32,985 8,461 352MB
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THOMAS'
CALCULUS Twelfth Edition
Based on the original work by
George B. Thomas, Jr. Massachusetts Institute of Technology as revised by
Maurice D. Weir Naval Postgraduate School Joel Hass University of California, Davis
AddisonWesley Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo
EdltoriDCblef: Deirdre Lynch Semor AcqailltiODl Editor: William Hoffinan Semor Project Editor: Rachel S. Reeve Associate Editor: Caroline Celano Associate Project Editor: Leah Goldberg Semor Managing Editor: Karen Wernhohn Semor Prodnctton Snpenlsor: Sheila Spinoey Senior Design Supervisor: Andrea Nix Digital Assets Manager: Mariaone Groth Media Producer: Lin Mahoney Software Development: Mary Dumwa1d and Bob Carroll EIecutive Marketing Manager: Jeff Weidenaar MarketingAsmlant: Kendra Bassi Senior Author Support/fec:hnology Specialist: Joe Vetere Senior Prepreill Supervilor: Caroline Fell Manufacturing Manager: Evelyn Beaton Production Coordinator: Kathy Diamond Composition: Nesbitt Graphics, Inc. lli••tratioDl: Karen Heyt, lllustraThch Cover Design: Rokusck Design Cover image: Forest Edge, Hokuto, Hokkaido, Japsn 2004 © Michael Kenna About the cover: The cover image of a tree line on a snowswept landscape, by the photographer Michael Kenna, was taken in Hokkaido, Japan. The artist was not thinking of calculus when he composed the image, but rather, of a visual haiku consisting of a few elements that would spaIl< the viewer's imagination. Similarly, the minima1 design of this text allows the central ideas of calculus developed in this book to unfold to igoite the learner's imagination.
For permission to use copyrighted material, grateful acknowledgment is made to the copyright holders on page CI, which is hereby made part of this copyright page. Many of the desigoations used by manufacturers and sellers to distinguish their products are claimed as tradenulrks. Where those designations appear in this book, and AddisonWesley was aware of a trademark claim, the designations have been printed in initial caps or all caps.
Ubrary of Congress CataloginginPublication Data Weir, Maurice D. Thomas' Calculus I Maurice D. Weir, Joel Hass, George B. Thomas.12th ed. p.em ISBN 9780321587992 1. CalculUl!Textbooks. I. Hass, Joel. II. Thoroas, George B. (George Brinton), 19142006. III. Thomas, George B. (George Brinton), 19142006. Calculus. Iv. Title V. Title: Calculus. QA303.2.W452009b 515dc22
2009023069
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CONTENTS
1
Preface
ix
I Functions
1 1.1 1.2 1.3 1.4
Functions and Their Graphs I CombiJring Functions; Shifting and Scaling Graphs Trigonometric Functions 22 Graphing with Calcolators and Computers 30 QuEsTIONS TO GUIDE YOUR REVIEW PRACTICE EXERCISES
34
35 37
AoDmONAL AND ADvANCED EXERCISES
2
Limits and Continuity 2.1 2.2 2.3 2.4 2.5 2.6
39
Rates of Change and Tangents to Curves 39 Limit of a Function and Limit Laws 46 The Precise Definition of a Limit 57 OneSided Limits 66 Continuity 73 Limits Involving Infinity; Asymptotes of Graphs QuEsTIONS TO GUIDE YOUR REVIEW PRACTICE EXERCISES
98
Differentiation 3.1 3.2 3.3 3.4 3.5 3.6
84
96
97
AoDmONAL AND ADvANCED EXERCISES
3
14
102 Tangents and the Derivative at a Point 102 The Derivative as a Function 106 Differentiation Rules 115 The Derivative as a Rate of Change 124 Derivatives of Trigonometric Functions 135 The Chain Rule 142
iii
iv
Contents
3.7 3.8 3.9
4
Applications of Derivatives 4.1 4.2 4.3 4.4 4.5 4.6 4.7
5
Extreme Values of Functions 184 The Mean Value Theorem 192 Monotonic Functions and the First Derivative Test Concavity and Curve Sketching 203 Applied Optimization 214 Newton's Method 225 Antiderivatives 230 QuESTIONS TO GUIDE YOUR REVIEW 239 PRACTICE EXERCISES 240 ADDITIONAL AND ADvANCED EXERCISES 243
184
198
I Integration
246 5.1 5.2 5.3 5.4 5.5 5.6
6
Implicit Differentiation 149 Related Rates 155 Linearization and Differentials 164 QuESTIONS TO GUIDE YOUR REVIEW 175 PRACTICE EXERCISES 176 ADDITIONAL AND ADvANCED EXERCISES 180
Area and Estimating with Finite Sums 246 Sigma Notation and Limits of Finite Sums 256 The Definite Integral 262 The Fundamental Theorem of Calculus 274 Indefinite Integrals and the Substitution Method 284 Substitution and Area Between Curves 291 QuESTIONS TO GUIDE YOUR REVIEW 300 PRACTICE EXERCISES 301 ADDITIONAL AND ADvANCED EXERCISES 304
Applications of Definite Integrals 6.1 6.2 6.3 6.4 6.5 6.6
Volumes Using CrossSections 308 Volumes Using Cylindrical Shells 319 Arc Length 326 Areas of Surfaces of Revolution 332 Work and Fluid Forces 337 Moments and Centers of Mass 346 QuESTIONS TO GUIDE YOUR REVIEW 357 PRACTICE EXERCISES 357 ADDITIONAL AND ADvANCED EXERCISES 359
308
Contents
7
Transcendental Functions 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8
8
9
Inverse Functions and Their Derivatives 361 Natural Logarithms 369 Exponential Functions 377 Exponential Change and Separable Differential Equations Indeterminate Forms and VHopitai's Rule 396 Inverse Trigonometric Functions 404 Hyperbolic Functions 416 Relative Rates of Growth 424 QuEsTIONS TO GUIDE YOUR REVIEW 429 PRACTICE EXERCISES 430 ADDmONAL AND ADvANCED EXERCISES 433
Integration by Parts 436 Trigonometric Integrals 444 Trigonometric Substitotions 449 Integration of Rational Functions by Partial Fractions Integral Tables and Computer Algebra Systems 463 Numerical Integration 468 Improper Integrals 478 QuEsTIONS TO GUIDE YOUR REVIEW 489 PRACTICE EXERCISES 489 ADDmONAL AND ADvANCED EXERCISES 491
453
496
Solutions, Slope Fields, and Euler's Method 496 FirstOrder Linear Equations 504 Applications 510 Graphical Solutions of Autonomous Equations 516 Systems of Equations and Phase Planes 523 QuEsTIONS TO GUIDE YOUR REVIEW 529 PRACTICE EXERCISES 529 ADDmONAL AND ADVANCED EXERCISES 530
Infinite Sequences and Series 10.1 10.2 10.3 lOA 10.5
387
435
FirstOrder Differential Equations 9.1 9.2 9.3 9.4 9.5
10
361
Techniques of Integration 8.1 8.2 8.3 8.4 8.5 8.6 8.7
V
Sequences 532 Infinite Series 544 The Integral Test 553 Comparison Tests 558 The Ratio and Root Thsts
532
563
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10.6 10.7 10.8 10.9 10.10
11
Parametric Equations and Polar Coordinates 11.1 11.2 11.3 11.4 11.5 11.6 11.7
12
660
ThreeDimensional Coordinate Systems 660 Vectors 665 The Dot Product 674 The Cross Product 682 Lines and Planes in Space 688 Cylinders and Quadric Surfaces 696 QuESTIONS TO GUIDE YOUR REVIEW 701 PRACTICE EXERCISES 702 ADDITIONAL AND ADvANCED EXERCISES 704
VectorValued Functions and Motion in Space 13.1 13.2 13.3 13.4 13.5 13.6
610
Parametrizations of Plane Curves 610 Calculus with Parametric Curves 618 Polar Coordinates 627 Graphing in Polar Coordinates 631 Areas and Lengths in Polar Coordinates 635 Conic Sections 639 Conics in Polar Coordinates 648 QuESTIONS TO GUIDE YOUR REVIEW 654 PRACTICE EXERCISES 655 ADDITIONAL AND ADvANCED EXERCISES 657
Vectors and the Geometry of Space 12.1 12.2 12.3 12.4 12.5 12.6
13
Alternating Series, Absolute and Conditional Convergence 568 Power Series 575 Taylor and Maclaurin Series 584 Convergence of Taylor Series 589 The Binomial Series and Applications of Taylor Series 596 QuESTIONS TO GUIDE YOUR REVIEW 605 PRACTICE EXERCISES 605 ADDITIONAL AND ADvANCED EXERCISES 607
Curves in Space and Their Tangents 707 Integrals of Vector Functions; Projectile Motion 715 Arc Length in Space 724 Curvatore and Normal Vectors ofa Curve 728 Tangential and Normal Components of Acceleration 734 Velocity and Acceleration in Polar Coordinates 739 QuESTIONS TO GUIDE YOUR REVIEW 742 PRACTICE EXERCISES 743 ADDITIONAL AND ADvANCED EXERCISES 745
707
Contents
14
Partial Derivatives 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8 14.9 14.10
15
16
747 Functions of Several Variables 747 Limits and Continuity in Higher Dimensions 755 Partial Derivatives 764 The Chain Rule 775 Directional Derivatives and Gradient Vectors 784 Tangent Planes and Differentials 791 Extreme Values and Saddle Points 802 Lagrange Multipliers 811 Taylor's Formula for Two Variables 820 Partial Derivatives with Constrained Variables 824 QUESTIONS TO GUIDE YOUR REVIEW 829 PRACTICE ExERCISES 829 ADomONAL AND ADvANCED EXERCISES 833
Multiple Integrals 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8
836 Double and Iterated Integrals over Rectangles 836 Double Integrals over General Regions 841 Area by Double Integration 850 Double Integrals in Polar Form 853 Triple Integrals in Rectangular Coordinates 859 Moments and Centers of Mass 868 Triple Integrals in Cylindrical and Spherical Coordinates Substitutions in Multiple Integrals 887 QUESTIONS TO GUIDE YOUR REVIEW 896 PRACTICE ExERCISES 896 ADomONAL AND ADvANCED EXERCISES 898
875
Integration in Vector Fields 16.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8
vii
Line Integrals 901 Vector Fields and Line Integrals: Work, Circulation, and Flux 907 Path Independence, Conservative Fields, and Potential Functions 920 Green's Theorem in the Plane 931 Surfaces and Area 943 Surface Integrals 953 Stokes' Theorem 962 The Divergence Theorem and a Unified Theory 972 QUESTIONS TO GUIDE YOUR REVIEW 983 PRACTICE ExERCISES 983 ADomONAL AND ADVANCED EXERCISES 986
901
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17
SecondOrder Differential Equations 17.1 17.2 17.3 17.4 17.5
online
SecondOrder Linear Equations Nonhomogeneous Linear Equations Applications Euler Equations Power Series Solutions
Appendices
AP1 A.I A.2 A.3 AA A.5 A.6 A.7 A.S A.9
Real Numbers and the Real Line API Mathematical Induction AP6 Lines, Circles, and Parabolas AP1O Proofs of Limit Theorems APIS Commonly Occurring Limits AP2I Theory of the Real Numbers AP23 Complex Numbers AP25 The Distributive Law for Vector Cross Products AP35 The Mixed Derivative Theorem and the Increment Theorem
AP36
Answers to OddNumbered Exercises
A1
Index
11
A Brief Table of Integrals
T1
Credits
C1
PREFACE We have significantly revised this edition of Thomas' Calculus to meet the changing needs of today's instructors and students. The result is a book with more examples, more midlevel exercises, more figures, better conceptual flow, and increased clarity and precision. As with previous editions, this new edition provides a modern introduction to calculus that supports conceptual understanding but retains the essential elements of a traditional course. These enhancements are closely tied to an expanded version for this text of MyMathLab® (discussed further on), providing additional support for students and flexibility for instructors. Many of our students were exposed to the terminology and computational aspects of calculus during high school. Despite this familiarity, students' algebra and trigonometry skills often hinder their success in the college calculus sequence. With this text, we have sought to balance the students' prior experience with calculus with the algebraic skill development they may still need, all without undermining or derailing their confidence. We have taken care to provide enough review material, fully steppedout solutions, and exercises to support complete understanding for students of all levels. We encourage students to think beyond memorizing formulas and to generalize concepts as they are introduced. Our hope is that after taking calculus, students will be confident in their problemsolving and reasoning abilities. Mastering a beautiful subject with practical applications to the world is its own reward, but the real gift is the ability to think and generalize. We intend this book to provide support and encouragement for both.
Changes for the TweLfth Edition CONTENT In preparing this edition we have maintained the basic structure of the Table of Contents from the eleventh edition. Yet we have paid attention to requests by current users and reviewers to postpone the introduction of parametric equations until we present polar coordinates, and to treat I'Hopital 's Rule after the transcendental functions have been studied. We have made numerous revisions to most of the chapters, detailed as follows.
• Functions We condensed this chapter even more to focus on reviewing function concepts. Prerequisite material covering real numbers, intervals, increments, straight lines, distances, circles, and parabolas is presented in Appendices 13. • Limits To improve the flow of this chapter, we combined the ideas of limits involving infinity and their associations with asymptotes to the graphs of functions, placing them together in the final chapter section. • Differentiation While we use rates of change and tangents to curves as motivation for studying the limit concept, we now merge the derivative concept into a single chapter. We reorganized and increased the number of related rates examples, and we added new examples and exercises on graphing rational functions.
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• Antiderivatives and Integration We maintain the orgaJrization of the eleventh edition in placing antiderivatives as the fInal topic of the chapter covering applications of derivatives. Our focus is on ''recovering a function from its derivative" as the solution to the simplest type offirstorder differential equation. Integrals, as "limits of Riemann sums," motivated primarily by the problem of rmding the areas of general regions with curved boundaries, are a new topic fanning the substance of Chapter 5. After carefully developing the integral concept, we turn our attention to its evaluation and connection to antiderivatives captured in the Fundamental Theorem of Calculus. The ensuing applications then define the various geometric ideas of area, volume, lengths of paths, and centroids all as limits of Riemann sums giving definite integrals, which can be evaluated by finding an antiderivative of the integrand. We return later to the topic of solving more complicated firstorder differential equations, after we derme and establish the transcendental functions and their properties. • Differential Equations Some universities prefer that this subject be treated in a course separate from calculus. Although we do cover solutions to separable differential equations when treating exponential growth and decay applications in the chapter on transcendental functions, we organize the bulk of our material into two chapters (which may be omitted for the calculus sequence). We give an introductory treatment of firstorder differential equations in Chapter 9, including a new section on systems and phase planes, with applications to the competitivehunter and predatorprey models. We present an introduction to secondorder differential equations in Chapter 17, which is included in MyMathLab as well as the Thomas' Calculus Web site, www.pearsonhighered.comlthomas. • Series We retain the organizational structure and content of the eleventh edition for the topics of sequences and series. We have added several new figures and exercises to the various sections, and we revised some of the proofs related to convergence of power series in order to improve the accessibility of the material for students. The request stated by one of our users as, "anything you can do to make this material easier for students will be welcomed by our faculty;' drove our thinking for revisions to this chapter. • Parametric Equations Several users requested that we move this topic into Chapter II, where we also cover polar coordinates and conic sections. We have done this, realizing that many departments choose to cover these topics at the beginning of Calculus m, in preparation for their coverage of vectors and multivariable calculus. • VectorValued Functions We streamlined the topics in this chapter to place more emphasis on the conceptual ideas supporting the later material on partial derivatives, the gradient vector, and line integrals. We condensed the discussions of the Frenet frame and Kepler's three laws of planetary motion. • Multivariahle Calculus We have further enhanced the art in these chapters, and we have added many new figures, examples, and exercises. We reorganized the opening material on double integrals, and combined the applications of double and triple integrals to masses and moments into a single section covering both two and threedimensional cases. This reorganization allows for better flow of the key mathematical concepts, together with their properties and computational aspects. As with the eleventh edition, we continue to make the connections of multivariable ideas with their ainglevariable analogues studied earlier in the book. • Vector Fields We devoted considerable effort to improving the clarity and mathematical precision of our treatment of vector integral calculus, including many additional examples, figures, and exercises. Important theorems and results are stated more clearly and completely, together with enhanced explanations of their hypotheses and mathematical consequences. The area of a surface is now organized into a single section, and surfaces defined implicitly or explicitly are treated as special cases of the more general parametric representation. Surface integrals and their applications then follow as a separate section. Stokes' Theorem and the Divergence Theorem are still presented as generalizations of Green's Theorem to three dimensions.
Preface
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EXERCISES AND EXAMPLES We know that the exercises and examples are critical components in learning calculus. Because of this importance, we have updated, improved, and increased the number of exercises in nearly every section of the book. There are over 700 new exercises in this edition. We continue our organization and grouping of exercises by topic as in earlier editions, progressing from computational problems to applied and theoretical problems. Exercises requiring the use of computer software systems (such as Maple® or Mathematica®) are placed at the end of each exercise section, labeled Computer Explorations. Most of the applied exercises have a subheading to indicate the kind of application addressed in the problem. Many sections include new examples to clarify or deepen the meaning of the topic being discussed, and to help students understand its mathematical consequences or applications to science and engineering. At the same time, we have removed examples that were a repetition of material already presented. ART Because of their importance to learning calculus, we have continued to improve existing figures in Thomas' Calculus and we have created a significant number of new ones. We continue to use color consistently and pedagogically to enhance the conceptual idea that is being illustrated. We have also taken a fresh look at all of the figure captions, paying considerable attention to clarity and precision in short statements.
y
No matter what
y=1
positive nwnber E is, the graph enters this band at x = ~ and stays.
FIGURE 2.50, page 85 The geometric explanation of a finite limit as x  ± 00.
FIGURE 16.9, page 908 A surface in a space occupied by a moving fluid.
MYMATHLAB AND MATH XL The increasing use of and demand for online homework systems has driven the changes to MyMathLab and MathXL® for Thomas' Calculus. The MyMathLab course now includes significantly more exercises of all types. New Java™ applets add to the already significant collection to help students visualize the concepts and generalize the material.
Continuing Features RIGOR The level of rigor is consistent with that of earlier editions. We continue to distinguish between formal and informal discussions, and to point out their differences. We think starting with a more intuitive, less formal approach helps students understand a new or difficult concept so they can then appreciate its full mathematical precision and outcomes. We pay attention to defining ideas carefully and to proving theorems appropriate for
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calculus students, while mentioning deeper or subtler issues they would study in a more advanced course. Our organization, and distinctions between informal and formal discussions, gives the instructor a degree of flexibility in the amount and depth of coverage of the various topics. For example, while we do not prove the Intermediate Value Theorem or the Extreme Value Theorem for continuous functions on a :5 x :5 b, we do state these theorems precisely, illustrate their meanings in numerous examples, and use them to prove other important results. Furthermore, for those instructors who desire greater depth of coverage, we discuss in Appendix 6 the reliance of the validity of these theorems on the completeness of the real numbers.
WRmNG EXERCISES Writing exercises placed throughout the text ask students to explore and explain a variety of calculus concepts and applications. In addition, the end of each chapter contains a list of questions for students to review and summarize what they have learned. Many of these exercises make good writing assignments. ENDOfCHAPTER REVIEWS AND PROJECTS In addition to problems appearing after each section, each chapter colminates with review questions, practice exercises covering the entire chapter, and a series of Additional and Advanced Exercises serving to include more challenging or synthesizing problems. Most chapters also include descriptions of several Technology Application Projects that can be worked by individual students, or groups of students, over a longer period of time. Thesc projects require the use of a computer, running Mathematica or Maple, and additional material that is available over the Internet at www.pearsonhigherecLcomlthomas and in MyMathLab. WRITING AND APPUCATIONS As always, this text continues to be easy to read, conversational, and mathematically rich. Each new topic is motivated by clear, easytounderstand examples and is then reinforced by its application to realworld problems of immediate interest to students. A hallmark of this book has been the application of calculus to science and engineering. These applied problems have been updated, improved, and extended continually over the last several editions. TECHNOLOGY In a course using the text, technology can be incorporated according to the taste of the instructor. Each section contains exerciscs requiring the use of technology; these are marked with a D if suitable for calculator or computer use or are labeled Computer Explorations if a computer algebra system (CAS, such as Maple or Mathematica) is required.
Text Versions THOMAS' CALCULUS, Twelfth Edition Complete (Chapters 116), ISBN 032158799519780321587992 Single Variable Calculus (Chapters 111), ISBN 032163742919780321637420 Multivariable Calculus (Chapters 1116), ISBN 0.32164369019780321643698
THOMAS' CALCULUS: EARLY TRANSCENDENTALS, Twelfth Edition Complete (Chapters 116), ISBN 032158876219780321588760 Single Variable Calculus (Chapters 111),032162883719780321628831 Multivariable Calculus (Chapters 1116), ISBN 0.32164369019780321643698 The early transcendentals version of Thomas' Calculus introduces and integrates transcendental functions (such as inverse trigonometric, exponential, and logsritlnnic functions) into the exposition, examples, and exercises of the early chapters alongside the algebraic functions. The Multivariable book for Thomas' Calculus: Early 1'rnnscendentals is the same text as Thomas' Calculus, Multivariable.
Preface
xiii
Instructor's Editions Thomas , Calculus, ISBN 032160075419780321600752 Thomas 'Calculus: Early Transcentientals, ISBN 032162718019780321627186 In addition to including all of the answers present in the stodent editions, the Instructor ~ Editions include evennumbered answers for Chapters l{j.
University Calculus (Early Transcendentals) University Calculus: Alternative Edition (Late Transcendentals) University Calculus: Elements with Early Transcendentals The University Calculus texts are based on Thomas' Calculus and featore a streamlined presentation of the contents of the calculus course. For more information about these titles, visit www.pearsonhighered.com.
Print Supplements INSTRUCTOR'S SOLUTIONS MANUAL Single Variable Calculus (Chapters 111), ISBN 032160807019780321608079 Multivariable Calculus (Chapters 1116), ISBN 032160072X 19780321600721 The Instructor ~ Solutions Manual by William Ardis, Collin County Community College, contains complete workedout solutions to all of the exercises in the text.
STUDENrs SOLUTIONS MANUAL Single Variable Calculus (Chapters 111), ISBN 032160070319780321600707 Multivariable Calculus (Chapters 1116), ISBN 032160071119780321600714 The Student~ SolutiollS Manual by William Ardis, Collin County Community College, is designed for the stodent and contains carefully workedout solutions to all the oddnumbered exercises in the text.
JUSTINTIME ALGEBRA AND TRIGONOMETRY FOR CALCULUS, Fourth Edition ISBN 032167104X 19780321671042 Sharp algebm and trigonometry skills are critical to mastering calculus, and JustinTIme Algebra and 1Hgonometry for Calculus by Guntram Mueller and Ronald I. Brent is designed to bolster these skills while stodents stody calculus. As stodents make their way through calculus, this text is with them every step of the way, showing them the necessary algebm or trigonometry topics and pointing out potential problem spots. The easytouse table of contents has algebm and trigonometry topics armnged in the order in which stodents will need them as they stody calculus.
CALCULUS REVIEW CARDS The Calculus Review Cards (one for Single Variable and another for Multivariable) are a stodent resource containing important formulas, functions, defmitions, and theorems that correspond precisely to Thomas' Calculus. These cards can work as a reference for completing homework assignments or as an aid in studying, and are available bundled with a new text. Contact your Pearson sales representative for more information.
Media and Online Supplements TECHNOLOGY RESOURCE MANUALS Maple Manual by James Stapleton, North Carolina State University Mathematica Manual by Marie Vanisko, Carroll College ITGraphing Calculator Manual by Elaine McDonaldNewman, Sonoma State University These manuals cover Maple 13, Mathematica 7, and the TI83 PlUS/TI84 Plus and TI89, respectively. Each manual provides detailed gnidance for integrating a specific software package or gmphing calculator throughout the course, including syntax and commands.
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These manuals are available to qualified instructors through the Pearson Instructor Resource Center, www.pearsonhigheredlirc, and MyMatbLab.
WEB SITE _w.pearsonhighered.com/thomas The Thomas' Calculus Web site contains the cbapter on SecondOrder Differential Equations, including oddnumbered answers, and provides the expanded historical biographies and essays referenced in the text. Also available is a collection of Maple and Mathematica modules, as well as the Technology Application Projects, which can be used as projects by individual students or groups of stodents.
MyMathLab Online Course (access code required) MyMathLab is a textspecific, easily customizable ouline course that integrates interactive multimedia inatruction with textbook content. MyMathLab gives you the tools you need to deliver all or a portion of your course ouline, whether your stodents are in a lab setting or working from home. o Interactive homework exercises, correlated to your textbook at the objective level, are algorithmically generated for unlimited practice and mastery. Most exercises are freeresponse and provide guided solutions, sample problems, and learning aids for extra help. o "Getting Ready" chapter includes hundreds of exercises that address prerequisite skills in algebra and trigonometry. Each student can receive remediation for just those skills he or she needs help with. o Personalized Study Plan, generated when students complete a test or quiz, indicates which topics have been mastered and links to tutorial exercises for topics stodents have not mastered. o Multimedia learning aids, such as video lectures, Java applets, animations, and a complete multimedia textbook, help stodents independently improve their understanding and performance. o
Assessment Manager lets you create ouline homework, quizzes, and tests that are automatically graded. Select just the right mix of questions from the MyMatbLab exercise bank and instructorcreated custom exercises.
o Gradebook, designed specifically for mathematics and statistics, automatically tracks students' results and gives you control over how to calculate f'mal grades. You can also add offiine (paperandpencil) grades to the gradebook. o MathXL Exercise Builder allows you to create static and algorithmic exercises for your ouline assignments. You can use the library of sample exercises as an easy starting point. o Pearson Tutor Center (www.pearsontutorservices.com) access is automatically included with MyMatbLab. The Tutor Center is staffed by qualified math instructors who provide textbookspecific tutoring for students via tollfree phone, fax, email, and interactive Web sessions. MyMathLab is powered by CourseCompass™, Pearson Education's ouline teaching and learning environment, and by MathXL, our ouline homework, tutorial, and assessment system. MyMatbLab is available to qualified adopters. For more information, visit www.mymathlab.com or contact your Pearson sales representative.
Video Lectures with Optional Captioning The Video Lectures with Optional Captioning feature an engaging team of mathematics instructors who present comprehensive coverage of topics in the text. The lectorers' presentations include examples and exercises from the text and support an approach that emphasizes visualization and problem solving. Available ouly through MyMathLab and MathXL.
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MathXL Online Course (access code required) MathXL is an online homework, tutorial, and assessment system that accompanies Pearson's textbooks in mathematics or statistics. • Interactive homework exercises, correlated to your textbook at the objective level, are algorithmically generated for unlimited practice and mastery. Most exercises are freeresponse and provide guided solutions, sample problems, and learning aids for extra help. • "Getting Ready" chapter includes hundreds of exercises that address prerequisite skills in algebra and trigonometry. Each student can receive remediation for just those skills he or she needs help with. • PersonaIized Study Plan, generated when students complete a test or qillz, indicates which topics have been mastered and links to tutorial exercises for topics students have not mastered. • Multimedia learning aids, such as video lectures, Java applets, and animations, help students independently improve their understanding and performance. • Gradebook, designed specifically for mathematics and statistics, automatically tracks students' results and gives you control over how to calculate fmal grades. • MathXL Exercise Builder allows you to create static and algorithmic exercises for your online assignments. You can use the library of sample exercises as an easy starting point. • Assessment Manager lets you create online homework, quizzes, and tests that are automatically graded. Select just the right mix of questions from the MathXL exercise bank, or instructurcreated custom exercises. MathXL is available to qualified adopters. For more information, visit our Web site at www.mathxl.com. or contact your Pearson sales representative.
TestGen® TestGen (www.pearsonhighered.comltestgen)enablesinstructurstobuild.edit.print. and administer tests using a computerized hank of questions developed to cover all the objectives of the text. TestGen is algorithmically based, allowing instructurs to create multiple but equivalent versions of the same question or test with the click of a button. Instructors can also modifY test hank questions or add new questions. Tests can be printed or administered online. The software and testbank are available for dowuload from Pearson Education's online catalog.
PowerPoint® Lecture Slides These classroom presentation slides are geared specifically to the sequence and philosophy of Thomas' Calculus series. Key graphics from the book are included to help bring the concepts alive in the classroom.These files are available to qualified instructors tbrough the Pearson Jnstructur Resource Center, www.pearsonhigheredlirc, and MyMathLab.
Acknowledgments We would like to express our thanks to the people who made many valuable contributions to this edition as it developed through its various stages:
Accuracy Checkers Blaise DeSesa Paul Lorczak Kathleen Pellissier Lauri Semarne Sarab Streett Holly Zullo
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Reviewers for the Twelfth Edition Meighan Dillon, Southern Polytechnic State University Anne Dougherty, University of Colorado Said Fariabi, San Antonio College Klaus Fischer, George Mason University Tim Flood, PiUsburg State University Rick Ford, California State UniversityChico Robert Gardner, East Tennessee State University Christopher Heil, Georgia Institute ofTechnology Joshua Brandon Holden, RoseHulman Institute ofTechnology Alexander Hulpke, Colorado State University Jacqueline Jensen, Sam Houston State University Jennifer M. Johnson, Princeton University Hideaki Kaneko, Old Dominion University Przemo Kranz, University ofMississippi Xin Li, University of Central Florida Maura Mast, University ofMassachusettsBoston Val Mohanakumar, Hillsborough Community CollegeDale Mabry Campus Aaron Montgomery, Central Washington University Cynthia Piez, University ofIdaho Brooke Quinlan, Hillsborough Community CollegeDale Mabry Campus Rebecca A. Segal, Virginia Commonwealth University Andrew V. Sills, Georgia Southern University Alex Smith, University ofWlSconsinEau Claire Mark A. Smith, Miami University Donald Solomon, University ofWISconsinMilwaukee Blake Thornton, Washington University in St. Louis David Walnut, George Mason University Adrian Wilson, University ofMontevallo Bobby Winters, PiUsburg State University Dennis Wortman, University ofMassachusettsBoston
1 FUNCTIONS OVERVIEW Functions are fundamental to the study of calculus. In this chapter we review what functions are and how they are pictured as graphs, how they are combined and transformed, and ways they can be classified. We review the trigonometric functions, and we discuss misrepresentations that can occur when using calculators and computers to obtain a function's graph. The real number system, Cartesian coordinates, straight lines, parabolas, and circles are reviewed in the Appendices. We treat inverse, exponential, and logarithmic functions in Chapter 7.
1.1
Functions and Their Graphs Functions are a tool for describing the real world in mathematical terms. A function can be represented by an equation, a graph, a numerical table, or a verbal description; we will use all four representations throughout this book. This section reviews these function ideas.
Functions; Domain and Range The temperature at which water boils depends on the elevation above sea level (the boiling point drops as you ascend). The interest paid on a cash investment depends on the length of time the investment is held. The area of a circle depends on the radius of the circle. The distance an object travels at constant speed along a straightline path depends on the elapsed time. In each case, the value of one variable quantity, say y, depends on the value of another variable quantity, which we might call x. We say that "y is a function of x" and write this symbolically as y
= f(x)
("y equals f of x").
In this notation, the symbol f represents the function, the letter x is the independent variable representing the input value of f, and y is the dependent variable or output value of
fatx.
DEFINITION
A function f from a set D to a set Y is a rule that assigns a unique E Y to each element XED.
(single) element f(x)
The set D of all possible input values is called the domain of the function. The set of all values of f(x) as x varies throughout D is called the range of the function. The range may not include every element in the set Y. The domain and range of a function can be any sets of objects, but often in calculus they are sets of real numbers interpreted as points of a coordinate line. (In Chapters 1316, we will encounter functions for which the elements of the sets are points in the coordinate plane or in space.)
1
2
Chapter 1: Functions
J
x :_.~
Input (donurin)
FIGURE 1.1
:"""":~~•• J(x)
Output
(nmge)
A diagram showing a
function as a kind of machine.
~
~J(a) D
= domain set
Vx
J(x)
Y = set containing tberange
A function from a setD to a set Yassigns a unique element of Y to each element in D. FIGURE 1.2
Often a function is given by a formula that describes how to calculate the output value from the input variable. For instance, the equation A = 1Tr2 is a rule that calculates the area A of a circle from its radius r (so r, interpreted as a length, can only be positive in this formula). When we derme a function y = f(x) with a formula and the domain is not stated explicitly or restricted by context, the domain is assumed to be the largest set of real xvalues for which the formula gives real yvalues, the socalled natura! domain. If we want to restrict the domain in some way, we must say so. The domain of y = x 2 is the entire set of real numbers. To restrict the domain of the function to, say, positive values of x, we would write "y = x 2 ,x > 0." Changing the domain to which we apply a formula usually changes the range as well. The range of y = x 2 is [0, 00). The range of y = x 2, X ;;,: 2, is the set of all numbers obtained by squaring numbers greater than or equal to 2. In set notation (see Appendix 1), the rangeis {x 2 Ix;;,: 2} or{Yly;;': 4} or [4, 00). When the range of a function is a set of real numbers, the function is said to be rea!valued. The domains and ranges of many realvalued functions of a rea! variable are intervals or combinations of interva1s. The interva1s may be open, closed, or half open, and may be finite or infinite. The range of a function is not a!ways eaay to find. A function f is like a machine that produces an output value f(x) in its range whenever we feed it an input value x from its domain (Figure 1.1). The function keys on a calculator give an example of a function as a machine. For instance, the key on a calculator gives an output value (the square root) whenever you enter a nonnegative number x and press the key. A function can also be pictured as an arrow diagram (Figure 1.2). Each arrow associates an element of the domain D with a unique or single element in the set Y. In Figure 1.2, the arrows indicate thatf(a) is associated with a, f(x) is associated with x, and so on. Notice that a function can have the same value at two different input elements in the domain (as occurs with f(a) in Figure 1.2), but each input element x is assigned a single output value f(x).
Vx
EXAMPLE 1 Let's verify the natural domains and associated ranges of some simple functions. The domains in each case are the values of x for which the formula makes sense. Function
Domain (x)
Range (y)
y = x2
(00,00) (00,0) U (0, 00) [0, 00) (00,4] [I, I]
[0, 00)
Y = I/x y=
Vx
y=~ y =
v'1=7
(00,0) U (0, 00) [0, 00) [0, 00)
[0, I]
Solution The formula y = x 2 gives a rea! yvalue for any real number x, so the domain is ( 00, 00). The range of y = x 2 is [0, 00) because the square of any real number is nonnegative and every nonnegative number y is the square of its own square root, y = (vY)2fory;;,: O. The formula y = l/x gives a real yvalue for every x except x = O. For consistency in the rules of arithmetic, we cannot divide any number by zero. The range of y = I/x, the set of reciprocals of all nonzero real numbers, is the set of all nonzero real numbers, since y = I/(I/y). That is, for y oF 0 the number x = I/y is the input assigned to the output valuey. The formula y = Vx gives a real yvalue only if x ;;,: O. The range of y = Vx is [0, 00) because every nonnegative number is some number's square root (namely, it is the square root of its own square). In y = '\I'4=x, the quantity 4  x cannot be negative. That is, 4  x ;;,: 0, or x :5 4. The formula gives real yvalues for all x :5 4. The range of '\I'4=x is [0, 00), the set of all nonnegative numbers.
1.1
Functions and Their Graphs
3
The formula y = ~ gives a real yvalue for every x in the closed interval from  1 to 1. Outside this domain. 1  x 2 is negative and its square root is not a real number. The values of 1  x 2 vary from 0 to 1 on the given domain, and the square roots of these values do the same. The range of ~ is [0. 1]. •
Graphs of Functions If j is a function with domain D, its graph consists of the points in the Cartesian plane whose coordinates are the inputoutput pairs for j. In set notation. the graph is {(x,j(x)) I XED}.
The graph of the function j(x) = x + 2 is the set of points with coordinates (x,y) for which y = x + 2. Its graph is the straight line sketched in Figure 1.3. The graph of a function j is a useful picture of its behavior. If (x. y) is a point on the graph, then y = j(x) is the height of the graph above the pointx. The height may be positive or negative, depending on the sign of j(x) (Figure 104).
y
X
y=xl
2 1 0 1
4 1 0 1
3
2
9 4
2
4
FIGURE 1.3 The graph of f(x) = x + 2 is the set of points (x,y) for whichy has the value x + 2.
FIGURE 1.4
f(x)
EXAMPLE 2
Graph the function y
=
If(x.y) lies on the graph of
f. then the value y = f(x) is the height of the graph above the poiut x (or below x if
is negative).
x 2 over the interval [2,2].
Solution Make a table ofxypairs that satisfy the equation y = x 2• Plot the points (x,y) whose coordinates appear in the table, and draw a smooth curve (labeled with its equation) through the plotted points (see Figure 1.5). •
y
How do we know that the graph of y = x 2 doesn't look like one of these curves? (2.4)
y
~~~~~~~X
2
1
FIGURE 1.5
Example 2.
0
2
Graph of the function in
y
4
Chapter 1: Functions
To find out, we could plot more points. But how would we then connect them? The basic question still remains: How do we know for sure what the graph looks like between the points we plot? Calculus answers this question, as we will see in Chapter 4. Meanwhile we will have to settle for plotting points and connecting them as best we can.
Representing a Function Numerically We have seen how a function may be represented algebraically by a formula (the area function) and visually by a graph (Example 2). Another way to represent a function is numerically, through a table of values. Numerical representations are often used by engineers and scientists. From an appropriate table of values, a graph of the function can be obtained using the method illustrated in Example 2, possibly with the aid of a computer. The graph consisting of only the points in the table is called a scatterplot.
EXAMPLE 3 Musical notes are pressure waves in the air. The data in Table l.l give recorded pressure displacement versus time in seconds of a musical note produced by a toning fork. The table provides a representation of the pressure function over time. If we first make a scatterplot and then connect approximately the data points (t, p) from the table, we obtain the graph shown in Figure 1.6. p (pressure)
TABLE 1.1 Tuning fork data Time
Pressure
Time
0.00091 0.00108 0.00125 0.00144 0.00162 0.00180 0.00198 0.00216 0.00234 0.00253 0.00271 0.00289
0.080
0.00362 0.00379 0.00398 0.00416 0.00435 0.00453 0.00471 0.00489 0.00507 0.00525 0.00543 0.00562
0.00307 0.00325 0.00344
0.200 0.480 0.693 0.816 0.844 0.771 0.603 0.368 0.099 0.141 0.309 0.348 0.248 0.041
0.00579 0.00598
Pressure 0.217 0.480 0.681 0.810 0.827 0.749 0.581 0.346 0.077 0.164 0.320 0.354 0.248 0.035
1.0 0.8 0.6 0.4 0.2
t (sec) 0.2 0.4 0.6
FIGURE 1.6 A smooth curve through the plotted points gives a graph of the pressure functioo represented by Table l.l (Example 3).
• The Vertical Line Test for a Function Not every curve in the coordinate plane can be the graph of a function. A function 1 can have ouly one value I(x) for each x in its domain, so 110 vertical line can intersect the graph of a function more than once. If a is in the domain of the function I, then the vertical line x = a will intersect the graph of 1 at the single point (a, I(a». A circle cannot be the graph of a function since some vertical lines intersect the circle twice. The circle in Figure 1.7a, however, does contain the graphs of two functions of x: the upper semicircle defmed by the function I(x) = ~ and the lower semicircle defined by the function g(x) =  ~ (Figures 1.7b and 1.7c).
1.1 Y
Functions and Their Graphs
Y
5
Y
I ~~~>x
I
0
~~~~x
o
(C)Y~~
FIGURE 1.7 (a) The circle is not the graph ofa function; itfails the vertical line test. (b) The upper semicircle is the graph ofa function f(x) = ~. (c) The lower semicircle is the graph ofa functiong(x) = ~.
y
PiecewiseDefined Functions Sometimes a function is described by using different formulas on different parts of its domain. One example is the absolute value funetion FIGURE 1.8 The absolute value function has domain (  00, 00) and range [0, 00). y
Ixl =
x;;"
0
0 is a positive constant and a oF 1, are called exponential fuoctions. All exponential functions have domain (  00, 00) and range (0, 00), so an exponential function never assumes the value O. We study exponential functions in Section 7.3. The graphs of some exponential functions are shown in Figure 1.22. y
1
0.5
y
12
12
10
10
8
8
6
6
4
4
2
2
0
0.5
1
x
(a)
FIGURE 1.22
Graphs of expooential functinos.
1
0.5
0 (b)
0.5
1.1
Functions and Their Graphs
11
Logarithmic Functions These are the functions t(x) = log" x, where the base a '" 1 is a positive constant. They are the inverse jUnctions of the exponential functions, and the calculus of these functions is studied in Chapter 7. Figure 1.23 shows the graphs of four logarithmic functions with various bases. In each case the domain is (0, 00) and the range is (00, 00). Y
y
~O~~ ~~~~=Y=~~IO~~XX 1 ,
FIGURE 1.23 functions.
Y = lOglOx
Graphs offour logaritlunic
FIGURE 1.24 Graph ofa cateoary or hanging cable. (The Latin word calena means "chain.")
Transcendental Functions These are functions that are not algebraic. They include the trigonometric, inverse trigonometric, exponential, and logarithmic functions, and many other functions as well. A particular example of a transcendental function is a catenary. Its graph has the shape of a cable, like a telephone line or electric cable, strung from one support to another and hanging freely under its own weight (Figure 1.24). The function defming the graph is discussed in Section 7.7.
Exercises 1.1 Functions
8...
Y
b.
y
In Exercises 1{j, find 1he domain and range of each function.
=
+ x' v'Sx +
=
4 3_ I
1. f(x) = I 3. F(x) S. f(t)
2. f(x) = 1 10
4. g(x)
=
6. G(t)
=
;Ix
v'x'  3x 2 t' _ 16 ~~X
In Exercises 7 and 8, which of1he graphs are graphs of functions oh, and which are not? Give reasons for your answers.
7.a.y
b.y
o
~+x
o
Finding Formulas for Functions 9. Express 1he area and perimeter of an equilateral triangle as a function of1he triangle's side leng1hx. 10. Express 1he side 1eng1h ofa square as a function of1he 1eng1h d of 1he square's diagonal. Then express 1he area as a function of 1he diagonalleng1h.
~~X
o
~+x
o
11. Express 1he edge leng1h ofa cube as a function of1he cube's diagonal 1eng1h d. Then express 1he surface area and volume of 1he cube as a function of1he diagonalleng1h.
12
Chapter 1: Functions
12. A point P in the first quadrant lies 00 the graph of the functioo I(x) ~ Express the coordinates of P as functions of the slope of the line joining P to the origin.
Vx.
31. a.
13. Consider the point (x, y) lying 00 the graph of the line a + 4y ~ 5. Let L he the distsnce from the point (x, y) to the origin (0, 0). WriteL as a functioo ofx. 14. Consider the point (x, y) lying on the graph of y = ~. Let L he the distsnce hetweeo the points (x,y) and (4, 0). WriteL as a function ofy.
Y
~x
17. g(x) =
a
16. I(x) = I 
'\J'Ix1
19. F(t) = t/ltl
2
20. G(t) =
!fit I
x' 
4 ~
h
(2,1)
Y
32. a.
b. Ix
A
The Greatest and Least Integer Functions 33. For what values oh is
9
x' + ,. x +4
+ Iyl = I
A
1:
b. rxl=O?
a.lxJ=O?
Iyl=x
Ixl
Y
1
n+~!'~x OTT
b. y'=x' 24. Graph the following equations and explain why they are not graphs of functioos of x. L
b.
o
23. Graph the following equations and explain why they are not graphs of functioos of x. L
 =ti  x
x'
x~.
21. Find the domain ofy = 22. Find the range of y
a 
18. g(x) =
Y
2
3
I
Functions and Graphs Find the domain and graph the functions in Exercises 1520.
15. I(x) = 5 
b.
34. What real nwnhers x satisfY the equation l x J = r xl? 35. Does r xl = lxJ for all real x? Give reasons foryour answer. 36. Graph the function
I(x) =
{f:~:
+ yl = I
x~o
x
< O.
Why is I(x) called the integer part ofx?
Piece>riseDefined Functions Graph the functions in Exercises 2528.
25. I(x) = {x, 2x, I  x 26. g(x) = { 2 _ x:
4 27. F(x) ~ { , x
28. G(x)
x'a'
+
~ {I/X, x,
Increasing and Decreasing Functions Graph the functioos in Exercises 3746. What synnnetries, if any, do the graphs have? SpecifY the intervals over which the function is increasing and the intervals where it is decreasing.
0 :5 x :5 I 1 :4
+ 10
+ 32x' + 10.
(b) The fannula is
Y =
!f(x)
=
!X 4 + 2x'  5.
•
Ellipses Although they are not the graphs of functions, circles can be stretched horizontally or vertically in the same way as the graphs of functions. The standard equation for a circle of radius r centered at the origin is x2 + y2 = r2. Substituting ex for x in the standard equation for a circle (Figure 1.36a) gives c2x2
y
+ y2
= r2.
(1)
y
y
r
~~4~x r r
r
o
r c
r
r
(a) circle
FIGURE 1.36
~~+~x
(b) ellipse, 0 < c
1
1.2 Combining Functions; Shifting and Scaling Graphs
If 0 < c < I, the graph of Equation (I) horizontally stretches the circle; if c > I the circle is compressed horizontally. In either case, the graph of Equation (I) is an ellipse (Figure 1.36). Notice in Figure 1.36 that theyintercepts of all three graphs are alwaysr and r. In Figure 1.36b, the line segment joining the points (±rlc, 0) is called the major axis of the ellipse; the minor axis is the line segment joining (0, ±r). The axes of the ellipse are reversed in Figure 1.36c: The major axis is the line segment joining the points (0, ±r), and the minor axis is the line segment joining the points (±rlc, 0). In both cases, the major axis is the longer line segment. Ifwe divide both sides of Equation (I) by r', we obtain
y
b
+~~~~x
a
19
a
b (2)
where a = ric andb = r. Ifa > b, the major axis is horizontal; if a < b, the major axis is vertical. The center of the ellipse given by Equation (2) is the origin (Figure 1.37). Substituting x  h for x, and y  k for y, in Equation (2) results in
FIGURE 1.37 Graph of tire ellipse 2
y2
+2 =
l,a > b,wherethemajor a b axis is horizontal. X
2
(x  h)' (y  kf 'a"'' + b'
=
(3)
I.
Equation (3) is the standard equation of an ellipse with center at (h, k). The geometric def"mition and properties of ellipses are reviewed in Section 11.6.
Exercises 1.2 Algebraic Combinations In Exercises I and 2, fmd tire domains and ranges of f, g, f
f·g· 1. I(x) = x, 2. I(x) =
+ g, and
v;=J Vx+J, g(x) = v;=J g(x) =
4. f(x)
=
g(x) = x 2
I, g(x)
=
I
+ I
+
Vx
+ 5 andg(x)
10. f(x)
=
I
• ,c;
vx
= x'  3, fmd the following.
a. l(g(O))
b. g(/(O))
c. I(g(x))
II. g(/(x))
•• 1(1(5))
r.
x+2
3 x' g(x)
go 1(I(x))
b. g(g(x))
+
x'
= x' +
I' h(x)
=
d.y=x6
•• y=2~
r.
y=
W=3
13. Copy and coroplere the following table.
I), fmd the following.
g(x)
I(x)
(/0 g)(x)
a. x  7
Vx
?
3x
?
a. f(g(1/2)) c. f(g(x)) e. 1(1(2))
b. g(/(1/2))
II. g(/(x))
c.
~
~
r.
b. g(g(x))
x x  I
?
go N(x))
x d.   I x
••
?
1+1 x
x
f.
I x
?
x
g(g(2))
In Exercises 710, write a formula for fog
8. I(x)
= x + I, = 3x + 4,
= 3x, g(x) = 2x
g(x)
h(x)
v2x
b. y = x'/2
c. y = x 9
b. x + 2
7. I(x)
.~
= x  3, g(x) = Vx, h(x) = x', and j(x) = 2x. Express each of the functions in Exercises 11 and 12 as a composite involving one or more of j,g, h, andj. 11••• y = Vx  3 b. y = 2Vx c. y = xl/4 d. y = 4x •• Y = y"(x:c )'" r. y = (2x  6)' 3
g(g(2))
6. If f(x) = x  I andg(x) = I/(x
I
+ I, g(x) = x + 4' h(x) = x
12••• y=2x3
Composites of Functions S. If I(x) = x
=
Let f(x)
In Exercises 3 and 4, fmd the domains and ranges of I, g, I/g, and g/f.
3. I(x) = 2,
9. f(x)
=4
 I, h(x)
0
h.
 x
= x2
?
20
Chapter 1: Functions 22. The accompanying figure shows the graph of y = x 2 shifted to two new positions. Write equations for the new graphs.
14. Copy and complete the following table.
g(x)
(j
f(x)
I x  I
L
0
g)(x)
?
Ixl
b.
?
x  I x
c.
?
Vx
d.
Vx
?
x x
+I Ixl Ixl
15. Evaluate each expression using the giveo table of values
2
I
0
I
2
f(x)
I
0
2
I
2
g(x)
2
I
0
I
0
x
23. Match the equations listed in parts (a)(d) to the graphs in the accompanying figure .
.. j(g(I))
g{j(O» •. g(j( 2))
b.
d. g(g{2»
.. y
c. j(j( I)) f.
j(g(I))
x g(x) = { x ~ I,
.. j(g(O» d. j(j(2»
g{j(3» •. g{j(O»
b.
=
Vx+l,
g(x)
18. f(x) = x 2 , g(x) = I 
=
f.
j(g(I/2)) 0
f and
i
Vx
x
0
d. Y = (x
c. g{g{I»
19. Let f(x) = ~2' Find a function y = g(x) so that (j
+2 + 3)2  2
b. Y = (x  2)2
2sx0
COMPUTER EXPLORATIONS In Exercises 6'J72, you will explore graphically the general sine
69. TheperiodB Set the constants A = 3, C = D = O. a. Plot f(x) for the values B = I, 3, 21f, 5" over the interval 41f :5 x :5 41f. Describe what happens to the graph of the general sine function as the period increases. b. What happens to the graph for negative values of B? Try it withB = 3 andB = 21f. 70. The horizontal shift C Set the constants A = 3, B = 6, D = O. a. Plot f(x) for the values C = 0, I, and 2 over the interval 41f :5 x :5 41f. Describe what happens to the graph of the general sine function as C increases through positive values. b. What happens to the graph for negative values of C? c. What smallest positive value should be assigned to C so the graph exhibits no horizontal shift? Confum your answer with a plot. 71. The vertical shift D Set the constants A = 3, B = 6, C = O. a. Plot f(x) for the values D = 0, I, and 3 over the interval 41f :5 x :5 41f. Describe what happens to the graph of the general sine function as D increases through positive values. b. What happens to the graph for negative values of D? 72. The amplitude A Set the constants B = 6, C = D = O. a. Describe what happens to the graph of the general sine function as A increases through positive values. Confirm your answer by plotting f(x) for the values A = I, 5, and 9. h. What happens to the graph for negative values of A?
functioo
f(x)
= A Sine: (x

C)) +
D
as you change the values of the constants A, B, C, andD. Use a CAS or coruputer grapher to perform the steps in the exercises.
1.4
Graphing with Calculators and Computers A graphing calculator or a computer with graphing software enables us to graph very complicated functions with high precision. Many of these functions could not otherwise be easily graphed. However, care must be taken when using such devices for graphing purposes, and in this section we address some of the issues involved. In Chapter 4 we will see how calculus helps us determine that we are accurately viewing all the important features of a function's graph.
Graphing Windows When using a graphing calculator or computer as a graphing tool, a portion of the graph is displayed in a rectangular display or viewing window. Often the default window gives an incomplete or misleading picture of the graph. We use the tenD square window when the units or scales on both axes are the same. Ibis tenD does not mean that the display window itself is square (usually it is rectangular), but ioslead it means that the xunit is the same as the yunit. When a graph is displayed io the default window, the xunit may differ from the yunit of scaling in order to fit the graph in the window. The viewing window is set by specifying an ioterval [a, b] for the xvalues and an ioterval [c, d] for the yvalues. The machine selects equally spaced xvalues in [a, b] and then plots the points (x, f(x». A poiot is plotted if and
1.4 Graphing with Calculators and Computers
31
only ifx lies in the domain of the function andf(x) lies within the interval [c, d]. A short line segment is then drawn between each plotted point and its next neighboring point. We now give illustrative examples of some common problems that may occur with this procedure. Graph the function f(x) = x 3 play or viewing windows:
EXAMPLE 1
(8) [10, 10] by [10, 10]
7x'

+ 28 in each of the following dis
(b) [4,4] by [50, 10]
(c) [4, 10] by [60,60]
Solution (8) We select a = 1O,b = lO,c = 1O,andd = 10 to specify the in!erval of xvalues and the range of yvalues for the window. The resulting graph is shown in Figure l.50a. It appears that the window is cutting off the bottom part of the graph and that the interval of xvalues is too large. Let's try the next window.
10
4 10
~~f+~' O. (Logarithmic and exponential functions are presented in Chapter 7.) To obtain the full picture showing both branches, we can graph the function
f(x)
=
1~1·lxll/'.
xl/'
This function equals except at x = 0 (where f is undefined, although 0 1/3 = 0). The graph of f is shown in Figure 1.55b. •
Exercises 1.4 Choosing a Viewing Window
o In Exercises 14, use a graphing calculator or coroputer to detennine which of the given viewing windows displays the most appropriate graph of the specified function. 1. f(xl = x'  7x 2 + 6x L
[I, I]by[I, I]
b. [2, 2] by [5, 5]
c. [10, 10] by [10, 10] d. [5,5] by [25, 15] = x'  4x 2  4x + 16
19. f(xl
[I, I] by [5,5] c. [5,5]by[1O,20] 3. f(xl = 5 + 12x  x' L [I, I]by[I, I] c. [4,4] by [20,20] 4. f(xl = '1'5 + 4x  x 2 L
[2,2] by [2,2]
c. [3, 7] by [0, 10]
b. [3, 3] by [10, 10]
22. f(xl
= 2
2 23. f(xl = 6x  15x + 6 4x 2  lOx
24. f(xl =   2
window for the given 3
6. f(xl =
+3 +2
Chapter
8. f(xl
2
x x T  "2 
16. Y
II
3
25. y
= sin 250x
26. y
= 3 cos 60x
27. y
= cos(;O)
28. y
=
29. y
=x +
_ 2 I 30. y  x + 50 cos IOOx
/OSin3Ox
110 sin (;0)
34. Graph two periods of the function f(x l = 3 cot
15. y = Ix 2


x
33. Graph four periods of the function f(xl =  tan 2x.
9. f(xl = x~ 11. y = 2x  3x 2/' 13. Y = 5x 2/'  2x x 17. y = x
x'  5x'
x2
d. [10, 10] by [10, 10]
4x'  x' 10. f(xl = x 2(6  x'l 12. y = x 1/'(x 2  8l 14. Y = x 2/'(5  xl
=
8 x  9
b. [2,6]by[1,4]
functioo and use it to display its graph.
7. f(xl
x +I
32. Graph the upper branch of the hyperbola y2  16x 2
o In Exercises 530, rmd an appropriate viewing + 15 + 10
1
31. Graph the lower half of the circle dermed by the equation x 2 +2x=4+4y y 2.
d. [4,5]by[15,25]
Finding a Viewing Window
5. f(xl = x'  4x'

21. f(x l = x2o"_x"_~6
d. [20, 20] by [100, 100] b. [5, 5] by [10, 10]
x2
= 2
2. f(xl L
2 x +2 x2 + I x  I
20. f(xl
=
18. y
=
2x
+I
=
Ix 2
=I

 x
xl I
+3
35. Graph the function f(xl
=
sin 2x
36. Graph the function f(xl
=
sin' x.
= I.
I + I.
+ cos 3x.
Graphing in Dot Mode
o Another
way to avoid incorrect conooctions when using a graphing device is through the use of a "dot mode:' which plots only the points. !fyour graphing utility allows that mode, use it to plot the functions in Exercises 3740. I 38. y = sin~ 37. y = x  3
39. y
=
xlxJ
40. y
x3

X

1 I
= 2
Questions to Guide Your Review
1. What is a function? What is its doroain? Its range? What is an arrow diagraro for a functioo? Give examples. 2. What is the graph of a realvalued functioo of a real varisble? What is the vertical line test?
3. What is a piecewi..,.dermed functioo? Give examples.
4. What are the importsnt types of functions frequently encountered in calculus? Give an example of each type.
Chapter 1 Practice Exercises
35
5. What is meant by an increasing function? A decreasing function? Give an example of each.
11. What is the stsndard equation of an ellipse with center (h, k)? What is its major axis? Its minor axis? Give examples.
6. What is an even function? An odd function? What symmetry properties do the graphs of such functions have? What advantage can we take of this? Give an example of a function that is neither even nor odd.
12. What is radian measure? How do you convert from radians to degrees? Degrees to radians?
7. If f and g are realvalued functions, how are the domains of f + g, f  g, fg, and f / g related to the domains of f and g? Give examples.
14. What is a periodic function? Give examples. What are the periods of the six basic trigonometric functions?
8. When is it possible to compose one function with another? Give examples of composites and their values at various points. Does the order in which functions are composed ever matter? 9. How do you change the equation y = f(x) to shift its graph vertically up or down by Ikl units? Horizontally to the left or right? Give examples.
10. How do you change the equation y = f(x) to compress or stretch the graph by a factor c > I? Reflect the graph across a coordinate axis? Give examples.
Chapter
a
13. Graph the six basic trigonometric functions. What symmetries do the graphs have?
15. Stsrting with the identity sin2 8 + cos2 8 = I and the formulas for cos (A + B) and sin (A + B), show how a variety of other trigonometric identities may be derived.
16. How does the formula for the general sine function f(x) = A sin «2'IT/B)(x  e)) + D relate to the shifting, stretching, compressing, and reflection of its graph? Give examples. Graph the general sine curve and identifY the constsnts A, B, e,andD. 17. Name three issues that arise when functions are graphed using a calculator or computer with graphing software. Give examples.
Practice Exercises
Functions and Graphs 1. Express the area and circumference of a circle as functions of the circle's radius. Then express the area as a function of the circumference.
2. Express the radius of a sphere as a function of the sphere's surface area. Then express the surface area as a function of the volume. 3. A point P in the frrst quadrant lies on the parabola y = x 2 • Express the coordinates of P as functions of the angle of inclination of the line joining P to the origin. 4. A hotair balloon rising straight up from a level field is traclred by a range fmder located 500 ft from the point ofliftoff. Express the balloon's height as a function of the angle the line from the range fmder to the balloon makes with the ground.
Iu Exercises 19. y =
11. y
= x2 + I = I  cosx
13• y
=~ 3
x  2x 15. Y = x + cosx
12. y
= x'  x 3 = sec x tsnx
14. Y
=
10. y
16. y
fmd the (a) domain and (b) range.
2
20. y = 2
= Vl6  x
23. y
= 2e~
25. y
=
27. y
= Iu (x
2
 3
2sin(3x
+ 'IT)
 3)
 I
+I
+~ +I
22. y =
32 >
24. y
=
tsn(2x  'IT)
26. y
=
x 2/'
28. y = I
+ ~2  x
29. State whether each function is increasing, decreasing, or neither. a. Volume of a sphere as a function of its radius b. The greatest integer function
c. Height above Earth's sea level as a function of atmospheric pressure (assumed nonzero) d. Kinetic energy as a function ofa particle~ velocity
Iu Exercises ~16, deterntine v.lrether the function is even, odd, or neith"" 9. y
Ixl 
21. y
Iu Exercises 51!, determine whether the graph of the function is sym
metric about the yaxis, the origin, or neither. 5. y = x'i' 6. y = x 2/' 7. y = x 2  2x  I 8. y = e"
1~28,
30. Find the largest interval on which the given function is increasing.
a. f(x)=lx21+ I c. g(x) = (3x  1)'/3
x
x  sinx
= xcosx
b. f(x)
= (x +
I)'
d. R(x)=~
PiecewiseDefined Functions Iu Exercises 31 and 32, fmd the (a) domain and (b) range.
17. Suppose that f and g are both odd functions defined on the entire real1ine. Which of the following (where defmed) are even? odd?
31
a. fg b. f3 c. f(sinx) d. g(sec x) •• Igl 18. If f(a  x) = f(a + x), show that g(x) = f(x + a) is an even function.
32.y=
_ .y
{h, Yx,
4 S x'" 0 0l
x2
+X x2
y x
+
X x2  X
2 •
Solution We cannot substitute x = I because it makes the denominator zero. We test the numerator to see if it, too, is zero at x = I. It is, so it has a factor of (x  I) in common with the denominator. Canceling the (x  I)'s gives a simpler fraction with the same values as the original for x oF I:
(a)
y=x+2

Eliminating Zero Denominators Algebraically
EXAMPLE 7
~ _~ 2 nO~+~x
51


(x  I)(x + 2) x(x  I)
2
x
x
+
2
x
'
if x oF I.
Using the simpler fraction, we find the limit of these values as x > I hy substitution:
3
lim x x+l
2
+x 
x2

X
2 = lim x x+l
+2 X
= I
+
2 = 3.
1
See Figure 2.11.

(b)
Using Calculators and Computers to Estimate Limits FIGURE 2.11 The graph of f(x) = (x' + x  2)/(x'  x) in part (a) is the same .. the graph of g(x) = (x + 2)/x in part (b) except at x = I, where f is undefmed. The functions have the same lintit .. x .... I (Example 7).
When we cannot use the Quotient Rule in Theorem I because the limit of the denon1inator is zero, we can try using a calculator or computer to guess the limit numerically as x gets closer and closer to c. We used this approach in Example I, but calculators and computers can sometimes give false values and misleading impressions for functions that are undefmed at a point or fail to have a limit there, as we now illustrate.
Solution Table 2.3 lists values of the function for several values near x = O. As x approaches 0 through the values ±I, ±0.5, ±O.IO, and ±0.01, the function seems to approach the number 0.05. As we take even smaller values of x, ±0.OO05, ±0.0001, ±0.00001, and ±O.OOOOOI, the function appears to approach the value O. Is the answer 0.05 or 0, or some other value? We resolve this question in the next example. _
52
Chapter 2: Limits and Continuity
~
Computer values of f(x)
TABLE 2.3
x
v'x' + 100 ,
10
x
near x
=
0
!(x)
±I ±0.5 ±O.I ±0.01
0.049876) 0.049969 h 0 05? 0.049999 approac es. . 0.050000
±0.0005 ±O.OOOI ±O.OOOOI ±O.OOOOOI
0.080000) 0.000000 h? 0.000000 approac es 0 . 0.000000
Using a computer or calculator may give ambiguous results, as in the last example. We cannot substitute x = 0 in the problem, and the numerator and denominator have no obvious common factors (as they did in Example 7). Sometimes, however, we can create a common factor algebraically.
EXAMPLE 9
Evaluate . Yx2 Iun
xo
+ 100  10 X
2
•
Solution This is the limit we considered in Example 8. We can create a common factor by multiplying both numerator and denominator by the conjugate radical expression Yx 2 + 100 + 10 (obtained by changing the sigu after the square root). The preliminary algebra rationalizes the numerator: Yx2 + 100  10
Yx 2 + 100  10. Yx 2 + 100
+ 10 v'x + 100 + 10 2
x 2 + 100  100 + 100 + 10)
X2(Yx2 X2(Yx2
x2 + 100 + 10)
1 Yx 2 + 100
+ 10'
Common fuctor i'
Cancel xl for x =F 0
Therefore, · Yx2 IlID x~o
+ 100  10 x2
=
I'lID r~~I=:~ ~ x~o Yx 2 + 100 + 10 1
Y02 ~
+ 100 + 10
Denominator not 0 at x = 0; substitute
1 20 = 0.05.
This calculation provides the correct answer, in contrast to the ambiguous computer results in Example 8. • We cannot always algebraically resolve the problem of rmding the limit of a quotient where the denominator becomes zero. In some cases the limit might then be found with the
2.2
Limit of a Function and Limit Laws
53
aid of some geometry applied to the problem (see the proof of Theorem 7 in Section 2.4), or through methods of calculus (illustrated in Section 7.5). The next theorem is also useful.
y
The Sandwich Theorem
~~~~X
o
c
The following theorem enables us to calculate a variety of limits. It is called the Sandwich Theorem because it refers to a function f whose values are sandwiched between the values of two other functions g and h that have the same limit L at a point c. Being trapped between the values of two functions that approach L, the values of f must also approach L (Figore 2.12). You will fmd a proof in Appendix 4.
FIGURE 2.12 The graph of f is saodwiched between the graphs of g aod h.
THEOREM 4The Sandwich Theorem
Suppose that g(x) :5 f(x) :5 h(x) for all x in some open interval containing c, except possibly at x = c itself. Suppose also that lim g(x) = lim h(x) = L.
xc
y
xc
Thenlim..... d(x) = L.
The Sandwich Theorem is also called the Squeeze Theorem or the Pincbing Theorem. y~
1
4x'
EXAMPLE 10
Given that 2
~~~+~>x
1
I 
0
FIGURE 2.13 Aoy functioo u(x) whose graph lies io the regioo betweeo y ~ I + (x'/2) andy = I  (x'/4) has limit I as x > 0 (Example 10).
4x
:5 u(x) :5 I
Solution
Since lim(1  (x2/4» = I
and
+
(x 2/2» = I,
•
The Sandwich Theorem helps us establish several important limit rules:
(a) lim sin 0 = 0 1r
lim(I
x~o
the Sandwich Theorem implies thatlimx~o u(x) = I (Figore 2.13).
EXAM PLE 11 ~~~~ 8
for all x oF 0,
fmd lim..... o u(x) , no matter how complicated u is.
x~o
y
x2
+2
(h) lim cos 0 = I
60
80
(c) For any function f, lim If(x) I = 0 implies lim f(x) = xc
xc
o.
Solution (a)
y
(a) In Section 1.3 we established that 101:5 sinO :5101 for all 0 (see Figure 2.14a). Since lim'hO ( I 0 I) = lim'hO 10 I = 0, we have lim sinO = O.
8~O
(h) From Section 1.3,0:5 I  cosO :5101 for all 0 (see Figore 2.14b), and we have Inn,,~o (I  cos 0) = 0 or
_~2~~1~~O~~1~2~~8 lim cos 0 = I.
(b)
FIGURE 2.14 The Sandwich Theorem
coofl1'!I1S the limits io Example II.
8~O
(c) Since If(x)l:5 f(x):5 If(x) I andlf(x)1 and If(x) I have limit 0 as x>c,it follows that limx~, f(x) = O. •
54
Chapter 2: Limits and Continuity
Another important property oflimits is given by the next theorem. A proof is given in the next section.
THEOREM 5 possibly at x then
TIf(x) :5 g(x) for ailxin some openintervai containingc, except x approaches c,
= c itself, and the limits of f and g both exist as
lim fix) :5 lim g(x).
x"'c
x"'c
The assertion resulting from replacing the less than or equal to (:5) inequality by the strict less than «) inequality in Theorem 5 is false. Figure 2.l4a shows that for 0 .. 0, 101 < sin 0 < 10 I, but in the limit as 0 > 0, equality holds.
Exercises 2.2 Limits from Graphs 1. For the function g(x) graphed here, fmd the following limits or explain why they do not exist.
a. lim g(x) xI
b. lim g(x) x2
c. lim g(x)
d.
%3
y
lim g(x)
x2.S
y
4. Which of the following statements about the function y graphed here are 1rue, and which are false?
a. b. c. d.
2. For the function f(/) graphed here, fmd the following limits or explain why they do not exist.
a. lim f(/)
b. lim f(/)
12
11
c. lim f(/)
d.
10
lim f(/)
lim f(x) does not exist.
x~2
lim f(x) = 2 lim f(x) does not exist. x~l lim f(x) exists at every pointxo in (I, I). x~2
x~
t{).S
...
e. lim f(x) exists at every pointxo in (I, 3).
s
x~
...
IL
ro 2
I
= f(x)
y
I
0
1
)
t
I
I
3. Which of the following statements about the function y = f(x) graphed here are 1rue, and which are false? a. lim f(x) exists. x~o
b. lim f(x) = 0 x~o
c. lim f(x) = I x~o
d. lim f(x) = I x~l
e. lim f(x) = 0 x~l
f. lim f(x) exists at every point Xo in (1, 1). x~
...
g. lim f(x) does not exist. x~l
Existence of Limits In Exercises 5 and 6, explain why the limits do not exist. 5lim~
.
x~o
Ixl
6. lim _11 xI x
7. Suppose that a function f(x) is defined for all real values ofx except x = Xo. Can anything be said about the existence of lim~" f(x)? Give reasons for your answer. 8. Suppose that a function f(x) is defmed for all x in [I, I]. Can anything be said about the existence of limx~o f(x)? Give rea
sons for your answer.
2.2
9. Iflim,.....1 f(x) ~ 5, must f be defmed at x ~ I? If it is, must f( I) ~ 5? Can we conclude anything about the values of f at x ~ I? Explain. 10. If f(1) ~ 5, must lim'~1 f(x) exist? If it does, thea must lim,.....1 f(x) ~ 5? Can we conclude anything about lim,.....1 f(x)? Explain.
Limit of a Function and Limit Laws
Using Limit Rules
51. Suppose limro f(x) ~ 1 and limz~o g(x) ~ 5. Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation. lim 2f(x)  g(x) ~ ,~o (j(x)
+ 7)2/3
%7
+ 7)2/3
(a)
14. lim
x~o
(b)
+ 7))2/3
2 lim f(x)  lim g(x)
z,,' + 4x + 8)
(x 3 
xo
( lim (J(x)
+ 5x  2)
%2
16
lim (j(x)
xo
12. lim (x'
13. lim 8(t  5)(t  7)
 g(x))
lim 2f(x)  lim g(x)
Find the limits in Exercises 1122.
+ 5)
1!'!l, (2f(x) x~o
Calculating Limits
11. lim (z"
55
xo
%+2
xa
(c)
16. lim 38(Zs  1) s2/3
17. lim 3(z"  I)'
18.lim
19. lim (5  y)4/3
20. lim (2z  8)1/3
%1
y3
21. lim
h~O
(2)(1)  (5) (1 + 7)'/3
y+2
y2 y2
+
5y
+
6
z~o
3
v'3h+1 + I
22. lim
v5h+4 
40
2
52. Let lim,.....1 h(x) ~ 5, limz~lp(X) ~ 1, and lim'~1 r(x) ~ 2. Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation.
h
Limits of quotients Find the limits in Exercises 2342.
lim
x~1
23.lim x  5 xs x 2  25
24. lim
x'+3x1O lim 25. %5 X + 5
' x2  7x + 10 26. lun 2 %2 x 2 28. lim t + 3t + 2 tl t 2  t  2
29. lim z"  4 %+2 x 3 + 2,x2 !  I
yo
39. lim
xI
Vx+3  2 W+12  4
x2
X 
41. lim 2 x3
2
w=s +
X
3
y'5lim h(x) ,~1
,
38. lim
W+8 
%+1
40. lim
X
+1
x
+2
53.
Supposelim,~,,(x) ~
+ 3g(x))
54. Suppose lim.~"(x) •• lim (g(x) + 3) %4
~
4x
Find the limits in Exercises
c. lim (g(x»2
44. lim sin2 x
45. lim sec x
46. lim tan x
47. lim 1 + x + sinx xo 3 cos x
48. lim (x'  1)(2  cosx)
,~o
xo
Vx+4 cos (x + 'IT)
,~o
xo
xo
SO. lim Y 7
xo
+ sec' X
~ 
3. Find
%4
r
d •
g(x)
x~ f(x}  1
55. Suppose lim'~b f(x) ~ 7 and lim'~b g (x) ~  3. Find b. lim f(x)' g(x) •• lim (j(x) + g(x)) xb
xb
c. lim 4g(x)
43. lim (2 sin x  I)
r f(x) • ,~ f(x)  g(x)
0 and lim.~4 g (x) b. lim xf(x)
%4
x~45~
2. Find
xc
d
,~,
3
~
b. lim 2f(x)g(x)
xc
c. lim (j(x)
xI
5 2
5andlim,.....,g(x)
•• lim f(x)g(x)
x~2~3
42.lim
(c)
xI
yI(5)(5)
X
4 
xI
(1)(4  2)
4350.
%'11"
xI
x+l
16
(b)
( lim p(x»)( lim (4  r(x»))
+ _1_
v2 v 4 
Limits with trigonometric functions
49. lim
,~1
xI
36. lim x x '~42Vx ,~1
vlim 5h(x)
( lim p(x»)( lim 4  lim r(x»)
34. lim if  8
37.lim
,~1
3y4  16y2 _1_ xI
(a)
lim (P(x)(4  r(x)))
p(x)(4  r(x))
+ 4x + 3
xo
x
xI
z~1
5y 3 + 8y' 30. lim ';cco
32. lim
31. lim xl
limY5h(x)
VShW
x+3
%+3 Xl
27.lim t'+t2 II t2  1
7 4
d. lim f(x)/g(x)
xb
xb
56. Suppose that limr2 p(x) lim,~_,,(x) ~ 3. Find
~
4,lim,....._2 r(x)
•• lim (P(x) + r(x) + ,(x» %2
b.
lim p(x)' r(x)' ,(x)
%+2
c. lim (4p(x) x2
+ 5r(x))l8(x)
~
0, and
56
Chapter 2: Limits and Continuity
Umits of Average Rates of Change
h. Support your conclusion in part 0 such that for all x 0< Ix  xol < Il
I/(x) 
LI
0 that places the open interval (xo  Il, Xo + Il) centered atxo inside the interval (a, b). The inequality I/(x)  L I < E will hold for all x oF Xo in this Ilinterval.
NOT TO SCALE
FIGURE 2.22 The function and intervals in Example 4.
EXAMPLE 5
Prove that limx~2 /(x) = 4 if
/(x) Solutton
{x2,
=
I,
Our task is to show that given E > 0 there exists all> 0 such that for all x 0< Ix  21 < Il
y
I/(x)  41
0,
m
>
0,
120/x,
29. !(x) ~ nIX Xo = 1/2,
~
• = c
m
Xo 11,
~ 0.1
E
~
I,
X'o
=
E
L
=
2m,
X'o = 2,
L
=
3m,
X'o
~
(m/2)
0,
L
~ I
48.
E
24,
=
~
x~o
0.1 I
I
3,
E
~
L = m
+ b,
+ b,
0.03
e=c>O
>0 m > 0,
4
if !(x)
~
2 {x , I,
X #2 x =2
' I 441 • llJlr. "2
I
xv3 x
1~ !(x) ~ 0
49. lim
~
=
# I x = I X
46.lim xI
~
I
3
x2  1 =2 xI x
E
E
0.4
38. lim (3x  7)
5
x~l
0.02
~ 0.1
E ~
4,
L ~ 3,
~
~
41. lim !(x) = I
43.
E
Xo ~ 23,
22. !(x) ~ x 2,
24. !(x)
E ~
Xo ~ 10,
L ~ 4, ~
om
2,
=
Xo ~ 0,
L ~ 3,
20. !(x) ~ ~,
E ~
X'o = 4,
6,
=
19. !(x) ~ ~,
21. !(x)
5,
=
• = 0.05 • = 0.5
E ~
x~.
Eacb of Exercises 1530 gives a function !(x) aod numbers L, XO, aod E > O. In each case, f'md an open interval about Xo on which the inequality I!(x)  L I < E holds. Then give a value for 8 > 0 such that forallxsatisfyingO < Ix  xol < 8 the inequality I!(x)  LI 1. y
The number L is the limit of f(x) as x approaches Xo if, given any E > 0, there exists a value ofx for which If(x)  LI
O,f"md an interval I = (4  S, 4), S > 0, such that if x lies in I, then ~ < •. What !intit is being verified and what is its value?
. h3h sm
,~o
33. lim sin (I  cos t) 10 1  cost
0 cot 48
.~o sin' 0 cot' 20
0' cot 30
Formal Definitions of OneSided Limits
Find the lintits in Exercises 2142.
x
73
sons for your answer.
(J
b. lim(t  ltJ)
xo
42. lim
tan 0
Continuity
.~o
Use the def"ntitions of righthand and lefthand lintits to prove the lintit statements in Exercises 49 and 50. 49. lim IXI = I xo
X
50. lim ( x2+ x
;1
=
I
51. Greatest integer function Find (a) lim,_.oo' l x J and (b) lim,_.oo l x J ; then use limit definitions to verilY your f"mdings. (0) Based on your conclusions in parts (a) and (b), can you say anything about lim,........ l x J ? Give reasons for your answer.
< 00 x> .
52. Onesided1intits Letj(x) = {:';in(I/X), x
vx,
Find (a) lim,.....o' j(x) and (b) lim,_o j(x); then use lintit definitions to verilY your f"mdings. (0) Based on your conclusions in parts (a) and (b), can you say anything about lim,~o j(x)? Give
reasons for your answer.
Continuity When we plot function values generated in a laboratory or collected in the field, we often connect the plotted points with an unbroken curve to show what the function's values are likely to have been at the times we did not measure (Figure 2.34). In doing so, we are assuming that we are working with a continuaus jUnction, so its outputs vary continuously with the inputs and do not jump from one value to another without taking on the values in between. The limit of a continuous function as x approaches c can be found simply by calculating the value of the function at c. (We found this to he true for polynomials in Theorem 2.) Intuitively, any function y = t(x) whose graph can be sketched over its domain in one continuous motion without lifting the pencil is an example of a continuous function. In this section we investigate more precisely what it means for a function to be continuous.
74
Chapter 2: Limits and Continuity
y
80
p
I Q
~9 f
We also study the properties of continuous functions, and see that many of the function types presented in Section 1.1 are continuous.
Continuity at a Point
Q /
To understand continuity, it helps to consider a function like that in Figure 2.35, whose limits we investigated in Example 2 in the last section.
Q 1/
I'"
..... o
5
10
Elapsed time (sec)
FIGURE 2.34 Connecting plotted points by an unbroken curve from experimental
data Q, , Q2, Q" ... for a falling object
EXAMPLE 1 Find the points at which the function 1 in Figure 2.35 is continuous and the points at which 1 is not continuous. The function 1 is continuous at every point in its domain [0,4] except at 2, and x = 4. At these points, there are breaks in the graph. Note the relationship between the limit of 1 and the value of 1 at each point of the function's domain.
Solution x = I, x
=
Points at which f is continuous:
y
2
Xf(X) • 1
2
3
=
0,
lim I(x) = 1(0). xo+
Atx
=
3,
AtO
x
Continuity Test A function f(x) is continuous at an interior point x = c of its domain if and only if it meets the following three conditions.
o
FIGURE 2.38 A function that has a jump discontinuity at the origin (Example 3).
Y
1. 2. 3.
3
Y~
2
LxJ
f(c) exists
(c lies in the doroain of f).
limx~,
f(x) exists
(f has a limit as x > c).
limx~,
f(x) = f( c)
(the limit equals the function value).
For onesided continuity and continuity at an endpoint, the limits in parts 2 and 3 of the test should be replaced by the appropriate onesided limits.

4
1
75
is rightcontinuous at a and continuous at a right endpoint b of its domain if it is leftcontinuous at b. A function is continuous at an interior point c of its doroain if and only if it is both rightcontinuous and leftcontinuous at c (Figure 2.36).
y
2
Continuity
EXAMPLE 4 Thefunctiony = lxJ introduced in Section l.l is graphed in Figure 2.39. It is discontinuous at every integer because the lefthand and righthand limits are not equal
........0
........0
asx~n:
limJx J = n  I
........0
and
x~n
x
1
I
0
2
3
4
2
FIGURE 2.39 The greatest integer function is continuous at every noninteger point It is rightcontinuous, but not leftcontinuous, at every integer point (Example 4).
Since l1l J = 11, the greatest integer function is rightcontinuous at every integer 11 (but not leftcontinuous). The greatest integer function is continuous at every real numher other than the integers. For example, lim lxJ = I = l1.5J.
x+1.5
In general, if 11

I
0, and we can remove the discontinuity by setting f(O) equal to this limit. The discontinuities in Figure 2.40d through f are more serious: lim..... o f(x) does not exist, and there is no way to improve the situation by changing f at O. The step function in Figure 2.40d has a jump discontinuity: The onesided limits exist but have different values. The function f(x) = l/x 2 in Figure 2.40. has an inUrnte discontinuity. The function in Figure 2.40f has an oscillating discontinuity: It oscillates too much to have a limit as x>O.
76
Chapter 2: Limits and Continuity y
y
y
y
1
x
Y ~ f(x)
0
(b)
(a)
x
(d)
(c)
y
J
Y~f(x)~l2
0
x
x x
(f)
(e)
FIGURE 2.40
The functioo in (a) is cootinuous at x
~
0; tire functioos in (b) through (f)
are not
Continuous Functions A function is continuous on an interval if and only if it is continuous at every point of the interval. For example, the semicircle function graphed in Figure 2.37 is continuous on the interval [  2, 2], which is its domain. A continuous function is one that is continuous at every point of its domain. A continuous function need not be continuous on every interval. Y
EXAMPLE 5 (a) The function y = l/x (Figure 2.41) is a continuous function because it is continuous at every point of its domain. It has a point of discontinuity at x = 0, however, because it is not defined there; that is, it is discontinuous on any interval containing x = O.
~t~x
\ "
The functioo y ~ l/x is continuous at every value of x except x = O. It has a point of discontinuity at x = 0 (Example 5). FIGURE 2.41
(h) The identity function I(x) = x and constant functions are continuous everywhere by Example 3, Section 2.3. • Algebraic combinations of continuous functions are continuous wherever they are defined.
THEOREM BProperties of Continuous Functions If the functions I and g are continuous at x = c, then the following combinations are continuous at x = c.
Differences:
I+g Ig
Constant multiples:
k· I, for any number k
Products:
I'g Ilg,
1. Sums:
2. 3. 4. 5. 6. 7.
Quotients: Powers:
r,
Roots:
"ifj
providedg(c) oF 0 n a positive integer provided it is deImed on an open interval
, containing c, where n is a positive integer
2.5
Continuity
77
Most of the results in Theorem 8 follow from the limit rules in Theorem I, Section 2.2. For instance, to prove the sum property we have
lim(j + g)(x)
xc
= lim(j(x) xc
= lim f(x) xc
= f(c) =
This shows that f
+ g(x)) + lim g(x),
Sum Rule, Theorem 1
xc
+ g(c)
Continuity of j. g at c
(j + g)(c).
+ g is continuous.
EXAMPLE 6 (a) Every polynomial P(x) = anx n + a._1Xn1 lim P(x) = P(c) by Theorem 2, Section 2.2.
+ ... + ao
is continuous because
x~c
(b) If P(x) and Q(x) are polynomials, then the rational function P(x)/Q(x) is continuous wherever it is dermed(Q(c) oF 0) by Theorem 3, Section 2.2. _
EXAMPLE 7 The function f(x) = Ixl is continuous at every value of x. If x > 0, we have f(x) = x, a polynomial. Ifx < 0, we have f(x) = x, another polynomial. Finally, at the origin, limx~o Ixl = 0 = 101. The functions y = sin x and y = cos x are continuous at x = 0 by Example 11 of Section 2.2. Both functions are, in fact, continuous everywhere (see Exercise 68). It follows from Theorem 8 that all six trigonometric functions are then continuous wherever they are dermed. For example, y = tanx is continuous on ... U (7r/2, 7r/2) U (7r/2, 37r/2) U ....
Composites All composites of continuous functions are continuous. The idea is that if f(x) is continuous atx = c andg(x) is continuous atx = f(c), theng 0 f is continuous atx = c (Figure 2.42). In this case, the limit as x > c is g(f(c».
Continuous at c g
f Continuous
•
c
FIGURE 2.42
ate
_
Continuous :;.o C_ _ _ _ atf(e) _ _ _ _=. . ~f(c) g(f(c»
Composites of continuous functions are continuous.
THEOREM 9Composite of Continuous Functions If f is continuous at c and g is continuous at f(c), then the composite g 0 f is continuous at c.
Intuitively, Theorem 9 is reasonable because if x is close to c, then f(x) is close to f(c), and since g is continuous at f(c), it follows thatg(f(x» is close to g(f(c». The continuity of composites holds for any finite number of functions. The only requirement is that each function be continuous where it is applied. For an outline of the proof of Theorem 9, see Exercise 6 in Appendix 4.
78
Chapter 2: Limits and Continuity
EXAMPLE 8 Show that the following functions are continuous everywhere on their respective domains. =
Yx 2
(e) y =
Ix2 _
(a) y
x 2/3
2x  5

+ x4
(b) y = I
21
x  2
(d) Y =
1:2S:~1
Solution
y
0.4 3
(a) The square root function is continuous on [0, 00) because it is a root of the continuous identity function fix) = x (part 7, Theorem 8). The given function is then the composite of the polynomial fix) = x 2  2x  5 with the square root function g(t) = Vi, and is continuous on its domain.
(b) The numerator is the cube root of the identity function squared; the denominator is an everywherepositive polynomial. Therefore, the quotient is continuous. (e) The quotient (x  2)/(x 2  2) is continuous for all x # ± v2, and the function is the composition of this quotient with the continuous absolute value function (Example 7).
FIGURE 2.43 The graph suggests 1hat y = I(x sinx)/(x 2 + 2) I is continuous
(Example 8d).
(d) Because the sine function is everywherecontinuous (Exercise 68), the numerator term x sin x is the product of continuous functions, and the denominator term x 2 + 2 is an everywherepositive polynomial. The given function is the composite of a quotient of continuous functions with the continuous absolute value function (Figure 2.43). • Theorem 9 is actually a consequence of a more general result which we now state and prove.
THEOREM lOLimits of Continuous Functions b and limx~d(x) = b, then lim~, g(f(x» =
Proof that
Let E
>
g(b)
If g is continuous at the point
= g(lim~, fix»~.
0 be given. Since g is continuous at b, there exists a nornber 8 , Ig(y)  g(b)1
1+, we have (x  I) ..... 0+ and I/(x  1) ..... oo.Asx ..... I, we have (x  I) ..... 0 and I/(x  I) .....
y
00.
} x 0
x.o+
No matter how high B is, the graph
EXAMPLE 10
•
Discuss the behavior of
goes higher.
~ x
I f(x) = 2" x x
FIGURE 2.59 The graph of f(x) io Example 10 approaches inf"mity as x ..... O.
x>
as
o.
As x approaches zero from either side, the values of l/x 2 are positive and become arbitrarily large (Figure 2.59). This means that
Solution
lim f(x) = lim
xo
1, =
x+Ox
00.
•
90
Chapter 2: Limits and Continuity The function y = I/x shows no consistent behavior as x > o. We have I/x > 00 if x>O+, but I/x> 00 ifx> 0. All we can say about lim.o (I/x) is that it does not exist. The function y = l/x 2 is different. Its values approach infinity as x approaches zero from either side, so we can say that limx _ o (l/x 2 ) = 00.
EXAMPLE 11 These examples illustrate that rational functions can behave in various ways near zeros of the denominator. I· (x  2)2 lim x  2 · (x  2f IlID 2 (a) x2 lID ( 2 = 0 x  4 = x2 x  2)(X + 2) = x2 X + (b) lim x  2 = lim .~2 X 2 
(c) y
4
.~2
x  2 = lim _1_ = ! (x  2)(x + 2) .~2 x + 2 4
lim x  3 = lim
.2+ X 2 
4
.2+ (X 
(d) lim x  3 = lim • 2 X 2 
4
.T
.2
(f) lim
B ~~r~~
;;If~;;;____c:\~x
Xo
+8
4
x  3 = 00 (x  2)(x + 2)
.2 .2
(e) lim x  3 = lim • 2 X 2 
x  3 = _ 00 2)(x + 2)
x  3 does not exist. (x  2)(x + 2)
The values are negative for x > 2, x near 2. The values are positive forx < 2, x near 2 . See parts (c) and (d).
.2
2x = lim (x2)=lim I =00 (x  2)3 (x  2)3 (x  2)2
In parts (a) and (b) the effect of the zero in the denominator at x = 2 is canceled because the numerator is zero there also. Thus a finite limit exists. This is not true in part (f), where cancellation still leaves a zero factor in the denominator. •
Predse Definitions of Infinite Limits FIGURE 2.60 For XQ  8 < x < xo + 8, the graph of f(x) lies above the lioe y = B. y
Instead of requiriog f(x) to lie arbitrarily close to a finite number L for all x sufficieotly close to xo, the defInitions of infinite limits require f(x) to lie arbitrarily far from zero. Except for this change, the language is very similar to what we have seeo before. Figures 2.60 and 2.61 accompany these definitions.
DEFINmONS 1. We say that f(x) approaches infinity as x approaches "0, and write lim f(x) = 00, XXo
if for every positive real number B there exists a correspoodiog Ii that for all x
>
0 such
f(x»B. o < Ix  xol < Ii 2. We say that f(x) approaches minus inf"mity as x approaches xo, and write lim f(x) = 00, XXo
FIGURE 2.61 For Xo  8 < x < XQ the graph of f(x) lies below the lioe y = B.
+ 8,
if for every negative real number  B there exists a correspondiog Ii that for all x f(x) < B. o < Ix  xol < Ii
>
0 such
The precise defmitions of ooesided infinite limits at Xo are similar and are stated in the exercises.
2.6
EXAMPLE 12 Solution
Prove that lim
xa
Given B
>
Limits Involving Infinity; Asymptotes of Graphs
~
91
= 00
X
>
0, we want to find 8
0 such that
1 implies 2>B. x
0;  4
y
For x numerically large, 2x
~ 4 is near O.
For x near 2, this term is very large.
If we want to know how f behaves, this is the way to find out. It behaves like y = (x/2) + I when x is numerically large and the contribution ofl/(2);  4) to the tota\ value of f is insignificant. It behaves like 1/(2);  4) when x is so close to 2 that
_~2~_~I~~Ott~~~2~X
5
1/(2);  4) makes the dominant contribution. We say that (x/2) + I dominates when x is numerically large, and we say that 1/(2);  4) dominates when x is near 2. Dominant terms like these help us predict a function's behavior.
Let f(x) = 3x 4  2>;3 + 3x 2  5x + 6 and g(x) = 3x4. Show that although f and g are quite different for numerically small values of x, they are virtually identical for Ix I very large, in the sense that their ratios approach I as x ..... 00 or
EXAMPLE 16
V'2
d. x+l
b. x+2+
d.x>2
e. What, if anything, can be said about the limit as x > O?
72. f(2)
+2
73. 74. 75.
t~/3) as
~
b. t+ 0
:/3
x+ 1+
.%0+
~
2, and
0, lim f(x) ~ 0, lim f(x) ~ x±oo
%1
~ 00,
I, f( I)
~
~ 00,
and lim f(x) .%1
0, lim f(x) x_OO
~
and lim f(x)
lim f(x) ~
.1'1+ ~ 00
0, lim f(x) .1'0+
~
00,
~ 00,
I
%_00
lim f(x)
~
0, lim f(x) ~
lim g(x)
~
0, lim g(x)
lim h(x)
~
I, lim h(x)
x_±OO x_±OO
.%2
.%3
00,
and lim f(x) ~ x2+
~ 00, ~
and lim g(x) ~ x3+
I, lim h(x)
.%_00
00
.%0
~
00
I, and
lim k(x) x_±OO
~
I, lim k(x) .%1
~ 00,
and lim k(x) xI+
~ 00
for your answer.
C:/5 + 7) as
t+ 0+
0, lim f(x)
79. How many horizootal asymptotes can the graph of a given rational func1ioo have? Give reasons for ynur answer.
d. x+ 1
Finding Limits of Differences when x> ±oo Find the limits in Exercises 8086. x~oo
81.
83.
b. x+O
x+ 1+
d. x+ 1
(Vx+9  Vx+4) lim (\I'x 2 + 25  w=J) lim (w+3 + x) x_oo lim (2x + yrc _c 4X"2+c3x 2)
80. lim
82.
~/3  (x  I I)4/') as x
a. x+o+ C.
~
~
77. Suppose that f(x) and g(x) are polynootials in x and that limx_oo (f(x)/g(x)) ~ 2. Can you conclude anything about limx_~ (f(x)/ g (x))? Give reasons for your answer.
t+ 0+
62. lim(
I, and
78. Suppose that f(x) and g(x) are polynootials in x. Can the graph of f(x)/g(x) have an asymptote if g(x) is never zero? Give reasons
x + (x  2 I)2/') as a. x+o+ b. x+OC.
~
X~ ...
Find the limits in Exercises 5'HJ2.
61. lim(
2, lim f(x)
lim h(x) ~ I
4x
e. What, if anything, can be said about the limit as x > O?
a.
I
2
.%_00
as 76.
60. lim
+
In Exercises 7376, fmd a func1ioo that satisfies the given cooditions and sketch its graph. (The answers here are not unique. Any func1ioo that satisfies the conditions is acceptable. Feel free to use formulas defmed in pieces if that will help.)
as
a. x+o+ c. x+2
a.
~
xo
c.x+l+
59. lim(2 
2x
X~OO
x_±OO
lim f(x)
a.x+2+

~
2, f( I)
0, lim f(x)
lim f(x)
1 56. hm 2x + 4 as
x3
~
xI+

x23x+2
~
0, f(l)
...
71. f(O) b. x+O
58. limx2  3x
~
lim f(x)
a. x+o+
r
68. Y ~ x
X~OO
70. f(O)
• un x 3 2x 2
x3
lim f(x) ~ I
c. x+ 1+
57
+3 +2
3
y~
Inventing Graphs and Functions In Exercises 61}72, sketch the graph of a func1ioo y ~ f(x) that satisfies the given oouditioos. No form:u1as are requiredjust label the coordinate axes and sketch an appropriate graph. (The answers are not unique, so your graphs may not be exactly like those in the answer sec1ioo.)
69. f(O)
a.x+l+
. x2
66.
65'Y~2x+4
54.lim+as x  I
55. lim(x; 
I 64'Y~x+1
I
lim
95
Limits Involving Infinity; Asymptotes of Graphs
X~OO
x_oo
84. lim (v'9x 2  x  3x) x~OO
96
Chapter 2: Limits and Continuity
85. lim (VX2
+ 11: 
86. lim (Vx2
+x
x~oo
X~OO
VX 2  2x)
Use the formal defmitioos from Exercise 93 to prove the lintit statemoots in Exercises 9498.
 ~)
94. lim xl =
Using the Formal Definitions Use the formal defmitioos of limits as x> ±oo to establish the limits in Exercises 87 and 88.
95. lim
00
%0+
96. lim _1_ =
2
x_2X 
00
00
97. lim _1_ =
2
x_2+X 
98. lim _1_2 =
00
00
%11  x
87. If j bas the coostant value j(x) = k, then lim j(x) = k.
1=
x_ox
X~OO
88. If j bas the coostant value j(x) = k, then lim j(x) = k. X~OO
Use formal defmitions to prove the lintit statements in Exercises 8c}'92. I 89. lim  2 = 00 90. lim _III = 00
xo
xo x
91. lim
x~3
2 (x  3)2
92. lim
=00
X~,
Oblique Asymptotes Graph the ratioua1 functioos in Exercises 9c}'104. Include the graphs and equations of the asymptotes. x2 x 2 +1 99. Y =   I 100. Y =   I
x
x2
X
(x
I
+ 5)2
= 00
93. Here is the defmition ofinfinite righthand limit.
x
4
x=T
102. Y
x2  1 103. y=x
104• Y
101. Y
=
=
x 2 1 2x + 4
=x3+1 X
2
Additional Graphing Exercises the curves in Exercises \05\08. Explain the relationship between the curve's formula and what you see.
D Graph We say that j(x) approaches infmity as x approaches Xo from the right, and write lim j(x) =
105.Y=~2 4  x
00,
xXQ+
if.
for every positive real number B. there exists a corresponding nomber 8 > 0 such that for all x
:%'0 o+? h. How does the graph behave as x > ± oo?
Modify the defmition to cover the following cases. L
lim_j(x)
= 00
c. How does the graph behave near x = I and x = I?
%%0
lim j(x)
= 00
Give reasons for your answers.
c. limj(x)
= 00
109. Y
h.
%xo+ %Xo
Chapter
~
=
r
&(x  ~
3(
X )2/3
1l0'Y=I x  I
Questions to Guide Your Review
1. What is the average rate of chaoge of the function g(t) over the interval from t = a to t = b? How is it related to a secant line?
6. What theorems are available for calculating lintits? Give examples of bow the theorems are used
2. What lintit must be calculated to fmd the rate of change of a function g(t) att = to?
7. How are onesided limits related to limits? How can this relationship sometimes be used to calculate a limit or prove it docs not exist? Give examples.
3. Give ao informal or inmitive defmition of the lintit lim j(x) = L. X~Xo
Why is the definition "informal''? Give examples. 4. Docs the existence aod value of the limit of a function j(x) as x approaches Xo ever depend on what happens at x = xo? Explain aod give examples. 5. What function behaviors might occur for which the limit may fail to exist? Give examples.
8. What is the value oflim._o «sinO)/O)? Docs it matter whcther 0 is measured in degrees or radiaos? Explain. 9. What exactly does lim~Xo j(x) = L mean? Give an example in which you fmd a 8 > 0 for a given j, L, Xo, aod. > 0 in the precise defmition oflintit.
10. Give precise defmitions of the following staterneots.
a. lim,....2 j(x) = 5 c. lim,....2 j(x) = 00
h. lim~2' j(x) = 5
d. lim,....2 j(x) = 
00
97
Chapter 2 Practice Exercises 11. What conditions must be satisfied by a function if it is to be continuous at an interior point of its domain? At an endpoint?
12. How can looking at the graph of a function help you tell where the fiwnctionis continuous? 13. What does it mean for a function to be rightcontinuous at a point? Leftcontinuous? How are continuity and onesided continuity related?
14. What does it mean for a fiwnction to be continuous on an interval? Give examples to illustrate the fact that a function that is not continuous on its entire domain may still be continuous on selected intervals within the domain.
15. What are the basic types of discontinuity? Give an example of each. What is a removable discontinuity? Give an example.
Chapter ~
16. What does it mean for a function to have the Intermediate Value Property? What conditions guarantee that a function has this property over an interval? What are the consequences for graphingandsolvingtheequationf(x) ~ O?
17. Under what circumstances can you extend a function f(x) to be
= c? Give an example.
continuous at a point x
18. What exact1y do limr _
oo
f(x) = L and limr~ f(x)
~
L mean?
Give examples.
19. What are limr_± k (k a constant) and lim~± (I/x)? How do you extend these results to other functions? Give examples. 20. How do you f"md the limit of a rational function as x + ± oo? Give examples. 21. What are horizontal and vertical asymptotes? Give examples.
Practice Exerdses
Lim;ts and ContinuUy 1. Graph the function
In Exercises 5 and 6, f"md the value that lim~og(x) must have if the given lintit statements hold.
I,
f(x) =
I
~:
x, I,
x:5
:'.
122
Chapter 3: Differentiation
I How to Read the Symbols for
If y" is differentiable, its derivative, y'" = dy" /dx = d'y/dx', is the third derivative ofy with respect to x. The names continue as you imagine, with
Derivatives
y' ''y prime" y. 'y double prime" d'y
dx' y~
yv.)
d"y
dx" D"
d d"y y(") = dxy("l) = dx" = D"y
"d squared y dx squared"
denoting the nth derivative of y with respect to x for any positive integer n. We can interpret the second derivative as the rate of change of the slope of the tangent to the graph of y = f{x) at each point. You will see in the next chapter that the second derivative reveals whether the graph bends upward or downward from the tangent line as we move off the point of tangency. In the next section, we interpret both the second and third derivatives in terms of motion along a straight line.
''y triple prime" ''y super n"
"d to the n of y by dx to the n" "Dtothen"
EXAMPLE 8
The
flISt
four derivatives of y
x'  3x' + 2 are
=
First derivative:
y' = 3x'  6x
Second derivative:
y"
= 6x 
Third derivative:
y.'
=
Fourth derivative:
y(4) =
6
6 O.
The function has derivatives of all orders, the fifth and later derivatives all being zero.
_
Exercises 3.3 Derivative Calculations
25. v
In Exercises 112, fmd the fITSt and second derivatives.
1. y = x' + 3
2. Y = x' + x + 8
3•• = 51'  31'
4. w
4x' 5. y=Tx
x3 x2 x 6·y=T+T+4
7. w = 3z2
8. •
}
= 3z'
 7z 3
+ 21z2
11.r=I~
3.'
=
X4
29. Y
=
10. Y
=4
31. Y
= (x
12. r
=
 2x  x,
12  ~ 8 8'
+ ~4 8
In Exercises 1316, find y' (a) by applying the Product Rule and (b) by multiplying the factors to produce a sum of simpler terms to differentiate.
37. w
=
Find the derivatives of the functions in Exercises 1728.
39. p =
2x + 5 17. Y = 3x  2 x2
19. g(x) = x
21. v
= (1
23. /(.)

