Second Course in Calculus

A SECOND COCi

1MY A

I B S

Rob

Constantstt ~ 3.141593 ir/2 « 1.570796 ir/3 « 1.047198 t / 4 « 0.785398 x / 6 « 0.523599 logio it « 0.497150 1 rad « 57.29578°1° » 0.0174533 rad e « 2.718282 e2 » 7.389056 1 /e « 0.367879 M = logio c « 0.4342945 1/A f = In 10 « 2.302585 logic M ~ 0 .637784 - 1 logio x = M In x V2 « 1.414214

V § « 1.732051 V 5 « 2.236068 30 m ph = 44 f t /s e c g « 980.62 c m /se c 2

« 32.173 f t /s e c 2

Definite Integralsr2ir r2r/ sin 2 nx dx = / cos2 nx dx = t (n > 1) Jo Jo

I0 J o

•2ir r2wsin mx sin nx dx — / cos mx cos nx dx = 0

Jo(m n)

sin mx cos nx dx = 03. f sii Jor*/2 /•?

4. / s in 2n xdx =Jo Jorir/2 /*ir/2

5. / s in 2n+1 x dx = / cos2n+1 x dx =Jo Jo (2n + l)!

6. [ e~x* dx = ^ V t t Jo 2

/2 (2n) ! * cos2n x ax = ~ ■■ - 22n(n !)2 2

2 2n(n !)2

Formulasax+1/ = axav (a&)x = axbx (ax)v = axv sin21 + cos2 £ = 1 1 + ta n 2 £ = sec2 £ 1 + co t2 2 = csc2 £

7T 7r 1— = cos - = —6 3 2

7T 7r 1-- = cos - = —4 4 2

7T 7r 1— = cos - = —3 6 2

1 /-

sin( — 0 = — sin £ cos( — 0 = cos t sin(x + y) = sin x cos y + cos(x + y) = cos x cos y — sin 2x = 2 sin x cos x cos 2x = cos2 x — sin 2 x sin2 x = ^(1 — cos 2x) cos2 x = i ( l + cos 2x) sinh t = £(e* — e~l) cosh t = + e“ 0 cosh2 £ — sinh2 £ = 1

cos x sin 2/ sin x sin 2/

Power Series

x| < 1r b ' X 1' 1

n =00

X x n— all x n\

n — 0

SX2n+1 ‘

n * 000

VA x2nc o s z = 2 / ( - 1) n (2^)! a l lx

n =0OO

ln ( l + *) = V ( - 1 ) " - 1— \x\ < 1nn = 1

arc tan x ■S'n = 0(- 1 )"

X 2n+1

2n -|" 1 1*1 < 1

Differentiation Rules

m II

cu cuU + V u + i)uv ui) + vuu vu — ui)V V2u[v(t)\ u[»(0]»(0

df . ,d f .= — x H---y.dx dyy

Differentiation Formulas

m £ - /«>

ctaela1

In t

sin £ cos t tan t cot t sec t csc t

arc sin t*

arc tan t

sinh t cosh t tanh t

0at“- 1 el(In a) a1 1 tcos t— sin t sec2 1— csc2 t ta n t sec t— cot t CSC t

1V i -

11 + 12 cosh t sinh t sech21

Polar Coordinates

{ x — r cos 0 y = r sin 0

Spherical Coordinates

Integrals (constant of integration omitted)

1. J u dv = uv — J v du

/ v - M ;

h4 [ d± = 1

J a2 — x2 2a

5. /

dx 1 (x—— - = - arc tan I - + x2 a \a,

x + a x — aIn

(a2 + x2)n (2 n — 2 )a2(a2 + x2)n_1 2n — 3 / dx< i-S [

~ 2) a 2 J (n > 1)(2n - 2)a2 J (a2 + x2)"-1

c f dx . (x \6. / — = arc sin 1 - JJ V a 2 - x2 W

7. J y/a2 — x2 dx = \/a2 — x2 + ~ arc sin

8.J V x 2 i a 2

y a/x2 ± a2 dx

= In |x + \/x2 ± a2

9.

= - V x 2 ± a2 ± In |x + \/x2 ± a2\

(continued inside back cover)

Vectors

a • b = dibi -f- (I2&2 "I- C&3&3

a x b =i j k

a 1 &2 as61 62 63

i j- dy+ k

dcte

V / = grad / = /,)

V • v = div v = ux + vy + w.

f x = p sin <t> cos 0 i j ks y = p sin <£ sin 0 V X v = curl v = d/dx d /d y d /azi 2 = p COS 0 U V w

A Second Course in Calculus

A Second Course in Calculus

Harley Flanders Tel Aviv University

Robert R. Korfhage Southern Methodist University

JllStin J. Price Purdue University

ACADEM IC PRESSNew York San Francisco LondonA Subsidiary of Harcourt Brace Jovanovich, Publishers

C o p y r i g h t © 1974, b y A c a d e m i c P r e s s , I n c .ALL RIGHTS RESERVED.NO PART OF THIS PUBLICATION MAY BE REPRODUCED OR TRANSMITTED IN ANY FORM OR BY ANY MEANS, ELECTRONIC OR MECHANICAL, INCLUDING PHOTOCOPY, RECORDING, OR ANY INFORMATION STORAGE AND RETRIEVAL SYSTEM, WITHOUT PERMISSION IN WRITING FROM THE PUBLISHER.

A C A D E M I C P R E S S , I N C .I l l Fifth Avenue, New York, New York 10003

United K ingdom Edition published by A C A D E M I C P R E S S , I N C . ( L O N D O N ) L T D .24 /28 Oval R oad, L ondon NW 1

Library of Congress Cataloging in Publication Data

Flanders, Harley.A second course in calculus.

1. Calculus. I. Korfhage, Robert R., joint author. II. Price, Justin J., joint author. III. Title.QA303.F5823 515 73-18943ISBN 0 - 1 2 - 2 5 9 6 6 2 - 5

PRINTED IN THE UNITED STATES OF AMERICA

Contents

Preface xi

1. INFINITE SERIES AND INTEGRALS

1. Infinite Series 12. Convergence and Divergence 43. Tests for Convergence 84. Series with Positive and Negative Terms5. Improper Integrals 166. Convergence and Divergence 237. Relation to Infinite Series 308 . Other Improper Integrals 359. Some Definite Integrals [optional] 40

10. Stirling’s Formula [optional] 43

2. TAYLOR APPROXIMATIONS

1. Introduction 522. Polynomials 523. Taylor Polynomials 564. Applications 635. Taylor Series 676. Derivation of Taylor’s Formula 69

3. POWER SERIES

1. Introduction 722. Ratio Test 783. Expansions of Functions 824. Further Techniques 90

vi Contents

5. Binomial Series 976. Alternating Series 1027. Applications to Definite Integrals [optional] 1058. Uniform Convergence [optional] 109

4. SOLID ANALYTIC GEOMETRY

1 . Coordinates and Vectors 1162. Length and Dot Product 1233. Lines and Planes 1304. Linear Systems and Intersections 1335. Cross Product 1416 . Applications 147

5. VECTOR CALCULUS

1. Vector Functions 1572. Space Curves 161

• 3. Curvature 1684. Velocity and Acceleration 1745. Integrals 1816. Polar Coordinates 1887. Polar Velocity and Acceleration [optional] 193

6 . FUNCTIONS OF SEVERAL VARIABLES

1. Introduction 1972. Domains 1993. Continuity 2054. Graphs 2095. Partial Derivatives 2136. Maxima and Minima 217

7. LINEAR FUNCTIONS AND MATRICES

1. Introduction 2242. Linear Transformations 2273. Matrix Calculations 232

Contents

4. Applications 2395. Quadratic Forms 2446 . Quadric Surfaces 2537. Inverses 2618. Characteristic Roots [optional] 269

8. SEVERAL VARIABLE DIFFERENTIAL CALCULUS

1 . Differentiable Functions 2812. Chain Rule 2873. Tangent Plane 2924. Gradient 297

« 5. Directional Derivative 3026. Applications 3067. Implicit Functions 3108 . Differentials 3169. Proof of the Chain Rule [optional] 319

9. HIGHER PARTIAL DERIVATIVES

1. Mixed Partials 3262. Higher Partials 3283. Taylor Polynomials 3334. Maxima and Minima 3395. Applications 3456. Three Variables 3497. Maxima with Constraints [optional] 3528 . Further Constraint Problems [optional] 360

10. DOUBLE INTEGRALS

1. Introduction 3692. Special Cases 3723. Iterated Integrals 3774. Applications 3825. General Domains 3876 . Polar Coordinates 399

11. MULTIPLE INTEGRALS

1 . Triple Integrals 4092 . Cylindrical Coordinates 419

viii Contents

3. Spherical Coordinates 4274. Center of Gravity 4405. Moments of Inertia 448

12. INTEGRATION THEORY

1. Introduction 4572. Step Functions 4583. The Riemann Integral 4634. Iteration 4725. Change of Variables 4826. Applications of Integration 4897. Improper Integrals [optional] 5028 . Numerical Integration [optional] 511

13. DIFFERENTIAL EQUATIONS

1. Introduction 5172. Separation of Variables 5203. Linear Differential Equations 5264. Homogeneous Equations 5305. Non-Homogeneous Equations 5326 . Applications 5407. Approximate Solutions 549

14. SECOND ORDER EQUATIONS AND SYSTEMS

1. Linear Equations 5562. Homogeneous Equations 5573. Particular Solutions 5624. Applications 5655. Power Series Solutions 5766. Matrix Power Series 5807. Systems 5858 . Uniqueness of Solutions [optional] 591

15. COMPLEX ANALYSIS

1. Introduction 5962. Complex Arithmetic 5983. Polar Form 602

Contents

4. Complex Exponentials 6105. Integration and Differentiation 6156. Applications to Differential Equations 6197. Applications to Power Series 622

Mathematical Tables

1. Trigonometric Functions 6292. Trigonometric Functions for Angles in Radians 6303. Four-Place Mantissas for Common Logarithms 6324. Antilogarithms 6345. Exponential Functions 636

Answers to Selected Exercises 643

Index 683

Preface

This text, designed for a second year calculus course, can follow any standard first year course in one-variable calculus. Its purpose is to cover the material most useful at this level, to maintain a balance between theory and practice, and to develop techniques and problem solving skills.

The topics fall into several categories:

Infinite series and integrals

Chapter 1 covers convergence and divergence of series and integrals. It contains proofs of basic convergence tests, relations between series and integrals, and manipulation with geometric, exponential, and related series. Chapter 2 covers approximation of functions by Taylor polynomials, with emphasis on numerical approximations and estimates of remainders. Chapter 3 deals with power series, including intervals of convergence, expansions of functions, and uniform convergence. It features calculations with series by algebraic operations, substitution, and term-by-term differentiation and integration.

Vector methods

Vector algebra is introduced in Chapter 4 and applied to solid analytic geometry. The calculus of one-variable vector functions and its applications to space curves and particle mechanics comprise Chapter 5.

Linear algebra

Chapter 7 contains a practical introduction to linear algebra in two and three dimensions. We do not attempt a complete treatment of foundations, but rather limit ourselves to those topics that have immediate application to calculus. The main topics are linear transformations in R2 and R3, their matrix representations, manipulation with matrices, linear systems, quadratic forms, and quadric surfaces.

Differential calculus of several variables

Chapter 6 contains preliminary material on sets in the plane and space, and the definition and basic properties of continuous functions. This is followed by partial derivatives with applications to maxima and minima. Chapter 8 continues with a careful treatment of differentiability and applications to tangent planes, gradients, directional derivatives, and differentials. Here ideas from linear algebra are used judiciously. Chapter 9 covers higher

xii Preface

order partial derivatives, Taylor polynomials, and second derivative tests for extrema.

Multiple integrals

In Chapters 10 and 11 we treat double and triple integrals intuitively, with emphasis on iteration, geometric and physical applications, and coordinate changes. In Chapter 12 we develop the theory of the Riemann integral starting with step functions. We continue with Jacobians and the change of variable formula, surface area, and Green’s Theorem.

Differential equations

Chapter 13 contains an elementary treatment of first order equations, with emphasis on linear equations, approximate solutions, and applications. Chapter 14 covers second order linear equations and first order linear systems, including matrix series solutions. These chapters can be taken up any time after Chapter 7.

Complex analysis

The final chapter moves quickly through basic complex algebra to complex power series, shortcuts using the complex exponential function, and applications to integration and differential equations.

Features

The key points of one-variable calculus are reviewed briefly as needed. Optional topics are scattered throughout, for example Stirling’s Formula,

characteristic roots and vectors, Lagrange multipliers, and Simpson’s Rule for double integrals.

Numerous worked examples teach practical skills and demonstrate the utility of the theory.

We emphasize simple line drawings that a student can learn to do himself.

Acknowledgments

We appreciate the invaluable assistance of our typists, Sara Marcus and Elizabeth Young, the high quality graphics of Vantage Art Inc., and the outstanding production job of the Academic Press staff.

1. Infinite Series and Integrals

1. INFIN ITE SER IESOne of the most important topics in mathematical analysis, both in theory

and applications, is infinite series. The basic problem is how to add up a sum with infinitely many terms. At first that seems impossible; life is too short. However, suppose we look at the sum

1 1 1x + 2 + i+ + 2 » +

and start adding up terms. We find 1, §, i , ^ • • •> numbers getting closer and closer to 2. The message is clear: in some limit sense the total of all the terms is 2.

If we try to add up terms of the sum

1 + 1 + 1 + •••,we find 1, 2, 3, 4, • • •, numbers becoming larger and larger. The situation is hopeless; there is no reasonable total.

Let us now consider in some detail two important infinite sums.

Geometric Series

A geometric series is an infinite sum in which the ratio of any two consecutive terms is always the same:

a + ar + ar2 + • • • + arn + • • • (a ^ 0, r 5 0).Let sn denote the sum of all terms up to arn,

sn = a + ar + ar2 + • • • + arn.If r = 1 , then sn = a + a + • • • + a = (n + 1) a, so sn-----» zi= <*>. Ifr ^ 1, there is a simple formula for sn:

/ l - rn+1\Sn = 0(1 + r + r2 + • • • + rn) = a ( — j .

(To check, multiply both sides by 1 — r.) If \r\ < 1 , then rn+1-----> 0 as nincreases. Hence a logical choice for the “ sum” of the geometric series is

2 1. INFINITE SERIES AND INTEGRALS

a/ (1 — r). But if \r\ > 1, then rn+1 grows beyond all bound, and the situation is hopeless. If r = —1, then sn is alternately a and 0. There is no reasonable sum in this case either.

An infinite geometric series

a + ar + ar2 + • • • + arn + —

has the sum a/ (1 — r) if \r\ < 1, but no sum if \r\ > 1.

Harmonic Series

The series1 1 1

1 + - + -+••• + -+•• •2 3 nis known as the harmonic series. It is not at all obvious, but the sums sn =1 + i + J + *# • + n~x increase beyond all bound, so the series has no sum. To see why, we observe that

Si = 1>i1 1 1 2

S2 = S i+ + - — - ,

/ I 1\ / I 1\ 2 1 3= 82 + (3 + i ) > S2 + ( l + l ) > 2 + 2 " 2 ’

/ l 1 1 1\ / l 1 1 1\ 3 1 4S8 = S4 + V5 + 6 + 7 + 8 ) >S4 + (8 + 8 + 8 + 8/) > 2 + 2 " 2 -

Similarly, su > 5/2, S32 > 6/2, • • •, s2« > (n + 1 )/2. Now the sequence of sums sn increases, and our estimates show sn eventually passes any given positive number. (This happens very slowly it is true; around 215 terms are needed before sn exceeds 10 and around 229 terms before it exceeds 20.)

Remark: Both the geometric series for 0 < r < 1 and the harmonic series have positive terms that decrease toward zero, yet one series has a sum and the other does not. This indicates the subtlety we must expect in our further study of infinite series.

EXERCISESFind the sum:

L1 + § + p+ — + ^ 2-1 _ ^+ l _ + h

1. Infinite Series 3

32 33 Sn+15. 3 + - + ^ H-----h — 6. 1 - y2 + y*~ -\------ + y*>

7. r1'2 + r + r3'2 H-----1- r4 8. (z + 1) + (x+ 1)H-----h (z + l )5.

Find the sum of the series:

9-1 - ? + ( 1 Y - (1 Y + — io. J - 7 + S - A + —5 W W ' 2 4 1 8 16

1L 2*0+^1 + ^2+ ••• 12-§ + ^ + i + ' “

13. — _--- 1---- \---- 1---- 1---- 1- ...2 + x2^ (2 + x2)2 ^ (2 + x2)3 ^

. . cos 6 , cos2 8 . cos3 6 .14 . ------ ------ h • • •.2 4 815. A certain rubber ball when dropped will bounce back to half the height from

which it is released. If the ball is dropped from 3 ft and continues to bounce indefinitely, find the total distance through which it moves.

16. Trains A and B are 60 miles apart on the same track and start moving toward each other at the rate of 30 mph. At the same time, a fly starts at train A and flies to train B at 60 mph. Then it returns to train A, then to B, etc. Use a geometric series to compute the total distance it flies until the trains meet.

17. (cont.) Do Ex. 16 without geometric series.18. A line segment of length L is drawn and its middle third is erased. Then (step 2)

the middle third of each of the two remaining segments is erased. Then (step 3) the middle third of each of the four remaining segments is erased, etc. After step n, what is the total length of all the segments deleted?

Interpret the repeating decimals as geometric series and find their sums:19. 0.11111-•• 20. 0.101010-••21. 0.434343- - - 22. 0.185185185- - -.Show that the series have no sums:

23. - + - + - + -+ ••• 24. 1 + - + - + - + • • • .2 4 6 8 3 5 725. Find n so large that

26. Aristotle summarized Zeno’s paradoxes as follows:I can’t go from here to the wall. For to do so, I must first cover half the distance, then half the remaining distance, then again half of what still remains. This process can always be continued and can never be completed.

Explain what is going on here.

2. CONVERGENCE AND DIVERGENCE

It is time to formulate the ideas of Section 1 more precisely.

4 I. INFINITE SERIES AND INTEGRALS

An infinite series is a formal sum

dl + 0,2 + ds +

Associated with each infinite series is its sequence {sn} of partial sums defined by

S i = d i , $2 = d i + 0 2 , • • • , sn = d i + a 2 + • • • + d n .

A series converges to the number S, or has sum £,' if limn_*« sn = S. A series diverges, or has no sum, if lim^*, sn does not exist.

A series that converges is called convergent; a series that diverges is called divergent.

Let us recall the meaning of the statement lin ing sn = S. Intuitively, it means that as N grows larger and larger, the greatest distance \sn — $|, for all n > N, becomes smaller and smaller. Precisely, for each e > 0, there is a positive integer N such that |sn — S | < e for all n > N. Let us rephrase the definition of convergence accordingly.

The infinite series ai + a2 + a3 + • • • converges to S if for each e > 0, there is a positive integer N such that

| (ai + a2 + • • • + an) — S\ < e

whenever n > N.

Thus, no matter how small e, you will get within e of S by adding up enough terms. For each e, the N tells how many terms are “ enough” . Naturally the smaller e is, the larger N will be. From the way convergence is defined, the study of infinite series is really the study of sequences of partial sums. Hence we may apply everything we know about sequences.

We know that inserting, deleting, or altering any finite number of elements of a sequence does not affect its convergence or divergence. The same holds for series. For instance, if we delete the first 10 terms of the series ai + a2 + a3 +• • •, then we decrease each partial sum sn (for n > 10) by the amount ai + a2 + • • • + ai0. If the original series diverges, then so does the modified series. If it converges to S, then the modified series converges to aS — (ai + a2 +• • • + aio).

W a r n in g : In problems where we must decide whether a given infinite series converges or diverges, we shall often, without prior notice, ignore or change a (finite) batch of terms at the beginning. This, we now know, does not affect convergence.

2. Convergence and Divergence 5

Notation

The first term of a series need not be a\. Often it is convenient to start with a0 or with some other a&.

It is also convenient to use summation notation and abbreviate ai + a2 + «3 + • • • by J2n=i an, and even simply J2 an- In summation notation, the partial sums sn of an infinite series ]Cn°°=i are given by

n

k= 1

Cauchy Criterion

Recall the Cauchy criterion for convergence of sequences:

A sequence {sn} converges if and only if for each e > 0, there is a positive integer N such that

whenever m,n > N.

Thus all elements of the sequence beyond a certain point must be within e of each other. The advantage of the Cauchy criterion is that it depends only on the elements of the sequence itself; you don’t have to know the limit of a sequence in order to show convergence. That’s a great help; sometimes it is very hard to find the exact limit of a sequence, whereas you may only need to know that the sequence does indeed converge to some limit.

Let us apply the Cauchy criterion to the partial sums of a series. We simply observe (for m > n) that

sm — sn = (ai + a2 + • • • + an + an+1 + • • • + am) —■ («i + a2 + • • • + an)

= &n+i + an+2 + • • • + am.

Cauchy Test An infinite series X an converges if and only if for each e > 0, there is a positive integer N such that

\a>n+l + &n+2 + ' • • + Om\ < €whenever m > n > N.

Thus beyond a certain point in the series, any block of consecutive terms, no matter how long, must have a very small sum.

In the last section we proved the harmonic series diverges by producing blocks of terms arbitrarily far out in the series whose sum exceeds J. In other words, we showed that the Cauchy test fails for € = J.

Suppose the Cauchy test is satisfied, and take m = n + 1. Then the block

consists of just one term am, so \am\ < e when m > N. In other words,dm * 0.

Necessary Condition for Convergence If the series X an converges, thenlim^oo dn = 0.

W a r n in g : This condition is not sufficient for convergence. The harmonic series 1 + \ + J + • • • diverges even though l/n -----» 0.

Positive Terms

Suppose an infinite series has only non-negative terms. Then its partial sums form an increasing sequence, S1 < S2 < S3 < S4 < • • •. Recall that an increasing sequence must be one of two types: Either (a) the sequence is bounded above, in which case it converges; or (b) it is not bounded above, and it marches off the map to + 00.

We deduce corresponding statements about series:


A series «i + &2 + «3 + • • • with an > 0 converges if and only if there exists a positive number M such that

d\ + a2 + • • • + dn < M for all n > 1 .

Using this fact, we can often establish the convergence or divergence of a given series by comparing it with a familiar series.

Comparison Test Suppose 2 «» and X bn are series with non-negative terms.

(1) If X) an converges and if bn < dn for all n > 1, then X bn also converges.

(2) If J2 diverges and if bn > dn for all n > 1, then X) bn also diverges.

Proof: Let sn and tn denote the partial sums of X) and X bn respectively. Then {sn} and {tn} are increasing sequences.

(1 ) Since X an converges, sn < X i°° dn = M for all n > 1 . Since bk < dk for all k, we have tn < s„ for all n. Hence tn < sn < M for all n > 1, so £ 6„ converges.

(2 ) Since X an diverges, the sequence {sn} is unbounded. Since bk > dk, we have tn > sn. Hence { n} is also unbounded, so X bn diverges.

N o t e : It is important to apply the Comparison Test correctly. Roughly speaking, ( 1 ) says that “ smaller than small is small” and (2) says that “ bigger than big is big” . However the phrases “ smaller than big” and “ bigger than small” contain little useful information.

EXAMPLE 2.1

Test for convergence or divergence:

'•> »> 2 * » X r f r ,Solution: (a) (sin2n)/3n < 1/3W. But £ l/3n converges, so the given

series converges.(b) 1 /y/n > 1 /n. But £ 1/n diverges, so the given series diverges.(c) Diverges because a„ = n/(2n + 1 )-----> 1 ^ 0 .

Answer: (a) converges (b) diverges (c) diverges.

p-Ser/es

The comparison test is useful provided you have a good supply of known series. An excellent class of series for comparisons are those of the form £ l/np.

The series £ l/np diverges if p < 1 and converges if p > 1.

Proof: If 0 l/n and the series diverges by comparison with the divergent series £ l /n .

If p > 1, we shall show that the partial sums of the series are bounded. We use an important trick: we interpret sn as an area and compare it with a region below the curve y = l/xp. See Fig. 2.1.

F ig . 2.1 Area under the curve exceeds the rectangular sum.

The combined areas of the rectangles shown is less than the area under the decreasing curve between x = 1 and x = n. Therefore

f ndx _ -1 1 n - L 1J l xp p — 1 xp~l 1 V - 1 Ik np-7

Since p — 1 > 0, the right side is a positive number, a little less than 1/ (p — 1) for all values of n. Hence

Sn = 1 + ( ~ + “ + * • • H— “ ) < 1 H----- :\2P 3P nv) p — 1for n > 1. Thus the partial sums are bounded if p > 1, so the series converges.

EXERCISESDetermine whether the series converges or diverges:

n 2 + 1 2 n\ / n

3 Y - 5 — 4 V ____l—L j 4n + 3 # (2n - 1

^ ny/n + 3 ^

r. V n2 s V ±/ j/ 2n4 7 / j/ In n

*2? “ X

)2

+

(n+ 1) (n + 3) (n + 5)11. Show that 2 ^ ! 1/n2 < 2. [Hint: See text.]12. Show that 1/w! < 3. Compare n! with 2n.]13. Prove that if £ and £ converge, then so does £ (an + bn), and find the

sum.14. If £ an and £ bn diverge, show by examples that £ (an + bn) may either con

verge or diverge.Let £ an be a convergent series of positive terms:15. Prove that £ an2 converges.16. Show by examples that £ y/om may either converge or diverge.

3. TESTS FOR CONVERGENCE

Suppose £ an is a given series, and c 5* 0. Then the two series £ an and £ can either both converge or both diverge. For the partial sums of the series are {sn} and {csn}. Clearly these sequences converge or diverge together.

3. Tests for Convergence 9

We can extend these remarks to a pair of series £ an and £ bn where the ratios bn/an are not constant, but restricted to a suitable range. Throughout the rest of this section all series will have only positive terms.

Let Yj an and £ bn be given series with positive terms. Suppose there exist positive numbers c and d such that

bn . ,c < — < d an

for all sufficiently large ft. Then the series both converge or both diverge.

Proof: If £ an converges, then so does £ dan. But bn < dan, so £ bn converges.

If £ an diverges, then so does £ can. But bn > can, so £ bn diverges. Done.

The conditions of the preceding test are automatically satisfied if the ratios bn/an actually approach a positive limit L. Then, by the definition of limit with e = JL, all ratios satisfy \L < bn/an < §L, except perhaps for a finite number of them.

Let £ an and £ bn have positive terms. If lim bn/an = L exists and if L > 0, then either both series converge or both series diverge.

EXAMPLE 3.1

Test for convergence or divergence;

w 2 ^ Solution: (a) When n is very large, n is much larger than y/n. This

suggests that the terms behave roughly like 1/n, so we apply the test with an = 1/n and bn = l/ (ft+ \/n ):

bn ft 1 1 ,— = ----------7=- = --------- ;---t=---------->-------- = 1 , as f t ---------> 00.an ft + V ft 1 + 1 /y/n 1 + 0

The ratios have a positive limit. Therefore £ bn diverges since £ 1/ft diverges.(b) When n is very large, the terms appear to behave like 4ft/3ft3 = 4/3ft2.

This suggests comparison with the convergent series £ V ^ 2- Let an = 1/ft2 and bn = (4ft + 1)/ (3ft3 — ft2 — 1). Then

bn (4ft + l)ft2 4 + 1/ft ______ 4~an = 3ft3 - ft2 - 1 = 3 - 1/ft - 1/ft3 * 3 ’

The ratios have a positive limit. Therefore £ bn converges because £ 1/n2 converges.

Answer: (a) diverges (b) converges.


The Ratio Test

In a geometric series, the ratio an+i/an is a constant, r. If |r| < 1, the series converges, basically because its terms decrease rapidly. B y analogy, we should expect convergence in general if the ratios are small, not necessarily constant.

Let £ an be a series of positive terms.

(1) The series converges if

an+l . 1-- < r < 1dn

from some point on, that is for n > N.(2) The series diverges if

an+l ^ ^

from some point on.

Proof: (1) Suppose an+i/an < r < 1 starting with n = N. Thenun+i < Gtfr, cln+2 < ctN+ir < aivT2,

and by induction, aw+k < clntk, that is, an < aNrn~N = (aNr~N)rn for all n > N. It follows that the series £ an converges by comparison with the convergent geometric series £ rn.

(2) From some point on, an+i > an. The terms increase, hence the series diverges.

W a rn in g : Note that the test for convergence requires an+i/an < r < 1, not just an+i/an < 1. The ratios must stay away from 1. If an+i/an < 1 butan+i/an-----» 1, we may have divergence. For example, take an = 1/n.Then an+1/an = n/(n + 1) = 1 — l/ (n + 1) < 1, but £ 1/n diverges.

It often happens that the ratios an+i/a,n approach a limit.

Ratio Test Let £ an be a series of positive terms. Suppose an+i/an ---------> r .

(1) The series converges if r < 1.(2) The series diverges if r > 1.(3) If r = 1, the test is inconclusive; the series may either converge or

diverge.

3. Tests for Convergence 11

Proof: (1) If r < 1, choose e so small that r + e < 1. B y definition ofthe statement an+i/an-----» r, there is a positive integer N such thatan+i/an < r + e < 1 for all n > N. Therefore the series converges by the preceding test.

(2) Similarly, if r > 1, then an+i/an > 1 from some point on. The series diverges.

(3) If r = 1, this test cannot distinguish between convergent and divergent series. For example, take an = 1 /nv. The series converges for p > 1, diverges for p < 1. But for all values of p,

On+ldn

nv(n + 1)p \n + 1/ \ n + 1/ (1 - 0)*’ = 1.

EXAMPLE 3.2

Test for convergence or divergence:

(a) l l <» 1 %.Solution: (a) Set an = n/2n. Then

CLn+ 1 _ n + 1 / n _ n + 1 _ 1 / 1\ an ~ 2n+1 / 2" “ 2n “ 2 \ n/

Since | < 1, the series converges by the ratio test.(b) Set an — 10n/n\. Then

Cln+l = 10n+1an 1*2* • «n(n + 1)

j 10”

/ 1*2•••n10n ->0.

Since 0 < 1, the series converges by the ratio test.

Answer; (a) converges (b) converges.

EXERCISESTest for convergence or divergence:

1' ■ I

' ■ I

1 2. V 1n2- 3 / \/2n3 — n

1 4. V* 5 + \/n4n — 1 A/ 1 + nn3n\ 6. H O "ne~n 8. Y 3-+1

/ j 5e” + n

12 J. INFINITE SERIES AND INTEGRALS

9 V 10 V —Z/ 3 » - n Z/ On n)"

i i V w! 12 V i*!Z .■ Z/ 1-3-5 ••• (2n — 1) • Z/ (2w) ! '

'ind all real numbers z for which the series converges:

X x2n 1 sin2 nxs' l i L ,—

15. ^ (3x)2n 16. ^ m:2".

17. (Root Test) If an > 0 and if y/cin < r < 1 for n > 1, show that £ an converges.

18*. Let an and £ bn be series with positive terms. If £ an converges, and if bn+i/bn < an+i/an, show that £ also converges.

19. Let £ an and £ 6n be series with positive terms. Suppose bn/an ---- » 0. Findan example where £ 6n converges while £ an diverges. Does this contradict the text?

4. SER IES WITH POSITIVE AND NEGATIVE TERM S

Infinite series with both positive and negative terms are generally more complicated than series with terms all of the same sign. In this section, we discuss two common types of mixed series that are manageable.

Alternating Series

An alternating series is one whose terms are alternately positive and negative. Examples:

1 1 11 _ 2 + 3 _ i + _ + ’

1 — x2 + x* — x6 H---- 1---• • • (alternating for all x ^ 0),rp2 ^.3 4

x — — + — — — H---- 1---- (alternating only for x > 0).

Such series have some extremely useful properties, two of which we now state.

(1) If the terms of an alternating series decrease in absolute value to zero, then the series converges.

(2) If such a series is broken off at the n-th term, then the remainder(in absolute value) is less than the absolute value of the in + l)-thterm.

4. Series with Positive and Negative Terms 13

These assertions provide a very simple convergence criterion and an immediate remainder estimate for alternating series. We shall not give a formal proof; rather we shall show geometrically that they make good sense. However, a proof is outlined in Exs. 15 and 16.

Suppose £ an is an alternating series whose terms decrease in absolute value to zero. (To be definite, assume a\ > 0.) Let sn = «i + a2 + • • • + an. Plot these partial sums (Fig. 4.1). The partial sums oscillate back and forth as shown. But since the terms decrease to zero, the oscillations become shorter and shorter. The odd partial sums decrease and the even ones increase, squeezing down on some number S. Thus, the series converges to S.

---------------------- |a2| ------------------------------------------------------------- \a3\ ------------------------------------- »

------------ M ----------- --*---- |a5| ---- -

— I---------- 1----1--1------- 1-------------- 1-------- 1—s 2 «4 se S s * s 3 Si = ai

F ig. 4.1 partial sums of an alternating series

If the series is broken off after n terms, the remainder is | S — sn|. But from Fig. 4.1,

\S Sn\ < Sn\ = |dn_j_i|.

Thus, the remainder is less than the absolute value of the (n + l)-th term.

EXAMPLE 4.1

Find all values of x for which the series

converges.

Solution: From the ratio test, it is easily seen that the series converges for |#| < 1 and diverges for |#| > 1. But what happens if \x\ = 1 ? At x = ±1, the series is

1 1 11 _ 1 + 2 ~ 3 + i ~ + " ' ’

an alternating series whose terms decrease in absolute value to zero. Such a series is guaranteed to converge by the statement above.

Answer: Converges for \x\ < 1.

N otation : A useful device for abbreviating alternating series is use of

the factor ( —l ) n, an automatic sign reverser. For example, the alternating harmonic series can be written £ (—1 )n-1/n.

EXAMPLE 4.2

It is known that the series 0 xn/n\ converges to ex for all real x. Use this series to estimate 1/e to 3-place accuracy.

Solution: Set x = — 1. Then e~l = £^=0 (~ 1 )n/n\. The signs alternate and the terms decrease in absolute value to 0. Therefore,

1 1 ( ~ l ) n . ,e = 1 — — + — ---h • • • H---- ;---h remainder,1 ! 2 ! n\where

| remainder | < -— 7-— .(n + 1 )!For 3-place accuracy, we need | remainder| < 5 X 10~4, so we want an n

for which

, I ■ , < 5 X 10-4, (n + 1)! > I X 104 = 2000.(n + 1)! 5

Now 6! = 720 and 7! = 5040. So we choose n + 1 = 7, that is, n = 6.Answer: l — l + i — -| + ^ r — 0.368.

Absolute Convergence

How is it that the harmonic series £ 1/n diverges but the alternating harmonic series £ ( — l ) n~1/n converges? Essentially the harmonic series diverges because its terms don’t decrease quite fast enough, like 1/n2 or 1/2W for example. Its partial sums consist of a lot of small terms which have a large total. The terms of £ l/ 2n, however, decrease so fast that the total of any large number of them is bounded.

The alternating harmonic series converges, not by smallness of its terms alone, but also because strategically placed minus signs cause lots of cancellation. Just look at two consecutive terms:

f - - 1 _ 1n n + 1 n(n + 1 )

Cancellation produces a term of a convergent series! Thus £ ( —1 )n~l/n converges because its terms get small and because a delicate balance of positive and negative terms produces important cancellations.

Some series with mixed terms converge by the smallness of their terms alone; they would converge even if all the signs were +. We say that a series £ an converges absolutely if £ \an\ converges. As we might expect, absolute convergence implies (is even stronger than) convergence.

4. Series with Positive and Negative Terms 15

If a series £ an converges absolutely, then it converges.

Proof: Suppose £ M converges. B y the Cauchy test, for each e > 0 there is an N such that

\an+i\ + \an+2\ + • • • + \dm\ < e, m > n > N.But

|f l » + l + d n + 2 + * * * + d m \ < |® n+ l | + * * * + \dm\ < €

by the triangle inequality. Therefore £ an converges by the Cauchy test.

R em ark : In studying series with mixed terms, it is a good idea to check first for absolute convergence. Just change all signs to +, then use any test for convergence of positive series.

EXAMPLE 4.3

Test for convergence and absolute convergence:

(a) 1 -j- — — — -f- — -f~ — — — -)- -f~ — • • * w 22 32 42 52 62

Solution: (a) The series of absolute values is £ V nS which converges. The series is absolutely convergent, hence convergent.

(b) The series of absolute values is £ 1 /y/n which diverges. Hence the given series does not converge absolutely. It does converge, nevertheless, because it satisfies the test for alternating series: the terms decrease in absolute value to zero.

Answer: (a) converges and converges absolutely, (b) converges but not absolutely.

EXERCISESTest the series for convergence and for absolute convergence:

1 . / { in n / j

In nL X ' - " ‘ c r . 2 - X3X <_1)"5^ 4'2 <- i ) " ,

7. V ( - 1 ) ” * 8. V (— 1 )B sin —/ ( n + inn n

9. y io. y ( - 1 )»<»+»/*, ' . ./_! n2 Z / Ve~” +Test for convergence or divergence:

iLi+H +i + H ++- - -12‘ 1 + l " § ~ 5 + + ------ •13. Estimate l/\/e to 4-place accuracy by using the technique of Example 4.2.

14*. Suppose the series £ anXn converges for the positive value x0. Show that the series converges absolutely for \x\ < x0.

15*. Suppose £ an is an alternating series whose terms decrease in absolute value toward 0. Suppose the first term is a\ > 0. If {sn} denotes the sequence of partial sums, show that the subsequence {s2n} is increasing and bounded above and the subsequence {s2n-i} is decreasing and bounded below.

16*. (cont.) Conclude that the two sequences coverge and have the same limit S.Show that £ an = S.

17*. Suppose £ an and £ bn2 both converge. Prove that £ anK converges absolutely. 18*. Suppose £ an converges, but not absolutely. Let and £ Ck be the series

made of the positive an’s and negative an’s respectively. Prove that both £ bj and £ Ck diverge.

5. IM PROPER INTEGRALS

In scientific problems, one frequently meets definite integrals in which one (or both) of the limits is infinite. Here is an example.

Imagine a particle P of mass m at the origin. Consider the gravitational potential at a point x = a due to P. This potential is the work required to move a unit mass from the point x = a to infinity, against the force exerted by P . According to Newton’s Law of Gravitation, the force is km/d2, where d is the distance between the two masses and k is a proportionality constant. The work done in moving the unit mass from x = a to x = b is

f b [ b km ( 1\ 16 J (force) dx = J —jd x = km ^---j | = km

c— »■

Let b -----> oo. Then 1/6-----> 0, hence

/ * * ! ? * ---------o ) - ^ .Ja X \a / a

Thus k m / a is the work required to move the mass from a to < » . It is convenient

5. Improper Integrals 17

to set

6 km km — ax = —

A definite integral whose upper limit is oo, whose lower limit is — oo, or both, is called an improper integral

In order to give a precise definition of infinite integrals, we must recall what is meant by an expression such as limx-*, F (x ) = L. The definition is patterned after the definition of the limit of a sequence.

Let F be defined for x > a, where a is some real number. Then

lim^oo F (x ) = L

if for each e > 0, there is a number b such that

\F(x) — L\ < e for all x > b.

For increasing and decreasing functions the basic fact is analogous to the one for sequences. (We say F is increasing provided F (x i) < F(x2) wheneverX i < x 2.)

Let F be an increasing function. Then lim *^ F (x ) exists if and only if F (x ) is bounded above, i.e., if and only if there exists a number M such that

F (x ) < M

for all x > a, that is, on the domain of F.

Similarly if F (x ) is a decreasing function, then lim*-* F (x ) exists if and only if F (x ) is bounded below.

We shall not give the proof; with some modification it is the same proof as for sequences.

Another important fact, also similar in spirit to one for sequences, is the Cauchy criterion:

Cauchy Criterion lim*-*, F (x) exists if and only if for each e > 0, there exists b such that

Limits

|F (x ) - F ( z )| <€whenever x > b and z > b.

There is a similar discussion for limits of the form lim*.*-*, F (x ) which is

hardly worth writing down. Obviously, if we set G(x) = F ( — x), then limx -oo F (x ) is equal to lim ^* G(x), so limits at — oo involve nothing new.

Definition of Improper Integrals

Suppose f(x ) is defined for x > a and is integrable on each interval a < x a. Set

provided the limit exists. If it does, the integral is said to converge, otherwise to diverge.

Similarly, define

Now linu oo F (b ) may or may not exist.

Define

provided both integrals on the right converge.

R em a rk : An integral from — oo to oo may be split at any convenient finite point just as well as at 0.

An improper integral need not converge. As an example, take

Since

and In b oo as b oo, the limit

does not exist; the integral diverges.

Remember that a definite integral of a positive function represents the area under a curve. We interpret the improper integral

[ f(x )dx, f (x )> 0,J a

as the area of the infinite region in Fig. 5.1. If the integral converges, the area is finite; if the integral diverges, the area is infinite.

y


F ig. 5.1 area of infinite region

At first it may seem unbelievable that a region of infinite extent can have finite area. But it can, and here is an example. Take the region under the curve y = 2~x to the right of the ?/-axis (Fig. 5.2). The rectangles shown in Fig. 5.2 have base 1 and heights 1, £, i, i, • • •• Their total area is

1 1 11 + 2 + i + 8 + = 2 ‘

Therefore, the shaded infinite region has finite area less than 2.

EXAMPLE 5.1

Compute the exact area of the shaded region in Fig. 5.2.

F ig. 5.2

Solution: The area is given by the improper integral

f 2rx dx = lim f 2~x dx.J 0 &-+00 J 0

An antiderivative of 2~x is — 2~x/\n 2. (Use 2~x = e~x ln2.) Hence


R em a rk : This answer is reasonable. A look at Fig. 5.2 shows that the area of the shaded region is between 1 and 2. A closer look shows that the area is slightly less than 1.5. Why?

Solution:b

= arc tan x = arc tan 6.o

Let b oo. Then arc tan b --------- » t/2. Hence

2 *

EXAMPLE 5.3

Evaluate e* dx.

Solution:


Let a • — oo. Then ea-----» 0. Hencers

fJ — *

ex dx = lim (es — ea) = e3.

: eV

The following integral arises in various applications such as electrical circuits, heat conduction, and vibrating membranes:

/ e~8Xf(x )dx . Jo

It is called the Laplace Transform of f (x ) .

EXAMPLE 5.4

f xEvaluate / e-’* cos x dx,

Jos > 0.

Solution: From integral tables or integration by parts,/*& S1—8X b

/ e-*'Jo

~3X cos x dx = ( —s cos x + sin x)?2 + 1

e~8b . s( — s cos 6 + sin b) +s2 + 1 + 1Now let b-----> oo :

cos x dx[o o rb/ e~8X cos x dx = lim / cos a: dx = 0 + 0 6->00 0 + 1

Answer;

EXAMPLE 5.5

f °°Evaluate /«/ —oo

dx3e2x + a”"2*

Solution: B y definition, the value of this integral is

f ° ° d x f ° dxJ 0 3e2x + e~2x + J_ „ 3e2x + e"2* ’

provided both improper integrals converge. From integral tables,rb dx

■ ' ”V 3 )


= [arc tan(e26\/3) — arc tan V^]-

Now arc tan(e26-\/3)-----> tt/2 as 6-----* oo. Note that arc tan \/S = 7r/3.Hence

f*3 dx . f 6 dx _ 1 T 7i- ?rl36** + e~2x = ™ J 0 3e2* + “ 2^/3 1.2 “ 3 j '

Similarly,

f .-- ^ — 7 = lim [ ° ^J - x 3e2x + e~2x 0„_M J a :3e2x + e~2x

- — [arc tan V 5 - arc tan 0] - ^ [ | - o] .

Thus both improper integrals converge. The answer is the sum of their values.

Answer:

R em ark : D o you prefer th is snappy calculation?

dx 1/ --~—^-arc tan (v^^2x)3e2* + e"2* 2 \ /3

1 1 7T 7T-- ~7=- (arc tan oo — arc tan 0) = -—7= - = -— j=-.2 \ /3 2 a /3 2 4 y/3

Warning. Try the same slick method onf " dx

J- x X2It fails. Why?

Evaluate:

EXERCISES

1.f 00 dx

/ 2 X3

2. I ° ° e~x dx

3.f °°/ xe~x dx h

4.[ - 1 dx

J - 0 0

5.[- 1 dx a dxJ- 00 1 + x2

0. J 4 x\/x

xer * 2 dx/ oo f ooe-1*1 dx 8. /

■00 J -00f 00 dx f 00

9 Ji W T + x 1 10- Jo e~'Xsinxdx (s > °)

/°° r r/rr /*00"4 (letw = :r2) 12. I xe~BX dx (s> 0)

/OO Too

x2e~9xdx (s > 0) 14. I xne~9x dx (s > 0)

r oo /*oo15. / e e~8x dx (s > a) 16. / xe e*™ dx (s > a)

/oo Coo

e~*x cosh x dx (s > 1) 18. / xe-** sin x dx (s > 0).

Is the area under the curve finite or infinite?19. 2/ — 1/x; from x = 5 to x = oo20. i/ = 1/x2; from x = 1 to x = oo

21. y = sin2 x; from x = 0 to x = oo22. y = (1.001 )” *; from X = 0 to x = oo .

Solve for b:oo f b , [ b dx /■« dx23 j.e *■ J, ’’ * j, r+?* j. r+ '25. Find 6 such that 99% of the area under y = e” * between x = 0 and z = oo is

contained between x = 0 and x = b.Denote the Laplace Transform of /(#) by L (/ ) (s), so

L (f )(s ) = f°° e-*f(x) dx.26*. Suppose /(x) is continuous for x > 0, and for some n and some constant c we have \f(x)\ < cxn for all x sufficiently large. Prove that L (f)(s ) exists for all s > 0.

27*. Suppose / has a continuous derivative /' for x > 0 and |/' (x)| < cxn for x sufficiently large. Prove for s > 0 that

L (f')(s ) = - f(0 ) + sL(f)(s).

28*. For the / in Ex. 26, set g(x) = / f(t) dt. Prove for s > 0 that

L ( g ) ( s ) = - L ( f ) ( s ) .

6. CONVERGENCE AND DIVERGENCEWhether an improper integral converges or diverges may be a subtle

matter. The following example illustrates this.

EXAMPLE 6.1

f 00 dxFor which positive numbers p does the integral / — converge? diverge? ^ ,7?

Solution: Suppose p ^ 1. Then

[ hdx 1 , 1 6 _____\_ / ____1 _ \J 1 Xv p — 1 ! p — 1 \ 6p-1/

As b •

and

Hence

bp-i

ftp-1

-> 0 if p — 1 > 0,

->oo if p — 1 < 0.

£P

exists if p > 1, does not exist if p < 1. That means the given integral converges if p > 1, diverges if p < 1.

If p = 1,

Idx = In 6- oo as b ■

the integral diverges.Answer: Converges if p > 1, diverges if p < 1.

R emark 1. Obviously the same is true of the integral

[*d xJ a X P

for any positive number a.R emark 2: Now, a subtle question. Why should this integral converge

if p > 1 but diverge if p < 1? (See Fig. 6.1.) The curves y = \/xv all decrease . as x increases. The key is in their rate of decrease. If p < 1, the curve decreases slowly enough that the shaded area (Fig. 6.1a) increases withoutbound as b ---------» oo. If p > 1, the curve decreases fast enough th a t theshaded area (Fig. 6.1b) is bounded by a fixed number, no matter how large b is.

F i g . 6 .1a

Convergence Criteria

We know that the integral00 d r .

(a > 0)

converges. Suppose that 0 < g(x) < 1/x2. Then

g (x) dx

fJ a

/ 'J a

also converges because the area under the curve y = g (x ) is even smaller than the area under y = 1/x2. This illustrates a general principle:

If/(# ) ^ OandifO < g(x) < f ( x ) f o r a < x < oo, then the convergence of/' oo roo/ f ( x ) dx implies the convergence of / g(x )dx.

J a J a

Proof: We imitate the proof of the comparison test for positive series.

Let

F Q>) = f / (* ) dx, G(b) = f g(x) dx.J a J a

Since f(x ) > 0 and g(x) > 0, both F(b ) and G(b) are increasing functions of b. Also g(x) < f(x ), so G(b) < F(b ). B y hypothesis lim^oo F (b) =Too r oo/ f(x ) dx exists, hence F (b ) < / f(x ) dx. It follows that G(b) < F (b ) <

J a J a

f(x ) dx, so limt oo G(b) exists, that is,f./■J a

g(x)dx converges.

EXAMPLE 6.2

Show that the integrals converge:f * dx Z*00 sin2 x

(a) i , ? T v i ; <b) J , (c) J , ~ d I-

Solution: Note that1 1

< “ I , 1 < X < oo,x2 + y/x x2

e - x< e~x, 0 < x < oo,

x + 1

sin2 x 1x° x°

Since the integralsdx [ °° dx

j , * • i '■**- ) , ? all converge, the given integrals converge by the preceding test.

A second important convergence criterion is this:

Suppose f(x ) > 0 and g(x) is bounded, i.e., \g(x)\ < M for some constant M. Then the convergence of

r oo r/ f(x )dx implies the convergence of / f(x)g(x)dx.

J a Jo ,

Proof: Let

F (b) = f f(x)dx, H (b) — I f(x ) g(x) dx.J a J a

We are given that lim*,.** F(b ) exists, and we must show that lim*.** H(b) exists.

By the Cauchy criterion, given e > 0, there exists B such that

- f J a

IF (c ) — F (b )I < — whenever c > b > B.M ~

It follows that

|H (c ) - H (b )| = i; r.f(x )g (x )dx\ < / |/(*)| |flr(ar)| dx

< M f(x ) dx = M [F {c ) - F (b ) J < M -j- = e J b M

whenever c > b > B. Therefore by the Cauchy criterion, limb-oo H(b) exists, that is,

/ "J af(x )g {x )d x converges.

EXAMPLE 6.3

Show that the integrals converge:f 0° In x(a) / r xsm 8xdx, b / — dx.

Jo J i Xs

Solution: Apply the above criterion,

(a) Since

/ e~x J o

dx

converges and |sin3 x\ < 1, the given integral converges,(b) Write

In x _ 1 In xx3 x2 x

The integralf«>dx

J i x2

converges and (\nx)/x is bounded. [The maximum of (In x)/x is l/e.~\ Hence the given integral converges.

R e m a r k : Both convergence criteria apply also to improper integrals of the forms

fJ — a

f(x )dx and/ ‘J —a

f ix ) dx.

A Divergence Criterion

Here is a simple criterion for divergence of an improper integral.

If f(x ) > 0 and if g (x) > f(x ) for a < x < oo, then the divergence of

/ f{x ) dxJ a

implies the divergence of

/ gix)dx.J a

Proof: This time

Gib) = f gix)dx > I fix )d x = F(b )J a J a

and F(b ) is unbounded, so G(b) is unbounded. Hence

g(x)dx diverges./ "J a

This criterion is obvious geometrically; since g(x) > /(» ), the region under y = g(x) contains the region under y = f(x ). If the second region has infinite area, so does the first.

EXAMPLE 6.4

Show that the integrals diverge:

(a) [ ” V x , . s [ x ___ dx_J i 1 + x J 2 -y/x - v 'z

f 00 Inx/ — dx.

Solution: Note that

1 1>

\ / x — y / x y / x ’

1 < X < 00,

2 < x < oo,

In x In 3 1-- > --- > - , 3 < x < oo .X X X

Since the integrals

dxI i s / x ’ J 3

all diverge, the given integrals diverge by the preceding criterion.

dx f°° dx f 00 dxJ x 1 + x ’ J2 y / x ’ Js x

EXERCISESDoes the integral converge or diverge?

h dxJo J + l

+ X 2

cosh x dx

’■ / ' T

7 f ° ° sin x7-00 i + v

, [ 'J i \ /^ (* + 4)

/■- z3 ,■ J,

f * fjj.13. Show that / —--

y 2 £ (In a;

9.

11

X2 + £

4. [ e~x* dx

, x dx6,

10

12

2J:

• /■“yo V52~+~3 f 00

. / sin a; dx

h dx ’ J 2 In x

■ / " i

)3Use the substitution w = In x.]

dx

+ x + ex *

converges if p > 1, diverges if p < 1.

14. Show thatf; converges if p > 1, diverges if p < 1.x In x[ln(ln x)]p15. Denote by the infinite region under y = 1/x to the right of x = 1. Suppose #

is rotated around the x-axis, forming an infinitely long horn. Show that the volume


of this horn is finite. Its surface area, however, is infinite (the surface area is certainly larger than the area of R). Here is an apparent paradox: You can fill the horn with paint, but you cannot paint it. Where is the fallacy?

Find all values of s for which the integral converges:r oo r oo g—sx

16. / e~sxex dx 17. / —.—5 dxJo Jo 1 + *2f 00 r 00 x8

18. / e~8xe~x2 dx 19. / ——,— dx.Jo J 1 (1 + x*)°

7. RELATION TO INFIN ITE SER IES

We have already seen a number of similarities between infinite series and infinite integrals. In this section we discuss a very useful connection between them which often enables us to establish the convergence or divergence of a series by studying a related integral. This is important, for usually it is easier to find the value of an integral than the sum of a series.

Consider the relation between the series

1 1 1— -j- — -j- • • • — -j- • • •22 32 ^ n2 ^

and the convergent integral

/00 dx

x2

(See Fig. 7.1.) The rectangles shown in Fig. 7.1 have areas 1/22, 1/32, •••. Obviously the sum of these areas is finite, being less than the finite area under the curve. Hence, the series converges. This illustrates a general principle:

F ig . 7.1 Note: x scale is 1/4 of y scale.

7. Relation to Infinite Series 31

Suppose/(#) is a positive decreasing function. Then the series

/ ( l ) + / ( 2 ) + ••• + / ( » ) + •••

converges if the integral

J f ( x ) dx

converges, and diverges if the integral diverges.

Proof: The argument given above for f(x ) = 1 /x2 holds for any positive decreasing function f(x ). Figure 7.1 indicates that

/(2 ) + / ( 3 ) + ••• + / ( » ) < J " f ( x ) dx.

If the infinite integral converges, then

* . = / ( ! ) + / ( 2 ) + ••• + / (» ) < /(I) + J " f ( x ) d x < / ( l ) + J f (x )dx.

Hence the increasing partial sums are bounded; the series converges.If the infinite integral diverges, the rectangles are drawn above the curve

(Fig. 7.2). Their areas a re/ (l), /(2), • • •. This time

j;

»* “ /(! ) + / (2 ) + ••• + / (» )> - r

f ( x ) dx.

But the integrals on the right are unbounded. Hence the increasing sequence {sn} is unbounded; the series diverges.

V = /(*)

1 2 3 4

F ig . 7.2 The rectangular sum exceeds the integral.

EXAMPLE 7.1

For which positive numbers p does the series

converge? diverge?Solution: The series can be written

/ ( l ) +/ (2) + . . . +/ (n) + ...,•where f(x ) = \/xv, a positive decreasing function. B y the preceding principle, the given series converges or diverges as

Convergence of Integrals

Sometimes we can turn the tables and use the convergence of a series to establish the convergence of an integral. If the integrand changes sign regularly, we may be able to compare the integral with an alternating series.

EXAMPLE 7.2

Solution: First sketch the graph of y = (sinjc)/rc. See Fig. 7.3. There

converges or diverges.

Answer: Converges if p > 1, diverges if p < 1.

is no trouble at x = 0 because (sin x)/x 1 as x 0.

y

F ig 7.3 graph of y = (sin x)/x

7. Relation to Infinite Series 33

The figure suggests that the integral is given by an alternating series. To be precise, let

’ (n-fl)TT ,

an = (- 1 )" /J nil

sin xdx.

Then an > 0; in fact an is the area of the n-th shaded region in Fig. 7.3. Now

/•(w-hDir |sin /’("+!)*• i T ian = / ---- dx 0. Furthermore

f w sin (x + mr) [*dn = ( “ 1)W / -- ------ ~dx = /J 0 x + mr J 0

so an > an+1 because

sin x x -\- mr dx,

sin x sin x>x + mr x + (n + 1)t ’

Therefore the alternating series

dO — di + d2 ~ ds +converges. But

0 < X < 7T.

do — di + a% — • • • + (

so we have the existence of

f(n+1)» - l ) nd n = /

Josin x

dx,

r ( n + l ) r g i n x

lim / --- dx,n-+oo J o

where n can take only integer values.If b is any positive real number, there is an integer n such that mr <b <

(n + l)7r. Then

f b sin x f (n+i)v sjn x r(/ = / ~ /J 0 x J o J b

dx.

If b-----» oo, then n -----» oo, so the first term on the right converges to alimit. The second term approaches 0 because

1 f (n+1,Tsinx f (n+1)ir sin x 7/ dx < / a# dn1 J b % ►0.

Therefore

converges.

[ * sin x _ f b sin # ,/ ------ a# = lim / ----- da:

Jo ® 6 ^ o c y 0 s

EXERCISES

Does the series converge or diverge?

L 1 + 7 i + 75 + ^ + -

2- - + ^ + T + T + " -e e2 e3 e4

3‘ 1 + ^+§3 + ^H---

4. , ..7 f + o r sk + 1 ' 11 + y/\ 2 -f- \ /2 3 -f- \ /3 4 -f- \/4

5 - ^ + ^ + 1 '2 In 2 3 In 3 4 In 4

6- ^ + 1 ' 1 O /l-~ O \rt 12 (In 2 )p 1 3 (In 3 )p 1 4(ln4)*>

7 . ^ + ^ + 1 ■1 + l2 ’ 1 + 22 ' 1 + 32 '

« L j . 3 5 7 l2 22 32 42

9. Show geometrically that the sum o fl + ^ + ;jj2“l- 2+ ’ *' is less than 2. See

Fig. 7.1. (It is known that the exact sum is 7r2/6, a startling fact.)10* Use the method of inscribing and circumscribing rectangles to show that

ln ( « + 1 ) < 1 + ^ + 5+ + ~ < 1 + lnn .Z O 71

Is 1 + + + • • • + more or less than 10?Z o 1 UUU

11. Estimate how many terms of the series l + + + 7 + * * * must be added before the sum exceeds 1000.

Does the series converge or diverge?

8. Other Improper Integrals 35

14

16*. Prove convergent:

n = 1 ‘ n = 2OO 00

I.

\ f n In r in=1 n=2

sin £ _ —7=- ax V X

17*. (cont.) Prove convergent:

sin x2 dx.

[Hint: Set x2 = u. ]18*. Let 0 < a < b. Prove convergent:

r°° cos ax — cos bxf: dx.

[Hint: Separate the difficulties a t 0 and 00 .]19*. Let 0 < a < b. Prove convergent:

1 arc tan bx — arc tan axJr dx.

8. OTHER IM PROPER INTEGRALS

A definite integral

Lbf(x ) dx, a and b finite,

is called improper if f(x ) “ blows up” at one or more points in the interval a < x < b. Examples are

ri° dxf s dx f 5 dx f 1Jo J 1 *2 J 6 ln(ir — 5)

The first integrand “ blows up” at x = 0, the second at x = 2, the third at x = 6. Such bad points are called singularities of the integrand.

We shall discuss integralsrb

f(x ) dxfJ a

where f(x ) has exactly one singularity which occurs either at x = a or at x — b. This is the most common case.

Consider the integralP dx

Jo V xwhose integrand has a singularity at x = 0. What meaning can we give to this integral?

Except at x = 0, the integrand is well-behaved. Hence if h is any positive number, no matter how small, the integral

[ ' dxA ' x

makes sense. Its value is easily computed: dxfJ hr = 2y/x

V *2 (V 3 - Vh)-

It is reasonable to definef3 dx fs dx/ —7= = lim / —7=. = 2a/3.

Jo V x h , J h V xNext, consider the integral

f 3 dxJo «

We try to “ sneak up” on the integral as before by computing

— = In 3 — In ft,Jh z

then letting h -----> 0. But In h -----> — oo as h -----» 0. Hence; dx

fJ hoo as h -----> 0.

xThere is no reasonable value for this integral.

Motivated by these examples, we make the following definitions:

Suppose/(#) has one singularity, at x = a, and that a < b. Define

f f(x ) dx = lim f f(x ) dx, (h > 0)J a h -+0 J a +h

provided the limit exists. If it does, the improper integral converges, otherwise, it diverges.

Similarly, if f(x ) has one singularity, at x = b, define rb rb-h/ f (x ) dx = lim / f(x ) dx j (h > 0)

J a h-+0 J a

provided the limit exists.

EXAMPLE 8.1f dxFor which positive numbers p does the improper integral / —

converge? diverge? * ° xP

Solution: The case p = 1 was just discussed; the integral diverges. Now assume p 9 1. By definition, the value of the integral is

limh-*0

provided the limit exists. Now

1 dx 1 1

f 3 dx im / — ,^0 Jh %p

fJ h xp p — 1 xp~l But, as h -----► 0,

1

vh p - 1 W - 1 3P-1/

hp~land

hp~l

-> 0 if p — 1 < 0,

00 if p — 1 > 0.

Hence the limit exists only if p < 1. In that case

f S - lim r ^Jo %p ™ Jh xp (p ~ l ) 3p_1

Answer: Converges if p < 1, diverges if p > 1.

R e m a r k 1: The answer applies as well tof b dx

Jo *p

for each positive number b since the upper limit plays no essential part in the discussion. Only the behavior of l/xp in the immediate neighborhood of x = 0 counts.

R e m a r k 2: If p > 1, the curve y — l/xp increases so fast as x -----> 0that the area of the shaded region (Fig. 8.1a) tends to infinity. If p < 1, the curve rises so slowly that the area of the shaded region (Fig. 8.1b) is bounded.

C a u t i o n : D o not confuse these results with those of Example 6.1 concerning

f*dxJ 1

Fig. 8.1 graph of y = l / xp

In fact,

f 1 dxJo xp

convergesdiverges

if p < 1, I*00 dx Jdiverges if p < 1,if P > 1, J i xp [converges if p > 1.

EXAMPLE 8.2

For which positive numbers p does the improper integralr4 dxfJ 3 (x - 3 )p

converge? diverge?Solution: Change variable. Let u = x — 3. Then

f 4 da: f 1 duJ 3 (x - 3)p J 0. '

But this integral was discussed above.Answer: Converges if p < 1, diverges if p > 1.

The techniques of the two preceding examples yield a general fact (for a and b finite) :

The integrals

f b dx f b dxJ a (s “ a)p ’ J a (b - x ) p

converge if p < 1, diverge if p > 1.

The convergence criteria given in Section 6 have analogues for improper integrals with finite limits. We state (without proof) just one of these.

Suppose that f (x ) > 0, and that f (x) has a singularity at x = a or at x = b. Suppose g(x) is a bounded function. Then the convergence of


A Convergence Criterion

f ( x ) dx implies the convergence of f ( x ) g ( x ) dx.'b

EXAMPLE 8.3

Show that the integrals converge:

Solution: Use the preceding criterion.

(a) Let f ( x ) = l / v ^ a n d g(%) = cos x.

(b) Let f ( x ) = l / y / 2 ~ = ~x and g(x) = l / \ / 2 + x.

EXERCISES

Does the integral converge or diverge?

40 1. INFIN ITE SERIES AND INTEGRALS

i5 - J !'J ^x d* COS X

9. SOME DEFINITE INTEGRALS [optional]

Sometimes it is important to know the exact value of an improper integral, not just that it converges. Certain (improper) definite integrals can be found exactly by tricks, even though the corresponding indefinite integrals are difficult, or even impossible to find. The most common type of trick involves a change of variable.

EXAMPLE 9.12 TT

Prove / In sin d dd = -- - In 2.Jo 2

Solution: Since 0 < sin d < 1 for 0 < d < we have — oo < In sin 0 < 0. The integrand approaches — oo as x --------> 0 + . To prove convergence, note that (2/ir)d < sin 6 < 1 for 0 < d < hence In 6 + In ( 2 / t ) < In sin d < 0. Therefore the integrand does not change sign and the integral converges by comparison with the convergent integral

r iIn d dd.I.o

Now setr*/ 2

I = / In sin 6 dd.Jo

Make the change of variable 6 = — a. Then

'dd.r o r*/2

I = — I In cos a da — In cos 6 < J t/2 J o' t /2

We now have two expressions for I ; average them:fr/2

(In sin 6 + In cos 6) dd= i r2 Jo

1 f r/2 1 ( sin 2 6 \ ,= - J In (sin 6 cos 6) dd = - J In ^—- —j dd

v/2(In sin 2d — In 2)«

. „ , it In 2 sin 2d d d ------ -— .

But[*12 j [ r/ In sin 26 dd = - I In sin 6 dd

Jo Jo

1 f T/2 . 1 f r = - / In sin 6 dd + - / In sin d dd

Jo yx/2

1 1 f *= - / + - / In sin d dd.

% -> Jir/2

Therefore

7r In 2

9. Some Definite Integrals 41

1 I f/ = - / + - / In sin 8 dd —

4 4 7 t/2

I t is obvious by symmetry (or by the transformation 8 = it — a) that

/ [ r/2In sin d dd = I In sin d dd — I,

■12 Johence we have

1 1 7T In 2

and the formula follows.

R e m a r k : I t is known that the indefinite integral

In sin d dd/cannot be expressed in terms of elementary functions (composite functions built with rational functions, radicals, exponentials, logs, and trigonometric functions).

In the next two examples we compute by tricks an improper integral whose corresponding indefinite integral can be worked out, but is complicated. Compare Exs. 9 and 10.

EXAMPLE 9.2

Prove that

f x dx f 00 x2 dxJ 0 X* + 1 Jo x4 + 1 *

Solution: Convergence is obvious. The problem suggests a change of variables. Try x = 1/u. Then

dx —du /u2 —u2dux4 + 1 (1/u 4) + 1 w4 + 1

Hence, if 0 < a < 6,

rb dx

Let a ■ * 0 and then let b ■

1/6 u2 du _ f 1,a u2 du1/o U*~+1 = J l/b u 4 + l

oo. Then 1 /a --------> oo and 1/6 ■

P da; _ _ f 1/b u2 du _ f 1,a

Ja + 1 Jl/a UA + I Jl/b

so the stated formula follows.

EXAMPLE 9.3 ; : / ;;Prove that

f°° dx f K x2 dx i r \ /2J0 x* + 1 J0 x4 + 1 4

Solution: The two integrals are equal by the previous example. This suggests averaging to increase the symmetry:

f 00 dx I / / * 00 & \ _ i p . r 2 + iJ 0 x 4 + l ~ 2 \ J 0 x 4 + i + J 0 x 4 + l ) ~ 2 J 0 x 4 + I

Now we need a really clever change of variable. Consider

u = x ---- .x

Clearly u • — oo as £

du

- 0 + and w ■ oo as .r • oo. Also

L _ + 1dx .r2 x2 ^

so u = w($) is a strictly increasing function taking the interval (0, oo ) onto (—oo, oo). Furthermore,

1 1 x4 -|- 1u2 = x2 - 2 + — , u2 + 2 = X 2 + — = ---- -—,

x2 x2 x2

hence

du (x2 + \ ) / x 2 . x2 + 1ax = :..d x ;

u2 + 2 (#4 + l ) / x 2 x4 + 1

1 f * + 1 [°° du V 2 fM dt2 J0 *< + 1 * ~ J ^ u > + 2 ~ 2 J _ x t* + 1 ’

V 2 arc tan £7 I " \ /2

EXERCISES

10. Stirling's Formula 43

Prove:f * 71-2 f i r f ir /2 f i r f ir/2 f ir/2

1. / x In sin x dx = — — In 2 [Hint: / = / + / = / + / (?).]> 2 > > >/2 > >

2iV/2

In tan x dx — 0

fx /2

3. / sin x (In sin x) dx = In 2 — 1 [Hint: Integrate by parts.]

dx — 7r

f:4' Jo (1 + x ) V x

5. J (In x)n dx = (—l)nw! for n = 1,2,3,

/ i 3 Toox ln(l — x)dx = — - 7. I

r°° dx _ f 00 x dx _ 27r\/38- Jo xM7! " Jo ^>+1 9“ •9. Start with x4 + 1 = ( 2 + V2# + 1) (z2 — \ / 2 x + 1) and obtain the partial

fraction decomposition1 _ j \ / 2 x + 2 i —\ \ / 2 x + |

dx _ 7T cosh x 2

+x4 + 1 x2 + y/*2x + 1 x2 — \ / 2 x + 1 * 10*. (cont.) Use this to obtain

7T\/2[°° dx _ f b dx J - oo X4 + 1 &-►oo J-b X4 + 1

(Compare Example 9.3.)

10. STIRLING’S FORMULA [optional]

In this section we obtain several useful and important results in analysis by exploiting the method of approximating integrals by sums.

Euler’s Constant

We know that the harmonic series 2] 1/n diverges. Now we show much more: that the partial sum sn is approximately Inn. Consider Fig. 10.1a. Comparing the area under the curve y = l / x between 1 and n with a sum of rectangles, we see that

1 1 f n dx1 + - + • • • + - = I -----b cn — In n + cn)

2 n — 1 Ji

where cn is the area of the shaded regions.


(b) A close-up of the error; it fits into the square of side one.

F ig . 10.1 approximation of sn by an integral

Now shift the shaded areas to the left as in Fig. 10.1b. They fit inside a square of side one! I t follows that {c„} is an increasing sequence bounded above by 1. Hence limn.**, cn = y exists, and we have

1 + i + ••• + - - l n n = - + cn --------* 7 -2 n n

There is a positive constant y such thatn

t o ( £ i - I n n ) - * .k =1

The number y is called Euler's constant. Its value is approximately0.57721566. In view of Fig. 10.1b, this value seems reasonable. For the curve y = 1/x is convex, so the combined shaded areas fill out a bit more than half of the square.

R e m a r k : In many computations, the exact value of y is not needed. What counts is that the difference between XI ^/n and In n is bounded,

so for large values of n, the approximation l / k ~ In n is often accurateenough.

Wallis’s Product

This is an old and remarkable formula:

Wallis’s Product

Hm 1 f 2-4 -6 - -»(2n) T _ *■ n™ 2n + 1 Ll-3-5---(2n - 1)J ~ 2

Its derivation is an exercise in integration. Set

/•»/2 rir/2J n = / cosn 0 dd = / cosn_1

J 0 J o0 cos 0 dd.

Integrate by parts with u = cos71-*1 0 and v = cos 0 :

r*i2J f*n = 0 + / (n — 1 ) cosn_2 0 sin2 0 dd.

J o

But

hence

Solve for J n:

cosn_2 0 sin2 0 = cosn”2 0(1 — cos2 0) = cosn_2 0 — cosn 0,

Jn = (n 1 ) («/n—2 Jn) •

j w ~ 1 ,n — «/ n—2‘

Now apply this reduction formula repeatedly and eventually reach either

r */2Jo = I dd = - or Ji = I cos 0 d0 = 1.

Clearly there are two cases, n even and n odd. The results are

I . 3 .5 . . . (2n — 1 ) tt

fr/2i - / ,

Jo

J 2n 2-4-6---(2n) 2 ’

2 -4-6 - -- (2ra)1-3-5---(2ra + 1) '

Since 0 < cos 6 < 1 for 0 < 6 < %ir, we have cos2"-1 6 > cos2n d > cos2n+16,

46 1. IN FIN ITE SERIES AND INTEGRALS

hence J 2n-i > Jin > Jin+i- Substitute, then rearrange:

2 -4 - . - (2 n ~ 2) . 1-3-• • (2w - 1 ) w > 2-4-•• (2 n)1*3* • • (2n — 1) 2*4* • • (2n) 2 1 -3-• • (2n + 1) ’

l~2*4-6- • • (2n — 2 )12 t f 2 . 4 - 6 . . . (2n) I2 1LI * 3 • 5 • • • (2n — 1)J 2 Ll-3 -5 -• • (2n - 1)J 2n + l '

For simplicity, introduce the quantities

1 f 2 - 4 . 6 . . . (2n) I2" 2n + 1 Ll .3.5. • • (2n - 1)J

Now divide the last inequality by H n :

2n -|- 1 ^ 7r ^2n > 2Wn > L

I t follows easily that x /2H n --------> 1, hence H n -------- > r as n --------> oo.Done.

Let us express H n in terms of factorials. The product of evens is easy:

2-4 -6 - --(2n) = 2n( l -2-3- - -n) = 2w(n!).

To get the product of odds, throw in the missing evens and compensate by dividing them right out again:

_ 1 -2-3-4 - - - (2n — 1) (2n) (2n) \1 -3 .0 . . . ( 2» 1 ) - 2 .4 .6 . . . (2») “ 2-»! •

I t follows that

1 p 2"(n!)2 THn ~ 2n + 1 L (2n ) ! J

and that Wallis's formula can be expressed in the form:

1 r 22n(n!)2T 7r ! Z 2 n + 1 L (2n)\ J = 2*

There are a number of interesting applications of this formula. Here is one concerning probability.

Suppose a coin is tossed 2n times, where n is large. Then the number of heads that can be expected is about n. Yet it seems unlikely that exactly n heads will appear. Just what is the probability of this event?

Let p n be the probability of n heads in 2n tosses. Then


Why? Because the probability that a given sequence of 2n heads and tails occurs is 2~2n, and there are (2 ) such sequences that contain n heads and n tails. (This is the number of ways to choose n positions for the heads from among 2n possible positions.)

Now, Wallis's formula may be rewritten by taking reciprocals:12

!” ( 2 n + 1 , k = w J - ; 'The quantity in brackets is p n. Hence, for large values of n,

(2n + 1 ) p n2 « - ,7r

that is,

\ ( 2n + l ) i r ~

Thus, for example, the probability of exactly 10,000 heads in 20,000 tosses of a coin is

pl000° ~ - v /T o W ~ ° '°0564'This is fairly small, roughly 1 in 180. Yet when you realize that there are 20,001 possibilities (no heads, one head, etc.), it is relatively quite large.

Stirling’s Formula

The numbers n\ occur frequently in applications of mathematics. They grow rapidly as n increases, and are tedious to compute. Still in many problems we need at least an estimate of their size. Such an estimate is provided by the remarkable formula of Stirling:

Stirling’s Formula

n\ t t \ Z 2 tu nne~n.

More precisely,

\ \ / 2 nne~nJ

Note that Stirling's formula does not give a close estimate in the usual sense, but rather an “order of magnitude" estimate. For n large, the difference between n\ and \ / 2 i rn nne~n is also large. However, this difference is small relative to n! In other words, for n large enough, \ / 2 ttu une~n approximates n! to within, say, 1%. But 1% of 100! is a huge number.

Proof: I t is easier to work with Inn! than n! itself. Accordingly,let

S n = In n! = In 1 + In 2 + In 3 + • • • + In n.

This is just the kind of sum we expect to be related to an integral. Let us con

sider a trapezoidal approximation to In x dx. See Fig. 10.2 . Comparing the

area of the trapezoids shown to the area under the curve, we see that

j In x dx t t \ In 1 + In 2 + In 3 + • • • + In (n — 1) + i In n

= S n — h In n.


Setf n

Tn = In x dx.

Then we have Tn t t S n — h In n, that is,

Sn t t Tn + I In n.

We shall show that this approximation is close, more precisely, that the difference of the two quantities approaches a constant. Therefore, we study the difference

An = Tn + i Inn - S n-

Our strategy is to prove the existence of l imAn by showing that {A n} is increasing and bounded. This we do in two steps. Then we complete the proof of Stirling's formula in two further steps.

Step 1: To prove Ai < A 2 < As < We note that

Ak ~ Ak-i — (Tk — Tk~i) + [[J In k — \ In (k — 1)] — (& — Sk-i).

But

so

- tv .i = rJ k- 1

In x dx and Sk — Sk-i = In kj

rkA k ~ Ak-1 = / In x dx — £[ln k + In (A; — 1)].

J k-1

From the trapezoidal approximation in Fig. 10.3a, we see that

f In x dx > J[ln k + In (k — 1 )].J k- 1

Therefore A k — A k~i > 0 ; the sequence increases.

(a) trapezoidal approximation (b) tangential approximation

Fig. 10.3 estimates of An

Step 2: To find an upper bound for {An}.Consider Fig. 10.3b. The tangent at (k, In A;) lies above the convex curve,

hence

for k > 2 . By summing, we conclude that

f n 1 /1 1 1\J In x dx < (In 2 + In 3 + • • • + In n) — - + - + • • • + - J ,

1 / 1 1 1\T n < S n ~ - ( - + - + ••• + - ) .

2 \2 3 n f

Therefore

A n = r , + | l n » - 5 . < | [ l n » - Q + i + ••• + .

But the quantity on the right is bounded, as we have seen in studying Euler's constant. Therefore { A n} is bounded above.

Step 3: The sequence { 4„} is increasing and bounded, hence there is a number C such that A n --------> C, that is,

T n + \ \ n n - S n --------*C.

But

so

x dx = n In n — n + 1,

>ec,

n In n — n + 1 + ^ In n — In n \ --------> C.

Take exponentials:

nne~ne y f u n\

that is,n\

(*) ---------7= -------->K,nne~ny / n

where K = el~c .

Step 4- Complete the proof by showing K = \Z 2r .We exploit Wallis's product, which provides a relation between (2n) \ and

(n!)2. From relation (*) above:

10. Stirling’s Formula 51

The quotient approaches K * / K = K . Divide and simplify:

—» K .[ 2 *-(»!)* ]

^ L ( 2n ) ! v ^ J

But the quantity in parentheses has limit as is seen from Wallis's product. Therefore K = \ / 2 t ; the proof of Stirling's formula is complete.

EXERCISES

1. Estimate the number of digits in 100!.2. Estimate the number of bridge hands (13 cards) that can be formed from a 52-card

deck.Find the limit:

n + 1 n + 2

1 | 1 i« + l 'n 2

f - ^ +

6. lim - (n!)1/n.n-+ oo 71

+

+ 2 ' ' 3n2

s)

i )

V .)

7*. Use the method of the text to prove the existence of

2. Taylor Approximations

1. INTRODUCTION

In this chapter we shall study approximations of functions by polynomials. Why approximate functions by polynomials? Because values of polynomials can be computed by addition and multiplication, simple operations well-suited for hand or machine computations.

Suppose, for example, you need a 6-place table of

f ( x ) = e3x2

at 1000 equally spaced values of x between —1 and 1. If possible, find a polynomial p (x ) such that

63x2 = e (X),

where |e(#)| is less than 5 X 10“7 for — 1 < x < 1. Then program a computer to tabulate the corresponding values of p( x) .

We shall discuss methods for finding polynomial approximations and obtain estimates for the errors in such approximations.

2. POLYNOMIALS

We begin with a basic algebraic property of polynomials: every polynomial can be expressed not only in powers of x , but also in powers of (x — a), where a is any number. This form of the polynomial is convenient for computations near x = a.

EXAMPLE 2.1

Express x2 + x + 2 in powers of x •— 1.

Solution: Set u = x — 1. Then x = u + 1, and

x2 + x + 2 = (w + 1 )2 + (u + 1 ) + 2

= (u2 + 2u + 1) + (u + 1) + 2 = u2 + 3u + 4.

Answer: (x — I )2 + 3(x — 1) + 4.

2. Polynomials 53

EXAMPLE 2.2

Express %z — §x2 + l l x — % in powers of x — 2 .

Solution: Set u = x — 2 . Then x = u + 2 , and

#3 — 6#2 + 11# — 6 = (u + 2 )3 — 6(u + 2 )2 + 11 (u + 2 ) — 6

= us — u.

Answer: (x — 2 )3 — (x — 2 ).

R emark: The answer reveals a symmetry about the point (2, 0 ) not evident in the original expression for the polynomial.

EXAMPLE 2.3

Express x4 in powers of x + 1.

Solution: x4 = [(# + 1) — l ] 4.

Answer: (x + l )4 — 4(.t + l )3 + 6(x + l )2 - 4 0 + 1 ) + 1.

The methods used in these examples is simple. To express

p (x ) = A 0 + Aix + A 2x2 + Azx* + • • • + A nxn

in powers of x — a, write u = x — a. Then substitute u + a for x:

p (x ) = Ao + A \ ( u + a) + A^iu + a ) 2 + • • • + A n(u + a )n.

Expand each of the powers by the Binomial Formula and collect like powers of u. The result is a polynomial in u = x — a, as desired.

This method is laborious when the degree of p ( x ) exceeds three or four. We now discuss a simpler, more systematic method.

Suppose p(x ) is a polynomial expressed in powers of x — a:

p ( x ) = A0 “h A\ {x — a) + A%(x — a ) 2 + • • • + A n(x — a ) n.

What are the coefficients A 0, Ai, A 2, • • •?There is an easy way to compute A 0. Just replace % by a. Then all terms on

the right vanish except the first:p ( a ) = Ao.

Now modify this trick to compute A\. Differentiate p(x ) :

p '{x) = Ai + 2A 2(x — a) + ••• + n A n{x — a )n_1.

Substitute x = a; again all terms vanish except the first:

p' (a) = Ai.

Differentiate again to find A 2:

p" (x) = 2A 2 + 3-2 A 3(x — a) + • • • + n( n — 1 ) A n(x — a ) n~2.

