

(0, 2)

One solution (3, 0)

(3, 0) x

x

Two solutions (0, 2)

A. You’ve just learned how to visualize possible solutions

Now try Exercises 7 through 12



B. Solving a Nonlinear System by Substitution Since graphical methods at best offer an estimate for the solution (points of intersection may not have integer values), we more often turn to algebraic methods. Recall the substitution method involves solving one of the equations for a variable or expression that can be substituted in the other equation to eliminate one of the variables. EXAMPLE 2



Solving a Nonlinear System Using Substitution Solve the system using substitution: e

Solution



y  x2  2x  3 . 2x  y  7

The first equation contains a single second-degree term in x, and is the equation of a parabola. The second equation is linear. Since the first equation is already written with y in terms of x, we can substitute x2  2x  3 for y in the second equation to solve. 2x  y  7 2x  1x2  2x  32  7 2x  x2  2x  3  7 x2  4x  3  7 x2  4x  4  0 1x  22 2  0

We find that x  2 is a repeated root.

second equation substitute x 2  2x  3 for y distribute simplify set equal to zero factor

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Since the second equation is simpler than the first, we substitute 2 for x in this equation and find y  3. The system has only one (repeated) solution at 12, 32, which is shown in the figure.

WORTHY OF NOTE Similar to our earlier work with repeated roots of polynomials, note the graph of y  x2  2x  3 “bounces” off (is tangent to) the line y  2x  7.

y 10

2x  y  7

y  x2  2x  3 10

10

x

(2, 3)

B. You’ve just learned how to solve nonlinear systems using substitution

10

Now try Exercises 13 through 18



C. Solving Nonlinear Systems by Elimination When both equations in the system have second-degree terms with like variables, it is generally easier to use the elimination method, rather than substitution. As in Chapter 8, watch for systems that have no solutions. EXAMPLE 3



Solving a Nonlinear System Using Elimination Solve the system using elimination: e

Solution



2y2  5x2  13 . 3x2  4y2  39

The first equation contains a difference of second-degree terms and is the equation of a central hyperbola. The second has a sum of second-degree terms with unequal coefficients, and represents a central ellipse. By mentally visualizing the possibilities, there could be zero, one, two, three, or four points of intersection (see Example 1). However, with both centered at (0, 0), we find there can only be zero, two, or four solutions. After writing the system so that x- and y-terms are in the same order, we find that multiplying the first equation by 2 will help eliminate the variable y: e

10x2  4y2  26 3x2  4y2  39

13x  0  13 x2  1 x  1 or x  1 2

rewrite first equation and multiply by 2 original second equation add divide by 13 square root property

Substituting x  1 and x  1 into the second equation we obtain: 3112 2  4y2  39 3  4y2  39 4y2  36 y2  9 y  3

3112 2  4y2  39 3  4y2  39 4y2  36 y2  9 y  3

Since 1 and 1 each generated two outputs, there are a total of four ordered pair solutions: 11, 32, (1, 3), 11, 32, and 11, 32. Once again, the graph shown supports these calculations.

y (1, 3)

5x2  2y2  13 (1, 3)

x 3x2  4y2  39

(1, 3)

(1, 3)

Now try Exercises 19 through 24



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Nonlinear systems may involve other relations as well, including power, polynomial, logarithmic, or exponential functions. These are solved using the same methods. EXAMPLE 4



Solving a System of Logarithmic Equations Solve the system using the method of your choice: e

Solution



WORTHY OF NOTE

5

10

Since both equations have y written in terms of x, substitution appears to be the better choice. The result is a logarithmic equation, which we can solve using the techniques from Chapter 4. log1x  42  1  log1x  72  2 log1x  42  log1x  72  1 log1x  421x  72  1 1x  421x  72  101 x2  11x  18  0 1x  921x  22  0 x  9  0 or x  2  0 x  9 or x  2

Since the domain of the functions are x 7 7 and x 7 4 respectively, x  9 cannot be a solution, as illustrated in the graphical representation of the system.

10

y  log1x  72  2 . y  log1x  42  1

substitute log1x  42  1 for y in first equation add log1x  72; subtract 1 product property of logarithms exponential form eliminate parentheses and set equal to zero factor zero factor theorem possible solutions

By inspection, we see that x  9 is not a solution, since log19  42 and log19  72 are not real numbers. Substituting 2 for x in the second equation we find one form of the (exact) solution is 12, log 2  12. If we substitute 2 for x in the first equation the exact solution is 12, log 5  22. Use a calculator to verify the answers are equivalent and approximately 12, 1.32.

5

C. You’ve just learned how to solve nonlinear systems using elimination

Now try Exercises 25 through 36



For practice solving more complex systems using a graphing calculator, see Exercises 37 through 42.

D. Solving Systems of Nonlinear Inequalities As with our previous work with inequalities in two variables, nonlinear inequalities can be solved by graphing the boundary given by the related equation, and checking the regions that result using a test point. For example, Figure 9.33 the inequality x2  4y2 6 25 is solved by graphing y 2 2 x  4y  25 [a central ellipse with vertices at 15, 02 and (5, 0)], deciding if the boundary is x2  4y2  25 included or excluded (in this case it is not), and using a test point from either the “outside” or “inside” region formed. The test point (0, 0) results in a true (5, 0) (5, 0) x statement since 102 2  4102 2 6 25, so the inside of the ellipse is shaded to indicate the solution region (Figure 9.33). For a system of nonlinear inequalities, we identify regions where the solution set for each inequality overlap, paying special attention to points of intersection. EXAMPLE 5



Solving a System of Nonlinear Inequalities Solve the system: e

x2  4y2 6 25 . x  4y  5

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Solution

y

We recognize the first inequality from Figure 9.33, an ellipse with vertices at 15, 02 and (5, 0), and a solution region in the interior. The second inequality is linear and after solving for x in the related equation we use a substitution to find points of intersection (if they exist). For x  4y
5, we have x
4y  5 and substitute 4y  5 for x in x2  4y2
25: x2  4y2
25 14y  52 2  4y2
25 20y2  40y  25
25 y2  2y
0 y1y  22
0 y
0 or y
2

x  4y
5 (3, 2) (5, 0)

x

x2  4y2
25

D. You’ve just learned how to solve nonlinear systems of inequalities

879

given substitute 4y  5 for x expand and simplify subtract 25; divide by 20 factor result

Back-substitution shows the graphs intersect at 15, 02 and (3, 2). Graphing a line through these points and using (0, 0) as a test point shows the upper half plane is the solution region for the linear inequality 3102  4102  5 is false]. The overlapping (solution) region for both inequalities is the elliptical sector shown. Note the points of intersection are graphed using “open dots,” (see figure) since points on the graph of the ellipse are excluded from the solution set. Now try Exercises 43 through 50




E. Solving Applications of Nonlinear Systems In the business world, a fast growing company can often reduce the average price of its products using what are called the economies of scale. These would include the ability to buy necessary materials in larger quantities, integrating new technology into the production process, and other means. However, there are also countering forces called the diseconomies of scale, which may include the need to hire additional employees, rent more production space, and the like. EXAMPLE 6




Solving an Application of Nonlinear Systems—Economies of Scale Suppose the cost to produce a new and inexpensive shoe made from molded plastic is modeled by the function C1x2
x2  5x  18, where C(x) represents the cost to produce x thousand of these shoes. The revenue from the sales of these shoes is modeled by R1x2
x2  10x  4. Use a break-even analysis to find the quantity of sales that will cause the company to break even.

Solution




Essentially we are asked to solve the system formed by the two equations: C1x2
x2  5x  18 . Since we want to know the point where the company e R1x2
x2  10x  4 breaks even, we set C1x2
R1x2 and solve. C1x2
R1x2 x  5x  18
x2  10x  4 2 2x  15x  22
0 12x  112 1x  22
0 11 or x
2 x
2 2

E. You’ve just learned how to solve applications of nonlinear systems

break even: cost
revenue substitute C 1x2 and R 1x2 for set equal to zero factored form result

With x in thousands, it appears the company will break even if either 2000 shoes or 5500 shoes are made and sold. Now try Exercises 53 through 56




For additional applications of nonlinear systems, see Exercises 57 through 62.

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CHAPTER 9 Analytical Geometry and the Conic Sections

9.5 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. Draw sketches showing the different ways each pair of relations can intersect and give one, two, three, and/or four points of intersection. If a given number of intersections is not possible, so state. a. circle and line b. parabola and line c. circle and parabola d. circle and hyperbola e. hyperbola and ellipse f. circle and ellipse

4. When both equations in the system have at least one -degree term, it is generally easier to use the method to find a solution.

2. By inspection only, identify the systems having no solutions and justify your choices. y2  x2  16 y  x2  4 a. e 2 b. e 2 2 x y 9 x  4y2  4 yx1 c. e 2 3x  4y2  12

6. Solve the system twice, once using elimination, then again using substitution. Compare/Contrast each process and comment on which is more 4x2  y2  25 . efficient in this case: e 2 2x  y  5

5. Suppose a nonlinear system contained a central hyperbola and an exponential function. Are three solutions possible? Are four solutions possible? Explain/Discuss.

3. The solution to a system of nonlinear inequalities is a(n) of the plane where the for each individual inequality overlap. 

DEVELOPING YOUR SKILLS

Identify each equation in the system as that of a line, parabola, circle, ellipse, or hyperbola, and solve the system by graphing.

8. e

2x  y  4 4x2  y2  16

y  x2  1 4x2  y2  100

10. e

x2  y2  25 x2  y  13

x2  y2  9 x2  y2  41

12. e

4x2  y2  36 y2  9x2  289

7. e

x2  y  6 xy4

9. e 11. e

Solve using substitution. [Hint: Substitute for x2 (not y) in Exercises 17 and 18.]

13. e

x2  y2  25 yx1

14. e

x  7y  50 x2  y2  100

15. e

x2  4y2  25 x  2y  7

16. e

x2  2y2  8 xy6

17. e

x2  y  13 9x2  y2  81

18. e

y  x2  10 4x2  y2  100

Solve using elimination.

19. e

x2  y2  25 2x2  3y2  5

20. e

y2  x2  12 x2  y2  20

21. e

x2  y  4 x2  y2  16

22. e

4x2  y2  13 x2  y2  1

23. e

5x2  2y2  75 2x2  3y2  125

24. e

3x2  7y2  20 4x2  9y2  45

Solve using the method of your choice. Answer in exact form.

25. e

y  5  log x y  6  log1x  32

26. e

y  log1x  42  1 y  2  log1x  72 log1x  1.12  y  3 y  4  log1x2 2

27. e

y  ln1x2 2  1 y  1  ln1x  122

28. e

29. e

y  9  e2x 3  y  7ex

30. e

y  2e2x  5 y  1  6ex

31. e

y  4x3 2 y  2x 3x  0

32. e

y  3x 2x  0 y  9x2 2

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33. e

x3  y  2x y  5x  6

34. e

y  x3  2 y  4  3x

35. e

x2  6x  y  4 y  2x  8

36. e

y  x  2 y  4x  x2

Solve each system using a graphing calculator. Round solutions to hundredths (as needed).

37. e 39. e

x2  y2  34 y2  1x  32 2  25

y  2x  3 y  2x2  9

1 2 1x  32 2 41. • 1x  32 2  y2  10

38. e

5x2  5y2  40 y  2x  x2  6

40. e

y  2 log1x  82 y  x3  4x  2

y



42.

Solve each system of inequalities

43. e

y  x2  1 xy3

44. e

x2  y2  25 x  2y  5

45. e

x2  y2 7 16 x2  y2  64

46. e

y  4  x2 x2  y2  34

47. e

y  x2  16 y2  x2 6 9

48. e

x2  y2  16 x  2y 7 10

49. e

y2  x2  25 x  1 7 y

50. e

y2  x2  4 xy 6 4

y2  x2  5 • 1 y 2 x1

WORKING WITH FORMULAS d 2 1a b a B

51. Tunnel clearance: h  b

The maximum rectangular clearance allowed by an elliptical tunnel can be found using the formula y2 x2 shown, where e 2  2  1 models the tunnel’s a b elliptical cross section and h is the height of the



881

Section 9.5 Nonlinear Systems of Equations and Inequalities

tunnel at a distance d from the center. If a  50 and b  30, find the maximum clearance at distances of d  20, 30, and 40 ft from center. 52. Manufacturing cylindrical vents: e

A  2rh V  r2h

In the manufacture of cylindrical vents, a rectangular piece of sheet metal is rolled, riveted, and sealed to form the vent. The radius and height required to form a vent with a specified volume, using a piece of sheet metal with a given area, can be found by solving the system shown. Use the system to find the radius and height if the volume required is 4071 cm3 and the area of the rectangular piece is 2714 cm2.

APPLICATIONS

Solve the following applications involving economies of scale.

53. Revenue from sales: Early in 2008, The Tata Company (India) unveiled the Tata Nano, the world’s most inexpensive car. With its low price and 54 miles per gallon, the car may prove to be very popular. Assume the cost to produce these cars is modeled by the function C1x2  2.5x2  120x  3500, where C(x) represents the cost to produce x thousand cars. Suppose the revenue from the sales of these cars is

modeled by R1x2  2x2  180x  500. Use a break-even analysis to find the quantity of sales (to the nearest hundred) that will cause the company to break even. 54. Document reproduction: In a world of technology, document reproduction has become a billion dollar business. With very stiff competition, the price of a single black and white copy has varied greatly is recent years. Suppose the cost to produce these copies is modeled by the function C1x2  0.1x2  1.2x  7, where C(x) represents

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the cost to produce x hundred thousand copies. If the revenue from the sales of these copies is modeled by R1x2  0.1x2  1.8x  2, use a break-even analysis to find the quantity of copies sold (to the nearest thousand) that will cause the copy company to break even. Market equilibrium: In a free-enterprise (supply and demand) economy, the amount buyers are willing to pay for an item and the number of these items manufacturers are willing to produce depend on the price of the item. As the price increases, demand for the item decreases since buyers are less willing to pay the higher price. On the other hand, an increase in price increases the supply of the item since manufacturers are now more willing to supply it. When the supply and demand curves are graphed, their point of intersection is called the market equilibrium for the item. 55. Suppose the monthly market demand D (in tenthousands of gallons) for a new synthetic oil is related to the price P in dollars by the equation 10P2  6D  144. For the market price P, assume the amount D that manufacturers are willing to supply is modeled by 8P2  8P  4D  12. (a) What is the minimum price at which manufacturers are willing to begin supplying the oil? (b) Use this information to create a system of nonlinear equations, then solve the system to find the market equilibrium price (per gallon) and the quantity of oil supplied and sold at this price. 56. The weekly demand D for organically grown carrots (in thousands of pounds) is related to the price per pound P by the equation 8P2  4D  84. At this market price, the amount that growers are willing to supply is modeled by the equation 8P2  6P  2D  48. (a) What is the minimum price at which growers are willing to supply the organically grown carrots? (b) Use this information 

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CHAPTER 9 Analytical Geometry and the Conic Sections

to create a system of nonlinear equations, then solve the system to find the market equilibrium price (per pound) and the quantity of carrots supplied and sold at this price. Solve by setting up and solving a system of nonlinear equations.

57. Dimensions of a flag: A large American flag has an area of 85 m2 and a perimeter of 37 m. Find the dimensions of the flag. 58. Dimensions of a sail: The sail on a boat is a right triangle with a perimeter of 36 ft and a hypotenuse of 15 ft. Find the height and width of the sail.

Exercise 58

59. Dimensions of a tract: The area of a rectangular tract of land is 45 km2. The length of a diagonal is 1106 km. Find the dimensions of the tract. 60. Dimensions of a deck: A rectangular deck has an area of 192 ft2 and the length of the diagonal is 20 ft. Find the dimensions of the deck. 61. Dimensions of a trailer: The surface area of a rectangular trailer with square ends is 928 ft2. If the sum of all edges of the trailer is 164 ft, find its dimensions. 62. Dimensions of cylindrical tank: The surface area of a closed cylindrical tank is 192 m2. Find the dimensions of the tank if the volume is 320 m3 and the radius is as small as possible.

EXTENDING THE CONCEPT

63. The area of a vertical parabolic segment is given by A  23BH, where B is the length of the H horizontal base of the segment and H is the height from the base to the vertex. Investigate how this B formula can be used to find the area of the solution region for the general system y  x2  bx  c of inequalities shown. e y  c  bx  x2 (Hint: Begin by investigating with b  6 and c  8, then use other values and try to generalize what you find.)

64. For what values of r will the volume of a sphere be numerically equal to its surface area? For what values of r will the volume of a cylinder be numerically equal to its lateral surface area? Can a similar relationship be found for the volume and lateral surface area of a cone? Why or why not? 65. A rectangular fish tank has a bottom and four sides made out of glass. Use a system of equations to help find the dimensions of the tank if the height is 18 in., surface area is 4806 in2, the tank must hold 108 gal 11 gal  231 in3 2 , and all three dimensions are integers.

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883

MAINTAINING YOUR SKILLS

66. (7.4) Determine the angle  between the vectors u  12i  35j and v  20i  21j. 67. (7.1) Estimate the length L to the base of the water tank, if the angle of elevation to the base of the tower is 10°, the angle of elevation to the base of the tank is 35°, and the tower makes a 112° angle with the hillside. W L 112

100 ft 10

68. (2.6/4.2) Sketch each transformation: 3 a. y  2x  3  1 b. y  1 x23 2 x1 3 c. y  1x  42  3 d. y  2 69. (2.3) In 2001, a small business purchased a copier for $4500. In 2004, the value of the copier had decreased to $3300. Assuming the depreciation is ¢value linear: (a) find the rate-of-change m  and ¢time discuss its meaning in this context; (b) find the depreciation equation; and (c) use the equation to predict the copier’s value in 2008. (d) If the copier is traded in for a new model when its value is less than $700, how long will the company use this copier?

9.6 Polar Coordinates, Equations, and Graphs Learning Objectives In Section 9.6 you will learn how to:

A. Plot points given in polar form

B. Convert from rectangular form to polar form

C. Convert from polar form to rectangular form

D. Sketch basic polar graphs using an r-value analysis

E. Use symmetry and families of curves to write a polar equation given a polar graph or information about the graph

Figure 9.34 N

4 mi 2 mi 60 30 Coast Guard (pole)

23 mi

shoreline

One of the most enduring goals of mathematics is to express relations with the greatest tan   cot   sin  cos, we would possible simplicity and ease of use. For tan2  cot2 definitely prefer working with sin  cos , although the expressions are equivalent. Similarly, we would prefer computing 13  13i2 6 in trigonometric form rather than algebraic form—and would quickly find the result is 1728. In just this way, many equations and graphs are easier to work with in polar form rather than rectangular form. In rectangular form, a circle of radius 2 centered at (0, 2) has the equation x2  1y  22 2  4. In polar form, the equation of the same circle is simply r  4 sin . As you’ll see, polar coordinates offer an alternative method for plotting points and graphing relations.

A. Plotting Points Using Polar Coordinates Suppose a Coast Guard station receives a distress call from a stranded boat. The boater could attempt to give the location in rectangular form, but this might require imposing an arbitrary coordinate grid on an uneven shoreline, using uncertain points of reference. However, if the radio message said, “We’re stranded 4 miles out, bearing 60°,” the Coast Guard could immediately locate the boat and send help. In polar coordinates, “4 miles out, bearing 60°” would simply be written 1r, 2  14, 30°2, with r representing the distance from the station and  7 0 measured from a horizontal axis in the counterclockwise direction as before (see Figure 9.34). If we placed the scenario on a rectangular grid (assuming a straight shoreline), the coordinates of the boat would be 1213, 22 using basic trigonometry. As you see, the polar coordinate system uses angles and distances to locate a point in the plane. In this example, the Coast Guard station would be considered the pole or origin, with the x-axis as the polar axis or axis of reference (Figure 9.35). A distinctive feature of polar coordinates is that we allow r to be negative, in which case P1r, 2 is the point r units from the pole in a direction opposite 1180°2 to that of  (Figure 9.36). For convenience, polar graph paper is often used when working with polar coordinates. It consists of a series of concentric circles that share the same center and have integer radii. The standard angles are marked off   15° depending on whether you’re working in radians or degrees in multiples of 12

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(Figure 9.37). To plot the point P1r, 2, go a distance of r at 0° then move ° counterclockwise along a circle of radius r. If r 7 0, plot a point at that location (you’re finished). If r 6 0, the point is plotted on a circle of the same radius, but 180° in the opposite direction. Figure 9.35

Figure 9.37

Figure 9.36 P(r, ␪) r0

120 135 150

r r Pole

␪ Pole

␪

105

165

Polar axis Polar axis

5 4 3 2 1

75

45 30 15

195

P(r, ␪) r0

60

345 330 315 300

210 225 240

255 285 Polar graph paper

EXAMPLE 1



Plotting Points in Polar Coordinates Plot each point P1r, 2 given A14, 45°2; B15, 135°2; C 13, 30°2; Da2, Ea5,

Solution



  b; and Fa3,  b. 3 6

2 b; 3

For A14, 45°2 go 4 units at 0°, then rotate 45° counterclockwise and plot point A. For B15, 135°2, move 5  5 units at 0°, rotate 135°, then actually plot point B 180° in the opposite direction, as shown. Point C13, 30°2 is plotted by moving 3  3 units at 0°, rotating 30°, then plotting point C 180° in the opposite 2  b, Ea5, b, and direction (since r 6 0). See Figure 9.38. The points Da2, 3 3  F a3,  b are plotted on the grid in Figure 9.39. 6 Figure 9.38 120 135 150 C (3, 30)

210 225 240

5 4 3 2 1

Figure 9.39 60 3␲ 4

45 A

30

(4, 45)

(5, 135)

2␲ 3

5␲ 6

(2, ) 2␲ 3

B

330 315 300

␲ 3

5 4 3 D 2 1

␲ 4 ␲ 6

F 3, ␲6

(

7␲ 6

E 5␲ 4

4␲ 3

)

␲ 3

(5, ) 5␲ 3

11␲ 6 7␲ 4

Now try Exercises 7 through 22



 While plotting the points B15, 135°2 and F a3,  b, you likely noticed that the 6 coordinates of a point in polar coordinates are not unique. For B15, 135°2 it appears  more natural to name the location 15, 315°2; while for F a3,  b, the expression 6

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a3,

11 b is just as reasonable. In fact, for any point P1r, 2 in polar coordinates, 6 P1r,  22 and P1r,  2 name the same location. See Exercises 23 through 36.

A. You’ve just learned how to plot points given in polar form

B. Converting from Rectangular Coordinates to Polar Coordinates Figure 9.40 Conversions between rectangular and polar coordiy nates is a simple application of skills from previous sections, and closely resembles the conversion from P(x, y) the rectangular form to the trigonometric form of a r ␪ y complex number. To make the connection, we first assume r 7 0 with  in Quadrant II (see Figure ␪r x 9.40). In rectangular form, the coordinates of the x point are simply (x, y), with the lengths of x and y forming the sides of a right triangle. The distance r from the origin to point P resembles the modulus of a complex number and is computed in the same way: y r  2x2  y2. As long as x 0, we have r  tan1a b, noting r is a reference x angle if the terminal side is not in Quadrant I. If needed, refer to Section 5.3 for a review of reference arcs and reference angles.

Converting from Rectangular to Polar Coordinates y

Any point P1x, y2 in rectangular coordinates can be represented as P1r, 2 in polar coordinates, y where r  2x2  y2 and r  tan1a b, x 0. x

P(x, y) r

y

␪ ␪r

x

x

EXAMPLE 2



Converting a Point from Rectangular Form to Polar Form Convert from rectangular to polar form, with r 7 0 and 0    360° (round values to one decimal place as needed). a. P15, 122 b. P13 12, 3122

Solution



a. Point P15, 122 is in Quadrant II.   tan1a

12 b 5  1169 r  67.4°  13   112.6° P15, 122 S P113, 112.6°2

r  2152 2  122

b. Point P13 12, 3122 is in Quadrant IV. r  313222 2  13222 2 B. You’ve just learned how to convert from rectangular form to polar form

 136 6

  tan1a

312 b 312

r  45°   315° P13 12, 3122 S P16, 315°2 Now try Exercises 37 through 44



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C. Converting from Polar Coordinates to Rectangular Coordinates The conversion from polar form to rectangular form is likewise straightforward. From Figure 9.41 y x we again note cos   and sin   , giving r r x  r cos  and y  r sin . The conversion simply consists of making these substitutions and simplifying.

Figure 9.41 y

P(x, y) r

y

␪ ␪r

x

x

Converting from Polar to Rectangular Coordinates Any point P1r, 2 in polar coordinates can be represented as P(x, y) in rectangular coordinates, where x  r cos  and y  r sin .

P(r cos ␪, r sin ␪) r

y y ␪

␪r

x

x

EXAMPLE 3



Converting a Point from Polar Form to Rectangular Form Convert from polar to rectangular form (round values to one decimal place as needed). 5 a. Pa12, b. P16, 240°2 b 3

Solution



C. You’ve just learned how to convert from polar form to rectangular form

5 b is in Quadrant IV. 3 x  r cos  y  r sin  5 5  12 cosa b  12 sina b 3 3 1 13  12a b b  12a 2 2 6  6 13 5 b S P16, 6132  P16, 10.42 Pa12, 3 b. Point P16, 240°2 is in Quadrant III. a. Point Pa12,

x  6 cos 240° y  6 sin 240° 1  23  6a b b  6a 2 2  3  3 13 P16, 240°2 S P13, 3132  P13, 5.22 Now try Exercises 45 through 52



Using the relationships x  r cos , y  r sin , and x2  y2  1, we can actually convert an equation given in polar form, to the equivalent equation in rectangular form. See Exercises 105 and 106.

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D. Basic Polar Graphs and r-Value Analysis To really understand polar graphs, an intuitive sense Figure 9.42 of how they’re developed is needed. Polar equations 5␲ 7␲ 2␲ 12 12 ␲ are generally stated in terms of r and trigonometric 5 3 3 ␲ 3␲ 4 4 4 functions of , with  being the input value and r ␲ 5␲ 3 being the output value. First, it helps to view the 6 6 ␲ 2 length r as the long second hand of a clock, but 11␲ 12 12 1 extending an equal distance in both directions from center (Figure 9.42). This “second hand” ticks 13␲ 23␲ around the face of the clock in the counterclockwise 12 12 7␲ direction, with the angular measure of each tick 11␲ 6 6 5␲  7␲ 4 radians  15°. As each angle “ticks by,” being 4 4␲ 5␲ 12 3 17␲ 19␲ 3 12 12 we locate a point somewhere along the radius, depending on whether r is positive or negative, and plot it on the face of the clock before going on to the next tick. For the purposes of this study, we will allow that all polar graphs are continuous and smooth curves, without presenting a formal proof.

EXAMPLE 4



Graphing Basic Polar Equations Graph the polar equations. a. r  4

Solution



b.  

 4

a. For r  4, we’re plotting all points of the form 14, 2 where r has a constant value and  varies. As the second hand “ticks around the polar grid,” we plot all points a distance of 4 units from the pole. As you might imagine, the graph is 5␲ 7␲ 2␲ 12 12 ␲ a circle with radius 4. r, 4 5 3 3 ␲ 3␲ 4 4 4  ␲ 5␲ b. For   , all points have the form 3 6 6 4 ␲ 2 11␲ 12 12   1 ar, b with constant and r varying. In 4 4 13␲ 23␲ this case, the “second hand” is frozen at 12 12 7␲  11␲ etc. 6 , and we plot any selection of r-values, 6 5␲ 4 7␲ 4 4 4␲ 5␲ producing the straight line shown in the 3 17␲ 19␲ 3 12 12 figure.

( )

Now try Exercises 53 through 56



To develop an “intuitive sense” that allows for the efficient graphing of more sophisticated equations, we use a technique called r-value analysis. This technique basically takes advantage of the predictable patterns in r  sin  and r  cos  taken from their graphs, including the zeros and maximum/minimum values. We begin with the r-value analysis for r  sin , using the graph shown in Figure 9.43. Note the analysis occurs in the four colored parts corresponding to Quadrants I, II, III, and IV, and that the maximum value of sin   1.

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Figure 9.43 r 1

(1)

(2) r  sin ␪ ␲

␲ 2

2␲

3␲ 2

(3)

␪

(4)

1

 , sin  is positive and sin  increases from 0 to 1. 2 1 for r  sin , r is increasing  2. As  moves from to , sin  is positive and sin  decreases from 1 to 0. 2 1 for r  sin , r is decreasing 3 3. As  moves from  to , sin  is negative and sin  increases from 0 to 1. 2 1 for r  sin , r is increasing 3 4. As  moves from to 2, sin  is negative and sin  decreases from 1 to 0. 2 1 for r  sin , r is decreasing 1. As  moves from 0 to

WORTHY OF NOTE It is important to remember that if r 6 0, the related point on the graph is r  units from center, 180° in the opposite direction: 1r, 2 S 1r,   180°2. In addition, students are encouraged not to use a table of values, a conversion to rectangular coordinates, or a graphing calculator until after the r-value analysis.

EXAMPLE 5

In summary, note that the value of r goes through four cycles, two where it is increasing from 0 to 1 (in red), and two where it is decreasing from 1 to 0 (in blue). 

Graphing Polar Equations Using an r-Value Analysis Sketch the graph of r  4 sin  using an r-value analysis.

Solution



Figure 9.44 3␲ 4

2␲ 3

5 4 3 2 1

5␲ 6

␲ 3

␲ 4 ␲ 6

(1)

7␲ 6

11␲ 6

5␲ 4 4␲ 3

5␲ 3

7␲ 4

Figure 9.45 3␲ 4

2␲ 3

5␲ 6

(2) and (4)

5 4 3 2 1

␲ 3

␲ 4 ␲ 6

(1) and (3) r  4 sin 

7␲ 6 5␲ 4

11␲ 6 4␲ 3

Begin by noting that r  0 at   0, and will increase from 0 to 4 as the clock    “ticks” from 0 to , since sin  is increasing from 0 to 1. (1) For   , , and 2 6 4   , r  2, r  2.8, and r  3.5, respectively (at   , r  4). See Figure 9.44. 3 2  (2) As  continues “ticking” from to , r decreases from 4 to 0, since sin  is 2 5 2 3 decreasing from 1 to 0. For   , , and , r  3.5, r  2.8, 3 4 6 3 , r and r  2, respectively (at   , r  0). See Figure 9.45. (3) From  to 2 increases from 0 to 4, but since r 6 0, this portion of the graph is reflected back 3  into Quadrant I, overlapping the portion already drawn from 0 to . (4) From 2 2  to 2, r decreases from 4 to 0, overlapping the portion drawn from to . We 2 conclude the graph is a closed figure limited to Quadrants I and II as shown in Figure 9.45. This is a circle with radius 2, centered at (0, 2). In summary:

5␲ 3

7␲ 4

 2



0 to

r

0 to 4

 to  2

 to

4 to 0

0 to 4

3 2

3 to 2 2 4 to 0

Now try Exercises 57 and 58



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Although it takes some effort, r-value analysis offers an efficient way to graph polar equations, and gives a better understanding of graphing in polar coordinates. In addition, it often enables you to sketch the graph with a minimum number of calculations and plotted points. As you continue using the technique, it will help to have Figure 9.43 in plain view for quick reference, as well as the corresponding analysis of y  cos  for polar graphs involving cosine (see Exercise 98). EXAMPLE 6



Graphing Polar Equations Using an r-Value Analysis Sketch the graph of r  2  2 sin  using an r-value analysis.

Solution



WORTHY OF NOTE While the same graph is obtained by simply plotting points, using an r-value analysis is often more efficient, particularly with more complex equations. 

r  4 sin 

0

0

30

2

45

212  2.8

60

213  3.5

90

4

120

213  3.5

135

2 12  2.8

150

2

180

0

Since the minimum value of sin  is 1, we note that r will always be greater than or equal to zero. At   0, r has a value of 2 (sin 0  0), and will increase from 2  to 4 as the clock “ticks” from 0 to (sin  is positive and sin  is increasing). 2  From to , r decreases from 4 to 2 (sin  is positive and sin  is decreasing). 2 3 From  to , r decreases from 2 to 0 (sin  is negative and sin  is increasing); 2 3 and from to 2, r increases from 0 to 2 (sin  is negative and sin  is 2 decreasing). We conclude the graph is a closed figure containing the points (2, 0), 5  3  a4, b, 12, 2 , and a0, b. Noting that   and   will produce 2 2 6 6  integer values, we evaluate r  2  2 sin  and obtain the additional points a3, b 6 5 b. Using these points and the r-value analysis produces the and a3, 6 graph shown here, called a cardioid (from the limaçon family of curves). In summary we have:

  2

2 to 4

 to  2

4 to 2

3 (3)  to 2

2 to 0

(1) 0 to (2)

D. You’ve just learned how to sketch basic polar graphs using an r-value analysis

r  2  2 sin 

(4)

3 to 2 2

3␲ 4 5␲ 6 11␲ 12

0 to 2

(2)

(3)

13␲ 12 7␲ 6

7␲ 2␲ 12 3

5␲ 4

4␲ 3 17␲ 12

5 4 3 2 1

5␲ 12

␲ 3

(1)

␲ 4

␲ 6 ␲ 12

(4)

19␲ 12

5␲ 3

7␲ 4

23␲ 12 11␲ 6

Now try Exercises 59 through 62



E. Symmetry and Families of Polar Graphs Even with a careful r-value analysis, some polar graphs require a good deal of effort to produce. In many cases, symmetry can be a big help, as can recognizing certain families of equations and their related graphs. As with other forms of graphing, gathering this information beforehand will enable you to graph relations with a smaller number of plotted points. Figures 9.46 to 9.49 offer some examples of symmetry for polar graphs.

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Figure 9.46 Vertical-axis symmetry: r  2  2 sin  5 4 3 2 1

Figure 9.47 Polar-axis symmetry: r  5 sin  5 4 3 2 1

Figure 9.48 Polar symmetry: r  5 sin 12 2 5 4 3 2 1

Figure 9.49 Polar symmetry: r 2  25 sin12 2 5 4 3 2 1

␪  ␲2

WORTHY OF NOTE In mathematics we refer to the tests for polar symmetry as sufficient but not necessary conditions. The tests are sufficient to show symmetry (if the test is satisfied, the graph must be symmetric), but the tests are not necessary to show symmetry (the graph may be symmetric even if the test is not satisfied).

The tests for symmetry in polar coordinates bear a strong resemblance to those for rectangular coordinates, but there is a major difference. Since there are many different ways to name a point in polar coordinates, a polar graph may actually exhibit a form of symmetry without satisfying the related test. In other words, the tests are sufficient to establish symmetry, but not necessary. The formal tests for symmetry are explored in Exercises 100 to 102. For our purposes, we’ll rely on a somewhat narrower view, one that is actually a synthesis of our observations here and our previous experience with the sine and cosine. Symmetry for Graphs of Certain Polar Equations Given the polar equation r  f 12 , 1. If f 12 represents an expression in terms of sine(s), the graph will be  symmetric to   : 1r, 2 and 1r,   2 are on the graph. 2 2. If f 12 represents an expression in terms of cosine(s), the graph will be symmetric to   0: 1r, 2 and (r, ) are on the graph. Figure 9.50 While the fundamental ideas from Examy ples 5 and 6 go a long way toward graphing 1 (1) other polar equations, our discussion would (2) (5) (6) not be complete without a review of the y  sin(2␪) period of sine and cosine. Many polar equa␲ 3␲ ␲ 2␲ ␪ tions have factors of sin1n2 or cos1n2 in 2 2 them, and it helps to recall the period formula (3) (4) (7) (8) 1 2 P . Comparing r  4 sin  from n Example 5 with r  4 sin122, we note the period of sine changes from P  2 to 2  , meaning there will be twice as many cycles and r will now go through P 2 eight cycles—four where sin122  is increasing from 0 to 1 (in red), and four where it is decreasing from 1 to 0 (in blue). See Figure 9.50.

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EXAMPLE 7



Sketching Polar Graphs Using Symmetry and r-Values Sketch the graph of r  4 sin122 using symmetry and an r-value analysis.

Solution



 Since r is expressed in terms of sine, the graph will be symmetric to   . We 2 n note that r  0 at   , where n is even, and the graph will go through the pole 2 at these points. This also tells us the graph will be a closed figure. From the graph 7  3 5 , of sin122 in Figure 9.50, we see sin122   1 at   , , and , so the 4 4 4 4  3 5 7 b, a4, b, and a4, b. Only the graph will include the points a4, b, a4, 4 4 4 4 analysis of the first four cycles is given next, since the remainder of the graph can be drawn using symmetry. Cycle

r-Value Analysis

Location of Graph

(1)

0 to

 4

r increases from 0 to 4

QI 1r 7 02

(2)

  to 4 2

r decreases from 4 to 0

QI 1r 7 02

(3)

 3 to 2 4

r increases from 0 to 4

QIV 1r 6 02

(4)

3 to  4

r decreases from 4 to 0

QIV 1r 6 02

Plotting the points and applying the r-value analysis with the symmetry involved produces the graph in the figure, called a four-leaf rose. At any time during this process, additional points can be calculated to “round-out” the graph. r  4 sin122   4   to 4 2  3 to 2 4 3 to  4

0 to

3␲ 4

|r| 0 to 4

7␲ 2␲ 12 3

5␲ 6 11␲ 12

5 4 3 2 1

5␲ 12

␲ 3

(2)

␲ 4

␲ 6

(1)

␲ 12

(4)

23␲ 12 11␲ 6

4 to 0 0 to 4 4 to 0

13␲ 12 7␲ 6 5␲ 4

(3) 4␲ 3 17␲ 12

5␲ 19␲ 3 12

7␲ 4

Now try Exercises 63 through 70



Graphing Polar Equations To assist the process of graphing polar equations: 1. Carefully note any symmetries you can use. 2. Have graphs of y  sin1n2 and y  cos1n2 in view for quick reference. 3. Use these graphs to analyze the value of r as the “clock ticks” around the polar grid: (a) determine the max/min r-values and write them in polar form, and (b) determine the polar-axis intercepts and write them in polar form. 4. Plot the points, then use the r-value analysis and any symmetries to complete the graph.

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Similar to polynomial graphs, polar graphs come in numerous shapes and varieties, yet many of them share common characteristics and can be organized into certain families. Some of the more common families are illustrated in Appendix VI, and give the general equation and related graph for common family members. Also included are characteristics of certain graphs that will enable you to develop the polar equation given its graph or information about its graph. For further investigations using a graphing calculator, see Exercises 71 through 76. EXAMPLE 8




Graphing a Limaçon Using Stated Conditions Find the equation of the polar curve satisfying the given conditions, then sketch the graph: limaçon, symmetric to

90°, with a
2 and b
3.

Solution




The general equation of a limaçon symmetric to

90° is r
a  b sin
, so our desired equation is r
2  3 sin
. Since a 6 b, the limaçon has an inner loop of length 3  2
1 and a maximum distance from the origin of 2  3
5. The polar-axis intercepts are (2, 0) and 12, 180°2 . With b 6 0, the graph is reflected across the polar axis (facing “downward”). The complete graph is shown in the figure.

(2, 180)

(2, 0) (1, 90)

(5, 270)

Now try Exercises 79 through 94

EXAMPLE 9







Modeling the Flight Path of a Scavenger Bird Scavenger birds sometimes fly over dead or dying animals (called carrion) in a “figure-eight” formation, closely resembling the graph of a lemniscate. Suppose the flight path of one of these birds was plotted and found to contain the polar coordinates (81, 0°) and (0, 45°). Find the equation of the lemniscate. If the bird lands at the point (r, 136°), how far is it from the carrion? Assume r is in yards.

Solution




Since (81, 0°) is a point on the graph, the lemniscate is symmetric to the polar axis and the general equation is r2
a2cos12
2 . The point (81, 0°) indicates a
81, hence the equation is r2
6561 cos12
2 . At

136° we have r2
6561 cos 272°, and the bird has landed r
15 yd away.

(15, 136) (81, 0)

Lemniscate

Now try Exercises 95 through 97

E. You’ve just learned how to use symmetry and families of curves to write a polar equation given a polar graph or information about the graph




You’ve likely been wondering how the different families of polar graphs were named. The roses are easy to figure as each graph has a flower-like appearance. The limaçon (pronounced li-ma-sawn) family takes its name from the Latin words limax or lamacis, meaning “snail.” With some imagination, these graphs do have the appearance of a snail shell. The cardioids are a subset of the limaçon family and are so named due to their obvious resemblance to the human heart. In fact, the name stems from the Greek kardia meaning heart, and many derivative words are still in common use (a cardiologist is one who specializes in a study of the heart). Finally, there is the lemniscate family, a name derived from the Latin lemniscus, which describes a certain kind of ribbon. Once again, a little creativity enables us to make the connection between ribbons, bows, and the shape of this graph.

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9.6 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. The point (r, ) is said to be written in coordinates. 2. In polar coordinates, the origin is called the and the horizontal axis is called the axis. 3. The point (4, 135°) is located in Q (4, 135°) is located in Q . 

, while

4. If a polar equation is given in terms of cosine, the graph will be symmetric to . 5. Write out the procedure for plotting points in polar coordinates, as though you were explaining the process to a friend. 6. Discuss the graph of r  6 cos  in terms of an r-value analysis, using y  cos  and a color-coded graph.

DEVELOPING YOUR SKILLS

Plot the following points using polar graph paper.

7. a4,

 b 2

8. a3,

3 b 2

 10. a4.5,  b 3

11. a5,

13. a3, 

 14. a4,  b 4

2 b 3

5 b 6

9. a2,

21.

5 b 4

12. a4,

16.

y

y

(0, 4)

(4, 4)

y

7 b 4

Express the points shown using polar coordinates with  in radians, 0    2 and r  0.

15.

22. y

(4, 0) x

x

x

x

(4√3, 4)

List three alternative ways the given points can be expressed in polar coordinates using r  0, r  0, and   [ 2, 2 ).

23. a3 12, 25. a2,

24. a4 13, 

3 b 4

11 b 6

26. a3, 

5 b 3

7 b 6

Match each (r, ) given to one of the points A, B, C, or D shown.

17.

18.

y

Exercise 27–36

y

(4, 4)

7␲ 2␲ 12 3

3␲ 4

x

x

5␲ 6 11␲ 12

(4, 4)

19.

20.

y

13␲ 12 7␲ 6

y

(4, 4√3) x

(4√3, 4)

27. a4, 

5 b 6

5␲ 12

␲ 3

␲ 4

B

␲ 6

C D 5␲ 4

x

A

5 4 3 2 1

4␲ 3 17␲ 12

5␲ 19␲ 3 12

28. a4, 

7␲ 4

␲ 12

23␲ 12 11␲ 6

5 b 4

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29. a4,

 b 6

31. a4,  33. a4,

5 b 4

13 b 6

35. a4, 

21 b 4

30. a4,

3 b 4

 32. a4,  b 4 34. a4,

36. a4, 

37. (8, 0)

35 b 6

38. (0, 7)

39. (4, 4)

40. 14 13, 42

41. (512, 5 12)

42. (6, 6 13)

43. (5, 12)

44. (3.5, 12)

Convert from polar coordinates to rectangular coordinates. A diagram may help.

45. (8, 45°)

46. (6, 60°)

47. a4,

48. a5,

3 b 4

49. a2,

7 b 6

51. 15, 135°2

5 b 6

50. a10,

4 b 3

52. 14, 30°2

53. r  5  6

54. r  6 56.   

3 4

57. r  4 cos 

58. r  2 sin 

59. r  3  3 sin 

60. r  2  2 cos 

61. r  2  4 sin 

62. r  1  2 cos 

63. r  5 cos122

64. r  3 sin142

65. r  4 sin 2

66. r  6 cos152

67. r  9 sin122

68. r2  16 cos122

 69. r  4 sina b 2

 70. r  6 cosa b 2

2

Use a graphing calculator in polar mode to produce the following polar graphs.

71. r  4 21  sin2, a hippopede 72. r  3  csc , a conchoid 73. r  2 cos  cot , a cissoid 74. r  cot , a kappa curve 75. r  8 sin  cos2, a bifoliate 76. r  8 cos 14 sin2  22 , a folium

WORKING WITH FORMULAS

77. The midpoint formula in polar coordinates: r cos   R cos  r sin   R sin  , b Ma 2 2 The midpoint of a line segment connecting the points (r, ) and (R, ) in polar coordinates can be found using the formula shown. Find the midpoint of the line segment between 1r, 2  16, 45°2 and 1R, 2  18, 30°2 , then convert these points to rectangular coordinates and find the midpoint using the “standard” formula. Do the results match? 

Sketch each polar graph using an r-value analysis (a table may help), symmetry, and any convenient points.

55.  

19 b 6

Convert from rectangular coordinates to polar coordinates. A diagram may help.



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CHAPTER 9 Analytical Geometry and the Conic Sections

78. The distance formula in polar coordinates: d  2R2  r2  2Rr cos(  ) Using the law of cosines, it can be shown that the distance between the points (R, ) and (r, ) in polar coordinates is given by the formula indicated. Use the formula to find the distance between 1R, 2  16, 45°2 and 1r, 2  18, 30°2 , then convert these to rectangular coordinates and compute the distance between them using the “standard” formula. Do the results match?

APPLICATIONS

Polar graphs: Find the equation of a polar graph satisfying the given conditions, then sketch the graph.

79. limaçon, symmetric to polar axis, a  4 and b  4

80. rose, four petals, two petals symmetric to the polar axis, a  6 81. rose, five petals, one petal symmetric to the polar axis, a  4

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82. limaçon, symmetric to  

 , a  2 and b  4 2

83. lemniscate, a  4 through (, 4) 84. lemniscate, a  8 through a8, 85. circle, symmetric to   containing a2,

 b 6

 b 4

  , center at a2, b, 2 2

86. circle, symmetric to polar axis, through 16, 2

a. c. e. g.

r  6 cos  r  6 cos142 r2  36 sin122 r  6 sin 

b. d. f. h.

895

r  3  3 sin  r2  36 cos122 r  2  4 sin  r  6 sin152

95. Figure eights: Waiting for help to arrive on foot, a light plane is circling over some stranded hikers using a “figure eight” formation, closely resembling the graph of a lemniscate. Suppose the flight path of the plane was plotted (using the hikers as the origin) and found to contain the polar coordinates (7200, 45°) and (0, 90°) with r in meters. Find the equation of the lemniscate.

Matching: Match each graph to its equation a through h, which follow. Justify your answers.

87.

88. 96. Animal territories: Territorial animals often prowl the borders of their territory, marking the boundaries with various bodily excretions. Suppose the territory of one such animal was limaçon shaped, with the pole representing the den of the animal. Find the polar equation defining the animal’s territory if markings are left at (750, 0°), (1000, 90°), and (750, 180°). Assume r is in meters.

6 6

89.

90. 6 6

91.

92. 6

93.

6

94. 6 6



97. Prop manufacturing: The propellers for a toy boat are manufactured by stamping out a rose with n petals and then bending each blade. If the manufacturer wants propellers with five blades and a radius of 15 mm, what two polar equations will satisfy these specifications?

98. Polar curves and cosine: Do a complete r-value analysis for graphing polar curves involving cosine. Include a color-coded graph showing the relationship between r and , similar to the analysis for sines that preceded Example 6.

EXTENDING THE CONCEPT

99. The polar graph r  a is called the Spiral of 1 Archimedes. Consider the spiral r  . As this 2 graph spirals around the origin, what is the distance between each positive, polar intercept? In QI, what is the distance between consecutive branches of the

 ? What is the 4 distance between consecutive branches of the spiral  at   ? What can you conclude? 2 spiral each time it intersects  

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As mentioned in the exposition, tests for symmetry of polar graphs are sufficient to show symmetry (if the test is satisfied, the graph must be symmetric), but the tests are not necessary to show symmetry (the graph may be symmetric even if the test is not satisfied). For r  f(), the formal tests for the symmetry are: (1) the graph will be symmetric to the polar axis if f()  f(); (2) the  graph will be symmetric to the line   if 2 f(  )  f (); and (3) the graph will be symmetric to the pole if f()  f().

100. Sketch the graph of r  4 sin122. Show the equation fails the first test, yet the graph is still symmetric to the polar axis. 101. Why is the graph of every lemniscate symmetric to the pole? 102. Verify that the graph of every limaçon of the form r  a  b cos  is symmetric to the polar axis. 

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CHAPTER 9 Analytical Geometry and the Conic Sections

103. The graphs of r  a sin1n2 and r  a cos1n2 are from the rose family of polar graphs. If n is odd, there are n petals in the rose, and if n is even, there are 2n petals. An interesting extension of this fact is that the n petals enclose exactly 25% of the area of the circumscribed circle, and the 2n petals enclose exactly 50%. Find the area within the boundaries of the rose defined by r  6 sin152. To develop an understanding of polar equations, we used the following facts x2  y2  r2, x  r cos , and y  r sin . Using these relationships, we can actually convert polar equations to rectangular equations and vice versa, showing that a particular equation can be graphed in either form. Use these relationships to write these polar equations in rectangular form. (Hint: Isolate the term kr (k a constant) on one side, then square.)

104. r 

1 1  sin 

105. r 

6 2  4 sin 

MAINTAINING YOUR SKILLS

106. (6.2) Verify the following is an identity: cos2x  sin2x  1  sin12x2 tan x. 107. (6.7) Solve for t  3 0, 22:  20  5  30 sin a2t  b. 6

109. (2.7) Graph the piecewise function shown and state its domain and range. x  2 5  x 6 1 f 1x2  • x 1  x 6 2 4 26x5

108. (1.3) Solve the absolute value inequality. Answer in interval notation: 32x  5  7 7 19

9.7 More on the Conic Sections: Rotation of Axes and Polar Form Learning Objectives In Section 9.7 you will learn how to:

A. Graph conic sections that have nonvertical and nonhorizontal axes (rotated conics)

B. Identify conics using the discriminant of the polynomial form—the invariant B2  4AC

C. Write the equation of a conic section in polar form

D. Solve applications involving the conic sections in polar form

Our study of conic sections would not be complete without considering conic sections whose graphs are not symmetric to a vertical or horizontal axis. The axis of symmetry still exists, but is rotated by some angle. We’ll first study these rotated conics using the equation in its polynomial form, then investigate some interesting applications of the polar form.

A. Rotated Conics and the Rotation of Axes It’s always easier to understand a new idea in terms of a known idea, so we begin our 1 study with a review of the reciprocal function y  . From the equation we note: x 1. The denominator is zero when x  0, and the y-axis is a vertical asymptote (the vertical line x  0). 2. Since the degree of the numerator is less than the degree of the denominator, the x-axis is a horizontal asymptote (the horizontal line y  0). 3. Since x 6 0 implies y 6 0 and x 7 0 implies y 7 0, the graph will have two branches—one in the first quadrant and one in the third.

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Section 9.7 More on the Conic Sections: Rotation of Axes and Polar Form

Note the polynomial form of this equation is xy  1. The resulting graph is shown in Figure 9.51, and is actually the graph of a hyperbola with a transverse axis of y  x. Using the 45-45-90 triangle indicated, we find the distance from the origin to each vertex is 12. If we rotated the hyperbola 45° clockwise, we would obtain a more “standard” graph with a horizontal transverse axis and vertices at 1a, 02 S 112, 02. The asymptotes would be b y  1x, and since y   x is the general form a we know b   12. This information can be used to find the equation of the rotated hyperbola.

Figure 9.51 y

yx

(1, 1) 1 1

(1, 1)

EXAMPLE 1



Finding the Equation of a Rotated Conic from Its Graph

Solution



Using the standard form

x

The hyperbola xy  1 is rotated clockwise 45°, with new vertices at 112, 02, asymptotes at y  1x and b   12. Find the equation and graph the hyperbola. y2 x2   1 and a2 b2 substituting  12 for a and b, the equation of the rotated hyperbola is y2 x2   1 or x2  y2  2 in polynomial 2 2 form. The resulting graph is the central hyperbola shown.

y

(√2, 0)

(√2, 0) x

Now try Exercises 7 and 8 Figure 9.52 y x  r cos ␣ y  r sin ␣

(x, y)

r

␣ x



It’s important to note the equation of the rotated hyperbola is devoid of the mixed “xy” term. In nondegenerate cases, the equation Ax2  Cy2  Dx  Ey  F  0 is the polynomial form of a conic with axes that are vertical/horizontal. However, the most general form of the equation is Ax2  Bxy  Cy2  Dx  Ey  F  0, and includes this Bxy term. As noted in Example 1, the inclusion of this term will rotate the graph through some angle . Based on these observations, we reason that one approach to graphing these conics is to find the angle of rotation  with respect to the xy-axes. We can then use  to rewrite the equation so that it corresponds to a new set of XY-axes, which are parallel to the axes of the conic. The mixed xy-term will be absent from the new equation and we can graph the Figure 9.53 conic on the new axes using the same ideas as before y Y (identifying a, b, foci, and so on). To find , recall that X  r cos(␣  ␤) Y  r sin(␣  ␤) a point (x, y) in the xy-plane can be written (X, Y) x  r cos , y  r sin , as in Figure 9.52. The diagram in Figure 9.53 shows the axes of a new XY-plane, rotated counterclockwise by angle . In r this new plane, the coordinates of the point (x, y) ␣ X become X  r cos1  2 and Y  r sin1  2 as ␣␤ shown. Using the difference identities for sine and ␤ x cosine and substituting x  r cos  and y  r sin  leads to

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X  r cos1  2  r1cos  cos   sin  sin 2  r cos  cos   r sin  sin   x cos   y sin 

Y  r sin1  2  r1sin  cos   cos  sin 2  r sin  cos   r cos  sin   y cos   x sin 

The last two equations can be written as a system, which we will use to solve for x and y in terms of X and Y. X  x cos   y sin  Y  y cos   x sin  X cos   x cos2  y sin  cos  e Y sin   y sin  cos   x sin2 X cos   Y sin   x cos2  x sin2 e

WORTHY OF NOTE

X cos   Y sin   x

If you are familiar with matrices, it may be easier to remember the rotation formulas in their matrix form, since the pattern of functions is the same, with only a difference in sign:

original system multiply first equation by cos  multiply second equation by sin  first equation  second equation factor out x 1cos2  sin2  12

Re-solving the system for y results in y  X sin   Y cos , yielding what are called the rotation of axes formulas (see Exercise 79). Rotation of Axes Formulas If the x- and y-axes of the xy-plane are rotated counterclockwise by the (acute) angle  to form the X- and Y-axes of an XY-plane, the coordinates of the points (x, y) and (X, Y ) are related by the formulas

x cos sin X c d  c dc d sin cos Y y X cos sin x c d  c dc d sin cos y Y See Exercises 86 and 87.

x  X cos   Y sin 

X  x cos   y sin 

y  X sin   Y cos 

Y  x sin   y cos 

EXAMPLE 2



Naming the Location of a Point After Rotating the Axes

Solution



Using the formulas with x  1, y  13, and   60°, we obtain

Given the point 11, 132 in the xy-plane, find the coordinates of this point in the XY-plane given the angle  between the xy-axes and the XY-axes is 60°. X  x cos   y sin   1 cos 60°  13 sin 60° 1 13  a b  13 a b 2 2 2

Y  x sin   y cos   1 sin 60°  13 cos 60° 13 13   2 2 0

The coordinates of P(X, Y) would be (2, 0). Now try Exercises 9 through 16



Figure 9.54 y : (2 ,0

)

X

pla

ne

Y

xy-plane: (1, √3)

XY -

The diagram in Figure 9.54 provides a more intuitive look at the rotation from Example 2. As you can see, a 30-60-90 triangle is formed with a hypotenuse of 2, giving coordinates (2, 0) in the XY-plane.

√3 60 1

x

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EXAMPLE 3



Writing the Equation of a Conic After Rotating the Axes The ellipse X2  4Y2  16 is rotated clockwise 45°. What is the corresponding equation in the xy-plane?

Solution



We proceed as before, using the rotation formulas X  x cos   y sin  and 12 Y  y cos   x sin . With   45° we have cos   sin   , yielding 2 X2  4Y2  16 2 1x cos   y sin 2  41y cos   x sin 2 2  16 use rotation formulas 12 12 2 12 12 2 12 a x yb  4a y xb  16 substitute for sin  and cos  2 2 2 2 2 1 1 1 1 a x2  xy  y2 b  4a x2  xy  y2 b  16 square binomials 2 2 2 2 1 2 1 x  xy  y2  2x2  4xy  2y2  16 distribute 2 2 5 2 5 x  3xy  y2  16 result 2 2 Now try Exercises 17 through 20



Note the equation of the conic in the standard xy-plane contains the “mixed” Bxyterm. In practice, we seek to reverse this procedure by starting in the xy-plane, and finding the angle  needed to eliminate the Bxy-term. Using the rotation formulas and the appropriate angle , the equation Ax2  Bxy  Cy2  Dx  Ey  F  0 becomes aX2  cY2  dX  eY  f  0, where the xy-term is absent. To find the angle , note that without loss of generality, we can assume D  E  0 since only the second-degree terms are used to identify a conic. Starting with the simplified equation Ax2  Bxy  Cy2  F  0 and using the rotation formulas we obtain 

Ax2

Bx

#



y

F0

Cy2

A1X cos   Y sin 2  B1X cos   Y sin 2 1X sin   Y cos 2  C1X sin   Y cos 2  F  0 2

2

Expanding this expression and collecting like terms (see Exercise 80), gives the following expressions for coefficients a, b, and c of the corresponding equation aX2  bXY  cY2  f  0: a S A cos2  B sin  cos   C sin2

a is the coefficient of X 2

b S 2A sin  cos   B1cos2  sin22  2C sin  cos 

b is the coefficient of XY

c S A sin   B sin  cos   C cos 

c is the coefficient of Y 2

fSF

f  F (the constant remains unchanged)

2

2

To accomplish our purpose, we require the coefficient b to be zero. While this expression looks daunting, the double-angle identities for sine and cosine simplify it very nicely: b S A12 sin  cos 2  B1cos2  sin22  C12 sin  cos 2  0 A sin122  B cos122  C sin(2)  0 1C  A2sin122  B cos122 B tan122  CA B tan122  ;AC AC

(1) (2) (3) (4)

(5)

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Note from line (3) that A  C would imply cos12
2  0, giving 2
 90° or
90°, with
 45° or
45° (for the sake of convenience, we select the angle in QI). B This fact can many times be used to great advantage. If A  C, tan12
2  and A
C we choose 2
between 0 and 180° so that
will be in the first quadrant 3 0 6
6 90° 4. The Equation of a Conic After Rotating the Axes For a conic defined by Ax2  Bxy  Cy2  Dx  Ey  F  0 and its graph in the B xy-plane, an angle
can be determined using tan12
2  and used in the A
C rotation formulas to find a polynomial aX2  cY2  dX  eY  f  0 in XY-plane, where the conic is either vertical or horizontal.

EXAMPLE 4




Rotating the Axes to Eliminate the Bxy-Term For x2
2 13xy  3y2
13x
y
16  0, eliminate the xy-term using a rotation of axes and identify the conic associated with the resulting equation. Then sketch the graph of the rotated conic in the XY-plane.

Solution




B
2 13 , giving tan12
2   13. A
C 1
3 13 This shows 2
 tan
1 13, yielding 2
 60° so
 30°. Using cos 30°  2 1 and sin 30°  along with the rotation formulas we obtain the following 2 XY-equation, with corresponding terms shown side-by-side for clarity: Since A  C, we find
using tan12
2 

Given Term in xy-Plane x2
2 13xy 3y2


13x
y
16

Corresponding Term in xy-Plane → a

13 1 2 3 1 13 X
Yb  X2
XY  Y2 2 2 4 2 4

13 3 2 1 1 13 3 2 →
213a 2 X
2 Yba 2 X  2 Yb 
2 X
13XY  2 Y 13 2 3 2 13 9 2 1 → 3a 2 x  2 Yb  4 X  3 2 XY  4 Y →
13a

13 1 3 13 X
Yb 
X  Y 2 2 2 2

13 1 13 1 →
a 2 X  2 Yb 
2 X
2 Y →
16

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Adding the like terms to the far right, the X2-terms (in red), the Y-terms (in bold), and the mixed XYterms (in blue) sum to zero, leaving the equation 2X  4Y2  16  0, which is the parabola defined by Y2  12 1X  82. This parabola is symmetric to the X-axis and opens to the right, with a vertex at (8, 0), Y-intercepts at (0, 2) and (0, 2), focus at 163 8 , 02 and directrix , 02 through 165 . The graph is shown in the 8 figure.

y Y X (0, 2) 30 x

(0, 2) (8, 0)

Now try Exercises 21 through 30

A. You’ve just learned how to graph conic sections that have nonvertical and nonhorizontal axes (rotated conics)



In Example 4, the angle  was a standard angle and easily found. In general, this is not the case and finding exact values of cos  and sin  for use in the rotation formulas sin122 , the corresponding (triangle) diagram, and the idenrequires using tan122  cos122 1  cos122 1  cos122 tities cos   and sin   . See Exercises 31, 32, 84, A B 2 2 and 85 for further study.

B. Identifying Conics Using the Discriminant In addition to rotating the axes, the inclusion of the “xy-term” makes it impossible to identify the conic section using the tests seen earlier. For example, having A  C no longer guarantees a circle, and A  0 or C  0 does not guarantee a parabola. Rather than continuing to look at what the mixed term and the resulting rotation changes, we now look at what the rotation does not change, called invariants of the transformation. These invariants can be used to double-check the algebra involved and to identify the conic using the discriminant. These are given here without proof. Invariants of a Rotation and Classification Using the Discriminant By rotating the coordinate axes through a predetermined angle , the equation Ax2  Bxy  Cy2  Dx  Ey  F  0 can be transformed into aX2  cY2  dX  eY  f  0 in which the xy-term is absent. This rotation has the following invariants: (1) F  f

(2) A  C  a  c

(3) B2  4AC  b2  4ac.

The discriminant of a conic equation in polynomial form is B2  4AC. Except in degenerate cases, the graph of the equation can be classified as follows: If B2  4AC  0, the graph will be a parabola. If B2  4AC 6 0, the graph will be a circle or an ellipse. If B2  4AC 7 0, the graph will be a hyperbola.

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EXAMPLE 5A



Verifying the Invariants of a Rotation of Axes Verify the invariants just given using the equations from Example 4. Also verify the discriminant test.

Solution



From the equation x2  2 13xy  3y2  13x  y  16  0, we have A  1, B  2 13, C  3, D   13, E  1, and F  16. After applying the rotation the equation became 2X  4Y2  16  0, with a  0, b  0, c  4, d  2, e  0, and f  16. Checking each invariant gives (1) 16  16✓, (2) 1  3  0  4✓, and (3) 12 132 2  4112 132  102 2  4102142 ✓. With B2  4AC  0, the discriminant test indicates the conic is a parabola ✓.

EXAMPLE 5B



Identifying the Equation of a Conic Using the Discriminant Use the discriminant to identify each equation as that of a circle, ellipse, parabola, or hyperbola, but do not graph the equation. a. 3x2  4xy  3y2  6x  12y  2  0 b. 4x2  9xy  4y2  8x  24y  9  0 c. 6x2  7xy  y2  5  0 d. x2  6xy  9y2  6x  0

Solution



a.

A  3; B  4; C  3

B  4AC  142  4132132 2

2

 20 circle or ellipse

c. A  6; B  7; C  1 B. You’ve just learned how to identify conics using the discriminant of the polynomial form—the invariant B2  4AC

B  4AC  172  4162112 2

2

 25

b. A  4; B  9; C  4

B2  4AC  192 2  4142142  17 hyperbola

d. A  1; B  6; C  9

B2  4AC  162 2  4112192 0

hyperbola

parabola Now try Exercises 33 through 36

Figure 9.55

C. Conic Equations in Polar Form

y P(r, ␪)

D

r

␪ F(0, 0)

A d

x B

You might recall that earlier in this chapter we defined ellipses and hyperbolas in terms of a distance between two points, but a parabola in terms of a distance between a point and a line (the focus and directrix). Actually, all conic sections can be defined using a focus/directrix development and written in polar form. This serves to unify and greatly simplify their study. We begin by revisiting the focus/directrix development of a parabola, using a directrix l and placing the focus at the origin. With the polar axis as the axis of symmetry and the point P1r, 2 in polar coordinates, we obtain the graph shown in Figure 9.55. Given D and A are points on l (with A on the polar axis), we note the following: (1) DP  FP (2) DP  AB

Directrix ᏸ



(3) FB  r cos  (4) AB  AF  FB

definition of a parabola equal line segments FB cos   r sum of line segments

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Using the preceding equations and representing the distance AF by the constant d, we obtain this sequence: AB  d  r cos 

substitute d for AF and r cos  for FB

FP  d  r cos  r  d  r cos 

substitute FP for AB since FP  DP  AB substitute r for FP

d , 1  cos  which is the equation of a parabola in polar form with its focus at the origin, vertex at d  3 a , b, and y-intercepts at ad, b and ad, b. Note the constant “1” in the denom2 2 2 inator is a key characteristic of polar equations, and helps define the standard form. Solving the last equation for r we have r  r cos   d, then r 

EXAMPLE 6A



Identifying a Conic from Its Polar Equation Verify the equation r 

6 represents a parabola, then describe and 3  3 cos 

sketch the graph. Solution



Figure 9.57 P3

D1 F P1

D1P3  2FP3 D2P4  2FP4

D2 ᏸ

P4

P2

Write the equation in standard form by dividing the numerator and denominator by 3, 2 obtaining r  . From this we see 1  cos  d  2 and the represents a parabola symmetric to the polar axis, with vertex at (1, ) and 3  b, as shown y-intercepts at a2, b and a2, 2 2 in the figure.

2␲ 3 3␲ 4

␲ 3

ᏸ

5␲ 6

4 3 (2, ␲2 )

␲ 4 ␲ 6

(1,␲)

(2, 3␲ 2 ) 7␲ 6 5␲ 4

11␲ 6 4␲ 3

5␲ 3

7␲ 4

The polar equation for a parabola depended on Figure 9.56 DP and FP being equal in length, with ratio FP  1. But what if this ratio is not equal to 1? DP2  2FP2 DP1  2FP1 DP D Similar to our introduction to conics in Section 9.1, P1 F P2 FP 1 Directrix  and investigate the graph that we assume ᏸ 2 DP results.Cross-multiplying gives 2FP  DP, which states that the distance from D to P is twice the distance from F to P. Note that we are able to locate two points P1 and P2 on the polar axis that satisfy this relation, rather than only one as in the case of the parabola. Figure 9.56 illustrates the location of these points. Using the focal chord for convenience, two additional points P3 and P4 can be located that also satisfy the stated condition (see Figure 9.57). In fact, we can locate an infinite number of these FP 1  , and the resulting graph appears to be an ellipse (and is definitely points using 2 DP not aparabola). These illustrations provide the basis for stating a general focus/directrix FP definition of the conic sections. The ratio is often represented by the letter e, DP and represents the eccentricity of the conic. Using FP  r and DP  d  r cos  from r FP   e, which enables us to state the general our initial development, d  r cos  DP

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de , 1  e cos  where the type of conic depends solely on e. Depending on the orientation of the conic, the general form may involve sine instead of cosine, and have a sum of terms in the denominator rather than a difference. Note once again that if e  1, the relation simplifies into the parabolic equation seen earlier. equation of a conic in polar form. Solving for r leads to the equation r 

The Standard Equation of a Conic in Polar Form Given a conic section with eccentricity e, one foci at the pole of the r-plane, and directrix l located d units from this focus. Then the polar equations r

de 1  e cos 

and

r

de 1  e sin 

represent one of the conic sections as determined by the value of e. • If e  1, the graph is a parabola. • If 0 6 e 6 1, the graph is an ellipse. • If e 7 1, the graph is a hyperbola. For the ellipse and hyperbola, the major axis and transverse axis (respectively) are both perpendicular to the directrix and contain the vertices and foci. Our earlier development of eccentricity can then be expressed in terms of a and c, as the c ratio e  . a As in our previous study of polar equations, if the equation involves cosine the  graph will be symmetric to the polar axis. If the graph involves sine, the line   is 2 the axis of symmetry. In addition, if the denominator contains a difference of terms (as in Example 6A), the graph will be above or to the right of the directrix (depending on whether the equation involves sine or cosine). If the denominator contains a sum of terms, the graph will be below or to the left of the directrix. EXAMPLE 6B



Using the Standard Equation to Graph a Conic in Polar Form 10 represents a parabola, ellipse, or 5  3 sin  hyperbola. Then describe and sketch the graph. Determine if the equation r 

Solution



C. You’ve just learned how to write the equation of a conic section in polar form

To write the equation in standard form, we divide 2␲ ␲ 3 (5, ␲2 ) 3 ␲ both numerator and denominator by 5, obtaining 3␲ 4 4 2 4 ␲ . From the standard 5␲ the equation r  6 6 3 3 1  sin  2 5 1 (2, 0) (2, ␲) 3 form we note e  so the equation represents an 5 (1.25, 3␲ 2 ) ellipse. With a difference of terms and the sine 11␲ 7␲ 6 function involved, the graph is symmetric to 6 ᏸ 5␲  7␲ 4 4   and is above the directrix. Given so much 4␲ 5␲ 2 3 3 information by the equation, we require very few points to sketch the graph and settle for those generated by  3  5 3   0, , , and , yielding the points (2, 0), a5, b, 12, 2, and a , b. The 2 2 2 4 2 graph is shown in the figure. Now try Exercises 37 through 56



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D. Applications of Conics in Polar Form For centuries it has been known that the orbits of the planets around the Sun are elliptical, with the Sun at one focus. In addition, comets may approach our Sun in an elliptical, hyperbolic, or parabolic path with the Sun again at the foci. This makes planetary studies a very natural application of the conic sections in polar form. To aid this study, it helps to know that in an elliptical orbit, the maximum distance of a planet from the Sun is called its aphelion, and the shortest distance is the perihelion (Figure 9.58). This means the length of the major axis is “aphelion  perihelion,” enabling us to find the value of c if the aphelion and peric helion are known (Figure 9.59). Using e  , we can a then find the eccentricity of the planet’s orbit.

EXAMPLE 7



Figure 9.58

Sun Perihelion

Aphelion

Figure 9.59 a Perihelion c

Determining the Eccentricity of a Planet’s Orbit In its elliptical orbit around the Sun, Mars has an aphelion of 154.9 million miles and a perihelion of 128.4 million miles. What is the eccentricity of its orbit?

Solution



The length of the major axes would be 2a  1154.9  128.42 mi, yielding a semimajor axis of a  141.65 million miles. Since a  c  perihelion (Figure 9.59), we have 141.65  c  128.4 so c  13.25. The eccentricity of the orbit is 13.25 c or about 0.0935. e  a 141.65 Now try Exercises 59 and 60



We can also find the perihelion and aphelion directly in terms of a (semimajor axis) and e (eccentricity) if these quantities are known. Using a  c  perihelion, we c obtain: perihelion  a  c. For e  , we have ea  c and by direct substitution we a obtain: perihelion  a  ea  a11  e2. For Example 8, recall that “AU” designates an astronomical unit, and represents the mean distance from the Earth to the Sun, approximately 92.96 million miles.

EXAMPLE 8



Determining the Perihelion of a Planet’s Orbit The orbit of the planet Jupiter has a semimajor axis of 5.2 AU (1 AU  92.96 million miles) and an eccentricity of 0.0489. What is the closest distance from Jupiter to the Sun?

Solution



With perihelion  a11  e2, we have 5.211  0.04892  4.946. At its closest approach, Jupiter is 4.946 AU from the Sun (about 460 million miles). Now try Exercises 61 through 64 To find the polar equation of a planetary orbit, it’s helpful to write the general polar equation in terms of the semimajor axis a, which is often known or easily found, rather than in terms of the distance d from directrix to focus, which is often unknown. Consider the diagram in Figure 9.60, which shows an elliptical orbit with the Sun at one focus, vertices



Figure 9.60 d D ᏸ

P1

F

a

C

P2

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P1 and P2 (perihelion and aphelion), and the center C of the ellipse. Assume the point FP1  e. From Example 8 we P used to define the conic sections is at position P1, giving DP1 have FP1  a11  e2. Substituting a11  e2 for FP1 and solving for DP1 gives a11  e2 . Using d  DP1  FP1, we obtain the following sequence: DP1  e d  DP1  FP1 a11  e2  a11  e2  e a11  e2 ae11  e2   e e a11  e211  e2  e a11  e2 2  e de  a11  e2 2

substitute

a 11  e2 e

for DP1 and a 11  e2 for FP1

common denominator combine terms, factor out a 11  e2 11  e211  e2  1  e 2 multiply by e

Substituting a11  e2 2 for de in the standard equation r 

de gives the 1  e cos  a11  e2 2 . equation of the orbit entirely in terms of a and e: r  1  e cos  EXAMPLE 9



Writing the Polar Equation of an Ellipse from Given Information At its aphelion, the dwarf planet Pluto is the most distant from the Sun at 4538 million miles. It has a perihelion of 2756 million miles. Use this information to find the polar equation Perihelion that models the orbit of Pluto, then find the length of the focal chord for this ellipse.

Solution



D. you’ve just learned how to solve applications involving the conic sections in polar form

a c Center

Aphelion

With all figures in millions of miles, the major axis is 2a  4538  2756  7294, so the semimajor axis has length a  3647. With a  c  perihelion, we obtain 891 3647  c  2756 or c  891. The eccentricity of the orbit is e   0.2443. 3647 136472 11  30.2444 2 2 The polar equation for the orbit of Pluto is r  or 1  3 0.24434 cos   3430 r . Substituting   (since the left-most focus is at the pole), 1  0.2443 cos  2 we obtain r  3430, so the length of the focal chord is 2134302  6860 million miles. Now try Exercises 65 through 70



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TECHNOLOGY HIGHLIGHT

Investigating the Eccentricity e Figure 9.61 One meaning of the word eccentric is “to deviate from a circular pattern.” In a very real sense, this is the role that eccentricity plays as it helps to describe the conic sections. For an ellipse we’ve learned that 0 6 e 6 1. If the eccentricity is near zero, there is little deviation and the ellipse appears nearly circular. If e is near 1, the ellipse is very elongated. To explore the eccentricity of an ellipse, enter the equation a11  e2 2 on the Y = screen, using a  2 (arbitrarily chosen) r 1  e cos  and ALPHA “E” for the eccentricity. The result is shown in Figure 9.61. We will enter and store values for E on the home screen and graph the resulting ellipse (see Exercise 2 for an alternative ALPHA method). Return to the home screen and enter 0.1 and graph the result on the ZOOM 4:ZDecimal screen. Repeat the procedure using e  0.25, 0.5, 0.75, and 0.9. The graphs for e  0.1 and e  0.9 are shown in Figures 9.62 and 9.63. As you can see, when e  0.1 the ellipse is nearly circular, while e  0.9 produces a graph that is cigar shaped. Figure 9.62

Figure 9.63

3

5

3

5

5

5

3

3

Exercise 1: Try entering a value of e  0, then use your graphing calculator and basic knowledge to verify the resulting graph is a circle. Exercise 2: Try the same exercise using the set/list option. In other words, enter the equation as shown 211  50.1, 0.25, 0.5, 0.75, 0.962 2 here, with the values of e in braces { }: r1  . This will enable you to 11  50.1, 0.25, 0.5, 0.75, 0.96cos122 view all five ellipses on the same \screen. Discuss the similarities and differences of this family of graphs.

9.7 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. The set of points (x, y) in the xy-plane are related to points (X, Y) in the XY-plane by the

formulas. To find the angle  between the original axes and the rotated axes, we use tan122  . .

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2. For a point P on the graph of a conic with focus F FP and D a point on the directrix, the ratio gives DP the ________ of the graph. For the eccentricity e, if e  1 the graph is a ________, if e 7 1 the graph is a ________, and if 0 6 e 6 1 the graph will be an ellipse. 3. Features or relationships that do not change when certain transformations are applied are called ________ of the transformation.

the ________ axis, and r 

de if symmetric 1  e sin 

to the line ________. 5. Discuss the advantages of graphing a rotated conic using the rotation of axes, over graphing by simply plotting points. 6. Discuss the primary advantages of using a11  e2 2 de r rather than r  to 1  e cos  1  e cos  develop the equation of planetary orbit.

4. The ________ form of the equation of a conic de is r  if the graph is symmetric to 1  e cos 



DEVELOPING YOUR SKILLS

The graph of a conic rotated in the xy-plane is given. Use the graph (not the rotation of axes formulas) to find the equation of the conic in the XY-plane.

7.

y Y

X

(2, 2)

x

(2, 2)

Y

14. 1 13, 32

15. (3, 4)

16. (12, 5)

17. X2  Y2  9; 60°

18. X2  Y  4; 60°

The conic sections whose equations are given in the xy-plane are rotated by the indicated angle. What is the corresponding equation in the XY-plane?

y X

19. 3x2  2xy  3y2  9; 45°

(3√3, 3)

20. x2  13xy  2y2  8; 60°

(√3, 3) 30 x

(3√3, 3)

13. 12, 2 132

The conic sections whose equations are given in the XY-plane are rotated clockwise by the indicated angle. Find the corresponding equation in the xy-plane.

45

8.

Given the point (X, Y) in the XY-plane, find the coordinates of this point in the xy-plane given the angle  between the xy-axes and the XY-axes is 30.

(√3, 3)

For the given conics in the xy-plane, (a) use a rotation of axes to find the corresponding equation in the XY-plane (clearly state the angle of rotation  ), and (b) sketch its graph. Be sure to indicate the characteristic features of each conic in the XY-plane.

21. x2  4xy  y2  2  0 Given the point (x, y) in the xy-plane, find the coordinates of this point in the XY-plane given the angle  between the xy-axes and the XY-axes is 45.

9. 1612, 62

11. (0, 5)

10. 14, 3 122 12. (8, 0)

22. x2  2xy  y2  12  0 23. 5x2  6xy  5y2  16 24. 5x2  26xy  5y2  72 25. x2  10 13xy  11y2  64 26. 37x2  42 13xy  79y2  400  0

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Section 9.7 More on the Conic Sections: Rotation of Axes and Polar Form

27. 3x2  2 13xy  y2  8x  813y  0

38.

␲ 2 2␲ ᏸ 3 3␲ 4

28. 6x  4 13xy  2y  2x  2 13y  0 2

2

29. 13x2  6 13xy  7y2  100  0

5␲ 6

30. x  4xy  y  12x  12y  11 2

2

(1.3, ␲)

Identify the graph of each equation using the discriminant, then find the value of cos122 using sin122 tan122  and the related triangle diagram. cos122 Finally, find sin  and cos  using the half-angle 1  cos122 and B 2 1  cos122 sin   . B 2

5␲ 4

␲ 3

2␲ 3

5␲ 6

For the following equations, (a) use the discriminant to identify the equation as that of a circle, ellipse, parabola, or hyperbola; (b) find the angle of rotation  and use it to find the corresponding equation in the XY-plane; and (c) verify all invariants of the transformation.

(4,

3␲ 2 )

4␲ 3

3␲ 4

ᏸ

36. 3x2  8 13xy  5y2  12y  2

5␲ 6

(1.5, ␲)

6 5 4 ␲ 3 (3, 2 ) 2 1

(3,

5␲ 6

␲ 4

4␲ 3

5␲ 4

3␲ 4

4␲ 3

(1.4, 0)

5␲ 3

2␲ 3

(3, ␲2 )

11␲ 6 5␲ 3

␲ 6

11␲ 6

41.

3␲ 2)

7␲ 4

␲ 4

7␲ 6

␲ 6

5

5␲ 4

␲ 3

2␲ 3

5␲ 6

7␲ 6

5␲ 3

4 3 ␲ 2 (0.84, 2 ) 1 (1.4, ␲)

Match each graph to its corresponding equation. Justify your answers (two equations have no match).

3␲ 4

␲ 6

11␲ 6

40.

␲ 3

␲ 4

5␲ 4

35. 3x  13xy  4y  4x  1

ᏸ

6 5 4 3 2 1 (0.8, ␲2 )

7␲ 6

2

2␲ 3

7␲ 4

5␲ 3

32. 25x2  840xy  16y2  400  0

37.

(4, 0)

4␲ 3

3␲ 4

31. 12x2  24xy  5y2  40x  30y  25

2

␲ 6

11␲ 6

39.

34. 2x2  3xy  2y2  0

␲ 4

7␲ 6

identities cos  

33. x2  2xy  y2  5  0

␲ 3

5 4 3 2 1

6 5 4 3 2 1

7␲ 4

␲ 3

␲ 4 ␲ 6

(3, ␲)

7␲ 4

(3,

7␲ 6

3␲ 2)

11␲ 6

5␲ 4 4␲ 3

5␲ 3

7␲ 4

909

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CHAPTER 9 Analytical Geometry and the Conic Sections

42. 3␲ 4

␲ 3

2␲ 3

6 5 4 3 2 (1, ␲2 ) 1 (2, 0)

5␲ 6

For the conic equations given, determine if the equation represents a parabola, ellipse, or hyperbola. Then describe and sketch the graphs using polar graph paper.

␲ 4 ␲ 6

ᏸ

7␲ 6

11␲ 6

5

5␲ 4 4␲ 3

a. r  c. r  e. r  g. r 

5 3 4 2

5␲ 3

10  5 sin  5.4  2 sin  12  6 sin  4  3 sin 

7␲ 4

b. r  d. r  f. r  h. r 

4 3 2 3

8  2 cos  4.2  2 sin  6  2 cos  9  6 cos 

43. r 

4 2  2 sin 

44. r 

10 5  5 sin 

45. r 

12 6  3 sin 

46. r 

6 4  3 cos 

47. r 

6 2  4 cos 

48. r 

2 2  3 sin 

49. r 

5 5  4 cos 

50. r 

2 4  5 sin 

Write the equation of a conic that satisfies the conditions given. Assume each has one focus at the pole.

51. ellipse, e  0.8, directrix to focus: d  4 52. hyperbola, e  1.25, directrix to focus: d  6 53. parabola, vertex at 12, 2

54. ellipse, e  0.35, vertex at (4, 0) 55. hyperbola, e  1.5, vertex at a3,

 b 2

56. parabola, directrix to focus: d  5.4



WORKING WITH FORMULAS

57. Equation of a line in polar form: C r A cos   B sin  For the line Ax  By  C in the xy-plane with C A slope m   and y-intercept a0, b, the B B corresponding equation in the r-plane is given by the formula shown. (a) Given the line 2x  3y  12 in the xy-plane, find the corresponding polar equation and (b) verify r1/22 A . that    B r102 

58. Polar form of an ellipse with center at the pole: a 2b 2 r2  2 2 a sin   b 2cos 2 If an ellipse in the r-plane has its center at the pole (with major axis parallel to the x-axis), its equation is given by the formula here, where 2a and 2b are the lengths of the major and minor axes, respectively. (a) Given an ellipse with center at the pole has a major axis of length 8 and a minor axis of length 4, find the equation of the ellipse in polar form and (b) graph the result on a calculator and verify that 2a  8 and 2b  4.

APPLICATIONS

Planetary motion: The perihelion, aphelion, and orbital period of the planets Jupiter, Saturn, Uranus, and Neptune are shown in the table. Use the information to answer or complete the following exercises. The formula L  220.51a2  b2 2 can be used to estimate the length of the orbital path. Recall for an ellipse, c2  a2  b2.

Planet

Perihelion (106 mi)

Aphelion (106 mi)

Period (yr)

Jupiter

460

507

11.9

Saturn

840

941

29.5

Uranus

1703

1866

84

Neptune

2762

2824

164.8

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Section 9.7 More on the Conic Sections: Rotation of Axes and Polar Form

59. Find the eccentricity of the planets Jupiter and Saturn. 60. Find the eccentricity of the planets Uranus and Neptune. 61. The orbit of Pluto (a dwarf planet) has a semimajor axis of 3647 million miles and an eccentricity of e
0.2443. Find the perihelion of Pluto. 62. The orbit of Ceres (a large asteroid) has a semimajor axis 257 million miles and an eccentricity of e
0.097. Find the perihelion of Ceres. 63. Which of the four planets in the table given has the greatest orbital eccentricity? 64. Which of these four planets has the greatest orbital velocity? 65. Find the polar equation modeling the orbit of Jupiter. 66. Find the polar equation modeling the orbit of Saturn. 67. Find the polar equation modeling the orbit of Uranus. 68. Find the polar equation modeling the orbit of Neptune. 69. Suppose all four major planets arrived at the focal  chord of their orbit a

b simultaneously. Use 2 the equations in Exercises 65 to 68 to determine the distance between each of the planets at this moment. 70. The polar equation for the orbit of Pluto (a dwarf planet) was developed in Example 9. From an earlier exercise, the polar equation for the orbit of 2793 Neptune is r
. Using the 1  0.0111 cos
TABLE of your graphing calculator, determine if Pluto is always the farthest planet from the Sun. If not, how much further from the Sun is Neptune than Pluto at their perihelion? Mirror manufacturing: A modern manufacturer of oval (elliptical) mirrors for consumer use has programmed the equipment to automatically cut the glass for each mirror (major axis horizontal). The most popular mirrors are those that fit within a golden rectangle (ratio of L to W is approximately 1 to 0.618). Find the polar equation the manufacturer should use to program the equipment for mirror orders of the following lengths. Recall that c2
a2  b2 c and e
and assume one focus is at the pole. a

71. L
4 ft

72. L
3.5 ft

73. L
1.5 m

911

74. L
0.5 m

75. Referring to Exercises 71 to 74, find the total cost of each mirror (to the consumer) if they sell for $75 per square foot ($807 per square meter). The area of an ellipse is given by A
ab. 76. Referring to Exercises 71 to 74, find the total cost of an elliptical frame for each mirror (to the consumer) if the frame sells for $12.50 per linear foot ($41.01 per meter). The circumference of an ellipse is approximated by C
221a2  b2 2. 77. Home location: Candice Exercise 77 is an enthusiastic golfer R d H and an avid swimmer. Home After being transferred to River a new city, she decides to buy a house that is an d equal distance from the
local golf course and the G B river running through the Golf course city. If the distance between the river and the golf course at the closest point is 3 mi, find the polar equation of the parabola that will trace through the possible locations for her new home. Assume the golf course is at the focus of the parabola. 78. Home location: Referring to Exercise 77, assume Candice finds the perfect dream house in a  subdivision located at a6, b. Does this home fit 3 the criteria (is it an equal distance from the river and golf course)? 79. Solve the system below for y to verify the rotation formula for y given on page 898. e

X
x cos   y sin  Y
y cos   x sin 

80. Rotation of a conic section: Expand the following, collect like terms, and simplify. Show the result is the equation aX2  bXY  cY2  f
0, where the coefficients a, b, c, and f are as given on page 899. A1X cos   Y sin 2 2  B1X cos   Y sin 2 1X sin   Y cos 2  C1X sin   Y cos 2 2  F
0

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CHAPTER 9 Analytical Geometry and the Conic Sections

EXTENDING THE CONCEPT

81. Using the rotation of axes formulas in the general equation Ax2
Bxy
Cy2
F  0 1D  E  02, we were able to obtain the equation aX2
bXY
cY2
f  0 (see page 899), where a S A cos2
B sin  cos 
C sin2 b S 2A sin  cos 
B1cos2  sin22
2C sin  cos  c S A sin2  B sin  cos 
C cos2 and f S F b. Use these to verify a
c  A
C.

a. Use these to verify b2  4ac  B2  4AC.

82. A short-period comet is one that orbits the Sun in 200 yr or less. Two of the best known are Halley’s Comet and Encke’s Comet. Using any of the resources available to you, find the perihelion and aphelion of each comet and use the information to find the lengths of the semimajor and semiminor axes. Also find the period of each comet. If the length of an elliptical (orbital) path is approximated by L  220.51a2
b2 2, find the approximate average speed of each comet in miles per hour. Finally, determine the polar equation of each orbit.

For the given conics in the xy-plane, use a rotation of axes to find the corresponding equation in the XY-plane. See Exercises 31 and 32.

84. 12x2
24xy
5y2  40x  30y  25 85. 25x2
840xy  16y2  400  0 86. A right triangle in the xy-plane had vertices at (0, 0), (8, 0), and (8,6). Use the matrix equation X cos  sin  # x c d  c d c d to find the y Y sin  cos  vertices in the XY-plane after the triangle is rotated 60°.

83. In the r
-plane, the equation of a circle having radius R, center at 1R, 2, and going through the pole is given by r  2R cos1
 2. Consider the circle defined by x2
y2  6 12x  6 12y  0 in the xy-plane. Verify this circle goes through the origin, then find the equation of the circle in polar form.




c. Explain why the invariant f  F must always hold.

87. A square in the XY-plane has vertices at (0, 0), 12 13, 22, 12 13  2, 2
2 132 and 12, 2 132. Use the matrix equation x cos  sin  # X c d  c d c d to find the Y y sin  cos  vertices in the xy-plane after the triangle is rotated 30°.

MAINTAINING YOUR SKILLS 91. (7.3) A ship is moving at 12 mph on a heading of 325°, with a 5 mph current flowing at a 100° heading. Find the true course and speed of the ship.

88. (8.2) Solve the system using elimination. x
2y  z  3 • 2x  6y
z  4 5x
4y  2z  3

N

89. (4.5) Solve for x (to the nearest tenth): 21.7  77.5e0.0052x  44.95 90. (5.5) Use the graph shown to write an equation of the form y  A sec1Bx
C2. Clearly state the values of A, B, and C.

12 mph ship

y 5

100 325 

 2

 2

5



3 2

x

5 mph current

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Precalculus—

9.8 Parametric Equations and Graphs Learning Objectives

A large portion of the mathematics curriculum is devoted to functions, due to their overall importance and widespread applicability. But there are a host of applications for which nonfunctions are a more natural fit. In this section, we show that many nonfunctions can be expressed as parametric equations, where each is actually a function. These equations can be appreciated for the diversity and versatility they bring to the mathematical spectrum.

In Section 9.8 you will learn how to:

A. Sketch the graph of a parametric equation

B. Write parametric equations in rectangular form

A. Sketching a Curve Defined Parametrically

C. Graph curves from

Suppose you were given the set of points in the table here, and asked to come up with an equation model for the data. To begin, you might plot the points to see if any patterns or clues emerge, but in this case the result seems to be a curve we’ve never seen before (see Figure 9.64).

the cycloid family

D. Solve applications involving parametric equations

Figure 9.64 1

y

1

1

x

1

Figure 9.65 1

1

y

Lissajous figure

1

x

1

x

0

13 2

13 2

0

y

1

13 2

1 2

0



13 2



13 2

0

1 2



13 2

1



You also might consider running a regression on the data, but it’s not possible since the graph is obviously not a function. However, a closer look at the data reveals the y-values could be modeled independently of the x-values by a cosine function, y  cos t for t  3 0,  4. This observation leads to a closer look at the x-values, which we find could be modeled by a sine function over the same interval, namely, x  sin 12t2 for t  30, 4. These two functions combine to name all points on this curve, and both use the independent variable t called a parameter. The functions x  sin 12t2 and y  cos t are called the parametric equations for this curve. The complete curve, shown in Figure 9.65, is called a Lissajous figure, or a closed graph (coincident beginning and ending points) that crosses itself to form two or more loops. Note that since the maximum value of x and y is 1 (the amplitude of each function), the entire figure will fit within a 1  1 rectangle centered at the origin. This observation can often be used to help sketch parametric graphs with trigonometric parameters. In general, parametric equations can take many forms, including polynomial, exponential, trigonometric, and other forms. Parametric Equations Given the set of points P(x, y) such that x  f 1t2 and y  g1t2, where f and g are both defined on an interval of the domain, the equations x  f 1t2 and y  g1t2 are called parametric equations, with parameter t.

EXAMPLE 1



Graphing a Parametric Curve Where f and g Are Algebraic Graph the curve defined by the parametric equations x  t2  3 and y  2t  1.

Solution

9-83



Begin by creating a table of values using t  33, 3 4. After plotting ordered pairs (x, y), the result appears to be a parabola, opening to the right.

913

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CHAPTER 9 Analytical Geometry and the Conic Sections

y

x  t2  3

t 3

x

y  2t  1

6

5

2

1

3

1

2

1

0

3

1

1

2

3

2

1

5

3

6

7

Now try Exercises 7 through 12, Part a



If the parameter is a trig function, we’ll often use standard angles as inputs to simplify calculations and the period of the function(s) to help sketch the resulting graph. Also note that successive values of t give rise to a directional evolution of the graph, meaning the curve is traced out in a direction dictated by the points that correspond to the next value of t. The arrows drawn along the graph illustrate this direction, also known as the orientation of the graph. EXAMPLE 2



Graphing a Parametric Curve Where f and g Are Trig Functions Graph the curve defined by the parametric equations x  2 cos t and y  4 sin t.

Solution



Using standard angle inputs and knowing the maximum value of any x- and y-coordinate will be 2 and 4, respectively, we begin computing and graphing a few points. After going from 0 to , we note the graph appears to be a vertical ellipse. This is verified using standard values from  to 2. Plotting the points and connecting them with a smooth curve produces the ellipse shown in the figure. y

x

t

x  2 cos t

y  4 sin t

0  6  3  2 2 3 5 6

2

0

13

2

1

2 13

0

4

1

2 13

13

2

2

0

 A. You’ve just learned how to sketch the graph of a parametric equation

Now try Exercises 13 through 18, Part a Note the ellipse has a counterclockwise orientation.



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Section 9.8 Parametric Equations and Graphs

915

B. Writing Parametric Equations in Rectangular Form When graphing parametric equations, there are sometimes alternatives to simply plotting points. One alternative is to try and eliminate the parameter, writing the parametric equations in standard, rectangular form. To accomplish this we use some connection that allows us to “rejoin” the parameterized equations, such as variable t itself, a trigonometric identity, or some other connection. EXAMPLE 3

Solution





Eliminating the Parameter to Obtain the Rectangular Form Eliminate the parameter from the equations in Example 1: x  t2  3 and y  2t  1. y1 , which we then substitute into Solving for t in the second equation gives t  2 y1 2 1 the first. The result is x  a b  3  1y  12 2  3. Notice this is indeed a 2 4 horizontal parabola, opening to the right, with vertex at 13, 12. Now try Exercises 7 through 12, Part b

EXAMPLE 4





Eliminating the Parameter to Obtain the Rectangular Form Eliminate the parameter from the equations in Example 2: x  2 cos t and y  4 sin t.

Solution



Instead of trying to solve for t, we note the parametrized equations involve sine and cosine functions with the same argument (t), and opt to use the identity cos2t  sin2t  1. Squaring both equations and solving for cos2t and sin2t yields y2 y2 x2 x2  cos2t and  sin2t. This shows cos2t  sin2t    1, and as we 4 16 4 16 suspected—the result is a vertical ellipse with vertices at 10, 42 and endpoints of the minor axis at 12, 02. Now try Exercises 13 through 16, Part b



It’s important to realize that a given curve can be represented parametrically in infinitely many ways. This flexibility sometimes enables us to simplify the given form, or to write a given polynomial form in an equivalent nonpolynomial form. The easiest way to write the function y  f 1x2 in parametric form is x  t; y  f 1t2, which is valid as long as t is in the domain of f(t). EXAMPLE 5



Writing an Equation in Terms of Various Parameters Write the equation y  41x  32 2  1 in three different parametric forms.

Solution



1. If we let x  t, we have y  41t  32 2  1. 2. Letting x  t  3 simplifies the related equation for y, and we begin to see some of the advantages of using a parameter: x  t  3; y  4t2  1. 3. As a third alternative, we can let x 

B. You’ve just learned how to write parametric equations in rectangular form

x

1 tan t  3, which gives 2

2 1 1 tan t  3; y  4 a tan tb  1  tan2t  1 or y  sec2t. 2 2

Now try Exercises 19 through 26



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CHAPTER 9 Analytical Geometry and the Conic Sections

C. Graphing Curves from the Cycloid Family The cycloids are an important family of curves, and are used extensively to solve what are called brachistochrone applications. The name comes from the Greek brakhus, meaning short, and khronos, meaning time, and deal with finding the path along which a weight will fall in the shortest time possible. Cycloids are an excellent example of why parametric equations are important, as it’s very difficult to name them in rectangular form. Consider a point fixed to the circumference of a wheel as it rolls from left to right. If we trace the path of the point as the wheel rolls, the resulting curve is a cycloid. Figure 9.66 shows the location of the point every one-quarter turn. Figure 9.66

By superimposing a coordinate grid on the diagram in Figure 9.66, we can construct parametric equations that will produce the graph. This is done by developing equations for the location of a point P(x, y) on the circumference of a circle with center (h, k), as the circle rotates through angle t. After a rotation of t rad, the x-coordinate of P(x, y) is x  h  a (Figure 9.67), and the y-coordinate is y  k  b. Using a right a b triangle with the radius as the hypotenuse, we find sin t  and cos t  , giving r r a  r sin t and b  r cos t. Substituting into x  h  a and y  k  b yields x  h  r sin t and y  k  r cos t. Since the circle has radius r, we know k  r (the “height” of the center is constantly k  r). The arc length subtended by t is the same as the distance h (see Figure 9.68), meaning h  rt (t in radians) Substituting rt for h and r for k in the equations x  h  r sin t and y  k  r cos t, gives the equation of the cycloid in parametric form: x  rt  r sin t and y  r  r cos t , sometimes written x  r 1t  sin t2 and y  r 11  cos t2. Figure 9.67

Figure 9.68

y

(h, k) r k

P(x, y)

t

b

a

h

x

Most graphers have a parametric MODE that enables you to enter the equations for x and y separately, and graph the resulting points as a single curve. After pressing the Y= key (in parametric mode), the screen in Figure 9.68 comes into view using a TI-84 Plus, and we enter the equation of the cycloid formed by a circle of radius r  3. To set the viewing window (including a frame), press WINDOW and set Ymin  1 and Ymax at slightly more than 6 (since r  3). Since the cycloid completes one cycle every 2r, we set Xmax at 2rn, where n is the number of cycles we’d like to see. In

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917

Section 9.8 Parametric Equations and Graphs

this case, we set it for four cycles 122132142  24 (Figure 9.69). With r  3 we conveniently set Xscl at 3122  6  18.8 to tick each cycle, and Xscl  3  9.4 to tick each half cycle (Figure 9.69). For parametric equations, we must also specify a range of values for t, which we set at Tmin  0, Tmax  8  25.1 for the four  cycles, and Tstep   0.52 (Tstep controls the number of points plotted and joined 6 to form the curve). The window settings and resulting graph are shown in Figure 9.70, which doesn’t look much like a cycloid because the current settings do not produce a square viewing window. Using ZOOM 5:ZSquare (and changing Yscl) produces the graph shown in Figure 9.71, which looks much more like the cycloid we expected.

Figure 9.69

Figure 9.70

Figure 9.71

7

28.9

␲

24␲

␲

1



EXAMPLE 6

24␲

22.9

Using Technology to Graph a Cycloid Use a graphing calculator to graph the curve defined by the equations x  3 cos3t and y  3 sin3t, called a hypocycloid with four cusps.



Solution Figure 9.72 r y

A hypocycloid is a curve traced out by the path of a point on the circumference of a circle as it rolls inside a larger circle of radius r (see Figure 9.72). Here r  3 and we set Xmax and Ymax accordingly. Knowing ahead of time the hypocycloid will have four cusps, we set Tmax  4122  25.13 to show all four. The window settings used and the resulting graph are shown in Figure 9.73 and 9.74. Figure 9.74

x

r

Figure 9.73

5

r

5

5

r

C. You’ve just learned how to graph curves from the cycloid family

5

Now try Exercises 27 through 35



D. Common Applications of Parametric Equations In Example 1 the parameter was simply the real number t, which enabled us to model the x- and y-values of an ordered pair (x, y) independently. In Examples 2 and 6, the parameter t represented an angle. Here we introduce yet another kind of parameter, that of time t. A projectile is any object thrown, dropped, or projected in some way with no continuing source of propulsion. The parabolic path traced out by the projectile (assuming negligible air resistance) will be fully developed in Section 7.4. It is stated here in

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CHAPTER 9 Analytical Geometry and the Conic Sections

parametric terms. For the projectile’s location P(x, y) and any time t in seconds, the x-coordinate (horizontal distance from point of projection) is given by x  v0t cos , where v0 is the initial velocity in feet per second and t is the time in seconds. The y-coordinate (vertical height) is y  v0t sin   16t2. EXAMPLE 7



Using Parametric Equations in Projectile Applications As part of a circus act, Karl the Human Cannonball is shot out of a specially designed cannon at an angle of 40° with an initial velocity of 120 ft/sec. Use a graphing calculator to graph the resulting parametric curve. Then use the graph to determine how high the Ring Master must place a circular ring for Karl to be shot through at the maximum height of his trajectory, and how far away the net must be placed to catch Karl.

Solution



The information given leads to the equations x  120t cos 40° and y  120t sin 40°  16t2. Enter these equations on the Y = screen of your calculator, remembering to reset the MODE to degrees (circus clowns may not know or understand radians). To set the window size, we can use trial and error, or estimate using   45° (instead of 40°) and an estimate for t (the time that Karl 12 b  360 12 for will stay aloft). With t  6 we get estimates of x  120162 a 2 the horizontal distance. To find a range for y, use t  3 since the maximum height of the parabolic path will occur halfway through the flight. This gives an estimate 12 b  16192  180 12  144 for y. The results are shown in Figures of 120132 a 2 9.75 and 9.76. Using the TRACE feature or 2nd GRAPH (TABLE) feature, we find the center of the net used to catch Karl should be set at a distance of about 450 ft from the cannon, and the ring should be located 220 ft from the cannon at a height of about 93 ft. Figure 9.76 Figure 9.75

0

0

509

100

Now try Exercises 46 through 49



It is well known that planets orbit the Sun in elliptical paths. While we’re able to model their orbits in both rectangular and polar form, neither of these forms can give a true picture of the direction they travel. This gives parametric forms a great advantage, in that they can model the shape of the orbit, while also indicating the direction of travel. We illustrate in Example 8 using a “planet” with a very simple orbit.

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EXAMPLE 8




Modeling Elliptical Orbits Parametrically The elliptical orbit of a certain planet is defined parametrically as x
4 sin t and y
3 cos t. Graph the orbit and verify that for increasing values of t, the planet orbits in a counterclockwise direction.

Solution




y2 x2 
1, 16 9 or the equation of an ellipse with center at (0, 0), major axis of length 8, and minor axis of length 6. The path of the planet is traced out by the ordered pairs (x, y) generated by the parametric equations, shown in the table for t
3 0,
4 . Starting at t
0, P(x, y)

begins at 10, 32 with x and y both increasing until t
. Then from t
to 2 2 t

, y continues to increase as x decreases, indicating a counterclockwise orbit in this case. The orbit is illustrated in the figure. Eliminating the parameter as in Example 4, we obtain the equation

y

(4, 0) x (3.46, 1.5) (0, 3)

(2, 2.6)

t

x  4 sin t

y  3 cos t

0
6
3
2 2
3 5
6

0

3

2

2.6

3.46

1.5

4

0

3.46

1.5

2

2.6

0

3




Now try Exercises 50 and 51

D. You’ve just learned how to solve applications involving parametric equations




Finally, you may recall from your previous work with linear 3  3 systems, that a dependent system occurs when one of the three equations is a linear combination of the other two. The result is a system with more variables than equations, with solutions expressed in terms of a parameter, or in parametric form. These solutions can be explored on a graphing calculator using ordered triples of the form (t, f(t), g(t)), where Y1
f 1t2 and Y2
g1t2 (see Exercises 52 through 55). For more information, see the Calculator Exploration and Discovery feature on page 931.

TECHNOLOGY HIGHLIGHT

Exploring Parametric Graphs Most graphing calculators have features that make it easy (and fun) to explore parametric equations. For example, the TI-84 Plus can use a circular cursor to trace the path of the plotted points, as they are generated by the equations. This can be used to illustrate the path of a projectile, the distance of a runner, or the orbit of a planet. Operations can also be applied to the parameter T to give the effect of “speed” (the points from one set of equations are plotted faster than the points of a second set). To help illustrate their use, consider again the simple, elliptical orbit of a planet in Example 8. Physics tells us the closer a planet is to the Sun, the faster its orbit. In fact, the orbital speed of Mercury is about twice that of Mars and about 10 times as fast as the dwarf planet Pluto (29.8, 15, and 2.9 mi/sec,

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respectively). With this information, we can explore a number of Figure 9.77 interesting questions. On the Y = screen, let the orbits of Planet 1 and Planet 2 be modeled parametrically by the equations shown in Figure 9.77. Since the orbit of Planet 1 is “smaller” (closer to the Sun), we have T-values growing at a rate that is four times as fast as for Planet 2. Notice to the far left of X1T, there is a symbol that looks like an old key “0.” By moving the cursor to the far left of the equation, you can change how the graph will look by repeatedly pressing ENTER . With this symbol in view, the calculator will trace out the curve with a circular cursor, which in this case represents the planets as they orbit (be sure you are in simultaneous MODE ). Setting the window as in Figure 9.78 and pressing GRAPH produces Figure 9.79, which displays their elliptical paths as they race around the Sun. Notice the inner planet has already completed one orbit while the outer planet has just completed one-fourth of an orbit.

Figure 9.79

Figure 9.78

10

10

10

10

Exercise 1: Verify that the inner planet completes four orbits for every single orbit of the outer planet. Exercise 2: Suppose that due to some cosmic interference, the orbit of the faster planet begins to decay at a rate of T0.84 (replace T with T0.84 in both equations for the inner planet). By observation, about how many orbits did the inner planet make for the first revolution of the outer planet? What is the ratio of orbits for the next complete orbit of the outer planet?

9.8 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section if needed.

1. When the coordinates of a point (x, y) are generated independently using x  f 1t2 and y  g1t2 , t is called a(n) .

2. The equations x  f 1t2 and y  g1t2 used to generate the ordered pairs (x, y) are called equations. 3. Parametric equations can both graph a curve and indicate the traveled by a point on the curve.

4. To write parametric equations in rectangular form, we must the parameter to write a single equation. 5. Discuss the connection between solutions to dependent systems and the parametric equations studied in this section. 6. In your own words, explain and illustrate the process used to develop the equation of a cycloid. Illustrate with a specific example.

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921

DEVELOPING YOUR SKILLS

For Exercises 7 through 18, (a) graph the curves defined by the parametric equations using the specified interval and identify the graph (if possible) and (b) eliminate the parameter (Exercises 7 to 16 only) and write the corresponding rectangular form.

7. x  t  2; t  3 3, 34 y  t2  1

27. x  8 cos t  2 cos14t2, y  8 sin t  2 sin14t2, hypocycloid (5-cusp)

8. x  t  3; t  3 5, 54 y  2  0.5t2

28. x  8 cos t  4 cos12t2, y  8 sin t  4 sin12t2, hypocycloid (3-cusp)

9. x  12  t2 2; t  3 0, 54 y  1t  32 2

29. x 

10. x  t3  3; t  3 2, 2.5 4 y  t2  1 5 11. x  , t  0; t  33.5, 3.5 4 t y  t2 3

t ; t  3 5, 5 4 10 y  t

12. x 

 3 d 16. x  4 cos12t2; t  c , 2 2 y  6 sin t 3 ; t  10, 2 17. x  tan t y  5 sin12t2  18. x  tan2t; t  , t  3 0,  4 2 y  3 cos t Write each function in three different parametric forms by altering the parameter. For Exercises 19–22 use at least one trigonometric form, restricting the domain as needed.

23. y  tan 1x  22  1 2

8 sin3t , cissoid of Diocles cos t

31. x  21cos t  t sin t2, y  21sin t  t cos t2, involute of a circle 32. 4x  116  362cos3t, 6y  116  362sin3t, evolute of an ellipse

35. x  23 3 cos t  cos13t2 4 , y  233 sin t  sin13t2 4, nephroid

15. x  4 sin12t2; t  3 0, 22 y  6 cos t

21. y  1x  32  1

30. x  8 sin2t, y 

34. x  t  3 sin t, y  1  3 cos t, prolate cycloid

14. x  2 sin t; t  3 0, 22 y  3 cos t

2

2 , y  8 sin t cos t, serpentine curve tan t

33. x  3t  sin t, y  3  cos t, curtate cycloid

13. x  4 cos t; t  3 0, 22 y  3 sin t

19. y  3x  2

The curves defined by the following parametric equations are from the cycloid family. (a) Use a graphing calculator or computer to draw the graph and (b) use the graph to approximate all x- and y-intercepts, and maximum and minimum values to one decimal place.

20. y  0.5x  6 22. y  21x  52 2  1 24. y  sin12x  12

25. Use a graphing calculator or computer to verify that the parametric equations from Example 5 all produce the same graph. 26. Use a graphing calculator or computer to verify that your parametric equations from Exercise 21 all produce the same graph.

Use a graphing calculator or computer to draw the following parametrically defined graphs, called Lissajous figures (Exercise 37 is a scaled version of the initial example from this section). Then find the dimensions of the rectangle necessary to frame the figure and state the number of times the graph crosses itself.

36. x  6 sin13t2 y  8 cos t

37. x  6 sin12t2 y  8 cos t

38. x  8 sin14t2 y  10 cos t

39. x  5 sin17t2 y  7 cos14t2

40. x  8 sin14t2 y  10 cos13t2

41. x  10 sin11.5t2 y  10 cos12.5t2

42. Use a graphing calculator to experiment with parametric equations of the form x  A sin1mt2 and y  B cos1nt2. Try different values of A, B, m, and n, then discuss their effect on the Lissajous figures. 43. Use a graphing calculator to experiment with a parametric equations of the form x  and tan t y  b sin t cos t. Try different values of a and b, then discuss their effect on the resulting graph, called a serpentine curve. Also see Exercise 29.

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44. The Folium of Descartes: 3kt 2 3kt ; y1t2  x1t2  1  t3 1  t3 The Folium of Descartes is a parametric curve developed by Descartes in order to test the ability of Fermat to find its maximum and minimum values. a. Graph the curve on a graphing calculator with k  1 using a reduced window ( ZOOM 4), with Tmin  6, Tmax  6, and Tstep  0.1. Locate the coordinates of the tip of the folium (the loop). b. This graph actually has a discontinuity (a break in the graph). At what value of t does this occur? c. Experiment with different values of k and generalize its effect on the basic graph.



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45. The Witch of Agnesi: x1t2  2kt; y1t2 

2k 1  t2

The Witch of Agnesi is a parametric curve named by Maria Agnesi in 1748. Some believe she confused the Italian word for witch (versiera), with a similar word that meant free to move. In any case, the name stuck. The curve can also be stated in trigonometric form: x1t2  2k cot t and y  2k sin2t. a. Graph the curve with k  1 on a calculator or computer on a reduced window ( ZOOM 4) using both of the forms shown with Tmin  6, Tmax  6, and Tstep  0.1. Try to determine the maximum value. b. Explain why the x-axis is a horizontal asymptote. c. Experiment with different values of k and generalize its effect on the basic graph.

APPLICATIONS

Model each application using parametric equations, then solve using the GRAPH and TRACE features of a graphing calculator.

46. Archery competition: At an archery contest, a large circular target 5 ft in diameter is laid flat on the ground with the bull’s-eye exactly 180 yd (540 ft) away from the archers. Marion draws her bow and shoots an arrow at an angle of 25° above horizontal with an initial velocity of 150 ft/sec (assume the archers are standing in a depression and the arrow is shot from ground level). (a) What was the maximum height of the arrow? (b) Does the arrow hit the target? (c) What is the distance between Marion’s arrow and the bull’s-eye after the arrow hits?

47. Football competition: As part of their contribution to charity, a group of college quarterbacks participate in a contest. The object is to throw a football through a hoop whose center is 30 ft high and 25 yd (75 ft) away, trying to hit a stationary (circular) target laid on the ground with the center

56 yd (168 ft) away. The hoop and target both have a diameter of 4 ft. On his turn, Lance throws the football at an angle of 36° with an initial velocity of 75 ft/sec. (a) Does the football make it through the hoop? (b) Does the ball hit the target? (c) What is the approximate distance between the football and the center of the target when the ball hits the ground? 48. Walk-off home run: It’s the bottom of the ninth, two outs, the count is full, and the bases are loaded with the opposing team ahead 5 to 2. The home team has Heavy Harley, their best hitter at the plate; the opposition has Raymond the Rocket on the mound. Here’s the pitch . . . it’s hit . . . a long fly ball to left-center field! If the ball left the bat at an angle of 30° with an initial velocity of 112 ft/sec, will it clear the home run fence, 9 ft high and 320 ft away? 49. Last-second win: It’s fourth-and-long, late in the fourth quarter of the homecoming football game, with the home team trailing 29 to 27. The coach elects to kick a field goal, even though the goal posts are 50 yd (150 ft) away from the spot of the kick. If the ball leaves the kicker’s foot at an angle

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of 29° with an initial velocity of 80 ft/sec, and the kick is “true,” will the home team win (does the ball clear the 10-ft high cross bar)?

50 yards (150 feet)

50. Particle motion: The motion of a particle is modeled by the parametric equations x  5t  2t2 . Between t  0 and t  1, is the y  3t  2 particle moving to the right or to the left? Is the particle moving upward or downward? e

51. Electron motion: The motion of an electron as it orbits the nucleus is modeled by the parametric x  6 cos t equations e with t in radians. Between y  2 sin t t  2 and t  3, is the electron moving to the right or to the left? Is the electron moving upward or downward? Systems applications: Solve the following systems using elimination. If the system is dependent, write the general solution in parametric form and use a calculator to generate several solutions.

2x  y  3z  3 52. • 3x  2y  z  4 8x  3y  z  5

x  5y  z  3 53. • 5x  y  7z  9 2x  3y  4z  6

5x  3z  1 54. • x  2y  2z  3 2x  6y  9z  10 x  y  5z  4 55. • 2y  3z  1 x  3y  z  3



923

56. Regressions and parameters: x y Draw a scatter-plot of the data 0 0 given in the table. Note that 12 0.25 connecting the points with a smooth curve will not result 2 2 in a function, so a standard 12 6.75 regression cannot be run on 0 16 the data. Now consider the  12 31.25 x-values alone—what do 2 54 you notice? Find a sinusoidal 85.75 12 model for the x-values, using T  0, 1, 2, 3, . . . , 8. Use the 0 128 same inputs to run some form of regression on the y-values, then use the results to form the “best-fit” parametric equations for this data (use L1 for T, L2 for the x-values, and L3 for the y-values). With your calculator in parametric MODE , enter the equations as X1T and Y1T, then graph these along with the scatterplot (L2, L3) to see the finished result. Use the TABLE feature of your calculator to comment on the accuracy of the model. 57. Regressions and parameters: x y Draw a scatter-plot of the data 1 0 given in the table, and connect 1.75 the points with a smooth curve. 1.2247 The result is a function, but no 3 1.5 standard regression seems to 3.75 1.8371 give an accurate model. The 4 2.25 x-values alone are actually 3.75 2.7557 generated by an exponential 3 3.375 function. Run a regression 4.1335 1.75 on these values using T  0, 1, 2, 3, . . . , 8 as inputs 5.0625 0 to find the exponential model. Then use the same inputs to run some form of regression on the y-values and use the results to form the “best-fit” parametric equations for this data (use L1 for T, L2 for the x-values, and L3 for the y-values). With your calculator in parametric MODE , enter the equations as X1T and Y1T, then graph these along with the scatterplot (L2, L3) to see the finished result. Use the TABLE feature of your calculator to comment on the accuracy of the model.

EXTENDING THE CONCEPT

58. What is the difference between an epicycloid, a hypercycloid, and a hypocycloid? Do a word study on the prefixes epi-, hyper-, and hypo-, and see

how their meanings match with the mathematical figures graphed in Exercises 27 to 35. To what other shapes or figures are these prefixes applied?

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59. The motion of a particle in a certain medium is x  6 sin14t2 . modeled by the parametric equations e y  8 cos t GRAPH (TABLE) Initially, use only the 2nd feature of your calculator (not the graph) to name the intervals for which the particle is moving (a) to the left and upward and (b) to the left and downward. Answer to the nearest tenth (set ¢Tbl  0.1). Is it possible for this particle to collide with another particle in this medium whose



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CHAPTER 9 Analytical Geometry and the Conic Sections

x  3 cos t  7 movement is modeled by e ? Discuss y  2 sin t  2 why or why not. 1 1x  32 2  1 in 2 parametric form using the substitution x  2 cos t  3 and the appropriate double-angle identity. Is the result equivalent to the original function? Why or why not?

60. Write the function y 

MAINTAINING YOUR SKILLS

61. (1.1) The price of a popular video game is reduced by 20% and is selling for $39.96. By what percentage must the sale price be increased to return the item to its original price? 62. (5.2) When the tip of the antenna atop the Eiffel Tower is viewed at a distance of 265 ft from its base, the angle of elevation is 76°. Is the Eiffel Tower taller or shorter than the Chrysler Building (New York City) at 1046 ft?

63. (3.4) Graph f 1x2  x3  2x2  5x  6 using information about end behavior, y-intercept, x-intercept(s), and midinterval points: 64. (6.6) The maximum height a projectile will attain depends on the angle it is projected and its initial velocity. This phenomena is modeled by the v2 sin2 , where v is the initial function H  64 velocity (in feet/sec) of the projectile and  is the angle of projection. Find the angle of projection if the projectile attained a maximum height of 151 ft, and the initial velocity was 120 ft/sec.

S U M M A RY A N D C O N C E P T R E V I E W SECTION 9.1

A Brief Introduction to Analytical Geometry

KEY CONCEPTS • The midpoint and distance formulas play an important role in the study of analytical geometry: x2  x1 y2  y1 , b distance: d  21x2  x1 2 2  1y2  y1 2 2 midpoint: 1x, y2  a 2 2

• The perpendicular distance from a point to a line is the length of a line segment perpendicular to a given line with the given point and the point of intersection as endpoints. • Using these tools, we can verify or construct relationships between points, lines, and curves in the plane; verify properties of geometric figures; prove theorems from Euclidean geometry; and construct relationships that define the conic sections.

EXERCISES 1. Verify the closed figure with vertices (3, 4), (5, 4), (3, 6), and (5, 2) is a square. 2. Find the equation of the circle that circumscribes the square in Exercise 1. 3. A theorem from Euclidean geometry states: If any two points are equidistant from the endpoints of a line segment, they are on the perpendicular bisector of the segment. Determine if the line through (3, 6) and (6, 9) is a perpendicular bisector of the segment through (5, 2) and (5, 4). 4. Four points are given below. Verify that the distance from each point to the line y  1 is the same as the distance from the given point to the fixed point (0, 1): (6, 9), (2, 1), (4, 4), and (8, 16).

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Summary and Concept Review

The Circle and the Ellipse

KEY CONCEPTS • The equation of a circle centered at (h, k) with radius r is 1x  h2 2  1y  k2 2  r2.

• Dividing both sides by r 2, we obtain the standard form vertical distance from center to graph is r.

1x  h2 2

1x  h2 2 r2

1y  k2 2

1y  k2 2 r2

 1, showing the horizontal and

 1. The center of the ellipse is (h, k), with a2 b2 horizontal distance a and vertical distance b from center to graph. y • Given two fixed points f1 and f2 in a plane (called the foci), an ellipse is the set of all points (x, y) such that the distance from the first focus to (x, y), plus the distance from (x, y) the second focus to (x, y), remains constant. d1 d2 (a, 0) (a, 0) For an ellipse, the distance a from center to vertex is greater than the distance c from • x (c, 0) (c, 0) center to one focus. d 1  d2  k • To find the foci of an ellipse: a2  b2  c2 (since a 7 c).

• The equation of an ellipse in standard form is





EXERCISES Sketch the graph of each equation in Exercises 5 through 9. 5. x2  y2  16 6. x2  4y2  36 7. 9x2  y2  18x  27  0 1y  22 2 1x  32 2  1 8. x2  y2  6x  4y  12  0 9. 16 9 10. Find the equation of the ellipse with minor axis of length 6 and foci at (4, 0) and (4, 0). 11. Find the equation of the ellipse with vertices at (a) (13, 0) and (13, 0), foci at (12, 0) and (12, 0); (b) foci at (0, 16) and (0, 16), major axis: 40 units. 12. Write the equation in standard form and sketch the graph, noting all of the characteristic features of the ellipse. 4x2  25y2  16x  50y  59  0

SECTION 9.3

The Hyperbola

KEY CONCEPTS 1x  h2 2 1y  k2 2   1. The center of the hyperbola • The equation of a horizontal hyperbola in standard form is a2 b2 is (h, k) with horizontal distance a from center to vertices and vertical distance b from center to the midpoint of one side of the central rectangle. y • Given two fixed points f1 and f2 in a plane (called the foci), a hyperbola is the set of all (x, y) d1 points (x, y) such that the distance from the first focus to point (x, y), less the distance d 2 (a, 0) (a, 0) from the second focus to (x, y), remains constant. (c, 0) (c, 0) x • For a hyperbola, the distance from center to one of the vertices is less than the distance from center to one focus. • To find the foci of a hyperbola: c2  a2  b2 (since c 7 a). d1  d2  k

EXERCISES Sketch the graph of each equation, indicating the center, vertices, and asymptotes. For Exercise 18, also give the equation of the hyperbola in standard form. 1y  32 2 1x  22 2 1y  12 2 1x  22 2  1  1 13. 4y2  25x2  100 14. 15. 16 9 9 4

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16. 9y2  x2  18y  72  0

17. x2  4y2  12x  8y  16  0

4 18. vertices at (3, 0) and (3, 0), asymptotes of y   x 3 19. Find the equation of the hyperbola with (a) vertices at (15, 0), foci at (17, 0), and (b) foci at (0, 5) with vertical dimension of central rectangle 8 units. 20. Write the equation in standard form and sketch the graph, noting all of the characteristic features of the hyperbola. 4x2  9y2  40x  36y  28  0

SECTION 9.4

The Analytic Parabola

KEY CONCEPTS • Horizontal parabolas have equations of the form x  ay2  by  c; a  0.

• A horizontal parabola will open to the right if a 7 0, and to the left if a 6 0. The axis of symmetry is y 

b , 2a

b or by completing the square and 2a y writing the equation in shifted form: x  a1y  k2 2  h. Given a fixed point f (called the focus) and fixed line D in the plane, a parabola is the set d1 (x, y) of all points (x, y) such that the distance from f to (x, y) is equal to the distance from (x, y) f d 1  d2 d2 to line D. Vertex x D The equation x2  4py describes a vertical parabola, opening upward if p 7 0, and opening downward if p 6 0. The equation y2  4px describes a horizontal parabola, opening to the right if p 7 0, and opening to the left if p 6 0. The focal chord of a parabola is a line segment that contains the focus and is parallel the directrix, with its endpoints on the graph. It has a total length of 4p, meaning the distance from the focus to a point of the graph is 2p. It is commonly used to assist in drawing a graph of the parabola. with the vertex (h, k) found by evaluating at y 

• • • •

EXERCISES For Exercises 21 and 22, find the vertex and x- and y-intercepts if they exist. Then sketch the graph using symmetry and a few points or by completing the square and shifting a parent function. 21. x  y2  4 22. x  y2  y  6 For Exercises 23 and 24, find the vertex, focus, and directrix for each parabola. Then sketch the graph using this information and the focal chord. Also graph the directrix. 23. x2  20y

SECTION 9.5

24. x2  8x  8y  16  0

Nonlinear Systems of Equations and Inequalities

KEY CONCEPTS • Nonlinear systems of equations can be solved using substitution or elimination. • First identify the graphs of the equations in the system to help determine the number of solutions possible. • For nonlinear systems of inequalities, graph the related equation for each inequality given, then use a test point to decide what region to shade as the solution. • The solution for the system is the overlapping region (if it exists) created by the areas shaded for the individual inequalities. • If the boundary is included, graph it using a solid line; if the boundary is not included (for strict inequalities) use a dashed line.

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927

EXERCISES Solve Exercises 25–30 using substitution or elimination. Identify the graph of each relation before you begin. x2  y2  25 y  x  1 x2  y  1 27. e 2 x  y2  7 25. e

29. e

y  x2  2 x2  y2  16

SECTION 9.6

x  y2  1 x  4y  5 x2  y2  10 28. e y  3x2  0 26. e

30. e

x2  y2 7 9 x2  y  3

Polar Coordinates, Equations, and Graphs

KEY CONCEPTS P(r, ␪) • In polar coordinates, the location of a point in the plane is denoted 1r, 2, where r is the r0 distance to the point from the origin or pole, and  is the angle between a stipulated polar axis and a ray containing P. r • In the polar coordinate system, the location 1r, 2 of a point is not unique for two reasons: (1) the angles  and   2n are coterminal (n an integer), and (2) r may be negative. ␪ Pole Polar axis The point can be converted to P(x, y) in rectangular coordinates where P1r, 2 x  r cos  • and y  r sin . • The point P(x, y) in rectangular coordinates can be converted to P1r, 2 in polar coordinates, where r  2x2  y2 y and r  tan1a b. x • To sketch a polar graph, we view the length r as being along the second hand of a clock, ticking in a  counterclockwise direction. Each “tick” is rad or 15°. For each tick we locate a point on the radius and plot it 12 on the face of the clock before going on. • For graphing, we also apply an “r-value” analysis, which looks where r is increasing, decreasing, zero, maximized, and/or minimized.  • If the polar equation is given in terms of sines, the graph will be symmetric to   . 2 • If the polar equation is given in terms of cosines, the graph will be symmetric to the polar axis. • The graphs of several common polar equations are given in Appendix V. EXERCISES Sketch using an r-value analysis (include a table), symmetry, and any convenient points. 31. r  5 sin  32. r  4  4 cos  33. r  2  4 cos  34. r  8 sin122

SECTION 9.7

More on the Conic Sections: Rotation of Axes and Polar Form

KEY CONCEPTS • Using a rotation, the conic equation Ax2  Bxy  Cy2  Dx  Ey  F  0 in the xy-plane can be transformed into aX2  cY2  dX  eY  f  0 in the XY-plane, in which the mixed xy-term is absent. B • The required angle of rotation  is found using tan122  ; 0 6 2 6 180°. AC • The change in coordinates from the xy-plane to the XY-plane is accomplished using the rotation formulas: x  X cos   Y sin 

y  X sin   Y cos 

• In the process of this conversion, certain quantities, called invariants, remain unchanged and can be used to

check that the conversion was correctly performed. These invariants are (1) F  f, (2) A  C  a  c, and (3) B2  4AC  b2  4ac.

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• The invariants B2  4AC  b2  4ac are called discriminants and can be used to classify the type of graph the

equation will give, except in degenerate cases: • If B2  4AC  0, the equation is that of a parabola. • If B2  4AC 6 0, the equation is that of a circle or an ellipse. • If B2  4AC 7 0, the equation is that of a hyperbola. • All conics (not only the parabola) can be stated in terms of a focus/directrix definition. This is done using the concept of eccentricity, symbolized by the letter e. FP • If F is a fixed point and l a fixed line in the plane with the point D on l, the set of all points P such that e DP (e a constant) is the graph of a conic section. If e  1, the graph is a parabola. If 0 6 e 6 1, the graph is an ellipse. If e 7 1, the graph is a hyperbola. • Given a conic section with eccentricity e, one focus at the pole of the r-plane, and directrix l located d units de de from this focus, then the polar equations r  and r  represent one of the conic sections 1  e cos  1  e sin  as determined by the value of e.

EXERCISES For the given conics in the xy-plane, use a rotation of axes to find the corresponding equation in the XY-plane, then sketch its graph. 35. 2x2  4xy  2y2  8 12y  24  0 36. x2  6 13xy  7y2  160  0 For the conic equations given, determine if the equation represents a parabola, ellipse, or hyperbola. Then describe and sketch the graphs using polar graph paper. 4 9 8 37. r  38. r  39. r  3  2 cos  4  6 cos  3  3 sin  40. Mars has a perihelion of 128.4 million miles and an aphelion of 154.9 million miles. Use this information to find a polar equation that models the elliptical orbit, then find the length of the focal chord.