18.
4
+ 0.5
 /)(1
= Vs Vs +
 4x)
4  3x z=,3x + x
20. /(1) =
22. w
I I
24 =5x+1 • u 2v'i
= (2x
 7,'(x
34
.
(8  1)(8'
,
+
8
e;/Z):'  lOx  5x'
27. Y
I + x  4v'i x
=
+ 5)
u'(O)
= 3,
v(O) = 1,
=0
v'(O) = 2.
Find the values of the following derivatives at x = O.
d d. dx (7v  2u)
3.3 42. Suppose u and v are differentiable functions of x and that
Differentiation Rules
52. Find all points (x,y) on the graph of f(x)
x' with tangent lines
passing through the point (3, 8).
u(i) = 2, u'(I) = 0, v(l) = 5, v'(I) = I. Find the values of the following derivatives at x = I.
d
=
123
f(x)
= xl
d d. dx (7v  2u)
a. dx (uv)
Slope. and Tangents 43. a. Normal to a curve Find an equation for the line perpendicular to the tangent to the curve y = x'  4x + I at the point (2, I). b. Smalle.t slope Wbat is the smallest slope on the curve? At what point on the curve does the curve have this slope?
c. Tangents having specified slope Find equations for the tangents to the curve at the points where the slope of the curve is 8. 44. a. Horizontal tangents Find equations for the horizontal tan3x  2. Also !md equations for gents to the curve y = the lines that are perpendicular to these tsngents at the points of tangency.
x' 
b. Smalle.t slope Wbat is the smallest slope on the curve? At what point on the curve does the curve have this slope? Find an equation for the line that is perpendicular to the curve~ tangent at this point. 45. Find the tangents to Newton's serpentine (graphed here) at the origin and the point (I, 2).
53. a. Find an equation for the line that is tangent to the curve y = xatthepoint(I,O).
x' 
D b.
Graph the curve and tsngent line together. The tsngent intersects the curve at another point. Use Zoom and Trace to estimate the point's coordiuates.
D c.
Conrmn your estimates of the coordinates of the secood intersection point by solving the equations for the curve and tsngent simultaneously (Solver key).
54. a. Find an equation for the line that is tsngent to the curve y = x 3  6x 2 + Sx at the origin.
D b.
Graph the curve and tsngent together. The tangent intersects
the curve at another point. Use Zoom and Trace to estimate
y
the point~ coordiuates.
D c.
Conf'mn your estimates of the coordinates of the second intersection point by solving the equations for the curve and tangent simultaneously (Solver key).
Theory and Example. 46. Find the tangent to the Witch ofAgnesi (graphed here) at the point
(2, I).
For Exercises 55 and 56 evaluate each limit by !lrSt coovertiog each to a derivative at a particular xvalue.
55. lim
xI
y y=_8_ x2 + 4
0
2 3
I
50
56.
"=xI
lim x'/9  I xl
X
+
1
57. Find the value of a that makes the following function differentiable for all xvalues.
x
47. Quadratic tangent to identity funetlon The curve y = + bx + c passes through the point (I, 2) and is tsngent to the line y = x at the origin. Find a, b, and c.
49. Find all points (x, y) on the graph of f(x) = 3x'  4x with tangent lines parallel to the line y = 8x + 5. SO. Find all points (x,y) on the graphofg(x) = tx'  ~x' tangent lines parallel to the line 8x  2y = I.
+
I with
51. Find all points (x, y) on the graph of y = x/(x  2) with tangent lines perpendicular to the line y = 2x + 3.
x:5
I 1
59. The general polynomial of degree n has the fonn
P(x} where a.
¢'
=
a,.x n + a,._tXnt
+ ... + a2x 2 + atX + ao
O. Find P' (x).
60. The body's re••tion to medicine The reaction of the body to • dose of medicine can sometimes be represented by an equation of the form
124
Chapter 3: Differentiation
where C is a positive constant and M is the amount of medicine absorbed in the blood. If the reaction is a change in blood pressure, R is measured in millimeters of mercury. If the reaction is a change in temperature, R is measured in degrees, and so on. Find dRj dM. This derivative, as a function of M, is called the sensitivity of the body to the medicine. In Section 4.5, we will see how to find the amount of medicine to which the body is most sensitive. 61. Suppose that the function v in the Derivative Product Rule has a constant value c. What does the Derivative Product Rule then say?
64. Power Rule for negative integers Use the Derivative Quotient Rule to prove the Power Rule for negative integers, that is, d (x m) dx
p=~_an2 V  nb
=
V2
'
in which a, b, n, and R are constants. Find dPjdV. (See accompanying figure.)
a. The Reciprocal Rule says that at any point where the function vex) is differentiable and different from zero,
(1)v
l
65. Cylinder pressure If gas in a cylinder is maintained at a constant temperature T, the pressure P is related to the volume Vby a formula of the form
62. The Reciprocal Rule
dx
mx  m
where m is a positive integer.
What does this say about the Derivative Constant Multiple Rule?
.E...
=
_l...dv v2dx'
Show that the Reciprocal Rule is a special case of the Derivative Quotient Rule. b. Show that the Reciprocal Rule and the Derivative Product Rule together imply the Derivative Quotient Rule.
63. Generalizing the Product Rule The Derivative Product Rule gives the formula d (uv) dx
=
dv udx
+ vdudx
66. The best quantity to order One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is
for the derivative of the product uv of two differentiable functions ofx.
A(q) =
a. What is the analogous formula for the derivative of the product uvw of three differentiable functions of x? b. What is the formula for the derivative of the product u 1 u2 u3 U4 offour differentiable functions of x? c. What is the formula for the derivative of a product Ul U2U3 ..• Un of a finite number n of differentiable functions of x?
3.4
km
hq
q + cm + 2'
where q is the quantity you order when things run low (shoes, radios, brooms, or whatever the item might be); k is the cost of placing an order (the same, no matter how often you order); c is the cost of one item (a constant); m is the number of items sold each week (a constant); and h is the weekly holding cost per item (a constant that takes into account things such as space, utilities, insurance, and security). Find dAjdq and d 2Ajdq2.
The Derivative as a Rate of Change In Section 2.1 we introduced average and instantaneous rates of change. In this section we study further applications in which derivatives model the rates at which things change. It is natural to think of a quantity changing with respect to time, but other variables can be treated in the same way. For example, an economist may want to study how the cost of producing steel varies with the number of tons produced, or an engineer may want to know how the power output of a generator varies with its temperature.
Instantaneous Rates of Change If we interpret the difference quotient (J(x + h)  !(x)) / h as the average rate of change in ! over the interval from x to x + h, we can interpret its limit as h ~ 0 as the rate at which! is changing at the point x.
3.4 The Derivative as a Rate of Change
DEFINmON
125
The instantaneous rate of change of / with respect to x at Xo is
the derivative /
'(x ) = lim /(xo
o
+ h)
 /(xo)
h
hO
'
provided the limit exists.
Thus, instantaneous rates are limits of average rates.
It is conventional to use the word instantaneous even when x does not represent time. The word is, however, frequently omitted. When we say rate of change, we mean instantaneous rate of change.
EXAM PLE 1
The area A of a circle is related to its diameter by the equation
A = TT D2 4 How fast does the area change with respect to the diameter when the diameter is 10m? Solution
The rate of change of the area with respect to the diameter is
~
= ; • 2D =
TTf.
When D = 10 m, the area is changing with respect to the diameter at the rate of (TT/2)10 = 5TT m2/m "" 15.71 m 2/m. •
Motion Along a Line: Displacement,. Velocity, Speed, Acceleration, and Jerk Suppose that an object is moving along a coordinate line (an saxis), usually horizontal or vertical, so that we know its position s on that line as a function of time t:
s Position at time t ...
I'
•
s~f(t)
and at time t
i1s >l'1
/(t).
The displacement of the object over the time interval from tto t
+ At
.
=
Ils = /(t
) S
s+l!.s~f(t+l!.t)
FIGURE 3.12 The positions ofa body moving along a coordinate line at time t aod shortly later at time t + b.t. Here the coordinate line is horizontal.
+ Ilt (Figure 3.12) is
+ Ilt)  /(t) ,
and the average velocity of the object over that time interval is
Vav
=
displacement Ils /(t travel time = Ilt =
+ tlt)  /(t) Ilt
To find the body's velocity at the exact instant t, we take the limit of the average velocity over the interval from t to t + Ilt as Ilt shrinks to zero. This limit is the derivative of / with respect to t.
DEFINmON Velocity (instantaneous velocity) is the derivative of position with respect to time. If a body's position at time tis s = /(t) , then the body's velocity at time t is
_ tis _
.
() vtdtInn &0
/(t + tlt)  /(t) Il t
126
Chapter 3: Differentiation Besides telling how fast an object is moving along the horizontal line in Figure 3.12, its velocity tells the direction of motion. When the object is moving forward (8 increasing), the velocity is positive; when the object is moving backward (. decreasing), the velocity is negative. If the coordinate line is vertical, the object moves upward for positive velocity and downward for negative velocity (Figure 3.13). s
)
,
o
s increasing:
s decreasing:
positive slope so
negative slope so moving downward
moving upward
FIGURE 3.13 For motion. ~ /(,) along a straight line (tire vertical axis), v = dsldt is positive when s increases and negative whens decreases. The blue curves represent position along lhe lioe over time; lhey do not portray lhe palh of motion, which lies along lhe .axis.
If we drive to a friend's house and back at 30 mph, say, the speedometer will show 30 on the way over but it will not show  30 on the way back, even though our distance from home is decreasing. The speedometer always shows .peed, which is the absolute value of velocity. Speed measures the rate of progress regardless of direction.
DEFINITION
Speed is the absolute value of velocity. Speed
~
IV(I)I
~ I~I
EXAMPLE 2 Figure 3.14 shows the graph of the velocity v = /'(1) ofa particle moving along a horizontal line (as opposed to showing a position function. = /(1) such as in Figure 3.13). In the graph of the velocity function, it's not the slope of the curve that tells us if the particle is moving forward or backward along the line (which is not shown in the figure), but rather the sign of the velocity. Looking at Figure 3.14, we see that the particle moves forward for the first 3 sec (when the velocity is positive), moves backward for the next 2 sec (the velocity is negative), stands motionless for a full second, and then moves forward again. The particle is speeding up when its positive velocity increases during the first second, moves at a steady speed during the next second, and then slows down as the velocity decreases to zero during the third second. It stops for an instant at 1 ~ 3 sec (when the velocity is zero) and reverses direction as the velocity starts to become negative. The particle is now moving backward and gaining in speed until 1 ~ 4 sec, at which time it achieves its greatest speed during its backward motion. Continuing its backward motion at time 1 ~ 4, the particle starts to slow down again until it finally stops at time 1 = 5 (when the velocity is once again zero). The particle now remains motionless for one full second, and then moves forward again at 1 ~ 6 sec, speeding up during the f"mal second of the forward motion indicated in the velocity graph. •
3.4 The Derivative as a Rate of Change
127
v
, MOVES FORWARD
,
I
(v> 0)
flQRWARD I
I
AGAIN
I
:
(v> 0)
:
Velocity v ~ 1'(')
,
I
I
I
I
"
I ,,
~ Speeds +: up ,I
Speeds~Steady~ Slows ~
up
:(v
= const) :
down
,
I
, ,I
I
"
rI
:
I
Stands
till
S
,"
~I
I (v = 0) I
I I
I I
0;;i"';c!2c"::i...~,+.5:6~'7~ ' ,(sec)
,
4 Speeds ~ Slows ~ I Up , down I I I I
I I MOVES BACKWARD
:
(v
< 0)
I I I
:
FIGURE 3.14 The velocity graph ofa particle moving aloog a horizoota1line, discussed in Example 2. HIsTORICAL BIOGRAPHY
Bernard Bolzano (17811848)
The rate at which a body's velocity changes is the body's acceleration. The acceleration measures how quickly the body picks up or loses speed. A sudden change in acceleration is called a jerk. When a ride in a car or a bus is jerky, it is not that the accelerations involved are necessarily large but that the changes in acceleration are abrupt.
DEFINmONS Acceleration is the derivative of velocity with respect to time. Ifa body's position at time 1 is s = /(1), then the body's acceleration at time 1 is a(l) =
~~ = ~:~.
Jerk is the derivative of acceleration with respect to time:
j(l) =
~~ = ~:~.
Near the surface of the Earth all bodies fall with the same constant acceleration. Galileo's experiments with free fall (see Section 2.1) lead to the equation
s = 19t2 2
'
where s is the distance fallen and g is the acceleration due to Earth's gravity. This equation holds in a vacuum, where there is no air resistance, and closely models the fall of dense, heavy objects, such as rocks or steel tools, for the first few seconds of their fall, before the effects of air resistance are sigoificant. The value of g in the equation s = (1/2)gI2 depends on the units used to measure 1and s. With 1in seconds (the usual unit), the value ofg determined by measurement at sea level is approximately 32 ft/sec 2 (feet per second squared) in English units, and g = 9.8 m/sec2
128
Chapter 3: Differentiation
t~O
t
=1
(meters per second squared) in metric units. (These gravitational constants depend on the distance from Earth's center of mass, and are slightly lower on top ofM!. Everest, for example.) The jerk associated with the constant acceleration of gravity (g = 32 ft/sec 2) is zero:
s (meters)
t (seconds) •
o
, ,
5
''
10 15
j =
20
30
EXAMPLE 3 Figure 3.15 shows the free fall of a heavy ball bearing released from rest at time I = Osee.
35
40
=3
~~:
(g) = O.
An object does not exhibit jerkiness during free fall.
25
t
!
(a) How many meters does the ball fall in the first 2 sec?
45
(b) What is its velocity, speed, and acceleration when t = 2?
FIGURE 3.15 A ball bearing falling from rest (Example 3).
Solution (a) The metric freefall equation is s = 4.91 2. During the first 2 sec, the ball falls s(2) = 4.9(2)2 = 19.6 m. (b) At any time I, velocity is the derivative of position:
s
v(l) = s'(I) =
!
(4.91 2) = 9.81.
At I = 2, the velocity is v(2) = 19.6 m/sec
in the downward (increasing s) direetion. The speed at I = 2 is speed = Iv(2)1 = 19.6m/sec. The acceleration at any time I is
a(l) = v'(I) = s"(I) = 9.8 m/sec2.
t
All = 2, the acceleration is 9.8 m/ sec2.
s~O s.m~
•
EXAMPLE 4
A dynamite blast blows a heavy rock straight up with a launch velocity of 160 ft/sec (about 109 mph) (Figure 3.16a). It reaches a height of s = 1601  161 2 ft after t sec.
(a)
s,V
s ~ 160t  16,z
(a) How bigh does the rock go? (b) What are the velocity and speed of the rock when it is 256 ft above the ground on the way up? On the way down?
(c) What is the acceleration of the rock at any time t during its flight (after the blast)? (d) When does the rock bit the ground again? 160
Solution (b)
FIGURE 3.16 (a) The rock in Example 4. (b) The graphs of s and v as functions of time; s is largest when v = ds/dt ~ O. The graph of s is not the path of the rock: It is a plot of height versus time. The slope of the plot is the rock's velocity, graphed here as a straight line.
(a) In the coordinate system we have chosen, s measures height from the ground up, so the velocity is positive on the way up and negative on the way down. The instant the rock is at its highest point is the one instant during the flight when the velocity is O. To find the maximum height, all we need to do is to find when v = 0 and evaluate s at this time. At any time I during the rock's motion, its velocity is V
tis d 2 = dt = dl (1601  161 ) = 160  321ft/sec.
3.4 The Derivative as a Rate of Change
129
The velocity is zero when
160  32t
=
0
or
t = 5 sec.
The rock's height at t = 5 sec is
s.... = s(5) = 160(5)  16(5)' = 800  400 = 400 ft. See Figure 3.16h.
(b) To find the rock's velocity at 256 ft on the way up and again on the way down, we rust rmd the two values of t for which s(t) = 160t  16t 2 = 256. To solve this equation, we write
16t2 16(t
+ 256 lOt + 16)
160t
2

= =
(t  2)(t  8) =
0 0 0
t = 2sec,t = 8 sec. The rock is 256 ft above the ground 2 sec after the explosion and again 8 sec after the explosion. The rock's velocities at these times are
v(2) = 160  32(2) = 160  64 = 96 ft/sec.
v(8) = 160  32(8) = 160  256 = 96 ft/sec. At both instants, the rock's speed is 96 ft/sec. Since v(2) > 0, the rock is moving upward (s is increasing) at t = 2 sec; it is moving downward (s is decreasing) at t = 8 because v(8) < O. (c) At any time during its flight following the explosion, the rock's acceleration is a constant
a _dv_d( dt  dt 160  32t)_  32 ft/2 sec. The acceleration is always downward. As the rock rises, it slows down; as it falls, it speeds up. (d) The rock hits the ground at the positive time t for which s = O. The equation 160t  16t 2 = 0 factors to give 16t(1O  t) = 0, so it has solutions t = 0 and t = 10. At t = 0, the blast occurred and the rock was thrown upward It retamed to the ground 10 sec later. _
Derivatives in Economics Engineers use the terms velocity and acceleration to refer to the derivatives of functions describing motion. Economists, too, have a specialized vocabulary for rates of change and derivatives. They call them marginals. In a manufacturing operation, the cost ofproduction c(x) is a function of x, the number of units produced. The marginal cost of production is the rate of change of cost with respect to level of production, so it is dc/dx. Suppose that c(x) represents the dollars needed to produce x tons of steel in one week. It costs more to produce x + h tons per week, and the cost difference, divided by h, is the average cost of producing each additional ton:
c(x
+ h)  c(x) h
average cost of each of the additional h tons of steel produced.
130
Chapter 3: Differentiation The limit of this ratio as h + 0 is the marginal cost of producing more steel per week when the current weekly production is x tons (Figure 3.17):
Cost y (dollars)
dc
ax ~~~~~x
o
x x+h Production (tons/week)
.
= hm
c(x
+ h)
h
h~O
 c(x). . = margmal cost of production.
Sometimes the marginal cost of production is loosely defined to be the extra cost of producing one additional unit: ac
FIGURE 3.17 Weekly steel production: c(x) is the cost of producing x tons per week. TIre cost of producing an additional h tons is c(x + h)  c(x). y
c(x
+
!>.X
I)  c(x) 1
which is approximated by the value of dc/ax at x. This approximation is acceptable if the slope of the graph of c does not change quickly near x. Then the difference quotient will be close to its limit dc/ax, which is the rise in the tangent line if!>.x = I (Figure 3.18). The approximation works best for large values of x. Economists often represent a total cost function by a cubic polynomial
c(x) = ax 3 + f3x 2
+ 1X + {J
where {J represents ['!Xed costs such as rent, heat, equipment capitalization, and management costs. The other terms represent variable costs such as the costs of raw materials, taxes, and labor. Fixed costs are independent of the number of units produced, whereas variable costs depend on the quantity produced. A cubic polynomial is usually adequate to capture the cost behavior on a realistic quantity interval.
EXAMPLE 5
Suppose that it costs
c(x) = x 3

6x 2
+
15x
~~L~~~x
o
x
x+ 1
FIGURE 3.18 Tlremarginalcostdc/dds approximately the extra cost ac of producing = I more unit.
ax
dollars to produce x radiators when 8 to 30 radiators are produced and that
1'(x) = x 3  3x 2
+
12x
gives the dollar revenue from selling x radiators. Your shop currently produces 10 radiators a day. About how much extra will it cost to produce one more radiator a day, and what is your estimated increase in revenue for selling II radiators a day?
Solution c'(IO):
The cost of producing one more radiator a day when 10 are produced is about
c'(x) = c'(IO)
:fx (x3 
6x 2 + 15x) = 3x 2

12x
+
IS
= 3(100)  12(10) + IS = 195.
The additional cost will be about $195. The marginal revenue is
r'(x) =
:fx (x 3 
3x 2 + 12x) = 3x 2  6x
+
12.
The marginal revenue function estimates the increase in revenue that will result from selling one additional unit. If you currently sell 10 radiators a day, you can expect your revenue to increase by about
r'(IO)
= 3(100)  6(10)
if you increase sales to II radiators a day.
+
12 = $252
•
3.4 The Derivative as a Rate of Change
131
EXAM PLE 6 To get some feel for the iangoage of marginal rates, consider marginal tax mtes. If your marginal income tax mte is 28% and your income increases by $1000, you can expect to pay an extra $280 in taxes. This does not mean that you pay 28% of your entire income in taxes. It just means that at your current income level I, the rate of increase of taxes Twith respect to income is dT/dI = 0.28. You will pay $0.28 in taxes out of every extra dollar you earn. Of course, if you earn a lot more, you may land in a higher tax bracket and your marginal mte will increase. _
Sensitivity to Change When a small change in x produces a large change in the value of a function f(x), we say that the function is relatively sensitive to changes in x. The derivative f' (x) is a measure of this sensitivity.
EXAMPLE 7
Genetic Data and Sensitivity to Change
The Austrian monk Gregor Johann Mendel (18221884), working with garden peas and other piants, provided the first scientific explanation ofbybri/dtwhen 8
d. f(g(x)),
r.
0
(x 11
x
~
0
+ f(x))2,
X
with A and b positive. The value of A is the amplitude of the motion, and b is the frequency (number of times the piston moves up and down each second). What efree! does doubliog the frequency have on the piston's velocity, acceleration, and jerk? (Once you fmd out, you will know why some machinery breaks when you ruo it too fast.)
~ I
x ~ 0
y
~ 3'1f/2 ifs ~ cos8andd8/dt ~ 5.
76. Find eIy/dt when x ~ I ify ~ x 2
+ 7x
 5andd>/dt ~ 1/3.
What happens if you can write a :function as a composite in different
77. Find dy/dx ify ~ x hy using the Chain Rule withy as a composite of
b. y ~ I
+7
and
u ~ 5x  35
+ (I/u)
and
u ~ I/(x  I).
y ~ u'
b. y~
Vu
and and
u~x'.
«x 
80. Find the tangeot to y ~
y'x 2
L
I)/(x 
X
+
1))2atx ~ O.
+ 7 at x ~
2.
Find the tangent to the curve y ~ 2 tan (=/4) at x ~ I.
b. Slopes on a tangent curve What is the smallest value the slope of the curve can ever have on the interval 2 < x < 2? Give reasons for your answer.
82. Slopes on sine curves L
25
ing when it is increasing at its fastest? y
60 H+++b.>"'"""'~ d+++++H fi:'
1
' 40 H
~
0

1 . . . .'
Find equations for the tangents to the curves y ~ sin 2x and ~ sin(x/2) at the origin. Is there anything special about how the tangents are related? Give reasons for your answer.
y
b. Can anything be said about the tangents to the curves y ~ sinmxandy ~ sin(x/m) at the origin (m a coostant 0# O)? Give reasons for your answer. c. For a giveo m, what are the largest values the slopes of the curvesy ~ sinmxandy ~ sin(x/m) can ever have? Give
reasons for your answer. d. The function y ~ sin x completes one period on the interval [0, 2'1f], the function y ~ sin 2x coropletes two periods, the functiony ~ sin(x/2) completes half a period, and so on. Is there any relation between the number of periods y ~ sin mx coropletes on [0, 2'1f] and the slope of the curve y ~ sinmx at the origin? Give reasons for your answer.
_
Hf71++++'"~+++++1 
   '~   ~
v
~
u ~ '\IX
79. Findthetangeottoy ~
81.
1+
and is graphed in the accompanying figure.
20
78. Find eIy/dx if y ~ x'/2 hy using the Chain Rule with y as a composite of L
101)
.. On what day is the temperature increasing the fastest?
ways? Do you get the same dctivative each time? The Chain Rule says you should. Try it with the functions in Exercises 77 and 78.
y ~ (u/5)
~ 37 sin [;;; (x 
b. About how many degrees per day is the temperature increas
Theory and Examples
L
A cos (2'1fbt) ,
84. Temperatures in Fairbanks, Alaska The graph in the accompanying figure shows the average Fahrenheit temperature in Fairbaoks, Alaaks, doring a typical 365day year. The equation that approsirnates the temperature on day x is
x ~ 0
b. f(x)g'(x),
L
~
85. Particle motion ordinate line is s =



 1'\   ~ I>x
V+
position of a particle moving along a co
1 41, with s in meters and t in seconds. Find the particle~ velocity and acceleration at t ~ 6 sec.
86. Constant acceleration Suppose that the velocity of a fa1liog body is v ~ kYs m/sec (k a constant) at the instant the body has fallen s m from its starliog point. Show that the body's accelera
tion is constant. 87. Falling meteorite The velocity of a heavy meteorite entering Earth's atmosphere is inversely proportional to Ys when it is s kIn from Earth's center. Show that the meteorite's acceleration is inversely proportional to 8 2 . 88. Particle acceleration A particle moves along the xaxis with velocity dx/dt ~ f(x). Show that the particle's acceleration is
f(x)/'(x). 89. Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period T and the length L of a simple pendulum with the equation
T
~ 2'1f~'
where g is the constant acceleration of gravity at the pendulum's
location. If we measure g in centimeters per second squared, we measure L in centimeters and T in seconds. If the pendulum is
3.7
made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to L. In symbols, with u being temperature and k the proportionality constant,
Implicit Differentiation
•• Graph dg/dt (where defmed) over ["" ".]. b. Findd//dt. c. Graph d// dt. Where does the approxinlation of dg/ dt by d//dt seem to be best? Least good? Approximations by 1rigonome1ric polynomia1s are important in the theories of
tiL du = kL.
heat and oscillation, but we must not expect too much of them, as we see in the next exercise.
Assuming Ibis to be the case, show that the rate at which the period changes with respect to temperature is kT/2. 90. Chain Rule
,
,
Ixl. Then the
Suppose that /(x) = x' and g(x) =
~
0
g)(x) =
lxi' =
x'
are both differentiable at x
and
(g
0
/)(x) =
Ix'i =
/"
x'
D 91.
The derivative of sin 2x Graph the function y = 2 cos 2x for 2 '" x s 3.5. Then, on the same screen, graph
y=
sin 2(x
+ h) 
sin 2x
h
for h = 1.0, 0.5, and 0.2. Expctiment with other values of h, including negative values. What do you see happening as h > O? Explain Ibis behavior.
D 92.
The derivative of cos (x') Graph y = 2x sin (x') 2 '" x s 3. Then, on the same screen, graph cos «x
y=
o
"
= 0 even though g itself is not differ
entiable at x = O. Does Ibis contradict the Chain Rule? Explain.
g(t)
'~f(t)
/
composites (j
149
"
94. (Continuation of Exercise 93.) In Exercise 93, the 1rigonome1ric polynomial /(t) that approximated the sawtooth function g(t) on [".,,,.] had a derivative that approximated the derivative of the sawtooth function. It is possible, however, for a trigonometric polyoomial to approximate a function in a reasonable way without its derivative approxinlating the function's derivative at all well. As a case in point, the "polynomial"
s = h(t) = 1.2732 sin 2t
+
0.4244 sin 6t
+ 0.25465 sin lOt
+ 0.18189 sin 14t + 0.14147 sin 18t
for
graphed in the accompanying figure approximates the step function s = k(t) shown there. Yet the derivative of h is nothing like the derivative of k.
+ h)')  oos (x') h
,
for h = 1.0,0.7, and 0.3. Expctiment with other values of h. What do you see happening as h > O? Explain this behavior.
s
COMPUTER EXPLORATIONS
I
Trigonometric Polynomials
"
93. As the accompanying figure shows, the 1rigonometric "polynomial"
k(t)
I. ~ /
1
2"
~
"2
0
h(t)
"
1
s = /(t) = 0.78540  0.63662 cos 21  0.07074cos6t  0.02546 cos lOt  0.01299 cos 14t
•• Graph dk/dt (where defmed) over ['IT, 'IT].
gives a good approximation of the sawtooth function s = g(t) on the interval ["" ".]. How well does the derivative of / approximate the derivative of g at the points where dg/ dt is defmed? To find out, carry out the following steps.
3.7
b. Find dh/ dt. c. Graph dh/ dt to see how badly the graph fits the graph of dk/dt. Comment on what you see.
Implicit Differentiation Most of the functions we have dealt with so far have been described by an equation of the form y = f(x) that expressesy explicitly in tenDs of the variable x. We have learned rules for differentiating functions defined in this way. Another situation occurs when we encounter equations like
x 3 + y3  9xy
= 0,
y'  x = 0,
or
x' + y'  25
= O.
150
Chapter 3: Differentiation (See Figures 3.26, 3.27, and 3.28.) These equations defme an implicit relation between the variables x andy. In some cases we may be able to solve such an equation for y as an explicit function (or even several functions) of x. When we cannot put an equation F(x, y) = 0 in the fonn y = f(x) to differentiate it in the usual way, we may still be able to fmd dyj ax by implicit differentiation. This section describes the tecbnique.
Y
Implicitly Defined Functions We begin with examples involving familiar equations that we can solve for y as a function ofx to calculate dyjax in the usual way. Then we differentiate the equations inlplicitly, and find the derivative to compare the two methods. Following the examples, we summarize the steps involved in the new method. In the examples and exercises, it is always assumed that the given equation determines y inJplicitly as a differentiable function of x so that dyj ax exists.
EXAMPLE 1 FIGURE 3.26 The curve + y'  9xy ~ 0 is not the graph of
x'
anyone function of x. The curve can,
however, be divided into separate arcs that are the graphs of functions ofx. This particular curve, called afolium, dates to Descartes in 1638.
Find dyjax ify2 = x.
The equation y2 = x defmes two differentiable functions of x that we can actoally find, namely Yl = VX and Y2 = VX (Figure 3.27). We know how to calculate the derivative of each of these for x > 0:
Solution
dYI ax
I 2VX
dY2 ax
and
1 2VX'
But suppose that we knew only that the equation y2 = x defmed y as one or more differentiable functions of x for x > 0 without knowing exactly what these functions were. Could we still find dy j ax? The answer is yes. To fmd dyj ax, we simply differentiate both sides of the equation y2 = x with respect to x, treating y = f(x) as a differentiable function of x:
y2
I I I I
= 1
2Y2
x
The Chain Rule gives
ax
Yx
dy ax
1
=
2Yx
FIGURE 3.27 The equationy2  x ~ 0, or y2 = X as it is usually written, dermes two differentiable functions of x on the interval x > O. Exarople 1 shows how to find the derivatives of these functions without solving the equation y2 = x for y.
! (y2)
=
d 2 dy dx [f(x)] ~ 2f(x)j'(x) ~ 2y dx'
2y dy = I
Q(x, \IX)
Y2~
/
Slope
=
I 2y'
This one fonnula gives the derivatives we calculated for both explicit solutions Yl = VX andY2 = VX:
dYI ax
EXAMPLE 2
I 2Yl
I 2VX
and
d)'2 ax
I 2Y2
I
2( vx)
I 2VX'
•
Find the slope of the circle x 2 + y2 = 25 at the point (3, 4).
The circle is not the graph of a single function of x. Rather it is the combined graphs of two differentiable functions, Yl = V25  x 2 and Y2 =  V25  x 2 (Figure 3.28). The point (3, 4) lies on the graph of)'2, so we can fmd the slope by calculating the derivative directly, using the Power Chain Rule:
Solution
dY21
ax x~3 =

I
2x 2V25  x 2 x~3
6 2v'25=9
3 4'
3.7 Implicit Differentiation y
We can solve this problem more easily by differentiating 1he given equation of the circle implicitly with respect to x:
A.. (x2) + A.. (y2) dx
dx
=
dy dx Y2~V25X2
The slope at (3, 4) is.x.1
Y
(3, 4).
A.. (25) dx
dy 2x+ 2y dx=O
_5~~Or~5~~x
FIGURE 3.28 The circle combines 1he graphs of two functions. The graph of Y2 is 1he lower semicircle and passes through
151
= (3, ....)
x
y.
~ = 1. 4 4
Notice that unlike the slope formula for dy2/ dx, which applies ouly to points below the xaxis, the formula dy/dx = x/yapplies everywhere the circle has a slope. Notice also that the derivative involves both variables x and y, not just the independent variable x. • To calculate the derivatives of other implicitly defmed functions, we proceed as in Examples 1 and 2: We treat y as a differentiable implicit function of x and apply the usual rules to differentiate both sides of the defming equation.
Implicit Differentiation 1. Differentiate both sides of the equation with respect to x, treating y as a differentiable function of x. 2. Collect the terms with dy/ dx on one side of the equation and solve for dy/ dx.
Y
EXAMPLE 3 Solution
Find dy/dx if y2 = x 2
(Figure 3.29).
We differentiate the equation implicitly. y2 =
x 2 + sinxy
:. (y2) = : . (x2)
FIGURE 3.29 The graph of y2 ~ x 2 + sin'9' in Example 3.
+ sinxy
+ :. (sinxy)
2y dy dx = 2x
+
d( ) (cosxy) dx xy
2y: = 2x
+
(cosxy)&
2Y:  (cosxy) (x : ) = 2x
+
(cosxy)y
dy (2y  xcosxy) dx = 2x dy dx
. .. treating Y as a function of and using the Chain Rule.
x
Treat xy as a product
Collect terms wilh dy/dx.
+ ycosxy
+ Y cosxy 2y  x cosxy
2x
+ x :)
Differentiate both sides with respect to x ...
Solve for dy/dx.
Notice that the formula for dy/dx applies everywhere that the implicitly defined curve has a slope. Notice again that the derivative involves both variables x andy, not just the independent variable x. •
152
Chapter 3: Differentiation
Derivatives of Higher Order Implicit differentiation can alao be used to find bigher derivatives.
EXAMPLE 4
Find d'y/dx 2 if 2x'  3y2 = 8.
To start, we differentiate both sides of the equation with respect to x in order to findy' = dy/dx.
Solution
JI (2x' dx
 3y2)
=
JI (8)
6Y.Y'
=
0
fu;2 
dx
x
Treaty as a function ob. 2
y' =y,
when y # 0
Solve for y'.
We now apply the Quotient Rule to f'md y' .
y"
=:.. (~) =2xy / 2y ' =~ _;~.y'
Finally, we substitute y' = x 2/y to express y" in terms ofx andy.
2 _ 2x x (X2) _ 2x x' y" Y y2 Y Y y"
Curve of lens smface
•
wheny # 0
Lenses, Tangents, and Normal Lines In the law that describes how light changes direction as it enters a lens, the important angles are the angles the light makes with the line perpendicular tu the surface of the lens at the point of entry (angles A and B in Figure 3.30). This line is called the normal tu the surface at the point of entry. In a profile view of a lens like the one in Figure 3.30, the normal is the line perpendicular to the tangent of the profile curve at the point of entry. FIGURE 3.30 TIre profile ofa leos, showing the beading (refraction) of a ray oflight as it passes through the leos surface.
y
EXAMPLE 5 Show that the point (2, 4) lies on the curve x' find the tangent and normal tu the curve there (Figure 3.31).
+ y'  9xy
= O. Then
Solution
The point (2, 4) lies on the curve because its coordinates satisfy the equation given for the curve: 2' + 4'  9(2)(4) = 8 + 64  72 = O. To find the slope of the curve at (2, 4), we first use implicit differentiation to f'md a fonnula for dy/ dx:
x'+y'9xy=0
JI (x') + JI (y') dx
dx

JI (9xy)
=
JI (0)
Y
+ Y dx) dx
=
0
Differentiate both sides with respect to x.
+ 3x 2  9y
=
0
TreatX)' as a product andy as a function of x.
dy  9 (xddx 3x2 + 3y2 dx dy (3y2  9x) dx
dx
dx
3(y2  3X): = 9y  3x 2 FIGURE 3.31 Example 5 shows how to fmd equations for the taogeot and normal to the folium of Descartes at (2, 4).
dy dx
3yx 2 y2  3x'
Solve for dy/d
X
(3, 4)
xV ~ 9,
(1,3) 32. y2  2x  4y  I ~ 0,
(2, I)
33. 6x 2 + 3xy + 2y2 + 17y  6 ~ 0, (1,0) 34. x 2  V3xy + 2y2 ~ 5, ( V3, 2 ) 35. 2xy
+ 'If siny
~ 2'1f,
36. xsin2y ~ ycos2x, 37. y
(I, 'If/2) ('If/4,'If/2)
~
2 sin ('IfX  y), (1,0) 2 38. x cos2y  siny ~ 0, (0,'If)
39. Parallel tangents Find the two points where the curve x 2 + xy + y2 ~ 7 crosses the xaxis, and show that the tangents to the curve at tlrese points are parallel. What is the common slope of these tangents? 40. Normals parallel to a lin. Find the normals to the curve xy + 2x  y ~ 0 that are parallel to the line 2x + y ~ O. 41. The eight curve Find the slopes of the curve y' ~ y2  x 2 at
(See Figure 3.26.) a. Find the slope of the folium of Descartes x'
44. The folium of Deseartes
c. Find the coordinates of the point A in Figure 3.26, where the folium has a vertical tsngent.
Theory and Examples 45. Intersecting normal The line that is normal to tire curve x 2 + 2xy  3y2 ~ 0 at (I, I) intersects the curve at what other point? 46. Power rule for rational ""ponents Let p aod q be integers with q > O. If y ~ xPi., differentiate the equivalent equation y' ~ xP implicitly aod show that, for y #' 0,
~ "'i.)1 . x
+ y'
at the points (4, 2) and (2, 4).
3.8
Related Rates
155
x' D 10 Exercises 51 aod 52, fmd both dy/dx (treatingy as a differentiable
48. Is there anything special about the tangents to the curves y' ~ and lx' + 3y' ~ 5 at the points (I, ± I)? Give reasons for your
answer. y
function of x) and dx/dy (treating x as a differentiable function ofy). How do dy/dx aod dx/dy seern to be related? Explain the relationship geometrically in terms of the graphs. 51. xy' +
x'y
~ 6
52. x' + y' ~ sin'y
COMPUTER EXPLORATIONS Use a CAS to perform the following steps in Exercises
5~.
a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point P satisfies the equation.
b. Using implicit differentiatioo, find a formula for the derivative dy/dx aod evaluate it at tlte given pnint P.
49. Verify that the following pairs of curves meet orthogoualiy.
a.
x' + y' ~ 4, x' ~ 3y'
b. x
3
50. The graph of y' ~ is called a semicubica! parabola aod is shown in the accoropaoying figure. Determine the coostant b so x + b meets this graph orthogoually. that the line y ~
l
53.
x' 
xy
+ y'
~ 7,
55. y
,
_2+x
+ Y  I _ x' P(O, I)
56. y' + cos xy ~ 57.
x',
59.
60.
P(1,O)
x+tan(~) ~2, P(I,;r) ~ I, 2y' + (xy)l/' ~ x' + 2,
58. xy' + tan (x + y)
3.8
P(2, I)
54. x' + y'x + yx' + y' ~ 4, P(I, I)
~ I  y', x ~ ly'
x'
c. Use tlte slope found in part (b) to fmd an equation for tlte tangent line to tlte curve at P. Then plot tlte implicit curve and tangent line togetlter no a single graph.
p(;r, 0) P(1, 1)
xv'1+2.Y + y ~ x', P(I,O)
Related Rates In this section we look at problems that ask for the mte at which some variable changes when it is known how the mte of some other related variable (or perhaps several variables) changes. The problem of finding a mte of change from other known rates of change is called a related rates problem.
Related Rates Equations Suppose we are pumping air into a spherical balloon. Both the volume and radius of the balloon are increasing over time. If V is the volume and r is the radius of the balloon at an instant of time, then
156
Chapter 3: Differentiation
Using the Chain Rule, we differentiate both sides with respect to t to rmd an equatiDJI relating the rates of change of V and r,
So if we know the radius r of the balloon and the rate dV/dt at which the volume is increasing at a given instant of time, then we can solve this last equation for dr/dt to fmd how fast the radius is increasing at that instant. Note that it is easier to directly measure the rate of increase of the volume (the rate at which air is being pumped into the balloon) than it is to measure the increase in the radius. The related rates equation allows us to calculate dr/d, from dV/dt. Very often the key to relating the variables in a related rates problem is drawing a picture that shows the geometric relations between them. as illustrated in the following example.
EXAMPLE 1 Water runs into a conical tank at the rate of 9 W/min. The tank stands point down and has a height of lOft and a base radius of 5 ft. How fast is the water level rising when the water is 6 ft deep? Solution
Figure 3.32 shows a partially filled conical tank. The variables in the problem are V = volume (W) of the water in the tank at time t (min)
x  radius (ft) of the surface of the water at time t y = depth (ft) of the water in the tank at time t. We assume that V, x, and y are differentiable functions of t. The constants arc the dimensions of the tank. We arc asked for dy/dt when FIGURE 3.32 The geometry of the conical tank and the rate at which water rills the tank dctmnine how fast the water level rises (Example 1).
y
~
6 ft
and
" d'v + ~~k';!~ d:x;' k d:x;k d'v
be differentiable for all values of x?
+"'+u dx '"
b. Discuss the geometry of the resulting graph of g.
19. Odd differentiable function. Is there aoytIting special about the derivative of an odd differentiable function of x1 Give reasoos for your answer.
The equations in parts (a) and (b) are special cases of the equation in part (c). Derive the equation in part (c) by mathematical induction, using
20. Even differentiable function.
m! m! (m)k + (m) \k + I ~ k!(m  k)! + (k + I)!(m  k 
Is there aoytIting special about the derivative of an eveo differentiable function of x1 Give reasons for your answer.
f and g are defmed throughout an open interval containing the point xo, that f is differentiable at:x:o, that f(xo) ~ 0, and that g is continuous at Xo. Show that the product fg is differentiable at Xo. This process shows, for example, that although Ixl is not differentiable at x ~ 0, the product x Ixl" differentiable at x ~ O.
21. Suppose that the functions
22. (Continuation of Exerc"e 21.) Use the result of Exercise 21 to show that the followiog functions are differentiable at x ~ O. L
Ixl sin x
d. h(x)
b. x'/' sinx
c."\Yx(l  cosx)
~ {x' sin (I/x), x
0# 0 x= 0
0,
23. Is the derivative of
h(x) ~ {x' sin (I/x), x 0# 0 0, x= 0 continuous atx ~ 01 How about the derivative ofk(x) Give reasons for your answers.
~
:x:h(x)1
24. Suppose that a function f satisfies the followiog cooditioos for all real values ofx andy:
i) f(x
+ y)
ii) f(x) ~ I
~
f(x)·f(y).
+ xg(x) , where linJ,....og(x)
~
I.
Show that the derivative I'(x) exists at every value oh and that I'(x) ~ f(x). 25. The generalized product rule Use mathematical induction to prove that if y = "1"2'" u" is a finite product of differentiable functions, then y is differentiable on their common domain and
du,
dy
ax
=
du,
dun
liX U2 "'u" + Ul(jX"'UII + ... + u t u2 "'ulI  l liX'
I)!'
27. The period of a dock pendulum
The period T of a clock pendulum (time for one full swing and back) is given by the formula T' ~ ~L/g, where Tis measured in seconds, g ~ 32.2 ft/sec', and L, the length of the pendulum, is measured in feet. Find approximately a. the length of a clock pendulum whose period is T ~ I sec.
b. the change dTin Tifthependulum in part (a) is lengthened
0.01 ft.
c. the amount the clock gains or loses in a day as a result of the period's changing by the amount dT found in part (b). 28. The melting ice cube Assume that an ice cube retains its cubical shape as it melts. Ifw. call its edge length s, its volume is V = 3 3 and its surface area is 63 2 . We assume that V and s are differentiable functions of time t. We assume also that the cube's volume decreases at a rate that is proportional to its surface area. (This latter assumption seems reasonable enough when we think that the melting takes place at the surface: Changing the amount of surface changes the amount of ice exposed to melt.) In mathematical terms, dV ~ k(6s') dt '
k> O.
The minus sign indicates that the volume is decreasing. We assume that the proportiona1ity factor k is coostant. (It probably depends on many things, such as the relative humidity of the surrounding air, the air temperatore, and the incidence or absence of sunlight, to nanre only a fow.) Assume a particular set of conditions in which the cube lost 1/4 of its volume during the first hour, and that the volume is Vo when t ~ O. How long will it take the ice cube to melt?
Chapter 3 Technology Application Projects
Chapter ID
183
Technology Application Projects
Mathematica,tMap1e Modules: Convergence ofSecant Slopes to the Derivative Function
You will visualize the secant line between successive points on a curve and observe what happens as the distance between them becomes small. The function, sample points, and secant lines are plotred on a single graph, while a second graph compares the slopes of the secant lines with the derivative function. DeriVlllives, Slopes, Tangent Lines, and Making Movies Part. Iill. You will visualize the derivative at a point, the linearization of a function, and the derivative of a function. You learn how to plot the function and selected tangeots on the same graph. Part IV (plotting Many Tangent.) Part V (Making Movies). Parts N and V of the module can he used to aoimate tangeot lines as one moves along the graph ofa function. Convergence ofSecant Slopes to the Derivative Function
You will visualize righthand and lefthand derivatives. MotWnAlollg tJ Stmight Line: Position .... Velocity .... AccelertJdDn
Observe dramatic animated visualizations of the derivative relations among the position, velocity, and acceleration functions. Figures in the text can he animated.
4 ApPLICATIONS OF DERIVATIVES OVERVIEW In this chapter we use derivatives to find extreme values of functions, to determine and analyze the shapes of graphs, and to find numerically where a function equals zero. We also introduce the idea of recovering a function from its derivative. The key to many of these applications is the Mean Value Theorem, which paves the way to integral calculus in Chapter 5.
Extreme Values of Functions
4.1
This section shows how to locate and identify extreme (maximum or minimum) values of a function from its derivative. Once we can do this, we can solve a variety of problems in which we find the optimal (best) way to do something in a given situation (see Section 4.5). Finding maximum and minimum values is one of the most important applications of the derivative.
DEFINITIONS Let f be a function with domain D. Then f has an absolute maximum value on D at a point c if
f(x)
:5
f(c)
for all x inD
and an absolute minimum value on D at c if y
f(x) y
=
f(c)
for all x in D .
sin x
++~+x
FIGURE 4.1 Absolute extrema for the sine and cosine functions on [7T/2, 7T/2]. These values can depend on the domain of a function.
184
2::
Maximum and minimum values are called extreme values of the function f. Absolute maxima or minima are also referred to as global maxima or minima. For example, on the closed interval [7T/2, 7T/2] the function f(x) = cos x takes on an absolute maximum value of 1 (once) and an absolute minimum value of 0 (twice). On the same interval, the function g(x) = sinx takes on a maximum value of 1 and a minimum value of 1 (Figure 4.1). Functions with the same defining rule or formula can have different extrema (maximum or minimum values), depending on the domain. We see this in the following example.
4.1
Extreme Values of Functions
185
EXAMPLE 1 The absolute extrema ofthe following functions on their domains can be seen in Figure 4.2. Notice that a function might not have a maxinnnn or minimmn if the domain is unbounded or fails to contain an endpoint
Function rule
DomainD
Absolute extrema on D
(a) y = x 2
(00,00)
No absolute maximmn. Absolute minimmn of 0 at x =
[0,2]
Absolute maximmn of 4 at x = 2. Absolute minimmn of 0 at x = o.
(c) y = x 2
(0,2]
Absolute maximmn of 4 at x = 2. No absolute minimum.
(d) y = x 2
(0,2)
No absolute extrema.
 "......:cx 2 (a) abs min only
FIGURE 4.2
HISWRICAL BIOGRAPHY
Daniel Bernoulli (170()1789)
o.
(b) y = x 2
 "......:c x 2
x  " 0, and a local maximum at the endpoint x = b if f(x) :5 f(b) for all x in some half·open interval (b  Ii, b J, Ii > O. The inequalities are reversed for local minimum values. In Figure 4.5, the function f has local maxima at c and d and local min· ima at a, e, and b. Local extrema are also called relative extrema. Some functions can have inrmitely many local extrema, even over a finite interval. One example is the func· tion f(x) = sin (l/x) on the interval (0, IJ. (We graphed this function in Figure 2.40.)
x
4.1
Extreme Values of Functions
187
Absolute maximum.
Local minimum No smaller value : off nearby. Absolute minimum. No smaller value of f anywhere. AlBo a I locaIminimum. : _ _ _ _ _ _ _ _ _ _ _ _ _ _ _LI
Local minimum. I No smaller value of :fnearby. : I
_ _ _ _ _ _~_ _ _ _ _ _LI
a
: : I
:
"; ' ~':_>x
e
d
b
FIGURE 4.5 How to ideotifY types of maxima and minima for a functioo with domain a ~ x ~ b.
An absolute maximum is also a local maximum. Being the largest value overall, it is also the largest value in its immediate neighborhood. Hence, a list ofall local maxima will automatically include the absolute maximum if there is one. Similarly, a list of all local minima will include the absolute minimum if there is one.
Finding Extrema Local maximum value
The next theorem explains why we usually need to investigate only a few values to find a function's extrema.
If f has a local maximum or minimum value at an interior point c of its domain, and if f' is defined at c, then
THEOREM 2The First Derivative Theorem for Local Extreme values
f'{c) = O. Secant slopes
2:
(never ,negative)
0
Secant slopes ::s 0 (never po~itive)
I I I I
~I~~~>,X
X
X
FIGURE 4.6 A curve with a local maximum value. The slope at c, simultaneously the limit of noopositive numbers and nonnegative numbers, is zero.
Proof To prove that f' (c) is zero at a local extremum, we show first that f' (c) cannot be positive and second that f' (c) cannot be negative. The only number that is neither positive nor negative is zero, so that is what f' (c) must be. To begin, suppose that f has a local maximum value at x = c (Figure 4.6) so that f{x)  f{c) ,;; 0 for all values of x near enough to c. Since c is an interior point of f's domain, f' (c) is derfied by the (w(}.sided limit
r
x~
f{x)  f{c) xc·
This means that the righthand and lefthand limits both exist at x = c and equal f'{c). When we examine these limits separately, we find that
f '{ C )
=
f '{ C )
=
l'
un
xc+
f{x)  f{c) xc
0
'
Because (x  c) > 0 and !(x) '" !(c)
(1)
Similarly,
l'
11D_
xc
f{x)  f{c) X
C
.
Because (x  c) < 0 and !(x) " !(c)
(2)
Together, Equations (1) and (2) imply f'{c) = O. This proves the theorem for local maximum values. To prove it for local minimum values, we simply use f{x) 2: f{c) , which reverses the inequalities in Equations (1) and (2). •
188
Chapter 4: Applications of Derivatives Theorem 2 says that a function's first derivative is always zero at an interior point where the function has a local extreme value and the derivative is defined. Hence the only places where a function 1 can possibly have an extreme value (local or global) are
1. 2.
interior points where f' = 0,
3.
endpoints of the domain of I.
interior points where f' is undefined,
The following definition helps us to summarize. (a)
y
~~~>x
(b)
FIGURE 4.7 Critical points wi1hout extrem.evalues. (a)y' = 3x 2 isOatx = 0, but y = x 3 has no extremum there. (b) y' = (1/3)x2/3 is undef"med atx = 0, but y = X 1/3 bas no extremum 1here.
DEFINITION An interior point of the domain of a function 1 where f' is zero or undef"med is a critical point of I.
Thus the only domain points where a function can assume extreme values are critical points and endpoints. However, be careful not to misinterpret what is being said here. A function may have a critical point at x = c without having a local extreme value there. For instance, both of the functions y = x' and y = x 1/' have critical points at the origin and a zero value there, but each function is positive to the right of the origin and negative to the left. So neither function has a local extreme value at the origin. Instead, each function has a point of inflection there (see Figure 4.7). We define and explore inflection points in Section 4.4. Most problems that ask for extreme values call for finding the absolute extrema of a continuous function on a closed and finite interval. Theorem 1 assures us that such values exist; Theorem 2 tells us that they are taken on only at critical points and endpoints. Often we can simply list these points and calculate the corresponding function values to fmd what the largest and smallest values are, and where they are located. Of course, if the interval is not closed or not fmite (such as a < x < b or a < x < 00), we have seen that absolute extrema need not exist. If an absolute maximum or minimum value does exist, it must occur at a critical point or at an included right or lefthand endpoint of the interval.
How to Find the Absolute Extrema of a Continuous Function f on a Finite Closed Interval 1. Evaluate 1 at all critical points and endpoints. 2. Take the largest and smallest of these values.
EXAMPLE 2 [2, I].
Find the absolute maximum and minimum values of I(x) = x 2 on
Solution
The function is differentiable over its entire domain, so the only critical point is where f' (x) = 2x = 0, namely x = O. We need to check the function's values at x = 0 and at the endpoints x = 2 and x = I: Critical point value: Endpoint values:
1(0) = 0
I( 2) = 4
1(1) = 1 The function has an absolute maximum value of 4 at x = 2 and an absolute minimum value of 0 atx = O. •
4.1
EXAMPLE 3 [2,1].
y 7
(1.7)
189
Extreme Values of Functions
Find the absolute maximum and minimum values of g(l) = 81  14 on
Solutton
The function is differentiable on its entire domain, so the only critical points occur where g'(I) = O. Solving this equation gives
y= 8t t 4
841'=0
1=\0/2>1,
or
a point not in the given domain. The function's absolute extrema therefore occur at the endpoints, g( 2) = 32 (absolute minimum), and g(l) = 7 (absolute maximum). See • Figure 4.8.
EXAMPLE 4
Find the absolute maximum and minimum values of I(x) = x2/' on the
interval [2,3]. FIGURE 4.8 The extreme values of g(t) = 8t  t 4 on [2, I] (Example 3).
y
Solutton
We evaluate the function at the critical points and endpoints and take the largest and smallest of the resulting values. The tITSt derivative
I'(x) y =x213 , 2sxs 3 Absolute maximum; also a local maximum
Loca1
I
0 ~ 1
2
3
Critical point value:
1(0)
Endpoint values:
I( 2) = (_2)2/3 =
Absolute minimum; also a local minimum
1(3)
FIGURE 4.9 The extreme values of f(x) = x 213 0n [2, 3] occuratx = oand x = 3 (Example 4).
~xI/3 = _2_ 3 3\Yx
has no zeros but is undefined at the interior point x = O. The values of I at this one critical point and at the endpoints are
~~~~~~~~X
2
=
=
0
= (3)'/3 =
V'4
'f9.
We can see from this list that the function's absolute maximum value is 'f9 .., 2.08, and it occurs at the right endpoint x = 3. The absolute minimum value is 0, and it occurs at the • interior point x = 0 where the graph has a cusp (Figure 4.9).
Exercises 4.1 finding Extrema from Graphs
3.
y
4.
y
In Exercises l{i, determine from the graph whether the function has any absolute ex1reme values on [a, b]. Then explain how your answer is consistent with Theorem 1.
1.
y
2.
y ~O~aL~c~b~X
~L~~+x
o
a
c
b
o
a
c
b
190 S.
Chapter 4: Applications of Derivatives 6.
Y
In Exercises 1520, sketch 1Ire graph of each function and determine whether the functioo has any absolute extreme values on its domain.
y
(~'
Explain how your answer is consistent with Theorem 1. 16. Y
~L~~+x
o
a
c
lxi, I 0 throughout an interval [a, b], then!, has at most one zero in [a, b]. What if/,,< 0 throughout [a, b] instead? 18. Show that a cubic polynontial can have at most three real zeros. Show that the functions in Exercises 1926 have exactly one zero in tire given interval. 19. f(x) = x'
20. f(x) = x 3 + 21. g(l) =
~
x
38. g'(x)
=
1, + lx, x
(;f, 0 )
39. 7'(0) = 8  csc'O, 40. 7'(1)
= secltanl 
I, P(O, 0)
F;nd;ng Po.;tion from Velodty or Acceleration Exercises 4144 give the velocity v = dsldl and initial position ofa body moving along a coordinate line. Find the body's position at time t. 41. v = 9.81 + 5,
.(0) = 10
42. v = 321  2,
.(0.5) = 4
(00,0)
44. v
(~)
24.7(0) = 20  oos2 0 + I 25. 7(0) = sec 0  03

8,
v2,
(0,00)
3.1,
(I, I)
(00,00) (00,00)
+ 5, (0, "'/2)
26. 7(0) = tanO  cotO  0,
(0, "'/2)
Find;ng Functions from Derivatives 27. Suppose that f( I) = 3 and that !,(x) = 0 for all x. Must f(x) = 3 for all x? Give reasons for your answer.
sec2 0
p(I, I)
+ 7,
4,
v8 
 I, P(O, 0)
43. v = sin ",I, .(0) = 0
~ I + v'l+t 
23. 7(0) = 0 + sin2
= lx
[2, I]
vt + v'l+t 
22. g(l) = I
37. !'(x)
I,
+ 3x +
c. y' =
+ cost
In Exercises 3740, fmd the function with the given dctivative whose graph passes through the point P.
33x + 216x = x(x  9)(x  24)
b. Use Rolle's Theorem to prove that between every two zeros of x" + Qn_tX" 1 + ... + QtX + ao there lies a zero of
v8
c. y' = 8in2t
I
Vx
=
2
21
7T cos W'
,
s('7J'""")
=
1
Exercises 4548 give the acceleration a = d 2.ldI 2 , initial velocity, and initial position of a body moving on a coordinste line. Find tire body's position at time I. 45. a = 32,
v(O) = 20,
46. a = 9.8,
v(O) = 3,
.(0) = 5 .(0) = 0
47. a = 4 sin 21,
v(O) = 2,
.(0) = 3
9 31 48. a = "" cos ""
v(O) = 0,
.(0) = I
Applkations 49. Temperatnre cbange It took 14 sec for a mercury thermometer to rise from 19'C to lOO'C when it was taken from a freezer and placed in boiling water. Show that sOO3ewhere along the way tire mercury was rising at the rate of 8.5°C/sec.
198
Chapter 4: Applications of Derivatives
50. A trucker handed in a ticket at a toll booth showing that in 2 hours she had covered 159 mi on a toll road with speed limit 65 mph. The trucker was cited for speeding. Why? 51. Classical accounts tell us that a 170oar trireme (aocient Greek or Roman WlIIship) once covered 184 sea miles in 24 hours. Explain why at some point during this feat the trireme's speed exceeded 7.5 koots (sea miles per hour). 52. A marathoner ran the 26.2mi New York City Marathon in 2.2 hours. Show that at least twice the marathoner was running at exactly II mph, assuming the initial and fmal speeds are zero.
60. Parallel tangents Assume that I and g are differentiable on [a, b j and that/(a) ~ g(a) aod/(b) ~ g(b). Show that there is at least one point between a and b where the taogents to the graphs of I and g are parallel or the same line. lllustrate with a sketch. 61. Suppose that f'(x):5 I for I :5 x :5 4. Show that 1(4)1(1) :5 3. 62. Suppose that 0 < f'(x) < 1/2 for all xvalues. Show that 1(1) < 1(1) < 2 + 1(1).
63. Show that Icou  II :5 Ix I for all xvalues. (Hint: Consider I(t) ~ cos t on [0, xj.)
53. Show that at some instaot during a 2hour automobile 1rip the car's speedometer reading will equal the average speed for the 1rip.
64. Show that for any numbers a and b, the sine inequality Isinb  sinal :5 Ib  al is true.
54. Free fall on the moon On our moon, the acceleration of gravity
65. If the graphs of two differentishle functions I(x) andg(x) start at the same point in the plane and the functions have the same rate of change at every point, do the graphs have to be identical? Give
is 1.6 m/sec2 • Ifa rock is dropped into a crevasse, how fast will it be going just before it hits bottom 30 sec later? Theory and Examples 55. The geometric mean of a and b The ~metric mean of two positive nombers a and b is the number Vab. Show that the value of c in the conclusion of the Meao Valoe Theorem for I(x) ~ I/x on an interval of positive numbers [a, b j is c ~ v;;J,.
56. The arithmetic mean of Il and b The arithmetic mean of two numbers a and b is the number (a + b)/2. Show that the value of c in the conclusion of the Meao Value Theorem for I(x) ~ x 2 on any interval [a,bj isc ~ (a + b)/2.
D 57.
reasons for your answer. 66. Ifl/(w)  I(x) I :5lw  xl for all values w and x and I is a differentiable function, show that I :5 f' (x) :5 I for all xvalues. 67. Assumethatl is differentiable ona :5 x:5 baodthat/(b) Show that f' is negative at some point between a and b.
68. Let/be a function defmed on ao interval [a, bj. What conditions could ynu place on I to gusrantee that
. I' < I(b)  I(a) < I' mm ba _max,
Graph the function
I(x) ~ sin x sin (x
+ 2)
 sin2 (x
+
I).
What does the graph do? Why does the function behave this way?
Give reasons for your answers. 58. RoUe's Theorem L
Construct a polynomial I(x) that has zeros atx I, and 2.
~
2, I, 0,
b. Graph I and its derivative f' together. How is what you see related to Rolle~ Theorem?
c. Do g(x) ~ sin x aod its derivative g' illustrate the same phenomenon as I and f'? 59. Unique solution Assume that I is continuous on [a, bj and differentiable on (a, b). Also assume that I(a) aod I(b) have opposite sigus and that f' # 0 between a aod b. Show that I(x) ~ 0 exactly once between a aod b.
4.3
< I(a).
where min f' and max f' refer to the minimum and maximum values off' on [a, b]? Give reasons for your aoswers.
D 69.
Use the inequalities in Exercise 68 to estimate 1(0.1) if f'(x) ~ 1/(1 + x'cosx) forO:5 x:5 0.1 aod/(O) ~ 1.
D 70.
Use the inequalities in Exercise 68 to estimate 1(0.1) if f'(x) ~ 1/(1  x 4 ) forO:5 x:5 0.1 aod/(O) ~ 2.
71. Let
1(1)
I
be differentishle at every value of x and suppose that I, that f' < 0 on (00, I), and thatf' > 0 on (1,00).
~
a. Show that I(x) '" I for all x. b. Mustf'(I) ~ O?Explain.
72. Let I(x) ~ px2 + qx + r be a quadratic function defined on a closed interval [a, b j. Show that there is exactly one point c in (a, b) at which I satisfies the conclusion of the Mean Value
Theorem.
Monotonic Functions and the First Derivative Test In sketching the graph of a dllferentiable function it is useful to know where it increases (rises from left to right) and where it decreases (falls from left to right) over an interval. This section gives a test to determine where it increases and where it decreases. We also show how to test the critical points of a function to identifY whether local extreme values are present.
4.3
Monotonic Functions and the First Derivative Test
199
Increasing Functions and Decreasing Functions All another corollary to the Mean Value Theorem, we show that functions with positive derivatives are increasing functions and functions with negative derivatives are decreasing functions. A function that is increasing or decreasing on an interval is said to be monotonic on the interval.
COROLLARY (a, h).
3
If I'(x)
Suppose that
I
is continuous on [a,
> 0 at each point x E (a, h), then I
Ifl'(x) < 0 at each point X
E
hI and differentiable on
is increasing on [a,
(a, h), then/is decreasing on [a,
Proof Let Xl andx2 be any two points in [a, applied to I on [Xh x21 says that
I(X2)  l(xI)
=
hI with Xl I'(C)(X2 
hI. hI.
< X2. The Mean Value Theorem
XI)
for some c between XI and X2. The sign of the righthand side of this equation is the same as the sign of I'(c) because X2  Xl is positive. Therefore, I(X2) > I(Xl) if I' is positive • on (a, h) and/(x2) < I(xtl if I' is negative on (a, h). Corollary 3 is valid for infinite as well as finite intervals. To find the intervals where a function I is increasing or decreasing, we first find all of the critical points of I. If a < h are two critical points for I, and if the derivative I' is continuous but never zero on the interval (a, h), then by the Intermediate Value Theorem applied to 1', the derivative must be everywhere positive on (a, h), or everywhere negative there. One way we can determine the sign of I' on (a, h) is simply by evaluating the derivative at a single point c in (a, h). Ifl'(c) > 0, thenl'(x) > 0 for all X in (a, h) so/isincreasing on [a, hI by Corollary 3; if I'(c) < 0, then I is decreasing on [a, hI. The next example illustrates how we use this procedore. y y~x'I2x  5
EXAMPLE 1 Find the critical points of I(x) = x3  12x  5 and identify the intervals on which I is increasing and on which I is decreasing.
20
Solution
The function I is everywhere continuous and differentiable. The f"trst derivative
I'(X)
=
3X2  12
= 3(x
FIGURE 4.20 The function /(x) ~ 12x  5 is monotonic on three separate intervals (Example I).
=
+ 2)(x
3(X2  4)  2)
is zero at X = 2 and X = 2. These critical points subdivide the domain of I to create nonoverlapping open intervals (00, 2), (2,2), and (2, 00) on which I' is either positive or negative. We determine the sign of I' by evaluating I' at a convenient point in each subinterval. The behavior of I is determined by then applying Corollary 3 to each subinterval. The results are summarized in the following table, and the greph of I is given in Fignre 4.20.
X3 
Intervnl /' evnluated Sign of/, Behaviorofl
00 < X o :
1' < 0
I I I
Local min : :
:
I I
I I
: /'= 0 I I
Absolute min
I
Local min
_ _ _ _ _ _ _ _ _ _ ~_ _ _ _ _ _ _ _ ~_ _ _ _ _ _ ~_ _ _ _ _ _ ~ ' _ _ _ _ _ _l ' _ _ _ _~_ _ _ _~'_ _ _ _ _ _ _ _ _ _~"
a
C2
b
FIGURE 4.21 The critical points of a functioo locate where it is increasing and where it is decreasing. The lust derivative changes sign at a critical point where a local extremum occurs.
These observations lead to a test for the presence and nature of local extreme values of differentiable functions.
First Derivative Test for Local Extrema Suppose that c is a critical point of a continuous function I, and that I is differentiable at every point in some interval containing c except possibly at c itself. Moving across this interval from left to right,
1. if f' changes from negative to positive at c, then I has a local minimum at c; 2. if f' changes from positive to negative at c, then I has a local maximum at c; 3. iff' does not change sign at c (that is, f' is positive on both sides of cor negative on both sides), then I has no local extremum at c. The test for local extrema at endpoints is similar, but there is ouly one side to consider.
Proof of the First Derivative Test Part (I). Since the sign of f' changes from negative
< c < b, f' < 0 on (a, c), and 0 on (c, b). IfxE (a, c), then/{c) < I{x) because f' < 0 implies that I is decreasing on [a, c]. If x E (c, b), then I{ c) < I{x) because f' > 0 implies that I is increasing on [c, b]. Therefore, I{x) ;", I{c) for every XE (a, b). By definition, I has a local minimumatc. Parts (2) and (3) are proved similarly. to positive at c, there are numbers a and b such that a
f' >
•
EXAMPLE 2
201
Monotonic Functions and the First Derivative Test
4.3
Find the critical points of
I(x) = xl/3(X  4) = x'/'  4XI/3. Identify the intervals on which 1 is increasing and decreasing. Find the function's local and absolute extreme values. Solution The function 1 is continuous at all x since it is the product of two continuous functions, x 1/' and (x  4). The first derivative
f'(x) =
:fx (X4/' 
_1
3 x
y
j x 1/'  j x2/'
2/'(x _ 1) _ 4(x3x
1)
2/'
is zero at x = 1 and undefined at x = O. There are no endpoints in the domain, so the critical points x = 0 and x = 1 are the only places where 1 might have an extreme value. The critical points partition the xaxis into intervals on which f' is either positive or negative. The sign pattern off' reveals the behavior of1 between and at the critical points, as summarized in the following table.
4
FIGURE 4.22 The functionj(x) = x 1/3 (x  4) decreases when x < I and increases when x > I (Example 2).
4XI/') =
Interval Sign ofr Behavior of I
x 2, 00
00
48. k(x) = x + 3x + 3x + I, 00 < x :5 0 49. j(x) = Y25  x 2 , 5 S x s 5 50. j(x) = Yx2  2:<  3, 3:5 x < 00 x2 51. g(x) = 2 , 0:5 x < I 3
V3 cos:< + sinx,
c. /'(x) > Oforx "" I;
b. Which of the extreme values, if any, are absolute?
x T 
0:5 x :5 2rr
56. j(x) =
+3
main, and say where they occur.
47. h(x) =
Graph the functioo and its derivative together. Connnent on the behavior of j in relation to the signs and values of /'.
18x
= 2: 0,
0+.
67. Discuss the extremevalue behavior of the function j(x) = x sin (I/x), x "" O. How many critical points does this function have? Where are they located 00 the xaxis? Does j have an
4.4 Concavity and Curve Sketching
69. Determine the values of constants a and b so that f(x) ax 2 + bx has an absolute maximum at the point (I, 2).
absolute minimum? An absolute maximum? (See Exercise 49 in Section 2.3.)
68. Find tire intervals on which the function f(x) ~ ax 2 + bx + c, a =F 0, is increasing and decreasing. Describe the reasoning behind your answer.
4.4
203 ~
70. Determine the values of constants a, h, c, and d so that f(x) ~ ax 3 + bx2 + ex + d has a local maximum at tire point (0,0) and a local minimum at the point (I, I).
Concavity and Curve Sketching We have seen how the first derivative tells us where a function is increasing, where it is decreasing, and whether a local maximum or local minimum occurs at a critical point. In this section we see that the second derivative gives us infonnation about how the graph of a differentiable function bends or turns. With this knowledge about the first and second derivatives, coupled with oUI previous understanding of asymptotic behavior and symmetry stodied in Sections 2.6 and 1.1, we can now draw an accurate graph of a function. By organizing all of these ideas into a coherent procedure, we give a method for sketching graphs and revealing visually the key featores of functions. Identifying and knowing the locations of these featores is of major importance in mathematics and its applications to science and engineering, especially in the graphical analysis and interpretation of data.
Concavity FIGURE 4.23 The graph of f(x) ~ x 3 is concave down on (  00, 0) and concave up on (0,00) (Example la).
As you can see in Figare 4.23, the curve y = x 3 rises as x increases, but the portions defmed on the intervals (00,0) and (0, 00) turn in different ways. As we approach the origin from the left along the curve, the curve turns to OUI right and falls below its tangents. The slopes of the tangents are decreasing on the interval (  00, 0) . As we move away from the origin along the curve to the right, the curve turns to OUI left and rises above its tangents. The slopes of the tangents are increasing on the interval (0, 00). This turning or bending behavior defmes the concavity of the curve.
DEFINmON
The graph of a differentiable function y ~ j(x) is
(a) concave up on an open interval I if j' is increasing on I;
(b) concave down on an open interval I if j' is decreasing on!.
If y
~
j(x) has a second derivative, we can apply Corollary 3 of the Mean Value Theorem to the first derivative function. We conclude that j' increases if f" > 0 on I, and decreases iff" < O.
The Second Derivative Test for Concavity Let y = j(x) be twicedifferentiable on an interval!.
1. If f"
0 on I, the graph of j over I is concave up.
2.
0 on I, the graph of j over I is concave down.
> If f"
O. (b) The curve y = x 2 (Figure 4.24) is concave up on (00, 00) because its second deriv_ ative y" = 2 is always positive. (8) The curve
EXAMPLE 2
y
=
Detennine the concavity of y = 3
+
sin x on [0, 21T].
Solution The first derivative of y = 3 + sin x is y' = cos x, and the second derivative is y" = sinx.Thegraphofy = 3 + sin x is concave down on (O,1T),wherey" = sinx is negative. It is concave up on (1T, 21T), where y" = sinx is positive (Figure 4.25). _ FIGURE 4.24
The graph of f(x) = x 2
is concave up on every interval (Example Ib).
The curve y = 3 + sin x in Example 2 changes concavity at the point (1T, 3). Since the first derivative y' = cos x exists for all x, we see that the curve has a taogent line of slope I at the point (1T, 3). This point is called a point of inflection of the curve. Notice from Figure 4.25 that the graph crosses its taogent line at this point and that the second derivative y" = sin x has value 0 when x = 1T. In general, we have the following definition.
y
y=3+sinx
2
Points of Inflection
: I
:~
DEFINITION A point where the graph of a function has a taogent line and where the concavity changes is a point of inflection.
~t.~~L~~~~~ x
0 ," / .. 1      . y" = sinx
2
Using the sign of y' to determine the concavity ofy (Example 2). FIGURE 4.25
We observed that the second derivative of f(x) = 3 + sin x is equal to zero at the inflection point (1T, 3). Generally, if the second derivative exists at a point of inflection (c, f(c», then f"(c) = O. This follows immediately from the Intermediate Value Theorem whenever f" is continuous over an interval containing x = c because the second derivative changes sign moving across this interval. Even if the continuity assumption is dropped, it is still true that f"(c) = 0, provided the second derivative exists (although a more advanced agrument is required in this noncontinuous case). Since a taogent line must exist at the point of inflection, either the first derivative f' (c) exists (is finite) or a vertical taogent exists at the point. At a vertical taogent neither the flISt nor second derivative exists. In summary, we conclude the following result.
At a point of inflection (c, f(c», either f"(c) = 0 or f"(c) fails to exist. y
The next example illustrates a function having a point of inflection where the flISt derivative exists, hut the second derivative fails to exist. EXAMPLE 3 The graph of f(x) = x'/3 has a horizontal taogent at the origin because f'(x) = (5/3)x2/3 = 0 when x = O. However, the second derivative
f"(x) = FIGURE 4.26 The graph of f(x) = x'/3 has a horizontal taogent at the origin where the coocavity chaoges, although f" does not exist at x = 0 (Example 3).
~G'X2/3)
=
10 x  1/ 3 9
failstoexistatx = O.Nevertheless,f"(x) < ofor x < Oandf"(x) > ofor x > O,sothe second derivative changes sign at x = 0 and there is a point of inflection at the origin. The graph is shown in Figure 4.26. _
4.4 Concavity and Curve Sketching y
205
Here is an example sbowing that an inflection point need not occur even though both derivatives exist and f" = O.
EXAMPLE 4 The curve y = x' has no inflection point at x = 0 (Figure 4.27). Even though the second derivative y" = 12x 2 is zero there, it does not change sign. •
FIGURE 4.27 The grapb of y = x' bas no inflection point at the origin, even though y" = 0 there (Example 4).
As our fmal illustration, we show a situation in which a point of inflection occurs at a vertical tangent to the curve where neither the first nor the second derivative exists. The graph of y = x 1/ 3 has a point of inflection at the origin because the second derivative is positive for x < 0 and negative for x > 0:
EXAMPLE 5
y Point of
inflection
~ ~~~x
FIGURE 4.28
A point of
inflection where y' and y" fail to exist (Example 5).
However, both y' = x 2/3/3 and y" fail to exist at x = 0, and there is a vertical tangent there. See Figure 4.28. • To study the motion of an o~ect moving along a line as a function of time, we often are interested in knowing when the object's acceleration, given by the second derivative, is positive or negative. The points of inflection on the gmph of the object's position function reveal where the acceleration changes sign.
EXAMPLE 6 A particle is moving along a horizontal coordinate line (positive to the right) with position function .(1)
=
21 3

