Loose-leaf Version for Calculus: Late Transcendentals [4 ed.] 1319055834, 9781319055837

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FOURTH EDITION

Jon Rogawski Colin Adams Robert Franzosa

Jon Rogawski • Cohn Adams • Robert Franzosa University of California, Los Angeles

Williams College

•h • freeman oko‘ w Macmillan Learning New York

The University of Maine

TO JULIE -Jon TO ALEXA AND COLTON -Colin TO MY FAMILY -Bob Vice President, STEM: Daryl Fox Program Director: Andy Dunaway Program Manager: Nikki Miller Dworsky Senior Marketing Manager: Nancy Bradshaw Marketing Assistant: Madeleine lnskeep Executive Development Editor: Katrina Mangold Development Editor: Tony Palermino Executive Media Editor: Catriona Kaplan Associate Editor: Andy Newton Editorial Assistant: Justin Jones Director, Content Management Enhancement: Tracey Kuehn Senior Managing Editor: Lisa Kinne Senior Content Project Manager: Kerry O'Shaughnessy Senior Workflow Project Manager: Paul Rohloff Director of Design, Content Management: Diana Blume Design Services Manager: Natasha Wolfe Cover Design Manager: John Callahan Interior 8( Cover Design: Lumina Datamatics, Inc. Senior Photo Editor: Sheena Goldstein Rights and Billing Associate: Alexis Gargin Illustration Coordinator: Janice Donnola Illustrations: Network Graphics Director of Digital Production: Keni deManigold Senior Media Project Manager: Alison Lorber Media Project Manager: Hanna Squire Composition: Lumina Datamatics, Inc. Printing and Binding: LSC Communications Cover Photo: Bettmann/Getty Images Library of Congress Control Number: 2018959763 Student Edition Hardcover: ISBN-13: 978-1-319-05073-3 ISBN-10: 1-319-05073-5 Student Edition Loose-leaf: ISBN-13: 978-1-319-05583-7 ISBN-10: 1-319-05583-4 Instructor Complimentary Copy: ISBN-13: 978-1-319-05579-0 ISBN-10: 1-319-05579-6 Copyright © 2019, 2015, 2012, 2008 by W. H. Freeman and Company All rights reserved Printed in the United States of America 1 2 3 4 5 6 23 22 21 20 19 18 W. H. Freeman and Company One New York Plaza Suite 4500 New York, NY 10004-1562 www.macmillanlearning.com

T THE AUTHORS Jon Rogawski

A

s a successful teacher for more than 30 years, Jon Rogawski listened and learned much from his own students. These valuable lessons made an impact on his thinking, his writing, and his shaping of a calculus text. Jon Rogawski received his undergraduate and master's degrees in mathematics simultaneously from Yale University, and he earned his PhD in mathematics from Princeton University, where he studied under Robert Langlands. Before joining the Department of Mathematics at UCLA in 1986, where he was a full professor, he held teaching and visiting positions at the Institute for Advanced Study, the University of Bonn, and the University of Paris at Jussieu and Orsay. Jon's areas of interest were number theory, automorphic forms, and harmonic analysis on semisimple groups. He published numerous research articles in leading mathematics journals, including the research monograph Automorphic Representations of Unitary Groups in Three Variables (Princeton University Press). He was the recipient of a Sloan Fellowship and an editor of the Pacific Journal of Mathematics and the Transactions of the AMS. Sadly, Jon Rogawski passed away in September 2011. Jon's commitment to presenting the beauty of calculus and the important role it plays in students' understanding of the wider world is the legacy that lives on in each new edition of Calculus.

Colin Adams

C

olin Adams is the Thomas T. Read professor of Mathematics at Williams College, where he has taught since 1985. Colin received his undergraduate degree from MIT and his PhD from the University of Wisconsin. His research is in the area of knot theory and low-dimensional topology. He has held various grants to support his research and written numerous research articles.

Colin is the author or co-author of The Knot Book, How to Ace Calculus: The Streetwise Guide, How to Ace the Rest of Calculus: The Streetwise Guide, Riot at the Cale Exam and Other Mathematically Bent Stories, Why Knot?, Introduction to Topology: Pure and Applied, and Zombies & Calculus. He cowrote and appears in the videos "The Great Pi vs. e Debate" and "Derivative vs. Integral: the Final Smackdown." He is a recipient of the Haimo National Distinguished Teaching Award from the Mathematical Association of America (MAA) in 1998, an MAA Polya Lecturer for 1998-2000, a Sigma Xi Distinguished Lecturer for 2000-2002, and the recipient of the Robert Foster Cherry Teaching Award in 2003. Colin has two children and one slightly crazy dog, who is great at providing the entertainment.

Robert Franzosa

R

obert (Bob) Franzosa is a professor of mathematics at the University of Maine where he has been on the faculty since 1983. Bob received a BS in mathematics from MIT in 1977 and a PhD in mathematics from the University of Wisconsin in 1984. His research has been in dynamical systems and in applications of topology in geographic information systems. He has been involved in mathematics education outreach in the state of Maine for most of his career. Bob is a co-author of Introduction to Topology: Pure and Applied and Algebraic Models in Our World. He was awarded the University of Maine's Presidential Outstanding Teaching award in 2003. Bob is married, has two children, three step-children, and one grandson.

III

SECTION 2.1 was largely rewritten so that its focus is on motivating the need for limits via the concepts of velocity and the tangent line. The content on rate of change that did not treat velocity was moved elsewhere.

Chapter 1: Precalculus Review 1 1.1 Real Numbers, Functions, 1.2 1.3 1.4 1.5

and Graphs 1 Linear and Quadratic Functions 14 The Basic Classes of Functions 22 Trigonometric Functions 27 Technology: Calculators and

Computers 35 Chapter Review Exercises 40

Chapter 2: Limits 43 -0 2.1 The Limit Idea: Instantaneous Velocity and Tangent Lines 43 2.2 Investigating Limits 49 2.3 Basic Limit Laws 58 2.4 Limits and Continuity 61 2.5 Indeterminate Forms 72 2.6 The Squeeze Theorem and Trigonometric Limits 77 2.7 Limits at Infinity 82 2.8 The Intermediate Value Theorem 88 2.9 The Formal Definition of a Limit 92 Chapter Review Exercises 99

Chapter 3: Differentiation 103

SECTION 4.1 was rewritten and reorganized to clarify the relationship between the different types of linear approximation. In particular, we wanted to reinforce the understanding that the various types of linear approximation are all based on the idea that the tangent line approximates the curve close to the point of tangency.

7.2 Inverse Functions 354

4.2 Extreme Values 186 4.3 The Mean Value Theorem and Monotonicity 196 4.4 The Second Derivative and Concavity 204 4.5 Analyzing and Sketching Graphs of Functions 210 4.6 Applied Optimization 218 4.7 Newton's Method 231 Chapter Review Exercises 236

Chapter 5: Integration 239

7.3 Logarithmic Functions and Their Derivatives 362 -0 7.4 Applications of Exponential and Logarithmic Functions 372 7.5 L'HOpital's Rule 380 7.6 Inverse Trigonometric Functions 388 7.7 Hyperbolic Functions 396 Chapter Review Exercises 403

Chapter 8: Techniques of Integration 407 8.1 Integration by Parts 407 8.2 Trigonometric Integrals 413

5.1 Approximating and Computing Area 239 5.2 The Definite Integral 251 5.3 The Indefinite Integral 263

8.3 Trigonometric Substitution 421 8.4 Integrals Involving Hyperbolic and Inverse Hyperbolic Functions 427

5.4 The Fundamental Theorem of Calculus, Part I 271 5.5 The Fundamental Theorem of

8.5 The Method of Partial

Calculus, Part II 277 5.6 Net Change as the Integral of a Rate of Change 285 5.7 The Substitution Method 291 Chapter Review Exercises 298

Chapter 6: Applications of the Integral 303

3.1 Definition of the Derivative 103 3.2 The Derivative as a Function 112 3.3 Product and Quotient Rules 125 3.4 Rates of Change 131 3.5 Higher Derivatives 141 3.6 Trigonometric Functions 146 3.7 The Chain Rule 151 3.8 Implicit Differentiation 158 3.9 Related Rates 165 Chapter Review Exercises 174

Cylindrical Shells 331 6.5 Work and Energy 338 Chapter Review Exercises 344

Chapter 4: Applications of the Derivative 177

Chapter 7: Exponential and Logarithmic Functions 347

4.1 Linear Approximation and o Applications 177

7.1 The Derivative of f(x) = 6' and the Number e 347

6.1 Area Between Two Curves 303 6.2 Setting Up Integrals: Volume, Density, Average Value 311 6.3 Volumes of Revolution: Disks and Washers 322 6.4 Volumes of Revolution:

SECTION 7.4 is new and contains some applications from the former Sections 7.4 and 7.5, along with applications that are new to this edition. Some material from the former Sections 7.4 and 7.5 has been moved elsewhere. For example, the material on differential equations and on exponential growth and decay was moved to the chapter on differential equations.

SECTION 9.1 The section on probability was moved from Section 8.8 to 9.1 so that it appears in the chapter on applications of thP integral rather than the chapte techniques of integration.

Fractions 432 8.6 Strategies for Integration 441 8.7 Improper Integrals 448 8.8 Numerical Integration 459 Chapter Review Exercises 469

Chapter 9: Further Applications of the Integral 473 9.1 Probability and Integration 473 9.2 Arc Length and Surface Area 479 9.3 Fluid Pressure and Force 486 9.4 Center of Mass 492 Chapter Review Exercises 502

Chapter 10: Introduction to Differential Equations 505 10.1 Solving Differential Equations 505 10.2 Models Involving y'= k(y - b) 515 10.3 Graphical and Numerical Methods 522 10.4 The Logistic Equation 529 10.5 First-Order Linear Equations 534 Chapter Review Exercises 540

SECTION 10.1 was rewritten to provide a more-straightforward introduction to differential equations and methods of solving them. Furthermore we wrote a few new examples that replaced a rather technical derivation to provide for a wider variety of simpler, more accessible application examples.

SECTION 15.4 The development of the concept of differentiability in Section 15.4 was rewritten to provide a clearer pathway from the basic idea of the existence of partial derivatives to the more-technical notion of differentiability. We dropped the concept of local linearity introduced in previous editions because it is redundant and adds an extra layer of technical detail that can be avoided.

Chapter 11: Infinite Series 543

14.2 Calculus of Vector-Valued

11.1 Sequences 543 11.2 Summing an Infinite Series 554 11.3 Convergence of Series with Positive Terms 566 11.4 Absolute and Conditional

Functions 750 14.3 Arc Length and Speed 760 14.4 Curvature 766

Convergence 575 11.5 The Ratio and Root Tests and Strategies for Choosing Tests 581 11.6 Power Series 586 11.7 Taylor Polynomials 598 11.8 Taylor Series 609 Chapter Review Exercises 621

Chapter 12: Parametric Equations, Polar Coordinates, and Conic Sections 625 12.1 Parametric Equations 625 12.2 Arc Length and Speed 637 12.3 Polar Coordinates 643 12.4 Area and Arc Length in Polar Coordinates 652 12.5 Conic Sections 657 Chapter Review Exercises 671

Chapter 13: Vector Geometry 675 13.1 Vectors in the Plane 675 13.2 Three-Dimensional Space: Surfaces, Vectors, and Curves 686 13.3 Dot Product and the Angle Between Two Vectors 696 13.4 The Cross Product 706 13.5 Planes in 3-Space 718 13.6 A Survey of Quadric Surfaces 725 13.7 Cylindrical and Spherical Coordinates 733 Chapter Review Exercises 740

Chapter 14: Calculus of VectorValued Functions 743 14.1 Vector-Valued Functions 743

14.5 Motion in 3-Space 778 14.6 Planetary Motion According to Kepler and Newton 786 Chapter Review Exercises 793

Chapter 15: Differentiation in Several Variables 795 15.1 Functions of Two or More Variables 795 15.2 Limits and Continuity in Several Variables 806 15.3 Partial Derivatives 814 -0 15.4 Differentiability, Tangent Planes, and Linear Approximation 824 15.5 The Gradient and Directional Derivatives 833 15.6 Multivariable Calculus Chain Rules 846 15.7 Optimization in Several Variables 856 15.8 Lagrange Multipliers: Optimizing with a Constraint 871 Chapter Review Exercises 881

Chapter 16: Multiple Integration 885 16.1 Integration in Two Variables 885 16.2 Double Integrals over More General Regions 897 16.3 Triple Integrals 911 16.4 Integration in Polar, Cylindrical, and Spherical Coordinates 923 16.5 Applications of Multiple Integrals 933 16.6 Change of Variables 945 Chapter Review Exercises 959

CALCULUS

CONTENTS Chapter 17: Line and Surface Integrals 963 17.1 Vector Fields 963 17.2 Line Integrals 973 17.3 Conservative Vector Fields 989 17.4 Parametrized Surfaces and Surface Integrals 1001 17.5 Surface Integrals of Vector Fields 1015 Chapter Review Exercises 1026

Chapter 18: Fundamental Theorems of Vector Analysis 1029 18.1 Green's Theorem 1029 18.2 Stokes' Theorem 1043 18.3 Divergence Theorem 1055 Chapter Review Exercises 1067

Appendices Al A. The Language of Mathematics Al B. Properties of Real Numbers A7 C. Induction and the Binomial Theorem Al2 D. Additional Proofs A16 ANSWERS TO ODD-NUMBERED EXERCISES ANSI REFERENCES R1 INDEX II Additional content can be accessed online at www.macmiI lanlearning.com/ ca1cu1us4e:

Additional Proofs: L'Hopital's Rule Error Bounds for Numerical Integration Comparison Test for Improper Integrals

Additional Content: Second-Order Differential Equations Complex Numbers

SECTION 11.1 We have chosen a somewhat traditional location for the section on Taylor polynomials, placing it directly before the section on Taylor series in Chapter 11. We feel that this placement is an improvement over the previous edition where the section was isolated in a chapter that primarily was about applications of the integral. The subject matter in the Taylor polynomials section works well as an initial step toward the important topic of Taylor series representations of specific functions. The Taylor polynomials section can serve as a follow-up to linear approximation in Section 4.1. Consequently, Taylor polynomials (except for Taylor's Theorem at the end of the section, which involves integration) can be covered at any point after Section 4.1.

CALCULUS, FOURTH EDITION On Teaching Mathematics

W

e consider ourselves very lucky to have careers as teachers and researchers of mathematics. Through many years (over 30 each) teaching and learning mathematics we have developed many ideas about how best to present mathematical concepts and to engage students working with and exploring them. We see teaching mathematics as a form of storytelling, both when we present in a classroom and when we write materials for exploration and learning. The goal is to explain to students in a captivating manner, at the right pace, and in as clear a way as possible, how mathematics works and what it can do for them. We find mathematics to be intriguing and immensely beautiful. We want students to feel that way, too.

On Writing a Calculus Text It has been an exciting challenge to author the recent editions of Jon Rogawslci's calculus book. We both had experience with the early editions of the text and had a lot of respect for Jon's approach to them. Jon's vision of what a calculus book could be fits very closely with our own. Jon believed that as math teachers, how we present material is as important as what we present. Although he insisted on rigor at all times, he also wanted a book that was clearly written, that could be read by a calculus student and would motivate them to engage in the material and learn more. Moreover, Jon strived to create a text in which exposition, graphics, and layout would work together to enhance all facets of a student's calculus experience. Jon paid special attention to certain aspects of the text: 1. Clear, accessible exposition that anticipates and addresses student difficulties. 2. Layout and figures that communicate the flow of ideas. 3. Highlighted features that emphasize concepts and mathematical reasoning including Conceptual Insight, Graphical Insight, Assumptions Matter, Reminder, and Historical Perspective. 4. A rich collection of examples and exercises of graduated difficulty that teach basic skills as well as problem-solving techniques, reinforce conceptual understanding, and motivate calculus through interesting applications. Each section also contains exercises that develop additional insights and challenge students to further develop their skills.

VI

Our approach to writing the recent editions has been to take the strong foundation that Jon provided and strengthen it in two ways: • To fine-tune it, while keeping with the book's original philosophy, by enhancing presentations, clarifying concepts, and emphasizing major points where we felt such adjustments would benefit the reader. • To expand it slightly, both in the mathematics presented and the applications covered. The expansion in mathematics content has largely been guided by input from users and reviewers who had good suggestions for valuable additions (for example, a section on how to decide which technique to employ on an integration problem). The original editions of the text had very strong coverage of applications in physics and engineering; consequently, we have chosen to add examples that provide applications in the life and climate sciences. We hope our experience as mathematicians and teachers enables us to make positive contributions to the continued development of this calculus book. As mathematicians, we want to ensure that the theorems, proofs, arguments, and derivations are correct and are presented with an appropriate level of rigor. As teachers, we want the material to be accessible and written at the level of a student who is new to the subject matter. Working from the strong foundation that Jon set, we have strived to maintain the level of quality of the previous editions while making the changes that we believe will bring the book to a new level.

What's New in the Fourth Edition In this edition we have continued the themes introduced in the third edition and have implemented a number of new changes.

A Focus on Concepts We have continued to emphasize conceptual understanding over the memorization of formulas. Memorization can never be completely avoided, but it should play a minor role in the process of learning calculus. Students will remember how to apply a procedure or technique if they see the logical progression of the steps in the proof that generates it. And they then understand the underlying concepts rather than seeing the topic

PREFACE

as a black box. To further support conceptual understanding of calculus, we have added a number of new Graphical and Conceptual Insights through the book. These include insights that discuss: • The differences between the expressions "undefined," "does not exist," and "indeterminate" in Section 2.5 on indeterminate forms, • How measuring angles in radians is preferred in calculus over measuring in degrees because the resulting derivafive formulas are simpler (in Section 3.6 on derivative rules of trigonometric functions), • How the Fundamental Theorem of Calculus (Part II) guarantees the existence of an antiderivative for continuous functions (in Section 5.5 on the Fundamental Theorem of Calculus, Part II), • How the volume-of-revolution formulas in Section 6.3 are special cases of the main volume-by-slices approach in Section 6.2, • The relationships between a curve, parametrizations of it, and arc length computed from a parametrization (in Section 12.2 on arc length and speed), • The relationship between linear approximation in multivariable calculus (in Section 15.4) and linear approximation for a function of one variable in Section 4.1.

Simplified Derivations We simplified a number of derivations of important calculus formulas. These include: • The Power Rule derivative formula in Section 3.2, • The formula for the area of a surface of revolution in Section 9.2, • The vector-based formulas for lines and planes in 3-space in Sections 13.2 and 13.5.

New Examples in the Life and Climate Sciences Expanding on the strong collection of applications in physics and engineering that were already in the book, we added a number of applications from other disciplines, particularly in the life and climate sciences. These include: • The rate of change of day length in Section 3.7 • The log wind profile in Section 7.4

vi

• A grid-connected energy system in Section 5.2 • A glacier height differential-equations model in Section 10.1 • A predator-prey interaction in Section 12.1 • Geostrophic wind flow in Section 15.5 • Gulf Stream heat flow in Section 16.1

An Introduction to Calculus In previous editions of the text, the first mathematics material that the reader encountered was a review of precalculus. We felt that a brief introduction to calculus would be a more meaningful start to this important body of mathematics. We hope that it provides the reader with a motivating glimpse ahead and a perspective on why a review of precalculus is a beneficial way to begin.

Additional Historical Content Historical Perspectives and margin notes have been a wellreceived feature of previous editions. We added to the historical content by including a few new margin notes about past and contemporary mathematicians throughout the book. For example, we added a margin note in Section 3.1 about the contributions of Sir Isaac Newton and Gottfried Wilhelm Leibniz to the development of calculus in the seventeenth century, and a margin note in Section 13.2 about recent Field's medalist Maryam Mirzakhani.

New Examples, Figures, and Exercises Numerous examples and accompanying figures have been added to expand on the variety of applications and to clarify concepts. Figures marked with a DF icon have been made dynamic and can be accessed via WebAssign Premium. A selection of these figures also includes brief tutorial videos explaining the concepts at work. A variety of exercises have also been added throughout the text, particularly following up on new examples in the sections. The comprehensive section exercise sets are closely coordinated with the text. These exercises vary in difficulty from routine to moderate as well as more challenging. Specialized exercises indicates problems are identified by icons. For example, There also student to give a written response. that require the of either graphingare icons for problems that require the use more advanced software such as calculator technology 'GU) or system a computer algebra (CAS)

CALCULUS, FOURTH EDITION offers an ideal balance of formal precision and dedicated conceptual focus, helping students build strong computational skills while continually reinforcing the relevance of calculus to their future studies and their lives.

FOCUS ON CONCEPTS CONCEPTUAL INSIGHTS encourage students to

HISTORICAL PERSPECTIVES are brief vignettes that

develop a conceptual understanding of calculus by

place key discoveries and conceptual advances

explaining important ideas clearly but informally.

in their historical context. They give students a glimpse into some of the accomplishments of

CONCEPTUAL INSIGHT In our work with functions and limits so far, we have encountered three expressions that are similar but have different meanings: undefined, does not exist, and indeterminate. It is important to understand the meanings of these expressions so that you can use them correctly to describe functions and limits.

great mathematicians and an appreciation for their significance.

• The word "undefined" is used for a mathematical expression that is not defined, such as 2/0 or InO. • The phrase "does not exist" means lirn (x) does not exist, that is. f (x) does not approach a particular numerical value as x approaches c. • The term "indeterminate" is used when, upon substitution, a function or limit has one of the indeterminate forms.

HISTORICAL PERSPECTIVE

1 1 1 4 S=T+—T+—T+•••=-TE—=—T For this and many other achievements. Archimedes is ranked together with Newton and Gauss atone of the greatest scientists of all time. The modern study of infinite series begun in the seventeenth centuly with Newton. Leihniz. and their mitemporaries. The diver-

GRAPHICAL INSIGHTS enhance students' visual understanding by making the crucial connections between graphical properties and the underlying concepts.

GRAPHICAL INSIGHT The formula (sin al = cos X seems reasonable when we compare the graphs in Figure I. The tangent lines to the graph of y = sin x have positive slope on the interval ( — and on this interval, the derivative y' = cos x is positive. The tangent lines have negative slope on the interval (S, In, where y' = cos x is negative. The tangent lines are horizontal at x = 1-` where cos x = 0.

Geometric series were toed as early as the third century 110E by Archimedes in a brilliant argument for determining the area Sofa "parabolic segment" (shaded region in Figure 3). Given two points A and C on a parabola. there is a point B between A and C' where the tangent line is parallel to AC (apparently, Archimedes was aware of the Mean Value Theorem mom than 2000 years before the invention of calculus). Let T be the area of triangle AABC Archimedes proved that if Ohs chosen in a similar fashion relative to AR and E is chosen relative to BC, then —T 4

area(AAD1D+ area(4BEC)

This construction of triangles can be continued. The next step would be to construct the four triangles on the segments AD. DB, BE. EC, of total area Op.

Then construct eight tri-

angles of total area (i)

T, and so ow In Mk

way, we obtain infinitely many triangles that completely fill up the parabola segment. By the 13 FIGURE 1 The graphs of y = sin A and its derivative y' = coax.

COSA

fonnula for the sum of a geometric series, we get

genre of

E I/. (called the harmonic series)

n= was known to the medieval scholar Nicole d'Orestne (1323-1382), but his proof was lust for centuries, and the result was rediscovered on more than one occasion. It was also known that the sum of the reciprocal squares

E 1/n2

converges, and in the 1640s. the

Italian Pietro Mengoli put forward the challenge of finding its sum. Despite the efforts of the best mathematicians of the day, inch ding Leibniz and the Bernoulli brothers Jakob and Johann. the problem resisted solution for nearly a century. In 1735, the great master Leonhard Euler (at the time, 28 years old) astonished his contemporaries by moving that I 12

I 22

1 32

1 42

1 52

1 62

.2 T

We examine the convergence of this series in Exercises 85 and 91 in Section 11.3.

FOCUS ON CLEAR, ACCESSIBLE EXPOSITION that anticipates and addresses student difficulties REMINDERS are margin notes

Using the trigonometric

that link the current discussion to important concepts introduced earlier in the text to give students a quick review and make

= cos' x

identities in the margin, we can also integrate cos2 x,

obtaining the following:

REMINDER Useful identities: — cos 20

j.

x 2

x dx

sin 2x 4

1(1 + cos 2x)

i

sin 2x = 2 sin x cos x

con2 x dx =

+

±c _x

- 2

I

2 sinx cos x + C

4x,c=1± si sinxcosx-1-C

cos 2x = cos' x —sin2 x

connections with related ideas.

CAUTION NOTES warn students of common pitfalls they may encounter in understanding the material.

ASSUMPTIONS MATTER uses short explanations and well-chosen counterexamples to help students

CAUTION When using L'IlOpital's Rule, be sure to take the derivative of the numerator and denominator separately,

Do not take the derivative of the function Y = 1(x)/ gin) as a quotient, for evampie using the Quotient Rule.

EXAMPLE 3 Assumptions Matter Show that the Product Law cannot be applied to lim f (x)g(x) if f (x) = x and g(x) = x-1. Solution For all x exists:

0. we have f(x)g(x)= x x-1 = I, so the limit of the product

limo f(x)g(x)= lim 1 =1 However, there is an issue with the product of the limits because linbx-1 does not exist (since g(x)= x-1 becomes infinite as x -> 0). Therefore, the Product Law cannot be applied and its conclusion does not hold even though the limit of the products does exist. Specifically, lies f(x)g(x).= I, but the product of the limits is not defined:

I (

lim f(x))(lim g(x)) =

x-50

x—.0

hm x)

x —s0

lim x-1)

x-50

Does not exist

SECTION SUMMARIES summarize a section's key points in a concise and useful way and emphasize for students what is most important in each section.

5-8

Solution Let fix) = x3 — Sand g(.0 ) = x° ± 2x — 20. Both f and g am differentiable and f g(x) is indeterminate of type 0/0 at a = 2 because f(2) = 5(2) = 0.

1411 f = tim l'('") g(x)

appreciate why hypotheses are needed in theorems.

EXAMPLE 1 Use I: Hopital '5 Rule to evaluate lim2

FOCUS ON EXERCISES AND EXAMPLES SECTION EXERCISE SETS offer a comprehensive set of exercises closely coordinated with the text. These exercises vary in difficulty from routine, to moderate, to more challenging. Also included are icons indicating problems that require the student to give a written response or require the use of technology. Each section offers PRELIMINARY QUESTIONS

Preliminary Questions 1. Assume that limf(x)

that test student understanding.

L

3. Sketch the graph of a function that approaches a limit as x does not approach a limit (either finite or infinite) use and

g(x)

4. What is the sign of a if f(x)= rix3 + o + I satisfies lim f(c) = cx.:?

Which of the following staten.nts are correct? (a) .t• = L is a vertical asymptote of g. (b)

S. What is the sign of the coefficient multiplying x' if f isa polynomial of degree 7 such that lim f L ) = to?

= L is a horizontal asymptote of g.

(c)a = L is a vertical asymptote off. (d) y = L is a horizontal asymptote off.

6. Explain why lire sin

teaches basic skills as

(r)

tim

Exercises I. What are the horizontal asymptotes of the function in Figure 6?

conceptual understanding, and motivates calculus

I 20

through interesting

CHALLENGES develop additional insights and

3. Sketch the graph of a function f with a single horizontal asymptote y = 3. 4. Sketch the graphs of functions f and g that have both y = -2 and y = 4w. horizontal asymptotes but Al;:lo f(x)0 .14, g(x).

techniques, reinforces

FURTHER INSIGHTS &

dues not exist. What is

.r 4

well as problem-solving

applications.

exists but lim sin

lint

2. What are the following limits? (a) lim (b) limo'

A main set of EXERCISES

oc but

40

(s0

80

FIGURE 6 2. Sketch the graph of a function f that has both y= -1 and y = S as horizontal asymptotes.

S. 1: :1 Investigate the asymptotic behavior of fix) = x + numerically anti graphically: (n) Make a table of valuts of fix) for x m +50. ±100. ±500. +1000. (h) Plot the graph off. (c) What are the horizontal asymptotes off? 6. C3 Investigate lim

I numerically and graphically:

(a) Make a table of values of f(x) x= ± 100. ±500, ±1000. *10.000.

I is 4- I s.r 43

1- 9

for the following:

Further Insights and Challenges 46. Every limit as x -s• oo can be expressed alternatively as a one-sided limit as 1 -+ 0+, where I = x-1. Settittg gliI = f we have ).!.tn. f

Show that

.

3x2 -

-

,

3 2 + 50 . and evaluate using the Quotient

Law.

l = ,11 1101+ g(

challenge students to further develop their skills.

47. Rtiverite the following as one-sided limits as in Exercise 46 awl ovalmite. 3- 12x3 (a) lim (bI .,U2)1 3V° x-)40 4x3 + 30 + I 1 (c) lim x sin -

48, Let GI b) = lint ( I + for b E 0. Investigate G(b) mimedz-sto) catty and graphically for h = 0.2. 0.8. 1 3, 5 land additional values if necessary). Then make a conjecture for the value of G(b) as a function of b. Draw a graph of y = G(b). Does G appear to he continuous? We will evaluate G(b) using L'HOpitalls Rule in Section 7.5 (see Exercise 69 there).

/-

RICH APPLICATIONS such as

EXAMPLE 3 A Glacial Thickness Model Let p = 917 kg/m3, g = 9.8 m/s2, and r = 75,000 N/m2 in Eq. (2). Use T(0) = 0 for an initial condition, and solve for r(x). 'Then use T(x) to determine the thickness of the glacier I km from its terminus.

this exercise on smart phone

Solution The differential equation that we need to solve is T

growth (below) and this

dT = 75.000 dx (917)(9.8)

example discussing glacier

It is a separable differential equation. We use the approximate value of 8.35 for the righthand side, and proceed as follows:

1 - its)

T dT =

thickness (left) reinforce

8.35 dx

the relevance of calculus

TerroinuN

I - 7'-= 8.355 + C 2 T(x.)= .116.7x +C

FIGURE 3 The glacier's thickness T is modeled as a function of distances from the terminus.

to students' lives and demonstrate the importance

Since T(0) = 0, we obtain T(x) = s/r6-.Tx (Figure 4). At a distance of 1 km from the terminus, the thickness is T(1000) = 129m.

ISO

of calculus in scientific •

research.

120 80 40 1

1 400

FIGURE 4

1

1 800

1

8. In 2009. 2012, and 2015, the number (in millions) of smart phones sold in the world was 172.4. 680,1, and 1423.9, respectively.

1200

T(s) =

(a) Let t represent time in years since 2009, and let S represent the number of smart phones sold in millions. Determine M, A. and k for a logistic model, 5(1) = , that fits the given data points. (b) What is the long-term expected maximum number of smart phones sold annually? That is, what is fliigo SW? (c) In what year does the model predict that smart-phone sales will reach 98% of the expected maximum?

CHAPTER REVIEW EXERCISES 1. The position of a particle at time f(s) is s(t) = rt : 1- I 111. Compute its average velocity over 12.51 and estimate its instantaneous velocity at t = 2. 2. A rock dropped from a state of rest at time I =0 on the planet Ginormon travels a distance Mt) = 15.2,2 to in t seconds. Estimate the instantaneous velocity at = 5. 3. For f (x)= ../rx compute the slopes of the secant lines from 16 to each of 16 ± 0.01, 16 ± 0.001.16 ± 0.01:01 and use those values to estimate the slope of the tangent line at x = 16. 4. Show that the slope of the secant line for f(x) = x3 - 2x over [5,x] is equal to x2 + 5x + 23. Use this to estimate the slope of the tangent line at x = 5. In Exercises 5-10, estimate the limit numerically to two decimal places or state that the limit does not exist. 5. lim

I - cos3(x)

5 _ x2 it. lim (3 +x 1 /2 )

12. lint

4 13. tim x3

14.

15. lim r-.9 t - 9

16. lim x-.3

X3

17. lim

1-59

21.

t-6 ,/t - 3

lim

6. lim

x- 2 xx - 4 8. lim 7. lim - 4 x-.2 x-.2 x2 - 4 35_9 7 3 10. lim 9. lim x-.2 5x - 25 ( I -x 7 -x 3 In Exercises 11-50, evaluate the limit if it exists. If not, determine whether the one-sided limits exist. For limits that don't exist indicate whether they can be expressed as = -00 or = oa

x+1

x3 -Zr 23. lim x-4.1 x 25. lim x-.0 27.

433 - 4x - 1

I lim Lx-.1.5 x

CHAPTER REVIEW EXERCISES offer a comprehensive set of exercises closely coordinated with the chapter material to provide additional problems for self-study or assignments.

18. lim

32 -I- 4x + I x+ + - 2 x- 3 2(a + /02 - 2a2

-

A-0

19. tim

-x

lim

4x +7

20. lim s-44)

I - yT

i

22. lim

3y2 +5y - 2 6y2 - 5y + I

24. lim a-.1,

a2 - 3ab + 2b2 a- b

26. lim 9-.0

sin 50 0

28. lim sect)

FOCUS ON MEDIA AND RESOURCES WebAssign putnatirt https://www.webassign.net/whfreeman WebAssign Premium offers course and assignment customization to extend and enhance the classroom experience for instructors and students. The fully customizable WebAssign Premium for Calculus integrates an interactive e-book, a powerful answer evaluator, algorithmically generated homework and quizzes, and Macmillan's acclaimed CalcTools: DYNAMIC FIGURES—Interactive versions of 165 text figures. Tutorial videos explain how to use select figures. CALCCLIPS—Step-by-step whiteboard tutorials explain key concepts from exercises in the book. E-BOOK Easy to navigate, with highlighting and note-taking features. LEARNINGCURVE—A powerful, self-paced assessment tool provides instant feedback tied to specific sections of the e-book. Difficulty level and topic selection adapt based on each student's performance. TUTORIAL QUESTIONS—This feature reviews difficult questions one segment at a time. A PERSONAL STUDY PLAN (PSP)—Lets each student use chapter and section assessments to gauge their mastery of the material and generate an individualized study plan.

FOR INSTRUCTORS available at https://www.webassign.net/whfreeman OVER 7,000 EXERCISES from the text, with detailed solutions available to students at your discretion. READY-TO-USE COURSE PACK ASSIGNMENTS drawn from the exercise bank to save you time. A SUITE OF INSTRUCTOR RESOURCES, including iClicker questions, Instructor's Manuals, PowerPoint lecture slides, a printable test bank, and more.

ADDITIONAL SUPPLEMENTS FOR INSTRUCTORS INSTRUCTOR'S SOLUTIONS MANUAL Worked-out solutions to all exercises in the text. ISBN: (SV) 978-1-319-25216-8; (MV) 978-1-319-25217-5 TEST BANK ISBN: 978-1-319-22128-7 INSTRUCTOR'S RESOURCE MANUAL ISBN: 978-1-319-22126-3 LECTURE SLIDES (customizable) IMAGE SLIDES (all text figures and tables)

iClicker iClicker's two-way radio-frequency classroom response solution was developed by educators for educators. To learn more about packaging iClicker with this textbook, please contact your local sales representative or visit www.iclicker.com. FOR STUDENTS STUDENT SOLUTIONS MANUAL Worked-out solutions to all odd-numbered exercises. ISBN: (SV) 978-1-319-25443-8; (MV) 978-1-319-25441-4 MAPLETM MANUAL ISBN: 978-1-319-23878-0 MATHEMATICA® MANUAL ISBN: 978-1-319-23879-7 7A11- WeBVVorK WEBWORK.MAA.ORG Macmillan Learning offers thousands of algorithmically generated questions (with full solutions) from this book through WeBWorK's free, open-source online homework system.

ACKNOWLEDGMENTS W

e are grateful to the many instructors from across the United States and Canada who have offered comments that assisted in the development and refinement of this book. These contributions included class testing, manuscript reviewing, exercise reviewing, and participating in surveys about the book and general course needs.

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Margolius, Cleveland State University; Tzu-Yi Alan Yang, Columbus State Community College; Greg S. Goodhart, Columbus State Community College; Kelly C. Stady, Cuyahoga Community College; Brian T. Van Pelt, Cuyahoga Community College; David Robert Ericson, Miami University; Frederick S. Gass, Miami University; Thomas Stacklin, Ohio Dominican University; Vitaly Bergelson, The Ohio State University; Robert Knight, Ohio University; John R. Pather, Ohio University, Eastern Campus; Teresa Contenza, Otterbein College; Ali Hajjafar, The University of Akron; Jianping Zhu, The University of Akron; Ian Clough, University of Cincinnati Clermont College; Atif Abueida, University of Dayton; Judith McCrory, The University at Findlay; Thomas Smotzer, Youngstown State University; Angela Spalsbury, Youngstown State University; James Osterburg, The University of Cincinnati; Mihaela A. Poplicher, University of Cincinnati; Frederick Thulin, University of Illinois at Chicago; Weimin Han, The Ohio State University; Crichton Ogle, The Ohio State University; Jackie Miller, The Ohio State University; Walter Mackey, Owens Community College; Jonathan Baker, Columbus State Community College; Vincent Graziano, Case Western Reserve University; Sailai Sally, Cleveland State University OKLAHOMA Christopher Francisco, Oklahoma State University; Michael McClendon, University of Central Oklahoma; Teri Jo Murphy, The University of Oklahoma; Kimberly Adams, University of Tulsa; Shirley Pomeranz, University of Tulsa OREGON Loma TenEyck, Chemeketa Community College; Angela Martinek, Linn-Benton Community College; Filix Maisch, Oregon State University; Tevian Dray, Oregon State University; Mark Ferguson, Chemekcaa Community College; Andrew Flight, Portland State University; Austina Fong, Portland State University; Jeanette R. Palmiter, Portland State University; Jean Nganou, University of Oregon; Juan Restrepo, Oregon State University PENNSYLVANIA John B. Polhill, Bloomsburg University of Pennsylvania; Russell C. Walker, Carnegie Mellon University; Jon A. Beal, Clarion University of Pennsylvania; Kathleen Kane, Community College of Allegheny County; David A. Santos, Community College of Philadelphia; David S. Richeson, Dickinson College; Christine Marie Cedzo, Gannon University; Monica Pierri-Galvao, Gannon University; John H. Ellison, Grove City College; Gary L. Thompson, Grove City College; Dale McIntyre, Grove City College; Dennis Benchaff, Harrisburg Area Community College; William A. Drumin, King's College; Denise Reboli, King's College; Chawne Kimber, Lafayette College; Elizabeth McMahon, Lafayette College; Lorenzo Traldi, Lafayette College; David L. Johnson, Lehigh University; Matthew Hyatt, Lehigh University; Zia Uddin, Lock Haven University of Pennsylvania; Donna A. Dietz, Mansfield University of Pennsylvania; Samuel

xv

Wilcock, Messiah College; Richard R. Kern, Montgomery County Community College; Michael Fraboni, Moravian College; Neena T. Chopra, The Pennsylvania State University; Boris A. Datskovsky, Temple University; Dennis M. DeTurck, University of Pennsylvania; Jacob Burbea, University of Pittsburgh; Mohammed Yandi, Ursinus College; Timothy Feeman, Villanova University; Douglas Norton, Villanova University; Robert Styer, Villanova University; Michael J. Fisher, West Chester University of Pennsylvania; Peter Brooksbank, Bucknell University; Larry Friesen, Butler County Community College; LisaAngelo, Bucks County College; Elaine Fitt, Bucks County College; Pauline Chow, Harrisburg Area Community College; Diane Benner, Harrisburg Area Community College; Erica Chauvet, Waynesburg University; Mark McKibben, West Chester University; Constance Ziemian, Bucknell University; Jeffrey Wheeler, University of Pittsburgh; Jason Aran, Drexel University; Nakia Rimmer, University of Pennsylvania; Nathan Ryan, Bucknell University; Bharath Narayanan, Pennsylvania State University RHODE ISLAND Thomas F. Banchoff, Brown University; Yajni Warnapala-Yehiya, Roger Williams University; Carol Gibbons, Salve Regina University; Joe Allen, Community College of Rhode Island; Michael Latina, Community College of Rhode Island SOUTH CAROLINA Stanley 0. Perrine, Charleston Southern University; Joan Hoffacker, Clemson University; Constance C. Edwards, Coastal Carolina University; Thomas L. Fitzkee, Francis Marion University; Richard West, Francis Marion University; John Harris, Furman University; Douglas B. Meade, University of South Carolina; GeorgeAndroulakis, University of South Carolina; Art Mark, University of South Carolina Aiken; Sherry Biggers, Clemson University; Mary Zachary Krohn, Clemson University; Andrew Incognito, Coastal Carolina University; Deanna Caveny, College of Charleston SOUTH DAKOTA Dan Kemp, South Dakota State University TENNESSEE Andrew Miller, Belmont University; Arthur A. Yanushka, Christian Brothers University; Laurie Plunk Dishman, Cumberland University; Maria Siopsis, Maryville College; Beth Long, Pellissippi State Technical Community College; Judith Fethe, Pellissippi State Technical Community College; Andrzej Gutek, Tennessee Technological University; Sabine Le Borne, Tennessee Technological University; Richard Le Borne, Tennessee Technological University; Maria E Bothelho, University of Memphis; Roberto Triggiani, University of Memphis; Jim Conant, The University of Tennessee; Pavlos Tzennias, The University of Tennessee; Luis Renato Abib Finotti, University of Tennessee, Knoxville; Jennifer Fowler, University of Tennessee, Knoxville; Jo Ann W. Staples, Vanderbilt University; Dave Vinson, Pellissippi State Community College; Jonathan Lamb, Pellissippi State Community College; Stella Thistlewaite, University of Tennessee, Knoxville TEXAS Sally Haas, Angelina College; Karl Havlak, Angelo State University; Michael Huff, Austin Community College; John M. Davis, Baylor University; Scott Wilde, Baylor University and The University of Texas at Arlington; Rob Eby, Blinn College; Tim Sever, Houston Community College—Central; Ernest Lowery, Houston Community College—Northwest; Brian Loft, Sam Houston State University; Jianzhong Wang, Sam Houston State University; Shirley Davis, South Plains College; Todd M. Steckler, South Texas College; Mary E. Wagner-ICrankel, St. Mary's University; Elise Z. Price, Tarrant County College, Southeast Campus; David Price, Tarrant County College, Southeast Campus; Runchang Lin, Texas A&M University; Michael Stecher, Texas A&M University; Philip B. Yasskin, Texas A&M University; Brock Williams, Texas Tech University; I. Wayne Lewis, Texas Tech University; Robert E. Byerly, Texas Tech University; Ellina Grigorieva, Texas Woman's University; Abraham Haje, Tomball College; Scott Chapman, Trinity University; Elias Y. Deeba, University of Houston Downtown; Jianping Zhu, The University of Texas at Arlington; Tuncay Aktosun, The University of Texas at Arlington; John E. Gilbert, The University of Texas at Austin; Jorge R. ViramontesOlivias, The University of Texas at El Paso; Fengxin Chen, University of Texas at San Antonio; Melanie Ledwig, The Victoria College; Gary L. Walls, West Texas A&M University; William Heierman, Wharton County Junior College; Lisa Rezac, University of St. Thomas; Raymond J. Cannon, Baylor University; Kathryn Flores, McMurry University; Jacqueline A. Jensen, Sam Houston State University; James Galloway, Collin County College; Raja Khoury, Collin County College; Annette Benbow, Tarrant County College—Northwest; Greta Harland, Tarrant County College—Northeast; Doug Smith, Tarrant County College—Northeast; Marcus McGuff, Austin Community College; Clarence McGuff, Austin Community College; Steve Roth, Austin Community College; Vicki Payne, Austin Community College; Anne Pradera, Austin Community College; Christy Babu, Laredo Community College; Deborah Hewitt, McLennan Community College; W. Duncan, McLennan Community College; Hugh Griffith, Mt. San Antonio College; Qin Sheng, Baylor University, My Linh Nguyen, University of Texas at Dallas; Lorenzo Sadun, University of Texas at Austin UTAH Ruth Trygstad, Salt Lake City Community College VIRGINIA Veme E. Leiningei; Bridgewater College; Brian Bradie, Christopher Newport University; Hongwei Chen, Christopher Newport University; John J. Avioli, Christopher Newport University; James H. Martin, Christopher Newport University; David Walnut, George Mason University; Mike Shirazi, Germanna Community College; Julie Clark, Hollins University; Ramon A. MateToledo, James Madison University; Adrian Riskin, Mary Baldwin College; Josephine Letts, Ocean Lakes High School; Przemyslaw Bogacki, Old Dominion University; Deborah Denvir, Randolph-Macon Woman's College; Linda Powers, Virginia Tech; Gregory Dresden, Washington and Lee University; Jacob A. Siehler, Washington and

xvi

ACKNOWLEDGMENTS

Lee University; Yuan-Jen Chiang, University of Mary Washington; Nicholas Hamblet, University of Virginia; Bernard Fulgham, University of Virginia; Manouchehr "Mike" Mohajeri, University of Virginia; Lester Frank Caudill, University of Richmond VERMONT David Dorman, Middlebury College; Rachel Repstad, Vermont Technical College WASHINGTON Jennifer Laveglia, Bellevue Community College; David Whittaker, Cascadia Community College; Sharon Saxton, Cascadia Community College; Aaron Montgomery, Central Washington University; Patrick Averbeck, Edmonds Community College; Tana Knudson, Heritage University; Kelly Brooks, Pierce College; Shana P. Calaway, Shoreline Community College; Abel Gage, Skagit Valley College; Scott MacDonald, Tacoma Community College; Jason Preszler, University of Puget Sound; Martha A. Gady, Whitworth College; Wayne L. Neidhardt, Edmonds Community College; Simrat Ghuman, Bellevue College; Jeff Eldridge, Edmonds Community College; Kris Kissel, Green River Community College; Laura Moore-Mueller, Green River Community College; David Stacy, Bellevue College; Eric Schultz, Walla Walla Community College; Julianne Sachs, Walla Walla Community College WEST VIRGINIA David Cusick, Marshall University; Ralph Oberste-Vorth, Marshall University; Suda Kunyosying, Shepard University; Nicholas Martin, Shepherd University; Rajeev Rajaram, Shepherd University; Xiaohong Zhang, West Virginia State University; Sam B. Nadler, West Virginia University WYOMING Claudia Stewart, Casper College; Pete Wildman, Casper College; Charles Newberg, Western Wyoming Community College; Lynne Ipina, University of Wyoming; John Spitler, University of Wyoming WISCONSIN Erik

R. Tou, Carthage College; Paul Bankston, Marquette University; Jane Nichols, Milwaukee School of Engineering; Yvonne Yaz, Milwaukee School of EngineeringiTh Simei Tong, University of Wisconsin—Eau Claire; Terry Nyman, University of Wisi, ) consin—Fox Valley; Robert L. Wilson, University of Wisconsin—Madison; Dietrich A. Uhlenbrock, University of Wisconsin—Madison; Paul Milewski, University of Wisconsin—Madison; Donald Solomon, University of Wisconsin—Milwaukee; Kandasamy Muthuvel, University of Wisconsin—Oshkosh; Sheryl Wills, University of Wisconsin—Platteville; Kathy A. Tomlinson, University of Wisconsin—River Falls; Cynthia L. McCabe, University of Wisconsin—Stevens Point; Matthew Welz, University of Wisconsin—Stevens Point; Joy Becker, University of Wisconsin-Stout; Jeganathan Sriskandarajah, Madison Area Tech College; Wayne Sigelko, Madison Area Tech College; James Walker, University of Wisconsin—Eau Claire CANADA Don St. Jean, George Brown College; Robert Dawson, St. Mary's University; Len Bos, University of Calgary; Tony Ware, University of Calgary; Peter David Papez, University of Calgary; John O'Conner, Grant MacEwan University; Michael P. Lamoureux, University of Calgary; Yousry Elsabrouty, University of Calgary; Darja Kalajdzievska, University of Manitoba; Andrew Skelton, University of Guelph; Douglas Farenick, University of Regina; Daniela Silvesan, Memorial University of Newfoundland; Beth Ann Austin, Memorial University; Brenda Davison, Simon Fraser University; Robert Steacy, University of Victoria; Dan Kucerovslcy, University of New Brunswick; Bernardo GalvaoSousa, University of Toronto; Hadi Zibaeenejad, University of Waterloo

The creation of this fourth edition could not have happened without the help of many people. First, we want to thank the primary individuals with whom we have worked over the course of the project. Katrina Mangold, Michele Mangelli, and Nikki Miller Dworsky have been our main contacts managing the flow of the work, doing all that they could to keep everything coming together within a reasonable schedule, and efficiently arranging the various contributions of review input that helped keep the project well informed. Their work was excellent, and that excellence in project management helped greatly in bringing this new edition of the book together. Tony Palermino has provided expert editorial help throughout the process. Tony's experience with the book since its beginning helped to keep the writing and focus consistent with the original structure and vision of the book. His eye for detail and knowledge of the subject matter helped to focus the writing to deliver its message as clearly and effectively as possible. Kerry 0' Shaughnessy kept the production process moving forward in a timely manner. Thanks to Ron Weickart at Network Graphics for his skilled and creative execution of the art program. Sarah Wales-McGrath (copyeditor) and Christine Sabooni (proofreader) both provided expert feedback. Our thanks are also due to Macmillan Learning's superb production team: Janice Donnola, Sheena Goldstein, Alexis Gargin, and Paul Rohloff. Many faculty gave critical feedback on the third edition and drafts of the fourth, and their names appear above. We are very grateful to them. We want to particularly thank all of the advisory board members who gave very valuable input on very specific questions about the approach to and the presentation of many important topics: John Davis (Baylor University), Judy Fethe (Pellissippi State Community College), Chris Francisco (Oklahoma State University), and Befit Givens (California State Polytechnic University, Pomona). The accuracy reviewers at Math Made Visible helped to bring the final version into the form in which it now appears. We also want to thank our colleagues in the departments where we work. We are fortunate to work in departments that are energized by mathematics, where many interesting projects take place, and clever pedagogical ideas are employed and debated. We would also like to thank our students who, over many years, have provided the energy, interest, and enthusiasm that help make teaching rewarding. Colin would like to thank his two children, Alexa and Colton. Bob would like to thank his family. They are the ones who keep us well grounded in the real world, especially when mathematics tries to steer us otherwise. This book is dedicated to them. Bob and Colin

INTRODUCTION TO CALCULUS

GRANGER / GRANGER—All rights reserved

W

Maria Gaetana Agnesi (1718-1799), an Italian mathematician and theologian, is credited with writing one of the first books about calculus, lnstituzioni analitiche ad uso della gioventa italiana. It was self-published and was written as a textbook for her brothers, who she was tutoring.

FIGURE 1

FIGURE 2 The approximation to the slope at P improves as Q approaches P.

e begin with a brief introduction to some key ideas in calculus. It is not an exaggeration to say that calculus is one of the great intellectual achievements of humankind Sending spacecraft to other planets, building computer systems for forecasting the weather, explaining the interactions between plants, insects, and animals, and understanding the structure of atoms are some of the countless scientific and technological advances that could not have been achieved without calculus. Moreover, calculus is a foundational part of the mathematical theory of analysis, a field that is under continuous development. The primary formulation of calculus dates back to independent theories of Sir Isaac Newton and Gottfried Wilhelm Liebnitz in the 1600s. However, their work only remotely resembles the topics presented in this book. Through a few centuries of development and expansion, calculus has grown into the theory we present here. Newton and Liebnitz would likely be quite impressed that their calculus has evolved into a theory that many thousands of students around the world study each year. There are two central concepts in calculus: the derivative and the integral. We introduce them next. The Derivative The derivative of a function is simply the slope of its graph; it represents the rate of change of the function. For a linear function y = 2.3x — 8.1, the slope 2.3 indicates that y changes by 2.3 for each one-unit change in x. How do we find the slope of a graph of a function that is not linear, such as the one in Figure 1? Imagine that this function represents the amount A of a drug in the bloodstream as a function of time t. Clearly, this situation is more complex than the linear case. The slope varies as we move along the curve. Initially positive because the amount of the drug in the bloodstream is increasing, the slope becomes negative as the drug is absorbed. Having an expression for the slope would enable us to know the time when the amount of the drug is a maximum (when the slope turns from positive to negative) or the time when the drug is leaving the bloodstream the fastest (a time to administer another dose). To define the slope for a function that is not linear, we adapt the notion of slope for linear relationships. Specifically, to estimate the slope at point P in Figure 2, we select a point Q on the curve and draw a line between P and Q. We can use the slope of this line to approximate the slope at P. To improve this approximation, we move Q closer to P and calculate the slope of the new line. As Q moves closer to P, this approximation gets more precise. Although we cannot allow P and Q to be the same point (because we could no longer compute a slope), we instead "take the limit" of these slopes. We develop the concept of the limit in Chapter 2. Then in Chapter 3, we show that the limiting value may be defined as the exact slope at P. The Definite Integral The definite integral, another key calculus topic, can be thought of as adding up infinitely many infinitesimally small pieces of a whole. It too is obtained through a limiting process. More precisely, it is a limit of sums over a domain that is divided into progressively more and more pieces. To explore this idea, consider a solid

X

X

xvii

INTRODUCTION TO CALCULUS

ball of volume 2 cm3 whose density (mass per unit volume) throughout is 1.5 g/cm3. The mass of this ball is the product of density and volume, (1.5)(2) = 3 grams. If the density is not the same throughout the ball (Figure 3), we can approximate its mass as follows: • Chop the ball into a number of pieces, • Assume the density is uniform on each piece and approximate the mass of each piece by multiplying density by volume, • Add the approximate masses of the pieces to estimate the total mass of the ball. FIGURE 3 For a ball of uniform density, mass is the product of density and volume. For a nonuniform ball, a limiting process needs to be used to determine the mass.

The irregular density of the moon presented a navigational challenge for spacecraft orbiting it. The first group of spacecraft (unmanned!) that circled the moon exhibited unexpected orbits. Space scientists realized that the density of the moon varied considerably and that the gravitational attraction of concentrations of mass (referred to as mascons) deflected the path of the spacecraft from the planned trajectory

We continually improve this approximation by chopping the ball into ever smaller pieces (Figure 4). Ultimately, an exact value is obtained by taking a limit of the approximate masses.

FIGURE 4

In Chapter 5, we define the definite integral in exactly this way; it is a limit of sums over an interval that is divided into progressively smaller subintervals. The Fundamental Theorem of Calculus Although the derivative and the definite integral are very different concepts, it turns out they are related through an important theorem called the Fundamental Theorem of Calculus presented in Chapter 5. This theorem demonstrates that the derivative and the definite integral are, to some extent, inverses of each other, a relationship that we will find beneficial in many ways.

1 PRECALCULUS REVIEW C

alculus builds on the foundation of algebra, analytic geometry, and trigonometry. In this chapter, therefore, we review some concepts, facts, and formulas from precalculus that are used throughout the text. In the last section, we discuss ways in which technology can be used to enhance your visual understanding of functions and their properties.

1.1 Real Numbers, Functions, and Graphs

Oscillatory phenomena, such as the extreme tides of the Bay of Fundy in Atlantic Canada, are modeled by the sine function from trigonometry.

We begin with a short discussion of real numbers. This gives us the opportunity to recall some basic properties and standard notation. A real number is a number represented by a decimal or "decimal expansion." There are three types of decimal expansions: finite, infinite repeating, and infinite but nonrepeating. For example, 3 — = 0.375, 8

1 = 0.142857142857 . . . = 0.142857 7

r

=

3.141592653589793 . . .

The number is represented by a finite decimal, whereas is represented by an infinite repeating decimal. The bar over 142857 indicates that this sequence repeats indefinitely. The decimal expansion of 7 is infinite but nonrepeating. The set of all real numbers is denoted by a boldface R. When there is no risk of confusion, we refer to a real number simply as a number. We also use the standard symbol E for the phrase "belongs to." Thus, a

E

R

"a belongs to R"

reads

The set of integers is commonly denoted by the letter Z (this choice comes from the German word Zahl, meaning "number"). Thus, Z = 1. . ,-2, —1, 0, 1, 2, . . . 1. A whole number is a nonnegative integer—that is, one of the numbers 0, 1, 2, . . . . A real number is called rational if it can be represented by a fraction plq, where p and q are integers with q 0 0. The set of rational numbers is denoted Q (for "quotient"). Numbers that are not rational, such as 7r and V-2-, are called irrational. We can tell whether a number is rational from its decimal expansion: Rational numbers have finite or infinite repeating decimal expansions, and irrational numbers have infinite, nonrepeating decimal expansions. Furthermore, the decimal expansion of a number is unique, apart from the following exception: Every finite decimal is equal to an infinite decimal in which the digit 9 repeats. For example, 1/5 = 0.5 = 0.499999 . . . Two algebraic properties of the real numbers are the commutative property of addition, a + b = b + a, and the distributive property of multiplication over addition, a(b + c)= ab + ac. A list of further properties can be found in Appendix B. Next, we present some properties of exponents that are used regularly when we work with exponential expressions and functions. Example

Rule =1

Exponent zero Products

Quotients

50 = 1

bx bY = bx+Y bY

= bx— Y 1 = — bx

Negative exponents

b

Power to a power

(bx

Roots

b11" =

= bxY

25 23 = 25+3 = 28 47 = 47-2 — 45 42 3-4 1 34 (32)4 = 32(4) = 38 51/2

=

N/3

CHAPTER

PRECALCULUS REVIEW

1

EXAMPLE 1 Rewrite as a whole number or fraction: (d)

(c) 416 • 4-18

(b) 272/3

(a) 16-1/2

93

Solution 1 (a) 16-1/2 = 161/2 (c) 416 • 4-18 = 4-2 =

1

1

V1 -6

4

(b) 272/3 = (271/3)2 = 32 = 9 93

=

(d)

(32)3

= y- =

36 =

3 1=

•

Another important algebraic relationship is the binomial expansion of (a + b)'1 . It is proved in Appendix C and is needed in the proof of the power law for derivatives in Section 3.2. Expanding (a + b)' 1 for n = 2, 3, 4, we obtain • (a + b)2 = (a + b)(a + b) = a2 + 2ab + b2 • (a + b)3 = (a + b)(a + b)2 = (a + b)(a2 +2ab + b2) = a3 + 3a2b + 3ab2 + b3 • (a + b)4 = (a + b)(a + b)3 = (a + b)(a3 +3a2b +3ab2 + b3) = a4 + 4a3b + 6a2b2 + 4ab3 + a4 Notice there are some patterns emerging here. In each case, the first and second terms are an and nan—l b, while the last two terms are nab' ' and bn. There is a general formula for the expansion, called the binomial expansion formula. It is expressed using summation notation as Imd REMINDER n-factorial is the number n! = n(n — 1)(n — 2) • • • (2)(1)

(a + br =

! E (n — np)! an-PbP p!

p=0

Thus,

1! = 1, 2! = (2)(1) = 2 3! = (3)(2)(1) = 6 By convention, we set 0! = 1.

We introduce summation notation in Section 5.1. For now, you can understand the formula as saying that (a + b)' 1 is a sum of terms (n _npl), p,an—PbP, with a term for each p going from 0 to n. So, for example, in (a + b)8, the first four terms are: a8 = a8, _8! -;a 5b3 = 56a5 b3. Working out the rest of the 7!1! b — 8a7 b ' 6!2! a6b2 = 28a6b2, and s terms, we find that: (a + b)8 = a8 + 8a7 b + 28a6b2 + 56a5b3 + 70a4b4 + 56a3b5 + 28a2b6 + 8ab7 + a8

—2

—1

0

1

2

FIGURE 1 The set of real numbers represented as a line.

In some texts, "larger than" is used synonymously with "greater than." We will avoid that usage in this text.

lal a

We visualize real numbers as points on a line (Figure 1), and we refer to that line as the real line. For this reason, real numbers are often called points. The point corresponding to 0 is called the origin. The real numbers are ordered, and we can view that ordering in terms of position on the real line: p is greater than q, written p > q, if p is to the right of q on the number line. p is less than q, written p 0, negative if x 0. Two other important terms we use, related to position on the real line, are "large" and "small." We say that p is large if p is distant from the origin, and p is small if p is close to the origin. While these definitions are somewhat vague, the meaning should be clear in the contexts in which they are used. The absolute value of a real number a, denoted la, is defined by (Figure 2):

0

FIGURE 2 I I is the distance from a to the origin.

laI = distance from the origin =

la

—a

if a > 0 if a < 0

SECTION 1.1

Real Numbers, Functions, and Graphs

3

For example, 11.21 = 1.2 and 1-8.351 = —( —8.35) = 8.35. The absolute value satisfies

—2

1

•

1

—1 a 0

lab! = lal Ibl

lal =

lb —al • • •

1

1

•-1 $

1

2

b

FIGURE 3 The distance between a and

b is

lb — al.

The distance between two real numbers a and b is lb — al, which is the length of the line segment joining a and b (Figure 3). Two real numbers a and b are close to each other if lb — al is small, and this is the case if their decimal expansions agree to many places. More precisely, if the decimal expansions of a and b agree to k places (to the right of the decimal point), then the distance lb — al is at most 10—k . Thus, the distance between a = 3.1415 and b = 3.1478 is at most 10-2 because a and b agree to two places. In fact, the distance is exactly 13.1478 — 3.14151 0.0063. Beware that la + bl is not equal to la 1 + 1bl unless a and b have the same sign or at least one of a and b is zero. If they have opposite signs, cancellation occurs in the sum a + b, and la + bl < 'al + Ibl. For example, 12 + 51 = 121+ 151 but 1-2 -I- 51 = 3, which is less than 1-21 +151 = 7. In any case, la + bl is never greater than lal ± bl, and this gives us the simple but important triangle inequality: la + bl

lal

1

Ibl

We use standard notation for intervals. Given real numbers a 0 • Absolute value: al = —a if a < 0 • Triangle inequality: la + bl lal + Ibl • Four intervals with endpoints a and b: (a, b),

[a, b],

[a, b),

(a, b]

• Writing open and closed intervals using absolute-value inequalities: (a, b) = Ix : lx — Cl < r}, [a, b] = Ix : lx —cl < r where c = (a + b) is the midpoint and r = 1(b — a) is the radius.

SECTION 1.1

Real Numbers, Functions, and Graphs

11

• Distanced between (xi, y) and (x2, y2):

=1(x2 — x) 2 +0,2 — yo2 • Equation of circle of radius r with center (a, b): (x — a)2

(y — b)2 = r2

• A zero or root of a function f is a number c such that f (c) = 0. • Vertical Line Test: A curve in the plane is the graph of a function of x if and only if each vertical line x = a intersects the curve in at most one point. Increasing: Nondecreasing: Decreasing: Nonincreasing:

f (xi) < f (x2) ifx1 f (xi) f (x2) if f (xi) > f (x2) if xi f (xi) f (x2) if xi

0 and satisfies 2—x p(a Ib) = p(b) — p(a). Prove that f (x) = p (-2 + x is an odd function. 70. State whether the function is increasing, decreasing, or neither

1 1 4 —3 —2 —1

I I I .X I I I I I I .X 1 2 3 —3 —2 —11— 1 2 3 — —2— —3 —

_

•

(a) Surface area of a sphere as a function of its radius (b) Temperature at a point on the equator as a function of time (c) Price of an airline ticket as a function of the price of oil (d) Pressure of the gas in a piston as a function of volume In Exercises 71-76, let f be the function shown in Figure 28. 71. Find the domain and range off. 72. Sketch the graphs of y = f (x + 2) and y = f (x) + 2.

13

(iv)

(v)

FIGURE 29 81. Sketch the graph of y = f (2x) and y = f (x), where f (x) = Ix I + 1 (Figure 29). 82. Find the function f whose graph is obtained by shifting the parabola y = x2 by 3 units to the right and 4 units down, as in Figure 30.

73. Sketch the graphs of y = f (2x), y = f (1x), and y = 2f (x). 74. Sketch the graphs of y = f (—x) and y = —f(—x). 75. Extend the graph of f to [-4, 4] so that it is an even function. 76. Extend the graph of f to [-4,4] so that it is an odd function.

FIGURE 30

1

2

3

4

FIGURE 28

83. Define f (x) to be the larger of x and 2 — x. Sketch the graph of f. What are its domain and range? Express f (x) in terms of the absolute value function. 84. For each curve in Figure 31, state whether it is symmetric with respect to the y-axis, the origin, both, or neither.

77. Suppose that f has domain [4,8] and range [2,6]. Find the domain and range of: (b) y = f (x + 3) (a) y = f (x) + 3 (d) y = 3 f (x) (c) y = f (3x) 78. Let f (x) = x2. Sketch the graph over [-2,21 of: (b) y = f (x) + 1 (a) y = f (x + 1) (d) y = 5f (x) (c) y = f (5x) 79. Suppose that the graph of f (x) = x4 — x2 is compressed horizontally by a factor of 2 and then shifted 5 units to the right. (a) What is the equation for the new graph? (b) What is the equation if you first shift by 5 and then compress by 2? (c) (GU) Verify your answers by plotting your equations. 80. Figure 29 shows the graph of f (x) = lx1+ 1. Match the functions (a)—(e) with their graphs (i)—(v). (c) y = —f(x) + 2 (b) y = — f (x) (a) Y = f(x — 1) + 1) = f (x (e) y 2 — 1) — (d) y = f(x

(B)

(A)

I

X

(D)

(C) FIGURE 31

85. Show that the sum of two even functions is even and the sum of two odd functions is odd.

14

PRECALCULUS REVIEW

CHAPTER 1

86. Suppose that f and g are both odd. Which of the following functions are even? Which are odd? (b) Y =fix)3 (a) y = f (x)g(x) (d) y =

(c) y = f (x) — g(x)

f (x) g(x)

Further Insights and Challenges 88. Prove the triangle inequality (la + bl inequalities: — Ial

a

— Ibl

la I + Ibl) by adding the two

2 . Note that Use this to find the decimal expansion of r = — 11 18

2

Ibi

b

=

= 11

89. Show that a fraction r = a / b in lowest terms has a finite decimal expansion if and only if b = 2n 5m

87. Prove that the only function whose graph is symmetric with respect to both the y-axis and the origin is the function f (x) = 0.

for some n,m > 0

Hint: Observe that r has a finite decimal expansion when 10N r is an integer for some N > 0 (and hence b divides 10N ).

102 -

1

91. El/ A function f is symmetric with respect to the vertical line x = a if f (a — x) = f (a +x). (a) Draw the graph of a function that is symmetric with respect to x = 2. (b) Show that if f is symmetric with respect to x = a, then g(x) = f (x + a) is even.

90. Let p = p1 . . . ps be an integer with digits pl, . . . .Ps. Show that 10s— 1

= 0.pi • • • Ps

92. ael Formulate a condition for f to be symmetric with respect to the point (a, 0) on the x-axis.

1.2 Linear and Quadratic Functions Linear functions are the simplest of all functions, and their graphs (lines) are the simplest of all curves. However, linear functions and lines play an enormously important role in calculus. For this reason, you should be thoroughly familiar with the basic properties of linear functions and the different ways of writing an equation of a line. Let's recall that a linear function is a function of the form f (x) = mx

b

(m and b constants)

The graph of f is a line of slope m, and since f(0) b, the graph intersects the y-axis at the point (0, b) (Figure 1). The number b is called the y-intercept. The slope-intercept form of the line with slope m and y-intercept b is given by y = mx + b

y-intercept--Xi

X2

FIGURE 1 The slope m is the ratio "rise over run."

The Greek letter A (delta) is commonly used to denote the change in a variable or function. Thus, letting Ax and Ay denote the change in x and y = f (x) over an interval [xi, x2], we have Ax = x2 — xi,

Ay = Y2 — Y1 =

f (x2) — f (xi)

The slope m of a line (Figure 1) is equal to the ratio Ay Ax

vertical change rise _ — honzontal change run

SECTION 1.2

Linear and Quadratic Functions

15

This follows from the formula y = mx + b: Ay y2 — yi — = Ax x2 —

(mx2 + b) — (mxi + b) x2 —

m(x2 — xi) —m x2 — xi

The slope m measures the rate of change of y with respect to x. In fact, by writing Ay = mAx we see that a 1-unit increase in x (i.e., Ax = 1) produces an m-unit change Ay in y. For example, if m = 5, then y increases by 5 units per unit increase in x. The rate-of-change interpretation of the slope is fundamental in calculus. Graphically, the slope m measures the steepness of the line y = mx + b. Figure 2(A) shows lines through a point of varying slope m. Note the following properties: • • • • • •

Steepness: The larger the absolute value Im I. the steeper the line. Positive slope: If m > 0, the line slants upward from left to right. Negative slope: If m 0 and decreasing if m 0. Show that the geometric mean is not larger than the arithmetic mean (a + b)12. Hint: Consider (a1/2 - b1/2)2. 53. If objects of weights x and W I are suspended from the balance in Figure 14(A), the cross-beam is horizontal if bx = aw l . If the lengths a and b are known, we may use this equation to determine an unknown weight x by selecting w such that the cross-beam is horizontal. If a and b are not known precisely, we might proceed as follows. First balance x by iv on the left, as in (A). Then switch places and balance x by w2 on the right, as in (B). The average i = (w1 + w2) gives an estimate for x. Show that is greater than or equal to the true weight x.

37. Find the roots of the quadratic polynomials: (b) f(x)= x2 - 2x - 1 (a) f (x)= 4x2 - 3x - 1 In Exercises 38-45, complete the square and find the minimum or maximum value of the quadratic function. 38. y = x2 + 2x + 5

39. y = x2 - 6x + 9

40. y = -9x2 +x

41. y =x 2 +6x +2

42. y = 2x2 - 4x - 7

43. y = -4x2 + 3x+ 8

44. y = 3x2 + 12x - 5

45. y = 4x - 12x2

(B)

(A) FIGURE 14

46. Sketch the graph of y = x2 - 6x + 8 by plotting the roots and the minimum point. 47. Sketch the graph of y = x2 + 4x + 6 by plotting the minimum point, the y-intercept, and one other point.

54. Find numbers x and y with sum 10 and product 24. Hint: Find a quadratic polynomial satisfied by x. 55. Find a pair of numbers whose sum and product are both equal to 8. 56. Show that the parabola y = x2 consists of all points P such that d1 = d2, where d1 is the distance from P to (0, /) and d2 is the distance from P to the line y = - / (Figure 15).

48. If the alleles A and B of the cystic fibrosis gene occur in a population with frequencies p and 1 - p (where p is between 0 and 1), then the frequency of heterozygous carriers (carriers with both alleles) is 2p(1 - p). Which value of p gives the largest frequency of heterozygous carriers? 49. For which values of c does f(x) = x2 + cx + 1 have a double root? No real roots? 50. Let f (x) = x2 + x - 1. (a) Show that the lines y =x + 3, y =x - 1, and y =x - 3 intersect the graph of f in two, one, and zero points, respectively. (b) Sketch the graph off and the three lines from (a). (c) 7.4 Describe the relationship between the graph of f and the lines y = x + c as c changes from -oo to oo.

FIGURE 15

Further Insights and Challenges 57. Show that if f and g are linear, then so is f +g. Is the same true of fg?

60. Complete the square and use the result to derive the quadratic formula for the roots of ax2 + bx + c = 0.

58. Show that if f and g are linear functions such that f(0) = g(0) and f(1) = g(1), then f = g.

61. Let a,c

59. Show that Ay/Ax for the function f(x)= x2 over the interval [xi , x2] is not a constant, but depends on the interval. Determine the exact dependence of Ayl Ax on xi and x2.

0. Show that the roots of

ax2 + bx + c = 0 are reciprocals of each other.

and

cx2 + bx + a = 0

22

CHAPTER 1

PRECALCULUS REVIEW

63. Prove Viete's Formulas: The quadratic polynomial with a and roots is x2 + bx + c, where b = —a — /3 and c = a/3.

62. Show, by completing the square, that the parabola y = ax2 + bx + c

0 as

can be obtained from y = ax2 by a vertical and horizontal translation.

1.3 The Basic Classes of Functions The primary condition on a function f is that it assigns to each element x of its domain a unique element f (x) in its range. There are no other restrictions on how that relation is defined. Usually we describe the relationship by a formula for the function, sometimes by a table or a graph, but the association could be quite complicated, not lending itself to any simple description. The possibilities for functions are endless. In calculus we make no attempt to deal with all possible functions. The techniques of calculus, powerful and general as they are, apply only to functions that are sufficiently "well behaved" (we will see what well behaved means when we study the derivative in Chapter 3). Fortunately, such functions are adequate for a vast range of applications. Most of the functions considered in this text are constructed from the following familiar classes of well-behaved functions: polynomials

rational functions

exponential functions logarithmic functions

algebraic functions

trigonometric functions inverse trigonometric functions

We shall refer to these as the basic functions. • Polynomials: For any real number m, f (x) =- xm is called the power function with exponent m. Power functions include f (x) = x3, f (x) = x 7, and f (x) -= x" . The base is the variable, and the exponent is a constant. For now, we are interested in power functions with exponents that are positive integers. A polynomial is a sum of multiples of power functions with exponents that are positive integers or zero (Figure 1): f (x) =- x5 — 5x3 + 4x, FIGURE 1 The polynomial function f (x) = x5 — 5x3 ± 4x.

g(t) -= 716 + t3 — 3t — 1,

h(x) = x9

Thus, the function f (x) = x + x-1 is not a polynomial because it includes a term x 1 with a negative exponent. The general polynomial P in the variable x may be written P(x) = anx" + an —ix"-1 + • • • + aix + ao

5

— — — —

The numbers ao, al,. , an are called coefficients. The degree of P is n (assuming that an 0 0). The coefficient a, is called the leading coefficient. The domain of P is R.

• A rational function is a quotient of two polynomials (Figure 2): f (x) = —2

(x) Q(x)

[P(x) and Q(x) polynomials]

The domain of f is the set of numbers x such that Q(x) 0 0. For example, —3

f(x) =

1

domain fx : x

01

domain ft :t

±11

X

FIGURE 2 The rational function x+1 f(x) = x3 — 3x + 2

h(t) =

7t6

t3 — 3t — 1 t2 — 1

Every polynomial is also a rational function [with Q(x) = 1].

SECT I ON 1.3

The Basic Classes of Functions

23

• An algebraic function is produced by taking sums, products, and quotients of roots of polynomials and rational functions (Figure 3): Z z-5/3 5z3 — A number x belongs to the domain of f if each term in the formula is defined and the result does not involve division by zero. For example, g(t) is defined if t > 0 and 0 2, so the domain of g is D = ft : t > 0 and t 41. • Exponential functions: The function f (x) = bx , where b > 0 and b 1, is called the exponential function with base b. Some examples are f (x) = / 1 +3x 2 —x4,

FIGURE 3 The algebraic function f(x) = 3x2 _ x4.

f (x) = 2x, Any function that is not algebraic is called transcendental. Exponential and trigonometric functions are examples, as are the Bessel and gamma functions that appear in engineering and statistics. The term "transcendental" goes back to the 1670s, when it was used by Gottfried Wilhelm Leibniz (1646-1716) to describe functions of this type.

g(t) =

g(t) = 10f,

— 2)-2,

h(x) =

h(z) =

,

p(t) =

Exponential functions and their inverses, the logarithmic functions, are treated in greater detail in Chapter 7. • Trigonometric functions are functions built from sin x and cos x. These functions are discussed in the next section.

Constructing New Functions Given functions f and g, we can construct new functions by forming the sum, difference, product, and quotient functions: (f

g)(x) = f (x)

g(x),

( f — g)(x) = f (x) — g(x) ) (x) =

(f g)(x) = f (x) g(x),

g

f (x) g(x)

(where g(x)

0)

For example, if f (x) = x2 and g(x) = sin x, then (f

g)(x) -= x2 + sin x,

(f — g)(x) = x2 — sin x 2

( -0(x) = x sin x

( f g)(x) = x2 sinx,

We can also multiply functions by constants. A function of the form h(x) = c f (x)

c2 g(x)

(c c2 constants)

is called a linear combination of f and g. Composition is another important way of constructing new functions. The composition of f and g is the function f o g defined by (f o g)(x) = f (g(x)). The domain of f o g is the set of values of x in the domain of g such that g(x) lies in the domain of f. EXAMPLE 1 Compute the composite functions f o g and g o f and discuss their domains, where f (x) = Example 1 shows that the composition of functions is not commutative: The functions f o g and g o f may be (and usually are) different.

g(x)

1—x

Solution We have ( f o g)(x) = f (g(x)) = f(1 — x) =

—x

— x is defined if 1 — x > 0, that is, for x < 1. Therefore, the domain The square root off o g is {x : x < 1}. On the other hand, (g o f)(x) = g( f (x)) = g(17x) = 1 — The domain of g o f is {x :x > 01. EXAMPLE 2 Surface Area and Volume function of its volume V.

• Express the surface area S of a cube as a

24

CHAPTER

1

PRECALCULUS REVIEW Solution We will derive a relationship via a composition of functions. The volume V and the length L of a side of a cube are related by V = L3 (Figure 4). Therefore, L = V1' 3. Thus, L(V) = V113 expresses the side length as a function of the volume. The surface area S is a function of the side length defined by S(L) = 6L2 (the cube has six sides, each with area L2). Thus, S depends on V by the composition S o L(V) = S(L(V)) = S(V113) = 6(V1I3)2 = 6V2I3 It follows that we can express surface area as a function of volume by S(V) = 6 V2/3 .

FIGURE 4 A cube of side length L.

A power law is a relationship in the form y = kx" for constants k and a. They are quite common in biology and ecology. We will introduce a number of examples in the text.

•

The simple geometric relationship derived in the previous example is the basis for a variety of theoretical power laws in biology and ecology in which an attribute proportional to an animal's surface area is related to an attribute proportional to its volume. For example, in a particular species, the mass M of an individual is proportional to its volume, and the mass F of its fur might be proportional to its surface area. Thus, the relationship between fur mass and animal mass could be modeled by a power law F = k M2I3 . Typically, scientists collect data to check the proposed relationship that either confirms this model or suggests adjustments or other factors that must be considered.

Elementary Functions As noted above, we can produce new functions by applying the operations of addition, subtraction, multiplication, division, and composition. It is convenient to refer to a function constructed in this way from the basic functions listed above as an elementary function. The following functions are elementary: f (x) = A/2x + sin x,

f (x) = 10'fi ,

f (x) =

1 + x-1 1 ± cos x

Piecewise-Defined Functions We can also create new functions by piecing together functions defined over limited domains, obtaining piecewise-defined functions. One example we have already seen is the absolute value function defined by Ix I =

I —x

when x < 0 when x > 0

EXAMPLE 3 Given the function f, determine its domain, range, and intervals where it is increasing or decreasing. y=x+1 f (x) -=

y= X x< 0

x>0

FIGURE 5 A function defined piecewise.

{1 x+1

when x < 0 when x > 0

Solution The graph of f appears in Figure 5. The function is defined for all values of x, so the domain is all real numbers. Now, for all x 0, the output covers all values greater than or equal to 1. Hence, the range of the function is fy : y > 1). The function is neither increasing nor decreasing for x 0. •

1.3 SUMMARY • For m a real number, f (x) = xin is called the power function with exponent m. A polynomial P is a sum of multiples of xm, where m is a whole number: P(x) = ax" + an—ixn-1 + • • • + ax

ao

S ECT I ON 1.3

• • • • •

•

The Basic Classes of Functions

25

This polynomial has degree n (assuming that an 0 0), and ay, is called the leading coefficient. A rational function is a quotient P/ Q of two polynomials [defined when Q(x) 01. An algebraic function is produced by taking sums, products, and quotients of roots of polynomials and rational functions. Exponential function: f(x) = bx , where b > 0 and b 1 (b is called the base). The composite function f o g is defined by (f o g)(x) = f(g(x)). The domain of f o g is the set of x in the domain of g such that g(x) belongs to the domain of f. The elementary functions are obtained by taking products, sums, differences, quotients, and compositions of the basic functions, which include polynomials, rational functions, algebraic functions, exponential functions, trigonometric functions, logarithmic functions, and inverse trigonometric functions. A piecewise-defined function is obtained by defining a function over two or more distinct domains.

1.3 EXERCISES Preliminary Questions are rational func-

1. Explain why both f (x) = x3 + 1 and g(x) = tions. 2.

Is y = lx1 a polynomial function? What about y =

1x 2 ±

11?

3. What is unusual about the domain of the composite function f o g for the functions f (x) = x 1/2 and g(x) = -1 - lx I? 4.

Explain why both f (x) = i -x7

and g(x) =

are algebraic x .N/T-x4

g(x) = x -2 ± 1 h(x) = 2x, and 5. We have f (x) = + 0 112, k(x) = x2 + 1. Identify which of the functions may be described by each of the following. (a) Transcendental (b) Polynomial (c) Rational but not polynomial (d) Algebraic but not rational

functions.

Exercises 1. 3.

f (x) = x 114

2.

g(t) = 12/ 3

f (x) = x3 + 3x - 4

4.

h(z) =

6.

f (x) =

x2 ±

8.

f (x) =

2

1

5.

g(t) = t

2

7.

G(u) -

9.

f (x) = x-4 + (x - 1)-3

1 U2 -

4

23. Is f (x) = 2x2 a transcendental function?

+ 1

4

.17x

x

-

10. F(s) = sin'

9 s s+1

In Exercises 11-22, identify each of the following functions as polynomial, rational, algebraic, or transcendental. 11. f (x) = 4x3 + 9x2 - 8

15. f (x) =

17. f (x) -

X

24. Show that f (x) = x2 + 3x-1 and g(x) = 3x3 - 9x + X -2 are rational functions-that is, quotients of polynomials. In Exercises 25-32, calculate the composite functions f o g and g o f, and determine their domains. 25. f (x) =

2

g(x) = X-4

26. f (x) = -1 , x

27. f (x)

g(x) = x + 1

,

12. f (x) = X -4 14. f (x) = Ji - x2

13. f (x) =

22. f (x) = sin(3x)

21. f (x) = x2 + 3x-1

In Exercises 1-10, determine the domain of the function.

-

1

g(x) = x 2

g(0) = sin 9

28. f (x) = lxl,

x + sin x 3x _ 9x-112

2x3 + 3x 9 - 7x2

19. f (x) = sin(x2)

29. f (0) = cos 0 ,

g(x) = x3 + x2

9 - 7x2 20. f(x) =

+1

30. f (x) -

1 x2 +

1,

g(x) = -7C -

2

26

CHAPTER 1

1 31. f (t) = — , 32. f (t) =

,

PRECALCULUS REVIEW

g(t) =

36. f(x) =j l

—t2

X2

g(t) = 1 — t3

37. f(x)=1_ x2

33. The volume V and surface area of a sphere [Figure 6(A)] are exand S(r)= 47rr2, respecpressed in terms of radius r by V(r)= tively. Determine r(V), the radius as a function of volume. Then determine S(V), the surface area as a function of volume, by computing the composite S o r(V).

lirr3

34. A tetrahedron is a polyhedron with four equilateral triangles as its faces [Figure 6(B)]. The volume V and surface area of a tetrahedron are ex.4L3 and pressed in terms of the side-length L of the triangles by V (L) = respectively. Determine L(V), the side length as a function S(L)= of volume. Then determine S(V), the surface area as a function of volume, by computing the composite S o L(V).

38. f(x)=

when x 0

xl

when x 0

2x — 2

when x < 0 when x > 0

39. Let f(x)= (a) What are the domain and range of f? (b) Sketch the graph off. (c) Express f as a piecewise-defined function where each of the "pieces" is a constant. 40. The Heaviside function (named after Oliver Heaviside, 1850-1925) is defined by: H(x)= I

0 1

when x < 0 when x > 0

The Heaviside function can be used to "turn on" another function at a specific value in the domain, as seen in the four examples here. For each of the following, sketch the graph off. (a) (b) (c) (d)

(A) FIGURE 6 A sphere (A) and tetrahedron (B). In Exercises 35-38, draw the graphs of each of the piecewise-defined functions. 3 35. f(x) = {,2 +3

when x 0

f (x) = f (x) = f(x) = f (x) =

H(x)x2 H(x)(1 — x2) H(x —1)x H(x + 2)x2

41. The population (in millions) of Calcedonia as a function of time t (years) is P(t)= 30 • 20.11. Show that the population doubles every 10 years. Show more generally that for any positive constants a and k, the function g(t)= a2kf doubles after 1/k years. 42. Find all values of c such that f (x) =

x +1 x2 ± 2cx + 4

has domain R.

Further Insights and Challenges In Exercises 43-49, we define the first difference 8f of a function f by elf (x) = f (x + 1) — f (x). 43. Show that if f(x)= x2, then 8f (x) = 2x + 1. Calculate 8f for f (x) = x and f (x) = x3. 44. Show that 8(1()x) = 9 • 10x and, more generally, that 8(bx)= (b —1)bx 45. Show that for any two functions f and g, 8(f + g)= elf + 3g and S(c f)= cS(f ), where c is any constant. 46. Suppose we can find a function P such that SP(x)= (x + 1)k and P(0) = 0. Prove that P(1) = lk, P(2) = ik + 211, and, more generally, for every whole number n, P(n)=

l k + 2k + • • • + n k

then BP = (x + 1). Then apply Exercise 46 to conclude that 1 + 2 +3 +• ••+n=

n(n + 1) 2

48. Calculate 8(x3), 3(x2), and 3(x). Then find a polynomial P of degree 3 such that SP = (x + 1)2 and P(0) = 0. Conclude that P(n)=12 +22 +. . .+n 2. 49. This exercise combined with Exercise 46 shows that for all whole numbers k, there exists a polynomial P satisfying Eq. (1). The solution requires the Binomial Theorem and proof by induction (see Appendix C). (a) Show that 3(x1 +1 ) = (k +1)xk +. . . ,where the dots indicate terms involving smaller powers of x. (b) Show by induction that there exists a polynomial of degree k + 1 with leading coefficient 1/(k ± 1): P(x)= —1 x k+ 1 + • • • k +1

47. Show that if P(x)=

x(x +1) 2

such that SP = (x + 1)11 and P(0) = 0.

SECTION 1.4

Trigonometric Functions

27

1.4 Trigonometric Functions We begin our trigonometric review by recalling the two systems of angle measurement: radians and degrees. They are best described using the relationship between angles and rotation. As is customary, we often use the lowercase Greek letter 9 (theta) to denote angles and rotations.

(A)

(B)

(C)

(D)

FIGURE 1 The radian measure 0 of a counterclockwise rotation is the length along the unit circle of the arc traversed by P as it rotates into Q.

FIGURE 2 On a circle of radius r, the arc traversed by a counterclockwise rotation of radians has length Or. TABLE 1 Rotation through

Radian measure

Two full circles Full circle Half circle circle One-sixth circle

47r 27 it 27r/4 = 7/2 27/6 = 7/3

Radians

Degrees

0 7r 6

00

4

30° 45° 60°

2

90°

Figure 1(A) shows a unit circle with radius OP rotating counterclockwise into radius 0 Q. The radian measure of this rotation is the length 9 of the circular arc traversed by P as it rotates into Q. On a circle of radius r, the arc traversed by a counterclockwise rotation of 9 radians has length Or (Figure 2). The unit circle has circumference 2n-. Therefore, a rotation through a full circle has radian measure 9 = 27 [Figure 1(B)]. The radian measure of a rotation through onequarter of a circle is 9 = 2n-/4 = n-I2 [Figure 1(C)] and, in general, the rotation through one-nth of a circle has radian measure 27/n (Table 1). A negative rotation (with 0 L].

c [and we write

In other words, as x approaches c, f(x) approaches L. See Figure 2 for the graphical interpretation. If the values of f(x) do not converge to any number L as x c, we say that lim f(x) does not exist. It is important to note that the value f (c) itself, which may

Y =f(x)

X-4 C

or may not be defined, plays no role in the limit. All that matters are the values of f(x) for x close to c. Furthermore, if f(x) approaches a limit as x c, then the limiting value L is unique.

X X

C -4- X

FIGURE 2 As x —> c, f(x) —> L.

EXAMPLE 1 Let f(x) = 5 and g(x) = 3x + 1. Use the definition above to verify the following limits: (a) lim f(x) = 5 x—).7

(b) lim g(x) = 13 x—>4

Solution (a) To show that lim f(x) = 5, we must show that 1f(x) — 51 becomes arbitrarily small x—>7

when x is sufficiently close (but not equal) to 7. But note that If (x) — 51 = 15 — 51 = 0 for all x, so what we are required to show is automatic. (b) To show that lim g(x) = 13, we must show that lg(x) — 131 becomes arbitrarily small when x is sufficiently close (but not equal) to 4. We have Ig(x) — 131 = 1(3x + 1) — 131 = 13x — 121 = 31x —41

SECTION 2.2 The concept of a limit was not fully clarified until the nineteenth century. The French mathematician Augustin-Louis Cauchy (1789-1857, pronounced Koh-shee) gave the following verbal definition: "When the values successively attributed to the same variable approach a fixed value indefinitely, in such a way as to end up differing from it by as little as one could wish, this last value is called the limit of all the others. So, for example, an irrational number is the limit of the various fractions which provide values that approximate it more and more closely" (Translated by J. Grabiner)

Investigating Limits

51

Because Ig(x) - 131 is a multiple of Ix - 41, we can make lex) - 131 arbitrarily small by taking x sufficiently close to 4. Reasoning as in Example 1 but with arbitrary constants, we obtain the following simple but important results: THEOREM 1

For any constants k and c,

(a) lim k = k,

(b) lim x = c. x-->c

To deal with more complicated limits and, especially, to provide mathematically rigorous proofs, the precise version of the above limit definition, presented in Section 2.9, is needed. There inequalities are used to pin down the exact meaning of the phrases "arbitrarily small" and "sufficiently close."

Graphical and Numerical Investigation

Utkin/Sputnik/The Ima

Our goal in the rest of this section is to develop a better intuitive understanding of limits by investigating them graphically and numerically.

Olga Ladyzhenskaya (1922-2004) was a Soviet-Russian mathematician who worked in the field of partial differential equations. A major contribution of hers was a proof of a theorem verifying that a numerical approximation method converged in the limit to solutions to the Navier—Stokes equations, an important model for fluid motion. Such verification is valuable because it ensures the accuracy of estimates made via the approximation method.

Graphical Investigation Use a graphing utility to produce a graph of f. The graph should give a visual impression of whether or not a limit exists. It can often be used to estimate the value of the limit. Numerical Investigation We write x -> c- to indicate that x approaches c through values less than c (i.e., from the left), and we write x -> c+ to indicate that x approaches c through values greater than c (i.e., from the right). To investigate lirn f (x), (i) Make a table of values of f (x) for x close to but less than c—that is, as x -> c- . (ii) Make a second table of values of f (x) for x close to but greater than c—that is, as x c+. (iii) If both tables indicate convergence to the same number L, we take L to be an estimate for the limit. The tables should contain enough values to reveal a clear trend of convergence to a value L. If f (x) approaches a limit, the successive values of f (x) will generally agree to more and more decimal places as x is taken closer to c. If no pattern emerges, then the limit may not exist. x- 9 graphically and numerically. x-0 - 3

EXAMPLE 2 Investigate lim

x- 9 is undefined at x = 9 because the formula for - 3 f(9) leads to the undefined expression 0/0. Therefore, the graph in Figure 3 has a gap at x -= 9. However, the graph suggests that f (x) approaches 6 as x approaches 9. For numerical evidence, we consider a table of values of f (x) for x approaching 9 from both the left and the right. Table 2 supports our impression that

Solution The function f (x) = Keep in mind that graphical and numerical investigations provide evidence for a limit, but they do not prove that the limit exists or has a given value. This is done using the Limit Laws established in the following sections.

lim

x- 9

=6

In Section 2.5, we will revisit this limit and show how we can use algebraic simplification • to prove that this limit is, in fact, 6.

52

CHAPTER 2

LIMITS

TABLE 2 x

x- 9

9-

v -3

x- 9

9+

6.01662 6.001666 6.000167 6.0000167

9.1 9.01 9.001 9.0001

5.98329 5.99833 5.99983 5.9999833

8.9 8.99 8.999 8.9999 FIGURE 3 Graph of f(x) =

x

x- 9 -3 .

EXAMPLE 3 Limit Equals Value of the Function

Investigate Ern x2.

Solution Figure 4 and Table 3 both suggest that lirn x2 = 16. Furthermore, note that x->4 f(x) = x2 is defined at x = 4 and f(4) = 16, so in this case, the limit is equal to the function value. This pleasant conclusion is valid whenever f is a continuous function, a • concept treated in Section 2.4.

TABLE 3 x

16

I , X

2

4

6

x2

x -> 4+

x2

15.21 15.9201 15.992001 15.99920001

4.1 4.01 4.001 4.0001

16.81 16.0801 16.008001 16.00080001

4-

3.9 3.99 3.999 3.9999

FIGURE 4 Graph of f(x) = x2. The limit is equal to the value of the function f(4)= 16.

EXAMPLE 4 Investigate lim (1 + x)1/x numerically and graphically. x->o Solution The function f(x) = (1 + x)14 is undefined at x = 0, but both Figure 5 and

Table 4 suggest that a limit exists and is approximately equal to 2.71828. In Chapter 7 we will see that this limit is equal to the important numerical value e, the base of the natural exponential and logarithmic functions. TABLE 4 FIGURE 5 f(0) is undefined but lim f(x) exists.

CAUTION Numerical investigations are often suggestive, but may be misleading in some cases. If, in Example 5, we had chosen to evaluate f(x) = sin at the values

7

x = 0.1,0.01,0.001,. .., we might have concluded incorrectly that f(x) approaches the limit 0 as x 0. The problem is that f(1O ) = sin(107r) = 0 for every whole number n, but f(x) itself does not approach any limit.

EXAMPLE 5 numerically.

x -> 0-

(1 + x)l /x

x

0+

(1 + x)lix

-0.01 -0.001 -0.0001 -0.00001 -0.000001

2.731999 2.719642 2.718418 2.718295 2.718283

0.01 0.001 0.0001 0.00001 0.000001

2.704814 2.716924 2.718146 2.718268 2.718280

A Limit That Does Not Exist

Investigate lim sin -7 graphically and x->13 x

Solution The function f(x) = sin Ix is not defined at x = 0, but Figure 6 suggests that it oscillates between +1 and -1 infinitely often as x 0. It appears, therefore, that Um sin zr- does not exist. This impression is confirmed by Table 5, which shows that the values of f(x) bounce around and do not tend toward any limit L as x

0.

•

SECTION 2.2

Investigating Limits

53

TABLE 5 The Function f (x) = sin 1ix Does Not Approach a Limit as x 0 x —2

2

- .0

sin — x

—0.1 —0.03 —0.007 —0.0009 —0.00065

0 0.866 —0.434 0.342 —0.935

x

X

0.1 0.03 0.007 0.0009 0.00065

sin — x —0.866 0.434 —0.342 0.935

FIGURE 6 Graph of f (x) = sin l. 7x

One-Sided Limits The limits discussed so far are two-sided. To show that lim f (x) = L, it is necessary to check that f (x) converges to L as x approaches c through values both greater than and less than c. In some instances, f (x) may approach L from one side of c without necessarily approaching it from the other side, or f (x) may be defined on only one side of c. For this reason, we define the one-sided limits lim f (x)

(left-hand limit),

lim f (x) x—>c+

(right-hand limit)

The limit itself exists if and only if both one-sided limits exist and are equal. Otherwise the limit does not exist. EXAMPLE 6

of f (x) = — as x lx 1

—3 —2 —1

2

3

0. Does lim f (x) exist? x—>o

Solution Figure 7 shows what is going on. For x 0, x f (x) = -f--= — = , Ix I x Therefore, lim f (x) = 1. These one-sided limits are not equal, so lim f (x) does not • exist. EXAMPLE 7 The function f in Figure 8 is not defined at c = 0, 2, 4. Investigate the one- and two-sided limits at these points. Solution • c =0: The left-hand limit lim f (x) does not seem to exist because f (x) appears x->0 to oscillate infinitely often to the left of x = 0. On the other hand, lim f (x) = 2. x • c = 2: The one-sided limits exist but are not equal: lim f (x) = 3

FIGURE 8

and

lim f (x) = 1 x—>2+

Therefore, lim f (x) does not exist. x—>.2 • c = 4: The one-sided limits exist and both have the value 2. Therefore, the two• sided limit exists and lirn f (x) = 2. x—>4

54

LIMITS

CHAPTER 2

Infinite Limits ai REMINDER Recall that "large" refers to distance from zero. So being negative and becoming arbitrarily large means becoming arbitrarily far from zero to the left of zero.

f(x) -

4-

-2

xl- 2

4

6

Asymptote x=2

-4

For some functions, f(x) tends to oo or -oo as x approaches a value c. It is important lim f(x) does not exist. to understand that oo and -oo are not numbers, and therefore x->c However, we say that f(x) has an infinite limit. More precisely, we write c. • lirn f(x) oo if f(x) is positive and becomes arbitrarily large as x x->c, • lirn f(x) = -oo if f(x) is negative and becomes arbitrarily large as x --> c. x-->c One-sided infinite limits are defined similarly. When f(x) approaches oo or -oo as x approaches c from one or both sides, the line x = c is called a vertical asymptote. In Figure 9, the line x =- 2 is a vertical asymptote in (A), and x = 0 is a vertical asymptote in (B). In the next example, the notation x -> c± is used to indicate that the left- and righthand limits are to be considered separately. EXAMPLE 8 Investigate the one-sided limits graphically: (a) lim

x-÷2± x —

(A)

1 (b) lim — x->o± x2

1 2

Solution (a) Figure 9(A) suggests that f(x) =

lim x->2-

Asymptote x=0

1 = - 2

oo,

lim

1

x->.2+ X —

2

= oo

The line x =- 2 is a vertical asymptote. Why are the one-sided limits different? Because 1 f(x) = is negative for x 2 (so the limit from the right is oo). 1 1 (b) Figure 9(B) suggests that lim - = oo. Indeed, f(x) = — is positive for all x 0 x->c) X2 and becomes arbitrarily large as x 0 from either side. The line x = 0 is a vertical asymptote. •

(B) FIGURE 9

CONCEPTUAL INSIGHT You should not think of an infinite limit as a true limit. The notation lirn f(x) = oo is merely a shorthand way of saying that f(x) is positive and arbitrarily large as x approaches c. The limit itself does not exist. We must be careful when using this notation because oo and -oo are not numbers, and contradictions can arise if we try to manipulate them as numbers. For example, if oo were a number, it would be larger than any finite number, and presumably, oo + 1 = oo. But then oo +1 = oo (c,o+ 1)- oo = oo - oo 1=0

(contradiction!)

To avoid errors like this, keep in mind the oo is not a number but rather a convenient shorthand notation.

2.2 SUMMARY - By definition, lim f(x) = L if ff(x) - LI can be made arbitrarily small by taking x sufficiently close (but not equal) to c. We say that

SECTION 2.2

Investigating Limits

55

- The limit of f (x) as x approaches c is L, or - f (x) approaches (or converges to) L as x approaches c. • If f (x) approaches a limit as x c, then the value of the limit L is unique. • If f (x) does not approach a limit as x -> c, we say that lim f (x) does not exist. x • The limit may exist even if f (c) is not defined. • One-sided limits: -

lim f (x) = L if f (x) converges to L as x approaches c through values less than c. x-÷c- lim f(x) = L if f (x) converges to L as x approaches c through values greater x-4c-F than c. • The limit exists if and only if both one-sided limits exist and are equal. • Infinite limits: lim f (x) = oo if f (x) is positive and becomes arbitrarily large as x approaches c, and lirn f (x) = -oo if f (x) is negative and becomes arbitrarily large as x approaches c. • In the case of a one- or two-sided infinite limit at c, the vertical line x = c is called a vertical asymptote.

2.2 EXERCISES Preliminary Questions 1. What is the limit of f (x) = 1 as x -> yr?

x

2. What is the limit of g(t) = t as t --> yr?

f (x)

3. Is lim 20 equal to 10 or 20? x-4113

0.9

0.99

0.999

1.001

1.01

1.1

7

25

4317

3.00011

3.0047

3.0126

6. Can you tell whether lim f (x) exists from a plot of f for x > 5? c if f (c) is undefined? If so, give an

4. Can f (x) approach a limit as x example.

5. What does the following table suggest about lim f (x) and lim f (x)? x- >1+ x

Explain. 7. If you know in advance that lim f (x) exists, can you determine its x->s value from a plot of f for all x > 5?

Exercises In Exercises 1-5, fill in the table and guess the value of the limit. x3 - 1 1. lim f (x), where f (x) = „ X4 -

x->1

x

f (x)

x 1.002

0.998

1.001

0.999

1.0005

0.9995

1.00001

0.99999

2.001

1.999

2.0001

1.9999

4. lim f(0), where f(0) = 0->o 0

±0.002

h(t) y2_ 3. hm f (y), where f (y) y-->2

y2

+y

y

2 - 6

±0.00005

sin 0 -

93 ±0.0001

±0.00005

±0.00001

f(0)

h(t) = h(-t).

±0.0001

f (y)

f (x)

cost - 1 . Note that h is even; that is, t2

±0.002

Y 1.998

5. lim f (t), where f (t) = 2. lim h(t), where h(t) =

f(y)

y 2.002

±0.00001

t

1 - cos 2t

f (t)

t

0.002

-0.002

0.001

-0.001

0.0005

-0.0005

0.00001

-0.00001

f (t)

, computing the values of sin x with x 6. Numerically investigate lim x->o x in degrees. Make an estimate of the limit accurate to 5 decimal places.

56

CHAPTER 2

LIMITS

7. Determine firn f (x) for f as in Figure 10.

39. lim IxI5

40. lim(1 +2r)

8. Determine lim g(x) for g as in Figure 11.

tan 0 - 2 sin cos 41. lim 19->rr/4 14

42.

r->0

x->0

x->0.5

x->0.5

. tan x - x -4 111 sin x -x

43. The greatest integer function, also known as the floor function, is defined by [x_1 = n, where n is the unique integer such that n < x 2.6

44. Determine the one-sided limits at c = 1, 2, and 4 of the function g shown in Figure 12, and state whether the limit exists at these points.

FIGURE 11

FIGURE 10 In Exercises 9-10, evaluate the limit. 9.

lim

x->21

10.

X

firn

11. Show, via illustration, that the limits lirn x and lim a are equal but x->a the functions in each limit are different.

FIGURE 12

12. Give examples of functions f and g such that lirn f (x) = lim g(x), x->o but f (x) g(x) for all x, including 0.

In Exercises 45-52, determine the one-sided limits numerically or graphically. If infinite, state whether the one-sided limits are oo or -oo, and describe the corresponding vertical asymptote. In Exercise 52, f (x) = is the greatest integer function defined in Exercise 43.

In Exercises 13-20, verify each limit using the limit definition. For example, in Exercise 13, show that I3x - 121 can be made as small as desired by taking x close to 4.

45.

sin x lim x--,4)± lxi

47.

lim x->o±

x - sin lx1 x3

49.

lirn x.-4.-2±

4x2 + 7 X3 + 8

13. lim 3x = 12 15. lim (5x + 2) = 17 x->3

17. lim x2 = 0 x-->0

19. lim (4x2 + 2x + 5) = 5 x-,4)

14. lim 3 = 3 16. lim (7x -4) = 10 x->2

18. lim (3x2 -9) = -9 x->o 20. lim (x3 + 12) = 12

In Exercises 21-42, estimate the limit numerically or state that the limit does not exist. If infinite, state whether the one-sided limits are oo or -Co. - 1

21. lim

x->1

X -

1

x2 + x - 6 23. lim x->2 x2 - x - 2 25. lim x->o

sin 2x

s n 3x 27. urn x->o 3x 29.

hm

i9->0

cos 61 - 1 U

31. lim 33. lim

(x - 4)3 x +3

x->--3 X2 + X -6

35. lim x->3+

x- 4 X2 - 9

1 37. lim sin h cos h-->0

22.

lim x->-4

2x2 - 32

46. lim 48. lim

x+1 4

x->4± X -

50.

x2

firn x->-3

X2 -

9

5

X + X - 2 51. lim i± x4 + x - 2

52. lim cos ( 1- (x - LxD)

53. Determine the one-sided limits at c = 2 and c = 4 of the function f in Figure 13. What are the vertical asymptotes off? 54. Determine the infinite one- and two-sided limits in Figure 14.

+

24. lim

x3 - 2x2 - 9 2x - 3

26. lim

sin 5x x

x->3 X 2 -

cos x 28. lim x o 3x 30. lim

x->0

sin x X2

3- x 32. lim x-›l- x-1 34.

x+1 lim x+2 3h

36.

lim h-->0

1 h

1 38. lim cos h->0 h

FIGURE 13

FIGURE 14

In Exercises 55-58, sketch the graph of a function with the given limits. 55. lim f (x) = 2,

lim f (x) = 0,

lim f (x) = 4 x->3+

56. lim f (x) = oo, x->1

57.

lim f (x) = 0,

x->3-

lim f (x)= f(2) = 3,

lim f (x) = -co

lim f (x)= - 1,

lim f (x) =2 0 f(4)

x

SECTION 2.2 58. urn f (x) = oo,

lim f (x) = 3,

urnf (x) = -oo

59. Determine the one-sided limits of the function f in Figure 15, at the points c = 1, 3, 5, 6.

Investigating Limits

57

(GU) In Exercises 61-66, plot the function and use the graph to estimate the value of the limit. sin 50 61. lim e-o3 sin 20

12x - 1 62. litn x->o 4x - 1

. 2x - cos x 63. hm x--q)

sin2 40 64. lim 9-4 cos& - 1

65. lim o->o

cos 70 - cos 50 02

66. hm o->o

s n2 20-0 sin 40 04

67. Let n be a positive integer. For which n are the two infinite one-sided limits lim 1/x" equal? x->o± n 1 68. Let L(n)=urni for n a positive integer. Investix-> i ( 1 - xn 1 - x) gate L(n) numerically for several values of n, and then guess the value of L(n) in general.

FIGURE 15 Graph of f . 60. Does either of the two oscillating functions in Figure 16 appear to approach a limit as x --> 0?

69. (GU) In some cases, numerical investigations can be misleading. Plot f (x)= cos I'x . (a) Does lim f(x) exist? x->o 1 1 1 (b) Show, by evaluating f (x) at x = ±- ,±- ' ±- '. , that you might be 2 4 6 able to trick your friends into believing that the limit exists and is equal to L = 1. (c) Which sequence of evaluations might trick them into believing that the limit is L = -1?

(B)

(A) FIGURE 16

Further Insights and Challenges 70. Light waves of frequency A passing through a slit of width a produce a Fraunhofer diffraction pattern of light and dark fringes (Figure 17). The intensity as a function of the angle 0 is

bX - 1 is less than 2 with b = 7 and is X Experiment with values of b to find an approxi8. than 2 with b = greater mate value of b for which the limit is 2.

(sin(R sin 0) 2 1(0) = im R sin0 ) where R = gap, and 1m is a constant. Show that the intensity function is not defined at 0 = 0. Then choose any two values for R and check numerically that 1(0) approaches Im as 0 -> 0.

x" - 1 for (m, n) equal to (2,1), (1,2), (2,3), and xm - 1 (3, 2). Then guess the value of the limit in general and check your guess for two additional pairs.

aT

72. Show numerically that lim

73. Investigate lim

74. Find by numerical experimentation the positive integers k such that s n(sin2 x) m exists. x-+O xk 75.

1-7 ,GU) Plot the graph of f (x) -

2x - 8 x 3.

(a) Zoom in on the graph to estimate L = lim f (x). x->.3

Incident light waves

(b) Explain why Slit

Viewing screen

Intensity pattern

FIGURE 17 Fraunhofer diffraction pattern. sin nO numerically for several positive integer values 71. Investigate lim e-o of n. Then guess the value in general.

f(2.99999) C

X ->C

(i) Sum Law: lim (f(x) + g(x)) exists and x-> c, lim g(x)

g(x)) = lim f(x)

lim (f(x)

X ->C

X ->C

X ->C

(ii) Constant Multiple Law: For any number k, lim kf (x) exists and x-> c. lim kf (x) = k lim f(x) X->C

(iii) Product Law: lim f (x)g(x) exists and lim f (x)g(x) = (lim f(x)) (Inn g(x)) x-›c• x->c —f(x) (iv) Quotient Law: If lim g(x) 0 0, then lim exists and x ->c g(x) x---c lim f (x) ,. f (x) x->c nm — = X -> C g(x) ail g(x) (v) Powers and Roots: If n is a positive integer, then lim[f(x)]'2 = (lim f(x)) , x->c x-*c

lirn ,'/f(x) = Oim f(x)

X ->C

X ->C

In the second limit, assume that lim f (x) > 0 if n is even. X - ->C

If p, q are integers with q

0, then lim [f ( )}P/ exists and x-÷c

lim [f (x)]Plq = (lim f(x))"'

X -4.0

X ->C

Assume that lim f(x) > 0 if q is even, and that lim f(x) x-y.c x->c

0 if p/q < 0.

Before proceeding to the examples, we make some useful remarks. • The Sum and Product Laws are valid for any number of functions. For example, lim (fi (x)

12(x)

f3(x)) = lim fi(x) X - >C

lim f2(x)

X--> C

lim f3(x)

X->C

• The Sum Law has a counterpart for differences: lirn ( f (x) - g(x)) = lim f(x) - lim g(x)

X->C

C

X ->C

This follows from the Sum and Constant Multiple Laws (with k = -1): lim (f(x) - g(x)) = lim f(x) + lim - g(x)) = Bin f(x) - lim g(x) X ->C X->C

X -+C

X->C

• Recall two basic limits from Theorem 1 in Section 2.2: lim k = k,

X ->C

lirn x = c

X -->C

X->C

SECTION 2.3

Basic Limit Laws

59

Applying Law (v) to f (x) = x, we obtain

iimxPiq = cP/q

1

--> C

for integers p and q such that q 0 0. Note, in Eq. (1) we need to assume that c > 0 if q is even and that c 0 if p / q 2

(b) lim(x3 + 5x + 7) x—>2

(c) lim Vx3 + 5x + 7 x—>2

Solution (a) By Eq. (1), Ern x3 = 2 3 = 8. x—>2 (b) lim (x3 + 5x + 7) = lim x3 + lim 5x + lim 7 x—>2 x—>2 x—>2 x—>2 =

x—>2

(Sum Law)

X3 ± 5 lim x + lim 7 x—>2

(Constant Multiple Law)

=8+5(2)+7=25 (c) By Law (v) for roots and (b), lim VX 3 x—>2 EXAMPLE 2 Evaluate

+

5x + 7 = .11im (x3 + 5x + 7) = Nr23 =- 5 x—>2 t +6 (a) lim t—>-1 2t4

and

•

(b) lirn t-1/4(t + 5)1/3. t—>3

Solution (a) Use the Quotient, Sum, and Constant Multiple Laws: You may have noticed that each of the limits in Examples 1 and 2 could have been evaluated by a simple substitution. For example, set t = —1 to evaluate t +6 2t4

-1+6 2(-1)4

5 2

Substitution is valid when the function is continuous, a concept we shall study in the next section.

lim

t +6 2t4

lim (t + 6) t-0-1 lirn 2t 4

lirn t + lim 6 2 lim t4

5 -1 + 6 - 2(-1)4 - 2

(b) Use the Product, Powers, and Sum Laws: lim t-1/4(t + 5)1/3 = (Ern C1/4) (Ern t—>3 t—>3

= (3-1/4) (3/lim t + 5) t—>3

2 = 3-1/4,0 ± 5 = 3-1/4 (2) = 31/4

•

The next example reminds us that the Basic Limit Laws apply only when the limits of both f (x) and g(x) exist. EXAMPLE 3 Assumptions Matter Show that the Product Law cannot be applied to Ern f (x)g(x) if f (x) = x and g(x) = x-1. Solution For all x exists:

0, we have f (x)g(x) = x • x-1 = 1, so the limit of the product

lim f (x)g(x) = lim 1 = 1 x-->o However, there is an issue with the product of the limits because lim x-1 does not exist (since g(x) = x-1 becomes infinite as x -÷ 0). Therefore, the Product Law cannot be

60

CHAPTER 2

LIMITS

applied and its conclusion does not hold even though the limit of the products does exist. Specifically, lirn f (x)g(x) = 1, but the product of the limits is not defined: 1=

lirn f (x))

lim g(x)) = x-+0

lim x) x,o

lirn x-1) Does

•

not exist

2.3 SUMMARY • The Basic Limit Laws: If lirn f (x) and Ern g(x) both exist, then x—>c x—>c g(x)) = Ern f (x) x—>c (ii) lirn kf (x) = k lirn f (x) x—>c x—>c lirn (f (x) (i) x—>c

lim f (x) g(x) = x—>c

Em g(x) x-+c

lim f (x))( lirn g(x)) x—>c x—>c

urnf (x) f(x) 0, then lim — = x —>c g(x) Ern g(x) x-÷c (v) If p, q are integers with q 0,

(iv) If Ern g(x) x—>c

lim [f(x)]Piq = Oim f(x)r iq x—).c For n a positive integer, limf(x)]" = (lim f (x)) , x—>c

Ern A/f(x) ' = x—>c

lim f (x) x—>c

• If lirn f (x) or lirn g(x) does not exist, then the Basic Limit Laws cannot be applied. x—>c

2.3 EXERCISES Preliminary Questions 1. State the Sum Law and Quotient Law. 2. Which of the following is a verbal version of the Product Law (assuming the limits exist)? (a) The product of two functions has a limit. (b) The limit of the product is the product of the limits.

(c) The product of a limit is a product of functions. (d) A limit produces a product of functions. 3. Which statement is correct? The Quotient Law does not hold if (a) The limit of the denominator is zero (b) The limit of the numerator is zero

Exercises In Exercises 1-26, evaluate the limit using the Basic Limit Laws and the limits lirn x = CPlq and lim k = k. x->c x->c 1. lim x 2. lim 14 x->9

x->-3

4. lirn

z->27

5. Ern t-1 t->2.

7. lim (3x + 4)

9. lirn (3x4 - 2x3 + 4x)

Z213

11. lim(x + 1)(3x2 -9) x--.2.

13. lim

x-> 7

1

t->4 t +4

15. urn

r.-).4

3t- 14 t+1

6. lirn x-2

17. lim (16y + 1)(2y1/2 + 1)

8. lim (3x3 + 2x2) x-).4

19. lim

10. lim(3x213 - 16x-1)

21. Ern

x->5

Y- >

y->4

12. limi (4x + 1)(6x - 1)

1 ,/6y + 1 ,

X

+ 4x

14. lim

3

z->0 z -

16. Ern

1

.ii

z->9 z -

2

18. lim x(x + 1)(x + 2) x->2

20. lirn

w - >7

22. lim

1 ,

-- 3 - 1 t2 +

1

+ 2)(t4 + 1)

SECTION 2.4 23. lim t->25 (t - 20)2

24. lim (18y2

25. lim (4t2 + 8t - 5)3/2

26. lim t->7 (t

4)4

± 2)1/2 1)20

27. Use the Quotient Law to prove that if lim f (x) exists and is nonzero, x->c

then

lim

1

f(x)

28. Assuming that lim f (x) = 4, compute: 1 (b) lim x->6 f (x)

(a) lim f(x)2 x->6

(c) Um x f (x)

29.

lirn f (x)g(x)

30. lim (2f (x)

x-> -4

• 31. urng(x) x->-4 X2 33.

35. Assume that if lim f (x) = L, then lirn sin f (x) = sin L. In each case x->a x->a evaluate the limit or indicate that the limit does not exist. (b) lim sin x (a) lirn sin x->0

( X -X 1)

3g(x))

f (x) 1 32. lim x-÷-4 3g(x) - 9

Can the Quotient Law be applied to evaluate lim sin x ? Explain. x-).0 X

x-*r/2

X

(d) lim x2 sin(7rx2) x->1

36. Assume that if lirn f (x) = L, then lim cos f(x)= cos L. In each x->a x->a case evaluate the limit or indicate that the limit does not exist. 2x cos x (a) urn cos (b) lim x (1 x-,7r/2 X

x->6

In Exercises 29-32, evaluate the limit assuming that lim f (x) = 3 and x->-4 lim g(x) -0 1.

61

34. Show that the Product Law cannot be used to evaluate the limit lirn - 2 tan x. x-,7r/2

3x (c) lim x-+1 sin(1 -x)

1

Limits and Continuity

1 -x 2

(c) lim x3 cos(1 - x)

(d) lim x->o 1 - cos(x2)

37. Give an example where lim ( f (x) x-0

g(x)) exists but neither lim f (x) x-o

1

nor lim g(x) exists.

38. Give an example where lim (f (x) • g(x)) exists but neither lim f (x) x-o x->1) nor lim g(x) exists. x-03 39. Give an example where lim - 'x) exists but neither lim f (x) nor x->o ex) Jim g(x) exists. x->0

Further Insights and Challenges 40. Show that if both lim f (x) g(x) and lim g(x) exist and x->c f (x) g(x) Jim g(x) 0, then lirn f (x) exists. Hint: Write f (x) = x->c x->c g(x) 41. Suppose that lim t g(t) = 12. Show that lim g(t) exists and t ->3 t->3 equals 4. = 5, then urnh(t) = 15. t->3

_ 42. Prove that if tlim >3 43. RI

Assuming that lim

= 1, which of the following statements

is necessarily true? Why? (b) lim f (x) = 0

(a) f(0) = 0 44. Prove that if lim f(x) = L Jim

0 and lim g(x) = 0, then the limit x->c

does not exist.

45. 21

Suppose that lim g(h) = L. h->() (a) Explain why Jim g(ah) = L for any constant a 0 0. (b) If we assume instead that lim g(h) = L, is it still necessarily true that h->1 lim g(ah) = L? h->1 (c) Illustrate (a) and (b) with the function f (x) = x2. ax - 1 exists for all a > 0. Assume also 46. Assume that L(a) = lim x->o x 1. that lim ax = x->() (a) Prove that L(ab) = L(a) + L(b) for a, b > 0. Hint: (ab)x - 1 = ax bx ax + ax - 1 = ax (bx - 1) ± (ax - 1). [This shows that L(a) behaves like a logarithm, in the sense that log(ab) = log(a) + log(b). In fact, it can be shown that L(a) is equal to what is known as the natural logarithm function.] (b) Verify numerically that L(12) = L(3) + L(4).

2.4 Limits and Continuity Y =f(x)

f(c)

FIGURE 1 f is continuous at x = c.

In everyday speech, the word "continuous" means having no breaks or interruptions. In calculus, continuity is used to describe functions whose graphs have no breaks. If we imagine the graph of a function f as a wavy metal wire, then f is continuous if its graph consists of a single piece of wire as in Figure 1. Many physical phenomena can be considered as continuous. Our position and velocity vary continuously with time. Barometric pressure varies continuously with altitude above the earth. The current in a simple circuit varies continuously with the voltage applied to it. Ultimately, when we determine the rate of change of a function as we do in the next chapter, we will need the function to be continuous for the mathematics to work properly.

62

CHAPTER 2

LIMITS

A break in the wire as in Figure 2 is called a discontinuity. Observe in Figure 2 that the break in the graph occurs because the left- and right-hand limits as x approaches c are not equal and thus liLnc g(x) does not exist. By contrast, in Figure 1, lim f (x) exists and

g(c)

y = g(x)

is equal to the function value f (c). This suggests the following definition of continuity in terms of limits. DEFINITION Continuity at a Point Assume that f (x) is defined on an open interval containing x = c. Then f is continuous at x = c if

al FIGURE 2 Discontinuity at x = c: The left- and right-hand limits as x —> c are not equal.

lim f (x) = f (c) x—>c. If the limit does not exist, or if it exists but is not equal to f (c), we say that f has a discontinuity (or is discontinuous) at x = c. Note that for f to be continuous at c, three conditions must hold: 1. f (c) is defined.

2. lim f (x) exists. x—>c

3. They are equal.

A function f may be continuous at some points and discontinuous at others. If f is continuous at all points in its domain, then f is simply called continuous. EXAMPLE 1 Show that the following functions are continuous: (a) f (x) = k

(k any constant)

(b) g(x) = xn

(n a whole number)

Solution

FIGURE 3 The function f (x) = k is continuous.

(a) We have lim f (x) = lim k = k and f (c) = k. The limit exists and is equal to the x—>c function value for all c, so f is continuous (Figure 3). (b) By Eq. (1) in Section 2.3, lim g(x) = xn = cn for all c. Also g(c) = c', so x—>c x—>c. again, the limit exists and is equal to the function value. Therefore, g is continuous. (Figure 4 illustrates the case n = 1.)

Examples of Discontinuities To understand continuity better, let's consider some ways in which a function can fail to be continuous. Keep in mind that continuity at a point x = c requires that: 1. f (c) is defined.

2. lim f (x) exists.

3. They are equal.

If lim f (x) exists, but either the limit is not equal to f (c), or f (c) is not defined, x—>c then we say that f has a removable discontinuity at x =- c. The function in Figure 5(A) has a removable discontinuity at c = 2 because FIGURE 4 The function g(x) = x is continuous.

f(2) = 10

but

lim f (x) = 5

x—>2

Limit exists but is not equal to function value

FIGURE 5 Removable discontinuity: The discontinuity can be removed by redefining f(2).

10

10

5

5

2 (A) Removable discontinuity at x = 2

2 (B) Function redefined at x = 2

SECTION 2.4

A removable discontinuity at x = c that occurs because f (c) is not defined is sometimes referred to as a removable singularity.

Limits and Continuity

63

Removable discontinuities are mild in the following sense: We can make f continuous at x = c by redefining f (c) [in the case lim f (x) f (c)] or defining f (c) [in x-).c the case f (c) is not defined] so that f (c) = lim f (x). In Figure 5(B), f(2) has been redefined as f(2) = 5, and this makes f continuous at x = 2. -8 has a removable discontinuity at x = 2. How x- 2 should g(2) be defined so that g is continuous at x = 2?

EXAMPLE 2 Show that g(x) =

Solution First note that g is not defined at x = 2 since evaluating g at 2 involves division by 0. Also, lim g(x) = lim

x—>2

x3 - 8

x-->2 X — 2

= urn

(x - 2)(x2 + 2x + 4)

x—>2

X —2

2

= lim (x + 2x + 4) = 12 x—>2

where the Basic Limit Laws are used to determine the value of the limit. Since firn g(x) x-->2

exists, but g(2) is not defined, g has a removable discontinuity at x = 2. If we define g(2) = 12, then g would be continuous at x = 2. • A worse type of discontinuity is a jump discontinuity, which occurs if the one-sided limits lim f (x) and lirn f (x) exist but are not equal. In this case f is not continuous x--)•cf at c because lirn f (x) does not exist. Figure 6 shows two functions with jump discontinuities at c = 2. Unlike the removable case, we cannot make f continuous simply by redefining f at the single point c.

FIGURE 6 These functions have jump discontinuities at x = 2.

2 (A)

2 (B)

In connection with jump discontinuities, it is convenient to define one-sided continuity. DEFINITION One-Sided Continuity

A function f is called

• Left-continuous at x = c if lim f (x) = f (c) x->c • Right-continuous at x = c if lim f (x) = f (c) x->c+ In Figure 6 above, the function in (A) is left-continuous at x = 2 but the function in (B) is neither left- nor right-continuous at x = 2. Many theorems in calculus apply to functions that are continuous on an interval, a concept that is defined as follows: DEFINITION Continuity on an Interval / Assume that / is an interval in the form (a, b), [a, b), (a, b], or [a, b]. Then f is continuous on / if f is continuous at each point in (a, b), f is right-continuous at a (if a is in /), and f is left-continuous at b (if b is in /). The next example explores one-sided continuity using a piecewise-defined function— that is, a function defined by different formulas on different intervals.

64

CHAPTER 2

LIMITS

EXAMPLE 3

Piecewise-Defined Function lx F(x) =

Discuss the continuity of

for x < 1

3

for 1 < x < 3

x

for x > 3

Solution The functions f (x) = x and g(x) = 3 are continuous, so F is also continuous, except possibly at the transition points x = 1 and x = 3, where the formula for F(x) changes (Figure 7). FIGURE 7 Piecewise-defined function F in

• At x = 1, the one-sided limits exist but are not equal: lim F(x) = lim x = 1,

Example 3.

CAUTION Piecewise-defined functions may

lim F(x) = lirn 3 = 3

Thus, F has a jump discontinuity at x = 1. However, the right-hand limit is equal to the function value F(1) = 3, so F is right-continuous at x = 1. • At x = 3, the left- and right-hand limits exist and both are equal to F(3), so F is continuous at x = 3:

or may not be continuous at points where

lirn F(x) = lim 3 = 3,

they are pieced together.

lim F(x) = urn x = 3

x—>3+

•

We say that f has an infinite discontinuity at x = c if one or both of the onesided limits are infinite [even if f (x) itself is not defined at x = c]. Like with a jump discontinuity, in this case f is not continuous at c because lim f (x) does not exist. Figure x-->c 8 illustrates three types of infinite discontinuities occurring at x = 2. Notice that x = 2 does not belong to the domain of the function in cases (A) and (B).

1r X

FIGURE 8 Functions with an infinite discontinuity at x = 2.

(A)

(B)

(C)

EXAMPLE 4 The Intensity of a Light Source A standard model for the intensity / of a light source at varying distances d from the light is an inverse-square law, 1(d) = k Id2 for a constant k > 0 depending on the light source (Figure 9). Show that / has an infinite discontinuity at d = 0.

FIGURE 9 Light intensity is inversely

proportional to d2. Solution Regardless of the value of k > 0, as d approaches 0 from the right, the values of 1(d) = k Id2 are positive and become arbitrarily large. Therefore lim 1(d) = co, and it follows that / has an infinite discontinuity at 0.

SECTION 2.4

Limits and Continuity

65

Note that this does not mean that the intensity of the light is actually unbounded as we get closer and closer to it. The relationship 1(d) = k / d2 is a model for the true behavior. The model does, however, properly reflect the fact that the intensity rises rapidly as we approach the source. • Finally, we note that some functions have more severe types of discontinuities than those discussed above. For example, f (x) = sin oscillates infinitely often between +1 and -1 as x 0 (Figure 10). Neither the left- nor the right-hand limit exists at x = 0, so this discontinuity is not a jump discontinuity. See Exercises 94 and 95 for even stranger examples.

Building Continuous Functions al FIGURE

10 Graph of y = sin 1- . The discontinuity at x = 0 is not a jump, removable, or infinite discontinuity.

Having studied some examples of discontinuities, we focus again on continuous functions. How can we show that a function is continuous? One way is to use the Laws of Continuity, which state, roughly speaking, that a function is continuous if it is built out of functions that are known to be continuous. THEOREM 1 Basic Laws of Continuity If f and g are continuous at x -= c, then the following functions are also continuous at x = c: (i) f + g and f - g (iii) f g (ii) kf for any constant k

(iv) f/ g if g(c)

0

Proof These laws follow directly from the corresponding Basic Limit Laws (Theorem 1, Section 2.3). We illustrate by proving the first part of (i) in detail. The remaining laws are proved similarly. By definition, we must show that lim (f (x) + g(x)) = f (c) + g(c). x-*C Because f and g are both continuous at x = c, we have Ern f (x) = f (c),

lim g(x) = g(c) x

The Sum Law for limits yields the desired result: •

lim (f (x) + g(x)) = lim f (x) + lim g(x) = f (c) + g(c) x ->c x-÷c x->c

In Section 2.3, we noted that the Basic Limit Laws for Sums and Products are valid for an arbitrary number of functions. The same is true for continuity; that is, if fi • • • , ft: are continuous, then so are the functions fi When a function f is defined and continuous for all values of x, we say that f is continuous on the real line.

1

4.10 REMINDER A rational function is a quotient of two polynomials P / Q .

+ f2 + • • • + f

ft, f2 • • • fn

The next two theorems assert that the indicated basic functions are continuous on their domains. Recall (Section 1.3) that the term "basic function" refers to polynomials, rational functions, nth-root and algebraic functions, trigonometric functions and their inverses, and exponential and logarithmic functions. THEOREM 2 Continuity of Polynomial and Rational Functions polynomials. Then:

Let P and Q be

• P and Q are continuous on the real line. • P / Q is continuous on its domain [at all values x = c such that Q(c)

0].

Proof The function f (x) = xi" is continuous for all whole numbers m by Example 1. By Continuity Law (ii), f (x) = axm is continuous for every constant a. A polynomial P(x) = ax n + an _ixn-1 + • • • +

ao

66

2

CHAPTER

LIMITS is a sum of continuous functions, so it too is continuous. By Continuity Law (iv), a quo• tient function P/ Q is continuous at x = c, provided that Q(c) 0. This result shows, for example, that f (x) = 3x4 — 2x3 + 8x is continuous for all x, and that g(x) =

x +3 X

2

—

1

±1. Note that if n is a positive integer, then f (x) = x- n is continis continuous for x = 1 / x' is a rational function. uous for x 0 because f (x) The continuity of the nth-root, sine, cosine, and exponential functions should not be surprising because their graphs have no visible breaks (Figure 11). However, complete proofs of continuity are somewhat technical and are omitted. THEOREM 3 Continuity of Some Basic Functions INNO REMINDER The domain of y = x11' is the real line ifn is odd and the half-line [0, co) if n is even.

• y = x1/ is continuous on its domain for n a natural number. • y =- sin x and y = cos x are continuous on the real line. • y = bx is continuous on the real line (for b > 0, b 0).

2

4

FIGURE 11 As the graphs suggest, these functions are continuous on their domains. Because f (x) = sin x and f (x) = cos x are continuous, the Continuity Law (iv) for Quotients implies that the other standard trigonometric functions are continuous on their domains, consisting of the values of x where the denominators, in the following quotient expressions for them, are nonzero: 1,x 37r 2

tan x =

sin x cos X

,

cos x cot x = . , sin x

1 cos X

,

1 csc x = . sin x

They have infinite discontinuities at points where the denominators are zero. For example, as illustrated in Figure 12, f (x) = tan x has infinite discontinuities at the points 7

FIGURE 12 Graph of y = tan x.

sec x =

x = ±— , 2

37r ±—, 2

5n±— 2

Finally, it is important to know that a composition of continuous functions is again continuous. The following theorem is proved in Appendix D. THEOREM 4 Continuity of Composite Functions If g is continuous at X = C, and f is continuous at x = g(c), then the composite function F(x) = f (g(x)) is continuous at x = C.

SECTION 2.4

Limits and Continuity

67

For example, F(x) = (x2 + 9)1/3 is continuous because it is the composite of the continuous functions f (x) =- x 1/3 and g(x) = x2 ± 9. Similarly, F(x) = cos(x 1) is continuous for all x 0. More generally, an elementary function is a function that is constructed out of basic functions using the operations of addition, subtraction, multiplication, division, and composition. Since the basic functions are continuous (on their domains), an elementary function is also continuous on its domain by the Laws of Continuity. An example of an elementary function is x2 ± cos4(2x ± 4) x+8 This function is continuous on its domain lx : x

8}.

Substitution: Evaluating Limits Using Continuity It is easy to evaluate a limit when the function in question is known to be continuous. In this case, by definition, the limit is equal to the function value: lim f (x) = f (c) x—>c We call this the Substitution Method because the limit is evaluated by substituting x = c in f (x).

EXAMPLE 5 Evaluate

(a)

lirn sin y

and

2x 113

(b) lim

x-*8 N/x + 1

Solution (a)

We can use substitution because f (y) = sin y is continuous. 7r

0

lim sin y = sin — = — 3 2 y(b) The function f (x) = 2x 1/3 /A/X + 1 is continuous at x = 8 because the numerator and denominator are both continuous at x = 8 and the denominator Vx ± 1 is nonzero at x = 8. Therefore, we can use substitution:

3

lim

2

i/ 3

x-->8 Nt7C + 1

-0

FIGURE 13 Graph of f (x)

= Lx j.

2(8)1/3

4

-\/8 + 1

3

The greatest integer function f (x) = [xj is the function defined by Lx] = n, where n is the unique integer such that n < x 2 • lim [xi = 1 and lim [xi = 2 x—>2+

68

CHAPTER 2

LIMITS

CONCEPTUAL INSIGHT Real-World Modeling by Continuous Functions Continuous functions are often used to model relationships between physical quantities such as position and time, temperature and altitude, and voltage and resistance. This reflects our everyday experience that change in the physical world tends to occur continuously rather than through abrupt transitions. However, mathematical models are approximations to reality and are based on assumptions about the phenomenon being studied. It is always important to be aware of a model's assumptions and the limitations they impose. In Figure 14, atmospheric temperature is represented as a continuous function of altitude. At such a large scale the assumption of continuity is reasonable because it is consistent with our experience. However, at smaller, less familiar scales the situation can be different. In fact, in 2002 scientists McGaughey and Ward reported observing temperature discontinuities at the surface of evaporating water droplets, suggesting that it may not be appropriate to assume temperature is continuous at such small scales. The size of a population is often treated as a continuous function of time. Strictly speaking, population size is a whole number that changes by ±1 when an individual is born or dies, so it really is not continuous. At the scale of the size of your family, it does not make sense to consider the number of people as a continuous variable. However, if a population is large, the effect of an individual birth or death is small, and at such a scale it is both reasonable and convenient to treat population as a continuous function of time. Ultimately, the test of a model is how well it enables us to understand and predict the behavior of the actual system. When it fails to do so, the assumptions need to be examined, and possibly adjusted, to try to find a better fit between model and reality.

Temperature (°C)

200

4.) C.

100 4.),

4.)

World population (millions)

C. t SE:4 (.1)

6,000 —

't4' )

L2'

C.

C.

4,000 —

0 —100 10

50

100

150

Altitude (km)

0 1700

1750

1800

1850

1900

1950

2000

Year

FIGURE 14 Atmospheric temperature and world population are modeled by continuous functions.

2.4 SUMMARY • Definition: f is continuous at x = c if lim f (x) = f (c). This means that f (c) exists, lim f (x) exists, and they are equal. x—> c. • If lim f (x) does not exist, or if it exists but does not equal f (c), then f is discontinx-*C uous at x = c. • If f is continuous at all points in its domain, f is simply called continuous. • Right-continuous at x = c: lim f (x) = f (c). x —>c+ • Left-continuous at x = c: lim f (x) = f (c). x-->c

SECTION 2.4

Limits and Continuity

69

• Three common types of discontinuities: — Removable discontinuity: lim f (x) exists, but either the limit does not equal f (c) x—> c. or f (c) is not defined. — Jump discontinuity: The one-sided limits both exist but are not equal. — Infinite discontinuity: The limit is infinite as x approaches c from one or both sides. • Laws of Continuity: Sums, products, multiples, inverses, and composites of continuous functions are continuous. The same holds for a quotient f g at points where g(x) O. • The following basic functions are continuous on their domains: polynomials, rational functions, nth-root and algebraic functions, trigonometric functions, and exponential functions. • Substitution Method: If f is known to be continuous at x = c, then the value of the limit lim f (x) is f (c).

2.4 EXERCISES Preliminary Questions 1. Which property of f (x) = x3 allows us to conclude that lim x3 = 8? x—>2 2. What can be said about f(3) if f is continuous and lim f (x) = x—>3 3. Suppose that f (x) 1 if x is negative. Can f be continuous at x = 0? 4. Is it possible to determine f(7) if f (x) = 3 for all x a exist and are equal. (b) f is continuous at x = a if the left- and right-hand limits of f (x) as x —> a exist and equal f (a). (c) If the left- and right-hand limits of f (x) as x —> a exist, then f has a removable discontinuity at x = a. (d) If f and g are continuous at x = a, then f + g is continuous at x =- a. (e) If f and g are continuous at x =a, then flg is continuous at x = a.

Exercises 1. Referring to Figure 15, state whether f is left- or right-continuous (or neither) at each point of discontinuity. Does f have any removable discontinuities?

5. In Figure 17, determine the one-sided limits at the points of discontinuity. Which discontinuity is removable and how should f be redefined to make it continuous at this point?

Exercises 2-4 refer to the function g whose graph appears in Figure 16. 2. State whether g is left- or right-continuous (or neither) at each of its points of discontinuity. 3. At which point c does g have a removable discontinuity? How should g(c) be redefined to make g continuous at x = c? 4. Find the point ci at which g has a jump discontinuity but is leftcontinuous. How should g(ci) be redefined to make g right-continuous at x = Cl? FIGURE 17 5—

5—

6. Suppose that f (x) = 2 for x < 3 and f (x) = —4 for x > 3.

4

4

(a) What is f(3) if f is left-continuous at x = 3?

3—

(b) What is f(3) if f is right-continuous at x = 3?

2

2— •

1—

1

1 2

I 3

In Exercises 7-16, use Theorems 1-4 to show that the function is continuous.

1— x 4

5

6

FIGURE 15 Graph of y = f (x).

x 1

2

3

4

5

6

FIGURE 16 Graph of y = g(x).

7. f (x) = x + sin x

8. f (x) = x sin x

9. f (x)

10. f (x) = 3x3 + 8x2 — 20x

3x + 4 sin x

70

LIMITS

CHAPTER 2

11. f (x) =

1

x2 ±

12. f (x) =

1

2

-

COS X

3 + cos x

14. f (x) = X 1i 3 cos 3x

13. f (x) = cos(x2) 15. f (x) = tan ( x2

X

1

±

16. tan

1)

ltX

2

1 + 2x2

In Exercises 17-38, determine the points of discontinuity. State the type of discontinuity (removable, jump, infinite, or none of these) and whether the function is left- or right-continuous. 1 17. f (x) = x 19. f (x) =

x- 2

lx - 11

18. f (x) =1x1

FIGURE 18

20. f (x) =1)d

54. Sawtooth Function Draw the graph of f (x) = x - Lx j. At which points is f discontinuous? Is it left- or right-continuous at those points?

22. g(t) =

21. f (x) =

23. h(x) =

1 2- Ix I

24. k(x) =

25. f (x) =

x+1 4x - 2

26. h(z) =

x- 2

31. f (x) =

1

x0 2

In Exercises 55-56, I-xl refers to the least integerfunction. It is defined by = n, where n is the unique integer such that n - 1 1

ix + 1 1

for x < 1 for x > 1

33. g(t) = tan 2t

34. f (x) -= csc(x2)

35. f (x) = tan(sin x)

36. f (x) = cos(7r [x])

37. f (x) = Lx + 3] + L2xJ

38. f (x) = 2Lx12] + 41_x/4_1

In Exercises 39-52, determine the domain of the function and prove that it is continuous on its domain using Theorems 1-4. 40. f (x) = • x2 + 9

39. f (x) = 2 sin x + 3 cos x

59.

f (x) =

x2 - 3x + 2 ix -21 0

x0 2 x =2

X3 ± 1 60. f (x) = 1 -x + 1 -x 2 +10x - 15

for -oo < x < 0 for 0 3

x-1 for x < -1 67. f (x) = I ax ± b for - 1 < x < x-1 for x >

74. f is right- but not left-continuous at x = 1, left- but not rightcontinuous at x = 2, and neither left- nor right-continuous at x = 3.

75. firn (2x3 - 4)

76. lim (5x - 12x-2) x-).2

x+2 77. lim x->3 X` ± 2x

78. urn sin( -x - 7t) x, 2

79. hill tan(3x)

1 80. lim x-4.71. cos x

81. lim x-512

82. lim ./x3 + 4x x-).2

83. firn (1 - 8x3)312 x -I

7x +2 \ 2/3 84. liM x-).2 ( 4 - x

68. Define Ix + 3 for x < -1 g(x)= cx for - 1 < x < 2 x + 2 for x > 2 Find a value of c such that g is (a) left-continuous In each case, sketch the graph of g.

71

In Exercises 75-86, evaluate using substitution.

for x < 5 for x > 5

{2x + 9x-1 -4x + c

Limits and Continuity

(b) right-continuous

t3 - 1 fort ±1. Answer the following questions, t2 - 1 using a plot if necessary. (a) Can g(1) be defined so that g is continuous at t = 1? If yes, how? (b) Can g(-1) be defined so that g is continuous at t = -1? If so, how? 69. Define g(t) =

70. Each of the following statements is false. For each statement, sketch the graph of a function that provides a counterexample. (a) If lim f (x) exists, then f is continuous at x = a. x-).a (b) If f has a jump discontinuity at x = a, then f (a) is equal to either lim f (x) or lim f (x). In Exercises 71-74, draw the graph of a function on [0,5] with the given properties. 71. f is not continuous at x = 1, but firn f (x) and lirn f (x) exist and are equal.

86.

85. lim 10'2-2x x-> 3

urn 3sin x

87. Suppose that f and g are discontinuous at x = c. Does it follow that f + g is discontinuous at x = c? If not, give a counterexample. Does this contradict Theorem 1(i)? 88. Prove that f (x) = Ixl is continuous for all x. Hint: To prove continuity at x = 0, consider the one-sided limits. 89. Use the result of Exercise 88 to prove that if g is continuous, then f (x) = Ig(x)1 is also continuous. 90. Which of the following quantities would be represented by continuous functions of time and which would have one or more discontinuities? (a) Velocity of an airplane during a flight (b) Temperature in a room under ordinary conditions (c) Value of a bank account with interest paid yearly (d) Salary of a teacher (e) Population of the world In 2017, the federal income tax T on income of x dollars (up to 91. $91,900) was determined by the formula

72. f is left-continuous but not continuous at x = 2, and right-continuous but not continuous at x = 3.

for 0 1

- X 12. lim x->3 x2 -9 14. lim

1 lim x-o+ (,Fx

x3 - 64x x- 8

X2

y->3

21. lim

x->4

km x-+8

h-).4

h-40

23. lim

8.