Substitute x — a :p" (a) = 2A 2.

Once again:

p/ f,(x) = 3-2A z + • • • + n(n - 1) (n - 2 ) A n(x - a )n~s,

p ' " (a ) = 3-2A z = 3! A z.

Continuing in this way yields

p (4) (a) = 4 \ A 4, p (5)(a) = 5!A 5, • • • , p (n)(a) = n ! A n.

(Here p {k) is the fc-th derivative.)

If p (x) is a polynomial of degree n and if a is a number, then

p( x ) = p (a) + p' ( a ) ( x - a) + ^ p " ( a ) ( x - a ) 2

+ V'" (a) (x - a)* + ••• + — p (n) (a) (x - a ) n.3! n!

EXAMPLE 2.4

Express p( x ) = xs — x2 + 1

(a) in powers of x — ~ , (b) in powers of x — 10.Jttd

Solution: Use the preceding formula with n = 3. Compute three derivatives:

p' (x) = Sx2 — 2x, p" (x) = 6^ — 2 , p ,,r(x) = 6 .

For (a), evaluate at x = J:

p(l)=-I-By the formula,

+£•«(*-I) •

For (b), evaluate at x = 10:

p(10) = 901, p '(10) = 280, p " ( 10) = 58, p '"(10) = 6 .

By the formula,

54 2. TAYLOR APPROXIMATIONS

2. Polynomials 55

Answer:

(b) 901 + 280(x - 10) + 2 9 (x - 10)2 + (x - 10)3.

The next example illustrates the computational advantages gained by expanding polynomials in powers of x — a.

EXAMPLE 2.5

Let p ( x ) ~ xz — x2 + 1. Compute p (0.50028) to 5 places.

Solution: Use answer (a) of the preceding example:

p (0.50028) = p 0 + 0.00028

7 1 1= - - - (0.00028) + - (0.00028 )2 + (0.00028 )3.

O t: JL

The last two terms on the right are smaller than 10~7. Therefore to 5 places, p (0.50028) agrees with

7 1 - - - (0.00028) = 0.87500 - 0.00007.

Answer: 0.87493.

Because we shall write polynomials frequently, it is convenient to use summation notation:

n

^ AiXx = Ao -f- A\x + A 2x2 + • • • + A nxn. i =0

The formula for an n-th degree polynomial in powers of x — a can be abbreviated:

P(») = V V ^ (x - a ) \1 = 0 t

Here p (i) (a) denotes the i-th derivative of p (x) evaluated at x = a, with the special convention p w (a) = p( a ) . (Also recall the convention 0 ! = 1.)

Expand in powers of x — a:1. x2 + 5x + 2; a = 1

EXERCISES

2. x8- 3x2 + 4x; a = 23. 2x3+ 5 x 2+ 1 3 x + 10; a = - 14. a:4 — 5x2 + x + 2; o = 25. 2x4+ 5 x * + 4 a : + 16; a = - 26. 3xs — 2x2 — 2x + 1; a = 17. 5a:5 + 4x4 — 3a:3 — 2x2 + x + 1; a = — 18. a:5 + 2a;4 + 3x2 + 4x + 5; a = — 29. a;4 - 7a:3 + 5x2 + 3x - 6 ; a = 0 .

Evaluate to 4 significant digits:10. x3 — 3a:2 4- 2x + 1; x = 1.00411. x6 + a:4 + a;3 + a;2 + a; + 1; x = 1.99412. 4a:4 - 3a:2 + lOx + 12; x = -0.989013. 10a;3 + 12x2 - 6x - 5; x = -3.042.

56. 2. TAYLOR APPROXIMATIONS

3. TAYLOR POLYNOMIALS(

Consider this problem: Given a function f (x ) and a number a, find a polynomial p(x) which approximates f (x ) for values of x near a.

One approach that seems reasonable is to construct a polynomial p n{x) of degree n so that

P n ( a ) = f ( a ) , pn' (a) = f ' ( a ) , pn" (a) = f" (a), •••, p„(n) (a) = / (B) (a).

Thus p n(x) mimics f ( x ) and its first n derivatives at x = a.Let us find p n(x) explicitly. We write

p n(x) = A o + A i ( x — a) + A 2 (x — a ) 2 + • • • + A n(x — a ) n

and choose the coefficients A k appropriately. But in the last section, we saw that Ak = p n(k) (a) /k\ . Since we want p n(k) (a) = / (A;)(a), we must choose Ak = /<*>(a)/fc!

The n-th degree Taylor polynomial of f (x) at x = a is

P n ( x ) = / (a) + / ' ( a ) ( x - a) + ^ f " { a ) ( x - a )2 + •••

+ ~ ; / (n)(a)(* - a)", w!

W hen/(#) is itself a polynomial of degree n, then

Vn(x) = / ( s ) .

This was shown in the last section; p n(x) is precisely the expression for f ( x )

*

in powers of x — a. Furthermore, in this case,

Pn(x) = Pn+l(x) = pn+2 (x) = •••.

(Why?) Thus for an n-th degree polynomial f ( x ) , the n-th degree and all higher Taylor polynomials equal f ( x ) .

Here are the first three Taylor polynomials explicitly:

pi (x) = f ( a ) + f ( a ) (x - a),

P2O) = / ( « ) + f ( a ) ( x - a) + ^ f ' ( a ) ( x - a ) 2,

Ps(x) = / (a ) +/ ' (<*) (x - a) + - a )2 + ^ f " ( a ) ( x - a)3.

The graph of the linear function y = Pi(x) is the tangent to the graph of V = f (x ) at (a, / (a ) ) . The graph of the quadratic function y = p2(x) is a parabola through (a, / (a ) ) , also with tangent y = pi (x) , and curved in the same direction as y = f ( x ) . See Fig. 3.1.

3. Taylor Polynomials 57

In general, each Taylor polynomial is derived from the preceding one by the addition of a single term :

Pn+i(x) = p n(x) + / (n+1)(a) (x - a ) n+l.(n + 1)!We anticipate that p n(x) is a good approximation to f ( x ) ; the

error is f (x ) — p n(x). We try to reduce this error by adding an additional term / (n+1)(a)(z — a )n+1/(n + 1)! to p n(x), thereby obtaining p n+i(x), an even better approximation (we hope). In Section 6 we shall justify the fol

lowing formula for the error:


Taylor’s Formula with Remainder S uppose/^) has derivatives up to and including /<n+1) (x) near x = a. Write

/ ( « ) = Pn( x ) + r n (x) ,

where p n (x) is the n-th degree Taylor polynomial at x = a and r n (x) is the remainder (or error). Then

Usually the integral expressing r n (x) cannot be computed exactly. Nevertheless, the integral can be estimated. Here is one important estimate:

Estimate of Remainder Suppose

f i x ) = Pn( x) + rn(x),

where p n (x) is the n-th Taylor polynomial at x = a. If

| / (w+1) 0c)| < M

in some interval including x = a, say b < x < c, then

This assertion is verified by a direct estimate of the integral:

EXAMPLE 3.1

Find the n-th degree Taylor polynomial of f ( x ) = ex at x = 0 . Estimate the error.

Solution:

/ ( 0 ) = 1, / ' ( 0 ) = 1, / " ( 0 ) = 1, • • • •

Hencen n

The (ft + l )- th derivative is / (n+1)(0 = e*. If x > 0, the largest value of f(n+1) for i between 0 and x is ex. By the remainder estimate with M = ex,

\rn(x)\ < ^ ^ xn+1 for X > 0.

If x < 0, the largest value of / (n+1)(0 between 0 and x is e° = 1. By the remainder estimate with M — 1,

I x I w~|r" (a:)l - ( » + ! ) ! •


^.3

Answer: p„ (*) = 1 + a; + — + — + • • • + —,,Z\ o ! ft!

\rn{%)\ < — “ 77 ; xn+1 for x > 0 , (n + 1 )!

j XIM # ) | < 7— 7 -7 7 . for x < 0 .

(n + 1 )!

R e m a r k : At first sight, the answer to Example 3.1 seems circular: the error estimate in approximating ex involves ex itself. However, if the ex in the remainder is replaced by something a little larger, say Sx, we still get a useful estimate of the remainder:

3Zk«(s)| < 7— ;—77 . xn+l for x > 0 .(n + 1 )!

EXAMPLE 3.2

Find the Taylor polynomials for sin x at x = 0 . Estimate the remainders.

Solution: Compute derivatives:

/(# ) = sin#, f ' ( x ) = cos#, /" (# ) = — sin#,

/" '(# ) = — cos#, f w (x) = sin#, •••,

repeating in cycles of four. At # = 0 , the values are

0 , 1, 0 , - 1, 0 , 1, 0 , - 1, 0 ,

Hence the ft-th degree Taylor polynomial of sin # is/y» 3 n* 5 /v» 7 /v»9

K . W - . - j i + j j - J j + j i - - ,

where the last term is dbxn/ n \ if n is odd and =txn~1/ (n — 1)! if ft is even.

For example,

ps (x) = Pi(x) = x - — ,

£ 5

p5(x) = p«(a:) = * - — + — ,

/y»3 rp o r p l

M * ) = M » ) +

Thus p 2m—i ( x ) = P2m(x) and the sign of the last term is plus if m is odd, minus if m is even:

' X3 XP * ^ ( X ) = V * m { x ) = x - - + - ------- + ( - ! ) - * (2w I-£)-,

m

- II =1

( - 1 ) - (2* - 1 )!

The remainder estimate is easy: | / (w+1)(jc)| ^ 1 because f (n+1)(x) = zfc sin x or dz cos x. Hence

knO)| <■ I n+1

( n + 1)!

Since p 2m - i ( % ) = P 2m ( x ) , it follows that

|r2m-l(^)| = k'2m(a0 | <.|2m+l

(2m + 1)!

Answer: sin# = p2m(%) + r2m(%),

P2m (a?)

m

I ( - D 1r2i—i

(2 i - 1 )! ’ - 1 (2m + 1)1

[Note: pim- i ( x ) = pim{x).'}

R em a rk : For the cosine, a similar argument showsn% 2 rv* 4 /y»6 v2wi

cos* = 1 - _ +

|r2m (jc)| = |r2m+i(x)| <. 12m+2

(2m + 2 )!

EXAMPLE 3.3

Find the n-th degree Taylor polynomial of In x at x = 5. Estimate the remainder if 4 < # < 6 .

Solution: Le t / (#) = In#. Then

x x2 #3

/(«(*) = ~ - 4 , •••, /<•)(*) = ~ , 1 ) ! .

The coefficient of (# — 5)* in the Taylor polynomial is

I / ( 0 (5 ) = ( _ i )i \ 5* i»5*

Therefore,

Pn( x) = In 5 + ^ (x - 5 ) - (x - 5 )2

+ (x ~ 5 ) 3 ~ ' ‘ * + ( - 1 )"-1 “ ^ r (x - 5 ) ”.3 • 5d n • 5 n

An error estimate in the range 4 < # < 6 is

M Mkn(a?)| < -7— —77; \x - 5 |n+1 < - — — — ( l ) n+1,

( n + 1)!where M is a bound for | /(n+1) (#)|. Now

( n + 1)!

n!r n+l

n !< — 7 for 4 < # < 6 .— 4*1+1 — —

Take this number as M. The resulting estimate is

I n ! 1M * )l < (n + 1)! 4n+1 (n + 1)4W+1

Answer: For 4 < x < 6 ,n

P n ( x ) = I n 5 + ^ (* ~ 5 ) i>t » l

1

1(n + l)4 n+1 *

Summary

T a y l o r 's F o r m u l a w it h R e m a i n d e r :

f ( x ) = Pn( x) + r n (x) ,

Pn{ x) = ^ (X ~ a )'> r"(a;) = ~ \ f ~ 0 n/ <n+1> ( 0t = 0 0

E st im a te o f t h e R e m a in d e r :

If | / (w+1) (01 < M for all t between a and x, then

M*)| < I* -

S pe c ia l F u n c t io n s :

n

ex = ) — + rn(x), \rn(x)\ < - i =0

exx(» + 1 )!

(n + 1 )!

if x > 0

if x < 0 .

1 V:------ = > ** + 71 — x L-! 1

x =0

.n-flif X 9 ^ 1 .

mYA (—l )*-1

sin a; = > — - ; a;2*-1 + r2m_i(a;), |r2m_i(a;)| < 1 = 1

. |2m+l

(2m + 1)!

cos

m

L , (« )! i = 0

x2i + r2m(x), Vim{x)\ <. 12m+2

(2m + 2 )!

EXERCISES

Find the Taylor polynomials at the given point; estimate the remainder:

1. f i x ) = sin 2 x ; x = 0 3. f i x ) = xex; x = 0

5. f { x ) = x2 ln x ; x = 1

7. f i x ) = xV"*; x = 0

9. f i x ) = x sin x; x — 0 11. f i x ) = sinx + cosx; x = 0

13. f i x ) = sinh x + sin x; x = 0

2. f i x ) = sin2x; x = tt/ 2

4. f i x ) = xex; x = 1 6. /(x) = x2 In x; x = e

8. f i x ) = x2e~x; x = 1 10. /(#) = xsinx; x = 7r/2 12. f i x ) = coshx; x = 0 14. /(:r) = 1 + ex + e2x; x = 0.

15. Let pOr) = x4 — 4x3 + 6x2 — 3x + 2. Estimate the error in the range f < x < f if this polynomial is approximated by its linear Taylor polynomial about x = 1.

16. Compare the Taylor polynomials for sin x at x = 7r/2 with those for cos x at x = 0.17. Let Pnix) be the n-th degree Taylor polynomial of f i x ) at x = a. Verify that

Pn (a) = / (i) ia) , for i = 0, 1, 2, • • •, n.

4. Applications 63

4. APPLICATIONS

Taylor's Formula with Remainder provides a practical method for approximating functions. First, it gives a simple procedure for obtaining a polynomial approximation. Second, it supplies an estimate of the error.

EXAMPLE 4.1

Find a polynomial p ( x ) such that \e* — p{x) \ < 0.001

(a) for all x in the interval —! < # < ! ,(b) for all x in the interval —2 < x < 2 .

Solution: By the answer to Example 3.1, a logical choice for p (x ) is one of the Taylor polynomials

7*2 ry>3 fy%n

p n(x) = l + x + - + - + . . . + - 2! 3! n\

We want to choose n so that

\ex — p n{x)| < 10~3.

To minimize computation and round-off error, we prefer n as small as possible.

(a) If — J < x < j , then

e1/2 / l \ n+1M ' ) I S ( » + 1) ! \ 2.

We choose n so thatpW

(n +

that is,

© »+ i i

< 1 0 0 0 ’

1 1 1< <(n + l ) !2 n+1 1000e1/2 1648

A few trials show

_1___ J _ J _ _ J L _4!24 “ 384 J 5!25 ~ 3840*

Hence we take n + 1 = 5, that is, n = 4. The desired polynomial is

. . X2 X3 X4M x ) + - + _ +

(b) If —2 < x < 2 , then

64 2. TA YLOR APPROXIMA TIONS

This time we choose n so that

(ft + 1)!

that is,

(ft + 1)! 7389'

A few trials show

210 1024 4 1____ — _______________ — - ___________ _____

10! 3,628,800 14,175 ~ 3544 ’

211 210 2 2 111! 10! * 11 ~ 38,981 ~ 19,500

Hence we take ft + 1 = 11, that is, n = 10. The desired polynomial is

method is approximation by Taylor polynomials. These should be of low degree (few terms) to limit the number of arithmetic operations necessary and to prevent accumulation of round-off errors. Note that we need tabulate sin # and cos x only for 0 < x < 7r/4 . (Why?)

Let us concentrate on sin#; a similar discussion applies to cos#. First we consider the third degree Taylor polynomial at # = 0,

#2 #3 #10

Tables of Sines and Cosines

Suppose we want to construct 5-place tables of sin # and cos #. A logical

From Section 3

For 5-place accuracy, the error must be less than 5 X 10~6. Therefore we want

4. Applications 65

An easy computation with slide rule (or 4-place log tables) shows this is the case if x < 0.22 rad « 12.5°. Thus up to about 12°, the third degree Taylor polynomial yields 5-place accuracy.

Next we try

Since

X1|sina: - p 5(a:)| < — ,

5-place accuracy is obtained if

x7- < 5 X 10-6, #7 < 0.0252.7!

Using the C.R.C. table of seventh powers, we find this is the case provided x < 0.59 rad t t 34°.

Next we try

The error is

/y»3 /y»5 /y* 7

X|sina: - pT(*)| < ^ .

For angles up to 45°, this error is at most

ifi ( i ) ' “ h (0-7854)’ < jjj <°-79>’ -

The C.R.C. table of ninth powers shows (79)9 is slightly less than 12 X 1016. Hence (0.79)9 is slightly less than 12 X 10-2. The C.R.C. tables show also that 1/9! « 0.276 X 10~5. Therefore

|error| < (12 X 10~2)(0.28 X 10~5) < 4 X 10“7.

Conclusion: For 0 < x < 7r/4, the Taylor polynomial Vi (x) yields 6-place accuracy in approximating sin#. Furthermore, since pi (x) involves only four terms, the probability of large accumulated round-off error is low. Thus pi (x) provides a practical way of constructing a table of sines.

I t is not necessary to limit ourselves to Taylor polynomials at x = 0. For example, if we are interested only in angles close to 45°, it is better to use Taylor polynomials at tt/4. They provide greater accuracy for the same amount of computation.

The third degree Taylor polynomial of sin x at tt/4 is

and

I sin x - p3(x)\ < - j ) •

If x differs from 45° by at most 0.1 rad « 5.7°, then the error is bounded by

j - (0.1)4 < 4.2 X lO"6.4!

Hence for x between 39.3° and 50.7°, p s ( x ) yields 5-place accuracy. Between 44° and 46°, the error is bounded by

(0.0175)4 < 4 X lO"9, (1° « 0.01745 rad)4!

and so p s ( x ) yields 8-place accuracy.We see that near tt/4, the Taylor polynomial p s ( x ) about x = 7r/4 gives

the same accuracy as does p i { x ) about x — 0. This is typical of Taylor polynomials: for values near x = a, the accuracy achieved by an n-th degree Taylor polynomial about x = 0 is matched by a lower degree Taylor polynomial about x = a. Generally the lower degree polynomial means less computation and less round-off error.

What is not typical is that every other coefficient is zero in the Taylor polynomials of sin x and cos x about x = 0. For computation, this is excellent; it means relatively little computation yields extraordinary accuracy. For example, the polynomial p i ( x ) of sin# actually involves only four terms, yet provides an approximation to within |#|9/9!. This is why the Taylor polynomials about points other than x = 0 are rarely used for sin x and cos x.

EXERCISES

1. Find a polynomial p ( x ) such that \e~x2 — p(#)| < 0.001 for all x in the interval - 1 < # < 1.

2. What degree Taylor polynomial about x = 0 is needed to approximate/{x) = cos x for — 7r/4 < x < 7r/4 to 5 decimal places?

3. Estimate the error in approximating f ( x ) = ln(l + x) for — | < x < | by its 10-th degree Taylor polynomial about x = 0.

4. Approximate f ( x ) = 1/(1 — x ) 2 for — j < x < J to 3 decimal places by a Taylor polynomial about x = 0.

5. Approximate sin2 x by its 4-th degree Taylor polynomial about x = 0. Estimate the error if \x\ < 0.1. [Hint: sin2 x = \ (1 — cos 2#).]

6. Show that for 100 < x < 101, the approximation y / x t t 10 + (x ~~ 100) correct to within 0.0002.

5. Taylor Series 67

7. Show that |sinx — pg(x)| < 5 X 10 6 for 0 < x < t / 2, where y$(x) is the 9-th degree Taylor polynomial for sin x at x = 0.

8. Find the smallest positive integer n such that for 0 < x <

1 — x— (1 + X + X2 + • • • + x n ) < 5 X 10"6.

9. Compute sin(57r/8 ) to 5-place accuracy. [Use Taylor polynomials of sinx, but not at x = 0.]

10. I f / ' (a) = /" (a) = • • • = f n~~l) (a) = 0, but / (n) (a) ^ 0, show that a reasonable(w) \

approximation to/(x) near a is/(x) t t f ( a ) + ’---- :— (x — a)n.n\

11. (cont.) Suppose n is even; show that / (x) has a maximum at x = a if / (w) (a) < 0, and a minimum at x = a if / (n)(a) > 0. Suppose n is odd; show that f ix ) has neither a maximum nor a minimum at x = a.

5. TAYLOR SERIES

Consider once again the Taylor polynomials and remainders of the exponential function at x = 0 :

n

6x = 2 / j \ + T n { x ) ’ i =0

where

M a;)| <g(x) |x|n+1 _ ( » + ! ) ! ’

g(x) = 1 for x < 0

g(x) = ex for x > 0 .

This estimate shows that no matter what x is, the error is very small if n is large enough. In other words, for fixed x,

g(x) H n+l( » + ! ) !

Here is the reason. The number x is fixed; set A = \x\. Pick m so that m > 10A. Then

<1 1

m + 2 " 10 ’ m + 3 < 10 ’

for any k. If n = m + k, then

|x|n _ A n A m A A n\ n\ m\ m + 1 m + 2 m + k

Now k ' ->oo as n ■

A m 1 1 A m 1^ ml 10 10 m\ 10fc

— > oo. Since g (x) Am/ m \ is fixed, the right-hand


-> 0 as n * oo. This means that rn (x)term -for each x , hence the infinite series S / lo x*/i\ converges to ex:

00

e ~ / / VT for all #.t =0

Similarly

Such representations of functions by what look like polynomials of infinite degree, are called Taylor series. A familiar one is the infinite geometric series:

In general, suppose/(.r) is a function that has derivatives of all orders. (We say that f ( x ) is infinitely differentiable.) Then for each n,

f { x ) = ^ (* - a Y + *■»(*)•t =0

If we can show that rn(x) ■ then we can write

0 as n •

' f {i) (fl)

oo for certain values of x}

f i x ) =i =0

This formula is called the expansion of f (x) in a Taylor series at x — a.In practice, it may be tedious to compute successive derivatives of / (x)

and difficult to determine whether rn( x ) --------» 0. In Chapter 3, we shall discuss the theory of representing functions by series and shall present various practical techniques for doing so.

EXERCISES

Find the Taylor series about the given point:1. f ix) = sin3z; x = 0 2. f(x) = cos^x; x = 0

3. f i x ) = sin2 x ; x = 05. f i x ) = e~2x; x = 07. /(* ) = e3*+2; x = - §

1

4. f i x ) = cos2 (2x — 1); x = J6. f i x ) = cosh x; x = 08. f i x ) = sin x + 2 cos x; x = 0

6. Derivation of Taylor's Formula 69

9. /(* ) =1 - Sx ’

1

10. f ( x ) =1

2 + x ; x = 0

12. An early model desk computer had an exponential pack, but no trigonometric one. Its program to compute the sine used the approximation

sin x : 2 (e* - e~x ) ~ l x i [ 1 + m ( x i + 7 ^ 0 *8) ] •Express the exact error as a power series.

13*. (cont.) Prove |error| < 10“8 for \x\ < \ i r. You may assume 15! > 1.3 X 1012 and( 7r)15 < 103.

6. DERIVATION OF TAYLOR’S FORMULA

We shall derive Taylor's Formula as stated in Section 3 :

f i x ) = Pnix) + rnix ),where

and

Pn(x) = h (x ~ i =o

a y

rn(x) = r (* - 0 ".n\ J a/("+»(o d<.

This is a consequence of the following assertion (actually a special case of Taylor’s Formula):

If

gia ) = g'(a) = g" (a) = ••• = gM (a) = 0 ,

then

1 [xg(x) = - - (x - t ) ng

n\ J a(n+1)(<)

If this assertion is* correct, how does Taylor’s Formula follow? Suppose / (:<:) is any function and p„ (x) is its n-th degree Taylor polynomial at x = a. Set

g(x ) = /(* ) - p„(«).

Now p n(x) agrees with f { x ) , and its first n derivatives agree with those of

f ( x ) at x = a. Therefore

g(a) = g ' ( a ) = g" (a) = ••• = g ™ ( a ) = 0 .

Thus g(x) satisfies the conditions of the assertion, so

g( x)l f x

= —. (x - t ) ng n\ J a

«"+»(<) dt.

But g(x) = f ( x ) — p„ (x) and g(n+1> (t) = / (n+1) (t) because the (n + l)- th derivative of the polynomial p n(x) is zero. Therefore,

f ( x ) - p n(x) = i rn \ J a

(x - <)n/ (n+1> (0 dt,

which is precisely Taylor's Formula.Let us return to the assertion and verify it for low values of ft. We integrate

by parts, noting that a and x are fixed.

C a s e n = 0:

ja ~ = fa M = 9 (Do not forget g ( a ) = 0.)

C a s e n = 1 : To evaluate

= g(x) .

i ra j. (x - t )g"(t) dt,

x r x

— / v( t )u ' ( t )d t .a J a

x r x r x

+ / g ' ( t ) d t = / g'(t) dt = g(x) ,a J a J a

set u( t ) = (x — t) and v(t ) = g'(t). Then

/ (x — t)g" (t) dt = / u d v = u( t )v ( t )J a J o

Therefore

I (x - t)g" (<) dt = (,x - t )g ' ( t )J a

by the previous case. (Do not forget g' (a) = 0 . )

C a s e n = 2: Set u(t) — i ( x — t )2 and v(t) = g"(t) . Then

^ I (x - t ) 2g'"(t) dt = f u d v-J • J a J a

= u (0» (0 " - J V ( i)u'( t) dt = 0 + I ( x - t)g" (<) dt = g{x) ,a J a &

by the previous case. Note that v(a) = 0 by the hypothesis g" (a) = 0.

6. Derivation of Taylor's Form ula 71

The general case is handled the same way. One integration by parts, with

u( t ) = ——T~~~~ and v(t ) = gM (t), n\

reduces

—j (z - t ) n9(n+1)(t) dt to -— [ (x - t ) n~ Y n)(t) dt.n\ Ja (tt - 1)! Ja

The latter is g(x) by the previous case.

3. Power Series

1. INTRODUCTION

In this chapter we study power series

do + ai{x — c) + d2(x — c)2 + • • • + an(x — c ) n +

and their applications. In most of our examples c = 0, but the discussion applies to c ^ 0 as well.

Power series serve two important purposes. First, they express known functions in a form particularly suitable for computation. Second, they define functions which are not simple to specify otherwise. Certainly nobody objects to defining a function by a polynomial,

provided, of course, that the series converges? We shall see, in fact, that in many ways power series resemble polynomials.

One useful power series is the geometric series

which converges to 1 /(1 — x) for \x\ < 1 (but diverges for |#| > 1).Other examples of power series were discussed in the preceding chapter:

These series were derived from Taylor's Formula with Remainder. Each

f ( x ) = a0 + aix + a2x2 + • • • + dnxn.

Then why not define a function by a power series,

f ( x ) = d0 + diX + d2x2 + • • •,

00

n =0

1. Introduction 73

converges for all values of x, whereas the geometric series converges only for\x\ < 1 .

Convergence and Divergence

For each fixed value of x , a power series is an infinite series of numbers. Therefore convergence is defined just as it was in Chapter 1.

A power series £ o ak(x — c)k converges at a point x if the sequence of partial sums

n

Sn(x) = ^ ak(x — c)k k=0

converges. The power series diverges at x if the sequence {sn( z ) } diverges.

You can think of a power series as infinitely many numerical series at once, one for each x. The series may converge at some points x and diverge at others. If it converges on the set of points D, then the sum of the series will generally vary as x varies in D; in other words, the sum is a function F(x) with domain D. We say that the power series converges to F(x) on D. Here is the precise definition:

The power series Xo°° ak(x — c ) k converges to the function F(x) on a set D, if given x in D and any e > 0, there exists a positive integer N such that

n^ ak(x - c ) k - F(x)

k =0< e for all n > N.

The number N tells how soon the partial sums are within e of F(x) , thus it measures “rapidity of convergence” . The series may converge at different rates for different points x ; so for e given, N may depend on x.

For example, take the geometric series xn, which converges to F(x) = 1 / ( 1 — x) in the interval \x\ < 1. Suppose e = 0.01. We ask how large n must be so that

n

F(x) — ^ x k k =0

For x = i , we require

/ l \ n+1 / 4 1 1 1\5 / / 5 < 100 J 5n < 400 J °n >

Clearly n > 4 will do, hence N = 4.

< e, that is,1 1 - # ” + l X n + l

1 — x I — X 1 — X< 0.01.

74 3. POWER SERIES

For x = 5, we require

©»+l / J

/ 2 < 100 ’This time we need n > 7, so iV = 7.

For ft = T9o, we require

2n > 100.

/ 9 V +1 / 1 J _ W / 10 < 100 ’ \ 10/ > 1000

Using logarithms, we find that we need n > 65; in this case N = 65.These results show that the geometric series converges less and less

rapidly (N increases) as x increases towards 1. Nevertheless it does converge for each x in the interval — 1 < x < 1.

Radius of Convergence

The set on which a power series converges turns out to be simple: it is either a single point, the whole line or an interval. The proof of this statement is based on the following fact.

Lemma If a power series XI anXn converges at x = xi, where xi j* 0, then it converges absolutely in the interval \x\ < \xi\.

Proof: If 2 anxin converges, its terms approach 0. Hence the terms are bounded, that is, there exists a positive number M such that |an#in| < M. Now

/ x \ n X= anxin [ - ) < MV i/ X i

\anxn\ =

If \x\ < |#i|, then \x/xi\ < 1 and the series converges absolutely by comparison with a convergent geometric series.

The proof of the next result will require a form of the basic completeness property of the field R of real numbers: Each set of real numbers that has an upper bound has a unique least upper bound (supremum) . This means that if S is a set of real numbers such that x < B for some B and all x in S, then there is a number L such that x < L for all x in S and no smaller number has this property.

diverges - ►< converges - diverges -

c — R c R

F ig. 1.1

/

1. Introduction 75

Theorem Given a power series X) an(x — c)n, precisely one of the following three cases holds:

(i) The series converges only for x = c.(ii) The series converges for all values of x.(iii) There is a positive number R such that the series converges for each x satisfying \x — c\ < R and diverges for each x satisfying |# — c\ > R. See Fig. 1.1.

Proof: We may take c = 0 to simplify the notation, that is, we simply replace x — c by x, a translation along the #-axis that takes c to 0. Now let D be the domain of convergence of X) anxn. Certainly 0 is in D because X) an0n converges no matter what the an are. We distinguish two cases.

Case 1: D is unbounded. Then there are numbers x\ with Yh anxin convergent and |#i| arbitrarily large. By the Lemma, D contains the interval \x\ < \xi\ in each such case. Since \xi\ can be taken arbitrarily large, this means that D contains all real numbers, Case (ii) in the Theorem.

Case 2: D is bounded. Then the set S of all numbers \xi\ where X anx in converges is bounded above. Let R be the supremum. Thus, if X) anXin converges, then |#i| < R, and R is the smallest number with this property.

I t follows that S anxn converges for some points x with |#| < R, and diverges for all points x with |#| > R. I t remains to show that the series converges for all x with \x\ < R. Suppose X) onx2n diverges; it is enough to show that R < |#21 • By the Lemma, the series must diverge for every x with \x2\ < |#| otherwise XI anx2n would converge. Therefore \x2\ is an upper bound for S, so R < \x2\ since R is the least upper bound.

If R > 0 we have (iii) of the Theorem. Otherwise R = 0 and we have (i). The proof is complete.

Case (i) is an extreme case, unimportant and uninteresting. I t occurs when the coefficients an grow so fast that the power series can converge only if all terms after the first vanish. An example is

1 + x + 22x2 + 3V + 44#4 + • • •.

For each non-zero x , note that \unxn\ = \nx\n -------- » oo as n --------» oo.Case (ii) is the opposite extreme and occurs when the coefficients a0, ai,

a2, • • • become small very rapidly. The series for ex is an example. Here the general term is xn/ n \ , which for each x tends to zero very quickly since the coefficients an = 1/n ! become small so fast.

Case (iii) lies in between. The coefficients do not increase so rapidly that the series never converges (except for x — c), nor do they decrease so rapidly that the series always converges. A typical example is the geometric series2 xn) where each an = 1. This series converges for |#| < 1 and diverges for \x\ > 1, hence R = 1.

Note that in Case (iii) nothing is said about the endpoints x = c + R and x = c — R. The series may or may not converge at either point.

The number R in Case (iii) is called the radius of convergence. By convention, R = 0 in Case (i), convergence for x = c only, and R = oo in Case(ii), convergence for all x. (The word “radius” will be clarified in Chapter 16, Section 7.)

The interval c — / 2 < £ < c + /2 in Case (iii) is called the interval of convergence. (Refer to Fig. 1.1.) By convention, the interval of convergence in Case (i) is the single point c; in Case (ii) it is the entire x-axis.

76 3. POWER SERIES

EXAMPLE 1.1

Find the sum of the power series and its radius of convergence R :

(a) 1 + - + —

(b) 1 + 5* + 5%2 + ------1- 5*xn + • •

Solution: Each is a geometric series

% y -n =0

ll - y ’ \y\ < !•

In (a), y = x/3; hence the series converges for \x/S\ < 1 and diverges for \x/3\ > 1, that is, converges for \x\ < 3 and diverges for \x\ > 3. Thus R = 3. In (b), y = ox, which implies convergence for |rc| i.

Note that the series with smaller coefficients has the larger radius of convergence.

EXAMPLE 1.2

Find the radius of convergence and the sum of the power series

Solution: The series has the form

y2 ys1 "l’ 2/ + 2! + 3! + ‘ **’ Wher6 V = ~ X2'

1. Introduction 77

Since the series converges to ev for all values of y, we may replace y by — ,r2 for any value of x.

Answer: R = « ; the series converges to e~z for all values of x.

EXERCISES

Find the sum of the series and its radius of convergence:1. 1 + ( x - 3 ) + ( x - 3)2 + (x — 3)3 + ••• + ( x - 3)»H-----

, , + ( l ± L » ) + ( £ ± i? )> + + . . . + ( £ + ! « ) • +

3. 1 — x2 -f- x4 — x6 -|- • • • -f- (— 1 )nx2n + •••

G K K ) ‘+~+(i)' +

E , I (x + 1) | (x + 1 )2 (x + 1 )3 (x + 1 )n5. 1 + ^ j — H------2!------1------3!— + - + — — + —

6 i - g + g ? _ g g + . . . + ( - 2^)"+ . . .1! 2! 3! T T n!

7. 1 + (5x)3 + (5*)6+ (5x)9H--------1- (5a:)3" H-----

_ „ 8x3 32x5 128a:7 . . , . (2a:)2"-1 .8. 2a:-------- ------------------- ! - • • • + (—1 )n_1 —-—-------- h • • •3! 5! 7! T - r y (2n - l ) ! T

„ , ( X - I Y , ( X ~ 1 )8 (x — 1 )12 , , (x - 1 )4n9. 1 -------- + _ ---------------------- ( _ l ) n

Find the sum of the series and all values of x for which the series converges :

10. - + -2+*li + H— * +X X2 X3 xn

11. 1 + ex + e2x + e3x + • • • + enx + • • •12. cos2 x + cos4 x + cos6 x + • • • + cos2n x + • • •

, „ „ sin Sx . sin2 Sx sin3 3x . , , ^. sinw 3x .13 1 ----------------------------------- U . . . 4- (— 1 )»--------- U . . .1! 2! 3! K } n!

„; „ . In x . In2 x , In3 x . . lnn x .14- 1 + T r + ^ r + ^ r + *, , + ^ r + " ’

15. 1 + 2 \ / x -f- 4x + 8x\ /x + 16x2 -)-••• + i^l\/x)n + • • •16. \nx + In (x112) + ln(z1/4) H--------h ln(z1/2n) H-----17. 1 + (x2 + a2) + (x2 + a2)2 + (x2 + a2)3 + • • •

78 3. POWER SERIES

~2/3 /v.4/3 /v.6/3 /v.2n/319 1 — 5__ U 5___ ______ |_ . . . _L ( _ 1 ------------ L

2! 4! 6 ! ^ 1 j (2n)!

2. RATIO TEST

For most power series that arise in practice, the radius of convergence can be found by the following criterion. Roughly speaking, if the coefficients of a power series behave very much like those of a geometric series with radius of convergence R , then the power series also has radius of convergence R. Such a geometric series is X) anxn = X) (1 / R ) nxn. Note that for this series an/a n+i = R •

Ratio Test Suppose the power series

a0 + ai(x — c) + a2(x — c)2 + • • • + an(x — c) n +

has non-zero coefficients. If

dndn+1

-> R as n ■

where R is 0, positive, or 00, then R is the radius of convergence.

Proof: For simplicity in notation, let us assume c = 0. Suppose Ircl < R. Then

\an+ix:n+i|

\anxn\&n+1

hence the series X) anXn converges (absolutely) by the ratio test for a series of constants (Chapter 1, Section 3).

Suppose 0 < R < 00 and \x\ > R. Then

\dn-\-lXn+II

|an£n| R > 1,

so for n sufficiently large, {\anxn\} is an increasing sequence. Therefore 2 anXn diverges because it terms do not approach 0. A slight modification of this argument applies if R = 0.

Thus the series converges for |#| < R and diverges for |#| > R , that is, its radius of convergence is R.

2. Ratio Test 79

EXAMPLE 2.1

Find the radius of convergence in each case:

x x2 xs xn(a) 1 + T + — + — + • • • H-------f* • ■ s

1 2 6 n

(b) (x — 5) — 4(# — 5 )2 + 9(# — o )3 — • • *

+ ( — l ) n~ln2(x - 5 ) n +

A» /y»2 /v»3 a<H(C) 1 + ^ - + - ^ — + - ^ — + . . . + ^ — + . . .w 2 + 1 2 * + 2 2* + 3 2" + n

ar ar(d) 1 - * + 5 ; - = - , + ••• + ( - l ) ” - n +

22 3* nn

/ y 3 /y»6 /y»9 a»3h(e) * + - ^ + . . . + ^ +

\ / 3 \ / 6 \ / 9 \ /3 ^

Solution: In each case apply the Ratio Test,(a) Here an = 1/n and

(In+l- I / 1 _ ” + 1

n / n + 1 n

Hence

ttn+l

so /2 = 1.

(b) an =

1 as n • -> 00,

an n2 if » \Cln+l (n + l )2 U + 1/

-> 1 as tt ■

so R = 1.

(c)

^n+1= - L - / _____ I_____=

2" + n / 2n+1 + n + 1

2 + in + 1) *2- n

2n+1 + n + 1 2" + n

2 + 0

as n ■

1 + n-2~n

» . Hence R = 2.

1 + 0= 2

80 3. POWER SERIES

(d)

= - / ------------n " / (« + l )n+1

I d n

I ^n+1

(» + 1)"+1_ = (n + 1} > w + i.

Hence \an/ a n+i \ --------> oo as n --------> oo, so R = oo ; the series converges forall values of x.

(e) The Ratio Test does not apply directly because “two-thirds” of the coefficients in this power series are zero. Nevertheless, the series may be written

v . y2 , y3 . , yn ,v ^ n + ' " ’

where y = xs. The Ratio Test does apply to the series in this form:

V 3 ( n + 1 ) = jn + 1

\ / S n * n1.

Hence the ^/-series converges for \y\ < 1 and diverges for \y\ > 1. Therefore, the original series converges for \xs\ < 1 and diverges for |#3| > 1, i.e., for |#| < 1 and |#| > 1, respectively. Hence R = 1.

Answer: (a) 1; (b) 1; (c) 2; (d) <»; (e) 1.

R e m a r k : The ratio test does not apply to all series. An example is

1 + 2x + x2 + 2xz + x4 + 2xb + • • •.

For this series

do 1 CLl 0*2 1 dz— = - , — = 2 , — = - , — = 2 , etc. di 2 a2 dz 2 a4

The ratios are alternately \ and 2 ; they do not have a limit and so the ratio test does not apply. Nevertheless, the given series is a perfectly decent one. I t is in fact the sum of two geometric series,

i1 + x2 + + • •

1 - X 2 ’

and

2#2x + 2xz + 2xb +

1 - x2 ’

2. Ratio Test 81

both of which converge for |#| < 1 and diverge for |#| > 1. Hence

Find the radius of convergence:1. 1 + x + 2x2 + 3rH --------\-nxn-\-----

2 3 n/y»2 /y»3 /y»4 rpTl

3.X + X- + ? r + ^ + - - - + * '3 5 1 7 1 2n - 1

/v»2 /v.3 /j*4A " I *V IV ■ ■ / « ^4 - * “ F + iT ~ 7 ^ + + (— 1) ;

x"5 9 17 v 7 2" + 1

/V» /Y»Z A»U /yiTl

5- l + ^ + ^7 + # ^ + ••• +1-2 2-4 ' 3 - 8 n-2n 1

6 ( # + i ) { (* + i )2 | ( x + i y | ( * + n *1-2-3 2-3-4 3-4-5 1 ' n ( n + l ) ( n + 2)

7. x + \ / 2 x 2 \ / S x s + - - - + y /nxn + - - -

8 . i _ 2 + 12_ * + . . . + (-* )" i2 24 29 2n!

9. (e - 1 )x + (e2 - 1 )x2 + (e3 - 1 )x3H---------1- (e» - 1 )x“ +

. o . - | + S - S + - + ( - D - (” + I ) ‘‘ -1* 1 23 33 v ' n3

„ 1 , 3 ( * - 5) , 32(x — 5)2 , , 3" (x — 5)" ,1L 4 -1------ 4?----^--------43-------r ----- h — ----H------

12. l + x + 2!x2 + 3!x3H------ . + n!x" H-------/y»2 /y»3 /v4 /ytW

10 *____ I___ ~____ L _______ L . # . J___ _____ L' 2 + In 2 3 + In 3 4 + In 4 n + l n n

, , 1 , 1*3 2 , 1-3-5 . , , l - 3 - 5 - - - ( 2 n - 1) .14. 1 + - X + — X2 + -■ X3 + • • • H----- -----— Xn +2 2*4 2-4*6 2*4*6*'*(2n)

15' 3 0 + (52 62) X + ( s 3 63) *2 "* *" (5n+l 6n+1) X"

16. 1 + x2 + x10 + x12 + x20 + x22 H--------h x10" + x10"+2 H-----17. 4-5x4 + 8-9x» H--------1- 4n(4n+ 1 )x4n -\-----18. 1 + (l + 2 )x + (l + 2 + 4)x2 + (l + 2 + 4 + 8 ) x 3H-----

+ (1 + 2 + 2M--------[- 2")xnH-----

+

EXERCISES

82 3. POWER SERIES

3. EXPANSIONS OF FUNCTIONS

For many applications, it is convenient to expand functions in power series. We have already done this for ex, sin x, and cos x by computing the coefficient of xn from the formula

Generally, however, computation of higher derivatives is extremely laborious. Try to find the seventh derivative of t an# or of 1/(1 + #4) and you will soon agree.

In this section we describe several techniques for deriving power series without tedious differentiation. They all depend on one basic principle:

Uniqueness of Power Series If f ( x ) has a power series expansion

convergent in some interval \x — c\ < R, then

Thus there is only one possible choice for the coefficients; the series is unique.

Once you find a power series for / (x) at c by any method, fair or foul, then you have it! There is no other series for/ (#).

Proof: The proof is based on a property of power series that will be discussed in Section 4: within its interval of convergence, a power series can be differentiated term-by-term, infinitely often. Having this, the rest is easy. If / (#) = 2 an(x — c )n, we differentiate n times, then set x = c:

/(n)( 0 )an = -----:—

This is exactly the same method we used in Chapter 2 , Section 2 to derive the coefficients in the Taylor polynomials of /(#) .

The first new technique we consider is addition and subtraction of power series.

3. Expansions of Functions 83

Two power series may be added or subtracted term-by-term within their common interval of convergence.

Proof: Suppose f ( x ) = an(x - x0)n and g(x) = Y o bn(x - x0)n.If sn (x ) and tn (x ) denote the partial sums of these series then sn ( x ) -------->f (x)and tn( x ) --------> g(x) for each x in the common interval of convergence. Butby a basic theorem on limits, sn(x) db tn(x) ■ -»/(#) =fc g(x) . That means

in other words, Y o (an ± bn) (x — #0)n = f ( x ) =h g{x).

EXAMPLE 3.1

Express cosh x in a power series at x = 0 .

Solution: cosh# = \ ( e x + e~x). But

x2 #3 ' - 1 + * + 2 i + 3 i + *

for all x,

X* X 6

e _ I = 2i ~ 3 !

Add these series and divide by 2 :

for all x.

/v» 2 <y»4- (e* + e- , ) = ! + - + - +

X* X*Answer: cosh # = 1 H— : + t : + • • • for all #.

2! 4!

The next technique is formal multiplication of series. To simplify notation we shall stick to c = 0 .

Theorem LetOO 00

/ (#) = ^ anxn and g(x) = ^ bnxnn=0 n=0

for |#| < R. Then the product f ( x ) g ( x ) has the power series expansion

also valid for I#I < R.

84 3. POWER SERIES

The theorem means that the power series for f (x )g (x) is obtained by multiplying each term atf* of f ( x ) by each term bjx3' of g (x) and collecting terms, just as in multiplying polynomials. To remember the rule, start with the lowest terms and work u p :

f ( x ) g ( x ) = (ao + aix + a2x2 + • • •) (&o + bix + b2x2 + • • •)= ciobo + (dobi + a\bo)x + (aob2 + aibi + a2bo)x2 + • • •.

The proof of this theorem for products is difficult and best postponed to an advanced course.

EXAMPLE 3.2

Compute the terms up to x* in the power series of x2ex sin 2x at x = 0 .

Solution:

Since only terms involving xe and lower powers are required, it suffices to compute the product

= x2 ^2x + 2x2 + 1 - 0 x* + 0 - 0 x* + • • • j .

Answer: For all .r,

x2ez sin 2x = 2xs + 2x4 — ~ x5 — x% +o

The next technique is substitution; it is used to find the power series for a composite function from the series for the component functions.

Theorem Let/ (z) = d o + CLiZ + a2z2 + • • •

converge in the interval w < R. Suppose g (x) is a function with domain D and such that \g(x)\ < R for all x in D. Then

(*) f l a ( x ) 2 = ao + axg{x) + a2\_g{x)J + a3[^(a; ) ]3 + ••••If in addition g(x) is a power series with g(0) = 0, that is,

g (x) = bix + b2x2 + &3#3 + • • •for |#| < r, then the series on the right side of (*) can be converted into a power series by formally squaring, cubing, etc., and collecting terms. The resulting power series is the valid expansion of f[_g(x)] for |x| < r.

The proof of this result is definitely beyond the scope of this course. R emark: Note that g(x) lacks a constant term, that is, b0 = 0.

Without this restriction, there are infinitely many terms contributing to each each coefficient of f[g(x)~], a tricky situation to handle.

The theorem often allows us to find series for various functions by simple modifications of known series. Here are a few everyday examples:

= 1 + x2 + x4 + x* + • • • (|#| < 1).1 - a?

1 = 1 + ( - 2 a * ) + ( - 2x3)2 + ( —2#3)3 +1 + 2x*

= 1 — 2x* + 4x* — Sx9 H---- -- • • (\x\ < l / \ / 2 ) .

( x2\ 1 ( x2\ 1 ( .r2\ 3

r ® x , / I I \= 1 _ 2~ + 2 ^ 2 ! _ 2^3! + ------- (aU X)'

Let us now consider a less simple example.

EXAMPLE 3.3

Find all terms up to the term in x% in the power series at x = 0 of the function

/ ( * ) = ;-------- 7 7~JZ ■1 — sm ( x )

Assume \x\ < 1.

Solution: If |#| < 1, then |sin(#2)| < 1 ; hence by (*),

f ( x ) = 1 + sin(x2) + [sin (a;2) ] 2 + [sin(x2) ] 3 +

Now, from the series for sin xy we have10

sin (x2) + +6 120

We substitute this series into the expression ior f ( x ) . We square, cube, etc., collecting powers of x up to the eighth power:

Answer: f i x ) = 1 + x2 + x* + f &’6 + f#8 +

OcM and Even Functions

The power series at x = 0 for the odd function sin x lacks x2, x4, xQ, Likewise the series at x = 0 for the even function cos x lacks x, xz, xb, These examples illustrate a useful principle:

86 3. POWER SERIES

If f i x ) is an odd function, f ( —x) = —f i x ) , then its power series at x = 0 has the form

f i x ) = clix + a*x3 + a5x5 + • • •.

If g(x) is an even function, g { —x) = g(x) , then its power series at x = 0 has the form

g(x) = a0 + a2x2 + a4£4 + • • •.

These statements are easy to remember: an odd (even) function involves only odd (even) powers of x at x = 0 .

The proofs of the principles depend in turn on another pair of important facts :

The derivative of an odd function is an even function; the derivative of an even function is an odd function.

Reason: Suppose/(x) is odd, f ( — x) = —f i x ) . Differentiate, using the Chain Rule on the left-hand side: —f i —x) = —f i x ) . Hence f ' i —x) = f i x ) , so f i x ) is even. Likewise, if g i ~ x ) = gix ) , then g' i ~ x ) = —g'ix) .

Now return to power series of odd and even functions. Suppose f i x ) is odd and at x = 0 has the power series

f i x ) = a0 + clix + a2x2 + • • •.

Since f i x ) is odd, /(0 ) = 0 . Hence a0 = 0 . The d e riv a tiv e / '^ ) is even and its derivative f ' { x ) is again odd. Hence /" (0 ) = 0 . But a2 = / " ( 0 ) / 2 !, so a2 = 0. Likewise aA = 0, a6 = 0 , • • •, so

f i x ) = dix + azx3 + a5#5 + • • • •

The statement about even functions can be proved similarly.

■


EXAMPLE 3.4

Find the power series for tan x a t x = 0 up to terms in x7.

Solution: The power series for tan x is obtained by long division: the series for sin x is divided by the series for cos x. Here is a systematic way to carry out the long division. Set

tan x = clix + ct3xz + a5&5 + cax1 + • • •,

valid because tan x is an odd function. Write the identity tan x cos x = sin x in terms of power series:

(diX + a3x 3 + asx5 + a7a;7 + • • •) ( l - ^ ^ ^

X 6 + 120 5040 +

Multiply the two series on the left, then equate coefficients.

x: a\ = 1;1 1a8 - - a 1 = - - ;

f l s _ r j + i fll = i F o ;

i i i ix ■ «7 — — —(- —— ei3 — —— ai =

2 24 720 5040

Solve these equations successively for a\, a3, as, (ii:

1 2 17a x - 1, a , - - , a6 - - , a , - — .

Alternate Solution: Write

1tan x = sin x •

cos x

Now

cos x - i ( - ^ \ _ 1 _\2 4! 6 ! " / *’

cos x 1 — z1 = 1 + z + z2 + z * +

88 3. POWER SERIES

Compute up to xe only:

x2 x4 xe Z = 2 ~~ 24 720

*! ■ (f ~ fi+ ’ (!)*(' - + '"J ‘ f - £ +

The quantities z4, 25, ••• can be ignored; they do not contain #6 or lower powers of x. Collect terms:

COS

Therefore

1tan x = sin x •

cos x

* ( * ■ i r + 4 ~ s S o + " ' X 1 + 1 + S * 1 + i o * 1 + " ■ )

■ I + ( “ S + 0 * ‘ + ( i ! o ~ i i + i i ) * ‘

/ 1 1 5 61 \ . ,_ 1_ ( — ---------- _J----------- — ------------- L ------- 1 /v.7 _J_ .

\ 5040 240 144 720/

1 2 17 ,

Answer:

1 3 2 5 i 17 7 .tan x = x + - x * + ~ x b + r r r x7 +o 15 olo

R e m a r k : The power series for tan x converges to tan x for \x\ < t / 2 . This is the largest interval about x = 0 in which the denominator cos x is non-zero.

EXAMPLE 3.5

What is the value of the seventh derivative of tan x at x = 0?

Solution: Denote tan x by f i x ) . The coefficient of x7 in its power series is / (7) (0)/7! But that coefficient is 17/315 from the answer to the preceding problem. Hence

/ (7) (0) = }7_ m . . . _ (17) (5040)7! 315’ 3 K ) 315

Answer: 272.

EXERCISES

Find the power series at x = 0 :

1. 2 . *21 — 5x 1 -f- x3

3. ex3 4. cos 2x

5. cosh y /x 6. x (sin x + sin Sx)

1 — cos x7. ( x - l)ex 8. x2

9 * 1Q s m x - x1 — X X 6

11. sin2x 12. sin (a;2).[Hint: Use a trigonometric identity.]

Compute the terms up to and including z6 in the power series at x = 0 :

13- ( 1 - 2 ,HI - 3,2) 14.e*sm(**)*

15. -— 16 11 — x2ex ’ 1 + x2 + x4

17. sin3 x 18. In cos x.

Find / (8) (0):

19. f ix) = -—-—i 20. f ix ) = zcosx 1 — x

21. f ix ) = 22. f ix ) = arc tan(z3).2* 3/2 6

23. Show that the series 1 - — + — — — + • • • converges to cos y /x for x > 0.2! 4! o!

24. (cont.) Show that it converges to cosh y/^^x for x < 0.

90 3. POWER SERIES

4. FURTHER TECHNIQUES

The methods of the last section enable us to derive new power series from known ones. Here is another important method for doing so. We postpone the proof until Section 7.

A power series may be differentiated or integrated term-by-term within its interval of convergence.

EXAMPLE 4.1

Find power series at x = 0 for (1 — x)~~2 and (1 — £)~3.

Solution: For |rc| < 1,

(1 - *)-2 = (1 _ *)-! ax

= — ( l + x + x2 + ••• + * » + •••) dx

= J - (1 ) + J - (x) + J - (X2) + • • • + J - (xn) + dx dx dx ax

— 1 + 2x + Sx2 + • • • + nxn~~l + • • •.

Differentiate again:

(1 — x)~z = “ ~ (1 — x)~2 = - -7- (1 + 2x + Sx2 + • • • + nxn~l + • • •)2 dx 2 dx

= i [2 + 6x2 + 12x* + • • • + n ( n - l ) x n~2 + • • •]•

Answer: For \x\ < 1,CD

(1 — x)~2 = 2 ^ nxn~l, (1 — x)~3n =1

CO

In = 2

n (n ~ 1 )- ...- .....1 xn~22

EXAMPLE 4.2

Find the power series at x = 0 for In (1 + x ).

4. Further Techniques 91

Solution: Notice that ln(l + x) is an antiderivative of (1 + x)~l —1 — x + x2 — x3 + • • •. Therefore

ln(l + x)r x dt

Jo 1 + ^%

= / [1 - t + t* - ••• + (-i)»<» + . . . ] d «Jo

= dt - t d t + t2 dt 1 )» j tn Jo Jo J l) Jo

v.n+1

Answer: For j#] < 1, ln(l + x)

+

y.n+1

+

EXAMPLE 4.3

< •3 /j*5

Sum the series x + V + zr + 3 5 + 2n +

Solution: By the ratio test, the series converges for \x\ < 1 to some function/(e). Write

+

Each term is the integral of a power of x. This suggests that f ( x ) is the integral of some simple function. Differentiate term-by-term:

f ( x ) = 1 + x2 + x4 + • • • + x2n~2 + 11 - X 2

Therefore, f ( x ) is an antiderivative of 1 /(1 — x2). Since / ( 0 ) = 0 , it follows that

f x dt 1 1 + t ~ J0 1 - t2 = 2 ln 1 - t1 1 + x

- 2 b m ,

1 . 1 + *Answer: - ln ------- .

2 1 - x

92 3. POWER SERIES

EXAMPLE 4.4

Sum the series x + 4a;2 + 9x3 + * • * + n2xn +

Solution: Write

f ( x ) = x + 4a;2 + 9a;3 + • • • + n2xn + • • • = %g{x),

whereg(x) = 1 + 22x + S2x2 + • • • + n2xn~l + • • •.

Now

g(x) = -7- (x + 2a;2 + 3a:3 + 4a;4 + • • • . + nxn + • • •) dx

= — \_x (1 + 2x + 3a;2 + 4a:3 + • • • + nxn~l + •••)]- dx

By Example 4.1,

1 + 2a; + 3a:2 + • • • + nxn~l + • • • = (1 — x)~2.Hence

so

, x d f x "1 1 + Xir=

f ( x ) = xg(x) =

x ) 3 ’

X + X 2

(1 - a;)3

. X + X 2Answer:

(1 — a:)3

Check: Evaluate the series at a: = 0.1:

1 4 9 16 25/((U) = io + w + To* + To*+ w +

According to the answer, the sum is

(0 .1 ) + (0 .1)2 0.11( 1 - 0 . 1 )3 (0.9)3

0.15089 16324.

I t is easy to make a convincing numeric check. Start with the first term and add successive terms: 0.1, 0.14, 0.149, 0.1506, 0.15085, 0.15088 6, 0.15089 09, 0.15089 154, 0.15089 1621, 0.15089 16310.

The power series for 1/ (x — 5) can be obtained from the geometric series. Just write

_L_________ 1 _ _ _ 1 1x — 5 5 — x 5 x ’

~ 5

and expand in a geometric series:


Partial Fractions

which converges if \x/5\ < 1, that is, |#| < 5.Combined with this trick, the method of partial fractions is useful in

finding power series for rational functions (quotients of polynomials).

EXAMPLE 4.5

1Find a power series at x = 0 for f ( x ) = 7 —-----— ---------7 .

(x — 2 ) (x ~ o)

Solution: By partial fractions

— ---------------- ----- - ( — ----------------- — ) .- 2 ) ( * - 5 ) 3 \a; — 5 x - 2 /(x

Expand each fraction by the preceding trick :<X)

l l VX - 2 2 L 4 \ 2 ) L j 2n+1

o

v a*Z / 2n+1

for \x\ < 2,

^ - ■ 5 2 x 0 f°r w<3'0 0

Therefore if |#| < 2, both series converge and

i p ___ M = -V P __ LV3 \ x — 5 x - 2 / 3 Zy \2 n+1 o"+7 ’

Answer: For |*| < 2,

i = I V ( — - — ^(x - 2) (* - 5) 3 A f \2 n+1 5»+7

n = 0

94 3. POWER SERIES

EXAMPLE 4.6

1Find the power series at x = 0 for

x2 + x — 1

Solution: The denominator can be factored:

x2 + x — 1 = (x — a) (x — 6),

where a and b are the roots of the equation

x2 + x — 1 = 0.

By the quadratic formula,

- 1 + y /B , - l - y / 5a = -------------- , b = ------- -------.2 2

Notice that ab = — 1 because the product of the roots of x2 + px + q = 0 is q. (Why?)

By partial fractions,

- 1x2 + x — 1 (#

— --------- = _ L _ ( _ J -------------— ^ .a)(x — b) a — b \ x — b x — a )

Suppose \x\ < | ( —1 + y /5 ) , the smaller of the numbers \a\ and |6|. Now expand:

i i / y xn y xn \ _ V * / 1 1 \; — 1 a — b \2* / an+1 L j bn+1) L j a — b \ a w+1 bn+1)x2 + x , . .

n =0 n =0 n =0

Note that

, - 1 + V 5 - 1 - V 5 a — b = ------------------------ ------- = y 5 ,

and (since ab = — 1)

1 , 1 + \ / 5 1 l - y / 5r - '

R e m a r k : It may not seem so, but the numbers cn are actually integers! The first few are 1, 1, 2, 3, 5, 8, 13, • • •; each is the sum of the previous two. See Ex. 17 below.

■SEXAMPLE 4.7

Find a power series for -----------—-----~ .(1 — #) (1 + #2)

Solution: By partial fractions,

R e m a r k : This example also can be done by multiplying together the series for (1 — x )~1 and (1 + x2)~l, also by multiplying the series for (1 — &4) -1 by 1 + x.

(1 - * ) ( 1 + * 2)1

— x6 H---- -- • •) + (x — xz + x5 — x7 H---- • • •)]

= 1 + x + x4 + x5 + x8 + x9 +

Answer: For |a?| < 1,

EXERCISESObtain the power series at x = 0:

x2 — 5x + 6 2. (2x + 1) (3a; + 4)1

3. 1 4. 1( x - l ) ( x - 2 ) ( x - 3) ( x - 3 ) ( x 2+ l )

1 + x2

Find the sum of the series:7. 4 + 5x + 6x2 + • • • + (n + 48. 1 + 4x + 9x2 + • • • + (n + 1

+ (n + 4 )xn + + (n + 1 )2xn +

[Hint: Differentiate.]

96 3. POWER SERIES

10 ____ _ H ) V + , . ,1-2 2*3 3*4 (ft — 1 )n

11. 2 + 3a; + 4x2 + • • • + (ft + 2)xn + • • • [Hint: Multiply by x and integrate.]

12. Evaluate l + ? + A + A + . . . + J L +

[Hint: Consider ^ ftzn 1.]n = 1

Verify by expressing both sides in power series:

13. y- (sina;) = #cosa; 14. ~ (e- *2) = — 2xe~x'2dx O/Xf

15. I x2 cos x dx = 2x cos x + (x2 — 2) sin x + Cs7, /16. I arc tan x dx = x arc tan x — ln(1 + a:2) + C,

17. Consider Example 4.6 from a different viewpoint. Suppose, a0, ai, a2, • • • is a sequence of integers satisfying a0 = a\ = 1 and an+2 = an + an+1 for ft > 1. Set f (x ) = a0 + aix + a2x2 + • • •. Show that xf(x) + x2f (x ) = f (x) — 1, and conclude that fix) = (1 — x — a;2)-1. Now obtain a formula for an. (This is the method of generating functions.)

18. Which power series is more efficient for computing ln(§), the one for ln(l + x)y or the one for ln[(l + x)/ (I — a;)]? See Examples 4.2 and 4.3. Compute ln (f) to 5 places.

19. Expand arc tan a; in a Taylor series at x = 0.[Hint: Integrate its derivative.] Conclude that

arc tan x = x — ^— = -+ ••• for \x\ < 1.o 5 7

It is known that the series in Ex. 19 is valid for x = 1, hence that

3 5 7 4

This is an interesting but poor formula for computing 7r, since its terms decrease veryslowly. Exercises 20-23 develop an efficient way to compute t .

20. Prove the formula arc tan x + arc tan y = arc tan for |a;| < 1 and»

/ j r + j A \1 — xy)

7T 1 1| y | < 1. Conclude that - = arc tan - + arc tan - .4 o

21. (cont.) Use this expression for 7r/4 and the power series in Ex. 19 to compute w to 4 places.

5. Binomial Series 97

22. (cont.) Show that the expression in Ex. 20 for 7r/4 can be modified to7T 1 1 . .- = 2 arc tan - + arc tan - . Why is this even better for computing 7r?

23. (cont.) Finally modify the expression in Ex. 22 to

j = 2 arc tan - + arc tan + 2 arc tan i . Now compute 7r to 7 places.4 D 1 o

5. BINOMIAL SERIES

The Binomial Theorem asserts that for each positive integer p,

/ t , , , P(P ~ 1 ) , , P(P ~ 1 ) (P — 2 )(1 + a:)p = 1 + pa; H------- —---- x2 H------------- —---------- a:3 +

, P(P ~ 1)(P ~ 2 ) - - - (p - n + 1) „ , , p! „H--------------------------------------------------a;" + • • • H----- -a:'’.n! p!

Standard notation for the coefficients in this identity is

. ( p \ _ j / p \ = P(P - 1 ) (P - 2)-«- (p - n + 1) 1 < n <\0 / ’ \ n j n\ ’ — n — P-

With this notation the expansion of (1 + x ) p can be abbreviated:

(i + * )p = £ ( £ ) * ”■ n =0

♦ A generalization of the Binomial Theorem is the binomial series for (1 + x ) p, where p is not necessarily a positive integer.

Binomial Series Suppose p is any number. Then

(1 + x ) p = ^ xn, —l < x < 1,n =0

where the coefficients in this series are

/ p \ = l / p \ = p(p - l ) (p - 2 ) - - - (p - n + 1) n > 1

\0 / ’ \ n j n\

R e m a r k : In case p happens to be a positive integer and n > p , then the coefficient

98 3. POWER SERIES

equals 0 because it has a factor (p — p). In this case, the series breaks off after the term in xp. The resulting formula,

(1 + x) p = xn,i =0

is the old Binomial Theorem again. But if p is not a positive integer or zero, then each coefficient is non-zero, so the series has infinitely many terms.

The binomial series is just the Taylor series for y(x) = (1 + x) p at x = 0. Notice that

y'(x) = p(l + * )p“l,

y" (x) = p ( p - 1)(1 + x )p~2,

y(n) (x) = p (p — l ) (p — 2) • • • (p — n + 1) (1 + x )p~n.

Therefore the coefficient of xn in the Taylor series is

0) p (p - l ) (p - 2)--« (p - n + 1)n\ ft!

The binomial series converges for |rc| < 1. When p is an integer this is obvious because the series terminates. When p is not an integer the ratio test applies:

&n+l - l O / U O l -ft + 1p — ft 1,

so the radius of convergence is R = 1.This, however, does not prove that the sum of the series is (1 + x) p.

That requires a delicate piece of analysis beyond the scope of this course.

EXAMPLE 5.1

Find the power series for(1 + * ) 2 ‘

Solution: Use the binomial series with p = — 2. The coefficient ofxn is

( ;2) -

( —2)( —3)( —4) - - . ( —2 - t t + l )nl

ft!

Therefore

(1n =0 n =0

I — 2x + Sx2 — 4#3 +

Answer: For \x\ < 1,(1 + x) i

n ~0

( - 1 )*(« + l ) x n.

C h e c k :

(1 +I — = - J - ( r ~ ) = - y (1 - X + x2 - x3 + - •••)

x)2 ax \1 + x / dx

= — ( — 1 + 2x — Sx2 H---- • • •).

EXAMPLE 5.2

Express•\/l + x

as a power series.

xn isSolution: Use the binomial series with p = — The coefficient of

(7) -----= ( - 1 ) ”

n\

1-3-5’ • • (2n - 1) 2"-n!

But

1 •3 •5 •••(2n - 1) =1-2-3-4---(2w ) _ (2n)!2 -4 -6 -8 ---(2n ) ~~ 2»•»!'

Hence the coefficient of a:" is

2n*n! 2n*n! (2w«n!)2

The following example will be used in Section 7.

100 3. POWER SERIES

EXAMPLE 5.3

Express y / \ — x as a power series.

Solution: Use the binomial series with p = J and replace x by — x. The constant term is

= 1.

The term in xn is

S X - S X -H - W( - * ) " = n\

{ ~ x ) n

, , 1 .3 -5 ---(2» - 3) ,= ( - 1 ) " - 1-------- -------------- - ( — 1)**".

2"*n! ( -1 ) "

But

1*3*5- • • (2n - 3) = :l*2-3*4*••(2n — 2)2-4-6-8--*(2w - 2)

(2n — 2)! (2n)!2n-t(n — 1)! (2”-n\) (2n — 1)

Therefore the term in xn is

(2»)! 1(2”>n!)(2n — 1) 2n>w!

a;" = —(2n)!

(2"-n!)2(2n - 1)x".

Answer: For |s| < 1,

a /1 — x = 1X (2”'

( 2 n ) t

nl)2(2n — 1)

EXAMPLE 5.4

Estimate a IOOI to 7 places.

Solution: Write^ lO O l = [1000(1 + lO"3) ]1'3 = 10(1 + 10~3)1/3.

Use the binomial series with x = 10-3 and p = §:

10(1 + 10-3)1'3 = 10 [1 + 0 (10-3) + Q (10-3)2 + • • •]

10-3 - ~ • 10—6 + y 1

The first three terms yield the estimate

A^lOO! « 10.0033322222-

The error in this estimate is precisely the remainder

10 (3) (io -3)3 + 0) (10- 8)< + . . . j .

Now each binomial coefficient above is less than 1 in absolute value:

1 2 5 3n—43 6 9 3w < 1.

1*2*3*•-n

Therefore, the error (in absolute value) is less than

io [( io -3)3 + (10-3)4 + . . . ] = io -8[ i + 10-3 + 10-® + 10-9 + • • • ] '

110-

1 - 10-310- 8.

Answer: To 7 places, y / 1001 ~ 10.0033322.

R emark: The estimate is actually accurate to at least 8 places. When we estimated each binomial coefficient by 1, we were too generous since

10\ \ V 2 - 5 = _ ,n/ 3-6-9 162’

n > 3.

(Why?) Therefore, the error estimate can be reduced by a factor of iVV, which guarantees at least 1 more place accuracy.

Expand in power series at x = 0:

11.

3.

(1 + x)3

1(1 - 4a;2)2

5. (1 + 2a^)1M

2. 1(1 - x ) 113

4. y / T — X

[Hint,

EXERCISES

1(3 - 4x2)2

102 3. POWER SERIES

Expand in power series at x = 1:

7. v T T x [Hint: Write l + z = 2 + ( z - l ) = 2 ^ 1 + .]

1(3 + x)2 '

Compute the power series at x = 0 up to and including the term in x4:

19. y / l + x2ex 10.

11. (sin 2z)\/3 + x 12.

cos 2xlo.

(1 + sin x)2

1\ / l + x + x2[Hint: Write x + x2 = u.~\

(i+i)‘Compute to 4-place accuracy using the Binomial Theorem:14. 15. ^82

1fi 1 (1.03 )5‘

17. Expand arc sin x in a Taylor series at x = 0.[Hint: Integrate its derivative.]

6. ALTERNATING SERIES

In Chapter 1, Section 4, we saw that alternating series whose terms approach 0 in absolute value are convenient to work with. In particular, remainder estimates are very simple and quite accurate: if you break off such a series, then the remainder is less than the absolute value of the first term omitted.

EXAMPLE 6.1

The power series at x = 0 for In (1 + x2) is broken off after n terms. Estimate the remainder.

Solution: The series is obtained by integrating from 0 to # the series for the derivative of ln (i + t2).

d- [ ln ( l + O ] = = 2t(l + ------- )

= 2{t - t3 + t* - f + - • • •), |<| < 1.

6. Alternating Series 103

I t follows that

. „ , f 2td t o A 2 ** \ ln( l + a;8) = / --------= 2 ( ------------- -------------- -----• • • )J 0 1 + <2 \2 4 6 8 /

a;4 a:6 a;8= * ~ 2 + 3 ~ I + ------- ’ W < L

This series alternates; since |rc| < 1, its terms decrease in absolute value to zero. Therefore, the remainder after n terms is less than the absolute value of the (n + l)- th term.

/ 2n-f2(remainder! < ------ - .

n + 1

Occasionally we encounter an alternating series whose terms ultimately decrease in magnitude, but whose first few terms do not. If the successive terms decrease starting at the A;-th term, the series will still converge, and the remainder estimate is still valid—beyond the A;-th term. The front end of the series, up to the (k — l)- th term, is a finite sum; it causes no trouble. The important part is the tail end, that is, the series starting with the A;-th term. I t is this series that we test for convergence.

EXAMPLE 6.2

The power series for e~x at x = 0 is broken off after n terms. Estimate the remainder for positive values of x.

Solution: If x > 0 the power series

x2 xse- . , i

alternates. If 0 < x < 1, the terms decrease to 0, and the above remainder estimate for alternating series applies. If x > 1, however, the first few terms may not decrease. (Take x = 6 for example:

e-6 = 1 _ 6 + | _ ^ + _ . . . = 1 _ 6 + 18 _ 36 + _

Nevertheless, for any fixed x, the terms do decrease ultimately.To see why, note that the ratio of successive terms is

rn+l

(tt + 1)

Take a fixed value of x. Then/ xn _ x

tt! tt + 1

104 3. POWER SERIES

as soon as n + 1 > x; from then on the terms decrease. Furthermore

x 1u -f- 1 2

as soon as n + 1 > 2x. From then on each term is less than one-half the preceding term. Hence, the terms decrease to 0.

Result: if the series is broken off at the ft-th term, the remainder estimate for an alternating series applies, provided n + 1 > x.

Answer: For n > x — 1,

|remainder| < :.....— .(ft + 1)!

As another application of this method, recall Example 5.4. There it was shown that three terms of the binomial series for

10(1 + lO"3)1'3

provide 7-place accuracy in computing \/l0 0 1 . But this series alternates and its terms decrease towards 0. Therefore, one deduces that the remainder after three terms is less than the fourth term, which is

so 8-place accuracy is assured. This estimate is both easier and more precise than the one in Example 5.4.

EXERCISES

X xn— converges inside the interval — 1 < x < 1. Show that it n

converges also at x = — 1.2. Give an example of an alternating power series that does not converge.3. Compute e-1/5 to 5-place accuracy.

jC 24. Find values of x for which the formula cos x t t 1 — — + — yields 5-place

accuracy.5. How many terms of the power series for ln x at x = 10 are needed to compute

ln(10.5) with 5-place accuracy? (Assume the value of ln 10 is known.)6. How many terms of the binomial series for y / l -f- x will yield 5-place accuracy

fm- o < x < 0.1 ?

7. APPLICATIONS TO DEFINITE INTEGRALS [optional]

Power series are used in approximating definite integrals which cannot be computed exactly.

EXAMPLE 7.1

Estimate / e~i2 dt to 6 places for \x\ < §.J o

7. Applications to Definite Integrals 105

Solution: A numerical integration formula such as Simpson’s Rule can be used, but a power series method is simpler. Expand the integrand in a power series:

t4 <6 - 1 - ,! + 2 ! - 3 ! +2! 3!«

Since this series converges for all x, it can be integrated term-by-term:

f i r s * ■)dt

X 6 X ° X 1 X y

“ x ~ 3" + 5^2! - 7^3! + 9T ! ~

Because of the large denominators, this series converges rapidly if x is fairly small. For |#| < J, the sixth term is at most

1< 4 X lO"7.

\ 2 / 11-5!

Since the series alternates, it follows that five terms provide 6-place accuracy.

Answer: For all x,

/ ■Joe~<2 dt = x X' ' X'

3 + 5-2! 7-3! +

For \x\ < J, five terms yield 6-place accuracy.

R e m a r k : The series converges for all values of x, but for large x it converges slowly. For example, it would be ridiculous to compute

I"Joe~l dt

by this method, since more than 100 terms at the beginning are greater than 1.

106 3. POWER SERIES

Elliptic Integrals

EXAMPLE 7.2

[*/2Express the integral / \ / l — k2 sin2 £ where /c2 < 1, as a

Jopower series in ft.

Solution: By Example 5.3,00

V 1 X = (1 ~ ^ )1/2 = 1 — ^ |#| < 1,n =1

where

(2»)l“ (2"-n!)2(2n - 1) '

Substitute ft2 sin21 for re. This is permissible because ft2 sin2 t < k2 < 1.00

(1 — ft2 sin2 £)1/2 = 1 — ^ anft2n sin2nn = 1

Now integrate term-by-term:

/•"/2 tt V f*'2/ (1 — /c2 sin21)112 dt = - — y ank2n / sin2n t dt.

n =1

The integrals on the right are evaluated by a reduction formula (see p. 45), and their values are listed inside the front cover:

fir/2/ (si

Jo/ • (2n) ! *sin x Y n dx = —---- — • - ,v ' (2n-n\)2 2

Therefore

r 2 • 2n,W, - f ( 2 a ) I I f ( 2 n ) l tt]a” io Sm I (2"-n!)2(2n — l)J|_(2"-n!)2 2J

_ r (2n)I ]2 1 *•L (2"*n!)2J (2n — 1)' 2

Substitute this* expression to obtain the answer.

[*! 2Answer: For k2 < 1, / \ / 1 — ft2 sin2 1 dt

J o

= ^ (i v r (2^ ) ; T fc2w |2 1 Z / [ (2 « -n ! )2J 2n - lj *

n =1

R e m a r k : This integral arises in computing the arc length of an ellipse. Suppose an ellipse is given in the parametric form

x = a cos 0, y = b sin 9,

where b > a. If s denotes arc length,

(I) - (I) + ( t ) - <-»“ *>’+= 62(sin2 9 + cos2 9) — (b2 — a2) sin20 = 62(1 — k2 sin2 9),

where k2 = (b2 — a2)/b2. The length of the ellipse is

r « / 2 / / 7 o \ r w 2 ___________________________

4 J j d9 = 46 J y / l — k2 sin2 9 d9.

7. Applications to Definite Integrals 107

EXAMPLE 7.3

u r - i -Estimate ~ I -x/1 — * sin2 x dx to 4-place accuracy.

Solution: By the last example with k2 = 5,

where

f (2n)! I2 p . 3 - 5 . . . ( 2 » - 1)T ” " L22"(n!)2J ~ L 2-4-6- - - (2n ) J ‘

Break off the series after the p-th term. The remainder (error) is

- - i * - t y rn = p + l

The problem: choose p large enough that |ep| < 5 X 10“5.The error is a complicated expression. I t can be enormously simplified

in two steps, each of which causes a certain amount of over estimation.First, each coefficient bn is less than 1 :

n .3 - 5 - - - ( 2 n - 1)T T1 3 5 2n - IT w “ L 2 • 4 • 6 • • • (2n) J _ l2 4 6 2n J < l;

hence

i i y n x rn=p + 1 ' * n = p + l

|6«,

Second, for n > p + 1,

108 3. POWER SERIES

1 1 1<

2n - 1 “ 2(p + 1) — 1 2p + 1 ’

hence replace l/(2 n — 1) by l/(2 p + 1 ) :

"■'< X feViXS'-sV, I (ITn —p -\-1 n =p +1

On the right is a geometric series; consequently

< 2?TT 0) [‘ + (0 + 0) +1 / l \ p+1 1 1

:T l \ 5 / 1 - i “ 4(2V +2p + 1 \ 5 / 1 - £ 4(2p + 1)5*'

To obtain 4-place accuracy, choose p so that

4(2p + 1)5- < 5 X

To find a suitable value of p, take reciprocals:

4(2p + 1)5P > 20000, (2p + 1)5P > 5000.

Trial and error shows

(2-3 + 1)53 = 875, (2*4 + 1)54 = 5625.

Accordingly, choose p = 4 and estimate the integral by

1 - ©'© - o w © - gs)w© - (mm1 Q 1 7

_ 1 _ __ _ ___ _ ___ _ _____ f) OztQO20 1600 6400 409600 ~

Answer: 0.9480.

EXERCISES

Express as a power series in x :* sin t . * 1 — cos t

8. Uniform Convergence 109

4. / sin (t2) dt.

Compute to 5-place accuracy:

/ 0.2e”*2 dx

r 1 / 4 _____________

7. / y/1 + x* dxCz. 01

8. I dx [Expand at a; = 3.]/ 3.01 ear

’’ /s.00 1 + 3J 3.00 1 + 39. Compute to 4-place accuracy the arc length of an ellipse with semi-axes 40 and

[Hint: Approximate the integral by the first significant term of its power series; use Ex. 3.]

11. (cont.) Refine your estimate of x to 4-place accuracy by taking the first two significant terms of the power series. Use Newton’s Method to solve approximately the resulting equation.

8. Uniform C onvergence [optional]

Any basic information we need about the convergence of an infinite series J2fn(x) of functions is carried in the corresponding sequence {$«(#)} of partial sums. In this section we shall study some important properties of sequences of functions, then apply the results to power series.

Definition Let {un(x)) be a sequence of functions, all with the same domain D. The sequence is said to converge uniformly on D to w(a;) if given any e > 0, there exists an N such that

for all n > N and all x in D.

The last four words are the key to this concept. We can control the degree of approximation of un(x) to u(x) independent of x. The next three results show the usefulness of uniform convergence.

Theorem 8.1 Let {un(x) } be a sequence of continuous functions on D and let un( x ) --------» u(x) uniformly on D. Then u(x) is continuous on D.

Proof: Let e > 0. Then there is an N such that \ u n ( x ) — u (x )| < i* for all x in D. Take any point c of D. Since un is continuous at c, there is 8 > 0 such that («) — w a t ( c ) | < |e for all x in D such that \x — c\ < 8.

41.10. Estimate the value of x for which

|Un(x) ~ U { x ) \ < €

If x is in D and \x — c\ < S, then

\u(x) — w(c)| = \u(x) — un (x) + u n (x ) — u n (c) + m ( c ) — u(c) |

< Iu(x) — Un {x ) I + \u n (x ) — Un {c)\ + |mat(c) — u(c)\

<§« + h + = «•This proves the continuity of u at c.

Theorem 8.2 Let {un( x) | be a sequence of continuous functions on aclosed interval [a, 6], and suppose un (x ) --------> u(x) uniformly on [a, 6],Then

110 3. POWER SERIES

r . r/ u{x) dx = lim / unJ a n-+ oo J a

(x ) dx.

Proof: The function u is continuous by the previous theorem, hence integrable. Let e > 0. Then there is an N such that \un (x) — u(x)\ < e/ (b — a) for all n > N and for all x in [a, 6].

If n > N, then

r f h i i r i/ Unix) dx — / u{x) dx = / [un(x) — u{x)~] dx\J a J a, \ \ J a '

f b f b e< / \un(x) — w(a;)| dx < / -------dx = e.

J a J a O a

Hencef b f b/ Un(x) dx -------->

J a J a

Cb

u{x) dx.