1412
+ 221 
5,
12: O.
Find the velocity and acceleration, and describe the motion of the particle.
Solutton
The velocity is
V(I)
= .'(t) = 6t 2
 28t + 22
= 2(t 
1)(3t  11),
and the acceleration is
a(t) = v'(t) = ."(t) = 12t  28 = 4(3t  7). When the function .(t) is increasing, the particle is moving to the right; when .(t) is de· creasing, the particle is moving to the left. Notice that the first derivative (v = .') is zero at the critical points 1 = 1 and 1 = 11/3.
Interval Sign ofv =.' Behavior of. Particle motion
0 O.
for
95. Sketch the graph of a twicediffereotiable function y ~ f(x) with the following properties. !.abel coordinates where possible.
x 74.
y
Graphing Rational Functions Graph the rational functions in Exereises 7592.
75 ~2x'+x1 • y x'  1 ~x4+1 77
x'
76.y~
x'49 x' + 5x  14 78 ~x'+4 .Y 2x 80. y
x' ~ ,
82. y
~
,x  2
83,y~x+1
84. y
~
X+T
~x'x+1 x I x'3x'+3x1 87.y~ x'+x2
86
~_x'x+1
88• y
~x'+x2 ,
79.
I
y~,
xI x2
2 81.y~x'_1 
x'
85
• y
Derivatives
x =
. During approximately what time intervals was the marginal revenue increasing? Decreasing?
101. Suppose the derivative of the functioo y = f(x) is
= (x
 I)'(x  2).
At what points, if aoy, does the graph of f have a local minimum, local maximum, or point of ioflectioo? (Hint: Draw the sigo pattern for y' .) 102. Suppose the derivative of the functioo y
y'
112. Find the values of coostaots a, b, aod c so that the graph of y = (x 2 + a)/(bx + c) has a local minimum at x = 3 aod a I()cal maximum at (I, 2). COMPUTER EXPLORATIONS 10 Exercises 113116, fmd the ioflectioo points (if aoy) 00 the graph of
y
y'
111. Find the values of coostaots a, b, aod c so that the graph of y = ax 3 + bx2 + ex has a local maximum at x = 3, local minimumatx = 1,aodioflectioopointat(l, 11).
= (x
=
f(x) is
 I)'(x  2)(x  4).
At what points, if aoy, does the graph of f have a local minimum, local maximum, or point of inflection?
the function aod the coordinates of the points on the graph where the functioo has a local maximum or local minimum value. Thcu graph the functiou in a regioo large coough to show all these points simuitaoously. Add to your picture the graphs of the function's fl!St and secood derivatives. How are the values at which these graphs intersect the xaxis related to the graph of the functiou? 10 what other ways are the grapha of the derivatives related to the graph of the functioo? 113. y = x'  5x'  240 114. Y = x 3  12x 2 115. Y
= ~x' + 16>;2 
116. Y
= "4  "3 
X4
x3
25
4x 2 + 12x + 20
117. Graph f(x) = 2x'  4x 2 + I and its frrst two derivatives lgether. Comment on the behavior off in relation to the sigos aod values of f' aod f" . 11S. Graph f(x) = x cos x aod its second derivative together for o ,;; x ,;; 2".. Comment 00 the behavior of the graph of fin relatioo to the sigos aod values of f" .
214
Chapter 4: AppLications of Derivatives
Applied Optimization
4.5
What are the dimensions of a rectangle with IlXCd perimeter having maximum area? What are the dimensions for the least expensive cylindrical can of a given volume? How many items should be produced for the most profitable production run? Each of these questions asks for the best, or optimal, value of a given function. In this section we use derivatives to solve a variety of optimization problems in business, mathematics, physics, and economics.
Solving Applied Optimization Problems 1. Read the problem. Read the problem until you understand it. What is given? What is the unknown quantity to be optimized? 2. Draw a picture. Label any part that may be important to the problem. 3. hltroduce variables. List every relation in the picture and in the problem as an equation or algebraic expression, and identify the unknown variable. 4. Write an equation for the unknown quantity. If you can, express the unknown as a function of a single variable or in two equations in two unknowns. This may require considerable manipulation. S. Test the critical points and endpoints in the domain ofthe Il1Iknown. Use what you know about the shape of the function's graph. Use the lust and second derivatives to identify and classify the function's critical points.
(.)
EXAMPLE 1 An opentop box is to be made by cutting small congruent squares from the comers of a 12in.by12in. sheet of tin and bending up the sides. How large should the squares cut from the corners be to make the box hold as much as possible? (b)
FIGURE 4.32 An open box made by cutting the corners from a square sheet of
Solution We start with a picture (Figure 4.32). In the figure, the corner squares are x in. on a side. The volume of the box is a function of this variable:
tin What size comers maximize the box's volmne (Example I)?
, y  x(12  'lx'jl,
O:Sx:S6
Since the sides of the sheet of tin are only 12 in. long, x :S 6 and the domain of Vis the interval 0 :S X :S 6. A graph of V (Figure 4.33) suggests a minimum value of 0 at x = 0 and x = 6 and amaximum near x = 2. To learn more, we examine the first derivative of V with respect
to x: min
o
V=hlw
:
~
144  9&
+ 12%' ~
12(12  8x
+ x') ~
12(2  x)(6  x).
/,
•
FIGURE 4.33 The volume of the box in Figure 4.32 graphed 18 a fimction of x.
Of the two zeros, x  2 and x  6, only x  2 lies in the interior of the function's domain
and makes the criticalpoint list. The values of Vat this one critical point and two endpoints are Criticalpoint value: ilDdpointvalucs:
1'(2)  128 1'(0)
~
0,
1'(6)
~
o.
The lll8Ximum volume is 128 in3 • The cutout squares should be 2 in. on a side.
•
4.5
AppLied Optimization
215
EXAMPLE 2 You have been asked to design a oneliter can shaped like a right circular cylinder (Figure 4.34). What dimensions will use the least material?
+2r~) 1
T h
1
SoLution Volume of can: If r and h are measured in centimeters, then the volume of the can in cubic centimeters is I liter = 1000 crn3
Surface area ofcan:
A = 27Tr2
+ 27Trh 'v'
FIGURE 4.34 This oneliter can uses the least material when h = 2r (Example 2).
circular cylindrical ends wall
How can we interpret the phrase "least material"? For a first approximation we can ignore the thickness of the material and the waste in manufacturing. Then we ask for dimensions r and h that make the total surface area as small as possible while satisfying the constraint 7Tr 2h = 1000. To express the surface area as a function of one variable, we solve for one of the variables in 7Tr 2h = 1000 and substitute that expression into the surface area formula. Solving for h is easier:
Thus,
Our goal is to find a value of r that such a value exists.
> 0 that minimizes the value of A. Figure 4.35 suggests
A
Tall and thin can Short and wide can A = 271"r2 Tall and thin
+
2000, r r
/> 0
r~~r
o
Short and wide
FIGURE 4.35
~
The graph of A = 27fr2
+ 2000/r is concave up.
Notice from the graph that for small r (a tall, thin cylindrical container), the term 2000/r dominates (see Section 2.6) and A is large. For large r (a short, wide cylindrical container), the term 27Tr2 dominates and A again is large.
216
Chapter 4: Applications of Derivatives Since A is differentiable on r > 0, an interval with no endpoints, it can have a minimum value only where its fITst derivative is zero.
dA dr
2000 r2
  = 4'ITr   
0= 4'ITr  2000 r2
SetdA/dr ~
4",,3 = 2000 r =
~5~0
o.
Multiply by r'.
'" 5.42
Solve forr.
What happens at r = "¢'500/'IT? The second derivative 2
d A = 4'11' + 4000 dr 2 r3 is positive throughout the domain of A. The graph is therefore everywhere concave up and the value of A at r = "¢'500/'IT is an absolute minimum. The corresponding value of h (after a little algebm) is
h = 1000 = 7f'r2
2~500
= 2r.
'1T
The oneliter can that uses the least material has height equal to twice the mdius, here with r '" 5.42 em andh '" 10.84 em. •
Examples from Mathematics and Physics EXAMPLE 3
A rectangle is to be inscribed in a semicircle of radius 2. What is the largest area the rectangle can have, and what are its dimensions?
y x'+y'~4
(x. V4  X2) 2 ~L~~~~~ x
2 x
0
x 2
FIGURE 4.36 The rectangle inscribed in the semicircle in Example 3.
Solution Let (x, ~) be the coordinates of the comer of the rectangle obtained by placing the circle and rectangle in the coordinate plane (Figure 4.36). The length, height, and area of the rectangle can then be expressed in terms of the position x of the lower righthand comer: Length: 2x,
Height:
V4 
x 2,
Area:2x~.
Notice that the values of x are to be found in the interval 0 :5 x :5 2, where the selected comer of the rectangle lies. Our goal is to find the absolute maximum value of the function
A(x)=2x~ on the domain [0, 2]. The derivative
dA dx
~+2~ 4  x2
is not defined when x = 2 and is equal to zero when
2x
2
~
+2~=0
2x 2 + 2(4  x 2 )
= 0 2 84x =0
x 2 = 20rx =
±v'2.
4.5
V2
V2,
Applied Optimization
217
V2
Of the two zeros, x = and x = only x = lies in the interior of A's domain and makes the criticalpoint list. The values of A at the endpoints and at this one critical point are
A( V2)
Criticalpoint value:
=
2V2v'4=2 =
A(O) = 0,
Endpoint values:
4
A(2) = O.
The area has a maximum value of 4 when the rectangle is 2x = 2 uuits long.
V2
V4 
x2 =
V2 units high and •
EXAMPLE 4 The speed of light depends on the medium through which it travels, and is generally slower in denser media. Fermat's principle in optics states that light travels from one point to another along a path for which the time of travel is a minimum. Describe the path that a my of light will follow in going from a point A in a medium where the speed of light is c, to a point B in a second medium where its speed is C2.
HiSTORICAL BIOGRAPHY
Willebrord Snell van Rayen
(158()"'1626)
y
Solution Since light traveling from A to B follows the quickest route, we look for a path that will minimize the travel time. We assume that A and B lie in the xyplane and that the line separating the two media is the xaxis (Figure 4.37). In a uuiform medium, where the speed of light remains constant, "shortest time" means "shortest path;' and the my oflight will follow a straight line. Thus the path from A to B will consist of a line segment from A to a boundary point P, followed by another line segment from P to B. Distance traveled equals rate times time, so
Medium I
ooot,"''Ir d\    ' Medium 2
dx B
Time = distance rate
FIGURE 4.37 Alightrayrefracted (deflected from its path) as it passes from
From Figure 4.37, the time required for light to travel from A to P is
one medium to a denser medium (Example 4).
t}
AP
= Ct
=
Va 2 + x 2 Ct
.
From P to B, the time is
PB
t2=C2=
Vb 2 + (d 
x)'
C2
The time from A toB is the sum of these: 1 = I,
+ 12
=
Va 2 + x 2 C,
+
Vb 2 + (d C2
x)'
This equation expresses 1 as a differentiable function of x whose domain is [0, d]. We want to find the absolute minimum value of 1 on this closed interval. We find the derivative
dx
dtIdx negative
dtIdx zero
(    : /,. + + + + + + + (
o
0
and observe that it is continuous. In terms of the angles 6, and 62 in Figure 4.37,
dtldx positive >x
dl dx
sin 6,
sin 62
d
FIGURE 4.38 The sign pattern of dt/dx in Example 4.
The function 1 has a negative derivative at x = 0 and a positive derivative at x = d. Since dl/dx is continuous over the interval [0, d], by the Intermediate Value Theorem for continuous functions (Section 2.5), there is a point Xo E [0, d] where dl/dx = 0 (Figure 4.38).
218
Chapter 4: Applications of Derivatives
There is only one such point because dt/ dx is an increasing function of x (Exercise 62). At this unique point we then have sin II,
sin 112 C2·
Ct
This equation is SneU's Law or the Law of Refraction, and is an important principle in the theory of optics. It describes the path the ray of light follows. •
Examples from Economics Suppose that r(x) = the revenue from selling x items c (x) = the cost of producing the x items p(x) = r(x)  c(x) = the profit from producing and selling x items. Although x is usually an integer in many applications, we can learn about the behavior of these functions by defining them for all nonzero real numbers and by assuming they are differentiable functions. Economists use the terms marginal revenue, marginal cost, and marginal profit to name the derivatives r'(x), c'(x), and p'(x) of the revenue, cost, and profit functions. Let's consider the relationship of the profit p to these derivatives. If r(x) and c(x) are differentiable for x in some interval of production possibilities, and if p(x) = r(x)  c(x) has a maximum vaIue there, it occurs at a critical point ofp(x) or at an endpoint of the interval. If it occurs at a critical point, then p'(x) = r'(x) c'(x) = 0 and we see that r'(x) = c'(x). In economic terms, this last equation means that
At a production level yielding maximum profit, marginal revenue equals marginal cost (Figure 4.39).
y
I
i Maximum profit, c'{x) = r'(x) I I I
I I
"
I
Local maximum. for loss (minimum profit), c'(x) = ,'(x)
~~ _ _L ______~ ____J~
o
____________>x
Items produced
FIGURE 4.39 The graph of a typical cost functioo starts coocave down aod larer turns coocave up. It crosses the revenue curve at the breakeven point B. To the left of B, the compaoy operates at a loss. To the right, the compaoy operates at a profit, with the maximmn profit occurriug where c' (x) = r' (x). Farther to the right, cost exceeds revenue (perllaps because of a combinatioo of rising labor aod material costs aod market saturatioo) aod productioo levels become unprofitable again.
4.5 Applied Optimization
219
EXAMPLE 5 Supposethatr(x) = 9xandc(x) = x 3  6>;2 + 15x, where x represents millions ofMP3 players produced. Is there a production level that maximizes profit? If so, what is it? y
Solution
Notice that r'(x) = 9 and c'(x) = 3x2

IZl: + 15.
2
3x  IZl: + 15 = 9 3x 2 1Zl:+6=0
Setc'(x) ~ r'(x).
The two solutions of the quadratic equation are
x,
= 12 6
V72
= 2 
Vz "" 0.586
X2
= 12 +6
V72
= 2 +
Vz "" 3.414.
: Maximum : for profit I I I
Local maximum for loss ~L~JJ~~c>x
o 2v2
2
2+v2
NOT TO SCALB
FIGURE 4.40 The cost and revenue curves for Example 5.
and
The possible production levels for maximum profit are x "" 0.586 million MP3 players or x"" 3.414 million. The second derivative of p(x) = r(x)  c(x) is p"(x) = c"(x) since r"(x) is everywhere zero. Thus, p"(x) = 6(2  x), which is negative atx = 2 + and positive at x = 2 By the Second Derivative Test, a maximum profit occurs at about x = 3.414 (where revenue exceeds costs) and maximum loss occurs at about x = 0.586. The graphs of r(x) and c(x) are shown in Figure 4.40. •
Vz
Vz.
Exercises 4.5 Mathematical Applications Whenever you are maximizing or minimizing a function of a single variable, we mge you to graph it over the domain that is appropriate to the problem you are solving. The graph will provide insight before you calculate and will furnish a visual context for understanding your aoswer. 1. Minimizing perimeter What is the smallest perimeter possible for a rectangle whose area is 16 in2 , and what are its dimensions? 2. Show that among all rectangles with an 8m perimeter, the one with largest area is a square. 3. The figure shows a rectangle inscribed in an isosceles right triangle whose hypotenuse is 2 units long. &.
Express the }"oordinate of P in terms of x. (Hint: Write an equation for the line AB.)
b. Express the area of the rectangle in terms or.. c. What is the largest area the rectangle can have, and what are its dimensions? y
4. A rectangle has its base on the xaxis and its upper two vertices on the parabola y ~ 12  x 2 • Whst is the largest area the rectangle can have, and what are its dimensions? 5. You are plauning to make an open rectangular box from an 8in.by15in. piece of cardboard by cutting congruent squares from the corners and folding up the sides. Whst are the dimensions of the box of largest volume you can make this way, and wbat is its volume?
6. You are planning to close off a comer of the fIrst quadrant with a line segment 20 units long running from (a, 0) to (0, b). Show that the area of the triangle enclosed by the segment is largest when
a = b. 7. The best fencing plan A rectangular plot offarmland will be bounded on one side by a river and on the other three sides by a singlestrand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its
dimensions? 8. The .hortest fence A 216 m 2 rectangu1ar pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. Whst dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed? 9. Designing. tank Your iron works bas contracted to design and build a 500 ft', squarebased, opentop, rectangular steel holding tank for a paper company. The tank is to be made by welding thin stainless steel plates together along their edges. As the production engineer, your job is to f"md dimensions for the base and height that will make the tank weigh as little as possible.
220
Chapter 4: Applications of Derivatives
a. What dimensions do you tell the shop to use? b. Briefly describe how you took weight into account.
oJ
~
u
10. Catching rainwater A 1125 ft3 opentop rectangular tank with a square base x ft on a side and y ft deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product xy.
'" l3
!3z
I 1
I+x+j
I+x+j T
IT
I
10"
Base
Lid
T
x .i
T
I+x+j
x .i
I+x+j IS"
l
I
a. If the total costis
c = 5(x 2
a. Write a formula V(x) for the volume of the box.
+ 4xy) + lOX)!,
b. Find the domain of V for the problem situation and graph V over this domain.
what values of x and y will minimize it?
c. Use a graphical method to find the maximum volume and the value of x that gives it.
b. Give a possible scenario for the cost function in part (a). 11. Designing a poster You are designing a rectangular poster to contain 50 in2 of printing with a 4in. margin at the top and bottom and a 2in. margin at each side. What overall dimensions will minimize the amount of paper used? 12. Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 3.
d. Confirm your result in part (c) analytically.
D 17.
Designing a suitcase A 24in.by36in. sheet of cardboard is folded in half to form a 24in.by18in. rectangle as shown in the accompanying figure. Then four congruent squares of side length x are cut from the corners of the folded rectangle. The sheet is unfolded, and the six tabs are folded up to form a box with sides and a lid. a. Write a formula V(x) for the volume of the box. b. Find the domain of V for the problem situation and graph V over this domain. c. Use a graphical method to find the maximum volume and the value of x that gives it. d. Confirm your result in part (c) analytically. e. Find a value of x that yields a volume of 1120 in 3 . f. Write a paragraph describing the issues that arise in part (b).
13. Two sides of a triangle have lengths a and b, and the angle between them is 8 . What value of 8 will maximize the triangle's area? (Hint: A = O/2)ab sin 8.) 14. Designing a can What are the dimensions of the lightest opentop right circular cylindrical can that will hold a volume of 1000 cm3 ? Compare the result here with the result in Example 2. 15. Designing a can You are designing a 1000 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. The total amount of aluminum used up by the can will therefore be
A = 8r 2 + 21Trh rather than the A = 21Tr2 + 21Trh in Example 2. In Example 2, the ratio of h to r for the most economical can was 2 to 1. What is the ratio now?
D 16.
Designing a box with a lid A piece of cardboard measures 10 in. by 15 in. Two equal squares are removed from the corners of a IOin. side as shown in the figure. Two equal rectangles are removed from the other corners so that the tabs can be folded to form a rectangular box with lid.
I
I I I I I I I I I I
24"
1
x
x
24"
x
x
:
I,
I
)1
36"
~ 18" ;.I
1
The sheet is then unfolded.
I 1 24"
I,
Base
36"
)1
18. A rectangle is to be inscribed under the arch of the curve y = 4 cos (0.5x) from x = 1T to x = 1T . What are the dimensions of the rectangle with largest area, and what is the largest area?
4.5
Applied Optimization
221
19. Find the dimensions of a right circular cylinder of maximum volume that can be inscribed in a sphere of radius 10 cm. What is the maximum volume?
cylindrical sidewall. Determine the dimensions to be used if the volume is fIxed and the cost of construction is to be kept to a minimum. Neglect the thickness of the silo and waste in construction.
20. a. The U.S. Postal Service will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 108 in. What dimensions will give a box with a square end the largest possible volume?
24. The trough in the figure is to be made to the dimensions shown. Only the angle 8 can be varied. What value of 8 will maximize the trough's volume?
Girth = distance
o b. Graph the volume of a 108in. box (length plus girth equals 108 in.) as a function of its length and compare what you see with your answer in part (a). 21. (Continuation ofExercise 20.)
25. Paper folding A rectangular sheet of 8.5in.byIIin. paper is placed on a flat surface. One of the comers is placed on the opposite longer edge, as shown in the figure, and held there as the paper is smoothed flat. The problem is to make the length of the crease as small as possible. Call the length L. Try it with paper. a. Show that L 2 = 2x 3/(2x  8.5) . b. What value of x minimizes L 2 ?
a. Suppose that instead of having a box with square ends you have a box with square sides so that its dimensions are h by h by wand the girth is 2h + 2w . What dimensions will give the box its largest volume now?
c. What is the minimum value of L?
Girth
Q (originally at A)
h
o b.
Graph the volume as a function of h and compare what you see with your answer in part (a).
22. A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only half as much light per unit area as clear glass does. The total perimeter is fIxed. Find the proportions of the window that will admit the most light Neglect the thickness of the frame.
26. Constructing cylinders Compare the answers to the following two construction problems. a. A rectangular sheet of perimeter 36 cm and dimensions x cm by y cm is to be rolled into a cylinder as shown in part (a) of the figure. What values ofx andy give the largest volume? b. The same sheet is to be revolved about one of the sides of length y to sweep out the cylinder as shown in part (b) of the figure. What values of x and y give the largest volume?
x y I
_1, /
23. A silo (base not included) is to be constructed in the form of a cylinder surmounted by a hemisphere. The cost of construction per square unit of surface area is twice as great for the hemisphere as it is for the
(a)
(b)
,
222
Chapter 4: AppLications of Derivatives
27. CoaJtructlDg conel A right triangle whose hypotcuuac ia m long is mvolwd about one of it! less to gcnmate a right circular OODC. Find the radiwJ, height, and vulumc of the OODC of ~t volume that can be made this way.
v'3
• 28. Find the point on the line ~
Building
,
+~=
1 that is closest to the origin
29. Find a positive mnnbcr for which the sum of it and its reciprocal ia the !lll8lJ.e!1: (least) possible. 30. Find a postitive mnnbcr for which the sum of it! reciprocal and four timca it! square is the smallest poYible. 31. A wire b m long is cut into two pieces. One piece is bent into an cquila.tcmJ. triangle and the other is bent into a circle. If the !IUlD. of the areas cnclo!Cd by each part ia a minjrnnm, what is the length of each part? 32.
~
and I in seconds. L
3
34. Determine the dimensions of the rectangle of largest area that can be inscribed in a semicircle ofradiWl 3. (See accompanying figure.)
35. What value of a makes f(x} = x'1.
•
I
~
b. a point of inflection at x  I 1
+ a;t2 + bx have
a local maximum at x   1 and a local mjnjrnmn at x  3 1
b. a local minimum. at x = 4 and a point of inflection at x = 11
Physical Applications 37. Ver11c.J. motloa The height above ground of an object moving vertically ia givm by
b. What is the farthest apart that the particles ever get?
c. When in the interval 0
:S t :S 2Tr is the distance between the particles changing the fastest?
41. Projectile motioa The runge R of a projectile flICd from the origin am: horizontal ground is the distance from the origin to the point of impact. If the projectile is flICd with an initial velocity VI) at an angle a with the horizonlal, then in Chapter 13 we find that
vJ.
+ 96t + 112,
o 43. Streqtb. of.
beam. The strength S of a rectangular wooden beam is proportional to it! width timca the square of its depth. (See the accompanying figm:.) L
Find the dimensions of the strongest beam that can be cut from a 12in..diamctcr cylindrical log.
b. GraphSas a function of the beam's width w, assuming the proportionality constant to be 1  1. Reconcile what you see with your answer in part (a).
c. On the same screen, graph S as a function of the beam's depth d, again taking 1  1. Compare the graphs with one mother and with your answer inpart (a). What would be the cifect of chansins to some other value of 11 Try it.
with a in feet and t in seconds. Find L
2Tr do the particles
wbcrc g is the downward acceleration due ttl gravity. Find the angle a for which the range R is the laIjest possible.
a local minimum. at x = 21
a  161 2
~
m/dx atx y(O)
54. Production level Prove that the productioo level (if any) at which average cost is smallest is a level at which the average cost equals marginal cost
~
oand x
~
Ltogetherwith
Htoshowthat
(have revenue equal cost). 56. Production level Suppose that e(x) ~ x'  2Ox 2 + 20,00Ox is the cost of manufacturing x iterus. Find a productioo level that
will minimize the average cost of making x items. 57. You are to construct an open rectangular box with a square base and a volume of48 ft'. Ifmaterial for the bottom costs $6/ft' and materisl for the sides costs S4/ft', what dimensions will result in
y
Landing path
the least expensive box? What is the minimum cost? 58. The 800room Mega Motel chain is filled to capacity when the room charge is $50 per night. For each $10 increase in room charge, 40 fewer rooms are filled each night. What charge per room will result in the maximum revenue per night?
H = Cruising altitude
Airport
~I'L~
Business and Economics 51. It costs you c dollars each to manufacture and distribute backpacks. If the backpacks sell at x dollars each, the number sold is given by
n~ x ~ e
+ b(IOO
Biology 59. Sensitivity to medicine (Continuation of Exercise 60, Section 3.3.) Find the amount of medicine to which the body is most sensitive by f'mding the value of M that maximizes the derivative dR/ dM, where
 x),
where a and b are positive constants. What selling price will bring a maximum profit?
and C is a constant 60. How we cough
52. You operate a tour service that offers the following rates:
.. When we cough, the trachea (windpipe) contracts to
$200 per persoo if 50 people (the minimum number to book the tour) go on the tour.
increase the velocity of the air going out. This raises the questions of how much it should contract to maximize the
For each additiooal persoo, up to a maximum of 80 people total, the rate per petliOO is reduced by $2.
velocity and whether it really cootraets that much when we cough.
Under reasonable assumptions about the elasticity of the
It costs $6000 (a fixed cost) plus $32 per petlion to cooduct the tour. How many people does it take to maximize your profit?
tracheal wall and about how the air near the wall is slowed by friction, the average flow velocity v can be modeled by the
53. Wilson lot size formula One of the formulas for inventory management says that the average weekly cost of ordering, paying for, and holding merchandise is
A(q) ~
km
equation
v ~ c(ro  r)r 2 em/sec,
hq
placing an order (the same, no matter how often you order), c is
L
Your job, as the inventory manager for your store, is to fmd the quantity that will minimize A(q). What is it? (The fonnula you get for the answer is called the Wilson lot size formula.)
b. Shipping costs sometimes depend 00 order size. When they do, it is more realistic to replace k by k + bq, the sum of k and a constant multiple of q. What is the most economical quantity to order now?
70,
length of the trachea. Show that v is greatest when r ~ (2/3)ro; that is, when the trachea is about 33% cootraeted. The remarkable fact is that Xray photographs conf'mn that the trachea contracts about this much during a cough.
where q is the quantity you order when things run low (shoes, radios, brooms, or whatever the item might be), k is the cost of
(a constant that takes into account things such as space, utilities, insurance, and security).
O. By coosidering the minimum, prove that f(x) '" 0 for all real x if, and ooly if, b 2  ac :s; O. 30. SchWlllZ'. inequolity
and deduce Schwarz's inequality: (al b l
+ a,b, + ... + a"b.)' '" (ai' + a,' + ... + a,,')(bl' + b,' + ... + b.').
b. Show that equality holds in Schwarz's inequality only ifthcre exists a real number x that makes a,x equal hi for every value of i from I to n.
•• In Exercise 29, let
Chapter 1:3
245
Technology Application Projects
Mathematica,tMap1e Modules: MOMIIAlolIg tJ Stmight Line: POSitioll .... Velocity .... Accelertllioll You will observe the shape of a graph through dramatic animated visualizatioos of the derivative relations among the position, velocity, and acceleration. Figures in the text can be animated Newtoll's Method: EstimtJte 'fr to How MtJlly PltJces? Plot a function, observe a root, pick a starting point near the root, and use Newton's Iteration Procedure to approximate the root to a desired accuracy. The numbers 'fr, e, and v2 are approsimated.
5 INTEGRATION OVERVIEW A great achievement of classical geometry was obtaining formulas for the areas and volumes of triangles, spheres, and cones. In this chapter we develop a method to calculate the areas and volumes of very general shapes. This method, called integration, is a tool for calculating much more than areas and volumes. The integral is of fundamental importance in statistics, the sciences, and engineering. We use it to calculate quantities ranging from probabilities and averages to energy consumption and the forces against a dam's floodgates. We study a variety of these applications in the next chapter, but in this chapter we focus on the integral concept and its use in computing areas of various regions with curved boundaries.
5.1
Area and Estimating with Finite Sums
y
~~~7~X
FIGURE 5.1
The area of the region R cannot be found by a simple formula.
246
The definite integral is the key tool in calculus for defining and calculating quantities important to mathematics and science, such as areas, volumes, lengths of curved paths, probabilities, and the weights of various objects, just to mention a few. The idea behind the integral is that we can effectively compute such quantities by breaking them into small pieces and then summing the contributions from each piece. We then consider what happens when more and more, smaller and smaller pieces are taken in the summation process. Finally, if the number of terms contributing to the sum approaches infinity and we take the limit of these sums in the way described in Section 5.3, the result is a definite integral. We prove in Section 5.4 that integrals are connected to antiderivatives, a connection that is one of the most important relationships in calculus. The basis for formulating definite integrals is the construction of appropriate finite sums. Although we need to define precisely what we mean by the area of a general region in the plane, or the average value of a function over a closed interval, we do have intuitive ideas of what these notions mean. So in this section we begin our approach to integration by approximating these quantities with finite sums. We also consider what happens when we take more and more terms in the summation process. In subsequent sections we look at taking the limit of these sums as the number of terms goes to infinity, which then leads to precise definitions of the quantities being approximated here.
Area Suppose we want to find the area of the shaded region R that lies above the xaxis, below the graph of y = 1  x 2 , and between the vertical lines x = 0 and x = 1 (Figure 5.1). Unfortunately, there is no simple geometric formula for calculating the areas of general shapes having curved boundaries like the region R. How, then, can we find the area of R? While we do not yet have a method for determining the exact area of R, we can approximate it in a simple way. Figure 5.2a shows two rectangles that together contain the region R. Each rectangle has width 1/2 and they have heights 1 and 3/4, moving from left to right. The height of each rectangle is the maximum value of the function j,
5.1
Area and Estimating with Finite Sums
y
247
y
y=1x 2 1+,(;;; 0'", 1)::,
1
0.5
(0, 1)
0.5 R
~~~~~X
o
o
0.5
0.25
0.5
(a)
0.75
(b)
FIGURE 5.2 (a) We get an upper estimate of the area of R by using two rectang1es containing R. (b) Four rectaogles give a better upper estimate. Both estimates oversboot the true value for the area by the amount shaded in light red.
obtained by evaluating f at the left endpoint of the subinterval of [0, I] forming the base of the rectangle. The total area of the two rectangles approximates the area A of the region R, A .., I·
2I + 43 . 2I = 87 =
0.875.
This estimate is larger than the true area A since the two rectangles contain R. We say that 0.875 is an upper sum because it is obtained by taking the height of each rectangle as the maximum (uppermost) value of f(x) for a point x in the base interval of the rectangle. In Figure 5.2b, we improve our estimate by using four thinner rectangles, each of width 1/4, which taken together contain the region R. These four rectangles give the approximation A .., 1·
1
15
1
3
1
7
1
4 + 16 . 4 + 4 . 4 + 16 . 4 =
25 32 = 0.78125,
which is still greater than A since the four rectangles contain R. Suppose instead we use four rectangles contained imide the region R to estimate the area, as in Figure 5.3a. Each rectangle has width 1/4 as before, but the rectangles are y
y
1
1
0.5
0.5
~~~~~~~~X
o
0.25
0.5
(a)
0.75
(U)
~~~L~~~~~~ x
o
0.25 0.5 0.75 1 0.125 0.375 0.625 0.875 (b)
FIGURE 5.3 (a) Rectang1es contained in R give an estimate for the area that uodershoots the true value by the amoout shaded in light blue. (b) The midpoint rule uses rectsngles whose height is the value of y = f(x) at the midpoints of their bases. The estimate appears closer to the true value of the area because the light red overshoot areas roughly balaoce the light blue undershoot areas.
248
Chapter 5: Integration shorter and lie entirely beneath the graph of j. The function f(x) = I  x 2 is decreasing on [0, 1], so the height of each of these rectangles is given by the value of f at the right endpoint of the subinterval fonning its base. The foorth rectangle has zero height and therefore contributes no area. Summing these rectangles with heights equal to the mini· mum value of f(x) for a point x in each base subinterva1 gives a lower sum approximation to the area, A ..,
y
15
1
3
1
1
1
17 32 = 0.53125.
This estimate is smaller than the area A since the rectangles all lie inside of the region R. The true value of A lies somewhere between these lower and upper sums:
1
0.53125
o (a)
. ~k2
c. ~k'
'1 13
>=1 13
I>. ~k2
c. ~k'
... 1 5 '1fk 22. ~T5
'1
Limits nf Riemann Sums For the functions in Exercises 3946, fmd a formula for the Riemann sum obtained by dividiog the ioterval [a, b] ioto n equal subintervals and using the righthand endpoint for each c•. Then tske a limit of these sums as n i' 00 to calculate the area under the curve over [a, b].
39. I(x) = I  x 2 over the interval [0, I]. 40. I(x)
=
2x over the ioterval[O, 3].
41. I(x) = x 2
42. I(x)
=
+
I over the interval [0, 3].
2
3x overtheioterval[O, I].
43. I(x) = x
+ x 2 0ver the interval [0, I].
44. I(x) = 3x + 2x 2 over the ioterval [0, I].
'1
•
24. ~(k2  5)
'1
45. I(x)
=
2x' over the ioterval [0, I] .
46. I(x) = x 2

x' over the interval [1,0].
The Definite Integral In Section 5.2 we investigated the limit of a finite sum for a function defined over a closed interval [a, b] usiog n subiotervals of equal width (or length), (b  a)/n. In this section we consider the limit of more general Riemann sums as the norm of the partitions of [a, b] approaches zero. For general Riemann sums the subintervals of the partitions need not have equal widths. The limiting process then leads to the definition of the definite integrol of a function over a closed interval [a, b].
Definition of the Definite Integral The definition of the defioite integral is based on the idea that for certain functions, as the norm of the partitions of [a, b] approaches zero, the values of the corresponding Riemann
5.3 The Definite Integral
263
sums approach a limiting value J. What we mean by this limit is that a Riemann sum will be close to tbe number J provided that tbe nann of its partition is sufficiently small (so that all of its subintervals have tbin enough widtha). We introduce tbe symbol E as a small positive number that specifies how close to J tbe Riemann sum must be, and tbe symbolll as a second small positive number that specifies how small tbe nann of a partition must be in order for convergence to happen. We now defme this limit precisely.
DEFINmON Let f(x) be a function defmed on a closed interval [a, hI. We say that a number J is tbe defmite integral of f over [a, bl and tbat J is tbe limit oftbe Riemann sums ~~=t!(c.) Ax. iftbe following condition is satisfied: Given any number E > 0 tbere is a corresponding number Il > 0 such tbat for every partition P = {xo, x" ... , xn } of [a, hI witb Ilpll < Il and any choice of in [Xk" xii, we have
c.
The defmition involves a limiting process in which tbe nann of tbe partition goes to zero. In tbe cases where tbe subintervals all have equal widtb ~x = (h  a)/n, we can fann
each Riemann sum as
•
• (h ; a ),
S. = ~ f(ck) ~Xk = ~ f(ck)
ax, = ax = (b  alln for all k
c.
where is chosen in tbe subinterval Ax•. If tbe limit of tbese Riemann sums as n > exists and is equal to J, tbenJ is tbe defmite integral of f over [a, hI, so J = lim
~f(ck)(h; a)
n + 00 (:1
=
lim
n+ 00
~f(ck) Ax.
{:1
00
ax = (b  all.
Leibniz introduced a notation for tbe defmite integral that captures its construction as a limit of Riemann sums. He envisioned tbe fmite sums ~~=1 f(Ck) Ax. becoming an infinite sum of function values f(x) multiplied by "infmitesimal" subinterval widtbs dx. The sum symbol ~ is replaced in tbe limit by tbe integral symbol j, whose origin is in tbe letter "S." The function values f(ck) are replaced by a continuous selection of function values f(x). The subinterval widtbs Ax. become tbe differential dx. It is as if we are summing all products of tbe fann f(x) • dx as x goes from a to h. While tbis notation captures tbe process of constructing an integral, it is Riemann's defmition that gives a precise meaning to tbe defmite integral. The symbol for tbe number J in tbe defmition of tbe definite integral is
tf(X)dx,
which is read as ''the integral from a to h off of x dee x" or sometimes as ''the integral from a to h off of x witb respect to x." The component parts in tbe integral symbol also have names: The function is the integrand. Upper limit of integration/
""f;(x) ,h~r"... . ~ Lower limit of mtegration
When you :find the value of the integral, you have Integral off from a to b ~ evaluated the integral. ,
y
264
Chapter 5: Integration When the condition in the dermition is satisfied, we say the Riemann swns of I on [a, b] converge to the definite integral J = I{x} ax and that I is integrable over [a, b]. We have many choices for a partition P with norm going to zero, and many choices of points CO for each partition. The definite integral exists when we always get the same limit J, no matter what choices are made. When the limit exists we write it as the definite integral
t
lim f/{Ck}
11111~O
.... ,
~k =
J =
l
a
b, {X} ax.
When each partition has n equal subintervals, each of width ~ = {b  a}Jn, we will also write n
lim ~/(ck} ~ = J =
11_ 00
1=1
lb a
I{x} ax.
The limit of any Riemann swn is always taken as the norm of the partitions approaches zero and the nwnber of subintervals goes to inImity. The value of the dermite integral of a function over any particular interval depends on the function, not on the letter we choose to represent its independent variable. If we decide to use t or u instead of x, we simply write the integral as instead of
or
[/{X}ax.
No matter how we write the integral, it is still the same nwnber that is dermed as a limit of Riemann swns. Since it does not matter what letter we use, the variable of integration is called a dummy variable.
Integrable and Nonintegrable Functions Not every function dermed over the closed interval [a, b] is integrable there, even if the function is bounded. That is, the Riemann swns for some functions may not converge to the same limiting value, or to any value at all. A full development of exactly which functions defined over [a, b] are integrable requires advanced mathematical analysis, but fortonately most functions that commonly occur in applications are integrable. In particular, every continuous function over [a, b] is integrable over this interval, and so is every function having no more than a finite nwnber of iwnp discontinuities on [a, b]. (The latter are called piecewisecontinuous /Unctions, and they are defined in Additional Exercises 1118 at the end of this chapter.) The following theorem, which is proved in more advanced courses, establishes these results.
THEOREM lIntegralrility of Continuous Functions If a function I is continuous over the interval [a, b], or if I has at most finitely many iwnp discontinuities there, then the dermite integral I{x} ax exists and I is integrable over [a, b].
t
The idea behind Theorem 1 for continuous functions is given in Exercises 86 and 87. Briefly, when I is continuous we can choose each Ck so that I{ Ck} gives the maximwn value of I on the subinterval [Xk), Xk], resulting in an upper swn. Likewise, we can choose CO to give the minimwn value of Ion [Xk), xii to obtain a lower swn. The upper and lower swns can be shown to converge to the same limiting value as the norm of the partition P tends to zero. Moreover, every Riemann swn is trapped between the values of the upper and lower swns, so every Riemann swn converges to the same limit as well. Therefore, the nwnber J in the definition of the definite integral exists, and the continuous function I is integmble over [a, b]. For integrability to fail, a function needs to be sufficiently discontinuous that the region between its graph and the xaxis cannot be approximated well by increasingly thin rectangles. The next example shows a function that is not integrable over a closed interval.
5.3 The Definite Integral
EXAMPLE 1
265
The function
{I,0,
f(x) =
if x is rational if x is irrational
has no Riemann iotegral over [0, I]. Underlying this is the fact that between any two numbers there is both a rational number and an irrational number. Thus the function jwnps up and down too erratically over [0, I] to allow the region beneath its graph and above the xaxis to be approximated by rectangles, no matter how thio they are. We show, io fact, that upper sum approximations and lower sum approximations converge to different limitiog values. If we pick a partition P of [0, I] and choose to be the poiot giving the maximum value for f on [x._ 10 x.tJ then the correspondiog Riemann sum is
c.
•
•
U = ~f(C.) Ax. = ~(1) ~x. = I,
sioee each subioterval [Xk 10 x.] contaios a rational number where f(c.) = 1. Note that the lengtha of the iotervals io the partition sum to I, ~f~l~X' = 1. So each such Riemann sum equals I, and a limit of Riemann sums using these choices equals 1. On the other hand, if we pick to be the poiot giving the minimum value for f on [x._1o x.tJ, then the Riemann sum is
c.
•
L = ~ f(c.)
Ax.
•
~(O) Ax.
=
=
0,
sioee each subioterval [x.1ox.tJ contaios an irrational number c. where f(C.) = 0. The limit of Riemann sums usiog these choices equals zero. Sioce the limit depends on the choices of c., the function f is not iotegrable. _ Theorem I says nothiog about how to calculate defioite integrals. A method of calculation will be developed io Section 5.4, through a connection to the proeess of taking antiderivatives.
Properties of Definite Integrals In defming tf(x) dx as a limit of sums ~f~!/(c.) Ax.. we moved from left to right across the ioterval [a, b]. What would happen if we iostead move right to left, starting with Xo = b and endiog at x. = a? Each Ax. io the Riemann sum would change its sign, with x.  XkI now negative iostead of positive. With the same choices of io each subioterval, the sign of any Riemann sum would change, as would the sign of the limit, the iotegral f(x) dx. Sioce we have not previously given a meaniog to iotegrating backward, we are led to defme
c.
[f(X)dx
= 
lb'
tf(X)dx.
Although we have ouly defmed the integral over an ioterval [a, b] when a < b, it is convenient to have adefmition for the iotegral over [a, b] when a = b, that is, for the iotegral over an ioterval of zero width. Sioce a = b gives ~x = 0, whenever f(a) exists we defioe
[f(X)dx
=
0.
Theorem 2 states basic properties of iotegrais, given as rules that they satisfy, iocludiog the two just discussed. These rules become very useful io the process of computing iotegrais. We will refer to them repeatedly to simplify our calculations. Rules 2 through 7 have geometric ioterpretations, shown io Figure 5.11. The graphs io these figures are of positive functions, but the rules apply to general iotegrable functions.
THEOREM 2 Whenf andg are iotegrable over the ioterval [a, b], the definite iotegral satisfies the rules io Table 5.4.
266
Chapter 5: Integration
TABLE 5.4
RuLes satisfied by definite integrals bf  l (x) ax
1.
Ordero/Integration: l"f(x) ax
2.
Zero Width Interval:
l
3.
Constant Multiple:
l\f(x)ax
4.
Sum and Difference:
l bU(x) ± g(x}} ax
5.
Additivity:
lbf(x) ax + l'f(x) ax
6.
MaxMin Inequality:
If f has maximum value max f and minimum value minf on [a, b], then
=
A Def"mition
a
A Def'mition when!(a) exists
f(x) ax  0
"
=
bf kl (x)ax =
Anycons_k
bf (x) ax ± l
l
b g(x) ax
l'f(x) ax
=
bf minf·(b  a) " ' l (x)ax '" maxf·(b  a). 7.
y
Domination:
f(x)
20
g(x) on [a, b] => l
f(x)
20
Oon[a,b] => l
y
O~~ a~ b~x
(a) Zero Width Interval:
(b) Constant Multiple: (k
a
[f(X)dx
~0
l ' kf(x) dx
y
bf (x)ax
20
l'
~k
~
2)
f(x) dx
o
l
0
(Special Case)
a
y
~ !(x)
y
~
y
~ !(x)
x
b
l'(f(x)
+ g(x))dx
~ l'f(x)dx + l'g(x)dx
y
min! y
[f(X)dx
+ [f(X)dx
~ [f(X)dx
~~~~x
o
a
b
(e) MaxMin Inequality:
mmf'(b  a)
"'1'
f(x)dx
'" maxf'(b  a) FIGURE 5.11
Geometric interpretations of Rules 27 in Table 5.4.
+ g(x)
g(x)
max!
(d) Additivity for definite integrals:
(x) ax
(c) Sum: (areas add)
y
O~~a~b~c~~x
bg
20
y
~~~x
o
bf (x) ax
~
g(x)
~~~~x
oa
b
(f) Domination:
f(x)
=>
20
l'
g(x) on [a, b]
f(x) dx
20
l'
g(x) dx
5.3 The Definite Integral
267
While Rules I and 2 are def'mitions, Rules 3 to 7 of Table 5.4 must be proved. The following is a proof of Rule 6. Similar proofs can be given to verify the other properties in Table 5.4.
Proof of Rule 6 Rule 6 says that the integral of I over [a, b] is never smaller than the minimum value of I times the length of the interval and never larger than the maximum value of I times the length of the interval. The reason is that for every partition of [a, b] and for every cboice of the points Ck, min/·(b  a) = mini'
• L dXk k=1 Constant Multiple Rule
min! S !(c,)
Constant Multiple Rule
maxi' (b  a).
=
In short, all Riemann sums for I on [a, b] satisfy the inequality
•
min I' (b  a) ,,; ~/(ck) dxk"; max I' (b  a).
•
Hence their limit, the integral, does too.
EXAMPLE 2
To illustrate some of the rules, we suppose that
[1(X)dx
j1
4/
= 5,
(X) dx
2,
=
th(x) dx
and
JI
= 7.
Then
1. [I(X) dx = 2.
3.
[[2 /
(X)
f'1(x) dx
11
EXAMPLE 3
j4/
(x) dx
+ 3h(x)] dx
= ( 2) =
=
2[I(X) dx
=
2(5)
+ 3[h(X) dx
+ 3(7)
j41/
Rulel
=
Rules 3 and 4
31
=
5
+ (2)
Show that the value of fa!.'/!
+
cosxdx is less than or equal to
=
+
2
tl(x) dx
11
(x) dx
=
3
•
Rule 5
V2.
The MaxMin Inequality for def'mite integrals (Rule 6) says that min I' (b  a) is a lower bound for the value of I(x) dx and that max I' (b  a) is an upper bound.
Solution
The maximum value of
VI
f:
+
11VI
cosx on [0, I] is
+
cosxdx";
v'1+l = V2, so
V2'(1 
0) =
V2.
•
268
Chapter 5: Integration
Area Under the Graph of a Nonnegative Function We now return to the problem that started this chapter, that of defining what we mean by the area of a region having a curved boundary. In Section 5.1 we approximated the area under the graph of a nonnegative continuous function using several types of finite sums of areas of rectangles capturing the regionupper sums, lower sums, and sums using the midpoints of each subinterva1all being cases of Riemann sums constructed in special ways. Theorem I guarantees that all of these Riemann sums converge to a single def'mite integral as the norm of the partitions approaches zero and the number of subintervals goes to inf'mity. As a result, we can now define the area under the graph of a nonnegative integrable function to be the value of that definite integral.
DEFINITION If Y = f{x) is nonnegative and integrable over a closed f{x> over [II, b] is the interval [a, b], then the area under the curve y integml of f from a to b,
=
For the first time we have a rigorous def'mition for the area of a region whose boundary is the graph of any continuous function. We now apply this to a simple example, the area under a straight line, where we can verifY that our new definition agrees with our previous notion of area.
EXAMPLE 4
Compute
J: ax x
and find the area A under
y=
x over the interva1
[O,b],b>O. Solution
~~~~+x
o
b
FIGURE 5.12 The region in Example 4 is a triangle.
The region of interest is a triangle (Figure 5.12). We compute the area in two ways.
To compute the def'mite integral as the limit of Riemann sums, we calculate lim lPl~o ~~,f{ck) dxk for partitions whose norms go to zero. Theorem 1 tells us that it does not matter how we choose the partitions or the points Ck as long as the norms approach zero. All choices give the exact same limit So we consider the partition P that subdivides the interval [0, b] into n subinterva1s of equal width dx = {b  O)ln = bin, and we choose Ck to be the right endpoint in each subinterval. The partition is p = { 0,
t 2:, 3:,,,. ,n: }andck
= ,,:. So
Constant Multiple Rule
=
b2 n2
n{n + 1) •
2
Sum of First n Integers
5.3 y
269
All n > 00 and Ilpll > 0, this last expression on the right has the limit b'/2. Therefore,
r
b
Jo xdx b a ~~L+~x
o
The Definite Integral
a I
h(x) dx
~ 4.
2f(x) dx
b.
1,'
[f(x)
+ h(x)] dx
c. 1,'[2f(x)  3h(x)] dx
d. J.' f (X) dx
e. l'f(x) dx
f. J.'[h(X)  f(x)] dx
11. Suppose that f.,'f(x) dx ~ 5. Find
l'
Y3f(z)
dz
a. l'f(U) du
b.
c. J,'f (l)dl
d.l'[f(X)]dx
t, g(l) dl ~ v'2. Find
l:
a. 1,'g(l) dl
b.
c·l°[
d.l° ~~
3
g(x)] dx
3
13. Suppose thatf is integrable and that
104 f(z) dz ~ 7. Find
a.
1,4f (z) dz
g (u) du
v2
dr
fa' f(z) dz ~ 3 and
b. /.' f(l) dl
14. Suppose that h is integrable and that f~, h(r) dr ~ 0 and
l, h(r) dr ~ 6. Find
a.l'h(r)dr
+
x'dx
_1,' h(U)dU
0
28. f(x)
on •• [1,0], b. [1,1]
Evaluattng Definite Integrals Use the results of Equations (I) and (3) to evaluate the integrals in Exercises 2940.
32.
i,
xdx
30.
sv'2 rdr
33.
J{'/2 v2
35. io
I'dl
l
v'3a xdx
1','
xdx
J.
34.
io{"'8' ds
w
05
{V'7
io x'dx 36. 1,w/2IP dO 39.
'W
31.
OdO
37·1"xdx
{~ x'dx
io
io(" x'dx
40.
Use the rules in Table 5.4 and Equations (1)(3) to evaluate the integrals in Exercises 4150.
41. 1,'7 dx
42.1,'sxdx
43.1,'(21  3) dl
44.
45.[(I+I)dz
46. 1,0(2z  3)
47·1'3U'dU
48. (' 24u' du
49.
1,\3x' + x 
{v2
5) dx
io
(I 
v'2) dl
dz
i'/2 50.1°(3x' + x 
5) dx
Finding Allla by Definite Integrals 10 Exercises 5154, use a definite integral to rmd the area of the region between the giveo curve and the xaxis on the interval [0, b].
51. Y ~ b.
>
on a. [2,2], b. [0,2]
38..
12. Suppose that
J~v'16 
26.1'3Id', O 00 and ax ~ (b  a)/n > O. 86. Suppose that f is continuous and nonnegative over [a, b], as in the accompanying figure. By inserting points XloX2,··· ,.:t,tloX} •••• , Xnl
as shown, divide [a, b] into n subintervals oflengths ax, ~ x,  a, dx2 = Xl  Xl, • •• , dA" = b  XII_I, which need not be equal. •• !fm, ~ min {f(x) for x in the kth subinterval}, explain the connection between the lower sum L = ml
axl + m2 dx2 + ... + mil 4x1l
and the shaded regions in the first part of the figure. b. !f M, ~ max {f(x) for x in the kth subinterval}. explain the
connection between the upper sum U~
M,ax, + M2ax2 + ... + M.ax.
and the shaded regions in the second part of the figure. c. Explain the connection between U  L and the shaded regions along the curve in the third part of the figure.
0
II
I,
~ ba
lj
,I
87. We say f is uniformly continuous on [a, b] if given any E > 0, thereisa8 > osuch thatifx"x2 are in [a, b] and lx,  x21 < 8, then If(x,)  f(X2) I < E. It can be shown that a continuous function on [a, b] is uniformly continuous. Use this and the figure for Exercise 86 to show that iff is continuous and E > 0 is given, it is possible to make U  L '" E' (b  a) by making the largest of the ax,'s sufficiently small. 88. If you average 30 mijh on a 150·mi trip and then return over the same 150 mi at the rate of 50 mi/h, what is your average speed for the trip? Give reasons for your answer. COMPUTER EXPLORATIONS
!fyour CAS can draw rectaugles associated with RieD2anD sums, use it to drsw rectangles associated with Riemann sums that converge to the integrals in Exercises 8g..94. Use n ~ 4, 10,20, and 50 subintervals of equal length in each case. 89.
[(I 
x)dx
~k
274 90.
Chapter 5: Integration
1' a.
(1)
For example, if I is nonnegative and x lies to the right of a, then F(x) is the area under the graph from a tox (Figure 5.18). The variable x is the upper limit of integration of an integra!, but F is just like any other realvalued function of a real variable. For each value of the input x, there is a welldermed numerical output, in this case the dermite integra! of I from a to x. Equation (1) gives a way to derme new functions (as we will see in Section 7.2), but its importance now is the connection it makes between integrals and derivatives. If I is any continuous function, then the Fundamental Theorem asserts that F is a differentiable function of x whose derivative is I itself. At every value of x, it asserts that d dx F(x) = I(x).
y
To gain some insight into why this result holds, we look at the geometry behind it. If I ;;,: 0 on [a, b], then the computation of F'(x) from the dermition of the derivative means taking the limit as h > 0 of the difference quotient F(x b
FIGURE 5.19
In Equation (I), F(x) is the area to the left of x. Also, F(x + h) is the area to the left of
+ h. The difference quotient [F(x + h)  F(x)l/h is then
x
approximately equal to f(x), the height of the rectangle shown here.
+ h)  F(x) h
For h > 0, the numerator is obtaioed by subtracting two areas, so it is the area under the graph of I from x to x + h (Figure 5.19). If h is small, this area is approximately equal to the area of the rectaogle ofheight/(x) and width h, which can be seen from Figure 5.19. That is, F(x
+ h)  F(x) '" h/(x).
Dividing both sides of this approximation by h and letting h > 0, it is reasonable to expect that F'(x) = lim F(x h~O
+ h)  F(x)
=
I(x).
h
This result is true even if the function I is not positive, and it forms the first part of the Fundamental Theorem of Calculus.
276
Chapter 5: Integration
THEOREM 4The Fundamental Theorem of Calculus, Part 1 If I is continuous on [a, b], thenF(x) = I(t) dt is continuous on [a, b] and differentiable on (a, b) and its derivative is I(x):
J:
X
(2)
F'(x) = ax d ia I(t) dt = I(x).
Before proving Theorem 4, we look at several examples to gain a better understanding of what it says. In each example, notice that the independent variable appears in a limit of integration, possibly in a formula.
EXAMPLE 2 (a) y =
Solution
i
Use the Fundamental Theorem to tmd dy/ax if
X
(t 3
+ 1) dt
(h) y = lS3tSintdt
(e) y = ] ' " costdt
We calculate the derivatives with respect to the independent variable x.
(a) dy = d ax ax
i a
X
(t 3
+
I)dt = x 3
+
Eq.(2)withf(') ~,'
1
dy = ax diS d(f' ) (h) ax x 3tsintdt = ax  Js 3tsintdt
+
I
Table 5.4, Rule I
rx
= 
d axJs 3tsintdt
= 
3x sin X
Eq. (2) with f(') ~ 3tsin'
(e) The upper limit of integration is not x but functions,
x 2• This
makes y a composite of the two
and
y = ] " costdt
We must therefore apply the Chain Rule when finding dy/ ax.
dy dy du ax=du'ax =
(~]" costdt). du 1
= cosu .
du ax
du ax
= cos(x 2 ). 2x =
2xcosx 2
•
Proof of Theorem 4 We prove the Fundamental Theorem, Part I, by applying the definition of the derivative directly to the function F(x), when x and x + h are in (a, b). This means writing out the difference quotient F(x + h)  F(x) h
(3)
5.4 The Fundamental Theorem of Calculus
277
and showing that its limit as h + 0 is the number I(x) for each x in (a, b). Thus,
F'(x)
Fi,,(x_+~h)'_Fi"(x') _
= lim h~O =
h
l~d [[+O/(t) dt l
= lim h h+O
l
[I(t) dt]
X
+hI(t)
x
dt
Table 5.4, Rule 5
According to the Mean Value Theorem for Definite Integrals, the value before taking the limit in the last expression is one of the values taken on by 1 in the interval between x and x + h. That is, for some number c in this interval,
11'H
Ii,
I(t) dt
=
I(c).
(4)
All h + 0, x + h approaches x, forcing c to approach x also (because c is trapped between x and x + h). Since 1 is continuous atx, I(c) approaches I(x): lim I(c) = I(x).
(5)
hO
In conclusion, we have
F'(x)
= lim hi hO
= lim
h~O
=
l
x
X +h
I(t) dt
I(c)
I(x).
Eq. (4) Eq. (5)
If x = a or b, then the limit of Equation (3) is interpreted as a onesided limit with h + 0+ or h + 0, respectively. Then Theorem I in Section 3.2 shows that F is continuous for every point in [a, b]. This concludes the proof. •
Fundamental Theorem, Part 2 (The Evaluation Theorem) We now come to the second part of the Fundamental Theorem of Calculus. This part describes how to evaluate definite integrals without having to calculate limits of Riemann sums. Instead we find and evaluate an antiderivative at the upper and lower limits of integration.
If1 is continuous at every point in [a, b] and F is any antiderivative of 1 on [a, b], then
THEOREM 4 (Continued)The Fundamental Theorem of Calcuius, Part 2
l b,
(x) dx
=
F(b)  F(a).
Proof Part I of the Fundamental Theorem tells us that an antiderivative of 1 exists, namely
G(x)
=
[I(t) dt.
Thus, if F is allY antiderivative of I, then F(x) = G(x) + C for some constant C for a < x < b (by Corollary 2 of the Mean Value Theorem for Derivatives, Section 4.2). Since both F and G are continuous on [a, b], we see that F(x) = G(x) + C also holds when x = a and x = b by taking onesided limits (as x + a + and x + r).
278
Chapter 5: Integration Evaluating F(b)  F(a), we have
F(b)  F(a) = [G(b)
+ C] 
[G(a)
+ C]
=
G(b)  G(a)
=
J.b/(t) dt  J."/(t) dt
=
J.b/(t) dt  0
=
J.b/(t) dt.
•
The Evaluation Theorem is important because it says that to calculate the definite integral of / over an interval [a, b] we need do only two things: 1.
Find an antiderivative F of /, and
2.
Calculate the number F(b)  F(a), which is equal to
1: /(x) dx.
This process is much easier than using a Rieroann sum computation. The power of the theorem follows from the realization that the definite integral, which is defmed by a complicated process involving all of the values of the function / over [a, b], can be found by knowing the values of any antiderivative F at only the two endpoints a and b. The usual notation for the difference F(b)  F(a) is
F(x)
1:
or
depending on whether F has one or more terms.
EXAMPLE 3 We calculate several definite integrals using the Evaluation Theorem, rather than by taking limits of Riemann sums. (a)
l
u
cosx dx = sin x
I
d
.
dx smx
=
cosx
d dxsecx
=
sec x tan x
= sin".  sinO = 0  0 = 0
(b)
r
Jw/4
sec x tan x dx = secx]O w/4 = secO sec (  ; ) = 1
(e)
l' (~Vi  x~)
dx =
[x
3/2
+
[(4)3/2 =
[8
Vi
iI + ~] [(1)3/2 +
+ 1]  [5]
=
t]
4.
•
Exercise 66 offers another proof of the Evaluation Theorem, bringing together the ideas of Riemann sums, the Mean Value Theorem, and the defmition of the definite integral.
The Integral of a Rate We can interpret Part 2 of the Fundaments! Theorem in another way. If F is any antiderivative of /, then F' = f. The equation in the theorem can then be rewritten as
J.bF'(X) dx = F(b)  F(a).
5.4 The Fundamental Theorem of Calculus
279
Now F'(x) represents the rate of change of the function F(x) with respect to x, so the integral of F' is just the net change in F as x changes from a to b. Fonnally, we have the following result.
THEOREM 5The Net Change Theorem
The net change in a function F(x) over an interval a :5 x :5 b is the integral of its rate of change:
F(b)  F(a)
EXAMPLE 4
=
l
b
F'(x) dx.
(6)
Here are several interpretations of the Net Change Theorem.
(a) If c(x) is the cost of producing x units of a certain commodity, then c'(x) is the marginal cost (Section 3.4). From Theorem 5, x
l..
,C'(X) dx
=
C(X2)  C(XI),
which is the cost of increasing production from XI units to X2 units.
(b) If an object with position function .(t) moves along a coordinate line, its velocity is v(t) = .'(t). Theorem 5 says that
l"
"v(t) dt = .(t2)  .(tl),
so the integral of velocity is the displacement over the time interval tl :5 t :5 t2' On the other hand, the integral of the speed Iv(t) I is the total distaoce traveled over the • time interval. This is consistent with our discussion in Section 5.1.
If we rearrange Equation (6) as
F(b)
=
F(a) + lbF'(X) dx,
we see that the Net Change Theorem also says that the f'mal value ofa functionF(x) over an interval [a, b] equals its initial value F(a) plus its net change over the interval. So ifv(t) represents the velocity function of an object moving along a coordinate line, this means that the object's final position .(t2) over a time interval tl :5 t :5 t2 is its initial position S(tl) plus its net change in position along the line (see Example 4b).
EXAMPLE 5 Consider again our analysis of a heavy rock blown straight up from the ground by a dynamite blast (Example 3, Section 5.1). The velocity of the rock at any time t during its motion was given as v(t) = 160  32t ft./sec. (a) Find the displacement of the rock during the time period 0 :5 t :5 8.
(b) Find the total distance traveled during this time period. Solution (a) From Example 4b, the displacement is the integral
10rv(t) dt = 10r(160 =
32t) dt = [160t  16t 2]8 0
(160)(8)  (16)(64) = 256.
This means that the height of the rock is 256 ft. above the ground 8 sec after the explosion, which agrees with our conclusion in Example 3, Section 5.1.
280
Chapter 5: Integration
(b) As we noted in Table 5.3, the velocity function v(t) is positive over the time interval [0,5] and negative over the interval [5, 8]. Therefore, from Example 4b, the total distance traveled is the integral
[lv(t)ldt
=
[lv(t)ldt + 18Iv(t)ldt
1 8
= 1'(160 =
32t) dt 
(160  32t) dt
[160t  16t2]~  [160t  16t 2]:
= [(160)(5)  (16)(25)]  [(160)(8)  (16)(64)  «160)(5)  (16)(25))] = 400  (144) = 544.
Again, this calculation agrees with our conclusion in Example 3, Section 5.1. That is, the total distance of 544 ft traveled by the rock during the time period 0 :5 t :5 8 is (i) the maximum height of 400 ft it reached over the time interval [0, 5] plus (ii) the additional distance of 144 ft the rock fell over the time interval [5, 8]. •
The Relationship between Integration and Differentiation The conclusions of the Fundamental Theorem tell us several things. Equation (2) can be
rewritten as
di
dx
a
X
j(t) dt
=
j(x),
which says that if you frrst integrate the function j and then differentiate the result, you get the function j back again. Likewise, replacing b by x and x by t in Equation (6) gives
i
X
F'(t) dt
=
F(x)  F(a),
so that if you first differentiate the function F and then integrate the result, you get the function F back (adjusted by an integration constant). In a sense, the processes of integration and differentiation are "inverses" of each other. The Fundamental Theorem also says that every continuous function j has an antiderivative F. It shows the importance of finding antiderivatives in order to evaluate definite integrals easily. Furthermore, it says that the differential equation dy/dx = j(x) has a solution (namely, any of the functions y = F(x) + C) for every continuous function j.
Total Area The Riemann sum contains terms such as j( Ck) ,lxk that give the area of a rectangle when j(ck) is positive. When j(ck) is negative, then the product j( Ck) ,lxk is the negative of the rectangle's area. When we add up such terms for a negative function we get the negative of the area between the curve and the xaxis. If we then take the absolute value, we obtain the correct positive area.
EXAMPLE 6 g(x) = 4 
Figure 5.20 shows the graph of j(x) = x 2  4 and its mirror image across the xaxis. For each function, compute
x 2 reflected
;
10. a. {' k d > ;
Jo
26. y
o
16 •
smw (3+2cosw)'
]4 2Vy(1 + Vy)' ely
1
28.
27. y
y = ~(COSX)(sin(1T + 1Tsinx»
y
298
Chapter 5: Integration
29.
36.
30.
Y
y
y
y=!sec 2 t 2
1
Y~ 1
0
"2
2
"
o
x
37.
38.
39.
40.
4 y
31. (2,8)
(2.8)
8
Y NOTTOSCAIE
32.
Y
y=~
3
o
(3, 5)
33.
Find the areas of the regions enclosed by the lines and curves in Exercises 4150. 41. Y = x 2
x2
42. Y = 1. 
34.
35. Y
Y
and y = 2
2

and y = 3
43.y=x' and y=8x 44.y=x2 1. and y=x 45. y=x2 and y= x 2 +4x 46. y=71. 2 and y=x2+4 47. y = x'  4x 2 + 4 and y = x 2 48. Y
=
49. Y
=
xv'a 2 
v'R
x 2,
a
>
0,
and
5y
=
x
and y
+6
=
0
(How many intersection points
are there?) 50. y y = _44
2
= Ix 2

41
and y
= (x 2/2) + 4
Find the areas of the regions enclosed by the lines and curves in Exercises 5158. 51. x = 2y 2,
X
= 0,
and y = 3
5.6
52. x ~ y'
and x ~ y
53. y'  4x ~ 4
and
54. x  y' ~ 0
and x
55. x ~ y'  y
and
56. x  y'/3 ~ 0 57. x ~ y'  I
+2
Sub,titution and Area Between (urves
299
y
4x  y ~ 16
+ 2y'
~ 3
x ~ 2y'  2y  6
and x
+ y'
~ 2
and x ~ lylv'i=Y'
58. x ~ y'  y'
and x ~ 2y
~~~~~X
o
2
Find the areas of the regions enclosed by the curves in Exercises 59{j2.
59. 4x'+y~4 60. x'y~O 61. x 62. x
and x4_y~ I
+ 4y' ~ 4 + y' ~ 3
79. The figure here shows triaogle AOC inscribed in the region cut from the parabola y ~ by the lioe y ~ Fiod the limit of the ratio of the area of the ttiangle to the area of the parabolic region as a approaches zero.
x'
3x'y~4
and and
x
and 4x
+ y' + y'
~ I,
for
x "" 0
~ 0
a'.
y
Find the areas of the regions eoclosed by the lines and curves in Exercises 6370.
63. Y
= 2 sin x and y = sin2x, 0
64. y ~ 8 cosx
and y ~ sec' x,
65. y ~ cos ('lTx/2) 66. y
~
sin ('lTx/2)
~
x
'IT/3:5 x :5 'IT/3
and y ~ I  x'
and y
~
x
67. y ~ sec' x, y ~ tan' x, x ~ 'IT/4, 68. x ~ tan'y
and
x ~ tan'y,
69. x ~ 3 siny v'cosy 70. y ~ sec' ('lTx/3)
~ 'IT
x ~ 'IT/4
'IT/4 s y '" 'IT/4
and x ~ 0,
and y ~ xli',
and
0:5 Y :5 'IT/2 I '" x :5 I
71. Find the area of the propellershaped region enclosed by the curve x  y' ~ o and the line x  y ~ O. 72. Find the area of the propellershaped region enclosed by the curves x  yl/3 ~ o and x  yl/' ~ O.
80. Suppose the area of the region betweeo the graph of a positive continuous function f and the xaxis from x = a to x = b is 4 square uoits. Find the area between the curves y ~ /(x) and y ~ 2/(x) from x ~ atox ~ b.
81. Show that the area of the shaded region equals 1/6 for all values ofz.
73. Find the area of the region in the f1I1!t quadrant bounded by the liney ~ x,the line x ~ 2,thecurvey ~ I/x',andthexaxis. 74. Find the area of the ''triangular'' region in the first quadrant bounded on the left by the yaxis and on the right by the curves y = sinx andy = cosx.
x'
75. The region bounded below by the parabola y ~ and above by the line y ~ 4 is to be partitioned into two subsections of equal area by cutting across it with the horizontal line y = c. ~ c across it that looks about right. In terms of c, what are the coordinates of the points where the line and parabola intersect? Add them to your figure.
a. Sketch the region and draw a line y
82. True, sometimes true, or never true? The area of the region between the graphs of the continuous functions y ~ /(x) and y ~ g(x)andtheverticsllinesx ~ a and x ~ b(a < b) is
b. Find c by integrating with respect to y. (This puts c in the limits ofintegration.)
c. Find c by integrating with respect to x. (This puts c into the integrand as well.) 76. Find the area of the region between the curve y ~ 3 line y ~ I by integrating with reopect to (a) x, (b) y.
x' and the
77. Find the area of the region in the fJrst quadraot bounded on the left by the yaxis, below by the line y ~ x/4, above left by the curvey ~ I + v'i,andaboverightbythecurvey ~ 2/v'i. 78. Find the area of the region in the fJrst quadraot bounded on the left by theyaxis, below by the curve x ~ 2Vy, above left by the curve x ~ (y  I)', and above right by the line x ~ 3  y.
l\/(x)  g(x)] dx. Give reasons for your answer. Theory and Examples 83. Suppose thatF(x) is an antiderivative of /(x) ~ (sinx)/x, x Express {' sin2x dx x
11 io terms ofF.
> O.
300
Chapter 5: Integration
84. Show that iff is continuous, then
l
lf
(X) dx =
l
lf(1 
y
x) dx.
85. Suppose that
Find tf(X)dx if (a) f is odd, (h) f is even. 86. L Show that if f is odd on [a, aJ, 1hen
89. Use a substitution to verify Equation (I). 90. For each of 1he following functions, graph f(x) over [a, bJ and f(x + c) over [a  c, b  cJ to convince yourself that Equation (1) is reasonable.
i:f(X)dx = O. b. Test1heresultinpar!(a)wi1hf(x) = sinxanda = '1f/2.
&.
87. Iff is a continuous function, fmd 1he value oftha integral ('
1=
Jo
b. f(x)
f(x) dx f(x) + f(a  x)
by making 1he substitution u integral to l.
=
c. f(x)
a  x and adding 1he resultiog
88. By using a substitution, prove 1hat for all positive numbers x andy,
t~'f(x + c) dx.
a. Plot the curves toge1her to see what they look like and how many points of intersection they have. b. Use 1he numerical equation solver in your CAS to fmd a111he points of intersection. c. lotegrate If(x)  g(x) I over consecutive pairs of intersection values. d. Sum toge1her 1he integrals found in par! (c).
(1)
The equation holds wbeoever f is integrable and defmed for 1he necessary values of x. For example, in 1he accompanying figure show that
r (x + 2)' dx = Jo(lx'dx J2 l
because the areas of the shaded regions are congruent.
Chapter
= x 2 , a = 0, b = I, c = I = sinx, a = 0, b = '1f, C = '1f/2 = v;=4, a = 4, b = 8, c = 5
COMPUTER EXPLORATIONS 10 Exercises 9194, you will find tha area between curves in tha plane when you caonot fmd 1heir points of intersection using siruple algebra. Use a CAS to perform 1he following steps:
The Sbift Property for Definite Integral. A basic property of definite integrals is thair invatiance under translation, as expressed by 1he equation [f(x) dx =
f(x)
3
2
91. f(x) =
x x T  2 
92. f(x)
=
2x' 
93. f(x)
=
x
94. f(x)
= x2
3x'
2x
+
1 + 3'
10,
+ sin (2x), cos x, g(x)
g(x)
g(x) =
g(x) = x  I
=
=
8  12x
x'
x'  x
Questions to Guide Your Review
1. How can you sometimes estimate quantities like distance traveled, area, and average value wi1h fmite sums? Why ntight you want to do so? 2. What is sigma notation? What advantage does it offer? Give examples. 3. What is a Riemann sum? Why ntight you want to consider such a S'lD1l?
4. What is 1he norm of a partition of a closed interval? 5. What is 1he defmite integral of a function f over a closed interval [a, bJ? When can you be sure it exists?
6. What is 1he relation between defmite integrals and area? Describe some other interpretations of def'mite integrals. 7. What is the average value of an integrable function over a closed interval? Must the function assume its average value? Explain. 8. Describe the rules for working wi1h defntite integrals (Table 5.4). Give examples. 9. What is 1he Fundamental Theorem of Calculus? Why is it so iruportant? Illustrate each part of the 1heorem with an example. 10. What is the Net Change Theorero? What does it say about the integral of velocity? The integral of margina1 cost?
301
Chapter 5 Practice Exercises 11. Discuss how the processes of inregration and differentiation can be considered as "inverses" of each other.
14. How can you sometimes evaluare indefutire integrals by substitution? Give examples.
12. How does the Fundamental Theorem provide a solution to the initial value problem dyldx = f(x), y(xo) = Yo, when f is continuous?
15. How does the method of substitution worl< for defutite inregrals? Give examples.
16. How do you defme and calculare the area of the region between
13. How is inregration by substitution relared to the Chain Rule?
Chapter
~
the graphs of two continuous functions? Give an example.
Practice Exerdses
Finite Sums and Estimates 1. The accompanying figure shows the graph of the velocity (fr/sec) of a model rocket for the first 8 sec after launch. The rocket accel· erared straight up for the fIrSt 2 sec and then coasred to reach its maximum height at t = 8 sec.
(5 )
10
c. ~(ak + bk  I)
10   b d. ~ k k I 2
k l
'0
20
4. Suppose that ~ak = 0 and ~bk = 7. Find the values of 20
'0
a. ~3ak
b. ~(ak
k l
k I
150 1 ¢l
+ bk)
'0
d. ~(ak  2)
~IOO
.~
~
Definite Integrals
50
o
2
6
4
8
Time after launch (sec)
a. Assuuting that the rocket was launched from ground level, about how high did it go? (This is the rocket in Section 3.3, Exercise 17, but you do not need to do Exercise 17 to do the exercise here.)
b. Sketch a graph of the rocket's height above ground as a function oftime for 0 :5 t :5 8. 2. a. The accompanying figure shows the velocity (mj sec) of a body moving along the s·axis duting the time in1llrva1 from t = 0 to t = 10 sec. About how far did the body travel duting those 10 sec?
b. Sketch a graph of s as a function of t for 0 ,;; t ,;; 10 assum· ings(O) = O.
v ....
5
I
\
6
8
10
Time (sec) 10
10
3. Suppose that ~ak = 2 and ~bk = 25. Find the value of iI
7. 8.
•
lim ~e, x,
'If/3:5 x :5 'If/3
29. Find the extren3e values of f(x) ~ x'  3x 2 and rmd the area of the region enclosed by the graph off and the xaxis. 30. Find the area of the regioo cut frOO3 the rlrst quadrant by the curve x 1/2 + yl/2 ~ a1/2. 31. Find the total area of the region enclosed by the curve x ~ y2/3 aod the lines x ~ y and y ~ I.
• ]0,
50.
{I (88'
Jo {27 Jl XX
Fulcrum at origin
The resulting system might balance, or it might not, depending on how large the masses are and how they are arranged along the xaxis. Each mass mk exerts a downward force mkg (the weight of mk) equal to the magnitude of the mass times the acceleration due to gravity. Each of these forces has a tendency to turn the axis about the origin, the way a child turns a seesaw. This turning effect, called a torque, is measured by multiplying the force mkg by the signed distance Xk from the point of application to the origin. Masses to the left of the origin exert negative (counterclockwise) torque. Masses to the right of the origin exert positive (clockwise) torque. The sum of the torques measures the tendency of a system to rotate about the origin. This sum is called the system torque. System torque = ml8X1
+
m28X2
+
m38X3
The system will balance if and only if its torque is zero. Ifwe factor out the g in Equation (I), we see that the system torque is g

.
(mlxl
+ m2X2 + m3x3).
a feature of the
a feature: of
environment
the system
(1)
6.6
Moments and Centers of Mass
347
Thus, the torque is the product of the gravitational acceleration g, which is a feature of the environment in which the system happens to reside, and the number (mlxl + m2X2 + m3x3), which is a feature of the system itself, a constant that stays the same no matter where the system is placed. The number (mlxl + m2X2 + m3x3) is called the moment of the system about the origin. It is the sum of the moments mlX!. m2X2, m3X3 of the individual masses.
Mo
=
Moment of system about origin = ~ mkXk
(We shift to sigma notation here to allow for sums with more terms.) We usually want to know where to place the fulcrum to make the system balance, that is, at what point x to place it to make the torques add to zero. Xl
x
X3
~
m3
X2
0
• ml
/\ • L__~~ m2
•
, X
Special location for balance
The torque of each mass about the fulcrum in this special location is distance) (downward) T orque 0 f mk ab out x  (Signed fi of mk fromx orce
When we write the equation that says that the sum of these torques is zero, we get an equation we can solve for x: ~ (Xk  x)mkg = 0
~ mkXk
_
x
Sum ofthe torques equals zero.
=
This last equation tells us to find the system's total mass:
~ mk .
Solved for x
x by dividing the system's moment about the origin by system moment about origin system mass
X=
(2)
The point x is called the system's center of mass.
Masses Distributed over a Plane Region FIGURE 6.44
A wrench gliding on
ice turning about its center of mass as the center glides in a verticalline. y
Suppose that we have a finite collection of masses located in the plane, with mass mk at the point (xt, Yk) (see Figure 6.45). The mass ofthe system is System mass: Each mass mk has a moment about each axis. Its moment about the xaxis is mkYk, and its moment about the yaxis is mkXk. The moments of the entire system about the two axes are Moment about xaxis:
Mx = ~ mkYk,
Moment about yaxis:
My = ~ mkXk.
The xcoordinate of the system's center of mass is defined to be Each mass mk has a moment about each axis. FIGURE 6.45
_
My M
x==
(3)
348
Chapter 6: Applications of Definite Integrals With this choice of X, as in the onedimensional case, the system balances about the line = x (Figure 6.46). The ycoordinate of the system's center of mass is dermed to be
y
x
_ Mx y= M=
L
mkYk Lmk'
(4)
With this choice of y, the system balances about the line y = y as well. The torques exerted by the masses about the line y = y cancel out Thus, as far as balance is concerned, the system behaves as if all its mass were at the single point (x, y). We call this point the system's center of mass.
o
Thin, Flat Plates FIGURE 6.46 A twodimensional array of masses balances on its center of mass.
y Strip of mass
.am
In many applications, we need to find the center of mass of a thin, flat plate: a disk of aluminum, say, or a triangular sheet of steel. In such cases, we assume the distribution of mass to be continuous, and the fonnulas we use to calculate x and y contain integrals instead of finite sums. The integrals arise in the following way. Imagine that the plate occupying a region in the .!J'plane is cut into thin strips parallel to one of the axes (in Figure 6.47, the yaxis). The center of mass of a typical strip is (x, y). We treat the strip's mass dm as if it were concentrated at (x, y). The moment of the strip about the yaxis is then dm. The moment of the strip about the xaxis is y dm. Equations (3) and (4) then become
x
_ My Lxdm x=M=Ldm' ~~~X
o
'i
FIGURE 6.47 A plate cut into thin sttips parallel to tire yaxis. The moment exerted by a typical sttip about each axis is the moment its mass .11m would exert if concentrated at the strip's center of mass
_
Mx
Y = M =
The sums are Riemann sums for integrals and approach these integrals as limiting values as the strips into which the plate is cut become narrower and narrower. We write these integrals symbolically as
x=
jxdm jdm
and
y=
jydm jdm'
(X',y).
Moments, Mass, aod Center of Mass of a Thin Plate Covering a Region in the xyPlaoe
I Density A material's density is its mass per unit area For wires, rods, and narrow s1rips, we use mass per unit length.
Moment about the xaxis:
Mx= J ydm
Moment about the yaxis:
My=Jxdm
Mass:
M=
J
(5)
dm
Center of mass:
The differential dm is the mass of the strip. Assuming the density Il of the plate to be a continuous function, the mass differential dm equals ilie product Il dA (mass per unit area times area). Here dA represents the area of the strip. To evaluate the integrals in Equations (5), we picture the plate in the coordinate plane and sketch a strip of mass parallel to one of the coordinate axes. We then express the strip's mass dm and the coordinates (x, y) of the strip's center of mass in tenos ofx or y. Finally, we integrate y dm, x dm, and dm between limits of integration determined by the plate's location in the plane.
6.6 y(em)
2
Moments and Centers of Mass
349
EXAMPLE 1 The triangular plate shown in Figure 6.48 has a constant density of Il = 3 g/=2. Find
(1,2)
(8) the plate's moment My about the yaxis.
(b) the plate's mass M.
(c) the xcoordinate of the plate's center of mass (c.m.).
Solution
Method 1: Vertical Strips (Figure 6.49)
(8) The moment My: The typical vertical strip has the following relevant data.
FIGURE 6.48
center of mass (c.m.): length: width: area: mass: distance of C.m. from yaxis:
The plate in Example 1.
(1,2)
My = Strip c.m. is halfway.
J
xdm = 116x 2 dx = 2x 3 ]: = 2g·=.
(b) The plate's mass:
lx, y) ~ (x, x)
M =
>~+I
dm =
_
Units in centimeters
My
x=M=
FIGURE 6.49 Modeling the plate in Example 1 with vertical strips.
6x dx = 3x 2 ]: = 3 g.
2g'cm 3g
2
=3='
By a similar computation, we could find Mx and y = Mx/M.
Method 2: Horizontal Strips
(Figure 6.50)
(8) The moment My: The ycoordinate of the center of mass of a typical horizontal strip is y (see the figore), so
y(em)
2
J 11
(c) The xcoordinate of the plate's center of mass:
~f,,~~~x
o
x
The moment of the strip about the yaxis is xdm = x'6xdx = 6x 2 dx. The moment of the plate about the yaxis is therefore
y 2
(x,y) = (x,x) 2x dx dA = 2x dx dm = IldA = 3·2xdx = 6xdx = x
y=y. (1,2) Strip c.m. is halfway.
__) (Y+2 (X,Y 4 ,Y) ~
The xcoordinate is the xcoordinate of the point halfway across the triangle. This makes it the average ofy/2 (the strip's lefthand xvalue) and 1 (the strip's righthand xvalue):
=
+ t
_
x =
(y/2) + 1 2
y 1 Y+2 = 4 + 2 = 4'
dy
o o+!~x (em)
FIGURE 6.50 Modeling the plate in Example 1 with horizootal strips.
We also have length:
1
width:
dy
area: mass:
2=
2  Y 2
2y
dA =dy 2
2y
dm = IldA = 3'dy 2
_
distance of c.m. to yaxis:
Y
y+2
x = 4'
350
Chapter 6: Applications of Definite Integrals
The moment of the sttip about the yaxis is
:tdm
=
y+2 4
0
3
2y 2dy
0
3 8(4  y2)dy.
=
The moment of the plate about the yaxis is
My
=
J :tdm
f~(4 
=
y2)dy
=
H4Y  y;]:
 y) dy
=
1
=
H~6) =
=
1(4  2)
2gocm.
(b) The plate's mass:
M
=
J
dm
=
( ' 1(2
102
[2 
y2]2
2Y 20
(c) The xcoordinate of the plate's center of mass: _ My 2g ocm x= M = 3g
2
=
3 g.
2
=3 cm.
By a similar computation, we could fmd Mx and y.
•
If the distribution of mass in a thin, flat plate has an axis of symmetry, the center of mass will lie on this axis. If there are two axes of symmetry, the center of mass will lie at their intersection. These facts often help to simplify our work. y 4
EXAMPLE 2
Find the center of mass ofa thin plate covering the region bounded above
by the parabola y = 4  x 2 and below by the xaxis (Figure 6.51). Assume the density of the plate at the point (x, y) is 8 = 2x 2 , which is twice the square of the distance from the point to the yaxis.
Solution The mass distribution is symmetric about the yaxis, so x = O. We model the distribution of mass with vertical strips since the density is given as a function of the varible x. The typical vertical strip (see Figure 6.51) has the following relevant data.
center of mass (c.m.): FIGURE 6051 Modeling tire plate in Example 2 with vertical sttips.
(:t, y)
=
(x, 4
~ x2)
length: 4  x 2 width: ax area: ciA = (4  x 2 ) ax mass: dm = 8 ciA = 8(4  x 2 ) ax _ 4  x2 distance from c.m. to xaxis: y=2The moment of the strip about the xaxis is
y dm
=
4  x2 8  2  08(4  x 2) ax = 2(4  x 2)2ax.
The moment of the plate about the xaxis is
Mx=JYdm= =
('06x 2
12

f2~(4_x2)'ax=
12
8>;4
+ x 6 ) ax
=
f\2(4x2),ax
12
2C::58 1
6.6
Moments and Centers of Mass
351
Therefore,
_ Y
Mx 2048 15 8 M = 105 • 256 = 7'
=
The plate's center of mass is
• Plates Bounded by Two Curves Suppose a piate covers a region that lies between two corves y = g(x) and y = f(x), where f(x) 2: g(x) and a :5 x :5 b. The typical vertical s1rip (see Figure 6.52) has
y
"y
=
g(x)
I I I
I
I I I
~L~tc~~x
o
a
dx
(X, y) = (x,! [[(x) + g(x)]) f(x)  g(x)
center of mass (c.m.): length: width: area: mass:
ax dA dm
= =
[[(x)  g(x)] ax Il dA = Il[[(x)  g(x)]
ax.
b
The moment of the plate about the y·axis is FIGURE 6.52 Modeling the plate bounded by _ curves with vertical s1rips. The s1rip
c.m. is halfway, so 'ji
=
t
[f(x)
My
J l xdm
=
+ g(x)].
b
=
xll[[(x)  g(x)]
ax,
and the moment about the x·axis is
Jy l l2 dm
Mx =
b
=
a
Il
b
=
t [[(x) + g(x)]'Il[[(x)  g(x)] ax
[f(x)  g2(x)]
ax.
These moments give the formulas
lib
X= M
a
Ilx [[(x)  g(x)]
ax
(6)
(7)
EXAMPLE 3 Find the center of mass for the thinpiate bounded by the curves g(x) = x/2 and f(x) = v;"O :5 x :5 I, (Figure 6.53) using Equations (6) and (7) with the density function Il(x) = x 2. Solutton We first compute the mass of the plate, where dm = Il[[(x)  g(x)] ax:
M
=
t 10
x2(v;, 
~) ax = t 2
10
(r'/2 _x2
3 )
ax
=
[2.x7/2 _ Ix,]! =~. 7 8 0 56
352
Chapter 6: Applications of Definite Integrals Then from Equations (6) and (7) we get X = 56
9
=
y
[x2.{Vx ~) (X7/2 _ ~4) dx
56 [ 9
= 56
dx
[2. x9/2 _
9 9
...Lx5]! 10
0
308
405'
and
~~~>x
o
1
252 FIGURE 6.53
405'
The region in Example 3.
•
The center of mass is shown in Figure 6.53.
Centroids When the density function is constant, it cancels out of the numerator and denominator of the formulas for x and y. Thus, when the density is constant, the location of the center of mass is a feature of the geometry of the object and not of the material from which it is made. In such cases, engineers may call the center of mass the centroid of the shape, as in "Find the centroid of a triangle or a solid cone." To do so, just set {; equal to 1 and proceed to fmd x and y as before, by dividing moments by masses. y A typical small segment of wire has
tim
~
a tis ~ aad8.
a
a (a)
EXAMPLE 4 Find the center of mass (centroid) of a thin wire of constant density {; shaped like a semicircle of radius a.
Va
length:
y
mass: distance ofc.m. tox·axis:
a c.m.
a
ds = a dO dm = {; tis = {;a dO y = a sin O.
Mass per unit length times length
Hence,
(0. ~a)
0
x
2 2 (Figure 6.54). The We model the wire with the semicircle y = distribution of mass is symmetric about the y·axis, so x = O. To find y, we imagine the wire divided into short subarc segments. If (:t, y) is the center of mass of a subarc and 0 is the angle between the x·axis and the radial line joining the origin to (:t, y), then y = a sin 0 is a function of the angle 0 measured in radians (see Figure 6.54a). The length ds of the subarc containing (:t, y) subtends an angle of dO radians, so tis = a dO. Thus a typical subarc segment has these relevant data for calculating y:
Solution
a
(b)
FIGURE 6.54 The semicircular wire in Example 4. e1ed by the ",,'s cen1roid during the revolution. If p is the distance from the axis of revolution to the centroid, then
s=
21TpL.
(11)
355
6.6 Moments and Centers of Mass
The proof we give assumes that we can model the axis of revolution as the xaxis and the arc as the graph of a continuously differentiable function of x.
y
Proof We draw the axis of revolution as the xaxis with the arc extending from x = a to x = b in the f"ITSt quadrant (Figure 6.59). The area of the surface generated by the arc is S=
o
l
x~b
lx~b
27T)1 tis = 2".
x a
Y tis.
(12)
x a
The ycoordinate of the arc's centroid is FIGURE 6.59
_ E:bytls E:bytls
Figure for proving
Pappus's Theorem for surface area. The arc lenglh differential tis is given by Equation (6) in Section 6.3.
y =
l
x b
tis
L
=
L
= J ds is the arc's
length and
y ~ y.
x=a
Hence
E:bytls
=
YL.
Substituting yL for the last integral in Equation (12) gives S = 2"'YL. With p equal to y, = 2".pL. •
we have S
EXAMPLE 8
Use Pappus's area theorem tof"md the surface area of the torus in Example 6.
Solution From Figure 6.57, the surface of the torus is generated by revolving a circle of radius a about the zaxis, and b ;" a is the distance from the centroid to the axis ofrevolution. The arc length of the smooth curve generating this surface of revolution is the circumference of the circle, so L = 2".a. Substituting these values into Equation (II), we f"md the surface area of the torus to be
• Exercises 6.6 10. a. The region cut from the frrst quadrant by 1he circle x 2
Thin Plates with Constant Density In Exercises 112, find the cenrer of mass ofa thin plate of coostant deosity B covering the given region.
1. The region bounded by the parabola y = x 2 and the line y = 4 2. The region bounded by the parabola y = 25  x 2 and the x·axis 3. The region bounded by the parabola y = x  x 2 and the line y = x 4. The region enclosed by the parabolas y = x 2
3 and y =  2x 2
+ y2
= 9
b. The region bounded by 1he x·axis and the semicircle
y=
vI9=7
Compare your answer in part (b) wi1h 1he answer in part (a). 11. The "ttiangular" region in 1he ftrst quadrant between the circle x 2 + y2 = 9 and 1he lines x = 3 and y = 3. (Hint: Use geometry to fmd the area.)
s.
12. The region bounded above by 1he curve y = I/x', below by the curve y = I/x', and on 1he left and right by 1he lines x = I and x = a > 1. Also, f"md lima_ oo j.
7. The region bounded by the x·axis and the curve y = cos x,
Thin Plates with varying Density 13. Find the center of mass of a thin plate covering the region between the x·axis and 1he curve y = 2/X2, I '" x s 2, if the

The region bounded by the yaxis and the curve x = y  y', O"'y'" I 6. The region bounded by the parabola x = y2  Y and the line y = x
"./2 '"
x"''''/2
8. The region betweeo 1he curve y = sec' x, "./4 s x s "./4 and the xaxis 9. The region bounded by the parabolas y = 2x 2
y=2xx 2