2 1_x 2 )

1 33. lim o-> (tan ° - 1

lim

16. lim

- 4 16

17. lim

x2 - 64 x->8 X - 9

31. lim r-*2.

26.

cot 0 28. lim s csc 0

1 29. lim x->i (1 -x

6.

10. ELM

. cotx 27. km

x->0 CSC X

In Exercises 5-34, evaluate the limit, if it exists. If not, determine whether the one-sided limits exist (finite or infinite). 5.

4

C

25. lim

36. The following limits all have the indeterminate form co/cxi One of the limits does not exist, one is equal to 0, and one is a nonzero limit. Evaluate each limit algebraically if you can or investigate it numerically if you cannot. x-4 3 cot x (a) lim (b) lim x->o 4 + x-1 x->I3 CSC X

1+ x2 (c) lim x ->o 1 + In Exercises 37 and 38, show that the limit is in an indeterminate form, then investigate the limit numerically to estimate the value. 37. lim o-.()

1 - cos() 02

38. lim

0 ->0

1 - cos2

02

SECTION 2.6

39. (GU) Use a plot of f (x)=

x- 4

to estimate lim f (x) to x->4

two decimal places. Compare with the answer obtained algebraically in Exercise 23. 4 to estimate lim f (x) 2 x- 4 numerically. Compare with the answer obtained algebraically in Exercise 25. 40. (GU' Use a plot of f (x) -

1

x3 - 8 x->2 X - 2

41. lim

43. lim x--+1 45. lim

x2 - 5x + 4 x3 -1

x4 -1 x- - 1

b3 = (a - b)(a2 + ab + b2) 42. lim

47. lim (2a + x) x-q)

48. lim (4ah + 7a)

49. lim (4t - 2at +3a)

50.

-

Si. lim

;/1 + h

-1. Hint: Set x = .`,Y1

function of x, and rewrite as a limit as x

x3 + 8 44. lim x-+-2 x2 ± 6x + 8

h, express h as a

1.

56. Evaluate lim '357-F h I Hint: Set x = h-0 h- 1 function of x, and rewrite as a limit as x 1.

x - 27 46. lim x-3.27 X., ' -3

4x2

x- a

1 1 a 54. lim h h- a

a)3 - a 3

55. Evaluate lim

-

(x

h->0

x->0

9

. x->a

52. lim

X - a

53. lim

x3 - 27

x->3 X 2 -

77

In Exercises 47-54, evaluate in terms of the constant a.

In Exercises 41-46, evaluate using the identity a3-

The Squeeze Theorem and Trigonometric Limits

express h as a

Further Insights and Challenges 61. For which sign, ± or -, does the following limit exist?

In Exercises 57-60, find all values of c such that the limit exists. x2 +3x +c x2 - 5x - 6 58. lim 57. lim x->c x-> X x- c

lim ( 1 ±

x-O

-

59. lim

1

60. lim

x-4.0

x->1 ( X - 1

X

1 x(x - 1))

1 ± cx2 - N/1 ± x2 x4

2.6 The Squeeze Theorem and Trigonometric Limits y = u(x)

FIGURE 1 f is trapped between / and u.

In our study of the derivative, we will need to evaluate certain limits involving transcendental functions such as sine and cosine. The algebraic techniques of the previous section are often ineffective for such functions, and other tools are required. In this section, we discuss one such tool-the Squeeze Theorem-and use it to evaluate the trigonometric limits needed in Section 3.6.

The Squeeze Theorem Consider a function f that is "trapped" between two functions 1, for lower bound, and u, for upper bound, on an interval I. In other words, 1(x)

y=u(x) Y =f (x) y =/(X)

u(x)

for all x e /

Thus, the graph off lies between the graphs of 1 and u (Figure 1). The Squeeze Theorem applies when f is not just trapped but squeezed at a point x = c (Figure 2). By this we mean that for all x c in some open interval containing c, 1(x)

113 FIGURE 2 f is squeezed by 1 and u at x = C.

f (x)

f (x)

u(x)

and

lirn 1(x) = firn u(x) = L x->c x->c

We do not require that f (x) be defined at x = c, but it is clear graphically that f (x) must approach the limit L, as stated in the next theorem. See Appendix D for a proof.

78

CHAPTER

2

LIMITS THEOREM " Squeeze Theorem taining c), 1(x)

f (x)

c (in some open interval con-

Assume that for x

u(x)

lirn 1(x) = firn u(x) = L x->c x->c

and

Then firn f (x) exists and lim f (x) = L. x-*c. EXAMPLE 1 Show that firn x sin I = O. x

x —>I3

Solution Although f (x) = x sin 1. - is a product of two functions, we cannot use the Product Law because lim sin I does not exist. However, the sine function takes on values y

between 1 and -1, and therefore I sin 11 < 1 for all x 0 0. Multiplying by Ix', we obtain x sin 11 < Ix I and conclude that (Figure 3) . 1 -Ix' < x sin - < Ix' x

I

Furthermore, we have 13 FIGURE

lim lxI = 0 x->o

3

and

lim(-Ixl) = 0

and therefore it follows that f (x) = x sin is squeezed between 1(x) = -Ix' and u(x) = lxi at x = 0. The Squeeze Theorem now applies, and we can conclude that lim x sin = 0. • x—>0

In Section 2.2, we found numerical and graphical evidence suggesting that the limit sin 0 lim is equal to 1. The Squeeze Theorem will allow us to prove this fact.

0-0

0

THEOREM 2 Important Trigonometric Limits Note that both ±'"9 si and c": -f are indeterminate at 9 = 0, so Theorem 2 cannot be proved by substitution.

sin hm — e-).13 0

=1

and

lim o-.3

1 - cos

=0

sin 0 To apply the Squeeze Theorem to prove that lim — = 1, we must find functions 61—>C1

sin 0 that squeeze — at 9 = 0. These are illustrated in Figure 4 and provided by the next theorem.

al FIGURE 4 Graph illustrating the inequalities of Theorem 3. THEOREM 3 cos 9< -

sin

0 to obtain sin 4h sin = lim4 h->0 h 0-40 0 11.m

( sin = 4 lim —) \O -*O 0

= 4(1) = 4

Note that the change of variables 6 = kx demonstrates that x -> 0. We use this limit to our advantage in the next example.

sin kx approaches 1 as kx

tan 3x EXAMPLE 3 Find lim x->o tan 2x Solution . tan 3x . sin 3x = hm lim x-40 tan 2x x-÷0 cos 3x

cos 2x . sin 3x = hm sin 2x x->o cos 3x

cos 2x sin 2x

x x

3 ( sin 3x ) ( 2x ) cos 2x 3 1 3 = lirn = • 1•1• T = I x->0 2 3x sin 2x cos 3x 2

•

2.6 SUMMARY • We say that a function f is squeezed at x -= c if there exist functions 1 and u such that 1(x) < f(x) < u(x) for all x c in an open interval I containing c, and lim 1(x) = lim u(x) = L x->c x->c The Squeeze Theorem states that in this case, lim f (x) = L. x->c

• Two important trigonometric limits: sin lim—=1 c+-0 0

and

lim e->o

1 - cos 0 =0 0

2.6 EXERCISES Preliminary Questions 1. Assume that —x4 < f (x) < x2. What is x—>o lim f (x)? Is there enough information to evaluate lhn f(x)? Explain. x—>1

sin5h it is a good idea to rewrite the limit 3h in terms of the variable (choose one):

2. State the Squeeze Theorem carefully.

(a) 0 = 5h

3. If you want to evaluate lim

h-.0

(b) 9=3h

(c)

5h = — 3

SECTION 2.6

The Squeeze Theorem and Trigonometric Limits

81

Exercises In Exercises 1-10, evaluate using the Squeeze Theorem.

15. State whether the inequality provides sufficient information to determine lirn f (x), and if so, find the limit. x-0.1 (a) 4x - 5 < f (x) 1 x- 1

4. lim (x2

1 5. lim(2t - 1) cos r

6. Ern ,rx 3c" (tri x) x->o+

16. (GU) Plot the graphs of u(x) = 1 ± Ix I and 1(x) = sin x on the same set of axes. What can you say about lim f (x) if f is squeezed by 1 and u at x = ?

8. lim tan x cos (sin! x->CI

In Exercises 17-26, evaluate using Theorem 2 as necessary.

7. lim(t2 - 4) cos t->2

x->3

1 t- 2

10.

9. Ern cos 0 cos(tan 0) 9->ir

9) x

3 31

lim (3t -1) sin2 (-1 ) t->o-

11. State precisely the hypothesis and conclusions of the Squeeze Theorem for the situation in Figure 6.

tan x

17. lim x->0

s n x sec x

18. lim x-,4)

X

9 sint

19. Ern t->.0

t

x2 21. lirn x-o3 sin2 x

20. lim

sin2 t

t->0

t

1 - cos t

22. lim r->

23. lim 0..0

sec - 1 0

24. Ern (9-q)

25. lim

sin t t

26. lim

1 - cos 0 sin COS t

-

COS2 t

. sin 1 lx using a substitution 0 =11x. 27. Evaluate lim FIGURE 6 12. In Figure 7, is f squeezed by u and / at x = 3? At x = 2?

. sin 7t . Hint: Multiply the numerator and denominator 28. Evaluate Inn . r-ri sin llt by (7)(11)t. In Exercises 29-48, evaluate the limit. 29. lim

sin 9h h

30.

31. lim

sin h 5h

32. lirn

h ->C1

1.5

h->0

6

7 FIGURE 8 14. Determine lim f (x) assuming that cos x < f (x) < 1.

sin 4h 4h sin 3x

sin 70 33. lim e-*() sin 30

tan 4x 34. lim x-op 9x

35. Em x csc 25x x-›-o

tan 4t 36. lim r->0 t sec t

FIGURE 7 lim f (x) if the limits 13. What does the Squeeze Theorem say about x->7 ittri 1(x) = lim u(x) = 6 and f ,u, and / are related as in Figure 8? The inx->7 x-,•7 equality f (x) < u(x) is not satisfied for all x. Does this affect the validity of your conclusion?

iitn h-->0

sin 2h sin 3h

37. Em

h->0

38. lim

h2

sin(-30) 39. lim 6.-->0 sin 40

sin(z/3) sin z

t n 4x 40. Inn x-o) tan 9x

41. lim

csc 8t csc 4t

sin 5x sin 2x 42. Ern ,x-o sin 3x sin 5x

43. lim

sin 3x sin 2x x sin 5x

44. lim h-).0

45. lim h->0

sin(2h)(1 - cos h) h2

46. lim t->0

47. lim 49-A)

cos 20 - cos 0

48.

1 - cos 2h 1 - cos 2t sin2 3t

lilll h--).

1 - cos 3h

49. Use the identity sin 20 = 2 sin 0 cos 6 to evaluate lirn e-o

sin 20 - 2 sin 0 02

82

CHAPTER 2

LIMITS

50. Use the identity sin 30 = 3 sin 0 - 4 sin3 0 to evaluate the limit, s n 30 - 3 sin 0 0 0 —> 03 51. Explain why lim (csc 0 - cot 0) involves an indeterminate form, and

54. (GU) Investigate lim

1 - cos h

h —+O

h2

prove that the limit is equal to

numerically or graphically. Then

Hint: See the proof of Theorem 2.

In Exercises 55-57, evaluate using the result of Exercise 54.

then prove that the limit equals 0.

55. Inn

cos 3h -1 h2

52. Explain why Elm (2 tan 0 - sec 0) involves an indeterminate form,

56. lim

cos 3h - 1 - 1

h-÷0

h—).0 cos 2h

and then evaluate the limit. 1 - cos 2h

53. (GU) Investigate lim

h—>0

numerically or graphically. Then

h2

evaluate the limit using the double angle formula cos 2h = 1 - 2 sin2 h.

57. lim

- cos t

t—>0

lim If(x)1 = 0, then 58. Use the Squeeze Theorem to prove that if x->c lim f (x) = 0.

Further Insights and Challenges 59. Use the result of Exercise 54 to prove that for m .m

coSMX

Conclude that sin2 0 < 2(1 - cos 0) < 02 and use this to give an alternative proof that the limit in Exercise 51 equals 0. Then give an alternative proof of the result in Exercise 54.

0,

— 1

X2

2 61. (a) Investigate the limit lim

x -->e

60. El/ Using a diagram of the unit circle and the Pythagorean Theorem, show that sin2 8 oo indicates that x increases without bound, and x —> —oo indicates that x decreases (through negative values) without bound. We write

10—

FIGURE 1 The graphed model P(t) fits recent world population data and is derived with the assumption that the population eventually levels off (see Section 7.4). In the model lim P(t) = 17.4, suggesting t->oo that the human carrying capacity of the earth is about 17.4 billion people. y = g(x)

• lim f (x) = L if f (x) gets closer and closer to L as x —> oo X—>00 •

Ilin

X-+ —co

f (x) = L if f (x) gets closer and closer to L as x —> —oo

As before, "closer and closer" means that I f (x) — LI becomes arbitrarily small. In either case, the line y =-- L is called a horizontal asymptote. We use the notation x —> ±oo to indicate that we are considering both infinite limits, as x oo and as x —oo. Infinite limits describe the asymptotic behavior of a function, which is determined by the behavior of the graph as we move out indefinitely to the right or the left. EXAMPLE 1 Discuss the asymptotic behavior in Figure 2. Solution The function g approaches L = 7 as we move to the right and it approaches L = 3 as we move to the left, so lim g(x) = 7

and

lirn g(x) = 3 x—>-00 Accordingly, the lines y = 7 and y = 3 are horizontal asymptotes of g. —400

—200

200

400

FIGURE 2 The lines y = 7 and y = 3 are horizontal asymptotes of g.

A function may approach an infinite limit as x lirn f (x) = oo

or

±co. We write lim f (x) = co

X—>-00

•

SECTION 2.7

(A)

Limits at Infinity

83

(B)

FIGURE 3

if f (x) is positive and becomes arbitrarily large as x oo or -oo. Similar notation is used if f (x) approaches -co as x ±oo. For example, we see in Figure 3(A) that lim 2x = cc —>00

and

lim 2x = 0 X—>-00

However, limits at infinity do not always exist. For example, f (x) = sin x oscillates indefinitely [Figure 3(B)], so the following limits do not exist: lim sin x

and

lim sin x

X—>00

X—>-00

The limits at infinity of the power functions f (x) = x" are easily determined. If n > 0, then x" increases without bound as x -> oo, so (Figure 4) lim xn = oo x->00

FIGURE 4

lim xn = oo (A) n even: lim xn = x--(X)

CAUTION lim x112 does not exist, since x—> the square root of a negative number is not a real number.

and

lim X—>00

1 firn — = 0

X —n =

X—>00 X n

• 1= (C) ltm

(B) n odd: lim xn = oo, lirn f=—oo

x-00

X-r 00

•

1 X =0 —

To describe the limits as x -> -oo, assume that n is a whole number so that X n is defined -oo, and if n is odd, for x 0, lim xn = oo .x,00

and

lim x- n = lim X—>00

X —>00

1 Xn

If n is a positive whole number, Jim

{cc-°°

if n is even if n is odd

and

lim X—>-00

1 lim — = 0 X—>-00

Note also that if p and q are positive integers and q is odd, then if p is even and

Xn

lim xPiq = oo

lim xP/q = -oo if p is odd. In the case that q is even, xP/q is not

defined for negative x, so it does not make sense to address the limit as x -> -oo.

84

CHAPTER 2

LIMITS

The Basic Limit Laws (Theorem 1 in Section 2.3) are valid for limits at infinity. For example, the Sum and Constant Multiple Laws yield lim (3 - 4x-3 + 5x-5) = lim 3 - 4 lim x-->oo

X-* 00

x->co

X -3 +

5 Bin

X-5

X-4 00

= 3+0+0 = 3

20x2 - 3x EXAMPLE 2 Calculate lim x->oo 3x5 - 4x2 + 5 Solution It would be nice if we could apply the Quotient Law directly, but this law is valid only if the denominator has a finite, nonzero limit. Our limit has the indeterminate form oo/oo because lim (20x2 - 3x) = oo

and

lim (3x5 - 4x2 + 5) = oo

X->00

X->00

The way around this difficulty is to divide the numerator and denominator by x5 (the highest power of x in the denominator): x-5(20x 2 — 3x) 20x2 - 3x 20x-3 — 3x-4 3x5 - 4x2 + 5 - x-5(3x5 - 4x2 + 5) - 3 - 4x-3 + 5x-5 Now we can use the Quotient Law: lim (20x-3 - 3x-4) 20x2 - 3x 0 x lim = = =0 x-,o0 3x5 -4x 2 +5 3 lim (3- 4x-3 +5x 5)

•

X-).00

In general, if f (x) =

anxn + an-if + • • • + ao bmxm + bm _ixm-1 + • • • + bo

where a, 0 0 and bm 0 0, divide the numerator and denominator by xm: anxn' + bm +

f (x)

= xn-m

+ • • • + aox' +. . . + box- m

(a n + bm +

+ • • • + aox-n +. . . + box- m )

The quotient in parentheses approaches the finite limit an I bm because lim (an + an-ix

X->00

lim (bm +

X-.>00

This also holds true for x

+ - • • + aox') = an + • • • + box') = kn

-oo, and therefore,

an + an _ix - 1 + • • • + aox- n lim f (x) = lim xn' lim x->-±co .x-).±00 .x.->±00 bm + + • • • + box- m

an

linl

bm ,c-4±00

x n-m

THEOREM 2 Limits at Infinity of a Rational Function The asymptotic behavior of a rational function depends only on the leading terms of its numerator and denominator. If an, b,n 0 0, then lim

.7c->-±00

an xn + an _ixn-1 + • • • + ao an . = — hm x' bmxm + bm-ixm-1 + • • • + bo bm x->±co

2.7

SECTION

Limits at Infinity

85

Here are some examples: • n = m:

xL imoo

3x4 - 7x + 9 3 i. 0 3 = - iim x = 7 x-÷0.0 7x4 - 4 7

• n inco 7 x->•00 7x4 - 4

• n > m, n - m odd:

urn x->-00

• n > m, n — m even:

Ern x—>.—co

3x8 - 7x + 9 7x3 - 4

=

3 lim x5 = -oo 7 x->-00

3x7 - 7x + 9 3 ,. — tim x4 = oo 7 x—>.-00 7x3 — 4

Our method can be adapted to noninteger exponents and algebraic functions. 3x7/2 ± 7x -1/2

EXAMPLE 3 Calculate the limits

(a) lim

x2

X->00

4x (b) lim x' ±°° A/x2 + 1

x1/2

Solution The Quotient Law is valid if lirn f(x)= co and lim g(x)= L, where L 0:

(a) As before, divide the numerator and denominator by x2, which is the highest power of x occurring in the denominator (this means multiply by x-2): 3x772

lim f(x) f (x) l oo if L > 0 lim — = = x—>c g(x) —oo if L < 0 lim g(x)

+ 7x-1/2

X 2 — X 1/2

x -2

3x712 4_ 7x-112

X -2

X2 — X 112

X-±C

3x7/2 + 7x -1/2

A similar result holds when lim f(x)= —oo.

x_ >. 00

x2

lim

(3x3/2

+

7x-5/2

1 — X -3 /2

7x -5/2) 00

x—>ao

x 1/2

3x312

00

1

hill (1 — X-3 /2) X—> co

oo. The key is to observe that the denominator of

(b) First, consider x

4x ,Vx2 + 1

behaves like x l =x: x2 +

1= Vx2(1 + x-2) = x

+x

(for x >0)

2

This suggests that we divide the numerator and denominator by x: 4x A/x2 ± 1 =

4x

4x

4

•Vx2+1

X A/1 + X-2

.s/1 ± X -2

Then apply the Quotient Law: 4 4x = -lim , lim , x->00 ,/i + x-2 x-÷00 Aix2 +1

lim 4 x ->00 hill A/1 + X-2

x->00

4 = - =4 1

For the limit as x -> -oo, one approach is to replace x with -t. Since x = -t and x -›- -oo, then t -)- oo. So we have X

FIGURE 5 There are horizontal asymptotes at y = ±4.

lim

4x

-00 ,Vx2 + 1

= lim

4(-t)

t—>oo V(_)2 ± 1

= lim

t—>c)o

-4t t2 ± 1

=

lim

4t

t—>oo ,Vt2 ±

1

=

4

where the last equality holds by the limit we previously calculated. 4x The limits in (b) indicate that the graph of f(x)= ,N,t F i has horizontal asymptotes at y = 4 and y = -4, which is confirmed in Figure 5.

•

2.7 SUMMARY • Limits at infinity: -

lim f(x)= L if I f (x) - Li becomes arbitrarily small as x increases without 00 bound.

86

CHAPTER 2

LIMITS

-

lirn f (x) = L if I f (x) - LI becomes arbitrarily small as x decreases without .7c--> -00 bound.

• A horizontal line y -= L is a horizontal asymptote of f if lim f (x) = L oo

Or

lirn f (x) = L x-4-00

A function can have 0, 1 or 2 horizontal asymptotes. • If n > 0, then lim x-n = 0. .x-).±00 • If n > 0 is a whole number, then lim • If f (x) =

anxn br„xm

= oo

and

lim x =

oo if n is even -oo if n is odd

an-ixn 1 ± • • • + ao with an, bm 0 0, then bm_ixm-1 ± • • • + bo an n-m lu rn f (x) = — lim x->±oo bn, x-±oo

• for 0 < a < 1, lim ax = oo and lim ax = for a> 1, lim ax = 0 and lim ax

oo

2.7 EXERCISES Preliminary Questions 3. Sketch the graph of a function that approaches a limit as x —> oo but does not approach a limit (either finite or infinite) as x —oo.

1. Assume that lirn f (x) = L

and

lim g(x) = oo

4. What is the sign of a if f (x) = ax3 lim f (x) = oo? X- -00

Which of the following statements are correct?

x + 1 satisfies

(a) x = L is a vertical asymptote of g. 5. What is the sign of the coefficient multiplying x7 if f is a polynomial of degree 7 such that lim f (x) = oo?

(b) y = L is a horizontal asymptote of g. (c) x = L is a vertical asymptote of f.

6. Explain why lim sin I exists but lim sin I does not exist. What is x->00 x lim sin 1 ? x

(d) y = L is a horizontal asymptote off. 2. What are the following limits? (a) lim x3

(b)

lim x3

(C)

iiM

X4

Exercises 1. What are the horizontal asymptotes of the function in Figure 6?

3. Sketch the graph of a function f with a single horizontal asymptote Y = 3. 4. Sketch the graphs of functions f and g that have both y = —2 and y = 4 as horizontal asymptotes but lim f (x) lim g(x).

Y =fix) 5.

-20

20

40

60

80

FIGURE 6 2. Sketch the graph of a function f that has both y = —1 and y = 5 as horizontal asymptotes.

(ou) Investigate

the

asymptotic

behavior

of

f (x) =

x2

x2 +

numerically and graphically: (a) Make a table of values of f (x) for x = ±50, ±100, ±500, ±1000. (b) Plot the graph off. (c) What are the horizontal asymptotes of f? 12x +1 6. (GU) Investigate urn00 numerically and graphically: x ,i4x2 + 9 12x + 1 (a) Make a table of values of f (x) = for the following: ,./4x2 + 9 x = ±100, ±500, ±1000, ±10,000.

S EC TION 2.7

(b) Plot the graph off. (c) What are the horizontal asymptotes of f?

37. 3x2 + 20x lim x->.00 4.x' + 9

7.

lim x->00 x + 9

8.

9.

3x2 + 20x lim x-*oc 2x4 3X3 - 29

10. lim x->oo

7x2 - 9 13. lirn x->--co 4x + 3 15.

lim x->-co

14.

3x3 - 10 x+4

16.

39.

lim (14x4 + 9x - 2x2)

38.

lim

40.

1 lim x-÷co(x

42.

4 + 521 lim t->-oc, 5 _ 53t

./c->co

4 x

5

9x2 - 2 12. lim x-)-0t) 6 - 29x

7x - 9 4x + 3

5x - 9 lim 4x3 + 2x + 7 lim

2x5 + 3x4 - 31x 8x4 - 31x2 + 12

In Exercises 17-24, find the horizontal asymptotes. 17. f (x) =

2x2 - 3x 8x2 + 8

18. f(x)=

19. f (x) =

V36x2 + 7 9x + 4

20. f (x) =

8X3 - X2

41.

lirn

-

2)

lx1+x X+1

lim (-19x3 + x -

X312)

1 x+2

43. al Let P(n) be the perimeter of an n-gon inscribed in a unit circle (Figure 7). (a) Explain, intuitively, why P(n) approaches 27r as n

oo.

(b) Show that P(n) = 2n sin (k). (c) Combine (a) and (b) to conclude that lim fl,--00

sin (I) = 1. n sin (d) Use this to give another argument that lim = 1. O-->0

7 + llx - 4x4 ,,/36x4 + 7 9x2 + 4 t1/3 (64t2 + 9)1/6

21. f(t)= 1 ± 3-t

22. f(t) =

10 23. g(t)= 1 ± 3-t

24. p(t)= 2-t2 n=

Eoll

The following statement is incorrect: "If f has a horizontal asymptote y = L at oo, then the graph of f approaches the line y = L as x gets greater and greater, but never touches it." In Exercises 25 and 26, determine lim f (x) and indicate how f demonstrates that the statement is incorrect. l 25. f (x) = 2x + ix 26. f (x) =

87

In Exercises 37-44, calculate the limit.

In Exercises 7-16, evaluate the limit.

11. lim

Limits at Infinity

sin x

6

n=

9

n =

12

FIGURE 7 44. Physicists have observed that Einstein's theory of special relativity reduces to Newtonian mechanics in the limit as c oo, where c is the speed of light. This is illustrated by a stone tossed up vertically from ground level so that it returns to Earth 1 s later. Using Newton's Laws, we find that the stone's maximum height is h = gI8 m (g = 9.8 m/s2). According to special relativity, the stone's mass depends on its velocity divided by c, and the maximum height is

In Exercises 27-34, evaluate the limit. 27.

lim x->co

29.

Him

/9.x4 + 3x + 2 4x3 + 1 8x 2

+

7x 173

,16X4

±

Ern x*oo

30.

4x - 3 lim x-"* -00 .../25x2 + 4x

6

t 4/3

31. 33.

lim firn

t 4/ 3 - 9t 1 /3

32.

firn t->oo (8t4 + 2)1/3

34.

4 + 6e2r lim t->---00 5 - 9e3t

(412/ 3 + 1)2

IX I + x x+1

+ 20x 10x - 2

28.

35. 174 Determine lim 5-1/`2 . Explain geometrically. t->00 36. Show that lim (Vx2 + 1 - x) = 0. Hint: Observe that X->00

X2 + 1

- X=

1

,./X 2 +1+ X

h(c) = c c2 I g2 ± 1/4 - c2 /g Prove that lim h(c) = g18. 45. According to the Michaelis-Menten equation, when an enzyme is combined with a substrate of concentration s (in millimolars), the reaction rate (in micromolars/min) is R(s) -

As K +s

(A, K constants)

(a) Show, by computing firn R(s), that A is the limiting reaction rate as the concentration s approaches oo. (b) Show that the reaction rate R(s) attains one-half of the limiting value A when s = K. (c) For a certain reaction, K = 1.25 mM and A = 0.1. For which concentration s is R(s) equal to 75% of its limiting value?

Further Insights and Challenges oo can be expressed alternatively as a one-sided 46. Every limit as x 0+, where t = x-1. Setting g(t) = f (t-1), we have limit as t Ern f(x) .= lirn g(t) t->0+

3- t 3x 2 - x ' and evaluate using the Quotient Show that urn = lim x->oo 2x' + 5 t-A3+ 2 + 5t 2 Law.

88

CHAPTER 2

LIMITS

47. Rewrite the following as one-sided limits as in Exercise 46 and evaluate. 3 — 12x3 (b) lim 31/x (a) Ern X-00 x ,00 4x3 + 3x + 1 (c) lirn x sin — x—>00

48. Let G(b) = urn (1 + bx )1 /x for b > 0. Investigate G(b) numerix—>00 cally and graphically for b = 0.2, 0.8, 2, 3, 5 (and additional values if necessary). Then make a conjecture for the value of G(b) as a function of b. Draw a graph of y = G(b). Does G appear to be continuous? We will evaluate G(b) using L'Hopital's Rule in Section 7.5 (see Exercise 69 there).

2.8 The Intermediate Value Theorem The Intermediate Value Theorem (IVT) says, roughly speaking, that a continuousfunction cannot skip values. Consider a plane that takes off and climbs from 0 to 10,000 m in 20 min. The plane must reach every altitude between 0 and 10,000 m during this 20-min interval. Thus, at some moment, the plane's altitude must have been exactly 8371 m. Of course, this assumes that the plane's motion is continuous, so its altitude cannot jump abruptly from, say, 8000 to 9000 m. To state this conclusion formally, let A(t) be the plane's altitude at time t. The IVT asserts that for every altitude M between 0 and 10,000, there is a time to between 0 and 20 min such that A(to) = M. In other words, the graph of A must intersect the horizontal line y = M [Figure 1(A)]. Altitude (m) 10,000 — 5000 —

to 20 (A) Altitude of plane A(t) Q

f(c) — M

a

i I c

I b

x

FIGURE 2 For every M between f (a) and f (b), there is a c between a and b such that f (c) = M.

y = sin x

0.3

FIGURE 1

By contrast, a discontinuous function can skip values. The greatest integer function f (x) = in Figure 1(B) satisfies L1 = 1 and [2] = 2, but it does not take on the value 1.5 (or any other value between 1 and 2).

f(a)

f(b) 0

y = 0.3

THEOREM 1 Intermediate Value Theorem If f is continuous on a closed interval [a, b], then for every value M, strictly between f (a) and f (b), there exists at least one value c E (a, b) such that f (c) = M. Graphically, as in Figure 2, the result appears obvious. For a continuous function, every horizontal line at height M between f (a) and f (b) is forced to hit the graph, and therefore there must be at least one value c in (a, b) such that f (c) = M. The proof appears in Appendix B. EXAMPLE 1 Prove that the equation sin x = 0.3 has at least one solution in the interval Solution We may apply the IVT because f (x) = sin x is continuous. The desired value 0.3 lies between the values of the function at the endpoints of the interval: sin 0 = 0

7r 2 FIGURE 3

(B) Graph of f(x) = LxJ

and

71"

sin — = 1 2

as illustrated in (Figure 3). The IVT tells us that sin x = 0.3 has at least one solution in (0, S) •

SECTION 2.8

The Intermediate Value Theorem

89

The IVT can be used to show the existence of zeros of functions. If f is continuous and takes on both nonpositive and nonnegative values (say, f (a) 0) then the IVT guarantees that f (c) = 0 for some c in [a, b]. This is extremely useful when we cannot explicitly solve for the zero but would like to know that there is one in the interval. COROLLARY 2 Existence of Zeros If f is continuous on [a, b], and if one of f (a) and f (b) is nonnegative and the other is nonpositive, then f has a zero in [a, b].

1

4.# REMINDER A zero or root of a function is a value c such that f (c) = 0.

We can locate zeros of functions to arbitrary accuracy using the Bisection Method. The idea is to find an interval [a, b] such that the function has opposite signs at the endpoints. Then Corollary 2 tells us that there is a zero on this interval. To find its location more precisely, we cut the interval into two equal subintervals. Then, check the signs at the endpoints of each of these intervals to see which one Corollary 2 tells us has a zero. (But keep in mind that there may be more than one zero, so both could contain a zero.) Next, we repeat the process on this smaller interval. Eventually, we narrow down on a zero. This is illustrated in the next example. EXAMPLE 2 The Bisection Method Show that f (x) = cos2 x — 2 sin 24 has a zero in (0,2). Then, using the Bisection Method, find a subinterval of (0,2) of length 1/8 that contains a zero of f. Solution To begin, we note that f is continuous on [0,2]. Calculating f(0) and f(2), we find that they have opposite signs: f(0) = 1 > 0,

FIGURE 4 Graph of f (x) = c0s2 x —2 sin "4 .

Computer algebra systems have built-in commands for finding roots of a function or solving an equation numerically. These systems use a variety of methods, including versions of the Bisection Method and Newton's Method, a process that we introduce in Section 4.7.

f(2)

—0.786 0)

+ h) — sin 1'6

sin(

+ h) — 0.5

This difference quotient represents the slope of a secant line through the graph at The resulting secant-line slopes are shown in Table 1. Note that Figure 11 indicates that the slope of the tangent line at -74- lies between the slopes of the secant lines for h > 0 and those for h 0. Explain.

for every slope m. Plot y = f(x) and y = L(x) on the same axes for several values of m until you find a value of m for which y = L(x) appears tangent to the graph of f. What is your estimate for f ' (1)? 69. (GU) Use a plot of f(x) = xx to estimate the value c such that f'(c)= 0. Find c to sufficient accuracy so that f (c + h) - f (c)

for

< 0.006

h = +0.001

70. (GU) Plot f(x) = xx and y = 2x + a on the same set of axes for several values of a until the line becomes tangent to the graph. Then estimate the value c such that f'(c) = 2. The vapor pressure of water at temperature T (in kelvins) is the atmospheric pressure P at which no net evaporation takes place. In Exercises 71-72, use the following table to estimate the indicated derivatives using the difference quotient approximation.

Y = f(x)

T (K)

293

303

313

323

343

333

353

P (atm) 0.0278 0.0482 0.0808 0.1311 0.2067 0.3173 0.4754 X

71. Estimate P'(T) for T = 293, 313, 333. (Include proper units on the derivative.)

(A)

72. Estimate P'(T) for T = 303,323, 343. (Include proper units on the derivative.)

FIGURE 17 66. E/1 Refer to the graph of f(x) = 2X in Figure 18. (a) Explain graphically why, for h > 0,

f(-h) -f(0) -h

i f (0)
0? Explain in terms of secant lines.

x

, X

(B) FIGURE 21

85. E4 Show that if f is a quadratic polynomial, then the SDQ at x = a (for any h 0) is equal to f'(a). Explain the graphical meaning of this result. 86. Let f (x) = x-2. Compute [(1) by taking the limit of the SDQs (with a = 1) ash O.

3.2 The Derivative as a Function In the previous section, we computed the derivative f'(a) for specific values of a. It is also useful to view the derivative as a function f' whose values f'(x) are defined by the limit definition of the derivative: f'(x) = lim

f (x

h->.O

If y = ,f (x), we also write y' or y'(x) for f'(x).

h) - f (x)

1

SECTION 3.2

Often, the domain off' is clear from the context. If so, we usually do not mention the domain explicitly

The Derivative as a Function

113

The domain of f' consists of all values of x in the domain of f for which the limit in Eq. (1) exists. We say that f is differentiable on (a, b) if f'(x) exists for all x in (a, b). When f'(x) exists for all x in the interval or intervals on which f (x) is defined, we say simply that f is differentiable. EXAMPLE 1 Prove that f (x) = x3 — 12x is differentiable. Compute f'(x) and find f'(-3), f' (0), f'(2), and (3). Solution We compute f'(x) in three steps as in the previous section. Step I. Write out the numerator of the difference quotient. f (x + h) — f (x) = ((x + h)3 — 12(x + h)) — (x3 — 12x) = (x3 + 3x2 h + 3xh2 + h3 — 12x — 12h) — (x3 — 12x) = 3x2h + 3xh2 + h3 — 12h = h(3x2 + 3x h + h2 — 12)

(factor out h)

Step 2. Divide by h and simplify. f (x + h) — f (x)

=

h(3x2 + 3x h + h2 — 12)

—

2

3x + 3x h + h2 — 12

(h

0)

Step 3. Compute the limit. f' (x) = lim

f (x + h) — f (x)

h —>0

= lim(3x2 + 3xh + h2 — 12) = 3x2 — 12 h—>0

h

In this limit, x is treated as a constant because it does not change as h —> 0. We see that the limit exists for all x, so f is differentiable and f'(x) = 3x2 — 12. Now evaluate: • • • •

f' ( —3) = 3(-3)2 — 12 = 15 PO) = 3(0)2 — 12 -= —12 f'(2) = 3(2)2 — 12 = 0 f'(3) = 3(3)2 — 12 = 15

These derivatives indicate the slope of the graph of f (and the tangent line to the • graph) at the corresponding points, as shown in Figure 1. 1 The derivative at each point is the slope of the tangent line.

FIGURE

In the previous example, we used the definition of f'(x) to find an equation for the derivative as a function of x. Using the formula for (x) we computed the derivative at specific values of x rather than computing each derivative separately using the limit definition. As we proceed with the development of the derivative, one goal will be to develop formulas and rules for the derivatives of functions so that we do not have to return to the limit definition for every derivative computation. EXAMPLE 2 Prove that y = X-2 is differentiable and calculate y'. Solution The domain of f (x) = x-2 is Ix : x 0), so assume that x f'(x) directly, without the separate steps of the previous example:

f'(x) =

urnf (x + h) — f (x) h—>0

=11m h—>0

h

x2 — (x + h)2 x2(x + h )2

= lim

1 (x + h)2

1

h—>-0

=lim h—>0

+ 1 (X 2 — h x2(x + h)2 )

0. We compute

114

CHAPTER 3

DIFFERENTIATION

= urn

2x + h

1 (-h(2x + h) = Ern h—>0 x2(x + h)2 )

x2(x ± h)2

h-->0 h

=

2x + 0 = X2'kX +0)2

(cancel h)

2x = -2x -3 x4

The limit exists for all x 0 0, so y = X-2 is differentiable and y' = -2x-3.

•

Leibniz Notation The "prime" notation y' and f'(x) was introduced by the French mathematician Joseph Louis Lagrange (1736-1813). There is another standard notation for the derivative that we owe to Gottfried Wilhelm Leibniz: dy df or dx dx In Example 2, we showed that the derivative of y = X-2 is y' = -2x-3. In Leibniz notation, we would write dy —=-2x -3 dx

or

d —x-2 = -2x-3 dx

To specify the value of the derivative for a fixed value of x, say, x = 4, we write df dx x=4

or

dy dx x=4

You should not think of dy/dx as the fraction "dy divided by dx." Separately, the expressions dy and dx are called differentials. They play a role in linear approximation (Section 4.1), and relationships between them are used as a guide for "substitutions" we do later when working with integrals.

1

We read dx and iy dx with respect to x."

as "the derivative of y

CONCEPTUAL INSIGHT Leibniz notation is widely used for several reasons. First, it reminds us that the derivative df/dx, although not itself a ratio, is in fact a limit of ratios Al./6,x. Second, the notation specifies the independent variable. This is useful when variables other than x are used. For example, if the independent variable is t, we write dfldt. Third, we often think of dldx as an "operator" that performs differentiation on functions. In other words, we apply the operator dldx to f to obtain the derivative df/dx. We will see other advantages of Leibniz notation when we discuss the Chain Rule in Section 3.7. Now we are ready to start assembling a collection of derivative rules and formulas that will enable us to compute derivatives of the most common functions in mathematics, the sciences, and engineering. We begin with two simple formulas that are consequences of Theorem 1 in the previous section: d —x = 1 dx

and

d — c = 0 for any constant c dx

The first indicates that the derivative with respect to x of x is 1, reflecting that the slope of the line y = x is 1. The second, known as the Constant Rule, indicates that the derivative of a constant is 0. This makes sense, of course, since a constant does not change and therefore has a rate of change of zero. As simple as the latter is, we will find it quite useful as we work with derivatives throughout the book. The next theorem will prove to be very valuable for differentiating polynomial functions and many other types of functions involving constant powers.

SECTION 3.2

THEOREM 1 The Power Rule

The Derivative as a Function

115

For all exponents n: d n —x dx

The Power Rule is valid for all exponents. We prove it here for positive integers n. See Exercise 93 for a proof for negative integers n and the marginal note on page 365 for arbitrary n. Proof We have f'(x) = lim

(x + h)n

—

xn

h—>0

To simplify the difference quotient, we need to expand the (x + On term. However, we do not need to write all the terms to work through the limit. The binomial expansion formula helps here (see Section 1.1); it indicates that expanding (x + On results in a n! xn— P hP , one for each p from 0 to n. For p = 0 we have xn , for sum of terms p!(n— p)! p = 1 we have nxn-l h, and for the rest of the terms, it is enough to observe that they are terms containing xq hP for p > 2. Thus, for our purposes, we can express the expansion of (x h)n as (x +

nXn-1 h

= Xn

+ [terms with xq hP , p > 2]

Now, we have f ,(x) = hip (x + h )n

— Xn

h—>0

= lim

(xn + nxn-1 h + [terms with xq hP , p > 2]) - xn

h—>0

= lim

nxn-1 h + [terms with xq hP , p > 2]

h—>0

= lim (nxn-1 + [terms with xq hP , p > 1]) h—>0

= nx n-1 0 because those At the last stage, the limit of the terms containing xq hP equal 0 as h terms include a factor of h to at least the first power. • This proves that f' (x) = nxn-1 for the case where n is a positive integer. We make a few remarks before proceeding: • The Power Rule in words: To differentiate x to a power, multiply by the power and reduce the power by one. _ x power = (power) xpowerdx • The Power Rule is valid for all exponents, whether positive, negative, fractional, or irrational: x"4 = = --x -8/5 — dx 5 dx • The Power Rule can be applied with any variable, not just x. For example, d d 1.20 = d 2 — r 1/2 = — r —1/2 20t19, — z = 2z, 2 dr dt dz

116

CHAPTER 3

DIFFERENTIATION

Next, we state the Linearity Rules for derivatives, which are analogous to the Linearity Laws for limits: THEOREM 2 Linearity Rules

Assume that f and g are differentiable. Then

Sum and Difference Rules:

f + g and f - g are differentiable, and =

(f

fl

g l,

( f

Constant Multiple Rule:

g/

= g) '

g) '

f'

For any constant c, cf is differentiable, and (cf)' = cf'

Proof To prove the Sum Rule, we use the definition (f + g)'(x) = lirn h->o

(f (x + h) + g(x + h)) - (f (x) + g(x))

This difference quotient is equal to a sum (h

0):

(f (x + h) + g(x + h)) - (f (x) + g(x))

f (x + h) - f (x)

g(x

h) - g(x)

Therefore, by the Sum Law for limits, (f + g) (x) = lim

f (x + h) - f (x)

h—*0

h

+ lim h-o3

g(x

h) - g(x)

= 1(x) + g'(x) as claimed. The Difference and Constant Multiple Rules are proved similarly.

•

In words, these rules state: • The derivative of a sum is the sum of the derivatives. • The derivative of a difference is the difference of the derivatives. • The derivative of a constant times a function is the constant times the derivative of the function.

EXAMPLE 3 Find the points on the graph of f (t) = t3 - 12t + 4 where the tangent line is horizontal. Solution We calculate the derivative: df dt f 40

±(t3_ Tt 12t + 4) d

d d t' - —(12t) + — 4 dt dt dt d 3 = — t — l2 —t +0 dt dt =

= 3t2 - 12

(Sum and Difference Rules) (Constant Multiple Rule and Constant Rule) (Power Rule)

The tangent line is horizontal at points where the slope f'(t) is zero, so we solve —40 FIGURE 2 Graph of f (t) = t3 — 12t + 4. Tangent lines at t = ±2 are horizontal.

3t2 —

12 = 0

=

t = ±2

Now f(2) = -12 and f(-2) -= 20. Hence, the tangent lines are horizontal at (2, -12) and (-2,20), as shown in Figure 2. •

SECTION 3.2

EXAMPLE 4 Calculate

dg dt

The Derivative as a Function

7 where g(t) = t 3

117

-

t=1

Solution We differentiate term-by-term using the Power Rule and the Linearity Rules. Writing ,N/i as t112, we have dg d = -(t-3 _,_ 2t1/2 t -4/S) — dt = -31.-4 + t -1/2 + 4 t-9/5 5 dg 4 6 So — = -3 + 1 + - = - 7 dt r =1 5 5

1 +2 2

t-1/2 —

t -9/5

•

EXAMPLE 5 A power-law model relating the pulse rate P (in beats per minute) in mammals to body mass m (in kilograms) is given by P = 200m-1/4 (see Figure 3). It is clear from the graph that, from species to species, as the mass increases, the pulse rate drops off. Furthermore the pulse rate drops off quickly for small mammals but relatively slowly for large mammals. Determine 131(m) at the mass of a guinea pig (1 kg) and at the mass of cattle (500 kg). Pulse (beats/min) . . Guinea.Oig...I. „ .

. ..

. .

.

..

..

... .

. ...

. .

.

i

.

..I

.. attle

200 : 100

.. i

.. 100

.. i

.... i

200

300

..

.. 400

500

Mass (kg)

FIGURE 3

Here tangent lines have negative slope. 16

Solution Applying the Power Rule and the Constant Multiple Rule to P(m) = 200m-1/4, we obtain P'(m) = -50m-5/4. So, • P'(1) = -50 beats per minute per kilogram • P(500) = -50(500-514) -0.02 beats per minute per kilogram These values confirm our observation that P decreases rapidly for small m and decreases • slowly for large m.

—16 (A) Graph of f(x) = x3 — 12x2 + 36x — 16

The Derivative and Behavior of the Graph The derivative f' gives us important information about the graph of f. For example, the sign of f'(x) tells us whether the tangent line has positive or negative slope. When the tangent line has positive slope, it slopes upward and the graph must be increasing. When the tangent line has negative slope, it slopes downward and the graph must be decreasing. The magnitude of f'(x) reveals how steep the slope is. EXAMPLE 6 f' and the Graph of f How is the graph of f (x) = x3 - 12x2 + 36x -16 related to the derivative f'(x) = 3x2 - 24x + 36?

(B) Graph of the derivative f'(x) = 3x2 — 24x + 36 FIGURE 4

Solution The derivative f'(x) = 3x2 - 24x + 36 = 3(x - 6)(x - 2) is negative for 2 6

Tangent has negative slope for 2 0. Show that the area of the triangle bounded by L and the coordinate axes does not depend on a. 69. In the setting of Exercise 68, show that the point of tangency is the midpoint of the segment of L lying in the first quadrant.

FIGURE 17

124

DIFFERENTIATION

CHAPTER 3

(GU) In Exercises 75-80, zoom in on a plot of f at the point (a, f (a)) and state whether or not f appears to be differentiable at x = a. If it is nondifferentiable, state whether the tangent line appears to be vertical or does not exist. 75. f (x) = (x — 1)Ixl,

a=0

76. f(x) = (x — 3)513,

77. f (x) = (x — 3)1/3,

a =3

78. f (x) = sin(x113),

79. f (x) = I sin

a =0

a=3

83. Calculate the subtangent of at x = 2

f (x) = x2 + 3x 84. Show that for n to cln.

0, the subtangent of f (x)= xn at x = c is equal

85. Prove in general that the subnormal at P is I f'(x)f (x)I.

80. f (x) =

— sinx I,

a=0

86. Show that PQ has length I f(x)I /1 + f (x)-2.

a=0

81. Find the coordinates of the point P in Figure 18 at which the tangent line passes through (5,0).

Y = f(x)

P = (x, f(x))

Tangent line

FIGURE 19 FIGURE 18 87. Prove the following theorem of Apollonius of Perga (the Greek mathematician born in 262 BCE who gave the parabola, ellipse, and hyperbola their names): The subtangent of the parabola y = x2 at x = a is equal to al2.

—1 for x > 0 and 82. (GU) Plot the derivative f' of f (x) = 2x3 — observe that / (x) > 0. What does the positivity of f'(x) tell us about the graph of f itself? Plot f and confirm this conclusion.

88. Show that the subtangent to y = x3 at x = a is equal to a/3.

Exercises 83-86 refer to Figure 19. Length QR is called the subtangent at P, and length RT is called the subnormal.

89. 17 y = x".

Formulate and prove a generalization of Exercises 87 and 88 for

Further Insights and Challenges 90. Two small arches have the shape of parabolas. The first is the graph of f (x) = 1 —x2 for —1 0+

f(c + h)— f(c)

exist but are not equal, then f is not differentiable at c, and the graph of f has a comer at c. Prove that f is continuous at c.

3.3 Product and Quotient Rules

1

1.10 REMINDER The product function fg is defined by (fg)(x)= f(x)g(x).

This section covers the Product Rule and Quotient Rule for computing derivatives. These two rules, together with the Chain Rule and implicit differentiation (covered in later sections), make up an extremely effective differentiation toolkit. THEOREM 1 Product Rule entiable and

If f and g are differentiable functions, then f g is differ-

(f g)/ (x) = f (x) g(x) + f (x)

(x)

It may be helpful to remember the Product Rule in words: The derivative of a product of terms is equal to the derivative of thefirst term times the second plus thefirst term times the derivative of the second: (first)' • second + first • (second)' Be careful when taking the derivative of products. The product rule is not (f g)' = f g'; that is, it does not say that the derivative of a product is the product of the derivatives. We prove the Product Rule after presenting some examples. EXAMPLE 1 Find the derivative of h(x)

x2(9x +2).

Solution This function is a product: First

Second

h(x) = x2 (9x + 2) By the Product Rule (in Leibniz notation), (First)'

. First

Second

(Second)'

d d h'(x) = —(x 2 ) (9x + 2) + (x2) —(9x + 2) dx dx = (2x)(9x + 2) + (x2)(9) = 27x2 + 4x EXAMPLE 2 Find the derivative of y = (2 ± x-1)(x372 + 1). Solution Use the Product Rule: (First)' • Second +First • (Second)' Note how the prime notation is used in the solution to Example 2. We write (x3/2 + 1)' to denote the derivative of X3/2 + 1.

yl = (2 ± x-1)' (x3/2 ± 1) ± (2 ± x-1) (x3/2 ± 1)'

3 x1/2 (compute the derivatives) = ( - x-2)(x3/2 ± 1) + (2 ± x -1) ( 2 , — 1 / 2 _ 1 x-1/2 _ x-2 ± 3x112 (simplify) , —x -1/2 _ x-2 ± 3x112 _,_ 2 '''' ' 2

— 2

126

CHAPTER 3

DIFFERENTIATION

In the previous two examples, we could have avoided the Product Rule by expanding the function. Thus, the result of Example 2 can be obtained as follows: y = (2 + x-i)(x3/2 + 1) =2x312 + 2 + x 1/2 +x 1 /2 y' -= — (2x3/2 + 2 + X 172 ± x-1 ) = 3x i/ 2 _.,_ 2""v -1 _ x -2 dx In many cases, the function cannot be expanded and we must use the Product Rule. One such function is f(x) = x cos x whose derivative we find in Section 3.6.

A(t)

EXAMPLE 3 Figure 1 depicts a rectangle whose length L(t) and width W(t) (measured in inches) are varying in time (t, in minutes). At t =- 5, the length is 8, the width is 5, and they are changing according to L'(5) = -4 and W'(5) = 3. Compute A'(5). W(t)

Solution Since the area is given by A(t) L(t)W(t), we can use the Product Rule to compute AV). We have A1(t) = Li(t)W(t) + L(t)W'(t). Therefore, L(t)

•

A'(5) = (-4)(5) + (8)(3) = 4

FIGURE 1 The length and width of the

It follows that the area of the rectangle in the example is increasing at a rate of 4 in.2/min at t = 5. This may appear counterintuitive, given that the length is decreasing at a faster rate than the width is increasing. What is important, as the Product Rule demonstrates, is that the decreasing length acts over a short width of 5, contributing -20 to the rate of change of area, while the increasing width acts over a long length of 8, contributing 24 to the rate of change of area, resulting in an increasing area.

rectangle change in time.

Proof of the Product Rule According to the limit definition of the derivative, (fa (x) = lim

f(x + h)(g(x + h) - g(x))

f(x)

f(x + h)

h->0

g(x)(f(x + h) - f(x))

el

f (x + h)g(x + h) - f (x)g(x)

We can interpret the numerator as the area of the shaded region in Figure 2: the area of the larger rectangle f(x + h)g(x + h) minus the area of the smaller rectangle f (x)g(x). This shaded region is the union of two rectangular strips, so we obtain the following identity [which we can also obtain algebraically by adding and subtracting the term f(x + h)g(x) from the left-hand side and then manipulating the result algebraically]: f(x + h)g(x + h) - f (x)g(x) = (f(x + h) - f (x))g(x) + f(x + h)(g(x + h) - g(x)) Now use this identity to write (fg)/(x) as a sum of two limits:

FIGURE 2

( h) - g(x) h) - f (x) g(x) + lim f (x + h)g x ± h h -o-0 h

f

g)/ (x)

h->0

We show that this equals f '(x)g(x).

I1

We show that this equals f (x)g'(x).

The use of the Sum Law is valid, provided that each limit on the right exists. To check that the first limit exists and to evaluate it, we note that f is differentiable. Thus, Jim f (x + h) - f (x) f (x + h) - f (x) iim g(x) g(x) = lim h h->0 h h-> 0

h-÷0

= f'(x) g(x)

2

The second limit is similar, but using the facts that f is continuous (because it is differentiable) and g is differentiable: lim f(x + h)

h->0

g(x + h) - g(x) h

= .

h-±

f(x + h)

.

h-->

g (x + h) - g(x)

= f(x)

(x) j3

Using Eq. (2) and Eq. (3) in Eq. (1), we conclude that f g is differentiable and that (f g)'(x) = f'(x)g(x) + f (x)g'(x) as claimed. •

SECTION 3.3

Product and Quotient Rules

127

CONCEPTUAL INSIGHT The Product Rule was first stated by the 29-year-old Leibniz in 1675, the year he developed some of his major ideas on calculus. To document his process of discovery for posterity, he recorded his thoughts and struggles, the moments of inspiration as well as the mistakes. In a manuscript dated November 11, 1675, Leibniz suggests incorrectly that (fg)' equals f g'. He then catches his error by taking f (x) = g(x) x and noticing that (f

(x) = (x2)' = 2x

is not equal to

f' (x)g'(x) = 1 • 1 = 1

Ten days later, on November 21, Leibniz writes down the correct Product Rule and comments, "Now this is a really noteworthy theorem." ad REMINDER

The quotient function f/ g

is defined by

(x) = f (x) g

The next theorem states the rule for differentiating quotients. Note, in particular, that (fIg)' is not equal to the quotient f' /g'; the derivative of the quotient is not the quotient of the derivatives.

g(x) THEOREM 2 Quotient Rule If f and g are differentiable functions, then f g is differentiable for all x such that g(x) 0, and )f i(x ) f (x)g'(x) ( 1 ) ' (x) = g g(x)2

The numerator in the Quotient Rule is the bottom times the derivative of the top minus the top times the derivative of the bottom. The denominator is the bottom squared: bottom • (top)' — top • (bottom)' bottom2 Proof of the Quotient Rule Let Q(x) = f (x)I g(x). Our goal is to find the formula for Q'(x). First, we multiply the equation for Q(x) through by g(x), then use the product rule on the result, and finally solve for Q'(x). So, multiplying by g(x) we have f (x) = Q(x) • g(x). Differentiating both sides, utilizing the Product Rule for the right side, we obtain f'(x) = Q'(x) • g(x) + Q(x) • g'(x). Solving for Q'(x), we obtain Q,(x)

f'(x) — Q(x) • g'(x) g(x)

f'(x) - 0 • ex)) g(x)

f'(x)g(x) — f (x)g'(x) g(x)2 •

as we wanted to show. An alternative proof appears in Exercises 66-68. EXAMPLE 4 Compute the derivative of f(x) —

1

Solution Apply the Quotient Rule: Bottom

f'(x) =-

(1 ± .X2)

(Top)'

Top

(Bottom)'

(x)' — (x) (1 + x2r (1 + x2)2

1 +. x2 — 2x2

1 — x2

(1 + x2 )2

(1 + x2)2

(1 + x2)(1) — (x)(2x) (1 + x2)2

•

128

CHAPTER

3

DIFFERENTIATION

EXAMPLE 5 Find the tangent line to the graph of f(x) = Solution

(4x3 + 1) (3x2 + x — 2)' — (3X 2 (4x 3 ± 1)2

(3x2 x — 4x3 + 1 dx

(Bottom)'

Top

(Top)'

Bottom

(x) =

3x2 +x —2 at x = 1. 4x3 + 1

(4x3 ± 1)(6x ± 1) — (3x2 (4x3 + 1) 2

+

X —

2) (4x3 + 1)'

x —2)(12x2)

(24x4 + 4x3 + 6x + 1) — (36x4 + 12x3 —24x2) (4x3 ± 1) 2 24X 2 +

—12X 4 — 8X 3

(4x 3 Note that it is not always the simplest method to apply the Quotient Rule. If we want to differentiate the function — X W(X) =

f(1) = 3 + 1 — 2 2 4+1 5 11 —12-8+24+6+1 =s f'(1) = 52

is easier to rewrite it

as w(x) = X -1/2 - X 2 and then differentiate it directly.

An equation of the tangent line at (1, i) is y—

FIGURE 3 Apparatus of resistance R attached to a battery of voltage V.

6x ± 1

1) 2

At x = 1,

3 , it

+

2 7 5

=

11 —(x — 1) 25

11 1 y = —x — — 25 25

or

•

EXAMPLE 6 Power Delivered by a Battery The power that a battery supplies to an apparatus such as a laptop depends on the internal resistance of the battery. For a battery of voltage V and internal resistance r, the total power delivered to an apparatus of resistance R (Figure 3) is V 2R P= (R + r)2 (a) Calculate dP1dR, assuming that V and r are constants. (b) Where, in the graph of P versus R, is the tangent line horizontal? Solution (a) Using the Constant Multiple Rule (V is a constant) and the Quotient Rule, we obtain dP 2 d =V dR dR

R 2 (R + =V (R + r)2 )

R f i,(R + r)2 (R + r)4

I 4

We have ; R R = 1, and ,d -FR-r = 0 because r is a constant. Therefore, d — (R + r)2 = dR dR d 2 R dR

+ 2r R + r2)

+

d d 2 2r — R + dR dR

= 2R + 2r + 0 = 2(R + r) Using Eq. (5) in Eq. (4), we obtain dP dR

112

(R + r)2 — 2R(R + r) = 2 (R + r) — 2R r—R V — V2 (R + r)4 (R + r)3 (R + r)3

6

(b) The tangent line is horizontal when the derivative is zero. We see from Eq. (6) that the derivative is zero when r — R = 0; that is, when R = r. •

SECTION 3.3

Product and Quotient Rules

129

GRAPHICAL INSIGHT Figure 4 shows that the point where the tangent line is horizontal is the maximum point on the graph. This proves an important result in circuit design: Maximum power is delivered when the resistance of the load (apparatus) is equal to the internal resistance of the battery.

3.3 SUMMARY FIGURE 4 Graph of power versus resistance: V2 R P(R + r)2

• Two basic rules of differentiation: ( fg

Product Rule:

(

Quotient Rule:

)'_

fg ) I

fl

g

+

f

g ,

2 gt g fgf

• Remember: The derivative of fg is not equal to f'g'. Similarly, the derivative of flg is not equal to f' g'

3.3 EXERCISES Preliminary Questions 1.

Are the following statements true or false? If false, state the correct version.

= f (4)g'(4) - g(4)f'(4) (d) -• (fg) dx x=4

(a) f g denotes the function whose value at x is f (g(x)).

= f'(0)g(0)+ f (0)g'(0) (e) -• (fg) dx -0

(b) f g denotes the function whose value at x is f (x)I g(x).

2.

Find (f/g)'(1) if f(1) = [co= g(1) = 2 and g'(1) ----- 4.

(c) The derivative of the product is the product of the derivatives.

3.

Find g(1) if f (1) = 0, f' (1) = 2, and (f g)/ (1) = 10.

Exercises In Exercises 1-6, use the Product Rule to calculate the derivative.

17. h(t) -

1.

f (x) = x3(2x2 +1)

2.

f (x) = (3x - 5)(2x2 - 3)

3.

f (x) = .1i(1 - x3)

4.

f (x) = (3x4 + 2x6)(x - 2)

5.

dh ds

6.

, h(s) = (s-112 + 2s)(7 s=4

y = (t - 8t-1)(t +t 2)

dg dt

, g(t) t=-2

11. g(x) -

t2 + 1 2_I

1 + X 312

1

18. g(x) =

t- 1

x3 + 2x2 + 3x-1

In Exercises 19-40, calculate the derivative. 19. f (x) = (x3 + 5)(x3 + x + 1)

20. f (x) =

dy 21. dx

22.

- x2) (x3 + 1)

)

In Exercises 7-12, use the Quotient Rule to calculate the derivative. x+4 8. f (x) 7. f (x) = X2 + X + x- 2 9.

t2 -

to.

dw dz

,

w=

z=9

Z2

s3/2 12. h(s) - s2 ± 1

In Exercises 13-18, calculate the derivative in two ways. First use the Product or Quotient Rule; then rewrite the function algebraically and directly calculate the derivative. X

X + 10

x=3

14. h(x) =

15. f (t) = (2t + 1)(t2 -2)

16. f (x) .= x2(3 +x-1)

,

z=

x=-2

9x512 1)(Nrx - 1)

23. f (x) =(./+ dy 25. dx ,

z,/.

dz dx

X

x=2

, x=1

29. h(t) =

Z -

4

x2 - 5

1

26. f (x) =

28. f (x) =

X3 + 1

3x2 + 1 2

24. f (x) =

A

-

x 4 + x -1 x+1 3x3 - x2 + 2 ,Nrx

(t +1)(t2 +1)

30. f (x) = X3/2 (2,4

3x ± x-1/2)

5

13. f (x) = X 3X-3

dz dx

31. f (x) = x2/3(x2 -1)

32. h(x) = 72(x - 1)

33. f (x) = (x + 3)(x - 1)(x -5)

34. h(s)= s(s + 4)(s2 + 1)

130

DIFFERENTIATION

CHAPTER 3

35. f (x)= 37. g(z)

X

3/2(x2 ± 1)

36. g(z)=

x+1

Hint: Simplify first.

40.

d (ax +E.) +d dx

2- x

x- 1 x +8

at x = a

passes through the origin (Figure 5).

(a, b constants)

38. - ((ax + b)(abx2 +1)) dx d Oct - 4 ) dt (\t

57. Find all values of a such that the tangent line to f (x)=

z2 - 1

z2 -4

39.

(z - 2)(z2 + 1)

-8

(x constant)

(a, b, c, d constants)

In Exercises 41-44, calculate f'(x) in terms of P(x), Q(x), and R(x), asP(x). suming that P'(x)= Q(x), Q'(x)= -R(x), and R' 41. f (x)= x R(x) + Q(x)

42. f (x)= Q(x)P(x)

P(x) - x 43. f(x)= Q(x)

44. f(x) -

Q(x)R(x) P(x)

In Exercises 45-48, calculate the derivative using the values: f(4)

f'(4)

g(4)

g'(4)

10

-2

5

-1

45. (fg)'(4) and (f/g)'(4) 46. F'(4), where F(x)= x2 f(x) 47. G'(4), where G(x)= (g(x))2

FIGURE 5 58. Current I (amperes), voltage V (volts), and resistance R (ohms) in a circuit are related by Ohm's Law, I =VIR. if V is constant with value V = 24.

(a) Calculate dR

R=6

dV (b) Calculate dR

R=6

if! is constant with value! = 4.

59. The revenue per month earned by the Couture clothing chain at time t is R(t)= N(t)S(t), where N(t) is the number of stores and S(t) is average revenue per store per month. Couture embarks on a two-part campaign: (A) to build new stores at a rate of five stores per month, and (B) to use advertising to increase average revenue per store at a rate of $10,000 per month. Assume that N(0) = 50 and S(0) = $150,000. (a) Show that total revenue will increase at the rate dR

48. H'(4), where H(x) =

dt =

g(x) f (x)

In Exercises 49 and 50, a rectangle's length L(t) and width W(t) (measured in inches) are varying in time (t, in minutes). Determine A'(t) in each case. Is the area increasing or decreasing at that time? 49. At t = 3, we have L(3) = 4, W(3) = 6, L'(3) = -4, and W'(3) = 5. 50. At t = 6, we have L(6) = 6, W(6) = 3, L'(6) = 5, and W'(6) = -2. 51. Calculate F'(0), where x9 + x8 + 4x5 -7x F(x) = x4 - 3x2 ± 2x + I Hint: Do not calculate F'(x). Instead, write F(x) = f (x)Ig(x) and express F'(0) directly in terms of f(0), f'(0), g(0), g'(0). 52. Proceed as in Exercise 51 to calculate F'(0), where F(x) = (1 + x + x4/3

5X4 ± 5x + 1 8x9 -7x 4 +1

+ x5/3) 3X5

55(t)± 10,000N(t)

Note that the two terms in the Product Rule correspond to the separate effects of increasing the number of stores and the average revenue per store. dR (b) Calculate dt f=0 (c) If Couture can implement only one leg (A or B) of its expansion at t = 0, which choice will grow revenue most rapidly? 60. The tip speed ratio of a turbine is the ratio R = T/ W, where T is the speed of the tip of a blade and W is the speed of the wind. (Engineers have found empirically that a turbine with n blades extracts maximum power from the wind when R = 27r/n.) Calculate dRldt (t in minutes) if W = 35 km/h and W decreases at a rate of 4 km/h per minute, and the tip speed has constant value T = 150 km/h. 61. The curve y = 1/(x2 + 1) is called the witch of Agnesi (Figure 6) after the Italian mathematician Maria Agnesi (1718-1799). This strange name is the result of a mistranslation of the Italian word la versiera, meaning "that which turns." Find equations of the tangent lines at x = ±1.

53. Verify the formula (x3)' = 3x2 by writing x3 = x • x • x and applying the Product Rule. 54. (--) Plot the derivative of f (x)= x1(x2 + 1) over [-4,4]. Use the graph to determine the intervals on which f'(x) > 0 and f'(x) < 0. Then plot f and describe how the sign of f'(x) is reflected in the graph off. 55. 12L. 11 Plot f(x)= x1(x2 - 1). Use the plot to determine whether f'(x) is positive or negative on its domain (x : x ±1}. Then compute f'(x) and confirm your conclusion algebraically. 56. Let P = V2 R /(R + r)2 as in Example 6. Calculate dPldr, assuming that r is variable and R is constant.

-

-2

I

-1

3

FIGURE 6 The witch of Agnesi. 62. Let f(x)= g(x)= x. Show that (fig)' 63. Use the Product Rule to show that (f 2)' 64. Show that (f 3)'

=

3 f2f'

f' I g'. =

2ff'.

X

SECTION 3.4

Rates of Change

131

Further Insights and Challenges 65. Let f, g,h be differentiable functions. Show that (fgh )'(x) is equal to f' (x)g(x)h(x) + f (x)g'(x)h(x)-1- f (x)g(x)h'(x)

Exercises 72 and 73: A basicfact of algebra states that c is a root of a polynomial f if and only if f (x) = (x — c)g(x) for some polynomial g. We say that c is a multiple root if f (x) = (x — c)2h(x), where h is a polynomial.

Hint: Write fgh as f (gh). 66. Prove the Quotient Rule using the limit definition of the derivative. 67. Derivative of the Reciprocal d dx

1 f (x)

72. Show that c is a multiple root off if and only if c is a root of both f and f'.

Use the limit definition to prove 73. Use Exercise 72 to determine whether c = —1 is a multiple root. (a) x5 + 2x4 — 4x3 — 8x2 — x + 2

f'(x) f 2(x)

Hint: Show that the difference quotient for 1/1(x) is equal to

(b) x4 ± x3 — 5x2 — 3x + 2

nr

f(x) — f (x +h) hf (x)f (x + h) 68. Prove the Quotient Rule using Eq. (7) and the Product Rule.

74. Figure 7 is the graph of a polynomial with roots at A, B, and C. Which of these is a multiple root? Explain your reasoning using Exercise 72.

69. Use the limit definition of the derivative to prove the following special case of the Product Rule: d — (xf (x)) = f (x) + xf'(x) dx 70. Use the limit definition of the derivative to prove the following special case of the Quotient Rule: d (f (x)) = xf'(x) — f (x) x2 dx x 71. The Power Rule Revisited If you are familiar with proof by induction, use induction to prove the Power Rule for all whole numbers n. Show that the Power Rule holds for n = 1; then write x" as x • xn —I and use the Product Rule.

FIGURE 7

3.4 Rates of Change In this section, we pause from building tools for computing the derivative and instead focus on the derivative as a rate of change, particularly in applied settings. Recall the notation for the average rate of change of a function y = f (x) over an interval [xo, xi]: Ay= change in y = f (xi) -

f (xo)

Ax = change in x = xi - xo AY average rate of change = — Ax We usually omit the word "instantaneous" and refer to the derivative simply as the rate of change. This is shorter and also more accurate when applied to general rates, because the term "instantaneous" would seem to refer only to rates with respect to time.

f (xi) — f (xo) xi - xo

In our prior discussion in Section 2.1, limits and derivatives had not yet been introduced. Now that we have them at our disposal, we can define the instantaneous rate of change of y with respect to x at x = xo: Ay Inn instantaneous rate of change = f(x 0) = urn— = X1-*X Ax->0 Ax

f (xi) — f (x0) X - X0

Keep in mind the geometric interpretations: The average rate of change is the slope of the secant line (Figure 1), and the instantaneous rate of change is the slope of the tangent line (Figure 2).

132

CHAPTER 3

DIFFERENTIATION

FIGURE 2 The instantaneous rate of change at xo is the slope of the tangent line.

FIGURE 1 The average rate of change over [xo, xi] is the slope of the secant line.

Mexey Stiop/Shutterstock

How important are units? In September 1999 the $125 million Mars Climate Orbiter spacecraft burned up in the Martian atmosphere before completing its scientific mission. According to Arthur Stephenson, NASA chairman of the Mars Climate Orbiter Mission Failure Investigation Board, 1999, "The 'root cause' of the loss of the spacecraft was the failed translation of English units into metric units in a segment of ground-based, navigation-related mission software."

Leibniz notation dy/dx is particularly convenient because it specifies that we are considering the rate of change of y with respect to the independent variable x. The rate dy/dx is measured in units of y per unit of x. For example, the rate of change of temperature with respect to time has units such as degrees per minute, whereas the rate of change of temperature with respect to altitude has units such as degrees per kilometer. In applications, it is important to be mindful of the units on rates of change and to interpret properly what the rate of change is communicating about the variables. EXAMPLE 1 Surface Temperatures During an Eclipse Since the moon has no atmosphere to help moderate the temperature, its surface experiences large extremes in temperature (typically between —150°C and 120°C). Furthermore, the primary source of heat is direct radiation from the sun, so a location experiences its highest temperatures when in direct sunlight and lowest when in darkness. With a "day" on the moon being 29.5 Earth days, a location on the moon will cycle between hottest and coldest temperatures over such a period of time. The only situation where this temperature-change cycle is disrupted is when the earth blocks the sun (that is, when a lunar eclipse is seen from Earth, as in Figure 3). During an eclipse, the temperature on the moon drops quickly, but then rebounds once the earth passes by the sun. Table 1 contains data on the temperature (T, in °C) at a location on the moon t minutes into an eclipse. TABLE 1

FIGURE 3 A time lapse of a lunar eclipse.

t

0 20 40 7

T 45 37

60 33

80 61

100 120 140 160 180 200 78 86 91 95 97 99

220 240 260 280 300 101 —84 —33 12 37

(a) Calculate the average rate of change of temperature T from the start of the eclipse to the time t* when the temperature was coldest and the average rate of change from t* to the end of the eclipse. (b) Use the difference quotient approximation to estimate the rate of change of the temperature 60, 160, and 260 min into the eclipse. Solution (a) From the table, we use t* = 220. The average rate of change from eclipse start to t* is —1oL-45 -0.66°C/mm . The average rate of change from t* to eclipse end is 37-(-101) 1.73°C/min. 300-220 (b) • The rate of change at t = 60 is approximately — 61—(-33) — _ 1.40°C/min. 20 • The rate of change at t = 160 is approximately —97-2V 5) -= —0.10°C/min.

• The rate of change at t = 260 is approximately 1213.33) = 2.25°C/min. For comparison, note that under normal circumstances if the moon heats from —150°C

to 120°C in one-half of a lunar day (21,240 min), then the average rate of change of 150) temperature is 120-(0.013°C/min. • 21,240

SECTION 3.4

Rates of Change

133

EXAMPLE 2 Let A = rr2 be the area of a circle of radius r. By Eq. (1), dA/dr is equal to the circumference 2irr.. We can explain this intuitively as follows: Up to a small error, the area AA of the band of width Ar in Figure 4 is equal to the circumference 27rr times the width Ar. Therefore, AA 27rr Ar and dA A — = lim = 2.7r dr r-.0 Ar

(a) Compute dAldr at r = 2 and r = 5. (b) Explain geometrically why dAldr is greater at r = 5 than at r = 2. Solution The rate of change of area with respect to radius is the derivative dA d 2 — = —(rr-) = 27r dr dr (a) We have dA = 27-/-(2) ••-•• 12.57 dr r=2

and

dA — = 27r(5) dr r=5

31.42

(b) The derivative dAldr measures how the area of the circle changes when r increases. Figure 4 shows that when the radius increases by Ar, the area increases by a band of thickness Ar. The area of the band is greater at r = 5 than at r = 2. Therefore, the derivative is larger (and the tangent line is steeper) at r = 5. In general, for a fixed Ar, the change in area AA is greater when r is larger. • Area Tangent at r

5

Tangent at r = 2

63 FIGURE 4 The pink bands represent the change in area when r is increased by Ar.

Marginal Cost in Economics Although C(x) is meaningful only when x is a whole number, economists often treat C(x) as a differentiable function of x so that the techniques of calculus can be applied. This is reasonable when the domain of C is large.

Let C(x) denote the dollar cost (including labor and parts) of producing x units of a particular product. The number x of units manufactured is called the production level. To study the relation between costs and production, economists define the marginal cost at production level xo as the cost of producing one additional unit: marginal cost = C(xo + 1) — C(xo) Note that if we use a difference quotient approximation for the derivative Ci(xo) with h = 1, we obtain C(xo + 1) — C(xo) = C(xo + 1) — C(x0) C'(x0) 1 and therefore we can use the derivative at xo as an approximation to the marginal cost. EXAMPLE 3 Cost of an Air Flight Company data suggest that when there are 50 or more passengers, the total dollar cost of a certain flight is approximately C(x) = 0.0005x3 — 0.38x2 + 120x, where x is the number of passengers (Figure 5).

15,000 -

(a) Estimate the marginal cost of an additional passenger if the flight already has 150 passengers. (b) Compare your estimate with the actual cost of an additional passenger. (c) Is it more expensive to add a passenger when x = 150 or when x = 200?

10,000 5,000

Solution The derivative is C'(x) = 0.0015x2 — 0.76x ± 120. I I I 50 100 150 200 250

X

FIGURE 5 Cost of an air flight. The slopes of the tangent lines decrease as x increases, so the marginal cost decreases as well.

(a) We estimate the marginal cost at x = 150 by the derivative C'(150) = 0.0015(150)2 — 0.76(150) ± 120 = 39.75 Thus, it costs approximately $39.75 to add one additional passenger. (b) The actual cost of adding one additional passenger is C(151) — C(150)

11,177.10— 11,137.50 = 39.60

Our estimate of $39.75 is close enough for practical purposes.

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DIFFERENTIATION

(c) The marginal cost at x = 200 is approximately C'(200) = 0.0015(200)2 — 0.76(200) + 120 = 28 Since 39.75 > 28, it is more expensive to add a passenger when x = 150 than when • x = 200.

In his famous textbook Lectures on Physics, Nobel laureate Richard Feynman (1918-1988) uses a dialogue to make a point about instantaneous velocity: Policeman: "My friend, you were going 75 miles an hour." Driver: "That's impossible, sir, I was traveling for only seven minutes."

Linear Motion Recall that linear motion is motion along a straight line. This includes horizontal motion along a straight highway and vertical motion of a falling object. Let s(t) denote the position on a line, relative to the origin, at time t. Velocity is the rate of change of position with respect to time: ds v(t) = velocity = — dt The sign of v(t) indicates the direction of motion. For example, if s(t) is the height above ground, then v(t) > 0 indicates that the object is rising. Speed is defined as the absolute value of velocity, I v(t)II. EXAMPLE 4 A truck enters the off-ramp of a highway at t = 0. Its position on the offramp after t seconds is s(t) = 25t — 0.3t3 m for 0 < t < 5. (a) How fast is the truck going at the moment it enters the off-ramp? (b) Is the truck speeding up or slowing down? Solution The truck's velocity at time t is v(t) = --c1--(25t — 0.3t3) = 25 — 0.9t 2. dt (a) The truck enters the off-ramp with velocity v(0) = 25 m/s. (b) Since v(t) = 25 — 0.9t2 is decreasing and positive (Figure 6), the speed is decreasing • and the truck is slowing down.

3020 10

1

2

3

4

5

FIGURE 6 Graph of velocity v(t) = 25 — 0.9t2.

When we say "speeding up" or "slowing down" we typically are referring to the speed of an object, not its velocity. The relationship between speed and velocity is simple: Speed is the absolute value of velocity. Use care to apply velocity and speed properly when describing an object's motion. For instance, as the next example demonstrates, an object's velocity can increase while its speed decreases. EXAMPLE 5 Velocity and Speed Figure 7 shows graphs of an object in linear motion whose position s is changing in time t in four different circumstances.

(A)

(B)

(C)

(D)

FIGURE 7

(a) In which cases is the velocity increasing? Decreasing? (b) In which cases is the speed increasing (so the object is speeding up)? Decreasing (so the object is slowing down)?

1

4=0 REMINDER "Larger" means farther from 0, while "smaller" means closer to 0.

Solution (a)

• In Figure 7(A), the slope is positive and getting larger, so the velocity (or rate change) is increasing. • In Figure 7(B), the slope is positive and getting smaller, so the velocity is decreasing.

SECTION 3.4

Rates of Change

135

• In Figure 7(C), the slope is negative and getting smaller; that is, getting closer to zero. Since the slope values are negative and approaching zero, the slope is increasing, and therefore the velocity is increasing. • In Figure 7(D), the slope is negative and is getting larger in the negative direction, so the velocity is decreasing. (b) Now we are considering the absolute value of the velocity; that is, the absolute value of the slope of the graph. It (and therefore speed) increases when the slope gets steeper, and that occurs in both Figures 7(A) and 7(D). Thus, in both of those cases the object is speeding up. On the other hand, in Figures 7(B) and 7(C), the slopes are getting less steep and therefore are getting smaller. Thus, in those cases the speed is decreasing and the object is slowing down. Notice that Figure 7(C) depicts a situation where the velocity is increasing but the object • is slowing down. Suppose s is the distance between a car and a wall during a crash test, and assume that during the test the car continued to speed up until it hit the wall. Which of the four graphs above best represents s(t)? This question, and others like it, is addressed in Exercises 15-16. EXAMPLE 6 Describe the motion and velocities of a shuttle train that runs on a straight track at the airport, ferrying passengers from Terminal 1 to Terminal 2 according to the graph given in Figure 8. Assume that s represents the distance from Terminal 1 in meters, t represents time in minutes, and the terminals are 800 m apart. Solution Note that the graph has portions resembling each of the four graphs in Example 5. Analyzing the motion: 2

• The train starts at rest, but then speeds up with increasing positive velocity for the first 2 min. • Over the interval [2,4], the velocity remains positive, but begins decreasing as the graph becomes less steep. The train is slowing down as it approaches Terminal 2. • In the interval [4,6], the graph is flat with slope 0. In this interval the train is stopped at Terminal 2. • The train speeds up again at t = 6, now with negative velocity since the distance to Terminal 1 is decreasing. Furthermore, since the graph has a negative slope and is getting steeper, the velocity is decreasing and getting larger, indicating that the train is speeding up. • Over the interval [8, 10], the velocity remains negative, but gets smaller as the graph become less steep. The train is slowing down as it approaches and arrives back at Terminal 1. •

Motion Under the Influence of Gravity Galileo discovered that the height s(t) and velocity v(t) at time t (seconds) of an object tossed vertically in the air near the earth's surface are accurately represented by the formulas Galileo's formulas are valid only when air resistance is negligible. We assume this to be the case in all examples.

s(t)= so + vot — I gt 2, 2

ds v(t) = — =

— gt

2

The constants so and vo are the initial values: • so = s(0), the position at time t = 0. • vo = v(0), the velocity at t = 0. • —g is the acceleration due to gravity on the surface of the earth (negative because the up direction is positive), where g ,=== 9.8 m/s2

or

g

32 ft/s2

136

CHAPTER 3

DIFFERENTIATION

Maximum height

A simple observation enables us to find the object's maximum height. Since velocity is positive as the object rises and negative as it falls back to Earth, the object reaches its maximum height at the moment of transition, when it is no longer rising and has not yet begun to fall. At that moment, its velocity is zero. In other words, the maximum height is attained when v(t) = 0. At this moment, the tangent line to the graph of s is horizontal (Figure 9). EXAMPLE 7 Finding the Maximum Height ground level with an initial velocity of 30 m/s.

FIGURE 9 Maximum height occurs when si(t) = v(t) = 0, where the tangent line is horizontal.

A projectile is launched upward from

(a) Find the velocity at t = 2 and at t = 4. Explain the change in sign. (b) What is the projectile's maximum height and when does it reach that height? Solution Apply Eq. (2) with so = 0, vo = 30, and g = 9.8: s(t) = 30t — 4.9t2,

Galileo's formulas: 1 s(t)= so ± vot — —gt 2 2 ds v(t) = — = vo — gt dt

v(t) = 30 — 9.8t

(a) Therefore, v(2) = 30 — 9.8(2) = 10.4 m/s,

v(4) = 30 — 9.8(4) = —9.2 m/s

At t = 2, the projectile is rising and its velocity v(2) is positive (Figure 9). At t = 4, the projectile is on the way down and its velocity v(4) is negative. (b) Maximum height is attained when the velocity is zero, so we solve 30 — 9.8t = 0

=

t=

150

— 3.06

The projectile reaches maximum height at t = 150/49 s. Its maximum height is s(150/49) = 30(150/49) — 4.9(150/49)2 •=z-', 45.92 m

FIGURE 10 To explain the motion of falling objects, Galileo studied the motion of balls on an inclined plane.

Galileo Galilei (1564-1642) discovered the laws of motion for falling objects on the earth's surface around 1600. This paved the way for Newton's general laws of motion. How did Galileo arrive at his formulas? The motion of a falling object is too rapid to measure directly, without modern photographic or electronic apparatus. To get around this difficulty, Galileo experimented with balls rolling down an incline (Figure 10). For a sufficiently flat incline, he was able to measure the motion with a water clock and found that the velocity of the rolling ball is proportional to Bettmann/Getty Images

Dorling Kindersley/Getty Images

HISTORICAL PERSPECTIVE

•

time. He then reasoned that motion in free-fall is just a faster version of motion down an incline and deduced the formula v(t) = —gt for falling objects (assuming zero initial velocity). Prior to Galileo, it had been assumed incorrectly that heavy objects fall more rapidly than lighter ones. Galileo realized that this was not true (as long as air resistance is negligible), and indeed, the formula v(t) = —gt shows that the velocity depends on time but not on the weight of the object. Interestingly, 300 years later, another great physicist, Albert Einstein, was deeply puzzled by Galileo's discovery that all objects fall at the same rate regardless of their weight. He called this the Principle of Equivalence and sought to understand why it was true. In 1916, after a decade of intensive work, Einstein developed the General Theory of Relativity, which finally gave a full explanation of the Principle of Equivalence in terms of the geometry of space and time.

SECTION 3.4

Rates of Change

137

3.4 SUMMARY • The (instantaneous) rate of change of y = f (x) with respect to x at x = xo is defined as the derivative AY f (xo) = lim — = lim Ax —>0 Ax xi—>xo

f (xi) — f (xo) — xo

• The rate dy ldx is measured in units of y per unit of x. • Marginal cost is the cost of producing one additional unit. If C(x) is the cost of producing x units, then the marginal cost at production level xo is C(xo + 1) — C(xo). The derivative Ci(x0) is often a good estimate for marginal cost. • For linear motion, velocity v(t) is the rate of change of position s(t) with respect to time—that is, v(t) = si(t). • Galileo's formulas for an object rising or falling under the influence of gravity near Earth's surface ignoring air resistance (so = initial position, vo = initial velocity): 1 s(t) = so ± vot — — 2 where g =•-•-•9.8 m/s2, or g

,

v(t) = vo — gt

32 ft/s2. Maximum height is attained when v(t) = 0.

3.4 EXERCISES Preliminary Questions 1. Which units might be used for each rate of change? (a) Pressure (in atmospheres) in a water tank with respect to depth (b) The rate of a chemical reaction (change in concentration with respect to time with concentration in moles per liter) 2. Two trains travel from New Orleans to Memphis in 4 h. The first train travels at a constant velocity of 90 mph, but the velocity of the second train varies. What was the second train's average velocity during the trip?

3. Discuss how it is possible to be speeding up with a velocity that is decreasing. 4. Sketch the graph of a function that has an average rate of change equal to zero over the interval [0, 1] but has instantaneous rates of change at 0 and 1 that are positive.

Exercises Distance (km) ,

In Exercises 1-8, find the rate of change. 1.

Area of a square with respect to its side s when s = 5

2.

Volume of a cube with respect to its side s when s = 5

150—........

3. Cube root .rx with respect to x when x = 1, 8, 27 4.

The reciprocal 1/x with respect to x when x = 1,2,3

5.

The diameter of a circle with respect to radius

6.

Surface area A of a sphere with respect to radius r (A = 4n r 2)

7. Volume V of a cylinder with respect to radius if the height is equal to the radius 8. Speed of sound v (in m/s) with respect to air temperature T (in kelvins), where v = 20,1f In Exercises 9-11, refer to Figure 11, the graph of distance s from the origin as a function of time for a car trip.

iI i , t(h) 0.5 1.0 1.5 2.0 2.5 3.0 FIGURE 11 Distance from the origin versus time for a car trip. 9.

Find the average velocity over each interval. (c) [1,1.5] (b) [0.5,1]

(a) [0,0.5]

(d) [1,2]

10. At what time is velocity at a maximum? 11. (i) (ii) (iii) (a)

Match the descriptions (i)—(iii) with the intervals (a)—(c) in Figure 11. Velocity increasing Velocity decreasing Velocity negative (c) [1.5,21 (b) [2.5,3] [0,0.51

138

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DIFFERENTIATION

Exercises 12 and 13 refer to the data in Example I. Approximate the derivative with the symmetric difference quotient (SDQ) approximation: T(t ± 20) — T(t — 20) T'(t) 40 12. (a) At what t does the SDQ approximation give the fastest rate of increase of temperature? What is the rate of change?

(b) At what t does the SDQ approximation give the fastest rate of decrease of temperature? What is the rate of change? 13. At what t does the SDQ approximation give the smallest (i.e., closest to 0) rate of change of temperature? What is the rate of change? Exercises 14-16 refer to the four graphs of s as a function of t in Figure 7. 14. Sketch s' for each of the four graphs of s. 15. Match each situation with the graph that best represents it. (a) Rocky slowed down his car as it approached the moose in the road. The distance from the car to the moose is s and the time since he spotted the moose is t.

20. At the start of the 27th century, the population of Zosania was approximately 40 million. Early-century prosperity saw the population nearly double in the first three decades, but the growth slowed in the 30s and 40s and then leveled off completely during the war years in the 50s. A postwar boom saw another rapid population increase, but that turned around in a major decline resulting from the great famine of the 70s. A slow end-ofcentury rebound resulted in an increase of the population to approximately 90 million at century's end. Let P represent the population in millions and t represent time in years since the start of the century. Make sketches of the graphs of P and P' as functions oft for Zosania's population during the century. 21. The velocity (in centimeters per second) of blood molecules flowing through a capillary of radius 0.008 cm is v = 6.4 x 10-8 — 0.001r2, where r is the distance from the molecule to the center of the capillary. Find the rate of change of velocity with respect to r when r = 0.004 cm. 22. Figure 13 displays the voltage V across a capacitor as a function of time while the capacitor is being charged. Estimate the rate of change of voltage at t = 20 s. Indicate the values in your calculation and include proper units. Does voltage change more quickly or more slowly as time goes on? Explain in terms of tangent lines.

(b) The rocket's speed increased after liftoff until the fuel was used up. The distance from the rocket to the launchpad is s and the time since liftoff is t.

V (V) 5— ,

(c) The increase in college costs slowed for the fourth year in a row. The cost of college is s and the time since the start of the 4-year period is t. 3—

16. Match each situation with the graph that best represents it.

2—

(a) Dusty's batting average increased over the first 10 games of the season but from game to game the amount of increase went down. Dusty's batting average is s and the time since the beginning of the season is t.

1—

(b) In performing a crash test, the car continued to speed up until it hit the

10

wall. The distance between the car and the wall is s and the time since the car started moving is t. (c) The hurricane strengthened at an increasing rate over the first day of its development. The strength of the hurricane is s, and the time since it started developing is t. 17. Sketch a graph of velocity as a function of time for the shuttle train in Example 6.

20

30

i ' t (s) 40

FIGURE 13 23. Use Figure 14 to estimate dT /dh at h = 30 and 70, where T is atmospheric temperature (in degrees Celsius) and h is altitude (in kilometers). Where is dTldh equal to zero? T (°C)

18. Figure 12 shows the height y of a mass oscillating at the end of a spring, through one cycle of the oscillation. Sketch the graph of velocity as a function of time.

250 — 200 — 150 — 8

'>'

100 — 50 — 0 —50 -10010

t i 50

l

l

I

I I 100

I

I

t

i f ,h(km) 150

FIGURE 12 FIGURE 14 Atmospheric temperature versus altitude. 19. Fred X has to make a book delivery from his warehouse, 15 mi north of the city, to the Amazing Book Store 10 mi south of the city. Traffic is usually congested within 5 mi of the city. He leaves at noon, traveling due south through the city, and arrives at the store at 12:50. After 15 mm at the store, he makes the return trip north to his warehouse, arriving at 2:00. Let s represent the distance from the warehouse in miles and t represent time in minutes since noon. Make sketches of the graphs of sands' as functions of t for Fred's trip.

24. The earth exerts a gravitational force of F(r)= (2.99 x 1016)/r 2 newtons on an object with a mass of 75 kg located r meters from the center of the earth. Find the rate of change of force with respect to distance r at the surface of the earth. 25. For the escape velocity relationship, vesc (2.82 x 107)r-1/2 m/s, calculate the rate of change of the escape velocity with respect to distance r from the center of the earth.

SECTION 3.4

26. The power delivered by a battery to an apparatus of resistance R (in ohms) is P = 2.25R /(R + 0.5)2 watts (W). Find the rate of change of power with respect to resistance for R = 3 Q and R =-- 5E2. 27. A particle moving along a line has position s(t) = t4 - 18t2 m at time t seconds. At which times does the particle pass through the origin? At which times is the particle instantaneously motionless (i.e., it has zero velocity)? 28. (GU Plot the position of the particle in Exercise 27. What is the farthest distance to the left of the origin attained by the particle? 29. A projectile is launched in the air from the ground with an initial velocity v0 = 60 m/s. What is the maximum height that the projectile reaches? 30. Find the velocity of an air conditioner accidentally dropped from a height of 300 m at the moment it hits the ground. 31. A ball tossed in the air vertically from ground level returns to Earth 4 s later. Find the initial velocity and maximum height of the ball. 32. Olivia is gazing out a window from the 10th floor of a building when a bucket (dropped by a window washer) passes by. She notes that it hits the ground 1.5 s later. Determine the floor from which the bucket was dropped if each floor is 5 m high and the window is in the middle of the 10th floor. Neglect air friction. 33. Show that for an object falling according to Galileo' s formula, the average velocity over any time interval [tr t2] is equal to the average of the instantaneous velocities at ti and t2. 34. Eel An object falls under the influence of gravity near the earth's surface. Which of the following statements is true? Explain. (a) Distance traveled increases by equal amounts in equal time intervals. (b) Velocity increases by equal amounts in equal time intervals. (c) The derivative of velocity increases with time. 35. By Faraday's Law, if a conducting wire of length meters moves at velocity v in/s perpendicular to a magnetic field of strength B (in teslas), a voltage of size V = -Hey is induced in the wire. Assume that B = 2 and t = 0.5. (a) Calculate dV ldv. (b) Find the rate of change of V with respect to time t if v(t) = 4t + 9. 36. The voltage V, current I, and resistance R in a circuit are related by Ohm's Law: V = I R, where the units are volts, amperes, and ohms. Assume that voltage is constant with V = 12 volts (V). Calculate (specifying units): (a) The average rate of change of I with respect to R for the interval from R = 8 to R = 8.1 (b) The rate of change of I with respect to R when R = 8 (c) The rate of change of R with respect to I when I = 1.5 37. El Ethan finds that with h hours of tutoring, he is able to answer correctly S(h) percent of the problems on a math exam. Which would you expect to be larger: Y(3) or Y(30)? Explain. 38. Suppose B(t) measures the angle between a clock's minute and hour hands. What is (At) at 3 o'clock? 39. To determine drug dosages, doctors estimate a person's body surface area (BSA) (in meters squared) using the formula BSA = fir; /60, where h is the height in centimeters and m the mass in kilograms. Calculate the rate of change of BSA with respect to mass for a person of constant height h = 180. What is this rate at m = 70 and m = 80? Express your result in the correct units. Does BSA increase more rapidly with respect to mass at lower or higher body mass?

Rates of Change

139

40. The atmospheric CO2 level A(t) at Mauna Loa, Hawaii, at time t (in parts per million by volume) is recorded by the Scripps Institution of Oceanography. Reading across, the annual values for the 4-year intervals are 1960 316.91

1964 319.20

1968 323.05

1972 327.45

1976 332.15

1980 338.69

1984 344.42

1988 351.48

1992 356.37

1996 362.64

2000 369.48

2004 377.38

2008 385.34

2012 393.87

(a) Estimate A1(t) in 1962, 1970, 1978, 1986, 1994, 2002, and 2010. (b) In which of the years from (a) did the approximation to Ai(t) take on its largest and smallest values? (c) In which of these years does the approximation suggest that the CO2 level was the most constant? 41. The tangent lines to the graph of f(x)= x2 grow steeper as x increases. At what rate do the slopes of the tangent lines increase? 42. According to Kleiber's Law, the metabolic rate P (in kilocalories per day) and body mass m (in kilograms) of an animal are related by a threequarter-power law P = 73.3m314. Estimate the increase in metabolic rate when body mass increases from 60 to 61 kg. 43. The dollar cost of producing x bagels is given by the function C(x) = 300 + 0.25x - 0.5(x/1000)3. Determine the cost of producing 2000 bagels and estimate the cost of the 2001st bagel. Compare your estimate with the actual cost of the 2001st bagel. 44. Suppose that for x > 1000, the dollar cost of producing x video cameras is C(x) = 500x - 0.003x2 + 10-8x3. (a) Estimate the marginal cost at production level x = 5000 and compare it with the actual cost C(5001) - C(5000). (b) Compare the marginal cost at x = 5000 with the average cost per camera, defined as C(x)/x. 45. E4 According to Stevens's Law in psychology, the perceived magnitude of a stimulus is proportional (approximately) to a power of the actual intensity I of the stimulus. Experiments show that the perceived brightness B of a light satisfies B =k1213, where I is the light intensity, whereas the perceived heaviness H of a weight W satisfies H = kW312 (k is a constant that is different in the two cases). Compute dB/dl and diI lcIW and state whether they are increasing or decreasing functions. Then explain the following statements: (a) An increase in light intensity is felt more strongly when I is small than when I is large. (b) An increase in load W is felt more strongly when W is large than when W is small. 46. Let M(t) be the mass (in kilograms) of a plant as a function of time (in years). Recent studies by Niklas and Enquist have suggested that a remarkably wide range of plants (from algae and grass to palm trees) obey a three-quarter-power growth law-that is, dM = CM3/4 dt

for some constant C

(a) If a tree has a growth rate of 6 kg/yr when M = 100 kg, what is its growth rate when M = 125 kg? (b) If M = 0.5 kg, how much more mass must the plant acquire to double its growth rate?

140

CHAPTER 3

DIFFERENTIATION

Further Insights and Challenges Exercises 47-49: The Lorenz curve y = F(r) is used by economists to study income distribution in a given country (see Figure 15). By definition, F(r) is the fraction of the total income that goes to the bottom rth part of the population, where 0 < r < 1. For example, if F(0.4) = 0.245, then the bottom 40% of households receive 24.5% of the total income. Note that F(0) 0 and F(1) = 1.

48. The following table provides values of F(r) for the United States in 2010. Assume that the national average income was A = $66,000.

F(r)

0 0

0.2 0.033

0.6 0.264

0.4 0.118

1 1

0.8 0.480

(a) What was the average income in the lowest 40% of households? (b) Show that the average income of the households belonging to the interval [0.4,0.6] was $48,180. (c) Estimate r(0.5). Estimate the income of households in the 50th percentile. Was it greater or less than the national average? 49. Use Exercise 47(c) to prove: (a) (r) is an increasing function of r. (b) Income is distributed equally (all households have the same income) if and only if F(r) = r for 0 < r < 1. In Exercises 50 and 51, the average cost per unit at production level x is defined as 0.2 0.4 0.6 0.8 1.0 (A) Lorenz curve for the United States in 2010

1.0

Cavg(x) =

C(x)

where C(x) is the cost of producing x units. Average cost is a measure of the efficiency of the production process. 50. The cost in dollars of producing alarm clocks is given by C(x) = 50x3 - 750x2 + 3740x + 3750

0.8

where x is in units of 1000. (a) Calculate the average cost at x = 4, 6, 8, and 10.

0.6

(b) Use the graphical interpretation of average cost to find the production level xo at which average cost is lowest. What is the relation between average cost and marginal cost at xo (see Figure 16)?

0.4 0.2

0.2 0.4 0.6 0.8 1.0 (B) Two Lorenz curves: The tangent lines at P and Q have slope 1.

15,000 ...

...

...

..

...

..

10,000 FIGURE 15 5000 47. Ell Our goal is to find an interpretation for F'(r). The average income for a group of households is the total income going to the group divided by the number of households in the group. The national average income is A = TIN, where N is the total number of households and T is the total income earned by the entire population. (a) Show that the average income among households in the bottom rth part is equal to (F(r)I r)A. (b) Show more generally that the average income of households belonging to an interval Er, r All is equal to (F(r

1 2 3 4 5 6 7 8 9 10

FIGURE 16 Cost function C(x) = 50x3 - 750x2 + 3740x + 3750. 51. Show that Cavg(x) is equal to the slope of the line through the origin and the point (x, C(x)) on the graph of y = C(x). Using this interpretation, determine whether average cost or marginal cost is greater at points A, B, C, D in Figure 17.

Ar) - F(r)) A Ar

(c) Let 0 < r < 1. A household belongs to the 100r th percentile if its income is greater than or equal to the income of 100r % of all households. Pass to the limit as Ar 0 in (b) to derive the following interpretation: A household in the 100rth percentile has income F'(r)A. In particular, a household in the 10Orth percentile receives more than the national average if F'(r) > 1 and less if F'(r) < 1. (d) For the Lorenz curves L1 and L2 in Figure 15(B), what percentage of households have above-average income?

x

Production level FIGURE 17 Graph of y = C(x).

SECTION

3.5

Higher Derivatives

141

3.5 Higher Derivatives Higher derivatives are obtained by repeatedly differentiating a function y = f(x). If f' is differentiable, then the second derivative, denoted f" or y", is the derivative f"(x) = — (f' (x)) dx For example, for f(x) =- x2 ±

± r x, we have

f (x) = 2x - x-2 + lx -112 2 1 f"(x) = 2+ 2x-3 4 The second derivative is the rate of change of (x), so it is the rate of change of the rate of change of f. The next example highlights the difference between the first and second derivatives. C (millions) 330 320 — 310 300 — 290 — 280 — 2702009

2010

EXAMPLE 1 Figure 1 and Table 1 show the number of cell phone subscribers C(t) in the United States in year t. Discuss C(t) and C"(t). TABLE 1

Number of Cell Phone Subscribers in the United States Year

2011

2012

FIGURE 1 Number of cell phone subscribers C in the United States in millions.

• dy/dx has units of y per unit of x. • d2y/dx2 has units of dy/dx per unit of x or units of y per unit of x squared.

Number in millions Change in C Change in the change in C

2009

2010

2011

2012

277

301 24

316 15 -9

326 10 -5

Solution We will show that C'(t) is positive but C"(t) is negative. According to Table 1, the number of cell phone subscribers each year was greater than the previous year, so the rate of change C1(t) is certainly positive. However, the amount of increase declined from 24 million in 2010 to 15 million in 2011 to 10 million in 2012. Thus, C1(t) is positive, but C'(t) decreases from one year to the next, and therefore its rate of change C"(t) is negative. Figure 1 supports this conclusion: The slopes of the segments in the graph are positive [C1(t) is positive], but the slopes decrease going from one segment to the next • [C"(t) is negative]. The process of differentiation can be continued, provided that the derivatives exist. The third derivative, denoted f'"(x) or f (3)(x), is the derivative of f"(x). More generally, the nth derivative f (x) is the derivative of the (n - 1)st derivative. We use parentheses on the superscript for the derivative to distinguish f ( i), the nth derivative of f, from f' , the nth power of f. We call f(x) the zeroth derivative and f' (x) the first derivative. In Leibniz notation, we write d3 f d4 f df d2 f dx'

dx2

dx3•

dX 4•• • •

EXAMPLE 2 Calculate f'"(-1) for f(x) = 3x5 - 2x2 + 7x-2. Solution We must calculate the first three derivatives: • x5 - 2x2 ± 7x-2) = 15x4 - 4x - 14x-3 f'(x) = —(3 dx • (15x4 - 4x - 14x-3) = 60x3 -4 + 42x-4 f"(x) = — dx • (60x3 - 4 + 42x-4) =- 180x2 - 168x-5 f m(x) = — dx At x = -1, f'"(-1) = 180+ 168 = 348.

•

142

CHAPTER 3

DIFFERENTIATION

Polynomials have a special property: Once n is large enough, the nth derivative is the zero function, and therefore so are all higher derivatives. More precisely, if f is a polynomial of degree k, then f 110(x) is zero for n > k. Table 2 illustrates this property for f (x) = x5. By contrast, the higher derivatives of a nonpolynomial function are never the zero function (see Exercise 85, Section 5.3). TABLE 2 Derivatives of x5 f (x)

f'(x)

f"(x)

f"(x)

f(4)(x)

f 5 (x)

x5

5x4

20x3

60x2

120x

120

f (6)(x)

EXAMPLE 3 Calculate the first four derivatives of y = x-1. Then find the pattern and determine a general formula for y("). Solution By the Power Rule, yl(x) = 4.1d REMINDER n-factorial is the number n!

n(n — 1)(n — 2) . . . (2)(1)

Thus,

1! = 1, 2! = (2)(1) = 2 3! = (3)(2)(1) = 6

—x-2,

y" =

2x-3,

)7" = _2(3)x-4,

y

(4) =

2(3)(4)x-5

First note that we have a leading negative sign on each odd derivative, but not the even derivatives. In a formula for the nth derivative, a factor of (-1)- will provide this alternating sign. Ignoring the negative sign, the coefficients can be seen to follow the pattern: 1, (1)(2), (1)(2)(3), (1)(2)(3)(4), and so on. This can be expressed with a factor of n! in the nth derivative. Finally, we see that the power on x is —n — 1 in the nth derivative. In general, therefore, y (x) = (-1)n n! • EXAMPLE 4 Find an equation of the tangent line toy r--- f'(x) at x = 4 where f (x) =

By convention, we set 0! = 1.

x3/2

It is not always possible to find a simple formula for the higher derivatives of a function. In most cases, they become increasingly complicated.

Solution The slope of the tangent line to y = f'(x) at x = 4 is the derivative f"(4). So we compute the first two derivatives and their values at x =- 4: f'(x) = 3 — x112, 2 3 f"(x) = — x-112, 4

f'(4) = 2 3 — (4)112 = 3 3 3 f"(4) = — (4)-1/2 = 4 8

Therefore, an equation of the tangent line is y

2

2.45

(s)

—

(4) = f”(4)(x — 4)

=

y

—3= 8 — 3 (x — 4)

In slope-intercept form, the equation is y = ix ± 3.

•

A second derivative that you might be familiar with is acceleration. An object that is in linear motion with position s(t) at time t has velocity v(t) = si(t) and acceleration a(t) = v'(t) =- s"(t). Thus, acceleration is the rate at which velocity changes and is measured in units of velocity per unit of time or "distance per time squared," such as

Velocity (m/s) 12

2.45 —12 (B) FIGURE 2 Height and velocity of a ball tossed vertically with initial velocity 12 m/s.

EXAMPLE 5 Acceleration Due to Gravity Find the acceleration a(t) of a ball tossed vertically in the air from ground level with an initial velocity of 12 m/s. How does a(t) describe the change in the ball's velocity as it rises and falls? Solution The ball's height at time t is s(t) = so ± vot — 4.9t2 m by Galileo's formula. In our case, so = 0 and vo = 12, so s(t) = 12t — 4.9t2 m [Figure 2(A)]. Therefore, v(t) s'(t) = 12 — 9.8t m/s and the ball's acceleration is a(t) = s"(t) =

—(12 — 9.8t) = —9.8 m/s2 dt

SECTION 3.5

Higher Derivatives

143

The acceleration is constant with value —9.8 = —g, where g (in m/s2) is the acceleration due to gravity, as introduced in Section 3.4. As the ball rises and falls, its velocity decreases from 12 to —12 m/s at the constant rate —g [Figure 2(B)]. • GRAPHICAL INSIGHT Can we visualize the rate represented by f"(x)? The second derivative is the rate at which f i(x) is changing, so f"(x) is large if the slopes of the tangent lines change rapidly, as in Figure 3(A). Similarly, f"(x) is small if the slopes of the tangent lines change slowly—in this case, the curve is relatively flat, as in Figure 3(B). If f is a linear function [Figure 3(C)], then the tangent line does not change at all and f"(x) = 0. Thus, f"(x) measures the "bending" or concavity of the graph.

(A) Large second derivative: Tangent lines turn rapidly.

(B) Smaller second derivative: Tangent lines turn slowly.

(C) Second derivative is zero: Tangent line does not change.

13 FIGURE 3

EXAMPLE 6 Identify curves I and II in Figure 4(B) as the graphs of f' or f" for the function f in Figure 4(A).

Slopes of tangent lines increasing

x a (A) Graph off

(B) Graph of first two derivatives

FIGURE 4

Solution The slopes of the tangent lines to the graph of f are increasing on the interval [a, b]. Therefore, f' is an increasing function and its graph must be II. Since f"(x) is the rate of change of f'(x), and f' (x) is increasing, f"(x) is positive and its graph • must be I. EXAMPLE 7 In a 1997 study, Boardman and Lave related the traffic speed S on a twolane road to traffic density Q (number of cars per mile of road) by the formula S = 2882Q-1 — 0.052Q + 31.73 for 60 < Q < 400 (Figure 5). Show that dSIdQ 0 and interpret each in relation to the situation being modeled.

144

CHAPTER 3

DIFFERENTIATION

S (mph) 70 60 50 40 — 30 — 20 — 10 — FIGURE 5 Speed as a function of traffic density.

Q (cars/mile) 100

200

300

400

Solution Taking the derivatives: dSIdQ = -2882Q-2 - 0.052 d2S/dQ2 = 5764Q-3 From these derivative formulas, it is clear that dSIdQ 0. Since dSIdQ is negative, it follows that as the traffic density increases, the traffic speed decreases, as we would expect. Furthermore, since d2SId Q2 is positive, it follows that dSIdQ is increasing. Given dSIdQ is negative and increasing and therefore is getting closer to zero, we conclude that dSIdQ is getting smaller. This implies that the speed decrease associated with increasing traffic density is larger at low density than at high density. This is to be expected as well— with low density and a high speed, the speed is going to drop off quickly as the density increases, but with higher density and an already low speed the speed is not going to drop off very much more as the density increases further. •

3.5 SUMMARY • The higher derivatives f', f",

",. . . are defined by successive differentiation:

d d2 f f"(x) = — f'(x) = dx dx2 '

d — f" (x) dx

d3 f dx3

,...

f The nth derivative is denoted f ° (x) = d" dx" ' • The second derivative plays an important role: It is the rate at which f' (x) changes. Graphically, f "(x) measures how fast the tangent lines change direction and thus measures the "bending" of the graph. • If s(t) is the position of an object at time t, then si(t) is velocity and sll(t) is acceleration.

3.5 EXERCISES Preliminary Questions 1. For each headline, rephrase as a statement

about first and second

derivatives and sketch a possible graph. • "Stocks Go Higher, Though the Pace of Their Gains Slows" • "Recent Rains Slow Roland Reservoir Water Level Drop" • "Asteroid Approaching Earth at Rapidly Increasing Rate!!" 2. Sketch a graph of position as a function of time for an object that is slowing down and has positive acceleration.

3. Sketch a graph of position as a function of time for an object that is speeding up and has negative acceleration. 4. True or false? The third derivative of position with respect to time is zero for an object falling to Earth under the influence of gravity. Explain.

5. Which type of polynomial satisfies f (x) = 0 for all x? 6. What is the millionth derivative of f (x) = ex ? 7. What are the seventh and eighth derivatives of f (x) = x7 ?

SECTION 3.5

Higher Derivatives

145

Exercises 37. (a) Find the acceleration at time t = 5 mm of a helicopter whose height is s(t) = 300t - 4t3 m. (b) Plot the acceleration s" for 0 < t < 6. Is the helicopter speeding up or slowing down during this time interval? Explain.

In Exercises 1-16, calculate y" and y' 1.

y

14x2

2.

y = 7 - 2x

3.

y

x4 - 25x2 + 2x

4.

y = 4t3 - 9t2 +7

6.

y= y = x-9I5

4

3

5.

y=

7.

y = 20t415 - 6t213

8.

9.

4 y= z- z

10. y = 5t-3 + 7t 813

-717

3

39. Figure 6 shows f, f', and f". Determine which is which.

12. y =(x2 + x)(x3 + 1)

11. y = 02(20 +7) 13. y =

38. Find an equation of the tangent line to the graph of y = f'(x) at x = 3, where f(x) = x4.

x- 4

14. y=

(A)

1 1- x

In Exercises 17-26, calculate the derivative indicated. 17. 19.

f(4)(1)

,

d 2y

dr 2

f( x ) =

,

(C)

FIGURE 6

16. y = (r112 + r)(1 - r)

15. y = s-112(s + 1)

(B)

40. The second derivative f" is shown in Figure 7. Which of (A) or (B) is the graph off and which is f'?

18. e"(-1), g(t)= -4t-5

x 4

y = 4t-3 + 3t2

t=1

20.

d4 f , dt4 1=1

21.

d4x dr4

f (t) = 6t9 - 2t5 (A)

f"(x)

, x = t -3/4

22. I n(4),

f (t) = 2t2 - t

24. f" (1),

f (t) =

26. g"(1),

g(s) =

FIGURE 7

t=16

23. f'"(-3),

25. hil(l),

12 3 f(x) = - - x

h(w) -

1 +1

t t+1

(B)

41. Figure 8 shows the graph of the position s of an object as a function of time t. Determine the intervals on which the acceleration is positive.

s+1

27. Calculate y(k)(0) for 0 < k < 5, where y = x4 + ax3 + bx2 + cx + d (with a, b, c, d the constants). 28. Which of the following satisfy f (k)(x) = 0 for all k > 6? (b) f (x) -= x3 - 2 (a) f (x) = 7x4 + 4 + x-1 (c) f(x) = ,Fx

(d) f (x) = 1 - x6

(e) f(x) = X9/5

(f) f(x) = 2x2 + 3x5

29. Use the result in Example 3 to find

d6

10

20

30

40

Time FIGURE 8

x '.

30. (a) Calculate the first five derivatives of f(x) = (b) Show that f (n)(x) is a multiple of X -n+1/2. (c) Show that f (")(x) alternates in sign as (-1)"-1 for n > 1.

42. Figure 9 shows the graph of the position s of an object as a function of time t. For each interval [0, 101, [10, 20], and so on, indicate whether the acceleration is negative, zero, or positive.

(d) Find a formula for f (n)(x) for n > 2. Hint: Verify that the coefficient 2n - 3 is ±1 • 3 5 2' • In Exercises 31-36, find a general formula for f

(x).

31. f(x) = x-2

32. f(x) = (x + 2)-1

33. f(x) = x-1/2

34. f(x) =

35. f(x) -

x+ 1 x4

36. f(x) =

10 x- 1 x

20

30

FIGURE 9

40

50

146

CHAPTER 3

DIFFERENTIATION

where s(t) is the distance a raindrop has fallen (in meters), D is the raindrop diameter, and g = 9.8 m/s2. Terminal velocity vterm is defined as the velocity at which the drop has zero acceleration (one can show that velocity approaches vterm as time proceeds). (a) Show that vtent, = ,/2000g D.

43. Find all values of n such that y = x" satisfies x2y" — 2xy' = 4y 44. Find all values of n such that y = xn satisfies

(b) Find therm for drops of diameter 10-3 and 10-4 m. (c) In this model, do raindrops accelerate more rapidly at higher or lower velocities?

x2y" — 12y = 45. According to one model that takes into account air resistance, the acceleration a(t) (in m/s2) of a skydiver of mass m in free-fall satisfies a(t) = —9.8 + where v(t) is velocity (negative since the object is falling) and k is a constant. Suppose that m = 75 kg and k = 0.24 kg/m. (a) What is the skydiver's velocity when a(t) = —4.9? (b) What is the skydiver's velocity when a(t) = 0? (This velocity is the terminal velocity, the velocity attained when air resistance balances gravity and the skydiver falls at a constant speed.)

I- 4

46. In contrast to Exercise 45, the size of a falling lightweight object may be more significant than its mass when taking into account air resistance. One model that takes such an approach for falling raindrops is d2s dt 2 =

g

0.0005 (ds D dt)

47. In a manufacturing process, a drill press automatically drills a hole into a sheet metal part on a conveyor. In the drilling operation the drill bit starts at rest directly above the part, descends quickly, drills a hole, and quickly returns to the start position. The maximum vertical speed of the drill bit is 4 in./s, and while drilling the hole, it must move no more than 2.6 in./s to avoid warping the metal. Let s(t) be the drill bit's height (in inches) above the part as a function of time t in seconds. Sketch possible graphs of the drill bit's velocity [s'(t)] and acceleration [s"(t)]. 48. (CAS) Use a computer algebra system to compute f (k)(x) for k = 1, 2, 3 for the following functions: (a) f (x) = (1 + x3)513

(b) f (x) —

49. CAS) Let f (x) =

x+2

1—x4 1 — 5x — 6x2

. Use a computer algebra system to com-

pute the f (k)(x) for 1 < k < 4. Can you find a general formula for f or)(x)?

Further Insights and Challenges 54. Compute

50. Find the 100th derivative of p(x) =

+x5

+ x7 )10

(l

+ x2)11(x3

+x 5 +x)

lim

51. What is p(99)(x) for p(x) in Exercise 50?

f (x + h) + f (x — h) — 2f (x)

h-).0

52. Use the Product Rule twice to find a formula for (fg)" in terms of f and g and their first and second derivatives.

h2

for the following functions:

53. Use the Product Rule to find a formula for (fg)" and compare your result with the expansion of (a + b)3. Then try to guess the general formula for (fg)(n).

(a) f (x) = x

(b) f (x) = x2

(c) f (x) = x3

Based on these examples, what do you think the limit represents?

3.6 Trigonometric Functions We can use the rules developed so far to differentiate functions involving powers of x, but we cannot yet handle the trigonometric functions. What is missing are the formulas for the derivatives of sin x and cos x. Fortunately, their derivatives are simple—each is the derivative of the other up to a sign. Recall our convention: Angles are measured in radians, unless otherwise specified. THEOREM 1 Derivative of Sine and Cosine are differentiable and

±c sin x = cos x dx

and

The functions y = sin x and y =- cos x

— cos x = — sin x dx

SECTION 3.6

Trigonometric Functions

147

Proof We must go back to the definition of the derivative: d sin(x + h) - sin x — sin x = lim dx 11-+ 0 *NO REMINDER Addition

formula for sin x:

sin(x + h)= sin x cos h + cos x sin h

1

We cannot cancel the h by rewriting the difference quotient, but we can use the addition formula (see marginal note) to write the numerator as a sum of two terms: sin(x + h) - sin x = sin x cos h ± cos x sin h - sin x

(addition formula)

= (sin x cos h - sin x) + cos x sin h = sin x(cos h - 1) ± cos x sin h This gives us sin(x + h) - sin x = sin x (cosh - 1) cos x sin h + h h h . sin(x + h) - sin x d . sin x (cosh - 1) . cosx sin h sin x = urn = lim + lim h--->0 dx h h-+0 h h->0 h = (sinx) lim

h-±0

cos h - 1 h

sin h h->0 h

+ (cos x) lim

2

We can take sin x and cos x outside the limits in Eq. (2) because they do not depend on the limiting variable h. The two limits are determined by Theorem 2 in Section 2.6, which indicates that lim

h->0

It follows that lim

h->0

1 - cos h =0 h

and

sin h =1 lim h->0 h

-2-1 = 0, and therefore, Eq. (2) reduces to

rcdx

sin x = cos x, as

cd cos x = - sin x is proved similarly (see Exercise 57). desired. The formula T, CONCEPTUAL INSIGHT One property of f (x) = sin x that makes its derivative formula . sin h so simple is that lint — = 1. The value of this limit depends on measuring angles h-*0 h in radians. The simplicity of this limit explains why we measure angles in radians when working with trigonometric functions in calculus. If instead we measure angles sin h . Nothing else in = in degrees, then, as we pointed out in Section 2.2, lim h-4-0 h the previous proof would change, and we would end up with the unwieldy derivative 1'80 cos x. formula -d d7 sin x = — EXAMPLE 1 For f (x) = sin x, compute f' at x = 0,

S, and

Solution We have f' (x) = cos x. Thus, f'(0) = cos(0) = 1, f' ( ) = cos Ce. = = cos ( ) = 9. f'( ) = cos (1) = 0, and f'

•

Note, in Example 6 in Section 3.1, for f (x) = sin x we estimated f' ( ) ;--z-2, 0.8660. Now we have the exact value =

4.

GRAPHICAL INSIGHT The formula (sin x)' = cos x seems reasonable when we compare the graphs in Figure 1. The tangent lines to the graph of y = sin x have positive slope S), and on this interval, the derivative y' = cos x is positive. The on the interval ( where y' = cos x is negative. tangent lines have negative slope on the interval where cos x = 0. , x = The tangent lines are horizontal at

(s, 4),

148

CHAPTER 3

DIFFERENTIATION

FIGURE 1 The graphs of y = sin x and its derivative y' = cos x.

EXAMPLE 2 Calculate f (x), where f(x) = x cos x . Solution By the Product Rule, f' (x) = x' cos x + x(cos x)' = cos x — x sin x f''(x) = (cos x — x sin x)/ = — sin x — (x/(sin x)

x (sin x )') •

= —2 sin x — x cos x

EXAMPLE 3 A projectile is shot from ground level at 100 ft/s at a launch angle of 0 that is between 0 and 7/2 (Figure 2). Assume that the projectile is acted on by gravity, but not air resistance. Then, in a simple projectile-motion model (see Section 14.5) it can be shown that the projectile lands at a distance R(0) = 625 sin 8 cos 9 ft from the launch point. What is the rate of change of the range with respect to the launch angle? For what angles does the range increase/decrease with increasing launch angle? What angle provides the maximum range, and what is that maximum range? FIGURE 2 The range of the projectile is R(0).

Solution Using the Product Rule and the derivative rules for sin 6 and cos 0, we have R'(9) = 625(sin 0)(— sin 8) + (cos 0)(cos 0) = 625(cos2

— sin2 0) = 625 cos 20

where the last equality results from the double angle formula for cos 20. Since cos 20 is positive for 0 0 g(x + h) — g(x)

It now follows that (f o g)'(x) =

(g(x))g'(x).

3.7 SUMMARY • The Chain Rule expresses (f o g)' in terms of f' and g': (f (g(x)))' = f'(g(x)) g'(x) dy dy du • In Leibniz notation: — = — —, where y = f (u) and u = g(x) dx du dx n-1 r • General Power Rule: — (g(x)r = n(g(x)) g (x) dx

•

156

DIFFERENTIATION

CHAPTER 3

3.7 EXERCISES Preliminary Questions 1. Identify the outside and inside functions for each of these composite functions. (b) y tan(x2 + 1) (a) y = ,V4x + 9x2 (d) (c) y = sec5 x y

(1

±

x

12)4

2. Which of the following can be differentiated without using the Chain Rule? (a) y = tan(7x2 +2) (b) = 1

(c) y =

A/.7 • sec x

(d) y = x,/Te-c7x (f) y = tan(4x)

(e) y = x sec ,rx 3.

Which is the derivative of f (5x)? (b) 5 f (5x)

(c) f'(5x)

(a) 5 f '(x)

4. Suppose that f'(4) = g(4) = g'(4) = 1. Do we have enough information to compute F'(4), where F(x) = f (g(x))? If not, what is missing?

Exercises In Exercises 25-28, compute the derivative off o g.

In Exercises 1-4, fill in a table of the following type:

g(x) = 2x + 1

25. f (u) = sin u, f (u)

f (g(x))

(x)

f (g(x))

(f o

(x) g(x) = sinx

26. f (u) = 2u + 1, 27. f (u) = u + 1.

f( u) = u3/2,

2.

f (u) = u3,

3.

f (u) = tan u,

4.

f (u) = u4 + u,

g(x)

= x4 ± 1

g(x) = 3x + 5

tan x

g(x) = csc x

In Exercises 29 and 30, find the derivatives of f (g(x)) and g(f (u)).

g(x) = x4

29. f (u) = cos u,

g(x) = cos x

In Exercises 5 and 6, write the function as a composite f (g(x)) and compute the derivative using the Chain Rule. 5.

g(x)

u 28. f (u) = , u- 1

y = (x + sin xrt

6.

y = cos(x3)

g(x) = x2 +1

1 g(x) = x +

30. f (u) = u3,

In Exercises 31-44, use the Chain Rule to find the derivative. 31. y = sin(x2)

32. y = sin2 x

d 7. Calculate - cos u for the following choices of u(x): dx (a) u(x) = 9 - x2 (b) u(x) = x-1 (c) u(x) = tan x

33. y = -Vt2 + 9

34. y =(t2 + 3t + 1)-5/2

d 2 8. Calculate - f (x +1) for the following choices of f (u): dx (a) f (u) = sin u (b) f (u) = 3u312 (c) f (u) = u2 - u

37. y =

9.

df df du Compute - if- = 2 and - =6. dx du dx

df 10. Compute - I if f (u) = u2, u(2) = -5, and u'(2) = -5. dx x=2 11. Let f (x) = (2x2 - 5)2. Compute f'(x) three different ways: 1) Multiplying out and then differentiating, 2) using the Product Rule, and 3) using the Chain Rule. Show that the results coincide.

35.

y

=

x 3 _ 1)2/3

(x4

1'\

(x+ -

)

36. y = (i7x

1-

1)3/2

38. y = c053(120)

1 39. y = sec -

40. y = tan(02 - 40)

41. y -= tan(0 + cos 0)

42. y = ./cot9

43. y = csc(9 - 202)

44. y = cot(,/0 - 1)

+I

In Exercises 45-74, compute the derivative using derivative rules that have been introduced so far.

12. Let f (x) = (x + sin x)-I . Compute f'(x) separately using the Quotient Rule and the Chain Rule. Show that the results coincide.

45. y = tan(x2 + 4x)

46. y = sin(x2 + 4x)

In Exercises 13-24, compute the derivative using derivative rules that have been introduced so far.

47. y = x cos(1 - 3x)

48. y = sin(x2) cos(x2)

49. y = (4t + 9)1/2

50. Y = (z + 1)4(2z - 1)3

51. y = (x3 + cos x)-4

52. y = sin(cos(sinx))

53. y = -Vsin x cos x

54. y = (9 - (5 -

13. y = (x4 + 5)3

14. y = (8x4 + 5)3

15. y = ,N/7x - 3

16. y = (4 - 2x - 3x2)5

17. y = (x2 + 9x)-2

18.

19. y = co54

20. y = cos(90 + 41)

55. Y = (cos 6x + sin x2) 2

56. y =

21. y = (2 cos 0 + 5 sin 0)9

22. y =

57. y = tan3 x + tan(x3)

58. y =

23. y = sin (./x2 + 2x + 9)

24. y = tan(4 - 3x) sec(3 - 4x)

24

7)3

y =(x3 + 3x + 9)-4/3

+ x + sin x

z- 1

(x + 1)112 x +2 - 3 cos x

60. y = (cos3 x + 3 cos x + 7)9

SECTION 3.7

61. y =

cos(1 +x) 1 + cos x

cos(1/x) 1 + x2

64. y =

65. y = (1 + c0t5(x4 + 1))9

66. y = -1cos 2x + sin 4x

67. y = (1 - csc2(1 - x3))6

68. y = sin Wm 0 + 1)

157

L (cm) 32 30 -

62. y = sec(/t2 - 9)

63. y = c0t7(x5)

The Chain Rule

20 10 -

69. y =

71. y

+ 1)-1/2

70. y = sec (1 + (4 + x)-3/2)

111 + 111 +

72. y =

73. y = (kx + b)- 113; 74. y -

1 +b

;

x +1+1

d2

4

6 8 t (year)

10 12

FIGURE 1 Average length of the species Clupea harengus. (a) How fast is the average length changing at age t = 6 years? (b) (GU) Use a plot of g'(t) to estimate the age tat which average length is changing at a rate of 5 cm/yr.

k and b constants 85. According to a 1999 study by Starkey and Scarnecchia, the average weight (in kilograms) at age t (in years) for channel catfish in the Lower Yellowstone River (Figure 2) is approximated by the function

k and b constants

In Exercises 75-78, compute the higher derivative. 75.

2

2

sin(x 2 )

d3

77 (9 - x)8 • dx3

d2

76.

W(t) = (0.14 + 0.115t - 0.002t2 + 0.000023114 2

5 +

9)

Find the rate at which the average weight is changing at t = 10 years. W (kg)

a'

78. Txs sin(2x)

79. Assume that the average molecular velocity v of a gas in a particular container is given by v(T) = 29,,ft mls, where T is the temperature in kelvins. The temperature is related to the pressure (in atmospheres) by dv T = 200P. Find dP P=1.5 80. The power P in a circuit is P = Ri2, where R is the resistance and i is the current. Find dPIdt at t = if R= 1000 S-2 and i varies according to i = sin(47rt) (time in seconds).

10 t (year)

15

20

FIGURE 2 Average weight of channel catfish at age t. 81. An expanding sphere has radius r = 0.4t cm at time t (in seconds). Let V be the sphere's volume. Find dV Idt when (a) r = 3 and (b) t = 3. r 0 models the length of a day 82. The function L(t) = 12 + 5.5 sin( ii-s from sunrise to sunset in Moscow, Russia, where t is the day in the year after the spring equinox on March 21. Determine L'(t), and use it to calculate the rate that the length of the days are changing on March 25, April 30, and June 10. Discuss what the results say about the changing day length in the spring in Moscow. 83. The function L(t) = 12 + 3.4 sin(t-t) models the length of a day from sunrise to sunset in Sapporo, Japan, where t is the day in the year after the spring equinox on March 21. Determine Li(t), and use it to calculate the rate that the length of the days are changing on December I, January 1, and February 1. Discuss what the results say about the changing day length in the late fall and winter in Sapporo.

86. Calculate Ml(0) in terms of the constants a, b, k, and m, where 1 M(t) = (a + (b - a) (1 + kmt + - (kmt)2))

87. Assume that f(1) = 4, f' (1) = -3, g(2) = 1, g'(2) = 3 Calculate the derivatives of the following functions at x = 2: (b) f (x/2) (c) g(2g(x)) (a) f (g(x)) 88. Assume that f (0) = 2, f'(0) = 3, h(0) = -1, h'(0) = 7

84. From a 2005 study by the Fisheries Research Services in Aberdeen, Scotland, we infer that the average length in centimeters of the species Clupea harengus (Atlantic herring) as a function of age t (in years) can be modeled by the function L(t) = 32 (1 - (1 + 0.37t + 0.068t2 + 0.0085t3 + 0.0009t4)- I ) for 0 < t < 13. See Figure 1.

Calculate the derivatives of the following functions at x = 0: (c) f (4x)h(5x) (b) f (7x) (a) (f(x))3 89. Compute the derivative of h(sin x) at x = h'(0.5) = 10.

assuming that

90. Let F(x) = f (g(x)), where the graphs of f and g are shown in Figure 3. Estimate g'(2) and f'(g(2)) and compute F'(2).

158

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96. LT:i1 Plot the "astroid" y = (4 - x2/3)312 for 0 1,

g(x) =

x

0

x=0

0

109. Chain Rule This exercise proves the Chain Rule without the special assumption made in the text. For any number b, define a new function F(u) =

f (u) - f (b) u- b

for all u

b

(a) Show that if we define F(b)= f'(b), then F is continuous at u = b. (b) Take b = g(a). Show that if x 0 a, then for all u, f(u)- f (g(a)) u - g(a) = F(u) x- a x- a

2

Note that both sides are zero if u = g(a). (c) Substitute u = g(x) in Eq. (2) to obtain

nir sinx = sin (x+-_) dxn Hint: Use the identity cos x = sin (x

1 x2 sin x

f (g(x)) - f (g(a))

F(g(x))

g(x) - g(a)

Derive the Chain Rule by computing the limit of both sides as x -> a.

3.8 Implicit Differentiation We have developed techniques for calculating a derivative dy/dx when y is given in terms of x by a formula-such as y = x3 +1; that is, when y is expressed explicitly as a function of x. But suppose that y is instead related to x by an equation such as

SECTION

y4

xy = x 3

— X±

3.8

Implicit Differentiation

159

1

2

In this case, we say that y is defined implicitly as a function of x. How can we find the slope of the tangent line at a point, such as (1, 1), on the graph (Figure 1)? Although it may be difficult or even impossible to solve for y explicitly as a function of x, we can find dy/dx using the method of implicit differentiation (see Example 2). To illustrate, we first compute the slope of the tangent line to the unit circle at (4 , ) (Figure 2). The equation of the unit circle is x2 FIGURE 1 Graph of y4

xy = x3 - x + 2.

y2

+

=

1

Compute dy/dx by taking the derivative of both sides of the equation:

rcdx (x2 + y2) = cr dx (1) ddx (x2) ± ddx (y2) = 0 d 2x ± r (y-) = 0 cx

L2

How do we handle the term rcdx (y2)? We use the Chain Rule. Think of y as a function y = y(x). Then y2 = (y(x))2 and by the Chain Rule, d —y 2 dx

1:111 FIGURE 2 The tangent line to the unit 3 circle x2 + y2 = 1 at P has slope — 4.

d =—

dx

dy (y(x))2 = 2y(x)— = 2y— dx dx

= 0, and we can solve for 4j.)-- if y

Equation (2) becomes 2x ± 2y

0:

dy dx

3

EXAMPLE 1 Use Eq. (3) to find the slope of the tangent line at the point P = the unit circle. Solution Set x =

(4, 4) on

in Eq. (3):

and y =

dy

X

3 4

dx

5

3 4

•

In this particular example, we could have computed dy/dx directly, without implicit differentiation. The upper semicircle is the graph of y = Ji - x2 and dy d Ji = dx dx v

x2

=

1 d (1 - x2 ) = (1 - x2)-112 — dx 2

/1_ x 2

This formula expresses dy/dx in terms of x alone, whereas Eq. (3) expresses dy/dx in terms of both x and y, as is typical when we use implicit differentiation. The two formulas - x2. agree because y = Before presenting additional examples, let's examine again how the factor dy/dx arises when we differentiate an expression involving y with respect to x. It would not appear if we were differentiating with respect to y. Thus, Notice what happens if we apply the Chain Rule to dy sin y. The extra derivative factor appears, but it is equal to 1: dy d sin y = (cos y)— = cosy dy dy —

d — sin y = cosy dy _

dy

y4

4y3

but

dy d — sin y = (cosy)— dx dx

but

4 4 3 dV —Y = Y — dx dx

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CHAPTER 3

DIFFERENTIATION

Similarly, the Product Rule applied to xy yields dx dx

dy +x— x dx

dy dx -; = +x -d

-d-

The Quotient Rule applied to t2/y yields =

d (2) dt y

)4

Tt = 1.2_ t 2 dy

2t y - t2 dt y2

y2

EXAMPLE 2 Find an equation of the tangent line at the point P = (1,1) on the curve in Figure 1 with equation 4 y +

xy = x 3

—x+

2

Solution We break up the calculation into two steps. Step I. Differentiate both sides of the equation with respect to x. Note that each occurrence of y in the original equation generates an additional -T ci upon differentiation. d

d Tc x:Y + dx 4

d dx

3

-x+2

3 dy dy 4y — + y+x— ) = 3x2 - 1 dx dx

4

dy dx Move the terms involving dy/dx in Eq. (4) to the left and place the remaining terms on the right: dy dy 4y' — x — =3x` - 1 - y dx dx

Step 2. Solve for

Then factor out dy/dx and divide: dy (4y' + x)— = 3x2 - 1 - y dx dy dx

3x2 - 1 - y 4y3 +x

5

To find the derivative at P = (1,1), apply Eq. (5) with x = 1 and y = 1: dy dx (1,1)

3 • 12 - 1 - 1 4 • 13 + 1

1 5

An equation of the tangent line is y - 1 =-- (x - 1) or y = x +

•

CONCEPTUAL INSIGHT The graph of an equation does not always define a function of x because there may be more than one y-value for a given value of x. Implicit differentiation works because the graph is generally made up of several pieces called branches, each of which does define a function (a proof of this fact relies on the Implicit Function Theorem from advanced calculus). For example, the branches of the unit circle x2 + y2 = 1 are the graphs of the functions y = N/1 - x2 and y = - x2. Similarly, the graph in Figure 3 has an upper and a lower branch. In most examples, the branches are differentiable except at certain exceptional points where the tangent line may be vertical.

SECTION 3.8

FIGURE 3 Each branch of the graph of y 4 xy = x 3 — x + 2 defines a function of x.

Upper branch

Implicit Differentiation

161

Lower branch

EXAMPLE 3 Calculate dyldx at the point (I, 7-1 ) on the curve Ni2- cos(x + y) = cos x — cosy Solution We follow the steps of the previous example, this time writing y' for dy Idx: d dx —

2 cos(x + y)) = d — x cos x — dx —cosy

sin(x + y) • (1

y') = — sin x + (sin y)y'

(Chain Rule)

— N/2 sin(x + y) — A,hy' sin(x + y) = — sin x + y' sin y — y' (sin y

sin(x + y)) = N/2 sin(x + y) — sin x Y

,

sin x —

sin(x

sin y +

sin(x + y)

(place y'-terms on left)

y)

71 , ÷ 21.) is The derivative at the point ( + sin

dy dx

— Nhsin (5- +

J/2— N/2

)

1

•

jr

EXAMPLE 4 Shortcut to Derivative at a Specific Point on the curve (Figure 4): P = (0,

dy Calculate — di

at the point

y cos(y + t + t2) = t3 Solution As before, differentiate both sides of the equation (we write y' for dy Idt): — y cos(y dt

t

d3 t 2 ) = —t dt

y' cos(y + t + t2) — y sin(y + t + t2)(y/ + 1 + 2t) =3t2 CI FIGURE 4 Graph of y cos(y + t + t2) = t3 The tangent line at P = (0, ; r-) has slope —1.

We could continue to solve for y' in terms oft and y, but that is not necessary since we are only interested in dyldt at the point P. Instead, we can substitute t = 0, y = directly in Eq. (6) and then solve for y': , (57r „ 9 — (-5TIT )sin G57 + 0 + 02) (y1 + 1 + 0) = 0 y cos — + + kr) 2

0- ( This gives us y' + 1 = 0 or y' = —1.

25r2

(1)(y/ + 1) = 0

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CHAPTER 3

DIFFERENTIATION

Finding Higher Order Derivatives Implicitly By using the implicit derivative process repeatedly, we can find higher order derivatives of a function that is defined implicitly. We do so in the next example. EXAMPLE 5 Find `dx2 12 7 for x2 + 4y2 = 7. Solution We differentiate with respect to x, writing y' for 2x ± 8yy' = 0 Solving for y', we obtain Y = — 4y Differentiating again with respect to x, we obtain Y

„

4y(-1) — (—x)(4y') = 16y2

—y xy' 4y2

Substituting in the fact that y' = =41, yields „ —y + x(—x/(4y)) y= 4y2

—4y2 — x2 16y3

—7 16y3

The last equality holds since x2 + 4y2 = 7

•

3.8 SUMMARY • Implicit differentiation is used to compute dyldx when x and y are related by an equation. Step I. Take the derivative of both sides of the equation with respect to x, treating y as a function of x. Step 2. Solve for dyldx by collecting the terms involving dyldx on one side and the remaining terms on the other side of the equation. • Remember to include the factor dyldx when differentiating expressions involving y with respect to x. For instance, d dy — sin y = (cos y) — dx dx

3.8 EXERCISES Preliminary Questions Which differentiation rule is used to show

dx

dy sin y = cos y — ? dx

One of (a)—(c) is incorrect. Find and correct the mistake. d — sin(y2) = 2y cos(y2 ) dy

3.

On an exam, Jason was asked to differentiate the equation x2 + 2xy + y3 = 7

Find the errors in Jason's answer: 2x + 2xy' + 3y2 = 0. 4.

Which of (a) or (b) is equal to —(x sin t)?

d — sin(x2) = 2x cos(x2) dx

(a) (x cos t)—

d — sin(y2) = 2y cos(y2 ) dx

5. Assume that a is a constant and that y is implicitly a function of x. Compute the derivative with respect to x of each of a2, x2, and y2.

dt dx

dt (b) (x cos t)— + sin t dx

SECTION 3.8

Implicit Differentiation

163

Exercises 1. Show that if you differentiate both sides of x2 + 2y3 = 6, the result is 2x + 6y2 `/Yx = 0. Then solve for dyldx and evaluate it at the point (2, 1). 2. Show that if you differentiate both sides of xy + 4x + 2y = 1, the result is (x + 2) .(Z. + y + 4 = 0. Then solve for dyldx and evaluate it at the point (1, -1). In Exercises 3-10, differentiate the expression with respect to x, assuming that y is implicitly a function of x. 3.

x2y3

4.

x3 2 Y

5.

(x2 ± y2)3/2

6.

.,/x + y

7.

x/7

8.

tan(xy)

y+1

. y 10. sin x

9.

In Exercises 11-28, calculate the derivative with respect to x of the other variable appearing in the equation. 12. y4 -2y =4x3 +x

11. 3y3 +x2 = 5 13. x2y

2x3y = x

14. xy2

y

16. x4 + z4 = 1

y x 17. - + - = 2y x y

18.

19. y-213 + x372 = 1

20. x172 + y2/3 = -4y

2

= y2 + y4,

(0,7r) (1,1)

41. Find the points on the graph of y2 = x3 - 3x + 1 (Figure 5) where the tangent line is horizontal, as follows: (a) First show that 2yy' = 3x2 - 3, where y' = dyldx. (b) Do not solve for y'. Rather, set y' = 0 and solve for x. This yields two values of x where the slope may be zero. (c) Show that the positive value of x does not correspond to a point on the graph. (d) The negative value corresponds to the two points on the graph where the tangent line is horizontal. Find their coordinates. 42. Show, by differentiating the equation, that if the tangent line at a point (x, y) on the curve x2y - 2x ± 8y = 2 is horizontal, then xy = 1. Then substitute y = x-1 in x2y -2x ± 8y = 2 to show that the tangent line is horizontal at the points (2, and ( - 4, -1). 43. Find all points on the graph of 3x2 + 4y2 + 3xy =- 24 where the tangent line is horizontal (Figure 6).

1 x

1 s

22. sin(xt) = t

±x

24. tan(x2y) = (x

23. sin(x + y) = x + cos y

26. x sin y - y cos x = 2

27. x + cos(3x - y) = xy

28. 2x2 _ x _

y =

FIGURE 5 Graph of y2 =x 3_ 3x ± 1.

y)3

25. tan(x + y) = tan x ± tan y

4,4 + y4

29. Show that x + yx 1 = 1 and y = x - x2 define the same curve [except that (0, 0) is not a solution of the first equation] and that implicit differentiation yields y' = yx -1 - x and y' = 1 - 2x. Explain why these formulas produce the same values for the derivative. 30. Use the method of Example 4 to compute flhi p at P = (2, 1) on the curve y2x3 + y3x4 - 10x + y = 5. In Exercises 31 and 32, find dyldx at the given point. 31. (x + 2)2 - 6(2y + 3)2 = 3, 32. sin2(3y) = x + y,

40. x +

X2

x2y5 - x3 = 3

15. x3 R5 = 1

21. y± -=x

39. sin(2x - y) =

FIGURE 6 Graph of + 3x2

4y2 =

24.