SECTION 9.8

Parametric Equations and Graphs

KEY CONCEPTS • If we consider the set of points P(x, y) such that the x-values are generated by f (t) and the y-values are generated by g(t) (assuming f and g are both defined on an interval of the domain), the equations x  f 1t2 and y  g1t2 are called parametric equations, with parameter t. • Parametric equations can be converted to rectangular form by eliminating the parameter. This can sometimes be done by solving for t in one equation and substituting in the other, or by using trigonometric forms. • A function can be written in parametric form many different ways, by altering the parameter or using trigonometric identities. • The cycloids are an important family of curves, with equations x  r1t  sin t2 and y  r11  cos t2. • The solutions to dependent systems of equations are often expression in parametric form, with the points P(x, y) given by the parametric equations generating solutions to the system. EXERCISES Graph the curves defined by the parametric equations over the specified interval and identify the graph. Then eliminate the parameter and write the corresponding rectangular form. 41. x  t  4: t  3 3, 3 4: 42. x  12  t2 2: t  30, 5 4: 43. x  3 sin t: t  30, 22: 2 y  1t  32 2 y  4 cos t y  2t  3 44. Write the function in three different forms by altering the parameter: y  21x  52 2  1 45. Use a graphing calculator to graph the Lissajous figure indicated, then state the size of the rectangle needed to frame it: x  4 sin15t2; y  8 cos t

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Mixed Review

929

MIXED REVIEW For Exercises 1 through 16, graph the conic section and locate the center, vertices, directrix, foci, focal chords, asymptotes, and other important features as these apply to a particular equation and conic.

a. 4.5

1. 9x2  9y2  54 2. 16x2  25y2  400 3. 9y2  25x2  225 1x  32 2 1y  12 2  1 4. 9 25 5. 41x  12 2  361y  22 2  144

b.

6. 161x  22  41y  12  64 2

2

7. y  2x2  10x  15 8. x  y2  8y  11

(4.5, 0)

9. x  y  2y  3 2

10. x  1y  22 2  3

11. x2  8x  8y  16  0 12. x2  24y 13. 4x2  25y2  24x  150y  289  0

c.

14. 4x  16y  12x  48y  19  0 2

2

15. 491x  22 2  1y  32 2  49

16. x2  y2  8x  12y  16  0

(4.5, 0)

17. Graph the curve defined by the parametric equations given, using the interval t  3 0, 10 4 . Then identify the graph: x  1t  22 2, y  1t  42 2 18. Plot the polar coordinates given, then convert to rectangular coordinates. 2 5 b b a. a3.5, b. a4, 3 4 19. Solve using elimination: 4x2  y2  9 a. e 2 x  3y2  79

b. e

4x2  9y2  36 x2  3y  6

20. Match each equation to its corresponding graph. Justify each response. (i) r  3.5  cos  (ii) r2  20.25 sin122 (iii) r  4.5 cos 

21. A go-cart travels around an elliptical track with a 100-m major axis that is horizontal. The minor axis measures 60 m. Write an equation model for the track in parametric form. 22. Except for small variations, a planet’s orbit around the Sun is elliptical, with the Sun at one focus. The perihelion or minimum distance from the planet Mercury to the Sun is about 46 million kilometers. Its aphelion or maximum distance from the Sun is approximately 70 million kilometers. Use this information to find the length of the major and minor axes, then determine the equation model for the orbit of Mercury in the standard form y2 x2   1. a2 b2

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CHAPTER 9 Analytical Geometry and the Conic Sections

23. The orbit of a comet can also be modeled by one of the conic sections, with the Sun at one focus. Assuming the equations given model a comet’s path, (1) determine if the path is circular, elliptic, hyperbolic, or parabolic; and (2) determine the closest distance the comet will come to the Sun (in millions of miles). 84 31 a. r  b. r  100  70 cos  5  5 sin  24. In the design of their corporate headquarters, Centurion Computing includes a seven-leaf rose in a large foyer, with a fountain in the center. Each of the

leaves is 5 m long (when measured from the center of the fountain), and will hold flower beds for carefully chosen perennials. The rose is to be symmetric to a vertical axis, with the leaf bisected  by   pointing directly to the elevators. Find the 2 equation of the rose in polar form. Y2 X2   1 in the 802 4002 XY-plane is rotated clockwise by 45°. What is the corresponding equation in the xy-plane?

25. The hyperbola defined by

PRACTICE TEST By inspection only (no graphing), match each equation to its correct description. 1. x2  y2  6x  4y  9  0

16. x  4 sin t y  5 cos t

2. 4y2  x2  4x  8y  20  0 3. x2  4y2  4x  12y  20  0 4. y  x2  4x  20  0 a. Parabola b. Hyperbola

c. Circle

d. Ellipse

Identify and then graph each of the following conic sections. State the center, vertices, foci, asymptotes, and other important points when applicable. 5. x2  y2  4x  10y  20  0 6. 251x  22 2  41y  12 2  100 12 10 8. r  5  4 cos  5  5 cos  1y  32 2 1x  22 2  1 9. 9 16 7. r 

10. 41x  12  251y  22  100 2

2

Use the equation 80x2  120xy  45y2  100y  44  0 to complete Exercises 11 and 12. 11. Use the discriminant B2  4AC to identify the B graph, and tan122  to find cos  and sin . AC 12. Find the equation in the xy-plane and use a rotation of axes to draw a neat sketch of the graph in the XY-plane. Graph each polar equation. 13. r  3  3 cos  15. r  6 sin122

For Exercises 16 and 17, identify and graph each conic section from the parametric equations given. Then remove the parameter and convert to rectangular form.

14. r  4  8 cos 

17. x  1t  32 2  1 yt2

18. Use a graphing calculator to graph the cycloid, then identify the maximum and minimum values, and the period. x  4T  4 sin T y  4  4 cos T 19. Solve each nonlinear system using the technique of your choice. 4x2  y2  16 4y2  x2  4 a. e b. e 2 yx2 x  y2  4 20. Halley’s comet has a perihelion of 54.5 million miles and an aphelion of 3253 million miles. Use this information to find a polar equation that models its elliptical orbit. How does its eccentricity compare with that of the planets in our solar system? 21. The soccer match is tied, with time running out. In a desperate attempt to win, the opposing coach pulls his goalie and substitutes a forward. Suddenly, Marques gets a break-away and has an open shot at the empty net, 165 ft away. If the kick is on-line and leaves his foot at an angle of 28° with an initial velocity of 80 ft/sec, is the ball likely to go in the net and score the winning goal? 22. The orbit of Mars around the Sun is elliptical, with the Sun at one foci. When the orbit is expressed as a central ellipse on the coordinate grid, its equation is y2 x2   1. Use this information to 1141.652 2 1141.032 2 find the aphelion of Mars and the perihelion of Mars in millions of miles.

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Calculator Exploration and Discovery

Determine the equation of each relation and state its domain and range. For the parabola and the ellipse, also give the location of the foci. 23.

24.

931

y (3, 4)

y (4, 1)

(6, 1) x

(1, 0)

(5, 2)

(3, 0) x

(0, 3)

25.

(2, 6)

(1, 4)

y

(5, 1)

(1, 1) x

(2, 4)

C A L C U L AT O R E X P L O R AT I O N A N D D I S C O V E RY Conic Rotations in Polar Form While all planets orbit around the sun in an Sun elliptical path, their ecliptic planes, or the planes containing the orbits, differ considerably. For example, using the ecliptic plane of the Earth for reference, the plane containing Mercury’s orbit is inclined by 7° and the plane of the dwarf planet Pluto by 17°! In addition, if we use the major axis of Earth’s orbit for reference, the major axes of the other planets, assuming they are transformed to the ecliptic plane, are rotated by some angle . We can gain a basic understanding of the rotations of an elliptical path (relative to some point of reference) using skills developed in this chapter. Here we’ve seen that the equation of a conic can be given in rectangular form, polar form, and parametric form. Each form seems to have its advantages. When it comes to the rotations of a conic section, it’s hard to match the ease and versatility of the polar form. To illustrate, recall that in polar form the general equation of a horizontal ellipse with one focus (the a11  e2 2 . The constant a gives Sun) at the origin is r  1  e cos  the length of the semimajor axis and e represents the eccentricity of the orbit. With the exception of Mercury and Pluto (a dwarf planet), the orbits of most planets are close to circular (e is very near zero). This makes the rota-

tions difficult to see. Instead we will explore the concept of axes rotation using “planets” with higher eccentricities. Consider the following planets and their orbital equations. The planet Agnesi has an eccentricity of e  0.5, while the planet Erdös is the most eccentric at e  0.75. 2.9 1  0.5 cos  5.75 Galois: 1  0.7 cos  7.875 Erdös: 1  0.75 cos  Agnesi:

We’ll investigate the Figure 9.80 concept of conic rotations in polar form by rotating these ellipses. With your calculator in polar MODE , enter these three equations on the Y = screen and use the settings shown in Figure 9.80 to set the window size (use max  7). The resulting graph is displayed in Figure 9.81, showing the very hypothetical case where all planets share the same major axis. To show a more realistic case

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Figure 9.81 where the planets 15 approach the Sun along orbits with differing major axes, we’ll use 32 Galois as a refer- 10 ence and rotate  Agnesi rad 4 15 clockwise and  Figure 9.82 Erdös rad counter12 clockwise. This is done by simply adjusting the argument of cosine in each equation, using  cosa  b for Agnesi 4  and cosa  b for Erdös. The adjusted Y = screen 12 is shown in Figure 9.82, and new graphs in Figure 9.83. Use these ideas to explore and investigate other rotations by completing the following exercises.

Figure 9.83 Exercise 1: What happens if the 15 angle of rotation is  ? Is the orbit identical if you 10 32 rotate by ? Exercise 2: If the denominator in the equation is 15 changed to a sum, what effect does it have on the graph? Exercise 3: If the sign in the numerator is changed, what effect does it have on how the graph is generated? Exercise 4: After resetting the orbits as originally given, use trial and error to approximate the smallest angle of rotation required for the orbit of Galois to intersect the orbit of Erdös. Exercise 5: What minimum rotation is required for the orbit of Galois to intersect the orbit of both Agnesi and Erdös? Exercise 6: What is the minimum rotation required for the orbit of Agnesi to intersect the orbit of Galois?

STRENGTHENING CORE SKILLS Simplifying and Streamlining Computations for the Rotation of Axes

Figure 9.85

While the calculations involved for eliminating the mixed xy-term require a good deal of concentration, there are a few things we can do to simplify the overall process. Basically this involves two things. First, in Figure 9.84 we’ve organized the process in flowchart form to help you “see” the sequence involved in finding cos  and sin  (for use in the rotation formulas). Second, calculating x2, y2, and xy (from the equations x  X cos   Y sin  and y  X sin   Y cos ) as single terms and apart from their actual substitution is somewhat less restrictive and seems to help to streamline the algebra. Illustration 1



For 2x2  12xy  3y2  42  0, use a rotation of axes to eliminate the xy-term, then identify the conic and its characteristic features. B , giving AC tan122  12 5 . Using the triangle shown in Figure 9.85 we find 5 . We then find the values of cos  and sin  (choosing cos122  13 2 in QII), using the double-angle identities as follows:

Solution  Since A  C, we find  using tan122  13

12

2␤ 5

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Strengthening Core Skills

Figure 9.84 Is A
C? (B  0)

cos



1  cos12
2 A 2 1




Q

5 13

2

18 13 S Q 2

3
113 1  cos12
2 sin

2 A 1



Q

2

5 13

8 13 S Q 2

2 113

yes



45

no Is

B AC 兹3?

1 or 兹3

yes



30 or

60

no Find sin(2
) and cos(2
) from the triangle corresponding to tan(2
), then cos  and sin  from the double-angle identities.

Use cos
and sin
in the rotation formulas to compute x 2, xy, and y 2, writing each as a single term.

Substitute, simplify, and use the invariants to double-check your work.

We now compute x2, xy, and y2 prior to substitution in the original equation, writing each as a single term:

• •

3 2 3X  2Y X Y
113 113 113 2 3X  2Y b x2
a 113

x





9X2  12XY  4Y2 13

• •

2 3 2X  3Y X Y
113 113 113 2 2X  3Y y2
a b 113 y





•

xy


13X  2Y212X  3Y2

113 6X2  5XY  6Y2
13

4X2  12XY  9Y2 13

Next, we substitute into the original equation, clearing denominators prior to using the distributive property. 42
2x2  12xy  3y2 9X2  12XY  4Y2 6X2  5XY  6Y2 4x2  12XY  9Y2 42
2a b  12a b  3a b 13 13 13 multiply both sides by 13, then distribute

546
18X2  24XY  8Y2  72X2  60XY  72Y2  12X2  36XY  27Y2 546
78X2  91Y2 combine like terms 2 2 42
6X  7Y simplify and check invariants: F
f ✓ A  C
a  c ✓ 2 2 B2  4AC
b2  4ac ✓ Y X  standard form 1
1 172 2 1 162 2

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The graph is a central hyperbola along the X-axis, with vertices at 117, 02 6 X. and asymptotes Y   A7 Exercise 1: Return to Section 9.6 and resolve Exercises 31 and 32 using these methods. Do the new ideas make a difference?

C U M U L AT I V E R E V I E W C H A P T E R S 1 – 9 Graph each relation. Include vertices, x- and y-intercepts, asymptotes, and other features.

Solve each equation.

1. 1x  2  2  13x  4

14. f 1x2  x  2  3

15. y  1x  3  1

16. g1x2  1x  321x  12 1x  42

2. x  6x  13  0 2

5. log381  x

x2 x2  9 19. f 1x2  log2 1x  12

7. 6 tan x  2 13

21. 41x  12 2  361y  22 2  144

  8. 25 sina x  b  3  15.5 3 6

22. y  2 cosax 

3. 4 # 2x1  x2

4. 3

1 8

17. h1x2 

7

6. log3x  log3 1x  22  1

9.

18. y  2x  3

20. x2  y2  10x  4y  20  0  b1 4

23. r  4 cos122

sin 27° sin x  18 35

10. Use De Moivre’s theorem to find the three cube roots of 8i. Write the roots in a  bi form.

24. x  2 sin t y  tan t

25. Use the dot product to find the angle between the vectors u  H4, 5I and v  H3, 7I.

11. The price of beef in Argentina varies directly with demand and inversely with supply. In the small town of Chascomus, the tender-cut lomito was selling for 18 pesos/kg last week. There were 1000 kg available, and 850 kg were bought. Next week there is a 3-day weekend, so the demand is expected to be closer to 1400 kg, but the butchers will only be able to supply 1200 kg. What will a kilogram of tendercut lomito cost next week?

Solve each system of equations.

13. A surveyor needs to estimate the width of a large rock formation in Canyonlands National Park. From her current position she is 540 yd from one edge of the formation and 850 yd from the other edge. If the included angle is 110°, how wide is the formation?

29. Decompose y 

12. Find the inverse of f 1x2  3 sin12x  12 .

540 yd

850 yd 110

4x  3y  13 26. • 9y  5z  19 x  4z  4 27. e

x2  y2  25 64x2  12y2  768

28. Find the equation of the parabola with vertex at (2, 3) and directrix x  0. 3x3  2x2  x  3 into partial x4  x2

fractions. 30. In the summer, Hollywood releases its big budget, big star, big money movies. Suppose the weekly summer revenue generated by ticket sales was modeled by the function R1w2  w4  25w3  200w2  560w  234, where R(w) represents the revenue generated in week w and 1  w  12. Use the remainder theorem to determine the amount of revenue generated in week 5.

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CONNECTIONS TO CALCULUS As with other relations and functions we’ve studied in precalculus, there is a high level of interest in finding rates of change in polar functions, and in solving systems of polar equations. From their use in architecture to their application in studies of planetary motion, both skills play a fundamental role in our continuing study.

Polar Graphs and Instantaneous Rates of Change Using the tools of calculus, it can be shown that the slope of the tangent line for many different polar graphs can be generated from a specific template. It can also be shown that under certain conditions, the zeroes of the numerator will give the location of horizontal tangent lines, while the zeroes of the denominator will given the location of vertical tangent lines. EXAMPLE 1



Finding Slopes of Tangent Lines For the cardioid r  1  sin , it can be shown that

cos  sin   11  sin 2cos  cos  cos   11  sin 2sin 

gives the slope of the tangent line at . a. Simplify the expression using double-angle identities.

b. Use the result to find the slope of the tangent line at   Then, determine the value(s) of  for which a tangent line is c. Horizontal. d. Vertical. Solution



2 . 3

a. After using the distributive property in the numerator and denominator, we have cos  sin   11  sin 2cos  cos  sin   cos   sin  cos   distribute cos  cos   11  sin 2sin  cos2  sin   sin2 2 sin  cos   cos   combine terms cos2  sin2  sin  sin122  cos   double angle identities cos122  sin 

2 1 4 13 b  cosa b   3 3 2 2  1.  4 2 1 13 cosa b  sina b   3 3 2 2 For the cardioid r  1  sin , the slope Figure C2C 9.1 2 of the tangent line at   is 1 3 5 (Figure C2C 9.1).

2 b. For   , the expression gives 3

sina

c. For horizontal tangents, set the numerator equal to zero and solve. sin122  cos   0 2 sin  cos   cos   0 cos  12 sin   12  0 cos   0 or

9–105

zeroes of numerator double angle identity factor cos 

sin   

1 2

4 3 2 1 54321 1 2 3 4 5

1 2 3 4 5

935

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Horizontal tangent lines will occur at

tangent does not occur at



 7 11 , , and . Note that a horizontal 2 6 6

3 , as this value also makes the denominator 2

of the expression equal to zero. Figure C2C 9.2

d. For vertical tangents, set the denominator equal to zero and solve.

3

cos12
2  sin

0 zeroes of denominator 2 2 cos
 sin2
 sin

0 double angle identity 1 1  sin2
 sin2
 sin

0 Pythagorean identity 2 1  sin
 2 sin

0 combine terms 3 2 1 11  2 sin
211  sin
2
0 factor 1 1 sin

or sin

1 2 2

1

2

3

3 5  and 6 6 3 (see enlargement in Figure C2C 9.2). A vertical tangent does not occur at

, 2 as this value makes both the numerator and denominator of the expression equal to zero.

Vertical tangent lines will occur at



Now try Exercises 1 through 4




Systems of Polar Equations In the “Connections” feature from Chapter 2, we investigated the area bounded by a given curve, the x-axis, and given vertical lines. In many applications, we need to find the area bounded between two curves, which requires that we find where the curves intersect. This can be accomplished by solving a system consisting of the two given r
f 1
2 functions. For polar equations, the system takes the form e . r
g1
2 EXAMPLE 2




Finding Intersections of Polar Curves To find the area of the region inside the circle r
4 sin
and outside the limacon r
1  2 sin
, we must find where the curves intersect. Set up and solve the required system (see figure).

Solution




For r
4 sin
and r
1  2 sin
, the system r
4 sin
is e . Substituting 4 sin
for r in r
1  2 sin
the second equation gives 4 sin

1  2 sin
, 2 sin

1 1 sin

2 1

sin1a b 2

substitute 4 sin
for r subtract 2 sin
divide by 2

apply inverse sine

5 4 3 2 1

54321 1 2 3 4 5

1 2 3 4 5

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937

 5 and   . It appears we’re finished, 6 6 but an inspection of the equations (or the graph) shows there must be a third point of intersection at the pole (0, 0). The reason we missed it is that in polar coordinates each point can be represented many different ways. Solutions using a system of polar equations finds only the intersections represented by the identical ordered pair, and a separate analysis is needed for the case where r  0. Solving the equation 7 0  4 sin  gives   k, while 0  1  2 sin  gives    2k and 6 11   2k. Although the ordered pairs are different, both represent the pole 6 (0, 0) and the graphs must intersect there. This shows the graphs intersect at  

Now try Exercises 5 through 10



Connections to Calculus Exercises For each polar function given, the expression for the slope of the tangent line is also given. (a) Simplify the expression. (b) Find the slope of the tangent line at the values of  shown. Then determine the value(s) of  in [0, 2) for which a tangent line is (c) horizontal, and (d) vertical.

1. r  1  cos  (cardioid), slope of the tangent line: cos   sin 1sin 2  cos  cos   11 ; , sin   2 cos 1sin 2 2 6 3. r  cos   sin  (circle), slope of the tangent line: cos  cos   sin 1sin 2  2 sin  cos  2 cos 1sin 2  3sin 1sin 2  cos  cos  4  3  , 4 4

2. r  6 cos  (circle), slope of the tangent line: 6 sin  sin   6 cos  cos   2 ; , 6 sin  cos   6 cos  sin  6 3 4. r  sin   cos  (circle), slope of the tangent line: cos  cos   sin 1sin 2  2 cos 1sin 2 2 sin 1cos 2  3cos  cos   sin 1sin 2 4  3  , 2 2

Solve each system. For Exercises 7 and 8, identify the graph of each function before you begin. Verify results using a graphing calculator.

r

3 2 5. • r  1  cos2

6. e

r  2 sin2 r  1  sin 

r  sin122 r  cos 

8. e

r  2 sin122 r  2 sin 

7. e

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9–108

Connections to Calculus

9. Find the points of intersection for the circle r
4 cos
and the curve r
4 sin12
2 , by solving r
4 cos
the system e . State your answer in r
8 sin
cos
both (r,
) form and (x, y) form. Note from the graph there are three points of intersection, and that a polar grid has been superimposed on a rectangular grid.

10. Find the points of intersection of the two four-leaf roses formed by r
4 sin 12
2 and r
4 cos 12
2 , r
4 sin12
2 by solving the system e (there are r
4 cos12
2 nine in all). Some of these points can be found by solving 4 cos 12
2
4 sin12
2 , others can be found using symmetry [since r can be positive or negative, we consider the possibility r1
r2, or 4 cos 12
2
4 sin 12
2 ]. Be sure to include an analysis of the case where r
0. Answer in (r,
) form.

Exercise 9 5 4 3 2 1 54321 1 2 3 4 5

1 2 3 4 5

Exercise 10 5 4 3 2 1 54321 1 2 3 4 5

1 2 3 4 5

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Precalculas—

10 CHAPTER CONNECTIONS

Additional Topics in Algebra CHAPTER OUTLINE 10.1

Sequences and Series 940

10.2

Arithmetic Sequences 949

10.3

Geometric Sequences 956

10.4

Mathematical Induction 966

10.5

Counting Techniques 975

10.6

Introduction to Probability 987

10.7

The Binomial Theorem 999

As part of a science experiment, a baseball, golf ball, superball, and bowling ball are dropped on a concrete sur face from a height of 2 m. Why do the golf ball and superball seem to bounce almost as high as the drop, while the bowling ball hardly bounces at all? The answer lies in a study of elastic rebound, and depends on the density and elasticity of the material used to make the ball. Questions like these are an important part of a fast growing field of study called sports science. Further inquiries seek to know the total distance traveled by the balls before they come to rest, as an additional measure of their suitability for use in sports. This application appears as Exercise 117 in Section 10.3.

While the notation and properties of summation were introduced as tools to work with finite sums, in Section 10.3 we will note that under certain conditions, consideration of an infinite sum is possible. The process will involve an ever larger sum of ever smaller pieces, a process further illustrated in the Connections to Calculus Connections to Calculus feature following Chapter 10. This feature also offers practice with the algebraic tools that make the ultimate conclusion possible. 939

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10.1 Sequences and Series Learning Objectives

A sequence can be thought of as a pattern of numbers listed in a prescribed order. A series is the sum of the numbers in a sequence. Sequences and series come in countless varieties, and we’ll introduce some general forms here. In following sections we’ll focus on two special types: arithmetic and geometric sequences. These are used in a number of different fields, with a wide variety of significant applications.

In Section 10.1 you will learn how to:

A. Write out the terms of a sequence given the general or nth term

B. Work with recursive

A. Finding the Terms of a Sequence Given the General Term

sequences and sequences involving a factorial

C. Find the partial sum of a series

D. Use summation notation to write and evaluate series

Year:

f (1) T

T

T

T

T

E. Use sequences to solve

Value:

$10,700

$11,449

$12,250

$13,108

$14,026 . . .

applied problems

f (2)

f (3)

f (4)

f (5) . . .

Note the relationship (year, value) is a function that pairs 1 with $10,700, 2 with $11,449, 3 with $12,250 and so on. This is an example of a sequence. To distinguish sequences from other algebraic functions, we commonly name the functions a instead of f, use the variable n instead of x, and employ a subscript notation. The function f 1x2  10,00011.072 x would then be written an  10,00011.072 n. Using this notation a1  10,700, a2  11,449, and so on. The values a1, a2, a3, a4, p are called the terms of the sequence. If the account were closed after a certain number of years (for example, after the fifth year) we have a finite sequence. If we let the investment grow indefinitely, the result is called an infinite sequence. The expression an that defines the sequence is called the general or nth term and the terms immediately preceding it are called the 1n  12st term, the 1n  22nd term, and so on.

WORTHY OF NOTE Sequences can actually start with any natural number. For instance, the sequence 2 must start at an  n1 n  2 to avoid division by zero. In addition, we will sometimes use a0 to indicate a preliminary or inaugural element, as in a0  $10,000 for the amount of money initially held, prior to investing it.

940

Suppose a person had $10,000 to invest, and decided to place the money in government bonds that guarantee an annual return of 7%. From our work in Chapter 4, we know the amount of money in the account after x years can be modeled by the function f 1x2  10,00011.072 x. If you reinvest your earnings each year, the amount in the account would be (rounded to the nearest dollar):

Sequences A finite sequence is a function an whose domain is the set of natural numbers from 1 to n. The terms of the sequence are labeled a1, a2, a3, p , ak, ak1, p , an1, an where ak represents an arbitrary “interior” term and an also represents the last term of the sequence. An infinite sequence is a function an whose domain is the set of all natural numbers.

EXAMPLE 1A



Solution



EXAMPLE 1B



Computing Specified Terms of a Sequence n1 For an  , find a1, a3, a6, and a7. n2 11 2 12 61 7  a6  2 36 6 a1 

31 4  2 9 3 71 8 a7   2 49 7 a3 

Computing the First k Terms of a Sequence Find the first four terms of the sequence an  112 n2n. Write the terms of the sequence as a list. 10-2

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10-3

Section 10.1 Sequences and Series

Solution



a2  112 222  4

a3  112 323  8

WORTHY OF NOTE

a4  112 424  16

The sequence can be written 2, 4, 8, 16, . . . , or more generally as 2, 4, 8, 16, . . . , 112 n2n, . . . to show how each term was generated.

When the terms of a sequence alternate in sign as in Example 1B, we call it an alternating sequence.

Now try Exercises 7 through 32

A. You’ve just learned how to write out the terms of a sequence given the general or nth term

EXAMPLE 2

a1  112 121  2

941





B. Recursive Sequences and Factorial Notation Sometimes the formula defining a sequence uses the preceding term or terms to generate those that follow. These are called recursive sequences and are particularly useful in writing computer programs. Because of how they are defined, recursive sequences must give an inaugural term or seed element, to begin the recursion process. Perhaps the most famous recursive sequence is associated with the work of Leonardo of Pisa (A.D. 1180–1250), better known to history as Fibonacci. In fact, it is commonly called the Fibonacci sequence in which each successive term is the sum of the previous two, beginning with 1, 1, . . . . Computing the Terms of a Recursive Sequence Write out the first eight terms of the recursive (Fibonacci) sequence defined by c1  1, c2  1, and cn  cn1  cn2.

Solution



The first two terms are given, so we begin with n  3. c3  c31  c32  c2  c1 11 2

c4  c41  c42  c3  c2 21 3

c5  c51  c52  c4  c3 32 5

At this point we can simply use the fact that each successive term is simply the sum of the preceding two, and find that c6  3  5  8, c7  5  8  13, and c8  13  8  21. The first eight terms are 1, 1, 2, 3, 5, 8, 13, and 21.

WORTHY OF NOTE One application of the Fibonacci sequence involves the Fibonacci spiral, found in the growth of many ferns and the spiral shell of many mollusks.

Now try Exercises 33 through 38



Sequences can also be defined using a factorial, which is the product of a given natural number with all those that precede it. The expression 5! is read, “five factorial,” and is evaluated as: 5!  5 # 4 # 3 # 2 # 1  120. Factorials For any natural number n,

n!  n # 1n  12 # 1n  22 # p # 3 # 2 # 1

Rewriting a factorial in equivalent forms often makes it easier to simplify certain expressions. For example, we can rewrite 5!  5 # 4! or 5!  5 # 4 # 3!. Consider Example 3. EXAMPLE 3



Simplifying Expressions Using Factorial Notation Simplify by writing the numerator in an equivalent form. 6! 9! 11! a. b. c. 7! 8!2! 3!5!

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Solution



WORTHY OF NOTE

a.

9! 9 # 8 # 7!  7! 7!

b.

9#8

Most calculators have a factorial option or key. On the TI-84 Plus it is located on a submenu of the MATH key: MATH

10-4

CHAPTER 10 Additional Topics in Algebra

 72

11! 11 # 10 # 9 # 8!  8!2! 8!2! 990  2  495

c.

6! 6 # 5!  3!5! 3!5! 6  6 1

Now try Exercises 39 through 44



PRB (option) 4: !

EXAMPLE 4



Computing a Specified Term from a Sequence Defined Using Factorials Find the third term of each sequence. n! a. an  n 2

Solution



B. You’ve just learned how to work with recursive sequences and sequences involving a factorial

3! 23 6 3   8 4

a. a3 

112 n 12n  12! b. cn  n! 112 3 32132  1 4 ! 3! 112 3 5 # 4 # 3!4 112152!   3! 3!  20

b. c3 

Now try Exercises 45 through 50 Figure 10.1



C. Series and Partial Sums Sometimes the terms of a sequence are dictated by context rather than a formula. Consider the stacking of large pipes in a storage yard. If there are 10 pipes in the bottom row, then 9 pipes, then 8 (see Figure 10.1), how many pipes are in the stack if there is a single pipe at the top? The sequence generated is 10, 9, 8, . . . , 3, 2, 1 and to answer the question we would have to compute the sum of all terms in the sequence. When the terms of a finite sequence are added, the result is called a finite series. Finite Series Given the sequence a1, a2, a3, a4, . . . , an, the sum of the terms is called a finite series or partial sum and is denoted Sn: Sn  a1  a2  a3  a4  p  an

EXAMPLE 5



Computing a Partial Sum Given an  2n, find the value of a. S4 and b. S7.

Solution



C. You’ve just learned how to find the partial sum of a series

Since we eventually need the sum of the first seven terms (for Part b), begin by writing out these terms: 2, 4, 6, 8, 10, 12, and 14. a. S4  a1  a2  a3  a4 2468  20 b. S7  a1  a2  a3  a4  a5  a6  a7  2  4  6  8  10  12  14  56 Now try Exercises 51 through 56



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Precalculas—

10-5

Section 10.1 Sequences and Series

943

D. Summation Notation When the general term of a sequence is known, the Greek letter sigma © can be used to write the related series as a formula. For instance, to indicate the sum of the first 4

four terms of an  3n  2, we write

 13i  22. This result is called summation or

i1

sigma notation and the letter i is called the index of summation. The letters j, k, l, and m are also used as index numbers, and the summation need not start at 1.

EXAMPLE 6



Computing a Partial Sum Compute each sum: 4

a.



i1

5

13i  22

b.

6

1 j1 j



c.

 112 k

k 2

k3

4

Solution



a.

 13i  22  13 # 1  22  13 # 2  22  13 # 3  22  13 # 4  22

i1 5

b.

 5  8  11  14  38

1

1

1

1

1

1

j 12345

j1



60 30 20 15 12 137      60 60 60 60 60 60

6

c.

 112 k

k 2

k3

 112 3 # 32  112 4 # 42  112 5 # 52  112 6 # 62  9  16  25  36  18

Now try Exercises 57 through 68



If a definite pattern is noted in a given series expansion, this process can be reversed, with the expanded form being expressed in summation notation using the nth term. EXAMPLE 7



Writing a Sum in Sigma Notation Write each of the following sums in summation (sigma) notation. a. 1  3  5  7  9 b. 6  9  12  p

Solution



a. The series has five terms and each term is an odd number, or 1 less than a 5

WORTHY OF NOTE By varying the function given and/or where the sum begins, more than one acceptable form is possible.

 13  3n2 q

For example 7(b)

n1

multiple of 2. The general term is an  2n  1, and the series is

 12n  12.

n1

b. This is an infinite sum whose terms are multiples of 3. The general term is q

an  3n, but the series starts at 2 and not 1. The series is

 3j.

j2

Now try Exercises 69 through 78



also works.

Since the commutative and associative laws hold for the addition of real numbers, summations have the following properties:

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10-6

CHAPTER 10 Additional Topics in Algebra

Properties of Summation Given any real number c and natural number n, n

(I)

 c  cn

i1

If you add a constant c “n” times the result is cn. n

(II)



n

cai  c

i1

a

i

i1

A constant can be factored out of a sum. n

(III)



i1

1ai  bi 2 

n



n

ai 

i1

b

i

i1

A summation can be distributed to two (or more) sequences. m

(IV)



n

i1

n



ai 

ai 

im1

a; 1  m 6 n i

i1

A summation is cumulative and can be written as a sum of smaller parts. The verification of property II depends solely on the distributive property. n

Proof:

 ca  ca i

1

 ca2  ca3  p  can

expand sum

i1

 c1a1  a2  a3  p  an 2

factor out c

n

a

c

write series in summation form

i

i1

The verification of properties III and IV simply uses the commutative and associative properties. You are asked to prove property III in Exercise 91. EXAMPLE 8



Computing a Sum Using Summation Properties 4

Recompute the sum

 13i  22 from Example 6(a) using summation properties.

i1 4

Solution





i1

13i  22 

4



4

2

property III

i  2

property II

3i 

i1 4

3

i1

D. You’ve just learned how to use summation notation to write and evaluate series

i1 4 i1

 31102  2142  38

1  2  3  4  10; property I result

Now try Exercises 79 through 82



E. Applications of Sequences To solve applications of sequences, (1) identify where the sequence begins (the initial term) and (2) write out the first few terms to help identify the nth term.

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10-7

Section 10.1 Sequences and Series

EXAMPLE 9



945

Solving an Application — Accumulation of Stock Hydra already owned 1420 shares of stock when her company began offering employees the opportunity to purchase 175 discounted shares per year. If she made no purchases other than these discounted shares each year, how many shares will she have 9 yr later? If this continued for the 25 yr she will work for the company, how many shares will she have at retirement?

Solution

E. You’ve just learned to use sequences to solve applied problems



To begin, it helps to simply write out the first few terms of the sequence. Since she already had 1420 shares before the company made this offer, we let a0  1420 be the inaugural element, showing a1  1595 (after 1 yr, she owns 1595 shares). The first few terms are 1595, 1770, 1945, 2120, and so on. This supports a general term of an  1595  1751n  12. After 9 years

After 25 years

a9  1595  175182  2995

a25  1595  1751242  5795

After 9 yr she would have 2995 shares. Upon retirement she would own 5795 shares of company stock. Now try Exercises 85 through 90

TECHNOLOGY HIGHLIGHT

Studying Sequences and Series To support a study of sequences and series, we can use a graphing calculator to generate the desired terms. This can be done either on the home screen or directly into the LIST feature of the calculator. On the TI-84 Plus this is accomplished using the “seq(” and “sum(” STAT commands, which are accessed using the keystrokes 2nd (LIST) and the screen shown in Figure 10.2. The “seq(” feature is option 5 under the OPS submenu (press 5) and the “sum(” feature is option 5 under the MATH submenu (press 5). To generate the first four terms of the sequence an  n2  1, and STAT to find the sum, CLEAR the home screen and press 2nd 5 to place “seq(” on the home screen. This command requires four inputs: an (the nth term), variable used (the calculator can work with any letter), initial term and the last term. For this example the screen reads “seq 1x2  1, x, 1, 42,” with the result being the four terms shown in Figure 10.3. To find the sum of these terms, we simply precede the seq 1x2  1, x, 1, 42 command by “sum(,” and two methods are shown in Figure 10.4. Each of the following sequences have some interesting properties or mathematical connections. Use your graphing calculator to generate the first 10 terms of each sequence and the sum of the first 10 terms. Next, generate the first 20 terms of each sequence and the sum of these terms. What conclusion (if any) can you reach about the sum of each sequence?

Exercise 1: an 

1 3n

Exercise 2: an 

2 n1n  12

Exercise 3: an 

Figure 10.2

Figure 10.3

Figure 10.4

1 12n  1212n  12



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10-8

CHAPTER 10 Additional Topics in Algebra

10.1 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. A sequence is a(n) specific . 2. A series is the given sequence.

5. Describe the characteristics of a recursive sequence and give one example.

of the numbers from a

3. When each term of a sequence is larger than the preceding term, the sequence is said to be 

4. When each term of a sequence is smaller than the preceding term, the sequence is said to be .

of numbers listed in a

.

6. Describe the characteristics of an alternating sequence and give one example.

DEVELOPING YOUR SKILLS

Find the first four terms, then find the 8th and 12th term for each nth term given.

7. an  2n  1

8. an  2n  3

9. an  3n  3

10. an  2n  12

2

3

11. an  112 nn 13. an 

12. an 

112 n n

1 n 14. an  a1  b n

n n1 n

n

1 15. an  a b 2

2 16. an  a b 3

1 17. an  n

1 18. an  2 n

112 n 19. an  n1n  12 21. an  112 n2n

20. an 

112 n1 2n2  1

22. an  112 n2n

Find the indicated term for each sequence.

23. an  n2  2; a9

24. an  1n  22 2; a9

25. an 

26. an 

112 n1 ; a5 n n1

1 27. an  2a b 2

112 n1 ; a5 2n  1 n1

; a7

1 28. an  3a b 3

; a7

1 n 29. an  a1  b ; a10 n

1 n 30. an  an  b ; a9 n

31. an 

1 ; a4 n12n  12

32. an 

1 ; a5 12n  12 12n  12

Find the first five terms of each recursive sequence.

33. e

a1  2 an  5an1  3

34. e

a1  3 an  2an1  3

35. e

a1  1 an  1an1 2 2  3

36. e

a1  2 an  an1  16

37. •

c1  64, c2  32 cn2  cn1 cn  2

38. e

c1  1, c2  2 cn  cn1  1cn2 2 2

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10-9

Section 10.1 Sequences and Series

Simplify each factorial expression.

39. 42.

8! 5!

40.

6! 3!3!

43.

7

12! 10!

41.

8! 2!6!

44.

9! 7!2! 10! 3!7!

Write out the first four terms in each sequence.

45. an 

n! 1n  12!

1n  12! 13n2! n n 49. an  n! 47. an 

46. an 

n! 1n  32!

1n  32! 12n2! n 2 50. an  n! 48. an 

Find the indicated partial sum for each sequence.

51. an  n; S5

52. an  n2; S7

53. an  2n  1; S8

54. an  3n  1; S6

1 55. an  ; S5 n

56. an 

n ; S4 n1

Expand and evaluate each series. 4

57.

 13i  52 5



k1 7

61.



k1 4

63.

58.

i1



 12i  32 5

60.

112 kk

62.

i2

j3

j

j3

112 k k3 k1k  22 8

67.

j

2



6

68.



112 k1

k2

k2  1

Write each sum using sigma notation. Answers are not necessarily unique.

69. 4  8  12  16  20 70. 5  10  15  20  25 71. 1  4  9  16  25  36 72. 1  8  27  64  125  216 For the given general term an, write the indicated sum using sigma notation.

73. an  n  3; S5 74. an 

n2  1 ; S4 n1

75. an 

n2 ; third partial sum 3

76. an  2n  1; sixth partial sum

i1

12k2  32

2



7

66.

2j

5

i1

59.

65.

947



1k2  12

5

112 k2k

k1



k1

77. an 

78. an  n2; sum for n  2 to 6 Compute each sum by applying properties of summation. 5

79.

i



i1

4

64.

n ; sum for n  3 to 7 2n

14i  52

6

80.

4

2

i2

81.

 13k

2

k1

 13  2i2

i1 4

 k2

82.

 12k

3

 52

k1

WORKING WITH FORMULAS

83. Sum of an  3n  2: Sn 

n(3n  1) 2

The sum of the first n terms of the sequence defined by an  3n  2  1, 4, 7, 10, . . . , 13n  22, . . . is given by the formula shown. Find S5 using the formula, then verify by direct calculation.

84. Sum of an  3n  1: Sn 

n(3n  1) 2

The sum of the first n terms of the sequence defined by an  3n  1  2, 5, 8, 11, . . . , 13n  12, . . . is given by the formula shown. Find S8 using the formula, then verify by direct calculation. Observing the results of Exercises 83 and 84, can you now state the sum formula for an  3n  0?

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APPLICATIONS

Use the information given in each exercise to determine the nth term an for the sequence described. Then use the nth term to list the specified number of terms.

85. Blue-book value: Steve’s car has a blue-book value of $6000. Each year it loses 20% of its value (its value each year is 80% of the year before). List the value of Steve’s car for the next 5 yr. (Hint: For a1
6000, we need the next five terms.) 86. Effects of inflation: Suppose inflation (an increase in value) will average 4% for the next 5 yr. List the growing cost (year by year) of a DVD that costs $15 right now. (Hint: For a1
15, we need the next five terms.) 87. Wage increases: Latisha gets $5.20 an hour for filling candy machines for Archtown Vending. Each year she receives a $0.50 hourly raise. List Latisha’s wage for the first 5 yr. How much will she make in the fifth year if she works 8 hr per day for 240 working days? 88. Average birth weight: The average birth weight of a certain animal species is 900 g, with the baby gaining 125 g each day for the first 10 days. List the infant’s weight for the first 10 days. How much does the infant weigh on the 10th day?


89. Stocking a lake: A local fishery stocks a large lake with 1500 bass and then adds an additional 100 mature bass per month until the lake nears maximum capacity. If the bass population grows at a rate of 5% per month through natural reproduction, the number of bass in the pond after n months is given by the recursive sequence b0
1500, bn
1.05bn1  100. How many bass will be in the lake after 6 months? 90. Species preservation: The Interior Department introduces 50 wolves (male and female) into a large wildlife area in an effort to preserve the species. Each year about 12 additional adult wolves are added from capture and relocation programs. If the wolf population grows at a rate of 10% per year through natural reproduction, the number of wolves in the area after n years is given by the recursive sequence w0
50, wn
1.10wn1  12. How many wolves are in the wildlife area after 6 years?

EXTENDING THE CONCEPT

91. Verify that a summation may be distributed to two (or more) sequences. That is, verify that the following statement is true: n

n

n


1ai  bi 2

ai 
bi.

i
1

i
1

i
1

Surprisingly, some of the most celebrated numbers in mathematics can be represented or approximated by a series expansion. Use your calculator to find the partial


10-10

CHAPTER 10 Additional Topics in Algebra

sums for n
4, n
8, and n
12 for the summations given, and attempt to name the number the summation approximates: n

1 k
0 k! n 1 94.
k k
1 2 92.


n

1 k k
1 3

93.


MAINTAINING YOUR SKILLS

95. (6.7) Solve csc x sin a


 xb
1 2

96. (2.5) Set up the difference quotient for f 1x2
1x, then rationalize the numerator.

97. (7.2) Given a triangle where a
0.4 m, b
0.3 m, and c
0.5 m, find the three corresponding angles. 98. (6.3) Solve the system using a matrix 25x  y  2z
14 equation. • 2x  y  z
40 7x  3y  z
13

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10.2 Arithmetic Sequences Learning Objectives

Similar to the way polynomials fall into certain groups or families (linear, quadratic, cubic, etc.), sequences and series with common characteristics are likewise grouped. In this section, we focus on sequences where each successive term is generated by adding a constant value, as in the sequence 1, 8, 15, 22, 29, . . . , where 7 is added to a given term in order to produce the next term.

In Section 10.2 you will learn how to:

A. Identify an arithmetic sequence and its common difference

B. Find the nth term of an

A. Identifying an Arithmetic Sequence and Finding the Common Difference

arithmetic sequence

C. Find the nth partial sum of an arithmetic sequence

An arithmetic sequence is one where each successive term is found by adding a fixed constant to the preceding term. For instance 3, 7, 11, 15, . . . is an arithmetic sequence, since adding 4 to any given term produces the next term. This also means if you take the difference of any two consecutive terms, the result will be 4 and in fact, 4 is called the common difference d for this sequence. Using the notation developed earlier, we can write d  ak1  ak, where ak represents any term of the sequence and ak1 represents the term that follows ak.

D. Solve applications involving arithmetic sequences

Arithmetic Sequences Given a sequence a1, a2, a3, . . . , ak, ak1, . . . , an, where k, n   and k 6 n, if there exists a common difference d such that ak1  ak  d, then the sequence is an arithmetic sequence. The difference of successive terms can be rewritten as ak1  ak  d (for k  12 to highlight that each following term is found by adding d to the previous term. EXAMPLE 1



Identifying an Arithmetic Sequence Determine if the given sequence is arithmetic. a. 2, 5, 8, 11, . . . b. 12, 56, 43, 76, p

Solution



a. Begin by looking for a common difference d  ak1  ak. Checking each pair of consecutive terms we have 523

853

11  8  3 and so on.

This is an arithmetic sequence with common difference d  3. b. Checking each pair of consecutive terms yields 5 1 5 3    6 2 6 6 2 1   6 3

5 8 5 4    3 6 6 6 3 1   6 2

Since the difference is not constant, this is not an arithmetic sequence. Now try Exercises 7 through 18

EXAMPLE 2





Writing the First k Terms of an Arithmetic Sequence Write the first five terms of the arithmetic sequence, given the first term a1 and the common difference d. a. a1  12 and d  4 b. a1  12 and d  13

10-11

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Solution



A. You’ve just learned how to identify an arithmetic sequence and its common difference

a. a1  12 and d  4. Starting at a1  12, add 4 to each new term to generate the sequence: 12, 8, 4, 0, 4 b. a1  12 and d  13. Starting at a1  12 and adding 13 to each new term will generate the sequence: 12, 56, 76, 32, 11 6 Now try Exercises 19 through 30



B. Finding the nth Term of an Arithmetic Sequence If the values a1 and d from an arithmetic sequence are known, we could generate the terms of the sequence by adding multiples of d to the first term, instead of adding d to each new term. For example, we can generate the sequence 3, 8, 13, 18, 23 by adding multiples of 5 to the first term a1  3: 3  3  1025

a1  a1  0d

13  3  1225

a3  a1  2d

8  3  1125

a2  a1  1d

18  3  1325

a4  a1  3d

23  3  1425

current term

initial term

S

a5  a1  4d S

950

coefficient of common difference

It’s helpful to note the coefficient of d is 1 less than the subscript of the current term (as shown): 5  1  4. This observation leads us to a formula for the nth term. The nth Term of an Arithmetic Sequence The nth term of an arithmetic sequence is given by

an  a1  1n  12d

where d is the common difference.

EXAMPLE 3



Finding a Specified Term in an Arithmetic Sequence Find the 24th term of the sequence 0.1, 0.4, 0.7, 1, . . . .

Solution



Instead of creating all terms up to the 24th, we determine the constant d and use the nth term formula. By inspection we note a1  0.1 and d  0.3. an  a1  1n  12d  0.1  1n  120.3  0.1  0.3n  0.3  0.3n  0.2

n th term formula substitute 0.1 for a1 and 0.3 for d eliminate parentheses simplify

To find the 24th term we substitute 24 for n: a24  0.31242  0.2  7.0

substitute 24 for n result

Now try Exercises 31 through 42

EXAMPLE 4





Finding the Number of Terms in an Arithmetic Sequence Find the number of terms in the arithmetic sequence 2, 5, 12, 19, . . . , 411.

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Solution



951

By inspection we see that a1  2 and d  7. As before, an  a1  1n  12d  2  1n  12 172  7n  9

n th term formula substitute 2 for a1 and 7 for d simplify

Although we don’t know the number of terms in the sequence, we do know the last or nth term is 411. Substituting 411 for an gives 411  7n  9 60  n

substitute 411 for an solve for n

There are 60 terms in this sequence. Now try Exercises 43 through 50



If the term a1 is unknown but a term ak is given, the nth term can be written an  ak  1n  k2d since n  k  1n  k2

(the subscript of the term ak and coefficient of d sum to n). EXAMPLE 5



Finding the First Term of an Arithmetic Sequence Given an arithmetic sequence where a6  0.55 and a13  0.9, find the common difference d and the value of a1.

Solution



At first it seems that not enough information is given, but recall we can express a13 as the sum of any earlier term and the appropriate multiple of d. Since a6 is known, we write a13  a6  7d (note 13  6  7 as required). a13  a6  7d 0.9  0.55  7d 0.35  7d d  0.05

a1 is unknown substitute 0.9 for a13 and 0.55 for a6 subtract 0.55 solve for d

Having found d, we can now solve for a1.

B. You’ve just learned how to find the nth term of an arithmetic sequence

a13  a1  12d 0.9  a1  1210.052 0.9  a1  0.6 a1  0.3

n th term formula for n  13 substitute 0.9 for a13 and 0.05 for d simplify solve for a1

The first term is a1  0.3 and the common difference is d  0.05. Now try Exercises 51 through 56



C. Finding the nth Partial Sum of an Arithmetic Sequence Using sequences and series to solve applications often requires computing the sum of a given number of terms. Consider the sequence a1, a2, a3, a4, . . . , an with common difference d. Use Sn to represent the sum of the first n terms and write the original series, then the series in reverse order underneath. Since one row increases at the same rate the other decreases, the sum of each column remains constant, and for simplicity’s sake we choose a1  an to represent this sum. a1  a2  a3  p  an2  an1  an Sn  add p Sn  an  an1  an2   a3  a2  a1 columns 2Sn  1a1  an 2  1a1  an 2  1a1  an 2  p  1a1  an 2  1a1  an 2  1a1  an 2 ↓ vertically

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To understand why each column adds to a1  an, consider the sum in the second column: a2  an1. From a2  a1  d and an1  an  d, we obtain a2  an1  1a1  d2  1an  d2 by direct substitution, which gives a result of a1  an. Since there are n columns, we end up with 2Sn  n1a1  an 2, and solving for Sn gives the formula for the first n terms of an arithmetic sequence. The nth Partial Sum of an Arithmetic Sequence Given an arithmetic sequence with first term a1, the nth partial sum is given by Sn  na

a1  an b. 2

In words: The sum of an arithmetic sequence is the number of terms times the average of the first and last term.

EXAMPLE 6



Computing the Sum of an Arithmetic Sequence 75

Find the sum of the first 75 positive, odd integers:

 12k  12.

k1

Solution



The initial terms of the sequence are 1, 3, 5, . . . and we note a1  1, d  2, and n  75. To use the sum formula, we need the value of an  a75. The nth term formula shows a75  a1  74d  1  74122, so a75  149. n1a1  an 2 2 751a1  a75 2 S75  2 7511  1492  2  5625 Sn 

C. You’ve just learned how to find the nth partial sum of an arithmetic sequence

sum formula

substitute 75 for n substitute 1 for a1, 149 for a75 result

The sum of the first 75 positive, odd integers is 5625. Now try Exercises 57 through 62



Figure 10.5

By substituting the nth term formula directly into the formula for partial sums, we’re able to find a partial sum without actually having to find the nth term: n1a1  an 2 2 n1a1  3a1  1n  12d 4 2  2 n  3 2a1  1n  12d 4 2

Sn  spiral fern

Figure 10.6

sum formula substitute a1  1n  12d for an alternative formula for the nth partial sum

See Exercises 63 through 68 for more on this alternative formula.

D. Applications nautilus

In the evolution of certain plants and shelled animals, sequences and series seem to have been one of nature’s favorite tools (see Figures 10.5 and 10.6). Sequences and series also provide a good mathematical model for a variety of other situations as well.

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EXAMPLE 7



Solving an Application of Arithmetic Sequences: Seating Capacity Cox Auditorium is an amphitheater that has 40 seats in the first row, 42 seats in the second row, 44 in the third, and so on. If there are 75 rows in the auditorium, what is the auditorium’s seating capacity?

Solution



D. You’ve just learned how to solve applications involving arithmetic sequences

The number of seats in each row gives the terms of an arithmetic sequence with a1  40, d  2, and n  75. To find the seating capacity, we need to find the total number of seats, which is the sum of this arithmetic sequence. Since the value of n a75 is unknown, we opt for the alternative formula Sn  3 2a1  1n  12d4 . 2 n Sn  3 2a1  1n  12d 4 2 75 S75  3 21402  175  12 122 4 2 75  12282 2  8550

sum formula

substitute 40 for a1, 2 for d and 75 for n

simplify result

The seating capacity for Cox Auditorium is 8550. Now try Exercises 71 through 76



10.2 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. Consecutive terms in an arithmetic sequence differ by a constant called the . 2. The sum of the first n terms of an arithmetic sequence is called the nth . 3. The formula for the nth partial sum of an arithmetic sequence is sn  the term.



, where an is

4. The nth term formula for an arithmetic sequence is term an  , where a1 is the and d is the . 5. Discuss how the terms of an arithmetic sequence can be written in various ways using the relationship an  ak  1n  k2d. 6. Describe how the formula for the nth partial sum was derived, and illustrate its application using a sequence from the exercise set.

DEVELOPING YOUR SKILLS

Determine if the sequence given is arithmetic. If yes, name the common difference. If not, try to determine the pattern that forms the sequence.

10. 1.2, 3.5, 5.8, 8.1, 10.4, . . . 11. 2, 3, 5, 7, 11, 13, 17, . . .

7. 5, 2, 1, 4, 7, 10, . . .

12. 1, 4, 8, 13, 19, 26, 34, . . .

8. 1, 2, 5, 8, 11, 14, . . .

13.

9. 0.5, 3, 5.5, 8, 10.5, . . .

1 1 1 1 5 24 , 12 , 8 , 6 , 24 ,

...

14.

1 1 1 1 1 12 , 15 , 20 , 30 , 60 ,

...

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15. 1, 4, 9, 16, 25, 36, . . .

45. a1  0.4, an  10.9, d  0.25

16. 125, 64, 27, 8, 1, . . .

46. a1  0.3, an  36, d  2.1

17. ,

5 2    7 3 5  , , , , , . . . 18. , , , , ,... 6 3 2 3 6 8 4 8 2

Write the first four terms of the arithmetic sequence with the given first term and common difference.

20. a1  8, d  3

21. a1  7, d  2

22. a1  60, d  12

23. a1  0.3, d  0.03

24. a1  0.5, d  0.25

25. a1  32, d  12

1 26. a1  15, d  10

27. a1  34, d  18

28. a1  16, d  13

29. a1  2, d  3

30. a1  4, d  4

Identify the first term and the common difference, then write the expression for the general term an and use it to find the 6th, 10th, and 12th terms of the sequence.

33. 5.10, 5.25, 5.40, . . . 35.

3 9 2, 4,

3,

15 4,

...

32. 7, 4, 1, 2, 5, . . . 34. 9.75, 9.40, 9.05, . . . 36.

5 3 7 , 14 ,

27,

11 14 ,

48. 3.4, 1.1, 1.2, 3.5, . . . , 38 49.

1 1 1 5 1 12 , 8 , 6 , 24 , 4 ,

...

51. a3  7, a7  19

50.

1 1 1 1 12 , 15 , 20 , 30 ,

. . . , 14

Find the indicated term using the information given.

37. a1  5, d  4; find a15 38. a1  9, d 2; find a17 1 39. a1  32, d  12 ; find a7 1 40. a1  12 25 , d  10 ; find a9

41. a1  0.025, d  0.05; find a50 42. a1  3.125, d  0.25; find a20 Find the number of terms in each sequence.

43. a1  2, an  22, d  3 44. a1  4, an  42, d  2

52. a5  17, a11  2

53. a2  1.025, a26  10.025 54. a6  12.9, a30  1.5 27 55. a10  13 18 , a24  2

56. a4  54, a8  94

Evaluate each sum. For Exercises 61 and 62, use the summation properties from Section 10.1. 30

57.

 13n  42

29

58.

n1 37

3 59. a n  2b n1 4



61.

 13  5n2

 14n  12

n1 20

60.

 a 2n  3b 5

n1 20

15

n4



. . . , 89

Find the common difference d and the value of a1 using the information given.

19. a1  2, d  3

31. 2, 7, 12, 17, . . .

47. 3, 0.5, 2, 4.5, 7, . . . , 47

62.

 17  2n2

n7

Use the alternative formula for the nth partial sum to compute the sums indicated.

63. The sum S15 for the sequence 12  19.52  172  14.52  p

64. The sum S20 for the sequence 92  72  52  32  p 65. The sum S30 for the sequence 0.003  0.173  0.343  0.513  p 66. The sum S50 for the sequence 122  172  1122  1172  p 67. The sum S20 for the sequence 12  2 12  3 12  4 12  p 68. The sum S10 for the sequence 12 13  10 13  8 13  6 13  p

WORKING WITH FORMULAS n1n  12 2 The sum of the first n natural numbers can be found using the formula shown, where n represents the number of terms in the sum. Verify the formula by adding the first six natural numbers by hand, and then evaluating S6. Then find the sum of the first 75 natural numbers.

69. Sum of the first n natural numbers: Sn 

70. Sum of the squares of the first n natural n1n  12 12n  12 numbers: Sn  6 If the first n natural numbers are squared, the sum of these squares can be found using the formula shown, where n represents the number of terms in the sum. Verify the formula by computing the sum of the squares of the first six natural numbers by hand, and then evaluating S6. Then find the sum of the squares of the first 20 natural numbers: 112  22  32  p  202 2.

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APPLICATIONS

71. Temperature fluctuation: At 5 P.M. in Coldwater, the temperature was a chilly 36°F. If the temperature decreased by 3°F every half-hour for the next 7 hr, at what time did the temperature hit 0°F?

has 88, the third row has 96, and so on. How many seats are in the 10th row? If there is room for 25 rows, how many chairs will be needed to set up the theater?

72. Arc of a baby swing: When Mackenzie’s baby swing is started, the first swing (one way) is a 30-in. arc. As the swing slows down, each successive arc is 3 2 in. less than the previous one. Find (a) the length of the tenth swing and (b) how far Mackenzie has traveled during the 10 swings.

75. Sales goals: At the time that I was newly hired, 100 sales per month was what I required. Each following month—the last plus 20 more, as I work for the goal of top sales award. When 2500 sales are thusly made, it’s Tahiti, Hawaii, and pina coladas in the shade. How many sales were made by this person in the seventh month? What were the total sales after the 12th month? Was the goal of 2500 total sales met after the 12th month?

73. Computer animations: The animation on a new computer game initially allows the hero of the game to jump a (screen) distance of 10 in. over booby traps and obstacles. Each successive jump is limited to 34 in. less than the previous one. Find (a) the length of the seventh jump and (b) the total distance covered after seven jumps. 74. Seating capacity: The Fox Theater creates a “theater in the round” when it shows any of Shakespeare’s plays. The first row has 80 seats, the second row



EXTENDING THE THOUGHT

77. From a study of numerical analysis, a function is known to be linear if its “first differences” (differences between each output) are constant. Likewise, a function is known to be quadratic if its “first differences” form an arithmetic sequence. Use this information to determine if the following sets of output come from a linear or quadratic function: a. 19, 11.8, 4.6, 2.6, 9.8, 17, 24.2, . . . b. 10.31, 10.94, 11.99, 13.46, 15.35, . . . 

76. Bequests to charity: At the time our mother left this Earth, she gave $9000 to her children of birth. This we kept and each year added $3000 more, as a lasting memorial from the children she bore. When $42,000 is thusly attained, all goes to charity that her memory be maintained. What was the balance in the sixth year? In what year was the goal of $42,000 met?

78. From elementary geometry it is known that the interior angles of a triangle sum to 180°, the interior angles of a quadrilateral sum to 360°, the interior angles of a pentagon sum to 540°, and so on. Use the pattern created by the relationship between the number of sides to the number of angles to develop a formula for the sum of the interior angles of an n-sided polygon. The interior angles of a decagon (10 sides) sum to how many degrees?

MAINTAINING YOUR SKILLS

79. (5.7) Identify the amplitude (A), period (P), horizontal shift (HS), vertical shift (VS) and endpoints of the primary interval (PI) for   f 1t2  7 sin a t  b  10. 3 6 80. (3.1) Graph by completing the square. Label all important features: y  x2  2x  3.

81. (2.3) In 2000, the deer population was 972. By 2005 it had grown to 1217. Assuming the growth is linear, find the function that models this data and use it to estimate the deer population in 2008. 82. (6.1) Verify identity.

sin x sin x  1 tan x   is an csc x cos x cot x

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10.3 Geometric Sequences Learning Objectives

Recall that arithmetic sequences are those where each term is found by adding a constant value to the preceding term. In this section, we consider geometric sequences, where each term is found by multiplying the preceding term by a constant value. Geometric sequences have many interesting applications, as do geometric series.

In Section 10.3 you will learn how to:

A. Identify a geometric sequence and its common ratio

A. Geometric Sequences

B. Find the nth term of a

A geometric sequence is one where each successive term is found by multiplying the preceding term by a fixed constant. Consider growth of a bacteria population, where a single cell splits in two every hour over a 24-hr period. Beginning with a single bacterium 1a0  12, after 1 hr there are 2, after 2 hr there are 4, and so on. Writing the number of bacteria as a sequence we have:

geometric sequence

C. Find the nth partial sum of a geometric sequence

D. Find the sum of an infinite geometric series

E. Solve application

bacteria:

problems involving geometric sequences and series

EXAMPLE 1

hours:

a1 T 2

a2 T 4

a3 T 8

a4 T 16

a5 T 32

... ...

The sequence 2, 4, 8, 16, 32, . . . is a geometric sequence since each term is found by multiplying the previous term by the constant factor 2. This also means that the ratio of any two consecutive terms must be 2 and in fact, 2 is called the common ratio r for ak1 , where ak this sequence. Using the notation from Section 10.1 we can write r  ak represents any term of the sequence and ak1 represents the term that follows ak. 

Testing a Sequence for a Common Ratio Determine if the given sequence is geometric. a. 1, 0.5, 0.25, 0.125, . . . b. 18, 14, 34, 3, 15, . . .

Solution



ak1 . ak a. For 1, 0.5, 0.25, 0.125, . . . , the ratio of consecutive terms gives 0.25 0.125 0.5 and so on.  0.5,  0.5,  0.5, 1 0.5 0.25 This is a geometric sequence with common ratio r  0.5. b. For 18, 14, 34, 3, 15, . . . , we have: 1 1 1 8 1 3 4 3 4 3 3   #   # and so on. 3  # 4 8 4 1 4 4 4 1 4 1 3 2 3 4

Apply the definition to check for a common ratio r 

Since the ratio is not constant, this is not a geometric sequence. Now try Exercises 7 through 24 EXAMPLE 2





Writing the Terms of a Geometric Sequence Write the first five terms of the geometric sequence, given the first term a1  16 and the common ratio r  0.25.

Solution



A. You’ve just learned how to identify a geometric sequence and its common ratio

Given a1  16 and r  0.25. Starting at a1  16, multiply each term by 0.25 to generate the sequence. a2  16 # 0.25  4 a4  1 # 0.25  0.25

a3  4 # 0.25  1 a5  0.25 # 0.25  0.0625

The first five terms of this sequence are 16, 4, 1, 0.25, and 0.0625. Now try Exercises 25 through 38

956



10-18

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957

B. Find the nth Term of a Geometric Sequence If the values a1 and r from a geometric sequence are known, we could generate the terms of the sequence by applying additional factors of r to the first term, instead of multiplying each new term by r. If a1  3 and r  2, we simply begin at a1, and continue applying additional factors of r for each successive term. 6  3 # 21

a2  a1r1

12  3 # 22

a3  a1r2

24  3 # 23

a4  a1r3

48  3 # 24

a5  a1r4

current term

initial term

S

a1  a1r0

S

3  3 # 20

exponent on common ratio

From this pattern, we note the exponent on r is always 1 less than the subscript of the current term: 5  1  4, which leads us to the formula for the nth term of a geometric sequence. The nth Term of a Geometric Sequence The nth term of a geometric sequence is given by an  a1r n1 where r is the common ratio. EXAMPLE 3



Finding a Specific Term in a Sequence Find the 10th term of the sequence 3, 6, 12, 24, . . . .

Solution



Instead of writing out all 10 terms, we determine the constant ratio r and use the nth term formula. By inspection we note that a1  3 and r  2. an  a1r n1  3122 n1

n th term formula substitute 3 for a1 and 2 for r

To find the 10th term we substitute n  10: a10  3122 101  3122 9  1536

substitute 10 for n simplify

Now try Exercises 39 through 46

EXAMPLE 4





Determining the Number of Terms in a Geometric Sequence 1 . Find the number of terms in the geometric sequence 4, 2, 1, . . . , 64

Solution



Observing that a1  4 and r  12, we have an  a1rn1 n th term formula 1 n1 1  4a b substitute 4 for a1 and for r 2 2 Although we don’t know the number of terms in the sequence, we do know the last 1 1 . Substituting an  64 or nth term is 64 gives 1 1 n1  4a b 64 2 1 1 n1 a b 256 2

substitute

1 for an 64

1 divide by 4 amultiply by b 4

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From our work in Chapter 4, we attempt to write both sides as exponentials with a like base, or apply logarithms. Since 256  28, we equate bases. 1 8 1 n1 a b a b 2 2 S8  n  1 9n

write

1 1 8 as a b 256 2

like bases imply exponents must be equal solve for n

This shows there are nine terms in the sequence. Now try Exercises 47 through 58



If the term a1 is unknown but a term ak is given, the nth term can be written an  akr nk, since n  k  1n  k2

(the subscript on the term ak and the exponent on r sum to n). EXAMPLE 5



Finding the First Term of a Geometric Sequence Given a geometric sequence where a4  0.075 and a7  0.009375, find the common ratio r and the value of a1.

Solution



Since a1 is not known, we express a7 as the product of a known term and the appropriate number of common ratios: a7  a4r 3 17  4  3, as required). a7  a4 # r 3 0.009375  0.075r3 0.125  r3 r  0.5

a1 is unknown substitute 0.009375 for a 7 and 0.075 for a4 divide by 0.075 solve for r

Having found r, we can now solve for a1

B. You’ve just learned how to find the nth term of a geometric sequence

a7  a1r6 0.009375  a1 10.52 6 0.009375  a1 10.0156252 a1  0.6

n th term formula substitute 0.009375 for a7 and 0.5 for r simplify solve for a1

The first term is a1  0.6 and the common ratio is r  0.5. Now try Exercises 59 through 64



C. Find the nth Partial Sum of a Geometric Sequence As with arithmetic series, applications of geometric series often involve computing a sum of consecutive terms. We can adapt the method for finding the sum of an arithmetic sequence to develop a formula for adding the first n terms of a geometric sequence. For the nth term an  a1r n1, we have Sn  a1  a1r  a1r2  a1r3  p  n1 . If we multiply Sn by r then add the original series, the “interior terms” sum a 1r to zero. 2  1a1r2  1a1r 3 2  p- 1a1rn1 2  1a1rn 2  rSn  a1r-----S --S --S -S S  Sn  a1  a1r  a1r 2  p  a1rn2  a1rn1

Sn  rSn 

a1 

0



0

 0 

0



We then have Sn  rSn  a1  a1rn, and can now solve for Sn: Sn  rSn  a1  a1r n Sn 11  r2  a1  a1r n a1  a1r n Sn  1r

difference of Sn and rSn factor out Sn solve for Sn

0

 1a1rn 2

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The result is a formula for the nth partial sum of a geometric sequence. The nth Partial Sum of a Geometric Sequence Given a geometric sequence with first term a1 and common ratio r, the nth partial sum (the sum of the first n terms) is a1 11  r n 2 a1  a1r n Sn   ,r1 1r 1r

In words: The sum of a geometric sequence is the difference of the first and 1n  12st term, divided by 1 minus the common ratio.

EXAMPLE 6



Computing a Partial Sum 9

Find the sum:

 3 (the first nine powers of 3). i

i1

Solution



The initial terms of this series are 3  9  27  p , and we note a1  3, r  3, and n  9. We could find the first nine terms and add, but using the partial sum formula gives a1 11  rn 2 1r 311  39 2 S9  13 3119,6822  2  29,523

Sn 

C. You’ve just learned how to find the nth partial sum of a geometric sequence

sum formula

substitute 3 for a1, 9 for n, and 3 for r

simplify result

Now try Exercises 65 through 88



D. The Sum of an Infinite Geometric Series To this point we’ve considered only partial sums of a geometric series. While it is impossible to add an infinite number of these terms, some of these “infinite sums” appear to have a limiting value. The sum appears to get ever closer to this value but never exceeds it— much like the asymptotic behavior of some graphs. We will define the sum of this infinite geometric series to be this limiting value, if it exists. Consider the illustration in Figure 10.7, where a standard sheet of typing paper is cut in half. One of the halves is again cut in half and the process is continued indefinitely, as 1 1 p , 32, with a1  12 and shown. Notice the “halves” create an infinite sequence 12, 14, 18, 16 1 1 1 1 1 1 r  2. The corresponding infinite series is 2  4  8  16  32  p  1n  p . 2

Figure 10.7 1 2

1 8 1 2

1 4

1 4 1 8

1 16

1 16

1 32 1 32

1 64

and so on

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CHAPTER 10 Additional Topics in Algebra

If we arrange one of the halves from each stage as shown in Figure 10.8, we would be rebuilding the original sheet of paper. As we add more and more of these halves together, we get closer and closer to the size of the original sheet. We gain an intuitive sense that this series must add to 1, because the pieces of the original sheet of paper must add to 1 whole sheet. To explore this idea further, consider what happens to 1 12 2 n as n becomes large.

Figure 10.8 1 64

1 16

1 32

1 4 1 8

1 4 n  4: a b  0.0625 2

1 2

The formula for the sum of an infinite geometric series can also be derived by noting that Sq  a1  a1r  a1r 2  a1r 3  p can be rewritten as Sq  a1  r 1a1  a1r  a1r 2  a1r 3  p 2  a1  rSq.

Infinite Geometric Series Given a geometric sequence with first term a1 and r  6 1, the sum of the related infinite series is given by

Sq  rSq  a1

Sq 11  r2  a1

Sq 

a1 . 1r

EXAMPLE 7

1 12 n  12: a b  0.0002 2

Further exploration with a calculator seems to support the idea that as n S q, 1 12 2 n S 0, although a definitive proof is left for a future course. In fact, it can be shown that for any r  6 1, r n becomes very close to zero as n becomes large. a1  a1r n a1 a1r n   , note that if In symbols: as n S q, r n S 0. For Sn  1r 1r 1r r  6 1 and “we sum an infinite number of terms,” the second term becomes zero, leaving only the first term. In other words, the limiting value (represented by Sq 2 is a1 Sq  . 1r

WORTHY OF NOTE

Sq 

1 8 n  8: a b  0.004 2

a1 ;r1 1r

If r  7 1, no finite sum exists.



Computing an Infinite Sum Find the limiting value of each infinite geometric series (if it exists). a. 1  2  4  8  p b. 3  2  43  89  p c. 0.185  0.000185  0.000000185  p

Solution



Begin by determining if the infinite series is geometric. If so, a1 use Sq  . 1r a. Since r  2 (by inspection), a finite sum does not exist. b. Using the ratio of consecutive terms we find r  23 and the infinite sum exists. With a1  3, we have Sq 

3 3 2  1  9 13 3

c. This series is equivalent to the repeating decimal 0.185185185 p  0.185. The common ratio is r  0.000185 0.185  0.001 and the infinite sum exists: D. You’ve just learned how to find the sum of an infinite geometric series

Sq 

0.185 5  1  0.001 27 Now try Exercises 89 through 104



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961

E. Applications Involving Geometric Sequences and Series Here are a few of the ways these ideas can be put to use. EXAMPLE 8



Solving an Application of Geometric Sequences: Pendulums A pendulum is any object attached to a fixed point and allowed to swing freely under the influence of gravity. Suppose each swing is 0.9 the length of the previous one. Gradually the swings become shorter and shorter and at some point the pendulum will appear to have stopped (although theoretically it never does). a. How far does the pendulum travel on its eighth swing, if the first swing was 2 m? b. What is the total distance traveled by the pendulum for these eight swings? c. How many swings until the length of each swing falls below 0.5 m? d. What total distance does the pendulum travel before coming to rest?

Solution



a. The lengths of each swing form the terms of a geometric sequence with a1  2 and r  0.9. The first few terms are 2, 1.8, 1.62, 1.458, and so on. For the 8th term we have: an  a1r n1 a8  210.92 81  0.956

n th term formula substitute 8 for n, 2 for a1, and 0.9 for r

The pendulum travels about 0.956 m on its 8th swing. b. For the total distance traveled after eight swings, we compute the value of S8. a1 11  r n 2 1r 211  0.98 2 S8  1  0.9  11.4 Sn 

n th partial sum formula substitute 2 for a1, 0.9 for r, and 8 for n

The pendulum has traveled about 11.4 m by the end of the 8th swing. c. To find the number of swings until the length of each swing is less than 0.5 m, we solve for n in the equation 0.5  210.92 n1. This yields 0.25  10.92 n1 ln 0.25  1n  12ln 0.9 ln 0.25 1n ln 0.9 14.16  n

divide by 2 take the natural log, apply power property solve for n (exact form) solve for n (approximate form)

After the 14th swing, each successive swing will be less than 0.5 m. d. For the total distance traveled before coming to rest, we consider the related infinite geometric series, with a1  2 and r  0.9. a1 1r 2 Sq  1  0.9  20

Sq 

E. You’ve just learned how to solve application problems involving geometric sequences and series

infinite sum formula substitute 2 for a1 and 0.9 for r result

The pendulum would travel 20 m before coming to rest. Now try Exercises 107 through 119



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CHAPTER 10 Additional Topics in Algebra

10.3 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. In a geometric sequence, each successive term is found by the preceding term by a fixed value r. 2. In a geometric sequence, the common ratio r can be found by computing the of any two consecutive terms. 3. The nth term of a geometric sequence is given by an  , for any n  1. 

4. For the general sequence a1, a2, a3, p , ak, p , the fifth partial sum is given by S5  . 5. Describe/Discuss how the formula for the nth partial sum is related to the formula for the sum of an infinite geometric series. 6. Describe the difference(s) between an arithmetic and a geometric sequence. How can a student prevent confusion between the formulas?

DEVELOPING YOUR SKILLS

Determine if the sequence given is geometric. If yes, name the common ratio. If not, try to determine the pattern that forms the sequence.

7. 4, 8, 16, 32, p 8. 2, 6, 18, 54, 162, p 9. 3, 6, 12, 24, 48, p 10. 128, 32, 8, 2, p 11. 2, 5, 10, 17, 26, p 12. 13, 9, 5, 1, 3, p 13. 3, 0.3, 0.03, 0.003, p

23. 240, 120, 40, 10, 2, p 24. 120, 60, 20, 5, 1, p Write the first four terms of the sequence, given a1 and r.

25. a1  5, r  2 27. a1  6, r 

26. a1  2, r  4 12

29. a1  4, r  13

30. a1  15, r  15

31. a1  0.1, r  0.1

32. a1  0.024, r  0.01

Find the indicated term for each sequence.

14. 12, 0.12, 0.0012, 0.000012, p

33. a1  24, r  12; find a7

15. 1, 3, 12, 60, 360, p

34. a1  48, r  13; find a6

16. 23, 2, 8, 40, 240, p 17. 25, 10, 4, 85, p 18. 36, 24, 16, 32 3,p 1 19. 12, 14, 18, 16 ,p 8 16 20. 23, 49, 27 , 81, p

21. 3,

12 48 192 ,p , , x x2 x3

22. 5,

10 20 40 , , ,p a a2 a3

28. a1  23, r  15

1 , r  5; find a4 35. a1  20

36. a1 

3 20 ,

r  4; find a5

37. a1  2, r  12; find a7 38. a1  13, r  13; find a8 Identify a1 and r, then write the expression for the nth term an  a1r n1 and use it to find a6, a10, and a12.

39.

1 27 ,

19, 13, 1, 3, p

41. 729, 243, 81, 27, 9, p 43. 12,

12 2 ,

1, 12, 2, p

40. 78, 74, 72, 7, 14, p 42. 625, 125, 25, 5, 1, p

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Section 10.3 Geometric Sequences

44. 36 13, 36, 12 13, 12, 413, p

 16  12  p ; find S7

76.

1 18

77.

4

5

45. 0.2, 0.08, 0.032, 0.0128, p 46. 0.5, 0.35, 0.245, 0.1715, p

10

j

47. a1  9, an  729, r  3

2 k1 5a b 3 k1

7



80.

10

48. a1  1, an  128, r  2

81.

1 49. a1  16, an  64 , r  12 1 50. a1  4, an  512 , r  12

1 i1 9a b 2 i4



83. a2  5, a5 

53. 2, 6, 18, 54, p , 4374

84. a3  1, a6  27; find S6

54. 3, 6, 12, 24, p , 6144

9 85. a3  49, a7  64 ; find S6

55. 64, 32 12, 32, 16 12, p , 1

2 86. a2  16 81 , a5  3 ; find S8

56. 243, 81 13, 81, 27 13, p , 1

87. a3  2 12, a6  8; find S7

57. 38, 34, 32, 3, p , 96

88. a2  3, a5  9 13; find S7

58.

5, p , 135

Find the common ratio r and the value of a1 using the information given (assume r  0).

1 25 ;

60. a5  6, a9  486

61. a4 

62. a2  16 81 , a5 

91. 9  3  1  p

a8 

63. a4  32 3 , a8  54

2 3

64. a3  16 25 , a7  25

Find the indicated sum. For Exercises 81 and 82, use the summation properties from Section 10.1.

65. a1  8, r  2; find S12 66. a1  2, r  3; find S8 67. a1  96, r  13; find S5 68. a1  12, r  12; find S8 69. a1  8, r  32; find S7 70. a1  1, r  32; find S10 71. 2  6  18  p ; find S6 72. 2  8  32  p ; find S7 73. 16  8  4  p ; find S8 74. 4  12  36  p ; find S8 1 75. 43  29  27  p ; find S9



Determine whether the infinite geometric series has a finite sum. If so, find the limiting value.

59. a3  324, a7  64 9 4

1 i1 5a b 4 i3

find S5

89. 3  6  12  24  p 90. 4  8  16  32  p

4 9,

 8

82.

52. a1  2, an  1458, r   13

53,

1 j1 3a b 5 j1

Find the indicated partial sum using the information given. Write all results in simplest form.

51. a1  1, an  1296, r  16

5 5 27 , 9,

k

k1

8

79.

2

78.

j1

Find the number of terms in each sequence.

963

92. 36  24  16  p 93. 25  10  4  85  p 2 94. 10  2  25  25 p 95. 6  3  32  34  p

96. 49  172  117 2  p 97. 6  3  32  34  p 98. 10  5  52  54  p 99. 0.3  0.03  0.003  p 100. 0.63  0.0063  0.000063  p 101.

3 2 k a b k1 4 3



102.

1 i 5a b 2 i1

103.

2 j 9a b 3 j1

104.

4 k 12a b 3 k1

q

q

q



 q



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WORKING WITH FORMULAS

105. Sum of the cubes of the first n natural numbers: n2(n  1)2 Sn  4 Compute 13  23  33  p  83 using the formula given. Then confirm the result by direct calculation.



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CHAPTER 10 Additional Topics in Algebra

106. Student loan payment: An  P(1  r)n If P dollars is borrowed at an annual interest rate r with interest compounded annually, the amount of money to be paid back after n years is given by the indicated formula. Find the total amount of money that the student must repay to clear the loan, if $8000 is borrowed at 4.5% interest and the loan is paid back in 10 yr.

APPLICATIONS

107. Pendulum movement: On each swing, a pendulum travels only 80% as far as it did on the previous swing. If the first swing is 24 ft, how far does the pendulum travel on the 7th swing? What total distance is traveled before the pendulum comes to rest? 108. Pendulum movement: Ernesto is swinging to and fro on his backyard tire swing. Using his legs and body, he pumps each swing until reaching a maximum height, then suddenly relaxes until the swing comes to a stop. With each swing, Ernesto travels 75% as far as he did on the previous swing. If the first arc (or swing) is 30 ft, find the distance Ernesto travels on the 5th arc. What total distance will he travel before coming to rest?

111. Equipment aging: Tests have shown that the pumping power of a heavy-duty oil pump decreases by 3% per month. If the pump can move 160 gallons per minute (gpm) new, how many gpm can the pump move 8 months later? If the pumping rate falls below 118 gpm, the pump must be replaced. How many months until this pump is replaced? 112. Equipment aging: At the local mill, a certain type of saw blade can saw approximately 2 log-feet/sec when it is new. As time goes on, the blade becomes worn, and loses 6% of its cutting speed each week. How many log-feet/sec can the saw blade cut after 6 weeks? If the cutting speed falls below 1.2 logfeet/sec, the blade must be replaced. During what week of operation will this blade be replaced? 113. Population growth: At the beginning of the year 2000, the population of the United States was approximately 277 million. If the population is growing at a rate of 2.3% per year, what will the population be in 2010, 10 yr later?

109. Depreciation: A certain new SUV depreciates in value about 20% per year (meaning it holds 80% of its value each year). If the SUV is purchased for $46,000, how much is it worth 4 yr later? How many years until its value is less than $5000? 110. Depreciation: A new photocopier under heavy use will depreciate about 25% per year (meaning it holds 75% of its value each year). If the copier is purchased for $7000, how much is it worth 4 yr later? How many years until its value is less than $1246?

114. Population growth: The population of the Zeta Colony on Mars is 1000 people. Determine the population of the Colony 20 yr from now, if the population is growing at a constant rate of 5% per year. 115. Population growth: A biologist finds that the population of a certain type of bacteria doubles each half-hour. If an initial culture has 50 bacteria, what is the population after 5 hr? How long will it take for the number of bacteria to reach 204,800? 116. Population growth: Suppose the population of a “boom town” in the old west doubled every 2 months after gold was discovered. If the initial population was 219, what was the population 8 months later? How many months until the population exceeds 28,000?

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117. Elastic rebound: Megan discovers that a rubber ball dropped from a height of 2 m rebounds four-fifths of the distance it has previously fallen. How high does it rebound on the 7th bounce? How far does the ball travel before coming to rest? 118. Elastic rebound: The screen saver on my computer is programmed to send a colored ball vertically down the middle of the screen so that it rebounds 95% of the distance it last traversed. If the ball always begins at the top and the screen is 36 cm tall,



965

how high does the ball bounce after its 8th rebound? How far does the ball travel before coming to rest (and a new screen saver starts)? 119. Creating a vacuum: To create a vacuum, a hand pump is used to remove the air from an air-tight cube with a volume of 462 in3. With each stroke of the pump, two-fifths of the air that remains in the cube is removed. How much air remains inside after the 5th stroke? How many strokes are required to remove all but 12.9 in3 of the air?

EXTENDING THE CONCEPT

120. As part of a science experiment, identical rubber balls are dropped from a certain height on these surfaces: slate, cement, and asphalt. When dropped on slate, the ball rebounds 80% of the height from which it last fell. On cement the figure is 75% and on asphalt the figure is 70%. The ball is dropped from 130 m on the slate, 175 m on the cement, and 200 m on the asphalt. Which ball has traveled the shortest total distance at the time of the fourth bounce? Which ball will travel farthest before coming to rest? 121. Consider the following situation. A person is hired at a salary of $40,000 per year, with a guaranteed raise of $1750 per year. At the same time, inflation is running about 4% per year. How many years until this person’s salary is overtaken and eaten up by the actual cost of living? 122. A standard piece of typing paper is approximately 0.001 in. thick. Suppose you were able to fold this 

Section 10.3 Geometric Sequences

piece of paper in half 26 times. How thick would the result be? (a) As tall as a hare, (b) as tall as a hen, (c) as tall as a horse, (d) as tall as a house, or (e) over 1 mi high? Find the actual height by computing the 27th term of a geometric sequence. Discuss what you find. 123. Find an alternative formula for the sum n

Sn 

 log k, that does not use the sigma notation.

k1

124. Verify the following statements: a. If a1, a2, a3, p , an is a geometric sequence with r and a1 greater than zero, then log a1, log a2, log a3, p , log an is an arithmetic sequence. b. If a1, a2, a3, p , an is an arithmetic sequence, then 10a1, 10a2, p , 10an, is a geometric sequence.

MAINTAINING YOUR SKILLS

125. (1.5) Find the zeroes of f using the quadratic formula: f 1x2  x2  5x  9. 126. (7.3) Find a unit vector in the same direction as 3i  7j. 127. (4.6) Graph the rational function: x2 h1x2  x1

128. (5.1) The cars on the Millenium Ferris Wheel are 100 ft from the center axle. If the top speed of the wheel is 1.5 revolutions per minute, find the linear velocity of a passenger in a car. Round your answer to the nearest whole number. Also, give the velocity in miles per hour.

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10.4 Mathematical Induction Learning Objectives

Since middle school (or even before) we have accepted that, “The product of two negative numbers is a positive number.” But have you ever been asked to prove it? It’s not as easy as it seems. We may think of several patterns that yield the result, analogies that indicate its truth, or even number line illustrations that lead us to believe the statement. But most of us have never seen a proof (see www.mhhe.com/coburn). In this section, we introduce one of mathematics’ most powerful tools for proving a statement, called proof by induction.

In Section 10.4 you will learn how to:

A. Use subscript notation to evaluate and compose functions

B. Apply the principle of mathematical induction to sum formulas involving natural numbers

A. Subscript Notation and Composition of Functions One of the challenges in understanding a proof by induction is working with the notation. Earlier in the chapter, we introduced subscript notation as an alternative to function notation, since it is more commonly used when the functions are defined by a sequence. But regardless of the notation used, the functions can still be simplified, evaluated, composed, and even graphed. Consider the function f 1x2  3x2  1 and the sequence defined by an  3n2  1. Both can be evaluated and graphed, with the only difference being that f(x) is continuous with domain x  , while an is discrete (made up of distinct points) with domain n  .

C. Apply the principle of mathematical induction to general statements involving natural numbers

EXAMPLE 1



Solution



A. You’ve just learned how to use subscript notation to evaluate and compose functions

Using Subscript Notation for a Composition

For f 1x2  3x2  1 and an  3n2  1, find f 1k  12 and ak1. f 1k  12  31k  12 2  1  31k2  2k  12  1  3k2  6k  2

ak1  31k  12 2  1  31k2  2k  12  1  3k2  6k  2 Now try Exercises 7 through 18



No matter which notation is used, every occurrence of the input variable is replaced by the new value or expression indicated by the composition.

B. Mathematical Induction Applied to Sums Consider the sum of odd numbers 1  3  5  7  9  11  13  p . The sum of the first four terms is 1  3  5  7  16, or S4  16. If we now add a5 (the next term in line), would we get the same answer as if we had simply computed S5? Common sense would say, “Yes!” since S5  1  3  5  7  9  25 and S4  a5  16  9  25✓. In diagram form, we have add next term a5  9 to S4

>

1  3  5  7  9  11  13  15  p c c S4 S5

sum of 4 terms sum of 5 terms

Our goal is to develop this same degree of clarity in the notational scheme of things. For a given series, if we find the kth partial sum Sk (shown next) and then add the next term ak1, would we get the same answer if we had simply computed Sk1? In other words, is Sk  ak1  Sk1 true?

966

10-28

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Section 10.4 Mathematical Induction

967

add next term ak1

>

a1  a2  a3  p  ak1  ak  ak1  p  an–1  an c c sum of k terms Sk sum of k  1 terms

Sk1

Now, let’s return to the sum 1  3  5  7  p  2n  1. This is an arithmetic series with a1  1, d  2, and nth term an  2n  1. Using the sum formula for an arithmetic sequence, an alternative formula for this sum can be established. Sn 

n1a1  an 2 2



n11  2n  12 2

substitute 1 for a1 and 2n 1 for an



n12n2 2

simplify

 n2

WORTHY OF NOTE No matter how distant the city or how many relay stations are involved, if the generating plant is working and the k th station relays to the (k  1)st station, the city will get its power.

summation formula for an arithmetic sequence

result

This shows that the sum of the first n positive odd integers is given by Sn  n2. As a check we compute S5  1  3  5  7  9  25 and compare to S5  52  25✓. We also note S6  62  36, and S5  a6  25  11  36, showing S6  S5  a6. For more on this relationship, see Exercises 19 through 24. While it may seem simplistic now, showing S5  a6  S6 and Sk  ak1  Sk1 (in general) is a critical component of a proof by induction. Unfortunately, general summation formulas for many sequences cannot be established from known formulas. In addition, just because a formula works for the first few values of n, we cannot assume that it will hold true for all values of n (there are infinitely many). As an illustration, the formula an  n2  n  41 yields a prime number for every natural number n from 1 to 40, but fails to yield a prime for n  41. This helps demonstrate the need for a more conclusive proof, particularly when a relationship appears to be true, and can be “verified” in a finite number of cases, but whether it is true in all cases remains in question. Proof by induction is based on a relatively simple idea. To help understand how it works, consider n relay stations that are used to transport electricity from (k  1)st kth Generating plant a generating plant to a distant relay relay city. If we know the generating plant is operating, and if we assume that the kth relay station (any station in the series) is making the transfer to the 1k  12st station (the next station in the series), then we’re sure the city will have electricity. This idea can be applied mathematically as follows. Consider the statement, “The sum of the first n positive, even integers is n2  n.” In other words, 2  4  6  8  p  2n  n2  n. We can certainly verify the statement for the first few even numbers: The first even number is 2 and . . . The sum of the first two even numbers is 2  4  6 and . . . The sum of the first three even numbers is 2  4  6  12 and . . . The sum of the first four even numbers is 2  4  6  8  20 and . . .

112 2  1  2 122 2  2  6

132 2  3  12 142 2  4  20

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CHAPTER 10 Additional Topics in Algebra

While we could continue this process for a very long time (or even use a computer), no finite number of checks can prove a statement is universally true. To prove the statement true for all positive integers, we use a reasoning similar to that applied in the relay stations example. If we are sure the formula works for n  1 (the generating station is operating), and assume that if the formula is true for n  k, it must also be true for n  k  1 [the kth relay station is transferring electricity to the 1k  12st station], then the statement is true for all n (the city will get its electricity). The case where n  1 is called the base case of an inductive proof, and the assumption that the formula is true for n  k is called the induction hypothesis. When the induction hypothesis is applied to a sum formula, we attempt to show that Sk  ak1  Sk1. Since k and k  1 are arbitrary, the statement must be true for all n. Mathematical Induction Applied to Sums WORTHY OF NOTE To satisfy our finite minds, it might help to show that Sn is true for the first few cases, prior to extending the ideas to the infinite case.

EXAMPLE 2



Let Sn be a sum formula involving positive integers. If 1. S1 is true, and 2. the truth of Sk implies that Sk1 is true, then Sn must be true for all positive integers n. Both parts 1 and 2 must be verified for the proof to be complete. Since the process requires the terms Sk, ak1, and Sk1, we will usually compute these first.

Proving a Statement Using Mathematical Induction Use induction to prove that the sum of the first n perfect squares is given by n1n  12 12n  12 . 1  4  9  16  25  p  n2  6

Solution



Given an  n2 and Sn 

n1n  1212n  12 , the needed components are . . . 6

For an  n2: ak  k2 For Sn 

and

ak1  1k  12 2

n1n  12 12n  12 k1k  1212k  12 1k  121k  2212k  32 and Sk1  : Sk  6 6 6

1. Show Sn is true for n  1. n1n  1212n  12 6 1122132 S1  6  1✓

Sn 

2. Assume Sk is true, 1  4  9  16  p  k2 

sum formula base case: n  1 result checks, the first term is 1

k1k  12 12k  12 6

and use it to show the truth of Sk1 follows. That is, ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

⎫ ⎬ ⎭

1  4  9  16  p  k2  1k  12 2  Sk

ak1

induction hypothesis: Sk is true

1k  121k  22 12k  32 6 Sk1

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Section 10.4 Mathematical Induction

⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭

Working with the left-hand side, we have 1  4  9  16  p  k2  1k  12 2 

  B. You’ve just learned how to apply the principle of mathematical induction to sum formulas involving natural numbers

 

k1k  12 12k  12  1k  12 2 6 k1k  12 12k  12  61k  12 2 6 1k  12 3 k12k  12  61k  12 4 6 1k  12 3 2k2  7k  64 6 1k  121k  22 12k  32 6

use the induction hypothesis: substitute k1k  12 12k  12 6

for 1  4  9  16  25  p  k2

common denominator

factor out k  1

multiply and combine terms

factor the trinomial, result is Sk1✓

Since the truth of Sk1 follows from Sk, the formula is true for all n. Now try Exercises 27 through 38



C. The General Principle of Mathematical Induction Proof by induction can be used to verify many other kinds of relationships involving a natural number n. In this regard, the basic principles remain the same but are stated more broadly. Rather than having Sn represent a sum, we take it to represent any statement or relationship we might wish to verify. This broadens the scope of the proof and makes it more widely applicable, while maintaining its value to the sum formulas verified earlier. The General Principle of Mathematical Induction Let Sn be a statement involving natural numbers. If 1. S1 is true, and 2. the truth of Sk implies that Sk1 is also true then Sn must be true for all natural numbers n.

EXAMPLE 3



Proving a Statement Using the General Principle of Mathematical Induction Use the general principle of mathematical induction to show the statement Sn is true for all natural numbers n. Sn: 2n  n  1

Solution



The statement Sn is defined as 2n  n  1. This means that Sk is represented by 2k  k  1 and Sk1 by 2k1  k  2. 1. Show Sn is true for n  1: Sn: S1:

2n  n  1 21  1  1 2  2✓

given statement base case: n  1 true

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Although not a part of the formal proof, a table of values can help to illustrate the relationship we’re trying to establish. It appears that the statement is true. n

1

2

3

4

5

n

2

4

8

16

32

2

3

4

5

6

2

n1

2. Assume that Sk is true, Sk: 2k  k  1

induction hypothesis

and use it to show that the truth of Sk1. That is, 2k1  k  2.

Sk1:

Begin by working with the left-hand side of the inequality, 2k1. 2k1  212k 2

 21k  12

 2k  2

properties of exponents induction hypothesis: substitute k  1 for 2k (symbol changes since k  1 is less than 2k) distribute

WORTHY OF NOTE

Since k is a positive integer, 2k  2  k  2,

Note there is no reference to an, ak, or ak1 in the statement of the general principle of mathematical induction.

showing 2k1  k  2.

EXAMPLE 4

Since the truth of Sk1 follows from Sk, the formula is true for all n. Now try Exercises 39 through 42



Proving Divisibility Using Mathematical Induction Let Sn be the statement, “4n  1 is divisible by 3 for all positive integers n.” Use mathematical induction to prove that Sn is true.

Solution



If a number is evenly divisible by three, it can be written as the product of 3 and some positive integer we will call p.

1. Show Sn is true for n  1: 4n  1  3p 4112  1  3p 3  3p✓

Sn: S1:

4n  1  3p, p   substitute 1 for n statement is true for n  1

2. Assume that Sk is true . . . Sk:

4k  1  3p

induction hypothesis

4  3p  1 k

and use it to show the truth of Sk1. That is, Sk1:

4k1  1  3q for q   is also true.

Beginning with the left-hand side we have: 4k1  1  4 # 4k  1

 4 # 13p  12  1  12p  3  314p  12  3q

properties of exponents induction hypothesis: substitute 3p  1 for 4k distribute and simplify factor



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The last step shows 4k1  1 is divisible by 3. Since the original statement is true for n  1, and the truth of Sk implies the truth of Sk1, the statement, “4n  1 is divisible by 3” is true for all positive integers n. Now try Exercises 43 through 47

C. You’ve just learned how to apply the principle of mathematical induction to general statements involving natural numbers



We close this section with some final notes. Although the base step of a proof by induction seems trivial, both the base step and the induction hypothesis are necessary 1 1 parts of the proof. For example, the statement n 6 is false for n  1, but true for 3 3n all other positive integers. Finally, for a fixed natural number p, some statements are false for all n 6 p, but true for all n  p. By modifying the base case to begin at p, we can use the induction hypothesis to prove the statement is true for all n greater than p. For example, n 6 13 n2 is false for n 6 4, but true for all n  4.

10.4 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. No statement

number of verifications can prove a true.

2. Showing a statement is true for n  1 is called the of an inductive proof. 3. Assuming that a statement/formula is true for . n  k is called the



4. The graph of a sequence is made up of distinct points.

, meaning it is

5. Explain the equation Sk  ak1  Sk1. Begin by saying, “Since the kth term is arbitrary . . .” (continue from here). 6. Discuss the similarities and differences between mathematical induction applied to sums and the general principle of mathematical induction.

DEVELOPING YOUR SKILLS

For the given nth term an, find a4, a5, ak, and ak1.

7. an  10n  6 9. an  n

8. an  6n  4 10. an  7n

11. an  2n1

12. an  213n1 2

For the given sum formula Sn, find S4, S5, Sk, and Sk1.

Verify that S4  a5  S5 for each exercise. These are identical to Exercises 13 through 18.

19. an  10n  6; Sn  n15n  12 20. an  6n  4; Sn  n13n  12 21. an  n; Sn 

n1n  12 2

13. Sn  n15n  12

14. Sn  n13n  12

n1n  12 15. Sn  2

7n1n  12 16. Sn  2

22. an  7n; Sn 

17. Sn  2  1

18. Sn  3  1

24. an  213n1 2; Sn  3n  1

n

n

7n1n  12 2

23. an  2n1; Sn  2n  1

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WORKING WITH FORMULAS

25. Sum of the first n cubes (alternative form): (1 + 2 + 3 + 4 + p + n)2

26. Powers of the imaginary unit: in 4  in, where i = 1 1

Earlier we noted the formula for the sum of the n2 1n  12 2 . An alternative is given first n cubes was 4 by the formula shown. a. Verify the formula for n  1, 5, and 9. b. Verify the formula using n1n  12 123pn . 2 

APPLICATIONS

Use mathematical induction to prove the indicated sum formula is true for all natural numbers n.

27. 2  4  6  8  10  p  2n; an  2n, Sn  n1n  12 28. 3  7  11  15  19  p  14n  12; an  4n  1, Sn  n12n  12 29. 5  10  15  20  25  p  5n; 5n1n  12 an  5n, Sn  2 30. 1  4  7  10  13  p  13n  22; n13n  12 an  3n  2, Sn  2 31. 5  9  13  17  p  14n  12; an  4n  1, Sn  n12n  32 32. 4  12  20  28  36  p  18n  42; an  8n  4, Sn  4n2 33. 3  9  27  81  243  p  3n; 313n  12 n an  3 , Sn  2 34. 5  25  125  625  p  5n; 515n  12 an  5n, Sn  4 35. 2  4  8  16  32  64  p  2n; an  2n, Sn  2n1  2 

Use a proof by induction to prove that powers of the imaginary unit are cyclic. That is, that they cycle through the numbers i, 1, i, and 1 for consecutive powers.

36. 1  8  27  64  125  216  p  n3; n2 1n  12 2 an  n3, Sn  4 37.

38.

1 1 1 1   p ; 1132 3152 5172 12n  1212n  12 1 n , Sn  an  12n  12 12n  12 2n  1 1 1 1 1   p ; 1122 2132 3142 n1n  12 1 n an  , Sn  n1n  12 n1

Use the principle of mathematical induction to prove that each statement is true for all natural numbers n.

39. 3n  2n  1

40. 2n  n  1

41. 3 # 4n1  4n  1

42. 4 # 5n1  5n  1

43. n2  7n is divisible by 2 44. n3  n  3 is divisible by 3 45. n3  3n2  2n is divisible by 3 46. 5n  1 is divisible by 4 47. 6n  1 is divisible by 5

EXTENDING THE THOUGHT

48. You may have noticed that the sum formula for the first n integers was quadratic, and the formula for the first n integer squares was cubic. Is the formula for the first n integer cubes, if it exists, a quartic (degree four) function? Use your calculator to run a quartic regression on the first five perfect cubes (enter 1

through 5 in L1 and the cumulative sums in L2). What did you find? Use proof by induction to show that the sum of the first n cubes is: n2 1n  12 2 n4  2n3  n2 1  8  27  p  n3   . 4 4

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Mid-Chapter Check

49. Use mathematical induction to prove that xn  1  11  x  x2  x3  p  xn1 2. x1



973

50. Use mathematical induction to prove that for 14  24  34  p  n4, where an  n4, n1n  12 12n  12 13n2  3n  12 . Sn  30

MAINTAINING YOUR SKILLS

51. (6.2) Verify the identity 1sin   cos 2 2  1sin   cos 2 2  2.

53. (2.1) State the equation of the circle whose graph is shown here.

52. (2.7) State the domain and range of the piecewise function shown here.

y 10 (1, 7) 8 6 (4, 3) 4 2

y 5 4 3 (1, 1) 2 1 54321 1 2 3 4 5

1 2 3

108642 2 4 6 8 10

5 x

(3, 2)

2 4 6 8 10 x

54. (7.4) Given p  H13, 1I and q  H1, 1I, find a. the dot product p # q b. the angle between the vectors

MID-CHAPTER CHECK c. sum of a finite geometric series d. summation formula for an arithmetic series e. nth term formula for a geometric series

In Exercises 1 to 3, the nth term is given. Write the first three terms of each sequence and find a9. 1. an  7n  4

2. an  n2  3

3. an  112 n 12n  12 4

4. Evaluate the sum

3

n1

n1

5. Rewrite using sigma notation. 1  4  7  10  13  16 Match each formula to its correct description.

11. Identify a1 and the common difference d. Then find an expression for the general term an. a. 2, 5, 8, 11, . . . b. 32, 94, 3, 15 4 , ... Find the number of terms in each series and then find the sum. 12. 2  5  8  11  p  74 1 2

 32  52  72  p  31 2

n1a1  an 2 6. Sn  2

13.

7. an  a1rn1

15. For a geometric series, a3  81 and a7  1. Find S10.

a1 8. Sq  1r

9. an  a1  1n  12d

a1 11  rn 2 1r a. sum of an infinite geometric series

10. Sn 

b. nth term formula for an arithmetic series

14. For an arithmetic series, a3  8 and a7  4. Find S10.

16. Identify a1 and the common ratio r. Then find an expression for the general term an. 1 a. 2, 6, 18, 54, . . . b. 21, 14, 18, 16 ,... 17. Find the number of terms in the series then compute the sum. 1 1 1 p  81 54  18  6  2

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18. Find the infinite sum (if it exists). 49  172  112 

117 2

p

19. Barrels of toxic waste are stacked at a storage facility in pyramid form, with 60 barrels in the first row, 59 in the second row, and so on, until there are 10 barrels in the top row. How many barrels are in the storage facility?

20. As part of a conditioning regimen, a drill sergeant orders her platoon to do 25 continuous standing broad jumps. The best of these recruits was able to jump 96% of the distance from the previous jump, with a first jump distance of 8 ft. Use a sequence/ series to determine the distance the recruit jumped on the 15th try, and the total distance traveled by the recruit after all 25 jumps.

REINFORCING BASIC CONCEPTS 12

Applications of Summation

IV:

The properties of summation play a large role in the development of key ideas in a first semester calculus course, and the following summation formulas are an integral part of these ideas. The first three formulas were verified in Section 10.4, while proof of the fourth was part of Exercise 48 on page 972. n n n1n  12 c  cn i (1) (2) 2 i1 i1 2 n n n1n  12 12n  12 n 1n  12 2 i2  i3  (3) (4) 6 4 i1 i1









To see the various ways they can be applied consider the following. Illustration 1  Over several years, the owner of Morgan’s LawnCare has noticed that the company’s monthly profits (in thousands) can be approximated by the sequence an  0.0625n3  1.25n2  6n, with the points plotted in Figure 10.9 (the continuous graph is shown for effect only). Find the company’s approximate annual profit. Figure 10.9 12

0

12

0

Solution  The most obvious approach would be to simply compute terms a1 through a12 (January through December) and find their sum: sum(seq(Y1, X, 1, 12) (see Section 10.1 Technology Highlight), which gives a result of 35.75 or $35,750. As an alternative, we could add the amount of profit earned by the company in the first 8 months, then add the amount the company lost (or broke even) during the last 4 months. In other words, we could apply summation property

a

8

n



12

a

n

a



n (see Figure 10.10), which gives

the same result: 42  16.252  35.75 or $35,750. Figure 10.10 i1

i1

i9

As a third option, we could use summation properties along with the appropriate summation formulas, and compute the result manually. Note the function is now written in terms of “i.” Distribute summations and factor out constants (properties II and III): 12

 10.0625i

3

 1.25i2  6i2 

i1

12

0.0625

i

i1

12

3

 1.25

i

i1

12

2

6

i

i1

Replace each summation with the appropriate summation formula, substituting 12 for n: n2 1n  12 2 n1n  1212n  12  0.0625 c d  1.25 c d 4 6 n1n  12  6c d 2 2 2 1122 11321252 1122 1132 d  1.25 c d  0.0625 c 4 6 1122 1132  6c d 2  0.0625160842  1.2516502  61782 or 35.75 As we expected, the result shows profit was $35,750. While some approaches seem “easier” than others, all have great value, are applied in different ways at different times, and are necessary to adequately develop key concepts in future classes. Exercise 1: Repeat Illustration 1 if the profit sequence is an  0.125x3  2.5x2  12x.

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10.5 Counting Techniques Learning Objectives In Section 10.5 you will learn how to:

A. Count possibilities using lists and tree diagrams

B. Count possibilities using the fundamental principle of counting

How long would it take to estimate the number of fans sitting shoulder-to-shoulder at a sold-out basketball game? Well, it depends. You could actually begin counting 1, 2, 3, 4, 5, . . . , which would take a very long time, or you could try to simplify the process by counting the number of fans in the first row and multiplying by the number of rows. Techniques for “quick-counting” the objects in a set or various subsets of a large set play an important role in a study of probability.

A. Counting by Listing and Tree Diagrams

C. Quick-count distinguishable permutations

Consider the simple spinner shown in Figure 10.11, which is divided into three equal parts. What are the different possible outcomes for two spins, spin 1 followed by spin 2? We might begin by organizing the possibilities using a tree diagram. As the name implies, each choice or possibility appears as the branch of a tree, with the total possibilities being equal to the number of (unique) paths from the beginning point to the end of a branch. Figure 10.12 shows how the spinner exercise would appear (possibilities for two spins). Moving from top to bottom we can trace nine possible paths: AA, AB, AC, BA, BB, BC, CA, CB, and CC.

D. Quick-count nondistinguishable permutations

E. Quick-count using combinations

Figure 10.12

Figure 10.11

Begin B A

C A A

EXAMPLE 1



B

B C

A

B

C C

A

B

C

Listing Possibilities Using a Tree Diagram A basketball player is fouled and awarded three free throws. Let H represent the possibility of a hit (basket is made), and M the possibility of a miss. Determine the possible outcomes for the three shots using a tree diagram.

Solution



Each shot has two possibilities, hit (H) or miss (M), so the tree will branch in two directions at each level. As illustrated in the figure, there are a total of eight possibilities: HHH, HHM, HMH, HMM, MHH, MHM, MMH, and MMM. Begin

H

M

H H

WORTHY OF NOTE Sample spaces may vary depending on how we define the experiment, and for simplicity’s sake we consider only those experiments having outcomes that are equally likely.

10-37

M M

H

H M

H

M M

H

M

Now try Exercises 7 through 10



To assist our discussion, an experiment is any task that can be done repeatedly and has a well-defined set of possible outcomes. Each repetition of the experiment is called a trial. A sample outcome is any potential outcome of a trial, and a sample space is a set of all possible outcomes. In our first illustration, the experiment was spinning a spinner, there were three sample outcomes (A, B, or C), the experiment had two trials (spin 1 and spin 2), and 975

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there were nine elements in the sample space. Note that after the first trial, each of the three sample outcomes will again have three possibilities (A, B, and C ). For two trials we have 32  9 possibilities, while three trials would yield a sample space with 33  27 possibilities. In general, we have A “Quick-Counting” Formula for a Sample Space If an experiment has N sample outcomes that are equally likely and the experiment is repeated t times, the number of elements in the sample space is N t. EXAMPLE 2



Counting the Outcomes in a Sample Space Many combination locks have the digits 0 through 39 arranged along a circular dial. Opening the lock requires stopping at a sequence of three numbers within this range, going counterclockwise to the first number, clockwise to the second, and counterclockwise to the third. How many three-number combinations are possible?

Solution



A. You’ve just learned how to count possibilities using lists and tree diagrams

35

5 10

30

25 There are 40 sample outcomes 1N  402 in this experiment, and 20 three trials 1t  32. The number of possible combinations is identical to the number of elements in the sample space. The quick-counting formula gives 403  64,000 possible combinations.

15

Now try Exercises 11 and 12



B. Fundamental Principle of Counting The number of possible outcomes may differ depending on how the event is defined. For example, some security systems, license plates, and telephone numbers exclude certain numbers. For example, phone numbers cannot begin with 0 or 1 because these are reserved for operator assistance, long distance, and international calls. Constructing a three-digit area code is like filling in three blanks with three digit digit digit

digits. Since the area code must start with a number between 2 and 9, there are eight choices for the first blank. Since there are 10 choices for the second digit and 10 choices for the third, there are 8 # 10 # 10  800 possibilities in the sample space.

EXAMPLE 3



Counting Possibilities for a Four-Digit Security Code A digital security system requires that you enter a four-digit PIN (personal identification number), using only the digits 1 through 9. How many codes are possible if a. Repetition of digits is allowed? b. Repetition is not allowed? c. The first digit must be even and repetitions are not allowed?

Solution



a. Consider filling in the four blanks

digit digit digit digit

with the number of

ways the digit can be chosen. If repetition is allowed, the experiment is similar to that of Example 2 and there are N t  94  6561 possible PINs. b. If repetition is not allowed, there are only eight possible choices for the second digit of the PIN, then seven for the third, and six for the fourth. The number of possible PIN numbers decreases to 9 # 8 # 7 # 6  3024. c. There are four choices for the first digit (2, 4, 6, 8). Once this choice has been made there are eight choices for the second digit, seven for the third, and six for the last: 4 # 8 # 7 # 6  1344 possible codes. Now try Exercises 13 through 20



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Given any experiment involving a sequence of tasks, if the first task can be completed in p possible ways, the second task has q possibilities, and the third task has r possibilities, a tree diagram will show that the number of possibilities in the sample space for task1–task2–task3 is p # q # r. Even though the examples we’ve considered to this point have varied a great deal, this idea was fundamental to counting all possibilities in a sample space and is, in fact, known as the fundamental principle of counting (FPC).

WORTHY OF NOTE In Example 4, we could also reason that since there are 6!  720 random seating arrangements and 240 of them consist of Bob and Carol sitting together [Example 4(a)], the remaining 720  240  480 must consist of Bob and Carol not sitting together. More will be said about this type of reasoning in Section 10.6.

Fundamental Principle of Counting (Applied to Three Tasks) Given any experiment with three defined tasks, if there are p possibilities for the first task, q possibilities for the second, and r possibilities for the third, the total number of ways the experiment can be completed is p # q # r. This fundamental principle can be extended to include any number of tasks.

EXAMPLE 4



Counting Possibilities for Seating Arrangements Adrienne, Bob, Carol, Dax, Earlene, and Fabian bought tickets to see The Marriage of Figaro. Assuming they sat together in a row of six seats, how many different seating arrangements are possible if a. Bob and Carol are sweethearts and must sit together? b. Bob and Carol are enemies and must not sit together?

Solution



Figure 10.13 Bob 1

Carol 2

3

4

5

6

1

Bob 2

Carol 3

4

5

6

1

2

Bob 3

Carol 4

5

6

1

2

3

Bob 4

Carol 5

6

1

2

3

4

Bob 5

Carol 6

B. You’ve just learned how to count possibilities using the fundamental principle of counting

a. Since a restriction has been placed on the seating arrangement, it will help to divide the experiment into a sequence of tasks: task 1: they sit together; task 2: either Bob is on the left or Bob is on the right; and task 3: the other four are seated. Bob and Carol can sit together in five different ways, as shown in Figure 10.13, so there are five possibilities for task 1. There are two ways they can be side-by-side: Bob on the left and Carol on the right, as shown, or Carol on the left and Bob on the right. The remaining four people can be seated randomly, so task 3 has 4!  24 possibilities. Under these conditions they can be seated 5 # 2 # 4!  240 ways. b. This is similar to Part (a), but now we have to count the number of ways they can be separated by at least one seat: task 1: Bob and Carol are in nonadjacent seats; task 2: either Bob is on the left or Bob is on the right; and task 3: the other four are seated. For task 1, be careful to note there is no multiplication involved, just a simple counting. If Bob sits in seat 1, there are four nonadjacent seats. If Bob sits in seat 2, there are three nonadjacent seats, and so on. This gives 4  3  2  1  10 possibilities for Bob and Carol not sitting together. Task 2 and task 3 have the same number of possibilities as in Part (a), giving 10 # 2 # 4!  480 possible seating arrangements. Now try Exercises 21 through 28



C. Distinguishable Permutations In the game of Scrabble® (Milton Bradley), players attempt to form words by rearranging letters. Suppose a player has the letters P, S, T, and O at the end of the game. These letters could be rearranged or permuted to form the words POTS, SPOT, TOPS, OPTS, POST, or STOP. These arrangements are called permutations of the four letters. A permutation is any new arrangement, listing, or sequence of objects obtained by changing an existing order. A distinguishable permutation is a permutation that produces a result different from the original. For example, a distinguishable permutation of the digits in the number 1989 is 8199. Example 4 considered six people, six seats, and the various ways they could be seated. But what if there were fewer seats than people? By the FPC, with six people

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and four seats there could be 6 # 5 # 4 # 3  360 different arrangements, with six people and three seats there are 6 # 5 # 4  120 different arrangements, and so on. These rearrangements are called distinguishable permutations. You may have noticed that for six people and six seats, we used all six factors of 6!, while for six people and four seats we used the first four, six people and three seats required only the first three, and so on. Generally, for n people and r seats, the first r factors of n! will be used. The notation and formula for distinguishable permutations of n objects taken r at a time is n! . By defining 0!  1, the formula includes the case where all n objects nPr  1n  r2! n! n! n!    n!. are selected, which of course results in nPn  1n  n2! 0! 1 Distinguishable Permutations: Unique Elements If r objects are selected from a set containing n unique elements 1r  n2 and placed in an ordered arrangement, the number of distinguishable permutations is n! or nPr  n1n  121n  22 # # # 1n  r  12 nPr  1n  r2! EXAMPLE 5



Computing a Permutation Compute each value of nPr using the methods just described. a. 7P4 b. 10P3

Solution



n! Begin by evaluating each expression using the formula nPr  , noting the 1n  r2! third line (in bold) gives the first r factors of n!. 7! 10! a. 7P4  b. 10P3  17  42! 110  32! 7 # 6 # 5 # 4 # 3! 10 # 9 # 8 # 7!   3! 7! # # # # 7 6 5 4  10 9 # 8  840  720 Now try Exercises 29 through 36

EXAMPLE 6





Counting the Possibilities for Finishing a Race As part of a sorority’s initiation process, the nine new inductees must participate in a 1-mi race. Assuming there are no ties, how many first- through fifth-place finishes are possible if it is well known that Mediocre Mary will finish fifth and Lightning Louise will finish first?

Solution



To help understand the situation, we can diagram the possibilities for finishing first through fifth. Since Louise will finish first, this slot can be filled in only one way, Louise by Louise herself. The same goes for Mary and her fifth-place finish: 1st Mary 2nd

C. You’ve just learned how to quick-count distinguishable permutations

3rd

4th

5th

. The remaining three slots can be filled in 7P3  7 # 6 # 5

different ways, indicating that under these conditions, there are 1 # 7 # 6 # 5 # 1  210 different ways to finish. Now try Exercises 37 through 42



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D. Nondistinguishable Permutations As the name implies, certain permutations are nondistinguishable, meaning you cannot tell one apart from another. Such is the case when the original set contains elements or sample outcomes that are identical. Consider a family with four children, Lyddell, Morgan, Michael, and Mitchell, who are at the photo studio for a family picture. Michael and Mitchell are identical twins and cannot be told apart. In how many ways can they be lined up for the picture? Since this is an ordered arrangement of four children taken from a group of four, there are 4P4  24 ways to line them up. A few of them are Lyddell Morgan Michael Mitchell Lyddell Michael Morgan Mitchell Michael Lyddell Morgan Mitchell

Lyddell Morgan Mitchell Michael Lyddell Mitchell Morgan Michael Mitchell Lyddell Morgan Michael

But of these six arrangements, half will appear to be the same picture, since the difference between Michael and Mitchell cannot be distinguished. In fact, of the 24 total permutations, every picture where Michael and Mitchell have switched places will be nondistinguishable. To find the distinguishable permutations, we need to take the total permutations (4P4) and divide by 2!, the number of ways the twins can be 24 4P4   12 distinguishable pictures. permuted: 122! 2 These ideas can be generalized and stated in the following way. WORTHY OF NOTE

Nondistinguishable Permutations: Nonunique Elements

In Example 7, if a Scrabble player is able to play all seven letters in one turn, he or she “bingos” and is awarded 50 extra points. The player in Example 7 did just that. Can you determine what word was played?

In a set containing n elements where one element is repeated p times, another is repeated q times, and another is repeated r times 1p  q  r  n2, the number of nondistinguishable permutations is n! nPn  p!q!r! p!q!r! The idea can be extended to include any number of repeated elements.

EXAMPLE 7



Counting Distinguishable Permutations A Scrabble player has the seven letters S, A, O, O, T, T, and T in his rack. How many distinguishable arrangements can be formed as he attempts to play a word?

Solution D. You’ve just learned how to quick-count nondistinguishable permutations



Essentially the exercise asks for the number of distinguishable permutations of the seven letters, given T is repeated three times and O is repeated twice. There are 7P 7  420 distinguishable permutations. 3!2! Now try Exercises 43 through 54



E. Combinations Similar to nondistinguishable permutations, there are other times the total number of permutations must be reduced to quick-count the elements of a desired subset. Consider a vending machine that offers a variety of 40¢ candies. If you have a quarter (Q), dime (D), and nickel (N), the machine wouldn’t care about the order the coins were deposited. Even though QDN, QND, DQN, DNQ, NQD, and NDQ give the 3P3  6 possible permutations, the machine considers them as equal and will vend your snack. Using sets, this is similar to saying the set A  5X, Y, Z6 has only one subset with three elements, since {X, Z, Y}, {Y, X, Z}, {Y, Z, X}, and so on, all represent the same set. Similarly, there are six, two-letter permutations of X, Y, and Z 1 3P2  62: XY, XZ, YX,

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YZ, ZX, and ZY, but only three two-letter subsets: {X, Y}, {X, Z} and {Y, Z}. When permutations having the same elements are considered identical, the result is the number of possible combinations and is denoted nCr. Since the r objects can be selected in r! nPr ways, we divide nPr by r! to “quick-count” the number of possibilities: nCr  , r! which can be thought of as the first r factors of n!, divided by r!. By substituting n! for nPr in this formula, we find an alternative method for computing nCr is 1n  r2! n! . Take special note that when r objects are selected from a set with n r!1n  r2! elements and the order they’re listed is unimportant (because you end up with the same subset), the result is a combination, not a permutation. Combinations The number of combinations of n objects taken r at a time is given by n! nPr or nCr  nCr  r! r!1n  r2!

EXAMPLE 8



Computing Combinations Using a Formula Compute each value of nCr given. a. 7C4 b. 8C3

Solution



7#6#5#4 4!  35

a. 7C4 

c. 5C2 8#7#6 3!  56

b. 8C3 

5#4 2!  10

c. 5C2 

Now try Exercises 55 through 64

EXAMPLE 9





Applications of Combinations-Lottery Results A small city is getting ready to draw five Ping-Pong balls of the nine they have numbered 1 through 9 to determine the winner(s) for its annual raffle. If a ticket holder has the same five numbers, they win. In how many ways can the winning numbers be drawn?

Solution



Since the winning numbers can be drawn in any order, we have a combination of 9 things taken 5 at a time. The five numbers can be 9#8#7#6#5  126 ways. drawn in 9C5  5! Now try Exercises 65 and 66



Somewhat surprisingly, there are many situations where the order things are listed is not important. Such situations include • The formation of committees, since the order people volunteer is unimportant • Card games with a standard deck, since the order cards are dealt is unimportant • Playing BINGO, since the order the numbers are called is unimportant When the order in which people or objects are selected from a group is unimportant, the number of possibilities is a combination, not a permutation. Another way to tell the difference between permutations and combinations is the following memory device: Permutations have Priority or Precedence; in other

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words, the Position of each element matters. By contrast, a Combination is like a Committee of Colleagues or Collection of Commoners; all members have equal rank. For permutations, a-b-c is different from b-a-c. For combinations, a-b-c is the same as b-a-c.

EXAMPLE 10



Applications of Quick-Counting — Committees and Government The Sociology Department of Lakeside Community College has 12 dedicated faculty members. (a) In how many ways can a three-member textbook selection committee be formed? (b) If the department is in need of a Department Chair, Curriculum Chair, and Technology Chair, in how many ways can the positions be filled?

Solution

E. You’ve just learned how to quick-count using combinations



a. Since textbook selection depends on a Committee of Colleagues, the order members are chosen is not important. This is a Combination of 12 people taken 3 at a time, and there are 12C3  220 ways the committee can be formed. b. Since those selected will have Position or Priority, this is a Permutation of 12 people taken 3 at a time, giving 12P3  1320 ways the positions can be filled. Now try Exercises 67 through 78



The Exercise Set contains a wide variety of additional applications. See Exercises 81 through 107.

TECHNOLOGY HIGHLIGHT

Calculating Permutations and Combinations Both the nPr and nCr functions are Figure 10.15 Figure 10.14 accessed using the MATH key and the PRB submenu (see Figure 10.14). To compute the permutations of 12 objects taken 9 at a time (12P9), clear the home screen and enter a 12, then press MATH 2:nPr to access the nPr operation, which is automatically pasted on the home screen after the 12. Now enter a 9, press ENTER and a result of 79833600 is displayed (Figure 10.15). Repeat the sequence to compute the value of 12C9 ( MATH 3:nCr). Note that the value of 12P9 is nPr much larger than 12C9 and that they differ by a factor of 9! since nCr  . r! Exercise 1: The Department of Humanities has nine faculty members who must serve on at least one committee per semester. How many different committees can be formed that have (a) two members, (b) three members, (c) four members, and (d) five members? Exercise 2: A certain state places 45 Ping-Pong balls numbered 1 through 45 in a container, then draws out five to form the winning lottery numbers. How many different ways can the five numbers be picked? Exercise 3: Dairy King maintains six different toppings at a self-service counter, so that customers can top their ice cream sundaes with as many as they like. How many different sundaes can be created if a customer were to select any three ingredients?

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10.5 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. A(n) has a(n)

is any task that can be repeated and set of possible outcomes.

2. If an experiment has N equally likely outcomes and is repeated t times, the number of elements in the sample space is given by . 3. When unique elements of a set are rearranged, the result is called a(n) permutation. 

4. If some elements of a group are identical, certain rearrangements are identical and the result is a(n) permutation. 5. A three-digit number is formed from digits 1 to 9. Explain how forming the number with repetition differs from forming it without repetition. 6. Discuss/Explain the difference between a permutation and a combination. Try to think of new ways to help remember the distinction.

DEVELOPING YOUR SKILLS 7. For the spinner shown here, (a) draw a tree diagram illustrating all possible outcomes for two spins and (b) create an ordered list showing all possible outcomes for two spins. Heads 8. For the fair coin shown here, (a) draw a tree diagram illustrating all possible outcomes for four flips and (b) create an ordered list showing the possible outcomes for four flips.

14. Repetition is not allowed? Z

W

Y

X

Tails

9. A fair coin is flipped five times. If you extend the tree diagram from Exercise 8, how many elements are in the sample space? 10. A spinner has the two equally likely outcomes A or B and is spun four times. How is this experiment related to the one in Exercise 8? How many elements are in the sample space? 11. An inexpensive lock uses the numbers 0 to 24 for a three-number combination. How many different combinations are possible? 12. Grades at a local college consist of A, B, C, D, F, and W. If four classes are taken, how many different report cards are possible? License plates. In a certain (English-speaking) country, license plates for automobiles consist of two letters followed by one of four symbols (■, ◆, ❍, or ●), followed by three digits. How many license plates are possible if

13. Repetition is allowed?

15. A remote access door opener requires a five-digit (1–9) sequence. How many sequences are possible if (a) repetition is allowed? (b) repetition is not allowed? 16. An instructor is qualified to teach Math 020, 030, 140, and 160. How many different four-course schedules are possible if (a) repetition is allowed? (b) repetition is not allowed? Use the fundamental principle of counting and other quick-counting techniques to respond.

17. Menu items: At Joe’s Diner, the manager is offering a dinner special that consists of one choice of entree (chicken, beef, soy meat, or pork), two vegetable servings (corn, carrots, green beans, peas, broccoli, or okra), and one choice of pasta, rice, or potatoes. How many different meals are possible? 18. Getting dressed: A frugal businessman has five shirts, seven ties, four pairs of dress pants, and three pairs of dress shoes. Assuming that all possible arrangements are appealing, how many different shirt-tie-pants-shoes outfits are possible? 19. Number combinations: How many four-digit numbers can be formed using the even digits 0, 2, 4, 6, 8, if (a) no repetitions are allowed; (b) repetitions are allowed; (c) repetitions are not allowed and the number must be less than 6000 and divisible by 10. 20. Number combinations: If I was born in March, April, or May, after the 19th but before the 30th,

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and after 1949 but before 1981, how many different MM–DD–YYYY dates are possible for my birthday? Seating arrangements: William, Xayden, York, and Zelda decide to sit together at the movies. How many ways can they be seated if

21. They sit in random order? 22. York must sit next to Zelda? 23. York and Zelda must be on the outside? 24. William must have the aisle seat? Course schedule: A college student is trying to set her schedule for the next semester and is planning to take five classes: English, art, math, fitness, and science. How many different schedules are possible if

25. The classes can be taken in any order. 26. She wants her science class to immediately follow her math class. 27. She wants her English class to be first and her fitness class to be last. 28. She can’t decide on the best order and simply takes the classes in alphabetical order. Find the value of nPr in two ways: (a) compute r factors n! of n! and (b) use the formula nPr  . 1n  r2!

29.

10P3

32. 5P3

30.

12P2

33. 8P7

35. T, R, and A

40. From a pool of 32 applicants, a board of directors must select a president, vice-president, labor relations liaison, and a director of personnel for the company’s day-to-day operations. Assuming all applicants are qualified and willing to take on any of these positions, how many ways can this be done? 41. A hugely popular chess tournament now has six finalists. Assuming there are no ties, (a) in how many ways can the finalists place in the final round? (b) In how many ways can they finish first, second, and third? (c) In how many ways can they finish if it’s sure that Roberta Fischer is going to win the tournament and that Geraldine Kasparov will come in sixth? 42. A field of 10 horses has just left the paddock area and is heading for the gate. Assuming there are no ties in the big race, (a) in how many ways can the horses place in the race? (b) In how many ways can they finish in the win, place, or show positions? (c) In how many ways can they finish if it’s sure that John Henry III is going to win, Seattle Slew III will come in second (place), and either Dumb Luck II or Calamity Jane I will come in tenth? Assuming all multiple births are identical and the children cannot be told apart, how many distinguishable photographs can be taken of a family of six, if they stand in a single row and there is

43. one set of twins 44. one set of triplets

31. 9P4

45. one set of twins and one set of triplets

34. 8P1

46. one set of quadruplets

Determine the number of three-letter permutations of the letters given, then use an organized list to write them all out. How many of them are actually words or common names?

983

47. How many distinguishable numbers can be made by rearranging the digits of 105,001? 48. How many distinguishable numbers can be made by rearranging the digits in the palindrome 1,234,321?

36. P, M, and A

37. The regional manager for an office supply store needs to replace the manager and assistant manager at the downtown store. In how many ways can this be done if she selects the personnel from a group of 10 qualified applicants? 38. The local chapter of Mu Alpha Theta will soon be electing a president, vice-president, and treasurer. In how many ways can the positions be filled if the chapter has 15 members? 39. The local school board is going to select a principal, vice-principal, and assistant viceprincipal from a pool of eight qualified candidates. In how many ways can this be done?

How many distinguishable permutations can be formed from the letters of the given word?

49. logic

50. leave

51. lotto

52. levee

A Scrabble player (see Example 7) has the six letters shown remaining in her rack. How many distinguishable, six-letter permutations can be formed? (If all six letters are played, what was the word?)

53. A, A, A, N, N, B 54. D, D, D, N, A, E

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nPr (r factors of n! Find the value of nCr: (a) using nCr  r! n! over r!) and (b) using nCr  r!1n  r2! .

55. 9C4

56.

58. 6C3

59. 6C6

10C3

57. 8C5 60. 6C0

Use a calculator to verify that each pair of combinations is equal.

61. 9C4, 9C5

62.