4x and
plate's deosity at 1he point (x,y) is B(x) = x 2.
14. Find the cenrer of mass of a thin plate coveriog the region bounded below by 1he parabola y = x 2 and above by 1he line y = x if1he plate's deosity at the point (x,y) is B(x) = l2x.
356
Chapter 6: Applications of Definite Integrals
±4/Yx
15. The region bounded by the curves y = and the lines x = I and x = 4 is revolved about the yaxis to generate • solid. L
25. Variable density Suppose that the density of the wire in Example 4 is 8 = hin6 (kconstant). Find the center of mass.
16. Variable density Suppose that the density of the wire in Exam
Find the volume of the solid.
b. Find the center of mass of. thin plate covering the region if the plate's density at the point (x, y) is 8(x) = I/x. c. Sketch the plate and show the center of mass in your sketch.
16. The region between the curve y = 2/x and the xaxis fium x to x = 4 is revolved about the xaxis to generate a solid
=I
a. Find the volume of the solid.
ple4is8 = I
Yx.
c. Sketch the plate aod show the center of mass in your sketch.
Centroids of Triangles 17. The centroid of a triangle lies at the intersection of the triangle's medians You may recall that the point inside a triaogie that lies acethird of the way from esch side toward the opposite vertex is the point where the triaog\e's three medians interse O.
18. (1,0), (I , 0), (0, 3) 19. (0,0), (I, 0), (0, I) 10. (0,0), (a, 0), (0, a) 21. (0,0), (a, 0), (0, b) 11. (0,0), (a , 0), (a/2 , b)
Thin Wires 23. Constant deoslty Find the moment about the xaxis of a wire of constant density that lies along the curve y = from x = 0 tox = 2.
Yx
14. Constant density Find the momeot about the xaxis of a wire of constant deosity that lies along the curve y
tox = I.
= x'
from x
=0
The Theorems of Pappus 33. The square region with vertices (0, 2), (2, 0), (4, 2), and (2, 4) is revolved about the xaxis to generate a solid. Fted the volume and surface area of the solid. 34. Use a theorem of Pappus to fmd the volume generated by revolving about the line x = 5 the triangular region bounded by the coordinate axe. and the line 2x + Y = 6 (see Exercise 17). 35. Find the volume of the torus generated by revolving the circle (x  2)' + y' = I about the yaxis.
Chapter 6 Practice Exercises 36. Use the theorems of Pappus to fmd the lateral surface area and the volume of a rightcircular cone. 37. Use Pappus's Theorem for sorface area and the fact that the surface area of a sphere of radius a is 4'1Ta 2 to fmd the centroid of the semicircle y ~ Va 2  x 2 • 38. As found in Exercise 37, the centroid of the semicircle y ~ Va 2  x 2 lies at the pcint (0, 2a/lr). Find the area of the sorface swept out by revolving the semicircle about the line y
~
a.
39. The area of the region R enclosed by the semiellipse y ~ (b/a)Va 2  x 2 and the xaxis is (I/2)'lTab, and the volume of the ellipsoid generated by revolving R about the xaxis is (4/3)'lTab 2 . Find the centroid of R. Notice that the location is independent of a. 40. As found in Example 7, the centroid of the region enclosed by the xaxis and the semicircle y = Va 2  x2 lies at the point (0, 4a/3'IT). Find the volume of the solid generated by revolving this region about the line y ~ a.
357
42. As found in Exercise 37, the centroid of the semicircle y ~ Va 2  x 2 lies at the pcint (0, 2a/'IT). Find the area of the sorface generated by revolving the semicircle about the line y = xa. In Exercises 43 and 44, use a theorem of Pappus to fmd the centroid
of the given triangle. Use the fact that the volume of a cone of radius r and height h is V ~ ~ 'lTr2h. 43.
y (0, b)
~~~~x (0,0) (a, 0)
44.
y
(a, c)
41. The region of Exercise 40 is revolved about the line y = x  a to generate a solid. Find the volume of the solid.
Chapter
!
Questions to Guide Your Review
1. How de you define and calculate the volumes of solids by the method of slicing? Give an example. 2. How are the disk and waaber methods for calculating volumes derived from the method of slicing? Give examples of volume calculations by these methods. 3. Describe the method of cylindrical sbells. Give an example. 4. How do you fmd the length of the graph of a smooth function over a closed interva1? Give an example. What about functions that do not have continuous flISt derivatives? 5. How do you defme and calculate the area of the sorface swept out by revolving the graphofa smooth functiony ~ f(x), a :5 x:5 b, about the xaxis? Give an example.
Chapter
[j
6. How de you defme and calculate the work done by a variable force directed along a pcrtion of the xaxis? How do you calculate the work it takes to pomp a liquid from a tank? Give examples. 7. How de you calculate the force exerted by a liquid against a portion of a flat vertical wall? Give an example. 8. What is a center of mass? a centroid? 9. How do you locate the center of mass ofa thin flat plate ofmatrial? Give an example. 10. How de you locate the center of mass of a thin plate bounded by _ curves y ~ f(x) andy ~ g(x) over a :5 x :5 b?
Practice Exercises
Volumes Find the volumes of the solids in Exercises 116.
the solid perpendicular to the xaxis are equilateral triangles whose bases stretch from the line to the curve.
1. The solid lies between planes perpendicular to the xaxis at x ~ 0 and x ~ 1. The crosssections perpendicu1ar to the xaxis between these planes are circular disks whose diameters run from the parabola y ~ x 2 to the parabola y ~ Yx.
3. The solid lies between planes perpendicular to the X f(XI) when X2 > Xl, is onetoone and has an inverse. Decreasing functions also have an inverse. Functions that are neither increasing nor decreasing may still be onetoone and have an inverse, as with the function f(x) = l/x for X oF 0 and f(O) = 0, defined on (  00, (0) and passing the horizontal line test.
Finding Inverses The graphs of a function and its inverse are closely related. To read the value of a function from its graph, we start at a point X on the xaxis, go vertically to the graph, and then move horizontally to the yaxis to read the value of y. The inverse function can be read from the grapb by reversing this process. Start with a point y on the yaxis, go horizontally to the graph of y = f(x), and then move vertically to the xaxis to read the value of x = r'(y) (Figure 7.2). y
y
~~~~.x
o
X DOMAINOF
o
f
X RANGE OF
11
r'
(b) The graph of is the graph off, but with x and y interchanged. To find the x that gave y. we start at y and go over to the curve and down to the xaxis. The domain off1 is the range off, The range off1 is the domain off.
(8) To find the value off at x, we start at X, go up to the curve, and then over to the yaxis.
y
y ~r'(x)
__1__~O~/~__~~______~y
o~~~~x
/
/ / / / /
/ /
(c) To draw the graph ofr' in the
(d) Then we interchange the letters x andy. We now have a normallooking graph ofjl as a function of x.
more usual way. we reflect the
system. across the line y = x.
r'
FIGURE 7.2 Detennining tlie graph ofy ~ (x) from tlie graph ofy = j(x). The graph of is obtaioed by reflecting the graph of j about the line y ~ x.
r'
r
We want to set up the graph of I so that its input values lie along the xaxis, as is usually done for functions, rather than on the yaxis. To achieve this we interchange the x and y axes by reflecting across the 45° line y = x. After this reflection we have a new graph that represents The value of r'(x) can now be read from the graph in the usual way, by starting with a point x on the xaxis, going vertically to the graph, and then horizontally
r'.
364
Chapter 7: Transcendental Functions to the yaxis to get the value of r'(x). Figure 7.2 indicates the relationship between the The graphs are interchanged by reflection through the line y = x. graphs of I and Although this geometric treatment is not a proof, it does make reasonable the fact that the inverse of a one·to·one continuous function defmed on an interval is also continuous. The process of passing from I to 1' can be summarized as a twc>step procedure.
r' .
y
1.
Solve the equation y = I(x) for x. This gives a formula x = r'(y) where x is ex· pressed as a function ofy.
2.
Interchange x and y, obtaining a formula y = r'(x) where r' is expressed in the conventional format with x as the independent variable and y as the dependent variable.
y~2x2
EXAMPLE 3
Find the inverse of y =
~ x + I, expressed as a function ofx.
Solution 1.
2. FIGURE 7.3 Graphingf(x) = (1/2)x + I and r'(x) = 2x  2 together shows the graphs' symmetry with respect to the line y = x (Example 3).
Solve for x in terms ofy:
Interchange x and y:
y =
y =
2y
=
x
=
1 2x +
x
I
+2
2y  2. 2x  2.
The inverse of the function I(x) = (1/2)x + I is the function r'(x) = 2x  2. (See Figure 7.3.) To check, we verify that both composites give the identity function:
r'(f(x»
=
2(~X + I)  2 =
I(r'(x»
=
~(2x
 2) + I
=
+2
x
 2 = x
x  I + I
x.
=
•
y
EXAMPLE 4
Find the inverse of the function y = x 2 , X
;;"
0, expressed as a function
ofx.
Solution
We flISt solve for x in terms of y:
y = x2
Ixl ~~~X
FIGURE 7.4 The functions y = Vi xl, X ~ 0, are inverses of one another (Example 4).
because x
2:
0
We then interchange x andy, obtaining
,0
and y =
= x
y=
Yx.
The inverse of the functiony O,isthefunctiony = Yx (Figure 7.4). Notice that the function y = x 2 , X ;;" 0, with domain restricted to the nonnegative real numbers, is one·to·one (Figure 7.4) and has an inverse. On the other hand, the func· tion y = x 2, with no domain restrictions, is not one·to·one (Figure 7.lb) and therefore has = x 2,x;;"
•
no inverse.
Derivatives of Inverses of Differentiable Functions If we calculate the derivatives of I(x) = (1/2)x + 1 and
its inverse r'(x) = 2x  2
from Example 3, we see that
~/(x)=~(~x+I)=~ ~r'(x)
=
~(2x
 2)
=
2.
365
7.1 Inverse Functions and Their Derivatives
The derivatives are reciprocals of one another, so the slope of one line is the reciprocal of the slope of its inverse line. (See Figure 7.3.) This is not a special case. Reflecting any nonhorizontal or nonverticalline across the line y = x always inverts the line's slope. If the original line has slope m oF 0, the reflected line has slope I/m.
The slopes are reciprocal: (r') '(b) ~ _I_or (f') '(b) ~ 1'(0)
IV
I '(b»
FIGURE 7.5 The graphs of inverse functions hsve reciprocal slopes at corresponding points.
r
The reciprocal relationship between the slopes of f and 1 holds for other functions as well, but we must be careful to compare slopes at corresponding points. If the slope of y = f(x) at the point (a,f(a»isj'(a) andj'(a) oF O,thentheslopeofy = (x) at the point (f(a), a) is the reciprocal 1/ j'(a) (Figure 7.5). Ifwe set b = f(a), then
r'
(r ' ),(b)
=
j'~a)
= j'(f \b))'
r'
If y = f(x) has a horizontal tangent line at (a, f(a» then the inverse function has a vertical tangent line at (f(a), a), and this wmite slope implies that is not differentiable at f(a). Theorem I gives the conditions under which is differentiable in its domain (which is the same as the range of f).
r'
r'
THEOREM IThe Derivative Rule for Inverses If f has an interval I as domain and j' (x) exists and is never zero on I, then 1 is differentiable at every point in its domain (the range of f). The value of(r y at a point b in the domain of ' isthereciprocalofthevalueofj' at the point a = r'(b):
r
r'
(f  1)'(b) _
I  j'(f '(b))
(i)
or
dr' I ax
x b 
_l_
dfl ax
x  r'(b)
Theorem 1 makes two assertions. The trrst of these has to do with the conditions under which f 1 is differentiable; the second assertion is a formula for the derivative of
366
Chapter 7: Transcendental Functions
I , when it exists. While we omit the proof of the first assertion, the second one is proved in the following way:
I(r'(x))
=
x
Inverse function relationship
fx I(r'(x))
=
I
Differentiating both sides
j'(r'(x))' ~ r'(x) dx d
dx
= 1
Chain Rule
1'( ) _
I
Solving for the derivative
x  j'(j '(x»"
Thefunction/(x) = x 2 ,x;" oand its inverse r'(x) = Yxhavederiva· tivesj'(x) = 2xand(r'),(x) = 1/(2Yx). Let's verify that Theorem I gives the same fannula for the derivative of r'(x):
EXAMPLE 5
(r'),(x)
j'(j \x))
=
1'(x) ~ 2xwithxreplaced by r1(x)
I
ccfL!~____!"~ x
o
1
2
3
4
FIGURE 7.6 The derivative of r'(x) = Vi at the point (4, 2) is the reciprocal of the derivative of f(x) = x 2 at (2, 4) (Example 5).
Theorem I gives a derivative that agrees with the known derivative of the square root function. Let's examine Theorem 1 at a specific point. We pick x = 2 (the number a) and 1(2) = 4 (the value b). Theorem I says that the derivative of I at 2, j'(2) = 4, and the derivative of at 1(2), )'(4), are reciprocals. It states that
r'
(r'
I _ I (/ ')'(4)  j'(j I '(4)) _ j'(2)  2x
I.2_4'I
•
See Figure 7.6.
y
6
y=x 3 
(2, 6)
2 Slope 3.1;2 ~ 3(2)2 ~ 12
We will use the procedure illustrated in Example 5 to calculate fannulas for the derivatives of many inverse functions throughout this chapter. Equation (I) sometimes enables us to find specific values of dr'/dx without knowing a formula for r'.
Let I(x) = x 3 finding a fannula for r'(x).
EXAMPLE 6
Reciprocal slope:
lz
L _ J    0, and suppose that j has a differentiable inverse, Revolve about the yaxis the regioo bounded by the graph of j and the lines x = a aod y = j(b) to geoerate a solid. Then the values of the integrals giveo by the washer and shell methods for the volume have ideotica1 values:
r' .
i
f
(') 1T«r'(y»'  a') t(y =
f(a)
l'
21TX(f(b)  j(x)) tIx.
To prove this equality, defme
i
e. Plot the functions j and g, the ideotity, the two tangoot lines, and the line segmoot joining the points (xo, j(xo» and (f(xo), xo). Discuss the symmetries you see across the main diagonal.
 3x+2 62 ,y2x11'
2< _ x O.
EXAMPLE 1
(2)
We use Equation (2) to fmd derivatives.
dId I I (8) dx In 2x = 2x dx (2x) = 2x (2) = x' (b) Equation (2) with u = x 2
iL ln (x 2 dx
+ 3)
=
x
>0
_1_.iL(x 2 x 2 +3 dx
+ 3)
+ 3 gives =
_1_. 2x =~. x 2 +3 x 2 +3
•
Notice the remarkable occurrence in Example la. The function y = In 2x has the same derivative as the function y = Inx. This is true of y = In bx for any constant b, provided that bx > 0:
dI d Inbx = '(bx) dx bxdx If x < 0 and b < 0, then bx b = I we get
> d
=
I (b) bx
I
=.
0 and Equation (3) still applies. In particular, if x
dx In (x)
=
1
X
for x
0 and x > 0, the natural logarithm satisfies the following rules:
b
1. Product Rule:
Inbx = Inb + Inx
2. Quotient Rule:
b Inx = Inb  Inx
3. Reciprocal Rule:
I Inx = Inx
Rule 2 with b
4. Power Rule:
Inx T = rlnx
For r rational
=
1
For now we consider ouly rational exponents in Rule 4. In Section 7.3 we will see that the rule bolds for all real exponents as well.
EXAMPLE 2 (a) 1n4 + In sin x = In (4 sin x)
Product
x + I (b) 1n2x _ 3 = In(x + I)  1n(2x  3)
Quotient
(e) Int = inS
Reciprocal
= In23 = 3ln2
•
Power
We now give the proof of Theorem 2. The properties are proved by applying Corollary 2 of the Mean Value Theorem to each of them. Proof that In bx = In b + In x have the same derivative:
The argument starts by observing that In bx and In x
d bid dx In (bx) = bx = x = dx In x. According to Corollary 2 of the Mean Value Theorem, the functions must differ by a constant, which means that
Inbx = Inx + C for some constant C. Since this last equation holds for all positive values of x, it must hold for x = I. Hence, In(b'l) = Inl + C Inb=O+C
lnl~O
C = Inb.
By substituting we conclude that
Inbx = Inb + Inx.
•
7.2
Proof that Inx'
=
Natural Logarithms
3 73
rlnx (assuming r rational) We use the samederivative argument
again. For all positive values of x,
l"
Eq. (2) with u
=
1 rl rx
General Power Rule for derivatives, r rational
=
r·k =
" In x T = T (x T ) dx x dx x'
=
x7
fx (rlnx).
Since In x' and r In x have the same derivative, Inx T = rlnx
+C
for some constant C. Taking x to be I identifies C as zero, and we're done. (Exercise 46 in Section 3.7 indicates a proof of the General Power Rule for derivatives when r is rational.) You are asked to prove Rule 2 in Exercise 86. Rule 3 is a special case of Rule 2, obtained by setting b = I and noting that In I = O. This covers all cases of Theorem 2. • y
We have not yet proved Rule 4 for r irrational; however, the rule does hold for all r, rational or irrational. We will show this in the next section after we def'me exponential functions and irrational exponents.
1 1 ~
+______
______
~~
__
2
O~+1~2~~x
1.
< xifx > 1.
71. Find the area between the curves y = Inx and y = In 2x from x=ltox=5.
72. Find the area between the curve y = tanx and the xaxis from x = 1T/4tox = 1T/3. 73. The region in the first quadrant bounded by the coordinate "'"'s, the line y = 3, and the curve x = 2/vy+! is revolved about
the yaxis to generate a solid. Find the volume of the solid. 74. The region between the curve y = ~ and the xaxis from x = 1T/6 to x = 1T/2 is revolved about the xaxis to generate a solid. Find the volume of the solid. 75. The region between the curve y = l/x2 and the xaxis from x = 1/2 to x = 2 is revolved about the yaxis to generate a solid. Find the volume of the solid.
76. In Section 6.2, Exercise 6, we revolved about the yaxis the region between the curve y = 9x/W+9 and the xaxis from x = 0 to x = 3 to generate a solid of volume 361T. What volume do you get if you revolve the region about the xaxis instead? (See Section 6.2, Exercise 6, for a graph.) 77. Find the lengths of the following curves.
a. y = (x 2/8)  Inx, 4 '" x '" 8 b. x = (y/4),  21n (y/4), 4 '" y '" 12
7.3 Exponential Functions 78. Find a curve through the point (I, 0) whose length from x = I to x
[~I + :,dx.
L =
D 79.
D 85.
2is
=
I to x
=
=
a. Derive the linearization 10 (1
2. Give the coordinates to
b. Sketcb the region and sbow the centroid in your sketch.
+ lox is oneroone.
D 87• ••
84.
d'y
dx' =
I
I
+ x'
D 88.
y(1)
=
3
sec' x, y(O)
=
0
aod y'(O)
+ sinx) for a
= 2,
I
Does the graph of y = Yx  lox, x> 0, have an inflection point? Try to answer this question (a) by graphing, (b) by using calculus.
Having developed the theory of the function In x, we introduce its inverse, the exponential function exp x = eX. We study its properties and compute its derivative and integral. We prove the power rule for derivatives involving general real exponents. Finally, we introduce general exponential functions, aX, and generallogaritbmic functions, log.,x.
y y = Inix
or 7
Graph y = sinxaod the curves y = In (a 4, 8, 20, and 50 together for 0 ,;; x ,;; 23.
Exponential Functions
7.3 8
=
O.
b. Why do the curves flatten as a increases? (Hint: Find an a 0
Chain Rule for e", Eq. (2)
=
In short, whenever x
For x < 0, if y
x"·!! x
Definition and derivative of 1n x
> 0,
=
x n, y' , and X,,l all exist, then
lnlyl
=
lnlxl'
n lnlx!
=
Using implicit differentiation (which assumes the existence of the derivative y') and Equation (4) in Section 7.2, we have
Y' y
'!
X'
Solving for the derivative, y'
=
y
nX
=
x'
nx
=
nx n t .
382
Chapter 7: Transcendental Functions It can be shown directly from the defInition of the derivative that the derivative equals 0 wben x = 0 and n ;;,: I. This completes the proof of the general version of the Power Rule for all values of x. •
EXAMPLE 4
> O.
Differentiate /(x) = xx, x
Solution We cannot apply the power rule here because the exponent is the variable x mther than being a constant value n (mtional or irrational). However, from the def"mition of the general exponential function we note !bat /(x) = XX = exlnx , and differentiation gives /,(x) =
~ (e xInX ) dx
=
e xlnx ~ (x In x) dx
=
e Xlnx ( Inx
=
xX(lnx + I).
Eq. (2) with u
~
.In.
+ x.:})
•
.>0
The Number e Expressed as a Limit We have def"med the number e as the number for which In e = I, or equivalently, the value exp (I). We see that e is an important constant for the logarithmic and exponential functions, but what is its numerical value? The next theorem shows one way to calculate e as a limit. The number e can be calculated as the
THEOREM 4The Number. as a Um;t
limit e = lim (I • 0
Proof If /(x)
=
+ x)l/x .
Inx, then /,(x) = I/x, so /,(1) = 1. But, by the def"mition of derivative,
/,(1) = lim /(1 hO
=
lim In (I
xo =
lim In (I
xo
+ h)  /(1)
=
lim /(1
h
xo
+ x)
 In I = lim
xo
X
+ x)l/x
=
In [lim(1
xo
+ x)
 /(1)
x
lin (I + x)
Inl~O
X
+ x)l/x]
In is continuous;
use Theorem lOin Chapter 2.
Because /,(1) = I, we have In [lim(1
.0
+ X)I/x]
=
I
Therefore, exponentiating both sides we get lim (I + x)l/x .0
=
e.
•
Approximating the limit in Theorem 4 by taking x very small gives approximations to e. Its value is e '" 2.718281828459045 to 15 decimal places as noted before.
The Derivative of aU To f"md tbis derivative, we start with the defIning equation aX = exlna . Then we have
383
7.3 Exponential Functions
We now see why eX is the exponential function preferred in calculus. If a = e, then In a = I and the derivative of aX simplifies to
With the Chain Rule, we get the following form for the derivative of the general exponential function.
If a > 0 and u is a differentiable function of x, then a" is a differentiable function ofx and
%x a"
= a"lna :,.
(3)
The integral equivalent of this last result gives the general antiderivative
y
=
lOx y
=
3x
y~ F ,....'
1
J
(4)
From Equation (3) with u = x, we see that the derivative of aX is positive if In a > 0, or a > I, and negative if In a < 0, or 0 < a < I. Thus, aX is an increasing function of x if a > I and a decreasing function of x if 0 < a < I. In each case, aX is onetoone. The second derivative d2 d (aX) = (axIna) = (InaJ2a x dx 2 dx
1
FIGURE 7.11 Exponential functions decrease if 0 < a < 1 and increase if a> 1.Asx+00,wehaveax+Oif O 1.As x+ 00, we have aX + 00 if 0 < a < 1 and aX + 0 if a > 1.
a" + C.
a"du = Ina
=2x
is positive for all x, so the graph of aX is concave up on every interval of the real line. Figure 7.11 displays the graphs of several exponential functions.
EXAMPLE 5 (a)
~3x
(b)
~rx
dx dx
We fmd derivatives and integrals using Equations (3) and (4).
= 3x ln3
=
rX(ln3)~(x) dx
=
rxln3
d· . d . (e) dx 30nx = 3,mx(ln3) dx (sinx) = 3,mx(ln3) cosx
(d) (e)
J J
2 x dx =
~x2 + C
2 .... cosxdx =
Eq.(3)wi1ha
~ 3,u ~x
... ,u
~
2"du =
2 sinx
= In2
+
C
~"2
+
C
u
~
x
~
,inx
Eq.(4)wi1ha
J
3,u
Eq.(3)wi1ha
~ 2,u ~ x
~ sinx,du ~
COSXtU,
u replaced by sin x
andEq. (4)
•
Logarithms with Base Q If a is any positive number other than I, the function aX is onetoone and has a nonzero derivative at every point. It therefore has a differentiable inverse. We call the inverse the logarithm ofx with base a and denote it by log.x.
384
Chapter 7: Transcendental Functions
DEFINITION
For any positive number a
'¢'
I,
log"x is the inverse function of aX.
~=;;of,fc:.~~~~~x
The graph of y = log"x can be obtained by reflecting the graph of y = aX across the 45° line y = x (Figure 7.12). When a = e, we have logex = inverse of eX = Inx. (The function 10glO x is sometimes written simply as log x and is called the common logarithm of x.) Since log. x and aX are inverses of one another, composing them in either order gives the identity function.
FIGURE 7.12 Thegraphof2x andits inverse, log2 x.
Inverse Equations for aX and log.x
a log.,x = x
(x> 0)
log" (aX) = x
(all x)
The function log" x is actually just a nmnerical multiple of In x. To see this, we let Y = log"x and then take the natural logarithm of both sides of the equivalent equation
a'
=
x to obtain Y In a
= In x. Solving for y gives
Inx log"x = Ina'
TABLE 7.2
Rules for base a
Logarithms For any nmnbers x
> 0 and
y> 0, 1.
2.
3.
Product Rule: log" xy = log" x
+ log" y
The algebraic rules satisfied by log. x are the same as the ones for In x. These rules, given in Table 7.2, can be proved using Equation (5) by dividing the corresponding rules for the natural logarithm function by In a. For example, Inxy = Inx
+ Iny
Inxy Inx Ina = Ina
+
log" xy = log" x
Iny Ina
+
log" y.
Rule 1 for natura110garitbms ... . .. djvided by In a ...
. .. gives Rule 1 for base a logarithms.
Quotient Rule: log" ~ = log" x  log" Y
Derivatives and Integrals Involving log. x
Reciprocal Rule:
To f"md derivatives or integrals involving base a logarithms, we convert them to natural logarithms. If u is a positive differentiable function of x, then
I log" y = log" Y 4.
(5)
Power Rule: log" x' = Y log" x
d d (lnu) I d I I du tU (Iog"u) = tU Ina = Ina tU (lnu) = lna'll tU'
d I I du (Iog"u) = .tU In a u tU
7.3
Exponential Functions
385
EXAMPLE 6 d
(a) dx log1O(3x
+
lid
1) = In 10 • 3x
+
r°'6)X dx = ~2J In/ dx
(b)
=
~2
J
U
I dx (3x
u
3
+
1) = (In 1O)(3x
1)
~ :;
log,x
du
+
~ lnx,
du
~ ~tU
I u2 I (lnxj2 (lnxj2 =ln2T+ C =ln22+ C = 2ln2 +C
•
Exercises 7.3 Solving Exponential Equations
35.
In Exercises 14, solve for I.
1. a.
e()·3t
=
~
2. a. e.....011
b.e kt
27
1000
b. e
tt
I
=
c.
e(lnO.2)t =
~
c.
e(Jn2)'
2
/0
~
0.4
t
37.
39. 41.
Finding Derivatives In Exercises 524, rmd the derivative of y with respect to x, I, or 8, as appropriate. 6. Y ~ e2>/3 5. y ~ 8. y = e(4v'i+x1) 7. y = e5  7:r:
e"
10. y ~ (1
9.y=xe x e x 11. Y ~ (x 2

2x + 2)e'
13. y ~ e'(sin 8 + cos 8) 15. Y ~ cos (e6')
+ 2x)e2>  fu + 2)e"
16. y ~
~ In (3tel)
18. Y
~ In (2e' sin I)
19. Y
~ In ("') 1 + eO
20. y
~ inC :~)
21. y
= e(coIHlnt)
22. y ~ e""(InI 2 + I)
23. y
~ /,'.. sin e' dl
In Exercises 2528, find 25. Iny
~
eY sinx
27. e2> ~ sin (x
+ 3y)
24. Y
~
l.
Inldl
26. In xy = e X +Y 28. !any
~
e'
+ Inx
Finding Integrals
31.
f l f
(e"
+ 5e~) fix
30.
8e«+I) fix
J1n2
34.
f
38'] e;;
21 e" dt
40.
f V/r f f e'l'
42.
7fix
44.
f f e~o(~+t) l f
dr
I'e(") dl eII" ;>fix
J.~j2 (1 + e""') csc2 8 d8
e sec frt sec 7ft tan wI dt
esc (."
+ I) cot (." + I) dl
2evcosevdv
48.
In ('11'/6)
49.
f f
fix
~/4
ln(wj2)
47.
44
/,v.;;,:
2 2 2xeX cos(eX )dx
0
e' I + e' dr
50.
f
fix I + e'
Initial Value Problems Solve the initial value problems in Exercises 5154.
51. :;;
~ e'sin(e' 
52. :;;
~ elsec2 (."e'),
d 2y
53. fix2 = 2e',
d 2y
2
dl
2), y(ln2)
0
y(ln4) = 2/."
y(O) = I
= I  e 2t ,
~
aod y'(O) ~ 0
y(l) = I
aod y'(I) ~ 0
(2e'  3e 2» fix
32. ('
1n2
33.
f
in,
e' fix
46.
54.
Evaluate the integrals in Exercises 2!150.
29.
45.
e
eV;
'b
e4Vi.
dy/ fix.
lln4
r /4 (1 + e""') see> 8 d8
lI'e29 cos 58
17. y
36. 10
43. 10
12. y ~ (9x 2
14. y ~ In (38e')
[m 16
{In' e,j2 fix
e~ fix
2e (2)1) fix
Differentiation In Exercises 55112, rmd the derivative of Y with respect to the given independent variable. 55.y~2'
56.y=3~
57. Y ~ 5V;
58. Y =
2(")
386
Chapter 7: Transcendental Functions Logarithmic Differentiation
59. y = x"
60. y = t l 
61. y = (cos8)v2
62. y = (ln8)"
63. y = 7sec 81n 7 65. y = 2sin3t
64. y = 3tan 8 In 3
67. y = IOg258
68. y = IOg3(l
66. y =
69. y = IOg4x + IOg4x2 71. y = x 3 10g lO x 73. y = log3 (
(~ ~
U
l n3)
e
In Exercises 111 118, use logarithmic differentiation to find the derivative of y with respect to the given independent variable. 111. y = (x + I)X 112. y = x 2 + x2x
5  cOS 21
+ 81n 3)
113. y = (VI)t
114. y = t Vi 116. y = x sinx 118. y = (lnx)lnx
70. y
=
IOg25 eX  IOg5vX
115. y = (sinx)X
72. y
=
log3 r· log9 r
117. y = sinxx
r;,;~
(3}~ 2 r 5
74. y = IOg5
Theory and Applications 119. Find the absolute maximum and minimum values of f(x) = eX  2x on [0, I].
75. y = 8 sin (lOg7 8)
76
77. y
858 78. y = 2  IOg5 8
10glO eX
=
= I
(sin8cos8) Og7 e 8 28
• y
79. y = 310g ,t
80. y = 3 logg (lOg2 t)
81. y = IOg2 (8t ln2 )
82. y = t IOg3 (e(Sin t)(ln 3))
120. Where does the periodic function f(x) = 2e sin (x/2) take on its extreme values and what are these values? y
~,
Integration Evaluate the integrals in Exercises 83 92.
83.
J
SX dx
84.
J ~x 3
121. Let f(x) = xe x.
a. Find all absolute extreme values for f.
3x dx
b. Find all inflection points for f. eX 122. Letf(x) = 2x' I +e a. Find all absolute extreme values for f.
(I3')
r,,/4
90.
Jo
92.
J
tan 1
b. Find all inflection points for f. sec2 t dt
where it is assumed.
x2x'
D 124.
,dx I + 2x
Evaluate the integrals in Exercises 93 106.
J 95.1 J i
3x v'3 dx
93.
94.
3
(0 + I)x v1dx
10glOx xdx
97.
103.
Jo
+ 2) + 2 dx 2 10glO (x + I) x + I dx
105.
x
r
xdx
126. Find the area of the "triangular" region in the first quadrant that is bounded above by the curve y = ex / 2 , below by the curve y = e  x/2 , and on the right by the line x = 2 In 2 .
21n 10 log 10 X x dx
127. Find a curve through the origin in the xyplane whose length from x = 0 to x = I is
i i 1 1 J I
r210g2 (x
Jo
J
10
102.
log 10 (lOx) x dx
L =
11)1 +
:teXdx .
1/10
104.
3 2 IOg2 (x  I) x _ I dx
2
106.
dx x log 10 X
Graph f(x) = (x  3)2e x and its first derivative together. Comment on the behavior of f in relation to the signs and values of f' . Identify significant points on the graphs with calculus, as necessary.
125. Find the area of the "triangular" region in the first quadrant that is bounded above by the curve y = e 2x , below by the curve y = eX, and on the right by the line x = In 3 .
410g2X e
100.
I
101.
x v2 1dx
I
41n210g2X x dx
99.
J
96·iexOn2)  ldx
98.
123. Find the absolute maximum value of f(x) = x 2 1n (I/x) and say
128. Find the area of the surface generated by revolving the curve x = (e Y + e Y )/2,0 ::s y ::s In 2, about the yaxis. eY + eY
dx x(logg x)2
X=
2
Evaluate the integrals in Exercises 107 110.
107. 109.
i i
lnx
I
I
l /X
I [dt,
x
I [dt,
x > 0
> I
108.
iex I I [dt
I 110. Ina
i
I
X
I [dt,
x> 0
x
387
7.4 Exponential Change and Separable Differential Equations In Exercises 129132, find the length of each curve. 129. y = t(e X
+ e~)fromx
130. y = In (eX  I) In(e'
137. Find the area of the region between the curve y = Zr/(l and the interval 2 ~ x ::s; 2 of the xaxis.
= Otox = I
+
I) from x = 1n2tox = 1n3
+ c.
graphing.
o 140. Could
Xln2 possibly be the same as 21n' for two functions and explain what you see.
b. Find the average value oflnx over [I, e].
134. Find the average value of f(x) = I/x on [I, 2]. 135. Thelinearizalion or.x at x = 0
a. Derive the linear approximation eX
~
1
+ x at x
= O.
x
eX
x :5 x :5 colors, if availahle. On what intervals does the approximation appear to overestimate eX? Underestimate eX?
136. The geometric, logarithmic, and arithmetic mean inequality
a. Show that the graph of e' is concave up over every interval of xvalues. b. Show, by reference to the accompanying figure, that if 0< a < bthen
e(ln.+lnb)/2'(lnb Ina)
O? Graph the
141. Thelinearizalion ofl'
o b. Estimate to five decimal places the magnitude of the error involved in replacing by I + on the interval [0, 0.2]. o c. Graph and I + together for  2 2. Use different eX
2
= 2' bas three sclutions: x = 2, x = 4, and one other. Estimate the third solution as accurately as you can by
132. y = In (esc x) from x = 'IT/6 to x = 'IT/4 133. a. ShowthatJlnxdx =xlnx x
138. Find the area of the region between the curve y = 2 1 x and the interval 1 :5 x :5 1 of the xaxis.
o 139. The equation x
131. y = In (cos x) from x = Otox = 'IT/4
+ x 2)
a. Find the linearization of f(x)
= 2' at x = O. Then round its coefficients to two decin2a1 places.
o b. Graph the linearization and function together for 3:5 3andl 1. x:5
:5 x:5
142. Thelinearizalion oflog, x
a. Find the linearization of f(x) = log, x at x its coefficients to two decimal places.
=
3. Then round
o b. Graph the linearization and function together in the
window
O:5x:5 8and2:5x :54.
o 143. Which is bigger, 'IT' or e ? W
Calculators have taken scme of the mystery out of this oncechallenging question. (Go ahead and check; you will see that it is a surprisingly close call.) You can answer the question without a calculator, though.
&.
Find an equation for the line through the origin tangent to the graph of y = In x.
E
B,
__~A~:Ina____~__~~__~~__~,x Ina+lnb
2
b. Give an argument hased on the grapha of y = In x and the tangent line to explain why In x < x/e for all positive x #' e.
c_ Show that In (x')
NorTOSCALB
c. Use the inequality in part (b) to conclude that
.rc
ba a+b vab < Inb  Ina < 2'
This inequality says that the geome1ric mean of two positive numbers is less than their logarithmic mean, which in turn is less than their arithmetic mean.
7.4
[3,6] by [3, 3]
d. Conclude that xt'
< x for all positive x #' e. < eX for all positive x =F e.
e. So which is bigger, ",e or e""?
o 144. A decimal representalion of.
Find e to as many decimal places as your calculator allows by sclving the equation Inx = I using Newton's method in Section 4.6.
Exponential Change and Separable Differential Equations Exponential functions increase or decrease very rapidly with changes in the independent variable. They describe growth or decay in many natural and industrial situations. The variety of models based on these functions partly accounts for their importance. We nOW investigate the basic proportionality assumption that leads to such exponential change.
388
Chapter 7: Transcendental Functions
Exponential Change In modeling many realworld situations, a quantity y increases or decreases at a rate proportional to its size at a given time I. Examples of such quantities include the amount of a decaying radioactive material, the size of a population, and the temperature difference between a hot object and its surrounding medium. Such quantities are said to undergo exponential change. If the amount present at time I = 0 is called Yo, then we can find y as a function of t by solving the following initial value problem:
dy dl
Differential equation:
y = Yo
Initial condition:
(la)
= ky
when
t
= O.
(lb)
Ify is positive and increasing, then It is positive, and we use Equation (Ia) to say that the rate of growth is proportional to what has already been accumulated. If y is positive and decreasing, then k is negative, and we use Equation (Ia) to say that the rate of decay is proportional to the amount still left. We see right away that the constant function y = 0 is a solution of Equation (Ia) if Yo = O. To find the nonzero solutions, we divide Equation (Ia) by y: Idy
y"O
y'dl=k
J~:dl=Jkdl In
Iyl Iyl Iyl
=
Integrate with respect to t;
kl + C
= ekt+c
J(1ju) du ~ In lui + C. Exponentiate.
= eC'e kt
= ±eCekt y = Ae kt .
y
If Iyl ~
T, tbeny
~
±T.
A is a shorter name for ±e c ,
By allowing A to take on the value 0 in addition to all possible values ±ec, we can include the solution y = 0 in the formula. We fmd the value of A for the initial value problem by solving for A when y = Yo and
1=0:
Yo
=
Ae k ' O = A.
The solution of the initial value problem is therefore y = yoe kt .
(2)
Quantities changing in this way are said to undergo exponential growth if It > 0, and exponential decay if k < O. The number k is called the rate constant of the change. The derivation of Equation (2) shows also that the only functions that are their own derivatives are constant multiples of the exponential function. Before presenting several examples of exponential change, let's consider the process we used to derive it.
Separable Differential Equations Exponential change is modeled by a differential equation of the form dy/dx = ky for some nonzero constant k. More generally, suppose we have a differential equation of the form
dy dx = f(x,y),
(3)
7.4 Exponential Change and Separable Differential Equations
389
where f is a function of both the independent and dependent variables. A solution of the equation is a differentiable function y = y(x) defined on an interval of xvalues (perhaps infmite) such that d
dx y(x) = f(x,y(x»
on that interval. 1bat is, wheny(x) and its derivative y'(x) are substituted into the differential equation, the resulting equation is true for all x in the solution interval. The general solution is a solutiony(x) that contains all possible solutions and it always contains an arlJitrary constant. Equation (3) is separable if f can be expressed as a product of a function of x and a function of y. The differential equation then has the fonn dy dx = g{x)H(y).
g is a function of x; H is a function of y.
When we rewrite this equation in the form dy dx
g(x) h(y) ,
I H(y) ~ hey)
its differential fonn allows us to collect all y terms with dy and all x terms with dx: h(y) dy = g(x) dx.
Now we simply integrate both sides of this equation:
J
h(y) dy =
J
(4)
g(x) dx.
After completing the integrations we obtain the solution y defmed implicitly as a function
ofx. The justification that we can simply integrate both sides in Equation (4) is based on the Substitution Rule (Section 5.5):
J
h(y) dy =
=
=
EXAMPLE 1
J J J
h(y(x)) : dx g(x) h(y(x)) h(y(x)) dx
ely
g(x)
;2
~ = yer' + 2vY er'
I
dy
20. fix=xy+3x2y6
27. Working underwater The intensity L(x) oflightx feet beneatlt the surface of the ocean satisfies the differential equation
tiL = eX  Y
+ eX +
0.6y
26. The inversion of sngar The processing of raw sugar has a step called ''inversion'' that changes the sugar's molecular s1ructw'e. Once the process has begun, the rate of chaoge of the amount of raw sugar is proportional to the ao30Unt of raw sugar remaining. If 1000 kg ofraw sugar reduces to 800 kg ofraw sugar during the 'IrSt 10 hours, how much raw sugar will remain after aoother 14 hours?
= e Y+';"
I
dy
Differential equation: Initial condition:
to express p in tenns of h. Determine tlte values of po aod k from tlte given altitudepressure data.
x>O 8. x"y' = xy  y2,
24. Atmospheric pressure The earth's atmospheric pressme p is often modeled by asson3ing that the rate dp/dh at which p chaoges witlt the altitode h above sea level is proportional to p.
candidate
2
6. y' = e"""  2xy 7. xy'
Solution
Initial condition
equation
theconditionthaty = 0.99yowhent = 1000 to find the value of k in the equation y = yoe it . Then use this value of k to answer the following questions.
b. In about how many years will humao teeth be 90% of their present size?
3
vT+7dt, y'
23. Human evolution continues The aoalysis of tooth shrinkage by C. Loring Brace and colleagues at the University of Michlgao's Musenm of Anthropology indicates that humao tooth size is continuing to decrease aod that tlte evolutionary process did not come to a halt some 30,000 years ago as many scientists contend In northern Emopeans, for example, tooth size reduction now has a rate of I % per 1000 years.
eY
+1
Applications and Examples
D The answers to most of the following exercises are in terms of logarithms aod exponentials. A calculator can be helpful, enabling you to express the aoswers in decimal form.
fix
=
AL.
As a diver, you know from experience that diving to 18 ft in tlte Caribbean Sea cuts tlte intensity in half. You caonot work witltout artificial light when the intensity falls below on.,.tenth of the surface value. About how deep can you expect to work witltout artificiallight?
7.4 Exponential Change and Separable Differential Equations 28. Voltage in • discharging capacitor Suppose that electricity is draiuiug from a capacitor at a rate that is proportional to the voltage V across its tenninals and that, if t is measured in seconds,
Solve this equation for V, using Vo to denote the value of V when t = O. How long will it take the voltage to drop to 10% of its original value? 29. Chol.ra b.ct.ria Suppose that the bactetia in a colony can grow unchecked, by the law of expouential cbange. The colony starts with I bacterium aod doubles every ba1fhour. How many bacteria will the colony contain at the end of24 hours? (Under fa
vorable laboratory conditions, the number of cholera bacteria can double every 30 min. In ao infected petlion, many bacteris are destroyed, but this example helps explain why a person whn feels well in the morning may be dangerously ill by evening.) 30. Growth of bacteria A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 hours there are 10,000 bacteria. At the end of 5 hours there are Ml,OOO. How many bacteria were present initially?
dis....
31. Th. incidence of. (Continuation of Example 3.) Suppose that in any given year the number of cases can be reduced by 25% instead of 20%.
a. How long will it take to reduce the number of cases to 1000? b. How long will it take to eradicate the disease, that is, reduce
the number of cases to less than I?
u.s.
32. Th. population The US. Census Bureau keeps a running clock totaling the US. popu1ation. On March 26, 2008, the tota1 was increasing at the rate of I petlion every 13 sec. The population figure for 2:31 P.M. EST on that day was 303,714,725.
a. Assuming exponential growth at a constaot rate, fmd the rate constaot for the population's growth (people per 365day year).
b. At this rate, what will the US. population be at 2:31 P.M. EST on March 26, 2015? 33. Oil depl.tion Suppose the amount of oil pumped from one ofthe canyon wells in Whittier, California, decreases at the contiouous rate of 10% per year. When will the well's output fall to on.,.fifth
ofits present value? 34. Continuous price discounting To encourage buyers to place 100unit ordetli, your firm's sales department applies a continuous discouot that makes the unit price a functionp(x) of the number of units x ordered. The discount decreases the price at the rate of $0.0 I per unit ordered. The price per unit for a lOOunit order is p( 100) = $20.09.
a. Find p(x) by solving the following initial value problem: Differential equation: Initial condition:
dp I dx = lOOP p( 100)
=
20.09.
b. Find the unit price p(1 0) for a 10unit order and the uoit price P(90) for a 90unit order. c. The sales department has asked you to find out if it is discouoting so much that the firm's revenue, r(x) = X' p(x) , will actually be less for a 100unit order than, say, for a
395
90unit order. Reassure them by showing that r has its maximum value atx = 100. d. Graph the reveoue function r(x) = xp(x) for 0 :5 x :5 200. 35. Plutonium239 The balflife of the plutooium isotope is 24,360 years. If 10 g of plutonium is released into the atmosphere by a nuclear accident, how many years will it take for 80% of the is()tope to decay? 36. Polonium210 The balflife of polonium is 139 days, but your sample will not be useful to you after 95% of the radioactive nuclei preseot on the day the sample arrives bas disintegrated. For about how many days after the sample arrives will you be able to
use the polonium? 37. Th. mean life of. radio.ctive nucleus Physicists using the radioactivity equation y = yoe kt call the number I/k the mean life
of a mdioactive nucleus. The mean life of a radon nucleus is about 1/0.18 = 5.6 days. The meao life ofa carbon14 nucleus is more than 8000 years. Show that 95% of the radioactive nuclei origiually preseot in a sample will disintegrate within three meao lifetimes, i.e., by time t = 3/k. Thus, the meonlife ofanucleus gives a quick way to eslimste how long the radioactivity of a sample will last. 38. Californiuno252 What costs $27 million per gram and can be used to treat brain cancer, analyze coal for its sulfur content, and detect explosives in luggage? The answer is californium252, a radioactive isotope so rare that only 8 g of it bave been made in the western world since its discovery by Gleno Seaborg in 1950. The ba1flife of the isotope is 2.645 yearsIong enough for a us.,. ful service life and short enough to bave a high radioactivity per unit mass. One microgram of the isotope releases 170 million
neutrons per minute. a. What is the value of k in the decay equation for this isotope?
b. What is the isotope's meao life? (See Exercise 37.) c. How long will it take 95% of a sample's radioactive nuclei to disintegrate? 39. Cooting .oup Suppose that a cup of soup cooled from 90'C to 60°C after 10 min in a room whose temperature was 20D e. Use Newton's law of cooling to aoswer the following questions. a. How much longer would it take the soup to cool to 35'C?
b. Instead ofbeing left to staod in the room, the cop of90'C soup is put in a freezer whose temperature is 15°C. How long will it take the soup to cool from 90'C to 35'C? 40. A beam of unknown temperatore An aluminum beam was brought from the outside cold into a machine shop where the temperature was held at 65'F. After 10 min, the beam warmed to 35'F and after another 10 min it was 50'F. Use Newton's law of cooling to estimate the beam's initial temperature. 41. Surrounding medium of unknown temperatore A pan of warm water (46'C) was put in a refiigerator. Ten minutes later, the water's temperature was 39'C; 10 min after that, it was 33'C. Use Newton's law of cooling to estimate how cold the refiigerator was. 42. SiIv.r cooling in air The temperature of an ingot of silver is 60'C above room temperature right now. Twenty minutes ago, it
was 70D e above room temperature. How far above room temperature will the silver be
•• 15 min from now?
b. 2 hours from now?
c. When will the silver be lODe above room temperature?
396
Chapter 7: Transcendental Functions
43. The age of Crater Lake The charcoal from a tree killed in the volcanic eruption that formed Crater Lake in Oregon cootained 44.5% of the carbon14 found in living matter. About how old is Crater Lake? 44. The sensitivity of carbon14 dating to measurement To see the effect of a relatively sma1l error in the estimate of the amount of carbon14 in a sample being dated, consider this hypothetical situation: L A fossilized bone found in centrallllinois in the year A.D. 2000 cootains 17% ofits original carbon14 cootent. Estimate the year the animal died.
45. Carbon14 The oldest known frozen humao mummy, discovered in the Schoalslal glacier of the Italian Alps in 1991 and called Olzi, was found wearing straw shoes and a leather coal with goat fur, and holdiog a copper ax and stone dagger. It was estimated that Otzi died 5000 years before he was discovered in the melting glacier. How much of the originsl carboo14 remained in Otzi al the time of his discovery? 46. Art forgery A painting attributed to Vermeer (16321675), which should cootain no more than 96.2% of its originsl carboo14, contains 99.5% instead. Aboulhow old is the forgery?
b. Repeat part (a) assuming 18% instead of 17%.
c. Repeat part (a) assuming 16% instead of 17%.
7.5 HISTORICAL
Indeterminate Forms and L'Hopital's Rule
BIOGRAPHY
Guillaume Fran~ois Antoine de I'H6pital (\6611704) Johano Bernoulli (16671748)
John (Joharm) Bernoulli discovered a rule using derivatives to calculate limits of fractions whose numerators and denominators both approach zero or + 00 . The rule is known today as I'HopitaI'. Rule, after Guillaume de I'HOpitai. He was a French nobleman who wrote the first introductory differential calculus text, where the rule 1mt appeared in print. Limits involving transcendental functions often require some use of the rule for their calculation.
Indeterminate Form 0/0 If we want to know how the function
F(x) = x  sinx x3 behaves near x = 0 (where it is undefined), we can examine the limit of F(x) as x + O. We carmol apply the Quotient Rule for limits (Theorem I of Chapter 2) because the limit of the denominator is O. Moreover, in this case, both the numerator and denominator approach 0, and % is unde1med. Such limits may or may not exist in general, but the limit does exist for the function F(x) under discussion by applying I'HOpital's Rule, as we will see in Example I d. If the continuous functions f(x) andg(x) are both zero at x = a, then lim f(x) x~a
g(x)
cannot be found by substitoting x = a. The substitotion produces 0/0, a meaningless expression, which we carmot evaluate. We use % as a notation for an expression known as an indeterminate form. Other meaningless expressions often occur, such as 00/00, 00 00, 0, 00  00, 0°, and 1 , which cannot be evaluated in a consistent way; these are called indeterminate forms as well. Sometimes, but not always, limits that lead to indetermiuate forms may be found by cancellation, rearrangement of terms, or other algebraic manipulations. This was our experience in Chapter 2. It took considerable analysis in Section 2.4 to find limx~o (sinx)/x. But we have had success with the limit
'( ) f a
=
Ii
m
x~a
f(x)  f(a) x a '
7.5
Indeterminate Forms and rH6pitars Rule
397
from which we calculate derivatives and which produces the indetenninant fann % when we substitute x = a. I:Hopitai's Rule enables us tu draw on our success with derivatives to evaluate limits that otherwise lead to indetenninate forms.
THEOREM 5 L'H6Jrital's Rule Suppose 1hatj(a) = g(a) = 0, thatj and g are differentiable on an open interval I containing a, and that g' (x) # 0 on !if x # a. Then lim j(x) g(x)
x~a
=
lim f'(x) x~a
g'(x) ,
assuming that the limit on the right side of this equation exists. We give a proof of Theorem 5 at the end of this section.
I
Caution
To apply I'H6pital's RnIe to Ilg, divide 1he derivative of I by 1he derivative of g. Do not fall into 1he trap of taking 1he detivative of IIg. The quotient to use is ng', not (f!g)'.
EXAMPLE 1 The following limits involve % indeterminate forms, so we apply I'HOpitai's Rule. In some cases, it must be applied repeatedly. (8) lim 3x  sinx = lim 3  cosx = 3  cosxl = Z xo X xo 1 1 x=o
1 (b) lim
xo
(c) lim
yIJ+x  I
=
x
yIJ+x 
lim
xo
zyIJ+x 1
I 2
1  x/2
o
2
xo
o
X
=
=
xr'/ xo 2x . (1/4)(1 + xr lim lim (I/Z)(I +
2

32 /
xa
2
I/Z
Still ~; differentiate again.
I 8
=
o
(d) lim x  sinx
xo
Not~; limit is found
o
x3
=
lim 1  cosx
xo =
Still Q
o
3x 2
lim sinx
xo 6x
Not~;limitisfound Here is a summary of the procedure we followed in Example 1. Using UHopital's Rule Tofmd
lim j(x) xa g(x)
by I'HOpital's Rule, continue to differentiate j and g, so long as we still get the form % at x = a. But as soon as one or the other of these derivatives is different from zero at x = a we stop differentiating. I:HOpitai's Rule does not apply
when either the numemtor or denominator has a fmite nonzero limit.
•
398
Chapter 7: Transcendental Functions
EXAMPLE 2
Be careful to apply I'Hopital's Rule correctly:
o o
lim I  cosx
xo =
X+X2
. sin x ;~ I + 2x
0
=
T = O.
Not ~; limit is found.
Up to now the calculation is correct, but if we continue to differentiate in an attempt to apply I'HOpital's Rule once more, we get
!
lim cosx =
xo 2
2'
wbich is not the correct limit. I:HOpital's Rule can ouly be applied to limits that give indeterminate forms, and 0/ I is not an indeterminate form. • I:HOpital's Rule applies to onesided limits as well.
EXAMPLE 3
In this example the onesided limits are different.
o o
I Recall that thing.
00
Positive for x > 0
and + 00 mean the same
o
. sin x (b) Iu n 2
o
xo x =
lim cosx xo 2x
Indeterminate Forms
=
00
•
Negative for x < 0
ot:J / ot:J, ot:J •
0,
ot:J 
ot:J
Sometimes when we try to evaluate a limit as x + a by substitoting x = a we get an indeterminant form like 00/00, 00·0, or 00  00, instead of 0/0. We first consider the form
00/00. In more advanced treatments of calculus it is proved that I'HOpital's Rule applies to the indeterminate form 00/00 as well as to 0/0. Iff(x) + ±oo andg(x) + ±oo asx + a, then lim fix) = lim f'(x) x~.
g(x)
x~.
g'(x)
provided the limit on the right exists. In the notation x + a, a may be either finite or infinite. Moreover, x + a may be replaced by the onesided limits x + a + or x + a .
EXAMPLE 4 (a)
lim
x~"/2 I
Find the limits of these 00/00 forms: secx
(b) lim
+ tanx
xOO
lnx
2Vx
• eX (c) lun 2'
xoo x
Solution O+
00 • 0
=
lim xO+
=
EXAMPLE 6
lim
xo+
I/x 1/2x 3/ 2
(2Vx)
I/x.
converted to
00/00
I'H8pital's Rule
•
= 0
Find the limit of this
00 
00
lim ( _ I xa sinx
Solution
~
form:
_1)
x·
Ifx+ 0+, then sinx+ 0+ and
I  ~oo I .smx x
00
Similarly, if x + 0, then sin x + 0 and
I  .I   + smx x
00 
(00) = 00
+
00.
Neither form reveals what happens in the limit. To find out, we flfst combine the fractions: I sin x
I

x
x  sinx x sin x
Common denominator is x sin x.
Then we apply I'HOpital's Rule to the result: lim
(_._1__ 1) =
xa smx
x
lim x  .sinx
xa x smx
= lim
I  cosx
xa sinx + xcosx sin x = Q= 0 x_o2cosxxsinx 2 .
= lim
o o Sti1I()
o
•
400
Chapter 7: Transcendental Functions
Indeterminate Powers Limits that lead to the indeterminate fonns 100, 0°, and 00° can sometimes be bandied by first taking the logarithm of the function. We use I'Hllpital's Rule to find the limit of the logarithm expression and then exponentiate the result to find the original function limit. This procedure is justified by the continuity of the exponential function and Theorem lOin Section 2.5, and it is fonnulated as follows. (The fonnula is also valid for onesided limits.)
Iflimx~a
In j(x) = L, then lim j(x) = lim elnf(x) = eL. x+a
xa
Here a may be either finite or infinite.
EXAMPLE 7
Apply I'Hllpital's Rule to show thatlimx~o+ (I
+ x) I/x
= e.
The limit leads to the indeterminate fonn 100. We let j(x) = (I find limx~o+ In j(x). Since
Solution
In j(x) = In (I
+ x)I/X
+ x)I/X
and
i In (1 + x),
=
I'Hllpital's Rule now applies to give
+ x)
lim In j(x) = lim In (I
xo+
xo+
X
0 0
I I +x Iim  xo+ 1
=+=1. Therefore, lim (I x+o+
EXAMPLE 8
+ x)l/x =
lim j(x)
x+o+
=
lim elnf~)
xo+
= el =
e.
•
Find limx~oo xl/x.
Solution
The limit leads to the indeterminate fonn 00°. We let j(x) = xl/x and f"md lim_oo In j(x). Since
~x,
Inj(x) = Inxl/x =
I 'Hllpital's Rule gives lim In j(x) = lim Inx x_OO x
x_OO
=
=
Therefore lim xl/x x+ 00
=
lim j(x) x+ 00
=
00
I/x lim
x+oo
¥
=
lim elnf(x) x 00
00
1
o. =
eO
= I.
•
Proof of L'Hiipital's Rule The proof of I'H6pital's Rule is based on Caucby's Mean Value Theorem, an extension of the Mean Value Theorem that involves two functions instead of one. We prove Caucby's Theorem first and then show how it leads to I'H6pital's Rule.
7.5
Indeterminate Forms and rH6pitars Rule
401
HisTORICAL BIOGRAPHY
AugustinLouis Cauchy (178!11857)
THEOREM 6Cauchy's Mean Value Theorem Suppose functions f and g are continuous on [a, bl and differentiable throughout (a, b) and also suppose g'(x) oF 0 throughout (a, b). Then there exists a number c in (a, b) at which
f'(c) g'(c)
f(b)  f(a) g(b)  g(a)"
Proof We apply the Mean Value Theorem of Section 4.2 twice. First we use it to show that g(a) oF g(b). For ifg(b) did equalg(a), then the Mean Value Theorem would give g'(c)
=
g(b)  g(a) ba
=
0
for some c between a and b, which cannot happen because g'(x) oF 0 in (a, b). We next apply the Mean Value Theorem to the function
F(x)
=
f(b)  f(a) f(x)  f(a)  g(b) _ g(a) [g(x)  g(a)].
This function is continuous and differentiable where f and g are, and F(b) = F(a) = O. Therefore, there is a number c between a and b for which F' (c) = O. When expressed in terms of f and g, this equation becomes
F'(c)
=
f'(c) 
:~:~ =~~~ [g'(c)1 =
0
so that
f'(c) g'(c)
y j '(c) slope ~,«c) 
.!
A (g(a).j(a)) ~~X
o
FIGURE 7.13 There is at least one point P on the curve C for which the slope of the tangent to the curve at P is the same as the slope of the secaot lioe joioing the poiots A(g(a), f(a)) aodB(g(b), f(b)).
f(b)  f(a) g(b)  g(a)"
•
Notice that the Mean Value Theorem in Section 4.2 is Theorem 6 with g(x) = x. Cauchy's Mean Value Theorem has a geometric interpretation for a general winding curve C in the plane joining the two points A = (g(a),/(a)) and B = (g(b), f(b)). In Chapter 11 you willleam how the curve C can be formulated so that there is at least one point P on the curve for which the tangent to the curve at P is parallel to the secant line joining the points A andB. The slope of that tangent line toms out to be the quotient f' /g' evaluated at the number c in the interval (a, b), as asserted in Theorem 6. Because the slope of the secant line joining A andB is
f(b)  f(a) g(b)  g(a)' the equation in Cauchy's Mean Value Theorem says that the slope of the tangent line equals the slope of the secant line. This geometric interpretation is shown in Figure 7.13. Notice from the figure that it is possible for more than one point on the curve C to have a tangent line that is parallel to the secant line joining A and B.
Proof of rHClpital's Rule We flfst establish the limit equation for the case x > a +. The method needs almost no change to apply to x > a , and the combination of these two cases establishes the result. Suppose that x lies to the right of a. Then g'(x) oF 0, and we can apply Cauchy's Mean Value Theorem to the closed interval from a to x. This step produces a number c between a and x such that f'(c) g'(c)
f(x)  f(a) g(x)  g(a)"
402
Chapter 7: Transcendental Functions But f(a) = g(a) = 0, so
j'(c) g'(c)
f(x) g(x)"
As x approaches a, c approaches a because it always lies between a and x. Therefore, lim f(x) = lim j'(c) = lim j'(x)
.ra+ g(x)
ca+ g'{c)
.ra+ g'(x) ,
which establishes I'HOpital's Rule for the case where x approaches a from above. The case where x approaches a from below is proved by applying Cauchy's Mean Value Theorem to the closed interval [x, aj, x < a. •
Exercises 7.5 Finding Limits in Two Ways In Exercises l{j, use I'HOpital's Ru1e to evaluare the limit. Then evaluare the limil using a method studied in Chapter 2.
. 3x  I 30. lun~1 xo~
31. lim In(x + I)
33. lim
3. lim 5x'  3x x_OO 7x 2 + 1
%0+
35. lim
5. lim I  cosx xo x 2
8.
t2
13

+
15 t  12
11. lim 5x'  2x x_OO 7x 3 + 3
,
.
13. lim sm 1 10
15.
t
8x '
lim=~.
xo cosX'
lim 20  'If · '~w/' cos (2'1f 0)
17
19
lim
I  sinO • 61T/2 1 + cos 26
,
10. lim
' 3"'"  I lun 27. ,~o 0
41
r
+ I))
(lnx)'
x~+ In (sinx) lim (_1___1_) xI Inx
• xI+
I'  I
xa
t 3
71'X
In (esc x) x~w/' (x  ('If/2))'
(3X + I__smX' .1_) X
42. lim (cscx  calx xo+
e' 
O
xo x smx
r 30 + 'If . '~'!!;/' sin (6 + ('If/3))
26.
y
38. lim (lnx  In sin x)
48. lun
xl
x? x_tInx  SIn
~a
43. lim cos 0  I 8_0efJ  81 1_00
18
(SInt
(x  ~2) sec x
•
36. lim
y~O
16. lim sinX'  x
24. lim
lim
39
14. lim sin 5/ 10 21
/(1  cos I) 23. lim .
x(7r/2)
25 5
12. lim x  8x' x_OO 12x2 + 5x
22.
25.
+
114t 3 
21 lim x • x~o In (sec x) 10
x' 
lim
x5 x
9. lim I'  41
y
X~OO
log, x
34. lim In (eX  I) xo+ Inx
+ 25  5
37. lim (In 2x  In(x
Use I'H6pital's rule 10 fmd the limits in Exercises 750. 7.lim x  2 x_2x2  4
Y5y
r
• x!'.?o log, (x + 3)
In(x' + 2x) In x
y~O
Applying l'H6pital's Rule
32
lo~x
x_OO
1. lim x+2 x_2x2  4

cost
(~2  x) Ian x
(1/2)"  I 6
57. lim  tanx
X
•• Estimate the value of
lim sin x olnx
%0+
xl
oc  oc Form
X~O+
68. lim
x_OO ~
+ " + sin bX) = O?
lim (tan 2x xa x3
Theory and Applications I:H6pital's Rule does not help with the limits in Exercises 6774. Try ityou just keep on eycling. Find the limits some other way. 67. lim
403
+
I)Yx + 2
1 hy graphing. Then confmn your estimate with I'H6pital's Rule. xI
X
84. This exercise explores the difference between the limit
a. lim x3 = lim 1...=1. x3 x 2  3 x3 2x 6 b. lim x3 =Q=o %3 x 2  3 6
and the limit
76. Which one is correct, and which one is wrong? Give reasons for
. (1+"I)X =e.
your answers.
lim
X~OO
2x  2 a. lim x '2x = l i m x_ox2  sinx xO 2x  cosx = lim x_o2 b.
2
+ sin x
•• Use I'H6pital's Rule to show that
lim x'  2x = lim 2x  2 xO 2x cosx _ sinx
2
.%_ox2
0  1 = 2
others wrong? Give reasons for your answers. a. lim xlnx = 0·(00) = 0
xo+ b. lim xlnx = 0'(00) = 00 xo+
lim Inx 00 c. x'r!1l+ X In X = x~o+ (I/x) = """00 =
lim
2+0
77. Only one of these calculations is correct Which one? Why are the
d.
. (1+"I)X =e.
= _2_ = I

1
X~OO
D b. Graph
f(x)
=
(I +:S
lim (I
.... 00
a. f(x) = x,
g(x) = x',
(a,b) = (2,0)
b. f(x) = x,
g(x)
(a, b) arbitrary
=
c. f(x) = x'/3  4x,
x',
g(x) = x',
(a, b) = (0,3)
79. Continuous OItension Find a value of c thst makes the function f(x) =
{
9X  3sin3x, x # 0 5x'
c,
x= 0
continuous at x = O. Explain why your value of c works.
(I +~)'
85. Show thst
lim x In x = lim In/x X~O+ (I x)
78. Find all values of c thst satisfY the conclusion of Cauchy's Mean Value Theoreru for the given functions and interval.
=
together for x '" O. How does the behavior off compare with thst of g? Estimate the value oflim~oo f(x). c. Confmn your estimate oflimx>oo f(x) by calculating it with I'H6pital's Rule.
X~O+
= lim (I/x) = lim (x) = 0 xo+ (1/x 2) xo+
and g(x)
86. Given that x a. Xl/x
>
1")'
+k
=
e'.
0, fmd the maximum value, if any, of
b. xl/i'
c. xlix' (n a positive integer) d. Show that 1imxoo xl/x· = 1 for every positive integer n. 87. Use limits to find horizontal asymptotes for each function.
•. y=xtan(~)
b. y
= { 0,
3x + e2x "'2x''+'e~3x"
x # 0
e11i'
88. Findf'(O) for f(x)
=
'
x
= O.
404
D 89.
Chapter 7: Transcendental Functions
D 90.
The continuous extension of(sinxf to [0, 'IT[
°
The function (sin x)... • (Continuation ofExercise 89.)
a. Graphf(x) ~ (sinxf on the interval :5 x:5 'IT. What value would you assign to f to make it continuous at x = O?
a. Graphf(x) ~ (sin x)"'" on the interva17 s x '" 7. How do you account for the gaps in the graph? How wide are the gaps?
b. VerifY your conclusion in part (a) by rmding lim,....o· f(x)
b. Now graph f on the interval :5 x '" 'IT. The function is not dermed at x ~ 'IT/2, hut the graph has no break at this point. What is going on? What value does the graph appear to give for f atx ~ 'IT/2? (Hint: Use I'H6pital~ Rule to rmd limf as x ..... ('IT/2) and x ..... ('IT/2)+.)
°
with I'H6pital~ Rule. c. Returning to the graph, estimate the maximwn value off on [0, 'IT]. About where is max f taken on? d. Sharpen your estimate in part (c) by graphing f' in the same window to see where its graph crosses the xaxis. To simplifY your work, you might want to delete the esponential factor from the expression for f' and graph just the factor that has a zero.
Inverse Trigonometric Functions
7.6
Inverse trigonometric functions arise when we want to calculate angles from side measurements in triangles. They also provide useful antiderivatives and appear frequently in the solutions of differential equations. This section shows how these functions are derIDed, graphed, and evaluated, how their derivatives are computed, and why they appear as important antiderivatives.
y x
= siny
2) VDomain: 'IT'
.
Y = sinIx
1';x,,1
Range: wl2 :S y :S 'lT12
_~
c. Continuing with the graphs in part (b), rmd max f and min f as accurately as you can and estimate the values of x at which they are taken on.
1
x
2
FIGURE 7.14 The graph ofy ~ sin1 x.
Defining the Inverses The six basic trigonometric functions are not oneroone (their values repeat periodically). However, we can restrict their domains to intervals on which they are oneroone. The sine function increases from I at x = 1f/2 to + I at x = 1f/2. By restricting its domain to the interval [1f/2, 1f/2], we make it on.,.toone, so that it has an inverse sin 1 x (Figure 7.14). Similar domain restrictions can be applied to all six trigonometric functions.
Domain restrictions that make the trigonometric functions onetoone Y
Y
sin x
= sinx Domain: ['IT/2, 'IT/2] Range: [1,1]
y
= cosx Domain: [0, 'IT] Range: [1, 1]
y
Y
2
i( ~ cotx Domain: (0, 'IT) Range: (00, 00)
= secx Domain: [0, 'IT/2) U ('IT/2, 'IT] Range: (00, 1] U [I, 00)
y
~ tao x Domain: ('IT/2, 'IT/2) Range: (00, 00)
y
sec x
o;;t,,;;;;;,,~x
y
tan x
;;,,;c;cotc,,;;~x
~
2
= cscx Domain: ['IT/2, 0) U (0, 'IT/2] Range: (00, 1] U [1, 00)
y
405
7.6 Inverse Trigonometric Functions
I
Since these restricted functions are now onetoone, they have inverses, which we denote by
The "Arc" in Arcsine and Arccosine The accompanying figure gives a
geometric interpretation ofy = sinI x andy = COSIX for radian angles in the frrst quadrant. For a unit circle, the = r8 becomes s = (J, so cen1ral angles and the arcs they subtend have the same measure. If x = sin y, then, in addition to being the angle whose sine is.:t, y is also the length of arc on the unit circle that subtends an angle whose sine is x. So we call y ''the arc
equation s
whose sine is .:t,"
y = sinIx
or
y = cos t x
or
y
= tanI
x
or
y
= coC I X
or
sec 1 x
or
Y = arccotx y = arcsecx
y = csc I x
or
y
y =
Y = arcsinx Y = arccosx y = arctanx
= arccscx
These equations are read "y equals the arcsine of x" or "y equals arcsin x' and so on. Caution The  1 in the expressions for the inverse means ''inverse.'' It does not mean reciprocal. For example, the reciprocal of sin x is (sinx)I = l/sinx = cscx.
y
x2
+ y2 = 1
Arc whose sine is x
~+
Angle whose sine is x ~~~L~t~x
o
x
1
The graphs of the six inverse trigonometric functions are shown in Figure 7.15. We can obtain these graphs by reflecting the graphs of the restricted trigonometric functions through the line y = x, as in Section 7.1. We now take a closer look at these functions and their derivatives. Domain: 1 =S x =S 1 Range' _1!:s yS 'I!
Domain: 1:S X:S 1 Range: 0 :S Y :S '1r
y
y
.
2
Angle whose
2
Domain: 00 < x < Range:
OQ
I 1 by applying Theorem 1 with f(x) = coshx andr ' (x) = cosh 1 x. Theorem I can be applied because the derivative of cosh x is positive for 0 < x.
(r' )'(x) = !'u I, (x)) 1 sinh (cosh
Theorem 1
1 x)
1 Ycosh2(cosh 1 x)  I
1
1'(u)
~
cosb2 U sinhu
=
sinh u 
sinh2 U
=
I,
v'cosh2 u  1
421
Hyperbolic Functions
7.7 The Chain Rule gives the final result:
HisTORICAL BIOGRAPHY
Sonya Kova1evsky (!8S()1891)
d ( sh 1 ) _ dx co u 
du
1
•
~dx'
With appropriate substitutions, the derivative formulas in Table 7.10 lead tu the integration formulas in Table 7.11. Each of the formulas in Table 7.11 can be verified by differentiating the expression on the righthand side. Integrals leading to inverse hyperbolic functions
TABLE 7.11
1.
1 1V2
du 2
Va + u
2.
3.
5.
a>O
du = sh1 (1l)+c a ' u  a 2 co
1 1V 1 V2 2

a
2
u =
a
u2
'
< a2
u2 > a2
(*) +
u
du =_1 a sechl(Il)+C a ' a2  u 2
O
Integration Fonnulas
1
VetifY the integration formulas in Exercises 3740. 37.
38. 39. 40.
J J J J J L
sechx dt = tanI (sinh x)
+C
b.
sechx dt = sin l (tanh x)
+C
xsechlxdt xcothlxdt tanhlxdt
Use the formulas in the box here to express the ntanhers in Exercises 61{i6 in tenns ofnaturallogatithms. 61. sinh l (5/12) 62. cosh l (5/3)
~ ~ sechlx  k~ + C ~ x' ~
I cothlx
+~+C
~ X tanhI x + kln(1
 x')
+C
63. tanhI (1/2)
64. coth l (5/4)
65. sechl (3/5) Evaluate the integrals in Exercises 6774 in tenns of
a. inverse hyperbolic functions. b. natural logarithms.
7.7
67.
69
•
J.
'v3
v4 +
o
l'
68.
""
dx
1 :7'==='" l/S
73. J.~
xVI 
80. Volume
6 dx
J. VI + 9x' 12 70. J.1 ~ o 1 x
x'
5/41x 2 3/13
71.
113
dx
.~
16>;'
cos x dx
•
l' 1
xVI
The region enclosed by the curve y = sech x, the
V3
81. An: length
Find the lengih of the graph of y = (1/2) cosh 2x
fromx = Otox = In
Vs.
82. Use the definitions of the hyperbolic funetions to fmd each of the following lintits.
1x~
74
(a) lim tanh
dx
lim tanhx
x~oo
+ (lnx)'
(c) lim sinhx
(d)
(e) lim sechx
(f) lim cotbx
x~oo
Appllcattons and Examples 75. Show that if a function f is defmed on an interval symme1ric ahout the origin (so that f is defmed at x whenever it is defined at x), then
_ f(x) + f( x) ( ) fx 2
+
f(x)  f( x) 2
423
xaxis, and the lines x = ± In is revolved about the xaxis to generate a solid. Find the volume of the solid.
0
72·1',=dx===
o VI + sin'x
Hyperbolic Functions
(1)
+ f( x))/2 is even and that (f(x)  f( x))/2 is odd.
Then show that (f(x)
W+i)
76. Derive the fornrula sinh 1 x = In (x + for all real x. Explain in your derivation why the plus sign is used with the square root instead of the minus sign. 77. Skydiving If a body of mass m falling from rest under the action of gravity encounters an air resistance proportional to the square of the velocity, then the body's velocity I sec into the fall satisfies the differential equation
x~oo
lim sinhx
x_oo x~oo
(g) lim coth x
xo+
(i)
lim csch x
x_oo
83. Hanging cable. ln3agine a cahle, like a telephone line or TV cable, strung from one support to another and hanging freely. The cahle's weight per unit lengih is a constant w and the horizontal tension at its lcmest point is a vector of length H. Ifwe choose a c0ordinate system for the plane of the cahle in which the xaxis is horizontal, the force of gravity is straight down, the positive yaxis points straight up, and the lowest point of the cahle lies at the point y = H/w on the yasis (see accompanying fignre), then it can be shown that the cahle lies along the graph of the hypetbolic cosine
H
w
y = wcoshJjx.
dv , mdt=mgkv, where k is a constant that depends on the body's aerodyuamic properties and the density of the air. CWe assume that the fall is short enough so that the variation in the air's density will not affect the outconre significantly.)
a. Showthat
satisfies the differential equation and the initial condition that v = o when I = O. b. Find the body's limiting velocity, Jim,....oo v. c. For a 160lb skydiver (mg = 160), with time in seconds and distance in feet, a typical value for k is 0.005. What is the diver~ limiting velocity'l
Such a curve is sometimes called a chain curve or a catenary, the latter deriving from the Latin catena, meaning "chain." a. LetP(x,y) denote an arbitrary point on the cable. The next accompanying fignre displays the tension at P as a vector of lengih (magnitude) T, as well as the tension H at the lowest pnintA. Show that the cable's slope atP is
dy . w tan = dx = sinhJjx.
78. Accelerations whose magnitndes IIl"O proportionlll to displacement Suppose that the position of a body moving along a coordinate line at time t is
y
y= l1cosh~x w H
a. s = acoskt + bsinkt. b. s = a coshkl + b sinhkl. Show in hoth cases that the acceleration d's/ dl' is proportional to s bot that in the frrst case it is directed toward the origin, whereas in the second case it is directed away from the origin. 79. Volume A region in the flIS! quadrant is hounded shove by the curve y = cosh x, below by the curve y = sinh x , and on the left and right by the yaxis and the line x = 2, respectively. Find the volume of the solid generated by revolving the region ahout the xaxis.
P(X,y)~.....
TcostfJ H
o
424
Chapter 7: Transcendental Functions
b. Using the result from part (a) aod the fact that the horizontal tension at P must equal H (the cable is not moving), show that
T = "y. Hence, the magnitude of the tension atP(x,y) is exactly equal to the weight of y units of cable. 84. (Continuation of Exercise 83.) TIre length of arc AP in the Exercise 83 figure is s = (1/a) sinh ax, where a = wlH. Show that the coordinates ofP may be expressed in terms of s as
x=
~Sinhl as,
y =
a. Show that the areaA(u) of sector AOP is A(u) = tCOshuSinhu
~S2 + :2"
85. Area Show that the area of the region in the fIrSt quadrant enclosed by the curve y = (1/a) cosh ax, the coordinate axes, and the line x = b is the sarae as the area of a rectangle of height 1/a and length s, where s is the length of the curve from x = 0 to x = b. Draw a figure illustrating this result. 86. The hyperbolic in hyperbolic functions Just as x = cos u and y = sin u are identified with points (x, y) on the unit circle, the functions x = cosh u and y = sinh u are identified with points (x, y) on the righthand branch of the unit hyperbola, X2 _
Another artalogy between hyperbolic and circular functions is that the variable u in the coordinates (cosh u, sinh u) for the points of the righthand branch of the hyperbola x 2  y2 = I is twice the area of the sector AOP pictured in the accompanying figure. To see why this is so, carry out the followiog steps.
1°O""~ fix.
b. Differentiate both sides of the equation in part (a) with respect to u to show that A'(u) =
t.
c. Solve this last equation for A(u). What is the value of A(O)? What is the value of the constant of integration C in your solution? With C determined, what does your solution say about the relationship of u to A(u)? y
y 2=1. y
y x2
+
r =1
P(cos U, sin u)
"...,.
~
t~~~x
Since cosh2 U  sinh2 U = I, the point (cosh u, sinh u) lies on the righthand branch of the hyperbola x 2  y2 = I for every value of u (Exercise 86).
7.8
One of the artalogies between hyperbolic and circular functions is revealed by these two diagrams (Exercise 86).
Relative Rates of Growth It is often important in mathematics, computer science, and engineering to compare the rates at which functions of x grow as x becomes large. Exponential functions are important in these comparisons because of their very fast growth, and logarithmic functions because of their very slow growth. In this section we introduce the liUleoh and bigoh notation used to describe the results of these comparisons. We restrict our attention to functions whose values eventoa1ly become and remain positive as x ..... 00 .
y
160 140 120 100
Growth Rates of Functions
80 60 40 20
d~~~~~~x o 1 2 3 4 5 6 7 FIGURE 7.27 andx2 .
The graphs of eX, 2x,
You may have noticed that exponential functions like 2" and eX seem to grow more rapidly as x gets large than do polynomials and rational functions. These exponentials certaiuly grow more rapidly than x itself, and you can see 2" outgrowing x 2 as x increases in Figure 7.27. In fact, as x ..... 00, the functions 2X and eX grow faster than any power of x, even x',ooo,ooo (Exercise 19). In contrast, logarithmic functions like y = log2x and y = lnx grow more slowly as x ..... 00 than any positive power of x (Exercise 21). To get a feeling fur how rapidly the values of y = eX grow with increasing x, think of graphing the function on a large blackboard, with the axes scaled in centimeters. At x = I cm,
7.8
425
the graph is e ' '" 3 em above the xaxis. At x = 6 em, the graph is e 6 '" 403 em '" 4 m high (it is about to go through the ceiliog if it hasn't done so already). At x = 10 em, the graph is e 'O '" 22,026 em '" 220 m high, higher than most buildings. At x = 24 cm, the graph is more than halfway to the moon, and at x = 43 em from the origin, the graph is high enough to reach past the sun's closest stellar neighbor, the red dwarf star Proxima Centauri. By contrast, with axes scaled in centimeters, you have to go nearly 5 lightyears out on the xaxis to find a point where the graph of y = lnx is even y = 43 em high. See Figure 7.28. These important comparisons of exponential, polynomial, and logarithmic functions can be made precise by defining what it means for a function I(x) to grow faster than another function g(x) as x + 00 .
y
y
Relative Rates of Growth
= eX
70 60 50 40 30 20
10 0
Y = lnx 10
20
30
40
50
60
FIGURE 7.28 Scale drawings of the graphs of e' and \0 x.
x
Rates of Growth as x +
DEFINmON
QC)
Let I(x) andg(x) be positive for x sufficiently large.
1.
1 grows faster than g as x + 00 if lim I(x) =
00
x_oo g(x)
or, equivalently, if
.
g(x)
lim I( X ) = O.
x+OO
2.
We also say that g grows slower than 1 as x + 00 . 1 and g grow at the same rate as x + 00 if lim I(x) = L x_oo
g(x)
where L is finite and positive.
According to these defmitions, y = 2x does not grow faster than y = x. The two functions grow at the same rate because lim
2x
x+OO X
=1im2=2 x+ OO
'
which is a finite, positive limit. The reason for this departure from more common usage is that we want "I grows faster than Ft' to mean that for large xvalues g is negligible when compared with f.
EXAMPLE 1
Let's compare the growth rates of several common functions.
(a) eX grows faster than x 2 as x +
e'
Iim 2 x
,.+00
lim eX 2x
=
xco
00/00 (b) 3' grows faster than
00
because
lim eX = 2
00.
x_OO
Using I'H6pital's Rule twice
00/00 2x
as x +
00
because
. 3x
. (3)X 2 =
Inn 2x = lim
xi>OO
xi>OO
00.
426
Chapter 7: Transcendental Functions (e) x 2 grows faster than In x as x >
00
because
2
lim x = lim k = lim 2>;2 = x+OO Inx x+OO l/x x+OO (d) In x grows slower than x '/n as X >
x_oo
j'H6pital's Rule
for any positive integer n because
00
lim Inx = lim x_oo xl/n
00
I/x (I/n)x(l/n)'
j'H6pital's Rule
= lim "= O. x_ co x l / 1I
11
is constant.
(e) As Part (b) suggests, exponential functions with different bases never grow at the same rate as X> 00. If a > b > 0, then a' grows faster than b'. Since (a/b) > I, lim
x_OO
ba: =
(ba )' =
lim
x_OO
00.
>
(f) In contrast to exponential functions, logarithmic functions with different bases a and b > I always grow at the same rate as x > 00 :
I
. log" x . In x/In a Inb hm=hm =10llbX x_OO Inx/lnb Ina'
x_OO
•
The limiting ratio is always finite and never zero.
x >
If 1 grows at the same rate as g as x > 00, and g grows at the same rate as h as 00, then 1 grows at the same rate as h as x > 00. The reason is that lim _gl = L,
x_OO
and
together imply
If L, andL 2 are fmite and nonzero, then so is L,L2 • Show that ~ and (2Vx
EXAMPLE 2

1)2 grow at the same rate asx>
00.
Solution We show that the functions grow at the same rate by showing that they both grow at the same rate as the function g(x) = x:
lim
x_OO
lim
~= lim~l+ x52 =1 ' x x_OO
(2Vx 
xoo
1)2 = lim
x
xoo
(2Vx Vi
1)2 = lim xoo
(2 __1_)2 = Vi
4.
•
Order and OhNotation The "littleoh" and ''bigoh'' notation was invented by number theorists a hundred years ago and is now commonplace in mathematical analysis and computer science.
DEFINITION
A function
1
is of smaller order than g as x >
lim g(/(X» = O. We indicate this by writing f
xOO
x
00
= o(g) ("I is littleoh of g'').
if
7.8
Notice that saying f
=
o(g) as x >
00
Relative Rates of Growth
427
is another way to say that I grows slower than gas
x~oo.
EXAMPLE 3 (a) lox
=
Here we use littleoh notation.
o(x) as x >
(b) x 2 = 0(x 3
00
because
+ I) as x > 00 because
2
lim _x_= 0
x_ oo x 3
+ 1
•
DEFINmON Let I(x) and g(x) be positive for x sufficiently large. Then I is of at most the order of g as x + 00 if there is a positive integer M for which I(x) < M g(x) ,
for x sufficiently large. We indicate this by writing /
EXAMPLE 4
= O(g) ("I is bigoh of g").
Here we use bigoh notation.
(a) x
+ sinx
(b) eX
+ x2
=
=
O(x) as x > O( eX) as x +
00
00
because because
x + sinx . x ,;; 2 for x sufficiently large. eX
+ x2 e
x
~lasx+oo
X. > 0 as x > 00 . • e If you look at the defmitions again, you will see that I = o(g) implies 1= O(g) forfunctions that are positive for x sufficiently large. Also, if I and g grow at the same rete, then I = O(g) and g = OU) (Exercise ll). (c) x = O( eX) as x >
00
because
Sequential vs. Binary Search Computer scientists often measure the efficiency of an algorithm by counting the number of steps a computer must take to execute the algorithm. There can be significant differences in how efficiently algorithms perform, even if they are designed to accomplish the saroe task. These differences are often described in bigoh notation. Here is an exarople. Websters International Dictionary lists about 26,000 words that begin with the letter a. One way to look up a word, or to learn if it is not there, is to read through the list one word at a time until you either fmd the word or determine that it is not there. This method, called sequential search, makes no particular use of the words' alphabetical armngement. You are sure to get an answer, but it might take 26,000 steps. Another way to find the word or to learn it is not there is to go straight to the middle of the list (give or take a few words). If you do not find the word, then go to the middle of the half that contains it and forget about the half that does not. (You know which half contains it because you know the list is ordered alphabetically.) This method, called a binary search, eliminates roughly 13,000 words in a single step. If you do not find the word on the second try, then jump to the middle of the half that contains it. Continue this way until you have either found the word or divided the list in half so many times there are no words left. How many times do you have to divide the list to find the word or learn that it is not there? At most 15, because (26,000/2 15 ) < 1. That certainly beats a possible 26,000 steps.
428
Chapter 7: Transcendental Functions For a list of length n, a sequential search algorithm takes on the order of n steps to find a word or determine that it is not in the list. A binary search, as the second algorithm is called, takes on the order of 10(U n steps. The reason is that if 2m  1 < n ,;; 2m , then m  I < 10(U n ,;; m, and the number of bisections required to narrow the list to one word will be at most m = 10(U n the integer ceiling for 10(U n. Bigoh notation provides a compact way to say all this. The number of steps in a sequential search of an ordered list is O(n); the number of steps in a binary search is O(iog2 n). In OUI example, there is a big difference between the two (26,000 ys. 15), and the difference can only increase with n because n grows faster than log2 n as n > 00.
r
1,
Exercises 7.8 Comparisons with the Exponential e' 1. Which of the following functions grow faster thao eX as x ..... 00 ? Which grow at the same rate as eX? Which grow slower?
3
b. x 3
c. Yx e. (3/2)'
d.4x
LX 
g. e
X
f.
/2
+ 8in2 x
ex/2
h. 10glOX
2. Which of the following functions grow faster thao eX as x ..... 00 ? Which grow at the same rate as eX? Which grow slower?
+ 30x +
L
lOx'
c.
vT+7
I
b. x Inx  x
6. Which of the following functions grow faster than In x as x .... oo? Which grow at the same rate as In x? Which grow slower?
a. logz (x 2 )
b. 10glO lOx
c. I/Yx e. x  2Inx
d. l/x 2 f. ex
g. In (In x)
b. In (2x
Ordering Functions by Growth Rates 7. Order the following functions from slowest growing to fastest growing as x +
d. (S/2)'
+ 5)
a.
00.
OX
c. (lnx)'
Comparisons with the Power i' 3. Which of the following functions grow faster than x 2 as x ..... 00 ? Which grow at the same rate as x 2 ? Which grow slower?
8. Order the following functions from slowest growing to fastest growing as x + 00. a. 2% b. x 2 d.
c. (ln2)'
OX
Bigoh and Litt1eoh; Order 9. True, or false? As x +
e. xInx g.
x 3ex
X
f. 2 h.
&x 2 2
4. Which of the following functions grow faster than x as x ..... 00 ? Which grow at the same rate as x 2 ? Which grow slower? L
x2
+ Yx
b. IOx 2
d. 10glO (x 2 ) f. (1/10)'
g. (1.1)'
h. x 2
+ IOOx
a. x = o(x) c. x = O(x + 5) e. 0% = 0(02%)
b.x=o(x+S)
g. Inx = 0(1n 2x)
b.
10. True, or false? As x +
x! = o(}) c. }  ;tI = o(})
a.
e. eX
5. Which of the following functions grow faster thao In x as x + oo? Which grow at the same rate as In x? Which grow slower? L
log,x
b. In 2x
c. InYx
d. Yx
e. x
f. Slnx
g.
l/x
h. eX
d. x = 0(2x) f. x
+
Inx = O(x)
W+5 =
O(x)
00,
3
2
Comparisons with the Logarithm In x
00,
+x
=
O(e X)
g. In (In x) = O(lnx)
d. 2
+ cosx
= 0(2)
f. x Inx = 0(x 2)
b. In (x) = 0(1n (x 2 + I))
11. Show that ifpositive functions f(x) and g(x) grow at the same rate as x"" 00, thenf= O(g) andg = O(f). 12. When is a polynomial f(x) of smaller order thao a polynomial g(x) as x + oo? Give reasons for your answer. 13. When is a polynomial f(x) of at most the order of a polynomial g(x) as x + oo? Give reasons for your answer.
Chapter 7 Questions to Guide Your Review 14. What do the conclusions we drew in Section 2.4 about the limits of rational functions tell us about the relative growth of polynomi· a1sasx~ oo?
429
Xl/l,ooo,ooo > Inx. You might start by observing that when x > 1 the equation Inx = X 1/1,000,000 is equivalent to the equation In (In x) ~ (lnx)/I,OOO,OOO.
o c. Even x takes a long time to overtake In x. Experiment with a calculator to fmd the value of x at which the graphs of x l/lO
o
Other Comparisons 15. Investigate
lim In (x + I) Inx
%_00
1/10
and
and Inx cross, or, equivalently, at which Inx ~ 10 In (lnx) . Bracket the crossing point between powers of 10 and then close in by successive halving.
. In (x + 999) lim In X .
.%_00
o d. (Continuation ofpart (c).) The value
ofx at which Inx ~ 10 In (lnx) is too far out for some graphers and root fmders to identify. Try it on the equipment available to you and see what happens. 22. The function In x grows slower than any polynomial Show that In x grows slower as x + 00 than any nonconstant polynomial.
Then use I'HOpital's Rule to explain what you fmd. 16. (Continuation ofExercise 15.) Show that the value of
lim In (x + a) Inx
];_00
is the same no matter what value you assign to the constant a. What does this say about the relative rates at which the functions f(x) ~ In (x + a)andg(x) ~ Inxgrow? 17. Show that v'lOx + 1 and Vx+l grow at the same rate as x > 00 by showing that they both grow at the same rate as Vi as
Algorithms and Searches 23. •• Suppose you have three different algorithms for solving the same problem and each algorithm takes a number of steps that is of the order of one of the functions listed here:
n lo~ n, n 3/2,
X~OO.
18. Show that ~ and v'x'  x' grow at the same rate as x + 00 by showing that they both grow at the same rate as x 2 as x + 00.
19. Show that eX grows faster as x + 00 than x" for any positive integer n, even x 1,000,000 . (Hint: What is the nth derivative of x"?) 20. The function e' outgrows any polynomial Show that e' grows faster as x > 00 than any polynomial
n(lo~ n? .
Which of the algorithms is the most efficient in the long ruo? Give reasons for your answer.
o b. Graph the functions in part (a) together to get a sense ofhow rapidly each one grows. 24. Repeat Exercise 23 for the functions n,
Vn log, n,
(log, n? .
o 25. Suppose you are looking for an item in an ordered list one million items long. How many steps might it take to fmd that item with a sequential search? A binary search?
21. a. Show that Inx grows slower as x > 00 than xii' for any positive integer n, even x 1/1,000,000 •
You are looking for an item in an ordered list 450,000 items long o b. Although the values ofxl/l,ooo,ooo eventually overtake the val o 26. (the length of Webster. Third New International Dictionary).
ues of In x, you have to go way out on the xaxis before this happens. Find a value of x greater than I for which
Chapter
,
How many steps might it take to fmd the item with a sequential search? A binary search?
Questions to Guide Your Review
1. What functions have inverses? How do you know if two functions f and g are inverses of one another? Give examples of functions that are (are not) inverses of one another. 2. How are the domains, raoges, and graphs of functions and their inverses related? Give an example.
3. How can you sometimes express the inverse of a function of x as a function ofx? 4. Under what circumstances can you be sure that the inverse of a function f is differentiable? How are the derivatives of f and l related?
r
5. What is the natural logarithm function? What are its domain, raoge, and derivative? What arithmetic properties does it have? Comment on its graph. 6. What is logarithmic differentiation? Give an example.
7. What integrals lead to logarithms? Give examples. What are the integrals of tan x and cot x? 8. How is the exponential function e' defined? What are its domain, range, and derivative? What laws of exponents does it obey? Comment on its graph. 9. How are the functions a' and log.x defmed? Are there any rs1rictions on a? How is the graph oflog. x related to the graph of In x? What troth is there in the statement that there is really only one exponential function and one logarithmic function? 10. How do you solve separable firstorder differential equations? 11. What is the law of exponential change? How can it be derived from an initial value problem? What are some of the applications of the law?
12. Describe I'HOpital's Rule. How do you know when to use the rule and when to stop? Give an example.
430
Chapter 7: Transcendental Functions
13. How can you sometimes handle limits that lead to indeterminate forms 00/00, 00' 0, and 00  001 Give examples.
14. How can you sometimes handle limits that lead to indeterminate forms rx>, 00, and 00 001 Give examples. 15. How are the inverse trigonometric functions deImed? How can you sometimes use right triangles to find values of these functions? Give examples. 16. What are the derivatives of the inverse trigonometric functions? How do the domains of the derivatives compare with the domains of the functions? 17. What integrals lead to inverse trigonometric functions? How do substitution and completing the square broaden the application of these integrals? 18. What are the six basic hyperbolic functions? Comment on their domains, ranges, and graphs. What are some of the identities relating them?
Chapter 'f)
1. y = LOe~/'
2. y
= VieW.
I. 3. Y _I"" "4xe  16 e.
4. y
=
=
In (sin2 0)
7. y = 10!!2 (x /2) 9. y = 81 11. y = 5x,·6
+ 2)"+2
15. y = sinI~, 16. y = sinI
x 221 e x
6. y = In (see> 0)
2
13. y = (x
X~OO?
23. What roles do the functions e% and In x play in growth comparisons?
24. Describe bigoh and littleoh notation. Give examples. 25. Which is more efficient
tan (In v) v dv
(lnx)' xfix (I + Inr) dr
42.
[~fix
1
~/6
cos I
",/2 1  sm t
dl
44.j~ vlnv 46.
j
ln(x  5) x 5 fix
48.
j
cos(Ilnv) v dv
49. jX3%' fix
SO. j 2"'" see> x fix
51. [ffix
52.
[2
;x fix
431
Chapter 7 Practice Exercises
53.
54.1 (~
l' (~+ ~)dx
55.1
56.1
1
e (>0+1) dx
/.ln5 e'(3e' + 1)'/2 dr
' l 1
+ 7lnxt l/'
dx
(In (v + I»)' v+1 dv
1 1
58.
60
"
65.
6 d6
1
'/4
'/4
6dx
4x 2
66.
68
J
dy yV4y2  I
74
dx V2xx 2 1
77 •
V
2
tO+
4  25x2
liS
~ 24dy
J I J
=r~==
yVy2  16
•
2tVs
•
dl
(I + I)VI 2 + 21  8
78 •
dy :::,~==
Iy IV5y2  3
dx Vx 2 +4x1
76 11
+ 4v + 5
1
J
3 dv 4v 2 + 4v + 4 dl
(31 + I)V912 + 61
Solving Equations In Exercises 75184, solve for y. 79. 3" = 2"+1
80. 4" = 3"+2
81. ge'" = x 2
82. 3"
83. In (y  I)
106. lim el/Y lny
= x + Iny
yo+
107. lim (eX + I x_OO
=
d. f(x)
=
e. f(x)
= csc
= 1n5x
tan~
xo x + smX'
· sinmx 90• Inn . xo smnx
89. Inn  ( 2)
xo tan x
l
x,
tanI X
=
g(x) = I/x g(x)
eX
=
.. f(x)
= 3~,
b. f(x) = 1n2x,
g(x)
= 2~
g(x)
=
Inx 2
c. f(x) = lOx' + 2x 2, g(x) = eX d. f(x) = tan l (1jx), g(x) = I/x
= sechx,
g(x) = l/x2 g(x) =
e~
.. ~2 + ~ = O(~) 2
b.
c. X = o(x + Inx) e. tanI x = 0(1)
d. In (lnx) = o(lnx) f. coshx = O(e X )
x
x4
x
~ + ~ = O(~) x 2 x4 x4
~= O(~+~) x4 x2 x4
c. Inx = o(x + I) e. secl x = 0(1)
d. 1n2x = O(lnx) f. sinhx
rl
x~o
G'
g(x)
x,
= O(e
X
)
113. Thefimctioof(x) = eX + x,beingdifferentishleaodonetoone, has a differentiable inverse (x). Find the value of /dx at the point f(ln 2).
xn/2
+ I  Vx2  xl
,)
lim _x_ _ _ x_ 21 x 2 +1
'x_oo
....
Theory and Applications
lim sec 7x cos 3x
93. lim (esc x  cotx)
96
(1+ ?)X
112. Tme, or false? Give reasons for your answers.
2 8in X
X
xo+
110. Does f grow faster, slower, or at the same rate as g as x ..... oo? Give reasons for your answers.
..
X~OO
108. lim
f. f(x) = sinh,
X
95. lim (VX2 +
)lnX
111. True, or false? Give reasons for your answers.
3 Inx
84. In (lOlny)
88. lim
91.
eX  1
Comparing Growth Rates of Functions 109. Does f grow faster, slower, or at the same rate as g as x ..... oo? Give reasons for your answers. .. f(x) = 10l!2 x, g(x) = log, x I b. f(x) = x, g(x) = x + X g(x) = xe~ c. f(x) = x/100,
f. f(x)
' x2+3x4 85. Inn I
•
+ 21)
I  In (1 2 t
e. f(x) = sinl (l/x),
Apply;ng L:H6pUal's Rule Use I'HOpital's Rule to find the limits in Exercises 85108. xI
100. lim Z.mx  I xo eXl
6dx
V.tVs
72.
J 75.1 ,~2~d~v_c: J 2
98. lim 3'  I 60 8
x
101. lim 5  5 cou x_oe x x 1 103. lim
dx
'8ln310g,6 d6 6
{'
70.
2/3 dy 71. /, :::r~= Yz/3IY IV9y2  I •
1)1/2 d6
• Jv'33 + 12
• _24+31
73
1 1V 1/5
:;:~~
'19 
l"Ix~
xo
99. lim 2'""  I x_oexl
dw
/.1n9 e'(e' _
64'1
6712~2 69 •
e2w
97. lim lOX  I
62. ],4(1 + 1n1)llnldl
81084 6
63.
0
In2
59.1' ~ (1 61.