44. Show that no point on the graph of x2 - 3xy + y2 = 1 has a horizontal tangent line. 45. Figure 1 shows the graph of y4 + xy = x3 - x ± 2. Find dyldx at the two points on the graph with x-coordinate 0 and find an equation of the tangent line at each of those points. 46. Folium of Descartes The curve x3 + y3 = 3xy (Figure 7) was first discussed in 1638 by the French philosopher-mathematician Rene Descartes, who called it the folium (meaning "lean. Descartes's scientific colleague Gilles de Roberval called it the jasmine flower. Both men believed incorrectly that the leaf shape in the first quadrant was repeated in each quadrant, giving the appearance of petals of a flower. Find an equation of the tangent line at the point (i ,

4).

(1,-i)

(2_ 7r 7r ) 4 4

In Exercises 33-40, find an equation of the tangent line at the given point. 33. xy + x2y2 = 6,

34. x213 + y213 = 2,

(2,1)

35. x2 + sin y -= xy2 + 1,

(1,0) rzt ,

36. sin(x - y) = x cos (y + 37. 2x 1 /2

+ 4Y-1/2 =

xy,

(1, 1)

(1,4)

ir

38.

x+1

+

y +1

-1,

(1,1)

FIGURE 7 Folium of Descartes: x3 + y3 = 3xy.

164

CHAPTER

3

DIFFERENTIATION

47. Find a point on the folium x3 + y3 = 3xy other than the origin at which the tangent line is horizontal.

2 (CAS) Plot (x2 + y2)2 = 12(x2 - y2) + 2 for x and y between - 4 and 4( using a computer algebra system. How many horizontal tangent lines does the curve appear to have? Find the points where these occur.

48. 1- 71 (.9L)) Plot x3 + y3 = 3xy + b for several values of b and de0. Then compute dyldx at the point scribe how the graph changes as b co? Do your plots confirm (b1/3, 0). How does this value change as b this conclusion?

53. Calculate dxIdy for the equation y4 ± 1 = y2 points on the graph where the tangent line is vertical.

49. Find the x-coordinates of the points where the tangent line is horizontal on the trident curve xy = x3 - 5x2 + 2x - 1, so named by Isaac Newton in his treatise on curves published in 1710 (Figure 8). Hint: 2x3 - 5x2 -I- 1 = (2x - 1)(x2 - 2x - 1).

± x2

and find the

54. Show that the tangent lines at x = 1 ± ,4 to the conchoid with equation (x - 1)2(x2 y 2‘ = 2x2 are vertical (Figure 11).

FIGURE 11 Conchoid: (x - 1)2(x2

_ 2. = 2x-2 . y

FIGURE 8 Trident curve: xy = x3 - 5x2 + 2x - 1. 50. Find an equation of the tangent line at each of the four points on the curve (x2 y2 - 4x)2 = 2(x2 + y2) where x = 1. This curve (Figure 9) is an example of a limacon of Pascal, named after the father of the French philosopher Blaise Pascal, who first described it in 1650.

55. CAS Use a computer algebra system to plot y2 = x3 - 4x for x and y between -4 and 4. Show that if dxIdy = 0, then y = 0. Conclude that the tangent line is vertical at the points where the curve intersects the x-axis. Does your plot confirm this conclusion? 56. Show that for all points P on the graph in Figure 12, the segments OP and PR have equal length.

y

FIGURE 9 Limacon: (x2 ± y2 - 4x)2 = 2(x2 + y2). 51. Find the derivative, dyldx, at the points where x = 1 on the folium y See Figure 10.

(x2 ± y2)2 = 42 x 2.

Tangent line

FIGURE 12 Graph of x2 _ y2 = a2.

In Exercises 57-58, first compute y' and y" by implicit differentiation. Then solve the given equation for y, and compute y' and y" by direct differentiation. Finally, show that the results obtained by each approach are the same. 57. xy = y - 2

58. xy3 = 8

In Exercises 59-62, use implicit differentiation to calculate higher derivatives. 59. Consider the equation y3 -

x2 = 1.

(a) Show that y' = x I y2 and differentiate again to show that

5 xy2. FIGURE 10 Folium curve: (x2 ± y2)2 = -2-4

„ y2 - 2xyy' Y =4 Y (b) Express y" in terms of x and y using part (a).

SECT 10 N 3.9

60. Use the method of the previous exercise to show that y" = —y-3 on the circle x2 + y2 = 1. 61. Calculate y" at the point (1, 1) on the curve xy + y — 2 = 0 by the following steps: (a) Find y' by implicit differentiation and calculate y' at the point (1, 1). (b) Differentiate the expression for y' found in (a). Then compute y" at (1, 1) by substituting x = 1, y = 1, and the value of y' found in (a). 62. Use the method of the previous exercise to compute y" at the point (1,1) on the curve x3 + y3 = 3x

y —2

Exercises 63 and 64 explore the radius of curvature of curves. There can be many circles that are tangent to a curve at a particular point, but there is one that provides a "best fit" (Figure 13). This circle is called an osculating circle of the curve. We define it formally in Section 14.4. The radius of the osculating circle is called the radius of curvature of the curve and can be computed using either of the formulas: r=

(1 + (dy1dx)2)312

or

r=

(1+ (dx Idy)2)312

1d2Y1dx2 1

Related Rates

165

63. Consider the ellipse x2 + 4y2 = 16. (a) Compute the radius of curvature in terms of x and y. (b) Compute the radius of curvature at (4, 0), (2, ,$ ), and (0, 2). Sketch the ellipse, plot these three points, and label them with the corresponding radius of curvature. 64. Consider the ellipse 9x2 ± y2 = 36. (a) Compute the radius of curvature in terms of x and y. (b) Compute the radius of curvature at (-2, 0), (1, and (0, 6). Sketch the ellipse, plot these three points, and label them with the corresponding radius of curvature. In Exercises 65-67, x and y are functions of a variable t. Use implicit differentiation to express dy Idt in terms of dx Idt, x, and y. 65. x2y = 3 66. x3 — 6xy2 = y

Id2x1dY2 1 67. y4 ± 2x2 = xy 68. The volume V and pressure P of gas in a piston (which vary in time t) satisfy PV312 = C, where C is a constant. Prove that

dPIdt dV1dt —

FIGURE 13 The osculating circle (solid red) at a point is the tangent circle that best fits the curve.

3P 2V

The ratio of the derivatives is negative. Could you have predicted this from the relation P V312 = C?

Further Insights and Challenges 69. Show that if P lies on the intersection of the two curves x2 — y2 = c and xy = d (c,d constants), then the tangents to the curves at P are perpendicular. 70. The lemniscate curve (x2 + y2)2 = 4(x2 — y2) was discovered by Jacob Bernoulli in 1694, who noted that it is "shaped like a figure 8, or a knot, or the bow of a ribbon." Find the coordinates of the four points at which the tangent line is horizontal (Figure 14).

71. Divide the curve in Figure 15 into five branches, each of which is the graph of a function. Sketch the branches.

2— 2

—2 — FIGURE 15 Graph of y5 — y = x2 y

x

1.

FIGURE 14 Lemniscate curve: (x2 -I- y2)2 = 4(x2 — y2).

3.9 Related Rates Suppose you are filling a bottle that has a cylindrical bottom and a top shaped like a cone. Water is flowing into the bottle from a faucet at a constant rate (Figure 1). Notice that the water level rises at a constant rate in the cylindrical part, but in the conical part, the level

166

CHAPTER 3

DIFFERENTIATION

Andreas Toulios/Sh

rises more rapidly. If you are not paying attention—expecting the water level to continue to rise at a constant rate—you might overfill the bottle. To investigate this situation, we build a related rates model, a relationship between the rates of change of different variables that themselves are related. In this case, the rates that are related are the rate of increase of the volume of water in the bottle and the rate of increase of the height of the water. We investigate these rates of increase of volume and height for the cylindrical bottom in Example 1 and separately for the cone-shaped top in Example 2.

FIGURE 1

EXAMPLE 1 Filling a Cylindrical Container Water flows into a cylindrical container at a rate of 5 in.3/s. Assume that the container has a height of 6 in. and a base radius of 2 in. At what rate is the water level rising in the container? Solution Let V represent the volume of the water in the container in in.3, and let h represent the height of the water in in. We draw a cylinder and label it with all the information we have (Figure 2). The goal is to find a relationship between the known rate, and the desired rate, . That is,

g_ ,

dh dV determine — given that — dt dt

Flow rate in = 5 in.3/s

5 in.3 /s.

First, we find a relationship between V and h and then, because we want a relationship between 4,`1 and , we differentiate the relationship between V and h with respect to t. The geometry is simple: The volume of the water in the container is the volume of a cylinder with height h and base radius 2. So V = 7t.22h = 4n-h. We differentiate with respect to t and obtain dV dt

dh 47r dt

We are given that ddvt = 5, so substituting that in the above equation and solving for we find FIGURE 2

dh dt

5 Ti7-r

0.40

Thus, the water level is rising in the container at a rate of approximately 0.40 in./s. Note that, as expected, the water level is changing at a constant rate. • We summarize the steps that we typically follow in developing a related rates model and solving an associated problem. Step 1. Identify variables and the rates that are related. Step 2. Find an equation relating the variables and differentiate it. Step 3. Use given information to solve the problem. EXAMPLE 2 Filling a Conical Container Water flows into a conical container at a rate of 5 in.3/s. Assume that the container has a height of 4 in. and a base radius of 2 in. Show that the rate that the water level is rising depends on the level of the water in the container, rising faster the higher the water level.

SECTION 3.9

Related Rates

167

Solution Flow rate in = 5 in.3/s

Step 1. Identify variables and the rates that are related. Let V represent the volume of the water in the container in in.3, and let h represent its height in in. Draw a cone with the given information (Figure 3). The goal: dV dh Find a relationship between — and — • dt dt

4

FIGURE 3

1.10 REMINDER The volume of a cone with base radius r and height h is inr2h.

Step 2. Find an equation relating the variables and differentiate it. We need to find a relationship between V and h that we can differentiate with respect to t. The geometry in this case is a little more involved than with the cylindrical container. To express a relationship between the volume of the water and height, we regard the shaded volume in Figure 4 as the difference between the volume of the conical container and the volume of the conical space in the container above the water. The volume of the conical container is 17(22)(4) = — 163n and the volume of the conical space is rrs2hs, where rs and h., are the base radius and height, respectively, of the conical space. Note that hs = 4 — h. Also note that there are similar triangles in Figure 4 from which we obtain th =- = 2. Thus, rs = = 4--2h . Now, putting together the terms, we have V=

167 3

1 (4 — h )2 (4 1 — h)3 n — h ) = 167 — —7r(4 3 12 3 2

Differentiating with respect to t, and using the Chain Rule on the (4 — h)3 term, we obtain dV = 0 dt

Water volume

02) (_ dh \

(3 70

2 dh h) dt

Space above water V= 1 7'r(rs2

Whole container v=

1 = 4.2“4

3

FIGURE 4

Step 3. Use the given information to solve the problem. the result is Substituting in 5 for `flif. and solving for

2- ,

dh dt

20 71-(4 — h)2

As we expected, the rate of change dh/dt of the water level depends on the level h itself. In fact, it increases as the level increases. For example, at h =1,2,3 in. we • find iddtz 0.71, 1.59, 6.37 in./s, respectively.

DIFFERENTIATION

CHAPTER 3

Next, we examine the "sliding ladder problem" where a ladder that is leaning against a wall has its bottom pulled away at constant velocity. The question is, "How fast does the top of the ladder move?" What is interesting and perhaps surprising is that the top and bottom travel at different speeds. Figure 5 shows this clearly: The bottom travels the same distance over each time interval, but the top travels farther during the second time interval than the first. In other words, the top is speeding up while the bottom moves at a constant speed. We will explore this further through a related rates model. EXAMPLE 3 Sliding Ladder Problem A 5-m ladder leans against a wall. The bottom of the ladder is 1.5 m from the wall at time t = 0 and slides away from the wall at a rate of 0.8 m/s. Find the velocity of the top of the ladder at time t = I. FIGURE 5 Positions of a ladder at times t = 0, 1,2.

Solution Step I. Identify variables and the rates that are related. Since we are considering how the top and bottom of the ladder change position, we use the variables x for the distance from the bottom of the ladder to the wall and h for the distance from the ground to the top of the ladder (Figure 6). The velocity of the bottom is dx/dt = 0.8 m/s. The unknown velocity of the top is dh/dt. Our goal is to dh dx compute — at t = 1, given that — = 0.8 m/s and x(0) = 1.5 m dt dt

FIGURE 6 The variables x and h.

Step 2. Find an equation relating the variables and differentiate it. We wish to find a relationship between dx/dt and dh/dt, and therefore we need an equation relating x and h (Figure 6). This is provided by the Pythagorean Theorem: x2 ± h2 =52 To calculate dh/dt, we differentiate both sides of this equation with respect to t: =-25 dt dt dt dx dh 2x —+2h— =0 dt dt dh x dx Therefore, — = dt h dt Step 3. Use the given information to solve the problem. Since L' dt — 0.8, we have

t

x

h

dh/dt

0 1 2 3

1.5 2.3 3.1 3.9

4.77 4.44 3.92 3.13

—0.25 —0.41 —0.63 —1.00

This table of values confirms that the top of the ladder is speeding up.

dh — dt

h

1

To use this equation to determine g at t = 1, we need to find x and h at that time. Since the bottom slides away from the wall at 0.8 m/s and we are given x(0) = 1.5, we have x(1) = 2.3 and h(1) = ,./52 _ 2.32 4.44. We obtain dh dt The negative value for the wall.

x(1) = —O. h1 — 8 ()

0.8

2.3 ••-•-• 0.41 m/s 4.44

reflects the fact that the top of the ladder is sliding down •

CONCEPTUAL INSIGHT A puzzling feature of Eq. (1) is that the velocity dh/dt, which is equal to —0.8x/h, becomes infinite as h 0 (as the top of the ladder gets close to the ground). Since this is impossible, our mathematical model must break down as h —> 0. In fact, using physics one can show that the ladder's top loses contact with the wall before reaching the bottom. From that moment on, the formula for dh/dt is no longer valid.

SECTION 3.9

Related Rates

169

EXAMPLE 4 Tracking a Rocket A spy uses a telescope to track a rocket launched vertically from a launch pad 6 km away, as in Figure 7. At the moment when the angle between the telescope and the ground is I, the angle is changing at a rate of 0.9 radians per minute. What is the rocket's velocity at that moment? Solution Step 1. Identify variables and the rates that are related. Let y be the height of the rocket at time t. We wish to determine the rocket's velocity dyldt when 6 = I. We are given that d9/dt = 0.9. Thus, our goal is to FIGURE 7 Tracking a rocket through a telescope.

dy d9 Jr compute — , given that — = 0.9 rad/min when 0 = dt 8=1 dt 3 Step 2. Find an equation relating the variables and differentiate it. We want a relationship between dyldt and dOldt; therefore, we need to find a relation between 0 and y. As we see in Figure 7, tan6 = — 6 Now differentiate with respect to time: sec-

dB 1 dy = dt 6th' dy 6 d0 dt cos2 0 dt

2

Step 3. Use the given information to solve the problem. At the given moment, 9 = and d9/dt = 0.9, so Eq. (2) yields dy dt

6 6 (0.9) = (0.5)2 (0.9) = 21.6kinimin COS2(n-13)

The rocket's velocity at 6 = I is 21.6 km/min, or approximately 1296 km/h.

•

EXAMPLE 5 Farmer John's tractor, traveling at 3 m/s, pulls a rope attached to a bale of hay through a pulley. With dimensions as indicated in Figure 8, how fast is the bale rising when x, the horizontal distance from the tractor to the hay bale, is 5 meters? Solution

FIGURE 8

Step I. Identify variables and the rates that are related. Let x be the horizontal distance from the tractor to the bale of hay, and let h be the height above ground of the top of the bale. We want to determine dh/dt when x = 5. We are given dx/dt = 3. Thus, our goal is to dx dh compute — , given that — = 3 m/s dt x=5 dt

CAUTION A common mistake is to substitute particular variable values (such as x = 5 here) prior to differentiating. Doing so changes the variable to a constant and might prevent you from finding the proper relationship between the rates of change. Wait until Step 3, after differentiating, to substitute particular values to answer the posed problem.

Step 2. Find an equation relating the variables and differentiate it. Let L be the total length of the rope. From Figure 8 (using the Pythagorean Theorem), L=

x2 + 4.52 + (6 — h)

Differentiating with respect to t, we obtain dL dt = .7dt (.%/ x2 ± 4.52 + (6

h)=

dx dt N/ x2 ± 4.52

dh dt

170

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DIFFERENTIATION

Now L, the length of the rope, is constant. Therefore, dLIdt = O. It follows from Eq. (3) that ,dx dt

dh

4

Step 3. Use the given information to solve the problem. Apply Eq. (4) with x = 5 and dxldt = 3. The bale is rising at the rate dh dt

(5)(3) ,s/52 + 4.52

2.23 m/s

•

3.9 SUMMARY • Related-rate models arise in situations where two or more variables are related and we wish to explore how the rate of change of one of the variables is related to the rates of change of the other variable(s). • The following steps can be helpful in developing a related-rates model and solving an associated problem. Step 1. Identify variables and the rates that are related. Step 2. Find an equation relating the variables and differentiate it. (A diagram is often helpful in identifying variables and relationships between them.) This gives us an equation relating the known and unknown derivatives. Remember not to substitute the specific values for the variables until after you have computed all derivatives. Step 3. Use the given information to solve the problem.

3.9 EXERCISES Preliminary Questions 1.

If

2.

If

= 3 and y = x2, what is

when x = —3, 2, 5?

2_ = 2 and y = x3, what is `/Yt when x = —4, 2, 6?

3. Assign variables and restate the following problem in terms of known and unknown derivatives (but do not solve it): How fast is the volume of a cube increasing if its side increases at a rate of 0.5 cm/s? 4.

What is the relation between dV Idt and dr/dt if V = (4)7rr3?

In Questions 5 and 6, water pours into a cylindrical glass of radius 4 cm. Let V and h denote the volume and water level, respectively, at time t. 5. Restate this question in terms of dV Idt and dhldt: How fast is the water level rising if water pours in at a rate of 2 cm3/min? 6. Restate this question in terms of dV Idt and dhldt: At what rate is water pouring in if the water level rises at a rate of 1 cm/min?

Exercises In Exercises 1 and 2, consider a rectangular bathtub whose base is 18ft2. 1. How fast is the water level rising if water is filling the tub at a rate of 0.7 ft3/min?

In Exercises 5-8, assume that the radius r of a sphere is expanding at a rate of 30 cm/min. The volume of a sphere is V = nr3 and its suiface area is 47 r2. Determine the given rate.

2. At what rate is water pouring into the tub if the water level rises at a rate of 0.8 ft/min?

5.

Volume with respect to time when r =15 cm

3. The radius of a circular oil slick expands at a rate of 2 m/min. (a) How fast is the area of the oil slick increasing when the radius is 25 m? (b) If the radius is 0 at time t = 0, how fast is the area increasing after 3 min?

7.

4. At what rate is the diagonal of a cube increasing if its edges are increasing at a rate of 2 cm/s?

8. Surface area with respect to time at t = 2 min, assuming that r = 10 at t = 0

6. Volume with respect to time at t = 2 min, assuming that r = 0 at t =0 Surface area with respect to time when r = 40 cm

SECTION 3.9

9. A conical tank (as in Example 2) has height 3 m and radius 2 m at the base. Water flows in at a rate of 2 m3/min. How fast is the water level rising when the level is 1 m and when the level is 2 m? 10. (GU) A conical tank (as in Example 2) has height 8 m and radius 4 m at the base. Water flows in at a rate of 3 m3/min. Determine Pit- as a function of h, and provide a graph of this relationship. In Exercises 11-14, refer to a 5-m ladder sliding down a wall, as in Figures Sand 6. The variable h is the height of the ladder's top at time t, and x is the distance from the wall to the ladder's bottom. 11. Assume the bottom slides away from the wall at a rate of 0.8 m/s. Find the velocity of the top of the ladder at t = 2 s if the bottom is 1.5 m from the wall at t = 0 s. 12. Suppose that the top is sliding down the wall at a rate of 1.2 m/s. Calculate dxldt when h -= 3 m. 13. Suppose that h(0) = 4 and the top slides down the wall at a rate of 1.2 m/s. Calculate x and dxldt at t = 2 s.

171

19. At a given moment, a plane passes directly above a radar station at an altitude of 6 km. (a) The plane's speed is 800 km/h. How fast is the distance between the plane and the station changing half a minute later? (b) How fast is the distance between the plane and the station changing when the plane passes directly above the station? 20. In the setting of Exercise 19, let 9 be the angle that the line through the radar station and the plane makes with the horizontal. How fast is changing 12 min after the plane passes over the radar station? 21. A hot air balloon rising vertically is tracked by an observer located 4 km from the liftoff point. At a certain moment, the angle between the observer's line of sight and the horizontal is i7 , and it is changing at a rate of 0.2 rad/min. How fast is the balloon rising at this moment? 22. A laser pointer is placed on a platform that rotates at a rate of 20 revolutions per minute. The beam hits a wall 8 m away, producing a dot of light that moves horizontally along the wall. Let 9 be the angle between the beam and the line through the searchlight perpendicular to the wall (Figure 11). How fast is this dot moving when 0 = Wall

14. What is the relation between h and x at the moment when the top and bottom of the ladder move at the same speed? 15. The radius r and height h of a circular cone change at a rate of 2 cm/s. How fast is the volume of the cone increasing when r = 10 and h = 20? 16. A road perpendicular to a highway leads to a farmhouse located 2 km away (Figure 9). An automobile travels past the farmhouse at a speed of 80 km/h. How fast is the distance between the automobile and the farmhouse increasing when the automobile is 6 km past the intersection of the highway and the road?

Related Rates

Laser FIGURE 11 23. A rocket travels vertically at a speed of 1200 km/h. The rocket is tracked through a telescope by an observer located 16 km from the launching pad. Find the rate at which the angle between the telescope and the ground is increasing 3 min after liftoff. 24. Using a telescope, you track a rocket that was launched 4 km away, recording the angle 0 between the telescope and the ground at halfsecond intervals. Estimate the velocity of the rocket if 0(10) = 0.205 and 0(10.5) = 0.225.

S.

80 km/h

Automobile

25. A police car traveling south toward Sioux Falls, Iowa, at 160 km/h pursues a truck traveling east away from Sioux Falls at 140 km/h (Figure 12). At time t = 0, the police car is 20 km north and the truck is 30 km east of Sioux Falls. Calculate the rate at which the distance between the vehicles is changing: (b) 5 min later (a) At time t = 0

FIGURE 9 17. A man of height 1.8 m walks away from a 5-m lamppost at a speed of 1.2 m/s (Figure 10). Find the rate at which his shadow is increasing in length.

1

160 km/h

Sioux Falls 140 km/h

5 FIGURE 12

FIGURE 10 18. As Claudia walks away from a 264-cm lamppost, the tip of her shadow moves twice as fast as she does. What is Claudia's height?

26. A car travels down a highway at 25 m/s. An observer stands 150 m from the highway. (a) How fast is the distance from the observer to the car increasing when the car passes in front of the observer? Explain your answer without making any calculations. (b) How fast is the distance increasing 20 s later?

172

CHAPTER 3

DIFFERENTIATION

27. In the setting of Example 5, at a certain moment, the tractor's speed is 3 m/s and the bale is rising at 2 m/s. How far is the tractor from the bale at this moment? 20 28. Placido pulls a rope attached to a wagon through a pulley at a rate of q m/s. With dimensions as in Figure 13: (a) Find a formula for the speed of the wagon in terms of q and the variable x in the figure. FIGURE 15

(b) Find the speed of the wagon when x = 0.6 if q = 0.5 m/s.

34. Two parallel paths 15 m apart run east—west through the woods. Brooke jogs east on one path at 10 km/h, while Jamail walks west on the other path at 6 km/h. If they pass each other at time t = 0, how far apart are they 3 s later, and how fast is the distance between them changing at that moment? 35. A particle travels along a curve y = f (x) as in Figure 16. Let L(t) be the particle's distance from the origin. FIGURE 13

29. Julian is jogging around a circular track of radius 50 m. In a coordinate system with its origin at the center of the track, Julian' s x-coordinate is changing at a rate of —1.25 m/s when his coordinates are (40, 30). Find dyldt at this moment. 30. A particle moves counterclockwise around the ellipse with equation 9x2 16y2 = 25 (Figure 14).

(a) Show that dL = (x + f (x)f' (x)) dx dt

x2

f (x)2

dt

if the particle's location at time t is P = (x, f(x)). (b) Calculate L'(t) when x = 1 and x = 2 if f(x)= N/3x2 — 8x + 9 and dx/dt =4.

(a) EZY In which of the four quadrants is dxIdt > 0? Explain. (b) Find a relation between dxIdt and dyldt. (c) At what rate is the x-coordinate changing when the particle passes the point (1, 1) if its y-coordinate is increasing at a rate of 6 m/s? (d) Find dyldt when the particle is at the top and bottom of the ellipse.

FIGURE 16

36. Let 6 be the angle in Figure 16, where P = (x, f (x)). In the setting of the previous exercise, show that FIGURE 14

d0

(x) — f (x)) dx ) dt

X2 ± f(X)2

In Exercises 31 and 32, assume that the pressure P (in kilopascals) and volume V (in cubic centimeters) of an expanding gas are related by PVb = C, where b and C are constants (this holds in an adiabatic expansion, without heat gain or loss). 31. Find dPIdt if b = 1.2, P = 8 kPa, V = 100 cm3, and dVIdt = 20 cm3/min. 32. Find b if P = 25 kPa, dPIdt = 12 kPa/min, V = 100 cm3, and dV Idt = 20 cm3/min. 33. The base x of the right triangle in Figure 15 increases at a rate of 5 cm/s, while the height remains constant at h = 20. How fast is the angle changing when x = 20?

Hint: Differentiate tanO = f(x)lx with respect to t, and observe that c0s6 = x/Vx2 f(x)2. Exercises 37 and 38 refer to the baseball diamond (a square of side 90 ft) in Figure 17. 37. A baseball player runs from home plate toward first base at 20 ft/s. How fast is the player's distance from second base changing when the player is halfway to first base? 38. Player 1 runs to first base at a speed of 20 ft/s, while player 2 runs from second base to third base at a speed of 15 ft/s. Let s be the distance between the two players. How fast is s changing when player 1 is 30 ft from home plate and player 2 is 60 ft from second base?

SECTION 3.9

Related Rates

173

the rate dhldt at which the water level changes at h = 0.3 m, assuming that k = 0.25 m/min. Second base 15 ft/s First base

20 Ws Home plate FIGURE 17

39. The conical watering pail in Figure 18 has a grid of holes. Water flows out through the holes at a rate of kA m3/min, where k is a constant and A is the surface area of the part of the cone in contact with the water. This surface area is A = nr.%/h2 + r2 and the volume is V = Irrr2h. Calculate

FIGURE 18

Further Insights and Challenges 40. E4 A bowl contains water that evaporates at a rate proportional to the surface area of water exposed to the air (Figure 19). Let A(h) be the cross-sectional area of the bowl at height h. (a) Explain why V(h + Ah)— V(h) A(h)Ah if Ah is small. dV (b) Use (a) to argue that h = A(h). (c) Show that the water level h decreases at a constant rate.

(b) What does this formula give for 9 = rr? 43. As the wheel of radius r cm in Figure 21 rotates, the rod of length L attached at point P drives a piston back and forth in a straight line. Let x be the distance from the origin to point Q at the end of the rod, as shown in the figure. (a) Use the Pythagorean Theorem to show that L2 = (x — r cos 0)2

V(h + Ah) — V(h) V(h)= volume up to height h Cross-sectional area A(h)

Ah

FIGURE 19

r2 sin2

5

(b) Differentiate Eq. (5) with respect to t to prove that

2(x — r cos 0)(

de dx + r sin 0 — dt dt

dB + 2r 2 sin 0 cos — = 0 dt

(c) Calculate the speed of the piston when 0 = I, assuming that r = 10 cm, L = 30 cm, and the wheel rotates at 4 revolutions per minute.

41. A roller coaster has the shape of the graph in Figure 20. Show that when the roller coaster passes the point (x, f (x)), the vertical velocity of the roller coaster is equal to f'(x) times its horizontal velocity.

Piston moves back and forth

FIGURE 21 FIGURE 20 Graph of f as a roller coaster track. 42. Two trains leave a station at t = 0 and travel with constant velocity v along straight tracks that make an angle 0. (a) Show that the trains are separating from each other at a rate v,/2 — 2 cos 0.

44. A spectator seated 300 m away from the center of a circular track of radius 100 m watches an athlete run laps at a speed of 5 m/s. How fast is the distance between the spectator and athlete changing when the runner is approaching the spectator and the distance between them is 250 m? Hint: The diagram for this problem is similar to Figure 21, with r = 100 and x = 300.

174

CHAPTER 3

DIFFERENTIATION

CHAPTER REVIEW EXERCISES In Exercises 1-4, refer to the function f whose graph is shown in Figure I. 1. Compute the average rate of change of f (x) over [0,21. What is the graphical interpretation of this average rate? f(0.7 + h) - f(0.7) equal to the slope of 2. For which value of h is and x = 1.1? where x = 0.7 the secant line between the points 3. Estimate

f(0.7 ± h)- f(0.7)

(I)

for h = 0.3. Is the value of this dif-

FIGURE 2 Graph off.

ference quotient greater than or less than f'(0.7)? 19. Is (A), (B), or (C) the graph of the derivative of the function f shown in Figure 3?

4. Estimate f'(0.7) and f (1.1).

0.5

1.0

1.5

2.0

(A)

FIGURE 1

(B)

(C)

FIGURE 3 20. Sketch the graph of f' if the graph off appears as in Figure 4.

In Exercises 5-8, compute f' (a) using the limit definition and find an equation of the tangent line to the graph off at x = a. 5. f (x) = x2 - x, a=l

6.

f (x) = 5 - 3x,

7. f (x) = x-1, a = 4

8.

f (x) = x3, a = -2

a=2

In Exercises 9-12, compute dy ldx using the limit definition. 9. y = 4 - x2

10. y = N/2x ± 1

1 - 2- X

12. y

1 (x - 1)2

In Exercises 13-16, express the limit as a derivative. 13. lirn

15. lirn -).7r

- 1 h sin t cos t t - 7F

14.

FIGURE 4 21. Sketch the graph of a continuous function f if the graph of f' appears as in Figure 5 and f(0) = 0.

x3 1 Jim x->-1 x + 1

16. lim (9-).7

cos 19 - sin ± 1 -7

17. Find f(4) and t (4) if the tangent line to the graph of f at x = 4 has equation y = 3x - 14. 18. Each graph in Figure 2 shows the graph of a function f and its derivative f'. Determine which is the function and which is the derivative.

FIGURE 5

Chapter Review Exercises 42. h(z) = (z + (z + 1)1/2)-3/2

22. Let N(t) be the percentage of a state population infected with a flu virus on week t of an epidemic. What percentage is likely to be infected in week 4 if N(3) = 8 and N'(t) = 1.2?

43. y = tan(t -3)

44. y = 4 cos(2 - 3x)

23. A girl's height h(t) (in centimeters) is measured at time t (in years) for 0 R)

where R = 3960 miles is the radius of the earth (Figure 10). (a) Show that the weight lost at altitude h miles above the earth's surface is approximately A W -(0.0005w)h. Hint: Use the Linear Approximation with dx = h. (b) Estimate the weight lost by a 200-lb football player flying in a jet at an altitude of 7 miles.

Linear Approximation and Applications

185

46. A golfer hits a golf ball at an angle of 0 = 23° with initial velocity v = 120 ft/s. (a) Estimate As if the ball is hit at the same velocity but the angle is increased by 3°. (b) Estimate As if the ball is hit at the same angle but the velocity is increased by 3 ft/s. 47. The radius of a spherical ball is measured at r = 25 cm. Estimate the maximum error in the volume and surface area if r is accurate to within 0.5 cm. 48. The dosage D of diphenhydramine for a dog of body mass w kg is D = 4.7w2/3 mg. Estimate the maximum allowable error in w for a cocker spaniel of mass w = 10 kg if the percentage error in D must be less than 3%. 49. The volume (in liters) and pressure P (in atmospheres) of a certain gas satisfy PV = 24. A measurement yields V = 4 with a possible error of ±0.3 L. Compute P and estimate the maximum error in this computation. 50. In the notation of Exercise 49, assume that a measurement yields V = 4. Estimate the maximum allowable error in V if P must have an error of less than 0.2 atm. 51. Approximate f(2) if the linearization of f (x) at a = 2 is L(x) = 2x +4.

FIGURE 10 The distance to the center of the earth is 3960 -I- h miles. 42. Using Exercise 41(a), estimate the altitude at which a 130-lb pilot would weigh 129.5 lb. 43. A stone tossed vertically into the air with initial velocity v cm/s reaches a maximum height of h = v2 /1960 cm. (a) Estimate Ah if v = 700 cm/s and A v = 1 cm/s. (b) Estimate Ah if v = 1000 cm/s and A v = 1 cm/s. (c) In general, does a 1-cm/s increase in v lead to a greater change in h at low or high initial velocities? Explain. 44. The side s of a square carpet is measured at 6 m. Estimate the maximum error in the area A of the carpet ifs is accurate to within 2 cm. In Exercises 45 and 46, use the followingfact derivedfrom Newton's Laws: An object released at an angle 8 with initial velocity v ft/s travels a horizontal distance 1 s = - v`, sin 20 ft (Figure 11) 32 45. A player located 18.1 ft from the basket launches a successful jump shot from a height of 10 ft (level with the rim of the basket), at an angle = 34° and initial velocity v = 25 ft/s. (a) Show that As 0.255A0 ft for a small change of AO. (b) Is it likely that the shot would have been successful if the angle had been off by 2°? (c) Estimate As if 0 = 34°, u = 25 ft/s, and Au = 2.

52. Compute the linearization of f (x) = 3x - 4 at a = 0 and a = 2. Prove more generally that a linear function coincides with its linearization at x = a for all a. 53. Estimate ,s/16.2 using the linearization L(x) of f (x) = at a = 16. Plot f and L on the same set of axes and from the plot indicate whether the estimate is greater than or less than the actual value. 54. (GU) Estimate 1/,s/13 using a suitable linearization of f (x)= 1/,,rx. Plot f and L on the same set of axes and from the plot indicate whether the estimate is greater than or less than the actual value. Use a calculator to compute the percentage error. In Exercises 55-63, approximate using linearization and use a calculator to compute the percentage error. 55.

1 ‘s/T.7

1 56. 101

1 (10.03)2

58. (17)114

59. (64.1)1/3

60. (1.2)5/3

61. tan(0.04)

3.1 62. cos ( -) 4

57.

63.

(3.1)/2 sin(3.1/2)

64. (GU) Compute the linearization L(x) of f (x)= x2 - x3/2 at a = 4. Then plot f - L and identify an interval I around a = 4 such that 1 f (x) - (x)I 0. Hint: Note that x = 0 is a root and apply Rolle's Theorem. 68. Prove that x = 4 is the greatest root of f (x) = x4 — 8x2 — 128. 69. The position of a mass oscillating at the end of a spring is s(t) = A sin wt, where A is the amplitude and w is the angular frequency. Show that the speed Iv(t)1 is at a maximum when the acceleration a(t) is zero and that la(t)1 is at a maximum when v(t) is zero.

[0,1]

52. y = sec2 x — 2 tan x,

70. The concentration C(t) (in milligrams per cubic centimeter) of a drug in a patient's bloodstream after t hours is

[—r/6,n73]

53. (GU, Plot f (x) = 42 on (0,5) and use the graph to explain why there is a minimum value, but no maximum value, of f on (0,5). Use calculus to find the minimum value. 4x _ _x2 on (0, 3) and use the graph to explain why 54. (GU) Plot f (x) = there is a maximum value, but no minimum value, of f on (0,3). Use calculus to find the maximum value. 55. Let f(0) = 2 sin 20 + sin 40. (a) Show that 0 is a critical point if cos 40 = — cos 20. (b) Show, using a unit circle, that cos 01 = — cos 02 if and only if 01 = ± 02 ± 271 for an integer k. (c) Show that cos 40 = — cos 20 if and only if 0 =

+ 7rk or

(d) Find the six critical points off on [0, 27] and find the extreme values of f on this interval. (e) (GU) Check your results against a graph off. 56. (GU) Find the critical points of f (x) = 2 cos 3x + 3 cos 2x in [0, 27]. Check your answer against a graph off. In Exercises 57-60, find the critical points and the extreme values on [0,4]. In Exercises 59 and 60, refer to Figure 21. 57. y = Ix — 21

63. f (x) =

[ ,2]

64. f (x) = sin2 x — cos2 x,

44. y = x + sin x,

48. y = cos 0 + sin 0,

FIGURE 21

[-2,2]

38. y = 2,Vx2 + 1 — x,

[0,1]

60. y = I cosx1

[4,6]

28. y = x3 — 6x2 + 8,

1

x2 +

35. y=

24. y = 2x2 + 4x + 5,

[ —2, 2]

23. y = 2x2 + 4x + 5,

59. y = 1x2 + 4x — 121

58. y =13x — 91

C(t) =

t2

0.016t 4t +4

Find the maximum concentration in the time interval [0, 8] and the time at which it occurs. 71. In 1919, physicist Alfred Betz argued that the maximum efficiency of a wind turbine is around 59%. If wind enters a turbine with speed vi and exits with speed v2, then the power extracted is the difference in kinetic energy per unit time: 1 2 1 P = — m v — — m v22 2 2

watts

where m is the mass of wind flowing through the rotor per unit time (Figure 22). Betz assumed that m = pA(vi + v2)/2, where p is the density of air and A is the area swept out by the rotor. Wind flowing undisturbed through the same area A would have mass per unit time pAvi and power Po = pA14. The fraction of power extracted by the turbine is F = F I Po. (a) Show that F depends only on the ratio r = v2/ vi and is equal to F(r) = (1 — r2)(1+ r), where 0 < r < 1. (b) Show that the maximum value of F, called the Betz Limit, is 16/27 0.59. (c) 17 Explain why Betz's formula for F is not meaningful for r close to zero. Hint: How much wind would pass through the turbine if v2 were zero? Is this realistic?

SECTION 4.2

Extreme Values

195

74. Find the maximum of y = x - xn on [0,1], where n > 1. 0.6 0.5 0.4 0.3 0.2 0.1

V2

75. Find the maximum of y =x'° - X b on [0, 1], where 0 0, where 0 takes its maximum value, if it exists, is called the resonant frequency. Show that co, = ico(1 - 2D2 if 0 < D < coo I .s,/1 and that no resonant frequency exists otherwise (Figure 24). 0 50

3 2-

co, (B)D = 0.2

2

84. A rainbow is produced by light rays that enter a raindrop (assumed spherical) and exit after being reflected internally as in Figure 25. The angle between the incoming and reflected rays is 0 = 4r - 2i, where the angle of incidence i and the angle of refraction rare related by Snell' s Law sin i = n sin r with n 1.33 (the index of refraction for air and water). dr cosi = (a) Use Snell's Law to show that di n cos r (b) Show that the maximum value Om. of 0 occurs when i satisfies c/0 n2 - 1 . Hint: Show that - = 0 if cosi = - cos r. Then use 2 di 3 Snell' s Law to eliminate r. 42.53°. (c) Show that Oma. cosi =

Incoming light ray Daniel Grill/iStockphoto.com

-2

Water droplet

Reflected ray

3 2 1 (C)D = 0.75 (no resonance)

FIGURE 24 Resonance curves with coo = 1.

FIGURE 25

196

CHAPTER 4

APPLICATIONS OF THE DERIVATIVE

Further Insights and Challenges

1 3

85. Show that the extreme values of f (x) = a sin x + b cos x are ±,Ia2 b2. 86. Show, by considering its minimum, that f (x) = x2 — 2x + 3 takes on only positive values. More generally, find the conditions on r and s under which the quadratic function f (x) = x2 + rx + s takes on only positive values. Give examples of r and s for which f takes on both positive and negative values.

1

(x) = — x ± —2 ax- + bx +c 3 that ensure f has neither a local min nor a local max. Hint: Apply Exercise 86 to f (x).

87. Show that if the quadratic polynomial f (x) = x2 + rx + s takes on both positive and negative values, then its minimum value occurs at the midpoint between the two roots. 88. Generalize Exercise 87: Show that if the horizontal line y = c intersects the graph of f (x) = x2 + rx + s at two points (xi, f (xi)) and +X2 (X2, f (x2)), then f takes its minimum value at the midpoint M = 2 (Figure 26).

(B)

(A) FIGURE 27 Cubic polynomials.

90. Find the min and max of f (x) = xP (1 — xr

on [0, 1]

where p,g >0.

FIGURE 26 89. A cubic polynomial may have a local min and max, or it may have neither (Figure 27). Find conditions on the coefficients a and b of

91. E4 Prove that if f is continuous and f (a) and f (b) are local minima where a < b, then there exists a value c between a and b such that f (c) is a local maximum. (Hint: Apply Theorem 1 to the interval [a ,b].) Show that continuity is a necessary hypothesis by sketching the graph of a function (necessarily discontinuous) with two local minima but no local maximum.

4.3 The Mean Value Theorem and Monotonicity Slope f '(c) Slope f(b) —f(a) b—a

FIGURE 1 By the MVT, there exists at least one tangent line parallel to the secant line.

We have taken for granted that if f (x) is positive, the function f is increasing, and if f (x) is negative, f is decreasing. In this section, we prove this rigorously using an important result called the Mean Value Theorem (MVT). Then we develop a method for "testing" critical points—that is, for determining whether they correspond to local maxima, local minima, or neither. The MVT says that a secant line between two points (a, f (a)) and (b, f (b)) on a graph is parallel to at least one tangent line in the interval (a, b) (Figure 1). Since the f (b) — f (a) secant line between (a, f (a)) and (b, f (b) has slope and since two lines b—a are parallel if they have the same slope, the MVT is claiming that there exists a point c between a and b such that f (b) — f (a) (c) b—a Slope of tangent line

Slope of secant line

THEOREM 1 The Mean Value Theorem Assume that f is continuous on the closed interval [a, b] and differentiable on (a, b). Then there exists at least one value c in (a, b) such that

f(c) =

f (b) — f (a) b—a

SECTION 4.3

The Mean Value Theorem and Monotonicity

197

Rolle's Theorem (Section 4.2) is the special case of the MVT in which f (a) = f (b). In this case, the conclusion is that f (c) = 0.

1

GRAPHICAL INSIGHT Imagine what happens when a secant line is moved parallel to itself. Eventually, it becomes a tangent line, as shown in Figure 2. This is the idea behind the MVT. We present a formal proof at the end of this section.

FIGURE 2 Move the secant line in a parallel fashion until it becomes tangent to the curve.

CONCEPTUAL INSIGHT

The conclusion of the MVT can be rewritten as f (b) — f (a) =

(c)(b — a)

We can think of this as a variation on the Linear Approximation, which says f (b) — f (a)

f (a)(b — a)

The MVT turns this approximation into an equality by replacing f (a) with f' (c) for a suitable choice of c in (a, b). EXAMPLE 1 Verify the MVT with f (x) =

, a = 1, and b = 9.

Solution First, compute the slope of the secant line (Figure 3):

Tangent line f(x)=4X (9,3) Secant line

f (b) — f (a) b—a

•N/§ — N/I

3—1

1

9—1

9—1

4

We must find c such that f (c) = 1/4. The derivative is f (x) = (1, 2

3

4

5

6

7

8

9 10 11 12

13 FIGURE 3 The tangent line at c = 4 is parallel to the secant line.

f (c) =

1 2.1-c;"

=

1

2

4

=4

The value c = 4 lies in (1,9) and satisfies f l(4) =

X -112,

and

c=4

This verifies the MVT.

•

As a first application, we prove that a function with zero derivative is constant. COROLLARY If f is differentiable and f (x) = 0 for all X on (a, b). In other words, f (x) = C for some constant C.

E

(a, b), then f is constant

Proof If al and bi are any two distinct points in (a, b), then, by the MVT, there exists c between al and bi such that

f (bi) — f(al) =

f f(c)(bi — ai) = 0

(since f (c) = 0)

Thus, f (bi) = f (ai). This says that f (x) is constant on (a, b). We say that f is "nondecreasing" if

f (xi) < f (x2) for xi "Nonincreasing" is defined similarly. In Theorem 2, if we assume that f /(x)> 0 (instead of > 0), then f is nondecreasing on (a, b). If f'(x) < 0, then f is nonincreasing on (a, b),

•

Increasing! Decreasing Behavior of Functions We prove now that the sign of the derivative determines whether a function f is increasing or decreasing. Recall that f is • Increasing on (a, b) if f (xi) • Decreasing on (a, b) if f (xi)

< f (x2) for all xi, X2 E (a, b) such that xi f (x2) for all xi , x2 E (a, b) such that x1 0 for X E (a, b), then f is increasing on (a, b). • If f' (x) 0 The inequality holds because f'(c) and (x2 — xi ) are both positive. Thus, f (x2) > f (xi), • as required. The case f'(x) 1 and negative for x < 1. By Theorem 2, f is decreasing on the interval (—oo, 1) and increasing on the interval (1, oo), as confirmed in Figure 6. •

Testing Critical Points

y = g(x) Sign change from + to —

\ -S\ No sign change Sign change from — to + FIGURE 7

There is a useful test for determining whether a critical point yields a min or max (or neither) based on the sign change of the derivative f'(x). To explain the term "sign change," suppose that a function g satisfies g(c) = 0. We say that g(x) changes from positive to negative at x = c if g(x) > 0 to the left of c and g(x) 0 for all x near but not equal to c, so f is increasing and has neither a local min nor a local max at c. A similar analysis holds when f' (c) does not exist and the possibilities for the sign of f' on either side of c are considered. As a result, we have the following theorem: THEOREM 3 First Derivative Test for Critical Points Then • f'(x) • f'(x)

changes from + to — at c changes from — to + at c

Let c be a critical point of f.

f(c) is a local maximum. f(c) is a local minimum.

SECTION 4.3

The Mean Value Theorem and Monotonicity

199

Y =f(x)

Neither a local mm nor max x

f '(x) = 3x2 — 27

Y = f'(x)

—3

• f'(x) does not change sign

f'(x) changes from — to +

f'(x) changes from + to —

(B)

(A) FIGURE 8

To carry out the First Derivative Test, we make a useful observation: f(x) can change sign at a critical point, but it cannot change sign on the interval between two consecutive critical points as long as the function is defined over the whole interval. In such a case, we can determine the sign of f'(x) on an interval between consecutive critical points by evaluating f'(x) at any test point xo inside the interval. The sign of f '(xo) is the sign of f'(x) on the entire interval. In a case where a function's domain is made up of separate intervals, this analysis of the sign of f' needs to be carried out individually on each of the intervals. EXAMPLE 4 Analyze the critical points of f(x)= x3 —27x — 20. Solution Our analysis will confirm the picture in Figure 8(A). Step 1. Find the critical points. We have f'(x)=3x2 — 27 = 3(x2 — 9). The critical points satisfy f i(c) , 0 and therefore are c = 13. Step 2. Find the sign of f(x) on the intervals between the critical points. The critical points c = 13 divide the real line into three intervals: (—oo, —3),

We chose the test points —4, 0, and 4 arbitrarily To find the sign of f'(x) on (—no, —3), we could just as well have computed f'(-5) or any other value off' in the interval (—Do, —3).

(-3,3),

(3, no)

To determine the sign of f'(x) on these intervals, we choose a test point inside each interval and evaluate. For example, in (—no, —3) we choose x = —4. Because f'(-4) = 21 > 0, f(x) is positive on the entire interval (-3, no). Taking this result, along with the results from test points at 0 and 4, we have f'(x)> 0

for all X E (-00, —3)

f'(0) = —27 0

f'(x)> 0

for all x E (3, oo)

f'(-4) = 21 > 0

This information is displayed in the following sign diagram: Behavior of Sign of f'(x) —3

0

200

CHAPTER 4

APPLICATIONS OF THE DERIVATIVE

Step 3. Use the First Derivative Test. • c = — 3: f'(x) changes from + to —= mum value. = • c = 3: f'(x) changes from — to + mum value.

f(-3) = 34 is a local maxif(3) = —74 is a local mini•

EXAMPLE 5 Analyze the critical points and the increase/decrease behavior of f (x) = cos2 x + sin x in (0,7r). Solution First, find the critical points: f'(x) = —2 cos x sin x + cos x = (cos x)(1 — 2 sin x) Therefore, the critical points are solutions to cos x = 0 or sin x = 1. Since we are just examining the interval (0, n-), the critical points of interest are e J . , , and . They divide (0, z) into four intervals: 7r Tr \ f7r 57r \ 5z (0, ), 6 We determine the sign of f'(x) by evaluating f'(x) at a test point inside each interval. Since 0.52, 1. 1.57, 2.62, and z 3.14, we can use the following test points: Interval (0, e")

Test

P(0.5) f'(1)

value

Sign of f'(x)

Behavior of f (x)

0.04 —0.37

(S,

f'(2) P.,- 0.34

( r 70

f'(3)

—0.71

Now apply the First Derivative Test: • Local max at c = 7-6r- and c = Graph of f (x) = cos2 x + sin x and its derivative.

• Local mm at c

FIGURE 9

because f'(x) changes from + to —.

because f'(x) changes from — to +.

The behavior of f (x) and f'(x) is reflected in the graphs in Figure 9.

•

EXAMPLE 6 Analyze the critical points and the increase/decrease behavior of f (x) = x

2

1

Solution Note that f is undefined at x = 0, so we need to analyze f separately on (—oo, 0) and (0, co). We have 2 f'(x) = 2x — The critical points are solutions to x — = 0; that is, to x4 — 1 = 0. They are c = ±1. Since we need to consider f separately on (—oo, 0) and (0, oo), there are four intervals on which we need to examine the sign of f'(x): (—oo, —1), (-1,0), (0,1), and (1, oo). We determine the sign of f'(x) by evaluating f'(x) at a test point inside each interval. Interval

FIGURE 10

Graph of f (x) = x2 +

(—oo, —1) (-1, 0) (0,1) (1, oo)

Test value = = f'(0.5) = f (2) =

f'(-2)

Sign of f'(x)

Behavior of f (x)

—3.75 15 —15 3.75

Applying the First Derivative Test, we see that both critical points are local minima. This is verified in the graph in Figure 10.

S ECT I ON 4.3

The Mean Value Theorem and Monotonicity

EXAMPLE 7 A Critical Point Where .r(x) of f (x) = (1 - x)213.

Is

Undefined

201

Analyze the critical points

Solution The derivative is f (x) = -4(1 - x)- I13 = 3 . The only critical point occurs at c = 1, when f (x) is undefined. For x < 1, f (x) is negative. For x > 1, f (x) is positive. So f (x) changes sign as we pass through c = 1, and by the First Derivative Test, f (c) is a local minimum. See Figure 11. • FIGURE 11 Graph of f (x) = (1 - x)213 .

EXAMPLE 8 Infinitely Many Critical Points, No Local Extrema points of f (x ) = x - sin x.

Analyze the critical

Solution We have f' (x) = 1 - cos x, and therefore critical points occur at solutions to cos x = 1; that is, at nit for all even integers n. At none of the critical points does the sign of f' change since f'(x) > 0 for all x. Therefore, none of the critical points are local extrema (Figure 12). • -15 -10 -5

5

10

15

5-

Proof of the MVT Let m = f (b)

f (a) be the slope of the secant line joining (a, f

b- a

-10-

and (b, f (b)). The secant line has equation consider the function

-15 -

y = mx + r

for some

r

(a))

(Figure 13). Now

G(x) = f (x) - (mx + r) FIGURE 12

a

Graph of f

= x-

sin x.

x

vertical distance between the graph and the secant line.

FIGURE 13 G(x) is the

As indicated in Figure 13, G(x) is the vertical distance between the graph and the secant line at x (it is negative at points where the graph of f lies below the secant line). This distance is zero at the endpoints, and therefore, G(a) = G(b) = 0. By Rolle's Theorem (Section 4.2), there exists a point c in (a, b) such that G'(c) = 0. But G'(x) = f (x) - m, • so G'(c) = f (c) - m = 0, and f' (c) = m as desired.

4.3 SUMMARY • The Mean Value Theorem (MVT): If f is continuous on [a, b] and differentiable on (a, b), then there exists at least one value c in (a, b) such that f(c)

.= f (b) - f (a) b- a

This conclusion can also be written f (b) - f (a) = f' (c)(b - a)

• Important corollary of the MVT: If

f (x) =

0 for all x

E (a, b),

then

f

is constant on

(a, b).

• The sign of

f (x)

determines whether

f (x) >

0 for x

E (a, b)

f (x)


- 4X -I

203

54. Ron's toll pass recorded him entering the tollway at mile 0 at 12:17 PM. He exited at mile 115 at 1:52 PM, and soon thereafter he was pulled over by the state police. "The speed limit on the tollway is 65 miles per hour," the trooper told Ron. "You exceeded that by more than five miles per hour this afternoon." "No way!" responded Ron. "I glance at the speedometer frequently, and not once did it read over 65!" How did the trooper use the Mean Value Theorem to support her claim that Ron must have gone more than 70 miles per hour at some point? 55. Two days after he bought a speedometer for his bicycle, Lance brought it back to the Yellow Jersey Bike Shop. "There is a problem with this speedometer," Lance complained to the clerk. "Yesterday I cycled the 22-mile Rogadzo Road Trail in 78 minutes, and not once did the speedometer read above 15 miles per hour!" "Yeah?" responded the clerk. "What's the problem?" How did Lance use the Mean Value Theorem to explain his complaint?

2x + 1 X2 + 1

56. Determine where f(x) = (1,000 - x)2 + x2 is decreasing. Use this to decide which is larger: 8002 ± 2002 or 6002 + 4002.

x3

57. Show that f(x) = 1 - lx1 satisfies the conclusion of the MVT on [a, b] if both a and b are positive or negative, but not if a 0. 58. Which values of c satisfy the conclusion of the MVT on the interval [a, b] if f is a linear function?

48. y =

so.

49. y = X I /3

x5/2

The Mean Value Theorem and Monotonicity

- 2 cos 8,

y =x 2/3

[0,2x]

-x 2

51. Show that f(x) = x2 + bx + c is decreasing on ( - no, no). creasing on ( -

and in-

52. Show that f(x) = x3 - 2x2 + 2x is an increasing function. Hint: Find the minimum value of f'. 53. Find conditions on a and b that ensure f(x) = x3 + ax + b is increasing on (-no, no).

59. Show that if f is any quadratic polynomial, then the midpoint c = a+b satisfies the conclusion of the MVT on [a, b] for any a and b. 2 60. Suppose that f(0) = 2 and f' (x) 3 for x > 0. Apply the MVT to the interval [0,4] to prove that f(4) 5. 14. Prove more generally that f(x) 0. 61. Show that if f(2) = -2 and f'(x)> 5 for x > 2, then f(4) > 8. 62. Show that if f(2) = 5 and f' (x) a 10 for x > 2, then f(x) a. 10x - 15 for all x >2.

Further Insights and Challenges c is increasing

68. Assume that f" exists and f"(x) = 0 for all x. Prove that f(x) = mx + b, where m = f'(0) and b = f(0).

64. Prove that if f(0) = g(0) and f'(x) g'(x) for x 0, then f(x) g(x) for all x > 0. Hint: Show that the function given by y = f(x) - g(x) is nonincreasing.

Define f(x) = x3 sin( ) for x 0 and f(0) = 0. 69. (a) Show that f' is continuous at x = 0 and that x = 0 is a critical point off. (b) (GU' Examine the graphs off and f'. Can the First Derivative Test be applied? (c) Show that f(0) is neither a local min nor a local max.

63. Show that a cubic function f(x) = x3 + ax2 on (-no, no) if b > a2 /3.

bx

65. Use Exercise 64 to prove that x < tan x for 0 < x < 17 and sin x < x for x >0. 66. Use Exercises 64 and 65 to prove the following assertions for all x > 0 (each assertion follows from the previous one): (a) cos x > 1 - x2 (b) sinx > x -

x3

70. Suppose that f(x) satisfies the following equation (an example of a differential equation): f"(x) = - f (x)

Ax4

(c) cos x < 1 - x2 + Can you guess the next inequality in the series? 67. Suppose that f(x) is a function such that f(0) = 1 and for all x, f' (x) = f(x) and f(x) > 0 (in Chapter 7, we will see that f(x) is the exponential function ex). Prove that for all x > 0 (each assertion follows from the previous one), (b) f (x) a 1 + x (a) f(x) a 1 1 (c) f (x) > 1 + x + - x2 2 Then prove by induction that for every whole number n and all x > 0, 1 1 + • • • + - xn f(x) 1 x + n ! 2!

1

(a) Show that f(x)2 + f'(x)2 = f(0)2 + f'(0)2 for all x. Hint: Show that the function on the left has zero derivative. (b) Verify that f(x) = sin x and f(x) = cos x satisfy Eq. (1), and deduce that sin2 x + cos2 x = 1. 71. Suppose that functions f and g satisfy Eq. (1) and have the same initial values-that is, f(0) = g(0) and f'(0) = g'(0). Prove that f(x) = g(x) for all x. Hint: Apply Exercise 70(a) to f - g. 72. Use Exercise 71 to prove f(x) = sin x is the unique solution of Eq. (1) such that f(0) = 0 and /(0) = 1; and g(x)= cos x is the unique solution such that g(0) = 1 and g'(0) = 0. This result can be used to develop all the properties of the trigonometric functions "analytically"-that is, without reference to triangles.

204

CHAPTER 4

APPLICATIONS OF THE DERIVATIVE

4.4 The Second Derivative and Concavity In the previous section, we studied the increasing/decreasing behavior of a function, as determined by the sign of the derivative. Another important property is concavity, which refers to the way the graph bends. Informally, a curve is concave up if it bends up and concave down if it bends down (Figure 1).

Concave up

Concave down

FIGURE 1

To analyze concavity in a precise fashion, let's examine how concavity is related to tangent lines and derivatives. Observe in Figure 2 that when f is concave up, f' is increasing (the slopes of the tangent lines increase as we move to the right). Similarly, when f is concave down, f' is decreasing. This suggests the following definition.

Concave up: Slopes of tangent lines are increasing.

Concave down: Slopes of tangent lines are decreasing.

FIGURE 2

DEFINITION Concavity Then

Let f be a differentiable function on an open interval (a, b).

• f is concave up on (a, b) if f' is increasing on (a, b). • f is concave down on (a, b) if f' is decreasing on (a, b). EXAMPLE 1 Concavity and Stock Prices The stocks of two companies, Arenot Industries (Al) and Blurbenthal Business Associates (BBA), went up in value, and both currently sell for $75 (Figure 3). However, one is clearly a better investment than the other, assuming these trends continue in the same manner. Explain in terms of concavity. Stock price

Stock price

75

75

25

25 Time Company Al

Time Company BBA

FIGURE 3

Solution The graph of Stock Al is concave down, so its growth rate (first derivative) is declining as time goes on. The graph of Stock BBA is concave up, so its growth rate is increasing. If these trends continue, Stock BBA is the better investment. • The concavity of a function is determined by the sign of its second derivative. Indeed, if f"(x) > 0, then f' is increasing and hence f is concave up. Similarly, if f"(x) 0 for x > 0 0

f(x) changes sign

, X

Concave down—local max

FIGURE 8

Concave up—local min

FIGURE 9 Concavity determines the type of the critical point. Mnemonic Device:

1(c) >0

local min

THEOREM 3 Second Derivative Test Let c be a critical point of f(x).If r(c) exists, then • f u(c) > 0 • f"(c) 0 f(x) = x5 - 5x4

=

f(4) is a local min

The Second Derivative Test fails at x = 0, so we fall back on the First Derivative Test. Choosing test points to the left and right of x = 0, we find f'(x) is positive on (-oo, 0)

f' (-1) = 5 + 20 = 25 > 0 f'(1) = 5-20 = -15 0 and f"(x) 2, and f"(x) > 0 for ix I 2. Show that f has at least one point of inflection if n is odd. Then give an example to show that f need not have a point of inflection if n is even. 66. Critical and Inflection Points If f'(c) = 0 and f (c) is neither a local min nor a local max, must x = c be a point of inflection? This is true for "reasonable" functions (including the functions studied in this text), but it is not true in general. Let

(a) Use the method of Exercise 62 to prove that the tangent line at x = c crosses the graph (Figure 21). Hint: Show that G(x) changes sign at x = c. by graphing f and (b) (GU) Verify this conclusion for f (x) = 32 +1 the tangent line at each inflection point on the same set of axes. 64. Let C(x) be the cost of producing x units of a certain good. Assume that the graph of C is concave up. (a) Show that the average cost A(x) = C(x)/x is minimized at the production level xo such that average cost equals marginal cost—that is, A(x0) = C'(xo). (b) Show that the line through (0,0) and (xo, C(xo)) is tangent to the graph of C.

f (x) =

x2 sin 0

for x 0 for x = 0

(a) Use the limit definition of the derivative to show that f'(0) exists and f '(0) = O. (b) Show that f(0) is neither a local min nor a local max. (c) Show that f'(x) changes sign infinitely often near x = 0. Conclude that x = 0 is not a point of inflection.

4.5 Analyzing and Sketching Graphs of Functions

+ Concave up \

r

— Concave down

f"

+ Increasing

— Decreasing

FIGURE 1 The four basic shapes.

In this section, our goal is to study graphs of functions f using the information provided by the first two derivatives f' and f". You will see that you can acquire a good understanding of the properties of a graph without plotting a large number of points. Evan though almost all graphs you may see are produced by computer (including, of course, the graphs in this textbook), the tools of calculus provide information beyond the image displayed on a computer. This information includes the exact locations of critical points and inflection points, the rates of increase and decrease over the function's domain, and the concavity of the function. Most graphs are made up of smaller arcs that have one of the four basic shapes, corresponding to the four possible sign combinations of f' and f" (Figure 1). Since f' and f" can each have sign + or —, the sign combinations are —

—

In this notation, the first sign refers to f' and the second sign to f". For instance, —+ indicates that f(x)< 0 and f''(x) > 0. We use a slanted arrow over the first sign to indicate whether the function is increasing or decreasing, and an upturned or downturned over the second sign to indicate the concavity.

SECTION 4.5

2 The graph off with transition points and sign combinations off' and f".

FIGURE

Analyzing and Sketching Graphs of Functions

211

In analyzing a graph, we focus on the transition points, where the basic shape changes due to a sign change in either f' (local mm or max) or f" (point of inflection). In this section, local extrema are indicated by solid dots, and points of inflection are indicated by green solid squares (Figure 2). In examining the properties of a function, it is often useful to investigate the asymptotic behavior—that is, the behavior of f (x) as x approaches either ±oo or a vertical asymptote. In the examples that follow, we use calculus to investigate the behavior of specific functions, and then we use the information we gather to construct a picture of the function's graph—that is, to "sketch the graph." The first three examples treat polynomials. Recall from Section 2.7 that the limits at infinity of a polynomial

f (x) = anxn + an_ixn-1 + • • + ax

ao

(assuming that an 0 0) are determined by urn f (x) = an lim xn x-+oo x-+00 In general, the graph of a polynomial oscillates up and down a finite number of times and tends to positive or negative infinity as x tends to positive or negative infinity. Typical examples appear in Figure 3.

FIGURE 3 Graphs of polynomials.

(C) Degree 5, a5 < 0

(B) Degree 4, a 4 > 0

(A) Degree 3, a3 > 0

EXAMPLE 1 Quadratic Polynomial and sketch its graph.

Investigate the behavior of f (x) =x 2 —4x +3

Solution Note that f (x) = (x — 1)(x — 3) so the graph intersects the x-axis at x = 1 and x = 3. We have f (x) = 2x — 4 = 2(x — 2). We can see directly that (x) is negative for x < 2 and positive for x > 2, but let's confirm this using test values, as in previous sections: Interval

Test value

(-00, 2) (2, oo)

f'(1) = —2 f'(3) = 2

Sign of f'

Furthermore, f "(x) = 2 is positive, so the graph is everywhere concave up. To sketch the graph, plot the local minimum (2, —1), the y-intercept, and the roots x = 1, 3. Since ±oo. This asymptotic behavior is the leading term of f is x2, f (x) tends to oo as x • noted by the arrows in Figure 4. FIGURE

4 Graph of f (x) = x2 — 4x + 3.

EXAMPLE 2 1 3 — 1 2 —

Cubic Polynomial Investigate the behavior of the cubic function f (x) = 2x + 3 and sketch the graph.

Solution Step /. Determine the signs of f' and f n. First, solve for the critical points: f'(x) = x2 — x — 2 -= (x + 1)(x — 2)

212

APPLICATIONS OF THE DERIVATIVE

CHAPTER 4

The critical points are c = -1,2, and they divide the x-axis into three intervals ' (-oo, -1), (-1,2), and (2, oo), on which we determine the sign off' by computing test values: Sign of f'

Test value

Interval (-co, -1) (-1,2) (2, oo)

IA-2) = 4 f'(0) = -2 f'(3) = 4

Next, f" (x) = 2x - 1, and therefore x = 1. is the only solution to f"(x) = 0. We have

v -+ I 0

! 2

FIGURE

Test value

(-oo,

f"(0) = -1

°°)

++

Sign of f"

f (1) = 1

X

I

Local max

Interval

Inflection point

Step 2. Note transition points and sign combinations. This step merges the information about f' and f" in a sign diagram (Figure 5). There are three transition points:

Local min

• c = -1: local max since f' changes from + to -. • c = corresponds to a point of inflection since f" changes sign.

5 Sign combinations of f' and f".

• c = 2: local mm since

f, changes from -

to +.

In Figure 6(A), we plot the transition points and, for added accuracy, the y-intercept f(0), using the values 25

f ( - 1) = -6 ,

(1 \ f

23 =

1 f (2) = - 3

f (0) = 3,

Step 3. Draw arcs of appropriate shape and asymptotic behavior. The leading term of f (x) is -x3. Therefore, lim f (x) = oo and lim f (x) = -oo. x-0.00 x-+-00 To create the sketch, it remains only to connect the transition points by arcs of the appropriate concavity and asymptotic behavior, as in Figure 6(B) and (C). • +-

-+

-

++

L'LL

_

+

)

3 23 \ a I k 2 ' 12) ++

k X

—3 FIGURE 6

f (x) =

—1

•

Graph of -

-

2x + 3.

I(

2, -1

(A)

3

)

(B)

(C)

EXAMPLE 3 Investigate the behavior of f (x) =- 3x4 - 8x3 + 6x2 + 1 and sketch its graph. Solution Step I. Determine the signs of f and f". First, solve for the transition points: f'(x) = 12x3 - 24x2 + 12x = 12x(x - 1)2, f"(x) = 36x2 - 48x + 12 = 12(x - 1)(3x - 1),

so f' = 0

=

so t

=

=0

x = 0,1 1

SECTION 4.5

Analyzing and Sketching Graphs of Functions

213

The signs of f' and f" are recorded in the following tables: Interval (-oo, 0)

(0,1) (1, oo) _ / Local

+ X

N Inflection

0

min

Inflection point

point

Test value f' (-1) =

Sign of f'

Interval

Test value

Sign of f"

-48

-

f"(0) = 12

±

4

+

( - °°, ) (1, oo)

f"(2) = 60

±

f'( ) = f'(2) = 24

(1.,1)

+

Step 2. Note transition points and sign combinations. The transition points c = 0, 1 divide the x-axis into four intervals (Figure 7). The type of sign change determines the nature of the transition point: • c = 0: local min since f' changes from - to +. • c= corresponds to a point of inflection since f" changes sign. • c =- 1: neither a local min nor a local max since f' does not change sign, but it is a point of inflection since f "(x) changes sign.

FIGURE 7

We plot the transition points c = 0, 1 in Figure 8(A) using function values and f (1) = 2. = (1) f 1, = f (0) Step 3. Draw arcs of appropriate shape and asymptotic behavior. Before drawing the arcs, we note that f (x) has leading term 3x4, so f (x) tends to oo -oo. We obtain Figure 8(B). • as x -> oo and as x

g,

++ 4

Points of inflection

FIGURE 8 f (x) = 3x4

(A)

- 8x3 + 6x 2 + 1.

over [0, in , and sketch its

EXAMPLE 4 Investigate the behavior of f (x) = cos x graph. Solution First, we find the transition points for x in [0, r]: 1 f' (x) = - sin x + - ' 2 f"(x) = - cos x,

so f (x) = 0

so f"(x) = 0

=

x=

7

57

-6

6

=

The sign combinations are shown in the following tables:

+-

Interval

Test value

(0, )

f'( -E) c-•-• 0.24

6-r )

0.5 X

7r 6

Jr

2

FIGURE 9 f (x) = cos x +

57r 6 x.

7r

fl (S) = f / (14) P--• 0.24

Sign of f

Interval

Test value

Sign of f"

(0,$) f"( )— (S,7)

f"(*) =

We record the sign changes and transition points in Figure 9 and sketch the graph using the values 57 0.44, f(7r) :=-•-•0.57 • 7 ) 0.79, f (- ) f(0) = 1, f ( 1 ) k" 1.13, f ( L 6 2 6

CHAPTER 4

APPLICATIONS OF THE DERIVATIVE

EXAMPLE 5 Investigate the behavior of f(x) =

3x + 2 and sketch its graph. 2x - 4

Solution The function f is not defined for all x. This plays a role in our analysis so we add a Step 0 to our procedure. Step 0. Determine the domain of f. Since f(x) is not defined for x = 2, the domain of f consists of the two intervals (-cc, 2) and (2, cc). We must analyze f on these intervals separately. Step 1. Determine the signs of f' and f". Calculation shows that f'(x) -

4 , (x - 2)2

f"(x) =

8 (x -2)

Although f'(x) is not defined at x = 2, it is not a critical point because x = 2 is not in the domain of f. In fact, f'(x) is negative for x 0 2, so f is decreasing and has no critical points. On the other hand, f"(x) > 0 for x > 2 and f"(x) 0

f(x) > 0

X

—1

0 f(x) 0, > 0, 0 for all x. Show that f (x) cannot be negative for all x. Hint: Show that f (b) 0 for some b and use the result of Exercise 62 in Section 4.4.

FIGURE 21

4.6 Applied Optimization D(v) (km) 200 150 100 50 10

20

30

I v (m/s) 40

FIGURE 1 Physiology and aerodynamics are applied to obtain a plausible formula for bird migration distance D as a function of velocity v. The optimal velocity corresponds to the maximum point on the graph (see Exercise 69).

Optimization plays a role in a wide range of disciplines, including the physical sciences, economics, and biology. For example, scientists have studied how migrating birds choose an optimal velocity v that maximizes the distance D they can travel without stopping, given the energy that can be stored as body fat (Figure 1). In many optimization problems, the first step is to write down the objective function. This is the function whose minimum or maximum we seek. Once we find the objective function, we can apply the techniques developed in this chapter. Our first examples require optimization on a closed interval [a, b]. Let's recall the steps for finding extrema developed in Section 4.2: (i) Find the critical points of f in [a, b]. (ii) Evaluate f (x) at the critical points and the endpoints a and b. (iii) The least and greatest values are the extreme values of f on [a, b]. EXAMPLE 1 A piece of wire of length L is bent into the shape of a rectangle (Figure 2). Which dimensions produce the rectangle of maximum area?

Y=

FIGURE 2

x

SECTION

An equation relating two or more independent variables in an optimization problem is called a constraint equation. The idea is that we cannot assume the variables take on any values we want; instead they are constrained to satisfy a specific equation. In Example 1, the constraint equation is 2x + 2y = L

4.6

A(x) = x

—L—x 2

1 Lx — x2 2

On which interval does the optimization take place? The sides of the rectangle are nonnegative, so we require both x > 0 and IL — x > 0. Thus, 0 < x 1L. Our problem is


0. FIGURE 19

39. Find the maximum area of a rectangle circumscribed around a rectangle of sides L and H. Hint: Express the area in terms of the angle (Figure 22).

30. Find the radius and height of a cylindrical can of total surface area A whose volume is as large as possible. Does there exist a cylinder of surface area A and minimal total volume? 31. A poster of area 6000 cm2 has blank margins of width 10 cm on the top and bottom and 6 cm on the sides. Find the dimensions that maximize the printed area. 32. According to postal regulations, a carton is classified as "oversized" if the sum of its height and girth (perimeter of its base) exceeds 108 in. Find the dimensions of a carton with a square base that is not oversized and has maximum volume. 33. Kepler's Wine Barrel Problem In his work Nova stereometria doliorum vinariorum (New Solid Geometry of a Wine Barrel), published in 1615, astronomer Johannes Kepler stated and solved the following problem: Find the dimensions of the cylinder of largest volume that can be

FIGURE 22

40. A contractor is engaged to build steps up the slope of a hill that has the x2(120 — x) shape of the graph of y — for 0 < x < 80 with x in meters 6400

SECT TO N 4.6

(Figure 23). What is the maximum vertical rise of a stair if each stair has a horizontal length of m?

FIGURE 23

Applied Optimization

227

47. Your task is to design a rectangular industrial warehouse consisting of three separate spaces of equal size as in Figure 26. The wall materials cost $500 per linear meter and your company allocates $2,400,000 for that part of the project involving the walls. (a) Which dimensions maximize the area of the warehouse? (b) What is the area of each compartment in this case?

FIGURE 26

41. Find the equation of the line through P = (4, 12) such that the triangle bounded by this line and the axes in the first quadrant has minimal area.

48. Suppose, in the previous exercise, that the warehouse consists of n separate spaces of equal size. Find a formula in terms of n for the maximum possible area of the warehouse.

42. Let P = (a, b) lie in the first quadrant. Find the slope of the line through P such that the triangle bounded by this line and the axes in the first quadrant has minimal area. Then show that P is the midpoint of the hypotenuse of this triangle.

49. According to a model developed by economists E. Heady and J. Pesek, if fertilizer made from N pounds of nitrogen and P lb of phosphate is used on an acre of farmland, then the yield of corn (in bushels per acre) is

43. Archimedes's Problem A spherical cap (Figure 24) of radius r and height h has volume V = irh2 (r — 1h) and surface area S = 2.7trh. Prove that the hemisphere encloses the largest volume among all spherical caps of fixed surface area S.

Y = 7.5 + 0.6N + 0.7P — 0.001N2 — 0.002P2

0.001NP

A farmer intends to spend $30/acre on fertilizer. If nitrogen costs 25 cents/lb and phosphate costs 20 cents/lb, which combination of N and P produces the highest yield of corn? 50. Experiments show that the quantities x of corn and y of soybean required to produce a hog of weight Q satisfy Q = 0.5x 1/2 y!/ 4. The unit of x, y, and Q is the cwt, an agricultural unit equal to 100 lb. Find the values of x and y that minimize the cost of a hog of weight Q = 2.5 cwt if corn costs $3/cwt and soy costs $7/cwt. 51. All units in a 100-unit apartment building are rented out when the monthly rent is set at r = $900/month. Suppose that one unit becomes vacant with each $10 increase in rent and that each occupied unit costs $80/mon in maintenance. Which rent r maximizes monthly profit?

FIGURE 24 44. Find the isosceles triangle of smallest area (Figure 25) that circumscribes a circle of radius 1 (from Thomas Simpson's The Doctrine and Application of Fluxions, a calculus text that appeared in 1750).

FIGURE 25

52. An 8-billion-bushel corn crop brings a price of $2.40/bushel. A commodity broker uses the rule of thumb: If the crop is reduced by x percent, then the price increases by 10x cents. Which crop size results in maximum revenue and what is the price per bushel? Hint: Revenue is equal to price times crop size. 53. The monthly output of a Spanish light bulb factory is P = 2LK2 (in millions), where L is the cost of labor and K is the cost of equipment (in millions of euros). The company needs to produce 1.7 million units per month. Which values of L and K would minimize the total cost L K? 54. The rectangular plot in Figure 27 has size 100 m x 200 m. Pipe is to be laid from A to a point P on side BC and from there to C. The cost of laying pipe along the side of the plot is $45/m and the cost through the plot is $80/m (since it is underground). (a) Let f (x) be the total cost, where x is the distance from P to B. Determine f (x), but note that f is discontinuous at x = 0 (when x = 0, the cost of the entire pipe is $45/m). (b) What is the most economical way to lay the pipe? What if the cost along the sides is $65/m?

45. A box of volume 72 m3 with a square bottom and no top is constructed out of two different materials. The cost of the bottom is $40/m2 and the cost of the sides is $30/m2. Find the dimensions of the box that minimize total cost. 46. Find the dimensions of a cylinder of volume 1 m3 of minimal cost if the top and bottom are made of material that costs twice as much as the material for the side.

FIGURE 27

228

APPLICATIONS OF THE DERIVATIVE

CHAPTER 4

, x„, find the value of x minimizing the sum

55. Brandon is on one side of a river that is 50 m wide and wants to reach a point 200 m downstream on the opposite side as quickly as possible by swimming diagonally across the river and then running the rest of the way. Find the best route if Brandon can swim at 1.5 m/s and run at 4 m/s.

61. Given n numbers xi , of the squares:

56. Snell's Law When a light beam travels from a point A above a swimming pool to a point B below the water (Figure 28), it chooses the path that takes the least time. Let v I be the velocity of light in air and v2 the velocity in water (it is known that vi > v2). Prove Snell's Law of Refraction: sin 6i = sin 02

First, solve for n = 2,3 and then try it for arbitrary n.

Vj

V2

- x02 +(x -

X2)2 + • • • ±

-

X0 2

62. A billboard of height b is mounted on the side of a building with its bottom edge at a distance h from the street as in Figure 31. At what distance x should an observer stand from the wall to maximize the angle of observation 0? 63. Solve Exercise 62 again using geometry rather than calculus. There is a unique circle passing through points B and C that is tangent to the street. Let R be the point of tangency. Note that the two angles labeled * in Figure 31 are equal because they subtend equal arcs on the circle. (a) Show that the maximum value of 0 is 0 = If. Hint: Show that = 0 + LPBA, where A is the intersection of the circle with PC.

A hi

(b) Prove that this agrees with the answer to Exercise 62. (c) Show that LQRB = ZRCQ for the maximal angle *.

FIGURE 28 57. Vascular Branching A small blood vessel of radius r branches off at an angle 0 from a larger vessel of radius R to supply blood along a path from A to B. According to Poiseuille's Law, the total resistance to blood flow is proportional to T

(a — bcot0 R4

b csc 0) r4

where a and b are as in Figure 29. Show that the total resistance is minimized when cos = (rIR)4.

FIGURE 29 In Exercises 58-59, a box (with no top) is to be constructedfrom a piece of cardboard with sides of lengths A and B by cutting out squares of length h from the corners andfolding up the sides (Figure 30). 58. Find the value of h that maximizes the volume of the box if A = 15 and B = 24. What are the dimensions of this box? 59. Which values of A and B maximize the volume of the box if h = 10 cm and AB = 900 cm2?

hi

--1

r-A FIGURE 30

60. Which value of h maximizes the volume of the box if A = B?

FIGURE 31 64. Optimal Delivery Schedule A gas station sells Q gallons of gasoline per year, which is delivered N times per year in equal shipments of QI N gallons. The cost of each delivery is d dollars and the yearly storage costs are sQT, where T is the length of time (a fraction of a year) between shipments and s is a constant. Show that costs are minimized for N= (Hint: T =11N.) Find the optimal number of deliveries if Q = 2 million gal, d = $8000, and s = 30 cents/gal-year. Your answer should be a whole number, so compare costs for the two integer values of N nearest the optimal value. 65. Victor Klee's Endpoint Maximum Problem Given 40 m of straight fence, your goal is to build a rectangular enclosure using 80 additional meters of fence that encompasses the greatest area. Let A(x) be the area of the enclosure, with x as in Figure 32. (a) Find the maximum value of A(x). (b) Which interval of x values is relevant to our problem? Find the maximum value of A(x) on this interval.

SECTION 4.6 40 20—x 1

I20 —x

Applied Optimization

229

70. The problem is to put a "roof' of side s on an attic room of height h and width b. Find the smallest length s for which this is possible if b = 27 and h = 8 (Figure 35). 71. Redo Exercise 70 for arbitrary b and h.

40 + x FIGURE 32 66. Let (a, b) be a fixed point in the first quadrant and let S(d) be the sum of the distances from (d, 0) to the points (0,0), (a, b), and (a, —b). (a) Find the value of d for which S(d) is minimal. The answer depends on whether b < ,N&L or b > ,N/a. Hint: Show that d = 0 when b > (b) (GU) Let a = 1. Plot S for b = 0.5, 13, 3 and describe the position of the minimum. 67. The force F (in Newtons) required to move a box of mass m kg in motion by pulling on an attached rope (Figure 33) is F(0) =

f mg cos 0 + f sin 0

where 0 is the angle between the rope and the horizontal, f is the coefficient of static friction, and g = 9.8 m/s2. Find the angle 0 that minimizes the required force F, assuming f = 0.4. Hint: Find the maximum value of cos 0 + f sine.

FIGURE 35

FIGURE 36

72. Find the maximum length of a pole that can be carried horizontally around a corner joining corridors of widths a = 24 and b = 3 (Figure 36). 73. Redo Exercise 72 for arbitrary widths a and b. 74. Find the minimum length t of a beam that can clear a fence of height h and touch a wall located b ft behind the fence (Figure 37).

FIGURE 33

t•

68. In the setting of Exercise 67, show that for any f the minimal force required is proportional to l/ \/1 f 2 . 69. Bird Migration Ornithologists have found that the power (in joules per second) consumed by a certain pigeon flying at velocity v m/s is described well by the function P(v) -= 17v-1 + 10-3 0 joules/s. Assume that the pigeon can store 5 x 104 joules of usable energy as body fat. (a) Show that at velocity v, a pigeon can fly a total distance of D(v) = (5 x 104)v/P(v) if it uses all of its stored energy. (b) Find the velocity vp that minimizes P. (c) Migrating birds are smart enough to fly at the velocity that maximizes distance traveled rather than minimizes power consumption. Show that the velocity vd which maximizes D(v) satisfies Pi(vd) = P(vd)/vd. Show that vd is obtained graphically as the velocity coordinate of the point where a line through the origin is tangent to the graph of P (Figure 34). (d) Find vd and the maximum distance D(vd).

FIGURE 37

75. 1- 7I A basketball player stands d feet from the basket. Let h and a be as in Figure 38. Using physics, one can show that if the player releases the ball at an angle 0, then the initial velocity required to make the ball go through the basket satisfies V

2

=

16d c0s2 0(tan 0 — tan a)

(a) Explain why this formula is meaningful only for a 1. 25. If you borrow L dollars for N years at a yearly interest rate r, your monthly payment of P dollars is calculated using the equation ( 1 _ b-i2N where b = 1 + -i-2L=P b- 1 ) (a) Find P if L = $5000, N = 3, and r = 0.08 (8%). (b) You are offered a loan of L = $5000 to be paid back over 3 years with monthly payments of P = $200. Use Newton's Method to compute b and find the implied interest rate r of this loan. Hint: Show that (Lip)b12N+1 _ (1 ± Li p)bi2N ± 1 = 0 26. If you deposit P dollars in a retirement fund every year for N years with the intention of then withdrawing Q dollars per year for M years, you must earn interest at a rate r satisfying P(bN - l)= Q(1 - bT41),

where b = 1 + r

Assume that $2000 is deposited each year for 30 years and the goal is to withdraw $10,000 per year for 25 years. Use Newton's Method to compute b and then find r. Note that b = 1 is a root, but you want the root satisfying b > 1.

FIGURE 8 21. (GU) Find the smallest positive value of x at which y = x and y = tan x intersect. Hint: Draw a plot.

27. There is no simple formula for the position at time t of a planet P in its orbit (an ellipse) around the sun. Introduce the auxiliary circle and angle in Figure 10 (note that P determines 8 because it is the central angle of point B on the circle). Let a = OA and e = OS/OA (the eccentricity of the orbit).

236

APPLICATIONS OF THE DERIVATIVE

CHAPTER 4

(a) Show that sector BSA has area (a2 /2)(0 - e sin 0). (b) By Kepler's Second Law, the area of sector BSA is proportional to the time t elapsed since the planet passed point A, and because the circle has area na2, BSA has area (ra2)(t1T), where T is the period of the orbit. Deduce Kepler's Equation: 27r t — = 0 e sin 0 T (c) The eccentricity of Mercury's orbit is approximately e = 0.2. Use Newton's Method to find 9 after a quarter of Mercury's year has elapsed (t = T/4). Convert 9 to degrees. Has Mercury covered more than a quarter of its orbit at t = T/4?

28. The roots of f(x) = x3 - 4x + 1 to three decimal places are -3.583, 0.251, and 3.332 (Figure 11). Determine the root to which Newton's Method converges for the initial choices xo = 1.85, 1.7, and 1.55. The answer shows that a small change in xo can have a significant effect on the outcome of Newton's Method.

Auxiliary circle

ArAi •

‘

\

\

.

Elliptical orbit .

/ .,/ /

`..

/

/

FIGURE 11 Graph of f (x) =

/

- 4x + 1.

29. What happens when you apply Newton's Method to find a zero of f (x) = x113? Note that x = 0 is the only zero. 30. What happens when you apply Newton's Method to the equation x3 - 20x = 0 with the unlucky initial guess xo = 2?

FIGURE 10

Further Insights and Challenges 31. Newton's Method can be used to compute reciprocals without performing division. Let c > 0 and set f (x) = x-1 - c. (a) Show that x - ( f (x)I (x)) = 2x - cx2 (b) Calculate the first three iterates of Newton's Method with c = 10.3 and the two initial guesses xo = 0.1 and xo = 0.5. (c) Explain graphically why xo = 0.5 does not yield a sequence converging to 1/10.3. In Exercises 32 and 33, consider a metal rod of length L fastened at both ends. If you cut the rod and weld on an additional segment of length m, leaving the ends fixed, the rod will bow up into a circular arc of radius R (unknown), as indicated in Figure 12.

32. Let h be the maximum vertical displacement of the rod. (a) Show that L = 2R sin 0 and conclude that

h-

(b) Show that L

L(1 - cos 0) 2 sin°

m = 2R0 and then prove sin 0 = L

2

m

33. Let L -= 3 and m = 1. Apply Newton's Method to Eq. (2) to estimate 0, and use this to estimate h. 34. Quadratic Convergence to Square Roots let en = xn - Jbe the error in xn •

Let f (x) = x 2 - c and

(a) Show that xn +i = (xn+c/x) and en +i = en2 /2xn•

FIGURE 12 The bold circular arc has length L

m.

(b) Show that if xo >

then xn >

(c) Show that if xo >

then en +1
0. 22. A function f has derivative f' (x) =

x4 + 1 val [1,4] does f take on its maximum value?

Where on the inter-

In Exercises 23-28, find the critical points and determine whether they are minima, maxima, or neither. 23. f (x) = x3 - 4x2 + 4x

24. s(t) = t4 - 8t2

25. f (x) = x2(x + 2)3

26. f (x) = x2/3(l - x)

27. g(0) = sin2 0 +0

28. h(0) = 2 cos 20 + cos 40

x (i) FIGURE 1 Graphs of the derivative. In Exercises 39-44, find the points of inflection. 39. y = x3 - 4x2 ± 4x 41. y =

40. y = x - 2 cos x

x2

42. y = (x2 _ 4)113

x2 + 4

x3 - x 43. f (x) = x2 + 1

44. f (x) = sin 2x - 4 cos x

In Exercises 45-54, sketch the graph, noting the transition points and asymptotic behavior. 45. y = 12x - 3x2

46. y = 8x2 - x4

47. y =x3 -2x 2 +3

48. y =4x

49. y = x3 ±

50. y = (x2 _ 4)2/3

51. y =

1

52. y = 12 - x3

Ix + 21 +

53. y = v sinx - cos x on [0,27r] 54. y = 2x - tan x on [0, 2711 55. Draw a curve y = f(x) for which f" and f" have signs as indicated in Figure 2.

In Exercises 29-36, find the extreme values on the interval. 29. f (x) = x(10 - x), [-1,31 30. f (x) = 6x4 - 4x6, [-2,2] 31. g(0) = sin2 0 - cos 0, [0, 2.71-] 32. R(t) =

t2

t+1

[0,3]

'

I

-2

0

I -4

1

3

++ I 4- -x 5

FIGURE 2 56. Find the dimensions of a cylindrical can with a bottom but no top of volume 4 m3 that uses the least amount of metal.

238

CHAPTER 4

APPLICATIONS OF THE DERIVATIVE

57. A rectangular open-topped box of height h with a square base of side b has volume V = 4 m3. Two of the side faces are made of material costing $40/m2. The remaining sides cost $20/m2. Which values of b and h minimize the cost of the box? 58. The corn yield on a certain farm is Y = —0.118x2 + 8.5x + 12.9

(bushels per acre)

where x is the number of corn plants per acre (in thousands). Assume that corn seed costs $1.25 (per thousand seeds) and that corn can be sold for $1.50/bushel. Let P(x) be the profit (revenue minus the cost of seeds) at planting level x. (a) Compute P(x0) for the value xo that maximizes yield Y. (b) Find the maximum value of P(x). Does maximum yield lead to maximum profit? 59. A quantity N(t) satisfies dNIdt =21t — 8/t2 for t > 4 (t in days). At which time is N increasing most rapidly? 60. A truck gets 10 miles per gallon (mpg) of diesel fuel traveling along an interstate highway at 50 mph. This mileage decreases by 0.15 mpg for each mile per hour increase above 50 mph.

FIGURE 4 64. A box of volume 8 m3 with a square top and bottom is constructed out of two types of metal. The metal for the top and bottom costs $50/m2 and the metal for the sides costs $30/m2. Find the dimensions of the box that minimize total cost. 65. Let f be a function whose graph does not pass through the x-axis and let Q = (a, 0). Let P = (x0, f(xo)) be the point on the graph closest to Q (Figure 5). Prove that PQ is perpendicular to the tangent line to the graph of xo. Hint: Find the minimum value of the square of the distance from (x, f (x)) to (a, 0).

(a) If the truck driver is paid $30/h and diesel fuel costs P = $3/gal, which speed v between 50 and 70 mph will minimize the cost of a trip along the highway? Notice that the actual cost depends on the length of the trip, but the optimal speed does not. (b) (= Plot cost as a function of v (choose the length arbitrarily) and verify your answer to part (a). (c) n Do you expect the optimal speed v to increase or decrease if fuel costs go down to P = $2/gal? Plot the graphs of cost as a function of v for P = 2 and P = 3 on the same axis and verify your conclusion. FIGURE 5 61. Find the maximum volume of a right-circular cone placed upsidedown in a right-circular cone of radius R = 3 and height H = 4 as in Figure 3. A cone of radius r and height h has volume 17t-r2h. 62. Redo Exercise 61 for arbitrary R and H.

66. Take a circular piece of paper of radius R, remove a sector of angle 09 (Figure 6), and fold the remaining piece into a cone-shaped cup. Which angle 8 produces the cup of largest volume?

FIGURE 6 FIGURE 3 63. Show that the maximum area of a parallelogram ADE F that is inscribed in a triangle ABC, as in Figure 4, is equal to one-half the area of AABC.

67. Use Newton's Method to estimate •:./T5 to four decimal places. 68. Use Newton's Method to find a root of f(x)= x2 — x — 1 to four decimal places.

f 4S_

a

-

P.44,4.4k.

5 INTEGRATION T

For many years, the elongated "s" was commonly used in writing and print. Here Gottfried Wilhelm Leibniz introduces it as a symbol for "summa," the mathematics concept he is developing that was to become the definite integral. Although the elongated "s" is no longer used in everyday writing, it became the iconic symbol representing the integral in mathematics. You will be introduced to the integral in this chapter, and the integral sign will become a familiar symbol that plays an important role in the mathematics that you learn and apply.

he last two chapters developed the derivative, one of the primary topics in calculus. The derivative was motivated by the basic problem of finding a line tangent to a curve at a given point. In addition, we saw that the derivative is much more significant than just a means for finding tangent lines. In this chapter, we introduce the next primary topic, the definite integral. It too is motivated by a basic problem: finding the area under a curve. And in this case as well, once you become familiar with the definite integral you will realize that its importance goes well beyond computing area. Computing tangent lines and areas may seem completely unrelated, and therefore the derivative and the definite integral may appear to be completely separate topics. They aren't. There is a deep connection between them that is revealed by the Fundamental Theorem of Calculus, discussed in Sections 5.4 and 5.5. This theorem expresses the "inverse" relationship between integration and differentiation. It plays a truly fundamental role in nearly all applications of calculus, both theoretical and practical.

5.1 Approximating and Computing Area Why might we be interested in the area under a graph? Consider an object moving in a straight line with constant positive velocity v. The distance traveled over a time interval [t1, t2] is equal to v At, where At = (t2 — t1) is the time elapsed. This is the well-known formula distance traveled = velocity x time elapsed vAt

Because v is constant, the graph of velocity is a horizontal line (Figure 1) and v At is equal to the area of the rectangular region under the graph of velocity over [4, t21. So we can write Eq. (1) as 2

distance traveled = area under the graph of velocity over [t1, /2]

There is, however, an important difference between these two equations: Eq. (1) makes sense only if velocity v is constant, whereas Eq. (2) is correct even if the velocity changes with time. We examine the relationship in Eq. (2) further in Section 5.6. Thus, the advantage of expressing distance traveled as an area is that it enables us to deal with much more general types of motion. To see why Eq. (2) might be true in general, let's consider the case where velocity changes over time but is constant on intervals. In other words, we assume that the object's velocity changes abruptly from one interval to the next as in Figure 2. The distance traveled over each time interval is equal to the area of the rectangle above that interval, so the total distance traveled is the sum of the areas of the rectangles. In Figure 2, distance traveled over [0,8] s -= 10+ 15 + 30 + 10 = 65 m Sum of areas of rectangles v (m/s) v (m/s)

Area = vAt 15 10

15 (s) I

IM =t2 -11

FIGURE 1 The rectangle has area vAt, which is equal to the distance traveled,

30

5

10

10

t (s) 1

3

4

5

6

7

8

FIGURE 2 Distance traveled equals the sum of the areas of the rectangles. 239

240

CHAPTER 5

INTEGRATION

Distance traveled is equal to the area under the graph. It is approximated by the sum of the areas of the rectangles.

FIGURE 3

Our strategy when velocity changes continuously (Figure 3) is to approximate the area under the graph by a sum of areas of rectangles. We can continually improve the approximation by using thinner and thinner rectangles, and then take a limit to obtain an exact value of distance traveled. This idea leads to the concept of an integral.

Approximating Area by Rectangles

Recall the two-step procedure for finding the slope of the tangent line (the derivative): First approximate the slope using secant lines and then compute the limit of these approximations. In integral calculus, there are also two steps: • First, approximate the area under the graph using rectangles. • Then compute the exact area (the integral) as the limit of these approximations.

Our goal is to compute the area under the graph of a function f. In this section, we assume that f is continuous and positive, so that the graph of f lies above the x-axis (Figure 4). The first step is to approximate the area using rectangles. To begin, choose a whole number N and divide [a, b] into N subintervals of equal width, as in Figure 4(A). The full interval [a, b] has width b — a, so each subinterval has width Ax = (b — a)I N . The right endpoints of the subintervals are xi = a ± Ax, x2 = a ± 26,x,

, xN_i = a + (N — 1).6a, xN = a ± NAx

Note that the last right endpoint is xN = b because a ± NAx = a ± N ((b — a)/ N) = b. The general term xi can be expressed as xi = a ± j Ax, indicating that it is obtained by adding j intervals of width Ax to a. Next, as in Figure 4(B), above each subinterval construct a rectangle whose height is the value of f (x) at the right endpoint of the subinterval. Height of second rectangle is f(x2). Height of first rectangle is f(x ).

a

X1

a

x2

(A) Divide [a, b] into N subintervals, each of width Ax.

x1

X2

...

(B) Construct right-endpoint rectangles.

FIGURE 4

The sum of the areas of these rectangles provides an approximation to the area under the graph. The first rectangle has base Ax and height f (xi), so its area is f(xi )Ax. Similarly, the second rectangle has height f (x2) and area f (x2) Ax, and so on. The sum of the areas of the rectangles is denoted RN and is called the Nth right-endpoint approximation: RN =

f (xi)Ax + f(x2)Ax +. . . + f (xN)Ax

Factoring out Ax, we obtain the formula RN

= Ax

(f (xi)

f (x2) + • • • + f(xN))

SECTION 5.1

To summarize: a = left endpoint of interval [a, b] b = right endpoint of interval [a, b] N = number of subintervals in [a, b] Ax =

b—a

Approximating and Computing Area

241

In words: RN is equal to /Ix times the sum of the function values at the right endpoints of the subintervals. EXAMPLE 1 Calculate R4 and

R6

for f (x) = x2 on the interval [1,3].

Solution Step I. Determine Ax and the right endpoints. To calculate Ra, divide [1,31 into four subintervals of width Ax = The right endpoints are the numbers xi = a + j = 1 + j(;) for j = 1, 2, 3, 4. They are spaced at intervals of beginning at ;, so, as we see in Figure 5(A), the right endpoints are Step 2. Calculate Ax times the sum of function values. R4 is Ax times the sum of the function values at the right endpoints:

3, 4, 3,

1 ( (2 \ 2 =2

2)

(4 ) 2

+ 2

(5

)2

1- 2

) 2 ± i

=

2

= 10.75

4

= 1-, and the right endpoints are spaced at intervals of 3 is similar: Ax = beginning at and ending at 3, as in Figure 5(B). Thus,

R6

/4\

1 / R6 =

/ 5\

V W

f

/ 6\ f

f ( -83) ± f G))

f

•

f(x)=x2

15

f(x)=x

15 10

10

3 2

1

5 2

(A) The approximation

R4

4 3

5 3

2

7 3

8 3

3

(B) The approximation

R6

13 FIGURE 5

The R4 and R6 values in the previous example are clearly overestimates of the area under y = x2 between 1 and 3, although the approximation improves going from R4 to R6. Later in the section, we show how to obtain a general expression for RN. With that, oo to obtain an exact value of 83 as the area (see we can then take the limit as N Exercise 55).

Summation Notation Summation notation is a standard notation for writing sums in compact form. The sum of numbers am . . . ..a (m 1)

For example,

E(j2 + j) = (32 + 3) + (42 + 5

+ (52

5) =62

j=3

can also be expressed as 5

j=3

j2 +Ej=02 +42 + +(3+4+ 5) = 50+ 12 = 62 5

j=3

The linearity properties can be used to write a single summation as a linear combination of several summations. For example, 100

100

100

100

k=0

k=0

k=0

k=0

E(7k2 _ 4k ± 9) = E7k2 + E(-4k) E 9 100 =7Ek k=0

100

100

2 —4Ek-F9E k=0

1 k=0

It is convenient to use summation notation when working with area approximations. For example, RN is a sum with general term RN

=

Ax[f (xi)

f (x2) + • + f (xN)]

SECTION 5.1

Approximating and Computing Area

243

The summation extends from j = 1 to j = N, so we can write RN concisely as

RN

=

Ax

E f(x1) j=1

We shall make use of two other rectangular approximations to area: the left-endpoint and the midpoint approximations. Divide [a, b] into N subintervals as before. In the leftendpoint approximation L N, the heights of the rectangles are the values of f(x) at the left endpoints [Figure 6(A)]. These left endpoints are

4=0 REMINDER LX =

b—a

xo = a, x1 = a + Ax, x2 = a + 2Ax,

xN_I = a + (N — 1)Ax

and the sum of the areas of the left-endpoint rectangles is L N = Ax f (x0) + f (xi) + f (x2) +

+ f(xN-1))

Note that both RN and LN have general term f (x j ), but the sum for LN runs from j = 0 to j = N — 1 rather than from j = 1 to j = N: N-1 LN = Ax

E f(xj)

J=0 In the midpoint approximation MN, the heights of the rectangles are the values of f(x) at the midpoints of the subintervals rather than at the endpoints. As we see in Figure 6(B), the midpoints are XN-1+ XN X0 + X1 X1+ X2 2 2 2 rectangles is midpoint areas of the sum of the The MN

= Ax ( f (xo

2

)

f ( xi + x2) + 2

+ xN)) 2 )

f(

In summation notation, N-1 MN =

E f (xi +2x +1 ) f (x° +2 xi) i(xl +2 x2)

,

xo +

2

(A) Left-endpoint rectangles

xt + x2

2

(B) Midpoint rectangles

FIGURE 6

EXAMPLE 2 Calculate R6, L6, and M6 for f(x) = x-1 on [2,4]. Solution In this case, Ax -= (b — a)IN = (4 — 2)/6 = A. For the six intervals, the right V, and 131, and 4, and the left endpoints are 2, endpoints are 3, ;, 3,

4,-4,3,

244

INTEGRATION

CHAPTER 5

Therefore (Figure 7), R6 =

=1 j-

Left-endpoint rectangle

7

1

i (f G) + f 7

L6 = 1 ( f

3

8

11 10 ± f (3) + f (-3—) + f (i) + f (4))

8 ()

-10

11

4

\ ( (2) + f (_7\ + f (_8 \ + f (3) + f (_10 \ 3) 3 ) +f 3 )) j 3)

= 1 (1 + 3 + 3 + 1 + 3 + 3 ) FIGURE 7 L6 and R6 for on [2,4].

-,•--,'0.737

f (x) = x -1 The general term in M6 is f (xi ± X j+i) 2 15 7 19 21 from j = 0 to 5, we In this case, the midpoints are 13 6 6 16 6 6 and -2i. Summing obtain (Figure 8)

+ f ( _21

+ f ( _) 15 + f

fri6 =

+ f (263)) )

= _1 3

2 13 1212312 21 23 4 6 6 6 6 6 6

FIGURE 8 M6 for

f (x) = x-1 on [2,4].

13

+ 6 15

+ 17

-19

21

23

•

0.692

GRAPHICAL INSIGHT Monotonic Functions Observe in Figure 7 that the left-endpoint rectangles for f (x) -= x -1 extend above the graph and the right-endpoint rectangles lie below it. The exact area A under the graph of f from 2 to 4 must lie between R6 and L6, and so, according to the previous example, 0.65 < A < 0.74. More generally, when f is monotonic (increasing or decreasing), the exact area lies between RN and LN (Figure 9):

• f increasing = • f decreasing =

LN RN

< area under graph < RN < area under graph < LN

Notice that M6 lies between R6 and L6. This is always the case for a monotonic function (see Exercise 91). In the upcoming sections, we will see how to determine the exact area in the example. It turns out to be 1n2 0.693, and thus, M6 provides the best estimate, which is understandable considering these observations.

Right-endpoint rectangle Left-endpoint rectangle

Computing Area as the Limit of Approximations FIGURE 9 When f is increasing, the left-endpoint rectangles lie below the graph and right-endpoint rectangles lie above it.

FIGURE 10 The error decreases as we use more rectangles.

Figure 10 shows several right-endpoint approximations. Notice that the error in computing the area, corresponding to the yellow region above the graph, gets smaller as the number of rectangles increases. In fact, it appears that we can make the error as small as we please by taking the number N of rectangles large enough. If so, it makes sense to consider the limit as N —> oo, as this should give us the exact area under the curve. The next theorem guarantees that the limit exists (see Theorem 7 in Appendix D for a proof and Exercise 91 for a special case).

N= 2

N= 4

N= 8

SECT! ON 5.1 In Theorem I, it is not assumed that f (x) > 0. If f (x) takes on negative values, the limit L no longer represents area under the graph, but we can interpret it as a "signed area," discussed in the next section.

Approximating and Computing Area

245

THEOREM 1 If f is continuous on [a, b], then the endpoint and midpoint approximations approach one and the same limit as N co. In other words, there is a value L such that lull RN = lim LN N —>oo

lirn MN = L N —>oo

=

If f (x) > 0 on [a, b], we define the area under the graph over [a, b] to be L.

CONCEPTUAL INSIGHT In calculus, limits are used to define basic quantities that otherwise would not have a precise meaning. Theorem 1 allows us to define area as a limit L in much the same way that we define the slope of a tangent line as the limit of slopes of secant lines. The next three examples illustrate Theorem 1 using formulas for power sums. The kth power sum is defined as the sum of the kth powers of the first N integers. We shall use the power sum formulas fork = 1, 2, 3. A method for proving power sum formulas is developed in Exercises 47-51 of Section 1.3. Formulas (3)-(5) can also be verified using the method of induction.

Power Sums N J =1

E j2 = 12 + 22

N(N + 1)

N2

2

2

N 2

N (N + 1)(2N + 1)

N2

6

j=1

E J3 = + 23 + • • •+ N3 = N"

3

N3

N

N2

4

— 3 + 2 +

N4 N3 N2 + 1)2 = 4 ± 2 ± 4 4

5

j= 1

For example, by Eq. (4),

E j2 = 12 + 6

22 +

32 +

42 +

52 +

i= 1

62 =

63

62

N3 N2 N

6

for

=91

N=6

As a first illustration, we verify this limit approach to area by computing the area of a right triangle, a figure whose area we can also compute geometrically. EXAMPLE 3 Find the area A under the graph of f(x) = x over [0,4] in three ways: (a) Using geometry

(c)

lim RN N—>co

lim LN N—>co

Solution The region under the graph is a right triangle with base b = 4 and height h = 4 (Figure 11).

4=0 REMINDER f (xi)

RN = AX N-I LN

(b)

=

f (xi )

Ax i=o

Ax =

b—a

xj =a+ j Ax

(a) By geometry, A = bh = ( )(4)(4) = 8. (b) We compute this area again as a limit. Since 6,x = (b — a)IN = f (xj) = f (a + j Ax) = f(0± j (— N 4

= 4Ni

N . RN = AXE f(x) -= — N Tvj=1 j=1

j=1

=

J

N and f (x) = x,

246

CHAPTER 5

INTEGRATION

al FIGURE 11

The right-endpoint approximations approach the area of the triangle. In the last equality, we factored out 4/N from the sum. This is valid because 4/N is a constant that does not depend on j. Now use formula (3): RN

16 =

E.; = 16

(N(N + 1) ) 2

J=1

8 ( N2 = — N2

+

N) = 8 + —8

Formula for power sum

The second term 8/N tends to zero as N approaches oo, so A = firn RN N-*00

In Eq. (6), we apply the formula

i=1

E./

i=1

16 N-1 LN

16

16

N v• -1

i=1 (N - 1)N 2

8 N

lim (8

—)

N-> co

=8

As expected, this limit yields the same value as the formula ,bh. (c) The left-endpoint approximation is similar, but the sum begins at j = 0 and ends at j -= N - 1:

N(N +1) = 2

with N - 1 in place of N: N-1

=

(N - 1)N 2

=8 -

8 iv

r

Note in the second step that we replaced the sum beginning at j = 0 with a sum beginning at j = 1. This is valid because the term for j = 0 is zero and may be dropped. Again, we find that A = lim LN = lim (8 - 8/N) =- 8. • In the next example, we compute the area under a curved graph. Unlike the previous example, it is not possible to compute this area directly using geometry. EXAMPLE 4 Let A be the area under the graph of f (x) (Figure 12). Compute A as the limit lirn RN •

2x2 - x + 3 over [2,4]

N->oo

Solution Step I. Express RN in terms of power sums. In this case, Ax = (4 - 2)/N = 2/N and RN

=

px,) = Ax

Ax

f (a ± j Ax)= 7 11--

f

+

)

J=.1 12 Area under the graph of f(x)=2x2 - x + 3 over [2,4].

FIGURE

Let's use algebra to simplify the general term. Since f(x)= 2x - x + 3, f (2 +

2j

N

- (2 + -2Nj )+3

8j N

4j 2 N2

2

) = 2 (2

(

2j N

A./ 2 ±

L4 j ± 9 N

SECTION 5.1

Approximating and Computing Area

247

Now we can express RN in terms of power sums:

J-1

J=1 16 = N3

•2 j=1

28 18 + N2 L-di+ N

J=1

J=1

Step 2. Use the formulas for the power sums. Using formulas (3) and (4) for the power sums in Eq. (7), we obtain RN

= 16 ( N3 + N2 + IsI ± 28 (N2 ± N3 3 2 6) N2 2 2) =

16

8 + -N- +

8

22

8

112

18 (N)

14 + (14 + -AT

+ 18

(112

8 ) 112 — = —

Step 3. Calculate the limit. A = lim RN = liM N-±oo N-4-co

3

22 — N

3N2

•

3

The area under the graph of any polynomial can be calculated using power sum formulas as in the examples above. For other functions, such as in the next example, the limit defining the area may be difficult or impossible to evaluate directly.

÷ 3,t,

EXAMPLE 5 Let A be the area under the graph of f (x) = sin x on the interval (Figure 13). Set up (but do not compute) the expression A = lim RN for determining N->co

the area.

E

E

j=1

j=1

II

The limit in the previous example is difficult to compute; however, it can be approximated by computing RN for large N (see Exercise 83). In Section 5.4, we will see that this area is straightforward to compute (and is ,./) using the definite integral and the Fundamental Theorem of Calculus.

HISTORICAL PERSPECTIVE

@ Bettmann/Getty Images

FIGURE 13 The area of this region is more difficult to compute as a limit of endpoint approximations.

Solution In this case, Ax = (37r/4 - 7r14)1N = 7r/(2N) and the area A is N N 7r sin (-7r ± -71 ) ,Ax f (a + j Ax) = lim — A = lim RN = lim 4 2N N--)-oo N->oo N->oo 2N

Jacob Bernoulli (1654-1705)

We used the formulas for the kth power sums for k = 1, 2, 3. Do similar formulas exist for all powers k? This problem was studied in the seventeenth century and eventually solved around 1690 by the great Swiss mathJacob ematician Bernoulli. Of this discovery, he wrote

With the help of [these formulas] it took me less than half of a quarter of an hour to find that the 10th powers of the first 1000 numbers being added together will yield the sum 91409924241424243424241924242500 Bernoulli's formula has the general form

1 k±i 1 k &_, k±ln + it - -1- —12n- "+• • •

-2 j=1 The dots indicate terms involving smaller powers of n whose coefficients are expressed in terms of the so-called Bernoulli numbers. For example,

E j4 =

1 1 c 1 A 3 n- -I- in - +

1 --0n

=1 These formulas are available on most computer algebra systems.

CHAPTER 5

INTEGRATION

5.1 SUMMARY • Approximations to the area under the graph of f over the interval [a, b] b—a x i =a+ jAx): (Ax

Power Sums

Ei= j=1 N 2_, J j=1

—

N 0 on [a, b], we take L as the definition of the area under the graph of y = f (x) over [a, b].

• If

5.1 EXERCISES Preliminary Questions 1. What are the right and left endpoints if [2,5] is divided into six subintervals? The interval [1,5] is divided into eight subintervals. (a) What is the left endpoint of the last subinterval? (b) What are the right endpoints of the first two subintervals?

4

(C)

2.

3.

Which of the following pairs of sums are not equal? 4

(a) Ei,

4

4

Et t=i

(b)

5

E

>k 2

j=1

k=2

Exercises 1. Figure 14

shows the velocity of an object over a 3-minute interval. Determine the distance traveled over the intervals [0,3] and [1,2.5] (remember to convert from kilometers per hour to kilometers per minute).

4. 5.

E J=. Explain:

5

4

(d)

—1)

E + 1),

i=2 100

100

i j= j=1

1=0

100 100 j but E 1 is not equal to E 1.

j=i

1=0

Explain why L toe > R100 for f (x) = x-2 on [3,7].

2. An ostrich (Figure 15) runs with velocity 20 km/hour for 2 minutes (min), 12 km/h for 3 mm , and 40 km/h for another minute. Compute the total distance traveled and indicate with a graph how this quantity can be interpreted as an area.

km/hour 30 20

min

FIGURE 14

FIGURE 15 Ostriches can reach speeds as high as 70 km/h.

SECTION 5.1

Approximating and Computing Area

249

3. A rainstorm hit Portland, Maine, in October 1996, resulting in record rainfall. The rainfall rate R(t) on October 21 is recorded, in centimeters per hour, in the following table, where t is the number of hours since midnight. Compute the total rainfall during this 24-hour period and indicate on a graph how this quantity can be interpreted as an area.

t (h)

0-2

2-4

4-9

9-12

12-20

20-24

R(t) (cm/h)

0.5

0.3

1.0

2.5

1.5

0.6 FIGURE 17

4. The velocity of an object is v(t) = 12t m/s. Use Eq. (2) and geometry to find the distance traveled over the time intervals [0,21 and [2,5]. 5.

Compute R5 and L5 over [0, 1] using the following values: x

0

0.2

0.4

0.6

0.8

1

f (x)

50

48

46

44

42

40

6. Compute R6, L6, and M3 to estimate the distance traveled over [0, 3] if the velocity at half-second intervals is as follows: t (s)

0

0.5

1

1.5

2

2.5

3

v (m/s)

0

12

18

25

20

14

20

7.

Let f (x) = 2x + 3. (a) Compute R6 and L6 over [0,3]. (b) Use geometry to find the exact value of the area A, and compute the errors IA — R6I and IA — L6I in the approximations. 8.

Repeat Exercise 7 for f (x) = 20 — 3x over [2,4].

9. Calculate R3 and L3 for f (x) = x2 — x + 4 over [1,4]. Then sketch the graph of f and the rectangles that make up each approximation. Is the area under the graph larger or smaller than R3? Is it larger or smaller than L3? 10. Let f (x) = x2 + 1 and Ax = 1. Sketch the graph of f and draw the right-endpoint rectangles whose area is represented by the sum

13. Let f (x) = x2. (a) Sketch the function over the interval [0, 2] and the rectangles corresponding to L4. Calculate the area contained within them. (b) Sketch the function over the interval [0, 2] again but with the rectangles corresponding to 124. Calculate the area contained within them. (c) Make a conclusion about the area under the curve f (x) = x2 over the interval [0,2]. 14. Let f (x) = .Nrx (a) Sketch the function over the interval [0,4] and the rectangles corresponding to L4. Calculate the area contained within them. (b) Sketch the function over the interval [0, 4] again but with the rectangles corresponding to R4. Calculate the area contained within them. over (c) Make a conclusion about the area under the curve f (x) = the interval [0,4]. In Exercises 15-22, calculate the approximation for the given function and interval. [0,2]

15. L4,

f (x) =

16. L6,

f (x) = .V6x + 2,

17. R6,

f (x)

18. R5,

f (x) = x2 + x , [-1,1]

19. M5,

f (x) —

20. M4,

f (x) = .N/7x,

21. La,

f (x) = cos2 x,

22. L6,

f (x) = x2 + 31x1, [ —2, 1]

2x — x2,

1 2 1, x +

[1,3] [0,2]

[1,5]

[3,5] ,

6 E

f(1 + Ax)Ax.

In Exercises 23-28, write the sum in summation notation. 11. Estimate R3, M3, and L6 over [0, 1.5] for the function in Figure 16.

23. 47 + 57 + 67 + 77 + 87 24. (22 + 2) + (32 + 3) + (42 + 4) + (52 + 5) 25. (22 + 2) + (23 + 2) + (24 + 2) + (25 + 2) 26. 1 +

+

4 +

2 1 27. 2 • 3 ± 3 • 4 +

+

116

+

• ± (n + 1)(n + 2)

28. sin(7) + sin(7/2) + sin(7/3) + • • • + sin(7/(n + 1)) 29. Calculate the SWIM 4

5

(a)

9

E

(c)

i= FIGURE 16 12. Calculate the area of the shaded rectangles in Figure 17. Which approximation do these rectangles represent?

k=2

30. Calculate the sums: 4

(a)

E i= 3

E

sin (j -T )

(b)

k3

250

INTEGRATION

CHAPTER 5

54. Show, for f (x) = 3x3 - x2 over [1,5], that

31. Let b1 = 4, b2 = 1, b3 = 2, and b4 -= -4. Calculate: (a)

3

2

4

Ebi

(b)

E(2bi b;)

(C)

Ekbk

i=2

32. Assume that at = -5,

20, and

i=1 (b)

28j N

i=1

In Exercises 55-62, find a formula for RN and compute the area under the graph as a limit.

to E(2ai - 3b1 )

Eai

(c)

56. f (x) = x2, [-1,5]

55. f (x) = x2, [1,3]

200

E

33. Calculate

128j2 N4

Then evaluate litn RN. N-,•oo

i=2

i=1

N3

E bi = 7. Calculate:

to

to (a) E(442; + 3)

(192 j 3 j

io

Ea,

4

RN =

k=1

j. Hint: Write as a difference of two sums and use

57. f (x) = 6x2 -4,

[0,2]

59. f (x) = x3 - x ,

j=101

formula (3).

58. f (x) = x2 + 7x, [6, 11]

[2,5]

60. f (x) = 2x3 + x2, [-2,2] 30

34. Calculate

E(2j

1)2. Hint: Expand and use formulas (3)-(4).

.1=1

62. f (x) = x2, [a, b] (a, b constants with a co N

36. E(4k - 3) k=1

E n2

E

38.

E i=1

65. lim N->oo N

k3

k=101

n=51 50

39.

E

J-0

66.

40.

- 1)

42. m=1

In Exercises 43-46, use formulas (3)-(5) to evaluate the limit. N

43. Inn

N

45.

lirn L.

N->oo

N

44. lim

E N2 i=1

.2

N-sco

.

+1

46.

N3

ii1T1 RN, N->oo

E N4

J=1

(i , N->co

f (x) = 9x, [0,2]

lim

N->oo

2N

E (2 +

4 -13 )

j=1

•)4

j=0

E sin ( 13 -

:1=1

4N

2N

67. RN,

f (x) = sin x over [0,7r]

68. RN,

f (x) = x-1 over [1,7]

69. LN,

f (x) = ,V2x + 1 over [7, 11]

70. LN,

f (x) = cos x over

N4

20) N

71. MN,

f (x) = tan x over [1, 1]

72. MN,

f (x) = x-2 over [3,5]

1 73. Evaluate urn -

E _ (_N )

2

by interpreting it as the area of

:1=1 part of a familiar geometric figure.

48. lim RN, N-+o

f (x) = 3x + 6, [1,4]

49.

lim LN, N

f (x) = x + 2, [0,4]

50. lim LN, N -,•co

f (x) = 4x - 2, [1,3]

51. lim MN, N-o•oo

f (x) = x, [0,2]

52.

f (x) = 12 - 4x,

lim MN, N->oo

64'

N

lilll ;7 3

.3

In Exercises 47-52, calculate the limit for the given function and interval. Verify your answer by using geometry. 47.

4

In Exercises 67-72, express the area under the graph as a limit using the approximation indicated (in summation notation), but do not evaluate.

30

41.

i

N-1

200

150

37.

[a, b] (a, b constants with a oo

8j\) 7,s7

1 3

1 2N

1 6N2,

MN -

3

12N2

Then, given that the area under the graph of y = x2 over [0,1] is 1, rank the three approximations RN, LN, and MN in order of increasing accuracy (use Exercise 74). 76. For each of RN, LN, and MN, find the smallest integer N for which the error is less than 0.001.

SECTION 5.2

In Exercises 77-82, use the Graphical Insight on page 244 to obtain bounds on the area. 77. Let A be the area under f (x)= ,17- c over [0, 1]. By computing R4 and L4, prove that 0.51 < A < 0.77. Explain your reasoning. 78. Use R5 and L5 to show that the area A under y = .X -2 over [10, 13] satisfies 0.0218 1. (b) By (a), the sequence {LN} is bounded, so it has a least upper bound L. By definition, L is the smallest number such that LN < L for all N. Show that L < Rm for all M.

xi_1 Midpoint xi

(c) According to (b), LN < L < RN for all N. Use Eq. (8) to show that lim LN = L and lim RN = L. N —>co

FIGURE 18

N—>oc

5.2 The Definite Integral In the previous section, we saw that if f is continuous on an interval [a, b], then the oo: endpoint and midpoint approximations approach a common limit L as N L = lim RN = lim LN = lim MN N—>oo N—>oo N—>oo

1

In a moment, we will state formally that L is the definite integral off over [a, b]. Before doing so, we introduce Riemann sums, sums that are structured similar to the approximating sums L N, RN, and MN, and that generalize them.

252

CHAPTER 5

INTEGRATION

Recall that RN, LN, and MN use rectangles of equal width Ax, whose heights are the values of f (x) at the endpoints or midpoints of the subintervals. In Riemann sums, we relax these requirements: The rectangles need not have equal width, f (x) can have any real-number value, and the height (which could be negative) may be any value of f (x) within the subinterval. To specify a Riemann sum, we choose a partition and a set of sample points: • Partition P of size N: a choice of points that divides [a, b] into N subintervals (not necessarily of equal width): P :a = C1 C2 I• I •

Xo = a

CN •I XN = b

x1


0, the signed area is positive and is equal to the area of the rectangle • If f (ci) 0 IIPII—>0

E f (ci)6,xi i=1

When this limit exists, we say that f is integrable over [a, b]. The definite integral is often called, more simply, the integral of f over [a, b]. The process of computing integrals is called integration and f (x) is called the integrand. The endpoints a and b of [a, b] are called the limits of integration. Finally, we remark that any variable may be used as a variable of integration. That is, that variable is a dummy variable. Thus, the following three integrals all denote the same quantity:

fa

Georg Friedrich Riemann (1826-1866)

fa

sin u du sin t dt fa The next theorem assures us that continuous functions (and even functions with finitely many jump discontinuities) are integrable (see Appendix D for a proof). In practice, we rely on this theorem rather than attempting to prove directly that a given function is integrable. sin x dx,

THEOREM 1 If f is continuous on [a , b], or if f is continuous except at finitely many jump discontinuities in [a, b], then f is integrable over [a, b]. A constant function f (x) = K is integrable over every interval [a, b]. The following theorem provides the value of the integral and indicates that when K is positive, the integral is the area of the rectangle in Figure 4.

254

CHAPTER 5

INTEGRATION

THEOREM 2 Integral of a Constant For any constant function f(x) = K, b l

a FIGURE 4]

a

Kdx =K (b — a).

Proof Let R(f,, P, C) =

E f(ci)Axi be a Riemann sum for f

4

over [a, b]. Then,

i=t f (ci ) = K for each ci, and therefore, R(f,, P ,C) =

E f (c)Axi = EKAxi= K E Axi = K(b — a)

i=i i=i i=i The latter equality holds because the sum of the Axi is the overall length b — a of the interval [a, b]. Since every Riemann sum has the value K(b — a), it follows that the f (x)dx = K(b — a).

integral does as well. Thus,

•

a

The Definite Integral and Signed Area We motivated the development of approximating sums, Riemann sums, and the definite integral with the problem of determining the area under the graph of a function. The definite integral provides an exact value for such areas. Specifically, if f is integrable b

and f (x) > 0 over [a, b], then the area under the graph off over [a, b] is

f (x)dx.

When the geometry of a region is simple (e.g., rectangles, triangles, circles) the definite integral corresponds to the area obtained by a geometric formula, as demonstrated for a rectangle in Theorem 2. In some instances, this correspondence enables us to compute a definite integral using geometric formulas (as in Example 2 below). Allowing the possibility that f (x) takes on both positive and negative values, we define the notion of signed area, where regions below the x-axis provide a negative contribution (Figure 5). Intuitively, the signed area of a region is the area above the x-axis minus the area below. The f(c)Lxi terms in a Riemann sum are signed areas of rectangles. Therefore, a signed area such as in Figure 5 can be approximated by Riemann sums and is given by a definite integral. Thus, for all integrable f over [a, b], b

The signed area between the graph of f and the x-axis over [a, b] is

f (x)dx. fa

EXAMPLE 2

Definite Intregrals Via Simple Geometry

fo

Calculate

5

Signed area is the area above the x-axis minus the area below the x-axis.

(3 — x)dx

FIGURE 5

and

f o5 13

dx

Solution The region between y = 3 — x and the x-axis consists of two triangles of areas and 2 [Figure 6(A)]. The triangle with area lies above the x-axis and therefore has signed area 3. The second triangle lies below the x-axis, so it has signed area —2. In the graph of y =13 — x 1, both triangles lie above the x-axis [Figure 6(B)]. It follows that

;

;

5

9 5 (3 — x)dx = — — 2 = 2 2

5

13—xldx=

9

+2=

13 2

•

In the next example, the first integral is geometrically simple but the second is not. For the second, we need to take a limit of Riemann sums to arrive at a value. In Section 5.4, we will start to develop tools that make the computation of integrals such as the latter much simpler.

SECTION 5.2

y=3 —

Area

The Definite Integral

255

x

9 2

Signed area —2 (A)

(B)

FIGURE 6

EXAMPLE 3 For b >0, calculate

(a)f

x dx

0

and

(b)

X 2 dx

f

0

Solution The integrals represent the shaded areas in Figure 7. (a) This integral represents the area of a triangle with base and height equal to b. Therefore,

L

b

1 12 x dx = — (b)(b) = -72 2

(b) We evaluate by taking a limit of Riemann sums. Because f (x) = x2 is continuous, it is also integrable by Theorem 1. It follows that the right-endpoint approximations RN b

converge to the integral. Therefore, we compute

x2 dx by computing lim RN . N-->oo We divide the interval [0, bl into N subintervals of width Ax = k T,(1 = t. f0

xl\±, J- 1 RN = AXE f(Xj)= b N I i=1 i=1 b3

= FIGURE 7

N3

2 i

b3 (N 3

= N3

b N AT)

N2

3 ± 2 + b

Taking the limit of RN as N --> oo, we obtain f0

( lb j=1 b3 b3 b3 = 3 + 2N ± 6N2

b3 x2 dx = — . 3

•

The results of the previous example will be helpful in other definite integral computations in this section Summarizing, if b > 0, then b

b2 x dx = — 2

b

and

b3 x2 dx = — 3 Jo

The integrals in Eq. (5) suggest a pattern. What do you think the value of

5 b

xn dx is? 0 The n = 3 case is considered in Exercise 50. We will compute this integral for general n in Section 5.4. In cases where we do not have an exact representation of a function, such as in the following example, the best we can do to compute a definite integral is to approximate it using a Riemann sum. 1

EXAMPLE 4 A Grid-Connected Energy System At their home, Malina and Hors have solar panels and an energy system that is grid connected. They use energy from their panels and from the Apollo Power Company (APC) for their household needs. When the

256

CHAPTER

5

INTEGRATION panels are supplying more energy than they need, the excess is fed into the grid and they receive energy credit from APC. We can view their daily APC energy use E (in kilowatts) as a function of time t (in hours). When E 0 over [a, b], then (as in Figure 13) the integral of f lies between the areas of two rectangles, one of height M enclosing the region associated with the integral, and one of height m enclosed in the region. 3 f 2 1 EXAMPLE 9 Prove the inequalities - < - dx < 3. 4 1/2 X Solution Because f (x) = x-1 is decreasing (Figure 14), its minimum value on [ , , 2] is m = f(2) = I2 and its maximum value is M = f( 2 ) = 2. By Eq. (8),

y = x-1

3

-

1 (

2-

2

1 )

f

—

1/2 x m(b—a)

dx

2 (2 - 1 2-) = 3 M(b—a)

2

fb

CONCEPTUAL INSIGHT FIGURE 14

•

Keep in mind that a definite integral

f (x)dx is defined as a

limit of Riemann sums over finer and finer partitions of [a, hi, and each Riemann sum R(f,, P, C) is a sum of terms f (c)Axi determined by a partition P. The situation that we saw with the derivative in Chapters 3 and 4 is being repeated in our development of the theory of the definite integral: • Definition: We define the definite integral via a limit definition Limits are needed in the definition in order to properly capture the concept. • Properties: We establish properties of the definite integral by proving theorems based on the limit definition. Some properties provide insight into the workings of the definite integral, and some provide computational tools.

260

CHAPTER 5

INTEGRATION

• Computation: Computing the definite integral directly from the limit definition is messy at best and generally very difficult. We establish rules that aid us significantly in carrying out definite integral computations. The most important is the first part of the Fundamental Theorem of Calculus that we introduce in Section 5.4. • Application: While most of the definite integral computations that we do are simplified by computation rules, we cannot lose sight of its definition. Knowing that a definite integral is a limit of sums defined over finer and finer partitions of the domain will help us identify when to use this tool in applications. We will see plenty of instances in the sections and chapters ahead.

5.2 SUMMARY • A Riemann sum R(f,, P, C) for the interval [a, b] is defined by choosing a partition P : a = xo < xi < x2 < • • • < xN = b and sample points C =

where ci E [Xi-1, X/ ]. Let Axi = xi — xi_i. Then R(f,, P ,C) =

E f (ci)Axi

• The maximum of the widths Axi is called the norm II P II of the partition. • The definite integral is the limit of the Riemann sums (if it exists):

f

f (x)dx = lim R(f , P ,C) 111'11—A

We say that f is integrable over [a, b] if the limit exists. • Theorem: If f is continuous on [a, b], then f is integrable over [a, b]. • The signed area of the region between the graph of f and the x-axis over [a, b] is b

f (x)dx b

• Using geometry: When the geometry of the corresponding region is simple, f

f (x) dx a

can be computed using geometric formulas to determine the signed areas involved. 1 b 1 ,,a . • I x dx = — b2 and f 3 x 2 ax = — b 2 3 • Properties of definite integrals:

IL fa

(f (x) + g(x)) dx = f

f (x)dx + f a

Cf (x)dx =C f

g(x)dx a

f (x)dx

for any constant C

a

fb a

f (x) dx = — f

F

fa

f (x)dx a

f (x)dx = 0

f (x)dx +f

f (x)dx = f

f (x)dx a

for all a, b, c

S EC TION 5.2

The Definite Integral

261

• Comparison Theorem: If f (x) < g(x) on [a, b], then

IL

f (x) dx < f

g(x)dx L1

• If m < f (x) < M on [a, b], then m(b — a)

f

f (x) dx

M(b — a)

a

5.2 EXERCISES Preliminary Questions 1.

5

What is f

(c) If f (x) 0, then 1 is the area between the graph and the x-axis over [2,7].

3.

Explain graphically: f

4.

Which is negative, f

cos

x dx = 0.

—5

— 8 dx or f

-1

8 dx? -5

Exercises ---

In Exercises 1-10, draw a graph of the signed area represented by the integral and compute it using geometry. 2.

2x dx

6

if(x)Idx

(2x + 4) dx

f

L3

3.

(b)f

f(x)dx

3

3

1.

4

14. Evaluate: (a) f

-2

4 dx

4.

j_2(3x + 4) dx

-2 37r/2

f 68

5.

sin x dx

6.

(7 — x)dx

7r/2 3

7.

8.

f o5 V25

dx

f

FIGURE 15 The two parts of the graph are semicircles.

-2

9.

f

10. f

(2 —Ix I) dx

5

(3

x — 21x1)dx

In Exercises 15 and 16, refer to Figure 16. 15. Evaluate

10 11. Calculate f

5

3

-2

Jo

g(t)dt and f

g(t)dt. 3

(8 — x)dx in two ways:

a

g(t)dt and f c g(t)dt are as large as

16. Find a, b, and c such that f (a) As the limit lim RN N—>oo

possible.

(b) By sketching the relevant signed area and using geometry y = g(t)

4

12. Calculate f

(4x — 8) dx in two ways:

(a) As the limit lim RN N—>oo (b) By using geometry

2

3

In Exercises 13 and 14, refer to Figure 15. 6

2

13. Evaluate: (a) f

f(x)dx

f(x)dx

(b) 0

FIGURE 16

4

5

INTEGRATION

CHAPTER 5

262

17. Describe the partition P and the set of sample points C for the Riemann sum shown in Figure 17. Compute the value of the Riemann sum.

In Exercises 45-48, calculate the integral, assuming that .105

f (x)dx = 5,

f o5 g(x)dx = 12

34.25

5

45.

.105(f

(x)+ g(x))dx

(2f (x) - -3 1- g(x)) dx

46. f 0

20 15

lo 47.

5 48. I (f (x) - x)dx 0

g(x)dx

49. Assume a 0, then f

b4

x3 dx = - . 4

51. Using the result of Exercise 50, prove that for all b (negative, zero, and positive),

C = {0.5, 2, 3, 4.5}

f ob

24. In Example 4, approximate the net APC energy use from noon to midnight. In Exercises 25-30, sketch the signed area represented by the integral. Indicate the regions of positive and negative area. 7r/4

25. f

(4x - x2)dx

26.

tan x dx

0

-7r/4

27r

27. f

sin x dx

28.

sin x dx 0

29.f

6

0

30. f

(t2 - 1)02 -4) dt -2

In Exercises 31-34, determine the sign of the integral without calculating it. Draw a graph if necessary.

31.f-

52. f 54.f 56.f

j :327

G1.1'

x sinx dx

1

34. (GU) f o

36. f

27

sin "y x dx

(4x + 7)dx

1

o

f

40. f

(9x - 4x2)dx -3

1

4

60. f

61. f

7

f(x)dx + f 0

f(x)dx 3

62. f

9

f(x)dx

f(x)dx

-f

9

63. f

5

f(x)dx - f 2

f(x)dx 2

a2

(x2

x)dx

44. f

x2 dx a

f(x)dx

1

f(x)dx 2

2

42. i

-3

-a

58. f

f(x)dx

4

f (x) dx = 7 1

2

f(x)dx

o

3

(7t2 + t ± 1)dt

13

9

(12y2 ± 6y)dy

f

In Exercises 61-64, express each integral as a single integral.

2

1

43.f

x2 dx

4

f (x)dx = 4,

1

59. f

0

41. f

f

4

57. f

1/2

(u2 - 2u)du

2

f (x) dx = 1,

3

5

38.

1

39. I

(2x3 - 3x2)dx

-3

9

37. f x2 dx JO

24x2 - 8x)dx

In Exercises 57-60, calculate the integral, assuming that

2

35. f (6t - 3)dt Jo

(x - x3)dx

1 55. j . (12x3

(2x3 - x ± 4) dx

fo

In Exercises 35-44, use properties of the integral and the formulas in the summa ry to calculate the integrals. 4

3

-2

2

33.

53.f

x3 dx

0

32. fl x 3 dx

x4 dx

3

2

2

(112- 4x1 -4) dx

9

In Exercises 52-56, evaluate the integral using the formulas in the summary and Eq. (9).

1

37r

b4

x3 d x = 4

23. In Example 4, approximate the net APC energy use from midnight to noon.

5

when x < 0 when x > 0

64.f

7

3

9

f (x)dx + f

f(x)dx 3

SECTION 5.3

In Exercises 65-66, prove the relationship for arbitrary a and b using the formulas in the summary. 65.

263

75. Use the Comparison Theorem to show that 1

Jo

b3 - a3 66. f a x2 dx = 3

b2 - a2 x dx = 2 a

The Indefinite Integral

1 x5 dx < I. x4 dx, Jo

2

2 X4 dx

_< f

x5 dx

6

- dx

67. E4 Explain the difference in graphical interpretation between

76. Prove that 1
0. a

Further Insights and Challenges 83. Explain graphically: If f is an odd function, then f a f (x)dx = 0. -a 1 84. Compute f (sin x)(sin2 x + 1) dx. 85. Let k and b be positive. Show, by comparing the right-endpoint approximations, that 1 Xk dx = bk+1 j x' dx Jo

86. E., 1 Suppose that f and g are continuous functions such that, for all a, a a f (x)dx = f g(x)dx -a Give an intuitive argument showing that f(0) = g(0). Explain your idea with a graph. 87. Theorem 4 remains true without the assumption a < b < c. Verify this for the cases b oo 6N

E sin j=1

15.

3

1

dt •N/4t + 12

,

u = 4t + 12

(x2 + 1) dx (x3 + 3x)4

42. f sec2(20)tan(20)c10,

E (10 + -3Nk)

43. I (20x4 - 9x3 - 2x) dx

(k > 0)

sin(0 - 8) dB

44.

2 (12x3 - 3x2)dx

0

18. f x9/4 dx 20. f cos(5 - 70)d0

46.

x2

49.f 51.f

3 dx

3

-3 4

1x2 - 4Idx

L2ti dt

(x7/3 - 2x1/4)dx

0

x5 + 3x4 47.

In Exercises 17-30, calculate the indefinite integral.

19.

u = tan(20)

1 45. f (2x2 - 3x)2 dx

1 k + 2k ± • • • ± Nk 16. lim N->oo Nk+1

u = cos x

In Exercises 43-70, evaluate the integral.

E J4 + 5jIN

17. f (4x3 - 2x2)dx

u = x3 + 3x

sin x cos4 x dx,

0

6N

lim = N-*Do N k0 = 5 lim = N->oo N

fo 40.f 41.f

39.

7r/6

+

N-1 14.

In Exercises 39-42, use the given substitution to evaluate the integral.

48.

1

r-4 dr

4 50. I 1(x - 1)(x - 3)Idx -2 2 52. I t - Ltj dt 0

300

INTEGRATION

CHAPTER 5

54.f

53. f(10t_7)1 dt

55.

(2x3 J (3x4

2

57. o5 15x../x + 4 dx 1

59.

7 - (t 3

58. f t2 N/t

Jo

f7/27 sin ( 59 6

FIGURE 5 ) dB

1)dt

62. f sin2 (30) cos(30) dB

75. For the function f illustrated in Figure 6 do the following:

sin cos20

64. f sin ON/4 — cos 0 de

66.f 68.f f_4 (x2+2)3 sec2 t dt

d0

(tan t — 1)2

67. f yi2y ± 3 dy

69.

8 dt

60.

63. f csc2(9 — 20)de

65. fJo

r (s)

2)) dt

61. f t2 sec2(9t3

g/3

y — 5 dy

—1 x dx 56. I' 3 (x2, + 5)2

3x) dx 9x2)5

f Jo cos(

3

1

A(x)

A(x

Ax) — A(x — Ax) 2Ax

(c) Plot the values of A.'(x) on a graph of f to demonstrate Af

8dt

f.

—2 12x dx

70

sec2(cos 0) sin 0d0

8 t2,,/t

(a) For x = 0, 1, 2, . . . , 10, approximate A(x) = I f (t)dt . 0 (b) For x = 1, 2, 3, .. . , 9, approximate Af(x) using Ax = 1 and the symmetric difference quotient approximation,

71. Combine to write as a single integral: 8

Jo

0 f (x)dx

f

—2

6 f (x)dx

f

8

f (x) dx

x 72. Let A(x) = I f (x)dx, where f is the function shown in 0 Figure 4. Identify the location of the local minima, the local maxima, and points of inflection of A on the interval [0, E], as well as the intervals where A is increasing, decreasing, concave up, or concave down. Where does the absolute maximum of A occur?

FIGURE 6

76. The sine integral function Si is an area function defined by Si(x) = f

0

x sin t —di' t

(a) Explain why Si has critical points at n7r for all nonzero integers n. (b) (CAS) Use Riemann sums to approximate Si(x) for x = 7r, 27, , 87r and sketch a graph of Si(x) for 0 < x < 87.

FIGURE 4

73. Find the local minima, the local maxima, and the inflection points x t dt of A(x) = / 3 0-2 + 1)2 ' 74. A particle starts at the origin at time t = 0 and moves with velocity v(t) as shown in Figure 5. (a) How many times does the particle return to the origin in the first 12 seconds? (b) What is the particle's maximum distance from the origin? (c) What is the particle's maximum distance to the left of the origin?

77. On a typical day, a city consumes water at the rate of r(t) = 100 ± 72t — 3t2 (in thousands of gallons per hour), where t is the number of hours past midnight. What is the daily water consumption? How much water is consumed between 6 PM and midnight? 78. The learning curve in a certain bicycle factory is L(x) = 12x-1/5 (in hours per bicycle), which means that it takes a bike mechanic L(n) hours to assemble the nth bicycle. If a mechanic has produced 24 bicycles, how long does it take her or him to produce a subsequent batch of 12? 79. Cost engineers at NASA have the task of projecting the cost P of major space projects. It has been found that the cost C of developing a projection increases with P at the rate dC1dP 21P-1165, where C is in thousands of dollars and P in millions of dollars. What is the cost of developing a projection for a project whose cost turns out to be P = $35 million?

Chapter Review Exercises

301

80. An astronomer estimates that in a certain constellation, the number of stars of magnitude m, per degree-squared of sky, is equal to A(m) = 2.4 x 10-6m74 (fainter stars have higher magnitudes). Estimate the total number of stars of magnitude between 6 and 15 in a 1-degree-squared region of sky. 8

81. Evaluate

x15 dx

f-8 3 + cos' x

, using the properties of odd functions.

1 82. Evaluate I f(x)dx, assuming that f is an even continuous 0 function such that

FIGURE 7

1

f(x)dx =5,

J —2 f(x)dx = 8

In Exercises 88-93, find the derivative.