Theorem 8.3 Let {un( x ) } be a sequence of continuous functions on an openinterval (a, b) and suppose un( x ) --------» u{x) for each x in (a, b). Assumealso that each un(x) is differentiable, that the derivatives un'{x) are continuous on (a, b), and that un' ( x ) --------> v{x) uniformly on (a, b). Thenu{x) is differentiable and u'(x) = v(x).

Proof: Fix c in (a, b) and let x be any point of this interval. Then

j r Un {t) d t --------> J' v it) dt

by the previous theorem. Hence

Unix) ~ Unic) --------> V it) dt.


But un(x) ■--------» u(x) and un(c) -------- > w(c), so

u(x) — u{c) = / v(t) dt.

By the Fundamental Theorem of Calculus, u is differentiable and ur = v.

R e m a r k : The last two results can be interpreted in terms of inter

changing operations. Theorem 8.2 says that lim^*, and can be inter

changed (under suitable hypotheses) and Theorem 8.3 says that limn-oo and d/dx can be interchanged. Note the key role played by uniform convergence in these results.

I t is easy to translate the results above into statements about infinite series of functions. The starting point is the elementary fact that everything reasonable holds for finite sums—that is what the partial sums of a series are.

Theorem 8.4 Let { f n( x) } be a sequence of continuous functions on a domain D and assume J2n=i fn (x ) converges uniformly on D to F ( x ) . Then

(1) F ( x ) is continuous on D.

(2) If D = [a, 6], then F ( x ) is integrable on [a, 6] and

(3) Suppose the functions f n (x ) have continuous derivatives and Y fn (#) converges uniformly to G(x) on an open interval D = (a, b). Then F(x) is differentiable and

Infinite Series

n =1

F ' ( x ) = G ( x ) = £ / . '(* )n =1

for each x in D.

Proof: Let sn (x) = 2 * = i /* («) • Then sn (x) --------> F (x) uniformlyon D. But sn(x) is continuous, being a finite sum of continuous functions; hence F(x) is continuous by Theorem 8.1. This proves (1).

112 3. POWER SERIES

Now F(x) is continuous, hence integrable on [a, 6]. By Theorem 8.2,

/ F(x) dx = lim / sn(x) dx = lim / / / * ( # ) dxJ a n-+ oo »/ o n-*oo J a J

k = 1

6

= lim V f f k(x) dx = V f fk(x) dx.n-+ oo ' T */ a a

fc = l k —1

This proves (2 ).By Theorem 8.3, the function F(x) is differentiable and

n

[ ^ / * ( ^ ) ] A- =1

F'(x) = G(x) = lim s'(a:) = lim -7-n-*oo d X

n co

= lim Y /* (* ) = V / t ( z ) -n ~*°° A: = 1 k = l

This proves (3).

R em ar k : Parts (2) and (3) of Theorem 8.4 are again results about interchanging the order of operations. They may be stated formally as follows:

00 00

(2) l f n(x) dx - u f n(x) dx,n =1

(3)n — 1 n =1

The M-test

In applying these results, the first step is always proving the uniform convergence of some series. Often we make use of the following criterion for uniform convergence, called the Weierstrass M-test.

Theorem 8.5 (M-test) Suppose f n(x) are given on a domain D and there are constants M n such that

(1) Y M n converges,(2 ) \ fn(x)\ < M n for all x in D and all n.

Then J2fn(x) converges uniformly on D.

Before reading the proof, it is advisable to review the Cauchy Test in Chapter 1, Section 2.

Proof: Let e > 0. By the Cauchy Test, there is an N such that

M n+ i + M n+2 + • • • + M m <

whenever m > n > N. Therefore

2 /*(*) < 2 I/* (* )!< 2 M * < |€/c=n+l A: =n+l k =n+l

for all x in D whenever m > n > N.By the Cauchy Test again (“if” part), for each x in D, the series X) /*(#)

converges to a number F(x). Now let --------> oo in the last displayed inequality:

' k =n-f*lfor all n > N and all x in D. An equivalent statement is

nF{x) - % f k(x)

k = l

that is,|F(«) - 8n(x)\ < e

for all n > N and all x in D, where sn(x) is the n-th partial sum of J^fk(x). This says that the series converges uniformly to F(x) on D.

Power Series

Let us derive the main facts about integration and differentiation of power series. First we need a result on the uniform convergence of power series.

Theorem 8.6 If a power series Y an(x — c)n converges in the interval \x — c\ < R, then it converges uniformly in each smaller interval \x — c\ < r, where 0 < r < R.

Proof: If \x — c\ < r, then

|an(x — c )n| < \anrn\.

Since r < R, the series Y arJ,n converges absolutely. Therefore the M-test applies with M n = \anrn\. Done.

Next we need uniform convergence of the series Y l nan{% — c)n_1 of term-by-term derivatives.

Theorem 8.7 If a power series Y an(x — c)n converges in the interval \x — c\ < R, then the series Y nan{x — c)n_1 converges uniformly in each smaller interval \x — c\ < r, where 0 < r < R.

Proof: By Theorem 8.6, it suffices to prove that the power series X nan(x — c)n_1 converges in \x — c\ < R.

Let \x — c\ < R. Choose r so that \x — c\ < r < R. Set b = \x — c\/r, so0 0 (by Lhospital’s rule for instance); henceribn~l < r for n sufficiently large. This implies

|n(x — c )n~l| < rn, |nan(x — c)n_1| < \anrn\

for n sufficiently large. But X) anTn converges absolutely since r < R. Therefore nan(x — c)n~~l converges by the M-test.

Now we can state the main results of this section.

114 3. POWER SERIES

Theorem 8.8 Let F(x) = X)an(rc — c)n, where the power series converges on the interval \x — c\ < R. Then

(1) F(x) is continuous on \x — c\ < R.(2) F(x) is integrable on \x — c\ < R and

oo

c>‘+'n = 0

for \x — c\ < R.(3) F(x) is differentiable on \x — c\ < R and

Ff (x) = ^ nan(x — c)n~l.n = 1

Proof: Except for a small technical detail, everything follows routinely from the previous results. We cannot conclude right off the bat that F(x) is continuous on \x — c\ < R because we do not know that Y an(x — c)n converges uniformly on all of the interval \x — c | < R. We know only that it converges uniformly on each closed interval \x — c\ < r, where r < R. But th a t is enough, for if £0 is any point such that |#0 — c\ < 72, we can choose r so that \x0 — c\ < r < R . Hence x0 belongs to the open interval \x — c\ < r. But Theorem 8.6 shows that the power series converges uniformly on \x — c\ < r, hence F(x) is continuous on \x — c\ < r. In particular F(x) is continuous at x0. (Of course we have used the obvious continuity of the summands—polynomials—an(x — c)n.) The rest of the theorem is proved similarly.

The last statement in the theorem can be applied to F', then to F';, etc. Thus F can be differentiated repeatedly, and

dkF V '—- = / n(n — 1)* • • (n — k + 1 )anxn~k for |.r — c\ < R.dxk

n —k

R emark: We have carefully avoided the endpoints of the interval of

convergence. If F(x) = 2 anxn has radius of convergence R, then the power series converges for \x\ < R, and we may not know what happens at x = dbR. But suppose the series converges at x = R. Then it can be proved that onxn converges uniformly on 0 < x < R, and hence that F(x) is continuous and can be integrated on 0 < x < R. This is a fairly deep result, and its proof is beyond our scope. I t is needed to prove such statements as

. * dx 1 1 1l n 2 * I +

7r

-jC r

■Jf rx , . d x 1 , 1 1 ,j - a r c t a n l - / —

2. Let fn (x) =

EXERCISES

1. Let/n(:r) = xn. Show that the sequence { /n(a0) converges uniformly on [0, but not on [0, 1].

r nx, 0 < x < 1/n

2 — nx, 1/n < x < 2/n

0 2/n < x.

Prove that /»(x ) --------» 0 for all x > 0, but the convergence is not uniform on[0, oo ).

3. Prove that xe~nx--------> 0 uniformly on [0, oo ).[Hint: Find the maximum value of xe~nx.]

4. Determine whether x2e~nx--------> 0 uniformly on [0, oo ).5. Prove that (sin nx)/n2 is continuous on R.6. Prove that 2 ” 1/ (1 + 3n) is continuous for x > 1.

00

7. Prove that f ( x ) = ^ e~nx sin nx is continuous for x > 0.n =1

8. (cont.) Justify the formula

/ ■ i t .n =1

f ( x ) d x = y I e~nx sin nx dx.

9. Justify the formula00 00

n ^ cos nxd |~V sin nxl _ V* <dx (_ n3 J

n =1

10*. In Theorem 8.3, suppose the hypothesis un (x )--------> u(x) for each x in (a, b) isreplaced by {un{c)) converges for some c in (a, 6). Prove that {un(x)} converges for each x follows anyhow.

11. If 2 an is an absolutely convergent series, prove that ^ an sin nx converges uniformly.

12. Suppose fn (x) --------> F (x) uniformly on [a, 6]. Does it follow that/n' (x) -------->F' (x) on [a, 6]?

4. Solid Analytic Geometry

1. COORDINATES AND VECTORS

In this chapter we shall develop tools useful for geometric applications of calculus and for the study of functions of several variables. Although we shall work in the euclidean three-space R3, we note that most of what we do applies as well to the euclidean plane R2.

First we introduce a rectangular coordinate system in R3. We select an origin 0 and three mutually perpendicular real axes through the origin (Fig. 1.1a).

F ig . 1.1

Once directions are fixed on the x- and y-axes, the direction on the 2-axis is determined by the right-hand rule: curl the fingers of your right hand from the positive ^-direction towards the positive ^/-direction; the thumb will point in the positive ^-direction (Fig. 1.1b). (In drawings of 3-space, it is convenient to think of the #-axis as pointing straight up from the paper).

We refer to the plane of the #-axis and y-axis as the x, 2/-cordinate plane or simply the x, y-plane, etc. (Fig. 1.2a).

Now take any point x in space. Pass planes through x parallel to each coordinate plane. Their intersections with the coordinate axes determine three

1. Coordinates and Vectors 117

(a) the coordinate planes (b) coordinates of a pointFig 1.2

numbers x, y, z, called the coordinates of x. See Fig. 1.2b. Conversely, each triple (x , y, z) of real numbers determines a unique point x in space. We shall write

x = (x, y, z).

A point (.x, y, z) is located by marking its projection (x, y, 0) in the Xj 2/-plane and going up or down the corresponding amount z. (From the habit of living in the x, y-plane for so long, we think of the ^-direction as “up” .) See Fig. 1.3a for some examples.

(a) locating points (b) dashes for hidden linesFig. 1.3

118 4. SOLID ANALYTIC GEOMETRY

The portion of space where x, y , and z are positive is called the first octant. (No one numbers the other seven octants.) Sometimes part of a figure which is not in the first octant is shown; dotted lines indicate it is behind the coordinate planes (Fig. 1.3b). The angle at which the #-axis is drawn in the y , 2-plane is up to you. Choose it so that your drawing is as uncluttered as possible. Actually it is perfectly alright to take a projection into other than the y , 2-plane, so that the y- and 2-axes are not drawn perpendicular.

Vectors

We now introduce the concept of a vector, then vector algebra and, later, vector analysis. Vectors are most useful for handling problems in space because (1) equations in vector form are independent of choice of coordinate axes, hence are well suited to describe physical situations; (2) each vector equation replaces three ordinary equations; and (3) several frequently occurring procedures can be summarized neatly in vector form.

Let the origin 0 be fixed once and for all. A vector in space is a directed line segment that begins at 0; it is completely determined by its terminal point. Denote vectors by bold-faced letters x, v, F, r, etc. (In written work use x or x instead of x.)

A point (x, y, z) in space is often identified with the vector x from the origin to the point.

A vector is determined by two quantities, length (or magnitude) and direction. Many physical quantities are vectors: force, velocity, acceleration, electric field intensity, etc.

Remember that the origin 0 is fixed, and that each vector starts at 0. We often draw vectors starting at other points, but in computations they all originate at 0. For example, if a force F is applied at a point x, we may draw Fig. 1.4a because it is suggestive. But the correct figure is Fig. 1.4b. One must specify both the force vector F (magnitude and direction) and its point of application x.

Fig. 1.4 drawing vectors

x

(a) picturesque (b) correct

With respect to coordinate axes, each vector x has three components (coordinates) x, y, and 2, which we indicate by the notation

X = (*, y , z).


See Fig. 1.5. Sometimes it is convenient to index the components, writingx = (#i, x2, £3) instead of x = (x, y ) z).

The zero vector (origin) will be written 0 = (0, 0, 0). For this vector only, direction is undefined.

F ig . 1.5 components of a vector

Addition of Vectors

The sum u + v of two vectors is defined by the parallelogram law (Fig.1.6). The points 0, u, v, u + v are vertices of a parallelogram with u + v opposite to 0.

F i g . 1 .6 parallelogram law of vector addition

Vectors are added numerically by adding their components:

(uh u2, u3) + (vh v2, Vs) = (Ui + Vi, U2 + V2, u3 + v3).

For example,

( - 1 ,3 ,2 ) + (1 ,1 ,4 ) = (0 ,4 ,6 ), (0 ,0 ,1 ) + ( - 1 ,0 ,1 ) = ( - 1 ,0 ,2 ) .

Let us prove that the sum of vectors, defined geometrically by the parallelogram law, can be computed algebraically by adding corresponding components. We pass planes P, Q, R through u, v, and w = u + v parallel to the #3, #i-plane (Fig. 1.7). They meet the #2-axis at u2, v2, and w2. Because vw and Ou are parallel, the directed distance from Q to R equals the directed distance from the xs, #i-plane to P. Hence w2 — v2 = u2, that is, w2 = u2 + v2. Similarly Wi = ui + V\ and Wz = Uz + vz.

Fig. 1.7 proof of componentwise addition

Multiplication b y a Scalar

Let v be a vector and let a be a number (scalar). We define the product av to be the vector whose length is |a| times the length of v and which points in the same direction as v if a > 0, in the.opposite direction if a < 0. If a = 0, then av = 0.

The physical idea behind this definition is simple. If a particle moving in a

2v

Fig. 1.8 scalar multiples

— v


certain direction doubles its speed, its velocity vector is doubled; if a horse pulling a cart in a certain direction triples its effort, the force vector triples. Figure 1.8 illustrates multiples of a vector.

Scalar multiples are computed in components by the following rule.

a(vh v2,vs) = (avh av2, avz) .

This rule is proved by similar triangles (Fig. 1.9). The triangle 0y2v is similar to 0w2w, hence w2 = av2, etc.

Fig. 1.9 proof of a(vi, v2, t>3) = (avi, av2, avz)

difference v — w of two vectors is defined by

V — W = V + ( — w).

The

v

0

(a)

— w

(b) (c)

Fig. 1.10 difference of vectors

v — w w


See Fig. 1.10. (The vector — w has the same length as w but points in the opposite direction.)

The segment from the tip of w to the tip of v (the dashed line in Fig. 1.10c) has the same length and direction as v — w. Hence if two points are represented by vectors v and w, the distance between them is the length of v — w.

The basic rules of vector algebra follow directly from the coordinate formulas for addition and multiplication by a scalar.

Rules of Vector Algebra

v + 0 = 0 + v = v v + ( - V ) = ( v) + v = 0

u + v = v + u U + (v + w) = (u + v) + w

Ov = 0 lv = v a (bw) = (ab)v

(a + J>)v = av + 6v a(v + w) = av + aw

EXERCISES

Draw axes as in Fig. 1.1a and locate each point accurately:1. (1,2,3), (1,3,4)* 2. (2,4,1), (2 ,-4 , 1)3. ( -1 , 2, 1), (2, 2, - 1 ) 4. (1, - 3 , 3), (3, 2, - 2 )5. (4, 6, - 1 ) , (—4, - 6 , 1) 6. (0, 0, - 3 ) , ( -2 , - 5 , - 3 ) .

Draw the parallelepiped with edges parallel to the axes and locate the vertices. The ends of a diagonal are:

7. (0,0,0), (2,3,1) 8. (4,2,3), (1,1,1).Are the points collinear?

9. (0, 0, 0), (1, 3, 2), (2, 6, 4) 10. (0, 0, 0), ( -1 , 3, - 4 ) , (2, - 5 , 8)11. (1,1,1), (0,1,2), ( - 1 , - 3 , - 5 ) 12. ( 1 , - 1 , - 2 ) , (-1 ,2 ,3 ), ( 3 , - 4 , - 7 ) .

Compute:13. (1, 2, - 3 ) + (4, 0, 7) 14. ( -1 , - 1 , 0) + (3, 5, 2)15. (4, 0, 7) - (1, 2, - 3 ) 16. (2, 1, 1) - (3, - 1 , - 2 )17. (1, 2, 3) - 6(0, 3, - 1 ) 18. 4 [ (1, - 2 , - 7 ) - (1, 1, 1)]19. 3(1, 4, 2) - 2(2, 1, 1) 20. 4(1, - 1 , 2) - 3(1, - 1 , 2).

Prove:21. u + v = v + u 22. u + (v + w) = (u + v ) + w23. (a + 6)v = av + 6v 24. a(v + w) = av + aw.25. Show that \ (v + w) is the midpoint of the segment from v to w. [Hint: Use the

parallelogram law.]26. (cont.) Use Ex. 25 to show that the segments joining the midpoints of opposite

sides of a (skew) quadrilateral bisect each other.[Hint: (u + v ) + (w + z ) = (v + w) + (u + z ) . ]

27. Show that J (u + v + w) is the intersection of the medians of the triangle with vertices u, v, and w.

2. Length and Dot Product 123

28*. In a tetrahedron, prove that the four lines joining each vertex to the centroid (intersection of the medians) of the opposite face are concurrent.

29*. Space billiards—no gravity. The astronaut cues a ball toward the corner of a rectangular room, with velocity v. The ball misses the corner, but rebounds off of each of the three adjacent walls. Find its returning velocity vector.

2. LENGTH AND DOT PRODUCT

The length |v| of a vector v = (vh v2, v3) is its distance from the origin. By regarding this distance as the length of the diagonal of a rectangular solid (Fig. 2.1a), we see that

V |2 = VI2 + V22 + Vz2.

(a) length of a vector: (b) distance from v to w ______ = |v - w|

iv|* = ( v v +

= l>i2 + Vi1 + v„2

Fig. 2.1 length and distance

Vector length has the following properties:

|0| = 0, |v| > 0 if v 5* 0,

|av| = |o|*|v|,

iv + w| < JvJ + |w| (triangle inequality).

The distance between two points v and w is equal to |v — w|. See Fig. 2.1b. If v = (vh v2, Vz) and w = (wi, wz), then v — w = (vi — wh v2 — w2)

Vz — Wz)- Therefore we have the distance formula:


Distance Formula The distance between two points v and w is |v — w|, where

|v — w|2 = ( 1 — W i ) 2 + (v2 — w2)2 + (Vz — W z ) 2.

Dot Product

There is another important vector operation, the inner product or dot product of two vectors. Let v and w be vectors, and let 0 be the angle between them (Fig. 2.2a). Define

Y-W = |V| • |w| COS0.

Since cos(—0) = cos 0, you can measure 0 from v to w or from w to v. Note (Fig. 2.2b, c) that |w| cos 0 is the (signed) projection of w on v, hence v w is |v| times the projection of w on v. If v = 0 or w = 0, we define v w = 0 even though 0 is not defined.

(a) (b) (c)

Fig. 2.2 dot product

Important: The dot product of two vectors is a scalar (number), not a vector.

The numerical rule for computing dot products is

V • W = V1W 1 + V2W 2 + VzWz,

an important formula. Let us prove it. See Fig. 2.3. By the Law of Cosines,

Hence

V#W = \ [jV ~ lv ~ wl2]

= 2 [l (Vl> V2’ V z I (Wl9 W2> W3 2 “ I “ Wh V* — ™2, 3 - |2J

= ^ £ (^l2 + V22 + VS2 ) + ( Wi 2 + W 22 + Ws2 )

— (Vi — Wi)2 — (v2 — W2)2 — (v3 — * ) ■ ]

= ViWi + V2W2 + VsWs.


The main algebraic properties of the inner product follow easily from the formula v*w = ViWi + v2w 2 + VsWs'.

v w = w*v (av)*w = v (aw) = a ( v w )

(u + v)*w = u*w + v w .

Two vectors v and w are perpendicular if 0 = w/2 , i.e., if cos 0 = 0 . This can be expressed very neatly as follows:

The condition for vectors v and w to be perpendicular is

v w = 0 .

(The vector 0 is considered perpendicular to every vector.)

For example, (1, 2 , 3) and ( — 1, —1, 1) are perpendicular because

(1, 2, 3 ) . ( — 1, - 1 , 1) = - 1 - 2 + 3 = 0.

There is a connection between lengths and dot products. The dot product of a vector v with itself is v v = Ivl2 cos 0 = Ivl2.

For any vector v,

V V = |v|2 = Vi2 + V22 + Vz2.

From the dot product can be found the angle 0 between any two non-zero vectors v and w. Indeed,

cos 0 =V* w

Ivl |w|

EXAMPLE 2.1

Find the angle between v = (1, 2, 1) and w = (3, —1, 1).

Solution:v w = 3 — 2 + 1 = 2,

| v |2 = 1 + 4 + 1 = 6, | w|2 = 9 + 1 + 1 = 11.Hence

cost/ =v W n

Answer: arc cos\/6 6

EXAMPLE 2.2

The point (1, 1, 2) is joined to the points (1, —1, —1 ) and (3, 0, 4) by lines L\ and L2. What is the angle 0 between these lines?

Solution: The vector

v = (1, - 1 , - 1 ) - (1 ,1 ,2 ) = (0, —2, —3)

is parallel to Li (but starts at 0). Likewise

w = ( 3 , 0 , 4 ) - (1, 1 ,2 ) = (2, - 1 ,2 )

is parallel to L2. Hence

v w 0 + 2 — 6 —4cost/ =

|vj |w| \ / 0 + 4 + 9 \ /4 + 1 + 4 V l 3 \ / 9 '

N ote: When we find cos# < 0 for an angle 6 between two lines, then


0 is not the smaller angle between the lines, but its supplement. The basic fact here is that cos(t — 0) = —cos 0.

Direction Cosines

I t is customary to use the notation

i = (1 ,0 ,0 ), j = (0 ,1 ,0 ), k = (0 ,0 ,1 )

for the three unit-length vectors along the positive coordinate axes (Fig. 2.4). If v is any vector, then

v = (vh v2, vz)

= Vi(l , 0, 0) + 02 (0, 1, 0) + f73 (0, 0, 1)

= Vi\ + v2] + 3k.

Fig. 2.4 the basic unit vectors

Thus v is the sum of three vectors yj, v2\ ) vzVi which lie along the three coordinate axes. The components vh v2, Vs can be interpreted as dot products:

v-i = (yh v2) Vs)• (1, 0, 0 ) = vi.

Similarly, v2 = v j and vs = v*k.Now suppose u is a unit vector, i.e., a vector of length one (Fig. 2.5a).

Let a be the angle from i to u. Define and y similarly. Then u*i = cos a, u • j = cos 0, and u • k = cos 7 . Hence

u = cos a i + cos /3 j + cos 7 k = (cos a , cos 13 cos 7 ).Since |u| = 1,

cos2 a + cos2 p + cos2 7 = 1.Unit vectors are direction indicators. Any non-zero vector v is a positive

multiple of a unit vector u in the same direction as v. In fact v = |v| u, so


k

Fig. 2.5 direction cosines

Each non-zero vector v can be expressed as

v = |v| u, u a unit vector,or as

v = |v| (cos a, COS 13, cos 7 ).

The numbers cos a, cos /3, cos 7 are called the direction cosines of v. They satisfy

cos2 a + cos213 + cos2 7 = 1.

If u is a unit vector (Fig. 2.5b) in the plane of i and j, then

u = cos a i + cos /3 j.

Since u is a unit vector, cos2 a + cos2 /3 = 1. But, as is seen in the figure, cos j3 = sin a. Therefore, the preceding equation simply says

cos2 a + sin2 a = 1.

Summary

A d d it io n o f V e c t o r s

(Vi, v2, Vs) + (tt>i, W2, Ws) = ( 1 + Wi, v 2 + w 2, Vs + Ws ) .

M u l t i p l i c a t i o n b y a S c a l a r :

a ( v i , v2) Vs) = ( a v h a v 2, a v s ) .

L e n g t h :

| V |2 = V V = V12 + V22 + Vs2.

v*w = |v| |w| cos 6 = ViWi + V2W2 + VsWs.

V e c t o r s i, j, k:

These are unit vectors in the direction of the positive #-axis, ?/-axis,2-axis, respectively. I f v = (vh v2, Vs), then v = v{\ + v2\ + v3k> where Vi = v i , v2 = v j , = v k .

D ir e c t io n C o s i n e s :

If u is a unit vector, then u = cos a i + cos @ j + cos 7 k, where a, 0, 7 are the angles to u from i, j, k, respectively. Furthermore cos2 a + cos2 P + cos2 7 = 1. Any non-zero vector v can be written asv = |v|u = |v| (cos a, cos 0 , cos 7 ). The numbers cos a, cos ft, cos 7 are the direction cosines of v.

D ot P ro duc t:

EXERCISES

Compute:1. (8, 2, 1). (3, 0, 5) 2. ( -1 , -1 , - 1 ) . (1, 2, 3)3. (1,0, 2).[(1,4, 1 )+ (2,0, - 3 ) ] 4. 1(2,-4,7)15. |3 l - j + k| 6. | (—1, —1, 0) — (3, 5, 2)|7. | W 3 (-1 , 1, 1)| 8. [3j - (1, 1, 2)3 • (4j - k).

Find the angle between the vectors:9. (4, 3, 0), (—3, 0, 4) 10. (1, 2, 2), ( -2 , 1, - 2 )

11. (6, 1, 5), ( -2 , - 3 , 3) 12. (-5 , 6, 1), (2, 3, - 8 )13. (1, 1 ,-1 ) , (2, 0, 4) 14. (2, 2, 2), ( -2 , 2, - 2 ) .Compute the distance between the points:15. (0, 1, 2), (5, - 3 , 1) 16. (1, 1, 1), (1, - 1 , 2)17. (7, 0, 0), (2, 3, 4) 18. (8, 5 ,- 1 ) , (7, 9, 3).Find the direction cosines:19. (1,0,1) 20. ( - 1 , - 1 , - 1 )21. (2, 1 ,- 3 ) 22. ( 4 , - 7 , - 4 ) .23. Find two non-collinear vectors perpendicular to (1, —1, 2).24. Find the angle between the line joining (0, 0, 0) to (1, 1, 1) and the line joining

(1, 0, 0) to (0, 1, 0).25. Prove the Cauchy-Schwarz inequality:

IV w| < | v| • Iw|.When does equality hold?

26. (cont.) Now prove the triangle inequality:|v + w| < |v| + |w|.

[Hint:|v + w|2 = | (v + w)« (v + w)| = I (v + w ) ' V + (v + w)»w|

< |v + w|- 1v| + |v + w|'|w| . ]

27. Prove the identity |v + w|2 — |v — w|2 = 4 v w .28. Let u be a unit vector. Show that the formula v = ( v u ) u — [v — ( v u ) u ]

expresses v as the sum of two vectors, one parallel to u, the other perpendicular to u, and is the only such expression.

29. Let u = (cos ai, cos a2, cos ck3) and v = (cos ft, cos ft, cos ft) be two unit vectors. Show that the angle 6 between them satisfies

cos 6 = cos ai cos ft + cos a2 cos ft + cos <23 cos ft.Interpret the formula when a3 = ft = 7r.

3. LINES AND PLANES

Take a non-zero vector v. The set of all scalar multiples x = tv of v is a line through the origin (Fig. 3.1a). If x0 is any point, then the set of all points x = x0 + £v is a line through x0 parallel to the first line (Fig. 3.1b). The equation x = x0 + tv is called a parametric vector equation for the line. For example

(jci, x2, xs) = (0, 1, - 2 ) + t (2, —1, 3)

is a parametric equation for the line through (0, 1, —2) parallel to (2, —1,3). This vector equation is equivalent to three parametric scalar equations:

xi = 2 1, x2 = — t + 1, X3 = 3t — 2 .

Fig. 3.1

(a) line through 0: X = tv

(b) line through x0 parallel to v

3. Lines and Planes 131

Given a point x0 of R3, and a non-zero vector v, the line through x0 parallelto v consists of all points

X = Xo + Jv,where — oo < t < o o .

EXAMPLE 3.1

Find a parametric vector equation for the line through (3, — 1, 2) and (4, 1, 1).

Solution: The line passes through x0 = (3, —1, 2) and is parallel to v = ( 4 , 1 , 1 ) - (3, - 1 ,2 ) = (1,2, — 1). Hence x = (3, - 1 ,2 ) + * (1 ,2 , - 1 ) is a parametric form for the line.

Answer: x *= (3, —1,2) + $ (1 ,2 , —1)= (3 + t, - 1 + 2t, 2 - t).

Equation of a Plane

Let P be a plane in R3. Draw the line L through 0 perpendicular to P and take one of the two unit vectors along L; call it n. See Fig. 3.2. The line L meets P in a point pn. If x is any point of P, then x — pn is perpendicular to n, therefore

(x — pn )* n = 0, x*n = pn*n = p.

L Fig. 3.2 normal form of a plane

Normal Form Each plane in R3 can be represented by an equationx n = p,

where n is a unit vector perpendicular to the plane and p is a real number.Conversely, each equation x*n = p represents a plane, the one through the point pn and perpendicular to n.

132 4 . SOLID ANALYTIC GEOMETRY

I t follows that each linear equation in 3 variables,

aiXi + a2x2 + azx3 = b (ai2 + a22 + a33 ^ 0),

is the equation of a plane. Just set a = (ah a2, a3) and n = a/|a|. Then the equation, in vector notation, is

ba *x = o, or n*x = — = p y|a|

the normal form of a plane.

R e m a r k 1: There really are two normal forms of a plane,

x*n = p, x* ( — n ) = —p.

R e m a r k 2: In applications it is often more convenient if n is not a unit vector. Certainly it is simpler to write

Oi, X2, x3)• (1, 1, 1) = 5

than the equivalent normal form

(xh x2, x3) - i V S ( l , 1, 1) = i h / 3 -

EXERCISES

Express in normal form:1. x\ — 2x2 + 2z3 = 1 2. 2x\ + 6 2 — 3z3 = 143. —&Ei + X2 — 4z3 = 27 4. Sxi — 2x2 — 6x3 = 45. x\ -j- x2 x3 — 3 6. x\ — X2 ~\~ x3— —12.7. Find a parametric form for the line through two distinct points x0 and xi.8. Let x = x0 + tu and x = xi + be two parallel lines. Prove that v = cu for

some c.9. Let x* n = p be a plane in normal form and x0 a point. Prove that the distance from

x0 to the plane is |x0«n — p\.10. (cont.) Prove that the point of the plane closest to x0 is z = x0 + (p — x0*n)n.11. Let x = x0 + be a line in parametric form and let x* n = p be a plane in normal

form. Prove that the line and plane are parallel if and only if v n = 0.12. (cont.) Prove that the line is on the plane if and only if v -n = 0 and x0*n = p.13. (cont.) Suppose v*n ^ 0. Prove that the point of intersection of the line and the

plane isXo*+ [ ( p — x0- n ) / ( v n ) ] v .

14. Let x« m = p and x* n = q be two non-parallel planes in normal form. Let 8 be one of their (dihedral) angles of intersection. Determine cos0.

15. Let x = x0 + £u be a line in parametric form, where u is a unit vector, and let yo be a point. Find the point on the line closest to y0.

16. (cont.) Prove that the distance D from yo to the line satisfies D2 = |xo — yo|2 — [ ( x 0- yo)*u]2.

4. Linear Systems and Intersections 133

4. LINEAR SYSTEMS AND INTERSECTIONS

Suppose we are given three planes

a r x = di, a2*x = d2) a3*x = d3.

In general, they will intersect in a single point x. How can we find this point? In coordinates, the problem is to solve simultaneously three linear equations

dix + biy + c\z = di

< a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

for x, y , z, where ah • • •, d3 are given constants.This is typical of problems in a subject called Linear Algebra, which we

shall take up in Chapter 7. While it is beyond the scope of this book to go into the general theory, we shall give a practical method of solution.

EXAMPLE 4.1

Find the solution (%, y, z) of the system

a? i + II

- 3y + 2z = - 1

—z = 3.

Solution: By the third equation, z = — 3. Substitute this into the first two equations. The result is a new system of two equations for x and y:

2x - y = 4 - ( - 3 ) = 7

32/ = — 1 — 2( —3) = 5.

By the second equation, y = f . Substitute this into the first equation; the result is a single equation for x :

2x = 7 + § =

Its solution is x = -V3--

Answer: ( ^ , f , — 3).

This example was very easy because we could solve for the unknowns one at a time. To solve a more general system, we reduce it to a system of this type by eliminating the unknowns one by one.

EXAMPLE 4.2

Find the solution of the system

Ii+1cl

*< 2x + 2y + 32 = 3

Ox — 9y — 2z — 17.

Solution: Eliminate x from the second and third equations. Subtract the first equation from the second, and subtract 3 times the first equation from the third; the result is an equivalent system of three equations (the first the same as before):

2 x — y + z = 4

Sy + 2z = — 1

— 6y — oz = 5.

Now eliminate y from the third equation. Add twice the second equation to the third, but keep the first two equations:

2 x — y + z = 4

3y + 2z = - 1

- 2 = 3.

This is the system in Example 4.1, which we can solve by elimination.

Answer: x = ^ y = f , z = —3.

The method of elimination works for a very simple reason: adding a constant multiple of one equation to another equation does not affect the solution of the system. To verify this, suppose a r x = di and a2*x = d2 are two linear equations. If c is any number, then the two systems

a r x = di ( a r x = diand \

a2*x = d2 [ (a2 + cai) *x = d2 + cdi

have exactly the same solutions. For if x satisfies a r x = d\ and a2* x = d2, then

(a2 + cai)*x = a2*x + c (a r x ) = d2 + cd\.

Conversely, if x satisfies a r x = di and (a2 + cai)*x = d2 + cdi, then

a2*x = (a2 + cai)*x — c (a r x ) = (d2 + cdi) — cdi = d2.

Thus the two systems are equivalent; they have precisely the same solutions.

P r a c t ic a l H i n t : When you apply the method of elimination, you do not have to eliminate first x and then y. Eliminate any two of the unknowns in an order that makes the computation easiest.

EXAMPLE 4.3

Solve the system

3a; + y — 2z = 4

-< - 5 x + 2z = 5

k — 7x — y + Sz = —2.

Solution: Since y is missing from the second equation, add the first to the third; then y is eliminated from two equations:

Sx + y — 2z = 4

< - 5 x + 2z = 5

— 4# + 2 = 2.

Now add —2 times the third to the second; this eliminates z :

Sx + y — 2z = 4

< 3* = 1

—4x + 2 = 2.

By the second equation, x = }. By the third equation, z = 2 + 4x = 2 + 4(A) = V - Finally,

y = 4 - 3* + 22 = 4 - 1 + ¥ = ¥ •

Answer: $ = f, y = 2 = ^

Certain systems of equations do not have precisely one solution. There may be either no solutions at all, or more than one solution. In the latter case it turns out that there are infinitely many. Both cases can be handled by elimination.

Inconsistent Systems

Sometimes the elimination process leads to an equation

Ox + Oy + Oz = d,

where d ^ 0. Obviously no choice of x, y, and z will satisfy this equation. Then the system simply has no solution, and it is called an inconsistent system.

An example with two unknowns will suffice to illustrate this. Look at thesystem

x + Sy = - 1

2x + 6y = 3.

Add — 2 times the first equation to the second. The result is the system

x + 3 y = - 1 .

0 = 5 .

There is no solution, for if there were, then 0 = 5 would be a correct statement.

Underdetermined Systems

Some systems have more than one solution. This happens when one of the equations is a consequence of the other two, so that really there are only two (or fewer) equations. Such systems are called underdetermined.

First consider an example in two unknowns :

x + y = 1

2x + 2y = 2.

The second equation is obviously twice the first. If we try to eliminate x by adding — 2 times the first to the second, the result is

x + y = 1

0 = 0 .

The second equation in this equivalent system gives no information whatever. Any solution of the first equation is a solution of the system. Thus there are infinitely many: each point (x, y) on the line x + y = 1 is a solution. I t is convenient to express these solutions in parametric form. A parametric representation of the line is (.x, y) = (t, 1 — t ), where — <*> < t < oo. Hence, the solution of the system is

x = t, y = 1 — t, — oo < t < oo.

Now consider the example

' x + y + z = 1

< x — 2y + 2z = 4

2x — y + 3z = 5.

Eliminate x from the second and third equations:

' x + y + z = 1

< - 3 y + z = 3

—3y + z = 3.

The last two equations both say the same thing. Therefore the system is equivalent to the system

x + y + z = 1

— 3y z = 3.

and no further elimination is possible. To get a parametric solution, set y = t. Then z = 3 + 3y = 3 + 3t, and

x = 1 - y - z = 1 - t - (3 + 30 = - 2 - 4t.

The most general solution is

(x, I/, 2) = ( - 2 - 4£, <, 3 + 30,

where — 00 < t < <*>. The set of solutions is a line, (x, y, z) = ( — 2, 0, 3) + t ( - 4, 1,3).

Finally, consider the system

" z + ?/ + z = 1

« 2z + 2y + 2z = 2

3x ~b 3y -|- 3z = 3.

This system is obviously equivalent to the single equation x + y + z = 1. Therefore the set of solutions (x, y, z) is a plane in space. For a parametric solution, set x = s and y = t. Then z = 1 — s — t. The general solution is

(x, y , z) = 0 , Z, 1 - s - t).

Intersections of Planes

Now we can close the books on intersection of planes. If a r x = di, a2*x = d2, a3*x = d3 are three planes in space, we find the points common to the three planes by the elimination method. Generally, there is a single common point (Fig. 4.1a). However, if the planes are parallel or if one is parallel to the intersection of the other two, then there is no common point (Fig. 4.1c). In this case, the corresponding system of equations is inconsistent. For

138 4. SOLID ANALYTIC GEOM ETRY

*example, the system

x + y + z = 1

- x + y + z = 2

3# — 2y + 42 = 7

is obviously inconsistent; the first two equations cannot both be satisfied. Geometrically, the first two planes are parallel.

parallel lines

(c) no intersection

Fig. 4.1 possible intersections of three distinct planes

The planes have more than one common point if they pass through a common line, or if two or all of them coincide. In this case, the corresponding system of equations is underdetermined. For example the system

x — 2y + Sz = 5

< 8x + 7y + z = 2

2x — 4y + ftz = 10

is underdetermined; the third equation is twice the first. Geometrically, the first and third planes coincide. The system represents two distinct planes that have a line in common (Fig. 4.1b).


Review of Determinants

An important computational tool in solving linear systems is the use of determinants. We recall from elementary algebra the definitions of determinants of orders two and three:

a i b\

CL2 2

d l 61 Cl

d2 b 2 C 2

dz b 3 Ci

— &1&2C3 ~t~ d 2bzCi “I" CLsblC2 — CLlbsCz — &2&1C3 — 0 362 1.

From the defining formulas: (1) if two rows (columns) are equal, the determinant is zero; (2) if two rows (columns) are transposed, the determinant changes sign; (3) if a multiple of one row (column) is added to another row (column), the determinant is unchanged; and (4) if all the terms in one row (column) are multiplied by a scalar, the determinant is multiplied by the same scalar.

Also the defining formulas imply various expansions by minors of a row (column), for instance

d\ b\ Ci62 C2 to Ci to d 2 62

d2 62 C2 = d l - h + Cl

bz Cz dz Cz dz bz

dz bz Cz

is the expansion by minors of the first row\ Here, for instance,

d2 C2

dz Cz

is the minor of bi. I t is the 2 X 2 determinant remaining after the row and the column containing bi are crossed off.

A system of equations

dix + biy + C\z = d \

d2X + b2y + C2Z = d2

, + bsy + c3z = dz

is both consistent (not inconsistent) and determined (not underdetermined)

if and only if the system determinant D ^ O , where


D =

ai bi ci

ci2 62 C2

dz bz Cz

When this is so, the system has a unique solution, given explicitly by Cramer’s Rule:

X ~ D

Cramer’s Rule will be derived in Chapter 7, Section 7.

di Cl ai di Cl dl b 1 di

C?2 62 c21

’ * - 5 &2 d/2 c2

-IQ11M d2 d2

3 63 Cz «3 dz C3 dz 63 dz

EXERCISES

Solve by elimination: \x + 2y = 1

1.

3.

5.

7.

9.

11.

Sy = 2x + 2y = 1

x + 3y = 2

2x — 3y = —1

Sx + by = 2

x + y + 2 = 0

2y — 3z = — 1

Sy + 5z = 2

£ + 2/ “ 2 = 0

x — y + z = 0

z + y + z = 0 2a: — y — 3z = 1

— x — 4?/ — 2z = 1

3x — y — 2 = 1

2.

4.

6.

8.

1 0 .

12

2x = 3

— x + y = 0

x - ^ y = a

x — y = b

' 2 x - 3 y = - I

— 3x + 62/ = 2 2x — y — 2 = 1

2y — 3z = — 1

— 32/ “h 52 = 2 2z + y + 32 = 1

—x + 4i/ + 2z = 0

3 z + y + 2 = —1 4z + 2y — 2 = 0

z + 32/ + 22 = 0

k x + y + 32 = 4.

5. Cross Product 141

Show that the system is inconsistent and interpret geometrically:[x = 2

13.

15.

Find all solutions of each underdetermined system:x + y = 0

17.

19.

18.

20.

x + y = 0

11* + 10y + 92 = 5

x + 2y + 3z = 1

3x + 2 y + 2 = 1 .21*. Suppose a 5 b, b 5* c, c a. Show that the system

x + y + 2 = d\

ax + by + cz = di

a?x + 62?/ + c22 = d 3

has a unique solution.22*. Suppose the three planes a^x = 0, a2*x = 0, a3*x = 0 have only 0 in common.

Show that the three planes ai*x = di, a2*x = d2, a3*x = d3 have exactly one point in common. [Hint: Apply the same elimination process to both systems.]

5. CROSS PRODUCT

Geometric Definition

Given a pair of vectors v and w, we define a new vector v X w.

The cross product of v and w, written

v X w,

is the vector whose direction is determined by the right-hand rule from the pair v, w, and whose magnitude is the area of the parallelogram based on v and w. See Fig. 5.1.

Note that v X w is a vector perpendicular both to v and to w. Note also that if v and w are collinear, then the parallelogram collapses, so v X w = 0.


In particular,

v X v = 0 for each vector v.

If v and w are interchanged, the thumb reverses direction, hence

w X v = — v X w.

For the basic vectors i, j, k, the cross products are simply

i X j = k, j X k = i, k X i = j.

R emark: This definition of cross product is motivated by physics. See the discussion of torque on p. 145.

Analytic Definition

Given a pair of vectors v = Vi\ + v2\ + ^k and w = wit + w2] + w3k, it seems reasonable to define their cross product using the cross products of the basic vectors i, j, k as follows:

v X w = (v!i + v2\ + v3k) X O ii + w 2j + w 3k)

= viw2i X j + v2wii X i + viwzi X k

+ vzWik X i + v2w3j X k + v3w 2k X j.

(We have used i X i = j X j = k X k = 0.) Now i X j = k = - j X i, etc. Hence, collecting terms, we have

V X W = (iV2Wz — VsW2)\ + (VSWi — ViWs)} + (viw2 — v2w i)k .

Note that each coefficient on the right is a 2 X 2 determinant.


Cross Product Let

v = (vi, v2) Vs) and w = (wh w2) W s ) .

Then

(v2 Vs Vs Vi Vi V2

V w2 Ws

1Ws W i

)W \ w2

v X w =

= ( v 2Ws — VsW2, VsWi — ViWs, ViW2 — V2W i ) .

Here are two numerical examples:

(4,3, - 1 ) X ( - 2 ,2 ,1 ) =

= (3 + 2, 2 - 4, 8 + 6) = (5, - 2 , 14).

3 - 1 - 1 4 4

CO

2 1>

1 - 2 1 to 2

(1 ,0 ,1 ) X (0, 1,1) =0 1 i i 1 0 \1 1 1 0 0 1 ) = (“ 1, - 1 ,1 ) .

A device for remembering the cross product is a symbolic determinant, to be expanded by the first row:

For the moment we shall take the formula for v X w as the definition of cross product. Our problem is to prove that it has the required geometric properties. We begin the proof with a formula interesting in itself:

U\ u2 Us

u* (v X w) = D ( u, v, w) = Vi V2 Vs

W i W 2 Ws


To prove it, simply expand the determinant D (u, v, w) by the first row:

U\ U 2 Uz

D(u, v, w) = Vi V2 Vz

Wi W 2 Wz

V2 Vz Vi Vz Vi V2

Ui — U 2 + Uz

W 2 Wz W \ Wz W i W 2

= u* (v X w).

As a consequence of the formula, v (v X w) = Z)(v, v, w) = 0 , because a determinant with two equal rows is zero. Hence v is perpendicular to v X w. Similarly so is w. We have proved

(1) v X w is perpendicular to v and to w.

This is one of the required geometric conditions. Next, we prove a formula for the length of v X w :

|v X w|2 = |v|2 |w|2 — (v w )2.

The left-hand side is(W 3 — V3W 2 ) 2 + ( v 3W i — VI W z ) 2 + (V\W2 — V2W 1 ) 2

and the right-hand side is

(Vi2 + V i + V32 ) ( W i 2 + W 22 + Wz2 ) — (ViWi + V2W 2 + VzWz) 2.

The first product in the right-hand side yields nine terms and the second product, six terms. The three terms like Vi2Wi2 cancel. There remain six terms like V\2w22 and three terms like — 2 V1V2W1W2, exactly what occur on the left- hand side.

From this last formula we deduce

(2) |v X w| is the area of the parallelogram determined by v and w.

For let 0 be the angle between v and w. Then v w = |v| • |w| cos 0, hence

|v X w|2 = |v|2 |w|2 — (v w )2 = |v|2 |w|2 (1 — cos20) = |v|2 |w|2sin20,|v X w| = |v| • |w[ sin0.

This last expression is precisely the required area (Fig. 5.2).Finally we must prove

(3) v, w, v X w is a right-handed system (provided v and w are not parallel).

In order to do so, we need some analytic way of deciding whether a given tripleu, v, w is a right-handed system or not.


w

Fig. 5.2 |v X w| = area of parallelogram

Now observe that u, v, w and v, u, w have opposite orientations, that is, one is a right-handed system and the other is left-handed. By analogy, the determinants D(u, v, w) and Z>(v, u, w) have opposite signs. This suggests that the sign of D (u, v, w) corresponds to the orientation of u, v, w. Since i, j, k is right-handed and D ( i, j, k) = 1, we suspect that u, v, w is right-handed if D ( u, v, w) > 0. This indeed is the case, but instead of proving it, we shall simply take the determinant criterion as the definition of right-handedness.

In view of this definition, we must prove that D (v, w, v X w) > 0. Now

Z)(v, w, v X w) = — Z)(v, v X w, w) = D ( v X w, v, w)

= (v X w)* (v X w) = |v X w|2 > 0.

This completes the proof that the analytic and geometric definitions of v X w coincide. We now summarize the main algebraic properties of the cross product. They follow readily from our discussion.

v X v = 0, w X v = — v X w ,

(au + 6v) X w = a(u X w) + 6(v X w),

u X (av + 6w) = a(u X v) + 6(u X w),

u • (v X w) = v (w X u) = w (u X v) = D (u, v, w),

v X w = 0 if and only if v and w are collinear.

R em ark : The quantity u* (v X w) = Z)(u, v, w) is sometimes written [u, v, w] and called the triple scalar product.

Torque

The original motivation for the cross product of vectors came from physics. Consider this situation.

A rigid body is free to turn about the origin. A force F acts at a point x of the body. As a result the body “wants” to rotate about an axis through 0 perpendicular to the plane of x and F (unless x and F are collinear; then there is no turning). See Fig. 5.3a. As usual, the force vector F is drawn at its point


of application x. But analytically it starts at 0. See Fig. 5.3b. The positiv axis of rotation is determined by the right-hand rule as applied to the pair x, F in that order: x first, F second.

Fig. 5.3 torque due to force F applied at x

In physics, one speaks of the torque (at the origin) resulting from the force F applied at x. Roughly speaking, torque is a measure of the tendency of a body to rotate under the action of forces. (Torque will be defined precisely in a moment.)

By experiment, if F is tripled in magnitude, the torque is tripled; if x is moved out twice as far along the same line and the same F is applied there, the torque is doubled. Hence the torque is proportional to the length of x and to the length of F. Therefore (Fig. 5.3b) the torque is proportional to the area of the parallelogram determined by x and F.

Resolve F into components F11 and F-*-, where F11 is parallel to x and F-1- is perpendicular to x. See Fig. 5.4a. Only produces torque; the amount of torque is the product | F-*-| |x| of the magnitude of F-*- by the length |x| of the lever arm. But this product is the area of the parallelogram determined by x and F. See Fig. 5.4b.

(a) (b)Fig. 5.4 Magnitude of torque equals area of parallelogram.

6. Applications 147

Therefore the torque about the origin is completely described by the one vector x X F. The length of x X F is the magnitude of the torque. The direction of x X F is the positive axis of rotation; with your right thumb along x X F, your fingers curl in the direction of turning. In physics, torque about the origin is defined to be the vector x X F.

EXERCISES

Find the cross product:1. ( - 2 ,2 , 1 )X (4, 3 , - 1 )3. (1, 2, 3) X (3, 2 ,1)5. ( - 2 , - 2 , - 2 ) X (1 ,1 ,0) 7. (0, 0, 0) X (1, 1, 2)9. (2, 1, 3) X (2, 2 , - 1 )

2. (1,0, 1 )X (1, 1,0)4. (3, 1 , - 1 ) X ( 3 , - 1 , - 1 )6. ( - 1 ,2 , 2) X ( 3 , - 1 ,2 )8. ( 1 , - 1 ,1 ) X ( - l , 1 , - 1 )

10. (1, 2, 3) X (4, 5, 6).

A force F is applied at point x Find its torque about the origin:11. F = ( - 1 , 1, 1), x = (10,0,0) 13. F - ( - 1 , 1 , 1), x = (2, 2, — 1)

Prove:15. u- ( v X w) = v ( w X u)17. (av) X w = a(v X w)

12. F = (3, 0, 0), 14. F = ( 2 , -1 ,5 ) ,

x = (0, 0, 1) x = ( -7 , 1, 0).

16. (u + v) X w = u X w + v X w18. v X (6w) = 6 ( v X w ) .

19*. Provep*u p ■ V

q-u q*v = (p X q ) ’ (u X w).

20. (cont.) Use this result for a new proof ofu X v s r|2 — (ll‘ v)2.

21. Prove the formula u X (v X w) = (u*w)v — (u- v)w.[Hint: By the linearity of each side and by symmetry, reduce to the cases where u, v, and w are chosen from i and j.]

22. (cont.) Prove the formula(a X b) X (u X v) = [(a X b ) -v ]u - [(a X b ) - u >

= [(u X v)*a]b — [ (u X v)*b]a.Hence show that the left-hand side is a vector along the line of intersection of the plane of a and b with the plane of u and v.

23. (cont.) Show that (a X b) X (a X c) is collinear with a.24. (cont.) Prove the Jacobi identity

u X (vX w) + v X (w X u) + w X (u X v) = 0.

6. APPLICATIONS

Volume

Two non-collinear vectors determine a parallelogram. Three non-coplanar vectors determine a parallelepiped (Fig. 6.1a) whose volume is given by the

formulaV = (area of base) • (height).

u X v

148 4. SOLID ANALYTIC GEOM ETRY

(a) parallelepiped determined (b) h = (projection of w on u X v)by three non-coplanar vectors

Fig. 6.1 volume of a parallelepiped

Suppose the base is the parallelogram determined by u and v; its area is |u X v|. Assume temporarily that w lies on the same side of the u, v-plane as u X v. Then the height is the projection of w on u X v. See Fig. 6.1b. Therefore

V = (projection w on u X v) • |u X v| = (u X v )-w = D ( u, v, w).

If w is on the other side of the u, v-plane, then V = — D (u , v, w). In any case:

The volume of the parallelepiped determined by three non-coplanar vectors u, v, w is given by the formula

7 = |Z>(u, v ,w ) | .

Intersection of Tw o Planes

Given two planes x*m = p and x*n = q, how can we find their line of intersection? We must assume the planes are not parallel, that is, m and n are not collinear. Then u = m X n is perpendicular to both m and n, so u is parallel to the line of intersection (Fig. 6.2).

If we can find a single point x0 on both planes, then the desired line is

x = x 0 + tu

6. Applications 149

Fig. 6.2 Intersection of two planes: m X n is parallel to their line of intersection. To find x0 on this line of intersection and in the plane of m and n.

in parametric form. The figure suggests a point in the plane of m and n, so let x0 = am + bn. Then the conditions x0*m = p and x0*n = q result in a system of linear equations for a and b :

am* m + 6m*n = p

a m*n + 6n*n = q.

The determinant of this system is

m • m m • n% = |m|2 |n|2 — (m*n)2 = |m X n|2 > 0.m*n n*n

Therefore there is a unique solution.

EXAMPLE 6.1

Find the line of intersection of the planes x + y + z = —1 and 2x + y — z = 3.

Solution: The equations of these planes are x • m = p and x • n = qy where

m = ( 1 ,1 ,1 ) , n = (2, 1, - 1 ) , p = - 1 , q = 3.Set

u = m X n = ( - 2 , 3, - 1 ) .

This vector is perpendicular to m and n, so it has the direction of the line of intersection.

To find a point on the line of intersection, set x0 = am + bn and choose a

150 4. SOLID AN ALYTIC GEOM ETRY

and b so that x0* m = — 1 and x0* n = 3. Now

|m|2 = 3, m*n = 2, |n|2 = 6,

and the equations Xo'm = —1 and x0*n = 3 become

3a -f~ 2b = — 1

2a -f- 66 = 3.

The solution is a = — f , b = t t ; therefore

Xo = — f (1 , 1, 1 ) + T T ( 2 , 1, — 1 ) = ( t t , — TT, — f f )•

Answer; x = - t r , — f f ) + t ( — 2, 3, - 1 ) .

Homogeneous Equations

Suppose we are given three planes through the origin,

x*u = 0, x*v = 0, x -w = 0.

In general, the planes will only have 0 in common. However, it may happen that they have a line in common, or even coincide. This occurs precisely when the three normal vectors u, v, w lie in the same plane. The situation can be described algebraically:

A system of three linear homogeneous equations

x * u = 0 , x*v = 0, x*w = 0

has a solution x0 ^ 0 (a non-trivial solution) if and only if

u* (v X w) = D (u , v, w) = 0.

If x0 is any solution, then £x0 is a solution for each t.

It is useful to restate this result in terms of determinants and linear equations.

A homogeneous system of linear equations

UiXi + U2X2 + u 3x 3 = 0

V1X1 + V2X2 + v3x 3 = 0

W1X1 + W2X2 + w 3x 3 = 0

has a solution (xi, x 2) x 3) (0, 0, 0) if and only if

U\ U2 u 3

Vi V2 v3 = 0.

W\ W2 w 3

If (xi, x 2, x 3) is any solution, then (tei, tx2, tx3) is a solution for each t.

6. Applications 151

EXAMPLE 6.2

Find a non-trivial solution of

x*u = x* v = x* w = 0,

where u = (1, — 2, 2), v = (3, 1, —2), and w = (5, —3, 2).

Solution: First

u • (v X w) = D (u, v, w) =

1 - 2 2

3 1 - 2

5 - 3 2

= 0.

(Note that w = 2u + v.) Therefore the vectors u, v, w are coplanar (the parallelepiped collapses) so the corresponding perpendicular planes have a line L in common. Certainly L contains the point 0 which is common to all three planes. Furthermore it contains u X v, since this vector starts at 0 and is parallel to L. Therefore L is the set of all multiples t(u X v), provided u X v ^ 0. A similar statement holds for v X w and w X u.

Note that u X v = (2, 8, 7), while v X w = ( — 4, —16, —14) and w X u = ( — 2, —8, —7), so these three vectors really are collinear.

R emark: Recall that Cramer's Rule (Section 4) guarantees a unique solution if Z)(u, v, w) ^ 0. Since (0 ,0 ,0 ) is obviously a solution to the homogeneous system, it is the only solution when D (u, v, w) ^ 0 . This proves again that for a homogeneous system to have a non-trivial solution, its determinant must be zero.

Skew Lines

Let x = x0 + su and x = y0 + tv be two lines in R3 that do not intersect and are not parallel, i.e., skew lines. We ask how far apart they are (Fig. (6.3a).

The vector u X v i s perpendicular to both lines, so n = (u X v ) / |u X v| is a unit vector perpendicular to both lines. From Fig. 6.3b we see that the required distance is the length of the projection of x0 — yo on n, that is,I (Xo - yo)*n|.

EXAMPLE 6.3

Find the distance between the lines

x = ( — i -f s , s, 2 + 2s) and x = (1 — t, 1 — i, 1 — t).

u X v

(a) skew lines (b) as seen from a direction that makes the lines appear to be parallel

F ig . 6.3

Solution: In this example

u = (1,1,2), V = ( - 1 , - 1 , - 1 ) , Xo= ( - 1 , 0 , 2 ) , y0 = (1,1,1).

Therefore u X v = (1, —1, 0) and

u X v 1

Finally,

n = r ^ n = o V 2 a , - i , o ) .u X V 2

(x0 - y0)*n = ( - 2 , - 1 , !)• J\ / 2 ( l , - 1 , 0) = - i \ / 2 .

Answer: i \ / 2 .

Parametric Form of a Plane

Two non-collinear vectors u and v in R3 determine a unique plane through 0 consisting of all points

X = su + tv,

where s and t are any real numbers. If x0 is any point of R3, then adding x0 to each point of the plane displaces it to a parallel plane through x0. See Fig. 6.4.

Given a point x0 of R3 and two non-collinear vectors u and v, the plane through x0 parallel to the plane of u and v consists of all points

x = x0 + su + tv,

where — oo < s < oo and — oc < t < oo.

6. Applications 153

The variables s and t are called parameters, and a plane presented in this fashion is said to be in parametric form.

Example:

Let x0 = ( — 1, 1, 2), u = (1, 0, 1), v = (1, 1, 0). Clearly neither u nor v is a multiple of the other, so they are not collinear. Then

x = Xo + su + tv = ( - 1 , 1, 2) + s ( l , 0, 1) + t ( l , 1, 0)

= ( — 1 + s + t, 1 + t, 2 + s).

In coordinates,

x = — 1 + s + t, y = 1 + t, z = 2 + s .

Given a plane in parametric form, how do we put it in normal form? We have x = x0 + su + tv, where u, v are linearly independent. A vector perpendicular to both u and v (hence to the plane) is u X v. This vector is guaranteed to be non-zero because u and v are not collinear. Set

n = (u X v )/ 1u X v|.

Then n is a unit vector perpendicular to the plane. Hence

x*n = Xo*n + su*n + t v n = x0*n = p.

This is a normal form.In the example above,

u X v = (1,0, 1) X (1 ,1 ,0) = ( - 1 , 1 , 1 ) ,

n = § V 3 ( - 1 , 1, 1), V = Xo-n = I V 3.

so a normal form is

% \ / 3 ( — Xi + X2 + £3) = i \ / S .


Plane through Three Points

Three non-collinear points x0, Xi, x2 determine a unique plane. If we want a normal form, we argue that u = Xi — x0 and v = x2 — x0 are parallel to the plane, hence m = u X v is perpendicular to it. Therefore x*m = x0*m is an equation of the plane.

If we want a parametric form, then x = x0 + su + £v does the job. Now

Xo + su + tV = Xo + s(X i - Xo) + t ( x 2 — Xo)

— (1 — s — t)x 0 + SXi + tx2,

so we have the alternative symmetric form:

The plane through three non-collinear points x0, Xi, x2 consists of all points

X = SoXo + SiXi + S2X2

where so, Si, s2 take on all real values subject to So + Si + s2 = 1.

Here is a physical interpretation. Put masses s0, si, s2 at x0, Xi, x2 respectively, where s0 + si + s2 = 1. Then x = s0x0 + SiXi + s2x2 is the center of gravity of the masses.

Equilibrium

Forces Fi, • • •, Fn are applied at points Xi, • • •, x n of a rigid body (Fig. 6.5). Now a rigid body is in equilibrium when both the sum of the forces vanishes and the sum of the turning moments (torques) of the forces about 0

6. Applications 155

vanishes. Thus the conditions for equilibrium are the two vector equations:

Fi + F2 + • • • + Fw = 0,

xi X Fx + x2 X F2 + • • • + xn X Fn = 0.

EXERCISES

Find the volume of the parallelepiped determined by1. (1, 1, 0), (0, 1, 1), (1, 0, 1) 2. (4, - 1 , 0), (3, 0, 2), (1, 1,1).

Find the line of intersection in parametric form of:4. x — y + z = 0, X'+ y + z = 3 6. x + y = 1, y + z = — 1.

3. x + 2y + 3z = 0, y — z = 1 5. x 2y z = 3, 2x — y + z = 4

Find a non-trivial solution, if it exists, of

|— 2x + 6y = 07.

[ 3x — 9y = 0

5x + 4y + 3z = 0

—x + 2y + z = 0

3x + y + z = 0

3x — + 2z = 0

11 5x + Qy = 0

x + 5y — 2 = 0

4x + 3y + 5z = 0

13. * —4* — 3y — 5z = 0

12z + 9i/ + 152 = 0

Find an equation for the parametric plane:15. x = (1, s, t) 16. x = (s, s + t, — 1 + 017. x = (2 + «, 1 + s + £, s — 0 18. x = (3s, 2s — t, 1 -j- 2£)*

Find an equation for the plane through the three points:19. (a, 0, 0), (0, b, 0), (0, 0, c), dbc 020. (1, 1, 0 ), (1, 0, 1 ), (0, 1, 1 ).

21. Provea\ bi c\ 2

8.

10.

12.

14.

— I2x + 4y = 0

3x — y = 0

— 2x + 2y + 42 = 0

— 3x — 8y — 52 = 0

— 3x — y + 22 = 0

3z + 3y + 2z = 0

7z + 5?/ + 122 = 0

x + 2y — 3z = 0

6x — 9i/ + 122 = 0

2x — 3y + 42 = 0

— 10x + 15 y — 20 2 = 0.

&2 &2 c2 < (ai2 + a22 + a32)(6i2 + 622 + 632)(d 2 + c22 + c32)

03 &3 C3by interpreting the determinant as a volume.

/


22. A seesaw with unequal arms of lengths a and b is in horizontal equilibrium. Find the relations between weights A and B at the ends and the upward reaction C at the fulcrum.

23. Unit vertical forces act downward at the points pi, • • •, pn of the horizontal x, 2/-plane. A force F acts at another point p of the plane so that the rigid system is in equilibrium. Find F and p.

24. A force F is applied at a point x. Its torque about a point p is (x — p) X F. Suppose Fi, • • •, Fn are applied at points Xi, • • •, xn of a rigid body and the body is in equilibrium. Show that the sum of the torques about p vanishes. (Here p is any point of space, not just 0.)

25. A couple consists of a pair of opposite forces F and — F applied at two different points p and q. Show that the total torque is unchanged if p and q are displaced the same amount, i.e., replaced by p + c and q + c. Interpret this total torque geometrically.

5. Vector Calculus

1. VECTOR FUNCTIONS

In this chapter we study functions whose values are vectors. For example, the position x of a moving particle at time t, or the gravitational force F on an orbiting satellite at time t are vector functions. To indicate that x is a function of time, we write

X = x(<);in components,

x(<) = ( x ( t ) , y ( t ) , z ( t ) ) .

Thus a vector function is a single expression for three ordinary (scalar) functions

x = x ( t ), y = y ( t ), z = z (t ) .

F ig . 1.1

What is the derivative of a vector function? Think of x = x (t) as tracing a path in space (Fig. 1.1). For h small, the difference vector

x ( t + h) — x( t )

represents the secant from x (t) to x (t + h). The difference quotient

x ( t + h) - x(Q h

158 5. VECTOR CALCULUS

represents this (short) secant divided by the small number h. The limit as h --------» 0 is called the derivative of the vector function:

. dx x( t + h ) - x ( t )x ( 0 = — = h m -----------------------

dt h

The derivative x(t ) is a vector in the direction of the tangent to the curve because the tangent is the limiting position of the secant.

To compute the derivative, express all vectors in components:

x ( t + h) - x ( 0 _ 1 + + ■ + _ 2 (<))]h h

= \ { x { t + h) - x ( t ) , y ( t + h) - y ( t ) , z ( t + h) - z { t ) ) h

x ( t + h ) — x ( t ) y ( t + h ) — y ( t ) z i t + h) - z ( t )h h h

It follows that

x ( t -f h) - x(<) lim ---------- ------------

x ( t + h) — x ( t ) z (t + h ) - z ( t ) lim ---------- ----------- , • • •, lim ----------- -----

h-*0

The result is

= ( l i m \h-*0

/ dx dy d z \= ’ dt ’ J t ) '

dx / dx dy d z \ dt ytt J dt ’ dt)

)•

)

dx / dx dy d z \ dt \ d t 1 dt ’ d t ) ’

The derivative of a vector function

x(f) = (x(t ) , y ( t ) , z { t ) )

is the vector function

dxdt

If a particle moves along a path x( t ) , its velocity is the vector function

dxv(<) - a -

The magnitude |v(2)| of the velocity is called the speed. It is a scalar (numerical) function. The direction of v(£) is tangential to the path of motion.

1. Vector Functions 159

EXAMPLE 1.1

The position of a moving particle at time t is (t, t2, tz). Find itsvelocity vector and its speed.

Solution: Let x ( t ) = (t, t2, ts). Then

v ( 0 = x ( 0 = (1, 2t, 3t2),

|v (0 |2 = 1 + (202 + (3<2)2 = 1 + 4<2 + 9<4.

Answer: v(<) = (1, 2t, 3<2),

-v/l + 4i2 + 9 i4.

EXAMPLE 1.2

If x (0 is a vector function whose derivative is zero, show that x (0 = c, a constant vector.

Solution:x ( 0 = ( x \ ( t ) y ± 2 ( t ) y X s ( t ) ) = ( 0 ,0 ,0 ) .

Hence ± i ( t ) = 0, and so X i ( t ) = d (constant) for i = 1, 2, 3. Therefore

x ( 0 = (ci> c2, c3) = c.

R e m a r k : Physically, this example simply says that an object with zero velocity is standing still.

Differentiation Formulas

The following formulas are essential for differentiating vector functions. Each can be verified by differentiating components.

|c/(0x(01 =/(0x(0 +/(0x(0,

dt [ x ( 0 + y (<)] = x(<) + y(t) ,

t , [ x ( 0 * y ( 0 J = x (0 * y (0 + x ( t ) - y ( t ) . dt

d ,— (v X w) = v X w + v X w .dt

7 x[ s ( 0 ] = -y (Chain Rule). dt ds dt

To establish the first formula, for example, write

Then

d_dt E /(0 x(0 ] = ( | c / ( 0 *i(0 l | c / ( 0 **(0 l | [ / ( 0 *.(0 ] )

= ( f ( t )x i ( t ) + f ( t ) x 1 (t) , f ( t )x2 (<) + f ( t ) x 2( t ) , f ( t ) x3(t) + f ( t ) x s (t))

= /(0 (*1 (0 » 2 (0 , *3 (<))+/(<) (*1 (0 , *2 (0 , *3 (0 )

= / ( 0x(0 +/(<)x(0-

EXAMPLE 1.3

Differentiate t2x{t ) , where x (0 = (cos 3£, sin 32, 0*

Solution: Apply the first formula above:

j [ < 2x ( 0 ] = 2 tx (t ) + H ( t ) at

= 2£(cos 32, sin 3£, t) + t2{ — 3 sin 3£, 3 cos St, 1).

Answer: (21 cos 32 — 312 sin 32, 22 sin 32 + 322cos 32, 322).

EXAMPLE 1.4

Suppose x (0 is a moving unit vector. Show that x (t) is always perpendicular to its velocity vector v(2).

Solution: Verify that x it) • v (2) = 0:

But xi2 + x22 + X32 = 1 for every t, since x is a unit vector. Hence x*v = 0.

Alternate Solution:

But by the third differentiation formula on the previous page,

x ( 0 * x ( 0 = |x(<)|2 = 1, | [ x ( 0 - x ( 0 ] = 0.

7 [x(0*x(0] = x(0*x(0 + x ( t ) ' x ( t ) = 2x ( t ) ' x ( t ) .at

Thus x (2) • x (2) = 0, that is, x (2) • v (2) = 0.

2. Space Curves 161

R e m a r k : This example makes sense geometrically. A moving unit vector represents a particle on the unit sphere |x| = 1. Its velocity vector is tangent to the sphere, i.e., perpendicular to the radius.

EXERCISES

Differentiate:1. x (0 = (e‘, e2*, e3t) 2. x(2) = (t\ 25, t«)3. x(0 = (2 + 1, 3 2 - 1, 41) 4. x(2) = (it2, 0, 23).

Find the velocity and the speed:5. x(2) = (t \ t*+ t \ 1) 6. x(2) = (22- 1, 32+ 1 , - 2 1+ 1)7. x(2) = (A coscot, A sinco2, Bt) 8. x(t) = (a + ^ + 62, ^2 + 63).9. Suppose that x = x(t) is a moving point such that x(2) is always perpendicular to

x(2). Show that x(t) moves on a sphere with center at 0.[Hint: Differentiate |x|2.]

d 110. Suppose x(t) 7* 0. Show that — |x(2)| = ^ x*x.

11. Prove the formula ^ Cx ( 0 + y (0 ] = x(2) + y(0-at

d12. Prove the formula — Cx (0 #y (0 ] = x*y + x*y.at

13. Prove the formula ^ -(x X y ) = x X y + x X y .at14. Suppose x(t) is a space curve which does not pass through 0, and that x(20) is the

point of the curve closest to 0. Show that x(fo)# x(2o) = 0.15. Suppose that x(t) and y(r) are two space curves which do not intersect. Suppose

the distance x(t) — y(r) is minimal at t = to and r = to. Show that the vector x(2o) — y(r0) is perpendicular to the tangents to the two curves at x(20) and y(r0), respectively.

2. SPACE CURVES

In this section we study the arc lengths and the tangents of curves in the plane and in space. To avoid analytic difficulties, we shall always assume that the vector functions under consideration have as many continuous derivatives as are needed. This applies to the following sections also.

Length o f a Curve

Let x = x (2) represent a curve in space. How long is the part of the curve between the points x(2o) and x(2i)? To answer this question, we need a reasonable definition of curve length. Intuitively, the velocity vector x(2) is

directed tangent to the curve (Fig. 2.1), and its length |x(£)l represents speed, the rate at which distance s along the curve increases with respect to time.

v«) = x(0

Fig. 2.1

This leads us to define arc length s by

Therefore,

Jt = |v(OI = |x(Ol> s(<o) = o.

( dsY I • /,\|2 \ ( dx dy dz\ I W ■ |X(W

- ( l ) ’+(I) '+(!)’•In terms of differentials,

* - M W +(!) *•This formula has a simple geometric interpretation. See Fig. 2.2. The tiny bit of arc length ds corresponds to three “displacements” dx, dy, and dz along the coordinate axes. By the Distance Formula,

(ds)2 = (dx)2 + (dy)2 + (dz)2.

Fig. 2.2

Divide by (dt)2 and take square roots to obtain

ds dt

This is the time derivative of arc length. Integrate it to obtain the arc length itself.

2. Space Curves 163

Suppose x ( t ) describes a curve in space. Let s( t) denote the length of the curve measured from a fixed initial point. Then

y - s / & + J/2 + 22. at

The length of the curve from x(t0) to x(ti) is

L = f \ / x 2 + y2 + z2 dt. j <0

For plane curves the formula is slightly simpler because z = 0.

Suppose x ( t ) = ( x ( t ) , y ( t ) ) describes a plane curve. The length of the curve from x(fo) to x(£i) is

L = I \ / x 2 + y2 dt.J to

If the curve is the graph of a function y = / ( x ), then its length from(x0, f ( x 0)) to (xi, f ( x i) ) is

l = [ X1 v r + T TJ XQ

)2dx.

The last formula is a special case of the preceding one. Indeed, set x = ty y = f (0 , where x0 < t < x\. Then x = 1 and y = /, so

- = y / W T t f = a A T ? = V I + ( / ' ) 2-

The formula for L follows.

EXAMPLE 2.1

Find the length of the parabola x( t ) = (t, t2), 0 < t < 1.

Solution: This plane curve is a parabola because

x = t} y = t2y

hence y = x2. Its length is

From integral tables,

J y / l + 4 t2 dt = + - ln (2 + \ / 5 ).

Answer: - [ 2 \ / 5 + In(2 + \ / 5 ) l t t 1.479.


EXAMPLE 2.2

Find the length of the curve y sin x for 0 < x < w.

Solution: The length L is given by

L = J yjl + (^ j dx = J y / l + cos2 x dx .

The exact evaluation of this integral is impossible. It can, however, be approximated by Simpson's Rule.

Answer: L I y / l + cos2 x dx t t 3.820. J o

EXAMPLE 2.3

Find the length of the curve x (t) = (t cos t, t sin t, 2t),0 < t < 4x. Sketch the curve.

2. Space Curves 165

Solution: Since

x2 + y2 = (t cos t )2 + {t sin t )2 = t2 = - z2.4

the curve lies on the right circular cone z2 = 4 ( x 2 + y2). As t increases, z steadily increases also, while the projection ( t cos t , t s i n t ) of x ( t ) on the x, £/-plane traces a spiral. Hence the space curve x = x(t) is a spiral on the surface of the cone (Fig. 2.3).

Compute ds/d t :

f d s V = ( d x V f d y V / d z Y \ d t j \ d t j \ d t j \ d t j

= (cos t — t sin t )2 + (sin t + t cos t )2 + (2)2

= 5 + t2.Hence

= T v sJo

1t2 dt — — [ t \ / 5 -|- t2 5 ln (t + \ / 5 “h" 2) ]

Answer: L = - 4wa + 5 ln(4x + a)

where a = y / 5 + 167r2. L

ln 5 j ,

R e m a r k : Suppose the same geometric curve has two different parametrizations. How do we know that we get the same length? We might have x = x ( t ) where to < t < th and x = x ( u ) where the corresponding interval on the w-axis is u0 0. The /-length and the ^-length of the curve are

- rJ to

dxdt

dt andr _ [ U1 \dx

JU " Ju o I dudu.

By the Chain Rule, d x /d u = (dx/dt) (dt /du) . The formula for change of variable in a definite integral implies

f ui I dx dt 7 f tl \dx u / I 7 ■» du — I I

Juo \dt du Jto \dtdt = L t.

This proves that the length of a curve is a geometric quantity, independent of the analytic representation of the curve.

Unit Tangent Vector

EXAMPLE 2.4

Plot the locus x(<) = ( t ° , n


Solution: The locus is the plane curve described by

x = t2, y = tz.Hence

x 3 = y2, y = d=#3/2.

The curve is defined only for x > 0. For each positive value of x, there are two values of y. (See Fig. 2.4.)

Fig. 2.4x

,*3)

Rem ark: The sharp point at the origin is called a cusp. At that point, a particle moving along the curve changes direction abruptly. Note that its velocity at the origin is zero since

v = x = (21, 3/2), v(0) = 0.

In fact, an abrupt change in direction can occur only when the velocity vector is zero. Physically, this seems plausible; a moving particle cannot change direction suddenly unless it slows down to an instantaneous stop at the “corner.”

To avoid such curves with cusps as shown in Fig. 2.4, we study only curves x( t ) for which the velocity x(t ) never equals zero.

Suppose a particle moves along a curve x ( t ) . Its velocity vector v ( t ) = dx/dt has length ds/dt and is directed along the tangent to the curve; hence

I

where T is a unit vector in the tangential direction. But by the Chain Rule,

_ dx _ dx ds dt ds dt

Compare these two expressions for v; the result is

ds dx ds ^ dt ds dt

Therefore dx/ds = T since ds/dt ^ 0 is assumed.

2. Space Curves 167

If x( t ) represents a space curve, then

is the unit tangent vector to the curve. In terms of the velocity vector v,

(It is assumed v ^ O . )

EXAMPLE 2.5

Find the unit tangent vector to the curve x ( t ) = (L t2. ts) at the point x ( l ) = (1, 1, 1).

Solution:

wherev =? x = (1, 2t, St2), |v|2 = 1 + 412 + 91\

Hence

T - - j r + w + m (1- 2‘- W ) -

Now substitute t = 1.

Answer: T = (1, 2, 3).

EXERCISES

1. Find the arc length of x (t) = (ait + bi, a2t + 62) azt + 63) for 0 < t < 1.2. Find the length of x(t) = (t2, ts) for 0 < t < a.

3. Find the length of x(t) = (t, sin t, cos 0 for 0 < t < 27r.4. Set up the length of y = x3 for — 1 < x < 1, but do not evaluate the integral.5. Set up the length of y = axn for xo < x < a*, but do not evaluate the integral.6. Set up the length of x(£) = (tm, 2n, 2r) for 0 < t < 6, but do not evaluate the

integral.7. Find the length of y = — z2 + 2z for —1 < x < 1.8. Carefully plot x(£) = (£2, t* + tb) for t near 0.9. Find the unit tangent T to the curve x(t) = (t, cos t, sin t) at t = 0.

10. Find the unit tangent T to the curve x(t) = (32 — 1, 4£, — 2£ + 1) at any point.11. Find the unit tangent T to the curve x(t) = (a\t + bi, a2t + b2, a3t + 63) at any

point.12. Find the unit tangent T to the curve x (t) = (t cos t, t sin t, 21) at any point.


3. CURVATURE

The curvature of a curve is a quantity which tells how fast the direction of the curve is changing relative to arc length.

The magnitude of the rate of change of the unit tangent T with respect to arc length is called the curvature of a curve, and is denoted by k :

k =dTds

Since the length of T is constant, T changes in direction only. Thus the curvature k measures its rate of change of direction. The curvature is a geometric quantity; it does not depend on how the curve is parametrized.

EXAMPLE 3.1

A curve has curvature zero. What is the curve?

Solution: A natural guess is a straight line. Let is prove this is so. We are given k = 0. Therefore,

hence

d l

d l

= 0,

= 0.

Consequently T is constant,

T — To — (t\, t2, tz).

3. Curvature 169

But dx/ds = T0, which means

dx dy dzT = f = t2, - = U.ds ds ds

*Integrating, we have

x = a + tis, y = b + t2s, z = c +

In vector notation,x = x0 + sT0.

But this is the vector equation of the line through x0 parallel to T0.

Answer: A straight line.

Computation of Curvature

The following three formulas are needed to compute curvature. The first two apply to curves given in parametric form, the third to the graph of a function.

If x = x( t ) is a space curve, then

_ [|x|2 lx |2 - (x« X)2] 1'2

1*1*

If x = (x( t ) j y { t ) ) is a plane curve, then

= xy - yx (x2 + y2y i 2 *

If a plane curve is the graph of a function y = f ( x ) , then

!/" (*)!/c =[1 + / ' ( ^ ) 2]3/2‘

Proof: By the Chain Rule,

ds dx ds ^dt ds dt ’ dt2 ' dt dtds dx ds _ .. d2s _ ds dT d2s / d s \2 dT

x = —-----= — T, x = — T H----------- = — T + I — ) — ,7' ’ 7J9 7' dt2 \ d t ) ds

Hence

ds d2sx* x = ,

dt dt2.•lo • - f dsYw - x ' x = w ■

.. .. / d2s \ 2 , / d s Y I d l 2 / d 2s V / d s V X*X W / (d t ) | ds ~ ( d l 2) \ d t ) ’ ’

*

Consequently

The first formula for k follows.If x = (x(t) , y ( t ) ) is a plane curve, then

t

|x|2 |x|2 - (x* x)2 = (x2 + y2) (x2 + y2) - (xx + y y ) 2

= (xy - y x ) 2y

so the second formula follows.Finally, if the plane curve is the graph of y = /(# ) , apply the second for

mula with t = x and x = ( t , f ( t ) ) = (x, f ( x ) ) . Thenx = 1, x = 0, y = f ' ( x ) , and y = f ” (x), so the third formula follows by direct substitution.

EXAMPLE 3.2

Find the curvature of a circle of radius a .

Solution: Let the equation of the circle be x2 + y2 = a2. Thus

y = zb a2 — x2.

(This equation describes either the upper or lower half of the circle depending on whether the positive or negative square root is chosen.) Differentiate:

x

y


y - xy'y2

y - x ( - x / y )

W

x2 + y2yZ y'

Now

Hence by the formula for curvature,

— a2/y* | a2 1(<a2/ y 2Y 12 a3 a

Alternate Solution: Write

x ( t ) = (a cos t, a sin t).