3x
1
2
57.
~) dx x2
8
drl
114. Find the inverse of the fimctioof(x) = I + (I/x),x # O.Then showthatr l (f(x» = f(r l (x» = x and that
drll dx
__1_ ft.) 
f'(x)"
432
Chapter 7: Transcendental Functions
In Exercises 115 and 116, fmd the absolure maximum and minimum values of each functioo 00 the given interval.
[i., fJ
115. y = xlo2x  x, 116. y = Hh:(2  Inx),
(0,
127. y.y' = secy' sec' x
.'l = 2(lnx)/x and the
Find the area between the curve y xaxis from x = 1 to x = e.
117. Area
118. a. Show that the area hetweoo the curve y = I/x and the xaxis from x = 10 to x = 20 is the same as the area between the curve and the xaxis from x = 1 to x = 2. b. Show that the area between the curve y = I/x and the xaxis from lea to kh is the same as the area between the curve and the xaxis from x = a tox = b (0 < a < b, k > 0).
119. A particle is traveling upward and to the right along the curve y = Inx. Its xcoordinare is increasing at the rate (fix/dt) = Vi mlsec. At what rare is the ycoordinare changiog at the point (.',2)? 120. A girl is sliding dowo a slide shaped like the curve y = 9. xl3 . Her ycoordinare is changiog at the rate dy/ dt = ( 1/4)~ Nsec. At approximarely what rate is her xcoordinate changiog when she reaches the bottom of the slide at x = 9 ft? (Take. 3 to be 20 and rouod your ans",", to the nearest sec.)
ftl
121. The rectangle shown here has one side
In Exercises 125128 solve the differential equatioo. dy _, 3y(x + I)' 125. fix = v'Ycos' vy 126. y' = y I
00
the positive yaxis,
one side on the positive xaxis, and its upper righthand vertex on the curve y = e r. What dimensions give the rectangle its largest area, and what is that area? y
128. y cos' x dy
+ sin x fix
= 0
In Exercises 129\32 solve the initial value problem.
.'Y',
129. :
=
dy 130. fix
=
131. xdy 
ylny
I
+x
"
y(O) = 2
y(0)
=•
(y + v'Y) fix =
132. y' dyfix =
,
0, y(l) = I
,f, y(O) = • +I
I
133. What is the age ofa sample of charcoal in which 90% of the carbon14 origioally present has decayed? 134. Cooling a pie A deepdish apple pie, whose inrerna\ temperature was 2200f when removed from the oven, was set out on a breezy 40°F porch to cool. Fifteen minures later, the pie's internal tcmperatore was 180°F. How long did it take the pie to cool from there to 70°F? 135. Locating a solar station You are under cootract to build a solar statioo at grouod level on tlte eastwest line between the two buildings shown here. How far from the taller building should you place the statioo to maximize the number of hours it will be in the sun 00 a day when the sun passes directly overhead? Begio by observing that
8
= 'If

1 X
cot
C1
60  co
50  x
~.
Then fmd the value of x that maximizes 8. ~~~~ x
o
60m 122. The rectangle shown here has one side
00
the positive yaxis,
one side on the positive xaxis, and its upper righthand vertex on the curve y = (lnx)/x'. What dimeosions give the rectangle its largest area, and what is that area? 0.2 0.1
t ,_ ..
o
D
~
~I
II
'x
123. Graph the following functioos and use what you see to locate and estimare the extreme values, identify the coordinares of the inflectioo points, and identifY the intervals 00 which the grapha
are concave up and concave down. Then confrrm your estimates by working with the functions' derivatives.
a. y
=
(lnx)/Vi
D 124. Graph f(x)
b. y = ...'
c. Y
=
(l
+ x).~
= x In x. Does the function appear to have an absolure minimum value? Cnnfmn your answer with calculus.
00 00 00 00
00 00
9
30m
~+~~~~x
o
x
SOm
136. A rouod underwater transmissioo cable consists of a core of copper wires surrounded by noocooducting insulatioo. 1f x denores the ratio of the radius of the core to the thickness of the insulation, it is known that the speed of the transmission signal is given by the equation v = x'in (I/x). If the radius of the core is I em, what insulatioo thickness h will allow the greatest transmissioo speed? Insulation
Chapter 7 Additional and Advanced Exercises
Chapter
f)
Additional and Advanced Exercises
Limits
16. Let g be a function that is differentiable throughout an open interval containing the origin. Suppose g has the following properties:
Find the limits in Exercises l{j.
[h dx bljo ~
2. lim} {' tanI I dl
1. lim
%_00
3. lim (cos Yx)l/x %0+
4. lim (x
Jo
• I) g(x
x
+ ex)2/x
+n+2 +... +~) 2n
.~o
_1_
iii) lim g(h)
1n (e1/1l + e2/11 + ... + e(II1)/1I + en/II)
volving the region about the xaxis. Find the following limits.
b. lim V(I)/A(t)
c. lim V(I)/A(I)
too
1_00
10+
8. Varying a logarithm's bas.
o
a. Findlimlog.2asa .... 0+, 1, 1+, and 00. b. Graph y ~ log. 2 as a function of a over the interval 0< a :5 4.
Theory and Examples 9. Find the areas between the curves y ~ 2(loll2x)/x and y ~ 2(log.,x)/x and the xaxis from x ~ I to x ~ e. What is the ratio of the larger area to the smaller?
0 10. Graph f(x) ~ tanI x + tanl(l/x) for 5
:5 x :5 5. Then use calculus to explain what you see. How would you expect f to behave beyond the interval [5,5]? Give reasons for your answer.
11. For what x
>
0 does x(x') ~ (xx)'? Give reasons for your
answer. 0
12. Graphf(x) ~ (sinx)"nxover[O, 31T]. Explain what you see. 13. Findj'(2) if f(x)
~
e«X) andg(x)
~ (X_t_, dl.
l,
~
0
~ I
h
a. Showthatg(O)
7. l.e!A(t) be the area of the region in the 'ITst quadrant enclosed by the coordinate axes, the curve y = e x, and the vertical line x ~ I, t > O. Let V(I) be the volume of the solid generated by rea. lim A(t)
g(x) + g(y) ~ I _ g(x)g(y) forallrealnurnbers x,y, and
+ y in the domain of g
hO 1I_00
+ y)
il) lim g(h)
%_00
5. lim (_1_ 11_00 n+ 1
6. lim
433
1+ t
14. a. Find df/dx if
~
O.
b. Showthatg'(x) ~ I
+ [g(x)]'.
c. Find g(x) by solving the differential equation in part (b).
17. Center of mas. Find the center of mass ofa thin plate of constant density covering the region in the first and fourth quadnmts enclosed by the curves y ~ 1/(1 + x') and y ~ 1/(1 + x') and by the lines x ~ 0 and x ~ 1. 18. Solid of revolution The region between the curve y ~ 1/(2Yx) and the xaxis from x ~ 1/4 to x ~ 4 is revolved about the xaxis to generate a solid. •• Find the volume of the solid.
b. Find the centroid of the region. 19. Th. best branching angl•• for blood v ••••l. and pipe. When a smaller pipe branches offfrom a larger one in a flow systero, we may want it to run off at an angle that is best from SOD2O energysaving point of view. We might require, for instance, that energy loss due to friction be minimized along the sectionAOB showo in the accompanying figure. In this diagram, B is a given point to be reached by the smaller pipe, A is a point in the larger pipe Ul'" stream from B, aod 0 is the point where the braochiog occurs. A law due to Poiseuille states that the loss of energy due to friction in nontorbulent flow is proportional to the length of the path aod inversely proportional to the fourth power of the radius. Thus, the loss along AO is (ledl)/R' and along OB is (Ied,)/r' , where /r; is a constant, dl is the length of AO, d, is the length of OB, R is the radius of the larger pipe, and r is the radius of the smaller pipe. The angle 8 is to be chosen to minimize the sum of these two losses:
f(x)~ ['2~ldt.
dl
d,
R
r
L~/r;4+/r;4·
b. Find f(O). c. What can you conclude about the graph off? Give reasons
for your answer. 15. Evenodd decompositions
a. Suppose that g is an even function of x and h is an odd function ofx. Show that if g(x) + h(x) ~ 0 for all x then g(x) ~ for aIIxandh(x) ~ for allx.
o
A
o
b. If f(x) ~ IE (x) + f 0 (x) is the sum of an even function h(x) and an odd function fo(x) , then show that lE(x) ~ f(x)
+ f( x) 2
and
fo(x) ~ f(x)  f( x)
2
c. What is the sigoificance of the result in part (b)?
In our model, we assume that AC Thus we have the relations
~
a and BC
d,sinO
~
b
~
b are fixed.
434
Chapter 7: Transcendental Functions
D 20.
so that d,=bcscO,
dl
=a
 d,cosO
=a
 beatO.
We can express the total loss L as a function of 0:
Urban gardening A vegetable gardeo 50 ft wide is to be growo between two buildiogs, which are 500 ft apart along an eastwest line. If the buildiogs are 200 ft and 350 ft ta1l, where should the garden be placed in order to receive the maximum llUll1ber ofbours of sunlight exposure? (Hint: Determine the value of x in the accompa
nying fIgure that maximizes sunlight exposure for the garden.)
L= k(a  bcotO + bCSCO). R4
L
74
00 00 00 00 00
Show that the critical value of 0 for which dL/ dO equals zero is
200 ft tall
b. If the ratio of the pipe radii isr/R = 5/6, estimate to the nearest degree the optimal branching angle given in part (a).
The mathematical analysis described here is also used to explain the angles at which arteries branch in an animal's body.
West
00 00 00 x
50
350 ft tall
East
8 TECHNIQUES OF INTEGRATION OVERVIEW The Fundamental Theorem tells us how to evaluate a definite integral once we have an antiderivative for the integrand function. Table 8.1 summarizes the forms of antiderivatives for many of the functions we have studied so far, and the substitution method helps us use the table to evaluate more complicated functions involving these basic ones. In this chapter we study a number of other important techniques for finding anti derivatives (or indefinite integrals) for many combinations of functions whose antiderivatives cannot be found using the methods presented before.
TABLE 8.1
1.
3.
4. 5. 6.
7. 8. 9. 10. 11.
Basic integration formulas
I
kdx = kx
+
I~ = In Ixl
I
eX dx = eX
lax
I I I I I I
dx =
+
C
+
12.
(n # 1)
13. 14.
C
C
15.
LIna + C
sinxdx = cosx
cosxdx = sinx 2
(any number k)
sec xdx = tanx
(a>O,a#l)
+
+ +
csc2 xdx = cotx
C
17.
C
18.
C
+
16.
19. C
20.
I I I I I I I v'ar: x (~) I ~ ~tan1 (~) I:!. I I v' tanxdx = In Isecxl
+
C
cotxdx = In Isinxl
+
C
secxdx
= In Isecx + tanxl + C
cscxdx
= In Icscx + cotxl + C
sinh x dx
= coshx + C
coshxdx
= sinhx + C 2 =
a2
x
secxtanxdx = secx
+
cscxcotxdx = cscx
C
+
x2 =
sin
1
+
+
dx =1 a sec1 a x 2  a2
C
C
+
C
(a> 0) C
(x> a
>
0)
435
436
Chapter 8: Techniques of Integration
8.1
Integration by Parts Integration by parts is a technique for simplifying integrals of the fann
J f(x)g(x) dx. It is useful when f can be differentiated repeatedly and g can be integrated repeatedly without difficulty. The integrals
and
JxCOSXdx
are such integrals because f(x) = x or f(x) = x 2 can be differentiated repeatedly to become zero, and g(x) = cos x or g(x) = eX can be integrated repeatedly without difficulty. Integration by parts also applies to integrals like and
J Inxdx
J
eX cosxdx.
In the rlISt case,j(x) = In x is easy to differentiate and g(x) = I easily integrates to x. In the second case, each part of the integrand appears again after repeated differentiation or integration.
Product Rule in Integral Form If f and g are differentiable functions of x, the Product Rule says that
d dx [f(x)g(x)] = f'(x)g(x)
+ f(x)g'(x).
In terms of indermite integrals, this equation becomes
J
fx [f(x)g(x)] dx
= J
[f'(x)g(x)
+
f(x)g'(x)] dx
or
J
fx [f(x)g(x)] dx
= J
f'(x)g(x) dx
+ J f(x)g'(x) dx.
Rearranging the terms of this last equation, we get
J f(x)g'(x) dx = J
fx [f(x)g(x)] dx 
J f'(x)g(x) dx,
leading to the integration by parts formula
J f(x)g'(x) dx = f(x)g(x)  J f'(x)g(x) dx
{l}
Sometimes it is easier to remember the formula if we write it in differential form. Let u = f(x} and v = g(x}. Then du = f'(x) dx and dv = g'(x) dx. Using the Substitution Rule, the integration by parts fannula becomes
8.1
437
Integration by Parts
Integration by Part. Formula
j udv
=
uv  j vdu
(2)
J
J
This formula expresses one integral, u dv, in terms of a second integral, v du. With a proper choice of u and v, the second integral may be easier to evaluate than the first. In using the formula, various choices may be available for u and dv. The next examples illustrate the technique. To avoid mistakes, we always list our choices for u and dv, then we add to the list our calculated new terms du and v, and finally we apply the formula in Equation (2).
EXAMPLE 1
Find
j We use the formula j
Solution
u dv
xcosxd!:.
=
uv  j v du with
dv = c08xdx, v = sinx.
u =x, du = d!:,
Simplest antiderivative of cos x
Then
j xcosxd!:
=
x sin x  j sinxd!:
=
x sin x
+ cosx +
C.
•
There are four choices available for u and dv in Example 1:
1. Letu = 1 anddv = xcosxd!:. Letu = xcosx and dv = d!:.
2. 4.
3.
Letu = xanddv = cosxd!:. Letu = cosxanddv = xd!:.
Choice 2 was used in Example I. The other three choices lead to integrals we don't know how to integrate. For instance, Choice 3 leads to the integral
j{xcosx  x 2 sinx)d!:.
J
The goal of integration by parts is to go from an integral u dv that we don't see how to evaluate to an integral v du that we can evaluate. Generally, you choose dv first to be as much of the integrand, including d!:, as you can readily integrate; u is the leftover part. When finding v from dv, any antiderivative will work and we usually pick the simplest one; no aIbitrary constant of integration is needed in v because it would simply cancel out of the righthand side of Equation (2).
J
EXAMPLE 2
Find
j Inxd!:. Solution
J udv = u du
Since
uv 
= lnx =
1
xd!:,
J In x d!: can J vduwith
be written as
Simplifies when differentiated
J In x • 1 d!:, dv
= d!:
v =x.
we use the formula Easy to integrate
Simplest antiderivative
438
Chapter 8: Techniques of Integration Then from Equation (2),
1
lnxdx
=
1
xlnx 
x·fdx
1
xlnx 
=
dx
=
xlnx  x + C.
•
Sometimes we have to use integration by parts more than once.
EXAMPLE 3
Evaluate
1
2
X
x e dx.
Solution
With u = x 2 , dv
= eX dx,
1
du = 2xdx, and v =
x 2e Xdx
x 2e x 
=
21
eX,
we have
xeXdx.
The new integral is less complicated than the original because the exponent on x is reduced by one. To evaluate the integral on the right, we integrate by parts again with u = x, dv = eX dx. Then du = tix, v = eX, and
1
xeXdx
=
xe X 
1
eXdx
=
xe X  eX +
c.
Using this last evaluation, we then obtain
1
x 2e Xdx
=
x 2e x 
= x 2e x 
21
xeXdx
2xe X
+
2e X
•
+ c.
J
The technique of Example 3 works for any integral x"e X dx in which n is a positive integer, because differentiating x" will eventually lead to zero and integrating eX is easy. Integrals like the one in the next example occur in electrical engineering. Their evaluation requires two integrations by parts, followed by solving for the unknown integral.
EXAMPLE 4
Evaluate
1
eXcosxdx.
Solution
Letu
=
eX and dv
1
=
cosxdx. Then du
eXcosxdx
=
eX sin x 
= eXdx,v
sinx,and
=
1
eXsinxdx.
The second integral is like the flISt except that it has sin x in place of cos x. To evaluate it, we use integration by parts with
dv Then
1
eX cosx dx
= sinxdx,
v =
du=exdx.
C08X,
1 1
=
eX sin x  (ex cosx 
=
eX sin x + eXcosx 
(cosx)(e Xdx»)
eXcosxdx.
Integration by Parts
8.1
439
The unknown integral now appears on both sides of the equation. Adding the integral to both sides and adding the constant of integration give
2J
e'cosxdx
=
e'sinx + e'cosx + CI.
,.
Dividing by 2 and renaming the constant of integration give
J
e , cosx dx  e
EXAMPLE 5
+' 2 e cosx +
SfiX
C.
•
Obtain a fonnula that expresses the integral
J
cos' xdx
in terms of an integral of a lower power of cos x. We may think of cos' x as cos· I x • cos x. Then we let
Solution
u
=
cosn
1X
dv
and
c08xdx,
=
so that
du = (n  I) cos· 2 x(sinx dx)
v = sinx.
and
Integration by parts then gives
J
cos'xdx = cos·Ixsinx
+
(n
1)J
sin2 xcos· 2 xdx
J
= cos· I x sin x
+ (n  1)
= cos·Ixsinx
+ (n
(1  cos2 x) cos· 2 X dx
1)J
cos· 2 xdx  (n 
1)J
cos·xdx.
Ifwe add (n
1)J
cos'xdx
to both sides of this equation, we obtain
J
n
cos' xdx = cos· I x sin x
+
J
(n  1)
We then divide through by n, and the final result is
J
.1' cos• x dx  cos nx smx
IJ
+~ n
cos· 2 xdx.
cosn2 x dx .
•
The fonnula found in Example 5 is called a reduction formula because it replaces an integral containing some power of a function with an integral of the same fonn having the power reduced. When n is a positive integer, we may apply the fonnula repeatedly until the remaining integral is easy to evaluate. For example, the result in Example 5 tells us that
J
3
2
cos xdx =
cos xsinx 3 2
2J
+3
= tcos xsinx
cosxdx
+ tsinx +
c.
440
Chapter 8: Techniques of Integration
Evaluating Definite Integrals by Parts The integration by parts fonnula in Equation (I) can be combined with Part 2 of the Fundamental Theorem in order to evaluate definite integrals by parts. Assuming that both f' andg' are continuous over the interval [a, b], Part 2 of the Fundamental Theorem gives
Integration by Parts Formula for Definite Integrals
tf{x)g'{x) dx
=
f{x)g(x)
l:  tf'{x)g(x) dx
(3)
In applying Equation (3), we normally use the u and v notation from Equation (2) because it is easier to remember. Here is an example. y
EXAMPLE 6 Find the area of the region bounded by the curve y = xex and the xaxis fromx = Otox = 4. Solution
0.5
The region is shaded in Figure 8.1. Its area is
1'xeX dx. 1
1
2
3
4
x
Letu
=
x,dv
exdx,v
=
ex,anddu = tix. Then,
1'xex dx
=
xe x l~
=
[4e4  (O)] +
=
4e4  ex]~
=
4e4  e4  (eO)
=
FIGURE 8.1 The region in Example 6.
14 (
e X) dx
14
eX dx
= I 
5e4 '" 0.91.
•
Tabular Integration
J
We have seen that integrals of the fonn f{x)g(x) dx, in which f can be differentiated repeatedly to become zero and g can be integrated repeatedly without difficulty, are natural candidates for integration by parts. However, if many repetitions are required, the calculations can be cumbersome; or, you choose substitutions for a repeated integration by parts that just ends up giving back the original integral you were trying to find. In situations like these, there is a way to organize the calculations that prevents these pitfalls and makes the work much easier. It is called tabular integration and is illustrated in the following examples.
EXAMPLE 7
Evaluate
J
2 x
x e dx.
Solution
With f{x) = x 2 and g{x) = eX, we list:
f(x) and its derivatives
g(x) and its integrals
Integration by Parts
8.1
441
We combine tbe products of tbe functions connected by tbe arrows according to tbe operation signs above tbe arrows to obtain
J
x'e' dx
x'e'  2xe'
=
+ 2e' + C.
•
Compare tbis witb tbe result in Example 3.
EXAMPLE 8
Evaluate
J
x 3 sinxdx.
Witb f(x) = x 3 and g(x) = sin x, we list:
Solution
f(x) and its derivatives
g(x) and its integrals
x3
(+)
sin x
3x'
()
cosx
(+) ()
.... sinx
6x

6

0

~
cos x
~
sin x
Again we combine tbe products of tbe functions connected by tbe arrows according to tbe operation signs above tbe arrows to obtain
J
x 3 sinxdx
= x 3 cosx
+ 3x'sinx + 6xcosx
 6 sin x
+ C.
•
The Additional Exercises at tbe eod of!bis chapter show how tabular integration can be used wheo neitber function f nor g can be differentiated repeatedly to become zero.
Exercises 8.1 Integration by Parts Evaluate the integrals in Exercises 124 using integmtion by parts.
15. / x'exdx
(x'  5x)e Xdx
1. / xsinIdx
2. /oeOS1rOdO
3. / t'eostdt
4. / x sinxdx
19. / x'exdx
6. /,e x31nx dx
21. /
5.
/,2
xlnxdx
7. /xeXdx 9.
/x e
11. /
2
X
2
8. dx
tan1ydy
13. / xsec2 xdx
17. /
10.
/xe
3x
12. /
23. / e lx cos 3x dx
dx
/(X 2 2x + l)e sin1 ydy
14. /4xsec 2 2xdx
e'sinOdO
2x
dx
16. / p
e" dp
4
18. /
(r' + r + l)e'dr
20. /
t'e 4t dt
22. /
eYeosydy
24. /
e2x sin2xdx
Using Substitution Evaluate the integrals in Exercises 2530 by using a substitutioo prior to integration by parts. 25. /
e v'hl9 tis
26.
J,lx~dx
442 27.
29.
Chapter 8: Techniques of Integration
J.
~/3
xtan2 xdx
0
j
28.
30.
sin (lnx) dx
j j
+ x 2)dx
In(x
d. What pattern do you see? What is the area b.eo the curve
and the xaxis for 2 (~)
z(lnz)2 dz
n an arbitrary positive integer? Give reasons for your answer.
Evaluating Integrals Evaluate the integrals in Exercises 3150. Some integrals do not require integration by parts.
31. jxsecX2 dx
32.
33. jXQnx)2dx
34.
45.
j ~: j j v?"+l j j e~ j Vx
47.
J.~/2 8 2 sin 28 d8
35. 37. 39. 41. 43.
49.
46.
48.
J.~/2 x 3 cos 2x dx
50.
J.
36.
x 3 e x'dx
38. dx
40.
sin 3x cos 2x dx
42.
sin eX fix
44.
dx
cos
tsecItdt
('
12;v3
(lnx)3
x
dx
x 2 sin x 3 dx sin 2x cos 4x dx
dx
e Vi dx
1;0.
0
~ 'IT.
55. Finding volume Find the volume of the solid generated by revolving the region in the fl1'St quadraot bounded by the coordinate axes and the curve y = cos x, 0 :5 x :s; 71'/2, about b. the line x = 1T/2. 56. Finding volume Find the volume of the solid generated by revolving the region bounded by the xaxis aod the curve y = xsinx,O:5 x:s; '11',about
a. the yaxis.
y=xsinx
5
o 5 52. Finding area Find the area of the region enclosed by the curve y = x cos x aod the xaxis (see the accompaoying figure) for
a. 1T/2 s x s 31T/2. 51T/2.
c. 51T/2 s x s 71T/2.
andx
=
e.
a. Find the area of the region. b. Find the volume of the solid formed by revolving this region
about the xaxis.
y
10
7T'.
(See Exercise 51 for a graph.)
nonnegative integer? Give reasons for your answer.
S
a. about the yaxis. b. about the line x = I.
57. Consider the region boonded by the graphs of y = In x, Y = 0,
d. What pattern do you see here? What is the area b.eo the curve and the xaxis for mr s x s (n + 1)1T, n an arbitnlry
'" x
54. Finding volume Find the volume of the solid generated by revolving the region in the fl1'St quadraot bounded by the coordinate axes, the curve y = e r, and the line x = 1
b. the line x =
1T'
b. 31T/2
53. Finding volume Find the volume of the solid generated by revolving the region in the fl1'St quadraot bounded by the coordinate axes, the curve y = eX. and the line x = In 2 about the line x = In2.
a. the yaxis. 2xsin1 (x 2 )dx
:s x :s 21T. c. 21T :s x :s 31T.
b.
10
xdx
51. Finding area Find the area of the region eoc1osed by the curve y = x sin x aod the xaxis (see the accompaoying figure) for ~
10
1_2 dx xQnx)
Theory and Examples
a. 0
y
jco~dx
j __ j j x' e" j j j :x j Vx
dx
x3
«~) 2 'fr,
< '1T_X_
c. Find the volume of the solid formed by revolving this region about the line x = 2. d. Find the centroid of the region. 58. Consider the region bounded by the graphs of y aodx = 1.
= tanI x, y = 0,
a. Find the area of the region. b. Find the volume of the solid formed by revolving this region about the yaxis. 59. Average value A retarding force, symbolized by the dashpot in the accompaoying fIgure, slows the motion of the weighted spring so that the mass's position at time t is
t '" O.
443
8.1 Integration by Parts Find the average value ofy aver the intervaJ. 0
:!;';
,
t
:!;';
2'11".
The idea is to take the most complicated part of the integral, in this case l(x), and simplify it fIrst For the integral ofln x, we get
r
J
lnxdx
~
J
,  lnx,
ye'dy
X _ 1I 7
dx  Il'rIy
+C x + C.
= ye'  II'
= xlnx 
o
,
For the integral of COlII X we get
f
I
C08l xdx  xoosIx
y
cosydy
=
coa1x
= xoosIx  siny + C
Use the formula
60. Avenge value In a massspringdashpot system l:ilcc the one in Exercise 59, the mass's position at time t is
y = ~t(sint  cost),
t:2!:
Find the average value ofy aver the interval 0
O.
Reduction Formulas
61.
61.
f x~
cosxdt = x"sinx 
1
1
x· sinxQ = x· cosx +
to cvaluate the integrals in Exercises 6770. Express your answers in terms ob.
69.
f
70·/IO~Xdx
secI x ttt
r
1
1
x· I COSXQ
If
(4)
fly) dy
Another way to integrate rl(x) (when course) is to use integration by parts with u rewrite the integral. of 1 as
X,,I sinxQ
If
J
r'(x) dx  xr'(x) 
:s: t :s: 2'11".
In Exercises 6164, use integration by parts to establish the reduction
.....
J
rl(x)dt
=
xr1(x)
I
r 1 is integrable, of r 1(x) and dv = dx to
x(!r1(x»)dt.
(5)
Exm:ises 71 and 72 compare the:results ofusing Equations (4) and (5). 64.
f
(lnx)"Q = x(lnx)· 
nf
71. Equations (4) and (5) give diffen:ul: form.ulas for the integral. of
(lnX),,I Q
cos1 x:
65. Show that
L
f
b.1 66. Use inqration by paI19 to obtain the formula
Integration by parts leads to a rule for integrating inverses that usually gives good results:
b.
J
~ yf(y) 
7 2
y> 7 x>1
10 • 12. 14.
23.
VI _
1 1 1Vy' 
dy, Y
2dx
1
X3
>~
25
y'
_~, 2  1
vx
5
15·1~dx 9 x 2
1 · W+4
17
x'dx
18.
1 1
0
27.
x'~
'/2'
29 1 Sdx • (4x' + I)'
x'dx x2  1
x> 1 33 1 v'dv • (1  v')'/2
x> 1
24.
26. 28.
1.' dx 0 (4 _ x')'/2 1
x'dx (x' _ 1
)'/2 '
x > 1
1(IX')I/2 4 dx x
6dt (9t' + I)'
30 1
·
32 1 xdx • 25 + 4x' 34.
1 (1  r')'/2 8 dr r
In Exercises 354S, use an appropriate substitution and then a 1rigonome1ric substitution to evaluate the integrals.
35 •
ln4
1. 0
[1/4
dx
22. 1 x Vx'  4dx
4x' dx (1 _ x')'/2
1 (1  x')'/2 6 dx x
31. 1
Assorted Integrattons
 x', d x 4+x
1.V3/2
dx
V9
>5
Vse any method to evaluate the integrals in Exercises 1534. Most will require 1rigonome1ric substitutions, but some can be evaluated by other methods. 16.
\00
+ 25x'
25.1 (x ,dx  1)
v'1=9t2dt 5dx x V25x'  9'
36
w~ W'dw
20. 1
37.
e
I dt
Ve u + 9 2 dt
JI/12 vi + 4tvl
36
• 38
lin
(4/')
In ('/4)
I dt
_"e~~
(1
['
· JI
+ eU )'/2
dy
yVI
+ (\ny)'
453
8.4 Integration of RationaL Functions by PartiaL Fractions
.. J J · '\17=1 J · V1+7 . . J)4 ;x '7. JVx~ 0, the improper integral can be interpreted as the (finite) area be• neath the curve and above the xaxis (Figure 8.15).
The Integral
l
""dx Ii X
1
The function y = l/x is the boundary between the convergent and divergent improper integrals with integrands of the form y = l/x p • Ai; the next example shows, the improper integral converges if p > 1 and diverges if p :5 1.
EXAMPLE 3
For what values ofp does the integral tegral does converge, what is its value?
Solution
Jt' dx/x
P
converge? When the in
If p "" 1,
I
1
bdx 
xP
=
x p+ 1
p
+1
]b 1
= _1_ bp+l 
1 p(
1 = 1 (1    1) ) 1  P b P 1
•
Thus,
=
b~oo [1  p (b
I
lim _1_ _ 1__ 1 P 1
= ) ]
p> 1
P  l' { co,
p
< 1
because lim _1_ = {O,
hOO bP  1
00,
p>1 p 0+:
Therefore the area under the curve from 0 to 1 is finite and is defined to be
DEFINmON Integrals of functions that become infmite at a point within the interval of integration are improper integrals of Type II. 1. If fix) is continuous on (a, bl and discontinuous at a, then b fix) dx = lim+ib fix) dx.
i i a
CQ
c
2. If fix) is continuous on [a, b) and discontinuous at b, then
a
b fix) dx = lim_ic fix) dx. cb
a
3. If fix) is discontinuous at e, where a [a, c) U (e, bl, then [
fix) dx = [f{X) dx
+
O.
For each positive;t. the number f{x) is the integral oftr1e ~ with respect to t from. 0 to 00. Figure 8.21 shDws the graph of f near the origin. You will sec how to calculate f(l/2) if you do Additional Exercise 23 in Chapter 14.
Chapter 8 Additional and Advanced Exercises y
493
As you will see if you do Exercise 104 in Section 10.1, Equation (4) leads to the approximation
~"'~.
(5)
o b. Compare your calculator's value for n! with the value given by Stirling's approximation for n = 10, 20, 30, ... , as far as your calculator can go.
o c. A refmoment of Equation (2) gives
r±+~+~~~~>X
o
2
3
1
f(x) =
2 3
(~Y l#e1/(l2x)(1
or f(x) '"
FIGURE 8.21 Euler's gamma function f(x) is a continuous function ofx whose value at each positive integer n + 1 is n!. The deftning integral fonnula for f is valid only for x > 0, but we can extend f to negative noninteger values of x with the fonnula f(x) = (f(x + 1»/x, which is the subject of Exercise 31. 31. H II is a nonnegative integer, f(1I
+ 1)
(6) Compare the values given for 1O! by your calculator, Stirling's approximation, and Equation (6).
Tabular Integratton The technique of tabular integration also applies to integrals of the f(x )g(x) fix when neither function can be differentiated reform peatedly to become zero. For example, to evaluate
J
= II!
b. Then apply integration by parts to the integral for f(x show that f(x + I) = xf(x). This gives
+
f(4) = 3f(3) = 6
~andits derivatives
= nf(n) = n!
(1)
Use mathematical induction to verify Equation (1) for every nonnegative integer n.
32. Stirling'. formula
Scottish mathematician James Stirling (16921770) showed that
,~('iYfIrf(X)=
integral.
e'"'~
cos x
2e2x~inX (+) ~ cosx
4e2¥
I
Ld;+R'~O dt l ,
0 < 6 < 1r/2
which is Equation (5) with V
Solve the initial value problems in Exercises 1520.
+ 2y
16. 1dy dl
~ 3,
c. Show that the value of the current when 1 ~ L/R is I/e. (The significaoce of this time is esplained in the next exercise.)
+ 2y _ 3 I, 1 > 0, y (2)
~
I
6> 0, Y(1r/2) ~ I
dy _ 3 18. 6 d6  2y  6 sec 6 tan 6,
6> 0,
dy
y(1r/3)
~
2
x'
19. (x
+ I) dx  2(x 2 + x)y ~ x e+ I' x > I, y(O) ~ 5
dy 20. dx
+ xy
~
O.
fall to half its original value?
y(O) ~ I
dy . 17. 6 d6 + Y ~ sm6,
~
a. Solve the equation to express i as a function of t. b. How loog after the switch is thrown will it take the current to
Solving In;tial Value Problems dy 15. dl
509
26. Current in an open RL circuit If the switch is thrown opeo after the current in an RL circuit has built up to its steadystate value I ~ VIR, the decaying current (see accompaoying figure) obeys the equatioo
0 < 6 < 1r/2
tan6,
FirstOrder Linear Equations
3!
x, y(O) ~ 6
R
21. Solve the expooeotial growth/decay initial value probl= for y as • functioo of 1 by thinking of the differential equation as a rITStorder linear equation with P(x) ~ kandQ(x) ~ 0:
dy dl
~
ky
(keonstant),
y(O)
~
Yo
22. Solve the following initial value probl= for u as a functioo of t.
~~ + ~u ~
0 (kaodm positive coostants), u(O)
~
Uo
27. TIme constants Engineers call the number L/R the time constanl of the RL circuit in Figure 9.9. The sigoificance of the time coostant is that the current will reach 95% of its final value within 3 time coostants of the time the switch is closed (Figure 9.9). Thus, the time coostant gives a builtin measure of how rapidly an individual circuit will reach equilibrium. •• Find the value of; in Equation (7) that correspoods to 1 ~ 3L/R and show that it is about 95% of the steadystate value I ~ V/ R.
b. Approximately what percentage of the steadystate current will be flowing in the circuit 2 time coostants after the switch is closed (i.e., wheo 1 ~ 2L/R)?
a. as a flIstorder linear equation. b. as a separable equatioo.
28. Derivation of Equation (7) in Example 4 a. Show that the solution of the equation
Theory and Examples 23. Is either of the following equatioos correct? Give reasoos for your
answers. b.
x/
tdx
~ xlnlxl + ex
24. Is either of the following equatioos correct? Give reasoos for your
answers. a.
co~x/ cosX' dx =
1/
b. cosX'
cosX' ax
=
is
; ~ l' + Ce (R/Ll' tanx
+C
tanx
+
C
cosX'
25. Current in a closed RL circuit How maoy seeonds after the switch in an RL circuit is closed will it take the current ; to reach half of its steadystate value? Notice that the time depeods 00 R and L and not 00 how much voltage is applied.
R
b. Theo use the initial cooditioo ;(0) ~ 0 to determine the value of C. This will complete the derivation ofEquatioo (7). c. Showthati ~ VIR is a solution of Equation (6) and that i = Ce {RILlt satisfies the equation
di dl
R.
+ I'
~
o.
510
Chapter 9: FirstOrder Differential Equations we have n ~ 2,
James Bernoulli
so thatu ~ yl2 ~ yl anddu/dx ~ y2dy/dx.Thendy/dx ~ y 2 du/dx ~ u2 du/dx.
(16541705)
Substitution into the original equation gives
HiSTORICAL BIOGRAPHY
A Bernoulli differential equation is of the form
dy dx
Observe that, if n
+ P(x)y
~
Q(x)y".
or, equivalently,
0 or I, the Bernoulli equation is lioear. For other values of n, the substitution u = y In transforms ~
du dx
+ u = ex.
the Bernoulli equation into the linear equation
: + (I 
n)P(x)u
~ (1 
This last equation is linear in the (unknown) dependent variable u.
n)Q(x).
Solve the Bernoulli equations in Exercises 2~32.
For example, in the equation dy  y dx
9.3
~ e~y2
29. y'
y~
_y2
30. y'  Y ~xy2
31. xy'
+y
y2
32.
~
x'y' + 2xy
~ y'
AppLications We now look at four applications of flfStorder differential equations. The flfSt application analyzes an object moving along a straight line while subject to a fOrce opposing its motion. The second is a model of population growth. The third application considers a curve or curves intersecting each curve in a second family of curves orthogonally (that is, at right angles). The fmal application analyzes chemical concentrations entering and leaving a container. The various models involve separable or liuear flfSt 0).
This is a separable differential equation representing exponential change. The solution to the equation with initial condition v ~ Vo at t = 0 is (Section 7.4) (1)
What can we learn from Equation (I)? For one thing, we can see that if m is something large, like the rnass of a 20,000ton ore boat in Lake Erie, it will take a long time for the velocity to approach zero (because t must be large in the exponent of the equation in order to make kt/m large enough for v to be small). We can learn even more ifwe integmte Equation (1) to find the position s as a function of time t. Suppose that a body is coasting to a stop and the only force acting on it is a resistance proportional to its speed. How far will it coast? To find out, we start with Equation (1) and solve the initial value problem ds _
dt  voe
(klm)t
,
s(O)
~
O.
9.3
Applications
511
Integrating with respect to I gives v~m e(klm)t
S = 
Substitoting S
~
0 when I
~
+ C.
0 gives
vom o ~+ k
C
vom k
C~
and
The body's position at time I is therefore
_ vom ("m)t ()st Te "
vom _ vom (
+ T  T Ie
(klm)t)
.
(2)
To find how far the body will coast, we fmd the limitofs(l) as 1> 00. Since (kim) < 0, we know that e (klm)t + 0 as I + 00 , so that
lim S(I) ~ lim vom (I t_OO k
e(klm),)
t_OO
~ v~m (I _ 0) ~ v~m .
Thus, Distance coasted
=
vom
T
(3)
The number vom/k is only an upper bound (albeit a useful one). It is true to life in one respect, at least: if m is large, the body will coast a long way.
l In the English system, where weight is measmed in pounds, mass is measured in slugs. Thus, Pounds
~
slugs X 32,
assuming the gravitational constant is 32 ft./sec'.
EXAMPLE 1
For a 192lb ice skater, the k in Equation (I) is about 1/3 slug/sec and
m = 192/32 = 6 slugs. How long will it take the skater to coast from 11 ftjsec (7.5 mph)
to I ft/sec? How far will the skater coast before coming to a complete stop? Solution We answer the flISt question by solving Equation (I) for I: lIetl18 ~ et118 ~
Eq. (I) with k ~ 1/3, m = 6, Vo = 11, v = 1
I
1/11
1/18
~
In (1/11)
~
I
~
18ln 11 '" 43 sec.
In 11
We answer the second question with Equation (3):
. D.stance coasted
vom
=
T
=
198 ft.
~
11·6 1/3
•
Inaccuracy of the Exponential Population Growth Model In Section 7.4 we modeled popolation growth with the Law of Exponential Change: dP iii =
P(O)
!P,
~
Po
whereP is the population at time t, k > 0 is a constant growth mte, and Po is the size of the population at time I ~ O. In Section 7.4 we found the solution P ~ Poe kt to this model. To assess the model, notice that the exponential growth differential equation says that dPt
dl
= k
(4)
512
Chapter 9: FirstOrder Differential Equations
TABLE 9.3 WorLd popuLation (midyear)
f
7000
World population (19802008)
( 5000
~O~~I~OAv
I
30
)
t
FIGURE 9.10 Notice that the value of the solutionP ~ 4454e°.0 17t is 7169 when t ~ 28, which is nearly 7% more than the actual population in 2008.
Orthogooal trajectory
Year
Population (millions)
1980 1981 1982 1983 1984 1985 1986 1987 1988 1989
4454 4530 4610 4690 4770 4851 4933 5018 5105 5190
76/4454 80/4530 80/4610 80/4690 81/4770 82/4851 85/4933 87/5018 85/5105
'" '" '" '" '" '" '" '" '"
0.0171 0.0177 0.0174 0.0171 0.0170 0.0169 0.0172 0.0173 0.0167
Source: u.s. Bureau of the Census (Sept., 2007): WWW.ceDSUS .gov/ipc/www/idb.
is constant. This rate is called the relative growth rate. Now, Table 9.3 gives the world population at midyear for the years 1980 to 1989. Taking dt = I and tiP '" l1P, we see from the table that the relative growth rate in Equation (4) is approximately the constant 0.017. Thus, based on the tabled data with t = 0 representing 1980, t = I representing 1981, and so forth, the world population could be modeled by the initial value problem,
tiP Iii =
FIGURE 9.11 An nrthogonal trajectory intersects the family of curves at right angles, or nrthogonally.
l1P/P
0.017P,
P(O) = 4454.
The solution to this initial value problem gives the population function P = 4454eo.017t.1n year 2008 (so t = 28), the solution predicts the world population in midyear to be about 7169 million, or 7.2 billion (Figure 9.10), which is more than the actoa1 population of 6707 million from the U.S. Bureau of the Census. A more realistic model would consider environmental and other factors affectiog the growth rate, which has been steadily decliniog to about 0.012 since 1987. We consider one such model in Section 9.4.
Orthogonal Trajectories y
An orthogonal trajectory of a family of curves is a curve that intersects each curve of the family at right angles, or orthogonally (Figure 9.11). For instance, each straight line through the origin is an orthogonal trajectory of the family of circles x2 + y2 = a 2 , centered at the origin (Figure 9.12). Such mutoa1ly orthogonal systems of curves are ofparticular importance in physical problems related to electrical potential, where the curves in one family correspond to strength of an electric field and those in the other family correspond to constant electric potential. They also occur in hydrodynamics and heatflow problems.
EXAMPLE 2 Find the orthogonal trajectories of the family of curves xy a .. 0 is an arhitrary constant.
FIGURE 9.12 Every straight line through the origin is orthogonal to the family of circles centered at the origin.
= a, where
Solution The curves xy = a form a family of hyperbolas having the coordinate axes as asymptotes. First we find the slopes of each curve in this family, or their dy/ ax values. Differentiatiog xy = a implicitly gives
dy
xax+y=O
or
dy
Y
ax
X'
9.3
Applications
513
Thus the slope of the tangent line at any point (x, y) on one of the hyperbolas xy = a is y' = y/x. On an orthogonal trajectory the slope of the tangent line at this same point must be the negative reciprocal, or x/yo Therefore, the orthogonal trajectories must satisfy the differential equation
dy dx
x y'
This differential equation is separable and we solve it as in Section 7.4:
ydy = xdx
Separate variables. Integrate both sides.
FIGURE 9.13 Each curve is orthogonal to every curve it meets in the other family
(Example 2).
(5)
where b = 2C is an arbitrary constant. The orthogonal trajectories are the family ofhyperbolas given by Equation (5) and sketched in Figure 9.13. •
Mixture Problems Suppose a chemical in a liquid solution (or dispersed in a gas) runs into a container holding the liquid (or the gas) with, possibly, a specified amount of the chemical dissolved as well. The nrixture is kept uniform by stirring and flows out of the container at a known mte. In this process, it is often important to know the concentration of the chemical in the container at any given time. The differential equation describing the process is based on the formula (mte at WhiCh) (mte at WhiCh) Rate of change of amount = chemical chemical in container arrives departs.
(6)
If y(t) is the amount of chemical in the container at time t and V(t) is the total volume of liquid in the container at time t, then the departure mte of the chemical at time t is y(t) Departuremte = V(t) • (outflow rate)
=
concentration in ) ( container at time t • (outflow rate) .
(7)
Accordingly, Equation (6) becomes
dy . . y(t) dt = (chenncal'samvalmte)  V(t) . (outflow rate).
(8)
If, say, y is measured in pounds, V in gallons, and t in minutes, the uuits in Equation (8) are pounds minutes
pounds pounds gallons minutes  gallons • minutes'
EXAMPLE 3 In an oil refinery, a storege tank contains 2000 gal of gasoline that initially has 100 lb of an additive dissolved in it. In preparation for winter weather, gasoline containing 2 lb of additive per gallon is pumped into the tank at a rate of 40 gal/min.
514
Chapter 9: FirstOrder Differential Equations
The wt:llmixed solution is pumped out at a rate of 45 gal/min. How much of the additive is in the taJ:Jk 20 min after the pumping process begins (Figure 9.14)? 40 gaIImin containing 21b1pl
• FIGURE 9.14 The storage tank in Example 3 mixes input liquid with stored liquid to produce an output liquid.
Lety be the amolDlt (in pounds) of additive in the taJ:Jk at time I. We know that y = 100 when I = O. The number of gallons of gasoline and additive in solution in the taJ:Jk at any time I is
Solution
~ 2000 gal + (40: 
V(I)
~
45 : ) (tmin)
(2000  51) gal.
Thcrefon:,
y(1)
Rate out
=
 (200: ~
Eq.(7)
Y(I)' outflow rate
Ootflawrate ia45 gaJjmin and v  2000  St.
51) 45
45y Ib 2000 51 min .
Also, Hate in
~ (2~)(4O:) ~801b. mm
The diffcrcn:tial equation modeling the
mnture process is
dy 45y dt ~ 80  2000 51
Eq. (8)
in pmmds per minute. To solve this dift"erent:i.al equation, wt: rll"St write it in standard linear form: dy dt
45
51 Y ~ 80.
+ 2000
Thus, P(I)  45/(2000  51) and Q(I)  80. Th. integnUing factnr is v(l)
=
eJPdt
=
eJ~dt
= e9In(2000St)
~ (2000 
5Ir'.
20005t> 0
9.3
Applications
515
Multiplying both sides of the standard equation by v(l) and integrating both sides gives 9 (2000  51r •
(2000  51r9
(t + 200~5_ Y)
t+
51
= 80(2000  51r
9
45(2000  51)10 Y = 80(2000  51r9
1
:1 [(2000  51)9y = 80(2000  51r9
(2000  51)9y =
J
80(2000  51)9 dl
9 (2000  51r' (2000  51) Y = 80· (8)(5)
+ c.
The general solution is
y
= 2(2000  51)
+
C(2000  51)9.
Because y = 100 when I = 0, we can detennine the value of C: 100 = 2(2000  0)
C=
+
C(2000  0)9
3900 (2000)9·
The particular solution of the initial value problem is
The amount of additive 20 min after the pumping begins is y(20) = 2[2000  5(20)1  ( 3900)9 [2000  5(20)1 9 .., 1342Ib. 2000
•
Exercises 9.3 Motion Along a Line 1. Coa.ting bicycle A 66·kg cyclist on a 7·kg bicycle starts coast· ing on level ground at 9 m/sec. The k in Equation (I) is about 3.9 kgfsec.
a. About how far will the cyclist coast before reaching a com· plete stop?
b. How long will it take the cyclist~ speed to drop to I m/sec? 2. Coa.ting battleship Suppose that an Iowa class battleship has rnass around 51,000 metric tons (51,000,000 kg) and a k value in
Equation (I) of about 59,000 kgfsec. Assume that the ship loses power when it is moving at a speed of9 mfsec. a. About how far will the ship coast before it is dead in the water?
b. About how long will it take the ship~ speed to drop to I mfsec? 3. The data in Table 9.4 were collected with a motion detector and a CBLTM by Valerie Sharritts, a mathematics teacher at St. Francis DeSales High School in Columbus, Ohio. The table shows the distance s (meters) coasted on in·line skates in t sec by her daughter Ashley when she was 10 years old. Find a model for Ashley's
516
Chapter 9: FirstOrder Differential Equations
position given by the data in Table 9.4 in the form of Equation (2). Her initial velocity was Vo ~ 2.75 mlsec, her mass m ~ 39.92 kg (she weighed 88Ib), and her total coasting distaoce was 4.91 m.
Orthogonal Trajectories In Exercises S10, fmd the orthogonal trajectories of the family of curves. Sketch several members of each family. 6.y~cx2
5.y~mx
7. k:x;2
TABLE 9.4 Ashley Sharritts skating data
t (sec)
s(m)
t (sec)
s(m)
t (sec)
s(m)
0 0.16 0.32 0.48 0.64 0.80 0.96 1.12 1.28 1.44 1.60 1.76 1.92 2.08
0 0.31 0.57 0.80 1.05 1.28 1.50 1.72 1.93 2.09 2.30 2.53 2.73 2.89
2.24 2.40 2.56 2.72 2.88 3.04 3.20 3.36 3.52 3.68 3.84 4.00 4.16 4.32
3.05 3.22 3.38 3.52 3.67 3.82 3.96 4.08 4.18 4.31 4.41 4.52 4.63 4.69
4.48 4.64 4.80 4.96 5.12 5.28 5.44 5.60 5.76 5.92 6.08 6.24 6.40 6.56
4.77 4.82 4.84 4.86 4.88 4.89 4.90 4.90 4.91 4.90 4.91 4.90 4.91 4.91
+ y2
~
I
11. Show that the
8.2x2+y2~c2
10. y
9. y = ce x curves 2x 2
+ 3y2
=
eb
~ S and y2 ~ x 3 are orthogonal.
12. Find the family of solutions of the given differential equation and the family of orthogonal trajectories. Sketch both families.
••
xdx+ydy~O
b.xdy2ydx~0
Mixture Problems 13. Salt miItnre A tack initially contains \00 gal of brine in which SO Ib of salt are dissolved. A brine containing 2 Ib/gal of salt runs into the tack at the rate of S gal/min. The mixture is kept uniform by stirring and flows out of the tack at the rate of4 gal/min. •• At what rate (pounda per minute) does salt enter the tack at time I?
b. What is the volume of brine in the tank at time I? •• At what rate (pounda per minute) does salt leave the tack at
time t? d. Write down and solve the initial value problem describing ilie mixing process. e. Find the concentration of salt in ilie tank 25 min afler the
process starts. 4. Coasting to ••Iop Thhle 9.S shows the distaoce s (meters) coasU:d on inline skates in terms of time t (seconds) by Kelly Schmilzer. Find a model for her position in the form of Equation (2). Her initial velocity was Vo ~ 0.80 mlsec, her mass m ~ 49.90 kg (110 Ib), and her total coasting distaoce was 1.32 m.
TABLE 9.5
•• At what time willilie tack be full?
b. At ilie time the tack is full, how many pounds of concentrate will it contain?
Kelly Schmitzer skating data
t (sec)
s(m)
t (sec)
s(m)
t (sec)
s(m)
0 0.1 0.3 0.5 0.7 0.9 1.1 1.3
0 0.07 0.22 0.36 0.49 0.60 0.71 0.81
1.5 1.7 1.9 2.1 2.3 2.5 2.7 2.9
0.89 0.97 1.05 1.11 1.17 1.22 1.25 1.28
3.1 3.3 3.5 3.7 3.9 4.1 4.3 4.5
1.30 1.31 1.32 1.32 1.32 1.32 1.32 1.32
9.4
14. MiItnre problem A 200gal tack is half full of distilled water. At time t ~ 0, a solution containing O.Slb/gal of concentrate enters ilie tack at the rate of 5 gal/min, and the wellstirred mixture is wiilidrawn at ilie rate 00 gal/min.
15. Fertilizer mhture A tack contains 100 gal offresh water. A solution containing I Ib/gal of soluble \awn fertilizer ruos into the tack at ilie rate of 1 gal/min, and ilie mixture is pumped out ofilie tack at the rate 00 gal/min Find ilie maxinuun amount of fertilizer in ilie tack and the time required to reach the maxinuun. 16. Carbon monmide pollution An executive conference room of a corporation contains 4S00 It' of air initially free of carbon monoxide. Starting at time t ~ 0, cigarette smoke containing 4% carbon monoxide is blown into the racon at the rate of 0.3 It'/min. A ceiling fan keeps the air in the room well circulated and the air leaves ilie room at the same rate of 0.3 It'/min. Find the time when the concentration of carbon monoxide in the room reaches 0.01 %.
Graphical Solutions of Autonomous Equations In Chapter 4 we learned that the sign of the first derivative tells where the graph of a function is increasing and where it is decreasing. The sign of the second derivative tells the concavity of the graph. We can build on our knowledge of how derivatives determine the shape of a graph to solve differential equations graphically. We will see that the ability to
9.4 Graphical Solutions of Autonomous Equations
517
discern physical behavior from gmphs is a powerful tool in understanding realworld systems. The starting ideas for a gmphical solution are the notions of phase line and equilibrium value. We arrive at these notions by investigating, from a point of view quite different from that studied in Chapter 4, what happens when !be derivative of a differentiable function is zero.
Equilibrium Values and Phase Lines When we differentiate implicitly the equation
I
Sin (5y  15)
= x
+ I,
we obtain
I( 5 )dY
S
5y  15
dx = I.
Solving for y' = dy/dx we find y' = 5y  15 = 5(y  3). In this case the derivative y' is a function ofy only (the dependent variable) and is zero when y = 3. A differential equation for which dy/ dx is a function ofy only is called an autonomous differential equation. Let's investigate what happens when the derivative in an autonomous equation equals zero. We assume any derivatives are continuous.
DEnNmON If dy/dx = g(y) is an autonomous differential equation, then the values of y for which dy/ dx = 0 are called equilihrium values or rest points.
Thus, equilibrium values are those at which no change occurs in the dependent variable, so y is at rest. The emphasis is on the value of y where dy/ dx = 0, not the value of x, as we studied in Chapter 4. For example, the equilibrium values for the autonomous differential equation dy dx = (y
+
I)(y  2)
arey = I andy = 2. To construct a graphical solution to an autonomous differential equation, we fIrst make a phase line for the equation, a plot on the yaxis that shows the equation's equilibrium values along with the intervals where dy/dx and d'y/dx 2 are positive and negative. Then we know where the solutions are increasing and decreasing, and the concavity of the solution curves. These are the essential features we found in Section 4.4, so we can determine the shapes of the solution curves without having to find formulas for them.
EXAMPLE 1
Draw a phase line for the equation dy dx = (y
+ I)(y  2)
and use it to sketch solutions to the equation.
Solution 1.
Draw a number line for y and mark the equilibrium values y =  I and y = 2, where dy/dx = O. 1@)i@~>, Y
1
2
518
Chapter 9: FirstOrder Differential Equations 2.
IdentifY and label the intervals where y' > 0 and y' < 0, This step resembles what we did in Section 4.3, only now we are marking the yaxis instead of the xaxis, I I I I
y'>o
I I I I
y' 2, we have y' > 0, so a solution with a yvalue greater than 2 will increase from there without bound, In short, solution curves below the horizontal line y =  I in the xyplane rise toward y =  L Solution curves between the lines y = I and y = 2 fall away from y = 2 toward Y =  L Solution curves above y = 2 rise away from y = 2 and keep going up,
3.
Calculate y' and mark the intervals where y' > 0 and y" entiate y' with respect to x, using implicit differentiation, y' = (y
+ I)(y  2)
= y2  Y 
2
O y">O
dx
differentiated implicitly with respect to x
2Y.)1'  y'
= (2y 
I)y'
= (2y 
I)(y
+
I)(y  2),
From this formula, we see that y" changes sign at y =  I, y = 1/2, and y = 2, We add the sign information to the phase line. y'>O y" o. Figure 9.17 adds this information to the phase line. The completed phase line shows that if the temperatore of the object is above the equilibrium value of 15°C, the graph of H(t) will be decreasing and concave upward. If the temperature is below 15°C (the temperature of the surrounding medium), the graph of H(t) will be increasing and concave downward. We use this information to sketch typical solution curves (Figure 9.18).
520
Chapter 9: FirstOrder Differential Equations From the upper solution curve in Figure 9.18, we see that as the object cools down, the rate at which it cools slows down because 1iH/ dt approaches zero. This observation is implicit in Newton's law of cooling and contained in the differential equation, but the flattening of the graph as time advances gives an immediate visual representation of the phenomenon.
A Falling Body Encountering Resistance Newton observed that the rate of change in momentum encountered by a moving equal to the net force applied to it. In mathematical terms,
F
=
d
o~ect
is
(2)
dt (mv),
where F is the net force acting on the object, and m and v are the object's mass and velocity. Ifm varies with time, as it will if the o~ect is a rocket burning fuel, the righthand side of Equation (2) expands to
using the Derivative Product Rule. In many situations, however, m is constant, dm/dt = 0, and Equation (2) takes the simpler form or
F=
'Ina,
(3)
s
known as Newton second law ofmotion (see Section 9.3). In free fall, the constant acceleration due to gravity is denoted by g and the one force acting downward on the falling body is
y
Fp FIGURE 9.19 An object falling under 1he influence of gravity wi1h a resistive force assumed to be proportional to 1he velocity.
=
mg,
the force due to gravity. If, however, we think of a real body falling through the air1!ay, a penny from a great height or a parachutist from an even greater heightwe know that at some point air resistance is a factor in the speed of the fall. A more realistic model of free fall would include air resistance, shown as a force F, in the schematic diagram in Figure 9.19. For low speeds well below the speed of sound, physical experiments have shown that F, is approximately proportional to the body's velocity. The net force on the falling body is therefore
F = Fp  F" giving
dv mdt=mgku dv k dt =g m V '
(4)
We can use a phase line to analyze the velocity functions that solve this differential equation. The equilibrium point, obtained by setting the righthand side of Equation (4) equal to

dv>O dt ____
I I I
dv v
{@~~
mg
zero, is
mg v=T'
k
FIGURE 9.20 Initial pbase line for 1he falling body encountering resistance.
If the body is initially moving faster than this, dv/dt is negative and the body slows down. If the body is moving at a velocity below mg/k, then dv/ dt > 0 and the body speeds up. These observations are captured in the initial phase line diagram in Figure 9.20.
9.4 Graphical Solutions of Autonomous Equations
dv>O dt 2 d v dt 2
0 dt
I I I I
• , @
dP 0 if 0 < P < M and dP/ dt < 0 if P > M. These observations are recorded on the phase line in Figure 9.23. We determine the concavity of the population curves by differentiating both sides of Equation (6) with respect to t:
(7)
If P = M/2, then d 2p/dt 2 = O. If P < M/2, then (M  2P) and dP/dt are positive and d 2p/dt 2 > O. If M/2 < P < M, then (M  2P) < 0, dP/dt > 0, and d 2p/dt 2 < O.
522
Chapter 9: FirstOrder Differential Equations
I I I
I
If P > M, then (M  2P) and dP/dt are both negative and d 2p/dt 2 > O. We add this information to the phase line (Figure 9.24). The lines P = M/2 and P = M divide the llfst quadrant of the tPplane into horizontal bands in which we know the signs of both dP/dt and d 2p/dt 2• In each band, we know how the solution curves rise and fall, and how they bend as time passes. The equilibrium lines P = 0 and P = M are both population curves. Population curves crossing the line P = M/2 have an inflection point there, giving them a sigmoid shape (curved in two directions like a letter S). Figure 9.25 displays typical population curves. Notice that each population curve approaches the limiting population M as t + 00.
dP >0 dt I
2
' 4X>o I dt 2
@l o
J'
2 : d P I
dt2
I,
8. y' = y3 _ y2
a
Models of Population Growth The autonomous diffi:rential equations in Exercises 912 represeot models for populatioo growth. For each exercise, use a phase line analysis to sketch solution curves for p(t), selecting different startiog values P(O). Which equilibria are stahle, aod which are uostable?
dP 9. dt = I  2P dP 11. dt
=
2P(P  3)
dP 10. dt
= P( I
and that the current populatioo Po is fairly close to the carrying capacity Mo. You ntight imagine a populatioo of fish living in a freshwater lake in a wilderuess area. Suddeoly a catastrophe such as the Mount St Helens volcauic eruption contaminates the lake and destroys a siguificant part of the food and oxygen 00 which the fish depend. The result is a new enviromnent with a carrying capacity Ml coosiderably less than Mo and, in fact, less than the current population Po. Startiug at some time before the catastrophe, slretch a ''beforeandafter'' curve that shows how the fish populatioo responds to the change in environment 14. Controlling a population The fish and game depar1ment in a certain state is planning to issue huntiog permits to cootrol the deer population (one deer per permit). It is knowo that if the deer populatioo falls below a certain level m, the deer will become extioct. It is also knowo that if the deer population rises above the carrying capacity M, the population will decrease back to M throngh disease and maIoutrition.
a. Discuss the reasonableness of the following model for the growth rate of the deer population as a function of time:
dP dt
= rP(M 
P)(P  m),
 2P)
12 dP = 3P(i _ P)(p • dt
_1) 2
13. Catastropbic change in logistic growth Suppose that a healthy population of some species is growing in a limited environment
where P is the population of the deer and r is a positive constant of proportionality. Ioclude a phase line.
b. Explain how this model differs from the logistic model dP/ dt = rP(M  Pl. Is it better or worse than the logistic model?
9.5
> Mforallt,thenlim,_ooP(t) What happens if P < m for all t?
c. ShowthatifP d.
~
M.
e. Discuss the solutions to the differential equation. What are the equilibrium points of the model? Explain the dependence of the steadystste value of P on the initial values of P. About how many pcrruits should he issued?
Applications and Examples 15. Skydiving If a body of mass m falling from rest under the ac· tion of gravity encouoters an air resistsoce proportional to the square of velocity, then the body's velocity t seconds into the fall
satisfies the equation dv 2 m=mgkv dt '
Systems of Equations and Phase Planes
523
a. Discuss the reasonableness of the model. b. Construct a phase line identifying the signs of X' and X".
c. Sketch representative solution curves. d. Predict the value of X for which the information is spreading most rapidly. How many people eventually receive the infor
mation? 19. Current in an RLcircuit The accoropanying diagram represents an electrical circuit whose total resistance is a constant R ohms and whose selfinductance, shown as a coil, is L henries, also a constant. There is a switch whose terminals at a and b can be closed to connect a constant electrical source of V volts. From Section 9.2, we have
k>O
di
.
Ldt+RI~V,
where k is a constsot that depends on the body's aerodynamic properties and the density of the air. CWe assume that the fall is too short to be affected by changes in the air's density.)
where i is the current in amperes and t is the time in seconds.
v
a. Draw a phase line for the equation.
+ d~ 
b. Sketch a typical velocity curve.
~b
c. For al60·lb skydiver (mg ~ 160) and with time in seconds and distsoce in feet, a typical value of k is 0.005. What is the diver's terminal velocity?
Switch
Vv
16. Resistance proportional to A body ofmassm is projected vertically downward with initial velocity Vo. Assume that the reo sisting force is proportional to the square root of the velocity and [mel the terruioal velocity from a graphical analysis. 17. Sailing A sailboat is running along a slraight course with the wind providing a constsot forward force of 50 lb. The only other force acting on the boat is resistance as the boat moves through the water. The resisting force is numerically equal to five times the boat's speed, and the initial velocity is 1 ft/sec. What is the maximum velocity in feet per second of the boat under this wind? 18. The spread of information Sociologists recognize a phenomenon called social diffUsion, which is the spreading of a piece of informa· tion, technological innovation, or cultoral fad among a population. The members of the population can he divided into tv. classes: those who have the information and those who do not In a fixed population whose size is known, it is reasonable to assume that the rate of diffusion is proportional to the number who have the infor· mation times the number yet to receive it IfX denotes the nwnhcr of individuals who have the information in a population of N people, then a mathematical model for social diffusion is given by
L
Use a phase line analysis to sketch the solution curve assuming that the switch in the RLcircuit is closed at time t ~ O. What happens to the current as t > oo? This value is called the steadystate solution. 20. A pearl in shampoo Suppose that a pearl is sinking in a thick fluid, like shampoo, subject to a frictional force opposing its fall and proportional to its velocity. Suppose that there is also a resistive buoyant force exerted by the shampoo. According to Archimedes'principle, the buoyant force equala the weight of the fluid displaced by the pearl. Using m for the mass of the pearl and P for the mass of the shampoo displaced by the pearl as it desceods, coroplete the following steps. a. Draw a schematic diagram showing the forces acting on the pearl as it sinks, as in Figore 9.19. b. Using v(t) for the pearl's velocity as a func1ion of time t, write a differential equation modeling the velocity of the pearl as a falling body. c. Construct a phase line displaying the signs of v' and v".
dX
dt ~ kX(N  X),
d. Sketch typical solution curves.
where t represents time in days and k is a positive constant.
9.5
R
e. What is the terminal velocity of the pearl?
Systems of Equations and Phase Planes In some situations we are led to consider not one, but several firstorder differential equations. Such a collection is called a system of differential equations. In this section we present an approach to understanding systems through a graphical procedure known as a phaseplane analysis. We present this analysis in the context of modeling the populations of trout and bass living in a common pond.
524
Chapter 9: FirstOrder Differential Equations
Phase Planes A general system of two firstorder clifferential equations may tske the form
dx dt = F(x,y), dy dt = G(x,y). Such a system of equations is called autonomous because dx/ dt and dy/ dt do not depend on the independent variable time t, but only on the dependent variables x andy. A solution of such a system consists of a pair of functions x(t) and yet) that satisfies both of the differential equations simultaneously for every t over some time interval (finite or infinite). We cannot look at just one of these equations in isolation to find solutions x(t) or y(t) since each derivative depends on both x andy. To gain insight into the solutions, we look at both dependent variables together by plotting the points (x(t), yet»~ in the xyplane starting at some specified point. Therefore the solution functions define a solution curve through the specified point, called a trajectnry of the system. The xyplane itself, in which these trajectories reside, is referred to as the phase plane. Thus we consider both solutions together and study the behavior of all the solution trajectories in the phase plane. It can be proved that two trajectories can never cross or touch each other. (Solution trajectories are examples of parametric curves, which are studied in detail in Chapter 11.)
A CompetitiveHunter Model Imagine two species of fish, say trout and bass, competing for the same limited resources (such as food and oxygen) in a certain pond. We let x(t) represent the number of trout and yet) the number of bass living in the pond at time t. In reality x(t) and y(t) are always integer valued, but we will approximate them with realvalued differentiable functions. This allows us to apply the methods of clifferential equations. Several factors affect the rates of change of these populations. As time passes, each species breeds, so we assume its population increases proportionally to its size. Taken by itself, this would lead to exponential growth in each of the two populations. However, there is a countervailing effect from the fact that the two species are in competition. A large number of hass tends to cause a decrease in the number of trout, and viceversa. Our model takes the size of this effect to be proportional to the frequency with which the two species interact, which in turn is proportional to xy, the product of the two populations. These considerations lead to the following model for the growth of the trout and hass in the pond:
dx dt = (a  by)x,
(la)
dy dt = (m  nx)y.
(lb)
Here x(t) represents the trout population, yet) the bass population, and a, b, m, n are positive constants. A solution of this system then consists of a pair of functions x(t) and yet) that gives the population of each fish species at time t. Each equation in (I) contains both of the unknown functions x and y, so we are unable to solve them individually. Instead, we will use a graphical analysis to study the solution trajectories of this competitivehunter model. We now examine the nature of the phase plane in the troutbass population model. We will be interested in the I st quadrant of the xyplane, where x ;" 0 and y ;" 0, since populations cannot be negative. First, we determine where the bass and trout populations are both constant. Noting that the (x(t),y(t» values remain unchanged when dx/dt = 0 and dy/dt = 0, Equations (Ia and Ib) then become
(a  by)x = 0, (m  nx)y =
o.
This pair of simultaneous equations has two solutions: (x, y) = (0, 0) and (x, y) = (m/n, a/b). At these (x,y) values, called equilibrium or rest points, the two populations
9.5
525
Systems of Equations and Phase Planes
remain at constant values over all time. The point (0, 0) represeots a pond containing no members of either fish species; the point (m/n, a/b) corresponds to a pond with an unchanging number of each fish species. Next, we note that ify = alb, then Equation (Ia) implies dx/dt = 0, so the trout popillation x(t) is constant. Similarly, if x = min, theo Equation (Ib) implies dy/dt = 0, and the bass popu1ation y(t) is constant. This information is recorded in Figure 9.26. y Bass
y Bass
y Bass I
!!
I, y(O) x
> 0,
+ 3y2) tho + Y dx = 0 (Hint: d(xy) = y dx + x tho) x tho + (3y  x2 cosx) dx = 0, x> 0
20. x tho
21 . .>:y'
+ (y 
+ (x
22. y dx
cosx)dx = 0,
 2)y
+ (3x
= h3e~,
 .>:y
y(l)
=
=
I
I
y(O) = I
amx
13. (l + eX) tho + (ye x + e~) dx = 0 14. e~ tho + (e~y  4x) dx = 0
15. (x
9. Why is the exponential model unrealistic for predicting longterm population grow1h? How does lbe logistic model correct for the deficiency in the exponential model for population grow1h? What is the logistic differential equation? What is the form of its solution? Desctibe the graph of the logistic solution.
In Exercises 1722 solve lbe initial value problem.
6. y' =
7. x(x  I) tho  ydx = 0 9. 2y'  y
8. How do you construct lbe phase line for an autonomous differential equation? How does lbe phase line help you produce a graph which qualitatively depicts a solution to the differential equation?
Practice Exercises
In Exercises 116 solve the differential equation.
3. secxtho
7. What is an autonomous differential equation? What are its equilibtium values? How do they differ from ctitical points? What is a stable equilibtiwn value? Unstable?
+ 2) dy
y(~) y(l)
= 0,
=
= 0
0
y(2) = I, y < 0
530
Chapter 9: FirstOrder Differential Equations
Euler's Method
Massm mgR2
In Exercises 23 and 24. use Euler's method to solve the initial value problem 00 the given interval starting at Xo with dx ~ 0.1.
0 23.
y' ~ y
+
cosx,
y(O) ~ 0;
0,;; x ,;; 2;
Xo
F~
,2
~
~ 0
0 24. y' ~ (2  y)(2x + 3), y(3) ~ I; 3:sx:s 1; Xo =3 In Exercises 25 and 26, use Euler's method with dx ~ 0.05 to estimate y(c) where y is the solution to the given initial value problem.
0 25. c ~ 3;
dy x2y dx ~ X+T' y(O) ~ I
0 26. c ~ 4;
dy x 2 2y+1 dx ~ x ,y(l) ~ I
In Exercises 27 and 28, use Euler's method to solve the initial value problem graphically, starting at Xo ~ 0 with L
0 27. dxdy
dx
~
0.1.
~
;
eX Y
+2'
b. dx
y(O)
~
~
0.1.
2gR.

v'2iii
30. y' ~ I/x
29.y'=x
32. y' ~ I/y
xy
Autonomous Differential Equations and Phase Lines In Exercises 33 and 34: IdentifY the equilibrium values. Which are stahle and which are unstable?
b. Construct a phase line. IdentifY the signs of y' and y'. c. Sketch a representative selection of solution curves.
dy
dy
33'dx~y21
34.dx~y_y2
Applications 35. Escape velocity
+ VOl
The gravitationsl attraction F exerted by an
airless moon on a body of mass m at a distance s:from the moon's center is given by the equation F = mg R 2s 2, where g is the acceleration of gravity at the moon's surface and R is the moon's radius (see accompanying figaro). The force F is negative because it acts in the direction of decreasing s.
Chapter
v'2iii,
The velocity Vo = is the moon's escape velocity. A body projected upward with this velocity or a greater one will escape from the moon's gravitational pull.
Slope Fields
L
2gR2 = .
Thus, the velocity remains positive as loog as Vo '"
In Esercises 2!132, sketch part of the equatioo's slope field. Then add to your sketch the solutioo curve that pssses through the point P( I, I). Use Euler~ method with Xo ~ I and dx ~ 0.2 to estimate y(2). Round your answers to four decimal places. Find the exact value of y(2) for comparison. ~
v2
2
dy x2 + Y 0 28. dx ~  eY + x' y(O) ~ 0
31. y'
a. If the body is projected vertically upward from the moon's surface with an initial velocity Vo at time t = O. use Newton's second law, F ~ rna, to show that the body's velocity at position s is given by the equation
b. Show that if Vo ~
v'2iii, then 3vo
.~RI+2Rt (
)2/3 .
36. Coasting to a stop
Table 9.6 shows the distance. (meters) coasted 00 inline skates in t sec by 10huathon Krueger. Find a model for his position in the form of Equatioo (2) ofSectioo 9.3. His initial velocity was Vo ~ 0.86 m/sec, his mass m ~ 30.84 kg (be weighed 68 Ib), and his total coasliog distance 0.97 m.
TABLE 9.6 Johnathon Krueger skating data
t (sec)
s(m)
t (sec)
s(m)
t (sec)
s(m)
0 0.13 0.27 0.40 0.53 0.67 0.80
0 0.08 0.19 0.28 0.36 0.45 0.53
0.93 1.06 1.20 1.33 1.46 1.60 1.73
0.61 0.68 0.74 0.79 0.83 0.87 0.90
1.86 2.00 2.13 2.26 2.39 2.53 2.66
0.93 0.94 0.95 0.96 0.96 0.97 0.97
Additional and Advanced Exercises
Theory and Applications 1. Transport through a cell membrane Under some conditions, the result of the movement ofa dissolved substance across a cell's membrane is described by the equation
dy A dt ~ kV(c  y). In this equation, y is the concentration of the substance inside the cell and dy/dtis the rate at whichy changes over time. The letter>
Chapter 9 Technology Application Projects k,A, V, and c stand for constants, kbeing the permeability coefficient (a property of the membrane), A the surface area of the membrane. V the cell's volume, and c the concentration of the
•• Vetify that y(x) problem
substance outside the cell. The equation says that the rate at which the cooceotratioo changes within the cell is proportional to the difference between it and the outside concentration.
b. For the integratiog factor v(x) = ef P«)"', show that
y'
In this equation, m is the mass of the system at time t, v is its velocity, and v + u is the velocity of the mass that is entering (or leaving) the aystern at the rate dm/ dt. Suppose that a rocket of initial mass mo starts from rest, but is driven upward by ftring some of its mass directly backward at the constant rate of dm/dt = b units per second and at constant speed relative to the rocket u = c. The ooly external force acting on the rocket is F = mg due to gravity. Under these assumptions, show that the height of the rocket above the ground at the end of t secoods (t smal\ compared to molb) is
_ [ mo  bt In mo  bt]_ 1 2 yct+ b mo 2gt· 3••• Assume that P(x) and Q(x) are continuous over the interval [a, b]. Use the Foodameotal Theorem of Calculus, Part I to show that aoy function y satisfYing the equation
v(x)y
=
J
v(x)Q(x) tb:
+C
for v(x) = ef p(.) '" is a solutioo to the tirstrder Hoear equstion
dy tb: + P(x)y = Q(x).
J:'
b. If C = yov(xo) v(t)Q(t) dt, then show that any solutioo y in part (a) satisfies the initial conditioo y(xo) = Yo.
4. (Continuation ofExercise 3.) Assume the hypotheses of Exercise 3, and assume that Yl (x) and Y2(X) are both solutions to the firstorder Hoear equation satisfYing the initial condition y(xo) = Yo.
Chapter
y(xo) = O.
o.
c. Fnnn part (a), we have Yl (xo)  Y2(XO) = O. Since v(x) > 0 fora < x < b, usc part (b) to estshlish that Yl(X)  Y:z(x) = 0 on the interwl (a, b). ThusYl(X) = Y2(X) for all a < x < b.
tim
+ (v + u)Tt·
= F
+ P(x)y = 0,
Conclude that v(x )[Yl (x)  Y2(X)] = constant.
2. Height of • rocket If an exteroa\ force F acts upon a aystem whose mass varies with time, Newton's law of motion is
d(mv)
Yl (x)  Y2(X) satisfies the initial value
d tb: (V(X)[yl(X)  Y2(X)]) =
•• Solve the equation for y(t), using Yo to deoote y(O). b. Find the steadystate cooceotratioo, lim,...ooy(t).
;It
=
531
Homogeneous Equations A rITstorder differeotiaI equation of the form
is called homogeneous. It can be transformed into an equation whose variables are separable by defIning the new variable v = y/x. Then, y = vxand dy dv tb:=v+xtb:. Substitution into the original differential equation and collecting terms with \ike vatiables then gives the separable equstioo tb:
dv
x + v _ F(v) =
O.
Afler solving this separable equation, the solution of the original equation is obtsined when we replace v by y/x .
Solve the branogeoeous equstioos in Exercises 510. First put the equation in the funn of a homogeneous equstioo. 5. (x 2 + y2)tb: + xydy = 0 6. x 2dy + (y2  xy)tb: = 0 7. (xeYi'
+ y)tb:  xdy
= 0
8. (x +y)dy+ (x y)tb: 9. y'
10.
Y
=i +
=
0
yx cos  x 
(x sin ~  ycos~) tb: + xcos ~ dy
Technology Application Projects
Mathematica/Maple Modules: Drllg Dosages: Are They Effective? Are They S"/e? Formulate and solve an initial value model for the absorption of a drug in the bloodstream. FirstOrtkr Differenlifll Eqllfltions and Slope Fields Plot slope fields and solution curves for various initial conditions to selected rrrstorder differeotiaI equations.
= 0
10 INFINITE SEQUENCES AND SERIES OVERVIEW Everyone knows how to add two numbers together, or even several. But how do you add infinitely many numbers together? In this chapter we answer this question, which is part of the theory of infinite sequences and series. An important application of this theory is a method for representing a known differentiable function f(x) as an infinite sum of powers of x, so it looks like a ''polynomial with infinitely many terms." Moreover, the method extends our knowledge of how to evaluate, differentiate, and integrate polynomials, so we can work with even more general functions than those encountered so far. These new functions are often solutions to important problems in science and engineering.
10.1 HrSlORICAL ESSAY
Sequences and Series
Sequences Sequences are fundamental to the study of infinite series and many applications of mathematics. We have already seen an example of a sequence when we studied Newton's Method in Section 4.6. There we produced a sequence of approximations Xn that became closer and closer to the root of a differentiable function. Now we will explore general sequences of numbers and the conditions under which they converge.
Representing Sequences A sequence is a list of numbers
in a given order. Each of aJ, a2, a3 and so on represents a number. These are the terms of the sequence. For example, the sequence 2,4,6,8, 10, 12, ... , 2n, ... has first term al = 2, second term a2 = 4, and nth term an = 2n. The integer n is called the index of an, and indicates where an occurs in the list. Order is important. The sequence 2, 4, 6, 8 ... is not the same as the sequence 4, 2, 6, 8 .... We can think of the sequence
as a function that sends 1 to al , 2 to a2, 3 to a3, and in general sends the positive integer n to the nth term an. More precisely, an infinite sequence of numbers is a function whose domain is the set of positive integers. The function associated with the sequence 2, 4, 6, 8, 10, 12, ... , 2n, ... sends 1 to al = 2, 2 to a2 = 4, and so on. The general behavior of this sequence is described by the formula an = 2n.
532
10.1 Sequences
533
We can equally well make the domain the integers larger than a given number no, and we allow sequences of!bis type also. For example, the sequence 12, 14, 16, 18,20,22 ... is described by the fonnula a" = 10 + 2n. It can also be described by the simpler fonnula bn = 2n, where the index n starts at 6 and increases. To allow such simpler fonnulas, we let the first index of the sequence be any integer. In the sequence above, {an} starts with a, while {b.} starts withb 6 • Sequences can be described by writing rules that specify their terms, such as • r an=vn,
bn = (_I)n+11 n'
dn = (_I)n+1,
n  I cn=n'
or by listing !enns:
{an}
=
{Vi, Vi, v'3, ... , Vn, ... }
{b n }
=
{1,!,t,i, .. ·,(I r1 k, ... }
{cn}
=
1234 nI } { 0'2'3'4'5'''''n''''
{dn }
= {I, I, I, I, I, I, ... ,
(I).+" ... }.
We also sometimes write
{a,,}
=
{Vn }:,.
Figure 10.1 shows two ways to represent sequences graphically. The fIrst marks the rrrs! few points from a" a2, a3,"" an,'" on the real axis. The second method shows the graph of the function defIning the sequence. The function is dermed only on integer inputs, and the graph consists of some points in the xyplane located at (I, ad, (2, a2), ... , (n, a,,), ....
+~~~I~I~ 'n
2
3
4
5
FIGURE 10.1 Sequences can be represented as peints on the real line or as points in the plane where the horizontal axis n is the index number of the term and the vertical axis all is its value.
Convergence and Divergence Sometimes the numbers in a sequence approach a single value as the index n increases. This happens in the sequence
534
Chapter 10: Infinite Sequences and Series whose terms approach 0 as n gets large, and in the sequence
{o,~, t, ~'~'"'' I