83. (GU) Plot the graph of f(x) = sin mx sin nx on [0,7r] for the pairs (m, n) = (2,4), (3,5) and in each case guess the value of ir I=f f(x)dx. Experiment with a few more values (including two . cases with m = n) and formulate a conjecture for when I is zero.

88. A'(x), where A(x) = f

84. Show that

90. — d f Y 3x dx dY —2

f

sin(t3)dt 3

89. Al (n.), where A(x) =

x cost dt f2 1 t

si n x

x f(x)dx = x F(x) — G(x)

t3 dt

91. 0x), where G(x) = f-2

where F'(x) = f(x) and 0x) = F(x). Use this to evaluate

f

x3 92. 02), where G(x) = f

x cosx dx.

85. Prove

2 f

2 2x dx

4

1

and

1

f 2

3—x dx < < 9 — 1

1

86. (GU) Plot the graph of f(x) = x -2 sin x, and show that

0.2 f

1

2

f (x)dx

93. I-01), where H(x) = f

9 1

— dt

4x 2

t

E/J

Explain with a graph: If f is increasing and concave up on 94. [a, b], then LN is more accurate than RN . Which is more accurate if f is increasing and concave down?

0.9.

1 f(x)dx, for y = f(x) in 87. Find upper and lower bounds for 0 Figure 7.

N/t ± 1 dt 0

95. r 7.

L

Explain with a graph: If f is linear on [a, b], then the 1

f(x)dx = —2 (RN

LN) for all N.

6 APPLICATIONS OF THE INTEGRAL I n the previous chapter, we used the integral to compute areas under curves and net change. In this chapter, we discuss some of the other quantities that are represented by integrals, including volume, average value, work, total mass, population, and fluid flow.

6.1 Area Between Two Curves

8

Sometimes we are interested in the area between two curves. Figure 1 shows projected electric power generation in the United States through renewable resources (wind, solar, biofuels, etc.) under two scenarios: with and without government stimulus spending. The area of the shaded region between the two graphs represents the additional energy projected to result from stimulus spending. How can we compute such an area? The scans of the cranium show slices through the brain. Each slice reveals information about the structure and health of the brain. The images are combined to show the functioning of the entire brain. In much the same way, a definite integral takes slice-by-slice information about a function and sums it to analyze the function over the entire domain. In this chapter, we will see how this accumulating property of the definite integral may be applied in many different settings.

U. S. Renewable Generating Capacity Forecast Through 2030 Gigawatts (GW) 160 —

160 gigawatts with stimulus spending 133 gigawatts without stimulus spending

120 100 — 80

2006 2010

2015

2020

2025

2030

Year

FIGURE 1 The area of the shaded region (which has units of power x time, or energy) represents the additional energy from renewable generating capacity projected to result from government stimulus spending in 2009-2010. Source: Energy Information Agency.

Now suppose that we are given two functions y = f (x) and y = g(x) such that f (x) > g(x) for all x in an interval [a, b] (Figure 2). We call such a region vertically simple since any vertical line that intersects the region does so in a single point or a single vertical line segment with its lower endpoint on the graph of y = g(x) and upper endpoint on the graph of y = f (x). Then the graph of y = f (x) lies above the graph of y = g(x), and the area between the graphs is equal to the area under the top function minus the area under the bottom function (Figure 3):

Vertical lines

Y = fix) y = g(x)

b

b

area between the graphs =

a

.a a

g(x)dx

f (x)dx —

f b

=1

FIGURE 2 A vertically simple region.

(f

(x) — g(x)) dx

1

a

Figure 3 illustrates this formula in the case that both graphs lie above the x-axis. We see that the region between the graphs is obtained by removing the region under y =- g(x) from the region under y = f (x). Y =f(x)

Y =f(x)

y = g(x)

y

x a

b

g(x)

a

Region between the graphs FIGURE 3 The area between the graphs is a difference of two areas. 303

304

CHAPTER 6

APPLICATIONS OF THE INTEGRAL

EXAMPLE 1 Find the area of the region between the graphs of the functions as shown in Figure 4: 1 g(x), even if f(x) and g(x) are not assumed to be positive. As in Figure 5, we can simply shift the two functions up by adding to each a constant C big enough so that both functions are positive over the interval [a, b]. This does not change the area between them. Then by our previous result, we have area between the graphs = f

((f(x) + C) — (g(x) + C)) dx a

=f

FIGURE 5 Shifting functions up to become positive-valued.

(f(x) — g(x)) dx

a

Writing ytop = f(x) for the top curve and ybot = g(x) for the bottom curve, we obtain

area between the graphs = f a

(Ytop — Ybot)dx = f

(f(x) — g(x)) dx

2

a

EXAMPLE 2 Find the area between the graphs of f(x) = x2 — 5x — 7 and — 12 over [-2,5].

g(x) = x

Solution First, we must determine which graph lies on top. Step 1. Sketch the region (especially, find any points of intersection). We know that y = f(x) is a parabola with y-intercept —7 and that y = g(x) is a line with y-intercept —12 (Figure 6). To determine where the graphs intersect, we look for values of x where f(x) = g(x), or equivalently, where f(x) — g(x) = 0: f (x) — g(x) = (x2 — 5x — 7) — (x — 12) = x2 — 6x + 5 = (x — 1)(x — 5) FIGURE 6

The graphs intersect where (x — 1)(x — 5) = 0, that is, at x = 1 and x = 5. Step 2. Set up the integrals and evaluate. We see that f(x) — g(x) > 0 for —2 < x < 1, and f(x) — g(x) < 0 for 1 < x < 5. Thus, In Example 2, we found the intersection points of y = f(x) and y = g(x) algebraically For more complicated functions, it may be necessary to use a computer algebra system.

f(x)

g(x) on [-2,1]

and

g(x)

f(x) on [1,5]

This tells us to subdivide our region into separate vertically simple regions, one over [-2, 1] and the other over [1, 5]. Therefore, we write the area as a sum of integrals over the two intervals:

SECTION 6.1 5 (Ytop f-2

1 Ybot)dx = f

f

—2

=-f

(g(x)— f(x))dx

1

((x2 — 5x —7) — (x — 12)) dx

-2

f =f

5 ((x — 12) — (x2 — 5x — 7)) dx

1

5

(x2 — 6x -I- 5) dx

1 = (- x3 — 3x2 3 (7

305

5 (f (x) — g(x))dx

-2

Area Between Two Curves

f

1

(—x2 + 6x — 5) dx 1

5x)

3

—2

(-74))

(25 3

3

5 x3 + 3x2 — 5x)

(-7)) 3

113

•

= 3

EXAMPLE 3 Calculating Area by Dividing the Region bounded by the graphs of y = 8/x2, y = 8x, and y = x.

Find the area of the region

Solution Step 1. Sketch the region (especially, find any points of intersection). The curve y = 8/x2 cuts off a region in the sector between the two lines y = 8x and y = x (Figure 7). We find the intersection of y = 8/x2 and y = 8x by solving 8 —=8x

x3 =1

=

=

x=1

and the intersection of y = 8/x2 and y = x by solving 8 —=x

X 3 =Q

=

=

x=2

Step 2. Set up the integrals and evaluate. Figure 7 shows that ybot = x, but ytop changes at x = 1 from ytop = 8x to ytop = 8/x2. So, the region is not vertically simple. Therefore, we break up the regions into two parts, A and B, each vertically simple, and compute their areas separately. area of A = f o (Ytop

Ybot) dx = f

2 area of B — f (Ytop — ybot) dx = — 1

0

(8x —x) dx = f

[2(8

2 --- x

The total area bounded by the curves is the sum

FIGURE 7 Area bounded by y = 8/x2, y = 8x, and y = x as a sum of two areas.

1 7 7 7x dx -= — x2 = — 2 2 0 o

dx = (_ )

+ ; = 6.

2 5 _ 1 x2 = — x 2 ) 1 2 •

CHAPTER 6

APPLICATIONS OF THE INTEGRAL

Integration Along the y-Axis Suppose we are given x as a function of y, say, x = g(y). What is the meaning of the d

integral

g(y)dy? This integral can be interpreted as signed area, where regions to fc

the right of the y-axis have positive area and regions to the left have negative area: g(y)dy = signed area between graph and y-axis for c < y < d In Figure 8(A), the part of the shaded region to the left of the y-axis has a negative signed area. The signed area of the entire region is 6

(y2

—

9) dy

f-6

6

1

= 36

= (—y3 — 9Y) 1

—6

Area to the right of y-axis minus area to the left of y-axis

Xleft =

h(Y)

Xright = g(y)

C

(A) Region between x = y2 — 9 and the y-axis

x = g(y)

Horizontal lines

FIGURE 9

(B) Region between x = h(y) and x = g(y)

FIGURE 8

More generally, if g(y) > h(y) as in Figure 8(B), then the graph of x = g(y) lies to the right of the graph of x = h(y). In this case, we write xright = g(y) and xieft = h(y). We call the region horizontally simple, since every horizontal line that intersects the region in more than a single point does so in a single line segment such that the left endpoint is on the curve x = h(y) and the right endpoint is on the curve x = g(y), as in Figure 9. The formula for the area corresponding to Eq. (2) is

A horizontally simple region. area between the graphs =

(xright — xieft)dY = f

(g(y) — h(y)) dy

EXAMPLE 4 Calculate the area enclosed by the graphs of h(y) = y2 — 1 and y2 i y4 ± 1. =

g(y)

Solution Figure 10 shows that the enclosed region stretches between the two points of intersection of the graphs and that g(y) > h(y) over the region. Therefore, the region is horizontally simple with Xright = g(y) and xieft = h(y). To set up the integral for the area, we need to determine the points of intersection. We do that by solving g(y) = h(y) for y: FIGURE 10

y2 _ _1 y4 ± 8

= y2 _

_1 y4 _ 2 = 0 8

y = 12

SECTION

6.1

Area Between Two Curves

307

Now, we have 1 xright - xieft = (y2 - -1 Y4 ± 1 - (y2 - 1) = 2 - - y8 8 It would be more difficult to calculate the area of the region in Figure 10 as an integral with respect to x because the curves are not graphs of functions of x.

The enclosed area is 2

2

f-2

—t (xngh

xieft)dy

=

-2 16

(

2

2

- - y4) dy = (2y 8 40 16"

-2

32

T CT) - T

•

For many regions, we have a choice of whether to find the area by integrating with respect to x or with respect to y. The decision is usually based on two factors: • How easy it is to obtain the curves as functions of one variable in terms of the other • How easy it is to subdivide the region into simple regions and to integrate the functions involved In the next example, we carry out the area computation, both integrating with respect to x and integrating with respect to y, demonstrating how these factors are involved. EXAMPLE 5 Find the area of the region that is bounded by the three curves y = x2, y = (x - 2)2, and y = O. x=2 —

x=

x=2+-5

y=

y = (x - 2)2

FIGURE 11 This region is horizontally simple, but it is easier to cut it into two vertically simple regions.

Solution The area appears in Figure 11. Notice immediately that it is not vertically simple, since the top function changes over the interval [0,21. It is horizontally simple, but to calculate the area using the fact it is a horizontally simple region takes some work to set up the expressions for xright and XleftWe first compute the area of the region by splitting it into two vertically simple regions. In this case, the area is given by 2

2

2 1 (x - 2)3 1 X3 1 = - + 0 -( - - ) = (x - 2)2 dx = — 3 3 3 Ji 3 0 3 1 Now, let's redo the problem using the fact the region is horizontally simple. To determine xright and xieft, we must invert the formulas for each of the parabolas. The left boundary of the region is the right side of the parabola given by y = x2. Solving for x, we have x = ±,g. The right side of the parabola corresponds to the positive choice; therefore, xieft = . .The right boundary of the region is the left side of the parabola y = (x - 2)2. We solve for x: x - 2 = ±.07

10

x2 dx

x =2± The left side of the parabola is obtained by choosing the minus sign, and therefore Xright = 2 Then the area is given by 1 2 =• dy = f (2 - 2y1/2) dy = (2y - y 3 / 2 ((2 ) o 3 3 0

1.

-

)-

)

6.1 SUMMARY • If f (x) > g(x) on [a, b], then the region between the graphs is vertically simple and we have area between the graphs -TT f a (ytop - ybot) dx - f (f (x) - g(x)) dx - a

CHAPTER 6

308

APPLICATIONS OF THE INTEGRAL

• To calculate the area between y = f (x) and y = g(x), sketch the region to find ytop. If necessary, find points of intersection by solving f (x) = g(x). d

• Integral along the y-axis: f

g(y) dy is equal to the signed area between the graph e and the y-axis for c < y < d. The signed area to the right of the y-axis is positive and the signed area to the left is negative. • If g(y) > h(y) on [c, d], then x = g(y) lies to the right of x = h(y) and the region is horizontally simple: area between the graphs = f

(xright — xieft) dy = f

(g(y) — h(y)) dy

6.1 EXERCISES Preliminary Questions 1. What is the area interpretation

b

of

f

(f(x) — g(x)) dx

if

4.

f (y)dy positive or negative?

f (x) 2.

g(x)? b Is f (1(x) — g(x)) dx equal to the area between the graphs of f a

and g if f (x) > 0 but g(x)

y = 3x2 + 12 and y = 4x + 4

over [-3,3] (Figure 12).

FIGURE 12

2. Find the area of the region between the graphs of 1(x) = 3x + 8 and g(x) = x2 + 2x + 2 over [0,2] (Figure 13).

FIGURE 13

11(x) — g(x)I dx represents.

Explain what i a

3. Suppose that f (x) g(x) on [0,31 and g(x) f (x) on [3,5]. Express the area between the graphs over [0, 5] as a sum of integrals.

Exercises 1. Find the area of the region between

b

5.

< 0?

Suppose that the graph of x = f (y) lies to the left of the y-axis. Is

6. Draw a region that is both vertically simple and horizontally simple.

3. Find the area of the region enclosed by the graphs of f (x) = x2 +2 and g(x) = 2x + 5 (Figure 14).

FIGURE 14 4. Find the area of the region enclosed by the graphs of f (x) = x3 — 10x and g(x) = 6x (Figure 15).

FIGURE 15

SECTION 6.1

In Exercises 5 and 6, sketch the region between y = sin x and y = cos x over the interval andfind its area. 5.

L4

6.

2J

Area Between Two Curves

309

19.

[0,7]

In Exercises 7 and 8, let f (x) = 20 + x — x2 and g(x) = x2 — 5x. 7. Sketch the region enclosed by the graphs of f and g, and compute its area. FIGURE 18 8. Sketch the region between the graphs of f and g over [4, 8], and compute its area as a sum of two integrals. 20. 9. (GU) Find the points of intersection of the graph of y = x(x2 — 1) and the graph of y = 1 — x2. Sketch the region enclosed by these curves over [—I, 1] and compute its area. 10. (GU Find the points of intersection of the graph of y = x(4 — x) and the graph of y = x2(4 — x). Sketch the region enclosed by these curves over [0, 4] and compute its area. 11. Sketch the region bounded by the line y = 2 and the graph of y = sec2 x for — 0 on [0, 11 and f(x) 0 for a < x < b. The solid obtained by rotating the region under the graph about the x-axis has a special feature: All vertical cross sections are circles (Figure 2). In fact, the vertical cross section at location x is a circle of radius R = f (x) and thus, area of the vertical cross section = 7R2

Th.f(x)2

We know from Section 6.2 that the total volume V is equal to the integral of crosssectional area. Therefore, V = n-

R2 dx =

n-f (x)2 dx . a

a

>x)

a

x

b

(A)

(B) Cross section is a circle of radius f(x).

(C) Solid of revolution

FIGURE 2

Volume of Revolution: Disk Method If f is continuous and f (x) > 0 on [a, b], then the solid obtained by rotating the region under the graph about the x-axis has volume R2 dx

f

1

Jr f a f (x)2 dx

a

EXAMPLE 1 Calculate the volume V of the solid obtained by rotating the region under y = x2 about the x-axis for 0 0 as in Figure 5(A). When this region is rotated about the x-axis, segment AB sweeps out the washer shown in Figure 5(B). The inner and outer radii of this washer (also called an annulus; see Figure 4) are Router —

FIGURE 3 Region under y = x2 rotated

Rinner = g(x)

The washer has area n Ro2ut„ —JtRner or 1r (f (x )2 — g(x)2), and the volume of the solid of revolution [Figure 5(C)] is the integral of this cross-sectional area:

about the x-axis. b V =

a

R outer — Ri2nner) dx

=f

(f(X)2

g(x)2) dx

324

CHAPTER 6

Area = n-(Ralter

APPLICATIONS OF THE INTEGRAL

nner)

f(x)

a

VIN

FIGURE 4 The region between two

Washer

concentric circles is called an annulus, or more informally, a washer.

(A)

(C)

(B)

FIGURE 5 AB generates a washer when rotated about the x-axis.

CAUTION When using the washer method, make sure you use (f (x)2 — g(x)2) in the integrand, not (f (x)— g(x))2.

EXAMPLE 2 Region Between Two Curves Find the volume V obtained by revolving the region between y = x2 + 4 and y -= 2 about the x-axis for 1 < x < 3. Solution The graph of y = x2 + 4 lies above the graph of y = 2 (Figure 6). Therefore, Router = x2 +4 and Rinner = 2. By Eq. (2), 3

f3

(Ro2uter — Ri2nner) dx =

V =

((x2 ± 4)2

22) dx

J1 3

=

f i

(x 4

8.X 2 ±

12)

dx = n-

± i x- ± 12x

2126 Jr 15

•

FIGURE 6 The area between y = x2 + 4 and y = 2 over [1,31 rotated about the x-axis.

In the next example, we calculate a volume of revolution about a horizontal axis parallel to the x-axis. EXAMPLE 3 Revolving About a Horizontal Axis Find the volume V of the "wedding band" [Figure 7(C)] obtained by rotating the region between the graphs of f (x) = x2 + 2 and g(x) = 4 — x2 about the horizontal line y = —3. When you set up the integral for a volume of revolution, visualize the cross sections. These cross sections are washers (or disks) whose inner and outer radii depend on the axis of rotation.

Solution First, let's find the points of intersection of the two graphs by solving

f (x) = g(x) =

x2 + 2 =

4 — x2

Figure 7(A) shows that g(x) > f (x) for —1 < x < 1.

X

2

= 1

=

X =

1

SECTION 6.3

Volumes of Revolution: Disks and Washers

325

f(x)= x2 + 2

Axis of rotation

(A)

(C)

(B) The inner and outer radii are 3 units longer.

FIGURE 7 If we wanted to revolve about the x-axis, we would use Eq. (2). Since we want to revolve around y = —3, we must determine how the radii are affected. Figure 7(B) shows that when we rotate about y = —3, AB generates a washer whose outer and inner radii are both 3 units longer, and therefore we have ' Router = g(x) (_3) = (4 _ x2) + 3 = 7 _ x2

f (x) - (-3) = (x2 + 2) + 3 = x2 + 5

• Ruiner

The volume of revolution (about y = —3) is equal to the integral of the area of this washer: We get Router by subtracting y = —3 from y = g(x) because vertical distance is the difference of the y-coordinates. Similarly, we subtract —3 from f (x) to get Rinser.

f (R.iuter — Ri2nner) dx v = J-1

=J =f 71,

_ x2)2

—1

= rr f

(x2 + 5)2) dx

1

—1

((49 — 14x2 ± x4) — (x4 + 10x2 ± 25)) dx (24 — 24x2) dx

rr (24x — 8x3)

—1

-= 327r

•

EXAMPLE 4 Find the volume obtained by rotating the graphs of f (x) = 9 — x2 and y = 12 for 0 0,

about y = 12

y = 10 — x2,

x > 0,

about x = —1

1

5 2

52. y2 = 4x,

about y-axis

about x-axis

45. y = 2,Fc,

49.

y = x,

329

about y-axis about y = 8

53. The bowl in Figure 14(A) is 21 cm high, obtained by rotating the curve in Figure 14(B) as indicated. Estimate the volume capacity of the bowl shown by taking the average of right- and left-endpoint approximations to the integral with N = 7. The inner radii (in centimeters) starting from the top are 0, 4, 7, 8, 10, 13, 14, 20.

FIGURE 12 26. Let R be the region enclosed by y = x2 + 2, y = (x — 2)2 and the axes x = 0 and y = 0. Compute the volume V obtained by rotating R about the x-axis. Hint: Express V as a sum of two integrals.

30 25 21 19 16 12 9

In Exercises 27-32, find the volume of the solid obtained by rotating region A in Figure 13 about the given axis. 27. x-axis

28. y = —2

29. y = 2

30. y-axis

31. x = —3

32. x = 2

21 cm

20 (B)

(A) FIGURE 14

54. The region between the graphs off and g over [0, 1] is revolved about the line y = —3. Use the midpoint approximation with values from the following table to estimate the volume V of the resulting solid: FIGURE 13 In Exercises 33-38, find the volume of the solid obtained by rotating region B in Figure 13 about the given axis. 33. x-axis

34. y = —2

35. y = 6

36. y-axis

Hint for Exercise 36: Express the volume as a sum of two integrals along the y-axis or use Exercise 30. 38. x = —3

37. x = 2

In Exercises 39-52, find the volume of the solid obtained by rotating the region enclosed by the graphs about the given axis. 39. y = x2,

y = 12 — x,

x = 0,

about y = —2

40. y = x2,

y = 12 — x,

x = 0,

about y = 15

x>0

x f (x) g(x)

0.3 7 3.5

0.1 8 2

0.5 6 4

0.7 7 3.5

0.9 8 2

In Exercises 55-56, you assist your grandfather Umberto who wants to know the volume of his wine barrels. Knowing that you are taking a calculus course, he thought that you might be able to help. So he measured the circumference around each barrel at regular intervals from the bottom to the top and provided the measurements to you. "Can you figure out from this how many gallons each holds?" he asked. 55. With the following barrel circumference measurements, estimate the volume of the barrel in gallons. Dist from Bottom (in.) Circumference (in.)

0 30

3 6 9 12 15 18 21 24 36 38 40 41 39 38 35 28

330

CHAPTER 6

APPLICATIONS OF THE INTEGRAL

56. With the following barrel circumference measurements, estimate the volume of the barrel in gallons. Dist from Bottom (in.) 0 4 8 12 16 20 24 28 32 36 62 70 75 79 83 83 77 74 68 62 Circumference (in.) 57. Find the volume of the cone obtained by rotating the region under the segment joining (0, h) and (r, 0) about the y-axis. 58. The torus (doughnut-shaped solid) in Figure 15 is obtained by rotating the circle (x — a)2 + y2 = b2 around the y-axis (assume that a > b). Show that it has volume 27r2ab2. Hint: After simplifying it, evaluate the integral by interpreting it as the area of a circle.

FIGURE 16 The hyperbola with equation y2 — x2 = 1. 61. A "bead" is formed by removing a cylinder of radius r from the center of a sphere of radius R (Figure 17). Find the volume of the bead with r = 1 and R = 2.

FIGURE 15 Torus obtained by rotating a circle about the y-axis.

59. (GU) Sketch the hypocycloid x2/3 + y2/3 = 1 and find the volume of the solid obtained by revolving it about the x-axis.

FIGURE 17 A bead is a sphere with a cylinder removed.

60. The solid generated by rotating the region between the branches of the hyperbola y2 — x2 = 1 about the x-axis is called a hyperboloid (Figure 16). Find the volume of the hyperboloid for —a < x < a.

Further Insights and Challenges 62. 1/1 Find the volume V of the bead (Figure 17) in terms of r and R. Then show that V = *h3, where h is the height of the bead. This formula has a surprising consequence: Since V can be expressed in terms of h alone, it follows that two beads of height 1 cm, one formed from a sphere the size of an orange and the other from a sphere the size of the earth, would have the same volume! Can you explain intuitively how this is possible? 63. The solid generated by rotating the region inside the ellipse with equation (i) 2 = 1 around the x-axis is called an ellipsoid. Show that the ellipsoid has volume rab2. What is the volume if the ellipse is rotated around the y-axis? 64. The curve y = f (x) in Figure 18, called a tractrix, has the following property: The tangent line at each point (x, y) on the curve has slope dy =

FIGURE 18 The tractrix. 65. Verify the formula t.x2

—y

-

X1XX

x2)

dx Let R be the shaded region under the graph of y = f (x) for 0 < x < a in Figure 18. Compute the volume V of the solid obtained by revolving R around the x-axis in terms of the constant c = f (a). Hint: Use the substitution u = f (x) to show that

V= f

1 UV1 -

u2 du

1

= — (-x — x2)3 6

3

Then prove that the solid obtained by rotating the shaded region in Figure 19 about the x-axis has volume V = '*BH2, with B and H as in the figure. Hint: Let xi and x2 be the roots of f (x) = ax + b — (mx + c)2, where xi no (Figure 3). This sum is also a right-endpoint approximation that converges b

xf (x) dx. Thus, we obtain Eq. (2) for the volume of the solid.

to 27 f .a

(B)

(A)

(C)

al FIGURE 3 Cylindrical shell approximations as N —> no.

Note: In the Shell Method, we integrate with respect to x when the region is rotated about the y-axis.

Volume of Revolution: The Shell Method The solid obtained by rotating the region under y = f (x) over the interval [a, b] about the y-axis has volume V = 2r f

(radius) (height of shell) dx = 2ir f

xf (x) dx

2

L

EXAMPLE 1 Find the volume V of the solid obtained by rotating the region under the graph of f (x) = 1 — 2x ± 3x2 — 2x3 over [0, 1] about the y-axis. Solution The solid is shown in Figure 4. By Eq. (2), V = 27r f

pl xf (x) dx = 27r f

= 2f

(x — 2x2 ( 1 x2 2

C3 FIGURE 4 The graph of f (x) = 1 — 2x + 3x2 — 2x3 rotated about the y-axis.

2 x3 3

x(1 — 2x ± 3x2 — 2x3)dx

3x3 — 2x4) dx 3 4 2 1 11 —x — — x5) = —7r 4 5 0 30

•

SECTION 6.4

Volumes of Revolution: Cylindrical Shells

333

CONCEPTUAL INSIGHT Shells Versus Disks and Washers • Shell Method: To calculate a volume, you must find the shell height, which is always parallel to the axis of rotation (Figure 5). • Disk and Washer Method: To calculate a volume, you must find the disk radius or washer radii, which are always perpendicular to the axis of rotation. Some volumes can be computed equally well using either the Shell Method or the Disk and Washer Method. In Example 1, however, the Shell Method is much easier because the shell height is f (x). Using the Disk Method would have been more challenging because we would need to find an expression for the radius of the disk perpendicular to the y-axis (Figure 5). This would require finding the inverse g(y) = f (y), and that could be difficult or impossible. In general: Use the Shell Method if finding the shell height is easier than finding the disk radius or washer radii. Use the Disk and Washer Method when finding the disk radius or washer radii is easier.

This segment generates a disk of radius R = g(y). This segment generates a shell of heightf(x). y = f(x) or x=g(y)

FIGURE 5 For rotation about the y-axis, the Shell Method uses y = f (x) but the Disk

When we rotate the region between the graphs of two functions f and g satisfying f (x) > g(x), the vertical segment at location x generates a cylindrical shell of radius x and height f (x) — g(x) (Figure 6). Therefore, the volume is

Method requires the inverse function x = g(y). V =

27r f

(radius) (height of shell) dx = 27r a

x (f (x) — g(x)) dx

3

a

FIGURE 6 The vertical segment at location x generates a shell of radius x and height f (x) — g(x).

EXAMPLE 2 Region Between Two Curves Find the volume V obtained by rotating the region enclosed by the graphs of f (x) = x(5 — x) and g(x) = 8 — x(5 — x) about the y-axis. Solution First, find the points of intersection by solving x(5 — x) = 8 — x(5 — x). We obtain 0 = x2 — 5x + 4 = (x — 1)(x —4), so the curves intersect at x = 1,4. Sketching the graphs (Figure 7), we see that f (x) > g(x) on the interval [1,4] and height of shell = f (x) — g(x) = x(5 — x) — (8— x(5 —x)) = 10x — 2x2 —8 4

4

V = 2r! (radius)(height of shell) dx = 27r f FIGURE 7 The reasoning in Example 3 shows that if we rotate the region under y = f(x) over [a, b] about the vertical line x = c, then the volume is

V= f V= f

(x — c)f(x) dx if c < a

a

a

(c — x)f (x) dx

ifc

b

1 10 = 27r (-3x3 — — x4 —

2

4

4x2)

1

=

x ( 10x — 2x — 8) dx = 457

(64 ▪

(—h))

•

EXAMPLE 3 Rotating About a Vertical Axis Use the Shell Method to calculate the volume V obtained by rotating the region under the graph of f (x) = .X -112 over 111, 411 about the axis x = —3. Solution If we were rotating this region about the y-axis (i.e., x = 0), we would use Eq. (3). To rotate it around the line x = —3, we must take into account that the radius of revolution is now 3 units longer.

334

CHAPTER 6

APPLICATIONS OF THE INTEGRAL

Figure 8 shows that the radius of the shell at x is now x — (-3) = x + 3. The height of the shell is still f(x) = x-1/ 2 , so 4 V = 27r f

(radius) (height of shell) dx

1

4

4

= 2.71- f

(x + 3)x-1/2 dx = 27r (-2 x3/2 ± 6,c1/2) 3 1 i

647C =

3

•

Axis x = —3

(11 FIGURE 8 Rotation about the axis x = —3.

—3

The method of cylindrical shells can be applied to rotations about horizontal axes, but in this case, the graph must be described in the form x = g(y). EXAMPLE 4 Rotating About the x-Axis Use the Shell Method to compute the volume V obtained by rotating the region in the first quadrant between y = 9 — x2 and the x-axis about the x-axis (Figure 9). Solution When we rotate about the x-axis, the cylindrical shells are generated by horizontal segments (AB in Figure 9) and the Shell Method gives us an integral with respect to y. The radius of the shell is y, the distance from the rotation axis to the segment. The length of AB is the height of the shell (we use the term "height" even though the shell is horizontal). The length of AB is given by the positive value of x on the parabola associated with y, that is, by x = 0 --77. The volume is then obtained as follows, where we use a substitution u = 9 — y, du = —dy, in the integral computation: 1=0 REMINDER After making the substitution u = 9 — y, the limits of integration must be changed. Since u(0)= 9 and u(9)= 0, we change fO 1 0 9 to

9 .

9 V = 27 f

0

9

(radius) (height of shell)dy = 27z f 0 9

-= 27T ( 6 /13/2 —

CI FIGURE 9 Shell generated by a horizontal segment in the region under the graph of y = 9— x2.

y19 — y dy

9

(9 — u).V71 du =- 27r f

= —27r f

0

(9u1/2 — u3/2)du 0

9 = 648

2 u5i2) 5 o

5

7

•

SECTION 6.4

Volumes of Revolution: Cylindrical Shells

335

6.4 SUMMARY • Shell Method: When you rotate the region between two graphs about an axis, the segments parallel to the axis generate cylindrical shells [Figure 10(A)]. The volume V of the solid of revolution is the integral of the surface areas of these shells: surface area of shell = 27r (radius) (height of shell) • Sketch the graphs to visualize the shells. • Figure 10(B): Region between y = f (x) (with f (x) > 0) and the y-axis, rotated about the y-axis: ph

V = 27r

(radius) (height of shell) dx = 27r f .a

xf (x)dx a

• Figure 10(C): Region between y = f (x) and y = g(x) (with f (x) rotated about the y-axis: V -= 27r f

(radius) (height of shell) dx = 27r f a

g(x) > 0),

x(f (x) — g(x))dx a

• Rotation about a vertical axis x = c. — Figure 10(D): c

a, radius of shell is (x — c): V = 27r f

(x — c)f (x)dx a

— Figure 10(E): c > a, radius of shell is (c — x): V = 27r f

(c — x)f (x)dx a

• Rotation about the x-axis using the Shell Method: Write the graph as x = V=

f

(radius)(height of shell) dy = 27r f

yg(y)dy

Surface area = 27rh (A)

Y = f(x)

Y = f(x)

Y x =c

Y = f(x)

y

y = g(x) a (B) FIGURE 10

x (C)

b

a

x (D)

b

x f(

a

x (E)

)

X

C

336

APPLICATIONS OF THE INTEGRAL

CHAPTER 6

6.4 EXERCISES Preliminary Questions 1. Consider the region under the graph of the constant function f (x)= h over the interval [0, r]. Give the height and the radius of the cylinder generated when the region is rotated about (b) The y-axis (a) The x-axis

(b) Does the Disk or Washer Method for computing V lead to an integral with respect to x or y? 3. If we rotate the region under the curve y = 8 between x = 2 and x = 3 about the x-axis, what answer should the Shell Method give us?

2. Let V be the volume of a solid of revolution about the y-axis. (a) Does the Shell Method for computing V lead to an integral with respect to x or y?

Exercises In Exercises 1-6, sketch the solid obtained by rotating the region underneath the graph of the function over the given interval about the y-axis, and find its volume. 1.

f (x) = x3,

3. 5.

2.

f (x) =

f (x) = x-1, [13]

4.

f (x) = 4 — x2,

[0,2]

f (x) =

6.

f (x) =

,

[0,1]

N/X 2 +

9,

[0,3]

y = 3x — 2,

8.

y = ,/7x,

9.

y = x2,

y = 6 — x,

N/1 +x3

[1,4]

y = 8 — x2, x = 0, y = 8 — 4x,

for x > 0

for x > 0

11. y = (x2 + 1)-2,

y = 2 — (x2 --I- 1)-2,

12. y = 1 — Ix — 1I,

y=0

13. y = 2 — x4,

y = x2,

14. y =

x = 0,

y = 0,

y=

x=2

for x > 0

x = 0,

y = sin(x2),

16. (GU) y = cos(x2),

x=4

y = x,

x>0 x=0

In Exercises 17-22, sketch the solid obtained by rotating the region underneath the graph off over the interval about the given axis, and calculate its volume using the Shell Method. 17. f (x) = x3 ,

[0,1],

18. f (x) = x3, [0,1],

about x = 2 about x = —2

19. f (x) = x-4, [-3,—i], 20. f (x) =

1

,

22. f (x) = 1 — x2, [-1,1], x = c

about x = —1 with c > 1

In Exercises 23-28, sketch the enclosed region and use the Shell Method to calculate the volume of rotation about the x-axis. 23. x = y, 24. x =

y = 0, + 1,

x=1

x = 3 — ly,

25. x = y(4 — y),

x=0

26. x = y(4 — y),

x = (y — 2)2

[0,2],

about x = 4 about x = 0

27. y = 4 — x2, x = 0, 28. y = X 1/3

In Exercises 15 and 16, use a graphing utility to find the points of intersection of the curves numerically and then compute the volume of rotation of the enclosed region about the y-axis. 15. (- 1

[0,a],

y=0

x =0

y = x2

10. y = 8 — x3,

with a >0,

[0,4]

In Exercises 7-14, use the Shell Method to compute the volume obtained by rotating the region enclosed by the graphs as indicated, about the y-axis. 7.

21. f (x) = a —x

—

2,

y = 0,

y=0 x = 27

29. Determine which of the following is the appropriate integrand needed to determine the volume of the solid obtained by rotating around the vertical axis given by x = —1 the area that is between the curves y = f (x) and y = g(x) over the interval [a, b], where a > 0 and f (x) > g(x) over that interval. (a) x(f (x) — g(x))

(b) (x + 1)(f (x) — g(x))

(c) x((f (x) — 1) — (g(x) — 1))

(d) (x —1)(f(x) — g(x))

(e) x(f (x + 1) — g(x

1))

30. Let y = f (x) be a decreasing function on [0, b], such that f (b)= 0. b S(0) Explain why 27r f xf (x)dx = n f (h(x))2 dx, where h denotes the Jo o inverse of f. 31. Use both the Shell and Disk Methods to calculate the volume obtained by rotating the region under the graph of f (x) = 8 — x3 for 0 < x < 2 about (a) The x-axis

(b) The y-axis

32. Sketch the solid of rotation about the y-axis for the region under the graph of the constant function f (x) = c (where c > 0) for 0 < x < r. (a) Find the volume without using integration. (b) Use the Shell Method to compute the volume.

SECTION 6.4

33. The graph in Figure 11(A) can be described by both y = f (x) and x = h(y), where h is the inverse of f. Let V be the volume obtained by rotating the region under the graph about the y-axis. (a) Describe the figures generated by rotating segments AB and CB about the y-axis. (b) Set up integrals that compute V by the Shell and Disk Methods.

Volumes of Revolution: Cylindrical Shells

43. y-axis

44. x = -3

45. x = 2

46. x-axis

47. y = -2

48. y = 8

337

In Exercises 49-56, use the most convenient method (Disk or Shell Method) to find the given volume of rotation. 49. Region between x = y(5 - y) and x = 0, rotated about the y-axis 50. Region between x = y(5 - y) and x = 0, rotated about the x-axis 51. Region bounded by y = x2 and x = y2, rotated about the y-axis

C' (B)

(A)

52. Region bounded by y = x2 and x = y2, rotated about x = 3 53. Region in Figure 13, rotated about the x-axis

FIGURE 11

54. Region in Figure 13, rotated about the y-axis 34. F74 Let W be the volume of the solid obtained by rotating the region under the graph in Figure 11(B) about the y-axis. (a) Describe the figures generated by rotating segments A'B' and A'C' about the y-axis. (b) Set up an integral that computes W by the Shell Method. (c) Explain the difficulty in computing W by the Washer Method. 35. Let R be the region under the graph of y = 9 - x2 for 0 < x < 2. Use the Shell Method to compute the volume of rotation of R about the x-axis as a sum of two integrals along the y-axis. Hint: The shells generated depend on whether y E [0,51 or y E [5,9]. 36. Let R be the region under the graph of y = 4x-1 for 1 < y < 4. Use the Shell Method to compute the volume of rotation of R about the y-axis as a sum of two integrals along the x-axis.

FIGURE 13

FIGURE 14

55. Region in Figure 14, rotated about x = 4 In Exercises 37-42, use the Shell Method to find the volume obtained by rotating region A in Figure 12 about the given axis.

56. Region in Figure 14, rotated about y = -2

37. y-axis

38. x = -3

39. x = 2

40. x-axis

In Exercises 57-60, use the Shell Method to find the given volume of rotation.

41. y -= -2

42. y = 6

57. A sphere of radius r 58. The "bead" formed by removing a cylinder of radius r from the center of a sphere of radius R (compare with Exercise 61 in Section 6.3) 59. The torus obtained by rotating the circle (x - a)2 + y2 = b2 about the y-axis, where a > b (compare with Exercise 58 in Section 6.3). Hint: Evaluate the integral by interpreting part of it as the area of a circle.

FIGURE 12 In Exercises 43-48, use the most convenient method (Disk or Shell Method) to find the volume obtained by rotating region B in Figure 12 about the given axis.

60. The "paraboloid" obtained by rotating the region between y = x2 and y = c (c > 0) about the y-axis 61. Given a and b, 0 < a 0, the region R becomes infinite but the volume V approaches 27r.

Y =f(x) FIGURE 15 The ellipse (--)2 'a!

= 1.

a

64. The bell-shaped curve y = f(x) in Figure 16 satisfies dyldx = —xy. Use the Shell Method and the substitution u = f(x) to show that the

FIGURE 16 The bell-shaped curve.

6.5 Work and Energy For those who want some proof that physicists are human, the proof is in the idiocy of all the different units which they use for measuring energy.

All physical tasks, from running up a hill to turning on a computer, require an expenditure of energy. When a force is applied to an object to move it, the energy expended is called work. When a constant force F is applied to move the object a distance d in the direction of the force, the work W is defined as "force times distance" (Figure 1):

—Richard Feynman, The Character of Physical Law

Force F

FIGURE 1 The work expended to move the object from A to B is W = F • d.

W= F•d

1

The International System (SI) unit of force is the newton (abbreviated N), defined as 1 kg-m/s2. Energy and work are both measured in units of the joule (J), equal to 1 N-m. In the British system, the unit of force is the pound, and both energy and work are measured in foot-pounds. Another unit of energy is the calorie. One ft-lb is approximately 1.356 J or 0.324 calories. To become familiar with the units, let's calculate the work W required to lift a 2-kg stone 3 m above the ground. Gravity acts on the stone of mass m with a force equal to —mg, where g = 9.8 m/s2. Therefore, lifting the stone requires an upward vertical force F = mg, and the work expended is W = (mg)h =- (2 kg)(9.8 rn/s2)(3 m) = 58.8 J F•d

The kilogram is a unit of mass in the SI system. In the British system, we typically work with weight, which is a force rather than a mass. Consequently, the factor g does not appear when work against gravity is computed in the British system because, essentially, this factor is incorporated in the weight. For example, the work required to lift a 2-lb stone 3 ft is W = (21b)(3 ft) -= 6 ft-lb Ed

a= x0

Xi_i

Xi

xN= b

FIGURE 2 The work to move an object from xi_1 to xi is approximately F(xi)6,x.

We are interested in the case where the force F(x) varies as the object moves from a to b along the x-axis. Eq. (1) does not apply directly, but we can break up the task into a large number of smaller tasks for which Eq. (1) gives a good approximation. Divide [a, b1 into N subintervals of length Ax = (b — a)I N as in Figure 2 and let Wi be the work required to move the object from xi_ j to xi. If Ax is small, then the force F(x) is nearly constant on the interval [xi _j, xi] with value F(x), so Wi F(x1 )6.x. Summing the contributions, we obtain W =

E i=1

E F(xi)Ax i=i Right-endpoint approximation

SECTION 6.5

Work and Energy

The sum on the right is a right-endpoint approximation that converges to

L

339

b F(x)dx.

a

This leads to the following definition.

DEFINITION Work The work performed in moving an object along the x-axis from a to b by applying a force F(x) is

[

,7W ;\ ; ; , 0,; t ,i: (,,,

A (i

[ ,

fli

,f'1( 1 , ,

t,

0 :I; ,1 0 if jff: 11-T

Restoring force —kx

LI'

I ! U-: (lHIiJiSOOAAA i I

x

a

,', ' '

Restoring force —kx

[

W= f

/. .

Equilibrium I 1 : position Ijal i I 1 hhi ir ii i LHI ,h

One typical calculation involves finding the work required to stretch or compress a spring. Assume that the free end of the spring has position x = 0 at equilibrium, when no force is acting (Figure 3). According to Hooke's Law, when the spring is stretched or compressed to position x, it exerts a restoring spring force F(x) = —kx, where k > 0 is the spring constant. If we want to stretch the spring from x = a to x = b, with 0 0 for all x. • The range of f (x) = bx is the set of all positive real numbers. • f (x) = bx is increasing if b > 1 and decreasing if 0 1, the exponential function f (x) = bx is not merely increasing but is, in a certain sense, rapidly increasing. Although the term "rapid increase" is perhaps subjective,

the following precise statement is true: For all n, if x is positive and large enough, then f (x) bx increases more rapidly than the power function g(x) = xi' (we will prove this in Section 7.5). For example, Figure 2 shows that f (x) = 3x eventually overtakes and

4-

Gordon Moore (1929—). Moore, who later became chairman of Intel Corporation, predicted that in the decades following 1965, the number of transistors per integrated circuit would grow "exponentially." This prediction has held up for nearly five decades and may well continue for several more years. Moore has said, "Moore's Law is a term that got applied to a curve I plotted in the mid-sixties showing the increase in complexity of integrated circuits versus time. It's been expanded to include a lot more than that, and I'm happy to take credit for all of it."

4

y

Y= -2

-1

1

I 2

-2

y = r is increasing

CP

-1

2

1

y = ( .)x is decreasing

FIGURE 1

Y f (x)= 3x g(x)= x5

150,000

150,000

150,000

10

10

13 FIGURE 2 Comparison of f (x) = 3' and power functions.

10

341

348

CHAPTER 7

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

increases faster than the power functions g(x) = x3, g(x) = x4, and g(x) = x5. Table 1 compares f (x) = 3x and g(x) = x5.

TABLE 1

1 5 10 15 25

x5

3x

1 3125 100,000 759,375 9,765,625

3 243 59,049 14,348,907 847,288,609,443

CONCEPTUAL INSIGHT The expressions "exponential growth" and "increases exponentially" (and related expressions) describe the growth in or modeled by an exponential function. A function whose graph increases more and more steeply does not necessarily increase exponentially. What distinguishes true exponential growth from other forms of growth is the precise way the function values increase. In exponential growth, an increase in x by 1 causes an increase in the function value by the same fixed percent, regardless of the value of x involved. In contrast, power functions do not exhibit such increase. For example, Table 2 demonstrates how f(x) = 3(2X) increases by 100% with each increase by 1 in x, but g(x) = 4x2 does not exhibit growth by a fixed percent.

TABLE 2 x

f (x) = 3(2x)

1 2 3 4 5

3(21) = 6 3(22) = 12 3(23) = 24 3(24) = 48 3(25) = 96

% increase

g(x)= 4x2

% increase

100 100 100 100

4(12) = 4 4(22) = 16 4(32) = 36 4(42) = 64 4(52) =- 100

300 125 78 56

The Derivative of f (x) = bx Next, we investigate the derivative of f (x) = bx . The rules of differentiation that we have developed so far are of no help because f (x) = bx is neither a product, quotient, nor composition of functions with known derivatives. We must go back to the limit definition of the derivative. The difference quotient (for h 0) is f (x + h) — f (x) Now, take the limit as h outside the limit:

—>

bx

bx bh

bx

bx (bh — 1)

0. The factor b' does not depend on h, so it may be taken

bx+h _ bx bx (bh — 1) d x —b = lim = lim = bx lim dx h.-4-0 h h--->0 h h—>0

(

bh — 1) h )

This last limit does not depend on x. We denote its value by m(b). What we have shown, then, is that the derivative of f (x) =- bx is proportional to bx:

—bx = m(b)bx , where m(b) = lim dx

h

)

1

At this point, this formula is not very helpful because we do not know much about m(b). In Section 7.3, we will learn that m(b) is equal to ln b, the natural logarithm of b. To proceed further, we investigate m(b) numerically for a few values of b. EXAMPLE 1 Estimate m(b) numerically for b = 2, 2.5, 3, and 10. Solution We create a table of values of difference quotients to estimate m(b):

The Derivative of f(x) = bx and the Number

SECTION 7.1

2h - 1

h 0.01 0.001 0.0001 0.00001

e

349

loh - 1

h

(2.5)h - 1 h

h

h

0.69556 0.69339 0.69317 0.69315

0.92050 0.91671 0.91633 0.916295

1.10467 1.09921 1.09867 1.09861

2.32930 2.30524 2.30285 2.30261

m(2) ',-:, 0.69

m(2.5) ,s-= 0.92

m(3) ,s•-'- 1.10

m(10) ', -,' 2.30 •

At least for the values of b considered, the results in Example 1 suggest that m(b) is defined (i.e., the limit exists) and that m(b) increases as b increases. In fact, the following properties of m(b) can be shown, but the proofs are somewhat technical and we do not present them: • m(b) is defined for all b > 0, • m is a continuous and increasing function. Given that m(2.5) ';=,' 0.92 and m(3) -,== 1.10, accurate to two decimal places, the Intermediate Value Theorem implies that there exists a value b between 2.5 and 3 such that m(b) = 1. Furthermore, since m is increasing, there is only one such value. This value is the number denoted by e. It can be shown that e ';'-' 2.718.

The Number e

There is a unique positive real number e with the property eh lim h->0 (

1)

h

-1

The value of e is approximately 2.718. Now, we define the (natural) exponential function to be the function f (x) = ex . By Eq. (1) and the definition of e, it follows that

Whenever we refer to the exponential function without specifying the base, the reference is to f(x) = ex. In some settings, ex is denoted exp(x).

d - ex = ex dx Thus, the exponential function f (x) = ex is equal to its own derivative! GRAPHICAL INSIGHT The graph of f (x) = bx passes through (0, 1) for all b > 0 because b° = 1 (Figure 3). The number m(b) is simply the slope of the tangent line at x = 0: d bx = m(b) • b° = m(b) dx x=0 These tangent lines become steeper as b increases and b = e is the unique value for which the tangent line has slope 1.

FIGURE 3 The tangent lines to y = bx at

x = 0 grow steeper as b increases.

EXAMPLE 2 Find an equation of the tangent line to the graph of f (x) = 3ex - 5x2 at x = 2.

350

CHAPTER 7

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Solution We compute both f'(2) and f(2): 5x2) = 3 d ex dx

f i(x) = dx

5 d x2 dx

=

3ex

10x

f'(2) = 3e2 - 10(2) ,^•-•-' 2.17 f(2) = 3e2 - 5(22) •=--, 2.17

f(x)= 3ex - 5x2

An equation of the tangent line is y = f(2) + f (2)(x - 2). Using these approximate values, we write the equation as (Figure 4) y = 2.17 + 2.17(x - 2)

or

•

y = 2.17x - 2.17

EXAMPLE 3 Calculate f'(0), where f (x) = ex cos x. Solution Use the Product Rule: f' (x) = ex • (cos x)/ ± cos x • (ex )' =- -e x sin x + cos x • ex = ex (cos x - sin x) •

Then f'(0) = e°(1 -0) = 1. FIGURE 4

To compute the derivative of a function of the form h(x) = eg(x), we recognize h(x) as a composite function h(x) = eg(x) = f (g(x)), where f (u) = eu , and we apply the Chain Rule: h'(x) =

(eg(x)) = [f (g(x))]' =

d

(g(x))g'(x) = eg(x)g'(x)

A special case is (ekx+b)' = kekx+b, where k and b are constants.

dx

=

ekx+b) = kekx+b

(x)eg(x),

(k, b constants)

dx

3

EXAMPLE 4 Differentiate: (a) f (x) =

e9x-5

(b) f (x) = ecos x

and

Solution Apply Eq. (3): (a) — e9x-5 dx

= 9e9x-5

and

(b) dx

EXAMPLE 5 Graph Sketching Involving ex interval [-4,21.

(ecosx ) = _ sin x (e' ) sx

•

Sketch the graph of f (x) = xex on the

Solution As usual, the first step is to determine the critical points. The derivative of f is d f' (x) = — x ex = x ex + ex = (x dx We need to solve (x x = -1 and

1)ex

1)ex = 0. Since ex > 0 for all x, there is a single critical point at

f' (x) = l < >0

for

x < -1

for

x > -1

Thus, f '(x) changes sign from - to ± at x = -1 and f (-1) is a local minimum. For the second derivative, we have f (x) = (x + 1) • (ex ) ± ex (x f(x) =

0

1)'

(x

1)ex + ex = (x

for

x < -2

for

x > -2

2)ex

SECTION 7.1

The Derivative of f(x)= bx and the Number e

351

Thus, x = —2 is a point of inflection, where the graph changes from concave down to concave up at x = —2. Figure 5 shows the graph with its local minimum and point of inflection. •

Integrals Involving ex The formula (ex y = ex says that the function f(x) = ex is its own derivative. But this means f(x) ex is also its own antiderivative. In other words, FIGURE 5 Graph of f(x)= xex . The sign combinations --, —±, ±± indicate the signs of f i and f".

f

ex dx = ex + C

More generally, for any constants b and k with k f

0,

1

ekx+b ax = — ekx+12 + k

We prove this formula by noting that

d dx

--(

1 _ k

ekx±b

=

ekx-Fb .

EXAMPLE 6 Evaluate: (a) f

e7x-5 dx

(b) f xe2x2 dx

(c) f

et dt 1 + 2et + e2t

Solution = — 1 e7x-5 ± C. 7 (b) Use the substitution u = 2x2, du = 4x dx:

(a)

f

e7x-5 dx

f

xe2x2 dx = — 4

eu du = —eu + C = —e`x + C 4 4

(c) We have 1 + 2et ± e2t = (1 + et )2. The substitution u = 1 + et, du = et dt gives et f 1+ 2et

e2t dt — f

_1

du — 14

—27

+ C = —( 1 +

et )

—1 + C

•

Other Approaches to Defining e Mathematicians first became aware of the special role played by the number e in the seventeenth century. The notation e was introduced around 1730 by Leonard Euler, who discovered many fundamental properties of this important number. In this section, we defined e as the unique value such that 11M h-±0 (

eh _1)

h

There are a number of other ways to define e, all of which either directly or indirectly involve a limit. Here are three other approaches to defining e. We will investigate each later in the text. • e is given by each of the following limits, where the second is obtained from the first by substituting 1/ t for x: e = lim(1 + x)1/x x—>13

and

1) t e = t j. ± rnoo I + — t

352

CHAPTER 7

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

• e is the limit as we progressively add terms to the sum 1 1 1 1 1 1 1+1+ — + — + — + — + — + • •• + — n! 2 6 24 120 720 (With infinitely many terms in the sum, the expression is known as an infinite series. We study infinite series in Chapter 11.) • e is the unique number such that the area of the region under the hyperbola y = 1/x, for 1 1 and decreasing if b < 1. • The derivative of f(x) = bx is proportional to bx: dx bh — 1 h • There is a unique number e

1).

bx = m(b)bx

where m(b) = lirn

h —*O

2.718 with the property m(e) = 1, so that d — ex = ex dx

• By the Chain Rule: d dx

= f(x)e

_d ekx-Fb =ke kx+b dx

,

(k,b constants)

• f ex dx = ex ± C. • f

ekx+b dx

1 = — ekx+b k

(k,b constants with k

0).

7.1 EXERCISES Preliminary Questions 1. To which of the following does the Power Rule apply? (a) f (x) = x2 (d) f(x) = ex

(b) f (x) = 2e (e) f(x) = xx

(c) f (x) = xe (f) f(x) = X -4 l 5

5. Which of the following statements is not true? (a) (ex)/ = ex 1

2. For which values of b does f(x) =- bx have a negative derivative?

(b) lim eh — h—>0 h

3. For which values of b is the graph of y = bx concave up?

(c) The tangent line to y = ex at x = 0 has slope e.

4. Which point lies on the graph of y = bx for all b?

(d) The tangent line to y = ex at x = 0 has slope 1.

—

Exercises In Exercises 1-4, determine the limit. 1. lim 4X 3. lim (-1) —x x—>0.0 4

7.

y

=

eX+2

x

o

=

2. lim x—>oo

In Exercises 9-30, find the derivative.

4. lirn ex —x2

9. f(x)= 7e2x

3e4x

X—. CXD

In Exercises 5-8, find the equation of the tangent line at the point indicated. 6. e4x , xo = 0 5. y = 4e , xo = 0 y

=

8. y = ex2,

xo = 1

10. f(x) = e-5x

11. f(x) = e x

12. f(x) = e3

13. f(x) = C

14. f(x) = 4e—x +7e—a

relx+9

SECTION 7.1 ex2 15. f(x) = -

The Derivative of f(x) = bx and the Number e

353

16. f(x) = X2 e2x 20

17. f (x) = (1 + ex)4

18. f(x) = (2e3x

19. f(x) =

20. f(x) =

2e-2x)4 10

ex2+2x-3

21. f(x) = esmx

22. f(x) = e(x2±2x+3)2

23. f(8) = sin(e8)

24. f (t) =

1

26. f (t) = cos(te-2t )

25. f (t) = 1 - e-3t

ex+t

51. Compute the linearization of f(x) = e-2x sin x at a = 0. 52. Compute the linearization of f(x) = xe6-3x at a = 2.

ex 27. f(x) = 3x + 1

29. f (x) =

FIGURE 8 Graphs of y = ex and y = 5x.

28. f(x) = tan(e5-6x)

53. Find the linearization of f(x) -= ex at a = 0 and use it to estimate e-0.1.

x 30. f(x) = eex

2ex - 1

54. Use the linear approximation to estimate f(1.03) - f(1), where y = x1/3e' .

In Exercises 31-36, calculate the derivative indicated. f(x) = e4x-3

31. f''(x); 33.

35.

d2 y

y = er sin t

d2 t t2 dt2

32. f m(x);

34.

36.

f(x) = e12-3x

d2y .

V

a't2 " d3 CO

= e-2t sin 3t

cos(e°)

In Exercises 37-42, find the critical points and determine whether they are local minima, maxima, or neither. 37. f (x) = ex - x

38. f (x) = x + e'

ex 39. f (x) = -

40. f(x) =

55. A 2005 study by the Fisheries Research Services in Aberdeen, Scotland, showed that the average length of the species Clupea Harengus (Atlantic herring) as a function of age t (in years) can be modeled by L(t) = 32(1 - e 37t ) cm for 0 < t < 13. (a) How fast is the average length changing at age t = 6 years? (b) At what age is the average length changing at a rate of 5 cm/year? (c) Calculate L = lim L(t). t->co 56. According to a 1999 study by Starkey and Scarnecchia, the average weight (in kg) at age t (in years) of channel catfish in the Lower Yellowstone River can be modeled by -0.03456t)34026 W(t) = (3.46293 - 3.32173e

for x > 0

X2ex

Find the rate at which weight is changing at age t = 10. 41. g(t) =

et

42. g(t) = (t3 - 2t)et

1.2 +

57. The functions in Exercises 55 and 56 are examples of the Von Bertalanffy growth function

In Exercises 43-48, find the critical points and points of inflection. Then sketch the graph. 44. y = e- x + ex

43. y = xe- x 45. y = e- x cos x 47. y = ex - x

on [ -

46. 48. y = x2e'

49. Find a > 0 such that the tangent line to the graph of f(x) = x2e- x at x = a passes through the origin (Figure 7).

f(x) = X2e-x

31(t) = (a ±

-aekm t) m

) biologist Karl Ludwig Von introduced in the 1930s by Austrian-born Bertalanffy. Calculate M'(0) in terms of the constants a, b, k, and m. 58. Find an approximation to m(4) using the limit definition and estimate the slope of the tangent line to y =- 4x at x = 0 and x = 2. 59. Find an approximation to m(1/2) using the limit definition and estimate the slope of the tangent line toy = (1/2)x at x = 0 and x = 1. 60. Find approximations to m(2.71) and m(2.72) using the limit definition. In Exercises 61-78, evaluate the integral. 61. f (ex + 2) dx

62.

f

e4x 6

I

63. f

dx

e-x /2 dx

64. f

e-3x dx

2 > x

65.i

3

e l -6t

dt

67. f (e4x + 1) dx 50. Use Newton's Method to find the two solutions of ex = 5x to three decimal places (Figure 8).

69.f

o

e4t-3 dt 2

0

FIGURE 7

3

66. i

68. f (ex + e')dx f2

1

xe

x-212

dx

ye3Y2 dy

70. 0

CHAPTER 7

354

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

71. f et .1et +1 dt f

73.

e 2x _

e4x

75. 1

ex

J ,N e' 77.

dx

ex 1

72. f (e-x - 4x) dx

This means that the probability that v lies between a and b is equal to the shaded area in Figure 9.

74. f ex cos(ex)dx

(a) Show that the probability that v E [0, b] is 1 (b) Calculate the probability that v E [2,5].

e —b2 /64 .

76. i ex (e2x +1)3 dx

dx

1 eVi dx j

78. i x_2/3e '3 dx

79. Find the area between y = ex and y = ex over [0,1]. 80. Find the area between y = ex and y = e- x over [0,2]. 81. Find the area bounded by y = e2, y = e, and x = 0. 82. Find the volume obtained by revolving y = ex about the x-axis for 0< x 1

holds fort > 0 because e > 1. Use (4) to prove that ex > 1 + x

for x > 0

Then prove, by successive integration, the following inequalities (for x > 0):

ex

1+x

+

1

1 + X + 2x 2

+ 1 — x3 6

89. Calculate the first three derivatives of f (x) = xex . Then guess the formula for f (?1)(x) (use induction to prove it if you are familiar with this method of proof). 90. Consider the equation ex = Ax, where A is a constant. (a) For which A does it have a unique solution? For intuition, draw a graph of y = ex and the line y =

91. Prove in two ways that the numbers m(a) satisfy 1

x2 + gx'

m(ab) = m(a) + m(b)

87. Generalize Exercise 86; that is, use induction (if you are familiar with this method of proof) to prove that for all n > 0, ex > 1

x—).cx) xn

(b) For which A does it have at least one solution?

-1 x-, 2

ex > 1+ x

ex x 88. Use Exercise 86 to show that — > - and conclude that x2 - 6 . ex = co. Then use Exercise 87 to prove more generally that x-,cx) x . ex — = co for all n.

+•••+

1

xn

(a) First method: Use the limit definition of m(b) and (ab)h

—1

, ( ah _ 1 ) =

h

bh _ h

(x>0) (b) Second method: Apply the Product Rule to ax bx = (ab)x

7.2 Inverse Functions

f-

I

FIGURE 1 A function and its inverse.

In this section, we discuss the concept of inverse functions and we develop an important theorem for computing their derivatives. In the next section, we will define logarithmic functions as inverses of exponential functions. The inverse off, denoted f —1 , is the function that reverses the effect off (Figure 1). For example, the inverse of f (x) = x3 is the cube root function f 1(x) = Given a table of function values for f, we obtain a table for f —1 by interchanging the x and y columns, assuming the resulting f' is a function:

SECTION 7.2

In general, f -1(x) L - . The expression f -1(x) is simply a notation for the inverse function, and the -1 does not represent an exponent.

Function

Inverse Functions

355

Inverse

f(x)=x3 -2 -1 0 1 2 3

-8 -1 0 1 8 27

(Interchange columns)

—8 -1 0 1 8 27

—2 —1 0 1 2 3

If we apply both f and f -1 to a number x in either order, we get back x. For instance, Apply f and then r

l:

2

Apply f -1 and then f:

8

(apply x3) (apply x113) --+

8 2

(apply x1/3) (apply x3)

2 8

This property is used in the formal definition of the inverse function: 4.0 REMINDER The domain is the set of numbers x such that f (x) is defined (the set of allowable inputs), and the range is the set of all values f (x) (the set of outputs).

DEFINITION Inverse Let f have domain D and range R. If there is a function g with domain R such that g(f(x)) = x

for x E D

and

f(g(x)) = x

for x E R

then f is said to be invertible. The function g is called the inverse function and is denoted f -1. EXAMPLE 1 Show that f(x) = 2x - 18 is invertible. What are the domain and range of f -1? Solution We show that f is invertible by computing the inverse function in two steps.

The variable y in f -1(y) = y + 9 is called a dummy variable. It is "local" to the equation, which means that changing y to a different symbol does not change the meaning of the relationship, as well as the mathematics that precedes and follows the relationship. We could write f -1(A) = A + 9 or f -1(DOG) = l(DOG)+ 9 and not affect the mathematics in Example 1. Generally, x is preferred as the independent variable when writing expressions for functions.

Step 1. Solve the equation y = f(x) for x in terms of y. y = 2x - 18 y ± 18 = 2x

x=

1

y+ 9

This gives us the inverse as a function of the variable y: f 1(y)= y + 9. Step 2. Interchange variables. We usually prefer to write the inverse as a function of x, so we interchange the roles of x and y: 2x + 9 f -1(x)= -1 Graphs of f and f -1 are shown in Figure 2. To check our calculation, let's verify that f -1(f(x))= x and f(f -1(x))= x: 2 (2x - 18) + 9 = (x -9) + 9 = x f -1(f(x)) = f -1(2x - 18) = -1 1 1 f (f -1(x)) = f (-x +9) = 2 (-x + 9) - 18 = (x + 18) - 18 = x 2 2

FIGURE 2

Because f -1 is a linear function, its domain and range are R.

•

356

CHAPTER 7

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

The inverse function, if it exists, is unique. However, some functions do not have Consider f (x) -= x2. When we interchange the columns in a table of values inverse. an (which should give us a table of values for f -1), the resulting table does not define a function: Inverse (?)

Function

f

f (x) = x2

'Another standard term for one-to-one is I injective.

—2 —1 0 1

4 1 0 1

2

4

(Interchange columns)

4 1 0 1 4

1( ) —2 —1 0 1 2

f -1(1) has two values: 1 and —1.

The problem is that every positive number occurs twice as an output of f (x) = x2. For example, 1 occurs twice as an output in the first table and therefore occurs twice as an input in the second table. So the second table gives us two possible values for f -1(1), namely f -1(1) = 1 and f 1(1) = —1. Neither value satisfies the inverse property. For instance, if we set f -1(1) = 1, then f -1(f(-1)) = f-1 0) = 1, but an inverse would have to satisfy f -1(f (-1)) = —1. So when does a function f have an inverse? The answer is: if f is one-to-one, which means that f takes on each value at most once (Figure 3). Here is the formal definition: DEFINITION One-to-One Function A function f is one-to-one on a domain D if, for every value c, the equation f (x) = c has at most one solution for x E D. Or, equivalently, if for all a, b E D, if a 0 b, then f (a) f (b).

FIGURE 3 A one-to-one function takes on each value at most once.

Think of a function as a device for "labeling" members of the range by members of the domain. When f is one-to-one, this labeling is unique and f -1 maps each number in the range back to its label.

One-to-one

Not one-to-one

f(x)= c has at most one solution for all c.

f(x)= c has two solutions: x = a and x = a'.

When f is one-to-one on its domain D, the inverse function f -I exists and its domain is equal to the range R of f (Figure 4). Indeed, for every c E R, there is precisely one element a E D such that f (a) = c and we may define f 1(c) = a. With this definition, f (f -1(c)) = f (a) = c and f -I ( f (a)) = f -1(c) -- a. This proves the following theorem: THEOREM 1 Existence of Inverses The inverse function f -I exists if and only if f is one-to-one on its domain D. Furthermore, • Domain of f = range of f -1 • Range of f =- domain of f -1

Domain off = range of

FIGURE 4 In passing from f to f —1 , the domain and range are interchanged.

Range off= domain off-'

SECTION 7.2

EXAMPLE 2 Show that f(x) = of f and f -1.

y=

357

3x + 2 is invertible. Determine the domain and range 5x - 1

Solution The domain off is D solve y = f(x) for x in terms of y:

5

Inverse Functions

:x

fl (Figure 5). Assume that x E D, and let's

3x + 2 5x - 1

y(5x - 1) = 3x ± 2 FIGURE 5 Graph of f (x) =

5xy - y = 3x ± 2

3x ± 2 5x — l •

5x y - 3x = y ± 2

(gather terms involving x)

x(5y - 3) = y + 2

(factor out x in order to solve for x)

1

(divide by 5y - 3)

2

x= Often, it is impossible to find a formula for the inverse because we cannot solve for x explicitly in the equation y = f(x). For example, the function f(x) = x + sin x has an inverse, but we must make do without an explicit formula for it.

y+2 5y - 3

The last step is valid if 5y - 3 0 0—that is, if y ;. But note that y = is not in the range of f. For if it were, Eq. (1) would yield the false equation 0 = + 2. Now, Eq. (2) shows that for all y ;, there is a unique value x such that f(x) = y. Therefore, f is one-to-one on its domain. By Theorem 1, f is invertible. The range of f is and R = ix : x f -1(x) =x+2 5x - 3 The inverse function has domain R and range D.

•

We can tell whether f is one-to-one from its graph. The horizontal line y = c intersects the graph of f at points (a, f (a)), where f (a) = c (Figure 6). There is at most one such point if f(x) = c has at most one solution. This gives us the following: Horizontal Line Test A function of x is one-to-one if and only if every horizontal line intersects the graph of the function in at most one point.

FIGURE 6 The line y = c intersects the graph at points where f (a) = c.

In Figure 7, we see that f(x) = x3 passes the Horizontal Line Test and therefore is one-to-one, whereas f(x) = x2 fails the test and is not one-to-one.

I

' X

X

(A) f(x) = x3 is one-to-one.

(B) f(x) = x2 is not one-to-one.

FIGURE 7 EXAMPLE 3 Increasing Functions Are One-to-One Show that increasing functions are one-to-one. Then show that f(x) = x5 + 4x + 3 is one-to-one. Solution An increasing function satisfies f (a) < f (b) if a 0, then f (x) = cxn is increasing. • A sum of increasing functions is increasing. Thus, g(x) = x5 and h(x) = 4x are increasing, and therefore so is the sum k(x) = x5 ± 4x. It follows that the function f (x) = x5 + 4x + 3 is increasing and thus is oneto-one (Figure 8). However, determining an explicit formula for its inverse would be • difficult.

FIGURE 8 The increasing function f (x) = x5 4x + 3 satisfies the Horizontal Line Test. One-to-one for x > 0

• -21

I -1

I

2

1

FIGURE 9 f (x) = x2 satisfies the Horizontal Line Test on the domain {x : x > 0}.

Note that using an argument like the one in the previous example, we can prove that decreasing functions are also one-to-one and therefore have inverses. We can make a function one-to-one by restricting its domain suitably. EXAMPLE 4 Restricting the Domain Find a domain on which f (x) one and determine its inverse on this domain.

x2 is one-to-

Solution The function f (x) = x2 is one-to-one on the domain D = : x > 01, for if a2 = b2, where a and b are both nonnegative, then a = b (Figure 9). The inverse of f on D is the positive square root f —1(x) = ,fx-. Alternatively, we may restrict f to the domain fx : x 01. We do not need a formula for g(x) to draw its graph. We simply reflect the graph of f through the line y = x, as in Figure 12. If desired, however, we can easily solve y = ,V4 — x to obtain x = 4 — y2 and thus g(x) = 4 — x2 with domain fx : x > 01. •

Derivatives of Inverse Functions Next, we derive a formula for the derivative of the inverse f -1, expressing how it is related to the derivative of f.

SECTION 7.2

Inverse Functions

359

THEOREM 2 Derivative of the Inverse Assume that f is differentiable and one-toone with inverse g(x) = f -1(x). If b belongs to the domain of g and f'(g(b)) 0 0, then g'(b) exists and g/ (b) =

1 f (g(b))

3

Proof The proof that g is differentiable if f'(g(x)) 0 is technical and therefore we omit it. To prove Eq. (3), note that f (g(x)) = x by definition of the inverse. Differentiate both sides of this equation, and apply the Chain Rule: d d — f (g(x)) = — x dx dx

f'(g(x))g'(x) = 1

=

=

g'(x) =

1 f'(g(x))

Set x = b to obtain Eq. (3).

•

GRAPHICAL INSIGHT The formula for the derivative of the inverse function has a clear graphical interpretation. Consider a line L of slope m and let L' be its reflection through y = x as in Figure 13(A). Then the slope of L' is 1/m. Indeed, if (a, b) and (c, d) are any two points on L, then (b, a) and (d, c) lie on L' and slope of L =

d- b

slope of L'= d c l ab

reciprocal slopes

Now recall that the graph of the inverse g is obtained by reflecting the graph of f through the line y = x. As we see in Figure 13(B), the tangent line to y = g(x) at x = b is the reflection of the tangent line to y = f (x) at x = a [where b = f (a)

and a = g(b)]. These tangent lines have reciprocal slopes, and thus g'(b) = 1If' (a) = 11f'(g(b)), as claimed in Theorem 2. y slope

ki „

y = g(x)

—x

slope m (d, c) (c, d) FIGURE 13 Graphical illustration of the formula g/(b) = 11f' (g(b)).

(A) If L has slope m, then its reflection L' has slope 1/m.

(B) The tangent line to the inverse y = g(x) is the reflection of the tangent line toy =f(x).

EXAMPLE 6 Using Equation (3) Calculate g/(x), where g is the inverse of the function f (x) = x4 + 10 on the domain Ix : x > 0). Solution Solve y = x4 + 10 for x to obtain x = (y - 10)1/4. Thus, g(x) = (x - 10)1/4. Since f (x) = 4x3, we have f'(g(x)) = 4g(x)3, and by Eq. (3), 1 1 1 1 = - (x - 10)-314 g'(x) 4 4(x - 10)3/4 4g(x)3 f'(g(x)) We obtain this same result by differentiating g(x) = (x - 10)1/4 directly.

•

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CHAPTER 7

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

EXAMPLE 7 Calculating g'(x) Without Solving for g(x) the inverse of f(x) = x ex .

f(x)= x + ex

Calculate g1(1), where g is

Solution In this case, we cannot solve for g(x) explicitly, but a formula for g(x) is not needed (Figure 14). All we need is the particular value g(1), which we can find by solving f(x) = 1. By inspection, x e = 1 has solution x = 0. Therefore, f(0) = 1 and, by definition of the inverse, g(1) = 0. Since f'(x) = 1 + ex , •

•

(0, 1)

g'(1) —

1 f' (g(1))

1

1 f '(0)

1

e°

1 — 2

•

(1,0)

FIGURE 14 inverse g.

Graph of f(x) = x

ex

and its

7.2 SUMMARY • A function f is one-to-one on a domain D if for every value c, the equation f(x) = c has at most one solution for x E D, or, equivalently, if for all a, b E D, f (a) f (b) unless a = b. • Let f have domain D and range R. The inverse f —1 (if it exists) is the unique function with domain R and range D satisfying f ( f —1(x)) = x and f —1( f (x)) = x. • The inverse of f exists if and only if f is one-to-one on its domain. • To find the inverse function, solve y = f(x) for x in terms of y to obtain x = g(y). The inverse is the function g. • Horizontal Line Test: f is one-to-one if and only if every horizontal line intersects the graph of f in at most one point. • The graph of f1 is obtained by reflecting the graph of f through the line y = x. • Derivative of the inverse: If f is differentiable and one-to-one with inverse g, then for x such that f' (g(x)) 0, g'(x) =

1

f' (g(x))

7.2 EXERCISES Preliminary Questions 1. Which of the following satisfy f —1(x) = f(x)?

Trenton

6:21

(a) f (x) = x

Hamilton Township

6:27

Princeton Junction

6:34

New Brunswick

6:38

(b) f (x) = 1 — x (c) f(x) = 1

(d) f (x) = .V7 (e) f(x) = Ix' 2. The graph of a function looks like the track of a roller coaster. Is the function one-to-one?

5. A homework problem asks for a sketch of the graph of the inverse of f(x) = x + cos x . Frank, after trying but failing to find a formula for f —1(x), says it's impossible to graph the inverse. Bianca hands in an accurate sketch without solving for f —1 . How did Bianca complete the problem?

3. The function f maps teenagers in the United States to their last names. Explain why the inverse function f —1 does not exist.

6. What is the slope of the line obtained by reflecting the line y = through the line y = x?

4. The following fragment of a train schedule for the New Jersey Transit System defines a function f from towns to times. Is f one-to-one? What is f'(6:27)?

7. Suppose that P = (2,4) lies on the graph off and that the slope of the tangent line through P is m = 3. Assuming that f exists, what is the slope of the tangent line to the graph of f —1 at the point Q = (4,2)?

(f) f (x) = x-1

Exercises 1. Show that f(x) = 7x — 4 is invertible and find its inverse. 2. Is f(x) = x2 + 2 one-to-one? If not, describe a domain on which it is one-to-one.

3. What is the largest interval containing zero on which f(x) = sinx is one-to-one? 4. Show that f(x) =

x —2 is invertible and find its inverse. x +3

SECTION 7.2

Inverse Functions

361

(a) What is the domain off? The range of f'? (b) What is the domain of f

? The range off?

5. Verify that f (x) = x3 + 3 and g(x) = (x - 3)1 /3 are inverses by showing that f (g(x)) = x and g(f (x)) = x. 6. Repeat Exercise 5 for f (t) =

t +1 t+1 and g(t) = . t- 1 t- 1

(A)

1 2G M R where G is the universal gravitational constant and M is the mass. Find the inverse of v expressing R in terms of v.

(B)

7. The escape velocity from a planet of radius R is v(R) =

Th

8. Show that the power law relationship P(Q) = kQr, for Q > 0 and r 0, has an inverse that is also a power law, Q(P) = m Ps, where m = k-l ir and s = 1/r.

, X

(C)

(D)

9. The volume V of a cone that has height equal to its radius r is given by V (r) = 3 r3 • Find the inverse of V(r), expressing r as a function of V. 10. The surface area S of a sphere of radius r is given by S(r) = 4nr2. Explain why, in the given context, S(r) has an inverse function. Find the inverse of S(r), expressing r as a function of S. In Exercises 11-17, find a domain on which f is one-to-one and a formula for the inverse off restricted to this domain. Sketch the graphs off and f_ 1 . 11. f (x) = 4 - x

1 12. f (x) = x + 1

1 13. f (x) = x _ 3

14. f(s) =

15. f (x) =

1 /x2 +1

FIGURE 16 20. Let n be a nonzero integer. Find a domain on which f (x) = (1 xn)1 I n coincides with its inverse. Hint: The answer depends on whether n is even or odd. 21. Let f (x) = x7 + x + 1. (a) Show that f' exists (but do not attempt to find it). Hint: Show that f is increasing. (b) What is the domain of f ? (c) Find f -1(3). 22. Show that f (x) = (x2 + 1)-1 is one-to-one on (-oo, 0], and find a formula for f -1 for this domain off.

1 5'

23. Let f (x) = x2 - 2x. Determine a domain on which f -1 exists, and find a formula for f -1 for this domain off.

16. f (z) = z3

24. Show that the inverse of f (x) = e- x exists (without finding it explicitly). What is the domain of f 1 ?

17. f (x) = • x3 + 9

25. Find the inverse g of f (x) = x2 +9 with domain x > 0 and calculate g'(x) in two ways: using Theorem 2 and by direct calculation.

18. For each function shown in Figure 15, sketch the graph of the inverse (restrict the function's domain if necessary).

26. Let g be the inverse of f (x) = x3 + 1. Find a formula for g(x) and calculate ex) in two ways: using Theorem 2 and then by direct calculation. In Exercises 27-32, use Theorem 2 to calculate g'(x), where g is the inverse of f.

(A)

(C)

(B)

27. f (x) = 7x + 6

28. f (x) =

29. f (x) = X -5

30. f (x) = 4x3 - 1

31. f (x) =

32. f (x) = 2 + x-1

x+1

33. Let g be the inverse of f (x) = x3 + 2x + 4. Calculate g(7) [without finding a formula for g(x)], and then calculate g'(7). X3

34. Find g'( - 1), where g is the inverse of f (x) = 2+ 1. In Exercises 35-40, calculate g(b) and g'(b), where g is the inverse off (in the given domain, if indicated).

, X

(D)

(E)

(F)

FIGURE 15

35. f (x) = x + cos x,

b=1

36. f (x) = 4x3 - 2x,

b = -2

37. f (x) =

x2 + 6x

for x > 0, for x < -6,

19. Which of the graphs in Figure 16 is the graph of a function satisfying f -1 = f?

38. f (x) = /x2 + 6x 1 1 39. f (x) = x+1' b=

b=4 b=4 40. f (x) = ex ,

b=e

362

CHAPTER 7

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

41. Let f (x) = x" and g(x) = x' ' . Compute gl(x) using Theorem 2 and check your answer using the Power Rule. 1- x 1 areinverses. Then comand g(x) 42. Show that f(x) = 1 +x pute (x) directly and verify that g'(x) = 1/1(g(x)). Use graphical reasoning to determine if the following state43. ments are true or false. If false, modify the statement to make it correct. (a) If f is increasing, then 1 -1 is increasing.

(b) If f is decreasing, then f -1 is decreasing. (c) If f is concave up, then f1 is concave up. (d) If f is concave down, then f -1 is concave down.

(e) Linear functions f (x) = ax + b (a 0 0) are always one-to-one. (f) Quadratic polynomials f(x)= ax2 + bx + c (a 0 0) are always oneto-one.

(g) f (x) = sin x is not one-to-one.

Further Insights and Challenges 44. Show that if f is odd and f -1 exists, then f -1 is odd. Show, on the other hand, that an even function does not have an inverse. 45. Let g be the inverse of a function f satisfying f '(x) = f (x). Show that g'(x) = x 1. [Note that this shows that the inverse of the exponential

function f (x) = ex is an antiderivative of x-1. That inverse is the natural logarithm function that we define in the next section.]

7.3 Logarithmic Functions and Their Derivatives Logarithmic functions are inverses of exponential functions. More precisely, if b > 0 and b 1, then the logarithm to the base b, denoted logs x, is the inverse of f (x) = bx . By definition, y = logs x if bY = x, so we have blogb x = x

and

logs(bx ) = x

In other words, logs x is the number to which b must be raised in order to get x. For example, log2(8) = 3

because

23 = 8

10g10(1) = 0

because

100 = 1

1 log3 ( 4) = —2

because

3-2 = 1

32

_

—4

The logarithm to the base e, denoted In x, plays a special role and is called the natural logarithm.

ln x = loge x In this text, the natural logarithm is denoted ln x. Other common notations are log x and Log x.

We use a calculator to evaluate logarithms numerically. For example, ln 17

2.83321

because

e2.83321

17

As in Figure 1, f (x) = ln x and g(x) = ex are inverse functions, so we have elnx = X

and

In(ex ) = x

Recall that the domain of f (x) = bx is R and its range is the set of positive real numbers fx : x > 0). Since the domain and range are reversed in the inverse function, • The domain of f (x) = logs x is {x :x >0). • The range of f (x) = logs x is the set of all real numbers R.

FIGURE 1 y = ln x is the inverse of y = ex.

If b> 1, then logs x is positive for x > 1 and negative for 0 0):

(a) f (x) = xx

and

(b) g(x) = xsi" x

Solution The graphs off and g are shown in Figure 4. We illustrate two different methods; either method could be used in either case.

4

(a) Method 1: Use the identity x -= eh x to rewrite f (x) as an exponential base e: 2

f (x) = xx = (e

f(x) = xx

)X = ex lnx

f (x) = (x ln x yex In x

= (1 + in X )ex

In x

— (1 + ln x)xx

,x 2

4

(b) Method 2: Apply Eq. (3) to ln(g(x)). Since ln(g(x)) = ln(xsinx) = (sin x) ln x, sin x d g'(x) + (cos x) ln x ln(g(x)) = — ((sin x) ln x) = = dx g(x) dx sin x sin x g/ (x) = (— + (cos x)ln x) g(x) = (— + (cos x) ln x) xsin x

FIGURE 4 Graphs of f (x) = xx

8,(x) = xsinx.

and

Note that since f (x) = xx has the variable x in both the base and the exponent, it is a rapidly growing function. It is evident from its graph that it decreases initially, reaches a minimum, and then increases from that point on. Where does that minimum occur and what is the minimum value? To answer this, we need to find the x where the slope of the graph of f is 0; that is, where f' (x) = 0. We solve: (1 + lnx)xx =0 1 + ln x = 0

(since xx is never 0)

ln x = —1 x = e-1 = 1/e So the minimum occurs at x = 1/e, and the minimum value is (1/e)1/e

0.6922.

Integrals Involving the Natural Logarithm In Chapter 5, we noted that the Power Rule for Integrals is valid for all exponents n

—1:

n+1

f xn dx =

n+1

+C

(n

—1)

This formula is not valid (or meaningful) for n = —1, so the question remained: What is the antiderivative of y = x-1? We can now give the answer: the natural logarithm.

368

CHAPTER 7

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Indeed, the formula (ln x)' = x-1 tells us that ln x is an antiderivative of y = x-1 for x >0: f dx — = ln x + C i x We would like to have an antiderivative of y = 1/x on its full domain, namely on the domain ix : x 0 0). To achieve this end, we extend F to an even function by setting F(x) = In jx1 (Figure 5). Then F(x) = F(-x), and by the Chain Rule, r(x) -F'(-x). For x 0 with b g(x) = bx ;

1, the logarithm function f(x) = logs x is the inverse of x

b

1, then logs x is positive for x > 1 and negative for 0 0. Hint: Use the substitution (a) a

t

ii

t

tla. (b) G(ab)= G(a) G(b). Hint: Break up the integral from 1 to ab into two integrals and use (a). (c) G(1) = 0 and G(a-1)= -G(a) for a >0. (d) G(an)= nG(a) for all a > 0 and integers n. U =

(e) G(a l /n) -=

G(a) for all a >0 and integers n

0.

(f) G(ar) = rG(a) for all a > 0 and rational numbers r. (g) G is increasing. Hint: Use FTC II. (h) There exists a number a such that G(a) > 1. Hint: Show that G(2) > 0 and take a =2- for m > 1/ G(2).

(i) lim G(x) = oo and lim G(x) = -co. x->oo x->o+ (j) There exists a unique number E such that G(E)= 1. (k) G(Er)= r for every rational number r. 117. Defining ex Use Exercise 116 to prove the following statements. (a) G has an inverse with domain R and range {x : x > 0}. Denote the inverse by F. (b) F(x + y) = F(x)F(y) for all x, y. Hint: It suffices to show that G(F(x)F(y))= G(F(x + y)). (c) F(r)= Er for all numbers. In particular, F(0) = 1. (d) F'(x)= F(x). Hint: Use the formula for the derivative of an inverse function. This shows that E = e and that F(x) is the function ex as defined in the text. 118. Defining bx Let b > 0 and let f (x) = F(xG(b)) with F as in Exercise 117. Use Exercise 116 (t) to prove that f (r) = br for every rational number r. This gives us a way of defining bx for irrational x, namely bx = f (x). With this definition, f (x)= bx is a differentiable function of x (because F is differentiable).

7.4 Applications of Exponential and Logarithmic Functions In this section, we explore some applications involving exponential and logarithmic functions. To begin, consider a quantity P(t) that depends exponentially on time: P(t)

Po

ek

If k > 0, then P(t) grows exponentially and k is called the growth constant. Note that P0 is the initial size (the size at t = 0): P(0)

poek.0

Po

We can change the base and also write P(t) = Pobt with b = ek , because bt = (e k)t = ekt A quantity that decreases exponentially is said to have exponential decay. In this case, we write P(t) = poe-kt with k > 0; k is then called the decay constant. Population is a typical example of a quantity that grows exponentially under suitable conditions, particularly in initial stages of growth. To understand why, consider a cell colony with initial population Po = 100 and assume that each cell divides into two cells after 1 hour. Then population P(t) doubles with each passing hour:

P(0)

= 100 (initial population)

P(1) = 2(100) = 200

(population doubles)

P(2) = 2(200) = 400

(population doubles again)

After t hours, P(t) = (100)21.. EXAMPLE 1 During the Ebola virus outbreak in West Africa, from 2014 to 2016, there were approximately 750 cases reported up to July 1, 2014, and the number N of reported cases was increasing exponentially with a growth constant of approximately k = 0.019 (with time measured in days). Use N(t) = 750e°.°19t as a model for the number of reported cases t days after July 1, 2014, and answer the following: (a) What did the model indicate the number of reported cases would be 2 weeks later on July 15?

S [CT I 0 N 7.4

Applications of Exponential and Logarithmic Functions

373

(b) According to the model, how many days would it take for the number of reported cases to double from 750 to 1500? Solution (a) July 15 corresponds tot =- 14. At that time, N(14) = 750e0.019.14 750e0266 reported cases. (b) The question asks for the time t such that N(t) = 1500, so we solve 750e0.019r = 1500

=.

979

e 0.019t = 1500 = 2 750

Taking the natural logarithm of both sides, we obtain ln (e°.019t) = ln 2, or 0.019t = 1n2 2000 — _ 1500

t=

ln 2 0.019

36.5

Therefore, the model indicates that N(t) would have reached 1500 reported cases approximately 36.5 days after July 1 (Figure 1). •

1000 — 500 — 10

20

30

40

50

The number of reported cases N doubled in approximately 36.5 days.

FIGURE 1

In exponential growth, the time it takes a quantity to double is called the doubling time. It depends only on the growth constant k, and does not depend on the initial quantity. In general, 1n2 doubling time = — k Similarly in exponential decay, the half-life is the time it takes a quantity to decrease by half. If k is the decay constant, then half-life =

ln 0.5

We discuss doubling time and half-life further in Section 10.1. Exponential growth cannot continue over long periods of time. The number of Ebola cases in West Africa continued to grow exponentially until about the end of 2014, but through 2015 the growth slowed, and the number of reported cases leveled off at approximately 28,500. with M, A, k all positive, is a function that The logistic function P(t) = is used to model phenomena that have an initial rapid increase but then level off toward some finite value (Figure 2). EXAMPLE 2 Logistic Functions and the Carrying Capacity of the Earth From 1950 to 1980 to 2010 the human population of the earth grew from 2.65 to 4.45 to 6.90 billion people. Let t represent time in years since 1950 and P represent the population in billions. Using the (t, P) values for 1950, 1980, and 2010, and employing a computer algebra system to solve for M, A, and k in P(t), we obtain the logistic world-population model: FIGURE 2 The logistic function

P(t) =

P(t) = I +111 Ae- ki

17.4 1 + 5.56e -0.022t

Find lim P(t). Solution Since lim

e -0.022 t =

0, then

t-00

lim P(t) = urn r- >co t--*00 1

17.4 17.4 17.4 5.56e-0.022r = 1 ± 0 =

374

CHAPTER 7

EXPONENTIAL AND LOGARITHMIC FUNCTIONS We can interpret this limiting value of 17.4 billion as a theoretical carrying capacity of the

earth. Opinions vary on the earth's actual human carrying capacity, and many approaches have been taken to develop estimates. This simply derived estimate is consistent with the carrying capacity models that are presented in a United Nations Environment Program 2012 review.

Convention: Time t is measured in years and interest rates are given as yearly rates, either as a decimal or as a percentage. Thus, r = 0.05 corresponds to an interest rate of 5% per year. In Exercises 33-35, we investigate periodically compounded interest and how the Continuously Compounded Interest formula arises as a limiting case of periodic interest where the compounding period shrinks to zero.

In Chapter 10, we further examine applications involving exponential growth and decay, as well as others involving the logistic function. In that setting, we demonstrate how the relationships arise via differential equation models. Exponential functions are used extensively in financial calculations. Two basic applications are compound interest and present value. An investment that earns continuously compounded interest at a fixed rate grows exponentially according to the following: Continuously Compounded Interest If Po dollars are deposited into an account earning interest at an annual rate r, compounded continuously, then the value of the account after t years is P(t) = Poe"

EXAMPLE 3 A principal of Po = $100,000 is deposited into an account paying 6% interest compounded continuously. Find the balance after 3 years. Solution After 3 years, the balance is 100,000e(0.06)3 -•••-•-•$119,722.

•

The concept of present value (PV) is used in business and finance to compare payments made at different times. Assume that there is an interest rate r (continuously compounded) at which an investor can lend or borrow money. By definition, the PV of P dollars to be received t years in the future is Pe': The PV of P dollars received at time t is Pe—rt. What is the reasoning behind this definition? When you invest at the rate r for t years, your principal increases by the factor et, so if you invest Pe st dollars, your principal grows to (Pe— rt )ert = P dollars at time t. The present value Pe' is the amount you would have to invest today in order to have P dollars at time t. Note: The mathematics of interest rates is the same for all currencies (dollars, euros, pesos, yen, etc.).

EXAMPLE 4 With an interest rate of r = 0.03, is it better to receive €2000 (euros) today or €2200 in 2 years? Does the answer change if the interest rate is r = 0.07 instead? Solution We compare €2000 today with the PV of €2200 received in 2 years. • If r = 0.03, the PV is 2200e-0•03)2 €2071.88. This is more than €2000, so a payment of €2200 in 2 years is preferable to a €2000 payment today. ) • If r = 0.07, the PV is 2200e —0.072 €1912.59. This PV is less than €2000, so it is better to receive €2000 today if r = 0.07. An income stream is a sequence of periodic payments that continue over an interval of T years. Consider an investment that produces income at a rate of $800/year for 5 years. A total of $4000 is paid out over 5 years, but the PV of the income stream is less. For instance, if r = 0.06 and payments are made at the end of the year, then the PV is 800e-0.06 + 800e— 01°6)2 ± 800e—ao6)3 + 800e—ao6)4 + 800e—(006)5 ',-•z,-' $3353.12 It is more convenient mathematically to assume that payments are made continuously at a rate of R(t) dollars per year. We can then calculate PV as an integral. Divide the time

SECTION 7.4

Applications of Exponential and Logarithmic Functions

375

interval [0, T] into N subintervals of length At = TIN. If At is small, the amount paid out between time t and t At is approximately R(t) x Rate

At

= R(t)At

Time interval

The PV of this payment is approximately e—rt R(t)At. Setting t, -= i At, we obtain the approximation

E e—rti R(ti)At

PV of income stream ==-•-•

T This is a Riemann sum whose value approaches i 0 In April 1720, Isaac Newton doubled his money by investing in the South Sea Company, an English company set up to conduct trade with the West Indies and South America. Having gained 7000 pounds, Newton invested a second time, but like many others, he did not realize that the company was built on fraud and manipulation. In what became known as the South Sea Bubble, the stock lost 80% of its value, and the famous scientist suffered a loss of 20,000 pounds.

R(t)e ' t dt as At —> 0.

PV of an Income Stream If the interest rate is r, the present value of an income stream paying out R(t) dollars per year continuously for T years is PV = f 0

R(t)e—rt dt

EXAMPLE 5 An investment pays out 800,000 Mexican pesos per year, continuously for 5 years. Find the PV of the investment for r = 0.04 and r = 0.06. Solution In this case, R(t) = 800,000. If r = 0.04, the PV of the income stream is equal (in pesos) to e-0.04t 5

5 800,000e-004 dt = —800 000

0.04 0

—16,374,615 — (-20,000,000) = 3,625,385

If r = 0.06, the PV is equal (in pesos) to 5 800,000e—u6t dt = —

800000 e-0.06t 5 0.06

—9,877,576 — (-13,333,333) = 3,455,757

FIGURE 3 Exponential functions grow rapidly; consequently, their inverses, logarithmic functions, grow very slowly.

The Moment Magnitude Scale was developed to improve on some shortcomings in the more familiar Richter Scale. It was defined to be relatively consistent with Richter-Scale values for medium-to-strong earthquakes and is the scale currently used by the U.S. Geological Survey when reporting earthquake magnitudes.

•

Applications of Logarithmic Functions Since logarithmic functions are inverses of exponential functions (which grow rapidly), logarithmic functions grow slowly, a property that is shown in Figure 3. Scientists exploit this slow growth of logarithmic functions to define logarithmic scales that measure phenomena that can have a very large range of values, such as the energy in earthquakes. One scale for earthquakes is called the Moment Magnitude Scale, a logarithmic scale defined by 2 M„, = — logic, E — 10.7 3 where Al„, is the unitless moment magnitude of an earthquake and E is the energy (in ergs) released by the earthquake, which is directly related to the size of the earthquake fault and the distance the fault moved. EXAMPLE 6 Earthquake Measurement The 2011 Tohoku earthquake in Japan, one of the strongest ever recorded, released 3.9 x 1029 ergs of energy, about 10 billion times

376

CHAPTER 7

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

as much energy as a mild earthquake that rattles the windows in a house. What are the magnitudes of the mild (1021 ergs of energy) and Tohoku earthquakes? Solution If E = 1021, then 2 2 — (21) — 10.7 = 3.3 Mu, = — 3 logy, 1021 — 10.7 = 3 For the Tohoku earthquake, E = 3.9 x 1029, so 2 2 M,„ = — logio(3.9 x 1029) — 10.7 = — (log10 3.9 + 29) — 10.7 3 3

9.0

•

The log wind profile is a formula for determining wind speeds at different heights near the surface of the earth. Over open agricultural land with few buildings and obstructions, it is expressed as = 160

ln(h/0.03) V° ln(h0/0.03)

where v (in m/s) is the wind speed at height h (in m) and vo is the known wind speed at reference height ho. Meteorologists believe this relationship is accurate for heights up to 200 m (Figure 4). The value 0.03 is called a surface roughness factor. On different surfaces, this factor takes on different values. For example, 0.0002 is used over open water, and 0.4 is used over villages, forests, and rough uneven terrain.

120 80 40

12

16

Graph of v(h) with vo = 10 and hp = 10. The independent variable his plotted on the vertical to depict how v is changing with height. The shape of the curve reflects the profile of the wind speed as the height increases. FIGURE 4

EXAMPLE 7 Suppose vo = 10 m/s at ho = 10 m above a large corn field (thus with surface roughness factor 0.03). Determine v and dv Idh at h = 60 (a typical height of a wind-turbine tower). Solution With vo = 10 at Ito -= 10, it follows that ln(h /0.03) 101n(10/0.03)

1.721n

— 0.03

= 1.72(1n h — in 0.03)

1.721n h + 6.03

Therefore, v(h) = 1.72In h + 6.03

and

dv 1.72 —= dh h

Thus, at h = 60 m, we have v k•-•-•13.07 nVs and dvldh 0.03 m/s per m. This rate of change indicates that at that height, the wind speed is increasing at three-hundredths of a meter per second for each meter increase in height. •

Gravity 1

Air resistance

5 A ball acted on by gravity and air resistance.

FIGURE

In the next example, the natural logarithm appears in a relationship for the maximum height attained by a projectile when air resistance is taken into consideration. The relationship itself is derived via a differential equation model in Section 10.2. EXAMPLE 8 Maximum Height Under Air Resistance A 1-kg bocce ball is launched straight upward at 30 m/s (Figure 5) and is acted on by gravity and an air resistance force. We assume that the latter force is in the form —kv(t), where v(t) is the ball's upward velocity and k is a positive constant reflecting the strength of the air resistance (the stronger the air resistance, the greater the value of k). The maximum height (in meters) that the ball reaches depends on the strength of the air resistance and is given by H(k)= 30k — 9.81n (4191 + 1) k2 Values of k between 0 and 1 are physically reasonable. Intuitively, it makes sense that the stronger the air resistance, the lower the maximum height attained by the ball

S EC TION 7.4

Applications of Exponential and Logarithmic Functions

377

(Figure 6). What is lim H(k)? In other words, what happens to the maximum height with k—>0

less and less air resistance, and ultimately with none at all?

60

Solution This limit has the indeterminate form 0/0. We cannot evaluate it by algebraic simplification, so we will estimate it here by examining it numerically. In the next section, we introduce L'Hopital's Rule, which enables us to obtain an exact value for the limit. In the meantime, consider the following table of values:

40 20 I

I 0.4

I

0.8

1 .k

TABLE 1

FIGURE 6 The maximum height decreases as the air resistance (and k) increases.

H(k)

0.1

0.01

0.001

0.0001

0.00001

0.000001

0.0000001

38.2785

45.0023

45.8249

45.9090

45.9174

45.9173

45.9184

The values in Table 1 suggest the limit is approximately 45.92. We can say that as the air resistance vanishes, the maximum height that the ball attains approaches approximately 45.92 m. • Note that the result in the previous example is consistent with the maximum projectile height obtained in Example 7 in Section 3.4, where the same initial velocity was used and air resistance was not taken into consideration.

7.4 SUMMARY

y.

• Exponential growth with growth constant k > 0: P(t) = Poekt , doubling time = • Exponential decay with decay constant k > 0: P(t) = Poe— kt , half-life = ln(k0.5). • Logistic function with M, A, k positive: P(t) = 1+AMe-k„ which models initial rapid oo. growth that eventually levels off to a limit of M as t • Account value with an interest rate r, compounded continuously: P(t) = poen' • The present value (PV) of P dollars (or other currency), to be paid t years in the future: PV = P e' t • Present value of an income stream paying R(t) dollars per year continuously for T years: PV = f

R(t)e ' t dt

2 • Moment Magnitude Scale for earthquake measurement: Mu, -- 3 log10 E — 10.7. ln(hlr) • The log wind profile: v = vo ln(ho/r) ' where r is a surface roughness factor.

7.4 EXERCISES Preliminary Questions 1. Two quantities increase exponentially with growth constants k = 1.2 and k = 3.4, respectively. Which quantity doubles more rapidly?

(b) The amount you would have to invest today in order to receive N dollars at time T

2. A cell population grows exponentially beginning with one cell. Which takes longer: increasing from one to two cells or increasing from 15 million to 20 million cells?

5. In one year, you will be paid $1. Will the PV increase or decrease if the interest rate goes up at the start of the year?

3. For the logistic function P(t)=

M

1+Ae- id

, with k > 0, what is

lim P(t)? t->oo 4. The PV of N dollars received at time T is (choose the correct answer): (a) The value at time T of N dollars invested today

6. For y = log10 x, if y increases by 2, then (choose the correct answer): (a) x increases by 20 (b) x decreases by 20 (c) x increases by a factor of 2 (d) x increases by a factor of 100

378

CHAPTER 7

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Exercises 1. A certain population P of bacteria obeys the exponential growth law P(t) = 2000e1.3f (tin hours). (a) How many bacteria are present initially? (b) At what time will there be 10,000 bacteria? 2. A quantity P obeys the exponential growth law P(t) = e5t (tin years). (a) At what time t is P = 10? (b) At what time t is P = 20? (c) At what time t is P = 40? (d) How long does it take for P to double? 3. Write f(t)= 5(7)` in the form f (t)= Poekt for some Po and k. 4. Write f (t) = 9e1.4 in the form f (t) = Pobt for some Po and b. 5. A quantity P obeys the exponential growth law P(t) = Cekt (t in years). Find the formula for P(t), assuming that P(0) = 100 and that it takes 5 years for P to double. 6. A 10-kg quantity of a radioactive isotope decays to 3 kg after 17 years. Find the decay constant of the isotope. 7. Assuming that population growth is approximately exponential, which of the following two sets of data is most likely to represent the population (in millions) of a city over a 5-year period? Year

2000

2001

2002

2003

2004

Set 1 SethI

3.14 3.14

3.36 3.24

3.60 3.54

3.85 4.04

4.11 4.74

8. In 2009, 2012, and 2015, the number (in millions) of smart phones sold in the world was 172.4, 680.1, and 1423.9, respectively. (a) (CAS Let t represent time in years since 2009, and let S represent the number of smart phones sold in millions. Determine M, A, and k for a M , that fits the given data points. logistic model, S(t) (b) What is the long-term expected maximum number of smart phones sold annually? That is, what is lim S(t)? t-oo (c) In what year does the model predict that smart-phone sales will reach 98% of the expected maximum?

14. Compute the PV of $5000 received in 3 years if the interest rate is (a) 6% and (b) 11%. What is the PV in these two cases if the sum is instead received in 5 years? 15. Is it better to receive $1000 today or $1300 in 4 years? Consider r = 0.08 and r = 0.03. 16. Find the PV of an investment that pays out continuously at a rate of $800/year for 5 years, assuming r = 0.08. 17. Find the PV of an income stream that pays out continuously at a rate R(t) = $5000e0 It/year for 7 years, assuming r = 0.05. 18. The decibel level for the intensity of a sound is a logarithmic scale defined by D = 10 log 10 / -I- 120, where I is the intensity of the sound in watts per square meter. (a) Express I as a function of D. (b) Show that when D increases by 20, the intensity increases by a factor of 100. (c) Compute dlIdD.

log io

E - 10.7, relating the moment 19. Consider the equation M. = magnitude of an earthquake and the energy E (in ergs) released by it. (a) Express E as a function of M. (b) Show that when M. increases by 1, the energy increases by a factor of approximately 31.6. (c) Compute dEldM.. 20. Over rough uneven terrain, the log wind profile is expressed as V - vo

ln(h/0.4) ln(h0 /0.4)

With vo = 10 m/s at ho = 10 m, determine v and dv/dh at h = 60. 21. Over open water, the log wind profile is expressed as = vo

ln(h /0.0002) ln(h0/0.0002)

With vo = 10 m/s at ho = 10 m, determine v and dvldh at h = 60. 22. The Palermo Technical Impact Hazard Scale P is used to quantify the risk associated with the impact of an asteroid colliding with the earth: (

9. Sam was 28 inches tall on her first birthday, 50 inches tall on her eighth, and 62 inches tall on her 14th. (a) CAS Let t represent Sam's age in years, and let h represent her height in inches. Determine M, A, and k for a logistic model, h(t) = 1+ Ae-k"

that fits the given height data.

(b) What is Sam's long-term expected height? That is, what is lim h(t)? (c) At what age does the model predict that Sam will reach 95% of her expected maximum height? 10. Suppose $500 is deposited into an account paying interest at a rate of 7%, continuously compounded. Find a formula for the value of the account at time t. What is the value of the account after 3 years? 11. How long will it take for $4000 to double in value if it is deposited in an account bearing 7% interest, continuously compounded? 12. How much must one invest today in order to receive $20,000 after 5 years if interest is compounded continuously at the rate r = 9%? 13. An investment increases in value at a continuously compounded rate of 9%. How large must the initial investment be in order to build up a value of $50,000 over a 7-year period?

P = log10

p

i

Ea8

0.03T

where P1 is the probability of impact, T is the number of years until impact, and E is the energy of impact (in megatons of TNT). The risk is greater than a random event of similar magnitude if P > 0. (a) Calculate dP/dT, assuming that pi = 2 x 10-5 and E = 2 megatons. (b) Use the derivative to estimate the change in P if T increases from 8 to 9 years. In Exercises 23 and 24, a ball is launched straight up in the air and is acted on by air resistance and gravity as in Example 8. The function H gives the maximum height that the ball attains as a function of the air-resistance parameter k. In each case, estimate the maximum height without air resistance by investigating limo H(k) numerically. 23. A ball with a mass of 1 kg is launched upward with an initial velocity of 60 m/s, and H(k) -

60k - 9.8 ln( 349°/`- + 1) k2

S ECT I 0 N 7.4

24. A ball with a mass of 500 g is launched upward with an initial velocity of 30 m/s, and H(k) =

Applications of Exponential and Logarithmic Functions

379

Exercises 28 and 29: The Gompertz differential equation dy iy = kyln dt

(:")/` + 1) 15k — 2.451n( 49 k2

25. The Beer—Lambert Law is used in spectroscopy to determine the molar absorptivity a > 0 or the concentration c > 0 of a compound dissolved in a solution at low concentrations (Figure 7). The law states that the intensity I of light as it passes through the solution satisfies Ingo / /) = acx, where /0 is the initial intensity and x is the distance traveled by the light. Show that I decays exponentially as a function of x. Intensity /

(where M and k are constants) was introduced in 1825 by the English mathematician Benjamin Gompertz and is still used today to model aging and mortality. 28. Show that y = Meaek' satisfies Eq. (2) for any constant a. 29. To model mortality in a population of 200 laboratory rats, a scientist assumes that the number P(t) of rats alive at time t (in months) satisfies Eq. (2) with M -= 204 and k = 0.15 (Figure 8). Find P(t) [note that P(0) = 2001 and determine the population after 20 months.

Rat population P(t) Distance

200

100 1 .0 0 FIGURE 7 Light of intensity /0 passing through a solution.

10 20 30 Time (months)

40

FIGURE 8

26. Consider the logistic world population model from Example 2: P(t) = 17.4 1+5.56,-0.022r •

(a) Determine the time t when the population is half of the expected longterm limit of 17.4 billion. (b) There is a single point of inflection for P. Determine it. [Note: You should find that the value of t is the same for both (a) and (b). This is not a coincidence; see the next exercise.] 27. Consider the general logistic function P(t)= 1 -1- M Ae—k" with A' 111, and k all positive. Show that mAk2,--kt me-kr_ (a) pl ( t ) M Ake kr and P"(t)— ' (1+ Ae- kt)3 ' (1+ Ae-1 )2 (b) lim P(t)= 0 and lim P(t) = M, and therefore P = 0 and P = M are horizontal asymptotes of P. (c) P is increasing for all t. (d) The only inflection point of P is at (it kA , tAn To the left of it, P is ). concave up, and to the right of it, P is concave down.

30. A company can earn additional profits of $500,000/year for 5 years by investing $2 million to upgrade its factory. Is the investment worthwhile if the interest rate is 6%? (Assume the savings are received as a lump sum at the end of each year.) 31. A new computer system costing $25,000 will reduce labor costs by $7000/year for 5 years. (a) Is it a good investment if r = 8%? (b) How much money will the company actually save? 7 Banker's Rule of 70 If you earn an interest rate of R per32. E4 cent, continuously compounded, your money doubles after approximately 70/R years. For example, at R = 5%, your money doubles after 70/5 or 14 years. Use the concept of doubling time to justify the Banker's Rule. (Note: Sometimes, the rule 72/R is used. It is less accurate but easier to apply because 72 is divisible by more numbers than 70.)

Further Insights and Challenges Periodically Compounded Interest: When interest is compounded in an account n times in a year (rather than continuously), the annual interest rate r is divided by n, and the amount paid each time that interest is compounded is A (0, where A is the amount that is in the account at the time the interest is compounded. 33. (a) Show that at an annual rate of r, if interest is compounded n times in a year, then the amount in the account at the end of the year is , where A was the amount in the account at the start of the A (1 + year. (b) Show that if Ao is initially invested in an account that pays annual interest r, compounded n times yearly, then the amount in the account after ;) t years is A(t) = Ao (1 + nt .

34. Suppose that $1000 is invested in an account that pays annual interest of 2.4%. How much is in the account after 1 year if the interest is compounded monthly? Continuously? 35. We will show that continuously compounded interest is a limiting case, as n oo, of periodically compounded interest. We first establish an important limit. (a) Show that ln(1 + r In) is the area under the graph of y = 1/x from 1 to 1 + dn. (b) Using rectangles with base length r n, prove that n < ln(1 + rIn)< tin 1 + tin

380

CHAPTER 7

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

(e) Prove that lirn (1 + r/n)' = er.

(c) Prove that < ln((l+ r/n)n) a g(x) Roughly speaking, L'Hopital's Rule states that when f (x)Ig(x) has an indeterminate form of type 0/0 or co I oo at x = a, then we can replace f (x)I g(x) by the quotient of the derivatives f (x)Ig'(x). THEOREM 1 L'Hapital's Rule interval containing a and that

Assume that f and g are differentiable on an open f (a) = g(a) = 0

Also assume that g'(x)

0 (except possibly at a). Then f (x) f '(x) hm—=lim x—>a g(x) x-->a g'(x)

if the limit on the right exists or is infinite (oo or —oo). This conclusion also holds if f and g are differentiable for x near (but not equal to) a and CAUTION When using L'Hopital's Rule, be sure to take the derivative of the numerator and denominator separately: f (x) f'(x) lim = lim x-->a g(x) x—>a (x) Do not take the derivative of the function y = f (x)/g(x) as a quotient, for example, using the Quotient Rule.

lim f (x) = ±oo

and

lim g(x) = ±oo

x-->a

Furthermore, this rule is valid for one-sided limits. X3 - 8

EXAMPLE 1 Use L'Hopital's Rule to evaluate lim

x-+2 x 4 + 2x — 20

Solution Let f (x) = x3 — 8 and g(x) = x4 + 2x — 20. Both f and g are differentiable and f (x)/ g(x) is indeterminate of type 0/0 at a = 2 because f(2) = g(2) = 0. Furthermore, gi (x) = 4x3 + 2 is nonzero near x = 2, so L'Hopitar s Rule applies. We may replace the numerator and denominator by their derivatives to obtain lim

X3 - 8

x->2 X4 ± 2X - 2

(x3 — 8)'

= lim

x--).2 (X4 + 2X - 2)'

• = lim

3x2

3(22)

L'Hopital's Rule COS2 X

EXAMPLE 2 Evaluate lirn x--›-7/12 1 — sin x Solution Again, the quotient is indeterminate of type 0/0 at x =C0S2 ( ±7 ) = 0,

2

12

6

4X3 + 2 — 4(23) -F. 2 — 34 = 17 II

1 — sin — = 1 — 1 = 0

2

since

SECT I ON 7.5

L'HOpital's Rule

381

The other hypotheses are satisfied, so we may apply L'Hopital's Rule: urn

2 COS X

(COS2 x)'

= lirn

x—>7/2 1 — sinx x —>n-/2 (1 — . -,,L'Hopital's Rule

sinx)'

-2 cos x sinx

= lim

x —>7r/2

— cosx

= lim (2 sin x) = 2 Simplified

-2 cos x sinx . is also indeterminate at x = 7r/2. We removed this - cos x indeterminacy by cancelling the factor - cos x. Note that the quotient

EXAMPLE 3

The Form 0 • oo

Evaluate lim x lnx. x->o+

Solution This limit is one-sided because f (x) = x ln x is not defined for x o. So f (x) presents an indeterminate form of type 0 • oo. To apply L'Hopital's Rule, we rewrite our function as f (x) = (ln x)/x-1 so that f (x) presents an indeterminate form of type -oo/oo. Then L'Hopital's Rule applies: lnx (ln x)' x-1 = x-44-\__ lim lim x ln x = lim — = lim = lim (-x) = 0 2/ x->o+ (x-1)' x-÷o+ x-1 x->o+ x—>0+ Simplified

L'Hopital's Rule

EXAMPLE 4

Using L'Hopital's Rule Twice

•

Evaluate lim

ex - x - 1 1

x—>0 cosx —

Solution The limit is in the indeterminate form 0/0 since at x = 0, we have ex -x-1=e

° -0-1

-=0,

cos x - 1 = cos 0 - 1 = 0

A first application of L'Hopital's Rule gives

1 - ex

e- 1

(ex - x - 1)' ex - x - 1 = lim = lim = lim urn x—>0 sinx x->o cosx - 1 x->o (cosx - 1)' x->o (- sinx )

This limit is again indeterminate of type 0/0, so we apply L'Hopital's Rule a second time: -ex -e° 1 ex = = lun = x—>0 sinx .)c-4 cos x cos() lim

1

It follows that ex - x - 1 lim =-1 x-4 cos x - 1

•

EXAMPLE 5 Maximum Height Under Air Resistance section, we introduced a function H (k) =

30k - 9.8 ln (1_50k 49

In Example 8 in the previous

_L 1) '

k2

that gives the maximum height attained by a 1-kg ball launched upward at 30 m/s with gravity and air resistance acting on it. (The function is derived in Section 10.2.) The variable k reflects the strength of the air resistance. We investigated what happens to the maximum height as the air resistance approaches zero; that is, we investigated lim H(k) k—>0

numerically. Show this limit can be evaluated using L'Hopital's Rule and find the limit.

382

CHAPTER 7

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Solution The quotient

30k —9.8 ln( W` +1)

has the indeterminate form 0/0. To evaluate the

k2

limit, we need to use L'Hopital's Rule twice: 4500/49

lirn

30k - 9.81n (150k 49

k—> 0

30

'

=19_ 1503k0+1

= lim

2k

k—> 0

k2

( 1r

= lim

k—> 0

+1)2

2

2250 49

45.92 •

This value of 45.92 m matches our numerical estimate in Example 8 in the previous section and the result we obtained separately in Example 7 in Section 3.4, where we considered the launched projectile's height, ignoring air resistance altogether.

EXAMPLE 6

Assumptions Matter

Can L'Hopital's Rule be applied to lirn

x2 +1 2x + 1

Solution The answer is no. The function does not have an indeterminate form because X

2

+ 1

2x + 1

12 +1

2

2 1+1

3

x2 + 1 2 This limit can be evaluated directly by substitution: lim = . An incorrect apx->1 2x + 1 3 plication of L'Hopital's Rule gives the wrong answer: lim

x-*1

EXAMPLE 7

(x2 + 1)' = lim =1 (2x + 1)' x—>1 2

(not equal to original limit)

•

1 1 Evaluate lim . - - . x-A) ( sin x x

The Form Do - oo

Solution Both 1/ sin x and 1/x become infinite at x = 0, so we have an indeterminate form of type oo - oo. We rewrite the function as 1

1

x - sin x

sin x

x

x sin x

to obtain an indeterminate form of type 0/0. Applying L'Hopitar s Rule twice yields lim x-o

1 sin x

1 x - sin x 1 - cos x - - ) = lim = lim x x sin x x cos x + sin x L'HOpites Rule

= lim

x—>0 —x

sin x 0 = =0 sin x + 2 cos x 2

L'Hopital's Rule again

The graph confirms that 1 1 y= . approaches 0 as x —> O. sin x x FIGURE 1

41.0 REMINDER The change-of-base formula, changing an exponential base a to base e, is a' = er Ina .

This value of the limit is confirmed graphically in Figure 1. Limits of functions of the form f (x)g(x) can lead to the indeterminate forms 00 , 100, or oo°. These are indeterminate since the limit can take on a variety of values, depending on the relative rates at which the base and exponent approach their limits. In evaluating these limits, we use the change-of-base formula to write f (x)g(x) , eg(x)ln f (x) and then we obtain lim f(x)g(x) =- lim eg(x)ln f (x)

x-± a

x-+.2

=

lim g(x)ln f(x)

The last equality is justified by the continuity of the exponential function.

7.5

SECTION

EXAMPLE 8

The Form 00

L'Hopitars Rule

383

Evaluate lirn xx.

Solution With x x = ex in x by the change-of-base formula, it will be enough to consider the limit of x In x. Example 3 showed litn x In x = 0. Therefore, x,o+ urnxx = urne x—>0+ x—>0+

lirn x In x lnx = ex-.± = e0 = ,

This value for the limit is confirmed graphically in Figure 2. FIGURE 2 The function y 1 as x

= xx

approaches

•

In Section 1 in this chapter, we pointed out that e is the value that (1 + x)1/x approaches as x approaches 0. This can be verified now by evaluating lim (1 + x)l /x using x—>0

L'Hopital's Rule. EXAMPLE 9 The Form 1°°

Find lim (1 ± x)14.

Solution This has the indeterminate form 1". We take the approach used in Example 8. Thus, we write (1 + x)14 = e

In(l+x)

and consider lim

x —A x

L'Hopital's Rule for the first equality) . x—>o

ln(1

x). We obtain (using

1n(1+x) • i-Fx = 1 — x —A 1 x

Therefore, ELM (1 + X )1/x = lim eTi ln(l+x)

urn

ex—>0

In(14-x) x =

e =e

•

r into lim (1 ± we obtain the limit in the pret = e. It is important to be familiar with these vious example. Therefore, lim (1 ± r—> 00 limits whose values are e: Note that if we substitute x =

e= lim (1 x—>0

x)1 x

and

e= lim (1

1 —)

They arise in limit evaluations that we will see subsequently in the text.

CONCEPTUAL INSIGHT Exponential Limit Forms Knowing that 0 • oo is an indeterminate n we can see why 00, 100 , and oo° form, and using the exponential identity a x = ex ma are indeterminate forms. A similar approach also shows why 0" is not indeterminate and corresponds to a limit that equals 0. 0. The Form 00: If lim f(x)g(x) is in the form 00, then f (x) —> 0 and g(x) x—o-a g(x) In f (x) has an expoTherefore, in the limit, the equivalent exponential expression nent in the indeterminate form 0(—oo) since g(x) —> 0 and In f (x) —> —oo [because f (x) —> 0]. Therefore, 00 is an indeterminate form. Similar arguments can be made to demonstrate that 1" and oo° are indeterminate forms (see Exercise 57). The Form 000: If lim f (x)g(x) is in the form 0", then f (x) -4- 0 and g(x) —> oo. x —>a l f (x) has an expoTherefore, in the limit, the equivalent exponential expression eg(x)n nent in the form (oo)(—oo). Since the limit of the exponent is —oo, it follows that the limit of eg(x)ln f(x) is 0, and therefore the limit of f (x)ex) is as well. Thus, the form 0" is not indeterminate but instead corresponds to a limit that is equal to 0.

384

CHAPTER 7

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

Comparing Growth of Functions Sometimes, we are interested in determining which of two given functions grows faster. For example, Quick Sort and Bubble Sort are two standard computer algorithms for sorting data (e.g., alphabetizing, ordering according to rank). The average time required to sort a list of size n is approxiamtely n ln n for Quick Sort and n2 for Bubble Sort. Which algorithm is faster when the size n is large? This problem amounts to comparing the oo. growth of Q(x) = x ln x and B(x) = x2 as x We say that f (x) grows faster than g(x) if f (x) lim — = oo x—*oo g(x)

or, equivalently,

g(x) lim =0 x—>00 f (x)

To indicate that f (x) grows faster than g(x), we use the notation g(x) oo x-›00 not exist. Do (a) and (b) contradict L'Hopital's Rule? Explain.

(x)/ g'(x) does

ln(1 bx) 68. Let H(b)= lim for b > 0. x-›00 x (a) Show that H(b) = ln b if b> 1. (b) Determine H(b) for 0 0. (a) Calculate lim f(x) and lim f (x). x->00 x-÷o+ (b) Find the maximum value off and determine the intervals on which f is increasing or decreasing. = c has 60. (a) Use the results of Exercise 59 to prove that x a unique solution if 0 < c < 1 or c = e h/e, has two solutions if 1 cc < elle, and has no solutions if c > elle. (b) COTO Plot the graph of f (x)= x i lx and verify that it confirms the conclusions of (a). 61. Determine whether f 0 and is a local maximum if f (c) 0 for x 0 a, so f (x) - f (c) must be positive for x near c if (x f (n)(c) > 0. (c) Use (a) to show that if n is odd, then f (c) is neither a local minimum nor a local maximum. 76. When a spring with natural frequency A/27r is driven with a sinusoidal force sin(wt) with co A, it oscillates according to y(t) = Let yo(t) = urn y(t).

1

(A sin(an) -

w sin(Aa))

388

CHAPTER 7

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

f (x) =

(a) Use L'Iropital's Rule to determine yo(t). (b) Show that yo(t) ceases to be periodic and that its amplitude I yo(t)I co (the system is said to be in resonance; eventually, tends to oo as t the spring is stretched beyond its structural tolerance).

- a).fi(x),

g(x)= (x - 0)81(x)

Use this to verify L'Hopital's Rule directly for lim f(x)Ig(x). 79. Patience Required Use L'Hopital's Rule to evaluate and check your answers numerically: ) I /x2 1 1 sin x (b) lim (a) lim — x-+) sin2 x x2 ) x->o+ x

(c) (CAS Plot y for A = 1 and to = 0.8, 0.9, 0.99, and 0.999. Do the graphs confirm your conclusion in (b)? in We expended a lot of effort to evaluate lim x->o x Chapter 2. Show that we could have evaluated it easily using L'Hopital's Rule. Then explain why this method would involve circular reasoning. 77. al

80. In the following cases, check that x = c is a critical point and use Exercise 75 to determine whether f (c) is a local minimum or a local maximum. (a) f (x) = x5 - 6x4 14x3 - 16x2 ± 9x + 12 (c= 1)

78. By a fact from algebra, if f, g are polynomials such that f(a)= g(a)= 0, then there are polynomials fi , gi such that

(b) f (x) = x6 - x3

(c = 0)

7.6 Inverse Trigonometric Functions

Do not confuse the inverse 5in-1 x with the reciprocal

(sin x)

1 = — = csc x sin x

The inverse functions sin-1 x, cos-I x,. . . are often denoted arcsinx, arccosx, and so on.

In this section, we introduce the inverse trigonometric functions and their derivatives, and we examine some integrals involving these functions. We have seen that an inverse function f -1 exists if and only if f is one-to-one on its domain. Because the trigonometric functions are not one-to-one, we must restrict their domains to define their inverses. First, consider the sine function. Figure 1 shows that f(9) = sin e is one-to-one on [—IL` 2 , 72 ] • With this interval as domain, the inverse is called the arcsine function and is denoted 0 = sin-1 x or 0 = arcsin x. By definition, 7 rr 0 = sin-1 x is the unique angle in [ - - , 2 2

such that sin 9 = x

9 sin 0 with restricted domain FIGURE 1

The range of f (x) = sin x is [-1, 1], so f gives some values of 0 = sin-1 x. Summary of inverse relation between the sine and arcsine functions: sin(sin—I x) = x sin—I (sin 0) =

TABLE 1 -1

for —1 < x < 1 7r

for-- 0). We let f(x) = cosh-1 x and f(x) = sech- I x denote the corresponding inverses (Figure 5). In reading the following table, keep in mind that the domain of the inverse is equal to the range of the function.

SECTION 7.7

Hyperbolic Functions

399

Inverse Hyperbolic Functions and Their Derivatives

The graphs of y = cosh-I x and y = sech-I x have a vertical tangent at the endpoint x = 1 of their domains and therefore the derivative is undefined there.

Function

Domain

y = sinh-1x

all x

y = cosh-1 x

x> 1

y = tanh 1 x

lx1 < 1

y = coth-1 x

jxj > 1

y = sech-I x

0 1 and b > 0,

First use Integration by Parts. Then assume f is increasing. Use the substitution u = f(x) to prove that f

1(a,b)=

xf'(x)dx is equal to the area of the

shaded region in Figure 1 and derive Eq. (8) a second time.

f' (x)2 dx

a b+1

1(a

1, b +1)

(d) Use (b) and (c) to calculate 1(1,1) and 1(3,2). (e) Show that 1(a, b) =

a! b! (a+b+1)!

97. Let In = f xn cos(x2) dx and J = fxn sin(x2)dx. (a) Find a reduction formula that expresses In in terms of Write xn cos(x2) as xn-1(x cos(x2)).

A_2.

Hint:

(b) 1- 7 .1 Use the result of (a) to show that I, can be evaluated explicitly if n is odd. FIGURE 1

(c) Evaluate /3.

8.2 Trigonometric Integrals In this section, we investigate integrals of various products of powers of trigonometric functions. We can often compute these integrals by combining substitution and Integration by Parts with trigonometric identities. In the section summary we expand on the

414

CHAPTER 8

TECHNIQUES OF INTEGRATION

table of integrals we have built so far, adding integral formulas employed or derived ir this section and some other formulas similar to them. We begin with integrals of the form

f

sin"' x cos' x dx

where m,n are whole numbers. The easier case is when at least one of m,n is odd. EXAMPLE 1 Odd Power of sin x

Evaluate fsin 3 x dx.

Solution We did this integral in Example 6 of the last section. However, we will use a different method that is more broadly applicable. Because sin3 x is an odd power, we split off one power of sin x and use the identity sin2 x = 1 — c0s2 x to convert the rest of the integrand into an expression in cos x: sin3 x = (sin2 x)(sin x) = (1 — c0s2 x) sin x We then use the substitution u = cos x, du = — sin x dx:

f

sin3 x dx = f (1 — cos2 x) sin x dx = — f (1 — u2) du u3 = --u+C=

COS3 X

•

cos x + C

3

3

The strategy of the previous example works when sin"' x appears with m odd, no matter what power of cos x is present. Similarly, if n is odd, we write cos" x as a power of (1 — sin2 x) times cos x. EXAMPLE 2

Odd Power of cos x

Evaluate

J

sin4 x c0s5 x dx.

Solution We take advantage of the fact that c0s5 x is an odd power to write sin4 x cos5 x = sin4 x cos4 x (cos x) = sin4 x(1 — sin2 x)2(cos x) (sin4 x — 2 sin6 x + sin8 x) cos x This allows us to use the substitution u = sin x, du = cos x dx:

f

sin4 x cos5 x dx =

(sin4 x — 2 sin6 x + sin8 x) cos x dx

= f (u4 — 2u6 + u8) du u5 5

2u7 7

+

u9 9

+C=

sin5 x

2 sin7 x

sin9 x

5

7

9

+C

We will need a different strategy when neither sin x nor cos x appears with an odd power. EXAMPLE 3 Evaluate f sin2 x dx Solution We utilize the trigonometric identity called the double angle formula sin2 x = I2 (1 — cos 2x). Then f sin2 x dx =

f

1 x —(1 — cos 2x)dx = 2 2

sin 2x +C 4

•

—

SECTION 8.2

4.00 REMINDER Useful identities:

cos

X =

415

Using the trigonometric identities in the margin, we can also integrate c0s2 x, obtaining the following:

1 sin2 x = —(1 — cos 2x) 2 2

Trigonometric Integrals

f

1 - (1 + cos 2x) 2

sin 2x = 2 sin x cos x

sin2 x dx = x 2

sin 2x x 1 + C = — — — sin x cos x + C 4 2 2

x , cos2 x ax =

sin 2x 4

2

x 1 C= — 2 + —2sin x cos x

C

2

cos 2x = cos2 x — sin2 x

EXAMPLE 4 Evaluate f sin4 x dx

As an alternative, the reduction formula for

Solution Double angle formulas will be of assistance to us here as well. Using sin2 x = (1 — cos 2x), we obtain

sin x dx, Eq.(5) in the previous J section, can be used to compute

f

f sin4 x dx = f (sin2 x)2 dx = f

sin2 x dx and f sin4 x dx.

Integrating sinm x cos' x

Applying the double angle formula to c0s2 2x, we get

Case 1:m odd

sin4 x dx =—J(1 — 2 cos 2x + c052 2x)dx 4 =

Either use the double angle formulas repeatedly or convert the integrand into an expression entirely in terms of sin x or cos x and then apply a reduction formula, either Eq. (5) or Eq. (6), in the integral table at the end of this section.

(1 — 2 cos 2x +

4

= 4

Case 2: n odd

Case 3: m, n both even

dx

1 = — f (1 — 2 cos 2x + cos2 2x) dx 4

Split off one power of sin x, and use sin2 x = 1 — c0s2 x to express the remaining powers of sin x in terms of cos x. Then substitute u = cos x, du = — sin x dx.

Split off one power of cos x, and use cos2 x = 1 — sin2 x to express the remaining powers of cos x in terms of sin x. Then substitute u = sin x, du = cos x dx.

2

1

—(1 — cos 2x) 2

f

2

1 + cos 4x ) dx 2

— 2 cos 2x + cc's 2

dx

By simple u-substitutions, this yields sin4 x dx =

3x sin 4x \ ( T. — sin 2x + 8 ) + C = 3x 8

sin 2x 4

sin 4x +C 32

•

As indicated in the above margin comment, the reduction formula for fsin" xdx can be used for the previous two examples. There is also a corresponding reduction formula for

cosn xdx. These two reduction formulas are (5) and (6) in the integral table

at the end of this section. We employ the reduction formula for fcosn xdx in the next example. EXAMPLE 5

Even Powers of sinx and cos x Evaluate

sin2 x cos4 x dx.

Solution Here, m =- 2 and n = 4. Since m 0. EXAMPLE 9 Evaluate Jt an2 xsec4 x dx. Solution In this case, since the derivative of tan x is sec2 x, we separate out sec2 x and convert the rest of the integrand into powers of tan x. Using the fact sec2 x = tan2 x + 1 yields

f

tan2 x sec4 x dx = f (tan2 x)(tan2 x + 1)(sec2 x)dx =

(tan4 x + tan2 x) sec2 x dx

Setting u = tan x and du = sec2 x dx, we have u5 u3 tan5 x f tan x sec4 x dx = f (u4 + u2)du = — + — + C = 5

3

5

tan3 x +C 3

•

The above method works to integrate tan"' x sec' x whenever n > 0 is even. The last case to consider is when m is even and n is odd. In that case we convert the integrand to be entirely in terms of powers of sec x and then apply reduction formulas. Examples of this case can be found in the exercises. CONCEPTUAL INSIGHT We previously mentioned that different methods for evaluating an indefinite integral may yield different expressions. Using identities and simplification, we can show that the results are equivalent. Keep in mind that

f

f(x)dx = F(x)+ C

and

f f(x)dx = G(x) + C

do not imply that F(x) = G(x). Instead, they indicate that the functions F and G differ by a constant. To verify that two such antiderivative results are equivalent, you need to show that F(x) = G(x)+ C for some constant C. For example, separately using the substitutions u = sin x and u = cos x, we obtain the results

f

2 sin x cos x dx = sin2 x + C

and

2 sin x cos x dx = — c052 x + C

Of course, sin2 x — c052 x. However, since sin2 x = — cos2 x + 1 it follows that the results are equivalent. Specifically, f 2 sin x cos x dx = sin2 x + C = — cos2 x + 1 + C = — cos2 x + C where, in the last equality, we express 1 + C as an arbitrary constant C.

418

CHAPTER 8

TECHNIQUES OF INTEGRATION

The formulas in Eqs. (15)—(17) in the integral table at the end of the section give the integrals of the products sin mx sin nx, cos mx cos nx, and sin mx cos nx. These integrals appear in the theory of Fourier Series, which is a fundamental technique used extensively in engineering and physics. 7

EXAMPLE 10 Integral of sin mx cos nx

Evaluate f

sin 4x cos3x dx.

Jo Solution Apply Eq. (16), with m = 4 and n = 3: Jr

fo

sin 4x cos 3x dx =

(

cos(4 + 3)x \ Jr 2(4 + 3) ) 0

cos(4 — 3)x 2(4 — 3) COS X

COS

2

7x

7

14 ) 0 ( 1 1)

1 \ ± 14 )

2

14 )

8

•

7

The following table of trigonometric integrals summarizes some of the integral formulas we have seen in this chapter and includes some other related formulas. TABLE OF TRIGONOMETRIC INTEGRALS f sin2 x dx =

x 2

sin 2x x 1 +C = — 2 — —2sin x cos x + C 4

-

f coS2 x dx = x + sin 2x + C = x + 1 sin x cos x + C —2 — 2 2 4 sinn-1 Jr cos x n—1f sinn- 2 x dx + n n

I sin' x dx = I cosn x dx =

f

tanm-1 x m—1

f

anm-2 x dx

8

r

cot x dx = — ln Icsc x I + C = ln 'sin x I + C m"1-11 x cot —

f cot m-2 x dx

11

tan x seen-2 x m—2 ± m—1 m—1

sec m-2 x dx

12 13

csc x dx = in Icsc x — cot x I + C

m —2 f cot x csc m-2 x 4csc m-2 x dx m—1 m—1 sin(m + n)x sin(m — n)x ± C (m 0 ±n) I sin mx sin nx dx = 2(m — n) 2(m + n)

I csc m x dx =

f

sin mx cos nx dx =

I cos mx cos nx dx =

cos(m — n)x 2(m — n)

9

10

sec x dx = ln I sec x + tan x I + C

I sec in x dx =

f

6

tan x dx = ln Isec x I + C =

f cot m x dx =

f

5

cosn-1 x sin x n-1 f cosn-2 x dx + n n

f tanm x dx =

f

4

cos(m + n)x +C 2(m +n)

sin(m + n)x sin(m — n)x +C + 2(m + n) 2(m — n)

(nz = ±n) (m 0 ±n)

14 15 16 17

SECTION 8.2

Trigonometric Integrals

419

8.2 SUMMARY • The integral f sin"' x cos' x dx can be evaluated as in the marginal note on page 415. • The integral f tanm x sec° x dx can be evaluated as in the marginal note on page 417.

8.2 EXERCISES Preliminary Questions 1. Describe the technique used to evaluate fsin 5 x dx.

4. Describe a way of evaluating fsin 6 x cos2 x dx. 5. Which integral requires more work to evaluate?

2. Describe a way of evaluating

sin6 x dx.

3. Are reduction formulas needed to evaluate why not?

f

sin7 x c0s2 x dx? Why or

sin798 x cos x dx

or

f sin4 x c054 x dx

Explain your answer.

Exercises In Exercises 1-6, evaluate the integral.

27.

1.

f c053 x dx

2.

i sin5 x dx

3.

fsin 3 0 c0s2 0 d0

4.

i sin5 x cos x dx

5.

fsin 3 t cos3 t dt

6.

fsin 2 x c0s5 x dx

7,

Compute the area under the graph of y = cos3 x from x = 0 to x = S

8.

Compute the area under the graph of y = sin5 x from x = 0 to x =

29.

f

COSS X

sin3 x

dx

f csc2(3 - 2x )dx

28.

f

sin7 x dx cos4 x

30. f csc3 x dx

31. f tan x sec2 x dx

32. f tan3 0 sec3 0 dB

33. f tan5 x sec4 x dx

34. f tan4 x sec x dx

35. f tan6 x sec4 x dx

36. f tan2 x sec3 x dx

37. f cot 5 x csc 5 x dx

38. f cot 2 x csc 4 x dx

39. f sin 2x cos 2x dx

40. f cos 4x cos 6x dx

41. f sin 2x cos3 x dx

42. f sin2 x sec4 x dx

14. f csc4 x dx

43. f t cos3(t2)dt

44.

16. f cot 5 X CSC x dx

45. f cos2(sin t) cost dt

46. f ex tan2(ex)dx

In Exercises 9-12, evaluate the integrals. 9.

f tan3 X sec x dx

11. f tan2 x sec4 x dx

10. f tan x sec3 x dx 12. f tan8 x sec2 x dx

In Exercises 13-16, evaluate using methods similar to those that apply to the integrals of tanm x seen x. 13. f 15. f

COt 3 x dx

COt 5 X CSC2 x dx

f tan3(In t)

17. Compute the area under the graph of y = tan2 x from x = 0 to x =

In Exercises 47-60, evaluate the definite integral.

18. Compute the area under the graph of y = sec4 x from x = 0 to x= .

47.

In Exercises 19-46, evaluate the integral. 19. f sin6 x dx 21. f cos5 x sin x dx 23. f c0s4(3x + 2) dx 25.

cos3(nO) sin4(70) dO

7r/2

27r

Jo

Jo

sin2 x dx

48.

sin5 x dx

50.

20. f sin2 x c0s2 x dx

dx cos x

dx 52. f 74. sin x

22. 24. 26.

f f

tan x dx

54. f

sec4 x dx

56.

J.

51. c053(2 - x) sin(2 - x)dx cos7 3x dx cos498 y sin3 y dy

53.f 55.f

cos3 x dx

n/2

7r/2

49.

dt

sin2 x c053 x dx

Jo

7r/4

n/3

tan5 x ax

0 3n/4

7r/4

-n/4

cot4 x csc2 x dx 7r/4

420

TECHNIQUES OF INTEGRATION

CHAPTER 8

f ir

7

57.

Jo

59. f

sin 3x cos 4x dx

sin x sin 3x dx

58.

(b) Prove that, form > 2, =

7r/4

7r/6 60. f Jo

sin 2x cos 4x dx

sin 7x cos 2x dx

M—1

(c) Use (a) and (b) to compute i n, for m = 2, 3, 4, 5. 7r

61. For n a positive integer, compute the area under the graph of y = sin' x cos 3 X for 0 1) (for Ix' < 1)

j 1 -x 2

dx = coth-1 x -F C 1 - x2 r

dx

J

X,N/1 — X2

(for ix] >1)

= - sech-1 x + C

dx + x2

CSCh-1 X +

C

(for 0 < x < 1) (for x

0)

8.4 EXERCISES Preliminary Questions 1. Which hyperbolic substitution can be used to evaluate the following integrals? (a)

f

dx

(b) 1

f

dx -Vx2 + 9

(c)

f

dx -V9x2 + 1

2. Which two of the hyperbolic integration formulas differ from their trigonometric counterparts by a minus sign? 3. Which antiderivative of y = (1 — x2)-1 should we use to evaluate the 5 integral f (1 — x2)-1 dx? 3

S E CT I 0 N 8.4

Integrals Involving Hyperbolic and Inverse Hyperbolic Functions

431

Exercises In Exercises 1-16, calculate the integral. 1.

32. Verify that tanh-1 x = 21 in i x sinh x2 dx

2.

f cosh(3x)dx

1+x for lx1 < 1. 1- x

33. Evaluate fA/x2 + 16 dx using trigonometric substitution. Then use Exercise 31 to verify that your answer agrees with the answer in Example 3.

3.

i x sinh x dx

4.

i sinh2 x cosh x dx

5.

f sech2(1 - 2x)dx

6.

f tanh(3x) sech(3x)dx

7.

9.

8.

ftanhx sech2 x dx

dx

cosh x dx sinh x

12.

f cosh x

J

sinh` x

dx

1 f cosh'? x dx = - coshn-1 x sinh x + n i n n

16. f tanh3 x dx

In Exercises 17-30, calculate the integral in terms of the inverse hyperbolic functions. 18.

17.

f

dx 'N/9x2 - 4

20.

19.

21.

f [

dx

J

+3x2

coshn-2 x dx

2

In Exercises 37-40, evaluate the integral. 37.

15. fsinh2 x cosh2 x dx

9 dx in two ways: using trigonometric substitution

36. Use Eq. (2) to evaluate Jcosh4 x dx.

14. f sinh3 x cosh6 x dx

13. f sinh2(4x - 9)dx

2 -

35. Prove the reduction formula for n > 2:

10. f x csch(x2) coth(x2)dx

i tanhx dx

11.

cosh x f 3 sinh x + 4

34. Evaluate f\/x

and using hyperbolic substitution. Then use Exercise 31 to verify that the two answers agree.

f

tanh-1 x dx x2 - 1

38.

39. f tanh-1 x dx

f

sinh-l x dx

40. f x tanh-1 x dx

41. (a) Compute the area under the graph of y = sinh x for 0 < x < 5. (b) Compute the area under the graph of y = sinh-1 x for 0 < x < sinh 5. (c) Show that the sum of the areas in (a) and (b) is equal to 5 sinh 5. (d) Refer to Figure 1 and explain why the sum of the areas in (a) and (b) is equal to 5 sinh 5.

22.

I X2 - 1 dx

y = sinh 23.

24.

25.

26. -1

27.

29.

dx

[0.8

dx

28.

I

x•Vx2 ± 16 f

./0.2

.17 J 1 - I dX

30. f

x2

9

X

-

dx x,vx4 ± 1

FIGURE 1 42. (a) Compute the area under the graph of y = tanhx for 0 < x < 4. (b) Compute the area under the graph of y = tanh-1 x for 0 < x < tanh 4. (c) Show that the sum of the areas in (a) and (b) is equal to 4 tanh 4. (d) Similar to Exercise 41(d), explain graphically why the sum of the areas in (a) and (b) is equal to 4 tanh 4.

31. Verify the formulas sinh-l x = Ink ± VX 2 cosh-1 x = ln Ix

± ii

(for x > 1)

Vx2 - 11

X

5

x2

Further Insights and Challenges In Exercises 44-46, evaluate using the substitution of Exercise 43.