\

(This describes the circle by its central angle t . ) Then

/ dsX2x = ( — a sin t, a cos t), |x|2 = f —J = ( — a sin t )2 + (a cos t )2 = a2,

3. Curvature 171

Hence

Differentiate:

. ds x = — = a.

1 1 dt

T = — = ( — sin cos 0-x

ds dT dT / .t ; V = ^7 = (~cos< , - s in 0- dt ds dt

Take lengths, substituting a = d s / d t :

= [ ( — cos 0 2 + (~ s in 0 2] 1/2 = 1,dsk =

dTds

1a

Rem ark: The curvature of a circle is the reciprocal of its radius. This is reasonable on two counts. First, the curvature is the same at all points of a circle. Second, it is small for large circles, since the larger the circle the more slowly its direction changes per unit of arc length.

The Unit Normal

The vector dT/ds has length k, the curvature. Therefore

dJds

= /cN,

F ig . 3.1

N


where N is a unit vector in the direction of dT/ds. (We assume k 7 0.) Since T is a unit vector, T is perpendicular to dT/ds; this was shown in Example 1.4. The vector N is called the unit normal vector to the curve (Fig. 3.1).

We summarize:

Let x( t ) represent a curve in space.

— = kN, k = k (s). ds

|T| = | N | = 1, T -N = 0.

The further study of space curves, not pursued here, begins with an analysis of dti /ds. That leads to another quantity, the torsion, which measures how fast the plane of T and N is turning around the tangent line.

EXAMPLE 3.3

Compute T, N and k for the circular spiral (helix) x (t) = (a cos ty a sin bt). Assume a > 0 and b > 0.

Solution: The projection of x (t) on the x, ?/-plane is (a cos t, a sin t, 0). As a particle describes the curve x (t), its projection describes a circle of radiusa. The third component of x ( t ) is bt; the particle moves upward at a steady rate. Thus, the curve is a spiral; it is circular but steadily rising (Fig. 3.2). Differentiate x = (a cos t, a sin t, bt) :

x = ( — a sin t, a cos t, b).

Introduce c > 0 by c2 = a2 + b2. Then

|x|2 = a2 + b2 = c2, ^ = |x| = Cy

and

1 . 1 /T = - x = - ( — a smty a cos t} b ). c c

dTTo find k and N, use the relation &N = — :

ds

3. Curvature 173

ak = — , N = ( — cos —sin t , 0).

c2

Since k > 0 and N is a unit vector,

Answer: &a2 + 62 ’

1\ / a 2 + b2

N = ( — cos t, — sin^O )

( — a sin a cos t , 6),

Rem ark: If 6 = 0, the spiral degenerates into a circle of radius a and the curvature k reduces to 1/a, which agrees with Example 3.2.

EXERCISES

Find the curvature:1. y = x2; at x = 12. x(0 = (£3, £2); at t = 13. x ( 0 = (it, *2, *3); at 2 = - 14. x(2) = (a\t a2t3, bit -f- b2t3, c\t + c2tz)', at t = 0.5. Let x = x(s) be a plane curve. Show that d t i /d s = —kT.

[Hint: Differentiate T*N = 0 and N-N = 1.]6. Find the point of the plane curve y = x2 where k is maximum.

7. Find the point of y = sin x where k is maximum, 0 < x < t .8. Find the curvature of y = xs at x = 0 and at x = 1.9. Show that the curvature of a plane curve at an inflection point is zero.

10. Let the tangent line of a plane curve intersect the z-axis with angle a. Show that k = \da/ds\.[.Hint: Write T = (cos a:, sin a:).]

11. A point moves along the curve y = ex at the rate of 3 in./sec. How fast is the tangent turning when the point is at (2, e2)?

12. Compute the maximum and minimum curvature of an ellipse with semimajor axis a and semiminor axis b. Check the case a = b.

13. From a graph, predict the behavior of the curvature of y = 1/x as x -------- > 0 andas x -------- > oo . Verify your prediction.

4. VELOCITY AND ACCELERATION

If x = x(t) represents the position of a moving particle, its velocity is

v ( 0 = x ( 0

and its acceleration is

a (t) = v (0 = x(<).

Velocity and acceleration are vectors, each having magnitude and direction. The direction of the velocity is the direction the particle is moving. The direction of the acceleration is the direction the particle is turning. The following example shows that the direction of the acceleration is not necessarily that of the velocity; it may even be perpendicular to the velocity.

EXAMPLE 4.1

The path of a particle moving around the circle x2 + y2 = r2 is given by x (t) = (r cos cot, r sin ut), where w is a constant. Find its velocity and acceleration vectors.

Solution: Differentiate to find v and a :

v(£) = x(t) = rco( —sin vt, cos cvt),

a (t) = v(t) = ra>2( — cos co£, — sin coO = —cv2x(t).

Answer: v(t) = ra>( — sin cot, cos wt),

a (t) = rco2( — coscot, —sin c o t).

Remark: The speed, |v| = rco, is constant; the motion is uniform circular motion. The velocity v(£) is perpendicular to the position vector x(t) since x (t) • v (t) = 0. This is expected since each tangent to a circle is perpendicular to the corresponding radius. But a (t) = — co2x( t ) , so the acceleration vector a (t) is directed opposite to the position vector x (t ). See Fig. 4.1. What is the physical meaning of this phenomenon?

4. Velocity and Acceleration 175

v(* + h) - v(0

x(t + h)

\ x ( 0

x

F ig . 4.1 F ig . 4 .2

Remember that a( t ) is the rate of change of the velocity vector. Observe the velocity vectors at t and an instant later at t + h. See Fig. 4.2. The difference y( t + h) — y ( t ) is nearly parallel, but oppositely directed, to x( t ) . Thus the velocity is changing in a direction opposite to x (t). It seems reasonable, therefore, that a (t) = —cx(t) , where c > 0.

But force and acceleration are vectors, both have magnitude and direction. Thus Newton's Law is a vector equation:

EXAMPLE 4.2

A particle of mass m is subject to zero force. What is its trajectory?

Solution: By Newton's Law,

Newton’s Law of Motion

This famous principle states that

force = mass X acceleration.

F = mx.

It is equivalent to three scalar equations for the components:

Fi = m x i , F 2 = m x 2, Fz = mx3.

m x = 0 , x = 0.Since x = v,

Integrate once; v is constant:


dxv = v0, — = v0.

dt

Integrate again:

The result is a straight line.

X = tv0 + Xo.

Answer: The trajectory is a straightline, traversed at constant speed.

Rem ark: Let us check the second integration in components. The equation

dx^ = Vo

means%1 — ^01> #2 = V02, #3 = 03,

where the v0j are constants. Integrating,

X \ = tV0i “H %0h # 2 = tv 02 + # 0 2 > # 3 = tV03 + #03-

Written as a vector equation, this is simply x = tv0 + x0.

EXAMPLE 4.3

A shell is fired at an angle a with the ground. What is its path? Neglect air resistance.

Fig. 4.3

Solution: Draw a figure, taking the axes as indicated (Fig. 4.3). Let v0 be the initial velocity vector, so v0 = ^o(cos a, sin a), where v0 is the initial speed. Let m denote the mass of the shell. The force of gravity at each point is

constant,F = (0, - m g ) .

d2xma = F, that is, — = (0, - g ) .


The equation of motion is

Integrate:

dx— = (0, —gt) + v0.

Integrate again, noting that x0 = 0 by the choice of axes:

x = ( ° ’ ~~ \ g t ) + <Vo'

Hence

( x ( t ) , y ( t ) ) = (o , - ^ g t 2 j + tv0(cos a, sin a)

= ^Vot cos a, v0t sin a — - g t .

To describe the path, eliminate t :

xx = Vot cos a, t =

Vo cos a

1 0y = Vot sin a - - gt2 = x tan a - — -----— x2.

2 2v02 cos2 a

The graph of this quadratic is a parabola.

Answer: x — (t>0 cos a)£, y = (^sina:)/ —

where is the initial speed. The path is a parabola:

02vq2 cos2 a ‘

EXAMPLE 4.4

In Example 4.3, what is the maximum range?

Solution: The shell hits ground when y = 0:

This equation has two roots. The root t = 0 indicates the initial point. We want the other root,

2v0 sin a

9

The range is the value of x at this tim e:


Clearly x is maximum when sin 2a = 1, or a = tt/4. The maximum range is v<?/g.

Rem ark: If the initial speed is doubled, the maximum range is quadrupled. Is this reasonable? (By what factor must the gunpowder be increased to double the initial speed?)

The arc length s, the unit tangent T, the unit normal N, and the curvature k are geometric properties of a curve. If a particle moves on the curve, it is useful to express its velocity and acceleration in terms of these quantities. We already know

which says that the motion is directed along the tangent with speed ds/dt.For further information, differentiate v with respect to time, using the

Chain Rule carefully:

Answer: The maximum range is v02/g- It is obtained by firing at 45°.

Components of Acceleration

This is an important equation in mechanics. It says that the acceleration is composed of two perpendicular components. The first is a tangential com-


ponent with magnitude s, the rate of change of the speed. The second is a normal component, directed along N with magnitude ks2. It is called the centripetal acceleration.

EXAMPLE 4.5

A particle moves along a circle. Find its velocity and acceleration.

Solution: Let r denote the radius and let 0 = 0{t) denote the central angle at time t. Place the circle in the x, 2/-plane with center at 0. Then the path is given by

x(t) = r (cos 0, sin 0).Differentiate:

v = x = r&( — sin0, cos 0).It follows that

dsT = ( — sin 0, cos 0), — = rd = rco(t).

dt

Here co(t) = 6(t) represents the instantaneous angular speed. Thus

v = ro?( — sin0, cos0) = rcoT.Differentiate:

a = v = rcjT + rcot = ro>( —sin0, cos0) + rcv2( — cos 0, — sin0)

= rcST + rco2N.

Answer: v == ro> T, a = ro?T + ro>2N, where w = 6 is the angular speed.

/0

(b)

(a)Fig. 4.4


R e m a r k : When the motion is uniform (co constant), then a = rco2N, so the acceleration is all centripetal, perpendicular to the direction of motion. This agrees with the answer to Example 4.1.

Angular Velocity

A rigid body rotates about an axis a through 0. See Fig. 4.4a. The central angle is 6 = 6 (t), so a> = 6 is the angular speed, the rate of rotation in radians per second.

The angular velocity is defined to be the vector co having magnitude 6 and pointing along the (positive) axis of rotation according to the right- hand rule (Fig. 4.4b).

Suppose the actual velocity v of a point x in the rigid body is required. How can it be expressed in terms of x and the angular velocity co ? See Fig. 4.5.

F ig . 4.5

Since the point x is rotating about the axis of co, its velocity vector v is perpendicular to the plane of co and x. By the right-hand rule, v points in the direction of co X x. The speed |v| is the product of the angular speed a> = |co| and the distance r of x from the axis of rotation. But r = |x| sin 0, hence

|v| = |co| • |x| sin 0 = |co X x|.Therefore:

The velocity of a point x in a rigid body rotating with angular velocity co is

v = co X x.

EXERCISES

1. A hill makes angle ft with the ground (Fig. 4.6). A shell is fired from the base of the hill at angle a with the ground. Show that the ^-component of the position where the shell strikes the hill is x = (2v^/g) (sin a cos a — tan ft cos2 a).

5. Integrals 181

xFig. 4.6

2. (cont.) Find the maximum of x as a function of a. Show that it occurs for

3. Let x(t) = (t, t2). Find v(0 and a (t).4. (cont.) Find the tangential and normal components of a at t = 0 and t = — 1.5. A particle moves along the curve y = sinx with constant speed 1. Find the tan

gential and normal components of a at x = 0 and at x = w/2.6. Find the tangential and normal components of acceleration

for x(t) = (cos t2, sin t2).7. Find the tangential and normal components of acceleration

for x (t) = (a cos wt, a sin cot, bt), where co is constant.8. A particle moves with constant speed 1 on the surface of the unit sphere |x| = 1.

Show that the normal component of the acceleration has magnitude at least 1.9. A particle moves on the surface z = x2 + y2 with constant speed 1. At a certain

instant t0 it passes through 0. Show that the tangential component of a is 0 and the normal component is (x(t0), jj(to), 2). Show also with xx + yy = 0 at t0.

10. Let x = x(t) be a space curve. Show that its curvature is k = |v X a|/|v|3.11. The earth turns on its axis with angular velocity 360° per day. Find the actual

speed (mph) of a point on the surface (a) at the equator, (b) at the 40-th parallel,(c) at the south pole. Approximate the earth by a sphere of radius 4000 miles.

5. INTEGRALS

Suppose a particle moves along a path x = x (t) from x(fo) to x(£i), acted on by a force F = F(£). How much work is done by the force?

From physics we learn that only the component of the force in the direction of motion does work, and that the amount of work done in a small displacement of length ds is

where Fa is the average component of force in the direction of motion.

d W = Fa ds,

F

Fig. 5.1

= x(0

Draw the unit tangent T, the force F, and a small portion of the path of length ds. See Fig. 5.1. Since T is a unit vector in the direction of motion, the component of the force F in the direction of motion is the dot product

HenceF -T .

dW = ( F - T )ds.

Replace ds by (<ds /d t ) dt. This makes good physical sense: The length ds traveled in a short time period dt is given by speed X time = (ds/dt) dt . The result is

. ds

But

hence

dW = F -T ^-dt. dt

ds dxdt dt ’

d W = ( F - v ) dt

We define the total work done by an integral that “adds up” these small bits of work from x (to) to x ( h ) :

Since (dx/dt) dt = dx, the integral can be writtenfx(ti)

W = F-dx.j X(fo)

This type of integral is called a line integral. It arises naturally in connection with work, but has many other practical applications in physics. The evaluation of a line integral involves nothing more than the evaluation of an ordinary integral.

5. Integrals 183

Suppose a particle moves on a curve x (t) from x (to) to x (ti) and is subject to a force F (t). Then the work done by the force is given by the line integral

J to \ dt / yx(*o)

In the integral on the right,

d t )

Let F(£) = (Fi(t) , F2(t), Fs (t)) . Then the line integral is evaluated as an ordinary integral:

EXAMPLE 5.1

rx( 3)Evaluate the line integral / F*dx, where F = (3, — 1, 2) and

Jx( o)

x (0 = (<, t \ tz).

Solution:dx = (1, 2t, 3t2) dt,

F-dx = (3, - 1 , 2)- (1, 2t, 3t2) dt = ( 3 - 2 1 + dt2) dt.

Therefore

f X(3)/*X(3) p/ _ F-dx = ( 3 - 2 1 + 6t2) dt = (31 - t2 + 2ts)

0)= (9 - 9 + 54).

Answer; 54.

EXAMPLE 5.2

Under the action of a force F (t), a particle moves on a path x (0 from x (to) to x (h). Let W denote the work done by the force. From Newton's Law, show that W = \ m |v (^i)|2 — \ m |v ( o)[2.

Solution:

W = I 1 F* x dt. J to

According to Newton's Law, F = mx, so

F*x = m x ' x .

But observe that

Therefore

so

. d . . 2x* x = — (x* x).

dt

- . 1 d -x 1 d . ioF* x = - m — (x* x) = ~ m — v 2,2 d r ' 2 d r 1

p i r fi i d i iT7 = / F* xd f = / ~ ? n - | v | 2d / = - m |v(^i)|2 - - m |v(^o)|2.

./«o J to * dt 2 2

1

Remark: The quantity Jm|v|2 is the kinetic energy of the particle. The result of this example is the Law of Conservation of Energy: work done equals change in kinetic energy.

Integrals of Vector Functions

Next we define the integral of a vector function as a componentwise operation.

Suppose u (t) = M O , u2(t), u3(t)) is defined for a < t < b. Then

la U (t)dt = ( l fa i: Uz (t) dt'j .

Notice that the integral of a vector function is a vector, whereas a line integral is a scalar.

EXAMPLE 5.3

Let u (0 = (1, t — 1, £2). Find J u ( t ) dt.

Solution:

t2) d tJ u (t) dt = J (1, t — 1,

i n « r ,

5. Integrals 185

The integral of a vector function u (t) is particularly easy to evaluate if an antiderivative of u (t) is known.

To prove this, simply check the three components; each is the integral of a derivative. Here is an example:

P f 2d/ (21, 3t \ 4*3) dt = - (t2, t \ t4) dt = (t2, *3,

^<4) = (4, 8, 16).

0

Momentum

The following applications show the importance in physics of vectorvalued integrals.

A particle of mass m has position x = x (t) at time t and moves under the action of a (variable) force F, so that

dxF = m x = m — .

dt

Integrate with respect to t on the interval to < t < t\:

[ 1 Fdt = f 1J <0 J to

dxm — dt ,

dt

hence,

[ tl/ F dt = mx( t \ ) — mx(to). J to

The quantity mx is the momentum of the particle, so the right-hand side is its change in momentum. The left-hand side is called the impulse of the force during the time from t0 to t\. This equation is a form of the Law of Conservation of Momentum: impulse equals change in momentum.

The angular momentum of the particle with respect to the origin 0 is defined as

mx X x.

since x X x = 0. Butmx = F

by Newton’s Law of Motion; hence

mx X x = x X mx = x X F,

which is the torque of F at x. The result is that

d[mx X x] = x X F.

dt

Integrating,

mx X x11 rti

= / x X<0 J <0

F dt.

This result, called the law of Conservation of Angular Momentum, asserts that the change in angular momentum during a motion is the time integral of the torque.

EXERCISES

Evaluate:/*x(i)

1. / (t, 2t, 3t)-dx; x(t) = (1, t, t2)Jx(0)

fx( T)2. / (cos t, sin t)*dx; x(t) = (sin 2, — cos 0

Jx(0)

/* X(2ir)3. / (0, 0, 3t)*dx; x(0 = (cos 2, sin 2, 2 + 1)

Jx(0)

f x (1)4. / (e*, e*, e*)'dx; x(t) = ( t I , t — I, t)

J x i - 1)

A 1.1.D5. / ( t ,—t2, t)»dx; straight path

y (o.o.o)

I* X(27r) / -- y £ \

6' i » , V + 7 ’ i ? + ? l ' h i/* x(i)

7. / xefc + i/<ft/; x(0 = (?, *3)Jx(o)

r (-1.0 4)8. / z dy + y dz + 2 dx; straight path,

y (1.1.2)9. Find the work done by the uniform gravitational field F = (0, 0, — <7) in moving

a particle from (0, 0, 1) to (1, 1, 0 ) along a straight path.

10. Find the work done by the central force field F = — x in moving a particle

from (2, 2, 2) to (1, 1, 1) along a straight path.

Evaluate:

11. j (1 -f- t, 1 -|- 2t, 1 -|- 3/) dt 12. J (cos t, sin t, 1) dt

u i ' 0 - ? . ? ) * •

5. Integrals 187

15*. Let x = x (0, where a < t < b, be a plane curve which does not passthrough (0, 0).

i1 /*x(6)Show that - (x dy — y dx) is the area in Fig. 5.2.2 Jx(a)

Show that t is related to the area A by t = 2A/ab. See Fig. 5.3. [Hint: Use Ex. 15.]

t, b sinh t)

Fig. 5.3

cosh


17. A circle of radius a rolls along the £-axis. Initially its center is (0, a). The point on the circle initially at (0, 0) traces a curve called the cycloid. Show that x(0) = aid — sin 0, 1 — cos 0) is a parametrization of the cycloid, where 0 is the angle at the center of the rolling circle, measured clockwise from the downward vertical to the moving point. Graph the curve.

18. (cont.) Show that the area under one arch of the cycloid is 3 times the area of the circle.[Hint: Express y dx in terms of 0 and integrate.]

19. (cont.) Find the arc length of one arch of the cycloid.20. Show that the “witch of Agnesi” y = 1/(1 + x2) is parameterized

by x = x(0) = 2(cot0, sin20). Sketch the curve.21. Show that the “folium of Descartes” x* + y* = 3axy is parameterized

by x = x(t) = (1 + t*)~l (3at, 3at2). Sketch the curve.22. The force F(2) = (1 — t, 1 — t2, 1 — t3) acts from t = 0 to t = 2. Find its im

pulse.23. The force F(t) = (e*, e2t, eu ) acts from t = —1 to t = 0. Find its impulse.24. An electron in a uniform magnetic field follows the spiral path

x(t) = (a cos t, a sin t, bt).Find its angular momentum with respect to 0.

25. A particle of unit mass moves on the unit sphere |x| = 1 with unit speed. Show that its angular momentum with respect to 0 is a unit vector.

6. POLAR COORDINATES

Review

The polar coordinates of a point x ^ 0 in the plane are the distance r = |x| of x from 0, and the angle 0 from the positive x-axis to the vector x, measured counterclockwise. The angle 0 is determined up to a multiple of 2ir.

We shall write polar coordinates {r, 0} with curly braces to distinguish them from rectangular coordinates ( x , y ) .

Any value of r is allowed, even r = 0 (the origin) and negative r; the point { — r, 0} is the reflection of {r, 0} through the origin and is identical with {r, 0 + 71-}. Note that 0 is undefined at the origin.

The rectangular coordinates (x, y ) of the point with polar coordinates {r, 0} are given by

x = r cos 0

y = r sin 0.

Conversely, given (x, y ) we find {r, 0} from

6. Polar Coordinates 189

(The single formula for the angle, tan 0 = y / x , is not adequate to distinguish quadrants. For example, tan 0 = tan t = 0.)

The number r is called the radius of the point x = {r, 6} and 0 is called the polar angle.

A curve may be presented by a relation between r and 0, often in the form r = f (6). For example, r = a is the equation of the circle of radius a with center 0. The graph of r = cos 0 is also a circle, but with center (J, 0) and radius §, because of the computation

r = cos 0, r2 = r cos 0, x2 + y2 = x,

(x2 - x ) + y2 = 0, ( x - 0 + y2 = ^ •

Length

Suppose a curve is given in the form

r = r( t) , 6 = 6 [t), U < t < t%.

What is the length? Write

x = (r cos 6, r sin 6) = r (cos 0, sin 6).Differentiate:

x = r(cos0, sin0) + r0( —sin0, cos0) = ru + r0w,

whereu = (cos 0, sin 0) and w = ( — sin 0, cos 0).

Now

= (ru + r0w)* (ru + r0w)

= r2u*u + 2rr0u*w + r202w*w.

But u*u = 1, w*w = 1, and u*w = 0. Hence,

(!) - ' ■ +

It follows that

Length = j y / f 2 + r202 dt.J to

Figure 6.1 provides an aid to memory. The “right triangle” has sides dr, r dd, and ds, so the Pythagorean Theorem suggests

(ds)2 = (dr)2 + r2 (dd)2.

y

Fig. 6.1

x

Suppose a curve is given by r = r (6). This is a special case of the previous situation with 0 = t and r = r(t ) . The length formula specializes to

Area

There are problems that require the area swept out by the segment joining 0 to a moving point on a curve (Fig. 6.2).

Suppose the curve is given by

r = r(t)y 0 = 0 (0 , t o < t < h .

In a small time interval dt, a thin triangle of base r dd and height r (ignoring

EXAMPLE 6.1

Find the length of the spiral r = 02, 0 < 0 < 2tt.

Solution:

8 8 Answer: L = - (ir2 + 1)3/2 — ~

o o

Fig. 6.2

negligible errors) is swept out (Fig. 6.3). Hence

1 1 dddA = - r 2 dd = - r2 — dt,

2 2 dt

Fig. 6.3

In the special case that the curve is given by r = r (6) for do < 0 < 0\, choose t = 0. The, formula specializes to

EXAMPLE 6.2

Find the area of the “four-petal rose” r = a cos 26.

Solution: Graph the curve carefully (Fig. 6.4). The portion on which0 < 0 < t / 2 is emphasized. Note that r < 0 for 7r/4 < 0 < ir/2. Because of symmetry it suffices to find the area of half of one petal. Thus

f r /4 [ * /4 f r / 2A = 8 / - (a cos 20)2 dd= 4a2 / cos2 20 J0 = 2a2 / cos2 1 dt.

J o 2 7 0

Answer; A7ra

Fig. 6.4

Summary

Suppose a curve is given in polar coordinates by

r = r( t) , 0 = 0 (0 , U < t < t \ r = r(0),

a r c l e n g t h :

< 0 < 0i.

a r e a :

H i 1 ,70 f d 1 1A - - * « , ) * > .

EXERCISES

Express in rectangular coordinates:1. { l ,i /2 } 2. { — 1, 3ir/2}

7. Polar Velocity and Acceleration 193

3. {2, tt/4} 5. {1, tt/6}

4. { -4 , 3tt/4} 6. {2, 5tt/6}.

Express in polar coordinates:7. ( 1 , - 1 )9. ( - V 3 , 1)

8. ( - 1 , - 1 )10. ( y/3/2).

11. Find the length of the “spiral of Archimedes” r = ad from 6 = 0 to 0 = 1.12. Set up an integral for the length of the “four-petal rose” r = a cos 20.13. Set up an integral for the length of the “three-petal rose” r = a cos 36.14. Find the length of the “one-petal rose” r = a sin 6. Precisely what is this curve?15. Find the area enclosed by the “three-petal rose” r = a cos 36.16. Find the area enclosed by the “ (2n + 1 )-petal rose” r = a cos (2n + 1 )6.17. Find the area enclosed by the “4n-petal rose” r = a cos 2nd.18. Find the area enclosed by the curve r = a cos2 2nd.19. Find the area enclosed by the “cardioid” r = a( 1 — cos6).20. Show that the “cissoid of Diodes” y2 = x3/ (a — x) can be expressed in polar

coordinates by r = a (sec 6 — cos 6); sketch the curve.21. Find the area enclosed by the figure eight, the “lemniscate of Bernoulli”

r2 = a2 cos 26. Sketch the curve.22. Sketch the “strophoid” r = a cos 26 sec 6. Find the area of the closed loop.23. Sketch the “limagon of Pascal” r = b + a cos 6 in the three cases 0 < a < 6,

0 < a = b, and 0 <. b < a. In the third case compute the area between the two

as the path of a particle. What are its velocity and acceleration vectors? In the above discussion of length, the perpendicular unit vectors

were introduced. In using polar coordinates, it is natural to express the velocity and acceleration in terms of these vectors (Fig. 7.1). Note that x = ru and

loops.

7. POLAR VELOCITY AND ACCELERATION [optional]

Let us think of a curve given by

r = r( 0 , 0 = 0(0

u = (cos 0, sin 0) and w = ( — sin 0, cos0)

thatu = 6w, w = — 0u.

Differentiate x = ru:

v = x = ru + rd w.


a = v = (ru + rd w) + (r0w + r0w — r02u).

Hencea = (r — rd2) u + (rd + 2r0)w.

Let us apply this formula to motion involving a central force

F = f ( t ) u .

At each instant, the force is directed toward or away from the origin. Since ma = F, the component of a in the direction of w is zero:

rd + 2rd = 0, that is, r2d + rfd = 0.


2

This is the same as

But

— ( - r2d J = 0, that is, - r2d = constant. dt \2 / 2

1 . dA- r2d = — ,2 dt

the rate at which central area is swept out by the curve. It follows that the same area is swept out in equal time anywhere along the path. This is Kepler’s Second Planetary Law; it is a case of the Law of Conservation of Angular Momentum.

Summary

Suppose a curve is given in polar coordinates by

r = r ( t) , d = d(t).

VELOCITY:

7. Polar Velocity and Acceleration 195

ACCELERATION I

a = f — - r ^—Y l u + Tr — + 2 — —lw[_dt2 \ d t ) J L dt2 dt d t j 1

where u = (cos 0, sin 0) and w = ( — sin 6, cos 8).

EXERCISES

This set of exercises develops Kepler’s First and Third Laws of Planetary Motion. Assume a particle of unit mass is moving in a central force field given by the inverse square law:

F - - I u .r2

J 2 11. Show that the equations of motion are r28 = J, r ------ = ---- -, where J is ar r2constant.

J 2 22. Show that r2 -\— r- = - + C, where C is a constant. This equation is essentially ther r

Law of Conservation of Energy.[Hint: Multiply the second equation in Ex. 1 by f and integrate.]

3. Set p = - . Show that + J 2p2 = 2p + C.r p

4. Imagine 6 = 8{t) solved for t as a function of 8 and this substituted into p = p(t). Thus p may be considered as a function of 8. Show that

■p (I)2 = p - and conclude that P [(I)2+p2] =2p + C.

d?p 15. Show that + p = -==. [Hint: Differentiate the previous relation.]du J 1

6. Show that p = A cos0+ Bsin8- \- l / J 2, where A and B are constants, is a solution of the preceding differential equation. (In Chapter 14 it is shown that every solution is of this type.)

7. Show that by a suitable choice of the z-axis, the solution may be written

- = -ji (1 — e cos 8), where e is a constant, the eccentricity of the orbit, e > 0.T J

8. By passing to rectangular coordinates, show that the orbit is a conic section.9. Suppose e = 0. Show that the orbit is a circle with center at 0, and that the speed

is constant.10. Suppose 6 = 1 . Show the orbit is a parabola with focus at the origin and opening

in the positive z-direction.

11. Suppose e > 1. Show that the orbit is a branch of a hyperbola with one focus at the origin.

(x c )2 y212. Suppose 0 < e < 1. Show that the orbit is the ellipse----^------ \- ^ = 1, where

J2 J2where a = ------ 5, b = . .... - , and c = ae.1 — eL v 1 — e2

[By Ex. 8-12, each closed orbit is an ellipse (or circle), Kepler’s First Law.]13. (cont.) Show that a2 = b2 + c2. Conclude that the foci of the ellipse are (0, 0)

and (2c, 0).14. (cont.) Let T denote the period of the orbit, the time necessary for a complete

revolution. Show that ~ T = t ab. [Hint: Use Kepler’s Second Law.] z

15. Conclude that T2 = 47r2a3. This is Kepler’s Third Law: The square of the period of a planetary orbit is proportional to the cube of its semimajor axis.

6 . Functions of Several Variables

1. INTRODUCTION

Elementary calculus is concerned with functions such as y = f ( x ) , where one quantity y depends on another quantity x. In all sorts of situations, however, a quantity may depend on several variables. Here are two examples:

(1) The speed v of sound in an ideal gas is

where D is the density of the gas, p is the pressure, and 7 is a constant characteristic of the gas. Then v depends on (is a function of) the two variables p and D. We may write

v = f ( p , D ) , or, v = v(p, D ) .

(2 ) The area of a triangle with sides x, y, z is

A = \ / s (s - *) (s — y ) (s - 2),

where s is the semiperimeter \ (x + y + z ). Then A depends on the three variables x, y, and z. We may write

Note that x, y, z are not three arbitrary numbers but must satisfy the inequalities x > 0 , y > 0 , z > 0 and z < x + 2/, x < y + z, y < z + x.

In Example (1), the quantity v is a function of p and D defined for a certain set of pairs (p, D) , which we can think of as a subset of the p, Z)-plane. In Example (2 ), the area A is a function of x, y, z defined for a certain set of points (x, y , z ) in space.

In general, let S be any subset of the plane R2 or space R3, and let / be a real-valued function defined on S. Thus / assigns a real number to each point in S. We write

and we say that S is the domain of /. Alternatively, we say that / is a function with domain S. Here R denotes as usual the set of all real numbers.

A = f ( x , y , z ) , or A = A ( x , y , z ) .

198 6. FUNCTIONS OF SEVERAL VARIABLES

We want to extend the concepts of one-variable calculus to functions of several variables, concepts such as continuity, derivative, and integral. Now in the one-variable situation, in order for these concepts to be meaningful, the domain of a function has to be a reasonably nice set, generally an interval or the union of several intervals. For example, the integral is usually defined for a function on a closed interval.

For a function of several variables, the nature of the domain is equally important. Therefore, we devote the next section to the study of properties of useful domains.

Set Notation

Let us review some standard notation that is useful in dealing with domains. The notation S = {# | P{ x ) } is read “S is the set of all x with property

P ( x ) ”. For example, an open interval is defined by

(a, 6) = [x | a < x < b}.

We even allow infinite open intervals such as

(a, oo ) = {x \ a < x}.

A closed interval is defined by

[a, 6] = {x \ a < x < b}.

The square brackets indicate that end points are included, the round brackets that they are excluded. Sometimes we used mixed types such as

[a, b) = {x \ a < x < b}.

Each of these sets is a subset of the reals R, and we write, for instance, (a, b) C R. In general, S C T . i s read “S is a subset of T”, and it means that each point of S is a point of T.

If x is a point of S, we write x 6 S and read “x belongs to S”.Given two sets S and T, we can form two other sets from them. First, their

intersection isS n T = | £ S and x G T}.

It consists of all points both in S and T. For example, let a < b < c < d. Then

(a, c) fl (b, d) = (b, c).

This means that both a < x < c and b < x < d if and only if b < x < c. The union (also called join) of S and T is

S U T = { # | £ £ S or x £ T or both}.

It consists of all the points of S and all the points of T thrown together. For example, the domain of F (x ) = y / x 2 — I is

{# | £ < — 1 or £ > ! } = ( — ° ° , — 1 ]U [ 1 , oo ).

2. Domains 199

Another example is( - o o , 1) U ( - 1 , oo) = ( - 0 0 , oo) = R.

In this case the two sets overlap, indeed, ( — » , 1) n ( — 1, oo) = ( —1,1). Still, each point of R is in either ( — oo, 1) or ( — 1, oo), some are in both.

In set theory there is something called the empty set, the set with no points at all. We have no real use for this here, so it will be understood, without further notice, that whenever we say “set” we mean a “non-empty set” , a set with at least one point.

2. DOMAINS

In this section, we discuss closed sets and open sets in the plane and in space. These are sets that share certain basic properties with closed and open intervals on the line.

A closed interval D = [a, 6] has this property: if x n G D and x n -------- » x,then x £ D. In other words if a sequence of points in D converges, it converges to a point in D. Not all intervals have this property. For example, take D = (0, 1) and x n = 1/n. Then x n £ D, the sequence {#n} converges, but lim x n D.

Convergence

We need the notion of convergence of a sequence of points in space. Henceforth, to avoid repetition, we shall use the word “space” to mean either two-space (plane) R2 or three-space R3.

The definition of convergence in space looks just like the definition on the line with the “nearness” of points x and y measured by |x — y|, the distance between the points.

Definition Let {xn} be a sequence of points in space and a another point. We say {xn} converges to a, and write

xn ------------> a or a = lim xnn - + oo

provided |xn — a|-------- > 0 as n -------- > oo.

From the given sequence {xn} and point a, we construct a new sequence { |xn — a | } of real numbers and we ask whether this sequence converges to 0.If yes, we say xn ---------> a. Just how close this definition is to the old definitionof convergence for sequences of real numbers is seen in the following result, which interprets convergence in terms of coordinates.

Theorem Let xn = (xn, y n, zn) and a = (a, b, c ). Then xn--------->aif and only if x n -------- » a, y n -------- * b, and zn -------- > c.


Proof: Suppose x n -------- > a, y n -------- > 6, and zn -------- > c. ThenIx n — a \ -------- > 0 so (;xn — a )2 -------- * 0. Similarly, (yn — b)2 -------- > 0 and(zn — c)2 -------- > 0. Therefore

|x„ - a|2 = (xn - a )2 + (yn ~ b)2 + (zn - c )2 -------- > 0,

and it follows that |xn — a | -------- » 0 so xn --------> a.Conversely, let xn-------- > a. Then |xn — a | -------- > 0. But

I Xn - a|2 = (xn - a ) 2 + (y n - b )2 + (zn - c )2 > (xn - a )2,

so |xn — a| > \xn — a\ > 0 and it follows that \xn — a \ -------- » 0, that is,x n -------- > a. Similarly, ?/n --------» 6 and -------- * c.

Closed Sets

Now that we know what it means for a sequence of points to converge, we can define closed sets.

Definition A set S in space is closed provided each limit of a sequence of points taken from S is itself in S. That is, if xn £ S for n — 1, 2, 3, • • • and if xn-------- > x, then x f S .

The intuitive idea of a closed set is a clump S of points in space which include all of its boundary points. Any point of space that you can “sneak up on” by points of S must be a point of S. The following examples should help.

EXAMPLE 2.1 (Closed Half-Plane)

Let a, by c be given with a2 + b2 ^ 0, and let

H = { (x, y ) | ax + by > c\.

Prove that H is a closed set.

(a) closed half-plane; (b) elliptical regionsee Example 2.1 see Example 2.2

Fig. 2.1

2. Domains 201

Solution: Suppose x „ f H and xn -------- >x. Prove th a t x 6 H. Nowxn = (#», 2/n)-------- > x = (#,?/) and a#n + byn > c. Since xn -------- > x, wehave #n -------- > x and yn -------- > y. Therefore a#n + byn -------- » ax + by.But axn + byn > c, hence ax + by > c, th a t is, (x, y) £ H. See Fig. 2.1a.

EXAMPLE 2.2

Let a > 0 and b > 0 and set

E = (*, y) - + - x and 2/n -------- » ?/, hence

&»* _____ ^z2 ^a2 62 * a2 D2 ’

B ut #n2/a 2 + yn2/b2 < 1. Therefore x2/a 2 + y2/b 2 < 1 so (x, y) G E. See Fig. 2.1b.

Let S and T be closed sets th a t have common points. Prove th a t S fl T is closed.

Solution: Let xn £ S n T and xn -------- > x. To prove: x £ S flT. Nowxn £ S and S is closed, so x £ S. Likewise x 6 T, so x £ S n T.

(a) intersection of five closed half-planes

(b) intersection oftwo closed half-planes and an elliptical lamina

(c) intersection of three elliptical laminas

F ig. 2.2 An intersection of closed sets— again a closed set.

R e m a r k : I t follows easily th a t the intersection of any number of closed sets, if non-empty, is another closed set. We can use this result to construct many closed sets. See Fig. 2.2 for some examples.

EXAMPLE 2.4

Let S and T be closed sets. Prove th a t S U T is a closed set.

Solution: Let xn £ S UT and xn-------- > x. To prove: x £ S UT. Noweither infinitely many of the points xn belong to S or infinitely many belong to T (or bo th). For otherwise there would be only a finite number of points xn. Suppose infinitely many belong to S. T hat means there is a subsequence{xni} of {xn} with xnj £ S. Since xn-------- > x, then also xnj -------- > x. B ut S isclosed, so x £ S. Hence x £ S U T.

Open Sets

Definition Let S be a set in space and x0 a point of S. Then x0 is called an interior point of S provided for some 8 > 0,

{x | |x — x0| < 5} C S.

Interior points are im portant because we have a fighting chance to define derivatives a t interior points of the domain of a function. See Fig. 2.3 for some plane examples th a t illustrate the concept. Note th a t if x0 is fixed, then {x | |x — x0| < 5} is the circular disk of radius 8 and center x0, without the boundary circle.

Fig. 2.3

D = {(«, y) | 0 < x < 1, 0 < y < 1}, unit square

{II, V, w are interior points

x, y are not

2. Domains 203

In the figure, u is an interior point of D because there is a circular disk centered a t u and entirely inside D. Likewise v and w are interior points of D. But x is not an interior point of D because any circular disk centered a t x has points outside of D. Likewise y is not an interior point. The interior points are precisely those points (x, y) where 0 < x < 1 and 0 < y < 1. The remaining points of D are not interior points. They are the points of D where x = 0 or x = 1 or y = 0 or ?/ = 1, th a t is, the points on the boundary of the square.

Now consider Fig. 2.4. The set D is the unit disk without its boundary. Every point of D is an interior point! For if |x0| < 1, then 8 = 1 — |x0| > 0, and if |x — x0| < 8, then

Ix| = I (x - Xo) + Xo| < |x - Xo| + |x01 < 8 + |x0| = 1,

so x 6 D.

D = {(x, y) \x 2 + y2 < 1}, open unit disk

Fig. 2.4 Every point of D is an interior point.

In R3, the set {x | |x — x0| < 8} is a ball of radius 8 and center x0, consisting of all points inside the spherical surface |x — x0| = 8.

Definition A set D is open if every point of D is an interior point.

The set in Fig. 2.4 is open; th a t in Fig. 2.3 is not open because the points on the boundary of the square are not interior points, bu t they are points of the set.

EXAMPLE 2.5

Let a > 0 and b > 0, and set

(*> y) - + f- < 1 a2 62

Solution: Suppose (x0, yo) £ D. To prove: (#o, 2/0) is an interior point of D, th a t is, th a t there is a positive 8 so small th a t (x, y) £ D whenever |(s , y) - (x<>, y0)I < S.

We have x02/a 2 + yo2/b 2 < 1. The idea is to choose 8 so small th a t x2/a 2 + y2/b 2 is very close to x02/a 2 + y02/b 2 whenever | (x , y) — (x0, yo)\ < 8. How close? Closer than e = 1 — (x02/a 2 + yo2/b 2). Then x2/a 2 + y2/b 2 must be less than 1.

Now f ( x ) = x2/a 2 and g(y) = y2/b 2 are continuous, so we can make 1/0*0 — f ( xo)l < h and |g(y) — g(yo)\ < \e by demanding th a t x be sufficiently near to x0 and y to y0, say |# — rc0| < 8 and |y — y0\ < 8. Then

i / x 2 y2\ _ , y f \I \ a 2 b2) \ a 2 b2 ) = I/(« ) - / Oo) + g(y) - g(yo)\

< I / (« ) - / ( « o)| + Ig (y) - g(y0)\

< h + h = «•

W a r n i n g : Despite what the words seem to suggest, not every set is either open or closed. For example, take a disk in the plane together with some, bu t not all, of its boundary points.

For a more unusual example let

S = {(x, y) \ x and y are rational numbers}.

Then S is very far from being either open or closed. Can you see why?

-> a and yn - -> c and xn - -» x and yn -

—> b. Prove that xn + yn ■ -> a. Prove that cnxn------

-> x. Prove |xn| • -» x and yn -----

-^y. Proveycn*yn - ------>1x1.

1. Let xn-2. Let cn —3. Let xn -4. Let xn -5. Let xn -

Prove each of the following sets is closed and sketch:6. { ( x , y ) \ y < x2\7. { { x , y ) \ y > x2}8. {(x, y) | x2/a 2 + y2/b2 > 1}

EXERCISES

► a b.

-> y. Prove xn Xy„*

x-y.

■^xXy.

3. Continuity 205

9. { (x, y ) | x2/a2 — y2/b2 < 1}10. {(x, y, z) I x2 + y1 + z2 < 1}.

11. Prove that R3 is a closed subset of R3.12. Show that {x | 0 < |x| < 1} is not a closed set.13. Prove that {(x, y) | 0 < x < 1} is an open set. Sketch it.14. Prove that S U T is open if S and T are.15. Suppose S and T are open and have common points. Prove that S fl T is open.

16*. Prove that S is closed if and only if its complement (the rest of space) is openor empty.

Is the set open or closed or neither?17. { {x, y ) \ 2 x + Z y > 1}

118. the domain of the function f ( x , y) = x — y19. {(x, y) | 0 < x < 4, 2 < y < 3}20. {(x, y) j 0 < y < ex\21. {(x, y ) | x is an integer}22. {(x, y) I neither x nor y is an integer}.

23*. Prove the Cauchy criterion: A sequence {xn} of points in R3 converges if and only if for each e > 0, there exists a positive integer N such that |xm — xn| < e whenever m, n > N. [Hint: Use the one variable case, p. 5.]

24*. Prove the Bolzano-Weierstrass Theorem: If {xn} is a bounded sequence of points in R3, then there is a convergent subsequence. You may presuppose the result for R.

3. CONTINUITY

For a function of several variables to be useful, it must have some reasonable properties. The most basic of such properties is continuity. Here is the formal definition, a direct generalization of the definition of continuity for a function of one variable.

Definition Let / : D -------- > R, where D, the domain o f / , is a subset of R2or R3. Let a be a point of D. We say / is continuous a t a if / ( x ) -------- ►/(a)as x --------- > a. Precisely, for each e > 0 there exists 8 > 0 such th a t| / ( x ) — / ( a ) | < e whenever x £ D and |x — a| < 8.We say / is continuous on D if / is continuous a t each point of D.

As for functions of one variable, this definition requires th a t a continuous function be “predictable” ; you should be able to predict the value of the function a t a from its values near a.

The elementary properties of continuous functions of one variable carry over to this case easily. In particular, sums, products, and quotients (with nonzero denominator) of continuous functions are continuous.


Obviously the functions defined by f (x , y) = x and g(x, y) = y are continuous. By forming products and sums we conclude th a t each 'polynomial is a continuous function on R2. (Of course the same holds on R3.) From this we deduce th a t each rational function is continuous wherever the denominator is not zero. (Recall th a t a rational function is a quotient of polynomials.)

I t is also not hard to see th a t if f ( x ) and g{y) are continuous functions of one variable, then h( x , y ) = f {x)g (y) is a continuous function of two variables. (See Ex. 2.) For example, the continuity of h ( x , y ) = x In y follows from th a t of f ( x ) = x and g(y) = In y.

Composite Functions

Suppose we wanted to prove th a t f (x , y) = yx is continuous on the domain x > 0, y > 0. We could write

f (x , y) = exlny = e0(x’v).

T h u s / (x, y) the composite of the continuous functions h (t) = el and t = x In y. I t seems reasonable th a t f (x , y) is continuous also.

Here is a more complicated type of example. Suppose we somehow manage to prove th a t

K(x, y , z ) = I (yz + t4) sin (zt2) dt Jo

is continuous in x , y, 2-space. We w ant to conclude th a t K (u v, vu, uv) is continuous on the domain u > 0, v > 0. W hat we need is the following theorem.

Theorem Let K(x , y, z) be continuous on a domain D in x, y, 2-space. Let / , g, h be continuous on a domain E of the u, y-plane, and suppose th a t ( f ( u , v ) , g ( u , v ) , h ( u , v ) ) £ D whenever (u, v) £ E. Then the composite function

k(u, v ) = K [ f ( u , v), g(u, v), h(u, v )]

is continuous on E.

Proof: Let (u , v ) -------- > (uo, v0). Then f (u, v ) -------- >/(wo, ^o),g(u, v ) -------- > g(uo, Vo), and h(u, v ) -------- » h(uo, v0). Hence

( / (« , v) ,g(u, v ) , h(u , v ) ) -------- > ( f ( u o, Vo),g(uo, v0), h(uo, v0)).

B ut K is continuous, so

k(u, v) = K [ f ( u , v), g(u, v), h(u, t>)]

-------- > Vo), g(uo, Vo), h (u0, «o)] = k(uo, v0),

therefore k is continuous.

N o t e : The theorem above is stated for a function of three variables, where each variable is replaced by a function of two variables. Clearly, there

3. Continuity 207

is nothing special about three and two, and the result may be modified as needed.

Maxima and Minima

One of the main concerns of calculus is maximum and minimum values of functions. Recall one of the basic facts about continuous functions of one variable:

If / is continuous on a closed interval [a, 6], then there exist points x0 and x\ in the interval such th a t

f i x o) < f i x ) < f i x i)for all x £ [a, 6].

The result says th a t

f(xo) = min{ f i x ) | a < x <b } ,

f i x i) = max{ f i x ) | a < x < b}.

If the interval is not closed, then / need not have a maximum or a minimum. For example, f i x ) — x has neither a maximum nor a minimum on the open interval (a, b). The same holds for any continuous increasing or decreasing function.

Furthermore, the result is not true on a domain which is unbounded, th a t is, contains points arbitrarily far from the origin. For example, on the domain [0, °°), the function f i x ) = e~x has a maximum but no minimum; on (0, oo ) it has neither.

The correct generalization of the preceding theorem requires a domain th a t is both closed and bounded.

Definition A subset S of space is bounded if there is a number B such th a t |x| < B for all x £ S.

Thus a set is bounded provided it is contained in some sphere of finite radius centered a t the origin.

Theorem Let S be a bounded and closed subset of space and let / have domain S. Then there exist points x0 and Xi of S such th a t

/(Xo) < f i x ) < / ( x :l)

for all x £ S.

A complete proof of this fundamental result is best postponed to a more advanced course; here is a brief sketch of a proof. Let M = sup { / (x) | x £ S }. (Possibly M = oo.) Choose xn = (xn, yn, zn) £ S so th a t / ( x n) -------- ► M .

The sequence {xn} is bounded (because each xn is in S and S is bounded). Therefore {xn} has a convergent subsequence. We may restrict attention to this subsequence only. Hence we may assume x n -------- > a.

Next we look a t {yn}- Again, passing to a subsequence, we may assumeyn -------- > b. Similarly we may assume zn -------- > c and hence xn -------- > a =(a, b, c). Then a £ S because S is closed, and f ( x n) -------- >f(a) because / iscontinuous. But / ( x n) -------- > M , therefore M = / ( a). A similar argumentapplies to the minimum.

Uniform Continuity

The final property of continuous functions we shall consider is the property of uniform continuity. I t is im portant in the theory of integration.

Definition Let / be defined on a set S. Then / is called uniformly continuous on S if for each e > 0 there is a 8 > 0 such th a t

l / (x ) - / ( z ) | < €

whenever x and z are in S and |x — z| < 8.

At first reading, this definition may appear to be the same as the definition of continuity. The point, however, is th a t given e, the same 8 works throughout the domain of /. In other words, the 8 given by the definition of continuity is independent of the point x.

I t is rather obvious th a t each uniformly continuous function is continuous. The converse is not true in general (see Ex. 5). The converse is true, however, provided the domain is closed and bounded.

Theorem Let S be a bounded and closed subset of space and let / be continuous on S. Then / is uniformly continuous on S.

The proof of this theorem, like the last one, is best postponed to a later course. However, for the brave we offer the following: Suppose for some e > 0 there is no 8 > 0 th a t fills the bill. Then we can choose xra £ S and yn £ Ssuch th a t |xn — yn\ -------- » 0 and | / ( x n) — / ( y n) | > e. As in the proof sketchedfor the previous theorem, by passing to a subsequence we may assumexn -------- >a £ S. Then yn-------- >a also, since yn = xn + (y„ — xn) and|y» - xn\ -------- > 0. Therefore / ( x n) -------- > /(a) and / ( yn) -------- ►/(a) since/ is continuous a t a. This contradicts | / ( x n) — f ( y n)\ > e for all n.

EXERCISES

1. Prove that the sum of two continuous functions is continuous.2. If f{x) and g(y) are continuous, prove that h(x,y) = f(x)g(y) is continuous.3. Let/ (x) = |x| for x £ R3. Prove th a t/(x ) is continuous using properties of length.

4. Graphs 209

4. (cont.) Do Ex. 3 using composite functions.5. Prove that f i x , y ) — 1 / (x + y) is continuous on the open first quadrant x > 0,

y > 0, but is not uniformly continuous.6*. (cont.) Prove that f ix, y) is uniformly continuous on the domain x > 1 , y > 1 .

Suppose/ (x, y ) is continuous on R2 and c is constant. Prove:

7. The set { ( x , y ) \ f ( x , y ) > c} is open (if non-empty).8. The set {(x, y) \f(x, y) = c} is closed (if non-empty).9. By what general principles do you know that sin § (x + y) has a minimum value

on the disk x2 + y2 < 1?10. Let S be a closed bounded set in R3 and let x0 be a point not in S. Prove there is a

point in S closest to x0. [Hint: Ex. 3.]11. (cont.) Show by example that there may be more than one point of S “closest” to x0.

12*. Let /(x ) be continuous for all x £ R3. Suppose that /(x ) = 0 for some x but/ ( 0) 7 0. Prove there is a point nearest to 0 where/(x) = 0.

13. Prove that f ( x , y ) = x2 — 6xy + 10y2 has a positive minimum value p on the circle x2 + y2 = 1 .

14. (cont.) If f ( x , y ) = ax2 + bxy + cy2 has a positive minimum p on x2 + y2 = 1 , prove that/Or, y) > 0 for all (x, y) except (0, 0). [Hint: Find the minimum of f(x, y) on x2 + y2 = r2.]

Let/(a;, y) — e“ (x2+1/2):15. Find the maximum of f ix, y ) on R2

16. Find the minimum of f {x, y) on R2

17. Is f{x, y) continuous on R2?18*. Is f ix, y) uniformly continuous on R2?

19. Let/Or, y) be continuous on R2 and let b be a real number. Prove that g{x) = f i x ,b) is continuous on R.

20. L et/(#, y) be continuous on R2. Prove that gix) = f ix, x) is continuous on R. 2 1*. Let git) have a continuous derivative for all t. Define f ix, y) by

x — y

J i x , x ) = gf ix).

Prove that f ix, y) is continuous on R2. [Hint: Use the Mean Value Theorem. It is stated on p. 285.]

4. GRAPHS

Suppose a function z = f i x , y) is defined for all points i x , y) in some domain in the plane. Its graph is the surface in space consisting of all points

(*, y , f ( x , ij)),

where ix, y) is in the domain o f/. Figure 4.1 illustrates the graph of a function defined for all points ix, y) in a circular disk.


To get an idea of the shape of the surface, draw several sections by planes perpendicular to the z-axis or the y-axis.

EXAMPLE 4.1

Graph the function z = f (x , y) = 1 — x2.

Solution: The function f ( x , y ) is independent of y. Its graph is a cylinder with generators parallel to the y-axis. To see this, first graph the parabola z = 1 — x2 in the x } z-plane (Fig. 4.2a). If (a, c) is any point on this parabola and b is any value of y whatsoever, then (a, b, c) is on the graph of 2 = f (x , y).

Answer: The graph is a parabolic cylinder with generators parallel to the ?/-axis. See Fig. 4.2b.

EXAMPLE 4.2

Graph the function z = x2 + y.

Solution: Each cross-section by a plane x = x0 is a straight line z = y + Xq2 of slope 1. The surface meets the x, 2-plane in the parabola z = x2. See Fig. 4.3a. The figure does not yet convey the shape of the surface, so look a t

4. Graphs 211

Now the picture is clearer.

Answer: The surface is a cylinder oblique to the x , z-plane;it intersects the x } 2-plane in the parabola z = x2.

F ig. 4.3a


EXAMPLE 4.3

Graph the function z = x2 + y2.

Solution: Each cross-section by a plane perpendicular to the rc-axis is a parabola z = x02 + y2. This is shown in Fig. 4.4a. B ut from this figure, it is hard to visualize the surface. You learn more studying the cross-sections of the graph by planes z = Zo perpendicular to the 2-axis. Each cross-section is a circle x2 + y2 = z0.

Answer: The surface is a paraboloid of revolution (Fig. 4.4b).

Fig. 4.4b

EXERCISES

Sketch the graphs:1. z = f(x, y) = 1 — 2x3. z = x + y5. z = (x — 2 )2 7. z — x3 + y9. z = x2 + x — y

11. z = xy

2. z = f (x , y ) = 3 + !/4. z = Sx + 2y6. z = (y + 1 )3 8. z = x2 - y2

10. z — x3 + yz12. z = x2 + xy + y2.

5. Partial Derivatives 213

13. Graph the function z = -s/9 — x2 — y2 for all points (x, t/) in the circle of radius 3 and center (0, 0).

14. Graph the function z = x4 + y* for all points (x, !/) in the domain x2 + ?/2 < 1, z > 0, y > 0.

15. Find the curves formed when planes parallel to the x, z-plane intersect the graph of z = 2x2+ 3 y2.

16. Find all points common to both the plane x = 3 and the graph of z = x2y.17. Let/(x, y) be a continuous function defined on a closed domain D. Prove that the

graph of / is a closed subset of R3.

18. Prove that the graph of /(x, y) is never an open subset of R3.

5. PARTIAL DERIVATIVES

Let z = f (x, y) be a function of two variables and a = (a, b) an interior point of its domain. Suppose we set y = b and allow only x to vary. Then f ( x , b ) is a function of the single variable x, defined a t least in some open interval including a. We define

p (a, b) = b)ox ax

This is called the partial derivative (or simply partial) of z with respect to x. (The “d” is a curly “d”.) I t measures the rate of change of z with respect to x while y is held constant.

Similarly, we define the partial derivative of z with respect to y:

dz . d— («, b) = — f ( a , y )

y=b

In like manner, given a function w = f (x, y, z) of three variables, we may define the three partial derivatives dw/dx, dw/dy, and dw/dz. For instance,

dw / 7 x d v— (a, b,c) = — /(a , y , c) dy dy y=b

Each of the partials is the derivative of w with respect to the variable in question, taken while all other variables are held fixed.

N o t e : Our definition of partial derivatives applies only a t interior points of the domain.

EXAMPLE 5.1

Let z = f ( x } y) = xy2. Find

dz r Q dZ f A— for y = 3, — for x = —4, dx dy

dz dz .— and — m general. dx dy


Solution: If y = 3, then z = 9x, and

dzdx y = 3

d= - (9®) = 9.

dx

Likewise, if x = —4, then z = —4y2, and

dz= J - (—4i/2) = -8 y .

dy

In general, to compute dz/dx just differentiate as usual, pretending y is a constant:

dz d{xy2) , d , ,— = — — = y2 — (x) = ^2. dx dx dx

To compute dz/dy , differentiate, pretending x is a constant:

dz d(xy2) d— = 7---- = « J " (y ) = 2xi/.2/ ^ dy

We consider two further examples of partial derivatives.

(1) The gas law for a fixed mass of n moles of an ideal gas is

TP = n R - ,

where R is the universal gas constant. Thus P is a function of the two variables T and V :

dP 1 dP _ T dT ~ n V ’ d V ~ n V 2 '

(2) The area A of a parallelogram of base b, slant height s, and angle a is A = sb sin a. The partial derivatives are

dA 1 . dA . dA— = b sin a, — = s sin a, — = sb cos a. ds db da

Geometric Interpretation

The graph of z = /(x , y) is a surface in three dimensions. A plane x = x0 cuts the graph in a plane curve x = x0, z — f ( x o, y). See Fig. 5.1a. If this

curve is projected straight back onto the y , 2-plane, the graph of the function z = f ( x o, y) is obtained (Fig. 5.1b). The partial derivative

5. Partial Derivatives 215

is the slope of this graph.

dldy (*o, y)

For example, suppose the graph of the function z = x + y2 is sliced by the plane x = x0. See Fig. 5.2a. The resulting curve is the parabola x = x0, z = y2 + x0. If this is projected onto the y, 2-plane, a parabola is obtained (Fig. 5.2b). Its slope is

dz— = 2 y. dy

Notation

Unfortunately there are several different notations for partial derivatives in common use. Become familiar w ith them ; they come up again and again in applications.

SupposeW = / ( X, y, z).

Common notations for dw/dz are:

fx, fx(x, y, z), wx, wx{x, y, z), Dxf.


For example, if

then

Fig. 5.2a

Fig. 5.2b

w = f (x , y , z) = x*y2 sin 2,

f x = 3x2y2 sin 2, wy = 2xzy sin 2, D zf = x*y2 cos z.

EXERCISES

dz , dz Find — and — :

dx dy

1. z = /(x, 1/) = x + 2i/

3. z = 3xi/

2x25. 2 = — —2/+ 1

7. z = x sin y

9 .2 = sin 2x + cos 3y

11. 2 = sin 2xy

2. 2 = /(x, 2/) = 3x + 42/

4. 2 = x2y

6. 2 = x2y + x?/2

8. z — y2 cos x

10. 2 = sin x — cos y

12. 2 = cos(2x + y)

5. Maxima and Minima 217

V

15. z = xey

17. 2 = exy19. z = e2x sin y

18. 2: = —3ex+y 20. z = e~y cos x.

21. Let z = x2y. Find dz/dx for y = 2, and dz/dy for x = — 1.22. Let z = y2/x. Find zx for y — 3.

23. Let w = xy2&. Find wx for y = 2 and 2 = 2 , find for x = 1 and 2 = 0 , and find wz for x = y.

r x -nv j dw i ^24. Let w = xy — xz — yz. Find - — - — h — .dx %

25. Show that 2 = (3x — y )2 satisfies dz/dx + 3 dz/dy = 0.

26. Show that 2 = f ( x ) + y2 satisfies dz/dy = 2y.

27. Show that 2 = x2 — y2 satisfies {dz/dx)2 — {dz/dy)2 = 42.28. Show that 2 = x6 — xby + 7x V satisfies x (dz/dx) + y (dz/dy) = 62.

6. MAXIMA AND MINIMA

Suppose

Suppose z takes on its minimum value a t (x, y) = (a, b). By holding y fixed a t y = b, the function z becomes a function of x alone with minimum a t x = a. Hence

These two relations are often enough to locate the points where a function takes on its minimum (or maximum) value. See Fig. 6.1.

Geometrically the idea is simple: If the graph of z(x, y) has a high (low) point a t (a, b, c), then so do its cross-sectional curves by planes through (a, b, c) parallel to the x , 2-plane and to the y, 2-plane. The slopes of these curves are dz/dx and dz/dy respectively.

Z = z(x, y)

is a function of two variables defined on the domain

x0 < x < xiy 2/0 < y < y 1.

Similarly


Fig. 6.1 In each case df/dx = df/dy = 0 at (a, b).

EXAMPLE 6.1

Find the minimum of z = x2 — xy + y2 + Sx.

Discussion: Is the question reasonable? For a fixed value of y, say y = b, the resulting function of x ,

z(x, b) = x2 + (3 — b)x + b2,

is a quadratic polynomial whose graph is a parabola turned upward; it has a minimum. Similarly for each fixed x = a, the function of y f

z(a , y) = y2 - ay + (a2 + 3a),

has a minimum.This is fairly good experimental evidence th a t the function z(Xjy) ought

to have a t least one minimum, probably no maximum.

Solution: The function is defined for all values of x and y. Begin by finding all points (x , y) a t which both

6. Maxima and M inima 219

Now

^ (x2 - xy + y2 + 3x) = 2x - y + 3, dx dx

dz d— = — ( X * _ X y -f- y 2 3 ^ ) = - ^ + 2y ,dy dy

so the conditions are

2x — y + 3 = 0

— x + 2y = 0 .Solve:

x = -2, y = - 1 .

The corresponding value of 2 is

* ( - 2 , - 1 ) = ( - 2 y - ( —2) ( — 1) + ( -1 )2 + 3( — 2)

= 4 — 2 + 1 — 6 = - 3 .

Is this value a maximum, a minimum, or neither? (Recall th a t the vanishing of the derivative of a function f ( x ) does not guarantee a maximum or minimum, e.g., f ( x ) = x* a t x = 0.)

In this case you can prove th a t the value z ( — 2, —1) = —3 is the minimum value of z by these algebraic steps. First set up a u, ^-coordinate system with its origin a t (x, y) = ( — 2, —1). Take

x — u — 2, y = v — 1.

Then

2 = (M - 2)2 - (ti - 2)(v - 1) + (v - l ) 2 + 3 (u - 2)

= u2 — uv + v2 — 3 .

Next, complete the square:

/ y \2 v2 / y \2 3* = ( . - - ) - ? + ^ - 3 = ( M - - ) + r > - 3 .

Since squares are nonnegative, z ( x , y ) > —3.

Answer; —3.

In general, to find possible maximum and minimum values of a function, locate points where all its partial derivatives are zero. W hether a particular one of these actually yields a maximum or a minimum may not be easy to determine. (Later we shall study a second derivative test which sometimes helps.)


EXAMPLE 6.2

Find the rectangular solid of maximum volume whose to tal edge length is a given constant.

Solution: As drawn in Fig. 6.2, the to tal length of the 12 edges is 4x + 4y + 4z. Thus 4x + 4y + 4z = 4/c, where & is a constant, so

x + y + z = k.The volume is

V = xyz = xy(k — x — y)

= kxy — x2y — xy2.

The conditions for a maximum,

W „ dV „— = 0, — = 0, dx dy

are

Iky — 2 xy — y2 = 0

kx — x2 — 2 xy = 0.

The nature of the geometric problem requires x > 0 and y > 0. Thus we may cancel y from the first equation and x from the second:

2 x + y = k

x + 2 y = k.

This pair of simultaneous linear equations has the unique solution

k k3 ’

Hence also z = k / 3; the solid is a cube.

* " 3 ’ V =

Answer: A cube.


EXAMPLE 6.3

W hat is the largest possible volume, and what are the dimensions of an open rectangular aquarium constructed from 12 ft2 of Plexiglas? Ignore the thickness of the plastic.

open top

y

Solution: See Fig. 6.3. The volume is V = xyz. The to tal surface area of the bottom and four sides is

xy + 2 yz + 2 zx = 12.

Solve for z, then substitute into the formula for V :

z =12 — xy

2 (x + y ) 9V =

(12 - xy)xy 2 (x + y)

Compute the partials carefully (only one computation is needed because of the symmetry in x and y ) :

dV _ y2( —x2 — 2xy + 12) dx 2(x + y ) 2

dV _ x2( —y2 — 2xy + 12) dy 2 {x + y ) 2

Now find all points (x, y) where both partials are zero. Such points must satisfy

y2( —x2 — 2 xy + 12) = 0

x2( - y 2 - 2xy + 12) = 0.

By the nature of the problem, both x and y are positive quantities. Therefore, the factors y2 and x2 may safely be canceled:

- x 2 - 2xy + 12 = 0

—y2 — 2 xy + 12 = 0.


I t is easy to solve these equations for x and y. Subtract: x2 — y2 = 0, hence y = ± x . Since x and y are both positive, only y = x applies. Now substitute y = x into either of the two equations:

— 3x2 -f* 12 = 0, x = ± 2 .

The only feasible choice is x = y = 2. I t follows th a t 2 = 1 . The optimal dimensions are 2 X 2 X 1; the maximal volume 4 ft3.

Alternate Solution: Previously we chose the side lengths as variables, because volume and surface area can be expressed in terms of these side lengths. Is this the only way? Perhaps we can use the face areas as variables. They are

u = yz, v = zx, w = xy.

The to tal surface area, in term s of u, v, w, is

w -J- 2u ~f- 2v — 12.

Can we express the volume in terms of u, v, and w l I t is unnecessary to solve the system

4 yz = u, zx = v, xy = w

for x , y, z in terms of u, v, w , and then substitute the resulting expressions inV = xyz. We can express V directly in u, v, w:

V2 = (xyz)2 = x2y2z2 = (yz) (zx) (xy) = uvw.

The maximum value of V corresponds to the maximum value of V2. Now we have formulated a reasonable problem: Maximize uvw subject to the constrain t 2u + 2v + w = 12.

Setg(u, v) = uv( 12 — 2u — 2v) = 12uv — 2u2v — 2uv2.

Take partials:gu = I2v — 4 uv — 2v2y

gv = 12m — 2 u2 — 4 uv.

These partials are 0 where the equations

12v — 4 uv — 2v2 = 0

12 u — 2 u2 — 4 uv = 0

are satisfied. Cancel the factor v in the first and u in the second (v = 0 or u = 0 doesn’t hold w ater). The equations reduce to

2 u -j- v — 6

u + 2v = 6.

Hence u = v = 2. I t follows th a t w = 4 and F 2 = uvw = 16. The answer, however, should be in terms of x, y , and z. Recall

yz = u = 2, zx = v = 2, xy = it? = 4.

Hence x = y and xy = #2 = w = 4. Therefore x = y = 2 and 2 = 1.

.Answer: The maximum volume is 4 ft3, achieved by a tank of base2 ft X 2 ft and height 1 ft.


R e m a r k : The theory of maxima and minima will be treated with more care in Chapter 9, Section 4.

EXERCISES

Find the possible maximum and minimum values:1. z = 4 — 2x2 — y2 2. z = x2 + y2 — 13 2 = ( z - 2 ) 2+ (y + 3)2 4. 0 = (* - 1 )2 + y2 + 35. z = x2 — 2xy + 2y2 + 4 6. 2 = xy — x2 — 2y2 + x + 2i/7. z = x — y2 — xs 8. 2 = 3x + 12i/ — a;3 — 2/3.

9. Find the rectangular solid of greatest volume whose total surface area is 24 ft2.10. Find the dimensions of an open-top rectangular box of given volume V, if the

surface area is to be a minimum.11. Find the dimensions of the cheapest open-top rectangular box of given volume 7,

if the base material costs 3 times as much per ft2 as the side material.

7. Linear Functions and Matrices

1. INTRODUCTION

In this chapter we shall discuss some parts of linear algebra th a t are useful in our further study of the calculus of functions of several variables.

Linear algebra can be looked a t as the study of one basic functional equation,

(1) F(au + bv) = aF( u) + bF (y ),

where a and b are scalars and u and v are vectors.This equation is fundamental in many branches of mathematics. We can

think of F as an operation th a t transforms an “ inpu t” u into an “ou tpu t” F(u) . Then equation (1) says th a t for input

a(inpu ti) + fr(input2),the output is

a (ou tpu ti) + 6 (outputs).

N ot every F has this property. Examples: F(x) = x2 for each real number x and F ( u) = |u| for each vector u in R3.

R em ark: If F satisfies (1), then it also satisfies

(2) F(au + by + cw) = aF( u) + bF(v) + cF( w)

(and similar equations for four or more summands). This equation follows in two steps from (1):

F (au + bv + cw) = F(au + bv) + cF ( w)

= [a F (u ) + 6 F (v )] + cF(w ).

An operation F satisfying (1) is called linear. In the most general setting, the inputs (domain) of F may come from any m athem atical system in which expressions au + bv make sense. The outputs may be in any other systems with the same property. Such systems are called abstract vector spaces; their study belongs to a course in linear algebra. We shall consider only linear operations with domain one of the spaces R, R2, R3 and with values also in one of these spaces, not necessarily the same one. We shall call such operations linear functions.

1. Introduction 225

Real-Valued Linear Functions

Let us consider a linear function F defined on three-space R3, with real values. Thus

F: R3--------- > R,

so F is a function of three variables of the type studied in Chapter 6 . However F is subject to an im portant restriction: it must satisfy (1 ).

Now what does F look like? From previous experience, we would guess th a t F(x, y, z) = px + qy + rz is a reasonable possibility. Let us verify th a t this function is indeed linear.

We set Xi = (#i, 2/1, zi) and x2 = (x2, 2/2, 22). Then

F(ax 1 + bx2) = F(ax 1 + bx2, ayi + by2, az\ + bz2)

= p(ax 1 + bx 2) + g(a2/i + by2) + r(azi + 622)

= a(px 1 + g?/i + r2i) + b(px2 + qy R

is given by

F(x, y, 2 ) = px + qy + rz,

where p, g, r are constants.

Proof: We use the decomposition of an arbitrary vector into a combination of the basic unit vectors:

(x, y , 2 ) = x\ + 2/j + 2k.

From (2 ) it follows th a t

F(x, y, 2 ) = F ( x i + 2/j + «k) = # F (i) + 2/F ( j) + z F ( k).

Set p = ^ ( i ) , q = F( j) , r = F (k ), three scalars which do not depend on (x, y, 2 ). We have proved th a t

F(x, y, z) = px + qy + rz.

R e m a r k : The proof illustrates an im portant principle: a linear functionF: R3-------- > R is completely determined by its effect on the three vectorsi, j, k.

As a corollary of the preceding theorem, we obtain a useful alternative representation for linear functions.

226 7. LINEAR FUNCTIONS AND MATRICES

Theorem Let F: R3----- — > R be linear. Then there is a vector p such th a t

F{x) = p*x.

Proof: Just observe th a t

F{x) = px + qy + rz = (p, q, r)* Or, y, z) = p-x,

where p = (p, q, r).

An interesting consequence is the following estimate on the order of m agnitude of a linear function. I t will be useful in the next chapter when we study composite functions.

Theorem Let F(x) = p*x. Then

I T O I < Ip I |x |.

Proof: Given x, let 0 be the angle between p and x. Then p*x =|p| |x| cos 0, hence

\F(x)\ = |p*x| = |p| |x| |cos0| < |p| |x|.

R e m a r k : In some discussions it is common to call a function of the form

f (x , y, z) = A x + By + Cz + D

a linear function (A, B , C, D are constants) or a linear polynomial. If D = 0, then f (x, y, z) is called a homogeneous linear function. In this chapter, we shall always mean homogeneous when we refer to linear functions.

EXERCISES

Let F : R3-------- > R be linear:1. Find F(x, y, z) if F ( 1 ,0 ,0) = 3, F(0, 1,0) = 2, F(0, 0, 1) = - 5 .2. Find F(x , y, z) if F( 1, 1, 0) = 0, F(0, 1, 1) = 1, F (l, 1, 1) = 3.3. Show that the set of x for which F ( x ) = 0 is a plane through 0, provided Ft* 0.4. Prove that F (x ) is determined by its values at i, i + j, and i + j + k.5. Prove that F (x ) is not determined by its values at i and j.

6. Prove that F(x) is continuous.

7. Prove that the sum of two linear functions is linear.8. Prove that cF is linear if F is linear and c is a scalar.9. Let g vary over continuous functions on [0, 1] and set

L(g) = J g(t) dt.

Show that L is linear.

2. Linear Transformations 227

10. Let g vary over polynomials and set

D(g) = g', the derivative. Show that D is linear.

11. Prove the identity

(ciiXi + a2x2 + )2 + (a\X2 — a2xi)2 + (a2xs — azx2 )2 + (a3x i — ai£3)2

= (ai2 + a22 + a32) fa2 + z22 + z32).

12. (cont.) Hence give another proof, free of cosines, that |a • x| < |a| |x|.

Let F: R3-------- > R. Call F an affine function if

F(au + 6v) = aF(u) + bF(v)

whenever a + b = 1.13. Prove that each non-homogeneous linear function is an affine function.

14*. Let F be an affine function. Prove that

F (au + 6v + cw) = aF (u) + bF (v) -f- cF (w)

if a + 6 + c = 1.15*. (cont.) Prove that F is a non-homogeneous linear function.

2. LINEAR TRANSFORMATIONS

In this section we study linear functions F: R3-------- > R3. (Linear functionstaking values in a space of dimension greater than one are often called linear transformations.) Let us begin by describing the most general function of R3 into R3, linear or not.

If F: R3-------- > R3, then for each x £ R3,

F(x) = (F , ( x ) ,F , ( x ) ,F , ( x ) ) .

Thus F is equivalent to a triple Fh F2, Fz of real-valued functions called coordinate functions associated with F.

Now, if F: R3-------- > R3 is linear, we can show th a t each coordinate function Fj\ R3-------- > R is linear. We write the basic equation F(au + 6v) =aF(u) + bF(v) in coordinates:

(Fi(au + by), F2(au + by), Fz(au + by))

= a(Fi(u) , F2(u), F 3(u )) + b(Fx(y), F2(v), f t ( v ) ) .

Now equate th e /- th coordinates:

Fj(au + by) = aFy(u) + 6Fy(v).

Therefore F}: R3-------- > R and Fj is linear. By the second theorem of Section 1,there is a vector p; such th a t F j (x) = py* x.

Theorem Let R3such th a t

-> R3 be linear. Then there are vectors pi, p2, P3

F(x) = (pi-x, p2 -x, p3* x).

A Bound for F

By analogy with our procedure in Section 1, we may use the preceding theorem to obtain a bound for |F (x )|.

Theorem L e tF iR 3 c such th a t

-> R3 be a linear function. Then there is a constant

|F (x )| < c |x | .

Proof: We have

F(x) = (pi-x, p2-x, p3* x)

and we know th a t |py#x| < |py| |x|. Therefore

|F (x )|2 = |Pl-x|2 + |p2-x|2 + |p3-x|2

< |pi|2 |x|2 + |p2|2 |x|2 + |p3|2 |x|2 = c2 |x|2,

where c2 = |pi|2 + |p2|2 + |p3|2. Hence |^ (x )| < c |x|.

The Matrix of F

Again consider a linear function F: R3 -------- > R3. Let

F (x ) = (pi*x, p2-x, p3*x ).

Each vector py has three components:

Pi = (pi, qi, r i)

P2 = (P2 , q%) r2)

. Pa = (ps, Qsf rz).

The function F is specified by the three vectors pi, p2, p3, and each of these is specified by its three components. Therefore F is determined by the square array of 9 numbers

*~Pi qi ri

A = p2 q2 r2

Jpz qz r3_


Such an array is called a 3 X 3 matrix. In this way each linear functionF: R3-------- > R3 gives rise to a 3 X 3 matrix A.

W e have seen th a t the rows of the matrix A are the vectors Pi, p2, p3 w ritten in components. There is also an interesting interpretation of the columns of A.

Recall th a t F(x ) = (pi*x, p2*x, p3*x ). In particular F ( i) = (pi*i, p2*i, p3*i). Now

Pi = pii + P2J + p3k, so p r i = pi.

Similarly, p2*i = p2 and p3*i = p3. Hence, F ( i) = (ph p2, p3). If this vector were written as a column

~Pi~

P2

_P3_

instead of a row, it would match the first column of A.Similarly we find th a t F (j) and F( k) correspond to

" V n

2 and r2

_ 3_ _r3_

Let F: R3-------- » R3 be a linear function. Then F determines a 3 X 3 matrixA whose columns represent the vectors F ( i), ^ (k ) respectively.

Conversely, each 3 X 3 matrix A determines a linear function F: R3-------- » R3 as follows:

If Ci, c2, c3 are the vectors represented by the columns of A, then F is determined by F(i) = Ci, F( j ) = c2, F ( k) = c3.

Geometric Interpretation

Let F: R3-------- > R3 be linear, so F assigns to each point x in 3-spaceanother point F (x ) in 3-space. Thus F can be looked a t geometrically as a mapping of space into itself.

EXAMPLE 2.1

Describe geometrically the transformations

(a) F { x , y , z ) = (x,y, 0)

(b) F(x, y, z) = (x, 2y, z).

In each case, find the associated matrix.


Solution: (a) The transform ation assigns to xi + y) + zk the vector .ri + y\. In other words F projects the vector x perpendicularly onto the x, 2/-plane. See Fig. 2.1a. Clearly F (i) = (1 ,0 ,0 ) , F ( j) = (0 ,1 ,0 ) and F (k) = (0, 0, 0). The corresponding column vectors are the columns of the associated matrix:

"1 0 0"

0 1 0 .

0 0 0

(Note position of axes.)

F ig . 2.1

(b) The transformation doubles the ^-component of x. Geometrically this amounts to a stretching of space by a factor of 2 in the (/-direction. See Fig. 2.1b. We note th a t F (i) = (1 ,0 ,0 ) , F ( j) = (0 ,2 ,0 ) , F (k ) = (0, 0, 1), so the associated matrix is

"1 0 0”

0 2 0 .

0 0 1

Answer: (a) (b) i

projection onto the x , y-plane; stretching parallel to the y -axis by a factor of 2;

iH- o o

1 1o0rH1

0 1 0 > 0 2 0 *

o o 0 1 0 H-* 1

EXERCISES1. Let F be linear with matrix A = [a;;]. Prove that |F (x)| < c |x| with c2 = ^ cuf.2. Let F and G be linear with respective matrices A and B. Prove that F + G is

linear and find its matrix.3. Let a be a vector. Prove that F(x) = a X x is linear.4. (cont.) Find the corresponding matrix.

Show that F is not linear:5. F(x, y, z) = (x, y, 1) 6. F(x, y, z) = (xy, 0, 0).

Find the matrix corresponding to the linear function:7. F(x, y, z) = (x,y,x + y + z) •8. F(x,y, z) — (2x — y — Sz, 4x + by + 6z, y — 2z)9. F ( i ) = 0, F( j) = i + j + 2k, F(k) = 4k

10. F(\) = l ,F( i ) = k , F(k) = \.For each matrix describe geometrically the corresponding linear function:

"1 0 0“ “0 0 0"

11. 0 1 0 12. 0 0 0

.0 0 1- -0 0 CL“1 0 0" "0 0 O'

13. 0 0 0 14. 0 0 0

-0 0 1- -0 0 1-—_ 1 0 0“ ■1 0 0“

15. 0 - 1 0 16. 0 - 1 0

_ 0 0 -1 _ _ 0 0 1-"1 0 0" “cos a: —sin a

17. 0 1 0 18. sin a cos a

-0 0 _ 1- 0 0


"2 0 0" “0 1 0“

19. 0 3 0 20. 1 0 0

-0 0 — 1. -0 0 1-

Let F: R3- > R3 be linear:21. If F (i), F (j), F (k) all lie in a plane P passing through 0, show that F (x) lies in P

for every x £ R3.22. If F(i), /'"(j), F(k) all lie on a line L passing through 0, show that F (x ) lies on L

for every x £ R3.

23. Let F: R3 -------- » R3 be linear and let a £ R3. Prove that x -------->a • F(x) islinear on R3 to R.

24. (cont.) Suppose F: R3 - -> R3 and x ■Prove that F is linear.

-> a • F (x) is linear for each a £ R3.

3. MATRIX CALCULATIONS

In Section 2, we saw the advantage of expressing vectors not only as row vectors,

a = (ai, a2, as),

bu t as column vectors

^ai

a' = a2

«3

We shall denote vectors in column form by primed letters.In this section we introduce various kinds of operations involving rows,

columns, and matrices. These facilitate the study of linear functions in Section 4.

Row-by-Column Multiplicationf—

The product of a row vector by a column vector is the scalar defined by

ab ' = (ai, a2, a3) — CLlbl + &2&2 ~h

I t is an alternative form of the dot product of two vectors. Note th a t this product is taken with the row first, then the column.