63. 8C5, 8C3

64. 7C2, 7C5

10C3, 10C7

65. A platoon leader needs to send four soldiers to do some reconnaissance work. There are 12 soldiers in the platoon and each soldier is assigned a number between 1 and 12. The numbers 1 through 12 are placed in a helmet and drawn randomly. If a soldier’s number is drawn, then that soldier goes on the mission. In how many ways can the reconnaissance team be chosen? 66. Seven colored balls (red, indigo, violet, yellow, green, blue, and orange) are placed in a bag and three are then withdrawn. In how many ways can the three colored balls be drawn? 67. When the company’s switchboard operators went on strike, the company president asked for three volunteers from among the managerial ranks to temporarily take their place. In how many ways can the three volunteers “step forward,” if there are 14 managers and assistant managers in all? 68. Becky has identified 12 books she wants to read this year and decides to take four with her to read while on vacation. She chooses Pastwatch by Orson Scott Card for sure, then decides to randomly choose any three of the remaining books. In how many ways can she select the four books she’ll end up taking? 69. A new garage band has built up their repertoire to 10 excellent songs that really rock. Next month they’ll be playing in a Battle of the Bands contest, with the 

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winner getting some guaranteed gigs at the city’s most popular hot spots. In how many ways can the band select 5 of their 10 songs to play at the contest? 70. Pierre de Guirré is an award-winning chef and has just developed 12 delectable, new main-course recipes for his restaurant. In how many ways can he select three of the recipes to be entered in an international culinary competition? For each exercise, determine whether a permutation, a combination, counting principles, or a determination of the number of subsets is the most appropriate tool for obtaining a solution, then solve. Some exercises can be completed using more than one method.

71. In how many ways can eight second-grade children line up for lunch? 72. If you flip a fair coin five times, how many different outcomes are possible? 73. Eight sprinters are competing for the gold, silver, and bronze medals. In how many ways can the medals be awarded? 74. Motorcycle license plates are made using two letters followed by three numbers. How many plates can be made if repetition of letters (only) is allowed? 75. A committee of five students is chosen from a class of 20 to attend a seminar. How many different ways can this be done? 76. If onions, cheese, pickles, and tomatoes are available to dress a hamburger, how many different hamburgers can be made? 77. A caterer offers eight kinds of fruit to make various fruit trays. How many different trays can be made using four different fruits? 78. Eighteen females try out for the basketball team, but the coach can only place 15 on her roster. How many different teams can be formed?

WORKING WITH FORMULAS

79. Stirling’s Formula: n!  12 # 1nn0.5 2 # en Values of n! grow very quickly as n gets larger (13! is already in the billions). For some applications, scientists find it useful to use the approximation for n! shown, called Stirling’s Formula. a. Compute the value of 7! on your calculator, then use Stirling’s Formula with n  7. By what percent does the approximate value differ from the true value? b. Compute the value of 10! on your calculator, then use Stirling’s Formula with n  10. By

what percent does the approximate value differ from the true value? 80. Factorial formulas: For n, k  , where n 7 k, n!  n1n  12 1n  22 p 1n  k  12 1n  k2! a. Verify the formula for n  7 and k  5. b. Verify the formula for n  9 and k  6.

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APPLICATIONS

81. Yahtzee: In the game of “Yahtzee”® (Milton Bradley) five dice are rolled simultaneously on the first turn in an attempt to obtain various arrangements (worth various point values). How many different arrangements are possible? 82. Twister: In the game of “Twister”® (Milton Bradley) a simple spinner is divided into four quadrants designated Left Foot (LF), Right Hand (RH), Right Foot (RF), and Left Hand (LH), with four different color possibilities in each quadrant (red, green, yellow, blue). Determine the number of possible outcomes for three spins. 83. Clue: In the game of “Clue”® (Parker Brothers) a crime is committed in one of nine rooms, with one of six implements, by one of six people. In how many different ways can the crime be committed? Phone numbers in North America have 10 digits: a threedigit area code, a three-digit exchange number, and the four final digits that make each phone number unique. Neither area codes nor exchange numbers can start with 0 or 1. Prior to 1994 the second digit of the area code had to be a 0 or 1. Sixteen area codes are reserved for special services (such as 911 and 411). In 1994, the last area code was used up and the rules were changed to allow the digits 2 through 9 as the middle digit in area codes.

84. How many different area codes were possible prior to 1994? 85. How many different exchange numbers were possible prior to 1994? 86. How many different phone numbers were possible prior to 1994? 87. How many different phone numbers were possible after 1994? Aircraft N-numbers: In the United States, private aircraft are identified by an “N-Number,” which is generally the letter “N” followed by five characters and includes these restrictions: (1) the N-Number can consist of five digits, four digits followed by one letter, or three digits followed by two letters; (2) the first digit cannot be a zero; (3) to avoid confusion with the numbers zero and one, the letters O and I cannot be used; and (4) repetition of digits and letters is allowed. How many unique N-Numbers can be formed

88. that have four digits and one letter? 89. that have three digits and two letters?

90. that have five digits? 91. that have three digits, two letters with no repetitions of any kind allowed? Seating arrangements: Eight people would like to be seated. Assuming some will have to stand, in how many ways can the seats be filled if the number of seats available is

92. eight

93. five

94. three

95. one

Seating arrangements: In how many different ways can eight people (six students and two teachers) sit in a row of eight seats if

96. the teachers must sit on the ends 97. the teachers must sit together Television station programming: A television station needs to fill eight half-hour slots for its Tuesday evening schedule with eight programs. In how many ways can this be done if

98. there are no constraints 99. Seinfeld must have the 8:00 P.M. slot 100. Seinfeld must have the 8:00 P.M. slot and The Drew Carey Show must be shown at 6:00 P.M. 101. Friends can be aired at 7:00 or 9:00 P.M. and Everybody Loves Raymond can be aired at 6:00 or 8:00 P.M. Scholarship awards: Fifteen students at Roosevelt Community College have applied for six available scholarship awards. How many ways can the awards be given if

102. there are six different awards given to six different students 103. there are six identical awards given to six different students Committee composition: The local city council has 10 members and is trying to decide if they want to be governed by a committee of three people or by a president, vicepresident, and secretary.

104. If they are to be governed by committee, how many unique committees can be formed? 105. How many different president, vice-president, and secretary possibilities are there?

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106. Team rosters: A soccer team has three goalies, eight defensive players, and eight forwards on its roster. How many different starting line-ups can be formed (one goalie, three defensive players, and three forwards)? 107. e-mail addresses: A business wants to standardize the e-mail addresses of its 

employees. To make them easier to remember and use, they consist of two letters and two digits (followed by @esmtb.com), with zero being excluded from use as the first digit and no repetition of letters or digits allowed. Will this provide enough unique addresses for their 53,000 employees worldwide?

EXTENDING THE CONCEPT

108. In Exercise 79, we learned that an approximation for n! can be found using Stirling’s Formula: n!  121nn0.5 2en. As with other approximations, mathematicians are very interested in whether the approximation gets better or worse for larger values of n (does their ratio get closer to 1 or farther from 1). Use your calculator to investigate and answer the question. 109. Verify that the following equations are true, then generalize the patterns and relationships noted to create your own equation. Afterward, write each of the four factors from Part (a) (the two combinations on each side) in expanded form and discuss/explain why the two sides are equal. a. 10C3 # 7C2  10C2 # 8C5 b. 9C3 # 6C2  9C2 # 7C4 

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c. 11C4 # 7C5  11C5 # 6C4 d. 8C3 # 5C2  8C2 # 6C3 110. Tic-Tac-Toe: In the game Tic-Tac-Toe, players alternately write an “X” or an “O” in one of nine squares on a 3  3 grid. If either player gets three in a row horizontally, vertically, or diagonally, that player wins. If all nine squares are played with neither person winning, the game is a draw. Assuming “X” always goes first, a. How many different “boards” are possible if the game ends after five plays? b. How many different “boards” are possible if the game ends after six plays?

MAINTAINING YOUR SKILLS

111. (5.4) Solve the given system of linear inequalities by graphing. Shade the feasible region. 2x  y 6 6 x  2y 6 6 μ x0 y0 12 , determine the other five 13 trig functions of the acute angle .

112. (5.2) Given sin  

113. (6.3) Rewrite cos122cos132  sin122sin132 as a single expression. 114. (7.3) Graph the hyperbola that is defined by 1x  22 2 1y  32 2   1. 4 9

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10.6 Introduction to Probability Learning Objectives

There are few areas of mathematics that give us a better view of the world than probability and statistics. Unlike statistics, which seeks to analyze and interpret data, probability (for our purposes) attempts to use observations and data to make statements concerning the likelihood of future events. Such predictions of what might happen have found widespread application in such diverse fields as politics, manufacturing, gambling, opinion polls, product life, and many others. In this section, we develop the basic elements of probability.

In Section 10.6 you will learn how to:

A. Define an event on a sample space

B. Compute elementary probabilities

C. Use certain properties of probability

A. Defining an Event

D. Compute probabilities

In Section 10.5 we defined the following terms: experiment and sample outcome. Flipping a coin twice in succession is an experiment, and two sample outcomes are HH and HT. An event E is any designated set of sample outcomes, and is a subset of the sample space. One event might be E1: (two heads occur), another possibility is E2: (at least one tail occurs).

using quick-counting techniques

E. Compute probabilities involving nonexclusive events

EXAMPLE 1



Stating a Sample Space and Defining an Event Consider the experiment of rolling one standard, six-sided die (plural is dice). State the sample space S and define any two events relative to S.

Solution



A. You’ve just learned how to define an event on a sample space

S is the set of all possible outcomes, so S  51, 2, 3, 4, 5, 66. Two possible events are E1: (a 5 is rolled) and E2: (an even number is rolled). Now try Exercises 7 through 10



WORTHY OF NOTE

B. Elementary Probability

Our study of probability will involve only those sample spaces with events that are equally likely.

When rolling the die, we know the result can be any of the six equally likely outcomes in the sample space, so the chance of E1:(a five is rolled) is 16. Since three of the elements in S are even numbers, the chance of E2:(an even number is rolled) is 36  12. This suggests the following definition. The Probability of an Event E Given S is a sample space of equally likely events and E is an event relative to S, the probability of E, written P(E), is computed as n1E2 P1E2  n1S2 where n(E) represents the number of elements in E, and n(S) represents the number of elements in S. A standard deck of playing cards consists of 52 cards divided in four groups or suits. There are 13 hearts (♥), 13 diamonds 12, 13 spades (♠), and 13 clubs (♣). As you can see in the illustration, each of the 13 cards in a suit is labeled 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, and A. Also notice that 26 of the cards are red (hearts and diamonds), 26 are black (spades and clubs) and 12 of the cards are “face cards” (J, Q, K of each suit).

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EXAMPLE 2



Stating a Sample Space and the Probability of a Single Outcome A single card is drawn from a well-shuffled deck. Define S and state the probability of any single outcome. Then define E as a King is drawn and find P(E).

Solution



Sample space: S  5the 52 cards6 . There are 52 equally likely outcomes, 1 so the probability of any one outcome is 52 . Since S has four Kings, n1E2 4  or about 0.077. P1E2  n1S2 52

Now try Exercises 11 through 14

EXAMPLE 3





Stating a Sample Space and the Probability of a Single Outcome A family of five has two girls and three boys named Sophie, Maria, Albert, Isaac, and Pythagoras. Their ages are 21, 19, 15, 13, and 9, respectively. One is to be selected randomly. Find the probability a teenager is chosen.

Solution



B. You’ve just learned how to compute elementary probabilities

The sample space is S  59, 13, 15, 19, 216. Three of the five are teenagers, meaning the probability is 35, 0.6, or 60%.

Now try Exercises 15 and 16



C. Properties of Probability A study of probability necessarily includes recognizing some basic and fundamental properties. For example, when a fair die is rolled, what is P(E) if E is defined as a 1, 2, 3, 4, 5, or 6 is rolled? The event E will occur 100% of the time, since 1, 2, 3, 4, 5, 6 are the only possibilities. In symbols we write P(outcome is in the sample space) or simply P1S2  1 (100%). What percent of the time will a result not in the sample space occur? Since the die has only the six sides numbered 1 through 6, the probability of rolling something else is zero. In symbols, P1outcome is not in sample space2  0 or simply P1~S2  0.

WORTHY OF NOTE In probability studies, the tilde “~” acts as a negation symbol. For any event E defined on the sample space, ~E means the event does not occur.

EXAMPLE 4

Properties of Probability Given sample space S and any event E defined relative to S. 1. P1S2  1



2. P1~S2  0

3. 0  P1E2  1

Determining the Probability of an Event A game is played using a spinner like the one shown. Determine the probability of the following events: E1: A nine is spun.

Solution



2

3 4

1

E2: An integer greater than 0 and less than 9 is spun.

8

5 7

6

The sample space consists of eight equally likely outcomes. P1E1 2 

0 0 8

P1E2 2 

8  1. 8

Technically, E1: A nine is spun is not an “event,” since it is not in the sample space and cannot occur, while E2 contains the entire sample space and must occur. Now try Exercises 17 and 18



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989

Because we know P1S2  1 and all sample outcomes are equally likely, the probabilities of all single events defined on the sample space must sum to 1. For the experiment of rolling a fair die, the sample space has six outcomes that are equally likely. Note that P112  P122  P132  P142  P152  P162  16, and 16  16  16  16  16  16  1. Probability and Sample Outcomes Given a sample space S with n equally likely sample outcomes s1, s2, s3, . . . , sn. n

 P1s 2  P1s 2  P1s 2  P1s 2  i

1

2

3

i1

# # #

 P1sn 2  1

The complement of an event E is the set of sample outcomes in S not contained in E. Symbolically, ~E is the complement of E. Probability and Complementary Events Given sample space S and any event E defined relative to S, the complement of E, written ~E, is the set of all outcomes not in E and: 1. P1E2  1  P1~E2

EXAMPLE 5



2. P1E2  P1~E2  1

Stating a Probability Using Complements Use complementary events to answer the following questions: a. A single card is drawn from a well-shuffled deck. What is the probability that it is not a diamond? b. A single letter is picked at random from the letters in the word “divisibility.” What is the probability it is not an “i”?

Solution



WORTHY OF NOTE Probabilities can be written in fraction form, decimal form, or as a percent. For P(E2) from Example 1, the probability is 3 4 , 0.75, or 75%.

EXAMPLE 6

a. Since there are 13 diamonds in a standard 52-card deck, there are 39 39 nondiamonds: P1~D2  1  P1D2  1  13 52  52  0.75. b. Of the 12 letters in d-i-v-i-s-i-b-i-l-i-t-y, 5 are “i’s.” This means 5 7 P1~i2  1  P1i2, or 1  12  12 . The probability of choosing a letter other than i is 0.583. Now try Exercises 19 through 22





Stating a Probability Using Complements Inter-Island Waterways has just opened hydrofoil service between several islands. The hydrofoil is powered by two engines, one forward and one aft, and will operate if either of its two engines is functioning. Due to testing and past experience, the company knows the probability of the aft engine failing is P1aft engine fails2  0.05, the probability of the forward engine failing is P1forward engine fails2  0.03, and the probability that both fail is P1both engines simultaneously fail2  0.012. What is the probability the hydrofoil completes its next trip?

Solution



Although the answer may seem complicated, note that P(trip is completed) and P(both engines simultaneously fail) are complements. P1trip is completed2  1  P1both engines simultaneously fail2  1  0.012  0.988 There is close to a 99% probability the trip will be completed. Now try Exercises 23 and 24



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The chart in Figure 10.16 shows all 36 possible outcomes (the sample space) from the experiment of rolling two fair dice. Figure 10.16

EXAMPLE 7



Stating a Probability Using Complements Two fair dice are rolled. What is the probability the sum of both dice is greater than or equal to 5, P1sum  52?

Solution



C. You’ve just learned how to use certain properties of probability

See Figure 10.16. For P 1sum  52 it may be easier to use complements as there are far fewer possibilities: P1sum  52  1  P1sum 6 52, which gives 6 1 5 1  1    0.83. 36 6 6 Now try Exercises 25 and 26



D. Probability and Quick-Counting Quick-counting techniques were introduced earlier to help count the number of elements in a large or more complex sample space, and the number of sample outcomes in an event. EXAMPLE 8A



Stating a Probability Using Combinations Five cards are drawn from a shuffled 52-card deck. Calculate the probability of E1:(all five cards are face cards) or E2:(all five cards are hearts)?

Solution



The sample space for both events consists of all five-card groups that can be formed from the 52 cards or 52C5. For E1 we are to select five face cards from the 12 that are available (three from each suit), or 12C5. The probability of five face n1E2 792 12C5 cards is  , which gives  0.0003. For E2 we are to select five n1S2 2,598,960 52C5 n1E2 13C5 hearts from the 13 available, or 13C5. The probability of five hearts is  , n1S2 52C5 1287  0.0005. which is 2,598,960



Stating a Probability Using Combinations and the Fundamental Principle of Counting

WORTHY OF NOTE It seems reasonable that the probability of 5 hearts is slightly higher, as 13 of the 52 cards are hearts, while only 12 are face cards.

EXAMPLE 8B

Of the 42 seniors at Jacoby High School, 23 are female and 19 are male. A group of five students is to be selected at random to attend a conference in Reno, Nevada. What is the probability the group will have exactly three females?

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Solution



D. You’ve just learned how to compute probabilities using quick-counting techniques

991

The sample space consists of all five-person groups that can be formed from the 42 seniors or 42C5. The event consists of selecting 3 females from the 23 available (23C3) and 2 males from the 19 available (19C2). Using the fundamental principle of counting n1E2  23C3 # 19C2 and the probability the group has 3 females is

# n1E2 302,841 23C3 19C2  , which gives  0.356. There is approximately a n1S2 850,668 42C5 35.6% probability the group will have exactly 3 females. Now try Exercises 27 through 34



E. Probability and Nonexclusive Events Figure 10.17 Sometimes the way events are defined S causes them to share sample outcomes. Using a standard deck of E1 E2 J♠ A♣ 2♣ Q♠ playing cards once again, if we define 3♣ J♣ the events E1:(a club is drawn) and J♦ K♠ 4♣ 5♣ Q♦ Q♣ E2:(a face card is drawn), they share 6♣ K♦ the outcomes J♣, Q♣, and K♣ as K♣ J♥ 8♣ 7♣ Q♥ shown in Figure 10.17. This overlap9♣ K♥ 10♣ ping region is the intersection of the events, or E1  E2. If we compute n1E1 ´ E2 2 as n1E1 2  n1E2 2 as before, this intersecting region gets counted twice! In cases where the events are nonexclusive (not mutually exclusive), we maintain the correct count by subtracting one of the two intersections, obtaining n1E1 ´ E2 2  n1E1 2  n1E2 2  n1E1  E2 2. This leads to the following calculation for the probability of nonexclusive events:

WORTHY OF NOTE This can be verified by simply counting the elements involved: n1E1 2  13 and n1E2 2  12 so n1E1 2  n1E2 2  25. However, there are only 22 possibilities—the J♣, Q♣, and K♣ got counted twice.

n1E1 2  n1E2 2  n1E1  E2 2 n1S2 n1E1 2 n1E1 2 n1E1  E2 2    n1S2 n1S2 n1S2  P1E1 2  P1E2 2  P1E1  E2 2

P1E1 ´ E2 2 

definition of probability

property of rational expressions definition of probability

Probability and Nonexclusive Events Given sample space S and nonexclusive events E1 and E2 defined relative to S, the probability of E1 or E2 is given by P1E1 ´ E2 2  P1E1 2  P1E2 2  P1E1  E2 2

EXAMPLE 9A



Stating the Probability of Nonexclusive Events What is the probability that a club or a face card is drawn from a standard deck of 52 well-shuffled cards?

Solution



As before, define the events E1:(a club is drawn) and E2:(a face card is drawn). 12 Since there are 13 clubs and 12 face cards, P1E1 2  13 52 and P1E2 2  52 . But three of 3 the face cards are clubs, so P1E1  E2 2  52 . This leads to P1E1 ´ E2 2  P1E1 2 13   52 22   52

 P1E2 2  P1E1  E2 2 12 3  52 52

0.423

nonexclusive events substitute

combine terms

There is about a 42% probability that a club or face card is drawn.

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EXAMPLE 9B



Stating the Probability of Nonexclusive Events A survey of 100 voters was taken to gather information on critical issues and the demographic information collected is shown in the table. One out of the 100 voters is to be drawn at random to be interviewed on the 5 O’Clock News. What is the probability the person is a woman (W) or a Republican (R)?

Solution



E. You’ve just learned how to compute probabilities involving nonexclusive events

Women

Men

Totals

Republican

17

20

37

Democrat

22

17

39

Independent

8

7

15

Green Party

4

1

5

Tax Reform Totals

2

2

4

53

47

100

Since there are 53 women and 37 Republicans, P1W2  0.53 and P1R2  0.37. The table shows 17 people are both female and Republican so P1W  R2  0.17. P1W ´ R2  P1W2  P1R2  P1W  R2  0.53  0.37  0.17  0.73

nonexclusive events substitute combine

There is a 73% probability the person is a woman or a Republican. Now try Exercises 35 through 48



Two events that have no common outcomes are called mutually exclusive events (one excludes the other and vice versa). For example, in rolling one die, E1:(a 2 is rolled) and E2:(an odd number is rolled) are mutually exclusive, since 2 is not an odd number. For the probability of E3:(a 2 is rolled or an odd number is rolled), we note that n1E1  E2 2  0 and the previous formula simply reduces to P1E1 2  P1E2 2 . See Exercises 49 and 50. There is a large variety of additional applications in the Exercise Set. See Exercises 53 through 68.

TECHNOLOGY HIGHLIGHT

Principles of Quick-Counting, Combinations, and Probability At this point you are likely using the Y = screen and tables ( , 2nd GRAPH TABLE, and so on) with relative ease. When probability calculations require a repeated use of permutations and combinations, these features can make the work more efficient and help to explore the patterns they generate. For choosing r children from a group of six children 1n  62, set the to AUTO, then press Y = and enter 6 nCr X as Y1 (Figure 10.18). Access the TABLE ( 2nd GRAPH ) and note that the calculator has automatically computed the value of 6C0, 6C1, 6C2, . . . , 6C6 (Figure 10.19) and the pattern of outputs is symmetric. For calculations similar to those required in Example 8B 1 23C3 # 19C2 2, enter

Figure 10.18

Figure 10.19

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993

Y1  23 nCr X, Y2  19 nCr 1X  12, and Y3  Y1 # Y2, or any variation of these. Use these ideas to work the following exercises. Exercise 1: Use your calculator to display the values of 5C0, 5C1, . . . , 5C5. Is the result a pattern similar to that for 6C0, 6C1, 6C2, . . . , 6C6? Repeat for 7Cr. Why are the “middle values” repeated for n  7 and n  5, but not for n  6? Exercise 2: A committee consists of 10 Republicans and eight Democrats. In how many ways can a committee of four Republicans and three Democrats be formed?

10.6 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

1. Given a sample space S and an event E defined relative to S, P1E2 



n1S2

.

4. The of an event E is the set of sample outcomes in S which are not contained in E.

2. In elementary probability, we consider all events in the sample space to be likely.

5. Discuss/Explain the difference between mutually exclusive events and nonmutually exclusive events. Give an example of each.

3. Given a sample space S and an event E defined relative to S: , P1S2  ,  P1E2  and P1~S2  .

6. A single die is rolled. With no calculations, explain why the probability of rolling an even number is greater than rolling a number greater than four.

DEVELOPING YOUR SKILLS

State the sample space S and the probability of a single outcome. Then define any two events E relative to S (many answers possible).

7. Two fair coins (heads and tails) are flipped.

Exercise 8

8. The simple spinner shown is spun.

1

2

4 3 9. The head coaches for six little league teams (the Patriots, Cougars, Angels, Sharks, Eagles, and Stars) have gathered to discuss new changes in the rule book. One of them is randomly chosen to ask the first question.

10. Experts on the planets Mercury, Venus, Mars, Jupiter, Saturn, Uranus, Neptune, and the dwarf planet Pluto have gathered at a space exploration conference. One group of experts is selected at random to speak first.

Find P(E) for the events defined.

11. Nine index cards numbered 1 through 9 are shuffled and placed in an envelope, then one of the cards is randomly drawn. Define event E as the number drawn is even. 12. Eight flash cards used for studying basic geometric shapes are shuffled and one of the cards is drawn at random. The eight cards include information on circles, squares, rectangles, kites, trapezoids, parallelograms, pentagons, and triangles. Define event E as a quadrilateral is drawn. 13. One card is drawn at random from a standard deck of 52 cards. What is the probability of a. drawing a Jack b. drawing a spade c. drawing a black card d. drawing a red three 14. Pinochle is a card game played with a deck of 48 cards consisting of 2 Aces, 2 Kings, 2 Queens, 2 Jacks, 2 Tens, and 2 Nines in each of the four

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standard suits [hearts (♥), diamonds (), spades (♠), and clubs (♣)]. If one card is drawn at random from this deck, what is the probability of a. drawing an Ace b. drawing a club c. drawing a red card d. drawing a face card (Jack, Queen, King) 15. A group of finalists on a game show consists of three males and five females. Hank has a score of 520 points, with Harry and Hester having 490 and 475 points, respectively. Madeline has 532 points, with Mackenzie, Morgan, Maggie, and Melanie having 495, 480, 472, and 470 points, respectively. One of the contestants is randomly selected to start the final round. Define E1 as Hester is chosen, E2 as a female is chosen, and E3 as a contestant with less than 500 points is chosen. Find the probability of each event. 16. Soccer coach Maddox needs to fill the last spot on his starting roster for the opening day of the season and has to choose between three forwards and five defenders. The forwards have jersey numbers 5, 12, and 17, while the defenders have jersey numbers 7, 10, 11, 14, and 18. Define E1 as a forward is chosen, E2 as a defender is chosen, and E3 as a player whose jersey number is greater than 10 is chosen. Find the probability of each event. 17. A game is played using a spinner like the one shown. For each spin, 1 2 a. What is the probability the arrow lands in a shaded region? 4 3 b. What is the probability your spin is less than 5? c. What is the probability you spin a 2? d. What is the probability the arrow lands on a prime number? 18. A game is played using a spinner like the one shown here. For each spin, 2 1 a. What is the probability the arrow lands in a lightly 3 6 shaded region? 5 4 b. What is the probability your spin is greater than 2? c. What is the probability the arrow lands in a shaded region? d. What is the probability you spin a 5? Use the complementary events to complete Exercises 19 through 22.

19. One card is drawn from a standard deck of 52. What is the probability it is not a club?

20. Four standard dice are rolled. What is the probability the sum is less than 24? 21. A single digit is randomly selected from among the digits of 10!. What is the probability the digit is not a 2? 22. A corporation will be moving its offices to Los Angeles, Miami, Atlanta, Dallas, or Phoenix. If the site is randomly selected, what is the probability Dallas is not chosen? 23. A large manufacturing plant can remain at full production as long as one of its two generators is functioning. Due to past experience and the age difference between the systems, the plant manager estimates the probability of the main generator failing is 0.05, the probability of the secondary generator failing is 0.01, and the probability of both failing is 0.009. What is the probability the plant remains in full production today? 24. A fire station gets an emergency call from a shopping mall in the mid-afternoon. From a study of traffic patterns, Chief Nozawa knows the probability the most direct route is clogged with traffic is 0.07, while the probability of the secondary route being clogged is 0.05. The probability both are clogged is 0.02. What is the probability they can respond to the call unimpeded using one of these routes? 25. Two fair dice are rolled (see Figure 10.16). What is the probability of a. a sum less than four b. a sum less than eleven c. the sum is not nine d. a roll is not a “double” (both dice the same) “Double-six” dominos is a game played with the 28 numbered tiles shown in the diagram.

26. The 28 dominos are placed in a bag, shuffled, and then one domino is randomly drawn. What is the probability the total number of dots on the domino a. is three or less b. is greater than three c. does not have a blank half d. is not a “double” (both sides the same)

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10-57 Find P(E) given the values for n(E) and n(S) shown.

27. 28. 29. 30.

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Section 10.6 Introduction to Probability

n1E2  6C3 # 4C2; n1S2  10C5 n1E2  12C9 # 8C7; n1S2  20C16 n1E2  9C6 # 5C3; n1S2  14C9 n1E2  7C6 # 3C2; n1S2  10C8

31. Five cards are drawn from a well-shuffled, standard deck of 52 cards. Which has the greater probability: (a) all five cards are red or (b) all five cards are numbered cards? How much greater? 32. Five cards are drawn from a well-shuffled pinochle deck of 48 cards (see Exercise 14). Which has the greater probability (a) all five cards are face cards (King, Queen, or Jack) or (b) all five cards are black? How much greater? 33. A dietetics class has 24 students. Of these, 9 are vegetarians and 15 are not. The instructor receives enough funding to send six students to a conference. If the students are selected randomly, what is the probability the group will have a. exactly two vegetarians b. exactly four nonvegetarians c. at least three vegetarians 34. A large law firm has a support staff of 15 employees: six paralegals and nine legal assistants. Due to recent changes in the law, the firm wants to send five of them to a forum on the new changes. If the selection is done randomly, what is the probability the group will have a. exactly three paralegals b. exactly two legal assistants c. at least two paralegals

a. b. c. d.

42. Eight Ball is a game played on a pool table with 15 balls numbered 1 through 15 and a cue ball that is solid white. Of the 15 numbered balls, 8 are a solid (nonwhite) color and numbered 1 through 8, and seven are striped balls numbered 9 through 15. The fifteen numbered pool balls (no cueball) are placed in a large bowl and mixed, then one is drawn out. What is the probability of drawing a. the eight ball b. a number greater than fifteen c. an even number d. a multiple of three e. a solid color and an even number f. a striped ball and an odd number g. an even number and a number divisible by three h. an odd number and a number divisible by 4 43. A survey of 50 veterans was taken to gather information on their service career and what life is like out of the military. A breakdown of those surveyed is shown in the table. One out of the 50 will be selected at random for an interview and a biographical sketch. What is the probability the person chosen is Women Private

Find the probability indicated using the information given.

35. Given P1E1 2  0.7, P1E2 2  0.5, and P1E1 ¨ E2 2  0.3, compute P1E1 ´ E2 2. 36. Given P1E1 2  0.6, P1E2 2  0.3, and P1E1 ¨ E2 2  0.2, compute P1E1 ´ E2 2.

37. Given P1E1 2  38, P1E2 2  34, and P1E1 ´ E2 2  15 18 ; compute P1E1 ¨ E2 2. 38. Given P1E1 2  12, P1E2 2  35, and P1E1 ´ E2 2  17 20 ; compute P1E1 ¨ E2 2. 39. Given P1E1 ´ E2 2  0.72, P1E2 2  0.56, and P1E1 ¨ E2 2  0.43; compute P(E1).

40. Given P1E1 ´ E2 2  0.85, P1E1 2  0.4, and P1E1 ¨ E2 2  0.21; compute P(E2).

41. Two fair dice are rolled. What is the probability the sum of the dice is

a multiple of 3 and an odd number a sum greater than 5 and a 3 on one die an even number and a number greater than 9 an odd number and a number less than 10

Totals

6

9

15

Corporal

10

8

18

Sergeant

4

5

9

Lieutenant

2

1

3

Captain Totals

a. b. c. d. e.

Men

2

3

5

24

26

50

a woman and a sergeant a man and a private a private and a sergeant a woman and an officer a person in the military

44. Referring to Exercise 43, what is the probability the person chosen is a. a woman or a sergeant b. a man or a private c. a woman or a man d. a woman or an officer e. a captain or a lieutenant

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A computer is asked to randomly generate a three-digit number. What is the probability the

45. ten’s digit is odd or the one’s digit is even 46. first digit is prime and the number is a multiple of 10 A computer is asked to randomly generate a four-digit number. What is the probability the number is

47. at least 4000 or a multiple of 5 48. less than 7000 and an odd number 49. Two fair dice are rolled. What is the probability of rolling a. boxcars (a sum of 12) or snake eyes (a sum of 2) b. a sum of 7 or a sum of 11 c. an even-numbered sum or a prime sum 

d. an odd-numbered sum or a sum that is a multiple of 4 e. a sum of 15 or a multiple of 12 f. a sum that is a prime number 50. Suppose all 16 balls from a game of pool (see Exercise 42) are placed in a large leather bag and mixed, then one is drawn out. Consider the cue ball as “0.” What is the probability of drawing a. a striped ball b. a solid-colored ball c. a polka-dotted ball d. the cue ball e. the cue ball or the eight ball f. a striped ball or a number less than five g. a solid color or a number greater than 12 h. an odd number or a number divisible by 4

WORKING WITH FORMULAS

51. Games involving a fair spinner (with numbers 1 through 4): P1n2  1 14 2 n Games that involve moving pieces around a board using a fair spinner are fairly common. If a fair spinner has the numbers 1 through 4, the probability that any one number is spun n times in succession is given by the formula shown, where n represents the number of spins. What is the probability (a) the first player spins a two? (b) all four players spin a two? (c) Discuss the graph of P(n) and explain the connection between the graph and the probability of consistently spinning a two. 

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CHAPTER 10 Additional Topics in Algebra

52. Games involving a fair coin (heads and tails): P1n2  1 12 2 n When a fair coin is flipped, the probability that heads (or tails) is flipped n times in a row is given by the formula shown, where n represents the number of flips. What is the probability (a) the first flip is heads? (b) the first four flips are heads? (c) Discuss the graph of P(n) and explain the connection between the graph and the probability of consistently flipping heads.

APPLICATIONS

53. To improve customer service, a company tracks the number of minutes a caller is “on hold” and waiting for a customer service representative. The table shows the probability that a caller will wait m minutes. Based on the table, what is the probability a caller waits a. at least 2 minutes b. less than 2 minutes c. 4 minutes or less d. over 4 minutes e. less than 2 or more than 4 minutes f. 3 or more minutes

Wait Time (minutes m)

Probability

0

0.07

0 6 m 6 1

0.28

1m 6 2

0.32

2m 6 3

0.25

3m 6 4

0.08

54. To study the impact of technology on American families, a researcher first determines the probability that a family has n computers at home. Based on the table, what is the probability a home

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a. b. c. d. e. f.

has at least one computer has two or more computers has less than four computers has five computers has one, two, or three computers does not have two computers Number of Computers

time you throw a dart. Assuming the probabilities are related to area, on the next dart that you throw what is the probability you a. score at least a 4? b. score at least a 6? c. hit the bull’s-eye? d. score exactly 4 points? 58. Three red balls, six blue balls, and four white balls are placed in a bag. What is the probability the first ball you draw out is a. red b. blue c. not white d. purple e. red or white f. red and white

Probability

0

9%

1

51%

2

28%

3

9%

4

3%

55. Jolene is an experienced markswoman and is able to hit a 10 in. by 20 in. target 100% of the time at a range of 100 yd. Assuming the 10 in. probability she hits a target is related to its area, what is the 20 in. probability she hits the shaded portions shown? a. isosceles triangle

b. right triangle

b. circle

c. isosceles trapezoid with b 

57. A circular dartboard has a total radius of 8 in., with circular bands that are 2 in. wide, as shown. You are skilled enough to hit this board 100% of the time so you always score at least two points each

59. Three red balls, six blue balls, and four white balls are placed in a bag, then two are drawn out and placed in a rack. What is the probability the balls drawn are a. first red, second blue b. first blue, second red c. both white d. first blue, second not red e. first white, second not blue f. first not red, second not blue 60. Consider the 210 discrete points found in the first and second quadrants where 10  x  10, 1  y  10, and x and y are integers. The coordinates of each point is written on a slip of paper and placed in a box. One of the slips is then randomly drawn. What is the probability the point (x, y) drawn a. is on the graph of y  x b. is on the graph of y  2x c. is on the graph of y  0.5x d. has coordinates 1x, y 7 22 e. has coordinates 1x  5, y 7 22 f. is between the branches of y  x2

c. equilateral triangle

56. a. square

997

B 2

2 4 6 8

61. Your instructor surprises you with a True/False quiz for which you are totally unprepared and must guess randomly. What is the probability you pass the quiz with an 80% or better if there are a. three questions b. four questions c. five questions 62. A robot is sent out to disarm a timed explosive device by randomly changing some switches from a neutral position to a positive flow or negative flow position. The problem is, the switches are independent and unmarked, and it is unknown

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which direction is positive and which direction is negative. The bomb is harmless if a majority of the switches yield a positive flow. All switches must be thrown. What is the probability the device is disarmed if there are a. three switches b. four switches c. five switches 63. A survey of 100 retirees was taken to gather information concerning how they viewed the Vietnam War back in the early 1970s. A breakdown of those surveyed is shown in the table. One out of the hundred will be selected at random for a personal, taped interview. What is the probability the person chosen had a a. career of any kind and opposed the war b. medical career and supported the war c. military career and opposed the war d. legal or business career and opposed the war e. academic or medical career and supported the war Career

Support

Opposed

T

Military Medical

9

3

12

8

16

24

Legal

15

12

27

Business

18

6

24

Academics Totals

3

10

13

53

47

100

64. Referring to Exercise 63, what is the probability the person chosen a. had a career of any kind or opposed the war b. had a medical career or supported the war c. supported the war or had a military career d. had a medical or a legal career e. supported or opposed the war 

65. The Board of Directors for a large hospital has 15 members. There are six doctors of nephrology (kidneys), five doctors of gastroenterology (stomach and intestines), and four doctors of endocrinology (hormones and glands). Eight of them will be selected to visit the nation’s premier hospitals on a 3-week, expenses-paid tour. What is the probability the group of eight selected consists of exactly a. four nephrologists and four gastroenterologists b. three endocrinologists and five nephrologists 66. A support group for hodophobics (an irrational fear of travel) has 32 members. There are 15 aviophobics (fear of air travel), eight siderodrophobics (fear of train travel), and nine thalassophobics (fear of ocean travel) in the group. Twelve of them will be randomly selected to participate in a new therapy. What is the probability the group of 12 selected consists of exactly a. two aviophobics, six siderodrophobics, and four thalassophobics b. five thalassophobics, four aviophobics, and three siderodrophobics 67. A trained chimpanzee is given a box containing eight wooden cubes with the letters p, a, r, a, l, l, e, l printed on them (one letter per block). Assuming the chimp can’t read or spell, what is the probability he draws the eight blocks in order and actually forms the word “parallel”? 68. A number is called a “perfect number” if the sum of its proper factors is equal to the number itself. Six is the first perfect number since the sum of its proper factors is six: 1  2  3  6. Twenty-eight is the second since: 1  2  4  7  14  28. A young child is given a box containing eight wooden blocks with the following numbers (one per block) printed on them: four 3’s, two 5’s, one 0, and one 6. What is the probability she draws the eight blocks in order and forms the fifth perfect number: 33,550,336?

EXTENDING THE CONCEPT

69. The function f 1x2  1 12 2 x gives the probability that x number of flips will all result in heads (or tails). Compute the probability that 20 flips results in 20 heads in a row, then use the Internet or some other resource to find the probability of winning a state lottery. Which is more likely to happen (which has the greater probability)? Were you surprised? 70. Recall that a function is a relation in which each element of the domain is paired with only one element from the range. Is the relation defined by C1x2  nCx (n is a constant) a function? To

investigate, plot the points generated by C1x2  6Cx for x  0 to x  6 and answer the following questions: a. Is the resulting graph continuous or discrete (made up of distinct points)? b. Does the resulting graph pass the vertical line test? c. Discuss the features of the relation and its graph, including the domain, range, maximum or minimum values, and symmetries observed.

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MAINTAINING YOUR SKILLS

71. (5.3) Given csc   3 and cos  6 0, find the values of the remaining five trig functions of . 72. (4.4) Complete the following logarithmic properties: logb b  __

logb 1  __

logb bn  __

blogbn  __

73. (6.4) Find exact values for sin122, cos122, and 21 tan122 given cos    and  is in Quadrant II. 29 74. (8.3) A rubber ball is dropped from a height of 25 ft onto a hard surface. With each bounce, it rebounds 60% of the height from which it last fell. Use sequences/series to find (a) the height of the sixth bounce, (b) the total distance traveled up to the sixth bounce, and (c) the distance the ball will travel before coming to rest.

10.7 The Binomial Theorem Learning Objectives In Section 10.7 you will learn how to:

A. Use Pascal’s triangle to find 1a  b2 n

B. Find binomial coefficients n using a b notation k C. Use the binomial theorem to find 1a  b2 n

D. Find a specific term of a binomial expansion

1 1 ?

1 1

?

Strictly speaking, a binomial is a polynomial with two terms. This limits us to terms with real number coefficients and whole number powers on variables. In this section, we will loosely regard a binomial as the sum or difference of any two terms. Hence 1 1 13 3x2  y4, 1x  4, x  , and   i are all “binomials.” Our goal is to develop x 2 2 an ability to raise a binomial to any natural number power, with the results having important applications in genetics, probability, polynomial theory, and other areas. The tool used for this purpose is called the binomial theorem.

A. Binomial Powers and Pascal’s Triangle

Much of our mathematical understanding comes from a study of patterns. One area where the study of patterns has been particularly fruitful is Pascal’s triangle (Figure 10.20), named after the French scientist Blaise Pascal Figure 10.20 (although the triangle was well known before his time). It begins 1 First row with a “1” at the vertex of the triangle, with 1’s extending diagonally downward to the left and right as shown. The entries on Second row 1 1 the interior of the triangle are found by adding the two entries Third row 1 2 directly above and to the left and right of each new position. There are a variety of patterns hidden within the triangle. In Fourth row 1 3 3 this section, we’ll use the horizontal rows of the triangle to help us raise a binomial to various powers. To begin, recall that 1 6 ? 1a  b2 0  1 and 1a  b2 1  1a  1b (unit coefficients are ? ? 1 ? included for emphasis). In our earlier work, we saw that a and so on binomial square (a binomial raised to the second power) always

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followed the pattern 1a  b2 2  1a2  2ab  1b2. Observe the overall pattern that is developing as we include 1a  b2 3: 1a  b2 0 1a  b2 1 1a  b2 2 1a  b2 3

1 1a  1b 1a2  2ab  1b2 1a3  3a2b  3ab2  1b3

row 1 row 2 row 3 row 4

Apparently the coefficients of 1a  b2 will occur in row n  1 of Pascal’s triangle. Also observe that in each term of the expansion, the exponent of the first term a decreases by 1 as the exponent on the second term b increases by 1, keeping the degree of each term constant (recall the degree of a term with more than one variable is the sum of the exponents). n

1a3b0  3a2b1  3a1b2  1a0b3 30 degree 3

21 degree 3

12 degree 3

03 degree 3

These observations help us to quickly expand a binomial power. EXAMPLE 1



Expanding a Binomial Using Pascal’s Triangle

Solution



Working step-by-step we have 1. The coefficients will be in the fifth row of Pascal’s triangle. 1 4 6 4 1 2. The exponents on x begin at 4 and decrease, while the exponents on 12 begin at 0 and increase.

Use Pascal’s triangle and the patterns noted to expand 1x  12 2 4.

1 0 1 1 1 2 1 3 1 4 1x4 a b  4x3 a b  6x2 a b  4x1 a b  1x0 a b 2 2 2 2 2 3. Simplify each term. 3 1 1 The result is x4  2x3  x2  x  . 2 2 16 Now try Exercises 7 through 10



If the exercise involves a difference rather than a sum, we simply rewrite the expression using algebraic addition and proceed as before. EXAMPLE 2



Raising a Complex Number to a Power Using Pascal’s Triangle Use Pascal’s triangle and the patterns noted to compute 13  2i2 5.

Begin by rewriting 13  2i2 5 as 33  12i2 4 5. 1. The coefficients will be in the sixth row of Pascal’s triangle. 1 5 10 10 5 1 2. The exponents on 3 begin at 5 and decrease, while the exponents on 12i2 begin at 0 and increase. 1135 212i2 0  5134 2 12i2 1  10133 212i2 2  10132 212i2 3  5131 2 12i2 4  1130 212i2 5 3. Simplify each term. A. You’ve just learned how 243  810i  1080  720i  240  32i to use Pascal’s triangle to find The result is 597  122i. Solution

1a  b2 n



Now try Exercises 11 and 12



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Expanding Binomial Powers 1a  b2 n 1. The coefficients will be in row n  1 of Pascal’s triangle. 2. The exponents on the first term begin at n and decrease, while the exponents on the second term begin at 0 and increase. 3. For any binomial difference 1a  b2 n, rewrite the base as 3a  1b2 4 n using algebraic addition and proceed as before, then simplify each term.

B. Binomial Coefficients and Factorials

Pascal’s triangle can easily be used to find the coefficients of 1a  b2 n, as long as the exponent is relatively small. If we needed to expand 1a  b2 25, writing out the first 26 rows of the triangle would be rather tedious. To overcome this limitation, we introduce a formula for the binomial coefficients that enables us to find the coefficients of any expansion. The Binomial Coefficients n For natural numbers n and r where n  r, the expression a b, read “n choose r,” r is called the binomial coefficient and evaluated as: n n! a b r!1n  r2! r In Example 1, we found the coefficients of 1a  b2 4 using the fifth or 1n  12st row of Pascal’s triangle. In Example 3, these coefficients are found using the formula for binomial coefficients. EXAMPLE 3



Computing Binomial Coefficients n n! Evaluate a b  as indicated: r r!1n  r2! 4 a. a b 1

Solution



4 b. a b 2

4 c. a b 3

4 # 3! 4 4! a. a b   4 1!14  12! 1!3! 1 4 # 3 # 2! 4#3 4 4! b. a b    6 2!14  22! 2!2! 2 2 4 # 3! 4 4! c. a b   4 3!14  32! 3!1! 3 Now try Exercises 13 through 20



4 4 4 Note a b  4, a b  6, and a b  4 give the interior entries in the fifth row of 1 2 3 Pascal’s triangle: 1 4 6 4 1. For consistency and symmetry, we define 0!  1, which enables the formula to generate all entries of the triangle, including the “1’s.” 4 4! a b 0 0!14  02! 4!  # 1 1 4!

apply formula

0!  1

4 4! 4! a b  # 4 4!14  42! 4! 0! 4!  # 1 4! 1

apply formula

0!  1

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n The formula for a b with 0  r  n now gives all coefficients in the 1n  12st r row. For n  5, we have 5 a b 0 1 EXAMPLE 4



5 a b 1 5

5 a b 2 10

5 a b 3 10

5 a b 4 5

5 a b 5 1

Computing Binomial Coefficients Compute the binomial coefficients: 9 a. a b 0

Solution



9 b. a b 1

6 c. a b 5

9 9! a. a b  0 0!19  02! 9!  1 9! 6 6! c. a b  5 5!16  52! 6!  6 5!

6 d. a b 6 9 9! b. a b  1 1!19  12! 9!  9 8! 6 6! d. a b  6 6!16  62! 6!  1 6! Now try Exercises 21 through 24

B. You’ve just learned how to find binomial coefficients n using a b notation k



n You may have noticed that the formula for a b is identical to that of nCr, and both r yield like results for given values of n and r. For future use, it will help to commit the n n n b  n, and general results from Example 4 to memory: a b  1, a b  n, a 0 1 n  1 n a b  1. n

C. The Binomial Theorem n Using a b notation and the observations made regarding binomial powers, we can now r state the binomial theorem. Binomial Theorem For any binomial 1a  b2 and natural number n, n n n 1a  b2 n  a banb0  a ban1b1  a ban2b2  p 0 1 2 n n a b a1bn1  a b a0bn n1 n The theorem can also be stated in summation form as n n 1a  b2 n  a b anrbr r0 r



The expansion actually looks overly impressive in this form, and it helps to summarize the process in words, as we did earlier. The exponents on the first term a begin at n and decrease, while the exponents on the second term b begin at 0 and increase,

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n keeping the degree of each term constant. The a b notation simply gives the coefficients r n of each term. As a final note, observe that the r in a b gives the exponent on b. r EXAMPLE 5



Solution



Expanding a Binomial Using the Binomial Theorem Expand 1a  b2 6 using the binomial theorem.

6 6 6 6 6 6 6 1a  b2 6  a b a6b0  a b a5b1  a b a4b2  a b a3b3  a b a2b4  a b a1b5  a b a0b6 0 1 2 3 4 5 6 6! 6 6! 5 1 6! 4 2 6! 3 3 6! 2 4 6! 1 5 6! 6 a  ab  ab  ab  ab  ab  b  0!6! 1!5! 2!4! 3!3! 4!2! 5!1! 6!0!  1a6  6a5b  15a4b2  20a3b3  15a2b4  6ab5  1b6 Now try Exercises 25 through 32

EXAMPLE 6



Using the Binomial Theorem to Find the Initial Terms of an Expansion

Solution



Use the binomial theorem with a  2x, b  y2, and n  10.



Find the first three terms of 12x  y2 2 10. 12x  y2 2 10  a

C. You’ve just learned how to use the binomial theorem to find 1a  b2 n

10 10 10 b 12x2 10 1y2 2 0  a b 12x2 9 1y2 2 1  a b 12x2 8 1y2 2 2  p 0 1 2

first three terms

 1121024x10  1102512x9y2 

a

 1024x10

10!  45 2!8!

 1024x10

10! 256x8y4  p 2!8!  5120x9y2  1452256x8y4  p  5120x9y2  11,520x8y4  p

10 10 b  1, a b  10 0 1

result

Now try Exercises 33 through 36



D. Finding a Specific Term of the Binomial Expansion In some applications of the binomial theorem, our main interest is a specific term of the expansion, rather than the expansion as a whole. To find a specified term, it helps to consider that the expansion of 1a  b2 n has n  1 terms: 1a  b2 0 has one term,

n 1a  b2 1 has two terms, 1a  b2 2 has three terms, and so on. Because the notation a b r always begins at r  0 for the first term, the value of r will be 1 less than the term we are seeking. In other words, for the seventh term of 1a  b2 9, we use r  6. The kth Term of a Binomial Expansion For the binomial expansion 1a  b2 n, the kth term is given by n a b anrbr, where r  k  1. r

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EXAMPLE 7



Solution



Finding a Specific Term of a Binomial Expansion Find the eighth term in the expansion of 1x  2y2 12.

By comparing 1x  2y2 12 to 1a  b2 n we have a  x, b  2y, and n  12. Since we want the eighth term, k  8 S r  7. The eighth term of the expansion is a

12 5 12! b x 12y2 7  128x5y7 7 7!5!  17922 1128x5y7 2  101,376x5y7

27  128 1 12 7 2  792 result

Now try Exercises 37 through 42



One application of the binomial theorem involves a binomial experiment and binomial probability. For binomial probabilities, the following must be true: (1) The experiment must have only two possible outcomes, typically called success and failure, and (2) if the experiment has n trials, the probability of success must be constant for n all n trials. If the probability of success for each trial is p, the formula a b 11  p2 nkpk k gives the probability that exactly k trials will be successful. Binomial Probability Given a binomial experiment with n trials, where the probability for success in each trial is p. The probability that exactly k trials are successful is given by n a b 11  p2 nk pk. k

EXAMPLE 8



Applying the Binomial Theorem–Binomial Probability Paula Rodrigues has a free-throw shooting average of 85%. On the last play of the game, with her team behind by three points, she is fouled at the three-point line, and is awarded two additional free throws via technical fouls on the opposing coach (a total of five free-throws). What is the probability she makes at least three (meaning they at least tie the game)?

Solution



Here we have p  0.85, 1  p  0.15, and n  5. The key idea is to recognize the phrase at least three means “3 or 4 or 5.” So P(at least 3)  P13 ´ 4 ´ 52. P1at least 32  P13 ´ 4 ´ 52  P132  P142  P152

“or” implies a union sum of probabilities (mutually exclusive events)

5 5 5  a b 10.152 2 10.852 3  a b 10.152 1 10.852 4  a b 10.152 0 10.852 5 4 5 3 D. You’ve just learned how to find a specific term of a binomial expansion

 0.1382  0.3915  0.4437  0.9734 Paula’s team has an excellent chance 197.3% 2 of at least tying the game. New try Exercises 45 and 46



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10.7 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.

4. In a binomial experiment with n trials, the probability there are exactly k successes is given

1. In any binomial expansion, there is always more term than the power applied.

2. In all terms in the expanded form of 1a  b2 n, the exponents on a and b must sum to .

by the formula

3. To expand a binomial difference such as 1a  2b2 , we rewrite the binomial as and proceed as before. 5



5. Discuss why the expansion of 1a  b2 n has n  1 terms. 6. For any defined binomial experiment, discuss the relationships between the phrases, “exactly k success,” and “at least k successes.”

DEVELOPING YOUR SKILLS

Use Pascal’s triangle and the patterns explored to write each expansion.

7. 1x  y2 5

10. 1x2  13 2 3

8. 1a  b2 6

11. 11  2i2 5

9. 12x  32 4

12. 12  5i2 4

Evaluate each of the following

7 13. a b 4 9 16. a b 5 40 19. a b 3 5 22. a b 0



.

8 14. a b 2 20 17. a b 17 45 20. a b 3 15 23. a b 15

5 15. a b 3 30 18. a b 26 6 21. a b 0 10 24. a b 10

Use the binomial theorem to expand each expression. Write the general form first, then simplify.

25. 1c  d2 5

26. 1v  w2 4

31. 11  2i2 3

32. 12  13i2 5

28. 1x  y2 7

29. 12x  32 4

27. 1a  b2 6

30. 1a  2b2 5

Use the binomial theorem to write the first three terms.

33. 1x  2y2 9

36. 1 12a  b2 2 10

34. 13p  q2 8

35. 1v2  12w2 12

Find the indicated term for each binomial expansion.

37. 1x  y2 7; 4th term

39. 1p  22 8; 7th term

41. 12x  y2 12; 11th term

38. 1m  n2 6; 5th term

40. 1a  32 14; 10th term

42. 13n  m2 9; 6th term

WORKING WITH FORMULAS

n 1 k 1 nk 43. Binomial probability: P1k2  a b a b a b 2 k 2 The theoretical probability of getting exactly k heads in n flips of a fair coin is given by the formula above. What is the probability that you would get 5 heads in 10 flips of the coin?

n 1 k 4 nk 44. Binomial probability: P1k2  a b a b a b 5 k 5 A multiple choice test has five options per question. The probability of guessing correctly k times out of n questions is found using the formula shown. What is the probability a person scores a 70% by guessing randomly (7 out of 10 questions correct)?

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APPLICATIONS

45. Batting averages: Tony Gwynn (San Diego Padres) had a lifetime batting average of 0.347, ranking him as one of the greatest hitters of all time. Suppose he came to bat five times in any given game. a. What is the probability that he will get exactly three hits? b. What is the probability that he will get at least three hits?

47. Late rental returns: The manager of Victor’s DVD Rentals knows that 6% of all DVDs rented are returned late. Of the eight videos rented in the last hour, what is the probability that a. exactly five are returned on time b. exactly six are returned on time c. at least six are returned on time d. none of them will be returned late

46. Pollution testing: Erin suspects that a nearby iron smelter is contaminating the drinking water over a large area. A statistical study reveals that 83% of the wells in this area are likely contaminated. If the figure is accurate, find the probability that if another 10 wells are tested a. exactly 8 are contaminated b. at least 8 are contaminated

48. Opinion polls: From past experience, a research firm knows that 20% of telephone respondents will agree to answer an opinion poll. If 20 people are contacted by phone, what is the probability that a. exactly 18 refuse to be polled b. exactly 19 refuse to be polled c. at least 18 refuse to be polled d. none of them agree to be polled



EXTENDING THE CONCEPT

49. Prior to calculators and computers, the binomial theorem was frequently used to approximate the value of compound interest given by the expression r nt a1  b by expanding the first three terms. For n example, if the interest rate were 8% 1r  0.082 and the interest was compounded quarterly 142152 1n  42 for 5 yr 1t  52, we have 11  0.08  4 2 20 11  0.022 . The first three terms of the expansion give a value of: 1  2010.022  19010.00042  1.476. a. Calculate the percent error: approximate value %error  actual value b. What is the percent error if only two terms are used.



50. If you sum the entries in each row of Pascal’s triangle, a pattern emerges. Find a formula that generalizes the result for any row of the triangle, and use it to find the sum of the entries in the 12th row of the triangle. n n b for n  6 and k  6. 51. Show that a b  a k nk

52. The derived polynomial of f (x) is f 1x  h2 or the original polynomial evaluated at x  h. Use Pascal’s triangle or the binomial theorem to find the derived polynomial for f 1x2  x3  3x2  5x  11. Simplify the result completely.

MAINTAINING YOUR SKILLS

53. (2.7) Graph the function shown and find x2 x2 f 132: f 1x2  e 2 1x  42 x 7 2 54. (5.4) Given the point (0.6, y) is a point on the unit circle in the third quadrant, find y.

55. (3.4) Graph the function g1x2  x3  x2  6x. Clearly indicate all intercepts and intervals where g1x2 7 0. 56. (6.5) Evaluate arcsin c sina

5 bd. 6

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S U M M A RY A N D C O N C E P T R E V I E W SECTION 10.1

Sequences and Series

KEY CONCEPTS • A finite sequence is a function an whose domain is the set of natural numbers from 1 to n. • The terms of the sequence are labeled a1, a2, a3, . . . , ak1, ak, ak1, p , an2, an1, an. • The expression an, which defines the sequence (generates the terms in order), is called the nth term. • An infinite sequence is a function whose domain is the set of natural numbers. • When each term of a sequence is larger than the preceding term, it is called an increasing sequence. • When each term of a sequence is smaller than the preceding term, it is called a decreasing sequence. • When successive terms of a sequence alternate in sign, it is called an alternating sequence. • When the terms of a sequence are generated using previous term(s), it is called a recursive sequence. • Sequences are sometimes defined using factorials, which are the product of a given natural number with all natural numbers that precede it: n!  n # 1n  12 # 1n  22 # p # 3 # 2 # 1. • Given the sequence a1, a2, a3, a4, . . . , an the sum is called a finite series and is denoted Sn. • Sn  a1  a2  a3  a4  p  an. The sum of the first n terms is called a partial sum. k

• In sigma notation, the expression

a  a i

1

 a2  p  ak represents a finite series.

i1

• When sigma notation is used, the letter “i” is called the index of summation. EXERCISES Write the first four terms that are defined and the value of a10. n1 1. an  5n  4 2. an  2 n 1 Find the general term an for each sequence, and the value of a6. 3. 1, 16, 81, 256, . . . 4. 17, 14, 11, 8, p Find the eighth partial sum (S8). 5.

1 1 1 2, 4, 8,

6. 21, 19, 17, p

p

Evaluate each sum. 7

7.



5

n2

8.

n1

 13n  22

n1

Write the first five terms that are defined. n! 9. an  1n  22!

10. e

a1  12 an1  2an  14

Write as a single summation and evaluate. 7

11.



7

n2 

n1

 13n  22

n1

SECTION 10.2

Arithmetic Sequences

KEY CONCEPTS • In an arithmetic sequence, successive terms are found by adding a fixed constant to the preceding term. • In a sequence, if there exists a number d, called the common difference, such that ak1  ak  d, then the sequence is arithmetic. Alternatively, ak1  ak  d for k  1.

1007

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• The nth term n of an arithmetic sequence is given by an  a1  1n  12d, where a1 is the first term and d is the

common difference. If • the initial term is unknown or is not a1 the nth term can be written an  ak  1n  k2d, where the subscript of the term ak and the coefficient of d sum to n. • For an arithmetic sequence with first term a1, the nth partial sum (the sum of the first n terms) is given by n1a1  an 2 Sn  . 2

EXERCISES Find the general term (an) for each arithmetic sequence. Then find the indicated term. 12. 2, 5, 8, 11, . . . ; find a40 13. 3, 1, 1, 3, p ; find a35 Find the sum of each series. 14. 1  3  7  11  p  75 16. 3  6  9  12  p ; S20

15. 1  4  7  10  p  88 17. 1  3  1  1  p ; S15 4

25

18.

 13n  42

19.

n1

SECTION 10.3

2

4

40

 14n  12

n1

Geometric Sequences

KEY CONCEPTS • In a geometric sequence, successive terms are found by multiplying the preceding term by a nonzero constant. ak1 • In other words, if there exists a number r, called the common ratio, such that a  r, then the sequence is k geometric. Alternatively, we can write ak1  akr for k  1. • The nth term an of a geometric sequence is given by an  a1rn1, where a1 is the first term and an represents the general term of a finite sequence. • If the initial term is unknown or is not a1, the nth term can be written an  akrnk, where the subscript of the term ak and the exponent on r sum to n. a1 11  rn 2 . • The nth partial sum of a geometric sequence is Sn  1r a1 . • If r  6 1, the sum of an infinite geometric series is Sq  1r EXERCISES Find the indicated term for each geometric sequence. 20. a1  5, r  3; find a7 21. a1  4, r  12; find a7

22. a1  17, r  17; find a8

Find the indicated sum, if it exists. 23. 16  8  4  p find S7 25. 4  2  1  1  p ; find S12

24. 2  6  18  p ; find S8 26. 4  8  12  24  p

27. 5  0.5  0.05  0.005  p

28. 6  3  32  34  p

5

5

5

8

29.

2 n 5a b 3 n1

10



4 n 12a b 3 n1 q

30.



1 n 5a b 2 n1 q

31.



32. Charlene began to work for Grayson Natural Gas in January of 1990 with an annual salary of $26,000. Her contract calls for a $1220 raise each year. Use a sequence/series to compute her salary after nine years, and her total earnings up to and including that year. (Hint: For a1  26,000, her salary after 9 yrs will be what term of the sequence?)

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33. Sumpter reservoir contains 121,500 ft3 of water and is being drained in the following way. Each day one-third of the water is drained (and not replaced). Use a sequence/series to compute how much water remains in the pond after 7 days. 34. Credit-hours taught at Cody Community College have been increasing at 7% per year since it opened in 2000 and taught 1225 credit-hours. For the new faculty, the college needs to predict the number of credit-hours that will be taught in 2009. Use a sequence/series to compute the credit-hours for 2009 and to find the total number of credit hours taught through the 2009 school year.

SECTION 10.4

Mathematical Induction

KEY CONCEPTS • Functions written in subscript notation can be evaluated, graphed, and composed with other functions. • A sum formula involving only natural numbers n as inputs can be proven valid using a proof by induction. Given that Sn represents a sum formula involving natural numbers, if (1) S1 is true and (2) Sk  ak1  Sk1, then Sn must be true for all natural numbers. • Proof by induction can also be used to validate other relationships, using a more general statement of the principle. Let Sn be a statement involving the natural numbers n. If (1) S1 is true (Sn for n  12 and (2) the truth of Sk implies that Sk1 is also true, then Sn must be true for all natural numbers n. EXERCISES Use the principle of mathematical induction to prove the indicated sum formula is true for all natural numbers n. 35. 1  2  3  4  5  p  n; 36. 1  4  9  16  25  36  p  n2; n1n  12 n1n  1212n  12 an  n and Sn  . an  n2 and Sn  . 2 6 Use the principle of mathematical induction to prove that each statement is true for all natural numbers n. 37. 4n  3n  1 38. 6 # 7n1  7n  1 39. 3n  1 is divisible by 2

SECTION 10.5

Counting Techniques

KEY CONCEPTS • An experiment is any task that can be repeated and has a well-defined set of possible outcomes. • Each repetition of an experiment is called a trial. • Any potential outcome of an experiment is called a sample outcome. • The set of all sample outcomes is called the sample space. • An experiment with N (equally likely) sample outcomes that is repeated t times, has a sample space with N t elements. • If a sample outcome can be used more than once, the counting is said to be with repetition. If a sample outcome can be used only once the counting is said to be without repetition. • The fundamental principle of counting states: If there are p possibilities for a first task, q possibilities for the second, and r possibilities for the third, the total number of ways the experiment can be completed is pqr. This fundamental principle can be extended to include any number of tasks. • If the elements of a sample space have precedence or priority (order or rank is important), the number of elements is counted using a permutation, denoted nPr and read, “the distinguishable permutations of n objects taken r at a time.” n! . • To expand nPr, we can write out the first r factors of n! or use the formula nPr  1n  r2! • If any of the sample outcomes are identical, certain permutations will be nondistinguishable. In a set containing n elements where one element is repeated p times, another is repeated q times, and another r times 1p  q  r  n2, n! nP n the number of distinguishable permutations is given by  . p!q!r! p!q!r!

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• If the elements of a set have no rank, order, or precedence (as in a committee of colleagues) permutations with the same elements are considered identical. The result is the number of combinations, nCr 

n! . r!1n  r2!

EXERCISES 40. Three slips of paper with the letters A, B, and C are placed in a box and randomly drawn one at a time. Show all possible ways they can be drawn using a tree diagram. 41. The combination for a certain bicycle lock consists of three digits. How many combinations are possible if (a) repetition of digits is not allowed and (b) repetition of digits is allowed. 42. Jethro has three work shirts, four pairs of work pants, and two pairs of work shoes. How many different ways can he dress himself (shirt, pants, shoes) for a day’s work? 43. From a field of 12 contestants in a pet show, three cats are chosen at random to be photographed for a publicity poster. In how many different ways can the cats be chosen? 44. How many subsets can be formed from the elements of this set: { , , , , }? 45. Compute the following values by hand, showing all work: c. 7C4 a. 7! b. 7P4 46. Six horses are competing in a race at the McClintock Ranch. Assuming there are no ties, (a) how many different ways can the horses finish the race? (b) How many different ways can the horses finish first, second, and third place? (c) How many finishes are possible if it is well known that Nellie-the-Nag will finish last and Sea Biscuit will finish first? 47. How many distinguishable permutations can be formed from the letters in the word “tomorrow”? 48. Quality Construction Company has 12 equally talented employees. (a) How many ways can a three-member crew be formed to complete a small job? (b) If the company is in need of a Foreman, Assistant Foreman, and Crew Chief, in how many ways can the positions be filled?

SECTION 10.6

Introduction to Probability

KEY CONCEPTS • An event E is any designated set of sample outcomes. • Given S is a sample space of equally likely sample outcomes and E is an event relative to S, the probability of E, n1E2 written P(E), is computed as P1E2  , where n(E) represents the number of elements in E, and n(S) n1S2 represents the number of elements in S. • The complement of an event E is the set of sample outcomes in S, but not in E and is denoted E. • Given sample space S and any event E defined relative to S: 112 P1S2  0, 122 0  P1E2  1, 132 P1S2  1, 142 P1E2  1  P1E2, and 152 P1E2  P1E2  1. • Two events that have no outcomes in common are said to be mutually exclusive. • If two events are not mutually exclusive, P1E1 or E2 2 S P1E1 ´ E2 2  P1E1 2  P1E2 2  P1E1 ¨ E2 2. • If two events are mutually exclusive, P1E1 or E2 2 S P1E1 ´ E2 2  P1E1 2  P1E2 2 . EXERCISES 49. One card is drawn from a standard deck. What is the probability the card is a ten or a face card? 50. One card is drawn from a standard deck. What is the probability the card is a Queen or a face card? 51. One die is rolled. What is the probability the result is not a three? 52. Given P1E1 2  38, P1E2 2  34, and P1E1 ´ E2 2  56, compute P1E1 ¨ E2 2. 53. Find P(E) given that n1E2  7C4 # 5C3 and n1S2  12C7.

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54. To determine if more physicians should be hired, a medical clinic tracks the number of days between a patient’s request for an appointment and the actual appointment date. The table given shows the probability that a patient must wait “d” days. Based on the table, what is the probability a patient must wait a. at least 20 days c. 40 days or less e. less than 40 and more than 10 days

SECTION 10.7

1011

Mixed Review

b. less than 20 days d. over 40 days f. 30 or more days

Wait (days d)

Probability

0

0.002

0 6 d 6 10

0.07

10  d 6 20

0.32

20  d 6 30

0.43

30  d 6 40

0.178

The Binomial Theorem

KEY CONCEPTS • To expand 1a  b2 n for n of “moderate size,” we can use Pascal’s triangle and observed patterns. n • For any natural numbers n and r, where n  r, the expression a b (read “n choose r”) is called the binomial r n n! coefficient and evaluated as a b  . r r!1n  r2! • If n is large, it is more efficient to expand using the binomial coefficients and binomial theorem. • The following binomial coefficients are useful/common and should be committed to memory: n n n n a b1 a bn a bn a b1 0 1 n1 n n n! 1 1    1. • We define 0!  1; for example a b  n n!1n  n2! 0! 1 n n n n n b a1bn1  a b a0bn. • The binomial theorem: 1a  b2 n  a b anb0  a b an1b1  a b an2b2  p  a 0 1 2 n1 n n • The kth term of 1a  b2 n can be found using the formula a b anrbr, where r  k  1. r EXERCISES 55. Evaluate each of the following: 7 8 a. a b b. a b 5 3 Use the binomial theorem to: 57. Write the first four terms of a. 1a  132 8 b. 15a  2b2 7

56. Use Pascal’s triangle to expand the binomials: a. 1x  y2 4

b. 11  2i2 5

58. Find the indicated term of each expansion. a. 1x  2y2 7; fourth b. 12a  b2 14; 10th

MIXED REVIEW 1. Identify each sequence as arithmetic, geometric, or neither. If neither, try to identify the pattern that forms the sequence. a. 120, 163, 206, 249, . . . b. 4, 4, 4, 4, 4, 4, . . . c. 1, 2, 6, 24, 120, 720, 5040, . . .

d. e. f. g. h. i.

2.00, 1.95, 1.90, 1.85, . . . ... 5.5, 6.05, 6.655, 7.3205, . . . 0.1, 0.2, 0.3, 0.4, . . . 525, 551.25, 578.8125, . . . 1 1 1 1 2, 4, 6, 8, . . . 5 5 5 5 8 , 64 , 512 , 4096 ,

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2. Compute by hand (show your work). a. 10!

b.

10! 6!

c.

10P4

d.

e.

10C6

f.

10C4

10P9

3. The call letters for a television station must consist of four letters and begin with either a K or a W. How many distinct call letters are possible if repeating any letter is not allowed? 4. Given a1  9 and r  13, write out the first five terms and the 15th term. 5. Given a1  0.1 and r  5, write out the first five terms and the 15th term. 6. One card is drawn from a well-shuffled deck of standard cards. What is the probability the card is a Queen or an Ace? 7. Two fair dice are rolled. What is the probability the result is not doubles (doubles  same number on both die)? 8. A house in a Boston suburb cost $185,000 in 1985. Each year its value increased by 8%. If this appreciation were to continue, find the value of the house in 2005 and 2015 using a sequence. 9. Evaluate each sum using summation formulas. 2 n a b n1 3

 5

c.

 19  2n2



b.

n1

14. The owner of an arts and crafts store makes specialty key rings by placing five colored beads on a nylon cord and tying it to the ring that will hold the keys. If there are eight different colors to choose from, (a) how many distinguishable key rings are possible if no colors are repeated? (b) How many distinguishable key rings are possible if a repetition of colors is allowed? 15. Donell bought 15 raffle tickets from the Inner City Children’s Music School, and five tickets from the Arbor Day Everyday raffle. The Music School sold a total of 2000 tickets and the Arbor Day foundation sold 550 tickets. For E1: Donell wins the Music School raffle and E2: Donell wins the Arbor Day raffle, find P(E1 or E2). Find the sum if it exists. 16. 13  23  1  43  p  20 3 17. 0.36  0.0036  0.000036  0.00000036  p Find the first five terms of the sequences in Exercises 18 and 19. 18. an 

n1 5

12n 

13. Use a proof by induction to show that 3n1n  12 3  6  9  p  3n  . 2

10

q

a.

10-74

CHAPTER 10 Additional Topics in Algebra



n1

152 

5



n2

n1

10. Expand each binomial using the binomial theorem. Simplify each term. a. 12x  52 5 b. 11  2i2 4 11. For 1a  b2 n, determine: a. the first three terms for n  20 b. the last three terms for n  20 c. the fifth term for n  35 d. the fifth term for n  35, a  0.2, and b  0.8

12. On average, bears older than 3 yr old increase their weight by 0.87% per day from July to November. If a bear weighed 110 kg on June 30th: (a) identify the type of sequence that gives the bear’s weight each day; (b) find the general term for the sequence; and (c) find the bear’s weight on July 1, July 2, July 3, July 31, August 31, and September 30.

19. e

12! 112  n2!

a1  10 an1  an 1 15 2

20. A random survey of 200 college students produces the data shown. One student from this group is randomly chosen for an interview. Use the data to find a. P(student works more than 10 hr) b. P(student takes less than 13 credit-hours) c. P(student works more than 20 hr and takes more than 12 credit-hours) d. P(student works between 11 and 20 hr or takes 6 to 12 credit-hours) 0-10 hr

11-20 hr

Over 20 hr

Total

1-5 credits

3

7

10

20

6-12 credits

21

55

48

124

8

28

20

56

32

90

78

200

over 13 credits Total:

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Practice Test

1013

PRACTICE TEST 1. The general term of a sequence is given. Find the first four terms, the 8th term, and the 12th term. 1n  22! 2n a. an  b. an  n3 n! a1  3 c. an  e an1  21an 2 2  1 2. Expand each series and evaluate. 6

a.

b.

3 j 122a b 4 j1

d.



k2 5

c.

6

12k2  32



 112 a j  1 b j

j

j2 q

1 k 7a b 2 k1



3. Identify the first term and the common difference or common ratio. Then find the general term an. a. 7, 4, 1, 2, . . . b. 8, 6, 4, 2, . . . c. 4, 8, 16, 32, . . . d. 10, 4, 85, 16 25 , . . . 4. Find the indicated value for each sequence. a. a1  4, d  5; find a40 b. a1  2, an  22, d  3; find n c. a1  24, r  12; find a6 d. a1  2, an  486, r  3; find n 5. Find the sum of each series. a. 7  10  13  p  100 37

b.

 13k  22

k1

c. For 4  12  36  108  p , find S7 d. 6  3  32  34  p 6. Each swing of a pendulum (in one direction) is 95% of the previous one. If the first swing is 12 ft, (a) find the length of the seventh swing and (b) determine the distance traveled by the pendulum for the first seven swings. 7. A rare coin that cost $3000 appreciates in value 7% per year. Find the value of after 12 yr. 8. A car that costs $50,000 decreases in value by 15% per year. Find the value of the car after 5 yr. 9. Use mathematical induction to prove that for 5n2  n an  5n  3, the sum formula Sn  is true 2 for all natural numbers n. 10. Use the principle of mathematical induction to prove that Sn: 2 # 3n1  3n  1 is true for all natural numbers n. 11. Three colored balls (Aqua, Brown, and Creme) are to be drawn without replacement from a bag. List all possible ways they can be drawn using (a) a tree diagram and (b) an organized list.

12. Suppose that license plates for motorcycles must consist of three numbers followed by two letters. How many license plates are possible if zero and “Z” cannot be used and no repetition is allowed? 13. How many subsets can be formed from the elements in this set: {,,,,,}. 14. Compute the following values by hand, showing all work: (a) 6! (b) 6P3 (c) 6C3 15. An English major has built a collection of rare books that includes two identical copies of The Canterbury Tales (Chaucer), three identical copies of Romeo and Juliet (Shakespeare), four identical copies of Faustus (Marlowe), and four identical copies of The Faerie Queen (Spenser). If these books are to be arranged on a shelf, how many distinguishable permutations are possible? 16. A company specializes in marketing various cornucopia (traditionally a curved horn overflowing with fruit, vegetables, gourds, and ears of grain) for Thanksgiving table settings. The company has seven fruit, six vegetable, five gourd, and four grain varieties available. If two from each group (without repetition) are used to fill the horn, how many different cornucopia are possible? 17. Use Pascal’s triangle to expand/simplify: a. 1x  2y2 4 b. 11  i2 4 18. Use the binomial theorem to write the first three terms of (a) 1x  122 10 and (b) 1a  2b3 2 8. 19. Michael and Mitchell are attempting to make a nonstop, 100-mi trip on a tandem bicycle. The probability that Michael cannot continue pedaling for the entire trip is 0.02. The probability that Mitchell cannot continue pedaling for the entire trip is 0.018. The probability that neither one can pedal the entire trip is 0.011. What is the probability that they complete the trip? 20. The spinner shown is spun once. 12 1 2 11 What is the probability of spinning 10 3 a. a striped wedge 9 4 b. a shaded wedge 8 5 7 6 c. a clear wedge d. an even number e. a two or an odd number f. a number greater than nine g. a shaded wedge or a number greater than 12 h. a shaded wedge and a number greater than 12

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CHAPTER 10 Additional Topics in Algebra

21. To improve customer Wait (days d ) Probability service, a cable 0 0.02 company tracks the 0 6 d 6 1 0.30 number of days a customer must wait 1d 6 2 0.60 until their cable service 2  d 6 3 0.05 is installed. The table 3d 6 4 0.03 shows the probability that a customer must wait d days. Based on the table, what is the probability a customer waits a. at least 2 days b. less than 2 days c. 4 days or less d. over 4 days e. less than 2 or at least 3 days f. three or more days 48 cm 22. An experienced archer can hit the rectangular target shown 100% of the time at a range of 75 m. Assuming the probability the target is hit is 64 cm related to its area, what is the probability the archer hits within the a. triangle b. circle c. circle but outside the triangle d. lower half-circle e. rectangle but outside the circle f. lower half-rectangle, outside the circle

23. A survey of 100 union workers was taken to register concerns to be raised at the next bargaining session. A breakdown of those surveyed is shown in the table

Expertise Level

Women

Men

Total

Apprentice

16

18

34

Technician

15

13

28

Craftsman

9

9

18

Journeyman

7

6

13

Contractor

3

4

7

50

50

100

Totals

in the right column. One out of the hundred will be selected at random for a personal interview. What is the probability the person chosen is a a. woman or a craftsman b. man or a contractor c. man and a technician d. journeyman or an apprentice 24. Cheddar is a 12-year-old male box turtle. Provolone is an 8-year-old female box turtle. The probability that Cheddar will live another 8 yr is 0.85. The probability that Provolone will live another 8 yr is 0.95. Find the probability that a. both turtles live for another 8 yr b. neither turtle lives for another 8 yr c. at least one of them will live another 8 yr 25. Use a proof by induction to show that the sum of the n1n  12 first n natural numbers is . That is, prove 2 n1n  12 123pn . 2

C A L C U L AT O R E X P L O R AT I O N A N D D I S C O V E RY Infinite Series, Finite Results Although there were many earlier flirtations with infinite processes, it may have been the paradoxes of Zeno of Elea (450 B.C.) that crystallized certain questions that simultaneously frustrated and fascinated early mathematicians. The first paradox, called the dichotomy paradox, can be summarized by the following question: How can one ever finish a race, seeing that one-half the distance must first be traversed, then one-half the remaining distance, then onehalf the distance that then remains, and so on an infinite number of times? Although we easily accept that races can be finished, the subtleties involved in this question stymied mathematicians for centuries and were not satisfactorily resolved until the eighteenth century. In modern notation, 1  p 6 1. This Zeno’s first paradox says 12  14  18  16 1 is a geometric series with a1  2 and r  12.

Illustration 1  For the geometric sequence with a1  12 1 1 and r  , the nth term is an  n . Use the “sum(” and 2 2 “seq(” features of your calculator to compute S5, S10, and S15 (see Technology Highlight from Section 10.1). Does the sum appear to be approaching some “limiting value”? If so, what is this value? Now compute S20, S25, and S30. Does there still appear to be a limit to the sum? What happens when you have the calculator compute S35? Solution  CLEAR the calculator and enter sum(seq (0.5^X, X, 1, 5)) on the home screen. Pressing ENTER gives

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Strengthening Core Skills

Figure 10.21 S50.96875 (Figure 10.21). Press 2nd ENTER to recall the expression and overwrite the 5, changing it to a 10. Pressing ENTER shows S10  0.9990234375. Repeating these steps gives S15  0.9999694824, and it seems that “1” may be a limiting value. Our conjecture receives further support as S20, S25, and S30 are closer and closer to 1, but do not exceed it. Note that the sum of additional terms will create a longer string of 9’s. That the sum of an infinite number of these terms is 1 can be understood by converting the repeating decimal 0.9 to its fractional form (as shown). For x  0.9, 10x  9.9 and it follows that

10x  9.9 x  0.9 9x  9 x 1

1015

For a geometric sequence, the result of an infinite sum a1 can be verified using Sq  . However, there are 1r many nongeometric, infinite series that also have a limiting value. In some cases these require many, many more terms before the limiting value can be observed. Use a calculator to write the first five terms and to find S5, S10, and S15. Decide if the sum appears to be approaching some limiting value, then compute S20 and S25. Do these sums support your conjecture? Exercise 1: a1  13 and r  13 Exercise 2: a1  0.2 and r  0.2 1 1n  12! Additional Insight: Zeno’s first paradox can also be “resolved” by observing that the “half-steps” needed to complete the race require increasingly shorter (infinitesimally short) amounts of time. Eventually the race is complete. Exercise 3: an 

STRENGTHENING CORE SKILLS Probability, Quick-Counting, and Card Games The card game known as Five Card Stud is often played for fun and relaxation, using toothpicks, beans, or pocket change as players attempt to develop a winning “hand” Five Card Hand

from the five cards dealt. The various “hands” are given here with the higher value hands listed first (e.g., a full house is a better/higher hand than a flush).

Description

Probability of Being Dealt

royal flush

five cards of the same suit in sequence from Ace to 10

0.000 001 540

straight flush

any five cards of the same suit in sequence (exclude royal)

0.000 013 900

four of a kind

four cards of the same rank, any fifth card

full house

three cards of the same rank, with one pair

flush

five cards of the same suit, no sequence required

straight

five cards in sequence, regardless of suit

three of a kind

three cards of the same rank, any two other cards

two pairs

two cards of the one rank, two of another rank, one other card

0.047 500

one pair

two cards of the same rank, any three others

0.422 600

0.001 970

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For this study, we will consider the hands that are based on suit (the flushes) and the sample space to be five cards dealt from a deck of 52, or 52C5. A flush consists of five cards in the same suit, a straight consists of five cards in sequence. Let’s consider that an Ace can be used as either a high card (as in 10, J, Q, K, A) or a low card (as in A, 2, 3, 4, 5). Since the dominant characteristic of a flush is its suit, we first consider choosing one suit of the four, then the number of ways that the straight can be formed (if needed). Illustration 1  What is the probability of being dealt a royal flush? Solution  For a royal flush, all cards must be of one suit. Since there are four suits, it can be chosen in 4C1 ways. A royal flush must have the cards A, K, Q, J, and 10 and once the suit has been decided, it can happen in only this (one) way or 1 C 1 . This means # 4C1 1C1 P 1royal flush2   0.000 001 540. 52C5

Illustration 2  What is the probability of being dealt a straight flush? Solution  Once again all cards must be of one suit, which can be chosen in 4C1 ways. A straight flush is any five cards in sequence and once the suit has been decided, this can happen in 10 ways (Ace on down, King on down, Queen on down, and so on). By the FCP, there are 4C1 # 10C1  40 ways this can happen, but four of these will be royal flushes that are of a higher value and must be subtracted from this total. So in the intended context we have # 4C1 10C1  4  0.000 013 900 P1straight flush2  52C5 Using these examples, determine the probability of being dealt Exercise 1: a simple flush (no royal or straight flushes) Exercise 2: three cards of the same suit and any two other (nonsuit) cards Exercise 3: four cards of the same suit and any one other (nonsuit) card Exercise 4: a flush having no face cards

C U M U L AT I V E R E V I E W C H A P T E R S 1 – 1 0 1. Robot Moe is assembling memory cards for computers. At 9:00 A.M., 52 cards had been assembled. At 11:00 A.M., a total of 98 had been made. Assuming the production rate is linear a. Find the slope of this line and explain what it means in this context. b. Determine how many boards Moe Table for can assemble in an eight-hour day. Exercise 3 c. Find a linear equation model for x y this data. 0 d. Determine the approximate time  that Moe began work this 6 morning. 2. When using a calculator to find 13 , yet sin 120°, you get 2 13 sin1a b 120°. Explain why. 2 3. Complete this table of special values for y  cos x without using a calculator.

 4  3  2 2 3 5 6 

4. Sketch the graph of y  1x  4  3 using transformations of a parent function. Label the x- and y-intercepts and state what transformations were used. 5. Solve using the quadratic formula: 3x2  5x  7  0. State your answer in exact and approximate form. 6. The orbit of Venus around the Sun is nearly circular, with a radius of 67 million miles. The planet completes one revolution in about 225 days. Calculate the planet’s (a) angular velocity in radians per hour and (b) the planet’s orbital velocity in miles per hour. 7. For the graph of g(x) shown, state where a. g1x2  0 b. g1x2 6 0 5 c. g1x2 7 0 d. g1x2c e. g1x2T f. local max g. local min h. g1x2  2 5 i. g(4) j. g112 k. as x S 1  , g1x2 S 5 l. as x S q, g1x2 S m. the domain of g(x)

y

g(x)

5 x

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Cumulative Review Chapters 1–10

8. Match each equation to its corresponding graph. a. y  sin1x2 b. y  sin1x  2  c. y  sin12x  2 d. y  sinax  b 2  e. y  sin12x2 f. y  sinax  b 2 (1) y (2) y 1

1

2

x

1

1

(3)

17. Solve for x. a. e2x1  217

2

x

y 1

1

1

2

x

1

1

1

2

x

(6)

1

y 1

1

2

x

1

b. log13x  22  1  4

18. Solve using matrices and row reduction: 2a  3b  6c  15 • 4a  6b  5c  35 3a  2b  5c  24 19. Solve using a calculator and inverse matrices. 0.7x  1.2y  3.2z  32.5 • 1.5x  2.7y  0.8z  7.5 2.8x  1.9y  2.1z  1.5 20. Find the equation of the hyperbola with foci at 16, 02 and (6, 0) and vertices at 14, 02 and (4, 0). 21. Write x2  4y2  24y  6x  29  0 by completing the square, then identify the center, vertices, and foci.

1

y

1

1

(4)

y

1

16. What interest rate is required to ensure that $2000 will double in 10 yr if interest is compounded continuously?

1

1

(5)

15. Write each expression in exponential form: a. 3  logx 11252 b. ln12x  12  5

1

1

1017

1

2

x

1

9. Graph the piecewise function and state the domain and range. 2 3  x  1 y  •x 1 6 x 6 2 x2 2x3 10. For u  3i  4j and v  12i  8j, find the resultant vector w  u  v, then use the dot product to compute the angle between u and w. 11. Compute the difference quotient for each function given. 1 a. f 1x2  2x2  3x b. h1x2  x2 12. Graph the polynomial function given. Clearly indicate all intercepts. f 1x2  x3  x2  4x  4

2x2  8 . Clearly x2  1 indicate all asymptotes and intercepts.

13. Graph the rational function h1x2 

14. Write each expression in logarithmic form: 1  34 a. x  10 y b. 81

22. Use properties of sequences to determine a20 and S20. a. 262144, 65536, 16384, 4096, . . . 7 27 19 11 b. , , , , . . . 8 40 40 40 23. Use the difference identity for cosine to  a. verify that cosa  b  sin  and 2 b. find the value of cos 15° in exact form. 24. Caleb’s grandparents live in a small town that lies 125 mi away at a heading of 110°. Having just received his pilot’s license, he sets out on a heading of 110° in a rented plane, traveling at 125 mph (total flight time 1 hr). Unfortunately, he forgets to account for the wind, which is coming from the northeast at 20 mph on a heading of 190°. (a) If Caleb starts out at coordinates (0, 0), what are the coordinates of his grandparent’s town? (b) What are the vector coordinates of the plane 1 hr later? (c) How many miles is he actually away from his grandparent’s town? 25. Empty 55-gal drums are stacked at a storage facility in the form of a pyramid with 52 barrels in the bottom row, 51 barrels in the next row, and so on, until there are 10 barrels in the top row. Use properties of sequences to determine how many barrels are in this stack.

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1018

CHAPTER 10 Additional Topics in Algebra

26. Three $20 bills, six $50 bills, and four $100 bills are placed in a large box and mixed thoroughly, then two bills are drawn out and placed in a savings account. What is the probability the bills drawn are a. first $20, second $50 b. first $50, second $20 c. both $100 d. first $100, second not $20 e. first $100, second not $50 f. first not $20, second $20 27. The manager of Tom’s Tool and Equipment Rentals knows that 4% of all tools rented are returned late. Of the 12 tools rented in the last hour, what is the probability that a. exactly ten will be returned on time b. at least eleven will be returned on time c. at least ten will be returned on time d. none of them will be returned on time

10-80

28. Use a proof by induction to show 3  7  11  15  p  14n  12  n12n  12 for all natural numbers n. 29. State the three double angle formulas for cosine. If 1 cos122  , what is the value of sin ? 2 30. A park ranger tracks the number of campers at a remote national park from January 1m  12 to December 1m  122 and collects the following data (month, number of campers): (3, 6), (5, 110), (7, 134), and (9, 78). Assuming the data is quadratic 1y  ax2  bx  c2, (a) select any of the three points and create a system of three equations in three variables to obtain a parabolic equation model for the data and (b) determine the month that brought the maximum number of campers. (c) What was this maximum number? (d) How many campers might be expected in September? (e) Based on your model, what month(s) is the park apparently closed to campers (number of campers is zero or negative)?

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Precalculus—

CONNECTIONS TO CALCULUS The properties of summation play an important role in the development of many key ideas in a first semester calculus course. Here, we’ll see how the summation properties (Section 10.1) are combined with the summation formulas (Section 10.4) and the concept of the area under a curve (Connections to Calculus Chapter 2), to produce some very interesting results. For convenience, the summation formulas are restated here: n

(1)

n

 c  cn

(2)

i1 n

(3)



i1

2n  3n  n 6 3

i2 

i1

i 

2

n2  n 2

n

(4)



i3 

i1

n4  2n3  n2 4

Applications of Summation For this study, consider the area under the line f 1x2  x  5, in the interval [0, 4] (between the y-axis and the vertical line x  4, Figure C2C 10.1). To find the total area, we could add the area of the rectangle and triangle shown (see Reinforcing Basic Concepts, page 974), or use f 102  5 and f 142  9 as the bases of a trapezoid h and simply apply the formula A  1b1  b2 2 . 2 Instead of being so direct, we’ll use this simple geometric shape to develop some powerful ideas, to be used when the desired area is much more complex. Using what we’ll call the rectangle method, we first approximate the total area by adding the area of the four rectangles shown in Figure C2C 10.2. Note this will give an area greater than the true value, but yields a reasonable estimate. Each rectangle will have a width of 1, since there are four rectangles in an interval four units wide 1 44  12, and the length of each rectangle will be f 1x2  x  5 for x  1, 2, 3, and 4. Using the formula A  LW, we can write this as

Figure C2C 10.1 y 10 9 8 7 6 5 4 3 2 1 1

2

3

4

5

6

x

5

6

x

5

6

x

Figure C2C 10.2 y 10 9 8 7 6 5 4 3 2 1 1

4

LW   f 1i2 112, where the function is written

2

3

4

in terms of i 3 f 1i2  i  5 4 for the sake of the summation notation. i1

4

 f 1i2 112  f 112 112  f 122 112  f 132 112  f 142 112

i1

 6  7  8  9  30 units2

To refine this estimate, consider the eight rectangles shown in Figure C2C 10.3. The original four-unit interval is now divided into eight parts, so each part is 48  12 unit wide. The length of each rectangle is still given by f 1x2  x  5, but we now evaluate the function in increments of 12:

Figure C2C 10.3 y 10 9 8 7 6 5 4 3 2 1 1

10–81

2

3

4

1019

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Precalculus—

1020

10–82

Connections to Calculus 8 rectangles

giving

one-half unit wide 8



LW 

 f a 2 iba 2 b 1

1

i1

evaluate f each 1/2 unit

for i  1, 2, 3, . . . , 8

Since the length of each rectangle is multiplied by W  12, we can factor out 12 and 1 1 8 f a ib for convenience. write the expression as 2 i1 2



EXAMPLE 1



Computing a Sum of Rectangular Areas Find the sum

Solution



1 1 8 f a ib by writing out each term. 2 i1 2



1 1 3 5 7 1 1 8 f a ib  c f a b  f 112  f a b  f 122  f a b  f 132  f a b  f 142 d 2 i1 2 2 2 2 2 2 1 11 13 15 17  a 6 7 8  9b 2 2 2 2 2 11 13 7 15 17 9  3   4   29 units2 4 4 2 4 4 2



Now try Exercises 1 and 2



y This approximation is still too large, but closer to the true value of 28 units2. At this point, 10 9 we could get an even better estimate using 16 rec- 8 4 1 tangles, with each width being 16  4 unit wide, 7 and the length still computed by f 1x2  x  5 6 but using increments of 14 . This would yield 5 4 16 1 1 1 16 1 LW  f a iba b  f a ib. As you 3 2 4 4 4 i1 4 i1 16 1 1 1 f a ib by might imagine, finding the sum 5 6 x 1 2 3 4 4 i1 4 writing out each term would be tedious and time consuming. Instead, the properties of summation and the summation formulas can be used to develop an expression where we can find the sum for n  16 directly. The key idea is to note that since 1 1 1 n 1 f a ib  i  5, we can write the summation as a i  5b and work as in 4 4 4 i1 4 Example 2.











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Connections to Calculus

EXAMPLE 2



1021

Computing a Sum of Rectangular Areas 1 n 1 a i  5b, use the summation properties and summation formulas 4 i1 4 to develop an expression where the sum can be found by substituting n  16 directly. For the sum

Solution





n 1 n 1 1 n 1 i a i  5b  a 5b 4 i1 4 4 i1 4 i1 n 1 1 n  a i 5b 4 4 i1 i1 1 1 n2  n b  5n d  c a 4 4 2 1 1 162  16 b  51162 d  c a 4 4 2 1 272 b  20  a 16 2  8.5  20  28.5









summation properties (distribute)



factor

1 from first summation 4

use summation formulas substitute 16 for n

distribute and simplify result

2

The approximate area is 28.5 units . Now try Exercises 3 and 4



The significance of this result cannot be overstated. As we’ll see in Chapter 11, similar calculations can be used if n rectangles are assumed, with the area easily calculated for any value of n. Further, you might imagine the amazement of the early students of mathematics, when they realized that these ideas could still be applied for areas where no elementary formula existed, as for the area under a nonlinear graph.

Connections to Calculus Exercises

1. For f 1x2  x  6 and x  3 0, 4 4 , (a) graph the function, then approximate the area under the graph in this interval using four rectangles of width 1 and length f(x) for x  1, 2, 3, and 4. Begin by developing the summation formula for

LW , then

expand the sum. (b) Repeat part (a) using eight rectangles of width one-half and length f(x). Draw both the graph and the rectangles in each case, and verify the increased number of rectangles gives a better estimate.

2. For f 1x2  x  4 and x  3 0, 6 4, (a) graph the function, then approximate the area under the graph in this interval using six rectangles of width 1 and length f (x) for x  1, 2, 3, 4, 5, and 6. Begin by developing the summation formula for

LW , then expand the sum. (b) Repeat part (a) using 12 rectangles of width one-half and length f(x). Draw both the graph and the rectangles in each case, and verify the increased number of rectangles gives a better estimate.

For Exercises 3 and 4, show all work requested using a methodical step-by-step process, similar to that illustrated in Example 2.

3. For f 1x2  x  6 and x  3 0, 44 from Exercise 1, (a) discuss how the area under the graph of f in this interval can be approximated using the rectangle method and n  32 rectangles without expanding the sum. (b) Find the area for n  32 by applying summation properties and formulas.

4. For f 1x2  x  4 and x  30, 6 4 from Exercise 2, (a) discuss how the area under the graph of f in this interval can be approximated using the rectangle method and n  36 rectangles without expanding the sum. (b) Find the area for n  36 by applying summation properties and formulas.

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PreCalculus—

11 CHAPTER CONNECTIONS

Bridges to Calculus: An Introduction to Limits CHAPTER OUTLINE 11.1 An Introduction to Limits Using Tables and Graphs 1023 11.2 The Properties of Limits 1034 11.3 Continuity and More on Limits 1044 11.4 Applications of Limits: Instantaneous Rates of Change and the Area under a Curve 1056

This chapter introduces the most central and power ful idea in calculus, the concept of a limit. While the study of limits isn’t a recent phenomenon (Archimedes inscribed a polygon within a circle and considered an ever increasing number of sides to estimate the circumference), a complete development of the idea had to wait for the talents of Augustin-Louis Cauchy (c. 1821) and a maturation of the notation needed to work with the concept effectively. Using a rational function to illustrate, essentially the concept involves recognizing the distinction between x 2  2x  3 the value of f 1x2
when x
1 x1 (the function is not defined at 1), and the limit of f (x) as x becomes ver y close to 1. This application appears as Exercise 33 in Section 11.1.

1023

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PreCalculus—

11.1 An Introduction to Limits Using Tables and Graphs Learning Objectives

The concept of a limit is the foundation of all calculus. If the concept is faulty, the conclusions of calculus can be neither supported nor sustained, and the whole becomes a great leap of faith. The inventors of calculus (Isaac Newton and Gottfried Leibniz) tried mightily to bridge the gap between the finite and the infinite, but it was only after a long evolution that the ideas would mature to a point where infinity could be tamed and calculus placed on an unassailable footing.

In Section 11.1 you will learn how to:

A. Distinguish between a limit and an approximation

B. Estimate limits using tables

A. Distinguishing between Limits and Approximations

C. Evaluate one-sided

To begin our study, consider the area of a regular polygon inscribed in a circle of radius r (inscribed: all vertices of the polygon are on the circumference). The radii drawn from the center to each vertex divide the polygons into triangles, enabling us to compute the area of each polygon by summing the area of these triangles (Figure 11.1).

limits

D. Determine when a limit does not exist

Figure 11.1

A3

A4

A5

A6

A8

If we let An represent the area of an inscribed polygon with n sides, we note that as the number of sides increases, the area of the polygon gives a better and better approximation for the area of the circle. In fact, it seems intuitive that as n becomes infinitely large, the area of the polygon becomes infinitely close to the area of the circle. Using the notation seen in previous chapters we would say, as n S q, An S r2. In a study of limits, limit notation is used, and the same statement is written lim An  r2

nSq

In words, “the limit of An as n becomes infinitely large is r2.” For practice using the new notation, see Exercises 7 through 12. In a study of limits it becomes extremely important to accept that no finite number of sides for our polygon will give the exact area of the circle (even with 100 sides we still have only a very good approximation). But that’s not what the concept and notation are saying. They only say that r2 is the limiting value, and that we can become as close as we like to this limit, by choosing n sufficiently large. EXAMPLE 1 WORTHY OF NOTE From the area formula 1 A  ab sin C, note that for 2 each triangle within the circle, the angle at the center C is 2 , and a  b  r. This shows n each triangle has area 2 1 A  r 2 sina b and there 2 n are n of these triangles.

1024



Finding the Area of an Inscribed Polygon If an n-sided regular polygon is inscribed in a circle of nr2 2 sina b. Assume radius r, its area is given by An  n 2 the circle has a radius of r  10 cm. Use a table of values to determine the number of sides needed (to the nearest 100) for the area of the polygon to approximate the area of the circle rounded to two decimal places ( A  314.16 units2 rounded).

x

An

100

313.95

200

314.11

300

314.14

400

314.15

500

314.15

600

314.15

700

314.16

y  314.155046835

11-2

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PreCalculus—

11-3

Section 11.1 An Introduction to Limits Using Tables and Graphs

Solution



A. You’ve just learned how to distinguish between a limit and an approximation

1025

2 b, where n represents the n number of sides. Evaluating the equation using multiples of 100 produces the table shown, and we note that when n  700, the area of the polygon approximates the area of the circle rounded to two decimal places. With r  10, the formula becomes An  50n sina

Now try Exercises 13 through 16



B. Estimating Limits Using Tables and Graphs

The limit in the discussion prior to Example 1 1 lim An 2 is called a limit at infinity, or nSq

a limit as n S q . These are discussed in more detail in Section 11.3. Here we turn our attention to limits at a constant c, or a limit as x S c (using the more common variable x). sin x Consider the function f 1x2  (x in radians), noting that x  0 is not in the x domain of f. However, the fact that we could not evaluate lim An for n  q in nSq

Example 1, did not stop us from evaluating An as n S q . Similarly, we cannot evaluate sin x f 1x2  at x  0, but we can evaluate f for values of x slightly less than zero, or x sin x slightly greater than zero, to investigate the limit of as x S 0. For additional x practice with the limit notation, see Exercises 17 through 22.

EXAMPLE 2



Determining Whether a Limiting Value Exists Use a table of values to evaluate f 1x2 

sin x as x S 0. If the function seems to x approach a limiting value, write the relationship in words and using the limit notation. Solution



For this exercise it seems appropriate to begin a short distance from zero, and approach it in small increments. Noting that 0 can be “approached” from either side, we use values less than zero (from the left-hand side) and values greater than zero (from the right-hand side). Beginning at x  0.5 and approaching 0 from the left produces Table 11.1. Starting at x  0.5 and approaching from the right yields Table 11.2. The tables seems to indicate that as x becomes very close to 0, sin x becomes very close to 1, or seems to have a limiting value of 1. In limit x sin x  1. notation we write lim xS0 x Table 11.1 x

f (x)

Table 11.2 x

f (x)

0.5

0.95885

0.5

0.95885

0.4

0.97355

0.4

0.97355

0.3

0.98507

0.3

0.98507

0.2

0.99335

0.2

0.99335

0.1

0.99833

0.1

0.99833

0.01

0.99998

0.01

0.99998

0.001

1

y  0.999999833333

0.001

1

y  0.999999833333

Now try Exercises 23 through 28



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PreCalculus—

1026

11-4

CHAPTER 11 Bridges to Calculus: An Introduction to Limits

sin x as close x as we like to 1 by using values of x that are sufficiently close to 0. This suggests the following definition of a limit, a definition, however, that will be refined and made more precise as we gain a greater understanding of the limit concept. Similar to our work with the polygon and circle, note that we can make

The Definition of a Limit For a fixed real number L and constant c,

lim f 1x2  L

xSc

means that values of f(x) can be made arbitrarily close (infinitely close) to L, by taking values of x sufficiently close to c, x  c. The graph of (1) f 1x2 

sin x is shown in Figure 11.2. As we did in Section 2.7, we x could create a piecewise function that defines f(x) at x  0, and two possibilities are listed here, followed by their graphs (Figures 11.3 and 11.4, respectively). sin x x0 122 f 1x2  • x 2 x0

sin x 132 f 1x2  • x 1

y

y

y

2.5

x0

Figure 11.4

Figure 11.3

Figure 11.2

x0

2.5

2.5

2

2

2

1.5

1.5

1.5

1

1

1

0.5

0.5

0.5

10 8 6 4 2 0.5

2

4

6

8 10

x

10 8 6 4 2 0.5

1

1

1.5

1.5

2

2

2.5

2.5

2

4

6

8 10

x

10 8 6 4 2 0.5

2

4

6

8 10

x

1 1.5 2 2.5

In (2), f(x) defines the function at (0, 2) but leaves a hole in the graph at (0, 1), while in (3), f(x) is a continuous function as we can now draw the entire graph without sin x 1 lifting our pencil (see Exercises 29 and 30). Note that in all three cases, lim x xS0 remains true, as the definition of a limit is still satisfied in every respect. In other words, the limit of a function as x S c can exist even if (2) the function is defined at c but f 1c2  L sin x , x0 sin x at c as in f 1x2  , as in f 1x2  • x , x 2 x0 sin x , x0 or (3) the function is defined at c and f 1c2  L as in f 1x2  • x . 1 x0

(1) the function is not defined

EXAMPLE 3



Determining Whether a Limiting Value Exists 1 2 2x  x Use a table of values to evaluate g1x2  , as x S 2. If the function seems to x2 approach a limiting value, write the relationship in words and using the limit notation.

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PreCalculus—

11-5

1027

Section 11.1 An Introduction to Limits Using Tables and Graphs 

Solution

Similar to Example 2, it seems appropriate to start a short distance from 2 on the left- and right-hand sides, then approach 2 in small increments. Starting from the left at 1.6 produces Table 11.3. Starting from the right at 2.4 yields Table 11.4.

Figure 11.5

Table 11.4

Table 11.3

y

x

5 4

g(x)

x

g(x)

1.6

0.8

2.4

1.2

1.7

0.85

2.3

1.15

1.8

0.9

22

11

1.9

0.95

2.1

1.05

1.99

0.995

2.01

1.005

⫺4

1.999

0.9995

2.001

1.0005

⫺5

1.9999

0.99995

2.0001

1.00005

3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

g(x) 1

2

3

4

5

x

⫺2 ⫺3

y  0.99995 B. You’ve just learned how to estimate limits using tables

y  1.00005

The tables seem to indicate that as x becomes very close to 2, g(x) becomes very close to 1, or seems to have a limiting value of 1. In limit notation, 1 2 2x  x  1. The graph is shown in Figure 11.5. lim xS2 x  2 Now try Exercises 31 through 36



C. One-Sided Limits In Examples 2 and 3 our discussion focused on what are called two-sided limits, as we approached the limiting value by selecting inputs to the left and to the right of c x2  2x ( x  c). But what of functions like y  12  x, y  and many others, 20 1x  2 that may only be defined to the left or right of c  2? Can these functions approach a limiting value as x approaches c from only one side?

EXAMPLE 4

Solution





Determining Whether a Limit Exists for a One-Sided Approach x2  2x Use a table of values to help determine if f 1x2  20 1x  2 x has a limiting value as x S 2 from the right. With the radical 1x  2 in the denominator, the domain of f is x  12, q 2 . For y  f 1x2 , we input values close to 2 but greater than 2, since 2 can only be “approached” from the right-hand side. The result is the table shown, and seems to indicate that as x becomes very close to 2 from the right, f(x) becomes very close to 0. Further, it appears we can make f(x) as close to zero as we like, by choosing values of x sufficiently close to 2 but greater than 2. This suggests the limit of f (x) as x approaches 2 from the right is 0.

f (x)

2.4

0.07589

2.3

0.06299

2.2

0.04919

2.1

0.0332

2.01

0.01005

2.001

0.00316

2.0001

0.001

y  0.00100005

Now try Exercises 37 through 42



For functions like those in Example 4, we can define a one-sided limit, using the general definition as a template and introducing a modified notation to represent the idea.

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PreCalculus—

1028

11-6

CHAPTER 11 Bridges to Calculus: An Introduction to Limits

Right-Hand Limits For a fixed real number L and constant c,

lim f 1x2  L

xSc

means that values of f(x) can be made arbitrarily close (infinitely close) to L, by taking values of x sufficiently close to c and to the right of c 1x 7 c2 . Left-Hand Limits For a fixed real number L and constant c,

lim f 1x2  L

xSc

means that values of f (x) can be made arbitrarily close (infinitely close) to L, by taking values of x sufficiently close to c and to the left of c 1x 6 c2 . Note how the superscript on c indicates the direction from which x is approaching c. For x S c, x is approaching c from the left, for x S c  , x is approaching c from the right (c can be any real number). For practice with this notation, see Exercises 43 through 46. EXAMPLE 5



Determining Left- and Right-Handed Limits x  2 For h1x2  , use a table to evaluate the expressions x2 a. lim h1x2 b. lim h1x2 xS2

xS2

If a limiting value appears to exist from either side, write the relationship in words and using the limit notation. Solution



Begin by noting the domain of h is x  , x  2. a. For lim h1x2 , we use a table to track the value of h as x approaches 2 from the xS2

left (x S 2). The result is shown in Table 11.5, and suggests lim xS2

In words, the limit of h(x) as x approaches 2 from the left is 1. Figure 11.6

Table 11.5

y

x

5 4 3 2

h(x)

1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1

2

3

4

5

x

⫺2 ⫺3 ⫺4 ⫺5

h(x)

C. You’ve just learned how to evaluate one-sided limits

Table 11.6 x

h(x)

1.6

1

2.4

1

1.7

1

2.3

1

1.8

1

2.2

1

1.9

1

2.1

1

1.99

1

2.01

1

1.999

1

2.001

1

1.9999

1

2.0001

1

y  1

x  2  1. x2

y1

x  2  1, or the limit of h(x) as x xS2 x2 approaches 2 from the right is 1. The graph of h is shown in Figure 11.6 and seems to support this conclusion.

b. Similarly, Table 11.6 suggests lim

Now try Exercises 47 through 54



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Section 11.1 An Introduction to Limits Using Tables and Graphs

1029

D. Limits That Fail to Exist Unlike the function in Example 4 (where x could approach 2 only from the right), the functions in Examples 3 and 5 allowed an approach from both the left and right sides. Note that the statement, “taking values of x sufficiently close to c” from the general definition of a limit implies a two-sided approach to c, and we must approach c from both sides wherever possible. Since this is not possible in Example 4, we use the notation x2  2x
0. For the function g(x) in Example 3, for one-sided limits to state: lim xS2 1x  2 a “two-sided” approach was possible and we saw the approach from each side was equal to 1. Using the general definition we were able to write lim g1x2
1 (the limit xS2

is L
1). For the function h(x) in Example 5, x could also approach c from both sides, but the left-hand limit was not equal to the right, and there is no single number L that h(x) approaches as x S 2. This leads us to the following statement: The Existence of a Limit For a fixed real number L and constant c, lim f 1x2
L if and only if

xSc

lim f 1x2
lim f 1x2
L.

xSc 

xSc

In words, “the limit of f(x) as x approaches c is L, if and only if the left-hand limit as x approaches c is equal to the right-hand limit as x approaches c, and both are also equal to L.”


EXAMPLE 6

Determining Whether a Limit Exists x2  3x

Use a table of values to evaluate the expression lim

. 2x2  6x  9 As a matter of choice we first evaluate the left-hand limit, and Table 11.7 suggests xS3




Solution

lim

xS3

lim

xS3 

x2  3x 2x2  6x  9 x2  3x 2x2  6x  9


3. For the right-hand limit, Table 11.8 suggests
3. Table 11.7

Table 11.8

x

x

y

y

2.5

2.5

3.4

3.4

2.6

2.6

3.3

3.3

2.7

2.7

3.2

3.2

2.8

2.8

3.1

3.1

2.9

2.9

3.01

3.01

7

2.99

2.99

3.001

3.001

6

2.999

2.999

3.0001

3.0001

Figure 11.7 y

5

y
3.0001

y
2.999

4 3 2

Since the left-hand limit  right-hand limit, lim

1 3 2 1 1 2 3

xS3

1

2

3

4

5

6

7

x

The graph of y


x2  3x 2x2  6x  9

x2  3x 2x2  6x  9

does not exist.

is shown in Figure 11.7. Now try Exercises 55 through 60




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There are actually other ways that a limit can fail to exist, and as a matter of order we will adopt the following notation for the case where the left-hand limit is not equal to the right-hand limit. Using dne as an abbreviation for “does not exist,” we will write lim f 1x2  A dne B . In Example 7 we note a second way that a limit can fail to exist. xSc

EXAMPLE 7



LHRH

Determining Whether a Limit Exists Given f 1x2 

Solution



x

9x , evaluate lim f 1x2 . xS2 1x  22 2

For the left-hand limit, we obtain the table shown and note as x S 2  , the value of f (x) increases without bound [for x  1.999, f(x) is almost 18,000,000]. A similar check shows that as x S 2  , the values of f(x) likewise become infinitely large. Since f(x) becomes infinitely large and positive as x S 2, the limit does not exist and we write lim f 1x2  adneb . xS2

f (x)

1.5

54

1.6

90

1.7

170

1.8

405

1.9

1710

1.99

179100

1.999

1.8E7

y  17991000

q

Now try Exercises 61 through 66



From our work with rational functions in Section 3.5, we know the graph of 9x f 1x2  has a vertical asymptote at x  2, verifying it is impossible for the 1x  22 2 function to have a limit as x S 2. Finally, we consider a third way that a limit can fail to exist, as when a function does not approach a fixed real number L, but does not grow infinitely large and positive, nor infinitely large and negative. EXAMPLE 8



Determining Whether a Limit Exists 1 For h1x2  2 ` sina b ` , try to find a fixed number L such that lim f 1x2  L. x xS0

Solution



WORTHY OF NOTE For students and instructors with a sense of humor, Example 8 illustrates what’s sometimes called the Christmas Case, because there’s No-el.

Using an approach from the right gives the result shown in the table. This time there is no apparent pattern in the output values, and a graph of the function (Figure 11.8) reveals why. As x S 0, the function begins to oscillate wildly between 0 and 2, with no approach to any fixed number L. A similar thing happens for a left-handed approach. As per our definition, the limit does not exist because L does not exist. In this case, we write lim h1x2  A dne B . xS2

L

Figure 11.8 y 4 3

h(x)

2 1

D. You’ve just learned how to determine when a limit does not exist

⫺0.4

⫺0.2

0.2 ⫺1

0.4

x

x

h(x)

0.4

1.1969

0.3

0.38114

0.2

1.9178

0.1

1.088

0.01

1.0127

0.001

1.6538

1E-4

0.61123

x  0.0001

Now try Exercises 67 through 70



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Section 11.1 An Introduction to Limits Using Tables and Graphs

1031

11.1 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

5. Discuss/Explain why lim g1x2  5 for g1x2 

1. The limit of f(x) as x S q , is called a limit at .

xS3

2. lim g1x2  L means that values of g(x) can be made xSc

to L, by taking values of sufficiently close to .

x2  x  6 , but g132  5. Can we redefine g(x) x3 to create a piecewise-defined function where lim g1x2  g132 ?

xS3

3. To evaluate the limit of a function as x S c, we must find the limit using values of x less than c, and the limit using values of x than c. 4. If a rational function v(x) has a vertical asymptote at x  c, the limit as x S c does not exist and we write lim v1x2  .

6. Discuss/Explain why lim f 1x2 does not exist for f 1x2 

x2  x

xS1

, even though the left-hand limit 21x  12 2 and the right-hand limit do exist.

xSc



DEVELOPING YOUR SKILLS determine the number of sides needed (to the nearest 100) for the area of the polygon to approximate the area of the circle correct to two decimal places when rounded. Compare your result with that of Example 1.

Write each of the following statements using limit notation.

4 7. As n approaches q , Vn approaches r3. 3 8. As n approaches q , An approaches a 3 b 2 x  x  cx  d . 3 2 9. As t approaches q , e f(t) approaches 0. 10. As t approaches q , sin[g(t)] approaches 1. 1 11. As x increases without bound, cosa b approaches 1. x 12. As x decreases without bound, approaches 52 .

5x2  3 2x2  x  1

Use a table of values for the following exercises.

13. If an n-sided regular polygon circumscribes a circle of radius r (see the figure), its area is given by  An  nr2tana b. For a circle n with radius r  10 cm,

A8 r

14. If an n-sided regular polygon is inscribed in a circle of radius r, its perimeter is given by  P  2nr sina b. For a circle with radius n r  50 mm, determine the number of sides needed (to the nearest 25) for the perimeter of the polygon to approximate the circumference of the circle correct to two decimal places when rounded. 15. If an n-sided regular polygon circumscribes a circle of radius r, its perimeter is given by  P  2nr tana b. For a circle with radius n r  50 mm, determine the number of sides needed (to the nearest 25) for the perimeter of the polygon to approximate the circumference of the circle correct to two decimal places when rounded. Compare your results with those in Exercise 13.

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16. One of the most famous and useful numbers in all of 1 x mathematics is the one defined as lim a1  b
e. x xSq Determine the smallest x to (the nearest 250) that can be used to approximate the value of e to three decimal places. Write each of the following statements using limit notation.

17. As t approaches 5, st approaches 5r. 18. As t approaches 3, dt approaches 29  r2.

19. As x approaches a, tan1 3 g1x2 4 approaches
3 . 20. As x approaches b, csc2[ f(x)] approaches 43. x3 21. As x S 3, 2 approaches 16. x 9 22. As x S
,

sin x approaches 2. x cosa b 2

Use a table of values to evaluate each function as x approaches the value indicated. If the function seems to approach a limiting value, write the relationship in words and using the limit notation.

23. p1x2
cos x  sina

3x b; x S
2

24. q1x2
tan x  3 sin 2x; x S


4


cosa b x 25. v1x2
;xS2 sin1
x2 26. w1x2


tan1x  22 ;xS2 x2

27. s1x2


2 cos x  2 ;xS0 x

28. t1x2


sin x  1
;xS sin12x2 2

The functions in Exercises 29 and 30 have a hole (discontinuity) in their graphs at x
2. Write a related piecewise-defined function that creates a continuous graph for each.

29. r1x2


2x2  7x  6 sin1x  22

30. s1x2


sin1x  4x  42 x2 2

Use a table of values to evaluate each function as x approaches the value indicated. If the function seems to approach a limiting value, write the relationship in words and using the limit notation.

31. f 1x2


x2  2x ;xS2 x2  4

32. v1x2


x2  3x  10 ; x S 2 2x  4

33. g1x2


x2  2x  3 ;xS1 x1

34. w1x2


x1x  22  8 ;xS4 41x  42

35. f 1x2
3x2  x  2; x S 1 36. g1x2
10x  x2; x S 2

See if a table of values suggests a limit exists for the functions and approaches indicated.

37. f 1x2
2x2  3x as x S 3 from the right. 1 38. g1x2
6x2  x as x S  from the right. 2 39. f 1x2


x3  8 as x S 2 from the left. 1 2x  1

40. g1x2


x4  1 as x S 1 from the left. x1

41. f 1x2


sin 1x as x S 0 from the right. 1x

42. g1x2


x as x S 0 from the right. ln x

Write each statement using limit notation.

43. As x approaches 3 from the left, Ix approaches 3 cos2 1R1  R2 2 .

44. As x approaches 1 from the right, Sx approaches log1  12 . 45. As x approaches m from the left, f approaches L. 46. As x approaches n from the right, g approaches L. Evaluate the following limits using a table of values.

47. Given f 1x2


sin x , find sin x a. lim f 1x2 b. lim f 1x2 xS


xS


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Section 11.1 An Introduction to Limits Using Tables and Graphs

48. Given g1x2   a.

sin 31x  12 4 , find x a. lim g1x2 b. lim g1x2

cos x , find cos x

lim g1x2

58. Given g1x2 

g1x2 b. lim 

xS 2 

xS 2

x  10x  24

xS0

2

49. Given f 1x2 

c. lim g1x2

, find

xS0

2x  12x  36 a. lim f 1x2 b. lim f 1x2 2

xS6

sin x

xS6

x  2x  3

59. Given f 1x2  μ tan x

2

50. Given g1x2  a.

, find

2x2  6x  9

lim  g1x2

b.

xS3

cos x

lim  g1x2

a. lim f 1x2

xS3

1 2 51. Given f 1x2   3x 2 , find x a. lim f 1x2 b. lim f 1x2

xS4

xS0

xS 4



c. lim f 1x2 xS 4

xSln 7.3

lim

xSln 7.3

60. Given g1x2  μ cot x

g1x2

x2  4 x 6 2 , find 2 sin1x  42 x  2 a. lim f 1x2 b. lim f 1x2

53. For f 1x2  e xS2

 3 tan c 1x  22 d 4 54. For g1x2  • 2x2  8

x  1 x 7 1

xS1

2x2  7 x 6 5 55. Given f 1x2  e , find 3  2x x  5 a. lim  f 1x2 b. lim  f 1x2 xS5

xS5

c. lim f 1x2

1x  12 x  10 , find log x x 7 10 lim  g1x2 b. lim  g1x2

56. Given g1x2  e xS10

xS10

c. lim g1x2 57. Given f 1x2 

a. lim f 1x2 xS

c. lim f 1x2 xS

b. lim g1x2 5

4

4

, find



xS



c. lim5 g1x2 xS

4

61. Given f 1x2  x2 

3 , find lim f 1x2 . x2 xS2

62. Given g1x2  2x2  3x  4x2, find lim g1x2 . xS0

 4 cosa xb 4 63. Given f 1x2  , find lim f 1x2 . x2 xS2 64. Given g1x2  csc1x2  17x  522 , find lim g1x2 .

xS5

xS10

g1x2 a. lim 5 xS

lim  g1x2

b.

xS1

a.

csc x

xS2

lim  g1x2

5 4 5 , find x 4 5 x 7 4

sec x x 6

x

a.

xS 4

xS4

52. Given g1x2  e  5, find lim  g1x2 a. b.

 4  x  , find 4  x 7 4 b. lim f 1x2  x 6

21  cos 12x2 12 sin x

, find

b. lim f 1x2 xS

xS13

2x 65. Given f 1x2  3 , find lim f 1x2 . xS6 x  216 66. Given g1x2 

6x2  x  1 , find lim1 g1x2 . 2x  1 xS 2

67. Given f 1x2  sin a

x1 b, find lim f 1x2 . x1 xS1

68. Given g1x2  cos a

x1 b, find lim g1x2 . x1 xS1

69. Given f 1x2  tan1cos x2 , find lim f 1x2 . xS 2

70. Given f 1x2  cos1tan x2 , find lim f 1x2 . xS 2

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11-121034

MAINTAINING YOUR SKILLS 71. (3.3) Use the rational zeroes theorem to write the polynomial in completely factored form: 3x4  19x3  15x2  27x  10.

73. (7.2) Use Heron’s formula to find the area of a triangle with sides a
5 in., b
8 in., c
9 in., rounded to two decimal places.

72. (1.4) Find the sum, difference, product, and quotient of 1  3i and 1  3i.

74. (5.7) Write a sinusoidal equation given the maximum is 100, the minimum is 20 and the period is 365.

11.2 The Properties of Limits Learning Objectives In Section 11.2 you will learn how to:

A. Distinguish between a declarative statement and a proof

B. Establish limit properties using a table

After the invention of calculus, but before the concept of a limit had been fully developed, the ideas of Newton (the method of fluxions) and Leibniz (the method of differentials) were often met with skepticism and sarcasm even within the academic world. In a tract called The Analyst (1794), mathematician and philosopher George Berkeley once said, “he who can digest a second or third fluxion . . . need not, methinks, be squeamish about any point in divinity.” In this section, we introduce the properties of limits, which helped to place the study of calculus on a sure footing. The plausibility of these properties will be illustrated using a table or graph, with their formal proof left for a future course.

C. Evaluate limits using the limit properties

A. Distinguishing between a Declarative Statement and a Proof Consider the function p1x2
x2  x  41. Evaluating p for the natural numbers 1 through 7 reveals that all outputs are prime numbers (Table 11.9). Made curious by this discovery, we investigate further and find that values of x from 8 through 14 also generate prime numbers (Table 11.10), as do the inputs 15 through 40. At this point it might seem reasonable to declare that p generates a prime number for all natural numbers n. This exercise points out the dangers of making a declarative statement without formal proof, because when x
41, the function returns a value of 1681, which is not a prime number (41 # 41
1681). Table 11.9 x

Table 11.10

p(x)

x

p(x)

1

41

8

97

2

43

9

113

3

47

10

131

4

53

11

151

5

61

12

173

6

71

13

197

83

14

7 x
7

223 x
14

Also, in Section 10.4 we noted 1  3  5  7
16, 1  3  5  7  9  11
36, 1  3  5  7  9  11  13  15  17  19  21  23
144 and it appeared the sum of the first n odd numbers was equal to n2. While this declaration turns out to be true, no number of finite checks can prove the statement is true for all n, and a formal proof had to be rendered using mathematical induction. In x2  9
6, using exactly the same way, we can view a table of values to “declare” lim xS3 x  3 values closer and closer to 3, then even closer still, but how do we prove the limit is 6?

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Section 11.2 The Properties of Limits

1035

It was this type of question that prompted additional sarcasm from George Berkeley. “And what are these fluxions? The velocities of evanescent increments? And what are these same increments? They are neither finite quantities, nor quantities infinitely small, nor yet nothing. May we not then call them the ghosts of departed quantities?” Example 1 demonstrates that in the absence of a formal proof, it appears Mr. Berkeley’s skepticism was justified. EXAMPLE 1




Evaluating Limits Using a Table Evaluate lim ax2  xS0

Solution




cos x  1b using a table of values. 1000

For the left- and right-hand limits we obtain Tables 11.11 and 11.12 respectively, and it appears we can declare that lim f 1x2
1. xS0

Table 11.11

Table 11.12

x

f(x)

x

0.6

1.361

0.6

1.361

0.5

1.251

0.5

1.251

0.4

1.161

0.4

1.161

0.3

1.091

0.3

1.091

0.2

1.041

0.2

1.041

0.1

1.011

0.1

1.011

0.01

1.0011

0.01

1.0011

y
1.00109999998 A. You’ve just learned how to distinguish between a declarative statement and a proof

f(x)

y
1.00109999998

However, if we continue the approach using values of x that are even closer to c
0, we find f 10.0012
1.001001, f 10.00012
1.00100001, and it turns out the limit is actually 1.001. Our declarative statement then has no standing and our work falls far short of a proof. Now try Exercises 7 through 14




B. Establishing the Limit Properties From Example 1 and our work in Section 11.1, it seems we need a more definitive method for evaluating limits. As it turns out, limits possess certain properties that, once established, enable us to evaluate even complex limits with complete confidence. The properties we’re interested in involve the various ways that limits can be combined using basic operations. For instance, from our study of operations on functions, we know 1 f  g21x2
f 1x2  g1x2 . Is it similarly true that lim 3 f 1x2  g1x2 4
lim f 1x2  lim g1x2 ? While a formal proof of the properties that xSc

xSc

xSc

follow will be offered in a first calculus course, for the time being we will accept them as true after demonstrating their plausibility using tables and our intuition (despite their shortcomings). EXAMPLE 2




Combining Limits to Establish Limit Properties Using a table of values and the techniques from Section 11.1, it can be shown that for f 1x2
x and g1x2
x2  7, lim f 1x2
3 and lim g1x2
2. Use these results to create a new function h1x2
1 f  g21x2 and investigate whether lim 3 f 1x2  g1x2 4
lim f 1x2  lim g1x2 . xS3

xSc

xSc

xSc

xS3

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CHAPTER 11 Bridges to Calculus: An Introduction to Limits

Solution



For f and g as given, we have h1x2  x  1x2  72 . This means we need to investigate lim h1x2  lim 3x  1x2  72 4 . The results are shown in Tables xS3

xS3

11.13 and 11.14 and suggest lim h1x2  1. xS3

Table 11.13

Table 11.14

x

x

h(x)

h(x)

2.6

2.84

3.4

1.16

2.7

2.41

3.3

0.59

2.8

1.96

3.2

0.04

2.9

1.49

3.1

0.49

2.99

1.0499

3.01

0.9499

2.999

1.005

3.001

0.995

2.9999

1.0005

3.0001

0.9995

y  1.00049999

y  0.99949999

It appears that such a property exists, as the tables suggest lim 3 f 1x2  g1x2 4  lim f 1x2  lim g1x2

xS3

xS3

1  3  122 11✓

xS3

Now try Exercises 15 through 34



Other investigations might involve whether limits have a “multiplicative” property. lim g1x2 g1x2 xSc In other words, is lim 3 f 1x2g1x2 4  lim f 1x2 lim g1x2 ? Is lim a true  xSc xSc xSc xSc f 1x2 lim f 1x2 xSc statement? We explore these questions in Example 3.

EXAMPLE 3



Combining Limits to Establish Limit Properties Using a table of values and the techniques from Section 11.1 suggests that for f 1x2  x2 and g1x2  3x, lim f 1x2  4 and lim g1x2  6. Note this yields xS2

lim f 1x2

xS2

xS2

lim g1x2

#

lim g1x2  4162  24 and

xS2

xS2

lim f 1x2



6 3  . Use this information to 4 2

xS2

a. Create a new function m1x2  f 1x2g1x2 to investigate whether lim 3 f 1x2g1x2 4  lim f 1x2 lim g1x2 . xS2

xS2

xS2

g1x2 b. Create a new function v1x2  to investigate f 1x2 lim g1x2 g1x2 xS2 whether lim .  xS2 f 1x2 lim f 1x2 xS2

Solution



a. For the product of f and g we have m1x2  1x2 2 13x2  3x3. Now using a single table to view the approach from the left and from the right (indicated by the arrows), Table 11.15 suggests lim m1x2  lim 13x3 2  24. xS2

xS2

Table 11.15 x

m(x)

1.9

20.577

1.99

23.642

1.999

23.964

2 2.001

24.036

2.01

24.362

2.1

27.783

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WORTHY OF NOTE While the limits illustrated in Tables 11.15 and 11.16 used functions that were defined at x
2, remember that the limit does not depend on the functions being defined there.

1037

Section 11.2 The Properties of Limits

b. For the quotient of g and f we have 3x 3 v1x2
2
 1x  02 . Table 11.16 indicates x x 3 3
. The tables once again seem lim v1x2
lim xS2 xS2 x 2 to indicate that such properties exist, as they suggest lim 3 f 1x2g1x2 4
lim f 1x2 # lim g1x2
24, and xS2

xS2

xS2

lim g1x2

g1x2 3 xS2 lim

. xS2 f 1x2 lim f 1x2 2

Table 11.16 x

v(x)

1.9

1.5789

1.99

1.5075

1.999

1.5008

2 2.001

1.4993

2.01

1.4925

2.1

1.4286

xS2

Now try Exercises 35 through 44 B. You’ve just learned how to establish limit properties using a table




The preceding observations seem to imply that the limit of a product can be found by taking the limit of each factor, and the limit of a quotient is equal to the quotient of the limits (if the limit of the denominator is not 0). In fact, each of these observations (and others) can be generalized and formally established using the definition of a limit.