43. Show that if u = tanh(x/2), then cosh x =

+ u2 1 - u2

sinh x -

2u

1-

dx 2

2du 1 - u2

44. f sech x dx

Hint: For the first relation, use the identities 46. sinh2

2

= -1 (cosh x - 1), 2

cosh2 ( ±) j = 1 (cosh x 2 2

1)

dx 1 - coshx

45.

[

dx 1 + cosh x

432

TECHNIQUES OF INTEGRATION

CHAPTER 8

Exercises 47-50 refer to the function gd(y) = tan- ' (sinti y), called the Gudermannian. In a map of the earth constructed by Mercator projection, points located y radial units from the equator correspond to points on the globe of latitude gd(y).

49. Let t(y)= sinh-I (tan y). Show that t(y) is the inverse of gd(y) for 0 y co 2 x3 R oo 2 2 R->oo 2 2 lim R-oo 1:13 FIGURE 3 The area over [2, oo) is equal to

1 \ 2R2 )

8

8

We conclude that the unbounded shaded region in Figure 3 has area

•

" dx EXAMPLE 2 Determine whether f — converges. J1000 x Solution Since the integration is from 1000 to oo, we take R > 1000 and compute the limit as R -> oo: R dx f 00 dx — = lim f — =urni ln lx I X R->oo moo x R->oo J1000 1000

=

lim ln R - ln 1000 = oo

R—>00

The limit is infinite, so the improper integral diverges. We conclude that the area of the unbounded region in Figure 4 is infinite. • 1000

2000

Note that in the previous example, even though f(x) = 1/x is less than 0.001 in FIGURE 4 The integral of

f(x) = x -1 over

[1000,00) is infinite,

(1000, oo) and is decreasing to 0, the area under the graph is infinite. Such is the somewhat perplexing nature of improper integrals.

CONCEPTUAL INSIGHT If you compare the unbounded shaded regions in Figures 3 and 4, you may wonder why one has finite area and the other has infinite area. Convergence of an improper integral depends on how rapidly f(x) tends to zero as x -> oo (or x -oo). The previous two examples show that f(x) = X -3 tends to zero quickly enough that the integral converges, whereas f(x) = x-1 does not. An improper integral of a power function f(x) = x- P is called ap-integral. Note that f(x) = xP decreases more rapidly as p gets larger. Interestingly, our next the—1 is the dividing line between convergence and orem shows that the exponent p divergence. THEOREM 1 The p-IntegraI over [a, oo)

Fora >0,

a l —p f cx) ia

dx XP

=

I p

if p

l

diverges

> 1

if p < 1

Proof Denote the p-integral by J. Then R1—

J= lim f R—).00

p-integrals are particularly important because they are often used to determine the convergence or divergence of more complicated improper integrals by means of the Comparison Test (see Example 10).

x- P dx =urn

R-->oo 1 — p a

=

R--oo (1 —p

a l —p 1 —p)

If p> 1, then 1 - p oo. In this case, J = If

p
0 and R1- P tends to oo. In this case, J diverges. If p = 1, then R

J diverges because lim

R--).00 fa

x-1 dx = lim (ln R - In a) = oo. R ->co

•

450

CHAPTER 8

TECHNIQUES OF INTEGRATION EXAMPLE 3 Gabriel's Horn is the surface obtained by rotating the graph of f(x) = for x > 1, about the x-axis (Figure 5). This surface is interesting because it contains a finite volume but has an infinite surface area. (We verify the latter fact in Section 9.2.) Compute the volume contained in Gabriel's Horn. Solution The volume contained in the horn is the volume obtained by rotating the area under the graph of f(x) = . , for x > 1, about the x-axis. We find the volume by the Disk Method from Section 6.3 where the radius of the disk is r = f(x) = . : volume =

7

=

FIGURE 5 Gabriel's Horn contains a finite volume but has an infinite surface area. Therefore—paradoxically—it can be filled with a finite volume of paint but requires an infinite amount of paint to cover its surface. (A resolution to this paradox is presented in Exercise 63 in Section 11.2.)

11\ °° -1 1 x

r2 dx = rt-

f1

lim —'r x R—> co

1

=-7r

R

2

dx = lim 7r R —>co

X -2

dx

J1

1 11M (— — 1) = R-+ oo R

•

Therefore, the volume contained in Gabriel's Horn is 7r.

A doubly infinite improper integral is defined as a sum (provided that both integrals on the right converge): 00 co 2 f(x)dx + f f(x)dx f(x)dx = f foo We can use some number other than 0 as a choice of where to split the integral, if it is more convenient to do so. EXAMPLE 4 Determine if

cx)

1

f_„ 1 ± x2

dx converges and, if so, compute its value.

Solution °°

L ,

0

1 1 + x2

fool+

dx =

x2

dx +

"

1 1 + x2 dx

Jo

assuming both of these integrals converge. For the second of these,

r Jo

DO

R

1

dx = Ern

,, dx = Ern tan-1 x

R —*co f

1 ± X2

1 R —*co

1 ± X'

= lim tan-1 R — 0 = o

R —>c°

2

Similarly, 1,0

1 dx = lim /2.---co j _co 1 ± X2

o

fR

1

0

, dx = lim tan-1 x R —>—co

1 + X'

=0—

lim tan-1 R =

R-÷ — co

Thus, since both integrals converge, f 70

j

1

1 + x2

dx =

1 f cx) 1 7T IT dx + dx = — + — = 7r 2 2 00 1 + x2 Jo 1 - I- x 2

•

Sometimes it is necessary to use L'Hopital's Rule to determine the limits that arise in improper integrals. 00 EXAMPLE 5

Using L'Hopital's Rule

Calculate f

xe —x dx.

Solution The integral corresponds to the area in Figure 6. First, we compute the associated indefinite integral using Integration by Parts with u = x and dv = dx:

The integral of f(x) = xe—x over [0, oo) is the shaded area. FIGURE 6

f fo

xe

dx = —xe—x + f e —x dx = —xe—x —

xe—x dx = —(x

1)e—x

= —(R 0

+ C = —(x

1)e—R ± 1 = 1

R

1)e —x + C 1

eR

SECTION 8.7

Improper Integrals

Now, compute the improper integral as a limit using L'Hopital's Rule: co R +1 1 xe' dx =1 — lim =1— lim — = 1—0=1 R—>00 eR R —> oo e R

fo

451

•

L'Hopitars Rule

Improper integrals arise in applications when it makes sense to treat certain large quantities as if they were infinite. In the next example, we determine the escape velocity of an object launched from Earth by assuming that the velocity is sufficient to take it "infinitely far" into space. In physics, we speak of moving an object "infinitely far away" In practice, this means "very far away" but it is more convenient to work with an improper integral.

4=4 REMINDER The mass of the earth is M,

5.98 • 1024 kg

EXAMPLE 6 Escape Velocity The earth exerts a gravitational force of magnitude F(r) = GMem/r2 on an object of mass m at distance r from the center of the earth. (a) Find the work required to move the object infinitely far from the earth. (b) Calculate the escape velocity vese on the earth's surface. Solution This amounts to computing a p-integral with p = 2. Recall that work is the integral of force as a function of distance (Section 6.5). (a) The work required to move an object from the earth's surface (r = re) to a distance R from the center is

The radius of the earth is re

f eR GMem fir = r2

6.37 • 106 m

The universal gravitational constant is

GMem R = Gmem r re

_1) joules R

re

The work moving the object "infinitely far away" is the improper integral

N_In21kg2 G ,'=1 6.67 • 10-n

GMem f

A newton is 1 kg-m/s2 and a joule is 1 N-m.

GMem ' dr (1 1) .--= lim GMem — — — = joules R—>oo re re .—rT R re

(b) By the principle of Conservation of Energy, an object launched with velocity vo will escape the earth's gravitational field if its kinetic energy ,/inq) is at least as large as the work required to move the object to infinity—that is, if 2 I Escape velocity in miles per hour is I approximately 25,000 mph.

In practice, the word "forever" means "a long but unspecified length of time." For example, if the investment pays out dividends for 100 years, then its present value is 100 6000e-6'64' dt

$147,253

The improper integral ($150,000) gives an approximation to this value.

GMem re

=

vo >

(2GMe ) 1/2 re

)

Using the values recalled in the marginal note, we find that vo > 11,200 m/s. The minimal • velocity is the escape velocity vesc = 11,200 m/s. EXAMPLE 7 Present Value of Future Income If an investment pays a dividend continuously at a rate of R(t) $/year and earns interest at rate r (in decimal form), then the present value of the dividend income, for the period from t = 0 to t = T, is given by PV = f

R(t)e

t dt.

We think of present value as the payment that we would need to receive at t = 0, instead of the dividend income, so that at time T the payment's value (with accumulated interest) would be the same as the amount accumulated from the dividend income (with its accumulated interest). It is essentially the present worth of the income we are about to receive up to time T. Assuming that the dividend rate is $6000/year, and the interest rate is 4%, compute the present value if the dividends continue forever. Solution Over an infinite time interval, 00 PV = f

6000e -0.04t T T —> oo 0 —0.04

6000e-0.04t dt = lim

6000 = $150,000 0.04

Although an infinite amount of money is paid out during the infinite time interval, the • total present value is finite.

452

CHAPTER 8

TECHNIQUES OF INTEGRATION

Unbounded Functions An integral over a finite interval [a, b] is improper if the integrand is unbounded. In this case, the region in question is unbounded in the vertical direction. For example,

9 dx

f Nrx is improper because the integrand f (x) = x- V2 tends to oo as x ---> 0+ (Figure 7). Improper integrals of this type are defined as one-sided limits.

If f is continuous on [a, b) and lim f (x) =

DEFINITION Unbounded Integrands ±00,

we define

7 By Example 8(a), the shaded region has area 6.

FIGURE

f(x)dx = lim f

f(x)dx

R —>b-

a

Similarly, if f is continuous on (a, b] and lirn f (x) = ±oo,

IL

f(x)dx = lim f R -->a+

f (x) dx R

In both cases, we say that the improper integral converges if the limit exists and that it diverges otherwise. Note that if there is a single point c in the interval [a, b] such that lim f (x) = ±oc, c

b

or lim f(x) = ±oo, and if f x —>c+ b

IL

a

c

f (x)dx = f

f (x)dx and f

f (x)dx ± f a

f (x)dx both converge, then we define c

b

f (x)dx . c

(a)f

EXAMPLE 8 Calculate:

9

/1/2

dx

Jo

and

x

(b)

dx x

Solution Both integrals are improper because the integrands have infinite discontinuities at x = 0. The first integral converges:

fo

9

9

9

dx

=

2x 1" 2

x — 1 /2 dX =

X

R—>0+

R

= lim (6 - 2R1/2) = 6 The second integral diverges: f 1/2 dx — = lim

Jo

X

R—>0+

1/2 R

dx 1 — = lim (ln - - ln R) X R—>0+ 2

1 = ln - - lim ln R 2 R->o+

fO

x

(x -

2 • 1) 3

Solution This integral is improper with an infinite discontinuity at x = 1 (Figure 8). Therefore, we write 2

The unbounded shaded region has area 6. FIGURE 8

•

dx

2

EXAMPLE 9 Calculate

oo

I

1

dx

—

2

2

dx

fo (x —

2 +

f

dx

(x —

2

SECTION 8.7

Improper Integrals

453

We consider each integral individually: 11

dx

2 - liM (x - 1)3 R-41- JO

R (x

dx -= lirn 3(x - 1)3 1)32 R-41-

0

= lim 3(R - 1)3 - 3(-1)3 = 3

f.112

dx

2

(x — 1)4

= lim R->1+

dx (x

= 3(1)3 -

2

2 = lim 3(x - 1)3

lim 3(R - 1)3 = 3 R-.+1+

Therefore, we obtain 12

(x —

J

Theorem 2 is valid for all exponents p. However, the integral is not improper if p 0, " dx fo =

1 P diverges

if p > 1

00 GRAPHICAL INSIGHT

but f

C0

oo

but fd 1

FIGURE 9

dx < X7/

x —=oo xq

The p-integrals i

1 x' dx and i

x" dx have opposite 1 o behavior for p 0 1. The first converges only for p > 1, and the second converges only for p < 1 (both diverge for p = 1). This is reflected in the graphs of y = x- P (with p > 1) and y = x' (with 0 < q < 1) in Figure 9. Since 0 < q < 1, the values of f(x) = x- q are arbitrarily large near x = 0 and decrease rapidly as x increases to 1, thereby resulting in the convergence of 1 x q dx. However, f(x) = x- q decreases slowly to 0 as x -> oo, resulting in the o 00 divergence of x q dx. 1 The graphs for q < 1 and p > 1 switch relative positions and behaviors at the point of intersection (1, 1). Since p > 1, the values of f(x) = x- P are arbitrarily large near x = 0, but 1 decrease slowly as x -›- 1, resulting in the divergence of f x- P dx. On the other o hand, with p> 1, f(x) = x- P decreases rapidly to 0 as x ->. oo, resulting in the 00 convergence of 1 x 1' dx. f

f

f

Comparing Integrals Sometimes we are interested in determining whether an improper integral converges, even if we cannot find its exact value. For instance, the integral r 00 e-x —dx ii x

454

CHAPTER 8

TECHNIQUES OF INTEGRATION cannot be evaluated explicitly. However, if x > 1, then 1 0< - 0 for x > a: 00 f (x)dx converges, then f g(x)dx also converges. • If f a a co co • If g(x)dx diverges, then f a f (x)dx also diverges. a The Comparison Test is also valid for improper integrals of unbounded functions on a finite interval.

EXAMPLE 10 Show that f i

dx ±

converges.

Solution We cannot evaluate this integral, but we can use the Comparison Test. To show convergence, we must compare the integrand (x3 + 1)-1/2 with a larger function whose integral we can compute. < A/x3 + 1. Therefore, It makes sense to compare with X-312 because 1 'S/X3 ± 1

< — 'N

= X -3/2

The integral of the larger function converges, so the integral of the smaller function also converges: dx fx

1 0°

converges

3/2

dx

converges

A/x3 + 1

p-integral with p > 1

Integral of smaller function 0.0

What the Comparison Test says (for nonnegative functions):

EXAMPLE 11

Choosing the Right Comparison

• If the integral of the larger function converges, then the integral of the smaller function also converges. • If the integral of the smaller function diverges, then the integral of the larger function also diverges.

Solution Since ,rx > 0, we have jic

dx e3x

converge?

e3x > e3x, and therefore, 1

N/TC

Does f

1 oo 3

- e

=

(converges)

•

SECTION 8.7

Improper Integrals

455

Our integral converges by the Comparison Test:

f

dx e

,,e3x dx

converges

3x

ft

Integral of larger function

also converges

Integral of smaller function

Had we not been thinking, we might have tried to use the inequality 1

< 1

+ e3x FIGURE 11 The divergence of the integral

of the larger function says nothing about the integral of the smaller function.

However,

VTX

dx

diverges (p-integral with p < 1), and this says nothing about the J 1 .Nrx • integral of the smaller function (Figure 11).

EXAMPLE 12

Endpoint Discontinuity

Does]

"

x8

dx + x2

converge?

1 Solution This integrand has a discontinuity at x = 0, since lim = -koo. x->0+ x8 + x2 We might try the comparison x

8

However, the p-integral f 0

+ X

2

> X

1

2

x8

1 < — x2 x2

c1.5 dx x diverges, so this says nothing about the integral involv-

ing the smaller function. But notice that if 0 1:

dx Ja

dx

p < 1: p = 1:

converges

and

diverges

diverges

and

converges

XP

a

XP

f a cx3

dx

and

both diverge

• The Comparison Test: Assume that f and g are continuous functions such that f (x) > g(x) > 0 for x > a. Then 00

IL If f If

00

f (x) dx converges,

then

g(x)dx diverges,

then

IL IL

g(x)dx converges.

00

00

a

f(x)dx diverges.

• Remember that the Comparison Test provides no information if the integral of the larger function diverges or if the integral of the smaller function converges. • The Comparison Test is also valid for improper integrals of functions with infinite discontinuities at an endpoint of the integral.

8.7 EXERCISES Preliminary Questions 1. State whether each of the following integrals converges or diverges:

(a) f

co X-3 dx

(c) f oe x-213 dx

(b) 1 31 X-3 dx

integral.

1 (d) f o x-2/3 dx

4. Which comparison would show that

7/12 2. Is f

f b

3. Find a value of b > 0 that makes

cot x dx an improper integral? Explain.

(d)f 2.

Let f(x)= x-413. R

(a) Evaluate f

f(x)dx. 1

(b) Evaluate] 1

00 f(x)dx by computing the limit urnf R-)•oo 1

f(x)dx

dx an improper

converges?

5. Explain why it is not possible to draw any conclusions about the condx vergence off — e-x dx by comparing with the integral f —. x x

Exercises 1. Which of the following integrals is improper? Explain your answer, but do not evaluate the integral. D. 2 dx " dx (c) f e-x dx (b) f i x0.2 (a) fo X113 -1 Ir. 1 (f) f sin x dx (e) f sec x dx e-x dx 0 0 13i 00 1 dx (i) i ln x dx (h) (g) f sinx dx ft ,./3 _ x2 1 o 3 (j) f In x dx o

J o x2 - 4

3.

co

Prove that f

X-213 dx diverges by showing that 1 lim R->oo

4.

f

R

x-213 dx = co

3 dx Determine whether f o (3 - ,03/2 converges by computing R

11111

R-0— f 0

dx (3 - )03/2

In Exercises 5-40, determine whether the improper integral converges if so, evaluate it. and, ' dx dx ' 5. f i. 6. 20/19 x19/ 20 f 4 7. -00

eo.000lt dt

8.

f" dt J20

t

SECT I ON 8.7

dx

5

9.

10. ./0.

x 20/ 19

./0

1.1. f o

dx 12. f 5 (x _ 5)3/2

,f4__ _. x

14.

13. f 2 D° x-3 dx "

17. il l

10

16. f

dx (T

+1)3

e

2'

dx

18.

x0.2

23.

25.

.12 m

x-113 dx

dx

dx

24.

e3x dx

f_4, +2, _00 °

dx 28. f 3 ,/,c _ 3

x dx (1 ± x2)2

6

31.J

32. J o

dx

x2

ri x- dx

34.

,./Z

-cc

x dx converges, thereby demonstrating that the

oo

f0

sin x dx

38.

/4

for -oo 1 (Figure 12). Use the Com2

parison Test to show that I e-x dx converges. Hint: It suffices (why?) Jo to make the comparison for x > 1 because

2 dx f i x In x

1

oo 41. Let / =

53. Sketch the region under the graph of the function f(x)=

7

tan x dx

40.f

1 lnx -dx 2 x

dx 52. Show that f I - converges if p < 1 and diverges if p > 1. 0XP

55. Show that

,./.7

36. f

37. f o I In x dx

eax dx converge? o

1 S i dx

0

fo

J

i c°

x2 ex dx

712

39.

Lf11 1> 11 R 0.3 R oe

X

0

30. f

35. f:

(b) Show that

dx

Xe-X2 dx

29. f o°3 xe-3x dx

'

x dx. - 00

definition of L . f(x)dx needs to be adhered to carefully.

26. f

"

33. fi

1 dx converges and, if so, to what. _ t x 113

R

0

Jo 1+ x

o ,,,/

dx (x2 + 1)3/2

(a) Show that it diverges.

dx 22. f i (x _ 1)2

dx

3

00 48' f_.

51. For which values of a does f

3

27. f o

, xe-x- dx

2

.ft ,I3 - x

'

f oo 47.

dx

4

e3x

46. f . e- lx1 dx

co

20. f

dx

0

.

x dx 1 + x2

50. Consider the integral i

e-3x

21.L

oo

f' j

45.

2

00 19. f

"

457

In Exercises 45-48, determine whether the doubly infinite improper integral converges and, if so, evaluate it. Use definition (2).

49. Determine whether f

dx

15. f_3 (x + 4)3/2 1

dx x19/20

6

dx

4

5

Improper Integrals

JO

In x -dx x2

1 e -X2

dx =

2

e'

00

dx + f

e-x-, dx

dx (x - 2)(x - 3Y

(a) Show that for R > 4, R

R- 3 dx (x - 2)(x - 3)- ln R - 2

(b) Then show that / = In 2. FIGURE 12 Comparison of y = e Ix' and y = 42. Evaluate the integral / =

43. Evaluate f 44. Evaluate

is" ii

dx x(2x + 5) •

dx or state that it diverges. o x(2x + 5)

L

dx or state that at it diverges. + 3)(x + 1)2 (x

58. Prove that

C° f

e'

2

2.

Co

dx converges by comparing with

(Figure 12). 59. Show that

e'

1 - sin x X2

dx converges.

e- lx1 dx

458

TECHNIQUES OF INTEGRATION

CHAPTER 8

xa = oo (by Exercise 63 in Section 7.5). 60. Let a > 0. Recall that lim x->co ln x (a) Show that xa > 21n x for all x sufficiently large.

83. When a capacitor of capacitance C is charged by a source of voltage V. the power expended at time t is

(b) Show that e- x" 1,

(x+

P1'2 1/2

dx

ja xi/2(x+1) - j, xi/2(x+1)

0

f (x)e-sx dx

Laplace transforms are widely used in physics and engineering. 88. Show that if f (x) = C, where C is a constant, then £f (s) = C Is for s >0. a 89. Show that if f (x) = sin ax, then Gf (s) = s 2 a 2 90. Compute £f (s), where f (x) = ex and s > a. 91. Compute ,C f(s), where f (x) = cos ax and s > 0.

Use the Comparison Test to show that the integral converges. 78. Determine whether] verges.

dx (defined as in Exercise 77) conx3/2(x + 1)

79. An investment pays a dividend of $250/year continuously forever. If the interest rate is 7%, what is the present value of the entire income stream generated by the investment? 80. An investment is expected to earn profits at a rate of 10,0000.mt dollars per year forever. Find the present value of the income stream if the interest rate is 4%. 81. Compute the present value of an investment that generates income at a rate of 5000te0 Olt dollars per year forever, assuming an interest rate of 6%.

92. 174 When a radioactive substance decays, the fraction of atoms present at time t is f (t)= e-kt , where k >0 is the decay constant. It can be shown that the average life of an atom (until it decays) is tf'(t)dt. Use Integration by Parts to show that A = f

A = -f

f (t)dt

and compute A. What is the average decay time of radon-222, whose halflife is 3.825 days? 93.

el Let

=f

e-° ' dx, where n

1 is an integer and a >0.

Prove that =a

A-1

and ./0 = 11a. Use this to compute J4. Show that J = n/an+1.

SECTION 8.8

for x 0 0 and f (0) = 0. xn eax _ I (a) Use L'Hopital's Rule to show that f is continuous at x = 0. 00 e —ax f (x) dx converges. Hint: Show that f (x) _< 2xn ' (b) Show that Jo if x is large enough. Then use the Comparison Test and Exercise 93. 94. Leta > 0 and n > 1. Define f(x) =

According to Planck's Radiation Law, the amount of electro95. magnetic energy with frequency between v and v + Av that is radiated by a so-called black body at temperature T is proportional to F(v) A v, where

F(v) =

87th

Numerical Integration

459

v3 1

where c,h,k are physical constants. Use Exercise 94 to show that the total radiated energy 00 E= I

F(v)dv

0

is finite. To derive his law, Planck introduced the quantum hypothesis in 1900, which marked the birth of quantum mechanics.

Further Insights and Challenges However, the integral is not absolutely convergent. It is known that I = The convergence depends on cancellation, as shown in Figure 13.

xP lnx dx.

96. Consider f

(a) Show that the integral diverges for p = —1. —1, then (b) Show that if p xp+1 1 +C ln x f xPlnxdx = p +1 (c) Use L'Hopitars Rule to show that the integral converges if p > —1 and diverges if p co

t

(1 —e—"„ r ) dt = lim —(r + 0e—f ir

— (r + R)e —R1r ) =r

0 as R —> no. It follows that the mean The last equality holds since (r + R)e—RIr of a random variable with the probability density function p(t) = .-e —t ir, over [0, no), • is r.

SECTION 9.1

Probability and Integration

475

EXAMPLE 3 Waiting Time The waiting time T between customer arrivals in a drivethrough fast-food restaurant is a random variable with exponential probability density. If the average waiting time is 60 seconds, what is the probability that a customer will arrive within 30 to 45 s after another customer? Solution If the average waiting time is 60 s, then r = 60 and p(t) = 6_10 e —t/60 because the mean of 4, .--"t/r is r by the previous example. Therefore, the probability of waiting between 30 and 45 s for the next customer is 120

30 45 60

180

13 FIGURE 3 Customer arrivals have an exponential distribution.

P(30 < T < 45) = i

45 1 45 _L e—t/60 = _ e —r/60 = —e-314 ± e-1/2 s•-' 30 30 60

This probability is the area of the shaded region in Figure 3.

0.134 is

The normal density functions, whose graphs are the familiar bell-shaped curves, appear in a surprisingly wide range of applications. The standard normal density is defined over (—oo, oo) by y 1

p(x) =

N/Yr

e

_x2/2

3

That p(x) satisfies Eq. (1) is not easy to show. One problem is that p does not have an elementary antiderivative. In Exercise 57 in Section 16.4 we will see an approach using multivariable calculus. More generally, we define the normal density function with mean kt and standard deviation a: FIGURE 4 Normal density functions.

p(x) =

6:00

7:00

8:00

9:00

FIGURE 5 Time that people leave for work.

1 a

The standard deviation a measures the spread; for larger values of a, the graph is more spread out about the mean p, (Figure 4). The standard normal density in Eq. (3) has kt = 0 and a = 1. A random variable with a normal density function is said to have a normal or Gaussian distribution. Examples of data whose distribution is modeled well by a normal distribution include current sale prices for houses in Denver, heights of female children of age 11 in Egypt, and systolic blood pressure readings for adults in Frigento, Italy. The normal distribution is ubiquitous in everyday life. For example, Figure 5 shows a bar graph of data from a survey on the time of day that workers in the United States leave for work. Note the approximate bell-shaped curve generated by the data. One difficulty with normal density functions is that they do not have elementary antiderivatives. As a result, we cannot evaluate the probabilities P(a 0 for x in J, and

f

j p(x)dx =1.

• Mean (or average) value of X: it = i

xp(x)dx I

• Exponential density function of mean r: p(x)= 71. e- xlr • Normal density of mean kt and standard deviation a: p(x) =

a

e

1 • Standard cumulative normal distribution function: F(z) = —,__.

_ 0 2/(2,2) 2 in

e - 1` dt

V271.

• If X has a normal distribution of mean .t and standard deviation a, then P(X

b) = F ( b

a

b

P (a < X < b) = F \

a

j

F (6I ; II )

9.1 EXERCISES Preliminary Questions 1.

The function p(x) = cos x satisfies f

p(x)dx = 1. Is pa proba-

3.

Which exponential probability density has mean js =-

8.

p(x) = Ce' e-e-'

bility density function on [-n12, 7r]? 2. Estimate P(2 < X -0.4)

20. Assume X has a normal distribution with u, = 0 and a = 5. Express each of the following probabilities in terms of F(z) and determine the value of each. (b) P(X > -0.4) (a) P(X < 1.2)

The smaller the value of a, the more tightly clustered are the values of the random variable X about the mean s. (The limits of integration need not be ±c)o if p is defined over a smaller domain.) 27. p(x) =

28. p(x) =

5 20 2 1 7r,/1 -

on [1, oo) on (-1, 1)

1 29. p(x)= - e'13 3

on [0, oo)

1 30. p(x) = - e'l r r

on [0, oo), where r > 0

SECTION 9.2

Arc Length and Surface Area

479

Further Insights and Challenges 31. F01 The time to decay of an atom in a radioactive substance is a random variable X. The law of radioactive decay states that if N atoms are present at time t = 0, then Nf(t) atoms will be present at time t, where f (t) = e —kt (k > 0 is the decay constant). Explain the following statements: (a) The fraction of atoms that decay in a small time interval [t,t + At] is approximately —/(t)At.

(b) The probability density function of X is y = — f'(t). (c) The average time to decay is 1/k. 32. The half-life of radon-222 is 3.825 days. Use Exercise 31 to compute:

(a) the average time to decay of a radon-222 atom. (b) the probability that a given atom will decay in the next 24 hours.

9.2 Arc Length and Surface Area We have seen that integrals are used to compute total amounts (such as distance traveled, total mass, total cost). Another such quantity is the length of a curve (also called arc length). We derive a formula for arc length using our standard procedure: approximation followed by passage to a limit. In this case, we approximate the curve by a path made up of line segments connecting points on the curve. It is easy to find the length of a collection of line segments. We improve the approximation by using more, but smaller, segments. Then we take the limit of the sum of their lengths as the number of line segments grows. To make this precise, consider the graph of y = f (x) over an interval [a, b]. Choose a partition P of [a, b] into N subintervals with endpoints P : a = xo 0, so the graph lies above the x-axis. We revolve the graph around the x-axis to obtain a surface of revolution R. Our goal is to determine the surface area S of R. To do so, we start by creating another surface of revolution R* by rotating a polygonal approximation to y = f (x) about the x-axis. The result is a surface R* that lies very close to R and whose surface area approximates S (Figure 8).

482

CHAPTER 9

FURTHER APPLICATIONS OF THE INTEGRAL

,

FIGURE 8 Rotating a polygonal approximation produces an approximation by slanted bands.

I

Axi

I

We will set up a Riemann sum that approximates the surface area of R* and therefore that also approximates the surface area of R. Taking a limit results in a definite-integral formula for S, the surface area of R. The surface R* is made up of slanted bands, as shown in Figure 8. Consider the slanted band corresponding to the interval [x1-1, xi 1. The segment Li along the slanted band is a segment in the polygonal approximation for y = f (x). As in the derivation of the arc length formula, the length of Li can be expressed as

1141 = \/1 + pco2Axi 141 f(b)

for some ci in [x1-1, x,] and Axi = xi — xi-1. Now, as Figure 9 illustrates, we can approximate the surface area of the single slanted band with the surface area of a cylinder of width I Li I and radius f(b1) for any bi in [xi _1, xi ]. Since we can use any bi in [xi _1, xi] for this approximation, we will use b, = c, to match the value used in the expression for IL I. The surface area of a cylinder of radius r and width w is 2n-r w; therefore, if we let bi = ci, the surface area of the cylinder in Figure 9 is 277/(ci)ILi = 27(141 +

The surface area of the slanted band is approximated by the surface area of the cylinder. FIGURE 9

The surface area of each slanted band in R* is approximated by the surface area of such a cylinder. We add up the surface areas of these cylinders to obtain an approximation to the surface area of R*: surface area of R*

27r

E f(c)

1 + f' (ci)2 Axi

i=1 As the norm of the partition goes to zero, the error in this approximation of the surface area of R* also goes to zero. Furthermore, the surface area of R* approaches the surface area of R. Therefore, the sum on the right-hand side of the approximation approaches S in the limit. That sum is a Riemann sum that converges to the integral in the following definition: Area of a Surface of Revolution Assume that f (x) > 0 and that f' exists and is continuous on the interval [a, b]. The surface area S of the surface obtained by rotating the graph of f about the x-axis for a < x < b is equal to

S= 2 a

f(x) l + f' (x)2 dx

2

SECT I ON 9.2

Arc Length and Surface Area

483

EXAMPLE 4 Calculate the surface area of a sphere of radius R. Solution The graph of f (x) = N/ R2 - x2 is a semicircle of radius R (Figure 10). We obtain a sphere by rotating it about the x-axis. We have X

f'(x) =

N/R 2 —

1+ X2

x2

f'(x)2 = 1 ± R2 _ x2

R2 = R2

x2

The surface area integral gives us the usual formula for the surface area of a sphere: f (x) 1 ± f' (x)2 dx = 27r

S =27 —R FIGURE 10 A sphere is obtained by revolving the red semicircle about the x-axis.

= 2m R f —R

f-R

A/R2

x2

,VR2

x2

dx

dx = 27 R(2R) = 4n- R2

•

EXAMPLE 5 Find the surface area of the surface, called a paraboloid, that is obtained by rotating the graph of f (x) = ,fx- about the x-axis for 0 < x < 1.

=

Solution The graph of f (x) = ,/.7 is the top half of a parabola opening along the x-axis, which becomes a paraboloid when rotated about the x-axis (Figure 11). Then f'(x) = 2.11i and hence, we obtain

X

1 S -=

2

f(x) 1 + f'(x)2 dx =

27t f

0 FIGURE 11 A paraboloid results when the top half of the parabola is revolved about the x-axis.

= 27r J o

,‘Fx

1 + (- 1

dx

0

N/4x + 1 dx =

A/ 4x

f

± 1 dx

0

=± 7 (4x ± 1)3/2 = -(53/2 - 1) 6 0 6

5.3304

•

EXAMPLE 6 A Corrugated Pipe A corrugated pipe is obtained by rotating the graph of f (x) = 1 + 0.1 sin(10x), for 0 1, about the x-axis. There we saw that the volume enclosed in the horn is Tr, and therefore is finite. Prove that the surface area of Gabriel's Horn is infinite. Solution With f (x) = I = x-1 , we have f '(x) = —x-2 and f '(x)2 = x -4. The surface area of Gabriel's Horn is 00 x _1 1+ x 4dx S = 2f

Gabriel's Horn is obtained by rotating the graph of f(x)= about the x-axis. FIGURE 13

Now, x-1s/1 + x-4 > x-1 over [1, oo), and f

co x-1 dx diverges (to infinity) by The-

orem 1 in Section 8.7. By the Comparison Test for Improper Integrals (Theorem 3 in • Section 8.7), it follows that S, the surface area of Gabriel's Horn, is infinite.

484

FURTHER APPLICATIONS OF THE INTEGRAL

CHAPTER 9

9.2 SUMMARY • The arc length of y = f (x) over the interval [a, b] is s =

(x)2 dx. 1+ .12b • Assume that f (x) > 0. The surface area of the surface obtained by rotating the graph of f about the x-axis for a •-•-•pf(xj) Ax. Furthermore, Mx,i = Tim j, where Ti is the y-coordinate of the COM of the strip. However, Ti f (x j) because the COM of a rectangle with uniform mass density is located at its center. Thus, 1 1 M,1 = m1357 '-.,' pf (x )Ax • i f (x j) = -i pf (x j)2 Ax My,

My

FIGURE 13 Center of mass of the lamina.

Mx = Q FIGURE 14 Because the shaded strip is nearly rectangular, its COM has an approximate height of f (x j).

N

N

i=i

j=1

E mx,.; i1 P E f(x1)2Ax

b This is a Riemann sum whose value approaches 1p a f (x)2 dx as N —> oo. The case f of a region between the graphs of functions fi and f2 where fi (x) > f2(x) > 0 is the difference of the moments corresponding to fi (x) and f2(x), by the principle of additivity of moments, so we obtain the following formulas for Mx for a lamina occupying a vertically simple region:

M1 =

1 2

b f

a

f(X)2 dX

or

The same idea holds for determining

Y

1p =_ 2

f

g

( y

)2

dy

or

Mx =-

My

1

-p 2

a

(fi(x)2 — ,f2(x)2) dx

3

in the horizontally simple case: 1 M = 2 Y

f

"1°

( y

)2

g2

( y

)2)

dy

4 I

S EC TION 9.4

Center of Mass

497

EXAMPLE 4 Find the centroid of the shaded region in Figure 15. Solution The centroid does not depend on p, so we may set p = 1 and apply Eqs. (1) and (3) with f (x) -= ex :

25

3

1 Mx -= —

1

3

3

1

e6 — e2

f (x)2 dx = — e2x dx = — e2x = 2 1 2 1 4 1

4

Using Integration by Parts, we get FIGURE 15 Region under the curve y = ex

3

between x = 1 and x = 3. My

=

3

3

xf(x)dx = f

f

2e3

xex dx = (x — 1)ex

3

The total mass is M = -=

FIGURE 16 The COM of a symmetric plate

lies on the axis of symmetry. 4m1 REMINDER A region is symmetric with respect to a line if reflection across the line sends each point of the region to another point of the region.

M

=

f1

ex dx = (e3 — e). The centroid has coordinates

2e3 3 e— e

e6

2.313,

51" =

e2

Mx = =•-•-' 5.701 M 4(e3 — e)

•

The symmetry properties of an object give information about its centroid (Figure 16). For instance, the centroid of a square or circular plate is located at its center. Here is a precise formulation (see Exercise 49): THEOREM 1 Symmetry Principle then its centroid lies on that line.

If a lamina is symmetric with respect to a line,

EXAMPLE 5 Using Symmetry Find the centroid of the half-disk of radius 3, between — x2, as shown in Figure 17. the x-axis and the graph of f(x) = Solution Symmetry cuts our work in half. The half-disk is symmetric with respect to the y-axis, so the centroid lies on the y-axis, and hence i = 0. It remains to calculate Mx and 57. By Eq. (3) with p = 1, 1 Mx =

3

f

2 _3

3 1f 3 1 (9 — x2)dx = — (9x — — 1 x3) f (x)2 dx = — = 9 — (-9) = 18 —3

The half-disk has area (and mass) equal to A = 7i-(32) = 97/2, so FIGURE 17 The half-disk under the graph of

f (x) =

18 _ Mx y= = 97.1

— x2 . EXAMPLE 6 Figure 18.

=

4

•

r=,-' 1.27

Using Additivity and Symmetry Find the centroid of the region R in

Solution We set p = 1 because we are computing a centroid. The region R is symmetric with respect to the y-axis, and therefore, i = 0. To find we compute the moment M. Step I. Use additivity of moments. Let Mxtriangle and Mxcircle be the x-moments of the triangle and the circle. Then m = mxtriangle x FIGURE 18 The moment of region R is the sum of the moments of the triangle and circle.

circle

Step 2. Moment of the circle. To save work, we use the fact that the centroid of the circle is located at the center (0,5) by symmetry. Thus, 7 1" 1e = 5 and we can solve for the moment: y-circie

mcircie mcircie x = 5 x mcircie — 4.

mVrcle

= 207

Here, the mass of the circle is its area Mcircle = 7(22) = 47r (since p = 1).

498

CHAPTER

9

FURTHER APPLICATIONS OF THE INTEGRAL

h —

y

Step 3. Moment of a .triangle. Let's compute Mxtriangle for an arbitrary triangle of height h and base b (Figure 19). Let i(y) be the width of the triangle at height y. By similar triangles, i(y) h—y

FIGURE 19 By similar triangles, i(y) b h —y = h

b h

By Eq. (2), mxtriangle = f

Ii

h

ye(y)dy

f

y

In our case, b = 4, h = 3, and Mtriangle

b b — —y) dy =

by2

by3 ) h = bh 2 — 6 0

32 = 4• =6. 6

Step 4. Computation of 53. Aix The triangle has mass 6 + 47 and

mtriangle

lu.exircle = 6 + 207r

• 4 • 3 = 6, and the circle has mass 47, so R has mass M = Mx =— M

6 + 207r 6 + 47r

•

3.71

We end this section with the Theorem of Pappus, attributed to Pappus of Alexandria, a mathematician of the fourth century BCE: THEOREM 2 Theorem of Pappus Let R be a region of area A in the plane. If we rotate R about an axis that is disjoint from R, then the volume of the resulting solid is the product of A with the distance traveled by the centroid of R.

Proof Since we assume a uniform density of 1, the area A of the region is equal to the mass M. We will prove the theorem only in the special case that we are rotating about the x-axis and that we have a region bounded by y = (x) and y = f2(x) for a < x < b and fi (x) > f2(x) > 0. In this case, we know that the volume is given by

(h(x)2 —

V = a

f2(x)2)dx =

1

(—f 2 a

— h(x)2)dx)

Mx

= 22rM = A • 27 — = A • 2753 A Thus V = A • 2703. This is the desired result because 53 is the distance from the rotation axis to the centroid (since we are rotating about the x-axis), and therefore, 2757 is the • distance traveled by the centroid. EXAMPLE 7 Find the formula for the volume of the solid torus obtained by rotating the disk of radius a centered at (b, 0) about the y-axis, where a n > 0, and find the COM of the region. Find a pair (n, m) such that the COM lies outside the region. 39. Find the formula for the volume of a right circular cone of height H and radius R using the Theorem of Pappus as applied to the triangle H, rotated - x bounded by the x-axis, the y-axis, and the line y = 4 about the y-axis. 40. Use the Theorem of Pappus to find the centroid of the half-disk bounded by y = .//22 — x2 and the x-axis. In Exercises 41-43, use the additivity of moments to find the COM of the region. 41. Isosceles triangle of height 2 on top of a rectangle of base 4 and height 3 (Figure 25)

FIGURE 27 45. Find the COM of the laminas in Figure 28 obtained by removing squares of side 2 from a square of side 8.

2

8

8 FIGURE 28

FIGURE 25

Further Insights and Challenges 46. A median of a triangle is a segment joining a vertex to the midpoint of the opposite side. Show that the centroid of a triangle lies on each of its medians, at a distance two-thirds down from the vertex. Then use this fact to prove that the three medians intersect at a single point. Hint: Simplify the calculation by assuming that one vertex lies at the origin and another on the x-axis.

47. Let P be the COM of a system of two weights with masses m and m2 separated by a distance d. Prove Archimedes's Law of the (weightless) Lever: P is the point on a line between the two weights such that miLi = m2L2, where Li is the distance from mass j to P.

502

FURTHER APPLICATIONS OF THE INTEGRAL

CHAPTER 9

48. Find the COM of a system of two weights of masses mi and m2 connected by a lever of length d whose mass density p is uniform. Hint: The moment of the system is the sum of the moments of the weights and the lever. 49. 17 Symmetry Principle Let R be the region under the graph of y = f(x) over the interval [-a, a], where f(x) > 0. Assume that R is symmetric with respect to the y-axis. (a) Explain why y = f(x) is even—that is, why f(x) = f(-x).

Hint: First apply the substitution y = f(x) to the integral on the left and observe that dx = g'(y) dy.. Then apply Integration by Parts. 51. Let R be a lamina of uniform density submerged in a fluid of density w (Figure 29). Prove the following law: The fluid force on one side of R is equal to the area of R times the fluid pressure on the centroid. Hint: Let g(y) be the horizontal width of R at depth y. Express both the fluid pressure [Eq. (2) in Section 9.3] and y-coordinate of the centroid in terms of g(Y).

(b) Show that y = xf (x) is an odd function. (c) Use (b) to prove that

My

=

Fluid level

0.

(d) Prove that the COM of R lies on the y-axis (a similar argument applies to symmetry with respect to the x-axis).

Ycm

50. Prove directly that Eqs. (2) and (3) are equivalent in the following situation. Let f be a positive decreasing function on [0, b] such that f (b) = 0. Set d = f(0) and g(y)= f -1(y). Show that 1f b 2 0 f(x) dx = f

(depth)

yg(y)dy

FIGURE 29

CHAPTER REVIEW EXERCISES 1.

Compute p(X < 1), where X is a continuous random variable 1 with probability density p(x) = 7r(x2 -I- 1)

2.

Show that p(x) = 4e-x I

±

x13 is a probability density over

12. 1 CAS Compute the trapezoidal approximation T5 to the arc lengths of y = tan x over [0, Td. In Exercises 13-16, calculate the surface area of the solid obtained by rotating the curve over the given interval about the x-axis.

the domain [0, no) and find its mean.

13. y = x

3. Find a constant C such that p(x) = Cx3e-x2 is a probability density over the domain [0, no) and compute P(0 < X < 1).

14.

2 2 y = - x3/4 - - x5/4, 3 5

[0, 1]

4. The interval between patient arrivals in an emergency department is a random variable with exponential density function p(t) = 0.125e -0.125t (tin minutes). What is the average time between pa-

2 1 15. y = - x3/2 - - x1/2, 3 2

[1,2]

tient arrivals? What is the probability of two patients arriving within 3 min of each other?

1,

[0,4]

1 2 16. y = - x , 2

[0,2]

17. Compute the total surface area of the coin obtained by rotating the region in Figure 1 about the x-axis. The top and bottom parts of the region are semicircles with a radius of 1 mm.

5. Calculate the following probabilities, assuming that X is normally distributed with mean p, = 40 and a = 5. (a) P(X > 45)

(b) P(0 < X 0.

The general solution with k > 0 is illustrated in Figure 8.

8

•

The function y = Dekt models exponential growth when k > 0, and it models exponential decay when k 0. (a) Set a = (g/k)'/2 and rewrite the differential equation as dv — dt

(b) Prove that the tractrix is the graph of

X =

y2

64. Find the family of curves satisfying y' = x ly and sketch several members of the family. Then find the differential equation for the orthogonal family (see Exercise 63), find its general solution, and add some members of this orthogonal family to your plot.

y2

=

2

2\ — V

Then solve using Separation of Variables with initial condition v(0) = 0. (b) Show that the terminal velocity th>nx, i v(t) is equal to —a. c

SECTION 10.2

67. If a bucket of water spins about a vertical axis with constant angular velocity co (in radians per second), the water climbs up the side of the bucket until it reaches an equilibrium position (Figure 16). Two forces act on a particle located at a distance x from the vertical axis: the gravitational force —mg acting downward and the force of the bucket on the particle (transmitted indirectly through the liquid) in the direction perpendicular to the surface of the water. These two forces must combine to supply a centripetal force mco2x, and this occurs if the diagonal of the rectangle in Figure 16 is normal to the water's surface (i.e., perpendicular to the tangent line). Prove that if y = f (x) is the equation of the curve obtained by taking a vertical cross section through the axis, then —1/y/ = g I(w2x). Show that y = f (x) is a parabola.

Models Involving y' = k(y — b)

515

y = f(x)

X

FIGURE 16

Further Insights and Challenges 68. 1.7 In Section 6.2, we computed the volume V of a solid as the integral of cross-sectional area. Explain this formula in terms of differential equations. Let V(y) be the volume of the solid up to height y, and let A(y) be the cross-sectional area at height y as in Figure 17. (a) Explain the following approximation for small Ay:

69. A basic theorem states that a linear differential equation of order n has a general solution that depends on n arbitrary constants. The following examples show that, in general, the theorem does not hold for nonlinear differential equations. In each case the differential equation is not linear because of the (y')2 term.

(a) Show that (y')2 + y2 = 0 is a first-order equation with only one soluV(y ± Ay) — V(y)

A(y) Ay

11

(b) Use Eq. (11) to justify the differential equation dV Idy = A(y). Then derive the formula

70. Show that y -= Ce" is a solution of y" + ay' + by = 0 if and only if r is a root of P(r) = r2 ar ± b. Verify directly that y = CI e3x C2e —x is a solution of y" — 2y' — 3y = 0 for any constants C1, C2.

A(y)dy

V = a

71. A spherical tank of radius R is half-filled with water. Suppose that water leaks through a hole in the bottom of area B. Let y(t) be the water level at time t (seconds). dy N/Y, (a) Show that = 7(2Ry — y2) . dt

b— Area of cross section is A(y)

y + Ay_

tion y = 0. (b) Show that (y')2 + y2 + 1 = 0 is a first-order equation with no solutions.

Volume of slice is V(y + Ay) — V(y)

A(y)Ay

(b) Show that for some constant C,

yo = a —

15B,/2-i;

(10Ry3/2 — 3y5R) = C — t

(c) Use the initial condition y(0) = R to compute C, and show that C = te, the time at which the tank is empty. FIGURE 17

(d) Show that te is proportional to R512 and inversely proportional to B.

10.2 Models Involving

= k(y — b)

In this section we examine the differential equation dy — = k(y — b) dt where k and b are constants. This differential equation describes a quantity y whose rate of change is proportional to the difference y — b. It arises in many different modeling situations. We will use it to model a cooling object in a fixed-temperature environment, vertical motion under the influence of gravity and air resistance, and the changing value of an annuity. c t = ky c and is called This differential equation can be written in the form Y linear because the relationship between C!It-- and y is linear. Furthermore, because the

516

CHAPTER 10

INTRODUCTION TO DIFFERENTIAL EQUATIONS

coefficient k and the term c are constant, the differential equation is said to have constant coefficients. Thus, the differential equations we examine in this section are first-order linear constant coefficient differential equations. In Section 10.5, we examine the general first-order linear differential equation where the terms k and c could be functions of t. We can use Separation of Variables to show that the general solution to Eq. (1) is 2

y(t) = b + Cekt

Alternatively, we may observe that (y - b)' = y' since b is a constant, so Eq. (1) may be rewritten —(y - b) = k(y - b) dt In other words, y - b satisfies the differential equation of an exponential function and thus y - b = cekt, or b + Cekt , as claimed. y

=

GRAPHICAL INSIGHT The behavior of the solution y(t) as t C and k are positive or negative:

FIGURE 1 Two solutions to yl = —2(y — 1) corresponding to C = 2 and C = —2.

Newton's Law of Cooling implies that the object cools quickly when it is much hotter than its surroundings (when y - To is large). The rate of cooling slows as y approaches To. When the object's initial temperature is less than To, y' is positive and Newton's Law models warming.

oo depends on whether

• When k > 0, et tends to oo and, therefore, y(t) tends to oo if C > 0 and y(t) tends to -oo if C 0. We take the upward direction to be positive, so v 0 for a rising object and -kv is a downward acting force. The force due to gravity on an object of mass m is -mg, where g is the acceleration due to gravity, so the total force is F = -mg - kv. By Newton's Second Law of Motion, F = ma =

(a = v/ is the acceleration)

Thus, my' = -mg - kv, which can be written as / 7.) =

I In this model, k has units of mass per I time, such as kilograms per second.

mg ) (v + — k

3

This equation has the form v' = -k(v - b) with k replaced by k I m and b = -mg/k. By Eq. (2), the general solution is v(t) = Since Ce-(k/ m)t tends to zero as t

mg

Ce-(klm)t

oo, v(t) tends to a limiting terminal velocity:

F-5

mg terminal velocity = lim v(t) = - — t-÷00

Without air resistance, the speed (absolute value of velocity) of a falling object would increase without bound until a sudden collision with the ground occurs. On the other hand, with air resistance the speed also increases, but levels off and approaches a limiting value of mg I k. EXAMPLE 2 An 80-kg skydiver steps out of an airplane. (a) What is her terminal velocity if k = 8 kg/s? (b) What is her velocity after 30 s? Solution (a) By Eq. (5), with k = 8 kg/s and g = 9.8 m/s2, the terminal velocity is

v (m/s) > t 110

20

30

_mg k

(80)9.8 = 8

98 m/s

(b) With tin seconds, we have, by Eq. (4), —50

—98

-(8/80)t v(t) = -98 + Ce-(ki" = -98 + Ce

Terminal velocity = —98 m/s

FIGURE 3 Velocity of 80-kg skydiver in free-fall with air resistance (k = 8 kg/s).

=

_ 98

ce-o.it

We assume that the skydiver leaves the airplane with no initial vertical velocity, so -98(1 _ e-o.it) (Figure 3). v(0) = -98 + C = 0, and C = 98. Thus, we have v(t) = The skydiver's velocity after 30 s is v(30) = -98(1 - e-13.1(313))

-93.1 m/s

•

CHAPTER 10

INTRODUCTION TO DIFFERENTIAL EQUATIONS

In Example 8 in Section 7.4 and Example 5 in Section 7.5, we investigated limits involving the maximum height attained by a 1-kg ball that is launched upward at 30 m/s and acted on by gravity and air resistance. We observed that the maximum height (in meters) depends on the strength of the air resistance (i.e., on the proportionality constant k), a relationship given by

H(k) =

30k — 9.8 ln (150k 49

j _ 1) '

k2

Using Eq. (4), we can now derive this equation. In the next example, we find equations for the projectile's velocity v(t) and height y(t). The maximum height H(k) is then found by determining the height when the velocity is zero (see Exercise 17). EXAMPLE 3 A 1-kg ball is launched from the ground at 30 m/s and is acted on by air resistance that is expressed in the form —kv and by gravity. Determine its velocity v(t) and height y(t). Solution By Eq. (4), the projectile's velocity is 9.8 v(t) = -+ Ce —kt k where C must be chosen so that v(0) = 30. That is, C must satisfy 30 =

9.8

C

Thus, C = 30 + 4 9 , and therefore, 9.8

v(t) =

(30 ±

k

9.8

e-kr

k

Next, we take the antiderivative of v(t) to find y(t). We have 9.8

y(t) = — 7 c-t —

1

30 +

e

+C

To satisfy y(0) = 0, we must have C =J (30 + 7). Using this equation to substitute for C in y(t), and simplifying, we find 9.8 30k + 9.8 ( y(t) = — — t + 1 k2

•

An annuity is an investment in which an amount of money /30, called the principal, is placed in an account that earns interest. Let P(t) be the balance in the annuity (in dollars) after t years. When interest on the balance is compounded continuously at rate r, the rate of growth of the balance is proportional to the balance, and the proportionality constant is r. That is, P'(t) = r P(t). If we withdraw from the annuity at a constant rate of N dollars per year, we can model the balance in the annuity by the differential equation P'(t) = Rate of change

r P(t) — Growth due to interest

N

6

Withdrawal rate

This equation has the form y' = k(y — b) with k = r and b = N1r, so by Eq. (2), the general solution is P(t) = — + Ce"

10.2

SECTION

Models Involving

= k( y — b)

519

Since P(0) =- Po, we know Po = - + C and, therefore, C is given by C = Po - - • r r Because e" tends to infinity as t no, the balance P(t) tends to no if C > 0. If C < 0, then P(t) tends to -no (i.e., the annuity eventually runs out of money). If C = 0, then P(t) remains constant with value Nit-. EXAMPLE 4 Does an Annuity Pay Out Forever? An annuity earns continuously compounded interest at the rate r = 0.07, and withdrawals are made continuously at a rate of N $500/year. (a) When will the annuity run out of money if the initial deposit is P(0) = $5000? (b) Show that the balance increases indefinitely if P(0) = $9000. Solution We have NI,- =

7143, so P(t)= 7143 + Ceu7t by Eq. (7).

(a) If P(0) = 5000 = 7143 + Ce°, then C = -2143 and P(t) = 7143 - 2143e0O7t The account runs out of money when P(t) = 7143 - 2143e0wt = 0, or e. 07t

7143 =

0.07t = In

2143

The annuity money runs out at time t = FIGURE 4 The balance in an annuity may increase indefinitely or decrease to zero (eventually becoming negative), depending on the size of initial deposit /30, the interest rate, and the rate of withdrawal.

(7143) 2143

1.2

17 years.

(b) If P(0) = 9000 = 7143 + Ce°, then C = 1857 and P(t) = 7143 + 1857e007t Since the coefficient C = 1857 is positive, the account never runs out of money. In fact, • P(t) increases indefinitely as t —> no. Figure 4 illustrates the two cases.

10.2 SUMMARY • The general solution of y' = k(y - b) is y = b + Cekt, where C is a constant. • The following table describes the solutions to y' = k(y - b) (see Figure 5): Equation (k > 0) = k(y — b)

y' = -k(y - b)

Behavior as t

Solution Y(t)= b + Cekt

lim y(t) = t—oo

y(t)= b + Ce-kt

Ern y(t) = b t-,00

no

no if C > 0 if C 0 C 0

520

CHAPTER 10

INTRODUCTION TO DIFFERENTIAL EQUATIONS

• Three applications: - Newton's Law of Cooling: y' -= -k(y - To), y(t)= temperature of the object, To = ambient temperature, k = cooling constant k mg - Free-fall with air resistance: vi = -- (v ± -), v(t) = velocity, m = mass, m . k k = air resistance constant, g = acceleration due to gravity N Continuous annuity: P' = r ( P - r interest rate, N = withdrawal rate

),

P(t) = balance in the annuity, r =

10.2 EXERCISES Preliminary Questions 1.

Write a solution to y' = 4(y - 5) that tends to -cc as t -> co.

2.

Does y' = -4(y - 5) have a solution that tends to oo as t --> cc?

4. As an object cools, its rate of cooling slows. Explain how this follows from Newton's Law of Cooling.

3. True or false? If k > 0, then all solutions of y' = -k(y - b) approach the same limit as t -> oo.

Exercises 1. Find the general solution of y' = 2(y - 10). Then find the two solutions satisfying y(0) = 25 and y(0) = 5, and sketch their graphs. 2. Verify directly that y = 12 + Ce-31 satisfies y' = -3(y - 12) for all C. Then find the two solutions satisfying y(0) = 20 and y(0) = 0, and sketch their graphs. 3. Solve y' = 4y ± 24 subject to y(0) = 5. 4. Solve y' + 6y = 12 subject to y(2) = 10. In Exercises 5-12, use Newton's Law of Cooling. 5. A hot anvil with cooling constant k = 0.02 s-1 is submerged in a large pool of water whose temperature is 10°C. Let y(t) be the anvil's temperature t seconds later. (a) What is the differential equation satisfied by y(t)? (b) Find a formula for y(t), assuming the object's initial temperature is 100°C. (c) How long does it take the object to cool down to 20°? 6. Frank's automobile engine runs at 100°C. On a day when the outside temperature is 21°C, he turns off the ignition and notes that 5 minutes later, the engine has cooled to 70°C. (a) Determine the engine's cooling constant k. (b) What is the formula for y(t)? (c) When will the engine cool to 40°C? 7. At 10:30 AM, detectives discover a dead body in a room and measure its temperature at 26°C. One hour later, the body's temperature had dropped to 24.8°C. Determine the time of death (when the body temperature was a normal 37°C), assuming that the temperature in the room was held constant at 20°C. 8. A cup of coffee with cooling constant k = 0.09 min- I is placed in a room at temperature 20°C. (a) How fast is the coffee cooling (in degrees per minute) when its temperature is T = 80°C? (b) Use the Linear Approximation to estimate the change in temperature over the next 6 s when T = 80°C. (c) If the coffee is served at 90°C, how long will it take to reach an optimal drinking temperature of 65°C?

9. A cold metal bar at -30°C is submerged in a pool maintained at a temperature of 40°C. Half a minute later, the temperature of the bar is 20°C. How long will it take for the bar to attain a temperature of 30°C? 10. When a hot object is placed in a water bath whose temperature is 25°C, it cools from 100°C to 50°C in 150 seconds. In another bath, the same cooling occurs in 120 s. Find the temperature of the second bath. 11. ( GU ) Objects A and B are placed in a warm bath at temperature To = 40°C. Object A has initial temperature -20°C and cooling constant k = 0.004 s- I . Object B has initial temperature 0°C and cooling constant k = 0.002 s- I . Plot the temperatures of A and B for 0 < t < 1000. After how many seconds will the objects have the same temperature? 12. In Newton's Law of Cooling, the constant r = 1/ lc is called the characteristic time. Show that r is the time required for the temperature difference (y - T0) to decrease by the factor e-1 0.37. For example, if Y(0) = 100°C and To = 0°C, then the object cools to 100/e 37°C in time r, to 100/e2 P--• 13.5°C in time 2r, and soon. In Exercises 13-16, use Eq. (3) as a model for free-fall with air resistance. 13. A 60-kg skydiver jumps out of an airplane. What is her terminal velocity, in meters per second, assuming that k = 10 kg/s for free-fall (no parachute)? 14. Find the terminal velocity of a skydiver of weight w = 192 pounds if k = 1.2 lb-s/ft. How long does it take him to reach half of his terminal velocity if his initial velocity is zero? Mass and weight are related by w = mg, and Eq. (3) becomes v' = -(kg1w)(v + w/ k) with g = 32 ft/s2. 15. An 80-kg skydiver jumps out of an airplane (with zero initial velocity). Assume that k = 12 kg/s with a closed parachute and k = 70 kg/s with an open parachute. What is the skydiver's velocity at t = 25 s if the parachute opens after 20 s of free-fall?

174

16. Does a heavier or a lighter skydiver reach terminal velocity more quickly?

SECTION 10.2

17. As in Example 3, a 1-kg ball is launched upward at 30 m/s and is acted on by gravity and air resistance. 1 (-30k (a) Show that the ball's velocity is zero at time t* = — In +1 . k 9.8 30k — 9.8 In (150k + i) 49 (thereby establishing k2 the formula for H(k) given prior to the example).

(b) Show that y(t*)=

18. A 500 g ball is launched upward at 60 m/s and is acted on by gravity and air resistance that can be expressed in the form —kv, where v is the ball's velocity. (a) Determine the ball's velocity v(t), expressed in terms of k. (b) Determine the ball's height y(t), expressed in terms of k. (c) Determine H(k), the maximum height reached by the ball, as a function of k. In Exercises 19(a)—(f), use Eqs. (6) and (7) that describe the balance in a continuous annuity. 19. (a) A continuous annuity with withdrawal rate N = $5000/year and interest rate r = 5% is funded by an initial deposit of Po = $50,000. i. What is the balance in the annuity after 10 years? ii. When will the annuity run out of funds? (b) Show that a continuous annuity with withdrawal rate N = $5000/year and interest rate r = 8%, funded by an initial deposit of Po -= $75,000, never runs out of money. (c) Find the minimum initial deposit Po that will allow an annuity to pay out $6000/year indefinitely if it earns interest at a rate of 5%. (d) Find the minimum initial deposit Po necessary to fund an annuity for 20 years if withdrawals are made at a rate of $10,000/year and interest is earned at a rate of 7%. (e) An initial deposit of 100,000 euros is placed in an annuity with a French bank. What is the minimum interest rate the annuity must earn to allow withdrawals at a rate of 8000 euros/year to continue indefinitely? (f) Show that a continuous annuity never runs out of money if the initial balance is greater than or equal to N1r, where N is the withdrawal rate and r the interest rate.

521

(b) How long will it take Sam to pay back the loan if N = $1200? (c) Will the loan ever be paid back if N = $800? 21. April borrows $18,000 at an interest rate of 5% to purchase a new automobile. At what rate (in dollars per year) must she pay back the loan, if the loan must be paid off in 5 years? Hint: Set up the differential equation as in Exercise 20. 22. Let N(t) be the fraction of the population who have heard a given piece of news t hours after its initial release. According to one model, the rate N'(t) at which the news spreads is equal to k times the fraction of the population that has not yet heard the news, for some constant k > 0. (a) Determine the differential equation satisfied by N(t). (b) Find the solution of this differential equation with the initial condition N(0) = 0 in terms of k. (c) Suppose that half of the population is aware of an earthquake 8 hours after it occurs. Use the model to calculate k and estimate the percentage that will know about the earthquake 12 h after it occurs. 23. Current in a Circuit When the circuit in Figure 6 (which consists of a battery of V volts, a resistor of R ohms, and an inductor of L henries) is connected, the current I(t) flowing in the circuit satisfies dI L — + RI = V dt with the initial condition /(0) = 0. (a) Find a formula for I(t) in terms of L, V, and R. (b) Show that lim I(t) = V I R. (c) Show that I(t) reaches approximately 63% of its maximum value at the characteristic time r = LIR.

Inductor

Sam borrows $10,000 from a bank at an interest rate of 9% and 20. pays back the loan continuously at a rate of N dollars per year. Let P(t) denote the amount still owed at time t. (a) Explain why P(t) satisfies the differential equation = 0.09y — N

Models Involving y' = k(y — b)

> R 4> Resistor

Battery 1—= V

FIGURE 6 Current flow approaches the level 'max = VIR.

Further Insights and Challenges 24. Show that the cooling constant of an object can be determined from t2 by the formula two temperature readings y(ti) and y(t2) at times t1 1 ln Y(tz) — To ) k= ti — t2 y(ti ) — To) 25. Show that by Newton's Law of Cooling, the time required to cool an object from temperature A to temperature B is t —

1 (A — To ) ln k B — To )

(a) Determine the velocity v(t). (b) Show that the projectile's velocity is zero at

k

—9.8 —

( vok ) — +1 9.8m

(c) Determine the height y(t). (d) The maximum height is y(t*). Show that

where To is the ambient temperature. 26. A projectile of mass m kg travels straight up from ground level with initial velocity vo m/s. Suppose that the velocity v satisfies v'(t) =

In

y(t*) —

mvok — 9.8m2 In ( v°k + 1) 9.8m k2

(e) If air resistance is negligible, then the maximum height reached by the projectile is lirn y(t*). Determine the limit. k-> 0

INTRODUCTION TO DIFFERENTIAL EQUATIONS

10

CHAPTER

10. Graphical and Numerical Methods In the previous two sections, we focused on finding solutions to differential equations. Differential equations cannot always be solved explicitly. Fortunately, there are techniques for analyzing the solutions that do not rely on explicit formulas. Even for differential equations that have exact solutions, these techniques are valuable because they provide us with extra insight into the behavior of the solutions. In this section, we discuss the method of slope fields, which provides us with a good visual understanding of firstorder equations. We also discuss Euler' s Method for finding numerical approximations to solutions. We use t as the independent variable. A first-order differential equation can then be written in the form dy = F(t,y) dt

11

2,

/

-

/

/

1

-

/

/

1

I

I

\

-

/

/

\

-

/

\

\-/

I

I

1-

-

1

\

I

I

1

1

-

1

1

-.

1

1

I /

/-

7

0 and A > 0): dy = k 1-Tit Y z)

Y=

A 1— e —ktIB'

or equivalently,

y—A

= Bek

• Two equilibrium (constant) solutions: — y = 0 is an unstable equilibrium. — y = A is a stable equilibrium. • If the initial value yo = y(0) satisfies yo > 0, then y(t) approaches the stable equilibrium y = A; that is, tlim y(t) = A.

10.4 EXERCISES Preliminary Questions 1. Which of the following differential equations is a logistic differential equation? dy dy y (a) — = 2y(1 — y2) (b) — = 2y (1 — — ) dt dt 3 dy dy (c) — = 2y (1 — (d) — 2y(1 — 3y) dt 4 dt

dy 2. What are the constant solutions to — = ky (1 — dt Al 3. Is the logistic equation separable?

S E C T10 N 10.4

Exercises 1. Find the general solution of the logistic equation

533

(b) Find the time t when yi (t) = 5. (c) When does y2(t) become infinite?

dy ( = 3y 1 — ) dt 5 Then find the particular solution satisfying y(0) = 2. dy 2. Find the solution of — = 2y(3 — y), y(0) = 10. dt In Exercises 3-4, for each of (a)—(c), give the solution y(t) satisfying the initial condition. (Note: the general solution formula for the logistic equation, Eq. (5), applies when a solution is not constant.) dy 3. = 3y(6 — y) dt (a) y(0) = 6

The Logistic Equation

12. A tissue culture grows until it has a maximum area of M square centimeters. The area A(t) of the culture at time t may be modeled by the differential equation 7 where k is a growth constant. (a) Show that if we set A = u2, then

(b) y(0) = 4

(c) y(4) = 0

du dt

lk 2

1 \ M

4. dY — y (2 — Then find the general solution using Separation of Variables. (a) y(0) = 6

(b) y(0) = 8

(c) y(0) = —2

5. A population of squirrels lives in a forest with a carrying capacity of 2000. Assume logistic growth with growth constant k = 0.6 yr . (a) Find a formula for the squirrel population P(t), assuming an initial population of 500 squirrels. (b) How long will it take for the squirrel population to double? 6. The population P(t) of mosquito larvae growing in a tree hole increases according to the logistic equation with growth constant k = 0.3 day-1 and carrying capacity A = 500. (a) Find a formula for the larvae population P(t), assuming an initial population of Po = 50 larvae. (b) After how many days will the larvae population reach 200? 7. Sunset Lake is stocked with 2000 rainbow trout, and after 1 year the population has grown to 4500. Assuming logistic growth with a carrying capacity of 20,000, find the growth constant k (specify the units) and determine when the population will increase to 10,000. 8. Spread of a Rumor A rumor spreads through a small town. Let y(t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1 — y that has not yet heard the rumor. (a) Write the differential equation satisfied by y in terms of a proportionality factor k. (b) Find k (in units of day '), assuming that 10% of the population knows the rumor at t = 0 and 40% knows it at t = 2 days. (c) Using the assumptions of part (b), determine when 75% of the population will know the rumor. 9. A rumor spreads through a school with 1000 students. At 8 AM, 80 students have heard the rumor, and by noon, half the school has heard it. Using the logistic model of Exercise 8, determine when 90% of the students will have heard the rumor. 10. (GU) A simpler model for the spread of a rumor assumes that the rate at which the rumor spreads is proportional (with factor k) to the fraction of the population that has not yet heard the rumor. (a) Compute the solutions to this model and the model of Exercise 8 with the values k = 0.9 and yo = 0.1. (b) Graph the two solutions on the same axis. (c) Which model seems more realistic? Why? 11. Let k = 1 and A =- tin the logistic equation. (a) Find the solutions satisfying yi (0) = 10 and y2(0) = —1.

(b) Show that the general solution to Eq. (7) is

A(t)= M

(Ce (k h hr"

— 1) 2

Ce(1 R/711)t +1

13. (GU) In the model of Exercise 12, let A(t) be the area at time t (hours) of a growing tissue culture with initial size A(0) = 1 cm2, assuming that the maximum area is M = 16 cm2 and the growth constant is k = 0.1. (a) Find a formula for A(t). Note: The initial condition is satisfied for two values of the constant C. Choose the value of C for which A(t) is increasing. (b) Determine the area of the culture at t = 10 hours. (c) (GU) Graph the solution using a graphing utility. 14. Show that if a tissue culture grows according to Eq. (7), then the growth rate reaches a maximum when A = M/3. 15. In 1751, Benjamin Franklin predicted that the U.S. population P(t) would increase with growth constant k = 0.028 year'. According to the census, the U.S. population was 5 million in 1800 and 76 million in 1900. Assuming logistic growth with k = 0.028, find the predicted carrying capacity for the U.S. population. Hint: Use Eqs. (3) and (4) to show that P(t) P(t) — A

Po kt e P0 — A

16. 174 Reverse Logistic Equation equation (with k,B > 0):

Consider the following logistic

dP = —kP(1- 1 4) dt

8

(a) Sketch the slope field of this equation. IC), where C is a nonzero (b) The general solution is P(t)= B/(1 — constant. Show that P(0) > B if C> 1 and 0 < P(0) < B if C B. (d) Show that P = 0 is a stable equilibrium and P = B is an unstable equilibrium.

534

CHAPTER 10

INTRODUCTION TO DIFFERENTIAL EQUATIONS

Further Insights and Challenges In Exercises 17 and 18, let y(t) be a solution of the logistic equation A dy = ky 1 — dt ( A

9 A 2

Inflection point

where A > 0 and k > 0. y(0)

17. (a) Differentiate Eq. (9) with respect to t and use the Chain Rule to show that

FIGURE 5 An inflection point occurs at y = A/2 in the logistic curve. d2y = k2y

y

( i

k

dt2

2y A)

A be the general nonequilibrium solution to 1 — e—kt/B Eq. (9). If y(t) has a vertical asymptote at t = tb, that is, if lim ±oo, we say that the solution "blows up" at t = tb.

18. Let y =

(b) Show that the graph of the function y is concave up if 0 < y < A/2 and concave down if A/2 < y A, then y blows up at a time tb, which is negative (and hence does not correspond to a real time). (c) Show that y blows up at some positive time tb if and only if y(0) co (a) lim (an + bn ) (b) lim a„3 n "

(i) cos nn

(c)

lim cos(gbn ) n-co

n! (ii) 2n (_ o n+1

In Exercises 15-28, use Theorem 1 to determine the limit of the sequence or state that the sequence diverges.

(d)

,f•

•••

(iv)

n +1

1 for n = 1, 2, 3, . . . . Write out the first three terms of 2n - 1 the following sequences. (a) bn = an+1 (b) cn = an+3 2.

Let an =

(c) d„ = an2

(d) e = 2an - an+i

In Exercises 3-12, calculate the first four terms of the sequence, starting wit/in = 1. 3n (2n - 1)! 3. cn = 4. bn = n! n! 5.

al = 2,

an+i = 2ari

6.

lit = 1,

6, = bn _i +

19. an = Gi

18. an

n

4 + n - 3n2 4n2 + 1

20. „ = ()

n

21. cn = 9n

22. Zn = 10-1/n

23. a„ -

24. an _

'1n2 +l ( 12n + 2 25. an = ln n + 4n)

27. zn =

-Vn3 + 1

26. rn = In n - ln(n2 + 1) 28. y„ = ne I In

en

In Exercises 29-32, use Theorem 4 to determine the limit of the sequence.

b_1

29. an =

bn = 5 + cosnn

9.

1 1 1 cn = 1 + i+- 3-+•• •+ 71

8.

cn = (- o2fl-F1

1 1 1 10. wn=l+ F -1-yf +• • •+j b2 = 3,

5n - 1 12n + 9

3

7.

11. b1 = 2,

bn

2anbn)

4 16. an = 20- -n2

15. an = 5 - 2n (d)

lim (a„2 -

bn = 2bn- + bn-2

Fn+i 12. an = where Fn is the nth Fibonacci number. Fn 13. Find a formula for the nth term of each sequence. 1 -1 1 2 3 4 (a) T' Fr • • • 6 7 8

31. an = cos-1

1 n

30. an = e4n/ (3n+9) \ n3 2n3 + 1)

32. an = tan ' (e") Fn

In Exercises 33-34 let an = --,+1 where fFn I is the Fibonacci sequence. Fn The sequence {an} has a limit. We do not prove this fact, but investigate the value of the limit in these exercises. 33. ( CAS ) Estimate lim an to five decimal places by computing an for n->oo

sufficiently large n.

34. Denote the limit of {a} by L. Given that the limit exists, we can determine L as follows: (a) Show that an+1 = 1+ 1a„ •

SECTION 11.1 1

1

N/n2 + 1

N/n2 + 2

(b) Given that {an} converges to L, it follows that {an±i } also converges to L (see Exercise 85). Show that L2 - L - 1 = 0 and solve this equation to determine L. (The value of L is known as the golden ratio. It arises in many different situations in mathematics.)

72. c,,

. Find a number M such that: n+1 11 < 0.001 for n > M. (a) Ian (b) Ian - 11 M. Then use the limit definition to prove that lim an = 1.

73. an = (2n + 3n )lin,

35. Let an =

74. an = (n + 10n )1/ n ,

.

(a) Find a value of M such that Ibn I < 10-5 for n > M. (b) Use the limit definition to prove that lim bn -= 0. n->00 37. Use the limit definition to prove that lim n-2 = 0. n-> oo 38. Use the limit definition to prove that nItno, n

76. Show that an =

n _,

In Exercises 39-66, use the appropriate limit laws and theorems to determine the limit of the sequence or show that it diverges. 39. a,,= 10 + (--

1)n

+3-

40. dn =

9

41. cn = 1.01"

42. bn = e l -n2

43. an = 21/ n

44. b,, =n "

9n 45. cn = n!

46. an =

47. an = 49. an =

cos n

50. c„ =

55. c,,

= in ( 2n + 1) +4)

57. Yn= 59. Yn -

(-1)n

54. bn = tan

60. bn =

5n

n) n vn

(-

1rn

3

. Jr 61. an -= n sin n 3_ 4n 63. bi, = 2+7•4n

3n3 + 4- n ! 62. bn = -n7rn 3_ 4n 64. a,, = 2+7•3n

65. a, = (1 + -1 ) n n

66. an = (1 + 1.2.) n n

In Exercises 67-70, find the limit of the sequence using L'Ilopital's Rule. 67. an =

(lnn)2

68. bn = N/Ti ln (1+ -1 ) 70. dn = n2 ( / n3 + 1 - n)

69. cn = n(Vn2 + 1 - n)

In Exercises 71-74, use the Squeeze Theorem to evaluate nlLnrcix, an by verifying the given inequality. 71. an =

1 / n4 + n8 '

1 -shn4

< an
co words, find a function f such that an = f (n) converges but X -00f (x) does not exist.

58. an =

en + (-3)n

1 is decreasing. 2n + I

82. Use the limit definition to prove that if {an} is a convergent sequence of integers with limit L, then there exists a number M such that an = L for all n > M.

.171

56. cn =

en

10 < an < (2 • 10n )l in

81. Using the limit definition, prove that if {an} converges and {bn} diverges, then {an + bn} diverges.

51. dn = ln 5n - inn! 52. dn = ln(n2 + 4) - ln(n2 -1) 4\ 1/3 ) 53. an = (2 +

n2 + 1 3 < an < (2 3n )lin =211n 3

80. Give an example of divergent sequences {an} and {b,,) such that {an + bn} converges.

n!

48. a„ =

1

+

79. Give an example of a divergent sequence {ad such that lim Ian I n->0.0 converges.

82n

3n2 +n+2 2n2 - 3

553

75. Ell Which of the following statements is equivalent to the assertion lirn an = L? Explain. fl-* 00 (a) For every c > 0, the interval (L - E, L e) contains at least one element of the sequence {an}. (b) For every E > 0, the interval (L - E, L 6) contains all but at most finitely many elements of the sequence Ian l•

ti->oo

36. Let b,, =

oo 86. Let {an } be a sequence such that lim Ian I exists and is nonzero. Show n->00 that lim an exists if and only if there exists an integer M such that the sign n-4.00 of an does not change for n > M. 87. Proceed as in Example 13 to show that the sequence is increasing and bounded above by M = 3. Then prove that the limit exists and find its value. 88. Let {an} be the sequence defined recursively by ao

0,

an±i =

+ an

a2 = 12 + 2, a3 = 2 + 12 + Thus, ai = (a) Show that if an 1. Hint: Show that Hn > f

(b)

(c) Prove that lim an exists. This limit, denoted y, is known as Euler's Constant. It appears in many areas of mathematics, including analysis and number theory, and has been calculated to more than 100 million decimal places, but it is still not known whether y is an irrational number. The first 10 digits are y 0.5772156649.

Geometric Arithmetic mean mean

a +,

1

lim cn . Use the Squeeze Theorem to determine n-)•oo

93. al

(a) Show that an < b,, for all n (Figure 14). (b) Show that {an} is increasing and {bn} is decreasing. bn - an (c) Show that b,,+1 - an+1 < 2 (d) Prove that both {an} and {bn} converge and have the same limit. This limit, denoted AGM(ai, b1), is called the arithmetic-geometric mean of ai and b1 . (e) Estimate AGM(1, ,12..) to three decimal places.

a,,

1

(a) Calculate cz, ca• (b) Use a comparison of rectangles with the area under y = x-1 over the interval [n, 2n] to prove that

90. Let b,, = (a) Show that In bn = -1 n

1

bn +

AGM(a l , b1) FIGURE 14

11.2 Summing an Infinite Series Many quantities that arise in mathematics and its applications cannot be computed exactly. We cannot write down an exact decimal expression for the number 7r or for values of the sine function such as sin 1. However, sometimes these quantities can be represented as infinite sums. For example, using Taylor series (Section 11.8), we can show that sin 1 = 1 -

1 —

1 — -

1 —

1 — -

1 —

•••

3! 5! 7! 9! 11! Infinite sums of this type are called infinite series. We think of them as having been obtained by adding up all of the terms in a sequence of numbers. But what precisely does Eq. (1) mean? How do we make sense of a sum of infinitely many terms? The idea is to examine finite sums of terms at the start of the series and see how they behave. We add progressively more terms and determine whether or not the sums approach a limiting value. More specifically, for the infinite series a + a2

a3

a4 ± • • • +

±•••

define the partial sums: Si = a S2 = al + az S3 =

SN =

al

al -I- a2

a2

a3

a3 + • • • ± aN

SECTION 11.2

Summing an Infinite Series

555

The idea then is to consider the sequence of values, S1, S2, 53, . . .,SN,. . ., and whether the limit of this sequence exists. For example, here are the first five partial sums of the infinite series for sin 1: Si = 1 , 1 S2 = 1— 1 = 1— g

S3 =

1—

54=1-7 S5 = 1 —

1

0.833

1

1

1

0.841667

+R=1-64-Tki 1 1 1 6 ± 120

1 g

1

+

120

;•"2, 0.841468

5040

0.8414709846

1 1 + 5040 362,880

Compare these values with the value obtained from a calculator: sin 1 ===, 0.8414709848079 We see that S5 differs from sin 1 by less than 10-9. This suggests that the partial sums converge to sin 1, and in fact, in Section 11.8 we will prove that sin 1 = fim SN

• Infinite series may begin with any value for the index. For example, 00 1 E — = — ± —± — +••• 4 5 n 3 n=3 When it is not necessary to specify the starting point, we write simply E an. • Any letter may be used for the index. Thus, we may write a., ak, a, and so on.

(see Example 2). It makes sense then to define the sum of an infinite series as a limit of partial sums. In general, an infinite series is an expression of the form 00

E n=1

= ai + a2

a3

±•••

where {an } is any sequence. For example, Sequence

General term

1 1 4' 277' • "

a_ n— — 3n

1 1 1 1 4' 5' Tg- • •

1 an = r7 2

1

1

Infinite series °

° i

1

/ 3n = 3 n=1

1 9 + 27 + 81 + • • •

00 1 _1+1+1 n=1 n2

I

4

9 ± 16 ± • • •

The Nth partial sum SN is the finite sum of the terms up to and including aN: SN =

E = al

a2 + a3 + • • • + aN

n=1

If the series begins at k, then SN = ak

ak+I ± • • • ± aN • 00

DEFINITION Convergence of an Infinite Series

An infinite series

E an converges

n=k to the sum S if the sequence of its partial sums {SA converges to S:

00

MIR SN = S N

In this case, we write S = n=k • If the limit does not exist, we say that the infinite series diverges. • If the limit is infinite, we say that the infinite series diverges to infinity.

556

INFINITE SERIES

CHAPTER 11

We can investigate series numerically by computing several partial sums SN. If the sequence of partial sums shows a trend of convergence to some number S, then we have evidence (but not proof) that the series converges to S. The next example treats a telescoping series, where the partial sums are particularly easy to evaluate. EXAMPLE 1

Investigate numerically:

Telescoping Series

n=1

1 1 1 1 1 n(n + 1) = 1(2) + 2(3) + 3(4) + 4(5) ± • • •

Then compute the sum of the series using the identity: 1 n

1 n(n + 1) Partial co 1 Sums for n(n + 1)

CD TABLE 1

E

Solution The values of the partial sums listed in Table I suggest convergence to S = 1. To prove this, we observe that because of the identity, each partial sum collapses down to just two terms:

n=1

= SN

10

50 100 200 300

0.90909 0.98039 0.990099 0.995025 0.996678

S2 = S3 =

1 1(2)

1 1 — — — 1 2

1 1(2)

2(3)

1

1(2)

2(3)

3(4)

)

3)

3

G-74)=1-

1

In general, 73 11)

SN

( 7

(1 =1

In most cases (apart from telescoping series and the geometric series introduced later in this section), there is no simple formula like Eq. (2) for the partial sum SN. Therefore, we shall develop techniques for evaluating infinite series that do not rely on formulas for SN.

1 n+1

+

1_

N±1)

1

2

N+1

The sum S is the limit of the sequence of partial sums: S= lim N

SN =

lim N

( 1

1 N+1)

=1

It is important to keep in mind the difference between a sequence Ian ) and an infinite 00 series n=1

EXAMPLE 2

Sequences Versus Series 1 an, where an = n(n 1) n=1

Discuss the difference between {an} and

E

Make sure you understand the difference between sequences and series. • With a sequence, we consider the limit of the individual terms an. • With a series, we are interested in the sum of the terms al +

a2 + a3 + • • •

Solution The sequence is the list of numbers verges to zero:

..

. . .This sequence con-

1 lima= lim =0 n—>00 n—>00 n(n + 1) The infinite series is the sum of the numbers an, defined as the limit of the sequence of partial sums. This sum is not zero. In fact, the sum is equal to 1 by Example 1:

1 E an n=1 E n(n + 1) = 1(2) ± 2(3) + 3(4) ± • • • = I n=1 00

which is defined as the limit of the sequence of partial sums.

27,

00

•

SECTION 11.2

Summing an Infinite Series

557

The next theorem shows that infinite series may be added or subtracted like ordinary sums, provided that the series converge. THEOREM 1 Linearity of Infinite Series

E

(an + bn ), Furthermore,

If E

an and E

bn converge, then

E(an — 10, and E can also converge, the latter for any constant C. E(an +b0)= a„ +E19„ E(an — bn ) a„ — L b, E can = cE

(c any constant)

an

Proof These rules follow from the corresponding linearity rules for limits. For example, oo E(a

n

bn ) = limE(an

n=1

bn ) = lim (E an ± N-4-oo

n=1 N

=

lim N —>oo

E an ± n=1

lim

N—> oo

n=1

E n=1 bn)

00

co

oo

n=1

n=1

n=1

E bn = E an ± E

bn

•

A main goal in this chapter is to develop techniques for determining whether a series converges or diverges. It is easy to give examples of series that diverge:

• E 1 diverges to infinity (the partial sums increase without bound): n=1 S1 = 1,

S2 =

1 ± 1 = 2,

S3

=1+1+1=3,

S4

=1+1+1+1=4,

...

co

• E(_1)n-1 diverges (the partial sums jump between 1 and 0): n=1 Si = 1,

S3 = 1 — 1 + 1 = 1,

S2 = 1 — 1 = 0,

S 4 = 1 — 1 + 1 — 1 = 0,

..

Next, we study geometric series, which converge or diverge depending on the common ratio r. A geometric series with common ratio r 0 is a series defined by a geometric sequence crn, where c 0. If the series begins at n = 0, then

.0 E crn

+ cr

cr2

cr3

cr4 + cr5 + • • •

n=0

For r = 21 and c = 1, we have the following series:

1 1 1 1 1 + ••• E — —2 + —4 + —8 + — 16 n=1

2n

Figure 1 demonstrates that adding successive terms in the series corresponds to moving stepwise from 0 to 1, where each step is a move to the right by half of the remaining distance. Thus it appears that the series converges to 1.

558

CHAPTER 11

INFINITE SERIES 1 2

0

I 1 2

I +1 2 4

0

I 1 i

1 3 ',"i

0

I 1

4 3

I 7 1

1 2

3 vei

7 15 i 75

1

oo

FIGURE 1 Partial sums of

I

1

_. E 2n 1

n=1

1

1

1

1

There is a simple formula for computing the partial sums of a geometric series: THEOREM 2 Partial Sums of a Geometric Series with r 0 1,

For the geometric series E nc3t o crn

SN = C + cr + cr2 ± cr 3 ± • • • + crN =

c(1 — r N+1)

1—r

3

Proof In the steps below, we start with the expression for SN, multiply each side by r, take the difference between the first two lines, and then simplify: SN = C + cr + cr2 ± cr3 + • • • ± crN rSN = cr + cr2 + cr3 + • • • ± crN + cr A1+1

SN — rSN = c — crN+1 SN(1 — r) = c(1 — r N+1)

Since r 0 1, we may divide by (1 — r) to obtain SN = Geometric series are important because they • arise often in applications. • can be evaluated explicitly. • are used to study other, nongeometric series (by comparison).

•

Now, the partial sum formula enables us to compute the sum of the geometric series

when Id < 1. THEOREM 3 Sum of a Geometric Series

00 crn = E n=0 In words, the sum of a geometric series is the first term divided by 1 minus the common ratio.

C(1 — r is1+1) 1—r

Let c 0 O. If 1r 1 < 1, then

c + cr ± cr2 ± cr3 ±

c

=

1—r

4

If VI > 1, then the geometric series diverges. Proof If r --= 1, then the series certainly diverges because the partial sums SN = Nc grow arbitrarily large. If r 0 1, then Eq. (3) yields Lim SN = liM A/—>oo isi oo

c(1 — r' +1) 1— r

=

c

1—r

c

liM r N+1

1 — r N—>oo

If Id < 1, then lim r AI +1 = 0 and we obtain Eq. (4). If Id > 1 and r 0 1, then N —>co liM N—>oo

r N+1 does not exist and the geometric series diverges.

li

SECTION 11.2

00 EXAMPLE 3 Evaluate E 5 n=o

Summing an Infinite Series

.

Solution This is a geometric series with common ratio r By Eq. (4), 00

1

1

1

5-1 and first term c = 1.

1 1_

n=0 oo

EXAMPLE 4 Evaluate

559

E 7 (- 43 ) n = 7 ( -34 n=3

3

5 =4

3 4

+ 7 (— 4

•

5

+ 7 (- -3-) + 4

Solution This is a geometric series with common ratio r = — 743 and first term 3 c = 7 (-4) . Therefore, it converges to

7(_)3 1—r

27 16

= I — (- 4)

•

EXAMPLE 5 Find a fraction that has repeated decimal expansion 0.212121 . . . . 100 213 Solution We can write this decimal as the series _12010 This is a geYou can check the result by dividing 7 by 33 on a calculator and seeing that the desired decimal expansion, 0.212121. , results.

ometric series with c = 4 0 and r = ffi. Thus, it converges to 21 100 1- r

1 - th

=

21 =7 —

99

33

•

EXAMPLE 6 A Probability Computation Nina and Brook are participating in an archery competition where they take turns shooting at a target. The first one to hit the bullseye wins. Nina's success rate hitting the bullseye is 45%, while Brook's is 52%. Nina pointed out this difference, arguing that she should go first. Brook agreed to give the first turn to Nina. Should he have? Solution We can answer this question by determining the probability that Nina wins the competition. It is done via a geometric series. Nina wins in each of the following cases.

Two events A and B are called independent if one of them occurring does not affect the probability of the other occurring. In such a case, the probability that A and B both occur is the product of the probabilities of each occurring individually. This idea applies to each case that leads to a win by Nina. For example, in the second case, the probability that Nina wins is the product of the probabilities of: Nina missing on Turn 1 (0.55), Brook missing on Turn 1 (0.48), and Nina hitting on Turn 2 (0.45).

• By hitting the bullseye on her first turn (which happens with probability 0.45), or • By having both players miss on their first turn and Nina hit on her second turn [which happens with probability (0.55)(0.48)(0.45)], or • By having both players miss on their first two turns and Nina hit on her third [which happens with probability (0.55)(0.48)(0.55)(0.48)(0.45)], and so on... There are infinitely many different cases that result in a win for Nina, and because they are distinct from each other (that is, no two of them can occur at the same time) the probability that some one of them occurs is the sum of each of the individual probabilities. That is, the probability that Nina hits the bullseye first is: 0.45 + (0.55)(0.48)(0.45) + (0.55)2(0.48)2(0.45) + • • • This is a geometric series with c = 0.45 and r = (0.55)(0.48) = 0.264. It follows that the probability that Nina wins is 1 %14 0.61. Thus, Brook would have been wise not to • let Nina go first.

560

CHAPTER 11

INFINITE SERIES

EXAMPLE 7 Evaluate

°° 2 + 3' E 5n •

n=0

Solution Write the series as a sum of two geometric series. This is valid by Theorem 1 because both geometric series converge: 00 = 2 2 =5 5n 5) 5n 5n La 5n

E +E

E E + n=0

n=0

n=0

n=0

5

n=0

Both geometric series converge.

•

CONCEPTUAL INSIGHT Assumptions Matter Knowing that a series converges, sometimes we can determine its sum through simple algebraic manipulation. For example, suppose we know that the geometric series with r = 1/2 and c = 1/2 converges. Let us say that the sum of the series is S, and we write 1 1 1 S= - +- +- +••• 2 4 8 1 1 1 2S = 1 + - + - + - + • • • = 1 + S 2 4 8 Thus, 2S = 1 + S, or S = 1. Therefore, the sum of the series is 1. Observe what happens when this approach is applied to a divergent series: S= 1 + 2 + 4 + 8 + 16 + • • • 2S = 2 + 4 + 8 + 16 + • • • = S - 1 This would yield 2S = S - 1, or S -1, which is absurd because the series diverges. Thus, without the assumption that a series converges, we cannot employ such algebraic techniques to determine its sum. 00

E 1 diverges because the Nth partial sum SN = N diverges to

The infinite series

k=1

infinity. It is less clear whether the following series converges or diverges: CO

+1 n E ( -1)n n=1

n

_1 1— 2

2

3

4

3 + 71. —

5 +

We now introduce a useful test that allows us to conclude that this series diverges. The idea is that if the terms are not shrinking to 0 in size, then the series will not converge. This is typically the first test one applies when attempting to determine whether a series diverges. The nth Term Divergence Test (also known as the Divergence Test) is often stated as follows: CO

If

Ean converges, then lim an = 0. n=1

n—> oo

n=1

diverges.

n

In practice, we use it to prove that a given series diverges. It is important to note that it does not say that if lirn a, = 0, then

E an necessarily converges. We will see n=1

1 that even though lim — = 0, the series n—>oo n I - diverges. n.= I

00 THEOREM 4 nth Term Divergence Test If lim an 0 0, then the series E an

Proof First, note that an = Sn - 5n-1 because = (ai + a2 + • • • + an-i) + an = Sn-1 ± an

00

If

E an converges with sum S, then n=1

lim an = lim (Sn — Sn—i ) = lim S0 n—+

n—> co

n—).Do

co

lim Sn — 1 = S — S = 0

n —> pc

oo

Therefore, if an does not converge to zero,

E, cannot converge. n =1

•

SECTION 11.2

Summing an Infinite Series

561

00 EXAMPLE 8 Prove the divergence of E 4n + 1 n=1 Solution We have 1 1 Jim a, = lim = Jim = n—>oo 4n + 1 n—>oo 4 ± 1/n 4 The nth term an does not converge to zero, so the series diverges by the nth Term Divergence Test (Theorem 4). • EXAMPLE 9 Determine the convergence or divergence of 00

n n+1

n=1

=

1

2

2

3

3 4 + — — — +. . . 4 5

Solution The general term a, = (- 1)n-1

does not approach a limit. Indeed, n n+1 tends to 1, so the odd terms a20+1 tend to 1, and the even terms azn tend to —1.

n+1 Because lim a, does not exist, the series diverges by the nth Term Divergence Test. n

oo

•

The nth Term Divergence Test tells only part of the story. If a, does not tend to zero, then a, certainly diverges. But what if an does tend to zero? In this case, the series may converge or it may diverge. In other words, nlitri a, = 0 is a necessary condition of

E

convergence, but it is not sufficient. As we show in the next example, it is possible for a series to diverge even though its terms tend to zero. EXAMPLE 10 divergence of

4-

E 1 = _1 _1 _1 +... Nif n=1 1 1 SN = — ± — ± • • • +

1

•

3

1 >

•

2 1

Prove the

Solution The general term 1/,/t tends to zero. However, because each term in the partial sum SN is greater than or equal to 1/1 g- -, we have

Partial sums, SN . • • • •

5

Sequence Tends to Zero, Yet the Series Diverges

1 +

•

±

Nt;rnis

Terms of sequence, a, • • • • • • • • • • 1 2 3 4 5 6 7 8 9 10 11 12

•

- N FIGURE 2 The partial sums of E n`xj_ i *diverge even though the terms an = 1 tend to zero.

1 +

z

N)

= A/TV

This shows that SN > ,N/TV. But ,N/Kr increases without bound (Figure 2). Therefore, SN • also increases without bound. This proves that the series diverges.

11.2 SUMMARY • An infinite series is an expression

E an = ai + az + a3 + a4 + • • •

n=1

We call a, the general term of the series. An infinite series can begin at n = k for any integer k.

2

CHAPTER 11

INFINITE SERIES • The Nth partial sum is the finite sum of the terms up to and including the Nth term:

E an = al ± a2

SN =

a3 ± • • • + aN

n=1 lim SN. If the limit exists, N —>o we say that the infinite series is convergent or converges to the sum S. If the limit does not exist, we say that the infinite series diverges. If the sequence of partial sums of a series increases without bound, we say that the series diverges to infinity. 00 an diverges. However, a series nth Term Divergence Test: If lim an 0 0, then n—>oo n=1 may diverge even if its general term a, tends to zero. Partial sum of a geometric series:

By definition, the sum of an infinite series is the limit S =

E

c + cr

cr2 ± cr3 ±• • •+ crN =

• Geometric series: Assume c

The geometric series diverges if Ir I

cr 2 + cr 3 +

=

1- r

1.

HISTORICAL PERSPECTIVE

triangles on the segments AD, DB, BE, EC,

Archimedes (287-212 BCE), who discovered the law of the lever, said, "Give me a place to stand on, and I can move the Earth" (quoted by Pappus of Alexandria c. 340 CE).

4T,

rN+1)

1- r

0. If In I < 1, then

00 ce = c + cr E n=0

FIGURE 3 Archimedes showed that the area S of the parabolic segment is where T is the area of AABC.

c(1 -

of total area ( 4 1 ) T. Then construct eight triangles of total area (1)3 T, and so on. In this way, we obtain infinitely many triangles that completely fill up the parabolic segment. By the formula for the sum of a geometric series, we get oo 1 1 1 4 S=T+—T+—T±• • •=TE—=—T 4 16 4n 3 n=0 Geometric series were used as early as the third century BCE by Archimedes in a brilliant argument for determining the area S of a "parabolic segment" (shaded region in Figure 3). Given two points A and C on a parabola, there is a point B between A and C where the tangent line is parallel to AC (apparently, Archimedes was aware of the Mean Value Theorem more than 2000 years before the invention of calculus). Let T be the area of triangle AABC. Archimedes proved that if D is chosen in a similar fashion relative to AB and E is chosen relative to BC, then 1 —T = area(AADB) 4

area(ABEC)

5

This construction of triangles can be continued. The next step would be to construct the four

For this and many other achievements, Archimedes is ranked together with Newton and Gauss as one of the greatest scientists of all time. The modern study of infinite series began in the seventeenth century with Newton, Leibniz, and their contemporaries. The divergence of

E

1/n (called the harmonic series) n=1 was known to the medieval scholar Nicole d' Oresme (1323-1382), but his proof was lost for centuries, and the result was rediscovered on more than one occasion. It was also known that the sum of the reciprocal squares CO

E

1/n2 converges, and in the 1640s, the n=1 Italian Pietro Mengoli put forward the challenge of finding its sum. Despite the efforts of the best

SECTION 11.2

mathematicians of the day, including L,eibniz and the Bernoulli brothers Jakob and Johann, the problem resisted solution for nearly a century. In 1735, the great master Leonhard Euler (at the time, 28 years old) astonished his contemporaries by proving that

1

1

Summing an Infinite Series

1

1

1

563

Jr 2

1 =

We examine the convergence of this series in Exercises 85 and 91 in Section 11.3.

11.2 EXERCISES Preliminary Questions 1. What role do partial sums play in defining the sum of an infinite series? 2. What is the sum of the following infinite series? 1 1 1 1 1 - +- + — + — + — +••• 16 32 64 4 8 3. What happens if you apply the formula for the sum of a geometric series to the following series? Is the formula valid?

5. Indicate whether or not the reasoning in the following statement is co correct: converges because n=1 Vn 1 lim =0 n —.0O vn

E _1,_

00

6. Find an N such that SN

>

25 for the series

E 2. n=1

1 + 3 + 32 + 33 +3+ • • • 4. Indicate whether or not the reasoning in the following statement is 00 1 1 0 because - 2 tends to zero. correct: n n2

E_ =

7. Does there exist an N such that SN Explain.

>

00

25 for the series n=1

8. Give an example of a divergent infinite series whose general term tends to zero.

Exercises 1. Find a formula for the general term a„ (not the partial sum) of the infinite series. 125 25 1 5 1 1 1 1 — +••• (b) - + - + — (a) 8 2 4 1 ±5± (c) (d)

1

44

33

22 -

2•1 + 3•2•1 1 2 + +

12 + 1

22 + 1

4•3• 2• 1 ± 1 2 +••• + 42 + 1 32 + 1

2. Write in summation notation: 1 1 1 (a) 1 + -4- + 7 — 16 + • • • 9

(C) - -3 + -5 -

1

In Exercises 9 and 10, use a computer algebra system to compute Sio, S10, S500, and Swoo for the series. Do these values suggest convergence to the given value? 9. (CAS ,r-3 4 10. CAS

1

1

7,4 1 1 1 — =1+ + + -471 + • • • 90

1

± •••

(b)

11. Calculate S3, S4, and S5 and then find the sum of the telescoping series

0. E (n + 1

- +••• 7

125 625 3125 15,625 (" 1 9 ± 16 ± 25 ± 36 ± In Exercises 3-6, compute the partial sums S2,

n+ 1 2)

n=1

00

Al

1 1 1 3. 1+ — 22 + — 32 + -42

1 + 8 • 9 • 10

1 1 + 6•7•8 4•5•6

1 2•3•4

12. Write S4,

and S6.

± •••

1 1 1 5. — + — + — + • • • 2•3 3•4 1•2 ) 7. The series 1 + (1) + (1)2 + ( 3 + • • • converges to . Calculate SN for N = 1, 2, . . . until you find an SN that approximates with an error less than 0.0001. 1 1 1 1 8. The series — - — + — - — + • • • is known to converge to e-1 0! 1! 2! 3! (recall that 0! = 1). Calculate SN for N = 1, 2, ... until you find an SN that approximates e-1 with an error less than 0.001.

E n(n

1

n=3

- 1)

as a telescoping series and find its sum. 00

E

13. Calculate 53, S4, and S5 and then find the sum 4n 2 identity n=1

—

1

using the

1 \ 1 1 1 4n2 - 1 - 2 \ 2n - 1 2n + 1 ) 00 1 as a telescoping series 14. Use partial fractions to rewrite n(n + 3) and find its sum. 1 1 1 15. Find the sum of — + — + — +... . 5•7 1•3 3•5 co and show that 16. Find a formula for the partial sum SN of

E n=,

E(-1)n-1 n=1

the series diverges.

564

INFINITE SERIES

CHAPTER 11

In Exercises 17-22, use the nth Term Divergence Test (Theorem 4) to prove that the following series diverge. 00

00

18.

17. E 10n± 12 n=1 0 1

oo 20. E(-1) n ri2 n=1

3 4

2 3

1 2

J2 + 1 n=1

00

1 1 1 21. cos - + cos - + cos - + • • • 4 3 2

22. E(/4n 2 + 1 - n) n=0

(c)

1 1 1 + 216 ± • • 6±

24.

43

44 ±

45 ±

7 7 7 7 25. -3 + -32 + -33 + -34 + • • • 26.

(')3

+()7 2

+(

7

n=5

E (a, + bn ) diverges. Hint: If not, derive a contradiction by writing n=1

0.

0.

E

00

b, = E (an + bn) - E an n.1 n=1 n=1

c° 9' + 2" 50. Prove the divergence of E 5n n=0 51. E4 Give a counterexample to show that each of the following statements is false.

)

+. . . o 7 (_3)n

28.

27. E (-3 1n 11 n=3

(d)

c° 1 48. Use the method of Example 10 to show that E -k1/3 diverges. k=.1 00 00 49. Prove that if E an converges and E b, diverges, then n=1 n=1

In Exercises 23-38, either use the formula for the sum of a geometric series to find the sum, or state that the series diverges. 23.

00

00 ri 2 2n n=0

t

5n

n=2

(a) If the general term an tends to zero, then E an = 0. n=1 (b) The Nth partial sum of the infinite series defined by {an} is aN • co

(C)

If an tends to zero, then E a, converges. n=1

29. n=-4 E

30.

-4 ) n 9

oo 31. Ee -n n=1

32.

+1 C° 8 ± 2 n 33. LJ 5n n=0

34.

5 4

35. 36. 37.

23

00

n=0

5 42 24

3(-2r - 5? 8n

5 43 25

7 49 343 7 .78 - 64 ± 512

26 2401 4096 4-

25 5 3 9 27 38. - + - + I + - + - + +••• 9 3 5 25 125 In Exercises 39-44, determine a reduced fraction that has this decimal expansion. 39. 0.222. . .

40. 0.454545. . .

41. 0.313131. . .

42. 0.217217217. . .

43. 0.123333333. . .

44. 0.808888888. . .

45. Verify that 0.999999. . . = 1 by expressing the left side as a geometric series and determining the sum of the series.

(d) If an tends to L, then E an = L. n=1 oo 52. Suppose that Ea, is an infinite series with partial sum n=1 2 SN = 5 10 16 (a) What are the values of E an and E a,? n=1 n=5 (b) What is the value of a3? (c) Find a general formula for an • 00 (d) Find the sum E an • n=1 53. Consider the archery competition in Example 6. (a) Assume that Nina goes first. Let p„ represent the probability that Brook wins on his nth turn. Give an expression for p, . (b) Use the result from (a) and a geometric series to determine the probability that Brook wins when Nina goes first. (c) Now assume that Brook goes first. Use a geometric series to compute the probability that Brook wins the competition. 54. Consider the archery competition in Example 6. Assume that Nina's probability of hitting the bullseye on a turn is 0.45 and that Brook's probability is p. Assume that Nina goes first. For what value of p do both players have a probability of 1/2 of winning the competition? 55. Compute the total area of the (infinitely many) triangles in Figure 4.

46. The repeating decimal 0.012345678901234567890123456789. . . can be expressed as a fraction with denominator 1,111,111,111. What is the numerator? 47. Which of the following are not geometric series? 00 7n co 1 (b) 7 (a) E n=0 n=3 n

I

I

168

I

4

FIGURE 4

SECTION 11.2

56. The winner of a lottery receives m dollars at the end of each year for N years. The present value (PV) of this prize in today's dollars is PV =

57. If a patient takes a dose of D units of a particular drug, the amount of the dosage that remains in the patient's bloodstream after t days is De — t, where k is a positive constant depending on the particular drug. (a) Show that if the patient takes a dose D every day for an extended De—k period, the amount of drug in the bloodstream approaches R = 1 — e—k •

(b) Show that if the patient takes a dose D once every t days for an extended period, the amount of drug in the bloodstream approaches De —kt R — 1 — e —kt • (c) Suppose that it is considered dangerous to have more than S units of the drug in the bloodstream. What is the minimal time between doses that is safe? Hint: D+ R < S. 58. In economics, the multiplier effect refers to the fact that when there is an injection of money to consumers, the consumers spend a certain percentage of it. That amount recirculates through the economy and adds additional income, which comes back to the consumers and of which they spend the same percentage. This process repeats indefinitely, circulating additional money through the economy. Suppose that in order to stimulate the economy, the government institutes a tax cut of $10 billion. If taxpayers are known to save 10% of any additional money they receive, and to spend 90%, how much total money will be circulated through the economy by that single $10 billion tax cut?

565

61. Show that if a is a positive integer, then

00 1

Emu + rr i , where r is the interest rate. Calculate PV if m =

i=1 $50,000, r = 0.06 (corresponding to 6%), and N = 20. What is PV if N = oo?

Summing an Infinite Series

1

n(n + a) =

1 ±2 —±

1 + a)

62. A ball dropped from a height of 10 ft begins to bounce vertically. Each time it strikes the ground, it returns to two-thirds of its previous height. What is the total vertical distance traveled by the ball if it bounces infinitely many times? 63. In this exercise, we resolve the paradox of Gabriel's Horn (Example 3 in Section 8.7 and Example 7 in Section 9.2). Recall that the horn is the surface formed by rotating y = for x > 1 around the x-axis. The surface encloses a finite volume and has an infinite surface area. Thus, apparently we can fill the surface with a finite volume of paint, but an infinite volume of paint is required to paint the surface. (a) Explain that if we can fill the horn with paint, then the paint must be Magic Paint that can be spread arbitrarily thin, thinner than the thickness of the molecules in normal paint. (b) Explain that if we use Magic Paint, then we can paint the surface of the horn with a finite volume of paint, in fact with just a milliliter of it. Hint: A geometric series helps here. Use half of a milliliter to paint that part of the surface between x = 1 and x = 2. 64. A unit square is cut into nine equal regions as in Figure 6(A). The central subsquare is painted red. Each of the unpainted squares is then cut into nine equal subsquares and the central square of each is painted red as in Figure 6(B). This procedure is repeated for each of the resulting unpainted squares. After continuing this process an infinite number of times, what fraction of the total area of the original square is painted?

59. Find the total length of the infinite zigzag path in Figure 5 (each zag occurs at an angle of 1÷, ).

(B)

(A) FIGURE 6

co

65. Let {bn} be a sequence and let an =

— bn_i. Show that

E an conn=1

verges if and only if lim bn exists. rz—>oo

FIGURE 5

Show, by giving counterexamples, that the oo oo bn are an and assertions of Theorem 1 are not valid if the series 66. Assumptions Matter

60. Evaluate that

E n=1

1 . Hint: Find constants A, B, and C such n(n +1)(n + 2)

E

E

n=0

n=0

not convergent.

B A 1 = + n+2 n(n +1)(n + 2) n n +1 oo 1 and use the result to evaluate n=l n(n +1)(n ±

E

Further Insights and Challenges In Exercises 67-69, use the formula 1±r±r2±. . .+r

x N iaa N

1

= 1 — rN 1—r

f' (a) =

L6

67. Professor George Andrews of Pennsylvania State University observed that we can use Eq. (6) to calculate the derivative of f(x) = x N (for N > 0). Assume that a 0 0 and let x = ra. Show that

iilim

x—>a

x

= a N-1 lim

rN

1

r —1

and evaluate the limit. 68. Pierre de Fermat used geometric series to compute the area under the graph of f (x)= x N over [0, Al. For 0 < r < 1, let F(r) be the sum of the

566

INFINITE SERIES

CHAPTER 11

areas of the infinitely many right-endpoint rectangles with endpoints Ar n , as in Figure 7. As r tends to 1, the rectangles become narrower and F(r) tends to the area under the graph. 1 —T (a) Show that F(r)= AN + 1 — rN+I . A

1 1 1 1 ) L (-4 + —8 + —16 + • • • + 2N+1

x N dx = lim F(r).

(b) Use Eq. (6) to evaluate

r-->1

0

each of these two sections (this stage removes L/8 of the table). Now four sections remain, each of length less than L/4. At Stage 3, remove the four central sections of length L/43, and so on. (a) Show that at the Nth stage, each remaining section has length less than L/2N and that the total amount of table removed is

(b) Show that in the limit as N --÷ co, precisely one-half of the table remains. This result is intriguing, because there are no nonzero intervals of table left (at each stage, the remaining sections have a length less than LI2N ). So, the table has "disappeared." However, we can place any object longer than L/4 on the table. The object will not fall through because it will not fit through any of the removed sections.

f(x)= xN

X

r3A r2A rA

L/4

L/16

A

LI16

FIGURE 7 69. Verify the Gregory—Leibniz formula in part (d) as follows. (a) Set r = —x2 in Eq. (6) and rearrange to show that 1 + X2

= 1_x2 + X4 —

+

(-1)N-1x2N-2

(_1)N x 2N

l+

x2

FIGURE 8

(b) Show, by integrating over [0, 11, that Jr

1 3

1 5

1 7

(-1)N-1 2N — 1 ± (-1)1v

f

71. The Koch snowflake (described in 1904 by Swedish mathematician Helge von Koch) is an infinitely jagged "fractal" curve obtained as a limit of polygonal curves (it is continuous but has no tangent line at any point). Begin with an equilateral triangle (Stage 0) and produce Stage 1 by replacing each edge with four edges of one-third the length, arranged as in Figure 9. Continue the process: At the nth stage, replace each edge with four edges of one-third the length of the edge from the (n — 1)st stage. (a) Show that the perimeter Pn of the polygon at the nth stage satisfies Pn =P n —I. Prove that lim Pn = oo. The snowflake has infinite length.

l x2Ndx

1 + x2

(c) Use the Comparison Theorem for integrals to prove that 1 x 2/V

0
0 for all n. We can visualize the terms of a positive series as rectangles of width 1 and height a„ (Figure 1). The partial sum

E

FIGURE 1 The partial sum SN is the sum of the areas of the N shaded rectangles.

SN = ai + az + • • • + aN is equal to the area of the first N rectangles.

SECTION 11.3

Convergence of Series with Positive Terms

567

The key feature of positive series is that their partial sums form an increasing sequence SAT < SN+1 for all N. This is because SN+1 is obtained from SN by adding a positive number: SN+1 = (al a2 ± • • • ± aN) aN+1 = aN+1 Positive

Recall that an increasing sequence converges if it is bounded above. Otherwise, it diverges (Theorem 6, Section 11.1). It follows that a positive series behaves in one of two ways. 00

THEOREM 1 Partial Sum Theorem for Positive Series If

Ea, is a positive series, n=1

then either

00

(i) The partial sums SN are bounded above. In this case,

E a, converges. Or, n=1

(ii) The partial sums SN are not bounded above. In this case,

00

E an diverges. n=1

• Theorem 1 remains true if a, > 0. It is not necessary to assume that a, > 0. • It also remains true if a, > 0 for all n > M for some M, because the convergence or divergence of a series is not affected by the first M terms. Assumptions Matter The theorem does not hold for nonpositive series. Consider 00 = 1 — 1+ 1 — 1 + 1 — 1+ • • • n=

The partial sums are bounded (because SN = 1 or 0), but the series diverges. Our first application of Theorem 1 is the following Integral Test. It is extremely useful because in many cases, integrals are easier to evaluate than series. THEOREM 2 Integral Test Let a, = f(n), where f is a positive, decreasing, and continuous function of x for x > 1. 00 oo an converges. f (x) dx converges, then (i) If f

E

n=1 oo

oo (ii)

If f

f (x)dx diverges, then

E a, diverges. n=1

FIGURE 2

Proof Because f is decreasing, the shaded rectangles in Figure 2 lie below the graph of f , and therefore for all N, oo f (x) dx f(x)dx f a2 ± • • • + aN Area of shaded rectangles in Figure 2

If the improper integral on the right converges, then the sums a2 + • • • + aN are bounded above. That is, the partial sums SN are bounded above, and therefore the infinite series converges by the Partial Sum Theorem for Positive Series (Theorem 1). This proves (i). On the other hand, the rectangles in Figure 3 lie above the graph of f, so

f (x) dx FIGURE 3

+ a2 + • • • + aN-1 Area of shaded rectangles in Figure 3

1

568

INFINITE SERIES

CHAPTER 11

N

oo

The Integral Test is valid for any series 00 f (n), provided that for some M > 0, n=k f is a positive, decreasing, and continuous function of x for x > M. The convergence of the series is determined by the convergence of imoo f (x) dx

f(x)dx increases without bound as N increases. The 1 1 inequality in (1) shows that SN also increases without bound, and therefore, the series • diverges. This proves (ii).

If f

f(x)dx diverges, then

00 Show that E I diverges.

EXAMPLE 1 The Harmonic Series Diverges

n=1

Solution Let f(x) =1. Then f(n) = and the Integral Test applies because f is positive, decreasing, and continuous for x > 1. The integral diverges: R dx " dx — = lirn ln R = oo —= lim f

The infinite series 1 n=1

n —

x

f1

R—>o0 1

R—>oo

X

oo 1 Therefore, the series E _ diverges.

•

n=1

is called the harmonic series.

00 EXAMPLE 2 Does n=1

= n 22 (n 2 + 1)2

52

+ - 3— + • • • converge? 102

Solution The function f(x) = (x2 + 1)2 is positive and continuous for x > 1. It is decreasing because f'(x) is negative: f'(x) =

1 - 3x2 1

Therefore, the Integral Test applies. Using the substitution u = x2 ± 1, du = 2x dx, we have dx = lim R—>00 f oo (x2 ± 1)2

(x2 ± 1)2

1 f R2+1 du dx = urn R—>oo 2 2 U2

R2+1 1 liM — = lim R-+00 2u 2 R oo \4 4 2(R21+ 1)) = 00 Thus, the integral converges, and therefore, E also converges by the Integral (n2 ± 1)2 n=1 =

Test.

The sum of the reciprocal powers n- P is called a p-series. As the next theorem shows, the convergence or divergence of these series is determined by the value of p. THEOREM 3 Convergence of p-Series

The infinite series

converges if

p> 1 and diverges otherwise.

Proof If p < 0, then the general term n- P does not tend to zero, so the series diverges by the nth Term Divergence Test. If p > 0, then f(x) = x- P is positive and decreasing for x > 1, so the Integral Test applies. According to Theorem 1 in Section 8.7, r

Ji 00 Therefore, E n=1

1

1 — dx =_- ip - 1 XP oo

if p> 1 if p < 1

1 _ converges for p> 1 and diverges for p < 1.

nP

S ECT I ON 11.3

Convergence of Series with Positive Terms

569

Here are two examples of p-series: 1

1

P=

n=1 3

1 +

1

1 +

+. . .=

=

00

00 ,1

diverges

converges

n=1 Another powerful method for determining convergence of positive series occurs via comparison with other series. Suppose that 0 < an < b„. Figure 4 suggests that if the larger sum b, converges, then the smaller sum an also converges. Similarly, if the smaller sum diverges, then the larger sum also diverges.

b2

bi

a2

a1

1

2

193

bN

a3

aN

3

E

n

E bn forces the convergence of E an •

E

THEOREM 4 Direct Comparison Test Assume that there exists M > 0 such that 0 1, we have 1

< 1

3n — 30

E 3—1n converges because it is a geometric series with r = 00

The larger series

< 1. By

n=1

00

FIGURE 5 The smaller series is contained in the smaller balloon.

the Direct Comparison Test, the smaller series

E ,sfii1 3n also converges.

n=1

•

570

CHAPTER 11

INFINITE SERIES 00

EXAMPLE 4 Does

E 012 ±13)1/3 converge?

n=2

Solution Let us show that 1 n

1 (n2 + 3)1/3

for n > 2

This inequality is equivalent to (n2 + 3) 0

for x > 2

The function f is increasing because its derivative f (x) = 3x2 — 2x = 3x — &) is positive for x > 2. Since f(2) = 1, it follows that f (x) > 1 for x > 2, and our original 00 1 inequality follows. We know that the smaller harmonic series diverges. Therefore,

E _n

n=2

00

the larger series

E (n2 +1 1)1/3 also diverges.

•

n=2

EXAMPLE 5 Determine the convergence of

00 E n(ln n)2 n=2

Solution We might be tempted to compare

oo 00 1 E n(ln1n)2 to the harmonic series E _

n=2

n=2

using the inequality (valid for n > 3 since ln 3> 1) 1 1 n(ln n)2 5." n

E _n1 diverges, and this says nothing about the smaller series E oo

However,

n=2

1 n)2

Fortunately, the Integral Test can be used. The substitution u = in x yields 1 00 dx f °° du 1 1) 1 0, then

SECTION 11.3

Convergence of Series with Positive Terms

571

E

Proof Assume first that L is finite (possibly zero) and that bn converges. Choose a positive number R > L. Then 0 < an Ibn < R for all n sufficiently large because an Ibn approaches L. Therefore, a, < Rbn. The series Rbn converges because it is a con-

E stant multiple of the convergent series E bn. Thus, E a, converges by the Direct Comparison Test. Next, suppose that L is nonzero (positive or infinite) and that E cin converges. Let K = Ern b,lan. Then either K = L-1 (if L is finite) or K = 0 (if L is infinite). In

either case, K is finite and we can apply the result of the previous paragraph with the roles of {ad and fbn l reversed to conclude that bn converges. •

E

CONCEPTUAL INSIGHT To remember the different cases of the Limit Comparison Test, you can think of it this way: If L > 0, then a, Lb, for large n. In other words, the an and bn are roughly multiples of each other, so one converges if and series only if the other converges. If L = oo, then a, is much larger than bn (for large n), so if bn certainly converges. Finally, if L = 0, then bn is much larger a, converges,

E

E E

E

than a, and the convergence of

E bn yields the convergence of E an •

00

EXAMPLE 6 Show that

E n4 - n2n -

n=2

1

converges.

Solution If we divide the numerator and denominator by n, we can conclude that for large n, 1 n2 n2 n4 - n - 1 To apply the Limit Comparison Test, we set a0 =

n2 n4 - n - 1

and

1 n2 b, = —

We observe that Ern 2,L.` exists and is positive: n -*00

n2 an L= lim — = lim n-oo bn n->co n4 - n - 1

E _12 converges, our series E n4 - n2n n=2

n-4 =1

00

00

Since

1 n2 = lim n-3 _ n->oo 1 1

n

n=2

also converges by Theorem 5.

•

1

00

EXAMPLE 7 Determine whether

E ,s/n21± 4 converges. 0=3

Solution Apply the Limit Comparison Test with an =

1 n2 +4

and bn

1 an =1 = lim L = lim — = lim fl-+00v,n2 + 4 0+00 bn n->00 1/i + 4/n2 00 00 1 also diverges. diverges and L > 0, the series Since n2 n ,../ + 4 n=3

E 1—

E

1 —. Then

•

n=3

In the Limit Comparison Test, when attempting to find an appropriate bn to compare with an , we typically keep only the largest power of n in the numerator and denominator of an, as we did in each of the previous examples.

572

CHAPTER 11

INFINITE SERIES

11.3 SUMMARY

E

an form an increasing sequence. • The partial sums SN of a positive series • Partial Sum Theorem for Positive Series: A positive series converges if its partial sums SN are bounded. Otherwise, it diverges. • Integral Test: Assume that f is positive, decreasing, and continuous for x > M. Set 00 00 f (x)dx dian converges, and if an = f (n). If f f (x)dx converges, then

E

E an diverges. 00 1 • p-Series: The series E _ converges if p> 1 and diverges if p < 1. nP verges, then

n=1

• Direct Comparison Test: Assume there exists M > 0 such that 0 < an < bn for all bn converges, then an converges, and if an diverges, then n > M. If bn diverges.

E

E

E

E

• Limit Comparison Test: Assume that {an } and fbn l are positive and that the following limit exists: an L = lim bn

E a, converges if and only if E bn converges. — If L = oo and Ean converges, then E bn converges. — If L = 0 and E bn converges, then E an converges. — If L > 0, then

11.3 EXERCISES Preliminary Questions 00 1. For the series E an, if the partial sums SN are increasing, then (choose n=1

the correct conclusion) (a) fa„) is an increasing sequence. (b) {ad is a positive sequence.

00

Er

4. Which test would you use to determine whether converges? n=1 5. Ralph hopes to investigate the convergence of 00

2. What are the hypotheses of the Integral Test? co

3. Which test would you use to determine whether

En-12

it with

1

+

00

-n E f —by comparing n=1

E _n . Is Ralph on the right track? 1

n=1

n=1

converges?

Exercises In Exercises 1-12, use the Integral Test to determine whether the infinite series is convergent. Do co v , 1 1 1. 2. (n ± 1)4 n +3 n=1 3.

00 En-I/3

co 4.

n=1

n2

E (n3 +

n=25 co

7.

1

9)5/ 2

co 11.

nE=2

—n2

12.

00

13. Show that

E

n3 ±

8n converges by using the Direct Comparison

n=1

E (n2 1)3/5

n=1

co

Test with

E n -3. n=1

co

oo

8.

14. Show that n 4 n 2 1—

E ne n= 1

1 n(ln n)3i2

00

6.

10.

n(n + 5)

1

n=5

co

5.

E

oo

CO

9.

l

E n=2

1

co

diverges by comparing with

E n-.. n=2

S ECT I ON 11.3 00

15.

For

E 1 , n=I n + „Ft

CO

verify that for

n > 1, 1 n+„Ft

„Ft

3n + 5

1

1

1

00

En2n

18.

E

n=1

n=1

E

n=1

23.

2n

&/3+

co

21.

20.

4

E m! + 4'n nt=1

E° °

24. E

k2

co

25. Y‘

27.

E n=1

26.

3

E

k2/9 k10/9

51.

1

(n ± 1)!

30.

(ln n) m°

32.

00

34.

Z:a 3n n=1

co

Show that

E n=1

1 sin n2

co sin(11n) E

inn n=2 (cos(1/n))/n. sin(l/n) > cos(1/n) >

Hint:

in ni

E co

(In n)1°

n=2 00

61. E

n5

n=3

La 2n n=1

00 E,1 12

E

(ln n)

n=2

n9/8

E n=1

4n 5? - 2n

60.

(ln n)4

1

n=1

E -1 n=1 nn 00 + Hon 65. E

Thus, sin(l/n) > 1/(2n) for

n >

2n 3n -

oo

62.

nlnn-n

oo

for x > 0.

E 00

1

E n=3

1

°° n 2 -

oo

2 + (-1) n

L-• n=1

n3/ 2

2 [because 00

J.

67.

E n=1

69.

1

sin n

°9 2n + 1

E

n=1

4n

oo

68.

E

. sin(l/n)

n=1

co

70.

,F1

1 E _TTz n=3e

n

4n3/2 n3

n=1

66.

n

nan n)2 -

64. E

n=1

E

1

n2 + sin n

co

58.

41 / k

63.

By Theorem 3 in Section 2.6,

n=2

n 1

co

56.

n3/2 inn

59. E

In Exercises 37-46, use the Limit Comparison Test to prove convergence or divergence of the infinite series. co 00 1 n2 38. n(n - 1) n4 - 1 n=2

1

k=1

converges. Hint: Use sin x < x

converge?

n3

n=1 3n

00

57.

n - cos n

1

E \--,

n=2

'n

00

52.

54.

(4/5)

co

1

00

N--, n

Does

1

n5 +1

55. E

co

12=1

11=1

36.

E m=2

n=1

35.

n2 -

00 E

using any

n2

n=1

n=5

n3 1 n=1

oo

Inn

E n3 . 31. E ni, 33.

4n +9

oo

n!

° n=1

50.

Na, °

53. E

Exercise 29-34: For all a > 0 and b > 1, the inequalities ltin < na , na < bn are true for n sufficiently large (this can be proved using L'Hopital's Rule). Use this, together with the Direct Comparison Test, to determine whether the series converges or diverges. 29.

Compare with the harmonic series.

9

c°

n=1

2 -k2

co

28.

n-2 .

n=1

48.

n2

n=1

k=1

1

co

co

E

co

1- ' 3 n + 3-n n=1 co

1 - ) Hint: Compare with n

1

nE=4

k=2

2

n 5/2

n=1

E(1 _2-1/n) Hint:

47.

49.

n-

n=4

cos

ln(n + 4)

44. E

n=1

,Vn 3 +2n -

22.

sink

k=1

1

E

n2

In Exercises 47-76, determine convergence or divergence method covered so far.

n3 n5 +4n +1

n=1

e2n n=1

co

46.

00

19.

+ 1

n2

-

n2 +

E _ n=1

In Exercises 17-28, use the Direct Comparison Test to determine whether the infinite series is convergent. 17.

+ 2n2

42.

2)

„Fi+lnn

co

45. n 2 + N/T1

E

n=1

? Explain. + 1

n3

00

Which of the following inequalities can be used to study the conver-

E n2

E n=2

n(n -1)(n -

43.

n=2

40.

573

1

Can either inequality be used to show that the series diverges? Show that 1 > for n > 1 and conclude that the series diverges. n + .171 2n

gence of

00

39. n=2 A/n3 + 1

1 1 < , n+,Ft n

16.

Convergence of Series with Positive Terms

574

INFINITE SERIES

CHAPTER 11 00

71.

E n2inn — 3n

n=1

n=4

00 73 .

n=2 oo

75.

E n=2

Is this consistent with Euler' s result, according to which this infinite series has sum

72.

86. CAS Use a CAS and the inequalities in (5) from Exercise 83 to 00 to within an error less than 10-4. Check determine the value of

00

1 nI/21nn

74. n=1

n3/2

—

En-6

(ln n)4

n=1

that your result is consistent with that of Euler, who proved that the sum is equal to 76 /945.

oo

4n2 + 15n 3n4 — 52 —17

76. E 4—n ±5 n.1

—fl

87. (CAS Use a CAS and the inequalities in (5) from Exercise 85 to co n-5 to within an error less than 10-4. determine the value of

E

00

E n(ln n)a converge?

77. For which a does

n=1

n=2

88. How far can a stack of identical books (of mass m and unit length) extend without tipping over? The stack will not tip over if the (n + 1)st book is placed at the bottom of the stack with its right edge located at or before the center of mass of the first n books (Figure 6). Let c,, be the center of mass of the first n books, measured along the x-axis, where we take the positive x-axis to the left of the origin as in Figure 7. Recall that if an object of mass mi has center of mass at xi and a second object of m2 has center of mass x2, then the center of mass of the system has x-coordinate

00

78. For which a does

E na 1Inn converge?

n=2

00

79. For which values of p does

E (n 3 n2 1)P converge? ±

n=1

oo

80. For which values of p does

E (1 +

n=1

e2x )P

m x + m2x2 mi +m2

converge?

Approximating Infinite Sums In Exercises 81-83, let a, = f (n), where f is a continuous, decreasing function such that f(x) 0 and oo f(x)dx converges.

f

81. Show that j •00

co

f(x)dx

00

E an

+f

f(x)dx

3

n=1 82. (CAS) Using the inequality in (3), show that oo 5 :5_ n=1

(a) Show that if the (n + 1)st book is placed with its right edge at c„, then its center of mass is located at c„ + (b) Consider the first n books as a single object of mass nm with center of mass at cn and the (n + 1)st book as a second object of mass m. Show that if the (n + 1)st book is placed with its right edge at c,„ then 1 . cn+1 = c,, + 2(n + 1) (c) Prove that lim c,, = co. Thus, by using enough books, the stack can n—>00 be extended as far as desired without tipping over.

1 6 n1•2 —

This series converges slowly. Use a computer algebra system to verify that S N 2 if n > C, where C = / (c) Show that the series converges.

than

E 11n 2? n=1

.

(c) (CAS, Compute

91. Kummer's Acceleration Method Suppose we wish to approximate oo S= 1/n2. There is a similar telescoping series whose value can be n=1 computed exactly (Example 2 in Section 11.2):

1000

E

n=1

n

Which is a better approximation to S, whose exact value is ir2 /6? oo

00

nE=i n(n +1)

92. CAS The sum S =

0. +E

n(n

1)

S

1+

n=i

Ek

3 has been computed to more than 100 k=1 million digits. The first 30 digits are

=1

(a) Verify that

s=E n=i

ioo

1 +E n2(n + 1)

S = 1.202056903159594285399738161511 Approximate S using Kummer's Acceleration Method of Exercise 91 with n2

n(n

1) )

E

(n(n + 1)(n + 2)r' and M = 500. According to n=1 Exercise 60 in Section 11.2, the similar series is a telescoping series with a sum of 14 .

the similar series

Thus for M large, 6 n=1

11 2 (11 +1)

11.4 Absolute and Conditional Convergence In the previous section, we studied positive series, but we still lack the tools to analyze series with both positive and negative terms. One of the keys to understanding such series is the concept of absolute convergence.

DEFINITION Absolute Convergence

E a,

The series E

an converges absolutely if

converges.

EXAMPLE 1 Verify that the series 00

n=1

converges absolutely.

(_1)n-1

n2

1

1

1

1

—2

22

32

42± • • •

576

CHAPTER 11

INFINITE SERIES

Solution This series converges absolutely because taking the absolute value of each term, we obtain a p-series with p = 2> 1:

0.

(convergent p-series)

12 1- 22 1- 32 1- 42 1- • • •

n2

L.s j n=1

The next theorem tells us that if the series of absolute values converges, then the original series also converges. THEOREM 1 Absolute Convergence Implies Convergence If then

E an also converges.

E

I converges,

Proof We have - Ian I < an < Ian I. By adding Ian I to all parts of the inequality, we get I also converges, and there0< la, I +a Ian I converges, then 2lan I. If

E2lan

E

fore, E(a n Ian I) converges by the Direct Comparison Test. Our original series converges because it is the difference of two convergent series:

EXAMPLE 2

E an = E(an + Ian') — Ian' 00 (_ on-1 Verify that E converges. n2

•

n=1

Solution We showed that Theorem 1,

0° E n=1

itself converges.

n2

EXAMPLE 3 Does

converges absolutely in Example 1. By •

E "

1 E_ N/T1 00

Solution The series of absolute values is co

diverges because p < 1. Therefore,

E n=1

N/-5

-

.

.

.

converge absolutely?

which is a p-series with p =

n=1 .111

does not converge absolutely.

The series in the previous example does not converge absolutely, but we still do not know whether or not it converges. A series an may converge without converging

E

absolutely. In this case, we say that

E a, is conditionally convergent.

DEFINITION Conditional Convergence tionally if

An infinite series

E an converges but E Ian I diverges.

E a, converges condi-

bi

b3 -b2

b5 —64

—b6

n

FIGURE 1 An alternating series with decreasing terms. The sum is the signed area, which is at most bi.

If a series is not absolutely convergent, how can we determine whether it is conditionally convergent? This is often a difficult question, because we cannot use the Integral Test or the Direct Comparison Test since they apply only to positive series. However, convergence is guaranteed in the particular case of an alternating series E (_ o n-i bn _ b1 - b2 + b3 - b4 ± • • n=1 where the terms bn are positive and decrease to zero (Figure 1).

Absolute and Conditional Convergence

SECTION 11.4 Assumptions Matter The Alternating Series Test is not valid if we drop the assumption that bn is decreasing (see Exercise 35).

THEOREM 2 Alternating Series Test decreasing and converges to 0:

577

Assume that fb,} is a positive sequence that is

bi > b2 > b3 > b4 > • • • > 0,

lim b, = 0 n-÷oo

Then the following alternating series converges:

= bi — b2

b3 — ba + • • •

n=I

co The Alternating Series Test is the only test for conditional convergence developed in this text. Other tests, such as Abel's Criterion and the Dirichlet Test, are discussed in textbooks on analysis.

Furthermore, if S =

(-1)'

'b,n then

n=1

0 < S 0 because b, is decreasing, and therefore, S4 < S6. In general, S2N + (b2N+1 — b2N+2) = S2N+2 where b2N+1 — b2N+2 > 0. Thus, S2N < S2N+2 and 0 < S2 1, choose r such that 1 < r < p. Then there exists a number M such that lan+i /an I > r for all n > M. Arguing as before with the inequalities reversed, we find that lam+, I > rn lam I. Since rn tends to oo, the terms am+, do not tend to zero, and consequently, E , diverges. Finally, Example 4 in this section shows that both convergence and divergence are possible when p =1, so the test is inconclusive in this case. •

582

CHAPTER 11

INFINITE SERIES oo 2n EXAMPLE 1 Prove that E — converges. n! n=1 2' Solution Compute the ratio and its limit with an = —. Note that (n + 1)! = (n + 1)n!. n! Thus, 2

2n n! 2+1 n! (n + 1)! 2n - 20 (n +1)!

an+1 an

n +1

We obtain lim p= n-> oo

a0+1 an

= lim n ->o. n

2 1

=0

oo 2' Since p < 1, the series E — converges by the Ratio Test. n! n=1 EXAMPLE 2 Does

to n 2 _ converge? 2n n=1

n2 Solution Apply the Ratio Test with an = — 2n (n + 1)2 2' 1 n2 + 2n + 1 2n Fl n2 2 n2 an

1

2 1

=

n

1 — n2

We obtain an -F1

p= nlimoo

=

an

lirn ( 1 + -2 n n

+12 n

21 •

Since p < 1, the series converges by the Ratio Test. 00 EXAMPLE 3 Does E (

n! converge? moon

n=0

Solution This series diverges by the Ratio Test because p > 1: p = lim n—> oo

(n+1)! 1000 0 n+1 = i = lim = oo n.-00 1000 n±>Ileo 1000 n +1 n!

an+1

an

In the next example, we demonstrate why the Ratio Test is inconclusive in the case where p =1. Show that both convergence and divergence oo oo are possible when p = 1 by considering E n2 and E n-2. n=1 n=1

EXAMPLE 4

Ratio Test Inconclusive

Solution For an = n2, we have p = lim oo

an+1 = liM n—> oo an

(n + 1)2 = n2

n2 ± 2n + 1 2 1 = (1 + + 17 2 ) = n2 n—> co 0lim00

Furthermore, for bn = n-2, p = lim n--*00

an bn-E i = lirn n->-oo bn an+1

1 urn n --> oo

=1 a.

SECTION 11.5

The Ratio and Root Tests and Strategies for Choosing Tests

583

00

Thus, p = 1 in both cases, but, in fact,

n=1

00

since Inn n2 = oc, and n—> co

En-2

E n2 diverges by the nth Term Divergence Test

Converges since it

is a p-series with p = 2 > 1. This

n=1

shows that both convergence and divergence are possible when p = 1.

•

Our next test is based on the limit of the nth roots Vlan I rather than the ratios a0+1 /an. Its proof, like that of the Ratio Test, is based on a comparison with a geometric series (see Exercise 63).

THEOREM 2 Root Test

Assume that the following limit exists: L= lim-Oanl n —> co

(i) If L < 1, then (ii) If L > 1, then

a, converges absolutely.

E a, diverges.

(iii) If L = 1, the test is inconclusive. ) 1.1

00

EXAMPLE 5 Does

E (2n + 3

converge?

n=1

Solution We have L = lim 4'AZ; = lim n —>co

n—>oo

verges by the Root Test.

2n ± 3

=

1 . Since L < 1, the series con2

Determining Which Test to Apply We end this section with a brief review of all of the tests we have introduced for determining convergence so far and how one decides which test to apply. 00 Let a, be given. Keep in mind that the series for which convergence or diver-

E

n=1

00

gence is known include the geometric series

1 p-series E _ , which converge for p > 1. nP 00

Earn, which converge for Ir I < 1, and the n=0

n=0

1. The nth Term Divergence Test Always check this test first. If lim an 0 0, then n--->co

the series diverges. But if lirn a, = 0, we do not know whether the series converges or n —> co

diverges, and hence we move on to the next step. 2. Positive Series If all terms in the series are positive, try one of the following tests: (a) The Direct Comparison Test Consider whether dropping terms in the numerator or denominator gives a series that we know either converges or diverges. If a larger series converges or a smaller series diverges, then the original series does the same. For 00 00 1 1 1 < — and example, converges because converges ,F1 n2 ± ,F2 n2 n2 n=1 n=1 (since it is a p-series with p = 2 > 1). On the other hand, this does not work for 00 00 1 1 while still converging, is 2 , j2 since then the comparison series n — n=1

E _1

E

E

n=2

E _,

584

CHAPTER 11

INFINITE SERIES

smaller than the original series, so we cannot compare the series with E and apply the Direct Comparison Test. In this case, we can often apply the Limit Comparison Test as follows. (b) The Limit Comparison Test Consider the dominant term in the numerator and denominator, and compare the original series to the ratio of those terms. For example, CC 1 for n2 is dominant over as it grows faster as n increases. So, we n2 — .1Ft ' n=2 1 let bn = — Then n2 1 n2 . an 11M — = lim —lim =1 n—>.00 bn n-->oo n2 00 The limit is a positive number, so the Limit Comparison Test applies. Since n2 n=i converges, so does the original series.

E

E _1

(c) The Ratio Test The Ratio Test is often effective in the presence of a factorial such as n! since in the ratio, the factorial disappears after cancellation. It is also effective when there are constants to the power n, such as 2n , since in the ratio, the power n oo 3n disappears after cancellation. For example, if the series is then applying the n=1 Ratio Test yields

E _n! ,

317+1

lim oo

an+1

(n+1)!

= lim n—> oo JL (n)! Therefore, the series converges. an

3 = lim =0oo n + 1

(d) The Root Test The Root Test is often effective when there is a term of the form 00 2n f (n)g(n) . For example, n2n is a good example since applying the Root Test yields n=1

E

lim

n—>oo

lati l l / n =

2121 nlim _> oc, ( n n

1/n =

/ /14 M0 0

---7‘.n2z.

=0< 1

Thus, the series converges. (e) The Integral Test When the other tests fail on a positive series, consider the Integral Test. If a, = f (n) is a decreasing function, then the series converges if and only 00 if the improper integral 1 f (x)dx converges. For example, the other tests do f 00 1 1 not easily apply to However, f (x) = is a decreasing function nlnn nlnn n=2 oo 00 1 and dx = ln(ln x) = oo. Thus, the integral diverges, implying that the 1 x ln x 2 series does as well.

E

.

3. Series That Are Not Positive Series 00

(a) Alternating Series Test If the series is alternating of the form

(_on -1 . n,

n show n =1 that 0 1.

- Then -

- The test is inconclusive if L = 1.

11.5 EXERCISES Preliminary Questions 1. Consider the geometric series

00

00

E crn

3. Is the Ratio Test conclusive for

n=0

n=1 n1

(a) In the Ratio Test, what do the terms -a''

equal?

(b) In the Root Test, what do the terms

equal?

4. Is the Root Test conclusive for oo

E n- P

E (1 +

n=1 (a) In the Ratio Test, what do the terms

n=1

n=1 n 4- 1

1? E_ 2n

n )

Is it conclusive for

n=1 ?

n

a F-l equal?

(b) What can be concluded from the Ratio Test?

Exercises In Exercises 1-20, apply the Ratio Test to determine convergence or divergence, or state that the Ratio Test is inconclusive. oo

1.

E 5n n=1

2.

E n=1

(-1)'

4.

6.

00 17.

'n

5n

E n=1

8.

9.

1 On 2_, 2n2 n=1

co

20.

E -Inn

n=2

E

nk

3-n converges for all exponents k.

n=1

En -

co

°° n3

22. Show that

E

n 2x'

converges if Ix' < 1.

n=1 co

23. Show that

E 2"xn converges if lxi
1. 64. Show that the Ratio Test does not apply, but verify convergence using the Direct Comparison Test for the series

65. Let

(-

58.

.=, co

60.

cos n=1

1 n

62.

1 be a positive series, and

n(Inn)

3

E 4-2n-1-1

57.

n!

E (2n)!

48.

00

61.

(a) Show that the series converges if L < 1. Hint: Choose R with L < R < 1 and show that a R" Rn for n sufficiently large. Then compare with

and n! < n".

n=1

n 2n + 1

En-0.8

n=1 oo

39.