3. Matrix Calculations 233

Now let"fen 6 12 &13

B = &21 &22 &23

_&31 &32 &33,

be a 3 X 3 matrix and a = (ai, a2, as) a row vector. We can multiply 5 on the left by a, taking the product of a with each column of B :

"bn bn bis

aB — (0 1 , ci2> <2 3) b2i b22 b23

_bsi bs2 bs3m

— (a i& ll + 02&21 + « 3&31, Glfrl2 + a2&22 + &3&32, ^1613 + « 2&23 + O s b s s ) •

Similarly we can multiply B on the right by a column vector c', taking the product of each row of B with c ':

bn & 12 bis Ci bnCi + bi2c2 + 6 1 3 C 3

B e ' = 6 2 1 b22 b2s c2 = b2iCi + b22c2 + b2sCs

_b 31 bs2 bss^ _ C 3_ J>3iCi + bs2c2 + bssCs_

The answer is a row vector in the first case, a column vector in the second. Examples:

1 2 0"

- 1 , 1 ) - 1 - 1 2 = 1(8, 5, - 3 ),

4 - 2 - 1 _

1 2 0“ 3“ " 1“

- 1 - 1 2 - 1 = 0

4 - 2 - 1 1 13 <

Sometimes it is convenient to abbreviate the notation. We can write B as a row of three column vectors:

"6 n“ 6 12 bis

B = (b /, b2', b3'), where b / = b2i , b2' = b22 , b3' = b2s

_bsi__ _b&_ J) 33_


Then aB is a row vector:

aB = a (b / , b2', b3') = (ab/, ab2', ab3').

Each a by' is a scalar, a row-by-column product. Similarly we can write B as a column of three row vectors:

b i = ( b n , &12, biz)

b2 = (£>21, &22> £>23)-

b3 = (&31, 32, £>33)

Then 5 c ' is a column vector:

"bi"

B = b2 , where •

_ b 3_

~br biC'-

B e = b2 c' = b2c'

b3_ b3c'_

f l o w X Matrix X Column

Let a be a row vector, B a matrix, and c' a column vector. Then B e' is a column vector, so we can form the scalar a (B e'). We can also form the scalar (aB )c '. It is an important fact (an associative law) that these two are equal:

a (B e') = (aB)c'.

To prove it, we simply compute the product in either order. With the notation above,

a (B e') = (0 1 , a2, a3)

^ bijCj

b%jCj

^ bsjCj

^ ^ bijCj ^ dibijCj. i=i y=i »*,/

The double sum contains 9 summands, and it does not depend on whether we


sum first on j or on i. If we sum first on i , then on j , we obtain

a (Be') = ^ CLibijCj = ^ ^ Cy

= (aB)c'.

If a is a row vector, c' is a column vector, and B is a matrix, then

a(Bc ' ) = (aB )c' = ^ a * b tycy.

Product of Matrices

I t is time to define the product of two 3 X 3 matrices A and B. As you might suspect by now, we shall define A B in terms of row-by-column products. Let fiy r2, r3 be the rows of A and let c / , c2', c3' be the columns of J5, so

A = and B = (c /, c2', c3') .

We define~rC ” r iC i' TiC2' TlC/

A B = r2 (Cl', Ci’ , c3') = r2Ci' r2c2' r2c3'

_r3_ _r3Ci' r3c2' r3c3'_

Thus the product A B is another 3 X 3 matrix. The number in the i} j -th position of A B is TiC/y the product of the z-th row of A by th e^-th column of B.

Example: C =

2 0 - 1 3 - 3 - 3 "( 2, o, - 1 )

- 1 3 - 2 1 2 1 = ( - 1 , 3, - 2 )

2 4 4_ _5 1 — 4_ _( 2, 4 , 4 )

- 3

2

1

- 3

1

- 4


Then

cu = (2,0, - 1 ) — 1, C32 — (2, 4, 4) 2

1

- 3

= 6, etc.

and we find

1 - 7 - 2 '

c = - 1 0 7 14

30 6 - 1 8

W a r n i n g : M ultiplication of matrices is not commutative'. In general, A B B A . For example, multiply the matrices of the example above in the opposite order:

"3 - 3 3" 2 0 - 1 “ “15 3 15“

1 2 1 - 1 3 - 2 = 2 10 - 1

_5 1 — 4_ 2 4 4_ _ 1 - 1 3 — 23_

The product is totally different.Despite the failure of the commutative law for multiplication of matrices,

the associative law does hold.

Associative Law If A , B, C are 3 X 3 matrices, then

A ( B C ) = {AB)C.

The proof is given in the next section.

Transpose

The transpose A' of a matrix A is obtained by flipping A across its main diagonal (mirror reflection in the main diagonal; interchange of rows and colum ns):

a b

c d

a n «12 O13/ an 021 O3I

/ a c=

bJ 021 &22 023 ~ aw 022 O32

_&31 O32 d33_ _dl3 023 d33_

We also define transpose for row and column vectors:


a a

b y b

_c_ _c_

In any case, if the transpose operation is performed twice in succession, we return to where we started:

( x ' ) ' = X, ( A ' ) ' = A , [ ( y = y' .

There is an im portant connection between matrix multiplication and transposing.

Let x and y be row vectors and A and B matrices. Then

xy' = yx', (xA)' = A'x', (Ay')' = yA',

xAy' = yA'x', (AB) ' = B 'A '.

The proofs of these relations are straightforward and are left as exercises.

Further Operations

Sometimes it is necessary to add two 3 X 3 matrices or multiply a matrix by a scalar. The definitions of these operations are quite natural:

an «12 dl3 bn £>12 £>13 an + £>n #12 + £>12 «13 + £>13

«21 «22 &23 + £>21 £>22 £>23 = a 21 + £>21 «22 + £>22 &23 “f" £>23

_&31 &32 «33_ _£>31 £>32 £>33_ _«31 + £>31 «32 + £>32 «33 + £>33_

an au «13 can can cau

«21 «22 «23 = ca2i ca22 ca23

_d31 «32 «33_ jca^i ca$2 caz 3_

W a r n i n g : In the second definition, each one of the 9 entries of the matrix is multiplied by c. This is different from the corresponding rule for determinants, where multiplying by c is equivalent to multiplying just one row or one column by c.

R e m a r k : I t should be clear th a t everything in this section applies not only to 3 X 3 matrices, bu t also to 2 X 2 or n X n matrices as well.


EXERCISES

Compute:

3. (2,0,8)

C I ]

1

1

LlJ

7.

0 1 1'

1 0 1

Ll 1 0J

9. (3,1,1)

1"

1

Ll.1 1 1‘

1 1 1

Ll 2 3J

- 1

- 1

L 2J

15.

2 1

- 1 2

L -2 - 2

1‘

-1

1J

0 3

1 1

L—1 - 1

17.

1 0 0"

0 0 0

LO 0 0.

CL\ 0*2 CLz

6l &2

Lei C2 C3.

4. (0,4,0)

1'

0

LlJ

V

j/J- 2 0 o'

0 3 0

L 0 0 4J

10. (2, 0, - 1)

L c J

- 1

1

L 1 1 - I J

" 2"

0

. - I -

14.

‘ 1

- 1

’ 3 0 1'

- 1 3 2

L—2 4 5J 1 1'

-1 - 1

1 0 5*

0 1 0

LO 0 U

0 0 0.

~b 1 0“

16.

18.

0 b 1

LO 0 bJ0 1 O'

1 0 0

LO 0 1.

“ ai 0*2 ciz

bi 62 h

-Cl C2 C3-

4. Applications 239

"0 a b~ 3 ~a 0 0“

19. 0 0 c 20. 0 b 0

-0 0 0- -0 0 C -

21. (x,y,z)

“ 1 0 O'

0 2 0

L O 0 3J

Compute A2 + 3A:

L z J

23 A =- 3

0 j

22. (x,y,z)

1 2 O'

2 3 0

LO 0 4J

24. A =

1 2 O'

0 - 3 1

Ll 1 - I J

25. Let 0 be the 3 X 3 matrix all of whose entries are zero. If A and B are 3 X 3 matrices and AB = 0, does it follow that either A = 0 or B = 0?

Prove

26. xy' = yx'28 (Ay')' = yA' 30. (AB)' = B 'A '

27. (xA)' = A'x'29. xAy' = yA'x'31. (A + B)' = A' + B'.

4. APPLICATIONS

We return to linear functions of R3 into R3 and apply vector and matrix techniques. I t will be convenient to write all vectors in columns. Thus we shall think of R3 as the set of all column vectors with three elements.

Suppose F: R3-------- > R3 is linear. According to Section 2, there existvectors pi, p2, P3 such th a t F(x) = (pi* x, p2* x, p3* x). B ut we were going to use column vectors instead of row vectors, so we re-write this formula as follows:

~Pix'~

F(x') = P2X

.P3X J

where pi, p2, P3 are row vectors. Recall th a t in Section 2 we associated with F the 3 X 3 matrix A whose rows are Pi, p2, p3. Therefore the last formula is simply F(x' ) = Ax'.

If F: R3 R3 is linear and if A is the matrix of F, then

F(x ' ) = Ax'

for each column vector x'.

R e m a r k : The bound |F (x')| < c |x'| th a t we derived before can now be written for the product Ax' of a matrix by a vector:

|Ax'| < c |x'|,

where c is a constant independent of x'.The formula F(x ' ) = Ax' provides a practical algorithm for computing the

values of a linear function.

EXAMPLE 4.1

Let F: R3-------- > R3 be linear and satisfy

"1“ 4" ~0~ “ 7" "0" “ - 1 "

F ( 0 ) = 5 > 1 ) = 12 , F( 0 ) = 2

_0_ _ —2_ _0_ _ 0_ _1_ 1_

Compute F (

Solution: Let A be the matrix of F. In Section 2, we showed th a t the columns of A are f

F (

"1"

i01 \01

0 ) . n i ) , n 0

_0 _ _o_ _1 _

(only there we used the notation F ( i), F ( j) , /^(k )). Consequently

4 7 - 1 “ 2”

A = 5 12 2 ,so F ( 7 ) =

— 2 0 1 _ - 3

4 7 - 1

5 12 2

- 2 0 1

2

7

- 3

60

4. Applications 241

60

A nswer: 88

7_

Composition of Linear Functions

Let F and G be linear functions with corresponding matrices A and B :

F(x ' ) = A x ' , <?(x') = Bx'.

As usual, the composite function F ° G is defined by

(F ° G) (x') = F [G (x ')] .

Theorem The composite of linear functions is again linear. If F(x ) = Ax' and G(x) = Bx', then

(F ° G) (x') = (AB)x' .

Proof: To prove th a t F ° G is linear, we go right back to the definition:

F[G(au' + 6 v ')] = F[_aG( u ') + « ? (v ') ] = aF[_G( u ') ] + 6F [(? (v ')],

the first equality because G is linear, the second because F is linear. I t follows th a t

{F ° G) (au' + by') = a ( F ° ( ? ) (u ') + b ( F ° G ) ( v ' ) ,

th a t is, F ° G is linear.We must compute the matrix corresponding to F ° G. We have

(F o G) (x') = F [G (x ') ] = F [B x 'l = A( Bx ' ) .

To finish the proof, we must establish the associativity relation

A ( B x ' ) = (AB)x' .

If ai denotes the t-th row of A, then the i-th element of the column vector A ( B x ' ) is a i (Bx' ) . B ut we know th a t a i (Bx' ) = (aiB)x' . However, a iB is the z-th row of the matrix A B . Consequently A (Bx' ) and (AB)x ' have the same elements, i.e., they are equal, A ( B x ' ) = (AB)x' . This completes the proof th a t ( F ° G ) ( x ' ) = (AB)x' .

Here is a concrete application of composition of linear functions, showing how matrix multiplication can be a labor saving device. Consider this problem in algebraic manipulation: Given

(1)

r = 3x + + 5z

s — 2x — u — z (2 )

t = 5x — 6y + 3z

express r, s, t in terms of u, v, w.

x — u + v +

y = 3u — 2v — w,

z = 6u + 5v + 6w

The obvious approach is to substitute equations (2) into equations (1), simplify and collect terms. But this is tedious, while matrix multiplication will do it for you automatically. Write

r X X u

s = A y y y = B V

t z Z wwhere

3 4 5“ "1 1 4"

A = 2 - 1 - 1 , B = 3 - 2 - 1

_5 - 6 3_ _6 5 6_Then

r u ■ 45 20 38“ u

s = (AB) V = - 7 - 1 3 V

_ t _ _w_ 5 32 44_ _w_Hence

r r = 4 5u + 20v + 38w

\ s = — 7u — v + Sw

J = 5u + 32v + 44w.

Associative Law

We can now prove, with a minimum of computation, the associative law for multiplication of matrices given in Section 3.

Let A, B, C be 3 X 3 matrices and let F , G, H be the corresponding linear functions:

F (x ') = Ax', Cr(x') = Bx', H (x') = Cx'.

We compute in ^wo ways. First, C?[77(x')] = ((? ° H ) (x'), so

F {G [tf(x ')]) = F [(G « ff ) (x ') ] = [ F* (<7«ff)](x ').

Second, F [G (y ')] = (F « 6 ) (y'). Replace y' by H(x ' ) :

F {G [ff(x ')]} = ( F 0( ? P ( x ' ) ] = [ ( F 0f f ) ^ ] ( x ' ) .

We conclude th a tF ° (G ° H) = (F ° G) ° H.

Now we translate this into matrices. The matrix corresponding to G ° H is

4. Applications 243

BC , hence the m atrix corresponding to F ° (G ° H) is A (B C ). Similarly the matrix corresponding to the linear function (F ° G) ° H is (A B ) C . Therefore A ( B C ) = (A £)C .

EXERCISES1. Let F : R3- R3 be linear. If

"1“ “6" ’0“ ’ - 4 " "0‘ ■ 8"

F( 0 ) = 2 F( 1 ) = 3 n 0 ) = - 1

- 0- -7_ - 0- - 0- _ 1_ - —3-compute

F(

’ 5'

- 1

2. Given a 3 X 3 matrix A, verify that x' • F: R3-------- > R3.

A x ' defines a linear function

Let F and G be linear functions R3- Describe F ° G:

R3 with associated matrices A and B.

’ 0 0 1 " “ 2 0 0 “

CO II 0 1 1

II

oq

3 0 0

. 0 0 0 - - 5 1 0 -

[cos 0 sin 0~1 cc

— sin 0 cos 0 J [_si

cos 0 — sin 0"; interpret geometrically.

-sin0 cos 0 j |_sin0 cos0

5. Use matrix and vector notation to write the linear system

a\x + b\y + c\z — d\

azx + b2y + c2z = d2

k dzX + bty + C3z = d3.

6. Let F: R3-------- > R3 be linear with matrix A. Given y', prove that there is aunique vector x' such that F ( x ') = y' if and only if the determinant of A is not zero.

7. Givenx — Su + 4v + 5w

y = u — v + 2w

z = 2u + 2v — w

express x, y, z in terms of r, s, t.

u = r — s — t

v — 2r + 6s — 7t

w = 3r — 2s + t}

8. Givenix = 2u — 3v u = 4r — 5s fr = 5p + 2q

y = r + 2s [s = — p + 3 q,[y = 6w + v

express x and y in terms of p and q.9. Do Ex. 4 of Section 2 in column notation. That is, prove

“ 0 “ 03 «2

a' X x' = Ax', where A = 0 — 01

_— &2 ai 0 .10. (cont.) Use this result to prove

a '- (b 'X c ' ) = — c' • X fit ).11. Use Ex. 9 to show that the matrix of

F(x ') = a 'X (b 'X x ') is - ( a ' 1•b')I + b'a

12. (cont.) Now prove the identity

a 'X (b 'X c ') = - (a '-b ') c ' + (a'-c')b'.

(Compare Ex. 21, p. 147.)13. Let A be a 3 X 3 matrix. We proved there is a constant c such that | Ax'\ < c |x '|

for all vectors x '. Prove that | A nx'| < cn |x '| for n = 1, 2, 3, • • •.14*. (cont.) Prove that

jn =0

n\A nx'

converges for each x'. [Hint: Use Ex. 23, p. 205.] 15*. (cont.) Prove that

■lh"n =0

is a linear function. (It defines a matrix called eA or exp A.)

5. QUADRATIC FORMS

This section paves the way for the study of maxima and minima of functions of several variables in Chapter 9. In one variable calculus, for f ( x ) to have a minimum at a point c where / '( c ) = 0, a sufficient condition is th a t / " ( c ) > 0. In several variables, the analogous second derivative test is more complicated, and requires some knowledge of quadratic forms.

A quadratic form is simply a homogeneous quadratic polynomial. The general quadratic form in two variables is

f ( x , y ) = ax2 + 2 bxy + cy2.

We note immediately a useful expression for / in terms of matrices:

5. Quadratic Forms 245

f i x , y) = (x, y)a b

b c

This expression justifies using 2b instead of b for the middle coefficient. The general quadratic form in three variables is

f {x, y, z ) = ix, y, z )

a b d

b c e y

d e / J

= ax2 + 2 bxy + cy2 + 2 dzx + 2 eyz + fz 2.

The matrices

a b

b cand

a b d

b e e

J ' e / _

have mirror symmetry in the main diagonal (top left to bottom right) and so are called symmetric matrices. This property is concisely stated in terms of transposes:

A matrix A is symmetric if and only if A ' = A .

To each symmetric matrix corresponds a unique quadratic form. Conversely, to each quadratic form corresponds a unique symmetric matrix. This converse statem ent may seem obvious, bu t it does require a proof, which we give for the two variable case.

Suppose

ai bi d2 b%Ax = and A 2 =

bi ci b 2 Ci

determine the same quadratic form. Then

aix2 + 2b\xy + c\y2 = a2x 2 + 2 b2xy + c2y2

for all x and y. The choices (x, y) = (1, 0), (0, 1), and (1, 1) yield

a\ — a2) C\ — c2, b\ = b2.

Hence A \ = A 2. A similar easy proof applies for three variables.


The general quadratic form may be w ritten as x.t x', where A is a unique symmetric matrix.

Examples of quadratic forms:

x2 + y2 = (x , y)1 0

0 1

2xy = (x, y )0 1

1 0

ax2 + by2 + cz1 = (x , y , z )

a 0 0

0 6 0

0 0 c.

0 c b

2cxy + 2ayz + 2bzx = (x , y, z) c 0 a

6 a 0

Positive Definite Matrices

Take a quadratic form in two variables,

f i x , y) = x.4x' = ax2 + 2 bxy + cy2,where

x = (x, y) and A =a b

b c

We want to find conditions on the coefficients a, b, c so th a t

f i x , y) > 0 for all {x, y) ^ (0, 0).When this is the case both / and A are called positive definite.

The simplest case of a positive definite quadratic form is f i x , y) = The associated positive definite matrix is

‘1 01A =

0 1Other examples:

1 1f i x , y) = x2 + 2xy + 5y2 = ix + y )2 + 4y2, A =

-1 51

f i x , y) = x2 — 6xy + 11 y2 = ix — 3y)2 + 2y2, A -- 3

x2 + y2.

-3

11

To check th a t (x — 3y )2 + 2y2 is positive definite, suppose (x } y) 9 (0, 0). If y 0, then

(x - 3y ) 2 + 2y2 > 2y2 > 0.

If y = 0, then x ^ 0 and

(x — 3y ) 2 + 2y2 = x2 > 0.

We are ready to state the basic facts about positive definite matrices. We begin with the 2 X 2 case:


Theorem 5.1 The symmetric matrix

a b

b c

is positive definite if and only if

a > 0 anda b

b c> 0.

Proof: Suppose the matrix is positive definite. Then

f ( x y y) = ax2 + 2bxy + cy2 > 0

whenever (x , y) ^ (0, 0). In particular a = / ( l , 0) > 0. Now complete the square:

f ( x , y ) = a ( x + V2-

T h e n /( — 6/a, 1) > 0, hence

ac — b2--------- > 0.

a

Therefore a > 0 and ac — b2 > 0.Conversely, suppose a > 0 and ac — b2 > 0. Then certainly

x / b V (ac - b 2\ f (x , y) = a l x + - y ) + I — - — J y2 > 0

for any (x, y). F u rth e rm o re ,/^ , y) = 0 only if each of the squared quantities is zero:

bx + - y = 0, y = 0,

a

hence only for (x, y) = (0, 0). This completes the proof.

There is a corresponding theorem for 3 X 3 symmetric matrices. A proof based on “completing the square” is possible, bu t it is long and tedious, so we shall omit it. In linear algebra courses other types of proofs are often developed. (Compare the remark after Theorem 8.8.)


a b d

b e e

_d e / _

is positive definite if and only if

a b d

a > 0,a b

b c> 0 , b e e

d e f

> 0.

Negative Definite Matrices

Let A be a symmetric matrix a n d /(x ) = xAx' the corresponding quadratic form. We say th a t A and / are negative definite if / ( x ) < 0 except forx = 0. I t is p re tty clear th a t / ( x ) < 0 is the same thing as — / ( x ) > 0, hence A is negative definite if and only if —A is positive definite, where — A denotes the matrix whose elements are the negatives of the elements of A (it is the matrix of —/) . A corresponding modification of Theorem 5.2 gives the following test for negative definiteness:


a b d

b e e

jd e f _

is negative definite if and only ifa b d

a ba < 0, > 0 , b e e < 0.

b cd e f

The corresponding conditions for a quadratic form ax2 + 2bxy + cy2 in two variables are simply a < 0, ac — b2 > 0.


A Lower Bound

The following result is essential in our discussion later of maxima and minima. We prove it for three variables, bu t an immediate modification covers the two variable case.

Theorem 5.4 L e t / ( x ) be a positive definite quadratic form. Then there exists a constant k > 0 such th a t

/ ( x ) > k |x|2for all x.

Proof: Let S = {x | |x| = 1} denote the unit sphere, a closed and bounded set. The function / is continuous on S, so it has a minimum on S. T hat is, there is a point x0 £ S such th a t / (x ) > / ( x 0) for all x £ S. We set Jc = f ( x 0). S in c e /is positive definite, we have k > 0.

Let x ^ O . Then x/|x | £ S, hence / (x / |x |) > k. B u t / ( x / |x |) = / ( x ) / |x |2 s in c e /is a quadratic form. Therefore/ (x ) /|x |2 > h, and fin a lly /(x ) > k |x|2.

Semi-definite Forms

Positive definiteness is defined by /(x ) > 0 for all x ^ 0. A quadratic fo rm /(x ) is called positive semi-definite if

/ (x ) > 0 for all x.

The property negative semi-definite is defined similarly. We use the same adjectives for the symmetric matrix corresponding to the fo rm /(x ) .

Theorem 5.5 The quadratic form /(x ) = ax2 + 2bxy + cy2 is positive semi-definite if and only if

a > 0, c > 0, ac — b2 > 0.

Proof: Suppose/(x) is positive semi-definite. Let h > 0. Then

/ ( x ) + h(x2 + y2) = (a + h)x2 + 2 bxy + (c + h)y2

is positive definite. By Theorem 5.1,

a + h > 0, (a + h) (c + h) — b2 > 0.Also c + h > 0 as we see, for instance, by interchanging x and y. Now let h -------- > 0. Then we conclude th a t

a > 0, c > 0, ac — b2 > 0.

Conversely, suppose these three inequalities hold. Let h > 0. Then a + h > 0, c + h > 0, and

(a + h)(c + h) — b2 = (ac — b2) + (a + c)h + h2 > h2 > 0.

Hence (a + h)x2 + 2bxy + (c + /i)?/2 is positive definite. Let (x, y) 9 0. Then

(a + h)x2 + 2bxy + (c + h)y2 > 0.Let h -------- > 0:

/ (x ) = arc2 + 2bxy + cy2 > 0.

T herefo re/(x ) is positive semi-definite.

R e m a r k 1: An alternate proof can be constructed by modifying the proof of Theorem 5.1.

R e m a r k 2 : The corresponding conditions for /(x ) to be negative semi- definite are

a < 0, c < 0, ac — b2 > 0.

We shall state without proof the corresponding results for three dimensions.


a b d

A = b e e

d e f .

is positive semi-definite if and only if all of its principal minors are nonnegative, th a t is,

a > 0, c > 0, / > 0,

ac - b 2 > 0, cf — e2 > 0, af - d2 > 0,

a b d

\A\ = b e e > 0 .

d e f

The conditions for A to be negative semi-definite are

a < 0, c < 0, / < 0,

ac — 62 > 0, cf — e2 > 0, af — d2 > 0

|A| < 0 .

For instance, consider

A =

1 1 1

1 1 1

1 1 1

Then a = c = f = I > 0, ac — b2 = cf — e2 = af — d2 = 0 > 0, and \A\ =0 > 0, so A is positive semi-definite. If x = (x, y, z), note that/(x) = xAx' = (x + y + z)2 > 0, confirming the diagnosis. (But A is not positive definite; for instance/(l, 1, — 2) = 0.)

Indefinite Forms

A quadratic form/(x) = xAx' is called indefinite if it is neither positive nor negative semi-definite, that is, if there are points Xi and x2 such that /(xi) < 0 and /(x2) > 0.

Theorem 5.7 A form /(x ) = <ax2 + 2bxy + cy2 in two variables is in-definite if and only if

a b< 0.

b c

Proof: First suppose ac — b2 < 0. By Theorem 5.5, /(x ) is not positive semi-definite. By the corresponding result mentioned in Remark 2 above, / (x) is not negative semi-definite. Therefore/(x) is indefinite.

Now suppose/(x ) is indefinite. We examine three cases: (1) a > 0, c > 0. Then ac — b2 < 0, otherwise /(x ) is positive semi-definite by Theorem 5.5.(2) a < 0, c < 0. Then ac — b2 < 0, otherwise/(x ) is negative semi-definite.(3) ac < 0. Then ac — b2 < ac < 0. This exhausts all possibilities; the proof is complete.

There is not such a simple criterion for indefiniteness .of a 3 X 3 matrix

a b d

A = b c e

J e f .

The best we can do is try the test in Theorem 5.6. If A fails to be positive semi-definite, and if it fails to be negative semi-definite, then it must be indefinite. Thus we first inspect a, c, /. If two have different signs, A is indefinite; if not we inspect the 2 X 2 principal minors

ac — 62, cf — e2, af — d2.

If one of these is negative, A is indefinite; if not we inspect the determinant \A\. If its sign differs from that of a, c, and e, then A is indefinite; if not, A is positive or negative semi-definite.

For instance, consider

A =

- 1 1 1

1 - 1 1

1

Then a = c = f = — 1 < 0, so we pass to 2 X 2 principal minors: ac — b2 = cf — e2 = af — d2 = 0, so we still do not know if A is or isn’t indefinite. But |A| = 4 > 0, so A is indefinite. If this is not a convincing argument, then here is an overwhelming one: set x = (1,0,0) and y = (1, 1, 1). Then xAx' = - 1 < 0 and yAy' = 3 > 0.

EXERCISES

Write out the quadratic form corresponding to the symmetric matrix:r 4 - i 1 ‘ - I 5

2.L - i 3J 5 —2"1 0 0" "1 2 3“

0 2 4 4. 2 4 6

O 00 1 _3 6 5-

3.

Find the symmetric matrix of the quadratic form:5. x2 — y2 6. xy7. (z + ?/)2 8. (x— 2y)29. (ax + by)2 10. (ax + by) (cx + dy)

11. (x + y)(y + z) 12. (x + 2y + 3z)213. (ax + by + cz)2 14. (ax + by + cz) (ax + fiy + yz).

Determine whether the quadratic form is positive definite:15. x2 + 4xy + 2y2 16 9x2 — 12xy + y217. x2 + fay + 10y218. 3:r2 + 2xy + y2 — 2zx + 3zy + z2.19. Let A be any 3X 3 matrix, not necessarily symmetric. Then/(x) = xAx' is ;

quadratic form. Find its matrix.20. (cont.) Do the special case

^0 0 0'

A = 2 0 0

-4 6 0.21. Prove that the sum of two positive definite quadratic forms is again positive

definite. Find the corresponding matrix.

6. Quadric Surfaces 253

22. Let A be symmetric, x = (xi, x2, £3), and/(x) = xAx'. Prove that

OXi d x 2 d x s

23. Let A be any 3X 3 matrix. Prove that J (A + A ' ) is symmetric.24. Let A be any 3X 3 matrix. Prove that A A' is symmetric.25. Suppose A and B are symmetric 3X 3 matrices that commute, that is, AB = BA.

Prove that AB is symmetric.26*. Suppose A and B are positive definite 2X 2 matrices. Is the quadratic form

/(x) = xABx' positive definite? (Note that AB need not be symmetric.)27. The form x2 + 2xy + 2y2 is positive definite so there is a k > 0 such that x2 +

2xy + 2y2 > k (x2 + y2) for all (x,y). Find the largest such k.28*. (cont.) Solve this problem in general for a positive definite

f i x ) = ax 2 + 2 bxy + cy2.

29. Prove the statement in Remark 2, p. 250.30. Carry out the proof suggested in Remark 1, p. 250.

31*. Let f ( x , y ) = a x 2 + 2 bxy + cy2 be an indefinite form. Prove that f i x , y ) is the product of two non-proportional linear functions. [Hint: Rotate coordinates.]

32. (cont.) Sketch in the plane the regions / > 0, / < 0, / = 0.

6. QUADRIC SURFACES

A quadric surface is the graph of an equation f(x, y , z) = 0, where f (x, y, z) is a quadratic polynomial. The most general quadratic polynomial / is the sum of a quadratic form and a linear function,

f (x, y ,z) = Q (x ) + px + qy + rz + k.

The quadratic form Q involves the pure terms x2, y2, z2 and the mixed terms xy, yz, zx. According to a fairly deep result in linear algebra, the mixed terms can be eliminated by rotating the coordinate system. We shall accept this result without proof. It means that we may confine our study of quadric surfaces to the cases where there are no mixed quadratic terms.

Thus we consider f(x, y, z) = 0 , where

f i x , y , z) = Ax2 + By2 + Cz2 + px + qy + rz + k.

Of course, if there are no quadratic terms, then / = 0 represents a plane, which doesn’t interest us here.

If A 9* 0, then a translation in the ^-direction eliminates the term px, and similarly if B 9* 0 or C 9* 0. This reduces our study to the following types of /:

(i) Ax2 + By2 + Cz2 + k

(ii) Ax2 + By2 + rz + k

(iii) Ax2 + qy + rz + k.

These include all possibilities, provided we are willing to permute the variables.

For instance, the polynomial Cz2 + px + qy + k becomes type (iii) when x and z are interchanged.

In type (ii), if 0, then a translation in the 2-direction eliminates k. In type (iii), a rotation in the y, z-plane, taken so that qy + rz = 0 is the new 2-axis, changes the function to the form ax2 + qy + k. Again, k can be eliminated by translation if q 9* 0.

This reduces our study to the following types of /:

(1) Ax2 + By2 + Cz2 + k

(2) Ax2 + By2 + rz (2') Ax2 + By2 + k

(3) Ax2 + qy (3') Ax2 + k.

We begin with (1) in case ABC 7* 0 and k 9* 0. Then / = 0 can be written in the form

x2 y2 z2 a1 bl cl

If all three signs are minus, the graph is empty. Otherwise the graph is symmetric in each coordinate plane because if (x , y, z) is on the graph, so are all eight points (d=£, ± y , ± z ) . Therefore it suffices to draw the first octant portion of the graph and determine the rest by symmetry.

Ellipsoids

Consider the graph of

X2 V2 z2+ t2 + — 1, a,b,c > 0. a2 b2 c2

Since squares are non-negative, each point of the graph satisfies

- < 1, - < 1, - < 1. a2 ~ ’ b2 ~ ’ c2 ~

This means the graph is confined to the box

— a < x < a, —b < y < bj —c < z < c.

Suppose — c < z0 < c. The intersection of the graph and the horizontal plane z = z0 consists of all points (x , y, z0) that satisfy

x2 y_ = z j a2 b2 c2 ’

This curve is an ellipse. It is as large as possible when z0 = 0, and it becomessmaller and smaller as Zo------- >c or z0--------> —c. Thus each such cross-section by a horizontal plane is an ellipse, except at the extremes zo = ±c, where it is a single point.


The same argument applies to plane sections parallel to the other coordinate planes. This gives us enough information for a sketch. The surface is called an ellipsoid (Fig. 6.1). In the special case a = b = c} it is a sphere.

(a) Horizontal cross-sections (b) the complete ellipsoidare ellipses.

Fig. 6.1 graph of ^ ^ ^ = 1a‘ o‘ &

Hyperboloids of One Sheet

Consider the graph ofv2r, 2•r tr _ * =a2 b2 c2

a , b , c > 0.

Each horizontal cross-section is an ellipse

z = Zo

& r i ,a2 b2 c2 ’

no matter what value z0 is. The smallest ellipse occurs for z0 = 0; as z0-or zo------- » — oo f the ellipses get larger and larger.

The surface meets the y, 2-plane in the hyperbola

t . - - = ib2 c2


and it meets the z, #-plane in the hyperbola

x* - i a2 c2

This information is enough to sketch the surface, called a hyperboloid of one sheet (Fig. 6.2a).

y = 0

?! _ — a2 c2

(a) one sheet:?! yl _ Z1 - ia2 b2 c2

_ ?! _ yl 4- zi = ia2 b2 ~t~ c2

Fig. 6.2 hyperboloids

Hyperboloids of Two Sheets

Consider the equation

_ _ _ ^ x _ = 1 a2 b2 c2

If (x, y, z) is a point on the surface, then

X‘

a , b , c > 0.

c2 a2 o2

hence z2 > c2. This means either z > c or z < —c, that is, there are no points of the surface between the horizontal planes z = c and z = —c.

If Zo2 > c2, the horizontal plane z = z0 meets the surface in the curve

Z = Zo

*2 + t = zl _ 1 > oa2 b2 c2 ’

an ellipse. Also the surface meets the y , 2-plane and the 2, rc-plane in the hyperbolas

“ w + 1 “ 1 and “ ~2 + 1 = 1 b2 c2 a2 c2

respectively. The surface breaks into two parts, and it is called a hyperboloid of two sheets (Fig. 6.2b).

Cones

Now we complete the study of type (1) on p. 254 for the case ABC 9* 0, but k = 0. Then / = 0 can be written

X2 V2 z 2 i — i — = 0 , a , b , c > 0 .

a2 b2 c2

If the signs are all the same, then (0, 0, 0) is the only point on the graph; not interesting. Otherwise two signs are equal, the other opposite. Changing signs if necessary, we are reduced to

2 x\ y 2 h ^22 = — + — , a,b > 0.a 2 b2

It is a surface with the following property. For each point x0 on the surface, the entire line x = £x0 lies on the surface. Such a surface is called a cone, and the lines x = tx0 are called generators of the cone.

To check that the graph of

2 x 2 iy2z = --------a2 b2

is a cone, we take any point x0 on the graph and check that t x 0 is also on the graph. If x0 = (%o, yo, 20), then t x 0 = ( t x 0, t y 0, t z 0) and

)2 + (tyo)2b2

Thus tx0 is on the graph.To sketch the cone, we note that it meets the horizontal plane z = 1 in


the ellipse

- + - = 1 .

a2 b2

For each point on this ellipse, we draw the line through the point and 0. See Fig. 6.3.

Paraboloids

Next we take up case (2) on p. 254 and study the graph of / = 0, where

f ( x , y , z) = Ax2 + By2 + rz, AB ^ 0, r ^ 0.

We may write the equation / = 0 in the form

x2 y2z = ± — ± — , a, b > 0.

a2 b2

If both signs are minus, replacing z by —z changes both signs to plus; therefore we are reduced to two cases.

The first case is the surface

X2 V2z = ~2 ^ h2 ’ a , b > 0 , a2 b2

called an elliptic paraboloid. It lies above the x, ?/-plane, and it is symmetric in the y, z- and z, ^-planes. Each horizontal cross-section

1 + U = 20 > 0a2 b2

is an ellipse, and these ellipses grow larger as z0 increases. The graph meets the y, z- and zy ^-planes in parabolas z = y2/b2 and z = x2/a2 respectively (Fig. 6.4a).


(a) elliptic paraboloid:3* £

a2 62(b) hyperbolic paraboloid:

* = - £ + £ a2 62Fig. 6.4 paraboloids

The second case is the hyperbolic paraboloid, the locus of

)‘2 + -r»2_ x _ r

a2 b2 'a, b > 0.

It is symmetric in the y, z- and z, ^-planes. The horizontal planes z = Zo > 0 meet it in hyperbolas whose branches open out in the y-direction. The horizontal planes z = Zo < 0 meet it in hyperbolas that open out in the rc-direction. The y , 2-plane meets the locus in the parabola z = y2/b2, which opens upwards;

and the z, #-plane meets it in the parabola, z = —x2/a2, which opens downwards. The best description is “saddle-shaped”. See Fig. 6.4b.

Cylinders

In cases (2') and (3) on p. 254, the variable z is missing. In general, when one variable is missing the locus is a cylinder. Take for example Ax2 + By2 + k = 0, where A > 0, B > 0, and k < 0. This can be written in the form

-2 + £ = a’ b > °- a2 b2

The surface meets each horizontal plane z = z0 in the same ellipse. If (#0,2/o, zo) is any point of the surface, the whole vertical line (#o, yo, z) for — oo < z < oo lies on the surface. The surface is an elliptic cylinder and these vertical lines that lie on the surface are called generators of the cylinder (Fig. 6.5). Any curve f(x, y) = 0 in the x, ?/-plane generates a cylinder in space consisting of all points (#0,2/o, z) for which f ( x 0, yo) = 0 . In particular, a circle generates a (right) circular cylinder, a hyperbola generates a hyperbolic cylinder, etc. Case (3) on p. 254 leads to a parabolic cylinder.

x2 y2F ig . 6.5 elliptic cylinder: — + — = 1ai bz

7. Inverses 261

In the final case (3') both y and z are missing. Depending on the signs of A and fc, the locus of Ax2 + k = 0 is empty or consists of one plane or two planes parallel to the y, 2-plane. In general, the locus in R3 of f (x) = 0 is a set of planes parallel to the y, 2-plane. For each zero x0 of f (x ) , the plane x = Xo is included in the locus.

EXERCISESSketch the first octant portion:

1. \x2 + y2 + \z2 = 1 2. z2 + \y2 + iz2= 13. x2 + y2 — z2 = 1 4. —x2 — y2 + z2 = 15. x2 — y2 + z2 = 1 6. —x2 + y2 — z2 = 17. z = x2 + y2 8. z — \x2 + y29. z = —x2 + y2 10. z = x2 — y2.

11. Identify the quadric surface z — x2 + 2x + y2. [Hint: Complete the square.]12. Identify the quadric surface z — xy. [Hint: Rotate 45° around the 2-axis.]Sketch the paraboloids:13. x = y2 + z2 14. y = x2 — z2.Sketch the surface in x, y, 2-space:15. x — z = 1 16. y = x217. xy = 1 18. - x 2 + y2= 119. x = 22 20. y2 + 422 = 121. z2= x2 + y2 22. z2 + 422 = 123. y2 = z2 + 4z2 24. z2 — xy.25. Suppose f(x, y) = 0, 2 = 1 is a curve on the plane 2 = 1. Find an equation for the

cone obtained by taking all points on all lines through 0 and points of the curve.26. (cont.) Test your result on x2 + y2 = 1, z = 1.27. Let f(y, z) = 0 be a curve in the y, 2-plane. Find an equation for the surface of

revolution obtained by revolving the curve around the y-axis.28. (cont.) Test your result on the curve y2 + 9z2 = 1.

7. INVERSES

The square matrix I with ones on the main diagonal and zeros everywhere else is called the identity matrix. The 2 X 2 and 3 X 3 identity matrices are

1 0

0 1

1 0 0

0 1 0

0 0 1

The matrix I behaves very much like the number 1. If x' is a column vector, then Ix' = x'; if A is a matrix (the same size as I ) then IA = A I = A.


Here is an important practical question. Given a 2 X 2 o r 3 X 3 matrix A, is there a matrix B (of the same size) such that AB = I and BA = /? Not necessarily; for example, if

1 0“and B =

'bu b 12

0 0 _&21 b 22then

bn b 12 'bn O '

AB = BA =

0 0 b 2i 0

Therefore, both equations AB = I and BA = I are impossible for any choice of B.

Let A be a square matrix for which there exists a matrix B satisfying AB = I and BA = I. Then we write B = A~l and call A~l the inverse of A. Thus

A A - 1 = A~'A = I .

R e m a r k : It is reasonable to say “the” inverse. For if A B = BA = 7, and also AC = CA = /, then

B = B I = B (A C) = CBA)C = IC = C.

Some matrices have inverses and some do not. How can we tell whether A~l exists? We shall derive a simple test using determinants.

Recall that to each square matrix A is associated a real number \A\, the determinant of A. For instance |/| = 1, as is easily checked. One of the most important properties of determinants concerns the determinant of a product of matrices.

Theorem If A and B are square matrices of the same size, then

________________________ 1^1 = |A[ IB|.________________________

The proof is not easy; we shall omit it.Suppose A has an inverse, B. Then AB = I, so by the theorem above

\AB\ = |/|, \A\ |2 ?| = 1 .

Conclusion: \A\ j* 0. Thus for A to have an inverse, it is necessary that \A\ 7* 0 . We are going to show that this condition is sufficient as well. It will then follow that the matrices possessing inverses are precisely those whose determinants are non-zero.

R e m a r k : The situation is similar to that for reciprocals of real numbers. The real number a has an inverse arl such that aar1 = arla = 1 if and only if a 0.

7. Inverses 263 \

Suppose \A\ 7* 0. We shall prove that A~l exists by actually computing it. The cofactor matrix cof A of a 2 X 2 matrix A is defined by

Cofactor Matrix

a b d - b, cof A =

c d_ —c a

For a 3 X 3 matrix, cof A is defined by

mn - m 21 rrizi

cof A = — m 12 rri22 — m32

miz — rri2z m33.

Here mty is the minor of a»y in A. Note that the signs alternate and that the i, /-element of cof A is not =tm*y. Precisely, cof A = \_dj] whereC {j = ( 1 )

The cofactor matrix cof A has an important relation to the original matrix A. We shall prove the basic formula:

A (cof A ) = (cof A ) A = \A\I.

This is easy for a 2 X 2 matrix:

A (cof A ) =a 6 d —6 ad — be 0

3 1e011 0 ad — bc_

w i ,

and similarly (cof A ) A = \A \ I. For a 3 X 3 matrix we have

mu -7/121 mzi an au aiz

(cof A ) A = — mi2 ni22 — niz2 «21 «22 a2z

mi3 - m 2z mSz_ _azi &32 azz_

Call the product [&»■/]• A typical element on the principal diagonal is

6 i i = ninCLn — Wi2i&2i + wizid&i.

This is nothing else but the expansion of \A\ by minors of the first column. Therefore bn = \A\.

A typical element off the diagonal is

&12 = VI11CL12 — VI21CI22 ^ ^ 31^ 32*

264 7. LINEAR FUNCTIONS AND MATRICESt

It is the expansion by minors (of the first column) of

&12 &12 &13

Cl 22 $22 ^23

&32 ^32 (I33

But this is the determinant of a matrix with two equal columns. Therefore6 1 2 — 0 .

We see that all elements on the principal diagonal of (cof A ) A are |4|, and all elements off are 0. Therefore (cofyl)il = |A| I. Using expansion by minors of rows, we can prove A (cof A ) = \A \ I similarly.

Example:'3 0 -1"

A =

cof *4 =

+

+

1 - 2

- 3 1

2 - 2

4 1

2 1

4 - 3

+

2 1

4 - 3

0 - 1

- 3 1

3 - 1

4 1

3 0

4 - 3

- 2

1

+

+

0 - 1

1 - 2

3 - 1

2 - 2

3 0

2 1

- 5 3 1

-1 0 7 4

-1 0 9 3

A (cof A ) = - 5 1 = (cof A)A , \A\ = -5 .

Let us return to the question of inverses. Assume |A| ^ 0. From the rule

A (cof A ) = (cof A )A = \A\ I,follows

A matrix A has an inverse if and only if \A\ 9* 0. If \A\ 9* 0, then

A~l = — cof A. \ A \

7. Inverses 265

Example 1:

A =5 3

- 3 1\A\ = —4 5* 0.

cof A =1 -3" i 1 -3"

— A - 1 = ----- 3 5 4 - 3 5

Check:

1

lCO1

1 - 3 ' 1

'o11 —

4 3 1 . - 3 5~ ~ 4

---------------------------------11o

—_—

1

= 7.

Example 2:

A =

1 0 1

2 - 1 1

- 2 - 2 1

\A\ — —5 0.

1 - 2 r 1 - 2 1“

cof A = - 4 3 1 A- i - - -

’ A " 5- 4 3 1

- “ 6 2 -1 _ — 6 2 1_

Check: A~lA =

1 - 2

- 4

- 6

3 1

2 - 1

1 0 1“ ”- 5 0 0”

2 - 1 1 1“ “ 5

0 - 5 0 = 7.

- 2 - 2 1 0 0 - 5

T e r m in o l o g y : A square matrix that has an inverse is called non-singular. A square matrix that does not have an inverse is called singular. Thus a square matrix A is non-singular if and only if \A\ ^ 0, and A is singular if and only if |A | = 0.

R e m a r k : Given A , if there is a B with AB = 7, then \A\ 0 so A has an inverse. This inverse must be B because,

A - 1 = A - 1 (I) = A - 1 (AB) = (A~1A ) B = IB = B.

Similarly if CA = /, then A has an inverse and A~l = C.


Linear Systems

A 3 X 3 linear system (three equations, three unknowns)

anXi + anX2 + «i3 3 = bi

* Cl2lX l -|- &22X2 ”1“ d 23%3 = 62

, (L31#1 + ^32^2 + «33^3 = 63

can be written neatly in matrix form as

Ax' = b',where

a 11 d l 2 d 13 X l "61"

02 1 Cl22 « 23 , x ' = X 2 , b ' = 62

_&31 ^32 &33_ _X3_

Suppose the matrix A is non-singular. Then we can multiply both sides of the equation by A-1:

A - 1 (Ax') = A~lb', Ix' = A~l b', x' = A~l b'.

Therefore, if there is a solution, it can only be the vector x' = A~lb'. But this vector is in fact a solution. Check:

Ax' = A (A ~ lb') = (AA~l)W = lb ' = b'.

A linear system Ax' = b' with A non-singular has the unique solution

x' = A~lb'.

R e m a r k : Note the formal similarity between the linear system Ax' = b' and the elementary equation ax = b. The first is solved by multiplying both sides by A~~l, the second by multiplying both sides by 1/a = arl.

EXAMPLE 7.1

Solve

"3 0 - 1 “ xi “ 15"

2 1 - 2 x% = 0

_4 - 3 1_ _ 5 _

Solution: Evaluate \A\, say by minors of the first row:

|il| = 3 ( -5 ) + 0 + ( —1) ( — 10) = -5 .

7. Inverses 267

Since |A| 9* 0, the matrix A is non-singular, and

■ - 5 3 1

A-1 = —77 cof A = — - - 1 0 7 4w 5

- 1 0 9 3Therefore the solution is

x' = A - 1 b'5

- 5 3 1

-1 0 7 4

-1 0 9 3

"15" "14"

0 = 26

_ 5_ _27_

Answer: xi = 14, = 26, #3 = 27.

The solution of Ax' = b' is

x' = A _1b' = t~t (cof A )b '.\M

From this formula follows Cramer’s Rule for the solution of a linear system. If rriij denotes the minor of an in A, then

(cof A) b' =

mu — ra21 m31

—mi2 m22 — m32

mi3 — m23 m33

W21161 — 77 2162 H" nisibs

— 777-1261 -| - 7722262 — 7723263

7ni36i — 7 2362 + m3363

Look carefully at the first element, mn6i — m2i62 + 7723163. It is the value of the determinant

6 1 a\2 « i3

6 2 & 2 2 ^23

6 2 &32 CZ33

obtained as the expansion by minors of the first column. A similar statement applies to the other elements. The result is a set of explicit formulas:

Let A = (ci', c2', c3') be a 3 X 3 non-singular matrix. Then the solution of Ax' = b' is given by [x\, x2, £3]', where

Xi =D ( b', c2', c3') Z)(ci', b', c3')

Xi = ------ rn------ > xsD (ci', c2', b')

|A| ’ ' | A | ' ‘ | A |

In other words, to obtain x i} replace the 2-th column of A by b', take the determinant, then divide by |A|.

EXERCISES

Find A”1 where A is:

Find A”1 where A is:’ 1 1 1" " 2 0 0”

11. 1 - 1 2 12. - 1 3 0

-1 1 4_ _ 1 2 - 1_— 2 2 6" " -1 2 - 9 “

13. 1 2 9 14. 1 3 - 2

_ 1 3 2_ - 2 2 6-“4 - 3 1“ "1 1 3“

15. 2 1 2 16. 2 4 7 •

.3 0 —1- 1 to 00 9-Solve by Cramer’s Rule:

“ 1 1 - 1“ “3“ "0 2 — 3” "0"

17. 1 - 1 1 x = 2

00▼■H 1 1 1 x = 1

_ - l 1 1- - 1- -0 3 5- -0-" 2 1 3 " " 1“

ICO11<N1

“1“

19. - 1 4 2 x = 0 20. (N1rHi x = 1

- 3 1 1 - - - 1- . 3 - 1 - 1- - 1_21. Let A and B have inverses. Prove that AB has an inverse and (AB) 1 = B lA \22. (cont.) Prove that cof (AB) = (cof B) (cof A ) if A and B are non-singular.

8. Characteristic Roots 269

23. Let A have an inverse. Prove that its transpose A' has an inverse and (A')_1 = (A*1)'.

24. Prove that a triangular matrix

A =

"an ai2 d\z

0 022 &23

0 0 a33

has an inverse if and only if ana22«33 9 0. If this condition is satisfied, prove that A”1 is also triangular.

The following exercises outline another method for inverting matrices. Set“0 1 0“ ’ 1 0 0“ “1 0 c~

Pl2 = 1 0 0 , A(c) = 0 1 0 , R*i(c) = 0 1 0

-0 0 1_ -0 0 C - _0 0 1_

25. Show that PuA is the result of interchanging row 1 and row 2 of A.26. Show that D3(c)A is the result of multiplying row 3 of A by c.27. Show that R3i (c)A is the result of adding c times row 3 of A to row 1.28. Write out P23, £>1 (c), and #12(0).

29*. Let A be non-singular. Show there are matrices Qi, Qz, • • •, Qn such that(1) QnQn—i* • * Q2Q1A = I,(2 ) each Q is a P»y or a Di(c) with c 9* 0 or an Rij(c).

30. (cont.) Interpret this statement as follows:Apply a sequence of elementary row operations to A (interchange rows, multiply a row by a non-zero constant, add a multiple of one row to another) so as to change A to I. Apply the same sequence of operations to I. The result is A-1.

Try the method of Ex. 30 on the A of:31. Ex. 1 32. Ex. 233. Ex. 11 34. Ex. 1235. Ex. 13 36. Ex. 16.

Let A = A (t) = [aty(0] be a matrix whose elements are differentiable functions. Define its derivative by A' = [_dn(t)~\. Prove:37. (A + BY = A* + B- 38. (/A)* = /*A +/A*39. (AB)' = A-B+ AB' 40. (A-1)* = — A-1 A’A-1.

8. CHARACTERISTIC ROOTS [optional]

Let A be a 2 X 2 matrix. A characteristic vector of A is a non-zero column vector x' such that Ax' is a multiple of x':

Ax' = Ax',where X is a scalar.

Given a vector x', the vector Ax' generally does not resemble x' at all, whereas for a characteristic vector, Ax' is simply a multiple of x'. Therefore a


characteristic vector is a rare animal, in fact, it is not obvious whether A has any characteristic vectors at all. One thing is clear though: if x' is a characteristic vector, then so is any non-zero multiple y' = ex'. For

Ay' = A (ex') = c (ix ') = c(Xx') = X(cx') = Xy'.

Let us investigate the existence of characteristic vectors. We write the condition Ax' = Xx' in the form

(XI - A)x ' = 0,

and ask whether there is a non-zero x' that satisfies this homogeneous system for some scalar X. Set

A =

The question we ask is whether there is a scalar X for which the homogeneous linear system

a b Xand x' =

c d JJ.

X — a b X " 0"

c \ — d JJ. _0

has a non-trivial solution. The answer is yes if and only if there is a X for which the determinant of the system is 0. Thus X must satisfy/(X) = 0, where

f i t ) = |tl - A\ =

= it

t — a — b

—c t — d

a) (t — d) — be = t2 — (a + d)t + (ad be).

The quadratic polynomial f(t) is called the characteristic polynomial of A, and its zeros are called the characteristic roots of A .

Conclusion: For each characteristic root X of A, there is a characteristic vector x' with Ax' = Xx'; conversely, each characteristic vector corresponds in this way to a characteristic root.

Terminology: Several adjectives are often used where we have used “characteristic”. The most common ones are “proper”, “eigen” (from German), and “latent”. Characteristic roots are also called characteristic “values”.

EXAMPLE 8.1

Find all characteristic roots and vectors of

'2 - 1 '.1

4 — 3•


Solution:

f i t ) = \ t l - A\ =t - 2 1

- 4 t + 3

= (£— !)(£ + 2 ).

= (t - 2)(* + 3) + 4 = t2 + t - 2

The characteristic roots are X = 1 and n = — 2 . The corresponding homogeneous systems (XI 7- A )x r = 0 and I — A ) y' = 0 are

y' = 0." -1 1“ "-4 1"

x' = 0 and

1— 1 - 4 1

Obvious non-trivial solutions are

Y Yx' = and y' =

1 4

Any non-zero multiple is also a characteristic vector.

Y YAnswer: X = 1, x' = a

1; m *= - 2, y' = b

4

The characteristic polynomial f i t ) of a 2 X 2 matrix A is quadratic. It has two complex roots which may or may not be real. Hence A will always have a characteristic vector if we admit complex scalars and vectors with complex entries. Otherwise, the discussion that follows applies only to matrices with real characteristic roots. (See Chapter 15 for complex numbers.)

Now f ( t ) = t2 — (a + d)t + (ad — be), so the two characteristic roots are real provided

A = (a + d)2 - 4 (ad - be) > 0.

The roots are distinct if A > 0, equal if A = 0.

Basic Structure Theorems

Suppose a certain problem involves a 2 X 2 matrix about which we know nothing. Now the general 2 X 2 matrix has four arbitrary elements, so is awkward to handle. The following results show that in many situations a general 2 X 2 matrix can be replaced by one of two simple types,

"X 0"or

X f

_0 _0 X

There are many practical applications.


Theorem 8.1 Let A be a 2 X 2 matrix with distinct characteristic roots X and /i. Then there is a non-singular matrix P such that

P- 'A P =X 0

0 ju

Proof: Let x' and y' be corresponding characteristic vectors. Then

Ax' = Xx', A y ' = ny',

x' 0, and y' ^ 0. Set P = (x', y' ) , considered as a 2 X 2 matrix. Then

A P = A ( x ' , y ' ) = (Ax', Ay ') = (Xx',»y') = (x' ,y')

If P is non-singular, then we multiply the last equation by P~l:

X o' 'x o'= p

0 M 0 M_

/ 'x o' \= P - ' ( A P ) = p-- l ( p

V 0 n /'x o' X 0" "X 0"

= (P~'P) = I =o M. 0 jU 0 n

It remains to prove that P is non-singular. If not, then the homogeneous system

a0

has a non-trivial solution. That is, there exist a and b, not both zero, such that ax' + by' = 0. Multiply by A :

aAx' + bAy' = 0, aXx' + buy' = 0.

Now eliminate x' from ax' + by' = 0 and aXx' + buy' = 0:b(X — n)y' = 0.

Since X ^ /x and y' ^ 0 , we have 6 = 0. Similarly a = 0, a contradiction.

EXAMPLE 8.2

Apply the theorem to the matrix of Example 8.1.

Solution: The matrix is

A =2 - 1

4 - 3


and we found the characteristic vectors

Y Yx' = and y' =

1 4

corresponding to X = 1 and n = — 2 . Set

P = (x', y') =1 1

1 4Then |P| = 3 9* 0 and

'2 - 1' 1 1" '1 — 2AP = = j

4 - 3 1 4 1 “ 8.

'1 O' "1 ll "1 0" "1 - 2

0 - 2 1 4 _0 - 2 1 - 8

'1 O' '1 O'

II p - iA p =

O 1 fcO 1 O 1 to

Theorem 8.2 Let A be a 2 X 2 matrix with equal characteristic roots X, X. Then there is a non-singular matrix P such that either

'\ O' ‘x 1'P - 'AP = or P~lA P =

0 X 0 X

Proof: Let x' be a characteristic vector, so x' ^ 0 and Ax' = Xx'. Choose any vector y' so that P = (x', y') is non-singular. Then x' and y' are non-collinear with 0, so the vector Ay' can be expressed as Ay' = ax' + fry'. In fact, if we set

then

p-iAy' =

Ay' = (PP_1) (Ay') = P(P~lAy') = P = ax' + by'.


Therefore

AP = A (x', y') = (Ax', Ay') = (Xx', ax' + by')

X a X a= (x', y') = P

0 b 0 b

hence P_1AP =X a

0 b

Now an important observation: P_1AP and A have the same characteristic polynomial. For

\tl - P_1AP| = |tP~lP - P_1AP|

= |P~l (tl - A)P| = IP-1! \tl - A| |P|

= It l - A\ IP-1! |P| = \tl - A| \P~lP\ = It l - A|.

L et/(0 denote this common characteristic polynomial. On the one hand, the zeros of f ( t ) are X, X by hypothesis. On the other hand,

f ( t ) = \tl - P - 'A P | =t — X —a

0 t - b

so the zeros are X, b. Hence X = b. Therefore

= (t - \ ) ( t - 6),

Ax' = Xx' X aand P~lA P —

Ay' = ax' + Xy' x-

If a = 0, we are done. If a 5* 0, we simply repeat the argument with x' replaced by ax'. Then

Ax' = Xx', Ay' = x' + Xy'

so, with P modified accordingly,

X 1]P~lA P =

0 XThis completes the proof.

EXAMPLE 8.3

Apply the theorem to

-1 0

3 - 1

Solution: The characteristic polynomial is / (t) = (< + l )2 so the characteristic roots are —1, —1. Let x' be a characteristic vector. Then

\ I — A = —I — A =0 0

-3 0so we must have

An obvious solution is

The easiest choice of y' is

Ty' =

o

o o

- 3 o

x' =

x' = 0.

■ < f

1

so P = (x', y') =

Then

Ax' = — x' and Ay' =

Therefore

- 1 0

3 - 1

0 1

1 0.

- 1

33x' - y'.

r .AP = A{x ' ,y ' ) = {Ax', Ay ' ) = ( -x '^ x ' - y')

"-1 3' ' - 1 3'(x', y') = P

0 - 1 0 - 1

We replace x' by 3x', that is, we now take

After these changes,

"O' 'l" 0 1'

II II ii

CO _o

----------1OCO

so

AP = P

’ - I "- x' and II II xn 1

3

" -1 1" - i rP~lAP =

0 - 1. 0 - 1

IiImII

We shall apply these results in Chapter 15, Section 6 to systems of differential equations.

3 X 3 Matrices

For a 3 X 3 matrix A, characteristic vectors and roots are defined just as * for 2 X 2 matrices. A characteristic vector of A is a column vector x' such that

Ax' = Xx' for some scalar X. Searching for characteristic vectors leads to the characteristic polynomial f{t) = \tl — A |, whose zeros are the characteristic roots of A .

Nowt - au - 0 1 2 013

m = “ 021 t — CJ22 023 — ts -1- bt2 -|- ct -f- d.

“ 031 “ 032 t — 033

a cubic polynomial with real coefficients. Since / has odd degree, there must be at least one real characteristic root. In general, the cubic has either one real and two complex zeros or three real zeros.

Theorem 8,3 Each 3 X 3 matrix has a real characteristic root.

Corresponding to the structure theorems above is a more complicated result which we state without proof.

Theorem 8.4 Let A be a 3 X 3 matrix, and let X, n, v be the characteristic roots of A y not necessarily all real or distinct. Then there is a non-singular matrix P such that P~lA P has one of the following forms:

X 0 0 X 1 0 X 1 o'

0 n 0 J 0 X 0 J 0 X 1 •

o o _0 0 n_

S<oO

R e m a r k : There is only one possible basic form if all three characteristic roots are distinct, but three basic forms if the three characteristic roots are equal, and two basic forms if two roots are equal and the third distinct.

Symmetric Matrices

All characteristic roots of a symmetric matrix are real. That is, if A is an n X n symmetric matrix (with real coefficients), then the characteristic polynomial/ ( 0 factors completely,

f i t ) = (t - X i)(* - X2) - - - ( * - Xn),

where Xi, • • •, Xn are real.The proof of this basic fact, while not hard, belongs to a course in linear


algebra. The 2 X 2 case is easy. For if

A =

then

f i t ) =

t — a — b

a b

b c

= (t — a) (t — c) — b2— b t — c

= t2 — (a + c)t + (ac — 62),

and the discriminant is

A = (a + c)2 - 4 (ac - b2) = (a - c)2 + 4b2 > 0.

Therefore the roots are real.We shall prove the 3 X 3 case in Chapter 9, Section 8 , but the analytic

proof there does not generalize to 4 X 4 or higher cases. We state the result formally.

Theorem 8.5 All characteristic roots o f a 2 X 2 o r 3 X 3 symmetric matrix are real.

The basic structure theorems take a particularly simple form for symmetric matrices. The possibilities

A 1

0 x

simply never occur! The following theorem summarizes the situation.

"x 1 0“ "A 1 0"

0 X 0 > 0 A 1

_0 0 _0 0 A_

Theorem 8.6 If A is a 2 X 2 symmetric matrix, then there is a non-singular matrix P such that

“A 0"P~lA P =

0 /i

I f Ai sa3 X 3 symmetric matrix, then there is a non-singular matrix P such that

"A 0 0"

P~lA P = 0 /x 0

0 0 P

The proof for 3 X 3 matrices is beyond the scope of this course. See Ex. 35 for the 2 X 2 case.

Now we come to an important relation between characteristic roots and definiteness.

Theorem 8.7 Let A be a symmetric matrix with characteristic roots Xi, X2, • • •. Then

(1) A is positive (negative) definite if and only if all X* > 0 (all X; < 0);

(2 ) A is positive (negative) semi-definite if and only if all Xt > 0 (allXi < 0 );

(3) A is indefinite if and only if some Xt > 0 and some Xy < 0.

We shall prove only a small part of this result because the complete proof, especially for 3 X 3 and higher, is hard. We shall show that if A is positive definite, then all Xi > 0. The negative definite case is similar.

Thus let A be positive definite and let X be any characteristic root. There is an x V 0 such that Ax' = Xx'. Therefore

xAx' = Xxx'.

But xAx' > 0 because A is positive definite and x' ^ 0; also xx' > 0 because x' 0. Therefore X > 0.

EXERCISESFind all characteristic roots and vectors:

“21.

3.

5.

7.

9.

11.

j

a

0

'1 2 "

4 3-3 1‘

-1 - 1

4.

6.

8.

10.

12.

CO io

r o 1_"1 1“

1 ■

o 1_“ 1 O'

1— 1 1 - 1- 2 2”

T"

•1 - 1.- 5 - r

r 1 4 2.”4 91_1 J

[-1 1 ]C -I]

‘ - n

r a

ca-

17. Suppose be > 0. Prove that A has real characteristic roots.18. (cont.) Under the same hypothesis, find when the roots coincide.19. Prove X + \x = a + d and X/x = ad — be.20. Prove A', the transpose, has the same roots as A.

Find P so that P~lAP

H ; a

■ c a

■ n

23. A

Find P so that P~lAP -C 3 '

- [ c . a- - u a

1r- 1 01 1r4 9125. 26.1L i - 1 J 1Li 2j

Find all characteristic roots and vectors:"3 0 0“ “2 1 0“

27. 0 1 0 28. 0 2 0

_0 0 2_ -0 0 1_“- 2 0 0" "3 1 0”

29. 1 - 2 0 30. 0 3 1

. 0 1 —2_ -0 0 3_’4 0 9“ ‘3 0 0“

31. 0 0 0 32. 0 1 2 •

_1 0 2. -0 1 2.

33*. Let A

c < n--ca have characteristic roots X < /x. Prove that X < a < n and X <


34*. Let A be a symmetric matrix and x' a characteristic vector. Let y' • x' = 0. Prove that (Ay')• x' = 0.

35*. (cont.) Prove Theorem 8.6 for 2 X 2 symmetric matrices.36. Let A be a 3 X 3 matrix with characteristic roots X, /z, v. Prove that |A| = \nv.

[Hint: Set t = 0 in the identity |tl — A\ = (t — X) (t — n) (t — y).]37*. (cont.) Prove X + ju + v = an + a22 + «33-38*. Let A be a 3 X 3 symmetric matrix. Prove that A = 0 if and only if 0 is the only

characteristic root of A.39. (cont.) Show this is false for some non-symmetric A.40. Let X be a characteristic root of A. Prove that X2 is a characteristic root of A2.

8. Several Variable Differential Calculus

1. DIFFERENTIABLE FUNCTIONS

It may be said that differential calculus is the study of functions which have a linear approximation at each point of their domains. We shall extend this point of view to functions of several variables, but first let us review the one variable situation.

Suppose/(#) has derivative/'(c) at x = c. Then for x near c, the approximation f (x ) ~ f { c ) + f '{c){x — c) is quite accurate. The term “accurate” is not well defined; let us express what we mean geometrically. We know that V = /(c) + f {c)(x — c) is the equation of the tangent line to the graph of y = f i x ) (c,/(c)). See Fig. 1.1. Geometrically, the tangent “hugs” the curve near the point of tangency. Therefore if e(x) is the vertical distance between the curve and the tangent line, e (x) ought to be small compared to \x — c|. In fact, the closer x is to c, the smaller the ratio e (x)/\x — c\ should be.

Let us now state these ideas more precisely.

Suppose / is differentiable at a point c of an open interval D. Set f i x ) = /(c) + f ' { c ) ( x - c) + e(x).

Then

e(x)1 -------- > 0 as x --------> c.\x — c|

This statement characterizes the number /'(c). It suggests that the derivative could have been defined by an approximation property:

Suppose there exists a number k such that

f i x ) = /(c) + k{x - c) + e(x), where

e (x )-------- - -------- > 0 as x ------- > c.\X - C\

Then / is differentiable at c and k — /' (c).

282 8. SEVERAL VARIABLE DIFFERENTIAL CALCULUS

This definition of derivative implies the usual one. For if the condition is satisfied, then

f (x ) f (.c) e(x)= k +x — c

[

-> k + 0 as X' -> c.x — c

Hence/'(c) = k.

Let us change our notation slightly. We shall always assume that c is an interior point of a domain D. Given such a c, there is a small interval (c — 8, c + 8) contained in D. Hence we can denote points in D near c by c + x where |#| < 8. Using this notation and writing e(x) instead of e(x + c), we may restate the basic facts about differentiability as follows:

Suppose/is differentiable at a point c of an open interval D. Set

/(c + x) = /(c) + f ( c ) x + e(x).

Then e(x ) / \ x \------- > 0 as x ------- » 0.

Conversely, suppose there exists a number k and 8 > 0 such that

/(c + x) = /(c) + kx + e(x) for |#| < 8,

where e (x) / \x \------- > 0. Then / is differentiable at c and k = /' (c).

For a function of one variable, the relation is clear: if the derivative exists, the linear approximation exists; if the linear approximation exists, the derivative exists.

For a function of several variables, the situation is not at all like this. It can happen that df/dx and df/dy both exist at a point (a, 6), but no linear approximation c + hx + ky exists!

1. Differentiable Functions 283

EXAMPLE 1.1

Define / by

§§jj m o) = o

(x, y) ^ (0 , 0 ).

Prove that df/dx and df/dy both exist at (0 , 0 ), but/ does not have a linear approximation at (0 , 0 ); indeed, / is not even continuous at (0 , 0 ).

Solution: Clearly/(#, 0) = 0 for all x. Hence

f (o ,o ) - < £ £ » )dx ax

= 0 .(0 ,0)

Similarly df/dy = 0 at (0, 0), so both partials exist and are zero. Now f ( x , y ) = 0 at each point of the x- and ?/-axes, so the only conceivable linear approximation is 0 + Ox + Oy = 0. But for x = y 9* 0,

f ( x , x ) =2x2

X2 + X2= 1.

Hence /(#, y) = 1 at all points of the line y = x except (0, 0) so /is not continuous at (0 , 0 ) and certainly 0 does not begin to approximate /.

R e m a r k : It is also possible to construct an example of a continuous function whose first partials exist, but which fails to have a linear approximation. See Exs. 1-3.

Differentiable Functions

As we have just seen, the mere existence of both partials of f(%,y) may not guarantee that f (x, y) is a reasonably behaved function. We shall therefore limit our attention to functions that have good linear approximations at each point.

Recall that each linear function L : R2------- > R is given by L(x) = k*xfor some fixed vector k.

L§t us now define differentiability for functions of two variables, that is,functions/: D--------> R, where D is a subset of R2. The discussion will extendeasily to functions of three or more variables.


Differentiable Function Let / be defined on a domain D of the plane R2 and let c 6 D. Then/is differentiable at c if there exist a linear function k*x and a function e(x) such that(i) / ( c + x) = / ( c ) + k*x + e(x) whenever c + x £ D,

p (x)(ii) - f - f ------- >0 a s x ------- > 0 .

The definition is completely analogous to the one for functions of a single variable. Just compare the relations

/ (c + x) = /(c) + kx + e(x) , e ( x ) / \ x \ ------- > 0 ,and

/ ( c + x) = / ( c ) + k*x + e(x), e (x ) / |x |--------»0.It is instructive to write out the definition in coordinate form:

A function / is differentiable at (a, 6 ) if there exists a linear function hx + ky such that

f (a + x ,b + y) = / (a , b) + hx + ky + e(x, y) ,where

e(x, y )

V x2 + y 2•0 as (*, j /)------- >(0 , 0 ).

An immediate consequence of differentiability is continuity.

Theorem 1.1 If / is differentiable at (a, b), then / is continuous at (a, b).

Proof: Let f (a + x , b + y ) = f(a, b) + hx + ky + e(x, y) as in thedefinition. If (x, y ) ------- » (0, 0), then x ------- » 0, y ------- > 0, ande(x, y ) ------- > 0. Consequently f (a + x, b + y ) ------- >/(«, b). This is continuity.

The mere existence of partial derivatives at (a, b) does not guarantee differentiability, as we noted in Example 1.1. However differentiability does guarantee the existence of partial derivatives.

Theorem 1.2 Let/be differentiable at an interior point (a, b) of its domain, and let

/(a + x,b + y) = /(a, b) + hx + ky + e(x, y) ,

where e(x, y ) / \ ( x , y ) \ ------- » 0 as 0x, y ) ------- » (0, 0). Then the firstpartials of / exist at (a, b) and

—■ (a, b) = h, j - (a, b) = k. dx dy

1. Differentiable Functions 285

Proof:

df , ^ v f (a + x ,b) - f (a ,b )— (a, b) = lim---------------------------to «

v hx + e (x ,0 ) e(z, 0)= lim --------------- = h + lim-------- = h.

x-> 0 X x -*0 X

Similarly f y (a, b) = k.

Again there is a strong analogy between the preceding formula and the corresponding one for functions of one variable. Just compare the relations

f (c + x) = /(c) + (- ~ } - x + e(x) ax

and

f (a + x ,b + y) = f(a, b) + — x + — y + e(x, y).dx dy

A Test for Differentiability

The next proof uses the Mean Value Theorem for functions of one variable. Let us review its content:

Mean Value Theorem Suppose / is continuous on a closed interval a < x < b and differentiable on the open interior a < x < b. Then there exists c such that a < c < x and

f ib ) - f ( a ) = f (c) (b - a).

We are prepared for a practical test for the differentiability of a function.

Theorem 1.3 Suppose / has domain D and (a, b) is an interior point of D. Suppose (1) df/dx and df/dy exist at each interior point of D and (2 ) these partials are continuous at (a, b). Then/ is differentiable at (a, b).

Proof: Write

/(a + x,b + y) - /(a, b)

= [ /(a + x, b + y) - f(a, b + y)~\ + [/(a , b + y) - f(a, 5)].

By the Mean Value Theorem, applied twice,

/(a + x, b + y) — /(a, b + y) = xfx (a + dx,b + y)

/(a, b + y) - f(a, b) = yfy{a, b + \y ) ,

where 0 < 0 < 1 and 0 < X < 1. (But 0 and X depend on x and y.) Set(1 ) /(a + x, b + y) = /(a, b) + xfx(a, b) + yfy (a, b) + e(x, y ), so that(2 ) e(s, 2/) = xg(x, y) + yh(xy y), where

d(x, y) = fx(a + dx, b + y) - f x(a, b)(3)

h(x9 y) = f y(a, b + \y ) - f v(a, b).Relation (1) implies that/is differentiable at (a, b), provided, we prove that

e (X>V) / x /n m ... ---------------------- >0 as ( x , y ) --------> (0 , 0 ).

+ yWe obviously have

(a + dx, b + y ) --------» (a, 6 ) and (a, b + \ y ) ------- > (a, b)as (x , y ) ------- » (0, 0). This implies

g(x>y)------- >0 and h ( x , y ) -------->0

as (x, y ) ------- > (0 , 0 ) by (3) and the assumption that/* and f y are continuousat (a, b).

Now divide both sides of (2 ) by y / x 2 + y2, take absolute values, and apply the triangle inequality:


e(x, y)y / x 2 + y2 < \g(x’ + / ifL 2y / x 2 + y2 y / x + y2

< !»(*, y )I + \h(x, y ) I-Clearly \g(x, y)\ + \h(x, y ) | --------> 0 as (x, y ) ------- > (0, 0). This completesthe proof.

Three Variables

For three or more variables, the definitions and theorems are easy extensions of those for two variables. We shall merely state the definition of differentiability for three variables and leave the extensions of Theorem 1.1-1.3 as exercises.

Differentiable Function Let /b e defined on a domain D of three-space R3, and let c £ D. Then / is called differentiable at c if there exist a linear function k - x and a function e(x) such that(i) / ( c + x) = / ( c ) + k*x + e(x) whenever c + x £ D,

p (x)(ii) j j - -------->0 as x ------ -»0.

IXIThe function / is called differentiable on D provided it is differentiable at each point of D.

2. Chain Rule 287

Exactly as for two variables, it follows from this definition that

k = (dl dl tf\|\dx ’ dy 9 dz) I x=c

The proof is left as an exercise.

EXERCISES

Show that f (x ,y) is differentiable at (a, b) by finding a linear function hx + ky that satisfies the definition of differentiability:

1- f(x, y) = xy2 at (0, 0 ) 2*. f i x , y) = 1 /xy at (1, 2 ).3. Define/on R2 by/(0, 0) = 0 and

f(x, y ) = for (X, y) ^ (0, 0 ).xL -+- yProve / is continuous on R2.

4*. (cont.) Prove that df/dx and df/dy exist on R2 and are continuous except at (0, 0).5. (cont.) Prove that/is not differentiable at (0, 0).6. Prove that the sum of two differentiable functions is differentiable.7. Prove that the product of differentiable functions is differentiable.8. Let g be differentiable at (a, b) and g(a,b) j* 0 . Prove that l/g is differentiable at

(a, 6).9. (cont.) Prove that the quotient of differentiable functions is differentiable when

ever the denominator is not zero.State and prove for three variables:10. Theorem 1.111. Theorem 1.2.

2. CHAIN RULE

The Chain Rule for functions of one variable gives the derivative of a composite function: if y = f i x ) where x = x(t), then

dy dy dx dt dx dt

The Chain Rule for functions of several variables really has two parts, a theoretical part, which says that a composite function built out of differentiable functions is itself differentiable, and a practical part, which is a formula for computing partial derivatives of such a composite function. We shall postpone a precise statement and proof until Section 9. First we shall work intuitively in order to gain a feeling for how the Chain Rule is used in practice.

Suppose z = f(x, y) where x = x{t) and y = y(t). Thus z is indirectly a

function of t. The Chain Rule asserts that

dz dz dx dz dy dt dx dt dy dt

This Chain Rule can be better understood in terms of vectors. The composite function z(t) = f[_x{t), y{ t )] can be thought of as z(t) = If t is time, then x (t) represents the path of a moving particle in the plane, and the composite function /[x (I)] assigns a number z to each value of t. The Chain Rule is a formula for the rate of change of/[x (£)] with respect to t. For instance /(x ) might be the temperature at position x. Then the Chain Rule tells how fast the temperature is changing as the particle moves along the curve x(£).

Chain Rule Let z = f (x, y), where x = x(t) and y = y{t). Thendz dz dx dz dy dt dx dt dy dt 1

dz dzwhere — and — are evaluated at (x(t)} In briefer notation,

dx dy

z = f xx + f vy.

In terms of vectors, 2 = /[x(£)],

^ = / » t * ( 0 > ( 0 + / » [ > ( < )] £ ( 0 -

Similar rules hold for functions of more than two variables. For instance, if w = f ( x , y , z )} where x = x ( t ), y = y(t), z = z(t), then

dw dw dx dw dy dw dz dt dx dt dy dt dz dt

EXAMPLE 2.1

Let w = f (x, y, z) = xy2zz, where x = t cos t, y = e\ and

z = ln (£2 + 2 ). Compute at t = 0 .dt

Solution: There is a direct but tedious way to do the problem. Write

w = (t cos t)e2t[ln(t2 + 2 )]3.

Differentiate, then set t = 0 . Thatfs quite a job! Use of the Chain Rule is much simpler:

2. Chain Rule 289

x0 = 0c(O), 2/(0 ), z(O)) = (0 , 1, In 2 ).

Since w = xy2z3,

dw dx

where the partial derivatives are evaluated at

= (In 2 )3,dw— y 2Z 3 = 2 xyz3

Xo X0 dy X0

= 0 ,

dwdz

= 3 xy2z2 = 0 ,

hence

Butw(0 ) = (In 2 )3x(0 ) + 0 + 0 .

x(0) = (cos t — t sin t ) |0 = 1,

therefore w(0) = (In 2 )3.Answer: (In 2 )3.

Another Version of the Chain Rule

Suppose z = f ( x , y ) , where this time x and y are functions of two variables, x = x(s, t) and y = y(s ,t ). Then indirectly, z is a function of the variables s and t. There is a chain rule for computing dz/ds and dz/dt:

Chain Rule If z = f ( x , y) is a function of two variables x and y, where x = x(s, t) and y = y(s, t)} then

dz dz dx ^ dz dy ds dx ds dy ds

dz dz dx dz dy dt ~ dx dt + dy dt ’

dz dzwhere — and — are evaluated at (x (s, t), y (s, t) ).

dx dy

This Chain Rule is a consequence of the previous one. For instance, to compute dz/ds, hold t fixed, making x(s, t) and y(s , t) effectively functions of the one variable s. Then apply the previous Chain Rule.

EXAMPLE 2.2

Let w = x2y , where x = s2 + t2 and y = cos st. Compute dwds


Solution:

dw dw dx dw dy ds dx ds dy ds

— (2xy)2s + x2( — t sin st)

= 2 (s2 + t2) (cos st)2s + (s2 + t2)2( — t sin st).

Answer: (s2 + £2)[4s cos st — t(s2 + t2) sin sQ*

The next example is important in physical applications.

EXAMPLE 2.3

If w = / ( x , y ) } where x = r cos 0 and y = r sin 0, show that

(»>\ + (»)' = + 1 . / \dy } \dr J r2 \d6 )

Solution: Use the Chain Rule to compute dw/dr and dw/dO:

dw dw dx dw dy dw dw . A— = ---------1---------= — cos 6 H------ sin 6;dr dx dr dy dr dx dy

dw dw dx dw dy dw . . , dw— ---------- b ----- - = — i — r sin 6) -----r cos 6d0 dx dd dy dd dx dy

( dw . dw \------ sin 6 + — cos 6 ) .

dx dy )From these formulas

Add:

( d w \2 ( dw\ dw dw . ( dw\ .( — ) = ( — J cos2 6 + 2 -------sin 0 cos 6 + ( — ) sm2 0,\dr / \ d x ) dx dy \dy /

1 / dw\ 2 / dw\ 2 dw dw . / dw\2- I — ) = [ — ) sin2 0 — 2 -------sm 0 cos 0 + ( — ) cos2 0.r2 \^0 / \dx / dx dy \dy /

2. Chain Rule 291

EXERCISESFind dz/dt by the Chain Rule:

1 . 2= exy; a; = St + 1, y = t2 2. z = #/?/; £ = + 1, y = — 13. z = x2 cos y — x; x = t2, y = 1/t 4. z = a?/y; * = cos t, y = 1 + £2.

Find dw/eft by the Chain Rule:5. w = xyz; x = t2, y — z — tA6. w = e* cos(i/ + 2); x = 1/ 2, ?/ = t2, z = —t7. w — e~xy2 sin 0; x = t, y = 2t, z = At8. w = (e~x sec )/?/2; x = t2, y = I t, z = tz.