C. Finding Limits Using the Limit Properties Building on our previous work, we might expect that limit properties exist for the other basic operations, and this is indeed the case: Properties of Limits Given that the limits lim f 1x2 and lim g1x2 exist, xSc

xSc

(I) lim 3 f 1x2  g1x2 4
lim f 1x2  lim g1x2 xSc

xSc

xSc

the limit of a sum is the sum of the limits (II) lim 3 f 1x2  g1x2 4
lim f 1x2  lim g1x2 xSc

xSc

xSc

the limit of a difference is the difference of the limits (III) lim 3 f 1x2g1x2 4
lim f 1x2 lim g1x2 xSc

xSc

xSc

the limit of a product is the product of the limits (IV) lim 3kg1x2 4
k lim g1x2 , k a constant xSc

xSc

the limit of a constant times a function is the constant times the limit of the function lim f 1x2 f 1x2 xSc
(V) lim , provided lim g1x2  0 xSc g1x2 lim g1x2 xSc xSc

WORTHY OF NOTE A formal proof of these properties, requiring a more sophisticated approach to limits, is offered in a calculus course.

the limit of a quotient is the quotient of the limits By repeatedly applying property III with f 1x2
g1x2 , we obtain the property related to the limit of a power. (VI) lim 3 f 1x2 4 n
3 lim f 1x2 4 n, for n a natural number xSc

xSc

the limit of a power is the power of the limit A similar property holds for the limit of an nth root.

(VII) lim 2f 1x2
2 lim f 1x2 , for n

, n 7 1 [if n is even, then f 1x2 7 0] n

xSc

n

xSc

the limit of an nth root is the nth root of the limit

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CHAPTER 11 Bridges to Calculus: An Introduction to Limits

Before we begin applying these properties to evaluate limits, we will accept the validity of the following basic limits, most of which are supported by the previous examples. These are used extensively in our application of the limit properties. Basic Limits 1. lim k  k

2. lim x  c

3. lim xn  cn, n a natural number

4. lim 1x  1c, n a natural number,

xSc

xSc

n

xSc

n

xSc

n 7 1 (if n is even, then c 7 0)

EXAMPLE 4



Using Limit Properties to Determine a Limit Find the following limits using the limit properties. State the property that justifies each step. x4  3x2  5 a. lim 1x3  3x  42 b. lim xS3 xS2 7  4x

Solution



a. lim 1x3  3x  42  lim x3  lim 3x  lim 4 xS3

xS3

xS3

 lim x3  3 lim x  lim 4 xS3 3

xS3

 132  3132  4

xS2

x4  3x2  5  7  4x 

basic limits 3, 2, and 1 result

lim 1x  3x  52 4

2

xS2

lim 17  4x2

property V

xS2 4

lim x  lim 3x2  lim 5

xS2

xS2

xS2

lim 7  lim 4x

xS2



property IV

xS3

 40 b. lim

properties I and II

xS3

properties I and II

xS2

lim x4  3 lim x2  lim 5

xS2

xS2

xS2

lim 7  4 lim x

xS2

2  312 2  5 7  4122 23   23 1 4

property IV

xS2

2



basic limits 3, 2, and 1

result

Note the general strategy is to “break-up” the expressions using limit properties, then use the basic limits to evaluate the result. Now try Exercises 45 through 56

EXAMPLE 5



Using the Limit Properties to Determine a Limit Evaluate the following limits using the limit properties. 9 2x2  4 3 a. lim 2x3  7x  28 b. lim xS4 xS0 x2  5



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Section 11.2 The Properties of Limits

Solution




3 3 3 a. lim 2 x  7x  28
2 lim 1x3  7x  282

xS4

property VII

xS4

3
2 lim x3  lim 7x  lim 28

properties I and II

3
2 lim x3  7 lim x  lim 28

property IV

3
2 142 3  7142  28

basic limits 3, 2, and 1

xS4

xS4

xS4

xS4

xS4

xS4


28
2 3

9 2x2  4
b. lim xS0 x2  5


1039

result

lim 19 2x2  42

xS0

lim 1x2  52

property V

xS0

9 lim 2x2  4 xS0

lim 1x2  52

property IV

xS0




92 lim 1x2  42 xS0 2

lim 1x  52

property VII

xS0




9 2 lim x2  lim 4 xS0

xS0

properties I and II

lim x2  lim 5

xS0

xS0

9 202  4 02  5 9122 18

5 5




basic limits 3, 2, and 1

result

Now try Exercises 57 through 64

EXAMPLE 6




Using Limit Properties to Determine Limits Find the following limits using the limit properties. If a limit does not exist, state why. x2  2x x2 lim a. lim 2 b. xS2 xS1 x  2x  1 2x2  4x  4 lim x2 x2 xS1
a. lim 2 xS1 x  2x  1 lim 1x2  2x  12

property V

xS1

lim lim x2


xS1

lim x2  lim 2x  lim 1

xS1




xS1

112

xS1

2

112  2112  1 1
0 2

property I

basic limits 3, 2, and 1

result

In this case, applying the limit properties results in an undefined expression, and we resort to a table of values to see what’s happening as x S 1.




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PreCalculus—

1040

11-18

CHAPTER 11 Bridges to Calculus: An Introduction to Limits

Table 11.17 x

Table 11.18

y

x

y

1.5

9

0.5

1

1.4

12.25

0.6

2.25

1.3

18.778

0.7

5.4444

1.2

36

0.8

16

1.1

121

0.9

81

1.01

10201

0.99

9801

1.001

1E6

0.999

998001

y  1002001

y  998001

Tables 11.17 and 11.18 show that as x S 1, function values grow infinitely x2 large and positive, and we conclude lim 2  A dne B . q xS1 x  2x  1 b. lim xS2

x2  2x 2x2  4x  4



lim 1x2  2x2

xS2

property V

lim 2x2  4x  4

xS2



lim lim x2  lim 2x

xS2xS2

xS2

2 lim 1x2  4x  42

properties I and VII

xS2

WORTHY OF NOTE Recall that the equation 1  x (with an undefined 0 expression) has no solution, 0 while the equation  x 0 (with an indeterminate expression) has infinitely many solutions—a unique solution cannot be “determined.”



lim lim x2  lim 2x

xS2xS2

2 lim x  lim 4x  lim 4 xS2



xS2 2

lim x B A xS2

property I

xS2

 2 lim x xS2

2 A lim x B  4 lim x  lim 4 2

xS2 2



xS2

2

xS2

properties VI and IV

xS2

122  2122



0

basic limits 3, 2, and 1

2122 2  4122  4 20 0  result 0 Here, the limit properties result in an indeterminate form, and we again resort to a table of values to check the behavior of the function as x S 2. Table 11.19

Table 11.20

x

y

x

2.5

2.5

1.5

1.5

2.4

2.4

1.6

1.6

2.3

2.3

1.7

1.7

2.2

2.2

1.8

1.8

2.1

2.1

1.9

1.9

2.01

2.01

1.99

1.99

2.001

2.001

1.999

1.999

y  2.001

y

y  1.999

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PreCalculus—

11-19

Section 11.2 The Properties of Limits

Tables 11.19 and 11.20 imply that lim  xS2

x2  2x

x2  2x

1041

 2 (the left-hand

2x2  4x  4

 2 (the right-hand limit is 2). 2x2  4x  4 Since these do not agree, the limit does not exist and we write x2  2x lim  A dne B . LHRH xS2 1x2  4x  4 limit is 2), while lim  xS2

C. You’ve just learned how to evaluate limits using the limit properties

Now try Exercises 65 through 76

11.2 EXERCISES 

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

1. The limit of a sum is the

of the

2. The limit of a product is the .

of the

3. The limit of an nth is the n n root of the limit: lim 2f 1x2  2 lim f 1x2 , xSc xSc provided if n is even.

.

4. The limit of a

is the quotient lim f 1x2 f 1x2 xSc , lim ,  xSc g1x2 lim g1x2

of the

xSc

provided

.

5. Discuss/Explain how applying the limit properties can result in an undefined or indeterminate expression. What should be done in these cases? 6. Discuss/Explain the relationship between limit properties VI and VII, and basic limits 3 and 4. How is basic limit 2 involved here?



DEVELOPING YOUR SKILLS

For Exercises 7 and 8, (a) complete the table given for each function and state the apparent limit as x S 0, then (b) continue the approach using values even closer to 0 and comment. 2

1000x3  1 8. f 1x2  1000

sin x 7. f 1x2  x  1000x 2

x

f (x)

x

f (x)

x

f (x)

x

0.5

0.5

0.5

0.5

0.4

0.4

0.4

0.4

0.3

0.3

0.3

0.3

0.2

0.2

0.2

0.2

0.1

0.1

0.1

0.1

0.01

0.01

0.01

0.01

0.001

0.001

0.001

0.001

f (x)



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CHAPTER 11 Bridges to Calculus: An Introduction to Limits

9. Complete the table shown (also see Exercise 11). yx

x

1x  2 10 x

2.7 2.8 2.9 3 3.1 3.2 3.3

10. Complete the table shown (also see Exercise 12).  2x 

x

x3 401x  32 2

1.7 1.8 1.9 2 2.1

13. a. Based on the table in Exercise 9, estimate 1x  2 b. lim ax  xS3 10x b. Based on the table in Exercise 11, what appears 1x  2 b? to be the actual value of lim ax  xS3 10x 14. a. Based on the table in Exercise 10, estimate x3 d. lim c 2x  xS2 401x  32 2 b. Based on the table in Exercise 12, what appears to x3 d? be the actual value of lim c 2x  xS2 401x  32 2 The following functions will be used for Exercises 15 through 44. a1x2
3  2x c1x2
1  x2

b1x2

x  2
d1x2
5x

f 1x2
1x  7

g1x2
cos a

2.2

h1x2
sin1x  32

2.3

k1x2

x2  16


11. Complete the table shown (also see Exercise 9). x

yx

1x  2 10 x

2.99 2.999


b 2x j1x2
log 3 15x2 2 4

l1x2
2x2  10x  25

For Exercises 15 through 24, use a table of values to evaluate each limit.

15. lim a1x2

16. lim b1x2

xS3

xS2

17. lim c1x2

18. lim d1x2

xS3

xS2

19. lim f 1x2

20. lim g1x2

3

xS3

xS2

3.0001

21. lim h1x2

22. lim j1x2

xS3

xS2

23. lim k1x2

24. lim l1x2

xS3

xS2

2.9999

3.001 3.01

12. Complete the table shown (also see Exercise 10). x 1.999 1.9999 1.99999 2 2.00001 2.0001 2.001

y  2x 

x3 401x  32 2

For Exercises 25 through 34, use a table of values to evaluate each limit. Compare each result with that of the corresponding Exercises in 15 through 24.

25. lim 3a1x2  c1x2 4 xS3

26. lim 3d1x2  g1x2 4

27. lim 3 f 1x2  h1x2  k1x2 4

xS2

xS3

28. lim 3b1x2  d1x2  j1x2 4 xS2

29. lim 3h1x2  c1x2 4 xS3

30. lim 3b1x2  g1x2 4

31. lim 3 f 1x2  k1x2  a1x2 4

xS2

xS3

32. lim 3 j1x2  d1x2  l1x2 4 xS2

33. lim 3k1x2  k1x2 4 xS3

34. lim 3d1x2  d1x2 4 xS2

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Section 11.2 The Properties of Limits

For Exercises 35 through 44, use a table of values to evaluate each limit. Compare each result with that of the corresponding Exercises in 15 through 24.

35. lim 3 c1x2 f 1x2 4

36. lim 3 j1x2d1x2 4

k1x2 37. lim xS3 f 1x2

d1x2 38. lim xS2 g1x2

xS3

xS2

xS3

39. lim 3 a1x2h1x2k1x2 4 41. lim 3 c1x2c1x2 4 xS3

a1x2c1x2 43. lim xS3 f 1x2

xS2

40. lim 3b1x2g1x2l1x2 4 42. lim 3d1x2d1x2 4 xS2

d1x2 44. lim xS2 g1x2j1x2

For Exercises 45 through 62, evaluate the limits using the limit properties.

45. lim 1x3  52 xS4

47. lim 12x2  x  72 xS4

49. lim 1x  51x  32

46. lim 1 1x  162

55. lim 1x2  32 3

56. lim 13x2  2x2 2

3 57. lim 5 22x  7

58.

59. lim 11x  7  7x2

60. lim 12x  14  x2

xS3

xS10

xS3

61. lim xS2

xS2

x3  2x  10 3 22 5x2  2x  3

2  3x x1 63. lim xS3 1x  2  1

65. lim xS2

67. lim

51. lim xS2

2x2  5 x3

1  x2 x1 53. lim xS1 1  2x2  x x


xS5

11  3x2

62. lim

2x2  3x  1

xS2

a

64. lim xS1

xS7

3x2  11x  4 x2  2x  8

66. lim xS1

68. lim

x2  5x  6 x2  4x  4

70. lim

69. lim

1 3 50. lim a x2  1 x  3b xS8 2

3 lim 2 2 x6

xS2

x2  5x  14 x7

2

xS4

xS2

2 x b  3x 2x  1

2x2  3  1

For Exercises 65 through 76, evaluate the limits using limit properties. If a limit does not exist, state why.

xS16

48. lim 1x4  2x  52

1043

xS2

2x2  x  3 x2  1

xS4

xS1

x2  7x  12 2x  8

x2  4x  3 x2  2x  1

2x  14x  49 2x2  x  3 72. lim xS7 xS1 2x2  2x  1 x2  8x  7 1x  32 2  9 1x  12 3  1 73. lim 74. lim x x xS0 xS0 2

52. lim xS3

x2  2x  3 x3

1  3x2 x2 54. lim xS3 2 1 x 6 x1

71. lim

75. lim xS1

x3  x2 2x2  2x  1

x3
x  1
xS1 x  1

76. lim

MAINTAINING YOUR SKILLS

77. (2.5) If p1x2
2x2  x  3, in what intervals is p1x2  0?

79. (6.2) Verify the identity: sec y  cos y
tan y sin y

78. (5.4) Solve the logarithmic equation log1x  22  log x
log 3.

80. (6.7) Use any appropriate method to solve in [0, 2
): sin 2x
cos x.

MID-CHAPTER CHECK 1. Express the following statement using limit notation: 6x2  3 As x decreases without bound, 12x3  x  1 approaches 0.

Use a table of values to complete Exercises 2 through 7. 2. Evaluate each limit. sin h cos h  1 a. lim b. lim hS0 h hS0 h

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CHAPTER 11 Bridges to Calculus: An Introduction to Limits

3. The formula for interest compounded n times per r nt year is A
Pa1  b , where P represents the n principal deposit, r is the annual interest rate, t is the number of years the funds are on deposit, and n is the number of compoundings per year. If we let n S q , this formula can be transformed into the formula for interest compounded continuously: A
Pert. Verify that $1000 invested at 3% compounded continuously for 1 yr grows to $1030.45. Then find the minimum number of compoundings per year needed to yield the same amount with the same $1000 principal.

6. Given h1x2
cota

7. a. Complete the table of
values shown. x y
cos a b x b. Based on your table, it 0.1 appears that 0.01
lim cosa b 0.001 x xS0 is approaching what 0 value? Graph 0.001
0.01 y
cosa b x 0.1 on a graphing calculator, and “Zoom In” a few times. Is your prior observation still valid?

4. The function f(x) shown has a pointwise discontinuity. Create a related piecewise-defined function F(x) that is continuous for x
. 6x2  19x  7 f 1x2
2x  7 x x 6 5 5. Given g1x2
• 2x  5 , find the cos1x  52 x  5 following: a. lim  g1x2 b. lim  g1x2 xS5

xS5

x2  4 b, find lim h1x2 . x2 xS2

For Exercises 8 through 10, evaluate each limit using the limit properties. If a limit does not exist, state why. 8. lim xS2

9. lim

4x2  x  5 x5  3x 2x  10

xS3

10. lim

c. lim g1x2

xS2

3 22 11  x2

2x2  4x  4 x2

xS5

11.3 Continuity and More on Limits Learning Objectives In Section 11.3 you will learn how to:

A. Determine if a function is continuous at c and find limits by direct substitution

Historically, the careful development of the limit concept also impacted the concept of continuity and the two seem to have matured simultaneously. In this section, we’ll see how limits help define the continuity of a function more precisely, and once established, how continuity makes finding certain limits an easier task. We’ll also see how the algebra skills developed in previous courses (adding rational expressions, simplifying complex fractions, multiplying by conjugates, etc.) are used to help determine some very important limits.

B. Evaluate limits using algebra and the properties of limits

C. Evaluate limits at infinity D. Use the definition of a limit to evaluate limits graphically

A. Continuity and Finding Limits by Direct Substitution In previous chapters, we stated that a function was continuous if you could draw the entire graph without lifting your pencil. We also noted there were several ways that continuity could be interrupted, such as the holes and gaps seen in a study of piecewise-defined functions (Section 2.7), and the vertical asymptotes in some rational and trigonometric graphs. These “interruptions” are called discontinuities and can occur with other types of functions as well. The graph of f (x) given in Figure 11.9 shows a vertical asymptote at x
1, which interrupts the continuity of the graph and indicates that f is not defined at 1.

Figure 11.9 y 4 3 2 1

5 4 3 2 1 1

x
1

1

2

3

4

5

x

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Section 11.3 Continuity and More on Limits

1045

This is called an asymptotic discontinuity. The graph also shows that continuity is interrupted at x
1 (there is a “hole” in the graph), even though f is still defined there: f 112
3. This is called a removable discontinuity, because the discontinuity can be removed/repaired by redefining the function as f 1x2
1 when x
1. The discontinuity at x
4 is also removable, even though the function is not defined at 4. A break in the graph also occurs at x
3, but this time it leaves a gap in the output values, and there is no way to redefine f to “fill the gap.” This is called a jump discontinuity or a nonremovable discontinuity. In fact, of all the points called out by this graph, the function is only continuous at x
2. By comparing the graph at x
2 with the graph of f at other points, we find the following conditions are required for continuity. Continuity A function f is continuous at c if 1. f is defined at c 2. lim f 1x2
lim f 1x2 xSc

xSc

3. lim f 1x2
f 1c2 xSc

In Figure 11.9, the discontinuities at x
1 and x
4 violate condition 1, the discontinuity at x
3 violates condition 2, and the discontinuity at x
1 violates condition 3. EXAMPLE 1




Analyzing the Continuity of a Function Graphically The graph of a function f is given. Use the graph to comment on the continuity of the function for all integer values from 2 to 3. If the function is discontinuous at any value, state which of the three conditions are violated.

WORTHY OF NOTE

y

The continuity of a function really boils down to condition 3, since if this condition is satisfied, conditions 1 and 2 are satisfied.

x⫽3

5 4 3 2 1 ⫺5 ⫺4 ⫺3 ⫺2 ⫺1 ⫺1

1

2

3

4

5

x

⫺2 ⫺3

Solution




x
2: discontinuous, condition 2 is violated x
0: discontinuous, condition 3 is violated x
2: discontinuous, condition 1 is violated

x
1: continuous x
1: continuous x
3: discontinuous, condition 1 is violated

Now try Exercises 7 through 16




The fact that condition 3, lim f 1x2
f 1c2 , is one of the requirements for continuity xSc offers a great advantage when evaluating the limit of many common functions. For example, using properties of limits it can be shown that polynomials are continuous for all real numbers, and rational, root, and trigonometric functions are continuous wherever they are defined. This means as long as c is in the domain of these functions, we can find the limit by direct substitution since lim f 1x2
f 1c2 can be assumed. xSc

Finding Limits Using Direct Substitution If f is a polynomial, rational, root, or trigonometric function and c is in the domain of f, then lim f 1x2
f 1c2

xSc

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CHAPTER 11 Bridges to Calculus: An Introduction to Limits

EXAMPLE 2



Finding Limits Using Direct Substitution Evaluate the following limits using direct substitution, if possible. x4  3x2  5 x2  3x  10 a. lim x3  3x  4 b. lim c. lim xS3 xS2 7  4x xS2 x2

Solution



a. Since x3  3x  4 is a polynomial, the limit can be found by direct substitution: lim x3  3x  4  132 3  3132  4

xS3

 27  192  4  40

substitute 3 for x simplify result

See Example 4 of Section 11.2 where this limit was found using limit properties. x4  3x2  5 b. Since lim is a rational function and 2 is in the domain, the limit xS2 7  4x can be found by direct substitution: lim xS2

A. You’ve just learned how to determine if a function is continuous at c and find limits by direct substitution

24  3122 2  5 x4  3x2  5  7  4x 7  4122 16  12  5  78  23

substitute 2 for x

simplify result

See Example 4 of Section 11.2 where this limit was found using limit properties. x2  3x  10 c. Since lim is a rational function and 2 is not in the domain, the xS2 x2 limit cannot be found by direct substitution. This doesn’t mean the limit doesn’t exist, only that direct substitution cannot be used. A table of values x2  3x  10 seems to indicate lim  7 (also see Example 3a). xS2 x2 Now try Exercises 17 through 26



B. Evaluating Limits Using Algebra and Limit Properties

Recall that for lim f 1x2  L, the difference between x and c is made very small by xSc

taking values of x very close to c but not equal to c. In many cases, knowing that xc  1 (even if x  c is very small), can help evaluate limits where a direct xc substitution is not initially possible. This is done by using algebra to rewrite the function prior to taking the limit.

EXAMPLE 3



Finding Limits Using Algebra and Limit Properties Evaluate the following limits. x2  3x  10 1x  4  3 a. lim b. lim xS2 x2 xS5 x5

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PreCalculus—

11-25

Section 11.3 Continuity and More on Limits

Solution



1047

a. Although the function given is rational, x  2 is not in the domain and the limit cannot be found using direct substitution. But noting the numerator is factorable, we can work as follows. lim xS2

1x  52 1x  22 x2  3x  10  lim x2 xS2 x2 1x  52 1x  22  lim xS2 x2  lim 1x  52

factor the numerator

factors reduce since x  2  0 result is a polynomial

xS2

257

evaluate by direct substitution

As in Example 2c, a table of values will support this result. b. Since x  5 is not in the domain, we again attempt to rewrite the expression in a form that would make direct substitution possible. Note that multiplying the numerator and denominator by 1x  4  3 (the conjugate of 1x  4  3) will eliminate the radical in the numerator since 1A  B21A  B2  A2  B2. This yields lim xS5

1 1x  4  321 1x  4  32 1x  4  3  lim x5 xS5 1x  521 1x  4  32 3 1x  42  9 4  lim xS5 1x  52 1 1x  4  32 1x  52  lim xS5 1x  521 1x  4  32 1  lim xS5 1x  4  3

multiply by

1x  4  3 1x  4  3

1A  B21A  B2  A 2  B 2 simplify numerator factors reduce since x  5  0

lim 1 

xS5

lim 1 1x  4  32

property V

xS5

1 15  4  3 1 1   6 19  3



evaluate by direct substitution

result

The result can be supported using a table of values. Now try Exercises 27 through 36



In Example 4, we need to find the limit as h S 0 rather than x S c, but the concepts h are identical a  1 as long as h is not 0b. If h is not in the domain, the limit cannot h be found by direct substitution, and we again attempt to write the expression in an alternative form.

EXAMPLE 4



Finding Limits Using Algebra and Limit Properties Evaluate the following limits. 3 1x  h2  31x  h2 4  1x  3x2 h 2

a. lim hS0

2

1 1  x xh b. lim hS0 h

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11-26

CHAPTER 11 Bridges to Calculus: An Introduction to Limits 

Solution

a. Here we begin by simplifying the numerator.

3 1x  h2 2  31x  h2 4  1x2  3x2 lim hS0 h x2  2xh  h2  3x  3h  x2  3x  lim hS0 h

2xh  h2  3h hS0 h h12x  h  32  lim hS0 h  lim 12x  h  32  lim

square binomial, distribute 3, distribute 1

combine like terms

factor out h factors reduce since h  0

hS0

 2x  3 evaluate by direct substitution b. Here we begin by simplifying the complex fraction, hoping to rewrite the expression in a form that allows a direct substitution.

B. You’ve just learned how to evaluate limits using algebra and the properties of limits

1x  h2 1 1 x   x xh x1x  h2 x1x  h2 lim  lim hS0 h hS0 h x  1x  h2 x1x  h2  lim hS0 h h x1x  h2  lim hS0 h h # 1  lim hS0 x1x  h2 h 1  lim hS0 x1x  h2 1  2 x

LCD for the numerator is x 1x  h2

combine terms in the numerator

simplify

invert and multiply

factors reduce since h  0

evaluate by direct substitution

Now try Exercises 37 through 44



C. Evaluating Limits at Infinity Figure 11.10 y 10 8 6 4 2 10 8 6 4 2 2 4

h(x) 2

4

6

8 10

x

The limit of a function as x S c necessarily has x moving infinitely close to a number c from the left or right. In contrast, limits at infinity are concerned with the value of a function as x moves infinitely away from an arbitrary number c to the left or right, meaning as x S q or x S q . 4 The graph of h1x2  2 is shown in Figure 11.10, and we note as x S q , x 1 values of h(x) become closer and closer to 0. Similar to our previous work, we could say that values of h(x) can be made arbitrarily close to 0, by taking values of x sufficiently large and positive. This suggests the following definition.

6 8 10

Limits at Positive Infinity For a function f defined on an interval (c, q ),

lim f 1x2  L

xSq

means that values of f (x) can be made arbitrarily close to L, by taking values of x sufficiently large and positive.

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1049

Section 11.3 Continuity and More on Limits

In the present case, we have lim xSq

EXAMPLE 5



4  0. x2  1

Evaluating Limits at Infinity x

3x2  x  12 For f 1x2  , use a table of values to 2x2  8 evaluate lim f 1x2 . xSq

Solution



From our work with rational functions in Chapter 3 3 (ratio of leading terms), we suspect lim f 1x2  . This xSq 2 is supported by the table shown.

y

250

1.498

500

1.499

750

1.4993

1000

1.4995

1250

1.4996

1500

1.4997

1750

1.4997

y  1.49971428534

Now try Exercises 45 through 52 WORTHY OF NOTE

The limit of a function as x S  q is likewise defined.

Since q is not a number, the notation x S q should no longer be read, “as x approaches infinity,” and instead we say, “as x becomes infinitely large,” or “as x increases without bound.”

EXAMPLE 6



Limits at Negative Infinity For a function f defined on an interval (q , c)

lim f 1x2  L

xSq

means that values of f (x) can be made arbitrarily close to L, by taking values of x sufficiently large and negative.



Finding Limits at Infinity 1 For f 1x2  , evaluate lim f 1x2 and lim f 1x2 . x xSq xSq

Solution



You might recognize this function from our work in Chapter 3, where we observed that as x becomes very 1 large, becomes very small and close to zero (see figure). x The result was a horizontal asymptote at y  0 (the x-axis). In fact, we can make f (x) as close to 0 as we like, by taking x to be a sufficiently large positive number, or a sufficiently large negative number. 1 1  0 and lim  0. This shows lim x xSq xSq x

y 5 4 3 2 1 5 4 3 2 1 1

1

2

3

4

5

x

2 3 4 5

Now try Exercises 53 through 60



From Example 6 we learn two important things. First, horizontal asymptotes can be defined in terms of limits at infinity. Horizontal Asymptotes The line y  L is a horizontal asymptote if lim f 1x2  L

xSq

or

lim f 1x2  L

xSq

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CHAPTER 11 Bridges to Calculus: An Introduction to Limits

Second, by repeatedly applying limit property VI (the limit properties also hold for limits at infinity), we can state the following result. Limits of Reciprocal Powers For any positive integer k, lim xSq

1 0 xk

and

lim xSq

1 0 xk

These limits are a great asset to finding limits at infinity, as we often need to rewrite an expression and apply this limit in order to find the desired limit. For rational functions, this involves dividing numerator and denominator by the highest power of x occurring in the denominator. 

EXAMPLE 7

Finding Limits at Infinity for a Rational Function Evaluate lim xSq



Solution

y 10 8 6 4 2 10 8 6 4 2 2 4 6 8 10

2

4

6

8 10

x

5x2  6x  3 . 4x2  9x  12

Begin by dividing the numerator and denominator by x2, the highest power of x in the denominator. 5x2 6x 3 2 2  2  2 5x  6x  3 x x x lim  lim 2 2 9x 12 xSq 4x  9x  12 xSq 4x  2  2 x2 x x 3 6 5  2 x x  lim 12 9 xSq 4  2 x x 3 6 lim a5   2 b x xSq x  12 9 lim a4   2 b x xSq x 6 3 lim 5  lim  lim 2 x xSq xSq xSq x  9 12 lim 4  lim  lim 2 xSq xSq x xSq x 500 5   400 4 A similar calculation for x S q shows lim

xSq

divide numerator and denominator by x 2

simplify

property V

properties I and II

lim

xSq

k 0 xn

5x2  6x  3 5 is also . The 4 4x2  9x  12

graph is shown in the figure. Now try Exercises 61 through 64



If the function contains a radical term in the numerator or denominator, we attempt to rewrite the expression as a single radical term. We can then apply limit property VII and treat the radicand as a rational function. When doing so, you must particularly note whether you’re taking the limit as x S q or as x S q , since this may affect the sign of the final result.

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Section 11.3 Continuity and More on Limits



EXAMPLE 8



Solution

1051

Finding Limits at Infinity for a Radical Function Evaluate each limit: 8x 8x a. lim b. lim 2 xSq 24x  1 xSq 24x2  1 a. Since x is becoming an infinitely large positive number, we know 8x can be written as  218x2 2   264x2 (signs added for emphasis). The original expression can then be rewritten as follows. lim xSq

8x 24x  1 2

 lim

264x2

xSq

24x  1 2

64x2 2 xSq B 4x  1

 lim 

lim a

B xSq

64x2 b 4x2  1

8x  264x 2

quotient property of radicals

property VII

We are now taking the limit of a rational expression, so we next divide the numerator and denominator by x2. 64x2 x2  lim ± 2 ≤ xSq 4x 1  2 b x2 x 

xSq

R 



C. You’ve just learned how to evaluate limits at infinity



64

lim

1 4 2 x

simplify

lim 64 xSq

1 lim a4  2 b R xSq x lim 64 xSq

property V

property I

1 lim 4  lim 2 R xSq xSq x 64 A4  0

 116  4

y

divide by x 2

lim xSq

k 0 xn

result

5 4 3 2 1 5 4 3 2 1 1 2 3 4 5

1

2

3

4

5

x

b. For part (b), x is becoming an infinitely large negative number, so the sign of 8x will be negative for all such values. For this reason 8x must be 24x2  1 written as 218x2 2   264x2. This is the only change to be made, and 8x after applying algebra and the limit properties we find lim  4. xSq 24x2  1 The graph is shown in the figure and shows that this function has two unique horizontal asymptotes. Now try Exercises 65 through 70



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CHAPTER 11 Bridges to Calculus: An Introduction to Limits

D. Evaluating Limits Graphically As a complement to the equations and tables we’ve used to explore the limit concept, we now look at how these concepts appear graphically. This will offer a heightened understanding of important ideas, and an excellent summary of the definitions, properties, and relationships involved. The graphs of two general functions f and g are given (Figures 11.11 and 11.12, respectively) for use in the following examples. Figure 11.12 Figure 11.11 y

y

5

5

g(x)

4

f (x)

3

2

2

1 5 4 3 2 1 1

EXAMPLE 9




4

3

1 1

2

3

4

5

5 4 3 2 1 1

x

2

2

3

3

4

4

5

5

1

2

3

4

5

x

Evaluating and Combining Function Values Using a Graph Use the graphs of f and g to perform the operations indicated. If any computation is not possible, state why. a. f 112  g142 b. g122  f 142 c. f 132  f 122 d. g122 g132

Solution




a. f 112  g142
3  2 b. g122  f 142
1  2
1
3 c. f 132  f 122
X d. g122 g132
112 102
0 not possible since f (2) is not defined

Now try Exercises 71 through 76 EXAMPLE 10







Evaluating One-Sided Limits Graphically Use the graphs of f and g to evaluate the one-sided limits necessary to complete each calculation. If any computation is not possible, state why. a. lim f 1x2  lim g1x2 b. lim f 1x2  lim  g1x2 xS1

c.

xS4

xS2

Solution




xS3

lim  g1x2 lim  f 1x2

d.

xS3

xS1

a. lim f 1x2  lim g1x2
0  2 xS1

c.

b. lim f 1x2 

xS4

lim  g1x2

xS2

xS5

lim  f 1x2  lim g1x2

xS3


2 lim  f 1x2
X

d.

xS3

xS5


4 lim  f 1x2  lim g1x2
X

xS1

not possible since lim f 1x2
A dne B q xS3 

lim g1x2
1  132

xS5 

xS5

not possible since 5 is not in the domain of g Now try Exercises 77 through 82

EXAMPLE 11




Evaluating Limits Graphically Use the graphs of f and g to evaluate the limits necessary to complete each calculation. If any computation is not possible, state why. a. lim 3 f 1x2  g1x2 4 b. lim 33f 1x2  g1x2 4 xS2

c. lim xS1

g1x2 f 1x2

xS1

d. lim 3 1 f 1x22 2  1g1x2 4 xS2




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Section 11.3 Continuity and More on Limits

Solution




lim 3 f 1x2  g1x2 4
lim f 1x2  lim g1x2

a.

xS2

xS2

xS1

xS1

result property II

xS1


3 lim f 1x2  lim g1x2

property IV


3122  3
3

result

xS1

c. lim

property I

xS2


13
4 b. lim 33f 1x2  g1x2 4
lim 3f 1x2  lim g1x2 xS1

1053

xS1

lim g1x2 g1x2 xS1
f 1x2 lim f 1x2

property V

xS1

not possible since lim f 1x2
0

limit of the denominator is 0

xS1

d. lim 3 1 f 1x2 2 2  1g1x2 4
lim 3 f 1x2 4 2  lim 1g1x2 xS2

xS2


S lim f 1x2 T 2  1 lim g1x2

D. You’ve just learned how to use the definition of a limit to evaluate limits graphically

property I

xS2

xS2 2


122  14
6

properties VI and VII

xS2

result

Now try Exercises 83 through 88




11.3 EXERCISES


CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

1. There are three types of discontinuities we may encounter. These are (1) , (2) , and (3) discontinuities. 2. In order for a function f (x) to be continuous at c, it must be at c. Furthermore, lim f 1x2 must with lim f 1x2


equal

xSc

.

3. If f(x) is continuous at c, we can evaluate lim f 1x2 xSc by . xSc




4. The notation x S q is read, “as x becomes ,” or “as x increases without

.”

5. Discuss/Explain the relationship between the horizontal asymptote of a rational function f, and lim f 1x2 . xSq

6. Discuss/Explain how the three different types of discontinuities appear on the graph of a function.

DEVELOPING YOUR SKILLS

For Exercises 7 through 16, use the given graphs to comment on the continuity of the function at the given x-values. If the function is discontinuous at any value, state which of the three conditions for continuity are violated. y 5 4 3 2 1 54321 1 2 3 4 5

y

f (x)

1 2 3 4 5 x

5 4 3 2 1 ⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

g(x)

1 2 3 4 5 x

7. f(x) at x
1 9. f(x) at x
1

8. f(x) at x
3 10. f(x) at x
2

11. f(x) at x
0

12. g(x) at x
1

13. g(x) at x
3

14. g(x) at x
1

15. g(x) at x
2

16. g(x) at x
0

Evaluate the following limits using direct substitution, if possible. If not possible, state why.

17. lim 2x2  5x  3 xS3

18. lim 3x2  5x  2 xS2

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19. lim 13x  19

20. lim 1x2  7x

xS5

21. lim xS2

xS6

x2 5x  2

23. lim xS1

22. lim xS8

x1 x2  1

2x  5 x2  5x

Use a table of values to evaluate the following limits as x increases without bound.

45. lim xSq

4x x2  5x  6 2

24. lim xS2

25. lim 2x  6x

26. lim 13x  5

2

xS5

xS6

x1 27. lim 2 xS1 x  1

4  x2 28. lim 2 xS2 x  5x  6

x4 29. lim xS4 1x  2

x  16 30. lim xS16 4  1x

31. lim xS3

2x2  3x  9 x3

1x  7  2 33. lim xS3 x3

32. lim xS2

3x2  7x  2 x2

12x  1  5 34. lim xS12 x  12

x3  8x2  16x 2x3  12x2  18x 35. lim 36. lim 2 xS4 x  7x  12 xS3 x2  7x  12 Evaluate the following limits. Write your answer in simplest form.

37. lim hS0

38. lim hS0

3 21x  h2 2  1x  h2 4  12x2  x2 h 3 31x  h2  1x  h2 2 4  13x  x2 2 h

3 3  xh2 x2 39. lim hS0 h

47. lim xSq

48. lim 49. lim xSq

51. lim xSq

53. 55.

hS0

44. lim hS0

50. lim

x 2  6x  9 2x3

52. lim

xSq

xSq

x2  1 2x  11 10  3x2 10  3x3

xSq

5x3  2 10x3  2x  1

54.

x2  1 2x  11

56.

6x2  x  2 xSq 2x2  1 lim

lim xSq

7x3 5x2  3x

10  3x x2  6x  9 58. lim 3 xSq 10  3x xSq 2x3 1 59. Given lim  0, find the smallest positive value xSq x 1 of x such that  0.001. x 1 60. Given lim  0, find the largest negative value xSq x 1 of x such that  0.01. x 2

57.

lim

For Exercises 61 through 64, evaluate the limits by dividing the numerator and denominator by the highest power of x occurring in the denominator.

xSq

2x  h  1  1x  1 h

1x  h  22  1x  22 h

7x3 5x2  3x

lim

63.

43. lim

3x  5x3 x3  2x2  3x  7

xSq

1x  h  2  1x  2 41. lim hS0 h

3

6x  x  2 2x2  1

lim

61. lim

hS0

xSq

8x  3x2 7x2  x  1

Use a table of values to evaluate the following limits as x decreases without bound.

2 2  xh1 x1 40. lim hS0 h

42. lim

46. lim

2

xSq

Evaluate the following limits by rewriting the given expression as needed.

5x3  2 10x3  2x  1

3

1x  h  42 3  1x  42 3 h

3x2  2x  1 8x2  5

lim xSq

2x2  1 x3  2x  12

62. lim xSq

64.

5x2  11x  3 12x2  7x  5

lim xSq

8x3  27 x4  1

For Exercises 65 through 70, evaluate each limit.

65. lim xSq

66.

236x2  11 3x

lim xSq

236x2  11 3x

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PreCalculus—

11-33

Section 11.3 Continuity and More on Limits

For Exercises 77 through 82, use the graphs of f and g to evaluate the one-sided limits necessary to complete each calculation. If any computation is not possible, state why.

212x2  6x  1 xSq 7x

67. lim 68.

212x2  6x  1 xSq 7x

77.

lim

3

79.

2

y 5 4 3 2 1

1 2 3 4 5 x

73. g142  f 122

75. g102  f 142

xS0

⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

xS2

81. lim f 1x2  lim  g1x2 xS5

xS2

82. lim g1x2  lim f 1x2 xS2

g(x) 1 2 3 4 5 x

For Exercises 71 through 76, use the graphs of f and g to perform the operations indicated. If any computation is not possible, state why.

71. f 102  g112

f 1x2

f 1x2  lim g1x2

lim

xS4

xS4

y 5 4 3 2 1

lim

xS2 

80. lim g1x2  lim  f 1x2

The graphs of two functions f and g are given here, for use in Exercises 71 through 88.

⫺5⫺4⫺3⫺2⫺1 ⫺1 ⫺2 ⫺3 ⫺4 ⫺5

xS0

xS4

3 2 216x  36x  6x  1 70. lim xSq 2x

f (x)

f 1x2  lim g1x2

lim

xS4 

78. lim g1x2 

2

3 2 216x  36x  6x  1 xSq 2x

69. lim

3

1055

72. f 152  g132 74. f 122  g122 76. g152  f 112

xS6

For Exercises 83 through 88, use the graphs of f and g to evaluate the limits necessary to complete each calculation. If any computation is not possible, state why.

83. lim 3 f 1x2  g1x2 4 xS0

85. lim c xS3

3f 1x2 d g1x2

86. lim e xS1

87.

84. lim 3g1x2  f 1x2 4 xS2

3 3g1x2 4 2 f f 1x2

3 lim 3 f 1x2  2 g1x2 4

xSq

88. lim 32g1x2f 1x2 4 xS3



MAINTAINING YOUR SKILLS

89. (1.3) Solve each equation: a. 3x2  4x  12  0 b. 13x  1  12x  1 3 1 2  2  c. x2 x3 x  5x  6  90. (6.6) Find cos1 c cosa b d . 6 91. (5.6) Solve the right triangle shown. Round answers to the nearest tenth. A

C

35⬚ 13.7 cm

B

92. (2.3) In 2005, a small business purchased a copier for $4500. By 2008, the value of the copier had decreased to $3300. Assuming the depreciation is ¢ value linear, (a) find the rate-of-change m  ¢ time and discuss its meaning in this context. (b) Find the depreciation equation and (c) use the equation to predict the copier’s value in 2012. (d) If the copier is traded in for a new model when its value is less than $700, how long will the company use this copier?

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PreCalculus—

11.4 Applications of Limits: Instantaneous Rates of Change and the Area under a Curve Learning Objectives In Section 11.4 you will learn how to:

A. Evaluate the limit of a difference quotient to find instantaneous rates of change

B. Evaluate the limit of a sum to find the area under a curve

At this point, the concept of rates of change is a familiar theme. From their first introy2  y1 ¢y
b to investigations of nonlinear functions duction in a linear context a x2  x1 ¢x using the difference quotient, the idea has proven powerful and effective. In Example 10 from Section 2.5, we used the difference quotient to approximate the velocity of a falling wrench with great accuracy. While this study certainly answers Galileo’s historic questions regarding the speed of a falling body, business, science, industry, education, and government present us with equally compelling questions that can likewise be answered using the difference quotient.

A. The Limit of a Difference Quotient In Example 8 of Section 2.5, the height of a soccer ball kicked straight up into the air was modeled by the function d1t2
16t2  64t. Using this function and the rated1t2 2  d1t1 2 ¢d
d , we found that between t
0.5 and of-change formula c ¢t t2  t1 t
1.0 sec, the average velocity of the ball was 40 ft/sec. Graphically, this was represented by the slope of the secant line through (0.5, 28) and (1.0, 48). For the time interval [1.0, 1.5], the average velocity had slowed to 24 ft/sec (the slope of this secant line is 24 1 ). Using smaller and smaller increments of time enables a better and better estimate of the velocity over a given time interval, but requires a new calculation for each interval. Instead, we consider applying the difference quotient, as it allows for a fixed point d1t  h2  d1t2 . t and a second point 1t  h2 that becomes arbitrarily close to t as h S 0: h Graphically, this means the two points used for the secant line are becoming arbitrarily close, and in fact, can be made as close as we like by applying the concept of a limit. Figure 11.13 In Figure 11.13, several secant lines are shown, each d(t) with a left endpoint at t
1. Selecting right endpoints that become closer and closer to 1 11  h as h S 02 , we obtain the sequence of secant lines shown in black, then 60 purple, then red. As 1  h gets infinitely close to 1, the 50 secant line reaches a “limiting position” (the yellow line) as the two points forming the line become infinitely 40 close. The end result is a tangent line to the graph of d at x
1 (tangent: touching the curve at only one point). 30 As the slope of each secant line gives the average rate of 20 change between two points, in this case the velocity of the ball, the slope of the tangent line gives the instanta- 10 neous velocity at precisely x
1. As the graph indicates, the slope of the tangent line is the limit of the slopes of t 0 1 2 3 4 the secant lines and we write, mtan
lim msec
lim hS0

hS0

d1t  h2  d1t2 h

For a general function f (x), we have the following result.

1056

11-34

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Section 11.4 Applications of Limits: Instantaneous Rates of Change and the Area under a Curve

1057

The Slope of a Tangent Line Given a function f(x) that is smooth and continuous over the interval containing x and x  h, the slope of a line drawn tangent to the graph of f at x is given by the function f (x), where f 1x  h2  f 1x2 f 1x2  lim hS0 h

EXAMPLE 1



Computing the Limit of a Difference Quotient For the function d1t2  16t2  64t (see page 146) a. Find the limit of the difference quotient for d, to obtain a function d (t) that represents the instantaneous velocity at time t. b. Use d (t) to find the velocity at t  1, t  2, and t  3.5.

Solution



a. For d1t2  16t2  64t, d1t  h2  161t  h2 2  641t  h2 . After squaring the binomial and applying the distributive property, we obtain d1t  h2  161t2  2th  h2 2  64t  64h  16t2  32th  16h2  64t  64h

To find d (t), we then have

d1t  h2  d1t2 16t 2  32th  16h2  64t  64h  116t 2  64t2 apply difference  lim lim hS0 h hS0 h quotient 16t2  32th  16h2  64t  64h  16t 2  64t  lim distribute 1 hS0 h 32th  16h2  64h  lim combine like terms hS0 h h 132t  16h  642  lim factor out h hS0 h h  lim 132t  16h  642  1, h  0 h

hS0

d 1t2  32t  64

result

b. For the velocity of the ball at t  1, t  2, and t  3.5, we evaluate d (t). t  1:

d 112  32112  64  32  64  32

substitute 1 for t simplify

At the moment t  1 sec, the velocity is 32 ft/sec. t  2:

d 122  32122  64  64  64  0

substitute 2 for t simplify

At the moment t  2 sec, the velocity is 0 ft/sec. This seems reasonable, as we expect the ball will slow as it approaches its maximum height, where it momentarily stops (velocity is 0), then begins its return to the ground. t  3.5: d 13.52  3213.52  64  112  64  48

substitute 3.5 for t simplify

At the moment t  3.5 sec, the velocity is 48 ft/sec. This indicates the ball is falling back to Earth (velocity is negative) at a rate of 48 ft/sec. Now try Exercises 7 through 12



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CHAPTER 11 Bridges to Calculus: An Introduction to Limits

In Example 10 of Section 2.5 (page 148), we modeled the distance a wrench had fallen from a skyscraper with the function d1t2
16t2. Using a very small time interval and the difference quotient, we obtained a close approximation for the velocity of the wrench in various intervals. Using the concept of a limit we can now find the instantaneous velocity at any time t. EXAMPLE 2




Finding an Instantaneous Rate of Change For the function d1t2
16t2, a. Find the limit of the difference quotient for d to obtain a function d (t) that represents the instantaneous velocity of the falling wrench. b. Use d (t) to find the velocity at t
2, t
7, and t
9.

Solution




a. In the referenced example we obtained the expression 32t  16h after applying the difference quotient. Using the ideas developed here, we know the velocity of the wrench will be d 1t2
lim 132t  16h2 hS0


32t by direct substitution. b. For the instantaneous velocity at t
2, t
7, and t
9, we evaluate d (t). t
2: d 122
32122
64

substitute 2 for t

t
7: d 172
32172
224

substitute 7 for t

t
9: d 192
32192
288

substitute 9 for t

At t
2 sec, the velocity of the wrench is 64 ft/sec. At t
7 sec, the velocity of the wrench is 224 ft/sec. At the instant t
9 sec, the velocity of the wrench has increased to 288 ft/sec. In fact, in the absence of air resistance the velocity of the wrench will continue to increase until it hits the ground (with a wham!). Now try Exercises 13 through 16




Mathematical models of real-world phenomena come in many different forms, and in a calculus course these ideas are applied to a very wide variety of functions. For practice with the algebra required to make these ideas work, see Exercises 17 through 20. The following applications demonstrate how these ideas are applied to root and rational function models. EXAMPLE 3




Computing a Rate of Change for the Distance to the Horizon Many new high-rise buildings have external elevators, allowing an unobstructed view of the city as it extends toward the horizon. The distance that Melody can see as the elevator takes her higher is modeled by the function d1x2
1.151x, where d(x) represents her sight distance in nautical miles, at a height of x feet. a. Find the limit of the difference quotient for d(x), to obtain a function d (x) that represents the instantaneous rate of change in the sight distance at x. b. Find the instantaneous rate of change at heights of 225, 400, and 625 ft.

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Section 11.4 Applications of Limits: Instantaneous Rates of Change and the Area under a Curve

Solution



lim hS0

1059

a. For d1x2  1.15 1x, d1x  h2  1.15 1x  h. To find d (x), we then have d1x  h2  d1x2 1.15 1x  h  1.15 1x  lim h hS0 h 1x  h  1x  1.15 lim hS0 h

apply difference quotient

factor out the constant

Noting that multiplying the numerator by 1x  h  1x will “free up” the h in the first radicand, we multiply numerator and denominator by 1x  h  1x.

1 1x  h  1x2 1 1x  h  1x2 hS0 h 1 1x  h  1x2 xhx  1.15 lim 1A  B 21A  B 2  A 2  B 2 hS0 h1 1x  h  1x2 h  1.15 lim combine like terms hS0 h1 1x  h  1x2 1 h  1.15 lim  1, h  0 h hS0 1x  h  1x 1  1.15a b apply limit 1x  1x 1.15 result d 1x2  2 1x b. To find the instantaneous rate of change at x  225, 400, and 625 ft, we evaluate d (t).  1.15 lim

d 12252 

1.15 1.15  21152 2 1225  0.038

substitute 225 for x

At x  225 ft, the sight distance is increasing at a rate of about 0.038 mi/ft (about 3.8 nautical miles per 100 ft). d 14002 

1.15 1.15  21202 2 1400  0.029

substitute 400 for x

At x  400 ft, the sight distance is increasing at a rate of about 0.029 mi/ft (about 2.9 nautical miles per 100 ft). d 16252 

1.15 1.15  21252 2 1625  0.023

substitute 625 for x

At x  625 ft, the sight distance is increasing at a rate of 0.023 mi/ft (2.3 nautical miles per 100 ft). Now try Exercises 21 through 24



Note that for square root functions, the algebra involves rationalizing the numerator to “free up” the constant h in hopes that the limit can eventually be found by direct substitution. For rational functions, the difference quotient yields a difference of rational expressions in the numerator, which we combine for the same reason.

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EXAMPLE 4



Computing Instantaneous Rates of Change for a Cost Function A county government finds the cost of removing pollutants from a stream can be 2x modeled by the function C1x2  , where C(x) represents the cost in millions 1x of dollars, to remove x percent of the pollutants 3C1x2 7 0, x is a decimal less than 1]. a. Find the limit of the difference quotient for C, to obtain a function C(x) that represents the instantaneous rate of change in the cost of removing pollutants. b. Find the rate the cost is increasing at the moment 60%, 70%, and 80% of the pollutants are removed. c. Use the information from Part (b) to estimate the rate the cost will be increasing at the moment 90% of the pollutants are removed, then compute the actual rate. Were you surprised?

Solution



21x  h2 2x , C1x  h2  . To find C(x) we then have 1x 1  1x  h2 21x  h2 2x  C1x  h2  C1x2 1  1x  h2 1x lim  lim apply difference quotient hS0 h hS0 h 1x  h2 x factor out 2, distribute 1  1xh 1x  2 lim hS0 h 1x  h211  x2  x11  x  h2 A C AD  BC   B D BD 11  x  h211  x2  2 lim hS0 h 2 x  x  h  xh  x  x2  xh F-O-I-L, distribute 11  x  h2 11  x2  2 lim hS0 h h 11  x  h211  x2  2 lim simplify numerator hS0 h h  2 lim invert and multiply hS0 11  x  h211  x2h 1 h  2 lim  1, h  0 h hS0 11  x  h211  x2 2 C 1x2  apply limit, multiply by 2 11  x2 2 a. For C1x2 

b. Evaluating C(x) for x  0.6, 0.7, 0.8 (60%, 70%, and 80%) yields C 10.62 

2 2 2  0.16 11  0.62  12.5

substitute 0.6 for x and simplify

At x  0.6, the cost is increasing at a rate of $12,500,000 for each additional percentage point of pollution removed. C 10.72 

2 2 2  0.09 11  0.72  22.2

substitute 0.7 for x and simplify

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Section 11.4 Applications of Limits: Instantaneous Rates of Change and the Area under a Curve

1061

At x  0.7, the cost is increasing at a rate of about $22,222,222 for each additional percentage point. C 10.82 

2 2 2  0.04 11  0.82  50

A. You’ve just learned how to evaluate the limit of a difference quotient to find instantaneous rates of change

substitute 0.8 for x and simplify

At x  0.8, the cost is increasing at a rate of $50,000,000 for each additional percentage point. c. Evaluating C(x) for x  0.9 shows the cost rises dramatically to a rate of $200,000,000 per percentage point. Removing anywhere near 100% of the pollutants would become prohibitively expensive. Now try Exercises 25 through 32



B. Limits and the Area under a Curve In the Connections to Calculus from Chapter 10, the area under the graph of f 1x2  x  5 for x in the interval [0, 4] was approximated using the rectangle method with n  4 and n  8 rectangles (see Example 1). While we agreed that using 16 rectangles would give a better estimate (see Figure 11.14), expanding the related 16 1 1 LW  f a iba b seemed too summation Figure 11.14 4 4 i1 y tedious and instead we opted to use the properties 10 of summation to write the expression in a form 9 where the sum could be found by substituting 8 n  16 directly. Here we extend this idea to a 76 situation where n rectangles are used, as we can 5 then take the limit of this sum as n S q . The key 4 to this extension is a result from Section 11.3, 3 where we noted for any positive integer k, 2 1 1 1 lim k  0 and lim k  0 (note the change 5 6 x 1 2 3 4 nSq n nSq n of variable from x to n, which is more commonly used for summations). The original 4 4-unit interval would then be divided into n parts, so each rectangle is units wide. n The length of each rectangle is still given by f 1x2  x  5, but we now evaluate f in n 4 4 4 4 4 increments of , giving . Since f a ib  i  5, we can LW  f a iba b n n n n n i1 n 4 4 a i  5b and proceed as before. write the sum as n i1 n











EXAMPLE 5



Finding the Limit of a Sum

n

Evaluate the following limit: lim

nSq

Solution



 a ni  5b n . 4

4

i1

Begin by applying summation properties and formulas to write the expression in a 4 form where the limit properties can be applied. Since is a constant, we begin by n factoring it out of the summation (summation property II).

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4 n

n

n 4 n 4 4 i a i  5b  a 5b n i1 n i1 n i1









4 4 n a i n n i1



summation properties (distribute)

n

 5b

factor

i1

4 4 n2  n  c a b  5n d n n 2 16 n2  n b  20 a 2 n2 16 n2  n  a b  20 2 n2 n2 n  8a 2  2 b  20 n n 8 1  8a1  b  20  28  n n 

4 from first summation n

apply summation formulas

distribute

4 n

rewrite denominators (commute)

decompose rational expression

result

We can now take the limit of this result and find 8 lim a28  b  28. n nSq This shows n 4 4 lim a i  5b  28, nSq i1 n n



and that the area under the graph of f 1x2  x  5, between x  0 and x  4, is exactly 28 units2! Now try Exercises 37 through 40



This result may seem unimpressive, as we could have quickly found the area using simple geometry. But as mentioned earlier, this method is easily applied to areas under a general function f (x). In the Connections to Calculus feature from Chapter 2, we noted how the area under a graph could represent the distance covered by a runner. Our work here now takes on added meaning, as the area under a general curve has many other real-world implications in a study of calculus. EXAMPLE 6



Using Limits to Find the Area under a Curve f(x)

During a training session for the Boston Marathon, 10 9 Georgiana plans to begin at a leisurely walk of 8 2 mi/hr and gradually increase her speed over a 7 5-hr period, finishing up at a clip of over 8 mi/hr. 6 Her speed might then be modeled by the function 5 4 1 2 f 1x2  x  2, where f(x) represents the velocity 3 4 2 in miles per hour, and x is the time in hours. The area 1 under the curve in the interval [0, 5] then represents the 2 1 1 2 3 4 5 6 x distance Georgiana has run in these 5 hr. a. Show that using the rectangle method results in the expression 5 n 1 5 2 c a ib  2 d . n i1 4 n b. Determine if Georgiana has run a “marathon distance” (about 26.2 mi) in these 5 hr, by applying summation properties and formulas, and taking the limit of this sum as n S q .



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Section 11.4 Applications of Limits: Instantaneous Rates of Change and the Area under a Curve

Solution



1063

a. The area described is shown in the figure. The given interval is 5 units wide, 5 and after dividing it into n parts, each rectangle will have a width of . The n 1 length of each one is given by f 1x2  x2  2, evaluated at each of 4 5 the subintervals. The total area is then n



n

LW 

5 5 5 f a ib a b  n n n i1



n

5 5 f a ib  n n i1



n



i1

c

1 5 2 a ib  2 d . 4 n

b. Proceed as in Example 5. 5 n 1 5 2 5 n 1 5 2 a ib  c a ib  2 d  c n i1 4 n n i1 4 n







n

 2d

i1 n

5 n 1 25 2 c a bi  n i1 4 n2



5 n 25 2  c i  n i1 4n2



 2d

i1

summation properties (distribute) 5 2 25 a ib  2 i 2 n n

n

 2d

multiply

i1



n 5 25 n 2 c a 2b i  2d n 4n i1 i1

factor



5 25 2n3  3n2  n c a b  2n d n 4n2 6

apply summation formulas





25 n2

from first summation

125 2n3  3n2  n 5 b  10 a distribute n 6 4n3 125 2n3  3n2  n  a b  10 rewrite denominators (commute) 24 n3 1 3 125 a2   2 b  10  decompose rational expression n 24 n We can now take the limit of this result, and after applying the needed limit properties we obtain 

1 125 3 lim a2   2b  lim 10 apply properties of limits n n 24 nSq nSq 125 b122  10  20.416 a 24 1 The area under the graph of f 1x2  x2  2 between x  0 and x  5 is 4 20.416 units2, which is numerically equivalent to the distance run. Georgiana is still about 5.8 mi short of a marathon. 

B. You’ve just learned how to evaluate the limit of a sum to find the area under a curve

Now try Exercises 41 through 44



Note: For Learning Objective B, the functions and intervals offered in the Exercise Set were deliberately chosen for their simplicity. In particular, all are polynomials where f 1x2 7 0 in the interval selected. In a calculus course, the same ideas are applied to a greater variety of functions, not all intervals begin at 0, and the function may vary from positive to negative within the interval. Regardless, the general ideas demonstrated here remain consistent.

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11-42

11.4 EXERCISES


CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.




1. The slope of a line drawn tangent to the graph of any function f (x) can be found by taking the limit of the quotient for f, as h approaches zero.

4. As we let the number of approximating rectangles increase without bound, the resulting limit gives a(n) value for the area under the curve.

2. The instantaneous rate of change of a function f (x) can be found by evaluating the expression .

5. Discuss/Explain the relationship between the difference quotient and the formula for finding the slope of a line given two points.

3. In order to obtain a better approximation for the area under a curve, we can use narrower (and thus more) in our computation.

6. Do some research on geometric shapes other than rectangles that can be used to approximate the area under a curve. Return to the graph on page 1061, and comment on why approximating rectangles consistently give an overestimate of the area.

DEVELOPING YOUR SKILLS

Use the following information to answer Exercises 7 through 12. Two model rockets are launched at a gathering of the National Association of Rocketry (NAR: www.nar.org). Frank’s Apollo II motor burns out at a height of 500 m, at which point the rocket has a velocity of 88.2 meters per second (m/sec). His rocket’s height in meters, t sec after engine burnout, is given by f 1t2
500  88.2t  4.9t2. Gwen’s Icarus Alpha motor burns out at a height of 600 m, at which point the rocket has a velocity of 78.4 m/sec. Her rocket’s height in meters, t sec after burnout, is given by g1t2
600  78.4t  4.9t 2.

11. Use the result from Exercise 7 to find the maximum height of Frank’s rocket. This occurs when v
0. 12. Use the result from Exercise 9 to find the maximum height of Gwen’s rocket. This occurs when v
0.

7. Find the limit of the difference quotient for f, to obtain a function f (t) that represents the instantaneous velocity at time t.

13. To replicate Galileo’s famous test as to whether the velocity of a falling body depends on its weight, a science class is dropping bowling balls of different weights from an 11-story building onto the lawn below. The ball’s height in meters, t sec after it is released, is modeled by d1t2
4.9t2  44.1. Find the limit of the difference quotient for d, to obtain a function d (t) that represents the instantaneous velocity of the bowling ball at time t.

8. Use the result from Exercise 7 to find the instantaneous velocity of Frank’s rocket at a. t
5 b. t
9 c. t
13

14. Use the results of Exercise 13 to find the instantaneous velocity of the bowling ball at a. t
1 b. t
2 c. t
3 (time of impact)

9. Find the limit of the difference quotient for g, to obtain a function g (t) that represents the instantaneous velocity at time t.

15. A rock climber’s carabineer falls off her harness 256 ft above the floor of the Grand Canyon. It’s height in feet, t sec after it falls, can be modeled by d1t2
16t2  256. Find the limit of the difference quotient for d, to obtain a function d (t) that represents the instantaneous velocity of the “biner” at time t.

10. Use the result from Exercise 9 to find the instantaneous velocity of Gwen’s rocket at a. t
2 b. t
8 c. t
10

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Section 11.4 Applications of Limits: Instantaneous Rates of Change and the Area under a Curve

16. Use the results of Exercise 15 to find the instantaneous velocity of the “biner” at a. t
1 b. t
3 c. t
4 (time of impact) Find the limit of the difference quotient for each function f(x) given, to obtain a function f (x) that represents the instantaneous rate of change at x for each function.

1 17. f 1x2
x  5 2

18. f 1x2
x2  3x

19. f 1x2
x3

20. f 1x2


1 x

21. The population of a small town can be modeled by the function p1t2
1.2 1t  40, where p is measured in thousands and t is the number of years after 2008. Find the limit of the difference quotient for p to obtain a function p (t) that represents the instantaneous rate of change of population at time t. 22. Use the results of Exercise 21 to find the instantaneous rate of change of population of the town in a. 2009 b. 2012 c. 2024 23. The number of bacteria in a person’s body, after they begin a regimen of antibiotics, can be modeled by the function b1t2
6  1t, where b is measured in tens of thousands and t is the number of hours after the first dose. Find the limit of the difference quotient for b to obtain a function b (t) that represents the instantaneous rate of change of number of bacteria at time t. 24. Use the results of Exercise 23 to find the instantaneous rate of change of number of bacteria after a. 1 hr b. 16 hr c. 25 hr For Exercises 25 through 28, find the limit of the difference quotient of the given function to obtain a function that represents the slope of a line drawn tangent to the curve at x.

25. f 1x2


2 x1

26. g1x2


3 x2

27. h1x2


5x x5

28. j1x2


x 21x  12

Use the results from the corresponding Exercises 25 through 28 for the following exercises.

29. Find the slope of the line drawn tangent to the graph of f (x) at x
3.

30. Find the slope of the line drawn tangent to the graph of g(x) at x
1. 31. Find the slope of the line drawn tangent to the graph of h(x) at x
0. 32. Find the slope of the line drawn tangent to the 1 graph of j(x) at x
 . 2 For Exercises 33 through 36, graph each function over the interval [0, 7]. Then use geometry to find the area of the region below the graph, and above the x-axis in the interval [0, 6].

1 33. q1x2
x 2

1 34. r1x2
x 3

1 35. t1x2
x  1 2

1 36. s1x2
x  2 3

For Exercises 37 through 40, evaluate the following limits and compare your result to the corresponding exercise in 33 through 36. n

37. lim

n

 a 2 # n ib n 1 6

6

38. lim

nSq i
1 n

39. lim

 a 3 # n ib n 1 6

6

nSq i
1

 a 2 # n i  1b n

6

1 6

nSq i
1 n

40. lim

 a 3 # n i  2b n

6

1 6

nSq i
1

y 41. For a new machine shop 12 11 employee, the rate of 10 production for a specialized 9 8 part is modeled by the 7 6 function 5 4 1 2 3 p1x2
x  3 2 2 1 (production increases 54321 1 2 3 4 5 x quickly with experience), where p(x) represents the number of parts completed per day after x days on the job. The area under this curve in the interval x
3 0, 44 then represents the total number of parts produced in the first 4 days. Using the rectangle method results in 4 n 1 4 2 c a ib  3 d . Find the total the expression n i
1 2 n number of parts produced by applying the summation properties/formulas and taking the limit as n S q .



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42. Water has been leaking undetected from a reservoir formed by an earthen dam. Due to the accelerating erosion, the rate of water loss can be modeled by 1 the function g1x2   x2  8, x 7 0, where g(x) 4 represents hundreds of gallons of water lost per hour. The area under the curve in the interval [0, 5] then represents the total water loss in 5 hr. Using the rectangle method results in the expression 1 5 2 5 n c  a ib  8 d . Find the total amount of n i1 4 n water lost by applying summation properties and formulas and taking the limit as n S q .





11-44

CHAPTER 11 Bridges to Calculus: An Introduction to Limits

MAINTAINING YOUR SKILLS

Exercise 42 y 12 11 10 9 8 7 6 5 4 3 2 1 ⫺6⫺5⫺4⫺3⫺2⫺1

1 2 3 4 5 6 x

Find the area under the curve for each function and interval given, using the rectangle method and n subintervals of equal width.

1 43. f 1x2   x2  4x, x  30, 6 4 2 1 44. f 1x2   x3  6x, x  30, 3 4 2 48. (5.2) Evaluate all six trigonometric functions at 2 t 3

45. (4.4) Solve for x: 350  211 e0.025x  450 46. (7.4) Given u  H2, 3I and v  H3, 5I, find 3u  v. 47. (1.5/1.6) Find all solutions by factoring: x3  5x2  3x  15  0.

S U M M A RY A N D C O N C E P T R E V I E W SECTION 11.1

An Introduction to Limits Using Tables and Graphs

KEY CONCEPTS • The concept of a limit involves the incremental approach of a variable x to a fixed constant c (the cause), with a focus on the behavior the function f during this approach (the effect). • The notation lim f 1x2  L means values of f(x) can be made arbitrarily close to L by taking values of xSc x sufficiently close to c, but not equal to c. L is called the limit of f as x approaches c. • Limits can be investigated using a table of values or the graph of a function. Choosing values of x that are closer and closer to c, we observe the resulting values of f(x) to see if it appears a limiting value exists. • The lim f 1x2 may exist even if f(x) is not defined at c, if f(x) is defined at c but f 1c2  L, or if f(x) is defined xSc

at c and f 1c2  L.

• We can also define a left-hand limit and a right-hand limit: lim f 1x2  L means values of f(x) can be made xSc

arbitrarily close to L by taking values of x sufficiently close to c and to the left of c. The right-hand limit is similarly defined. • A limit can fail to exist in one of three ways. dne B , (1) the left-hand limit is not equal to the right-hand limit: lim f 1x2  A LHRH (2) as x S c, f (x) grows without bound: lim f 1x2  A dne B , or

xSc

q

(3) as x S c, there is no fixed number L that f(x) approaches: lim f 1x2  A dne B . xSc

xSc

L

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Summary and Concept Review

1067

EXERCISES 1. Estimate the following limits (if they exist) using the tables shown. a. lim f 1x2 b. lim f 1x2 c. lim f 1x2 xS2

xS2

xS2

x

f(x)

x

1.5

2.5

2.5

5.5

1.6

2.6

2.4

5.4

1.7

2.7

2.3

5.3

1.8

2.8

2.2

5.2

1.9

2.9

2.1

5.1

1.99

2.99

2.01

5.01

2.001

5.001

1.999 2.999 y  2.999

f(x)

y  5.001

Use a table of values to evaluate each limit. x2  2x  15 2. lim 2 xS5 x  6x  5

SECTION 11.2

x4 x  2 4. lim f 1x2, f 1x2  • x xS2 x2  1 x 7 2

sin12x  32 3. lim3 xS 2 2x  3

The Properties of Limits

KEY CONCEPTS • When evaluating limits using a table, great care must be exercised as results can sometimes be misleading. Using a graph and some common sense will sometimes help. • Limits possess certain properties that seem plausible using a table of values. These include the sum, difference, product, and quotient properties, as well as the power and root properties. For a review of these properties, see page 1037. • Once the limit properties have been established, they can be used to evaluate the limit of many different kinds of functions. • These four basic limits play an important role in the evaluation of general limits. k k (1) lim k  k (2) lim x  c (3) lim xk  ck (4) lim 1x  1c, if c  0 when k is even xSc

xSc

xSc

xSc

• If the limit properties lead to an undefined or indeterminate expression, a limit may still exist and other means should be used to investigate.

EXERCISES Determine the following limits using limit properties. 5.

lim 2x3  5x  1

xS3

SECTION 11.3

6. lim xS7

111  x

7. lim

2x  3x  1 3

2

xS1

1x 1x2  12x  92 5

8. lim xS1

22  x  1 1x

Continuity and More on Limits

KEY CONCEPTS • Generally, a function is continuous if you can draw the entire graph without lifting your pencil. • Specifically, a function f is continuous at c if these three conditions are met: (1) c is in the domain of f (2) lim f 1x2  lim f 1x2 (3) lim f 1x2  f 1c2 . xSc

xSc

xSc

• The continuity of a function can be interupted by (1) asymptotic, (2) jump (non-removable), or (3) removable discontinuities. • If a function is continuous at c, the limit lim f 1x2 can be found by direct substitution. xSc

• Polynomial functions are continuous for x  , while rational, root, and trigonometric functions are continuous everywhere they are defined.

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CHAPTER 11 Bridges to Calculus: An Introduction to Limits

• Using algebraic methods, a limit that cannot be evaluated by direct substitution can sometimes be rewritten in a form where a direct substitution becomes possible. • Limits can also occur as x becomes infinitely large: lim f 1x2  L, provided f (x) can be made arbitrarily close xS  q to L, by taking values of x sufficiently large. f (x)

EXERCISES 9. Name the x-value(s) where the graph has a. an asymptotic discontinuity b. a jump discontinuity c. a removable discontinuity Use the graph of f to evaluate the following limits. 10. lim f 1x2 11. lim f 1x2 12. lim f 1x2 xSq

13.

lim

xS3

f 1x2

SECTION 11.4

xS3

14. lim f 1x2 xS3

xS3

15. lim f 1x2

y 5 4 3 2 1

54321 1 2 3 4 5

1 2 3 4 5 x

xS1

Applications of Limits: Instantaneous Rates of Change and the Area under a Curve

KEY CONCEPTS • The rate of change of f (x) in the interval [x1, x2] can be approximated using the rate-of-change formula: f 1x2 2  f 1x1 2 ¢f .  x2  x1 ¢x • The rate of change of f (x) in the arbitrary interval [x, x  h] can be approximated using the difference quotient: f 1x  h2  f 1x2 . h • The instantaneous rate of change of f at point x is the limit of the difference quotient (if it exists): f 1x  h2  f 1x2 . f 1x2  lim hS0 h • Graphically, the slope of the secant line gives a better and better approximation for the instantaneous rate of change as h S 0, and we define the instantaneous rate of change as the limiting value of this slope: f 1x  h2  f 1x2 . mtan  lim msec  lim hS0 hS0 h • For f 1x2 7 0 in the first quadrant, the area under the graph of f in the interval [0, b] can be found using the b rectangle method and n rectangles of equal width. The width of each rectangle will be , the height of each is n b found by evaluating f in increments of : n n n b b LW  lim f a ib . A  lim n n nSq i1 nSq i1





EXERCISES Find the limit of the difference quotient as h S 0 to determine the instantaneous rate of change for the functions given. 1 16. f 1x2  x2  5x  2 17. g1x2  12x  1 18. v1x2  x3 19. Take the limit of a difference quotient to find 20. Use the rectangle method to find the area in QI, the slope of a line drawn tangent to under the graph of f 1x2  x2  6x  9 in the 2 f 1x2  x  3x at x  4. interval [0, 3].

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Mixed Review

1069

MIXED REVIEW 1. Express the following statements using limit notation: a. As x approaches a from the right, f(x) approaches b. b. As x increases without bound, f (x) approaches b. c. As x approaches a from both sides, f (x) approaches b. d. As x approaches a from the left, f (x) increases without bound.

For Exercises 14 and 15, use the graphs shown to evaluate each limit. If any computation is not possible, state why.

54321 1 2 3 4 5

Use a table of values to evaluate the following limits. 2x2  7x 3. lim xS0 sin x

sin x 2. lim xS0 x  tan x

4. Determine the limiting value of g1x2 

x3  8 x2

as x S 2 from the right.

5. Given f 1x2  sin csc x, find lim f 1x2 . xS

For Exercises 6 and 7, use the following information to evaluate the limits: lim f 1x2  5

xSc

6. lim xSc

f 1x2  g1x2 h1x2

2 3 6f 1x2 7. lim g1x2  xSc A h1x2

lim g1x2  13

xSc

lim h1x2 

xSc

y

y 5 4 3 2 1

5 4 3 2 1 1 2 3 4 5 x

f (x)

14.

54321 1 2 3 4 5

lim f 1x2  lim g1x2 

xS2 

15. lim c xS2

f 1x2

1g1x2

xS1

1 2 3 4 5 x

g(x)

lim 2f 1x2

xS2 

d

16. A little boy throws a rock straight down off a bridge into a river. The rock’s height in feet, t sec after it is released, is modeled by the function d1t2  16t2  10t  85. Compute the limit of the difference quotient for d, to obtain a function d (t) that gives the instantaneous velocity of the rock at time t. 17. What is the rock’s instantaneous velocity at a. t  1 sec? b. t  2 sec? c. t  0 sec (when it was released)?

Evaluate each limit using the limit properties. If a limit does not exist, state why. 1  x2 3x 2 8. lim1 1x  6x  22 9. lim xS 2 xS2 5x2  x 2 x 6 10. lim 2 xS16 x  6

18. The spreading of a particularly juicy bit of gossip at a local college can be modeled by the function p1t2  0.1 110t, where p is the percentage of faculty/staff “in the know” after t days. Find the limit of the difference quotient for p, to obtain a function p (t) that represents the instantaneous rate of change of percentage of faculty who have heard the gossip at time t.

Evaluate the following limits.

19. Referring to Exercise 18, what is the instantaneous rate of change in this percentage after 2 a. day? b. 1.6 days? c. 5 days 5

x 8 2 xS2 x  4 1x  1  3 12. Evaluate lim xS10 x  10 3

11. Evaluate lim

13. Evaluate the limits by dividing the numerator and denominator by the highest power of x occurring in x3  2x  1 the denominator: lim xSq 2x3  5

20. Find the area in QI, under the graph of g1x2  10x  2x2 in the interval [0, 5]. Use the rectangle method and n subintervals of equal width.

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11-48

CHAPTER 11 Bridges to Calculus: An Introduction to Limits

PRACTICE TEST 1. Write the expression in words, as though you were reading it out loud: lim f 1x2  10 xS5

2. Fill in the blanks with the correct word: lim f 1x2  L, xSc

means that values of _____ can be made arbitrarily close to _____, by taking values of x _______ close to _____. State true or false and justify your answer. x2  1 , lim f 1x2 cannot exist since 3. For f 1x2  x  1 xS1 x  1 is not in the domain. x  1, x  1 , lim f 1x2 must exist x  1, x 6 1 xS1 since x  1 is in the domain.

4. For f 1x2  e

5. For g1x2  1x  1  2, state which of the following one-sided limits exist. Justify your answer and then find the limit. a. lim 1 1x  1  22 xS1

b. lim 1 1x  1  22 xS1

6. Match each expression to its conclusion. 1 a. lim cosa b I. A dne q B x xS0 5x b. lim 2 II. A dne L B xS2 x  4x  4 sin x dne B c. lim III. A LHRH x xS0 2x2  2x d. lim IV. 1 xS1 2x2  2x  1 Determine the following using the graphs of f (x) and g(x) shown. 7. a.

lim

xS1

f 1x2

8. a. lim g1x2 xS4

9. a. f 102  g102 b. g152 f 142 10. a. [g(1)]

2

b. lim f 1x2 xS3

b. lim g1x2 xS3

f 142 b. g132

11. a. lim 3f 1x2  g1x2 4 xS4

b. lim f 1x2g1x2 xS1

y 5 4 3 2 1

f (x)

54321 1 2 3 4 5

1 2 3 4 5 x

y

g(x)

5 4 3 2 1

54321 1 2 3 4 5

1 2 3 4 5 x

12. a.

lim 1g1x2

xS5 

b.

lim 3g1x2 4 2

xS1

13. Evaluate the following limits using a table. If the limit does not exist, state why. cos x  1 x3  1 a. lim b. lim x xS0 xS1 21x  12 2 sin13x2 sin x c. lim d. lim xS0 2x  tan x xS0 3x 14. Each of the following limits does not exist. State why. x2  1 1 a. lim b. lim tana b x xS1 x  1 xS0 2 x 1 c. lim  d. lim 1x  4 xS2 x2 xS3 Evaluate each limit. 15. lim xS1

17. lim

x3  1 x1

16.

1x  5 x  25

18. lim

lim xSq

3x2  5x  4 2x2  x  3

cos12x2

cos2 x 19. A controlled explosion at a strip mine causes a huge shower of dirt and rock. The height of any debris that has been projected vertically upward by the blast can be modeled by the function d1t2  16t2  224t, where d(t) is the height in feet after t sec. (a) Compute the limit of the difference quotient for d to find a function d (t) that gives the instantaneous velocity of the debris at time t. (b) Find the velocity at t  2, 6, 7, and 11, and explain the significance of each result. xS25

xS0

20. In preparation for another sold-out performance by the Danish pianist and comedian Victor Borge (Phonetic Punctuation), stage hands move a large grand piano 15 ft across the main stage and then carefully into position. The force applied can be modeled by the function f 1x2  225  x2 (the force is much greater as they start than when they finish). Since Work  Force # Distance, the area under the curve from 0 to x represents the amount of work done in foot-pounds. Use the rectangle method to (a) find an expression representing the amount of work done in moving the piano the first 6 ft, and (b) find the amount of work done in the interval [0, 6].

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11-49

Cumulative Review Chapters 1–11

1071

C U M U L AT I V E R E V I E W C H A P T E R S 1 – 1 1 1. Find all zeroes, real and complex: f 1x2  2x3  3x2  9  6x

2. Graph the function f 1x2  2x2  12x  15 by completing the square. 3. Solve the system using elimination. x  3y  2z  6 • 2x  y  z  2 3x  4y  2z  3 4. Write as a sum/difference of logarithmic terms. 2

x3 2x2  3 ln c d 12x  12 4 5. Standing 134 ft from the St. Louis Arch, the angle of elevation to the top of the arch is 78°. How tall is the arch?

14. An airline pilot is having to fight a stiff crosswind to stay on course. If the plane is flying at 250 mph on a heading of 20°, and the wind is from the west at 50 mph, what would be the course and heading of the plane if no course corrections are made? 15. Graph the piecewise-defined function given, then state whether its graph is continuous. 1x  22 2 x  0 f 1x2  e x 7 0 x2 16. Solve the following system using inverse matrices and your calculator. w  x  y  z  2 2w  x  3y  z  9 μ w  x  y  z  4 3w  2x  2y  3z  19

6. Verify that a

17. Solve for x: 3e2x1  28.08.

7. Find the sum of the infinite series 1 1 1 1    p 3 9 27 81

19. Evaluate the limit:

8 15 ,  b is a point on the unit circle, 17 17 then use it to find the value of sin , cos , and tan .

8. Use the Guidelines for Graphing Rational Functions x . to graph v1x2  2 x 9 9. Write f 1x2  x3  2x2  5x  6 in completely factored form, then graph the function. 10. Evaluate the following limits: 0x  1 0 0x  1 0 a. lim 2 and lim 2 xS1 xS1 x 1 x 1 0x  1 0 b. lim 2 xS1 x  1 11. For z  2  2i, use DeMoivre’s theorem to find z6  12  2i2 6.

12. Solve the equation 2  13 tan 1  12  3. Find all solutions in [0, 2). 13. I am a heavy pencil user. My new pencils are 19 cm long, but after 30 days of heavy use they now average only 9 cm in length. Assuming their length decreases linearly, (a) find a function L(x) that models the length of my pencils after x days of use. (b) Use the model to find the length of a pencil I’ve been using for 15 days. (c) If one of my pencils is 11 cm long, how many days has it been in use?

18. Graph the function y  2 cos ax 

lim

 b. 4

6x

xSq 24x2

5 20. Verify the following is an identity: sin4x  cos4x  2 sin2x  1 21. Decompose the rational expression

x2  x  4 into x2  x  2

a sum of partial fractions. 22. Write each expression in exponential form: a. 3  log 0.001 b. ln x  2 23. Find all real solutions for  3500  7000 sin a4  b  10. 6 24. Identify and sketch the graph of the relation defined 8 . by r  4  3 sin  25. Find the equation of the ellipse with vertices at (4, 1) and (6, 1), and foci at (3, 1) and (5, 1). 26. Graph the hyperbola by completing the square in x and y, then writing the equation in standard form: x2  4y2  24y  6x  43  0.

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CHAPTER 11 Bridges to Calculus: An Introduction to Limits

Determine the following using the graph of f (x) shown: 27. a. b.

lim

xS3

lim

xS3

f 1x2

f 1x2

c. lim f 1x2 xS2

d. lim f 1x2 xS5

28. a. f 132 f 102 b. f 122  f 142 c. f 112  f 132 f 102 d. f 122

y

f (x)

5 4 3 2 1

54321 1 2 3 4 5

1 2 3 4 5 x

11-50

29. Use the rectangle method to find the area under the 1 graph of f 1x2   x2  9 for x  30, 4 4 . 4 30. Hi’ilani has to reset her five-digit PIN number. How many PINs are possible if no repetitions are allowed and two common vowels must be followed by three nonzero digits?

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Precalculus—

APPENDIX I

A Review of Basic Concepts and Skills APPENDIX OUTLINE A

The Language, Notation, and Numbers of Mathematics A-1

B

Algebraic Expressions and the Properties of Real Numbers A-8

C

Exponents, Scientific Notation, and a Review of Polynomials A-14

D

Factoring Polynomials A-24

E

Rational Expressions A-31

F

Radicals and Rational Exponents A-40

A The Language, Notation, and Numbers of Mathematics In Section A you will review:

A. Sets of numbers, graphing real numbers, and set notation

B. Inequality symbols and order relations

C. The absolute value of a real number

D. The order of operations

A. Sets of Numbers, Graphing Real Numbers, and Set Notation To effectively use mathematics as a problem-solving tool, we must first be familiar with the sets of numbers used to quantify (give a numeric value to) the things we investigate. Only then can we make comparisons and develop the equation models that lead to informed decisions. Our primary focus here is the set of real numbers, symbolized by the letter  written in the special font shown. This set is made up of several smaller sets called subsets, which are described in Table AI.1. Note the members or elements of a set are grouped within braces “{ },” and separated by commas. The three dots “. . .” seen in some subsets is called an ellipsis, and indicates the list of elements is infinite. Table AI.1 Subsets of  Subset

Description

Additional Examples

Natural numbers 

numbers used to count physical objects: {1, 2, 3, 4, 5, . . . }

19, 52, 1089

Whole numbers 

natural numbers with the 0 element: {0, 1, 2, 3, 4, 5, . . . }

0, 31, 107, 3215

Integers 

whole numbers and their opposites: { . . . , 3, 2, 1, 0, 1, 2, 3, . . . }

725, 48, 7, 0, 7, 48, 725

Rational numbers 

numbers that can be wriiten in fraction form, as the ratio of two p integers , q  0 q

Irrational numbers 

numbers with a nonrepeating and nonterminating decimal form; numbers that cannot be written as a ratio of two integers



19 1 , 7, 0.4, , 29 2 3

  3.141529 p 13   0.866025 p 2 0.02002000200002 p

A-1

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Appendix I A Review of Basic Concepts and Skills

EXAMPLE 1



Graphing Rational and Irrational Numbers Graph the each number by writing the decimal form and estimating its location between two integers (round to hundredths as needed). 2 3 a.  b. c. 119 d.  3 2

Solution



2 a.   0.67 3

b.

3  1.5 2

c. 119  4.36

0.67 . . . 1

0

 19

1.5 1

d.   3.14

2

3

5 . . .

4

Now try Exercises 1 through 8



Figure AI.1 illustrates the relationship between these sets of numbers. Notice how each subset appears “nested” in a larger set. R (real): All rational and irrational numbers Q (rational): {qp, where p, q  z and q  0}

H (irrational): Numbers that cannot be written as the ratio of two integers; a real number that is not rational. 2, 7, 10, 0.070070007... and so on.

Z (integer): {. . . , 2, 1, 0, 1, 2, . . .} W (whole): {0, 1, 2, 3, . . .} N (natural): {1, 2, 3, . . .}

Figure AI.1

EXAMPLE 2



Solution



A. You’ve just reviewed sets of numbers, graphing real numbers, and set notation

Identifying Numbers

List the numbers in set A  52, 0, 5, 17, 12, 23, 4.5, 121, , 0.756 that belong to a.  b.  c.  d.  a. 2, 0, 5, 12, 23, 4.5, 0.75   c. 0, 5, 12  

b. 17, 121,    d. 2, 0, 5, 12   Now try Exercises 9 through 12



B. Inequality Symbols and Order Relations We compare numbers of different size using inequality notation, known as the greater than 172 and less than 162 symbols. Note that 4 6 3 is the same as saying 4 is to the left of 3 on the number line. In fact, on a number line, any given number is smaller than any number to the right of it (see Figure AI.2). 4 3 2 1

Figure AI.2

a

0 1 4  3

ab

2

3

b

4

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A-3

Order Property of Real Numbers Given any two real numbers a and b. 1. a 6 b if a is to the left of b on the number line. 2. a 7 b if a is to the right of b on the number line. Inequality notation is used with numbers and variables to write mathematical statements. A variable is a symbol, commonly a letter of the alphabet, used to represent an unknown quantity. Over the years x, y, and n have become most common, although any letter (or symbol) can be used. Often we’ll use variables that remind us of the quantities they represent, like L for length, and D for distance. EXAMPLE 3



Writing Mathematical Models Using Inequalities Use a variable and an inequality symbol to represent the statement: “To hit a home run out of Jacobi Park, the ball must travel over three hundred twenty-five feet.”

Solution



Let D represent distance: D 7 325 ft. Now try Exercises 13 through 16



In Example 3, note the number 325 itself is not a possible value for D. If the ball traveled exactly 325 ft, it would hit the fence and stay in play. Numbers that mark the limit or boundary of an inequality are called endpoints. If the endpoint(s) are not included, the less than 162 or greater than 172 symbols are used. When the endpoints are included, the less than or equal to symbol 1 2 or the greater than or equal to symbol 12 is used. The decision to include or exclude an endpoint is often an important one, and many mathematical decisions (and real-life decisions) depend on a clear understanding of the distinction.

B. You’ve just reviewed inequality symbols and order relations

C. The Absolute Value of a Real Number In some applications, our primary interest is simply the size of n, rather than its sign. This is called the absolute value of n, denoted n, and can be thought of as its distance from zero on the number line, regardless of the direction (see Figure AI.3). Since distance is always positive or zero, n  0. 4  4

Figure AI.3

EXAMPLE 4



4 3 2 1

3  3 0

1

2

3

4

Absolute Value Reading and Reasoning In the table shown, the absolute value of a number is given in column 1. Complete the remaining columns.

Solution



Column 1 (In Symbols)

Column 2 (Spoken)

Column 3 (Result)

Column 4 (Reason)

7.5

“the absolute value of seven and five-tenths”

7.5

the distance between 7.5 and 0 is 7.5 units

2

“the absolute value of negative two”

2

the distance between 2 and 0 is 2 units

6

“the opposite of the absolute value of negative six”

6

the distance between 6 and 0 is 6 units, the opposite of 6 is 6

Now try Exercises 17 through 20



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Appendix I A Review of Basic Concepts and Skills

Example 4 shows the absolute value of a positive number is the number itself, while the absolute value of a negative number is the opposite of that number (recall that n is positive if n itself is negative). For this reason the formal definition of absolute value is stated as follows. Absolute Value For any real number n, 0n 0  e

n n

if if

n0 n 6 0

The concept of absolute value is often used to find the distance between two numbers on a number line. For instance, we know the distance between 2 and 8 is 6. Using absolute values, we write 08  2 0  06 0  6, or 0 2  8 0  0 6 0  6. Generally, if a and b are two numbers on the real number line, the distance between them is 0a  b 0 or 0b  a 0 . EXAMPLE 5



Using Absolute Value to Find the Distance between Points Find the distance between 5 and 3 on the number line.

Solution



0 5  3 0  08 0  8

0 3  152 0  08 0  8.

or

Now try Exercises 21 through 30

C. You’ve just reviewed the absolute value of a real number



D. The Order of Operations When a number is repeatedly multiplied by itself as in (10)(10)(10)(10), we write it using exponential notation as 104. The number used for repeated multiplication (in this case 10) is called the base, and the superscript number is called an exponent. The exponent tells how many times the base occurs as a factor, and we say 104 is written in exponential form. Numbers that result from squaring an integer are called perfect squares, while numbers that result from cubing an integer are called perfect cubes (see Table AI.2). Table AI.2 Perfect Squares

EXAMPLE 6



2

Perfect Cubes

N

2

N

N

N3

N

N

1

1

7

49

1

1

2

4

8

64

2

8

3

9

9

81

3

27

4

16

10

100

4

64

5

25

11

121

5

125

6

36

12

144

6

216

Evaluating Numbers in Exponential Form Write each exponential in expanded form, then determine its value. a. 43 b. 162 2 c. 62

Solution



a. 43  4 # 4 # 4  64 c. 62  16 # 62  36

d. 1 23 2 3

b. 162 2  162 # 162  36 3 8 d. A 23 B  23 # 23 # 23  27

Now try Exercises 31 and 32



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Section A The Language, Notation, and Numbers of Mathematics

A-5

Examples 6(b) and 6(c) illustrate an important distinction. The expression 162 2 is read, “the square of negative six” and the negative sign is included in both factors. The expression 62 is read, “the opposite of six squared,” and the square of six is calculated first, then made negative.

Square Roots and Cube Roots Index

兹A 3

2 For the square root operation, either the 1 or 1 notation can be used. The 1 symbol is called a radical, the number under the radical is called the radicand, and the small case number used is called the index. The index tells how many factors are needed to obtain the radicand. In general, 1a  b only if b2  a. All numbers greater than zero have one positive and one negative square root. The positive or principal square root of 49 is 7 1 149  72 since 72  49. The negative square root of 49 is 3 7 1 149  7). The cube root of a number has the form 1 a  b, where 3 3 3 3 b  a. This means 1 27  3 since 3  27, and 1 8  2 since 122 3  8. The cube root of a real number has one unique real value. In general, we have the following:

Radical

Radicand

Square Roots

Cube Roots

1a  b if b2  a

3 1 a  b if b3  a

This indicates that

This indicates that

1a # 1a  a

3 3 3 1 a# 1 a# 1 aa

1a  02

or 1 1a2 2  a EXAMPLE 7



1a  2

3 or 1 1a2 3  a

Evaluating Square Roots and Cube Roots Determine the value of each expression. 3 9 a. 149 b. 1 c. 216 125

Solution



d. 116

e. 125

a. 7 since 7 # 7  49 b. 5 since 5 # 5 # 5  125 3 # 3 9 3 c. 4 since 4 4  16 d. 4 since 116  4 # e. not a real number since 5 5  152152  25 Now try Exercises 33 through 38



When basic operations are combined into a larger mathematical expression, we use a specified priority or order of operations to evaluate them. The Order of Operations 1. Simplify within grouping symbols (parentheses, brackets, braces, etc.). If there are “nested” symbols of grouping, begin with the innermost group. If a fraction bar is used, simplify the numerator and denominator separately. 2. Evaluate all exponents and roots. 3. Compute all multiplications or divisions in the order they occur from left to right. 4. Compute all additions or subtractions in the order they occur from left to right.

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Appendix I A Review of Basic Concepts and Skills

EXAMPLE 8



Evaluating Expressions Using the Order of Operations Simplify using the order of operations: a. 5 2 # 3 # 0.075 12 15 c. 7500 a1

b 12

Solution



a. 5 2 # 3  5 6  11

b. 8 36  4112  32 2 4.5182  3 d. 3 1125 23

multiplication before addition result

b. 8 36  4112  3 2  8 36  4112  92  8 36  4132  8 9132  8 27  35 2

d.

4.5182  3

D. You’ve just reviewed the order of operations

3 2 125 23 36  3  5 8 39  13  3

12  9  3 division before multiplication multiply result

#

0.075 12 15 b 12 #  750011.006252 12 15  750011.006252 180  7500(3.069451727)  23,020.89

c. 7500 a1

simplify within parentheses

original expression simplify within the parenthesis (division before addition) simplify the exponent exponents before multiplication result (rounded to hundredths) original expression

simplify the numerator and denominator

combine terms result

Now try Exercises 39 through 54



A EXERCISES 

DEVELOPING YOUR SKILLS

Convert to decimal form and graph by estimating the number’s location between two integers.

1.

4 3

2. 78

3. 259

4. 156

Use a calculator to find the principal square root of each number (round to hundredths as needed). Then graph each number by estimating its location between two integers.

5. 7

6.

75 4

7. 3

8.

25 2

For the sets in Exercises 9 through 12:

a. List all numbers that are elements of (i) , (ii) , (iii) , (iv) , (v) , and (vi) .

b. Reorder the elements of each set from smallest to largest. c. Graph the elements of each set on a number line.

9. 51, 8, 0.75, 92, 5.6, 7, 35, 66

10. 57, 2.1, 5.73, 356, 0, 1.12, 78 6

11. 55, 149, 2, 3, 6, 1, 13, 0, 4, 6

12. 58, 5, 235, 1.75, 22, 0.6, , 72,2646 Use a descriptive variable and an inequality symbol 1, , , 2 to write a model for each statement.

13. To spend the night at a friend’s house, Kylie must be at least 6 yr old.

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Precalculus—

Section A The Language, Notation, and Numbers of Mathematics

14. Monty can spend at most $2500 on the purchase of a used automobile. 15. If Jerod gets no more than two words incorrect on his spelling test he can play in the soccer game this weekend. 16. Andy must weigh less than 112 lb to be allowed to wrestle in his weight class at the meet. Evaluate/simplify each expression.

18. 7.24

19. 4

20. 6

24. Write the statement two ways, then simplify. “The distance between 1325 and 235 is . . .” 25. What two numbers on the number line are five units from negative three? 26. What two numbers on the number line are three units from two? 27. If n is positive, then n is

.

28. If n is negative, then n is

. .



25 B 49

34. 

3 35. 1 8

3 36. 1 64

39. 24  1312

40. 45  1542

41. 456 112 2 43.

4 5

42. 118 134 2

 182

44. 15  12

45. 23  16 21

46. 34  78

Evaluate without a calculator, using the order of operations.

47. 12  10  2 5 132 2 48. 15  22 2  16  4 # 2  1 49. 51.

9 3 5 2  #a b B 16 5 3

3 2 9 25 50. a b  a b  2 4 B 64

4172  62

52.

6  149

5162  32 9  164

. Evaluate using a calculator (round to hundredths).

Without computing the actual answer, state whether the result will be positive or negative. Be careful to note what power is used and whether the negative sign is included in parentheses.

31. a. 172 2 c. 172 5

121 B 36

33. 

Perform the operation indicated without the aid of a calculator.

22. 4  7

23. Write the statement two ways, then simplify. “The distance between 7.5 and 2.5 is . . .”

30. If n 7 0, then 0 n 0 

Evaluate without the aid of a calculator.

38. What perfect cube is closest to 71?

Use the concept of absolute value to complete Exercises 23 to 30.

29. If n 6 0, then 0 n 0 

b. 73 d. 74

37. What perfect square is closest to 78?

17. 2.75 21. 3  162 

32. a. 172 3 c. 172 4

A-7

b. 72 d. 75

#

0.06 4 10 53. 2475a1

b 4 #

0.078 52 20 b 54. 5100 a1

52

APPLICATIONS

Use positive and negative numbers to model each situation, then compute.

55. At 6:00 P.M., the temperature was 50°F. A cold front moves through that causes the temperature to drop 3°F each hour until midnight. What is the temperature at midnight? 56. Most air conditioning systems are designed to create a 2° drop in the air temperature each hour. How long would it take to reduce the air temperature from 86° to 71°?

57. The state of California holds the record for the greatest temperature swing between a record high and a record low. The record high was 134°F and the record low was 45°F. How many degrees difference are there between the record high and the record low? 58. In Juneau, Alaska, the temperature was 17°F early one morning. A cold front later moved in and the temperature dropped 32°F by lunch time. What was the temperature at lunch time?

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Appendix I A Review of Basic Concepts and Skills

EXTENDING THE CONCEPT

59. Here are some historical approximations for . Which one is closest to the true value? Archimedes: 317 Aryabhata: 62,832 20,000

60. If A 7 0 and B 6 0, is the product A # 1B2 positive or negative? 61. If A 6 0 and B 6 0, is the quotient 1A  B2 positive or negative?

355 Tsu Ch’ung-chih: 113 Brahmagupta: 110

B Algebraic Expressions and the Properties of Real Numbers A. Algebraic Expressions and Mathematical Models

In Section B you will review how to:

A. Identify algebraic

An algebraic term is a collection of factors that may include numbers, variables, or expressions within parentheses. Here are some examples:

expressions and create mathematical models

(1) 3

B. Evaluate algebraic expressions

C. Identify and use properties of real numbers

D. Simplify algebraic expressions

EXAMPLE 1



(2) 6P

(3) 5xy

(4) 8n2

(5) n

(6) 21x 32

If a term consists of a single nonvariable number, it is called a constant term. In (1), 3 is a constant term. Any term that contains a variable is called a variable term. We call the constant factor of a term the numerical coefficient or simply the coefficient. An algebraic expression can be a single term or a sum or difference of terms. To avoid confusion when identifying the coefficient of each term, the expression can be rewritten using algebraic addition if desired: A  B  A 1B2. To identify the coefficient of a rational term, it sometimes helps to decompose the term, rewriting it 2  15 1n  22 and 2x  12x. using a unit fraction: n  5 The key to solving many applied problems is finding an algebraic expression that accurately models the situation given. First, we assign a variable to represent an unknown quantity, then build related expressions using words that suggest a mathematical operation. Translating English Phrases into Algebraic Expressions Assign a variable to the unknown number, then translate each phrase into an algebraic expression. a. twice a number, increased by five b. six less than three times the width

Solution

A. You’ve just reviewed how to identify algebraic expressions and how to create mathematical models



a. Let n represent the number. Then 2n represents twice the number, and 2n 5 represents twice a number, increased by five. b. Let W represent the width. Then 3W represents three times the width, and 3W  6 represents six less than three times the width. Now try Exercises 1 through 14



Identifying and translating such phrases when they occur in context is an important problem-solving skill. Note how this is done in Example 2.

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Precalculus—

Section B Algebraic Expressions and the Properties of Real Numbers

EXAMPLE 2



A-9

Creating a Mathematical Model The cost for a rental car is $35 plus 15 cents per mile. Express the cost of renting a car in terms of the number of miles driven.

Solution



Let m represent the number of miles driven. Then 0.15m represents the cost for each mile and C  35 0.15m represents the total cost for renting the car. Now try Exercises 15 through 26



B. Evaluating Algebraic Expressions We often need to evaluate expressions to investigate patterns and note relationships.

WORTHY OF NOTE In Example 3, note the importance of the first step in the evaluation process: replace each variable with open parentheses. Skipping this step could easily lead to confusion as we try to evaluate the squared term, since 32  9, while 132 2  9.

EXAMPLE 3

Evaluating a Mathematical Expression 1. Replace each variable with open parentheses ( ). 2. Substitute the given values for each variable. 3. Simplify using the order of operations. In this evaluation, it’s best to use a vertical format, with the original expression written first, the substitutions shown next, followed by the simplified forms and the final result. The numbers substituted or “plugged into” the expression are often called the input values, with the resulting values called outputs. 

Evaluating an Algebraic Expression Evaluate the expression x3  2x2 5 for x  3.

Solution



For x  3:

x3  2x2 5  132 3  2132 2 5  27  2192 5  27  18 5  40

substitute 3 for x

simplify: 132 3  27, 132 2  9 simplify: 2192  18 result

Now try Exercises 27 through 42 B. You’ve just reviewed how to evaluate algebraic expressions



For exercises that combine the skills from Examples 2 and 3, see Exercises 59 to 62.

C. Properties of Real Numbers While the phrase, “an unknown number times five,” is accurately modeled by the expression n5 for some number n, in algebra we prefer to write numerical coefficients before variable factors. When we reorder the factors as 5n, we’re using the commutative property of multiplication. A reordering of terms involves the commutative property of addition. The Commutative Properties Given that a and b represent real numbers: ADDITION:

a bb a

Terms can be combined in any order without changing the sum.

MULTIPLICATION:

a#bb#a

Factors can be multiplied in any order without changing the product.

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Appendix I A Review of Basic Concepts and Skills

Each property can be extended to include any number of terms or factors. While the commutative property implies a reordering or movement of terms (to commute implies back-and-forth movement), the associative property implies a regrouping or reassociation of terms. For example, the sum A 34 35 B 25 is easier to compute if we regroup the addends as 34 A 35 25 B . This illustrates the associative property of addition. Multiplication is also associative. The Associative Properties Given that a, b, and c represent real numbers: ADDITION:

MULTIPLICATION:

Terms can be regrouped.

Factors can be regrouped.

1a b2 c  a 1b c2

EXAMPLE 4



1a # b2 # c  a # 1b # c2

Simplifying Expressions Using Properties of Real Numbers Use the commutative and associative properties to simplify each calculation. a. 38  19 58 b. 32.5 # 11.22 4 # 10

Solution



a.

3 8

 19 58  19 38

5 8 3 5 8 8

2  19 1  19 1  18 b. 3 2.5 # 11.22 4 # 10  2.5 # 3 11.22 # 10 4  2.5 # 1122  30

commutative property associative property simplify result associative property simplify result

Now try Exercises 43 and 44



For any real number x, x 0  x and 0 is called the additive identity since the original number was returned or “identified.” Similarly, 1 is called the multiplicative identity since 1 # x  x. For any real number x, there is a real number x such that x 1x2  0. The number x is called the additive inverse of x, since their sum results in the additive identity. Similarly, the multiplicative inverse of any nonzero number x is 1x , since p q x # 1x  1 (the multiplicative identity). This property can also be stated as q # p  1 q p p 1p, q  02 for any rational number q. Note that q and p are reciprocals. The distributive property of multiplication over addition is widely used in a study of algebra, because it enables us to rewrite a product as an equivalent sum and vice versa. The Distributive Property of Multiplication over Addition Given that a, b, and c represent real numbers: a1b c2  ab ac A factor outside a sum can be distributed to each addend in the sum.

ab ac  a1b c2 A factor common to each addend in a sum can be “undistributed” and written outside a group.

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Precalculus—

Section B Algebraic Expressions and the Properties of Real Numbers

EXAMPLE 5



A-11

Simplifying Expressions Using the Distributive Property Apply the distributive property as appropriate. Simplify if possible. a. 71p 5.22

Solution



b. 12.5  x2

a. 71p 5.22  7p 715.22  7p 36.4

c. 7x3  x3

d.

1 5 n n 2 2

b. 12.5  x2  112.5  x2  112.52  1121x2  2.5 x 5 1 5 1 d. n na bn 2 2 2 2 6 a bn 2  3n

c. 7x3  x3  7x3  1x3  17  12x3  6x3

Now try Exercises 45 through 50

C. You’ve just reviewed how to identify and use properties of real numbers



D. Simplifying Algebraic Expressions Two terms are like terms only if they have the same variable factors (the coefficient is not used to identify like terms). For instance, 3x2 and 17x2 are like terms, while 5x3 and 5x2 are not. We simplify expressions by combining like terms using the distributive property, along with the commutative and associative properties.

EXAMPLE 6



Combining Like Terms Using the Distributive Property

Solution



712p2 12  11p2 32  14p2 7  1p2  3  114p2  1p2 2 17  32  114  12p2 4  13p2 4

Simplify the expression completely: 712p2 12  1p2 32 . original expression; note coefficient of 1 distributive property commutative and associative properties (collect like terms) distributive property result

Now try Exercises 51 through 58

The steps for simplifying an algebraic expression are summarized here: To Simplify an Expression D. You’ve just reviewed how to simplify algebraic expressions

1. Eliminate parentheses by applying the distributive property. 2. Use the commutative and associative properties to group like terms. 3. Use the distributive property to combine like terms.



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Appendix I A Review of Basic Concepts and Skills

B EXERCISES 

DEVELOPING YOUR SKILLS

Translate each phrase into an algebraic expression.

1. seven fewer than a number 2. x decreased by six 3. the sum of a number and four 4. a number increased by nine 5. the difference between a number and five is squared 6. the sum of a number and two is cubed 7. thirteen less than twice a number 8. five less than double a number 9. a number squared plus the number doubled 10. a number cubed less the number tripled 11. five fewer than two-thirds of a number 12. fourteen more than one-half of a number 13. three times the sum of a number and five, decreased by seven 14. five times the difference of a number and two, increased by six Create a mathematical model using descriptive variables.

15. The length of the rectangle is three meters less than twice the width. 16. The height of the triangle is six centimeters less than three times the base. 17. The speed of the car was fifteen miles per hour more than the speed of the bus. 18. It took Romulus three minutes more time than Remus to finish the race. 19. Hovering altitude: The helicopter was hovering 150 ft above the top of the building. Express the altitude of the helicopter in terms of the building’s height.

20. Stacks on a cruise liner: The smoke stacks of the luxury liner cleared the bridge by 25 ft as it passed beneath it. Express the height of the stacks in terms of the bridge’s height. 21. Dimensions of a city park: The length of a rectangular city park is 20 m more than twice its width. Express the length of the park in terms of the width. 22. Dimensions of a parking lot: In order to meet the city code while using the available space, a contractor planned to construct a parking lot with a length that was 50 ft less than three times its width. Express the length of the lot in terms of the width. 23. Cost of milk: In 2008, a gallon of milk cost two and one-half times what it did in 1990. Express the cost of a gallon of milk in 2008 in terms of the 1990 cost. 24. Cost of gas: In 2008, a gallon of gasoline cost one and one-half times what it did in 1990. Express the cost of a gallon of gas in 2008 in terms of the 1990 cost. 25. Pest control: In her pest control business, Judy charges $50 per call plus $12.50 per gallon of insecticide for the control of spiders and other insects. Express the total charge in terms of the number of gallons of insecticide used. 26. Computer repairs: As his reputation and referral business grew, Keith began to charge $75 per service call plus an hourly rate of $50 for the repair and maintenance of home computers. Express the cost of a service call in terms of the number of hours spent on the call. Evaluate each algebraic expression given x  2 and y  3.

27. 4x  2y 29. 2x 3y 2

28. 5x  3y 2

31. 2y2 5y  3 33. 13x  2y2 2 35.

12y 5 3x 1

37. 112y # 4

30. 5x2 4y2 32. 3x2 2x  5 34. 12x  3y2 2 36.

12x 132 3y 1

38. 7 # 127y

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Precalculus—

Section B Algebraic Expressions and the Properties of Real Numbers

Evaluate each expression for integers from 3 to 3 inclusive. What input(s) give an output of zero?

39. x2  3x  4

40. x2  2x  3

41. x3  6x 4

42. x3 5x 18

Rewrite each expression using the given property and simplify if possible.

43. Commutative property of addition a. 5 7 b. 2 n c. 4.2 a 13.6 d. 7 x  7 44. Associative property of multiplication a. 2 # 13 # 62 b. 3 # 14 # b2 c. 1.5 # 16 # a2 d. 6 # 156 # x2 Simplify by removing all grouping symbols (as needed) and combining like terms.

45. 51x  2.62 

A-13

47. 23 115p 92

2 48. 56 115 q 242

49. 23x 34x 50.

5 12 y

 38y

51. 31a2 3a2  15a2 7a2 52. 21b2 5b2  16b2 9b2 53. x2  13x  5x2 2

54. n2  15n  4n2 2

55. 13a 2b  5c2  1a  b  7c2

56. 1x  4y 8z2  18x  5y  2z2 57. 34 15n  42 58 1n 162 58. 23 12x  92 34 1x 122

46. 121v  3.22

APPLICATIONS

Translate each key phrase into an algebraic expression, then evaluate as indicated.

59. Cruising speed: A turbo-prop airliner has a cruising speed that is one-half the cruising speed of a 767 jet aircraft. (a) Express the speed of the turbo-prop in terms of the speed of the jet, and (b) determine the speed of the airliner if the cruising speed of the jet is 550 mph. 60. Softball toss: Macklyn can throw a softball twothirds as far as her father. (a) Express the distance that Macklyn can throw a softball in terms of the distance her father can throw. (b) If her father can throw the ball 210 ft, how far can Macklyn throw the ball? 61. Dimensions of a lawn: The length of a rectangular lawn is 3 ft more than twice its width. (a) Express the length of the lawn in terms of the width. (b) If the width is 52 ft, what is the length? 62. Pitch of a roof: To obtain the proper pitch, the crossbeam for a roof truss must be 2 ft less than three-halves the rafter. (a) Express the length of the crossbeam in terms of the rafter. (b) If the rafter is 18 ft, how long is the crossbeam? 63. Postage costs: In 2010, a first class stamp cost 27¢ more than it did in 1978. Express the cost of a 2010 stamp in terms of the 1978 cost. If a stamp cost 15¢ in 1978, what was the cost in 2004?

64. Minimum wage: In 2009, the federal minimum wage was $4.95 per hour more than it was in 1976. Express the 2009 wage in terms of the 1976 wage. If the hourly wage in 1976 was $2.30, what was it in 2009? 65. Repair costs: The TV repairman charges a flat fee of $43.50 to come to your house and $25 per hour for labor. Express the cost of repairing a TV in terms of the time it takes to repair it. If the repair took 1.5 hr, what was the total cost? 66. Repair costs: At the local car dealership, shop charges are $79.50 to diagnose the problem and $85 per shop hour for labor. Express the cost of a repair in terms of the labor involved. If a repair takes 3.5 hr, how much will it cost? 67. If C must be a positive odd integer and D must be a negative even integer, then C2 D2 must be a: a. positive odd integer. b. positive even integer. c. negative odd integer. d. negative even integer. e. cannot be determined. 68. Historically, several attempts have been made to create metric time using factors of 10, but our current system won out. If 1 day was 10 metric hours, 1 metric hour was 10 metric minutes, and 1 metric minute was 10 metric seconds, what time would it really be if a metric clock read 4:3:5? Assume that each new day starts at midnight.

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Precalculus—

C Exponents, Scientific Notation, and a Review of Polynomials A. The Properties of Exponents

In Section C you will review how to:

A. Apply properties of

The Product and Power Properties

exponents

Recall that bn indicates the number b is multiplied by itself n times. There are two properties that follow immediately from this definition. When b3 is multiplied by b2, we have an uninterrupted string of five factors: b3 # b2  1b # b # b2 # 1b # b2, which can be written as b5. This is an example of the product property of exponents.

B. Perform operations in scientific notation

C. Identify and classify polynomial expressions

D. Add and subtract

Product Property Of Exponents

polynomials

E. Compute the product of

For any base b and positive integers m and n:

two polynomials

bm # bn  bmn

F. Compute special products

A special application of the product property uses repeated factors of the same exponential term, as in 1x2 2 3. Using the product property, we have 1x2 21x2 21x2 2  x6. Notice the same result can be found more quickly by multiplying the inner exponent # by the outer exponent: 1x2 2 3  x2 3  x6. We generalize this idea to state the power property of exponents.

WORTHY OF NOTE In this statement of the product property and the exponential properties that follow, it is assumed that for any expression of the form 0m, m 7 0 hence 0m  0.

EXAMPLE 1

Power Property of Exponents For any base b and positive integers m and n:

1bm 2 n  bm n #



Multiplying Terms Using Exponential Properties Compute each product. a. 4x3 # 12x2 b. 1 p3 2 2 # 1 p4 2 2

Solution



a. 4x3

b. 1 p3 2 2

# 12x2  14 # 12 21x3 # x2 2

#

 122 1x32 2  2 x5 4 2 1 p 2  p6 # p8  p68  p14

commutative and associative properties product property; simplify result power property product property result

Now try Exercises 1 through 6



When the power property is extended to include more than one factor within the parentheses, we obtain the product to a power property. We can also raise a quotient of exponential terms to a power. The result is called the quotient to a power property. Product to a Power Property For any bases a and b, and positive integers m, n, and p: 1ambn 2 p  amp # bnp

A-14

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Precalculus—

A-15

Section C Exponents, Scientific Notation, and a Review of Polynomials

Quotient to a Power Property For any bases a and b  0, and positive integers m, n, and p: a

EXAMPLE 2



am p amp b  np bn b

Simplifying Terms Using the Power Properties Simplify using the power property (if possible): 2 5 a. 13a2 2 b. a a3bb 2

Solution



a. 13a2 2  132 2 # 1a1 2 2  9a2

5 3 2 5 2 32 2 b. a a bb  a b 1a 2 b 2 2 

25 6 2 ab 4

Now try Exercises 7 through 12



Applications of exponents sometimes involve linking one exponential term with another using a substitution. The result is then simplified using exponential properties. EXAMPLE 3



Applying the Power Property after a Substitution The formula for the volume of a cube is V  S3, where S is the length of one edge. If the length of each edge is 2x2: a. Find a formula for volume in terms of x. b. Find the volume if x  2.

Solution



a. V  S3 ↓2

 12x 2  8x6

S  2 x2 3

2x2 2x2

2x2 b. For V  8x6, V  8122 6 substitute 2 for x  8 # 64 or 512 122 6  64 The volume of the cube would be 512 units3.

Now try Exercises 13 and 14



The Quotient Property x5 x#x#x#x#x  or x3, the exponent of the final result appears to be the difference x#x x2 between the exponent in the numerator and the exponent in the denominator. This seems reasonable since the subtraction would indicate a removal of the factors that reduce to 1. Regardless of how many factors are used, we can generalize the idea and state the quotient property of exponents. For

Quotient Property of Exponents For any base b  0 and positive integers m and n:

bm  bmn bn

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Appendix I A Review of Basic Concepts and Skills

Zero and Negative Numbers as Exponents If the exponent of the denominator is greater than the exponent in the numerator, the x2 quotient property yields a negative exponent: 5  x25  x3. To help understand x what a negative exponent means, let’s look at the expanded form of the expression: x2 x#x 1  # # # #  3 # A negative exponent can literally be interpreted as “write 5 x x x x x x x the factors as a reciprocal.” A good way to remember this is three factors of 2

!

!

23

written as a reciprocal

1 1 23  3 1 8 2

Since the result would be similar regardless of the base used, we can generalize this idea and state the property of negative exponents. WORTHY OF NOTE

Property of Negative Exponents

The use of zero as an exponent should not strike you as strange or odd; it’s simply a way of saying that no factors of the base remain, since all terms have 23 been reduced to 1. For 3 , 2

For any base b  0 and integer n:

1

1

bn 

1 bn

1 bn

 bn

a n b n a b a b ;a0 a b

x3 x3 by division, and  1  x33  x0 using the x3 x3 quotient property, we conclude that x0  1 as long as x  0. We can also generalize this observation and state the meaning of zero as an exponent. In words the property says, any nonzero quantity raised to an exponent of zero is equal to 1. Finally, when we consider that

1

8 2# 2 # 2 we have  1, or  1, 8 2#2#2 33 0 or 2  2  1.

Zero Exponent Property For any base b  0: b0  1

EXAMPLE 4

Solution





Simplifying Expressions Using Exponential Properties Simplify using exponential properties. Answer using positive exponents only. 2a3 2 a. a 2 b b. 13hk2 2 3 16h2k3 2 2 b 12m2n3 2 5 c. 13x2 0  3x0  32 d. 14mn2 2 3 3 2 2 2 2a b a. a 2 b  a 3 b property of negative exponents b 2a 

1b2 2 2

22 1a3 2 2 b4  6 4a

power property

result

b. 13hk2 2 3 16h2k3 2 2  133h3k6 2 162h4k6 2  33 # 62 # h34 # k66 27h7k0  36 3h7  4

power property product property simplify a62  result 1k 0  12

1 62



1 b 36

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Precalculus—

Section C Exponents, Scientific Notation, and a Review of Polynomials

WORTHY OF NOTE Notice in Example 4(c), we have 13x2 0  13 # x2 0  1, while 3x0  3 # x0  3112. This is another example of operations and grouping symbols working together: 13x2 0  1 because any quantity to the zero power is 1. However, for 3x0 there are no grouping symbols, so the exponent 0 acts only on the x and not the 3.

c. 13x2 0  3x0  32  1  3112  4

d.

12m2n3 2 5 14mn2 2 3



1 32

1 9

1 37 4  9 9 122 5 1m2 2 5 1n3 2 5

43m3 1n2 2 3 32m10n15  64m3n6 m7n9  2

A-17

zero exponent property; property of negative exponents

simplify

result

power property

simplify

quotient property

Now try Exercises 15 through 46



Summary of Exponential Properties

A. You’ve just reviewed how to apply properties of exponents

For real numbers a and b, and integers m, n, and p (excluding 0 raised to a nonpositive power) bm # bn  bmn Product property: # Power property: 1bm 2 n  bm n Product to a power: 1ambn 2 p  amp # bnp am p amp Quotient to a power: a n b  np 1b  02 b b m b Quotient property:  bmn 1b  02 bn Zero exponents: b0  1 1b  02 bn 1 1 a n b n Negative exponents:  n , n  bn, a b  a b 1a, b  02 bn  a 1 b b b

B. Exponents and Scientific Notation In many technical and scientific applications, we encounter numbers that are either extremely large or very, very small. For example, the mass of the moon is over 73 quintillion kilograms (73 followed by 18 zeroes), while the constant for universal gravitation contains 10 zeroes before the first nonzero digit. When computing with numbers of this size, scientific notation has a distinct advantage over the common decimal notation (base-10 place values).

WORTHY OF NOTE Recall that multiplying by 10’s (or multiplying by 10k, k 7 02 shifts the decimal to the right k places, making the number larger. Dividing by 10’s (or multiplying by 10k, k 7 0) shifts the decimal to the left k places, making the number smaller.

Scientific Notation A nonzero number written in scientific notation has the form N  10k

where 1  0 N 0 6 10 and k is an integer. To convert a number from decimal notation into scientific notation, move the decimal point to the immediate right of the first nonzero digit (creating a number less than 10 but greater than or equal to 1) and multiply by 10k. Here k represents the number of decimal places needed to return the decimal to its original position.

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Appendix I A Review of Basic Concepts and Skills

EXAMPLE 5



Converting from Decimal Notation to Scientific Notation The mass of the moon is about 73,000,000,000,000,000,000 kg. Write this number in scientific notation.

Solution



Place decimal to the right of first nonzero digit (7) and multiply by 10k. 73,000,000,000,000,000,000  7.3  10k To return the decimal to its original position would require 19 shifts to the right, so k must be positive 19. 73,000,000,000,000,000,000  7.3  1019 The mass of the moon is 7.3  1019 kg. Now try Exercises 47 and 48



Converting a number from scientific notation to decimal notation is simply an application of multiplication or division with powers of 10. EXAMPLE 6



Converting from Scientific Notation to Decimal Notation The constant of gravitation is 6.67  1011. Write this number in common decimal form.

Solution



Since the exponent is negative 11, shift the decimal 11 places to the left, using placeholder zeroes as needed to return the decimal to its original position: 6.67  1011  0.000 000 000 066 7

B. You’ve just reviewed how to perform operations in scientific notation

Now try Exercises 49 through 52



C. Identifying and Classifying Polynomial Expressions A monomial is a term using only whole number exponents on variables, with no variables in the denominator. A polynomial is a monomial or any sum or difference of monomial terms. For instance, 12x2  5x  6 is a polynomial, while 3n2  2n  7 is not. We classify polynomials according to their degree and number of terms. The degree of a polynomial in one variable is the largest exponent occurring on the variable. The degree of a polynomial in more than one variable is the largest sum of exponents in any one term. A polynomial with two terms is called a binomial (bi means two) and a polynomial with three terms is called a trinomial (tri means three). There are special names for polynomials with four or more terms, but for these, we simply use the general name polynomial (poly means many). EXAMPLE 7



Classifying and Describing Polynomials For each expression: a. Classify as a monomial, binomial, trinomial, or polynomial. b. State the degree of the polynomial. c. Name the coefficient of each term.

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Section C Exponents, Scientific Notation, and a Review of Polynomials

Solution



Expression

Classification

Degree

5x y  2xy

binomial

three

5, 2

x2  0.81

binomial

two

1, 0.81

z3  3z2  9z  27

polynomial (four terms)

three

1, 3, 9, 27

binomial

one

trinomial

two

2

3 4 x 2

5

2x  x  3

A-19

Coefficients

3 4 ,

5

2, 1, 3

Now try Exercises 53 through 58



A polynomial expression is in standard form when the terms of the polynomial are written in descending order of degree, beginning with the highest-degree term. The coefficient of the highest-degree term is called the leading coefficient. EXAMPLE 8



Writing Polynomials in Standard Form Write each polynomial in standard form, then identify the leading coefficient.

Solution



Polynomial

Standard Form

9x

x  9

2

2

5z  7z  3z  27 2

3

1

3z  7z  5z  27 3

2

2  34 x

C. You’ve just reviewed how to identify and classify polynomial expressions

Leading Coefficient

3  2x2  x

3 4 x 2

2

3 3 4

2x  x  3

2

Now try Exercises 59 through 64



D. Adding and Subtracting Polynomials To add polynomials, use the distributive, commutative, and associative properties to combine like terms. As with real numbers, the subtraction of polynomials involves adding the opposite of the second polynomial using algebraic addition. EXAMPLE 9



Adding and Subtracting Polynomials Perform the indicated operations:

10.7n3  4n2  82  10.5n3  n2  6n2  13n2  7n  102.

Solution D. You’ve just reviewed how to add and subtract polynomials



0.7n3  4n2  8  0.5n3  n2  6n  3n2  7n  10  0.7n3  0.5n3  4n2  1n2  3n2  6n  7n  8  10  1.2n3  13n  18

eliminate parentheses (distributive property) use real number properties to collect like terms combine like terms

Now try Exercises 65 through 70



E. The Product of Two Polynomials To multiply polynomials, we use the distributive property together with the product property of exponents.

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Appendix I A Review of Basic Concepts and Skills

EXAMPLE 10 Solution





Multiplying Polynomials

Find the product: a. 2a2 1a2  2a  12 2

2

2

2

Now try Exercises 71 through 78

WORTHY OF NOTE Consider the product 1x  32 1x  22 in the context of area. If we view x  3 as the length of a rectangle (an unknown length plus 3 units), and x  2 as its width (the same unknown length plus 2 units), a diagram of the total area would look like the following, with the result x2  5x  6 clearly visible.

By observing the product of two binomials in Example 10(b), we note a pattern that can make the process more efficient. We illustrate here using the product 12x  12 13x  22. The F-O-I-L Method for Multiplying Binomials The product of two binomials can quickly be computed by multiplying:

6

S

S

6x2  x  2

The first term of the result will always be the product of the first terms from each binomial, and the last term of the result is the product of their last terms. We also note that here, the middle term is found by adding the outermost product with the innermost product.

(x  3)(x  2)  x2  5x  6

EXAMPLE 11

and combining like terms

Inner Outer



Multiplying Binomials Using F-O-I-L Compute each product mentally: a. 15n  121n  22 b. 12b  32 15b  62 a. 15n  121n  22:

5n2  9n  2

product of first two terms

sum of outer and inner

S



S

Solution

10n  (1n)  9n

S

2x

S

S

2

S

3x

S

12x  12 13x  22 S

x2

6x2  4x  3x  2 First Outer Inner Last

Last First

3

x



The F-O-I-L Method

S

x

b. 12z  121z  22

a. 2a 1a  2a  12  2a 1a 2  12a 212a1 2  12a2 2112 distribute simplify  2a4  4a3  2a2 b. 12z  12 1z  22  2z1z  22  11z  22 distribute to every term in the first binomial eliminate parentheses (distribute again)  2z2  4z  1z  2  2z2  3z  2 simplify 2

product of last two terms

12b  15b  3b

E. You’ve just reviewed how to compute the product of two polynomials

product of first two terms

sum of outer and inner

S

S

S

b. 12b  3215b  62: 10b2  3b  18 product of last two terms

Now try Exercises 79 through 86



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Section C Exponents, Scientific Notation, and a Review of Polynomials

A-21

F. Special Polynomial Products Certain polynomial products are considered “special” for two reasons: (1) their products follow a predictable pattern, and (2) their results are frequently used to simplify expressions, graph functions, solve equations, and/or develop other skills. Recall that for an expression of the form A  B, the conjugate of the expression is A  B. Special Products Binomial Conjugates

1A  B2 1A  B2  A  B 2

2

Binomial Squares

1A  B2 2  A2  2AB  B2 1A  B2 2  A2  2AB  B2

These special products can be verified using the F-O-I-L method, and should be committed to memory. EXAMPLE 12



Using Patterns for Special Products Compute each special product using the patterns outlined previously. 2 2 a. 12x  52 12x  52 b. ax  b ax  b 5 5 c. 1a  92 2 d. 13x  52 2

Solution

a. 12x  5y2 12x  5y2  12x2 2  15y2 2  4x2  25y2 2 2 2 2 b. ax  b ax  b  x2  a b 5 5 5 4  x2  25 c. 1a  92 2  a2  21a # 92  92  a2  18a  81 2 d. 13x  52  13x2 2  213x # 52  52  9x2  30x  25



F. You’ve just reviewed how to compute special products

CAUTION

1A  B2 1A  B2  A2  B 2

result 1A  B2 1A  B2  A 2  B 2 result special product A 2  2AB  B 2 result special product A 2  2AB  B 2 result

Now try Exercises 87 through 102





Note the square of a binomial always results in a trinomial (three terms). Specifically 1A  B2 2 Z A2  B2.

C EXERCISES 

DEVELOPING YOUR SKILLS

Determine each product using the product and/or power properties.

1.

2 2# n 21n5 3

3. 16p2q2 12p3q3 2

5. 1a2 2 4 # 1a3 2 2 # b2 # b5

3 2. 24g5 # g9 8 4. 11.5vy2 2 18v4y2

6. d2 # d 4 # 1c5 2 2 # 1c3 2 2

Simplify each expression using the product to a power property.

7. 16pq2 2 3

9. 10.7c4 2 2 110c3d2 2 2 11.

A 34x3y B 2

8. 13p2q2 2

10. 12.5a3 2 2 13a2b2 2 3 12.

A 45x3 B 2

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Appendix I A Review of Basic Concepts and Skills

13. Volume of a cube: The formula 3x2 for the volume of a cube is V  S3, where S is the length of one edge. If the length of each edge is 3x2, 3x2 a. Find a formula for volume in terms of the variable x. b. Find the volume of the cube if x  2. 14. Area of a circle: The formula for the area of a circle is A  r2, where r is the length of the radius. If the radius is given as 5x3, a. Find a formula for area in terms of the variable x. b. Find the area of the circle if x  2.

37. 3x2

6w5 2w2

17.

12a b 4a2b4

16.

3 5

19. 21.

A 23 B 3

20.

2

h

22.

3

23. 122

3

45. 5x0  15x2 0

5x3

2p4 q3

b

2

5m n 10mn2

0.2x2 3 27. a b 0.3y3 29. a

5m2n3 2 b 2r4

3 m2

2

5v4 2 b 7w3

0.5a3 2 28. a b 0.4b2 30. a

4p3 2

3x y

b

3

Use properties of exponents to simplify the following. Write the answer using positive exponents only.

31.

9p6q4 12p4q6

20h2 33. 12h5

5m5n2 32. 10m5n 5k3 34. 20k2

35.

ab 1a2 b3 2 4

1p q 2

4 8 2

3 2

36.

5 2

pq

40. 132 0  172 0 42. 41  81

44. 22  21  20 46. 2n0  12n2 0

47. In mid-2007, the U.S. Census Bureau estimated the world population at nearly 6,600,000,000 people. 48. The mass of a proton is generally given as 0.000 000 000 000 000 000 000 000 001 670 kg. Convert the following numbers to decimal notation.

49. As of 2006, the smallest microprocessors in common use measured 6.5  109 m across. 50. In 2007, the estimated net worth of Bill Gates, the founder of Microsoft, was 5.6  1010 dollars.