(2")"

r+

n

n=1 co

oo

36. n=0 E _ion

2 n2 =

n=1

n=1 nP

1

2n2

oo . E sin n

47.

n=1

oo

v ,

45.

00

E3nan cc 33. E 31.

c'c' E _ diverges. Hint: Use

n)

n=4

n =1 00

n=1

41.

In Exercises 43-62, determine convergence or divergence using any method covered in the text so far. co 00 3 4n n 43. 44. 7n n!

00

n=1

n2

00

1 -n

n=1

co

28. Does

00 40.

1

1

1

1

±•••

E en!, where c is a constant. nn

n=1

(a) Prove that the series converges absolutely if id < e and diverges if Icl > e. en! (b) It is known that lim = Verify this numerically. n (c) Use the Limit Comparison Test to prove that the series diverges for c = e.

11.6 Power Series With series we can make sense of the idea of a polynomial of infinite degree: 00

F(x) =

Eanxn = cto

aix

a2x2

a3x3 ± • • •

n=0

Specifically, a power series with center c is an infinite series 00 F(x) = ai (x - c) a2(x - c)2

an(x

n=0

=.0 +

a3(x - c)3 ± • • •

SECTION 11.6

Power Series

587

where x is a variable. For example, F(x) = 1 — x +x2 —x3 + • • • 1

G(x) = 1 + (x — 2) + 2(x — 2)2 + 3(x — 2)3 + • • •

Many functions that arise in applications can be represented as power series. This includes not only the familiar trigonometric, exponential, logarithm, and root functions, but also the host of "special functions" of physics and engineering such as Bessel functions and elliptic functions.

are power series where F(x) has center c = 0 and G(x) has center c = 2. 00 A power series F(x) = an(x — c)° converges for some values of x and may

E

n---0

diverge for others. For example, if we set x = the infinite series

in the power series of Eq. (1), we obtain 2

9 G(7 4

3

= 1 + ( 9 — 2) + 2 ( 7 9 — 2) + 3 (-9 — 2) + • • • 4

4

4

(41)2 + 3 (1)3 +

1 4

+ n 71\n

= 1+ (- ) + 2

±

This converges by the Ratio Test: lim

fl-f 00

an+1

an

= urn

n—>oo

n+1 4n+I 4n

= hill 1 (n+1\ 1/n lim 1 (1 + 1/n) 1 -= — n—>oo 4 1 4 n ) l/n = n- 0.0 4

On the other hand, the power series in Eq. (1) diverges for x = 3 by the nth Term Divergence Test: G(3) = 1 + (3 — 2) + 2(3 — 2)2 + 3(3 — 2)3 +. . . = 1

1 ±2±3±•

••

There is a surprisingly simple way to describe the set of values x at which a power series F(x) converges. According to our next theorem, either F(x) converges absolutely

for all values of x or there is a radius of convergence R such that F(x) converges absolutely when Ix — c R. This means that F(x) converges for x in an interval of convergence consisting of the open interval (c — R,c + R) and possibly one or both of the endpoints c — R and c + R (Figure 1). Note that F(x) automatically converges at x = c because F (c) = cio ± al (c — c)

a2(c — c)2

a3(c — c) + • • • = ao

We set R = 0 if F(x) converges only for x = c, and we set R = oo if F(x) converges for all values of x. Diverges

Converges absolutely Ix cl 0) or infinity (R = oo). If R is finite, F(x) converges absolutely when Ix — cl R. If R = oo, then F(x) converges absolutely for all x.

588

CHAPTER 11

INFINITE SERIES Proof We assume that c = 0 to simplify the notation. If F(x) converges only at x =- 0, then R = 0. Otherwise, F(x) converges for some nonzero value x = B. We claim that F(x) must then converge absolutely for all Ix I < I B I. To prove this, note that because 00 a, B° converges, the general term a0 B0 tends to zero. In particular, there F(B) =

E

n=0

exists M > 0 such that Ian Bn1 < M for all n. Therefore, 00

oo

00

Eicinfi=EianBnI n=0

Least Upper Bound Properly: If S is a set of real numbers with an upper bound M (i.e., x < M for all x E S), then S has a least upper bound L. See Appendix B.

0 (see marginal note). In this case, there exist numbers B E S smaller than but arbitrarily close to L, and thus, S contains (-B, B) for all 0 < B n

n=1 13 FIGURE 3 The power series 00 , , \ " n (x 5)0 has an interval of 4nn n=1 convergence (1,9].

cc (-1)E

5)0 _

n=1

9

Converges

if

n=1

We conclude that F(x) converges for x in the half-open interval (1,9] shown in Figure 3. • Some power series contain only even powers or only odd powers of x. The Ratio Test can still be used to find the radius of convergence. 00 x2n An Even Power Series Where does — converge? EXAMPLE 3 1---i (2n)! n=0

Solution Although this power series has only even powers of x, we can still apply the Ratio Test with an = x2' /(2n)!. We have = an+1

X 2(n+1)

x2n+2

(2(n+ 1))! = (2n + 2)!

n+21)! = (2n ± 2)(2n + 1)(2n)!, so Furthermore, (2na+ x2n+2

p = lim

n--).co

n—0

= nL irn oo

(2n)!

(2n ± 2)! x2"

= ix 12 lim

n—>oo

1 (2n + 2)(2n + 1)

=

0

Thus, p = 0 for all x, and F(x) converges for all x. The radius of convergence • is R = no. When a function f is represented by a power series on an interval 1, we refer to it as the power series expansion off on I.

0. Geometric series are important examples of power series. Recall the formula x in place of r, we obtain a power seE rn = 1/(1 — r), valid for VI < 1. Writing

ries expansion with radius of convergence R = 1: 1 1—x

00 = E xn

for lx1 < 1

2

n=0

The next two examples show that we can modify this formula to find the power series expansions of other functions.

590

CHAPTER 11

INFINITE SERIES EXAMPLE 4

Geometric Series

Prove that oo

E2nx0 1 — 2x — 1

for Ix I
0

69. Use Exercise 68 to show that for x > 0 and all n, ex

X2 — -1- • • •

x

2!

In Exercises 71-75, we estimate integrals using Taylor polynomials. Exercise 72 is used to estimate the error. 71. Find the fourth Maclaurin polynomial late I =

T4(x)dx as an estimate for Jo

72. Approximating Integrals Let L > 0. Show that if two functions f and g satisfy If (x) — g(x)I < L for all x c [a, b], then f (x)dx — f

73. Let T4 be the fourth Maclaurin polynomial for f (x) = cos x. (a) Show that ()6 )1 I COSS — T4(S,,
0, I f (k)(x)1 < ec+R for x E (c — R, c + R). Applying Theorem 2 with K = ec+R , we conclude that T (x) converges to f(x) for all x E (c — R, c + R). Since R is arbitrary, the Taylor expansion holds for all x. For c = 0, we obtain the standard Maclaurin series 2 X ex = 1 + X ± — 21

X3

•

+—+ 31

Shortcuts to Finding Taylor Series There are several methods for generating new Taylor series from known ones. First of all, we can differentiate and integrate Taylor series term by term within its interval of convergence, by Theorem 2 of Section 11.6. We can also multiply two Taylor series or substitute one Taylor series into another (we omit the proofs of these facts). EXAMPLE 4 Find the Maclaurin series for f(x) = X2 ex . Solution Multiply the known Maclaurin series for ex by x2: In Example 4, we can also write the Maclaurin series as

X 2 ex

E00 xn+2 n=0

n!

EXAMPLE 5

= x2

X2 (1 + X + —

= X2'

x4 Xj + — 2!

2!

Substitution

X3 —

X4 —

3!

x5

4!

x6

X5 — + •••)

5!

oo

x7

+ — + — + — ±••• = 3! 4! 5!

E

xn

•

n=2

Find the Maclaurin series for f(x) = e—X2

Solution Substitute —x2 for x in the Maclaurin series for ex: 00 (_ .x.2 e —x2

=

E

n=0

n!

=

00 ( _i n x2n )!

E n=0

n

4

X6

_ _ x- + _2! _ _3! + _4! _ • • • 2

X

X8

The Taylor expansion of ex is valid for all x, so this expansion is also valid for all x. EXAMPLE 6

Integration

Find the Maclaurin series for f(x) = 1n(1

•

x).

Solution We integrate the geometric series with common ratio —x (valid for Ix I < 1): 1 = 1 — x + x 2 — x3 + • • • 1+ x

ln(1 + x) =

dx x2 X3 =A+x--+---+• ,1 1 + x 2 3

X4

4

00 ‘n xn • •=A+E(-1) --- -

The constant of integration A on the right is zero because ln(1

n=1

x) = 0 for x = 0, so

ln(1 +x) =E( -1)n-1 Xn n=1

This expansion is valid for Ix I < 1. It also holds for x = 1 (see Exercise 92).

•

In many cases, there is no convenient formula for the coefficients of a Taylor series for a given function, but we can still compute as many coefficients as desired, as the next example demonstrates.

SECTION 11.8

EXAMPLE 7 Multiplying Taylor Series Maclaurin series for f (x) = ex cos x.

Taylor Series

Write out the terms up to degree 5 in

613

the

Solution We multiply the fifth-order Maclaurin polynomials of ex and cos x together, dropping the terms of degree greater than 5: (

x2

X3

X4

X5 —)

2

6

24

120

(

X 4)

X2

,

2

24

Distributing the term on the left (and ignoring products that result in terms of degree greater than 5), we obtain x2

(1 + X +

+

X3

X4

X5

X2

X3

+ 724 + -120) — (1 + x + — 2 X3

X

4

— 6 X

X2

—) 2

+ (1 +

x)

5

We conclude that the Maclaurin series for f (x) = ex cos x (with the terms up to degree 5) appears as X3

X4

X5

1 + x — — — — — — +. . . 3 6 30

ex COSX =

•

In the next example, we express a definite integral of sin(x2) as an infinite series. This is useful because the definite integral cannot be evaluated directly by finding an antiderivative of sin(x2). sin(x2) dx.

EXAMPLE 8 Let J = f

(a) Express J as an infinite series. (b) Determine J to within an error less than 10-4. Solution (a) The Maclaurin expansion for f (x) = sin x is valid for all x, so we have

E (2n + 1)! (3°

sinx

(— lr

E (2n(—l) ± 1)! oo

X 2n+1

sin(x2) =

1

X

4n+2

n=0

n=0

We obtain an infinite series for J by integration: 1 J=f 0

1 3

co 00 E (—on ( 1 ) x,n+2dx = (-1) E sin(x2) dx = (2n ± 1)! \ 4n ± 3 J h n

n=0

1 42

f 1

(2n + 1)!

n=0

1

1 1320

2

75,600

(b) The infinite series for J is an alternating series with decreasing terms, so the sum of the first N terms is accurate to within an error that is less than the (N 1)st term. The absolute value of the fourth term 1/75,600 is smaller than 10-4, so we obtain the desired accuracy using the first three terms of the series for J: J

1 3

1 1 ± 42 1320

0.31028

The error satisfies /1 3

1 42

1 \ 1320 )

1 75,600

1.3 x 10-5

The percentage error is less than 0.005% with just three terms. The next example demonstrates how power series can be used to assist in the evaluation of limits.

614

CHAPTER 11

INFINITE SERIES

x - sin x x-“) x3 cos x

EXAMPLE 9 Determine lirn

0 Solution This limit is of indeterminate form -, so we could use L'Hopital's Rule 0 the Maclaurin series. We have with work will instead, we However, repeatedly. X

3

X

5

sin x = x - — + — - • • • 5! 3! 2 X4 x — - ••• COS X = 1 - — 4! 2! Hence, the limit becomes x - (x -

sin x

x

lim - lim x—>0 x-*O x3 cosx

X3

5

+ •1 - • • • )

_ . . .)

x3(1

x3

7

= m

-5T

•

Binomial Series Isaac Newton discovered an important generalization of the Binomial Theorem around 1665. For any number a (integer or not) and integer n > 0, we define the binomial coefficient: a(a- 1)(a - 2) • • • (a - n + 1) (a) =1 n! 0 For example, (6 W

=

6.5.4 = 20 3 • 2. 1 '

_ 4 • ( i) (4) = 4 3) 3.2.1 81

Let f (x) = (1 + x)a The Binomial Theorem of algebra (see Appendix C) states that for any whole number a, (r + sr = ra + Setting r =- 1 and s

a ra-1s 1

a 2

r a-2s 2

a rsa-1 a- 1

sa

x, we obtain the expansion of f (x):

+ x)a = 1+ ( a)x + ( 61).x2 + \1 \21

+

(

a

a )

a_1 +x

We derive Newton's generalization by computing the Maclaurin series of f (x) without assuming that a is a whole number. Observe that the derivatives follow a pattern: f (x) = (1 + x)a

f (0) = 1

f (x) = a(1 + x)a-1

f'(0) = a

f''(x) = a(a - 1)(1 + xr -2

f"(0) = a(a - 1)

= a(a - 1)(a - 2)(1 + x)a-•3

f m(0) = a(a - 1)(a - 2)

SECTION 11.8

In

general, f

(0) = a(a - 1)(a - 2) • • • (a

Taylor Series

615

1) and

—n

f

When a is a positive whole number,

(n a)

is

zero for n > a, and in this case, the binomial series breaks off at degree n. The binomial series is an infinite series when a is not a positive whole number.

(0) a(a — 1)(a — 2) • • •(a - n + 1) (a) n! n! n) Hence, the Maclaurin series for f (x) = (1 ± x)a is the binomial series

E (a)xn

+

ax x + x +•••+ +••• 2! 3! n n=0 The Ratio Test shows that this series has radius of convergence R = 1 (Exercise 94), and an additional argument (developed in Exercise 95) shows that it converges to (1 + x)a for 1x1 < 1 .

THEOREM 3 The Binomial Series For any exponent a and for 1x1 oo. Let / = (c — R, c + R) with R > 0. Suppose that there exists K > 0 such that I f (k)(x)I < K for all x E I and all k. Then f is represented by its Taylor series on I; that is, f (x) = T (x) for x E I. A good way to find the Taylor series of a function is to start with known Taylor series and apply one of the following operations: differentiation, integration, multiplication, or substitution. For any exponent a, the binomial expansion is valid for Ix < 1: (1 + x)a = 1 + ax +

a(a — 1) 2 a(a — 1)(a — 2) x3 + x + 2! 3!

a) xn

11.8 EXERCISES Preliminary Questions 1. Determine f(0) and f'"(0) for a function f

with Maclaurin series

T(x) = 3 + 2x + 12x2 +5x3 + • • • 2.

Determine f(-2) and f (4)(-2) for a function with Taylor series T(x) = 3(x +2) + (x + 2)2 — 4(x + 2)3 + 2(x + 2)4 + • • •

3. What is the easiest way to find the Maclaurin series for the function f (x) = sin(x2)? 4. Find the Taylor series for f centered at c = 3 if f(3) = 4 and 1(x) has a Taylor expansion 00 f'(x) =

E(x-3) (x n=1

SECTION 11.8

5. Let T(x) be the Maclaurin series of f (x). Which of the following guarantees that f(2) = T(2)? (a) T (x) converges for x = 2.

Exercises 1. Write out the first four terms of the Maclaurin series of f (x) if (0) = 3,

f(0) = 2,

f"(0) = 4,

f'"(0) = 12

2. Write out the first four terms of the Taylor series of f (x) centered at c = 3 if f(3) = 1,

f'(3) = 2,

f"(3) = 12,

f'"(3) = 3

42. Show that for ix I < 1, x3 x5 tanh I x = x + - + - + • • • 3 5 d 1 Hint: Recall that - tanh-I x = dx 1 - X2 43. Use the Maclaurin series for 1n(1 + x) and ln(1 - x) to show that

1

f (x) =

f (x) = cos 3x

6.

f (x) = sin(2x)

7.

f (x) = sin(x2)

8.

f (x) = e4x

9.

f (x) = ln(1 - x2)

10. f (x) = (1 - x)-112

5.

1 + 10x

12. f (x) = x2ex2

13. f(x) = e-2

14. f (x) =

15. f (x) = 1n(1 - 5x)

16. f (x) = (x2 ± 2x)ex

17. f (x) = sinh x

18. f (x) = cosh x

sin-I x = x +

1 - cos x

22. f (x) =

1

5

5

± •••

3 • 5 • • • (2n - 1) x 2n+ I 2 • 4 • 6 • • (2n) 2n + 1

•

46. Use the first five terms of the Maclaurin series in Exercise 45 to approximate sin-1 1. Compare the result with the calculator value. 47. How many terms of the Maclaurin series of f (x) = ln(1 + x) are needed to compute In 1.2 to within an error of at most 0.0001? Make the computation and compare the result with the calculator value. 48. Show that

20. f (x) = ex ln(1 - x)

sin x 1- x

E n=1

In Exercises 19-30, find the terms through degree 4 of the Maclaurin series of f (x). Use multiplication and substitution as necessary.

21. f (x) -

3

X -

1 44. Differentiate the Maclaurin series for twice to find the Maclau1- x 1 rin series of (1 - x)3 • 1 , that 45. Show, by integrating the Maclaurin series for f (x) - X2 for 'xi 00. (c) The remainder Rk(2) approaches zero as k oo.

In Exercises 3-18, find the Maclaurin series andfind the interval on which the expansion is valid. 3.

Taylor Series

1 sinx

7r 3

7 5

77

converges to zero. How many terms must be computed to get within 0.01 of zero?

23. f (x) = (1 + x)1I4

24. f (x) = (1 + X)-312

25. f (x) = ex tan-I x

26. f (x) = sin (x3 -x)

27. f (x) = es'"

28. f (x) = e(ex)

49. Use the Maclaurin expansion for e-t2 to express the function e -t2 dt as an alternating power series in x (Figure 3). F(x) = Jo (a) How many terms of the Maclaurin series are needed to approximate the integral for x = 1 to within an error of at most 0.001?

29. f (x) = cosh(x2)

30. f (x) = sinh(x) cosh(x)

(b) (CAS) Carry out the computation and check your answer using a computer algebra system.

I

In Exercises 31-40, find the Taylor series centered at c and the interval on which the expansion is valid. 31. f(x) =

1

32. f (x) = e3x ,

c=1

C = -1

X

33. f (x) = -1

1 - x

34. f (x) = sin x,

c=5

35. f (x) = x4 + 3x - 1,

c=2

36. f (x) = x4 -I- 3x - 1,

c=0

1 37. f (x) = - 2- , 39. f (x) -

y = F(x)

2

1

FIGURE 3 The Maclaurin polynomial T15(x) for F(x) = f c=4

1 1-

c= 2

, X2

c- 3

38. f (x) = A/7, 40. f (x) =

c=4

1 3x - 2'

50. Let F(x) =c = -1

41. Use the identity c052 x = 1(1 + cos 2x) to find the Maclaurin series for f (x) = co52 x.

ix sin t dt

h

t

F(x) = x -

. Show that x3

x5 ± 5. 5!

Evaluate F(1) to three decimal places.

x7 ± 7 • 7!

x 0

,2 e- ` dt .

CHAPTER 11

620

INFINITE SERIES

In Exercises 51-54, express the definite integral as an infinite series and find its value to within an error of at most 10-4. 51. f

cos(x2) dx

52.f

e -x 3 dx

54.

V I(t) = (- ) (1 - e-RtIL ) R

tan-1(x2)dx

0 1

I

53. f

Jo

Vt Expand 1(t) in a Maclaurin series. Show that 1(t)-% - for small t.

dx

•,/x4 +

74. Use the result of Exercise 73 and your knowledge of alternating series to show that

In Exercises 55-58, express the integral as an infinite series. 55. 57.

f f

ox 1 - cos t

dt,

ox ln(1 + t 2 )dt,

f ox t - sin t

for all x

56.

for Ix! < 1

58. fo

dt

dt,

E(_

, for Ix! < 1

2x'?

n=0 60. Which function has the following Maclaurin series?

E (-ok +1 (x 00

3)k

For which values of x is the expansion valid? 61. Using Maclaurin series, determine to exactly what value the following series converges: E( -1)n " (2n)! n=0 62. Using Maclaurin series, determine to exactly what value the following series converges: 00 (ln 5)n L-' n! n=0 In Exercises 61-64, use Theorem 2 to prove that the f (x) is represented by its Maclaurin seriesfor all x. 63. f (x) = sin(x/2) + cos(x/3)

64. f (x) = e- x

65. f (x) = sinh x

66. f (x) = (1 + x)1°°

In Exercises 67-70, find the functions with the following Maclaurin series (refer to Table 2 prior to the section summary). x9

x 12

53x3 55x5 + 3! 5!

57x7

7!

x x20 x28 x 12 '2 +- +••• 3 5 7 In Exercises 71 and 72, let 1 (1 - x)(1 - 2x)

2 1 - 2x

1 1- x

72. Find the Taylor series for f (x) at c = 2. Hint: Rewrite the identity of Exercise 71 as f (x) =

2 -3 - 2(x - 2)

76. Find f (7)(0) and f ( (0) for f (x) = tan-1 x using the Maclaurin series. Use substitution to find the first three terms of the Maclau-

77.

rin series for f (x) = exm . How does the result show that f (k)(0) = 0 for 1 < k < 19? - x2.

79. Does the Maclaurin series for f (x) = (1 + .0314 converge to f (x) at x = 2? Give numerical evidence to support your answer. Explain the steps required to verify that the Maclaurin series for 80. f (x) = ex converges to f (x) for all x. 81. (2.(1) Let f (x) = (a) Use a graphing calculator to compare the graph of f with the graphs of the first five Taylor polynomials for f. What do they suggest about the interval of convergence of the Taylor series? (b) Investigate numerically whether or not the Taylor expansion for f is valid for x = 1 and x = -1. 82. Use the first five terms of the Maclaurin series for the elliptic integral E(k) to estimate the period T of a 1-m pendulum released at an angle 0 = (see Example 12). 83. Use Example 12 and the approximation sin x X to show that the period T of a pendulum released at an angle 0 has the following second-order approximation: L 2) T 1 + 4.g

cos x - 1 +

85. urnsinx - x

x4

1 -1 - (x - 2)

3

T x

x5

tan 1 x - x cos x - -61 x 3 86. lim x5

71. Find the Maclaurin series of f (x) using the identity f (x) =

75. Find the Maclaurin series for f (x) = cos(x3) and use it to determine f (6)(0).

84. lim x->o

70. x4

f (x) =

(for all t)

In Exercises 84-87, the limits can be done using multiple L'Hopital's Rule steps. Power series provide an alternative approach. In each case substitute in the Maclaurin seriesfor the trig function or the inverse trigfunction involved, simplify, and compute the limit.

+- + +••• L! 3! 4! 68. 1 - 4x + 42X2 43x3 44x4 45x5 69. l -

< 1(t) < V t R 2L) L

78. Use the binomial series to find f(NO) for f (x) =

k=0

x6

V t (1 L

for all x

00

59. Which function has Maclaurin series

67. 1 + x' +

73. When a voltage V is applied to a series circuit consisting of a resistor R and an inductor L, the current at time t is

87. l im x->o (

sin(x2) x4

cos x x2

88. Use Euler's Formula to express each of the following in a + bi form. (a)

(b) 4e* i

(c) ie -21

89. Use Euler's Formula to express each of the following in a + bi form. (a) -e* i

(b) e21r i

(c) 3ie+ri

In Exercises 90-91, use Euler's Formula to prove that the identity holds. Note the similarity between these relationships and the definitions of the hyperbolic sine and cosine functions. 90. cos z -

eiz

e- iz 2

91. sin z -

eiz

Chapter Review Exercises

621

Further Insights and Challenges 92. In this exercise, we show that the Maclaurin expansion of f(x) = ln(1 + x) is valid for x = 1. (a) Show that for all x -1,

+x = E(-1)nxn 1

n=0

(-0N-F1xN+1 1+x

97. Assume that a oo an

E (n3 ±

26. Define an+ i = Van ± 6 with a l = 2. (a) Compute an for n = 2, 3, 4, 5.

E

(b) Show that (an ) is increasing and is bounded by 3. (c) Prove that lim an exists and find its value. oo

27. Calculate the partial sums S4 and S7 of the series

E n2n -

n=1

2

+2n

•

31. Use series to determine a reduced fraction that has decimal expansion 0.108108108 • • • . oo 32. Find the sum n=2

n=-1

E (b n=1

44.

\---, cx) n2 -I- 1

46.

48. n=2 Vn5 +5

E

E

1 3n - 2"

n=1

00 n 10 ± 10n il + 11"

50.

+

1 n 1 n - In n

n=1

47.

49.

E n=1 oo

00 n20 + 21"

E

n=1 n21

20"

oo

51. Determine the convergence of

E 2"+n 3" -2 using the Limit Cornn=1

2n+3

parison Test with bn =

3n

(-2 ) n .

3

Inn E _1.5 oo

oo

34. Show that

00 E 1)2 n=1 (n+

nLa =2 n" - 2 oo

30. Use series to determine a reduced fraction that has decimal expansion 0.121212 • • • .

00

43.

45.

4 8 16 32 29. Find the sum - + - + - + ±••• . 9 27 81 243

E

E

In Exercises 43-50, use the Direct Comparison or Limit Comparison Test to determine whether the infinite series converges.

1 1 1 28. Find the sum 1 - - ± - - - + • • • . 4 42 43

33. Find the sum

In Exercises 39-42, use the Integral Test to determine whether the infinite series converges. 00 00 n2 n2 40. 39. 1)1M 71_ 1 n3 + 1 n=1 oo CO 3 1 n 41. 42. (n + 2)(1n(n ± 2))3 n=1 e n=1

tan-1 n2) diverges if b

--7r 2-

52. Determine the convergence of ison Test with bn =-

1.4" •

n=1

using the Limit Compar-

Chapter Review Exercises 00

53. Leta, = 1 - \11 - 4.. Show that nlirn an =0 and that 00

E (n2 ±

+

57.

E n1.1 ln(n + 1)

c° cos a 59. E n =1

n-=1

+ 27 n)

N/72

z

1 72. E (n) n! 4

n!

n=1

75.

oo

76

E( 43n )

n=1

77. Ec° ( _2 ) n 3) 79.

78.

k=0

Ee

80.

n=1

n=1

n=10 (_ on

n=2

n=2

1 E (an + _,) nz'

(b)

oo

(C)

1

(d)

E 1 ± a,2,

Ec-iyian oo E lani

n=1

n=1

91.

93.

n

mine whether the following series converge or diverge: 00 00 (c) 30 an 2an (b) (a) n=1

co

E

E .J--in

n=1

n=1

In Exercises 65-72, apply the Ratio Test to determine convergence or divergence, or state that the Ratio Test is inconclusive. 65.

E -5n °°

n=1

n5

66.

°° N/n + 1 E n8

n=1

nt

nn +inn

E

(1

90.

00

95.

E

1

E ,Z/12(1 1+ N/T1) . 94. E ( In n - 1n(n + 1)) 00

92.

n=1

,/n + 1

)

n=1

96.

n=1

E nln1 n n=2 oo

99.

E sin2 -7n

n=1

00 cos(nn) E n=2

co

oo

97.

E enn!

n=1

1

NiTz

.Vn3/2 + 1

oo

E (5n n+ 2 )) n 0. 1 v ,

n=1

64. Let {an } be a positive sequence such that n limoo 4'/Tn = 1. . Deter-

E

n=1

n=1

n=1

n=1

88.

ni2

00

89.

oo

00

(a)

E cos n

n3 - 2n2 + n - 4 2n4 + 3n3 - 4n2 - 1

DO

87.

E an be an absolutely convergent series. Determine whether

the following series are convergent or divergent:

1 n(ln n)3i2

n! (2n)!

In n 86.

n=1

n=1

E ne-0.02n oo

83.

oo

63. Let

7n

e8n

n=1

E

E

n3

oo

-0.02n

co

E

00 7r E -

82. 00

an and 62. Give an example of conditionally convergent series 00 oo n=1 (an + bn ) converges absolutely. bn such that

)

n=1

oo

n=1

"k

(a) How many terms of the series are needed to calculate K with an error of less than 10-6? (b) ( CASs Carry out the calculation.

COS

n=1

81.

E (2k + 1)2

E

In Exercises 77-100, determine convergence or divergence using any method covered in the text.

.0

n

E (_n2 ) n=1

\n

n=1

co

61. Catalan's constant is defined by K =

74. 00

to within an error of at most 10-5.

n3 ± n=1

n=4

n=1

to approximate

60. (CAS) Use a computer algebra system

co

Inn E n3/2

co

fl

n=1 7n)

oo

73.

( i)

cos (

n=1

In Exercises 73-76, apply the Root Test to determine convergence or divergence, or state that the Root Test is inconclusive.

In Exercises 56-59, determine whether the series converges absolutely. If it does not, determine whether it converges conditionally.

+ 2n

4

n Z-v n!

oo

n=1 1)2

(b) (CAS) Use the inequality in (4) from Exercise 83 of Section 11.3 with M = 99 to approximate the sum of the series. What is the maximum size of the error?

58.

70.

71.

n=1

n=1

°° 2n2 n!

n=1

oo

(a) Show that the series converges.

00 E 56.

68.

n=1

00

55. Consider

00

E n20 + n3

69. E converges.

n2

n=2

oo

67.

n=1

1 diverges. Hint: Show that an > -• 2n oo 1 54. Determine whether I - -

E

>a0

623

98.

1

E 1n3 n

n=2

00 22n n=0

624

INFINITE SERIES

CHAPTER 11

In Exercises 101-106, find the interval of convergence of the power series. n 102. 101. n+1 n-=0 n6

3r

(x

103.

104.

E nxn n=0

12. 0 n8 + 1 oo 105. Etitxr n=0

106.

cs-.) x •-•, (2x - 3)n n=2

nlnn

2

as a power series centered at c = 0. De4- 3x termine the values of x for which the series converges.

107. Expand f (x) =

108. Prove that

cp. e x ne-nx = E (1 - e- x)2 n=0

123. Find n such that le - Tn (1)I < 10-8, where Tn is the nth Maclaurin polynomial for f (x) = ex . 124. Let T4 be the Taylor polynomial for f (x) = x ln x at a = I computed in Exercise 115. Use the Error Bound to find a bound for If (12) - 7•4(1 •2)1. 125. Verify that T(x) = 1 + x + x2 + • • • + xn is the nth Maclaurin polynomial of f (x) = 1/(1 - x). Show using substitution that the nth Maclaurin polynomial for f (x) = 1/(1 - x/4) is

2k • k!. k=0 (a) Show that F(x) has infinite radius of convergence. (b) Show that y = F(x) is a solution of y(0) = 1,

y" = xy/ + y,

)/(0) = 0

(c) (CAS) Plot the partial sums SN for N = 1, 3, 5, 7 on the same set of axes. 00 110. Find a power series P(x) = that satisfies the Laguerre differential equation n=0

E anxn

xy” + (1 - x)y' - y = 0 with initial condition satisfying P(0) = 1.

5 and let ak be the coefficient of xk in 4 + 3x - x2 the Maclaurin polynomial Tn for k < n.

(a) Show that f (x) =

112. Use power series to evaluate lim x-o- 0

x2(1 - ln(x + 1)) sin x - x •

In Exercises 113-118, find the Taylor polynomial at x = a for the given function. a =1

114. f (x) = 3(x + 2)3 - 5(x + 2),

T3,

a = -2

1/4 (1 - x /4

In Exercises 127-136, find the Taylor series centered at c. =0

127. f (x) 129. f (x) = x4,

130. f (x) = x3 - x,

133. f (x) = 134. f (x) =

1 1 - 2x (1 -

c= ,

2.02 ,

1

117. f (x) = xe-x2 ,

a=0

T3,

a =0

119. Find the nth Maclaurin polynomial for f (x) = e3x. 120. Use the fifth Maclaurin polynomial of f (x) = ex to approximate Use a calculator to determine the error.

132. f (x) = eX

,

c = -1

c = -2

135. f (x)

ln

2

'

c=2

136. f (x) = x ln (1 ± -), c -= 0 2 In Exercises 137-140, find the first three terms of the Maclaurin series of f (x) and use it to calculate f(3)(0).

139. f (x) =

118. f (x) --= ln(cos x),

7r

c = -2

T3,

T4,

c = -1

c = -2

116. f (x) = (3x + 2)113 ,

a=2

128. f (x) = e2x,

c=2

137. f (x) = (x2 - x)ex2

T4,

1

(c) Compute T3.

a=1

115. f (x) = x ln(x),

1 1 + x) •

(b) Use Exercise 125 to show that ak =

131. f (x) = sin x,

111. Use power series to evaluate lim x-> 0 cos x - 1

T3,

1 1 2 -x + • • • + - xn 4n 42

1 ? What is the nth Maclaurin polynomial for g(x) =1+x

x2ex

113. f (x) = x3,

1 -x 4

126. Let f (x) =

x 2k

109. Let F(x) =

122. Let T4 be the Taylor polynomial for f (x) = ji at a = 16. Use the Error Bound to find the maximum possible size of If (17) - 7•40 7/1.

Tn (x) = 1

Hint: Express the left-hand side as the derivative of a geometric series. 00

121. Use the third Taylor polynomial of f (x) = tan-1 x at a = 1 to approximate f (1.1). Use a calculator to determine the error.

1 1 + tan x

3 5 it 7r 71141. Calculate - + 2 233! 255!

138. f (x) = tan-1(x2 - x) 140. f (x) = (sin x),V171- x 77 277! ± •

•

142. Find the Maclaurin series of the function F(x) = f Jo

x et - 1 t dt.

12 PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTI 1NS We can study the interaction between two animal species with populations q(t) and p(t), where each population is a function of time, to investigate how the two populations change. Combining the functions in the form (q(t), p(t)) yields a parametric representation of a curve in the qp-plane. Tracing this curve as t changes creates a story about this interaction and its impact on population size.

T

his chapter introduces two important new tools. First, we consider parametric equations, which describe curves in a form that is particularly useful for analyzing motion and is indispensable in fields such as computer graphics and computer-aided design. We then study polar coordinates, an alternative to rectangular coordinates that simplifies computations in many applications. The chapter closes with a discussion of the conic sections (ellipses, hyperbolas, and parabolas).

12.1 Parametric Equations We use the term "particle" when we treat an object as a moving point, ignoring its internal structure. Position at time t

Imagine a particle moving along a curve C in the plane as in Figure 1. We would like to be able to describe the particle's motion along the curve. To express this motion mathematically, we consider how its coordinates x and y are changing in time, that is, how they depend on a time variable t. Thus, x and y are both functions of time, t, and the location of the particle at t is given by

y(t))

c(t) = (x(t), y(t))

- -Curve C t=0

t=4

FIGURE 1 Particle moving along a curve C in the plane.

This representation of the curve C is called a parametrization with parameter t, and C is called a parametric curve. In a parametrization, we often use t for the parameter, thinking of the dependent variables as changing in time, but we are free to use any other variable (such as s or 0). In plots of parametric curves, the direction of motion is often indicated by an arrow as in Figure 1. Specific equations defining a parametrization, such as x = 2t — 4 and y = 3 + t2 in the next example, are called parametric equations. EXAMPLE 1 Sketch the curve with parametric equations x = 2t — 4,

y =

1

3 + t2

Solution First compute the x- and y-coordinates for several values of t as in Table 1, and plot the corresponding points (x, y) as in Figure 2. Then join the points by a smooth • curve, indicating the direction of motion (direction of increasing t) with an arrow. =4 (4, 19)

TABLE 1 4

x —2 2 4 Graphing calculators or CAS software can be used to sketch and examine parametric curves.

—8 —4 4

y = 3 - t2 7 3 7 19

=2 (0, 7) (-43) -8

—4

41.

X

M FIGURE 2 The parametric curve x = 2t —4, y = 3 ± t2.

CONCEPTUAL INSIGHT The graph of y = x2 can be parametrized in a simple way. We take x = t and y = t2. Then, since y = t2 and t = x, it follows that y = x2. Therefore, the parabola y = x2 is parametrized by c(t) = (t, t2). More generally, we can 625

626

CHAPTER 12

PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS

parametrize the graph of y = f (x) by taking x = t and y = f (t). Therefore, c(t) = (t, f (t)) parametrizes the graph of y = f (x). For another example, the graph of y -= ex is parametrized by c(t) = (t, et). An advantage of parametric equations is that they enable us to describe curves that are not graphs of functions. For example, the curve in Figure 3 is not of the form y = f (x) but it can be expressed parametrically. As we have just noted, a parametric curve c(t) need not be the graph of a function. If it is, however, it may be possible to find the function f by "eliminating the parameter" as in the next example. EXAMPLE 2 FIGURE

Eliminating the Parameter

3 The parametric curve

Describe the parametric curve

c(t) = (2t — 4,3 + t2)

x = 5 cos(3t)cos(i sin(5t)),

of the previous example in the form y = f (x).

y = 4 sin(3t)cos(i sin(5t)).

Solution We eliminate the parameter by solving for y as a function of x. First, express t in terms of x: Since x = 2t — 4, we have t = x + 2. Then substitute Y=3+t

2

1 x+ 2

=3

2

1

= 7 + 2x -P —x2 4

Thus, c(t) traces out the graph of f (x) = 7 + 2x + lx2 shown in Figure 2.

•

EXAMPLE 3 A model rocket follows the trajectory c(t) = (80t, 200t — 4.9t2) until it hits the ground, with t in seconds and distance in meters (Figure 4). Find: (a) The rocket's height at t = 5 s.

(b) Its maximum height.

Solution The height of the rocket at time t is y(t) = 200t — 4.9t2. (a) The height at t = 5 s is 1,000

2,000

3,000

FIGURE 4 Trajectory of rocket. CAUTION The graph of height versus time for an object tossed in the air is a parabola (by Galileo's formula). But keep in mind that Figure 4 is not a graph of height versus time. It shows the actual path of the rocket (which has both a vertical and a horizontal displacement).

y(5) = 200(5) — 4.9(52) = 877.5 m (b) The maximum height occurs at the critical point of y(t) found as follows: y'(t) = --61--(200t — 4.9t2) = 200 — 9.8t dt Thus, y' = 0 when 200 — 9.8t = 0, that is, for t=

200 ,=-= 20.4 s 9.8

The rocket's maximum height is y(20.4) = 200(20.4) — 4.9(20.4)2

2041 m.

•

We now discuss parametrizations of lines and circles. They will appear frequently in later chapters. To begin, note that the parametric equations x = t,

y = mt

— 00 0) c(t) = (t —

t t tanh — , f sech —

has the following property: For all t, the segment from c(t) to (t, 0) is tangent to the curve and has length (Figure 28).

FIGURE 29 The parameter 0 on the ellipse n

y 2 2 ± (—) = 1. a b

109. Show that the parametrization of the ellipse by the angle 9 is ab cos X =

a2 sin2 9 + b2 cos2 ab sin s/a2 sin2 9 + b2 c0s2 FIGURE 28 The tractrix c(t) =

—

tanh

t sech — ).

SECT I ON 12.2

Arc Length and Speed

637

12.2 Arc Length and Speed We now derive a formula for the arc length s of a curve in parametric form. Recall that in Section 9.2, arc length of a curve C was defined as the limit of the lengths of polygonal approximations of C (Figure 1).

P3

FIGURE 1 Polygonal approximations to a curve C for N = 5 and N = 10.

134

, X

, X

N=5

N =10

To compute the length of C via a parametrization, we need to assume that the parametrization directly traverses C, that is, the path traces C from one end to the other without changing direction along the way. Thus, assume that c(t) = (x(t), y(t)) is a parametrization that directly traverses C for a < t < b. We construct a polygonal approximation L consisting of the N segments obtained by joining points P1 = c((i ),

Po = c(to),

,

PN = c(tN)

corresponding to a choice of values to = a 0: 1

PF1 + PF2 = K We assume always that K is greater than the distance F1F2 between the foci, because the ellipse reduces to the line segment F 1F2 if K = F1F2, and it has no points at all if K < F1F2.

The points F1 and F2 are called the foci (plural of "focus") of the ellipse. Note that if the foci coincide, then Eq. (1) reduces to 2P F1 = K and we obtain a circle of radius 1K centered at F1. We use the following terminology: • The midpoint of F1 F2 is the center of the ellipse. • The line through the foci is the focal axis. • The line through the center perpendicular to the focal axis is the conjugate axis. Conjugate axis B = (0, b) 4111111ft

= (x, y)

Focal axis A'= (-a, 0) (-c, 0)

Semiminor axis

(c, 0) A = (a, 0)

Center

B' = (0, -6)

Semimajor axis (A) The ellipse consists of all points P

(B)

Ellipse in standard position:

such that PF1 + PF2 = K

w2

(i )2 =

FIGURE 2

The ellipse is said to be in standard position (and is called a standard ellipse) if the focal and conjugate axes are the x- and y-axes, as shown in Figure 2(B). In this case, the foci have coordinates Fi = (c, 0) and F2 = (—C, 0) for some c > 0. Let us prove that the equation of this ellipse has the particularly simple form f.D2

ki nb)2 = 1

2

a)

where a = K/2 and b = ,s/ a2 e2. By the distance formula, P = (x, y) lies on the ellipse in Figure 2(B) if

PFI + PF2 --1(x —

+ y2 -1-1(x + c)2 + y2 = 2a

3

In the second equation, move the first term on the left over to the right and square both sides: (x Strictly speaking, it is necessary to show that if P = (x, y) satisfies Eq. (4), then it also satisfies Eq. (3). When we begin with Eq. (4) and reverse the algebraic steps, the process of taking square roots leads to the relation 02 + y2 ± ‘ Ax ± 02 ± y2 =

±2,

c)2 + y2 = 4a2 — 4a \ (x — c)2 +

4a1(x — C)2 ± y2 = 4a2

c)2 — (X +

4.

(x

02

= 4a2 —

±

y2

4cx

Now divide by 4, square, and simplify: a2(x2 — 2cx + e2 + (a2

e2)x2

x2 a2

However, because a > c this equation has no solutions unless both signs are positive.

(x

y2

a2y2 =_

Y a2

y2) = a4

- 2a2CX

a4-

C2X2

a2c2= a2(a2

e2)

2 e2 =

This is Eq. (2) with b2 = a2 — c2 as claimed.

1

4

SECTION 12.5

Conic Sections

659

The ellipse intersects the axes in four points A, A', B, B' called vertices. Vertices A and A' along the focal axis are called the focal vertices. Following common usage, the numbers a and b are referred to as the semimajor axis and the semiminor axis (even though they are numbers rather than axes). THEOREM 1 Ellipse in Standard Position Let a > b > 0, and set c = N/ a2 b2. The ellipse P P F2 = 2a with foci F1 = (c, 0) and F2 = (—c, 0) has equation &a .x)2

(Y )2 = 1

Furthermore, the ellipse has • semimajor axis a, semiminor axis b. • focal vertices (±a, 0), minor vertices (0, ±b). If b > a > 0, then Eq. (5) defines an ellipse with foci (0, ±c), where c = ,/b2 _ az. EXAMPLE 1 Find the equation of the ellipse with foci (I N/1i, 0) and semimajor axis a = 6. Then find the semiminor axis and sketch the graph. Solution The foci are (±c, 0) with c = fa., and the semimajor axis is a = 6, so we can use the relation c = A/a2 — b2 to find b: b=5

b2 = a2 — c2 = 62 — (4 -1)2 = 25

()x 2 ())2 = 1. Thus, the semiminor axis is b =- 5 and the ellipse has equation -6 + — 5 To sketch this ellipse, plot the vertices (±6,0) and (0, ±5) and connect them as in • Figure 3.

0)

(0,5)

(0,-5)

0)

(6, e

(-6, 0)

(6,0)

(-6,0)

(0,5)

(0,-5)

FIGURE 3

To write the equation of an ellipse with axes parallel to the x- and y-axes and center translated to the point C = (h, k), replace x by x — h and y by y — k in the equation (Figure 4): —ah) 2 +

(y — b k\) 2 =

(x EXAMPLE 2 Translating an Ellipse Find an equation of the ellipse with center at C =- (6,7), vertical focal axis, semimajor axis 5, and semiminor axis 3. Where are the foci located?

660

PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS

CHAPTER 12

Solution Since the focal axis is vertical, we have a = 3 and b = 5, so that a < b (Fig) ure 4). The ellipse centered at the origin would have equation ( 2 ± (i)2= 1. When the center is translated to (h, k) = (6,7), the equation becomes ix — 6 \ 2 3 )

)2

( y —

5

= 1

Furthermore, c = N/b2 _ a 2 ,‘ /52 32 = 4, so the foci are located ±4 vertical units • above and below the center—that is, F1 = (6, 11) and F2 = (6,3). A hyperbola is the set of all points P such that the difference of the distances from P to two foci F1 and F2 is ±K: P

FIGURE 4 An ellipse with a vertical major axis and center at the origin, and a translation of it to an ellipse with center C = (6,7). Conjugate axis

F1

Focal axis'

6

— P F2 = ±K

We assume that K is less than the distance F1F2 between the foci (the hyperbola has no points if K > F1 F2). Note that a hyperbola consists of two branches corresponding to the choices of sign ± (Figure 5). As before, the midpoint of F1F2 is the center of the hyperbola, the line through F1 and F2 is called the focal axis, and the line through the center perpendicular to the focal axis is called the conjugate axis. The vertices are the points where the focal axis intersects the hyperbola; they are labeled A and A' in Figure 5. The hyperbola is said to be in standard position (and is called a standard hyperbola) when the focal and conjugate axes are the x- and y-axes, respectively, as in Figure 6. The next theorem can be proved in much the same way as Theorem 1. THEOREM 2 Hyperbola in Standard Position Let a > 0 and b > 0, and set c = b2. The hyperbola P — P F2 = ±2a with foci F1 =(c, 0) and F2 = ( — c, 0) has equation

,1a 2

(X\2 k a)

FIGURE 5 A hyperbola with center (0,0).

(Y\ 2

1

b)

A hyperbola has two asymptotes that we claim are the lines y = ± 161x. Geometrically, these are the lines through opposite corners of the rectangle whose sides pass through (±a, 0) and (0, ±b) as in Figure 6. To prove the claim, consider a point (x, y) on

Conjugate axis

b 17 -1 \

the hyperbola in the first quadrant. By Eq. (7),

\ \\

\

F2=(

a

•

- Focal axis

Fi = (c, 0)

-b

13 FIGURE 6 Hyperbola in standard position.

=

b2 _7x 2

b2 =

a4

a2

a

The following limit shows that a point (x, y) on the hyperbola approaches the line y =!ix as x ----> oo: lim x—>oo

— — b x) = — b lim (Vx2 — a2 — x) a

a x-00 4-

b = — lim a x -›-co

( x2 — a2 — x

= 11 lim ax->cx)(,/x2

) (A

—a2 a2

x

2 —a2

) =0 N/x2

The asymptotic behavior in the remaining quadrants is similar.

x\

SECTION 12.5

Conic Sections

661

EXAMPLE 3 Find the foci of the hyperbola 9x2 - 4y2 = 36. Sketch its graph and asymptotes. Solution First, divide by 36 to write the equation in standard form: X2

y2

=1 Thus, a = 2, b = 3, and c = -Va2

b2 =

=

FIGURE 7 The hyperbola 9x2

—

4y2 = 36.

)2 — (y)2

Ix

or

=

0),

= 1

The foci are

F2 =

To sketch the graph, we draw the rectangle through the points (±2,0) and (0, ±3) as in Figure 7. The diagonals of the rectangle are the asymptotes y = ±;.x. The hyperbola passes through the vertices (±2,0) and approaches the asymptotes. • Unlike the ellipse and hyperbola, which are defined in terms of two foci, a parabola is the set of points P equidistant from a focus F and a line D called the directrix:

Axis PF F = (0, c)

, X

2

Vertex V -c

Directrix V

y = -c

8 Parabola with focus (0, c) and directrix y = -c.

al FIGURE

PD

8

Here, when we speak of the distance from a point P to a line D, we mean the distance from P to the point Q on D closest to P, obtained by dropping a perpendicular from P to D (Figure 8). We denote this distance by PD. The line through the focus F perpendicular to D is called the axis of the parabola. The vertex is the point where the parabola intersects its axis. We say that the parabola is in standard position (and is a standard parabola) if, for some c, the focus is F = (0, c) and the directrix is y = -c, as shown in Figure 8. We prove in Exercise 75 that the vertex is then located at the origin and the equation of the parabola is y = x2/ 4c. If c 0 and downward if c 1. How the eccentricity determines the shape of a conic is summarized in Figure 10. Consider the ratio bla of the semiminor axis to the semimajor axis of an ellipse. The ellipse is nearly circular if bla is close to 1, whereas it is elongated and flat if bla is small. Now b

N/a2 _

a

a

c2 =

1—

a2 =

e2

This shows that bla gets smaller (and the ellipse gets flatter) as e The most round ellipse is the circle, with e = 0. Similarly, for a hyperbola,

1 [Figure 10(B)].

— = /1+ e2 a The ratios ±bla are the slopes of the asymptotes, so the asymptotes get steeper as e ---> oo [Figure 10(C)].

Circle

Parabola

0

1

e Ellipses

(A) Eccentricity e

0

FIGURE 10

Hyperbolas

(B) Ellipse flattens as e --+1.

(C) Asymptotes of the hyperbola get steeper as e—>oo.

SECTION 12.5

Conic Sections

663

CONCEPTUAL INSIGHT There is a more precise way to explain how eccentricity determines the shape of a conic. We can prove that if two conics C1 and C2 have the same eccentricity e, then there is a change of scale that makes C1 congruent to C2. Changing the scale means changing the units along the x- and y-axes by a common positive factor. A curve scaled by a factor of 10 has the same shape but is 10 times as large. By "congruent," we mean that after scaling, it is possible to move C1 by a rigid motion (involving rotation and translation, but no stretching or bending) so that it lies directly on top of C2. All circles (e = 0) have the same shape because scaling by a factor r > 0 transforms a circle of radius R into a circle of radius rR. Similarly, any two parabolas (e = 1) become congruent after suitable scaling. However, an ellipse of eccentricity e = 0.5 cannot be made congruent to an ellipse of eccentricity e = 0.8 by scaling (see Exercise 76). Directrix x=a

Eccentricity can be used to give a unified focus-directrix definition of the conic sections with e > 0. Given a point F (the focus), a line D (the directrix), and a number e > 0, we consider the set of all points P such that 10

PF = ePD

For e = 1, this is our definition of a parabola. According to the next theorem, Eq. (10) defines a conic section of eccentricity e for all e > 0 (Figures 11 and 12). THEOREM 5 Focus-Directrix Relationship Ellipse 31 FIGURE 11 The ellipse consists of points P such that PF = ePD.

• If 0 b > 0 and c = N/a2 — b2, then the ellipse (: )2

=

satisfies Eq. (10) with F = (c,0), e =

and vertical directrix x =

Hyperbola FIGURE 12 The hyperbola consists of points P such that PF = ePD.

• If e > 1, then the set of points satisfying Eq. (10) is a hyperbola, and xy-coordinate axes can be chosen, and a,b defined, so that the hyperbola has eccentricity e and is in standard position with equation 1

/D 2

'a! • Conversely, if a,b > 0 and c = N/a2 /..1\ 2

a)

-

b) b2, the hyperbola fl y

_ 1

- kb)

satisfies Eq. (10) with F = (c, 0), e = ca , and vertical directrix x Proof We prove the first part of the hyperbola relationship. The remaining parts of the theorem are proven in Exercises 65, 66, and 68. We are assuming we have a set of points satisfying Eq. (10) with e > 1. Let d denote the distance between the focus and directrix. We want to set up our x- and y-axes and define a,b,c so that the focus is located at (c, 0)

664

CHAPTER 12

PARAMETRIC EQUATIONS, POLAR COORDINATES, AND CONIC SECTIONS

a Directrix D x = -e-

and the directrix is x = that follow) is to set

ice

c= F = (c,

as in Figure 13. What works (and we will see why in the steps

b

a = —, e

1 — e-2 '

.1c2

a2

With the coordinate axes set up so that the focus is located at (c, 0), the directrix is then the line x=c—d=c—c(1—e-2) = c e-2 -= a

FIGURE 13

which is what we wanted. 2 ( y )2 -= 1 Now, we need to show that the equation P F = ePD is in the form ela in the coordinate system. This establishes that the resulting curve is a hyperbola with an equation in the standard-position form in the coordinate system. A point P = (x, y) that satisfies P F = e PD then satisfies 1/(x — e)2 -I- y2 =

—

(a Ie))2

PD

PF

Algebraic manipulation yields (x — c)2

+

y2 = e2

— (a1e))2

x2 — 2ex + c2 + y2 = e2x2 — 2aex x2

2 ete-x-

a 2e2 ± y2

e2x2

Y

a2

a 2(e2

a2

X —2e+ a2

(e2 — 1)x2 — y2 = a2(e2 _ 1) x2

(square)

(use c = ae) (rearrange)

2 (divide)

— 1

This is the desired equation because a2(e2 —

1)

c2

a 2 = b2.

•

Note that Theorem 5 indicates that for every e > 0, the solution to Eq. (10) produces a conic section in standard position. Also, all conic sections in standard position, except for the circle, can be obtained via the focus-directrix relationship in Eq. (10). EXAMPLE 5 Find the equation, foci, and directrix of the standard ellipse with eccentricity e = 0.8 and focal vertices (±10, 0). Solution Directrix x= 12.5

The vertices are (±a, 0), with a = 10 (Figure 14). By Theorem 5, b =_

c = ae = 10(0.8) = 8,

az _

= .1102

82 =

6

Thus, our ellipse has equation

(-

) 2

(6-)2

=1

The foci are (±c, 0) = (±8,0) and the directrix is x = Ellipse of eccentricity e = 0.8 with focus at (8,0).

FIGURE 14

= 4 ( 1 = 12.5.

•

In Section 14.6, we discuss the famous law of Johannes Kepler stating that the orbit of a planet around the sun is an ellipse with one focus at the sun. In this discussion, we will need to write the equation of an ellipse in polar coordinates. To derive the polar equations of the conic sections, it is convenient to use the focus-directrix definition with

SECTI 0 N 12.5

Conic Sections

665

focus F at the origin 0 and vertical line x = d as directrix D (Figure 15). Note from the figure that if P =- (r, 8), then P F -= r,

PD = d — r cos

Thus, the focus-directrix equation of the ellipse P F = ePD becomes r e(d — r cos 0), or r(1 e cos 0) =- ed. This proves the following result, which is also valid for the parabola and hyperbola (see Exercise 69).

FIGURE 15 Focus-directrix definition of the ellipse in polar coordinates.

THEOREM 6 Polar Equation of a Conic Section The conic section of eccentricity e > 0 with focus at the origin and directrix x = d has polar equation r=

ed 1

11

e cos 0

EXAMPLE 6 Find the eccentricity, directrix, and focus of the conic section r=

24 4 ± 3 cos 0

Solution First, we write the equation in the standard form: r=

24 4 + 3 cos

6 1 + 24 cos 0

Comparing with Eq. (11), we see that e = and ed -= 6. Therefore, d = 8. Since e 2 FIGURE 19 34.

X2 +

y2 + z2 = 9, with x, y, z > 0

35.

x2 +

y2

=

7, with lz < 7

56. Find a parametrization of the line through P = (4, 9, 8) perpendicular to the yz-plane. 57. Show that ri (t) and r2(t) define the same line, where

36. x2 + y2 = 4, with y, z > 0 ri(t) = (3, -1,4) + t (8,12, -6) In Exercises 37-42, give an equation for the indicated surface. 37. The sphere of radius 3 centered at (0,0, -3) 38. The sphere centered at the origin passing through (1,2, -3) 39. The sphere centered at (6, -3, 11) passing through (0, 1, -4)

r2(t) = (11, 11, -2) +t (4, 6, -3) Hint: Show that r2(t) passes through (3,-i, 4) and that the direction vectors for ri(t) and r2(t) are parallel. 58. Show that ri(t) and r2(t) define the same line, where 1.1(0 = t (2,1, 3) ,

r2(t) = (-6, -3, -9) + t (8,4,12)

40. The sphere with diameter P Q where P = (1, 1, -3) and Q = (1, 7, 1)

59. Find two different vector parametrizations of the line through P = (5, 5, 2) with direction vector v = (0,-2, 1).

41. The cylinder passing through the origin with the vertical line through (1, -1,0) as its central axis

60. Find the point of intersection of the lines r(t) = (1, 0, 0) + t (-3,1, 0) and s(t) = (0, 1, 1) +t (2, 0, 1).

SECTION 13.2

61. Show that the lines ri (t) = (-1, 2, 2) + t (4, -2, 1) and r2(t) = (0,1, 1) + t (2, 0, 1) do not intersect.

Three-Dimensional Space: Surfaces, Vectors, and Curves

695

70. Find the components of the vector w whose tail is C and head is the midpoint of AB in Figure 20.

62. Determine whether the lines r1 (t) = (2, 1, 1) + t (-4,0,1> and r2(s) = (-4, 1,5> + s (2, 1, -2) intersect, and if so, find the point of intersection. 63. Determine whether the lines ri(t) = (0, 1, 1) t (1, 1,2) and r2(s) = (2, 0, 3) + s (1, 4, 4) intersect, and if so, find the point of intersection.

A = (1,0, 1)

64. Find the intersection of the lines ri (t) = (-1,1) + t (2,4) and r2(s) -= (2,1) s (-1, 6) in the plane. 65. A meteor follows a trajectory r(t) = (2, 1,4> ± t (3,2, -1) km with t in minutes, near the surface of the earth, which is represented by the xyplane. Determine at what time the meteor hits the ground. 66. A laser's beam shines along the ray given by ri (t) = (1, 2, 4) ± t(2, 1, -1) for t > 0. A second laser's beam shines along the ray given by r2(s) = (6,3, -1) + s(-5, 2, c) for s > 0, where the value of c allows for the adjustment of the z-coordinate of its direction vector. Find the value of c that will make the two beams intersect. 67. The line with vector parametrization r(t) = (3, 1, -4) + t (-2, -2,3) intersects the sphere (x - 1)2 + (y + 3)2 + z2 = 8 in two points. Find them. Hint: Determine t such that the point (x(t), Y(t), z(t)) satisfies the equation of the sphere, and then find the corresponding points on the line.

B = (1,1,0) FIGURE 20 71. A box that weighs 1000 kg is hanging from a crane at the dock. The crane has a square 20 m by 20 m framework as in Figure 21, with four cables, each of the same length, supporting the box. The box hangs 10 m below the level of the framework. Find the magnitude of the force acting on each cable. (-10, -10,0)

(-10, 10,0)

68. Show that the line with vector parametrization r(t) = (3, 5, 6) + t (1, -2, -1) does not intersect the sphere of radius 5 centered at the origin. 69. Find the components of the vector v whose tail and head are the midpoints of segments AC and BC in Figure 20. [Note that the midpoint of (a l bi a2 + b2 a3 b3 (ai , a2, a3) and (bi , b2, b3) is 2 2 2

(10, -10,0)

(10, 10,0)

(0,0, -10) FIGURE 21

Further Insights and Challenges In Exercises 72-78, we consider the equations of a line in symmetric form, when a 0 0, b 0, c 0. x - xo Y YO Z - Zo 10 a 72. Let r be the line through Po = (xo, yo, zo) with direction vector v (a, b, c). Show that r is defined by the symmetric equations (10). Hint: Use the vector parametrization to show that every point on r satisfies (10). 73. Find the symmetric equations of the line through Po = (-2,3,3) with direction vector v = (2,4, 3). 74. Find the symmetric equations of the line through P = (1, 1,2) and Q = (-2,4,0).

79. A median of a triangle is a segment joining a vertex to the midpoint of the opposite side. Referring to Figure 22(A), prove that three medians of triangle ABC intersect at the terminal point P of the vector .1 (u + v + w). The point P is the centroid of the triangle. Hint: Show, by parametrizing the segment AA', that P lies two-thirds of the way from A to A'. It will follow similarly that P lies on the other two medians. 80. A median of a tetrahedron is a segment joining a vertex to the centroid of the opposite face. The tetrahedron in Figure 22(B) has vertices at the origin and at the terminal points of vectors u, v, and w. Show that the medians intersect at the terminal point of t (u + v + w).

75. Find the symmetric equations of the line x = 3 ± 2t,

y = 4 - 9t,

z = 12t

76. Find a vector parametrization for the line x- 5 9 -

y+3 =z 7

10

77. Find a vector parametrization for the line ;-1 =

=

78. Show that the line in the plane through (xo, yo) of slope m has symmetric equations Y

0 (A)

(B)

YO FIGURE 22

696

CHAPTER 13

VECTOR GEOMETRY

13.3 Dot Product and the Angle Between Two Vectors Wind parallel to bridge

FIGURE 1 A bridge oriented 32 degrees east

of north with a 60 km/h westerly wind.

Operations on vectors are widely used in engineering and other scientific disciplines. For example, an operation called the dot product can be applied to analyze the wind blowing toward a bridge. Suppose a 60 km/h wind w is blowing from the west toward a bridge that is oriented 32 degrees east of north (Figure 1). A civil engineer needs to compute how much of the wind is blowing directly at the bridge to determine whether large trucks will be permitted on the bridge under such wind conditions. To analyze this problem we express w as a sum of vectors, one parallel to the bridge and a second perpendicular to it. We can compute the parallel part by projecting w onto a vector parallel to the bridge. Finding projections of one vector onto another is easily done with the dot product, a concept we introduce in this section. The dot product is one of two important products that we define on pairs of vectors. The other, cross product, is introduced in the next section. Dot product is defined as follows: DEFINITION Dot Product

The dot product v • w of two vectors

v= (vi, v2, v3) ,

w

(W1, W27 W3)

is the scalar defined by V•W=

Important concepts in mathematics often have multiple names or notations either for historical reasons or because they arise in more than one context. The dot product is also called the scalar product or inner product, and in many texts, v • w is denoted (v, w) or (v, w).

The dot product appears in a very wide range of applications. For example, to determine how closely a Web document matches a search input, a dot product is used to develop a numerical score by which candidate documents can be ranked. (See The Anatomy of a Large-Scale Hypertextual Web Search Engine by Sergey Brin and Lawrence Page.)

V2W2

V3W3

In words, to compute the dot product, multiply the corresponding components and add. For example, (2, 3, 1) • (-4, 2, 5) = 2(-4) + 3(2) + 1(5) = —8 + 6 + 5 = 3 The dot product of vectors v = (v1, v2) and w = (wi, w2) in R2 is defined similarly: V • w = viwi

v2w2

We will see in a moment that the dot product is closely related to the angle between v and w. Before getting to this, we describe some elementary properties of dot products. First, the dot product is commutative: v • w = w • v, because the components can be multiplied in either order. Second, the dot product of a vector with itself is the square of the length: If v = (vi, v2, v3), then v • v = vi vi + v2v2 + v3v3

vi. +

+ vi = 111'112

The dot product also satisfies the Distributive Law and a scalar-multiplication property as summarized in the next theorem (see Exercises 94 and 95). THEOREM 1 Properties of the Dot Product (i) 0 v = v • 0 = 0 (ii) Commutativity: v-w=w•v (iii) Pulling out scalars: (Av) • w = v • (Avv) = X(v • w) (iv) Distributive Law: u • (v + w) =u•v+u•w (v+w)•u=v•u+w•u (v) Relation with length:

v•v

EXAMPLE 1 Verify the Distributive Law u • (v + w) =u•v+u•w for u= (4, 3, 3) ,

v= (1, 2, 2) ,

w= (3, —2, 5)

SECTION 13.3

Dot Product and the Angle Between Two Vectors

697

Solution We compute both sides and check that they are equal: u • (v + w) = (4,3,3) • ( (1,2,2) + (3, —2,5) ) = (4, 3, 3) • (4,0,7) = 4(4) + 3(0) + 3(7) = 37 u•v+u•w= (4, 3, 3) • (1, 2, 2) ± (4, 3, 3) • (3, —2, 5) = (4(1) + 3(2) ± 3(2)) + (4(3)

3(-2)

3(5))

= 16 + 21 = 37

By convention, the angle between two vectors is chosen so that 0 ----> --> the vectors QP and QR (or RP and RQ) to find a normal vector to P.

PQ = (2,2,1) — (1,0,—i) = (1,2,2) PR = (4, 1, 2) — (1, 0, —1) = (3,1,3) i --> —> n= PQ x PR = 1 3

j 2 1

k 2 = 4i 4 3j — 5k = (4,3, —5) 3

By Eq. (3), P has equation 4x + 3y — 5z = d for some d. Step 2. Choose a point on the plane and compute d. Now, choose any one of the three points—say, P = (1,0, —1)—and compute CAUTION When you find a normal vector to the plane containing points P, Q, R, be sure to compute a cross product such as PQ A common mistake is to use a cross product such as OP x or --> OP x OR, which need not be normal to the plane.

A.

06

d=n• OP= (4, 3, —5) • (1, 0, —1) = 9 We conclude that P has equation 4x + 3y — 5z = 9.

•

To check our result, simply verify that the three points we were given satisfy this equation of the plane. Example 3 shows that three noncollinear points determine a plane. A plane P also can be determined by any of the following: • Two lines that intersect in a single point; • A line and a point not on the line; • Two distinct parallel lines. For each of these situations, think about how to determine a normal vector to the plane and a point on the plane. We examine these cases in the exercises.

SECTION 13.5

Planes in 3-Space

721

EXAMPLE 4 Intersection of a Plane and a Line Find the point P where the plane 3x — 9y + 2z = 7 and the line r(t) = (1,2, 1) + t (-2,0, 1) intersect. Solution The line has parametric equations x

1 — 2t,

y = 2,

z=1

t

Substitute in the equation of the plane and solve for t: 3x — 9y + 2z = 3(1 — 2t) — 9(2) + 2(1 + t) = 7 If we think oft representing time on a path on the line, then t = —5 is the time that the path meets the plane. The point P = (11,2, —4) is the location on the path at that time.

Simplification yields —4t — 13 = 7 or t = —5. Therefore, P has coordinates x= 1 — 2(-5) = 11,

y = 2,

z=

The plane and line intersect at the point P = (11,2, —4). The intersection of a plane P with a coordinate plane or a plane parallel to a coordinate plane is called a trace. The trace is a line unless P is parallel to the coordinate plane (in which case, the trace is empty or is P itself). EXAMPLE 5 Traces of the Plane its traces in the coordinate planes.

Graph the plane —2x + 3y + z = 6 and then find

Solution To draw the plane, we determine its intersections with the coordinate axes. To find where it intersects the x-axis, we set y = z = 0 and obtain —2x = 6

so

x = —3

It intersects the y-axis when x = z = 0, giving 3y = 6

so

y=2

It intersects the z-axis when x = y = 0, giving z=6 Thus, the plane appears as in Figure 7. We obtain the trace in the xy-plane by setting z = 0 in the equation of the plane. Therefore, the trace is the line —2x + 3y = 6 in the xy-plane (Figure 7). Similarly, the trace in the xz-plane is obtained by setting y = 0, which gives the line • —2x + z = 6 in the xz-plane. Finally, the trace in the yz-plane is 3y z = 6.

FIGURE 7 The three blue lines are the traces of the plane —2x + 3y + z = 6 in the coordinate planes.

In most cases, you can picture a plane by determining where it intersects the coordinate axes. However, if the plane is parallel to one of the axes and does not intersect it, or the plane passes through the origin, and therefore intersects all three axes there, we can use the normal vector to determine how the plane is oriented.

13.5 SUMMARY • Plane through Po = (xo, yo, zo) with normal vector n = (a, b, c): — Geometrically: The tips of all vectors based at Po that are perpendicular to n — Algebraically: n • (x, y, z) = d

Vector form: Scalar forms:

a(x — x0)

b(Y — YO)

c(z — zu) = 0

ax + by + cz = d where d = n • (xo, yo, zo) = axo + byo + CZ.

P2

CHAPTER 13

VECTOR GEOMETRY

• The family of parallel planes with given normal vector n = (a, b, c) consists of all planes with equation ax + by + cz = d for some d. • A plane is determined in each of the following cases. In each case a point on the plane and a normal vector to the plane can be determined in order to find an equation for the plane. — — — —

Three noncollinear points Two lines that intersect in a point A line and a point not on it Two distinct parallel lines

• The intersection of a plane P with a coordinate plane or a plane parallel to a coordinate plane is called a trace. The trace in the yz-plane is obtained by setting x = 0 in the equation of the plane (and similarly for the traces in the xz- and xy-planes).

13.5 EXERCISES Preliminary Questions 1. What is the equation of the plane parallel to 3x + 4y — z = 5 passing through the origin? 2. The vector k is normal to which of the following planes? (a) x = 1 (b) y = 1 (c) z = 1 3. Which of the following planes is not parallel to the plane x+y+z=1? (a) 2x + 2y + 2z = 1 (b) x+y+z=3 (c) x — y + z = 0

4.

To which coordinate plane is the plane y = 1 parallel?

5.

Which of the following planes contains the z-axis?

(a) z = 1

(b) x + y = 1

(c) x + y = 0

6. Suppose that a plane P with normal vector n and a line .0 with direction vector v both pass through the origin and that n • v = 0. Which of the following statements is correct? (a) is contained in P. (b) C is orthogonal to P.

Exercises In Exercises 1-8, write the equation of the plane with normal vector n passing through the given point in the scalar form ax + by + cz = d. 1.

n= (1,3,2),

3.

n = (-1, 2, 1),

5.

n=

7.

n = k,

9.

Write the equation of any plane through the origin.

(4,-1,1)

2.

n = (-1, 2, 1),

(3, 1, 9)

(4,1,5)

4.

n = (2, —4,1),

( ,4,i)

(3, 1, —9)

6.

n =j,

(6,7,2)

8.

n = i — k,

(-5, (4, 2, —8)

10. Write the equations of any two distinct planes with normal vector n = (3,2, 1) that do not pass through the origin. 11. Which of the following statements are true of a plane that is parallel to the yz-plane? (a) n = (0,0, 1) is a normal vector. (b) n = (1, 0, 0) is a normal vector.

(A)

(B)

(C)

FIGURE 8 In Exercises 13-16, find a vector normal to the plane with the given equation. 13. 9x — 4y — llz = 2

14. x — z = 0

15. 3(x — 4) — 8(y — 1) + llz = 0 16. x = 1

(c) The equation has the form ay + bz = d.

In Exercises 17-20, find the equation of the plane with the given description.

(d) The equation has the form x = d.

17. Passes through 0 and is parallel to 4x — 9y + z = 3

12. Find a normal vector n and an equation for the planes in Figures 8 (A)—(C).

18. Passes through (4,1, 9) and is parallel to x + y + z = 3 19. Passes through (4, 1,9) and is parallel to x = 3

S EC TION 13.5

20. Passes through P = (3,5, —9) and is parallel to the xz-plane

Planes in 3-Space

723

39. Find an equation of the plane P in Figure 9.

In Exercises 21-24, find an equation of the plane passing through the three points given. 21. P = (2, —1,4),

Q = (1,1,1),

R = (3, 1, —2)

22. P = (5, 1, 1),

Q =- (1,1,2),

R = (2,1,1)

23. P = (1,0,0),

Q = (0,1,1),

R = (2, 0,1)

24. P = (2, 0, 0),

Q = (0, 4, 0),

R = (0,0,2)

E/J

25. In each case, describe how to find a normal vector to the plane: (a) Three noncollinear points are given. The plane contains all three points. (b) Two lines are given that intersect in a point. The plane contains the lines. 26. al In each case, describe how to find a normal vector to the plane: (a) A line and a point that is not on the line are given. The plane contains the line and the point. (b) Two lines are given that are parallel and distinct. The plane contains the lines. 27. In each case, determine whether or not the lines have a single point of intersection. If they do, give an equation of a plane containing them. (a) ri(t) = (t, 2t — 1, t —3) and r2(t) = (4,2t — 1,—i) (b) ri (t) = (3t ,2t + 1, t — 5) and r2(t) = (4t, 4t — 3, —1) 28. In each case, determine whether or not the lines have a single point of intersection. If they do, give an equation of a plane containing them. (a) r1(t) = (5t, 2t — 1, 2t —2) and r2(t) = (t — 5, —t + 4, t —7) (b) ri (t) = (3,-2t + 1, t —3) and r2(t) = (2t — 1, —t, —t —1)

FIGURE 9 40. Verify that the plane x — y +5z = 10 and the line r(t) = (1, 0, 1) + t (-2, 1, 1) intersect at P = (-3,2,3). In Exercises 41-44, find the intersection of the line and the plane. 41. x + y + z = 14, 42. 2x + y = 3, 43. z = 12,

r(t) = (1,1,0) + t (0,2, 4)

r(t) = (2, —1, —1) + t (1, 2, —4)

r(t) = t (-6,9,36)

44. x — z = 6,

r(t) = (1,0,-1) + t (4,9,2)

In Exercises 45-50, find the trace of the plane in the given coordinate plane. 46. 3x — 9y + 4z = 5, xz

29. In each case, determine whether or not the point lies on the line. If it does not, give an equation of a plane containing the point and the line. (a) (2, 2, —1) and r(t) = (4t, 6t — 1,—i) (b) (3, —3,2) and r(t) = (4t + 3, 4t — 3, t + 1)

45. 3x — 9y + 4z = 5,

30. In each case, determine whether or not the point lies on the line. If it does not, give an equation of a plane containing the point and the line. (a) (-7, 10, —3) and r(t) = (1 — 4t, 6t — 5, t — 5) (b) (-1, 5, 9) and r(t) -= (4t + 3, t + 6, 5 — 4t)

51. Does the plane x = 5 have a trace in the yz-plane? Explain.

31. In each case, determine whether or not the lines are distinct parallel lines. If they are, give an equation of a plane containing them. (a) ri (t) = (t ,2t — 1, t —3) and r2(t) = (3t — 3, 6t — 1, 3t —1) (b) ri (t) = (3t ,2t + 1, t — 5) and r2(t) = (-6t, I — 4t, 2t — 3)

47. 3x +4z = —2,

yz

xy

49. —x + y = 4, xz

48. 3x +4z = —2, xz 50. —x+y= 4,

yz

52. Give equations for two distinct planes whose trace in the xy-plane has equation 4x + 3y = 8. 53. Give equations for two distinct planes whose trace in the yz-plane has equation y = 4z. 54. Find parametric equations for the line through /30 = (3,—i, 1) perpendicular to the plane 3x + 5y — 7z = 29.

32. In each case, determine whether or not the lines are distinct parallel lines. If they are, give an equation of a plane containing them. (a) ri = (2t + 1, —2t — 1, 3t — 7) and r2(t) = (7 — 6t,6t — 7,2 — 9t) (b) ri = (-4t, 2t + 1, 8t +5) and r2(t) = (2t —2, —t +4,5 — 4t)

55. Find all planes in R3 whose intersection with the xz-plane is the line with equation 3x + 2z = 5.

In Exercises 33-37, draw the plane given by the equation.

56. Find all planes in R3 whose intersection with the xy-plane is the line r(t) = t (2,1,0).

33. x+y+z=4

34. 3x + 2y — 6z = 12

In Exercises 57-62, compute the angle between the two planes, defined as the angle 0 (between 0 and v) between their normal vectors (Figure 10).

35. 12x — 6y + 4z = 6

36. x + 2y = 6

57. Planes with normals ni = (1, 0, 1), n2 = (-1, 1, 1)

37. x+y+z= 0 38. Let a, b, c be constants. Which two of the following equations define the plane passing through (a, 0,0), (0, b, 0), (0,0, c)? (b) bcx + acy + abz = abc (a) ax + by + cz = 1 z x y (d) — + — + — = 1 (c) bx + cy + az = 1 c a b

58. Planes with normals ni = (1, 2, 1), n2 = (4, 1, 3) 59. 2x + 3y + 7z = 2 and 4x — 2y + 2z =4 60. x — 3y + z = 3 and 2x — 3z = 4 61. 3(x — 1) — 5y + 2(z — 12) = 0 and the plane with normal n = (1, 0, 1)

724

CHAPTER 13

VECTOR GEOMETRY

62. The plane through (1,0,0), (0,1,0), and (0,0, 1) and the yz-plane

70. The plane x

y

z

intersects the x-, y-, and z-axes in points P, Q, and R. Find the area of the triangle APQR.

ae

71. In this exercise, we show that the orthogonal distance D from the plane P with equation ax ± by ± cz = d to the origin 0 is equal to (Figure 11) D= FIGURE 10 By definition, the angle between two planes is the angle between their normal vectors. 63. Find an equation of a plane making an angle of 3x + y — 4z = 2.

with the plane

64. L.T Let Pi and P2 be planes with normal vectors n1 and n2. Assume that the planes are not parallel, and let L be their intersection (a line). Show that ni x 112 is a direction vector for L.

Id' ,‘ /a2

b2

e2

Let n = (a, b, c), and let P be the point where the line through the origin with direction vector n intersects P. By definition, the orthogonal distance from P to 0 is the distance from P to 0. d (a) Show that P is the terminal point of v = n. n•n (b) Show that the distance from P to 0 is D. 72. Use Exercise 71 to compute the orthogonal distance from the plane x + 2y + 3z = 5 to the origin.

65. Find a plane that is perpendicular to the two planes x + y = 3 and x + 2y — z = 4. 66. Let L be the line of intersection of the planes x + y + z = 1 and x + 2y + 3z = 1. Use Exercise 64 to find a direction vector for L. Then find a point P on L by inspection, and write down the parametric equations for .C. 67. Let L denote the line of intersection of the planes x — y — z = 1 and 2x + 3y + z = 2. Find parametric equations for the line L. Hint: To find a point on L, substitute an arbitrary value for z (say, z = 2) and then solve the resulting pair of equations for x and y. 68. Find parametric equations for the line of intersection of the planes 2x + y — 3z = 0 and x + y = 1. 69. Vectors v and w, each of length 12, lie in the plane x + 2y — 2z = 0. The angle between v and w is 7r/6. This information determines v x w up to a sign ±1. What are the two possible values of v x w?

FIGURE 11

Further Insights and Challenges In Exercises 73 and 74, let P be a plane with equation

73. Show that the point P on P closest to Q is determined by the equation

ax + by + cz = d and normal vector n = (a, b, c). For any point Q, there is a unique point P onP that is closest to Q, and is such that P Q is orthogonal toP (Figure 12).

0P=0 Q+

d— 0Q•n n•n

5

74. By definition, the distance from Q = (xi, yi,zi) to the plane P is the distance to the point P on P closest to Q. Prove distance from Q to P =

lax' + byi + cz —dl

111111

6

75. Use Eq. (5) to find the point P nearest to Q = (2, 1,2) on the plane x + y + z = I. 76. Find the point P nearest to Q = (-1,3, —1) on the plane x — 4z = 2 77. Use Eq. (6) to find the distance from Q = (1, 1, 1) to the plane 2x + y + 5z = 2. 78. Find the distance from Q = (1, 2, 2) to the plane n • (x, y, z) = 3, where n = FIGURE 12

SECT 10 N 13.6

79. What is the distance from Q = (a, b, c) to the plane x = 0? Visualize your answer geometrically and explain without computation. Then verify that Eq. (6) yields the same answer.

A Survey of Quadric Surfaces

725

80. The equation of a plane n • (x, y, z) = d is said to be in normal form if n is a unit vector. Show that in this case, Id I is the distance from the plane to the origin. Write the equation of the plane 4x — 2y + 4z = 24 in normal form.

13.6 A Survey of Quadric Surfaces

To ensure that Eq. (1) is genuinely quadratic, we assume that the degree-2 coefficients A, B,C,D,E,F are not all zero.

Quadric surfaces are the surface analogs of conic sections. Recall that a conic section is a curve in R2 defined by a quadratic equation in two variables. A quadric surface is defined by a quadratic equation in three variables: Ax2

By2

Cz2 + Dxy

Eyz+ Fzx + ax + by ± cz+ d =0

1

Like conic sections, quadric surfaces are classified into a small number of types. When the coordinate axes are chosen to coincide with the axes of the quadric, the equation of the quadric has a simple form. The quadric is then said to be in standard position. In standard position, the coefficients D, E, F are all zero. In this short survey of quadric surfaces, we restrict our attention to quadrics in standard position. The idea here is not to memorize the formulas for the various quadric surfaces, but rather to be able to recognize and graph them using cross sections obtained by slicing the surface with certain planes, as we describe in this section. The surface analogs of ellipses are the egg-shaped ellipsoids (Figure 1). In standard form, an ellipsoid has the equation

Ellipsoid

FIGURE 1 Ellipsoid with equation (z)2 (y)2 (x)2

=1

..x)2

(— a

y)2

— b

(z)2

1

— c

For a = b = c, this equation is equivalent to x2 + y2 ± z2 = a2 and the ellipsoid is a sphere of radius a. Surfaces are often represented graphically by a mesh of curves called traces, obtained by intersecting the surface with planes parallel to one of the coordinate planes (Figure 2), yielding certain cross sections of the surface. Algebraically, this corresponds to holding one of the three variables constant. For example, the intersection of the horizontal plane z = zo with the surface is a horizontal trace curve. EXAMPLE 1

The Traces of an Ellipsoid

( 3x 2 ) ±

y)2

Describe the traces of the ellipsoid (z) 2

=1

Solution First, we observe that the traces in the coordinate planes are ellipses [Figure 3(A)1: Trace curve FIGURE 2 The intersection of the plane z = zo with an ellipsoid is an ellipse.

(X)2

±

()2

xy-trace (set z = 0, blue in figure):

=1 ( 02 + ( D2 = 1

yz-trace (set x = 0, green in figure): xz-trace (set y = 0, red in figure):

( i .) 2 ± (D 2

=1

726

CHAPTER 13

VECTOR GEOMETRY

(C) Vertical traces

(B) Horizontal traces FIGURE 3 The ellipsoid

(X)2

3

±

()2 ( y )2 z , +

4 = 1.

In fact, all the traces of an ellipsoid are ellipses (or just single points). For example, the horizontal trace defined by setting z = zo is the ellipse [Figure 3(B)] trace at height zo:

/X\ 2 W

/ 41 \ 2

/y\ 2 + W

+(

or

=1

)

x2 y2 — + — =

25

49

1 —

2 Zo

=81

A constant

The trace at height zo = 9 is the single point (0, 0, 9) because x2/25 + y2 /49 = 0 has only one solution: x = 0, y = 0. Similarly, for zo = —9, the trace is the point (0,0, —9). If Izo I > 9, then 1 — 4/81 < 0 and the plane z = zo lies above or below the ellipsoid. The trace has no points in this case. The traces in the vertical planes x = xo and y = yo have a similar description [Figure 3(C)]. • The surface analogs of the hyperbolas are the hyperboloids, which come in two types, depending on whether the surface has one or two components that we refer to as sheets (Figure 4). Their equations in standard position are Hyperboloids

One sheet: Two sheets:

2

y 2

z 2

na

± (—) b = (—) c 2 ) 2 z 2

Ca-)

± (—) b = (—) c

+1

2 —1

Notice that a hyperboloid of two sheets does not contain any points whose z-coordinate satisfies —c 0 such that every point on the surface lies at a distance of at most M from the origin. Which of the quadric surfaces are bounded?

True or false? All traces of a hyperboloid are hyperbolas.

3. Which quadric surfaces have both hyperbolas and parabolas as traces? 4.

6.

What is the definition of a parabolic cylinder?

Is there any quadric surface whose traces are all parabolas?

Exercises In Exercises 1-8, state whether the given equation defines an ellipsoid or hyperboloid, and if a hyperboloid, whether it is of one or two sheets.

1. (

+()

3.

x 2 ± 3y2 ▪

5.

x2

7.

iz\2

iy•2

x‘2 -2 )

2.

=1

9z2

=

(D2

(z)2

5)2 (y

(D 2_

9z2 =

▪

3y2

(-5 )

=1

(i)2

± ( D2 =

4. 6.

1

8.

x2 + y2 = 4 - 4z2

x2 _ 3y2

9z2 =

z=

FIGURE 14

11. z = x \2

( x) 2

2

(X 4- ) 2 + (i) 2

10. Z

=

4 +

(y)2

12. 4z = 9x2 + 5y2

Y \2 VE)

14. 3x2

15.

16. y =3x2 -4z 2

4z2

26. Describe the surface that is obtained when, in the equation ±8x2 ± 3y2 z2 = 1, we choose (a) all plus signs, (b) one minus sign, and (c) two minus signs. 27. What is the equation of the surface obtained when the elliptic paraboloid z = (jc- )2 + ( 1 ) 2 is rotated about the x-axis by 90°? Refer 2 4 to Figure 15.

7y2 = 14z2

13. 3x2 - 7y2 = z y2 =5x2

(C)

x2 + 3y2 = 9 + z2

In Exercises 9-14, state whether the given equation defines an elliptic paraboloid, a hyperbolic paraboloid, or an elliptic cone. 9.

( 3)

(A)

In Exercises 17-24, state the type of the quadric surface and describe the trace obtained by intersecting with the given plane. 17. x2 -1- (i . ) 2 + z2 = 1,

y

0

18.

X2 ± (-1 ) 2 ± Z2 =

1,

y=5

19.

X 2 4- ( 1 ) 2 + Z2 =

1,

1 Z=4

4

x) 2 20. (2

4

( y) 2 -5.z` = 1, 5

FIGURE 15 x

21. ( .)2 + (-) 2 - 5z2 = 1, 5 3

y= 1

28. Describe the intersection of the horizontal plane z = h and the hyperboloid -x 2 - 4y2 + 4z2 = 1. For which values of h is the intersection empty?

( 1 ) 2 - 2z2 = -1, 3

z=1

In Exercises 29-42, sketch the given surface. 29. x 2 ± y2 _ z2 = 1

22. 4x2

23. y = 3x2,

z = 27

30. ( x4 )2 + ( 24. y = 3x2,

25. Match each of the ellipsoids in Figure 14 with the correct equation: (a) x 2 ± 4 y2 4z2 = 16 (b) 4x2 + y2 + 4z2 = 16 (c) 4x2 + 4y2 + z2 = 16

) 2 + ( iz2)2 = 1

y = 27 31. z = (x4)2 + (28.,. ) 2 33. z2 =

(i)2

± (D2

32.

Z =

4

34. y = -x2

-

(1 )2 8

732

CHAPTER 13

35.

_ x2 _ y2 + 9z2 =

VECTOR GEOMETRY

9

36. x2 + 36y2 = 1

37. xy = 1

38. x = 2y2 — z2

39. x = 1 + y2 + z2

ao.

41. x2 + 9y2 + 4z2 = 36

42. y 2 —

x2 — 4)/2 = Z A

2 _

Z

2 =

4

43. Find the equation of the ellipsoid passing through the points marked in Figure 16(A). 44. Find the equation of the elliptic cylinder passing through the points marked in Figure 16(B).

(B)

(A)

45. Find the equation of the hyperboloid shown in Figure 17(A).

FIGURE 17

46. Find the equation of the quadric surface shown in Figure 17(B). 47. Determine the vertical traces of elliptic and parabolic cylinders in standard form. 48. What is the equation of a hyperboloid of one or two sheets in standard form if every horizontal trace is a circle? 49. Let C be an ellipse in a horizonal plane lying above the xy-plane. Which type of quadric surface is made up of all lines passing through the origin and a point on C? 50. The eccentricity of a conic section is defined in Section 12.5. Show that the horizontal traces of the ellipsoid t rx: \ 2 + (17 \ 2

(A)

(B) FIGURE 16

a)

b)

= 1

c)

+

are ellipses of the same eccentricity (apart from the traces at height h = ±c, which reduce to a single point). Find the eccentricity.

Further Insights and Challenges 51. Let S be the hyperboloid x2 + y2 = z2 + 1 and let P = (a, 0) be a point on S in the (x, y)-plane. Show that there are precisely two lines through P entirely contained in S (Figure 18). Hint: Consider the line r(t) = (a + at, + bt,t) through P. Show that r(t) is contained in S if (a, b) is one of the two points on the unit circle obtained by rotating (a, p) through ± This proves that a hyperboloid of one sheet is a doubly ruled surface, which means that it can be swept out by moving a line in space in two different ways.

Cone on ellipse C

Cone on parabola C (half of cone shown) FIGURE 19

FIGURE 18

SECTION 13.7

In Exercises 52 and 53, let C be a curve in R3 not passing through the origin. The cone on C is the surface consisting of all lines passing through the origin and a point on C [Figure 19(A)].

a)

733

53. Let a and c be nonzero constants and let C be the parabola at height c consisting of all points (x,ax2, c) [Figure 19(B)]. Let S be the cone consisting of all lines passing through the origin and a point on C. This exercise shows that S is also an elliptic cone. (a) Show that S has equation yz = acx2. (b) Show that under the change of variables y = u + v and z = u — v, this equation becomes acx2 = u2 — v2 or u2 = acx2 + v2 (the equation of an elliptic cone in the variables x, v, u).

= ( X \ 2 ± ( .1. ) 2 is, in fact, 52. Show that the elliptic cone ( 4 c k a ). \b a cone on the ellipse C consisting of all points (x, y, c) such that (D 2 ± ( n2

Cylindrical and Spherical Coordinates

= 1.

)

13.7 Cylindrical and Spherical Coordinates Wire

This section introduces two generalizations of polar coordinates to R3: cylindrical and spherical coordinates. These coordinate systems are commonly used in problems having symmetry about an axis or rotational symmetry. For example, the magnetic field generated by a current flowing in a long, straight wire is conveniently expressed in cylindrical coordinates (Figure 1). We will also see the benefits of cylindrical and spherical coordinates when we study change of variables for multiple integrals.

Magnetic field

I

Current flow

FIGURE 1 The magnetic field generated by a current flowing in a long, straight wire is conveniently expressed in cylindrical coordinates.

Cylindrical Coordinates In cylindrical coordinates, we replace the x- and y-coordinates of a point P = (x, y, z) by polar coordinates. Thus, the cylindrical coordinates of P are (r, 6, z), where (r, 0) are polar coordinates of the projection Q = (x, y, 0) of P onto the xy-plane (Figure 2). Note that the points at fixed distance r from the z-axis make up a cylinder; hence, the name cylindrical coordinates. We convert between rectangular and cylindrical coordinates using the rectangularpolar formulas of Section 12.3. In cylindrical coordinates, we usually assume r > 0. Cylindrical to rectangular

Rectangular to cylindrical

x =COS

r = 1 x2 ± y2

y = r sin 0

tan 6+ = 27x

Z = Z

Z = Z

EXAMPLE 1 Converting from Cylindrical to Rectangular Coordinates Find the rectangular coordinates of the point P with cylindrical coordinates (r,0,z) = (2, I4`1- , 5). al FIGURE 2 P has cylindrical coordinates (r, 0,z).

Solution Converting to rectangular coordinates is straightforward (Figure 3): x = r cos

Cylindrical (2, ———

37r = 2 cos — = 2 (-4) 4

2

—

+, ,5)

Rectangular

.4, 5)

•

The z-coordinate is unchanged, so (x, y, z) = (—v.N/2, 5). Q=(-4,V2,0)

EXAMPLE 2 Converting from Rectangular to Cylindrical Coordinates Find cylindrical coordinates for the point with rectangular coordinates (x,y,z) -= (—IA —3,5). Solution We have r = .1x2 ± y2 = 1/(___3,/j)2 + (_3)2 = 6. The angle 0 satisfies tan 0 =

13 FIGURE 3

y

x

-----

1 —3 _ = --,, /3 —3,

7

0= —

Or

77r --

The correct choice is 0 = 2k, because the projection Q = (-3,A —3,0) lies in the third • quadrant (Figure 4). The cylindrical coordinates are (r, 9, z) = (6, 2i, 5) .

734

CHAPTER 13

VECTOR GEOMETRY

FIGURE 4 The projection Q lies in the third quadrant. Therefore, 0 = 4. Level Surfaces in Cylindrical Coordinates: r =R

Cylinder of radius R with the z-axis as axis of symmetry

0 = 00 Half-plane through the z-axis making an angle 00 with the xz-plane z=c

0 FIGURE 5 Level surfaces in cylindrical coordinates.

The level surfaces of a coordinate system are the surfaces obtained by setting one of the coordinates equal to a constant. In rectangular coordinates, the level surfaces are the planes x = xo, Y = Yo, and z = zo. In cylindrical coordinates, the level surfaces come in three types (Figure 5). The surface r = R is the cylinder of radius R consisting of all points located a distance R from the z-axis. The equation 0 = 00 defines the half-plane of all points that project onto the ray 0 = 90 in the (x, y)-plane. Finally, z = c is the horizontal plane at height c.

Horizontal plane at height c EXAMPLE 3 Equations in Cylindrical Coordinates Find an equation of the form z = f(r, 0) for the surfaces: (a) x 2 + y2 + z2 = 9, with z > 0 (b) x + y ± z = 1 Solution We use the formulas x2 ± y2 =

r2,

X = r COS 9 ,

y = r sin 9

(a) The equation x2 +y2 +z2 = 9 becomes r2 +z2 = 9, or z = -V9 — r2 (since z >0). This is the upper half of a sphere of radius 3. (b) The plane x±y+z= I becomes z=1—x—y=l—rcos0—rsin0

or

z = 1 — r(cos 0 + sin 0)

s

EXAMPLE 4 Graphing Equations in Cylindrical Coordinates Graph the surface corresponding to the equation in cylindrical coordinates given by z = r2. Y

FIGURE 6 The surface z = r2 is the paraboloid z = x2 ± y2 and can also be seen to be the result of rotating the curve z = y2 around the z-axis.

Solution We consider two straightforward ways of picturing the surface. First, convert to rectangular coordinates to obtain z = x2 + y2. The resulting surface is the paraboloid illustrated in Figure 6. Alternatively, note that since the equation of the surface does not depend on 0 we can graph its intersection with any plane containing the z-axis and rotate the resulting curve around the z-axis to obtain the surface. In the yz-plane, where x = 0, z = r2 corresponds with the parabola z -= y2. When we rotate this parabola around the z-axis, we obtain the circular paraboloid in Figure 6. •

Spherical Coordinates Spherical coordinates make use of the fact that a point P on a sphere of radius p is determined by two angular coordinates 0 and 0 (Figure 7): • 0 is the polar angle of the projection Q of P onto the xy-plane. • 0 is the angle of declination, which measures how much the ray through P declines from the vertical.

SECTION 13.7

Cylindrical and Spherical Coordinates

735

Spherical Coordinates: p = distance from origin 0 = polar angle in the xy-plane = angle of declination from the vertical In some textbooks, 0 is referred to as the azimuthal angle and 0 as the polar angle.

FIGURE 7 Spherical coordinates

1:3 FIGURE 8

(p,0,4).

• The symbol 0 (usually pronounced fee, but sometimes pronounced fie) is the 21st letter of the Greek alphabet. • We use p (written out as "rho" and pronounced row) for the radial coordinate, although r is also used to denote distance from the origin in other contexts.

Thus, P is determined by the triple (p,0,0), which are called spherical coordinates. Typically, we restrict the coordinates so that p > 0 and 0 < < Suppose that P = (x, y, z) in rectangular coordinates. Since p is the distance from P to the origin, p

=

1/ x2 + y2 + z 2

On the other hand, we see in Figure 8 that cos

tan 0 =

= -

The radial coordinate r of Q = (x, y, 0) is r = p sin 0, and therefore x = r cos° = p sin

cos 0,

y = r sin 0 =- p sin 0 sine,

z = p cos 0

Spherical to rectangular

Rectangular to spherical

x = p sin 0 cos 0

p=

y = p sin

2 ± y2 ± z2

tan0 = x z cos goo = -

sin 0

z = p cos 0

EXAMPLE 5 From Spherical to Rectangular Coordinates Find the rectangular coordinates of P = (p,0,0)= (3, 1,-), 7 and find the radial coordinate r of its projection Q onto the xy-plane. Solution By the formulas discussed, x = p sin 0 cos 0 = 3 sin 1' cos - 3 4 3 Jr y = p sin (/) sin 0 = 3 sin - sin -

4

3

=

Jr ,./1 3,./2 z = p cos (/) = 3 cos - = 3 — = 2 2 4

2 '‘/2

1 = 2

2 1 2 -

4 4

736

CHAPTER 13

VECTOR GEOMETRY

3N4/ , 0)

Now consider the projection Q = (x, y, 0) = dinate r of Q satisfies

(

2

r2

= x2 +

1 = ( 3

y2

2

9

3N/6 4 )

+

4

(Figure 9). The radial coor-

=

•

Therefore, r =

Find the spherical coordi-

EXAMPLE 6 From Rectangular to Spherical Coordinates nates of the point P = (x,y,z)= (2, —IA 3). FIGURE 9 Point with spherical coordinates

i,1).

Solution The radial coordinate is coordinate 8 satisfies tan 0 =

y

=

2

p

=

1 /22

±

(__2,/j)2

±

=

=

32

27r 3

=

25 = 5. The angular 57r or —

Since the point (x, y) = (2, —2,./j) lies in the fourth quadrant, the correct choice is 0.93. Therefore, P (Figure 10). Finally, cos 4) = = and so 0 = cos-1 = • 0.93). has spherical coordinates (5, Figure 11 shows the three types of level surfaces in spherical coordinates. Notice that if 0 0 0, i7 or 7r, then the level surface 0 = 00 is the right circular cone consisting of points P such that OP makes an angle 00 with the z-axis. There are three exceptional cases: 0 = defines the xy-plane, = 0 is the positive z-axis, and 0 = 7 is the negative z-axis. 13 FIGURE 10 Point with rectangular

coordinates (2, —IA 3).

p=R Sphere of radius R

= 00 Vertical half-plane

= 00 Right circular cone

FIGURE 11

EXAMPLE 7 Finding an Equation in Spherical Coordinates form p = f (0 , (/)) for the following surfaces: (a) x2 + y2 + z2 = 9

Find an equation of the

(b) z = x2 — y2

Solution (a) The equation x2 + y2 + z2 = 9 defines the sphere of radius 3 centered at the origin. Since p2 = x2 + y2 + z2, the equation in spherical coordinates is p = 3. (b) To convert z = x2 — y2 to spherical coordinates, we substitute the formulas for x, y, and z in terms of p, 8, and 0: x

p cos

= (p sin

Y

2

cos 0)2 — (p sin 4) sin 0)2

cos 0 = p sin2 0(cos2 0 — sin2 8)

(divide by p and factor)

cos 0 = p sin2 0 cos 28

(since c0s2 8 — sin2 8 = cos 20)

SECTION 13.7

Solving for p, we obtain p = 0 0 7r/4, 37/4, 57r/4, 77r/4.

Cylindrical and Spherical Coordinates

137

,cos , which is valid for 0 0 0,7r, and when sin' 0 cos 20 •

The angular coordinates (0,0) on a sphere of fixed radius are closely related to the longitude-latitude system used to identify points on the surface of the earth (Figure 12). By convention, in this system, we use degrees rather than radians. • A longitude is a half-circle stretching from the North to the South Pole (Figure 13). The axes are chosen so that 0 = 0 passes through Greenwich, England (this longitude is called the prime meridian). We designate the longitude by an angle between 0 and 1800 together with a label E or W, according to whether it lies to the east or west of the prime meridian. • The set of points on the sphere satisfying 0 = 0)0 is a horizontal circle called a latitude. We measure latitudes from the equator and use the label N or S to specify the Northern or Southern Hemisphere. Thus, in the upper hemisphere 0 < 00 < 90°, and a spherical coordinate 00 corresponds to the latitude (90° — 00) N. In the lower hemisphere 90° < yho < 180°, and 00 corresponds to the latitude (00 — 90°) S.

0,0)

Latitude 90 —

FIGURE 12 Longitude and latitude provide spherical coordinates on the surface of the

FIGURE 13 Latitude is measured from the equator and is labeled N (north) in the

earth.

upper hemisphere, and S (south) in the lower hemisphere.

EXAMPLE 8 Spherical Coordinates via Longitude and Latitude Find the angles (9,0) for Nairobi (1.17° S,36.48° E) and Ottawa (45.27° N, 75.42° W). Solution For Nairobi, 0 = 36.48° since the longitude lies to the east of Greenwich. Nairobi's latitude is south of the equator, so 1.17 = 00 — 90 and 4)0 = 91.17°. For Ottawa, we have 9 = 360 — 75.42 = 284.58° because 75.42° W refers to 75.42° in the negative 0 direction. Since the latitude of Ottawa is north of the equator, 45.27 = • 90 — 00 and 00 = 44.73°. EXAMPLE 9 Graphing Equations in Spherical Coordinates Graph the surface corresponding to the equation in spherical coordinates given by p = sec 0. Solution We could plug in values for 0, obtain the corresponding values for p, and then plot points, but it would be difficult to obtain an accurate representation of the surface in this way. Instead, notice that we can rewrite the equation: p=

1 cos 0

p cos0 = FIGURE 14

p = sec '.

This plane is the graph of

1

From our conversion equations, we see that this is z = 1. Hence, our surface is sim• ply the horizontal plane at height z = 1, as in Figure 14.

738

CHAPTER 13

VECTOR GEOMETRY

13.7 SUMMARY Cylindrical (r, 0, z) P Rectangular (x, y, z)

• Conversion from rectangular to cylindrical (r, 0, z) and spherical (p , , 0) coordinates (Figures 15 and 16): Rectangular to cylindrical

Q = (x, y, 0)

r = -1.x2

FIGURE 15 Cylindrical coordinates (r,0 , z).

y2

Rectangular to spherical p

=

A/ x2 + y2 ± z2

tan

=x , z cos dp = P

tan° = z=z

The angles are chosen so that 0 < 0 to.

We can visualize the limit of a vector-valued function as a vector r(t) moving toward the limit vector u (Figure 1). According to the next theorem, vector limits may be computed componentwise.

SECTION 14.2

Calculus of Vector-Valued Functions

751

THEOREM 1 Vector-Valued Limits Are Computed Componentwise A vector-valued function r(t) = (x(t), Y(t), z(t)) approaches a limit as t to if and only if each component approaches a limit, and in this case, lirn r(t) = firn x(t), lim y(t), urn z(t)) t --*to t->to t->to

1

Proof Let u = (a, b, c) and consider the square of the length 111'(t) - u112 = (x(t) -

± (At)

b)2

2

- 02

The term on the left approaches zero if and only if each term on the right approaches zero (because these terms are nonnegative). It follows that II r(t) - till approaches zero if and only if lx(t) - al, ly(t) - hi, and lz(t) - cl tend to zero. Therefore, r(t) approaches a limit u as t ->- to if and only if x(t), y(t), and z(t) converge to the components a, b and c, respectively. • EXAMPLE 1 Calculate lim r(t), where r(t) = (t2, 1- t, t-1). t—>3

The Limit Laws of scalar functions remain valid in the vector-valued case. They are verified by applying the Limit Laws to the components.

Solution By Theorem 1, lim r(t) = lim(t2, 1 — t t-1) -= t2, ilM(1 — t), limt-1 = t—.>3 t—>3 t—>3 t—>3 t—>3

9, -2, -I 3

•

Continuity of vector-valued functions is defined in the same way as in the scalar case. A vector-valued function r(t) -= (x(t), y(t), z(t)) is continuous at to if lim r(t) = r(to) t-* to By Theorem 1, r(t) is continuous at to if and only if the components x(t), y(t), z(t) are continuous at to. We define the derivative of r(t) as the limit of the difference quotient: (t) = — r(t) = lim h-÷o dt

r(t + h) - r(t)

3

In Leibniz notation, the derivative is written drldt. We say that r(t) is differentiable at t if the limit in Eq. (3) exists, and we say thet r is differentiable if it is differentiable at all tin its domain. Notice that the components of the difference quotient are themselves difference quotients: lim

r(t

h--+0

h) - r(t) = h

lim

(x(t + h) - x(t) y(t

h) - y(t) z(t

h) - z(t))

h—>0

and by Theorem 1, r(t) is differentiable if and only if the components are differentiable. In this case, 11(0 is equal to the vector of derivatives (x'(t), y'(t), z'(t)). THEOREM 2 Vector-Valued Derivatives Are Computed Componentwise A vectorvalued function r(t) -= (x(t), y(t), z(t)) is differentiable if and only if each component is differentiable. In this case, r'(t) = — r(t) = (x'(t), (t), (t)) dt By Theorems 1 and 2, vector-valued limits and derivatives are computed componentwise, so they are no more difficult to compute than ordinary limits and derivatives.

Here are some vector-valued derivatives, computed componentwise: d — (t2 t3, sin t) = (2t, 3t2, cos t), dt

_(cos t, —1, e2t) = (— sin t, 0, 2e2 )

CHAPTER 14

CALCULUS OF VECTOR-VALUED FUNCTIONS

Higher order derivatives are defined by repeated differentiation: r"(t) = — rt(t), dt

r"(t) = — r"(t), dt

...

EXAMPLE 2 Calculate r"(3), where r(t) = (ln t, t, t2). Solution We perform the differentiation componentwise: d r'(t) =— (1nt,t,t2) = (C I , 1,2t) dt d r"(t) = — „ , 1, 2t) = (-t -2, 0,2) dt Therefore, r"(3) = (-

0, 2).

•

The differentiation rules of single-variable calculus carry over to the vector setting. Differentiation Rules

Assume that r(t), ri(t), and r2(t) are differentiable. Then

• Sum Rule: (ri(t)± r2(t))' = rii(t)+ r'2 (t) • Constant Multiple Rule: For any constant c, (c r(t))' = c ri(t). • Scalar Product Rule: For any differentiable scalar-valued function f, -crdt (f (t)r(t)) = f'(t)r(t) + f (Or/ (t) • Chain Rule: For any differentiable scalar-valued function g, d —t r(g(t)) =

rt(g(t))gt(t)

Proof Each rule is proved by applying the single-variable differentiation rules to the components. For example, to prove the Scalar Product Rule (we consider vector-valued functions in the plane, to keep the notation simple), we write f (t)r(t) = f (t) (x(t), y(t)) = ( f (t)x(t), f (t)y(t)) Now apply the Product Rule to each component: dt

f (t)r(t) =

(-dri f (t)x(t), Tit f(t)y(t))

= (f t(t)x(t) + f (t)xt(t), f t(t)y(t) + f (t)y'(t)) (f' (t)x(t), .f i (t)Y(t)) + ( f (t)xt(t), f (t)yt(t)) = f'(0 (x(t), y(t)) + f (t)(xt(t), yt(t)) = f t(t)r(t) + f (Or'(t) The remaining proofs are left as exercises (Exercises 73-74). EXAMPLE 3 Let r(t) = (t2, 5t, 1) and f (t) = d (a) — f (t)r(t) dt

e3t .

Calculate:

d (b) — r(f(t)) dt

Solution We have ri(t) = (2t, 5,0) and f '(t) = 3e3`. (a) By the Scalar Product Rule, d — f (t)r(t) = f t(t)r(t) + f (t)rt(t) = 3e3" (t2,5t , 1) + e3 (2t, 5,0) dt = ((3t2 + 2t)e3r, (15t + 5)e3`,3e3')

•

Calculus of Vector-Valued Functions

SECTION 14.2

753

t e3r), and Note that we could have first found f (t)r(t) = e3t (1.2, 5t, 1) = (e3t t2 , e35t, then differentiated to obtain the same answer. In that case, we would have needed to use the single-variable product rule when differentiating the x- and y-components. (b) By the Chain Rule, —r(f (t)) = r'(f (0).00 = r'(e3t)3e3' = (2e3t, 5, 0)3e3t = (6e6t, 15e3t, 0) dt

•

In addition to the derivative rule for the product of a scalar function f and a vectorvalued function r stated earlier in this section, there are product rules for the dot and cross products. These rules are very important in applications, as we will see.

THEOREM 3 Product Rules for Dot and Cross Products are differentiable. Then d , —(ri(t) • r2(t)) =

Dot Product Rule: Cross Product Rule: CAUTION Order is important in the Cross Product Rule. The first term in Eq. (5) must be written as

(t) x r2(t) not r2(t) x r (t). Remember, cross product is not commutative. Similarly, the second term is ri(t) x r(t). Why is order not a concern for dot products?

Assume that ri(t) and r2(t)

(t) • r2(t)

ri(t) • 112(0

— (ri(t) x r2(t)) = (r(t) x r2(t)) dt

4

(ri(t) x 1)2(0) 5

We have seen three product rules involving vector-valued functions: the Scalar Product Rule, the Dot Product Rule, and the Cross Product Rule. Each has the same form as the single-variable product rule: The derivative of the first term "times" the second, plus the first term "times" the derivative of the second. The type of product referred to by "times" in each case is different—scalar multiplication in the first case, dot product in the second, cross product in the third. Proof We prove Eq. (4) for vector-valued functions in the plane. If ri(t) = (xi (t), Yf (0) and r2(t) = (x2(t), Y2(t)), then

r2(t))

d (xi(t)x2(t) + yi(t)Y2(t)) dt = x'1 (t)x2(t) + xi(t)x(t) +

(t)y r(t)

(t)y2(t)) + (xi (t),y(t) + (t)A(t))

= (x'1(t)x2(t)

= 1-'1 (0 • r2(t)

(0)72(0 +

ri(t) • 1)2(t)

The proof of Eq. (5) is left as an exercise (Exercise 75).

•

In the next example and throughout this chapter, all vector-valued functions are assumed differentiable, unless otherwise stated. d EXAMPLE 4 Prove that — dt

(t)) = r(t) x r"(t).

x

Solution By the Cross Product Rule, d — dt

x i(t)) = r'(t) x r'(t) + r(t) x r"(t) = r(t) x r"(t) Equals 0

Here, r'(t) x 1.'(t) = 0 because the cross product of a vector with itself is zero.

•

754

CHAPTER 14

CALCULUS OF VECTOR -VALUED FUNCTIONS

The Derivative as a Tangent Vector The derivative vector e(to) has an important geometric property: It points in the direction tangent to the path traced by r(t) at t = to. To understand why, consider the difference quotient, where Ar = r(to h) — r(t0) and At = h with h > 0: Ar

r(to + h) — r(to)

6

At —

The vector Ar points from the head of r(to) to the head of r(to h) as in Figure 2(A). The difference quotient Ar/At is a positive scalar multiple of Ar and therefore points in the same direction [Figure 2(B)].

C13 FIGURE 2 The difference quotient

points in the direction of Ar = r(to h)— r(to). If we think of r(t) as indicating the position of a particle moving along a curve, then 1.'(t) gives the rate of change of position with respect to time, which is the velocity of the particle. Since the velocity vector is tangent to the curve, it indicates the (instantaneous) direction of motion of the particle. In the next section we will see that it also indicates the particle's speed.

(A)

(B)

As h = At tends to zero, Ar also tends to zero but the quotient Ar/ At approaches a vector e(to) (assuming it exists), which, if nonzero, points in the direction tangent to the curve. Figure 3 illustrates the limiting process. We refer to r i(to) as the tangent vector or the velocity vector at r(to).

h tending to zero

Limit ash —)0

(B)

(C)

(A)

Cti FIGURE 3

rIto)

The difference quotient converges to a vector e(to), tangent to the curve.

The tangent vector ri(to) (if it exists and is nonzero) is a direction vector for the tangent line to the curve. The tangent line then is the line with vector parametrization: Tangent line at r(to):

L(t) = r(to)

t ri(t0)

7

EXAMPLE 5

Plotting Tangent Vectors CASJ Plot r(t) = (cos t, sin t, 4 cos2 t) together with its tangent vectors at t = and 4-r . Find a parametrization of the tangent line at t =

Solution The derivative is i(t) = (— sin t, cost, —8 cost sin t), and thus the tangent vectors at t = and are , r Ul-)

2

2

(-3+ r ) = (1,0,0)

SECTION 14.2

Calculus of Vector-Valued Functions

755

Figure 4 shows a plot of r(t) with e( ) based at r(i) and r' ( 4) based at r(f). At t = 4 r(-71-) 4 = (""L' 2 fl 2 " 2) and thus the tangent line is parametrized by 7r L(t) = r( 7 )-1-tr ( 73 =( 2 , 2 ,2) -Ft(

There are some important differences between vector- and scalar-valued derivatives. The tangent line to a plane curve y = f (x) is horizontal at xo exactly when f (x0) = 0. But in a vector parametrization of a plane curve, the tangent vector e(to) = (x'(to), Y'(to)) is horizontal and nonzero if y'(to) = 0 but x/(to) 0 0. In the case where a vector-valued function r(t) describes a particle moving along a curve, if ri(to) = 0, the particle has momentarily stopped at to. It could subsequently continue to move in the same direction, or move in any other direction from that point, including reversing itself and returning along the path upon which it arrived. We will see such instantaneous-stop behavior in the next example.

FIGURE 4 Tangent vectors to r(t) = (cos t, sin t, 4 cos2 t) at t -

2 , 2 , 4

i74 and 2 •

EXAMPLE 6 Tangent Vectors on the Cycloid Recall from Example 7 in Section 12.1 that a cycloid is a curve traced out by a point on the rim of a rolling wheel as the center of the wheel moves horizontally. We assume the point begins on the ground and the center of the wheel is moving to the right at a speed of 1. If the radius of the wheel is 1, then the resulting cycloid is traced out by r(t) = (t -

sin t, 1 - cos t) ,

fort >0

Find the points where: (a)

11(0 is horizontal and nonzero.

(b) r'(t) is the zero vector.

Solution The tangent vector is 11(0 = (1 - cos t, sin t). The y-component of ri(t) is zero if sin t = 0-that is, if t = 0, 7r, 27r, . . . . Therefore (see Figure 5), r'(7r) horizontal r'(27r)

0

X

47r FIGURE 5 Points on the cycloid where r' is 0 or horizontal.

• At r(0) = (0,0), we have r'(0) = (1 - cos 0, sin 0) = (0,0), so r' is the zero vector. • At r(7r) = (7r, 2), we have r/(7r) = (1 - cos 7r, sin 7r) = (2,0), so r' is horizontal and nonzero. By periodicity, we conclude that e(t) is nonzero and horizontal for t = 71", 3n-, 57, . . . [and therefore at (7r, 2), (3r,2), (57r, 2), . ] and ri(t) = 0 fort = 0, 27r, 47r, . . . [and therefore at (0,0), (27r, 0), (47r, 0),. 1. Note that, while the center of the wheel moves with a constant speed of 1, the point on the rim that traces out the cycloid has speed 2 at t = Tr, 37r, 57r, . . and has speed 0 at t = 0i, 2n-, 47r, . . . •

CONCEPTUAL INSIGHT The cycloid in Figure 5 has sharp points called cusps at points where x = 0, 27r, 47r, . . . . If we represent the cycloid as the graph of a function y = f (x), then f (x) does not exist at these points. By contrast, the vector derivative r i(t) = (1 - cost, sin t) exists for alit, but 1-'(t) = Oat the cusps. In general, e(t) is a direction vector for the tangent line whenever r'(t) exists and is nonzero. If 'J(t) = 0, then either the curve does not have a tangent line or the curve has a tangent line and r'(t) (being the zero vector) is not a direction vector for it. The next example establishes an important property of vector-valued functions that will be used in later sections in this chapter.

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EXAMPLE 7 Orthogonality of r and r' when r Has Constant Length Prove that if r(t) and ri(t) are nonzero and r(t) has constant length, then r(t) is orthogonal to r' (t).

1

4.4 REMINDER Ilr(t)II represents the length of the vector r(t).

Solution We prove this by considering -`,T 1. dr(t)112. On one hand, this derivative is equal to 0 because II r(t) is constant. On the other hand, by the Dot Product Rule, d d lIr(t)112 = — r(t)) = '1(0 • r(t) + r(t) • r'(t) = 2r'(t) • r(t) d —t dt (r(t) • It follows that ri(t) • r(t) = 0, and r(t) is orthogonal to 11(4

•

GRAPHICAL INSIGHT The result of Example 7 has a geometric explanation. A vector parametrization r(t) consisting of vectors of constant length R traces a curve on the surface of a sphere of radius R with its center at the origin (Figure 6). Thus, ri(t) is tangent to this sphere. But any line that is tangent to a sphere at a point P is orthogonal to the radial vector through P. and thus r(t) is orthogonal to ri(t).

Vector-Valued Integration FIGURE 6 r' (t) is orthogonal

has fixed length.

to r(t) if r(t)

The integral of a vector-valued function can be defined in terms of Riemann sums as in Chapter 5. We will define it more simply via componentwise integration (the two definitions are equivalent). In other words,

IL

r(t)dt =(f

x(t)dt, f a

y(t)dt, f a

z(t)dt) a

The integral exists if each of the components x(t), y(t), z(t) is integrable. For example, 7T 7r 1 t,sint) dt = f ldt, f tdt, f ir sintdt = 2 ' 2 0 0 0

f(1,

Vector-valued integrals obey the same linearity rules as scalar-valued integrals (see Exercise 76). An antiderivative of r(t) is a vector-valued function R(t) such that R'(t) = r(t). In the single-variable case, two functions fi and f2 with the same derivative differ by a constant. Similarly, two vector-valued functions with the same derivative differ by a constant vector (i.e., a vector that does not depend on t). This is proved by applying the scalar result to each component of r(t). THEOREM 4

If R1(t) and R2(t) are differentiable and lei (t) = R'2 (t), then R1(t) = R2(t) + c

for some constant vector c. The general antiderivative of r(t) is written

f

r(t)dt =R(t)±c

where c = (ct, c2, c3) is an arbitrary constant vector. For example, f(1,

t, sin t)

dt = (t,

1

,

cos t)

= (t

ci

1 , '

+ c2, — cost

EXAMPLE 8 Finding Position via Vector-Valued Differential Equations a particle satisfies 1 dr = (1 — 6 sin3t, —t) dt 5 Find the particle's location at t = 4 if r(0) = (4, 1).

c

The path of

SECTION 14.2

Calculus of Vector-Valued Functions

757

Solution The general solution is obtained by integration: 1 1 , r(t) = f (1 — 6 sin 3t, —t) dt = (t -I- 2 cos 3t, 5

c

From the general solution and from the initial condition, we have two expressions for r(0): r(0) = (2, 0) + c

3 (7.69, 2.6) t= 4 1

(4, 1) t=0 0

2

4

r(0) = (4, 1)

Therefore, we have c = (2, 1), and it follows that r(t) =

6

and

9

1 + 2 cos 3t, —t -, )± (2,1) = 10

1 , ± 1) ± 2 cos 3t + 2, —t` 10

The path is illustrated in Figure 7. The particle's position at t = 4 is FIGURE

7 Particle path (t ±

1 r(4) = (4 + 2 cos 12 ± 2, —(41 ± 1) •c,•,- (7.69,2.6) 10

2 cos 3t +2, -t2 + 1)

•

The Fundamental Theorem of Calculus from single-variable calculus naturally carries over to vector-valued functions: Fundamental Theorem of Calculus for Vector-Valued Functions Part I: If r(t) is continuous on [a, II], and R(t) is an antiderivative of r(t), then

I;

r(t) dt = R(b) — R(a)

Part II: Assume that r(t) is continuous on an open interval I and let a be in I. Then d t — f r(s)ds = r(t) dt a

14.2 SUMMARY • Limits, differentiation, and integration of vector-valued functions are performed componentwise. • Differentation rules: — Sum Rule: (ri(t)

r2(t))/ =

(t)

1.'2(0

— Constant Multiple Rule: (c r(t))' = c (t) d — Chain Rule: — r(g(t)) = g'(t)e(g(t)) dt • Product Rules: d — Scalar times vector: — f (t)r(t)) = f'(t)r(t) f (t)r'(t) dt d , — Dot product: — lri(t) • r2(t)) -= 111(0 • r2(t) ri(t) • e2(t) dt — Cross product: — (ri(t) x r2(t)) dt

(t) x r2(t))

(ri(t) x r'2(t))

• The derivative rl(to) is called the tangent vector or velocity vector. • If ri(to) is nonzero, then it points in the direction tangent to the curve at r(to). The tangent line at r(to) has vector parametrization L(t) = r(to)

re(to)

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• If (t) = le2(t), then R1(t) = R2(t) c for some constant vector c. • The Fundamental Theorem of Calculus for vector-valued functions: If r(t) is continuous, then — if le(t) = r(t), then f

r(t) dt = R(b) — R(a), a

— — d

ja

r(s) ds = r(t)

14.2 EXERCISES Preliminary Questions 1. State the three forms of the Product Rule for vector-valued functions. In Questions 2-6, indicate whether the statement is true or false, and if it is false, provide a correct statement. 2. The derivative of a vector-valued function is defined as the limit of a difference quotient, just as in the scalar-valued case. 3. The integral of a vector-valued function is obtained by integrating each component. 4. The terms "velocity vector" and "tangent vector" for a path r(t) mean the same thing.

5. The derivative of a vector-valued function is the slope of the tangent line, just as in the scalar case. 6. The derivative of the cross product is the cross product of the derivatives. 7. State whether the following derivatives of vector-valued functions ri(t) and r2(t) are scalars or vectors: d (a) — ri(t) dt

(b) — (ri (t) • r2(t)) dt

(c) — (ri(t) x r2(t)) dt

Exercises 18. Determine the values oft between 0 and 27r such that the tangent vector to the cycloid r(t) = (t — sin t ,1 — cos t) is a unit vector.

In Exercises 1-6, evaluate the limit. 1.

2. lim sin 2ti + cos tj + tan 4tk

(12,41, — 1)

-17

4. lim

3. lim e2t + ln(t + 1)j + 4k t—).0

5. Evaluate lim

r(t + h) — r(t)

In Exercises 19-22, evaluate the derivative by using the appropriate Product Rule, where

et —1 1 ,41) t +1 t

for r(t) = (t-1, sin t, 4).

ri(t) = (t2,t3,t),

6. Evaluate L.an

r(t) t

for r(t) = (sin t, 1 — cost, —2t).

In Exercises 7-12, compute the derivative.

-(ri(t) • r2(t)) dt

21.

-(ri(t) x r2(t)) dt

8. r(t) = (2t2,/

7. r(t) = (t, t2, t3) 9. r(s)

22. T,dt (1.(x) • 1.1(0 ,2 — t-2)

— s, ln(1 — s))

1)-1) 10. b(t) = (e3t-4,e6—t, (t +

= (e3t , e2t , et)

d 4 20. — (t 1%(t)) dt

19.

h —>CI

r2(t)

t=2

, assuming that

r(2) = (2, 1, 0) ,

(2)

(1, 4, 3)

In Exercises 23 and 24, let 11. c(t)

t —i i _ ezt k

12. a(61) = (cos 30)1+ (sin2 0)j + (tan 0)k 13. Calculate 11(0 and r"(t) for r(t) = (t, t2, t3). 14. Sketch the curve parametrized by r(t) = (1 — t2, t) for —1 0 but aT is positive or negative, depending on whether the object is speeding up or slowing down along the curve. EXAMPLE 6 For r(t) = (t2, 2t, ln t), determine the acceleration a(t). At t = decompose the acceleration vector into tangential and normal components, and find the curvature of the path (Figure 8). Solution First, we compute the tangential components T and ar. We have v(t) = 1-'(t) = (2t, 2, CI ),

a(t) = r"(t) = (2,0, -t-2)

At t =

2

a = r /,

1

1

1

v=r' (-1)

2

2 (i)_

2 )

= (2o. —

= (2,0,-4)

el

FIGURE 8 The vectors T, N, and a at t = on the curve given by r(t) = (t2, 2t, In t).

Thus, T

= livli =

(1, 2, 2) /1 2 2 \ 12 + 22 + 22 = \j'

and by Eq. (6), aT = a • T = (2,0, —4)

= _2

Next, we use Eq. (7):

Summary of steps in Example 6:

This vector has length 64 16 aN = IlaNNII = iv +

T= — iivii aT = N=

a•T

N

— aN N aN

=4

and thus,

a — aTT

aN = liaN NII

64

N

aNN = aN

1,-1) /2 4

1 \3 3

2\ 3/

Finally, we obtain the decomposition

a = (2,0, —4) = aTT

aNN = —2T + 4N

Now, since aN = 4 at t = 1, and we know aN = K v2 from Eq. (5), to obtain the curvature at t = divide 4 by the square of the speed. With v = (1, 2, 2) at t = , we have v2 -= 9, and therefore K(1/2) = 4/9. • EXAMPLE 7 Nonuniform Circular Motion Figure 9 shows the acceleration vectors of three particles moving counterclockwise around a circle. In each case, state whether the particle's speed v around the circle is increasing, decreasing, or momentarily constant.

SECT 10 N 14.5

FIGURE 9 Acceleration vectors of particles moving counterclockwise (in the direction of T) around a circle.

(A)

Motion in 3-Space

783

(C)

(B)

Solution The rate of change of speed depends on the angle 8 between a and T: v' =

= a • T = hail IITII cos 0 = Hail cos

Here, the first equality follows from Eq. (5), the second from Eq. (6), the third from the geometric interpretation of the dot product, and the last since T is a unit vector. • In (A), 8 is obtuse, so cos 8 < 0 and v' 0. The particle's speed is increasing. •

14.5 SUMMARY • For an object whose path is described by a vector-valued function r(t), v(t) = r'(t),

v(t) = Ilv(t)II,

a(t) = r"(t)

• The velocity vector v(t) points in the direction of motion. Its length v(t) = II v(011 is the object's speed. • The acceleration vector a is the sum of a tangential component (reflecting change in speed along the path) and a normal component (reflecting change in direction): a(t) = aT(t)T(t)+ aN(t)N(t) Unit tangent vector

T(t) =

Unit normal vector

N(t) =

Tangential component

v(t) II v(011 T'(t)

11r(t)II

= vf(t) = a • T = a•v arT = (—) v v•v

Normal component

a•v

hlvil

= K(t)v(t)2 = .111a112 — laTI2 a•v aN N = a— aTT = a — (—) v v•v

14.5 EXERCISES Preliminary Questions 1. If a particle travels with constant speed, must its acceleration vector be zero? Explain.

2. For a particle in uniform circular motion around a circle, which of the vectors v(t) or a(t) always points toward the center of the circle?

3. Two objects travel to the right along the parabola y = x2 with nonzero speed. Which of the following statements must be true? (a) Their velocity vectors point in the same direction. (b) Their velocity vectors have the same length. (c) Their acceleration vectors point in the same direction.

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CALCULUS OF VECTOR -VALUED FUNCTIONS

CHAPTER 14

4. Use the decomposition of acceleration into tangential and normal components to explain the following statement: If the speed is constant, then the acceleration and velocity vectors are orthogonal. 5. If a particle travels along a straight line, then the acceleration and velocity vectors are (choose the correct description): (b) parallel. (a) orthogonal.

6. What is the length of the acceleration vector of a particle traveling around a circle of radius 2 cm with constant speed 4 cm/s? 7. Two cars are racing around a circular track. If, at a certain moment, both of their speedometers read 110 mph, then the two cars have the same (choose one): (a) a'r

(b) aN

Exercises 1. Use the table to calculate the difference quotients

r(1 + h) — r(1)

for

h = —0.2, —0.1, 0.1,0.2. Then estimate the velocity and speed at t = 1.

In Exercises 15-18, find r(t) and v(t) given a(t) and the initial velocity and position. 15. a(t) = (t,4),

(1.557,2.459, —1.970) (1.559,2.634, —1.740) (1.540,2.841, —1.443) (1.499,3.078, —1.035) (1.435,3.342, —0.428)

r(0.8) r(0.9) r(1) r(1.1) r(1.2)

16. a(t) = (el, 2t, t + 1), 17. a(t) = tk,

v(0) = 8,

18. a(t) = cos tk, r(2 + h ) — r(2)

2. Draw the vectors r(2 + h) — r(2) and

v(0) = (3,-2),

for h =- 0.5 for

the path in Figure 10. Draw v(2) (using a rough estimate for its length).

r(0) = (0,0)

v(0) = (1, 0, 1),

r(0) = (2, 1, 1)

r(0) =j

v(0) = i — j,

r(0) = i

19. A projectile is launched from the ground at an angle of 45°. What initial speed must the projectile have in order to hit the top of a 120-m tower located 180 m away? 20. Find the initial velocity vector vo of a projectile released with initial speed 100 m/s that reaches a maximum height of 300 m. 21. Assume that astronaut Alan Shepard hit his golf shot on the moon (acceleration due to gravity = 1.6 m/s2) with a modest initial speed of 35 m/s at an angle of 30°. How far did the ball travel?

r(2)

0

22. Golfer Judy Robinson hit a golf ball on the planet Priplanus with an initial speed of 50 m/s at an angle of 40°. It landed exactly 2 km away. What is the acceleration due to gravity on Priplanus?

FIGURE 10 In Exercises 3-6, calculate the velocity and acceleration vectors and the speed at the time indicated. 3. r(t) = (t3, 1 — t,4t2),

t=1

5. 1.(9) = (sin 0, cos 8, cos 30), 6. r(s) -=

+1 s2

±S s2

4. r(t) = er j — cos(2t)k,

t=0

9=

23. Show that a projectile launched at an angle 9 with initial speed vo travels a distance (nil g) sin 20 before hitting the ground. Conclude that the maximum distance (for a given vo) is attained at 9 = . 24. Show that a projectile launched at an angle 9 will hit the top of an h-meter tower located d meters away if its initial speed is

, s=2 77.d sec 0

7. Find a(t) for a particle moving around a circle of radius 8 cm at a constant speed of v = 4 cm/s (see Example 4). Draw the path, and on it, draw the acceleration vector at t = 8. Sketch the path r(t) = (1 — t2, 1 — t) for —2 < t < 2, indicating the direction of motion. Draw the velocity and acceleration vectors at t =- 0 and t = 1. 9. Sketch the path r(t) = (t2, t3) together with the velocity and acceleration vectors at t = 1. 10. The paths r(t) = (t2, t3) and 1.1(0 = (et, t6) trace the same curve, and ri (1) = r(1). Do you expect either the velocity vectors or the acceleration vectors of these paths at t = 1 to point in the same direction? Explain. Compute these vectors and draw them on a single plot of the curve. In Exercises 11-14, find v(t) given a(t) and the initial velocity. 11. a(t) = (t,4),

v(0) =

12. a(t) = (e1,0,t + 1), 13. a(t) = k,

v(0) = i

—2) v(0) --= (1, —3, ,%/2) 14. a(t) = t2k,

v(0) = i — j

VO=

tant9 — h

25. A quarterback throws a football while standing at the very center of the field on the 50-yard line. The ball leaves his hand at a height of 5 ft and has initial velocity vo = 40i + 35j + 32k ft/s. Assume an acceleration of 32 ft/s2 due to gravity and that the i vector points down the field toward the endzone and the j vector points to the sideline. The field is 150 ft in width and 300 ft in length. (a) Determine the position function that gives the position of the ball t seconds after it is thrown. (b) The ball is caught by a player 5 ft above the ground. Is the player in bounds or out of bounds when he receives the ball? Assume the player is standing vertically with both toes on the ground at the time of reception. 26. A soccer ball is kicked from ground level with (x, y)-coordinates (85, 20) on the soccer field shown in Figure 11 and with an initial velocity vo = 10i — 5j + 25k ft/s. Assume an acceleration of 32 ft/s2 due to gravity and that the goal net has a height of 8 ft and a total width of 24 ft. (a) Determine the position function that gives the position of the ball t seconds after it is hit. (b) Does the ball go in the goal before hitting the ground? Explain why or why not.

SECTION 14.5

a(t) =

_

12 , _ 12

10 5 ft 165 11

16511

785

44. Let r(t) = (t2, 4t —3). Find T(t) and N(t), and show that the decomposition of a(t) into tangential and normal components is

,.

10 5 ft

Motion in 3-Space

G t2+ 2t 4

T

(

4

)

N/t2 + 4

45. Find the components alr and aN of the acceleration vector of a particle moving along a circular path of radius R = 100 cm with constant speed v0 = 5 cm/s. 46. In the notation of Example 5, find the acceleration vector for a person seated in a car at (a) the highest point of the Ferris wheel and (b) the two points level with the center of the wheel. 47. Suppose that the Ferris wheel in Example 5 is rotating clockwise and that the point P at angle 45° has acceleration vector a = (0, —50) m/min2 pointing down, as in Figure 12. Determine the speed and tangential component of the acceleration of the Ferris wheel.

FIGURE 11

27. A constant force F = (5,2) (in newtons) acts on a 10-kg mass. Find the position of the mass at t = 10 seconds if it is located at the origin at t = 0 and has initial velocity vo = (2, —3) (in meters per second). 28. A force F = (24t, 16 — 8t) (in newtons) acts on a 4-kg mass. Find the position of the mass at t = 3 s if it is located at (10, 12) at t = 0 and has zero initial velocity. 29. A particle follows a path r(t) for 0 < t

T, beginning at the origin 0.

1f

if (t) dt is called the average velocity vector. SupThe vector V = — T 0 pose that V = 0. Answer and explain the following: (a) Where is the particle located at time T if V = 0? (b) Is the particle's average speed necessarily equal to zero? 30. At a certain moment, a moving particle has velocity v = (2,2, —1) and acceleration a = (0,4,3). Find T, N, and the decomposition of a into tangential and normal components. 31. At a certain moment, a particle moving along a path has velocity v = (12, 20, 20) and acceleration a = (2,1, —3). Is the particle speeding up or slowing down? In Exercises 32-35, use Eq. (6) to find the coefficients aT and aN as a function oft (or at the specified value oft).

FIGURE 12 48. At time to, a moving particle has velocity vector v = 2i and acceleration vector a = 38 + 18k. Determine the curvature K(to) of the particle's path at time to. 49. A satellite orbits the earth at an altitude 400 km above the earth's surface, with constant speed v = 28,000 km/h. Find the magnitude of the satellite's acceleration (in kilometers per square hour), assuming that the radius of the earth is 6378 km (Figure 13).

33. r(t) -= (t, cos t, sin t)

32. r(t) = (t2, t3) t =1

34. r(t) = (t —1 , In t, t2),

t=0

35. r(t) = (e2t , t,

In Exercises 36-43, find the decomposition of a(t) into tangential and normal components at the point indicated, as in Example 6. 36. r(t) = (es, 1 — t),

t =0

37. r(t) = (1t3, 1 — 3t), 38. r(t) = (t 39. r(t) = (t,

t2, t3),

t = —I t =

P 3),

t =4

40. r(t) = (4 — t, t + 1, t2), 41. r(t) = (t, ei ,t ef ),

1

t=2

t =0

42. r(B) -= (cos 0, sin 8, (9),

9=0

43. r(t) = (t, cos t, t sin t),

t=

FIGURE 13 Satellite orbit. 50. A car proceeds along a circular path of radius R = 300 m centered at the origin. Starting at rest, its speed increases at a rate of t m/s2. Find the acceleration vector a at time t = 3 s and determine its decomposition into normal and tangential components. 51. A particle follows a path ri (t) on the helical curve with parametrization r(8) = (cos 0, sin 0,0). When it is at position r( ), its speed is 3 m/s and its speed is increasing at a rate of m/s2. Find its acceleration vector a at this moment. Note: The particle's acceleration vector does not coincide with r"(0).

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14

CALCULUS OF VECTOR -VALUED FUNCTIONS

E/J

00

Explain why the vector w in Figure 14 cannot be the accelera52. tion vector of a particle moving along the circle. Hint: Consider the sign of w • N.

(A)

(B)

(C)

FIGURE 15 FIGURE 14

54. Prove that aN =

53.121 Figure 15 shows the acceleration vectors of a particle moving clockwise around a circle. In each case, state whether the particle is speeding up, slowing down, or momentarily at constant speed. Explain.

x v11 11v11

•

55. Suppose that r(t) lies on a sphere of radius R for all t. Let J = r x r'. Show that r' = (J x r)/11r112. Hint: Observe that r and r' are perpendicular.

Further Insights and Challenges

174

56. The orbit of a planet is an ellipse with the sun at one focus. The sun's gravitational force acts along the radial line from the planet to the sun (the dashed lines in Figure 16), and by Newton's Second Law, the acceleration vector points in the same direction. Assuming that the orbit has positive eccentricity (the orbit is not a circle), explain why the planet must slow down in the upper half of the orbit (as it moves away from the sun) and speed up in the lower half. Kepler's Second Law, discussed in the next section, is a precise version of this qualitative conclusion. Hint: Consider the decomposition of a into normal and tangential components. planetary motion

Note that braking (v' 0, the level curve is an ellipse. • For c = 0, the level curve is just the point (0,0) because x2 + 3y2 = 0 only for (x, y) = (0,0). • There is no level curve for c < 0 because f (x, y) is never negative.

FIGURE 11 f(x, y) =- x2 ± 3y2. Contour interval m = 10.

The graph of f (x, y) is an elliptic paraboloid (Figure 11). As we move away from the origin, f (x, y) increases more rapidly. The graph gets steeper, and the level curves • become closer together.

800

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES

EXAMPLE 6 The hyperbolic paraboloid in Figure 12 is often called a "saddle" or "saddle-shaped surface."

Hyperbolic Paraboloid

ad REMINDER

Sketch the contour map of

g(x, y) = x2

3y2

Solution The level curves have equation g(x, y) = c, or x2

3y2 = c

• For c 0, the level curve is the hyperbola x2 — 3y2 = c. • For c = 0, the level curve consists of the two lines x = ±Nijy because the equation g(x, y) = 0 factors as follows: x2

—

3y2

=

(x — ,Vjy)(x

•Ij'y) = 0

The graph of g(x, y) is a hyperbolic paraboloid (Figure 12). When you stand at the origin, g(x, y) increases as you move along the x-axis in either direction and decreases as you move along the y-axis in either direction. Furthermore, the graph gets steeper as you • move out from the origin, so the level curves grow closer together. EXAMPLE 7

Contour Map of a Linear Function

Sketch the graph of

f (x , y) = 12 — 2x -- 3y and the associated contour map with contour interval m = 4.

g(x, y) decreasing

c = 30 g (x, y) increasing

x g(x, y) increasing

c = —30 g (x, y) Y decreasing

FIGURE 12 g(x, y) = x2 — 3y2. Contour interval m = 10.

Solution Note that if we set z = f (x, y), we can write the equation as 2x + 3y + z= 12. As we discussed in Section 13.5, this is the equation of a plane. To plot the graph, we find the intercepts of the plane with the axes (Figure 13). The graph intercepts the z-axis at z = f(0, 0) = 12. To find the x-intercept, we set y = z = 0 to obtain 12 — 2x — 3(0) = 0, or x = 6. Similarly, solving 12 — 3y = 0 gives the y-intercept y = 4. The graph is the plane determined by the three intercepts. In general, the level curves of a linear function f (x, y) = qx ry s are the lines with equation qx ± Ty ± s = c. Therefore, the contour map of a linear function consists of equally spaced parallel lines. In our case, the level curves are the lines • 12 — 2x — 3y = c, or 2x ± 3y = 12— c (Figure 13). How can we measure steepness of the graph of a function quantitatively? Let's imagine the surface given by z = f (x, y) as a mountain (Figure 14). We place the xy-plane at sea level, so that f (a, b) is the height (also called altitude or elevation) of the mountain above sea level at the point (a, b) in the plane. Figure 14(A) shows two points P and Q in the xy-plane, together with the points and Q on the graph that lie above them. We define the average rate of change:

average rate of change from P to Q =

where A altitude = change in the height from from P to Q.

F and

A altitude A horizontal

6, and A horizontal = distance

CONCEPTUAL INSIGHT We will discuss the idea that rates of change depend on direction when we come to directional derivatives in Section 15.5. In single-variable calculus, we measure the rate of change by the derivative f l(a). In the multivariable case, there is no single rate of change because the change in f (x , y) depends on the direction: The rate is zero along a level curve [because f (x, y) is constant along level curves], and the rate is nonzero in directions pointing from one level curve to the next [Figure 14(B)].

SECTION 15.1

Functions of Two or More Variables

801

c = 20 c=16 c = 12 c=8 c=4 - c =0 2100

4 6

c = 20 c=16 c=12 c=8 c=4 c=0 c = —4

Function does not change along the level curve Contour interval: 0.8 km Horizontal scale: 2 km

(Interval m = 4) (A)

FIGURE 13 Graph and contour map of f (x, y) = 12— 2x — 3y.

A contour map is like a topographic map that hikers would use to help understand the terrain that they encounter. They are both two-dimensional representations of the features of three-dimensional structures.

A A

200

B 400

Contour interval: 100 m Horizontal scale: 200 m (B)

FIGURE 14

EXAMPLE 8 Average Rate of Change Depends on Direction Compute the average rate of change from A to the points B, C, and D in Figure 14(B). Solution The contour interval in Figure 14(B) ism = 100 m. Segments AB and AC both span two level curves, so the change in altitude is 200 m in both cases. The horizontal scale shows that AB corresponds to a horizontal change of 200 m, and AC corresponds to a horizontal change of 400 m. On the other hand, there is no change in altitude from A to D. Therefore,

average rate of change from A to B =

A altitude 200 = = 1.0 A horizontal 200

average rate of change from A to C =

A altitude 200 = — = 0.5 A horizontal 400

average rate of change from A to D =

A altitude = 0 A horizontal

We see here explicitly that the average rate varies according to the direction.

A path of steepest descent is the same as a path of steepest ascent traversed in the opposite direction. Water flowing down a mountain approximately follows a path of steepest descent.