Find dz/ds and dz/d2 by the Chain Rule:9. z = z3/V; x = s2 — t, y = 2st

10. z = (x + ?/2)4; x = se*, y =11. z = \ / i + z2 + y*\ x — st2> y — 1 + si

13. The radius r and height A of a conical tank# increase at rates r = 0.3 in./hr and h = 0.5 in./hr. Find the rate of increase V of the volume when r = 6 ft and h = 30 ft.

14. Prove that

-(dX J gh(x)

F(t) dt — F[h(x)lh'(x) - F\ j ,(xW(x).q{x)

15. Given F (x, y), show that

~ F(u + v, u — v) + F(u + v, u — v) = 2Fx(u + v. u — v). ou ov

A function w = f ix , y, z) is homogeneous of degree n if f(tx, ty, tz) = tnf (x , y, z) for all t > 0. The condition of homogeneity can be written vectorially:

f(tx) = tnf (x).Show that the function is homogeneous: What degree?16. x2 + yz 17. x — y + 2z18. x3 + ys + z3 - 3xyz 19. x2e~ylz

20 . a . 1x4 + y* + z4 a: + y *

22. Suppose/and <7 are homogeneous of degree m and n respectively. Show that fg is homogeneous of degree mn.

23. Let f(x, y, z) be homogeneous of degree n. Show that f x is homogeneous of degree n — 1. (Exception: n = 0 and / constant.)

24. Let f(x, y, z) be homogeneous of degree n. Prove Euler’s Relation: xfx + yfv + zfz = nf.[Hint: Differentiate f(tx, ty} tz) = tnf(x, y, z) with respect to t, using the Chain Rule; then set t — 1.]

25. Verify Euler’s Relation for the functions in Exs. 18 and 19.26*. (Converse of Euler’s Relation) Let /(x ) be differentiable for x ^ 0, and suppose

x • grad / = nf. Prove / is homogeneous of degree n.[Hint: Show that d[t~nf (tx)~\/dt = 0.]

3. TANGENT PLANE

The graph of2 = f ix , y)

is a surface. Given a point x0 = (x0, y0, z0) on this surface, we are going to describe the plane tangent to the surface at x0.

Let 2 = f(x, y) represent a surface and let x0 = (x0} y0) z0) be a fixed point on it. Consider all curves lying on the surface and passing through x0. Their velocity vectors fill out a plane through the origin. The parallel plane through x0 is called the tangent plane at x0.

Let us see why this is so. Any curve on the surface (Fig. 3.1) is given by

x(<) = ixit), y{t), 2 (0 ) = ix it), y it), f i x (0 , 2/(0 ])-

Its velocity vector is found by the Chain Rule:

v it) = x(<) = ixit), y i t ) , f xx + f vy).

Let us assume time is measured so that x(0 ) = Xo, that is,

z (0 ) = x o , 2/(0) = 2/0, 2(0) = 20.

3. Tangent Plane 293

Thenv(0 ) = (£ (0 ), y (0 ), f x(x0, yo)x(0) + f y(x0) 2/0)2/ (0 ))

= x ( 0 ) (1 , 0, fx(x0, y0)) + 2/(0) (0, l , fy(x0, yo))-The vectors

Wi = (1, 0 , f x(x0, y0)) and w2 = (0 , l , f y(x0, yo))

depend only on the function / and the point x0, not on the particular curve (Fig. 3.2).

Suppose a curve lies on the surface z = f(x, y) and passes through

x0 = (x0, y0) zo).

Then its velocity vector has the form

v = awi + 6w2,where

Wi = (1, 0,fx(xo, y0)), W2 = (0, l , f y(x0, yo)), and a and b are constants, a = x (0 ) and b = 2/(0 ).

Given any pair of numbers a and b, there are curves x(t) which lie on the surface, pass through x0 when t = 0 , and have x(0 ) = a and 2/(0 ) = b. One such is

x(t) = (xo + cit, 2/0 + bt,f(xo + at, yo + bt)).


Its velocity vector is awx + frw2. Therefore, all vectors awi + 6w2 actually occur as velocity vectors. It follows that the velocity vectors fill out a plane through the origin. The parallel plane through x0 is the tangent plane to the surface at x0. It consists of all points

x = x0 + awi + frw2, a and b arbitrary.

EXAMPLE 3.1

Find the tangent plane to 2 = x2 + y &t (1, 1, 2 ).

Solution:

Wl = (1, 0 , ^ (1, 1)) = (1, 0 , 2 ), w2 = (0 , 1, ^ (1, 1)) = (0 , 1, 1 ). dx dy

Hence the typical velocity vector is

v = awi + &w2 = a(l, 0 , 2 ) + 6 (0 , 1, 1) = (a, b, 2a + b).

The tangent plane consists of all points

(1, 1, 2 ) + (a, b, 2a + b) = (a + 1, 6 + 1, 2a + b + 2 ),

where a and 6 are arbitrary.

Answer: All points (a + 1, b + 1, 2a + b + 2).

R e m a r k : The typical point on this tangent plane is

x = (xy y, z) = (a + 1, b + 1, 2a + b + 2 ).Thus

a = x — 1, b = y — 1,and

2 = 2a + b + 2 = 2{x - 1) + (y - 1) + 2 = 2x + y - 1.

Consequently an equation for the tangent plane is

2 = 2# + y — 1.

Tfce Normal

The vectorn = ( f x j ~ f y , 1 )

is perpendicular to both tangent vectors Wi and w 2:

n * W i = ( — f x , — f y , 1 ) * (1 , 0 , f x ) = — f x + 0 + f x = 0 ,

n - w 2 = ( - f x , - f y , 1 ) * (0 , 1 , f y ) = 0 - / „ + / „ = 0 .

Consequently n is perpendicular to each vector a w i + &w2:

i t ( a w i + 6 w 2) = a ( n * Wi ) + & ( n * w 2) = 0 .

3. Tangent Plane 295

In other words, n is perpendicular to the tangent plane at x0, that is, n is a normal, so we may write a normal form of the plane.

The tangent plane to the surface z = f(x, y) at the point x0 = (#o, yo, zQ) is given by

- f x - (x ~ Xo ) - f y - (y - yo) + (z - z0) = 0,

where/x and/^ are evaluated at (#0, yo)- The vector form of this equation is

(x - x0)-n = 0 , n = ( - /„ - f y) 1).

EXAMPLE 3.2

Find an equation for the tangent plane to z = x2 + y at (1, 1, 2).

Solution: Write f(x, y) = x2 + y. Then

fx(x, y) = 2x, f y(x, y) = 1.At (1, 1, 2 ),

f x = 2 , f y = 1.Hence the equation is

— 2(x - 1) - (y - 1) + (z - 2 ) = 0 .

Answer: 2x + y — z = 1.

The vector (—f xy —f V) 1) is perpendicular to the tangent plane and has a positive z-component (it is the upward normal rather than the downward normal). We introduce the unit vector in the same direction.

The unit normal at x0 to the surface z = f (x , y) is the vector

N = V J T T W n ( ~ fx’ ~ fv’ 1}’where f x and/y are evaluated at (#0, yo)-

EXAMPLE 3.3

Find the unit normal at (1, 1, 2) to the surface z = x2 + y.

Solution: We have (— f x, —f y, 1) = ( — 2, —1, 1), hence

N = ;..- 1 = ( -2 , - 2 , 1).a /22 + l2 + l2

Answer: N = -^= ( — 2, —1, 1).

EXAMPLE 3.4

How far is 0 from the tangent plane to z = x2 + y at (1, 1, 2 )?


Solution: The distance is the length of the projection of x0 = (1, 1, 2) on N. See Fig. 3 .3 .

0Fig. 3.3

Distance = |xo*N|

(1, 1, 2 ) - - ^ ( - 2 , - 1, 1)

Answer:V 6

EXERCISES

Give the equation z — ax + by + c of the tangent plane to the surface at the indicated point:

1. z = x2 — y2; x = 0, y = 0 2. z = x2 — y2; x = 1, y = — 13. z = x2 + Ay2) x = 2, y = 1 4. z — x2ey; x = — 1, y = 25. z = x2y + ys; x = — 1, y = 2 6. z = x cos y + y cos z; x = 0, y = 0 .

Find the unit normal to the surface at the indicated point:7. z — x + x2yz + 2/; x = 0, y = 08. 2 = z3+ 2/3; * = 1, 2/= - 19. z = x2 + xi/ + 2/2; x = — 1, 2/ = 2

10. 2 = v 1 - a:2 — s/2; x = i , y = $.

4. Gradient 297

11. Show that the tangent plane to the hyperbolic paraboloid z = x2 — y2 at (0, 0, 0 )intersects the surface in a pair of straight lines.

12. (cont J Show that the conclusion is valid for the tangent plane at any point of the surface.

13*. (cont.) Show that the property of the tangent planes in Ex. 12 is also valid for the hyperboloid of one sheet z2 = x2 + y2 — 1.

4. GRADIENT

The gradient field of a function / on a region is the assignment of a certain vector, called grad /, to each point of that region.

Suppose z = z(x, y) is defined on a region D of the y-plane. The gradient of / is the vector

grad/ = (/*,/„).

Likewise, if w = f(x, y, z) is defined on a region D of space, the gradient of / is the vector

grad/ = (/*,/„,/*).

The gradient field of a function / is the assignment of the vector grad / to each point of the region D.

For example, if

f (x, y) = .r2 + y, grad/ = (2®, 1);if

/(«, y, z) = |x|2 = x2 + y2 + 22, grad/ = (2x, 2y, 2z) = 2x.

In this section, we discuss several uses of the gradient. The first of these concerns notation. Certain formulas are simplified if expressed in terms of gradients. An example is the Chain Rule, which asserts

z = f xx + f uy

if z = f(x, y) and x = x(t), y = y(t). But grad/ = (/.,/„ ) and x = (x, y). Therefore, in vector notation

z = (grad/) • x.

Level Curves

Imagine a surface z = f (x, y) above a portion of the x, y-plane. In the plane we can draw a contour map of the surface by indicating curves of constant altitude. These are called contour lines or level curves. Figure 4.1


shows a contour map with two hills and a pass between them. Level curves are obtained by slicing the surface z = f{x, y) with planes z = c for various constants c. Each plane z = c intersects the surface in a plane curve (Fig. 4.2). The projection of this curve onto the x, y-plane is the level curve at level c. It is the graph of f (x, y) = c. Where level curves are close together the surface is steep; where they are far apart it is relatively flat.

peak

An important relation exists between the gradient field and the level curves of a function.

The gradient field of / is orthogonal (perpendicular) to the level curves of /.

4. Gradient 299

For suppose a particle moves along the level curve f ( x y y) = c. Let its position at time t be x(t) = (x(t), y(t)). Then

f lx ( t ) , ?/(0] = C.

Therefore the time derivative is zero :

fxX + f yy = 0 , (grad /) • x = 0 .

Hence grad / is orthogonal to the tangent to the level curve.Suppose we want an explicit equation for the tangent to a level curve

f (x, y) = c at a point x0 = (x0, y0). We know a point, x0, on the tangent and we know a normal, grad/(x0), so the answer is

(x - x0)*grad/(xo) = 0 .

EXAMPLE 4.1

Find the tangent to the cubic x2 = y3 at ( — 1, 1).

Solution: Set f (x, y) = x2 — y3. Then the cubic is the level curve/ = 0 . We have

grad/(xo) = (2x, — 32/2) | = (“ 2, - 3 ) .Hence the tangent is

[ ( * , y ) - ( — 1, 1 )D* (— 2, —3) = 0,

— 2(x + 1) - 3 (y - 1) = 0.

Answer: 2x + 32/= 1.

Level Surfaces

We cannot graph a function of three variables

w = f (x , y, *)

(the graph would be four-dimensional). We can, however, learn a good deal about the function by plotting in three-space the level surfaces

f ( x , y , z ) = constant.

For example, the level surfaces of

f (x, y , z) = x2 + y2 + z2

are the spheresx2 + y2 + z2 = c2

centered at the origin.

The gradient field of / is orthogonal to the level surfaces of /.


The proof of this statement is practically identical to the proof of the analogous statement in two variables.

Just as for curves, the gradient provides a convenient tool for finding explicitly tangents to level surfaces. Suppose we want the tangent plane at a point x0 of the level surface/(x) = c. The answer is

(x - x0)*grad/(x0) = 0 .

Note that this method fails at points where grad / = 0. At such points there usually is not a clearly defined tangent plane anyhow.

EXAMPLE 4.2

Let A be a non-zero 3 X 3 symmetric matrix and let x0 be a point on the quadric surface

xAxf = 1.

Find the tangent plane to the surface at Xo.

Solution: Let /(x) = xAx'. Then the quadric is the level surface / = 1, and since x0 is on the surface, x0Ax0' = 1.

Our problem is to compute grad/. To be definite, set

A =

a b d

b e e

A e /_

The terms involving x in xAx' are ax2 + 2bxy + 2dzx, so df/dx = 2 (ax + by + dz). Similarly df/dy = 2 (bx + cy + ez) and df/dz = 2 (dx + ey + fz). It follows that

grad/ = 2xA.

Therefore, the tangent plane at x0 is given by

(x - x0)* (2x0A) = 0.

(Note that x0A ^ 0 since x0Ax0' = 1.) By the symmetry of A, the equation can be written

(x — x0)(x0A)' = 0, (x — x0)Ax0' = 0, xAx0' = x0Ax0' = 1.

Answer: xAx0' ~ 1.

R e m a r k : This answer is particularly simple. You just replace one of the x’s in xAx' = 1 by x0. For instance, the tangent plane to the hyperboloid 2x2 + 3y1 — 4z2 = 1 at a point (x0, yo, Zo) on the surface is 2##o + Syy0 — 4zz0 = 1, by inspection.

4. Gradient 301

EXERCISES

Plot the level curves and the gradient field in the region \x\ < 3, \y\ < 3:

9. For each function in Exs. 5-8, find the gradient field.10. Suppose z = f(r, 0) is given in terms of polar coordinates. Show that

where u = (cos0, sin0) and w = (— sin0, cos0).11. Find grad (r~2 cos 20).

Find the tangent line to12. x2 — 3xy + y2 = —1 at (1, 2 )13. x + z3?/4 — y = 0 at (0, 0)14. y + sin xy = 1 at (0, 1).

Find the tangent plane to15. z = z 2- 2/2at (1,0,1)16. xyz = 1 at (1, 1, 1).17. Let a be a constant vector. Find the gradient of /(x ) = a • x.

18*. Let x0 be a point of the quadric xAx' + 2a x' = 1, where A is a 3 X 3 symmetric matrix and a a vector. Assuming x0A + a ^ 0, prove that the tangent plane at x0 is

Apply this to z = xy .Let u = (u, v, w ) be a vector field (p. 306) on an open domain of R3 with differentiable coordinate functions. Define the divergence and curl of u by

19. div(/u ) = (grad/) *u + /(d iv u )20. div(/(p)x) = [p3/(p)]'/p2, where p2 = x2 + y2 + z221. curl(/u) = (grad/) X u + /(cu rlu )22. curl (grad/) = 023. curl[/(p)x] = 0, where p2 = x2 + y2 + z224. div (curl u ) = 025. curl (a X x) = 2a

26*. div[grad/(p)] = [p/(p)]"/p, where p2 = x2 + y2 + z2.

1. z = x — 2y3. z = x2 + y

2. z = x2 — y2 4. z = x2 + Ay2.

Describe the level surfaces:5. w = x y z7. w — xyz

6. w = x2 + 4y2 + 9z28. w = x2 + y2 — z2.

gradz = fr u + - fe w,r

xAx o' + a(x' + x0') = 1.

Prove:


5. DIRECTIONAL DERIVATIVE

Given a function w = f (x , y, z ) and a point j^ in space, we may ask how fast the function is changing at x in various directions. (A direction is indicated by a unit vector u.)

The directional derivative of f ( x , y, z ) at a point x in the direction u is

A i / 0 0 = l / ( x + t o)dt (=0

Think of a directional derivative this way. Imagine a particle moving along a straight line with constant velocity u, passing through the point x when t = 0. See Fig. 5.1. To each point x + tu of its path is assigned the number

w ( t ) = /(x + tu).

Fig. 5.1

ThenD u f ( x ) = w ' ( 0),

the rate of change of w ( t ) as the particle moves through the point x. For example, suppose f (x , y, z ) is the steady temperature at each point (x , y, z) of a fluid. Suppose a particle moves with unit speed through a point x in the direction u. Then D uf ( x ) measures the time rate of change of the particle’s temperature.

5. Directional Derivative 303

p(f) = x + tu.

Then w i t ) = /C p(0], a composite function. By the Chain Rule,

There is a handy formula for directional derivatives. Let

D uf { x ) = w( 0) = ^ / [ P ( 0 ] = (grad /)*p(0 ) = (grad/)*u.

Since u is a unit vector, (grad/)*u is simply the projection of grad/ on u.

The derivative of / in the direction u is the projection of grad / on u :

D u /(x ) = (grad/)*u.

In particular if u = i = (1, 0, 0), then

A / ( x ) = (g ra d /) - (1, 0, 0 ) = (1’ ° ’ °) = r -\ d x dy d z / dx

A similar situation holds for j = (0, 1, 0) and k = (0, 0, 1).

The directional derivatives of f ( x , y, z ) in the directions i, j, and k are the partial derivatives:

df df df A / - J - ’ D ; f = T ’ Z)^ = r -dx dy dz

EXAMPLE 5.1

Compute the directional derivatives of f ( x , y , z ) = x y 2zz at (3, 2, 1), in the direction of the vectors

(a) ( - 2 , - 1 , 0 ) , (b) (5 ,4 ,1 ).

Solution:D uf ( x ) = (grad/)*u,

where u is a unit vector in the desired direction, and grad / is evaluated at (3, 2, 1). Since

df df df— = y 2z3, — = 2 xyz3, and — = 3 xy2z2, dx dy dz

g rad / = (4,12,36).(3,2,1)


Thus£>u/(x ) = (4, 12,36) -u.

(a) u = - ^ ( - 2 , - l , 0 ) ,

D uf ( x ) = (4, 12, 3 6 ) ~ ( - 2 , - 1 , 0) = .

(k) u =

Du/(x ) = (4, 12, 36)* 7= (5, 4, 1) = 104V 42 V 42

Answer: (a) ; (b) '

Question: In what direction is a given function / increasing fastest? In other words, at a fixed point x in space, for which unit vector u is

D umlargest? Now

D u /(* ) = (grad/) • u = |g ra d / |c o s0,

where 0 is the angle between g rad / and u. Therefore, the largest value of Z)u/(x ) is |grad/|, taken where cos 0 = 1, that is, 0 = 0.

The direction of most rapid increase of f (x , y, z ) at a point x is the direction of the gradient. The derivative in that direction is |grad/|.

The direction of most rapid decrease is opposite to the direction of the gradient.

EXAMPLE 5.2

Find the direction of most rapid increase of the function

f ( x y y , z ) * #2 + yz

at (1, 1, 1) and give the rate of increase in this direction.

Solution:

grad/ = (2x , 2, y ) = (2, 1, 1).(i.i.i) (i,i,D

The most rapid increase is

D u f = | g rad /| = a /2 2 + l 2 + l2 = y / 6 ,

5. Directional Derivative 305

where u is the direction of grad/:

= g rad / _ _1_ (2 i i )I grad /| a/ 6

Answer: D u f(1,14 )

\ / 6 for uV 6

(2, 1, 1).

We conclude this section with two examples involving directions of most rapid change of a function.

Consider water running down a hill from a spring at a point P. See Fig. 5.2. The water descends as quickly as possible. What is the path of the stream?

From physics, change in kinetic energy [ |m (speed)2] equals change in potential energy [height]. Hence the speed of a water particle depends only on how far it has descended (its altitude). Since the speed at a given time does not depend on direction, the particle “chooses” the direction of steepest descent (most rapid change of altitude). Let the hill be represented by the surface z = f ( x , y ) . Then water will flow in the direction of —grad/, that is, perpendicular to the level curves.

Next, consider a function z = f ( x , y ) and all curves x (t) = ( x ( t ) , y ( t ) ) in the x, y-plane which pass through a fixed point x0 with speed 1. Along which of these is f (x , y ) increasing the fastest at x0 ?

To find the direction of most rapid increase, write

z i t ) = / [ x ( 0 ] .By the Chain Rule,

z = (grad / ) • x = (grad / ) • v,


where v is the velocity vector. But v is a unit vector; hence z is the directional derivative in the direction of v (tangential to the curve x ( t ) ) . Thus2 is greatest for those curves whose tangents at x0 point in the direction of g rad /; such curves are orthogonal to the level curves.

EXERCISES

Find the directional derivative of f(x, y, z) at Xo in the directions of vi, V2, and V3:1. / = x + y + 2; x0 = 0, vi = (1, 0, 0), v2 = (0, 1, 0), v3 = (0, 0, 1)2. / = xy + yz + zx; Xo = (1,1,1), Vi= (1 ,1 ,1), v2 = (1, —1,1),

v3 = (—1, —1, —1)3 . f = x y z ; * - ( 1 , - 1 , 2 ) , vx= (1,1,0), v2 = (1,0,1), v ,= (0,1,1)4. / = xhfz*; Xo= ( - 1 , - 1 , - 1 ) , * - ( 1 , 2 , 3 ) , v2 = (1, 1, 0), v3= (3,2 ,1).

Find the largest directional derivative oi f (x , y , z) at x0:5 . f = x 3 + y2 + z; x0 = 0 6. f = xyz; Xo = (— 1, — 1, — 1)7. f = x * + y2 + z2; x0 = (1 ,2 ,2 ) 8. / = z2 - y2 + \ z \ x„ = ( - 1 , - 1 , 1 ) .

6. APPLICATIONS

A vector field is the assignment of a vector F (x) to each point x of a region D in space. (The gradient of a function is one example.)

Let F(x) be a vector field on D and suppose x (?) is a path in D from x0 = x (l0) to Xi = x( t i ) . The line integral

LXl

F*dxX0

is defined over this path and is computed by the ordinary integral

1

fJ toF [x (0 ]# x(0 dt.

Its value generally depends on the path x( t ) connecting x0 and Xi.

EXAMPLE 6.1

Let F(x) be the vector field F(x) = (x, y, x + y + z) . Let x0 = (0, 0, 0 ) and Xi = (1, 1, 1). Compute the line integral

[ X i

/ Fvfa over the paths J X o

(a) x(t) = ( t , t , t ) , (b) x ( t ) = ( t , P , P ) .

Solution: Notice that for both paths, x0 = x(0 ) and Xi = x ( l ) .

(a) On this path x = (£, t} t) and x{ t ) = (1, 1, 1).

f* F'dx = [ F[x(<)]*x(<) dt = [ ( t , t , 3 t ) ' ( l , 1 , 1 ) dt = [ 5 t d t = ~ . xo Jo Jo Jo 2

(b) This time x = (t, t2, ts) and x( t ) = (1, 2t, St2).

6. Applications 307

In an important special case, the line integral does not depend on the path x( t ) , but only on the initial and terminal points x0 and Xi:

If the vector field F(x) is the gradient of some function/,

Thus the value of the line integral is independent of the path connecting x0 and Xi.

This assertion is easily verified by means of the Chain Rule,

^ ( t + 5 t > + W + 3 t > ) d t = ± + 5 + | + I = |

5 57Answer: (a) - ; (b) — .

F = grad/,then

~ /E x (0 ] = (grad/)* x.

If F = grad/, then


R e m a r k : Not every vector field is the gradient of some function. For example, the field F(x) = (x, y, x + y + z ) is not a gradient;

r F-dxJ Xo

is not independent of the path (see Example 6.1).

Application to Physics

If F = grad / is a force, the net work done by this force in moving a particle from x0 to Xi is/(x i) — /(x 0), independent of the path. The function/, which is unique up to an additive constant, is called the potential of the force.

An important example is that of a central force subject to the inverse square law, for instance, the electric force E on a unit charge at x due to a unit charge of the same sign at the origin. The magnitude of the vector E is inversely proportional to |x|2. Its direction is the same as that of x. See Fig. 6.1.

The unit vector in the direction of x is

x x— = - , p = X . 1*1 P

Therefore, expressed in suitable units,

_ 1 x x p2 p p3

The force field E is defined at all points of space except the origin. We shall prove that E is the gradient of a function, in fact that

E = grad/, where f ( x , y, z ) = - - =p \ / x 2 + y 2 + z2 '

6. Applications 309

Let us compute the gradient of / = — 1/p:

df x

and similarly

Therefore

dx (x2 + y 2 + z2)312 p3 5

df_ = y_ df = z_ dy p3 ’ dz p3

Since E = grad/, it follows from our discussion of line integrals that

L E-dx = /(xO - /(xo) = 777 - ITT •Xo |Xo| | X i|

The right-hand side is the potential difference or voltage. I t represents the work done by the electric force when a unit charge moves from x0 to Xi along any path.

If Xi is far out, then l/|x i| is small, so

LX, I

E' d x -— - . |xo|

As Xi moves farther out, the approximation improves. In mathematical shorthand,

Lxi i

E * d x -------->7—: as |xi|X . Xo

or

E *dx = 7~ Xo/ • J xo

Physical conservation laws are usually derived by identifying an appropriate vector field with the gradient of a function and then evaluating a line integral.

EXERCISES

f (0,-1)1. Compute / (xy, 1 + 2y )*dx

y (o,i)(a) along the straight path,(b) along the semicircular path passing through (—1, 0 ).


[ Q X D2. Let F = (3x2y2z, 2x*yz, x*y2). Show that / F• dx is independent of the path, and

evaluate it. J(o,o,o)[(a,b,c)

3. Let F = (x2 + yz, y2 + zx, z2 + xy). Show that / ¥*dx is independent of the path, and evaluate it. y (o.o.o)

— v x \2 , 2 , 2 , 2 ) . x y x y j

5- Find

rr—2/ dx + x dy .—— — over the circle x = a.

' + y2

6. Let F = x/|x|5 and suppose 0. Show that I F*dx, taken along any path

from a which does not pass through 0 and goes out indefinitely, depends only on a = |a|. Evaluate the integral.

7. (cont.) Do the same for F = x/|x|n, for any n > 2,8. (cont.) Show that F = x/|x|2 is a gradient.

7. IMPLICIT FUNCTIONS

Often a function y = g ( x) is defined only as the root of an equation

f ( x , y ) = 0,

which may be hard or impossible to solve explicitly. In such a case, the equation is said to define an implicit function y = g (x ) . For example, Fig. 7.1 shows part of the graph of

?/6 + y + xy — x = 0 .Near the origin, this equation defines y as an implicit function of x. (It is hopeless to express y as an explicit function of x. )

Fig. 7.1

7. Implicit Functions 311

What is the derivative of an implicit function y = g ( x ) defined by f (x , y ) = 0? Substitute y = g( x) :

f i x , g{x)~\ = o.

Differentiate with respect to x using the Chain Rule:

/■fx + fy-g' - 0, g' - —f y

If y is an implicit function of x defined by

f ( x , y ) = 0 ,then

dy

dxfx(x9 y )

fy(x, V )

at each point (x , y ) where f (x , y ) = 0 and/^Or, y ) ^ 0 .

Differentiation of implicit functions is called implicit differentiation.

R e m a r k : The minus sign in this formula may appear puzzling. I t seems to contradict the natural procedure of “canceling” differentials. For instance, if g = g[_y(x)~\, the Chain Rule says dg /dx = (d g / d y ) (<d y / d x ). Hence

dy dg /dx

dx dg /d y ’

so dg appears to cancel out.The reason for the minus sign is this. When we w r ite / (x, y ) = 0, we have

“taken y to the other side” . The equation y = g( x) is equivalent to f ( x , y ) = y ~~ 9 ( x ) = 0. Now “canceling” differentials fails because

fx _ - g ' j x ) _ _ ,

fy

There is the minus sign!

EXAMPLE 7.1

Findax

where y = g ( x ) is defined by y* + y + xy — x = 0 .<0 ,0)

Solution:f ( p > y ) = y* + y + xy -

fx = y — l, f v — 6y5 + l + X.

Hence

dy f xdx f y 6i/5 + 1 + x


At (0, 0),

dy_dx

- 1= 1.

Alternate Solution: Differentiate the equationy* + y + xy - x = 0,

treating ? /a sa function of x :

a b dy dy dy+ + + y - 1 = 0 ,

dx dx dx

y - 1dy = ____________dx 6t/5 + 1 + x

Answer: = 1.dx (0,0)

R e m a r k : The technique in the alternate solution is equivalent to use of the rule

dy = _ f xdx f y

because the rule was derived by that very technique.

EXAMPLE 7.2

Let y = y / l — x2. Compute y ' and y" by differentiating implicitly x2 + y2 — 1 = 0 .

Solution: Differentiate:

2x + 2yy' = 0, y' = - - .y


y - x

y" = ~

since x2 + y 2 = 1.

y - xy (-3 + X2 1

r

Answer:

—xy f - -

y

%1*—1 >

l - 1y * ”

y3 (1 - X2)3/2 '


Applications

Implicit differentiation is useful when a function must be maximized or minimized subject to certain restrictions.

EXAMPLE 7.3

A cylindrical container (right circular) is required to have a given volume V. The material on the top and bottom is k times as expensive as the material on the sides. What are the proportions of the most economical container?

Solution: The cost C of the container is proportional to

(area of side) + /c(area of top + area of bottom).

Let r and h denote the radius and height of the container. In the proper units,

C = 2t rh + k(2irr2),

into the equation for C. Then C is an explicit function of r which can be minimized.

I t is simpler, however, not to make the substitution, but to consider C as a function of r anyway (as if the substitution had been made). Differentiate implicitly:

I t is easily verified that C is minimal for h = 2kr. Since t r2h is constant, h is large if r is small and decreases as r increases. Therefore (2kr — h) increases from negative to positive as r increases. Thus d C / d r satisfies the conditions for C to have a minimum at h = 2kr.

We must minimize C subject to this restriction.One approach is obvious: solve the last equation for h and substitute

Now differentiate the equation for V with respect to r:

Substitute this value of dh/dr into the preceding equation:

dr L \ r /

Hence

dC r / - 2/A 1— = 2t r I ----- ) + h + 2kr = 2n(2kr — h).

when h = 2kr.

Answer: height = 2k X radius.


R e m a r k : The special case k = 1 is interesting. All parts of the cylinder are equally expensive; the cheapest cylinder is the one with least surface area. Conclusion: Of all cylinders with fixed volume, the one with least surface area is the one whose height is twice its radius.

EXAMPLE 7.4

Find the greatest distance between the origin and a point of the curve x4 + y 4 = 1.

Solution: Draw a graph (Fig. 7.2). Because of symmetry we need consider only x > 0 and y > 0 . Since the curve lies outside of the circle x2 + y2 = 1, the maximum distance is greater than 1 and occurs at some point (x, y ) where x > 0 and y > 0 .

The square of the distance from any point (#, y ) to the origin is x2 + y2. Hence, we must maximize

L 2 = x2 + y 2

subject tox4 + y 4 — 1 = 0.

Differentiate both relations with respect to x :

j— (L2) = 2x + 2 yy', 4z3 + 4 y 3y' = 0.

dx

I t follows that

This derivative vanishes in the first quadrant only for x = y. Hence the maximum distance occurs at the point (x , y ) of the curve in the first quadrant for which x = y. Thus

x4 + y 4 = 1, x = y,from which

" = 2/ = ^ l ’and

A ns we r: \ / 2 .

EXERCISESFind dy/dx:

1. x + y = x sin y 2. x2 + ys = xy3. exy — 3xy2 4. x4 — y4 = 3x2y*5. ex sin y = ey cos x 6. x*y* = x2 — y2 + 17. x4 + Sye = 1 8. x5 + yb = xy + 1.

9*. Find the maximum and minimum values of f ix, y, z) = x4 + y4 + z4 on the surface of the unit sphere x2 + y2 + z2 = 1.

10*. (cont.) Deduce that (x2 + y2 + z2)2 < x4 + y4 + z4 < (x2 + y2 + 22)2 for any (x, V,z)-

11*. (cont.) Find the corresponding inequalities for n > 3 relating xn + yn + zn and x2 + y2 + z2. (Assume x > 0, y > 0, z > 0 if w is odd.)

12*. (cont.) Find the largest A and the smallest B so thatA (x4 + y4 + z4)3 < ( 6 + 2/6 + 26)2 prove

dz_ Fx[xf y , z (x , y ) ' ] dz = Fy[x, y, z(x, y)J dx Fz[_x, y, z(x, y)~\ 7 dy Fz[x, y, z(x, y)~\ *

\_Hint: Use the Chain Rule.]14*. (cont.) Suppose z ( x , y ) and w ( x , y ) are defined implicitly by

lF(x, y, z , w ) = 0

[G(x, y } z , w ) = 0.

Find formulas for dz/dx, dz/dy, dw/dx, dw/dy under suitable hypotheses.


8. DIFFERENTIALS

We shall re-examine our basic approximation formula

/ ( c + x ) = / ( c ) + k -x + e (x )

with error e (x). We assume that the domain D of / is an open set and that / is differentiable at every point c of D. Hence

Suppose we allow both c and x to vary. I t is customary to replace c by x and x by dx = (dx, dy, dz) , a new quantity. In this notation,

/ ( x + dx) = / ( x ) + k *dx

The boxed quantity is particularly significant. For each x it is a linear function of dx. We therefore give it a name, df, the differential of /, and write

, , df , , df df df = — dx + — dy + — dz.

dx dy dz

The differential is really a function of the six variables x, y, z, dx, dy, dz, where x is confined to D and dx is arbitrary. We could even write df = d f (x, dx) but this is a cumbersome notation.

The differential has elementary algebraic properties that correspond to analogous results for derivatives:

d ( f + g) = d f + dg d ( a f ) = a d f

d ( f g ) = W ) g + f dg d ( f / g ) = 2 g ^ 0.

Substitution

The differential has a useful formal property of remaining unchanged when the variables are replaced by functions of other variables. This is another form of the Chain Rule.

For instance, suppose / = f (x , y, z ) so

df = f x d x + f y dy + fz dz.

Now suppose x = x ( u , v ) , y = y ( u , v ) , z = z ( u , v ) . We can look at the composite function f [ x (u, v)^\ and compute its differential, another “d f f :

d f = f u du + f v dv.

8. Differentials 317

Is this different? No, because by the Chain Rule

fu du + f v dv = ( f xxu + f yy u + f zZu) du + ( f xx v + A?/* + /*z.) dv

,/x du I x v d v ) I /y (2/w I y v d v ) -j- f % (Zu du I z ® d v )

= f x d x + f y dy + fz d z .

This may appear as simply a consequence of sloppy notation for composite functions. Still there is something more to it.

Suppose we have a function / of independent variables u, v, but ther^are some intermediate variables in the way. After some computation we arrive at an expression

(1) df = M d u + N d v .

Then we know automatically that M — f u and N = / , no matter how we obtained (1).

For example, suppose z = z(x, y ) , and we know that

x2 + y 2 + z2 = 1.

Then, since the differential of a constant function is 0,

x dx + y dy + z dz — 0,hence

dz = — - dx — - dy. z z

We conclude that dz /dx = —x / z and d z / d y = —y/ z .More generally, if z = z (x, y ) is constrained by a relation

F(x, y , z ) = 0,then

Fx dx -f- Fy dy -f- F z dz = 0,which implies

7 7 Fydz = ~ y T Vy

whenever F z ^ 0 at (x , y, z (x, y ) ) . Therefore

dz Fx dz _ Fy dx F z ’ dy F z

Here is a situation where an intermediate variable is in the way. Suppose z = z(x, y ) is subject to two constraints,

F(x, y, z, u ) = 0

G(x, y, u ) = 0 .

If we could solve the second relation for u and substitute the solution in the first, the result would be a relation H (x, y, z ) = 0. But that might be impractical to carry out, so we proceed indirectly by forming differentials:

Fx dx -|- Fy dy -j- F z dz -|- Fu du = 0

Gx dx -f- Gy dy -f- Gu du = 0 .

Now we eliminate du:

(GUFX — FuGx) dx -f- (GuFy — F uGy) dy + GUF z dz = 0 .

At points where Gu ^ 0 and F z ^ 0, we have

dz GUFX — FUGX dz GuFy — FuGy dx GUF Z dy GUF z

Numerical Approximations

The approximationf ( x + dx) & f ( x ) + df

can supply quick numerical estimates.

EXAMPLE 8.1

Estimate (2.01 )0-98.

Solution: Set f (x , y ) = x v. Then In / = y In x

— = - dx + (In x) dy. f x

Set x = 2, y = 1, dx = 0.01, dy = —0 .02. Also use the estimate In 2 ^ 0.69. Then / (2, 1) = 2 and

5 (0.01) - (0.69)(0.02) = -0.0088,

d f « -0.0176, f t t 2 + d f t t 1.9824.

Answer: 1,9824,

R e m a r k : By 4-place logs, (2.01 ) 0-98 « 1.982.

EXERCISES


Compute dz :

5. Proof of the Chain Rule 319

3. x = z2 - y2 4. ze~xy + xe~yz = 0.

5. For x = r cos 0, y — r sin 0, prove r dr = x dx + y dy and x dy — y dx = r2 dd.6. Let z = z ( x , y ) , p = dz/dx, q = dz/dy, r = d2z/dx2, s = d2z /d x d y , 2 = d2z/dy2.

Express dz, dp, dq in terms of dx, dy.7. Let / be homogeneous of degree n, that is, f(tx, ty, tz) — tnf(x, y , z ) for t > 0-

Compute “d” of both sides of this relation and equate coefficients of dx, dy, dz, dt- What results? (Compare Ex. 24, p. 291.)

8. Let z — z(x, y ) be defined byx2 + y2 + z2 = t2

y = tx.

Find zx and zy.

If v = v(s, t) = (f(s, t), g(s, t), h(s, t ) ) is a vector function of several variables, then we define dv by dy = (df, dg, dh) = vs ds + v t dt.

9. Consider the vectors x = (x, y) , u = (cos 6, sin6), w = (— sin0, cos0) of Chapter 5, Section 7, p. 193. Prove that

du = w dd, dw = — u dd,

dx = u dr + rw dB.

Estimate using differentials:10. 5.1 X 7.1 X 9.9 11. V (5.99 )2 - (3.02 )3.

-» R3. In column

9. PROOF OF THE CHAIN RULE [optional]

Vector-valued Functions

In Chapter 7 we discussed linear functions L: R3- vector notation, each linear L has the form

L (x ') = Ax' ,

where A is a 3 X 3 matrix. I t will be convenient throughout this section to think of R3 as the space of column vectors, even when we deal with more general functions, not necessarily linear.

We shall consider functions F: D --------> R3, where D C R3. Such a functionassigns to each x' in the domain D a column vector F (x ') in R3. We may write

7 i ( x T

F( x ' ) = / 2(x')

L/3(x' ) J


Thus the vector-valued function F is equivalent to a triple /i, / 2, fz of realvalued functions, the components of F.

Our first chore is to say when a vector-valued function is differentiable.

Differentiable Vector-valued Function Let F: D --------- > R3, where D C R3.Then F is differentiable at a point c' of D if there are a matrix A and a vector-valued function E such that

(i) F ( c' + x') = F ( c') + Ax' + E { x ' ) whenever c' + x' £ D,

(ii) r ~ E (x ') -------->0 as x '-------->0.I x |

I t is understood in (ii) that c' + x' £ D when x '--------> 0.We want to identify the matrix A in (i). If we compare this situation with

that of a differentiable/: R3--------- » R, as discussed in Section 1, we suspectthat A must involve partial derivatives. That this is so will fall out of the following characterization of differentiability.

Theorem Let F: D --------- > R3 and let /i , / 2, fz be the components of F, sof i i D --------- > R. Then F is differentiable at a point c' if and only if /i, / 2,and/3 are differentiable at c'.

Proof: Suppose F is differentiable. Write

7 i ( x ' ) “ ai ~ei (x')~

(1) F (X ') = /s (x ') , a = a2 , E ( x ' ) = e2(x')

_/3(x')_ _ a 3_ _e3(x')_

where ai, a2, a3 are the rows of A . From

(2 ) F( c ' + x') = F ( c ' ) + Ax ' + E ( x ' )

we have

(3) f t (c' + x') = f i ( c ' ) + atx' + 6t (x ').

The vector function |x'|~1 J57 (x ') -------->0 as x '--------> 0, hence each of itscomponents approaches 0, that is, |x'|_1 e ; (x ')--------> 0. Hence by (3), each /*is differentiable.

Conversely, if each / t- is differentiable at c', then (3) holds and|x'|_1 e i { x ' ) -------->0 for i = 1 ,2 ,3 . Consequently (2 ) holds with A andE { x ' ) as in (1); furthermore jx'|_1 E (x ') ----— > 0 as x '--------> 0 because eachcomponent of Ix'!-1 l£ (x ') approaches 0. Therefore F is differentiable.


Theorem Let F be differentiable at an interior point c' of its domain andlet

F ( c' + x') = F (c ') + Ax' + E (x ')

where Ix'h1 E (x') ■ 0 as x'

A =

—>0. Then

dA & d_hdXi dXi dx3

d_h d_h dfrdx\ dx2 dx3

dh dh dfs_dXi dXi dXi

where fi, f 2, fs are the components of F and all partials are evaluated at c.

This is an immediate consequence of the corresponding result for the functions /»•: R3--------- > R and the previous theorem.

The matrix A written above is called the Jacobian matrix of F. I t is important in the study of transformations of regions of space and in change of variables in multiple integrals.

Chain Rule

We first consider a composite function h = f ° G, where G is a differentiable function on a domain S of u-space into x-space a n d /is a differentiable function on a domain D of x-space into R. We assume G sends S into D. We want to prove that h is differentiable and to compute its partials. See Fig. 9.1.

Fig. 9.1 composite vector function

Let us experiment with the special case in which both / and G are non- homogeneous linear:

/ ( x ' ) = d + kx', Cr(u') = b' + Au'.Then

h( u') = f [ G ( u ' ) ] = d + kb' + kAu'.For any c',

h( c' + u') = d + kb' + kA(c' + u') = h ( c') + kAu'.

Therefore h is also non-homogeneous linear and

/ dh_ dh_ d h \ = kA = df_ \ d l t i ’ du2 1 du z j \ d # i ’ dx2 ’ dx j \_dU j \

With this easy case as a guide, we are ready to formulate the general result.


Chain Rule Let / : D ---------> R be defined on a domain D in x-space,differentiable at a point a'. Let G: S --------> R3 be defined on a domain Sof u-space into x-space, differentiable at a point c' such that G ( c') = a'. Assume G (u' ) £ D whenever u' £ S.

Then the composite function h = f ° G is differentiable at c'.

If/ ( a ' + x') ^ / ( a ' ) + kx' and G ( c' + u ') ^ G (c ' ) + Au',

thenh ( c' + u') h ( c') + kAu'.

Finally, if a' and c' are interior points of D and S respectively, then

dh(c ' ) = y a /(a ') a ^ ( c ')

i=i

for j = 1, 2, 3, where the gi are the components of G.

Proof: By hypothesis,

(1) / ( a ' + x') = / ( a ' ) + kx' + e (x ')

(2 ) G ( c' + u') = G(c') + Au' + E ( u'),

where fx'

(3) y T p --------»0 as x '-------->0lX I

(4) -------->0 as u '-------->0.I** I

Substitute:

h(c ' + u') = /[(? (c' + u ') ]

= / [G (c ' ) + Au' + £ ( u ' ) ]

= / [ ^ ( c ' ) ] + k [Au' + t f (u ' ) ] + e[Au' + # ( u ' ) ] ,hence

(5) h ( c ' + u ') = h(c ' ) + kAu' + {kE'(u') + e[_Au' + i?(u ')]} .

We must prove that the term in braces goes to zero faster than u', that is,

kE (u') + e[_Au' + E ( u ' ) ]* 0 as u' ■ -+0.

For then (5) says that h is differentiable at c' and that h (c' + u') ^ h (c') + kAu'.

Clearly

k E ( u ' )

by (4). I t remains to prove that

e\_Au' + -E(u')]

— E ( u ' ) lu'l K ’

0 as u'

->0

0.

This is the crux of the proof, and it requires some care and patience.First, by the triangle inequality

\Au' + E ( u ' ) \ < \Au'\ + \E(u') \ .

We proved in Chapter 7, Section 4, that |Au'| < bi |u'| where bi is a constant.W ealsohave \E(u') \ < |u'| for |u'| sufficiently small since |u'|-1l£ ( u ' ) --------->0by (2 ). Therefore

\Au' + E ( u ' ) \ < 6 lu'l,

where b = bi + 1 is a constant.Now write e(x ') = |x'| e i(x '), where by (3) we have ei(x ') -

x '--------- > 0. Then

|e[Au' + E ( u')] | = |Au' + E ( u ' ) \ ^ [ A u ' + E { u')] |

< 6 |u'| \e , tAu' + E ( u ' ) J .

But Au' + E ( u ') --------> 0 as u ' --------> 0, hence

e[Au' + E(u')~\

-» 0 as

 0

as u ' ---------> 0. This completes the proof of the first part of the Chain Rule.

If c' and a ' are interior points, then

^ 1w ’ at-2:’ <32-3/ 1

and A =

dg 1 dg iawi dU2 a^3

ICD

1 <Hj2 dg%awi du2 du*

tty? dg* dg*awi dU2 dl ls c '

We have proved that

h(c ' + u ') = ft(c') + kAu' + E (u ')

where |u'|~1E ( u ' ) --------- >0 as u '--------- > 0. Hence the elements of the rowvector kA are the partials of h at c'. But the ^-th element of kA is precisely

as required. Xd l d Q idXi dUj

i = l

Composite of Vector-valued Functions

We next consider a more general form of the Chain Rule. I t turns out to be an easy consequence of the previous one.

Theorem Let F : D --------> R3 and G : S --------» R3, where G(u') £ D foreach u' £ S. Suppose that G ( c') = a', that G is differentiable at c', and that F is differentiable at a'. Then the composite function H = F ° G is differentiable at c'. If

F( x ' ) « F ( a ') + Ax' and <?(u') « G(c') + £u ',then

ff(u ') (c') + (Afi)u'.

If a' and c' are interior points, then

hi (O') = y a /,(a ') agy(c')a ^ -w a#/ at/fc

j =i

Proof: Write

" M u ' f 7 i(x ') "

H ( u ') = /l2(u') and F (x ') = /*(x ')

Ja(x ')_


Then hi = f i ° G because H = F ° G. Since F is differentiable, each /» is differentiable. Hence, by the previous theorem, each hi is differentiable and

3dhj V dft dgj

dUk L j dXj dUk j= l

in the case of interior points. But this implies H is differentiable since a vector-valued function is differentiable if and only if its component functions are differentiable.

R e m a r k : The theorem asserts that the Jacobian matrix of the composite function F 0 G built from F and G is the matrix product A B of their respective Jacobian matrices A and B.

Other Cham Rules

The definitions, theorems, and proofs in this section can be modified with little trouble to cover all useful versions of the Chain Rule. The most general case might be indicated schematically as follows

GR m--------> R n

\\

H = F ° G \\

F

Rp.

Here F and G are differentiable on suitable domains. Then H is differentiable and

nd h i V s d h i d g j

dUk L j dXj dUk i =1

at interior points. The long proof starting on page 322 covers the case p = 1 almost verbatim, and the last proof above extends to arbitrary p.

9. Higher Partial Derivatives

1. MIXED PARTIALS

Differentiate a function f ( x , y ) of two variables. There are two first derivatives,

f x ( z , y ) and f y { x , y ) ,

each itself a function of two variables. Each in turn has two first partial derivatives; these four new functions are the second derivatives of f { x , y ) . Figure 1.1 shows their evolution:

fa j d

dx dy 1_a__|____ d_dx dy

f x x f x y f y x f y y

F ig . 1.1

The pure second partials

f x x and fyy

represent nothing really new. Each is found by holding one variable constant and differentiating twice with respect to the other variable.

Alternate notation:

_ ay ayJxx d x 2 , Jyy d y 2 -

For example, if f (x , y ) = x3y 4 + cos oy,

f x = 3 x Y , f y = - 5 sin oy,

f x x = 6 x y \ f y y = ! 2 x Y - 25 cos 5y.

a | adx dy

1. Mixed Partials 327

u „±(*±),*l anddy \ d x / dy dx dx \ d y / dx dy

are new. The mixed partial f xy measures the rate of change in the ^/-direction of the rate of change of / in the ^-direction. The other mixed partial f yx measures the rate of change in the x-direction of the rate of change of / in the y - direction. I t is not easy to see how, if at all, the two mixed partials are related to each other.

Let us compute the mixed partials of the function / (x, y ) = x*yA + cos 5y:

f x = 3 x2y \ fxy = 3 -4z y .

f y = 4x*y3 — 5 sin 5 y , f yx = 4 -3 # y + 0.

The mixed partials are equal! This is not an accident but a special case of a general phenomenon, true for functions normally encountered in applications.

The mixed second partials

Theorem Let /(# , y ) be defined on an open set D. If f xy and f yx exist at each)point of D and are continuous at (a, 6), then

W at (a, 6).dx dy dy dx

Proof: We consider the mixed second difference

(1) A = [ / ( a + h , b + k ) - / ( a + h, 6) ] - [ / ( a , b + k ) — /(a , 6)]

and its alternate form

(2) A = I f (a + h , b + k ) - /(a , b + &)] — [ / ( a + A, b) - /(a , &)].

We shall apply the Mean Value Theorem (twice) to (1). To do this, we set

g ( x ) = f ( x , b + k ) - f ( x , b ).Then

A = g ( a + h) - g ( a ) = hg'{xi),

where xi is between a and a + h. Next,

g' 0 0 = f x (xi, b + fc) - f x (xi, b) = kfxy(xi, yi ) ,>

where yi is between b and b + k. Hence

(3) A = hkfxy(xi, y x).

We apply similar reasoning to the second expression for A to obtain

(4) A = hkfyX(x2, 2/2 )>

where x is between a and a + h and y 2 is between b and b + k.

328 9. HIGHER PARTIAL DERIVATIVES

Now we take h = k 5* 0. By (3) and (4),

(5) f x y ( % l , y i ) = y % ) •

Let h --------> 0. Then (#i, y \ ) --------» (a, b ) and (%2, y i ) --------* (a , b ) . Sincef xy and f yx are continuous at (a, b ), we deduce from (5) that

f x y b ) = f y x ( d , b ).

EXERCISES

d2/ d2/Compute -—— and -—— ; verify that they are equal:ax ay ay ax

1. x*yb 2. xy6 3. x /y2 4. x + z32/ + yA5. sin (x + y ) 6. cos (xy)7. ex,y 8. arc tan (a; + 2?/)9. xw2/n 10. g(x) + /&(?/)

11. xy 12. y*

13. 14. 2X2/x — y x2 + y2

x — y 1 + xy

15. ~~—— 16. (x — y ) ( x — 2y) (x — Sy)

1 x2 +17. -7------- T 7----- s - t 18. ^(x — i/)(x — 2y) * t/2 + z19. ax2 + 2bxy + cy2 20. sinh(x + y2).

21. Show that each function of the form/(a;, y) = g(x y) + h(x — y) satisfies thed2f d2fpartial differential equation —2 — —2 = 0.

22. Show that each function of the form f ( x , y ) — g(x) + h(y) satisfies the partiald2fdifferential equation -—— = 0.

ax ay23. Prove the converse of Ex. 22: each solution of fxy = 0 has the form f ( x , y ) =

g(x) + h(y).

2. HIGHER PARTIALS

A function z = f ( x , y ) has two distinct first partials:

2. Higher Partials 329

and three distinct second partials:

dh d2z d*zdx2 y dx dy ’ dy2

Because the mixed second partials are equal, so are certain mixed third partials. For example,

/ d z \ _ d_ / d h \ J dy \ d x 2) ’

( d z \ _ d_ l d _ / d A ] _ d_ f d _ / d z \ l _ f d z \ ] _ d_ / & z \\ d y ) dx [dx \ d y ) J dx |_dy \ d x ) J dy [_d£ \ d x ) J dy \ d x 2)

dx2 \ d y /

since

dx2 '

Thus there are precisely four distinct third partials:

dh d3z d3z dhdx3 * dx2 dy 1 dx dy2 1 dy3

In general there are n + 1 distinct partials of order n :

dnzdxk dyn~

k = 0, 1, 2, • • •, n.

EXAMPLE 2.1

Find all functions z = f (x , y ) which satisfy the system of partial

d2z d2z d2zdifferential equations —- = 0 , ------ = 0 , —- = 0 .

dx2 dx dy dy2

Solution: If both first partials of a function are 0 everywhere, then the function is constant. (Why?) This applies to the function dz/dx since

0 and U * ) - dx \ d x / dx2 d\J \ d x /

d2Z d2Z= 0 .

dy dx dx dy

Hence

— = A. dx

Integrate with respect to x, holding y constant:

z = A x + g ( y ) ,

where the “constant of integration” is an arbitrary function of y. Sinced h / d y 2 = 0,

Consequently g ( y ) is a linear function,g (y) = B y + C.


Answer: z = A x + B y + C.

EXAMPLE 2.2

Find all functions z = f ( x , y ) whose third partials are all 0.

Solution: The second partials of dz /dx are all 0. By the last example,

Integrate:

But

— = A x + B y + C. dx

z = - A x 2 + B x y + Cx + g ( y ) .Zi

0 = — = ^dyz dy z

Consequently g ( y ) is a quadratic polynomial in y.

Answer: Any quadratic polynomial in x and y.

EXAMPLE 2.3

d2zFind all functions z = f ( x , y ) that satisfy — = 0 .

Solution: Write the equation

— ( — ) = 0, dx \ d x /

then integrate:

dzdx = g ( y ) -

Integrate again:2 = g ( y ) x + h ( y ) .

Answer: z {x , y ) = g ( y ) x + h ( y ) , where g i y ) and h (y) are arbitrary functions of y.

EXAMPLE 2.4

Find all functions z = f ( x , y ) that satisfyd2z

dx dy

2. Higher Partials 331

Solution: Write the equation

then integrate:

Integrate again:

— = 0 , dy W

dz e ^— = P \x ) . dx

Z = a M + h ( y ) ,

where g(x ) is an antiderivative of p ( x ) . Note that g (x) is an arbitrary function of x since p ( x ) is.

Answer: z (x, y ) = g( x) + h { y ) , where g(x) and h(y ) are arbitrary differentiable functions of one variable.

C h e c k :

d2dx dy [0(3) + h ( y ) 2 = + h ( y ) ] j = £ ; [ > '( j / ) ] = 0.

EXAMPLE 2.5

Find all functions z = f ( x , y ) that satisfy the system of partial

differential equations — = y, — = 1.dx dy

Solution: Integrate the first equation:

z = xy + g ( y ) .

Substitute this into the second equation:

— Ij cy + g ( y ) l = 1, x + g ' ( y ) = 1, g ' ( y ) = 1 - x. dy

This is impossible since the left-hand side is a function of y alone.

Answer: No solution.

This example illustrates an important point. A system of partial differential equations may have no solution at all! Could we have forseen this catastrophe for the system above? Yes; for suppose there were a function/ (x, y ) satisfying


Then

j E L , J L _ ± ( 1 ) . „ .d?/ d# d?/ ^ dy dx

so the mixed partials would be unequal, a contradiction.

If the system of equations — = p ( x , y ) , — = q(x, y )dx dy

has a solution, then — = — .dy dx

Indeed,

dp _ d_ / d z \ _ d_ / d z \ _ dq

dy dy \ d x j dx \ d y / dx

More Variables

All that has been said applies to functions of three or more variables. For example, suppose w — f ( x , y, z ) . Then w has three first partials:

dw dw dw

dx ’ dy 1 dz

The nine possible second partials may be written in matrix form:

d2W d2W d2W

dx2 dx dy dx dz

d2W d2W d2W

dy dx d y 2 d y dz

d2W d2W d2wdz dx dz dy dz2

This matrix is symmetric since the mixed second partials are equal in pairs:

d2W d2W d2W d2W d2W d2Wdy dx dx dy ’ dx dz dz dx 1 dz dy dy dz

EXERCISES


L dx dy2, d*f j d3/ Compute t i t - and

dar d y

1. x*y*3. x2yA 5. cos (xy)7. e^sinz

9. xlly

11.

2. xAyb4. xmyn6. sin (x2y)8. xy

x2 + ;

10.

12.

13. Find all functions f(x, y) such that aydx2 dy

x — y x + y

xy x2 + y2

= 0.

d*f d*f14. Find all functions f ( x , y ) such that _ 9 _ = 0 and -—-r-z = 0.dx2, dy dx dy2

15. Find all functions f(x, y) whose 4-th partial derivatives all equal 0.a4/16. Find all functions / (a;, y) such that • = 0.

dx2 dy2

Find all functions f(x, y) that satisfy the system of partial differential equations:

17 dl = n dl = hdx ’ dy

1Q df 2 df19. — = y2, — = X2dx dy 20- s - v ' I - 3*’9’-

22. xy + yz + zx 24. x2 + yz.

Write the matrix of 9 second partials:21. xmynzp23. sin (a; + 2y + Sz)

25. How many distinct third partials does/(a ;, y, z) have?26. (cont.) Find an explicit function for which they really are distinct.

d3/27. Find all functions / (x, y, z) satisfying ^ ^ ^ = 0.

28. How many distinct second partials does a function / (x, y, z ,w ) of 4 variables have? How many distinct third partials?

3. TAYLOR POLYNOMIALS

Let us recall some facts about Taylor polynomials. (See Chapter 2, Section 3.) If y = f ( x ) , then

f ( x ) = / (a) + f ' ( a ) (x - a ) + n(a;),

and

f ( x ) = f ( a ) + f ' ( a ) ( x - a ) + ^ f " ( a ) ( x - a ) 2 + r2(x),

where

kiO*OI < (x - a )2, |r2(a:)| < j * \x - a |3,

and where M 2 and M s are bounds for | f ' { x )| and | f " ( x )| respectively.The Taylor polynomial

P i0*0 = f i f l ) + f ( a ) ( x - a ) -

is constructed so that p i ( a ) = /(a ) and p i (a) = / '( a ) . The Taylor polynomial

P2 (x) = / (a ) + f ' ( a ) ( x - a ) + ^ f ” ( a ) ( x - a )2

is constructed so that p 2(a) = /(a ) , p 2 (a) = / ' (a), p 2" (a) = f " (a).In a similar way, one can construct linear and quadratic polynomials in

two variables approximating a given function of two variables.

Taylor polynomials Let f (x , y ) have continuous first and second partials on an open domain D. The first degree and second degree Taylor polynomials of / at (a, b ) are

P ifc y ) = /(«> &) + /* • 0 “ «) + / y* (:V ~ 6),

P2^ , y) = PiO, 2/) + hlfxx- (x - a )2 + 2f xy- (x - a ) ( y - b)

+ fyy9 (y “ &)2],where all the partials are evaluated at (a, b).

I t is easy to check that p i ( a , b) = / (a, b) and that the first partials of pi agree with those of / a t (a, b). Similarly, p 2(a, b) = f (a, b) and all first and second partials of p 2 agree with the corresponding partials of / a t (a, b).

Now we ask how closely these Taylor polynomials approximate f ( x , y ) for (x , y ) near (a, b). In other words, we want estimates for the errors in the approximations f ( x , y ) Pi(x, y ) and f {x , y ) P2(x} y) . We shall obtain error estimates subject to the mild condition that / is defined on a convex domain. A set S in the plane or space is called CQXivex if it contains the whole segment joining any two of its points.

Theorem L e t/b e defined on a convex open domain D in R2 and let a £ D.

(1) Suppose / has continuous first and second derivatives on D and the second derivatives satisfy

I fxx\ < M 2, | fxy\ < M 2, I fyy\ < M2, for all x £ D.

Let pi be the first degree Taylor polynomial of / at a. Then

f ( x ) = p i(x) + ri(x),

where |ri(x)| < M2 |x — a|2.

(2) Suppose / has continuous first, second, and third derivatives on D and the third derivatives satisfy

| fxxx\ < Mz, I f X X y \ < Mz, I fxyy\ < M z , | /y*/j/| < ^3-

Let p2 be the second degree Taylor polynomial of / at a. Then

f i x ) = p2(x) + r2(x),

a/2where |r2(x)| < — - M z |x — a|3.

o

Proof : The idea is to interpret the problem in such a way that we can use the error estimates on the previous page for a function of one variable.

Assume at first that a = 0; this will simplify the notation considerably. Now fix a point x £ D. By convexity, D contains the entire line segment connecting 0 and x, that is, D contains all points Zx for 0 < t < 1. (Actually because it is open, D contains a slightly larger open segment.) %

Set g ( t ) = f ( t x ) . Then g ( t ) is a function of one variable defined for 0 < t < 1 and

9 (0 ) = / ( 0), 0(1) = / 00 - Let us compute the first and second degree Taylor polynomials of g at t = 0. By the Chain Rule,

g'( 0 = fx ( t* )x + f y ( t x ) y ,g ' ( 0) = f x ( 0 ) x + f y (0)y ,

g " ( t ) = f x x ( t * ) x 2 + 2f Xy ( t x ) x y + f y y ( t x ) y 2,

g " ( 0) = f x x ( 0 ) x 2 + 2 f Xy ( 0 ) x y + /„ (0 )i/2,

^"'(0 = f x x x ( t x ) x3 + 3 f XXy ( t x ) x 2y + 3 f Xy y ( t x ) x y 2 + f y y y ( t x ) y 3.

From these calculations we see that

Pl(x) = ff(0) + ff'(0 ), p,(x) = !7(0 ) + p '(0 ) + ^ " ( 0 ) .

Thus p-i(x) and p«(x) are the first and second degree Taylor polynomials of g ( t ) evaluated at t = 1. Therefore

max \g” (t)\


K (x ) | = | / ( x ) - P l(x)| <

|r2(*)l = l / ( * ) - Pa(*)I <

21

max \g"' (<)|3!

I t remains to estimate \ g " ( t ) \ and \ g " ' ( t ) \ . We have

W (01 < Mi |*|2 + 2Mi |*| \y\ + M2 \y\2 = M 2(\x\ + \y\)2,

\g"'(t)\ < M 3 |*|3 + 3M s |*|2 \y\ + 3M 3 \x\ \y\2 + M 3 |*/|3 = M 3(\x\ + |t/|)3.

Now we modify these estimates slightly as follows. From (|*| — \y \ )2 > 0 we have 2 |*| jz/| < |*|2 + |?/|2, hence

(1*1 + \ y\ )2 = 1*1* + 2 1*1 \v\ + \y\2 < 2 (l*l* + \y\2) = 2IXI2.

Take the § power:(1*1 + \y \ )3 < 23/2|x|3.

Therefore,\g" (01 < 2M 2 |x |2, \ g " ' ( t ) \ < 2 y / 2 M 3 |x |3.

Combining results we obtain

, , „ ^ 2 M 2 |x |2 i , „ ^ 2 y / 2 M 3 |x |3 l* i(x)l < 2 , , | r2( x ) | < — — -------- .

This completes the proof assuming a = 0. In the general case, we define g{ t ) = /[a + £(x — a)]. The proof proceeds as before, except that (#, y ) is replaced by (x — a, y — b) and the partials of / are all evaluated at a.

R e m a r k : There are Taylor polynomials of higher degree and corresponding error estimates. The notation for these polynomials is complicated, and since we shall not need them, we leave their study to an advanced calculus course.

EXAMPLE 3.1

Compute the Taylor polynomials pi (x , y ) and P2 &, y ) of the funct io n /^ , y ) = \ A 2 + V2 (3> 4).

Solution:

df______ x d f______ ydx y / x 2 + y 2 ’ dy \ / x 2 + y 2 ’

d2f y 2 d2f —xy d2f ______ x2dx2 (X2 + y 2y 12 ’ dx dy (X2 + y2)3/2 ’ dy2 (;X2 + y 2)*12

At (3, 4),

ctf _ 3 £ f _ 4 ay _ _ie_ j y _ = _ } 2 _ a2/ _ 9ax _ 5 ’ a ^ s ’ dx2 ~ 125 ’ ax ay - 125 ’ a?/2 125 '

Therefore

p2(*> 2/) = 5 + ^ (a: — 3) + - (y — 4)

1 T 1 fi 24 9+ 2 [ l 2 5 < I - 3)' - i r 5 <* - 3 ) ( » - 4) + l 2 5 (!' - 4)!J '

The polynomial pi (x, y ) is just the linear (first degree) part of p2 (a:, y ) .

Answer:

Pi(x, y ) = l l 25 + 3(x - 3) + 4(y - 4 )];5

Pafo 2/) = Pi(x, y ) + ~ [16 (a; - 3 )2

— 24(a; — 3) (2/ — 4) + 9 (2/ — 4)2].

EXAMPLE 3.2

Estimate V ( 3 - l ) 2 + (4.02)2 by(a) p i ( % ,y ) , (b) p2(x, y ) ,

the Taylor polynomials in the last example.

Solution: Let f ( x , y ) = \ / x 2 + y 2.

(a) Near (3, 4),

/(* , y ) » I [25 + 3(* - 3) + 4(2/ - 4)],

y (3.1, 4.02) « - [25 + 3(0.1) + 4(0.02)] = 5.076.5

(b)

f ( x , y ) « pi (x, y ) + ~ [16 (a; - 3 )2 — 24(a: - 3) ( y - 4) + 9 ( y - 4)2],

/(3.1, 4.02) « p i(3.1, 4.02) + [16 (0.1 )2 - 24(0.1) (0.02) + 9(0.02)2]250

= 5.076 + = 5.0764624.

Answer: (a) Approximately 5.076;(b) Approximately 5.0764624.(Actual value to 7 places: 5.0764555.)

EXAMPLE 3.3

If \x\ <0 . 1 and \y\ < 0.1, prove that

\ex sin Or + y ) — (x + y ) \ < 0.05.

Solution: Let f ( x , y ) = ex sin(x + y ) . Then as is easily checked, x + y is simply pi (x, y ) , the first degree Taylor polynomial of f (x , y ) at (0, 0). Thus we are asked to verify that |ri(x)| < 0.05 for points x = (x, y ) with |#| < 0.1 and \y\ < 0.1. Such points satisfy |x|2 = (0.1 )2 + (0.1 )2, so we may restrict the domain o f f to the open disk |x|2 < 0.02.

Our error estimate yields|ri(x)| < M 2 \x\2 = (0.02)M2,

where M 2 is a bound for \fxx\i | fxy\) I f w \ • To find a suitable value for M 2y compute the second partials:

fxx = 2ex cos (x + y ) , f xy = 6x[cos(a; + y ) - sin (a; + y)],

fyy = ~ e x sin (x + y ) .

The elementary estimates |sin(a; + y )\ < 1 and |cos(# + y) \ < 1 show that

| fxx I < 26X, | f xy\ < 2eX, | fyy | < 6*.

Now \x\ < 0.1, soex < 60*1 < 1.11,

as is seen from a table. Therefore we take M2 = 2(1.11) = 2.22, and it follows that

|n (x )| < (0.02) (2.22) = 0.0444 < 0.05.

EXERCISES

Compute the Taylor polynomials p i (x, y ) and p2 (x, y):1. x2y2; at (1, 1) 2. x4ys; at (2 ,-1 )3. sin (xy); at (0,0) 4. exy; at (0,0)5. xy\ at (1,0) 6. xy; at (1,1)7. cos{x + y); at (0, t / 2 ) 8. 1 + x y ; at (1, 1)9. In Or+ 2y); at (^,i) 10. x2ey; at (1,0).

Estimate using the second degree Taylor polynomial. Carry your work to 5 significant figures:11. (1.1 J1*2 12. [ (1.2 )2 + 7.2]1/3

4. Maxima and M inim a 339

13. /(1.01, 2.01), where f(x, y) = x*y2 — 2xy4 + yb14. V (1.99)2+ (3.01)2+ (6.01 )2.

15. Let pi(x, y) be the first degree Taylor polynomial of f ( x , y ) at (a, 6). Identify the graph of z = p i (x , y ) .

16. Let p i ( x , y , z ) be the first degree Taylor polynomial of f ( x , y , z ) at (a, b, c). Assume grad / 9 0 at (a, b, c). Identify the graph of the equation pi(x, y, z) = /(a, b, c).

Prove the inequality, given \x\ < 0.1 and \y\ < 0.1:17. |\/1 + x + 2y — (1 + \ x + 2/)I < 0.0418. \ex sin (a; + y) — (1 + z)(l + y)\ < 0.01

(even <0.005 with more careful estimates).

4. MAXIMA AND MINIMA

In this section we shall develop second derivative tests for maxima and minima.

We shall assume here and in the rest of this chapter that all functions have continuous first, second, and third partial derivatives. With this assumption, the Taylor approximations of the previous section apply.

Let us begin with a brief review of the one-variable case. We consider a function g( t ) and a point c where g' (c) = 0. Suppose g" (c) > 0. We want to conclude that g(c) is a relative minimum of g. For this purpose, an excellent tool is the second degree Taylor approximation of g at c :

g(t) = g(c) + ^g"{c){ t - c) + r»(t),

1 2(01 5= & V ~ cl3> k > 0.

I t follows that

gi t ) ~ g(c) = ^ g"{c)( t - c)2 + r2(<) > ~ g " (c) it - c)2 - k\ t - c\3

= it - c )2 |^ 5 f" (c) - k \t - c|J .

Since g"(c) > 0, the quantity on the right is positive if 0 < \t — c\ < ^"(c)/A;. Thus there is a positive number 8 = \ g " ( c ) / k such that g{ t ) — g(c) > 0 when 0 < \t — c\ < 8. In other words, g(c ) is smaller than any other value of g in an interval of radius 8 and center c. Hence g(c) is a relative minimum of g.

Hessian Matrix

Let us try to generalize these ideas to the two variable case. Given /(# , y ), suppose (a, b) is an interior point of the domain of / where f x (a, b) =

fy(a, b) = 0. We would like a condition on the second derivatives of/, analogous to g" (c) > 0, guaranteeing th a t/(a , b) is a relative minimum of /. Now the condition g" (c) > 0 can be interpreted as meaning that

g ' ' ( c ) ( t - c ) 2

is a positive definite quadratic form in the one variable t — c. I t is natural, therefore, to examine the quadratic part of the second degree Taylor polynomial o i f ( x , y ) at (a, b). If it is positive definite, we suspect th a t/(a , b) is a minimum of /.

The quadratic part in question is

^ [/**• (* - «)2 + 2f*y' (x - a ) ( x - b ) + f yv• (2/ - b)2].

In matrix-vector notation, this can be expressed as

- (x - a ) H f (x - a )',

where

H f =fxx fxy

Jyx fyy.

The matrix H f is called the Hessian matrix of/. Our assumption on the partial derivatives of / ensure that f xy = f yx. Hence H f is a symmetric matrix.

N o t e : I t is more accurate to write H f ( x , y ) since the Hessian matrix generally depends on the point (x, y ) . When the meaning is clear from the context, however, we shall use the simpler notation H f .

Second Derivative Test

Theorem 4.1 Let a = (a, 6) be an interior point of the domain of /. Sup-pose that

f x { a , b ) = 0, f y { a , b ) = 0,and that the Hessian matrix H / ( a , b ) is positive definite.

Then / (a, b ) is a relative minimum of / . That is, there is 8 > 0 such that

/ ( * , V ) > f ( a , b )

whenever0 < |x — a < S.

Proof: To make the notation simple, let us take (a, b) = (0,0). Then the second degree Taylor approximation of / a t (0, 0) is

f f (%, y ) = / ( 0, 0) + \ \_Ax2 + 2B x y + Cy21 + r(x, y ) ,

where

H f { 0,0) =' A B fxx fxy

B C_ Jv x fyy_ (0 ,0)

and h is a positive constant. Sinee H f is positive definite, we can apply Theorem 5.4, p. 249; there is another positive constant k such that

A x 2 + 2Bx y + Cy2 > k |x|2.Consequently

/(* , V) “ /(0 , 0) > \ k |x|2 - h |x|3 = |x|2 (J* - h |x|).

This implies f ( x , y ) — f ( 0,0) > 0, that is f ( x } y ) > f ( 0,0), provided 0 < |x| < \ k /h . Finally, for 8 we may choose any number such that first0 < 8 < \ k / h and second 8 is so small that the disk |x| < 8 is contained in the domain D.

R e m a r k : The essential point in the proof is that for |x | small, the positive definite quadratic form A x 2 + 2B x y + Cy2 is much larger than the remainder r(x, y ) because r is of third order: |r(x, y) \ < h |x|3.

By Theorem 5.1, p. 247, the matrix H f (a, b) is positive definite if and only if

f x x f x y

that is,

f x x ( a , b ) > 0 ,

f y x f y y

f x x iP', b ) > 0 , ( f x x f y y / Xy)

> 0 ,(a,b)

> 0.(a ,6)

Suppose we want a maximum instead of a minimum. We reason that finding a maximum of / is the same as finding a minimum of —/. Therefore the following conditions are sufficient fo r/(a , b) to be a relative maximum o f / at a point of its (open) domain:

fx(a, b) = 0, f v (a, b) = 0,

fxx (a, b) < 0 , (fxx fyy ~ f l y ) > 0.(a ,6)

The two inequalities make up a necessary and sufficient condition for the Hessian matrix of —/,

- f x x - f x y

H - f =_ fyx fyy_

to be positive definite, namely, —f Xx > 0 and det H - f > 0. Note particularly that det H - S = det H f , not the negative.

(

Let us review our procedures. We start with / on an open domain D. We find a point (a, b ) where /* = 0 and f y = 0. We compute the Hessian matrix

H f =f x x fxx

. f y x f y y _

at (a, b). There are three possibilities for det H f = f x x f yy — f \

(1) det H f > 0,

In case (1), we have

(2) det H f = 0, (3) det H f < 0.

so fxx 7* 0. Either f xx > 0 and f ( a , b ) is a relative minimum, or f xx < 0 and /(a , b) is a relative maximum.

In case (2), no conclusion can be drawn. For example take the functions f ( x , y ) = x3y s and g { x } y ) = xAy*. Then f x = f y = 0 and H f = 0 at (0, 0), and similarly for g. Yet g has a minimum at (0, 0) whereas / has neither a minimum nor a maximum.

In case (3) there is useful information and we state it as a formal theorem.

Theorem 4.2 Suppose f x ( a , b ) = 0 and f y ( a , b ) = 0 at an interior point of the domain of /. Suppose

f x x f x y

f y x f y y

< 0.(a ,6)

Then f ( a , b ) is neither a relative maximum nor a relative minimum of /. That is, there are points (x, y ) arbitrarily close to (a, b) where /(# , y ) < f (a, b) and there are points (x, y ) arbitrarily close to (a, b) w here /(x, y ) > /(a , b).

Proof: We take (a, b) = (0, 0) for simplicity and consider the Taylor approximation

/(«> y ) = /(0 , 0) + ^ (Ax2 + 2Bx y + Cy2) + r(x, y ), \r(x, y) \ < h |x|3.

SinceA B f x x fx y

B C fy x fyy (0,0)

the quadratic form Q(x, y ) = A x 2 + 2B x y + Cy 2 is indefinite by Theorem 5.7, p. 251. Therefore there are points (x\, y i ) and (x2, y 2) such that Q ( x i, y \ ) < 0 and Q (x 2, y 2) > 0.

4. Maxima and M inim a 343

(b) level curves

Fig. 4.1 graph of z = y2 — x2

Consider (x, y ) = (tei, ty i ) , where t > 0 and small. Then

-Q(tei , fr/i) = ^Q(#i, 2/i) 2 = kt2, k < 0 ,

|r(tei, tyi)\ < h(x\ + y \ ) l,2t3 = W 8,hence

/(tei, tyi) - /(0 , 0) < kt2 + hit* = t2(k + W ).

Now for t sufficiently small, t2(k + hit) < 0, hence

/(tei, tyi) < / (0 , 0)

for points (tei, tyi) as close to (0, 0) as we please.Sim ilarly/(te2, ty2) > /(0 , 0) for points (te2, ty2) as close to (0, 0) as we

please. Therefore / (0 , 0) is neither a max nor a min of /.

In the case we have just discussed, / is said to have a saddle point at (a, b ) ; the surface z = f ( x , y ) is shaped like a saddle near (a, b, f (a, b ) ) . The tangent plane is horizontal; the surface rises in some directions, falls in others, so that it crosses its tangent plane. This is like the summit of a mountain pass. See Fig. 4.1 for an example.

Summary Let (a, b) be an interior point of the domain of / and suppose

f x ( a , b) = f v (a , b) = 0.

(1) If f xx > 0 and f xxf yy — f l y > 0, then / (a, b) is a relative min.

(2) If f xx < 0 and f x x f yy — f xy > 0, then /(a, b) is a relative max.

(3) If f xxfyy — f l y < 0, then f ( a , b ) is a saddle point, neither a relative max nor min.

(4) If fxxfyy — f l y = 0, no conclusion can be drawn from this informationalone.

Does the function z = f ( x , y ) have Try the first and second derivative

1. z = x23. z = x2 - 4y25. z = —x2 + 2xy — y27. z = x4 + y49. z = x3 + y3

EXERCISES

a possible maximum or minimum at the origin? tests. Sketch a few level curves near the origin:

2. z = xy4. z = x2 + 2xy + y26. z = x4 + y28. z = —x2y2

10. z = x2 + yb.

11. Suppose f ( x , y ) is a function of two variables and x = x( t ) and y = y( t ) are functions of time. Form the composite function g(t) = / [ z (t), y 0)]. By the Chain

5. Applications 345

Rule, g(t) = fx[x(t) , y ( t )]£(*) + fv[x(t) , y(t)~\y(t). Show that

g = fxxX2 + 2fxyXy + fyyi/2 + fxX + fyjj.

12. (cont.) Suppose also that fx(0, 0) = fy (0,0) = 0, and that only curves x(t) = ( x ( t ) , y ( t ) ) are allowed which pass through (0, 0) with speed 1 at t = 0. Suppose for each such curve g(0) > 0. Show that g(0) = 0 and fxx(0, 0) > 0 and/w(0» 0) > 0. /

13. (cont.) Show also th a t/2x(0, 0)/^(0, 0) — fxy(0, 0)2 > 0.14. (cont.) Find a formula for cPg/dt*.15. (cont.) Suppose g( x ,y ) = Ax2 + 2Bxy + Cy2 and AC — B2 = 0. Conclude that

g(x, y) = -b {ax + by)2.

5. APPLICATIONS

EXAMPLE 5.1

Show t h a t / (#, y ) = 4 x2 — xy + y2 has a minimum at (0 , 0 ) .

Solution:

f x = 8x - y , f y = - x + 2 y .

Both partials are 0 at (0 , 0 ) . Furthermore,

f x x = 8 , f x y = 1, f y y = 2.Hence,

f x x > 0 , f x x f y y - f xy2 = 15 > 0 .

These conditions ensure a minimum at the origin.

EXAMPLE 5.2

Of all triangles of fixed perimeter, find the one with maximum area.

Solution: If the sides are a, b, c, then the area A is found by Heron’s formula:

A 2 = s ( s — a) (s — b) (s — c),where s = + b + c) is the semiperimeter. Thus

2s = a + b c, s — c = a + b — s,

A 2 = s ( s — a) (s — b) (a + b — s) .

Here s is constant and the variables are a and b. In order to maximize A , it suffices to maximize

/(a , b) = (s — a) (s — b) (a + b — s) .Now

f a = (s — b) (2s — 2a — b), f h = (s — a) (2s — a — 26),

f a a — —2( s — b), f a b — — 3s + 2a + 26, f bb — —2 ( s — a) .

The equations f a = 0 and f b = 0 imply

2a -j- b = 2s, a 26 = 2s,

since s = a or s = b is impossible in a triangle. (Why?) I t follows that

For these values of a and 6,

/ a o — f bb — 3 < f a b ~ ^ ,

4S2 g2 g2 f a a f b b ~ f a b2 = “ ~ = ~ > 0.

These conditions ensure a maximum. Now compute c :

2 2 2 c = 2s — a — b = 2s — ~ s — - s = - s .

O o O

Hence a = b = c.

Answer: The equilateral triangle.

EXAMPLE 5.3

x2 y2Find the point on the paraboloid z = ~

closest to i = (1, 0, 0).

Solution: Let x = (#, y, z ) be a point on the surface. Then

|x — i|2 = (x — l ) 2 + y2 + 22

= (x - l ) 2 + y2 + ^ + 0 = f ( x , y ) .

The function/ (x, y ) must be minimized. Now

/ ■ - 2 < * - > ) + * ( f + f ) ,

The condition /j, = 0 is satisfied if either

y = 0 or 1 + ? ^ + * V o .9 \4 9 /

The latter is impossible. Therefore f y = 0 implies y = 0. But if y = 0, the condition f x = 0 means

32 (x — 1) + — = 0, that is, x* + Sx — 8 = 0.

A rough sketch shows this cubic has only one real root, and the root is near x = 1. By Newton’s Method iterated twice, x 0.9068.

Thus f ( x , y ) has a possible minimum only at the point (a, 0), where a3 + 8a — 8 = 0. Test the second derivatives at (a, 0):

/** = 2 + | a2 > 0, f xy = 0, fyy = 2 + i a2 > 0,

/«e/w - /*y2 > 0.

Therefore the minimum does occur at (a, 0).