51. The average distance between the Earth and the planet Jupiter is 465,000,000 mi. How many hours would it take a satellite to reach the planet if it traveled an average speed of 17,500 mi per hour? How many days? Round to the nearest whole.

A 56 B 1

26. a

18n3 813n2 2 3

Compute using scientific notation. Show all work.

Simplify each expression using the quotient to a power property.

25. a

38.

Convert the following numbers to scientific notation.

8z7 16z5

24. 142

39. 40  50 43. 30  31  32

3 5

18.

10x2

41. 21  51

Simplify using the quotient property or the property of negative exponents. Write answers using positive exponents only.

15.

612x3 2 2

52. In fiscal terms, a nation’s debt-per-capita is the ratio of its total debt to its total population. In the year 2007, the total U.S. debt was estimated at $9,010,000,000,000, while the population was estimated at 303,000,000. What was the U.S. debtper-capita ratio for 2007? Round to the nearest whole dollar. Identify each expression as a polynomial or nonpolynomial (if a nonpolynomial, state why); classify each as a monomial, binomial, trinomial, or none of these; and state the degree of the polynomial.

53. 35w3  2w2  112w2  14 54. 2x3  23x2  12x  1.2 55. 5n2  4n  117 57. p3 

2 5

4  2.7r2  r  1 r3 58. q3  2q2  5q 56.

Write the polynomial in standard form and name the leading coefficient.

59. 7w  8.2  w3  3w2 60. 2k2  12  k 61. c3  6  2c2  3c

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Section C Exponents, Scientific Notation, and a Review of Polynomials

62. 3v3  14  2v2  112v2

77. 1b2  3b  282 1b  22

64. 8  2n2  7n

79. 17v  42 13v  52

63. 12  23x2

Find the indicated sum or difference.

65. 13p3  4p2  2p  72  1p2  2p  52 66. 15q2  3q  42  13q2  3q  42

67. 15.75b2  2.6b  1.92  12.1b2  3.2b2

68. 10.4n2  5n  0.52  10.3n2  2n  0.752 69. 1 34x2  5x  22  1 12x2  3x  42

70. 1 59n2  4n  12 2  1 23n2  2n  34 2 Compute each product.

71. 3x1x  x  62

78. 12h2  3h  82 1h  12 81. 13  m213  m2 83. 1m 

3 4 2 1m



3 42

85. 12x2  521x2  32

A-23

80. 16w  1212w  52 82. 15  n215  n2 84. 1n  25 21n  25 2

86. 13y2  2212y2  12

For each binomial, determine its conjugate and then find the product of the binomial with its conjugate.

87. 4m  3

88. 6n  5

89. 7x  10

90. c  3

91. 6  5k

92. 11  3r

93. x  16

94. p  12

2

72. 2v2 1v2  2v  152 73. 13r  52 1r  22 74. 1s  32 15s  42

75. 1x  32 1x  3x  92 2

76. 1z  52 1z  5z  252 2



Find each binomial square.

95. 1x  42 2

97. 14g  32 2

99. 14p  3q2 2

101. 14  1x2 2

96. 1a  32 2

98. 15x  32 2

100. 15c  6d2 2 102. 1 1x  72 2

APPLICATIONS

103. Attraction between particles: In electrical theory, the force of attraction between two particles P and kPQ Q with opposite charges is modeled by F  2 , d where d is the distance between them and k is a constant that depends on certain conditions. This is known as Coulomb’s law. Rewrite the formula using a negative exponent. 104. Intensity of light: The intensity of illumination from a light source depends on the distance from k the source according to I  2 , where I is the d intensity measured in footcandles, d is the distance from the source in feet, and k is a constant that depends on the conditions. Rewrite the formula using a negative exponent. Rewriting an expression: In advanced mathematics, negative exponents are widely used because they are easier to work with than rational expressions. 3 2 5 105. Rewrite the expression 3  2  1  4 using x x x negative exponents.

106. Rewrite the expression

2 1 6 4  2  a using a a

negative exponents. 107. Maximizing revenue: A sporting goods store finds that if they price their video games at $20, they make 200 sales per day. For each decrease of $1, 20 additional video games are sold. This means the store’s revenue can be modeled by the formula R  120  1x2 1200  20x2, where x is the number of $1 decreases. Multiply out the binomials and use a table of values to determine what price will give the most revenue. 108. Maximizing revenue: Due to past experience, a jeweler knows that if they price jade rings at $60, they will sell 120 each day. For each decrease of $2, five additional sales will be made. This means the jeweler’s revenue can be modeled by the formula R  160  2x21120  5x2, where x is the number of $2 decreases. Multiply out the binomials and use a table of values to determine what price will give the most revenue.

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Appendix I A Review of Basic Concepts and Skills

EXTENDING THE CONCEPT

109. If 13x2  kx  12  1kx2  5x  72  12x2  4x  k2  x2  3x  2, what is the value of k?

1 2 1 b  5, then the expression 4x2  2 2x 4x is equal to what number?

110. If a2x 

D Factoring Polynomials A. The Greatest Common Factor and Factoring by Grouping

In Section D you will review:

A. Factoring out the greatest common factor and factoring by grouping

To factor an expression means to rewrite the expression as an equivalent product. The distributive property is an example of factoring in action. To factor 2x2  6x, first rewrite each term using the common factor 2x: 2x2  6x  2x # x  2x # 3, then apply the distributive property to obtain 2x1x  32. The greatest common factor (or GCF) is the largest factor common to all terms in the polynomial.

B. Factoring quadratic polynomials

C. Factoring special forms and quadratic forms

EXAMPLE 1



Factoring Polynomials Factor each polynomial: a. 12x2  18xy  30y

Solution



c. 1x  32x2  1x  325

b. x5  x2

a. 6 is common to all three terms: 12x2  18xy  30y mentally: 6 # 2x2  6 # 3xy  6 # 5y 2  612x  3xy  5y2 2 b. x is common to both terms: x5  x2 mentally: x2 # x3  x2 # 1 2 3  x 1x  12 c. 1x  32 is common to both terms: 1x  32x2  1x  325  1x  32 1x2  52 Now try Exercises 1 through 4



One application of removing a binomial factor involves factoring by grouping. At first glance it appears, the expression 3t3  15t2  6t  30  31t3  5t2  2t  102 cannot be factored further. But by grouping the terms (applying the associative property), we can remove a monomial factor from each subgroup, which then reveals a common binomial factor. EXAMPLE 2



Factoring by Grouping Factor 3t3  15t2  6t  30.

Solution



A. You’ve just reviewed how to factor out the greatest common factor and factor by grouping

Notice that all four terms have a common factor of 3. Begin by factoring it out. 3t3  15t2  6t  30  31t3  5t2  2t  102  31t3  5t2  2t  102  33 t2 1t  52  21t  52 4  31t  52 1t2  22

original polynomial factor out 3 group remaining terms factor common monomial factor common binomial

Now try Exercises 5 and 6



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Section D Factoring Polynomials

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When asked to factor an expression, first look for common factors. The resulting expression will be easier to work with and help ensure the final answer is written in completely factored form. If a four-term polynomial cannot be factored as written, try rearranging the terms to find a combination that enables factoring by grouping.

B. Factoring Quadratic Polynomials A quadratic polynomial is one that can be written in the form ax2  bx  c, where a, b, c  , and a  0. One common form of factoring involves quadratic trinomials such as x2  7x  10 and 2x2  13x  15. For this study, it will help to place the trinomials in two families—those with a leading coefficient of 1 and those with a leading coefficient other than 1.

WORTHY OF NOTE Similarly, a cubic polynomial is one of the form ax3  bx2  cx  d. It’s helpful to note that a cubic polynomial can be factored by grouping only when ad  bc, where a, b, c, and d are the coefficients shown. This is easily seen in Example 2, where 1321302  1152 162 gives 90  90✓.

ax2  bx  c, where a  1

When a  1, the only factor pair for x2 (other than 1 # x2 2 is x # x and the first term in each binomial will be x: (x )(x ). The following observation helps guide us to the complete factorization. Consider the product 1x  b21x  a2: 1x  m21x  n2  x2  nx  mx  mn  x2  1m  n2x  mn

F-O-I-L distributive property

This shows that to factor the expression x  bx  c, we’re looking for two numbers m and n that multiply to c 1mn  c2 and add to b1m  n  b2. It is also helpful to note that if the constant term is positive, the binomials will have like signs, since only the product of like signs is positive. If the constant term is negative, the binomials will have unlike signs, since only the product of unlike signs is negative. This means we can use the sign of the linear term (the term with degree 1) to guide our choice of factors. 2

Template for Factoring Trinomials with a Leading Coefficient of 1 If the constant term is positive, the binomials will have like signs: 1x  2 1x  2 or 1x  21x  2 ,

to match the sign of the linear (middle) term. If the constant term is negative, the binomials will have unlike signs: 1x  21x  2,

with the larger factor placed in the binomial whose sign matches the linear (middle) term.

EXAMPLE 3



Factoring Trinomials Factor these expressions: a. x2  11x  24

Solution



b. x2  10  3x

a. First rewrite the trinomial in standard form as 11x2  11x  242. For x2  11x  24, the constant term is positive so the binomials will have like signs. Since the linear term is negative,

11x2  11x  242  11x  21x  2 like signs, both negative  11x  821x  32 182 132  24; 8  132  11 b. First rewrite the trinomial in standard form as x2  3x  10. The constant term is negative so the binomials will have unlike signs. Since the linear term is negative, x2  3x  10  1x  2 1x  2  1x  22 1x  52

unlike signs, one positive and one negative 5 7 2, 5 is placed in the second binomial; 122 152  10; 2  152  3

Now try Exercises 7 and 8



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Appendix I A Review of Basic Concepts and Skills

Sometimes we encounter prime polynomials, such as x2  9x  15. The factor pairs of 15 are 1 # 15 and 3 # 5, with neither pair having a sum of 9.

ax2  bx  c, where a  1 If the leading coefficient is not one, the possible combinations of outers and inners are more numerous and we generally employ a trial-and-error approach. To factor 2x2  13x  15, note the constant term is positive so the binomials must have like signs. The negative linear term indicates these signs will be negative. We then list possible factors for the first and last terms of each binomial, then sum the outer and inner products in a search for the correct linear term.

Possible First and Last Terms for 2x2 and 15

Sum of Outers and Inners

2. 12x  1521x  12

2x  15x  17x

1. 12x  121x  152

30x  1x  31x

3. 12x  321x  52

10x  3x  13x

4. 12x  521x  32

d

6x  5x  11x

As you can see, only possibility 3 yields a linear term of 13x, and the correct factorization is then 12x  321x  52. With practice, this trial-and-error process can be completed very quickly. If the constant term is negative, the number of possibilities can be reduced by finding a factor pair with a sum or difference equal to the absolute value of the linear coefficient, as we can then arrange the sign of each binomial to obtain the needed result (see Example 4).

EXAMPLE 4



Factoring a Trinomial Using Trial and Error Factor 6z2  11z  35.

Solution

B. You’ve just reviewed how to factor quadratic polynomials



Note the constant term is negative (binomials will have unlike signs), 11  11, and the factors of 35 are 1 # 35 and 5 # 7. Two possible first terms are: (6z )(z ) and (3z )(2z ), and we begin with 5 and 7 as factors of 35. (6z

)(z

)

Outers/Inners

(3z

)(2z

)

Sum

Diff

1. (6z

5)(z

7)

47z

2. (6z

7)(z

5)

37z

Outers/Inners Sum

Diff

37z

3. (3z

5)(2z

7)

31z

11z d

23z

4. (3z

7)(2z

5)

29z

1z

Since possibility 3 yields the linear term of 11z, we need not consider other factors of 35 and write the factored form as 6z2  11z  35  13z 5212z 72. The signs can then be arranged to obtain a middle term of 11z: 13z  5212z  72, 21z  10z  11z ✓. Now try Exercises 9 and 10



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Section D Factoring Polynomials

C. Factoring Special Forms and Quadratic Forms

WORTHY OF NOTE

Multiplying and factoring are reverse processes. This means that to factor the special products from Section C, we simply view the multiplication process “in reverse.” When doing so, it may help to rewrite certain terms in the original expression as a perfect square: ( )2.

In an attempt to factor a sum of two perfect squares, say v2  49, let’s list all possible binomial factors. These are (1) 1v  72 1v  72, (2) 1v  72 1v  72, and (3) 1v  72 1v  72. Note that (1) and (2) are the binomial 2 2 squares 1v  72 and 1v  72 , with each product resulting in a “middle” term, whereas (3) is a binomial times its conjugate, resulting in a difference of squares: v2  49. With all possibilities exhausted, we conclude that the sum of two squares is prime!

EXAMPLE 5

Factoring Special Forms Difference of Squares

Perfect Square Trinomials

A  B  1A  B2 1A  B2 2

A2  2AB  B2  1A  B2 2

2

A2  2AB  B2  1A  B2 2

Note that the sum of two perfect squares A2  B2 cannot be factored using real numbers (the expression is prime). As a reminder, always check for a common factor first and be sure to write all results in completely factored form. See Example 5(c). 

Factoring a Difference of Squares Factor each expression completely. a. 4w2  81 b. v2  49 c. 3n2  48

Solution



a. 4w2  81  12w2 2  92  12w  9212w  92 2 b. v  49 is prime. c. 3n2  48  31n2  162  3n2  142 2  31n  42 1n  42

d. z4 

    e. x 2  7   1 81

1z2 2 2  1 19 2 2 1z2  19 2 1z2  19 2 1z2  19 2z2  1 13 2 2 1z2  19 2 1z  13 2 1z  13 2 1x2 2  1 172 2 1x  172 1x  172

1 d. z4  81

e. x2  7

write as a difference of squares A2  B 2  1A  B2 1A  B2 factor out 3 write as a difference of squares A2  B 2  1A  B2 1A  B2 write as a difference of squares A 2  B 2  1A  B2 1A  B2 write as a difference of squares result write as a difference of squares A 2  B 2  1A  B2 1A  B2

Now try Exercises 11 and 12 EXAMPLE 6





Factoring a Perfect Square Trinomial Factor 12m3  12m2  3m.

Solution



12m3  12m2  3m  3m14m2  4m  12

check for common factors: GCF  3m factor out 3m

For the remaining trinomial 4m2  4m  1 . . . 1. Are the first and last terms perfect squares?

4m2  12m2 2 and 1  112 2 ✓ 2. Is the linear term twice the product of 2m and 1? 2 # 2m # 1  4m ✓

Yes.

Yes.

Factor as a binomial square: 4m  4m  1  12m  12 2 2

This shows 12m3  12m2  3m  3m12m  12 2.

Now try Exercises 13 and 14



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Appendix I A Review of Basic Concepts and Skills

CAUTION



As shown in Example 6, be sure to include the GCF in your final answer. It is a common error to “leave the GCF behind.”

In actual practice, the tests for a perfect square trinomial are performed mentally, with only the factored form being written down. There is one additional special form that merits attention, and that is factoring the sum or difference of two cubes. Factoring the Sum or Difference of Two Cubes: A3  B3 A3  B3  1A  B2 1A2  AB  B2 2 A3  B3  1A  B2 1A2  AB  B2 2 EXAMPLE 7



Factoring the Sum and Difference of Two Cubes Factor completely: a. x3  125

Solution



b. 5m3n  40n4

write terms as perfect cubes x3  125  x3  53 3 3 2 2 Use A  B  1A  B21A  AB  B 2 factoring template x3  53  1x  52 1x2  5x  252 A S x and B S 5 check for common 5m3n  40n4  5n1m3  8n3 2 b. factors 1GCF  5n2

a.

 5n 3 m3  12n2 3 4 Use A  B  1A  B2 1A2  AB  B2 2 m3  12n2 3  1m  2n2 3 m2  m12n2  12n2 2 4  1m  2n21m2  2mn  4n2 2 1 5m3n  40n4  5n1m  2n21m2  2mn  4n2 2. 3

3

write terms as perfect cubes factoring template

A S m and B S 2n simplify factored form

The results for parts (a) and (b) can be checked using multiplication. Now try Exercises 15 and 16



Using u-Substitution to Factor Quadratic Forms For any quadratic expression ax2  bx  c in standard form, the degree of the leading term is twice the degree of the middle term. Generally, a trinomial is in quadratic form if it can be written as a1 __ 2 2  b1 __ 2  c, where the parentheses “hold” the same factors. The equation x4  13x2  36  0 is in quadratic form since 1x2 2 2  131x2 2  36  0. In many cases, we can factor these expressions using a placeholder substitution that transforms these expressions into a more recognizable form. In a study of algebra, the letter “u” often plays this role. If we let u represent x2, the expression 1x2 2 2  131x2 2  36 becomes u2  13u  36, which can be factored into 1u  921u  42. After “unsubstituting” (replace u with x2), we have 1x2  921x2  42  1x  32 1x  321x  22 1x  22.

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Precalculus—

Section D Factoring Polynomials

EXAMPLE 8



Factoring a Quadratic Form

Solution



Expanding the binomials would produce a fourth-degree polynomial that would be very difficult to factor. Instead we note the expression is in quadratic form. Letting u represent x2  2x (the variable part of the “middle” term), 1x2  2x2 2  21x2  2x2  3 becomes u2  2u  3.

A-29

Write in completely factored form: 1x2  2x2 2  21x2  2x2  3.

u2  2u  3  1u  321u  12

factor

 1x2  2x  32 1x2  2x  12

substitute x2  2x for u

To finish up, write the expression in terms of x, substituting x2  2x for u. The resulting trinomials can be further factored.

 1x  32 1x  121x  12 2

C. You’ve just reviewed how to factor special forms and quadratic forms

x2  2x  1  1x  12 2

Now try Exercises 17 through 20

D EXERCISES 

DEVELOPING YOUR SKILLS

Factor each expression using the method indicated. Greatest Common Factor

1. a. 17x  51 b. 21b  14b  56b c. 3a4  9a2  6a3 2

3

2

2. a. 13n2  52 b. 9p2  27p3  18p4 5 4 c. 6g  12g  9g3 Common Binomial Factor

3. a. 2a1a  22  31a  22 b. 1b2  323b  1b2  322 c. 4m1n  72  111n  72 4. a. 5x1x  32  21x  32 b. 1v  522v  1v  523 c. 3p1q2  52  71q2  52 Grouping

5. a. 9q3  6q2  15q  10 b. h5  12h4  3h  36 c. k5  7k3  5k2  35 6. a. 6h3  9h2  2h  3 b. 4k3  6k2  2k  3 c. 3x2  xy  6x  2y

Trinomial Factoring where |a|  1

7. a. p2  5p  14 c. n2  20  9n

b. q2  4q  45

8. a. m2  13m  42 c. v2  10v  15

b. x2  12  13x

Trinomial Factoring where a  1

9. a. 3p2  13p  10 c. 10u2  19u  15 10. a. 6v2  v  35 c. 15z2  22z  48

b. 4q2  7q  15 b. 20x2  53x  18

Difference of Perfect Squares

11. a. 4s2  25 c. 50x2  72 e. b2  5

b. 9x2  49 d. 121h2  144

1 12. a. 9v2  25 4 c. v  1 e. x2  17

1 b. 25w2  49 4 d. 16z  81

Perfect Square Trinomials

13. a. a2  6a  9 c. 4m2  20m  25

b. b2  10b  25 d. 9n2  42n  49



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Appendix I A Review of Basic Concepts and Skills

14. a. x2  12x  36 b. z2  18z  81 c. 25p2  60p  36 d. 16q2  40q  25 Sum/Difference of Perfect Cubes

15. a. 8p  27 c. g3  0.027

b. m  d. 2t4  54t

16. a. 27q3  125 c. b3  0.125

8 b. n3  27 d. 3r4  24r

3

3

1 8

u-Substitution

17. a. x4  10x2  9 c. x6  7x3  8

b. x4  13x2  36

18. a. x6  26x3  27 b. 31n  52 2  12n  102  21 c. 21z  32 2  13z  92  54 19. Completely factor each of the following (recall that “1” is its own perfect square and perfect cube). a. n2  1 c. n3  1

b. n3  1 d. 28x3  7x

20. Carefully factor each of the following trinomials, if possible. Note differences and similarities. a. x2  x  6 

c. x2  x  6 e. x2  5x  6

d. x2  x  6 f. x2  5x  6

Factor each expression completely, if possible. Rewrite the expression in standard form (factor out “1” if needed) and factor out the GCF if one exists. If you believe the expression will not factor, write “prime.”

21. a2  7a  10

22. b2  9b  20

23. 2x2  24x  40

24. 10z2  140z  450

25. 64  9m2

26. 25  16n2

27. 9r  r2  18

28. 28  s2  11s

29. 2h2  7h  6

30. 3k2  10k  8

31. 9k2  24k  16

32. 4p2  20p  25

33. 6x3  39x2  63x

34. 28z3  16z2  80z

35. 12m2  40m  4m3

36. 30n  4n2  2n3

37. a2  7a  60

38. b2  9b  36

39. 8x3  125

40. 27r3  64

41. m2  9m  24

42. n2  14n  36

43. x3  5x2  9x  45 44. x3  3x2  4x  12

b. x2  x  6

APPLICATIONS

In many cases, factoring an expression can make it easier to evaluate as in the following applications.

45. Conical shells: The volume of a conical shell (like the shell of an ice cream cone) is given by the 1 1 formula V  R2h  r2h, where R is the 3 3 outer radius and r is the inner radius of the cone. Write the formula in completely factored form, then find the volume of a shell when R  5.1 cm, r  4.9 cm, and h  9 cm. Answer in exact form and in approximate form rounded to the nearest tenth. 46. Spherical shells: The volume of a spherical shell (like the outer r shell of a cherry cordial) is given R by the formula V  43R3  43r3, where R is the outer radius and r is the inner radius of the shell. Write the right-hand side in completely factored form, then find the volume of a shell where R  1.8 cm and r  1.5 cm.

47. Volume of a box: The volume of a rectangular box x inches in height is given by the relationship V  x3  8x2  15x. Factor the right-hand side to determine: (a) The number of inches that the width exceeds the height, (b) the number of inches the length exceeds the height, and (c) the volume given the height is 2 ft. 48. Shipping textbooks: A publisher ships paperback books stacked x copies high in a box. The total number of books shipped per box is given by the relationship B  x3  13x2  42x. Factor the right-hand side to determine (a) how many more or fewer books fit the width of the box (than the height), (b) how many more or fewer books fit the length of the box (than the height), and (c) the number of books shipped per box if they are stacked 10 high in the box. 49. Space-Time relationships: Due to the work of Albert Einstein and other physicists who labored on space-time relationships, it is known that the faster an object moves the shorter it appears to become. This phenomenon is modeled by the

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Section E Rational Expressions

than at the sides, where the tube exerts a backward drag. Poiseuille’s law gives the velocity of the flow G 2 1R  r2 2, at any point of the cross section: v  4 where R is the inner radius of the tube, r is the distance from the center of the tube to a point in the flow, G represents what is called the pressure gradient, and  is a constant that depends on the viscosity of the fluid. Factor the right-hand side and find v given R  0.5 cm, r  0.3 cm, G  15, and   0.25.

v 2 1a b, c B where L0 is the length of the object at rest, L is the relative length when the object is moving at velocity v, and c is the speed of light. Factor the radicand and use the result to determine the relative length of a 12-in. ruler if it is shot past a stationary observer at 0.75 times the speed of light 1v  0.75c2 . Lorentz transformation L  L0

50. Tubular fluid flow: As a fluid flows through a tube, it is flowing faster at the center of the tube 

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EXTENDING THE CONCEPT

51. Factor out a constant that leaves integer coefficients for each term: a. 12x4  18x3  34x2  4 b. 23b5  16b3  49b2  1 52. If x  2 is substituted into 2x3  hx  8, the result is zero. What is the value of h? 53. Factor the expression: 192x3  164x2  270x. 54. As an alternative to evaluating polynomials by direct substitution, nested factoring can be used. The method has the advantage of using only products and sums—no powers. For P  x3  3x2  1x  5, we begin by grouping all variable terms

and factoring x: P  3 x3  3x2  1x 4  5  x 3 x2  3x  14  5. Then we group the inner terms with x and factor again: P  x3 x2  3x  14  5  x 3 x1x  32  1 4  5. The expression can now be evaluated using any input and the order of operations. If x  2, we quickly find that P  27. Use this method to evaluate H  x3  2x2  5x  9 for x  3. Factor each expression completely.

55. x4  81

56. 16n4  1

57. p6  1

58. m6  64

59. q4  28q2  75

60. a4  18a2  32

E Rational Expressions In Section E you will learn how to:

A. Write a rational expression in simplest form

B. Multiply and divide rational expressions

C. Add and subtract rational expressions

D. Simplify compound

A. Writing a Rational Expression in Simplest Form A rational number is one that can be written as the quotient of two integers. Similarly, a rational expression is one that can be written as the quotient of two polynomials. We can apply the skills developed in a study of fractions (how to reduce, add, subtract, multiply, and divide) to rational expressions, sometimes called algebraic fractions. A rational expression is in simplest form when the numerator and denominator have no common factors (other than 1). After factoring the numerator and denominator, we apply the fundamental property of rational expressions.

fractions

E. Rewrite formulas and algebraic models

Fundamental Property of Rational Expressions If P, Q, and R are polynomials, with Q, R  0, (1)

P#R P  # Q R Q

and

(2)

P#R P  # Q Q R

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Appendix I A Review of Basic Concepts and Skills

1. A rational expression can be simplified by canceling common factors in the numerator and denominator. 2. An equivalent expression can be formed by multiplying numerator and denominator by the same nonzero polynomial.

EXAMPLE 1



Simplifying a Rational Expression Write the expression in simplest form:

Solution



1x  12 1x  12 x2  1  2 1x  12 1x  22 x  3x  2 1x  121x  12  1x  121x  22 x1  x2

x2  1 . x  3x  2 2

factor numerator and denominator

cancel common factors

simplest form

Now try Exercises 1 through 4

CAUTION





When reducing rational numbers or expressions, only common factors can be reduced. 6  423 It is incorrect to reduce (or divide out) individual terms:  3  423, and 2 x1 1  (except for x  0) x2 2

When simplifying rational expressions, we sometimes encounter expressions ab ab of the form . If we factor 1 from the numerator, we see that  ba ba 11b  a2  1. ba EXAMPLE 2



Simplifying a Rational Expression Write the expression in simplest form:

Solution

A. You’ve just reviewed how to write a rational expression in simplest form



16  2x2

213  x2 1x  321x  32 x 9 122 112  x3 2  x3 2



16  2x2 x2  9

.

factor numerator and denominator

reduce:

13  x2 1x  32

 1

simplest form

Now try Exercises 5 through 10



B. Multiplication and Division of Rational Expressions Operations on rational expressions use the factoring skills reviewed earlier, along with much of what we know about rational numbers.

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Section E Rational Expressions

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Multiplying Rational Expressions Given that P, Q, R, and S are polynomials with Q, S  0, PR P #R  Q S QS 1. Factor all numerators and denominators completely. 2. Reduce common factors. 3. Multiply numerator  numerator and denominator  denominator. EXAMPLE 3



Multiplying Rational Expressions Compute the product:

Solution



2a  2 # 3a2  a  2 . 3a  3a2 9a2  4

21a  12 13a  221a  12 2a  2 # 3a2  a  2 #  2 2 3a11  a2 13a  2213a  22 3a  3a 9a  4

factor

112

21a  12 13a  221a  1 2 #  3a11  a2 13a  2213a  22

reduce:

21a  12  3a13a  22

simplest form

1

1

a1  1 1a

1

Now try Exercises 11 through 14



To divide fractions, we multiply the first expression by the reciprocal of the second. The quotient of two rational expressions is computed in the same way. Dividing Rational Expressions Given that P, Q, R, and S are polynomials with Q, R, S  0, R P S PS P   #  Q S Q R QR Invert the divisor and multiply. EXAMPLE 4



Dividing Rational Expressions Compute the quotient:

Solution



4m3  12m2  9m 10m2  15m  . m2  49 m2  4m  21

4m3  12m2  9m 10m2  15m  m2  49 m2  4m  21 4m3  12m2  9m # m2  4m  21  m2  49 10m2  15m 2 m14m  12m  92 1m  72 1m  32 #  1m  72 1m  72 5m12m  32

m 12m  32 12m  32 1m  72 1m  32 #  1m  72 1m  72 5m 12m  32 1

1

1

12m  321m  32  51m  72

invert and multiply

factor

1

1

factor and reduce

1

lowest terms

Note that we sometimes refer to simplest form as lowest terms. Now try Exercises 15 through 30



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Appendix I A Review of Basic Concepts and Skills

CAUTION



1w  721w  72 1w  22 # , it is a common mistake to think that all 1w  721w  22 1w  72 factors “cancel,” leaving an answer of zero. Actually, all factors reduce to 1, and the result is a value of 1 for all inputs where the product is defined. For products like

1w  72 1w  72 1w  22 1

B. You’ve just reviewed how to multiply and divide rational expressions

1

1

#

1w  72 1w  22 1w  72 1

1

1

1

C. Addition and Subtraction of Rational Expressions Recall that the addition and subtraction of fractions requires finding the lowest common denominator (LCD) and building equivalent fractions. The sum or difference of the numerators is then placed over this denominator. The procedure for the addition and subtraction of rational expressions is very much the same. Addition and Subtraction of Rational Expressions 1. 2. 3. 4.

EXAMPLE 5



Find the LCD of all rational expressions. Build equivalent expressions using the LCD. Add or subtract numerators as indicated. Write the result in lowest terms.

Adding and Subtracting Rational Expressions Compute as indicated: 7 3 a.  10x 25x2

Solution



b.

10x 5  x3 x 9 2

a. The LCD for 10x and 25x2 is 50x2.

7 3 7 # 15x2 3 # 122   2  10x 10x 15x2 25x 25x2 122 35x 6  2  50x 50x2 35x  6  50x2

find the LCD write equivalent expressions

simplify add the numerators and write the result over the LCD

The result is in simplest form. b. The LCD for x2  9 and x  3 is 1x  32 1x  32.

10x 10x 5 5 # 1x  32    x3 1x  321x  32 1x  32 1x  32 x 9 10x  51x  32  1x  32 1x  32 10x  5x  15  1x  321x  32 5x  15  1x  321x  32 2

find the LCD write equivalent expressions subtract numerators, write the result over the LCD distribute

combine like terms

1

51x  32 5   1x  32 1x  32 x3

factor and reduce

1

Now try Exercises 31 through 36



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Precalculus—

Section E Rational Expressions

EXAMPLE 6



Adding and Subtracting Rational Expressions Perform the operations indicated: 5 n3 a.  2 n2 n 4

Solution



b.

c b2 2  a 4a

a. The LCD for n  2 and n2  4 is 1n  221n  22.

5 5 n3 n3 # 1n  22    2 n2 1n  22 1n  22 1n  22 1n  22 n 4 51n  22  1n  32  1n  221n  22 5n  10  n  3  1n  221n  22 4n  7  1n  221n  22 b2 b2 c c 14a2 2 2    # b. The LCD for a and 4a is 4a : 2 2 a a 14a2 4a 4a b2 4ac 2  4a 4a2 2 b  4ac  4a2 

C. You’ve just reviewed how to add and subtract rational expressions

write equivalent expressions subtract numerators, write the result over the LCD distribute

result write equivalent expressions

simplify subtract numerators, write the result over the LCD

Now try Exercises 37 through 48

CAUTION



A-35



When the second term in a subtraction has a binomial numerator as in Example 6(a), be sure the subtraction is applied to both terms. It is a common error to write 51n  22 n3 5n  10  n  3 X in which the subtraction is applied   1n  221n  22 1n  221n  22 1n  221n  22 to the first term only. This is incorrect!

D. Simplifying Compound Fractions Rational expressions whose numerator or denominator contain a fraction are called 3 2  3m 2 compound fractions. The expression is a compound fraction with a 3 1  4m 3m2 3 1 2 3  and a denominator of  . The two methods commonly numerator of 3m 2 4m 3m2 used to simplify compound fractions are summarized in the following boxes. Simplifying Compound Fractions (Method I) 1. Add/subtract fractions in the numerator, writing them as a single expression. 2. Add/subtract fractions in the denominator, also writing them as a single expression. 3. Multiply the numerator by the reciprocal of the denominator and simplify if possible.

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Appendix I A Review of Basic Concepts and Skills

Simplifying Compound Fractions (Method II) 1. Find the LCD of all fractions in the numerator and denominator. 2. Multiply the numerator and denominator by this LCD and simplify. 3. Simplify further if possible. Method II is illustrated in Example 7 (for Method I, see Example 9). EXAMPLE 7



Simplifying a Compound Fraction Simplify the compound fraction: 3 2  3m 2 3 1  4m 3m2

Solution



The LCD for all fractions is 12m2. 2 3 3 12m2 2  a  ba b 3m 2 3m 2 1  3 12m2 3 1 1  a  b a b 4m 4m 1 3m2 3m2 2 12m2 3 12m2 a ba ba ba b 3m 1 2 1  3 12m2 12m2 1 a ba b  a 2b a b 4m 1 1 3m 8m  18m2  9m  4

multiply numerator and denominator by 12m 2 

12m 2 1

distribute

simplify

1

2m14  9m2   2m 9m  4

D. You’ve just reviewed how to simplify compound fractions

factor and write in lowest terms

Now try Exercises 49 through 58



E. Rewriting Formulas and Algebraic Models In many fields of study, formulas and algebraic models involve rational expressions and we often need to write them in an alternative form. EXAMPLE 8



Rewriting a Formula In an electrical circuit with two resistors in parallel, the total resistance R is related 1 1 1  . Rewrite the right-hand side as to resistors R1 and R2 by the formula  R R1 R2 a single term.

Solution



1 1 1   R R1 R2 R2 R1   R1R2 R1R2 R2  R1  R1R2

LCD for the right-hand side is R1R2

build equivalent expressions using LCD

write as a single expression

Now try Exercises 59 and 60



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Precalculus—

Section E Rational Expressions

EXAMPLE 9



A-37

Simplifying an Algebraic Model When studying rational expressions and rates of change, we encounter the 1 1  x xh . Simplify the compound fraction. expression h

Solution



E. You’ve just reviewed how to rewrite formulas and algebraic models

Using Method I gives:

1 1 1 # x  1 # 1x  h2  x x 1x  h2 xh xh x  h h x  1x  h2 x1x  h2  h h x1x  h2  h h # 1  x1x  h2 h 1  x1x  h2

LCD for the numerator is x (x  h)

write numerator as a single expression

simplify

invert and multiply

result

Now try Exercises 61 through 64



E EXERCISES 

DEVELOPING YOUR SKILLS

Reduce to lowest terms.

1. a.

a7 3a  21

b.

2x  6 4x2  8x

2. a.

x4 7x  28

b.

3x  18 6x2  12x

3. a.

x2  5x  14 x2  6x  7

b.

a2  3a  28 a2  49

4. a.

r2  3r  10 r2  r  6

b.

m2  3m  4 m2  4m

x7 5. a. 7x

5x b. x5

v  3v  28 6. a. 49  v2

u  10u  25 b. 25  u2

2

12a b 4a2b4 y2  9 c. 3y

3 5

7. a.

7x  21 63 3 m n  m3 d. 4 m  m4n

5v  20 25 4 w  w4v d. 3 w v  w3

2n3  n2  3n n3  n2 x3  8 c. 2 x  2x  4

6x2  x  15 4x2  9 mn2  n2  4m  4 d. mn  n  2m  2

9. a.

10. a. c.

2

b.

5m3n5 10mn2 n2  4 c. 2n

8. a.

x3  4x2  5x x3  x 2 12y  13y  3 27y3  1

b.

b.

b.

5p2  14p  3

5p2  11p  2 ax2  5x2  3a  15 d. ax  5x  5a  25

Compute as indicated. Write final results in lowest terms.

11.

a2  4a  4 a2  2a  3 # a2  9 a2  4

12.

b b2  5b  24 # 2 2 b  6b  9 b  64

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Appendix I A Review of Basic Concepts and Skills

13.

x2  7x  18 # 2x2  7x  3 x2  6x  27 2x2  5x  2

14.

6v2  23v  21 4v2  25 # 3v  7 4v2  4v  15

31.

3 5 2  2x 8x

32.

15 7  2 16y 2y

p3  64

33.

7 1 2 3  4x y 8xy4

34.

5 3 3  6a b 9ab3

15.

p3  p2



p2  4p  16

Compute as indicated. Write answers in lowest terms [recall that a  b  1(b  a)].

p2  5p  4

4p



2 p6

3q

a2  3a  28 a3  4a2  3 16. 2 a  5a  14 a 8

35.

3x 3x  9  17. 4x  12 5x  15

37.

2b 5b  10  18. 7b  28 5b  20

39.

8 # 1a2  2a  352 19. 2 a  25

40.

3 4n  4n  20 n  5n

m2  5m 20. 1m2  162 # 2 m  m  20

41.

a 1  2 a4 a  a  20

42.

x5 2x  1  2 x  3x  4 x  3x  4

21.

xy  3x  2y  6 x  3x  10 2

xy  3x xy  5y



22.

ab  2a 2a  ab  7b  14  2 ab  7a b  14b  49

23.

m  2m  8 m  16  2 m  2m m2 2

24.

2

18  6x 2x2  18  3 2 x  25 x  2x2  25x  50

25.

x2  0.49 x2  0.10x  0.21  x2  0.5x  0.14 x2  0.09

26.

x  0.8x  0.15 x  0.25  x  0.1x  0.2 x2  0.16 2

2

2

4 4 4 n  n2  n  9 3 9 27.  13 2 1 n2  n  n2  15 15 25 2

q2  28. q2  29. 30.

9 25

3 1 q 10 10

17 3 q 20 20 1 q2  16

q2  

3a3  24a2  12a  96 6a2  24  a2  11a  24 3a3  81 p3  p2  49p  49 p2  6p  7



p2  p  1 p3  1

43.

p  36 2

36.

4 m  4m m  16

38.

2

y1 y  y  30 2



q  49 2



3 2q  14

2 5 m7

2 y6

2

2

3y  4 y  2y  1 2



2y  5 y  2y  1 2

44.

2 7  2 3a  12 a  4a

45.

m5 2  2 m2  9 m  6m  9

46.

m6 m2  2 m2  25 m  10m  25

Write each term as a rational expression. Then compute the sum or difference indicated. 47. a. p2  5p1

48. a. 3a1  12a2 1

b. x2  2x3

b. 2y1  13y2 1

Simplify each compound rational expression. Use either method. 8 1 5 1  3  a 4 27 x 49. 50. 1 25 1 2  2  x 16 3 a 1 3 p 1 p2 y6 51. 52. 1 9 1 y p2 y6

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Precalculus—

Section E Rational Expressions

2 3  3x x3 53. 5 4  x x3 2 2 y  y  20 55. 4 3  y4 y5

1 2  y5 5y 54. 3 2  y y5 2 2 x  3x  10 56. 4 6  x2 x5

Rewrite each expression as a compound fraction. Then simplify using either method. 57. a. 

1  3m1 1  3m1

b.

1  2x2 1  2x2

4  9a2 3a2

b.

3  2n1 5n2

Rewrite each expression as a single term.

1 1  f1 f2 a a  x xh 61. h 1 1 2  21x  h2 2x2 63. h 59.

60.

1 1 1   w x y

a a  x hx 62. h a a 2  2 1x  h2 x 64. h

APPLICATIONS

65. Stock prices: When a hot new stock hits the market, its price will often rise dramatically and then taper off over time. The equation 5017d2  102 models the price P d3  50 of stock XYZ d days after it has “hit the market.” Create a table of values showing the price of the stock for the first 10 days and comment on what you notice. Find the opening price of the stock— does the stock ever return to its original price? 66. Population growth: The Department of Wildlife introduces 60 elk into a new game reserve. It is projected that the size of the herd will grow 1016  3t2 , where according to the equation N  1  0.05t N is the number of elk and t is the time in years. Approximate the population of elk after 14 yr.



58. a.

A-39

67. Typing speed: The number of words per minute that a beginner can type is approximated by the 60t  120 , where N is the number equation N  t of words per minute after t weeks, 2 6 t 6 12. Use a table to determine how many weeks it takes for a student to be typing an average of forty-five words per minute. 68. Memory retention: A group of students is asked to memorize 50 Russian words that are unfamiliar to them. The number N of these words that the average student remembers D days later is modeled by the 5D  35 1D  12. How many words equation N  D are remembered after (a) 1 day? (b) 5 days? (c) 12 days? (d) 35 days? (e) 100 days? According to this model, is there a certain number of words that the average student never forgets? How many?

EXTENDING THE CONCEPT

69. One of these expressions is not equal to the others. Identify which and explain why. 20n a. b. 20 # n  10 # n 10n 1 20 # n c. 20n # d. 10n 10 n

70. The average of A and B is x. The average of C, D, and E is y. The average of A, B, C, D, and E is 2x  3y 3x  2y a. b. 5 5 21x  y2 31x  y2 c. d. 5 5

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Precalculus—

F Radicals and Rational Exponents In Section F you will learn how to:

Square roots and cube roots come from a much larger family called radical expressions. Expressions containing radicals can be found in virtually every field of mathematical study, and are an invaluable tool for modeling many real-world phenomena.

A. Simplify radical expresn

sions of the form 1 an

B. Rewrite and simplify

n

A. Simplifying Radical Expressions of the Form 1 an

radical expressions using rational exponents

In Section A we noted 1a  b only if b2  a. The expression 116 does not represent a real number because there is no number b such that b2  16 1 1a is a real number only if a  02. Of particular interest to us now is an inverse operation for a2. In other words, what operation can be applied to a2 to return a? Consider the following.

C. Use properties of radicals to simplify radical expressions

D. Add and subtract radical expressions

E. Multiply and divide EXAMPLE 1

radical expressions; write a radical expression in simplest form



Evaluating a Radical Expression

Evaluate 2a2 for the values given: a. a  3 b. a  5 c. a  6

F. Evaluate formulas involving radicals

Solution



a. 232  19 3

b. 252  125 5

c. 2162 2  136 6

Now try Exercises 1 and 2



The pattern seemed to indicate that 2a2  a and that our search for an inverse operation was complete—until Example 1(c), where we found that 2162 2  6. Using the absolute value concept, we can repair this apparent discrepancy and state a general rule for simplifying these expressions: 2a2  a. For expressions like 249x2 and 2y6, the radicands can be rewritten as perfect squares and simplified in the same manner: 249x2  217x2 2  7x and 2y6  21y3 2 2  y3. The Square Root of a2: 2a2 For any real number a, 2a2  a. EXAMPLE 2



Simplifying Square Root Expressions Simplify each expression. a. 2169x2 b. 2x2  10x  25

Solution



a. 2169x2  13x  13x b. 2x  10x  25  21x  52  x  5 2

since x could be negative 2

since x  5 could be negative

Now try Exercises 3 and 4

CAUTION





In Section C, we noted that 1A  B2 2  A2  B2. In a similar way, 2A2  B2  A  B,

and you cannot take the square root of individual terms. There is a big difference between the expressions 2A2  B2 and 21A  B2 2  A  B. Try evaluating each when A  3 and B  4.

A-40

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Precalculus—

Section F Radicals and Rational Exponents

A-41

3 3 3 3 To investigate expressions like 2 x , note the radicand in both 1 8 and 1 64 can be written as a perfect cube. From our earlier definition of cube roots we know 3 3 3 3 1 8 2 122 3  2, 1 64  2 142 3  4, and that every real number has only one real cube root. For this reason, absolute value notation is not used or needed when taking cube roots. 3 3 The Cube Root of a3: 2 a 3 3 For any real number a, 2 a  a.

We can extend these ideas to fourth roots, fifth roots, and so on. For example, the 5 fifth root of a is b only if b5  a. In symbols, 1 a  b implies b5  a. Since an odd number of negative factors is always negative: 122 5  32, and an even number of negative factors is always positive: 122 4  16, we must take the index into account n when evaluating expressions like 1 an. If n is even and the radicand is unknown, absolute value notation must be used.

WORTHY OF NOTE 2

Just as 116 is not a real 4 6 number, 1 16 or 1 16 do not represent real numbers. An even number of repeated factors is always positive!

n

The nth Root of an: 2an For any real number a, n 1. 1 an  a when n is even.

EXAMPLE 3



n

2. 1 an  a when n is odd.

Simplifying Radical Expressions Simplify each expression. 3 3 a. 2 b. 2 27x3 64n6 4 e. 2 16m4

Solution

A. You’ve just reviewed how to simplify radical n n expressions of the form 1 a



4 c. 1 81

6 g. 2 1m  52 6

5 f. 2 32p5

3 3 2 27x3  2 13x2 3  3x 4 1 81  3 c. 4 4 e. 2 16m4  2 12m2 4  2m or 2m 6 g. 2 1m  52 6  m  5

4 d. 1 81

7 h. 2 1x  22 7

3 3 b. 2 64n6  2 14n2 2 3  4n2 4 d. 181 is not a real number 5 5 f. 232p5  2 12p2 5  2p 7 h. 2 1x  22 7  x  2

a.

Now try Exercises 5 through 8



B. Radical Expressions and Rational Exponents As an alternative to radical notation, a rational (fractional) exponent can be used, along 3 with the power property of exponents. For 2a3  a, notice that an exponent of onethird can replace the cube root notation and produce the same result: 1 3 3 3 2 a  1a3 2 3  a3  a. In the same way, an exponent of one-half can replace the 1 2 square root notation: 2a2  1a2 2 2  a2  a. In general, we have the following: Rational Exponents If a is a real number and n is an integer greater than 1, n

n

1

then 1 a  2a1  an n

provided 1 a represents a real number.

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Precalculus—

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Appendix I A Review of Basic Concepts and Skills

EXAMPLE 4

Solution





Simplifying Radical Expressions Using Rational Exponents Simplify by rewriting each radicand as a perfect nth power and converting to rational exponent notation. 3 4 a. 2125 b. 216x20 3 3 3 4 4 a. 2125  2 152 b. 216x20   2 12x5 2 4  12x5 2 44 4  12x5 2 4  2x5

1

1

 152 33 3  152 3  5

Now try Exercises 9 and 10 n

Any rational number can be decomposed into the product of a unit fraction and m 1 an integer:  # m. n n

WORTHY OF NOTE While1 the expression 3 182 3  18 represents the real number 2, the expres-

Rational Exponents

sion 182  1 1 82 is not a real number, even though 1 2  . Note that the second 3 6 exponent is not in lowest terms. 6

2

EXAMPLE 5

1

When a rational exponent is used, as in 1a  2a1  an, the m an denominator of the exponent represents the index number, while n m the numerator of the exponent represents the original power on (a ) a. This is true even when the exponent on a is something other 4 Figure AI.4 than one! In other words, the radical expression 2163 can be 1 3 3 4 4 rewritten as A 16 B  16 . This is further illustrated in Figure AI.4 where we see the rational exponent has the form, “power over root.” To evaluate this expres3 1 sion without the aid of a calculator, we use the commutative property to rewrite A 161 B 4 1 3 1 3 as A 164 B 1 and begin with the fourth root of 16: A 164 B 1  23  8. m In general, if m and n have no common factors (other than 1) the expression a n can be interpreted in the following two ways.

WORTHY OF NOTE

2 6

n



If

m n

is a rational number expressed in lowest terms with n  2, then (1) a n  A 2a B m m

n

m

n

(2) a n  2am

or

n

(compute 1 a, then take the mth power) (compute am, then take the nth root) n provided 1 a represents a real number.



Simplifying Expressions with Rational Exponents Find the value of each expression without a calculator, by rewriting the exponent as the product of a unit fraction and an integer. 5 2 4 3 4x6 2 3 3 b a. 27 b. 182 c. a d. 492 9

Solution

B. You’ve just reviewed how to rewrite and simplify radical expressions using rational exponents



2

1

#

a. 273  273 2 1  A 273 B 2  32 or 9 1# 6 52 4x 4x6 2 5 c. a b a b 9 9 1 4x6 2 5 b d  ca 9 2x3 5 32x15  c d  3 243

b. 182 3  182 3 4 1  182 34    24  16 4

1

#

d. 492   A 492 B 3  1 1492 3  172 3 or 343 3

1

Now try Exercises 11 through 16



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Precalculus—

Section F Radicals and Rational Exponents

A-43

C. Properties of Radicals and Simplifying Radical Expressions The properties used to simplify radical expressions are closely connected to the properties of exponents. For instance, the product to a power property: 1xy2 n  xnyn 1 1 1 holds true, even when n is a rational number. This means 1xy2 2 is equal to x2 y2 and 1 1 1 14 # 252 2 is equal to 42 # 252. When this statement is expressed in radical form, we have 14 # 25  14 # 125, with both having a value of 10. It is likewise true that 100 1100  , which suggests the following properties of radicals. These can be A 25 125 extended to include cube roots, fourth roots, and so on. Properties of Radicals n

n

If 1 A and 1 B represent real valued expressions, and B  0, n

n

n

n

(1) 1 AB  1 A # 1 B; n

n

(2)

n

1 A # 1 B  1 AB

CAUTION

1A A  n ; BB 1B n 1A A  n n 1B B B n

Note that this property applies only to a product of two terms, not to a sum or difference. In other words, while 29x2  3x, 29  x2  |3  x| !

One application of the product property is to simplify radical expressions. In n general, the expression 1a is in simplified form if a has no factors (other than 1) that are perfect nth roots. EXAMPLE 6



Simplifying Radical Expressions Write each expression in simplest form using the product property. 3 3 3 a. 118 b. 5 2125x4 c. 1.2 216n4 24n5

Solution



a. 118  19 # 2  1912  3 12

3 3 b. 5 2 125x4  5 # 2 125 # x4 3 3 3 # 3 1 These steps can 5# 2 125 # 2 x 2x be done mentally. e 3  5 # 5 # x # 1x 3  25x 1 x

3

3 3 c. 1.2 216n4 2 4n5  1.2 264 # n9 product property Since the index is 3, we look for perfect cube factors in the radicand.

WORTHY OF NOTE Rational exponents also could have been used to simplify the expression from Example 9, since 9 3 3 1.2 1 64 2 n9  1.2142n3  3 4.8n . Also see Example 8.

3 3 9  1.2 2 64 2 n 3 3  1.2 264 2 1n3 2 3  1.2142n3  4.8n3

product property rewrite n9 as a perfect cube simplify result

Now try Exercises 17 through 20



When radicals are combined using the product property, the result may contain a perfect nth root, which should be simplified. Note that the index numbers must be the same in order to use this property.

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Precalculus—

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Appendix I A Review of Basic Concepts and Skills

EXAMPLE 7



Simplifying Radical Expressions Simplify each expression: 218a5 a. 22a

Solution



a.

218a5 22a



b.

18a5 B 2a

b.

 29a4  3a2

81 A 125x3 3

3 81 1 81  3 3 A 125x 2125x3 3 1 27 # 3  5x 3 31 3  5x 3

Now try Exercises 21 and 22



Radical expressions can also be simplified using rational exponents. EXAMPLE 8



Using Rational Exponents to Simplify Radical Expressions Simplify using rational exponents: 3 4 a. 236p4q5 b. v 2 v

Solution



a. 236p4q5  136p4q5 2 2 1 4 5  362p2q2 4 1  6p2q12  2 2 1  6p2q2q2  6p2q2 1q 1

1

3 3 c. 2 1x  2x2 1 1  1x2 2 3 1 #1  x2 3 1 6  x6 or 1 x

C. You’ve just reviewed how to use properties of radicals to simplify radical expressions

3 c. 2 1x

3 d. 1 m1m 4

3 4 b. v 2 v  v1 # v3 3 4  v3 # v3 7  v3 6 1  v3v3 3  v2 1 v 1

1

3 d. 1 m1m  m3m2 1 1  m3  2 5  m6 6  2m5

Now try Exercises 23 and 24



D. Addition and Subtraction of Radical Expressions 3 Since 3x and 5x are like terms, we know 3x  5x  8x. If x  1 7, the sum becomes 3 3 3 3 1 7  5 1 7  8 1 7, illustrating how like radical expressions can be combined. Like radicals are those that have the same index and radicand. In some cases, we can identify like radicals only after radical terms have been simplified.

EXAMPLE 9



Adding and Subtracting Radical Expressions Simplify and combine (if possible). 3 3 a. 145  2 120 b. 2 16x5  x 2 54x2

Solution

D. You’ve just reviewed how to add and subtract radical expressions



a. 145  2 120  3 15  212 152 simplify radicals: 245  29 # 5; 220   3 15  4 15 like radicals  7 15 result 3 3 3 3 5 2 3 # 2 # # b. 216x  x 254x  28 2 x x  x 2 27 # 2 # x2 3 3 2 2  2x 22x  3x 22x simplify radicals 3 2  x 22x result

24 # 5

Now try Exercises 25 through 28



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Precalculus—

Section F Radicals and Rational Exponents

A-45

E. Multiplication and Division of Radical Expressions; Radical Expressions in Simplest Form Multiplying radical expressions is simply an extension of our earlier work. The multiplication can take various forms, from the distributive property to any of the special products reviewed in Section C. For instance, 1A  B2 2  A2  2AB  B2, even if A or B is a radical term. EXAMPLE 10



Evaluating a Quadratic Expression Show that when x2  4x  1 is evaluated at x  2  13, the result is zero.

Solution



x2  4x  1 original expression 12  132  412  132  1 substitute 2  23 for x 4  4 13  3  8  413  1 multiply 14  3  8  12  14 13  4 132 commutative and associative properties 0✓ 2

Now try Exercises 29 through 36



When we applied the quotient property in Example 7, we obtained a denominator free of radicals. Sometimes the denominator is not automatically free of radicals, and the need to write radical expressions in simplest form comes into play. This process is called rationalizing the denominator. Radical Expressions in Simplest Form A radical expression is in simplest form if: 1. The radicand has no perfect nth root factors. 2. The radicand contains no fractions. 3. No radicals occur in a denominator. As with other types of simplification, the desired form can be achieved in various ways. If the denominator is a single radical term, we multiply the numerator and denominator by the factors required to eliminate the radical in the denominator [see Example 11(a)]. If the radicand is a rational expression, it is generally easier to build an equivalent fraction within the radical having perfect nth root factors in the denominator [see Example 11(b)]. In some applications, the denominator may be a sum or difference containing a radical term. In this case, we multiply by a conjugate since 1A  B21A  B2  A2  B2. If either A or B is a square root, the result will be a denominator free of radicals.

EXAMPLE 11



Simplifying Radical Expressions Simplify by rationalizing the denominator. Assume a  0. 2 1 3 3 a. b. c. 4 A 4a 16  12 5 23

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Precalculus—

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Appendix I A Review of Basic Concepts and Skills

Solution



a.

2

2

#

23

5 23 23 223 2 23   2 15 51 232 3 3 2a2 b. 3 4  3 4 # 2 A 4a B 4a 2a 6a2  3 6 B 8a 3 26a2  2a2 c.

E. You’ve just reviewed how to multiply and divide radical expressions and write a radical expression in simplest form

5 23



multiply numerator and denominator by 23

simplify—denominator is now rational 4 # 2  8 is the smallest perfect cube with 4 as a factor; a4 # a2  a6 is the smallest perfect cube with a4 as a factor

the denominator is now a perfect cube—simplify

result

1 1 # 1 16  122  16  12 16  12 1 16  122 16  12  1 162 2  1 122 2 16  12  62 

16  12 4 Now try Exercises 37 through 42



F. Formulas and Radicals Hypotenuse

Leg 90

Leg

One application that often involves radical terms is the Pythagorean theorem. A right triangle is one that has a 90° angle. The longest side (opposite the right angle) is called the hypotenuse, while the other two sides are simply called “legs.” The Pythagorean theorem is a formula that says if you add the square of each leg, the result will be equal to the square of the hypotenuse. Furthermore, we note the converse of this theorem is also true. Pythagorean Theorem 1. For any right triangle with legs a, b and hypotenuse c, a2  b2  c2 2. For any triangle with sides a, b, and c, if a2  b2  c2, then the triangle is a right triangle. A geometric interpretation of the theorem is given in the figure, which shows 32  42  52.

Area 16 in2

4

ea Ar in2 25

25

13

5

c

7

5

3

Area 9 in2

12

24

  25  144  169  52

122

132

b

  49  576  625 72

242

252

 b2  c2 general case

a2

a

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Precalculus—

Section F Radicals and Rational Exponents

EXAMPLE 12



A-47

Applying the Pythagorean Theorem An extension ladder is placed 9 ft from the base of a building in an effort to reach a third-story window that is 27 ft high. What is the minimum length of the ladder required? Answer in exact form using radicals, and approximate form by rounding to one decimal place.

Solution



We can assume the building makes a 90° angle with the ground, and use the Pythagorean theorem to find the required length. Let c represent this length. c2  a2  b2 c2  192 2  1272 2 c2  81  729 c2  810 c  2810 c  9 210 c  28.5 ft

27 ft

c

Pythagorean theorem substitute 9 for a and 27 for b 92  81, 272  729 add definition of square root; c 7 0 exact form: 2810  281 # 10  9210 approximate form

The ladder must be at least 28.5 ft tall. Now try Exercises 43 and 44 9 ft F. You’ve just reviewed how to evaluate formulas involving radicals

F EXERCISES 

DEVELOPING YOUR SKILLS

Evaluate the expression 2x2 for the values given.

1. a. x  9

b. x  10

2. a. x  7

b. x  8

Simplify each expression, assuming that variables can represent any real number.

3. a. 249p2 c. 281m4

b. 21x  32 2 d. 2x2  6x  9

4. a. 225n2 c. 2v10

b. 21y  22 2 d. 24a2  12a  9

5. a. 1 64

b. 2125x 3 3 v d. B 8

3

3 c. 2216z12

6. a. 1 8 3

3 c. 227q9

3

3

6 b. 1 64 5 d. 2243x5 6 f. 2 1h  22 6

3 9. a. 1125

4 b. 281n12 49v10 d. B 36

4 8. a. 1 216 5 c. 21024z15 5 e. 2 1q  92 5

c. 236 3 10. a. 1216

c. 2121 3

4 b. 1216 5 d. 21024z20 6 f. 2 1p  42 6

4 b. 216m24 25x6 d. B 4

b. a

2

b. 2125p 3 3 w d. B 64 3

6 7. a. 1 64 5 c. 2243x10 5 e. 2 1k  32 5

11. a. 83 4 2 c. a b 25 3

d. a

3

16 2 b 25 27p6 8q3

2 3

b



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Appendix I A Review of Basic Concepts and Skills 3

3

12. a. 92 c. a

34

16 b 81

4 2 b. a b 9 2 125v9 3 b d. a 27w6

22. a.

227y7

23y 20 c. B 4x4

b.

3 2 72b5

3 2 3b2 125 d. 9 3 A 27x6

3

3 2

13. a. 144

c. 1272

23

4 2 b. a b 25 4 27x3 3 b d. a 64

23. a. 232x10y15 5

14. a. 100

c. 11252

23

4

16 3

4 3 c. 31 b

d.

26

e. 2b 1b 4

3

3 2

b. x 2x5

49 2 b. a b 36 4 x9 3 d. a b 8

4 24. a. 2 81a12b16 4 c. 32a

5 6 b. a 2 a

d.

3 1 3 4 2 3

3 4 e. 1 c1c

Use properties of exponents to simplify. Answer in exponential form without negative exponents.

15. a. A 2n2p5 B 5 2

16. a. a

24x

3 8

1 2

4x

b

2

b. a

3 4

8y

3 2

64y

b

1 3

b. A 2x4y4 B 4 1

3

Simplify each expression. Assume all variables represent nonnegative real numbers.

17. a. 218m2 3 3 c. 264m3n5 8 6  228 e. 2 18. a. 28x6

b. 2 2125p3q7 3

d. 232p3q6 f.

d. 254m6n8

e.

12  248 8

f.

19. a. 2.5 218a22a3 x3y 4x5y c. B 3 B 12y 20. a. 5.1 22p232p5 c. 21. a.

ab2 25ab4 B 3 B 27 28m5

22m 45 c. B 16x2

d. 29v u23u v 3

5 2

4 b.  25q220q3 5 3 3 d. 25cd2 125cd

b.

3 2 108n4

3 2 4n 3 81 d. 12 A 8z9

3280  2 2125 5 212  2 227 3 212x  5 275x 3 240q  9 210q

3 3 28. a. 5 254m3  2m216m3 b. 110b  1200b  120  140

2 b.  23b212b2 3 2

26. a. b. c. d.

c. 272x3  150  17x  127

20  232 4

3

12 272  9 298 8 248  3 2108 7 218m  250m 2 228p  3 263p

b. 14  13x  112x  145

3 b. 3 2128a4b2

2 3 227a2b6 9

25. a. b. c. d.

3 3 27. a. 3x 154x  5 216x4

27  272 6

c.

Simplify and add (if possible).

c. 275r3  132  127r  138 Use a substitution to verify the solutions to the quadratic equation given.

29. x2  4x  1  0 a. x  2  13

b. x  2  13

30. x2  10x  18  0 a. x  5  17

b. x  5  17

31. x  2x  9  0 a. x  1  110 b. x  1  110 2

32. x2  14x  29  0 a. x  7  215

b. x  7  215

Compute each product and simplify the result.

33. a. 17 122 2 c. 1n  152 1n  152

b. 131 15  172 d. 16  132 2

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Section F Radicals and Rational Exponents

34. a. 10.3 152 2 c. 14  132 14  132

b. 151 16  122 d. 12  152 2

38. a.

35. a. 13  2172 13  2 172 b. 1 15  11421 12  1132

c.

c. 12 12  6 16213110  172

Rationalize each expression by building perfect nth root factors for each denominator. Assume all variables represent positive quantities.

c.



3 112

b.

20 B 27x3

27 B 50b

d.

1 A 4p

b.

125 B 12n3

5 B 12x

d.

3 3 A 2m2

e.

8 3 31 5

Simplify the following expressions by rationalizing the denominators. Where possible, state results in exact form and approximate form, rounded to hundredths.

36. a. 15  41102 11  2 1102 b. 1 13  1221 110  1112 c. 13 15  4 1221 115  162

37. a.

4 120

3

e.

39. a.

8 3  111

b.

1 2  13

40. a.

7 17  3

b.

12 1x  13

41. a.

110  3 13  12

b.

7  16 3  3 12

42. a.

1  12 16  114

b.

1  16 5  2 13

5 1a 3

APPLICATIONS

43. Length of a cable: A radio tower is secured by cables that are anchored in the ground 8 m from its base. If the cables are attached to the tower 24 m above the ground, what is the length of each cable? Answer in (a) exact form using radicals, and (b) approximate form by rounding to one decimal place.

24 m

c

8m 44. Height of a kite: Benjamin Franklin is flying his kite in a storm once again. John Adams has walked to a position directly under the kite and is 75 m from Ben. If the kite is 50 m above John Adams’ head, how much string S has Ben let out? Answer in (a) exact form using radicals, and (b) approximate form by rounding to one decimal place.

S

75 m

50 m

The time T (in days) required for a planet to make one revolution around the sun is modeled by the function 3 T  0.407R2, where R is the maximum radius of the planet’s orbit (in millions of miles). This is known as Kepler’s third law of planetary motion. Use the equation given to approximate the number of days required for one complete orbit of each planet, given its maximum orbital radius.

45. a. Earth: 93 million mi b. Mars: 142 million mi c. Mercury: 36 million mi 46. a. Venus: 67 million mi b. Jupiter: 480 million mi c. Saturn: 890 million mi 47. Accident investigation: After an accident, police officers will try to determine the approximate velocity V that a car was traveling using the formula V  2 26L, where L is the length of the skid marks in feet and V is the velocity in miles per hour. (a) If the skid marks were 54 ft long, how fast was the car traveling? (b) Approximate the speed of the car if the skid marks were 90 ft long. 48. Wind-powered energy: If a wind-powered generator is delivering P units of power, the velocity V of the wind (in miles per hour) can be 3 P , where k is a constant determined using V  Ak

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A-50

Appendix I A Review of Basic Concepts and Skills

that depends on the size and efficiency of the generator. Rationalize the radical expression and use the new version to find the velocity of the wind if k  0.004 and the generator is putting out 13.5 units of power.

of the cone. Find the surface area of a cone that has a radius of 6 m and a height of 10 m. Answer in simplest form. 50. Surface area: The lateral surface a area S of a frustum (a truncated cone) is given by the formula h 2 2 S  1a  b2 2h  1b  a2 , b where a is the radius of the upper base, b is the radius of the lower base, and h is the height. Find the surface area of a frustum where a  6 m, b  8 m, and h  10 m. Answer in simplest form. The expression x2  7 is not factorable using integer values. But the expression can be written in the form x2  1 272 2, enabling us to factor it as a “binomial” and its conjugate: 1x  2721x  272. Use this idea to factor the following expressions.

49. Surface area: The lateral surface area (surface area excluding the base) S of a cone is given by the formula S  r 2r2  h2, where r is the radius of the base and h is the height 

h

51. a. x2  5

b. n2  19

52. a. 4v2  11

b. 9w2  11

r

EXTENDING THE CONCEPT

53. The following terms 23x  29x  227x  . . . form a pattern that continues until the sixth term is found. (a) Compute the sum of all six terms; (b) develop a system (investigate the pattern further) that will enable you to find the sum of 12 such terms without actually writing out the terms.

1 1 1 1 9 55. If A x2  x2 B 2  , find the value of x2  x2. 2

56. Rewrite by rationalizing the numerator: 1x  h  1x h

54. Find a quick way to simplify the expression without the aid of a calculator. 3 4

2 5

aaaa

a a a a a3 b 5 6

4 5

3 2

10 3

OVERVIEW OF APPENDIX I Important Definitions, Properties, Formulas, and Relationships A Notation and Relations concept • Set notation:

notation {members}

• Is an element of • Empty set • Is a proper subset of • Defining a set

 {} ( 5x| x . . .6

description braces enclose the members of a set

indicates membership in a set a set having no elements indicates the elements of one set are entirely contained in another the set of all x, such that x . . .

example set of even whole numbers A  50, 2, 4, 6, 8, p6 14  A odd numbers in A S  50, 6, 12, 18, 24, p6 S ( A S  5x |x  6n for n  6

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Precalculus—

Overview of Appendix I

A Sets of Numbers

• Natural:   51, 2, 3, 4, p6

• Integers:   5. . . , 3, 2, 1, 0, 1, 2, 3, . . .6 • Irrational:   {numbers with a nonterminating, nonrepeating decimal form}

Absolute Value of a Number 0n 0  e

n n

if n  0 if n 6 0

• Whole:   50, 1, 2, 3, p6 p • Rational:   e , where p, q  ; q  0 f q • Real:   {all rational and irrational numbers}

Distance Between a and b on a Number Line a  b or b  a

B Properties of Real Numbers: For real numbers a, b, and c, Commutative Property • Addition: a  b  b  a • Multiplication: a # b  b # a Identities • Additive: 0  a  a • Multiplicative: 1 # a  a

Associative Property • Addition: 1a  b2  c  a  1b  c2 • Multiplication: 1a # b2 # c  a # 1b # c2 Inverses • Additive: a  1a2  0 p q • Multiplicative: #  1; p, q  0 q p

C Properties of Exponents: For real numbers a and b, and integers m, n, and p (excluding 0 raised to a nonpositive power), m

• Product property: b

# bn  bmn

• Product to a power: 1ambn 2 p  amp # bnp bm • Quotient property: n  bmn 1b  02 b 1 a n b n • Negative exponents: bn  n ; a b  a b a b b 1a, b  02

Special Products

• 1A  B2 1A  B2  A2  B2 • 1A  B2 1A2  AB  B2 2  A3  B3

D Special Factorizations

• A2  B2  1A  B21A  B2 • A3  B3  1A  B21A2  AB  B2 2

• Power property: 1bm 2 n  bmn am p amp • Quotient to a power: a n b  np 1b  02 b b 0 • Zero exponents: b  1 1b  02 • Scientific notation: N 10k; 1 N 6 10, k  

• 1A  B2 2  A2  2AB  B2; 1A  B2 2  A2  2AB  B2 • 1A  B2 1A2  AB  B2 2  A3  B3 • A2  2AB  B2  1A  B2 2 • A3  B3  1A  B2 1A2  AB  B2 2

E Rational Expressions: For polynomials P, Q, R, and S with no denominator of zero, P#R P P#R P • Lowest terms: #  • Equivalence:  # Q R Q Q Q R P #R P#R PR R P S PS P • Multiplication: • Division:   #   #  Q S Q S QS Q S Q R QR • Addition:

Q PQ P   R R R

• Subtraction:

Q PQ P   R R R

A-51

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Precalculus—

A-52

Appendix I A Review of Basic Concepts and Skills

• Addition/subtraction with unlike denominators: 1. Find the LCD of all rational expressions. 2. Build equivalent expressions using LCD. 3. Add/subtract numerators as indicated. 4. Write the result in lowest terms.

F Properties of Radicals • 1a is a real number only for a  0 n • 1a  b, only if bn  a n

• For any real number a, 1an  a when n is even • If a is a real number and n is an integer greater than 1, 1 n n then 1a  an provided 1a represents a real number n

n

• If 1A and 1B represent real numbers, n n n 1 AB  1 A # 1 B

• A radical expression is in simplest form when:

• 2a  b, only if b2  a n • If n is even, 1a represents a real number only if a  0 n • For any real number a, 1an  a when n is odd m • If is a rational number written in lowest terms with n m m n n n  2, then a n  ( 1a)m and a n  1am provided n 2a represents a real number. n n • If 1A and 1B represent real numbers and B  0, n 1A n A  n BB 1B 1. the radicand has no factors that are perfect nth roots, 2. the radicand contains no fractions, and 3. no radicals occur in a denominator.

Pythagorean Theorem • For any triangle with sides a, b, and c, if a2  b2  c2, then the triangle is a right triangle.

• For any right triangle with legs a and b and hypotenuse c: a2  b2  c2.

PRACTICE TEST 1. State true or false. If false, state why. a.  (  b.  (  1 c. 22   d.   2

6. State the value of each expression, if possible. a. 0  6 b. 6  0 7. State the number of terms in each expression and identify the coefficient of each. c2 a. 2v2  6v  5 b. c 3

2. State the value of each expression. 3 a. 2121 b. 1 125 c. 236 d. 2400

8. Evaluate each expression given x  0.5 and y  2. Round to hundredths as needed.

3. Evaluate each expression: 7 1 5 1 a.  a b b.   8 4 3 6 c. 0.7  1.2 d. 1.3  15.92

a. 2x  3y2

b. 22  x14  x2 2 

y x

9. Translate each phrase into an algebraic expression. a. Nine less than twice a number is subtracted from the number cubed. b. Three times the square of half a number is subtracted from twice the number.

4. Evaluate each expression: 1 a. 142a2 b b. 10.6211.52 3 2.8 c. d. 4.2  10.62 0.7

12 10 5. Evaluate using a calculator: 200011  0.08 12 2

#

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Precalculus—

Practice Test

10. Create a mathematical model using descriptive variables. a. The radius of the planet Jupiter is approximately 119 mi less than 11 times the radius of the Earth. Express the radius of Jupiter in terms of the Earth’s radius. b. Last year, Video Venue Inc. earned $1.2 million more than four times what it earned this year. Express last year’s earnings of Video Venue Inc. in terms of this year’s earnings. 11. Simplify by combining like terms. a. 8v2  4v  7  v2  v b. 413b  22  5b c. 4x  1x  2x2 2  x13  x2 12. Factor each expression completely. a. 9x2  16 b. 4v3  12v2  9v c. x3  5x2  9x  45 13. Simplify using the properties of exponents. 5 a. 3 b. 12a3 2 2 1a2b4 2 3 b 5p2q3r4 2 m2 3 c. a b d. a b 2n 2pq2r4 14. Simplify using the properties of exponents. 12a3b5 a. 3a2b4 b. 13.2 1017 2 12.0 1015 2 a3 # b 4 c. a 2 b d. 7x0  17x2 0 c 15. Compute each product. a. 13x2  5y213x2  5y2 b. 12a  3b2 2 16. Add or subtract as indicated. a. 15a3  4a2  32  17a4  4a2  3a  152 b. 12x2  4x  92  17x4  2x2  x  92

A-53

Simplify or compute as indicated. 17. a. c. e. f. 18. a. c. e. g.

x5 4  n2 b. 2 5x n  4n  4 3 x  27 3x2  13x  10 d. 2 x  3x  9 9x2  4 x2  25 x2  x  20  3x2  11x  4 x2  8x  16 m3 2  2 51m  42 m  m  12 8 b. 3 21x  112 2 A 27v3 3 25 2 4  232 a b d. 16 8 7 240  290 f. 1x  2521x  252 2 8 h. B 5x 26  22

19. Maximizing revenue: Due to past experience, the manager of a video store knows that if a popular video game is priced at $30, the store will sell 40 each day. For each decrease of $0.50, one additional sale will be made. The formula for the store’s revenue is then R  130  0.5x2140  x2, where x represents the number of times the price is decreased. Multiply the binomials and use a table of values to determine (a) the number of 50¢ decreases that will give the most revenue and (b) the maximum amount of revenue. 20. Use the Pythagorean theorem to determine the length of the diagonal of the rectangular prism shown in the figure. (Hint: First find the diagonal of the base.)

42 cm

32 cm

24 cm

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Precalculus—

APPENDIX II

More on Synthetic Division As the name implies, synthetic division simulates the long division process, but in a condensed and more efficient form. It’s based on a few simple observations of long division, as noted in the division (x3  2x2  13x  172  1x  52 shown in Figure AII.1. Figure AII.1

Figure AII.2

x  3x  2 x  5 x  2x2  13x  17 2

x  5 1

3

 1x  5x 2 3

2

3 13

2 17

5

↓

3x  13x 2

1 2

 13x  15x2 2

3 15

↓

2x  17

2

 12x  102

10

7

7

remainder

remainder

A careful observation reveals a great deal of repetition, as any term in red is a duplicate of the term above it. In addition, since the dividend and divisor must be written in decreasing order of degree, the variable part of each term is unnecessary as we can let the position of each coefficient indicate the degree of the term. In other words, we’ll agree that 1

2

13

17

represents the polynomial 1x3  2x2  13x  17.

Finally, we know in advance that we’ll be subtracting each partial product, so we can “distribute the negative,” shown at each stage. Removing the repeated terms and variable factors, then distributing the negative to the remaining terms produces Figure AII.2. The entire process can now be condensed by vertically compressing the rows of the division so that a minimum of space is used (Figure AII.3). Figure AII.3 1 3

x  5 1

2

quotient

Figure AII.4 1 3 2 2 13 17

2

13

17

dividend

5

15

10

products

↓

5

15

3

2

7

sums

1

3

2

x  5 1

10 7

dividend products remainder

quotient

Further, if we include the lead coefficient in the bottom row (Figure AII.4), the coefficients in the top row (in blue) are duplicated and no longer necessary, since the quotient and remainder now appear in the last row. Finally, note all entries in the product row (in red) are five times the sum from the prior column. There is a direct connection between this multiplication by 5 and the divisor x  5, and in fact, it is the zero of the divisor that is used in synthetic division ( x  5 from x  5  0). A simple change in format makes this method of division easier to use, and highlights the location of the divisor and remainder (the blue brackets in Figure AII.5). Note the process begins by “dropping the lead coefficient into place” (shown in bold). The full process of synthetic division is shown in Figure AII.6 for the same exercise. A-54

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Appendix II More on Synthetic Division

Figure AII.5 2 13

divisor (use 5, not 5) →

5 1 ↓ drop lead coefficient into place → 1

17

A-55

← coefficients of dividend ← quotient and remainder appear in this row

We then multiply this coefficient by the “divisor,” place the result in the next column and add. In a sense, we “multiply in the diagonal direction,” and “add in the vertical direction.” Continue the process until the division is complete.

multiply by 5

↓ 1

The result is x2  3x  2 

2 5 3

↓

5 1

Figure AII.6 13 17 15 10 2 7

← coefficients of dividend ← quotient and remainder appear in this row

7 , read from the last row. x5

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Precalculus—

APPENDIX III

More on Matrices Reduced Row-Echelon Form A matrix is in reduced row-echelon form if it satisfies the following conditions: 1. All null rows (zeroes for all entries) occur at the bottom of the matrix. 2. The first non-zero entry of any row must be a 1. 3. For any two consecutive, nonzero rows, the leading 1 in the higher row is to the left of the 1 in the lower row. 4. Every column with a leading 1 has zeroes for all other entries in the column. Matrices A through D are in reduced row-echelon form. 0 A  £0 0

1 0 0

0 0 0

0 1 0

0 0 1

5 3§ 2

1 B  £0 0

0 1 0

0 0 0

5 3§ 0

1 C  £0 0

0 1 0

0 3 0

5 2 § 0

1 D  £0 0

0 1 0

5 2 0

0 0§ 1

Where Gaussian elimination places a matrix in row-echelon form (satisfying the first three conditions), Gauss-Jordan elimination places a matrix in reduced rowechelon form. To obtain this form, continue applying row operations to the matrix until the fourth condition above is also satisfied. For a 3  3 system having a unique solution, the diagonal entries of the coefficient matrix will be 1’s, with 0’s for all other entries. To illustrate, we’ll extend Example 4 from Section 8.5 until reduced rowechelon form is obtained. EXAMPLE 4



Solving Systems Using the Augmented Matrix 2x  y  2z  7 Solve using Gauss-Jordan elimination: • x  y  z  1 2y  z  3 2x  y  2z  7 • x  y  z  1 2y  z  3 1 1 1 1 £2 1 2 7 § 0 2 1 3 1 1 1 1 £0 1 4 5§ 0 2 1 3 1 1 4 5§ 1 1

1 £0 0

1 1 0

R 1 4 R2

x  y  z  1 • 2x  y  2z  7 2y  z  3

matrix form →

1 1 1 1 £2 1 2 7 § 0 2 1 3

2R1  R2 S R2

1 1 1 1 £ 0 1 4 5 § 0 2 1 3

1R2 → R2

1 1 1 1 £0 1 4 5§ 0 2 1 3

1 1 4 5§ 7 7

2R2  R3 S R3

1 £0 0

1 1 0

R2  R1 S R1

1 £0 0

0 3 6 1 4 5§ 0 1 1

R3 S R3 7

3R3  R1 S R1 4R3  R2 S R2

1 £0 0

1 1 0

1 1 4 5§ 1 1

1 £0 0

0 1 0

0 3 0 1§ 1 1

The final matrix is in reduced row-echelon form with solution (3, 1, 1) just as in Section 8.5.

A-56

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Precalculus—

A-57

Appendix III More on Matrices

The Determinant of a General Matrix To compute the determinant of a general square matrix, we introduce the idea of a cofactor. For an n  n matrix A, Aij  112 ij  Mij  is the cofactor of matrix element aij, where Mij represents the determinant of the corresponding minor matrix. Note that i  j is the sum of the row and column of the entry, and if this sum is even, 112 ij  1, while if the sum is odd, 112 ij  1 (this is how the sign table for a 3  3 determinant was generated). To compute the determinant of an n  n matrix, multiply each element in any row or column by its cofactor and add. The result is a tier-like process in which the determinant of a larger matrix requires computing the determinant of smaller matrices. In the case of a 4  4 matrix, each of the minor matrices will be size 3  3, whose determinant then requires the computation of other 2  2 determinants. In the following illustration, two of the entries in the first row are zero for convenience. For 2 1 A ≥ 3 0 we have: det1A2  2 # 112

11

2 † 1 3

0 2 1 3 0 4 2

3 0 4 2

0 2 ¥, 1 1

2 1 13 1 †  132 # 112 †3 1 0

2 1 3

2 1† 1

Computing the first 3  3 determinant gives 16, the second 3  3 determinant is 14. This gives: det1A2  21162  31142  74

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Precalculus—

APPENDIX IV

Deriving the Equation of a Conic The Equation of an Ellipse In Section 9.2, the equation 21x  c2 2  y2  21x  c2 2  y2  2a was developed using the distance formula and the definition of an ellipse. To find the standard form of the equation, we treat this result as a radical equation, isolating one of the radicals and squaring both sides. 21x  c2 2  y2  2a  21x  c2 2  y2

1x  c2 2  y2  4a2  4a 21x  c2 2  y2  1x  c2 2  y2

isolate one radical square both sides

We continue by simplifying the equation, isolating the remaining radical, and squaring again. x2  2cx  c2  y2  4a2  4a 21x  c2 2  y2  x2  2cx  c2  y2 4cx  4a  4a 21x  c2  y 2

2

2

simplify

a 21x  c2  y  a  cx 2

2

2

isolate radical; divide by 4

a 3 1x  c2  y 4  a  2a cx  c x 2

2

2

4

2

2 2

square both sides

a x  2a cx  a c  a y  a  2a cx  c x 2 2

2

2 2

2 2

4

2

expand binomials

2 2

expand and distribute a 2 on left

a2x2  c2x2  a2y2  a4  a2c2

add 2a 2cx and rewrite equation

x2 1a2  c2 2  a2y2  a2 1a2  c2 2 y2 x2  1 a2 a2  c2

factor divide by a 2 1a 2  c 2 2

Since a 7 c, we know a2 7 c2 and a2  c2 7 0. For convenience, we let b2  a2  c2 and it also follows that a2 7 b2 and a 7 b (since c 7 0). Substituting b2 for a2  c2 we obtain the standard form of the equation of an ellipse (major axis y2 x2 horizontal, since we stipulated a 7 b): 2  2  1. Note once again the x-intercepts a b are (a, 0), while the y-intercepts are (0, b).

The Equation of a Hyperbola Similar to the development of the equation of an ellipse, the equation 21x  c2 2  y2  21x  c2 2  y2  2a could have been developed using the distance formula and the definition of a hyperbola. To find the standard form of this equation, we apply the same procedures as before. 21x  c2 2  y2  2a  21x  c2 2  y2

isolate one radical

1x  c2  y  4a  4a 21x  c2  y  1x  c2 2  y2 2

2

2

2

2

x  2cx  c  y  4a  4a 21x  c2  y  x  2cx  c 2

2

2

2

2

2

4cx  4a2  4a 21x  c2 2  y2 cx  a  a 21x  c2  y 2

2

2

c x  2a cx  a  a 3 1x  c2  y 4 2 2

2

4

2

2

2

4

2 2

c2x2  a2x2  a2y2  a2c2  a4

2

x2 1c2  a2 2  a2y2  a2 1c2  a2 2 2

A-58

2

y x 1 2  2 a c  a2

2

square both sides expand  y2 binomials

simplify isolate radical; divide by 4

2

square both sides

c x  2a cx  a  a x  2a cx  a c  a y 2 2

2

2 2

2 2

expand and distribute a 2 on the right add 2a 2cx and rewrite equation factor divide by a 2 1c 2  a 2 2

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Precalculus—

Appendix IV Deriving the Equation of a Conic

A-59

From the definition of a hyperbola we have 0 6 a 6 c, showing c2 7 a2 and c2  a2 7 0. For convenience, we let b2  c2  a2 and substitute to obtain the stany2 x2 dard form of the equation of a hyperbola (transverse axis horizontal): 2  2  1. a b Note the x-intercepts are (0, a) and there are no y-intercepts.

The Asymptotes of a Central Hyperbola From our work in Section 9.3, a central hyperbola with a horizontal axis will have b asymptotes at y   x. To understand why, recall that for asymptotic behavior we a investigate what happens to the relation for large values of x, meaning as x S q . y2 x2 Starting with 2  2  1, we have a b b2x2  a2y2  a2b2

clear denominators

ay bx ab 2 2

2 2

2 2

a2y2  b2x2 a1  y2 

a b x2

b2 2 a2 2 x a1  2 b a x

b a2 y x 1 2 a B x As x S q,

isolate term with y

2

factor out b 2x 2 from right side

divide by a 2

square root both sides

b a2 2 S 0, and we find that for large values of x, y   a x. x

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Precalculus—

APPENDIX V

Selected Proofs Proof of the Complex Conjugates Corollary Given p(x) is a polynomial with real number coefficients, complex solutions must occur in conjugate pairs. If a  bi, b  0 is a solution, then a  bi must also be a solution. To prove this for polynomials of degree n 7 2, we let z1  a  bi and z2  c  di be complex numbers, and z1  a  bi and z2  c  di represent their conjugates, and observe the following properties: 1. The conjugate of a sum is equal to the sum of the conjugates. sum: z1  z2 1a  bi2  1c  di2 1a  c2  1b  d2i

sum of conjugates: z1  z2 1a  bi2  1c  di2 1a  c2  1b  d2i ✓

S conjugate of sum S

2. The conjugate of a product is equal to the product of the conjugates. product of conjugates: z1 # z2 1a  bi2 # 1c  di2

product: z1 # z2 1a  bi2 # 1c  di2

ac  adi  bci  bdi2 1ac  bd2  1ad  bc2i

ac  adi  bci  bdi2 1ac  bd2  1ad  bc2i ✓

S conjugate of product S

Since polynomials involve only sums and products, and the complex conjugate of any real number is the number itself, we have the following: Given polynomial p1x2  anxn  an1xn1  p  a1x1  a0, where an, an1, . . . , a1, a0 are real numbers and z  a  bi is a zero of p, we must show that z  a  bi is also a zero. anzn  an1zn1  p  a1z1  a0  p1z2 evaluate p(x) at z anzn  an1zn1  p  a1z1  a0  0 anzn  an1zn1  p  a1z1  a0  0 an zn  an1zn1  p  a1z1  a0  0 an 1zn 2  an1 1zn1 2  p  a1 1z1 2  a0  0 an 1zn 2  an1 1zn1 2  p  a1 1z1 2  a0  0 p1z2  0 ✓

p 1z2  0 given

conjugate both sides property 1 property 2 conjugate of a real number is the number result

Proof of the Determinant Formula for the Area of a Parallelogram 1 Since the area of a triangle is A  ab sin , the area of the corresponding parallelogram 2 is twice as large: A  ab sin . In terms of the vectors u  Ha, bI and v  Hc, dI, we have A  uv sin , and it follows that

A-60

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Precalculus—

Appendix V Selected Proofs

Area  uv 21  cos2,  uv

u ␪

v

 uv

B

1

Pythagorean identity

1u # v2 2

substitute for cos 

u2v2

u2v2  1u # v2 2 B u2v2

 2u2v2  1u # v2 2

common denominator simplify

 21a  b 2 1c  d 2  1ac  bd2 2

2

2

2

2

substitute

 2a c  a d  b c  b d  1a c  2acbd  b d 2 2 2

2 2

2 2

2 2

2 2

2 2

expand

 2a d  2acbd  b c

simplify

 21ad  bc2

factor

2 2

A-61

2 2

2

 ad  bc

result

Proof of Heron’s Formula Using Algebra Note that 2a2  d2  h  2c2  e2. It follows that a d

h

2a2  d2  2c2  e2

c b

a2  d2  c2  e2

e

a2  1b  e2 2  c2  e2

a2  b2  2be  e2  c2  e2 a2  b2  c2  2be b2  c2  a2 e 2b This shows:

1 A  bh 2 1  b 2c2  e2 2 1 b2  c2  a2 2 b  b c2  a 2 B 2b 1 b4  2b2c2  2a2b2  2a2c2  a4  c4  b c2  a b 2 B 4b2 1 4b2c2 b4  2b2c2  2a2b2  2a2c2  a4  c4  b  a b 2 B 4b2 4b2 1 4b2c2  b4  2b2c2  2a2b2  2a2c2  a4  c4  b 2 B 4b2 1  22a2b2  2a2c2  2b2c2  a4  b4  c4 4 1  2 3 1a  b2 2  c2 4 3 c2  1a  b2 2 4 4 1  21a  b  c2 1a  b  c21c  a  b21c  a  b2 4

From this point, the conclusion of the proof is the same as the trigonometric development found on page 652.

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Precalculus—

A-62

Appendix V Selected Proofs

Proof that u  v  compv u  |v| z

Consider the vectors in the figure shown, which form a triangle. Applying the Law of Cosines to this triangle yields: u–v

v ␪ x

u y

0 u  v 0 2  0 u 0 2  0 v 0 2  2 0 u 0 0v 0 cos 

Using properties of the dot product (page 680), we can rewrite the left-hand side as follows: 0 u  v 0 2  1u  v2 # 1u  v2

u#uu#vv#uv#v  0u 0 2  2 u # v  0v 0 2

Substituting the last expression for 0u  v 0 2 from the Law of Cosines gives 0 u 0 2  2 u # v  0v 0 2  0 u 0 2  0 v 0 2  2 0 u 0 0 v 0 cos  2 u # v  2 0 u 0 0v 0 cos  u # v  0u 0 0 v 0 cos 

 0 u 0 cos   0v 0 .

Substituting compv u for 0 u 0 cos  completes the proof: u # v  compv u  0 v 0

Proof of DeMoivre’s Theorem: (cos x  i sin x)n  cos(nx)  i sin(nx) For n 7 0, we proceed using mathematical induction. 1. Show the statement is true for n  1 (base case):

1cos x  i sin x2 1  cos11x2  i sin11x2 cos x  i sin x  cos x  i sin x ✓

2. Assume the statement is true for n  k (induction hypothesis): 1cos x  i sin x2 k  cos1kx2  i sin1kx2

3. Show the statement is true for n  k  1:

1cos x  i sin x2 k1  1cos x  i sin x2 k 1cos x  i sin x2 1  3cos1kx2  i sin1kx2 4 1cos x  i sin x2 induction hypothesis  cos1kx2cos x  sin1kx2sin x  i 3 cos1kx2sin x  sin1kx2cos x 4 F-O-I-L  cos 3 1k  12x 4  i sin 3 1k  12x 4 ✓ sum/difference identities By the principle of mathematical induction, the statement is true for all positive integers. For n 6 0 (the theorem is obviously true for n  0), consider a positive integer m, where n  m. 1cos x  i sin x2 n  1cos x  i sin x2 m 1  1cos x  i sin x2 m 1  cos1mx2  i sin1mx2  cos1mx2  i sin1mx2  cos1mx2  i sin1mx2  cos1nx2  i sin1nx2

negative exponent property

DeMoivre’s theorem for n 7 0 multiply numerator and denom by cos1mx2  i sin1mx2 and simplify even/odd identities n  m

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Precalculus—

APPENDIX VI

Families of Polar Curves Circles and Spiral Curves

r

a

a

Circle Circle r  a cos ␪ r  a sin ␪ a represents the diameter of the circle

Circle ra 2

Spiral r  k␪

Roses: r  a sin(n) (illustrated here) and r  a cos(n) a

Four-petal rose r  a sin(2␪)

a

a

Three-petal rose Eight-petal rose r  a sin(3␪) r  a sin(4␪) If n is odd S there are n petals, if n is even S there are 2n petals. |a| represents the maximum distance from the origin (the radius of a circumscribed circle)

a

Five-petal rose r  a sin(5␪)

Limaçons: r  a  b sin  (illustrated here) and r  a  b cos 

|a|  |b| |b|  |a|

|a|  |b|

|a|  |b| |a|  |b|

a

a

|a|  |b| |a|  |b|

a

a

Cardioid Apple Eye-ball Limaçon (limaçon where |a|  |b|) (limaçon where |a| |b|) (limaçon where |a|  2|b|) (inner loop if |a| |b|) r  a  b sin ␪ r  a  b sin ␪ r  a  b sin ␪ r  a  b sin ␪ |a|  |b| represents the maximum distance from the origin (along the axis of symmetry)

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Precalculus—

A-64

Appendix VI Families of Polar Curves

Lemniscates: r 2  a2sin(2) and r 2  a2cos(2)

a a

Lemniscate Lemniscate r2 ⫽ a2 cos(2␪) r2 ⫽ a2 sin(2␪) a represents the maximum distance from the origin (the radius of a circumscribed circle)

cob19421_es.indd Page Sec1:2 10/7/09 4:05:00 PM user-s180

▼

Special Constants

e ⬇ 2.7183

▼

▼

␲ ⬇ 1.0472 3

␲ ⬇ 1.5708 2

␲ ⬇ 3.1416

▼

/Volumes/204/MH00823_r1/0073519421%0/cob19421_pagefiles

12 ⬇ 1.4142

13 ⬇ 1.7321

␲ ⬇ 0.2618 12 13 ⬇ 0.5774 3

1x ⫹ a21x ⫹ b2 ⫽ x2 ⫹ 1a ⫹ b2x ⫹ ab

1a ⫹ b21a ⫺ b2 ⫽ a2 ⫺ b2

1a ⫹ b2 2 ⫽ a2 ⫹ 2ab ⫹ b2

1a ⫺ b2 2 ⫽ a2 ⫺ 2ab ⫹ b2

1a ⫹ b2 3 ⫽ a3 ⫹ 3a2b ⫹ 3ab2 ⫹ b3

1a ⫺ b2 3 ⫽ a3 ⫺ 3a2b ⫹ 3ab2 ⫺ b3

a2 ⫺ b2 ⫽ 1a ⫹ b21a ⫺ b2

a2 ⫹ 2ab ⫹ b2 ⫽ 1a ⫹ b2 2

a2 ⫺ 2ab ⫹ b2 ⫽ 1a ⫺ b2 2

a ⫺ b ⫽ 1a ⫺ b2 1a ⫹ ab ⫹ b 2 3

2

a ⫹ b ⫽ 1a ⫹ b21a ⫺ ab ⫹ b 2

2

3

3

2

2

▼

Formulas from Plane Geometry: P S perimeter, C S circumference, A S area Rectangle

Square

w

P ⫽ 2l ⫹ 2w

Regular Polygon s

P ⫽ 4s

l

A ⫽ s2

A ⫽ lw

Parallelogram

Trapezoid

h

A ⫽ bh

A⫽

b

Triangle Sum of angles

C

A

a2 ⫹ b2 ⫽ c2

Ellipse

A⫽

b

c

1 bh 2

b

a

C ⬇ ␲ 221a ⫹ b 2

b

2

4 A ⫽ ab 3

a

▼

Rectangular Solid

Cube

Right Circular Cylinder

V ⫽ lwh

V ⫽ s3

V ⫽ ␲r2h

S ⫽ lw ⫹ lh ⫹ wh

S ⫽ 6s2

S ⫽ 2␲r 1r ⫹ h2

Right Circular Cone

Right Square Pyramid

Sphere

1 V ⫽ ␲r2h 3

1 V ⫽ b2h 3

4 V ⫽ ␲r3 3

S ⫽ ␲r 1r ⫹ s2

S ⫽ b2 ⫹ b2b2 ⫹ 4h2

S ⫽ 4␲r2

Front endsheets Color: 5 Pages: 4, 2, 5 3

y ⫺ y1 ⫽ m1x ⫺ x1 2

y ⫽ mx ⫹ b, where b ⫽ y1 ⫺ mx1

Parallel Lines

Perpendicular Lines

Slopes Are Equal: m1 ⫽ m2

Slopes Have a Product of ⫺1: m1m2 ⫽ ⫺1

Intersecting Lines

Dependent (Coincident) Lines

Slopes Are Unequal: m1 ⫽ m2

Slopes and y-Intercepts Are Equal: m1 ⫽ m2, b1 ⫽ b2

Logarithms and Logarithmic Properties y ⫽ logb x 3 b y ⫽ x

logb b ⫽ 1

logb bx ⫽ x

blogb x ⫽ x logb

logb 1 ⫽ 0 logc x ⫽

M ⫽ logb M ⫺ logb N N

logb x logb c

logb MP ⫽ P # logb M

Applications of Exponentials and Logarithms A S amount accumulated

P S initial deposit, p S periodic payment

n S compounding periods/year

r S interest rate per year

r R S interest rate per time period a b n

t S time in years

Interest Compounded Continuously

nt

A ⫽ Pert

Payments Required to Accumulate Amount A p⫽

AR 11 ⫹ R2 nt ⫺ 1

Sequences and Series: a1 S 1st term, an S nth term, Sn S sum of n terms, d S common difference, r S common ratio

Formulas from Solid Geometry: S S surface area, V S volume

ISBN: 0-07-351942-1 Author: John W. Coburn Title: College Precalculus, 2e

Slope-Intercept Form (slope m, y-intercept b)

p A ⫽ 冤11 ⫹ R2 nt ⫺ 1冥 R

b ▼

Point-Slope Form

Accumulated Value of an Annuity

Right Parabolic Segment

A ⫽ ␲ab

Equation of Line Containing P1 and P2

r A ⫽ P a1 ⫹ b n

C ⫽ 2␲r ⫽ ␲d

¢y y2 ⫺ y1 ⫽ x2 ⫺ x 1 ¢x

Equation of Line Containing P1 and P2

Interest Compounded n Times per Year

r

A ⫽ ␲r2 b

m⫽

logb MN ⫽ logb M ⫹ logb N

h

Circle

a

Pythagorean Theorem

A ⫹ B ⫹ C ⫽ 180°

2

h

Right Triangle

B

a

▼

Triangle

a

h 1a ⫹ b2 2

s

P ⫽ ns a A⫽ P 2

Slope of Line Containing P1 and P2

d ⫽ 21x2 ⫺ x1 2 2 ⫹ 1y2 ⫺ y1 2 2

Special Factorizations x2 ⫹ 1a ⫹ b2x ⫹ ab ⫽ 1x ⫹ a21x ⫹ b2

Formulas from Analytical Geometry: P1 S (x1, y1), P2 S (x2, y2) Distance between P1 and P2

Special Products

3

▼

␲ ⬇ 0.5236 6 13 ⬇ 0.8660 2

␲ ⬇ 0.7854 4 12 ⬇ 0.7071 2

Arithmetic Sequences

Geometric Sequences

a1, a2 ⫽ a1 ⫹ d, a3 ⫽ a1 ⫹ 2d, . . . , an ⫽ a1 ⫹ 1n ⫺ 12d

a1, a2 ⫽ a1r, a3 ⫽ a1r2, . . . , an ⫽ a1rn⫺1

Sn ⫽ Sn ⫽ ▼

a1 ⫺ a1rn 1⫺r a1 Sq ⫽ ; 冟r冟 6 1 1⫺r

n 1a1 ⫹ an 2 2

Sn ⫽

n 冤2a1 ⫹ 1n ⫺ 12d冥 2

Binomial Theorem n n n 1a ⫹ b2 n ⫽ a b anb0 ⫹ a b an⫺1b1 ⫹ a b an⫺2b2 ⫹ 0 1 2 n! ⫽ n1n ⫺ 121n ⫺ 22

# # # 132122112

# # #

⫹a

n n ba 1bn⫺1 ⫹ a ba0bn n⫺1 n

n n! a b⫽ ; k k!1n ⫺ k2!

0! ⫽ 1

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Student Answer Appendix 43. x  32, 52 ;

CHAPTER 1

[

Exercises 1.1, pp. 10–13 3. literal; two 5. Answers will vary. 7. x  3 27 6 9. v  11 11. b  13. b  15 15. m   17. x  12 5 4 19. x  12 21. p  56 23. a  3.6 25. v  0.5 20 12 27. n  29. p  31. contradiction; { } 21 5 11 P 33. conditional; n   35. identity; 5x |x  6 37. C  10 1M 2Sn T1P2V2 C 3V 39. r  41. T2  43. h  45. n  2 P1V1 a1  an 4r2 21S  B2 20 A C 16 x 47. P  49. y   x  51. y  S B B 9 3 4 x  5 55. a  3; b  2; c  19; x  7 53. y  5 57. a  6; b  1; c  33; x  16 3 59. a  7; b  13; c  27; x  2 61. h  17 cm 63. 510 ft 65. 56 in. 67. 3084 ft 69. 48; 50 71. 5; 7 73. 11: 30 A.M. 75. 36 min 77. 4 quarts; 50% O.J. 79. 16/lb; $1.80/lb 81. 12 lb 83. 16 lb 85. Answers will vary 87. 69 89. 3 91. a. 12x  3212x  32 b. 1x  321x2  3x  92

1. identity; unknown

1

2

)

3

5. Answers will vary.

4

13. 4 3 2 1 0 1 2 3 4 [5 6 15. 4 3 2 1 0 )1) 2 3 4 5 6

3 2 1

0

25. 5n |n  16; 0

[

1

2

[

27. 5x| x 6

1

21. 5x | 2  x  16; 32, 1 4

3

4

2

32 5 6;

5

; a  32, q 2

3

; n  3 1, q 2

)

10 9 8 7 6 5 43 2 1 0

29. 35. 37. 39. 41.

; x  1q, 32 5 2

{ } 31. 5x | x  6 33. 5x | x  6 526; 53, 2, 1, 0, 1, 2, 3, 4, 6, 86 56; 53, 2, 1, 0, 1, 2, 3, 4, 5, 6, 76 54, 66; 52, 4, 5, 6, 7, 86 x  1q, 22 ´ 11, q2;

)

4 3 2 1

0

)

1

2

2

3

3

)

4

5

6

3

4

45. no solution 47. x  1q, q 2 ; 4 3 2 1

49. x  35, 0 4;

0

1

2

[

[

6 5 4 3 2 1

0

1 51. x  1 1 3 , 4 2;

1

0.5

)(

0

0.5

53. x  1q, q 2; 4 3 2 1

55. x  34, 12;

0

1

[

6 5 4 3 2 1

2

1

59. x  116, 82;

1

3

4

2

3

)

0

57. x  31.4, 0.8 4; 1.4

[

2

1

1

0.8

[

0

0

1

4

2

)

8

12

61. m  1q, 02 ´ 10, q 2 63. y  1q, 72 ´ 17, q2 65. a  1q, 12 2 ´ 1 12 , q 2 67. x  1q, 42 ´ 14, q2 69. x  3 2, q 2 71. n  3 4, q 2 73. b  3 43 , q 2 75. y  1q, 24 BH2 77. a. W  b. W 6 177.34 lb 79. x  81% 81. b  $2000 704 83. 0 6 w 6 7.5 m 85. 7.2° 6 C 6 29.4° 87. h 7 6 89. Answers may vary. 91.  93.  95.  97. 99. 2n  8 101. 17 18 x  5

Exercises 1.3, pp. 29–31

17. 43 21 0 1 2) 3 4 5) 6 7 19. 5x| x  26; 3 2, q 2 23. 5a |a  26;

1

20 1612 8 4

1. set; interval 3. intersection; union 7. w  45 9. 250 6 T 6 450 0

0

[

Exercises 1.2, pp. 20–23

11. 3 2 1

3 2 1

5. no solution; answers will vary. 7. 54, 66 1 1 8 9. 52, 126 11. 53.35, 0.856 13. e  , 2 f 15. e  , f 7 2 2 17. { } 19. 510, 66 21. {3.5, 11.5} 23. 51.6, 1.66 3 8 14 d 25. 35, 9 4 27. 29. a1, b 31. 15, 32 33. c , 5 3 3 7 35. 37. c  , 0 d 39. 1q, 102 ´ 14, q2 4 41. 1q, 3 4 ´ 3 3, q 2 43. aq,  7 d ´ c 7 , qb 3 3 45. aq, 3 d ´ 3 1, q 2 47. 1q, q 2 49. 1q, 02 ´ 15, q2 . 7

1. reverse

3. 7; 7

51. 1q, 0.75 4 ´ 3 3.25, q 2

53. aq,  7 b ´ 11, q2 15 55. 45  d  51 in. 57. in feet: [32,500, 37,600]; yes 59. in feet: d 6 210 or d 7 578 61. a. s  37.58  3.35 b. [34.23, 40.93] 63. a. s  125  23 b. [102, 148] 65. a. | d  42.7 | 6 0.03 b. |d  73.78 | 6 1.01 c. |d  57.150 | 6 0.127 d. | d  2171.05 | 6 12.05

SA1

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SA2

Student Answer Appendix 4 67. a. x  4 b. c , 4 d 3

e. golf: t  0.0014

e. { } 69. 3x 12x  5213x  42

71.

3 c. x  0 d. aq, d 5

3  13  0.21 6

Mid-Chapter Check, pp. 31–32 1. e. 3. 4.

50 a. r  9 b. x  6 c. identity; m   d. y  13 H  16t2 contradiction: { } f. x  5.5 2. v0  t S x B 12  y2 a. x  1 or x  2

[

4 3 2 1

0

[

1

2

3

b. 16 6 x  19 0

15

(

[

20

5 5 17 5. a. x  aq, b ´ a , qb b. x  aq, d 2 2 6 6. a. 54, 146 b. { } 7. a. q  18, 02 b. 566 23 19 8. a. d  1q, 04 ´ 34, q2 b. y  aq,  b ´ a , q b 2 2 c. k  1q, q 2 9. 1 hr, 20 min 10. w  38, 26 4 ; yes

Reinforcing Basic Concepts, p. 32 Exercise 1: x  3 or x  7 Exercise 2: x  3 5, 3 4 Exercise 3: x  1q, 1 4 ´ 3 4, q 2

Exercises 1.4, pp. 39–42 1. 3  2i 3. 2; 3 12 5. (b) is correct. 7. a. 4i b. 7i c. 3 13 d. 612 9. a. 3i 12 b. 5i 12 c. 15i d. 6i 11. a. i 119 3 12 2 13 b. i131 c. i d. i 13. a. 1  i; a  1, b  1 5 8 b. 2  13i; a  2, b  13 15. a. 4  2i; a  4, b  2 b. 2  12i; a  2, b   12 17. a. 5  0i; a  5, b  0 b. 0  3i; a  0, b  3 19. a. 18i; a  0, b  18 12 12 b. i; a  0, b  21. a. 4  5 12i; a  4, b  5 12 2 2 b. 5  313i; a  5, b  3 13 7 7 12 7 712 1 110 1 110  i; a  , b  i; a  , b  b.  4 8 4 8 2 2 2 2 25. a. 19  i b. 2  4i c. 9  10 13i 27. a. 3  2i b. 8 1 1 c. 2  8i 29. a. 2.7  0.2i b. 15  i c. 2  i 12 8 31. a. 15 b. 16 33. a. 21  35i b. 42  18i 35. a. 12  5i b. 1  5i 37. a. 4  5i; 41 b. 3  i 12; 11 39. a. 7i; 49 b. 21  23 i; 25 41. a. 41 b. 74 43. a. 11 b. 17 36 36 45. a. 5  12i b. 7  24i 47. a. 21  20i b. 7  6 12i 49. no 51. yes 53. yes 55. yes 57. yes 59. Answers will vary. 4 2 i 61. a. 1 b. 1 c. i d. i 63. a. i b. 7 5 21 14 10 15 3 2  i b.  i 67. a. 1  i b. 1  i 65. a. 13 13 13 13 4 3 69. a. 113 b. 5 c. 111 71. A  B  10 AB  40 73. 7  5i  75. 25  5i V 77. 74  i  79. a. 1x  6i2 1x  6i2 b. 1m  i1321m  i 132 c. 1n  2i 1321n  2i 132 d. 12x  7i212x  7i2 81. 8  6i 83. a. P  4s; A  s2 23. a.

bh b. P  2L  2W; A  LW c. P  a  b  c; A  2 2 d. C  d; A  r 85. John

Exercises 1.5, pp. 52–56 1. descending; 0 3. quadratic; 1 5. GCF factoring: x  0, x  54 7. a  1; b  2; c  15 9. not quadratic 11. a  14 ; b  6; c  0 13. a  2; b  0; c  7 15. not quadratic 17. a  1; b  1; c  5 19. x  5 or x  3 21. m  4 23. p  0 or p  2 25. h  0 or h  1 27. a  3 or a  3 2 29. g  9 31. m  5 or m  3 or m  3 33. c  3 or c  15 35. r  8 or r  3 37. t  13 or t  2 39. x  5 or x  3 41. w  12 or w  3 43. m  4 45. y  217; y  5.29 47. no real solutions 49. x  121 51. n  9; n  3 4 ; x  1.15 53. w  5 13; w  3.27 or w  6.73 55. no real solutions 57. m  2 3 12 59. 9; 1x  32 2 7 ; m  2.61 or m  1.39 3 2 1 1 2 9 61. 4 ; 1n  2 2 63. 9 ; 1p  3 2 65. x  1; x  5 67. p  3 16; p  5.45 or p  0.55 69. p  3 15; p  0.76 or p  5.24 113 71. m  3 2 2 ; m  0.30 or m  3.30 73. n  52 3 15 2 ; n  5.85 or n  0.85 75. x  12 or x  4 77. n  3 or n  3 2 79. p  38 141 8 ; p  1.18 or p  0.43 81. m  72 133 2 ; m  6.37 or m  0.63 83. x  6 or x  3 85. m  52 87. n  2 2 15 ; n  2.12 or n  0.12 89. w  23 or w  1 2 91. m  32 16 93. n  32 2 i; m  1.5 1.12i 95. w  4 97. a  16 123 5 or w  2 6 i; a  0.16 0.80i 99. p  3 52 16 ; p  1.58 or p  0.38 101. w  1 10121 ; w  0.56 or w  0.36 103. a  34 131 4 i; a  0.75 1.39i 105. p  1 3 12 2 i; p  1 2.12i 12 107. w  1 3 3 ; w  0.14 or w  0.80 109. a  6 2 3 12 ; a  0.88 or a  5.12 111. p  4 61394 ; p  3.97 or p  2.64 113. two rational; factorable 115. two complex 117. two rational; factorable 119. two complex 121. two irrational 123. one repeated; factorable 125. x  32 12 i 127. x  12 i 13 2 129. x  54 3i 417 133. t 

6  1138 2 2

131. t  v

2v2  64h 32

sec, t  8.87 sec

135. 30,000 ovens

137. a. P  x  120x  2000 b. 10,000 139. t  2.5 sec, 6.5 sec 141. x  13.5, or the year 2008 143. 36 ft, 78 ft 145. a. 7x2  6x  16  0 b. 6x2  5x  14  0 3 c. 5x2  x  6  0 147. x  2i; x  5i 149. x  i; x  2i 4 151. x  1  i; x  13  i 153. a. P  2L  2W, A  LW b. P  2r, A  r2 c. A  12 h1b1  b2 2, P  c  h  b1  b2 d. A  12 bh, P  a  b  c 155. 700 $30 tickets; 200 $20 tickets

Exercises 1.6, pp. 65–70 1. excluded 3. extraneous 5. Answers will vary. 7. x  2, x  0, x  11 9. x  3, x  0, x  23 11. x  32 , x  0, x  3 13. x  0, x  2, x  1 i13 15. x  2, x  5 17. x  3, x  2i 19. x  15, x  6 21. x  0, x  7, x  2i 23. x  3, x  3i 25. x  4, x  4i 27. x  12, x  1, i 29. x  1, x  2, x  1 i 13 1 i 13 31. x  12 i 13 33. x  1 35. a  32 2 , x  2 2 , x  1 37. y  12 39. x  3; x  7 is extraneous 41. n  7 f1 f2 E  IR E 43. a  1, a  8 45. f  47. r  or  R f1  f2 I I

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Student Answer Appendix 3V 3V 51. r3  53. a. x  14 3 4 r2 b. x  8, x  1 is extraneous 55. a. m  3 b. x  5 c. m  64 d. x  16 57. a. x  25 b. x  7; x  2 is extraneous c. x  2, x  18 d. x  6; x  0 is extraneous 59. x  32 61. x  9 63. x  32, x  22 65. x  27, 125 67. x  5, x  i 69. x  1, 2 71. x  1, 14 73. x  13 , 12 75. x  4, 45 S 2 77. x  6; x  74 79. a. h  a b  r2 9 is extraneous B r b. S  12134 m2 81. x  3, x  2 83. x  2, 4, 6 or x  2, 0, 2 85. 11 in. by 13 in. 87. r  3 m; r  0 and r  12 m do not fit the context 89. either $50 or $30 91. a. 32 ft, 1h  322 b. 11 sec c. pebble is at canyon’s rim 93. 12 min 95. v  6 mph 97. P  52.1% 99. a. 36 million mi b. 67 million mi c. 93 million mi d. 142 million mi e. 484 million mi f. 887 million mi 101. The constant “3” was not multiplied by the LCD, 3x1x  328x  x  3; x  1, 1 103. x  1, 22 ´ 12, q2 105. a. x  5, 3, 5, 7 b. x  2, 1, 6, 7 c. x  2, 1, 3 d. x  4, 2, 3 e. x  1, 1, 7 f. x  1, 1, 2, 7 107. 2111 cm 109. 1 6 x 6 5; 49. h 

)

2 1

) 0

1

2

3

4

5

6

Summary and Concept Review, pp. 70–74 3. n  4 4. m  1 V 5. x  6. no solution 7. g  10 8. h  2 9. L  P 2 2W r 10. x  c a b 11. y  23 x  2 12. 8 gal 13. 12  98  ft2  15.5 ft2 14. 23 hr  40 min 15. a  35 16. a 6 2 17. s  65 18. c  1200 19. 15, q2 20. 110, q2 21. 1  q, 2 4 23 22. 19, 94 23. 1  6, q 2 24. A q, 8 5 B ´ A 10 , q B 3 3 25. a. 1q, 32 ´ 13, q2 b. A q, 2 B ´ A 2 , q B c. 3 5, q2 d. 1q, 6 4 26. x  96% 27. {4, 10} 28. 57, 36 29. 55, 86 30. 54, 16 31. 1q, 62 ´ 12, q 2 32. [4, 32] 33. { } 34. { } 35. 1q, q 2 36. 3 2, 6 4 37. 1q, 2 4 ´ 3 10 3 , q2 38. a. r  2.5  1.7 b. highest: 4.2 in., lowest: 0.8 in. 39. 6 12i 40. 2413i 41. 2  12i 42. 312i 43. i 44. 21  20i 45. 2  i 46. 5  7i 47. 13 48. 20  12i 49. 15i2 2  9  34 15i2 2  9  34 2 25i  9  34 25i2  9  34 25  9  34 ✓ 25  9  34 ✓ 50. 12  i 152 2  412  i 152  9  0 4  4i15  5i2  8  4i 15  9  0 5  152  0 ✓ 12  i152 2  412  i 152  9  0 4  4i15  5i2  8  4i 15  9  0 5  152  0 ✓ 51. a. 2x2  3  0; a  2, b  0, c  3 b. not quadratic c. x2  8x  99  0; a  1, b  8, c  99 d. x2  16  0; a  1, b  0, c  16 52. a. x  5 or x  2 b. x  5 or x  5 c. x  53 or x  3 d. x  2 or x  2 or x  3 53. a. x  3 b. x  2 15 c. x  15i d. x  5 54. a. x  3 or x  5 b. x  8 or x  2 c. x  1 110 d. x  2 or x  13 2 ; x  2.58 or x  0.58 55. a. x  2 15i; x  2 2.24i b. x  3 2 12 ; x  2.21 or x  0.79 c. x  32 12 i 56. a. 1.3 sec b. 4.67 sec c. 6 sec 57. a. 0.8 sec b. 3.2 sec c. 5 sec 58. $3.75; 3000 59. 6 hr 60. x  13, x  7 61. x  2, x  0, x  13 62. x  0, x  2, x  1 13i 63. x  12 , x  12 i 64. x  1 2 65. h  53 , h  2 66. n  13; n  2 is extraneous 67. x  3; x  3 68. x  4; x  5

1. a. yes 1 6

b. yes

c. yes

2. b  6

SA3

69. x  1; x  7 is extraneous 70. x  52 71. x  5.8; x  5 72. x  2, x  1, x  4, x  5 73. x  3, x  3, x  i 12, x  i 12 74. a. 12,000 kilocalories b. 810 kg 75. width, 6 in.; length, 9 in. 76. 1 sec; 244 ft; 8 sec 77. $24 per load; $42 per load

Mixed Review, p. 75

4 1. a. x  18, q 2 b. x  1q, 4 3. a. x  2, x  5i 3 2 ´ 1 3 , q2 5 5i 13 b. x  0, x  5, x  2 2 c. x  73 , x  53 d. 1q, 3 4 ´ 327, q 2 e. v  27 f. x  80 5. y  3 4 x  3 7. a. x  2 b. n  5 9. x  7, 11 11. x   16, 16 13. x  45 15. x  15, i 15 17. a. v  6, 2 is extraneous b. x  5; x  4 c. x  2; x  18 is extraneous 19. 6¿10–

Practice Test, pp. 75–76 P 1. a. x  27 b. x  2 c. C  1  d. x  4, x  1 k 2. 30 gal 3. a. x 7 30 b. 5  x 6 4 c. x   d. x  2, x  4 e. x 6 4 or x 7 2 4. S  177 5. z  3, z  10 6. x  5i 7. x  1 i13 8. x  1, x  4 9. x  23 , x  6 10. x  2, x  32 11. x  6, x  2 is extraneous 12. x  32 , x  2 13. x  16, x  4 is extraneous 14. x  11, x  5 15. a. $4.50 per tin b. 90 tins 16. a. t  5 (May) b. t  9 (Sept.) c. July; $3000 more 17. 43 i 15 18. i 19. a. 1 b. i13 3 c. 1 20. 32  32 i 21. 34 22. 12  3i2 2  412  3i2  13  0 5  12i  8  12i  13  0 0  0✓ 23. a. x  5 12 b. x  54 i 17 24. a. x  3 3 13 2 4 b. x  1 3i 25. a. F  64.8 g b. W  256 g

Strengthening Core Skills, pp. 77–78 7 5 b 7# 7 c  112    ✓  ✓ 112  2 2 a 2 2 a 2  3 12 2  3 12 4 b 2  3 12 # 2  312    Exercise 2: ✓ 2 2 2 a 2 2 14 7 c    ✓ 4 2 a b Exercise 3: 15  213i2  15  2 13i2  10  ✓ a c 15  2 13i2 15  213i2  25  12  37  ✓ a Exercise 1:

Connections to Calculus Exercises, p. 82 1. x 

3 6 , 2x   4.24 in. 12 12

3. t 

6 3 sec, t  sec 5 2

3 is very small, then the difference 2 3 between h(x) and 6 is very small: if ` x  ` 6 , then h1x2  6 6 . 2 5. If the difference between x and

7. If the difference between x and 2 is very small, then the difference between w(x) and 14 is very small: if x  2 6 , then w1x2  14 6 .

CHAPTER 2 Exercises 2.1, pp. 93–96 1. first, second 3. radius, center 5. Answers will vary. 7. Year in D  51, 2, 3, 4, 56 college GPA R  52.75, 3.00, 3.25, 3.50, 3.756 1

2.75

2

3.00

3

3.25

4

3.50

5

3.75

9. D  51, 3, 5, 7, 96; R  52, 4, 6, 8, 106 11. D  54, 1, 2, 36; R  50, 5, 4, 2, 36

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Student Answer Appendix

13.

10 8 6 4 2 108642 2 4 6 8 10

15.

y

2 4 6 8 10 x

x

x

y

5

2

0

3

3

0

2, 2

0

1

1

3, 3

3

1

3

5, 5

6

3

6

8, 8

8

13 3

7

9, 9

15 12 9 6 3 54321 3 6 9 12 15

19.

y

1 2 3 4 5 x

x

10 8 6 4 2

8

4

3

3

3

4

0

1

0

5

2

3

2

121

3

8

3

4

4

15

4

3

108642 2 4 6 8 10

23.

2 4 6 8 10 x

x

y

10

5 4 3 2 1 108642 1 2 3 4 5

108642 2 4 6 8 10

9

2

5

2, 2

2

1

4

13, 13

1

0

1, 1

0

1

0.5, 0.5

4

3 1 5

0

7

108642 2 4 6 8 10

2 4 6 8 10 x

49. 1x  42 2  1y  32 2  4 10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

59. 1x  32 2  1y  42 2  41 15 12 9 6 3 1512963 3 6 9 12 15

y

3 6 9 12 15 x

63. (2, 3), r  2, x  30, 44 , y  31, 5 4 10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

67. 14, 02, r  9, x  313, 5 4 , y  39, 9 4

69. 1x  52 2  1y  62 2  57, (5, 6), r  157

71. 1x  52 2  1y  22 2  25, 15, 22, r  5

108642 2 4 6 8 10

15 12 9 6 3

15 12 9 6 3

73. x2  1y  32 2  14, 10, 32, r  114

75. 1x  22 2  1y  52 2  11, 12, 52, r  111

51. 1x  72 2  1y  42 2  7 10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

10 8 6 4 2 108642 2 4 6 8 10

2

10 8 6 4 2

10 8 6 4 2

1512963 3 6 9 12 15

(3, 1) 27. (0.7, 0.3) 29. A B 31. (0, 1) (1, 0) 35. 2 134 37. 10 39. not a right triangle not a right triangle 43. right triangle 47. 1x  52 2  y2  3 x2  y2  9 y

2 4 6 8 10 x

2 4 6 8 10 x

108642 2 4 6 8 10

1 1 20 , 24

10 8 6 4 2

y

108642 2 4 6 8 10

y

65. 11, 22, r  2 13, x  31  2 13, 1  2 134, y  32  2 13, 2  2 134

2 4 6 8 10 x

3, 3

1

4 8 12 16 20 x

10 8 6 4 2

1512963 3 6 9 12 15

y

2

y

55. 1x  42 2  1y  52 2  12

y

x

1.25

2 4 6 8 10 x

61. 1x  52 2  1y  42 2  9

y

2

y

2016 1284 4 8 12 16 20

2 4 6 8 10 x

3

10 8 6 4 2

20 16 12 8 4

10 8 6 4 2

x

y

57. 1x  72 2  1y  12 2  100

y

108642 2 4 6 8 10

y

21.

25. 33. 41. 45.

108642 2 4 6 8 10

6

17.

10 8 6 4 2

2 4 6 8 10 x

108642 2 4 6 8 10

y

53. 1x  12 2  1y  22 2  9

y

10 8 6 4 2

10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

y

3 6 9 12 15 x

y

3 6 9 12 15 x

y

2 4 6 8 10 x

y

2 4 6 8 10 x

y

2 4 6 8 10 x

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Student Answer Appendix 77. 1x  72 2  y2  37, 17, 02, r  137

y

15 12 9 6 3

y

15 12 9 6 3

3 6 9 12 15 x

1512963 3 6 9 12 15

81. a. (1, 71.5), (2, 84), (3, 96.5), (5, 121.5), (7, 146.5); yes b. $159 c. 2011 d. s 155 145 135 125 115 105 95 85 75 65

10 8 6 4 (6, 0)2

27.

10 8 6 4 (5, 0) 2

29.

y

108642 2 2 4 6 8 10 x (0, 2) 4 6 (5, 4) 8 10

33. m  1; 12, 42 and 11, 32 8

8

4

4

8

4

3 6 9 12 15 x

89. Answers will vary. 91. a. center: 16, 22; r  0 (degenerate case) b. center: (1, 4); r  5 c. r2  1; degenerate case 93. a. 0 b. not possible c. 0.3; many answers possible d. not possible e. not possible f. 13; many answers possible 95. n  1 is a solution, n  2 is extraneous

Exercises 2.2, pp. 106–110 3. negative, downward 5.

y

6

6

3

4

0

2

3

0

15.

10 8

y

108642 2 4 6 8 10

108642 2 4 6 8 10

yes m1 m2 9.

4

8

x

80

31.

y

10 8 6 4 (0, 4) 2 (3, 0)

(4, 6) (2, 3) (0, 0) 2 4 6 8 10 x

108642 2 2 4 6 8 10 x 4 6 (6, 4) 8 10

(10, 3)

4

x

8

4

4 4

8

x

(4, 5)

8

y

(3, 7)

y

2 4 6 8 10 x

y

2

1

0

4

13.

7

4

10

10 (4, 10) 8 (2, 7) 6 4 (0, 4) (2, 1) 2 108642 2 4 6 8 10

2 4 6 8 10 x

y

10 8 6 (0, 6) 4 (1, 3) 2 (2, 0) 2 4 6 8 10 x

108642 2 4 6 8 10

17.

2

y

y

10 8 (2, 8) 6 4 (T, 0) 2 (0, 3) 108642 2 4 6 8 10

2 4 6 8 10 x

19.

y

(2, 0)

10 8 6 4 (4, R) 2

108642 2 2 4 6 8 10 x  4 6 (0, R) 8 10

y

8

(3, 6) 4

4 8

39. m  4 7 ; 110, 102 and 111, 22

8

4

4

8

4 8

x

(1, 8)

8

4

(4, 2) 4

8

4

x

8

41. a. m  125, cost increased $125,000 per 1000 sq ft b. $375,000 43. a. m  22.5, distance increases 22.5 mph b. about 186 mi 45. a. m  23 6 , a person weighs 23 lb more for each additional 6 in. in height b. 3.8 47. In inches: (0, 6) and (576, 18): m  1 48 . The sewer line is 1 in. deeper for each 48 in. in length. 49. 51. y y 10 8

10 8 6 4 2

4 2

108642 2 (3, 4) 4 6 8 10

no m1 # m2 1

x

2 4 6 8 10 x

 19 4 ✓ 10 8 6 4 2

(0, 25)

y

37. m  15 4 ; 11, 82 and 11, 12 2

(3, 0)

6 (6, 6) 4 (3, 4) 2 (0, 2) (3, 0)

11. 0.5  32 132  4 0.5  92  4 0.5  0.5 ✓ 19 3 1 4  2 12 2  4 19 3 4  4  4 19 4

40

4 40

8

(3, 5) 6

x

(q, 0) 8

y

8

6 3

7.

108642 2 4 6 8 10

y 80

(1, 71.5)

83. a. 1x  52 2  1y  122 2  625 b. no 85. Red: 1x  22 2  1y  22 2  4; Blue: 1x  22 2  y2  16; Area blue  12 units2 y 87. No, distance between centers is less than sum of 15 12 radii. 9

1. 0, 0

10 8 6 4 2

(4, 6) (3, 5)

4

25.

y

10 8 6 4 2 (0, 0) (4, 2) 108642 2 2 4 6 8 10 x 4 6 (2, 1) 8 10

35. m  43 ; 17, 12 and 11, 92

y

(7, 146.5)

0 1 2 3 4 5 6 7 8 9 10 t

1512963 3 6 9 12 15

23.

y

108642 2 2 4 6 8 10 x 4 (0, 4) (3, 6) 6 8 10

3 6 9 12 15 x

1512963 3 6 9 12 15

79. 1x  32 2  1y  52 2  32, 13, 52, r  4 12

21.

2 4 6 8 10 x

108642 2 4 6 8 10

(2, 4) (2, 0) 2 4 6 8 10 x

(2, 6)

53. L1: x  2; L2: y  4; point of intersection (2, 4) 55. a. For any two points chosen m  0, indicating Justices there has been no increase or decrease in the number 109 8 of supreme court justices. b. For any two points 7 6 1 Nonwhite, nonmale 5 chosen m  10 , which indicates that over the last 5 4 decades, one nonwhite or nonfemale justice has been 32 1 added to the court every 10 yr. 57. parallel 10 20 30 40 50 60 Time t 59. neither 61. parallel 63. not a right triangle 65. not a right triangle 67. right triangle 69. a. 76.4 yr b. 2010 71. v  1250t  8500 a. $3500 b. 5 yr 73. h  3t  300 a. 273 in. b. 20 months 30 75. Yes they will meet, the two roads are not parallel: 38 12 9.5 . 77. a. $3789 b. 2012 79. a. 23% b. 2005 81. a  6 83. a. 142 b. 83 c. 9 d. 27 2 85. perimeter of a rectangle, volume of a rectangular prism, volume of a right circular cylinder, circumference of a circle 87. 2 hr

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Student Answer Appendix

Exercises 2.3, pp. 118–122 1. 7 3. 2.5 5. Answers will vary. 4 ; (0, 3) 7. y  4 9. y  2x  7 11. y  5 5 x  2 3 x  5 x 5 2 0 1 3

x 5 2 0 1 3

y 6 18 5

2 6 5 2 5

y 3 3 7 9 13

x y 5 10 3 2 5 3 0 5 20 1 3 3 10

5 13. y  2x  3: 2, 3 15. y  5 3 x  7: 3 , 7 35 35 17. y  6 x  4: 6 , 4

19.

y

10 8 6 (0, 5) 4 2 (3, 1) 108642 2 2 4 6 8 10 x 4 (6, 3) 6 8 10

21.

y

10 8 6 4 2

23.

10 8

(6, 5) 6 4 2 2 4 6 8 10 x

108642 2 4 6 8 10

108642 2 4 6 8 10

(0, 2) (2, 5) (4, 8)

y (0, 4) (6, 3) 2 4 6 8 10 x

Exercises 2.4, pp. 132–137

25. a. 3 b. y  3 c. The coeff. of x is the slope and the con4 4 x  3 stant is the y-intercept. 27. a. 25 b. y  25 x  2 c. The coeff. of x is the slope and the constant is the y-intercept. 29. a. 45 b. y  45 x  3 c. The coeff. of x is the slope and the constant 2 is the y-intercept. 31. y  2 3 x  2, m  3 , y-intercept (0, 2) 5 5 33. y  4 x  5, m  4 , y-intercept (0, 5) 35. y  13 x, m  13 , 3 y-intercept (0, 0) 37. y  3 4 x  3, m  4 , y-intercept (0, 3) 2 39. y  3 x  1 41. y  3x  3 43. y  3x  2 75 45. y  250x  500 47. y  2 x  150 49. y  2x  13 3 y 51. y  5 x  4 53. y  23 x  5 55. 10 y

10 8 6 4 (0, 4) 2 108642 2 4 6 8 10

57.