2"

•

When we walk up a mountain, the incline at each moment depends on the path we choose. If we walk around the mountain, our altitude does not change at all. On the other hand, at each point there is a steepest direction in which the altitude increases most rapidly. On a contour map, the steepest direction is approximately the direction that takes us to the closest point on the next highest level curve [Figure 15(A)]. We say "approximately" because the terrain may vary between level curves. A path of steepest ascent is a path that begins at a point P and, everywhere along the way, points in the steepest direction. We can approximate the path of steepest ascent by drawing a sequence of

802

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES

15(B) segments that move as directly as possible from one level curve to the next. Figure dashed the but ascent, steepest of path a path is solid shows two paths from P to Q. The shortest path is not, because it does not move from one level curve to the next along the possible segment.

Approximate path of steepest ascent starting at P

Not a path of steepest ascent (B)

(A) Vectors pointing approximately in the direction of steepest ascent FIGURE 15

More Than Two Variables There are many modeling situations where it is necessary to use a function of more than two variables. For instance, we might want to keep track of temperature at the various points in a room using a function T(x, y, z) that depends on the three variables corresponding to the coordinates of each point. In making quantitative models of the economy, functions often depend on more than 100 variables. Unfortunately, it is not possible to draw the graph of a function of more than two variables. The graph of a function f(x, y,z) would consist of the set of points (x, y, z, f(x, y, z)) in four-dimensional space R4. However, just as we can use contour maps to visualize a three-dimensional mountain using curves on a two-dimensional plane, it is possible to draw the level surfaces of a function of three variables f (x, y, z). These are the surfaces with equation f (x, y, z) = c for different values of c. For example, the level surfaces of f(x, y, z) = x2 ± y2 + z2 c (Figure 16). In the case of a function are the spheres with equation x2 ± T(x, y, z) that represents temperature of points in space, we call the level surfaces corresponding to T(x, y, z) = k the isotherms. These are the collections of points, all of which have the same temperature k. For functions of four or more variables, we can no longer visualize the graph or the level surfaces. We must rely on intuition developed through the study of functions of two and three variables. y2

FIGURE 16 The level surfaces of f (x,y,z) = x2 + y2 ± z2 are spheres.

±

Z2

=

EXAMPLE 9 Describe the level surfaces of g(x, y, z) .= x2 + y2 _ z2. Solution The level surface for c = 0 is the cone x2 -I- y2 — z2 = 0. For c 0 0, the level surfaces are the hyperboloids x2 + y2 _ z2 = c. The hyperboloid has one sheet if c > 0 and it lies outside the cone. The hyperboloid has two sheets if c 0) FIGURE

g(x, y, z) = 0

Functions of Two or More Variables

803

g(x, y, z)= c (c < 0)

17 Level surfaces of g(x, y, z) = x2 + y2 — z2.

15.1 SUMMARY • The domain D of a function f(xi, xn ) of n variables is the set of n-tuples (ai, an) in Rn for which f (ai, . • • , an) is defined. The range of f is the set of values taken on by f. • The graph of a continuous real-valued function f (x, y) is the surface in R3 consisting of the points (a, b, f (a, b)) for (a, b) in the domain D of f. • A vertical trace is a curve obtained by intersecting the graph with a vertical plane x = a or y = b. • A level curve is a curve in the xy-plane defined by an equation f(x, y) = c. The level curve f (x, y) -= c is the projection onto the xy-plane of the horizontal trace curve, obtained by intersecting the graph with the horizontal plane z = c. • A contour map shows the level curves f (x , y) =- c for equally spaced values of c. The spacing m is called the contour interval. • When reading a contour map, keep in mind: — Your altitude does not change when you hike along a level curve. — Your altitude increases or decreases by m (the contour interval) when you hike from one level curve to the next. • The spacing of the level curves indicates steepness: They are closer together where the graph is steeper. • The average rate of change from P to Q is the ratio

altitude

Ahorizontal • A direction of steepest ascent at a point P is a direction along which f (x, y) increases most rapidly. The steepest direction is obtained (approximately) by drawing the segment from P to the nearest point on the next level curve. • Level surfaces can be used to understand a function f (x, y, z). In the case where the function represents temperature, we call the level surfaces isotherms.

15.1 EXERCISES Preliminary Questions 1. What is the difference between a horizontal trace and a level curve?

4.

Describe the contour map of f (x, y) = x with contour interval I.

How are they related?

5.

How will the contour maps of

2.

Describe the trace of f (x, y) = x2 — sin(x3y) in the xz-plane.

3. Is it possible for two different level curves of a function to intersect? Explain.

f (x, y) = x

and

with contour interval 1 look different?

g(x, y) = 2x

804

DIFFERENTIATION IN SEVERAL VARIABLES

CHAPTER 15

Exercises In Exercises 1-6, at each point evaluate the function or indicate that the function is undefined there. 1.

f (x, y) = x + yx3,

(1, 2), (-1, 6), (e, tr)

2.

g(x, y) = x2 _

(1,3), (3, —3), (,/, 2)

3.

h(x, y) =

4.

k(x,y) = xe—Y ,

5.

h(x,y,z)

6.

w(r, s,t) =

y2

'

_ v2, x—y

—S

FIGURE 19

(3, 7, —2), (3,2, ), (4, —4, 0) (7,7r, tt), ( — 2, 2, lk,"

,

In Exercises 7-14, sketch the domain of the function. 7.

f (x, y) = N/81 — x2

f (x , y) = 12x — 5y

8.

f (x, y) = ln(4x2 — y)

1 10. h(x,t) = x+t

21. Match the functions (a)—(f) with their graphs (A)—(F) in Figure 20.

(a) f (x, y) = lx1 + ly1 (b) f (x, y) = cos(x — y) —1 (c) f (x, y) = 1 + 92 y2 (d)

9.

11. g(Y, Z) =

(B)

(A)

(1,0), (3, —3), (0,12)

xyz —2,

sin t

(20,2), (1,-2), (1, 1)

1

f (x, y) = cos(y2)e0

2

2)

—1 1 + 9x2 9y2 2±y2 ) (f) f (x, y) = cos(x2 y2 )e —o.1(x (e) f (x, y) —

12. f (x, y) = sin

z + y2

13. F(1, R) =

14. f (x, y) = c0s-1(x + y)

In Exercises 15-18, describe the domain and range of the function. 15. f (x, y, z) = xz

eY

16. f (x, y, z) =

17. P(r ,s,t) = \/16 _ r2 52t2

18. g(r,$) = cos—I (rs)

19. Match graphs (A) and (B) in Figure 18 with the functions: (i) f (x, y) = —x + y2

(A)

(B)

(ii) g(x, y) = x + y2

(C)

(A)

(B) FIGURE 18

20. Match each of graphs (A) and (B) in Figure 19 with one of the following functions: (i) f (x, y) = (cos x)(cos y) g(x, y) = cos(x2 + y2)

(E)

(F) FIGURE 20

SECTION 15.1

Functions of Two or More Variables

805

22. Match the functions (a)-(d) with their contour maps (A)-(D) in Figure 21. (a) f (x, y) = 3x + 4y

(b) g(x, y) = x3 - y

(c) h(x, y) = 4x - 3y

(d) k(x, y) = x2 - y

10

FIGURE 22 Contour map with contour interval m = 6. 40. Use the contour map in Figure 23 to calculate the average rate of change: (a) from A to B. (b) from A to C. (A)

-10-1' 1 -10 -5

0

5

10

-10 -5 (D)

(C)

FIGURE 23

FIGURE 21

Exercises 41-43 refer to the map in Figure 24.

In Exercises 23-28, sketch the graph and draw several vertical and horizontal traces. 23. f (x, y) = 12 - 3x - 4y

24. f (x, y) = \/4 - x2 - y2

25. f (x, y) = x2 + 4y2

26. f (x, y) = y2

27. f (x, y) = sin(x - y)

1 28. f (x, y) = x2 ±y2 + 1

41. (a) At which of A-C is pressure increasing in the northern direction? (b) At which of A-C is pressure increasing in the westerly direction? 42. For each of A-C indicate in which of the four cardinal directions, N, S. E, or W, pressure is increasing the greatest. 43. Rank the following states in order from greatest change in pressure across the state to least: Arkansas, Colorado, North Dakota, Wisconsin.

29. Sketch contour maps of f (x, y) = x + y with contour intervals m = 1 and 2. 30. Sketch the contour map of f (x,y) = x2 + y2 with level curves c = 0, 4, 8, 12, 16. In Exercises 31-38, draw a contour map of f (x,y) with an appropriate contour interval, showing at least six level curves. 31. f (x, y) = x2 - y

32. f (x, y) =

33. f (x, y) =

34. f (x, y) = xy

35. f (x,y) = x2 + 4y2

36. f (x, y) = x + 2y - 1

37. f (x,y) = x2

38. f (x,Y)

=

3x2

FIGURE 24 Atmospheric pressure (in millibars) over North America on March 26, 2009.

y2

Find the linear function whose contour map (with contour in39. terval m = 6) is shown in Figure 22. What is the linear function if m = 3 (and the curve labeled c = 6 is relabeled c = 3)?

In Exercises 44-47, let T (x, y, z) denote temperature at each point in space. Draw level surfaces (also called isotherms) corresponding to the fixed temperatures given. 44. T(x,y,z) = 2x + 3y - z,T = 0,1,2 45. T(x, y, z) = x - y + 2z, T = 0, 1, 2

806

DIFFERENTIATION IN SEVERAL VARIABLES

CHAPTER 15

46. T(x, y, z) =x2 +

y2

z,T = 0, 1, 2

In Exercises 52-55, refer to Figure 26.

47. T(x, y, z) = x2 — y2 + z2, T = 0,1,2, —1,-2

52. Find the change in seawater density from A to B.

In Exercises 48-51, p(S,T) is seawater density (kilograms per cubic meter) as a function of salinity S (parts per thousand) and temperature T (degrees Celsius). Refer to the contour map in Figure 25.

53. Estimate the average rate of change from A to B and from A to C.

48. Calculate the average rate of change of p with respect to T from B to A.

54. Estimate the average rate of change from A to points i, ii, and iii. 55. Sketch the path of steepest ascent beginning at D.

49. Calculate the average rate of change of p with respect to S from B to C. 50. At a fixed level of salinity, is seawater density an increasing or a decreasing function of temperature? 51. Does water density appear to be more sensitive to a change in temperature at point A or point B?

Temperature T (°C)

25 20

Contour interval =20 m

1 FIGURE 26 10 56. Let temperature in 3-space be given by T(x,y,z) = x2 + y2 — z. Draw isotherms corresponding to temperatures T = —2, —1, 0, 1, 2.

5 0 31.5

57. Let temperature in 3-space be given by T(x,y,z)= Draw isotherms corresponding to temperatures T = 0, 1,2.

7" 77B 32.0

32.5 33.0 33.5 Salinity (ppt)

34.0

34.5

FIGURE 25 Contour map of seawater density p(S,T) (kilograms per cubic meter).

+

58. Let temperature in 3-space be given by T(x,y,z) = x 2 Draw isotherms corresponding to temperatures T = —1,0, 1.

+ z2.

_ y2 _ z.

59. Let temperature in 3-space be given by T(x,y,z) = x 2 _ y2 Draw isotherms corresponding to temperatures T = —2, —1,0,1,2.

_ z2.

Further Insights and Challenges 60. E4 The function f(x,t)= , whose graph is shown in Figure 27, models the temperature along a metal bar after an intense burst of heat is applied at its center point.

Temperature

T

(a) Sketch the vertical traces at times t = 1, 2, 3. What do these traces tell us about the way heat diffuses through the bar?

Time t

(b) Sketch the vertical traces x = c for c = ±0.2, ±0.4. Describe how temperature varies in time at points near the center. 2

61. Let

f(x,y)=

Vx2 ± y2

for (x, y)

(0, 0)

Write f as a function f (r ,0) in polar coordinates, and use this to find the level curves of f.

Metal bar

FIGURE 27 Graph of f(x,t)=1.-112e—x2 / r beginning shortly after t =0.

15.2 Limits and Continuity in Several Variables This section develops limits and continuity in the multivariable setting. We focus on functions of two variables, but similar definitions and results apply to functions of three or more variables.

S EC TION 15.2

Limits and Continuity in Several Variables

807

Recall that on the real number line, a number x is close to a if the distance lx — al is small. In the plane, a point (x, y) is close to another point P = (a, b) if the distance D*(P, r) excludes P

P (x, y) Open disk D(P, r)

The open disk D(P,r) consists of points (x, y) at distance < r from P. It does not include the boundary circle.

FIGURE 1

d((x, y), (a, b)) = (x — a)2 (y — b)2 between them is small. Note that if we take all the points that are a distance of less than r from P = (a, b), as in Figure 1, this is a disk D(P,r) centered at P that does not include its boundary. If we insist also that d((x, y), (a, b)) 0, then we get a punctured disk that does not include P and that we denote D*(P,r). Now assume that f (x, y) is defined near P but not necessarily at P itself. In other words, f (x, y) is defined for all (x, y) in some punctured disk D*(P,r) with r > 0. We say that f (x, y) approaches the limit L as (x, y) approaches P = (a, b) if If (x, y) — LI becomes arbitrarily small for (x, y) sufficiently close to P = (a, b) [Figure 2(A)]. In this case, we write firn

(x,y)-÷ P

f (x, y) =

lim

(x,y)—>.(a,b)

f (x, y) = L

Here is the formal definition. DEFINITION Limit

Assume that f (x, y) is defined near P = (a, b). Then firn

(x,y)—> P

f (x, y) = L

if, for any c > 0, there exists 8 > 0 such that if (x, y) satisfies 0< d((x, y), (a, b)) (a ,b)

y = b.

Solution Let P = (a, b). To verify (a), let f (x, y) = x and L = a. We must show that for any c > 0, we can find 6 > 0 such that If 0 < d((x, y), (a, b)) < 8,

then

If(x,Y)— LI = Ix — al 0, if 0 < d((x, y), (a, b)) < (a). The limit (b) is similar (see Figure 3).

_ D*(P, 3)

(x,y)->(a,b)

-

al
P

g(x, y)) =

(f(x, y)

lim

(x,y)-> P

lirn

f (x, y)

P

g(x, y)

(ii) Constant Multiple Law: For any number k, lirn

(x,y)-> P

kf (x, y)

k

lim

(x,y)--> P

f (x, y)

(iii) Product Law: lirn

(x,y)-> P

f (x, y) g(x, y) =

(iv) Quotient Law: If

lim g(x, y)

(x,y)-> P

lim

(x ,y)-> P

lim

P

f (x, y))

Elm

(x,y)-> P

g(x, y))

0, then lim

Groo-,

f (x, y) g(x, y)

urn

=-

f (x, y) P

(x,y)-> P

g(x, y)

As in the single-variable case, we say that f is continuous at P = (a, b) if f (x, y) approaches the value of the function f (a, b) as (x, y) (a, b). DEFINITION Continuity

A function f of two variables is continuous at P =- (a, b) if urn

(x,y)-) (a,b)

f (x, y) = f (a, b)

We say that f is continuous if it is continuous at each point (a, b) in its domain. The Limit Laws tell us that all sums, multiples, and products of continuous functions are continuous. When we apply them to f (x, y) = x and g(x, y) = y, which are continuous by Example 1, we find that the power functions f (x, y) = xmy n are continuous for all whole numbers m, n and that all polynomials are continuous. Furthermore, a rational function h(x, y)/ g(x, y), where h and g are polynomials, is continuous at all points (a, b) where g(a, b) 0. As in the single-variable case, we can evaluate limits of continuous functions using substitution. EXAMPLE 2 Evaluating Limits by Substitution FIGURE 4 Top view of the graph 3x y f (x' Y)= x2 +

y2 + 1 •

3x

f (x, Y) = x2 + is continuous (Figure 4). Then evaluate

lim

Show that y

y2 +

(x,y)--> (1,2)

f (x, y).

SECTION 15.2

Limits and Continuity in Several Variables

809

Solution The function f is continuous at all points (a, b) because it is a rational function whose denominator Q(x, y) = x2 + y2 + 1 is never zero. Therefore, we can evaluate the limit by substitution: lim(1,2) x2

3x + y

(., 0)

3(1) + 2

+ y2 + 1

12 + '+2 I- 1

5 - 6

•

If f is a product f (x, y) -= h(x)g(y), where h(x) and g(y) are continuous, then the limit is a product of limits by the Product Law: lirn

(x,y)-->(a,b)

EXAMPLE 3

f (x , y) =

lirn

(x,y)—>(a,b)

Product Functions

Solution Since lim x3 and lim

y-4.0

h(x)g(y) = (lim h(x)) —>a

Evaluate

Y

lirn

(x,y)—> (3,0)

lim g(y))

y—> b

x3 sin Y y

3' both exist, the desired limit can be expressed as a

product of limits: lim x (x,y)—>.(3,o)

3 sin y

y

=

(. 3) ( sin y\ x . = (33 )(1) = 27 \x-+3 y—>0 y

•

Composition is another important way to build functions. If f is a function of two variables and G(u) a function of one variable, then the composite function G o f is the function of two variables given by G(f (x, y)). According to the next theorem, a composition of continuous functions is continuous. THEOREM 2 A Composition of Continuous Functions Is Continuous If a function of two variables f is continuous at (a, b) and a function of one variable G is continuous at c f (a, b), then the composite function G(f (x, y)) is continuous at (a, b). EXAMPLE 4 Write H(x , y) = e —x2 +2Y as a composite function and evaluate lim

(x ,y)—> (1,2)

H(x, y)

Solution We have H (x, y) = G(f (x, y)), where G(u) = e" and f (x, y) =- —x2 + 2y. Both f and G are continuous, so H is also continuous and lfin

(x,y)—*(1,2)

H(x , y) =

lim

(x,y)—>(1,2)

As we indicated previously, if a limit

e —,2±2) = e_(1)2 +2(2) = e'2 lim

(x,y)—*(a,b)

•

f (x , y) exists and equals L, then

f (x, y) tends to L as (x, y) approaches (a, b) along any path. In the next example, we prove that a limit does not exist by showing that f (x, y) approaches different limits when (0,0) is approached along different lines through the origin. We use three different methods on the problem to demonstrate a variety of approaches one can take. EXAMPLE 5

Showing a Limit Does Not Exist

Examine

X2

numerilirn (x,y)-0.(0,0) x2 + y2

cally. Then prove that the limit does not exist. Solution If the limit existed, we would expect the values of f (x, y) in Table 1 to get closer to a limiting value L as (x, y) gets close to (0,0). However, the table suggests that: • As (x, y) approaches (0,0) along the x-axis, f (x, y) approaches 1. • As (x, y) approaches (0,0) along the y-axis, f (x, y) approaches 0. • As (x, y) approaches (0,0) along the line y = x, f (x, y) approaches 0.5.

810

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES

Therefore, f(x, y) does not seem to approach any fixed value L as (x, y)

Values of f(x, y) =

TABLE 1

X

(0,0).

2

x2 + y2 0.1

0.3

0.2

0.4

0.5

0.39

0.5

-0.5

-0.4

-0.3

-0.1

0

0.5

0.5

0.39

0.265 0.138 0.038

0

0.038 0.138 0.265

0.4

0.61

0.5

0.36

0.059

0

0.059

0.2

0.36

0.5

0.61

0.5

0.64

0.735

0.692

0.8

0.862

0.3 0.2 0.1

-0.2

0.2

0.735

0.64

0.5

0.308

0.1

0

0.1

0.308

0.862

0.8

0.692

0.5

0.2

0

0.2

0.5

0.962 0.941 1

0

0.9

0.8

0.5

1

1

1

0.9

0.8

0.5

0.692

0.5

0.2

1

-0.1

0.962 0.941

-0.2

0.862

0.8

0.5

0.8

0.9

1

1

1

0

0.5

0.8

0.9

0

0.2

0.5

0.692 0.5 0.36

0

-0.3

0.735 0.640

0.5

0.308

0.1

0

0.1

0.308

-0.4

0.610

0.5

0.360

0.2

0.059

0

0.059

0.2

-0.5

0.5

0.39

0.265 0.138 0.038

0

0.038 0.138 0.265

0.941 0.962 1

1

0.941 0.962 0.8

0.862

0.640 0.735 0.5

0.61

0.390

0.5

Now, let's prove that the limit does not exist. We demonstrate three different methods. First Method We show that f(x, y) approaches different limits along the x- and y-axes

(Figure 5): x2

Limit along x-axis:

firn f(x, 0) = lim = lim 1 = 1 x--0 x->13 x2 + 02 x->0

Limit along y-axis:

lim f(0, y) = firn ,

y--).0 0'

These two limits are different, and hence,

x

/ / c =/ 1 , 0.9 0.7 0.5 0.3 0.1

1

0

Y

, = lim 0 = 0 +y" y->0

f(x, y) does not exist.

Second Method If we set y = mx, we have restricted ourselves to the line through the origin with slope m. Then the limit becomes lim f (x, mx) -= lim

FIGURE 5 Graph and contour map of X2 f (x, y) = x2 + ,2•

lim

(x,y)-+(0,0)

02

x->o

X

2

1

, = 1 + m2 x--oD x' ± (mx)2

This clearly depends on the slope m, and therefore gives different values when the origin is approached along lines of differing slope. For instance, when m = 0, so that we

are approaching along the x-axis, we have a limit of 1. But when m = 1, so that we are approaching along the line y = x, the limit is 1. Hence, the overall limit does not exist. The contour map in Figure 5 shows the variety of limits that occur as we approach the origin along different lines. Third Method We convert to polar coordinates, setting x = r cos 0 and y = r sin O. Then for any path that approaches (0,0), it must be the case that r approaches 0. Different linear paths can be considered by fixing 0 at various values and having r approach 0. Hence, we can consider x2 (r cos 0)2 2 - lim cos 0 r1.imo x2 + y2 = rut ilo (r cos 0)2 + (r sin 0)2 r-0)

This result depends on O. For instance, fixing 0 at 0 would mean we are approaching (0,0) along the positive x-axis, and that gives a limit of 1. Fixing 0 at 7r/2 would mean we are approaching (0,0) along the positive y-axis, and that gives a limit of 0. Since different values of 9 yield different results, the overall limit does not exist. •

SECTION 15.2

EXAMPLE 6 Verifying a Limit for (x, y)

Calculate

Limits and Continuity in Several Variables

lim

(x,y)—>(0,0)

811

f (x, y), where f (x , y) is defined

(0, 0) by xy2 y2

f(x,y)= x 2 + as in Figure 6.

Solution Since substitution yields the indeterminate form of type alternate method. We convert to polar coordinates: X = r COS 0 ,

xy 2

FIGURE 6 Graph of f (x,Y) = x2

g, we need to try an

y = r sin

Keep in mind that for any path approaching (0,0), r approaches 0. Then x2 + y2 = r2 and for r 0,

+y2 0
(o,0) lx I + I I lim

ex 2

12 6

0

y2

(x ,y)-> (2,1)

(B) Contour map of g(x, y)

(A) Contour map of f(x, y)

1 x2

30

3

x2 ± y2

(x ,y)-> (0,0)

(x

45. 21 the limit

FIGURE 7

y2)

Further Insights and Challenges 46. Evaluate

lim

(1 +x)

(x,y)->(0,2)

49. Prove that the function

.

47. Is the following function continuous? { x2 + y2

f(x , Y) =

1

if x2 ± y2

f (x,y)= {

(2' - 1)(sin y) xy ln 2

if xy

0

if xy = 0

1

if x2 ± y2 >

48. CAS EleJ The function f(x, y) = sin(xy)/xy is defined for xy 0 0. (a) Is it possible to extend the domain of f to all of R2 so that the result is a continuous function? (b) Use a computer algebra system to plot f. Does the result support your conclusion in (a)?

is continuous at (0,0). 50. Prove that if f(x) is continuous at x = a and g(y) is continuous at y = b, then F (x y) = f (x)g(y) is continuous at (a, b). x3 y x6 + 2y2. (a) Show that as (x, y) -> (0,0) along any line y = mx, the limit equals 0. (b) Show that as (x, y) -> (0,0) along the curve y = x3, the limit does f (x , y) does not exist. firn not equal 0, and hence, (x,y)->(0,0) 51. Consider the function f(x, y) =

814

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES

15.3 Partial Derivatives We have stressed that a function f of two or more variables does not have a unique rate of change because each variable may affect f in different ways. For example, the current I in a circuit is a function of both voltage V and resistance R given by Ohm's Law: V I (V , R) = — R The current I is increasing as a function of V (when R is fixed) but decreasing as a function of R (when V is fixed). The partial derivatives are the rates of change with respect to each variable separately. A function f (x, y) of two variables has two partial derivatives, denoted fx and f y, defined by the following limits (if they exist): . f (a + h,b) — f (a, b) , fx(a,b) = urn h--+0

The partial derivative symbol a is a rounded "d." It is used to distinguish derivatives of a function of multiple variables from derivatives of functions of one variable where the straight "d" is used.

h

f y(a,b) = .

f (a, b

k) — f (a, b)

Thus, fx is the derivative of f (x,b) as a function of x alone, and f y is the derivative of f (a, y) as a function of y alone. We refer to fx as the partial derivative of f with respect to x or the x -derivative of f. We refer to f y similarly. The Leibniz notation for partial derivatives is

0f

= fx,

af ax (a,b)

= f x(a, b),

= f y(a,b) (a,b)

If Z = f (x, y), then we also write az/ax and az/ay. Partial derivatives are computed just like ordinary derivatives in one variable with this difference: To compute fx, treat y as a constant and take the derivative of f with respect to x, and to compute f y, treat x as a constant and take the derivative of f with respect to y. EXAMPLE 1 Compute the partial derivatives of f (x, y) = x2y5. Solution

af =

y5 Tx a (x2) = y5(2x) = 2xy5 Treat y5 as a constant.

ay = __:.(x2y5)= x 2 5; ay (y5) =

x2(5y4) = 5x2y4

•

Treat x2 as a constant.

GRAPHICAL INSIGHT The partial derivatives at P = (a, b) are the slopes of the tangent lines to the trace curves through the graph of f (x, y) at the point (a, b, f (a, b)) in Figure 1(A). To compute f( a, b), we set y = band differentiate in the x -direction. This gives us the slope of the tangent line to the trace curve in the plane y = b [Figure 1(B)]. Similarly, f y(a, b) is the slope of the trace curve in the plane x = a [Figure 1(C)]. The differentiation rules from calculus of one variable (the Product, Quotient, and Chain Rules) are valid for partial derivatives.

SECTION 15.3

Partial Derivatives

815

Slope fx(a, b) The trace curve (x, b, f(x, b))

Plane y = b

(B)

(A)

FIGURE 1 The partial derivatives are the slopes of the trace curves.

Y

EXAMPLE 2 Calculate g(1, 3) and g(1, 3), where g(x, Y)

2

(1 + X2)3

Solution To calculate gx, treat y (and therefore y2) as a constant and differentiate with respect to x:

a ( 2 +X2)3) ax (1 y

gx(X, y) =

gx(1, FIGURE 2 The slopes of the tangent lines to the trace curves are g(1, 3) and g( l, 3).

—6xy2 (1 + x2)4

—

4

To calculate gy, treat x [and therefore (1 + x2)3] as a constant and differentiate with respect to y:

a CAUTION It is not necessary to use the Quotient Rule to compute the partial derivative in Eq. (1). The denominator does not depend on y, so we treat it as a constant when differentiating with respect toy.

+x2)-3 =

2"87

6+01)23)2

-=

r a x (1

gy(x, y) =

g(1, 3) =

ay

y2 i +x

2(3) 1+ 12)3

1

a

3 ) — (1 +x 2)3 ay

,2

2y — (1 +x 2) 3

1

3 4

These partial derivatives are the slopes of the trace curves through the point P = (1,3, 3-) shown in Figure 2. Note that in the figure, g is decreasing as x increases through P, consistent with our determination that gx (1, 3) 0. The Chain Rule was Used in Example 2 to compute gx(x, y). We use the Chain Rule to compute partial derivatives of a composite function like f(x, y) = sin(3x2 + 4y) in the same way the Chain Rule is applied in the single variable case. For example, to compute the partial derivative of f with respect to x, we take the derivative of the outside function at the inside function [yielding cos(3x2 + 4y)] and multiply by the derivative (with respect to x) of the inside function; that is, by 6x. Therefore,

a ax

a (3x,- + 4y) = 6x cos(3x2 + 4y) — sin(3x2 + 4y) = cos(3x2 + 4y) — ax

In multivariable calculus, there are a number of different ways compositions of functions can arise. Consequently, there are multiple possibilities for chain-rule derivative computations. We examine other possibilities for multivariable chain rules in Sections 15.5 and 15.6.

816

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES

a

Compute — ln(xy — 2y2). ay

EXAMPLE 3 Chain Rule for Partial Derivatives Solution Using the Chain Rule, we have:

a — ln(xy ay

2y2)

1

a

)

1

) (x (xy - 2y2) - ( xy - 2y2 ay ,

- (xy - 2y2

4y) =

x — 4y xy - 2y2

Chain Rule

EXAMPLE 4 Wind Chill The wind-chill temperature W(T, v) (in °C) measures how cold people feel (based on the rate of heat loss from exposed skin) when the outside temperature is T°C (with T < 10) and the wind velocity is v m/s (with v > 2): W = 13.1267 ± 0.6215T - 13.9470.16 + 0.486T v°16 and -BZ . Show that at a fixed wind speed, the impact on wind chill of a Calculate temperature does not depend on the temperature, but the impact of a changing changing wind speed is larger the colder the temperature. Solution Computing the partial derivatives:

aw aT aw aV

—

=

-13.947(0.16)v-0.84

= 0.6125 + 0.4860 16

0.486T(0.16)v - 0.84

g

=

2.2315v - 0.84 ± 0.0778Tv —0.84

Note that does not depend on T, so that at a fixed wind speed g is the same, no matter the temperature. For example, at 20 m/s, 0.6125 ± 0.486(20)0.16 1.3877 (°C per °C) at all values of T. On the other hand, at a fixed wind speed, decreases as T decreases. For example (in units of °C per m/s),

=

aw av 1(5,10)

-0.2663

aw

(-5,10) =-- -0.3788

aw ay 1(-15,10) ,---- -0.4912

Therefore at a fixed wind speed, an increase in wind speed has a larger cooling effect at colder temperatures. • Partial derivatives are defined for functions of any number of variables. We compute the partial derivative with respect to any one of the variables by differentiating with respect to that variable while holding the remaining variables constant. EXAMPLE 5

More Than Two Variables

Calculate L(0,0,1, 1), where

f(x, y,z, w) =

exz+Y Z2 ± W

In Example 5, the calculation

a _ex-Y = xex-Y az

Solution Use the Quotient Rule, treating x, y, and w as constants and differentiating with respect to z:

a fz(x,

aZ

w

follows from the Chain Rule, just like (Z2

_ e4z+2 = 4e4z+2 dz

(z2

exz-FY

W)Xexz+Y — 2Zexz+Y (Z2 + 0 2

f(0,0, 1,1) =

—2e0 (12 ± 1)2

=

1 2

az-

exz+y

k(z2 w)

(z2 + (Z2X

WX — 2Z)exz+Y (Z2 ±

•

SECTION 15.3

Partial Derivatives

817

In the next example, we estimate a partial derivative numerically. Since A and f y are limits of difference quotients, we have the following approximations when Ax and Ay are small: Af Ax

fx(a,b) These approximation formulas are multivariable versions of the difference quotient approximation introduced in Section 3.1.

=

f (a ± Ax,b) — f (a,b) Ax

Af f (a, b + Ay) — f (a, b) —A-- = y Ay Similar approximations are valid in any number of variables. EXAMPLE 6 Estimating Partial Derivatives Using Contour Maps Seawater density depends on salinity and temperature and can be expressed as a function p(S, T), where p is in kilograms per cubic meter, salinty S is in parts per thousand, and temperature T is in degrees Celsius. Use the contour map of seawater density appearing in Figure 3 to estimate Op/OT and Op/OS at A -= (33, 15).

25

Solution We estimate Op/OT at A in two steps. Step I. Choose AT, and estimate or evaluate p(33, 15 + AT). With S held constant at 33, a change in T moves us vertically on the contour map from the point A. Any choice of small AT can be used to make our estimate. We choose AT = 2 because the corresponding point (B on the contour map) lies on a level curve near A, and at B we can evaluate p, rather than estimate it. With AT = 2, we have p(33, 15 + AT) = p(33, 17) = 1.0240.

5

0

32.5

34.0 33.0 33.5 Salinity S (ppt)

34.5

Contour map of seawater density as a function of temperature and salinity.

FIGURE 3

Step 2. Compute the difference quotient and make the approximation.

Op OT

(33,15)

p(33, 17) — p(33, 15) 2

1.0240 — 1.0245 2

—0.0005 2

= —0.00025 kg-m-3/°C We estimate Op/OS in a similar way, using AS of p on the contour map. We obtain

Op as (33,15)

p(33.7, 15) — p(33, 15) 0.7 0.0007 kg-m-3/ppt

0.7 to put us at point C on a level curve 1.0250 — 1.0245 0.7

0.0005 0.7 •

Higher Order Partial Derivatives The higher order partial derivatives are the derivatives of derivatives. The second-order partial derivatives of f are the partial derivatives of A and f y. We write f x., for the x-derivative of A and Ay for the y-derivative of f y:

a (af ax —, xa) We also have the mixed partials:

a (af)

fxy — ay ax , The process can be continued. For example, Ayx is the x -derivative of f xy, and fxyy is the y-derivative of Ay (perform the differentiation in the order of the subscripts from left to right). The Leibniz notation for higher order partial derivatives is

818

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES

a2f fxx = ax2

a2 f

a2f

a2 f

fYY = ay2

aXay'

fYx

fxY = ayaX

Higher order partial derivatives are defined for functions of three or more variables in a similar manner. EXAMPLE 7 Calculate the second-order partial derivatives of f (x, y) = x3 + y2ex Solution First, we compute the first-order partial derivatives: a f

x

(x

,

y)

=

(x3 + y2ex) 3x2

a (x3 fy(x, y) = — ay

y2ex,

ax

y2 ex) =

2yex

Then we can compute the second-order partial derivatives:

a

fxx(x, y) = ax fx

= a

(3x

2

y2 ex )

a 2yex — ay

fy

= 2ex

= 6x + y2ex, afx

a

fyy(x, y)

a

2

f xy(X,Y) = T y = --

a 2yex f yx(x, y) = afY = Tx

y2ex)

= 2yex ,

•

= 2yex

It is not a coincidence that in the previous example. This result is an example of a general theorem that we present after the next example. Ay

Remember how the subscripts are used in partial derivatives. The notation fx„ indicates that we first differentiate with respect to x and then differentiate twice with respect to y.

EXAMPLE 8 Calculate

fxyy

Ax

for f (x, y) = x3 + y2ex .

Solution By the previous example, f xy = 2yex . Therefore,

fxyy =

a

=

-a6aT 2yex = 2ex

•

The next theorem, named for the French mathematician Alexis Clairaut (Figure 4), indicates that in a mixed partial derivative, the order in which the derivatives are taken does not matter, provided that the mixed partial derivatives are continuous. A proof of the theorem is provided in Appendix D. The hypothesis of Clairaut's Theorem, that fxy and f yx are continuous, is almost always satisfied in practice, but see Exercise 80 for an example where the mixed partial derivatives are not equal.

THEOREM 1 Clairaut's Theorem: Equality of Mixed Partials If and f yx both exist and are continuous on a disk D, then f xy(a, b) = fyx(a,b) for all (a, b) E D. Therefore, on D, Ay

a2f a2f ax ay - ay ax EXAMPLE 9 Check that

a2w auaT

a2w =

a au

for W

eu/ T.

Solution We compute both mixed partial derivatives and observe that they are equal:

aw = eUIT a (U = -uT-2eulT, aT aw = _ T-2eul T _ UT-3eUIT au aT

aw = eU/T a u 7-1 6,u/T au -ari a aw T-2euir _uT-3eulT • aT au = -

SECTION

15.3

Partial Derivatives

819

Although Clairaut's Theorem is stated for f xy and f yx, it implies more generally that partial differentiation may be carried out in any order, provided that the derivatives in question are continuous (see Exercise 71). For example, we can compute f by differentiating f twice with respect to x and twice with respect to y, in any order. Thus, fuxy

fxxyy

fyyxx

fyxyx

EXAMPLE 10 Choosing the Order Wisely where g(x, y,z, w)

fuyx

fyxxy

Calculate the partial derivative gzz wx,

xy

= x 3 w2z2

+ sin — . z2

Solution Let's take advantage of the fact that the derivatives may be calculated in any order. If we differentiate with respect to w first, the second term disappears because it does not depend on w: gw =

a

3

— aw

2 2 w z +

sin (x•Y )) = n

3 wz 2

Next, differentiate twice with respect to z and once with respect to x: 4 Alexis Clairaut (1713-1765) was a brilliant French mathematician who presented his first paper to the Paris Academy of Sciences at the age of 13. In 1752, Clairaut won a prize for an essay on lunar motion that Euler praised (surely an exaggeration) as "the most important and profound discovery that has ever been made in mathematics."

FIGURE

gwz -=

a —2x3tvz2 = 4x3wz az a = 4x3w

gwzz = —4x-'wz az 3 a gwzzx = — 4x w

ax

= 12x 2 w

We conclude that gzz wx = A partial differential equation (PDE) is a differential equation involving functions of several variables and their partial derivatives. A solution to a PDE is a function that satisfies the equation. The heat equation in the next example is a PDE that models temperature as heat spreads through an object. There are infinitely many solutions, depending on the initial temperature distribution in the object. The particular function in the example describes temperature at times t > 0 along a metal rod when the center point is given a burst of heat at t = 0 (Figure 5). Temperature T

Time t

FIGURE 5 The plot of 1 u(x t) —

e —(x214t)

illustrates the diffusion of a burst of heat

Metal bar

over time.

EXAMPLE 11 The Heat Equation

Show that u(x, t) —

t > 0, satisfies the heat equation

au a2u at = ax2

1

(

1), defined for

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES a2u

Solution We write u(x, t) -

au ax

2 1 t-1/2e-(x214t) . We first compute —: ax2 N/Tr

a ( 1 ciRe_(x2140) ax 2.,/Tr a 1 Xt-3/2e-(x2

a2u

1

4.,Fr

ax2

=

Xt

_3 /2 ' e _( 214o)

t -3/2e-(x214t) +

140) =

1

x 2t-5/2e-(x214t)

ax

Then compute au/at and observe that it equals a2 u/ax2 as required:

au at

=

a ( 1 t_112e-(x2140) at

1

t -3/2e-(x214t)

HISTORICAL PERSPECTIVE

1 16 Joseph Fourier (1768-1830)

1

2 t —5/2 _—(x2/4t)

8,/Tr

4N/71-

Adolf Fick (1829-1901)

The general heat equation, of which Eq. (2) is a special case, was first introduced in 1807 by French mathematician Jean Baptiste Joseph Fourier. As a young man, Fourier was unsure whether to enter the priesthood or pursue mathematics, but he must have been very ambitious. He wrote in a letter, "Yesterday was my 21st birthday, at that age Newton and Pascal had already acquired many claims to immortality." In his twenties, Fourier got involved in the French Revolution and was imprisoned briefly in 1794 over an incident involving different factions. In 1798 he was summoned, along with more than 150 other scientists, to join Napoleon on his unsuccessful campaign in Egypt.

15.3 SUMMARY • The partial derivatives of f (x, y) are defined as the limits af fx(a,b). —

ax

= lim

af fy(a, b) = —

ay

= (a,b)

f (a

h, b) - f (a , b)

h—>0

(a,b)

e

•

Fourier' s true impact, however, lay in his mathematical contributions. The heat equation is applied throughout the physical sciences and engineering, from the study of heat flow through the earth's oceans and atmosphere to the use of heat probes to destroy tumors and treat heart disease. Fourier also introduced a striking new technique—known as the Fourier transform—for solving his equation, based on the idea that a periodic function can be expressed as a (possibly infinite) sum of sines and cosines. Leading mathematicians of the day, including Lagrange and Laplace, initially raised objections because this technique was not easy to justify rigorously. Nevertheless, the Fourier transform turned out to be one of the most important mathematical discoveries of the nineteenth century. A Web search on the term "Fourier transform" reveals its vast range of modem applications. In 1855, the German physiologist Adolf Fick showed that the heat equation describes not only heat conduction but also a wide range of diffusion processes, such as osmosis, ion transport at the cellular level, and the motion of pollutants through air or water. The heat equation thus became a basic tool in chemistry, molecular biology, and environmental science, where it is often called Fick's Second Law.

NLM/Science Source

820

. k —>-O

f (a, b

k) - f (a, b)

SECTION 15.3

Partial Derivatives

821

• Compute f x by holding y constant and differentiating with respect to x, and compute f y by holding x constant and differentiating with respect to y. fx(a, b) is the slope at x = a of the tangent line to the trace curve z = f (x,b). Simi• larly, f y(a,b) is the slope at y = b of the tangent line to the trace curve z = f (a, y). • Approximating partial derivatives: For small Ax and Ay, fx(a,b)

Af f (a + Ax, b) - f (a, b) =. Ax Ax

Af f (a, b + Ay) - f (a, b) fy(ab) , c•-' — Ay = Ay Similar approximations are valid in any number of variables. • The second-order partial derivatives are

82

.92

82

ay ax

ax ay

82

ay f = fyy

• Clairaut's Theorem states that mixed partials are equal—that is, f xy = f yx provided that fxy and f yx are continuous. • More generally, higher order partial derivatives may be computed in any order. For example, f xyyz = fyxzy if f is a function of x, y, z whose fourth-order partial derivatives are continuous.

— 15.3 EXERCISES Preliminary Questions 1. Patricia derived the following incorrect formula by misapplying the Product Rule:

_a (x2y2) = x2(2y)± y2(2x) Ox

3. Which of the following partial derivatives should be evaluated without using the Quotient Rule?

a xy ax y2 + 1

4.

What is f x, where f

What was her mistake and what is the correct calculation? 2. pute

Explain why it is not necessary to use the Quotient Rule to com-

a

x +1

Ox

y+1

. Should the Quotient Rule be used to compute

a tx+y , ay Y + 1 ) •

a 2 2 —xy =y, Ox 2.

a xy2 — ay

2xy

Use the limit definition of the partial derivative to verify the formulas

a (.1\ = 1 ax y y'

a ay

(b) f yyx

Use the Product Rule to compute f x and f y for f

fxyy

(C)

a

y

5.

Use the Quotient Rule to compute

6.

a 2 Use the Chain Rule to compute — ln(u

7.

Calculate f(2,3, 1), where f

au

ay x + y

(d)

fyxx

.

uv).

(x, y,z) = xyz.

— sin(cx) = c cos(cx), dx

4.

ax y2 + 1

8. F -71 Explain the relation between the following two formulas (c is a constant):

= -x y y2

3. Use the Product Rule to compute f x and f y (x2 - y)(x — y2).

y2

5. Assuming the hypotheses of Clairaut's Theorem are satisfied, which of the following partial derivatives are equal to f rxy?

(a) fxyx

Use the limit definition of the partial derivative to verify the formulas

(C)

(x, y, z) = (sin yz)ez3-z-1,5 ,9

Exercises 1.

a

a xy (b) ay y2 +

(a)

for f(x, y)

(x y) = xyex sin y.

a sin(xy) = y cos(xy) — Ox

9. The plane y = 1 intersects the surface z = x4 + 6xy — y4 in a certain curve. Find the slope of the tangent line to this curve at the point P = (1, 1,6).

822

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES

10. Determine whether the partial derivatives af/ax and 8f/ 8y are positive or negative at the point P on the graph in Figure 6.

29. W = er±s

30. Q = re°

31. z = exY

32. R = e-u2lk

33. z = e-x2- Y2

Fz2 34. P = eiT2T-

e-rt 35. U = r

36. z = yx

37. z = sinh(x2y)

38.

39. w = xy2z3

40. w =

41. Q =

42. w= (x2 + y2

e-Ltim

cosh(t - cos x)

Z =

y+z z2)312

In Exercises 43-46, compute the given partial derivatives. .fx (1,2)

43. f (x,y) = 3x2y + 4x3y2 - 7xy5, 44. f(x, y) = sin(x2 - y), FIGURE 6

46. h(x,z) = exz-x2z3

11. Estimate f x and f y at point A. 12. Is f positive or negative at B? 13. Starting at point B, in which compass direction (N, NE, SW, etc.) does f increase most rapidly? 14. At which of A, B, or C is f y the least?

4-

49. Use the contour map of f(x, y) in Figure 8 to explain the following statements: (a) f y is larger at P than at Q, and L is more negative at P than at Q. (b) fx(x, y) is decreasing as a function of y; that is, for any fixed value x = a, f x (a, y) is decreasing in y.

-30 10 0• 10

l(T, H) = 45.33 +0.6845T ± 5.758H - 0.00365T2 - 0.1565HT +0.001HT2 (a) Calculate I at (T, H) = (95, 50). (b) Which partial derivative tells us the increase in I per degree increase in T when (T, H) = (95,50)? Calculate this partial derivative. 48. Calculate apiaT and apla V, where pressure P. volume V. and temperature T are related by the Ideal Gas Law, PV = nRT (R and n are constants).

70 50 30 / / /

0-

h(3, O)

47. The heat index I is a measure of how hot it feels when the relative humidity is H (as a percentage) and the actual air temperature is T (in degrees Fahrenheit). An approximate formula for the heat index that is valid for (T, H) near (90,40) is

In Exercises 11-14, refer to Figure 7.

2-

f y(0,n-)

45. g(u , v) = u ln(u + v), g(1,2)

-10

:- 57g

-2

0

2

4

FIGURE 7 Contour map of f (x, y). In Exercises 15-42, compute the first-order partial derivatives. 15. z = x2 4 _ y2 16. z = x4y3 17. z = x 4y

xy -2

19. z = y 21. z =

20. z = - x2 - y2

23. z = (sin x)(cos y) 25. z = cos

18. V = irr2h

1- x

27. w = ln(x2 - y2)

22. z _

FIGURE 8 50. Estimate the partial derivatives at P of the function whose contour map is shown in Figure 9.

X

2118 15 12 9

x- y 4 A/ x 2 ± y2

26. 0 = tan-I (xy2) 28. P = sin(2s

-

3

2

24. z = tan(uv3)

3t)

1

1

2

1

1

1

4 FIGURE 9

I

I

I

I

, X

SECTION 15.3

51. Over most of the earth, a magnetic compass does not point to true (geographic) north; instead, it points at some angle east or west of true north. The angle D between magnetic north and true north is called the magnetic declination. Use Figure 10 to determine which of the following statements is true: 3D 3D aD „ 3D (b) >0 (c) >0 (0) > ax °Y A °Y aY Note that the horizontal axis increases from right to left because of the way longitude is measured.

57. f (x, y) = x ln(y2), 58. g(x,y) = xe'Y ,

Partial Derivatives

823

f( 2,3) g (-3,2)

59. Compute f xyxzy for

f (x,y, z) = y sin(xz) sin(x + z) + (x ± z2) tan y + x tan (

z + z -I y - Y-1

Hint: Use a well-chosen order of differentiation on each term. 60. Let

Magnetic Declination for the U.S.

f(x,y,u,v) =

50°N

x 2 ey v 3y2 + ln(2 + u2)

What is the fastest way to show that fuvxyvu,-, fx y, u, v) = 0 for all (x, y,u, v)? In Exercises 61-68, compute the derivative indicated.

I 40°N

61. f (u, v) = cos(u

v2),

62. g(x, y, z) = X4 Y5Z6

30°N

gXXYZ

63. F(r, s, t) = r(s2 + t2), 120°W

110°W

100°W

90°W

80°W

70°W

64. u(x, t)

=

fuuv

FrSt

t -112 e -(X 214t) ,

uxx

65. F(0 , u, v) = sinh(uv + 02),

FIGURE 10 Contour interval 1 0 .

66. R(u, v, w) 52. Refer to Table 1. (a) Using difference quotients, approximate Op/ 3T and 8p/DS at the points (S, T) = (30,2), (32,6), and (35, 10).

v

w

Fuue

, Rut

67. g(x ,y,z) = Il ycz + y2 + 4

gxyz

(b) For fixed salinity S = 32, determine whether the quotients Ap/ A T are increasing or decreasing as T increases. What can you conclude about the sign of 32p/8 T2 and the concavity of p as a function of T?

68. u(x, t) = sech2(x - t),

1 Seawater Density p as a Function of Temperature T and Salinity S

70. 17 Prove that there does not exist any function f (x, y) such that = x2. Hint: Consider Clairaut's Theorem. -a/ = xy and -af ay ax

TABLE 7-

30

31

32

33

34

35

36

12

22.75

23.51

24.27

25.07

25.82

26.6

27.36

10

23.07

23.85

24.62

25.42

26.17

26.99

27.73

8

23.36

24.15

24.93

25.73

26.5

27.28

29.09

6

23.62

24.44

25.22

26

26.77

27.55

28.35

4

23.85

24.62

25.42

26.23

27

27.8

28.61

26.38

27.18

28.01

28.78

26.5

27.34

28.12

28.91

5

2

24

24.78

25.61

0

24.11

24.92

25.72

uxxx

af af 69. Find a function such that - = 2xy and - = x 2. ay aX

71. Assume that f xy and fyx are continuous and that f yxx exists. Show that f xyx also exists and that fyxx = fxyx • 72. Show that u(x, t) = sin(nx)e-n2t satisfies the heat equation for any constant n: au a 2 u 3 at = 3x2 73. Find all values of A and B such that f (x,t) = eAx+Bt satisfies Eq. (3). 74. The function

In Exercises 53-58, compute the derivatives indicated. a2 f 53. f (x, y) = 3x2 y - 6xy4,

-

54. g(x,y) =

55. h(u, v) =

u +4v

3x2 and 3y2

02g a x ay

xy x- y ' ,

hyv(u,v)

56. h(x, y) = ln(x3 ± y3),

a2f

hxy(x, Y)

f (x,t) -

1

e

_x214t

describes the temperature profile along a metal rod at time t > 0 when a burst of heat is applied at the origin (see Example 11). A small bug sitting on the rod at distance x from the origin feels the temperature rise and fall as heat diffuses through the bar. Show that the bug feels the maximum temperature at time t = x2. In Exercises 75-78, the Laplace operator A is defined by Af = fxx .fyy• A function u(x, y) satisfying the Laplace equation Au = 0 is called harmonic.

824

DIFFERENTIATION IN SEVERAL VARIABLES

CHAPTER 15

78. Find all constants a, b such that u(x, y) = cos(ax)ebY is harmonic.

75. Show that the following functions are harmonic: (b) u(x, y) = ex cos y (a) u(x, y) = x (d) u(x, y) = ln(x2 + y2) = tan—I u(x, y) (c) 76. Find all harmonic polynomials u(x, y) of degree 3, that is, u(x, y) = ax3 + bx2 y cxy2 dy3.

79. Show that u(x, t) = sech2(x — t) satisfies the Korteweg—deVries equation (which arises in the study of water waves): 4ut + uxxx + 12uux = 0

77. Show that if u(x, y) is harmonic, then the partial derivatives au/ax and au/ay are harmonic.

Further Insights and Challenges 80. Assumptions Matter This exercise shows that the hypotheses of Clairaut's Theorem are needed. Let x2

y2

f (x, y) = xy x2

± y2

(b) Use the limit definition of the partial derivative to show that f x (0,0) = f(0, 0) = 0 and that f( O, 0) and f(0, 0) both exist but are not equal. (c) Show that for (x, y) 0 (0,0):

for (x, y) (0, 0) and f (0,0) = 0. (a) Verify for (x, y) 0 (0,0): y(x4

iy(x, Y) —

ixy(x, Y) = .fyx(x, Y) = Show that

4x2y2 — y4) (x2 ± y2)2

fxy

x6 + 9x4y2 — 9x2y4 — y6 (x2 + y2)3

is not continuous at (0,0). Hint: Show that Inn f xy(h, 0) 0 h—>0

lim f xy(0,h).

h—).0

x(x4 — 4x2y2 — y4) (x2 +y2)2

(d) Explain why the result of part (b) does not contradict Clairaut's Theorem.

15.4 Differentiability, Tangent Planes, and Linear Approximation

=ftx,

Tangent plane at P

FIGURE 1 Tangent plane to the graph of f (x, y).

In this section, we explore the important concept of differentiability for functions of more than one variable, along with the related ideas of the tangent plane and linear approximation. In single-variable calculus, a function f is differentiable if the derivative f' exists. By extension, one might expect that a function f (x, y) would be differentiable if the partial derivatives f x (x , y) and f y(x , y) exist. Unfortunately, the existence of partial derivatives is not a strong enough condition for differentiability. First, we will show why the existence of the partial derivatives is not sufficient. Differentiability of f (x, y) at (a, b) should ensure that there is a tangent plane to the graph of f (x, y) at P = (a, b, f (a, b)) as illustrated in Figure 1. If f (x , y) has partial derivatives f x(a,b) and f y(a,b) at (a, b), then these derivatives determine lines that are tangent to the graph of f (x, y) at P. Figure 2(A) shows that one

Slope f x(a, b)

Plane determined by f x and f y Slope f y (a, b)

Tangent line for f

=fix, y)

Plane y = b

Tangent line for f3

b, 0)

Plane x = a (A)

FIGURE 2 Is the plane determined by fx and f y tangent to the graph?

(B)

(C)

SECTION 15.4

FIGURE 3 The vectors v and w are parallel to the plane determined by fx and fy.

Differentiability, Tangent Planes, and Linear Approximation

825

of these tangent lines lies in the plane y = b, and the other lies in the plane x = a . We refer to these lines as the tangent line for A and the tangent line for f y, respectively. These two tangent lines determine a plane that is certainly a good candidate for a tangent plane to the graph [Figure 2(B)]. We refer to this plane as the plane determined by A and f y. Unfortunately, this plane might not be fully tangent to the graph at P because other lines through P in this plane might not be tangent to the graph as in Figure 2(C). We will give an example of just such a situation later in the section. To identify a condition that guarantees that the plane determined by A and f y is tangent to the graph, we first need an equation of this plane. (We will also use this condition to define differentiability.) We begin by finding a normal vector to the plane determined by A and f y. To do that, we find direction vectors for the tangent lines for A and f y (which are parallel to the plane) and then take their cross product (which results in a vector normal to the plane). Consider first the tangent line for f x, which lies in the plane y = b. In that plane, the line has slope f x(a, b). Therefore, if we move on the line 1 unit in the positive xdirection from P, then we move f x (a, b) units in the z-direction. It follows that the vector v = (1,0, f( a, b)) is a direction vector for this line, as in Figure 3. Similarly, the vector w = (0, 1, f y(a,b)) is a direction vector for the tangent line for f y . As indicated previously, a normal vector to the plane determined by A and f y is obtained by taking a cross product of the vectors v and w. It is convenient to compute this cross product as w x v: i W X V= 0 1

j 1 0

k f y(a,b) = (fx(a,b), fy(a, b), —1) f x(a,b)

Now, P = (a, b, f (a, b)) lies on the plane determined by A and f y, and the vector (fx(a,b), fy(a,b), —1) is normal to it. Therefore, the plane has equation 1.10 REMINDER A plane through the point P = (xo, yo, zo) with normal vector n (A, B, C) has equation A(x — xo) + B(Y — Yo) + C(z — zo) = 0

b)(x — a) + fy(a,b)(Y — b) — (z —

f (a, b)) = 0

or z = f (a, b) + f x(a,b)(x — a) + f y(a,b)(y — b) Next, we use this equation for the plane determined by A and f y to identify a condition that ensures this plane is fully tangent to the graph of f (x, y) [and thus that f (x,y) is differentiable]. Let

e(x, y) = f(x, y)— L(x, y) = f(x,

z= L(x, y) (a, b, 0)

FIGURE 4 The error approximating f (x, y) with L(x, y).

L(x, y) =

f

+ f x(a,b)(x — a) + f y(a,b)(y — b)

We refer to L(x, y) as the linearization of f (x,y) centered at (a, b). The linearization can be used to approximate f (x, y) near (a, b). (Later in this section, we will develop this idea further and present some examples.) The graph of L(x, y) is the plane determined by A and f y. Figure 4 demonstrates that the difference f (x, y) — L(x, y) is the error e(x, y) obtained when approximating f (x, y) by L(x, y). As (x, y) approaches (a, b), this error approaches zero because the two functions are continuous and agree at (a, b). If the error goes to zero fast enough that the graph of f (x, y) flattens and becomes approximately a plane, then the plane determined by Ix and f y is an actual tangent plane. (We will explain why when we introduce directional derivatives in the next section.)

826

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES

What suffices for "fast enough" is that as (x, y) goes to (a, b), the error goes to zero faster than the distance from (x, y) to (a, b). (In Section 15.6 we will show why this is so.) That is, lim

e(x, y)

=

(x,y) (a,b) ./(x — 02 + (y — b)2

lim

f (x, y) — L(x, y)

=0

(x,y)—>(a,b) V(x — (2)2 ± (y — b)2

This now brings us to definitions of differentiability and the tangent plane: DEFINITION Differentiability and the Tangent Plane Assume that f (x , y) is defined in a disk D containing (a, b) and that f( a, b) and f y(a, b) exist. Let L(x, y) =

f (a, b) +

fx(a, b)(x — a) + fy(a, b)(y — b)

• f (x, y) is differentiable at (a, b) if

lim

f (x , y) — L(x, y)

=0

(x,y)—>(a,b) V(x — a)2 ± (y — b)2

• If f(x,y) is differentiable at (a, b), then the tangent plane to the graph at (a, b, f (a, b)) is the plane with equation z = L(x, y). Explicitly, the equation of the tangent plane is z = f (a, b) + f x(a, b)(x — a) ± fy(a, b)(y — b)

The definition of differentiability extends to functions of n-variables, and Theorem 1 holds in this setting: If all of the partial derivatives of f (xi_ . , xn ) exist and are continuous on an open domain D, then f (xi, . . . , xn ) is differentiable on D.

1

If f (x, y) is differentiable at all points in a domain D, we say that f (x, y) is differentiable on D. To prove that a particular function is differentiable, we need to prove that the limit in the definition holds. That can be tedious to verify (see Exercise 43), but fortunately, this is rarely necessary. The following theorem provides conditions that imply differentiability and are easy to verify. It assures us that most functions arising in practice are differentiable on their domains. See Appendix D for a proof.

THEOREM 1 Confirming Differentiability If f x (x, y) and f y(x, y) exist and are continuous on an open disk D, then f (x, y) is differentiable on D.

EXAMPLE 1 Show that f(x, y) = 5x ± 4y2 is differentiable on its domain, R2. Solution The partial derivatives are fx(x, y) = 5,

f(x,y) = 8y

These are continuous functions over all of R2. It follows by Theorem 1 that f (x , y) is differentiable for all (x, y). • EXAMPLE 2 Let f(x,y) = x2 ± 2y2 _ y —4. Find an equation of the tangent plane to the graph off at P = (1,2, f(1, 2)).

SECTION 15.4

Differentiability, Tangent Planes, and Linear Approximation

827

Solution We have f x(x y) = 2x and f y(x, y) =- 4y — 1. From that, fx(1,

=2

and

f y(1, 2) = 7

These values, along with f (1, 2) = 3, enable us to determine the equation of the tangent plane: z=

3 + 2(x — 1) + 7(y — 2)

= —13 + 2x + 7y

f(a, b)+ fx(a, b)(x — a)± f y(a, b)(y — b) The tangent plane through P = (1, 2, 3) has equation z -= —13 + 2x ± '7y (Figure 5). • The graph of f (x, y) = x2 + 2y2 —y — 4 and the tangent plane at P = (1, 2, 3). FIGURE 5

EXAMPLE 3 Find an equation of the tangent plane to the graph of f (x, y) = xy3 ± x2 at (2, —2, f(2, —2)). Solution The partial derivatives of f (x, y) are f x(x, y) = y3 + 2x and f y(x, y) -= 3xy2 With f(2, —2) = —12, f(2, —2) = —4, and f y(2, —2) = 24, the tangent plane through (2, —2, —12) has equation z = —12 — 4(x — 2) + 24(y + 2) This can be rewritten as z = 44 — 4x + 24y.

•

In single-variable calculus, a function f that is differentiable at a has a graph that, as you continuously zoom in at (a, f (a)), appears more and more like the tangent line to the graph at (a, f (a)). A similar situation holds for differentiable functions of two variables. Specifically, if f (x, y) is differentiable at (a, b), then as you continuously zoom in on the graph at P = (a, b, f (a, b)), it appears more and more like the tangent plane to the graph at P (Figure 6).

FIGURE 6 The graph looks more and more

like the tangent plane at P as we zoom in.

As Figure 6 suggests, differentiability at (a, b) implies that in a small region around P, the graph of f (x , y) is nearly indistinguishable from the tangent plane at P. That is to say, we can approximate f (x, y) near (a, b) by the linearization L(x, y). We have obtained the multivariable version of approximation by linearization: Approximating f (x, y) by Its Linearization If f (x, y) is differentiable at (a, b), and (x, y) is close to (a, b), then f (x , y) L(x , y). Thus,

f (x, y)

f(a, b)

fx(a,b)(x — a) + f y(a,b)(y — b)

2

828

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES

I.10 REMINDER The percentage error is equal to error actual value

Compare the approxi. EXAMPLE 4 Use linearization to approximate ((2.92)2) A/4 percentage error. estimate the value and calculator a mation with Solution We use the linearization of f (x, y) = (x2) ,,Fy centered at (3,4). We have

x2

x 100%

f y(x, y) = 1\5

and

fx(x, Y) -= (2x),5

Then with f (3, 4) = 18, f x (3, 4) = 12, and f(3,4) = 9/4 = 2.25, the linearization centered at (3,4) is L(x, y) = 18+ 12(x — 3) + 2.25(y — 4) Therefore, 18 + 12(2.92 — 3) + 2.25(4.08 — 4) = 17.24

(2.92)2) ,N/4.08

A calculator yields ((2.92)2) N/4.08 -,== 17.2225 rounded to four decimal places. The percentage error is 17.24 — 17.2225 x 100% 17.2225

•

0.10%

Recall in Examples 5 and 6 in Section 13.3, we examined a problem where we had a parallelogram of side lengths a and b and diagonal lengths x and y (Figure 7). We straightened the parallelogram into a rectangle that also had side lengths a and b and asked the question "How are the diagonal lengths in the rectangle related to the original diagonal lengths x and y?" Using vector geometry, we were able to answer that the x2

y2

± rectangle diagonal lengths are 2 . We went on to explore the situation where a carpenter, wanting to "square-up" a parallelogram with diagonal lengths x and y, uses the simpler expression -x-2E2 for the target diagonal length. We are now in a position to show that the carpenter's formula is

a

2

2

obtained via linearization of R(x, y) = A/X ±2 Y for positive values of x and y close to each other.

a FIGURE 7

EXAMPLE 5 Let R(x, y) = A/ X2 + 2 "Y2 . Use linearization to show that R(x, y) positive values of x and y that are close to each other.

x+ 2Y

for

Solution Carrying out an approximation for x and y close to each other means that we are considering (x, y) close to (a, a) for some a. Furthermore, we are assuming that a is positive since x, y > 0. To determine the linearization of R(x, y) centered at (a, a), compute the partial derivative with respect to x: R(x R , y) =

1 (x2

—1/2

± y2

2

2x

x

2

= 2

)

2

=

x2 + Y2

V2(x2 + y2)

The partial derivative with respect to y is obtained similarly: R(x, y) =

,

2(x2 + y2)

Since the linearization is centered at (a, a), we need R(x, y), Rx, and R v evaluated there: a2 _Fa2

R(a, a) =

Rx(a, a) =

=

2 a

-12(a2

a2)

=

1 2

=a

since a >0

and similarly,

Ry(a,

a) =

1

S EC TION 15.4

Differentiability, Tangent Planes, and Linear Approximation

829

Therefore, we have by Eq. (2):

1

R(x, y)

1

a ± —(x — a) + —(y —a) = 2 2

x

y

•

2

Like functions of a single variable, the approximation by linearization has associated linear approximation formulas for the change in f. If we let Ax and Ay represent small changes in x and y, and set x = a + Ax and y = b ± Ay, then from the linearization formula, Eq. (2), we obtain the Linear Approximation:

f(a ± Ax, b

f (61, b)

AY)

fy(a, b).6,y

f x(a,b)Ax

3

We can also write the Linear Approximation in terms of the change in f: Af = f (x, y) — f (a, b)

Af

BMI is one factor used to assess the risk of certain diseases such as diabetes and high blood pressure. The range 18.5 < / 0) = fy(o,o) = o

then lim (x,Y)—>(0,0)

f(x,y)

7

—o

x 2 ± y2

72. This exercise shows that there exists a function that is not differentiable at (0,0) even though all directional derivatives at (0,0) exist. Define f(x, y) = x2Y1(x2 + 3,2) for (x, y) 0 and f(0, 0) = 0. (a) Use the limit definition to show that D, f (0, 0) exists for all vectors v. Show that fx (0, 0) = f y (0, 0) = 0. (b) Prove that f is not differentiable at (0,0) by showing that Eq. (7) does not hold. 73. Prove that if f(x, y) is differentiable and Vf(x,y) = 0 for all (x, y), then f is constant.

FIGURE 16 The path r(t) is orthogonal to the level curves of f(x, y) = 2x2 + 8y2.

74. Prove the following Quotient Rule, where f, g are differentiable: f

gVf — f Vg

g2

77. (CAS) Find the curve y = g(x) passing through (0, 1) that crosses each level curve of f (x, y) = y sin x at a right angle. Using a computer algebra system, graph y = g(x) together with the level curves off.

15.6 Multivariable Calculus Chain Rules We have seen a few different chain rule formulas for functions involving multiple variables. In this section, we show how they all fall under a general scheme for identifying the structure of a composite function and determining the type of chain rule formula from the structure. To begin, we return to the Chain Rule for Paths and prove it. In the proof, we use the limit condition in the definition of differentiability, demonstrating why that limit is a necessary and important part of the concept of differentiability.

SECT ION 15.6

1.0 REMINDER We regard r(t) as representing both a vector, (x(t), y(t)) in the plane or (x(t),y(t),z(t)) in 3-space, and a point (x(t), y(t)) or (x(t), y(t),z(t)).

Multivariable Calculus Chain Rules

847

The Chain Rule for Paths applies to compositions f (r(t)), where f and r are differentiable. We primarily consider the cases where f is a function of x and y, and r(t) is a path in the plane, or f is a function of x, y, and z, and r(t) is a path in 3-space. THEOREM 1 Chain Rule for Paths

If f and r(t) are differentiable, then

— f (r(0) = Vfr(t) • r'(t) dt

In the cases of two and three variables, this chain rule states:

af dx

af dy

a x dt

ay dt

0f 8f 8f af dx \ • (x'(t), y'(t), z'(t)) = f(r(t)) = (Tx , Tz, / ax dt dt

af dy ay dt

1f

I af (x'(t) '4)) (r(t)) = \Tx'Tyl • Y

af dz az dt

We prove the theorem for the two-variable case. Proof By definition,

dt

f (r(t))

. f( x(t + h), y(t h rn

h)) — f (x(t), y(t))

To calculate this derivative, set Af = f (x(t

h), y(t

Ax = x(t + h) — x(t), e(x(t

h), y(t

h)) = f (x(t + h), y(t

h)) — f (x(t), y(t)) Ay = y(t + h) — y(t) h)) — (f (x(t), y(t))

f x(x(t), y(t))Ax

+ f y(x(t), y(t))Ay) The last term is the error, as in Section 15.4, in approximating f with its linearization centered at (x(t), y(t)). Putting these terms together, we have Af =- fx(x(t), Y(t))Ax + fy(x(t), Y(t))AY + e(x(t

h), y(t + h))

Now, set h = At and divide by At: e(x(t + At), y(t + At)) Ay Ax Af At + f y(x(t), y(t)) At + = fx(x(t), y(t))— At At We show below that the last term tends to zero as At desired result:

0. Given that, we obtain the

Af urn— Ar—>0 At

— f (r(t)) dt

Ax lim — = ix(x(t), At)) At-0 At

fy(x(t), y(t))

dy dx + fy(x(t), y(t))— = fx(x(t), y(t))— dt dt =

fr(f) • rf(t)

Ay . — At-0 At

848

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES

We verify that the last term tends to zero as follows: hill e(x(t + At), y(t + At)) . e(x(t + At), y(t + At)) = hm V(Ax)2 ± (Ay)2 At-÷0 At

At —o3

=- ( lim At -±0 ,

e(x(t + At), y(t + At))

lim

At-0

A/0,02 ± (60)2 Zero

(

V(Ax)2 + (Ay)2 At

( Ax 2 Ay 2 At ) ±( At ) ) = °

The first limit is zero, as indicated, because f is differentiable. The second limit is equal • to ,/x/(t)2 + y/(t)2, a finite value, and therefore the product is zero. d , where EXAMPLE 1 Calculate — f (r(t)) dt t=7112 and

f(x, y, z) = xy + z2

r(t) = (cost, sin 1, t)

Solution We have r( ) = (cos I, sin -'i, I) = (0, 1, i). Compute the gradient:

af af af ay az

Vf = (—ax, —, —) = (y, x, 2z) ,

Vfror /2) = Vf (0, 1, -7( ) = (1,0,70 2

Then compute the tangent vector:

r'( )

1)(0 = (- sin t, cos t, 1),

2

= (- sin -3T-, cos -7r-, 1) = (-1, 0, 1) 2 2

By the Chain Rule, d — f(r(t)) = Vfr( r/2) • r' (-7r ) = (1, 0, Tr) • (-1,0, 1) = IT — 1 dt 2 t=7r/2

•

Next, let's consider the case of more general composite functions. Suppose, for example, that x, y, z are differentiable functions of s and t—say, x = x(s, t), Y = Y(s, t), and z = z(s, t). The composition

f (x(s,t), As, t), z(s, 0)

1i

is then a function of s and t. We refer to s and t as the independent variables. EXAMPLE 2 Given that f(x, y, z) = xy + z and x = s2, y = Si', z -= t2, find the composite function. Solution We can keep track of which variable depends on which other variable by using a chart as in Figure 1. The composite function is given by

f (x(s,t), As, t), z(s, 0) = xy + z = (s2)(st) + t2 = s3 t + r2

FIGURE 1 Keeping track of the relationships between the variables.

•

The Chain Rule expresses the derivatives of f with respect to the independent variables. For example, the partial derivatives of f(x(s,t), y(s,t), z(s,t)) are

af as af dt

_ — — —

0f ax af ay af az a•x as ± ay as ± a•zTy af ax a•x at ± ay at ± a•z -a

2 3

SECTION 15.6

Note that we can obtain the formula for

af as

Multivariable Calculus Chain Rules

849

by labeling each edge in Figure 1 with

the partial derivative of the top variable with respect to the bottom variable as in Figure 2. af Then to obtain the formula for — , we consider each of the paths along the edges down as from f to s: the first through x, the second through y, and the third through z. Each path contributes a term to the formula, and those terms are added together. The first term, through x, is the product of the partial derivatives labeling the path's edges, giving af ax af ay az --. Similarly, the second term is — — , and the third term is — . Thus, we obtain FIGURE 2 Keeping track of the relationships between the variables.

ax as

ay as

az as

the formula

af = af ax as ax as

af ay ay as

af az az as

We obtain the formula for —in a similar manner. To prove these formulas, we

at

af

observe that — when evaluated at a point (so, to), is equal to the derivative with respect as s on the path obtained by fixing t = to and letting s vary. That path is r(s) = (x(s, to), As, to), z(s, to)) Fixing t = to and taking the derivative with respect to s, we obtain

d —(so, to) = — f(r(s)) as ds s=s0 The tangent vector is ax ay az e(s)= (— (s, to), —(s, to), _( st O)) as as as Therefore, by the Chain Rule for Paths,

af as

d = — f(r(s)) (so,to) ds

s=so

af ax af ay = Vf • e(so) = — — + — — + ax as

ay as

af az az as

The derivatives on the right are evaluated at (so, to). This proves Eq. (2). A similar argu,x,), where the ment proves Eq. (3), as well as the general case of a function f (xi, ,t . variables x, depend on independent variables ti, THEOREM 2 General Version of the Chain Rule Let f(xi,. . .,xn) be a differen, xn is a differtiable function of n variables. Suppose that each of the variables xi, . Then, for k . . . independent .t, . = variables of m 1, • • • , m, function entiable

af _ af axi atk axi at

af ax2 ax2 at,. -F. •

af ax„ ax„ atk

4

We keep track of the dependencies between the variables as in Figure 3. As an aid to remembering the Chain Rule, we will refer to

FIGURE 3 Keeping track of the dependencies between the variables. The term "primary derivative" is not standard. We use it in this section only, to clarify the structure of the Chain Rule.

af ax,

af ax„

as the primary derivatives. They are the components of the gradient V'f. By Eq. (4), the derivative of f with respect to the independent variable tk is equal to a sum of n terms: jth term:

af a ax;at,,

---h-

for j = 1,2, . . . ,n

850

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES

Note that we can write Eq. (4) as a dot product: af

I af

af

af

atk

\axi

ax2

axn

af atk

vf .

( axi ax2

ark

axn atk

(axi 8x2 atk ark

atk

axn ) atk

5

af Let f(x,y,z) = xy + z. Calculate — , where as x = s2, y = st, z = t 2

EXAMPLE 3 Using the Chain Rule

Solution We keep track of the dependencies of the variables as in Figure 2. Step /. Compute the primary derivatives.

af ay

x'

af —1 az

Step 2. Apply the Chain Rule. a 2 a a af = af ax ±af ay + af az az -a—s- = y Ts (s ) + x Ts. (st)+ .5;42) as ax as ay as = (y)(2s) + (x)(t) +0 = 2sy + xt This expresses the derivative in terms of both sets of variables. If desired, we can substitute x s2 and y = st to write the derivative in terms of s and t: af as = 2ys + xt = 2(st)s

(52)t = 3s2 t

To check this result, recall that in Example 2, we computed the composite function:

f(x(s,O,Y(s,t),z(s,t))= ps2,st,t2) =

S3t

t2

af From this, we see directly that — =3s2t, confirming our result. as

•

Lf

EXAMPLE 4 Evaluating the Derivative Let f(x, y) = exY . Evaluate at at (s,t,o= (2, 3, —1), where x st, y = s — ut2. Solution We keep track of the dependencies of the variables as in Figure 4. We can use either Eq. (4) or Eq. (5). We'll use the dot product form in Eq. (5). We have FIGURE 4 Keeping track of the dependencies between the variables.

a af Vf = (fTx , Ty ) = (yexY ,xexY) ,

ax ay \at , at = (s, —2ut) I \

and the Chain Rule gives us af ax ay — = Vf • —) = (ye , xexY) • (s, —2ut) at at at =- yexY(s)+ xexY (-2ut) =- (ys — 2xut)exY af To finish the problem, we do not have to rewrite — in terms of s, t, u. For (s,t,u) = at (2,3, —1), we obtain x = st = 2(3) = 6,

y -= s — ut2 = 2— (-1)(32) = 11

SECTION 15.6

Multivariable Calculus Chain Rules

851

With (s,t , u) = (2, 3, —1) and (x, y) = (6, 11), we have

af at

((11)(2) — 2(6)(— i)(3)) e6(11) = 58e66

= (ys — 2xut)exY (2,3,-1)

•

(2,3,-1)

EXAMPLE 5 Polar Coordinates (r, 0) be polar coordinates.

Let f (x, y) be a function of two variables, and let

a af af —f in terms of — and — ae ax ay af (b) Evaluate — at (x, y) = (1, 1) for f (x, y) = x 2 y. ae (a) Express

Solution (a) Since x = r cos

and y = r sin 0, ay — = r cos e

ax — = —r sin 0,

ae

ae

By the Chain Rule,

af af ax af ay ae = ax ae ± ay ae If you have studied quantum mechanics, you may recognize the right-hand side of Eq. (6) as the angular momentum operator (with respect to the z-axis) applied to the function f.

af

= —r sin0 Tx

af

r cos 0 Ty

af Since x = r cos° and y = r sin 0, we can write — in terms of x and y alone:

af

af

ae af TK

6

(b) Apply Eq. (6) to f (x, y) = x2y:

af 80 af de

a a = x 0y(x2 y) — y —(x2y) = x3 — 2xy2 ax --

1 3 — 2(1)(12 ) = — 1

•

(x,y)=(1,1)

Notice that the General Version of the Chain Rule encompasses the Chain Rule for Paths. In the Chain Rule for Paths, there is just one independent variable, which is the parameter for the path.

Implicit Differentiation In single-variable calculus, we used implicit differentiation to compute dyldx when y is defined implicitly as a function of x through an equation f (x, y) = 0. This method also works for functions of several variables. Suppose that z is defined implicitly by an equation F(x, y,z) = 0 Here we are treating z as a dependent variable with independent variables x and y. We could switch the roles of the variables and similarly work with y(x,z), where y is dependent while x and z are independent. Likewise, we could work with x(y, z).

Thus, z = z(x, y) is a function of x and y. We may not be able to solve explicitly for z(x, y), but we can treat F(x, y, z) as a composite function with x and y as independent variables and use the Chain Rule to differentiate implicitly with respect to x:

aF ax ax ax

0F 8y ay ax

aF az 0z 8x =°

We have ax/ax = 1 and also ay/ax = 0, since y does not depend on x. Thus,

aF — ax

0F — 8z = Fx + az = 0 ax a ax

852

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES 0, we may solve for az/az (we compute az/ay similarly):

If F-

az ay

Fy Fz

7

EXAMPLE 6 Calculate az/ax and az/ay at P = (1,1, 1), where F (x, y, z) = x2 ± y2 _ 2z2 ± 12x — 8z —4 = 0 What is the graphical interpretation of these partial derivatives? Solution We have Fx = 2x + 12, Hence, az Fx — = ——

ax

F,

Fy

=

2y,,

2x + 12 4z + 8

= —4z — 8

az

Fy

ay

Fz

2y 4z + 8

The derivatives at P = (1,1, 1) are az

0x FIGURE 5 The surface x2 ± y2 — 2z2 + 12x — 8z — 4 = O. A small patch of the surface around P can be represented as the graph of a function of x and y.

(1,1,1) -

2(1) + 12 14 7 = = 4(1) + 8 12 6'

az

aY

(1,1,1)

2 1 2(1) 4(1) + 8 = 12 =

Figure 5 shows the surface F(x,y,z) = 0. The surface as a whole is not the graph of a function f (x, y) because it fails the Vertical Line Test [that is, for some (x, y) there is more than one point (x, y, z) on the surface]. However, a small patch near P may be az az represented as a graph of a function z = f y), and the partial derivatives — and — x ay are equal to f x and f y. Implicit differentiation has enabled us to compute these partial derivatives without finding f (x, y) explicitly. • Assumptions Matter Implicit differentiation is based on the assumption that we can solve the equation F(x, y, z) = 0 for z in the form z = f (x, y). Otherwise, the partial az az derivatives — and — would have no meaning. The Implicit Function Theorem of

ax

ay

advanced calculus guarantees that this can be done (at least near a point P) if F has continuous partial derivatives and F(P) 0 0. Why is this condition necessary? Recall that the gradient vector V Fp = (Fx(P), F(P), F(P)) is normal to the surface at P, so F(P) = 0 means that the tangent plane at P is vertical. To see what can go wrong, consider the cylinder (shown in Figure 6):

FIGURE 6 Graph of the cylinder x2 + y2 _I = 0.

F(x, y, z) = x2 + y2 —1 = 0 In this particular case, F., is 0 for all (x, y, z). The z-coordinate on the cylinder does not depend on x or y, so it is impossible to represent the cylinder as a graph z = f(x, y) and az az the derivatives — and — do not exist.

ax

ay

15.6 SUMMARY • If f (x, y, z) is a function of x, y, z, and if x, y, z depend on two other variables, say, s and t, then

f (x, y, z) = f(x(s,t),y(s,t),z(s,t)) is a composite function of s and t. We refer to s and t as the independent variables. • The Chain Rule expresses the partial derivatives with respect to the independent variables s and t in terms of the primary derivatives:

af ax'

af ay'

af az

SECTION 15.6

Multivariable Calculus Chain Rules

853

Namely, af _ af ax af ay af az as ax as + ay as ±

af _ af ax af ay af az at ax at ± ay at ± az at

• In general, if f (xi, , xn ) is a function of n variables and if xi, independent variables ti,. , tm, then

0f _ af axi atk axi atk

, xn depend on the

af axn • • + axn atk

af ax2 ax2 atk

• The Chain Rule can be expressed as a dot product:

af at,,

af \axi ax2

af fax ax2 ax/ \atk ' atk

I af

ax„ —) • ' atk

Vf

• Implicit differentiation is used to find the partial derivatives az/ax and az/ay when z is defined implicitly by an equation F(x, y, z) = 0: az

Fx

ax

az ay =

Fy

15.6 EXERCISES Preliminary Questions , h1

1. Let f (x, y) = xy, where x = uv and y = u +v. (a) What are the primary derivatives off? (b) What are the independent variables? In Exercises 2 and 3, suppose that f(u,v)= ueu, where u = rs and V = r + S. 2. The composite function f (u, v) is equal to: (b) res (a) rser+s 3.

(c) rse"

What is the value of f(u, v) at (r , s) = (1,1)?

4. According to the Chain Rule, 8f/&r is equal to (choose the correct answer):

af ax

af ax

(a) — — + — — ax ar ax as

Exercises 1. Let f(x,y,z) = x2y3 +z4 and x = s2, y = st2, and z = 52 t.

‘c1

af ax af ay 8x 8r + 8y 8r 0f 8r ar ax

8f as

as ax

5. Suppose that x, y,z are functions of the independent variables u, v, w. Which of the following terms appear in the Chain Rule expression for Of/Ow? f h\ Of Ow af az (a) -a— Ow ax 8v av (c) Tz Ow 6. With notation as in the previous exercise, does 0x/8v appear in the Chain Rule expression for 8fl0u?

In Exercises 3-10, use the Chain Rule to calculate the partial derivatives. Express the answer in terms of the independent variables.

af af af (a) Calculate the primary derivatives 8x ay az Ox ay az (b) Calculate —, Os Os Os af using the Chain Rule: (C) Compute

Ts

af ay af az Of _ af ax Os ax as + ay as + az vTs

3.

Of Of „ _; j (x,y, z) = Os Or

4.

af af T at ; f(x,y,z) = xy + z2, x = r + s — 2t, y = 3rt , z = s2 r,—

5.

ag g Tx , Ty ; g(0, 0) = tan(0 + 0),

6.

8R aR R(x,y) = (x — 2y)3, x = Ov Ow

7.

aF —; F(u, v) = eu+v , u = x2, v = xy ay

8.

_af px,y)= x2 ±y2,x =e u+v, au

xy

z2, X =

s2, y = 2rs, z = r2

=x+y

= xy,

Express the answer in terms of the independent variables s, t. 2.

Let f (x, y) = x cos(y) and x

= u2

u2

andy

= U — V.

8f (a) Calculate the primary derivatives 8f ax ay (b) Use the Chain Rule to calculate Of/0v. Leave the answer in terms of both the dependent and the independent variables. (c) Determine (x, y) for (u, v) = (2, 1) and evaluate af/0v at (u, v) = (2,1).

y

w2,

=

y

u

=

vW

v

854

9.

DIFFERENTIATION IN SEVERAL VARIABLES

CHAPTER 15

an

—; h(x, y) = — , x = t2

a

tit2,

y=

20. Jessica and Matthew are running toward the point P along the straight paths that make a fixed angle of 6 (Figure 8). Suppose that Matthew runs with velocity va meters per second and Jessica with velocity vb meters per second. Let f (x, y) be the distance from Matthew to Jessica when Matthew is x meters from P and Jessica is y meters from P.

/2

af 10. — ; f (x, y, z) = xy — z2, x = r cos°, y = cos2 , z = r 00 In Exercises 11-16, use the Chain Rule to evaluate the partial derivative at the point specified. 11. Of/au and af/av at (u, v) = (-1,-1), where f (x, y, z) = x 3 = U2 v, y = u v 2 , z = uv

±

yz2,

(a) Show that f (x, y) = x 2 ± y2 — 2xy cos°. (b) Assume that 6 = 7/3. Use the Chain Rule to determine the rate at which the distance between Matthew and Jessica is changing when x = 30, y = 20, va = 4 m/s, and vb = 3 m/s.

12. Of/Os at (r , s) = (1, 0), where f (x,y) = ln(xy), x = 3r + 2s, y= 5r + 3s where

(r , 0) = (2,4, I), 13. ag/00 at x = r cos 8, y = r sin 14. dglds at s = 4, where g(x,

=

x2

g(x, y) =11(x + y2),

y2, x = s 2

Y)

y = 1 — 2s

15. ag/a u at (u, v) = (0, 1), where g(x, y) = x2 — y2, x = eu cos v, y = eu sin v ah 16. — at (q, r) = (3, 2), where h(u, v) = ueu , u = q3, v = qr2 aq 17. Given f (x,y) and y = y(x), we can define a composite function g(x) = f (x, y(x)). (a) Show that g'(x) = ia

Ox

+ ia y' By

(b) Let f (x, y) = x3 — xy2 and y(x) = 1 — x. With g(x) = f (x, y(x)), use the formula in (a) to determine g'(x), expressing the result in terms of x only. (c) With f (x, y) and y(x) as in (b), give an expression for g(x) in terms of x. Then compute g'(x) from g(x), and show that the result coincides with the one from (b). 18. Let

f (x, y) = 4 — x2y2 + e2x

and y(x) =

f(x,y(x)).

5.

Define g(x) =

(a) Use the derivative formula from Exercise 17(a) to prove that g'(x) = 0 and therefore that g is a constant function. (b) Express g(x) directly in terms of x, and simplify to show that g is indeed a constant function.

FIGURE 8 21. Two spacecraft are following paths in space given by ri = (sin t, t, t2) and r2 = (cost, 1 — t, t3). If the temperature for points in space is given by T(x, y, z) = x2y(1 — z), use the Chain Rule to determine the rate of change of the difference D in the temperatures the two spacecraft experience at time t = 7r. 22. The Law of Cosines states that c2 = a2 b2 — 2ab cos 0, where a, b,c are the sides of a triangle and 9 is the angle opposite the side of length c. (a) Compute ao/aa, mob, and a0/.9c using implicit differentiation. (b) Suppose that a = 10, b = 16, c = 22. Estimate the change in 0 if a and b are increased by 1 and c is increased by 2. 23. Let u = u(x, y), and let (r,0) be polar coordinates. Verify the relation 8

= 19. A baseball player hits the ball and then runs down the first base line at 20 ft/s. The first baseman fields the ball and then runs toward first base along the second base line at 18 ft/s as in Figure 7.

Hint: Compute the right-hand side by expressing uo and Ur in terms of u, and uy. 24. Let u(r,, 0) = r2 c052 O. Use Eq. (8) to compute II Vu112. Then compute Vu112 directly by observing that u(x, y) = x2, and compare. 25. Let x = s + t and y = s — t. Show that for any differentiable function

f(x>y), af \ 2

af )2 af af (ay = as at

26. Express the derivatives

Of

Of Of

Op'

ae '

in terms of

Of — Ox

Of Of ,— ,— ayOz

where (p, 0, 0) are spherical coordinates. FIGURE 7

Determine how fast the distance between the two players is changing at a moment when the hitter is 8 ft from first base and the first baseman is 6 ft from first base.

27. Suppose that z is defined implicitly as a function of x and y by the equation F(x, y, z) = xz2 + y2z + xy — 1= 0. (a) Calculate F,, Fy, F.

Oz

(b) Use Eq. (7) to calculate Tr and

Oz

SECTION 15.6

Multivariable Calculus Chain Rules

855

28. Calculate az/ax and az/ay at the points (3,2, 1) and (3,2, -1), where z is defined implicitly by the equation e t + z2x2 - y - 8 = 0. In Exercises 29-34, calculate the partial derivative using implicit differentiation. io

2

aX

am 30. — az az 31. —, ay

ar

x2y+y2z+xz

x`w + w3

aw , ay

2

+

3y z = 0

exY + sin(xz) + y = 0

at

32. — and —, ar at 33.

+ WZ

= 10

r

2

=

FIGURE 9 Graph of x2 + y2 - z2 - 12x - 8z -4 = 0.

te s/r

1

1

w2 +x 2

w2 ± y 2

40. For all x > 0, there is a unique value y = r(x) that solves the equation y3 + 4xy = 16. (a) Show that dyldx = -4y/(3y2 + 4x). (b) Let g(x) = f (x,r(x)), where f (x, y) is a function satisfying

= 1 at (x, y, w) = (1, 1, 1)

34. awn' and may, (T U - V)2 1n(W - UV) = 1n2 at (T,U,V,W) ,---- (1,1,2,4)

f(1,2) = 8,

35. Let r = (x,y, z) and er = r/lIr II. Show that if a function f (x, y, z) = F(r) depends only on the distance from the origin r = lirli = y + z2, then ,/x2 -2 9

Vf = Fr(r)er

Use the Chain Rule to calculate g'(1). Note that r(1) = 2 because (x, y) (1,2) satisfies y3 + 4xy = 16. 41. The pressure P. volume V, and temperature T of a van der Waals gas with n molecules (n constant) are related by the equation (

36. Let f (x, y z) = e- x2- Y2- z2 = e- r2 , with r as in Exercise 35. Compute Vf directly and using Eq. (9).

f( 1, 2) = 10

P+

an 2

)(V - nb) = nRT

where a, b, and R are constant. Calculate a P laT and av/a P. 42. When x, y, and z are related by an equation F(x, y, z) = 0, we sometimes write (0z/ax) in place of az/ax to indicate that in the differentiation, z is treated as a function of x with y held constant (and similarly for the other variables). (a) Use Eq. (7) to prove the cyclic relation

37. Use Eq. (9) to compute V (-1 ). r 38. Use Eq. (9) to compute V(ln r). 39. Figure 9 shows the graph of the equation F(x , y, z)

= x 2 + y2

z2

10

_ 12x - 8z -4 = 0

(a) Use the quadratic formula to solve for z as a function of x and y. This gives two formulas, depending on the choice of sign. (b) Which formula defines the portion of the surface satisfying z > -4? Which formula defines the portion satisfying z < -4? (c) Calculate az/ax using the formula z = f (x, y) (for both choices of sign) and again via implicit differentiation. Verify that the two answers agree.

("M

(2 - ) z ( Z)

x =

(b) Verify Eq. (10) for F(x, y, z) = x y + z =0. (c) Verify the cyclic relation for the variables P. V ,T in the Ideal Gas Law PV - nRT = 0 (n and R are constants). 43. Show that if_f(x)-is differentiable and c 0 0 is a constant, then -rt)-=- f(-x-T-- ct) satisfies the so-called advection equation au au — + c— = 0

at

ax

Further Insights and Challenges In Exercises 44-47, a function f (x, y, z) is called homogeneous of degree n if f (Xx, , Xz) = Xn f (x, y, z) for all X E R.

45. Prove that if f (x, y,z) is homogeneous of degree n, then f x(x, y,z) is homogeneous of degree n - 1. Hint: Either use the limit definition or apply the Chain Rule to f (Ax, AY, Az).

44. Show that the following functions are homogeneous and determine their degree:

46. Prove that if f (x,y,z) is homogeneous of degree n, then

(a) f (x, Y,z)

x2y + xyz

(c) f (x,y,z) = In ( -x-7Y,-) z`

(b) f (x, y,z) = 3x + 2y - 8z (d)

af Z— af = af +y— + x— ax ay az

nj

11

f(x,y,z) = z4 Hint: Let F(t) = f (tx,ty,tz) and calculate F'(1) using the Chain Rule.

856

DIFFERENTIATION IN SEVERAL VARIABLES

CHAPTER 15

harmonic. A function f(x,y) is called radial if f (X, y) = g(r), where = X2 -I- y2.

47. Verify Eq. (11) for the functions in Exercise 44. 48. Suppose that f is a function of x and y, where x = g(t,$),y = h(t,$). Show that f it is equal to

ax ) 2 + 2 fxy

frx

ax\

( (— at ) 82 y

a2x

)2

=-

4

a2g gr,

aXi

1 1 foe + —fr rr

13

53. Verify that f(x, y) = x and f (x y) = y are harmonic using both the rectangular and polar expressions for Af .

x? + • • • + x, and let g(r) be a function of r. Prove the for-

ag — „ = dXi

frr

52. Use Eq. (13) to show that f(x, y) = in r is harmonic.

at2

fY

fx at 2

49. Let r = mulas

ay

)+f(

51. Use Eq. (12) to prove that in polar coordinates (r,0),

= r 2 grr ±

54. Verify that f (x, y) = tan-1 Ix is harmonic using both the rectangular and polar expressions for Af..

r 2 - x•I2 gr 3

55. Use the Product Rule to show that 50. Prove that if g(r) is a function of r as in Exercise 49, then 82 g

a

2

= grr

+••

frr + — fr =

n -1

g

a ar

f

( —ra ) ar

gr

In Exercises 51-55, the Laplace operator is defined by if = fxx + fryA function f(x,y) satisfying the Laplace equation if = 0 is called

Use this formula to show that if f is a radial harmonic function, then rf, = C for some constant C. Conclude that f (x, y) = Gin r + b for some constant b.

15.7 Optimization in Several Variables Local and global / maximum Local maximum

Local and global minimum

Recall that optimization is the process of finding the extreme values of a function. This amounts to finding the highest and lowest points on the graph over a given domain. As we saw in the one-variable case, it is important to distinguish between local and global extreme values. A local extreme value is a value f (a, b) that is a maximum or minimum in some small open disk around (a, b) (Figure 1).

DEFINITION Local Extreme Values A function f (x, y) has a local extremum at P = (a, b) if there exists an open disk D(P, r) such that • Local maximum: f (x, y) < f (a, b) for all (x, y) E D(P, r) • Local minimum: f (X, y) > f (a , b) for all (x, y) E D(P, r)

Disk D(P, FIGURE 1 f (x, y) has a local maximum at P.

Fermat' s Theorem for functions of one variable states that if f (a) is a local extreme value, then a is a critical point and thus the tangent line (if it exists) is horizontal at x = a. A similar result holds for functions of two variables, but in this case, it is the tangent plane that must be horizontal (Figure 2). The tangent plane to z = f (x, y) at P (a, b) has equation

Imd REMINDER The term "extremum" (the plural is "extrema") means a minimum or maximum value.

z = f (a, b) + fx(a,b)(x — a) + fy(a, b)(y — b) Thus, the tangent plane is horizontal if f x (a, b) = f y(a, b) = 0—that is, if the equation reduces to z = f (a, b). This leads to the following definition of a critical point, where we take into account the possibility that one or both partial derivatives do not exist.

• More generally, (a1,.. .,a„) is a critical point of f(xi .. . . ..x) if each partial derivative satisfies

fx, (al , • • • , an ) = 0 or does not exist.

DEFINITION Critical Point critical point if:

• fx(a, b) =

A point P = (a, b) in the domain of f (x, y) is called a

0 or f x(a, b) does not exist, and • f y(a,b) = 0 or f y(a,b) does not exist.

SECTION 15.7

Optimization in Several Variables

Local maximum

Local maximum

FIGURE 2 The tangent line or plane is horizontal at a local extremum.

857

(B)

(A)

As in the single-variable case, we have the following: • Theorem 1 holds in any number of variables: Local extrema occur at critical points.

THEOREM 1 Fermat's Theorem If f (x, y) has a local minimum or maximum at P = (a, b), then (a, b) is a critical point of f (x , y). Proof If f (x , y) has a local minimum at P = (a, b), then f (x , y) > f (a, b) for all (x, y) near (a, b). In particular, there exists r > 0 such that f (x , > f (a, b) if lx —al < r. In other words, g(x) = f (x, b) has a local minimum at x = a. By Fermat's Theorem for functions of one variable, either g'(a) = 0 or g'(a) does not exist. Since e(a) = f x (a, b), we conclude that either f x(a,b) = 0 or f x (a,b) does not exist. Similarly, f y(a,b) = 0 or f y(a,b) does not exist. Therefore, P = (a, b) is a critical point. The case of a local • maximum is similar. In most cases, the partial derivatives exist for the functions f (x, y) we encounter. In such cases, finding the critical points amounts to solving the simultaneous equations f x (x , y) = 0 and f y(x, y) = 0. EXAMPLE 1 Show that f (x, y) = 11x2 — 2xy 2y2 + 3y has one critical point. Use Figure 3 to determine whether it corresponds to a local minimum or maximum. Solution The partial derivatives are

fx(x, y) = 22x

—2y,

f y(x, y) = —2x + 4y + 3

Set the partial derivatives equal to zero and solve: 22x — 2y = 0 D FIGURE 3 Graph of f(x, y) = 11x2 — 2xy

—2x + 4y + 3 = 0

2y2 +3y.

By the first equation, y

11x. Substituting y = 1 lx in the second equation gives —2x + 4(11x) + 3 = 42x + 3 =0

Thus, x = — and y = — -Pi . There is just one critical point, P = (— T1:4 , — 41 ). Figure 3 • shows that f (x, y) has a local minimum at P (that is, in fact, a global minimum). As the next example demonstrates, computational software can be of assistance in finding critical points. EXAMPLE 2

LCASJ

Determine the critical points of x —y

FIGURE 4 Graph of x —y

f(x,)7) =

2x2

8y2 ± 3 •

f (x, y) = 2x2 + 8y2 + 3 Are they local minima or maxima? Refer to Figure 4.

858

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES Solution We use a CAS to compute the partial derivatives, obtaining fx(x, Y) =

—2x2 + 8y2 + 4xy + 3 (2x2

8y2 + 3)2

f y(x, y) =

—2x2 + 8y2 — 16xy — 3 (2x2

8y2 + 3)2

To determine where the partial derivatives are zero, we set the numerators equal to zero: —2x2 + 8y2 + 4xy + 3 = 0 —2x2 + 8y2 — 16xy — 3 = 0 Figure 4 suggests that f (x, y) has a local max with x > 0 and a local min with x < 0. Using a CAS to solve the resulting system of equations, we have solutions at ( \ /i,

-V-3,-,-

and (—j,3 .140 3 ). The former is the local maximum we see in the figure; the latter is the local minimum •

I The discriminant is also referred to as the I "Hessian determinant."

We know that in one variable, a function f may have a point of inflection rather than a local extremum at a critical point. A similar phenomenon occurs in several variables. Each of the functions in Figure 5 has a critical point at (0,0). However, the function in Figure 5(C) has a saddle point, a critical point that is neither a local minimum nor a local maximum. If you stand at the saddle point and begin walking, some directions such as the +j or —j directions take you uphill and other directions such as the +i or —i directions take you downhill. As in the one-variable case, there is a Second Derivative Test determining the type of a critical point (a, b) of a function f (x, y) in two variables. This test relies on the sign of the discriminant D = D(a,b), defined as follows: D = D(a,b) = fxx(cl, b)fyy(a,b) — fiy (a, b) We can remember the formula for the discriminant by recognizing it as a determinant: =

(A) Local maximum

f„(a,b) f y,(a,b)

f xy(a,b) fyy(a,b)

(B) Local minimum

(C) Saddle

FIGURE 5 THEOREM 2 Second Derivative Test for f (x, y) Let P = (a, b) be a critical point of f (x, y). Assume that fxx, fyy, fxy are continuous near P. Then If D >0, then f xx (a, b) and f yy(a, b) must have the same sign, so the sign of f yy(a, b) also determines whether f (a, b) is a local minimum or a local maximum in the D > 0 case.

(i) If D (ii) If D (iii) If D (iv) If D

> 0 and f1( a, b) > 0, then f (a, b) is a local minimum. > 0 and f (a, b) A P II

Show that V.I.A(P) = ea], Note that we can derive this result without calculation: Because VfA (P) points in the direction of maximal increase, it must point directly away from A at P, and because the distance f A (x, y) increases at a rate of 1 as you move away from A along the line through A and P, Vf A(P) must be a unit vector.

b

that minimizes the sum of the squares (Figure 26): Distance fA(x, y) E(m,b) =

(y j

P =

— f(x j))2

(x, y)

j=1 = (a, b) Show that the minimum value of E occurs for m and b satisfying the two equations

(E)C1) J=i n

bn =

n

rnEX j=i

? + bEX

J

i=1

E y; n

j• — —

E xi y j j=1

FIGURE 27 The distance from A to P increases most rapidly in the direction eA p .

SECTION 15.8

Lagrange Multipliers: Optimizing with a Constraint

871

Further Insights and Challenges

E/J

63. In this exercise, we prove that for all x, y > 0:

64. The following problem was posed by Pierre de Fermat: Given three points A = (ai , a2), B = (b1, b2), and C = (ci, c2) in the plane, find the point P = (x, y) that minimizes the sum of the distances

1 1 A > xy — xa + a

f(x,y)= AP + BP +CP

where a > 1 and /3 > I are numbers such that a-1 + /3-1 = 1. To do this, we prove that the function f (x, y) = a-I xa + 13-1 yfl —xy satisfies f (x, y) > 0 for all x, y > 0. (a) Show that the set of critical points of f(x, y) is the curve y = xa-1 (Figure 28). Note that this curve can also be described as x = yfi-1. What is the value of f(x, y) at points on this curve? (b) Verify that the Second Derivative Test fails. Show, however, that for fixed b > 0, the function g(x) = f(x, b) is concave up with a critical point at x =b' '. (e) Conclude that for all x > 0, f (x, b) > f (bfi-1, b) = 0.

(V- 1, b) inc

Let e, f, g be the unit vectors pointing from P to the points A, B,C as in Figure 29. (a) Use Exercise 62 to show that the condition Vf (P) = 0 is equivalent to e+f+g= 0

(b) Show that f(x, y) is differentiable except at points A, B, C. Conclude that the minimum of f(x, y) occurs either at a point P satisfying Eq. (3) or at one of the points A, B, or C. (c) Prove that Eq. (3) holds if and only if P is the Fermat point, defined as the point P for which the angles between the segments AP, BP, CP are all 1200 (Figure 29). (d) Show that the Fermat point does not exist if one of the angles in 6,ABC is greater than 1200 . Where does the minimum occur in this case?

inc

------------------------------

Critical points of z =f(x, y)

FIGURE 28 The critical points of f (x,y) = form a curve y x"-1.

Constraint g(x, y) = 2x+3y-6=0 /

Point on the line closest to the origin

3

xa

15-1/ 3 — xy

(B) The Fermat point does not exist.

(A) P is the Fermat point (the angles between e, f, and g are all 1200).

FIGURE 29

15.8 Lagrange Multipliers: Optimizing with a Constraint Some optimization problems involve finding the extreme values of a function f(x, y) subject to a constraint g(x, y) = 0. Suppose that we want to find the point on the line closest to the origin (Figure 1). The distance from (x, y) to the origin is f(x, y) = x2 ± y2, so our problem is

2x + 3y =- 6

Minimize f (x, y) 13 FIGURE 1 Finding the minimum of f (x, y) = .1x2 ± y2 on the line 2x + 3y =6.

N/x2 +y 2

subject to

g(x, y) = 2x

+3y - 6 = 0

We are not seeking the minimum value of f(x, y) (which is 0), but rather the minimum among all points (x, y) that lie on the line. The method of Lagrange multipliers is a general procedure for solving optimization problems with a constraint. Here is a description of the main idea.

872

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES

GRAPHICAL INSIGHT Imagine standing at point Q in Figure 2(A). We want to increase the value of f while remaining on the constraint curve g(x, y) = 0. The gradient vector VfQ points in the direction of maximum increase, but we cannot move in the gradient direction because that would take us off the constraint curve. However, the gradient points to the right, so we can still increase f somewhat by moving to the right along the constraint curve. We keep moving to the right until we arrive at the point F, where Vfp is orthogonal to the constraint curve [Figure 2(B)1. Once at P, we cannot increase f further by moving either to the right or to the left along the constraint curve. Thus, f(P) is a local maximum subject to the constraint. Now, the vector Vgp is also orthogonal to the constraint curve because it is the gradient of g(x, y) at P and therefore is orthogonal to the level curve through P. Thus, Vfp and Vgp are parallel. In other words, Vfp = AVgp for some scalar A (called a Lagrange multiplier). Graphically, this means that a local max subject to the constraint occurs at points P where the level curves of f and g are tangent. The same holds for a local min subject to a constraint. Y

'

Tangent line at P

VfQ Level curves of z =f(x, )7) 4 3 2 1 -- Constraint curve g(x, y)= 0

(B)

(A) f increases as we move to the right along the constraint curve.

The local maximum off on the constraint curve occurs where Vf p and Vgp are parallel.

1:13 FIGURE 2

THEOREM 1 Lagrange Multipliers Assume that f(x,y) and g(x, y) are differentiable functions. If f (x, y) has a local minimum or a local maximum on the constraint curve g(x, y) = 0 at P = (a, b), and if Vgp 0 0, then there is a scalar A such that In Theorem 1, the assumption Vgp 0 guarantees (by the Implicit Function Theorem of advanced calculus) that we can parametrize the curve g(x, y) = 0 near P by a path r(t) such that r(0) = P and

ri(0) 0 O.

Vfp = XVgp

Proof Let r(t) be a parametrization of the constraint curve g(x, y) = 0 near P, chosen so that r(0) = P and 1.1(0) 0 0. Then f(r(0)) = f (P), and by assumption, f(r(t)) has a local min or max at t = 0. Thus, t = 0 is a critical point of f(r(t)) and -f(r(t))

= Vfp • r'(0) = 0 t=0

Chain Rule

This shows that Vfp is orthogonal to the tangent vector 1.1(0) to the curve g(x, y) = 0. The gradient Vgp is also orthogonal to e(0) [because Vgp is orthogonal to the level curve g(x, y) -= 0 at P]. We conclude that Vfp and Vgp are parallel, and hence Vfp is a multiple of Vgp as claimed. • We refer to Eq. (1) as the Lagrange condition. When we write this condition in terms of components, we obtain the Lagrange equations: f x(a,b) = Xgx(a,b) f y(a, b) =

y (a , b)

SECTION 15.8

Lagrange Multipliers: Optimizing with a Constraint

873

A point P = (a, b) satisfying these equations is called a critical point for the optimization problem with constraint and f (a, b) is called a critical value. EXAMPLE 1 Find the extreme values of f(x, y) = 2x ± 5y on the ellipse x) 2

(y ) 2

=1

Solution Step /. Write out the Lagrange equations. The constraint curve is g(x, y) -= 0, where g(x, y) = (x / 4)2 ± (y/3)2 — 1. We have Vf = (2,5) ,

Vg =

Ix

2y) — 9

The Lagrange equations Vfp = AVgp are Ax 2— — 8

Ix 2y ( —) 8 9

5=

X(2y) 9

2

Step 2. Solve for A in terms of x and y. Equation (2) gives us two equations for X: 16

X

45 X= — 2y

x

3

To justify dividing by x and y, note that x and y must be nonzero, because x = 0 or y = 0 would violate Eq. (2). Step 3. Solve for x and y using the constraint. 16 45 45 — or y = — x. Now The two expressions for X must be equal, so we obtain — 32 x 2y substitute this in the constraint equation and solve for x:

Level curve of f(x, y) = 2x + 5y Constraint curve g(x, y) = 0

31 ( 732 + (4-1

2

=

17 x2

±

16

where a level curve of f is tangent to the constraint curve g(x, y) = 0.

x2 (/ 289\ 289 )

1024 )_= 11

32 = ± — , and since y = — , the critical points are P = 32 17 289 45 32 and Q = (- 77,- 77).

Thus, x = al FIGURE 3 The min and max occur

225\ 1024)

' 1024

32 45 — j7) 17

Step 4. Calculate the critical values.

f(P)

f

45 32 732 45 ) = 17 T7 , 71 7-) = 2 (-17 ) + 5 (

and f (Q) = —17. We conclude that the maximum of f (x, y) on the ellipse is 17 and • the minimum is —17 (Figure 3). Assumptions Matter According to Theorem 3 in Section 15.7, a continuous function on a closed, bounded domain takes on extreme values. This tells us that if the constraint curve is closed and bounded (as in the previous example, where the constraint curve is an ellipse), then every continuous function f(x,y) takes on both a minimum and a maximum value subject to the constraint. Be aware, however, that extreme values need not exist if the constraint curve is not bounded. For example, the constraint x — y = 0

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES

is a line which is unbounded. The function f (x, y) = xy2 has neither a minimum nor a maximum subject to x — y = 0 because every point (a, a) satisfies the constraint, yet f (a, a) = a3 can be arbitrarily large positive (so there is no maximum) and arbitrarily large negative (so there is no minimum). EXAMPLE 2 Cobb—Douglas Production Function By investing x units of labor and y units of capital, a watch manufacturer can produce P(x, y) = 50x°A y" watches. Find the maximum number of watches that can be produced on a budget of $20,000 if labor costs $100 per unit and capital costs $200 per unit. y (capital) Increasing output 120

Solution The total cost of x units of labor and y units of capital is 100x + 200y. Our task is to maximize the function P(x, y) = 50x0.4 y0.6 subject to the following budget constraint (Figure 4): g(x, y) = 100x +200y —20,000 = 0

Budget constraint

60

Step I. Write out the Lagrange equations.

/

P(x, ,Y) = Agx(x, y) : 20x —°.6y0.6 x (labor)

40

80

Contour plot of the Cobb—Douglas production function 50x0.4y0.6. The level curves of a production function are called isoquants. y)

P(x,

120

FIGURE 4

4

Xg

y)

y

y)

:

30x

OA

)r

= 100A

OA

=

200A.

Step 2. Solve for A in terms of x and y. These equations yield two expressions for A that must be equal:

=

1 y )0.6 5 x

P(x,

3 (

y )

—0.4

5

20 x

Step 3. Solve for x and y using the constraint. Multiply Eq. (5) by 5(y/x)0.4 to obtain y/x = 15/20, or y = ix. Then substitute in Eq. (4): 3 100x + 200y = 100x + 200 (-4 x) = 20,000

Global maximum

Q Maximum on the constraint curve

z = f(x, y)

Constrained max occurs here g(x, y)= 0 y (A)

(B) FIGURE 5

=

250x = 20,000

20,000 = 80 and y = ix = 60. The critical point is A = (80, 60). 250 Step 4. Calculate the critical values. Since P(x, y) is increasing as a function of x and y, VP points to the northeast, and it is clear that P(x, y) takes on a maximum value at A (Figure 4). The maximum is P(80, 60) = 50(80)04(60)0.6 = 3365.87, or roughly 3365 watches, with a cost per 20,000 watch of or about $5.94. • 3365 We obtain x =

GRAPHICAL INSIGHT In an ordinary optimization problem without constraint, the global maximum value is the height of the highest point on the surface z = f (x, y) [point Q in Figure 5(A)]. When a constraint is given, we restrict our attention to the curve on the surface lying above the constraint curve g(x, y) = 0 . The maximum value subject to the constraint is the height of the highest point on this curve. Figure 5(B) shows the optimization problem solved in Example 1. The method of Lagrange multipliers is valid in any number of variables. Imagine, for instance, that we are trying to find the maximum temperature f (x, y, z) for points on a surface S in 3-space given by g(x, y, z) = 0, as in Figure 6. This surface is a level surface for the function g, and therefore, Vgp is perpendicular to the tangent planes to this surface at every point P on the surface. Consider the level surfaces for temperature, which we have called the isotherms. They appear as surfaces in 3-space, and their intersections with S yield the level sets of temperature on S. If, as in the figure, the temperature increases as we move to the right on the surface, then it is apparent that the maximum temperature

SECTION 15.8

Lagrange Multipliers: Optimizing with a Constraint

875

for the surface occurs when the last isotherm intersects the surface in just a single point and hence that isotherm is tangent to the surface. That is to say, the last isotherm and the surface share the same tangent plane at their single point of intersection. However, as we know, Vfp is always perpendicular to the tangent plane to the level surfaces for f at each point P on a level surface. So at the hottest point on the surface, Vgp and Vfp are both perpendicular to the same tangent plane. Hence, they must be parallel, and one must be a multiple of the other. Thus, at that point, Vfp = XVgp. A similar argument holds for the minimum temperature on the surface. Vf Vf=0 \

Vf Vg

Vf Vg

g =0

Vf

Vg

g=0

Vf Vg

f=4

=2

f=3

Vg f= 1

f=4 f

f =2 f=2

As we move to the right, temperature increases, attaining a maximum on the surface off = 4 at P.

FIGURE 6

3

f =3

As we move to the right, temperature increases and then decreases.

CI FIGURE 7

There is one other situation to consider. Imagine that as we move left to right across our surface, temperature first increases to f = 4 and then it decreases again, as in Figure 7. There is a collection of points with the maximal temperature of f = 4. In this case, Vf must point to the right on isotherms that are to the left of f = 4 since this is the direction of increasing temperature, and Vf must point to the left on isotherms that are to the right of f = 4 since this is the direction of increasing temperature. Hence, in order for the right-pointing gradient vectors to become left-pointing gradient vectors in a continuous manner, they must be equal to 0 on the f = 4 isotherm. This makes sense, since on that isotherm there is no direction of increasing temperature. So for all of the points on the surface with maximal temperature, of which there are many, the equation Vf = ?Vg is satisfied, but by taking = 0. In the next example, we consider a problem in three variables. EXAMPLE 3 Lagrange Multipliers in Three Variables Find the point on the plane x y z — + — + — = 1 closest to the origin in R3. 2 4 4 subject to the conSolution Our task is to minimize the distance d = V x 2 + z . x y stramt — + — + — = 1. But, finding the minimum distance d is the same as finding the 4 2 4 minimum square of the distance d2, so our problem can be stated: y2

Minimize f(x, y, z) = x 2 ±

y2 ± z2

subject to

Vf

- 2 Z

z x y g(x, y, z) = — + — + — — 1 = 0 2 4 4

The Lagrange condition is (2x , 2y, 2z) =

±

1 1 l\ 2 4 4 Vg

876

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES

This yields = 4x =- 8y = 8z

=

Substituting in the constraint equation, we obtain x (0, 4, 0)

FIGURE 8 Point P closest to the origin on the plane.

y

z

2z

z

z

3z

,

z=

2

Thus, x = 2z = and y = z =4. This critical point must correspond to the minimum of f. There is no maximum of f on the plane since there are points on the plane that are arbitrarily far from the origin. Hence, the point on the plane closest to the origin is • P = 3 ' 3' 3 (Figure 8). The method of Lagrange multipliers can be used when there is more than one constraint equation, but we must add another multiplier for each additional constraint. For example, if the problem is to minimize f(x, y, z) subject to constraints g(x,y, z) = 0 and h(x, y, z) = 0, then the Lagrange condition is Vf = AVg

I The intersection of a sphere with a plane I through its center is called a great circle.

µ,Vh

EXAMPLE 4 Lagrange Multipliers with Multiple Constraints The intersection of the plane x + + z= 0 with the unit sphere x2 + y2 + z2 = 1 is a great circle (Figure 9). Find the point on this great circle with the greatest x-coordinate. Solution Our task is to maximize the function f (x, y, z) = x subject to the two constraint equations 1 1 g(x,y,z) = x + —y + —z = 0, 2 3

h(x,y,z)

=

1 =

x2 ± y2 ± z2

0

The Lagrange condition is Vf = AV'g + µVh (1,0,0)

2' 3

Note that µ cannot be zero, since, if it were, the Lagrange condition would become (1, 0, 0) = A(1, 1), and this equation is not satisfed for any value of A. Now, the Lagrange condition gives us three equations: 1 —A + 2ity = 0, 2

A + 2ptx = 1,

1 — ± 2,ttz = 0 3

The last two equations yield A = —4,uy and A = —6ptz. Because tt 0 0, FIGURE 9 The plane intersects the sphere in a great circle. Q is the point on this great circle with the greatest x-coordinate.

—4y= —6itz

=

y=

3 —z

2

Now use this relation in the first constraint equation: 1 1 1 /3 \ 1 x+-2 Y+3 z=x+ n -2 z) +-3z=13

13 x= --12 z

Finally, we can substitute in the second constraint equation: 13 z 2 + x2 ±

y2 ± Z2 — 1 = (— — 12

to obtain .1447 z2 = 1 or z = P=

18 7

7N/ff

12

3z 2 ± 2

2

=

0

s• •

13 Since x = — — 12 z and y = 23 z, the critical points are

12

Q=

(il -j

7

18

12 )

SECTION 15.8

Lagrange Multipliers: Optimizing with a Constraint

877

The critical point with the greatest x-coordinate [the maximum value of f(x, y, z)] is Q V-13 • 0.515. with x -coordinate 7

15.8 SUMMARY • Method of Lagrange multipliers: The local extreme values of f(x, y) subject to a constraint g(x, y) = 0 occur at points P (called critical points) satisfying the Lagrange condition Vfp = WVgp. This condition is equivalent to the Lagrange equations fy(X,y) = Agy(x, y)

f(x, y) = Agx(x Y),

• If the constraint curve g(x, y) = 0 is bounded [e.g., if g(x, y) = 0 is a circle or ellipse], then global minimum and maximum values of f subject to the constraint exist. • Lagrange condition for a function of three variables f(x, y, z) subject to two constraints g(x, y, z) = 0 and h(x, y, z) = 0: Vf = XVg

p,Vh

15.8 EXERCISES Preliminary Questions 1. Suppose that the maximum

of f(x,y) subject to the constraint

g(x,y)= 0 occurs at a point P = (a, b) such that Vfp 0 0. Which of the

following statements is true? (a) Vfp is tangent to g(x,y)= Oat P. (b) Vfp is orthogonal to g(x,y)--= Oat P.

3. On the contour map in Figure 11: (a) Identify the points where Vf = AVg for some scalar A. (b) Identify the minimum and maximum values of f(x, y) subject to g(x,y) = 0.

2. Figure 10 shows a constraint g(x, y)= 0 and the level curves of a function f. In each case, determine whether f has a local minimum, a local maximum, or neither at the labeled point.

g(x, y)= 0

g(x, y)= 0

FIGURE 10

Contour plot of f(x, y) (contour interval 2) FIGURE 11

Exercises In this exercise set, use the method of Lagrange multipliers unless otherwise stated. 1. Find the extreme values of the function f(x, y) = 2x + 4y subject to the constraint g(x, y) = x 2 ± y2 5 = 0. (a) Show that the Lagrange equation Vf = XVg gives Ax = 1 and Ay = 2. (b) Show that these equations imply A 0 and y = 2x. (c) Use the constraint equation to determine the possible critical points (x, y). (d) Evaluate f(x, y) at the critical points and determine the minimum and maximum values.

2. Find the extreme values of f(x,y)= x2 + 2y2 subject to the constraint g(x,)')= 4x — 6y = 25. (a) Show that the Lagrange equations yield 2x = 4A, 4y = —6A. (b) Show that if x = 0 or y = 0, then the Lagrange equations give x = y = 0. Since (0,0) does not satisfy the constraint, you may assume that x and y are nonzero. (c) Use the Lagrange equations to show that y -= —

ix.

(d) Substitute in the constraint equation to show that there is a unique critical point P.

DIFFERENTIATION IN SEVERAL VARIABLES

CHAPTER 15

878

(e) Does P correspond to a minimum or maximum value of f? Refer to Figure 12 to justify your answer. Hint: Do the values of f (x, y) increase or decrease as (x, y) moves away from P along the line g(x, y) = 0?

4—

FIGURE 13 Contour map of f (x, y) = x3 + xy + y3 and graph of the constraint g(x, y) = x3 — xy + y3 = 1.

—4 — x —4

0

8

FIGURE 12 Level curves of f (x, y) = x2 + 2y2 and graph of the constraint g(x, y) = 4x — 6y — 25 = 0.

3. Apply the method of Lagrange multipliers to the function f (x, y) = (x2 + 1)y subject to the constraint x2 + y2 = 5. Hint: First show that y 0; then treat the cases x = 0 and x 0 separately. In Exercises 4-15, find the minimum and maximum values of the function subject to the given constraint. 4. 5.

x2 + y2,

2x + 3y =6

(c) Does a cone with given volume V and maximum surface area exist?

f

(X

,

y)

=

f (x, y) = xy,

8.

f (x y) = x2 y +x + y,

12. 13.

(a) Determine the ratio h I r for the cone with given surface area S and maximum volume V. (b) What is the ratio h r for a cone with given volume V and minimum surface area S?

7.

11.

19. The surface area of a right-circular cone of radius r and height h is S= h2, and its volume is V = itr2h.

x2 + y2 = 4

f (x, y) = 4x2 + 9y2, xy = 4

10.

18. Find the rectangular box of maximum volume if the sum of the lengths of the edges is 300 cm.

f(x, y) = 2x + 3y,

6.

9.

17. Find the point (a, b) on the graph of y = ex where the value ab is the least.

f

(x

,

y)

=

y)

=

4x2 + 9y2 = 32

x2 + y2, x2y4,

xy = 4 x4 + y4 = 1

=

f (x, y, z) =

x2

—y

xy + 2z,

z,

f(x,y,Z) = xy

16.

Let

x2 + 6y2 + 3xy = 40

X2 +2y2

x2

6z2 = 1

with the greatest x-coordinate (Figure 14).

y2 + z = 0

x2 ± y2 ± z2 =36

14. f (x, y, z) = x2 + y2 + z2, 15.

21. Find the point on the ellipse

x2 + 2y2 = 6

f (x, y, z) = 3x + 2y + 4z, z)

20. In Example 1, we found the maximum of f (x, y) = 2x + 5y on the ellipse (x/4)2 + (y/3)2 = 1. Solve this problem again without using Lagrange multipliers. First, show that the ellipse is parametrized by x = 4 cos t, y = 3 sin t. Then find the maximum value off (4 cos t, 3 sin t) using single-variable calculus. Is one method easier than the other?

X + 3y + 2z =36

=4 xz, X2+y2+z2

f (x, y) = X3 + xy

y3,

g(x, y) = x3 —xy +y3

(a) Show that there is a unique point P = (a, b) on g(x, y) = I where Vfp = AVgp for some scalar A. (b) Refer to Figure 13 to determine whether f (P) is a local minimum or a local maximum of f subject to the constraint. (c) Does Figure 13 suggest that f (P) is a global extremum subject to the constraint?

FIGURE 14 Graph of x2 + 6y2 + 3xy = 40.

22. Use Lagrange multipliers to find the maximum area of a rectangle inscribed in the ellipse (Figure 15): X2

y2 ±

=1

SECT I ON 15.8

Lagrange Multipliers: Optimizing with a Constraint

879

32. E,4 In a contest, a runner starting at A must touch a point P along a river and then run to B in the shortest time possible (Figure 17). The runner should choose the point P that minimizes the total length of the path. (a) Define a function f (x, y) = AP + P B,

where P = (x, y)

Rephrase the runner's problem as a constrained optimization problem, assuming that the river is given by an equation g(x,y) = 0. FIGURE 15 Rectangle inscribed in the ellipse

23. Find the point (xo, yo) on the line 4x origin.

x2

y2 + —2- = 1. a2 b

9y = 12 that is closest to the

(1)) Explain why the level curves of f (x,y) are ellipses. (c) Use Lagrange multipliers to justify the following statement: The ellipse through the point P minimizing the length of the path is tangent to the river. (d) Identify the point on the river in Figure 17 for which the length is minimal.

24. Show that the point (xo, yo) closest to the origin on the line ax + by = c has coordinates xo =-- a2

ac

River

bc

b2 ,

= a2

b2

25. Find the maximum value of f (x, y) = x' y' for x > 0, y > 0 on the line x + y = 1, where a, b > 0 are constants. 26. Show that the maximum value of f (x, y) = x2y3 on the unit circle is 6/i 5 27. Find the maximum value of f (x, y) = xa yb for x > 0, y > 0 on the unit circle, where a, b > 0 are constants.

FIGURE 17

28. Find the maximum value of f (x, y, z) = xa yb zc for x, y,z > 0 on the unit sphere, where a, b, c > 0 are constants.

In Exercises 33 and 34, let V be the volume of a can of radius r and height h, and let S be its surface area (including the top and bottom).

29. Show that the minimum distance from the origin to a point on the plane ax + by + cz = d is

33. Find r and h that minimize S subject to the constraint V = 54m.

,s/ a 2± b2

17

Show that for both of the following two problems, P = (r ,h) is 34. a Lagrange critical point if h = 2r:

Idl e2

30. Antonio has $5.00 to spend on a lunch consisting of hamburgers ($1.50 each) and french fries ($1.00 per order). Antonio's satisfaction from eating xi hamburgers and x2 orders of french fries is measured by a funcHow much of each type of food should he purtion U(xi, x2) = chase to maximize his satisfaction? (Assume that fractional amounts of each food can be purchased.) 31. E4 Let Q be the point on an ellipse closest to a given point P outside the ellipse. It was known to the Greek mathematician Apollonius (third century BCE) that P Q is perpendicular to the tangent to the ellipse at Q (Figure 16). Explain in words why this conclusion is a consequence of the method of Lagrange multipliers. Hint: The circles centered at P are level curves of the function to be minimized

• Minimize surface area S for fixed volume V. • Maximize volume V for fixed surface area S. Then use the contour plots in Figure 18 to explain why S has a minimum for fixed V but no maximum and, similarly, V has a maximum for fixed S but no minimum.

Increasing S Level curves of S

Critical point P = (r, h) Level curve of V Increasing V r FIGURE 18

FIGURE 16

35. Figure 19 depicts a tetrahedron whose faces lie in the coordinate planes and in the plane with equation — + — + — = 1 (a, b, c > 0). The c a b volume of the tetrahedron is given by V = abc. Find the minimum value of V among all planes passing through the point P = (1,1,1).

880

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES

FIGURE 20 FIGURE 19

40. Find the maximum value of f (x y, z) = xy + xz + yz — xyz subject to the constraint x + y + z = 1, for x > 0, y > 0, z > 0.

36. With the same set-up as in the previous problem, find the plane that minimizes V if the plane is constrained to pass through a point P = (a, y) with a,I3,y >0.

41. Find the minimum of f (x,y,z) = x2 ± y2 constraints x + y + z = 1 and x + 2y + 3z = 6.

37. Show that the Lagrange equations for f (x, y) = x + y subject to the constraint g(x, y) = x + 2y = 0 have no solution. What can you conclude about the minimum and maximum values off subject to g = 0? Show this directly.

42. Find the maximum of f(x, y, z) = z subject to the two constraints x2 +y2 = 1 and x +y +z = 1.

Show that the Lagrange equations for f (x, y) = 2x + y subject 38. to the constraint g(x, y) = x2 y2 = 1 have a solution but that f has no min or max on the constraint curve. Does this contradict Theorem 1? 39. Let L be the minimum length of a ladder that can reach over a fence of height h to a wall located a distance b behind the wall. (a) Use Lagrange multipliers to show that L = (h2/3 4. b213)312 (Figure 20). Hint: Show that the problem amounts to minimizing f (x, y) = (x + b)2 + (y + h)2 subject to ylb = hlx or xy = bh. (b) Show that the value of L is also equal to the radius of the circle with center (—b, —h) that is tangent to the graph of xy = bh.

± z2

subject to the two

43. Find the point lying on the intersection of the plane x + y + and the sphere x2 + y2 + z2 = 9 with the greatest z-coordinate.

=0

44. Find the maximum of f (x, y, z) =x+y+z subject to the two constraints x2 + y2 + z2 = 9 and x2 + 1)12 + 4z2 = 9. 45. The cylinder x2 + y2 = 1 intersects the plane x + z = 1 in an ellipse. Find the point on such an ellipse that is farthest from the origin. 46. Find the minimum and maximum of f (x, y, z) = y + 2z subject to two constraints, 2x + z = 4 and x2 + y2 = 1. 47. Find the minimum value of f (x, y, z) = x2 + y2 + z2 subject to two constraints, x + 2y + z = 3 and x — y = 4.

Further Insights and Challenges 48. Suppose that both f (x, y) and the constraint function g(x, y) are linear. Use contour maps to explain why f (x , y) does not have a maximum subject to g(x, y) = 0 unless g = af + b for some constants a, b. 49. Assumptions Matter Consider the problem of minimizing f (x, y) = x subject to g(x, y) = (x — 1)3 — y2 = 0. (a) Show, without using calculus, that the minimum occurs at P = (1,0). (b) Show that the Lagrange condition V fp = AVgp is not satisfied for any value of A. (c) Does this contradict Theorem 1? 50. Marginal Utility Goods 1 and 2 are available at dollar prices of p1 per unit of Good 1 and p2 per unit of Good 2. A utility function U(xi, x2) is a function representing the utility or benefit of consuming xj units of good j. The marginal utility of the jth good is a U/ax j, the rate of increase in utility per unit increase in the jth good. Prove the following law of economics: Given a budget of L dollars, utility is maximized at the consumption level (a, b) where the ratio of marginal utility is equal to the ratio of prices:

(b) Calculate the value of the Lagrange multiplier A occurring in (a). (c) Prove the following interpretation: A is the rate of increase in utility per unit increase in total budget c. 52. This exercise shows that the multiplier A may be interpreted as a rate of change in general. Assume that the maximum of f (x, y) subject to g(x, y) = c occurs at a point P. Then P depends on the value of c, so we may write P = (x(c),y(c)) and we have g(x(c), y(c)) = c. (a) Show that Vg(x(c), y(c)) • (x'(c),y'(c)) = 1 Hint: Differentiate the equation g(x(c), y(c)) = c with respect to c using the Chain Rule. (b) Use the Chain Rule and the Lagrange condition Vfp = AVgp to show that

marginal utility of Good 1 Uxl (a,b) _ Pi marginal utility of Good 2 — Ux2 (a, b) — P2 51. Consider the utility function U(xl x2) = xi x2 with budget constraint

pixi + p2x2 = C.

(a) Show that the maximum of U(xi, x2) subject to the budget constraint is equal to c2 A4P i P2).

dc

f (x(c), y(c)) =

(c) Conclude that A is the rate of increase in f per unit increase in the "budget level" c.

Chapter Review Exercises

Show that there is a constant p, such that xi = A-1e4Ei for i = 1,2,3, where A = N-1(epE,

53. Let B > 0. Show that the maximum of

eiLE3).

XIX2 • • • Xn

f(Xl, • • • ,Xn)

subject to the constraints xi + • • • + xn = B and xi > 0 for j = 1, . . . , n occurs for xi = • • • = xn = Bln. Use this to conclude that (aia2 • • • a„)I /n < al + •

56. Boltzmann Distribution Generalize Exercise 55 to n variables: Show that there is a constant µ such that the maximum of

+ an

for all positive numbers ai, • • • an •

S = xi ln xi + • • • + xn ln xn subject to the constraints

54. Let B > O. Show that the maximum of f (xi • • • xn) = xi + • • • + xn subject to xf + • • • + xn2 = B2 is la. Conclude that ± • • • + an2)I/2 lai I + • • • + Ian I < ,Fn(a?

xi + • • • + xn = N,

E2, E3,

subject to two constraints: xi + X2

+ X3 =

Eixi + E2x2 + E3x3 = E

N,

A = N-i (ep.Ei + • • • + eikEn)

consider the maximum of

x2, x3) = xi In xi + x2 ln x2 + x3 1n x3

Eixi + • • • + Enxn = E

occurs for xi = A-I elLEi , where

for all numbers al, • • • , an • 55. Given constants E, E1,

This result lies at the heart of statistical mechanics. It is used to determine the distribution of velocities of gas molecules at temperature T; xi is the number of molecules with kinetic energy Ei; = -(k T) ', where k is Boltzmann's constant. The quantity S is called the entropy.

CHAPTER REVIEW EXERCISES ,./x2 _ y2 1. Given f (x, y) =

x+3 (a) Sketch the domain off.

•

(b) Calculate f(3, 1) and f(-5, -3). (c) Find a point satisfying f (x, y) = 1. 2.

Find the domain and range of:

(a)

f(x,y,z) =

+

(c) What are the signs of

A and f y at D?

(d) At which of the labeled points are both 7. Describe the level curves of: e4x-y (a) f (x, Y)=

(c)

f (x,y)

=

3x

2

4y

A and f y negative?

(b) f (x, y) = ln(4x - y) (d) f (x, y) = x + y2

2

—z

(b) f (x, y) = ln(4x2 —y) 3. Sketch the graph f (x, y) = x2 - y + 1 and describe its vertical and horizontal traces. 4. I, CAS Use a graphing utility to draw the graph of the function cos(x2 y2 ),1_ ' in the domains [-1,11 x [-1,11, [-2,2] x [-2,2], and [-3,3] x [-3,3], and explain its behavior.

(B)

(A)

5. Match the functions (a)-(d) with their graphs in Figure 1. (a) f (x, y) = x2 + y (b)

f (x, Y) = x2 +

(c)

f (x, y) = sin(4xy)e-x2- Y2

4y2

(d) f (x, y) = sin(4x)e-x2 6.

881

-y2

Referring to the contour map in Figure 2:

(a) Estimate the average rate of change of elevation from A to B and from A to D. (b) Estimate the directional derivative at A in the direction of v.

(D)

(C) FIGURE 1

882

CHAPTER 15

DIFFERENTIATION IN SEVERAL VARIABLES

In Exercises 17-20, compute f x and f y. 17. f (x, y) = 2x + y2

18. f (x, y) = 4xy3

19. f (x, y) = sin(xy)e—x-

20. f (x, y) = ln(x2 + xy2)

21. Calculate f xxy, for f (x, y, z) = y sin(x + z). 22. Fix c> 0. Show that for any constants a, 13, the function u(t, x) = sin(act + 13) sin(ax) satisfies the wave equation Contour interval =50 m

0

2 km

FIGURE 2

.9 2u

2 a2u

at2

c ax2

23. Find an equation of the tangent plane to the graph of f (x,y) = xy2 — xy + 3x3y at P = (1,3).

8. Match each function (a)—(c) with its contour graph (i)—(iii) in Figure 3:

(a) f (x, y) = xy

24. Suppose that f (4, 4) = 3 and f x (4, 4) = f y (4, 4) = —1. Use the Linear Approximation to estimate f (4.1, 4) and f(3.88, 4.03). 25. Use a Linear Approximation of f (x, y, z) = x2 + y2 + z to estimate V7.12 + 4.92 + 69.5. Compare with a calculator value.

(b) f (x, y) = exY

(c) f (x, y) = sin(xy)

26. The plane z = 2x — y — 1 is tangent to the graph of z = f (x, y) at P = (5,3). (a) Determine f(5, 3), f(5, 3), and .6(5, 3). (b) Approximate f(5.2, 2.9). 27. Figure 4 shows the contour map of a function f (x, y) together with a path r(t) in the counterclockwise direction. The points r(1), r(2), and r(3) are indicated on the path. Let g(t) = f (r(t)). Which of statements (i)—(iv) are true? Explain. (i) g/(1) > 0. (ii) g(t) has a local minimum for some 1 < t < 2. (iii) g'(2) = 0. (iv) gl(3) = 0.

0 FIGURE 3

In Exercises 9-14, evaluate the limit or state that it does not exist. 9.

lim (xy + y 2 ) (x,y)—>(1,-3)

10.

lim ln(3x (x,y)—>(1,-3)

11.

xy xy2 lim (x,y)—>(0,0) x2 + y2

12.

lim (x,y)-4(0,0)

13.

lira (2x + y)e—x+Y (x,y)—>(1,-3)

14.

lim (x,y)—> (0,2)

x3y2

y) x2y3

x4 + y4

(ex — 1)(eY — 1)

15. Let

FIGURE 4

f (x, y) =

I

(xy)P x 4 ± ya 0

(x, Y) 0 (0,0) (x, y) = (0,0)

Use polar coordinates to show that f (x , y) is continuous at all (x, y) if p> 2 but is discontinuous at (0,0) if p 0 on a domain D in the xy-plane, the integral represents the volume of the solid region between the graph of f (x , y) and the xy-plane (Figure 1). More generally, the integral represents a signed volume, where positive contributions arise from regions above the xy-plane and negative contributions from regions below. There are many similarities between double integrals and single integrals: • Double integrals are defined as limits of Riemann sums. • Double integrals are evaluated using the Fundamental Theorem of Calculus (but we have to use it twice—see the discussion of iterated integrals below).

FIGURE 1 The double integral of f (x, y) over the domain D yields the volume of

the solid region between the graph of f(x, y) and the xy-plane over D.

An important difference, however, is that the domains of integration of double integrals are often more complicated. In one variable, the domain of integration is simply an interval [a, b]. In two variables, the domain D is a plane region whose boundary can be made up of a number of different curves and segments (e.g., D in Figure 1 and 7Z in Figure 2). In this section, we focus on the simplest case where the domain is a rectangle, leaving more general domains for Section 16.2. Let 'R. = [a , b] x [c, d] denote the rectangle in the plane (Figure 2) consisting of all points (x, y) such that R.:

a < x
.

gradient

function

F

vx curl

vector field

v.

____*

G

g

div

vector field

function

One basic fact is that the result of two consecutive operations in this diagram is zero: curl(gradient(f)) = 0, V x (V f) = 0,

div(curl(F)) = 0 V • (V x F)

0

The first identity follows from Theorem 1 of Section 17.1. The second identity appeared as Exercise 33 in Section 17.1. An interesting question is whether every vector field satisfying curl(F) = 0 is necessarily conservative—that is, F = Vf for some function f. The answer is yes, but only if the domain D is simply connected. For example, in R2 the vortex field satisfies curls (F) = 0 and yet cannot be conservative because its circulation around the unit circle is nonzero (which is not possible for conservative vector fields since their circulation around a closed path must be zero). However, the domain of the vortex vector field is R2 with the origin removed, and this domain is not simply connected. The situation for vector potentials is similar. Can every vector field G satisfying div(G) = 0 be written in the form G = curl(A) for some vector potential A? Again, the answer is yes—provided that the domain is a region W in R3 that has no "holes," a region like a ball, a solid cube, or all of R3. The inverse-square field FIs = /r2 plays the role of the vortex field in this setting: Although div(Fis) = 0, Fis cannot have a vector potential over its whole domain because, as shown in Theorem 2, its flux through the unit sphere is nonzero (which is not possible for a vector field with a vector potential since Stokes' Theorem implies that the flux of such a vector field over a closed surface must be zero). In this case, the domain of F15 = er /r2 is R3 with the origin removed, which has a hole. These properties of the vortex and inverse-square vector fields are significant because they relate line and surface integrals to topological properties of the domain, such as whether the domain is simply connected or has holes. They are a first hint of the important and fascinating connections between vector analysis and the area of mathematics called topology.

18.3 SUMMARY • The Divergence Theorem: If W is a region in R3 whose boundary oriented by normal vectors pointing outside 142, then fl aw F • dS = iff

aw is a surface,

div(F)dV

• Corollary: If div(F) = 0, then F has zero flux through the boundary a W of any W contained in the domain of F. • The divergence div(F) is interpreted as flux per unit volume, which means that the flux through a small closed surface containing a point P is approximately equal to div(F)(P) times the enclosed volume. • Basic operations on functions and vector fields: div

curl

function

vector field

vector field

function

1064

CHAPTER 18

FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS • In these cases, the result of two consecutive operations is zero: curl(Vf) = 0,

div(curl(F)) = 0

0, satisfies div(Fis) = 0. The • The inverse-square field FIs = er /r2, defined for r contains the origin and is zero flux of Fis through a closed surface S is 47r if S otherwise.

HISTORICAL PERSPECTIVE Vector analysis was developed in the nineteenth century, in large part, to express the laws of electricity and magnetism. Electromagnetism was studied intensively in the —James Clerk Maxwell period 1750-1890, (1831-1879) culminating in the famous Maxwell Equations, which provide a unified understanding in terms of two vector fields: the electric field E and the magnetic field B. In a region of empty space (where there are no charged particles), the Maxwell Equations are

This is not just mathematical elegance. . . but beauty. It is so simple and yet it describes something so complex. Francis Collins (1950-), leading geneticist and former director of the Human Genome Project, speaking of the Maxwell Equations.

Magnetic field B

Direction of wave motion

Electric field E

0B curl(curl(E)) = curl (- —

at

a = — —curl(B) at

Then apply Maxwell's fourth equation to obtain

aE\ ( at VID"— at

a curl(curl(E)) = - —

82E = - 11060 -87 2

8

Finally, let us define the Laplacian of a vector field F=

F2, F3)

by applying the Laplacian A to each component, AF = AF1, A F2, A F3). Then the following identity holds (see Exercise 36):

div(B) = 0

div(E) = 0, aB curl(E) = - — ,

at

curl(curl(F)) = V(div(F)) - AF

aE

curl(B) = ItoE0 — at

where AO and E0 are experimentally determined constants. In SI units,

Applying this identity to E, we obtain curl(curl(E)) = -AE because div(E) = 0 by Maxwell's first equation. Thus, Eq. (8) yields 82E AE = act0E0 — at2

kto = 47 x 10-7 henries/m EO

We will show that the components of E satisfy this wave equation. Take the curl of both sides of Maxwell's third equation:

8.85 x 10-12 farads/m

These equations led Maxwell to make two predictions of fundamental importance: (1) that electromagnetic waves exist (this was confirmed by H. Hertz in 1887), and (2) that light is an electromagnetic wave. How do the Maxwell Equations suggest that electromagnetic waves exist? And why did Maxwell conclude that light is an electromagnetic wave? It was known to mathematicians in the eighteenth century that waves traveling with velocity c may be described by functions co(x, y, z, t) that satisfy the wave equation 1 82c,

FIGURE 17 The E and B fields of an electromagnetic wave along an axis of motion.