k, .. ·}
whose terms approach I. On the other hand, sequences like
{VI, Yz, V3, ... , V;;, ... } LE
o
L L+E
•• ••• • (• • I )
~ L+E
L
have tenDs that get larger than any number as n increases, and sequences like
(n,a,,) ~  . 
LE
{I, I, I, I, I, I, ... , O.
DEFINmONS The sequence {a,} converges to the number L iffor every positive number E there corresponds an integer N such that for all n,
~~~~~~~~n
o
2 3
N
n
la,  LI
N
In the representation ofa sequence as points in the plane, all ~ L if y = L is a horizontal asymptote of the sequence of points (in, a,)}. In this figure,
FIGURE 10.2
E.
Ifno such number L exists, we say that {an} diverges. If {an} converges to L, we write lim,~oo an = L, or simply a, + L, and call L the limit of the sequence (Figure 10.2).
all the a,/s after aN lie within E of L. The definition is very similar to the definition of the limit ofa function fix) asx tends to 00 (limx~oof 0)
n~OO
S.
lim 11_ 00
(1+'")"=e n
4. limxn=o n_OO
(lxl
odefine a sequence that, when it converges, gives a solution to the equation sin x  x 2 = o. •
Bounded Monotonic Sequences Two concepts that play a key role in determining the convergence of a sequence are those of a bounded sequence and a monotonic sequence.
DEFINmONS A sequence {a.} is bounded Crom above if there exists a number M such that a. ,,; M for all n. The number M is an upper bound for {a.}. If M is an upper bound for {a.} but no number less than M is an upper bound for {a.}, then M is the least upper bound for {a,,}. A sequence {a,,} is bounded from below if there exists a number m such that a. ;" m for all n. The number m is a lower bound for {a,,}. If m is a lower bound for {a,,} but no number greater than m is a lower bound for {a.}, then m is the greatest lower bound for {a,,}. If {a.} is bounded from above and below, the {a.} is bounded. If {a.} is not bounded, then we say that {a.} is an unbounded sequence. EXAMPLE 11 (a) The sequence 1,2,3, ... , n, ... has no upper bound since it eventually surpasses every number M. However, it is bounded below by every real number less than or equal to I. The number m = 1 is the greatest lower bound of the sequence.
123
n
.
(b) Thesequence Z'3'4"'" n + 1"" lSbounded above by every real number greater than or equal to 1. The upper bound M = 1 is the least upper bound (Exercise 125). The sequence is also bounded below by every number less than or equal to its greatest lower bound
t,
which is •
540
Chapter 10: Infinite Sequences and Series
I
Convergent .equences are bounded
If a sequence {a,,} converges to the number L, then by detmition there is a number N such that Ia"  L I < 1 if n > N. That is,
Ll N and n > N
~
I .... 
a" 1

133. Sequence. generated by N_on'. method Newton's method, applied to a differentiable functioo f(x), begins with a starting value Xo and coostructs from it a sequence of numbers {x,} that under favorable circumstances converges to a zero of f. The
recursion formula for the sequence is Xn+l
I
Which of the sequences in Exercises 115124 converge, and which di· 115.a,~I1i
a n +l =
I)!
114. a" ~ 2  Ii  2'
I
1,
al =
125. The sequence {II/(II + I)} bas a lea.t upper bound of 1 Show that if M is a number less thao I, then the terms of {n/(n + I)} eventually exceed M. That is, if M < I there is ao inreger N such that n/(n + I) > M whenever n > N. Since n/(n + 1) < I for every n, this proves that I is a least upper bouodfor {n/(n + I)}.
132. Prove that a sequence {a,} converges to 0 ifaod only if the quence of absolure values {I an I} cooverges to O.
In Exercises 111114, detennine if the sequence is monotooic aod if it ~
124.
131. For a sequence {a,,} the tenDs of even index are denored by au and the terms of odd index by au+ I. Prove that if au + L and a2k+1 + L, then a" + L.
107. Prove that lim.~oo 'C",; ~ 1.
111. a,
4"+ 1 + 3" 4'
n+ I 122. an =  n 
129. Uniqueness of limit. Prove that limits of sequences are uoique. That is, show that if LI aod L, are numbers such that On + LI and all + L2, then Ll = L,..
if c is any positive constant.
109. Prove Theorero 2.
~ I +~
123. a" ~
+ I)}.
max {x.. cos (n
127. Is it true that a sequence {a,} of positive numbers must cooverge if it is bouoded from above? Give reasons for your answer.
vi,;! ~ ~ for large values of n.
D b.
I)
=X
f(xn) ll 
!'(x
)'
lI
.. Show that the recursioo formula for f(x) ~ x 2 caobewrittelIaSxn+l ~ (xn + a/x,)/2.
D b.

a, a
> 0,
Starting with Xo ~ I aod a ~ 3, calculate successive tenDs of the sequence uotil the display begins to repeat. What number is being apprnsimared? Explain.
544
Chapter 10: Infinite Sequences and Series
D 134. Arecursi.. defioi1ionof"./2
Ifyou start with Xl = I anddefme the subsequent terms of {x,} by the rule x, = X,_I + COSX,_I, you geoerate a sequeoce that converges rapidly to 1f/2. (a) Try it. (b) Use the accompaoying figure to explain why the conver
a. Calculate and theo plot the frrst 25 terms of the sequeoce. Does the sequeoce appear to be bouoded from above or be
gence is so rapid.
b. If the sequeoce converges, find an integer N such that Ia"  L I :5 0.01 for n '" N. How far in the sequeoce do you have to get for the terms to lie within 0.0001 of L?
low? Does it appear to converge or diverge? If it does cooverge, what is the limit L?
y
135. a, =
COS~_l
o
I
138.
an+l = an
at = =
1,
(I + 0~5)"
= an + 5"
QII+I
+ (2)11 140. a"
sinn
sinn
COMPUTER EXPLORATIONS Use a CAS to perform the followiog steps for the sequeoces in Exercises 135146.
10.2
136. a" =
137. at = I,
139. an
~¥~~J>x
%
=
n sin
k
Inn
141. a, = n
142. a" = "...
143. a, = (0.9999)'
144. a" = (123456)1/,
145. an
=
n 41
8' I
146. a" = 19"
n.
Infinite Series An infinite series is the sum of an infmite sequence of numbers
al + a2 + a, + ... + a, + ... The goal of this section is to understand the meaning of such an infinite sum and to develop methods to calculate it. Since there are infmitely many terms to add in an infinite series, we cannot just keep adding to see what comes out. Instead we look at the result of summing the first n terms of the sequence and stopping. The sum of the first n terms Sn =
al
+
a2
+ a3 + ... + all
is an ordinary finite sum and can be calculated by normal addition. It is called the nth partial sum. As n geta larger, we expect the partial sums to get closer and closer to a limiting value in the same sense that the terms of a sequence approach a limit, as discussed in Section 10.1. For example, to assign meaning to an expression like
1+
1
1
1
2 + 4 + "8 +
1
16 + ...
we add the terms one at a time from the beginning and look for a pattern in how these partial sums grow.
Partial sum
Value
Suggestive expression for partial sum
First:
Sl =
1
1
2  1
Second:
s2=1+ 21
Third:
1 1 s,=I++2 4
3 2 7 4
1 22 1 24
nth:
1 1 1 s, =1+++···+2 4 2,1
2'  1 2n  1
2 n11 2 
10.2
Infinite Series
545
Indeed there is a pattern. The partial sums form a sequence whose nth tenn is SrI
=
1 2  2,,1.
This sequence of partial sums converges to 2 because 1im,,~00(I/2·1) = O. We say "the sum of the infinite series 1
+ !2 + !4 + ... + _1_ + ... 2. 1
is 2."
Is the sum of any fmite numberoftenns in this series equal to 2?No. Can we actually add an infinite number oftenns one by one? No. But we can still define their sum by defming it to be the limit of the sequence of partial sums as n > 00, in this case 2 (Figure 10.8). Our knowledge of sequences and limits enables us to break away from the confines of finite sums.
1
114
~
o
112
FIGURE 10.8 As the lengths 1, ,/2, '/.,
1/8
2
'I" ... are added one by one, the sum
approaches 2.
HISTORICAL BIOGRAPHY
Blaise Pascal (16231662)
DEFINmONS
Given a sequence of numbers {a.}, an expression of the form
al+a2+ a3+ .. ·+ a.+ .. · is an inf"mite series. The number a. is the 11th term of the series. The sequence {s.} defmed by
is the sequence of partial sums of the series, the number s. being the 11th partial sum. If the sequence of partial sums converges to a limit L, we say that the series converges and that its sum is L. In this case, we also write 00 al + az + ... + a. + ... = ~ a. = L . • 1
If the sequence of partial sums of the series does not converge, we say that the series diverges.
When we begin to study a given series al + az + ... + an + .. " we might not know whether it converges or diverges. In either case, it is convenient to use sigma notation to write the series as or
A useful shor1hand when smnmation from 1 to 00 is
understood
546
Chapter 10: Infinite Sequences and Series
Geometric Series Geometric series are series of the form 00
a + aT + ar2 + ... + ar n t + ...
=
L ar n 1 11=1
in which a and r are fixed real numbers and a oF ~:o ar". The ratio r can be positive, as in
o. The
I I
(1)"1 + ...
1t+~···+
(tf' + ....
1+++···+ 2
4
2
series can also be written as
r~I/2,a~1
'
or negative, as in r~ 1/3,a~
1
If r = I, the nth partial sum of the geometric series is
s"
= a + a(l) + a(I)2 + ... + a(I)"1 = na,
and the series diverges because lim"~oo s" = ± 00, depending on the sign of a. If r = I, the series diverges because the nth partial sums alternate between a and O. If Ir I oF 1, we can determine the convergence or divergence of the series in the following way:
rSn
=
+ aT + ar2 + ... + arl'll aT + ar2 + ... + arn 1 + ar1!
Sn  rSn
=
a  ar"
Sn =
a
s.(1  r) = a(1  r") 8n =
Multiply s. by r. Subtract rSn from Sn. Most of the terms on the right cancel.
Factor.
a(1  r") 1 r '
(r oF 1).
We can solve for 311 if r =F 1.
If Irl < I, then rn > 0 as n > 00 (as in Section 10.1) and s" > a/(I  r). Iflrl then Ir"I> 00 and the series diverges.
Iflrl < 1, the geometric series a to a/(I  r):
+ ar + ar2 + ... + ar"l + ...
> I,
converges
00
L ar n  1 =
11=1
_a_, 1 r
Irl
< I.
If Ir I ;;,: I, the series diverges.
We have determined when a geometric series converges or diverges, and to what value. Often we can determine that a series converges without knowing the value to which it converges, as we will see in the next several sections. The formula a/ (I  r) for the sum of a geometric series applies only when the summation index begins with n = I in the expression ~:"l ar"l (or with the index n = 0 if we write the series as ~:o ar").
EXAMPLE 1
The geometric series with a
I
I
I
= 1/9 and r = 1/3 is
I (I )"'
9 + 27 + Sf + ... = ~ 9 3 EXAMPLE 2
00
1/9 I  (1/3)
The series 00
(I)n5 4"
5
5
5
4
16
64
~=5++···
n~O
I 6·
•
10.2
is a geometric series with a
a
547
= 5 and r = 1/4. It converges to 5
a
+ (l/4)
1 r
ar  
Infinite Series
=
•
4.
EXAMPLE 3 You drop a ball from a meters above a flat surface. Each time the ball hits the surface after falling a distance h, it rebounds a distance rh, where r is positive but less than 1. Find the total distance the ball travels up and down (Figure 10.9).
_( _ _ _
ar2 __ _
Solution
The total distance is
ar 3   S
2ar = a + 2ar + 2ar 2 + 2ar 3 + ... = a +  = all+r lr r This sum is 2ar/ (I  r) .
If a = 6 m and r = 2/3, for instance, the distance is (a)
1
S
EXAMPLE 4
~ o o
Solution
o o a
+ (2/3)
(5/3)
= 6 1 _ (2/3) = 6 1/3
•
= 30 m.
Express the repeating decimal 5.232323 ... as the ratio of two integers.
From the definition of a decimal number, we get a geometric series
5.232323 ... = 5
+ 12030 + ~ + ~ + .. .
o
(l00)
(l00)
1
( 1 100
23 (
Q
= 5 + 100 1 + 100 + o
)2 + ... )
a = 1, r = 1/ 100
1/ (1  0.01)
00. n=1
(b)
00,,+1. ,,+1 L ,,diverges because  , ,  > I. n=1
(c)
L ( 1).+1 diverges because lim.~oo(  1).+1 does not exist.
00
.1 00
n
(d) ~ 2n
+
EXAMPLE 8
.
.
n
5 diverges because lim.~oo 2n
+5
1
= 
2 oF o.
•
The series
IIIIII II I 1+++++++···+++···++··· 224444 2'2' 2' ~
2 terms
4 terms
2" terms
diverges because the terms can be grouped into infinitely many clusters each of which adds to I, so the partial sums increase without bound. However, the terms of the series form a sequence that converges to O. Example 1 of Section 10.3 shows that the harmonic series also behaves in this manner. _
10.2 Infinite Series
549
Combining Series Whenever we have two convergent series, we can add them term by term, subtract them term by term, or multiply them by constants to make new convergent series.
If ~an = A and ~bn = B are convergent series, then
THEOREM 8
~(an + bn) = ~a. + ~bn = A + B
1. Sum Rule:
2. Difference Rule: 3. Constant Multiple Rule:
(any number k).
Proof The three rules for series follow from the analogous rules for sequences in Theorem 1, Section lO.1. To prove the Sum Rule for series, let ~=~+_+
...
+~
~=b,+~+···+~.
Then the partial sums of ~(an + bn) are Sn = (a, + b,) + (a2 + b2) + ... + (an + bn) = (a, + ... + an) + (b, + ... + bn) =
An
+ Bn·
Sinee An + A and Bn + B, we have Sn + A + B by the Sum Rule for sequences. The proof of the Difference Rule is similar. To prove the Constant Multiple Rule for series, observe that the partial sums of ~kan form the sequence Sn = ka, + ka2 + ... + kan = kia, + a2 + ... + an) = kAn.
•
which converges to kA by the Constant Multiple Rule for sequences. All corollaries of Theorem 8, we have the following results. We omit the proofs.
1. Every nonzero constant multiple of a divergent series diverges. 2. If ~an converges and ~bn diverges, then ~(an + bn) and ~(a. diverge.
 bn) both
Remember that ~(an + bn) can converge when ~a. and ~bn both diverge. = I + I + I + "'and~bn = (I) + (I) + (1) +"'diverge, whereas ~(an + bn) = 0 + 0 + 0 + ... converges to O. Caution
Forexample,~an
EXAMPLE 9 00
(8)
L n l
Find the sums of the following series.
3n' _ 1 6n ,
00
=
L n l 00
=
L
11=1
(
1 1 ) 2n'  6n'
1
00
L
I
2n'  11=1 6n' I 1
1  (1/2)
=2 Q =± 5
5
1  (1/6)
Difference Rule Geometric series with a ~ 1 andr ~ 1/2,1/6
550
Chapter 10: Infinite Sequences and Series
(b)
Cons_ Multiple Rule
=
4C  ~1/2»)
Geometric series with a
I, r
=
=
1/2
•
=8
Adding or Deleting Terms We can add a finite number of terms to a series or delete a rmite number of terms without altering the series' convergence or divergence, although in the case of convergence this will usually change the sum. If ~:;"~1 a, converges, then ~:;;'ka. converges for any k> land 00
Lan n=1
00
+
= at
a2
+ ... +
Conversely, if ~:;"ka. converges for any It
>
akt
+
Lan" 1I=k
I, then ~:::'1 a, converges. Thus,
and
HISTORICAL BIOGRAPHY
Richard Dedekind (18311916)
Reindexing As long as we preserve the order of its terms, we can reindex any series without altering its convergence. To raise the starting value of the index h units, replace the 11 in the fonnula fora, by 11  h: 00
00
L an = 11L1+11: anh = 11  1
at
+
a2
+ a3 + ....
To lower the starting value of the index h units, replace the 11 in the fonnula for a. by 11 00
+ h:
00
:L an = 11=111: :L an+h = 11=1
at
+
a2
+
Q3
+ ....
We saw this reindexing in starting a geometric series with the index 11 = 0 instead of the index 11 = 1, but we can use any other starting index value as well. We usually give preference to indexings that lead to simple expressions.
EXAMPLE 10
We can write the geometric series
1
00
~ 2,1
1
=
I
1
+ 2 + 4 + ...
as 00
I
~2n'
00
1
~ 2,5' 11=5
00
or even
1
n~2n+4'
The partial sums remain the same no matter what indexing we choose.
•
10.2 Infinite Series
551
Exercises 10.2 Finding nth Partial Sums
Using the nthTenn Test
Iu Exercises l{j, fmd a formula for the nth partial sum of each series and use it to fmd the series' sum if the series converges.
Iu Exercises 2734, use the nthTerm Test for divergence to show that
1. 2+~+~+l+ ... +~+ ... 3 9 27 3,1 2 ~ + _9_ + _9_ + ... + _9_ + ... • 100 100' 100' 100' 3.1_
1 + 1 _ 1 +"'+(_1),1_1_+ ... 2
4
8
2,1
4.1 2 +4  8 + ... + (1),12,1 + ...
the series is divergent, or state that the test is inconclusive. 00 n ~ n(n + I) 27. ~ n + 10 28. ,I (n + 2)(n + 3) 00 00 I 30. ~+29. ~~4 11""0 n 11 1 n + 3 00 , 00 I 31. ~ cos" 32. ~++ 11"'0 e n
,I
34. ~ cosmT
I I I I 5. 2. 3 + 3· 4 + 4· 5 + ... + (n + I)(n + 2) + ... 5 5 5 5 6. 1'2 + 2'3 + 3'4 + ... + n(n + I) + ...
Series with Geometric Terms Iu Exercises 714, write out the fITst few tenDs of each series to show how the series starts. Then fmd the sum of the series. 00 (I)'
7. ~4'
, 0
00 7
(5 + I) (I + sn (I)') ~
m (~y +
+
(~)' + (~)' + .. .
1& (~2y + (~2)' + (~2)' + (~2)' + (~2)6 + ... Repeating Decimals Express each of the numbers in Exercises 1926 as the ratio of two integers. 19. 0.23 = 0.23 23 23 ... 20. 0.234 = 0.234 234 234 ... 21. 0.7 = 0.7777 .. . 22. O.d = O.dddd ... , where d is a digit
= 0.06666 .. .
24. 1.414
(I"~I I) n
00 37. ~ (In v;;+!
,I
(t) + (t)' + (t)' + (t)' + (t)' + ...
23. 0.06
11 1
00 40. ~
211
16. 1+ (3) + (3)2 + (3)' + (3)' + .. . 17.
00 35. ~

36.~(3
3) (n + I)'
'1~'
In Vn)
311
In Exercises 1518, determine if the geometric series converges or diverges. If a series converges. fmd its sum.
15. I +
Iu Exercises 3540, fmd a formula for the nth partial sum of the series and use it to detennine if the series converges or diverges. If a series converges, ftnd its sum.
11=1
00 11. ~ 211 00
Telescoping Series
38. ~ (tan (n)  tan (n  I))
9. ,I ~4'
13.
11=0
1.414414414 ...
=
25. 1.24123
=
26. 3.142857
1.24123 123 123 .. .
= 3.142857 142857 .. .
(Vn+4  Vn+3)
Find the sum of each series in Exercises 4148. 41
~.
4
. ::1 (4n  3)(4n + I)
43
~ 40n • ,~1 (2n  1)'(2n + I)'
45.
00(1 I) ~ Vn  v;;+!
47
~.
'.;:1
(
42
~
. ::1 (2n
6
 1)(2n + I)
44.~2n+1 ,~1 n'(n
46.
00(1 ~ 21/,
+ I)' I)  21/(,+ 1)
I _ I ) In(n+2) In(n+l)
48. ~(tanl (n)  taoI (n + I))
,I
Convergence or Divergence Which series in Exercises 49{j8 converge, and which diverge? Give reasons for your answers. If a series converges, f"md its sum.
00(1)' 49. ~ • r.:: 11=0 v2 51.
~ (1)'+1
11=1
00 53. ~ cosmT 11=0
3, 2
00 SO. ~(V2r n=O
52. ~ ( 1)'+1n n=l
00 54. ~ cos:'11' n=O 5
552
Chapter 10: Infinite Sequences and Series
00
,I
59.
61.
~~
11=1
83. Show by example that ~(aJb,) may diverge even though and ~b, converge and no b, equals O.
56. ~ In 3'
00
57.
I
00
e'" ,0
55. ~
00
58.
10"
00
2"_ 1
00
1
84. Find convergent geometric series A ~ ~a, and B ~ ~b, that illustrate the fact that ~a"b. may converge without being equal toAB.
I
~"' Ixl> 11=0 X
I
I)'
~3"
00 ( 60.~ IIi
~ I~O'
00
•
00
21'.
85. Show by example that ~(a,/b,) may converge to something other than A/Beven when A ~ ~a., B ~ ~b. #' 0, and no b, equals O.
62. ~"1Ft n!
2"+ 3" 63. ~11=1 4" 00
64.
86. If ~a, converges and a, > 0 for all n, can anything be said about ~(l/a,,)? Give reasons for your answer.
+ 4"
~311+411
87. What happens if you add a fmite number of terms to a divergent series or delete a ftnite number of tenns :from a divergent series? Give reasons for your answer.
65.~1n(n :1) 11=1
Geometric Series with a Variable x In each of the geometric series in Exercises 6972, write out the fl!St few terms of the series to fmd a and r, and fmd the sum of the series. Then express the inequality Ir I < I in terms of x and fmd the values of x for which the inequality holds and the series converges.
69. ~(I)'x'
70. ~(I)'x'"
.0
(I)' (
~2 3 + 00
88. If ~a" converges and ~b. diverges, can anything be said about their termbyterm sum ~(a, + b,)? Give reasons for your answer. 89. Make up a geometric series ~ar"l that converges to the number 5if a. a
=2
b. a ~ 13/2.
90. Find the value of b for which
1 + eb + e2JJ + e3b + ... = 9. 91. For what values of r dues the infinite series
,0
72.
~a"
I
sinx
)' converge? Find the sum of the series when it converges.
10 Exercises 7378, find the values of x for which the given geometric senes converges. Also, fmd the sum of the series (as a function ofx) for those values of x. 00
00
73. ~2'x'
74. ~(!)y'"
,0
,0
00
92. Show that the error (L  so) obtained by replacing a convergent geometric series with one of its partial sums s" is ar"/( 1  r). 93. The accompanying figure shows the first five of a sequence of squares. The outermost square has an area of 4 m2 • Each of the other squares is obtained by joining the midpoints of the sides of the squares before it Find the sum of the areas of all the squares.
I)"(x  3)' 00 (  76. ~
75. ~(I)'(x + I)' 11=0
11=0
00
00
.
2
78. ~(lnx)'
77. ~ sin"x
Theory and Examples 79. The series in Exercise 5 can also be written as
~
I
,~I (n + I)(n + 2)
d ~ an ,~_I (n
Write it as a sum beginning with (a) n (0) n ~ 5.
~
I
+ 3)(n + 4) . 2, (b) n
~
94. Helga von Kocb's snowflake CIIl"W Helga von Koch's snowflake is • curve of infmite length that encloses a region of fmite area. To see why thls is so, suppose the curve is generated by starting with an equilateral triangle whose sides have length 1.
0,
80. The series in Exercise 6 can also be written as 00
5
~ n(n + I)
00
and
~ (n +
5 !)(n
Write it as a sum beginning with (a) n (0) n ~ 20.
+ 2) . ~
I, (b) n
~
3,
81. Make up an infinite series of nonzero terms whose sum. is L I b . 3 c. O. 82. (Continuation o/Exercise 81.) Can you make an write series of nonzero terms that converges to any number you want? Explain.
.. Find the length L, of the nth curve C, and show that limn. . . oo L" = 00. b. Find the area An of the region enclosed by C, and show that lim,~ooA. ~ (8/5)A , .
LOO O C2
c,
10.3 The Integral Test
10.3
553
The Integral Test Given a series, we want to know whether it converges or not. In this section and the next two, we study series with nonnegative terms. Such a series cooverges if its sequence of partial sums is bounded. If we establish that a given series does converge, we generally do not have a formula available for its sum, so we investigate methods to approximate the sum instead.
Nondecreasing Partial Sums Suppose that ~~, O.
the series converges by the Integral Test. We emphasize that the sum of the pseries is not I/(p  I). The series converges, but we don't know the value it converges to. Ifp < I,thenlp > oand
roo ~dx = x
11
P
_1_ lim (blp  I) = 1  Pb_OO
00.
The series diverges by the Integral Test. If p = I, we have the (divergent) hannonic series I I I 1+++···++··· 2 3 n We have convergence for p
>
I but divergence for all other values ofp.
•
The pseries with p = I is the harmonic series (Example I). The pSeries Test shows that the hannonic series is just barely divergent; if we increase p to 1.000000001, for in
stance, the series converges! The slowness with which the partial sums of the hannonic series approach infinity is impressive. For instance, it takes more than 178 million terms of the hannonic series to move the partial sums beyond 20. (See also Exercise 43b.)
2+ I)) is not apseries, but it converges by the
The series ~::'~I (1/(n Integral Test. The function j(x) = 1/(x 2 x ~ 1,and
EXAMPLE 4
+
I) is positive, continuous, and decreasing for
{"" _1_ dx = lim [arctan x2 + I b~oo
11
=
x]b 1
lim [arctan b  arctan I]
b~oo
1T
1T
1T
2
4
4'

Again we emphasize that 1T/4 is not the sum of the series. The series converges, but we do not know the value of its sum. •
Error Estimation If a series ~a. is shown to be convergent by the Integral Test, we may want to estimate the size of the remainder R. between the total sum S of the series and its nth partial sum s•. That is, we wish to estimate Rn =
S
9"
= an+l +
tln+2
+
an+3
+
556
Chapter 10: Infinite Sequences and Series To get a lower bound for the remainder, we compare the sum of the areas of the rectangles with the area under the curve y = f(x) for x ;;,: n (see Figure 1O.lla). We see that
1
00
R. = a.+1
+
a.+2
+
a,,+3
+ ... ;;,:
f(x) fix .
• +1
Similarly, from Figure 10.llb, we fmd an upper bound with
1
00
R.
= a,,+1
+
a.+2
+
a,,+3
+ ... ,,;;
f(x) fix.
These comparisons prove the following result giving bounds on the size of the remainder.
Bounds for the Remainder in the Integral Test Suppose {at} is a sequence of positive terms with at = f(!), wbere f is a continuous positive decreasing function of x for all x ;;,: n, and that l:a" converges to S. Then the remainder R. = S  s. satisfies the inequalities
1
00
1
00
f(x) fix ,,;; R. ,,;;
11+1
f(x) fix.
(1)
11
Ifwe add the partial sum s. to each side of the inequalities in (I), we get
1
00
s.
+
1
00
f(x) fix ,,;; S ,,;; s.
+
n+l
f(x) fix
(2)
n
since s. + R. = s. The inequalities in (2) are useful for estimating the error in approximating the sum of a convergent series. The error can be no larger than the length of the interval containing S, as given by (2).
EXAMPLE 5 n = 10.
Estimate the sum of the series l:(I/n 2 ) using the inequalities in (2) and
Solution We have that
Using this result with the inequalities in (2), we get SIO
Taking S10 = I equalities give
I
1000?
558
Chapter 10: Infinite Sequences and Series
48. ~:= 1 (l/n') converges a. Use the accompanying graph to determine the error if S30 = ~~~1 (I/n') is used to estimate the value Of~:::'l (I/n'). y
f(x)
29
x
31
I
00
I I 1++···+2 n
= 1,
30
I
00
c. ~, d. ~, ,2 nln(n ) ,2 n(lnn) 57. Euler's constant Graphs like those in Figure 10.11 suggest that as n increases there is little change in the difference between the sum
and the integral 32
... 33
x
Inn
["I dx x .
= Jl
To explore this idea, carry out the followiog steps. b. Find n so that the partial sum s, = ~~1 (I/i') estimates the value Of~~l (l/n') with an error of at most 0.000001.
a. Bytakingf(x) In(n
49. Estimatethevalueof~:=l (I/n') to within 0.01 of its exact value. 50. Estimate the value of ~~2 (l/(n 2 act value.
+ 4)) to within 0.1
of its ex
51. How many terms of the convergent series ~:1 (l/nl.l) should be used to estimate its value with error at most 0.00001?
54. Use the Cauchy coodensatioo test from Exercise 53 to show that 00
I
a. ~   diverges; n:z2nlnn b.
~.~ cooverges if p
::1.
> I and diverges if p
'" I.
I
1,00 x(:x)P cooverges if and ooIy if p
an
1+
=
"2I + ... + IiI  Inn
is bounded from below and from above. h. Showthat
1'+11x
I I 1.
(_n_)2 > + n
I
2= I
1.
In both cases, p = I, yet the first serles diverges, whereas the second converges.
_
The Ratio Test is often effective when the tenns of a series contain factorials of expressions involving n or expressions raised to a power involving n.
EXAMPLE 1 (a)
Investigate the convergence of the following serles.
~ 2'3~ 5
Solution
(2n)! n!n!
00
(b)
~
00
~
(c)
4' , ,
(2:)!'
We apply the Ratio Test to each series.
(a) For the serles ~~o (2' 1
a.+1
(2.+ (2'
a.
+
+ 5)/3.+ + 5)/3'
1
5)/3', =
1. 3
2.+ 1 + 5 = 1. (2 2' + 5 3 I
+ 5'2) >1. l + 5' 23 I
=
l 3.
The series converges because p = 2/3 is less than 1. This does not mean that 2/3 is the sum of the series. In fact, 00 2' + 5 00 (2)' ~ . = ~ 3 .0 3
.~O
5
+ .~ n _03 00
=
I I  (2/3)
5
+ "'coI  (1/3)
21 2'
10.5 The Ratio and Root Tests
565
(2n)! (2n + 2)! (b) If a, = n.n. ,, ' then a,+1 = ( n + 1)'( . n + I)'. and n!n!(2n + 2)(2n + 1)(2n)! (n + I)!(n + 1)!(2n)!
a,+1 a,
(2n + 2)(2n + I) 4n + 2 = >4 (n + I)(n + I) n + I .
=
The series diverges because p = 4 is greater than I. (c) If a, = 4'n!n!/(2n)!, then
4'+I(n + I)!(n + I)! (2n)! (2n + 2)(2n + 1)(2n)!· 4'n!n!
a.+l
a. =
4(n + I)(n + I) 2(n + I) (2n + 2)(2n + I) = 2n + I ..... I.
Because the limit is p = I, we cannot decide from the Ratio Test whether the series converges. When we notice that a.+1/a, = (2n + 2)/(2n + I), we conclude that a.+l is always greater than a. because (2n + 2)/(2n + 1) is always greater than I. Therefore, all terms are greater than or equal to a, = 2, and the nth term does not ap_ proach zero as n ..... 00. The series diverges.
The Root Test The convergence tests we have so far for ~a, work best when the fonnula for a, is relatively simple. However, consider the series with the terms
n/2', a, = { 1/2',
nodd
n even.
To investigate convergence we write out several terms of the series:
fa=l+l+~+l+~+l+2+ ... 3 5 6
,  1'
2'
22
113
2
24
2
15
2
27
17
= 2 + 4 + 8 + 16 + 32 + 64 + 128 + ... Clearly, this is not a geometric series. The nth term approaches zero as n ..... 00, so the nthTenn Test does not tell us if the series diverges. The Integral Test does not look promising. The Ratio Test produces
a,+1 =
a.
12~'
n+ I 2 '
n odd n even.
All n ..... 00, the ratio is alternately small and large and has no limit. However, we will see that the following test establishes that the series converges.
THEOREM 13The Root Test and suppose that
Let
~a,
be a series with
a. ;",
Then (8) the series converges if p < I, (b) the series diverges if p finite, (0) the test is inconclusive if p = I.
0 for n ;", N,
>
lor P is in
566
Chapter 10: Infinite Sequences and Series
Proof (8) P < 1. Choose an e > 0 so small that p + e < I. Since %;; > p, the terms %;; eventually get closer than e to p. In other words, there exists an index M 2: N such that
%;; I. _
. the senes · · WI t h terms an = CODSl'der agam
EXAMPLE 2
{n/2" /.
I 2,
nodd
n even.
Does ~a. converge?
Solution We apply the Root Test, finding that
%: = {"'I;/n/2, a.
nodd n even.
1/2,
Therefore,
t ,; %;; ,; fn· Since "'I;/n > 1 (Section 10.1, Theorem 5), we have Iim"~",, %;; = 1/2 by the Sandwich _ Theorem. The limit is less than I, so the series converges by the Root Test.
EXAMPLE 3 ""
(a)
2
~ ~.
(b)
.1
Solution
Which of the followiog series converge, and which diverge?
~ 2: n
"" (
~
(c)
n=1
1
1
+n
).
We apply the Root Thst to each series.
"" n 2
(a) ~ 2' converges because _1
"" 2' .
(b) ~ , diverges because ,,=1 n
""(I)'
(c) ~ +I ,,=1 n
1f.n W ("'I;/ny 12 2' = _.r.:;. = 2 > 2 < I. 2
v~
~2' , n
2
r)3
= ( •• Vn
converges because
2
> ,
1
> I.
~(1)' 1In > 0 < I. +I = + n

10.5 The Ratio and Root Tests
567
Exercises 10.5 Using the Ratio Test In Exercises 18, use the Ratio Test to determine if each series con
verges or diverges. 00 2' 1. /I  I n. ~ (n  I)! 3. ~ 2 ,~I (n + I) S. 7.
2.
~
in~2 00
4.
4
6.
~
37.
3
11",1
~:II 00
+ 2)!
8
2"+1 n3',1
f. 3Inn i (2n + 3)n5'In (n + I) 11
2
+
Using the Root Test In Exercises 916, use the Root Thst to detennine if each series con· verges or diverges. 00 7 00 4' 10. (3 )" 9. 11  1 (2n + 5 )' 11"'1 n
L
L
3n  5
11  1
13 ~ • ,;';:1 (3 15.
i
,I
+
(I 
8 (1In))2n
14.
i i
(In(e sin'
2
+
k) r'
(~)
k)'"
(Hint: lim (I + xln)' ,~OO
= e")
I
00
16.
12.
"5'.
~n
1+,
Determining Convergence or Divergence In Exercises 1744, use any method to detennine if tire series con·
verges or diverges. Give reasons for your answer. 00 V2 00 17.
"5' n ,
:.:1
00
00
19. Ln!e' 11=1
21.
L InO' ,I 00
23.
~
2
+ (I)' 1.25'
00 ( 25'LI
,I
27.
i
~ Inri,
22.
i (n; 2)' ,I
3)'
n
(2)' , 24. ,I 3 00
L
OO(
I)'
26'LI11"'1
3n
(Inn)' L,n 00
In;
28.
11  1 n
(I I)
II  I
(I I )'
29. ~ li n 2
30. ~ Ii  n 2
~. ~n
~_ n~n
00
31. 33.
32.
,;":1 ~ ~
/I  I
(n
00
+ I)(n + 2) , n.
,;":1
+ I)!
3'n.,
00
,
~n 00
L L
40
L n ,~2 (In n )(,/2)
42.
3' L';; n2
•
00
11 1
(n!? (2n)'.
00
43.
~
n2'(n
11 _
38. "5' n;
I)!
,,=2
L
,,"" I
~
44.~ ,I
(2n + 3)(2' + 3) 3'+2
Recunively Def"med Terms Which of the series ~;l an defmed by the formulas in Exercises 4554 converge, and which diverge? Give reasons for your answers. all +l =
al =
1,
a,,+1 =
a, =
I
47.
3'
""+I =
48.
al
49.
al =
2,
SO.
a, =
5,
51.
a,
= 3,
1 + sinn n
all
1 + tanI n
46.
n
a"
3n  I
2n + 5 "" n
an+1 = n
+ 1 a"
n
+ Inn
= I,
I
2'
=
a,,+1 =
a, t, ""+1 54. a, t, ""+I 53.
,
20.
I.
00
,
~ (2n:
52. al
18. Yn 2e
~ 2
oo
36.
(Inn )' n 00 'In 41 n. n . ,_I n(n + 2)!
39.
• ,~I
3211
11. i (4n + 3)'
+ 3)! 3!n!3"
(n
00
11=2
n2 (n n!
~ 00
L, 00
00
35.
n
+ 10
an
~
=
=
=
= (",,),+1
Convergence or Divergence Which of the series in Exercises 5562 converge, and which diverge? Give reasons for your answers. 00 2' , , 00 (3n)! 55 L~ 56. ~ n!(n + I)!(n + 2)! • ,~I (2n)! 00 (n!)" 57. II I n ')2
L( 00
59. Y
,
n( ')
:.:1 2 00
IJ
1.3 .... '(2n  1) 4"2IJn !
61.
~
62.
~ [2 4 ...• (2n)](3' + I)
00 00
13 ... '(2n I)
568
Chapter 10: Infinite Sequences and Series
Theory and Examples 63. Neither the Ratio Test nor the Root Thst helps with pseries. Try them an
n/2',
65. Let a, = { 1/2', Does
I ~11=1 nP 00
~an
if n is a prime number otherwise.
converge? Give reasons for your answer.
66. Show that ~:_ I 2(.i')/n! diverges. Recall from the Laws ofExponents that 2('~ = (2')'.
and show that both tests fail to provide information shout convergence. 64. Show that neither the Ratio Test nor the Root Test provides information about the convergence of
(p constant).
10.6
Alternating Series, Absolute and Conditional Convergence A series in which the terms are alternately positive and negative is an alternating series. Here are three examples:
1
1
1
1
2 + 3  4 + 5  ... +
I 
(_1),+1
n
+ ...
(1)
1 1 1 (1)'4 2 + 1  2 + 4  8 + ... + z'" + ... 1 2
(2)
+ 3  4 + 5  6 + ... + (I)·+l n + ...
(3)
We see from these examples that the nth term of an alternating series is of the fonn
an
= (_1).+l u•
or
a. = (I)·u.
where u. = Ia.1 is a positive number. Series (1), called the alternating harmonic series, converges, as we will see in a moment Series (2), a geometric series with ratio r = 1/2, converges to 2/[1 + (1/2)] = 4/3. Series (3) diverges because the nth term does not approach zero. We prove the convergence of the alternating barmonic series by applying the Alternating Series Test. The Test is for convergence of an alternating series and cannot be used to conclude that such a series diverges.
THEOREM 14The Alternating Series Test (Leibniz's Test)
The series
00
~(_1)'+IU. = UI  U2
+ U3

U4
+ ...
n=1
converges if all three of the following conditions are satisfied: 1. The u. 's are all positive. 2. The positive un's are (eventuslly) nonincreasing: u. ;;" U.+I for all n ;;" N, for some integer N.
3. u...... O. Proof Assume N = I. If n is an even integer, say n = 2m, then the sum of the {lISt n tenns is S2m =
(UI  U2)
+
(U3 
= UI  (U2  U3) 
U4) (U4 
+ ... +
(U2m1 
us)  ... 
U2m)
(U2m2 
U2mI) 
U2m.
10.6 Alternating Series, Absolute and Conditional Convergence
569
The first equality shows that S2m is the sum of m nonnegative terms since each tenn in parentheses is positive or zero. Hence S2m+2 ;;,: S2m, and the sequence {S2m} is nondecreasing. The second equality shows that S2m ,;; U1. Since {S2m} is nondecreasing and bounded from above, it has a limit, say (4)
If n is an odd integer, say n = 2m S2m+l
=
82m
+
I, then the sum of the first n terms is
+ U2m+l. Sinceun~O,
and,asmi>00,
+
S2m+l = 82m
U2m+l + L
+0=
Iim,,~oo Sn ~
Combining the results of Equations (4) and (5) gives Exercise 131).
EXAMPLE 1
(S)
L.
L (Section 10.1, _
The alternating harmonic series 00
L(l)n+11 =
n~l
I
n
_1 + 1_1 + ... 2
3
4
clearly satisfies the three requirements of Theorem 14 with N verges.
~
I; it therefore con

Rather !ban directly verifying the defInition Un ;;,: Un+1o a second way to show that the sequence {un} is nonincreasing is to define a differentiable function j(x) satisfying j{n) = Un. That is, the values of j match the values of the sequence at every positive integer n. If j' (x) ,;; 0 for all x greater than or equal to some positive integer N, then j(x) is nonincreasing for x ;;,: N. It follows that j(n) ;;,: j(" + I), or Un ;;,: Un+ 10 for" ;;,: N.
EXAMPLE 2 Consider the sequence where Un ~ IOn/(,,2 + 16). Detme IOx/(x 2 + 16). Then from the Derivative Quotient Rule, j '() X
j(x) ~
2
~
1O{16  x ) 1, the series converges absolutely. If 0 tionally. Conditional convergence: Absolute convergence:
p>O
< P ,;; 1, the series converges condi
1 __1_+_1___1_+ ...
v'2
v'3
v'4
1 __1_+_1___1_+ ... 23(2
33/2
4 3/ 2

572
Chapter 10: Infinite Sequences and Series
Rearranging Series We can always rearrange the terms of a finite sum. The same result is true for an inimite series that is absolutely convergent (see Exercise 68 for an outline of the proof).
THEOREM 17The Reammgement Theorem for Absolutely Convergent Series If ~:::, a. converges absolutely, and bj, b2, ... , b., ... is any arrangement of the sequence {a.} , then ~b. converges absolutely and 00
00
Lb. = n=1 La.·
11=1
If we rearrange the terms of a conditionally convergent series, we get different results. In fact, it can be proved that for any real number r, a given conditionally convergent series can be rearranged so its sum is equal to r. (We omit the proof of this fact.) Here's an example of summing the terms of a conditionally convergent series with different orderings, with each ordering giving a different value for the sum.
EXAMPLE 6 We know that the alternating harmonic series ~:::, (I).+I/n converges to some number L. Moreover, by Theorem 15, L lies between the successive partial sums 82 = 1/2 and 83 = 5/6, so L '¢' O. If we multiply the series by 2 we obtain
2L=2~
=2(1_1+1_1+1_1+1_1+1_~+~_ ...) .1(_1).+1 n 2 3 4 5 6 7 8 9 10 11 212121212 =21 +32+53+74+95+0 .. ·.
Now we change the order of this last sum by grouping each pair of tenns with the same odd denominator, but leaving the negative terms with the even denominators as they are placed (so the denominators are the positive integers in their natural order). This rearrangement gives
(2  1) 
! + (t t)  t + (t t)  i + (%  t)  i + ... (1 ! + t  t + t  i + t  i + ~  1~ + 1\ _ .. ) 00
L 11=1
(1).+1 n
= L.
So by rearranging the terms of the conditionally convergent series ~:::,2( _I).+I/ n, the series becomes ~:;O~I ( 1)'+ '/ n, which is the alternating harmonic series itself. If the two • series are the same, it would imply that 2L = L, which is clearly false since L '¢' O. Example 6 shows that we cannot rearrange the terms of a conditionally convergent series and expect the new series to be the same as the original one. When we are using a conditionally convergent series, the terms must be added together in the order they are given to obtain a correct result. On the other hand, Theorem 17 guarantees that the terms of an absolutely convergent series can be summed in any order without affecting the result.
Summary of Tests We have developed a variety of tests to determine convergence or divergence for an infinite series of constants. There are other tests we have not presented which are sometimes given in more advanced courses. Here is a summary of the tests we have considered.
10.6 Alternating Series, Absolute and Conditional Convergence
573
1. The 11thTerm Test: Unless a, > 0, the series diverges. 2_ Geometric series: ~ar' converges if Ir I < I; otherwise it diverges. 3. pseries: ~ I/n" converges if p > I; otherwise it diverges.
4. Series with nonnegative terms: Try the Integral Test, Ratio Test, or Root Test. Try comparing to a known series with the Comparison Test or the Limit Comparison Test. ~la,1 converge? If yes, so does ~a, since absolute convergence implies convergence.
5. Series with some negative terms: Does 6. Alternating series:
~a. converges if the series satisfies the conditions of the
Alternating Series Test.
Exercises 10.6 Determining Convergence or Divergence
00
In Exercises 114, detennine if the alternating series converges or
27. L( 1)'n'(2/3)'
diverges. Some of the series do not satisfY the conditions of the Alternating Series Test.
29.
,~I
::1
4. L (  I ) '  )' ( 11=2 In n 00 '5 6. L(I),+I
;±n + 4
11=1
10"
(Xl
8.
,&1 (I)' (n + I)!
10. i(l)n+I_I
Inn
11"'2
12. i(l)nln(1
14.
~(I)'+I
+
k)
3v;;:t:l" Vn+1
~
00
33.
00
37.
00
19.
L ( 1),+1
1Ft 00
21.
++ n
1
I
(I)' ,~I I + Vn
n; n
5
+n +n
00 (2),+1 24·L +S' 11"'1 n
25. i(_I),+11
~n
26. L(  I)n+I("'\Ylo)
23. i(I)'+l 3 11  1
11=1
n
00
,I
11
(_1),1
~,c'='1 n + 2n + 1
00
36. ,~_. CO~n1T ~
(I)'(n + I)' (2n)n
00
38.
~
(I),+I(n!)' (2n)!
00 (nl)'3' 40. ~(I)' (2n.+ I)!
00
00
11=1
11=1
41. L(I)'(v;;:t:l"  Vn)42. L(I)'(~  n) 00
43. L(I)'(Vn
+ Vn  Vn)
11=1
44.
L _r 11""1
(I)' _ r:c
00
45. L(I)' sechn
,I
vn+ vn+l
00
46. L(I)'cschn
,I I
I
I
I
I
I
"4  6 + "8  10 + IT  14 + ...
48. I +
1111111 2S + 36  49  64 + ...
"4  "9  16 +
•
22. L(I)n sm,n 11=1
~
L
2
11=1 00
~(I)' n + 3
I
00
20. L(  I)n+l
00
34.
00 ,(2n)! 39. ~(I) 2"n!n
47.
18. i
~
35. ~_ cos mr ~ nVn
nI
17. i(I),_I,~I Vn
32. Y(st'
(100)' I n.
00
Which of the series in Exercises 1548 converge absolutely, which converge, and which diverge? Give reasons for your answers. 00 00 (0 I)' 15. L(_I),+I(O.I)' 16. L(I)'+Ij,
n
~ I
00
Absolute and Conditional Convergence
~1
L /I _ I
Inn
n  Inn
00
31. Y(I)'
4
00
3. i(I)'+I.!.". 1Ft n3
11 1
00
2. i(I)'+I+ 11=1 n /2
Vn
30. i(I)'
1
n
nlnn
11=2
i( I)' ~I+ n
II I
1. i(_I),+I_I
28. i(_I),+I_I
,I
Errnr Estimation In Exercises 4g...S2, estimate the magoitude of the error involved in using the sum of the first four tenns to approximate the sum of the entire series.
574
Chapter 10: Infinite Sequences and Series is the same as the sum of the ftrSt n terms of the series
N. you will see in Section 10.7, the sum is In(l.Ol).
52. I
~t~
i.(
I
I)'t',
0< t
N.
r
_______  _____
Idl
R
.
lei
•
0
lei
____ 
R
~
x
Idl
FIGURE 10.16 Convergence of ~a"x' at x = c implies absolute convergence on the interval I c I < x < IcI; divergence at x ~ d implies divergence forlxl > Idl. The corollary to Theorem 18 asserts the
existence of a radius of convergence R '" O.
Since Ixlc I < 1, it follows that the geometric series ~~o Ixl c converges. By the Comparison Test (Theorem 10), the series ~~o la,llx'l converges, so the original power series ~~o a,x' converges absolutely for I c I < x < Ie I as claimed by the theorem. (See Figure 10.16.) Now suppose that the series ~::'~o a,x' diverges at x ~ d. If x is a number with Ixl > Idl and the series converges at x, then the first half of the theorem shows that the series also converges at d, contrary to our assumption. So the series diverges for all x with Ixl > Idl· • To simplify the notation, Theorem 18 deals with the convergence of series of the form  a)' we can replace x  a by x' and apply the results to the series ~a.(x')'. ~a,x'. For series of the form ~a,(x
The Radius of Convergence of a Power Series The theorem we have just proved and the examples we have studied lead to the conclusion that a power series ~c.(x  a)' behaves in one of three possible ways. It might converge only at x ~ a, or converge everywhere, or converge on some interval of radius R centered at x ~ a. We prove this as a Corollary to Theorem 18.
COROLLARY TO THEOREM 18 The convergence of the series ~c,(x  a)' is described by one of the following three cases: 1. There is a positive number R such that the series diverges for x with Ix  a I > R but converges absolutely for x with Ix  a I < R. The series may or may not converge at either of the endpoints x ~ a  R and x ~ a + R. 2. The series converges absolutely for every x (R ~ 00). 3. The series converges at x ~ a and diverges elsewhere (R = 0).
10.7
Power Series
579
Proof We first consider the case where a = 0, so that we have a power series ~:o c"x' centered at O. If the series converges everywhere we are in Case 2. If it converges only at x = 0 then we are in Case 3. Otherwise there is a nonzero number d such that ~:;"~o c.d· diverges. Let S be the set of values of x for which ~:o c"x' converges. The set S does not include any x with Ixl > since Theorem 18 implies the series diverges at all such values. So the set S is bounded. By the Completeness Property of the Real Numbers (Appendix 7) S has a least upper bound R. (This is the smallest number with the property that all elements of S are less than or equal to R.) Since we are not in Case 3, the series converges at some number b oF 0 and, by Theorem 18, also on the open interval (Ibl, Ibl). ThereforeR > O. Iflxl < R then there is a number c in S with Ixl < c < R, since otherwise R would not be the least upper bound for S. The series converges at c since c E S, so by Theorem 18 the series converges absolutely at x. Now suppose Ixl > R. If the series converges at x, then Theorem 18 implies it converges absolutely on the open interval (lxi, Ixl), so that S contains this interval. Since R is an upper bound for S, it follows that Ix I :5 R, which is a contradiction. So if Ix I > R then the series diverges. This proves the theorem for power series centered at a = O. For a power series centered at an arbitrary point x = a, set x' = x  a and repeat the argument above replacing x with x' . Since x' = 0 when x = a, convergence of the series ~:olc'(x') I' on a radiusR open interval centered at x' = 0 corresponds to convergence oftheseries~:olc.(x  a)I' onaradiusR open interval centeredatx = a. _
14
R is called the radius of convergence of the power series, and the interval of radius R centered at x = a is called the interval of convergence. The interval of convergence may be open, closed, or halfopen, depending on the particular series. At points x with Ix  a I < R, the series converges absolutely. If the series converges for all values of x, we say its radius of convergence is infinite. If it converges only at x = a, we say its radius of convergence is zero.
How to Test a Power Series for Convergence
1. Use the Ratio Test (or Root Test) to find the interval where the series converges absolutely. Ordinarily, this is an open interval Ix  al
I
Test each endpoint ofthe (rmite)
2.
interval of convergence.
R (it does not even converge conditiona1ly) because the nth term does not approach zero for those values of x.
Operations on Power Series On the intersection of their intervals of convergence, two power series can be added and subtracted term by term just like series of constants (Theorem 8). They can be multiplied just as we multiply polynomials, but we often limit the computation of the product to the 1rrst few terms, which are the most important. The following result gives a formula for the coefficients in the product, but we omit the proof.
580
Chapter 10: Infinite Sequences and Series
THEOREM
A(x)
=
19The Series Multiplication Theorem for Power Series
~::'~o a.x· and B(x) = ~:;;,o b.x· converge absolutely for Ixl
c. = aob.
If
< R, and
•
= ~akb..,
+ a,b., + a2b.2 + ... + an,b, + anbo
k~O
then ~::'~o c.x· converges absolutely to A(x)B(x) for Ixl
< R:
Finding the general coefficient c. in the product of two power series can be very tedious and the term may be unwieldy. The following computation provides an illustration of a product where we find the first few terms by multiplying the terms of the second series by each term of the fITst series:
00 ) • (00 .+1 ) ~(l)'_x( ~x' ~o ~o n + I Multiply second series ...
(x 
f + X;  .. ) +
(x2 
by!
~ + X;  .. ) +
t
(x3 
by.
+
X;  .. ) + ...
by.' and gather the IllS! four powers.
We can also substitute a function j(x) for x in a convergent power series.
If ~:;;'o a.x· converges absolutely for Ix I < R, then ~:o an (j(x))" converges absolutely for any continuous function j on Ij(x) I < R.
THEOREM
20
Since I/(i  x) = ~:;;'ox' converges absolutely for Ixl < I, it follows from Theorem 20 that 1/(1  4x 2 ) = ~::'_o(4x2)'convergesabsolutelyforI4x21 < 1 orlxl < 1/2. A theorem from advanced calculus says that a power series can be differentiated term by term at each interior point of its interval of convergence. THEOREM 21The TermbyTerm Differentiation Theorem
radius of convergence R
>
If ~cn(x  a)' has
0, it defines a function
00
j(x)
= ~cn(x
 a)'
on the interval
a  R
0 has a Taylor series that converges to the function at every point of (R, R). Show this by showing that the Taylor selies generated by f(x) = ~::'o a.x· is the series ~:o a"x" itself. An immediate consequence of this is that series like •
2
6
8
5!
7!
x4 + x  x + ...
3!
approximation with an error less than 102 ? b. What is the maximum error we could expect ifwe replace the function by each approximation over the specified interval? Using a CAS, perform the following steps to aid in answering questions :2 21
4
' Y  taoI Y 33. lun ,
yo
dl,
(a) [0,0.5]
(b) [0, I]
35. lim x 2(e 1/ x'

x~oo
1n(1 + x ) 37. lim """"'"''"xa 1 cosX' 2
sinB  B + (11'/6) 32. 90 lim "' tr '
34. lun
yo
y
I)
3'x'
+ 41  61 +'" X S + x 7  x 9 + XII  •••
51. I +2>;3x 2 +4x'5x 4 + ...
Theory and Examples 53. Replace x by x in the Taylor series for In (I + x) to ob1ain a series for In (I  x). Then subtract this from the Thylor series for In (I + x) to showthatforlxl < I,
54. How many lem3s of the Taylor series for In (I + x) should you
2
t
34x4
+ ...
add to be sure of calculatiog In (1.1) with ao error of magnitude
less than 108? Give reasons for your answer.
I  cos 1  (1 2/2)
10
2'
2 3 22.t4 23x 5 2 4x 6 50. x 2>; +21"""31+41'"
(b) [0, I]
Indeterminate Forms
xa
3"7!
In I + _ x =2 ( x+ x' + x' + . . .) . I x 3 5
Use series to evaluate the limits in Exercises 2940. eX  (I + x) 29. lim
3"5!
"3  3"3 + 3"5  3"7 x 3 + X4 + x 5 + x 6 + ...
=
I
_1_ _ _1_ + ... 3'2' 4'24
1r  r + 1f'5   1T'7  + ...
. 3
48.1
10 Exercises 2528, fmd a polynomial that will approximate F(x) throughout the given interval with ao error of magnitude less than
/.X sin 12 dl, [0, I] 26. F(x) /,xI2e t'dl, [0, I] 27. F(x) = /.X taoI dl, (a) [0,0.5]
. 2
47.
0.1
s~x dx
0
1. __1_ +
44
4S
1922 with an error of magnitude less than 108 .
19.
+ x') sinx 2
41. I + I + 2! + 3! + 4! + ...
D 10 Exercises 1518, use series to estimate the integrals' values with an
/.°
X'
Using Table 10.1 10 Exercises 4152, use Table 10.1 to fmd the sum of each series.
+ x 2)'
12. (1
. 3
sm x 1 cos 2x
603
_tao__'~y__sin ____ y
, y cosy
36. lim (x + I) sin _+1 I x_OO x
55. According to the Alternatiog Series Estimation The0ren3, how many lem3s of the Thylor series for taoI I ""uld you have to add to be sure of fmding 'If/4 with ao error of magnitude less than
1031 Give reasons for your answer. 56. Show that the Thylor series for !(x) Ixl> I.
D 57.
= taoI x
diverges for
Estimating Pi About how many tenns of the Taylor series for taoI x v.uld you have to use to evaluate each term on the righthand side of the equation
_ 48
'If 
1
tao
I + 18
32
1
tao
I 20 1 I 57 tao 239
with ao error of magnitude less than IO.... ? 10 contrast, the convergence of ~:~I(I/n2) to ~/6 is so slow that even 50 terms will not yield twoplace accoracy.
604
Chapter 10: Infinite Sequences and Series
58. Integrate the rlrst three nonzero terms of the Taylor series for tan t from 0 to x to obtain the fIrst three nonzero terms of the Taylor series for In sec x.
59.
L
by integrating the series
Use the binomial series and the fact that
in the first case from x to tox. to generate the fIrst four nonzero terms of the Taylor series
for sin1 x. What is the radius of convergence? b. Series for cos1 x Use your result in part (a) to rmd the rrrst five nonzero terms of the Taylor series for 008 1 x. 60.
L
Seri.. for sinh1 x Taylor series for
Find the rrrst four nonzero terms of the
sinb1 x
=
r
dt
Jo \IT+7
+ x)'
from the series for
+ x).
62. Use the Taylor series for 1/(1  x') to obtain a series for 22/(1  x')'.
D 63.
00
Euler's Identity 67. Use Equation (4) to write the followiog powers of e in the form a + hi. a. el'll' 68. Use Equation (4) to show that
69. Establish the equations in Exercise 68 by combining the formal Taylor series for e ifJ and eifJ • 70. Show that
a. coshiO = cosO,
estimation error. 61. Obtain the Taylor series for 1/(1
and in the second case from 
.
D b. Use the rrrst three terms of the series in part (a) to estimate sinb1 0.25. Give an upper bound for the magnitude of the
1/(1
00
Estimating Pi The English mathernaticiao Wallis discovered the fom2Ula ~
2,4,4,6,6,8, .. .
4
3,3,5,5,7,7, ... .
b. sinhi8
= isin8.
71. By multiplying the Taylor series for eX and sin x, rmd the terms through x 5 of the Taylor series for eX sin x. This series is the imaginary part of the series for eXoeix
=
e(l+i~.
Use this fact to check your anawer. For what values of x should the series for eX sin x converge? 72. Wheo a aod b are real, we derme e(a+ib).< with the equation e(a+ib)x
=
etIXoeibx
=
etIX(cosbx
+ isinbx).
Find ~ to two decin2al places with this formula. Differentiate the righthand side of this equation to show that
64. Use the followiog steps to prove Equation (I). L
:fx
Differentiate the series f(x)
=
I
+
~ (:)x'
e(a+ib).
a" > Oforalln,abowthat~:lln(1  an)eonverges.
Does the value of
.
n2 •
(2n  I)' Ul ~ I, Un+l ~ (2n)(2n + I) Un·
converges absolutely.
hm
j(n)
n
Apply Raabe's Test to determine whether the series converges.
00
L
+ f: +
where Ij(n)l < K for n '" N, then ~::;'1 Un converges ifC > I and diverges if C '" I. Show that the results of Raabe's Test agree with what you know about the series ~:;" 1 (I/n') and ~:;" 1 (I/n).
I   , dx. l+x
18. Find all values of x for which
D 19.
I
26. (Continuation ofExercise 25. ) Suppose that the terms of ~:;" 1 Un are defmed recursively by the formulas
17. Evaluate
~
~
(Hint: First show that Iln(1  an) I '" a"f(1  a,,).)
eos(a/n»)n a constant, n •
appear to depend on the value of a? If so, how?
I
b. Does the value of . ( cos (a/n»)n hm 1bn ' 11_00
a and b constant, b
Prove Nicole Oresme's Theorem that
29. Nicole Oresme's Theorem
¢'
0,
+ 1. 2
2
+ 1 . 3 + ... + "+ ... ~ n 1 4
2

4.
(Hint: Differentiate both sides of the equatioo 1/(1  x) 1 + ~:_ IX".)
~
30. a. Show that
appear to depend on the value of b? If so, how? c. Use calculus to confirm your fmdings in parts (a) and (b). 20. Show that if ~:;"~1 a" converges, then
00
~I n
n(n
+ I) Xn
2>;'
~(x  I )'
for Ix I > I by differentiating the identity 00
LXn+
,
1
IFI
converges.
=_x_ 1  x
twice, multiplying the result by x, and then replacing x by I/x.
21. Find a value for the eonstant b that will make the radius of con
vergence of the power series
b. Use part (a) to fmd the real solmioo greater than I of the equation
x~fn(n;l). 11=1
equal to 5.
x
31. Quality control
22. How do yoo know that the functioos sinx, lnx, and e' are not polyoomia\s? Give reasons for your answer.
23. Find the value of a for which the limit lim sin (ax)  sinx  x
xa
x3
is fmite and evaluate the limit. 24. Find values of a and b for which . cos(ax)b hm , 2x
xo
~l.
25. Raahe'. (or Gau••'.) Te.t The fo\lowiog test, which we state without proof, is an extension of the Ratio Test.
a. Differentiate the series _1_= 1 Ix
+X+X2+ ... +X"+'"
toobtainaseriesforl/(lx)'.
b. In ooe throw of two dice, the probability of gettiog a roll of7 is p ~ 1/6. If you throw the dice repeatedly, the probability that a 7 will appear for the fll'St time at the nth throw is qnlp , where q ~ I  p ~ 5/6. The expected number of throws until a 7 frrst appears is ~:=1 nq"Ip . Find the sum of this series. c. As an engineer applying statistical control to an industrial operation, yoo inspect items taken at randem from the assembly line. You classifY each sampled item as either "good" or
Chapter 10 Technology Application Projects ''bad.'' If the probability of an item's beiog good is p and of an item's beiog bad is q ~ I  p, the probability that the fust bad item found is the nth one inspected is p,,lq. The average number iospected up to and iocluding the fust bad itero found is ~:lnplllq.Evaluatethis sum, assuming 0 < P < 1. 32. EIpected value Suppose that a random variable X may assome the values 1, 2, 3, ... , with probabilities Pt,P2,P3, ... , where Pi is the probability that X equals k (k ~ I, 2, 3, ... ). Suppose also that Pk '" 0 and that ~:'Pk ~ 1. The OIpected value of X, denoted by E(X), is the number ~:,kp1' provided the series converges. In each of the following cases, show that ~~'Pk ~ I and fmdE(X) ifit exists. (Hint: See Exercise 31.)
C. P1~
D 33.
c. Ifk ~ 0.01 h 1 and to ~ IOh, fmd the smallestn such that
R. > (1/2)R. (Source: Prescribing Safe and Efficnve Dosage, B. Horelick and S. Kooot, COMAP, Inc., Lexington, MA.)
34. Time b_een drug do... (ContinUfJtion of Exercise 33.) If a drug is known to be ineffective below a concentration CL and harmful above some higher concentration CH , one needs to fmd values of Co and to that will produce a concentration that is sare (not above CH ) but effective (not below CL ). See the accompanyiog figure. We therefore want to fmd values for Co and to for which
R
511
~
CL
and
Co
+R
b.Pk~1
a.Pk=2j;
6
I
I
"§ :0 C H
.S
Safe and effective dosage The concentration io the blood resultiog from a siogle dose of a drug nonnaily decreases with time as the drug is eliotioated from the body. Doses may therefore need to be repeated periodically to keep the conccn1ration from dreppiog below some particular level. One model for the effect of repeated doses gives the residual concentration just before the (n + I)stdoseas
Coekto
+ Coe2kto + ... + Coenkto ,
where Co ~ the change io conceotratioo achievable by a single dose (mgjmL), k ~ the elimination constant (h1) , and to ~ time between doses (h). See the accompanying figure.
~
CH.
C
I
k(k+ 1) ~k k+ I
RII =
609
.g•
~ ~
Highest safe !evel
t
Co
CL
.j,
I
I
Lowest effective level
~ to~ 0
Time
Thus Co ~ CH  CL . Wheo !bese values are substituted io !be equatioo for R ohtsioed io part (a) of Exercise 33, the resultiog equatioo simplifies to
C
To reach an effective level rapidly, one might administer a "Ioadiog" dose that would produce a conccn1ration of CH mg/mL. This could be followed every to hours by a dose that raises the conceotratioo by Co ~ CH  CL mg/mL. a. VerifY the preceding equation for to.
Time (h)
a. Write R" in closed form as a single fraction, and fmd R =
litnn_oo R".
b. CaicuiateR, and RIO for Co ~ I mg/mL, k ~ 0.1 h 1 , and to ~ 10 h. How good an estimate of R is RIO?
Chapter
2J!)
b. If k ~ 0.05 h 1 and the highest safe conccn1ration is e times the lowest effective concen1ration, fmd the leogth of time b _ doses that will assure safe and effective concentrations. c. GivenCH~ 2mg/mL,CL ~ O.5mg/mL,andk~ 0.02h 1 , determine a scheroe for administeriog the drug. d. Suppose that k ~ 0.2 h1 and that the sma11est effective coocen1ratioo is 0.03 mgjmL. A siogle dose that produces a coocen1ratioo of 0.1 mg/mL is administered. About how long will the drug reroaio effective?
Technology Application Projects
Mathematica/Maple Modules: Bouncing BaH The model predicts the height of a bounciog ball, and the time until it stops bounciog. ~Ior Po/ynomiDIApproximll/ion.
ofII Function
A graphical animation shows the convergence of the Taylor polyoomials to functioos baviog derivatives of all orders over an ioterval io their domaios.
11 PARAMETRIC EQUATIONS AND POLAR COORDINATES OVERVIEW In this chapter we study new ways to define curves in the plane. Instead of thinking of a curve as the graph of a function or equation, we consider a more general way of thinking of a curve as the path of a moving particle whose position is changing over time. Then each of the x and ycoordinates of the particle's position becomes a function of a third variable t. We can also change the way in which points in the plane themselves are described by using polar coordinates rather than the rectangular or Cartesian system. Both of these new tools are useful for describing motion, like that of planets and satellites, or projectiles moving in the plane or space. In addition, we review the geometric definitions and standard equations of parabolas, ellipses, and hyperbolas. These curves are called conic sections, or conics, and model the paths traveled by projectiles, planets, or any other object moving under the sole influence of a gravitational or electromagnetic force.
11.1
Parametrizations of PLane Curves In previous chapters, we have studied curves as the graphs of functions or equations involving the two variables x and y. We are now going to introduce another way to describe a curve by expressing both coordinates as functions of a third variable t.
Parametric Equations Figure 11.1 shows the path of a moving particle in the xyplane. Notice that the path fails the vertical line test, so it cannot be described as the graph of a function of the variable x. However, we can sometimes describe the path by a pair of equations, x = f(t) and y = g(t), where f and g are continuous functions. When studying motion, t usually denotes time. Equations like these describe more general curves than those like y = f(x) and provide not only the graph of the path traced out but also the location of the particle (x, y) = (f(t), g(t» at any time t.
DEFINITION
Ifx andy are given as functions
x = f(t),
FIGURE 11.1 The curve or path traced by a particle moving in the xyplane is not always the graph of a function or single equation.
610
y = g(t)
over an interval loftvalues, then the set of points (x, y) = (f(t), g(t» defined by these equations is a parametric curve. The equations are parametric equations for the curve.
The variable t is a parameter for the curve, and its domain I is the parameter interval. If I is a closed interval, a :5 t :5 b, the point (f(a), g(a») is the initial point ofthe curve and (f(b), g(b) is the terminal point. When we give parametric equations and a parameter
11.1
611
Parametrizations ofplan. Curves
interval for a curve, we say that we have parametrized the curve. The equations and interval together constitute a parametrization of the curve. A given curve can be represented by different sets of parametric equations. (See Exercises 19 and 20.)
EXAMPLE 1
Sketch the curve defined by the parametric equations
y = I
+ I,
00 < 1 O. The bead's velocity at any later point (x, y) on the cycloid is v =
V2g(y 
YO) =
V'2ga(ccoSIoCosc I).
y ~ a(l 
cost)
616
Chapter 11: Parametric Equations and Polar Coordinates Accordingly, the time required for the bead to slide from (xo, YO) down to B is
T=
l
w
a'(2  2 cos I) ~lw dl= 2ga(cos lo  cost) g ~
~
_~r  "{gi"
w
I
2 sin'(t/2) dl '1(2 cos' (10/2)  I)  (2 cos' (t/2)  I)
fc; rw
=
V~il'
fc;l' V~
sin (1/2) dl Vcos'(to/2)  cos'(t/2)
1=1,
u 2 du c
2du
=w
=
1  cos I dl cosio  cosl
Ve' 
u'
= cos (t/2) = sin (t/2) dt = cos (to/2)
. _lU]'=W _~[ g sm C 1=10
~
fc; [ . 1
cos (t/2) ]W cos (to/2) I,
=
"V g
=
~ (sin1 0 + sin1 1) = '1T~.
sm
y
This is precisely the time it takes the bead to slide to B from O. It takes the bead the same amount of time to reach B no matter where it starts. Beads starting simultaneously from 0, A, and C in Figure 11.11, for instance, will all reachB at the same time. This is the reason that Huygens' pendulum clock is independent of the amplitode of the swing.
FIGURE 11.11 Beads released simultaneously on the upside·down eycloid at 0, A, and C will reach B at the same time.
Exercises 11.1 Finding Cartesian from Parametric Equations
15. x = sec'l  I,
Exercises 118 give parametric equations and parameter intervals for the motion of a particle in the xyplane. Identify the particle's path by fmding a Cartesian equation for it. Graph the Cartesian equation. (The graphs will vary with the equation used.) Indicate the portion of the graph traced by the particle and the direetion of motion.
17. x
1. x 2. x
= 31, Y = Vi,
= 9t 2 ,
00
y = I,
I '"
0 0'" 0 '" '1r/2
2/(1  cosO), '1r/2 '" 0 s
'1r
~ a
cos 8,
'1r/2 s 8 s '1r/2
32. r = /(9) .s. r = 2/(9) Can anything be said about the relative lengths of the curves r ~ /(8), 8 '" (3, and r ~ 2/(8), a ~ (J ~ f3? Give reasons for your answer.
a'"
Conic Sections In this section we detme and review parabolas, ellipses, and hyperbolas geometrically and
derive their standard Cartesian equations. These curves are called conic sections or conics b.,. cause they are formed by cutting a double cone with a plane (Figure 11.36). This geometry method was the only WH:f they could be described by Greek mathematicians who did not have our tools of Cartesian or polar coordinates. In the next section we express the conics in polar coordinates.
Parabolas DEFINITIONS A set that consists of all the points in a plane equidistant from a given tlXed point and a given fixed line in the plane is a parabola. The fixed point is the focus of the parabola. The fixed line is the directrix.
640
Chapter 11: Parametric Equations and PoLar Coordinates
Circle: plane perpendicular to cone axis
Ellipse: plane oblique to cone axis
Hyperbola: plane cuts both halves of cone
Parabola: plane parallel to side of cone (a)
Point: plane through cone vertex only
Pair of intersecting lines
Single line: plane tangent to cone (b)
FIGURE 11.36 The standard conic sections (a) are the curves in which a plane cuts a double cone. Hyperbolas come in two parts, called branches. The point and lines obtained by passing the plane through the cone's vertex (b) are degenerate conic sections.
If the focus F lies on the directrix L, the parabola is the line through F perpendicular to y
L. We consider this to be a degenerate case and assume henceforth that F does not lie on L. A parabola has its simplest equation when its focus and directrix straddle one of the coordinate axes. For example, suppose that the focus lies at the point F(O, p) on the positive yaxis and that the directrix is the line y = p (Figure 11.37). In the notation of the figure, a point P(x, y) lies on the parabola if and only if PF = PQ. From the distance formula,
The vertex lies ~"""~~++x halfway between directrix and focus. P L Directrix: y = P Q(x, p)
PF
=
PQ
=
Y(x  of + (y  pf = Yx 2 + (y  p)2 Y(x  xf + (y  (p)f = Y(y + p)2.
When we equate these expressions, square, and simplify, we get FIGURE 11.37 The standard form of the parabola x 2 = 4py, p > o.
x2
y = 4p
or
x2
=
4py.
Standard form
(1)
These equations reveal the parabola's symmetry about the yaxis. We call the yaxis the axis of the parabola (short for "axis of symmetry"). The point where a parabola crosses its axis is the vertex. The vertex of the parabola x 2 = 4py lies at the origin (Figure 11.37). The positive number p is the parabola's focal length.
11.6
y
~
Conic Sections
641
If the parabola opens downward, with its focus at (0, p) and its directrix the line p, then Equations (I) become 2
x y=
and
4p
x 2 ~ 4py.
By interchanging the variables x and y, we obtain similar equations for parabolas opening to the right or to the left (Figure 11.38). y
y
Directtix
Directtix
x=p
x=p
Vertex
Vertex
(b)
(a)
FIGURE 11.38 (a) The parabola y2 ~ 4px. (b) The parabolay 2 ~ 4px.
EXAMPLE 1 Solution
Find the focus and directrix of the parabolay2 ~ lOx.
We rmd the value ofp in the standard equation y2 ~ 4px:
4p = 10,
so
Then we find the focus and directrix for this value ofp: Vertex
Vertex
Focus: Directrix: FIGURE 11.39 Points on the focal axis of an ellipse.
(p, 0)
x
~ (~, 0 )
~p
or
x
~
5
2'
•
Ellipses
y b
FIGURE 11.40 The ellipse defined by the equation PFl + PF2 ~ 2a is the grapb of the equation (x 2/a 2) + (y2/b 2) ~ I, where b 2 = a 2  c 2•
DEFINmONS An ellipse is the set of points in a plane whose distances from two fixed points in the plane have a constant sum. The two fixed points are the foci of the ellipse. The line through the foci of an ellipse is the ellipse's focal ans. The point on the axis halfway between the foci is the center. The points where the focal axis and ellipse cross are the ellipse's vertices (Figure 11.39).
If the foci are Fl( c, 0) andF2(c, 0) (Figure 11.40), and PFI + PF2 is denoted by 2a, then the coordinates of a point P on the ellipse satisfY the equation
Y(x + c)2 + y2 + Y(x  c)2 + y2 ~ 2a.
642
Chapter 11: Parametric Equations and Polar Coordinates To simplify this equation, we move the second radical to the righthand side, square, isolate the remaming radical, and square agam, obtaining
x2
2 + a
y2 2
a  c
2=
1.
(2)
Since PFI + PF2 is greater than the length FIF2 (by the triangle inequality for triangle PFIF2), the number 2a is greater than 2c. Accordingly, a > c and the number a2  c 2 in Equation (2) is positive. The algebraic steps leading to Equation (2) can be reversed to show that every point P whose coordinates satisfy an equation of this form with 0 < c < a also satisfies the equation PFI + PF2 = 2a. A point therefore lies on the ellipse if and only if its coordinates satisfy Equation (2). If
Va 2  c2 ,
b = then a
2

c
2
=
(3)
2
b and Equation (2) takes the form x 2 y2
,+,=1.
(4) b Equation (4) reveals that this ellipse is symmetric with respect to the origin and both coordinate axes. It lies inside the rectangle bounded by the lines x = ±a and y = ±b. It crosses the axes at the points (±a, 0) and (0, ±b). The tangents at these points are perpendicular to the axes because
a
dy
b 2x
ax
a 2y
Obtained from Eq. (4) by implicit differentiation
is zero if x = 0 and inf'Inite if Y = O. The major axis of the ellipse in Equation (4) is the line segment oflength 2a joining the points (±a, 0). The minor axis is the line segment of length 2b joining the points (0, ±b). The number a itself is the semimajor axis, the number b the semiminor axis. The number c, found from Equation (3) as
c
=
Va 2 
b2 ,
is the centertoCocus distance of the ellipse. If a = b, the ellipse is a circle. 2 y2 Y 1 I6+9~ (0,3)
x
EXAMPLE 2
The ellipse
Vertex
(5)
(4,0)
(Figure HAl) has
a=
Sernimajor axis:
v'i6 = 4,
Centertofocus distance: (0, 3)
FIGURE 11.41 An ellipse with its major axis horizontal (Example 2).
Foci:
(±c, 0)
Vertices: Center:
=
c =
Serniminor axis:
b
=
v'9 = 3
v'l6=9 = 0
(± 0,0)
(±a,O)
= (±4,0)
(0,0).
•
Ifwe interchange x andy in Equation (5), we have the equation
x2
y2
9+16=1.
(6)
The major axis of this ellipse is now vertical instead of horizontal, with the foci and vertices on the yaxis. There is no confusion in analyzing Equations (5) and (6). Ifwe f'md the intercepts on the coordinate axes, we will know which way the major axis runs because it is the longer of the two axes.
11.6
Conic Sections
643
StandardForm Equations for Ellipses Centered at the Origin 2
Foci on the xaxis:
X, a
2
+ y2
=
b
I
(a
> b)
Centertofocus distance:
2
X, b
Va 2 
b2
=
Va 2 
b2
(±a, 0) 2
+ y2 a
=
I
(a
> b)
Centertofocus distance: Foci:
=
(±e, 0)
Foci:
Vertices:
Foci on theyaxis:
e
e
(0, ±e) (0, ±a)
Vertices:
In each case, a is the semimajor axis and b is the semiminor axis.
Hyperbolas DEFINmONS A hyperbola is the set of points in a plane whose distances from two fixed points in the pIane have a constant difference. The two f'lxed points are the foci of the hyperbola. The line through the foci of a hyperbola is the focal axis. The point on the axis halfway between the foci is the hyperbola's center. The points where the focal axis and hyperbola cross are the vertices (Figure 11.42).
FIGURE 11.42 Poiots 00 the focal axis of a hyperbola.
If the foci are F,( e, 0) and F2(c, 0) (Figure 11.43) and the constant difference is 2a, then a point (x, y) lies on the hyperbola if and only if (7)
y
x=a
x=a
To simplifY this equation, we move the second radical to the rightband side, square, isolate the remaining radical, and square again, obtainiog (8)
FIGURE 11.43 Hyperbolas have _ branches. For poiots 00 the righthand branch of the hyperbola shown here, PF,  PF2 = 2a. For poiots on the lefthand branch, PF,  PF, = 2a. We then leth = Ve'  a'.
So far, this looks just like the equation for an ellipse. But now a 2  e 2 is negative because 2a, being the difference of two sides of triangle PF,F2 , is less than 2e, the third side. The algebraic steps leading to Equation (8) can be reversed to show that every point P whose coordinates satisfy an equation of this form with 0 < a < e also satisf'les Equation (7). A point therefore lies on the hyperbola if and only if its coordinates satisfy Equation (8). Ifwe let b denote the positive square root of e 2  a 2 ,
b = then a 2

Ve 2 
a2 ,
(9)
e 2 = b 2 and Equation (8) takes the more compact form
x2 y2 a2  b 2 = I.
(10)
644
Chapter 11: Parametric Equations and Polar Coordinates The differences between Equation (10) and the equation for an ellipse (Equation 4) are the minus sign and the new relation FromEq. (9)
Like the ellipse, the hyperbola is symmetric with respect to the origin and coordinate axes. It crosses the xaxis at the points (±a, 0). The tangents at these points are vertical because
b 2x
dy dx
Obtained from Eq. (10) by implicit differentiation
a'y
is iufinite when y = O. The hyperbola has no yintercepts; in fact, no part of the curve lies between the lines x = a and x = a. The lines
y =
±%x
are the two asymptotes of the hyperbola deimed by Equation (10). The fastest way to find the equations of the asymptotes is to replace the I in Equation (10) by 0 and solve the new equation for y:
x2
y2
2
2
=
a
b
I >
x2
ofor I
hyperbola
EXAMPLE 3
y y
b 0 > y = ±a x .
y2
;;I  b2 =
asymptotes
The equation
~ v'5.
x2
2
y2
45=
(11)
I
is Equation (10) with a 2 = 4 and b 2 = 5 (Figure 11.44). We have Centertofocus distance: Foci: Center:
b2
Vertices:
= V4+5 =
3
(±a,O) = (±2, 0)
(0,0)
x2
TIre hyperbola and its asymptotes in Example 3.
= Va 2 +
(±c,O) = (±3,0),
Asymptotes: FIGURE 11.44
c
y2
4  5
= 0
or y =
Y5
•
±T x .
If we interchange x and y in Equation (II), the foci and vertices of the resulting hyperbola will lie along the yaxis. We still find the asymptotes in the same way as before, but now their equations will be y = ±2x/Y5.
StandardForm Equations for Hyperbolas Centered at the Origin Foci on the xaxis:
x2 y2 2  2 = I a b
Centertofocus distance: Foci:
(±c,O)
Vertices:
(±a,O)
c =
Va 2 + b 2
x2
2
b2
a
Centertofocus distance: Foci:
I
c =
Va 2 + b 2
(0, ±c)
Vertices:
(0, ±a)
x2 y2 b  2   2 = 0 or y = ±ax Asymptotes: a b Notice the difference in the asymptote equations (b/a in the imt, a/b in the second). Asymptotes:
y2
=
Foci on the yaxis:
y2
x2
 2   2 = 0 a b
or y =
±"x b
11.6 Conic Sections
We shift conics using the principles reviewed in Section 1.2, replacing x by x ybyy + k. Show that the equation x 2  4y2 perbola. Find its center, asymptotes, and foci.
EXAMPLE 4
Solution follows:
+ 2x +
645 + h and
8y  7 = 0 represents a by
We reduce the equation to standard form by completing the square in x and y as (x 2 (x 2
+
2x
+ I) 
+ 2x)
 4(y2  2y) = 7
4(y2  2y (X
+ 1)2 4
+ 1)
=
7 + I  4
 (y  1)2 = I.
This is the standard form Equation (lO) of a hyperbola with x replaced by x + 1 and y replaced by y  I. The hyperbola is shifted one unit to the left and one unit upward, and it has center x + 1 = oandy  I = O,orx = 1 andy = I. Moreover,
so the asymptotes are the two lines
x2 + 1  (y  1) = 0 The shifted foci have coordinates ( 1 ±
and
x ~ 1 + (y _ 1)
=
O.
v'5, 1).
•
Exercises 11.6 Identity;ng Graphs Match the parabolas in Exercises 14 with the following equations:
x' = 2y, x' = 6y, y'
= 8x,
y'
Match each conic section in Exercises 58 with one of these equations:
= 4x.
Then f'md each parabola's focus and direc1rix.
1.
Y
y
2.
Then fmd the conic section's foci and vertices. If the conic section is a hyperbola, f'md its asymptotes as well. S. Y 6. Y
~f>x
3.
Y
4.
Y +++~X
 ....~>x
646
Chapter 11: Parametric Equations and Polar Coordinates
Shifting Conic Sections
8.
7.
You may wish to review Section 1.2 before solving Exercises 31}56.
y
y
39. The parabola y2 = 8x is shifted down 2 units aod right I unit to generatetheparabola(y + 2)2 = 8(x  I) . •• Find the new parabola's vertex, focus, aod direc1rix. b. Plot the new vertex, focus, and directrix, aod sketch in the parabola.
40. The parabola x 2 = 4y is shifted left I uoit aod up 3 units to generate the parabola (x + I)" = 4(y  3) . •• Find the new parabola's vertex, focus, aod direc1rix.
Parabolas Exercises 1}16 give equations of parabolas. Find each parabola's focus and directrix. Then sketch the parabola. Ioclude the focus aod direc1rix in your sketch. 9. y2 = 121: 12. y2 = 21: 15. x
10. x 2 = 6y
11. x 2 = 8y
= 4x 2
14. Y = 8x 2
13. Y
= 3y2
b. Plot the new vertex, focus, and directrix, aod sketch in the parabola. 41. The ellipse (x 2/16) + (y2/9) = 1 is shifted 4 units to the right aod 3 units up to generate the ellipse (x  4)2 16
16. x = 2y2
+
(y  3)2 9
= 1.
a. Find the foci, vertices, and center of the new ellipse.
EIUpses Exercises 1724 give equations for ellipses. Put each equation in staodard form. Then sketch the ellipse. Ioclude the foci in your sketch. 17. lfu;2 + 25y2 = 400 19. 21:2 +y2=2
18. 7x 2 + 16y2 = 112 20. 21: 2 +y2=4
21. 3x2
= 6
22. 9x 2
= 54
24.
23. fu;2
+ 2y2 + 9y2
b. Plot the new foci, vertices, aod center, aod sketch in the new ellipse. 42. The ellipse (x 2/9) + (y2/25) = I is shifted 3 units to 1he left aod 2 units down to generate 1he ellipse
(x + 3)2
+ IOy2 = 90 169x 2 + 25y2 = 4225
9
+
(y
+
2)2 = 1.
25
a. Find the foci, vertices, and center of the new ellipse. b. Plot the new foci, vertices, and center, and sketch in the new
Exercises 25 aod 26 give information about the foci and vertices of ellipses centered at the origin of the xyplane. 10 each case, f"md the ellipse's staodardform equation from the given information.
ellipse. 43. The hyperbola (x 2/16)  (y2/9) right to generate the hyperbola
=I
is shifted 2 units to the
25. Foci: (± 0,0) Vertices: (±2, 0) 26. Foci: (0, ±4) Vertices: (0, ±5)
27. x 2  y2 = I
28. 9x 2  16y2 = 144
29.y2_ X2=8
30.y2_X2=4

y2
16
"9 =
1.
a. Find the center, foci, vertices, and asymptotes of the new
Hyperbolas Exercises 2734 give equations for hyperbolas. Put each equation in staodard form aod f"md the hyperbola's asymptotes. Then sketch the hyperbola. Ioclude the asymptotes aod foci in your sketch.
31. 8x2
(x  2)2
2y2
=
hyperbola.
b. Plot the new center, foci, vertices, aod asymptotes, aod sketch in the hyperbola. 44. The hyperbola (y2/4)  (x 2/5) = 1 is shifted 2 units down to generate the hyperbola
32.y2_3x 2 =3
16
(y
4
34. 64x 2  36y2 = 2304
33. 8y2  21: 2 = 16
+ 2)'
x2
"5
=
1.
a. Find the center, foci, vertices, and asymptotes of the new Exercises 3538 give information about the fuci, vertices, aod asymptotes of hyperbolas centered at 1he origin of 1he xyplane. 10 each case, f"md 1he hyperbola~ staodardfurm equation from the information giveo. 35. Foci:
36. Foci:
(0, ±0)
Asymptotes: 37. Vertices:
y = ±x
(±3, 0)
Asymptotes:
y
=
(±2,0)
Asymptotes: 38. Vertices:
±~ x 3
y = ±_I x
v3
(0, ±2)
Asymptotes:
y
=
hyperbola.
b. Plot the new center, foci, vertices, aod asymptotes, aod sketch in the hyperbola.
Exercises 4548 give equations for parabolas aod tell how many units up or down and to the right or left each parabola is to be shifted. Find ao equation for the new parabola, aod f"md 1he new vertex, focus, and directrix. 45. y2
±l2 x
= 4x,
47. x 2 = 8y,
= 121:,
left 2, down 3
46. y2
right I, down 7
48. x 2 = 6y,
right 4, up 3
left 3, down 2
11.6 Exercises 4~52 give equations for ellipses and tell how many units up or down and to the right or left each ellipse is to be shifted. Find an
equation for the new ellipse, and fmd the new foci, vertices, and center. x2 49. (;
y2
+9
2
50.
x T + y2
51.
3 +T
52.
16 + 25
x2
y2
Xl
y2
~
left 2, down I
I,
~ I,
right 3, up4
~
right2,up3
I,
~ I,
x2
y2
16  9
~
left 4, down 5
I,
left 2, down I
55. y2  x 2 ~ I,
left I, down I
y2
56.
3 
x 2 ~ I,
647
70. Suspension bridge cable. hang in parabolas The suspension bridge cable shown in the accorupanying figure supports a uniform load of w pounds per horizontal foot It can be shown that if H is the horizontal tension of the cable at the origin, then the curve of the cable satisfies the equation
Show that the cable hangs in a parabola by solving this differential equation subject to the initial condition that y ~ 0 when x ~ O.
Exercises 5356 give equations for hyperbclas and tell how many units up or down and to the right or left each hyperbcla is to be shifted. Find an equation for the new hyperbola, and fmd the new center, foci, vertices, and asymptotes. x 2 y2 53. 4  5 ~ I, right 2, up2 54.
Conic Sections
l~ 71. The width of a parabola .t the fucu. Show that the number 4p is the width of the parabola x 2 ~ 4py (p > 0) at the focus by showing that the line y ~ p cuts the parabola at points that are 4p units apart
72. The asymptote. of (x'Ia')  (yl Iii') = 1 Show that the vertical distsnce between the line y ~ (bla}x and the upper half of the righthand branch y ~ (bla)Vx 2  a 2 of the hyperbola (x 2/a 2)  (y2/b 2) ~ I approaches 0 by showing that
right I, up 3
Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections in Exercises 5768. 57. x 2 + 4x + y2 ~ 12 58. 2x2 + 2y2  28x
+
12y
+
Sintilar results hold for the remaining portions of the hyperbola andthelinesy ~ ±(b/a}x.
114 ~ 0
59.X2+2x+4y3~0
60. y2  4y  8x  12 ~ 0
61. x 2 + 5y2
62. 9x 2 + 6y2
+ 4x
~ I
+
36y ~ 0
63. x 2 + 2y2  2x  4y ~ I 64. 4x 2 + y2 65.
+
8x  2y ~I
x2_y2_2x+4y~4
67. 2x2

y2
+ 6y
~ 3
66. x2_y2+4x6y~6
68. y2  4x 2
+
16x ~ 24
Theory and Examples 69. lflines are drawn parallel to the coordinate axes through a point P on the parabola y2 ~ Irx, /r; > 0, the parabola partitions the rectsngular region bounded by these lines and the coordinate axes into two arualler regions, A and B. a. If the two arualler regions are revolved about the yaxis, show that they generate solids whose volumes have the ratio 4: 1. b. What is the ratio of the volumes generated by revolving the regions about the xaxis? y p
B ~~~~X
o
73. Area Find the dimensions of the rectsng1e of largest area that can be insctibed in the ellipse x 2 + 4y2 ~ 4 with its sides parallel to the coordinate axes. What is the area of the rectangle? 74. Volume Find the volUD2e of the solid generated by revolving the region enclosed by the ellipse 9x' + 4yl ~ 36 about the 0
Polar Coordinates 17.
L
Find an equation in polar coordinates for the curve
x = e 2 tcost. y = e2t sint;
00
 0 from (I), tanI/J
= tan(t/>
tant/>  tanO
 0)
= I + tant/>tanO'
Furthermore,
p
dy
dy/dO
tant/> = 4 The exterior of the sphere x 2 + y2 + z2 = 4. 2 (d) x + y2 + z2 = 4, z :5 0 The lower hemisphere cut from the sphere x 2 +
y2
+ z2
= 4 by the xyplane (the plane
z = 0) .
•
Just as polar coordinates give another way to locate points in the xyplane (Section 11.3), alternative coordinate systems, different from the Cartesian coordinate system developed here, exist for threedimensional space. We examine two of these coordinate systems in Section 15.7.
Exercises 12.1 Geometric Interpretations of Equations
Geometric Interpretations of Inequalities and Equations
In Exercises 116, give a geometric description of the set of points in
In Exercises 1724, describe the sets of points in space whose coordinates satisfY the given inequalities or combinatioos of equatioos aod inequalities.
space whose coordinates satisfY the given pairs of equations. 1. x = 2, Y = 3 2. x = 1, z = 0
= 0, 2 x + y2
3. y
z = 0
S.
=
2
4,
z = 0
+z 2 =4, y=O 9. x 2 + y2 + z2 = I, x = 0
7.
X
11. 12. 13. 14. 15.
16.
+ y2 = 4, z = 8. y2 + z2 = I, x =
6. xl
+ y2 + z2 = 25, Y = 4 x 2 + y2 + (z + 3)' = 25, z = 0 x 2 + (y  1)2 + z2 = 4, Y = 0 x 2 + y2 = 4, z = Y x 2 + y2 + z2 = 4, Y = x Y = x 2, z = 0 z = y2, X = 1
10. x 2
17. a. x
4. x = I, Y = 0 2 0
~
0, y
0,
~
Z
18.•. 0 '" x '" 1 c. 0 ~ x ~ 1, 0
~
y
= 0
~
b. x
1,
~
+ y2 + z2
:s 1
20. a. x 2
+ y2 :s 1,
Z
+ y2:5
norestrictiononz
a.
C. x 2
1,
= 0
~
b. 0 '" x '" I, 0 ~z~ 1
x2
19.
0, y
b.
Xl
b. x 2
0,
Z
= 0
0 '" Y s 1
+ y2 + z2 >
1
+ y2 :s
=3
1,
Z
21. a. 1 :sx 2 +y2+z2 :S4 b. x 2
+ y2 + z2 :s
22. a. x = y,
23. a. y ~ x 2,
z = 0 Z
24. a. z = 1  y, b. z =
y3,
X
~ 0
I, z ~ 0 b. x
=
y,
b. x
S
y2,
no restriction onx
=2
no restriction on z
0
S Z
s 2
662
Chapter 12: Vectors and the Geometry of Space
Distance and Spheres in Space
Z
The formula for the distance between two points in the xyplane extends to points in space.
The Distance Between P1(XhYh Zl) and P 2(X2,Y2, Z2) is
+
IPI P21 = V(X2  XI)2
x
FIGURE 12.5 We find the distance between PI and P 2 by applying the Pythagorean theorem to the right triangles PIAB and PIBP2.
(Y2  YI)2
+
(Z2  ZI)2
Proof We construct a rectangular box with faces parallel to the coordinate planes and the points PI and P2 at opposite comers of the box (Figure 12.5). If A(X2, YI, zd and B(X2, Y2, Zl) are the vertices of the box indicated in the figure, then the three box edges P I A, AB, and BP2 have lengths
Because triangles PIBP2 and PIAB are both rightangled, two applications of the Pythagorean theorem give IPIP212 = IPIBI 2 + IBP21 2
and
(see Figure 12.5). So IPIP212 = IPIBI 2 + IBP21 2 IPIA 12
+
IABI2
+
Substitute IP]BI 2 = IP]AI 2 + IABI2 .
IBP212
IX2  xl1 2 + IY2  YI1 2 + IZ2  zI1 2 =
(X2  XI)2
+
(Y2  Ylf
+
(Z2  zlf
Therefore
• EXAMPLE 3
The distance between P I (2, 1,5) and P2( 2,3,0) is IPIP21 = V(2  2f
Z
Po(Xo, Yo, zo)
\
P(x,Y,z)
al
I
=
V16
=
V45
+4+ I"::j
+
(3  1)2
+
(0  5)2
25
•
6.708.
We can use the distance formula to write equations for spheres in space (Figure 12.6). A point P(x, y, z) lies on the sphere of radius a centered at Po(xo,Yo, zo) precisely when IPoPI = a or
I
 I f"
,,/ 1
The Standard Equation for the Sphere of Radius a and Center (xo,Yo, zo) (x  xof
+ (y 
YO)2
+
(z  zO)2
= a2
Y x
FIGURE 12.6 The sphere of radius a centered at the point (xo,Yo, zo).
EXAMPLE 4
Find the center and radius of the sphere x2
+ y2 + z2 +
3x  4z
+ 1 = O.
SoLution We find the center and radius of a sphere the way we find the center and radius of a circle: Complete the squares on the X, Y, and zterms as necessary and write each
664
Chapter 12: Vectors and the Geometry of Space
In Exercises 2534, describe the given set with a single equatioo or
with a pair of equations. 25. The plaoe perpendicular to the L
xaxis at (3, 0, 0)
b. yaxis
c. zaxis
27. The plaoe througb the point (3, I, I) parallel to the xyplane
b. yzplane
c. xzplane
28. The circle of radius 2 centered at (0, 0, 0) and lying in the L
xyplane
b. yzplane
c. xzplane
29. The circle of radius 2 centered at (0, 2, 0) and lying in the L
xyplane
b. yzplane
parallel to the xyplane
b. yzplane
Sphel'l!s Find the centers and radii of the spheres in Exercises 4750.
+2)' + +(z  2)' = 8 48. I)' +&+tY +(z +3)' 25 49. (x  Vz)' +{y  Vz)2 +(z +Vz)' 50. x+ &+tY +(ztY ~ 47. (x
y2
(x 
=
2
b. yaxis
2
=
Find equations for the spheres whose centers aud radii are given in Exercises 5154.
c. xzplaoe
Center
Radius
31. The line through the point (I, 3, I) parallel to the
LXaxis
=
c. planey = 2
30. The circle of radius I centered at (3,4, I) and lying in a plane L
P2(2, 2, 2)
46. P I (5, 3, 2), P2(0, 0, 0)
26. The plaoe througb the point (3, 1,2) perpendicular to the
L
P 2(2, 3, 4)
45. PI(O, 0, 0), b. yaxis at (0, I, 0)
c. zaxis at (0, 0, 2) LXaxis
44. P I (3, 4, 5),
c. zaxis
32. The set of points in space equidistant from the origin and the point (0, 2, 0)
53.
33. The circle in whicb the plane through the point (I, 1,3) perpendicular to 1Ire zaxis meets the sphere of radius 5 centered at the origin 34. The set of points in space that lie 2 units from 1Ire point (0, 0, I) and, at 1Ire saroe time, 2 units from the point (0, 0, I) Inequalities to Describe Sets of Points Write inequalities to describe the sets in Exercises 3540.
35. The slab bounded by the planes z = 0 and z = I (planes
Vi4
51. (1,2,3) 52. (0, 1,5)
2
(I,t,t)
4 9
54. (0, 7,0)
7
Find the centers and radii of the spheres in Exercises 5558. 55. x 2 y2 z2 4x  4z = 0
+++ 56. ++ + 57. + + +++ 9 58. 3x + + + 9 x2
lx
y2
2 2
z2 
8z = 0
6y
2y2
2z2
X
3y2
3z2
2y 
Y
z =
2z =
included) 36. The solid cube in the frrst octant bounded by 1Ire coordioate plaoes and 1Ire planes x = 2, y = 2, and z = 2
37. The halfspace consisting of the points 00 and below the xyplane 38. The upper hemisphere of the sphere of radius I centered at the origin
39. The (a) interior and (b) exterior of the sphere of radius I centered at the point (I, I, I) 40. The closed regioo bounded by the spheres of radius I and radius 2 centered at the origin. (Closed means the spheres are to be included. Had we wanted the spheres left out, we would have asked for the open regioo bounded by the spheres. This is analogous to the way we use closed and open to describe intervals: closed means endpoints included, open means endpoints left out. Closed sets include boundaries; open sets leave them out)
Distance In Exercises 4146, fmd the distsnce between points PI and P2. 41. Pl(l, I, I),
P 2(3, 3, 0)
42. Pl( I, 1,5), P2(2, 5, 0) 43. PI(I, 4, 5),
P2(4, 2,7)
Theory and Examples 59. Find a formula for the distance from the point P(x, y, z) to the
a. xaxis
c. zaxis
b. yaxis
60. Find a formula for the distance from the point P(x, y, z) to the
a. xyplane
c. xzplane
b. yzplane
61. Find the perimeter of the triangle with vertices A( 1,2, I), B(I, I, 3), and C(3, 4, 5). 62. Show that the point p(3, I, 2) is equidistant from the points A(2, 1,3) andB(4, 3, I). 63. Find an equation for the set of all points equidistant from the planesy = 3 andy = \. 64. Find an equation for the set of all points equidistant from the point (0, 0, 2) and the xyplane. 65. Find the point on the sphere x 2 nearest •• thexyplane.
+(y  + + 3)'
(z
5)' = 4
b. thepoint(O, 7, 5).
66. Find the point equidistant from the points (0, 0, 0), (0,4, 0), (3, 0, 0), and (2, 2, 3).
12.2
12.2
Vectors
665
I~ ~_ ct_ or_ s __________________________________ Some of the things we measure are determined simply by their magnitudes. To record mass, length, or time, for example, we need only write down a number and name an appropriate unit of measure. We need more information to describe a force, displacement, or velocity. To describe a force, we need to record the direction in which it acts as well as how large it is. To describe a body's displacement, we have to say in what direction it moved as well as how far. To describe a body's velocity, we have to know where the body is headed as well as how fast it is going. In this section we show how to represent thiogs that have both magnitude and direction in the plane or in space. Terminal
FIGURE 12.7
The directed line segment
AB is called a vector.
Component Form A quantity such as force, displacement, or velocity is called a vector and is represented by a directed line segment (Fignre 12.7). The arrow points in the direction of the action and its length gives the magnitude of the action in terms of a suitably chosen uuit. For example, a force vector points in the direction in which the force acts and its length is a measure of the force's strength; a velocity vector points in the direction of motion and its length is the speed of the moving object. Figure 12.8 displays the velocity vector v at a specific location for a particle moving along a path in the plane or in space. (This application of vectors is studied in Chapter 13.) y
y B _______ D
A
C p ~~~~x
o
FIGURE 12.9 The four arrows io the plane (directed lioe segments) shown here have the same length and direction. They therefore represeot the same vector, aodwewriteAB = cD = OP = EF.
~~~~~y
o
F
E
o
~~X
x
(a) two dimensions
(b) three dllnensions
FIGURE 12.8 The velocity vector of a particle moving aloog a path o. The crosssections in planes perpendicular to the zaxis above and below the xyplane are hyperbolas. The crosssections in planes perpendicular to the other axes are parabolas.
Near the origin, the surface is shaped like a saddle or mountain pass. To a person traveling along the surface in the yzplane the origin looks like a minimum. To a person traveling the xzplane the origin looks like a maximum. Such a point is called a saddle point of a surface. We will say more about saddle points in Section 14.7. _ Table 12.1 shows graphs of the six basic types of quadric surfaces. Each surface shown is symmetric with respect to the zaxis, but other coordinate axes can serve as well (with appropriate changes to the equation).
12.6
699
Cylinder.; and Quadric Surfaces
TABLE 12.1 Graphs of Quadric Surfaces
,
, Tho PEIobolu _ ~l:3
ill Ibo ",,"pImo
*~~+,, T', ,, ,,
" ......
, 'Ibe p,nbolu _
'~/
~:I
____
;"'1110 1f'1Ilmo
,
,
• ELLIPSOID
ELLIPTICAL PARABOLOID
•
,/
)fr"
HYPERBOLOID OF ONE SHEET
ELLIPTICAL CONE
,
. .,.
'Ibe e1Hpee 
+ _1
.'f  cV'i
,
,
. HYPERBOLOID OF 1WO SHEETS
HYPERBOLIC PARABOLOID
% C'
.>0
700
Chapter 12: Vector.; and the Geometry of Space
Exercises 12.6 Matchtng Equations with Surfaces In Exercises 112, match the equation with the surface it defines.
k.
,
L
Also, identify each surface by type (paraboloid, ellipsoid, etc.) The swfaces arc labeled (a){I).
1. x:1 +y:1 + 4z2 = 10 3. 9y2 +z:1 = 16 5. x _ y 2_ z 2 7.x2 +2z 2 _ S
2.z:1+4y 2_4x:1=4 4.y:1+ z 2=X 2 6.x _ _ 2_ 2 z y 8.z 2 +x:1 y 2 _ 1 10. z = _4.1: 2 _ y2
9. X=z2_ y :1 11. x:1 + 4z:1 = y2
12. 9x:1
+ 4y:1 + 2z2
Drawtng
Sketch the surfaces in Exercises 1344.
= 36
,
b.
L
,
CYUNDERS 13. X2 +y2=4 15. x2+~ _ 16
14.z=y21 16. 4x 2 + y2 _ 36
ELUPSOIDS 17. 9;t2 + y2 + z2 = 9 19. 4x 2 + 9y2 + ~ _ 36
18. 4x:1 20. 9;t2
+ 4y:1 + z:1 = 16 + 4y2 + 36z2 _ 36
PARABOLOIDS AND CONES
,
,.
,
d.
:n.z=S:x2 y:1
21. z=x:1+4y:1 23. x=44y:1~ 25. X:1+ y 2=z:1
24.y=I:x 2 z2 26. 4x:1 + ~ = 9y2
HYPERBOLOIDS 28.y:1+ z 2_ x 2=1 30. (y'/4)  (.'/4) 
27.x:1+ y 2 z :1=1 29. z2_ X:1_ y 2= 1
y
z'
~
1
HYPERBOUC PARABOLOIDS
••
,
f•
31. y2:x:1 _ z
,
ASSORTED
y
..
,
h•
x2
38.x2+~ _ y
40. 16y2 + 9z:1 _ 4x 2 42.y:1_ x 2_ z 2=1 44.x 2 +y:1=z
+z2
_ 1 41. z = _(X2 + y2) 43.4y 2+ z 2_4x:1=4
39.
y
34. 4x:1 + 4y:1 = z2 36. 1&2 + 4y2 _ 1
33. z= 1 +y2_X:1 35. Y _ _ (x:1 + z2) 37. x 2 +y2_z2 _ 4
Theory and Examples 45. L Express the area.4. of the crosssection cut from. the ellipsoid
, ,
x:1+L+~=1 4 9
L
,
J.
,
by the plane z  c as a function of c. (The area of an ellipse with semiaxesa andbis7rab.) b. Use slices perpendicular to the zaxis to fmd the volume of
the ellipsoid in part (a). c. Now find the volume of the ellipsoid
x2
y2
z:1
++ 1. a:1 b2 c2 y
Does your formula give the volume of a sphere of radius a if a  b  c?
Chapter 12
701
h. Express your answer in part (a) in terms of h and the areas Ao and Ah of the regions cut by the hyperboloid from the planes z = 0 andz = h.
46. The barrel shown here is shaped like an ellipsoid with equal pieces cut from the ends by planes perpendicular to the zaxis. The crosssections perpendicular to the zaxis are circular. The barrel is 2h units high. its midsection radius is R, and its end radii are both r. Find a formula for the barrel's volume. Then check two things. First, suppose the sides of the barrel are straightened to turn the barrel into a cylinder of radius R and height 2h. Does your formula give the cylinder's volume? Second, suppose r = 0 and h = R so the barrel is a sphere. Does your formula give the sphere's volume?
z
Questions to Guide Your Review
c. Show that the volume in part (a) is also given by the formula V=
"6h (Ao + 4Am + Ah ),
where Am is the area of the region cut by the hyperboloid from the plane z = h/2.
Viewing Surfaces
D Plot the surfaces in Exercises 4952 over the indicated domains.
If
you can, rotate the surface into different viewing positions.
2 ~ x ~ 2,
49. z = y2,
2 ~ x ~ 2,
50. z = 1  y2, y
51. z = x 2 52. z = x 2
47. Show that the volume ofthe segment cut from the paraboloid x2 y2 z 2+  = a b2 c
by the plane z = h equals half the segment's base times its altitude. 48. a. Find the volume of the solid bounded by the hyperboloid x2 y2 z2 2+ 2  = 1 a b c2
and the planes z = 0 and z = h, h
Chapter
>
~
c. 2 d. 2
~
x
~
x
2 ~ y ~ 2
x ~ 3,
3 ~ y ~ 3
~
3,
3
~
y
~
3
~x~
1, 2,
2
~
y
~
3
~
2
~
y
~
2
~
2,
1
~y~
1
x
COMPUTER EXPLORATIONS Use a CAS to plot the surfaces in Exercises 5358. Identify the type of quadric surface from your graph. x2
53.
y2
9 + 36
55. 5x 2
o.
+ y2, 3 ~ + 2y2 over
a. 3 h. 1
0.5 ~ y ~ 2
=
9 
1  25
z2  3y2
x2
57.
z2 =
y2
1
=
54.
56. z2
16 + 2
x2
z2
9  9 y2
16
58. y 
=
1
=
1
x2
y2
16
9 +z
V4 
z2 = 0
Questions to Guide Your Review
1. When do directed line segments in the plane represent the same vector? 2. How are vectors added and subtracted geometrically? Algebraically? 3. How do you find a vector's magnitude and direction? 4. If a vector is multiplied by a positive scalar, how is the result related to the original vector? What if the scalar is zero? Negative? 5. Define the dot product (scalar product) of two vectors. Which algebraic laws are satisfied by dot products? Give examples. When is the dot product of two vectors equal to zero? 6. What geometric interpretation does the dot product have? Give examples. 7. What is the vector projection of a vector u onto a vector v? Give an example of a useful application of a vector projection. 8. Define the cross product (vector product) of two vectors. Which algebraic laws are satisfied by cross products, and which are not? Give examples. When is the cross product of two vectors equal to zero? 9. What geometric or physical interpretations do cross products have? Give examples.
10. What is the determinant formula for calculating the cross product of two vectors relative to the Cartesian i, j, kcoordinate system? Use it in an example. 11. How do you find equations for lines, line segments, and planes in space? Give examples. Can you express a line in space by a single equation? A plane? 12. How do you find the distance from a point to a line in space? From a point to a plane? Give examples. 13. What are box products? What significance do they have? How are they evaluated? Give an example. 14. How do you find equations for spheres in space? Give examples. 15. How do you find the intersection of two lines in space? A line and a plane? Two planes? Give examples. 16. What is a cylinder? Give examples of equations that define cylinders in Cartesian coordinates. 17. What are quadric surfaces? Give examples of different kinds of ellipsoids, paraboloids, cones, and hyperboloids (equations and sketches).
702
Chapter 12: Vectors and the Geometry of Space
Chapter
Practice Exercises
Vedor Calculations in Two Dimensions In Exercises 14, let u = (3,4) and v = (2, 5). Find (a) the componeot form of the vector aod (b) its magnitude.
In Exercises 25 and 26, fmd (a) the area of the parallelogram determined by vectors u and v and (b) the volume of the parallelepiped determined by the vectors u, v, and w.
1.3u4v
2.u+v
2S.u=i+jk,
3. 2u
4. 5v
26.u=i+j,
In Exercises 58, fmd the component form of the vector.
5. The veetor obtained by rotating (0, I) througb an angle of21f/3
radians 6. The unit vector that makes an angle of 1f/6 radian with the positivexaxis 7. The vector 2 units loog in the directioo 4i  j
v=2i+j+k,
v=j,
w=i+j+k
Lines, Planes, and Distances 27. Suppose that n is DOnnai to a plane and that v is parallel to the plane. Describe how you would fmd a veetor n that is both perpendicular to v and parallel to the plane. 28. Find a veetor in the plane parallel to the line ax + by = c. In Exercises 29 and 30, fmd the distance from the point to the line.
8. The veetor 5 units loog in the directioo opposite to the directioo of(3/5)i + (4/5)j
29. (2,2,0); x
=
t, Y
=
t,
Z =
Express the vectors in Exercises 912 in terms of their lengtha aod directioos.
12. Velocity vector v = (etcost  etsint)i + (etsint + etcost)j wheol = In 2.
Vedor Calculations in Three Dimensions Express the veetors in Exercises 13 and 14 in terms of their leogtha and directions.
+ 6k
14. i
Z
= I
31. Parametrize the line that passes througb the point (I, 2, 3) parallel to the vector v =  3i + 7k. 32. Parametrize the line segment joining the points P(I, 2, 0) and
10. i  j
11. Velocityveetorv = (2sinl)i + (2cosI)jwheol = 1f/2.
13. 2i  3j
+t
1
30. (0,4, I); x = 2 + I, y = 2 + I,
9. v2i + v2j
w=i2j+3k
+ 2j  k
15. Find a vector 2 units long in the direction ofv = 4i  j + 4k. 16. Find a vector 5 units loog in the direction opposite to the directioo ofv = (3/5) i + (4/5)k.
Q(I,3, I). In Esercises 33 and 34, fmd the distance from the point to the plane.
33. (6,0, 6),
x  y = 4
34. (3,0, 10),
2x
+
3y
+Z
= 2
35. Find an equation for the plane that passes through the point (3, 2, 1) normal to the veetor n = 2i + j + k. 36. Find an equatioo for the plane that passes through the point (1,6,0) perpendicular to the line x = I + I,y = 6  21, Z = 31. In Exercises 37 and 38, fmd an equatioo for the plane througb points
P,Q,andR. In Esercises 17 and 18, fmd lvi, lui, V'u, U'v, v X u, u X v,
37. P(i, 1,2),
Iv X u I, the angle between v and u, the scalar compooent of u in the direction afv, and the vector projection ofo onto v.
38. P(i, 0, 0),
17. v = i + j
18. v = i + j + 2k
u=2i+j2k
u=ik
In Exercises 19 and 20, fmd proj. u.
19. v=2i+jk u=i+j5k
Q(2, 1,3),
R( 1,2,  1)
Q(O, 1,0), R(O, 0, I)
39. Find the points in which the line x = I + 21, y = I  I, z = 3t meets the three coordinate planes. 40. Find the point in which the line througb the origin perpendicular to the plane 2x  Y  Z = 4 meets the plane 3x  5y + 2z = 6. 41. Find the acute angle between the planes x = 7 and x + y = 3.
v2z
20. u=i2j
v=i+j+k
+
42. Find the acute angle betweeo the planes x + y = I and
y+z= 1. In Exercises 21 and 22, draw coordinate axes and then sketch u, v, and
u X v as vectors at the origin. 21. u=i.,
v=i+j
22. u=ij,
v=i+j
23. !flvl = 2, Iwl = 3,andthe angie between v andw is 1f/3,fmd Iv  2wl· 24. For what value or values of a will the vectors u = 2i + 4j  5k and v = 4i  8j + ak be parallel?
43. Find pararne1ric equations for the line in which the planes x + 2y + Z = I and x  y + 2z = 8 intersect. 44. Show that the line in which the planes
x+2y2z=5
and
5x2yz=0
intersect is parallel to the line
x = 3 + 21, y = 31, z = I + 41.
Chapter 12 45. The planes 3x L
+ 6z = I and 2x + 2y  z = 3 intcncct in a Iinc.
Shaw that the planes are orthogonal.
L (21  3J + 3k)' b. x  3t,
b. Find equatiOlli for the line of intersection.
046. Find an equation for the plane that pasSCll through the point (1,2, 3)paraIleltou  2i + 3j + kandv  i  j + 2k. 47. Is v = 2i  4j + k related in any !pCcial way to the plane 2x + Y = 5? Give reasons for your answer. 48. The equation.' P;P = 0 represents ~plane through Po normal to D. What set docs the inequality D' PoP> 0 tcpICscn.t?
e. (x
«x + 2)1 + (y 
y  Ilt,
+ 2) + l1(y
Practice Exercises
703
I)J + zk)  0
z  23t
 1)  3z
d. (2i  3j + 3k) X «x + 2)i + (y  I)j + zk) = 0 e. (2i  j + 3k) X (3i + k)·«x + 2)i + (y  I)j + zk) ~O
6l. The puallelogram !hawn here has vertices at A(2, 1,4), B(I, 0, I), C(I, 2, 3), andD. Find
,
49. Find the distance from the point P(I, 4, 0) to the plane through A(O. o. 0). B(2. o. 1). "'" C(2. 1.0).
D),
50. Find the distance from the point (2, 2, 3) to the plane 2x + 3y + 5z  O. 51. Find a vector parallel to the plane 2x  y  z  4 and orthogonaltot+J+k. A(2,l,4)
51. Find a unit vector orthogonal to A in the plane of B and C if A  2i  j + k,B  i+ 2j + k,andC  i + j  2k.
C(l.2, 3)
53. Find a vector of magnitude 2 parallel to the line of intcncction of theplane8x + 2y +z  I = and x y + 2z+ 7 = O.
o
54. Find the point in which the line through the origin perpendicular totheplane2x  y  z = 4mcctsthep1mc3x  5y +22' = 6.
y B(l, 0. 1)
55. Find the point in which the line through p(3, 2, I) normal ttl the plane 2x  y + 2z = 2 meets the plane. 56. What angle doc! the line of intcncction of the planca 2x + Y  z = 0 and x + Y + 2z = 0 make with the positive
xaxis?
L the coordinates of D, b. the cosine of the interior angle atB,
e. the vector projection of BA. onto iiC,
57. The line
d. the area of the parallelogram, L:
x=3+2t,
y=2t,
z=t
in1eDcctll the plane x + 3y  z = 4 in a point P. Find the coonIin.atcs of P and find equations for the line in the plane through P perpendicular ttl L.
58. Shaw that for every real number k the plane
x  2y+ z + 3 + k(2x  y z + I) = 0
contains the line of intmscction of the planes
c. an equatioo for the plane of the parallelogram. f. the areas of the orthogonal projections of the panillelogram
on the three coord.i.na.tJ; planes. 63. Distmte betweea lineI Find the distance bctwccn the line L, through the points .4(1,0,1) and B(I,I,O) and the line L2 through the points C(3, I, I) andD(4, 5, 2). The distance is ttl be meuured along the line perpendicular In the two lines. FltBt fmd • vector D perpendicular to both lines. Thm project AC onto D. 64. (Continuatio1l o/Exerci.!e 63.) Find the distance between the line through A(4, 0, 2) and B(2, 4, 1) and the line through C(t, 3, 2)
""'D(2. 2. 4). x  2y + z + 3 = 0 and 2x  y  z + I = O. Quadric SumCt!!s 59. Find an equation for the plane through A(2,O, 3) and B(I, 2, I) that lies parallel to the line through
Identify and sketch the surfaces in Exercises 6576.
C( 2. 13/'. 26/') ""'D(16/'. 13/'. 0). 60. Is the line x = 1 + 2t, y = 2 + 31, z = 5t related in any way to the plane 4x  6y + 10z  91 Give reasoos for your
67.4x 2 +4y 2+ Z l=4 69. z = _(x 2 + y2)
""""'.
61. Which of the following are equations for the plane through the points P(I, I, I), Q(3, 0, 2), andR(2, I, O)?
65. x 2 + y2 + z2 = 4
71. X2 +y2=z2 73. X2 +y2_z2=4 75. yl_x2_zl= 1
66. x 2 + (y  1)2 + z2 = I 68. 3fu:l + 91 2 + 4z2 = 36
70. Y = _(Xl
+ zZ)
71. Xl+zZ=yl
74. 4yl+zZob:l =4 76. z2 _x 2 _ y2 = I
704
Chapter 12: Vectors and the Geometry of Space
Chapter
Additional and Advanced Exercises
1. Submarine hunting Two surface ships on maneuvers are trying to determine a submarine's course and speed to prepare for an aircraft intercept. As shown here, ship A is located at (4, 0, 0), whereas ship B is located at (0, 5, 0). All coordinates are given in thousands of feet. Ship A locates the submarine in the direction of the vector 2i + 3j  (I/3)k, and ship B locates it in the direction of the vector 18i  6j  k. Four minutes ago, the submarine was located at (2, 1, 1/3). The aircraft is due in 20 min. Assuming that the submarine moves in a straight line at a constant speed, to what position should the surface ships direct the aircraft?
clockwise when we look toward the origin fromA. Find the velocity v of the point of the body that is at the position B(l, 3, 2).
z
IA~(:: l,~l,~l!)
_I.
B
I
:~: : .J!___ v: / I
z