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

2 4 6 8 10 x

69. 77. 79. 81. 83.

65. y 

8 6 4 2

2 4 6 8 10 x

108642 2 4 6 8 10

(0, 5)

59.

y

63. y  25 x  4

y

10 8 6 4 2

y

10 8 6 4 2

2 4 6 8 10 x

108642 2 4 6 8 10

5 3 x

7

61.

10 8 6 4 2 108642 2 4 6 8 10

29 67. y  12 5 x  5

y  5 71. perpendicular 73. neither 75. neither 5 a. y  3 b. y  43 x  20 4 x  2 3 4 31 3 a. y  9 x  9 b. y  9 4 x  4 1 a. y  2 x  2 b. y  2x  2 85. y  4  38 1x  32 y  5  21x  22 10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

87. y  3.1  0.51x  1.82 10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

10 8 6 4 2 8642 2 4 6 8 10

y

2 4 6 8 10 12 x

2 4 6 8 10 x

108642 2 4 6 8 10

89. y  2  65 1x  42 ; For each 5000 additional sales, income rises $6000. 91. y  100  20 1 1x  0.52 ; For every hour of television, a student’s 1 final grade falls 20%. 93. y  10  35 2 1x  2 2 ; Every 2 in. of rainfall increases the number of cattle raised per acre by 35. 95. C 97. A c 99. B 101. D 103. m  a a. m  3 b , y-intercept  b 4 , y-intercept 5 (0, 2) b. m  2 , y-intercept ( 0, 3 ) c. m  , 5 6 y-intercept (0, 2) d. m  53 , y-intercept (0, 3) 105. a. As the temperature increases 5°C, the velocity of sound waves increases 3 m/s. At a temperature of 0°C, the velocity is 331 m/s. b. 343 m/s c. 50°C 107. a. V  20 b. Every 3 t  150 3 yr the value of the coin increases by $20; the initial value was $150. c. $223.33 d. 15 years, in 2013 e. 3 yr 109. a. N  7t  9 b. Every 1 yr the number of homes with Internet access increases by 7 million. c. 1993 d. 86 million e. 13 yr f. 2010 111. a. P  58,000t  740,000 b. Each year, the prison population increases by 58,000. c. 1,726,000 113. Answers will vary. 115. (1) d (2) a (3) c (4) b (5) f (6) h 117. x  5 23 113 ; x  0.74 or x  4.07 119. 113.10 yd2

y

2 4 6 8 10 x

1. first 3. range 5. Answers will vary. 7. function 9. Not a function. The Shaq is paired with two heights. 11. Not a function; 4 is paired with 2 and 5. 13. function 15. function 17. Not a function; 2 is paired with 3 and 4. 19. function 21. function 23. Not a function; 0 is paired with 4 and 4. 25. function 27. Not a function; 5 is paired with 1 and 1. 29. function 31. function 33. function y y 5 4 3 2 1

54321 1 2 3 4 5

7 6 5 4 3 2 1

1 2 3 4 5 x

7654321 1 2 3

1 2 3 x

35. function, x  3 4, 54 y  3 2, 3 4 37. function, x  34, q2 y  34, q 2 39. function, x  3 4, 4 4 y  35, 1 4 41. function, x  1q, q2 y  1q, q 2 43. Not a function, x  33, 5 4 y  33, 3 4 45. Not a function, x  1q, 3 4 y 1q, q2 47. x  1q, 52 ´ 15, q 2 49. x  3 5 3 , q2 51. x  1q, 52 ´ 15, 52 ´ 15, q 2 53. v  1q, 3 122 ´ 13 12, 3 122 ´ 13 12, q2 55. x  1q, q 2 57. x  1q, q2 59. x  1q, q2 61. x  1q, 22 ´ 12, 52 ´ 15, q 2 63. x  32, 52 2 ´ 1 52 , q 2 65. x  12, q2

7 1 69. f 1c  12  c  2 2 3 73. h1a  22  a2

67. x  14, q2

71. f 1c  12  3c2  2c  1 75. h1a  22  5a

|a  2 | a2

b

3 77. g142  8, ga b  3, g12c2  4c, g1c  32  21c  32 2 3 9 79. g142  16, ga b  , g12c2  4c2, g1c  32  1c2  6c  92 2 4 3 81. p152  213, pa b  26, p13a2  26a  3, p 1a  12  22a  1 2 3 7 27a2  5 14 3a2  6a  2 83. p152  , pa b  , p 13a2  , p1a  12  2 2 5 2 9 9a a  2a  1 85. a. D  51, 0, 1, 2, 3, 4, 56 b. R  52, 1, 0, 1, 2, 3, 46 c. 1 d. 1 87. a. D  35, 5 4 b. y  3 3, 4 4 c. 2 d. 4 and 0 89. a. D  33, q 2 b. y  1q, 4 4 c. 2 d. 2 and 2 1 91. a. 186.5 lb b. 37 lb 93. A  182  22  1  25 units2 2 95. a. N1g2  2.5g b. g  30, 5 4 ; N  3 0, 12.54 97. a. 30, q ) b. 750 c. 800 99. a. c1t2  42.50t  50 b. $156.25 c. 5 hr

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Student Answer Appendix d. t  3 0, 10.6 4 ; c  3 0, 500 4 101. a. Yes. Each x is paired with exactly one y. b. 10 P.M. c. 0.9 m d. 7 P.M. and 1 A.M. 1 103. a. ¢fertility ¢time  20 , negative, fertility is decreasing by one child every 0.8 20 yr b. 1940 to 1950: ¢f ¢t  10 ; positive, fertility is increasing by less ¢f than one child every 10 yr c. 1940 to 1950: ¢t  0.8 10 ; 1980 to 1990: ¢f 0.2 , the fertility rate was increasing four times as fast from 1940  ¢t 10 to 1950. 105. negative outputs become positive y  x

10 8 6 4 2

108642 2 4 6 8 10

y

2 4 6 8 10 x

108642 2 4 6 8 10

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

10 8 6 4 2

 3 107. a. x  1q, 22 ´ 12, q 2 ; x  2y 1  y ; y  1q, 12 ´ 11, q 2 b. x   x  1y  3; y  33, q2 109. a. 19 16 b. 1 111. a. 1x  321x  521x  52 b. 12x  321x  82 c. 12x  5214x2  10x  252

108642 2 4 6 8 10

10 8 6 4 2

y

10 8 6 4 (6, 4) 2 (3, 0) 108642 2 2 4 6 8 10 x 4 (0, 4) 6 8 10

108642 2 4 6 8 10

18 7

3. positive, loss is decreasing (profit is increasing); m  32 , yes; 1.5 1 , each year Data.com’s loss decreases by 1.5 million. y 4. y  32 x  52 10 2.

8 6 4 2 108642 2 4 6 8 10

15 12 9 6 3 1512963 3 6 9 12 15

2 4 6 8 10 x

5. x  3; no; input 3 is paired with more than one output. 4 6. y  x  4; yes 7. a. 0 b. x  3 3, 5 4 c. 1 3 d. y  34, 5 4 8. from x  1 to x  2; steeper line S greater slope 9. F1p2  34 p  54 , For every 4000 pheasants, the fox population increases by 300: 1625. 10. a. x  53, 2, 1, 0, 1, 2, 3, 46 y  53, 2, 1, 0, 1, 2, 3, 46 b. x  3 3, 4 4 y  3 3, 4 4 c. x  1q, q 2 y  1q, q2

108642 2 4 6 8 10

10 8 6 4 2 108642 2 4 6 8 10

7 2

y

2 4 6 8 10 x

c. y  1 2 x  6

y

3 6 9 12 15 x

5

5 x

5

13.

y 10

2 4 6 8 10 x

2 4 6 8 10 x

c. y  3 4 x 

5

y

y

2 4 6 8 10 x

1. linear; bounce 3. increasing 5. Answers will vary. y 7. 9. even 11. even

c. y  13 x  5

2. a. 7 b. y  9  7 3 , decreasing 3 1x  02 d. 7x  3y  27 e. (0, 9), (22 7 , 0)

y

Exercises 2.5, pp. 150–156

Reinforcing Basic Concepts, p. 138

10 8 6 4 2

c. y  34 x  14

6. a. 1 b. y  7  1 2 , decreasing 2 1x  22 d. x  2y  12 e. (0, 6), (12, 0)

(1, 4) (4, 2)

1. a. 13 , increasing b. y  5  13 1x  02 d. x  3y  15 e. (0, 5), (15, 0)

2 4 6 8 10 x

5. a. 3 b. y  5  3 4 , decreasing 4 1x  22 d. 3x  4y  14 e. (0, 72 ), (14 3 , 0)

Mid-Chapter Check, p. 137 1.

c. y  12 x  12

y

4. a. 34 , increasing b. y  4  34 1x  52 1 d. 3x  4y  1 e. (0, 1 4 ), (3 , 0)

y

10 x 8 6 y x 4 2

yx

3. a. 12 , increasing b. y  2  12 1x  32 d. x  2y  1 e. (0, 12 ), (1, 0)

10

c. y  7 3 x  9

10 x

10

15. odd 17. not odd 19. neither 21. odd 23. neither 25. x  31, 1 4 ´ 33, q 2 27. x  1q, 12 ´ 11, 12 ´ 11, q2 29. p1x2  0 for x  3 2, q 2 31. f 1x2  0 for x  1q, 24 33. V1x2c: x  13, 12 ´ 14, 62 V1x2T: x  1q, 32 ´ 11, 42 constant: none 35. f 1x2c: x  11, 42 f 1x2T: x  12, 12 ´ 14, q2 constant: x  1q, 22 37. a. p1x2c: x  1q, q 2 p1x2T: none b. down, up 39. a. f 1x2c: x  13, 02 ´ 13, q2 f 1x2T: x  1q, 32 ´ 10, 32 b. up, up

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Student Answer Appendix

41. a. x  1q, q 2, y  1q, 52 b. x  1, 3 c. H1x2  0: x  3 1, 3 4 H1x2  0: x  1q, 14 ´ 33, q 2 d. H1x2c: x  1q, 22 H1x2T: x  12, q 2 e. local max: y  5 at (2, 5) 43. a. x  1q, q 2, y  1q, q 2 b. x  1, 5 c. g1x2  0: x  31, q2 g1x2  0: x  1q, 1 4 ´ 30, 3.5 4 d. g1x2c: x  1q, 12 ´ 15, q 2 g1x2T: x  11, 52 e. local max: y  6 at (1, 6); local min: y  0 at (5, 0) 45. a. x  3 4, q 2, y  1q, 34 b. x  4, 2 c. Y1  0: x  3 4, 2 4 Y1  0: x  32, q 4 d. Y1c: x  14, 22 Y1T: x  12, q2 e. local max: y  3 at (2, 3) 47. a. x  , y   b. x  4 c. p1x2  0: x  34, q 2; p1x2  0: x  1q, 4 4 d. p1x2c: x  1q, 32 ´ 13, q 2; p1x2T: never decreasing e. local max: none; local min: none 49. a. x  1q, 3 4 ´ 3 3, q2 , y  3 0, q 2 b. 13, 02 , (3, 0) c. f 1x2c: x  13, q 2 f 1x2T: x  1q, 32 d. even 51. a. x  3 0, 2604, y  3 0, 804 b. 80 ft c. 120 ft d. yes e. (0, 120) f. (120, 260) 53. a. x  1q, q2; y  31, q 2 b. (1, 0), (1, 0) c. f 1x2  0: x  1q, 14 ´ 3 1, q2; f 1x2 6 0: x  11, 12; d. f 1x2c: x  10, q 2, f 1x2T: x  1q, 02 e. min: (0, 1) 55. a. t  372, 96 4 , I  37.25, 16 4 b. I1t2c: t  172, 742 ´ 177, 812 ´ 183, 842 ´ 193, 942 I1t2T: t  174, 752 ´ 181, 832 ´ 184, 862 ´ 190, 932 ´ 194, 952 I(t) constant: t  175, 772 ´ 186, 902 ´ 195, 962 c. max: (74, 9.25), (81, 16) (global max), (84, 13), (94, 8.5), min: (72, 7.5), (83, 12.75), (93, 7.25) d. Increase: 80 to 81; Decrease: 82 to 83 or 85 to 86 y 57. zeroes: (8, 0), (4, 0), (0, 0), (4, 0); 8 min: (2, 1), (4, 0); max: (6, 2), (2, 2) 4 8

4

4

8

4

x

8

59. a. 7 b. 7 c. They are the same. d. Slopes are equal.

10 8 6 4 (1, 1) 2 54321 2 4 6 (2, 8) 8 10

y (2, 8) (1, 1)

c.

48 42 36 30 24 18 12 6 6 12

As height increases you can see farther, the sight distance is increasing much slower.

y

x 100 200 300 400 500

79. no; no; Answers will vary. 83. x  2, x  10

Exercises 2.6, pp. 166–171 1. stretch; compression 3. (5, 9); upward 5. Answers will vary. 7. a. quadratic; b. up/up, (2, 4), x  2, (4, 0), (0, 0), (0, 0); c. D: x  , R: y  3 4, q 2 9. a. quadratic; b. up/up, (1, 4), x  1, (1, 0), (3, 0), (0, 3); c. D: x  , R: y  3 4, q2 11. a. quadratic; b. up/up, (2, 9), x  2, (1, 0), (5, 0), (0, 5); c. D: x  , R: y  3 9, q2 13. a. square root; b. up to the right, (4, 22 , (3, 0), (0, 2); c. D: x  34, q 2 , R: y  32, q2 15. a. square root; b. down to the left, (4, 3), (3, 0), (0, 3); c. D: x  1q, 4 4 , R: y  1q, 3 4 17. a. square root; b. up to the left, (4, 0), (4, 0), (0, 4); c. D: x  1q, 4 4 , R: y  30, q2 19. a. absolute value; b. up/up, (1, 4), x  1, (3, 0), (1, 0), (0, 2); c. D: x  , R: y  34, q 2 21. a. absolute value; b. down/down, (1, 6), x  1, (4, 0), (2, 0), (0, 4); c. D: x  , R: y  1q, 6 4 23. a. absolute value; b. down/down, (0, 6), x  0, (2, 0), (2, 0), (0, 6); c. D: x  , R: y  1q, 6 4 25. a. cubic; b. up/down, (1, 0), (1, 0), (0, 1); c. D: x  , R: y   27. a. cubic; b. down/up, (0, 1), (1, 0), (0, 1); c. D: x  , R: y   29. a. cube root; b. down/up, (1, 1), (2, 0), (0, 2); c. D: x  , R: y   31. square root function; y-int (0, 2); x-int (3, 0); initial point (4, 2); up on right; D: x  3 4, q 2, R: y  32, q 2 33. cubic function; y-int (0, 2); x-int (2, 0); inflection point (1, 1); up, down; D: x  , R: y   35. 37. 39. y y y

1 2 3 4 5 x

8

8

4

f(x)

4

4

h(x)

4

8

x

8

43. q(x)

4

4

8

4

8

x

75. a. d.

¢x

 3x  3xh  h

20 16 12 8 4 54321 4 8 12 16 20

¢g

8

x

8

y

b.

53.

x

8

4

4

8

x

4

8

x

8

x

4 8

51.

y 8 4

4

4

8

4

x

8

4 4

8

55.

y

8

8

8

4

4 4

8

4

x

8

59.

61.

y

4 4

4 4 8

p(x) q(x)

4

4

8

x

10 8 6 4 2 108642 2 4 6 8 10

y

8

Y1

8 Y2 4

x

8

8

8

57.

r(x)

4

8

¢d b.  0.05 ¢h

y

8

4

1 2 3 4 5 x

¢d 77. a.  0.25 ¢h

Y2

¢g

 12.61 c.  0.49 ¢x ¢x Both lines have a positive slope, but the line at x  2 is much steeper. 2

8

4 4

4

2

4

y

4

x

Y1

4

8

4

8

y 8

4

8

¢g

45.

4

49.

4 4

8

y

8

4

p(x)

4

8

1 2 3 4 x

8

8

y

8

654321 1 2 3

4

x

8

4

47.

8

8

y

4 3 2 1

4 4

8

8

8

p(x) q(x)

4

8

41.

r(x)

8

g(x)

4

61. a. 176 ft b. 320 ft c. 144 ft/sec d. 144 ft/sec; The arrow is going down. 63. a. 17.89 ft/sec; 25.30 ft/sec b. 30.98 ft/sec; 35.78 ft/sec c. Between 5 and 10. d. 1.482 ft/sec, 0.96 ft/sec 65. 2 2 67. 2x  h 69. 2x  2  h 71. x1x  h2 ¢g ¢g ¢g 73. a.  2x  2  h b.  3.9 c.  3.01 ¢x ¢x ¢x y d. The rates of change have opposite sign, with the 7 6 secant line to the left being slightly more steep. 5

81. Answers will vary. 2 85. y  x  1 3

8

4 4 8

y

2 4 6 8 10 x

Y3 4

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Student Answer Appendix 63. g 65. i 67. e 75. left 2, down 1

69. j

71. l

y 8 4 8

101. f 1x2  45 x  4

73. c 77. left 3, reflected across x-axis, down 2

4

4

100 80 60 40 20

8

x

8

1

4

8

8

4

4

4

105.

x

8

(3, 2)

79. left 3, down 1

3 2 1 20 40 60 80 100 x

8

4

107.

4

(3, 1)

a. compressed vertically, b. 216 W, c.  15.6, 161.5, power increases dramatically at higher windspeeds

P(v)

1000 800

4

x

8

4

8

4

(1, 2)

8

x

4 8 4 (0, 1)

600 400 200

8

4

83. left 3, reflected across x-axis, down 2

85. left 1, reflected across x-axis, stretched vertically, down 3

y

4

x

8

4 (3, 2)

8

4

8

12

1

4

x

8

4 (1, 3)

16

20

2

111.

3

y

4

89. left 1, reflected across y-axis, reflected across x-axis, stretched vertically, up 3,

y 8 4 8

4

4

(2, 1)4

x

8

91. right 3, compressed vertically, up 1

(2, 4) 4 (3, 1)

8

4

4

x

8

8

8

8

c.

y

x

8

8

4

8

y

4

8

x

8

4

95. a.

8

4

4

4

4

8

8

4

4

8

4

x

f(x) f(x)  x2  4 1 2 3 4 5 x

5 4 3 2 1 5 4321 1 2 3 4 5

F(x) F(x)  x2  4 1 2 3 4 5 x

10 8 6 4 2

8

4

4 4

8

108642 2 4 6 8 10

x

8

99. p1x2  1.5 1x  3

15. D: x  32, q 2 ; R: y  34, q2

8

x

2 4 6 8 10 x

y 8 4 8

4

4 4 8

4 8

4

97. f 1x2  1x  22 2

8

4

8

8

y

4

x

x

y

8

4

d.

c.

y

8

4

x

8

8

b.

y

4 4

8

8

1. continuous 3. smooth 5. Each piece must be continuous on the corresponding interval, and the function values at the endpoints of each interval must be equal. Answers will vary. x2  6x  10 0  x  5 7. a. f 1x2  e 3 b. y  31, 11 4 5 5 6 x9 2x  2 1 9. 2, 2, 2 , 0, 2.999, 5 11. 5, 5, 0, 4, 5, 11 13. D: x  36, q 2 ; R: y  34, q2 y

4 4

g(x) 4

Exercises 2.7, pp. 180–185

8

4 4

4

8

d.

y

4

x

8

x  10, 42 ; yes, x  14, q 2 ; yes

115. p  140 in. A  1168 in2 117. f 1x2 T : x  1q, 42 f 1x2 c : x  14, q 2

(0, 4) (1, 2) (5, 2)

8

4

8

5 4321 1 2 3 4 5

4 4

8

8

4

4

b.

5 4 3 2 1

8

4

t

113. Any points in Quadrants III and IV will reflect across the x-axis and move to Quadrants I and II.

y

8

4

a. vertical stretch by a factor of 2, b. 12.5 ft, c. 5, 13, distance fallen by unit time increases very fast

8

93. a.

y

4 4

1 x

987654321 1 2 3 4 5

8

8

y

5 4 (1, 3) 3 2 1

28 v

f(x)

8

8

87. left 2, reflected across x-axis, compressed vertically down 1,

24

d(t)  2t2

4 4

8

d(t) 14 12 10 8 6 4 2

8

4 8

109.

y

8

r

compressed vertically, 2.25 sec

T(x)

y

8

3

4

81. left 1, down 2

y

2

5

8

4

 4.2, about 65 units2, 65.4 units3, yes

V(r)  4.2r3 120

y

(2, 1) 4

8

103.

8

x

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Student Answer Appendix

y

8

8 4

4

4

8

4

x

8

4

4

8

23. x  1q, q 2; y  1q, 62 ´ 16, q2 ; discontinuity at x  3, redefine f 1x2  6 at x  3; c  6

0 2 45. C1a2  μ 5 7 5

8 4 4

4

8

x

4

2

c. 80¢, y

2 2

4

x

4

y

51. Y1 has a removable discontinuity at x  2; Y2 has a discontinuity at x  2 53. x  7, x  4 55. a. 4 15 cm b. 1615 cm2 c. V  320 15 cm3

8 4 4

Cost of call in cents

a 6 2 10 8 2  a 6 13 6 13  a 6 20 4 20  a 6 65 2 a  65 0 10 20 30 40 50 60 70 Age in years $38 47. a. C1w  12  173 w  14  80, b. 0 6 w  13; d. 165¢, e. 165¢, f. 165¢, g. 182¢ 5 x  3 4 49. yes; h1x2  • 2x  1 3 6 x 6 2 2 5 x2

8

8

(30, 99) 10 20 30 40 50 60

y

4

25. x  1q, q2; y  3 0.75, q 2; discontinuity at x  1, redefine f 1x2  3 at x  1; c  3

200

Duration of call in min

8

8

250

50

x

8

4

300

100

y

4

350

150

21. x  1q, q 2; y  30, q 2

8

3.3m 0  m  30 ; 7m  111 m 7 30 $2.11

1 2 3 4 5 x

54321 1 2 3 4 5

19. x  1q, 92; y  3 2, q 2

43. c1m2  e

y

5 4 3 2 1

Cost of admission

17. D: x  1q, q 2 ; R: y  1q, 32, ´ 13, q2

4

x

8

4 8

Exercises 2.8, pp. 196–202  1 4  x 6 2 x2  2x  3 27. f 1x2  e 29. p1x2  e 3x  6 x  2 x1 y 31. Graph is discontinuous at 8 x  0; f 1x2  1 for x 7 0; f 1x2  1 4 for x 6 0. 1 2x

8

4

4

8

4

x1 x 7 1

x

8

33. a. S 1t2  e 35. a.

t2  6t 5

Year (0 S 1950) 5 15 25 35 45 55

37. C1h2  e

0t5 b. S1t2  3 0, 9 4 t 7 5 b. Each piece gives a slightly different value due to rounding of coeffiPercent cients in each model. At t  30, we 7.33 use the “first” piece: 14.13 P1302  13.08. 14.93 22.65 41.55 60.45

0.09h 0  h  1000 0.18h  90 h 7 1000 C 112002  $126

200 180 160 140 120 100 80 60 40 20

C(h)

h 200

39. C1t2  e

0.75t 0  t  25 1.5t  18.75 t 7 25 C 1452  $48.75

60 55 50 45 40 35 30 25 20 15 10 5

Spending (in billions of dollars)

450 250 150 50 2

6 10 14 18 22

t (years since 1980)

$498 billion, $653 billion, $782 billion

30

40

1400

t 20

1.35t2  31.9t  152; 0  t  12 2.5t2  80.6t  950; 12 6 t  22 (12, 340)

350

1000

C(t)

10

41. S1t2  e

600

50

1. ( f  g)(x); A  B 3. intersection; g(x) 5. Answers will vary. 7. a. x   b. f 122  g122  13 9. a. h 1x2  x2  6x  3 b. h122  13 c. they are identical 11. a. x  33, q 2 b. h1x2  1x  3  2x3  54 c. h142  75, 2 is not in the domain of h. 13. a. x  3 5, 3 4 b. r 1x2  1x  5  13  x c. 2172  17  1, 4 is not in the domain of r. 15. a. x  34, q2 b. h 1x2  1x  412x  32 c. h142  0, h1212  225 17. a. x  31, 7 4 b. r 1x2  1x2  6x  7 c. 15 is not in the domain of r, r 132  4 19. a. x  1q, 42 ´ 14, q2 b. h1x2  x  4, x 4 21. a. x  1q, 42 ´ 14, q2 b. h1x2  x2  2, x 4 23. a. x  1q, 12 ´ 11, q2 b. h1x2  x2  6x, x 1 25. a. x  1q, 52 ´ 15, q2 2x  3 x1 b. h1x2  , x 5 27. a. x  1q, 22 b. r 1x2  x5 12  x 15 c. 6 is not in the domain of r. r 162   29. a. x  15, q 2 2 x5 b. r 1x2  c. r 162  1; 6 is not in the domain of r. 1x  5 x2  36 13 31. a. x  a , q b b. r 1x2  c. r 162  0, r 162  0 2 12x  13 2x  4 33. a. h1x2  b. x  1q, 32 ´ 13, q2 c. x 2, x 0 x3 35. sum: 3x  1, x  1q, q 2 ; difference: x  5, x  1q, q2 ; product: 2x  3 2x2  x  6, x  1q, q2 ; quotient: , x  1q, 22 ´ 12, q2 x2 2 37. sum: x  3x  5, x  1q, q2 ; difference: x2  3x  9, x  1q, q 2 ; product: 3x3  2x2  21x  14, x  1q, q 2 ; x2  7 2 2 quotient: , x  aq, b ´ a , q b 3x  2 3 3 39. sum: x2  3x  4, x  1q, q2 ; difference: x2  x  2, x  1q, q2 ; product: x3  x2  5x  3, x  1q, q2 ; quotient: x  3, x  1q, 12 ´ 11, q 2 41. sum: 3x  1  1x  3, x  33, q 2 ; difference: 3x  1  1x  3, x  33, q2 ; product: 13x  12 1x  3, x  33, q 2 ; 3x  1 quotient: , x  13, q 2 43. sum: 2x2  1x  1, x  3 1, q2 ; 1x  3

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Student Answer Appendix difference: 2x2  1x  1, x  3 1, q2 ; product: 2x2 2x2 1x  1, x  3 1, q 2 ; quotient: , x  11, q2 1x  1 7x  11 45. sum: , x  1q, 22 ´ 12, 32 ´ 13, q 2 ; 1x  321x  22 3x  19 difference: , x  1q, 22 ´ 12, 32 ´ 13, q 2 ; 1x  321x  22 10 product: 2 , x  1q, 22 ´ 12, 32 ´ 13, q 2 ; 1x  x  62 2x  4 quotient: , x  1q, 22 ´ 12, 32 ´ 13, q2 15x  152 47. 0; 0; 4a2  10a  14 a2  9a 49. a. h1x2  12x  2 b. H1x2  2 1x  3  5 c. D of h(x): x  3 1, q2 ; D of H(x): x  33, q 2 51. a. h1x2  13x  1 b. H1x2  3 1x  3  4 c. D of h(x): x  3 13 , q2 D of H(x): x  3 3, q 2 53. a. h1x2  x2  x  2 b. H1x2  x2  3x  2 c. D of h(x): x  1q, q2 D of H(x): x  1q, q2 55. a. h1x2  x2  7x  8 b. H1x2  x2  x  1 c. D of h(x): x  1q, q2 D of H(x): x  1q, q2 57. a. h1x2  3x  1  5 b. H1x2  3x  16 c. D of h(x): x  1q, q2 D of H(x): x  1q, q2 59. a. 1 f  g21x2 : For g(x) to be defined, x 0. 2g1x2 5 For f 3 g1x2 4  , g1x2 3 so x  . g1x2  3 3 5 domain: e x | x 0, x  f 3 b. 1g  f 21x2 : For f(x) to be defined, x 3. 5 For g3 f 1x2 4  , f 1x2 0 so x 0. f 1x2 domain: {x | x 0, x 3} 10 5x  15 c. 1 f  g21x2  ; ; 1g  f 21x2  5  3x 2x the domain of a composition cannot always be determined from the composed form 61. a. 1 f  g21x2 : For g(x) to be defined, x 5. 4 For f 3g1x2 4  , g1x2 0 and g(x) is never zero g1x2 domain: {x| x 5} b. 1g  f 21x2 : For f(x) to be defined, x 0. 4 1 For g 3 f 1x2 4  , f 1x2 5 so x . f 1x2  5 5 4 domain: e x | x 0, x f 5 x c. 1 f  g21x2  4x  20; 1g  f 21x2  ; the domain of a composition 4  5x cannot always be determined from the composed form 63. a. 41 b. 41 65. g1x2  1x  2  1, f 1x2  x3  5 67. p1x2  21x  42 2  3, q1x2  12x  72 2  1 69. a. 6000 b. 3000 c. 8000 d. C(9)  T(9); 4000 71. a. $1 billion b. $5 billion c. 2003, 2007, 2010 d. t  12000, 20032 ´ 12007, 20102 e. t  12003, 20072 f. R152  C152 ; $4 billion 73. a. 4 b. 0 c. 2 1 d. 3 e.  f. 6 g. 3 h. 1 i. 1 j. undefined k. 0.5 l. 2 3 2 75. h1x2   x  4 77. h1x2  4x  x2 3 79. A  2r 120  r2; f 1r2  2r, g1r2  20  r; A152  250 units2 81. a. P1x2  12,000x  108,000; b. nine boats must be sold 83. a. P1n2  11.45n  0.1n2 b. $123 c. $327 d. C11152 7 R11152 85. h1x2  x  2.5; 10.5 87. a. 4160 b. 45,344 c. M1x2  453.44x; yes 89. a. 6 ft b. 36 ft2 c. A1t2  9t2; yes 91. a. 1995 to 1996; 1999 to 2004 b. 30; 1995

c. 20 seats; 1997 d. The total number in the senate (50); the number of additional seats held by the majority 93. Answers will vary. 95. no, yes x f(x) g(x) ( f  g)(x) ( f  g)(x) 2

27

15

12

1

18

11

7

0

11

7

4

1

6

3

3

2

3

1

4

3

2

5

7

4

3

1

4

5

6

3

3

6

11

7

4

7

18

11

7

8

27

15

12

97. a.

b.

y f(x)

4

4

4

x

8

4

8

5

x

g(x)

4

4

8

4

8

8

y

3 99. y   x 2

8

x

h(x)

4 8

2 1

8

4

c.

5

y

8

8

10

4

4

8

4

x

8

Summary and Concept Review, pp. 202–209 1. x  57, 4, 0, 3, 56 y  52, 0, 1, 3, 86 7

3

4

2

5

1

3

0

0

8

2. x  35, 5 4 y  30, 5 4

5 4 3 2 1 54321 1 2 3 4 5

3. 65 mi

4. 1 52 , 32

5.

5 4 3 2 1 54321 1 2 3 4 5

7. 1x  1.52 2  1y  22 2  6.25 y 8. a. b. 10 8 6 4 2

x y 0 5 3 4 2 121  4.58 0 5 2 121  4.58 4 3 5 0

y

1 2 3 4 5 x

6.

y

1 2 3 4 5 x

10 8 6 9 4 2

5 108642 2 2 4 6 8 10 x 4 9 6 8 10

108642 2 4 6 8 10

5 , 114, 72 9

1 3,

10, 32

y

3 2 4 6 8 10 x

5 4 3 2 1 54321 1 2 3 4 5

y

1 2 3 4 5 x

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Student Answer Appendix

9. a. parallel 10. a.

b. perpendicular y b. 10 8 6 4 2

10 8 6 4 2 (0, 1)

(1, 1)

108642 2 2 4 6 8 10 x 4 (0, 2) 6 8 10

11. a.

y

y

10 8 6 (0, 2) 4 2 (3, 0) 108642 2 2 4 6 8 10 x 4 6 8 10

12. a. vertical b. horizontal c. neither

108642 2 4 6 8 10

b.

10 8 6 4 2

2 4 6 8 10 x

(2, 2)

y

(w, 0)

108642 2 2 4 6 8 10 x 4 (0, 2) 6 8 10

5

33. squaring function a. up on left/up on the right; b. x-intercepts: (4, 0), (0, 0); y-intercept: (0, 0) c. vertex (2, 4) d. x  1q, q 2, y  34, q 2 34. square root function a. down on the right; b. x-intercept: (0,0); y-intercept: (0, 0) c. initial point (1, 2); d. x  31, q 2, y  1q, 2 4 35. cubing function a. down on left/up on the right b. x-intercepts: (2, 0), (1, 0), (4, 0); y-intercept: (0, 2) c. inflection point: (1, 0) d. x  1q, q 2, y  1q, q2 36. absolute value function a. down on left/down on the right b. x-intercepts: (1, 0), (3, 0); y-intercept: (0, 1) c. vertex: (1, 2); d. x  3 q, q 2, y  1q, 2 4 37. cube root a. up on left, down on right b. x-intercept: (1, 0); y-intercept: (0, 1) c. inflection point: (1,0) d. x  1q, q 2, y  1q, q 2 38. quadratic 39. absolute value y

y

8

y

4

x5

2y  x  5

(0, ) 5 2

4

(3, 0) 8

(5, 0)

5

8

4

4

x

8

(2, 5) 4

5x

8

4

4

x

8

4

8

8

y  4 5

10 8 6 4 2

108642 2 4 6 8 10

17. a.

10 8 6 4 2 108642 2 4 6 8 10

10 8 6 4 2

2 4 6 8 10 x

108642 2 4 6 8 10

b.

y

2 4 6 8 10 x

10 8 6 4 2 108642 2 4 6 8 10

4

4

4 8

8

(7.5, 2)

x

41. square root y

2 4 6 8 10 x

8

8

4

4

y 8

(9, 4)

4

(2, 0)

(5, 2) 8

4

4

4

(0, 1)

x

8

8

y

4

4

x

8

4

8

1 2 3 4 5 x

54321 1 2 3 4 5

54321 1 2 3 4 5

(0.5, 1)

8

x

c. 5 4 3 2 1

(3, 3)

4 8

b. y

5 4 3 2 1

4 4

8

43. a. (2, 3)

2 4 6 8 10 x

42. cube root

y

8

11 18. y  5, x  2; y  5 19. y  3 20. f 1x2  43 x 4 x  4 2 2 21. m  5 , y-intercept (0, 2), y  5 x  2. When the rabbit population increases by 500, the wolf population increases by 200. 15 22. a. y  90  15 x  105 2 1x  22 b. (14, 0), (0, 105) c. f 1x2  2 5 d. f 1202  45, x  12 23. a. x  3 4 , q2 2 b. x  1q, 22 ´ 12, 32 ´ 13, q 2 24. 14; 26 25. It is 9 ; 18a  9a a function. 26. I. a. D  51, 0, 1, 2, 3, 4, 56 R  52, 1, 0, 1, 2, 3, 46 b. 1 c. 2 II. a. x  1q, q2 y  1q, q2 b. 1 c. 3 III. a. x  33, q y  3 4, q 2 b. 1 c. 3 or 3 27. D: x  1q, q 2 R: y  3 5, q 2 f 1x2c: x  12, q2 f 1x2T: x  1q, 22 f 1x2 7 0: x  1q, 12 ´ 15, q2 f 1x2 6 0: x  11, 52 28. D: x  3 3, q 2 R: y  1q,02 f 1x2c: none f 1x2T: x  13, q2 f 1x2 7 0: none f 1x2 6 0: x  13, q2 29. D: x  1q, q2, R: y  1q, q 2 f 1x2c: x  1q, 32 ´ 11,q 2 f 1x2T: x  13, 12 f 1x2 7 0: x  15, 12 ´ 14, q 2 f 1x2 6 0: x  1q, 52 ´ 11, 42 1 30. a. odd b. even c. neither d. odd 31. a. ; the graph is rising 4 to the right. b. 2x  1  h; 3.01 y 32. zeroes: (6, 02 , (0, 0), (6, 0) (9, 0) 8 (10, 6) min: 13, 82, (7.5 2) (3, 4) 4 max: (6, 0), (3, 4) 8

40. cubic

y

5 4 3 2 1

4, w 1 2 3 4 5 x

54321 1 2 3 4 5

y

1, w 1 2 3 4 5 x

5 x  3 44. a. f 1x2  • x  1 3 6 x  3 b. R: y  3 2, q2 3 1x  3  1 x 7 3 45. D: x  1q, q 2, y R: y  1q, 82 ´ 18, q2 , 8 discontinuity at x  3; 4 h(x) define h1x2  8 at x  3 8

4

(3, 6)

4

x

8

4 8

46. 4, 4, 4.5, 4.99, 3 13  9, 3 13.5  9 47. D: x  1q, q 2 R: y  34, q2 y

8 4 8

4

4 4

8

x

8

20x x2 48. • 30x  20 2 6 x  4 40x  60 x 7 4 For 5 hr the total cost is $140.

160 120

Cost

13. yes 14. m  23 , y-intercept (0, 2); when the rodent population increases by 3000, the hawk population increases by 200. 4 15. a. y  4 b. y  53 x  5, m  53 , 3 x  4, m  3 , y-intercept (0, 4) y-intercept 10, 52 16. a. b. y y

80 40 0

49. 52. 55. 56. 58.

2

4

6

Hours

8

a2  7a  2 50. 147 51. x  1q, 23 2 ´ 1 23 , q 2 4x2  8x  3 53. 99 54. x; x f 1x2  1x  1; g1x2  3x  2 1 f 1x2  x2  3x  10; g1x2  x3 57. A1t2  12t  32 2 1 a. 4 b. 7 c. 6 d. 5 e. 14

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Student Answer Appendix

Mixed Review, pp. 209–210 3. a. 1q, 12 ´ 11, 42 ´ 14, q 2

4 1. y   x  4 3 3 5. y   x  2 2 9. y 5 4 3 2 1

54321 1

y  3

14. 3 b. a , q b 2

x  5

108642 2 4 6 8 10

4

4

4

(3, 2)

(2, 3) 8

x

8

4 8

8

4

4

x

8

4 8

16. a. V1t2  43 1 1t2 3 b. 36 in3 17. a. D: x  34, q2 ; R: y  33, q 2 b. f 112  2.2 c. f 1x2 6 0: x  14, 32 f 1x2 7 0: x  13, q2 d. f 1x2c: x  14, q2 f 1x2T: none e. f 1x2  31x  4  3 18. a. 4, 4, 6.25 b. 19. 20. y y y

1 2 3 4 5 x

10 8 6 4 2

y 8

(0, 5) 4

7. (2, 2); 1x  22 2  1y  22 2  50

3 4 5

11. a.

15.

y 8

y

2 4 6 8 10 x

rate of change is positive in 32, 1 4 since p ¢y 14 is increasing in 1q, 22; less; in  ¢x 1 ¢y 2  in [1, 2] 3 2, 1 4 ; ¢x 1

5 4 3 2 1

8 4 8

(2, 4)

4

4

(2, 4) 4

8

x

8

5 4321 1 2 3 4 5

5 4 3 2 1

1 2 3 4 5 x

5 4321 1 2 3 4 5

1 2 3 4 5 x

Strengthening Core Skills, p. 213

¢A  200.1 b. In the interval [15, 15.01], ¢t 1 1 1 ; aq, b ´ a , 1b ´ 11, q 2 13. 3 3 3x2  4x  1 ¢f ¢g 3  2x  h,  3; For small h, 2x  h  3 when x  . 15. ¢x ¢x 2 17. D: x  1q, 6 4; R : y  1q, 34 g1x2c: x  1q, 62 ´ 13, 62 g1x2T: x  13, 32 g(x) constant: x  16, 32 g1x2 7 0: x  17, 12 g1x2 6 0: x  1q, 72 ´ 11, 62 max: y  3 for x  16, 32; y  0 at 16, 02 min: y  3 at 13, 32 19. f 1x2  2x2  x  3

Exercise 1: h1x2  x2  28; x  4 217 Exercise 2: h1x2  x2  1; x  2 i 5 13 Exercise 3: h1x2  2x2  32 ; x  2 2

Cumulative Review Chapters 1–2, p. 214

8

4

3 2 1

x  4

1 2 3 4 5 x

8

x

8

1. a. a and c are nonfunctions, they do not pass the vertical line test 2. neither 3. 4. 12, 32; r  4 y y

54321 1 2 3 4 5

4 4

Practice Test, pp. 211–212 5 4 3 2 y  11

3. 29.45 cm 5. x  1 7. a. 1 b. 35 3 1 7 # y  2 x  2 11. 1f g21x2  3x3  12x2  12x; y 8 f a b1x2  3x, x 2; 1g  f 2  22 4 g

1. x2  2 9.

321 1 2 3 4 5 6 7

1 2 3 4 5 6 7 x

6 2 5. y   x  6. a. (7.5, 1.5), b.  61.27 mi 5 5 7. L1: x  3 L2: y  4 8. a. x  54, 2, 0, 2, 4, 66 y  52, 1, 0, 1, 2, 36 b. x  3 2, 6 4 y  31, 4 4 9. a. 300 b. 30 c. W1h2  25 d. Wages are $12.50 per hr. 2 h e. h  3 0, 404 ; w  3 0, 5004 10. I. a. square root b. x  3 4, q2, y  3 3, q2 c. 12, 02, 10, 12 d. up on right e. x  12, q 2 f. x  3 4, 22 II. a. cubic b. x  1q, q2 y  1q, q2 c. 12, 02, 10, 12 d. down on left, up on right e. x  12, q2 f. x  1q, 22 III. a. absolute value b. x  1q, q2 y  1q, 4 4 c. 11, 02, 13, 02, 10, 22 d. down/down e. x  11, 32 f. x  1q, 12 ´ 13, q2 IV. a. quadratic b. x  1q, q 2 ; y  35.5, q 2 c. 10, 02, 15, 02 d. up/up e. x  1q, 02 ´ 15, q 2 f. x  10, 52 31 7 a2  6a  7 8 11. a. b. 2 c.   i 2 25 25 a  6a  9 1 12. 3x  1; x  3 3 , q2 13. a. No, new company and sales should be growing b. 19 for [5, 6]; 23 for [6, 7] ¢s c.  4t  3  2h. For small h, sales volume is approximately ¢t 93,000 units 37,000 units 69,000 units in month 10, in in month 18, and 1 mo 1mo 1 mo month 24

13. a. D: x  1q, 84 , R: y  3 4, q 2 b. 5, 3, 3, 1, 2 c. 12, 02 d. f 1x2 6 0: x  12, 22 f 1x2 7 0: x  1q, 22 ´ 3 2, 84 e. min: (0, 4), max: (8, 7) f. f 1x2c: x  10, 82 f 1x2T: x  1q, 02 y

8 4 8

4

4 4

8

x

8

x7 b2  4ac b. 1x  52 1x  22 4a2 17. a. False;  X  b. False;  X  c. True d. False;  X  12 19. x  5 ; x  5.707; x  4.293 2 5 21. W  31 cm, L  47 cm 23. a. x  4 b. x  5, 13, 13 3 ,2 25. p  15  197 units  24.8 units. No, it is not a right triangle. 52  1 1972 2 102 15. a.

Connections to Calculus Exercises, pp. 217–218 1. a. 3 b. as h S 0, 3 remains constant 3. a. 2x  3  h b. as h S 0, 2122  3  h S 1 1 1 1 5. a. b. as h S 0, S 1x  h2x 4 12  h2 2 2122  h 2x  h 1 cS 7. a. 2 b. as h S 0, 8 2x 1x  h2 2 2122 2 12  h2 2 9. 10 v(t)  3 11. v(t)  t  4  7 10 9 8 7 6 5 4 3 2 1

9 8 7 6 5 4 3 2 1

1 2 3 4 5 6 7 8 9 10 t

A  24, distance  24 ft

1 2 3 4 5 6 7 8 9 10 t

A  40, distance  40 ft

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Student Answer Appendix

CHAPTER 3 Exercises 3.1, pp. 226–230 1. 25 3. 0; f (x) 2 7. left 2, down 9

5. Answers will vary. 9. reflected across x-axis; right 1, up 4

y

y

8

8

4

(5, 0) 8

(0, 3) 4 (1, 0)

(1, 0) 4

4 4

8

8

x

(1, 4) (3, 0)

4

4

8

x

4

(0, 5)

(2, 9) 8

iii. x  4  114 iv. x  2  12 v. t  2.7, t  1.3 2 vi. t  1.4, t  2.6 41. a. 10, 66,0002; when no cars are produced, there is a loss of $66,000. b. (20, 0), (330, 0); no profit will be made if less than 20 or more than 330 cars are produced. c. 175 d. $240,250 43. a. 6 mi b. 3600 ft c. 3200 ft d. 12 mi 45. a. 10, 33002; if no appliances are sold, the loss will be $3300. b. (20, 0), (330, 0); if less than 20 or more than 330 appliances are made and sold, there will be no profit. c. 0  x  200; maximum capacity is 200 d. 175, $12,012.50 47. a. 288 ft b. c. 484 ft; 5.5 sec d. 11 sec h(t) 600

8

500 400

11. stretched vertically, left 1, down 8

13. stretched vertically reflected across x-axis; right 2, up 15

y

(0, 7) 8 (.7, 0)

(.6, 0)

4

4 4

8

8

x

17. reflected across x-axis; left 76 , up 121 12 ; stretched vertically

y

(C, 121 ) 12

(0, 3)

4

4

8

8

(3, 0)

x

8

7  25 , 8 4

)

4

y

8

8

4 (0, 2) 4

4

(4.6, 0)

(3.6, 0)

8

8

4 8

(e,



x

(1.6, 0)

4

4 4

)

*

(1, 7)

8

23. reflected across x-axis; right 2, up 6 y

x

y 8

(0, 2) 4

3, e

(4.4, 0) 4

8

(0, 6)

25. left 3, up 52 ; compressed vertically

8 (2, 6)

8

x

21. left 1, down 7

y

4

8

8

(.4, 0)

4

8

8

x

(0, 7)

4

4

4

(.4, 0) 4

4

8

8

8

x

27. reflected across x-axis; right 25 , 29. right 32 , down 6; stretched up 11 vertically 2 ; stretched vertically y 8

y

4

4

(0, 3)

(4.2, 0)

4

(0, 7)

8

e, y

(0.8, 0) 8

4

8

8

x

4

(2.7, 0) 4

4

(0.3, 0) 4

8

8

8

x

w, 6

31. left 3, down 19 2 ; compressed vertically y 8 4

(7.4, 0) 8

(1.4, 0) 4

3, p

8 10 12 t

49. a. h1t2  16t  32t  5 b. (i) 17 ft (ii) 17 ft c. it must occur between t  0.5 and t  1.5 d. t  1 sec e. h112  21 ft f. 2 sec 51. 155,000; $16,625 53. a. 96 ft  48 ft b. 32 ft  48 ft 55. f 1x2  x2  4x  13 57. a. radicand will be negative—two complex zeroes. b. radicand will be positive—two real zeroes. c. radicand is zero—one real zero. d. two real, rational zeroes. e. two real, irrational zeroes. x2 59. 61. x  33, 23 4 x5

Exercises 3.2, pp. 238–242

(s, 0)

4 4

19. right 25 , down 17 4

8

6

(0, 6) 4

(3, 0)

(

y

12

4

8

x

16

8

4

8

8

15. right 74 , down 25 8 , stretched vertically

8

4

4

2

(4.7, 0)

4

(0, 5)

(1, 8) 8

(q, 0)

100 2

(2, 15)

16

4 8

200

y

8

(2.6, 0)

300

4 8 4 (0, 5)

x

8

33. y  11x  22 2  1 3 37. y   1x  22 2  3 2

35. y  11x  22 2  4 39. i. x  3  15

ii. x  4  13

1. synthetic; zero 3. P1c2 ; remainder 5. Answers will vary. 7. x3  5x2  4x  21  1x  221x2  3x  102  3 9. 2x3  5x2  4x  17  1x  3212x2  x  72  4 11. x3  8x2  11x  20  1x  521x2  3x  42  0 2x2  5x  3 0 13. a.  12x  12  x3 x3 b. 2x2  5x  3  1x  3212x  12  0 x3  3x2  14x  8 0 15. a.  1x2  5x  42  x2 x2 b. x3  3x2  14x  8  1x  221x2  5x  42  0 x3  5x2  4x  23 3  1x2  3x  102  17. a. x2 x2 b. x3  5x2  4x  23  1x  221x2  3x  102  3 2x3  5x2  11x  17 13  12x2  3x  12  19. a. x4 x4 b. 2x3  5x2  11x  17  1x  4212x2  3x  12  13 21. x3  5x2  7  1x  121x2  4x  42  11 23. x3  13x  12  1x  421x2  4x  32  0 25. 3x3  8x  12  1x  1213x2  3x  52  7 27. n3  27  1n  32 1n2  3n  92  0 29. x4  3x3  16x  8  1x  221x3  5x2  10x  42  0 4x  3 7x  5 31. 12x  72  2 33. 1x2  42  2 x 3 x 1 35. a. 30 b. 12 37. a. 2 b. 22 39. a. 1 b. 3 41. a. 31 b. 0 43. a. 10 b. 0 45. a. yes b. yes 47. a. no b. yes 49. a. yes b. yes 51. 3 1 2 5 6 53. 2 1 0 7 6 3 6 2 4 6 3 1 1 2 0 1 2 3 0 55. 23 9 18 4 8 6 16 8 9 24 12 0 57. P1x2  1x  221x  321x  52, P1x2  x3  4x2  11x  30 59. P1x2  1x  221x  2321x  232, P1x2  x3  2x2  3x  6 61. P1x2  1x  521x  2 2321x  2232, P1x2  x3  5x2  12x  60 63. P1x2  1x  121x  221x  2102 1x  2102, P1x2  x4  x3  12x2  10x  20 65. P1x2  1x  221x  321x  42 67. p1x2  1x  32 2 1x  321x  12 69. f 1x2  21x  32 2 1x  221x  52

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Student Answer Appendix 71. p1x2  1x  321x  32 2 73. p1x2  1x  22 3 75. p1x2  1x  321x  32 3 77. p1x2  1x  321x  32 2 1x  42 2 79. 4-in. squares; 16 in.  10 in.  4 in. 81. a. week 10, 22.5 thousand b. one week before closing, 36 thousand c. week 9 83. a. 198 ft3 b. 2 ft c. about 7 ft 85. k  10 87. k  3 89. The theorems also apply to complex zeroes of polynomials. 91. S3  36; S5  225 93. yes, John wins. 95. G1t2  1400t  5000

b. C1z2  1z  9i2 1z  42 1z  12 c. C1z2  1z  3i2 1z  1  2i2 1z  1  2i2 d. C1z2  1z  i21z  2  5i2 1z  2  5i2 e. C1z2  1z  6i2 1z  1  13 i21z  1  13 i2 f. C1z2  1z  4i2 1z  3  12 i2 1z  3  12 i2 g. C1z2  1z  2  i21z  3i2 1z  i2 h. C1z2  1z  2  3i2 1z  5i2 1z  2i2 117. a. w  150 ft, l  300; b. A  15,000 ft2 119. r 1x2  21x  4  2

Exercises 3.3, pp. 252–257

Exercises 3.4, pp. 267–270

1. coefficients 3. a  bi 5. b; 4 is not a factor of 6 7. P1x2  1x  221x  221x  3i21x  3i2 x  2, x  2, x  3i, x  3i 9. Q1x2  1x  221x  221x  2i21x  2i2 x  2, x  2, x  2i, x  2i 11. P1x2  1x  12 1x  12 1x  12 x  1, x  1, x  1 13. P1x2  1x  521x  52 1x  52 x  5, x  5, x  5 15. 1x  52 3 1x  92 2; x  5, multiplicity 3; x  9, multiplicity 2 17. 1x  72 2 1x  22 2 1x  72; x  7, multiplicity 2; x  2, multiplicity 2; x  7, multiplicity 1 19. P1x2  x3  3x2  4x  12 21. P1x2  x4  x3  x2  x  2 23. P1x2  x4  6x3  13x2  24x  36 25. P1x2  x4  2x2  8x  5 27. P1x2  x4  4x3  27 29. a. yes b. yes 31. a. yes b. yes 3 5 1 15 3 5 33. 51, 15, 3, 5, 14 ,  15 4 , 4 , 4 , 2 ,  2 , 2 , 2 6 3 5 35. 51, 15, 3, 5, 12 , 15 ,  ,  6 2 2 2 1 7 2 7 1 7 28 4 37. 51, 28, 2, 14, 4, 7, 16 , 14 3 , 3 , 3 , 3 , 6 , 2 , 2 ,  3 , 3 6 1 1 1 1 1 3 3 39. 51, 3, 32 , 2 , 16 , 4 , 8 , 32 , 32 , 16 , 34 , 38 6 41. 1x  421x  121x  32, x  4, 1, 3 43. 1x  321x  221x  52, x  3, 2, 5 45. 1x  321x  121x  42, x  3, 1, 4 47. 1x  221x  321x  52, x  2, 3, 5 49. 1x  421x  121x  221x  32, x  4, 1, 2, 3 51. 1x  721x  221x  121x  32, x  7, 2, 1, 3 53. 12x  3212x  121x  12; x  32 , 12 , 1 55. 12x  32 2 1x  12; x  32 , 1 57. 1x  221x  1212x  52; x  2, 1, 52 59. 1x  1212x  121x  1521x  152; x  1, 12 , 15,  15 61. 1x  1213x  221x  2i21x  2i2; x  1, 23 , 2i, 2i 63. x  1, 2, 3, 3 65. x  2, 1, 2 67. x  2, 32 , 4 2 3 69. x  3, 1, 53 71. x  1, 2, 3, 17 i 73. x  2, 23 , 1, 13 i 75. x  1, 2, 4, 2 77. x  3, 1,  12 79. x  1, 32 ,  13 i 1 81. x  2 , 1, 2,  13 i 83. a. possible roots: 51, 8, 2, 46 ; b. neither 1 nor 1 is a root; c. 3 or 1 positive roots, 1 negative root; d. roots must lie between 2 and 2 85. a. possible roots: 51, 26 ; b. 1 is a root; c. 2 or 0 positive roots, 3 or 1 negative roots; d. roots must lie between 3 and 2 87. a. possible roots: 51, 12, 2, 6, 3, 46 ; b. x  1 and x  1 are roots; c. 4, 2, or 0 positive roots, 1 negative root; d. roots must lie between 1 and 4 89. a. possible roots: 1, 20, 2, 10, 4, 5, 12 , 52 ; b. x  1 is a root; c. 1 positive root, 1 negative root; d. roots must lie between 2 and 1 91. 1x  4212x  3212x  32; x  4, 32 , 32 93. 12x  1213x  221x  122; x  12 , 23 , 12 95. 1x  2212x  1212x  121x  122; x  2, 12 , 12 , 12 97. a. 5 b. 13 c. 2 99. yes 101. yes 103. a. 4 cm  4 cm  4 cm b. 5 cm  5 cm  5 cm 105. length 10 in., width 5 in., height 3 in. 107. 1994, 1998, 2002, about 5 yr 109. a. 8.97 m, 11.29 m, 12.05 m, 12.94 m; b. 9.7 m,  3.7 111. a. yes, b. no, c. about 14.88 113A. a. 1x  5i21x  5i2 b. 1x  3i21x  3i2 c. 1x  i 1721x  i 172 113B. a. x   17, 17 b. x  213, 2 13 c. x  3 12, 3 12 115. a. C1z2  1z  4i21z  321z  22

1. zero; m 3. bounce; flatter 5. Answers will vary. 7. polynomial, degree 3 9. not a polynomial, sharp turns 11. polynomial, degree 2 13. up/down 15. down/down 17. down/up; 10, 22 19. down/down; 10, 62 21. up/down; 10, 62 23. a. even b. 3 odd, 1 even, 3 odd c. f 1x2  1x  32 1x  12 2 1x  32, deg 4 d. x  , y  3 9, q2 25. a. even b. 3 odd, 1 odd, 2 odd, 4 odd c. f 1x2  1x  321x  12 1x  22 1x  42, deg 4 d. x  , y  1q, 25 4 27. a. odd b. 1 even, 3 odd c. f 1x2  1x  12 2 1x  32, deg 3 d. x  , y   29. degree 6; up/up; 10, 122 31. degree 5; up/down; 10, 242 33. degree 6; up/up; 10, 1922 35. degree 5; up/down; 10, 22 37. b 39. e 41. c 43. 45. 47. y y y 10 8 6 4 2

108642 2 4 6 8 10

49.

10 8 6 4 2 108642 2 4 6 8 10

55.

10 8 6 4 2 108642 2 4 6 8 10

61.

10 8 6 4 2 108642 2 4 6 8 10

67.

10 8 6 4 2 54321 2 4 6 8 10

73.

40 32 24 16 8 54321 8 16 24 32 40

10 8 6 4 2

108642 2 4 6 8 10

2 4 6 8 10 x

y

51.

108642 2 4 6 8 10

2 4 6 8 10 x

y

57.

63.

69.

1 2 3 4 5 x

200 160 120 80 40 54321 40 80 120 160 200

1 2 3 4 5 x

y

20 16 12 8 4 54321 4 8 12 16 20

2 4 6 8 10 x

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

y

10 8 6 4 2

75.

30 24 18 12 6 54321 6 12 18 24 30

10 8 6 4 2

108642 2 4 6 8 10

2 4 6 8 10 x

y

53.

108642 4 8 12 16 20

2 4 6 8 10 x

y

59.

65.

1 2 3 4 5 x

20 16 12 8 4 54321 4 8 12 16 20

1 2 3 4 5 x

y

20 16 12 8 4 108642 4 8 12 16 20

2 4 6 8 10 x

y

20 16 12 8 4

71.

40 32 24 16 8 54321 8 16 24 32 40

y

1 2 3 4 5 x

77. h1x2  1x  421x  232 1x  232 1x  23i21x  23i2 79. f 1x2  21x  52 21x  222 1x  222 1x  2321x  232

2 4 6 8 10 x

y

2 4 6 8 10 x

y

2 4 6 8 10 x

y

1 2 3 4 5 x

y

1 2 3 4 5 x

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Student Answer Appendix

81. P1x2  16 1x  421x  121x  32, P1x2  16 1x3  13x  122 83. P1x2  x4  2x3  13x2  14x  24 85. a. 280 vehicles above average, 216 vehicles below average, 154 vehicles below average b. 6:00 A.M. 1t  02, 10:00 A.M. 1T  42 , 3:00 P.M. 1t  92 , 6:00 P.M. 1t  122 c. max: about 300 vehicles above average at 7:30 A.M.; min: about 220 vehicles below average at 12 noon 300 240 180 120 60 60 120 180 240 300

43. (0, 0) cross, (3, 0) cross 45. 14, 02 cross, (0, 4) 47. (0, 0) cross, (3, 0) bounce 49. 51. 53. y y 10 8 6 4 2

108642 2 4 6 8 10

y

55.

10 8 6 4 2

2 4 6 8 10 12 14 16 x

87. a. 3 b. 5 c. B1x2  14 x1x  421x  92, $80,000 89. a. f 1x2 S q, f 1x2q b. g1x2 S q, g1x2 S q; x4 0 for all x 1  2x 1 91. verified 93. h1x2  ; D : x  5x | x  06; H1x2  2 ; x2 x  2x D : x  5x| x  0, x  26 95. a. x  2 b. x  8 c. x  4, x  6

Mid-Chapter Check, p. 271

1. a. x3  8x2  7x  14  1x2  6x  521x  22  4 x3  8x2  7x  14 4  x2  6x  5  b. x2 x2 2. f 1x2  12x  321x  12 1x  121x  22 3. f 122  7 4. f 1x2  x3  2x  4 5. g122  8 and g132  5 have opposite signs 6. f 1x2  1x  221x  121x  221x  42 7. x  2, x  1, x  1  3i 8. 9. y y 15 12 9 6 3

108642 3 6 9 12 15

10 8 6 4 2

2 4 6 8 10 x

108642 2 4 6 8 10

108642 2 4 6 8 10

61.

10 8 6 4 2 108642 2 4 6 8 10

67. f 1x2 

10 8 6 4 2

108642 2 4 6 8 10

2 4 6 8 10 x

57.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

63.

y

2 4 6 8 10 x

1x  421x  12

10 8 6 4 2 108642 2 4 6 8 10

y

108642 2 4 6 8 10

2 4 6 8 10 x

59.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

65.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

69. f 1x2 

10 8 6 4 2

2 4 6 8 10 x

y

2 4 6 8 10 x

y

2 4 6 8 10 x

x2  4

1x  221x  32 9  x2 71. a. Population density approaches zero far from town. b. 10 mi, 20 mi c. 4.5 mi, 704 people per square mi 73. a. $20,000, $80,000, $320,000; cost increases dramatically b. c. as p S 100  , C S q Cost ($1000) 900 700 500 300 100

2 4 6 8 10 x

10. a. degree 4; three turning points b. 2 sec c. A1t2  1t  12 2 1t  321t  52 A1t2  t4  10t3  32t2  38t  15 A122  3; altitude is 300 ft above hard-deck A142  9; altitude is 900 ft below hard-deck

Percent 10 30 50 70 90

75. a. 5 hr; about 0.28 b. 0.019, 0.005; As the number of hours increases, the rate of change decreases. c. h S q, C S 0  ; horizontal asymptote 77. a. 2; 10 b. 10; 20 79. a. y 0.9 W(t) c. On average, 50 (1, 46) 0.7 6 words will be 40 0.5 remembered for life. 30 0.3

20

Reinforcing Basic Concepts, pp. 271–272 Exercise 1: 1.532 Exercise 2: 2.152, 1.765

Exercises 3.5, pp. 283–289 1. as x S q, y S 2 3. denominator; numerator 5. about x  98 7. a. as x S q, y S 2 9. a. as x S q, y S 1 as x S q, y S 2 as x S q, y S 1 b. as x S 1  , y S q b. as x S 2  , y S q as x S 1  , y S q as x S 2  , y S q 1 11. reciprocal quadratic, S1x2  2 1x  12 2 1 13. reciprocal function, Q1x2  2 x1 1 15. reciprocal quadratic, v 1x2  5 1x  22 2 17. S 2 19. S q 21. 1;  q 23. x  3, x 1q, 32 ´ 13, q2 25. x  3, x  3, x  1q, 32 ´ 13, 32 ´ 13, q 2 5 5 27. x  5 2 , x  1, x  1q, 2 2 ´ 12 , 12 ´ 11, q 2 29. No V.A., x  1q, q2 31. x  3, yes; x  2, yes 33. x  3, no 35. x  2, yes; x  2, no 37. y  0, crosses at 1 32 , 02 39. y  4, crosses at 121 41. y  3, does not cross 4 , 42

0.1

10

70 140 210 280 350x 10 20 30 40 50 t

b. 35%; 62.5%; 160 gal; c. 160 gal; 200 gal; d. 70%; 75% 81. a. $225; $175 b. 2000 heaters c. 4000 d. The horizontal asymptote at y  125 means the average cost approaches $125 as monthly production gets very large. Due to limitations on production (maximum of 5000 heaters) the average cost will never fall below A150002  135. 83. a. 5 b. 18 c. The horizontal asymptote at y  95 means her average grade will approach 95 as the number of tests taken increases; no d. 6 85. a. 16.0 28.7 65.8 277.8 b. 12.7, 37.1, 212.0 c. a. 22.4, 40.2, 92.1, 388.9 b. 17.8, 51.9, 296.8; answers will vary. 87. a. q1x2  3, horizontal asymptote at y  3; r1x2  7x  10, graph 10 crosses HA at x  b. q1x2  2, horizontal asymptote at 7 y  2; r1x2  7, no zeroes—graph will not cross 4 1 89. y  91. 39, 32 , 1 x 3 3

Exercises 3.6, pp. 298–302 1. nonremovable

3. two

x2  4 7. F1x2  • x  2 4

5. Answers will vary.

x  2 x  2

10 8 6 4 2 108642 2 (2, 4) 4 6 8 10

y

2 4 6 8 10 x

cob19421_saa_SA14-SA22.qxd 01/09/2009 02:22 PM Page SA17 ssen 1 HD 049:Desktop Folder:MHDQ103:MHDQ103-SAA:

Student Answer Appendix x2  2x  3 9. G1x2  • x  1 4

3x  2x2 2x  3 11. H1x2  μ 3 2

x 8 x2 13. P1x2  • 12

x  1

10 8 6 4 2

x  1

3 2 3 x 2

x2

y

(2, 12)

4

8

x

x  3x  x  3 3

19.

10 8 6 4 (2, 0) 2 108642 2 4 6 8 10

25. 10 8 6 4 (1, 0) 2

(2, 0) 2 4 6 8 10 x

y yx2

31.

27. 10 8 6 4 2

108642 2 4 6 8 10

2 4 6 8 10 x

33.

y

10 8 yx5 6 4 (0.8, 0) 2 (1, 0) 108642 2 4 6 8 10

37. 10

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

(4.8, 0)

39.

y

108642 2 4 6 8 10

43.

yx

(4, 0)

10 8 6 4 2

2 4 6 8 10 x

(3, 0)

(4, 0)

108642 2 2 4 6 8 10 x 4 6 (0, 0) 8 10

x3

10 8 6 4 (2, 0) 2 108642 2 4 6 8 10

(0, 0.2)

yx

yx3

y yx

(2, 0)

yx1

108642 2 4 6 8 10

10

x1 8 6 4 (2, 0) 2

(1, 0)

y

10 8 6 4 (2, 0) 2

x3

41.

2 4 6 8 10 x

2 4 6 8 10 x

35.

2 4 6 8 10 x

yx

(1, 0)

108642 2 4 6 8 10

2 4 6 8 10 x

y x3

Sheds made 12 16 20

2x3  48 x c. S1x2 is asymptotic to y  2x2. d. x  2 ft 3.5 in.; y  2 ft 3.5 in. 2x  55 59. a. A1x, y2  xy; R1x, y2  1x  2.521y  22 b. y  x  2.5 2x2  55x A1x2  c. A(x) is asymptotic to y  2x  60 x  2.5 2V V d. x  11.16 in.; y  8.93 in. 61. a. h  b. S  2r2  2 r r 2r3  2V c. S  d. r  5.76 cm, h  11.51 cm; S  625.13 cm2 r r3  2V ; r  3.1 in., h  3 in. 63. Answers will vary. 65. S  r 3 3 67. y  4 x  4, m  4 , 10, 42 69. a. P  30 cm, b. CD  60 13 cm,

108642 2 4 6 8 10

47.

2 4 6 8 10 x

y

yx x1 (2, 0)

2 4 6 8 10 x

(0, 0)

y 8 7 6 5 4 3 2 1 54321 1 2

b. S1x2 

Exercises 3.7, pp. 311–315

10 8 6 4 2

(1, 0)

y yx1

8

2 4 6 8 10 x

29.

y

4

4320 2 2 c. 30 cm2, d. A  750 169 cm , and A  169 cm

108642 2 4 6 8 10

x1

y

y

10 8 6 4 2

108642 2 2 4 6 8 10 x 4 (0, 4) 6 8 10

45.

2 4 6 8 10 x

23.

y

6 4 (0, 1) 2 yx1

(1, 2)

108642 2 4 6 8 10

10 8 yx1 6 4 (2, 0) 2 (2, 0)

x  1 8

y

10 8 6 (3, 2) 4 2

108642 2 2 4 6 8 10 x 4 6 8 y  x 10

(2, 0)

108642 2 4 6 8 10

1 2 3 4 5 x

y

200

57. a. S1x, y2  2x2  4xy; V1x, y2  x2y

x  3 x1

10 8 6 4 (3, 0) 2

yx

c. 8, $116.25

100

x  1

y

x  3, x  1

21.

y

6 x

300

x  1

2

x2  2x  3 17. R1x2  μ 2 2

4

400

54321 2 (1, 4) 4 6 8 10

4 4 2

2 4 6 8 10 x

10 8 6 4 2

8

2

53. a. a  5, y  3a  15 b. 60.5 c. 10 4x2  53x  250 55. a. A1x2  ; x  0, g1x2  4x  53 x b. cost: $307, $372, $445, Avg. cost: $307, $186, $148.33 d. 500 Cost

x3  7x  6 x1 15. q1x2  • 4

x2

16 12 8 4

6 4 2 4 8 12 16

w, w

108642 2 4 6 8 10

16

8

y  14  x2

y

10 8 6 4 2

51. 119.1

y

108642 2 2 4 6 8 10 x 4 (1, 4) 6 8 10

x

3

12

49.

y

SA17

y  x2  1 1 2 3 4 5 x

1. vertical; multiplicity 3. empty 5. Answers will vary. 7 7. x  10, 42 9. x  1q, 5 4 ´ 31, q 2 11. x  11, 2 2 3 133 3 133 13. x  3  17, 174 15. x  32  2 , 2  2 4 17. x  1q, 53 4 ´ 31, q2 19. x  1q, q2 21. { } 23. x  1q, 52 ´ 15, q 2 25. { } 27. x  1q, q2 29. x  1q, q2 31. x  1q, 5 4 ´ 35, q2 33. x  1q, 04 ´ 35, q 2 35. { } 37. x  13, 52 39. x  34, q2 ´ 516 41. x  1q, 2 4 ´ 526 ´ 3 4, q2 43. x  12  13, 2  132 45. x  3 q, 3 4 ´ 516 47. x  13, 12 ´ 12, q2 49. x  1q, 32 ´ 11, 12 ´ 13, q2 51. x  1q, 22 ´ 12, 12 ´ 13, q2 53. x  31, 1 4 ´ 536 55. x  33, 22 57. x  1q, 22 ´ 12, 12 59. x  1q, 22 ´ 3 2, 32 61. x  1q, 52 ´ 10, 12 ´ 12, q 2 63. x  14, 2 4 ´ 11, 2 4 ´ 13, q 2 65. x  17, 32 ´ 12, q 2 67. x  1q, 2 4 ´ 10, 22 69. x  1q, 172 ´ 12, 12 ´ 17, q2 7 71. x  13, 4 4 ´ 12, q 2 73. x  12, q2 75. x  11, q2 77. 1q, 32 ´ 13, q 2 79. x  1q, 34 ´ 35, q2 81. x  3 3, 0 4 ´ 33, q2 83. x  1q, 22 ´ 12, 32 85. x  1q, 2 4 ´ 11, 12 ´ 3 3, q 2 87. b 89. b 91. a. verified 1 b. D  41p  34 2 1p  32 2, p  3, q  2; p  3 4 ,q  4 c. 1q, 32 ´ 13, 3 d. verified 2 4 93. d1x2  k1x3  192x  10242 a. x  15, 84 b. 320 units c. x  30, 32 d. 2 ft 95. a. verified b. horizontal: r2  20, as r1 increases, r2 decreases to maintain R  40 vertical: r1  20, as r1 decreases, r2 increases to maintain R  40 c. r1  120, 402 97. R1t2  0.01t2  0.1t  30 a. 3 0°, 30°2 b. 120°, q 2 c. 150°, q2 99. a. n 4 b. n  9 c. 13 101. a. yes, x2 0

b. yes,

x2

x2  1

0

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Student Answer Appendix

103. x1x  221x  12 2 7 0;

x1x  22

1x  12 105. R1x2 6 0 for x  12, 82 ´ 18, 142 107. F1x2  e

f 1x2 6

x  4 x  4

3 2 1 54321 1 2 3 4 5 6 7

109.

2

43. E  0.5mv2; 612.50 J 45. cube root family; answers will vary; 0.054 or 5.4%

7 0

y

Amount A

Rate R

1.0

0.000

1.05

0.016

1.10

0.032

1.15

0.048

1.20

0.063

1.25

0.077

1 2 3 4 5x

2

Exercises 3.8, pp. 321–326 1. constant

3. y 

k x2

7. d  kr

5. Answers will vary.

9. F  ka

x

y

500

12.5

650

16.25

750

18.75

13. w  9.18; h $321.30; the hourly wage; k  $9.18/hr 192 15. a. k  192 47 S  47 h b. 360 c. 330 stairs

d. S  331; yes

Number of stairs

320 280 240

s(t) 

1 48 ; 32 volunteers 49. M  E; 41.7 kg V 6 51. D  21.6 1S; 144.9 ft 53. C  8.5LD; $76.50 p1p2 55. C  14.4  104 2 2 ; about 223 calls 57. a. about 23.39 cm3, d b. about 191% 59. a. M  kwh2 1 L1 2 b. 180 lb ¢y ¢y 10 110   ; less; for both f and g, as 61. For f: For g: ¢x 3 ¢x 9 x S q, y S 0 63. a. about 3.5 ft b. about 6.9 ft 65. x  0, x  2  2i 67. y 47. T 

11. y  0.025 x

8

192 47 h

(3, 5)

4

200 160 120

8

80

4

4

x

8

4

40

8

10 20 30 40 50 60 70 80 90100

Height h (in meters)

17. A  kS2 19. P  kc2 21. k  0.112; p  0.112 q2

Summary and Concept Review, pp. 326–331 q

1.

p

2.

y

226.8

45 55

338.8

8

548.8

70

8

23. k  6, A  6s2; 55,303,776 m2 25. a. k  16 d  16t2 b. 360 Height h in feet

320

d(t)  16t

280

2

240 200 160 120 80 40 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Time t (in sec)

c. about 3.5 sec 31. Y 

d. 3.5 sec; yes

12,321

Z

Z2

3,072,000,000

r2 37. A  kh1B  b2 6.75R 41. C  S2

9

74

2.25

; 48 kg

27. F 

Y

37

111 33. w 

e. 2.75 sec

1 35. l  krt

39. V  ktr2 R

S

C

120

6

22.5

200

12.5

350

15

8.64 10.5

4

k 2 d

29. S 

k L

8

(0, 3) 4

4

(2, 1)

(3, 0)

(4, 1)4

y

8

4

(5, 0)

3.

y

8

4

8

x

8

4

4

(0.3, 0) 8

x

8

4 4

8

8

(0, 5)

(2.7, 0) 4

4

8

x

w, 6

4. a. 0 ft b. 108 ft c. 2.25 sec d. 144 ft, t  3 sec 5. q1x2  x2  6x  7; R  8 6. q1x2  x  1; R  3x  4 7. 7 2 13 6 9 14 7 7 14 14 2 1 2 0 1 Since R  0, 7 is a root and x  7 is a factor. 8. x3  4x  5  1x  221x2  2x2  5 9. 1x  421x  121x  32 10. h1x2  1x  12 1x  42 1x2  2x  22 11. 12 4 8 3 1 2 5 1 4 10 2 0 Since R  0, 12 is a root and 1x  12 2 is a factor. 12. 3i 1 9 2 18 3i 18 9  6i 1 2  3i 0 6i Since R  0, 3i is a zero 13. 7 1 9 13 10 7 7 14 1 2 1 3 h172  3 14. P1x2  x3  x2  5x  5 15. C1x2  x4  2x3  5x2  8x  4 16. a. C102  350 customers b. more at 2 P.M., 170 c. busier at 1 P.M., 760 7 710 17. 51, 12 , 14 , 5, 10, 52 , 54 , 26 18. x  12 , 2, 52 19. p1x2  12x  32 1x  421x  12 20. only possibilities are 1, 3, none give a remainder of zero 21. [1, 2], [4, 5]; verified

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Student Answer Appendix 22. one sign change for g1x2 S 1 positive zero; three sign changes for g1x2 S 3 or 1 negative zeroes; 1 positive, 3 negative, 0 complex, or 1 positive, 1 negative, 2 complex; verified 23. degree 5; up/down; 10, 42 24. degree 4; up/up; 10, 82 25. 26. 27. y y y 10 8 6 4 2

54321 2 4 6 8 10

20 16 12 8 4

15 12 9 6 3

54321 4 8 12 16 20

1 2 3 4 5 x

54321 3 6 9 12 15

1 2 3 4 5 x

1 2 3 4 5 x

108642 2 4 6 8 10

33. V1x2 

108642 2 4 6 8 10

x2  x  12

8 6 4 2

108642 2 4 6 8 10

108642 4 8 12 16 20

38.

y

10 8 6 4 2

10 8 6 4 2

108642 2 4 6 8 10

2 4 6 8 10 x

y

2 4 6 8 10 x

2 4 6 8 10 x

0

1

y 10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

2

2

42.

Neg 0

1x  221x  12 x1x  22

Pos

Pos 2

1

0

1

Pos

x  2, 02 ´ 1, 22

196

7

2.88

38.75

1.25

17.856

24

0.6

48

108642 4 8 12 16 20

2 4 6 8 10 x

Practice Test, pp. 332–333

b. g1x2  12 1x  42 2  8 y

y (5, 9) (8, 0) (4, 8)

18 16 (0, 16) 14 12 10 8 6 4 2

108642

2 4 6 8 10 x

2. 12, 02, y  2x  4x 3. a. 40 ft, 48 ft b. 49 ft c. 14 sec 14x  3 2 4. x  5  2 5. x2  2x  9  x2 x  2x  1 0 15 10 24 6. 3 1 3 9 18 24 1 3 6 8 0 R0✓ 7. 1 8. P1x2  x 3  2x2  9x  18 9. Q1x2  1x  22 2 1x  12 2 1x  12, 2 mult 2, 1 mult 2, 1 mult 1 10. a. 1, 18, 2, 9, 3, 6 b. 1 positive zero, 3 or 1 negative zeroes; 2 or 0 complex zeroes c. C1x2  1x  221x  121x  3i21x  3i2 11. a. 1992, 1994, 1998 b. 4 yr c. surplus of $2.5 million 12. 13. 14. y y y 500

20 16 12 8 4

300

2 4 6 8 10 x

108642 4 8 12 16 20

10 8 6 4 2

2 4 6 8 10 x

15. a. removal of 100% of the contaminants dramatic increase c. 88% 16. a. b. y y 50 40 30 20 10

outputs are negative or zero for

2

z

15. x  1q, 32 ´ 12, 22 17. a. V1x2  124  2x2 116  2x21x2  4x3  80x2  384x b. 512  4x3  80x2  384x 0  x3  20x2  96x  128 c. for 0 6 x 6 8, possible rational zeroes are 1, 2, and 4 d. x  4 e. x  8  422  2.34 in. 19. R  kL1 A1 2

500

outputs are positive or zero for x    2, 22 ´ 5, q2

0

When x  1 Neg Pos Neg

2 4 6 8 10 x

300

5

2

w

20 16 12 8 4

100

When x  0 Pos

v

46. 4.5 sec

108642 100

1x  521x  22 x2  3x  10 41.  0 x2 x2 Neg

w2

2

b. about 2450 favors c. about $2.90 ea. 40. factored form 1x  421x  121x  22 7 0 outputs are positive for When x  0 Neg Pos Neg Pos x  14, 12 ´ 12, q 2 4

157.5

42 4 2 4 6 8 10 12 14 16 x 8 (2, 0) 12 16 (0, 16) 20

39. a.

y

729

20 16 12 8 4

108642 2 4 6 8 10

20 16 12 8 4

12.25

1. a. f 1x2  1x  52 2  9

2 4 6 8 10 x

2

37.

0.343

108642 1 2 3 4 5

x2  x  6 34. a. y  15; as x S q A1x2 S 15. As production increases, average cost decreases and approaches 15. b. x 7 2000 y 35. removable discontinuity at 12, 52 ; 10

x  1

105

5 4 3 2 1

2 4 6 8 10 x

x  1

216

0.72v

1. y  21x  12 2 2  92 3. 80 GB, $40.00 5. q1x2  x3  2x2  x  3; R  7 7. a. P112  42 b. P112  26 c. P152  6 9. a. x  9; x  83 b. P1x2  1x  22 1x  121x2  92; x  2, x  1, x  3i, x  3i 11. 13. y y

; V102  2

x  3x  4 x1 36. H1x2  • 5

y

44. k  0.72; z 

Mixed Review, pp. 331–332

10 8 6 4 2

2 4 6 8 10 x

x

45. t  160

28. a. even b. x  2, odd; x  1, even; x  1, odd c. deg 6: P1x2  1x  221x  12 2 1x  12 3 29. a. 5xx  ; x  1, 46 b. HA: y  1; VA: x  1, x  4 c. V102  94 (y-intercept); x  3, 3 (x-intercepts) d. V112  43 30. No—even multiplicity; yes—odd multiplicity 31. 32. y y 10 8 6 4 2

3 43. k  17.5; y  17.5 1 x

108642 10 20 30 40 50

108642 2 4 6 8 10

2 4 6 8 10 x

b. $500,000; $3,000,000;

10 8 6 4 2

2 4 6 8 10 x

108642 2 4 6 8 10

2 4 6 8 10 x

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Student Answer Appendix

18. a. x  1q, 3 4 ´ 3 1, 4 4 b. x  1q, 42 ´ 10, 22 3 y b. h   1 55; no c. 28.6% 29.6% 0.8 d. 11.7 hr e. 4 hr 43.7% 0.4 f. The amount of the chemical in the blood4 0 4 8 12 16 20x stream becomes neglible. 20. 520 lb

17. 800 19. a.

1.2

Strengthening Core Skills, pp. 334–335 Exercise 1: Exercise 2: Exercise 3: Exercise 4: Exercise 5: Exercise 6:

x  1q, 3 4 x  12, 12 ´ 12, q 2 x  1q, 42 ´ 11, 32 x  3 2, q 2 x  1q, 22 ´ 12, q2 x  3 3, 14 ´ 3 3, q2

Cumulative Review Chapters 1–3, pp. 335–336 R1R2 3. a. 1x  121x2  x  12 R1  R2 b. 1x  32 1x  221x  22 5. all reals 7. verified 11 1009 9. y  x  ; 39 min, driving time increases 11 min every 60 days 60 60 x3  3 11. Month 9 13. f 1 1x2  2 15. 17. X  63 19. y y 1. R 

5

10 8 6 4 2

5

8

54321 2 4 6 8 10

5 x

5

paired with y  1 17. not one-to-one, y  7 is paired with x  2 and x  2 19. one-to-one 21. one-to-one 23. not one-to-one; p1t2 7 5, corresponds to two x-values 25. one-to-one 27. one-to-one 29. f 1 1x2  5 11, 22, 14, 12, 15, 02, 19, 22, 115, 526 31. v1 1x2  513, 42, 12, 32, 11, 02, 10, 52, 11, 122, 12, 212, 13, 3226 5 x3 33. f 1 1x2  x  5 35. p1 1x2  x 37. f 1 1x2  4 4 3 3 39. Y1 41. f 1 1x2  x3  2 43. f 1 1x2  2 x1 1  x  4 8 x 1 1 45. f 1x2   2 47. f 1x2  49. a. x  5, y  0 x 1x 1 b. f 1x2  1x  5, x  0, y  5 51. a. x 7 3, y 7 0 8  3, x 7 0, y 7 3 53 a. x  4, y  2 b. v 1 1x2  Ax 1 b. p 1x2  1x  2  4, x  2, y  4 55. 1 f  g2 1x2  x, 1g  f 21x2  x 57. 1 f  g21x2  x, 1g  f 21x2  x 59. 1 f  g2 1x2  x, 1g  f 21x2  x 61. 1 f  g21x2  x, 1g  f 21x2  x x5 63. f 1 1x2  65. f 1 1x2  2x  5 67. f 1 1x2  2x  6 3 x3  1 3 3 69. f 1 1x2  2x  3 71. f 1 1x2  73. f 1 1x2  2 2x  1 2 x2  2 2 , x  0; y  c  , q b 75. f 1 1x2  3 3 x2  3, x  0; y  33, q 2 77. p1 1x2  4 1 79. v 1x2  1x  3, x  3; y  30, q 2 81. 83. y y

1 2 3 4 5 x 8

8

f(x) f 1(x)

4 4

4

8

f 1(x)

4 8

x

f(x)

4

4

4

4

8

8

8

x

Connections to Calculus Exercises, pp. 339–340 1. x  1q, 34 , as x S q, y S q, (0, 0); (0, 0), (3, 0) 5 4 3 2 1 54321 1 2 3 4 5

3. x  3 3, 3 4 , as x S 3, y S 3, as x S 3, y S 3; (0, 0); (3, 0), (0, 0), (3, 0)

y f (x) 1 2 3 4 5 x

5. x  1q, q 2, down/up, 10, 62, 13, 02, 11, 02, (2, 0)

5 4 3 2 1 54321 1 2 3 4 5

10 8 6 4 2 54321 2 4 6 8 10

85.

8

y h(x)

1 2 3 4 5 x

89.

p(x)

91. 1 2 3 4 5 x

2  119 2  119 , pa bb or about 3 3 12.11, 4.062. The skater has a maximum anxiety level of near 4, 2  119 2  119 , pa bb or about 2 min before starting his routine; a 3 3 about (0.79, 8.21). The skater has a minimum anxiety level of near 8, shortly after starting his routine. 11. a

CHAPTER 4 Exercises 4.1, pp. 350–354

1. second; one 3. 111, 22 , (5, 0), (1, 2), (19, 4) 5. False, answers will vary. 7. one-to-one 9. one-to-one 11. not a function 13. one-toone 15. not one-to-one, fails horizontal line test: x  3 and x  3 are

4

8

4

x

8

4 4

8

8

5 4 3 2 1

5 4 3 2 1

5 4 3 2 1 54321 1 2 3 4 5

y

f(x) f 1(x) 4

4

54321 1 2 3 4 5

93.

y 8

f(x)

4

54321 1 2 3 4 5

y

87.

f 1(x)

4

7. max y  2 at x  2 9. max at y  4.5 at x  1.5 12; min at y  4.5 at x  1.512

y 8

8

x

D: x  3 0, q 2 , R: y  32, q2 ; D: x  32, q 2 , R: y  30, q 2

1 2 3 4 5 x

y

D: x  10, q 2 , R: y  1q, q2 ; D: x  1q, q 2 , R: y  10, q 2

1 2 3 4 5 x

y

D: x  1q, 4 4 , R: y  1q, 4 4 ; D: x  1q, 4 4 , R: y  1q, 44

1 2 3 4 5 x

95. a. 31.5 cm b. The result is 80 cm. It gives the distance of the projector from the screen. 97. a. 63.5°F b. f 1 1x2  2 7 1x  592; independent: temperature, dependent: altitude c. 22,000 ft 99. a. 144 ft 1x b. f 1 1x2  , independent: distance fallen, dependent: time fallen 4 3 3x , independent: volume, c. 7 sec 101. a. 28,260 ft3 b. f 1 1x2  B

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Student Answer Appendix dependent: height c. 9 ft 103. Answers will vary. 105. d 107. x  31, 2 4 109. a. P  2l  2w b. A  r2 c. V  r2h d. V  13 r2h e. C  2r f. A  12 bh g. A  12 1b1  b2 2h h. V  43 r3 i. a2  b2  c2

Exercises 4.2, pp. 362–366 1. bx; b; b; x 3. a; 1 5. False; for b 6 1 and x2 7 x1, bx2 6 bx1, so function is decreasing 7. 40,000; 5000; 20,000; 27,589.162 9. 500; 1.581; 2.321; 221.168 11. 10,000; 1975.309; 1487.206; 1316.872 13. increasing 15. decreasing

75. no, they will have to wait about 10 min 77. a. $100,000 b. 3 yr 79. a.  $86,806 b. 3 yr 81. a. $40 million b. 7 yr 83. 32% transparent 85. 17% transparent 87.  $32,578 89. a. 8 g 3 b. 48 min 91. 9.5  107; answers will vary 93. 9 95. 2 ¢y  0.3842, 0.056, 0.011, 0.003; the rate of growth seems to 97. a. ¢x be approaching zero b. 16,608 c. yes, the secant lines are becoming virtually horizontal 99. x  32, q2, y  31, q 2 y 10 8

y0

y

10 8 6 4 2

10

(0, 1)

y0

2 4 6 8 10 x

108642 2 4 6 8 10

2 4

2 4 6 8 10 x

decreasing

y2

108642 2 4 6 8 10

y0

2 4 6 8 10 x

21. reflect across y-axis

108642 2 4 6 8 10

108642 2 4 6 8 10

y  3

(3, 1)

108642 2 4 6 8 10

2 4 6 8 10 x

108642 2 4 6 8 10

2 4 6 8 10 x

2 4 6 8 10 x

(2, 10) 10

y1

108642 2 4 6 8 10

2 4 6 8 10 x

(1, 2)

6 4 2 108642 2 4 6 8 10

14 12 10 8 6 4 2

(2, 7)

(2, 1) 2 4 6 8 10 x

33. e 35. a 37. 45. 4.113250 47. y

y  2 4 2 2 4 6

b

10 8 6 4 (2.3, 0) 2

2 4 6 8 10 x

39. 2.718282 49.

41. 7.389056

10 8 6 4 2

2 4 6 8 10 x

53. 3 55. 32 57. 13 59. 4 69. 2 71. 3 73. a. 1732, 3000, 5196, 9000 b. yes c. as t S q, P S q

1 2 3 4 x

61. 3 50,000

y

2

4

43. 4.481689 y

10 8 6 (0, 6.39) 4 2 (2, 0) 108642 2 4 6 8 10

63. 3

x

(0, 1)

(0, 1)

654321 1 2 3 4 5 6

8 10 x

65. 2

10 8 6 4 2

(1, 3) 2 4 6 8 10 x

108642 2 4 6 8 10

(3, 3) 2 4 6 8 10 x

65. reflect across x-axis, shift left 1 y

y

10 8 6 4 2 (0, 0) 108642 2 2 4 6 8 10 x 4 6 8 10

67. II 69. VI 71. V 73. x  1q, 12 ´ 13, q2 75. x  1 32 , q 2 77. x  13, 32 79. pH  4.1; acid 81. a. 4.7 b. 4.9 83. about 3.2 times 85. a. 2.4 b. 1.2 87. a. 20 dB b. 120 dB 89. about 3162 times 91. 6,194 m 93. a. about 5434 m b. 4000 m 95. a. 2225 items b. 2732 items c. $117,000 d. verified 97. a. about 58.6 cfm b. about 1605 ft2 99. a. 95% b. 67% c. 39% 101.  4.3; acid 103. Answers will vary. a. 0 dB 2 b. 90 dB c. 15 dB d. 120 dB e. 100 dB f. 140 dB 105. a. 3 3 5 b. c. 107. D: x   R: y   y 10 2 2 8 6 4 2

2 4 6 8 10 x

67. 2

y

108642 2 4 6 8 10

2 4 6 8 10 x

109. x  1q, 52; f 1x2  1x  52 1x  42 2  x3  3x2  24x  80

Mid-Chapter Check, pp. 379–380

p

5

40,000 30,000 20,000 10,000

1

c. volume of a

5; answers will vary 7. 23  8 15. 21  2 17. 72  49 log464  3 25. log3 19  2 1 log 1000  3 33. log100  2 1 39. 1 41. 2 43. 1 45. 2 53. 0.4700 55. 5.4161 57. 0.7841 61. shift right 2, up 3

y

108642 2 2 4 6 8 10 x 4 6 (0, 0) 8 10

51.

y

4 3 2 1

51. 1.6990

63. shift left 1

y

31. down 2 y 10 8 (0, 9)

47. 2 49. 2 59. shift up 3

108642 2 4 6 8 10

8 6 4 2 (0, 2)

(2, 5)

37. log4 18  3 2

3 2

10 8 6 4 2

27. up 1

y

6

1. logb x; b; b; greater 3. (1, 0); 0 5. 1 9. 71  17 11. 90  1 13. 83  2 19. 102  100 21. e4  54.598 23. 27. 0  loge1 29. log13 27  3 31. 35. log48 

y

29. right 2

108642 2 4 6 8 10

8 6 4 2

10 8 (2, 7) 6 4 (0, 4) y3 2

25. left 1, down 3 10 8 6 4 2

y

23. reflect across y-axis, up 3

y

10 (3, 8) 8 6 4 2 (0, 1)

4

4

Exercises 4.3, pp. 375–379 (1, 9)10

10 8 6 (1, 5) 4 (0, 3) 2

2

2

101. a. volume of a sphere b. area of a triangle rectangular prism d. Pythagorean theorem

19. left 3 y

y0

4

6 4 2 (0, 1)

increasing

17. up 2

y0

6

(2, 9) 8

(2, 9)

108642 2 4 6 8 10

y

2

3 4 (days)

5t

1. a. 23  log279 b. 54  log81243 2. a. 83  32 b. 12960.25  6 3. a. x  5 b. b  54 4. a. x  3 b. b  5 5. a. $71,191.41 b. 6 yr 6. F1x2  4 # 5x3  2 7. f 1 1x2  1x  12 2  3, D: x  3 1, q2 ; R: y  3 3, q 2 ; verified 8. a. 4  log381, verified 2 b. 4  ln 54.598, verified 9. a. 273  9, verified b. e1.4  4.0552, verified 10. 7.9 times more intense

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Student Answer Appendix

Reinforcing Basic Concepts, p. 380

Exercises 4.5, pp. 405–410

Exercise 1: about 158 times Exercise 2: about 501 times Exercise 3: about 12,589 times

1. Compound 3. Q0ert 5. Answers will vary. 7. $4896 9. 250% 11. $2152.47 13. 5.25 yr 15. 80% 17. 4 yr 19. 16 yr 21. $7561.33 23. about 5 yr 25. 7.5 yr 27. no 29. a. no b. 9.12% 31. 7.9 yr 33. 7.5 yr 35. a. no b. 9.4% 37. a. no b. approx 13,609 euros 39. No; $234,612.01 41. about 7 yr A AP 43. 23 yr 45. a. no b. $302.25 47. a. t  b. p  pr 1  rt A ln a b Q1t2 p nt A  1b b. t  49. a. r  n a 51. a. Q0  rt Bp e r n ln a1  b n Q1t2 lna b Q0 b. t  53. $709.74 55. a. 5.78% b. 91.67 hr 57. 0.65 g r 59. 816 yr 61. about 12.4% 63. $17,027,502.21 65. 7.93% 67. 2548.8 m 69. P1x2  x4  4x3  6x2  4x  15

Exercise 4: about 398 times Exercise 5: about 39,811 times

Exercises 4.4, pp. 392–396 7. x  29.964 9. x  1.778 65 11. x  2.200 13. x  1.260 15. x  ln  2, x  4.7881 4 ln 2.32 17. x  log1782  5, x  3.1079 19. x   , x  1.1221 0.75 8 1.25 3 21. x  e  4, x  10.3919 23. x  5  10 , x  12.7828 e0.4  5 25. x  , x  1.7541 27. ln12x2  14x2 29. log1x2  12 2 x5 x b 35. ln a b 37. ln1x  22 39. log242 31. log34 33. loga x1 x 41. log5 1x  22 43. 1x  22 log 8 45. 12x  12 ln 5 47. 12 log 22 49. 4 log5 3 51. 3 log a  log b 53. ln x  14 ln y 55. 2 ln x  ln y 57. 12 3 log1x  22  log x 4 59. ln 7  ln x  12 ln13  4x2  ln 2  3 ln1x  12 ln 152 ln 60 61. ; 2.104076884 63. ; 3.121512475 ln 7 ln 5 log 1.73205 log 0.125 65. ; 0.499999576 67. ;3 log 3 log 0.5 log1x2 ; f 152  1.4650; f 1152  2.4650; f 1452  3.4650; 69. f 1x2  log132 outputs increase by 1; f 133 # 52  4.465 log1x2 ; h122  0.3155; h142  0.6309; h182  0.9464; 71. h1x2  log192 outputs are multiples of 0.3155; h124 2  410.31552  1.2619 73. x  32 75. x  6.4 77. x  20, 5 is extraneous 79. x  2, 52 is extraneous 81. x  0 83. x  52 85. x  23 e2  63 87. x  32 89. x  19 91. x  9 9 93. x  2; 9 is extraneous 95. x  3e3  12 ; x  59.75661077 97. no solution 99. t  12 ; 4 is extraneous 101. x  2  13, x  2  13 is extraneous ln 128,965 ln 231 2  2; x  0.7968 105. x   ; x  3.1038 103. x  ln 7 3 ln 5 3 ln 2 ln 9  ln 5 ; x  1.7095 109. x  ; x  0.5753 107. x  ln 3  ln 2 2 ln 5  ln 9 C p  1 ln a b a , t  55.45 111. x  46.2 113. t  k 115. a. 30 fish b. about 37 months 117. about 3.2 cmHg 119. about 50.2 min 121. $15,641 123. 6 hr, 18.0% 125. Mf  52.76 tons 127. a. 26 planes b. 9 days 129. a. log34  log35  2.7268 b. log34  log35  0.203 c. 2 log35  2.9298 131. a. d b. e c. b d. f e. a f. c 133. x  0.69314718 135. a. 1 f  g21x2  31log3x22 2  3log3x  x; 1g  f 2 1x2  log3 13x2 2  2  x  2  2  x b. 1 f  g2 1x2  e1ln x12 1  eln x  x; 1g  f 2 1x2  ln ex1  1  x  1  1  x x 137. a. y  ex ln 2  eln 2  2x; x ln y y  2 1 ln y  x ln 2, e  ex ln 2 1 y  ex ln 2 b. y  bx, ln y  x ln b, eln y  ex ln b, y  exr for r  ln b 139. Answers will vary. 141. b 143. y 1. e

3. extraneous

5. 2.316566275

10 8

(2, 0)

108642 2 4 6 8 10

1. no

2. no

3. yes

(2, 0) 2 4 6 8 10 x

x1

4. f 1 1x2 

x2 5. f 1 1x2  1x  2 3 8. f 1x2: D: x  1q, q 2, R: y  1q, q 2; f 1 1x2: D: 1q, q 2, R: y  1q, q2

6. f 1 1x2  x2  1; x 0 7. f 1x2: D: x  34, q2, R: y  3 0, q2; f 1 1x2: D: x  30, q2, R: y  34, q 2 y

f(x)

y 5 4 3 2 1

5 4 3 2 1

54321 1 2 3 4 5

54321 1

1 2 3 4 5 x

1 2 3 4 5 x

2 f(x) 3 4 5

9. f 1x2: D: x  1q, q2, R: y  10, q2; f 1 1x2: D: x  10, q2, R: y  1q, q2

10. a. $3.05 11.

10 8 6 4 2

y 5 4 3 2 f(x) 1 54321 1 2 3 4 5

1 2 3 4 5 x

 2 b. f 1 1t2  t 0.15 , f 1 13.052  7 c. 12 days 12. 13. y y 10 8 6 4 2

y3

108642 2 4 6 8 10

2 4 6 8 10 x

108642 y  1 2 4 6 8 10

2 4 6 8 10 x

10 8 6 4 2

108642 2 4 6 8 10

y

2 4 6 8 10 x

y  2

1 14. 2 15. 2 16. 52 17. 12.1 yr 18. 32  9 19. 53  125 3.7612 20. e  43 21. log525  2 22. ln 0.7788  0.25 23. log381  4 24. 5 25. 1 26. 12 27. 28. 29. y y y x  3 10 8 6 4 2

108642 2 4 6 8 10

10 8 6 4 2

2 4 6 8 10 x

30. x  1q, 02 ´ 16, q2 33. a. x  e32

(0, 4) 6 4 2

Summary and Concept Review, pp. 410–414

34. a. x 

108642 2 4 6 8 10

2 4 6 8 10 x

31. x  132 , q 2

b. x  102.38

ln 4 , x  2.7726 0.5

10 8 6 4 2

c. x  ln 9.8 b. x 

108642 2 4 6 8 10

x1

2 4 6 8 10 x

32. a. 4.79 b. 107.3 I0 1 d. x  log 7 2

ln 19 , x  6.3938 0.2

103 , x  33.3333 d. x  e2.75, x  0.0639 35. a. ln 42 3 3 d. log1x2  x2 36. a. 2 log59 b. 2 log74 b. log930 c. ln 1 xx   12 c. x 

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Student Answer Appendix c. 12x  12ln 5 d. 13x  22ln 10 37. a. ln x  14 ln y b. 13 ln p  ln q c. 53 log x  43 log y  52 log x  32 log y log 45 d. log 4  53 log p  43 log q  32 log p  log q 38. a.  2.215 log 6 log 128 ln 124 ln 0.42 b.  4.417 c.  6.954 d.  0.539 log 3 ln 2 ln 5 ln 7 2 ln 5 39. x  ln 2 40. x  ln 3  1 41. x  1ln 3 42. x  6.389 43. x  5; 2 is extraneous 44. x  4.25 45. a. 17.77% b. 23.98 days 46. 38.6 cmHg 47. 18.5% 48. Almost, she needs $42.15 more. 49. a. no b. $268.93 50. 55.0%

Mixed Review, pp. 414–415 log 30

 4.9069 b. 1.5 log 2 c. 1x  32 ln 2 5. 7. y

1. a.

y0

10 8 6 4 2

108642 2 4 6 8 10

2 4 6 8 10 x

1 3

c.

10 8 6 4 2

108642 2 4 6 8 10

3. a. 2 log1020

b. 0.05x

9. a. 54  625

y

11. x  3, x  2 1multiplicity 22; x  4 15. f 1 1x2 

5x  3 2

10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

Connections to Calculus Exercises, p. 423 3 1 1. 5x  ln x 3. 5 log x  3 log y  log z 2 2 1 5. A1x2  x2 2 y yt

x0

x

A(x)

1129 4

x

0

t

7. verified (factor out 12 , then combine like terms and simplify)

vary. 13. 6  log 2 17. I  6.3  1017  19. 1.6 m, 1.28 m, 1.02 m, 0.82 m, 0.66 m, 0.52 m

9. t 

Practice Test, pp. 415–416

CHAPTER 5

1. 34  81 2. log255  12 3. 52 logbx  3 logby  logbz 5 m2n3 4. logb 5. x  10 6. x  7. 2.68 8. 1.24 3 1p 9. 10. 11. a. 4.19 b. 0.81 y y

Exercises 5.1, pp. 435–440

10 8 6 4 2

108642 2 4 6 8 10

y3 2 4 6 8 10 x

10 8 6 4 2

108642 2 4 6 8 10

x2

2 4 6 8 10 x

12. f is a parabola (hence not one-to-one), x  , y  3 3, q2 ; vertex is at 12, 32 , so restricted domain could be x  3 2, q 2 to create a one-to-one function; f 1 1x2  1x  3  2, x  33, q2, y  3 2, q 2 . 13. x  1  ln 3 14. x  1, x  5 is extraneous 15. 5 yr 16.  8.7 yr 17. 19.1 months 18. 7% compounded semi-annually 19. a. no b. $54.09 20. a. 10.2 lb b. 19 weeks ln 89

Strengthening Core Skills, p. 418 Exercise 1: Exercise 2: Exercise 3: Exercise 4:

Answers will vary. a. log1x2  3x2 b. ln1x2  42 c. logx Answers will vary. a. x log 3 b. 5 ln x c. 13x  12 ln 2

x  3

Cumulative Review Chapters 1–4, pp. 418–419

1. x  2  7i 3. 14  5i2 2  814  5i2  41  0 9  40i  32  40i  41  0 0  0✓ 5. f 1g1x22  x g1f 1x2 2  x Since 1f  g21x2  1g  f21x2, they are inverse functions. ¢T 455 7. a. T1t2  455t  2645 11991 S year 12 b. , triple births  ¢t 1 increase by 455 each year c. T162  5375 sets of triplets, T 1172  10,380 sets of triplets 9. D: x  310, q2, R: y  3 9, q2 y 10 8 h1x2c: x  12, 02 ´ 13, q 2 h1x2T: x  10, 32 6 4 2

108642 2 4 6 8 10

2 4 6 8 10 x

2V 13. 2a b

17. x  5, x  6 is an extraneous root 19. a. 88 hp for sport wagon, ~81 hp for minivan b. 3294 rpm c. minivan, 208 hp at 5800 rpm

2 4 6 8 10 x

b. e0.45  0.15x c. 107  0.1  108 11. a. x  31, q2, y  3 2, q2 b. g1 1x2  1x  22 2  1, x  32, q2, y  3 1, q2 c. Answers will 9 15. 4

SA23

x1 1 lna b 2 1x

1 2 1. Complementary; 180; less; greater 3. r, 2r , radians 5. Answers will vary. 7. a. 77.5° b. 30.8° 9. 53° 11. 42.5° 13. 67.555° 15. 285.0025° 17. 45.7625° 19. 20° 15¿ 00– 21. 67° 18¿ 25.2– 23. 275° 19¿ 48– 25. 5° 27¿ 9– 27. No, 19  16 6 40 29. 69° 31. 25° 33. 62.5 m 35. 41 12 ft  58 ft  10 ft  68 ft 37. 645°, 285°, 435°, 795° 39. 765°, 405°, 315°, 675° 41. s  980 m 43.   0.75 rad 8 mi 49.   0.2575 rad 45. r  1760 yd 47. s  3 51. r  9.4 km 53. A  115.6 km2 55.   0.6 rad 57. r  3 m 59.   1.5 rad; s  7.5 cm; r  5 cm; A  18.75 cm2 61.   4.3 rad s  43 m; r  10 m; A  215 m2 63.   3 rad; A  864 mm2; s  72 mm; r  24 mm  7 2 rad 69. 65. 2 rad 67. rad 71. rad 73. 0.4712 rad 4 6 3 75. 3.9776 rad 77. 60° 79. 30° 81. 120° 83. 720° 85. 165° 87. 186.4° 89. 171.9° 91. 143.2° 93. h  7.06 cm; m  3.76 cm; n  13.24 cm 95. 960.7 mi apart 97. a. 50.3 m2 b. 80° c. 17 m 99. a. 1.5 rad/sec b. about 15 mi/hr  101. a. 40 rad/min b. ft/sec  0.52 ft/sec c. about 11.5 sec 6 103. a. 1000 m b. 1000 m c. 100012 m  1414.2 m 45

1000 m

√2 (1000) m

45 1000 m

105. 50 12 or about 10.7 mi apart 107. a. 50.3°/day; 0.8788 rad/day b. 0.0366 rad/hr c. 6.67 mi/sec 109. Answers will vary. 111. a. 192 yd b. 86.6 rpm 1 113. 8.14% 115. y  1x  22 2  4 4

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SA24

Student Answer Appendix

Exercises 5.2, pp. 449–454

y 1. x, y, origin 3. x, y, x , sec t, csc t, cot t 5. Answers will vary. 5 12 111 5  111 15 7. 10.6, 0.82 9. a , b 11. a , b 13. a , b 13 13 6 6 4 4 15. 10.9769, 0.21372 17. 10.9928, 0.11982 13 1 13 1 13 1 19. a , b, a , b, a , b 2 2 2 2 2 2  111 5 111 5 111 5 , b, a , b, a , b 21. a 6 6 6 6 6 6 23. 10.3325, 0.94312, 10.3325, 0.94312, 10.3325, 0.94312 25. (0.9937, 0.1121), 10.9937, 0.11212, 10.9937, 0.11212 1 13   12  12 27. a , b is on unit circle 29. ; a , b 2 2 4 2 2  13 1  13 1   12 12 31. ; a , b 33. ; a , b 35. ; a , b 6 2 2 4 2 2 6 2 2 12 12  12  12 12  12 12 b. c. d. e. f. g. 2 2 2 2 2 2 2 12 13  13 h. 39. a. 1 b. 1 c. 1 d. 0 41. a. b. 2 2 2 13 13 13 13  13 13 c. d. e. f. g. h. 2 2 2 2 2 2 43. a. 0 b. 0 c. undefined d. undefined 45. sin t  0.6, cos t  0.8, tan t  0.75, csc t  1.6, sec t  1.25, cot t  1.3 5 12 13 13 47. sin t  12 13 , cos t  13 , tan t  5 , csc t  12 , sec t   5 , 5 cot t  12 6 111 5 111 6 111 49. sin t  , cos t  , tan t  , csc t  , sec t  , 6 6 5 11 5 5111 cot t  11 121 2  121 5 121 , cos t  , tan t  , csc t  , 51. sin t  5 5 2 21 5 2 121 sec t  , cot t  2 21 2 12 1 3 12 , cos t  , tan t  2 12, csc t  , 53. sin t  3 3 4 12 sec t  3, cot t  4 1 13 2 13 55. sin t  , cos t  , tan t  13, csc t  , sec t  2, 2 2 3 13 cot t  3  12 12 57. sin t  , cos t  , tan t  1, csc t  12, sec t   12, 2 2 cot t  1 59. QI, 0.7 61. QIV, 0.7 63. QI, 1 65. QII, 1.1 67. QII, 0.4 2 7 2  3 5 69. QIV, 3.1 71. 73. 75. 77. 79. , 3 6 3 2 4 4  3 3 5 3 4 3 4 81. , 83. 85. 0,  87. a. 1 4 , 5 2 b. 1 4 , 5 2 , 2 2 4 4 89. 2.3416 91. 1.7832 93. 3.5416 5 12 5 2 25 144 169 12 2 95. a. 1 13 , 13 , 12, 1 13 2  1 12 13 2  169  169  169  1; sin t  13 , 5 12 13 13 5 cos t  13 , tan t  5 , csc t  12 , sec t  5 , cot t  12 7 24 7 2 49 576 625 24 2 b. 1 25 , 25 , 12, 1 25 2  1 24 25 2  625  625  625  1; sin t  25 , 7 25 25 7 cos t  25 , tan t  24 , csc t  , sec t  , cot t  7 24 7 24 35 12 2 35 2 144 1225 1369 35 c. 1 12 37 , 37 , 12, 1 37 2  1 37 2  1369  1369  1369  1; sin t  37 , 12 35 37 37 12 cos t  37 , tan t  12 , csc t  35 , sec t  12 , cot t  35 9 40 9 2 81 1600 1681 40 2 d. 1 41 , 41 , 12, 1 41 2  1 40 41 2  1681  1681  1681  1; sin t  41 , 9 41 41 9 cot t  cos t  41 , tan t  40 , csc t  , sec t  , 9 40 9 40 97. a. 5 rad b. 30 rad 99. a. 5 dm b.  6.28 dm 101. a. 2.5 AU b.  6.28 AU 103. yes 105. range of sin t and cos t is [1, 1] 107. a. 2t  2.2 b. QI c. cos t  0.5 d. No 109. a. d  10 3 b. midpoint: (1, 1) c. m  111. a. x  6, 4 b. x  24 4 37. a.

Exercises 5.3, pp. 466–470

1. increasing 3. 1q, q2 ; 31, 1 4 7. t y  cos t  1 13 2 12  2 1  2

7 6 5 4 4 3 3 2 5 3 7 4 11 6 2 9. a. II 11.



0 1 2 12 2 13 2 1 b. V

c. IV y

d. I 13.

e. III y

1

3

2

5. Answers will vary.



1



 2

2

t

1



 2

2

1

15. A  3, P  2

17. A  2, P  2

y

y 2

3 

2



t

3

2

t

2

19. A  12 , P  2

21. A  1, P  

y

y

Q

1



2

 2

t

Q



t

1

23. A  0.8, P  

25. A  4, P  4

y

y

0.8

4

 2



2

t

0.8

4

t

4

27. A  3, P  12

29. A  4, P  65

y

y 4

3

~

Q

t

E

3

T

t

2

t

4

1 31. A  2, P  128

33.

y

y

2

3

1 256

2

2 t

3 2



1 128



t 3

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Student Answer Appendix

35.

37. A  2, P 

y

 ,k 2

25.

y  2 tan t

y y  tan t

3  2 2 1 2 2 3 4 5

 2

5 4 3 2 1

2 

2

t

2

39. 0 A 0  3, P  , f

41. P  4, h

1 47. A  4, P  72 ,d

45. P  14 , j

43. 0 A 0  34 , P  5, b

49. y  34 cos18t2 2 51. y  0.2 csc1 12 t2 53. y  6 cosa tb 3 3 7 55. red: y  cos x; blue: y  sin x; x  , 4 4 57. red: y  2 cos x; blue: y  2 sin13x2; 3 3 7 11 7 15 x , , , , , 8 4 8 8 4 8 112 59. cos t  113 , (15, 112, 113)  61. a. 3 ft b. 80 mi c. h  1.5 cos a xb 40  63. a. D  4 cos a tb b. D  3.86 c. 72° 12 65. a. D  15 cos1t2 b. at center c. Swimming leisurely. One complete cycle in 2 sec! 67. a. Graph a b. 76 days c. 96 days 69. a. 480 nm S blue b. 620 nm S orange 71. I  30 sin150t2, I  21.2 amps 73. Since m t 0  2  3 2 2

 M, 0; y avg. value  3; shifted up 3 units; 3 avg. value  1; amplitude is “centered” on average value. 5

29.

1 3

75. g(t) has the shortest period; y 1

1 2954

b. 1  6i



3 2 2

t

 8

 4

t

3 2

3

t

1 2

1

t

y

5

37.

 4

t

39.

y

y

3

2

1

1 4

2 4

1 2

t

43. y  2 cot a

1 41. y  3 tana tb 2

2 tb 3

45.

 3 , 8 8

47. about 137.8 ft  xb; P  12; asymptotes at x  6  12k, k  ; using 12 (3, 5.2), A  5.2; at x  2, model gives y  3.002; at x  2, model gives y  3.002; answers will vary. 51. Answers will vary; y  11.95 tan ; P  180°; asymptotes at   90°  180°k; A  11.95 from 130°, 6.9 cm2; pen is  12 cm long 53. a. 20 cm  62.8 cm b. 80 cm; it is a square c. n P y  5.2 tan a

64.984 63.354 63.063 62.853

getting close to 20

  tan 

2.5 2

3  7i d. 2

cos t 5. a. use sin t

1 , 1, 13, und. 13 9. 1.6, 0.8, 0.5, 1.4, 0.7, 1.2 11. a. 1 b. 13 c. 1 d. 13 7 7 5 3 1 b. c. d. 15. und., 13, 1, ,0 13. a. 4 6 3 4 13 13 35 59  17. , , 19. 1.6, 4.6, 7.8 21.  k, k  Z 24 24 24 10  23.  k, k  Z 12 b. use reciprocals of tan t 7. 0,

 2

t

35.

 8

55.

Exercises 5.4, pp. 479–484 3. odd; f 1t2 ; 0.268

y h(t)

y

2

1.5 1 0.5 0

 1. ; P  B

 2

y

10 20 30 100

c. 7  3i

3  2 2 1 2 2 3 4 5

t

31.

 4

33.

1

5 4 3 2 1

1



4

1

79. a. 3  4i

y  cot x

1 

1 t 1477

200 77. distance  yd  115.5 yd 13

3 2 2

y

2

49.

3



27.

SA25

10

20

30

40

50

60 

a. no; 35° b. 1.05 c. Angles will be greater than 68.2°; soft rubber on sandstone 57. a. 5.67 units b. 86.5° c. Yes. Range of tan  is (q, q ). d. The closer  gets to 90°, the longer the line segment gets. 59. 3 2, 34 S  7.1 m/sec; 3 3, 3.5 4 S  26.1 m/sec; 33.5, 3.8 4 S  128 m/sec. The velocity of the beam is increasing dramatically, 33.9, 3.99 4 S  12,733 m/sec 61. a. x-intercepts: (0, 0), (3, 0); y-intercepts: (0, 0); vertical 3 asymptotes: x  2, x  2; horizontal asymptote: y  2 b. x-intercept: (1, 02 ; y-intercept: none; vertical asymptotes: x  0, x  4; horizontal asymptote: y  0 1 c. x-intercepts: (1, 02, (1, 0); y-intercepts: a0,  b; vertical asymptote: 2 x  2 slant asymptote: y  x  2 63. 7.37 hr

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Student Answer Appendix

Mid-Chapter Check, pp. 484–485

21. P1t2  250 cos c

1. a. 36.11°N, 115.08°W b. 2495.7 mi. 2.   4.3; A  860 cm2 1  12 3. a. b. 2 13 4. a. 1.0353 b. 8.9152 2 2  15 3 3 ; sin   , csc   , cos   , sec   5. y  , 3 3 2 3 15 2 15 tan   , cot   6. 221.8°, 3.8711 2 15 7. asymptotes: x  5, 3, 1, 1, 3, 5 8. | A |  3, P  4

23. 25. 27. 29. 31. 33. 35.

y

y

39.

3

3

2 2 1t  2.752 d  950; P1t2  250 sina tb  950 11 11 A  120; P  24; HS: 6 units right; VS: (none); PI: 6  t 6 30 A  1; P  12; HS: 2 units right; VS: (none); PI: 2  t 6 14 A  1; P  8; HS: 32 unit right; VS: (none); PI: 23  t 6 26 3 A  24.5; P  20; HS: 2.5 units right; VS: 15.5 units up; PI: 2.5  t 6 22.5 A  28; P  12; HS: 25 units right; VS: 92 units up; PI: 52  t 6 29 2 A  2500; P  8; HS: 13 unit left; 1 23 VS: 3150 units up; PI: 3  t 6 3    y  250 sina tb  350 37. y  5 sina t  b  13 12 50 2   y  4 sina t  b  7 41. 100 y 180 4 80 60

6

4

2

1 2 3 4 5 6

4

t

2

4

t

40 20

3

3

9. a. QIV

2

2

43.

b. 2  5.94  0.343 c. sin t, tan t 3 8 b. f 1t2  6 cos a tb 4 3

4

6

8

10 t

3 2

10. a. A  6, P  c. f 12  3

4

y

1 0 1 2

 4

 2

3 2

t

3

Reinforcing Basic Concepts, pp. 485–486 1. 112 ,

23 1 23 2 2, cos t  2 , sin t  2 5 2. t  , negative since x 6 0 6 3. QIV, negative since y 6 0  4. QI, cos t  12 , sin t  23 2 ,t  3

4

2 2 1 1 2 ,B ;f ,P ;B  2f. B P P f 1/f A sin1Bt2  A sin 3 12f 2t 4 1 47. a. P  4 sec, f  cycle/sec b. 4.24 cm, moving away 4 c. 4.24 cm, moving toward d. about 1.76 cm. avg. vel.  3.52 cm/sec, greater, still gaining speed 5 49. d1t2  15 cosa tb 51. red S D3; blue S A#3 4 53. D3: y  sin 3146.84 12t2 4; P  0.0068 sec; G4: y  sin 3 392 12t2 4 ; P  0.00255 sec 55. a. Caracas: 11.4 hr, Tokyo: 9.9 hr b. (i) Same # of hours on 79th day & 261st day (ii) Caracas: 81 days, Tokyo: 158 days 45. P 

Exercises 5.5, pp. 494–499 y  A sin1Bt  C2  D; y  A cos1Bt  C2  D 0  Bt  C 6 2 5. Answers will vary. a. A  50, P  24 b.  25 c. [1.6, 10.4] a.  A   200, P  3 b. 175 c. [1.75, 2.75] y 100  11. y  40 sina tb  60 80 15 60

1. 3. 7. 9.

40 20 6

13. y  8 sina

 tb  12 180

20

12

18

24

13

30 t

y

16 12 8

0

4 60

15. a. y  5 sina

 tb  34 12

b.

120 180

40

365

240 300 360 t

y

32 24

c.  1:30 A.M., 10:30 A.M.

11

16 8 5

10

15

20

25 t

 17. a. y  6.4 cosa tb  12.4 6 b. 20 y c. 134 days 16 12 8

57. a. Adds 12 hr. The sinusoidal behavior is actually based on hours more/less than an average of 12 hr of light. b. Means 12 hr of light and dark on March 20, day 79 (Solstice!). c. Additional hours of deviation from average. In the north, the planet is tilted closer toward the Sun or farther from Sun, depending on date. Variations will be greater! 59. QIII; 3.7    0.5584

4 2

4

6

8

10

c. max  1200, min  700 d. about 2 yr.

19. a. P  11 yr b. 1500 P 1200 900 600 300 2

4

6

61. sum: 2, difference: 2i 25, product: 6, quotient:

12 t

8

10

12 t

2 i25  3 3

Exercises 5.6, pp. 506–511 1.   tan1x 3. opposite; hypotenuse 5. To find the measure of all three angles and all three sides. 13 13 12 5 7. sin   12 13 , csc   12 , sec   5 , tan   5 , cot   12

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Student Answer Appendix 85 13 84 85 9. cos   13 85 , sec   13 , cot   84 , sin   85 , csc   84 11 5 15 2 11. sin   511 , tan   , csc   , cos   , sec   5 15 2 11 5 15 2 15 13. 15. Angles Sides Angles Sides A  30° a  98 cm A  45° a  9.9 mm

17.

B  60° C  90°

b  98 13 cm c  196 cm

Angles A  22° B  68° C  90°

Sides a  14 m b  34.65 m c  37.37 m

B  45° C  90° 19.

verified 21.

23. 35. 47. 55. 59. 65. 71. 77. 79. 83.

Angles A  65° B  25° C  90°

Angles A  32° B  58° C  90°

b  9.9 mm c  9.9 12 mm Sides a  5.6 mi b  8.96 mi c  10.57 mi

verified Sides a  625 mm b  291.44 mm c  689.61 mm

875 m

85. approx. 450 ft 87. a. approx. 20.2 cm for each side b. approx. 35.3° x 89. cot u  h x  h cot u xd cot v  h h cot u  d cot v  h h cot v  h cot u  d d  h cot u  h cot v d h cot u  cot v 91. a. approx. 3055.6 mi b. approx. 9012.8 mi c. approx. 7 hr, 13 min 93. a. local max: 15, 22, (2, 3); local min: 12, 12, 17, 22, 16, 32 b. zeroes: x  6, 3, 1, 4 c. T1x2T: x  15, 22 ´ 12, 62; T1x2c: x  17, 52 ´ 12, 22 d. T1x2 7 0: x  16, 32 ´ 11, 42; T1x2 6 0: x  17, 62 ´ 13, 12 ´ 14, 62 95. d  53.74 in. D  65.82 in.

Exercises 5.7, pp. 519–522 1. origin; x-axis 3. positive; clockwise 5. Answers will vary. 7. slope  13, equation: y  13x, 13 1 sin 60°  , cos 60°  , tan 60°  13 2 2

QI/III;

y

5 4 3 2 1 54321 1 2 3 4 5

1 2 3 4 5 x

3 3 (4, 3): sin   ; (4, 3): sin    5 5 4 4 cos   cos    5 5 3 3 tan   tan   4 4 y 11. QII/QIV; 5 4 3 2 1

54321 1 2 3 4 5

0.4540 25. 0.8391 27. 1.3230 29. 0.9063 31. 27° 33. 40° 40.9° 37. 65° 39. 44.7° 41. 20.2° 43. 18.4° 45. 46.2° 61.6° 49. 21.98 mm 51. 3.04 mi 53. 177.48 furlongs They have like values. 57. They have like values. 1 13 13 13 1 2 13 , , , , 13, 2, , 13 43° 61. 21° 63. , 2 2 3 2 2 3 6  2 23 67. 7  423 69.   11.0°,   23.9°,  145.1° approx. 300.6 m 73. approx. 1483.8 ft 75. approx. 118.1 mph a. approx. 250.0 yd b. approx. 351.0 yd c. approx. 23.1 yd approx. 1815.2 ft; approx. 665.3 ft 81. approx. 386.0 a. 875 m b. 1200 m c. 1485 m; 36.1°

 1200 m

9.

1 2 3 4 5 x

1 1 (3, 13): sin   ; 13,  132: sin    2 2 13 13 cos    cos   2 2 1 1 tan    tan    13 13 17 8 17 15 15 13. sin   , csc   , cos   , sec   , tan   , 17 15 17 8 8 8 cot   15 21 29 20 15. sin   , csc   , cos   , 29 21 29 29 21 20 sec   , tan   , cot   20 20 21  12 2 12 17. sin   , csc   , , cos   2 2 12 2 sec   , tan   1, cot   1 12 1 13 19. sin   , csc   2, cos   , 2 2 2 1 sec   , tan   , cot   13 13 13 4 117 1 21. sin   , csc   , cos   , 4 117 117 1 sec   117, tan   4, cot   4 2  113 3 23. sin   , csc   , cos   , 2 113 113  113 2 3 sec   , tan   , cot   3 3 2 6 161 5 25. sin   , csc   , , cos   6 161 161  161 6 5 sec   , tan   , cot   5 5 6 2 15 121 1 27. sin   , , csc    , cos   121 2 15 121 1 sec   121, tan   2 15, cot   2 15 29. x  0, y  k; k 7 0; r  k; k 0 k sin 90°  , cos 90°  , tan 90°  , k k 0 sin 90°  1, cos 90°  0, tan 90° undefined csc 90°  1, sec 90° undefined cot 90°  0

SA27

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Student Answer Appendix

31. 60° 33. 45° 35. 45° 37. 68° 39. 40° 41. 11.6° 43. QII 45. QII 1 13 1 47. sin    ; cos   ; tan    2 2 13  12 12 49. sin   ; cos   ; tan   1 2 2  13 1 51. sin   ; cos   ; tan   13 2 2  13 1 1 ; tan   53. sin    ; cos   2 2 13 3 5 4 55. x  4, y  3, r  5; QIV; sin   , csc   , cos   , 5 3 5 5 3 4 sec   , tan   , cot   4 4 3 35 37 57. x  12, y  35, r  37; QIII; sin   , csc   , 37 35 12 37 35 12 cos   , sec   , tan   , cot   37 12 12 35 1 212 , 59. x  212, y  1, r  3; QI; sin   , csc   3, cos   3 3 3 1 sec   , tan   , cot   2 12 2 12 2 12 8 7 61. x   115, y  7, r  8; QIII; sin   , csc   , 8 7 115 8 7  115 , tan   , cot   , sec    cos   7 115 115 8 63. 52°  360°k 65. 87.5°  360°k 67. 225°  360°k 13 1 1 13 1 , ,  13 73.  , , 69. 107°  360°k 71. 2 2 2 2 13  13 1 , cos   , tan   13 75. sin   2 2  13 1 , cos    , tan   13 77. sin   2 2 1  13 1 , cos   , tan   79. sin   2 2 13 1  13 1 , cos   , tan    1 81. sin   2 2 13 83. QIV, neg., 0.0175 85. QIV, neg., 1.6643 87. QIV, neg., 1.5890 89. QI, pos., 0.0872 91. a. approx. 144.78 units2 b. 53° c. The parallelogram is a rectangle ab sin  whose area is A  ab. d. A  2 93.   60°  360°k and   300°  360°k 95.   240°  360°k and   300°  360°k 97.   61.1°  360°k and   118.9°  360°k 99.   113.0°  360°k and   293.0°  360°k 101. 1890°; 90°  360°k 103. head first; 900° 105. approx. 701.6° 107. 343.12 in2 109. Answers will vary. 111. a. 12,960° b. 125.66 in. c. 15,080 in. d. 85.68 mph 113. about 555.4 ft 115. y  54 x  2

14. sin t  

17 4 , csc t   , 4 17

3 4 17 3 cos t  , sec t  , tan t   , cot t   4 3 3 17  2 and 15. 16. t  2.44 17. a. approx. 19.6667 rad b. 25 rad 3 3 18. A  3, P  2 19. P  2 y

4 3 2 1

y 10 8 6 4 2  2

1 2 3 4

2 t

3 2





 2

 2 2 4 6 8 10

20. A  1, P  



t

3 2

 2

21. A  1.7, P 

y

2

y

1 1

0.5  2

 2

0.5

0

t

 8

1

1

 4

2

22. A  2, P 

1 2

23. A  3, P 

y

4

1 199

y

2 2

1 1

1

4

8

1 8

1

1 4

0

t

1 398

2

2

1 199

t

4

24. y  0.75 sin16t2 25. y  4 csc13t2 26. green; red 2 7  1 2 ; 27. tan a b  1; cot a b  28.   4 3 3 3 13 29.

30.

y 12 6 2 



6

1 0.51 2 3 4 5

t

2

5 4 3 2 1

12

y

0.5

1

t

31. 1.55  k radians; k  Z 32. 3.5860 33.  151.14 m  34. y  5.2 tana xb; period  12; A  5.2; asymptotes x  6, x  6 12 35. a. A  240, P  12, HS: 3 units right, VS: 520 units up b. 760 y 520

280 0

3

6

9

15 t

12

36. a. A  3.2, P  8, HS: 6 units left, VS: 6.4 units up y b. 8 6

Summary and Concept Review, pp. 522–530 3. 10.125  13.5  16.875 7 1 7. approx. 4.97 units 8.  4. approx. 692.82 yd 5. 120° 6. 6 2 9. s  25.5 cm. A  191.25 cm2 10. r  41.74 in., A  2003.48 in2 11.   4.75 rad, s  38 m 12. a. approx. 9.4248 rad/sec b. approx. 3.9 ft/sec c. about 15.4 sec 6 113 6 113 6 113 6 13. y   , a , b, a ,  b, and a , b 7 7 7 7 7 7 7

t 2

3 8

4

1. 147.613 2. 32°52¿12–

2 8

6

4

2

t 2

37. A  125, P  24, HS: 3 units right, VS: 175 units up,  y  125 cos c 1t  32 d  175 12 3 38. A  75, P  , HS: (none), 8 16 VS: 105 units up, y  75 sin a tb  105 3

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Student Answer Appendix  39. a. P1t2  0.91 sin a tb  1.35 b. August: 1.81 in., Dec: 0.44 in. 6 40. a. A  0.80, b. A  64.3° 41. a. cot 32.6°, b. cos170°29¿45– 2 42.