AC° =

at2

7

where A is the Laplace operator also known as the Laplacian) 82c, Ay) =

8x2

82c, 82(p -8---- + T y2 z2

In other words, each component of the electric field satisfies the wave equation (7), with c = (A0E0)-1/2. This tells us that the E-field (and similarly the B-field) can propagate through space like a wave, giving rise to electromagnetic radiation (Figure 17). Maxwell computed the velocity c of an electromagnetic wave: c = (j1,00)-112

3 x 108 m/s

and observed that the value is suspiciously close to the velocity of light (first measured by Olaf Romer in 1676). This had to be more than a coincidence, as Maxwell wrote in 1862: "We can scarcely avoid the conclusion that light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena." Needless to say, the wireless technologies that drive our modern society rely on the unseen electromagnetic radiation whose existence Maxwell first predicted on mathematical grounds.

Divergence Theorem

SECTION 18.3

1065

18.3 EXERCISES Preliminary Questions 1. What is the flux of F = (1, 0, 0) through a closed surface? 2. Justify the following statement: The flux of F = (x3, y3, Z3) through every closed surface is positive. 3. Which of the following expressions are meaningful (where F is a vector field and f is a function)? Of those that are meaningful, which are automatically zero? (c) Vcurl(f ) (b) curl(Vf) (a) div(Vf ) (f) V(div(F)) (e) curl(div(F)) (d) div(curl(F))

4. Which of the following statements is correct (where F is a continuously differentiable vector field defined everywhere)? (a) The flux of curl(F) through all surfaces is zero. (b) If F = Vcp, then the flux of F through all surfaces is zero. (c) The flux of curl(F) through all closed surfaces is zero. 5. How does the Divergence Theorem imply that the flux of the vector field F = (x2, y — ez, y — 2zx) through a closed surface is equal to the enclosed volume?

Exercises In Exercises 1-4, verify the Divergence Theorem for the vector field and region. 1. F(x, y, z) =

x, Y),

2. F(x, y, z) = (y,x, z),

the box [0, 4] x [0,21 x [0,3] the region x2 + y2

3. F(x, y, z) = (2x, 3z, 3y), 4. F(x, y, z) = (x, 0, 0),

4

± z2

the region x2 + y2 < 1,0 4, then x2 >16. 13. If m2 is divisible by 3, then m is divisible by 3. 14. If x2 = 2, then x is irrational.

26. (a) (b) (c)

State the inverses of these implications: If X is a mouse, then X is a rodent. If you sleep late, you will miss class. If a star revolves around the sun, then it's a planet.

In Exercise 15-18, give a counterexample to show that the converse of the statement is false.

27. 1-7

15. If m is odd, then 2m ± 1 is also odd.

28.

16. If A ABC is equilateral, then it is an isosceles triangle.

29. Theorem 1 in Section 2.4 states the following: "If f and g are continuous functions, then f g is continuous." Does it follow logically that if f and g are not continuous, then f + g is not continuous?

17. If m is divisible by 9 and 4, then m is divisible by 12. 18. If m is odd, then I113

-

m

is divisible by 3.

In Exercise 19-22, determine whether the converse of the statement is false. 19. If x > 4 and y > 4, then x + y > 8. 20. If x > 4, then x2 > 16.

Explain why the inverse is equivalent to the converse.

1-371 State the inverse of the Pythagorean Theorem. Is it true?

30. Write out a proof by contradiction for this fact: There is no smallest positive rational number. Base your proof on the fact that if r > 0, then 0 < r/2 < r. 31. Use proof by contradiction to prove that if x y > 1 (or both).

y > 2, then x > 1 or

21. If Ix' > 4, then x2 >16.

In Exercises 32-35, use proof by contradiction to show that the number is irrational.

22. If m and n are even, then mn is even.

32.

In Exercises 23 and 24, state the contrapositive and converse (it is not necessary to know what these statements mean).

36. An isosceles triangle is a triangle with two equal sides. The following theorem holds: If A is a triangle with two equal angles, then A is an isosceles triangle. (a) What is the hypothesis? (b) Show by providing a counterexample that the hypothesis is necessary. (c) What is the contrapositive? (d) What is the converse? Is it true?

23. If f and g are differentiable, then f g is differentiable. 24. If the force field is radial and decreases as the inverse square of the distance, then all closed orbits are ellipses. In Exercises 25-28, the inverse of A

B is the implication

25. Which of the following is the inverse of the implication, "If she jumped in the lake, then she got wet"? (a) If she did not get wet, then she did not jump in the lake. (b) If she did not jump in the lake, then she did not get wet. Is the inverse true?

33.

34.

35.

37. Consider the following theorem: Let f be a quadratic polynomial with a positive leading coefficient. Then f has a minimum value. (a) What are the hypotheses? (b) What is the contrapositive? (c) What is the converse? Is it true?

Further Insights and Challenges 38. Let a, b, and c be the sides of a triangle and let 0 be the angle opposite c. Use the Law of Cosines (Theorem 1 in Section 1.4) to prove the converse of the Pythagorean Theorem. 39. Carry out the details of the following proof by contradiction that Nif is irrational (this proof is due to R. Palais). If N/2 is rational, then nlf, is a whole number for some whole number n. Let n be the smallest such whole number and let m = n N/2 - n. (a) Prove that m 1, define the dth truncation of a = n.a1a2a3a4 . . . to be the finite decimal a(d) =- a.aia2 ad obtained by truncating at the dth place. To form the sum a + b, assume that both a and b are infinite (possibly ending with repeated nines). This eliminates any possible ambiguity in the expansion. Then the nth digit of a + b is equal to the nth digit of a(d)± b(d) ford sufficiently large [from a certain point onward, the nth digit of a(d)H- b(d) no longer changes, and this value is the nth digit of a ± b]. Multiplication is defined similarly. Furthermore, the Commutative, Associative, and Distributive Laws hold (Table 1). TABLE 1 Algebraic Laws a+b=b+a, ab=ba (a + b) + c = a + (b + c), (ab)c = a(bc) a(b + c)= ab + ac

Commutative Laws: Associative Laws: Distributive Law:

Every real number x has an additive inverse -x such that x (-x) = 0, and every nonzero real number x has a multiplicative inverse x -1 such that x(x -1) = 1. We do not regard subtraction and division as separate algebraic operations because they are defined in terms of inverses. By definition, the difference x - y is equal to x (-y), and the quotient x I y is equal to x(y ) for y 0. In addition to the algebraic operations, there is an order relation on R: For any two real numbers a and b, precisely one of the following is true: Either TABLE 2 If a If a If a If a

Order Properties

< b and b (-3)5

The algebraic and order properties of real numbers are certainly familiar We now discuss the less familiar Least Upper Bound (LUB) Property of the real numbers. This property is one way of expressing the so-called completeness of the real numbers. There are other ways of formulating completeness (such as the so-called nested interval property discussed in any book on analysis) that are equivalent to the LUB Property and serve the same purpose. Completeness is used in calculus to construct rigorous proofs of basic Al

APPENDIX B

A8

PROPERTIES OF REAL NUMBERS

theorems about continuous functions, such as the Intermediate Value Theorem, (IVT) or the existence of extreme values on a closed interval. The underlying idea is that the real number line "has no holes." We elaborate on this idea below. First, we introduce the necessary definitions. Suppose that S is a nonempty set of real numbers. A number M is called an upper bound for S if x L. For example (Figure 2),

M I

I

for all x

3

2 M = 3 is an upper bound for the set S -= (-2, 1). The LUB is L = 1.

FIGURE

• M = 3 is an upper bound for the open interval S = (-2, 1). • L =1 is the LUB for S = (-2, 1). We now state the LUB Property of the real numbers. THEOREM 1 Existence of a Least Upper Bound Let S be a nonempty set of real numbers that is bounded above. Then S has an LUB. In a similar fashion, we say that a number B is a lower bound for S if x > B for all x E S. We say that S is bounded below if S has a lower bound. A greatest lower bound (GLB) is a lower bound M such that every other lower bound B satisfies B < M. The set of real numbers also has the GLB Property: If S is a nonempty set of real numbers that is bounded below, then S has a GLB. This may be deduced immediately from Theorem 1. For any nonempty set of real numbers S, let —S be the set of numbers of the form —x for x E S. Then —S has an upper bound if S has a lower bound. Consequently, —S has an LUB L by Theorem 1, and —L is a GLB for S.

4

1 -3

-2 -1

0

1 01 1I 2

,x 3

FIGURE 3 The rational numbers have a "hole" at the location Ari.

CONCEPTUAL INSIGHT Theorem 1 may appear quite reasonable, but perhaps it is not clear why it is useful. We suggested above that the LUB Property expresses the idea that R is "complete" or "has no holes." To illustrate this idea, let's compare R to the set of rational numbers, denoted Q. Intuitively, Q is not complete because the irrational numbers are missing. For example, Q has a "hole" where the irrational number A/2 should be located (Figure 3). This hole divides Q into two halves that are not connected to each other (the half to the left and the half to the right of A/2). Furthermore, the half on the left is bounded above but no rational number is an LUB, and the half on the right is bounded below but no rational number is a GLB. The LUB and GLB are both equal to the irrational number A/2, which exists in only R but not Q. So unlike R, the rational numbers Q do not have the LUB property. EXAMPLE 1 Show that 2 has a square root by applying the LUB Property to the set S = Ix : x2 2, then x satisfies x2 > 4, and hence x does not belong to S. By the LUB Property, S has a least upper bound. Call it L. We claim that L = Nn, or, equivalently, that L2 = 2. We prove this by showing that L2 > 2 and L2 < 2. If L2 0. Then b2 = L2 + 2Lh + h2 = L2 + h(2L + h)

1

We can make the quantity h(2L + h) as small as desired by choosing h > 0 small enough. In particular, we may choose a positive h so that h(2L + h) < 2 — L2. For this choice, b2 L since h > 0, and thus

APPENDIX B

PROPERTIES OF REAL NUMBERS

A9

L is not an upper bound for S, in contradiction to our hypothesis on L. We conclude that L2 > 2. If L2 > 2, let b = L — h, where h > 0. Then b2 = L2 — 2Lh

h2 = L2 — h(2L — h)

Now choose h positive but small enough so that 0 < h(2L — h) L2 (L2 — 2) = 2. But b < L, so b is a smaller lower bound for S. Indeed, if x > b, then x2 > b2 > 2, and x does not belong to S. This contradicts our hypothesis that L is the LUB. We conclude that L2 2, we • have L2 = 2 as claimed. We now prove three important theorems, the third of which is used in the proof of the LUB Property below. THEOREM 2 Bolzano—Weierstrass Theorem Let S be a bounded, infinite set of real numbers. Then there exists a sequence of distinct elements {an} in S such that the limit L = lim a, exists. n-+ 00

Proof For simplicity of notation, we assume that S is contained in the unit interval [0, 1] (a similar proof works in general). If 1(1, k2, ,k, is a sequence of n digits (i.e., each ki is a whole number and 0 < ki oo

•

We use the Bolzano—Weierstrass Theorem to prove two important results about sequences {a,}. Recall that an upper bound for {an ) is a number M such that a1 < M for all j. If an upper bound exists, {an} is said to be bounded from above. Lower bounds are defined similarly and {an} is said to be bounded from below if a lower bound exists. A sequence is bounded if it is bounded from above and below. A subsequence of fad is a sequence of elements a,,, a02 , an,, . . , where ni oo terms form a convergent subsequence. This proves the next result. I Section

11.1

THEOREM 3

Every bounded sequence has a convergent subsequence.

THEOREM 4 Bounded Monotonic Sequences Converge • If {an} is increasing and a, < M for all n, then {an} converges and lim a, < M. n-4 00 • If (ad is decreasing and a, > m for all n, then (an ) converges and lim a, > m. n—> 00

Al 0

APPENDIX B

PROPERTIES OF REAL NUMBERS Proof Suppose that {an} is increasing and bounded above by M. Then {a} is automatically bounded below by m = ai since a l < az < a3 • • • . Hence, {an} is bounded, and by Theorem 3, we may choose a convergent subsequence a 1 , an2, . . Let L = lim

k—>0. aflk

Observe that an L for some n and then an, > an > L for all k such that nk > n. But this contradicts that an, —> L. Now, by definition, for any > 0, there exists N, > 0 such that

Choose m such that nn, >

lank — LI < E if nk > NE . If n > n., then an,, no E = (L — M)/2 and choose N so that Ia—Ll N

Then a, > L —€-=M+ E. This contradicts our assumption that M is an upper bound for {a}. Therefore, L a(d), then

E

S. If x

E

S

a(d) < x(d) < M Thus, by the remark of the previous paragraph, there are at most finitely many values of x(d) in S(d) larger than a(d). The largest of these is the maximum element in S(d). For d = 1, 2, . . . , choose an element xd such that xd(d) is the maximum element in S(d). By construction, {xd(d)} is an increasing sequence (since the largest dth truncation cannot get smaller as d increases). Furthermore, xd(d) < M for all d. We now apply Theorem 4 to conclude that {xd(d)} converges to a limit L. We claim that L is the LUB of S. Observe first that L is an upper bound for S. Indeed, if x E S, then x(d) < L for all d and thus x no. EXAMPLE 1 Prove that 1 + 3 + • • + (2n — 1) = n2 for all natural numbers n. Solution As above, we let P(n) denote the equality P(n) :

1 + 3 + • • • + (2n — 1) = n2

Step 1. Initial step: Show that P(1) is true. We checked this above. P(1) is the equality 1 = 12. Step 2. Induction step: Show that if P(n) is true for n = k, then P(n) is also true for n = k 1. Assume that P(k) is true. Then I + 3 + • • + (2k — 1) = k2 Add 2k + 1 to both sides: [1 + 3 + • • • + (2k — 1)] + (2k + 1) = k2 + 2k + 1 = (k + 1)2 1 + 3 +• • •+(2k+1)=(k+ 1)2 This is precisely the statement P(k ± 1). Thus, P(k By the Principle of Induction, P(k) is true for all k.

1) is true whenever P(k) is true. •

The intuition behind the Principle of Induction is the following. If P(n) were not true for all n, then there would exist a smallest natural number k such that P(k) is false. Al2

APPENDIX C

INDUCTION AND THE BINOMIAL THEOREM

A13

Furthermore, k> 1 since P(1) is true. Thus, P(k - 1) is true [otherwise, P(k) would not be the smallest "counterexample"]. On the other hand, if P(k - 1) is true, then P(k) is also true by the induction step. This is a contradiction. So P(k) must be true for all k. EXAMPLE 2 Use Induction and the Product Rule to prove that for all whole numbers n, dx

xn = nxn-1

d Solution Let P(n) be the formula — = nxn-1. dx Step I. Initial step: Show that P(1) is true. We use the limit definition to verify P(1): —x = lim dx h—>0

(x

h) - x h = lim - = lim 1 = 1 h h—>0 h h—>0

Step 2. Induction step: Show that if P(n) is true for n = k, then P(n) is also true for n = k +1. d To carry out the induction step, assume that —x' = kx", where k> 1. Then, by dx the Product Rule,

In Pascal's Triangle, the nth row displays the coefficients in the expansion of (a ± :

d d d k+1 = — —x (x • xk) = x— xk dx dx dx kxk This shows that P(k

1 1 1 1

1 3

1

4

6 4 1 10 10 5 1 6 15 20 15 6 1 1

1

1)xk

1) is true. •

1 2

3

xk

By the Principle of Induction, P(n) is true for all n > 1.

1 2 3 4 5 6

Xk = (k

xk — d x = x(kxk-1) dx

5

As another application of induction, we prove the Binomial Theorem, which describes the expansion of the binomial (a + by1. The first few expansions are familiar: (a + b)1 = a (a ± b)2

The triangle is constructed as follows: Each entry is the sum of the two entries above it in the previous line. For example, the entry 15 in linen = 6 is the sum 10+5 of the entries above it in linen = 5. The recursion relation guarantees that the entries in the triangle are the binomial coefficients.

b

a2 + 2ab

(a ± b)3 = a3

b2

3a2b

3ab2

b3

In general, we have an expansion (a ± b)n = a" ± ()a 1

-l b ± n ) an-2b2 2

( n) an-3b3 3

_n1)

abn-1

bn

n , is called the binomial coefficient. Note (k that the first term in Eq. (2) corresponds to k = 0 and the last term to k = n; thus, in\ = ( n) = 1. In summation notation, V))

where the coefficient of an-kb', denoted

(a ±

= En ( n)an-k bk k=0

A14

APPENDIX C

INDUCTION AND THE BINOMIAL THEOREM

Pascal's Triangle (described in the marginal note on page A13) can be used to computt binomial coefficients if n and k are not too large. The Binomial Theorem provides the following general formula: n(n — 1)(n — 2) - • • (n — k +1) k(k — 1)(k — 2) • • • 2 • 1

n! k!(n — k)!

ft )

3

Before proving this formula, we prove a recursion relation for binom'al coefficients. Note, however, that Eq. (3) is certainly correct for k = 0 and k = n (recall that by convention, 0! = 1): (n) 1:1)

n! n! = 1, (n — 0)! 0! = n!

n! = (n — n)!n!

(nn

n! n!

THEOREM 2 Recursion Relation for Binomial Coefficients

k) =

— k

k

1 1)

for 1 2n for n > 4.

15. Expand (x +x 1 )4. 16. What is the coefficient of x9 in (x3 ± x)5?

Let {Fn} be the Fibonacci sequence, defined by the recursion formula Fn = Fn _i

Pn _2,

F1=F2=1

The first few terms are 1,1,2, 3,5, 8, 13, . . .. In Exercises 7-10, use induction to prove the identity.

17. Let S(n) =

E (k).

k=0 (a) Use Pascal's Triangle to compute S(n) for n = 1, 2, 3, 4. (b) Prove that S(n) = 2" for all n > 1. Hint: Expand (a ± by, and evaluate at a = b =1.

7. Fi + F2 ± • • • + Fn = Fn+2 - 1 8. Fi2 ± 9. Fn -

± • • • + Fn2 = Fn+1Fn Rn - Rn

, where R± =

1 2

18. Let T(n)=E(-1) k ( n). k=0 (a) Use Pascal's Triangle to compute T(n) for n =1, 2, 3, 4. (b) Prove that T(n)= 0 for all n > 1. Hint: Expand (a ± b)n and evaluate at a = 1, b = -1.

D ADDITIONAL PROOFS In this appendix, we provide proofs of several theorems that were stated or used in the text. I Section 2.3

THEOREM 1 Basic Limit Laws

Assume that lim f (x) and lim g(x) exist. Then: X -->C

(i) lim (f (x) g(x)) = lim f (x) lim g(x) x—>c x—>c (ii) For any number k, urn kf (x) = k lim f (x) (iii) lim f (x)g(x) = (lim f (x)) (lim g(x)) X ->C

X ->C

(iv) If lim g(x)

0, then

X ->C

lim f (x) f (x) x—>c lim — = x-+c g(x) xltg(x)

Proof Let L = lirn f (x) and M = lim g(x). The Sum Law (i) was proved in SecX -+C X ->C tion 2.9. Observe that (ii) is a special case of (iii), where g(x) = k is a constant function. Thus, it will suffice to prove the Product Law (iii). We write f (x)g(x) — LM = f (x)(g(x) — M) -F M(f(x)— L) and apply the Triangle Inequality to obtain

if(x)g(x) —

LMI

If (x)(g(x) — MA ± IM( f (x) —

By the limit definition, we may choose 3 > 0 so that if 0 < Ix — cl < 3, then

If(x)- Li < 1

If follows that I f(x)I < ILI + 1 for 0 < Ix — c I 0. Applying the limit definition again, we see that by choosing a smaller 8 if necessary, we may also ensure that if 0 < Ix — cl < 8, then E If(X) — LI

2(011+1)

and

I g(x)

MI

E

2(14+ 1)

Using Eq. (1), we see that if 0 < Ix — c I < 3, then

If(x)ex)- Lml

If(x)1 lex)- MI + IMl If(x)- LI OLP + 1)

2(ILI +1)

+ IMI

2(Im +1)

Since E is arbitrary, this proves that lim f (x)g(x) = LM. To prove the Quotient Law (iv), it suffices to verify that if M

0, then 1 1 lim = 7A1 x _> c g(x)

Al6

2

ADDITIONAL PROOFS

APPENDIX D

All

For if Eq. (2) holds, then we may apply the Product Law to f(x) and g(x)-1 to obtain the Quotient Law: lim

f(x) = lim f(x) 1 = lirn f(x)) lim 1 ) x—>c g(x) x—>c g(x) x—>c g(x) L L (-41 1—= TI

We now verify Eq. (2). Since g(x) approaches M and M 0, we may choose 3 > 0 so — cl < 8, then Ig(x)I > IMI /2. Now choose any number E > 0. By choosthat if 0 < ing a smaller 8 if necessary, we may also ensure that for 0
c x —>c contradiction. If L > M, let E = (L — M). By the formal definition of limits, we may choose 3 > 0 so that the following two conditions are satisfied: If Ix — cl c

and

Assume that the following limits exist:

M=

lim f(x)

x->L

Then lint f (g(x)) = M.

Proof Let E

>

•

0 be given. By the limit definition, there exists 81 > 0 such that if 0 < Ix — LI c Section 2.4

I

•

THEOREM 4 Continuity of Composite Functions Let F(x) = f (g(x)) be a composite function. If g is continuous at x = c and f is continuous at x = g(c), then F is continuous at x = c. Proof By definition of continuity, lim g(x) = g(c)

and

lim f (x) = f (g(c)) x g(c)

Therefore, we may apply Theorem 3 to obtain lirn f (g(x)) = f (g(c)) This proves that F(x) = f (g(x)) is continuous at x = c. I Section 2.6

THEOREM 5 Squeeze Theorem taining c), 1(x) < f (x)

Assume that for x

u(x)

c (in some open interval con-

lim 1(x) = lim u(x) = L x—>c x—>c

and

Then lim f (x) exists and Inn f (x) = L. x—>c Proof Let c > 0 be given. We may choose 8 > 0 such that if 0 < Ix — cl 1, and therefore, we may choose a point bn in [a, b] such that 1 M — — < f (bn) < M n Again by Theorem 3 in Appendix B, there exists a subsequence of elements On, bn2, • • • I in 01, b2,. I that converges to a limit—say, lim bn„ = c

k—>oo

Furthermore, this limit c belongs to [a, b] because [a, b] is closed. Let E > 0. Since f is continuous, we may choose k so large that the following two conditions are satisfied: f (c) — f (bnk )I < e /2 and nk > 2/e. Then

I f (c) — MI

If (c)

f (bn,)1 + I f (bnk)

I

=e

Thus, I f (c) — MI is smaller than E for all positive numbers e. But this is not possible unless I f (c) — MI = 0. Thus, f (c) = M, as desired. •

I Section 5.2

THEOREM 7 Continuous Functions Are Integrable If f is continuous on [a, b], or if f is continuous except at finitely many jump discontinuities in [a, b], then f is integrable over [a, b]. Proof We shall make the simplifying assumption that f is differentiable and that its derivative f' is bounded. In other words, we assume that I f (x)I < K for some constant K. This assumption is used to show that f cannot vary too much in a small interval. More precisely, let us prove that if [ao, bo] is any closed interval contained in [a, b] and if m and M are the minimum and maximum values off on [a0, bp], then

IM — ml

0 such that

if 0< Ix — L I 0

h2

. = Inn f xy(ai b1) = f xy(a,b) h—>0

The last equality follows from the continuity of fxy since (al, b1) approaches (a, b) as h 0. To prove that L = f yx(a,b), repeat the argument using the function F(y) = f (a + h, y) — f (a, y), with the roles of x and y reversed. • I Section 15.4

THEOREM 10 Confirming Differentiability If f x(x, y) and fy(x, y) exist and are continuous on an open disk D, then f (x, y) is differentiable on D. Proof Let (a, b)

E

D and set

L(x, y) = f (a, b) + f x(a,b)(x — a) + f y (a, b)(y — b) It is convenient to switch to the variables h and k, where x = a + h and y = b + k. Set Af = f (a + h,b + k) — f (a, b) Then L(x, y) =

f(a, b) + fx(a, b)h

f y (a,b)k

and we may define the function e(h,k) = f (x, y) — L(x, y) = Af — (fx(a,b)h + f y (a,b)k) To prove that f (x, y) is differentiable, we must show that e(h,k) lim =0 (h,k)--0(O,o) ,/h2 k2 To do this, we write Af as a sum of two terms: Af = (f (a + h, b + k) — f (a, b + k)) + (f (a, b + k) — f (a , b)) and apply the MVT to each term separately. We find that there exist al between a and a ± h, and b1 between b and b k, such that f (a + h,b + k) — f (a , b + k) = hf x (al, b +k) f (a , b + k) — f (a , b) = kfy(a, bi) Therefore, e(h, k) = h(fx(ai, b + k) — f x(a , b)) and for (h, k)

k(fy(a,

— fy(a,b))

(0,0),

e(h,k)

h(fx (ai, b + k) — f x(a,b))+ k( f y(a, b1) — f y(a,b))

h2 + k2

h2 + k2 h(fx (ai,b + k) — fx(a,b)) h2 k2 =

k( fy(a, b1) — fy(a,b)) h2 + k2

b +k) — f x (a,b)l + f y(a, bi) — f y(a,b)1

In the second line, we use the Triangle Inequality [see Eq. (1) in Section 1.11, and we may pass to the third line because hR/h2 + k2 1 and 1k//h2 + k2 1 are both less than 1. Both terms in the last line tend to zero as (h, k) (0,0) because fx and f y are assumed to be continuous. This completes the proof that f (x, y) is differentiable. •

ANSWERS TO ODD-NUMBERED EXERCISES Chapter 1

57. Zeros: ±2; Increasing: x > 0; Decreasing: x 0 satisfy lal = a and I - al = a. The numbers a < 0 satisfy la I = -a. 3.

a = -3 and b = 1

4. No 5. (9, -4)

6. (a) First quadrant (b) Second quadrant (c) Fourth quadrant (d) Third quadrant 7.

59. Zeros: 0, ±2; Symmetry: f (-x) = -f (x), so origin symmetry

3 8. (b) 9. Symmetry with respect to the origin

10. The only function that is both even and odd is the constant function f (x) = O.

Section 1.1 Exercises 1. (a) Correct (b) Correct (c) Incorrect (d) Correct 3.

128- 448x + 672x2 - 560x3 + 280x4 - 84x5 + 14x6 - x7

61. This is an x-axis reflection of y = x3 translated up 2 units. There is one zero at x =

7. Ix I 0 and < 0 so < 33. la + b - 131 = 1(a -5) + (b 35. (a) 11

(b) 1

8)1

la - 51+ lb -81
01

0}; R: {y : y > 0} 53. On the interval (-1,00)

55. On the interval (0, co)

77. (a) D: [4, 8], R: [5,9] (b) D: [1,5], R: [2,6] R: [2, 6] (d) D: [4, 8], R: [6, 18]

(c) D: [4,

ANSI

ANS2

ANSWERS TO ODD -NUMBERED EXERCISES Large positive a

79. (a) f (x) = (2(x - 5))4 - (2(x - 5))2 (b) f (x) = (2x - 5)4 - (2x - 5)2 (c)

Small positive a

Y = f(x)

2

-2

4

Small negative a

6

Large negative a

81. 6

6

4

4

1

-3 -2 -1

2

3

-3 -2

f(2x)

b b2 Open up with intercepts at 0 and -b. Vertex at (- 7, - 71-). For b 0, vertex in third quadrant.

8.

2

-1

Small b positive

3

y

b = 0 Small b \ negative

f (x 2)

83.

Large b positive

1

2

D: all reals; R: fy I Y

Large b negative

3 1); f (x) = Ix -fl + 1

85. Even: (f + g)(-x) = f (-x) + g(-x) evg l f (x) + g(x) = (f + g)(x) Odd: (1+ g)(-x) = f (-x) + g(-x) °Li-d - f (x) + -g(x) = -(f + g)(x) 87. If f is symmetric with respect to the y-axis, then f (-x) = f (x). If f is also symmetric with respect to the origin, then f (-x) = - f (x). Thus f (x) = - f (x) or 2f (x) 0 or f (x) = 0. 91. (a) There are many possibilities, one of which is

Section 1.2 Exercises 1. m = 3; y = 12; x 5.

m=3

13.

y =

-2

7. m = 15.

21. y = -2x + 9

= -4

--4

y =

3. m =

-4; y =

x=

9. y = 3x + 8 11. y = 3x - 12

3x - 2

17. 5x - 3y = 1 19. y = 4

23. 3x + 4y = 12

25. (a) c = (b) c = -2 slope equal to 0 (d) c =0

(c) No value for c that will make this

27. (a) N(P) = -5P + 15,000; (b) Slope = -5 computers/ dollar. For every dollar increase in price, five fewer computers are sold. (c) AN = -500 29. (a) Slope = -70 students/week. Enrollment dropped by 70 students per week during Fall 2017. (b) Slope = 3.5 dollars/person. The rental cost increases by 3.5 dollars for each person attending. 31. (a) 40.0248 cm

(b) 64.9597 cm

(c) L = 65(1 + a(T - 100))

33. b = 4 35. No, because the slopes between consecutive data points are not equal. 37. (a) 1 or -

(b) 1 ±

39. Minimum value is 0 41. Minimum value is -7 value is IN 45. Maximum value is

A

(b) Let g(x) = f (x + a). Then g(-x) = f (-x + a) = f (a - x) = f (a + x) = g(x)

43. Maximum

47. 10 8 6 4 2

Section 1.2 Preliminary Questions 1.

-4

3.

Parallel to the y-axis when b = 0; parallel to the x-axis when a = 0

4.

Ay = 9

) X

2. No

5. -4

6. (x - 0)2 ± 1

7. Fora 0 0, vertex at (0, -1). Large a 0: opens up, more narrow.

49. A double root occurs when c = ±2. There are no real roots when -2 1, no points of intersection; if Ic I = 1, one point of intersection; if Icl < 1, two points of intersection.

sinO = 0.918, cos 0 = 0.3965, and tan9 =

0.918 = 2.3153 0.3965

45.

sin 0 = 0.3965, cos = -0.918, and 0.3965 = -0.4319 -0.918

tan 0 =

sin 0 = -0.918, cos 0 = -0.3965, and - 0.918 = 2.3153 tan 0 = -03965 • The point in the fourth quadrant: sin 0 = -0.3965, cos 0 = 0.918, and -0.3965 = -0.4319 0.918

tan 0 =

49. By the corresponding double-angle formula, sin2 = 1(1 - cos 2q)) = 151. cos(0 + 7r) = cos 0 cos 7 - sin 0 sin n- = c056(-1) = - cos 53. Using Exercises 50 and 51, sin(71- - 0) - sin(-0) sin tan(n- - 0) = = = cos(7 - 0) cos(-0) - cos

• The point in the third quadrant:

59. cos2

cos

0.918 tan9 = - 2.3153 0.3965

sin 0 = 0.918, cos 0 = -0.3965, and =

0.918 -0.3965

2.3153

• The point in the third quadrant:

tan 8=

- 0.918 - 2.3153 -0.3965

= -0.918, cos 0 = 0.3965, and tan 0 =

--

4

f-6

cot* =

and cos 74 =

7

> 0, so cos -8- =

11

=-+-

2

4

Nri

+ 7 4-

61. 16.928

Section 1.5 Preliminary Questions 2. (a) The screen will display nothing. (b) The screen will display the portion of the parabola between the points (0, 3) and (1, 4).

• The point in the fourth quadrant:

33. sin 77( 12 35. v

=

1. It is best to experiment.

sin 0 = -0.918, cos 0 = -0.3965, and

31. cos* = 0.3, sin*

cos 8

1 + cos 2 Jr 1+ -2 1 + cos 4 = 2 2

2

65. Using the distances labeled in Figure 28(A), we see that the slope of the line is given by the ratio rls. The tangent of the angle 8 is given by the same ratio. Therefore, m = tan 0.

• The point in the second quadrant:

tan

=

8 Jr

sin 0 = 0.918, cos 0 = 0.3965, and

tan 0

sin 2x 2 sin x cos x 2 sin x cos x sin x = tan x 1 + cos 2x = 1 + 2 cos2 x - 1 2 cos2 x - cos x sin(0 + - sin 0 57. tan(0 + 7) = = tan 0, and cos(0 + 7r) - cos 0 cos(0 + 7) - cos cot(0 + 7) = = . = cot°. Thus, both tan 0 and sin(0 + 7r) -sinO cot 8 are periodic with period 7.

2

• The point in the first quadrant:

=

55.

Now consider the four points in Figure 22(B).

sin

- sin sin 0 = cos2 0 - sin2 6 =

47. cos 20 = cos(0 +9) = cos 0 cos 2 cos2 9 - 1

• The point in the second quadrant:

3.

No

4. Experiment with the viewing window to zoom in on the lowest point on the graph of the function. The y-coordinate of the lowest point on the graph is the minimum value of the function.

-0.918 0.3965

2.3153

Section 1.5 Exercises

1 , and csc

=

1.

‘ /-2

37.

x =- -3,x = -1.5,x = 1, and x = 2 3.

Two positive solutions

7.

Nothing; an appropriate viewing window: [50, 150] by [1000, 2000]

5. There are no solutions.

ANSWERS TO ODD -NUMBERED EXERCISES

ANS5

19. The table and graphs below suggest that as n gets large, f (n) tends toward oo.

9.

11.

t I

I 3

I I 3.8 3.9

I 4

n

(1 + ) n2

10 102 103 104 105 106

13780.61234 1.635828711 x 1043 1.195306603 x 10434 5.341783312 x 104342 1.702333054 x 1043429 1.839738749 x 10434294

x

1 x 1o43

10000

0

2

4

6

8

10

0

20

40

60

80 100

0.4

21. The table and graphs below suggest that as x gets large, f (x) approaches 1.

0.2 I

-ilJ

I

x

I

I

176 3.78 3.8 3.82 3.84 -0.2

13. The maximum is approximately 0.604, occurring at x

-0.716.

15. N R4, 161

808 806 804 802 800 798 796 794 792

x

(x tan 1)x

10 102 103 104 105 106

1.033975758 1.003338973 1.000333389 1.000033334 L000003333 1.000000333

1.5 1.4 1.3 1.2 1.1

160

161

162

163

164

I

I

I

10

15

20

IX

20

40

60

80 100

23.

17. The table and graphs below suggest that as n gets large, n lin approaches 1.

10 102 103 104 105 106

(A, B)= (1, 2)

(A, B)= (1, 1)

n un 1.258925412 1.047128548 1.006931669 1.000921458 1.000115136 1.000013816

(A, B)= (3, 4)

25. x 27.

0

2

4

6

8

10

0

200 400 600 800 1000

E

(-2, 0) U (3, co) 1 (i

1 2

)

x2 + 6x + 1 4(x + 1)

x x2 +6x+i 4(x + 1)

6x + 1 4(x + 1)

x2

x4 + 28x3 + 70x2 + 28x + 1 8(1 + x)(1 + 6x + X2)

NY,

ANSWERS TO ODD -NUMBERED EXERCISES 39.

and

fs(x) -

1+ 120x + 1820x2 +8008x3 + 12870x4 + 8008x5 + 1820x6 + 120x7 +x 8 16(1 + x)(1 + 6x + x2)(1 + 28x + 70x2 + 28x3 + x4)

It appears as if the f n are asymptotic to .1.V. 41. Let g(x) = f 04

Chapter 1 Review 1. (a) No match (b)

Then

(c) (i)

3.

ix : Ix - 71 < 3} 5. [-5, -1] U [3, 7]

7.

(x, 0) with x > 0; (0, y) with y -11; R: {y : y > 0} 15. D: {x : x 03}; R: {y : y 17. (a) Decreasing 19. 2x - 3y = -14

(c) Neither

(d) Increasing

y=5

3.

(a) 63 mi/h

(b) 42 rah

(c) 0 mi/h

4. The slope of the line tangent to the graph of position as a function of time at t = to

21. 6x - y = 53

25. x

23. x + 3y = 5

0}

(b) Neither

The graph of position as a function of time

2. No. Instantaneous velocity is defined as the limit of average velocity as time elapsed shrinks to zero.

27. Yes

29. (a) C (P) - 2250P + 1,225,000

Section 2.1 Exercises

(b) Slope = -2250 customers/dollar. For every dollar increase in the monthly price, there are 2250 fewer customers. (c) 225,000 customers.

1.

31. Roots: x = -2, x = 0 and x = 2; decreasing: x < -1.4 and 0 i+ No because lim f(x) may not be equal to lim f(x)

7.

Yes

5.

oo.

3x + 9. The tangent line slope

31. (a) With 2 subintervals, the total rectangle area = 9/16. With 3 subintervals, the total rectangle area = 4/9. With 5 subintervals, the total rectangle area = 9/25. With 10 subintervals, the total rectangle area = 0.3025. (b) A(2) = 9/16; A(3) = 4/9; A(5) = 9/25; A(10) = 0.3025 (c) A(100) = 0.255025; A(1000) = 0.25050025; A(10, 000) = 0.2500500025; conjecture A = 1/4

1. 1 2.

4+, f(x)

x.•5-

49.

lim f(x) = -1, lim f(x) = 1 1 lirn f(x) =- no, Inn f(x) = 6 4x2 + 7 4x2 +7 =-no, lirn = no lim

x-).--2-

X3 + 8

x--2 + x3 + 8

lim x5+x-2 = 2 x-›i± x2 +x -2 53. • lim f(x) = no and lim f(x) = no. 51.

• lirn f(x) = -no and lim f(x) = 10. The vertical asymptotes are the vertical lines x = 2 and x = 4. 57. Y

Section 2.2 Exercises 1.

x f(x)

1.499625

1.499250

1.498501

0.99999

3

1.499993

2

0.9995

0.999

0.998

x

1.00001

1.0005

1.001

1.002

f(x)

1.500008

1.500375

1.500750

1.501500

The limit as x -> 1 is 3. 3.

Y

1.998

1.999

1.9999

f (y)

0.59984

0.59992

0.599992

Y f (y)

2.0001

2.001

2.002

0.600008

0.60008

0.60016

7.

t 0.002 0.001 0.0005 0.00001 Limit guess: 0 1.5 9. 21

f (t) 0.004 0.002 0.001 0.00002

• lirn f (x) = 4 x->3+ • lim f (x) = 2 •

x-÷5lim x->5+

f(x) = -3

• lirn f (x) = urn f (x) = 00

The limit as y -> 2 is ;. 5.

59. • lim f(x) = lirn f(x) = 3 x->i• lim f(x) = -no

t -0.002 -0.001 -0.0005 -0.00001

f (t) -0.004 -0.002 -0.001 -0.00002

61. 75 2

2.44 2.42

ANS8

ANSWERS TO ODD -NUMBERED EXERCISES (d) True

63. 0.693 (The exact answer is In 2.)

(e) False. The correct statement is "If f and g are continuous at x = a and g(a) 0, then f/g is continuous at x = a."

0.6940-

Section 2.4 Exercises

0.6935 0.6930

Y= 0.6925 0.6920-

65. -12

1. • The function f is discontinuous at x = 1; it is left-continuous there. • The function f is discontinuous at x = 3; it is neither left-continuous nor right-continuous there. • The function f is discontinuous at x = 5; it is left-continuous there. None of these discontinuities are removable. 3.

x = 3; redefine g(3) = 4

5.

The function f is discontinuous at x = 0, at which lim f (x) = oo

and lirn urn f (x) = 2. The function f is also discontinuous at x = 2, at x->o+ which lim f (x) = 6 and lim f (x) = 6. The discontinuity at x = 2 is

-11.4

x-•2-

x-+2+

removable. Assigning f(2) = 6 makes f continuous at x = 2. 7. y = x and y = sin x are continuous; so is f (x) = x + sin x by Continuity Law (i). 67. For n even = 1 for all integers n 69. (a) No (b) f 1,1, . ., the value of f (x) is always -1. (c) At x = 1, n xn - 1 sin tv9 = n 73 1 2, 32' 32 ' 71. - 1 = m x->1 (b) L = 5.545 75. (a) • 5.565 5.555 5.545

15. *

5.535

continuous on (0, 1], so f (x) = tan ( ± 14 2 ) is continuous by Theorem 4.

5.525 -

17. Discontinuous at x = 0, at which there is an infinite discontinuity. The function is neither left- nor right-continuous at x = 0.

Section 2.3 Preliminary Questions Suppose lirn f (x) and lirn g(x) both exist. The Sum Law states that lim (f (x)

X->•C

Provided lirn g(x) x->c

g (x)) = urn f (x) + Ern g(x)

(b)

X -->C

0, the Quotient Law states that . f (x) hrn x-4c g(x)

2.

limx4, f (x) hmx, g(x)

5 I 21.

7 4.6 23. (ian i)

9 1 11. 9 13. 1/8 15. 25. 64

I = 29. 3 31. 33. No hal (f (x))) f (x) 35. (a) 0 (b) 2/7r (c) The limit does not exist. (d) 0 37. f (x) = 1/x and g(x) = -1/x 39. f(x) = 1/x and g(x) -1/x 41. Write g(t) = . 43. (b) 27 lina

f (i)

=

Section 2.4 Preliminary Questions 1.

19. Discontinuous at x = 1, at which there is an infinite discontinuity. The function is neither left- nor right-continuous at x = 1. 21. Discontinuous at even integers, at which there are jump discontinuities. The function is right-continuous at the even integers but not left-continuous. 23. Infinite discontinuities at x = ±2. h is neither left- nor right-continuous at both of these points. 25. Discontinuous at x = 1, at which there is an infinite discontinuity. The function is neither left- nor right-continuous at x = 29. Jump discontinuity at x = 2. The function is left-continuous at x = 2 but not right-continuous.

Section 2.3 Exercises 9 3 17. 10 19.

is continuous and takes values in the interval (0, 1] and tan x is

27. Continuous for all x

3. (a)

1.

11. Since y = x is continuous, so is y = x2 by Continuity Law (iii). Recall that constant functions, such as 1, are continuous. Thus, 1 y = x2 + 1 is continuous by Continuity Law (i). Finally, f (x) = x2 + 1 is continuous by Continuity Law (iv) because x2 + 1 is never 0. 13. The function f is a composite of two continuous functions: y = cos x and y -= x2, so f is continuous by Theorem 5.

2-8

x=3

1.

9. Since y = x and y = sin x are continuous, so are y = 3x and y = 4 sin x by Continuity Law (ii). Thus, f (x) = 3x + 4 sin x is continuous by Continuity Law (i).

Continuity 2. f(3) = 3. No 4. No; yes 5. (a) False. The correct statement is "f is continuous at x = a if the left- and right-hand limits of f (x) as x -4- a exist and equal f (a)." (b) True (c) False. The correct statement is "If the left- and right-hand limits of f (x) as x -• a are equal but not equal to f (a), then f has a removable discontinuity at x = a."

31. Removable discontinuity at x = -5. f is neither left- nor right-continuous at x = -5. 33. Discontinuous whenever t = (2n 1)n- where n is an integer. At every such value of t, there is an infinite discontinuity. The function is neither left- nor right-continuous at any of these points of discontinuity. 35. Continuous everywhere 37. Continuous everywhere 39. The domain is all real numbers. Both y = sin x and y = cos x are continuous on this domain, so f (x) = 2 sin x + 3 cos x is continuous by Continuity Laws (i) and (ii). 41. Domain is x > 0. Since y = „rx and y = sin x are continuous, so is f (x) = f. x-- sin x by Continuity Law (iii). 43. Domain is x > 0. Because x2 and 3x 1 /2 are continuous, so is f (x) = x2 - 3x 112 by Theorem 1(i). 45. Domain is x 0. Because the function y = .X413 is continuous and not equal to zero for x 0, f (x) = .X -413 is continuous for x 0 by Continuity Law (iv).

ANSWERS TO ODD -NUMBERED EXERCISES ±(2n - 1)7/2, where n is a positive integer. 47. Domain is all x Because y = tan x is continuous on this domain, it follows from Continuity Law (iii) that f (x) = tan2 x is also continuous on this domain. is all real numbers. Because 49. Domain of f (x) = (x4 + y = X3/2 and the polynomial y = x4 ± 1 are both continuous, so is the composite function f (x) = (x4 + 1)3/2. 1)3/2

ANS9

f is continuous in (b) and (d). f has a discontinuity at x = 0 in (a). f has a discontinuity at x = 2 in (c).

3

65. c=

67. a = 2 and

69. (a) g(1) = 3/2 71.

=1

(b) No. 73.

Y

51. Domain is all x ±1. Because the functions y = cos x and y = x2 are continuous on this domain, so is the composite function y = cos(x2). Finally, because the polynomial y = x2 - 1 is continuous and not equal c_ _2 xos(x21) is continuous by to zero for x ±1, the function f (x) = Continuity Law (iv). 53. Right-hand limit at x = 1: 9; left-hand limit at x = 1: 4; both rightand left-hand limits at x = 2: 8; f is right-continuous at x = 1; f is continuous at x = 2. 55.

77. -1

75. -6

79. -1

83. 27

81.

85. 1000

87. No. Take f (x) = -x -1 and g(x) = x-1. 89. f(x) = ig(x)1 is a composition of the continuous functions g and Y = Ixl• 91. No

15,000 10,000 5,000

f has a jump discontinuity at x = n for all integers n. At each discontinuity, f is left-continuous.

40,000 20,000

80,000 I x

60,000

57. The function f is continuous everywhere. 93. f (x) = 3 and g(x) = 95. In this case, y = f (x)2 is the constant function 1.

Section 2.5 Preliminary Questions x2 _ 1

1 .

3- 2

2.

(a) f (x) =

(b) f(x) =

59. The function f is neither left- nor right-continuous at x = 2.

Section 2.5 Exercises 1. lin, X 2 - 36 = lim (x x--46

3. 0 19. 61* lim

x->-4 x

6 - xlim (x -

4

(c) f (x) =

3. The simplify and plug-in strategy is based on simplifying a function that is indeterminate to a continuous function. Once the simplification has been made, the limit of the remaining continuous function is obtained by evaluation.

x-6

5.

x-4-6

7. -1

6)(x +6) = x- 6

9.

11M(x +

11. 2 13. 1

6) = 12 15. 2

17.

21. Limit does not exist

4) = 8 0 10 = f (4)

63.

• As h

0+, ‘Th ± 2 - 2

• Ash

0 , ' 1/717-1 2

23. 2

25 I

27 1

2

29 - I

oo. 00.

31 9

33.

35. -1, does not exist; 0 I X

(A)

37. 1/2 39. lim f (x) "-=', 2.00; to two decimal places, this matches the value of 2 x->.4

obtained in Exercise 23. 2.001 2.000 1.999 1.998 1.997 1.996 (C)

(D)

> x 3.6

3.8

4.0

4.2

4.4

ANSWERS TO ODD -NUMBERED EXERCISES

ANSI°

43. -1

41. 12

47. 2a

49. -4 + 5a

Here, c = -741- and cos c =

51.

59. c = 3

57. c = -1 and c = 6

55.

53. 3(12

45.

61. +

Section 2.6 Preliminary Questions 1. 2.

lim f (x) = 0: no

x->0

c (in some open interval containing c),

Assume that for x

1(x)

f (x)

u(x)

9.

lim (2! - 1) cos

= 0 7. lim (t2 - 4) cos

lim f (x) = 2 x-±

(b) lim

21. 1 23. 0 25. 4s 2", 37. 6 39. -4 41.

- cos t 57. lim -o4 t does not exist 61. (a)

sin x - sin c X-

C

= cos C

c - .01

c - .001

c + .001

c + .01

-0.411593

-0.415692

-0.416601

-0.420686

Here, c = 2 and cos c = cos 2

-0.416147

x

c - 0.01

c - 0.001

c 4- 0.001

c + 0.01

sin x - sin c x- c

0.863511

0.865775

0.866275

0.868511

Here, c =

and cos c =

0.866025

(b)

x3 = -co

3.

x->7

15. (a) Not sufficient information (c) r-• I f =3

55. -

-0.005000

2. (a) lim, x3 = no (c) x4 = no

13. lim f (x) = 6; no

49. 0

-0.000500

Section 2.7 Preliminary Questions 1. (a) Correct (b) Not correct (c) Not correct (d) Correct

Therefore, by the Squeeze Theorem,

A-

0.000500

x

=0

lim t(x) = lim u(x) = 2 x->

19. 3 35.

0.005000

=0

1 on the open interval (0, 2) containing x = 1, u(x). Moreover,

17. 1 33. 4.

c + .01

sin x - sin c x- C

lim cos 0 cos(tan ) =0 o->

11. For all x i(x) < f (x)

c + .001

(c)

(a)

t->0

c - .001

x-sc

Section 2.6 Exercises 1. lin) x2 cos =0 3. lim (x - 1) sin x 5.

c - .01

Here, c = -1 and cos c = 0

lirn f (x) = L x->c 3.

x sin x - sin c x -c

(b) lim

and that lim 1(x) = lim u(x) = L. Then lirn f (x) exists and

0.707107

2

2

; lim t--o-

f (x) = 1

27. 11 43.

- cos t t

29. 9 31. 45. 0 47. 0

2

x

c - 0.01

c - 0.001

c + 0.001

c + 0.01

sin x - sin c x -c

0.999983

0.99999983

0.99999983

0.999983

4.

Negative 5. Negative

6.

As x -> oo,

-> 0, so 1 lim sin - = sin 0 = 0 x->00 X

On the other hand, -> ±oo as x -> 0, and as oscillates infinitely often.

-> ±c>o, sin

Here, c = 0 and cos c = 1 x sin x - sin c X -

C

c -0.01

c - 0.001

c ± 0.001

c + 0.01

0.868511

0.866275

0.865775

0.863511

1.

Here, c = i7- and cos c =

y = 1 and y = 2

3.

0.866025

x

c - 0.01

c - 0.001

c + 0.001

c + 0.01

sin x - sin c x -c

0.504322

0.500433

0.499567

0.495662

Here, c =

Section 2.7 Exercises

5.

(a) From the table below, it appears that

and cos c =

tim

x

c -0.01

c - 0.001

c + 0.001

c + 0.01

sin x - sin c x- c

0.710631

0.707460

0.706753

0.703559

x2X+2 1 =1

x

±50

±100

±500

±1000

f (x)

0.999600

0.999900

0.999996

0.999999

ANSWERS TO ODD -NUMBERED EXERCISES

ANS11

11. For each positive integer k, let f (x) = X k - COS X . Observe that f is continuous on [0, -72E] with f(0) = -1 and f( ) = (1.)k > 0. Therefore, by the IVT, there is a c E [0, I] such that f (c) = Ck - COS(C) = 0. 13. Let f (x) = 2X + 3X - 4X. Observe that!is continuous on [0, 2] with

(b) From the graph below, it also appears that x2 =1 Lim x-).±0.0 X-,, +1

f(0) = 1 > 0 and f(2) = -3 0, and f (- 116) = 21, 10 - 10 + 4 < 0.

f (-1) = 271 +

Therefore, by the IVT, there is a c E (-1, - TO such that

f (c) = 2c + + 4 = 0, so that 2c +

= -4.

(c) The horizontal asymptote of f is y = 1. 19. y = 13. -co 15. no 17. y = 7. 1 9. 0 11. and y = -3 21. y = 0 23. y = 0 and y = 10 25. lim f (x) = 3. The expression "never touches it" is incorrect. Here

17. Apply Corollary 2 to the Intermediate Value Theorem. f is a polynomial and continuous everywhere. f(-3) = 170, f(-2) = -25, f( -l) = 2, f(0) = -1, f(l) = 2, f(2) = -25, and f(3) = 170. Thus, f has a zero in each of these intervals: (-3, -2), (-2, -1), (-1,0), (0, 1), (1, 2), (2, 3), and f must have six distinct solutions.

the graph of f coincides with the horizontal asymptote y = 3 for all 33.0 x > 0. 27.0 29.2 31.

19. The IVT does not apply since g is not continuous on the interval [-I, 1]. However, this function does take on all values between g(-1) = land g(1) = 2.

4

A-

35. 1; the graph of y = 5-1/ t2 has a horizontal asymptote at y = 1 37. 0 39. oo 41. 0 A As = A. 45. (a) lim R(s) = lim = lim s->co s-oo K + s s-+00 1 + A AK AK = = -2- half of the limiting value. (b) R(K) = K +K (c) 3.75 mM

21. (a) f(l) = 1; f(1.5) = 21.5 - (1.5)3 < 3 - 3.375 0 and f(1.5) 0. Let 8 = 6/8 and suppose Ix -41

: We are done. • (I f (x) - LI < 1): This means < L < 3. In this case, let x= f (x) = -1 so 3 < L - f (x).

13. 3 = 0.05

In either case, there exists an x such that Ix' < 8, but I f (x) - LI > 29. Let c > 0 and let 8 = min(1, I). Then, whenever Ix - 11 0 for all t, the slope of the tangent line to g is always nonnegative. (c) t = 0, 22r, 4rr 53. f'(x) = sec2 x = cosi x • Note that f'(x) x has numerator 1; the equation f'(x) = 0 therefore has no solution. The least slope for a tangent line to tan x is 1. Here is a graph of f ' .

5.

4(x + sin x)3 (1 + cosx)

7.

(a) 2x sin(9 - x2)

9.

12

I -4 55.

dR = (LI. dB g

(COS2 19 -

I -2

31. 2x cos (x2)

\7_1 2

I 4

x

2v2

sin2 0) =

- 0- cos(20). g g

(2q, ) sin(rr/4) cos(rr/4) = ( T., 21 g f,(x)= iim cos(x + h) h->0

= 0 when

_ vP (Nh---2-) (4)

cosx

33.

t ../t2 +9

35. 7 2, (x4 - x3 - 11 1/3 (4x3 - 3x2) 37. 8(1 + x)3 3 (1 - x)5 sec (1/x) tan (1/x) 39. 41. (1 - sin 0) sec2 (0 + cos 0) x2 43. 40 cot(202 - 9) csc(202 - 9) 45. (2x + 4) 5ec2 (x2 + 4x) 47. cos(1 - 3x) + 3x sin(1 - 3x) 49. 2(4t + 9)-1/2 51. 4(sinx - 3x2)(x3 + cos x)-5 cos 2x x cos(x2) - 3 sin 6x 53. 55. sin 2x N/cos 6x + sin(x2) 57. 3(tan2 x sec2 x + X2 sec2 (x3))

59.

h

,. cos x cosh - sin x sin h -cosx = um h->0 = lim ((- sinx) h->0

sin h

+ (cos x)

cosh - 1 h )

= (- sin x) • 1 + (cosx) • 0 = - sinx

Section 3.7 Preliminary Questions 1.

d - f (g(x)) = - sin(x2 + 1)(2x) = -2x sin(x2 + 1) dx - g(f (u)) = -2 sin u cos u du

cos(20) = 0, and for 0 < 0 < 7r/2 that occurs only when 0 = rt/4. The corresponding range is

57.

(c) - sec2 x sin(tanx)

11. Multiplying out, f (x) = 4x4 - 20x2 + 25, so f'(x) = 16x3 - 40x = 8x(2x2 - 5). To use the Product Rule, write f (x) = (2x2 - 5)(2x2 - 5). We have f'(x) = (4x)(2x2 - 5) + (2x2 - 5)(4x) = 8x(2x2 - 5). Using the Chain Rule, f'(x) = 2(2x2 - 5)(4x) = 8x(2x2 - 5). 7 13. 12x3(x4 + 5)2 15. 2,./7x - 3 17. -2(x2 + 9x)-3(2x + 9) 19. -4 cos3 0 sin 0 21. 9(2 cos 0 + 5 sin 0)8(5 cos 0 - 2 sine) 23. (x + 1)(x2 + 2x + 9)-1/2 cos ,k/./c2 + 2x + 9 25. 2 cos(2x + 1) 27. 5ec2 x - csc2 x 29.

14 12 10 8 6 4

sin(x ) x2

(b)

sin(-1) - sin(1 + x) 61. (1 + cos x)2

+ 1 (z - 1)3/2

63. -35x4 cot6 (x5) csc2 (x5)

65. -180x3 cot4 (x4 + 1) csc2 (x4 -- 1) (1 + cot5 (x4 + 1))8 67. -36x2 (1 - csc2(1 - x3))5 csc2(1 - x3) cot(1 -x 3) 69.

1 - X2 ( x. 2X2

1 X

)3/2

71.

(a) The outer function is ,rx, and the inner function is 4x + 9x2.

(b) The outer function is tan x, and the inner function is x2 + 1. (c) The outer function is x5, and the inner function is sec x.

73. - 3- (kx + b)-4/3

(d) The outer function is x4, and the inner function is 1 + x 12.

77. -336(9 - x)5

75. 2 cos (x2) - 4x2 sin (x2)

79.

dv dP

= p

=i .s

2913. 3

m s • atm

ANSWERS TO ODD -NUMBERED EXERCISES dV = 1.67r (3)2 45.24 cm3/s. 81. (a) When r = 3, dt (IA When t = 3, we have r = 1.2. Hence,

y

21. y' =

= 1.67(0.2)2 P-1 7.24 cm3 /s.

dt 83. LAO = f8r cos (A-r5 t). December 1: L'(255)

25. y'

-0.019 h/day -L1 min/day. January 1: L'(286) 0.012 h/day 0.7 min/day. February 1: 0317) 0.04 h/day P.' 2.4 min/day. The lengths of the days are decreasing in late fall, but then are increasing once the winter starts. As the winter progresses, the rate of increase of the lengths of the days is increasing. 0.3600 kg/year 87. (a) -9 (b) -3/2 (c) 18 89. 5,N/ dollars dP 1 - 0.727 91. 12 93. -1- 95. year 6 dt r=3 dP 4.08569 x 10-13 (288.14- 0.00649h)4256 97. dh = 99. 0.0973 kelvins/year 101. f" (g(x))(g' (x))2 f' (g(x))g''(x)

85. W'(10)

df dv du

df dv dv dx

2xy2)

y2

19.

, 23. y =

1

=

9 x1/2y5/3 4

1 - cos(x + y) y) + sin y

COS(X

tan2 (x +y) - tan2(x) tan2 (x + y) - tan2 (y)

27. y' =

y + 3 sin(3x - y) -1 sin(3x - y) - x

29. Multiplying both sides of x yx-1 = 1 by x gives x2 + y = x, so the two define the same curve except when x =- 0. Since y = x - x2, X = 1 - 2x. differentiating the first form gives y' = 7 - x = X-xX2 1 31. -

1 33. y = - - x + 2

39. y4 = 2x +7r

35. y = -2x + 2

12

37.

32

2

41. The tangent is horizontal at the points (-1, ,I3-) and (-1, --V5). 43. The tangent line is horizontal at

(

(g(h(x))g' (h(x))h' (x)

dv du dx =

x2)

x2

(2x + 1)y2

-

103. Let u = h(x), v = g(u), and w = f (v). Then dw dx

(y2

17. y'= x(y2

ANS19

2,N/T3 13

13

2.At 13

and

13

45. At (0, 21/4), (1 /), = -22 /14/Z1 , and the tangent line has equation

107. For n = 1, we find

y = ( -22 1,4,Z1 ) x

Jr)

- sin x = cos x = sin (x ± 2 dx

dk

)

dx

=

,

and the tangent line

41. ) x - 21/4

has equation y = ( 1-

as required. Now, suppose that for some positive integer k, y x-f, sin x = sin (x

21/4. At (0, -21/4),

1 49. x = - , 1 ± 2 1 51. • At (1,2), y' = 3

47. (21/3,22/3)

Then

• At (1, -2),

knd dk+1 sin x = - sin (x + - ) dx 2 dxk+1

• At (1, 1),

=

11

11 • At (1, -1-),y' = --E

(k + 1)7r lor = cos (x ± - ) = sin (x + 2 2 )

53. There are vertical tangent lines at six points (+1, 0) and

(± 2 ' ± 2 ) -

Section 3.8 Preliminary Questions 1.

The Chain Rule

2.

(a) This is correct.

2y = 0 when y = 0, so the tangent • it follows that dx = dy dy 3x2 - 4 ' line to this curve is vertical at the points where the curve intersects the x-axis.

55. (b) This is correct.

(c) This is incorrect. Because the differentiation is with respect to the variable x, the Chain Rule is needed to obtain

dx

d - sin(y2 ) = 2y cos(y2)-4 dx dx 3. There are two mistakes in Jason's answer. First, Jason should have applied the Product Rule to the second term to obtain dy -dx (2xy) = 2x -dx + 2y Second, he should have applied the General Power Rule to the third term to obtain 2 dY - y = 3y dx

4. (b)

5. ±C ,2 = 0, _d x2 dx dx

=

dy d 2x, - y2 = 2y dx dx

57. By implicit differentiation, y' = T÷ T, and y" = 0 2Yx)2 . Solving the

Section 3.8 Exercises 1. 5. 7.

dy (2, 1), dx dl ( x2

+

2 d ( 3. x`y-') = 3x2 y2 y' + 2xy3 -dx 3 2)3/2) = 3 (x x2 ± y2

xy-2/3

yy

9.

2xy + 6x2y - 1 , 13. Y 1 - X2 - 2X3

i )

d y dx y +1 15.

=

1, I

'

2x 11. y 9y2 = 1)2 (y+ 3R 5x

x . Differentiating, we obtain original equation for y, we obtain y = 4x)3 . If we substitute y = 2 x into the 2x)2 and _ )7/ = expressions for y' and y" found by implicit differentiation, we obtain the expressions obtained by direct differentiation. 59. y" = (y2 - 2xyy')/Y4 = (322 - 2xYx/Y2)/Y4 = (y3 - 2x2)/y5 61. (a) y' = -y 2 /(2xy + 1); y'l(1, 1) = -1/3 (b) y" = (2y2 - 2xy2y' - 2yy')/(2xy + 1)2; y"I(1, 1) = 10/27 (16y2 + x2)3/2 63. (a) r= 64

ANS20

ANSWERS TO ODD -NUMBERED EXERCISES 133/2

(b) At (4, 0), r = 1. At (2, 0), r =

".% 5.86. At (0, 2), r = 8.

8

Section 3.9 Preliminary Questions 1. When x = -3,

= -18. When x = 2, di = 12. When x = 5,

= 30.

dt

When x = -4, id = 96. When x =2, i`

2.

= 24. When x = 6,

= 216.

dt

3. Let s and V denote the length of the side and the corresponding if a = 0.5 cm/s. volume of a cube, respectively. Determine dr dV if dV - 2 cm3/min, 5. Determine dh = 47rr2 4. dt dt dt dt dV if (2 = 1 cm/min. 6. Determine

g

y - 4x dx dy dx dy 67. - = 65. - = -6x-3dt dt 4y3 -x dt dt 69. The derivatives of these two curves are y' =xly and y' = -y/x, respectively. So at any point (x, y) satisfying both equations, the slopes of one tangent line will be the negative reciprocal of the other slope. That is, the tangent lines will be perpendicular. 71. • Upper branch:

Section 3.9 Exercises 1. 0.039 ft/min (a) 1007r

5.

270007r cm3 /min

9.

When h = 1, dJ.= 1.43 m/min.

7. 96007r cm2 /min

td=

2

11. -0.632 m/s 13. x -4

2

-2

21. 1.22 km/min

• Lower part of lower left curve:

100 13

25. (a)

-2

0.36 m/min. When h =2,

4.737 m; 4,4i

-I

23. 1200 241

0.405 m/s

4.98 rad/h

- 27.735 km/h

4.025 m

27. ,,/1-0 -3

cz%

10007r 15. 1047.20 cm3 /s 17. 0.675 m/s 3 (a) 594.6 km/h (b) 0 km/h 19.

4

-2

-4

(b) 247r =•--' 75.40 m2 /min

314.16 m2 /min

3.

1 33. - - rad/s 8

5 29. - m/s 3

(b) 112.963 km/h 31. -1.92 kPa/min

16 35. (b) when x = 1, L'(t) = 0; when x = 2, L'(t) = 3 37. -4,/3 -8.94 ft/s 39. -0.79 m/min

-2 -

• Upper part of lower left curve:

41. Let the equation y = f (x) describe the shape of the roller coaster -/T . track. Taking " Tit of both sides of this equation yields `i = f' (x)c!x 43. (a) The distance formula gives L =. .1(x - r cos 0)2 + (-r sin 0)2 -1 -

Thus, L2 = (x - r cos 61)2

r 2 sin2

(b) From (a), we have • Upper part of lower right curve:

dx dO dB 0 = 2 (x - r cos 0) (- + r sini9-) + 2r2 sin0 cos 0 dt dt dt (c) -807r ',-,' -251.33 cm/min

1

7 I X

45. (c)

250

',-=- •• 0.027 m/min

Chapter 3 Review

-2

1. 3; the slope of the secant line through the points (2, 7) and (0, 1) on the graph of f (x) 7 - ; the value of the difference quotient is larger than the value of 3. 3 the derivative 1 1 1 5. f'(1) = 1; y = x - 1 7.

• Lower part of lower right curve:

7

• x

1 13. f' (1), where f (x) = (2 - x)2 15. f (7), where f (t) = sin t cost 17. f(4) = -2; f'(4) = 3 9.

-2

-2x

11.

19. (C) is the graph of f' (x).

ANSWERS TO ODD -NUMBERED EXERCISES 19. L(x) = 4x - 3; f(0.96) 0.84 (1.147r) 21. L(x) =x - -7r + -; v0.5785 1 f 4 2 1 23. L(x) = --x 1; f(0.08) 0.96 2 1 25. L(x) =- -e(x ± 1); f(0.85) 2.5144 2 27. Ay -0.007 29. Ay -0.026667 31. f(4.03) ••-, 2.01

21.

23. (a) 8.05 cm/year (c) h'(3)

(b) Larger over the first half 6.0 cm/year (using the difference

7.8 cm/year; h'(8)

quotient approximation h t(t)

h(t+1? -11(t) )

25. At (t) measures the rate of change in automobile production in the United States; A'(1971) 0.25 million automobiles/year; (using the difference quotient approximation A' (t)

A(t+1?-A(t) )

33. A/2717 - A/2 is larger than •%/

_-

35. R(9) = 25110 euros; if p is raised by 0.5 euros, then AR euros; on the other hand, if p is lowered by 0.5 euros, then AR euros. 37. AL 39. P

-0.00171 cm 5.5 + 0.5 • (-.87) = 5.065 kilopascals

A'(1974) would be negative. 27. 15x4 - 14x

29. -7.3t 83

31.

41. (a) AW

1 - 2x - x2 (x2 ± 1)2

41.

x+ A/Tc )

(x+

1/2 ( 1 + -2 (x +

2w R2 h = R3

43. (a) Ah R=, 0.71 cm (b) Ah 1.02 cm effect at higher velocities. 45. (a) If 8 = 340 (i.e., t = ,73 7r), then 2(1+

2wh

0.0005 w h

(c) There is a bigger

x _1/2)) 2

43. -3/-4 5ec2(t-3) 45. -6 sin2 x c0s2 x + 2 c0s4 x 8 c5c2 1 + sect - t sect tan t 49. 47. (1 + cot 0)2 (1+ sec t)2 51. y' = -100x99 sin(x 57 57. -18 59. (-1, -1) and (3,7) 53. -27 55. - 16 1 il. a = - 63. 72x - 10 65. -(2x + 3)-3/2 6 69. -2ci =

67. 8x2 sec2(x2) tan(x2) + 2 sec2 (x2)

41P(R)Ax =

-0.7 pounds (lb)

(b) AW

33. 6(4x3 - 9)(x4 - 9x)5 35. 27x(2 + 9x2)1/2 3 2- z 39. 2x 37. 2(1 - Z)312

585 -585

625 (17 st(t)At = cos -457r) At 16

As

= 625 cos ( .1_7 7r Ao• -180 16 45 )

(b) If AO = 2°, this gives As 0.51 ft, in which case the shot would not have been successful, having been off half a foot. (c) As 47. AV

2.897 ft 47(25)2(0.5)

49. P = 6 atm; AP

cos(x y) dy = y2 ± 4x dy 73. 1 - 2xy dx = 1 - cos(x y) dx 4y2 x2 .x dzy dy 16y3 4y' dx2 dx 77. For the plot on the left, the red, green, and blue curves, respectively, are the graphs off, f', and f". For the plot on the right, the green, red, and blue curves, respectively, are the graphs off, f', and f". -11n640 0.00567 cm/s -0.407 cm/min 81. 79. 6)2

71.

0.255A0

3927 cm3; AS

±0.45 atm

87(25)(0.5) ,' JJ 314.2 cm2

51. f(2) = 8

53. ./16.2 •-•-1 L(16.2) = 4.025. Graphs off and L are shown below. Because the graph of L lies above the graph of f, we expect that the estimate from the Linear Approximation is too large.

83. 0.284 m/s 1 L(17) 0.24219; the percentage error is 0.14%. ,s/17 1 57. L(10.03) = 0.00994; the percentage error is 0.0027%. (10.03)2

s.

Chapter 4 Section 4.1 Preliminary Questions 1. True 2. g(1.2) - g(1)

0.8

3. f (2.1)

1.3

4. The Linear Approximation tells us that up to a small error, the change in output Af is directly proportional to the change in input Ax when Ax is small.

Section 4.1 Exercises 1.

3. Af

-0.00222

5. Af P.' 0.003333

Af

0.12

7.

Af

0.0074074

9.

Af

0.05; error is 0.000610; the percentage error is 1.24%. -0.03; error is 0.0054717; the percentage error is 22.31%.

11. Af

13. Af P.' 0.1; f(26) R.-, 5.1; error 0.00098 -0.0005; f(101) 0.0995; error 0.000004 15. Af 1 0.08333; f(9) 2.08333; error 0.00325 17. Af 12

59. (64.1)1/3 0.000019%.

L(64.1) 'A% 4.002083; the percentage error is

61. tan(0.04) L(0.04) = 0.04; the percentage error is .053%. 3.1/2 63. L(3.1/2) = 1.55; the percentage error is 0.0216%. sin(3.1 /2) 65. Let f (x) = .7x. Then f(9) = 3, f t (x) = lx-1/2 and f(9) = Therefore, by the Linear Approximation,

f (9 ± h) - f (9) = 9 + h - 3

- h 6

SO f"(x)I = X-3/2. Because this is a Moreover, f "(x) = decreasing function, it follows that for x > 9, 1 oo as x approaches 0 from the right, there is no maximum value. Using calculus, the minimum is 2,./ occurring at x = 55. (d)

,

, and +` ; the maximum value is

y and the minimum value is

f( ) = f( 2 ) = J (16s.) = f(1167 =

(e) We can see that there are six points where the graph has a horizontal tangent line on the graph between 0 and 27r, as predicted. There are four local extrema, and two points at (1, 0) and 0) where the graph has neither a local maximum nor a local minimum.

1.

(a) 3, 5, 7 (b) Maximum: 6, minimum: 1 (c) Local maximum 5 at x = 5, local minima: 3 at x = 3, and 1 at x = 7 (d) [2, 6] is an example. (e) (0, 2) is an example. (f) (4, 6) is an example. 3. x = 1 5. x = -3 and x = 6 7. x = 2 9. x = ±1 x = ±1 15. 0=! 11. t = 3 and t = -1 13. x = 0, x = 2 17. (a) Critical point at x = 2; f(2) = -1 (b) Minimum: -1, maximum: 17 (c) Minimum: 1, maximum: 71 7r maximum value: ,,n; minimum value: 1 19. x 4 21. Maximum = 5, minimum = 3. Maximum 4 3 Minimum 2

57. Critical point: x = 2; minimum value: f(2) = 0, maximum: f(0) = f(4) = 2 59. Critical point: x = 2; minimum value: f(2) = 0, maximum: f (4) = 20 15 63. c= — 4 65. f(0) 0, so there is at least one root by the Intermediate Value Theorem; there cannot be another root because f (x) > 4 for all x. 61. c = 1

67. There cannot be a root c > 0 because f ' (x) > 4 for all x > 0. 23. Minimum: f(-1) = 3, maximum: f(2) = 21 25. Minimum: f(0) = 0, maximum: f(3) = 9 27. Minimum: f(4) = -24, maximum: f(6) = 8 29. Minimum = -19, maximum =3. 31. Minimum: f(1) = 5, maximum: f(2) = 28 33. Minimum: f(2) = -128, maximum: f(-2) = 128 35. Minimum: f(6) = 18.5, maximum: f(5) = 26 37. Minimum: f(1) = -1, maximum: f(0) = f(3) = 0

71. (a) F =

— 2

(1 ±

v2)

(b) F (r) achieves its maximum

V1

value when r = 1/3. (c) If 1)2 were 0, then no air would be passing through the turbine, which is not realistic. 75. • The maximum value off on [0, 1] is 0 f

39. Minimum: f(0) = 2,/g '-'.• 4.9, maximum: f (2) = 4,..h 'r ,' 5.66 41. Minimum: f

2 VI

2

P.,- -0.589980, maximum: f(4) '- --• 0.472136 4

1/0 - 1

=

ryl(b—a)

r

) bl(b—a)

ANSWERS TO ODD -NUMBERED EXERCISES 77. Critical points: x = 1,x = 4, and x = ;; maximum value: minimum value: f(-5) = f(1) = f(4) =

ANS23

• The function graphed here is discontinuous at x = 0.

1.2 1

2

-4 -3 -2 -1

3

Section 4.3 Preliminary Questions

4

79. (a) There are therefore four points at which the derivative is zero: (-1,

(1, -

(-1,

1.

m =3

2. (c)

3. Yes. The figure below displays a function that takes on only negative values but has a positive derivative.

),(1, ,./2)

There are also critical points where the derivative does not exist: 127, 0) (0, 0), (±-4 (b) The curve 27x2 = (x2 + y2)3 and its horizontal tangents are plotted here.

83.

81. 10

4

(a) f (c) must be a local maximum f (x) = sec x and f (x) = csc x

6.

f(X) = I sin x I is an example.

2

3. c=

3n4

or

85. If f (x) = a sin x + b cos x, then f' (x) = a cos x - b sin x, so f' (x) = 0 implies a cos x - b sin x = O. This implies tan x = .Then sin x =

a2 + b2

and

cos x =

±b a2 + b2

5. c = ±V7

1.

c=4

7.

c=0

9.

The slope of the secant line between x = 0 and x = 1 is f(1)- f (0) =1 1- 0

0

±a

(b) No

Section 4.3 Exercises

3 -10

4. 5.

1 Since f' (x) = 2x, solving 2c = 1 gives c = - . A graph off and the 2 tangent line appears here:

Therefore, f (x) = a sin x + b cos x = a

±a ,/ a2 +b2 a2

-

+b

±b V a2

b2

= ±.1a2 + 62 ., /a2 + b2

87. Let f (x) = x2 + rx + s and suppose that f (x) takes on both positive and negative values. This will guarantee that f has two real roots. By the quadratic formula, the roots of f are x=

-r

r2 - 4s

2 Observe that the midpoint between these roots is 1 (- r 2

r2 - 4s 2

1.4 1.2 1.0 0.8 0.6 0.4 0.2

b2

-r -

r2 - 4s) 2

2

and, because the graph off is Next, f' (x) = 2x + r -= 0 when x = an upward-opening parabola, it follows that f (-L,) is a minimum. 89. b> ,t a2 91. • Let f be a continuous function with f (a) and f (b) local minima on the interval [a, b]. By Theorem 1, f (x) must take on both a minimum and a maximum on [a, b]. Since local minima occur at f (a) and f (b), the maximum must occur at some other point in the interval, call it c, where f (c) is a local maximum.

-0.2

11. c =

0.2 0.4 0.6 0.8 1.0 1.2

ANSWERS TO ODD -NUMBERED EXERCISES

ANS24

37. c = (D2/5

13. The slope of the secant line between x = 0 and x = I is 2—0

f (1) — f (0)

=2 = 1—0 It appears that the x-coordinate of the point of tangency is approximately 0.62.

0.5

f' f

— \

0.56

0.6

0.64

15. The derivative is positive on the intervals (—no, 1) U (3, 5) and negative on the intervals (1, 3) U (5, 6). 17. f(2) is a local maximum; f(4) is a local minimum.

o m

,cc)

+ 7

0

(0, co)

f'

+

o

—

f

/

m

\

43. c = 0

21. 8

10 -

( ( i )2/5

(—no, 0)

x

0.3 0.52

Y

32/5

41. c = 0

0.4

y = 2 x - 0.764

19.

(0, (3)2/5)

39. Critical points: x = —1 (local maximum); x = 1 (local minimum); (—co, —1) increasing, (-1, 0) decreasing, (0, 1) decreasing, (1, co) increasing

0.6

5 + X2

.1' =

X

x

(-00, 0)

0

(0, co)

f'

+

o

+

f

/

—

i f

6

8-

45. c =

and c = 7r

6-

20

1

2

3

4

27. c = 3x

(—cc, )

7/2

( , oo)

F

+

0

—

i

i t

i.

(1,7r)

7r

(7r, 27r)

f'

+

0

—

0

+

f

/

lvi

N

m

/

m

(—oo, 0)

0

(0,8)

8

(8, no)

f'

+

0

—

0

+

f

/

m

N

m

/

1

(1, co)

f'

—

0

+

0

—

0

+

f

N

m

ir

M

\

m

7

33. c = —2, —1 x

(—co, —2)

—2

(-2, —1)

—1

(-1, co)

f'

+

o

—

0

+

f

/

M

\

m

/

(—co, 0)

0

(0, co)

f'

+

0

+

/

0

/

Hz

f' f

+ 7

0

0

+

M

m

i v

\

(

117r 6

( 117r 1,,,,) 6 ' '-'"

—

0

+

\

m

/

2 '

0 m

'' t'

3,r\

n 7

_37r 2

'Jr 77r\ l7 , —6- /

( 77r

7r \ l' , 7/

117r ) 6

49. c = 0; f' is positive on (—cc, 0) and on (0, cc) and is undefined at 0; f is increasing on (—cc, 0) and (0, cc); x = 0 is not a local minimum or maximum.

31. c = —2, —I, 1 (—co, —2) —2 (-2,—i) —1 (-1, 1)

37r 'and 2

(II

x

f

x

x

72T 6

f'

N

35. c = 0

47. c =

X

29. c = 0, 8

f

(0, -J i)

5

23. Critical point: c = 3; since the derivative changes sign from + to —, this is a point of local maximum. 25. Critical points: c = —2 and c = 0. Since f'(x) changes sign from + to — at c = —2, this is a point of local maximum. Since f'(x) changes sign from — to + at c = 0, this is a point of local minimum.

x

x

4

3

4-

x

(—co, 0)

0

(0, 00)

f'

+

o

+

f

/

—

/

53. a > 0 69. (b) First Derivative Test does not apply

Section 4.4 Preliminary Questions 1. (a) increasing 2. f (c) is a local maximum. 3. False 4. False. For example, with f (x) = x4, we have f"(0) = 0, but there is not an inflection point at x = 0 since the concavity does not change there. 5. No. An inflection point is a point on the graph of a function where the concavity changes. Since f is not defined at x = 0, there is no point on the graph of f at x = 0 and therefore no inflection point corresponding to the change in concavity that occurs going from negative to positive values of x. 6. Yes. For example, for f (x) = x3, there is a critical point at x =- 0, and (0, 0) is an inflection point.

ANSWERS TO ODD -NUMBERED EXERCISES

x (— 0., 3) 3 (300.) f" _ o +

Section 4.4 Exercises 1.

(a) In C, we have f"(x) 0 for all x. (d) In D, f"(x) goes from — to +. Point of inflection is (- 43 '

5.

Concave up everywhere; no points of inflection

27 )

, 3-; concave down for 7. Concave up for x < —0 and for 0 0, which means there are no inflection points.

15. Concave up for Ix' > 1; concave down for Ix' < 1; point of inflection at both x = —1 and x = 1

21. Near the point of inflection, the curve is roughly a straight line going through (55, 200) and (35, 100), so the rate of change is roughly 2°0 55 i3i5oo = 5 cm/day. So when the growth rate starts to slow down, the height is growing at about 5 cm/day. Plots of the first and second erivatives are

—

o

f"

13. Concave up on (0, 1); concave down on (—co, 0) U (1, no); point of inflection at both x = 0 and x = 1

17. (—oo, —1) concave up, (-1,0) concave down, (0, oo) concave up; inflection point: (0, 0)

I

(— 00,o)

9. Concave up for 0 0. In B, f is increasing and concave up, so f t > 0 and f" > 0. In C, f is increasing and concave down, so f' > 0 and f" 0 and f" 0

Now, if h is sufficiently small but negative, then f'(c + h) must also be negative [so that the ratio f'(c + h)I h will be positive] and c h < c. On the other hand, if h is sufficiently small but positive, then f'(c + h) must also be positive and c h > c. Thus, there exists an open interval (a, b) containing c such that f'(x) 2 and decreasing for -2 oo 8 27 189 + - • 222 59. RN = 2 + -6- + - ; 2 57. RN = 222 + N2 N N N2 (b - a)2 ; (b2 + b) - (a2 + a) N 61. RN = (b - a)(2a + 1) + (b - a)2 +

f

3 2x dx = 9 -9 = 0

63. The area between the graph of f (x) = x4 and the x-axis over the interval [0, 1] 65. The area between the graph of y = x4 and the x-axis over the interval [-2, 3] 67.

69.

E sin

. 7r lim RN = lim N-roo N

N-roo

lim

N =

ii111 -

3. The region bounded by the graph of y = 3x + 4 and the x-axis over the interval [-2, 1] consists of two right triangles. One has area 1( below the axis, and the other has area I2 ( 2)(7) = -4Z above 2 2 3 )(2) = 3 the axis. Hence,

k=1 N-1

E 1/15

4 lim LN = lim N-roo N-roo N

71.

kr ) -

E tan

1

1 1 ( - +2 2N

1))

49 L5 J-2(3x + 4) dx = - = 6 3 2

73. Represents the area between the graph of y = f (x) = N1 /- -x2 and the x-axis over the interval [0, 1]. This is the portion of the circular disk x2 + y2 < 1 that lies in the first quadrant. Accordingly, its area is 75. Of the three approximations, RN is the least accurate, and then LN and finally MN are the most accurate. 77. The area A under the curve is somewhere between •'-% 0.768.

L4

0.518 and

79. f is increasing over the interval [0, r/2], so 0.79 L4 < A < R4 1.18. 81. L100 = 0.793988; R100 = 0.80399; L200 = 0.797074; Rak = 0.802075; thus, A = 0.80 to two decimal places. 83. R100

1.4142 85. R100

5. The region bounded by the graph of y = 7 - x and the x-axis over the interval [6, 8] consists of two right triangles. One triangle has area (1)(1) =- above the axis, and the other has area (1)(1) = below the axis. Hence,

0.9946; guess: area = 1

87. Graph off(x)

1

Right endpt approx, n=1

Right endpt approx. n=2

8

1 6.

0.8 0.6

0.5 -

(7

1 1 - x)dx = - - - =0 2 2

0.5

0.4 0.2 1 1 1 0.2 0.4 0.6 0.8

0 0.5

0

05

89. When f' is large, the graph of f is steeper and hence there is more gap between f and LN or RN. 93. N > 30,000

Section 5.2 Preliminary Questions 1. -

2

2. (a) False. f c:i f (x) dx is the signed area between the graph and the x-axis. (b) True (c) True 3. Because cos(r - x) = - cos x, the "negative" area between the graph of y = cos x and the x-axis over [11, r] exactly cancels the "positive" area between the graph and the x-axis over [0, 11]. -5

4.

LI

8 dx

7. The region bounded by the graph of y = ,./25 - x2 and the x-axis over the interval [0, 5] is one-quarter of a circle of radius 5. Hence,

f125

- x2 dx = 1 71-(5)2 = 257( 4 4

ANS34

ANSWERS TO ODD -NUMBERED EXERCISES

9. The region bounded by the graph of y = 2 - Ix I and the x-axis over the interval [-2, 2] is a triangle above the axis with base 4 and height 2. Consequently, f 2 J -2 (2

11. (a) Ern RN

=

12

At = 1, f

E(t)dt

E(0)At + E(1)At + • • • + E(11)At

(1.1)(1) + (1)(1) + (1)(1) + (1.1)(1) + (1.3)(1) + (1.8)(1) + (2.5)(1) + (2)(1) + (0.3)(1) + (0.5)(1) + (0.4)(1) + (-0.8)(1) kilowatt-hours 25. r 27. 4

- Ix I) dx =

50 Ern (30 - -)

23. Using a left-hand Riemann sum approximation with

12.2

29.

= 30

(b) The region bounded by the graph of y = 8 - x and the x-axis over the interval [0, 10] consists of two right triangles. One triangle has area 1(8)(8) = 32 above the axis, and the other has area (2)(2) = 2 below the axis. Hence, 10

fo

13. (a) -

71.

(b)

(8- x)dx = 32-2 = 30

37 -F

5 3 g(t)dt = - ;f g(t)dt = 0 2 0 17. The partition P is defined by 3

15.

xo = 0


0 and c > 0) is an exponential decay relationship.

(b) lim f (x) = lim x(2 + sin x) > lim x = no and X 00 x->00 lim g(x) = lim (x2 + 1) = no, but x->00

0.02; 136 rats after 20 months

29. P(t) = 204eae0.15t with a

f i(x) x (cos x) + (2 + sin x) = lim gf(x) x->co 2x

lim

31. (a) The present value of the reduced labor costs is e-0.16

7000(e-008

e-0.4,) = $27,708.50

e-0.24 ±e -0.32

does not exist since cos x oscillates. This does not violate L'Hopital's Rule since the theorem clearly states

This is more than the $25,000 cost of the computer system, so the computer system should be purchased. (b) The present value of the savings is

or co c' c)

6=0.25

2. No

4 3 2

1

1 5

10

It is a continuous function.

L'Hopital's Rule does not apply. L'Hopital's Rule does not apply.

5.

L'Hopital's Rule does not apply.

7.

L'Hopitar s Rule does not apply.

51. )(m-n)/2

COS nx

0 (_1)(171

53. (a) no

(b) 1

m, n even m even, n odd m odd, n even m, n odd

does not exist, -

9000 55. 79-

-1)

/ 2

0; therefore,

lim f (x) = 0; lim f (x) = e° = 1 x->co x-o+ (b) f is increasing for 0 e, and has a maximum at x = e. The maximum value is f (e) = el le 1.444668.

59. (a)

, = lim 1 = lim 63. urnx- >CX) a X OO axa -, x->co x' 67. (a) 1 < 2 + sin x < 3, so

61. Neither

3x

x(2 + sin x) -

x2 +l

It follows by the Squeeze Theorem that hill x->co

10

5

15

t -

=

k

t->00 et2

=

0

by Exercise 68. 73. For x Assume

(x) = e-1lx2 (3). Here, P(x) = 2 and r = 3.

0,

f (k) (X)

=

P(x)xer I

.

Then

f (k+i) (x) = e _1 x2 (x 3 P'(x) + (2- rx2)P(x)) xr+3 which is of the form desired. Moreover, from Exercise 70, f'(0) = 0. Suppose f 1(k)(0) =

f(k)(x)

f (k+i)(0) =

xl

-

(k) (0)

= 0. Then

Px) ( xre+-11/x2

= P(0) lim f (x, = 0 x-*0

(b) Here, f (x) -> no, so ln f (x) -> no, and g(x) g(x) ln f (x) has the form 0 • no.

x2 + 1

X

115

lim x_,•0 xk e i/x2

183.7m

0, and g(x) -> Do; therefore, 57. (a) Here, f (x) -+ 1, so ln f (x) g(x) In f (x) has the indeterminate form no • 0.

x2 + 1 -

12=3.0

= limo xkellix2 . Let t = 1/x. As x -> 0, t -> Do. Thus,

71. limo

9 0 11. Quotient of the form 22 ; -" 2 13. Quotient of the form ; 0 15. Quotient of the form ÷:3) ; 0 9 7 5 3 27. 1 29. 2 31. -1 23. 7 25. 17. g 19. - 3 21. 7 7 2 1 39. 1 41. Does not exist 43. 0 35. 0 37. - 33. 7r 45. Ina 47. e 49. e-312

lim

6 5 4 3 2

15

X

10

9.

x-r/2

10

5

4 3 2

3.

cos MX

15

b=2.0

Section 7.5 Exercises

b=0.5

I

4 3 2

4. The function does not have an indeterminate form at x = 0. It has the form 0', corresponding to a limit that equals 0.

1.

Thus, G(b) = 1 if

(b)

You separately 3. You do not apply the quotient rule on differentiate the numerator and denominator, and work with the new rational expression obtained this way.

5.

lim (x) x->oo g, (x)

69. (a) Using Exercise 66, we see that G(b) = e". 0 < b < land G(b) = b if b > 1.

Section 7.5 Preliminary Questions g

lim f (x) x->oc g(x)

"provided the limit on the right exists."

$27,708.50 - $25,000 = $2708.50

1. Not of the form

ANS45

x(2 + sin x) =0 X2 + 1

77. lim sinx = lim = 1. To use L'Hopital's Rule to evaluate x->0 x->o lim -th1 "1 , we must know that the derivative of sin x is cos x, but to x-13 /.

x

determine the derivative of sin x, we must be able to evaluate lim x->o x 79. (a) e-1 /6 0.846481724 x

(

=0

sinx) 1/X2 -

1

0.1

0.01

0.841471

0.846435

0.846481

x

(b) 1/3 x 1 2 sin x

1 --- x

±1

±0.1

±0.01

0.412283

0.334001

0.333340

ANSWERS TO ODD -NUMBERED EXERCISES

ANS46

Section 7.6 Preliminary Questions 1. (b) and (c) 2. 0 = 37r; no 3. The sum of the two non-right angles in a right triangle is 1. It tells us that the derivatives of the two functions are negatives of each other (since their sum is zero). 4.

5. x = 4u

b=

to x = a horizontally and from y = 0 to y = Ina vertically and then subtracting the area of the shaded region. This yields a

Ina

In x dx = a ln a

In a

eY dy = eY

Section 7.6 Exercises 5.

Jo

7. 2.

-2x

35. ,s

43. 49. dy dx

X4

2

13. n-

21. N/3 23. 3 N/ Ix2- 1 7 29. 5/4 31. 33. -ff A/1 _49x2 ex 41. 39. 37. tan-1 x + x2 x± , - e2x

15. Not defined 17. Nr-1 x 25. An-

11. -

9.

a

19.

lnxdx = a ln a - (a -1) = a lna - a + 1 1- t

(d) Based on these results, it appears that

J

./l - t 2

d , -1 cos- (ln x) = 45. 0 47. dt x2 +l x 1 - 1n2 x cos y = x, so implicit differentiation gives - sin(y)y/ = 1, or -1 = - csc y = _ x2

tan y = ±A/x2 - 1. If 51. Since 1 + tan2 y = sec2 y, and sec y = x, tan y >0. If x 1, arcsec x = y E [0, arcsec x = y E (7r/2, 7] tan y = -,/x 2 - 1 0 for all x. Now, sinh x + cosh x =

ex - e-x 2

ex +e' 2

51. Local minimum at x = 1; local maximum at x = e-2; point of inflection at x = ; lirn x(logx)2 = 0; lim x(logx)2 = oo. x-ArE

= ex 7 2).

so x = In(sinhx + cosh x). Finally, sinh-I t = ln(t + 43. Let A = tanh-I t. Then t = tanh A =

1.0

sinh A eA - e-A cosh A - eA + e- A

0.8

Solving for A yields 0.6

1 1+t A = tanh-I t = - In 2

0.4

45. (a) By Galileo's law, w = 500 + 10 = 510 m/s. Using Einstein's law, w = c • tanh(1.7 x 10-6) 510 m/s. (b) By Galileo's law, u + v = 107 + 106 = 1.1 x 107 m/s. By Einstein's law, w c • tanh(0.036679) r-z% 1.09988 x 107 m/s. 49. (d)

t io,17.1,

limv(t)=, k

150(78,545.5) =

icto2

Vm

=17.1=."7

0.2

53. 55.

s = y(M) - y(0)

63. + C - (a cosh 0 + C) = a cosh (-) a

(b)

da dL =

ia il

(2 sinh (4) -

(b) (1n 2)2x

3.

f -I (x) =

7.

A follows upon

1}; range: {y : y

; domain {x : x

1)

17. (4 - 2t)e4t - t2 19. cot ex+inx 21. eexxiztax 23. ( 1 + 27. - 21n 7 • 7-2x 31.

25. -21-Y 1n2

29. v ,s4„.2 1

Ix' csc-I x•‘ /. -1 37. 2t cosh(t2) 39.

33. 1 + ins

35. (sin21 t)(21n sin t + 2t cot t)

45. y = 0 and y = 14.72 47. Global minimum at x = ln(2); no maximum. 49. Local minimum at x = e-I ; no points of inflection; lirn x lnx = 0; lim x lnx = oo. „x->00 x->o+

41. y =

+6

eb(a-u))2

=7

>0

(lnx)3 + C

67. - sin-1(ex) + C

69. tan-1(1n t) + C 73. l• cos(e

)+C

71. -1e9-2x + C 75.

85.

sin-I x2 + C

91.

(sin-I x)2 + C

77.

(e9 - e) 81.

87. 2(e _

2)

2

_2

89.

;•-- 3.63

83. 1 1n2

1n2

93. 4 - tan ' (4/)

95. g %---•12.27 hr-1-K-1; approximate change ink when T is raised from 500 to 510: 122.7 hr- I . 97. P'(t) = .05P(t) - 2,000

(d) (i) 9. le 13. _e---(x+1) 15. ins

(a)(iii) (b) (iv) (c)

11. - 36e-4x

beNa- u)

79. sin-1 (3) - sin- ' (1)

x-(x-1) =

g(g(x))=

f' (u)=

1.38 x 1028 at

1 (b) u = a + - ln 2 65.

Review

1.

(a)

R-=

a

al'I cosh (-11a-f• )) -1

,ct:,i • A.

5.

-

sinh (11 a-) - 1

The formula for (c) By the Chain Rule, A, = substituting the results from parts (a) and (b). Chapter 7

= 4e(x-1)2 e(x-3)2 (x -2)

g

51.

(a ) ds 1) da = COSh( 4

8 ) 4x-2

57. v = e3x(x(ffF1)22)2 ( 3 ± x}1) 59. v R% 29.2 m/s, 0.068 m/s per m 61 dE 1.38 x 1021 at Mzo = 3; g

1,178.18 lbs/mi

= a cosh (-) a

3 (4x-2)2 \x+1

-

99.

(a) lim tan-1 x = and lim tan-I x = imply that x->oo lim R(x) = 0 and lim R(x) = M and therefore R = 0 and R = M

are horizontal asymptotes of R. (b) R' (x) - 7, (1m +x2) > 0 for all x. Therefore R is increasing. (c) R"(x) - ,r(-12±frI A 2 . R"(x) > 0 for x ,,(21c -1 lim Cs' (s sin(at) + a cos(at)) -a 89. Lf (s) = 2 S 93. = 91. 4_ 71,

s, = (.; 1) =

95. E = 42 87 f oCC

dv. Because a > 0 and 87111c3 is a constant, we know E is finite by Exercise 92. ' dt x dt - > lnx. Thus, > 97. Because t > ln t fort > 2, F(x) = i2 In t I2 t F(x) -> oo as x -> no. Moreover, F(x) 1 lim is of the form lim G(x) X->00 1/x X->00 X->00 x 00 G(x) no/no, and L'Hopitars Rule applies. Finally, lnx F(x) lim - 1. L = Um = urn In x x->0.0 111 . x->00 lux - 1 X-00 G(x) (In x)2 99. The integral is absolutely convergent. Use the Comparison Test with

ro2 - a2) s(b - a) = f f (x)dx 2 a 63. (a) This result follows because the even-numbered interior endpoints overlap:

E

(N-2)/2

= b6

i=0

a , L(Yo + 4yi

y2) + (y2 + 4y3 ± y4) ± ...1

b- a =1..Yo + 43'1 + 2)72 +4y3 2),4 ± • • • + 4yN-i + ynd = SN 6 (b) If f (x) is a quadratic polynomial, then by part (a), we have f (x) dx a 65. Let f (x) = ax3 bx2 cx + d, with a 0 0, be any cubic polynomial. Then f (4) (X) = 0, so we can take K4 = 0. This yields is exact for all cubic error(SN) 1810)N4 - 0. In other words, SN polynomials for all N. SN =

+

+ • • • + 4 1-2 =

ANS52

ANSWERS TO ODD -NUMBERED EXERCISES

Section 9.1 Exercises

Chapter 8 Review

1. c = 2; P(0 < X < 1) =

1. (a) (v) (b) (iv) (c) (iii) (d) (i) (e) (ii) c sin: 10 1 7 tan Osec3B _L tan %ec ± ,16 In 1 sece + tan 01+ C 5. tan Osec50 6 24 7. sec-l x + C 9. 2tan-1,rx + C ,Vx 2 -1 tanx-ix ln 11. ln (1 + x2) + C 03026e cosNe c .2. _5_ 4 - 1 72 60 ' " 32 e - 8• 50 15. 17. 51n lx - 11± ln Ix + 11 C 19. tn 36 ± tan 0 + C 21. ln 8 - 1 cos55e 2c03s3e cos 8 + C 25. 23. 4 27. 3 (tan z)3/2 + C 29. ± sj 6, - si4 9 + C 31. -1 u3 + C = - cot3 x + C 3.

sin990

33. j: //: cot2 (x)csc3 (x) dx = A ln 3 + I 7

t+4 35. 49 1,m t-3 dx

+C

1n % - 2

C

a 0 +C

C=

y

1 ' 2 see- ' 2 + C

1 j_ t-3 - 2 ta =

ln (V-2 - 1) +

1 for any function f, we have 4 length of graph of y = f (x) over [1,4] = f

1 + f' (x)2 dx

>f

4 1 dx -= 3

Section 9.2 Exercises 1. 7.

= f26

+ 16x6 dx

-..1.7-(22,/-n - 13,11 )

3 13 5. 3,7-111 ' 12 9. e2 1n22

11. f i2

+x6 dx a3.957736

15. f 13

+ idx = 2.29896

13. f i2 17.

±

320.0

dx

19. 6

1.132123 21. 7.6337

23. a = sinh-1(5) = ln(5 + ,T2T5) 27. Let s denote the arc length. Then s= + 4a2 + ln 1,/1 + 4a2 + 2al. Thus, when a = 1, s= + + 2) 1.478943. 1+e-1 v,1+e2.+1 37. 167r 39.

Chapter 9

29. N/1 +e2a ±

Section 9.1 Preliminary Questions

35. 1.552248

1. No, p(x) > 0 fails. 3. p(x) = 4e -4x

41• IT32 (132,717 - In (4 + J17))

ln

+

ln

(1453/2 - 1) 53.23

31. ln(1 +

-

ANSWERS TO ODD -NUMBERED EXERCISES 43.

45.

3rr

(e4 _ y)

49. 22z- /02 e-x2 /2 V1 3. 472hr 57.

55.

261.13

47. 27 f i3

x2e- x2 dx

(1n(

+ x-4 dx

8.222696

7.60306

7.

A sketch of the lamina is shown here.

51. 27 In 2 +

45 + 12) + 12,11-47) R-J 77.33

59. 27b2 +

8

sin-1 (•Va2 -b2

27rba2

6

a

A./;i-171

4 2

Section 9.3 Preliminary Questions

I

0

1. Pressure is defined as force per unit area. 2. The factor of proportionality is the weight density of the fluid, w = pg.

(a)

=

M

y

0.5

1

1.5

F

2

i

= 243

(b) Area = 9 cm2; center of mass: 0, E)

3. Fluid force acts in the direction perpendicular to the side of the submerged object.

9.

4. Pressure depends only on depth and does not change horizontally at a given depth.

11. (a) M = 24 (b) M = 12, so ycn, = 2; center of mass: (0,2)

5. When a plate is submerged vertically, the pressure is not constant along the plate, so the fluid force is not equal to the pressure times the area.

13. (4, 4)

Section 9.3 Exercises 1.

x

2.5 3

Mx =

+ 6 16 ;

My

ap; center of mass: 0, 16\

=

15. (N, 1) .`5

17. 0, 150

( 1 - 5e-4

1 - e-8 19. 21. (5-, i) 1 - e-4 4(1 - e-4) 23. A sketch of the region is shown here.

(a) Top: F = 176,400 newtons; bottom: F = 705,600 newtons

(b) F

5

8

E pg3y) Ay

(c) F = f 2 pg3y dy

4

.1=1

3

(d) F = 882,000 newtons

2

3. Difference = force when top edge is 1 meter below the water - force when top edge is level with water = 19,600 newtons 5. (a) The width of the triangle varies linearly from 0 at a depth of y = 3 m to 1 at a depth of y = 5 m. Thus, f (y) = 4(y - 3).

1

b) The area of the strip at depth y is 4 (y - 3),Ay, and the pressure at depth y is pgy, where p = 103 kg/m3 and g = 9.8. Thus, the fluid force acting on the strip at depth y is approximately equal to pgy(y - 3),Ay. (c) F (d) F =

E pg 127,400

0

• x 2

The region is clearly symmetric about the line y = 3, so we expect the centroid of the region to lie along this line. We find Mx = 24, My = centroid: 3).

9\ A)

9, 25. ( A

27. (

(y - 3) Ay ->f pgly(y - 3) dy 29. newtons

19100 r 3 newtons

11.5

01.5

1 e2 - 3 2(e - 2) 4(e - 2) 1 )

4W2 - 1)

4(4-

31. A sketch of the region is shown here. Centroid: (0, 4)

7.

(b) F = -

9.

F = 200 r3 + 49007mr2 newtons

328,224,000 lb

11. F

815,360

newtons 13. 333,200 newtons 15. F = 17. F 6153.18 newtons 19. F 5652.4 newtons 21. F = 940,800 newtons 23. 5.4604 x 1011 newtons 25. F = (15b + 30a)h2 lb H 3;

27. Front and back: F -

slanted sides: F =

621 4 t H2

1. m, =

My

=

0

2. Mx = 21

3. M, = 5; My

=

10

4. Because a rectangle is symmetric with respect to both the vertical line and the horizontal line through the center of the rectangle, the Symmetry Principle guarantees that the centroid of the rectangle must lie along both these lines. The only point in common to both lines of symmetry is the center of the rectangle, so the centroid of the rectangle must be the center of the rectangle. 5. If the plate looks like a ring, then the center of mass doesn't occur at any point on the plate.

37. (0,

(a) -

3.

(a) M = 4m; My

(b) ( 44, PO

(b) =

9m; center of mass:

(

, 1)

i(r2-h2)3/2 )

r2

(0, 47,3_'43.,

)

;

4: with r = 1 and h =

(0,0.71) 41. (0, PO

39. V =

43. (_ 947,

)

45. For the square on the left: (4, 4); for the square on the right: (4, 24)

Chapter 9 Review 1.

3ection 9.4 Exercises 1.

35. (t,

33. (0, #)

Section 9.4 Preliminary Questions

3. C = 2; p(0 < X < 1) = 1 (b) 0.49997

5.

(a) 0.1587

7.

+1 09

9. 4,./. 17

13. 247,4

15.

677 36

17. 12.7r ± 4.72 19. 176,400 newtons 21. Fluid force on a triangular face: 183,750, + 306,250 newtons; + 294,000 newtons fluid force on a slanted rectangular edge: 122,500

ANSWERS TO ODD -NUMBERED EXERCISES

ANS54

23. Mx = 20,480; My

25,600; center of mass: (2, g)

=

25.

63. y = Cx3 and y = ±.1A 65. (b) v(t) = -9.8t + 100(1n(50) - ln(50 - 4.75t)); v(10) = -98 + 100(1n(50) - ln(2.5)) 201.573 in/s 14n R5/2 71. (c) C = ise,/2,T

Chapter 10 Section 10.1 Preliminary Questions 1. 2.

(d) Order 2

4. b, c, d, e

3. a, c, d, f

b, d, e

(c) Order 3

(b) First order

(a) First order

Let y = 4x2. Then y' = 8x and y' - 8x = 8x - 8x = 0.

3. Let y = 25e-2x2 . Then y' = -100xe-2x2 and y' + 4xy = -100xe-2X2 4x(25e-2x2 ) = 0. 5. Let y = 4x4 - 12x2 + 3. Then y" - 2xy' + 8y =(48x2 - 24) - 2x(16x3 - 24x) + 8(4x4 - 12x2 +3) = 48x2 - 24 - 32x4 + 48x2 +32x4 -96x2 +24 = 0 7.

y(t) = -1 and y(t)= 0.001x - 1.001

9.

(d) y =

-

Section 10.2 Preliminary Questions 1.

Section 10.1 Exercises 1.

(b) Since -1 (ye-b ) = 0 [by part (a)] the corollary implies that ye-(t = D for some constant D. Therefore, y

y(t) = 5 - ee4t for any positive constant c 2. No 3. True 4. The difference in temperature between a cooling object and the ambient temperature is decreasing. Hence, the rate of cooling, which is proportional to this difference, is also decreasing in magnitude.

Section 10.2 Exercises 1. General solution: y(t)= 10 + ce2t; solution satisfying y(0) = 25: y(t)= 10 + 15e2t; solution satisfying y(0) = 5: y(t) =10 - 5e2I 0.5

800

+

600

11. y = (8x3 + C)114, where C is an arbitrary constant.

400

13. y = Ce-x3/3, where C is an arbitrary constant.

200

y(0) =25

15. y = in (4t5 + C), where C is an arbitrary constant.

I

0.5

1

/ X

-50 -100 -150 -200 -250

1

1.5

y(0) = 5

1.5

17. y = CC-5-' 12 + 4, where C is an arbitrary constant. 19. y =

2, where C is an arbitrary constant.

21. y = ±N/x2 + C, where C is an arbitrary constant. 23. x =- tan(It2 + t + C), where C is an arbitrary constant. 25. y = sin-I (1x2 + C), where C is an arbitrary constant.

3.

y = -6 + 11e4x

5. (c) 7. 11.

(a) y' = -0.02(y -10) (b) y = 10 + 90e- 46` 1001n3 s 109.8s P.', 5:50 AM 9. 0.77 min = 46.6s 5001nisP-• 203 s= 3 min 23 s

27. y = C sec t, where C is an arbitrary constant. 29.

75e-2x

31. y = -On (x2 + e4) 39. 35. y = tan (x2/2) 37. y

=

y

=

t

y

33. y = 2 + 2e x(x-2)/2 1

=

41. y = sin-1 (fr) 43. (a) With an approximation of the numerical values involved in the differential equation and solution, T(x) = 0.091x. (b)

13. -58.8 m/s 17. (a)

80

15. -11.8 in/s 9.8 („ , 9.8) _kt* 0=- e JO k k 9.8 = (30k + 9.8)e-kt*

60 40

ekt* = 301c + 1 9.8 = in (30k + 1 \ kt* 9.8 )

20 I

400

I

I

I

1 X

800

45. (a) ' 51145 s or 19.1 min (b) "=.%3910 s or 65.2 min 47. y = 8- (8 + 0.000221502/3; te -z," 66000 s or 18 hr, 20 min 51. After 5 h, amount 184 g; After 10 h, amount 68 g 53. (a) g = kN, N(0) = 11.3, N(10) = 11.7; solution: N(t)= 11.3e°•0035` (b) Approximately 2.3 million 55. (a) t = 41 (b) Doubles in approximately 0.46 h; triples in approximately 0.73 h; increases tenfold in approximately 1.54 h

= k (b) y(t*)=

=

57. (a) q(t) = CV (1 - e'I RC) (c) lim q(t)= lim CV (1- e- `11 )= Ern CV (1 -0) = CV t->oo t->co (d) q(RC) = CV (1 - e-1) (0.63) CV 59. cubic; V = (kt13 + C)3, V increases roughly with the cube of time. 61. (a) (ye-kr) =

(e-kt) + y(-ke-kr ) = ky(e-kI ) - ky(e-la ) =0

=

in (30k + 1 \ 9.8 )

9.8 30k 1 9.8 ln (+ 1) + (30+ k2 9.8

(1

9.8 ( 30k + 1 \ + 1 ( 30 + 9.8\ ( k2 111 9.8 ) k k )

_e

1

+1))

9.8 \ 30k + 9.8)

9.8 (30k 1 (30k + 9.8 \ (30k30k _) \ ln +1 + k2 9.8 k k ) \, + 9.8) 9.8 (30k ) 30 in -- + 1 + k2 9.8 k 30k - 9.8 ln(W w` + 1)

k2

ANSWERS TO ODD -NUMBERED EXERCISES

_

19. (a) i. $17,563.94 ii. approximately 13.86 years or about 13 years 10 months (c) $120,000 (d) $107,629.00 (e) 8% 21. $4068.73 per year

23. (a) 1(t) =

1.

7

2. y = ±../1

t

3. (b)

Section 10.4 Exercises 5

1.

Y=

3.

(a) y(t)= 6

5.

2000 (a) P(t) = 1+ 3e-0.6t

(1 - e- ( 7- )t)

Section 10.3 Preliminary Questions

7.

4. 20

9.

5 1+ (3/2)e-3t

and y =

1- e-3t IC

(b) y(t)=

(c) y(t) =0

i+0.65e- '8, (b) t =

A In 3 '--z-- 1.83 years

k = In 14 -- -%0.96 year-1; t = 21nr911131 P,-, 2.29 years After t = 7.6 h, or at :36P2m(t) =

11. (a) y1(() =

Section 10.3 Exercises

A[N'SJ -3

10 9e- '

10

3and y

1 - 12e-1

(b) t = ln ;

(c) t = In 2 13. (a) A(t)= 16(1 _ 3er/40)2/0 (b) A(10) (c)

5 2

2.1

Y 16

A(t)

I

0

15.

5.

(a)

3et/40)2

100

r X

200

943 million 17. (d) t =

(ln yo - In (A - ye))

7. 3 2

"MAIM ';; .,

0

2 - --• 1 - \

i-

0

,„,_ASIdsiiii

-1

0

2

-- / - --s. • s

/ / / / ' ...- / / - -- / \ \ • s. - .., \ \ \ • -

/ / / / /

\

\

\

•

s. -

- \ \ -2 - I I

I I 1 I

\ \ \ I

\ \ \ \

\ \ \ \

• \ \ .

3

-2 -1

0

/

1 1 / / /

1 / / /

I 1 1 /

/

1

--• / / / /

\

-1

/ / / / /

/

/

2

9. For y' = t, y' depends only on t. The isoclines of any slope c will be the vertical lines t = c. 2

•Wh4:QF,,3; -

-2 -2

-1

(b)

3. P(x)=x '

4. P(x) =1

Section 10.5 Exercises

+

1.

(c) y =

5.

y= x+

(d) y =

-

7. y = -ix -I + Cx1/3

9. y = x2 + + Cx-3 11. y = -x lnx + Cx 13. y = + Ce- ' 15. y = x cos x + C cos x 17. y = Ce2x - le' 19. y = xx + Cxxe- x 21. y = -l ezx - e-3x 23. y = 1+: jj x(xl+i) + x+ 51 25. y = - cos x + sin x 27. y = tanh x + 3sech x

2

29. The differential equation is first-order linear with p(x) = 0 and Q(x)= x. Constant functions are antiderivatives of p(x), and therefore we can take ce(x) = cc, for any C, as an integrating factor. By Theorem 1 then, the solution to the differential equation is y =4 f ec x dx = f xdx. From here, obtaining the solution amounts to directly integrating the right side of the differential equation. 31. Form -n: y = y = (x + C)e- nx

11. (i) C (ii) B (iii) F (iv) D (v) A (vi) E 13. (a) 2

2.

/

ss - -- / / • s. - ...- / \ • s. - ..\ \• s- 1

Section 10.5 Preliminary Questions 1. (a) Yes (b) No (c) Yes (d) No

33. (a) y' - 4000

ssss,.------- "Z;Z

:::=:.---;:;;•;;;

e" + Ce';

sof+Yaor ; Y

for m = -n:

1000 412 ±21010`2-5F 125

(b) 40 g/L 35. 50 g/L

0

11,,11!15T7(411(

i';

37. (a)

;; • '-----""-------

-1

15. (a) yi = 3.1

0

1

-5 and V(t) = 201n(1 + t) - 5t + 100

(c) Estimate empty at

-------..... ------:--,::::\ K

-2 -2

=

(b) The maximum value is V(3) = 201n4 - 15 + 100 r=.-: 112.726. 34 min

2

(b) y2 = 3.231

(c) y = 3.3919, y4 = 3.58171, y5 = 3.799539, y6 = 4.0445851 (d) y(2.2) r-z-: 3.231, y(2.5)

3.799539

17. y(0.5) •--z-% 1.7210 19. y(3.3) --=1 3.3364 -

25. y(0.5)

1.794894

27. y(0.25)

21. y(2) P-• 2.8838

1.094871

Section 10.4 Preliminary Questions

39. 1(0 =

1.

(a) No

41. (a)

2.

y(t)= 0 and y(t) = A

(b) Yes

(c) No

(d) Yes

3. Yes

(1 _ e -201) = V -

il e-(RIL)t

43. (b) c1 (t) = 10e-06

(c) Approximately 0.0184 s

ANSWERS TO ODD -NUMBERED EXERCISES

ANS56

Chapter 10 Review

Section 11.1 Exercises

1. (a) No, first order (b) Yes, first order (c) No, order 3

1.

(a) (iv)

(d) Yes, second order

3.

ci = 3; c2 = ;; c3 = 1; c4. = at = 2; a2 = 5; a3 = 47; a4 = 4415

3.

y = ± (4t3 + C)1/4, where C is an arbitrary constant.

y = Cx2 - ;, where C is an arbitrary constant. 4 2, 9' Y = 13-I2x2 7. y = 1(x + I sin 2x) + ÷1 Y 13. 11. y . .

5.

2 1 0 -1

I X

I % - • •

\ X • \'•,--

/

/

1

1

1

• / •

/ /

I I

I I

/

/// . / /

/

/ /

I I 1 -

- ....h --

•

•

•

•

•

X

X

X 1 -

••-/ • -- -

/r../ -/

/ 1

/

'

/-•-• /

/••••

2

"')1/ V/3

/ •

\VAI

X 1 1 1

-2 -2 -2 -1

0

1

2

-6 -4 -2 0 2 4 6

15. y(t) = tan t

5.

(b) (i)

/

/

i /

/ /

/

/

/

1

I

/

/

/

/

/

/

/

/

/

/

/

/ //

/

/

/

/

/

2 0 -2

. /

/

4,

7.

bi = 4; b2 = 6; b3 = 4; b4. = 6 ci = 1; c2 = ;; c3 = c4 = 2; b2 = 3; b3 = 8; b4 = 19

11. bi =

13. (a) an =

/

/

/

/

/

/

/

/

/

/

/

/

/

/

/

/

/

/

/

/

/

/

- / /

/ /

/ /

/ /

/

/

/

/

/ -,

/

/

/

/

/

/

-2 -1

0

1

17. „limco I2n+9 = 12

21. The sequence diverges.

27. w = tan (lc lnx + 29. y = - cos x + -sixnx

-

29. lim

27. Limit = 0

+

n

ln 3

=2

33. Limit

31. urn cos- 1 (2n'3'3+ 1) n->oo

1.61803

(b) M = 99,999 41. The sequence diverges.

1/3 1 -= 2I/3 53. lim (2 + 72-) n->oo 57. The sequence diverges. 59. lim

51. The sequence diverges. en+1) = 3n+4

where C is an arbitrary constant. ; solution satisfying

3

e" 5. (-3)" -- 0

61. lirn n sin

63.

3-4" 11111 2+7.4. -

67.

lim(In n)2 = 0 n-> n

t2 + 15

31. Solution satisfying y(0) = 3: y(t) = 4 y(0) = 4: y(t) = 4

12n +2) 25. lim ln ( -9+4n

+1 - 1 23. nLimo. r nn2--

2

25. y = 1 -

19. Diverges

43. lim 2I / n -= 1 45. lim -9n = 0 n-soo n! • 3n2 +n+2 _ 3 49. lim COS fl = 0 / 2n2 -3 - 2 • oo n

17. y(0.1) r--.• 1.1; y(0.2) =•••-' 1.209890; y(0.3) •c-• • 1.329919 e-x _ 4 e-2x 19. y = 4x2 21. y = 23. y = ; sin 2x - 2 cos x

15. Diverges

(b) an =

39. 1i.rno3 (10 + (-4) n ) = 10 , -4

//#' / .- /

(d) (ii)

9.

35. (a) M = 999 . /

(c) (iii)

= 7r

65. lim (1 +

7

55. lim

n

co

=e

69. lim n(Vn2 + 1 - n) = co

71.

lirn 1 - 0 73. lim (2' + 3n)i ln = 3 75. (b) n->co n-> 77. Any number greater than or equal to 3 is an upper bound.

79. Example: an = (-1)" 91. (e) AGM (1, ./2)

83. Example: f (x) = sin rrx

1.198

y(0)=4 3

Section 11.2 Preliminary Questions

2

1. The sum of an infinite series is defined as the limit of the sequence of partial sums. If the limit of this sequence does not exist, the series is said to diverge.

1

0.5 1

1.5

.x

33. (a) 12 (b) oo, if y (0) > 12; 12, if y(0) = 12; -oo, if y(0) < 12 (c) -3 35. 400,000- 200,000e°•25 R-•- $143,194.91 37. $400,000 39. (a) P (t) = 500e0.262r (b) Approximately 2.65 h + Ce-Af and hin y = t di. dy -7,/1113, dt (30y + 8100) ' t = 3225.88 s or 51 min 56 s 45. 2 47. t = 51n441 30.45 days L = -ct 51. (a) (b) c1 (t) = 8e(-2/5)( g/L

2.

S=

3. The result is negative, so the result is not valid: A series with all positive terms cannot have a negative sum. The formula is not valid because a geometric series with Ir I > 1 diverges. 4.

No

7.

No, SN is increasing and converges to 1, so SN < 1 for all N.

8.

Example:

5. No

41. Solutions are of the form y =

E

n=1

Section 11.2 Exercises 1. (a) an = -k,

Chapter 11 Section 11.1 Preliminary Questions 1. a4 = 12 2. (c) 3. 17-4 lim a, = ../1 4. (b) 00 5. (a) False; counterexample: an = cos n n (b) True (c) False; counterexample: an = (-1)"

6. N = 13

(b) an =

(c) a n = (-1)n±1 nn';

(d) an = I

3.

s6 = ±i 6:)90 ,4

52 =

S4 =

3;

(;;;; 11+1

S4 = '4 ; S6 = 7. S6 = 1.24992 5. S2 = 9. S10 = 0.03535167962; S100 = 0.03539810274; S500 = 0.03539816290; S1000 = 0.03539816334; yes.

ANSWERS TO ODD -NUMBERED EXERCISES

11.

co

S3 = 3 ., c = 31 ,• S5 -

5

+1

n=1 oo `3. 53 =

4; S4 =

t ; S5 =

5

11. f20° xonlx>2 dx converges, so the series converges.

( n 1 - n + 2 \_i- 2

V'

n=1

1 13' n3 +8n _< n13 , so the series converges.

4n21_ =

17. n 2I"

19. n1/31+ 2,,

15. S= 1

17' n->oo lim lOnn+ 12 = 110 ° (z_1) 1,1.7+1 19. urn (_ ) does not exist. n->00 21. lirn an = urncos 7,7IF-T = 1 0 23. fl - 00 n-,-00 25. ; 27. The series diverges. 29. S = 59.049 3328

31. S = W

33. S = T 35

21.

m! + 4 4. 23. 0 < sikn72 k 25.

35. S =- 4 37. S =

-A

47. (b) and (c)

+ RT, 9 ±

39. 2

A + • • • = jom _ =

51. (a) Counterexample.

00

n= 53. (a) (.55)n(.48)"(.52) (b) E,711 (.55)" (.48)n-1(.52) = 0 39 0.286 1-0.264 • (.52)((.48)(.55))n-1 = (c)

n , so the series converges. (-) „ m 4 , so the series converges. ki2 , so the series converges. 2 (1)n, so the series converges.

3n

(

n E(I) 2 -

E

E cos 2n- n

the series converges.

31. (1nn)l00 < i09 for n sufficiently large, so the series converges. -52 ) n for n > 1, so the series converges. 37. The series converges. 39. The series diverges. 41. The series converges. 43. The series diverges. 45. The series converges. 47. The series converges. 49. The series diverges. 51. The series converges.

n=1 (b) Counterexample: If a = 1, then SN = N co (c) Counterexample: diverges n=1 (d) Counterexample:

n, so

1 27' , so the series converges. (n+1)1 n 29. ln "7 for n 1, so the series converges.

43. 300 3370

45. 0.999999 . . . =

()

1

53. The series diverges.

55. The series converges.

57. The series diverges.

59. The series diverges. 63. The series converges. 67. The series diverges.

61. The series diverges. 65. The series diverges. (.55)(.52)((.55)(.48))

=

69. The series converges.

71. The series converges.

73. The series diverges.

0.71

7.2264

The total area is

75. The series converges. 77. The series converges for a > 1 and diverges for a 1. 79. The series converges for p > 1 and diverges for p 1. co 87. n-5 1.0369277551 n=1 100 1000 1.6448848903. The = 1.6439345667 and 1 + n 2 (ni +1) n=1 n=1

E

- De- k 7. (a) De-k + De- 2k + De-3k + - 1 - e-k _ De- ki (b) De-kr ± De-2kt ± De- St T--7k7 ln(1 (c) t >

91. E

59. The total length of the path is 2 + oo the diameter of the horn goes to zero. If we can fill 63. (a) As x the horn all of the way to the end, then the paint must be capable of being spread thin enough to fit all of the way into the horn. (b) Use a volume of 4.r milliliters of paint to paint the portion of the horn between x = n 1 and x -= n + 1. Overall we use a volume of + milliliter of paint to paint the surface.

+ + + ••• =

E

jr second sum is a better approximation to w

1.6449340668.

Section 11.4 Preliminary Questions 1. Example: 4.

IS - Stool

EL-if

2. (b)

3. No

< 10-3, and S is larger than Sloo•

Section 11.3 Preliminary Questions

Section 11.4 Exercises

1. (b)

3. Converges conditionally 5. Converges absolutely 7. Converges absolutely 9. Converges conditionally

2. A function f such that an = f (n) must be positive, decreasing, and continuous for x > 1. 3. Convergence of p-series or integral test 4.

Comparison Test

5.

1 No; 2 - diverges, but since n n=1

00

-„T e-"
1, the Comparison 00

Test tells us nothing about the convergence of ‘-‘2_, n=1

•

Section 11.3 Exercises

a

0, +' 04 dx converges, so the series converges.

Acx) x-113 dx = no, so the series diverges. 2

7.

dx converges, so the series converges. 2 (x, J25 (x3 +x 9)5,2 foe x2d+x 1 converges, so the series converges.

9.

converges. fx0:1 x(x +5) dx converges, so the series

5.

11.

n 1 2 3 4 5

S, 1 0.875 0.912037037 0.896412037 0.904412037

n 6 7 8 9 10

Sn 0.899782407 0.902697859 0.900744734 0.902116476 0.901116476

13. S5 = 0.947 15. S44 = 0.06567457397 17. Converges (by geometric series) 19. Converges (by Limit Comparison Test) 21. Converges (by Limit Comparison Test) 23. Diverges (by Limit Comparison Test) 25. Converges (by geometric series and linearity) 27. Converges absolutely (by Integral Test) 29. Converges (by Alternating Series Test) 31. Converges (by Integral Test) 33. Converges conditionally

ANSWERS TO ODD -NUMBERED EXERCISES

ANS58

Section 11.5 Preliminary Questions 1. 3. 4.

2. (,,\ ( n ) (a) VI (b) ''' n+ i Yes in the first case, no in the second. Yes in the first case, no in the second. IcIlin Id

41. g(x) = P

(b) Nothing

Section 11.5 Exercises 1. Converges absolutely 3. Converges absolutely 5. 9. 13. 17. 21.

The Ratio Test is inconclusive 7. Diverges Converges absolutely 11. Converges absolutely Diverges 15. The Ratio Test is inconclusive Converges absolutely 19. Converges absolutely 23. p = 21x1 25. p = InI p=
11Moo

-F 1 an n=1 (n ± 1) lx I3(n I" = IX 13 . It follows that the radius of convergence is nIx13" n->"Co 1. (b) Multiplying term-by-term, we find that (f (x))2 = 1 + x3 +x6 +x9 +x3 +x6 +x9 +x6 +x9 +x9 +.. • = 1 + 2x3 + 3x6 + 4x9 + • • • . Expanding out h(x) up to n = 4, we have h(x) = 1 + 2x3 +3x6 +4x9 + • • • .

43. (a) h(x) =

47.

Ec-iy-Fi(x - 5)n on the interval (4,6) n=0

51. (c) S4 = 6To and IS - S4I 53. R = 1

55.

E =2

0.000386 < a5 = 1920 57. F(x) =

; c3X2

35.

61.

=e- P C1411100 1-r? ) = ect = 1. lim " = lim n- Pin = lime n-400 n Do n-> oo Therefore the root test is inconclusive. 37. Converges absolutely 39. Converges absolutely 41. Converges absolutely 43. Converges (by geometric series and linearity)

n=0 63. N must be at least 5; S5 = 0.3680555556 oo 1.3.5.• • (2n - 3) x 2n ; R = 65. P(x) = 1 - X2 no (2n)! n=2

45. 47. 49. 51. 53. 55.

Diverges (by the Divergence Test) Converges (by the Direct Comparison Test) Diverges (by the Direct Comparison Test) Converges (by the Ratio Test) Converges (by the Limit Comparison Test) Diverges (by p-series) 57. Converges (by geometric series)

59. Converges (by Limit Comparison Test) 61. Diverges (by Divergence Test) 65. (b) Ar2Tr 2.50663

E

Section 11.7 Preliminary Questions 1. T3 (X) = 9 + 8(x - 3) + 2(x - 3)2 + 2(x - 3)3 2.

e n! 1000 1500 2000 2500 3000

The polynomial graphed on the right is a Maclaurin polynomial.

3.

A Maclaurin polynomial gives the value of f(0) exactly.

4.

The correct statement is (b): I T3(2) - f (2)1 5_ 4.

Section 11.7 Exercises 1.

T2(x)=x; T3(x)=x -

3. T3 (X) =

0-1-1/2

2.506837 2.506768 2.506733 2.506712 2.506698

-

(x- 2) + 7 (x - 2)2 -

(x - 2)3

5. T2(x) = 75 + 106(x - 3) + 54(x - 3)2; T3 (X) = 75+ 106(x - 3) + 54(x _3)2 + 12(x - 3)3 7. T2(x) = 1 + 1(x - 1) - (x - 1)2, T3(x) = 1 ± 1(x - 1) - (.7C - 1)2 + - 1)3

4

Section 11.6 Preliminary Questions

9.

1. Yes. The series must converge for both x = 4 and x = -3.

11. T2(x) = 2 - 3x + V; T3 (X) = 2 - 3x +

2.

13. T2(x) T3(X) =

4.

(a), (c) F'(x) =

3. R = 4 oo

En2xn-1; R = 1 n=1

Section 11.6 Exercises 1. R = 2. It does not converge at the endpoints. 3. 9. 17. 27.

R = 3 for all three series. (-1, 1) 11. [-42, jf] 13. [-1, 1] 15. (-oo, oo) (-oo, oo) 19. (-1, 1] 21. (-1, 1) 23. [-1, 1) 25. (2,4) (6,8) 29. [ 31. (-oo, oo) -) 00 33. (2- ,2+ 35. on the interval ( - 4, ) n=0 00 37. on the interval (-3, 3) n=0

E3nxn

39.

59. -1 oo

fl-5 00

11. lim 21/ n2 = 1

lim (,./n ± 5 - ,s,/n + 2) = 0

n ->co

ti-+oo

15. lirn tan- 1 n-> co

n2 +1) =

n-> oo

5 -4

(n + 2) _ n+

11-

3m

=e3

urn 1 + 1 m m->oo( oo

29.

lirn an + n-.-00

.0 E

3

27.

an

4

(4) n

41 S4 = 11, S7 = 630

31. S =

0°

33.

3n

IT3(1.1)

x3

00

127. e4x =

E

xn

n=0

35. an =

()

n ± 1 - 2n , 17„

=

2n - 1

39. The series diverges.

37. S = g

41. fr(x+ 2)(11n(x+2))3 dx = 20n1(3))2' so the series converges. , so the series converges. 43. (n +1 1)2 < co 45. converges, so the series converges.

E '

n=1

47. 49.

, n

n

V n5 +5

E

(n)

1 , so the series converges.

converges, so the series converges.

n

129. x4 = 16 + 32(x - 2) + 24(x - 2)2 + 8(x - 2)3 + (x - 2)4 co 131. sin x =

E (-1);'n'+x17)2"+'

n=0 oo

133. 1-2x _

E „2:

n=0

00 (x +

139. i+ta l 141.

n.

= 1 - x + x2 -

- k 3f +

-

±r/2"x-2Y'

3) (0) = -6 4x3 + • • • , so f (3) (0) = -8

2_

n=0

E

n=1

E tx2n-F oo

137. (x2 - x)ex2 =

135. In x _

2)n

+

• • • =

)

so f (

sin SI = 1

n=0

51. Converges 55. (b) 0.3971162690 0

91.

A

97.

1

t=2

=

99. -25

_21 512 101. 32

t-3 =

Section 12.2 Preliminary Questions (c) The predator has its peaks shortly after the prey does. This makes sense because when the predator population is large, the prey population should be increasing due to the abundance of prey. After the prey reaches its peak and begins to decline, the predator will soon reach its peak and begin to decline in response to the decrease in available prey.

1.

S = f ab \Ix'(t)2 y'(t)2 dt 2. No. They are equal when the curve traced by c(t) is a line segment from the initial to the final point, and c(t) is a one-to-one function. 3. The speed at time t 4. Displacement: 5; no 5. L = 180 cm 6.

47r

ANSWERS TO ODD -NUMBERED EXERCISES

ANS62

Section 12.2 Exercises

5. 5

5

if 3-dt = 4 -3-. The path is a line JO segment from (-1, 2) to (14, —8), and the length of the segment is 3. S=

1.

+ (-2)2dt =

I

(653/2 — 53/2) P•,- 256.43

5. S =

2.34

9. S = —8 (4-2: — 1)

= 4,./1T) ••-- 12.65 m/s 17.

ds -d7

t=2

= 1 mis t=o J4.89Rz,' 2.21 21. (fst )rnin

=

6.4 m/s

t=9

(A): 0 r 3,7r 5_ 0 < 27, (B): 0 < r < 3, 37 (iii), (d) —> (i) 61. —5i — 3j

OR =/

7. PQ

s 5.1 11. P = (-2, 6,0)

13. (a) Parallel and same direction (b) Not parallel (c) Parallel and opposite directions (d) Not parallel

(-0.22, 0.97)

53. A = ±—L.

PQ = (1, 1, —1)

59. 9i + 7j

63.

15. Not equivalent 17. Not equivalent 19. (-8, —18, —2) 21. (-2, —2, 3) 25. Not parallel

23. (16, —1, 9)

27. Not parallel

,A)

w ) es, = (4' — 4' 31' e = (A , 33. The top half of a sphere of radius 2, centered at (0, 0, 2) 29.

X

1 X

65. u= 2v — w 35. The part of the cylinder of radius .0, with the z-axis as central axis, lying between z = —7 and z = 7

67. The force on cable 1 is 207 newtons 69. 230 km/h

71. r

444 newtons, and force on cable 2 is

(6.45, 0.38) 37.

(4, 3, 2)

5.

Infinitely many direction vectors

8.

(0, 0, 0), (0, 0, —1), (1, —1, 1), (1, 1, 0)

2. (3, 2, 1)

Section 13.2 Exercises 1.

11v11=

3. (a)

4. (c) 6. True

y2

+

(z

+

3)2

=

9

39. (x _6)2 + (y + 3)2 + (z — 11)2 = 277 41. (x — 1)2 + (y + 1)2 = 2 43. r(t) = (1 + 2t, 2 + t, —8 + 3t) 45. r(t) = (4 + 7t, 0, 8 + 4t)

Section 13.2 Preliminary Questions 1.

x2 ±

7. ,s,/3

47. r(t) = (1 + 2t, 1 — 6t, 1 + t) 49. r(t) = (4t, t, t) 51. r(t) = (0, 0, t) 53. r(t) = (—t, —2t, 4— 2r) 55. (c) 59. ri (t) = (5, 5, 2) + t (0, —2, 1); r2(t) = (5, 5, 2) + r (0, —20, 10) 63. The lines intersect at (3, 4, 7). 65. 4 min

67. (-1, —3, 2) and 0-4, — }:17 , —

ANS66

ANSWERS TO ODD -NUMBERED EXERCISES

69. (0, 1,

71. 2450 newtons

-)

73. -xV = 2 13• =

z-3 3

1. -5 3. -15 5. -8 7. 0 9. (1, 2, -5) 11. (6, 0, -8) 13. (0.02, -0.01,0) 15. -j + i 17. i + j + k 19. (-1, -1,0) 21. (-2, -2, -2) 23. (4, 4, 0) 25. v x = cj - bk; v x j = -ci + ak; v x k = bi - aj

77 r (t) = (2t, 7t,8t)

75.

Section 13.3 Preliminary Questions 1. Scalar 2. Obtuse 3. Distributive Law 4.

27. -u

(a) 5. (b); (c) 6. (c)

29. (0, 3, 3)

33.

e'

35. (a) 0.0067 N (b) 0.0067 N 37. Fi

Section 13.3 Exercises 1. 15 3. 2 5. 5 7. 0 9. 1 11. 0 13. Obtuse 15. Orthogonal 17. Acute 19. 0 21. *-(--) 23. 7/4 25. P--0.615 27. 27/3 29. cos-I -)171 •cz--' 0.641 31. (a) b=-

Section 13.4 Exercises

(c) 0 (d) 0.0095 N

41. 2.N/

43. The volume is 4.

(b) b = 0 or b =

37. five 33. vi = (0, 1, 0) , v2 = (3, 2, 2) 35. 39. IlvII2 - IIwII2 41. 8 43. 2 45. 7 47. (b) 7 51. 51.91° (b) 1111v 53. (a)

vn. 57. (-4, 0, -i)

55. (4,

65. Nil7 67. a=( . 69. a = (0,

1)+(

59. -4k

61. ai

63.

45.

5.92

47.

, -1)

Q= (0, 3.3)

_)+(4,

Y) 75. ',- z-.35° 77. AD 79. Diagonal length = 02,359.625 •=z-- 228.82 in.; split difference approximation = 228 in. 81. c,--109.5° 85. The wind vector is w = (0, -45). It is expressed as w = wIl wi, where wo is parallel to the bridge and equals -45 sin 58° (cos 58°, sin 58°) (-20.22, -32.36) , while wi is perpendicular to the bridge and equals (0, -45) +45 sin 58° (cos 58°, sin 58°) (20.22, -12.64). The magnitude of the perpendicular term 23.85 km/h. 87. P,-68.07 newtons 93. Consider the parallelogram shown in Figure 8. Assume that the diagonals are perpendicular. Then x • y = 0. Since x = a+ b and y = a - b, it follows that 0 =(a+b) •(a-b)=a•a-b•b=

11a112 - 11b112

Therefore, Hall = Ilb II, implying that the parallelogram is a rhombus. Thus, if the parallelogram has perpendicular diagonals, then it is a rhombus. Conversely, assume that the parallelogram is a rhombus. Then Ilall = 11b11. This implies that 0 = 11a112 - 11b112 =a•a-b•b= (a + b) • (a - b) = x • y. Therefore, the diagonals are perpendicular. Thus, if the parallelogram is a rhombus, then its diagonals are perpendicular.

P = (3, 3, 0)

The area of the triangle is y

7.8.

49. 3 51. 57. (a) i x j = 1 0 i k x j= 0 0

k 0 = (0)i + (0)j + (1)k = k 0

0 1 j 0 1

k 1 = (-1)1+ (0)j + (0)k = 0

(b) The length of i x j is dill Iill sin 90° = 1 since i and j are unit vectors and the angle between them is 90°. Also, curling from i to j, the right-hand rule yields the positive z-direction. Therefore, i x j is a vector of length 1, pointing in the positive z-direction; that is, i x j = k. The length of k x j is Ilk II MI sin 90° = 1 since k and j are unit vectors and the angle between them is 90°. Also, curling from k to j, the right-hand rule yields the negative x-direction. Therefore k x j is a vector of length 1, pointing in the negative x-direction; that is, k x j = 63. X = (a, a, a +1) 67. r = 250 sin 125° k 204.79 k N-m

105. 2x + 2y - 2z = 1

Section 13.4 Preliminary Questions -5 -1 2. Ile x fll = 3. u x w = (-2, -2, -1) 4 0 4. (a) 0 (b) 0 5. ix j=k and i x k = -j 6. v x w = 0 if either v or w (or both) is the zero vector or v and w are parallel vectors. 7. (a) Not meaningful because you cannot find the cross product of a scalar and a vector (b) Meaningful because it represents the dot product of two vectors (c) Meaningful because it is the product of two scalars (d) Meaningful because it is a scalar multiple of a vector 8. (b) 1.

Section 13.5 Preliminary Questions 1. 5.

3x +4y -z = 0 2. (c): z = 1 3. Plane (c) (c): x + y = 0 6. Statement (a)

4. xz-plane

Section 13.5 Exercises 1.

x+3y+2z= 3 3. -x+2y+z= 3 5. x=3 7. z = 2 9. x = 0 11. Statements (b) and (d) 13. (9, -4, -11) 15. (3, -8, 11) 17. 4x - 9y + z =0 19. x = 4 21. 6x + 9y + 4z = 19 23. x + 2y - z = 1

25. (a) Let the points be P, Q, and R. Thenn = PQ x PR is a normal vector to the plane. (b) Let vi and v2 be direction vectors for the lines. Then n=vi xv2 is a normal vector to the plane.

ANSWERS TO ODD -NUMBERED EXERCISES 27. (a) Do not intersect (b) Have a single point of intersection; an equation for the plane containing the lines is x - y - z = 4.

29.

1NS67

31.

29. (a) The point is on the line. (b) The point is not on the line. An equation for the plane containing the point and the line is x - y = 6. 31. (a) They are distinct parallel lines. An equation for the plane containing the lines is 4x - 5y + 6z = -13. (b) The lines are not parallel. 33.

35.

Graph of x2 + y2 - z2 = 1

33.

35.

37.

37.

39.

39. 10x + 15y + 6z = 30 41. (1,5, 8) 43. (-2, 3, 12) 49. x = -4 45. -9y + 4z = 5 47. x = 51. The two planes have no common points. 53. y - 4z = 0 x+y- 4z= 0 55. 59. 63. 67. 75.

(3A)x + by + (2.A.)z = 5X, A 0 0 57. 8=r/2 0 R-‘-' 1.143 radians or 0 Ps-1 65.49° 61. 8 55.0° x+y+z= 1 65. x-y-z=dla - 3t, z = 2 + 5t 69. ±24 (1, 2, -2) x = ; + 2t, y = (2 _1 2) 77. 6 %`•,' 1.095 79. lal ‘VT3 3 3

41.

'

Section 13.6 Preliminary Questions 1.

True, mostly, except at x = ±a, y = ±b, or z = ±c

2.

False

6.

All vertical lines passing through a parabola c in the xy-plane

3. Hyperbolic paraboloid

4. No

5. Ellipsoid

Section 13.6 Exercises 5. Hyperboloid of one sheet

1.

Ellipsoid

3. Ellipsoid

7.

Ellipsoid

9. Elliptic paraboloid 11. Hyperbolic paraboloid

13. Hyperbolic paraboloid

15. Elliptic cone

43. (i) ± (0 2 ±

=1

45.

(i) 2

±(

47. One or two vertical lines, or an empty set

)

2

2

(T

) =1

49. An elliptic cone

Section 13.7 Preliminary Questions 1. Cylinder of radius R whose axis is the z-axis, sphere of radius R centered at the origin 2.

(b)

3. (a)

4. 0 =0, n

5.

= I; the xy-plane

17. Ellipsoid; the trace is a circle on the xz-plane. 19. Ellipsoid; the trace is an ellipse parallel to the xy-plane. 21. Hyperboloid of one sheet; the trace is a hyperbola. 23. Parabolic cylinder; the trace is the parabola y = 3x2. 25. (a) 4-->- Figure b; (b) Figure c; (c) Figure a 27.

= (i) 2 + ( )2

Section 13.7 Exercises 1. (-4, 0, 4) 3. (0, 0, 1) 5. (12, .7,fr , 1) 7. (2, 9.

(5,

2)

15. r2 < 9,

-t 57

11. r2 < 3 13. r2