x
(l, 3, 2)
I
/
I
1/
I~
I /
_
I
1//
~y
//
ShipB
Ship A
(4,0,0)
(0,5,0)
y
x
5. Consider the weight suspended by two wires in each diagram. Find the magnitudes and components of vectors F I and F 2, and angles a and {3.
a.
\ Submarine
Nor TO SCALE
2. A helicopter rescue Two helicopters, HI and H 2, are traveling together. At time t = 0, they separate and follow different straightline paths given by
HI:
x = 6 + 40t,
Y = 3 + lOt,
z = 3 + 2t
H2:
x = 6 + 11 Ot,
Y = 3 + 4t,
z = 3 + t.
Time t is measured in hours and all coordinates are measured in miles. Due to system malfunctions, H2 stops its flight at (446, 13, 1) and, in a negligible amount of time, lands at (446, 13,0). Two hours later, HI is advised of this fact and heads toward H2 at 150 mph. How long will it take HI to reach H2? 3. Torque The operator's manual for the Toro® 21 in. lawnmower says "tighten the spark plug to 15 ftlb (20.4 N· m)." If you are installing the plug with a lO.5in. socket wrench that places the center of your hand 9 in. from the axis of the spark plug, about how hard should you pull? Answer in pounds.
b.
(Hint: This triangle is a right triangle.)
6. Consider a weight of w N suspended by two wires in the diagram, where T I and T 2 are force vectors directed along the wires.
a. Find the vectors T I and T 2 and show that their magnitudes are w cos (3 ITII = sin (a + (3) 4. Rotating body The line through the ongm and the point A( 1, 1, 1) is the axis of rotation of a right body rotating with a constant angular speed of3/2 rad/sec. The rotation appears to be
and T

121
wcosa sin (a+{3)
705
Chapter 12 Additional and Advanced Exercises b. For a fIxed {j determine the value of a which minimizes the
magnitude IT ,I. c. For a fIxed a determine the value of (j which minimizes the magnitude IT21· 7. Determinants and planes
X
X2 
X
X3 
x
Y'  Y Y2  Y y,  Y
Zt 
z
Z2 
Z
Z3 
z
d
b. What set of points in space is described by the equation
Y
Z
Xl
Yl
Zl
X2
Y2
Z2
X3
Y3
Z3
8. Determinant. and line.
lax, + by,

cl
~ ~';=.c===.
0, there exists a corresponding number II
LI
'1T/4
t1>'1T/4
+
+
tk, then
(lim sin t)j tw/4
+
lim t)k ( lw/4
=V2·+V2·+1T k J 2'
2
•
4'
We derme continuity for vector functions the same functions.
WH:j
we derme continuity for scalar
DEFINmON A vector function r{t) is continuous at a point t = to in its domain if lim,....~ r(t) = r(to). The function is continuous if it is continuous at every point in its domain. From Equation (3), we see that r(t) is continuous at t = to if and only if each component function is continuous there (Exercise 31).
EXAMPLE 3 (a) All the space curves shown in Figures 13.2 and 13.4 are continuous because their component functions are continuous at every value of t in (  00, 00).
(b) The function g(t) = (cost)i
+
(sint)j
+
ltJk
is discontinuous at every integer, where the greatest integer function discontinuous.
l t J is •
Derivatives and Motion Suppose that r(t) = I(t)i + g(t)j + h(t)k is the position vector of a particle moving along a curve in space and that I, g, and h are differentiable functions of t. Then the difference between the particle's positions at time t and time t + !J.t is ~r =
r(t
+ !J.t)  r(t)
710
Chapter 13: VectorValued Functions and Motion in Space
(Figure 13.5a). In terms of components, ar = r(t =
= ~~y
(a)
+
M)  r(t)
+ at)i + g(t + M)j + h(t + M)k]  [/(t)i + g{t)j + h(t)k] [j(t + at)  I(t)]i + [g(t + M)  g{t)]j + [j(t
[h(t
+
M)  h(t)]k.
As at approaches zero. three things seem to happen simultaneously. First. Q approaches P along the curve. Second, the secant line PQ seems to approach a limiting position tangent to the curve atP. Third, the quotient arl M (Figure 13.5b) approaches the limit
lim arat [lim I(t + at)at 
I(t)]i
=
.1.,0
[lim g{t + at)at  g(t)]J.
+
.l.tO
+
[lim h(t
+ at)  h(t)]k
.1.,0 =
.l.tO
M
g [d/]i + [d ].J + [dh]k dt' dt dt
We are therefore led to the following definition. ~~y
(b) x
FIGURE 13.5 As M .... O. thepointQ approaches the point P along the curve C. In the limit, the vector PQ/ at becomes the tangent vector r'(t).
DEFINmON The vector function r(t) = I(t)i + g{t)j + h(t)k has a derivative (is differentiable) at t if I, g, and h have derivatives at t. The derivative is the vector function '(t) r
= dr =
l' r(t dt,J!!!o
+ at)
 r(t)
at
dl. dt I
+ dg. + dt J
dh k dt'
A vector function r is differentiable if it is differentiable at every point of its domain. The curve traced by r is smootb if dr/dt is continuous and never 0, that is, if I, g, and h have continuous first derivatives that are not simultaneously O. The geometric significance of the definition of derivative is shown in ~ure 13.5. The points P and Q have position vectors r(t) and r(t + at), and the vector PQ is represented by r(t + at)  r(t). For at > 0, the scalar multiple (1/at)(r(t + at)  r(t» points in the same direction as the vector As at ..... 0, this vector approaches a vector that is tangent to the curve atP (Figure l3.5b). The vector r'(t), when different from 0, is defined to be the vector tangent to the curve at P. The tangent line to the curve at a point U(to), g{to), h(to» is defined to be the line through the point parallel to r'(to). We require dr/dt "" 0 for a smooth curve to make sure the curve has a continuously turoing tangent at each point. On a smooth curve, there are no sharp corners or cusps. A curve that is made up of a finite number of smooth curves pieced together in a continuous fashion is called piecewise smooth (Figure 13.6). Look once again at Figure 13.5. We drew the figure for at positive, so ar points forward, in the direction of the motion. The vector arl at, having the same direction as ar, points forward too. Had M been negative, ar would have pointed backward, against the direction of motion. The quotient arI at, however, being a negative scalar multiple of ar, would once again have pointed forward. No matter how ar points, arl at points forward and we expect the vector dr/dt = lim&+o arl M, when different from O. to do the same. This means that the derivative drI dt, which is the rate of change of position with respect to time, always points in the direction of motion. For a smooth curve, drI dt is never zero; the particle does not stop or reverse direction.
PQ.
FIGURE 13.6 A piecewise smooth curve made up offive smooth curves connected end to end in • continuous fashion. The curve here is not smooth at the points joining the five smooth curves.
=
13.1 Curves in Space and Their Tangents
711
DEFINmONS If r is the position vector of a particle moving along a smooth curve in space, then dr V(I) = dl is the particle's velocity vector, tangent to the curve. At any time I, the direction of v is the direction of motion, the magnitude of v is the particle's speed, and the derivative 8 = dv/dl, when it exists, is the particle's acceleration vector. In
summary, dr v = dl'
1. Velocity is the derivative of position:
2. Speed is the magnitude of velocity:
Speed = Ivl.
3. Acceleration is the derivative of velocity: 4. The unit vectorv/lvl is the direction of motion at time I.
EXAMPLE 4 Find the velocity, speed, and acceleration of a particle whose motion in space is given by the position vector r( I) = 2 cos I i + 2 sin I j + 5 cos2 I k. Sketch the velocity vector V(77T/4). Solution
The velocity and acceleration vectors at time I are V(I) = r'(I) = 2sinli + 2coslj  10 cos I sinlk
z
= 2 sin I i + 2 cos I j  5 sin 21 k, 8(1) = r"(I) = 2cosli  2sinlj  10cos21k, and the speed is
y t
=
7," 4
IV(I) I = Y(2sinl)2 + (2 cos 1)2 + (5 sin 21)2 = Y4 + 25si02 21. When 1= 77T/4, we have
x
FIGURE 13.7 The curve and ilie velocity vector when t = 7'IT/4foriliemotion given in Example 4.
V(7:)=Vzi+ Vz i+ 5 k, A sketch of the curve of motion, and the velocity vector when 1= 77T/4, can be seen in Figure 13.7. • We can express the velocity of a moving particle as the product of its speed and direction: Velocity =
IVIC~I) =
(speed)(direction).
Differentiation Rules Because the derivatives of vector functions may be computed component by component, the rules for differentiating vector functions have the same form as the rules for differentiating scalar functions.
712
Chapter 13: VectorValued Functions and Motion in Space
Differentiation Rules for Vector Functions Let u and v be clifferentiable vector functions of t, C a constant vector, c any scalar, and / any differentiable scalar function. d
1. Constant Function Rule:
dtC
2. Scalar Multiple Rules:
:t [cu(t)]
=
0
cu'(t)
=
:t [f(t)u(t)]
I When you use the Cross Product Rule,
remember to preserve the order of the factors. Ifu comes fIrst on the left side of the equation, it must also come first on the right or the signs will be wrong.
!,(t)u(t) + /(t)u'(t)
=
3. Sum Rule:
:t [u(t) + v(t)]
4. Difference Rule:
dt [u(t)  v(t)]
S. Dot Product Rule:
:t [u(t)· v(t)]
6. Cross Product Rule:
dt [u(t) X v(t)]
7. Chain Rule:
:t [u(f(t))]
d
=
d
=
=
u'(t) + v'(t)
=
u'(t)  v'(t)
u'(t)· v(t) + u(t)· v'(t) =
u'(t) X v(t) + u(t) X v'(t)
!'(t)u'(f(t))
We will prove the product rules and Chain Rule but leave the rules for constants, scalar multiples, sums, and differences as exercises. Proof of the Dot Product Rule
Suppose that
u = ul(t)i
+ U2(t)j + u3(t)k
and Then d
dt(u'v)
=
d
dt(U1Vl + U2 V2 + U3 V3)
= U1VI
+ ulv2 +
U3V3
+ UtVl + U2V2 + U3V3.
u' ·v
u·v'
•
Proof of the Cross Product Rule We model the proof after the proof of the Product Rule for scalar functions. According to the detmition of derivative,
E. (
)_ .
dtUXV~
u(t + h) X v(t + h)  u(t) X v(t) h
To cbange this fraction into an equivalent one that contains the difference quotients for the derivatives ofu and v, we subtract and add u(t) X v(t + h) in the numerator. Then
:t
(u X v) = 1~~u(~t_+~h~)_X~v(~t_+~h~)_~u(~~~X~v~(t_+~h)~+~u~(t~)_x_v~(t~+~h~)_~u~(t~)_x_v~(~t)
~o
.
= lun
h~O
=
h [U(t
+ h)h  u(t) Xvt+h ( ) () +ut
X
v(t
+ h)h  V(t)]
I~ u(t + h)  u(t) X I~ v(t + h) + I~ u(t) X I~ v(t + h)  v(t) hO
h
hO
hO
hO
h
13.1
713
Curves in Space and Their Tangents
The last of these equalities holds because the limit of the cross product of two vector functions is the cross product of their limits if the latter exist (Exercise 32). As h approaches zero, vet + h) approaches v(t) because v, being differentiable at t, is continuous at t (Exercise 33). The two fractions approach the values of du/dt and dv/dt at t. In short,
d (
tit
IAs an a1~braic convenience, we
sometimes write the product of a scalar c and a vector v as vc instead. of cv. Thill permi11l us, for instance, to write the Chain Rule in a familiar form.:
~
)
du Xv oX dv =tit + dt·
•
Proof of the Chain Rule Suppose that o(s) = a(s)i + b(s)j + c(s)k is a differentiable vector function ofs and thats = J(t) is a differentiable scalar function oft. Then a, b, and c are differentiable functions of t, and the Chain Rule for differentiable realvalued functions gives
~[u(')l ~
du da ds (ji  didt'
whore,
oXv
:1 + 0 = (cosht)i  (sinht)j + tk
11. r(t) = (e'cost)i
12. r(t) 13. r(t) 14. r(t) 15. r(t)
16. r(t)
More on Curvature 17. Show that the parabola y = ax', a #' 0, has its largest curvature at its vertex and has no minimmn curvature. (Note: Since the curvature ofa curve remains the same if the curve is translated orrotated, this result is true for any parabola.)
734
Chapter 13: VectorValued Functions and Motion in Space
18. Show that the ellipse x = a cos I, y = b sin I, a > b > 0, has its largest cmvature on its major axis and its smallest curvature on its minor axis. (As in Exercise 17, the same is true for any ellipse.) 19. Muimizing the curvature of. helli In Example 5, we found the curvature of the helix r(l) = (a cos I)i + (a sin I)j + blk (a, b '" 0) to be " = a/(a' + b'). What is the largest value" can have for a given value of b? Give reasons for your answer. 20. Total curvature We find the tutal curvature of the portion of a smooth curve that runs from s = So to s = 91 > So by integrating K from So to Sl • If the curve has some other parameter, say t, then the total curvature is
where to and t, correspond to So and L
s, .
Find the tota! curvatures of
The portion of the helix r(l) = (3 cos I)i 0:5 t:5 41T.
h. The parabola y = x 2 ,
00
0
tcost)i,
in the form a = aTT + lINN. (The path of the motion is the involute of the circle in Figure 13.27. See also Section 13.3, Exercise 19.)
Solution
We use the rlISt of Equations (2) to rmd aT: v =
~;
=
(sint + sint + tcost)i + (cost  cost + tsint)i
= (tcost)i
+ (tsint)j
Ivl =
Vt 2 cos2 t + t 2 sin2 t
aT =
dt
d
d Ivl = dt(t)
=
=
W
It I = t
=
t> 0 Eq. (2)
1.
Knowing aT, we use Equation (3) to find aN:
FIGURE 13.27 The tangential and normal components of the acceleration of
a = (cost  tsint)i
laI 2 =t2 +1 aN = Vlal 2
themotionr(l) = (cos I + I sin I)i + (sin I  I cos I)j, for I > O. If a string
After some algebra
~~~
wound around a ftxed circle is unwound while held taut in the plaoe of the circle, its endP traces an involute of the circle (Example I).
+ (sint + tcost)i
=
V(t2
al

+ 1)  (I)
=
W
t.
=
We then use Equation (I) to rmd a: a = aTT
+ lINN
=
(I)T + (t)N = T + tN.
•
Torsion How does dB/lis behave in relation to T, N, and B? From the rule for differentiating a cross product, we have
dB = d(T X N) = dT X N
lis
lis
+ T X dN
lis
ds'
Since N is the direction of dT/Iis, (dT/ds) X N = 0 and
dB = 0
lis
+T
X dN = T X dN.
ds
ds
From this we see that dB/lis is orthogonal to T since a cross product is orthogonal to its factors. Since dB/lis is also orthogonal to B (the latter bas constant length), it follows that dB/lis is orthogonal to the plane of B and T. In other words, dB/ds is parallel to N, so dB/lis is a scalar multiple ofN.1n symbols,
dB
ds
= TN.
The negative sign in this equation is traditional. The scalar 'I' is called the torsion along the curve. Notice that
dB
ds • N
= TN' N = '1'(1) =
We use this equation for our next definition.
'I'.
13.5
I
Rectifying
plane
Binormal
DEFINmON
Let B
=
Tangential and Normal Components of Acceleration
737
T X N. The torsion function of a smooth curve is
Normal plane
T=
Principal normal
< A
Unit tangent
FIGURE 13.28 The names of the three planes determined by T. N, and B.
dB dS·N.
(4)
Unlike the curvature K, which is never negative, the torsion T may be positive, negative, or zero. The three planes determined by T, N, and B are named and shown in Figure 13.28. The curvature K = IdT / ds Ican be thought of as the rete at which the normal plane turns as the point P moves along its path. Similarly, the torsion T = (dB/ds)' N is the rete at which the osculating plane turns about T as P moves along the curve. Torsion measures how the curve twists. Look at Figure 13.29. If P is a !min climbing up a curved track, the rete at which the headlight turns from side to side per unit distance is the curvature of the tmck. The rete at which the engine tends to twist out of the plane formed by T and N is the torsion. In a more advanced course it can be shown that a space curve is a helix if and only if it has constant nonzero curvature and constant nonzero torsion.
The tmsion
I ~ ~ l(dT/d,)I· "
/
The curvature at P
s increases
~
p
I
8=0
FIGURE 13.29 Every moving body travels with a TNB frame that characterizes the geometty of its path of motion.
Computational Formulas The most widely used formula for torsion, derived in more advanced texts, is
x x T=
y )i
x' y
Iv X al
z z 'z' 2
(ifv X a .. 0).
(5)
The dots in Equation (5) denote differentiation with respect to t, one derivative for each dot. Thus, x ("x dot") means dx/dt, x ("x double dof') means d 2x/dt 2 , and x ("x triple dot'') means d 3x/dt 3 • Similarly, y = dy/dt, and so on. There is also an easytouse formula for curvature, as given in the following summary table (see Exercise 21).
738
Chapter 13: VectorValued Functions and Motion in Space
Computation Formulas for Curves in Space v
[Vj
Unit tangeot vector:
T
Principal unit normal vector:
dT/dt N = IdT/dtl
Bioormal vector:
B=TXN
=
Curvatore:
Torsion:
T =
Tangeotial and normal scalar componeots of acceleration:
a
dB
tis
d aT=dt
aN =
Y
z
x
ji "ji
z
x ·Z· 2 Iv x al
'N =
= aTT
i:
+ aNN
lvl
Klvl2 = Vlal 2
al
Exercises 13.5 Finding Tangential and Normal Components In Exercises 1 and 2, write a in the form • finding T and N. 2. r(l) = (I
+ 31)i + (I 
4. r(l) = S. r(l) = 6. r(l) =
+
14. r(l) =
(sin I)j
+
II 0
Physical Applications
+ 21j + 12k, I = I (I cos I)i + (I sin I)j + 12k, I = 0 12; + (I + (I/3)1')j + (I  (I/3)1')k, (e' cos I)i + (e' sin I)j + Y2e'l in terms of r and riJ by evaluating the dot produets v· i and v· j.
r
x
b. Express and r iI in terms of and j> by evaluating the dot products v • Ur and v· lie.
6. Express the curvatore of a twicedifferentiable curve r = 1(0) in the polar coordinate plane in terms of 1 and its derivatives. 7. A slender rod through the origin of the polar coordinate plane r0tates (in the plane) about the origin at the rate of 3 rad/min. A beetle starting from the point (2, 0) crawls along the rod toward the origin at the rate of I in/min.
Positive zaxis . points down.
1 z
2. Suppose the curve in Exercise I is replaced by the conical helix r = aO, z = bO shown in the accompanying figure.
a. Express the angular velocity dOjdt as a function ofO. h. Express the distance the particle travels along the helix as a function of o.
•• Find the beetle's acceleration and velocity in polar form when it is halfway to (I in. froru) the origin.
D b.
To the nearest tenth of an inch, what will be the length of the path the beetle has traveled by the time it reaches the origin? 8. Arc length in cylindrical coordinate. a. Show that when you express ds' = dx' + ely' + dz'in terms of cylindrical coordinates, you get ds' = dr' + r'do' + dz'. b. Interpret this result geometrically in terms of the edges and a
diagonal of a box. Sketch the box. c. Use the result in part (a) to fmd the length of the curve r = e 6,z = e 6,O::S O::s 8InS. Conical helix T
1
= a8,z =
be
Conez= ~r
Positive zaxis points down.
z
9. Unit vecton for position and motion in cylindrical coordinates When the position of a particle moving in space is given in cylindrical coordinates, the unit vectors we use to describe its position and motion are
u, = (cosO)i + (sinO)j,
... = (sinOli +
(cosO)j,
and k (see accompanying figure). The particle's position vector is then r = rUr + zk, where r is the positive polar distance coordinate of the particle's position.
14 PARTIAL DERIVATIVES OVERVIEW Many functions depend on more than one independent variable. For instance, the volume of a right circular cylinder is a function V = 7Tr 2h of its radius and its height, so it is a function V(r, h) of two variables rand h. In this chapter we extend the basic ideas of single variable calculus to functions of several variables. Their derivatives are more varied and interesting because of the different ways the variables can interact. The applications of these derivatives are also more varied than for singlevariable calculus, and in the next chapter we will see that the same is true for integrals involving several variables.
14.1
Functions of SeveraL VariabLes In this section we define functions of more than one independent variable and discuss
ways to graph them. Realvalued functions of several independent real variables are defined similarly to functions in the singlevariable case. Points in the domain are ordered pairs (triples, quadruples, ntuples) of real numbers, and values in the range are real numbers as we have worked with all along.
DEFINITIONS Suppose D is a set ofntuples of real numbers (x], X2, " " xn). A realvalued function f on D is a rule that assigns a unique (single) real number w =
f(Xl , X2, " . ,
xn)
to each element in D. The set D is the function's domain. The set of wvalues taken on by f is the function's range. The symbol w is the dependent variable of f, and f is said to be a function of the n independent variables Xl to X n . We also call the x/ s the function's input variables and call w the function's output variable.
If f is a function of two independent variables, we usually call the independent variables X and y and the dependent variable z, and we picture the domain of f as a region in the xyplane (Figure 14.1). Iff is a function of three independent variables, we call the independent variables x, y, and z and the dependent variable w, and we picture the domain as a region in space. In applications, we tend to use letters that remind us of what the variables stand for. To say that the volume of a right circular cylinder is a function of its radius and height, we might write V = f(r , h). To be more specific, we might replace the notation f(r, h) by the formula that calculates the value of V from the values of rand h, and write V = 7Tr 2h. In either case, rand h would be the independent variables and V the dependent variable of the function.
747
748
Chapter 14: Partial Derivatives
f f(a, b) ~~~~~~~Z
o
FIGURE 14.1
f(x,y)
An arrow diagram for the functionz = j(x, y).
As usual, we evaluate functions defined by fannulas by substituting the values of the independent variables in the fannula and calculating the corresponding value of the 2 + y2 + z2 at the point dependent variable. For example, the value of f(x,y,z) = (3,0,4) is
Yx
f(3,0,4) =
Y(W + (0)2 + (4)2 = V25 =
5.
Domains and Ranges In defining a function of more than one variable, we follow the usual practice of excluding x 2,y cannot inputs that lead to complex numbers or division by zero. If f(x,y) = 2 be less than x . If f(x,y) = I/(xy),xy cannot be zero. The domain of a function is assumed to be the largest set for which the defining rule generates real numbers, unless the domain is otherwise specified explicitly. The range consists of the set of output values for the dependent variable.
Yy 
EXAMPLE 1 (a) These are functions of two variables. Note the restrictions that may apply to their domains in order to obtain a real value for the dependent variable z. Function z =
Yy 
x2
I z=xy z
=
sinxy
Domain
Range
y ~ x2
[0, 00)
xy '" 0
(00,0) U (0, 00)
Entire plane
[1,1]
(b) These are functions of three variables with restrictions on some of their domains.
Function w=
Yx 2 + y2 + z2
w=
~'o~
W
I
x2
+ y2 + z2
= xyInz
Domain
Range
Entire space
[0, 00)
(x,y,z) '" (0,0,0)
(0, 00)
Halfspace z
>
0
(00,00)
•
Functions of Two Variables Regions in the plane can have interior points and boundary points just like intervals on the real line. Closed intervals [a, b] include their boundary points, open intervals (a, b) don't include their boundary points, and intervals such as [a, b) are neither open nor closed.
14,1 Functions of Several Variables
749
DEFINmONS A point (xo, YO) in a region (set) R in Ibe .!J'plane is an interior point of R if it is Ibe center of a disk of positive radius !bat lies entirely in R (Figure 14.2). A point (xo, YO) is a boundary point of R if every disk centered at (xo, YO) contains points !bat lie outside of R as well as points !bat lie in R. (The boundary point itself need not belong to R.) The interior points of a region, as a set, make up Ibe interior of Ibe region. The region's boundary points make up its boundary. A region is open ifit consists entirely of interior points. A region is closed if it contains all its boundary points (Figure 14.3).
(a) Interior point
y
y
y
, 1
R
«.' yol
o
/
o
(b) Boundary point
FIGURE 14,2 Interior points and boundary points of a plane regioo R, An interior point is necessarily a point of R. A
Ix
((x,y) 2 + y2< 1) Open unit disk. Every point an interior point
{(x,Y)lx2+y2~ 1)
Boundary of unit disk. (The unit circle.)
Ix
((x,y) 2 + y2,,;; 1) Closed unit disk. Contains all boundary points.
boundary point of R need not belong to R. FIGURE 14.3
Interior points and boundary points of 1he unit disk in 1he plane.
As wilb a halfopen interval of real numbers [a, b), some regions in Ibe plane are neiIber open nor closed. If you start wilb Ibe open disk in Figure 14.3 and add to it some of but not all its boundary points, Ibe resulting set is neilber open nor closed. The boundary points !bat are Ibere keep Ibe set from being open. The absence of Ibe remaining boundary points keeps Ibe set from being closed.
DEFINmONS A region in Ibe plane is bounded if it lies inside a disk offIxed radius. A region is unbounded if it is not bounded.
y Interior points, wherey  x 2 > 0
/
Examples of bounded sets in Ibe plane include line segments, triangles, interiors of triangles, rectangles, circles, and disks. Examples of unbounded sets in Ibe plane include lines, coordinate axes, Ibe graphs of functions defined on infinite intervals, quadrants, halfplanes, and Ibe plane itself.
EXAMPLE 2
Describe Ibe domain oflbe function f(x,y) =
Yy 
x 2•
Since f is defmed only where y  x 2 ;" 0, Ibe domain is Ibe closed, unbounded region shown in Figure 14.4. The parabola y = x 2 is Ibe boundary oflbe domain. The points above Ibe parabola make up Ibe domain's interior. •
Solutton
~~~~~~x
1
0
FIGURE 14,4 The domain of f(x, y) in Example 2 consists of1he shaded regioo and its bounding parabola.
Graphs, Level Curves, and Contours of Functions of Two Variables There are two standard ways to picture Ibe values of a function fix, y). One is to draw and label curves in Ibe domain on which f has a constant value. The olber is to sketch Ibe surface z = fix, y) in space.
750
Chapter 14: Partial Derivatives
DEFINITIONS The set of points in the plane where a function f(x, y) has a constant value f(x, y) = c is called a level curve of f. The set of all points (X, y, f(x, y) in space, for (X, y) in the domain of f, is called the graph of f. The graph of f is also called the surface z f(x,y) .
=
The surface z =f(x,y) = 100  x 2 _ y2 is the graph off.
f(x, y) = 75
X'
X
f(x,y) = 51 (a typical level curve in the function's domain)
y
EXAMPLE 3 Graph f(x , y) = 100  x 2  y2 and plot the level curves f(x,y) f(x, y) = 51, and f(x, y) = 75 in the domain of f in the plane.
SoLution The domain of f is the entire xyplane, and the range of f is the set of real numbers less than or equal to 100. The graph is the paraboloid z = 100  x 2  y2, the positive portion of which is shown in Figure 14.5. The level curve f(x, y) = 0 is the set of points in the xyplane at which f(x,y)
The contour curve f(x , y) = 100  x 2  y2 = 75 is the circle x 2 + y2 = 25 in the plane z = 75.
f1~"
\
= 75
~
P~
z = 100  x 2 _ y2
> 100, then the values of f(x, y) are negative. For example, the circle x 2 + y2 = 144, which is the circle centered at the origin with radius 12, gives the constant value f(x, y) = 44 and is a level curve of f. • The curve in space in which the plane z = c cuts a surface z = f(x, y) is made up of the points that represent the function value f(x, y) = c. It is called the contour curve f(x, y) = c to distinguish it from the level curve f(x, y) = c in the domain of f. Figure 14.6 shows the contour curve f(x,y) = 75 on the surface z = 100  x 2  y2 defined by the function f(x, y) = 100  X2  Y 2. The contour curve lies directly above the circle x 2 + y2 = 25, which is the level curve f(x,y) = 75 in the function's domain. Not everyone makes this distinction, however, and you may wish to call both kinds of curves by a single name and rely on context to convey which one you have in mind. On most maps, for example, the curves that represent constant elevation (height above sea level) are called contours, not level curves (Figure 14.7).
Functions of Three Variables
/
x The level curvef(x, y) = 100  x 2  y2 = 75 is the circle x 2 + y2 = 25 in the xyplane.
FIGURE 14.6 A plane z = c parallel to the xyplane intersecting a surface z = f(x, y) produces a contour curve.
In the plane, the points where a function of two independent variables has a constant value f(x,y) = c make a curve in the function's domain. In space, the points where a function of three independent variables has a constant value f(x, y, z) = c make a surface in the function's domain. DEFINITION The set of points (x, y, z) in space where a function of three independent variables has a constant value f(x, y, z) = c is called a level surface of f.
Since the graphs of functions of three variables consist of points (x,y, z, f(x,y, z)) lying in a fourdimensional space, we cannot sketch them effectively in our threedimensional frame of reference. We can see how the function behaves, however, by looking at its threedimensional level surfaces. EXAMPLE 4
Describe the level surfaces of the function
f(x,y, z) = Yx 2
+ y2 + z2 .
14.1
Functions of Several Variables
751
FIGURE 14.7 Contours on Mt. Washington in New Hampshire. (Reproduced by permission from the Appalachian Mountain Club.) FIGURE 14.8 The level surfaces of f(x,y,z) = Yx 2 + y2 + z2 are concentric spheres (Example 4).
(a) Interior point
(b) Boundary point
FIGURE 14.9 Interior points and boundary points of a region in space. As with regions in the plane, a boundary point need not belong to the space region R.
Solution The value of f is the distance from the origin to the point (x, y, z). Each level surface x 2 + Y 2 + Z 2 = c, c > 0, is a sphere of radius c centered at the origin. Figure 14.8 2 + y2 + z2 = 0 shows a cutaway view of three of these spheres. The level surface consists of the origin alone. We are not graphing the function here; we are looking at level surfaces in the function's domain. The level surfaces show how the function's values change as we move through its domain. If we remain on a sphere of radius c centered at the origin, the function maintains a constant value, namely c. If we move from a point on one sphere to a point on another, the function's value changes. It increases if we move away from the origin and decreases if we move toward the origin. The way the values change depends on the direction we take. The dependence of change on direction is important. We return to it in Section 14.5. •
Y
Yx
The definitions of interior, boundary, open, closed, bounded, and unbounded for regions in space are similar to those for regions in the plane. To accommodate the extra dimension, we use solid balls of positive radius instead of disks.
DEFINITIONS A point (xo, Yo, zo) in a region R in space is an interior point of R if it is the center of a solid ball that lies entirely in R (Figure 14.9a). A point (xo, Yo, zo) is a boundary point of R if every solid ball centered at (xo, Yo , zo) contains points that lie outside of R as well as points that lie inside R (Figure 14.9b). The interior of R is the set of interior points of R. The boundary of R is the set of boundary points of R. A region is open if it consists entirely of interior points. A region is closed if it contains its entire boundary.
Examples of open sets in space include the interior of a sphere, the open halfspace z > 0, the first octant (where x, y, and z are all positive), and space itself. Examples of closed sets in space include lines, planes, and the closed halfspace z 2: O. A solid sphere
752
Chapter 14: PartiaL Derivatives with part of its boundary removed or a solid cube with a missing face, edge, or comer point is neither open nor closed. Functions of more than three independent variables are also important. For example, the temperature on a surface in space may depend not only on the location of the point p(x, y, z) on the swface but also on the time 1 when it is visited, 80 we would write T = /(x, y, z, t).
computer Graphing Threedimensional graphing programs for computers and calculators make it possible to graph functions of two variables with only a few keystrokes. We can often get information more quickly from a graph than from a formula. w
EXAMPLE 5 The temperature w beneath the Earth's surface is a function of the depth x beneath the swface and the time t of the year. If we measure x in feet and t as the number of days elapsed from the expected date of the yearly highest surface temperature, we can model the variation in temperature with the function w = 008(1.7 X 1021  O.2x)eo·~.
FIGURE 14.10 Thisgrapbshowsthe seasonal variation of the temperatuIe below ground as a fraction of surface
tmnpenrtmc (Exampl' 5).
(The temperature at 0 ft is scaled to vary from +1 to 1, so that the variation atx feet can be interpreted as a fraction of the variation at the surface.) Figure 14.10 shows a graph of the function. At a depth of 15 ft, the variation (change in vertical amplitude in the figure) is about 5% of the surface variation. At 25 ft, there is almost no variation during the year. The graph also shows that the temperature 15 ft below the surface is about half a year out of phase with the surface temperature. When the temperature is lowt:st on the surface (late January, say), it is at its highest 15 ft below. Fifteen feet below the ground, the seasons are reversed. _ Figure 14.11 shows computergenerated graphs of a number of functions of two variables together with their level curves.
y
x
(a) z  sin x
FIGURE 14.11
+ 2 sin]
(b) l  (4x2
+ ]2.}o!rr
Computergenerated graphs and level curves of typical functions of two variables.
14.1 Functions of Several Variables
753
Exercises 14.1 Domain, Range, and Level Curves
21. f(x,y)
~
23. f(x,y)
~ \I'
In Exercises 14, find the specific function values.
+ xy'
1. f(x,y) ~ x'
b. f(I, I)
a. f(O,O) c. f(2,3) 2. f(x,y) ~ sin (xy)
d. f(3,2)
16 
25. f(x,y) ~ In (x' 27. f(x,y)
f( 3, ;;) d. f( f, 7)
22. f(x,y) ~ y/x'
xy I
x' _ y'
+ y')
~ sin1 (y 
x)
+ y' 
24. f(x,y)
~ \1'9  x'  y'
26. f(x,y) ~ .""",'+Y') 28. f(x,y)
~ tanI (~)
b.
29. f(x,y) ~ In (x'
a. f(3, 1,2)
b.
Exercises 3136 show level curves for the functions graphed in (aHf) on the following page. Match each set of curves with the appropriate function.
c. f(O,t,O)
d. f(2,2, 100)
a. f(2,');) c.
f(,."t)
3. f(x,y,z)
~
1) 30. f(x,y) ~ In (9  x'  y')
Matching Surfaces with Level Curves
xy ,, y +z
f(l,t, t)
31.
32. y
4. f(x,y,z) ~ \1'49  x'  y' 
a. f(O,O,O)
z'
I(~t
b. f(2,3,6)
c. f(1,2,3)
d.
( 4 5 6)
f \1'2' \1'2' \1'2
~¥~'
In Exercises 512, fmd and sketch the domain for each function.
5. f(x,y) ~ \l'y  x  2 6. f(x,y) ~ In (x'
+ y'
(x  I)(y
 4)
+ 2)
7. f(x,y) ~ (y _ x)(y _ x') sin (xy)
8. f(x,y)
~,
9. f(x,y)
~ cos 1 (y
x
+y
10. f(x,y) ~ In(xy
,
33.
34. y
 25
 x')
+x  y 
1)
11. f(x,y) ~ \I'(x'  4)(y'  9)
I
12. f(x,y) ~ In (4 _ x' _ y') In Exercises 1316, fmd and sketch the level curves f(x, y) ~ c on the same set of coordinate axes for the given values of c. We refer to these level curves as a contour map.
13. f(x,y)
~
x
+y
 I, c
~
3, 2, I, 0, 1,2,3
14. f(x,y) ~ x' + y', c ~ 0, 1,4,9, 16,25 15. f(x,y)
~
xy,
c
~
9, 4, I, 0, 1,4,9
16. f(x,y) ~ \1'25  x'  y',
c ~ 0, 1,2,3,4
In Exercises 1730, (a) fmd the function's domain, (b) fmd the function's range, (oj describe the function's level curves, (dJ fmd the boondary of the function's don3ain, (eJ determine if the don3ain is an open reo gion, a closed region, or neither, and (f) decide if the don3ain is boonded or unbounded
17. f(x,y)
~
y  x
19. f(x,y) ~ 4x'
+ 9y'
18. f(x,y) ~ ~ 20. f(x,y) ~ x'  y'
35.
36. y
1111
IIlIJi'
754
Chapter 14: PartiaL Derivatives
L
L
b.
,.
Functions of Two variables Display the values of the :functions in Exercises 3748 in two ways: (a) by sketching the surfacez = f(",y) and (b) by drawing an assortment oflevel curves in the function's dmnain. Label each level curve with its function value.
Vx
37. f(x,y) = yl
38. f(x,y) =
39. f(x,y) _ xl + yl 41. f(x,y) _ xl  y
40. f(x,y) _ V"x,cc+cy" 42. f(x,y) _ 4  "l _ yl
43. f(x,y) _ 4"l ... f(x.y)  1
47. f(",y) =
+ yl
44. f(x,y)  6  2x  3y
Iyl
... f(x.y)  1
yGx ,i'+'y,OC+04
48. f(x,y) =
Ixl Iyl
y,,2 + yl
4
FInding Level Curves d.
In Exercises 4952, fmd an equation for and sketch the graph of the level curve of the function f(X, y) that passes through the given point.
,
4•• !(x.y) = 16  x'  y'.
(20.0)
50. !(x.y) 
Vx'=l.
(1.0)
51. f(",y) =
y" + yl
3,
52. f(x,y)  x
(3,1)
2y "
+ Y + I' (1, I)
y
Sketching Level Surfaces In Exercises 5360, sketch a typical level surface for the :functioo..
••
+ yl + z2 54. f(x,y,z) _ 10(,,2 + yl + z2) 56. f(x,y,z) = z S5. f(x,y,z) =" + z 58. f(x,y,z) _ y2 + z2 57. f(x,y,z) _,,2 + yl 53. f(x,y,z) _ ,,2
59. f(x,y,z)  z  x2 _ y2 60. f(x,y,z) = (x 2/25) + (y2/16)
+ (z2/9)
FInding Level Surfaces In Exercises 6164, fmd an equation for the level surface of the function through the given point.
61. f(",y,z) = ~  loz, 62. f(x,y,z) _
1o(x 2
(3, 1, I)
+ y+z2), (1,2,1)
14.2 63. g(x,y,z)
= Yx 2 + y2 + Z2, (I, I, v'2)
64. g(x,y, z)
= ,_
Limits and Continuity in Higher Dimensions
71. f(x, y) = sin (x + 2 cos y),  2". '" X '" 2"., 2". "'y '" 2"., P(".,,,.) 72. f(x,y) = e(z'Ly) sin (x 2 + y2), 0'" X '" 2".,
xy+z + yz ,(1,0, 2)
.LA.
2". '" Y ",,,.,
In Exercises 65{j8, rmd and sketch 1he domain of f. Then find an equation for 1he level curve or surface of1he function passing through 1he given point.
65. f(x,y)
=
it. ~Y. 00
66.g(x,y,z)=~
(x
1 1~ dO
vlOZ
Y
68. g(x,y,z) =
%
I
+t
(0, I)
+ {' ~, (0, I, \13) Jo 4  OZ
COMPUTER EXPLORATIONS Use a CAS to perfonn 1he following steps for each of 1he functioos in Exercises 6!172. L
Plot 1he surface over the given rectangle.
c. Plot 1he level curve of f through 1he given point.
69. f(x, y) = x sin ~
+ Y sin 2>:,
0 '" x '" 5".,
0 '" Y '" 5".,
14.2
 (cosy)Yx 2
0 '" x '" 5".,
+ z2
=
2
some parameter interval I, you can sometimes describe surfaces in
u,
0:5 u:S; 2,
78.x=ucosv, y=usinv, z=v, o :5 v :s; 2'1T
0::;; u:s; 2,
=
u cos v, y :s 21T
o :5 v =
=
u sin v, z
(2 + cosu) cos v, y
o :s; u :s; 2'17",
70. f(x,y) = (sinx)(cosy)e Vbl"/', o '" Y '" 5"., P(41T, 4".)
I
space wi1h a1riple of equations x = f(u, v),y = g(u, v),z = h(u, v) defined on some parameter rectangle a ::;; u :s; h, c :s; v :s; d. Many computer algebra systeros permit you to plot such surfaces in parametric mode. (parametrized surfaces are discussed in detail in Sectioo 16.5.) Use a CAS to plot 1he surfaces in Exercises 7780. Also plot sevetallevel curves in 1he xyplaoe.
79. x
P(3"., 3".)
(~)
74. x 2 + z2 = I
I
=
Parametrized Surfaces Just as you describe curves in 1he plaoe parametrically wi1h a pair of equations x = f(t),y = g(t) dermed on
77. x
b. Plot several level cutves in 1he rectangle.
+ z2)
+ y2  3z 2 =
76. sin
(ln4,ln9,2)
_ r.:;,'
%
75. x
+ y)' I"
y
67. f(x, y) =
(1,2)
P("., ".)
Use a CAS to plot 1he implicitly dermed level surfaces in Exercises 7376. 73. 4ln (x 2 + y2
n.z
11=0
755
O:s;
V
=
=
(2 + cosu) sin v, z
=
sinu,
:s; 2'17"
80. x = 2cosucosv, y = 2 cos u sin v, z o :5 u :s; 2'1T, 0:5 v :s; '11'
=
28inU,
Limits and Continuity in Higher Dimensions This section treats limits and continuity for multivariable functions. These ideas are analogous to limits and continuity for singlevariable functions, but including more independent variables leads to additional complexity and important differences requiring some new ideas.
Limits for Functions of Two Variables If the values of f(x, y) lie arbitrnrily close to a f"lxed real number L for all points (x, y) suff"lciently close to a point (xo, yo), we say that f approaches the limit L as (x, y) approaches (xo, yo). This is similar to the informal definition for the limit of a function of a single variable. Notice, however, that if (xo, Yo) lies in the interior of j's domain, (x, y) can approach (xo, yo) from any direction. For the limit to exist, the same limiting value must be obtained whatever direction of approach is taken. We illustrate this issue in several examples following the detmition.
756
Chapter 14: Partial Derivatives
DEFINITION We say that a function /(x, y) approaches the limit L as (x, y) approaches (xo, yo), and write lim
(x, y)(x. Yo)
/(x,y) = L
if, for every number E > 0, there exists a corresponding number 6 for all (x, y) in the domain of /,
O.
While we won't prove Theorem I here, we give an informal discussion ofwby it's true. If (x, y) is sufficiently close to (XO,yo), then f(x, y) is close to Land g(x, y) is close to M (from the informa! interpretation of limits). It is then reasonable that f(x, y) + g(x, y) is closetoL + M;f(x,y)  g(x,y) is close to L  M; kf(x,y) is close to kL; f(x,y)g(x,y) is close toLM; and f(x,y)/g(x,y) is close to L/Mif M # O. When we apply Theorem I to polynomials and rational functions, we obtsin the useful result that the limits of these functions as (x, y) > (xo, Yo) can be ca!culated by evaluating the functions at (xo, yo). The only requirement is that the rational functions be def"med at (xo, yo).
EXAMPLE 1 In this example, we can combiae the three simple results following the limit def"mition with the results in Theorem I to calculate the limits. We simply substitule the x and y values of the point being approached into the functional expression to f"md the limiting value. ( ) a
(h)
x  xy + 3
lim
(x,y)~(O,l) x2y + 5xy  y3 lim
(x,y)~(3,
EXAMPLE 2
4)
Vx 2
+ y2
=
0  (0)(1) + 3 = 3 (0)2(1) + 5(0)(1)  (1)3
V(3)2
+ (4)2
=
v'25 =

xy
Find
x2 lim
(x,y)(O,O)
.Vr .r· x  VY
5
•
758
Chapter 14: Partial Derivatives
Vx  vY
Solution Since the denominator approaches 0 as (x,y) + (0, 0), we cannot use the Q1!otient Rule from Theorem 1. If we multiply numerator and denominator by + Vy, however, we produce an equivalent fraction whose limit we can find:
Vx
(x 2  .!JI)( Vx + vY) (x, y)~(O,O) Vx  vY (x,y)~(O,O) (Vx  vY)(Vx + vY) x2
lim

n,
lim
J
x(x lim
Y)(Vx + vY) x
(x,y)(O,O)
=
lim
(x,y)~(O,O)
Algebra
y
Cancel the nonzero fa 0 be given, but arbitrary. We want to rmd all> 0 such that 4xy2
1x2 + y2

01
;Y  yz (temperature in degrees Celsius, distance in feet) at P(I, I, I) is 3°C/It? Give reasons for your answer.
In Exercises 2528, sketch the curve I(x, y) = c together with VI and the tangent line at the given point Then write an equation for the tangent line.
35. The derivative of I(x,y) at Po(I,2) in the direction ofi + j is 2V2 and in the direction of 2j is 3 . What is the derivative of I in the directioo of i  2j? Give reasons for yoor answer.
x' + y' = 4, (v'2, v'2) 26.x'y=l, (v'2,I)
36. The derivative of I(x, y, z) at a point P is greatest in the direction of v = i + j  k. In this direction, the value of the derivative is
25.
27. >;)' = 4,
2v'3.
(2, 2)
28.x'>;),+y'=7,
a. What is Vf at P? Give reasons for your answer.
(1,2)
b. What is the derivative of I at P in the direction ofi
Theory and Examples 29. Let I(x,y) = x'  >;)' + y'  y. Find the directions u and the values of D. 1(1, I) for which
b.
c. Du/(1, I) e. Du/(1, I)
d. D./(!,I)
30. Let I(x,y) =
D./(H)
= =
0 3
~~ ~ ~~. Find the directions u
=
4
derivative of a differentiable function I(x,y, z) at a point Po in the direction of a unit vector u related to the scalar component of (V/)p, in the direction ofn? Give reasoos foryoor answer.
+~) is largest
b. D.
c. Du/(
+~) = 0
d.
38. Directional derivative. and partial derivative. Assuming that the necessary derivatives of I(x, y, z) are dermed, how are DI/, DJ/, and D.I related to I., I y , and I.? Give reasons for your
answer.
and the values of
39. Line. in the xyplane Show that A(x  xo) + B(y  Yo) = 0 is an equation for the line in the xyplane through the point (xo,Yo) norma1 to the vectorN = Ai + Bj.
for which
a. Du/(
e.
37. Directional derivatives and scalar components How is the
Do I(!, I) is smallest
a. Du 1(1, I) is largest
1(+ ~)
is smallest
D./(t,~) =2
40. The algebra rule. for gradient. gradients
Du/(+~) = I
14.6
+ j?
VI
= al i + al. + al k, ax
ayJ
az
Given a constant k and the
ag ag ag Vg=i+j+k, ax ay az
establish the algebra rules for gradients.
Tangent Planes and Differentials In this section we derme the tangent plane at a point on a smooth surface in space. Then we show how to calculate an equation of the tangent plane from the partial derivatives of
the function defining the surface. This idea is similar to the dermition of the tangent line at a point on a curve in the coordinate plane for singlevariable functions (Section 3.1). We then study the total differential and linearization of functions of several variables.
Tangent Planes and Normal Lines If r = g(t)i + h(t)j + k(t)k is a smooth curve on the level surface f(x, y, z) = c of a differentiable function f, then f(g(t), h(t), k(t)) = c. Differentiating both sides of this
792
Chapter 14: Partial Derivatives
equation with respect to t leads to
d dtf(g(t), h(t), k(t))
=
d dt (c)
+ af dh + af dk
=
0
af dg ax dt
FIGURE 14.32 The gradient VI is orthogonal to the velocity vector of every smooth curve in the surface through Po. The velocity vectors at Po therefore lie in a common plane, which we call the tangent plane at Po.
f ( a~ i
+ af. + af ~J
'
~
ay dt
az
Chain Rule
dt
k)' (d~g i + dh. + dk k) O. ~J ~ =
~
(1)
' ~
VI
dr/dt
At every point along the curve, Vf is orthogonal to the curve's velocity vector. Now let us restrict our attention to the curves that pass through Po (Figure 14.32). All the velocity vectors at Po are orthogonal to Vf at Po, so the curves' tangent lines all lie in the plane through Po normal to Vf. We now define this plane.
DEFINITIONS The tangent plane at the point Po(xo, Yo, zo) on the level surface f(x, y, z) = c of a differentiable function f is the plane through Po normal to Vflpo' The normal line of the surface at Po is the line through Po parallel to Vf
IPo'
From Section 12.5, the tangent plane and normal line have the following equations:
Tangent Plane to f(x,y, z)
fx(Po)(x  xo)
=
cat Po(xo,Yo, zo)
+ fy(Po)(y  Yo) + fz(Po)(z  zo) =
0
(2)
Normal Line to f(x,y, z) = cat Po(xo,Yo, zo)
x z
Po(l, 2, 4)
1
:/
EXAMPLE 1
The surface
xZ+r+z
9 =O
I
=
Xo
+ fiPo)t,
z = Zo
+ fAPo)t
(3)
Find the tangent plane and normal line of the surface
f(x,y, z)
=
x 2 + y2
+z
 9 = 0
A circular paraboloid
at the point Po(1, 2, 4).
I
Nonnalline
SoLution The surface is shown in Figure 14.33. The tangent plane is the plane through Po perpendicular to the gradient of f at Po . The gradient is
Vflpo = (2xi
+
2yj
+
k)(1 ,2,4)
= 2i + 4j + k.
The tangent plane is therefore the plane x
/
FIGURE 14.33 The tangent plane and normal line to this surface at Po (Example I).
2(x  1)
+ 4(y
 2)
+
(z  4) = 0,
or
2x
+ 4y + z =
14.
The line normal to the surface at Po is
x = 1
+
2t,
y = 2
+ 4t,
z = 4
+
t.
•
To find an equation for the plane tangent to a smooth surface z = f(x, y) at a point = f(xo, Yo), we first observe that the equation z = f(x, y) is
Po(xo, Yo, zo) where Zo
793
14.6 Tangent Planes and Differentials
equivalent to f(x,y)  z = O. The surface z = f(x,y) is therefore the zero level surface of the function F(x, y, z) = f(x, y)  z. The partial derivatives of Fare
o
Ix 
F. = ox (f(x,y)  z) =
0 =
Ix
o
Fy = oy (f(x,y)  z) = /y  0 = /y
o
Fz = oz(f(x,y)  z) = 0  1 = 1.
The formula F'(Po)(x  xo)
+ Fy(Po)(y  Yo) + F.{Po)(z  zo)
=
0
for the plane tangent to the level surface at Po therefore reduces to f.(xo,yo)(x  xo)
+ fy(xo,yo)(Y  Yo)  (z  zo)
=
O.
Plane Tangent to a Surface z = f(x,y) at (Xo,Yo, fVco,yo» The plane tangent to the surface z = f(x, y) of a differentiable function f at the poiot Po(xo,Yo, zo) = (xo,Yo, f(xo,yo)) is f.(xo,yo)(x  xo)
EXAMPLE 2
+
fY(xo,yo)(Y  Yo)  (z  zo) = O.
(4)
Fiod the plane tangent to the surface z = xcosy  ye' at (0,0,0).
Solution We calculate the partial derivatives of f(x, y) = x cos y  ye' and use Equation (4): f.(O,O) = (cosy  ye')(o.o) = 1  0·1 = 1
/y(O,O)
=
(x sioy  e')(o.o)
0 1
=
1.
=
The tangent plane is therefore 1'(x  0)  1'(Y  0)  (z  0) = 0,
or x  y  z =
EXAMPLE 3
Eq.(4)
o.
•
The surfaces f(x,y,z) = x 2
+ y2
 2 = 0
A cylinder
and g(x,y,z) = x
+z 
4 = 0
ApJane
meet in an ellipse E (Figure 14.34). Fiod parametric equations for the lioe tangent to E at the poiot Po(I, 1,3). Solution The tangent lioe is orthogonal to both Vf and Vg at Po, and therefore parallel tov = Vf X Vg. The components ofv and the coordinates of Po give us equations for the lioe. We bave Vfl(l.l~) = (2xi
Vgl (l.l~)
FIGURE 14.34 This cylinder and plane intersect in an ellipse E (Example 3).
=
(i
+ 2yj)(1.1.3)
+ k)(1.1.3)
v = (2i
=
i
=
2i
+ 2j
+k
+ 2j) X (i + k)
=
2 1
j 2 0
k 0 = 2i  2j  2k. 1
794
Chapter 14: Partial Derivatives The tangent line is
y=I2t,
x=I+2t,
•
z=32t.
Estimating Change in a Specific Direction The directional derivative plays the role of an ordinary derivative when we want to estimate how much the value of a function / changes if we move a small distance tis from a point Po to another point nearby. If / were a function of a single variable, we would have
d/
=
f'(Po) tis.
Ordinary derivative x increment
For a function of two or more variables, we use the formula
d/
=
(V/lpo"u) tis,
Directional derivative x increment
where u is the direction of the motion away from Po.
Estimating the Change in / in a Direction u To estimate the change in the value of a differentiable function / when we move a small distance tis from a point Po in a particular direction u, use the formula
d/
=
(V/lpo" u)
tis
Directional Distance derivative increment
EXAMPLE 4
Estimate how much the value of
/(x,y,z) = ysinx + 2yz will change if the point P(x,y, z) moves 0.1 unit from Po(O, 1,0) straight toward Pt(2, 2, 2).
Solution We first fmd the derivative of / at Po in the direction of the vector P;;Pt = 2i + i  2k. The direction of this vector is
The gradient of / at Po is
V/I(O.1.0)
=
((ycosx)i
+
(sinx
+ 2zli + 2yk)(O.1.0)
= i
+ 2k.
Therefore,
V/lpo·u = (i
+
2k)· (t + ti  tk) t t to i
=
=
The change d/ in / that results from moving tis = 0.1 unit away from Po in the direction of u is approximately
d/
=
(V/lpo"u)(tis) =
(t)tO.I) '"
0.067unit.
•
How to Linearize a Function of Two Variables Functions of two variables can be complicated, and we sometimes need to approximate them with simpler ones that give the accuracy required for specific applications without being so difficult to work with. We do this in a way that is similar to the way we fmd linear replacements for functions of a single variable (Section 3.9).
14,6 Tangent Planes and Differentials
A point _ near (xQ. Yo)
(x, y)
Suppose the function we wish to approximate is z = I(x, y) near a point (xo, Yo) at which we know the values of I, Ix> and Iy and at which I is differentiable. Ifwe move from (xo, Yo) to any nearby point (x, y) by increments Ax = x  Xo and ~y = y  Yo (see Figure 14.35), then the defmition of differentiability from Section 14.3 gives the change
I(x,y)  I(xo,yo) A point where f is differentiable
795
=
fx(xo,Yo)Ax
+ ly(xo,Yo)~y + EtAx +
E2~y,
where E1, E2 > 0 as Ax, ~y > O. If the increments Ax and ~y are small, the products EJAx and E2~y will eventoally be smaller still and we have the approximation
/
I(x,y) "" I(xo,yo)
+ fx(xo,Yo)(x  xo) + /y(xo,Yo)(y  yo). L(x,y)
If! is differentiable at (xo, yo), then the value of! at aoy point (x, y) nearby is approximately !(xo,Yo) + fx(xo,Yo)/!x + !y(xO,yo)4y. FIGURE 14.35
In other words, as long as as the linear function L.
DEFINmONS
I
~x
and ~y are small, I will have approximately the same value
The linearization of a function I(x, y) at a point (xo, Yo) where
is differentiable is the function
L(x,y)
I(xo,yo) + fx(xo,Yo)(x  xo) + /y(xo,Yo)(y  yo).
=
(5)
The approximation
I(x,y) "" L(x,y) is the standard linear approximation of I at (xo, yo).
From Equation (4), we fmd that the plane z = L(x,y) is tangent to the surface z = I(x, y) at the point (xo, yo). Thus, the linearization of a function of two variables is a tangentplane approximation in the same way that the linearization of a function of a single variable is a tangentline approximation. (See Exercise 63.)
EXAMPLE 5
Find the linearization of
I(x,y)
=
x 2  xy
+ ty2 + 3
at the point (3, 2).
Solutton
We first evaluate 1./.. and Iy at the point (xo, Yo) = (3,2):
1(3,2)
(x2  xy
=
fx(3,2)
= :
/y(3,2)
=
x
+ 1 y2 +
(x2  xy
aa (x2  xy Y
2
3)
+ 21 y2 +
= (3,2)
3)(3,2)
+ 21 y2 + 3)
(3,2)
8 =
(2x  y)(3,2)
=
(x
+ Y)(3,2)
=
4
=
I,
giving
L(x, y)
=
I(xo, Yo) + fx(xo, Yo)(x  xo) + /y(xo, Yo)(Y  Yo)
=
8
+ (4)(x  3) + (I)(y  2)
The linearization of I at (3, 2) is L(x,y) = 4x  y  2.
=
4x  y  2.
•
When approximating a differentiable function I(x, y) by its linearization L(x, y) at (xo, Yo), an important question is how accurate the approximation might be.
796
Chapter 14: Partial Derivatives
y
If we can f'md a common upper bound M for II", I, IIyy I, and II", I on a rectangle R centered at (xo,Yo) (Figure 14.36), then we can bound the error E throughout R by using a simple formula (derived in Section 14.9). The error is def'med by E(x,y) =
I(x,y)  L(x,y).
R ~~X
o
FIGURE 14.36 The rectangular region R: Ix .tol '" h, Iy  yol '" kinthe xy·plane.
The Error in the Standard Linear Approximation If I has continuous first and second partial derivatives throughout an open set containing a rectangle R centered at (xo, YO) and if M is any upper bcund for the values of 1/",1, I/yyl, and 1/",1 on R, then the error E(x, y) incurred in replacing I(x, y) on R by its linearization
L(x, y)
=
I(xo, YO) + fx(xo, yo)(x  xo) + /y(xo, YO)(y  YO)
satisfies the inequality
IE(x,y) I'" t M(lx  xol + Iy  yoll'·
To makeIE(x,y) I small for a givenM, we just make Ix  xol andly  yol small.
EXAMPLE 6 Find an upper bound for the error in the approximation I(x, y) '" L(x, y) in Example 5 over the rectangle R:
Ix  31 '" 0.1,
Iy  21'" 0.1.
Express the upper bound as a percentage of 1(3, 2), the value of I at the center of the rectangle. Solution
We use the inequality
IE(x,y) I'" t M(lx  xol + Iy  yoll'· To f'md a suitable value for M, we calculate I ",,j"', and I yy, f'mding, after a routine differentiation, that all three derivatives are constant, with values 1/",1 = 121 = 2,
I/yyl = III = I.
1/",1 = III = I,
The largest of these is 2, so we may safely takeMto be 2. With (xo, Yo) = (3,2), we then know that, throughout R,
IE(x,y) I '" t(2)(lx  31
+ Iy
 21)2 = (Ix  31
+ Iy
 21)'.
Finally, since Ix  31 '" 0.1 andly  21 '" 0.1 onR, we have
IE(x,y) I '" (0.1 + 0.1)2
= 0.04.
As a percentage of 1(3,2) = 8, the error is no greater than
0.~4
X 100 = 0.5%.
•
Differentials Recall from Section 3.9 that for a function of a single variable, y = I(x), we defined the change in I as x changes from a to a + Oat(a,b).  Ix'? < oat (a, b).
iv) the test is inconclusive at (a, b) if 1",lyy  Ix'? = 0 at (a, b). In this case, we must fmd some other way to determine the behavior of I at (a, b).
(a)
The expression Ixxlyy  Ix'? is called the discriminant or Hessian of I. It is sometimes easier to remember it in determinant form,
Ixxlyy
!xl =
I~: ~:I·
Theorem 11 says that if the discriminant is positive at the point (a, b), then the surface curves the same way in all directions: downward if Ixx < 0, giving rise to a local maximum, and upward if Ixx > 0, giving a local minimum. On the other hand, if the discriminant is negative at (a, b), then the surface curves up in some directions and down in others, so we have a saddle point. (b)
(a) The origin is a saddle point of the function j(x, y) = y2  x 2. There are no local extreme values (Example 2). (b) Level curves for the function j io Example 2. FIGURE 14.44
EXAMPLE 3
Find the local extreme values of the function
I(x,y) = xy  x 2
y2  2x  2y

+ 4.
Solution The function is defined and differentiable for all x and y and its domain has no boundary points. The function therefore has extreme values only at the points where Ix and Iy are simultaneously zero. This leads to Ix = y  2x  2 = 0,
Iy = x  2y  2 = 0,
or
x =y
=
2.
Therefore, the point ( 2, 2) is the only point where I may take on an extreme value. To see if it does so, we calculate
Ixx = 2,
Iyy = 2,
Ixy = I.
The discriminant of I at (a, b) = (2, 2) is
Ixxlyy  Ixy2 = (2)(2)  (1)2 = 4  1 = 3. The combination
Ixx tells us that
I
I( 2, 2)
8.
=
EXAMPLE 4 Solution
0
imply that (18,18) gives a maximum volume. The dimensions of the package are x = 108  2(18)  2(18) = 36 in.,y = 18 in., andz = 18 in. The maximum volume is V = (36)(18)(18) = 11,664 in3, or 6.75 W. • Despite the power of Theorem II, we urge you to remember its limitations. It does not apply to boundary points ofa function's domain, where it is possible for a function to have extreme values along with nonzero derivatives. Also, it does not apply to points where either Ix or Iy fails to exist.
Summary of MaxMin Tests The extreme values of I(x, y) can occur only at i) boundary points of the domain of I ii) critical points (interior points where Ix = Iy = 0 or points where Ix or Iy fails to exist). If the first and secondorder partial derivatives of I are continuous throughout a disk centered at a point (a, b) and Ix(a, b) = Iy(a, b) = 0, the natore of I(a, b) can be tested with the Second Derivative Test: i) ii) iii)
iv)
I., < 0 and/.,lyy  Ix/ > 0 at (a, b) => local maximum I., > 0 and/.,lyy  Ix/ > 0 at (a, b) => local minimum 1.,lyy  Ix/ < 0 at (a, b) => saddle point 1.,1yy  Ix/ = 0 at (a, b) => test is inconclusive
Exercises 14.7 finding Local Extrema Find all the local maxima, local minima, and saddle points of the functions in Exercises 130.
1. I(x,y) 2. I(x,y) 3. I(x,y) 4. I(x,y)
+ :x:y + y2 + 3x  3y + 4 = 2xy  5x2  2y2 + 4x + 4y 2 = x + :x:y + 3x + 2y + 5 = 5:x:y  7x 2 + 3x  6y + 2 =
x2
4
I(x,y) = 2xy  x 2  2y2 + 3x + 4 6. I(x,y) = x 2  4:x:y + y2 + 6y + 2
s.
7. I(x,y) = 2x 2 + 3:x:y + 4y2  5x + 2y 8. I(x,y) = x 2  2xy + 2y2  2x + 2y + I 9. I(x,y) = x 2  y2  2x + 4y + 6 10. I(x,y) = x 2 + 2xy
Exbeme Values and Saddle Points
14.7
11. f(x,y) = "';56.1:2
15. f(x,y) 16. f(x,y) 17. f(x,y) 18. f(x,y)
16.1:
31
+I
39. Find two munbcrs a and b with a s b !Nch that
 8x
[b(6 x
V'x2 + y2
11. f(x,y) _ I 13. f(x,y)  x 3 14. f(x,y) _ x 3
8y2

y3  2xy
+6
+ 3xy + y3 2x 3 + 3y2 + 6xy 3 = x + y3 + 3.1: 2  3y2  8 = x 3 + 3xy2  15x + y3  15y  2x 3 + 2y 3  9x 2 + 3y2  12y
l
2 X
1
1
2
 1
23. f(x,y) = ysinx 25. f(x,y) 27. f(x,y) _ 29. f(x,y) = 38. f(x,y) =
22. f(x,y) = X
1 + xy + Y
11:x.y) = x2
24. f(x, y) = e2.: cosy
f(x.y) ~ xy in the open flISt quadnurt (x OIl. a miIrimum thcI'C:.
finding Absolute Extrema 31. f(x,y) = 2x 2  4x + y2  4y + I on the closed triangular plate bounded by the lines x = 0, y = 2, Y = 2x in the fll'St quodnrot  xy + y2 + Ion the closed triangularplatcin the fmt quadnntbounded by the lines x = O,y = 4,y = x 33. f(x,y)  x 2 + y2 on the cl.o&cd triangular plate bounded by the Iin.ca x = 0, y = 0, y + 2x = 2 in the fmt quadIant
31. D(x,y) = x2
on
+2
the
dr
rectangular
+ 2y2
 x.
42. Find the critical point of
In Exerciscs 3138, fmd the absolute maxima and minima of the :functions on the given d.omain9.
34. 11:x, y)  x + xy + y2  6x o :!;'; x :!;'; 5, 3 :!;'; y :!;'; 3 35. 11:x, y)  x 2 + xy + y2  6x o :!;'; x :!;'; 5, 3 :!;'; y :!;'; 0
X 2 )1/3
Find the tcmpcraturcs at the hottest and coldest points on the plate.
er+r.v 26. f(x.y)  "  y;2  2y2 + 3, 3/2'" x'" 3/2,  3/2 ,;; Y ,;; 3/2 73. I(x,y) = 5x' + lax'  3Ox' + 3Oxy2  12Ox', 4 :s x :s 3, 2:s y :s 2 (x,y) # (0,0) 74. I(x,y) = {~:In(x2 + y2), (x,y) = (0,0)' 2 :s x :5 2, 2:s y :5 2
IrLa~g~~_n= ge__ M_ UL_ ti~ Pl_ ie_~________________________________
HIsTORICAL BIOGRAPHY
Joseph Louis Lagrange (17361813)
Sometimes we need to find the extreme values of a function whose domain is constrained to lie within some particular subset of the planea disk, for example, a closed triangular region, or along a curve. In this section, we explore a powerful method for rmding extreme values of constrained functions: the method of Lagrange multipliers.
Constrained Maxima and Minima We first consider a problem where a constrained minimum can be found by eliminating a variable.
EXAMPLE 1
FindthepointP(x,y,z)ontheplane2x
+ y z 
5 =
othat is closest
to the origin.
Solution
The problem asks us to rmd the minimum value of the function
IoFl = Y(x  0)2 + (y = Yx2 + y2 + z2
 0)2
+
(z  0)2
subject to the constraint that
2x+yz5=0.
oF I has a minimum value wherever the function
Since I
f(x,y, z) = x 2
+ y2 + z2
has a minimum value, we may solve the problem by finding the minimum value of f(x, y, z) subject to the constraint 2x + y  z  5 = 0 (thus avoiding square roots). Ifwe regard x and y as the independent variables in this equation and write z as z=2x+y5, our problem reduces to one of finding the points (x,y) at which the function
h(x, y) = f(x, y, 2x
+y
 5) = x 2
+ y2 +
(2x
+y
 5)'
812
Chapter 14: Partial Derivatives has its minimum value or values. Since the domain of h is the entire xyplane, the First Derivative Test of Section 14.7 tells us that any minima that h might have must occur at points where hy = 2y + 2(2x + y  5) = O.
hx = 2x + 2(2x + y  5)(2) = 0, This leads to
lOx + 4y = 20,
4x
+ 4y
=
10,
and the solution
x
5
=
3'
We may apply a geometric argument together with the Second Derivative Thst to show that these values minimize h. The z·coordinate of the corresponding point on the plane
z=2x+y5is 5 6' Therefore, the point we seek is Closest point:
•
The distance fromPto the origin is 5jV6 '" 2.04.
Attempts to solve a constrained maximum or minimum problem by substitotion, as we might call the method of Example 1, do not always go smoothly. This is one of the rea· sons for learning the new method of this section.
z
EXAMPLE 2
Find the points on the hyperbolic cylinder x 2  z2  I = 0 that are clos·
est to the origin. Solution 1 The cylinder is sbowo in Figure 14.49. We seek the points on the cylinder closest to the origin. These are the points whose coordinates minimize the value of the function
/(x,y,z) = x 2 + y2
+
z2
Square of the distance
subject to the constraint that x 2  z2  I = O. Ifwe regard x andy as independent vari· abIes in the constraint equation, then
and the values of /(x,y, z) = x 2
+ y2 + z2 on the cylinder are given by the function
h(x, y) = x 2 + y2 + (x 2  I) = 2x 2 + y2  1. To f'md the points on the cylinder whose coordinates minimize /, we look for the points in the :ry·plane whose coordinates minimize h. The only extreme value of h occurs where FIGURE 14.49
x'  z' 
The hyperbolic cylinder
I = 0 in Example 2.
hx =4x=0
and
hy =2y=0,
that is, at the point (0, 0). But there are no points on the cylinder where both x and y are zero. What went wrong? What happened was that the First Derivative Test found (as it sbonld have) the point in the domain of h where h has a minimum value. We, on the other hand, want the points on the cylinder where h has a minimum value. Although the domain of h is the entire
14.8 Lagrange Multipliers The hyperbolic cylinder x 2 On this part,
x= v?+l
 Z2 =
1
On this part,
z x=v?+l
xyplane, the domain from which we can select the first two coordinates of the points (x, y, z) on the cylinder is restricted to the "shadow" ofthe cylinder on the xyplane; it does not include the band between the lines x = 1 and x = 1 (Figure 14.50). We can avoid this problem if we treat y and z as independent variables (instead of x and y) and express x in terms of y and z as
+
x 2 = z2
k(y, z)
1.
+ y2 + z2 becomes
With this substitution, I(x,y, z) = x 2
FIGURE 14.50 The region in the ;ryplane from which the first two coordinates of the points (x,y, z) on the hyperbolic cylinder x 2  z2 = 1 are selected excludes the band 1 < x < 1 in the xyplane (Example 2).
813
= (z2 + 1) + y2 + z2 = 1 + y2 + 2z2
and we look for the points where k takes on its smallest value. The domain of k in the yzplane now matches the domain from which we select the y and zcoordinates of the points (x, y, z) on the cylinder. Hence, the points that minimize k in the plane will have corresponding points on the cylinder. The smallest values of k occur where
ky = 2y = 0 or where y
kz
and
=
4z = 0,
= z = O. This leads to x2
= z2 + 1 = 1,
x = ±1.
The corresponding points on the cylinder are (± 1,0,0). We can see from the inequality
k(y, z)
= 1 + y2 + 2z2
~ 1
that the points (± 1, 0, 0) give a minimum value for k. We can also see that the minimum distance from the origin to a point on the cylinder is 1 unit.
\
Solution 2 Another way to find the points on the cylinder closest to the origin is to imagine a small sphere centered at the origin expanding like a soap bubble until it just touches the cylinder (Figure 14.51). At each point of contact, the cylinder and sphere have the same tangent plane and normal line. Therefore, if the sphere and cylinder are represented as the level surfaces obtained by setting and
x
/
FIGURE 14.51 A sphere expanding like a soap bubble centered at the origin until it just touches the hyperbolic cylinder x 2  z2  1 = 0 (Example 2).
g(x,y,z)
= x 2  z2  1
equal to 0, then the gradients VI and Vg will be parallel where the surfaces touch. At any point of contact, we should therefore be able to find a scalar A ("lambda") such that
VI = AVg, or
2xi
+
2yj
+
2zk = A(2xi  2zk).
Thus, the coordinates x, y, and z of any point of tangency will have to satisfy the three scalar equations
2x
= 2Ax,
2y
= 0,
2z
= 2Az.
For what values of A will a point (x, y, z) whose coordinates satisfy these scalar equations also lie on the surface x 2  z2  1 = O? To answer this question, we use our knowledge that no point on the surface has a zero xcoordinate to conclude that x O. Hence, 2x = 2Ax only if
*'
2 = 2A,
or
A=1.
For A = 1, the equation 2z = 2Az becomes 2z = 2z. If this equation is to be satisfied as well, z must be zero. Since y = 0 also (from the equation 2y = 0), we conclude that the points we seek all have coordinates of the form
(x, 0, 0).
814
Chapter 14: Partial Derivatives What points on the surface x 2 points (x, 0, 0) for which
x2