Angles

Sides

A  49° B  41° C  90°

a  89 in. b  77.37 in. c  117.93 in.

43.

Angles

Sides

A  43.6° B  46.4° C  90°

a  20 m b  21 m c  29 m

44. approx. 5.18 m 45. a. approx. 239.32 m b. approx. 240.68 m apart 46. approx. 54.5° and 35.5° 47. 207°  360°k; answers will vary. 48. 28°, 19°, 30° 35 37 12 37 49. a. sin   , csc   , cos   , sec   , 37 35 37 12 35 12 tan   , cot   12 35 3  113 2 113 b. sin   , csc   , cos   , sec   , 3 2 113 113 3 2 tan   , cot   2 3 3 5 50. a. x  4, y  3, r  5; QIV; sin    , csc    , 5 3 4 5 3 4 cos   , sec   , tan   , cot   5 4 4 3 12 13 b. x  5, y  12, r  13; QIV; sin   , csc   , 13 12 5 13 12 5 , cot   cos   , sec   , tan   13 5 5 12 51. a.   135°  180°k b.   30°  360°k or   330°  360°k c.   76.0°  180°k d.   27.0°  360°k or   207.0°  360°k

Mixed Review, pp. 530–531 1. a. A  10 b. D  15 c. P  6 d. f142  20 2 4 3. t  and t  5. 220°48¿50– 3 3 7. 1212 in.; 60 12  84.9 in. 28 112 9. arc length:   29.3 units; area:  117.3 units2 11. 86.915° 3 3 8 17 15 13. sin   , sec   , cos   , 17 15 17 17 8 15 csc   , tan   , cot   8 15 8 15. 60° 17. a. 6 rad/sec b. 20162 cm/sec  377 cm/sec 7 19. a. A  5; P  ; HS: (none); b. A  ; P  4; HS: 1 unit right; 2 VS: 8 units down; PI: 0  t 6  VS: (none); PI: 1  t 6 5 15 12 9 6 3 3 6 9 12 15

y

4 3 2 1  2



3 2

2

t

c. A: NA; P  4; HS: none; VS: none; PI: (2, 2) 10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

1 2 3 4

y

Practice Test, pp. 532–533

 1. complement: 55°; supplement: 145° 2. a. 45° b. 30° c. 6 3. 30°  360°k; k  Z 4. a. 100.755° b. 48°12¿45– 5. a. 430 mi b. 215 13  372 mi 6.

t

sin t

cos t

tan t

csc t

0

0

1

0

undefined

1

undefined

2 3

13 2



1 2

 13

213 3

2

 13 3

7 6



cot t

1 2



13 2

13 3

2

2 13 3

13



12 2

1

 12

 12

1

2

13 3

213 3

13

5 4



12 2

5 3



13 2

13 6

sec t

 3

d.

1 2

1 2

 13

13 2

13 3



2 13 3 2

5 5  121  121 , , tan   , csc   7. sec   , sin   121 2 5 2 2 cot   121 2 12 2 1 1 2 8 8. a b  a b   1 3 3 9 9 9. a. 225.8 ft or 225 ft 9.6 in.

b.

23  0.1505 rad/sec 480

c. 11.29 ft/sec  7.7 mph 10. Angles Sides A  33° B  57° C  90°

a  8.2 cm b  12.6 cm c  15.0 cm

11. about 67 cm, 49.6°

12. 57.9 m

13. a.

7 6

b.

11 6

c.

3 4

 tb  34.1 12 b. 433,000 gal; 249,000 gal 15. a. D: t  R, R: y  32, 2 4 , P  10, A  2;

14. a. W1t2  18.4 sina

2

y

1 2

4

6

8

10 t

1 2

 12k  12 for k  Z, 2 R: y  1q, 1 4 ´ 31,  q2, P  2   c. D: t  12k  12 for k  Z, R: y  R, P  6 3 b. D: t 

2

4

6

8

10 t

 to the 2 right; VS: none; PI: (, )

d. A: NA; P  2; HS:

10 8 6 4 2 6543 21 2 4 6

8

y

1

2 4

3

4 8

 3

2 3



4 3

5 3

2 t

y

2

1

y

1 2 3 4 5 6 x

3

 2



3 2

2 t

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Student Answer Appendix

 2 , HS: units right, VS: 19 units up 3 12 , 3  y  4 31 32 t 6 PI: 12 4 24

hyp  89;   64°; 90    26°

16. A  12, P 

3  17  17 4 , tan t  , sec t  , 5. cos t  , sin t  4 4 3 3 4 4 17 3 17 3 csc t   , cot t   7 7 17 17

16 8

,  7 12 7  6

 3

 2

2 3

5 3



3 7. a. D: x  c , q b, R: y  30, q2 2 b. D: x  1q, 72 ´ 17, 72 ´ 17, q 2, R: y  1q, q 2 9. a. max: (2, 4), endpoint max: (4, 0) min: (2, 4), endpoint min: (4, 0) b. f 1x2 0: x  3 4, 02 ´ 546 f 1x2 6 0: x  10, 42 c. f 1x2c: x  14, 22 ´ 12, 42 f 1x2T: x  12, 22 d. function is odd: f 1x2  f 1x2 y 11.  114.3 ft 13. 5

t

17. 1260° 18. a. D: t 

  12k  12, k  Z; R: y  R; P  ; 4 2

y

8 4

 2

4

3 2



2 t

8

b. D: t  2k, k  Z; R: y  R, P  2,

y

4

x  1 (0, 1)

5

2

5 x

y  2  2

2



3 2

2 t

5

4

  19. y  7.5 sina t  b  12.5 20. a.  4 6 2

b.  2.3

Strengthening Core Skills, pp. 535–536 Exercise 1:

t

0

 6

 4

 3

 2

sin t  y

0

1 2

12 2

13 2

1

cos t  x

1

13 2

12 2

1 2

0

y x

0

13 3

1

13

—

2 3

3 4

5 6



7 6

5 4

13 2

12 2

1 2

0

1 2

 12 2

1 2

 12 2

13 2

1

13 2

12 2

1

 13 3

0

13 3

tan t 

13 Exercise 2: 4 5 a. t  , 3 3

b. t 

 7 , 4 4

c. t 

 7 , 6 6

Exercise 3: a. no solution b. t  1.2310, t  5.0522 d. t  1.9823, t  4.3009

d. t 

1. 5 6 x 6 3 3. 89

80

 39

f(x)

3 2 1

54321 1 2 3 4 5

1 2 3 4 5 x

f 1 (x)

25. about 6.85%

Connections to Calculus Exercises, pp. 541–542 1.

hyp  x  4 4 2x2  8x sin   , cos   x4 x4 x4 x4 csc   , sec   4 2x2  8x 4 2x2  8x tan   , cot   4 2x2  8x

3. x2  16

1

 7 , 4 4

c. t  6.0382, t  2.8966

Cumulative Review Chapters 1–5, pp. 536–537

15. x  9, y  40, r  41, QII; 9 40 40 41 41 cos   , sin   , tan   , sec   , csc   , 41 41 9 9 40 9 cot   ,   102.7° 40 3  1 17. S  18 m; A  135 m2 19. y  sina4t  b  2 2 2 3 3 y 21. 23. y  x  2 m  , y-intercept (0, 2) 5 4 4 , 4

␪ 4

cot  

x

sin  

x

, cos  

4 x

13 2u2  169 2 , sec   7. r  u 13 sin  3 9. r  11. x2  y2  5y  0; circle sin   2 cos  13. 3x  2y  6; line 5. cot  

4

2x  16 2x  16 2x2  16 2x2  16 csc   , sec   x 4 2

2

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Student Answer Appendix

CHAPTER 6 1  sin2x 1. sin ; sec ; cos  3. one; false 5. ; Answers will vary. sin x sec x 7. Answers may vary; sec x sin x sec x 1 sec x 1 sin x tan x  ;  ;  ;  csc x cos x csc x cot x csc x cot x cos x 9. 1  sec2x  tan2x; tan2x  sec2x  1; 1  1sec x  tan x21sec x  tan x2; tan x   2sec2x  1 cos x 11. sin x cot x  sin x  cos x sin x 2 1 cos x 1 13. sec2x cot2x    csc2x cos2x sin2x sin2x 15. cosx 1sec x  cos x2  cos x sec x  cos2x  1 cos x  cos2x  1  cos2x  sin2x cos x 17. sin x1csc x  sin x2  1  sin2x  cos2x 19. tan x1csc x  cot x2  tan x csc x  tan x cot x  sin x cos x 1 sin x 1    1  sec x  1 cos x sin x cos x sin x cos x 2 2 2 2 21. tan x csc x  tan x  tan x 1csc2x  12  1; tan2x 1cot2x2  1 sin x 1cos x  12 sin x cos x  sin x sin x  tan x;  tan x  23. cos x 11  cos x2 cos x cos x  cos2x 11211  sin x2 1  sin x 1    sec x 25. cos x  cos x sin x 1cos x211  sin x2 cos x sinx 1tan x  12 sin x tan x  sin x sin x    27. tan x11  tan x2 tan x tan x  tan2x sin x sin x cos x   cos x sin x/cos x sin x 2 1sin x  cos x2 sin2x  2 sin x cos x  cos2x   29. cos x cos x 2 2 sin x  cos x  2 sin x cos x 1  2 sin x cos x   cos x cos x 1 2 sin x cos x   sec x  2 sin x cos x cos x 31. 11  sin x2 31  sin1x2 4  11  sin x211  sin x2  1  sin2x  cos2x 1csc x  cot x21csc x  cot x2 csc2x  cot2x 1    cot x 33. tan x tan x tan x 2 2 2 cos x sin x cos x  sin x 1     csc x 35. sin x 1 sin x sin x sin x cos x 1 tan x sin x tan x cos x  sin x csc x cos x sin x  1     37. csc x cos x csc x cos x 1 cot x cos x sin x 39.

sec x csc x sec2x  sin x csc x sec2x  1 tan2x      tan x sin x sec x sin x sec x 1 tan x sin x cos x

 21  sin2x B cot x sin x 21  sin2x 21 21 29 29 20 47. sin   , tan    , sec    , csc   , cot    29 20 20 21 21 8 15 17 17 8 49. cos    , sin    , sec    , csc    , cot   17 17 8 15 15 5 5 x , sin   , tan   , 51. cos   x 2x2  25 2x2  25 41.

sin x

43. 

1

2

1

45.

2x2  25 2x2  25 , csc   x 5 1120 2 130 7 13  , tan   , sec    , 53. cos    13 13 2 130 2 130 2 130 13 csc    , cot   7 7

sec  

4 12 7 4 12 , cos    , tan    , 9 9 7 9 7 57. Answers will vary. csc   , cot    4 12 4 12 59. Answers will vary. 61. Answers will vary. 4 18 m2 2  nx2  63. a. A  cot a b b. A  cot a b  64 m2 # 1  64 m2 4 n 4 4 c. A  119.62 in2 65. cos3x  1cos x21cos2x2  1cos x2 11  sin2x2 67. tan x  tan3x  1tan x211  tan2x2  1tan x21sec2x2 69. tan2x sec x  4 tan2x  1tan2x21sec x  42  1sec x  421tan2x2  1sec x  421sec2  12  1sec x  421sec x  12 1sec x  12 71. cos2x sin x  cos2x  1cos2x21sin x  12  11  sin2x21sin x  12  11  sin x211  sin x2 1sin x 12  11 sin x211 sin x2 112 11 sin x2  112 11  sin x2 11  sin x2 2   73. a. A  nr2tan a b b. A  4 # 42 tan a b  64 m2 n 4 1  m1m2 2 c. A  51.45 m 75. tan   77.   45° m2  m1 55. sin  

Exercises 6.1, pp. 548–551

SA31

79. 0, 83.

  2 3 , , , , 3 2 3 2

81. about 1148 ft

y 2

y = 2 sin(2t)

1

t 1

2

3 2

2

2

Exercises 6.2, pp. 555–558 1. substituted 3. complicated; simplify; build 5. Because we don’t know if the equation is true. 1  cos x 9. cos x 11. sin x sin2x 2 2 13. cos x tan x  cos2x cos2x  sin2x  1  cos2x sin x cos x 15. tan x  cot x   cos x sin x sin2x  cos2x  cos x sin x 1  cos x sin x 1 1  cos x sin x  sec x csc x 1 17. csc x  sin x   sin x sin x 1  sin2x  sin x cos2x  sin x cos x  sin x cos x cos x  tan x

7.

1  sin x cos x

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Student Answer Appendix

19. sec   tan    

1 sin   cos  cos  1  sin  cos  11  sin 211  sin 2

cos 11  sin 2 1  sin2  cos 11  sin 2 cos2  cos 11  sin 2 cos   1  sin 

21.

23.

25.

27.

29.

11  sin x211  sin x2 1  sin x  cos x cos x 11  sin x2 1  sin2x  cos x 11  sin x2 cos2x  cos x11  sin x2 cos x  1  sin x csc x cos x csc2x  cos2x   cos x csc x cos x csc x csc2x  11  sin2x2  1 cos x sin x csc2x  1  sin2x  cot x cot2x  sin2x  cot x sin x 11  sin x2  sin x 11  sin x2 sin x sin x   1  sin x 1  sin x 11  sin x211  sin x2 sin x  sin2x  sin x  sin2x  1  sin2x 2 2 sin x  cos2x  2 tan2x cot x 11  csc x2  cot x 11  csc x2 cot x cot x   1  csc x 1  csc x 11  csc x211  csc x2 cot x  cot x csc x  cot x  cot x csc x  1  csc2x 2 cot x csc x  cot2x 2 csc x  cot x 1 2 sin x  cos x sin x 2  cos x  2 sec x sec2x sec2x  1  cot2x csc2x 1 



cos2x 1 sin2x sin2x

cos2x  tan2x

31. sin2x1cot2x  csc2x2  sin2x cot2x  sin2x csc2x cos2x 1  sin2x 2  sin2x 2 sin x sin x  cos2x  1  sin2x cos x  sin x 33. cos x cot x  sin x  cos x sin x 2 cos x   sin x sin x 2 cos x  sin2x  sin x 1  sin x  csc x 1 1sin x21cos x2 sec x cos x  35. cos x sin x cot x  tan x a  b 1sin x21cos x2 sin x cos x sin x  cos2x  sin2x sin x  1  sin x sin x  csc x sin x csc x   37. csc x csc x csc x  sin2x  1  cos2x 1 sin x 1 39.  csc x  sin x 1csc x  sin x2 sin x sin x  1  sin2x sin x  cos2x sin x 1  cos x cos x  tan x sec x 11  sin x2 11  sin x2 1  sin x 41.  1  sin x 11  sin x2 11  sin x2  

1  2 sin x  sin2x 1  sin2x 1  2 sin x  sin2x

cos2x 1 sin2x sin x 1   2 2 cos x cos x cos x cos2x  sec2x  2 tan x sec x  tan2x  1sec x  tan x2 2

 1tan x  sec x2 2 1cos x  sin x2 1cos x  sin x2 cos x  sin x  43. 1  tan x 11  tan x2 1cos x  sin x2 1cos x  sin x2 1cos x  sin x2  sin2x cos x  sin x  sin x  cos x 1cos x  sin x21cos x  sin x2  sin2x cos x a1  b cos2x 1cos x  sin x21cos x  sin x2  cos x 11  tan2x2 1cos x  sin x21cos x  sin x2  cos x 11  tan x2 11  tan x2 

1cos x  sin x21cos x  sin x2

1cos x  sin x211  tan x2 cos x  sin x  1  tan x

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Student Answer Appendix 1tan x  cot x21tan x  cot x2 tan2x  cot2x  tan x  cot x 1tan x  cot x2  tan x  cot x sin x cos x   cos x sin x sin2x  cos2x  cos x sin x 1  cos x sin x 1 1  cos x sin x  sec x csc x  csc x sec x cos x 1cos x21sin x2 cot x sin x 47.  cot x  tan x cos x sin x 1cos x21sin x2  sin x cos x cos2x  cos2x  sin2x cos2x  1  1  sin2x 1sec2x  tan2x21sec2x  tan2x2 sec4x  tan4x 49.  2 2 sec x  tan x 1sec2x  tan2x2 2  sec x  tan2x 1 1cos2x  sin2x21cos2x  sin2x2 cos4x  sin4x 51.  2 cos x cos2x 2 2 1cos x  sin x2112  cos2x cos2x sin2x   cos2x cos2x  1  tan2x  1  1sec2x  12  1  sec2x  1  2  sec2x 53. 1sec x  tan x2 2  sec2x  2 sec x tan x  tan2x 1 2 sin x sin2x    2 2 cos x cos x cos2x 2 1  2 sin x  sin x  cos2x 11  sin x2 2  cos2x 1sin x  12 2  cos2x cos2x sec x  sin2x sec x  csc x sin x cos x cos x sin x csc x    55. sin x cos x sec x sin x cos x sec x sec x1cos2x  sin2x2  112cos x  sin x112 sec x  cos x  sin x 1sin2x  cos2x21sin2x  cos2x2 sin4x  cos4x  57. sin3x  cos3x 1sin x  cos x21sin2x  sin x cos x  cos2x2 1121sin x  cos x21sin x  cos x2  1sin x  cos x21sin2x  cos2x  sin x cos x2 sin x  cos x  1  sin x cos x 59. a. d2  120  x cos 2 2  120  x sin 2 2  400  40x cos   x2cos2  400  40 x sin   x2sin2  800  40x1cos   sin 2  x2 1cos2  sin22  800  40x1cos   sin 2  x2 b. 42.2 ft 45.

SA33

61. a. h  1cot x  tan x; h  3.76 sin x cos x b. cot x  tan x   sin x cos x cos2x  sin2x  sin x cos x 1  sin x cos x  csc x sec x; h  1csc x sec x h  3.76; yes 63. D2  400  40x cos   x2 D  40.5 ft 65. sin   cos  67. Answers will vary. 69. 1sin2x  cos2x2 2  112 2  1 17 3 3 71. sin t  , cos t  , tan t  4 4 17 73. y 6 5 4 3 2 1

4 321 1 1 1 2 3 4 5 6 x 2 3 4

Exercises 6.3, pp. 563–567 1. false; QII 3. repeat; opposite 12  16 12  16 7. 9. 4 4

5. Answers will vary.

11. a. cos145°  30°2  cos 45° cos 30°  sin 45° sin 30° 

26  22 4

b. cos1120°  45°2  cos 120° cos 45°  sin 120° sin 45°   22  26 26  22  4 4  13 16 13. cos152 15. 17. 19. sin 33° 21. cot a b 2 65 12  23. cos a  b 25. sin(8x) 27. tan 132 29. 1 31. 13 3 16  12 304 304 16  12 33. a. b. 35. 37. 425 297 4 4 1 13  39.  41.  13 3 13 16  12 43. a. sin145°  30°2  sin 45° cos 30°  cos 45° sin 30°  4 b. sin1135°  120°2  sin 135° cos 120°  cos 135° sin 120° 12 1  12 13 a b a b  a ba b 2 2 2 2  12 16   4 4 16  12  4 319  12  16 319 480 45. 47. a. b. c.  4 481 481 360 3416 3416 1767 49. a. b. c. 4505 4505 2937 12 13  5 12  513 12  5 13 51. a. b. c. 26 26 12 13  5 247 96 247 53. 190°  2    190°  2  180° a. b. c. 265 265 96 55. sin1  2  sin  cos   cos  sin   0  112sin   sin 

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Student Answer Appendix

   b  cos x cosa b  sin x sina b  4 4 4 12 12 12 cos xa b  sin xa b 1cos x  sin x2 2 2 2  tan x  tana b 4 tan x  1 1  tan x  59. tanax  b    4  1  tan x 1  tan x 1  tan x tana b 4 57. cosax 

61. cos1  2  cos1  2  cos  cos   sin  sin   cos  cos   sin  sin   2 cos  cos  63. cos12t2  cos1t  t2  cos t cos t  sin t sin t  cos2t  sin2t 65. sin13t2  sin12t  t2  sin12t2 cos t  cos12t2 sin t  2 sin t cos t cos t  1cos2t  sin2t2 sin t  2 sin t cos2t  sin t cos2t  sin3t  3 sin t cos2t  sin3t  3 sin t11  sin2t2  sin3t  3 sin t  3 sin3t  sin3t  4 sin3t  3 sin t    67. cosax  b  cos x cosa b  sin x sina b 4 4 4 12 12 b  sin xa b  cos xa 2 2 12  1cos x  sin x2 2 Wk 1  13 69. F  c 1  13 cos s cos t 71. R  C sin1s  t2 cos s cos t  C1sin s cos t  cos s sin t2 1 cos s cos t cos s cos t  1 C1sin s cos t  cos s sin t2 cos s cos t 1  sin s cos t cos s sin t b C a  cos s cos t cos s cos t 1  C1tan s  tan t2 sin  cos190°  2 A  73. B cos  sin190°  2 sin  1cos 90° cos   sin 90° sin 2 A  B cos  1sin 90° cos   cos 90° sin 2 sin 10  sin 2  cos  1cos   02 sin2  cos2  tan2 75. verified using sum identity for sine sin1x  h2  sin x f 1x  h2  f 1x2  77. h h sin x cos h  cos x sin h  sin x sin x cos h  sin x  cos x sin h   h h sin x1cos h  12  cos x sin h cos h  1 sin h   sin x  cos x h h h 12 1 79. 81. 2 2

83. D  d, so D2  d2, and D2  1cos   cos 2 2  1sin   sin 2 2  cos2  2 cos  cos   cos2  sin2  2 sin  sin   sin2  2  2 cos  cos   2 sin  sin  d2  sin2 1  2  3cos1  2  1 4 2  sin2 1  2  cos2 1  2  2 cos1  2  1  2  2 cos1  2 D2  d2 so 2  2 cos  cos   2 sin  sin   2  2 cos1  2 2 cos1  2 2 cos  cos   2 sin  sin   2 2 cos  cos   sin  sin   cos1  2 85. a. P  16,

b. P 

 2

87. about 19.3 ft

Exercises 6.4, pp. 576–580 1. sum;   

3. 2x; x

5. Answers will vary

120 120 119 , cos122  , tan122  7. sin122  169 169 119 720 720 1519 , cos122  , tan122  9. sin122  1681 1681 1519 2184 2184 6887 , cos122  , tan122  11. sin122  7225 7225 6887 5280 5280 721 , cos122  , tan122  13. sin122  5329 5329 721 24 24 7 , cos122  , tan122  15. sin122  25 25 7 4 4 3 17. sin   , cos   , tan   5 5 3 21 20 21 19. sin   , cos   , tan   29 29 20 21. sin132  sin12  2  sin122cos   cos122sin   12 sin  cos 2cos   11  2 sin22sin   2 sin  cos2  sin   2 sin3  2 sin 11  sin22  sin   2 sin3  2 sin   2 sin3  sin   2 sin3  3 sin   4 sin3 1 12 1 1 23. 25. 27. 1 29. 4.5 sin(6x) 31.  cos14x2 4 2 8 8 3 3 9 33.  cos12x2  cos14x2 8 2 8 5 7 3 1 35.  cos12x2  cos14x2  cos12x2cos14x2 8 8 8 8 12  12 12  12 , cos   , tan   12  1 37. sin   2 2 12  13 12  13 , cos   , tan   2  13 39. sin   2 2 12  12 12  12 , cos   , tan   12  1 41. sin   2 2 12  12 12  12 , cos   , tan   12  1 43. sin   2 2 12  12  13 12  12  12 45. 47. 49. cos 15° 2 2 51. tan 2 53. tan x  3  2  3 , cos a b  , tan a b  55. sin a b  2 2 2 2 113 113  3  1  , cos a b  , tan a b  3 57. sin a b  2 2 2 110 110 7  5  7  , cos a b  , tan a b  59. sin a b  2 2 2 5 174 174

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Student Answer Appendix 1  15  1  61. sin a b  , cos a b  , tan a b  2 2 2 15 1226 1226  5  2  5 63. sin a b  , cos a b  , tan a b   2 2 2 2 129 129 1 65. 3cos1122  cos142 4 67. cos12t2  cos15t2 2 1  13 1 69. cos11540t2  cos12418t2 71. 73. 2 4 55 27 x 75. 2 sin a kbcos a kb 77. 2 sin x sin a b 2 2 6 2061 357  12 tb cos a tb 81. 79. 2 cos a 2 2 2 sin12x2 2 sin x cos x 83.  cos12x2 cos2x  sin2x  tan12x2 85. 1sin x  cos x2 2  sin2x  2 sin x cos x  cos2x  sin2x  cos2x  2 sin x cos x  1  2 sin x cos x  1  sin12x2 87. cos182  cos12 # 42  cos2 142  sin2 142 cos 122 cos2  sin2  89. 2 sin  sin2 2 cos  sin2   2 sin  sin2 2  cot   1 2 tan  91. tan 122  1  tan2 1 12 tan 2 tan   1 11  tan22 tan  2  1  tan  tan  2  cot   tan  2 93. 2 csc 12x2  sin 12x2 2  2 sin x cos x 1  sin x cos x sin2x  cos2x  sin x cos x sin2x cos2x   sin x cos x sin x cos x sin x cos x   cos x sin x  tan x  cot x x x x 95. cos2 a b  sin2 a b  cos a2 # b 2 2 2  cos x 97. 1  4 sin2  4 sin4  11  2 sin22 2  3 cos 122 4 2  cos2 122  1  sin2 122 2 sin 1100t2cos 120t2 sin 1120t2  sin 180t2 99.  cos 1120t2  cos 180t2 2 sin 1100t2sin 120t2 cos 120t2  sin 120t2  cot 120t2

SA35

101. sin2  11  cos 2 2  sin2  1  2 cos   cos2  sin2  cos2  1  2 cos   1  1  2 cos   2  2 cos  1  cos    2  2 11  cos 2  4 a b  4 sin2 a b  c 2 sin a b d 2 2 2 103. sin 122  sin 1  2  sin  cos   cos  sin   sin  cos   sin  cos   2 sin  cos  tan 1  2  tan 1  2 tan   tan   1  tan  tan  2 tan   tan 122  1  tan2 1 105. 3 cos 1  2  cos 1  2 4  sin  sin  2 2 107. a. M  , M  3.9 22  13 2 b. M  , M  2.6 c.   60° 22  12 109. a. 288  144 12 ft  84.3 ft b. 288  144 12 ft  84.3 ft 111. cos 32112092t 4  cos 3 219412t 4; the * key t b` 60 1 t  ` 6 sin a # b ` 2 30

113. d 1t2  ` 6 sin a

† °  6 

1  cosa Q

2

t b 30 ¢ †

t 1  cosa b 30 6 R 2 t 1  cosa b 30  36 R 2 t  18 c 1  cos a b d A 30 115. a. sin 12  90°2  1  sin 122cos 90°  cos 122sin 90°  1  0  cos 122  1  1  cos 122 b. 2 sin2  sin2  sin2  1  cos2  sin2  1  1cos2  sin22  1  cos122 c. 1  sin2  cos2  1  1cos2  sin22  1  cos122 d. 1  cos122  1  cos122 117. a.  0.9659;  0.9659 12  13 2 16  12 2 b a b b. a 2 4 2  13 6  2 112  2  4 16 2  13 8  4 13  4 16 2  13 2  13  4 4 119. Must be a unit circle with  in radians. Must use a right triangle opposite side  sin   . definition of tangent: tangent a b  2 adjacent side 1  cos  121. x  1; x  2; x   16; x  16 3969 4225 63 2 256 16 2    1, 123. a b  a b  65 65 4224 4225 4225

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Student Answer Appendix 63 65 ; sec   16 16 65 2 63 2 1a b a b 16 16 3969 4225 1  256 256 3969 4225 256   256 256 256

tan  

Mid-Chapter Check, pp. 580–581

1. sin x 3 csc x  sin x 4  sin x csc x  sin2x 1  sin x  sin2x sin x  1  sin2x  cos2x cos2x 2. cos2x  cot2x  cos2x  sin2x 1  cos2xa1  b sin2x 2 2  cos x11  csc x2  cos2x1cot2x2  cos2x cot2x 2 sin x cos x 2 sin x csc x  cos x sec x   3. sec x csc x sec x csc x 2112  1  sec x csc x 1  sec x csc x  cos x sin x 4. 1  sec2x  tan2x 1  sec20  tan20 1  12  02 110 2  0 False 1sin x  cos x21sin2x  sin x cos x  cos2x2 sin3x  cos3x  5. a. sin x  cos x 1sin x  cos x2  1sin2x  cos2x  sin x cos x2  1  sin x cos x 1 1 1  sec x 1  cos x cos x 1  cos x    b. csc x cot x 1 cos x sin x sin x sin x sin x  asin x  ba  sin xb cos x cos x sin x sin x   sin x  sin x  cos x cos x 0 sec2x sec2x  tan2x tan2x 6. a.   2 2 sec x sec x sec2x 2 sin x 1

3193 456 c. 5785 5767 7  2413 24  7 13 7  24 13 , cos A  , tan A  8. sin A  50 50 24  7 13  4  1 9. sin a b  , cos a b  2 2 117 117 336 527 336 , cos 122  , tan 122  10. sin 122  625 625 527 7. a.

b.

Reinforcing Basic Concepts, pp. 581–582 1. sin2x  cos2x  1 sin2x cos2x 1   2 sin x sin2x sin2x 1  cot2x  csc2x ✓ sin2x  cos2x  1 sin2x cos2x 1   2 cos x cos2x cos2x tan2x  1  sec2x ✓ 2. cos 1  2  cos 1  2  cos  cos   sin  sin   cos2  sin2  cos2  11  cos22  2 cos2  1  cos2  sin2  11  sin22  sin2  1  2 sin2

Exercises 6.5, pp. 593–598   3. 31, 1 4 ; c  , d 5. cos1 1 15 2 2 2 1     7. 0; ;  ;  9. 11. 13. 1.0956, 62.8° 2 6 2 4 2 12  15. 0.3876, 22.2° 17. 19. 21. 45° 23. 0.8205 2 3 13  25. 0; ; 120°;  27. 29.  31. 1.4352; 82.2° 2 3  12 3 33. 0.7297; 41.8° 35. 37. 0.5560 39.  41. 4 2 4    43. 0;  13; 30°; 13; 45.  47. 49. 1.1170, 64.0° 3 6 3  13 51. 0.9441, 54.1° 53.  55. 57. 12 59. 30° 6 3  61. cannot evaluate tan a b 2  63. csc  12 7 1, not in domain of sin1x. 4 3 4 3 65. sin   , cos   , tan   5 5 4 2x2  36 6 2x2  36 67. sin   , cos   , tan   x x 6 1. horizontal; line; one; one

69.

cos2x 1

cos2x  1  sin2x  cos2x cot x  tan x cot x tan x b.   csc x sec x csc x sec x csc x sec x cos x sin x sin x cos x   1 # 1 1 # 1 sin x cos x sin x cos x cos2x sin x sin2x cos x   sin x cos x  cos2x  sin2x

456 5785

24 25

y

71.

25 24 ␪ 7

73.

15 3

5 √25  9x2

3

√5

␪ 2

x

225  9x2 3x

y

3x

75.

12 B 12  x2

x

√12  x2

x

√12

  81. 83. 80.1° 85. 67.8° 6 6 87. a. FN  2.13 N; FN  1.56 N b.   63° for FN  1 N,   24.9° for FN  2 N 89. 30° 91.   72.3°; straight line distance; 157.5 yd 77. 0; 2; 30°; 1; 

79.

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Student Answer Appendix 4 1.5 93. a.   tan1a b,   tan1a b x x 4 1.5 b.   tan1a b  tan1a b c.   27.0° at x  2.5 ft x x 70 1 94 95. a.   tan a b  tan1a b b.   8.4° at d  81.1 ft x x 97. a.   15.5°;   0.2705 rad b. 29 mi 99. a. 413.6 ft away 84 b. 503 ft c. 651.2 ft 101. sin122  85 103. x  1q, 3 4 ´ 3 0, 34

Exercises 6.6, pp. 606–610   3  3 ; ; ;  2k;  2k 4 4 4 4 4 7. a. QIV b. 2 roots 9. a. QIV b. 2 roots

1. principal; 3 0, 22 ; real 5. Answers will vary. 11.



sin 

cos 

tan 

0

0

1

0

 6

1 2

13 2

13 3

 3

13 2

1 2

13

 2

1

0

und.

2 3

13 2



5 6

1 2



0

7 6 4 3 13. 29. 37. 43. 45. 47. 49. 53. 57. 63. 67. 69. 71. 73. 75. 79.

3.

 

1 2

13 2



1 2

13 2

1 

 13 

13 3 0

13 2

13 3

1 2

13



    5   15.  17. 19.  21.  23. 25. 27. 6 3 3 6 6 4 4 2 5  5  5 7 11 3 7 31. 33. 35. , , , , , , 6 6 3 3 6 6 6 6 4 4  2 4 5  3 5 7  3 , , , , , 39. , 41. , 3 3 3 3 4 4 4 4 2 2   1.2310  2k or 5.0522  2k 5   x   k or  2k or  2k 2 6 6 4 2 x  2k or  2k or 1.4455  2k or 4.8377  2k 3 3 7  5 5 x   k or  k 51. x   2k or  2k 6 6 4 4 3 3 5 x  2k or  2k 55. x   k 4 4 4 3  2  x   k or  k 59. x   k 61. x  3  6k 3 3 8 2  5    x   k 65. x   k or  k or  k 2 6 3 12 12 a. x  1.2310 b. x  1.2310  2k, 5.0522  2k a. x  1.2094 b. x  1.2094  2k, 5.0738  2k a.   0.3649 b.   0.3649  k, 1.2059  k a.   0.8861 b.   0.8861  k, 2.2555  k  5 2 4 4 10 x   k or  k 77. x   k or  k 6 6 9 3 9 3    k 81.   0.3398  2k or 2.8018  2k 83. x  0.7290 2

85. x  2.6649 87. x  0.4566 89. 22.1° and 67.9° 91. 0°; the ramp is horizontal. 93. 30.7°; smaller 95.   35°,   25.5° 97. k  1.36,   20.6° 99. a. 7 in.   k, explanations b. 1.05 in. and 5.24 in. 101. 1.1547 103. 2 will vary. 1 12 105. f 12  i2  12  i2 2  4 12  i2  5 107. a.  b.  2 13 2  4  4i  i  8  4i  5  4  4i  1  8  4i  5  0

Exercises 6.7, pp. 615–618 1. sin2x  cos2x  1; 1  tan2x  sec2x; 1  cot2x  csc2x  5 , 3. factor; grouping 5. Answers will vary. 7. 9. 0 12 12  5  5  3 5 7 , , , , , 0.8411, 11. 0.4456, 1.1252 13. , 15. , 4 4 6 6 4 4 4 4 5.4421 2  5  3 5 7 , , , , 0.7297, 2.4119 21. 3 19. , 4 4 4 4 6 6  2 5 2  3 5 7  k,  k; k  0, 1, 2 25. , , , , 23. 9 3 9 3 4 4 4 4 27. P  12; x  3; x  11 29. P  24; x  0.4909, x  5.5091  17 31. 12 , 12 33. 0.3747, 5.9085, 2.7669, 3.5163  3 3 7  5 13 17 a is extraneousb 37. , , , , 35. 39. 2 2 4 4 12 12 12 12  112.5  x cosa b 4 5 5  41. I. a. a , b b. D  112.5,   , y  2 2 4  sina b 4 c. verified 215  x cos 1.1071 II. a. (2, 4) b. D  2 15,   1.1071, y  sin 1.1071 c. verified  2  x cosa b 3  III. a. (1, 13) b. D  2,   , y  c. verified 3  sina b 3 17.

43. a. 2500 ft3  7853.98 ft3 b. 7824.09 ft3 c.   78.5° 45. a. 78.53 m3/sec b. during the months of August, September, October, and November 47. a. $3554.52 b. during the months of May, June, July, and August 49. a. 12.67 in. b. during the months of April, May, June, July, and August 51. a. 8.39 gal b. approx. day 214 to day 333 53. a. 68 bpm b. 176.2 bpm c. from about 4.6 min to 7.4 min 2  xb  29 55. a. y  19 cosa  xb  53 b. y  21 sina 6 365 57. a. L  25.5 cm. b.   38.9° or 33.4°, depending on what side you consider the base. y 59. 11, 02, (0, 0), (2, 0) (multiplicity 2): up/up; 3

3

3 3

61.   4.56°

Summary and Concept Review, pp. 619–623 1. sin x 1csc x  sin x2  sin x csc x  sin x sin x 1  sin x  sin2x sin x  1  sin2x  cos2x

x

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2.

Student Answer Appendix

tan2x csc x  csc x 2

sec x

 

csc x 1tan2x  12 2

sec x csc x sec2x

sec2x  csc x 1sec x  tan x21sec x  tan x2 sec2x  sec x tan x  sec x tan x  tan2x  3. csc x csc x sec2x  tan2x  csc x 1  tan2x  tan2x  csc x 1  csc x  sin x sec2x sec2x  sin x csc x  sin x  4. csc x csc x sec2x  1  csc x tan2x  csc x 35 37 12 35 37 , csc   , cot   , tan   , sec   5. sin   37 35 35 12 12 4 16 25 23 4 16 , csc   , cot   , tan    , 6. sin   25 23 4 16 4 16 23 cos   25 1  cos x 7. ; answers will vary. 8. sec x  tan x; answers will vary. sin x 2 csc x 11  cos2x2 csc2x sin2x  9. 2 tan x tan2x 1  tan2x  cot2x csc x cot x 1   cot x 10.  cot x csc x sec x tan x sec x  cot x cos x  cot x csc x  cot x1cos x  csc x2 1sin2x  cos2x21sin2x  cos2x2 sin4x  cos4x  11. sin x cos x sin x cos x 1sin2x  cos2x2112  sin x cos x sin x sin x cos x cos x   sin x cos x sin x cos x sin x cos x   cos x sin x  tan x  cot x 1sin x  cos x2 2 sin2x  2 sin x cos x  cos2x  12. sin x cos x sin x cos x sin2x  cos2x 2 sin x cos x   sin x cos x sin x cos x 1  2 sin x cos x  csc x sec x  2 16  12 13. a. cos 75°  4 1 13  12 2  13  1   2  13 b. tan a b  12 2 1  13 1 13  12 2 13  1   2  13 14. a. tan 15°  2 1  13 13  12  16 b b. sin a 15. a. cos 180°  1 b. sin 120°  12 4 2 5x 16. a. cos x b. sina b 17. a. cos 1170°  cos 90°  0 8  12 57 b  sin a b  b. sin a 4 4 2

x  x 7   xb 18. a. cos a b  sin a  b b. sin a x  b  cos a 8 2 8 12 12 tan 45°  tan 30° 19. tan145°  30°2  1  tan 45° tan 30° 13 3  13 13 1 1 3 3 3    13 13 3  13 11# 1 3 3 3 131 13  12 3 3  13 3  13 13  1     3 3  13 3  13 131 13  12 13  1 tan 135°  tan 120° tan 1135°  120°2  1  tan 135° tan 120° 1  13 13  1 13  1    1  112 1 132 1  13 13  1   20. cos ax  b  cos ax  b  13 cos x 6 6      cos x cos a b  sin x sin a b  cos x cos a b  sin x sin a b 6 6 6 6 13  b  13 cos x  2 cos x cos a b  0  2 cos x a 6 2 84 13 2184 21. a. sin122  2 a ba b  85 85 7225 13 2 84 2 6887 cos122  a b  a b  85 85 7225 7225 2184 2184 a b tan122  7225 6887 6887 20 21 840 ba b b. sin122  2 a 29 29 841 21 2 20 2 441  400 41 cos122  a b a b   29 29 841 841 20 2a b 21 840 tan122   41 20 2 1a b 21 21 20 21 , tan    , 22. a. sin   , cos   29 29 20 7 24 24 7 7 , tan   b. sin   or sin   , cos   or cos   25 25 25 25 24 24 or tan   7 12  13 23. a. cos 45°  b. cosa b  2 6 2 12 1 2 2  12 1  cos 135°   24. a. sin 67.5  A 2 R 2 B 4 22  12  2 12 1 1  cos 135° 2 2  12 cos 67.5    A 2 R 2 B 4 22  12  2 5 12 1  cosa b 1 4 5 2 2  12   b. sin a b  8 R 2 R 2 B 4 22  12  2 5 12 1  cosa b 1 4 2 2  12 5   cos a b   8 R 2 R 2 B 4 22  12  2

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Student Answer Appendix  1  24/25 25  24 1  25. a. sin a b     , in QII 2 A 2 A 50 5 12 2 1  24/25  25  24 cos a b    2 A 2 A 50 49 7    , in QII A 50 5 12 2  1  56/65 65  56 b. sin a b    2 A 2 A 130 3  9  , in QIV  A 130 1130 2  1  56/65 65  56 121 11  cosa b     , in QIV 2 A 2 A 130 A 130 1130 2 2 sin122 sin  cos132  cos  26.  cos132  cos  2 cos122 cos  2 sin2 2 sin2   2 2 2 cos   sin  sin   cos2 2 tan2 2 sin2   1  2 cos2 sec2  2 27. cos13x2  cos x  0 S 2 cos12x2cos x  0   cos12x2  0: x   k; k   4 2  cos x  0: x   k; k   2 30° 30° 1  cos 30° 1  cos 30° 2 28. a. A  12 sin a b cos a b  144 2 2 A 2 A 2 13 13 1 1 2 2 2  13 2  13  144  144 R 2 2 4 4 R B B 14414  3  36 cm2; yes  4   2 b. x sina b cosa b 2 2  1 1 Let u  , then  x 2 sin u cos u  x 2 12sin u cos u2  x 2 sin 12u2 2 2 2 1 1 1  x2 sin; A  1122 2 sin130°2  72a b  36 cm2; yes 2 2 2 5   29. or 45° 30. or 30° 31. or 150° 32. 1.3431 or 77.0° 4 6 6  1 36. 33. 1.0956 or 62.8° 34. 0.5054 or 29.0° 35. 2 4 3 37. undefined 38. 1.0245 39. 60° 40. 4 41.

42.

y

37

35

7

␪ 12

␪ x

√49  9x2 3x

x 44.   cos1a b 5

y

1

√8 ␪

2

x

cot 

45.   sec1a

x b 7 13

  3  3 b. , c. x   2k or  2k, k   4 4 4 4 4 2 2 4 2 4 48. a. b. c. ,  2k or  2k, k   3 3 3 3 3 2  2 5 49. a.  b. c. ,  k, k   3 3 3 3 50. a. 1.1102 b. 1.1102, 5.1729 c. 1.1102  2k or 5.1729  2k, k   51. a. 0.3376 b. 0.3376, 1.2332, 3.4792, 4.3748 c. 0.3376  k or 1.2332  k, k   52. a. 0.3614 b. 0.3614, 2.7802 c. 0.3614  2k or 2.7802  2k, k    5  5 53.   1.1547 54. x  , 55. x  0.7297, 2.4119; x  , 12 12 6 6  5 11  , 56. x  , 57. x  58. P  12; x  2.6931, x  9.3069 6 6 6 2 9 59. P  6; x  0, x  60. a.  43110002  $43,000 2 b. April through August 47. a.

Mixed Review, pp. 623–624 6 2 9 6  1117 , tan    , cos   , sec   , 9 9 3 1117 1117 3 1117 , cot   csc   3. 13  2 6 2 x y tan   5. 2100  x2 1. sin  

10

x

␪ √100  x2

x

7. x  0.4103; x  4.9230 1 2 b a bv2 sin  cos  32 1 1 a bv2 2 sin  cos  32 1 a bv2 sin 2 32 b. sin122  sin 3 2190  2 4  sin 1180°  22  sin122   13 11. cos c 2a b d  cosa b  12 6 2 1cos t  sin t2 2 cos2t  2 cos t sin t  sin2t  13. tan t tan t 1  2 cos t sin t  tan t 1 cos t   2 cos t sin t tan t sin t  cot t  2 cos2t 3  x 15. or 135° 17. or 60° 19.   tan1a b 4 3 10 21. a. 6 ft: 2 A.M., 2 P.M., 10 A.M., 10 P.M. 10 ft: 6 A.M., 6 P.M. 13 22  12  22  12 b. about 8.9 ft 23. a. b. 2 2 12  13 1  12 25. a. b. 4 4 9. a. a

Practice Test, pp. 625–626

x 9

x

249  9x2 tan  3x

35 sin  37 43.

y

SA39

x

9 x

x  46.   sin1a b  4 6

1.

1csc x  cot x2 1csc x  cot x2 sec x

  

csc2x  csc x cot x  csc x cot x  cot2x sec x csc2x  cot2x scc x 11  cot2x2  cot2x

1 sec x  cos x 

sec x

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2.

3. 4. 7.

8.

Student Answer Appendix

1sin x  cos x21sin2x  sin x cos x  cos2x2 sin3x  cos3x  1  cos x sin x 1  cos x sin x 1sin x  cos x211  sin x cos x2  1  cos x sin x  sin x  cos x 55 73 48 55 73 sin   , sec   , cot   , tan   , csc   73 48 55 48 55 13  1 12  12 5. 6. 2 2 13  1   sin ax  b  sin ax  b 4 4      sin x cosa b  cos x sina b  sin x cosa b  cos x sina b 4 4 4 4    sin a b cos x  sin a b cos x 4 4   2 sina bcos x 4 12 2 cos x 2  12 cos x 15 8 15  13 1 6 sin   , cos   , tan   9. 10. ; 17 17 8 2 137 137

16  12 16  12  0.2588;  0.9659 4 4 1 13. a. y  30° b. f1x2  c. y  30° 2 157 rad 14. a. y  0.8523 rad or y  48.8° b. y  78.5° or 360 7 rad or 52.5° c. y  24 y 33 x y cos   cot   15. 16. 65 5 5

56

√2

65

12.

x2

11. 20 22  12

␪

5

Strengthening Core Skills p. 627 Exercise 1: Exercise 2: Exercise 3: Exercise 4:

x x x x

10.6025, 2.53912 3 0, 0.79454 ´ 34.4415, 2 4 30, 2.6154 4 ´ 39.3847, 12 4 167.3927, 202.60732

Cumulative Review Chapters 1–6, pp. 628–629 85 13 85 1. sin   84 85 , csc   84 , cos   85 , sec   13 , 84 13 tan   13 , cot   84 3. g12  132  12  132 2  412  132  1  4  4 13  3  8  4 13  1 0 5. about 474 ft 7. y 5 4 3 2 1

54321 1 2 3 4 5

1 2 3 4 5 x

9 11 1 , d 13. a. y   x  31 b. every 2 2 2 2 years, the amount of emissions decreases by 1 million tons. c. 23.5 million tons; 11 million tons 15. x  11, 52 17. $7 cos x1sec x  12 99 cos x 19. 21.  sec x  1 1sec x  121sec x  12 101 1  cos x  sec2x  1 1  cos x  tan2x  2 23. a. y  5.4 sin a x  b  27.1 b. from early May until late 6 3 August 25. a. volume of a cylinder b. volume of a rectangular solid c. circumference of a circle d. area of a triangle 9. 50.89 km/hr

11. x  c

␪ 33

x

x

x

 12 3 3 5 b , b. x  2 4 4 4 3 5   11  2k or  2k, k   II. a. c. x  b. x  , 4 4 6 6 6  11  2k, k   18. I. a. x  0.1922 c. x   2k or 6 6 b. x  0.1922, 1.3786, 3.3338, 4.5202 c. x  0.1922  k or 1.3786  k, k   II. a. x  0.9204 b. x  0.9204, 2.2212, 4.0620, 5.3628 c. x  0.9204  k or 2.2212  k, k   19. a. x  1.6875, 0.3413, 1.1321, 2.8967 b. x  0.9671, 2.6110, 7 11 7 11 19 23 , , , , 3.4538 20. a. x  0, , b. x  6 6 12 12 12 12  3 5 11 ; x  3.3090, 6.1157 22. x  , 21. x  , 2 2 6 6 23. a. 6 or $6,000 b. January through July 24. a. y  35.223 sin10.576x  2.5892  6.120 b. 25. cos12418t2  cos11540t2 Month Low (Jan S 1) Temp. (F ) 17. I. a. cos1a

1

26

3

21

5

16

7

41

9

25

11

14

Connections to Calculus Exercises, p. 632 1. a.  csc ,

b. y  csc c tan1a

3. 4 tan  sec 

5. sin 

 3 5 7 , , 9. x  , 4 4 4 4

x b d c. verified 13 2 4 , , 7. 0, 3 3

CHAPTER 7 Exercises 7.1, pp. 641–646 1. ambiguous 3. I; II 5. Answers will vary. 7. a  8.98 9. C  49.2° 11. C  21.4° 13. C  78°, b  109.5 cm, c  119.2 cm 15. C  90°, a  10 in., c  20 in. 17. 19. C  90°, a  15 mi, b  15 mi 33

19 in. 102

21. A  57°, b  49.5 km, c  17.1 km 23.

112

25. a. 10 cm

b. 0

c. 2

d. 1

0.8 cm 56

27. not possible 29. B  60°, C  90°, b  12.913 mi 31. A  39°, B  82°, a  42.6 mi or A  23°, B  98°, a  26.4 mi 33. A  39°, B  82°, a  42.6 ft or A  23°, B  98°, a  26.4 ft 35. not possible 37. A  80.0°, B  38.0°, b  1.8  1025 mi 39. A1  19.3°, A2  160.7°, 48°  160.7° 7 180°; no second solution possible 41. C1  71.3°, C2  108.7°, 57°  108.7° 6 180°; two

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Student Answer Appendix 12 2 49. a. No b.  3.9 mi

solutions possible 47. 51. 53. 55.

57.

43. not possible, sin A 7 1

45.

34.6 million miles or 119.7 million miles V 4 S  41.7 km, V 4 P  80.8 km a. No b. about 201.5 ft c. 15 sec Two triangles Angles Angles Sides A2  138.9° A1  41.1° a  12 cm B  26° B  26° b  8 cm C2  15.1° C1  112.9° c1  16.8 cm Angles A1  47.0° B1  109.0° C  24°

Angles A2  133.0° B2  23.0° C  24°

Sides a  9 cm b1  11.6 cm c  5 cm

Sides a  12 cm b  8 cm c2  4.8 cm Sides a  9 cm b2  4.8 cm c  5 cm

59. a  33.7 ft, c  22.3 ft 61. Rhymes to Tarryson: 61.7 km, Sexton to Tarryson: 52.6 km 63.  3.2 mi 65. h  161.9 yd 67. angle  90°; sides  9.8 cm, 11 cm; diameter  11 cm; it is a right triangle. 69. a. about 3187 m b. about 2613 m c. about 2368 m 71. sin 60° sin 90° 60 ; 12  13  20.4 cm sin 30° sin 45° 10.2 cm

A  139.7°, B  23.7°, C  16.6° C  86.3° 37. about 1688 mi 39. P  27.7°; heading 297.7° It cannot be constructed (available length  10,703.6 ft) 1678.2 mi 45. P  22.4 cm, A  135°, B  23.2°, C  21.8° A  20.6°, B  15.3°, C  144.1° 49. 58.78 cm a  13 A  133.2° 53. 33.7º; 150 ft2 b5 B  16.3° c  182 C  30.5° 55. a. 0.65  65% b. $1,950,000 57. about 483,529 km2 59. 387  502  889 6 902 61. (1) a2  b2  c2  2bc cos A (2) b2  a2  c2  2ac cos B, use substitution for a2 and (2) becomes b2  1b2  c2  2bc cos A2  c2  2ac cos B. Then 0  2c2  2bc cos A  2ac cos B, 2bc cos A  2ac cos B  2c2, b cos A  a cos B  c 63. 2 13 12 13 5 65. sin x  , csc x  , cos x  , sec x  , 13 5 13 12 5 12 tan x  , cot x  12 5 33. 35. 41. 43. 47. 51.

Exercises 7.3, pp. 669–673 1. scalar 3. directed; line 7. 9. 12 knots

5. Answers will vary. 11.

250 N

V1

30

M

9 knots

10.2(√3) cm

73. A  19°, B  31°, C  130°, a  45 cm, b  71.2 cm, c  105.8 cm

25 J 210 N

V2

6 knots

V3

V4

75. 12,564 mph 77. tan2x  sin2x   

sin2x cos2x sin2x 2

 sin2x 

13.

sin2x cos2x (3, 2)

2

cos x cos x sin2x  sin2x cos2x

321 1 2 3 4 5

cos2x sin x11  cos2x2



y

15.

(4, 5) H7, 3I

cos2x 2 sin x sin2x

H6, 5I

2 ( 3)  5

4 8

1 2 3 4 5 6 7 x

8

8

H8, 3I QI

4

4

t

8

4

y

5

(1, 9) v r 55

u

mi 10 25 2.3 

 4.1

10

5 x

u

(5, 3) 5 x

5

d. H8, 9I y

6

(0.5, 15) 5

1.5v 10

r

2v

x

r 10

2u 5

B

(8, 9)

5 x

33. a. H8, 4I

b. H6, 8I y

y

5

(8, 4) 2

r

u

x

5

v 2

u 5

C

v r

u

C

27. A  137.9°, B  15.6°, C  26.5° 29. A  119.3°, B  41.5°, C  19.2° 31. A

43.5

y

5

y

mi

x

b. 173 c. 20.6° b. 129 c. 68.2° 25. H10.9, 5.1I 27. H106, 92.2I 29. H9.7, 2.6I 31. a. H1, 9I b. H5, 3I

20

25

4

4

c. H0.5, 15I

260.9 mm

i 33.2

2 2

H2, 5I

538 mm

2.9  10 25 m

2

8

5

103.3

QIII

2

4

1. cosines 3. Pythagorean 5. B  33.1°, C  129.9°, a  19.8 m; law of sines 7. yes 9. no 11. yes 13. verified 15. B  41.4° 17. a  7.24 19. A  41.6° 21. A  120.4°, B  21.6°, c  53.5 cm 23. A  23.8°, C  126.2°, b  16 mi 25. B

29 465 mm

y

4

4

Exercises 7.2, pp. 653–658

91.2

8 x (5, 3)

4 4

8

cos x sin2x  sin2x cos2x 2  sin x tan2x 5 2 79. a. y  x  b. 1106 units 9 9

A

(1, 2)

4

4  (3)  7 523

2

59.8

y 8 1 5  6

17. Terminal point: 15, 12 , magnitude: 153 19. Terminal point: 11, 12 , magnitude: 134 y 21. a. 23. a.

2



5 4 3 2 1

5

x

5

r

v (6, 8)

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Student Answer Appendix

c. H15.5, 5I y

5

d. H5, 14I y

(15.5, 5)

3

r

r 20 x

10

63.

x

u

1.5v

2v

10

2u

67.

(5, 14)

5

35. a. H3, 6I

71.

b. H5, 2I

y

75.

y 5

(3, 6) 5

v

r

5

u 5 3

r

d. H6, 6I

y

(6.5, 10)

3 x

(5, 2)

5 x

c. H6.5, 10I

8 3 , i  H5.46, 2.05I 79. 14.4 81. 24.3° 173 173 hor. comp.  79.9 ft/sec; vert. comp.  60.2 ft/sec 83. 85. heading 68.2° at 266.7 mph 87. 182.10 cm, 22.00 cm2 89. 1Ha, bI  H1a, 1bI  Ha, bI 91. Ha, bI  Hc, dI  Ha  c, b  dI  Ha  1c2, b  1d2I  Ha, bI  Hc, dI  Ha, bI  1Hc, dI  u  11v2 93. 1ck2u  Hcka, ckbI  cHka, kbI  c1ku2 c1ku2  Hcka, ckbI  Hkca, kcbI  kHca, cbI  k1cu2 95. u  1u2  Ha, bI  Ha, bI  Ha  a, b  bI  H0, 0I 97. 1c  k2u  1c  k2Ha, bI  H1c  k2a, 1c  k2bI  Hca  ka, cb  kbI  Hca, cbI  Hka, kbI  cu  ku 99. H1, 3I  H3, 3I  H4, 1I  H2, 4I  H4, 3I  H6, 2I  H0, 0I 101. Answers will vary, one possibility: 0°, 81.4°, 34° 103. a. not a real number b. not possible c. not a real number x  0, 17; see graph 105. y 77. 5.83 h

u

v

5

y

10 3

1.5v

u 5

5

r

x

3

v r 2u

5

5

37. True 39. False 43. u  v  H8, 6I u  v  H6, 2I

41. True

y

5

(6, 6)

x

45. u  v  H9, 6I u  v  H7, 0I y

10 8 6 4 2

5

uv

uv 5

uv

5

x

5

5 5

47. u  v  H3, 6I u  v  H7, 0I

49. u  8i  15j u  17 16

u x

5

uv 5

2

8 x 4

51. p  3.2i  5.7j p  6.54

53. a.

4

y

r  16

y 10

8 x

v 5

5

5 x

b. v  H11.5, 3.3I c. v  11.5i, 3.3j

p 5

55. a.

10

5

b. w  H2.5, 9.2I c. w  2.5i  9.2j

y

w

57. a. b. c. d. 59. a. b. c. d. 61. a. b.

Reinforcing Basic Concepts, pp. 673–674 1.

Angles A  35° B  81.5° C  63.5°

Sides a  11.6 cm a  20 cm c  18 cm

2. For A  35°, a  10.3 For A  50°, a  14.2 For A  70°, a  19.1; yes, very close

Very close.

r  74.5 5

a2  c2  b2 b sin A 2. cos B  2ac a 3. a  129 m, B  86.5°, C  62.5° 4. A  42.3°, B  81.5°, C  56.2° 5. A  44° a  2.1 km 6. A  18.5° a  70 yd B  68.1° b  2.8 km B  134.5° b  157.1 yd C  67.9° c  2.8 km C  27° c  100 yd or A  44° a  2.1 km B  23.9° b  1.2 km C  112.1° c  2.8 km 7. about 60.7 ft 8. 169 m 9.   49.6°;   92.2°;   38.2° 10. 9.4 mi 1. sin B 

5

uv

1 2 3 4 5 x

Mid-Chapter Check, p. 673

y

y

4

54321 2 4 6 8 10

x

uv

5

5

c. r  18i  8j; r  19.7,   24.0° d. s  20i  4j; s  20.4,   11.3° 20 21 7 24 h , i , verified 65. h , i , verified 25 25 29 29 20 21 24 7 i  j, verified 69. i  j, verified 29 29 25 25 13 3 11 6 h , i , verified 73. i j, verified 1178 1178 1157 1157 5 2 4.48 h , i  H4.16, 1.66I 129 129

5 x

p  2i  2j; p  2 12,   135° q  6i  8j; q  10,   306.9° r  2i  1.5j; r  2.5,   143.1° s  10i  13j; s  16.4,   307.6° p  2 12i  2j; p  3.5,   35.3° q  8 12i  12j; q  16.5,   46.7° r  5.5 12i  6.5j; r  10.1,   39.9° s  11 12i  17j; s  23.0,   47.5° p  8i  4j; p  8.9,   26.6° q  16i  4j; q  16.5,   14.0°

Exercises 7.4, pp. 683–687 1. equilibrium; zero 3. orthogonal 5. Answers will vary. 7. H6, 8I 9. H5, 10I 11. 6i  8j 13. 2.2i  0.4j 15. H11.48, 9.16I 17. H24, 27I 19. F3  3336.8;   268.5° 21. 37.16 kg 23. 644.49 lb 25. 2606.74 kg 27. approx. 286.79 lb 29. approx. 43.8° 31. 1125 N-m 33. approx. 957.0 ft 35. approx. 64,951.90 ft-lb 37. approx. 451.72 lb 39. approx. 2819.08 N-m 41. 800 ft-lb 43. 118 ft-lb 45. verified 47. verified 49. a. 29 b. 45° 51. a. 0 b. 90° 53. a. 1 b. 89.4° 55. yes 57. no 59. yes 61. 3.68 63. 4 65. 3.17 67. a. H3.73, 1.40I b. u1  H3.73, 1.40I, u2  H1.73, 4.60I 69. a. H0.65, 0.11I b. u1  H0.65, 0.11I, u2  H1.35, 8.11I 71. a. 10.54i  1.76j

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Student Answer Appendix b. u1  10.54i  1.76j, u2  0.54i  3.24j 73. a. projectile is about 375 ft away, and 505.52 ft high b. approx. 1.27 sec and 12.26 sec 75. a. projectile is about 424.26 ft away, and 280.26 ft high b. approx. 2.44 sec and 6.40 sec 77. about 74.84 ft; t  3.9  1.2  2.7 sec 79. w # 1u  v2  He, f I # Ha  c, b  dI  e1a  c2  f 1b  d 2  ea  ec  fb  fd  1ea  fb2  1ec  fd2  He, f I # Ha, bI  He, f I # Hc, dI w#uw#v 81. 0 # u  H0, 0I # Ha, bI  01a2  01b2  0 u # 0  Ha, bI # H0, 0I  a102  b102  0 83.   56.9°; answers will vary. 85. x  20 87. a  138.4, B  106.8° C  41.2°; P  560.4 m, A  11,394.3 m2

Exercises 7.5, pp. 695–698

4 2

41. z1z2 43. z1z2 45. z1z2

49. z1z2

51. verified; verified, u2  v2  w2  uv  uw  vw 11  4 13i2  197  2013i2  139  60 13i2

yi

z2

6

z3 z1

 117  1213i2  13  16 13i2  145  56 13i2,

x z3

59  84 13i  59  84 13i

7 8 x

z1 z2

13. 101cos 210°  i sin 210°2

11. 2121cos 225°  i sin 225°2

3 3 11 11 15. 6 c cosa b  i sina b d 17. 8 c cosa b  i sina bd 4 4 6 6 6 19. 10 cis c tan1a b d ; 10 cis 36.9° 8 12 21. 13 cis c 180°  tan1a b d ; 13 cis 247.4° 5 1 17.5 23. 18.5 cis c tan a b d ; 18.5 cis 1.2405 6 5 25. 2134 cis c   tan1a b d ; 2 134 cis 2.1112 3 yi  4 27. r  2,   3 4 2  (√2, √2) z  2 cisa b 1 2 4

4  12  12i 2 1 1 2 x

53. a. V1t2  170 sin1120t2 b. t V(t)

55. 59. 61.

1

 3  z  4 13 cisa b 3  2 13  6i

63.

yi

29. r  413,  

(2√3, 6)

65.

5

4√3

3

67.

5

5 x 2

31. r  17,   tan1a

15 b 8 15 z  17 cis c tan1a b d 8 8 15  17a  ib  8  15i 17 17

33. r  6,     tan1a

z1 4 4 13   i z2 3 3 z1 13 1  2113  21i,   i z2 7 7 z1  10.84  12.04i,  1.55  4.76i z2 z1 5 13 5  0  40i,   i z2 4 4 z1 5  10  10 13i,   0i z2 2 z1  2.93  8.5i,  2.29  3.28i z2

39. z1z2  24  0i,

47. z1z2

1. modulus; argument 3. multiply; add 5. 21cos 240°  i sin 240°2, z is in QIII 7. z2  z1  z3 9. z2  z1  z3 yi

35. r1  212, r2  312, 1  135°, 2  45°; z  z1z2  12  0i 1 r  12,   180°; r1r2  21213 122  12✓ 1  2  135°  45°  180°✓ 37. r1  2, r2  2, 1  30°, 2  60°; r1 z1 2 13 1 z   i 1 r  1,   30°;   1✓ r2 2 z2 2 2 1  2  30°  60°  30°✓

20

69.

yi (8, 15)

10



0 62.6

0.002

116.4

0.003

153.8

0.004

169.7

0.005

161.7

0.006

131.0

0.007

81.9

0.008

21.3

c. t  0.00257 sec a. 17 cis 28.1° b. 51 V 57. a. 8.60 cis 324.5° b. 15.48 V a. 13 cis 22.6° b. 22.1 V I  2 cis 30°; Z  5 12 cis 45°; V  10 12 cis 75° 17 17113 I  113 cis 326.3°; Z  cis 61.9°; V  cis 28.2° 4 4 12 V  4 cis 60°; Z  4 12 cis 315°; I  cis 105° 2 10 V  5 cis 306.9°; Z  8.5 cis 61.9°; I  cis 245° 17 165 cis 29.7° 71. verified 4 24 7 24 7 5 13 29 37 z2   i, z3    i 75. , , , 5 5 5 5 24 24 24 24

77.

17

y

 

15 tan1 8

2

5 b 111 5 d z  6 cis c   tan1 111 111 5  6a  ib   111  5i 6 6

73.

0 0.001

x

5

3 1

yi (√11, 5)

5

6 5   tan1√11 

5

3 x 3

1 2 3 4

x

SA43

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Student Answer Appendix

Exercises 7.6, pp. 703–705

11.

r5 3 cos152  i sin152 4 ; De Moivre’s 3. complex z5  2 cis 366°  2 cis 6°, z6  2 cis 438°  2 cis 78°,  2 cis 510°  2 cis 150°; Answers will vary. r  312; n  4;   45°; 324 9. r  2; n  3;   120°; 8 13 1 11. r  1; n  5;   60°;  i 13. r  1; n  6;   45°; i 2 2 15. r  4; n  3;   330°; 64i 12 1 1 17. r  ; n  5;   135°;  i 2 8 8 19. verified 21. verified 23. verified 25. verified 27. r  1; n  5;   0°; roots: 1, 0.3090 0.9511i, 0.8090 0.5878i 29. r  243; n  5;   0°; roots: 3, 0.9271 2.8532i, 2.4271 1.7634i 3 13 3 3 13 3 31. r  27; n  3;   270°; roots: 3i,  i,  i 2 2 2 2 33. 2, 0.6180 1.9021i, 1.6180 1.1756i 313 3 3 13 3 35.  i,   i, 3i 2 2 2 2 37. 1.1346  0.1797i, 0.1797  1.1346i, 1.0235  0.5215i,  0.8123  0.8123i, 0.5215  1.0235i 1 13 39. x  1,  i. These are the same results as in Example 3. 2 2 41. r  16; n  4;   120°; roots: 13  i, 1  13i,  13  i, 1  13i 43. r  7 12; n  4;   225°; roots: 0.9855  1.4749i, 1.4749  0.9855i, 0.9855  1.4749i, 1.4749i  0.9855i 1 1 1 45. D  4, z0  86 cis 45°, z1  86 cis 165°, z2  86 cis 285°, 1 1 1 z0  86 cis 75°, z1  86 cis 195°, z2  86 cis 315° 47. verified 4 49. a. numerator: 117  44j, denominator: 21  72j b. 1  j 3 c. verified 51. Answers will vary. 53. 7  24i 55. z  2.7320, z  0.7320, z  2. Note: Using sum and difference identities, all three solutions can actually be given in exact form: 1  13, 1  13, 2. 1. 5. z7 7.

tan2x sec2x  1  sec x  1 sec x  1 1sec x  121sec x  12  sec x  1  sec x  1 1 cos x   cos x cos x 1  cos x  cos x 12 4 59. y   x  5 5 57.

12. 8i  3j; u  8.54;   159.4°

y

5 10.30 29.1 9 x

13. horiz. comp.  11.08, vertical comp.  14.18 14. H4, 2I; 2u  v  4.47,   206.6°

7 12 i j 1193 1193 16. QII; since the x-component is negative and the y-component is positive. 17. 16 mi 18. approx. 19.7° 19. H25, 123I 20. approx. 0.87 21. 4 22. p # q  6;   97.9° 23. 4340 ft-lb 24. approx. 417.81 lb 25. approx. 8156.77 ft-lb 26. a. x  269.97 ft; y  285.74 ft b. approx. 0.74 sec 27. 21cos 240°  i sin 240°2 28. 3  3i 5 z1  29. 30. z1z2  16 cisa b;  4 cisa b yi 12 z2 12 15.

\, e 5√3

3 2

30 4 5 x

32. Z  10.44;   16.7°, 10.44 cis 16.7° 5 13 5 513 5 33. 16  16 13i 34. verified 35.  i,   i, 5i 2 2 2 2 36. 6, 3 3i 13 37. 2  2i, 2 2i 38. 1 2i, 1 2i 39. verified 31. 213  2j

Mixed Review pp. 709–710 1.

Angles Sides A  41° a  13.44 in. B  27° b  9.30 in. , C  112° c  19 in. Area  57.9 in2 3. x  16.09, y  13.50 5. approx. 176.15 ft heading 28.2° 9. One solution possible since side a 7 side b Angles A  31° B  20.1° C  128.9°

7. approx. 793.70 mph;

Sides a  36 m b  24 m c  54.4 m

11. No; barely touches (“tangent”) at 30° 13. a. 4 121cos 315°  i sin 315°2 yi b. 3  3 13 i

yi 6 120

Summary and Concept Review pp. 705–709 45 4

1.

2.

Angles A  36° B  21° C  123°

Sides a  205.35 cm b  125.20 cm c  293 cm

Angles A  28° B  10° C  142°

Sides a  140.59 yd b  52 yd c  184.36 yd

4. approx. 20.2° and 159.8° 5. Angles Sides A  35° a  67 cm B1  64.0° b  105 cm C1  81.0° c1  115.37 cm

x

x

4

15. 13.1° 3. approx. 41.84 ft

17. compvu  0.87, projvu 

38 26  i 53 53

16 12  12 16  i, z1   i, 2 2 2 2  16 12 12 16 z2   i, z3   i 2 2 2 2

19. z0 

Practice Test pp. 710–712 Angles A  35° B2  116.0° C2  29.0°

6. no; 36° 7. approx. 36.9° 8. approx. 385.5 m 9. 133.2°, 30.1°, and 16.7° 10. 85,570.7 m2

Sides a  67 cm b  105 cm c2  56.63 cm

1. 6.58 mi 2. 137.18 ft 3. Angles Sides (in.) A1  58.8° a  15 B  20° b6 C1  101.2° c1  17.21

Angles A2  121.2° B  20° C2  38.8°

Sides (in.) a  15 b6 c2  11.0

4. a. No b. 2.66 mi 5. a. No b. 1 c. 8.43 sec 6. a. 2.30 mi b. 7516.5 ft 7. A  438, 795 mi2, P  61.7°, B  61.2°, M  57.1° 8. speed  73.36 mph, bearing  47.8°

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Student Answer Appendix 9.   36.5° 10. 63.48 cm to the right and 130.05 cm down from the initial point on the ceiling 11. F3  212.94 N,   251.2° 12. a.   42.5° b. projv u  H2.4, 7.2I c. u1  H2.4, 7.2I, u2  H6.6, 2.2I  13. 104.53 ft; 3.27 sec 14. 2 cisa b 15. 48 12 cis 75°; verified 24 513 5 5 13 5 16. 8  8 13i 17. verified 18.  i,   i, 5i 2 2 2 2 2 19. 2.3039  1.5192i, 2.3039  1.5192i 20.  2, 414,300 mi

Strengthening Core Skills p. 713 Exercise 1: Exercise 3:

664.46 lb, 640.86 lb yes

Exercise 2:

106.07 lb, 106.07 lb

Cumulative Review Chapters 1–7 pp. 713–714 1 2A  1r2 2  4 3 5 5. QIV sin   3 5 ; cos   5 ; tan   4 ; csc   3 ; 5 4 sec   4 ; cot   3 7. x  4  16 5 5 9. cos  19°  0.94, cos 125°  0.58 11. a. about $66,825 y2  y1 b. 13, 13, 7 12; A  59.5 mi2 13. a. m  x2  x1 x2  x1 y2  y1 b  2b2  4ac b. a , b c. x  2 2 2a 2 d. d  21x2  x1 2  1y2  y1 2 2 e. A  Pert 15. A  37°, a  33 cm, B  34.4°, b  31 cm, C  108.6°, c  52 cm 17. about 422.5 lb 19. y 1. 20 23; 40; 60°; 90°

7.

1800 ft D

d 525 ft

d

Post

a. D  25252  18002  10002  2125 ft b. d  H525, 1800, 0I, D  H525, 1800, 1000I H525, 1800, 0I  H525, 1800, 1000I cos   118752 121252 15   cos1a b  28.07°; verified 17 9. a.     

3. R 

7 18  tan1a b  tan1a b x x 1 7 1 18 b.   tan a b  tan a b 46 46  30.02°

18 ft ␤ ␣

25 ft x 7 ft

15 ft 8 ft

CHAPTER 8 Exercises 8.1, pp. 727–731 1. inconsistent 3. consistent; independent 5. Multiply the first equation by 6 and the second equation by 10. 7. y  74 x  6, y  4 9. y  x  2 11. x  3y  3 3 x  5 13. y  x  2, x  3y  3 15. yes 17. yes 19. 21. y y 10 8 6 4 2

8 4 54321 4

1 2 3 4 5 x

108642 2 4 6 8 10

8

x1q, 12 ´ 12, 32 21. 128  128i 13

23. about 3.6 yr

25. A  2, B  1, C 

 4

Connections to Calculus Exercises, pp. 717–718 a.   tan1a

1. Plane

b.   35°

9000 b d

9000 ft ␪ d

Light

a. d  2242  322  21242 1322cos  b. d  54.13 m c. about 106.1°

3.

1000 ft

␪

32 m ␪

10 8 6 4 2

108642 2 2 4 6 8 10 x 4 (1.1, 4.6) 6 8 10

2 4 6 8 10 x

(6, 3)

23. 14, 12

25. 13, 52

33. 13, 12

35. 12, 32

27. second equation, y, 14, 32

29. second equation, x, 110, 12 43. 16, 122

31. second equation, x, 1 52 , 74 2

37. 1 11 2 , 22

39. 12, 32

45. (2, 8); consistent/independent

41. 13, 42

47. ; inconsistent

49. 5 1x, y2 |6x  y  226; consistent/dependent 51. (4, 1); consistent/independent 53. 13, 42; consistent/independent

4 55. 1 1 57. 12, 52 2 59. 12, 12 2 , 3 2; consistent/independent 61. 1 mph 4 mph 63. 2318 adult tickets; 1482 child tickets 65. premium: $3.97, regular: $3.87 67. nursing student $6500; science major $3500 69. 150 quarters, 75 dimes 71. a. 100 lawns/mo, b. $11,500/mo 73. a. 1.6 billion bu, 3 billion bu, yes; b. 2.7 billion bu, 2.25 billion bu, yes; c. $6.65, 2.43 billion bu 75. a. 3 mph, b. 5 mph 77. a. 3.6 ft/sec, b. 4.4 ft/sec 79. 1776; 1865 81. Tahiti: 402 mi2, Tonga: 290 mi2 83. m1  m2; consistent/independent 85. $6552 at 8.5%; $11,551 at 6% 87. 472°, 832°, 248°, 608° 89. verified

24 m

Exercises 8.2, pp. 740–743 5.

64 ft v

18 ft

␪2 u u

48 ft

␪1

a. v  2482  642  182  82 ft b. u  H48, 64, 0I, v  H48, 64, 18I H48, 64, 0I  H48, 64, 18I cos 2  1802 1822 40 2  cos1a b  12.68°; verified 41

1. triple 3. equivalent; systems 5. z  5 7. Answers will vary. 9. Answers will vary. 11. yes; no 13. (5, 7, 4) 15. (2, 4, 3) 17. (1, 1, 2) 19. (4, 0, 3) 21. (3, 4, 5) 23. (1, 6, 9) 25. no solution, inconsistent 27. 1p, 2  p, 2  p2 5 2 29. a p  , p  2, pb, other solutions possible 3 3 31. 1p, 2p, p  12 33. 1p  9, p  4, p2 35. 5 1x, y, z2 | x  6y  12z  56 1 5 37. (1, 1, 2) 39. e 1x, y, z2 |x  y  2z  3 f 41. a2, 1, b 2 3

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Student Answer Appendix

43. 1p  5, p  2, p2

45. 118, 6, 102

47. a

11 10 7 , , b 3 3 3

19.

49. 11, 2, 32 55. 57. 59. 61. 63. 67.

1 1 51. a , , 3b 53. 3.464 units 2 3 Monet $1,900,000; Picasso $1,100,000; van Gogh $4,000,000 elephant, 650 days; rhino, 464 days; camel, 406 days Albatross: 3.6 m, Condor: 3.0 m, Quetzalcoatlus: 12.0 m 175 $5 gold pieces; 50 $10 gold pieces; 25 $20 gold pieces A  1, B  1, C  2; verified 65. x2  y2  4x  6y  9  0 H11, 5I; H6, 43 69. x  1 2I

108642 2 4 6 8 10

25.

47.

1 10



49.

1 10



1 100

1 P 100  10P P 10  10 P 4 49 62 4 3   51. 53. 55. 50 125 ln x  2 ln x  1 1ln x  12 2 57. Factor out 1 from the denominator: 1 x2 x  2 3 x2     1x  1211  x2 1x  121121x  12 x1 1x  12 2 1x  12 2 1 59. 2x  1  2 61. Verified x x6

1. half; planes 3. solution 5. The feasible region may be bordered by three or more oblique lines, with two of them intersecting outside and away from the feasible region. 7. No, No, No, No 9. No, Yes, Yes, No 11. 13. y y

108642 2 4 6 8 10

10 8 6 4 2

108642 2 4 6 8 10

2 4 6 8 10 x

15. No, No, No, Yes

17.

10 8 6 4 2 108642 2 4 6 8 10

37.

10 8 6 4 2 108642 2 4 6 8 10

10 8 6 4 2 108642 2 4 6 8 10

21.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

27.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

33.

y

5 4 3 2 1 54321 1 2 3 4 5

2 4 6 8 10 x

39. e

y

y

23.

108642 2 4 6 8 10

2 4 6 8 10 x

y

29.

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

y

10 8 6 4 2

35.

1 2 3 4 5 x

10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

y

2 4 6 8 10 x

y

2 4 6 8 10 x

yx1 xy 7 3

2 4 6 8 10 x

yx1 41. • x  y 6 3 43. (5, 3) 45. (12, 11) 47. (2, 2) y0 49. (4, 3) 51. 5 6 H 6 10 53. 10 55. 300 acres of corn; 200 acres of soybeans 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10

A (in ten thousands)

J  A  50,000 J  20,000 A  25,000 57. 240 sheet metal screws; 480 wood screws 59. 65 traditionals, 30 Double-T’s 61. 220,000 gallons from Tulsa to Colorado; 100,000 gal from Tulsa to Mississippi; 0 thousand gal from Houston to Colorado; 150,000 gal from Houston to Mississippi y 63. 65. a. 35 b. 54 c. 34 67. 324 10 8 6 4 2

Exercises 8.4, pp. 764–767

10 8 6 4 2

31.

J (in ten thousands)

1 100

10 8 6 4 2 108642 2 4 6 8 10

Exercises 8.3, pp. 753–755 1. template 3. repeated linear 5. Answers will vary B B A A 7.  9.  x3 x2 x1 1x  12 2 A B C B C A 11.   13.   x1 x2 x3 x x3 x1 A B C A B C 15.   17.  2 x5 x x2 1x  52 2 1x  52 3 x B C D A Bx  C A 19.  2  21.  2 x x5 x3 x 1x  52 2 x  5x  7 A Bx  C 4 3 Dx  E 23. 25.  2   2 x1 2x  5 x3 x 2 1x  22 2 7 2 1 4 5 1 27.  29.    x x1 x1 x x1 1x  12 2 5 x1 2 3x  1 31. 2 33.  2  2 x2 x 1 x 2 x 3 x1 5x  2 2 1 1 35.  2 37.  2  2 x x 1x  12 2 x x x3 3 5 4 2 39. 41.   2x x3 4  2x  x2 1x  32 2 2x  1 3 3x 2 43. 2 45.   2 x1 x 1 1x  12 2 1x  12 3

10 8 6 4 2

108642 2 4 6 8 10

(3, 3) 2 4 6 8 10 x

(3, 3); optimal solutions occur at vertices

Mid-Chapter Check, p. 768 2 4 6 8 10 x

y

2 4 6 8 10 x

1. (1, 1) consistent 2. (5, 3) consistent 3. 20 oz 4. No 5. The second equation is a multiple of the first equation. 6. (1, 2, 3) 2 3 1 7. (1, 2, 3) 8. 9. Mozart  8 yr;   x1 x2 1x  22 2 Morphy  13 yr; Pascal  16 yr 10. 2 table candles, 9 holiday candles

Reinforcing Basic Concepts, pp. 768–769 Exercise 1. Premium: $4.17/gal, Regular: $4.07/gal 15.3R  35.7P  211.14 e P  R  0.10 Exercise 2. Verified

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Student Answer Appendix

Exercises 8.5, pp. 776–779 1. square 3. 2; 3; 1 5. Multiply R1 by 2 and add that result to R2. This sum will be the new R2. 7. 3  2, 5.8 9. 4  3, 1 1 2 1 1 0 1 3 § ; diagonal entries 1, 0, 1 11. £ 1 2 1 1 3 x  2y  z  0 x  4y  5 1 S 13, 2 13. e 15. • y  2z  2 S 111, 4, 32 2 y  12 z3 x  3y  4z  29 1 6 2 y  32 z  21 d 17. • 19. c 2 S 14, 15, 32 0 28 6 z3 1 3 3 2 3 1 1 8 23 12 15 § 23. £ 0 3 3 6 § 21. £ 0 2 1 0 4 0 10 13 34 25. 2R1  R2 S R2 27. 5R1  R2 S R2 4R1  R3 S R3 3R1  R3 S R3 29. (20, 10) 31. (1, 6, 9) 33. (1, 1, 2) 35. (1, 1, 1) 37. 11, 3 2 , 22 39. linear dependence (p  4, 2p  8, p) 41. coincident dependence {(x, y, z)|3x  4y  2z  2} 43. no solution 45. linear dependence, 154 p  3, 18 p  12 , p2 47. 28.5 units2 49. Heat: 95, Mavericks: 92 51. Poe, $12,500; Baum, $62,500; Wouk, $25,000 53. A  35, B  45, C  100 55. $.4 million at 4%; $.6 million at 7%; $1.5 million at 8% 57. x  84; y  25 5 13 5 59. a. z1  110 cis 3   tan1 132 4 b. z2    i 2 2 61. C  30,000 in the year 2011 (t  6.39)

Exercises 8.6, pp. 787–791 1. aij; bij 3. scalar 5. Answers will vary. 7. 2  2, a12  3, a21  5 9. 2  3, a12  3, a23  6, a22  5 11. 3  3, a12  1, a23  1, a31  5 13. true 15. conditional, c  2, a  4, b  3 10 0 d 19. different orders, sum not possible 17. c 0 10 5 1 2 0 1 0 2 20 15 1 § 25. £ 0 1 d 23. £ 0 7 2 § 21. c 2 25 10 3 2 6 4 3 6 2 1 0 6 3 9 12 24 90 d d 29. c d 31. c 27. c 0 1 12 0 6 6 15 57 79 33. c 50

30 d 19

0.71 37. c 1.78

0.65 d 3.55

1 41. c 0 49. c

0 d 1

1.75 7.5

43. 2.5 d 13

42 18 60 d 35. c 12 42 36 1 1.25 0.25 2.13 § 39. £ 0.5 0.63 3.75 3.69 5.94 1 0 0 0 3 4 57 1 £ 0 1 0 § 45. c 19 d £ 47. 2 1 5 19 57 1 0 0 1 4 0.26 0.32 0.07 d 51. c 0.63 0.30 0.10

55. verified 57. P  21.448 cm; T S T S 3820 1960 S 4220 V  D £ 2460 1240 § M  D £ 2960 P 1540 920 P 1640 b. 3900 more by Minsk c. d. 8361.6 3972.8 2038.4 £ 5636.8 V  £ 2558.4 1289.6 § 3307.2 1601.6 956.8 4388.8 3078.4 M  £ 3078.4 3369.6 § 1705.6 852.8

53. verified 59. a.

Female 32.4 Male c 29.9

10.3 9.6

21.3 d, 19.5

the approximate number of females expected to join the writing club 2n1 0 2n1 n £ 2  1 1 2n  1 § 2n1 0 2n1 69. a  2, b  1, c  3, d  2 71. 0.3211 73. x2  2x  5 67.

Exercises 8.7, pp. 801–805 1. diagonal; zeroes 3. identity 5. Answers will vary. 7. verified 9. verified 11. verified 13. verified 2 1 5 1.5 9 9 d 19. verified 21. verified 15. c 1 17. c 5 d 2 0.5 9 18 23. £

2 39 1 3 4 39

1 29. £ 1 2

31.

33. 41. 51.

1 13

0 2 13

10 39 1 3 § 19 39

9 80

1 25. £ 80

1 20

31 400 41 400 1 100

27 400 3 400 § 17 100

27. c

2 5

3 x 9 dc d  c d 7 y 8

2 1 x 1 0 1 § £y§  £3§ 1 1 z 3

2 1 4 5 w 3 2 5 1 3 x 4 ≥ ¥≥ ¥  ≥ ¥ 3 1 6 1 y 1 1 4 5 1 z 9 (4, 5) 35. (12, 12) 37. no solution 39. 11.5, 0.5, 1.52 no solution 43. 11, 0.5, 1.5, 0.52 45. 1, yes 47. 0, no 49. 1 singular matrix 53. singular matrix 55. 34 57. 7

59. det1A2  5; (1, 6, 9)

1

61. det1A2  0

13 63. A1  c 2 13

3 4 3 8 11 16

A  27.7269 cm2 S 2960 3240 § 820 5116.8 4659.2 § 1809.6

61. 322,000 19,000 23,500 14,000 4 ; total profit North: $22,000 South: $19,000 East: $23,500 West: $14,000 63. a. $108.20 b. $101 Science 100 101 119 c. c d Math 108.2 107 129.5 First row, total cost for science from each restaurant; Second row, total cost for math from each restaurant. 65. a. 10 b. 20 c. Spanish Chess Writing

1 4 1 8 § 1 16

5 13 3 d 13

65. singular 67. 31 behemoth, 52 gargantuan, 78 mammoth, 30 jumbo 69. Jumpin’ Jack Flash: 3.75 min Tumbling Dice: 3.75 min You Can’t Always Get: 7.5 min Wild Horses: 5.75 min 71. 30 of clock A; 20 of clock B; 40 of clock C; 12 of clock D 73. p1  72.25°, p2  74.75°, p3  80.25°, p4  82.75° 75. y  x3  2x2  9x  10 77. 2 oz food I, 1 oz Food II, 4 oz Food III 79. Answers will vary. 81. a. 45, b. 52 c. 19 d. 4 2 9 1 83. A  125, period  85. x  aq,  d ´ c  , q b 3 2 2

Exercises 8.8, pp. 811–814 1. a11a22  a21a12 3. constant 5. Answers will vary. 2 5 7 5 2 7 7. D  ` ` ; Dx  ` ` ; Dy  ` ` 3 4 1 4 3 1 26 25 9. 15, 92 11. a , b 13. no solution 3 3

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Student Answer Appendix

4 1 2 5 1 2 15. a. D  † 3 2 1 † , Dx  † 8 2 1 † , 1 5 3 3 5 3 4 5 2 4 1 5 Dy  † 3 8 1 † , Dz  † 3 2 8† 1 3 3 1 5 3 D  22, solutions possible b. D  0, Cramer’s rule cannot be used; The sum of the first two rows of D gives row 3 17. (1, 2, 1) 3 5 1 19. a , , b 21. 10, 1, 2, 32 23. 320  32  420.5 in2 4 3 3 25. 8 cm2 27. 27 ft2 29. 19 m3 31. V  96 in3 33. yes 35. no 37. yes yes yes verified x2 2 1 1 3 39.  2 41.   x x1 x3 1x  12 2 1x  32 3

15,000x  25,000y  2900 ; 6%, 8% 25,000x  15,000y  2700 2x  2y  10z  3.26 45. • 3x  2y  7z  2.98; apples, 29¢/lb kiwi, 39¢/lb pears, 19¢/lb 2x  3y  6z  2.89 47. 10 lb of $1.90, 8 lb of $2.25, 6 lb of $3.50 49. Answers will vary. 51. x2  y2  4x  6y  12  0 43. e

53.

10 8 6 4 2 54321 2 4 6 8 10

55. B  76.3°, C  54.7°, side c  9.4 in.

y

28.

Maximum of 270 occurs at both (0, 6) and (3, 4)

y 10 8 6 4 2

(3, 4)

6 4 2 2 4 6

2 4 6 8 10 x

29. 50 cows, 425 chickens

33. c

30. 12, 42

32. 12, 7, 1, 82

31. (1, 6, 9) 7.25 0.875

5.25 d 2.875

34. c

6.75 1.125

6.75 d 35. not possible 1.125 1 0 4 2 6 1 0 36. c d 37. c d 38. £ 5.5 1 1 § 1 7 0 1 10 2.9 7 3 6 4 8 12 0 39. £ 4.5 3 1 § 40. not possible 41. £ 2 4 4 § 2 3.1 3 16 0.4 20 15.5 6.4 17 42. £ 9 17 2§ 18.5 20.8 13 43. D 44. It’s an identity. 45. It’s the inverse of B. 46. E 47. It’s an identity matrix. 48. It’s the inverse of F. 49. matrix multiplication is not generally commutative 50. 18, 62 51. 12, 0, 32 37 36 31 19 25 , b 53. a , , b 35 14 19 19 19 5 2x  1 91 55.  2 units2 56. x2 2 x 3 52. a

1 2 3 4 5 x

54. 11, 1, 22

Summary and Concept Review, pp. 814–819

Mixed Review, pp. 819–820

1.

1. consistent/dependent; consistent; inconsistent

5

10 8 6 4 2

(4, 4)

108642 2 4 6 8 10

3.

2.

y 3x  2y  4

10 8 6 4 x  3y  8 2

54321 2

2 4 6 8 10 x

x  0.3y  1.4 4

6 8 10

(4, 4)

y

3 12, 32 5. A 1, 12 , 2 B 7. 110, 122 3 8 16 2 9.  11. a. c 12 0 x2 x3

0.2x  0.5y  1.4 1 2 3 4 5 x

(q, 3)

1 12 , 32

13. 19, 3, 22 17.

y

2x  y  2 4

6 1

9 7

7 d 2

19. 7 unicycles; 9 bicycles; 5 tricycles

y

6 4

1 2 3 4 5 x

(U, T)

A 85, 6 5 B

9. A B ; consistent 10. Sears Tower is 1450 ft; Hancock Building is 1127 ft. 11. (0, 3, 2) 12. (1, 1, 1) 13. no solution, inconsistent 14. 72 nickels, 85 dimes, 60 quarters 15. y  x2  10x  9 4 2 2 2 7 1 16. 17. 18.    2 x3 x5 x3 2x  1 x4 x 1 x2 x1 3 3 19. 20.  2  2 x5 x3 x 3 x  3x  9 5 x2 1 3 5 21. 22.  2   x1 x2 x3 x x1 1x  22 2 1 3 2 2 x5 23. 24. 2    2 x1 x5 1x  12 2 x 3 1x  32 2 25. 26. 27. y y y 11 1 4, 6

10 8 6 4 2

10 8 6 4 2

2 4 6 8 10 x

(1.5, 4)

2

4. no solution; inconsistent 5. 15, 12; consistent 6. (7, 2); consistent 7. 13, 12; consistent 8. (2, 2); consistent

108642 2 4 6 8 10

10 57 A 33 31 , 31 , 31 B

b. c

8

3 2 1 54321 1 2 x  2y  4 3 4 5

10

15.

10 d 6

108642 2 4 6 8 10

10 8 6 4 2

2 4 6 8 10 x

108642 2 4 6 8 10

2 4 6 8 10 x

2

4

6

8

10 x

Practice Test, pp. 820–821 1.

y

10 3x  2y  12 8 x  4y  10 6 4 2 (2, 3) 108642 2 4 6 8 10

(2, 3)

5 1.2 d b. c 9 1.2 1 d e. 2 1.5 0.1 0 0.6 0 § 0.8 0.9

0.31 c. £ 0.01 0.39

0.13 0.05 0.52

7. a2, 1,

3. 13, 22

4. 12, 1, 42

2 4 6 8 10 x

6 8 2 d. c 2.5 0 6. a. £ 0.5 0.2 5. a. c

2 4 2. a , b 5 5

1 b 3

0.08 0.02 § 0.02

8. 13, 2, 32

1.2 d 2

c. c

3 3

0.3 b. £ 0.06 0.18 d. 9. a

40 17 40 £ 17 35 17

1 d 5

0.06 0.06 0.24 0 10 5

97 18 , b 34 17

10 17 10 17 § 30 17

0.12 0 § 0.48 e.

10. (1, 6, 9)

17 500

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Student Answer Appendix 11. 21.59 cm by 35.56 cm 12. Tahiti 402 mi2; Tonga 290 mi2 13. Corn 25¢ Beans 20¢ Peas 29¢ 14. $15,000 at 7% $8000 at 5% $7000 at 9% 15. h1t2  16t2  144 16. 144 ft, 3 sec y 17. 18. (5, 0) 19. 50 y 30 plain; 10 x  y  50 8 40 20 deluxe 6 30

4 2 108642 2 4 6 8 10

20.

(30, 20)

20

2x  3y  20

10

2 4 6 8 10 x

10

20

30

40

1 2 x 29. 4x2  3y2  48 16 31. Verified, verified 33. 3x2  y2  3 4 12 30 35. a. a ,  b, b. a2,  b 37. Verified (both add to 8) 7 7 3 y 1x  321x  32 4 10 39. x  41. h1x2  8 6 3 1x  221x  22 25. Verified

27. y  

4 2

50 x

y1

8642 2 4 6 8

1 3x  2  2 x3 x  3x  9

2 4 6 8 x

x  2

x2

Exercises 9.2, pp. 847–852 Strengthening Core Skills, p. 824

1. c2  | a2  b2 | 3. 2a; 2b 5. answers will vary. 7. x2  y2  49 9. 1x  52 2  y2  3 11. 1x  12 2  1y  52 2  25 13. 1x  62 2  1y  52 2  9 15. 1x  22 2  1y  52 2  25 center: (6, 5), r  3 center: 12, 52, r  5

1. Exercise 1 11, 4, 12

Cumulative Review Chapters 1–8, pp. 825–826 2 b. x  0, 7 c. x  5, i 12 3 1 y 3. R  2A  1r2 2 5.  6 1. a. x 

d. x  1, 0, 4

10

y

10

(6, 5) 10

10

10 x

r5

4 10

2

4

6

6

b 2b2  4ac c. x  2a

15. H3, 18I

19.

y 8

2

17. a.

2121  x2 11

b.

x  1q, 12 ´ 12, 32 23. about 3.6 yr

4 54321 4

b. 1a  bi21a  bi2  a2  1bi2 2 113 13 , cos   11. sin   , 4 4 y1  y y x2  x1 2 1 , b b. a x1 2 2

d. d  21x2  x1 2  1y2  y1 2

e. A  Pert

10

17. 1x  32 2  y2  14 center: (3, 0), r  114

8 10 12 x

4

7. a. 1a  bi2  1a  bi2  2a 115  a2  b2 9. x  12 3 y2  139 tan   13. a. m  3 x2 

10 x

(2, 5)

2 2 2

y

r3

2

10

r  3.7 10

10 x

(3, 0) 10

19.

10 8 6 4 2

x 29  x2

21. 128  128i 13  25. A  2, B  1, C  4

1 2 3 4 5 x

8

y

108642 2 4 6 8 10

21.

y

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

y2 x2   1, (0, 0), a  4, b  2 16 4 b. (4, 0), (4, 0), (0, 2), (0, 2) c.

25. a.

CHAPTER 9

10 8 6 4 2 108642 2 4 6 8 10

1. a.

27. a.

y2 x2   1, (0, 0), a  3, b  4 9 16

y2 x2   1, (0, 0), a  15, b  12 5 2 b. ( 15, 0), ( 15, 0), (0,  12), (0, 12) c.

13 2 25 2 17. ax  b  1y  92 2  a b 2 2 19. a. d  13; B, C, E, G; b. 113, 3  4132, 114, 82; Many others 8 15 21. Verified, d  23. a. B, C, E; b. Answers will vary. 5

2 4 6 8 10 x

2 4 6 8 10 x

108642 2 4 6 8 10

29. a.

y

y

10 8 6 4 2

b. (0, 4), (0, 4), 13, 02 , (3, 0) c.

Exercises 9.1, pp. 837–839 1. geometry, algebra 3. perpendicular 5. point, intersecting 7. 12, 22; verified 9. 12, 22; verified 11. 1 13 2 , 92; verified 13. 1x  22 2  1y  22 2  52 15. 1x  22 2  1y  22 2  52

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

Connections to Calculus Exercises, p. 829 k A B   ; 1A  B2x  1AB  Ba2  k 1x  a21x  b2 xa xb b. B  A 1 Ab  1Aa2  k; A1b  a2  k k 16 2 2 c. A  ;   b  a 1x  321x  52 x3 x5 1 1 1 2 2 2    3. 5. 7. x3 x3 x7 x4 x8 x5 9. A  114 units2 11. A  36 units2 13. V  240 units3 V  LWH; 152162182  240 units3

23.

y

5

y

2 4 6 8 10 x

y

5

5 x

5

31. ellipse

10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

33. circle

10 8 6 4 2 108642 2 4 6 8 10

y

2 4 6 8 10 x

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Student Answer Appendix

35. ellipse

1. transverse 3. midway 7. 9. y

108642 2 4 6 8 10

37. x2 

1y  32

2 4 6 8 10 x

(3, 0)

1

39.

5 x

41.



4 10

2

10

1

43.

1x  32 2 25 10

10

10 x

(5, 5) (3, 5)

10

6 4 (0, 0) 2

(2, 1)

108642 2 4 6 8 10

1y  22 2 10

19.

10

(3, 5  √10)

45. 20 47. 20 49. a. (2, 1) b. (3, 1) and (7, 1) d. (2, 3) and (2, 1) e.

108642 2 2 4 6 8 10 x 4 6 8 (0, 3) 10

(3, 2  √10)

23. 25. 27.

10

31.

51. a. (4, 3) b. (4, 2) and (4, 8) d. (0, 3) and (8, 3) e.

c. (4, 0) and (4, 6) 10

10

10

10

39.

y (4, 3)

10

10

(√6, 0) 10

10

1y  22 1x  32 y x   1 57.  1 36 20 9 25 2 2 2 1x  32 1y  12 2 y x 59.   1, 1 17, 02 61.   1, 16 9 4 16 2

2

2

55.

13, 1 2132

67. 73. 75. 77. 79. 83.

63. A  12 units2 65. 27  2.65 ft; 2.25 ft y2 y2 x2 x2 8.9 ft; 17.9 ft 69.  2  1; 6.4 ft 71.  1 2 2 8 15 36 135.252 2 a  142 million miles, b  141 million miles, orbit time  686 days 90,000 yd2 L  8 units; 13 15, 42, 1315, 42, 13 15, 42, 13 15, 42; verified kL Verified 81. R  2 k  0.003 250 d 261.8 mph, heading 26.2°

10

41.

y

10

(0, 0) 10

10 x

10 x

(0, 2) 10

45.

y

y (0, 2)

(3, 0)

10

4 5, 3

10

10

47.

y

10

(8, 2)

y

(√6, 0) 10 x

10

(1, 8) 14

(0, 0)

(0, 0)

10 x

10 x (1, 3)

(2, 1)

(3, 0)

10

10

10

10 x

43.

y

10 x

(2, 1)

10

53. a. (2, 2) b. (5, 2) and (1, 2) c. (2  13, 2) and (2  13, 2) d. (2, 2  16) and (2, 2  16) e. y

6

(1, 2)

4 5, 3 10

35.

y

10

(6, 1) 10 x

10 x

x

(3, 2)

33.

y

(5, 1)

37.

2

(0, 6)

10

10

10

10

(0, 6) (0, 0) 2 4 6 8 10 x

(3, 2)

5, 1 7

10

(0, 2√3)

(9, 2)

(0, 3)

10

y

2 4 6 8 10 x

y

5, 1 7

10

(0, 0)

10

10 x (0, 1) 10

10 x

y

10 8 6 4 (0, 2√3) 2

14, 22, 12, 22, y  2, 11, 22, x  1 14, 12, 14, 32, x  4, 14, 12, y  1 29. y y

10

y

(0, 3)

108642 2 4 6 8 10

(0, 1)

10

2 4 6 8 10 x

108642 2 4 6 8 10

2 4 6 8 10 x

10 8 6 4 2

(0, 0)

y

(0, 3) (0, 0)

10864 6 2 2 4 6 8 10

21.

y

c. (2  121, 1) and (2  121, 1)

10

(6, 0)

17.

y

10 8 6 4 2

2 4 6 8 10 x

10 8 (0, 3) 6 4 2

1

10 x (8, 2)

(2, 2)

10

(2, 1)



108642 2 4 6 8 10

10

(6, 0) 8

(3, 2  √10) (3, 2)

(3, 5  √10)

(2, 0)

15.

y

(7, 0)

4 (0, 0) 2

10 x

10

13.

y

10 8

(7, 0) 6

10

10 x

1

10 x

y

(1, 5)

4

10

1y  52

10

1y  12

(2, 0)

(3, 0)

10

(2, 1)

(0, 5)

2

(0, 0)

2

y

10

(1, 3)

(0, 3) 5



16

(2, 3) (6, 1)

(0, 1)

1x  32

1x  22

2

10

(1, 3)

10

(0, 0)

y

5

5. Answers will vary. 11. y

10

2

4 5

Exercises 9.3, pp. 862–865

y

10 8 6 4 2

10

10 x

(2, 2) 10

(5, 2)

10

10 x (4, 3)

(0, 3) 10

(2, 3)

49. circle 51. circle 53. hyperbola 55. hyperbola 57. circle 59. ellipse 61. 8, 2a  8, 2b  6 63. 12, 2a  16, 2b  12 65. a. (3, 4) b. (0, 4) and (6, 4) c. 13  113, 42 and 13  113, 42 d. 2a  6, 2b  4 e. y 10 8 6 4 2

108642 2 4 6 8 10

2 4 6 8 10 x

67. a. (0, 3) b. 12, 32 and (2, 3) d. 2a  4, 2b  8 e. y

c. 12 15, 3 and 12 15, 32

14 12 10 8 6 4 2

108642 2 4 6

2 4 6 8 10 x

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Student Answer Appendix 69. a. 13,22 b. 11, 22 and 15, 22 d. 2a  4, 2b  4 13 e. y

c. 11, 22 and 17, 22

27. x  3 21, q 2, y  1q, q 2

10 8 6 4 2

75. 79. 81. 85.

87. a.

1x  42 1 4

89. a

91.

 1y  22 2  0

1x  22 2 43



1y  32 2 9

25 15 5 3 6 9 12 15

2 4 6 8 10 x

b. 1x  22 2  1

1x  42

1. horizontal; right; a 6 0 3. (p, 0), x  p 5. Answers will vary. 7. x  1q, q 2, y  3 4, q 2 9. x  1q, q 2, y  318, q 2

108642 2 (0, 3) 4 6 8 10

43.

8

(5, 0)

16 8

2 4 6 8 10 x

8

16

13. x  3 4, q 2, y  1q, q 2 y

12

8

6

4

(3.5, 0)

(1, 0)

12 6

6 12 6 (0, 7)

12 (1.25, 10.125)

15. x  1q, 164 , y  1q, q2

8

4

4

(3, 0) 4 (0, 1)

8

(0, 7)

4

(16, 3)

(7, 0)

8 16 4 (0, 1)

x

8

16

x

4

8

8

19. x  39, q 2, y  1q, q2 y

y

10 8 6 4 (0, 2) (4, 0) 2

(9, 3) 8 4 (0, 6)

(0, 0) 4

21. x  3 4, q 2, y  1q, q 2

4

8

x

4 8

23. x  1q, 04 , y  1q, q 2

108642 2 4 6 8 10

2 4 6 8 10 x

(0, 2)

108642 2 4

2 4 6 8 10 x

(6.25, 0.5) 6 (0, 3) 8 10

10 (0, 6)

45.

y x1

9

x  2

(0, 0)

3

51. (4, 0) 9

y 2

9 ( 2,

108642 2 4 6 8 10

2 4 6 8 10 x

57.

20 (0, 6) 16 12 8 (5, 6) 4

10 8 6 (1, 3) 4 2

5

10 8 6 4 5,  2 0 2 108642 2 4 6 8 10

(0, 0)

53

y

108642 2 4 6 8 10

2 4 6 8 10 x

0)

2 4 6 8 10 x

10 8 6 (7, 4) 4 2 25 15 5 2 4 6 8 10

15 x

47

y

10 8 6 4 2

15

25 x

10 8 6 4 2 10 6

(7, 2) y 8

59

y

3

y  2

y 5

x2 (0, 0) 2 4 6 8 10 x

y (4, 2)

2

6

10 x

y 4

(4, 3)

y

(2, 3)

108642 3 6 9 12 15

x7

2 4 6 8 10 x

(6, 5)

x2  8y 63. y2  16x 65. x2  20y 1y  22 2  121x  22 69. 1x  42 2  121y  72 1x  32 2  81y  22 73. y2  81x  12 vertex (1, 0) focus (1, 0) 1y  22 2  81x  22 directrix: x  0 endpoints (4, 4) and (4, 0) 16 units2 y

(4, 2.5)

,  25 64 0 1

1

2

3

4

5 x

(4, 2.5)

81. 85. 87. 91. 93.

2 2 4 6 8 10

2 4 6 8 10 x

15 12 (5, 5) 9 6 3

x3

2 4 6 8 10 x

10 8 6 0, 3  2 4 2

108642 2 4 6 8 10

20 x

25

y

10 8 6 (0, 2) 4 2

(0, 0) 12

x  64

25. x  3 6.25, q 2, y  1q, q 2

y

10 8 6 4 2 (0, 1) 108642 2 2 4 6 8 10 x 4 (1, 0) 6 8 10

61. 67. 71. 75. 77. 79.

4

6

3 2 4 6 8 10

y

2 (0, 0)

y  2

108642 4 8 12 16 20

4 (0, 4)

16 8

(11, 0)

y6

20 12 4 2

2 4 6 8 10 x

y

x  5

8

(16, 0)

(2, 3) 3 6 9 12 15 x

41.

y

10 6

y

15 9

55.

39.

10 8 6 (4, 2) 4 2

y

8

y

(19, 0)

10 8 6 4 (1, 0) 2

8

y



4 8 12 16 20 x

y

49.

x

17. x  1q, 0 4 , y  1q, q 2

15 12963 2 4 6 8 10

10 8 6 4 (0, 2) 2

(0, 3)

(4, 1) x

(11, 2) 4  22 2

(1, 3)

108642 2 4 6 8 10

(2, 18)

y

8

x

(0, 10) 8

(1, 4)

11. x  1q, q 2, y  3 10.125, q 2

16 8

16

0,

y

108642 2 (0, 0) 4 6 8 10

y

(1, 0)

10 8 6 4 2

(2, 3)

20 16 1284 2 4 6 8 10

37.

(3, 0)

(3, 0) 3 6 9 12 15 x

33. x  32, q 2, y  1q, q 2

2 4 6 8 10 x

10 8 6 4 2

Exercises 9.4, pp. 871–874

(1, 0)

3 15 12963 3 6 9 12 15

25 x

35. x  31, q2, y  1q, q2

0

16



y

108642 2 4 6 8 10

93. 700 cos 65°  295.8, yes

y

15

y

4  22 12 9 2 6

(4, 0)

10 8 (0, 7) 6 4 2

2

1 5

5

31. x  1q, q 2, y  3 3, q 2

95. b and c

10 8 6 4 2



0,

9 6 3 (0, 5  21)

y2 y2 1x  22 2 x2   1 73.  1 36 28 9 9 2 2 1y  12 2 1x  22 2 y x   1, 1 113, 02 77.   1, 4 by 2 15 4 9 4 5 2 2 2 a. y  3 2x  9 b. x  1q, 3 4 ´ 3 3, q 2 c. y  2 3 2x  9 40 yd 83. 40 ft y2 x2   1, about 124.1, 602 or 124.1, 602 225 2275 2

15

(21, 5) 12

108642 2 4 6 8 10

71.

29. x  1q, 114 , y  1q, q 2

y 15 (0, 5  21)

6 in.; (13.5, 0) 83. 14.97 ft, (0, 41.75) y2  5x or x2  5y, 1.25 cm 1x  22 2  12 1y  82; p  18 ; 12, 82 89. 18 units2 2, 2, 1  23i, 1  23i, 1  23i, 1  23i about 120 days

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Student Answer Appendix

Mid-Chapter Check, p. 874 1.

2.

y

d. 0, 1, 2, 3, and 4 solutions possible 3.

y 8

4 4

4

8

12

x

4

4

4

8

12

16 x

321 1 2 3 4 5

4

8

8

12

12

4.

5.

y

8 7 6 5 4 3 2 1

y

5 4 3 2 1

4

1 2 3 4 5 6 7 x

e. 0, 1, 2, 3, and 4 solutions possible

6.

y

y 4

12 8

8

12 8

654321 1 2

4

1 2 3 4 x

1x  32 2

4

4

8

x

4

4 4

x

4

8

1y  12 2

 1; D: x  3 5, 1 4 ; R: y  3 3, 5 4 4 16 2 2 b. 1x  32  1y  22  16; D: x  3 1, 7 4 ; R: y  32, 6 4 c. y  1x  32 2  4 D: x  1q, q2; R: y  3 4, q 2 y2 x2   1 10. yes, distance d  49 mi y 8. 9. 12 16 4 7. a.

(6, 4)

10 8 6 4 2

108642 2 4 6 8



f. 0, 1, 2, 3, and 4 solutions possible

12

(6, 4)

3. region; solutions

10 8 6 (1, 5) 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

5. Answers will vary.

7. first: parabola; second: line y

(2, 2) 2 4 6 8 10 x

9. first: parabola; second: ellipse

Reinforcing Basic Concepts, p. 875 Exercise 1.

Exercise 2.

Exercise 3.

1x  22

1y  32

2

12 2 b a 5 1x  12 2

5 17 2 b a 14 1x  32 2 7 2 a b 2



2 2 a b 3 1y  22 2



5 13 2 b a 12 1y  12 2



6 2 a b 5

10 8 6 (3, 2.2) 4 2

(6.5, 1)

108642 2 (3, 0.2)4 6 8 10

Exercise 4.

1x  32 2

4 15 2 a b 3

108642 2 4 6 8 10

1a

2 12 ,b 5 3 5117 5 13 ,b 14 12

6 7  1; a  , b  2 5

(0.5, 1) 2 4 6 8 10 x

9 2 a b 2

10 8 6 4 2

1a

y

1y  12 2



(3, 1)

(6, 1)

2

10 (3, 8) 8 6 4 2 108642 2 4 6 8 10

(3, 8) (5, 4) 2 4 6 8 10 x

19. 23. 27. 31. 35. 39. 43.

108642 2 4 6 8 10

49.

1. a. 0, 1, and 2 solutions possible

c. 0, 1, 2, 3, and 4 solutions possible

5 4 3 2 1 54321 1 2 3 4 5

Exercises 9.5, pp. 880–883 b. 0, 1, and 2 solutions possible

(5, 4) 2 4 6 8 10 x

(5, 4)

(4, 3), 14, 32, 14, 32, 14, 32 21. no solutions 15, 52, (5, 5), 15, 52, 15, 52 25. (5, log 5  5) (3, ln 9  1), (4, ln 16  1) 29. (0, 10), (ln 6, 45) (3, 1), (2, 1024) 33. (3, 21), (1, 1), (2, 4) (2, 4), (6, 4) 37. (3, 5), (3, 5) 12.43, 2.812, (2, 1) 41. (0.72, 2.19), (2, 3), (4, 3), (5.28, 2.19) y y 45. 47. no solution 10 10 8 6 4 2 2 4 6 8 10 x

y

2 4 6 8 10 x

108642 2 (5, 4) 4 6 8 10

y

3 15. a4, b, (3, 2) 2 17. 1 110, 32, 1 110, 32, 15, 122, 15, 122

415 9  3, b  3 2

(0, 1)

10 8 6 4 2

13. 14, 32 , (3, 4)

8 6 4 2

 1; a 

11. first: hyperbola; second: circle

y

y

108642 2 4 6 8 10

20 16 12 8 4

2 4 6 8 10 x

2016 1284 4 8 12 16 20

y

4 8 12 16 20 x

51. h  27.5 ft; h  24 ft; h  18 ft

1 2 3 4 5 x

53. The company breaks even if either 18,400 or 48,200 Nanos are sold. 55. $1.83; $3 90,000 gal 10P2  6D  144 e 2 8P  8P  4D  12 57. 8.5 m  10 m 59. 5 km, 9 km 61. 8  8  25 ft 63. Answers will vary. 65. 18 in. by 18 in. by 77 in. 67. h  62 ft 69. a. m  400 1 , the copier depreciates by $400 a year. b. y  400x  4500 c. $1700 d. 9.5 yr

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Student Answer Appendix

Exercises 9.6, pp. 893–896

1. polar 3. II; IV 5. To plot the point 1r, 2 start at the origin or pole and move r units out along the polar axis. Then move counterclockwise an angle measure of . You should be r units straight out from the pole in a direction of  from the positive polar axis. If r is negative, final resting place for the point 1r, 2 will be 180° from . 2␲ 2␲ 2␲ ␲ ␲ ␲ 7. 9. 11. 3 3 3 3 3 3 5␲ 6

7␲ 6

3␲ 4

␲ 4

5

P 4 3 2 1

5␲ 6

3␲ 4

5␲ 6

11␲ 6

5␲ 4 4␲ 3

13.

␲ 6

7␲ 6

7␲ 5␲ 4 3

2␲ 3

5 4 3 2 1

␲ 3 ␲ 4

P

3␲ 4

␲ 4

5 4 3 2 1

P

␲ 6

5␲ 6

11␲ 6

5␲ 4 4␲ 3

7␲ 6

7␲ 5␲ 4 3

3␲ 4

␲ 4

5 4 3 2 1

P 5␲ 4 4␲ 3

␲ 6

11␲ 6

5␲ 6

7␲ 6

7␲ 5␲ 4 3

3␲ 4

2␲ 3

␲ 3 ␲ 4

10 8 6 4 2

5␲ 6

␲ 6

11␲ 6

5␲ 4 4␲ 3

5␲ 6

3␲ 4

2␲ 3

3␲ 4

2␲ 3

5 4 72 3 2 1

144

␲ 3 ␲ 4

␲ 6

4 7␲ 6

7␲ 5␲ 4 3

83. r2  16 cos122

  b 17. a4 12, b 2 4 2 3 19. a8, b 21. a412, b 3 4 15. a4,

␲ 6

413  312 4  3 12 , b; 1312, 3 122; 1413, 42; yes 2 2 3 12  413 3 12  4 , b Ma 2 2 79. r  4  4 cos  81. r  4 cos152 77. a

␲ 3 ␲ 4

5 4 3 2 1

216

288

5␲ 4 4␲ 3

11␲ 6

7␲ 5␲ 4 3

85. r  4 sin  4

␲ 6

␲ 2

␲ 6

2 7␲ 6

11␲ 6

5␲ 4 4␲ 3

23. a312, 25. a2,

5 7 11  b, a3 12, b, a3 12, b, a3 12, b 4 4 4 4

7 17  5 b, a2, b, a2, b, a2, b 6 6 6 6

33. B 35. D

45. 1412, 4 122 5␲ 6

7␲ 6

5␲ 6

3␲ 4

5 4 3 2 1

2␲ 3

5␲ 6

3␲ 4

2␲ 3

5␲ 6

3␲ 4

2␲ 3

55. ␲ 6

5␲ 6

11␲ 6

5 4 3 2 1

7␲ 6

5 4 3 2 1

␲ 3 ␲ 4

␲ 6

7␲ 6

11␲ 6

7␲ 6

3␲ 4

6

␲ 3 ␲ 4

5 4 3 2 1

3␲ 4

␲ 6

7␲ 6

␲ 6

5␲ 6

7␲ 6

␲ 6

Open dot 7␲ 6

5␲ 4 4␲ 3

75. 5␲ 6

7␲ 6

11␲ 6 7␲ 5␲ 4 3

3␲ 4

2␲ 3

5␲ 4 4␲ 3

5 4 3 2 1

␲ 3 ␲ 4

7␲ 6

5␲ 4 4␲ 3

␲ 3 ␲ 4

5 4 3 2 1

3␲ 4

␲ 6

11␲ 6

7␲ 5␲ 4 3

2␲ 3

5 4 3 2 1

␲ 3

11␲ 6 7␲ 5␲ 4 3

5␲ 4 4␲ 3

3␲ 4

2␲ 3

5␲ 4 4␲ 3

11␲ 6

5␲ 4 4␲ 3

7␲ 5␲ 4 3

87. a; this is a circle through 16, 0°2 symmetric about the polar axis   89. g; this is a circle through a6, b symmetric about   . 2 2  91. f; this is a limaçon symmetric about   with an inner loop. 2 Thus a 6 b. 3  b. 93. b; this is a cardioid symmetric about   through a6, 2 2 95. r2  72002sin122 97. r  15 cos152 or r  15 sin152 99. ; ; ; Answers will vary. 101. Consider r  a1cos122 and r  a 1cos122 ; both satisfy r2  a2cos122. Thus, 1r, 2 and 1r, 2 will both be on the curve. The same is true with a1sin122 and a 1sin122. 103. 9 units2 105. 3y2  x2  12y  9  0 2 5 , , 107. t  0, 3 3 109. D: x  35, 22 ´ 12, 5 4 y 5 R: y  3 3, 22 ´ 546 4 3 2 1

␲ 4

54321 1 2 3 4 5

␲ 6

(2.5, 30)

(2.5, 60)

69.

␲ 3 ␲ 4

2␲ 3

5␲ 4 4␲ 3

5␲ 6

11␲ 6

5 4 3 2 1

5 12 512 , b 2 2

(2.5, 150)

7␲ 5␲ 4 3 2␲ 3

3␲ 4

63.

7␲ 5␲ 4 3

5␲ 4 4␲ 3

5␲ 6

7␲ 6

11␲ 6

2␲ 3

73. ␲ 6

5 4 3 2 1

5␲ 4 4␲ 3

5␲ 6

11␲ 6

␲ 3 ␲ 4

3␲ 4

5␲ 6

11␲ 6

␲ 3 ␲ 4

51. a

57. ␲ 6

7␲ 5␲ 4 3

2␲ 3

67. ␲ 6

␲ 3 ␲ 4

5 4 3 2 1

5␲ 4 4␲ 3

5␲ 6

7␲ 5␲ 4 3

5 4 3 2 1

2␲ 3

3␲ 4

61.

␲ 3 ␲ 4

7␲ 5␲ 4 3

5␲ 4 4␲ 3

71.

47. 12 12, 2122 49. 1 13, 12 ␲ 3 ␲ 4

 b 4

43. 113, 247.4°2 or 113, 4.31762

7␲ 5␲ 4 3

5␲ 4 4␲ 3

65.

7␲ 6

2␲ 3

5␲ 4 4␲ 3

59.

7␲ 6

3␲ 4

 b 4

27. C 29. C 31. D

39. 14 12, 45°2 or a412,

37. (8, 180°) or 18, 2

41. 110, 45°2 or a10,

53.

7␲ 6

7␲ 5␲ 4 3

1 2 3 4 5 x

11␲ 6

(5, 90) 5 4 3 2 1

7␲ 5␲ 4 3 ␲ 3 ␲ 4

␲ 6

Exercises 9.7, pp. 907–912 1. rotation of axes; 2

B AC

3. invariants

5. Answers will vary.

2

Y X   1 9. 6  3 12  X, 6  3 12  Y 8 8 5 12 5 12  X,  Y 13. 0  x, 4  y 11. 2 2 y2 x2 3 13 3  2  x;  2 13  y 17.  xy 13  9 15. 2 2 2 2 2 2 19. 4X  2Y  9 21. a. 3X2  Y2  2 23. a. 4X2  Y2  8 b. b. y y 7.

11␲ 6

7␲ 5␲ 4 3

Y

X

Y

45 ␲ 6

vertices: √6, 0 3 foci: 2√10, 0 3 asymptotes: Y  √3X

11␲ 6 7␲ 5␲ 4 3

X

45

x

vertices: (0, 2√2) foci: (0, √6) minor axis endpoints: ( √6, 0)

x

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Student Answer Appendix

25. a. Y2  4X2  16 b. y

X

X

Y

Y

y

x

60

(5, 1)

directrix: X  1

5

ᏸ

(5, 1) 2

29. a. X  4Y  25 b. y

60

vertices: ( 5, 0)

x

7␲ 2␲ 12 3␲ 3 4 5␲ 6 11␲ (2 2 , 5␲ ) 3 6 12

␲ 12 23␲ 12

(4, 7␲ 6) 5␲ 4 4␲ 3 17␲ 12

13␲ 12 7␲ 6

11␲ 7␲ 6 5␲ 4 19␲ 3 12

13␲ 12 7␲ 6

5␲ 4 4␲ 3 17␲ 12

5␲ 12 ␲ 3 ␲ 4 ␲ 6

(1, ␲2 ) ( 59 , 0)

5␲ 12 ␲ 3 ␲ 4 ␲ 6

(2, 0) (1 13 , 3␲ 2 )

5␲ 4 4␲ 3 17␲ 12

49. ellipse 7␲ 2␲ 12 3␲ 3 4 5␲ 6 11␲ (5, ) 12

17.

y 10

10

y

3 7 4 ;  cos ;  sin  25 5 5 33. a. parabola b.   45°; 2Y2  5 c. verified 9 5Y2 35. a. circle or ellipse b.   60°; X2   2X  2 13Y  1 2 2 (ellipse) c. verified 37. f 39. g 41. h 43. parabola 45. ellipse 47. hyperbola

13␲ 12 7␲ 6

x

10 x

6 10

31. 336 7 0; hyperbola; cos122 

(2, 0)

(4, 0)

(0, 3)

15. a. Lissajous figure 1 x b. y  6 cos c sin1a b d 2 4

foci: 5√3, 0 2 minor axis 0, 5 endpoints: 2

(1, ␲2 )

(4, 0)

5 x

X

Y

5␲ 12 ␲ 3 ␲ 4 ␲ 6

2

6

2

7␲ 2␲ 12 3␲ 3 4

(0, 3)

foci: (1, 0)

asymptotes: Y  2X

5␲ 6 11␲ 12

(2, 6.25) 6

vertex: (0, 0)

foci: ( 2√5, 0)

y

(2, 6.25)

x

60

vertices: ( 4, 0)

2

11. a. power function with p  2 13. a. ellipse y2 25 x2  1 b. y  2 , x  0 b. 16 9 x

27. a. Y2  4X  0 b. y

51. r 

7␲ 2␲ 12 3␲ 3 4

␲ 12 23␲ 12

11␲ 7␲ 6 5␲ 4 19␲ 3 12

5␲ 6 11␲ 12 13␲ 12 7␲ 6

5␲ 12 ␲ 3 ␲ 4 ␲ 6

␲ 12

(⫺3, ␲)

(1, 0)

23␲ 12 11␲ 7␲ 6 5␲ 4 19␲ 3 12

(1.5, 5␲ 3)

5␲ 4 4␲ 3 17␲ 12

3.2 1  0.8 cos 

4

4

x

6

1 19. x  t, y  3t  2; x  t, y  t  2; x  cos t, y  3 cos t  2 3 21. x  t, y  1t  32 2  1; x  t  3, y  t2  1; x  tan t  3, 12k  12 y  sec2t, t  ,k 2  2 23. x  t, y  tan 1t  22  1, t  k   2, k  ; x  t  2, 2 1 y  sec2t, t  ak  b, k  ; x  tan1t  2, y  t2  1 2 25. verified 27. a.

␲ 12

23␲ 12 11␲ 7␲ 6 4 5␲ 19␲ 3 12

4 7.5 55. r  1  cos  1  1.5 sin  r 1/22 12 2 A 2 57. a. r  b.  and   2 cos   3 sin  r 102 3 B 3 59. Jupiter: e  0.0486, Saturn: e  0.0567 61. about 2757.1 million miles 63. Saturn: e  0.0567 482.36 1780.77 65. r  67. r  1  0.0486 cos 1  0.0457 cos 69. In millions of miles (approx): JS: 405.3, JU: 1298.4, JN: 2310.3, SU: 893.1, SN: 1905.0, UN: 1011.9 0.7638 0.2864 71. r  73. r  1 0.7862 cos  1 0.7862 cos  3 75. $582.45; $445.94; $881.32; $97.92 77. y  1  cos  79. verified 81. Answers will vary  83. r  12 cosa  b  612 1cos   sin 2 4 85. 425X2  416Y2  400  0 87. (0, 0), (4, 0), (4, 4), (0, 4) 89. x  29.0 91. 9.2 mph at heading 347.7° 53. r 

1. parameter 3. direction 5. Answers will vary. 7. a. parabola with vertex at (2, 1) 9. a. parabola b. y  x2  4x  3 b. y  x 2 1x  1 y

3

3

1

3

x

29. a.

b. x-intercepts none, y-intercepts none; no minimum or maximum x-values; minimum y-value is 4 and maximum y-value is 4 31. a.

Exercises 9.8, pp. 920–924

y

b. x-intercepts: t  0, x  10, y  0 and t  , x  6, y  0; y-intercepts: t  1.757, x  0, y  6.5 and t  4.527, x  0, y  6.5; minimum x-value is 8.1; maximum x-value is 10; minimum y-value is 9.5; the maximum y-value is 9.5

x

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Student Answer Appendix b. x-intercepts: t  0, x  2, y  0 and t  4.493, x  9.2, y  0; infinitely many others; y-intercepts: t  2.798, x  0, y  5.9 and t  6.121, x  0, y  12.4; infinitely many others; no minimum or maximum values for x or y 33. a.

43. The maximum value (as the graph swells to a peak) is at b 1x, y2  aa, b. The minimum value (as the graph dips to the valley) 2 b is at 1x, y2  aa, b. 2 3 3 45. a. The curve is approaching y  2 as t approaches , but cot a b is 2 2 undefined, and the trig form seems to indicate a hole at 3 t , x  0, y  2. The algebraic form does not have this 2 problem and shows a maximum defined at t  0, x  0, y  2. b. As t S q, y1t2 S 0 c. The maximum value occurs at (0, 2k). 47. a. Yes

b. no x-intercepts; y-intercept is t  0, x  0, y  2; no minimum or maximum x-values; minimum y-value is 2; maximum y-value is 4 35. a.

c.  0.82 ft

b. Yes

49. No, the kick is short.

51. The electron is moving left and downward. 6t 6 13t 21 53. at,  ,  b 55. Inconsistent, no solutions 17 17 17 17 57. x  1.22475t y  0.25t2  2t

y 1 5 x

The parametric equations fit the data very well. b. x-intercepts: t  0, x  4, y  0 and t  , x  4, y  0; 3  , x  0, y  8; y-intercepts: t  , x  0, y  8 and t  2 2 minimum and maximum x-values are approx. 5.657; minimum and maximum y-values are 8

59. Answers will vary.

5

61. by 25%

63.

y

10 8 6 4 2 654321 2 4 6 8 10

37.

1 2 3 4 x

Summary and Concept Review, pp. 924–928 1. verified (segments are perpendicular and equal length) 2. x2  1y  12 2  34 3. verified 5. 6. y 10 8 6 4 2

width 12 and length 16; including the endpoint t  2, the graph crosses itself two times from 0 to 2.

10 8 6 4 2

108642 2 4 6 8 10

39. 8.

41.

9.

y

5 4 3 2 1 54321 1 2 3 4 5

width 10 and length 14; including the endpoint t  2, the graph crosses itself nine times from 0 to 2.

10 8 6 4 2 108642 2 4 6 8 10

1 2 3 4 5 x

(3, 1)

14.

20

y2 x2  1 400 144

b. y 8 4

(2  21, 1) (2, 1)

8

4

15.

15

16.

y

9

4 12

20 x

x

15

9

3 3

20

y

12 (0, 1)

3 4

8

8

1 2 3 4 5 6 7 x

y

4 4

(7, 1)

12

20 12 4 4

2 4 6 8 10 x

2 4 6 8 10 x

y

2 1

321 1 2 3 4 5

y

y

y2 y2 x2 x2   1 11. a.  1 25 9 169 25 1y  12 2 1x  22 2   1 13. 12. 25 4 5 4

10 8 6 4 2 108642 2 4 6 8 10

2 4 6 8 10 x

10.