5. Applications 347

Answer: ( a , 0, j a*) , where

a3 + 8a — 8 = 0, so a « 0.9068.

The following example shows that you must be careful when the variables are restricted.

EXAMPLE 5.4

y 2 z2Find the points on the ellipsoid x2 + •— + — = 1 nearest to and

y 4

farthest from the origin.

Solution: The square of the distance from (x , y, z ) to (0, 0, 0) is

f {x , y ) = x2 + y2 + z2 = x2 + y2 + 4 - x2 - = 4 - dx2 + ^ y2.

Since

there is a natural restriction on x and y :

Thus the domain o f / (x, y ) is the closed set bounded by the ellipse x2 + \ y 2 = 1. The first partials are

These vanish at (0, 0) only. However, at (0, 0),

f x x f y y — f x y 2 = — 6 •“ 0 < 0 .

Therefore f ( x , y ) has neither a maximum nor a minimum at (0, 0). There seems to be no possible maximum or minimum.

We have forgotten the boundary! The continuous function / has both a maximum and a minimum in its bounded closed domain, and since they do not occur inside the domain, they must occur on the boundary curve (Fig. 5.1).

Notice that if (x , y ) is on the ellipse, then

V2x2 + — = 1 and 2 = 0.

Thus the nearest and farthest points from the origin are actually points on the ellipse. By inspection (Fig. 5.1), they are the points (±1 , 0, 0) and (0, ± 3 , 0). [The points (0, 0, ± 2 ) are saddle points for f (x , y ).]

Answer: Nearest: ( ± 1 , 0 , 0 ) ; farthest: (0, ± 3 , 0).

EXERCISES

Find the point nearest the origin in the first octant and on the surface:1. xyz = 1 2. xy2z — 1.

6. Three Variables 349

3. Find the maximum and minimum of sin (xy) for 0 < x} y < tt.4. Find the points on the hyperboloid x2 + y2/ 4 — z2/ 9 = 1 nearest the origin.5. Given n points Pi, • • •, P n of the plane, find the point P such that "=i P P / is

least.6. For each point (x, y , z ) , let f(x, y, z) denote the sum of the squares of the distances

of (x, y, z) to the three coordinate axes. Find the maximum of f(x, y, z) on the unit sphere x2 + y2 + z2 = 1.

7. Find the least distance between the line x(t) = (£ + 1, t, — t — 3) and the line y( r ) = (—2r + 3, r + 1, —r — 2 ).

8. Give an alternate solution to Example 5.3. By inspection, find the point closest toi on each cross-section z = constant.

9. Approximate to 3 places the minimum first octant distance from the origin to the surface xeyz = 1.

6. THREE VARIABLES

The extension to three variables of the second derivative test for maxima and minima is straightforward.

Theorem Let a be an interior point of the domain of / and suppose

g rad / = 0.

Let

H f ( a) =

f x x f x y f x z

f y x f y y f y z

_ / zx f z y f z z _

denote the Hessian matrix of / at a.

(1) If H f (a) is positive definite, then / ( a ) is a relative minimum of /, that is, there is 8 > 0 such that / (x) > / ( a) whenever 0 < |x — a| < 8.

(2) If H f (a) is negative definite, t hen / ( a ) is a relative maximum of /, that is, there is 8 > 0 such that / (x) < / ( a) whenever 0 < |x — a| < 8.

(3) If H f (a) is indefinite, then / ( a ) is neither a relative max nor min.

(4) Except in these cases, no conclusion can be deduced from H f ( a) alone.

The proof of (1) is essentially the same as the proof of Theorem 4.1. As we know, (2) follows by applying (1) to —/. The proof of (3) is essentially the proof of Theorem 4.2.

We can use the practical tests in Chapter 7. For instance Theorem 5.2, p. 248, tells us that H f is positive definite if and only if

f x x > 0 ,f x x f x \

> 0, and \Hf \ > 0.f y x f y y

Let us take a very simple example. Suppose

f ( p , y , z ) = + by2 + cz2,

where a, b, c are nonzero. Then

grad/ = (2ax, 2by, 2cz),

so grad/(x) = 0 only at x = 0. The matrix A in this case is

f x x f x y f x z 2a 0 0 “

A = fyx fyy fyi = 0 2b 0

_ f z x fry _0 0 2c_

According to the test, / (0) is a minimum if the three determinants are positive:

2a 0 0

2a > 0,2 a 0

0 2b> 0 , 0 2b 0

0 0 2c

> 0 ,

that is, if2a > 0, 4ab > 0, 8abc > 0.

This is so precisely if a > 0, b > 0, and c > 0.Similarly/(0) is a maximum if

2a < 0, 4ab > 0, 8abc < 0,

which is so precisely if a < 0, b < 0, and c < 0.These results agree with common sense:

/ ( x ) = ax2 + by2 + cz2

so obviously / (0) = 0. If a , b, c are all positive and x ^ 0, then / ( x ) > 0. If a, b , c are all negative and x ^ 0 , then/ (x) < 0. (If a, b, c are not all of the same sign, then /(0 ) is neither a maximum nor a minimum. Why?)

EXAMPLE 6.1

Find all maxima and minima of the function

/ (x) = x2 + 3 y 2 + 4 z2 — 2 xy — 2 yz + 2zx.

6. Three Variables 351

Solution:

grad/ = (2x — 2y + 2z, —2x + 6y — 2z, 2x — 2y + 8z).

The vector equation g rad / = 0 amounts to the system of scalar equations (divide by 2 )

x — y + z = 0

-x + Sy — z = 0

x — y + 42 = 0 ,

whose only solution is x = 0.The matrix A is

f x x f x y f x z 2 - 2 2"

f y x f y y f y* = - 2 6 - 2

_ f z x f z y fzz__ 2 - 2 8_

so the three relevant determinants are

2,

2 - 2

- 2 6and

2 - 2 2

- 2 6 - 2

2 - 2 8

= 48.

All are positive. Hence by the test above, / ( 0 ) is a minimum of /, the only one; there are no maxima.

Answer: The only extreme value of / is the m inim um /(0 ) = 0 .

EXERCISES

The second derivative test is inconclusive at (0, 0, 0) for the given function. Determine nonetheless if the function has a maximum, a minimum, or neither at the origin:

1. x2 + y2 + z4 2. x2 + y2z23. x2 + y2 4. x4+ y2 — z65. x2 + y4 + 26 6. x*y*£7. x4 + 2/323 8. x4y4 — 25.9. x4 + y2z2 10. xs + y3 + 2*

11. x ^ z 3 12. x2y2z2.

Find the extreme values:13. —2x2 — y2 — 322 + 2xy — 2 xz


14. x2 + 2y2 + z2 + 2xy — tyz15. 2x2 + y2 + 2z2 + 2xy + 2yz + 2zx + x — 3z16. £2 + 3^2/ + y2 — z2 — x — 2y + 2 + 3.Determine if the function has a maximum, a minimum, or neither at the origin:17. x2 + y2 + z2 + xy + yz + zx 18. x2 + 4y2 + 9z2 — xy — 2yz19. - x 2 - 2y2 - z2 + yz 20. x2 + y2 + 2z2 - 10yz21. x2 — y2 + 3z2 + 12xy 22. Sx2 + y2 + 4z2 — xy — yz — zx.

23. A surface x = x(u,v) and a curve p = p(t) are given. Suppose the distance \x(u, v) — p(0 | of a point on the surface to a point of the curve is minimized (or maximized) for x0 on the surface and po on the curve. Show that the segment from x0 to po is normal to both the surface and the curve. Find the one exception to this assertion.

24. (cont.) Formulate the corresponding statement for two surfaces. (This is a four- variable problem!)

7- MAXIMA WITH CONSTRAINTS [optional]

Here are several problems that have a common feature.(a) Of all rectangles with perimeter one, which has the shortest diagonal?

That is, minimize (;x2 + y2)112 subject to 2x + 2y = 1.(b) Of all right triangles with perimeter one, which has largest area?

That is, maximize xy/ 2 subject to x + y + (x2 + y2) 1'2 = 1.(c) Find the largest value of x + 2y + 3z for points (x, y, z) on the unit

sphere x2 + y2 + z2 = 1.(d) Of all rectangular boxes with fixed surface area, which has greatest

volume? That is, maximize xyz subject to xy + yz + zx = c.

Each of these problems asks for the maximum (or minimum) of a function of several variables, where the variables must satisfy an equation (constraint).

7. Maxima with Constraints 353

For example, in (a) you are asked to minimize

f (x, y ) = x2 + y2,

where x and y must satisfy

g(x, y ) = 2x + 2y - 1 = 0.

Such problems may be analyzed geometrically. Suppose you are asked to maximize a function f ( x , y ) , subject to a constraint g(x, y) = 0. On the same graph plot g(x, y) = 0 and several level curves of f ( x , y ) , noting the direction of increase of the level (Fig. 7.1). To find the largest value of f (x, y) on the curve g ( x , y ) = 0 , find the highest level curve that intersects g = 0 . If there is a highest one and the intersection does not take place at an end point, this level curve and the graph g = 0 are tangent.

Suppose f (x, y ) = M is a level curve tangent to g(x, y) = 0 at a point ( x , y ) . See Fig. 7.2. Since the two graphs are tangent at (x , y ) , their normals at (x, y ) are parallel. But the vectors

grad f i x , y) , grad g ( x , y )

point in the respective normal directions (see p. 298), hence one is a multiple of the other:

grad/ (x, y ) = X grad g(x, y)

for some number X. (The argument presupposes that grad g 9 0 at the point in question.)

This geometric argument yields a practical rule for locating points on 9 (x, y ) = 0 where f ( x , y ) may have a maximum or minimum. Note that where the condition of tangency is satisfied, there may be a maximum, a minimum, or neither (Fig. 7.3).


To maximize or minimize a function f ( x , y ) subject to a constraint g(x, y) = 0 , solve the system of equations

( f x j y ) = * ( g , , gy), g ( x , y ) = 0

in the three unknowns x , y , X. Each resulting point (x , y) is a candidate.

The number X is called a Lagrange multiplier, or simply multiplier.

To apply this rule, three simultaneous equations

fx(x, y ) = \g x(x, y )

* f v ( x > y ) = 2/)

, g(x, y ) = o,

must be solved for three unknowns x, y, X.

EXAMPLE 7.1

Find the largest and smallest values of f (x, y ) = x + 2y on the circle x2 + y2 — 1.

Solution: Draw a figure (Fig. 7.4). As seen from the figure, / takes its maximum at a point in the first quadrant, and its minimum at a point in the


third quadrant. Heref (x, y ) = x + 2y, g(x, y ) = x2 + y2 - 1 ; grad/ = (1, 2 ), grad# = (2x, 2y ).

The conditions= Xfo,, &,), y) = 0

become

Thus(1, 2 ) = \ (2 x , 2y) , x2 + y2 = 1.

1

By the third equation,5 1 _

A* = - , X = ± - V 5 -

The value X = i \ / 5 yields1 2

X~VE’ y~ V s ’the value X = — § \ /5 yields

* = - V 5 ’ y = ~ v t ’

f (x, y) = = \ /5 ;


Answer: Largest \ /5 ; smallest —y/5.

EXAMPLE 7.2

Find the largest and smallest values of xy on the segment 2x + y = 2 , x > 0 , y > 0 .

Solution: Draw a graph (Fig. 7.5). Evidently the smallest value of xy is 0 , taken at either end point.

To find the largest value, use the multiplier technique with

f i x , y ) = xy, g(x, y ) = 2x + y - 2 .

The relevant system of equations is

(y , x) = A (2, 1)

*

2x + y - 2 = 0.


Thus x = X, y = 2X, and

2X + 2A - 2 = 0, X = ^ .

Therefore

(*,2/) = Q ’ 1 ) ’ / ( i , 1 ) = 2 ’

Answer; Largest - ; smallest 0.

EXAMPLE 7.3

Find the largest and smallest values of x2 + y2, where xA + yA = 1.

Solution: Graph the curve x* + yA = 1 and the level curves of f ( x> y) = x2 + y2. See Fig. 7.6. By drawing xA + yA = 1 accurately, you see that the graph is quite flat where it crosses the axes and most sharply curved

where it crosses the 45° lines y — ± x . I t is closest to the origin (jx2 + y2 is least) at (dbl, 0 ) and (0 , ± 1 ), and farthest where y = ± x .

The analysis confirms this. Use the multiplier technique with

f (x, y ) = x2 + y2, g(x , y) = x* + y* - 1.

The relevant equations are(2x, 2y) = X(4#3, 4y*)

#4 + yA = l*

Obvious solutions are

x = 0 , y = dbl, * = V = 0 , « = ± 1, x = ^*

Thus the points (0, ± 1 ) and (± 1 ,0 ) are candidates for the maximum or minimum. At each of these p o in ts /(x, y ) = 1.

Suppose both x ^ 0 and y ^ 0 . From

2x = 4X#3, 2 y — 4 X?/3follows

#2 = y2

Hence X = 1/ (2x2) > 0. From xA + y* = 1 follows

2X

GO + (i) = X2“i ’ x =V 2 ’

V 2

Consequently, the four points

( * ^ 2 ’ ± ^ )are candidates for the maximum or minimum. At each of these points f (x, y ) = x2 + y2 = 2/ V 2 = y /2 .

Answer: Largest \ / 2 ; smallest 1.

EXAMPLE 7.4

Maximize 2y — x on the curve y = sin x for 0 < x < 2ir.

Solution: Here f ( x , y ) = 2 y — x and g ( x , y ) — y — sin#. The equations are

( — 1, 2) = X( — cos Xj 1)

y = sin x ,

from which follow X = 2, cos x = x = 7r/3 or 57r /3 . The maximum must occur at

/ 7T a/ 3 \ / V jA\ 3 ’ “ 2 / ’ \ 3 ’ 2 / ’

or at one of the end points (0, 0) and (2t , 0). But

./7T \ / 3 \ 777 7T \ / 3 \ AT 57T

f in , 0) - 0, / (2 t , 0) = —2».

Of these numbers, \ / S — 7r/3 is the largest, being the only positive one.

Answer: \ / 3 — " .3

R e m a r k : The preceding example is illustrated in Fig. 7.7. From the figure, can you tell where the maximum occurs if x is restricted to the interval tt/2 < X < 2tt?

F ig. 7.7

EXERCISES

1. Find the maximum and minimum of x + y on the ellipse (z2/4) + (y2/ 9) = 1.2. Find the extreme values of x — y on the branch x > 0 of the hyperbola (x2/9) —

(2/74) = 1.3. Find the extreme values of x — y on the branch x > 0 of the hyperbola (z2/4) —

(y2/ 9) = 1. Explain.4. Find the extreme values of x — y on the branch x > 0 of the hyperbola x2 — y2 = 1.

Explain.5. Find the maximum and minimum of xy on the circle x2 + y2 = 1.6. Find the maximum and minimum of xy on the ellipse (x2/4 ) + (y2/ 9 ) = 1.7. Find the rectangle of perimeter 1 with shortest diagonal.

8 . Find the right triangle of perimeter 1 with greatest area.[Hint: Eliminate X.]

9. Find the right circular cone of fixed lateral area with maximum volume.10. Find the right circular cone of fixed total surface area with maximum volume.11. Find the right circular cylinder of fixed lateral area with maximum volume.12. Find the right circular cylinder of fixed total surface area with maximum volume.13. Let 0 0 and y > 0.14. (cont.) Let 0 0, y > 0. Show that

xp + ypJ ,p < f xq + yqyJ lq

15. Find the maximum and minimuni of x2y on the short arc of circle x2 + y2 = 1 between (^a/3, i ) and

8- FURTHER CONSTRAINT PROBLEMS [optional]

In space, the problem is to maximize (minimize) a function f ( x , y , z ) subject to a constraint g(xy y, z) = 0 . One seeks level surfaces of f (x, y, z) tangent to the surface g(x9y , z ) = 0. See Fig. 8.1. Each point of tangency is a candidate for a maximum or minimum. At a point of tangency, the normals to the two surfaces are parallel (Fig. 8 .2 ). But the two vectors

grad /(x ), grad <7 (x)

point in the respective normal directions, so the first must be a multiple of the

8. Further Constraint Problems 361

9 = 0 f = M

common direction of normals

F ig . 8.2

second, provided grad g(x) 0 . This observation leads to a practical method for locating possible maxima and minima (proof om itted).

To maximize (minimize) a function f ( x , y , z ) subject to a constraint y> z) = 0 , solve the system of equations

( fxyfvj f z ) = X (gx, gv, gz), ^ (x) = 0

in the four unknowns x, y , z , X. Each resulting point (x , y, z) is a candidate.

In applications, the usual precautions concerning the boundary must be observed.

EXAMPLE 8.1

Find the longest and shortest distance from the origin to the

ellipsoid x2 + — + ~ = 1. 9 4

Solution: Set /(# , y, z) equal to the square of the distance,

f ( x y y, z) = x2 + y2 + z2,

and set

Then

grad/ = (2x, 2y, 2z)


The required system of equations is

(2x, 2y, 2z) = X

If x 9 0, then by the first equation X = 1; consequently y = z = 0, so by the second equation x2 = 1, x = ± 1 . Thus two candidates are the points ( ± 1, 0, 0). Similarly, there are four other candidates, namely

R e m a r k : Compare this procedure with the previous solution of the same problem in Section 5, p. 347. The advantage of the present method will be crystal clear.

EXAMPLE 8.2

Find the volume of the largest rectangular solid with sides parallel to the coordinate axes that can be inscribed in the ellipsoid

Solution: One-eighth of the volume is

f ( x , y , z ) = xyz,

where a; > 0, y > 0, z > 0 . See Fig. 8.3. The constraint is g(x, y, z) = 0, where

X = 9: (0, ± 3 ,0 ) ; X = 4: (0, 0, ± 2 ).Since

/ ( ± 1 , 0, 0) = 1, /(0 , ± 3 , 0) = 9, /(0 , 0, ± 2 ) = 4,

the maximum distance is \ / 9 and the minimum distance is 1.

Answer: Longest 3; shortest 1.

v2 z2g(x, y, z) = x2 + - + - - 1.

Set g rad / = X grad g and g = 0:

8. Further Constraint Problems

that is,yz = 2\ x

2 y2 z2

xy = -Xz

Multiply the first two equations and cancel x y :

Likewise

a:2 = - X2, y2 = X2.

Substitute in the fourth equation:

1 1 1- x2 + - X2 + - X2 = 1, X2 = 3, 9 9 9

3 ’y2 = 3, z2 =

Hence

/(* , y, z)2 = a;2J/2z2 = ^ , / m« =


Answer: 82 y / 3 16 vs.

Two Constraints

Fig. 8.4. two constraints

Suppose the problem is to maximize (minimize) f (x, y, z) , where (x, y , z) is subject to two constraints, g ( x , y , z ) = 0 and h ( x , y , z ) = 0. Each constraint defines a surface, and these two surfaces in general have a curve of intersection (Fig. 8.4). A candidate for a maximum or minimum of /(x ) is a point x where a level surface of / is tangent to this curve of intersection (Fig. 8.5). The vector g rad /(x ) is normal to the level surface at x, hence normal to

the curve. But the vectors grad <7 (x) and grad h(x) determine the normal plane to the curve at x. Hence for some constants X and ju,

R e m a r k : The existence of such multipliers X and ju presupposes that grad g 0, grad h ^ 0, and that neither is a multiple of the other.

To maximize (minimize) a function f (x, y, z) subject to two constraints g(x, y, z) = 0 and h(x, y, z) = 0 , solve the system of five equations

( fx) fyy f z ) = X (gX) gy, gz ) H (hx, ky, kz)

g(x) = 0 , h(x) = 0

in five unknowns x, y, z, X, ju. Each resulting point (x , y, z) is a candidate.

EXAMPLE 8.3

Find the maximum and minimum of f (x, y, z) = x + 2y + z on the ellipse x2 + y2 = 1, y + z = 1.

grad /(x ) = X grad g(x) + ju grad h(x) .

F i g . 8.6

Solution: Here

g(x , y, z) = x2 + y2 - 1, h(x, y, z) = y + z - 1 ;

Fig. 8.6 shows the curve of intersection of the surfaces g(x) = 0 and h(x) = 0,

a plane section of a right circular cylinder. The equations to be solved are

1(1, 2, 1) = \ ( 2 x , 2 y , 0 ) + „ (0 , 1, 1 )

X 2 + j/2 = l , y + Z = 1 .

1 1

Hence

1 = 2\x , 2 = 2 \y + n, 1 = n; x =2X ’ * 2X ’

The solution X = \ /2 /2 leads to the point

( y /2 y /2 _ y / 2 \1 \ 2 ’ 2 ’ 2 j ’

and the solution X = — y /2 /2 leads to the point

These are the only candidates for maxima or minima. But / ( x i ) = 1 + y /2 , / (X 2) = 1 - y /2 .

Answer : Maximum 1 + \ / 2 ; minimum 1 — V 2 -

Quadratic Forms

In Chapter 7, we postponed part of the proof of Theorem 8.5 that a 3 X 3 symmetric matrix has three real characteristic roots. We fill this gap now using Lagrange multipliers.

Let A be a 3 X 3 symmetric matrix. We consider the problem of finding the minimum of / (x) = x^4x' on the surface of the unit sphere. We set g (x) = xx' — 1, so the problem is to minimize/(x) subject to g(x) = 0 .

Since the (surface of the) unit sphere is a closed bounded set, the continuous function /(x ) has a minimum value at some point Xo. By Lagrange multiplier theory, g(*o) = 0 , that is, XoXo = 1, and

grad/(x0) = X grad g (xo), for some X. A direct computation shows that

grad/(x) = 2Ax' and grad <7 (x) = 2x', hence the condition is

Axo = Xxo.Thus Xo is a characteristic vector of A and X is the corresponding characteristic root. Furthermore,

f ( x 0) = Xo Ax{) = Xx0Xq = X,

so X actually is the minimum value. I t follows that X is the smallest real characteristic root of A.

Similarly, /(x ) has a maximum value v on the unit sphere; this value is the largest real characteristic root of A .

I t may be that X = v. If so, then /(x ) has the constant value X on the sphere. H ence/(x) = Xxx' and it follows that A = X/. Thus the characteristic roots of A are X, X, X, all real.

If X < v, then A has two distinct real characteristic roots. I t follows that the third characteristic root must also be real. For the characteristic polynomial of A is a cubic, and if a cubic has two real zeros, the third zero is also real, since complex zeros of polynomials with real coefficients come in conjugate pairs.

A Second Derivative Test

Suppose we wish to minimize f ( x , y , z ) subject to the constraint d(x, y> z ) = 0- Suppose we have found a solution (#, y, 2, X) to the system of equations

g rad / = X grad g

0 = 0 .

I t would be nice to have a test for a minimum analogous to the second derivative test in Section 6 . There is such a test, but its proof is beyond the scope of this course.

Theorem Let x0, X be a solution of

g rad / = grad g

where Xo is interior to the common domain of / and g, and (grad g) (xo) ^ 0. If the matrix

( Hf - \ H 0)

is positive definite, then there is a 8 > 0 such that / ( x ) > / ( x 0) for all x satisfying g(x) = 0 and 0 < |x — x0| < 8.

EXERCISES

1. Assume a , b , c > 0. Find the volume of the largest rectangular solid (with sides parallel to the coordinate planes) inscribed in the ellipsoid

_j_ t . - l t = 1 a2 ft2 c2

2. Find the triangle of largest area with fixed perimeter.[Hint: Use Heron’s formula A2 = s(s — x) (s — y ) ( s — z), where x, y, z are the sides and s is the semiperimeter.]

3. Find the maximum and minimum of the function x + 2y + 3z on the sphere x2 + y2 + z2 = 1.

4. Find the rectangular solid of fixed surface area with maximum volume.5. Find the rectangular solid of fixed total edge length with maximum surface area.6. Find the rectangular solid of fixed total edge length with maximum volume.7. Maximize xyz onx + y z = 1.

8. (cont.) Conclude that y/xyz < x for x > 0, y > 0, and z > 0.o9. Let 0 0, y > 0, and z > 0.10. (cont.) Let 0 0, y > 0, z > 0. Show that

/ xp + yp + zp\ 1/p ^ / x q + y q + zq

\ 3 ) ~ \ 311. Show that xy + yz + zx < x2 + y2 + z2 for x > 0, y > 0, and 2 > 0.12. Find the minimum of x2 + y2 + z2 on the line x + y + z = 1, x + 2y + 3z = 1.13. Find the maximum and minimum volumes of a rectangular solid whose total edge

length is 24 ft and whose surface area is 22 ft2.14. Find the maximum and minimum of x + y + z on the first octant portion of the

curve xyz — 1, x2 + y2 + z2 = .[Hint: Show that x, y, z are roots of the same quadratic equation; conclude that two of them are equal.]

15* Let A and B be 3 X 3 matrices with B positive definite. Show that each Lagrange multiplier X in the problem of maximizing xAx' on the ellipsoid xBx' = 1 is a characteristic root of AB~~l. Assume A and B are symmetric.

16. Let f ( x , y , z ) = x2 + y2 + z — z2 and g(x,y, z) = z . Prove that the minimum of / on g = 0 is 0 = f(0, 0, 0 ).

17. (cont.) Show however that the second derivative test at the end of the section is inconclusive for this example.

10. Double Integrals

1. INTRODUCTION

A geometric motivation for the definite integral (also called simple integral)

f f ( x ) dx J a

is the problem of finding the area under a curve y = /(# ). Suppose we consider instead the problem of finding the volume under a surface z = f {x, y ) y where f { x , y ) is a continuous (positive) function defined on a rectangular domain

a < x < b, c < x < d.

This leads us to a new kind of integral called a double integral. Now the theory behind the double integral (and the triple integral of the next chapter) is rather technical and lengthy, so we shall postpone it until Chapter 12. Meanwhile we shall proceed intuitively, taking a lot of things for granted. Our present aim is to develop a working knowledge of how to set up, evaluate, and apply double and triple integrals (called collectively multiple integrals).

370 10. DOUBLE INTEGRALS

Volume

Consider the prismatic solid (Fig. 1.1) bounded above by the surface s = f (x, y ) and below by the rectangular domain a < x < 6, c < y < d.

Intuitively it seems obvious that the solid has a well-defined volume. For we can carve the solid out of a homogeneous material like clay, plaster, or steel and then weigh it or submerge it in a tank of water to find its volume. This is fine, but how do we compute the volume? We shall use a procedure analogous to rectangular approximations for plane areas.

We shall approximate the given solid by a large number of long, thin rectangular solids, and add up their volumes to approximate the desired volume.

Partition the base of the solid into mn equal small rectangles by drawing segments parallel to the x- and y -axes that divide [a, 6] into m equal parts and [c, d~\ into n equal parts (see Fig. 1.2).

d = y n

Vi 2/i—1

c = y o

a = Xo Xi-i Xi x m = b

Fig. 1.2 division of the base into mn equal rectangles

Thus

a = xo < x\ < < • • • < xm = b, Xi — Xi-i =b — a

c = yo < yi < y 2 < • • • < y n = d, y j - y j- i =d — c

n

The part of the solid above the i , j -th small rectangle is approximately a thin rectangular column of height f (xi , y j ), where (xi} yj ) is the midpoint of the base (Fig. 1.3).

1. Introduction 371

M M -The volume of the column is

f ( x i , y j )

Add to approximate the total volume:

i=i y=i

Now partition the base into more and more rectangles by letting and n both increase to infinity, so (b — a ) / m and (d — c ) / n both decrease to zero. Then under reasonable conditions the double sum converges to a lim it:

m n

m • -> oo, n ■z=i j =i

One reasonable condition is that / be continuous, as will be shown in Chapter12. The double sum expression suggests the double integral notation

J J f (x, y ) dx dy = J J f (x, y ) dx dy,a<x<bc^y^d

where D denotes the rectangular domain of /. Thus the full definition of the double integral is

where

D n —><» i = 1 , • • - ,mj =1, • • -,n

The formula for #t- simply says that Xi is the midpoint (average) of

Xi-i = a + (i — 1) ( ----- - J and = a + i\ m /

Likewise yj = HVj- i + 2/i)-The definition does not require that / be positive. If / takes both positive

and negative values, the double integral represents an algebraic volume rather than a geometric volume. The volume between the surface z = f (x, y ) and the x , y-plane counts positively where / > 0 and negatively where / < 0 .

Assuming the double integral exists, the sum formulas

2 2 kf (*i , Vi) = k ^ £ / ( * < , J/y),

[ / ( * < » 2/y) + 9 (*<> Vi) ] = ^ ^ /(*•'. yi) + Y, Y , 9 ^ ’

imply the relations

(— )■ \ m /

f f kf{x, y ) d x d y - k J f t i f , y ) dxdy ,

J J I f i x , y ) + g{x,y)~\ d x d y = J J f { x , y ) d x d y + J J g { x , y ) d x d y .D D D

Once the double integral has been defined, the immediate problem is how to evaluate it. Sections 2 and 3 provide a solution to this problem.

2. SPECIAL CASES

In certain cases where f (x, y ) has a particularly simple form, it is easy to evaluate the double integral

J I f i x , y ) dx dy.D

As before, D denotes the rectangle a < x < b y c < y < d.

2. Special Cases 373

Suppose first that f ( x , y ) = h , a constant. Then the double integral represents the volume of a rectangular solid of height h. This volume is h times the area of the base:

/ / hd x dy = h X (area D) = h(b — a) (d — c).

Suppose next that f (x, y) is a function of x alone,

f i x , y ) = gix) .

Then each cross-section of the solid by a plane parallel to the 2, #-plane is an identical plane region (Fig. 2. la, b ). The volume is the area of this region times the y-length of the rectangle:

J J g ix) dx dy = ( J j g i x) d x j id - c).

(b) area of the section

F ig . 2.1

Similarly if f i x , y ) = h{y) ,

J J h{y) d x d y = ( J h{y ) dy \ (b - a).D °

These formulas are special cases of the following one:

J J gi x) h i y ) d x d y = ( J g{x) dx X f hi y ) d y )

This formula gives a double integral as the product of two ordinary (single) integrals. I t applies to such functions as

x5y7, ex~v, ex cos y , etc.


Fig. 2.2 cross-sectional slab

Let us try to justify the formula intuitively. The volume of a solid (the left-hand side) is evaluated by slicing the solid into slabs, then adding up (Fig. 2.2a). Fix y , then slice the solid by the planes parallel to the z , rc-plane at y and at y + dy. The result is a slab of thickness dy. The projection (Fig. 2 .2b) of this slab on the z , #-plane is the region in that plane under the curve z = z(x) = h(y)g(x) . (Remember y is fixed!) Hence its area is

Lb r b r b

z ( x ) d x = / h(y) g(x) dx = h(y) / g ( x ) dx .J a J a

The volume of the slab is its area times its thickness:

dV = ( h ( y ) J g(x) d x ) dy = (^J g(x) d x ) h ( y ) dy.

On the right-hand side, the first factor is a constant, therefore adding the slabs yields

g 0*0 d x ) h(y) dy = ( ^ j gi x) dx)(^j^ h( y ) d y ) .

This is the desired formula.

EXAMPLE 2.1

Find J J x4y6 dx dy.0^x,y£l

Solution:

J J x y d x d y = (J xl dx)(^J^ i f d y ) = g-~ .0 £x, y 1

2. Special Cases 375

Answer:35

EXAMPLE 2.2

Find I I0 < x £ l

ir /2 <>y< tt

cos y dxdy.

ThenSolution: Let D denote the rectangle, 0 < x < 1 and t / 2 < y < w.

I f = (/ eda:X/ cosydy) = (eI| )(sin| ) = (e ~ 1)(~1)-

Q u e s t i o n : The answer is negative. Why?Answer: 1 —6.

EXAMPLE 2.3

Find / / ex~~y dx dy.0 < x £ l -2 <,y^

Solution: Since ex~v = exe~v,

J J ex~v dx dy = ^ J ex e~v dy'jD °

= ( e x | ^ = (e - l ) ( e 2 - e ).

Answer: e(e — l ) 2.

EXAMPLE 2.4

Find I I (x2y — 3xy2) dx dy.l£x£2

- 1 < y < \

xy2 dx dy.

Solution: Use the linear property of the double integral:

J J ( x 2y — 3x y 2 ) d x d y = J J x 2y d x d y — 3 J JD D D

Evaluate these two integrals separately:

J J x 2y d x d y = ^ J x 2 d x ^ J y d y ^ j = 0;D 1

J J xy2 dx dy = ( J x d x ) ( ^ J y2 dy) = = 1.

Answer: — 3.

EXERCISES

Evaluate:

1. II (3x — 1) dx dy;■//

> • / /

‘ • / f

/ /

•//. J J (x17 - J/17)

•/// / / /I I (x ~ y ^ dxdy’

14. J J ( x — y ) 7bd x d y ;

15 i f (x ~^y ^ dxdy ’’

2 J J X

3. x2y2dxdy\

4. x2y2 dx dy ;

5. f f x3y3dxdy;

y ) dxdy ;

7. II (x2— y2)dxdy;

7) efcdt/;

ey2) dx dy;

10. ^ cos x cos ydxdy;

11. J J (x2+ y 2)dxdy;

12. [ f ( x — y)2dxdy;

13

- 1 < * < 2 , 0 < y < 5

— l < x < 1, 0 < y < In 2

- 1 < s < 0, 0 < i / < 1

0 < x , y < 1

0 < x, y < 1

-1 < 2/ < 1

■ 1 < x, y < 1

-1 < V < 1

■ l < y < i

-1 < 2/ < 1

3. Iterated Integrals 377

3 M ' ' • / M

/ /

16. 11 p dx dy; 1 < x < 2, 1 < y < 4

17. j / —x —2 dx d y ; 0 < x < 2, 0 < ?/< 1

18. J J xy In x dxdy; 1 < x < 4, — 1 < 2/ < 2

19. / / x ln(x?/) dx cfa/; 2 < x < 3, 1 < 2/ ^ 2

> • / /20. e*-H/ cos 2^ dy; 0 < x < 7r, 1 < y < 2.

3. ITERATED INTEGRALS

The formula in the last section for integrating products g ( x ) h ( y ) is a useful one, but it is inadequate for many functions, such as f ( x , y ) = 1/ (x + y ) and/(x, y ) = y cos (#2/)- In this section we derive the most general method for evaluating double integrals, the method of iterated integration. This method includes the previous rule for products as a special case.

The problem is to compute

- f fv = / / f (x, y ) dx dy.a <b c < y <d

Consider the integral as a volume to be found by slicing.Fix a value of y and slice the solid by the corresponding plane parallel

to the Xj 2-plane (Fig. 3 .1).

F ig . 3.1

The resulting cross-section has area A (y). Denote the volume to the left of this slice by V( y ) . (Thus V(c) = 0 , V( d ) = the desired volume.) The fundamental fact needed here is that

d V A ( \y ~ = A ( y ) .dy

Intuitively this is fairly clear. The derivative d V / d y is the limit as h --------» 0of

V ( y + h) - V j y ) h

For h very small, the numerator is the volume of a slab of width h and cross- sectional area approximately A ( y ) . Hence the quotient is approximatelyA (y).

Now integrate d V / d y = A (y) to find V :

- rJ cA (y ) dy.

But A (y), the area of the cross-section (Fig. 3.2), can be expressed as a simple integral. Indeed, A ( y ) is the area under the curve z = f ( x , y ) from x = a to x = b. (Remember y is fixed in this process.) Thus

A ( y ) = f /(« , y) dx ,J a

where y is treated as

Fig. 3.2

The final result is the following pair of formulas.

Iteration Formulas

I f / ( * , „ ) * # . / ( j f ( x , y ) d x j d y

= fa ^ ^ X’ V ^ dX'

a <b c <2/

a constant in computing the integral.

(The second form is obtained by reversing the roles of x and y in the argument. )

Our intuitive justification of the formulas will be replaced by an accurate proof in Chapter 12 for continuous, and even somewhat more general, integrands.

Rem ark: The expressions on the right are iterated integrals, i.e., repetitions of simple integrals. They are often written this w ay:

EXAMPLE 3.1

l < <£2

Solution:

Now for fixed x ,

f ; = In (a; + y )x + y

ln (2 + x) — ln (l + x ).y = 1

Hence

[In (2 + x) — ln (l + # )] dx.

But

J In u du = u In u — u + C,

therefore.

i= [ ( 2 + x) In (2 + x) - (2 + x) - (1 + x) ln (l + x) + (1 + # )]

o= 3 In 3 - 2 In 2 - 2 In 2 = 3 In 3 - 4 In 2.

i 27Answer: In — 16

An important feature of the Iteration Formulas above is that the iteration may be done in either order. Sometimes the computation is difficult in one order but relatively easy in the opposite order.

EXAMPLE 3.2

I IFind / / y cos (xy) dx dy.O^x^l0£y<>T

Solution: Here is one set-up:

i i - a i ; y cos (xy) d y^ dx.

The inner integral,

y cos (xy) dy,Jo

while not impossible to integrate (by parts), is tricky. The alternate procedure is iteration in the other order:

a - i : a : y cos (xy) dy.

Since y is constant in the inner integration, this can be rewritten as

I f ■ f » ( f cos (xy) dx^ dy.Now

f 1 / x 7 i • / \ I sin y/ cos (xy) dx = - sin(xy) = ------.Jo y

HenceI x—0

f [ = r y ™ y d y - f * sin y dy = 2.J J Jo y Jo

Alternate writing of the solution:

f f y cos (xy) d x d y = J y dy f cos(xy) dxD

. f \ ( * « | V . -2 . 'y0 \ y I o/

Answer: 2.

EXERCISES

Evaluate:

- f f ^ - r « • * > •

2. J J ^ d x d y ; 0 < x < 1, 1 < y < 5

> ' / /~2d x d y ; — 1 < a: < 1, 1 < 2/ < 3

4 . / / . - * < * - . < » , „ < < >

5. i/2sin (a??/) d x d y ; 0 < x < 27r, 0 < y < 1

6. (1 — 2x) sin(i/2) d x d y ; 0 < x , y < 1

7. y j sin (x/y) dx dy ; —7r/2 < £ < 7r/2, 1 < 1/ < 2

s. / y d - x + 2y)2 dxdy; 3 < a; < 4, 1 < 1/ < 2* 7 / ° -

> • / /9. J J (l + x + y)(3 + x — y ) dxdy ; 2 < x , y < 3

10. y^* sin (a; + 2/) dx dy ; 0 < a;, y < 7r/2

11. / / < * + i/)w dx dy; 0 < a; < 1, 1 < 2/ < 2

12. J J (x — ?/)n dx dy ; 0 < a?, y.< 1.

13. Suppose/(—a;, — y) = —/(as, 2/)- Prove J J fix, y) dx dy = 0 ; — 1 < *, y < 1.

14. Suppose / (a;, — y ) = —f(x, y) . Prove J J f(x, y) dx dy = 0 ; — 1 < x, y < 1.

15. Find the constant A that best approximates f(x, y) on the square 0 < x, y < 1 in the least squares sense. In other words, minimize

I f [/(«> 2/) “ dx dy; 0 < x, y < 1.

16. (cont.) Find'the least squares linear approximation

A + Bx + Cy

to f(x, y) = xy on the square 0 < x, y < 1.17. (cont.) Show that the coefficients of the least squares linear approximation

A + Bx + Cy to / (x, y) on the square 0 < x, y < 1 satisfy

A + \ B + l c = J J f dx dy

+ + ~ J J xf dx dy

\ a + \ b + 1- C - J J yfdxdy.

4. APPLICATIONS

Ariass and Density

Suppose a sheet of non-homogeneous material covers the rectangle a < x < b and c < y < d. See Fig. 4.1. At each point (x, y) , let p(x, y) denote the density of the material, i.e., the mass per unit area. (Dimen- sionally, planar density is mass divided by length squared. Common units are gm/cm2 and lb /ft2.)

The mass of a small rectangular portion of the sheet (Fig. 4.2) isd M p(x, y) dx dy.

Therefore the total mass of the sheet is

y

c -

a b x

Fig. 4.1 noil-homogeneous sheet

a <x <b c —d

The density (lb/ft2) at each point of a one-foot square of plastic is the product of the four distances of the point from the sides of the square. Find the total mass.

EXAMPLE 4.1

4. Applications 383

Solution: Take the square in the position 0 < x, y < 1. Then

p(x, y ) = a;(l - x ) y ( 1 - y) ,

M = J J p(x, y ) d x d y

» ( * - » ) * ) “ H - s -

Answer: — lb. 36

Moment and Center of Gravity

Suppose gravity (perpendicular to the plane of the figure) acts on the rectangular sheet of Fig. 4.1. The sheet is to be suspended by a single point so that it balances parallel to the floor. This point of balance is the center of gravity of the sheet and is denoted x = (x,y).

The center of gravity x is found in three steps:(1) Find the mass M .(2) Find the moment of the sheet with respect to the origin. This is

the vector

Here D denotes the region the sheet occupies, a < x < b and c < y < dy and p = p(x, y ) is the density.

(3) Divide the moment by the mass to obtain the center of gravity:


1

This formula will be derived after two examples.

EXAMPLE 4.2

Find the center of gravity of a homogeneous rectangular sheet.

Solution: “Homogeneous” means the density p is constant. Take the sheet in the position 0 < x < a and 0 < y < b. The mass is M = pab, and the moment is

This is the midpoint (intersection of the diagonals) of the rectangle. (Of course the rectangle balances on its midpoint; no one needs calculus for this, but it is reassuring that the analytic method gives the right answer.)

The center of gravity is

Answer: The midpoint of the rectangle.

EXAMPLE 4.3

A rectangular sheet over the region 1 < x < 2 and 1 < y < 3 has density p ( x , y ) ~ xy. Find its center of gravity.

Solution: The mass is

The moment is

4. Applications 385

m = J J xy x ^ = i ^ J J x2y J J xy2 dx= ( J * x 2 dx f * y dy , J * X d x J ' y> d y )

( 1 3 26\ 1 .- ( s ' 4 - 2 - j ) - 3 <28' 3 9 ) -

Therefore

■ ( I t )Note that x is inside the rectangle, a little northeast of center. Could you have predicted this?

Answer: x\ 9 ’ 6 /

The formula for center of gravity is derived by balancing the rectangular sheet on various knife edges.

Suppose a knife edge passes through x and the sheet balances (Fig. 4.3). Divide the sheet into many small rectangles. The turning moments of these pieces about the knife edge must add up to zero.

Let n be a unit vector in the plane of the rectangle perpendicular to the knife edge. A small rectangle with sides dx and dy located at x has (signed) distance (x — x)*n from the knife edge and has mass p d x d y . Hence its turning moment is

p(x — x)*n dx dy.

The sum of all such turning moments must be zero:

//■p(x — x)*n dx dy = 0.

Since x and n are constant, this relation may be written

n • J J p x d x d y = (n*x) J J p d x d y ,

orri* m = Mn* x.

This equation of balance is true for each choice of the knife edge (each choice of the unit vector n ). Hence,

n* (m — Mx) = 0

for each unit vector n. This means the component of m — Mx in each direction is zero. Therefore,

m — Mx = 0,

m = Mx.

EXERCISESFind the volume under the surface and over the portion of the x, f/-plane indicated. Draw a figure in each case:

1. z = 2 — (x2 + y2); —I < x , y < 12. z = 1 — xy\ 0 < x, y < 13. z = x2 fy2; . 0 < z < 2 , 0 < 2/ < l4. z = sin x sin y; 0 < x, y < tt5. z = x2y + y2x; l ^ z ; < 2, 2 < y < 36. 2 = (1 + x*)y2; — 1 < x, y < 1.

A sheet of non-homogeneous material of density p gm/cm2 covers the indicated rectangle. Find the mass of the sheet, assuming lengths are measured in centimeters:7. p = 3(1 + x) (1 + y); 0 < x, y < 18. p = 1 — 0.2 xy; 0 < x < 1, 1 < y < 1.59. p = 3 + O.lz; 2 < a; < 3, — 1 < 2/ < 1

10. p = 4e*+* - 2; 0 < * < 1, 0 < y < 0.5.Find the center of gravity of each rectangular sheet, density as given:11. p = (1 — x) ( l — y) + 1; 0 < x, y < 112. p = sin x; 0 < x < 7r, 0 < y < 1

5. General Domains 387

13. p = sinz(l — sin y) ;14. p = 10 -15. p = 1 + z2 + i/2;16. p = 2 + xV ;

7 r/2 < X, y < 7T

0 < x , y < 1 - 1 < X < 1, 1 < 2/ < 4— 1 < x < 1, 0 < 2/ < l .

5. GENERAL DOMAINS

Suppose we want the volume of a solid (Fig. 5.1) bounded on top by a surface z = f (x, y) defined over a non-rectangular domain D. Here the bottom of the solid is D itself, and the side of the solid is the cylindrical wall with generator parallel to the 2-axis and base the boundary of D. The solid is the set

{(*, V, z) | (x, y ) e D and 0 < 2 < /(* , y ) \ .

LZ

Fig. 5.1 solid under a surface

y

We shall suppose the volume is

V - J J /(« , y ) dx dy,D

where this double integral over D has properties consistent with our intuitive ideas about volume. Let us consider some examples to see how we can compute such volumes and what properties the double integral ought to have.

EXAMPLE 5.1

Find the volume under the surface z = e~(x+y), and over the domain of the y -plane bounded by the x-axis, the line y — x, and the lines x = f and x = 1.


Solution: Draw figures of the domain (Fig. 5.2a) and the solid (Fig. J 5.2b). The problem is solved by the slicing method. Slice the solid into slabs by planes perpendicular to the #-axis. Let the area of the slab at x be denoted by A (x). See Fig. 5.3a. The volume of the slab is

so the total volume isd V = A (x ) dx,

V = I A ( x ) dx.J 1/2

Now carefully draw the cross-section (Fig. 5.3b). Its base has length x, and it is bounded above by the curve z = e~xe~y. Note that x is constant; in this plane section 2 is a function of y alone. This is the absolute crux of the matter I

(a) (b)

Fig. 5.3

The area of the cross-section at x is

A (x) = I e~xe~y dy = e~x j e~y dy — —e~x(e~y)Jo Jo

y= x

y—0


Thus

V = [ (e~x — e~2x) dx = ( - e~2x — e~x\ = - e~2 — — e~l + e~112.J 1/2 \ 2 / 1 1/2 2 2

N ote: The solution can be set up as follows:

F = / ( / e ~ i x + v ) d y ) d x '

In the inner integral the variable of integration is y , while x is treated like a constant, both in the integrand and in the upper limit. But once the inner integral is completely evaluated, x becomes the variable of integration for the outer integral.

1Answer: ~e~2(l — 3e + 2e3/2).

EXAMPLE 5.2

Find the volume under the surface z = 1 — x2 — y2, lying over the square with vertices ( ± 1, 0 ) and (0 , ± 1).

Solution: First draw the square (Fig. 5.4a). Observe that by symmetry, it suffices to find the volume over the triangular portion in the first quadrant, and then to quadruple it.

(a) F ig. 5.4 (b)


Draw the corresponding portion of the solid (Fig. 5.4b). Slice by planes perpendicular to the x-axis. For each x, the plane cuts the solid in a cross- section (Fig. 5.5) whose area is

A ( x )- rJo

(1 - x2 - y2) dy.

Fig. 5.5

The volume of the solid is

V = 4 [ A (x) dx= 4 J A (x) dx = 4 J (1 — x2 — y2) dy^j dx

- 4 f

= 4 Jo [ _ ~ ~~ ~ ^ _

= 4 J | \ — x — x2 + a;3 — ^ (1 — a;)3j dx

T 1 1 1 1 1 4 4= 4 1 -------------1----------- ---- 4 — = - .

L 2 3 4 12j 12 3

dx

Answer:

Find the volume under the plane z =* 1 + x + y> and over the domain bounded by the curves x = J, x = 1, y = x2, y = 2x2.

EXAMPLE 5.3

Solution: The domain and the solid are drawn in Fig. 5.6. The cross- section by a plane perpendicular to the x-axis at x is a trapezoid (Fig. 5.7).

5. General Domains

z = 1 + x + y

Note the range of y } namely .t2 < y < 2x2. Thus

V = f A ( x ) dx ,J 1/2

where

" ( *i

(1 + x + y ) dy = ( y + xy + - y2A (x) = IJ xi

= (2x2 + 2x3 + 2x4) — ^#2 + x3 + - x =

y=2x2

X 2 + X s +

Finally,

to i c

o

Answer:4960

Iteration

With these examples behind us, we are prepared for the statement and solution of the double integration problem. Suppose a domain D in the x , y - plane is bounded by lines x = a and x = 6, and by two curves y = g ( x ) and y = f ( x ) . Assume a < b and g ( x ) < f ( x ) for each x . See Fig. 5.8. Suppose

a surface z = H(x, y) is given, defined over D. See Fig. 5.9. The volume of the solid column over D and under the surface is then

V = / / « < « , y) dxdy .

(As is usual with integrals, the portion where H < 0 is counted negative.)To evaluate the double integral, consider a slab parallel to the y, 2-plane

at x. Its face area is

/•/(*>A ( x ) = / H(x, y) dy.

* Oix)

See Fig. 5.10. In this integration, x is constant. Notice that y, the variable of integration, disappears when the definite integral A (x) is evaluated.

Conclusion: the volume integral is

r-f)D

f H ( x , y ) d x d y = J A ( x ) d x = j COf f (x) \f H ( x , y ) d y ) d x .0(x) /

This is called the iteration formula. In Chapter 12, we shall prove that it holds for any continuous function, not necessarily positive.

EXAMPLE 5.4

Find the volume under the surface z = xy, and over the domain D bounded by y = x and y = x2.

Solution: The line and parabola intersect at (0 , 0 ) and (1, 1). See Fig. 5.11. For each value of x, the range of y is

x2 < y < x.

Hence

V = J J xy d x d y = J J xy d y ) dx = J Q xy21 ^ dxD O x 0 y x

J o 2 2 \4 6/ 24

Alternate Solution: The domain D may be thought of as bounded by the curves x = y (below) and x = \ / y (above), where 0 < y < 1. See Fig. 5.12. For each y y the range of x is y < x < \ / y . Therefore the set-up of the iteration is

- i h y' - y,)dy- l ( l - i ) ~ k -

The iteration method does not apply directly to every example. The boundary of D may be too complicated, in which case you must break the domain into several smaller regions, and deal with each as a separate problem. In Fig. 5.13 two examples are shown. The set-up for the domain in Fig. 5.13a is

I fd 2

J J H ( x , y ) d x d y = J J H { x , y ) d x d y + 11 H (x, y) dx dyD Di

/ l(z) \ f b / f

H ( x , y ) d y j d x + ( / H ( x , y ) d y/ J c g(x)


Fig. 5.13 cut up the domain

The set-up for the domain in Fig. 5.13b is

J J H ( x , y ) d x d y

— J J H ( x , y ) d x d y + J J H(x, y ) d x d y + J J H ( x , y ) d x d yDi D2 Da

m/l(*) \ fc / f h ( x ) \H ( x , y ) d y ) d x + I / H (x, y ) d y ) dx

2(x) / J e \ J qi(x) /e \ J m(x)

+ H r H ( x , y ) d y ) d x .J c 02(x) /

EXAMPLE 5.5

Without evaluating, set up the calculation for

dx dyII x2 -I- y2

over the region indicated in Fig. 5.14.

Solution: Draw a vertical line from (1,0) to (1, 2 ). Call the left-hand part Di and the right-hand part D2. See Fig. 5.15. Then

C f d x d y = f f d x d y C fJ J x 2 + y 2 J J x 2 + y 2 J JD Di D2

d x d y■2 _|_ yi _r J J xi _|_ yi

D2


Area

I t is often convenient to compute areas by double integrals, since*

J J I d x d y = area(D).D

EXAMPLE 5.6

Find the total area A bounded by y = x2 and y = x \

A - J f d , d y - £ ( £ < * ) * - £ | . ± .


Solution: Draw the domain (Fig. 5.16). Then

Alternate Solution:

ri / r< n

Answer: A = — .15

EXERCISES

Find the volume under the surface z = f (x, y), and over the indicated domain of the x, y-plane:

1 . * = l ; Fig. 5.172. z = y; Fig. 5.173. 2 = x; Fig. 5.174. z = l + x + y, Fig. 5.17

5. z = x2; Fig. 5.186. z = xy; Fig. 5.187. 2 = y2; Fig. 5.188. 2 = (x - y)2; Fig. 5.18

(Exs. 1-4) F ig . 5.18 (Exs. 5-8)F ig . 5.17


9. 2 = 1; Fig. 5.1910. 2 = 1 + x; Fig. 5.1911. z = y, Fig. 5.19

1 2 . 2= 1 ; Fig. 5.2013. 2 = 1 + x; Fig. 5.2014. 2 = x2y; Fig. 5.2015. 2 = y3; Fig. 5.2016. z = x2y2; Fig. 5.20

17 . 2= 1 ; Fig. 5.21 18. 2 = 1 + *; Fig. 5.21.

F ig. 5.21 (Exs. 17-18)

19. Describe each of the domains in Figs. 5.17-5.21 by inequalities between x and y.

Compute the double integral over the domain whose boundary curves are indicated; in each case draw a figure.


20. J J xexy dx dy; x = 1, x = 3, xy — 1, xy = 2

21. J J x2y d xdy ; y = 0, x = 0, x = (y •— l )2

22. J J (x3 + i/3) dx dy; x2 + y2 = 1

23. J J (x + y)2 dxdy; x + y = 0, y = x2 + x

24. ^ (1 + xy) dx dy; y = 0, y = x, y = 1 — x

25. / / y2dx dy; y = =tx, y = |x + 3.

26. Find J J (1 + xy2) dx dy over the domain bounded by the parabola x = — y2 andD

the segments from (2, 0 ) to (—1, ± 1).

6. POLAR COORDINATES

Suppose a domain (Fig. 6 .1a) in the x , y-plane is described in polar coordinates b y r0 < r < ri and do < 6 < 0i. Suppose a surface (Fig. 6.1b) is given by a function z = /(r , 6) over this domain. What is the volume of the solid between the surface and the domain?

F ig. 6.1

Split the interval r0 < r < ri into small pieces. Do the same for 60 < 6 < di. The result is a decomposition of the plane domain into many small, almost rectangular, regions (Fig. 6 .2 ). Each has dimensions dr and rdO, hence area dA = r dr dd. The portion of the solid lying over this elementary “rectangle” has height z = /(r , 0), hence its volume is d V = /(r , 6) r dr dd. See Fig. 6.3.

i y

Fig. 6.2 elementary polar area

The total volume is

1Fig. 6.3 “rectangular” column

v = f j K r’ e)rdrde = i:u: f i r , 6 ) r d r j d dD *° r°

= J ' l r ( J ° ' f ( r , 6) do) dr.

EXAMPLE 6.1

Find the volume under the cone z = r, and over the domain0 < r < a and 0 < $ < tt/2 .

Solution: The solid in question is shown in Fig. 6.4. Its volume is

*12 m a3 7r3 ------- ,

3 2

Answer: V = —6

Mass and Center of Gravity in Polar Coordinates

The mass and the center of gravity of a non-homogeneous sheet over the domain in Fig. 6.2 can be computed by double integrals. Since the element of area is

dA = r dr dB,

the elementary mass is

d M = p(r, 0 )r dr dd.

Note that p must be expressed in polar coordinates. Thus

To find the moment, express x in polar coordinates:

x = (x> y) = (r cos 0, t sin 6).

Then x = m/Af, where

m = JJ x d M = JJ x p r d r d d — J J (r cos 6, r sin 0)pr dr <D D r o^r ^n

eo^e^ei

Find the center of gravity of a homogeneous sheet in the shape of the semicircle 0 < r < a and 0 < 6 < r.

EXAMPLE 6.2

Solution: Here p is constant. The mass is

M — - ira2p.2

The moment is

m = 11 xpr dr dd= JJ xprdrdd = JJ (r cos d, r sin d)pr dr dd = p JJ r2 cos 6 dr dd, JJ r2 sin 6 dr dd'j = p ^J r2 dr J cos 6 dd, J r2 dr J sin d dd'j = p 0, a j .

Therefore

1 2x = — m = — -

M ttcl2 ( ° -1 “*) ■

Answer: x (*£)■EXAMPLE 6.3

Find the center of gravity of a sheet with density p = r covering the quarter circle 0 < r < a and 0 < d < w/2.

Solution: The mass is

M = J J pr dr dd = J J r2 dr dd = J r2 dr J dd = — .

The moment is

= J J xpr dr dd = J J (r cos d, r sin d)r2 dr ddm

fa [ic/2= I r3 dr I

Jo Jo(cosd, sind) dd = — (1, 1).

Therefore

1x = i m = - 3 TM 7r a3 4

Answer: x/ 3a 3o \

= \2 ^ ’ 2^ /

Could you have predicted that x lies on the line y = x and |x| > a/2?

y

Fig. 6.5 domain between two polar curves

Other Domains

Now consider (Fig. 6.5a) a domain D defined by inequalities

g(fl) < r < /(0 ) , do < 0 < 0i,

where/(0) and g(d) are continuous functions and g{6) < /(0). The iteration formula for such a domain is

J J H(x, y ) dx dy = J J H (r cos 0, r sin 0) r dr ddD D

- A fJ df) \« / 7*=

fr=/(0)(r cos 0, r sin 0) r dr

! r=g(0 ) >

EXAMPLE 6.4

r r dx dySet up in polar coordinates the integral / / - - ■ ..... - over the

JJ x2 + y2D

domain D of Fig. 5.14. Do not evaluate.

Solution: The domain D splits naturally into two regions, one the reflection of the other in the line y = x; the function is symmetric in this line (Fig. 6.6). In polar coordinates, the line x = 2 is r cos 6 = 2 or r — 2 sec 6. Consequently, the lower region Di is determined by 0 < 6 < 7r/4 and 1 < r <2 sec 6. The set-up is

f C d x d y f f r dr ddJ J x2 + y2 J J r2D Di

C r !4 / [ 2 seed d \ p / 4

= 2 J —j d0 = 2 J In (2 sec 0) dO.

EXAMPLE 6.5

Compute xy dx dy over the domain D defined by r = sin 20:

r = sin 26

Fig. 6.7

Solution: Draw a figure (Fig. 6.7). In polar coordinates,


hence

J J xy dx dy = J J - r2 sin 2dr dr dO = ^ J (sin 26) ^ J r3 dr j ddD D 0 0

1 f T/2 / I \ 1 f r/2 1= ~ (sin 20) I - sin4 20 J d6 = - / sin5 26 d6 = — / sin5 a da

2 A \4 / 8 70 16 J0

1 f r /2 1 2*4 1(by tables).

Answer:15

EXAMPLE 6.6

Find the volume common to the three solid right circular cylinders bounded by x2 + y2 = 1, y2 + z2 = 1, + x2 = 1.

(a) (b)Fig. 6.8 three intersecting cylinders

Solution: The first problem is drawing the solid. By symmetry, only the portion in the first octant need be shown. In Fig. 6.8a the common part of y2 + z2 = 1 and z2 + x2 = 1 is shown. The base of the third cylinder x2 + y2 = 1 is dotted. In Fig. 6.8b this third cylinder is drawn, cutting off the desired solid. One sees that the solid is symmetric in the plane x — y, hence only the half to the left of this plane is needed (Fig. 6.9a). This is the solid under the surface z = y / \ — x2 and over the wedge-shaped domain D shown in Fig. 6.9b.


(a)Fig. 6.9

(b)

V

Compute the volume using polar coordinates:

= 16 j j y / 1 — x2 dx dy = 16 j j y / l — r2 cos2 6 r dr dd

= 16 j j j ry / 1 — r2 cos2 6 d rj dd

nU - f f

| r=0-l 3 J 0— sin3 6 cos2 6

dd.

Now

1 — sin3 6 1 — sin3 6 1 + sin 6 + sin2 6 . 1= sm 6 +

cos2 0 1 — sin2 0

From tables,

1 + sin 0 1 + sin 6

f d O ( i r 0 \/ / ------- ;— = — tan I ------- ) + C,1 J l + sin0 \4 2 /

hence

16 f / tt 0 \ H t/4 16 T y / 2 7r 1■ T L 1 _ T “ t an8 + 1J3 L 2 2 + V 2J

Answer: 8(2 — y / 2 ) .

N o t e : The expression for ta n (71-/8) comes from the half-angle formula ta n (0/2) = sin 0/(1 + cos 0).

Find the volume over the region given in polar coordinates and under the indicated surface. Sketch:

4. Use polar coordinates to find the volume of a hemisphere of radius a.5. Find the volume of the region bounded by the two paraboloids of revolution

z = x2 + y2 and z = 4 — 3 (x2 + y2).6. Find the volume of the lens-shaped region common to the sphere of radius 1

centered at (0, 0, 0) and the sphere of radius 1 centered at (0, 0, 1).7. A drill of radius b bores on center through a sphere of radius a, where a > 6.

How much material is removed?8. Find the center of gravity of a sheet with uniform density in the shape of a quarter

circle 0 < r < a and 0 < 0 < t / 2 .

9. (cont.) Now do it the easy way, using the result of Example 6.2.10. Find the center of gravity of a homogeneous wedge of pie in the position 0 < r < a

and 0 < 0 < a.11. (cont.) Hold the radius a of Ex. 10 fixed, but let a --------> 0. What is the limiting

position of the center of gravity?12. A quarter-circular sheet in the position 0 < r < a and 0 < 0 < t / 2 has density

p = a2 — r2. Find its center of gravity.13. Find the volume under the surface z = x4?/4, and over the circle x2 + y2 < 1.14. Find the volume under the surface z = r3, and over the quarter circle 0 < r < 1,

0 < 0 < t / 2 .

15. Find the volume under the cone z = 5r, and over the rose petal with boundary r = sin 30, 0 < 0 < t / S .

16. Find the area bounded by the spiral r = 0, 0 < 0 < 2t , and the x-axis from 0 to 2 t .

17. Evaluate over the unit disk 0 < r < 1:

EXERCISES

1 . z = x2 + y2; l < r < 2 , 0 < 0 < t / 2

2 . z = x ) 0 < r < l , - 7 r /2 < 0 < 7 r /23. z = xy; 1 < r < 2, tt/4 < 0 < t / 2 .

/ / n S r ? - / / — * * •21. Find the volume common to two right circular cylinders of radius 1 intersecting at

right angles on center.22. (cont.) Find the volume of the region in space consisting of all points (x, y , z ),

where one or more of the following three conditions hold:(1) x2+ y 2 < 1, - 2 < z < 2(2) y2 + z2< 1, - 2 < z < 2(3) z2 + x2< l , - 2 < y < 2.

Use the results of Ex. 21 and Example 6.6; after that, its pure logic, not calculus.] The game of “jacks” has objects more or less in the given shape.

23. Evaluate the integral in Example 6.6 using rectangular coordinates.

11. Multiple Integrals

1. TRIPLE INTEGRALS

In this chapter we shall learn how to set up and apply triple integrals. As with double integrals, we shall work intuitively, and postpone the theory behind multiple integration until Chapter 12.

Let us consider the following problem. Suppose we have a bounded domain D in space filled by a non-homogeneous solid. At each point x the density of the solid is 8 (x) gm/cm3. What is the total mass?

Our previous experience with problems of this type on the line and in the plane leads us to expect an answer of the form

- / / /D

M = 8(x) dx dy dz ,

where the triple integral is defined by decomposing D into many small boxes, forming a suitable triple sum

Y X X 5 (*•■> (~Xi - Xi- 1) - y s - 0 ~

and taking limits. The theory says that this process does indeed define the triple integral, and that in practice the triple integral can be evaluated by iteration.

Here is how iteration works when the domain is the part of a cylinder bounded between two surfaces, each the graph of a function. Precisely, suppose two surfaces z = g ( x , y ) and z = f (x, y ) are defined over a domain S in the x , y-plane, and that g(x, y ) < f(x, y ) . See Fig. 1.1. These surfaces may be considered as the top and bottom of a domain D in the cylinder over S. Thus D consists of all points (x, y, z) where (#, y) is in S and

g( x, y ) < z < f ix, y ) .

In this situation the iteration formula is

f f f 8( x) dx dy dz = f [ ( [ f>(x,y, z) dz \ dx dy.J J J J J \ J z=g(x,y) '

D

410 U . MULTIPLE INTEGRALS

R e m a r k : Some prefer the notation

r r r / ( x , y )

/ / dx dy / 8(x, y, z) dz .JJ J g(x,y)

The reason for the iteration formula is illustrated in Fig. 1.2. First, the little pieces of mass 8 (x, y , z ) dx dy dz in one column are added up by an integral in the vertical direction (x and y fixed, z variable). The result is

8 (x, y, z) dz J dx dy.0(x, y) /

Then the masses of these individual columns are totaled by a double integral over S.

EXAMPLE 1.1

Find j j j (,x2 + y2) dx dy dz, taken over the block 1 < x < 2,

0 < y < 1 , 2 < z < 5 .

Solution: The upper and lower boundaries are the planes z = 5 and

1. Triple Integrals 411

/(*, y) rs(x) dx dy dz

9(x, y)

z = 2. Therefore

I f f (x* + y ) dy dz = m i : (x2 + y) dz'j dx dy ,s

where S is the rectangle 1 < x < 2 and 0 < ?/ < 1. Now x and y are constant in the inner integral:

fJ 2(a;2 + y) dz = 3 (a;2 + 2/),

hence

JJJ (x2 + y) dx dy dz = j j 3 (#2 + y) dx dys

Answer:17

EXAMPLE 1.2

Compute JJJ xzy2z dx dy dz over the domain D bounded by

x = 1, x = 2; y = 0, y = #2; and 2 = 0, 2 = 1/jc.

412 U . M ULTIPLE INTEGRALS

Solution: The domain D is the portion between the surfaces z = 0 and z = 1/x of a solid cylinder parallel to the 2-axis. The cylinder has base S in the x , y-plane, where S is shown in Fig. 1.3. The solid D itself is sketched in Fig. 1.4. (A rough sketch showing the general shape is quite satisfactory.)

A*±

Q /O

oy == x2/

«

1 /1

0 | !

Fig. 1.3I ^

Fig. 1.4

The iteration is

rZy2z dx dy dzJ J J xzy2z dx dy dz = J J ( J xzy2z dz \ dx dyD S °

= J J x3y2 0 z21 ^ d x d y = ^ J J xy2 dx dy

48 v ’ 48 16

F ig . 1.5


Alternate Solution: The domain may be considered as the portion between the surfaces y — 0 and y = x2 of a solid cylinder parallel to the y-axis. The cylinder has base T in the z, x-plane (Fig. 1.5). From this viewpoint, the first integration is with respect to y ; the iteration is

J J J xzy2z dx dy dz = J J xzz ^ J y2 dy j dx dz = J J ^D T T

x9z dx dz

i y*2 / ri/x \ 1 f 2 1

’ • ) * — » } , ' “ s ' * - 1 ' - a -

Answer:8516

R e m a r k : I t is bad technique to consider the region as a solid cylinder parallel to the x-axis, because the projection of the solid into the y, 2-plane breaks into four parts. Therefore, the solid D itself must be decomposed into four parts, and the triple integral correspondingly expressed as a sum of four triple integrals (Fig. 1.6). The resulting computation is much longer than that in either of the previous solutions.

1 / /f

/

F ig . 1.6

In practice, try to pick an order of iteration which decomposes the required triple integral into as few summands as possible, hopefully only one. The typical summand has the form

rh(x) / f f ( x ,y)

p(x, y, z ) dz 'rb r rh(x) / /•/(*,*)

J a L J k(x) g(x, y)z ) d y \ dx.

(Possibly the variables are in some other order.) Once the integralrhx.v)/ p(x, y, z) dz

J o(x,y)

is evaluated, the result is a function of x and y alone; z does not appear. Likewise, once the integral

CHx) / r f ( x ,

J k(x) \ J g(x , y

V) \

p(x, y, z) d z ) dyy) /

is evaluated, the result is a function of x alone; y does not appear.Remember there are six possible orders of iteration for triple integrals.

If you encounter an integrand you cannot find in tables, try a different order of iteration.

Domains and Inequalities

A domain in the plane or in space is frequently specified by a system of inequalities. A single inequality f ( x , y , z ) > 0 determines a domain whose boundary is f ( x , y , z ) = 0. To find the domain described by several such inequalities, draw the domain each describes, then form their intersection.

EXAMPLE 1.3

Draw the plane domain given by x + y < 0, y > x2 + 2x.

Solution: The first inequality determines the domain below (and on) the line x + y = 0. See Fig. 1.7a. The second inequality determines the domain above (and on) the parabola y = x2 + 2x. See Fig. 1.7b. The line and parabola intersect at (0,0) and ( — 3,3). The domain satisfying both inequalities is shown in Fig. 1.7c.

= 0

(b) F ig . 1.7

y = x 2 + 2x

(c)

jy = x 2 + 2x

EXAMPLE 1.4

Describe the plane domain S defined by 0 < x < y < a. Write

the integral j j f { x , y ) dx dy as an iterated integral in both orders, s

Solution: The region is described by the three inequalities

x > 0, y > x, y < a.

Draw the corresponding domains and take their intersection, a triangle (Fig. 1.8).

i V

x > 0

X

a

y < ax

To integrate first on x, arrange the inequalities describing the triangle in his order:

0 < y < a, 0 < x < y.

Think of y as fixed in the interval [0, a]. Then x runs from 0 to y, leading to

/ /(« , y) dx.Jo

'Jow sum these quantities for y running from 0 to a. The result is

f(x, y ) dx j dy.

To integrate first on y , describe the region in the other order:

0 < # < a, x < y < a.


Now the result is

j j f ix , y ) d x d y = m f i x , y ) d y ) dx.S O x

Answer: The triangle with vertices (0,0), (0, a), (a, a);tv: f i x , y ) d x j dy = m f i x , y ) d y ) dx.

R e m a r k : A special case is interesting. Suppose/(x, y) = g(x) , a function of x alone. Then the right-hand side is

l ( ^ l g(x) d y ) dx = J g(x) (^J d y ) dx = J (a - x)g(x ) dx.

The following formula is a consequence:

I (^1 g(x) d x ) dy = J (a - x )g (x) dx.

Thus the iterated integral of a function of one variable can be expressed as a simple integral.

Tetrahedra

If a domain D is specified by inequalities, it may be possible to arrange the inequalities so that limits of integration can be set up automatically. For example, suppose the inequalities can be arranged in this form:

a < x < b, h(x) < y < k(x), g ( x , y ) < z < f (x, y) .

Thenr r r rb r m / r m , v) x -i

8 (x, y, z) dx dy dz = / I / 8 (x, y, z) d z j d y l dx.J J J J a - J h(x) \ J g(x,y) / J

D

Tetrahedral domains can be expressed by such inequalities, and they occur frequently enough that it is useful to practice setting up integrals over them.

EXAMPLE 1.5

A tetrahedron T has vertices at (0 ,0 ,0 ), (a, 0 ,0 ), (0 ,6 ,0 ),

(0, 0, c), where ayb ,c > 0. Set up J j j 8 (x , y , z) dx dy dz as anT

iterated integral.

x- + l + z- = i. a b c


Solution: The slanted surface (Fig. 1.9) has equation

The domain is defined by the inequalities

0 < 0 < ?/, 0 < 2, — h 7 H— ^ 1.a b c

Any order of iteration is satisfactory; for instance, choose the order of integration

/ [ / ( / p(x, y, z ) d z j efo/J dx.

To find the limits of integration, arrange the inequalities in this form:

a < x < by h(x) < y < k(x), g (xf y ) < z < f ( x , y ).

If x and y are fixed, then

If x is fixed, then

. < sr < i ( l — | < 6 ( l — .

Finally,

0 < x < a ^1 — ^ — -^ < a.

The resulting system of inequalities (equivalent to the original system) is

0 < £ < a, 0 < y < b ^1 — , 0 < z < c ^1 — - — .

The corresponding iteration is r r r r» r r K 1 ~ I) / A 1 ~ I - 5) \ -j

JJJ p(x, y, z) dx dy dz = J J y j p(x, y, z) dzJ d y J cte.


EXERCISES

Evaluate the triple integral over the indicated domain:

1. JJJ ~ z) dx dy dz; solid cylinder under z = 2 — x2, based on the triangle

with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0)

2. f j j y t e l , * ; solid cylinder under , - * + V, b»*d „„ the .rinngle in E , 1

’ / / / dx dy dz; pyramid with apex (0, 0, 1), based on the square with vertices

(dzl, ± 1 )4. The same, except the square base has vertices (dzl, 0), (0, d=l)

5. [ [ f a z dx dy dz; the domain lies in the first octant and is bounded by x = 0,

z = 0 , y = x2, and z = 1 — y2

6. J J J (3x2 — z2) dx dy dz; the domain in the slab 0 < y < 1 bounded by z = y2,

z = —y2, and y = x, y = —x.

7. A solid cube has side a. Its density at each point is k times the product of the 6 distances of the point to the faces of the cube, where k is constant. Find the mass.

8. Charge is distributed over the tetrahedron with vertices 0, i, j, k. The charge density at each point is a constant k times the product of the 4 distances from the point to the faces of the tetrahedron. Find the total charge.

9. Find ( i ( (x + y + z)2 dx dy dz over the domain in the first octant bounded by

the coordinate planes, the plane x + y + z = 2, and the 3 planes x = 1, y — 1, 2 = 1 .

2. Cylindrical Coordinates 419

( l l dx dy dz10. Find / / / t —|-----;— rz over the domain in the first octant between the planesJ J J ( x + y + z)2

x + y + z = 1 and x + y + z = 4. [Hint: Think!]Sketch the domain:11. x2 + y2 < 1, y + x2 > 012. x2 + y2 < 1, —x2 < y < x213. x2 + y2 > 1, (X - 2)2 + y2 < 914. x > 3, y < —5, y — x > —1015. x + y < 0, xy < 1, (x — y)2 < 116. (x + y)2 < 1 , (x — y)2 < 1.

Express the double integral J J f (x ,y ) dx dy over the specified domain as the sum of

one or more iterated integrals in which y is the first variable integrated:17. x2 + y2 < 1, x2 + (y - 1 )2 < 118. y > (x + 1 )2, y + 2x < 319. x > 0, 0 < y < t , x < sin y20. x > 0, x2 — y2 > 1, x2 + y2 < 9.

21. Describe the domain D :0< x < y < z < 1. Iterate III p (x, y, z) dx dy dz in theI I I '

orders x, y, z; z, y, x; and x, z,y. D

22. Repeat Ex. 21 for a < x < y < z < 6.23. Repeat Ex. 21 for 0 < x < 2y < 3z < 6.24. Express a triple integral over the domain determined by (x — 1 )2 + y2 < 4,

0 < z < y, (x + 1 )2 + y2 < 4 as one iterated integral (not a sum of two or more). Sketch the domain.

25. Find the integral of x over the tetrahedron with vertices (5,6,3), (4,6,3), (5,5,3), (5,6,2).

26. Set up the integral of 5(x, y, z) over the tetrahedron with vertices (5, —5, 1), ( 5 , - 5 , - 2 ) , (5, 1,1), ( - 2 , - 5 ,1 ) .

27. Express

g (x) dxj dyj dz

as a simple integral.28. Take four vertices of a unit cube, no two adjacent. Find the volume of the tetra

hedron with these points as vertices.29. (cont.) Now take the tetrahedron whose vertices are the remaining four vertices

of the cube. The two tetrahedra intersect in a certain polyhedron. Describe it and find its volume.

\

2. CYLINDRICAL COORDINATES

Cylindrical coordinates are designed to fit situations with rotational (axial) symmetry about an axis.

The cylindrical coordinates of a point x = ( x , y t z) are {r, 6, z] , where


[ry 6) are the polar coordinates of (xy y ) and z is the third rectangular coordinate (Fig. 2.1a). Each surface r = constant is a right circular cylinder, hence the name, cylindrical coordinates (Fig. 2.1b).

(a) (b)F ig . 2.1

Through each point x (not on the 2-axis) pass three surfaces, r = constant, 6 = constant, z = constant (Fig. 2.2). Each is orthogonal (perpendicular) to the other two at their common intersection x.

F ig . 2.2


The relations between the rectangular coordinates (x, y, z) and the cylindrical coordinates {r, 0, z) of a point are

x = r cos i

y = r sin 6

z = z

r2 = x2 + y2

z = z.

The origin in the plane is given in polar coordinates by r = 0; the angle 0 is undefined. Similarly, a point on the 2-axis is given in cylindrical coordinates by r = 0, z — constant; 0 is undefined.

EXAMPLE 2.1

Graph the surfaces (i) z — 2r, (ii) z — r2.

Solution: Both are surfaces of revolution about the 2-axis, as is any surface z = f ( r ) . Since z depends only on r, not on 6, the height of the surface is constant above each circle r = c in the x, y-plane. Thus the level curves are circles in the x, y-plane centered at the origin.

In (i), the surface meets the first quadrant of the y, 2-plane in the line z = 2y. (In the first quadrant of the y, 2-plane, x = 0 and y > 0. Since r2 = x2 + y2 = y2, it follows that r = y. ) Rotated about the 2-axis, this line spans a cone with apex at 0. See Fig. 2.3a.

(a)Fig. 2.3

In (ii), the surface meets the y, 2-plane in the parabola z = y2. Rotated about the 2-axis, this parabola generates a paraboloid of revolution (Fig. 2.3b).


EXAMPLE 2.2

Find the level surfaces of the function /(r , 0, z) = r.

Solution: Each surface is defined by r = c. This is a right circular cylinder whose axis is the 2-axis (Fig. 2.4).

Answer: Concentric right circular cylinders about the 2-axis.

The Natural Frame

I t is convenient to fit a frame of three mutually perpendicular vectors to cylindrical coordinates just as the frame i, j, k fits rectangular coordinates. At each point {r, 6, z } of space attach three mutually perpendicular unit

F ig . 2.5

\2. Cylindrical Coordinates 423

vectors u, w, k chosen so

u ( \ r

- w points in the direction of increasing d

k w z j

Thus (Fig. 2.5)

The vectors u, w, k form a right-hand system:

u X w = k, w X k = u, k X u = w.

Note that u and w depend on d alone, while k is a constant vector, our old friend from the trio i, j, k. Note also

dU aw— = W, — = —u.dd dd

In situations with axial symmetry, it is frequently better to express vectors in terms of u, w, k rather than i, j, k.

Integrals

If a solid has axial symmetry, it is often convenient to place the 2-axis on the axis of symmetry, and use cylindrical coordinates {r, d, z } for the computation of integrals.

In polar coordinates {r, 0}, the element of area is r dr dd. Hence the element of volume in cylindrical coordinates {r, 6, z } is

dV = r dr dd dz.

To gain further insight, let us argue intuitively for a moment. Given x = {r, d, z }, how is this point displaced if we give small increments dr , dd, dz to its three coordinates? Figure 2.6 shows that the displacement in the direction of u is dr u. The displacements in the directions of w and k are r dd w and dz k. The displacement of x is approximately the sum of these three small vectors.

Accordingly we write


dx = dr u + r dd w + dz k.

Since the displacements are mutually perpendicular vectors, they span a small “rectangular box” of volume (dr) (r dd) (dz) . Thus we arrive again at the formula d V = r dr dd dz.

EXAMPLE 2.3

Evaluate j j j (x2 + y2) ll2z dx dy dz} taken over the cone with

apex (0, 0, 1) and semicircular base bounded by the x-axis and y = \ / 4 — x2.

Solution: Write the integral in cylindrical coordinates:

/ = J J J (rz)r dr dO dz.

The lateral surface of the cone (Fig. 2.7) is given by


12 = 1 — 2 r>

so the solid is described by

O < 0 < T , 0 < r < 2, 0 < z < 1 - ^r .

Therefore

' - r’( f *)* - 'H O “ r h= “ / r2 ( 1 — r + - r2) dr = ^ / (r4 — 4r3 + 4r2) dr = — .

2 Jo \ 4 / 8 y0 15

Answer: — , 15

EXAMPLE 2.4

A region D in space is generated by revolving the plane region bounded by 2 = 2x2, the x-axis, and # = 1 about the 2-axis. Mass is distributed in D so that the density at each point is proportional to the distance of the point from the plane 2 = — 1, and to the square of the distance of the point from the 2-axis. Compute the total mass.

Solution: The density is

8 = k(x2 + y2) (2 + 1) = kr2(z + 1 ) ,

where k is a constant. The portion of the solid in the first octant is shown in Fig. 2.8. In cylindrical coordinates, the solid is described by the inequalities

O < 0 < 2 t t , 0 < r < 1, 0 < 2 < 2r2.

Therefore its mass is

426 11. M ULTIPLE INTEGRALS

EXERCISES

1. Find the rectangular coordinates of {2, 57t/4, —3}, {l, w, 2}.2. Find the cylindrical coordinates of (0, —2, —2), (5, —5, 0), (—1, — \/3 , “ !)•3. A space curve is given by r = r(0 , 0 = 0(O> 2 == z(0* Show that its arc length

satisfies s = (r2 + r202 + £2)1/2.4. (cont.) Find the length of the spiral r = a, 6 = bt, z = ct for 0 < t < 1.5. Prove analytically the formula

dx = dr u + r dB w + dz k.6. Prove

~du~

dw =

_dk_

0

L 0

dd O'

0 0

0 0.

u

w

Lk.7*. Letf = f ( r,0, z) . Prove

g rad /. 2 u + I | w + ^ k.dr r dd dz

[Hint: Use Ex. 5.]