(0)'
z2 =

=
I have coordinates oflhis fonn? The answer is the
x2
I,
=
I,
or
x
=
± I.
The points on the cylinder closest to the origin are the points (± 1,0,0).
•
The Method of Lagrange Multipliers In Solution 2 of Example 2, we used the method of Lagrange multipliers. The method says that the extreme values of a function I(x, y, z) whose variables are subject to a constraint g(x,y,z) = 0 are to be found on the surface g = 0 among the points where VI = AVg for some scalar A (called a Lagrange multiplier). To explore the method further and see why it works, we first make the following observation, which we state as a theorem.
THEOREM 12The Orthogonal Gradient Theorem
Suppose that I(x, y, z) is differentiable in a region whose interior contains a smooth curve C:
r(t)
=
g(t)i
+ h(t)i + k(t)k.
If Po is a point on C where I has a local maximum or minimum relative to its values on C, then VI is orthogonal to C at Po.
Proof We show that VI is orthogonal to the curve's velocity vector at Po. The values of I on C are given by the composite I(g(t), h(t), k(t», whose derivative with respect to tis
dl dt At any point Po where curve, dl/dt = 0, so
I
=
al dg ax dt
al dh
al dk
+ ay dt + az dt
= VI' v.
has a local maximum or minimum relative to its values on the
V/'Y
=
o.
•
By dropping the ztenns in Theorem 12, we obtain a similar result for functions of two variables.
COROLLARY OF THEOREM 12 At the points on a smooth curve r(t) = g(t)i + h(t)j where a differentiable function I(x, y) takes on its local maxima and minima relative to its values on the curve, VI' Y = 0, where Y = dr/dt.
Theorem 12 is the key to the method of Lagrange multipliers. Suppose that I(x, y, z) and g(x, y, z) are differentiable and that Po is a point on the surface g(x, y, z) = 0 where I has a local maximum or minimum value relative to its other values on the surface. We assume also that Vg '" 0 at points on the surface g(x,y,z) = o. Then I takes on a local maximum or minimum at Po relative to its values on every differentiable curve through Po on the surface g(x,y, z) = o. Therefore, Viis orthogonal to the velocity vector of every such differentiable curve through Po. So is Vg, moreover (because Vg is orthogonal to the level surface g = 0, as we saw in Section 14.5). Therefure, at Po, VI is some scalar multiple A ofVg.
14.8
815
Lagrange Multipliers
The Method of Lagrange Multipliers Suppose that I(x, y, z) and g(x, y, z) are clifferentiable and Vg '" 0 when g(x,y,z) = O. To find the local maximum and minimum values of I subject to the constraint g(x, y, z) = 0 (if these exist), f"md the values of x, y, z, and A that simultaneously satisfY the equations
VI
AVg
=
g(x,y, z)
and
o.
=
(1)
For functions of two independent variables, the condition is similar, but without the variable z.
Some care must be used in applying this method. An extreme value may not actually exist (Exercise 41). y
EXAMPLE 3 X2
Y2
8+
Find the greatest and smallest values that the function
y2
I(x,y)
2~1
=
xy
takes on the ellipse (Figure 14.52) x
0
2Y2 We want to find the extreme values of I(x, y)
Solution FIGURE 14.52 Example 3 shows how to f"md the largest and smallest values of the product"" on this ellipse.
g(x,y)
=
x'
y'
"8 + "2  I
xy subject to the constraint
=
=
o.
To do so, we first find the values ofx,y, and A for which
VI
=
AVg
g(x,y)
and
=
O.
The gradient equation in Equations (1) gives yi
+ xj
=
Ay,
=
~ xi + Ayj,
from which we find y
xy=2
xy=2
x
so that y
~~i~;~~~~~~
=
0 or A = ±2. We now consider these two cases.
Case 1: If y = 0, then x = y = O. But (0, 0) is not on the ellipse. Hence, y '" O. Case2: If y '" 0, then A = ±2 and x = ±2y. Substituting this in the equation g(x, y) = 0 gives
x
(±2y)' 8xy=2/
FIGURE 14.53
eonstraiot g(x, y)
and
y'
+ "2
=
I,
4y' + 4y'
=
8
and
y
=
±l.
The function I(x,y) = xy therefore takes on its extreme values on the ellipse at the four points (±2, 1), (±2, I). The extreme values are xy = 2 andxy = 2.
xy=2
When subjected to the = x'/8
+ y'/2  I
=
0, The Geometry of the Solution The level curves of the function I(x, y) = xy are the hyperbolas xy = c (Figure 14.53). The farther the hyperbolas lie from the origin, the larger the absolule value of I. We want to f"md the es1reme values of I(x, y), given that the point (x, y) also lies on the ellipse x' + 4y' = 8. Which hyperbolas intersecting the ellipse lie farthest from the origin? The hyperlJolas that just graze the ellipse, the ones that are tangent to it, are
the functioo f(x, y) = "" takes 00 extreme values at the four points (±2, ± I ~ These are the points 00 the ellipse when Vf (red) is a scalar multiple ofVg (blue) (Example 3).
816
Chapter 14: Partial Derivatives farthest. At these points, any vector nonna! to the hyperbola is nonna! to the ellipse, so VI = yi + xj is a multiple (A ~ ±2) ofVg ~ (x/4)i + yj. At the point (2, I), for example, VI
~
i
+
Vg ="2' 1. + J, .
2j,
aod
VI = 2Vg.
aod
VI
At the point (  2, I), VI = i  2j,
EXAMPLE 4 3x
Vg =
ti +
j,
~
2Vg.
•
Find the maximum aod minimum values of the function I(x,y) = ~ 1.
+ 4y on the circle x 2 + y2
Solution
We model this as a Lagraoge multiplier pmblem with
g(x,y) ~ x 2 + y2  I
+ 4y,
I(x,y) ~ 3x
aod look for the values ofx,y, aod A that satisfy the equations VI ~ AVg:
g(x,y) ~
+ 4j = 2xAi + 2yAj 0: x 2 + y2  I = O. 3i
The gradient equation in Equations (I) implies that A "" 0 aod gives
x
3 = 2A'
Y
~
2
X'
These equations tell us, among other things, that x aod y have the same sign. With these values for x aod y, the equation g(x, y) ~ 0 gives
(iAY + (iY 
I
~ 0,
so
9
+ 16
= 4A2,
4A2 = 25,
5 A ~ ±"2'
aod
Thus,
y
3
3
x~2A=±S'
2
4
Y ~ X = ±S'
aod/(x,y) = 3x + 4yhas extreme values at (x,y) ~ ±(3/5,4/5). By calculating the value of 3x + 4y at the points ±(3/5, 4/5), we see that its maximum aod minimum values on the circle x 2 + y2 ~ I are 3x+4y=5
aod 3x
+ 4y =5
FIGURE 14.54 The functionj(x,y) ~ 3x + 4y takes 00 its largest value 00 the unitcircleg(x,y) ~ x 2 + y2  I ~ Oat the point (3/5, 4/5) and its smallest value at the point (3/5, 4/5) (Example 4). At each of these points, Vj is a scalar multiple ofVg. The figure shows the gradients at the Il!St point but not the second.
The Geometry of the Solution The level curves of I(x, y) ~ 3x + 4y are the lines 3x + 4y ~ c (Figure 14.54). The farther the lines lie from the origin, the larger the absolute value of f. We want to find the extreme values of I(x,y) given that the point (x,y) also lies on the circle x 2 + y2 ~ I. Which lines intersecting the circle lie farthest from the origin? The lines taogent to the circle are farthest. At the points oftaogency, aoy vector nonna! to the line is normal to the circle, so the gradient VI ~ 3i + 4j is a multiple (A ~ ±5/2) of the gradient Vg ~ 2xi + 2yj. At the point (3/5, 4/5), for example, VI = 3i
+
4j,
Vg
~
6. + 8. S' SJ,
aod
5
VI = "2Vg.
•
14.8 Lagrange Multipliers
817
Lagrange Multipliers with Two Constraints Many problems require us to find the extreme values of a differentiable function I(x, y, z) whose variables are subject to two constraints. If the constraints are and
c
\
and gl and g2 are differentiable, with Vg 1 not parallel to Vg2, we find the constrained local maxima and minima of I by introducing two Lagrange multipliers A and JL (mu, pronounced "mew"). That is, we locate the points P(;x, y , z) where I takes on its constrained extreme values by finding the values of x, y , z, A, and JL that simultaneously satisfy the equations
gl = 0
FIGURE 14.55 The vectors Vg 1 and Vg2 lie in a plane perpendicular to the curve C because Vg 1 is normal to the surface gl = 0 and Vg2 is normal to the surface
g2 = O.
(2)
°
°
Equations (2) have a nice geometric interpretation. The surfaces gl = and g2 = (usually) intersect in a smooth curve, say C (Figure 14.55). Along this curve we seek the points where I has local maximum and minimum values relative to its other values on the curve. These are the points where VI is normal to C, as we saw in Theorem 12. But Vg 1 and Vg2 are also normal to C at these points because C lies in the surfaces gl = and g2 = 0. Therefore, VI lies in the plane determined by Vg 1 and Vg2, which means that VI = AVgl + JL Vg2 for some A and JL . Since the points we seek also lie in both surfaces, their coordinates must satisfy the equations gl(X, y, z) = and g2(X,y, z) = 0, which are the remaining requirements in Equations (2).
°
°
EXAMPLE 5 The plane x + y + z = 1 cuts the cylinder x 2 + y2 = 1 in an ellipse (Figure 14.56). Find the points on the ellipse that lie closest to and farthest from the origin. Solution
We find the extreme values of
I(x,y, z)
=
x 2 + y2
+ z2
(the square of the distance from (x, y , z) to the origin) subject to the constraints
gl(X, y , Z)
=
x 2 + y2  1
g2(X, y, z)
=
x
=
°
(3)
+ Y + z  1 = 0.
(4)
The gradient equation in Equations (2) then gives
= AVg 1 + JLVg2 2xi + 2yj + 2zk = A(2xi + 2yj) + JLO + j + k) 2xi + 2yj + 2zk = (2Ax + JL)i + (2Ay + JL)j + JLk VI
or 2x = 2Ax FIGURE 14.56 On the ellipse where the plane and cylinder meet, we find the points closest to and farthest from the origin. (Example 5).
+ JL,
2y
=
2Ay
+ JL,
2z = JL .
(5)
The scalar equations in Equations (5) yield
+ 2z~(1  A)x = z, 2Ay + 2z~(1  A)y = z.
2x = 2Ax
2y
=
(6)
°
Equations (6) are satisfied simultaneously if either A = 1 and z = or A*1 and x = y = z/(1  A). If z = 0, then solving Equations (3) and (4) simultaneously to find the corresponding points on the ellipse gives the two points (1, 0, 0) and (0, 1,0). This makes sense when you look at Figure 14.56.
818
Chapter 14: Partial Derivatives If x = y, then Equations (3) and (4) give X2 +X 2
x+x+z 1=0
1=0 2>;2 = I
z=I2>;
x=±v'Z 2
z= I 'f
v'Z.
The corresponding points on the ellipse are PI
=
('?, '?,
1
v'Z)
and
Here we need to be careful, however. Although PI and P2 both give local maxima of f on the ellipse, P2 is farther from the origin than PI. The points on the ellipse closest to the origin are (I, 0, 0) and (0, 1,0). The point on the ellipse farthest from the origin is P2 • •
Exercises 14.8 Two Independent Variables with One Constraint 1. Extrema on an ellipse Find the points 00 the ellipse x 2 + 2y2 = I where f(x, y) = xy has its extreme values. 2. Extrema on a circle Find the extreme values of f(x, y) = xy subject to the constraint g(x, y) = x 2 + y2  10 = 0. 3. Maximum on a line Find the maximum value of f(x, y) = 49  x 2  y2 00 the line x + 3y = 10.
4. Extrema on a line Find the local extreme values of f(x, y) 00 the line x + y = 3.
= x7
S. Constrained minimum nearest the origin.
Find the points on the curve xy2 = 54
6. Constrained minimum
Find the points
00
the curve x7 = 2
nearest the origin. 7. Use the method of Lagrange multipliers to find L
Minimum on a hyperbola The minimum value of x subject to the coostraints xy = 16, x > 0, Y > 0
+ y,
b. Muimum on a line The maximum value ofxy, subject to the constraint x + y = 16.
Comment on the geometry of each solution. 8. EItrema on a curve Find the points on the curve x + xy + y2 = I in the xyplane that are nearest to aod farthest from the origio. 2
9. Minimum .urface area with ("ned volume Find the dimensioos of the closed right circular cylindrical cao of smallest surface area whose volume is 16'7T cm.3• 10. Cylinder in a .phere Find the radius aod height of the open right cireolar cylinder oflargest surface area that cao be inscribed in a sphere of radius a. What is the largest surface area? 11. Rectaogle of greate.t area in an ellipse Use the method of Lagrange multipliers to f"md the dimensions of the rectsng1e of greatest area that can be inscribed in the ellipse x 2/16 + y2/9 = I with sides parallel to the courdinate ases.
ll. Rectaogle of longest perimeter in an ellipse Find the dimensioos of the rectangle oflargest perimeter that cao be inscribed in
the ellipse x 2/a 2 + y2/b 2 = I with sides parallel to the coordinate ases. What is the largest perimeter? 13. Extrema on a circle Find the maximum aod minimum values of x 2 + y2 subject to the coostraint x 2  2x + y2  4y = O. 14. Extrema on a circle Find the maximum and minimum values of 3x  y + 6 sui!ject to the constraint x 2 + y2 = 4. 15. Ant on a metal plate TIre temperature at a point (x, y) on a metal plate is T(x, y) = 4x 2  4xy + y2. An aot on the plate wa1ks arouod the circle of radius 5 centered at the origio. What are the highest aod lowest temperatures encoontered by the aot? 16. Cheape.t storage taok Yoor f"mn has been asked to design a storage tauk for liquid petroleum gas. The customer's specifications call for a cylindrical tank with hemispherical ends, aod the tank is to hold 8000 m' of gas. The customer also wants to use the smallest arooont of materisl possible in building the tank. What radius and height do you recommend for the cylindrical portion of the tank?
Thllle Independent Variables with One Cons!ra;nt 17. Minimnm distance to a point Find the point x + 2y + 3z = 13 closest to the point (I, I, I).
00
the plane
18. Maximnm distaoce to a point Find the point on the sphere x 2 + y2 + z2 = 4 farthest from the point (I, I, I). 19. Minimnm di.laoce to the origin
from the surface x 2
Find the minimum distance
z2 = 1 to the origin.
 y2 
20. Minimnm distance to the origin z = xy + 1 nearest the origin.
Find the point on the surface
21. Minimum di.laoce to the origin Find the points on the surface z2 = xy + 4 closest to the origin. 22. Minimnm distaoce to the origin face xyz = I closest to the origio.
Find the point(s) on the sur
23. Extrema on a sphere Find the maximoro aod minimum values of
f(x,y,z) on the sphere
x2
=
+ y2 + z2
x  2y = 30.
+ 5z
819
14.8 Lagrange Multipliers 24. Extrema on a sphere Find the points on the sphere x 2 + y2 + z2 ~ 25 where I(x,y, z) ~ x + 2y + 3z has its
maximum and minimum values.
34. Minimize the function I(x,y,z) ~ x 2 + y2 + z2 subject to the constraints x + 2y + 3z ~ 6 and x + 3y + 9z ~ 9.
25. Minjmjzing a sum of squares Find three real numbers whose sum. is 9 and the sum of whose squares is as small as possible.
35. Minimum distance to the origin Find the point closest to the origin on the line of intersection of the planes y + 2z ~ 12 and x + y ~ 6.
26. Maximizing a product Find the largest product the positive numbersx,y,andz can have if x + y + z2 = 16.
36. Muimum value on line of intersection Find the maximum
27. Rectangular bOJ[ of largest volume in a sphere Find the di· mensions of the closed rectangular box with maximum volume that can be inseribed in the unit sphere.
37. EItrema on a curve of intersection Find the extreme values of
28.
])OJ[ with vertex on a
plane Find the volume of the largest closed rectangular box in the fIrSt oclaot having three faces in the coordinareplanesandavertexontheplanex/a + y/b + z/c ~ I,where a> O,b > o,andc > O.
29. Hottest point on a space probe A space probe in the shape of the ellipsoid 4x 2 + y2 + 4z2 ~ 16 enters Earth's atmosphere and its surface begins to heat After I hour, the temperature at the point (x,y, z) on the probe's surface is
T(x,y,z) ~
ax 2 + 4yz 
16z + 600.
Find the hotrest point on the probe's surface.
30. Extreme temperatores on a sphere Suppose that the Celsius temperature at the point (x, y, z) on the sphere x 2 + y2 + z2 ~ I is T ~ 40Oxyz2 . Locare the highest and lowest temperatures on the sphere. 31. Maximizing a utility function: an example from economics In economics, the usefulness or utility of amounts x and y of two capital goods G1 and G2 is sometimes measured by a function U(x, y). For example, Gl and (h might be two chemicals a pharmaceutical coropany needs to have on hand and U(x, y) the gain from manufactoring a product whose synthesis requires different amounts of the chemicals depending on the process used. If Gl costs a dollars per kilogram, (h costs b dollars per kilogram, and the total amount allocared for the purchase of G 1 and G2 together is c dollars, then the company's managers want to maximize U(x, y) given that ax + by ~ c. Thus, they need to solve a typical Lagrange multiplier problem. Suppose that
U(x,y) ~ xy and that the equation ax
+
by
2x + Y
~
~
+ 2x
c simplifies to
30.
Find the maximum value of U and the corresponding values of x and y subject to this latter constraint
32. Locating a radio telescope You are in charge of erecting a radio telescope on a newly discovered planet. To minimize interference, you waot to place it where the magnetic field of the plauet is weakest. The plauet is spherical, with a radius of 6 units. Based on a coordinare system whose origin is at the center of the plauet, the streugth of the magnetic field is given by M(x, y, z) ~ 6x  y2 + xz + 60. Where should you locare the radio relescope?
Extreme value. Subject to Two Constraints
33. Maximize the function I(x,y, z) ~ x 2 + 2y  z2 subject to the constraints 2x  y ~ 0 aod y + z ~ O.
value that I(x, y, z) ~ x 2 + 2y  z2 can have on the line of intersection of the planes 2x  Y ~ 0 and y + z ~ O.
I(x, y, z) ~ x'yz
+
1 on the intersection of the plane z ~ I with
the sphere x 2 + y2 + z2 = 10. 38. a. Muimum on line of intersection Find the maximum value of w ~ xyz on the line of intersection of the two planes x + y + z ~ 40 and x + y  z ~ O. b. Give a geometric argument to support your claim that you have found a maximum, and not a minimum, value of w.
39. EItrema on a circle ofintenection Find the extreme values of the function/(x,y, z) ~ xy + z2 on the circle in which the plane y  x = 0 intersects the sphere x 2 + y2 + z2 = 4.
40. Minimum distance to the origin Find the point closest to the origin on the curve of intersection of the plane 2y the cone z2 = 4x 2 + 4y2.
+ 4z
~
5 and
Theory and Examples 41. The condition VI = AVgisnotsnllicient Although VI ~ AVg is a necessary condition for the occurrence of an extreme value of I(x, y) subject to the conditions g(x, y) ~ 0 and Vg #' 0, it does
not in itself guarantee that one exists. As a case in point, try using the method of Lagrange multipliers to fmd a maximum value of I(x, y) ~ x + y subject to the constraint that xy ~ 16. The method will identifY the two points (4, 4) and (4, 4) as candidares for the location of extreme values. Yet the sum (x + y) has no maximUD2 value on the hyperbola xy ~ 16. The farther you go from the origin on this hyperbola in the first quadrant, the larger the sUD2/(x,y) ~ x + ybecomes.
42. A lea.t squa...s plane The plaue z ~ Ax "fitted" to the following points (x" Yb z,):
(0,0,0),
(0, I, I),
+ By + C
is to be
(1,0, I).
(I, I, I),
Find the values of A, B, and C that minimize
f1• (Ax, + By, + C 
z,)"
the sum of the squares of the deviations. 43. a. Muimum on a sphere Show that the maximum value of a 2b 2c 2 on a sphere of radius r centered at the origin of a Cartesian abccoordinare systero is (r 2/3)3 b. Geometric and arithmetic means Using part (a), show that for nonnegative numbers a, b, and c,
a (abc ) 1/3 < 
+b +c. 3
'
that is, the geometric mean of three nonnegative numbers is less than or equal to their arithmetic mean. 44. Sum of products Let at. a2, ... , an be n positive numbers. Find the maximum of l:?=1 a,x, subject to the constraint l:1=1 x? = 1.
820
Chapter 14: Partial Derivatives
COMPUTER EXPLORATIONS
46. Minimize J(x, y, z) = :x:yz subject to the constraints x 2 + y2  1 = 0 and x  z = O. 47. Maximize J(x, y, z) = x 2 + y2 + z2 sui!ject to the constraints 2y + 4z  5 = 0 and4x 2 + 4y2  z2 = O.
In Exercises 4550, use a CAS to perform the following steps implementing tire method of Lagrange multipliers for fmding constrained extrema: L
Formthefunctionh = J  AlgI  A2g2,whereJ is the function to optimize subject to tire constraints gl = 0 and g2 = O.
48. Minimize J(x,y,z) = x 2 + y2 + z2 subject to the constraints x 2  :x:y + y2 _ z2  I = 0 andx 2 + y2  I = O. 49. Minimize J(x,y,z, w) = x 2 + y2 + z2 +w 2 subject to the
h. Determine all the f!rst partial derivatives of h, including the partials with respect to Al andA2, and set them equal to O.
2:x:y+zwI=O
constraints
and
x+yz+
w  I = O.
c. Solve the system of equations found in part (b) for all the unknowns, including Al and A2.
50. Determine the distance from the line y = x + I to the parabola y2 = x. (Hint: Let (x, y) be a point on tire line and (w, z) a point on the parabola. You want to minimize (x  w)' + (y  z)'.)
eI. Evaluate J at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the
exercise. 45. Minimize J(x,y,z) = :x:y + yz sui!ject to x 2 + y2  2 = 0 andx 2 + z2  2 = O.
14.9
rl
the
constraints
_~~y~Lo_(_s_F_o_ rm_u_La__ ro_r_Tw _ o__ Va_n_'a_b_Le_s______________In this section we use Taylor's formula to derive the Second Derivative Test for local extreme values (Section 14.7) and the error formula for linearizations of functions of two independent variables (Section 14.6). The use of Taylor's formula in these derivations leads to an extension of the formula that provides polynomial approximations of all orders for functions of two independent variables.
Derivation of the Second Derivative Test t
~
1 S(a
+ h,b + k)
Parametrized
Let I(x, y) have continuous partial derivatives in an open region R containing a point P( a, b) where Ix = Iy = 0 (Figure 14.57). Let h and k be increments small enough to put the
point S(a + h, b segment PS as
+ k) and the line segment joining it to P inside R. We parametrize the x=a+th,
If F(t) = I(a
O:5t:51.
y=b+tk,
+ th, b + tk), the Chain Rule gives
,
t= 0
F (t)
P(a, b)
=
ax
Ix dt
dy
+ Iy dt
=
hlx
+ k/y.
Part of open region R
FIGURE 14.57 We begin the derivation of the Second Derivative Test atp(a, b) by parametrizing a typical line segment from P to a point S nearby.
Since Ix and Iy are differentiable (they have continuous partial derivatives), F' is a differentiable function of t and
F"
=
aF' ax + aF'dy axdt aydt
= h 21xx
=
~(hf + kf)'h + ~(hf + kf)'k ax x y ay x y f", ~ fyx
+ 2hklxy + elyy .
Since F and F' are continuous on [0, I] and F' is differentiable on (0, 1), we can apply Taylor's formula with n = 2 and a = 0 to obtain
F(I)
=
F(O) + F'(O)(1  0) + F"(c)
F(I)
=
F(O) + F'(O)
t
+ F"(c)
(1
2
0)' (1)
14.9 Taylor's Formula for Two Variables
821
for some c between 0 and 1. Writing Equation (1) in terms of I gives I(a
+ h, b + k)
= I(a, b)
+ hlx(a, b) +
kly(a, b)
+ ~ (h 2lxx + 2hkl"1 + k 2lyy) I
.
(2)
(a+ch, b+cl)
Since Ix(a, b) = I,(a, b) = 0, this reduces to I(a
I(a
+ h,b + k)
 I(a,b) =
~(h2Ixx + 2hkl"1 + k 2lyy) I(a+ch, b+ck).
(3)
The presence of an extremum of I at (a, b) is determined by the sigo of k)  I(a, b). By Equation (3), !bis is the same as the sigo of
+ h, b +
Q(c) = (h 2lxx
+ 2hkl"1 + k2Iyy)l(a+do.b+ck)'
Now, if Q(O) oF 0, the sigo of Q(c) will be the same as the sigo of Q(O) for sufficiently small values of h and k. We can predict the sigo of
Q(O)
2
= h Ixx(a, b)
+ 2hkl"1(a, b) + k 2Iyy(a, b)
(4)
from the signs of Ixx and fulyy  1"12 at (a, b). Multiply both sides of Equation (4) by Ixx and rearrange the rightband side to get (5)
From Equation (5) we see that
1. If Ixx < 0 and Ixxlyy  1"12 > 0 at (a, b), then Q(O) < 0 for all sufficiently small nonzero values of h and Ie, and I has a local maximum value at (a, b). 2. If Ixx > 0 and Ixxlyy  1"12 > 0 at (a, b), then Q(O) > 0 for all sufficiently small nonzero values of h and Ie, and I has a local minimum value at (a, b). 3. If Ixxlyy  1"12 < 0 at (a, b), there are combinations of arbitrarily small nonzero values of h and k for which Q( 0) > 0, and other values for which Q( 0) < O. Arbitrarily close to the point Po(a, b, I(a, b)) on the surface z = I(x,y) there are points above Po and points below Po, so I has a saddle point at (a, b). 4. If I xxlyy  1"12 = 0, another test is needed. The possibility that Q(O) equals zero prevents us from drawing conclusions about the sigo of Q(c).
The Error Formula for Linear Approximations We want to show that the difference E(x, y), between the values of a function I(x, y), and its linearization L(x, y) at (xo, Yo) satisfies the inequality IE(x,yli:5
~M(lx 
xol
+ Iy
 yoll'·
The function I is assumed to have continuous second partial derivatives throughout an open set containing a closed rectangular region R centered at (xo, yo). The number M is an upperbouod forl/xxl, I/yyl, and 1/"11 onR. The inequality we want comes from Equation (2). We substitute Xo and Yo for a and b, and x  Xo and y  Yo for h and k, respectively, and rearrange the result as I(x,y) = l(xo,Yo)
+ fx{xO,yo)(x
 xo)
+ Iy(xo,yo)(y
 Yo)
linearization L(x, y) 1
+ 2 (x
 Xol'lxx
+ 2(x
 xo)(y  yo)I"1
+ (y
error E(x, y)
 Yol'lyy)l(Xo + c (XXo.Yo ) + c (YYo » .
822
Chapter 14: Partial Derivatives
TIris equation reveals that
IEI:5 !Ox  xo1 2 1/",1 + 21x  xolly  yoll/.,,1 + Iy 
YOI21/yy!).
Hence, if Mis an upper bound for the values ofl I", I, 1/.,,1, andl/yyl onR,
IEI:5 !Ox 
xol 2M
!M; ~ 4x  "  xy 2
", 0
~ 6  2y,
(4)
x O
The volume of the entire solid is therefore
Volum, ~
y
,
,
\ 1.'.' ,·0
A(y) _
A(y) dy ~
1',0·'
(6  2y) dy ~ [6y  y']: ~ 5,
in agreement with our earlier calculation. Again, we may give a formula for the volume as an iterated integral by writing
(4x)')d.t
FIGURE 15.5 To obtaW. the crosssectional. area A(y), we hold y fDWd and integrate with respect to x.
1,0,·,
Volume =
11102.(4  x 
y) duly.
The expression on the right says we can find the volume by intcgIating 4  x  y with respect tox from x  0 tax  2 as in Equation (4) and integrating the result with respect to y from y  0 to Y  1. In this. iterated integral, the order of integration is fll'S1: x and then y, the reverse of the order in Equation (3). What do these two volume calculations with iterated integrals have to do with the double integral
jj(4  x  y) dA R H:rsWRICAL BIOGRAPHY
Guido Fubini (18791943)
over the rectangle R: 0 ~ x ~ 2, 0 ~ y ~ I? The answer is that both iterated integrals give the value of the double integral. This is what we would reasonably expect, since the double integml measures the volume of the smne region as the two iterated integrals. A theorem published in 1907 by Guido Fubini says that the double integral of any continuous functioo aver a rectangle can be calculated as an itemted integral in either order of integration. (Fubini proved his theorem in greater generality, but this is what it says in our setting.)
THEOREM lFubini's Theorem ;dy ~ 1'1' /(x,y) dy
d>;,
R
Fubini's Theorem says that double integrals over rectangles can be calculated as iterated integrals. Thus, we can evaluate a double integral by integrating with respect to one variable at a time. Fubini's Theorem also says that we may calculate the double integral by integrating in either order, a genuine convenience. When we calculate a volume by slicing, we may use either planes perpendicular to the xaxis or planes perpendicular to the yaxis.
EXAMPLE 1
Calculate .O'R/(x,y)
f(x,y) ~ 100 
6x'y
dA for and
R:
O::Sx::S2,
1::Sy::sl.
840
Chapter 15: Multiple Integrals
,
.: 100 6.1:7
Solution
Figwe 15.6 displays the volume beneath the surface. By Fubini's Theorem,
~ 1\200 ,  16y) dy ~
[201ly  8y']:,
~ 400.
Reversing the order of integration gives the same answer: x
R
,
1 ,
{' {\100 
Jo Jl
6x'y)dydx
~ ('[lOlly  3'V~::, dx
Jo
~ 1,\(100 
FIGURE 15.6 The double integral f(x, y) cIA. gives the volume under this surface aver the n:ctangular region R (Example 1).
DR
3x')  (100  3x')] dx
£2
200 dx _ 400.
EXAM PLE 2
Find the volume of the region bounded above by the ellipitical paraboloid
z  10 + x2 + 3y 2 andbelowbytherectangleR:0::S; x S 1,0
Sy::s;
2.
Solution The surface and volume a:re shown in Figure 15.7. The volume is given by the double integral
R
ff + l' [lOy
FIGURE 15.7 The double inWgral f(x, y) gives the volume under this smfaoc aver the n:ctangular region R
DR
•
(10
V=
dA
x 2 + 3y2)dA =
11 1
2
(10
+ x 2 + 3y2) dydx
R
+ x'y + y'~:: dx
(Example 2).
=
Jor (20 + 2x
2
+ 8)dx = [2Ox + ~X3 + 8x]1 3
•
= 86.
3
0
Exercises 15.1 ~~tingH8~In~~
11.
In Exercises 112, evaluate the iterated integral.
1.
J.'110f'
3.1:1: la la 3
5.
2. ].' {' (x  y)dydr.
2xydydr.
(x
+y +
l
(4 
l)dxdy
y~dydx
Y 7. ].0 '].' 0 1 +xydxdy
9. ].0 "'11[b' e2>:+y dy dx
011 x' + Y') 4. ]. '].' (1  dr.dy o 0
12·1 la" lfr
y sin;t dx dy
(sin;t
+ cosy)dxdy
Evaluating Double Integra~ over Rectangles In Exercises 1320, evaluate the double integral. aver t1u: given regionR.
2
13.
f' (x'y 012
6. ].'
1,'],."'
2xy) dy dr.
jj(6y2 2x)cIA.,
R:
O:!;';;t:!;'; I,
0
:!;';y:!;'; 2
R
14.
jj (v:)cIA., R
15.
R:
jjxycosYdA, R: R
0 S;t S 4,
1 Sy S 2
Y
IS;tSI,
OSys",.
15.2 Double Integrals over General Regions
16.
JJ JJ JJ
y sin (x
+ y) dA,
R:
1T':S
x :s 0,
O:s y ::s
1T'
R
17.
.'Y dA,
0 '" x '" In 2,
R:
0 '" Y '" In 2
R
18.
xyexi' dA,
0 '" x '" 2, 0 '" Y '" I
R:
R
19.
ff
,xy3
}} x + I
dA,
0 '" x '" I,
R:
0 '" Y '" 2
R
20.
JJ X'y: +
I dA,
R:
0 '" x '" I,
0 '" Y '" I
R
In Exercises 21 and 22, integrate f over the given region.
21. Square f(x,y) = l/(xy) over the square I '" x'" 2, I'" Y '" 2 22. Rectangle f(x,y) = y cosxy over the rectangle 0 '" x '" 'fr,
O"'y"'l
841
Volume Beneath a Surface 23. Find the volume of the region bounded above by the paraboloid z = x' + y' and below by the square R: I '" x '" I, I "'y"'i. 24. Find the volume of the region bounded above by the ellipiticai paraboloid z = 16  x'  y' and below by the square R: 0 :s x :s 2,0 :s y :s 2. 25. Find the volume of the region bounded above by the plane z = 2  x  y and below by the square R: 0 '" x '" 1,
o '" Y '" i. 26. Find the volume of the region bounded above by the plane z = y/2 and below by the rectangle R: 0 '" x '" 4,0 '" Y '" 2. 27. Find the volume of the region bounded above by the surfaco z = 2 sin x cos y and below by the rectangle R: 0 '" x '" 'fr/2, o '" Y '" 'fr/4. 28. Find the volume of the region bounded above by the surfaco z = 4  y' and below by the rectangle R: 0 '" x '" 1, o '" Y '" 2.
z= f(x,y)
Double Integrals over General Regions
15.2
In this section we detme and evaluate double integrals over bounded regions in the plane which are more general than rectangles. These double integrals are also evaluated as iterated integrals, with the main practical problem being that of determining the limits of integration. Since the region of integration may have boundaries other than line segments parallel to the coordinate axes, the limits of integration often involve variables, not just constants.
Double Integrals over Bounded, Nonrectangular Regions
"
0, the width and height of each enclosed rectangle goes to zero and their number goes to intmity. If f(x, y) is a continuous function, then these Riemann sums converge to a limiting value, not dependent on any of the choices we made. This limit is called the double integral of f(x,y) over R:
842
Chapter 15: Multiple Integrals
The nature of the boundary of R introduces issues not found in integrals over an interval. When R has a curved boundary, the n rectangles of a partition lie inside R but do not cover all of R. In order for a partition to approximate R well, the parts of R covered by small rectangles lying partly outside R must become negligible as the norm of the partition approaches zero. This property of being nearly filled in by a partition of small norm is satisfied by all the regions that we will encounter. There is no problem with boundaries made from polygons, circles, ellipses, and from continuous graphs over an interval, joined end to end. A curve with a "fractal" type of shape would be problematic, but such curves arise rarely in most applications. A careful discussion of which type of regions R can be used for computing double integrals is left to a more advanced text.
Volumes If I(x, y) is positive and continuous over R, we define the volume of the solid region betweenR and the surface z = I(x,y) to be JJRI(x,y) dA, as before (Figure 15.9). If R is a region like the one shown in the xyplane in Figure 15.10, bounded "above" and ''below'' by the curves y = g2(X) and y = gl (x) and on the sides by the lines x = a, x = b, we may again calculate the volume by the method of slicing. We first calculate the crosssectional area A(x) =
l
y =g2(X)
I(x,y) dy
y=gl(X)
and then integrate A(x) from x
V=
l
=
a to x = b to get the volume as an iterated integral:
bA(x)dx= lblg2(X) I(x,y)dydx.
a
a
(1)
gl(X)
z z =f(x,y)
o
y
x
Volume
=
lim !fiXk, Yk) aAk =
If f(x, y) dA R
FIGURE 15.9 We define the volumes of solids with curved bases as a limit of approximating rectangular boxes.
FIGURE 15.10 The area of the vertical slice shown here is A(x). To calculate the volume of the solid, we integrate this area from x = a to x = b:
l
a
b
A(x)dx =
lblg2(X) a
g,(x)
f(x,y)dydx.
15.2 Double Integrals over General Regions
843
Similarly, if R is a region like the one shown in Figure 15.11, bounded by the curves x = h2(y) and x = hl(y) and the lines y = c and y = d, then the volume calculated by slicing is given by the iterated integral
z z =f(x,y)
x
Il d
Volume =
y
hb )
(2)
I(x,y) dxdy.
h,(y)
c
That the iterated integrals in Equations (1) and (2) both give the volume that we defined to be the double integral of lover R is a consequence of the following stronger form of Fubini 's Theorem.
FIGURE 15.11 shown here is
I
c
The volume of the solid
dA(y)dy = Idlh2(v) j(x,y)dxdy. h,(v) c
TH EOREM 2Fubini's Theorem (Stronger Form)
Let I(x, y) be continuous on a
regionR .
1. If R is defined by a on [a , b] , then
:5
x
:5
b, gl (x)
Y
:5
:5
r iirrI(x, y) dA = laig,(x)
For a given solid, Theorem 2 says we can calculate the volume as in Figure 15.10, or in the way shown here. Both calculations have the same result.
g2
b
g2(X) , with gl and g2 continuous
I(x, y) dy dx.
(x)
R
2. If R is defined by c on [c, d], then
:5
Y
:5
d, hl(y)
iJif I(x,y) dA = I
x
:5
:5
h2(y), with hI and h2 continuous
dlh2(y)
I(x,y) dxdy. hICy)
c
R
EXAMPLE 1 Find the volume of the prism whose base is the triangle in the xyplane bounded by the xaxis and the lines y = x and x = 1 and whose top lies in the plane z = I(x, y) = 3  x  y.
Solution See Figure 15.12. For any x between 0 and l,y may vary from y = 0 to y = x (Figure 15.12b). Hence, V=
=
11 1
x
o
(3  x  y) dy dx
11 [
=
0
1
1 (
o
3y  xy  y2]Y=X dx 2 y =O
0
3
3x  ~2) dx = 2
[3~ 2

2
:!.3 ]X=I 2
= 1.
x=o
When the order of integration is reversed (Figure 15.12c), the integral for the volume is
V =
= =
11 [ 1111 11 (3  ~ 11 (%  ~ 2) [% o
(3  x  y) dx dy
3x  ~2  xy ]X=I dy
=
0
y
y  3y
4y
+
Y
x=y
+ Y22 + y2) dy
dy =
The two integrals are equal, as they should be.
y  2y2
+ ~3 I:~ =
1.
•
844
Chapter 15: Multiple Integrals z y
(3,0,0)
FIGURE 15.12 (a) Prism with a triangular base in the xyplane. The volume of this prism is defined as a double integral over R. To evaluate it as an iterated integral, we may integrate first with respect to y and then with respect to x, or the other way around (Example I). (b) Integration limits of
l
x=11Y=x f(x, y) dy dx. x=o y=o
If we integrate first with respect to y, we integrate along a vertical line through R and then integrate from left to right to include all the vertical lines in R. (c) Integration limits of
l
Y=11X=1 f(x,y) dx dy. y=o X=Y
If we integrate first with respect to x, we integrate along a horizontal line through R and then integrate from bottom to top to include all the horizontal lines in R.
Although Fubini's Theorem assures us that a double integral may be calculated as an iterated integral in either order of integration, the value of one integral may be easier to find than the value of the other. The next example shows how this can happen.
EXAMPLE 2
Calculate
11 si~x
dA,
R
where R is the triangle in the xyplane bounded by the xaxis, the line y = x, and the line x = 1.
15.2 y x
Double Integrals over General Regions
845
Solution The region of integration is shown in Figure 15.13. Ifwe integrate IlISt with respect to y and then with respect to x, we Imd
=1
l' (lS~XdY)dx l' &s~x IJdx =
= cos (I) ~~~~X
o
+
I
RJ
=
1'SinXdx
0.46.
If we reverse the order of integration and attempt to calculate
J.
o
FIGURE 15.13 The region of integration in Example 2.
111. y
s~x dx dy,
we run into a problem because J«sinx)/x) dx cannot be expressed in terms of elementary functions (there is no simple antiderivative). There is no general rule for predicting which order of integration will be the good one in circumstances like these. If the order you IlISt choose doesn't work, try the other. Some_ times neither order will work, and then we need to use numerical approximations.
y
Finding Limits of Integration We now give a procedure for finding limits of integration that applies for many regions in the plane. Regions that are more complicated, and for which this procedure fails, can often be split up into pieces on which the procedure works. Using Vertical Crosssections When faced with evaluating lfR f(x,y)dA, integrating IlISt with respect to y and then with respect to x, do the following three steps:
(a)
y
/
Leaves at y~~ Enters at y=lx
1.
Sketch. Sketch the region of integration and label the bounding curves (Figure 15.14a).
2.
Find the ylimits of integration. Imagine a vertical line L cutting through R in the direction of increasing y. Mark the yvalues where L enters and leaves. These are the ylimits of integration and are usually functions of x (instead of constants) (Figure 15.14b).
o
3. Find the xlimits of integration. Choose xlimits that include all the vertical lines x
throughR. The integral shown here (see Figure 15.14c) is
ff
(b)
f(x,y)dA
=
E~IL~~ f(x,y)dydx.
R
y
Leaves at
/y~~ Enters at
y=lx
Using Horizontal Crosssections To evaluate the same double integral as an iterated integral with the order of integration reversed, use horizontal lines instead of vertical lines in Steps 2 and 3 (see Figure 15.15). The integral is
JJrr f(x,y)dA R
=
J.lkr~ f(x,y)dxdy. 0
y
~t~~~~x
l
x
/
Smallest x
Largest x
isx=O
isx=l
Largesty Y isy = 1 ~1
Enters at
x=ly
(c)
FIGURE 15.14 Finding the limits of integration when integrating first with respect to y and then with respect to x.
Y r~~~~
Smallest y isy ~ 0
"
~ Leaves at x
~
v'l=Y'
;;ct~~~x
o
FIGURE 15.15 Finding lhe limits of integration when integrating first with respect to x and then with respect to y.
846
Chapter 15: Multiple Integrals
EXAMPLE 3
Sketch the region of integration for the integral
1212x(4x + 2)dydx and write an equivalent integral with the order of integration reversed.
Solution The region of integration is given by the inequalities x 2 :5 y :5 2x and o :5 x :5 2. It is therefore the region bounded by the curves y = x 2 and y = 2x between x = 0 and x = 2 (Figure 15.16a). To find limits for integrating in the reverse order, we imagine a horizontal line passing from left to right through the region. It enters at x = y/2 and leaves at x = To include all such lines, we lety run from y = 0 to Y = 4 (Figure 15.16b). The integral is
...;y.
y
r
rv,.(4x + 2) dx dy. 101Y/2
•
The common value of these integrals is 8.
Properties of Double Integrals ~~~~X
Like single integrals, double integrals of continuous functions have algebraic properties that are useful in computations and applications.
If f(x, y) and g(x, y) are continuous on the bounded region R, then the following properties hold
FIGURE 15.16 Region of integration for Example 3.
1. constantMUltiPle:ff cf(x,y)dA = cff f(x,y)dA R
(any number c)
R
2. Sum and Difference: ffU(X,Y) ± g(x,y)) dA
ff f(x,y) dA ± ff g(x,y) dA
=
R
R
R
3. Domination: y
(a)
if
ff f(x,y) dA ;" 0
f(x,y) ;" 0 onR
R
(b)
ff f(x,y)dA;" ff g(x,y)dA R
o
II f(x, R
y) dA
II
II
~ f(x, y) dA + f(x, y) dA RI
f(x,y) ;" g(x,y) onR
R
4. Additivity: ff f(x,y) dA ~*~>x
if
R
=
ff f(x,y) dA + ff f(x,y) dA Rl
•
if R is the union of two nonoverlapping regions R, and R2
R2
FIGURE 15.17 The Additivity Property for rectangular regioos holds for regions bounded by continuous curves.
Property 4 assumes that the region of integration R is decomposed into nonoverlap· ping regions R, and R2 with boundaries consisting of a finite number of line segments or smooth curves. Figure 15.17 illustrates an example of this property.
15.2 Double Integrals over General Regions
847
The idea behind these properties is that integrals behave like swns. If the function f(x, y) is replaced by its constant multiple cf(x, y), then a Riemann swn for f n
Sn =
z
L f(xk, Yk) flAk k=!
is replaced by a Riemann swn for cf n
L
k=!
c
=
L
k=!
f(xk, Yk) flAk
",, \
\... I I
(a)
y
y
=
4x 2
2
~r~~L~x
(b)
FIGURE 15.18 (a) The solid "wedgelike" region whose volume is found in Example 4. (b) The region of integration R showing the order dx dy.
Find the volwne of the wedgelike solid that lies beneath the surface z = 16  x 2  y2 and above the region R bounded by the curve Y = 2Vx, the line Y = 4x  2, and the xaxis. Solution Figure 15.18a shows the surface and the "wedgelike" solid whose volwne we want to calculate. Figure 15.18b shows the region of integration in thexyplane. If we integrate in the order dy ax (first with respect to y and then with respect to x), two integrations will be required because y varies from y = 0 to y = 2vX for 0 :::; x :::; 0.5, and then varies from y = 4x  2 to Y = 2vX for 0.5 :::; x :::; 1. So we choose to integrate in the order ax dy, which requires only one double integral whose limits of integration are indicated in Figure 15 .18b. The volwne is then calculated as the iterated integral:
11
(16  x 2  y2) dA
= =
1 1
2 1(Y+2)!4
o
2 [
o
(16  x 2  y2)dxdy
y./4 x3 ]X=(Y+2)!4 16x    xy2 dx 3 x=y./4
t [
= Jo
4(y
_ [191Y
+
2) 
63y2
 z4 + ~ 
(y + 2)3 3 . 64 145y 3 ~

(y
+ 2)y2  4y2
4
49y4 768
+
y5 20
+
y6 3' 64
+ o

Sketching Regions of Integration In Exercises 1 8, sketch the described regions of integration. 1. 0 :::; x :::; 3, 0:::; y ~ 2x
xI ~ y ~ x 2
7. 0 ~ y ~ 1,
0 ~ x ~ sin I Y
8. 0 ~ y ~ 8,
i
3. 2 ~ y ~ 2, y2 ~ X ~ 4 ~
y
~
1, y
~
x
~
2y
y
~ x ~ Y 1/3
y4]
+ 4 dy
7 ]2 _ 20803
Y 1344
Exercises 15.2
4. 0
cSn .
EXAMPLE 4
R
2. 1 ~ x ~ 2,
=
Taking limits as n ~ 00 shows that c limn>:,y = x, aodx =
15. 16. 17.
°
18. Bounded by y = x 2 andy = x
+2
r
20. Jo Jo 21.
1, 1vG' VI=7 Jl Jo
45.
1,'1"
y dy dx
Jtr'r' Jo e +Y dx dy X
(' p'
25. Quadrilateral I(x, y) = x/y over the region in the flIst quadrant bounded by the lines y = x, y = 2>:, x = I, and x = 2
26. Triaogle I(x, y) = x2 + y2 over the ttiaogular region with vertices (0, 0), (I, 0), aod (0, I)
27. Triaogle I(u, v) = v  Vu over the ttiangular region cut from the flISt quadraot of the uvplane by the line u + v = I 28. Curved region I(s, t) = e' In t over the region in the flISt quadraot of the stplane that lies above the curve s = In t from
t=llot=2 Each of Exercises 21}32 gives an integral over a region in a Cartesian coordinate plane. Sketch the region and evaluate the integral.
3ydxdy
42.
ydxdy
1v'k'
2
44.
1,o .v'k' ~/6L1/2
1,o
6x dydx
xy 2 dydx
sinr
{V3 {,m,"
+ y) dx dy
46. Jo
Jo
v;Jdxdy
In Exercises 4756, sketch the region of integration, reverse the order of integration, aod evaluate the integral.
r Jxrsiny y dy dx
48.
47. Jo
1,111 2y'j;31y'j;3 51. 1, ,/2
x 2e"" dx dy
o
50.
1,2122y2 sinxy dy dx 2j,4'X' xe ~
1,o
4 _
0
52. {' {' Jo
ex' dx dy
JWi
Y
dy dx
.r dy dx
53. {1/16 {1/2 cos (I&nx') dx dy
Jo
23. Jo Jo 3y'e"" dx dy In Exercises 2528, integrate 1 over the giveo region.
dy dx
dx dy
{' r·r
xy dy dx (x
J"
40. Jo Jo
{' {Inx
43.
49.
{.;nx
1x
16x dy dx
Jo
o
finding Regions of Integration and Double Integrals In Exercises 11}24, sketch the region of integration and evaluate the integral.
0
dx dy
I,I.il'X'
1
41.
13. Bounded by y 14.
y2
38. Jo
{'/2 {"4x' 39. Jo
0
{1n2 {'
{' {" 37. Jo
1,2:£0
54.
),1/4
l,'h o
2
dydx +I
~ y4
55. Square region lfR (y  2>:2) dA where R is the region boundedbythesquarelxl + Iyl = I 56. Triaogular region
thelinesy
= ;t,Y =
lfRXY dA where R is the region hounded by +y = 2
a,andx
Volume Beneath a Surface z
=
/(x, y)
57. Find the volume of the region bounded above by the paraboloid z = x2 + y2 and below by the ttiangle enclosed by the lines y = x,x = 0, and x + y = 2 in thexyplane. 58. Find the volume of the solid that is hounded above by the cylinder z = x 2 aod below by the region enclosed by the parabola y = 2  x 2 and the line y = x in the xyplane.
15.2 Double Integrals over General Regions 59. Find the volume of the solid whose base is the region in the xyplane that is bounded by the parabola y ~ 4  x 2 and the line y = 3x, while the top of the solid is bounded by the plane z = x + 4. 60. Find the volume of the solid in the fIrSt octant bounded by the coordinate planes, the cylinder x 2 + y2 = 4, aod the plaoe z + y = 3. 61. Find the volume of the solid in the flIst octant bounded by the coordinate planes, the plaoe x = 3, aod the parabolic cylinder z ~ 4  y2. 62. Find the volume of the solid cot froro the fIrSt octant by the surfacez = 4  x 2  y.
63. Find the volume of the wedge cut from the fIrSt octaot by the cylinder z ~ 12  3y2 and the plaoe x + y ~ 2. 64. Find the volume of the solid cot froro the square colunm Ixl + Iyl'" I by the planes z = Oaod3x + z = 3. 65. Find the volume of the solid that is bounded 00 the froot aod back by the plaoes x ~ 2 aod x = I, 00 the sides by the cylioders y = ± I/x, aod above aod below by the plaoes z = x + I aod z ~ O. 66. Find the volume of the solid bounded 00 the froot aod back by the planes x = ±'1T/3, on the sides by the cylinders y = ±secx, above by the cylinder z ~ I + y2, aod below by the xyplaoe. In Exercises 67 and 68, sketch the region of integratioo aod the solid whose volume is giveo by the double integral. 67.
/.J
/'
':£'116 v'25 /.o y Y
68.
(I  tx 
2r
v'16
'
l
Integrals over Unbounded Regions Improper double integrals cao often be coroputed similarly to improper integrals of ooe vatiable. The flIst iteratioo of the following improper integrals is cooducted just as if they were proper integrals. One thea evaluates ao improper integral of a single vatiable by taking appropriate limits, as in Sectioo 8.7. Evaluate the improper integrals in Exercises 6')..72 as iterated integrals. 69.
1
00'1. '
I
70.
74. f(x,y) = x + 2y over the regioo R inside the circle (x  2)' + (y  3)' = I using the partition x = 1,3/2,2,5/2, 3 aody ~ 2, 5/2, 3, 7/2, 4 with (Xk,Yk) the center (centroid) in the kth suhrectangle (provided the suhrectaogle lies within R)
Theory and Examples
v'
75. Cin:ular sector Integrate f(x, y) ~ 4  x 2 over the smaller 2 sector cut from the disk x + y2 S 4 by the rays 8 = '1T/6 aod 8 = '1T/2. 76. Unbounded region Integrate f(x,y) ~ 1/[(x 2  x)(y  1)2/3]
aver the infmite rectangle 2 ::s x
1
89.
3\IJ 
x 2  y2dydx
x
1:1.
' _+1 dydx lOX Y
1
2].4Y' 0
eX)! dx dy
2~s y'
I v'x2
+ y2
dxdy
0
Area by Double Integration In this section we show how to use double integrals to calculate the areas of bounded regions in the plane, and to find the average value of a function of two variables.
Areas of Bounded Regions in the Plane If we take !(x, y) = I in the defmition of the double integral over a region R in the preceding section, the Riemarm sums reduce to
S. =
• • L !(XhY.) aA. = L aA•. i I i I
(1)
11ris is simply the swn of the areas of the small rectangles in the partition of R, and approximates what we would like to call the area of R. As the norm of a partition of R approaches zero, the height and width of all rectangles in the partition approach zero, and the coverage of R becomes increasingly complete (Figure 15.8). We define the area of R to be the limit lim . aA. = 1IJ1I~o~
DEFINITION
iJif dA.
(2)
R
The area of a closed, bounded plane region R is
As with the other defmitions in this chapter, the definition here applies to a greater variety of regions than does the earlier singlevariable definition of area, but it agrees with the earlier defmition on regions to which they both apply. To evaluate the integral in the definition of area, we integrate the constant function !(x, y) = I over R.
EXAMPLE 1 quadrant.
Find the area of the region R bounded by Y = x and y = x 2 in the frrst
15.3
851
Area by Double Integration
Solution
y
We sketch the region (Figure IS.19), noting where the two curves intersect at the origin and (1, 1), and calcolate the area as
A = =
!oJ dyax 10f' (x 
x
2 )
=
[~]:ax
ax = [; 2
3]'0
~
=
1 6'
ax,
FIGURE 15.19
The region in Example I.
Notice that the singlevariable integral Jo' (x  x 2 ) obtained from evaluating the inside iterated integral, is the integral for the area between these two curves using the method of Section S.6. • Find the area of the regionR enclosed by the parabolay = x 2 and the line
EXAMPLE 2 y = x + 2. Solution
Ifwe divide culate the area as
y
Rinto the regions R, and R2 shown in Figure IS.20a, we may cal
On the other hand, reversing the order of integration (Figure IS.20b) gives A =
=~~~x o
o
x
(b)
FIGURE 15.20 Calculating this area ta1res (a) two double integrals if the rust integration is with respect to x, but (b) only one if the ftrSt integration is with respect to y (Example 2).
The average value of an integrable function of one variable on a closed interval is the integral of the function over the interval divided by the length of the interval. For an integrable function of two variables defmed on a bounded region in the plane, the average value is the integral over the region divided by the area of the region. This can be visualized by thinking of the function as giving the height at one instant of some water sloshing around in a tank whose vertical walls lie over the boundary of the region. The average height of the water in the tank can be found by letting the water settle down to a constant height. The height is then equal to the volume of water in the tank divided by the area of R. We are led to define the average value of an integrable function f over a region R as follows:
10fRJJ IdA.
Averagevalueof/overR = area
(3)
R
If I is the temperatore of a thin piate covering R, then the double integral of lover R divided by the area of R is the plate's average temperature. If I(x, y) is the distance from the point (x, y) to a flXed point P, then the average value of lover R is the average distance of points inR fromP.
852
Chapter 15: Multiple Integrals
EXAMPLE 3 R: 0 :5 x :5
Find the average value of 'IT,
0
:5
f(x,y)
= xcosxy
over the rectsngle
f
= sinxy + C
Y :5 1.
The value of the integral of f over R is
Solution
1"1' o
xcosxy dydx =
0
1"[
sinxy
0
]Yl dx
xcosxydy
y=O
= l"(SinX  0)
I
dx = cosx
= I
+
I = 2.
•
The area of R is 7f. The average value of f over R is 2/7f.
Exercises 15.3 Area by Double Integrals In Exercises 112, sketch 1he region bounded by 1he given lines and curves. Then express the region's area as an iterated double integral and evaluate 1he integral.
1. The coordinate axes and 1he line x
+y
= 2
2. The lines x = O,y = 2x,andy = 4
3. The parabola x
_y2 and 1he line y
=
x
=
+2
4. The parabola x = y  y2 and 1he line y = x 5. The curve y = eX and 1he lines y = 0, x = 0, and x = In 2 6. Thecurvesy = lox andy = 2loxand1helinex = e,in1hefrrst quadrant
Find;ng Average Values 19. Find 1he average value of f(x, y) = sin (x
a. the rectangle 0 ::s x :s
1r.
b. 1he rectangle 0 ,;; x ,;; 7f,
O:s y :s
+ y) over 1r.
0,;; y ,;; 7f/2.
20. Which do you think will be larger, 1he average value of f(x,y) = xy over 1he square 0 '" x '" 1,0 '" Y '" I, or 1he average value of f over 1he quarter circle x 2 + y2 ,;; I in 1he first quadrant? Calculate 1hern to fmd out 21. Find 1he average height of 1he paraboloid z = x 2 + y2 over 1he
square 0 :s x :s 2, 0 :s y :s 2. 22. Find 1he average value of f(x,y) = I/(xy) over 1he square 102'" x '" 2102,102'" Y '" 2102.
7. The parabolas x = y2 and x = 2y _ y2 Theory and Examples 8. The parabolas x = y2  I and x = 2y2  2 9. The linesy = x,y = x/3, andy = 2 10. The lines y = I  x and y = 2 and 1he curve y = eX 11. Thelinesy = 2x,y = x/2,andy = 3  x 12. The lines y = x  2 and y = x and 1he curve y =
Vx
Identlfytng the Region of Integration The integrals and sums of integrals in Exercises 1318 give 1he areas of regions in 1he xyplane. Sketch each region, label each bounding curve with its equati~ and give the coordinates of the points where 1he curves intersect Then find 1he area of1he region.
11 1"1'1. 1°1 '6
II
o
2Y
K
dxtty
ylj3
0
o
0%
16.
ttydx
sinx X
17.
1
18.
ttydx
+
Ox 2 4
12:£'x ttydx x/2 0
2;(
12:£0
x Y+2
00
15.
131X(2X) ttydx
dydx
+
14:£Vx 00
ttydx
 y. 1~
dx tty
23. Baeterium population If f(x,y) = (IO,OOOe Y )/(\ + IxV2) represents 1he "population density" of a certain bactetiun3 on 1he xyplane, where x and y are measured in centimeters, fmd the tota! population of bactetia wi1hin 1he rectangle 5 '" x '" 5 and 2 '" Y '" O. 24. Regional population If f(x, y) = 100 (y + 1) represents 1he population density of a planar region on Earth, where x and y are measured in miles, find the number of people in the region bouoded by 1he curves x = y2 aod x = 2y _ y2. 25. Average temperature in Teus According to 1he Texas Almanac, Texas has 254 counties and a National Weather Service station in each county. Assume that at time to, each of the 254 wea1her stations recorded 1he local teroperature. Find a formula that would give a reasonable approximation of the average temperature in Texas at time to. Your answer should involve information that you would expect to be readily available in 1he Texas Almanac. 26. If Y = f(x) is a nonnegative contiouous function over 1he closed interval a '" x '" b, show that 1he double integral defInition of area for 1he closed plane region bounded by the graph of f, 1he vertical lines x = a and x = b, and the xaxis agrees with the defmition for area beuea1h 1he curve in Section 5.3.
853
15.4 Double Integrals in Polar Form
15.4
~I_D_ou_b_Le__ In_~~g~~_~_l_ 'n_P_o_ ~_ r_ ~_r_ m___________________________ Integrals are sometimes easier to evaluate if we change to polar coordinates. This section shows how to accomplish the change and how to evaluate integrals over regions whose boundaries are given by polar equations.
Integrals in Polar Coordinates When we dermed the double integral of a function over a region R in the xyplane, we began by cutting R into rectangles whose sides were parallel to the coordinate axes. These were the natoralshapes to use because their sides have either constant xvalues or constant yvalues. In polar coordinates, the natural shape is a ''polar rectangle" whose sides have constant r and Ovalues. Suppose that a function I(r, 0) is dermed over a region R that is bounded by the rays = a and 0 = {3 and by the continuous curves r = gM) and r = g2(0). Suppose also that 0 :5 gl(O) :5 g2(O) :5 a for every value of 0 between a and {3. Then R lies in a fanshaped region Q dermed by the inequalities 0 :5 r :5 a and a :5 0 :5 {3. See Figure 15.21.
o
r=a
o~
,,"''''''''~o ~
o
0
FIGURE 15.21 TheregionR:g,(8) '" r s g2(8),a s 8 s ,B,is contained in the fanshaped region Q: 0 :5 r S a, a s 8 s ,B. The partition of Q by circular arcs aod rays induces a partition of R.
We cover Q by a grid of circular arcs and rays. The arcs are cut from circles centered at the origin, with radii !J.r, 2!J.r, ... , m!J.r, where !J.r = aim. The rays are given by () =
Q,
0= a + M,
0= a + 2M,
... ,
o=
a + m'!J.O
= {3,
where !J.O = ({3  a)lm'. The arcs and rays partition Q into small patches called ''polar rectangles." We number the polar rectangles that lie inside R (the order does not matter), calling their areas !J.A" !J.A 2, ... , !J.A •. We let (rk, Ok) be any point in the polar rectangle whose area is !J.Ak. We then form the sum
If I is continuous throughout R, this sum will approach a limit as we refine the grid to make !J.r and !J.O go to zero. The limit is called the double integral of lover R. In symbols, lim S. =
n~OO
JJRrr I(r, 0) dA.
854
Chapter 15: Multiple Integrals To evaluate this limit, we first have to write the sum S. in a way that expresses aA. in terms of!!J.r and!!J.lI. For convenience we choose r. to be the average of the radii of the inner and outer arcs bounding the kth polar rectangle aA•. The radius of the inner arc bounding aA. is then r.  (!!J.r/2) (Figure 15.22). The radius of the outer arc is r. + (!!J.r/2). The area of a wedgeshaped sector of a circle having radius r and angle II is
A = Large sector
o FIGURE 15.22 aA k =
The observation lbat
1 11 • r2 2
'
as can be seen by multiplying 7Tr 2 , the area of the circle, by 1I/27T, the fraction of the circle's area contained in the wedge. So the areas of the circular sectors subtended by these arcs at the origin are
Car::e:!m) (sn:::!or)
leads to lbe formula aA, = r, !!J.r !!J.O.
)2 M
Inner mdius:
!!J. "21 ( r.f
Outer radius:
"21 ( r. + T!!J.r)2 !!J.II.
Therefore,
aA.
= area oflarge sector  area of small sector =
y
~II [ (r. + ~r)' 
(r. 
~)'] = ~O (2r. !!J.r) =
r. ar M.
Combining this result with the sum defming S. gives
•
S. = ~ j(r., lI.h ar M.
~~>x
As n > integral
00
and the values of ar and ao approach zero, these sums converge to the double
o
lim S. =
(a)
n_OO
rr
JJ R
y
Leaves at r
~
2 r sin 8 : : ~ r =
j(r, II) r dr dll.
A version of Fubini's Theorem says that the limit approached by these sums can be evaluated by repeated single integmtions with respect to r and II as
2
=
L
R
V2 t.....,,\t~
Y2 esc 9
Enters atr
=
V2 esc 8
8
~~~>x
o
Largest 6 is
I'
L
2
/
// Y
=x
V2 I+....:/~ / / / /
/
R
1
,~plr~"'(.)
8=a
j(r,lI) r dr dll. r=g,(8)
Finding Limits of Integration The procedure for finding limits of integration in rectangular coordinates also works for polar coordinates. To evaluate lfR j(r, II) dA over a region R in polar coordinates, integmting fITst with respect to r and then with respect to II, take the following steps.
(b) y /
iJif
j(r,lI) dA =
~ Smallest 6 is ~.
1. Sketch. Sketch the region and label the bounding curves (Figure 15.23a). 2. Find the rlimits of integration. Imagine a my L from the origin cutting through R in the direction of increasing r. Mark the rvalues where L enters and leaves R. These are the rlimits of integmtion. They usually depend on the angle II that L makes with the positive xaxis (Figure 15.23b).
/
~~L>x
o
(c)
FIGURE 15.23 Finding !be limits of integration in polar coordinates.
3.
Find the IIlimits of integration. Find the smallest and largest IIvalues that bound R. These are the IIlimits of integmtion (Figure 15.23c). The polar iterated integral is
iJif R
j(r, II) dA =
1
·~"/21r~2
9='1fj4
r=V2csc6
j(r, II) r dr dll.
15.4 Double Integrals in Polar Form
855
EXAMPLE 1 Find the limits of integration for integrating fir, 0) over the region R that lies inside the cardioid r = I + cos 0 and outside the circle r = I.
y
Solutton
O=I
L
Enters at r=1
Leaves at
1. We flfst sketch the region and label the bounding curves (Figure 15.24). 2. Next we fmd the rlimits of integration. A typical ray from the origin enters R where r = I and leaves where r = I + cos O. 3. Finally we find the Olimits ofintegration. The rays from the origin that intersectR run from 0 = 7f/2 to 0 = 7f/2. The integral is
r=I+0089
1
fTj2
FIGURE 15.24 Finding the limits of
integratioo in polar coordinates for the regioo in Example 1.
~/211+'0"
•
fir, 0) r dr dO.
1
If fir, 0) is the constant function whose value is I, then the integral of f over R is the areaofR.
Area in Polar Coordinates
I Area Differential in Polar Coordinate.
The area of a closed and bounded region R in the polar coordinate plane is
dA = rdrd6
A = IfrdrdO. R
This fonnula for area is consistent with all earlier fonnulas, although we do not prove this fact. y
Leaves at r=
V4 co, 28
" I. ~~~+~X
Enters at r= 0
r 2 =400829
To integrate over the shaded regioo, we run r from 0 to y'4 cos 26 and 6 from 0 to ",/4 (Exarople 2). FIGURE 15.25
EXAMPLE 2
Find the area enclosed by the lemniscate r2
=
4 cos 20.
Solutton We graph the lemniscate to determine the limits of integration (Figure 15.25) and see from the symmetry of the region that the tota1 area is 4 times the firstquadrant portion.
J.
A=4
~/4:£V4"'26
J.~/4
00
=
J.
4
0
[r2]'=V4'0'2'
rdrdO=4"2 0
~/4
2 cos 20 dO
]~/4
=
4 sin 20
0
dO
r=O
=
4.
•
Changing Cartesian Integrals into Polar Integrals The procedure for changing a Cartesian integralifR fix, y) dx dy into a polar integral has two steps. First substitute x = r cos 0 and y = r sin 0, and replace dx dy by r dr dO in the Cartesian integral. Then supply polar limits of integration for the boundary of R. The Cartesian integral then becomes
II R
fix, y) dx dy =
II
fir cos 0, r sin 0) r dr dO,
G
where G denotes the same region of integration now described in polar coordinates. This is like the substitution method in Chapter 5 except that there are now two variables to substitute for instead of one. Notice that the area differential dx dy is not replaced by dr dO but by r dr dO. A more general discussion of changes of variables (substitutions) in multiple integrals is given in Section 15.8.
856
Chapter 15: Multiple Integrals
EXAMPLE 3
Evaluate
II
e
x2 y2
+ dy dx,
R
where R is the semicircular region bounded by the xaxis and the curve y = ~ (Figure 15.26). y
y=~
SoLution In Cartesian coordinates, the integral in question is a nonelementary integral and there is no direct way to integrate e x2 +y2 with respect to either x or y. Yet this integral and others like it are important in mathematicsin statistics, for exampleand we need to find a way to evaluate it. Polar coordinates save the day. Substituting x = r cos e, y = r sin e and replacing dy dx by r dr de enables us to evaluate the integral as
0=0 ~~~~~
__ x
= Jot'l"2 (e  1) de
FIGURE 15.26 The semicircular region in Example 3 is the region
o :s r :s
1,
o :s ()
:s
7T.
= ;
(e  1).
The r in the r dr de was just what we needed to integrate e r2 • Without it, we would have been unable to find an antiderivative for the first (innermost) iterated integral. _
EXAMPLE 4
SoLution
Evaluate the integral
Integration with respect to y gives
t (x2~+
Jo
(1  X2) 3/2) 3 dx,
an integral difficult to evaluate without tables. Things go better if we change the original integral to polar coordinates. The region of integration in Cartesian coordinates is given by the inequalites 0 ::; y ::; ~ and o ::; x ::; 1, which correspond to the interior of the unit quarter circle x 2 + y2 = 1 in the first quadrant. (See Figure 15.26, first quadrant.) Substituting the polar coordinates x = r cos e, y = r sin e, 0 ::; e ::; 7T/2 and 0 ::; r ::; 1, and replacing dx dy by r dr de in the double integral, we get
z
_1

o
7T /
2
_1
[r4 ]r=1 4 de 
r=O
7T /
2
!4 de _ 1T8·
0
Why is the polar coordinate transformation so effective here? One reason is that x 2 simplifies to r2. Another is that the limits of integration become constants.
EXAMPLE 5 z x
FIGURE 15.27 Example 5.
_
Find the volume of the solid region bounded above by the paraboloid
= 9  x 2  y2 and below by the unit circle in the xyplane.
The region of integration R is the unit circle x 2 + y2 = 1, which is described in polar coordinates by r = 1,0 ::; e ::; 21T. The solid region is shown in Figure 15.27. The volume is given by the double integral
SoLution The solid region in
+ y2
15.4 Double Integrals in Polar Form
=
10f'''10{' (9r 
r
3
)
857
dr dO
• EXAMPLE 6 Using polar integration, find the area of the regionR in the xyplane eoclosed by the circle x 2 + y2 + 4, above the line y = 1, and below the line y = 0x. Solutton
A sketch of the region R is showo in Figure 15.28. First we note that the line y = 0x has slope 0 = tan 6, so 6 = 7T/3. Next we observe that the line y = 1 intersects the circle x 2 + y2 = 4 wheo x 2 + 1 = 4, or x = 0. Moreover, the radial line from the origin through the point (0,1) has slope 1/0 = tan 0, giving its angle of inclination as 6 = 7T/6. This information is showo in Figure 15.28. Now, for the regionR, as 6 varies from 7T/6 to 7T/3, the polar coordinate r varies from the horizontal line y = 1 to the circle x 2 + y2 = 4. Substituting r sin 0 for Y in the equation for the horizontal line, we have r sin 6 = 1, or r = esc 6, which is the polar equation of the line. The polar equation for the circle is r = 2. So in polar coordinates, for 7T/6 :5 6 :5 7T/3, r varies from r = esc 0 to r = 2. It follows that the iterated integral for the area theo gives
y
~~~~~~X
o
2
FIGURE 15.28 Example 6.
1 2
/3
{{ dA = {" Jrr/6
JJR
The region R in
 {"/3  Jrr/6
rdrd6
esc (J
[1 r2l'~2 2
d6
r csc9
,,/3 1
csc20] dO
=
1
=
21 [46 + cot 6 ],,/ "/6
=
1 (47T +
,,/6
2 [4 
3
_1_) _1(47T + 0)
23026
= 7T 
Exercises 15.4 Regions in Polar Coordinates
4.
3.
In Exercises 11l, describe the given region in polar coordinates.
L
y
1 y ~~~~>x
1
4+,> x 9
+~4>x
o
4
0
1
0
3'
•
858
Chapter 15: Multiple Integrals Area in Polar Coordinates
6.
5. y
27. Find the area of the region cut from the fIrst quadrant by the curve r = 2(2  sin 26)112.
y
28. Cardioid overlapping a clrcle Find the area of the region that lies inside the cardioid r = 1 + cos 0 and outside the circle r = 1.
2t     ,
29. One leaf of a rose r = 12 cos 36.
30. SnaR shen Find the area of the region eoclosed by the positive xaxis aod spiral r = 46/3,0 '" 6 '" 21T. The region looks like a anall shell.
2
7. The region enclosed by the circle x 2
8.
+ y2 = 21:. Theregionenclosedbythesemicirclex 2 + y2 =
2y,y '" O.
In Exercises 922, change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral.
1 11,~
11. Jo Jo
I
2
~ 1
+
~
Vd'x'
22
YX
+ y2
/,v22/,'v4=Y'
Find the awrage height of
Ya 2  x2 
y2 above the disk
34. Average height ofa cone Find the average height of the (single) dxdy
dydx
conez = Yx 2
+ y2 above the diskx 2 + y2
'" a 2 inthexyplane.
35. Average distance from interior of disk to center Find the average distance from a point P(x, y) in the disk x 2 + y2 '" a 2 to the origin.
2
2 X + y2f
+
dydx
°
1 1Viv'I=Yl =Y'
f(r, 6) r dr d6.
R
33. Awrage height of a hemisphere
1n2:£V(ln2)'" • ~ • vx'+,' dx dy
1,°
JJ
36. Average distance squared from a point in a disk to a point in
~ (I
1
20.
1
16.
dydx
18.111~ 19.
12.
Are~(R)
a1~ dydx
the hemispherical surface z = x 2 + y2 ~ a 2 in the xyplane.
17.1°1° 1
+ y2) dx dy
+ y2) dx dy
Average values In polar coordinates, the average value of a function over a region R (Section 15.3) is giveo by
0
x dx dy
V3r Jl
1
(x 2
0
(x
13. 1,61,' 1
1,11,Vi=Y'
0
{'{VH
15.
10.
dy dx
1
31. Cardioid in the r"st quadrant Find the area of the region cut from the frrst quadrant by the cardioid r = 1 + sin 6. 32. Overlapping cardioids Find the area of the region conunon to the interiors of the cardioids r = I + cos 6 and r = I  cos 6.
Evaluating Polar Integrals
9.
Find the area enclosed by one leaf of the rose
its houndary Find the average value of the square of the distance from the point p(x, y) in the disk x 2 + y2 '" I to the boundary pointA(I, 0).
Theory and Examples In (x 2
+ y2 + I) dx dy
1
37. Converting to a polar integral Integrate f(x,y) = [In (x 2 + y2)]fYx 2 + y2 OWl the region 1 '" x 2 + y2 S e.
21.1,11~ (x + 2y)dydx
38. Converting to a polar integral Integrate f(x, y) = [In (x 2 + y2)]/(x 2 + y2) 0'"' the region I '" x 2 + y2 S .2.
22. 121,~
39. Volume of noncircular right cylinder The region that lies inside the cardioid r = 1 + cos 8 and outside the circle r = I is the base of a solid right cylinder. The top of the cylioder lies in the plane z = x. Find the cylinder's volume.
1
0
1 (X 2
+ y2)2
dydx
In Exercises 2326, sketch the region of integration and convert each polar integral or sum of integrals to a Cartesian integral or sum of in· tegrals. Do not evaluate the integrals. 12 {' 23. Jo Jo r3 sin 6 cos 6 dr d6
r
24.
r121='
r2 cos 6 dr d6
i1r/6 I /4 {''''''' 25. Jo Jo
r
40. Volume of noncircular right cylinder The region enclosed by the leoutiscate r2 = 2 cos 26 is the base of a solid tight cylinder whose top is bounded by the sphere z = ~. Find the cylinder's volume. 41. Converting to polar integrals
a. The usual way to evaluate the improper integral 00 I = e ;f1 dx is frrst to calculate its square:
10
r' sin2 6 dr d6 Evaluate the last integral using polar coordinates and solve the resulting equation for 1.
15.5 Triple Integrals in Rectangular Coordinates b. Evaluate lim erf(x) =
xHX)
lim
!O0
X
2  /' • e/ dt.
XH)()
46. Area Suppose that the area of a region in the polar coordinate plane is
1
3Tr/412sin6
A =
V 11'
Tr/4
42. Converting to a polar integral
Evaluate the integral
43. Existence Integrate the function j(x, y) = 1/(1  x 2  y2) over the disk x 2 + y2 ~ 3/4. Does the integral of j(x, y) over
rdrdO. esc 6
Sketch the region and find its area. COMPUTER EXPLORATIONS In Exercises 47 50, use a CAS to change the Cartesian integrals into an equivalent polar integral and evaluate the polar integral. Perform the following steps in each exercise.
a. Plot the Cartesian region of integration in the xyplane.
the disk x 2 + y2 ~ I exist? Give reasons for your answer.
b. Change each boundary curve of the Cartesian region in part (a) to its polar representation by solving its Cartesian equation for r andO.
44. Area formula in polar coordinates Use the double integral in polar coordinates to derive the formula
1
859
.6 1
2
c. Using the results in part (b), plot the polar region of integration in the rOplane.
for the area of the fanshaped region between the origin and polar curve r = j(O), a ~ 0 ~ {3.
d. Change the integrand from Cartesian to polar coordinates. Determine the limits of integration from your plot in part (c) and evaluate the polar integral using the CAS integration utility.
A =
a
 r 2 dO
45. Average distance to a given point inside a disk Let Po be a point inside a circle of radius a and let h denote the distance from Po to the center of the circle. Let d denote the distance from an arbitrary point P to Po. Find the average value of d 2 over the region enclosed by the circle. (Hint: Simplify your work by placing the center of the circle at the origin and Po on the xaxis.)
15.5
47.
49.
1111 1l o
x
1
o
Y
 2   2 dy
+Y
x
Y 3 /
y/3
dx
Y
Yx 2 + y2
48.
l1 1
o
x 2 x / dydx
0
x 2 + y2
dxdy
Triple Integrals in Rectangular Coordinates Just as double integrals allow us to deal with more general situations than could be handled by single integrals, triple integrals enable us to solve still more general problems. We use triple integrals to calculate the volumes of threedimensional shapes and the average value of a function over a threedimensional region. Triple integrals also arise in the study of vector fields and fluid flow in three dimensions, as we will see in Chapter 16.
Triple Integrals If F(x, y, z) is a function defined on a closed, bounded region D in space, such as the region occupied by a solid ball or a lump of clay, then the integral of F over D may be defined in the following way. We partition a rectangular boxlike region containing D into rectangular cells by planes parallel to the coordinate axes (Figure 15.29). We number the cells that lie completely inside D from 1 to n in some order, the kth cell having dimensions ~Xk by ~Yk by ~Zk and volume ~ Vk = ~k~Yk~Zk. We choose a point (Xk, Yk, Zk) in each cell and form the sum n
Sn = ~ F(xk, Yk , Zk) ~ Vk.
(1)
k=l
FIGURE 15.29 Partitioning a solid with rectangular cells of volume ~ Vk •
We are interested in what happens as D is partitioned by smaller and smaller cells, so that ~k, ~Yk, ~Zk and the norm of the partition Ilpll, the largest value among ~k, ~Yk, ~Zk, all approach zero. When a single limiting value is attained, no matter how the partitions and points (Xk, Yk, Zk) are chosen, we say that F is integrable over D. As before, it can be
860
Chapter 15: Multiple Integrals shown that when F is continuous and the bounding surface of D is fonned from tmitely many smooth surfaces joined together along tmitely many smooth curves, then F is integrable. As Ilpll > 0 and the number of cells n goes to 00, the sums S. approach a limit. We call this limit the triple integral of F over D and write
•
~S. =
iiiF(x,y,Z)dV
lim S. =
or
IWII~O
D
rrr F(x,y,z)
111 D
dxdy dz .
The regions D over which continuous functions are integrable are those having ''reasonably smooth" boundaries.
Volume of a Region in Space If Fis the constant function whose value is 1, then the sums in Equation (1) reduce to
As dxk, t.Yk, and t.Zk approach zero, the cells t. V. become smaller and more numerous and ml up more and more of D. We therefore define the volume of D to be the triple integral
DEFINITION
The volume of a closed, bounded region D in space is
V= iff dV. D
TIris detmition is in agreement with our previous detmitions of volume, although we omit the verification of this fact. As we see in a moment, this integral enables us to calculate the volumes of solids enclosed by curved surfaces.
Finding Limits of Integration in the Order dz dy dx We evaluate a triple integral by applying a threedimensional version of Fubini's Theorem (Section 15.2) to evaluate it by three repeated single integrations. As with double integrals, there is a geometric procedure for finding the limits of integration for these single integrals. To evaluate
iff F(x,y,z)dV D
over a region D, integrate first with respect to z, then with respect to y, and finally with respect to x. (You might choose a different order of integration, but the procedure is similar, as we illustrate in Example 2.)
1.
Sketch. Sketch the region D along with its "shadow" R (vertical projection) in the xyplane. Label the upper and lower bounding surfaces of D and the upper and lower bounding curves of R.
15.5 Triple Integrals in Rectangular Coordinates
861
z z = fz(x, y)
D
y
x
2.
Find the zlimits a/integration . Draw a line M passing through a typical point (x,y) in R parallel to the zaxis. As z increases, M enters D at z = f 1(x, y) and leaves at z = h(x,y). These are the zlimits of integration. z
D
y
a b
 
x
(x,y)
3.
Find the ylimits a/integration. Draw a line L through (x,y) parallel to the yaxis. As y increases, L enters Rat y = gl(X) and leaves at y = g2(X). These are the ylimits of integration. z
D
Enters at y = g,(x) a
y
x b
x
Leaves at y = g2(x)
862
Chapter 15: Multiple Integrals
4.
Find the xlimits of integration. Choose xlimits that include all lines through R parallel to the yaxis (x = a and x = b in the preceding figure). These are the xlimits of
integration. The integral is
l
x=bly=g2(X)lz= h(x,y)
x=a
F(x, y, z) dz dy dx. y=gl(X)
z= !J(x, y)
Follow similar procedures if you change the order of integration. The "shadow" of region D lies in the plane of the last two variables with respect to which the iterated integration takes place. The preceding procedure applies whenever a solid region D is bounded above and below by a surface, and when the "shadow" region R is bounded by a lower and upper curve. It does not apply to regions with complicated holes through them, although sometimes such regions can be subdivided into simpler regions for which the procedure does apply.
EXAMPLE 1
Find the volume of the region D enclosed by the surfaces z = x 2
+
3y2
and z = 8  x 2  y2. SoLution
The volume is
v=
111
dzdydx,
D
the integral of F(x, y, z) = lover D. To find the limits of integration for evaluating the integral, we first sketch the region. The surfaces (Figure 15.30) intersect on the elliptical cylinder x 2 + 3y2 = 8  x 2  y2 or x 2 + 2y2 = 4, z > O. The boundary of the region R, the projection of D onto the xyplane, is an ellipse with the same equation: x 2 + 2y2 = 4. The ''upper'' boundary of R is the curve y = V(4  x 2)/2. The lower boundary is the curve y = V(4  x 2 )/2. Now we find the zlimits of integration. The line M passing through a typical point (x,y) in R parallel to the zaxis enters D at z = x 2 + 3y2 and leaves at z = 8  x 2  y2. M
The curve of intersection
'(2, 0,4)
I I
Enters at
z =x 2 +
3y2
I
I
J
 I I , I '.,,1
Enters at~=_ y = V(4  x 2 )/2
x
R
Leavesat
~
L
y
y = V(4  x 2 )/2
FIGURE 15.30 The volume of the region enclosed by two paraboloids, calculated in Example 1.
15.5 Triple Integrals in Rectangular Coordinates
863
Next we find the ylimits of integration. The line L through (x, y) parallel to the yaxis enters Rat y =  Y(4  x 2 )/2 and leaves at y = Y(4  x 2 )/2. Finally we find the xlimits of integration. As L sweeps across R, the value of x varies from x = 2 at (2,0,0) to x = 2 at (2, 0, 0). The volume of Dis
v=
JJJ
dzdydx
D
1 11 1
21V(4  X2)/ 218  X2  y2
=
 2  V(4  x 2)/ 2 x2 +3y2
V (4  X2)/ 2
2
=
dzdydx
(8  2x2  4y2) dy dx
 2  V(4  x 2)/ 2 2 [
=
2
=
4 ]y=V(4  X2)/ 2 (8  2x 2)y  _y3 dx 3 y=  V(4  x 2)/ 2
81TY2.
•
After integration with the substitution x = 2 sin u
In the next example, we project D onto the xzplane instead of the xyplane, to show how to use a different order of integration. z __________ (0, 1,1)
EXAMPLE 2 Set up the limits of integration for evaluating the triple integral of a function F(x, y, z) over the tetrahedron D with vertices (0, 0, 0), (1 , 1,0), (0, 1, 0), and (0, 1, 1). Use the order of integration dy dz dx. Solution
We sketch D along with its "shadow" R in the xzplane (Figure 15.31). The upper (righthand) bounding surface of D lies in the plane y = 1. The lower (lefthand) bounding surface lies in the plane y = x + z. The upper boundary of R is the line z = 1  x. The lower boundary is the line z = 0. First we find the ylimits of integration. The line through a typical point (x, z) in R parallel to the yaxis enters D at y = x + z and leaves at y = 1. Next we find the zlimits of integration. The line L through (x, z) parallel to the zaxis enters R at z = and leaves at z = 1  x. Finally we find the xlimits of integration. As L sweeps across R, the value of x varies from x = to x = 1. The integral is
°
(1,1,0)
°
ll i
x
Finding the limits of integration for evaluating the triple integral of a function defined over the tetrahedron D (Examples 2 and 3).
l
o
FIGURE 15.31
l x 
0
l
•
F(x,y, z) dy dz dx.
x+ z
EXAM PLE 3 Integrate F(x, y , z) = 1 over the tetrahedron D in Example 2 in the order dz dy dx, and then integrate in the order dy dz dx. Solution
First we find the zlimits of integration. A line M parallel to the zaxis through a typical point (x, y) in the xyplane "shadow" enters the tetrahedron at z = and exits through the upper plane where z = y  x (Figure 15.32). Next we find the ylimits of integration. On the xyplane, where z = 0, the sloped side of the tetrahedron crosses the plane along the line y = x . A line L through (x, y) parallel to the yaxis enters the shadow in the xyplane at y = x and exits at y = 1 (Figure 15.32).
°
864
Chapter 15: Multiple Integrals
z
Finally we find the xlimits of integration. As the line L parallel to the yaxis in the previous step sweeps out the shadow, the value of x varies from x = 0 to x = 1 at the point (1, 1,0) (see Figure 15.32). The integral is
ri
Jo
x
(yx
l
Jo
F(x,y,z) dzdydx.
For example, if F(x,y, z) = 1, we would find the volume of the tetrahedron to be
v=
1
lillY Xdzdydx l
= l i\yx)dY dx x
FIGURE 15.32 The tetrahedron in Example 3 showing how the limits of integration are found for the order dz dy dx.
1
6' We get the same result by integrating with the order dy dz dx. From Example 2,
V
=
l i o
l l X l dydzdx l 0
r r
x+z
x
=
Jo Jo
=
1 (1  x)3
6
(1  x  z) dz dx
]1
•
0
Average Value of a Function in Space The average value of a function F over a region D in space is defined by the formula Average value of F over D =
vol~e of D
Jff
F dV.
(2)
D
For example, if F(x,y, z) = Yx 2 + y2 + z2, then the average value of F over D is the average distance of points inD from the origin. If F(x,y, z) is the temperature at (x,y, z) on a solid that occupies a region D in space, then the average value of F over D is the average temperature of the solid.
865
15.5 Triple Integrals in Rectangular Coordinates
EXAMPLE 4 Find the average value of F(x,y, z) = xyz throughout the cubical region D bounded by the coordinate planes and the planes x = 2, Y = 2, and z = 2 in the tIrst octant. Solution We sketch the cube with enough detail to show the limits of integration (Figure 15.33). We then use Equation (2) to calculate the average value of F over the cube. The volume of the region Dis (2)(2)(2) = 8. The value of the integral of F over the cube is
D
121212
xyzdxdydz =
000
FIGURE 15.33 in Example 4.
The region of integration
1212 [2 12 12
~ yz ]X~2 dydz
00
=
[]Y~2 y2z dz=
o
= 12122yzdydz
x O
y=O
00
[]2
4zdz= 2z2
0
=8.
0
With these values, Equation (2) gives / Average value of = lYz over the cube vo ume
rrr
111 cube
(81 )(8)
xyz dV =
=
I.
In evaluating the integral, we chose the order dx dy dz, but any of the other five possible orders would have done as well. •
Properties of Triple Integrals Triple integrals have the same algebraic properties as double and single integrals. Simply replace the double integrals in the four properties given in Section 15.2, page 846, with triple integrals.
Exercises 15.5 Triple Integrals In Different Iteration Orders 1. Evaluate 1he integral in Example 2 taking F(x,y, z) = 1 to Imd 1he volmne of1he tetrahedron in 1he order dz dx dy. 2. Volume of rectangular .olid Write six different iterated triple
Evaluating Triple Iterated Integrals Evaluate 1he integrals in Exercises 720.
7. 1, '1, '1,
'·1" r 0 10
or;>
171'1~
T
!
dx dt dv
24. The regicm. in the flrst octant bounded by the coordinate planes andthep1anesx + z  l,y + 2z  2
(tuxspace)
1 tip dq fir
,
(pqrspace)
Finding Equivalent Iterated Integrals 21. Here is the region of integration of the integral
1'1, '1.'' r
1
,
dzdydx.
0
25. The region in the flrst octant bounded by the coontinate planes, the plane y + z = 2, and the cylinder x = 4 _ y2
,
Top:y+z=1 1
/
;...."> (1, 1,0)
,
1
x
,,
(I, I, 0)
Rewrite the integral as an equivalent iterated integm in the order L dydzdx b. dydxdz c, dxdydz d. dxdzdy e, dzdxdy. 22. Here is the regicm. of integraticm. of the integral
1. ,11f'1.T o
26. The wedge cut from the cylinder x 2
z  yandz  0
x
1
z _ ,'1.
27. The tetrahedron in the first octantboundedbythe coordinate planes and the plane passing through (1, 0, 0), (0, 2, 0), and (0, 0, 3)
,
o , (1,1,0)
 1 by the planes
,
,
(0,1, I)
+ y2
,
dzdydx.
0
(1,1, I)
.
1 x
Rewrite the integral. as an equivalent iterated integral. in the order dydzdx b. dydxdz c. dxdydz d. dxdzdy e. dzdxdy.
(0,2,0)
,
L
Finding Volumes UsIng TrIple Integrals Find the VOiUlllCS ofb regions in Exercises 2336. 23. The region between the cylinder z = y' and the xyplane that is bounded by the planes x = O,x = l,y = 1,y = 1
(1,0,0) ,/
x 28. The region in the fmt octant bounded by the coordinate planes, b plane Y = 1  x, and the surface z = cos ('/I"xj2), Osxsl
,
,
, x
15.5 Triple Integrals in Rectangular Coordinates 29. The region common to the interiors of the cylinders x 2 + y2 = 1 and x 2 + z2 = 1, oneeighth of which is shown in the accompanying figure
33. The region between the planes x z = 4 in the first octant
+ y + 2z
34. The finite region bounded by the planes z y = 8, andz = 0
=
2 and 2x
867 + 2y +
= x, x + z = 8, z = y,
35. The region cut from the solid elliptical cylinder x 2 + 4y2 ::::; 4 by the xyplane and the plane z = x + 2
xZ+r=l
36. The region bounded in back by the plane x = 0, on the front and sides by the parabolic cylinder x = 1  y2, on the top by the paraboloid z = x 2 + y2, and on the bottom by the xyplane
Average Values In Exercises 3740, find the average value of F(x, y, z) over the given region. 37. F(x,y, z) = x 2 + 9 over the cube in the first octant bounded by
the coordinate planes and the planes x
x
30. The region in the first octant bounded by the coordinate planes and the surface z = 4  x 2  Y
= 2, y = 2, and z = 2
38. F(x, y, z) = x + Y  z over the rectangular solid in the first octant bounded by the coordinate planes and the planes x = l,y = 1, andz = 2 39. F(x,y,z) = x 2 + y2 + z2 over the cube in the first octant bounded by the coordinate planes and the planes x = l,y = 1, andz = 1 40. F(x, y, z) = xyz over the cube in the first octant bounded by the coordinate planes and the planes x = 2, y = 2, and z = 2
z
Changing the Order of Integration Evaluate the integrals in Exercises 4144 by changing the order of integration in an appropriate way.
10 12 1o r r1112xzezy2 Jo Jo 4
41.
0
y
42.
x
1
4 cos (x • r 2 Yz
2y
1hI x2l
2 )
dxdydz
dy dx dz
x2
31. The region in the first octant bounded by the coordinate planes, the plane x + y = 4, and the cylinder y2 + 4z2 = 16
z
43. 44.
1
o
{2
~
ln3
'Tfe
2x
0
r r si~2z
Jo Jo Jo
4
sin 7TJ'2
Y
2
dxdydz
dy dz dx
z
Theory and Examples 45. Finding an upper limit of an iterated integral y
114ax214X2y
1 o