(2  21, 1) 3

width 20 and length 20; including the endpoint t  4, the graph crosses itself 23 times from 0 to 4.

108642 2 4 6 8 10

2 4 6 8 10 x

4. verified 7.

y

4 3

9

15 x

20 12 4 4

12

9

12

20

15

20

4

12

20 x

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Student Answer Appendix

17.

15 9

18.

y

10 8 6 4 2

(6, 1)

3 25 15 5 3

5

15

108642 2 4 6 8 10

25 x

9 15

2

2

19. a.

y x  1 225 64

1x  52 2

20.

9 y

10 8 6 (2, 2) 4 2

b.

1y  22 2



2



4 3

(4, 0)

16 3

10 8 6 4 2

23.

10

6 5 4 3 2 1

10

(4, 2)

y  2

(4, 0)

(8, 2) 15 x

10

5432 1 2 3 4 5

1 2 3 4 5 6 x

33.

8 6 4 2

4 2 6 4 2 2

2 4 6 8

1 2 3 4 5

34. 6

2

4

8 6 42 2 4 6 8

6

4 6

35. Y2  2Y  2X  6  0 7 1Y  12 2  2aX  b 2

2 4 6 8

36. 5X2  Y2  80  0 X2 Y2  1 16 80 y X

X

Y

Y 60

45

x 1

foci: ( 4√6, 0) asymptotes: Y  √5X

38. hyperbola, e  32 ;

4 3 2 1 1 2 3 4

(√6, 0) 1 2 3 4 x

2

4

6

8

108642 2 4 6 8 10

10864 2 4 6 8 10

2 4 6 8 10

39. parabola, e  1;

y

2

2

y

y

5x 3

10 8 6 4 (0, 5) 2

5432 1 2 3 4 5

y

5x 3

7. parabola; p  0.125 vertex 12.5, 27.52; focus 12.5, 27.3752; directrix y  27.625; y-intercepts: approx. 16.2, 02 , (1.2, 0)

y 4

y ,x  ,5 8 4

(5, 2)

y  ,x  ,7

2

4

2

8

12 x (7, 2)

4 (1, 2) 6 8

y  27.625 30

y

27 24 21 18 15 (0, 15) 12 9 6 (6.2, 0) 3 (1.2, 0) 876543213 1 2 3 4 x 6

y 4 3 2 1

x  1.75 (3, 0)

21 1 2 3 4 5 6

11. parabola; p  2 vertex (4, 0); focus (4, 2); directrix y  2; x-intercept (4, 0)

1 2 3 4 5 6 7 8 x

(2.25, 1)

y

5

(4, 2) (4, 0)

4 2

2 4 6 8 10 12 x

y  2 1 2 3 4 5

de with e  0.0935 and d  1501.1; focal cord: 1  e cos  280.82 million miles

40. r 

x

8

3. hyperbola; center (0, 0); a  5, b  3, c  134; vertices (0, 5), 10, 52; foci 10, 1342, 10,  1342; asymptotes y  53 x, y  53 x

5. hyperbola; center 11, 22; a  6, b  2, c  2110 vertices 15, 22, 17, 22; foci 11  2 110, 22, 11  2 110, 22; asymptotes: y  13 x  73 , y  13 x  53

5 4 3 2 1

10 8 6 4 2 2 4 6 8 10

4

4

10 x

(0, 2) 10 8 6 4 2

2 2

54321 1 2 3 4 5 x 2 4 (0, 5) 6 8 10

x

vertices: ( 4, 0)

vertex: 7 , 1 2 foci: (3, 1) y-intercepts: (0, √7) and (0, √7  1)

(√6, 0)

4 3 2 1

9. parabola; p  0.25 vertex (2, 1); focus (2.25, 1); directrix x  1.75; y-intercepts: none

ᏸ



2

y

5 4 3 2 1

654321 1 2 3 4 5 6

8 6 4 2

Y

2

1. circle, center: (0, 0); r  16

y

(0, 2)

(0, 5) (10, 5)

6 5 4 3 2 1

37. ellipse, e  23 ;

2

8

25. circle, line, (4, 3), (3, 4) 26. parabola, line, (3, 2) 27. parabola, circle, 1 13, 22, 1 13, 22 28. circle, parabola, (1, 3), (1, 3) 29. Parabola, circle 30. Circle, parabola 31. y y

y

2

4

15

10 x

(10, 5)

10

8 6 42 2 4 6 8

4

4

24.

10

2 4 6 8 10 x

32.

4

4

y5

8 (0, 3)

1 2 3 4 5 6 x

2

6

(0, 2)

(0, 0)

(6.25, 0.5) 6

654321 1 2 3 4 5 6

y 4

44. Answers will vary. 45. x  34, 4 4: y  3 8, 84

y

y

10

(0, 2)

108642 2 4

2 x

2 4 6 8 10 12 14 16

y

8

Mixed Review, pp. 929–930

y

10 8 6 (6, 0) 4 2

4 2

10

y2 x2  1 9 16

43.

108642 2 2 4 6 8 10 x 4 6 (0, 2) 8 10

2 4 6 8 101214 x 2

6

x 1 9

(8, 2)

y  3 x 

8

42. y  11 1x2 2

y

4 2

2 4 6 8 10 x

 1 21.

4

y 3x

y

16

41. y  21x  42 2  3

2

(5, 2)

642 2 4 6 8 10

22.

2

y2 x2  1 9 16

y

13. hyperbola; center (3, 3); a  5, b  2, c  129 vertices (8, 3), 12, 32 ; foci 13  129, 32, 13  129, 32,  12.39, 32, 18.39, 32 ; asymptotes y  25 x  95 , y  25 x  21 5

y  2x5  215

y

8 7 6 5 4 (≈2.4, 3) 3 2 1 (3, 3)

y  2x5  95

(≈8.4, 3)

129631 3 6 9 12 15 18 x 2

4

x

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Student Answer Appendix

17. parabola

4

2

11. parabola;   36.87°; cos   45 , sin   35

y

(2, 10)

10 8 6 4 2

3 11 2 12. Y  25 16 X  4 X  20

2

2 4 6

4

y

) ,0 0.4 (

15. ellipse; center (2, 3); a  1, b  7, c  4 13 vertices 12, 42, 12, 102 ; endpoints of minor axis 13, 32, 11, 32 ; foci 12, 3  4132, 12, 3  4132,  12, 3.932,  12, 9.932

x

(2, 4)

(0.24, 0.64)

y

13.

14.

15.

4 2 6 4 2

2 4 6 8 1012 14 16 x

2

2

4

2

2

16. ellipse;

y

54321 1

1 2 3 4 5 x

21 1 2 3 4 5 6 7 8 9 10 x 2

10

x

(2, 2)

4

(5, 5)

(2, 5)

2

y

19. a. 1313 , 513 2, 12, 02

(2, 5.6) (2, 1)

(4, 1)

(0, 1) x

(2, 3.6) (2, 4)

10 8 6 4 2 108642 2 4 6 8 10

8. parabola; vertex 11.2, 02; focus (0, 0); directrix at y  2.4

2 4 6 8 10

10. hyperbola; center 11, 22; vertices 14, 22, 16, 22; foci 11  129, 22, 11  129, 22,  14.39, 22, 16.39, 22; asymptotes: y  25 x  85 , y  25 x  12 5

48

72

x

b. a

12 8 12 8 , b, a , b A 5 A5 A5 A5 2 165411  0.967 2 8 12 8 12 a , b, a , b 20. r  A 5 A5 A5 A5 1  0.967 cos  e is very close to 1. This makes its orbit a very elongated ellipse, where the orbit of most planets is nearly circular. 21. The ball is 0.43 ft above the ground at x  165 ft, and will likely go into the goal. 22. Perihelion: 128.41 million miles Aphelion: 154.89 million miles 23. y  1x  12 2  4; D: x  R; R: y  34, q 2 ; focus: (1, 3.75) 24. 1x  12 2  1y  12 2  25; D: x  3 4, 64 ; R: y  34, 6 4 1x  22 2 1y  12 2 25.   1; D: x  35, 1 4 ; R: y  34, 6 4 9 25

Strengthening Core Skills, pp. 932–934

10 8 6 4 2 108642 2 4 6 8 10

9. hyperbola; center: 12, 32; vertices: (2, 0), 12, 62; foci: 12, 82, (2, 2); asymptotes: 3 y  34 x  92 , y  3 4 x  2

24

2 4

(2, 8) (2, 6) Focal chord

y

8 6

3 (1, 5)

6

y

18. max: y  8; min: y  0; P  8

y

4

6

10 9 8 7 6 5 4 3 2 1

2 3 4 5

1. c 2. d 3. b 4. a 5. circle; center 12, 52; radius 3

2

4

17. parabola; x  1y  52 2  1

y x  1 16 25 5 4 3 2 1

Practice Test, pp. 930–931

2 6 4 2 2

4 8 12

4 8 12

6

2 2 b. (0, 2), a2 12,  b, a2 12,  b 21. x  50 cos t; y  30 sin t 3 3 23. a. elliptic; 0.494 million miles b. parabolic; 3.1 million miles 25. 12x2  26xy  12y2  160,000

4

12 8 4

6

4

19. a. (2, 5), 12, 52, 12, 52, 12, 52

6

12 8 4

6

4

7. ellipse; center 1 40 9 , 02; vertices 80 1 10 9 , 02 , (10, 0); foci (0, 0), 1 9 , 02

x

(0.88, 0) X

16 14 12 10 8 6 4 2 42 2

6. ellipse; center 12, 12; vertices 12, 42, 12, 62; foci 12, 1  1212, 12, 1  1212

Y

5

2 4 6 8 10

Exercise 1: Yes, the calculations are much “cleaner” with the right triangle definitions.

Cumulative Review Chapters 1–9, p. 934 y

6 4 2 8642 2 4 6 8 1012 x 2 4 6 8 10 12 14

y 4 3 2 1 12 8 10 642 1 2 4 6 8 1012 x 2 3 4 5 6 7 8

1. x  7, x  1 is extraneous 3. x  6 5. x  4 5 7.  k, k  Z 6 9. x  61.98°  360°k; k  Z x  118.02°  360°k; k  Z 11. about 24.7 pesos/kg 13. The formation is 1152.4 yd wide 15. 17. horizontal asymptote: y  0 y 8 vertical asymptotes: x  3, x  3, 6 x-intercept: (2, 0); y-intercept: 10, 29 2 4 (7, 3)

x  3

2 (3, 1) 2

4

6

8

6

y x3

4

10 x

2

2 6 4 2

2 4 6

(2, 0) 2

4

6 x

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Student Answer Appendix 21. center 11, 22 ;

y

x  1

foci 11  2 110, 22  17.32, 22,

4

11  2110, 22  15.32, 22 ; asymptotes

2

(0, 0)

2

2

4

6

8 x

2

y  13 x  73 , y  13 x  53

4

y 2 16 8

8

16

x

2 4

(1, 2)

6

23.

27. 13, 42 , (3, 4),

25. 61.9° 5 4 3 2 1

13, 42, 13, 42

29.

54321 1 2 3 4 5 2 3 4 5

1 3 2x  1  2 2 x x x 1

Connections to Calculus Exercises, pp. 937–938 1. a.  3. a.

cos 122  cos  sin 122  sin 

b. 1, 1

cos122  sin122

b. 1, 1

sin122  cos122

 5 c. , 3 3

2 4 d. 0, , 3 3

 5 9 13 , , , 8 8 8 8

c.

 3 5 7 , , , ; verified 4 4 4 4   5 3  7. four-leave rose, circle; , , , ; verified 9. (r, ): a0, b, 6 2 6 2 2  5 3 a213, b, a2 13, b, a0, b; (x, y): (0, 0), 13, 132 , 13,  132 6 6 2 d.

3 7 11 15 , , , 8 8 8 8

5.

CHAPTER 10 Exercises 10.1, pp. 946–948 1. pattern; order 3. increasing 5. formula defining the sequence uses the preceding term(s); answers will vary. 7. 1, 3, 5, 7; a8  15; a12  23 9. 0, 9, 24, 45; a8  189; a12  429 11. 1, 2, 3, 4; 1 2 3 4 8 12 a8  8; a12  12 13. , , , ; a8  ; a12  2 3 4 5 9 13 1 1 1 1 1 1 1 1 1 1 1 15. , , , ; a8  ; a12  17. 1, , , ; a8  ; a12  2 4 8 16 256 4096 2 3 4 8 12 1 1 1 1 1 1 , , , ; a8  ; a12  19. 2 6 12 20 72 156 1 21. 2, 4, 8, 16; a8  256; a12  4096 23. 79 25. 51 27. 32 11 10 1 29. 1 10 2 31. 36 33. 2, 7, 32, 157, 782 35. 1, 4, 19, 364, 132, 499 37. 64, 32, 16, 8, 4 39. 336 41. 36 43. 28 45. 12 , 13 , 14 , 15 1 1 47. 13 , 120 49. 1, 2, 92 , 32 51. 15 53. 64 55. 137 , 15,1120 , 3,991,680 3 60 57. 10

59. 95

61. 4

63. 15

65. 50

67.

27 112

5

69.

 14n2

n1 6

7 n2 n 112 n 73. 1n  32 75. 71. 77. 79. 35 n n1 n1 n1 3 n3 2 n1 81. 100 83. 35 85. an  600010.82 ; 6000, 4800, 3840, 3072, 2457.60, 1966.08 87. 5.20, 5.70, 6.20, 6.70, 7.20, $13,824 89. 2690 3 7 , 91. verified 93. approaches 12 95. 4 4 97. A  53.1°, B  90°, C  36.9°



5

n 2

3







Exercises 10.2, pp. 953–955 1. common; difference

3.

n1a1  an 2

; nth 5. Answers will vary. 2 7. arithmetic; d  3 9. arithmetic; d  2.5 1 11. not arithmetic; all prime 13. arithmetic; d  24  2 15. not arithmetic; an  n 17. arithmetic; d  6 19. 2, 5, 8, 11 21. 7, 5, 3, 1 23. 0.3, 0.33, 0.36, 0.39 25. 32 , 2, 52 , 3 27. 34 , 58 , 12 , 38

29. 2, 5, 8, 11 31. a1  2, d  5, an  5n  3, a6  27, a10  47, a12  57 33. a1  5.10, d  0.15, an  0.15n  4.95, a6  5.85, a10  6.45, a12  6.75 33 39 35. a1  32 , d  34 , an  34 n  34 , a6  21 4 , a10  4 , a12  4 37. 61 39. 1 41. 2.425 43. 9 45. 43 47. 21 49. 26 472 51. d  3, a1  1 53. d  0.375, a1  0.65 55. d  115 126 , a1  63 57. 1275 59. 601.25 61. 534 63. 82.5 65. 74.04 67. 21022 69. S6  21; S75  2850 71. at 11 P.M. 73. 5.5 in.; 54.25 in. 75. 220; 2520; yes 77. a. linear function b. quadratic 1 1 13 79. A  7, P  6, HS: unit right, VS: 10 units up, PI:  t 6 . 2 2 2 81. f 1x2  4ax  972, 1364

Exercises 10.3, pp. 962–965 1. multiplying 3. a1r n1 5. Answers will vary. 7. r  2 9. r  2 11. an  n2  1 13. r  0.1 15. not geometric; ratio of terms decreases by 1 17. r  25 19. r  12 21. r  4x 240 3 25. 5, 10, 20, 40 27. 6, 3, 3 2 ,4 n! 29. 4, 4 13, 12, 12 13 31. 0.1, 0.01, 0.001, 0.0001 33. 38 35. 25 4 1 1 37. 16 39. a1  27 , r  3, an  27 132 n1, a6  9, a10  729, a12  6561 1 1 41. a1  729, r  13 , an  7291 13 2 n1, a6  3, a10  27 , a12  243 1 1 n1 43. a1  2 , r  12, an  2 1 122 , a6  212, a10  8 12, a12  1612 45. a1  0.2, r  0.4, an  0.210.42 n1 a6  0.002048, a10  0.0000524288, a12  0.000008388608 47. 5 49. 11 51. 9 53. 8 55. 13 57. 9 59. r  23 , a1  729 32 61. r  32 , a1  243 63. r  32 , a1  256 65. 10,920 81 3872 67. 27  143.41 69. 2059 71. 728 73. 85 8  257.375 8  10.625 75. 1.60 77. 1364 79. 31,525 81. 387 2187  14.41 512  0.76 83. 521 85. 3367 87. 14  15 12 89. no 91. 27 93. 125 25 1296 7 3 95. 12 97. 4 99. 31 101. 32 103. 18 105. 1296 5 107. about 6.3 ft; 120 ft 109. $18,841.60; 10 yr 111. 125.4 gpm; 10 months 113. about 347.7 million 115. 51,200 bacteria; 12 half-hours later (6 hr) 117. 0.42 m; 8 m 119. 35.9 in3; 7 strokes 121. 6 yr 123. Sn  log n! 5 111 125. x   i 127. y 10 2 2 8 23. not geometric; an 

6 4 2

108642 2 4 6 8 10

2 4 6 8 10 x

Exercises 10.4, pp. 971–973 1. finite; universally 3. induction; hypothesis 5. Answers will vary. 7. an  10n  6 a4  10142  6  40  6  34; a5  10152  6  50  6  44; ak  10k  6; ak1  101k  12  6  10k  10  6  10k  4 9. an  n a4  4; a5  5 ; ak  k; ak1  k  1 11. an  2n1 a4  241  23  8; a5  251  24  16; ak  2k1; ak1  2k11  2k 13. Sn  n15n  12 S4  415142  12  4120  12  41192  76; S5  515152  12  5125  12  51242  120; Sk  k15k  12; Sk1  1k  12151k  12  12  1k  1215k  5  12  1k  1215k  42

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Student Answer Appendix

15. Sn  S4  S5  Sk 

n1n  12 2 414  12 2 515  12 2 k1k  12

Sk1 

 

4152 2 5162 2

 10;

5  10  15 

 15;

; 2 1k  121k  1  12 2

29. 1. Show Sn is true for n  1. 511211  12 5122 S1   5 2 2 2. Assume Sk is true:



1k  121k  22

2 17. Sn  2n  1 S4  24  1  16  1  15; S5  25  1  32  1  31; Sk  2k  1; Sk1  2k1  1 19. an  10n  6; Sn  n15n  12 S4  415142  12  4120  12  41192  76; a5  10152  6  50  6  44; S5  515152  12  5125  12  51242  120; S4  a5  S5 76  44  120 120  120 Verified n1n  12 21. an  n; Sn  2 414  12 4152 S4    10; 2 2 a5  5 ; 515  12 5162 S5    15; 2 2 S4  a5  S5 10  5  15 15  15 Verified 23. an  2n1; Sn  2n  1 S4  24  1  16  1  15; a5  251  24  16; S5  25  1  32  1  31; S4  a5  S5 15  16  31 31  31 Verified 25. an  n3; Sn  11  2  3  4  # # #  n2 2 S1  12  13 S5  11  2  3  4  52 2  152  225 1  8  27  64  125  225 S9  11  2  # # #  92 2  452  2025 1  8  # # #  729  2025 n1n  12 2 n2 1n  12 2 c d  2 4 27. 1. Show Sn is true for n  1. S1  111  12  1122  2 Verified 2. Assume Sk is true: 2  4  6  8  10  # # #  2k  k1k  12 and use it to show the truth of Sk1 follows. That is: 2  4  6  # # #  2k  21k  12  1k  121k  22 Sk  ak1  Sk1 Working with the left hand side: 2  4  6  # # #  2k  21k  12  k1k  12  21k  12  k2  k  2k  2  k2  3k  2  1k  121k  22  Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n.

31. 1. 2.

33. 1.

2.

35. 1.

# # #

 5k 

SA59

Verified 5k1k  12

2 and use it to show the truth of Sk1 follows. That is: 51k  121k  1  12 5  10  15  # # #  5k  51k  12  2 Sk  ak1  Sk1 Working with the left hand side: 5  10  15  # # #  5k  51k  12 5k1k  12   51k  12 2 5k1k  12  101k  12  2 1k  1215k  102  2 51k  121k  22  2  Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n. Show Sn is true for n  1. S1  112112  32  5 Verified Assume Sk is true: 5  9  13  17  # # #  4k  1  k12k  32 and use it to show the truth of Sk1 follows. That is: 5  9  13  17  # # #  4k  1  41k  12  1  1k  12121k  12  32 Sk  ak1  Sk1 Working with the left hand side: 5  9  13  17  # # #  4k  1  4k  5  k12k  32  4k  5  2k2  3k  4k  5  2k2  7k  5  1k  1212k  52  Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n. Show Sn is true for n  1. 3131  12 313  12 3122    3 Verified S1  2 2 2 Assume Sk is true: 313k  12 3  9  27  # # #  3k  2 and use it to show the truth of Sk1 follows. That is: 3  9  27  # # #  3k  3k1 313k1  12  2 Sk  ak1  Sk1 Working with the left hand side: 3  9  27  # # #  3k  3k1 313k  12   3k1 2 313k  12  213k1 2  2 3k1  3  213k1 2  2 313k1 2  3  2 313k1  12  2  Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n. Show Sn is true for n  1. Sn  2n1  2 S1  211  2  22  2  4  2  2 Verified

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Student Answer Appendix

2. Assume Sk is true: 2  4  8  # # #  2k  2k1  2 and use it to show the truth of Sk1 follows. That is: 2  4  8  # # #  2k  2k1  2k1  2 Sk  ak1  Sk1 Working with the left hand side: 2  4  8  # # #  2k  2k1  2k1  2  2k1  212k1 2  2  2k2  2  Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n. 37. 1. Show Sn is true for n  1. n Sn  2n  1 1 1 1 S1    Verified 2112  1 21 3 2. Assume Sk is true: 1 1 1 1 k    # # #   3 15 35 12k  1212k  12 2k  1 and use it to show the truth of Sk1 follows. That is: 1 1 1 1    # # #  3 15 35 12k  1212k  12 1 k1   121k  12  12121k  12  12 21k  12  1 Sk  ak1  Sk1 Working with the left hand side: 1 1 1 1 1    # # #   3 15 35 12k  1212k  12 12k  12 12k  32 k 1   2k  1 12k  1212k  32 k12k  32  1  12k  1212k  32 2k2  3k  1  12k  1212k  32 12k  121k  12  12k  1212k  32 k1  2k  3  Sk1 Since the truth of Sk1 follows from Sk, the formula is true for all n. 39. 1. Show Sn is true for n  1. S1 : 31  2112  1 321 3  3 Verified 2. Assume Sk : 3k  2k  1 is true and use it to show the truth of Sk1 follows. That is: 3k1  2k  3. Working with the left hand side: 3k1  313k 2  312k  12  6k  3 Since k is a positive integer, 6k  3  2k  3 Showing Sk1 : 3k1  2k  3 Verified 41. 1. Show Sn is true for n  1. S1 : 3 # 411  41  1 3 # 40  4  1 3#13 3  3 Verified 2. Assume Sk : 3 # 4k1  4k  1 is true and use it to show the truth of Sk1 follows. That is: 3 # 4k  4k1  1. Working with the left hand side: 3 # 4k  3 # 414k1 2  4 # 314k1 2  414k  12  4k1  4 Since k is a positive integer, 4k1  4  4k1  1 Showing that 3 # 4k  4k1  1

43. n2  7n is divisible by 2 1. Show Sn is true for n  1. Sn : n2  7n  2m S1 : 112 2  7112  2m 1  7  2m  6  2m Verified 2. Assume Sk : k2  7k  2m for m  Z and use it to show the truth of Sk1 follows. That is: 1k  12 2  71k  12  2p for p  Z . Working with the left hand side:  1k  12 2  71k  12  k2  2k  1  7k  7  k2  7k  2k  6  2m  2k  6  21m  k  32 is divisible by 2. 45. n3  3n2  2n is divisible by 3 1. Show Sn is true for n  1. Sn : n3  3n2  2n  3m S1 : 112 3  3112 2  2112  3m 1  3  2  3m 6  3m 2m Verified 2. Assume Sk : k3  3k2  2k  3m for m  Z and use it to show the truth of Sk1 follows. That is: Sk1 : 1k  12 3  31k  12 2  21k  12  3p for p  Z . Working with the left hand side: 1k  12 3  31k  12 2  21k  12 is true.  k3  3k2  3k  1  31k2  2k  12  2k  2  k3  3k2  2k  31k2  2k  12  3k  3  k3  3k2  2k  31k2  2k  12  31k  12  3m  31k2  2k  12  31k  12 is divisible by 3. 47. 6n  1 is divisible by 5 1. Show Sn is true for n  1. Sn : 6n  1  5m S1 : 61  1  5m 6  1  5m 5  5m 1  m Verified 2. Assume Sk : 6k  1  5m for m  Z and use it to show the truth of Sk1 follows. That is: Sk1 : 6k1  1  5p for p  Z . Working with the left hand side:  6k  1  616k 2  1  615m  12  1  30m  6  1  30m  5  516m  12 is divisible by 5, Verified 49. Verified 51. Verified 53. 1x  32 2  1x  42 2  25

Mid-Chapter Check, pp. 973–974 1. 3, 10, 17, a9  59

2. 4, 7, 12, a9  84

3. 1, 3, 5, a9  17

6

4. 360

5.

 13k  22

6. d

7. e

8. a

9. b

10. c

k1

11. a. a1  2, d  3, an  3n  1 b. a1  d  an  34 n  34 12. n  25, S25  950 13. n  16, S16  128 14. S10  5 29,524 15. S10  16. a. a1  2, r  3, an  2132 n1 27 b. a1  12 , r  12 , an  1 12 2 n 17. n  8, S8  1640 18. 343 19. 1785 27 6 20. 4.5 ft; 127.9 ft 3 2,

3 4,

Reinforcing Basic Concepts, p. 974 Exercise 1: $71,500

Exercises 10.5, pp. 982–986 1. experiment; well-defined 7. a. 16 possible

3. distinguishable

W

W

X

X

Y

Z

W

5. Answers will vary.

Begin

X

Y

Y

Z

W

X

Z

Y

Z

W

X

Y

Z

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Student Answer Appendix b. WW, WX, WY, WZ, XW, XX, XY, XZ, YW, YX, YY, YZ, ZW, ZX, ZY, ZZ 9. 32 11. 15,625 13. 2,704,000 15. a. 59,049 b. 15,120 17. 360 if double veggies are not allowed, 432 if double veggies are allowed. 19. a. 120 b. 625 c. 12 21. 24 23. 4 25. 120 27. 6 29. 720 31. 3024 33. 40,320 35. 6; 3 37. 90 39. 336 41. a. 720 b. 120 c. 24 43. 360 45. 60 47. 60 49. 120 51. 30 53. 60, BANANA 55. 126 57. 56 59. 1 61. verified 63. verified 65. 495 67. 364 69. 252 71. 40,320 73. 336 75. 15,504 77. 70 79. a. 1.2% b. 0.83% 81. 7776 83. 324 85. 800 87. 6,272,000,000 89. 518,400 91. 357,696 93. 6720 95. 8 97. 10,080 99. 5040 101. 2880 103. 5005 105. 720 10! 9! 11! 8! 107. 52,650, no 109. a. b. c. d. 2!3!5! 2!3!4! 4!5!2! 2!3!3! 111. 113. cos 152 y 10 8 6 4 2

108642 2 4 6 8 10

2 4 6 8 10 x

Exercises 10.6, pp. 993–999 1. n1E2 3. 0; 1; 1; 0 5. Answers will vary. 7. S  5HH, HT, TH, TT6, 14 9. S  {coach of Patriots, Cougars, Angels, Sharks, Eagles, Stars}, 16 1 1 11. P1E2  49 13. a. 13 b. 14 c. 21 d. 26 1 5 3 15. P1E1 2  8 , P1E2 2  8 , P1E3 2  4 17. a. 43 b. 1 c. 41 d. 12 1 19. 34 21. 67 23. 0.991 25. a. 12 b. 11 c. 98 d. 56 27. 10 12 21 60 29. 143 31. b, about 12% 33. a. 0.3651 b. 0.3651 c. 0.3969 7 7 35. 0.9 37. 24 39. 0.59 41. a. 61 b. 36 c. 91 d. 49 2 9 2 43. a. 25 b. 50 c. 0 d. 25 e. 1 45. 34 47. 11 15 1 1 5 1 49. a. 18 b. 29 c. 98 d. 34 e. 36 f. 12 51. 14 ; 256 ; answers will vary. 53. a. 0.33 b. 0.67 c. 1 d. 0 e. 0.67 f. 0.08 1 1 9 1 5 55. a. 2 b. 2 c. 0.2165 57. a. 16 b. 14 c. 16 d. 16 3 3 1 9 2 11 1 1 3 59. a. 26 b. 26 c. 13 d. 26 e. 13 f. 26 61. a. 8 b. 16 c. 16 47 2 3 9 11 5 8 1 63. a. 100 b. 25 c. 100 d. 50 e. 100 65. a. 429 b. 2145 67. 3360 1 69. 1,048,576 ; answers will vary; 20 heads in a row. 1 3 2 12 1 71. sin   , cos    , tan    , sec    , 3 3 2 12 2 12 840 41 840 cot   2 12 73. sin 122   , cos 122  , tan 122   841 841 41

Exercises 10.7, pp. 1005–1006 5 1. one 3. 1a  12b22 5. Answers will vary. 5 4 3 2 7. x  5x y  10x y  10x2y3  5xy4  y5 9. 16x4  96x3  216x2  216x  81 11. 41  38i 13. 35 17. 1140 19. 9880 21. 1 23. 1 25. c5  5c4d  10c3d2  10c2d3  5cd4  d5 27. a6  6a5b  15a4b2  20a3b3  15a2b4  6ab5  b6 29. 16x4  96x3  216x2  216x  81 31. 11  2i 33. x9  18x8y  144x7y2  p

15. 10

20 2 35. v24  6v22w  33 37. 35x4y3 39. 1792p2 41. 264x2y10 2v w 43. 0.25 45 a. 17.8% b. 23.0% 47 a.  0.88% b.  6.9% c.  99.0% d.  61.0% 49. a. 99.33% b. 94.22% 6 6 6 6 6 6 6 51. a b  a b  1; a b  a b  6; a b  a b  15; a b  20 6 0 5 1 4 2 3 53. 55. g1x2 7 0: x  12, 02 ´ 13, q2 y 10 8 6 4 2

108642 2 4 6 8 10

f(3)  1

2 4 6 8 10 x

10 8 6 4 (2, 0) 2 108642 2 4 6 8 10

y

(0, 0) (3, 0) 2 4 6 8 10 x

Summary and Concept Review, pp. 1007–1011 5 11 1. 1, 6, 11, 16; a10  46 2. 1, 35 , 25 , 17 ; a10  101 3. an  n4; a6  1296 4. an  17  1n  12132; a6  2 6. 112 7. 140 8. 35 9. not defined, 2, 6, 12, 20, 30

5.

255 256

7

10. 12 , 34 , 54 , 94 , 17 4

11.

 1n

2

 3n  22; 210

n1

an  2  31n  12; 119 13. an  3  1221n  12; 65 740 15. 1335 16. 630 17. 11.25 18. 875 19. 3240 3645 21. 32 22. 2401 23. 10.75 24. 6560 25. 819 512 does not exist 27. 50 28. 4 29. 63,050 30. does not exist 9 6561 5 32. a9  $36,980; S9  $314,900 33. 7111.1 ft3 a9  2105 credit hrs; S9  14,673 credit hours 35. verified 111  12 (1) Show Sn is true for n  1: S1   1✓ 2 (2) Assume Sk is true: k1k  12 1  2  3 p k  2 Use it to show the truth of Sk1: 1k  121k  22 1  2  3  p  k  1k  12  2 left-hand side: 1  2  3  p  k  1k  12 12. 14. 20. 26. 31. 34.

 

k1k  12

21k  12 k1k  12  21k  12   2 2 2 1k  121k  22 2

36. verified (1) Show Sn is true n  1: S1  (2) Assume Sk is true: 1  4  9  p  k2 

1 32112  1 4 11  12 6

k12k  12 1k  12 6

Use it to show the truth of Sk1: 1  4  9  p  k2  1k  12 2 

1k  1212k  321k  22

left-hand side: 1  4  9  p  k2  1k  12 2 

k1k  12 12k  12



 1✓

61k  12 2



6

1k  12 3 12k2  k  6k  64

6 6 6 1k  1212k2  7k  62 1k  1212k  321k  22   6 6 37. verified (1) Show Sn is true for n  1: S1: 41  3112  1✓ (2) Assume Sk is true: 4k  3k  1 Use it to show the truth of Sk1: 4k1  31k  12  1  3k  4 left-hand side: 4k1  414k 2  413k  12  12k  4 Since k is a positive integer, 12k  4  3k  4 showing 4k1  3k  4 38. verified (1) Show Sn is true for n  1: S1: 6 # 711  71  1✓ (2) Assume Sk is true: 6 # 7k1  7k  1 Use it to show the truth of Sk1: 6 # 7k  7k1  1 left-hand side: 6 # 7k  7 # 6 # 7k1  7 # 7k  1  7k1  1 39. verified (1) Show Sn is true for n  1: S1: 31  1  2 or 2112 ✓ (2) Assume Sk is true: 3k  1  2p for p  Z Use it to show the truth of Sk1: 3k1  1  2q for q  Z left-hand side: 3k1  1  3 # 3k  1  3 # 2p  213p2  2q is divisible by 2

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Student Answer Appendix

40. 6 ways

Cumulative Review Chapters 1–10, pp. 1016–1018

Begin

A

B

1. a. 23 cards are assembled each hour. c. y  23x  155 d.  6:45 A.M.

C

3. B

C

A

C

A

B

C

B

C

A

B

A

41. 720; 1000 42. 24 43. 220 44. 32 45. a. 5040 b. 840 c. 35 46. a. 720 b. 120 c. 24 47. 3360 48. a. 220 b. 1320 4 3 7 49. 13 50. 13 51. 56 52. 24 53. 175 54. a. 0.608 b. 0.392 396 c. 1 d. 0 e. 0.928 f. 0.178 55. a. 21 b. 56 56. a. x4  4x3y  6x2y2  4xy3  y4 b. 41  38i 57. a. a8  813a7  84a6  16813a5 b. 78,125a7  218,750a6b  262,500a5b2  175,000a4b3 58. a. 280x4y3 b. 64,064a5b9

b. an  4

7. d. g. k. 9.

A

B

C

B

C

A

C

A

B

C

B

C

A

B

A

b. ABC, ACB, BAC, BCA, CAB, CBA 12. 302,400 13. 64 14. 720, 120, 20 15. 900,900 16. 302,400 17. a. x4  8x3y  24x2y2  32xy3  16y4 b. 4 18. a. x10  1012 x9  90x8 b. a8  16a7b3  112a6b6 5 5 7 19. 0.989 20. a. 41 b. 12 c. 31 d. 12 e. 12 f. 14 g. 12 h. 0 21. a. 0.08 b. 0.92 c. 1 d. 0 e. 0.95 f. 0.03 22. a. 0.1875 59 b. 0.589 c. 0.4015 d. 0.2945 e. 0.4110 f. 0.2055 23. a. 100 53 13 47 b. 100 c. 100 d. 100 24. a. 0.8075 b. 0.0075 c. 0.9925 25. verified

Strengthening Core Skills, pp. 1015–1016 Exercise 1. Exercise 2. Exercise 3. Exercise 4.

4C1

# 13C5  40 52C5

4#

13C3

# 39C2

52C5

4

# 13C4 # 39C1 52C5

4#

10C5

52C5

 0.001 970

 0.326 170  0.042 917

 0.000 388

 6

 4

 3

 2

2 3

y

1

13 2

12 2

1 2

0



1 2

5 6 

13 2

 1

D: x  3 3, 3 4 R: y  32, 24 ´ 11, 22 ´ 34, 94

y

10 8 6 4 2 54321 2 4 6 8 10

1 2 3 4 5 x

11. a. 4x  2h  3 13.

15 12 9 6 3 108642 3 6 9 12 15

Practice Test, pp. 1013–1014 8 1. a. 12 , 45 , 1, 87 ; a8  16 b. 6, 12, 20, 30; a8  90, a12  182 11 , a12  5 c. 3, 212, 17, 16; a8  12, a12  i 12 311 2. a. 165 b. 420 c. 2343 d. 7 512 3. a. a1  7, d  3, an  10  3n b. a1  8, d  2, an  2n  10 c. a1  4, r  2, an  4122 n1 d. a1  10, r  25 , an  101 25 2 n1 4. a. 199 b. 9 c. 34 d. 6 5. a. 1712 b. 2183 c. 2188 d. 12 6. a. 8.82 ft b. 72.4 ft 7. $6756.57 8. $22,185.27 9. verified 10. verified 11. a. Begin

0

5  1109 ; x  0.91; x  2.57 6 a. x  0 b. x  11, 02 c. x  1q, 12 ´ 10, q2 x  1q, 12 ´ 11, 12 e. x  11, q 2 f. y  3 at (1, 3) none h. x  2.3, 0.4, 2 i. g142  0.25 j. does not exist q l. 0 m. x  1q, 12 ´ 11, q 2

c. an  n!

d. arithmetic e. geometric 1 f. geometric g. arithmetic h. geometric i. an  3. 27,600 2n 5 5. 0.1, 0.5, 2.5, 12.5, 62.5; a15  610,351,562.5 7. 6 9. a. 2 b. 200 c. 210 11. a. a20  20a19b  190a18b2 b. 190a2b18  20ab19  b20 4 c. 52,360a31b4 d. 4.6  1018 13. verified 15. 0.01659 17. 11 2 2 2 19. 10, 2, 5 , 25 , 125

x

5. x 

Mixed Review, pp. 1011–1012 1. a. arithmetic

b. 184 cards

b.

1 1x  h  221x  22

y

2 4 6 8 10 x

15. a. x3  125 b. e5  2x  1 17. a. x  3.19 b. x  334 19. (5, 10, 15) 21. (3, 3); (7, 3), (1, 3); (3  213, 3), (3  213, 3) 16  12 4 25. 1333 27. a. 7.0% b. 91.9% 23. a. verified

d. a

b.

c. 98.9%

12 b10.042 12 10.962 0; virtually nil 0

29. cos122  cos2  sin2  1  2 sin2  2 cos2  1; 1  1  2 sin2 2 1  sin2 4 1   sin  2  5 7 13 , ,  , 6 6 6 6

Connections to Calculus Exercises, p. 1021 4

4

i1

i1

LW   f 1i2 112  1  f 1i2

1. a. y 7 6 5 4 3 2 1

4

1

sum of rectangles

 f 1i2  1 3 f 112  f 122  f 132  f 142 4

i1

definition of summation

 5  4  3  2 evaluate  14 units2 result 8 1 1 1 1 8 LW  f a ib a b  f a ib sum of rectangles b. 2 2 2 i1 2 i1 y 7 1 1 8 1 1 3 5 6 f a ib  c f a b  f 112  f a b  f 122  f a b 5 2 i1 2 2 2 2 2 4 3 2 7 definition of 1  f 132  f a b  f 142 d summation 1 2 3 4 5 6 7 x 2 1 2 3 4 5 6 7 x









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Student Answer Appendix 1 11 9 7 5  c  5   4   3   2d evaluate 2 2 2 2 2 1 2  3 30 4  15 units result 2 2 True area is 16 units , more rectangles S better estimate. 3. a. Since the interval [0, 4] is 4 units wide and we’re using 32 subintervals of equal length, the width of each interval (the width of each rectan4 gle) will be 32  18 . The length of each rectangle is determined by a point of the graph of f 1x2  x  6, so the length of the first rectangle is f 1 18 2, the second length is f 1 28 2, the third is f 1 38 2, and so on up to the 32nd rectangle. Since A  LW, we multiply each length f 1 18 i2 by width 18 and sum the areas of all such rectangles. Using i for a counter, this can be written as 32 32 1 1 1 A LW  f a iba b. Since all lengths are multiplied by 8 8 i1 i1 8 (the counter i does not affect the constant 18), we can factor out this term 1 1 32 1 a i  6b. and evaluate f 1x2  x  6 at x  i. The result is 8 8 i1 8



1. 3. 5. 7.



32







Exercises 11.1, pp. 1031–1034 1. infinity

3. left-hand; right-hand; greater 5. Answers will vary. 4 7. lim Vn  r3 9. lim e f 1t2  0 nSq 3 tSq 1 11. lim cos a b  0 13. 500 sides xSq x 15. 425 sides 17. lim st  5r tS5  x3 1 19. lim tan1 3 g1x2 4  21. lim 2  xSa 3 xS3 x  9 6 23. As x approaches , p(x) approaches 2: lim p1x2  2 xS 25. As x approaches 2, v(x) approaches 1 1 : lim v1x2  4 xS2 4 27. As x approaches 0, s(x) approaches 0: lim s1x2  0 xS0 2x2  7x  6 x2 29. R1x2  • sin1x  22 1 x2 1 1 31. As x approaches 2, f (x) approaches : lim f 1x2  2 xS2 2 33. As x approaches 1, g(x) approaches 4: lim g1x2  4 xS1

35. As x approaches 1, f (x) approaches 0: lim f 1x2  0 xS1 37. 27 39. 24 41. 1 43. lim Ix  3 cos2 1R1  R2 2 45. lim f  L 47. a. 1 b. 1 xS3

xSm

11 11 b.  53. a. 0 b. 0 2 2 dne dne B 55. a. 43 b. 13 c. A LHRH B 57. a. 1 b. 1 c. A LHRH 51. a. 

12 12 12 59. a. b. c. 2 2 2 L

69. 0

f (x)

x

f (x)

0.2510

0.5

0.2510

0.4

0.1610

0.4

0.1610

0.3

0.0910

0.3

0.0910

0.2

0.0410

0.2

0.0410

0.1

0.0110

0.1

0.0110

0.01

0.0011

0.01

0.0011

0.001

0.001001

0.001

0.001001 11.

61. A dne q B

63. 

65. A dne q B

71. 1x  521x  221x  12 13x  12

73. 19.90 in2

yx

x



CHAPTER 11

49. a. 2 b. 2

x

9.



32 1 1 32 1 c i 6 d factor  from first summation 8 8 i1 8 i1 1 1 322  32 b  61322d apply summation formulas  c a 8 8 2 1 1056  a b 24 distribute and simplify 64 2  8.25  24  15.75 result For n  32 the approximate area under the graph is 15.75 units2, even closer to the known area of 16 units2.



b. limit is actually 0.001

0.5

32

1 1 1 1 a i  6b  c a ib  6d summation properties (distribute) 8 i1 8 8 i1 8 i1

67. A dne B

sum; limits root; n th; f 1x2 7 0 Answers will vary. a. limit appears to be 0



32

b.

Exercises 11.2, pp. 1041–1043

1x  2 10 x

x

yx

1x  2 10 x

2.7

2.7017

2.99

2.991

2.8

2.8014

2.999

3

2.9

2.9012

2.9999

3.0009

3

—

3

—

3.1

3.1008

3.0001

3.0011

3.2

3.2007

3.001

3.002

3.3

3.3006

3.01

3.011

b. 3.001 15. 9 17. 8 19. 2 21. 0 23. 7 25. 1 7 27. 9 29. 8 31. 0 33. 14 35. 16 37. 39. 0 41. 64 43. 36 2 3 1 45. 69 47. 29 49. 9 51. 53.  55. 216 57. 15 59. 23 5 4 17 dne B 73. 6 61. 1 63.  65. A dne 67. 9 69. A dne 71. A LHRH q B q B 4 3 dne B 77. 31, 4 79. verified 75. A LHRH 2 13. a. 3

Mid-Chapter Check, pp. 1043–1044 6x2  3

 0 2. a. 1 b. 0 3. 49 per year (almost weekly) 12x3  x  1 2 6x  19x  7 7 x 2x  7 2 dne B 6. dne 4. F1x2  µ 5. a. 1 b. 1 c. A LHRH AqB 7 23 x 2 2 1.

lim

xSq

7. a.

b. Answers may vary. x

 y  cos a b x

0.1

1

0.01

1

0.001

1

0

—

0.001

1

0.01

1

0.1

1

8.

1 2

3 9. 1 4

10.

dne B A LHRH

Exercises 11.3, pp. 1053–1055 1. asymptotic; removable; jump 3. direct substitution 5. Answers may vary. 7. not continuous, condition 1 is violated 9. continuous 11. not continuous, condition 2 is violated 13. continuous 15. not continuous, condition 3 is violated 17. 36 19. Direct substitution not possible, 5 is not in the domain.

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21.

1 2

Student Answer Appendix

23. Direct substitution not possible, 1 is not in the domain.

25. 155 39. 49. 61. 73. 79.

27. 

3

1 2

29. 4

31. 9

1 4

33.

37. 4x  1

35. 0

1 1 43. 31x  22 2 45. 47. 3 2 2 1x  2 1 51. 0 53. 55. A dne 57. 0 59. x  1000 A dne q B q B 2 213 3 63. 0 65. 2 67. 69. 3 71. 3 8 7 not possible since f(2) not defined 75. 3 77. 0 1 81. not possible, lim  g1x2 does not exist 1x  22 2

85. 3

xS0

87. 3

2  2 110 b. x  0, x  8 c. x  2 3 91. A  55°, C  90°, b  9.6 cm, c  16.7 cm 89. a. x 

1. difference 3. rectangles 5. Answers may vary. 7. f 1t2  88.2  9.8t 9. g 1t2  78.4  9.8t 11. 896.9 m 13. d 1t2  9.8t 15. d 1t2  32t 1 0.6 1 17. f 1x2  19. f 1x2  3x2 21. p 1t2  23. b 1t2   2 1t 2 1t 2 1 25 25. f 1x2  27. h 1x2  29.  2 1x  12 2 1x  52 2 31. 1 33. 9 units2 35. 15 units2 37. 9 units2 39. 15 units2 4 n 1 4 2 c a ib  3 d n i1 2 n n 4 n 1 4 2  c 3 d summation properties (distribute) a ib  n i1 2 n i1 n n 4 1 16 2  c i  3 d simplify n 2 i1 n2 i1











4 16 n 2 i  c n 2n2 i1



n



3d

factor

i1 2

16 n2

from first summation

4 8 2n3  3n  n c a b  3n d apply summation formula n n2 6 4 32 2n3  3n2  n  c 3a b  12 d distribute 6 n n 32 2n3  3n2  n  c a b  12 d rewrite denominators 6 n3 

16 3 1  a2   2 b  12 decompose rational expression 3 n n 3 16 1 lim a2   2 b  lim 12, and Applying the limit properties gives 3 nSq n nSq n the area under the curve is a

16 68 b 122  12  units2. The new employee 3 3 has produced 22 complete parts.



LW

 f a n ib a n b 6

6

area formula, rectangle method

i1 n



6 6 f a ib n i1 n



6 n 1 6 2 6 c  a ib  4a ib d n i1 2 n n



n 6 6 1 n 6 2 a ib  4 a ib d c n 2 i1 n i1 n



n 6 1 n 36 2 6 c i 4 id 2 n 2 i1 n i1 n



factor

6 n



 





3

n

6 evaluate f at i n distribute summation

simplify

36 n2

and

6 n

2n3  3n2  n 144 n2  n b 2 a b 6 2 n

distribute

6 n

108 2n3  3n2  n 144 n2  n a b a b rewrite 3 6 2 n n2 denominators decompose rational 3 1 1  18 a2   2 b  72 a1  b expression n n n 3 1 As n S q, , 2 S 0, and the area is 1182 122  72  36 units2. n n 45. x  29.87 47. x  5,  13i

Summary and Concept Review, pp. 1066–1068 dne B A LH RH

2. 2

3. 1

4.

dne B A LH RH

2 1 1 5. 38 6. 7.  8. 3 32 2 9. a. x  3, b. x  2, 1, 3, 4, c. x  1, 2 10. 3 11. 1 12. 2 13. A dne B 14. A dne B 15. 2 16. f 1x2  2x  5 q

LH RH

1 1 18. v 1x2  12x  1 1x  32 2 19. mtan  2x  3; at x  4, mtan  5 20. 9 units2 17. g 1x2 

Mixed Review, p. 1069

1. a. lim f 1x2  b b. lim f 1x2  b xSa

xSq

c. lim f 1x2  b d. lim f 1x2  1dne 2 xSa

3. 7 17. a. 19. a. c.

xSa

q

1 1 12 5. 1dne 2 7. 4 12 9.  11. 3 13. 15.  L 6 2 2 42 ft/sec b. 74 ft/sec c. 10 ft/sec 25% of the staff per day b. 12.5% of the staff per day 12  7% of the staff per day 20

Practice Test, p. 1070 1. The limit of f(x) as x approaches 5 is 10. 2. f(x); L; sufficiently; c 3. False, a limit can exist even if c is not in the domain. 4. False, a limit can fail to exists even if a function is defined at c. 5. As the domain of g is x  1, the limit in b. exists and the limit in a. does not. lim 1 1x  1  22  2. xS1

6. a. S II b. S I 7. a. 2

b. A dne B

c. S IV

8. a. 1

b. 0

d. S III 9. a. 3

q

12. a. 2

i1 n

a

108

factor



10. a. 4

n



summation 18 2n3  3n2  n 24 n2  n 6 c 2 a b a b d formulas n 6 n 2 n

1. a. 3, b. 5, c.

Exercises 11.4, pp. 1064–1066

43. A 







xS2



6 36 24 n c  2 i2  id n n i1 2n

41.

83. not possible, lim g1x2 does not exist

41.



14. a.

b. undefined, g132  0 11. a. 4 b. 9

dne B A LH RH

b. 12 b. 4

1 13. a. 0 b. A dne B , c. d. 1 LH RH 3 b. A dne B , c. A dne B d. 3 is not in the domain L

q

3 1 15. 3 16. 17. 18. 1 2 10 19. a. d 1t2  32t  224 b. d 122  160, the debris is rising at a velocity of 160 ft/sec; d 162  32, the upward velocity of the debris has slowed to 32 ft/sec; d 172  0, the debris has reached its maximum height (velocity is 0 ft/sec); d 1112  128, the velocity of the debris in now in the downward direction (v 6 0) at 128 ft/sec. n n 6 6 6 n 6 2 20. lim LW  lim f a ib  lim c225  a ib d ; 1278 ft-lb xSq i1 nSq i1 n n nSq n i1 n







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Student Answer Appendix

Cumulative Review Chapters 1–11, pp. 1071–1072 3 1. x  , x   i 13 3. (1, 1, 1) 5. about 630 ft 2 9. f 1x2  1x  321x  121x  22 10 8 6 4 2 54321 2 4 6 8 10

1 7. 2

y

1 2 3 4 5 x

1 13. a. L1x2   x  19 3 15. no, lim f 1x2 lim f 1x2 11. 512i

xS0

b. 14 cm

xS0

5 4 3 2 1

54321 1 2 3 4 5

c. 24 days

y

1 2 3 4 5 x

ln 9.36  1 , x  1.618 19. 3 2 2 2 1  x2 x1       k;  k 12 2 4 2 1y  12 2 1x  12 2  1 25 9 a. 1 b. 0 c. 3 d. 5 not in domain

17. x  21. 23. 25. 27.

29. 30.6 units2

1. 1.3 0

1

2

3

4

0

1

2

3

4

0

1

2

3

2

3

4

5

3. 2.5

5. 2.65 2

1

7. 1.73 0

1

55. 32°F

57. 179°F

Exercises AI-B, pp. A-12–A-13

1. n  7 3. n  4 5. 1n  52 2 7. 2n  13 9. n2  2n 11. 23 n  5 13. 31n  52  7 15. Let w represent the width. Then 2w represents twice the width and 2w  3 represents three meters less than twice the width. 17. Let b represent the speed of the bus. Then b  15 represents 15 mph more than the speed of the bus. 19. h  b  150 21. L  2W  20 23. M  2.5N 25. T  12.50g  50 27. 14 29. 19 31. 0 33. 144 35. 41 37. 24 5 39. 41. x Output x Output 3 5 14 3 2 8 6 2 1 9 0 1 0 4 0 4 1 1 1 6 2 0 2 6 3 13 3 4 2 has an output of 0. 1 has an output of 0. 43. a. 7  152  2 b. n  122 c. a  14.22  13.6  a  9.4 2 d. x  7  7  x 45. 5x  13 47. 15 p  6 49. 17 12 x 51. 2a2  2a 53. 6x2  3x 55. 2a  3b  2c 57. 35 8n  7 59. a. t  12 j b. t  225 mph 61. a. L  2W  3 b. 107 ft 63. t  c  27; 42¢ 65. C  25t  43.50; $81 67. a. positive odd integer

9. a. i. {8, 7, 6} ii. {8, 7, 6} iii. 11, 8, 7, 62 iv. 51, 8, 0, 75, 92 , 5, 6, 7, 35 , 66 v. { } vi. 51, 8, 0.75, 92 , 5.6, 7, 35 , 66 b. 51, 35 , 0.75, 92 , 5.6, 6, 7, 86 c.

5.6 E t 0.75 2 1 0 1 2 3 4 5 6 7 8

11. a. i. 5 249, 2, 6, 46 ii. 5 249, 2, 6, 0, 46 iii. 55, 249, 2, 3, 6, 1, 0, 46 iv. 55, 249, 2, 3, 6, 1, 0, 46 v. 5 23, 6 vi. 55, 249, 2, 3, 6, 1, 23, 0, 4, 6 b. 55, 3, 1, 0, 23, 2, , 4, 6, 2496 c. 3 p 49 65 4 3 2 1 0 1 2 3 4 5 6 7

13. Let a represent Kylie’s age: a  6 years. 15. Let n represent the number of incorrect words: n  2 incorrect. 17. 2.75 19. 4 21. 9 23. 10 25. 8, 2 27. negative 29. n 11 31. a. positive b. negative c. negative d. negative 33.  6 1 35. 2 37. 92  81 is closest 39. 7 41. 413 43. 10 45. 78

3. 12p5q4

9. 49c14d 4

9 6 2 11. 16 xy 27 6 3 3 13. a. V  27x b. 1728 units 15. 3w 17. 3ab 19. 8 8 2 6 4 6 4p 3p 1 8x 25m n 21. 2h3 23. 25. 6 27. 29. 31. 9 8 8 27y 4r 4q2 q 11 5 a 12 7 33. 35. 14 37. 39. 2 41. 10 43. 13 45. 4 9 3h7 b 5x4 47. 6.6  109 49. 0.000 000 006 5 51. 26,571 hr; 1,107 days 53. polynomial, none of these, degree 3 55. nonpolynomial because exponents are not whole numbers, NA, NA 57. polynomial, binomial, degree 3 59. w3  3w2  7w  8.2; 1 61. c3  2c2  3c  6; 1 2 2 63. 2 65. 3p3  3p2  12 67. 7.85b2  0.6b  1.9 3 x  12; 3 1 2 69. 4 x  8x  6 71. 3x3  3x2  18x 73. 3r2  11r  10 75. x3  27 77. b3  b2  34b  56 79. 21v2  47v  20 9 81. 9  m2 83. m2  16 85. 2x4  x2  15 87. 4m  3; 16m2  9 89. 7x  10; 49x2  100 91. 6  5k; 36  25k2 93. x  16; x2  6 95. x2  8x  16 97. 16g2  24g  9 99. 16p2  24pq  9q2 101. 16  8 1x  x 103. F  kPQd2 105. 5x3  3x2  2x1  4 107. $15 109. 6

1. 14n7

Exercises AI-A, pp. A-6–A-8

1

49.

Exercises AI-C, pp. A-21–A-24

APPENDIX I

1

11 51. 64 53. 4489.70 12 59. Tsu Ch’ung-chih: 355 61. negative 113 47. 4

5. a14b7

7. 216p3q6

Exercises AI-D, pp. A-29–A-31

1. a. 171x2  32 b. 7b13b2  2b  82 c. 3a2 1a2  2a  32 3. a. 1a  2212a  32 b. 1b2  32 13b  22 c. 1n  7214m  112 5. a. 13q  2213q2  52 b. 1h  122 1h4  32 c. 1k2  721k3  52 7. a. 11p  72 1p  22 b. 1q  921q  52 c. 1n  421n  52 9. a. 13p  221p  52 b. 14q  52 1q  32 c. 15u  3212u  52 11. a. 12s  52 12s  52 b. 13x  7213x  72 c. 215x  6215x  62 d. 111h  122 111h  122 e. 1b  2521b  252 13. a. 1a  32 2 b. 1b  52 2 c. 12m  52 2 d. 13n  72 2 15. a. 12p  3214p2  6p  92 b. 1m  12 21m2  12 m  14 2 c. 1g  0.321g2  0.3g  0.092 d. 2t1t  321t 2  3t  92 17. a. 1x  32 1x  321x  12 1x  12 b. 1x2  921x2  42 c. 1x  22 1x2  2x  421x  121x2  x  12 19. a. 1n  121n  12 b. 1n  12 1n2  n  12 c. 1n  12 1n2  n  12

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Student Answer Appendix

7x12x  1212x  12 21. 1a  521a  22 23. 21x  22 1x  102 113m  8213m  82 27. 1r  321r  62 29. 12h  32 1h  22 13k  42 2 33. 3x12x  721x  32 35. 4m1m  52 1m  22 1a  521a  122 39. 12x  5214x2  10x  252 41. prime 1 43. 1x  521x  321x  32 45. V  h1R  r21R  r2; 3 3 3 6 cm ; 18.8 cm d. 25. 31. 37.

47. V  x1x  521x  32 a. 3 in. b. 5 in. c. V  2412921272  18,792 in3 v v 49. L  L0 a1  ba1  b , L  12211  0.752 11  0.752 B c c  3 27 in.  7.94 in. 51. a. 18 14x4  x3  6x2  322 1 b. 18 112b5  3b3  8b2  182 53. 2x116x  27216x  52 55. 1x  321x  321x2  92 57. 1p  121p2  p  121p  121p2  p  12 59. 1q  521q  521q  2321q  232

Exercises AI-E, pp. A-37–A-39 x3 2x1x  22

1. a. 

1 3

b. 1

7. a. 3ab9

b.

b.

a4 5. a. 1 a7 x3 1 c. 11y  32 d. 9 m 1a  221a  12 c. x  2 d. n  2 11. 1a  321a  22 81a  72 y 15 m 19. 21. 23. 4 a5 x m4 3. a. simplified

b.

2n  3 3x  5 b. n 2x  3 1p  42 2 13. 1 15. 17. p2 1 n 31a2  3a  92 5 3  20x x  0.3 25. 27. 29. 31. 2 x  0.2 2 8x2 n 3 14y  x y  11 2 3m  16 33. 35. 37. 39. 2 4 p  6 1m  421m  42 1y  62 1y  52 8x y m2  6m  21 2a  5 1 41. 43. 45. 1a  421a  52 y1 1m  32 2 1m  32 1 5 1  5p 2 x2 1 4a 47. a. 2  ; b. 2  3 ; 49. 51. p  1 p a  20 p p2 x x x3 2 3 1 2 2 1 m m3 x 2 x x 2 53. 55. 57. a. ; b. ; 3 m3 2 x2  2 9x  12 y  31 1 1 2 m x 12x  h2 f2  f1 a 59. 61. 63. f1 f2 x1x  h2 2x2 1x  h2 2 65. Price rises rapidly for Day Price first four days, then begins a 0 10 gradual decrease. Yes, on the 1 16.67 35th day of trading. 2 32.76 3 47.40 4 53.51 5 52.86 6 49.25 7 44.91 8 40.75 9 37.03 10 33.81 9. a.

67. t  8 weeks

69. b. 20 # n  10 # n  2n2, all others equal 2

Exercises AI-F, pp. A-47–A-50 1. a. 9

b. 10

5. a. 4

b. 5x

3. a. 7p c. 6z

4

b. x  3 c. 9m2 d. x  3 v d. 7. a. 2 b. not a real number 2

e. k  3 f. h  2 9. a. 5 b. 3n3 7v5 9p4 64 125 c. not a real number d. 11. a. 4 b. c. d. 6 125 8 4q2 1 256 13. a. 1728 b. not a real number c. d. 9 81x4 10 32n 1 3 3 2 3 15. a. b. 17. a. 3m 12 b. 10pq 1q c. mn 2n2 1 2 p2 2y4 9 d. 4pq3 12p e. 3  17 f.  12 19. a. 15a2 b. 4b 1b 2 3 x4 1y 3 15 181 3 3 c. d. 3u2v1v 21. a. 2m2 b. 3n c. d. 3 3 4x z 6 5 3 1 2 6 4 12 23. a. 2x2y3 b. x2 2x c. 1 b d. 6  e. b4 6 16 25. a. 9 12 b. 14 13 c. 16 12m d. 5 17p 3 27. a. x 12x b. 2  13x  315 c. 6x22x  522  27x  323 29. Verified 31. Verified 33. a. 98 b. 115  121 c. n2  5 d. 39  1213 35. a. 19 b. 110  165  217  1182 c. 12 15  2114  36115  6142 3 3 2 2 2p2 23 2 215x 3 26b 52 a 37. a. b. c. d. e. 2 10b 2p a 9x2 39. a. 12  4 111; 1.27 b. 2  13 41. a. 130  215  313  312; 0.05 7  7 12  16  213 ; 7.60 b. 3 43. a. 8 210 m; b. about 25.3 m 45. a. 365.02 days b. 688.69 days c. 87.91 days 47. a. 36 mph b. 46.5 mph 49. 12 234  219.82 m2 51. a. 1x  152 1x  152 b. 1n  11921n  1192 322 53. a. 1313x  391x b. Answers will vary. 55. 2 c. 3x2

d. 3x

Practice Test, pp. A-52–A-53 1. a. True b. True c. False; 12 cannot be expressed as a ratio of two integers. d. True 2. a. 11 b. 5 c. not a real number d. 20 3. a. 89 b. 7 c. 0.5 d. 4.6 4. a. 28 b. 0.9 c. 4 d. 7 6 3 5.  4439.28 6. a. 0 b. undefined 7. a. 3;  2, 6, 5 b. 2; 13 , 1 n 2 8. a. 13 b.  7.29 9. a. x3  12x  92 b. 2n  3a b 2 10. a. Let r represent Earth’s radius. Then 11r  119 represents Jupiter’s radius. b. Let e represent this year’s earnings. Then 4e  1.2 million represents last year’s earnings. 11. a. 9v2  3v  7 b. 7b  8 c. x2  6x 12. a. 13x  42 13x  42 b. v12v  32 2 25 2 2 m6 c. 1x  52 1x  321x  32 13. a. 5b3 b. 4a12b12 c. d. p q 3 4 8n 12 a 14. a. 4ab b. 6.4  102  0.064 c. 4 8 d. 6 bc 15. a. 9x4  25y2 b. 4a2  12ab  9b2 16. a. 7a4  5a3  8a2  3a  18 b. 7x4  4x2  5x x5 2n x5 17. a. 1 b. c. x  3 d. e. 2n 3x  2 3x  1 31m  72 2 64 1 12 f. 18. a. x  11 b. c. d.   51m  42 1m  32 3v 125 2 2 110x 2 e. 11 110 f. x  5 g. h. 21 16  122 5x 19. 0.5x2  10x  1200; a. 10 decreases of 0.50 or $5.00 b. Maximum revenue is $1250. 20. 58 cm

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Index A Absolute value applications of, 28 of complex numbers, 41, 254 explanation of, 24 on graphing calculators, 28, 81 multiplicative property of, 25 of real numbers, A–3 – A–4 Absolute value equations on graphing calculator, 28 method to solve, 24–25, 72 property of, 24 Absolute value function characteristics of, 157 explanation of, 182 Absolute value inequalities applications of, 28 distance and, 32 explanation of, 26 on graphing calculator, 28 limits and, 80–81 methods to solve, 26–27 Accumulated value, 399, 401 Acute angles explanation of, 427 functions of, 499–500 Addition associative property of, A–10 commutative property of, A–9 – A–11 of complex numbers, 35 distributive property of multiplication over, A–10 – A–11 of matrices, 781–782 of polynomials, A–19 of radical expressions, A–44 Additive identity, A–10 Additive inverse, A–10 Additive property of equality, 2, 3 of inequality, 15, 16 Agnesi, Maria, 922 Algebra to evaluate limits, 1046–1048 of functions, 186, 194–195, 199, 208–209 fundamental theorem of, 242–245 of matrices, 780–787, 817 of vectors, 664–665 to verify identities, 545–546 Algebraic equations. See Equations Algebraic expressions evaluation of, 76–77, A–9 explanation of, 2, A–8 mathematical models and, A–8 – A–9 simplification of, A–11 Algebraic fractions. See Rational expressions

Algebraic methods to find inverse functions, 344–346 trigonometric equations and, 610–611 Algorithms, division, 231 Allowable values, 18–20 Alpha (), 426 Alternating current, 41, 693–695 Altitude, 373 Ambiguous case applications of, 639–640 explanation of, 637 method to avoid, 648 scaled drawings and, 637–638, 673–674 solution to, 638–639 Amplitude on graphing calculator, 465, 614, 626 of sine and cosine functions, 460–461 Analytical geometry, 832, 924 Analytic parabolas, 870, 926 Angle reduction formulas, 567 Angles acute, 427, 499–500 central, 430 complementary, 426 coterminal, 430, 518 of elevation/depression, 504–505 explanation of, 426 measurement of, 426–427, 429–433, 522–523 negative, 429, 430 obtuse, 427 positive, 429, 430 quadrantal, 430 reference, 444, 515–517, 535 right, 426 standard, 430, 535 straight, 426 supplementary, 426 viewing, 591–592 Annuities, 401–402 Aphelion, 905 Approximations, limits vs., 1024–1025 Arc length explanation of, 431 formula for, 431, 433 of right parabolic segment, 872 Area. See also Surface area of circle, A–22 of circular sector, 431–432 under curve, 216–217, 1061–1063, 1068 of ellipse, 849 formulas for, 651, 652, 766, 789 of inscribed square, 95 of Norman window, 812 of polygon, 550, 551, 1024–1025

of right parabolic segment, 872 of triangle, 299, 557, 706, 778, 809 Area functions, 421–422 Argument of z, 689 Arithmetic sequences applications of, 952–953 explanation of, 949, 1007–1008 finding nth partial sum of, 951–952 finding nth term of, 950–951 identifying and finding common difference in, 949–950 Associated minor matrices, 797 Associative properties, A–10 Asymptotes horizontal, 274, 277–278, 1049–1050 of hyperbolas, 854, A–59 nonlinear, 279, 291 oblique, 291–293 vertical, 275–277, 289 Asymptotic discontinuities, 1045 Augmented matrices matrix inverses and, 794, 823–824 solving systems using, A–56 of system of equations, 770–771 triangularizing, 771–772 Average rate of change applications of, 148–149 difference quotient and, 146, 147 explanation of, 130–132, 145 formula for, 215 projectile velocity and, 145–146 Average value, of function, 460, 486 Axes conjugate, 853 parabolas with horizontal, 866–867 polar, 883 rotation of, 832–834, 898–902 transverse, 853 vertical, 732 Axis of symmetry, 139

B Back-substitution, 3 Base, A–4 Base case of inductive proof, 968 Base-e exponential functions, 357–359, 389 Beats, 626–627 Berkeley, George, 1034 Binomial coefficients, 1001–1002 Binomial cubes, 41 Binomial experiment, 1004 Binomial powers explanation of, 1001 Pascal’s triangle and, 999–1000 Binomial probability, 1004, 1005

I-1

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I-2

Index

Binomials expansion of, 1000, 1003–1004 explanation of, A–18, A–19 F-O-I-L method to multiply, A–20 Pascal’s triangle and, 999–1001 square of, A–21 Binomial theorem explanation of, 698, 1002–1003 key concepts on, 1011 Bisection, 271–272 Body mass index formula, 22 Boundary, 756 Brachistochrone applications, 916 Branches, of hyperbola, 852 Break-even analysis, 726 Brewster’s law, 566

C Calculators. See Graphing calculators Capacitive reactance, 694 Cardano, Girolomo, 41 Cardano’s formula, 256 Cauchy, Augustin-Louis, 1023 Cayley-Hamilton theorem, 813 Ceiling function, 178 Celsius temperature, 135 Centimeters of mercury, 373 Central angle, 430 Central circles, 430, 441 Central rectangles, 855 Change-of-base formula, 386–387 Characteristic polynomials, 813 Circles area of, A–22 center of, 440 central, 430, 441 circumscribed, 643, 833 equation of, 87–90, 839–840 explanation of, 89, 440, 835, 925 on graphing calculator, 92 graphs of, 90–92, 840, A–63 involute of, 521 radius of, 89, 91, 440 unit, 440–449, 524 Circular functions, 445. See also Trigonometric functions Coefficient matric, 770 Coefficients explanation of, A–8 of friction, 483 leading, A–19, A–25 polynomial equations with complex, 55 tangent and cotangent functions and, 474–477 Cofactors, 797, A–57 Cofunction identities, 561 Cofunctions explanation of, 503 summary of, 504 to write equivalent expressions, 503–504 Coincident dependence, 733, 738 Coincident lines, 724 Collinear points, 810

Combinations explanation of, 979–981 stating probability using, 990–991 Combined variation, 320–321 Common difference d, of arithmetic sequence, 949 Common logarithms, 368 Common ratio r, 956 Commutative properties, A–9 – A–11 Complementary angles, 426 Complements stating probability using, 989–990 to write equivalent expressions, 503–504 Completing the square applications of, 42 explanation of, 45 to graph ellipse, 846 to graph hyperbola, 856–859 to graph parabola, 867 to graph quadratic functions, 220–221 quadratic equations solved by, 45–46 Complex conjugates product of, 36–37 use of, 38, 49 Complex conjugates corollary explanation of, 244 proof of, A–60 Complex numbers absolute value of, 41, 254 addition and subtraction of, 35 applications of, 693–695 converted from rectangular to trigonometric form, 689–690 converted from trigonometric form to rectangular form, 690 cube of, 696 De Moivre’s theorem and, 699, 709 division of, 38 explanation of, 34, 72 graphs of, 687–688 historical background of, 33 identifying and simplifying, 33–35 multiplication of, 36–38 in rectangular form, 688 standard form of, 34–35 in trigonometric form, 688–690, 708 Complex plane, 688 Complex polynomial functions, 242 Complex roots, 37 Complex zeroes, 334 Composition of functions, 190–195, 208–209 inverse trigonometric functions to evaluate, 588–589 transformations via, 213 Compound annual growth, 200 Compound fractions, A–35 – A–36 Compound inequalities explanation of, 16–17 method to solve, 17–18 Compound interest, 200 Compound interest formula, 200, 399 Conditional equations, 4

Cones characteristics of, 834 volume of, 616–617 Conic equations after rotating axes, 899, 900 explanation of, 857 in polar form, 902–905 Conic sections. See also Circles; Ellipses; Hyperbolas; Parabolas characteristics of, 834–836 degenerate cases of, 865 eccentricity of, 903–905 explanation of, 832 formulas for, 152 identified by their equations, 857 in polar form, 928, 931–932 rotated, 896–901, 927, 931–932 use of discriminant to identify, 901–902 use of focal chord to graph, 851 Conjugate axis, 853 Conjugates complex, 36–37, 49 verifying identities by multiplying, 554 Consecutive integers, 62 Consistent systems, 724 Constant, A–8 Constant of variation, 316 Constraints, 758 Continuous functions, 1044, 1045 Continuous graphs, 86, 258 Continuous intervals, 510 Continuously compounded interest, 399–400 Contradictions, 4, 736 Convenient values decomposition using, 747–749 explanation of, 747 Coordinate grid, 85 Coordinate plane, trigonometry and, 512–518, 529–530 Cosecant function characteristics of, 464 graphs of, 463–464 inverse, 589 Cosine law of, 646–650, 706 sum and difference identities for, 558–560 table of values for, 459 Cosine function amplitude of, 460–461 applications of, 488, 492, 493 characteristics of, 457 graphs of, 459–462 inverse, 585–587 Cotangent, table of values for, 473 Cotangent function applications of, 477–478 characteristics of, 474 graphs of, 473–477 inverse, 589, 590 period of, 475–476 Coterminal angles, 430, 518

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Index Counting techniques combinations and, 979–981 distinguishable permutations and, 977–978 fundamental principle of, 976–977 key concepts on, 1009–1010 using tree diagram, 975–976 nondistinguishable permutations and, 979 Cramer’s rule, 807–809, 819 Cube root function, characteristics of, 158 Cube roots. See also Radical expressions explanation of, A–5 notation for, A–5 of unity, 701 Cubes binomial, 41 factoring sum or difference of, A–28 perfect, A–4 sum of, 964, 972 volume of, A–22 Cubic equations, 704 Cubic functions, 158, 475 Cubic polynomials, A–25 Current, 694 Curves area under, 216–217, 1061–1063, 1068 from cycloid family, 916–917 parametric, 913–914 vertical distance between, 199 Cusp, 258 Cycloids, 916–917 Cylinders surface area of, 11, 54, 200, 299 volume of, 616 Cylindrical vents, 881

D Decay rate constant, 402 Decimal degrees, 427 Decimal notation, conversions between scientific notation and, A–18 Declarative statement, 1034–1035 Decomposition of composite functions, 193 explanation of, A–8 of rational numbers, A–42 of special forms, 827 using convenient values, 747–749 using matrices and technology for, 810 using system of equations, 750–751 Decomposition template, 743–747 Decreasing functions, 142–144 Degenerate cases, 96, 865 Degrees as angle measurement, 426–427 conversion between radians and, 432–433 decimal, 427 Delta (), 99 Delta/epsilon form, 81–82 Demand curves, 882 Demographics, 354 De Moivre’s theorem checking solutions to polynomial equations and, 699–700 explanation of, 698–699, 709

multiple angle identities and, 704 nth root theorem and, 700–702 proof of, A–62 Denominators, rationalizing, A–45 – A–46 Dependent systems explanation of, 724, 733 of linear equations in three variables, 733, 737–738 of linear equations in two variables, 724–725 matrices and, 774 Dependent variables, 84 Depression, angles of, 504–505 Descartes, René, 33 Descartes rule of signs, 249–250 Determinant formula for area of parallelogram, A–60 – A–61 Determinants of column rotation, 798–799 of general matrix, A–57 geometry of, 828 singular matrices and, 796–800 solving systems using, 806–810 Difference identity for cosine, 559. See also Sum and difference identities Difference quotient application of, 148–149 explanation of, 146–147 limit of, 1056–1061 rates of change and, 146, 147, 215–216 Directed line segments, 658–659 Directrix explanation of, 834 of hyperbola, 838 of parabola, 868–869 Direct substitution, 1045–1046 Direct variation, 316–319, 331 Discontinuities asymptotic, 1045 explanation of, 1044 jump, 1045 nonremovable, 276, 1045 removable, 176, 289–290, 297, 1045 Discontinuous functions, 175–176 Discriminant of cubic equations, 704 explanation of, 48 on graphing calculator, 51–52 identifying conics using, 901–902 of quadratic formula, 48–49 of reduced cubic equation, 313 Diseconomies of scale, 879 Disjoint intervals, 18 Distance absolute value equations and inequalities and, 32 between point and line, 833–834 number line, A–4 perpendicular, 833 uniform motion, 8 vertical, between curves, 199 Distance formulas, 88–89, 482, 557, 646, 741, 832 Distinguishable permutations, 977–978

I-3

Distributive property of multiplication over addition, A–10 – A–11 Division of complex numbers, 38 long, 230–231, 279 of polynomials, 231, 233, 234 of radical expressions, A–45 – A–46 of rational expressions, A–33, A–34 synthetic, 232–235, 237, 327, A–54 – A–55 Division algorithm, 231 Divisor, nonlinear, 234 Domain of expression, 18–19 of functions, 125–126, 130 functions of real numbers and, 447 implied, 127–128 of logarithmic functions, 371 of piecewise-defined functions, 172–173 of relation, 84 restricting, 345 of trigonometric functions, 447 Dominant term, 259 Dot products angle between two vectors and, 678–680 explanation of, 678, 707–708 Double-angle formula, 569–570 Double-angle identities explanation of, 568, 582 use of, 569–570 Dynamic trigonometry, 512

E e (the number), 357 Eccentricity of conics, 903–905 on graphing calculator, 907 Ecliptic planes, 931 Economies of scale, 879 Elementary row operations, 771–772 Elements, of set, A–1 Elevation, angles of, 504–505 Elimination explanation of, 722 Gaussian, 771, A–56 Gauss-Jordan, 773, 774, A–56 to solve systems of equations in three variables, 734–736 to solve systems of equations in two variables, 722–724 to solve systems of nonlinear equations, 877–878 Ellipses area of, 849 definition of, 844 equation of, 840–843, 846–847, 906, A–58 explanation of, 835–836, 841, 925, A–1 foci of, 843–847 graphs of, 842–847 perimeter of, 849 polar form of, 910 with rational/irrational values of a and b, 875

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I-4

Index

End behavior of graphs, 143, 144 of polynomial graphs, 258–261 of quadratic functions, 259 of rational functions, 273 Endpoint maximum, 144 Endpoints, A–3 Equality additive property of, 2, 3 of matrices, 780–781 multiplicative property of, 2, 3 power property of, 59, 61 square root property of, 44–45 Equation form, relations in, 85 Equations. See also Identities; Linear equations; Systems of linear equations; Systems of linear equations in three variables; Systems of linear equations in two variables; Systems of nonlinear equations; specific types of equations absolute value, 24–28, 32 of circle, 87–90, 839–840 conditional, 4 of conic sections, 857, 899, 902–905 cubic, 704 of ellipse, 840–843, 846–847, 906, A–58 equivalent, 2 explanation of, 2 exponential, 381, 389–390 families of, 2 of functions, 165 of hyperbolas, 152, 852–857, A–58 – A–59 literal, 4–6 logarithmic, 381–383, 387–388, 878 logistics, 416–417, 754 matrix, 795–796 of parabola, 868–870 parametric, 313, 913–920, 928 polar, 887–891, 896, 903 polynomial, 56–57, 699–700 quadratic, 42–51, 77–78 in quadratic form, 61–62 of quadratic function, 223 radical, 59–61 rational, 57–59 of rational functions, 281 of semi-hyperbola, 864 solutions of, 2 that are not identities, 547–548 trigonometric, 599–605, 610–614, 622, 623 written from graphs, 464 Equilateral triangles in complex plane, 696 height of, 438 Equilibriant, 675 Equilibrium explanation of, 491 force required to maintain, 566 static, 713 vectors and, 674–675 Equivalent equations, 2 Equivalent systems of equations, 734 Equivalent vectors, 660

Euclid, 426 Even functions explanation of, 459 symmetry and, 139 Even multiplicity, 244 Events explanation of, 987 nonexclusive, 991–992 ex, 422 Existence theorem, 245 Expansion by minors, 798 Experiment binomial, 1004 explanation of, 975 Exponential decay, 402–404 Exponential equations explanation of, 381 on graphing calculator, 361–362 logarithmic form and, 366–368 methods to solve, 382, 389–390 uniqueness property to solve, 359–361 Exponential form, A–4 – A–5 Exponential functions applications of, 361, 390 base-e, 357–359 evaluation of, 354–355 explanation of, 354, 411 graphs of, 355–357, 359 natural, 358 power functions vs., 359 Exponential growth, 402–403 Exponential growth formula, 402 Exponential notation, A–4 – A–5 Exponents explanation of, A–4 properties of, A–14 – A–17, A–51 rational, 61 scientific notation and, A–17 – A–18 zero and negative numbers as, A–16 – A–17 Extraneous roots, 58, 63–64, 387 Extreme values on graphing calculator, 295 quadratic functions and, 223–225

F Factorial formulas, 984 Factorials, 941–942 Factors/factoring greatest common, A–24 by grouping, A–24 –A–25 least powers, 80 quadratic polynomials, A–25 – A–27 to solve polynomial equations, 56–57 to solve trigonometric equations, 611 special forms and quadratic forms, A–27 – A–29 Factor theorem explanation of, 235 polynomials and, 236–237 proof of, 235–236 Fahrenheit temperature, 135 Family of curves, 835. See also Conic sections Family of equations, 2

Family of identities, 544–545 Family of polar graphs, 889, 892 Fibonacci (Leonardo of Pisa), 941 Fibonacci sequence, 941 Finite sequences, 940 Finite series, 942 Fixed costs, 725 Floor function, 178–179 Fluid motion formula, 170 Focal chord for graphing conic sections, 851 of hyperbola, 864, 873 of parabola, 869 Foci applications of, 847 of ellipse, 843–847 explanation of, 835 of hyperbola, 838, 857–859 of parabola, 868–869 Foci formula for ellipses, 846 for hyperbolas, 858 Focus-directrix form, of equation of parabola, 868–870 F-O-I-L method, 36, A–20 Folium of Descartes, 313, 922 Force components of, 676–677 normal, 595 Force vectors, 666–667, 678 Formulas absolute value of complex number, 41, 254 alternative form for law of cosines, 655 angle between intersecting lines, 520 angle reduction, 567 arc length, 431, 433 area, 651, 652, 766, 789 area of circular sector, 431–432 area of ellipse, 849 area of inscribed square, 95 area of Norman window, 812 area of parallelogram, 520 area of regular polygon, 550, 551 area of right parabolic segment, 872 area of triangle, 299, 557, 778 barometric pressure, 373, 391 average rate of change of function, 215 binomial cubes, 41 binomial probability, 1005 body mass index, 22 Brewster’s law, 566 Celsius to Fahrenheit conversion, 135 change-of-base, 386–387 compound annual growth, 200 compound interest, 200, 399 conic sections, 152 continuously compounded interest, 400 cost of removing pollutants, 285 cube of complex number, 696 dimensions of rectangular solid, 741 discriminant of reduced cubic, 313 discriminant of reduced cubic equation, 704 distance, 88–89, 482, 557, 646, 741, 832 distance in polar coordinates, 894

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Index double-angle, 569–570 equation of line in polar form, 910 equation of line in trigonometric form, 616 equation of semi-hyperbola, 864 equilateral triangles in complex plane, 696 explanation of, 4 factorial, 984 fine-tuning golf swing, 608 fluid motion, 170 Folium of Descartes, 313, 922 force between charged particles, 324 force normal to object on inclined plane, 595 force required to maintain equilibrium, 566 forensics - estimating time of death, 394 games, 364, 996 general linear equation, 120 general summation, 967 growth of bacteria population, 364 half-angle, 571–572 heat flow on cylindrical pipe, 596 height of building, 511 height of equilateral triangle, 438 height of projected image, 352 height of projectile, 54 Heron’s, 652–653 human life expectancy, 108 hydrostatics, surface tension, and contact angles, 468 ideal weight for males, 135 illuminance of point on surface, 550 illumination of surface, 508 intercept/intercept form of linear equation, 120 interest earnings, 108 inverse of 2 x 2 matrix, 803 lateral surface area of cone, 67 logistic growth, 394 logistics equations and population size, 754 magnitude of vector in three dimensions, 671 Malus’s law, 578 manufacturing cylindrical vents, 881 midpoint, 87–88, 832, 894 number of daylight hours, 497 painted area on canvas, 67 perimeter, 789 perimeter of ellipse, 849 perimeter of trapezoid, 655 perpendicular distance from point to line, 837 pH level, 376 Pick’s theorem, 135 population density, 285 position of image reflected from spherical lens, 482 power of imaginary unit, 972 power reduction, 570 Pythagorean theorem in trigonometric form, 468 Pythagorean triples, 452 quadratic, 47–49 radius of circumscribed circle, 643 radius of sphere, 352

range of projectile, 608, 685 relationship between coefficient, frequency and period, 497 relationships in right triangle, 438 required interest rate, 324 revenue, 64 root tests for quartic polynomials, 269 sine of angle between sides of triangle, 508 solving for specified variable in, 59 spending on Internet media, 95 spring oscillation, 30 square root of complex numbers, 254 Stirling’s, 984 sum of cubes, 964, 968 sum of first n term of sequence, 947 sum of infinite geometric series, 960 sum of n natural numbers, 954 supersonic speeds, sound barrier, and Mach numbers, 578 surface area of cylinder, 11, 54, 200, 299 surface area of rectangular box with square ends, 227 time required to double investment, 376 trigonometric graphs, 153 triple angle formula for sine, 642 tunnel clearance, 879 uniform motion, 729 vertex, 222–223 vertex/intercept, 227 volume, 766 volume of cube, A–22 volume of open box, 240 volume of sphere, 170 written in alternative form, A–36 – A–37 Four-leaf rose, 891 45-45-90 triangles, 428, 429, 499 Fractions compound, A–35 – A–36 partial, 743, 751–752 Frequency, 492 Frequency variations, 163 Function families explanation of, 157 on graphing calculator, 166 Functions. See also Trigonometric functions absolute value, 157, 182 of acute angles, 499–500 algebra of, 186, 194–195, 199, 208–209 area, 421–422 ceiling, 178 composite, 193, 196 composition of, 190–195, 208–209 continuous, 1044, 1045 discontinuous, 175–176 domain of, 125–126, 130 equations of, 165 evaluation of, 128–129 even, 139, 459 explanation of, 122–123, 205 exponential, 354–361, 390, 411 evaluating, 128, 759 floor, 178–179 graphical view of operations on, 193–194 graphs of, 123, 124, 138–149, 205, 206

I-5

greatest integer, 178 growth, 357 identification of, 123–124 input/output nature of, 189 intervals and increasing or decreasing, 142–144 intervals and positive or negative, 141–142 inverse, 343–349 linear, 955 logarithmic, 367–370, 412 maximum and minimum value of, 144–145, 337–338 monotonic, 356 notation for, 128–130, 190 odd, 140–141, 463 one-to-one, 342–343, 410, 587 periodic, 456–457 piecewise-defined, 172–180, 208 polynomial, 215, 242, 251, 257–266, 328 products and quotients of, 187–189 quadratic, 220–225, 242, 326, 489, 955 radical, 216, 337, 1051 range of, 126–127, 130 rates of change and, 145–149 rational, 272–282, 289–297, 329, 330, 742, 1050 reciprocal, 273, 896–897 relations as, 123 sand dune, 182 step, 178–179 toolbox, 157–158, 207 transcendental, 366, 604 two variable, 296, 759 vertical line test for, 124–125 Fundamental identities. See also Identities algebra to verify, 545–546 explanation of, 544, 581, 619 trigonometric functions and, 546–547 Fundamental principle of counting (FPC), 976–977 Fundamental properties of logarithms, 381 Fundamental property of rational expressions, A–31 Fundamental theorem of algebra, 242–245

G Games, 996 Gamma (), 426 Gauss, Carl Friedrich, 33, 242, 773 Gaussian elimination, 771, 773, A–56 Gauss-Jordan elimination, 774, A–56 General function, 164, 165 General matrix, A–57 General solutions, literal equations and, 5–6 General summation formulas, 967 Geometric sequences applications of, 961 explanation of, 401, 956, 1008 finding nth partial sum of, 958–959 finding nth term of, 957–958 Geometric series applications of, 961 infinite, 959–960

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I-6

Index

Geometry analytical, 832, 924 applications of, 62–63 of determinants, 828 of vectors, 828 Global maximum, 144 Graphing calculator features parametric mode on, 916–917 split screen viewing, 252 TABLE, 9–10, 668 window size, 105, 768–769 Graphing calculators absolute value equations and inequalities on, 28, 81 amplitude on, 465, 514, 626 asymptotes on, 478–479 composite functions on, 196 cycloids on, 916–917 discriminant on, 51–52 eccentricity on, 907 equations and identities on, 548 evaluating expressions and looking for patterns using, 76–77 exponential equations on, 361–362 extreme values of function on, 295 function families on, 166 graphs of circles on, 92 hyperbolas on, 861 inequalities on, 334 infinite series on, 1014–1015 interest on, 399, 404–405 intermediate value theorem on, 252 inverse functions on, 349, 592 as investigative tools, 9–10 irrational zeros on, 225–226 linear equations on, 105 linear programming on, 822–823 logarithms on, 368–369, 371, 389 logistic equations on, 416–417 matrices on, 775–776, 785–786, 795, 800, 810 maximums and minimums on, 149–150 parameterized solutions on, 739 parametric equations on, 918–920 permutations and combinations on, 981 piecewise-defined functions on, 179–180 polynomial graphs on, 237 polynomial inequalities on, 305, 306, 310–311 projectile motion on, 712 rational functions on, 282 rational inequalities on, 307, 308, 310–311 removable discontinuities on, 297 roots on, 494 sequences on, 945 series on, 945 systems of linear equations on, 727, 739 systems of linear inequalities on, 763–764 transformations on, 212–213 trigonometric functions on, 465, 478–479, 491, 584, 585 vectors on, 668–669 velocity on, 575

x-intercepts on, 494 zeroes on, 149–150, 251, 334, 478–479, 494 Graphs. See also Trigonometric graphs absolute value, 125 analysis of function, 144 characteristics of linear, 97 of circle, 90–92, 840, A–63 continuous, 86, 258 of cosecant function, 463–464 of cosine function, 469–462 of cotangent function, 473–477 of curves from cycloid family, 916–917 of discontinuous functions, 175–176 of ellipse, 842–847 to estimate limits, 1025–1027, 1035, 1066 of exponential functions, 355–357 of function and its inverse, 346–348 of functions, 123, 124, 138–149, 205, 206 horizontal and vertical lines on, 101 of hyperbolas, 852–859, 897 importance of reading, 129–130 of linear equations, 97–98, 104, 203 of lines, 111–115 of logarithmic functions, 369–370, 372 maximum and minimum values on, 337–338 one-dimensional, 732 orientation of, 914 of parabolas, 141, 164 parallel and perpendicular lines on, 102–104 of piecewise-defined functions, 173–177 polar, 887–892, 935–936, A–63 – A–64 of polynomial functions, 257–266 of quadratic functions, 220–224 of rational functions, 279–281, 290–294 of reflections, 161, 162 of relations, 85–87 of secant function, 463–464 of sine function, 454–458, 460–462 sinusoidal, 487 smooth, 258 to solve inequalities, 141–142 to solve system of equations, 720–721 stretches and compressions in, 162–163 symmetry and, 138–141 of tangent function, 471–477 of toolbox functions, 157, 158 transformations of, 158–159, 164, 357 translation of, 160–161 of trigonometric functions, 627 two-dimensional, 732 used to evaluate limits, 1052–1053 of variation, 317–318 Greater than, A–2 – A–3 Greatest common factors (GCF), A–24 Greatest integer function, 178 Grid lines, 85 Grouping, factoring by, A–24 –A–25 Grouping symbols, 187 Growth functions, 357 Growth rate constant, 402

H Half-angle formulas, 571–572 Half-angle identities, 571–572, 582 Half-life of radioactive substances, 403–404 Half planes, 756 Harmonic motion explanation of, 491 sound waves and, 492–493 springs and, 491–492 Heron’s formula explanation of, 652–653 proof of, A–61 Horizontal asymptotes limits at infinity and, 1049–1050 of rational functions, 274, 277–278 Horizontal boundary lines, 126–127 Horizontal component, of vector, 660, 662, 664–669 Horizontal hyperbolas, 853 Horizontal lines of orientation, 504 slope of, 101 Horizontal line test, 342 Horizontal parabolas, 866–867 Horizontal shifts, 159–160, 489–490 Horizontal translations, 159–160, 488–491 Horizontal unit vectors, 664, 665 Human life expectancy, 108 Hyperbolas applications of, 859–860 central, 853, A–59 definition of, 858 equation of, 152, 852–857, A–58 – A–59 explanation of, 852, 925 focal chord of, 864 foci of, 838, 857–859 on graphing calculators, 861 graphs of, 852–859, 897 horizontal, 853 with rational/irrational values of a and b, 875 vertical, 854 Hyperbolic trigonometric functions, 558 Hypocycloid with four cusps, 917 Hypotenuse, 499, A–46

I Identities. See also specific identities additive, A–10 applications of, 574–575 cofunction, 561 constructing and verifying, 545–546, 552–555, 619 creating, 552–553 double-angle, 468–570, 582 due to symmetry, 544 equations that are not, 547–548 explanation of, 4, 544, 581–582, 620–621 fundamental, 544, 581, 619 on graphing calculator, 548 half-angle, 571–572, 582 multiplicative, A–10 power reduction, 570–571, 582

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Index product-to-sum, 572–573 Pythagorean, 544, 545, 581 ratio, 544, 581 reciprocal, 544, 581 to solve trigonometric equations, 611–612 sum and difference, 558–563, 582, 620 sum-to-product, 573–574 trigonometric, 604 Identity function, characteristics of, 157 Imaginary numbers historical background of, 33, 41 identifying and simplifying, 33–34 Imaginary unit, 33, 972 Impedance, 694 Implied domain, 127–128 Inconsistent systems explanation of, 724 of linear equations in three variables, 736–737 of linear equations in two variables, 724–725 matrices and, 774 Increasing functions, 142–144 Independent systems, 724 Independent variables, 84 Indeterminate case, 1040, 1067 Index of refraction, 608 Index of summation, 943 Induction. See Mathematical induction Induction hypothesis, 968 Inductive reactance, 694 Inequalities. See also Linear inequalities absolute value, 26–28, 32, 80–82 additive property of, 15, 16 applications of, 18–20, 309–310 compound, 16–18 on graphing calculator, 334 interval tests to solve function, 309 linear in one variable, 14–18 linear in two variables, 755–757 method to solve, 16 multiplicative property of, 15 polynomial, 304–306, 310–311, 330 push principle to solve, 334–335 quadratic, 303–304 rational, 307–308, 310–311, 330 symbols for, 14, 15, A–2 – A–3 systems of nonlinear, 878–879 using graphs to solve, 141–142 writing mathematical models using, A–3 Infinite geometric series, 959–960 Infinite sequences, 940 Infinity, limits at, 1025, 1048–1051 Infinity symbol, 14 Initial point, 659 Initial side, 429 Input value, A–9 Instantaneous rates of change, 1058–1060, 1068 Instantaneous velocity, 1056 Integers, A–1 Intercept/intercept form of linear equations, 120 Intercept method, 98

Interest applications of, 401–402 compound, 200, 397–399, 413 continuously compounded, 399–400, 413 on graphing calculators, 399, 404–405 required, 200 simple, 397, 413 Interest earnings, 108 Intermediate value theorem (IVT) explanation of, 245–246 on graphing calculators, 252 Intersection, of sets, 17 Interval notation, 14 Intervals where function is increasing or decreasing, 142–145 where function is positive or negative, 141–142 Interval test method, 309 Invariants, 901, 902 Inverse additive, A–10 of matrices, 793–794, 803, 823–824 multiplicative, A–10 Inverse cosecant functions, 589 Inverse cosine functions, 585–587 Inverse cotangent functions, 589, 590 Inverse functions algebraic method to find, 344–346 applications of, 348–349 explanation of, 343–344 on graphing calculator, 349, 592 graphs of, 346–348 ordered pairs and, 344 Inverse secant functions, 589–590 Inverse sine functions, 582–585 Inverse tangent functions, 587, 590, 591 Inverse trigonometric functions cosecant, 589 cosine, 585–587 cotangent, 589, 590 on graphing calculator, 584, 588, 592 key concepts on, 621–622 principal roots and, 600 secant, 589–590 sine, 582–585 tangent, 587, 590, 591 used to evaluate compositions, 588–589 Inverse variation, 319–320, 331 Involute of circle, 521 Irrational numbers, 358, A–1 – A–2 Irrational zeros, 225–226 Irreducible quadratic factors, 244

J Joint variation, 320–321 Jump discontinuity, 1045

L Lateral surface area of cone, 67 Latitude, 433 Latus rectum, 873

I-7

Law of cosines alternative form for, 655 applications using, 649–650 evaluation of, 648 explanation of, 647, 706 scaled drawings and, 673–674 side-angle-side triangles and, 646–648 side-side-side triangles and, 648–649 Law of sines ambiguous case of, 637–639 applications of, 639–640 explanation of, 635 scaled drawings and, 637–638, 673–674 unique solutions and, 634–636 Leading coefficients explanation of, A–19, A–25 quadratic equation with, 46 Least common denominator (LCD), to clear fractions, 3 Leibniz, Gottfried von, 1024, 1034 Lemniscates, A–64 Leonardo of Pisa, 941 Less than, A–2 – A–3 Life expectancy, 108 Lift capacity formula, 22 Like terms, combining, A–11 Limaçons, graphs of, 892, A–63 Limiting value, 1024 Limits absolute value inequalities and, 80–81 approximations vs., 1024–1025 area under curve and, 1061–1063, 1068 definition of, 1026 of difference quotient, 1056–1061 existence of, 1029–1030 at infinity, 1025, 1048–1051 notation for, 1024 one-sided, 1027–1028 properties of, 1034–1041, 1046–1048, 1067 tables and graphs to estimate, 1025–1027, 1035, 1066 two-sided, 1027 use of algebra and limit properties to find, 1046–1048 use of direct substitution to find, 1044–1046 using graphs to evaluate, 1052–1053 Linear dependence, 733 Linear equations. See also Equations applications of, 104–105, 116–117 explanation of, 97 forms of, 138 general, 120 on graphing calculators, 105 graphs of, 97–98, 104, 203 identities and contradictions as, 4 intercept/intercept form of, 120 methods to solve, 3 in one variable, 2, 70 in point-slope form, 115 problem-solving guide for, 6–9 properties of equality to solve, 2–3 slope-intercept form and, 110–111

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I-8

Index

Linear equations—Cont. specified variables in literal equations and, 4–6 standard form of, 138 in two variables, 97–98 Linear factorization theorem explanation of, 242–243 proof of, 243 Linear functions, 955. See also Functions Linear inequalities. See also Inequalities; Systems of linear inequalities applications of, 18–20 compound, 16–18 explanation of, 755 method to solve, 15–16 in one variable, 14–18, 71 solutions sets and, 14 systems of, 757–759, 763–764, 816 in two variables, 755–757 Linear programming explanation of, 759–760 on graphing calculators, 822–823 optimal solutions and, 822–823 solutions to problems in, 760–763 Linear scale, 380 Linear systems. See Systems of linear equations; Systems of linear equations in three variables; Systems of linear equations in two variables Lines angle between intersecting, 520 characteristics of, 138 coincident, 724 distance between point and, 833–834 equation of, 616 finding equation of, 112 horizontal, 101 parallel, 102–104, 113–114 perpendicular, 103–104, 114–115 secant, 114, 119 slope-intercept form and graph of, 111–115 slope of, 98–100, 935–936 tangent, 1056 vertical, 101 vertical boundary, 125–126 Line segments directed, 658–659 midpoint of, 87 Lissajous figure, 913 Literal equations explanation of, 4 solving for specified variable in, 4–6 Local maximum, 144 Logarithmic equations explanation of, 381 method to solve, 381, 383, 387–388 solutions to systems of, 878 Logarithmic functions applications of, 390–392 domain of, 371 explanation of, 367–368, 411–412 graphs of, 369–370 Logarithmic scales, 380

Logarithms applications for, 372–375 base-b, 381 change-of-base formula and, 386–387 common, 368 fundamental properties of, 381 on graphing calculator, 368–369, 372, 389 natural, 368 properties of, 383–387, 412–413, 418, 421 uniqueness property of, 387–388 Logistic equations, on graphing calculator, 416–417 Logistic growth, 390, 394 Long division explanation of, 230–231 nonvertical asymptotes and, 279 Longitude, 433 LORAN (long distance radio navigation system), 857 Lorentz transformation, A–31 Lower bound, 250

M Mach numbers, 570 Magnitude of earthquakes, 372 explanation of, 372 of vector, 659, 661 of velocity, 659 Major axis, of ellipse, 841 Malus’s law, 578 Mapping notation, 84, 191 Market equilibrium, 882 Mathematical induction applied to sums, 966–969 explanation of, 1009 general principle of, 969–971 subscript notation and composition of functions and, 966 Mathematical models algebraic expressions and, A–8 – A–9 using inequalities to write, A–3 Matrices addition and subtraction of, 781–782 algebra of, 780–787, 817 applications using, 774–775 augmented, 770–772, 794, 823–824, A–56 coefficient, 770 of constants, 770 decomposing rational expressions using, 810 determinants and singular, 796–800 determinants of general, A–57 entries of, 769 equality of, 780–781 explanation of, 769–770 on graphing calculator, 775–776, 785–786, 795, 800 identity, 791–793 inverse of, 793–794, 823–824 multiplication and, 782–787, 791–792 order of, 780

reduced row-echelon form of, 773, A–56 row-echelon form of, 773 solving systems using, 771–774, 817 Matrix equations, 795–796, 818 Matrix multiplication explanation of, 783–786 properties of, 786–787 Maximum values explanation of, 144 of functions, 144–145, 337–338 on graphing calculator, 149–150 method to find, 631 Midinterval points, 263, 265 Midpoint, of line segment, 87 Midpoint formula, 87–88, 832 Minimum values explanation of, 144 of functions, 144–145, 337–338 on graphing calculator, 149–150 method to find, 631 Minor axis, of ellipse, 841 Minors, 798–799. See also Associated minor matrices Minutes, 427 Mixture problems, 8–9, 725 Models of systems of equations, 725–726 of tides, 534–535 Modulus, 41, 689 Monomials, A–18 Monotonic functions, 356 Motion projectile, 574–575, 712 tidal, 534–535 uniform, 8–9, 726, 729 Multiple angle identities, 704 Multiplication. See also Product associative property of, A–10 commutative property of, A–9 – A–11 of complex numbers, 36–38 matrix, 783–787, 791–793 of radical expressions, A–45 – A–46 of rational expressions, A–33, A–34 scalar, 659, 782–783 Multiplicative identity, A–10 Multiplicative property of absolute value, 25 of equality, 2, 3 of inequality, 15 Multiplicity roots of, 141 vertical asymptotes and, 276–277 zeroes of, 244, 261–264

N Nappe, 834 Natural exponential functions, 358 Natural logarithms, 368 Natural numbers, 954, A–1 Navigation, 655 Negative angles, 429, 430 Negative exponents, A–16, A–17 Negative functions, 141–142

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Index Negative numbers as exponents, A–16, A–17 square root of, 36 Negative slope, 100 Newton, Isaac, 1024, 1034 Newton’s law of cooling, 360 Nondistinguishable permutations, 979 Nonexclusive events, 991–992 Nonlinear asymptotes, 279, 291 Nonlinear systems of equations. See Systems of nonlinear equations Nonremovable discontinuities, 276 Nonrigid transformations, 164 Normal force, 595 Notation/symbols angle, 426 degree, 426 delta (), 99 exponential, A–4 – A–5 function, 128–130, 190 grouping, 187 identity, 559 inequality, 14, 15, 17, 26, A–2 – A–3 interval, 14 inverse trigonometric functions, 584 limit, 1024 mapping, 84, 191 scientific, A–17 – A–18 set, 14, 17, A–1 square and cube root, A–5 subscript, 966 summation, 943–944 variable, A–3 vector, 658–659 nth roots, method to find, 701–702 nth roots theorem, 700–701 nth term of arithmetic sequence, 950–951 of geometric sequence, 957–958 of sequence, 940 Number line, 14, A–4 Number puzzles, 62 Numbers. See also Complex numbers imaginary, 33–34, 41 irrational, 358, A–1 – A–2 natural, 954, A–1 negative, 36, A–16, A–17 rational, A–1 – A–2, A–42 real, A–1, A–9 – A–11 sets of, A–1 – A–2 whole, A–1 Numerical coefficients, A–8. See also Coefficients

O Object variables, 4–5 Oblique asymptotes, 291–293 Oblique triangles, 634, 705 Obtuse angles, 427 Odd functions, 140–141, 463 Odd multiplicity, 244 One-dimensional graphs, 732 One-sided limits, 1027–1028 One-to-one functions, 342–343, 410, 587

Ordered pairs, 85, 344 Ordered triples, 732–733 Order of operations evaluating expressions using, A–5 – A–6 explanation of, A–4 – A–5 Origin explanation of, 85 symmetry to, 140 Orthogonal components of vectors, 675, 680–682 Oscillation, spring, 30 Outputs, A–9 Output value, 339

P Painted area on canvas formula, 67 Parabolas analytic, 870, 926 applications of, 870 equation of, 868–870 explanation of, 835 focal chord of, 869, 873 focus-directrix form of equation of, 868–870 graphs of, 141, 164 horizontal, 866–867 with horizontal axis, 866–867 polar equation for, 903 vertical, 86, 868 Parabolic segment, right, 872 Parallel lines explanation of, 102–104 finding equation for, 113–114 slope of, 102, 103 Parallelogram method, 668 Parallelograms formula for area of, 520 proof of determinant formula for area of, A–60 – A–61 Parameter, 724, 739, 913 Parametric curves, 913–914 Parametric equations applications of, 917–919 explanation of, 313, 913, 928 on graphing calculators, 918–920 graphing curves and, 916–917 in rectangular form, 915 sketching curves defined parametrically and, 913–914 Parent function, 164 Partial fraction decomposition. See Decomposition Partial fractions explanation of, 743 telescoping sums and, 751–752 Partial sum, 942, 951–952, 958–959 Pascal, Blaise, 999 Pascal’s triangle, 999–1001 Perfect cubes, A–4 Perfect squares, A–4, A–27 Perfect square trinomials, A–27 – A–28 Perihelion, 905

Perimeter of ellipse, 849 of rectangle, 789 of trapezoid, 655 Period formula for sine and cosine, 462 for tangent and cotangent, 475–476 Periodic functions, 456–457 Periods of cosecant and secant functions, 463 on graphing calculator, 465 of sine and cosine functions, 456, 457, 461, 462 of tangent and cotangent functions, 475–476 Permutations distinguishable, 977–978 nondistinguishable, 979 Perpendicular distance from point to line, 837 Perpendicular lines explanation of, 103 finding equation for, 114–115 slope of, 103, 104 Phase angle, 694 pH level, 376 Pick’s theorem, 135 Piecewise-defined functions applications of, 177–179 domain of, 172–173 explanation of, 172, 208 on graphing calculator, 179–180 graphs of, 173–177 , 582 Planes ecliptic, 931 half, 756 in space, 732 Point of inflection, 126 Points distance between line and, 833–834 test for collinear, 810 trigonometric ratios and, 512–515 Point-slope form explanation of, 115, 138 linear equations in, 115 used to find equation model, 116–117 Poiseuille’s law, A–31 Polar axis, 883 Polar-axis symmetry, 890 Polar coordinates converted to rectangular coordinates, 541, 886 converting from rectangular coordinates to, 540–541, 885 distance formula in, 894 explanation of, 883, 927 midpoint formula in, 894 plotting points using, 883–885 Polar equations explanation of, 887–891, 896, 903, 906, 927 systems of, 936–937

I-9

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Index

Polar form applications of conics in, 905–906 conic equations in, 902–905 conic rotations in, 931–932 of ellipse, 910 of equation of line, 910 use of, 883 Polar graphs development of, 887 instantaneous rates of change and, 935–936 r-value analysis and, 887–889, 891 symmetry and families of, 889–892, A–63 – A–64 use of, 927 Polar symmetry tests, 890, 896 Polygons area of, 550, 551, 1024–1025 perimeter of, 551 Polynomial equations checking solutions to, 699–700 with complex coefficients, 55 explanation of, 56 higher-degree, 56–57 Polynomial form, equation of hyperbola in, 856–857 Polynomial functions applications of, 251, 266 applying difference quotient to, 215 complex, 242 graphs of, 257–266, 328 (See also Polynomial graphs) zeroes of, 242–250, 327–328 Polynomial graphs end behavior of, 258–261 guidelines for, 264–266, 328 identification of, 257–258 turning points and, 258 zeroes of multiplicity and, 261–264 Polynomial inequalities explanation of, 304, 330 on graphing calculator, 305, 306, 310–311 method to solve, 304–306, 309 Polynomials addition and subtraction of, A–19 applications of, 251, 266 characteristic, 813 cubic, A–25 degree of, A–18 division of, 231, 233, 234 explanation of, A–18 factoring, A–24 – A–29 identifying and classifying, A–18 – A–19 prime, A–26 as product of linear factors, 243–244 products of, A–19 – A–21 quadratic, A–25 – A–27 quartic, 269 rational zeroes of, 247–249 in standard form, A–19 use of remainder theorem to evaluate, 235 Population density, 285 Position vectors, 660 Positive angles, 429, 430 Positive functions, 141–142

Positive slope, 100 Power functions, 359 Power property after substitution, A–15 of equality, 59, 61 of exponents, A–14, A–17 of logarithms, 384, 385 proof of, 384 quotient to, A–14 – A–15 simplifying terms using, A–15 Power reduction formula, 570 Power reduction identities, 570–571, 582 Pressure wave, 492 Primary interval, for sinusoidal graph, 490 Prime polynomials, A–26 Principal roots explanation of, 599 trigonometric equations and, 599–603 Principal square roots, A–5 Probability binomial, 1004, 1005 elementary, 987–988 explanation of, 987, 1010 on graphing calculator, 992–993 nonexclusive events and, 991–992 properties of, 988–990 quick-counting and, 990–991 Problem solving explanation of, 6 guide for, 7–9 trigonometry and, 715–716 Product property of exponents, A–14, A–17 of logarithms, 384 proof of, 384 Products. See also Multiplication of complex conjugates, 36–37 of complex numbers, 691 of functions, 187–188 of polynomials, A–19 – A–21 to power property, A–14, A–17 in trigonometric form, 691–692 Product-to-sum identities, 572–573 Projectiles explanation of, 598, 917–918 height of, 54 motion of, 574–575, 682–683, 712 range of, 608, 685 Projectile velocity, 145–146 Proof by induction, 966 Property of negative exponents, A–16, A–17 Property of radicals, A–43 – A–44 Protractors, 426 Push principle, 334–335 Pythagorean identities explanation of, 544, 545, 581 use of, 651 verifying, 551 Pythagorean theorem explanation of, 428, A–46 in trigonometric form, 468 use of, 501, 502, 646, A–47 Pythagorean triples, 452

Q Quadrantal angles, 430 Quadrantal points, 441 Quadrant analysis, 517, 547 Quadrants, 85, 442 Quadratic, 43 Quadratic equations checking solutions to, 77–78 completing the square to solve, 45–46 explanation of, 42, 43, 73 with leading coefficient, 46 methods to solve, 50, 51–52 quadratic formula to solve, 47–49 square root property of equality to solve, 44–45 standard form of, 42, 47 zero property product to solve, 43–44 Quadratic factors decomposition template for, 746 irreducible, 244 Quadratic form equations in, 61–62 explanation of, A–28 factoring, A–27 – A–29 Quadratic formula applications of, 50–51 discriminant of, 48–49 explanation of, 47 solving quadratic equations by using, 47–49 Quadratic functions completing the square to graph, 220–221 end behavior of, 259 explanation of, 220, 242, 326, 489, 955 extreme values and, 223–225 finding equation of, 223 graphs of, 220–224 in standard form, 220, 224 vertex formula to graph, 222–223 Quadratic inequalities, 303–304 Quadratic polynomials explanation of, A–25 factoring, A–25 – A–27 Quartic polynomials, root tests for, 269 Quick-counting techniques, 990–991 Quotient property of exponents, A–15, A–17 of logarithms, 384 Quotients of functions, 187–189 to power property, A–14 – A–15

R Radians angle measure in, 431 arc length in, 431 area in, 431–432 conversion between degrees and, 432–433 explanation of, 430–431 Radical equations explanation of, 59 methods to solve, 59–60, 79–80 power property of equality and, 59, 61

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Index Radical expressions. See also Cube roots; Square roots addition and subtraction of, A–44 evaluation of, A–40 explanation of, A–40 multiplication and division of, A–45 – A–46 rational exponents and, A–41 – A–42 simplification of, A–40 – A–41, A–43 – A–46 Radical functions applying difference quotient to, 216 graphs of, 337 limits at infinity for, 1051 Radicals explanation of, A–5 properties of, A–43 – A–44, A–52 Radioactive elements, 403–404 Radius of circle, 89, 91, 550 of circumscribed circle, 643 of sphere, 352 Range explanation of, 84 of functions, 126–127, 130 of projectile, 608 Rates of change average, 130–132, 145–149 difference quotient and, 146–149, 215–216 explanation of, 99, 204, A–37 limits and instantaneous, 1058–1060, 1068 polar graphs and instantaneous, 935–936 slope as, 99–100 Ratio identities, 544, 581 Rational equations explanation of, 57 methods to solve, 57–58 Rational exponents explanation of, A–41 – A–42 simplifying radical expressions with, A–42, A–44 solving equations with, 61 Rational expressions addition and substraction of, A–34 – A–35 compound, A–35 – A–36 explanation of, A–31, A–51 – A–52 formulas and algebraic models and, A–36 – A–37 fundamental property of, A–31 multiplication and division of, A–32 – A–34 in simplest form, A–31 – A–32 Rational functions applications of, 281–282, 294–297 end behavior of, 273–274 equation of, 281 explanation of, 272–273, 329, 330 on graphing calculator, 282 graphs of, 279–281, 290–294, 329 horizontal asymptotes of, 277–278 limits at infinity for, 1050 with oblique and nonlinear asymptotes, 291–293 removable discontinuities and, 289–290

rewriting, 742 vertical asymptotes of, 275–277 Rational inequalities explanation of, 307, 330 on graphing calculator, 307, 308, 310–311 method to solve, 307–308 Rationalizing the denominator, A–45 – A–46 Rational numbers, A–1 – A–2, A–42 Rational zeroes theorem, 246–249 Real numbers absolute value of, A–3 – A–4 explanation of, 242, A–1 order property of, A–3 properties of, A–9 – A–11, A–51 subsets of, 242 trigonometry of, 440, 446–449, 485 Real polynomials, intermediate value theorem and, 245–246 Real roots, 599, 602–603 Reciprocal function, 273, 896–897 Reciprocal identities, 544, 581 Reciprocal powers, 1050 Reciprocals, A–10 Reciprocal square functions, 273–274 Rectangles central, 855 reference, 458, 460, 462 Rectangular box, surface area of, 227 Rectangular coordinates converted to polar coordinates, 540–541, 885 converting from polar coordinates to, 541, 886 explanation of, 85 vectors and, 660–662 Rectangular form complex numbers in, 688 parametric equations in, 915 working in, 883 writing polar equations in, 896 Rectangular solid, 741 Recursive sequences, 941–942 Reduced row-echelon form, of matrices, 773, A–56 Reference angles explanation of, 444 trigonometric functions and, 515–517 Reference arc, 446 Reference intensity, 372 Reference rectangles explanation of, 458 to graph cosine, 460, 462 Reflections, 161, 162 Relations domain of, 84 explanation of, 84, 202 as functions, 123 graphs of, 85–87 Remainder theorem, 234–235 Removable discontinuities, 176, 289–290, 297, 1045 explanation of, 176 on graphing calculators, 297 rational functions and, 289–290

I-11

Repeated zeroes, 334 Required interest rate, 324 Resistance, 694 Resultant vectors, 662 Revenue models, 64–65 Right angles, 426 Right parabolic segment, 872 Right triangles applications of, 505 explanation of, A–46 relationships in, 438, 539–541 solutions to, 501–503 trigonometry of, 499–505, 528–529 Rigid transformations, 164 Roots. See also Cube roots; Square roots complex, 37 of equations, 2 extraneous, 58, 63–64, 387 principal, 599 real, 599, 602–603 Roots of multiplicity, 141 Root tests for quartic polynomials, 269 Roses, graphs of, A–63 Rotated conics, 896–901, 931–932 Rotation of axes, 898–902, 932–934 Rotation of axes formulas, 898 Row-echelon form, 773, A–56 Rule of fourths, 458, 461, 487, 490 r-value analysis, 887–889, 891

S Sample outcomes, 975, 976 Sample space, 975, 976, 987, 988 Sand dune function, 182 Scalar, 658 Scalar measurements, 658 Scalar multiplication explanation of, 659 matrices and, 782–783 Scalar quantities, 658 Scaled drawings law of cosines and, 673–674 law of sines and, 637–638, 673–674 Scientific notation, A–17 – A–18 Secant function characteristics of, 464 graphs of, 463–464 inverse, 589–590 Secant line explanation of, 114, 119 slope of, 145 Seconds, 427 Seed element, 941 Semicircles, 87 Semi-hyperbolas, equation of, 864 Sequences. See also Series; specific types of sequences applications of, 944–945 arithmetic, 949–953, 1007–1008 explanation of, 940, 1007 finding terms of, 940–941 finite, 940 geometric, 401, 956–961, 1008 on graphing calculator, 945

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Index

Sequences—Cont. infinite, 940 nth term of, 940 recursive, 941–942 terms of, 940 Series. See also Sequences explanation of, 940, 1007 finite, 942 geometric, 956, 959–961 on graphing calculator, 945 infinite, 1014–1015 Set notation, 14 Sets intersection of, 17 of numbers, A–1 – A–2 Shifted form, of trigonometric functions, 489, 490 Shifts horizontal, 159–160, 489–490 vertical, 159 Side-angle-side (SAS) triangles, 646–648 Side-side-angle (SSA) triangles, 636–640 Side-side-side (SSS) triangles, 648–649 Sigma notation. See Summation notation Sign analysis, 517 Similar triangles, 428 Simplest form, rational expressions in, A–31 Sine law of, 634–640 sum and difference identities for, 560–562 table of values for, 459 triple angle formula for, 642 Sine function amplitude of, 460–461 applications of, 487, 492, 493 characteristics of, 457 graphs of, 454–458, 460–462 inverse, 582–585 Singular matrices, 796–800 Sinking fund, 402 Sinusoidal graphs, 487 Sinusoidal models, 490 Sinusoidal pattern, 486 Slope explanation of, 98–99 of horizontal and vertical lines, 101–102 of parallel lines, 102, 103 of perpendicular lines, 103, 104 positive and negative, 100 as rate of change, 99–100 of secant line, 145 of tangent lines, 935–936 Slope formula explanation of, 99, 138 as rate of change, 99–100 Slope-intercept form explanation of, 111, 138 graph of line using, 113–115 linear equations and, 110–111 method to find, 112 Smooth graphs, 258 Snell’s law, 608 Solution region, 756 Solutions, 2, 733

Solution sets, 14 Sound energy, 492 Sound waves, 492–493, 575, 626–627 Spheres, volume of, 170 Spiral curves, A–63 Spiral of Archimedes, 521, 895 Spring oscillation, 30 Springs, 491–492 Square root function, characteristics of, 157 Square root property of equality, 44–45 Square roots. See also Radical expressions explanation of, A–5 of negative numbers, 36 notation for, A–5 principal, 33, A–5 simplification of, A–40 – A–41 Squares. See also Completing the square of binomials, A–21 factoring difference of, A–27 perfect, A–4, A–27 Square systems, 737. See also Systems of linear equations in three variables; Systems of linear equations in two variables Squaring function, characteristics of, 157 Standard angles, 535 Standard form complex numbers in, 34–35 equation of circle in, 89–91, 840 equation of ellipse in, 841 equation of hyperbola in, 854–856 linear equations in, 138 polynomials in, A–19 quadratic equations in, 42, 47 quadratic functions in, 220, 224 trigonometric functions in, 489, 490 Standard position, of angles, 430 Statement of proof, 1034–1035 Static equilibrium, 713 Static trigonometry, 512 Statistics, 987 Step functions, 178–179 Stirling’s formula, 984 Straight angles, 426 Stretches, 162–163 Subscript notation, 966 Subsets, 242, A–1, A–2 Substitution to check solutions, 3, 58 direct, 1045–1046 explanation of, 721 to solve systems of equations, 721–722 to solve systems of nonlinear equations, 876–877 in trigonometry, 630 u-, A–28 – A–29 Subsystems, 735 Subtends, 430 Subtraction of complex numbers, 35 of matrices, 781–782 of polynomials, A–19 of radical expressions, A–44 of rational expressions, A–34 – A–35 of vectors, 663

Sum and difference identities for cosine, 558–560 explanation of, 582, 620 for sine and tangent, 560–562 to verify other identities, 562–563 Sum identity for cosine, 559. See also Sum and difference identities Summation applications of, 974, 1019–1021 index of, 943 properties of, 944 Summation notation, 943–944 Sum-to-product identities, 573–574 Supplementary angles, 426 Supply curves, 882 Surface area. See also Area of cylinder, 11, 54, 200, 299 of rectangular box, 227 Symbols. See Notation/symbols Symmetry axis of, 139 even functions and, 139 to find function values, 455–456 graphs and, 138–141 identities due to, 544 odd functions and, 140–141 to origin, 140 in polar graphs, 889–892, 896 in polynomial functions, 264 unit circles and, 441–442 Synthetic division explanation of, 232–233, 327, A–54 – A–55 factor theorem and, 237 polynomials and, 233 remainder theorem and, 234, 235 Systems of equations decomposition using, 750–751 determinants to solve, 806–810 elimination to solve, 722–724, 734–736 equivalent, 734 explanation of, 720, 814, 815 on graphing calculators, 739 graphs to solve, 720–721 inconsistent and dependent, 724–725, 736–738 matrices and, 769–774, 817, 818 modeling and, 725–726 solutions to, 720, 732–734 substitution to solve, 721–722 Systems of linear equations in three variables applications to, 738–739 Cramer’s rule and, 808–809 elimination to solve, 734–736 explanation of, 815 inconsistent and dependent, 736–738 solutions to, 732–733 visualizing solutions to, 732 Systems of linear equations in two variables Cramer’s rule and, 807–808 elimination to solve, 722–724 explanation of, 720, 814 on graphing calculator, 727 graphs to solve, 720–721

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Index inconsistent and dependent, 724–725 modeling and, 725–726 substitution to solve, 721–722 Systems of linear inequalities applications of, 758–759 explanation of, 757, 816 on graphing calculators, 763–764 linear programming and, 759–763 solutions to systems of, 757–758 Systems of nonlinear equations applications of, 879 elimination to solve, 877–878 explanation of, 875, 926 solutions to, 875–876 substitution to solve, 876–877 Systems of nonlinear inequalities, 878–879, 926 Systems of polar equations, 936–937

T Tables, to estimate limits, 1025–1027, 1035, 1066 Tangent meaning of term, 483, 551 sum and difference identities for, 560–562 table of values for, 472–473 Tangent function applications of, 477–478 characteristics of, 474 graphs of, 471–477 inverse, 587, 590, 591 period of, 475–476 Tangent line, 1056 Telescoping sums, 751–752 Terminal point, 659 Terminal side, 429 Theta (), 426 30-60-90 triangles, 428, 429, 499 3 x 3 systems. See Systems of linear equations in three variables Tidal motion, 534–535 Toolbox functions direct variation and, 316–319 explanation of, 157–158, 207 graphs of, 157, 158 transformations and, 212–213 Transcendental functions, 366. See also Logarithmic functions; Trigonometric functions explanation of, 604 Transformations area under curve and, 216–217 explanation of, 158, 527 of general function, 163–165 to graph logarithmic functions, 370 of graphs, 158–159, 164, 357 horizontal reflections and, 162 nonrigid, 164 of parent function, 164 rigid, 164 solving equations involving, 612–613 of trigonometric graphs, 457–458, 487, 490 vertical reflection and, 161 via composition, 213

Transformed function, 164 Translations horizontal, 159–160, 488–491 vertical, 159, 486–488 Transverse axis, 853 Tree diagrams, 975 Trial-and-error process, A–26 Trials, 975 Triangles area of, 299, 557, 650–653, 706, 809 equilateral, 438, 696 explanation of, 427 45-45-90, 428, 429, 499 law of sines to solve, 635–639 oblique, 634, 705 properties of, 427–429, 635 right, 438, 499–505, 528–529, 539–541, A–46 side-angle-side, 646–648 side-side-angle, 636–640 side-side-side, 648–649 similar, 428 30-60-90, 428, 429, 499 unit circle and special, 442–445 Triangular form of matrix, 771, 772 Triangularizing, augmented matrix, 771–772 Trigonometric equations algebraic methods and, 610–611 applications of, 613–614 explanation of, 599 of form A sin (Bx  C)  D = k, 612–613 on graphing calculators, 604–605 identities and, 604, 611–612 methods to solve, 600–603, 622, 623 real roots and, 602–603 roots of, 599–600 Trigonometric form complex numbers in, 688–690, 693–695, 708 converting to rectangular form to, 690 equation of line in, 616 products and quotients in, 691–692 Trigonometric functions. See also specific trigonometric functions of acute angles, 500 of any angle, 535 applications of, 517–518 domains of, 447 evaluation of, 446, 512–515 fundamental identities and, 544–546 on graphing calculator, 465, 478–479, 491, 584, 585, 588 graphs of, 627 hyperbolic, 558 inverse, 582–592 points on unit circle and, 445–449 reference angles and, 515–517 written in terms of another, 546–547 Trigonometric graphs applications of, 491–493, 527–528 of cosecant function, 463–464 of cosine function, 459–462 of cotangent function, 473–477 explanation of, 153, 525, 526, 601

of secant function, 463–464 of sine function, 454–458, 460–462 of tangent function, 471–477 transformations of, 457–458 value in studying, 153 writing equations from, 464 Trigonometric identities, 604 Trigonometric ratios points and, 512–515 values of, 499–500 Trigonometry area of triangle and, 650–653 coordinate plane and, 512–518, 529–530 dynamic, 512 historical background of, 426, 427 problem solving and, 715–716 of real numbers, 440, 446–449, 485 of right triangles, 499–505, 528–529 static, 512 substitutions in, 630 Trinomials explanation of, A–18, A–19 factoring, A–25 – A–26 perfect square, A–27–A–28 quadratic, A–25, A–28 Triple angle formula, for sine, 642 Tunnel clearance, 879 Turning points, 258 Two-dimensional graphs, 732 Two-sided limits, 1027 2 x 2 systems. See Systems of linear equations in two variables

U Uniform motion, 8–9, 727, 729 Union, 17 Uniqueness property, 359–361, 387–388 Unique solutions, 733 Unit circles explanation of, 441, 451, 524 points on, 441–442 special triangles and, 442–445 symmetry and, 441–442 trigonometric functions and points on, 445–448 Unit vectors, 665 Upper and lower bounds property, 249–250 Upper bound, 250 u-substitution to factor expressions in quadratic form, 61, 62, A–28 – A–29 to solve trigonometric equations, 603

V Variable amplitudes, 534–535 Variable costs, 725 Variables dependent, 84 explanation of, A–3 independent, 84 object, 4–5 Variable terms, A–8

I-13

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Index

Variation constant of, 316 direct, 316–319, 331 inverse, 319–320, 331 joint or combined, 320–321 Vector projections, 675, 680–682 Vector quantities, 658 Vectors algebraic, 664–665 applications of, 659, 666–668, 677–678, 707–708 component of u along v, 675–677 dot products and angle between two, 678–680 equilibrium and, 674–675 equivalent, 660 explanation of, 659, 706–707 force, 666–667, 678 geometry of, 828 on graphing calculators, 668–669 horizontal and vertical components of, 660, 662, 664–669 horizontal unit, 664, 665 magnitude of, 659, 661, 671 notation and geometry of, 658–659 operations on, 662–664 orthogonal components of, 675, 680–682 position, 660 projectile motion and, 682–683, 712 projvu and, 680–681 properties of, 664 rectangular coordinate system and, 660–662 resultant, 662 subtraction of, 663 in three dimensions, 716

unit, 665 vertical unit, 664, 665 Velocity angular, 434–435 average rate of change applied to projectile, 145–146 explanation of, 434 on graphing calculator, 575 instantaneous, 1056 linear, 217, 434–435 Vertex of absolute value graph, 125 of angle, 426 of cone, 834 of ellipse, 841 of hyperbola, 853, 854 Vertex formula, 222–223 Vertex/intercept formula, 227 Vertical asymptotes multiplicities and, 276–277 of rational functions, 275–277, 289 Vertical axis, 732 Vertical-axis symmetry, 890 Vertical boundary lines, 125–126 Vertical change, 98–99 Vertical component of vectors, 660, 661, 664–669 Vertical format, A–9 Vertical hyperbolas, 854 Vertical lines, slope of, 101 Vertical line test for functions, 124–125 Vertical parabolas, 86, 868 Vertical shifts, 159 Vertical translations, 159, 486–488 Vertical unit vectors, 664, 665 Voltage, 694

Volume of cone, 616–617 of cube, A–22 of cylinder, 616 formulas for, 766 of open box, 240 of sphere, 170 surface area of cylinder with fixed, 299

W Whole numbers, A–1 Witch of Agnesi, 922 Work, 677–678 Wrapping function, 485

X x-intercept, 98, 224 xy-plane, 85

Y y-axis, 139, 140 y-intercept, 98

Z Zeroes approximating real, 271–272 as exponents, A–16, A–17 of function, 141 on graphing calculator, 149–150 irrational, 225–226 of multiplicity, 244, 261–264 of polynomial functions, 242–250, 327–328 use of factor theorem to find, 237 Zero placeholder, 233 Zero product property, 43–44

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