3. Spherical Coordinates 427

8. Use cylindrical coordinates to find the volume of a right circular cone of base radius a and height h.

9. Find J J J xyz dx dy dz over the quarter cylinder 0 < r < a, 0 < 0 < ^tt, 0 <

z < h.

10. Find / / / ' dx dy dz over the domain common to the sphere x2 + y2 + z2 < a2

and the cylinder r < b, where b < a.

11. Find ( I L dx dy dz over the cylindrical wedge 0 < z < y, r < a, 0 < 0 < 7r.

12. Sketch the domain r2 < z < 2r2, r < 1. Now find / / / • z dx dy dz over this domain.

13. Sketch the domain \z\ < r2, r < a, x > 0, y > 0. Now find dx dy dz over the

domain.14*. Sketch the domain 0 < r < cos 20, — 7r < 0 < £71*, 0 < 2 < 1 — r2. Now find

///■fz dx dy dz. (Use the definite integral table inside the front cover to complete

the solution.)

3. SPHERICAL COORDINATES

Spherical coordinates are designed to fit situations with central symmetry. The spherical coordinates [p, 0, 0] of a point x are its distance p = |x| from the origin, its elevation angle 0, and its azimuth angle 0. (Often 0 is called the longitude and 0 the co-latitude.) See Fig. 3.1.

F ig . 3.1

x

Relations between the rectangular coordinates (x , y, z) of a point and its spherical coordinates may be read from Fig. 3.2. They are

Note that 0 is not determined on the 2-axis, so points of this axis are usually avoided. In general 0 is determined up to a multiple of 2 t , and 0 < 0 < r.

The level surfaces

p = constant concentric spheres about 0

j 0 = constant ► are right circular cones, apex 0 *

0 = constant y planes through the 2-axis

At each point x the three level surfaces intersect orthogonally (Fig. 3.3).

The Natural Frame

Select unit vectors \ , [i, v at each point x of space (not on the 2-axis) such that

P

points in the direction of increasing 0 >

V 0 y

See Fig. 3.4. Then \ , |x, v is a right-hand system.Our immediate problem is to express \ , |A, v in terms of p, 0, 0. Here is a

short cut for doing so:x = p (sin 0 cos 0, sin 0 sin 0, cos 0),

dx = (sin 0 cos 0, sin 0 sin 0, cos 0) dp

+ p (cos 0 cos 0, cos 0 sin 0, — sin 0) d<t>

+ p (— sin 0 sin 6, sin 0 cos 0, 0) dd

= \ d p + |i p d(j> + v p sin 0 dd.

I t is easy to check that \i, v are unit vectors. Furthermore, they point in the directions of increasing p, 0, 0 respectively. Precisely, if only p increases, then d<t> = dd = 0, hence dx = \ dp. Similarly, if only 0 increases, thendx = |x p d0,


F ig . 3.3

Fig. 3.4

and if only 6 increases, then dx = vp sin 0 dd. Conclusion:

\ = (sin 4> cos 0, sin <£ sin 0, cos <t>)

\ [x —— (cos <f> cos 0, cos 0 sin 0, — sin <£)

{ v = (— sin 0, cos 0, 0),

dx = \ dp + |x p d<t> + v p sin <t> dd.

Integrals

If a solid has central symmetry, it is often convenient to place the origin at the center of symmetry and use spherical coordinates [p, 0, 0] for the computation of integrals.

The displacement of a point of space, in terms of the frame \ , |x , v natural to spherical coordinates, is

dx = dp \ + p d<t> | i + p sin 0 dd v .

See Fig. 3.5. The element of volume of the small “rectangular” solid is the product of its sides:

d V = (dp) (p d<l>) (p sin 0 dd) = p2 sin 0 dp d<t> dd.

EXAMPLE 3.1

Use spherical coordinates to find the volume of a sphere of radius a.

Solution:

V ! sin 0 dp dcf) dd■ / / / ' “

Answer: - %a?.O

EXAMPLE 3.2

Find the volume of the portion of the unit sphere which lies in the right circular cone having its apex at the origin and making angle a with the positive 2-axis.

Solution: The cone is specified by 0 < 0 < a, so the portion of the sphere is determined by 0 < 0 < 2ir, 0 < 0 < a, and 0 < p < 1. See Fig. 3.6. Hence the volume is

( / d 6 ) d o sin ^ ) ( / p2 rfp) = ^ - c°s Q ) •

R e m a r k : As a check, let a --------> 7r. Then the volume should approachthe volume of a sphere of radius 1. Does it?

EXAMPLE 3.3

A solid fills the region between concentric spheres of radii a and b} where 0 < a < b. The density at each point is inversely proportional to its distance from the center. Find the total mass.

Solution: The solid is specified by a dp d<t> dd

= k ( J * * d e ) ( J * sin * P dp) = (27rk) (2) ■

Answer: 2 irk(b2 — a2).

R e m a r k : As a --------> 0, the solid tends to the whole sphere, with infinitedensity at the center. But M --------> 2irkb2, which is finite.

EXAMPLE 3.4

A cylindrical hole of radius J is bored through a sphere of radius 1. The surface of the hole passes through the center of the sphere. How much material is removed?

Solution: Center the sphere at 0 and let the cylinder (hole) be parallel to the z-axis, with axis through (0, J). See Fig. 3.7. By symmetry the volume

Fig. 3.7

is four times that in the first octant. The equation of the cylindrical surface is x2 + (y - J )2 = I, or

x2 + y2 = y.

But in spherical coordinates, x2 + y2 = p2 sin2 0 and y = p sin 0 sin 0. Hence the equation of the cylinder is p2 sin2 0 = p sin 0 sin 0, or

sin 0

sm 0

The cylinder intersects the sphere in the curve (Fig. 3.8)

i sinfl i P = T — = 1.sm 0

In the first octant this is the curve

p = 1, 0 = 0.

The first-octant portion of the volume naturally splits into two parts, separated by the cone 0 = 6 of segments joining 0 to the curve of intersection. In the lower part, each radius from the origin ends at the cylinder, so the limits are

7r 7r sin 6O < 0 < ~ , 0 < 0 < - , 0 < p < — .

2 2 sm 0

In the upper part, each radius from the origin ends on the sphere, so the

limits are

O < 0 < - , 0 < 4 < e , 0 /'7r/2 /'7r/2V i = dO sin <t>d<t) p2dp = - sin3 0 dd / . o -

Jo A Jo 3 J 0 Je sm2 0

! /*T/2 . 1 /^/2 . 1 = - / sm3 0 cot 0 dO = - / sm2 0 cos 0 dO = - ,

3 Jo 3 70 9

/•"V2 re ri j /-ir/2 j / \F 2 = J dO J sin 4)d<t) J p2 dp = - J (1 — cos 0) d0 = - — ly .

^ = 4 ^ - + - - - ^ = —\ 9 6 3/ 3 9

and

Therefore

Answer;2tt __ 83 9

Spherical Area

Suppose a point moves on the surface of the sphere p = a. Then dp = 0, so the formula for displacement specializes to

dx = a d<t> |x + a sin <f> dOv.

The area of the small “rectangular” region on the surface of the sphere corresponding to changes d<t> and dd is

dA = (a d<f>) (a sin <j> dd) = a2 sin <£ d<j) dd.

See Fig. 3.9. The integral of this expression over a region is the area of that region.


EXAMPLE 3.5

Find the area of the polar cap, all points of co-latitude a or less on the unit sphere.

Solution: See Fig. 3.6. The region is defined on the sphere p = 1 by0 < </> < oi, hence

A = J ^ j sin <t)d<^jdd = 2tt(1 — cos a).

Answer: 2tt(1 — cos a).

R e m a r k : Suppose S is a region on the unit sphere. The totality of infinite rays starting at 0 and passing through points of S is a cone which is called a solid angle (Fig. 3.10). A solid angle is measured by the area of the base region S. Since S lies on the unit sphere, its area is measured in square radians, a dimensionless unit. Thus the solid angle of the whole sphere has

436 11. MULTIPLE INTEGRALS

z

y

Fig. 3.10

measure 47r rad2, or simply 47r. The solid angle determined by the first octant has measure 47r/8 = t t/ 2 . The solid angle of the polar cap in Example 3.5 has measure 2t (1 — cos a). Incidentally, in the metric (SI) system, the unit for solid angles is the steradian (sr).

EXERCISES

1. Convert to rectangular coordinates: [1, 3tt/4, 37t/4], [3, t / 2 , 5t / 4], [2, 27r/3, 7t/3]2. Convert to spherical coordinates:

(1, 1, 1 ), (1, - 1, 1 ), ( - 1, - 1, - 1), ( - 1, 1, - 1).3. If x has spherical coordinates [p, 0, 0], find the spherical coordinates of — x and

of 3x.4. Suppose a space curve is given byp = p (t), <t> = <t>(t),d = 0 (t). Show that its arc

length satisfiess2 = p2 + p202 + p2 (sin2 <t>)62.

5. (cont.) A rhumb line on a sphere of radius a is a curve that intersects each meridian at the same angle a. (Follow a constant compass setting.) Find the length of a rhumb line from the equator to the north pole.

6*. Let / = f(r, <£, 6). Prove1, , d f . . l d f ,

grad/ + • , .adp pd(j) p sin <f> dddf

V.

7. Prove" d \ ~

d[L =

- d V -

0

— d<f)

cUf)

0

V

LV J_—sin <l>dd —cos <t>dd 08. Let v be a radial vector field of the form v = g(p)\ . Prove that v is a gradient

field. [Hint: Use Ex. 6.]

9. Suppose /[p, $, 0] is homogeneous of degree n with respect to rectangular coordinates, that is, f(tx) = tnf (x) for all t > 0. Prove/[p, <t>, 0] = pn (</>, 0), where g(4>, 0 )= /[ l,< M ].

10. Find / / / ' n dx dy dz over the sphere p < a, where n > 0.

11. Find J J J (1 — p)n dx dy dz over the sphere p < 1, where n > 0.

Find J j j 1 dx dy dz over the first octant portion of the sphere p < a.

.3. F i n d / / / p"~2 dx dy dz over the domain z > 0, a < p < b, where 0 < a < 6.

14. Find ///■ dx dy dz over the domain 1 < z, p < f \/3 .

15. Two parallel planes at distance /i intersect a sphere of radius a. Find the surface area of the spherical zone between the planes.

16. In Example 3.4, how much of the surface area of the sphere is removed?

Additional Volume Problems

Use any method you please to set up the integral. You may not be able to evaluate it, but at least try to reduce the answer to a simple integral, not a double or triple one.

17. Find the volume of the (solid) right circular torus obtained by revolving the circle (y — A )2 + z2 = a2, x = 0 about the z-axis. Assume 0 < a < A.

18. Suppose in Ex. 17 that 0 < A < a. Let the portion of the circle to the right of the z-axis generate volume V\ and the portion to the left generate volume Vi. Find V i — V 2.

19*. A sphere of radius a touches the sides of a cone of semi-apex angle a. See Fig. 3.11. Find the volume of the portion of the cone above the sphere.


front view side viewF ig . 3.11 (Ex. 19)


20*. A circular hole is bored through a right circular cone, dimensions as indicated in Fig. 3.12. The axes are perpendicular and the hole just fits. How much material is removed?

21*. (cont.) How much of the cone remains above the hole?

front view side viewF ig . 3.12 (Exs. 20, 21)

F ig . 3.13 (Ex. 25)


F ig . 3.15 (Ex. 27)

front side

22*. A cylindrical hole of radius a is bored through a solid cylinder of radius 2a; the hole is perpendicular to the solid cylinder and just touches a generator. Find the volume removed.

23. A circular hole of radius a is bored through a solid right circular cylinder of radius b. Assume the axis of the hole intersects the axis of the solid at a right angle and that a < 6. Show that the volume of material removed is

/a _____ _____ f r /2 _________y /aL — x2\ / b z — x2 dx = 8a2 / (sin2 0)\/&2 — a2 cos2 6 dB.

24. Do Ex. 23 assuming that the axes meet obliquely at angle a. Set up, but do not evaluate the integral. Can you evaluate it when a = b?

25. A square hole is bored through a right circular cylinder (Fig. 3.13). How much material is removed?

26*. A square hole is bored through a right circular cylinder (Fig. 3.14). How much material is removed?

27*. A square hole is bored through a right circular cone (Fig. 3.15). How much material is removed? (Do not evaluate the integral.)

4. CENTER OF GRAVITY

Suppose a solid D has density 8 (x) at each point x. Then its mass is

M = J J J 8 (x) dx dy dz.D

Define its moment about the origin

m = i i i y dz>and its center of gravity

1x = — m.

M

Note that m and x are vectors. I t is often convenient to express them in terms of components:

m = (mx, my, ?nz), x = (x, y, z).

The center of gravity may be considered as a sort of weighted average of the points of the solid. Recall in this connection that the center of gravity of a system of point-masses M i, • • •, M n located at Xi, • • •, xn is

x = (MiXi + M2x2 + • • • + M nxn),

where M = M i + • • • + M n.If a solid is symmetric in a coordinate plane, then the center of gravity

lies on that coordinate plane. For example, suppose D is symmetric in the x , y-plane. This means that whenever a point (x, y , z ) is in the solid, then

4. Center of Gravity 441

(,xj y, —z) is in the solid, and 8(x, y, z) = 8(x} y, —z ). The contribution to m z at (x, y, z) is

8 (x , 2/, 2 ) 2 dx dy dz;

it is cancelled by the contribution

Vi —z) (—z ) dx dy dz = —8(x, y , z ) z d x dy dz

at (x, y, —z). Hence m2 = 0 and 2 = 0.Similarly, if D is symmetric in a coordinate axis, then the center of gravity

lies on that axis. Finally, if D is symmetric in the origin, then x = 0.To compute the center of gravity of a solid, exploit any symmetry it has

by choosing an appropriate coordinate system. Of course, express the element of volume dV = d x d y dz in the coordinate system chosen.

EXAMPLE 4.1

Find the center of gravity of a uniform hemisphere of radius a and mass M.

(a) (b)F ig. 4.1

Solution: To exploit symmetry, choose spherical coordinates, with the hemisphere defined by p < a and 0 < <£ < t / 2 . See Fig. 4.1a. The density 8 is constant, so the mass is 8 times the volume:

1 /4 \ 2M = — I — 7ra3) 8 = — 7ra35.

2 \3 / 3

Because the hemisphere is symmetric in the y, 2-plane, mx = 0. Likewise rny = 0; only m z requires computation:


Hencev i z 7r5a4/4 3

2 = H = 2x5a3/3 = 8 a '

Answer; The center of gravity lies on the axis of the hemisphere, f of the distance from the equatorial plane to the pole (Fig. 4.1b).

R e m a r k : The answer is independent of 8. For uniform solids in general, the constant 8 cancels when you divide m by M , so x is a purely geometric quantity. I t is then called the center of gravity or centroid of the geometric region (rather than the material solid). For uniform solids, from now on take 5 = 1 and M = V, the volume.

EXAMPLE 4.2

Find the center of gravity of a uniform right circular cone.

(a) (b)

Fig. 4.2

Solution: Choose cylindrical coordinates with the apex of the cone at0 and the base of radius a centered at (0, 0, h). See Fig. 4.2a. The lateral surface of the cone is

h az = - r, that is, r — - z,

a h

and the volume of the cone is

By symmetry mx = my = 0. Compute m z\

m z = J J J zr dr dd dz = ^ [J* ( j ” '* r dr'j z dzj dd


Hence■ f , ' 1* I ! l ( i :' ) ' 42 - (2'> (!■ ) I ? ' * -

m z Td2h2/4: 3

7Ta 2 h 4 7TT T 'T = 7 fl2/i2. h2 4 4

— — h.V 7ra2h/3 4

Answer: The center of gravity is on the cone’s axis, J of the distance from the base to the apex (Fig. 4.2b).

EXAMPLE 4.3

The solid 0 < a; < 1, 0 < y < 2 , 0 < 2 < 3 has density xyz gm/cm3. Find its center of gravity.

Solution:

M = j j j xyz dx dy dz = j^ x dx j y dy j z d z = ^ Sm*

m = j j j (x , y, z) xyz dx dy dz

- ( I l l x'yzdxdydz, J J J x f z dx dy dz, J J J xyz'dxdydz)Ofi r 2 rz r l /*2 rz

x2 dx / y dy I zdz , / x d x y2 dy I z dz,

0 0 «/o »/0 */0 •/()

f 1xdx f 2 y dy [° z* dz) J o J o J o /

/ I 4 9 1 8 9 1 4 27\ 4-9 „ nx = ( 3 2 2 ’ 2*3*2 ’ 2*2"1T/ = 3*2*2 = gm-cm.

Hence

2 /2 4 \= 9 (3) (1, 2’ 3) = \3 ’ 3 ’ /

cm.

Answer: xe - i - o

cm.

The definitions of moment and center of gravity for solids can be modified in an obvious way to fit plane regions with a mass distribution.

Plane Regions

EXAMPLE 4.4

Find the center of gravity of a uniform semicircular disk.

Fig. 4.3

Solution: Choose polar coordinates, taking the disk in the position0 < r < a, O < 0 < 7 r . See Fig. 4.3a. Take 5 = 1; the mass equals the area

A = - (tta2).

By symmetry mx = 0. Compute my:

my = j j y r dr dd = j j r2 sin 6 dr dd = j sin 6 dd j r2 dr = - a3.

Hence

- i - A (o , | «)■

A nsw er: x See Fig. 4.3b.

There is a useful connection between the centers of gravity of plane regions and volumes of revolution.

First Pappus Theorem Suppose a region D in the x, y-plane, to the right of the y-axis, is revolved about the y-axis. Then the volume of the resulting solid is

where A is the area of the plane region D and x is the ^-coordinate of its center of gravity (Fig. 4.4).

V = 2trxA,

In words, the volume is the area times the length of the circle traced by the center of gravity. Proof: A small portion dx dy at x revolves into a thin ring of volume

dV = 2 t t x dx dy.Hence

- / /But mx = xA, hence V = 2 t t x A .

2 t t x dx dy = 2 t mx

Wires

A non-homogeneous wire is described by its position, a space curve x = x (s) where a < s < b, and its density 8 = 8(s). (Here s denotes arc length.) Its mass is

its moment is

m

8 (s) ds,

x ( s )8 ( s ) ds,

and its center of gravity is

If the wire is uniform, then 8(s) is a constant. In this case, the center of gravity is independent of 8, hence it is a property of the curve x = x (s) alone; you can take <5 = 1 and replace M by L, the length.

EXAMPLE 4.5

Find the center of gravity of the uniform semicircle r = a, y > 0.

m = J x d s = J (a cos d, a sin d) a dd


Solution: The length is L = ira. The moment is

« . / •Jo

(cos 6, sin0) dd = a2(0, 2).

Hence

x = ^ m = — a2(0, 2) = - (0, 2). L ira 7r

-Answer

Suppose a plane curve is revolved about an axis in its plane, generating a surface of revolution. There is a useful relation between the center of gravity of the curve and the area of the surface.

Second Pappus Theorem Suppose a curve in the x, y -plane to the right of the y -axis is revolved about the y-axis. Then the area of the resulting surface is

A = 2ttxL j

where L is the length of the curve and x is the ^-coordinate of the center of gravity (Fig. 4.5).

In words, the area is the length of the curve times the length of the circle traced by the center of gravity. Proof: A short segment of length ds of the curve at the point x(s) revolves into the frustum of a cone with lateral area

rb rb

dA = 2t x ds. See Fig. 4.6. Hencerb rb

A — ) 2irx ds = 2t I x ds = 2Trmx = 2irxL.J a J a

There is a useful aid to problem solving which we state for solids. With obvious modifications, it applies to wires or laminas.

radius x

Addition Law Suppose a solid D of mass M and center of gravity x is cut into two pieces D0 and Di, of masses M 0 and M h and centers of gravity x0 and Xi. Then

M = M o + M i , x — ( M qXq + M i X i ) .

The first formula is obvious. The second is simply a decomposition of the moment integral:

Mx = j j j 5(x) x d V = j j j + J j j = M0X0 + M,X1.Do Di

EXERCISES

1. Find the center of gravity of the first octant portion of the uniform sphere p < a.2. Suppose the density of the hemisphere p < a , z > 0 i s 5 = a — p. Find the center

of gravity.3. Find the center of gravity of the uniform spherical cone p < a , 0 <</>< a.4. Find the center of gravity of the uniform hemispherical shell a 0.5. Find the center of gravity of a uniform circular wedge (sector).6. The plane region bounded by z = 1 and the parabola z = y2 is revolved about

the z-axis. Find the center of gravity of the resulting uniform solid.7. Find the center of gravity of a uniform wire in the shape of a quarter circle.

8. Find the center of gravity of the uniform spiral x(t) = (a cos t, a sin t, bt), 0 < t < t 0.

9. A copper wire in the shape of a semicircle of radius 100 cm is steadily tapered from 0.1 to 0.5 cm in diameter. Find its center of gravity.

10. Verify the First Pappus Theorem for a semicircle revolved about its diameter.11. Use the First Pappus Theorem to find the volume of a right circular torus.12. Verify the First Pappus Theorem for a right triangle revolved about a leg.13. Use the Second Pappus Theorem to find the surface area of a right circular torus.14. Use the Second Pappus Theorem to obtain another solution of Example 4.5.15. Find the center of gravity of the uniform spherical cap (surface)

p = a , 0 < 0 < a .

16. Find the center of gravity of the uniform spherical cap (solid) p < a , a — h < z < a.

17. Suppose a solid of density 5(x) is acted upon by a uniform (constant) gravitational field f so the force on a small portion at x is [5 (x) rf7]f. Show that the solid is in equilibrium if a single force — M f is applied at x.

18. Find the center of gravity of the uniform triangle with vertices a, b, c.19. Find the center of gravity of the uniform tetrahedron with vertices a, b, c, d.

Recall that the kinetic energy of a moving particle of mass m and speed v is K = \mvl. The kinetic energy of a system of moving particles is the sum of their individual kinetic energies. To define the kinetic energy of a moving solid body D, decompose the body into elementary masses 6(x) d V , where

5. MOMENTS OF INERTIA

z

y / x 1 + y 2

xF ig . 5.1

5. Moments of Inertia 449

K = l j f j 5(x) lv(x- ONF.D

Here v(x, t) is the velocity of the point x of the body at time t. Thus K varies with time.

We shall compute the kinetic energy of a rigid body rotating with angular velocity <o about an axis through the origin.

First consider rotation about the 2-axis with angular speed co, so the angular velocity is co = cok. See Fig. 5.1. An elementary volume d V = dx dy dz at x has mass d M = 8(x) dV and speed co\/x2 + y2, since y / x2 + y2 is the distance from x to the 2-axis. Hence its kinetic energy is

1

5(x) is the density at x, and form the integral

d K = - ( u y / x 2 + y2) 2 8 (x) d V = - u 2 8 ( x ) ( x 2 + y2) dV. z z

To find the kinetic energy of the entire solid D, integrate, obtaining

1K = - / 2*co2,

where

■ m 8 ( x ) ( x 2 + y2) dV

is the moment of inertia of D about the 2-axis. Similarly define

■ I I I 8 ( x ) ( y 2 + z2) d V and - I f f 8(x) (z2 + x2) dV.

From these moments of inertia one can compute the kinetic energy, provided the body rotates about one of the coordinate axes:

1K = - I x (rotation about the z-axis),

K = \ l (rotation about the y-axis).

Products of Inertia

For rotation about more general axes, however, three other quantities called products of inertia (also mixed moments of inertia) are needed:

8 (x )y z d V , I 2X = I xz = — J JD

= - J J J S (x )x ydV .

8 ( x ) z x dV ,

Ixy— lyx

The three moments and three products of inertia together form the symmetric matrix of inertia

I XX I x y I x z

I = I y x l y y Iy.

J « I zy I z z .

The quadratic form associated with this matrix gives the kinetic energy of a solid rotating about an axis through 0 .

Suppose D rotates about an axis through 0 with angular velocity o> =

(co*, co„, a)z) . Then its kinetic energy is

I X X I x y N1

C0X

K = - co/co' = - (cOx, J U y , U > z )

A zI yx l y y l y z CCy

J ~ I zy I z z - _ 0 > Z _

Proof: Let u be a unit vector along the axis of rotation. Then a> = coil, where co2 = <o*a> = wx2 + coy2 + coz2. If x 6 D, the projection of x on the axis of rotation is x* u. Therefore the distance of x from the axis is [|x |2 — (x* u )2] 1/2. See Fig. 5.2. Hence the speed at x is w[|x|2 — (x* u )2] 1/2, and the kinetic energy

of an elementary mass at x is

clK = ^ 5(x)co2[jx|2 — (x*u)2] d V = ^ 5(x)[co2 |x|2 — (x*<o)2] dV.Z A

But

CO2 |x|2 — (x* CO)2 = CO

> | 2 0 0 X2 xy xz

0 |x|2 0 (O' - (0 yx yi yz

_ 0 0 |x|2_ _zx zy z2_

CO

hence

d K = - « ( x )<d

-xz

CO ' d V .

y2 + z2 —xy

— yx z2 + x2 —yz

—zx —zy x2 + y2_

The formula for K follows by integrating over D.

Applications

Geometric symmetries simplify calculation of the moments and products of inertia. For example, suppose that a solid is symmetric in the x, y-plane. Recall this means that whenever a point (x, y , z) is in the solid, then so is (x, y, —z), and 5(z, y , —z) = 8(x, y } z) . Then I lx = 0, because the contribution

y, z) zx dV

at each point (x, y, z) above the x, y-plane is cancelled by the contribution

f>(x> Vi ~ z ) ( ~ z ) (x) dV = — 8(x, y , z)zx dV

at the symmetric point (x, y, —z) below the x> y-plane. Likewise Iy, = 0 under the same symmetry condition. For example, a uniform right circular cone with axis the 2-axis has all products of inertia 0, since it is symmetric in two coordinate planes.

EXAMPLE 5.1

Compute the moments and products of inertia for a uniform sphere of radius a and mass M with center at the origin. Measure length in cm and mass in gm.

Solution: Let 8 denote the constant density; then M = 47ra35/3 gm. Because the sphere is symmetric in each coordinate plane, all products of inertia are 0.

By symmetry, I xx = I yy = I zz. I t appears most natural to use spherical

coordinates and compute I zz:

I zz = 8 J J J ( x2 + y 2) = 6 (p2 sin2 0 cos2 0 + p2 sin2 0 sin2 0) dF

(p2 sin2 0) p2 sin 0 dp d<f) dd


- // /= 6 / p4 dp f sin3 0 d0 f

J o »/o */osin3 <t> d<t> I * I Ma*.

15 5

Answer: I** = = Izx = 0,

iza; = I y y = / = T M<Z2 gm-CHl2.5

EXAMPLE 5.2

Find the products of inertia of the first octant portion of the sphere of Example 5.1.

Solution: By symmetry I xy = I yz = I zx. Choose spherical coordinates and compute that product of inertia whose formula seems the most symmetric; this is Ixyj since the 2-axis is special in spherical coordinates:

Ixy = —8 J J J xy dx dy dz

= —8 J J J (p2 sin2 0 cos 6 sin 6) p2 sin 0 dp d<j> dd

r a ric/2 rir/2

= —8 p4 dp sin30d0 / cos 0 sin 0 ./0 ^0 ^0

But

M = - ( - Tra?8 ) = - tta35. 8 \3 / 6

Hence I xy = — 2Ma2/5ir.

Answer: I xy = I yz = I zx5w

Ma2 gm-cm2.

N ote o n u n i t s : The unit of work, or energy, in the CGS system is1 erg = 1 dyne-cm. Remember 1 dyne = 1 gm-cm/sec2.

EXAMPLE 5.3

The solid of Example 5.2 rotates with angular velocity = (<o/\/3) (1, 1, 1). Find its kinetic energy.

Solution: By symmetry, the moments of inertia are f those of the full sphere, and the mass M is | th a t of the full sphere. By Example 5.1

2Ix x = I yy ~ I zz = ~ M a 2,5

where M denotes the mass of the first octant portion of the sphere. Since cox = o)y = co 2 = co /\/3 , the formula for kinetic energy yields

K = \ (3 J„ + 6 /„ )

- K i X ? " -

Parallel Axis Theorem

Take an axis 0 anywhere in space (not necessarily through the origin). Suppose a rigid body D rotates about this axis with angular speed co. See Fig. 5.3. The speed a t each point x is coBx, where B x is the distance from x to the axis 0. Hence the kinetic energy is

K = - Ip co2, where Ip = J J J Bx28(x) dV.D

This defines Ip, the moment of inertia of D about the axis (3. The Parallel Axis Theorem allows us to compute Ip in terms of the moment of inertia about a parallel axis through the center of gravity.

Parallel Axis Theorem If a is an axis through the center of gravity of D, and (3 is an axis parallel to a, then

Ip = I a + M d 2,

where d is the distance between the axes and M is the mass of D.

^ Mo2] = i (1 - M a W .57T / 5 \ 7T/

Answer: K = ~ M a2co2 erg.


The proof of this result in the general case is not hard, bu t involves some technicalities. So let us content ourselves with the special case in which x = 0 and a is the 2-axis. Suppose ft passes through (a, b, 0). See Fig. 5.4.

F ig . 5.4

For each point x of D, the distance A x of x from a is given by

A x* = X1 + y1 = I (x, y ) I2,

and the distance Bx of x from 0 is given by

B x2 = (x - a )2 + (y - b )2 = | (x, y) - (a, 6 )|2

= IO , y) |2 - 2(o, b) • (x, 2/) + I (a, 6 )|2

= A x2 — 2 (a, 6) • (s, 2/) + d2.

M ultiply by the element of mass, 5(x) dV, and integrate. Result:

Ip = I a — 2 (a, b) • (mx, my) + M d 2.

But mx = my = 0 since x = 0. Hence Ip = I a + Md2.

EXAMPLE 5.6

Find the moment of inertia of a uniform sphere of radius a and mass M about an axis tangent to the sphere.

Solution: From Example 5.1, the moment of inertia about any axis through the center (e.g.) is 2M a 2/5. The distance from a tangent axis /3 to the center is a, so the Parallel Axis Theorem implies

2 7Ip = - M a 2 + M a 2 = - M a 2.

5 5

Ansiver: ~ M a 2. 5

EXERCISES1. A uniform cylinder of mass M gm occupies the region — h < z < /&, x2 + 1/2 < a2,

distance in cm. Find its moments and products of inertia. [Hint: Use cylindrical coordinates.]

2. (cont.) The cylinder rotates with angular speed co rad/sec about an axis through 0 and (1, 1, 1). Find its kinetic energy.

3. Find the moments and products of inertia of the uniform hemisphere p < a, z > 0 of mass M.

4. The axis of a uniform right circular cone of mass M is the 2-axis; its apex is at 0 and its base of radius a is at z — h > 0. Find its moments and products of inertia.

5. Find the moments and products of inertia of the uniform rectangular solid of mass M bounded by the planes x = ± a , y = ± b , z = =tc.

6. Find the moments and products of inertia of the uniform rectangular solid of mass M bounded by the coordinate planes and the planes x = a, y = b, z = c.

7. The circular disk y = 0, (x — A )2 + z2 < a2, 0 < a < A is revolved about the 2-axis. Suppose the resulting anchor ring (solid torus) is a uniform solid of mass M. Find I zz.

8. In Example 5.1, use symmetry to prove, without integrating, that

- i n p2 dV. Then evaluate the integral.

9. Find I zz for the uniform solid paraboloid of revolution of mass M bounded by az = x2 + y2 and z = h.

10*. Find the moments of inertia of the uniform solid ellipsoid of mass M ,

i 2 _l t . i zl = i a2_r 62_r c2

[Hint: Stretch a suitable amount in each direction until the solid becomes a sphere; set x = ua, y = vb, and z = wc.]

11*. Find I zz for the uniform solid elliptic paraboloid of mass M bounded byx2 . y2z ------- b 7" , z = h.a b

[Hint: Use the result of Ex. 9 and the hint of Ex. 10.]12*. Find I zz for the uniform solid of mass M bounded by the hyperboloid of revolution

z2 1— = — {x2 y2) — 1 and the planes z = dzh.cl or

13. Find the moments of inertia of the uniform solid of mass M in the region— a < x, y, z < a, x2 + y2 + z2 > a2.

14. Find the moment of inertia of a uniform solid right circular cylinder about a generator of its lateral surface. (This is important in problems concerning rolling.)

15*. Suppose a rigid body R is rotating about an axis through 0 with angular velocity

co. Show that the angular momentum, J = / / / [5 (x)x X v] dV (v is velocity)

is given by

J = (co*, coy, u z)

Ixx Ixy Ix

lyx I yy

— zx I zy

The definitions of moments and products of inertia can be easily modified to apply to wires and laminas rather than solids. Such moments can be computed directly, or sometimes by limit arguments.

16. Find the moments of inertia of the uniform circular disk x2 + y2 < a2, z = 0 of mass M. (The units are gm and cm.) [Hint: Let h ---------> 0 in Ex. 1.]

17. A uniform wire of mass M lies along the 2-axis from z = — h to z = h. Find its moments of inertia. (The units are gm and cm.) [Hint: Let a -------- > 0 in Ex. 1.]

18. A uniform rod of length L and mass M lies along the positive z-axis on the interval a < x < a + L. Find Iyy.

19. Find the moments of inertia of a uniform spherical shell (surface) of radius a and mass M about an axis through its center.

20. A uniform circular wire hoop x2 + y2 = a2, z = 0 has mass M . Find its moments of inertia.

21. Find I zz for the toroidal shell, the surface of the solid torus in Ex. 7.22. Find the moments of inertia of the uniform cylindrical shell x2 + y2 = a2,

—h < z < h.

12. Integration Theory

1. INTRODUCTION

In this chapter we shall give the theoretical background for the techniques of multiple integration developed in the last two chapters. We shall concentrate mainly on the double integral, both because it is easier to visualize the plane than space, and because no really new ideas are involved in doing the three- dimensional theory once we m aster the two-dimensional case.

First we must give an accurate definition of the integral, one broad enough to apply at least to continuous functions. The idea will be to define the integral for step functions in an obvious way, then to obtain the integral of a more general function by squeezing the function between step functions.

Let us review briefly how we do this on the line. We work on a fixed closed interval I = [a, 6]. A partition of I consists of a finite increasing sequence of division points (not necessarily equally spaced):

A step function is a function s(x) on [a, 6] that is constant on the open intervals of a partition (Fig. 1.1). Thus

a = xo < Xi < < • • • < xn = b.

s(x) = Bi for Xi-1 < x < Xi.

y

F ig . 1.1x

a — Xo X] X'i xz xAxh = b

The integral of s(x) is defined byn

s(x) d x = s(x) dx = Bi(xi — Xi-i)\ J a

Now let f (x ) be any function with domain I, arbitrary except we insist th a t / is bounded, \ f ( x ) \ < M . We call / integrable if for each e > 0 there

458 12. INTEGRATION THEORY

exist step functions s and S on I such th a t

(1) s(x) < f ( x ) < S(x ) for all x £ I

(2) / S(x) dx — s(x) dx < e.J | J j

Next we show th a t if / is integrable, then there exists a unique number, called

/ f (x) dx,

such th a t

J s(x) dx < j f (x) dx < J S(x) dx

for all step functions s(x) and S(x) th a t satisfy (1).We prove th a t a continuous function is integrable, using its known uniform

continuity to approximate it by step functions.This then is the way integrals are developed on the line; we shall use the

same approach in the plane. I t is by no means the only way to achieve a theory of integration; we like it because it goes quickly, by easy natural steps. Also, the definitions, theorems, and proofs, with only minor modifications, cover the one-variable case. Therefore you do not have to know the theory of integration on the line to read this chapter.

2. STEP FUNCTIONS

We fix a closed rectangle I in the plane,

I = { (x, y) \ a < x < b, c < y < d)

and denote its area by 111. Thus

|l| = (b — a) (d — c).

A partition of I is a decomposition of I into a finite number of sub-rectangles by lines parallel to the axes. Thus we have partitions of [a, 6] and [c, d ],

a = x0 < Xi < x2 < • • • < xm = b,

c = 2/o < Vi < 2/2 < • • • < yn = d,

defining the partition of I. See Fig. 2.1. We denote the individual rectangles of the partition by

I jk = { ( z , y) I Xj- 1 < x < Xj, yk-x < y < yk),

j = 1, 2, •• •, m, k = 1, 2, • • •, n.

2. Step Functions 459

2/» = d

2/n-l

y*2/i

Jl2_

«u In

a = x o £2

F ig . 2.1 a partition of I

Xm-l

Step Function A step function on I is a real-valued function s(x, y) with domain I such th a t(1) s is bounded;(2) there is a partition of I such th a t s is constant on the open interior

of each rectangle I# of the partition. T hat is, s ( x , y ) = Bjk for xj-i < x < xj and yk. i < y < yk.

Note th a t s is unrestricted on the division lines of the partition (except th a t s m ust be bounded).

If s(x, y) is a step function relative to some partition, then clearly it is a step function relative to any finer partition, a partition with more division lines (Fig. 2.2).

(a) partition (b) finer partition (refinement)

F ig . 2.2


We want to handle sums, products, and other combinations of step functions. B ut two different step functions m ay be associated with different partitions, so what do we do? We superpose the partitions (Fig. 2.3); then both functions are step functions relative to the new finer partition.

+

(a) partition (b) second partition (c) superposed(common refinement)

F ig. 2.3

We can use this construction to prove routinely the following result (see Exs. 1-4).

Theorem 2.1 Let Si and s2 be step functions on I and k a constant. Then each of the following is a step function:

Si + s2, ksij | Si |, S1S2

S(x, y) = max{si(z, y), s2(x, y ) ) ,

s ( x , y ) = m in{si(z, y), s2(x, y ) ) .

We are now ready to define the integral of a step function.

Integrals of Step Functions

Integral of a Step Function Let s be a step function on I relative to a partition of I into rectangles !jk. Suppose s(x, y) = B jk for (z, y) in the interior of I jk. Define

J J s d x d y = ^ Bjk |lyfc|.I j,k

(Recall th a t 11 1 denotes the area of I#.)This definition looks innocent, yet it contains two subtle points. The first is

th a t the same step function s(x, y) may be associated with two different partitions, Pi and P 2. Do we get the same answer when we compute the integral relative to Pi and relative to P 2? Let us show th a t we do.

Suppose th a t J is a sub-rectangle of Pi and th a t s(z, y) takes the constant value Bj on J. If P 3 is the superposed partition (common refinement) of Pi and P 2, then J decomposes into a number of sub-rectangles J »•/ of P3, and their areas add up to the area of J. On each, s(x, y) has the same constant value Bj.

r i r1 1 1i 1 t1 1 11 1 1 i i 1 1 1 1

= Il

- 1—-1— 1—

2. Step Functions 461

Therefore,

J J s(x, y) dxdy = ^ Bj |J| = ^ Bj ^ |J<3|( f t ) j

J i,i ( f t )By the same reasoning,

J J s(x, y) dxdy = J J s(x, (ft) (ft)

?/) dx d?/.

Hence,

J J s(x, y) dx dy = J J s(x, y) dx dy.(p i) ' ' ' (p 2)

The second subtle point is this. Is there anything to be gained by using more general partitions of I into sub-rectangles than those arising from partitions of [a, 6] and [c, d~\l The answer is no, as Fig. 2.4 indicates.

Now we state the main elementary properties of the integral of a step function. Except for the last two, their proofs are easy and are left as exercises.

Theorem 2.2 Let sh s2, s be step functions on I. Then:

(1) J J (si + s2) dx dy = J J Si dx dy + J J s2 dx dy. i i i

(2) J J ks dxdy = k J J s dx dy (k any constant), i i

(3) If s > 0 on I, then J J s dx dy > 0.i

(4) If Si < s2 on I, then J J sx dx dy < J J s2 dx dy.i i

(5) If I is partitioned into sub-rectangles I#, then

J J s d x d y = I f f s dx dy.I j,k I jk

(6) max | J J si dx dy , J J s2 dx dy J < J J max{si, s2} dx dy.i i i

J J minfsi, s2} dx dy < min I^JJ sx dx dy, J J s2 dx dy j .(7)

(a) more general partition (b) the segments extendedF ig . 2.4

To prove (6), note th a t Si < max{$i, s2} on I, hence

/ / - / / max{si, s2}

by (4). Likewise

consequently

max

J J Si < J J max {Si, S2J ;

The proof of (7) is similar.

R em ark: Note the meaning of (5). The given partition of I is not assumed to be related to a partition attached to the given step function.

Iteration

We complete the story of step functions with the iteration formula.

Theorem 2.3 Let s(x, y) be a step function on 1, attached to the partition

a = x0 < xi < • • • < xm = b, c = y0 < yi < • • • < y n = d.

Define t(x) by ,f d/ s(x, y) dy for x xj

t(x) = < J c

arbitrary for x = Xj.

Then t(x) is a step function on [a, b] and

J J s dx dy =1

= J td x - J ^ J s(x, y) d y j dx.

Proof: As usual, let

I/* = { (x, y) | Xj. i < a; < xjy y k_x < y < y k)

be the sub-rectangles of the partition and let s(x, y) = B jk for (x, y) in the open interior of ly*. For xy_i < x < Xj we have

3. The Riemann Integral 463

n

= J' s(x, y) dy = ^ B jk(yk - yk- 1).t(x)k = 1

This proves th a t <(x) is constant on the open interval (zy_i, Xy), hence is a step function on [a, 6]. By definition,

m n

J t(x) dx = V ( V B jk(yk - 2/t-i)^ (Xj - Xj.° ;=1 A: =1

= ^ ^ Bjk I lyjbj = J J s dx dy.

i)

EXERCISESProve in detail Theorem 2.1 for1. Si + S23. |Sl|

Prove Theorem 2.2:5. part (1)7. part (3)9. part (5).

3. THE RIEMANN INTEGRAL

Let / be a bounded real-valued function with domain I. We shall try to squeeze / between step functions whose integrals are arbitrarily close to each other. This is not possible for every function, bu t when it is, we are able to define the integral of /.

Since / is bounded, there exist step functions s and S such th a t s < / < S on I. Indeed, if | f\ 0, there exist step functions s and S on I such th a t

(1) s < f < S on I,

(2) J J S dx dy — J J s dx dy < e.

The first thing we m ust do is show th a t if / is an integrable function, we can assign t o / a unique number, its integral.

Theorem 3.1 Let / be integrable on I. Consider all step functions s and S such th a t

s < f < S on I.Then

sup J J s dx dy = inf J J S dx dy.

The common value of the sup and inf is called the integral of / on I, and is written

I fi / dx dy.

Proof: If s < / < S on I, then

J J s dx dy < J J S dx dyi i

by Theorem 2.2, part (4). Hold S fixed and let s vary. I t follows th a t

sup J J s dxdy < J J S dx dy.

The left side of this inequality is independent of S. Now let S vary; it follows th a t

(1) sup / / s dx dy < inf J J S dx dy.

This inequality holds for any bounded function. Next we show th a t if / is an integrable function, then the reverse inequality also holds, so (1) becomes an equality.

Let / be integrable. Given e > 0, there exist step functions s0 and S 0 such th a t So < f < So and

J J Sodxdy < e + J J s0 dx dy.

inf J J S dxdy < J J So dx dy < e + J J s0 dx dy

< e + sup J J s dx dy.

Therefore

i

This is true for all e > 0, hence

(2) inf J J S dxd y < sup J J s dx dy.

By combining (1) and (2) we have the desired equality.

R e m a r k 1: The theorem implies th a t if / is integrable and s < f < S,

then J J s < J J f < J J S. This relation determines J J f. In fact, J J f is

the only number between all pairs J J s and / / *

R em a r k 2: The integral we have defined is called the Riemann (double) integral, and integrable functions are also called R-integrable. I t is possible to define integrals th a t apply to more functions—the most im portant is the Lebesgue integral—but their theories are more advanced.


Theorem 3.2 Let / be a step function on I. Then / is integrable and the i?-integral of / is the integral defined in Section 2.

In particular, if / = B is constant on I, then / / f — B |l|.

Proof: Given a step function / and e > 0, take S = s = f in the defini

tion of integrable function. Then s < f < S and / / • - / / S, etc.

Continuous Functions

I t is an im portant practical m atter th a t all continuous functions are integrable. This is certainly not obvious from the definition.

Theorem 3.3 Each continuous function on I is integrable.

Proof: Let / be continuous on I. By the theorems in Chapter 6, Section3, / is bounded and is uniformly continuous on I, since I is a bounded and closed set. Because / is bounded, we are allowed to ask whether or not it is integrable. Thus let e > 0. Since / is uniformly continuous, there is 5 > 0 such th a t | / (x ) — / ( z)| < e/\\\ whenever |x — z| < 8.

Choose a partition with sub-rectangles I # so small th a t |x — z| < 8 whenever x and z lie in the same I#. Set

mjk = in f{ /(x ) | x 6 Ijk), M jk = su p { /(x ) | x £ ly*}.

Then mjk < /(x ) < M jk for all x 6 Iy*, and M jk — mjk < €/|l|.

Now define step functions s and S by requiring th a t s(x) = mjk and S (x) = Mjk for x in the open interior of I#, and s(x) = S(x) = f (x ) for all points x on the division lines of the partition. Then s < / < S on I and

J J S dx dy — J J s dx dy = J J (S — s) dx dy

hi i i

< I / j ^ d x d y = Tjj-|l| = €. r

Therefore / satisfies the definition of an integrable function.

Sums and Multiples

Theorem 3.4 Suppose / i and / 2 are integrable on I and Ci and c2 are constants. Then Ci /i + c2 / 2 is integrable and

J J (cifi + Cift) dx dy = ci J J fi dx dy + c2 J J / 2 dx dy. i i i

Proof: We divide the proof into several steps.

(1) If / is integrable, then —/ is integrable and J J ( —/) = — J J f.

For let e > 0. Choose s and S so s < / < S and / / - / / s < €. Then

— S < —f < —s and ( - S ) = / / - / / s < e, etc.

(2) If / is integrable and c > 0, then cf is integrable and J J (cf) — c J J f.

This time choose s and S so s < / < S and J J S — J J s < e/c. Then

cs < cf < cS and J J cS — J J cs = c ( / / - / / • ) < e, etc.

(3) If / is integrable and c is a constant, then cf is integrable and

/ / < < « - « / / / .

We have just disposed of c > 0. If c < 0, then cf = ( — c) ( —/) does it in two steps. The case c = 0 is obvious.

(4) If fi and / 2 are integrable, then fi + / 2 is integrable and

/ / ( / ■ + /O - / / ' ■ + / / / -

Given e > 0, choose step functions satisfying

si < jfi < Si, $2 < J2 < S2,

Then Si s2 < /1 -(- / 2 < Si + S2 and

/ / (A + S.) - j f <* + »,) - ( / / S. - / / « ) + (' J j S , - j j a )

€ , €

2 2 = * ’

etc. Now given integrable functions/1 a n d /2, it follows from (3) th a t C1/1 and (hfi are integrable, and from (4) th a t Ci/i + c2/ 2 is integrable. Again using(3) and (4) we have

/ / ( * /> + <*/’ ) = c> / / / . + « / / &

Inequalities

Theorem 3.5 Suppose / and g are integrable on I.

(1) If / > 0 on I, then J J f dx dy > 0.1

(2) If / < g on I, then J J f dx dy < J J g dx dy.1 1

(3) If m < f < M on I, then m |l| < J J f dx dy < M |l|.1

Proof: For (1), simply observe th a t s0 = 0 is a step function and s0 < f

on I. By definition, J J s < J J f for any step function s satisfying s < /, in

particular for s0. Now (2) follows by applying (1) to g — f. Finally, (3) is immediate from the choice of s = m and S = M o n l.

For the next theorem we need a simple fact, which we state as a lemma.

Lemma Let Ai, A 2, Bi, B2 be four real numbers. Then

max{i?i, B2\ — max{Ai, A 2) < max{l?i — A h B2 — A 2).

If in addition Bi > Ai and B2 > A 2, then

max{J5i, B2) — m axjAi, A 2\ < (Bi — Ai) + (B2 — A 2).

Proof: For i = 1, 2,

Bi = Ai + (Bi — At) < max{Ai, A 2) + m ax{£i — Ai, B 2 — A 2}.

Here max{i?i, B2) < (right-hand side), and the first inequality follows.If Bi > A i and B 2 > A 2) then Bi — Ai > 0, so B 2 — A 2 < (Bi — A x) +

(JB2 — A 2). Likewise Bi — A\ < (B\ — A\) + (B 2 — A 2), so

maxfjBi — Ai, B 2 — A 2\ < (Bi — Ai) + (B2 — A 2).

The second inequality now follows from the first.

Theorem 3.6 Suppose fi and f 2 are integrable on I. Then max{ f h f 2} and min{ f i y f 2} are integrable and

(1) max I f f f i dx dy , J J f 2 dx dy\ < J J max{ f h f 2} dx dy ,i i i

(2) J J min{ f h f 2) dx dy < min ^ J J fi dx dy , J J f 2 d x d y .

Proof: Let e > 0 and choose step functions satisfying Si < f i < S i9

< f 2 < S 2y J J Si — J J Si < and J J S2 — J J s2 < %e. Set S =

max {Si, S2] and s = max{«i, s2}, both step functions. Clearly s < m axj / i , / 2} < S.

We shall prove first th a t J j S — J J s < e. This will imply th a t max {/i,/2

is integrable. By the lemma,

S - s < (Si - si) + (S2 - * ) .

By (4) of Theorem 2.2,

J J (S — s) < J J (Si — Si) + (S2 — s2) < \e ~b he =

Hence F = max) /i, / 2} is integrable. Since fi < F and / 2 < F, we have

/ / » « / / F and / / - / / F by (2) of Theorem 3.5. Relation (1) fol

lows immediately.Relation (2) is equivalent to (1) by the equation

min { fi, /2} = - m a x { - / i , —/2}.


Proof: To prove | / | is integrable, we use the relation (easily verified)

| / | = maxj /, 0} - min{ /, 0}

and the last theorem. Since / < | / | and — / < | / | , by Theorem 3.5 we have

j j f< j j l / l a n d - / / / < / / I / I H e n c e \jjf\< j j | / | .R e m a r k : W hy not prove this theorem directly, starting with s < f < S,

etc.? Because even though we find easily th a t | / 1 < m axj |s|, \S \}, it is hard to find a step function below | / |. Certainly min{|s|, \S\} < \ f | is wrong!

Additivity of the Integral

Theorem 3.8 Suppose / i s a bounded function on I, and suppose I is partitioned into sub-rectangles I#. Then / is integrable on I if and only if / is integrable on all I#. If so, then

/ / f d* dv - X I f f dx dy.j ,k I jk

Proof: The theorem is true when / is a step function, by Theorem 2.2(5). We shall use this fact.

First suppose / is integrable on each of the mn sub-rectangles I#. Let c > 0. Choose step functions satisfying sjk < / < Sjk on I jk and

(1)

Define S and s on I by

S = s = f

f i Sjk I I S i k < m n ■Ijk I jk

S = Sjk and s = Sjk

on the dividing lines,

on the interior of I/*.

Then s < / < S on I and

» I I > - l l - 111- I I I -I I j , k I jk j ,k I jk

■ i n - i n -j ,k I jk j ,k I jk

■ m > - n - ) ‘ U -j ,k I jk I jk j ,k

Therefore / is integrable on I, and

(3) f f s < f f , < f f S .I I

B ut s < / < S on each I so

» a - - i n “ i a - ‘ i a - i i -1 j ,k I jk j ,k \ jk j ,k \ jk I

Compare the results of (2), (3), and (4):

/ / • ^ / / « / / ‘ - ’ i i jk iI t follows readily th a t

\ U ’ - i i i ' \ < - i i jk

Since e is arbitrary, (6) implies

n > - i i bI I jk

Now to prove the rest of the theorem, suppose / is integrable on I. I t is enough to prove / is integrable on each I#, for then the first part of the proof will give the sum formula.

Let e > 0. Choose s and S so s < / < S on I and 8 - S < €./ / - ■ / /

Then s < f < S on \Jk and

f f s - f f . z f j v - . x .\ j k Iji I

Hence / is integrable on I#, which completes the proof.

Products


Theorem 3.9 If / and g are integrable on I, then fg is integrable.

Proof: The identity

fg = \ [(/ + sO2 - ( /- ?)2]shows th a t it is enough to prove th a t the square of an integrable function is integrable. Let / be integrable. Then f 2 = | f \2 and | / | is integrable. Hence we may assume f > 0. Also / is bounded, so 0 < / < B.

I fI f -Let 6 > 0. Choose step functions 0 < s < f < S < B and

s < c/2B. Then s2 < f 2 < S 2 and

S 2 - s2 = ( S + s) (S - s) < 2 B ( S - s),

hence J J S2 — J J s2 < 2B J J (S — s) < c. This p roves/2 is integrable and

completes the proof.

R e m a r k : A much more general theorem is true, bu t its proof is very h a rd : If H(z, w) is continuous for all z, w and if / and g are integrable on I, then the composite function

h(x, y) = H [ f O , y), g(x, y ) ]

is integrable. The special case H ( z } iv) = zw implies Theorem 3.9.

EXERCISES

1. Suppose a function / on I has the following property: if c > 0, there exists a step function s such that

| f(x, y) — s(x, y )| < c for all (*, y) € I.

Prove / is integrable.

2. Suppose g(x) is integrable on [a, &]. Set / (x,y) = g {x) on I: a < x < b, c < y < d. Prove/is integrable on I.

3. Define / by f (x ,y ) = 1 if £ and t/ are both rational, / = 0 otherwise. Prove th a t / is not integrable on any rectangle.

4. Suppose / is integrable on a < x < b, c < y < d. Define g(x,y) = f (y ,x) .

Prove that g is integrable on c < x < d, a < y < b. Find JJ&-5. (cont.) Suppose / is integrable on 0 < x, y < 1 and f ( x , y ) - { - f ( x , y ) = 0. Prove

l f = o.


/ / '6*. For each n, divide I into n2 equal rectangles IC h o o se any point i/yjfc) 6 ly*

and set

Prove An -------- > J j f if / is integrable. [Hint: First prove it for step functions.]

7. Let f(x, y) = 0 on a rectangle I except at a single point p. Show th a t/is integrable

and J J f dx dy = 0.I

8*. Let/(:r, y) be defined in the square I with vertices (0, 0), (0, 1), (1, 1), (1, 0) by

f f a y ) =

f f a h yjk). i,k

2) 3) 4>

0 otherwise.

Prove that / is integrable and f dx dy = 0./ / .I

If-9. Suppose/(x, y) > 0. Exercises 7 and 8 show that JJ f dx dy = 0 does not implyI

that f(x, y) = 0 at each point of I. Prove, however, that if / is continuous, then

/ / f dx dy = 0 does imply that f (x , y ) = 0 at each point of I.

4. ITERATION

Here is the main theorem on iteration of integrals over a rectangle I = { y) | a ^ x ^ b, c < y < d).

4. Iteration 473

Theorem 4.1 Let / be a bounded function on the domain

I = { (x, ?/) \ a < x < b, c < y < d).Suppose:(1) / i s integrable on I.

(2) For each x £ [a, 6], with possibly a finite set of exceptions, the integral

g(x) = J' f ix , y) dy

exists.

Then g is integrable on [a, 6] and

f bgd x = f ( f / (* ,» )J a J a \ J c

< * ) * - / / / dx dy.

Proof: Let « > 0. Choose step functions satisfying s < f < S and

/ / - / / s < e. Define step functions

t ix) = f ' s i x , y ) d y , Tix) = J S i x , y ) d y .

Then for each x, except perhaps the finitely m any exceptions, the inequality 8 < f < S implies

t{x) < g(x) < T{x) .By Theorem 2.3,

J T(x) dx — J t(x) dx = J J S dx dy — j j s dx dy < c.

This says th a t g{x) is integrable on [a, 6]. I t also says th a t

J J s dx dy < J g(x) dx < J J S dx dy

for each choice of s and S. B ut the only number th a t satisfies these inequalities

i s / / / .for all s and S is /. I t follows th a t

completing the proof.

J gix) dx = J J f d x d y ,


R e m a r k : Hypothesis (1 ) is not sufficient for this theorem. There are examples of functions th a t satisfy (1) bu t not (2), so g(x) is not even defined, let alone integrable.

Corollary If / is continuous on I, then

j j f d x d y = a i : f (x , y) d y j dx = M f ix , y) dy.

Proof: By Theorem 3.3, / is integrable on I. Because/ is continuous on I, for each x0 the function f ( x 0, y) is continuous on [c, d], hence integrable. Thus the hypotheses of Theorem 4.1 are satisfied, so the first equality follows. By symmetry so does the second.

Non-rectangular Domains

This completes the theory of the double integral over a rectangle. Our next job is to extend this theory to functions with non-rectangular domains.

The first problem is to decide which sets we shall allow as domains. This is not an easy question; the complete theory is beyond the scope of this course. We want a theory th a t includes a t least the domains th a t arise in practice, b u t not so many more th a t the technical details become oppressive.

The kind of domain D we want can be described accurately this way.(1) I t is a closed bounded subset of R2. (2) I t is connected, i.e., any two points of D can be connected by a curve in D. (3) I t has m any interior points (that is, each point of D is either an interior point or a limit of a sequence of interior points). (4) Its boundary consists of a finite number of arcs th a t have continuously turning tangents and a finite number of corners (Fig. 4 .1a).

(a) smooth boundary (b) convex domain (c) jigsaw puzzleexcept for a piecefew corners

F ig . 4.1 domains of integration

M any of the domains we deal with are convex (Fig. 4.1b). A domain is convex if it contains the line segment connecting any two of its points. Probably the most complicated domains arising in practice are shaped like the pieces of a jigsaw puzzle (Fig. 4.1c). Still, we can define the double integral

4. Iteration 475

of a continuous function on such a domain. We first cut the domain into a finite number of simple domains, then use iteration and the techniques developed in Chapter 10 to evaluate the integral on each of these.

The double integral can be defined on domains th a t have area in a sense we define now.

Area Let D be a closed subset of the rectangle I. Define the characteristic function kD of D by

The set D is R-measurable if kD is integrable on I. If so, the area of D is defined by

R em a rk 1: Note th a t an /^-measurable domain is automatically a closed set by this definition. Nevertheless, we shall stress this closure in the following theorems.

R em a r k 2: I t will be left as an exercise to prove th a t if I and J are two rectangles, both containing D, then |D| is the same whether computed in I or in J. This is an easy, but necessary, result in order th a t |D| be properly defined.

The definition makes good sense. First it gives the right answer for any rectangle J in I (with sides parallel to the axes). Second, it is additive. This means th a t if two sets D and E are R-measurable and they do not intersect, then their union D U E is R-measurable and | DUE| = |D| + |E|. This statem ent is contained in the next theorem.

N o t a t io n : At this point it is convenient to introduce the symbol 0 for the empty set, the subset of I with no points. This set has the characteristic

function k 0 = 0, it is R-measurable, and | 0 | = J J 0 = 0. The statem ent

“ D and E have no common points” is abbreviated simply “D n E = 0 ” .

Theorem 4.2 Suppose D and E are closed R-measurable subsets of I.

(1) D D E is R-measurable.(2) D U E is R-measurable.(3 ) |D U E| = |D| + | E| - |D fl E|.

Area

1 if (x, y) £ D( x , y ) =

0 if (x, y) $ D.

Proof: First we observe th a t kDf]E = kDkE. For kDnE(x) = 1 precisely when x f D n E, th a t is, x £ D and x f E, which happens precisely when fcD(x) = 1 and feE(x) = 1, th a t is, fcD(x)fcE(x) = 1.

Now (1) follows because /cDflE is the product of integrable functions, hence integrable.

For (2) we need the relation

&dUE ~ — 1DflE*

I t is easily checked, because the right-hand side is 1 precisely when x is in D and not in E (1 + 0 — 0), or x is in E and not in D (0 + 1 — 0), or x is in both (1 + 1 — 1), th a t is, when x f D U E.

Now (3) follows from

|D U E| = J J /cDUE = J J kD + J J kE — J J kDf]E

= |D| + | E| — |D fl E|.

The next theorem includes a familiar situation, the area under a curve (the case c = 0).

Theorem 4.3 Let h(x) be a continuous function on [a, w ith c < h(x) < d. Define D by

D = {(x, y) | a < x < by c < y < h(x ) }.

Then D is R-measurable and'b

- rJ a[h(x) — c] dx.

F ig . 4.2 The set (shaded) below the graph of h is D.

4. Iteration 477

Proof: First suppose h is a step function on [a, &]. Then kD is a step function on I. See Fig. 4.2. Hence kD is integrable and

J J kD = J ( J kD(x, y ) dy'j dx.

For fixed x, we have kD(x, y) = 1 if c < y < h(x) and kD(x, y) = 0 if h(x) < y < d. Hence

rakf)(x, y) dy = h(x) - c,rJ c

SO

f h = i : \_h(x) — c] dx.

Now suppose h is continuous on [a, 6]. Then D is closed (why?) and we m ust prove D is immeasurable. Given e > 0, there are step functions s and S on [a, 6] such th a t

c < s < h < S < d and f ( S — s) dx < e.J a

To estimate kD, we construct the sets

C = { (x, y) | a < x < 6, c < y < s ( x ) }and

E = { (x, y) \ a < x < b, c < y < S(x) } .

Clearly C C D C E, so kc < kD < kE. As we have seen, kc and kE are step functions, and

[aS(x) — c ] dx — J [ s ( z ) — c ] dx

= f [ S ( x ) — s ( x ) ' ] d x < e.J a

This proves th a t kD is integrable on I, hence D is R-measurable, and also th a t

J [« (« ) — c ] d x < J J kD d x d y < J [ S ( x ) — c] dx.

Since J J kD = |D|, this relation says th a t

for all s and S such th a t s < h < S. But the only number satisfying these in

e q u a litie s is J (h — c), hence |D| = J (h — c).

As a corollary, we obtain the area of a region bounded above and below by the graphs of two continuous functions.


Theorem 4.4 Suppose g(x) and h(x) are continuous functions on [a, 6] with c < g(x) < h(x) < d. Define D by

o = { (x, y) I a < X < b, g(x) < y < h ( x ) }.

Then D is R-measurable and

- f J a\_h(x) — g(x)~\ dx.

Proof: This is a consequence of the last theorem, applied twice, and the relation (Fig. 4.3)

D = { {x, y) | c < y < h ( x ) } n {(*, y) \ g{x) < y < d ) .

R e m a r k : The preceding results are adequate for proving th a t standard geometric figures, like polygons and circles, have area. W ith a little patience one can now derive the usual area formulas for triangles, parallelograms, etc.

Integrals

Let D be a closed subset of a rectangle I and let / be a bounded realvalued function with domain D. We seek conditions under which

J J f d x d yD

can be defined. A first modest requirement is th a t D should be R-measurable.

y - h(x)

4. Iteration 479

Then a t least

J J 1 dx dy = J J kD dx dyD I

makes sense. We now need a reasonable definition of integrability on non- rectangular closed domains.

Integrable Function Let D be a closed R-measurable subset of I, and let / have domain D. Define F on I by

F = f on D, F = 0 elsewhere.

We say / is integrable on D if F is integrable on I, and we define

J j f dxdy = J J F dx dy.D I

I t can be shown th a t this definition is independent of the rectangle I.

Theorem 4.5 Suppose D is a closed R-measurable subset of I, and suppose / is integrable on the rectangle I. Then / is integrable on D.

*Proof: The product F = kDf of two functions integrable on I is itself integrable. But clearly F = f on D and F = 0 outside D, so this integrable function F is exactly the F of the definition. H e n ce /is integrable on D.

The General Iteration Theorem

Theorem 4.6 Let g(x) and h(x) be continuous functions on [a, 6], with c < g(x) < h(x) < d. Define D as the subset of I bounded by the graphs of g and h:

D = { (x, y) | a < x < b, g(x) < y < h(x) }.

Let f(x, y) be continuous on D. Then / is integrable on D and

/ / ^ x ' v) d x d y = m : f ( x , y) d y j dx.

Proof: We begin by constructing a continuous function /* on I such th a t /* = / on D. The idea is to make /* constant on the vertical segments above and below D. Thus we set

f*(x,y) =

fix, g(x)2

f(x, y) for

f£x, h(x)~\

c < y < g(x)

g{x) < y < h(x)

„ h(x) < y < d.

By definition, /* agrees with / on D. The proof th a t /* is continuous is routine and is left as an exercise. By Theorem 3.3, /* is integrable on I, so Theorem 4.5 implies / is integrable on D.

Let F = f on D and F = 0 outside D. By definition,

J J f dx dy = J J F dx dy.

For each fixed x,

j: F(x, y) dy

exists because for fixed x,

F(xj y) =

0, c < y < g(x)

f (x, y)> g(x) < y < H x )

0, h(x) < y < d,

a step function on the two end intervals [c, g(x)~\ and [h(x), d] and a continuous function on the middle interval \_g(x) , h(x) ]. Therefore the hypotheses of Theorem 4.1 are both satisfied by /, so we conclude

j j f d x d y = f i F d x d y = I ' M ' F d y j dx.

Butrd rg{x) rh{x) rd rh(x)/ F dy = I F d y + F d y + F d y = f (x, y) dy ,

J c J c J g(x) J h(x) J g(x)

The theorem follows.

Theorem 4.7 Suppose D is a closed R-measurable subset of I, and suppose / is a continuous function with domain D. Then / is integrable on D.

A complete proof of this theorem, with the tools we have developed, is possible bu t arduous. Instead we shall give a short elegant proof, for which we borrow an im portant and plausible result from an advanced course on real functions, the Tietze Extension Theorem: There exists a continuous function /* with domain I such th a t /* = / on D. W ith this assumed, the proof of Theorem 4.7 is very short indeed: Since /* is continuous on I, it is integrable. Apply Theorem 4.5. Done!

R e m a r k : The Tietze Theorem stated precisely is this. Let D be a closed subset of R2 (also for R3, or Rn in general). L e t /b e a continuous real-valued function on D. Then there is a continuous function /* on R2 such th a t /* = / on D.

4. Iteration 481

Additivity

We close this section with a broad generalization of Theorem 3.8. F irst we require a more general definition of partition.

Partition Let D be R-measurable. A partition of D consists of R-measurable sets Di, • • •, Dn such tha t

(1) D = Di UD2 U • •• UDn,(2) |D» n Dj\ = 0 if i j .

Theorem 4.8 Suppose D is partitioned into Di, • • •, Dn, and su p p o se /is a bounded function on D. Then / is integrable on D if and only if / is integrable on each Dy. If so, then

// f dxdy = I f f f dx dy./ = 1 Dy

Proof: Let the characteristic function of Dy be /cy, an integrable function by hypothesis. Suppose / is integrable on D. Then fkj is integrable on D, which means / is integrable on Dy since Dy C D.

Now we compute Clearlym^ kj = kD + e,

3=1where the value of the error function e = e(x) is one less than the number of sets Dy containing x. Since e(x) > 0 only if x belongs to two or more of the Dy, and |D* fl Dy| = 0, we have

j j e < (m — 1) ^ |D{ n Dy| = 0.D ij*j

j j fe < (m ax /) j j e = 0, hence

Therefore

D D

But on D,n

f = fkD = ^ fkj + fe,j - 1

so

j = 1 Dy ; = 1 D

Now suppose / is integrable on each Dy. Then the same relation, / = 'Efk j + fe, implies / is integrable on D. This completes the proof.

1. In the proof of Theorem 4.1, we omitted to prove that g (x) is bounded. Supply this missing step.

2. Prove the assertion about two rectangles in Remark 2 on p. 475.3. Prove that the area of a line segment is 0.4. Prove, on the basis of Theorem 4.4, that a triangle with vertices 0, {x\, yi), and

(%2, 2/2), where x\ < 0 < X2, has area \ \x1y2 — £22/11-5. (cont.) Prove that a triangle with vertices {x\, yi), (z2, 2/2), (£3, 2/3) has area

Let us begin with a review of the one-variable situation.

Theorem 5.1 Suppose <t>(u) is continuously differentiable on [a, b] and suppose 4> (u) > 0. Let f (x) be integrable on [0 (a ) , 0 (b )]. Then /[0 (w )]0 '(w ) is integrable on [a, b] and

We shall not give a detailed proof of this theorem. We simply note th a t it is easily verified when / is a step function. The usual approximation—with some technical details—then proves it for / integrable.

Let us rather interpret Theorem 5.1 in such a way th a t its analogue in more dimensions, Theorem 5.2, will seem natural.

We write x = 4>(u), where u runs over the interval D = [a, b] and x runs over the interval E = [0 (a ), 0 (b )]. Since > 0, it follows th a t x increases as u increases. Thus 0 is a one-one mapping on the set D onto the set E, and we may write E = 0 (D ) . W ith this notation, the formula of Theorem 5.1 can be written

EXERCISES

Xi 2/1 1

12 X2 2 /2 1

Xz 2/3 16. (cont.) Prove A — bh for a parallelogram.

7*. Prove that the function/* in the proof of Theorem 4.6 is continuous.

5. CHANGE OF VARIABLES

5. Change of Variables 483

or with x = x(u) instead of x = 0(w)>

J f (x) dx = J f[_x(u)~\^du.

A similar formula holds in the plane and in space. There 0 is a one-one mapping of some domain D of, say, R2 onto another domain E. We shall need some information about such mappings, in particular, w hat replaces the derivative dx/du.

Jacobians

Let us note some facts about mappings. Let S be an open set in the u, v- plane and 0 a continuously differentiable mapping of S into the x, y-plane. W rite

x = x(u, v)0 :

y = y(u, v).

Thus the four partials dx/du , • • •, dy/dv exist and are continuous on S. We define the Jacobian of 0 to be the determ inant

I t is a continuous function on S.Suppose 0 maps S into an open set T and \p in tu rn is a continuously

differentiable mapping of T into the z , w-plane. The composite \p 0 0 is a mapping of S into the z, w-plane. Thus

z = z(x, y) = z [x (u } v), y(u, v)]

w = w ( x , y) = w[x(u, v), y ( u } v)"]-

By the Chain Rule, we have the m atrix relation

~dz dz dz dz dx dxdu dv dx dy du dv

dw dw dw dw dy dy__ du dv _ _dx dy _ Jdu dv_

Recall th a t the determ inant of a product of matrices is the product of their

determinants. Therefore


d(z, w) d(z> w) d(x, y) d(u, v) d(x, y) d(u, v)

Suppose in particular th a t <t> has an inverse and = <t>~l. Then </> is one-one on S onto T ; also \p is one-one on T onto S, and \p ° 0 = identity. Then

so

u = u(x, y)

v = v(x, 2/),

d(u, v) d(x, y) a(x, y) d(u, v)

u = u[x(u , v), y(u, y )]

v = v[x(u, v), y(u, v )],

1 0

0 1= 1.

A continuously differentiable mapping </> th a t has a continuously differentiable inverse is called a regular transformation. We have proved:

The Jacobian of a regular transform ation is never 0.

Suppose the domain S of 0, in addition to being open, is connected. T hat is, any two of its points can be connected by a curve in S. Then the Jacobian of a regular transformation has constant sign in S, either always positive or always negative. For otherwise it would have to be zero a t some point along a curve joining points where it had opposite signs. We shall concentrate on the case where the sign is always positive. We shall call a regular transform ation with everywhere positive Jacobian a proper transformation.

Examples

Example 1. Translation:

x = u v+ a d(x, y) 1 0

y = v + b d{u, v) 0 1

The transform ation is proper on the whole plane.

Example 2. Linear transformation:

x = an u + a-uV d(x, y) an a i2

y = <kiu + 022 d(u, v) 021 22

If we assume the determ inant is positive, then </> is a proper transform ation on the whole plane. If D is the unit square in the u, y-plane, then E = </>( D) is a parallelogram (Fig. 5.1). More generally, any rectangle in the u, u-plane is mapped to a parallelogram.


Example 3. Polar coordinates:

x = r cos 6 d(x, y ) cos 6 — r sin 6

y = r sin 6 d(r ,d) sin 6 r cos 0

This is a proper transformation on any domain D of the r, 0-plane th a t avoids the 0-axis. For instance, the rectangle D in Fig. 5.2 is mapped to the circular region E.

x = r cos 6F ig . 5.2

y = r sind

The last two examples illustrate the most common reason for changing variables, to change the domain of integration to a rectangle. Here is a further such example, but in 3-space.

Example 4. Spherical coordinates:/ #

x = p sin 0 cos 0

0 : y = p sin 0 sin 0

djx, y, z ) d(p, <t>, 6)

2 = p COS 0,

sin 0 cos 0 p cos 0 cos 0 — p sin 0 sin 0

sin 0 sin 0 p cos 0 sin 0 p sin 0 cos 0

cos 0 — p sin 0 0

sin 0 cos 0 cos 0 cos 0 — sin 0

p2 sin 0 = p2 sin 0.

cos 0 — sin 0 0

(The determ inant can be expanded easily by minors of the third row.) A rectangular solid D: 0 < p0 i < 0 < 0O <0 < 0i < 2r in p, 0, 0-space maps to a curved solid in x, y , 2-space.

77ie Main Theorem

Theorem 5.2 Let S be an open subset of the u, y-plane and 0 a proper transformation on S into the x, ?/-plane. Let D be an R-measurable subset of S and E = 0 (D ) the image of D under 0. Then:

(1) E is R-measurable.

d (x, 2/)(2) If / is integrable on E, then the product [ ( / ° 0) (u, v) ] —r1—- is in-

d(u , v)

tegrable on D and

J J /(* , y) d x d y = J J f [x( u , v), y(u, v) ] j du dv-E D

A proof is hard and beyond the scope of this course. We shall note one case where the theorem is quite plausible, th a t of the linear transformation on a rectangle (Example 2 above). First suppose / = 1 on E, and look carefully again a t Fig. 5.1. The formula in Theorem 5.2 boils down to

| E| = | D| (#11022 — O12O21) == O11O22 — 012 21 •

This is certainly correct, since the area of the parallelogram E, by vector algebra, is

I E| = | (an, 021) X (ai2, 022) | = an022 — 012021.

In fact any rectangle Dx in the u , v-plane is mapped to a parallelogram Ex with | Ei| = |Di| (011022 — O12O21). Therefore if f (x , y) is a step function on E relative to a division of E by lines parallel to the sides of E, then / is a step function on D and (2) in Theorem 5.2 is again true. This suggests an approximation procedure for proving (2) for any integrable function on E, but we m ust omit the details.

Once Theorem 5.2 is known for proper linear transformations on rectangles, it can be proved in general by another, deeper, approximation technique. This is to approximate the general proper transformation by a piecewise linear transformation; the Jacobian plays the role of an area distortion factor. The complete proof is quite formidable.

Polar Coordinates

The polar coordinate mapping

x = r cos 0 djx, y)• a d(r, 0)y — r sin 0 \ /

= r

is a proper transformation in the open set

S = {(r, 0) I 0 < r, 0 < 0 < 2 r ) ,

and (Fig. 5.3) it maps S onto

T = {(x, y) | y 9 0 or y — 0 and x < 0}.

F ig . 5.3 polar coordinate transformation

There is a problem with r = 0 because m any typical integration problems include r = 0. See Fig. 5.4.

F ig . 5.4 including the origin

The problem is solved by deleting a small circular sector and taking limits:

y) dx dyJ J f (x , y) dx dy = lim J JE €"*° E, r >«

-1“ I fD, r > e

f dx dy

f ( r cos 0, r sin 0) r dr dd = J J f r dr dd.

The process works because the excluded region (circular sector in E, rectangular strip in D) has area th a t goes to 0 as e -------- > 0.

Similarly we can take care of a domain E th a t goes completely around the origin, so th a t it hits both horizontal boundaries of S in Fig. 5.3. In summary, the formula

fff d x d y - jf. d(x, y)d(r, e) -ff>f ( r cos 6, r sin 6) r dr dddr dd

holds without restriction on the R-measurable domain D.Analogous considerations apply to the transform ation from spherical

to rectangular coordinates:

x — p sin 0 cos 6

y = p sin 0 sin 6

2 = p cos <t>

a(s, y, g) d(p, , 6)

p2 sin 0.

6. Applications of Integration 489

EXERCISES

Compute the Jacobian and prove that the mapping is a proper transformation:1. x = 2u + v, y = u + 2v, all (u , v)2. x = u v, y = uv, u > v3. x = u + v, y = u2 + v2, v > u4. x — u, y = uv, u > 05. x = u, y — uv, z = u > 0, v > 06. x = vw, y = wu, z — uv, u > 0, > 0, w > 0.7. The transformation of inversion in the plane is defined by x -------- > x/|x|2 for

x 9 0. Find its Jacobian.8. (cont.) Generalize to R3.9. What domain in p, <j>, 0-space corresponds to the solid sphere |x| < e in x, y, 2-space?

Show that its volume goes to 0 as €-------- > 0.

6. APPLICATIONS OF INTEGRATION

Leibniz Rule

In this section we take up a number of im portant applications of multiple integration. The first concerns differentiation under the integral sign. Consider a function defined by a definite integral,

= I f ( x> 0 dx.J a

Think of t as a param eter. When the variable x is “integrated out” , there remains a function of t. Problem: find the derivative F'( t) . The answer is called the Leibniz Rule, or the rule for differentiating under the integral sign.

Leibniz Rule Suppose f ( x , t) and the partial derivative f t (x, t) are con-tinuous on a rectangle

a < x < b, c < t < d.Then

^ J f (x , t) dx = J f t(x, t) dx

for c < t < d.

Proof: For each t 6 [c, d~\ let Dt denote the rectangle

= { (x, s) | a < x < b, c < s < t)

in the x, s-plane. The idea is to evaluate

G{t) = J J M x , . ) d x d sDt

in two different ways, then to compute G'(t). On the one hand,

( 1 ) G(t) = ( J f . (x, s) d s ) dx = J [ f ( x , t) - f (x , c )] dx,

where the inner integral is evaluated by the Fundam ental Theorem of Calculus. On the other hand,

(2) • G(t) = f„(x, s) d x ) ds.

From (1),

d d t b d f b d f h d t = d t j ^ X’ ^ dX ~ d t j ^ X’ ° dX = d t j ^ X> ® dX‘

From (2 ),

G® = J tJ c ( / ^ x> ^ dx) ds = J ^ x> ^ dx’The Leibniz Rule follows upon equating these expressions for G'(t).


EXAMPLE 6.1

. , d f w sin tx „ 1Find - / -------dx at

dt J 0 x 2

Solution:

d_ d t .

When t = J, the value is (sin %t )/% = 2 .

d /’7r sin tx [* d /s in tx \ [*— / ------ dx = I — I --------J dx = / cos tx dx =d t j 0 x J 0 d t \ x / J 0

sin 7rt t

A?iswer: 2.

R e m a r k : I t is known th a t F (t) - J [(sin tx) / x ] dx cannot be expressed

in terms of (a finite number of) the usual functions of calculus. In other words, you won’t find it in a table of integrals, except as an infinite series. Nevertheless, F(t) is a perfectly good differentiable function. B ut to compute its derivative, you need the Leibniz Rule.

Volume of an Ellipsoid

EXAMPLE 6.2

Find the volume enclosed by the ellipsoid

+ “ +a2 ¥1, ay b , c > 0 .

Solution: Define sets

D = { (u, v, w) | u2 + v2 + w2 < 1},

E = \ (x , y , z) - 2 + £ + * < l \ f a2 b2 c2

and the mapping

0: x = au, y = bv, z = cw.

Then 0 takes D onto E, and 0 is a proper transform ation (non-singular linear actually), so

- / / / * > * * - / / / E D

- / / /

d(x, y, z)du dv dw

d (u, v, w)

abc du dv dw = abc |D|.

But D is the solid unit sphere, so |D| = iw.Answer: iirabc.

Surface Area

Our aim is to give a reasonable definition of surface area in R3.Up to now, a surface has always m eant the graph of a function z = f ( x , y ) .

A surface represented this way is called a non-parametric surface. There is, however, a more flexible concept called a parametric surface. Such a surface is the image under a nice mapping of a domain in the plane, just as a parametric curve is the image of an interval.

We start with a reasonable domain D in the u, y-plane (param eter plane) and a differentiable function x = x(w, v) on D with values in R3. We assume th a t x maps D one-to-one on a set S C R3- Then S is a parametric surface.

Each point of S is determined by the unique point of D from which it comes. Thus the point x(w, v) can be assigned the coordinates (u ,v ) . In general, the grid of coordinate lines u = const., v = const, is mapped onto a grid of coordinate curves on S. See Fig. 6.1.

The horizontal coordinate line through a point (a, b) of D can be w ritten u = a + t, v = b. The corresponding curve on the surface is

x (0 = x(o + ty b).

The velocity vector of this curve is

dx

Thus xw is tangent to the coordinate curve, hence tangent to the surface. A similar statem ent holds for x .


F ig . 6.1 parametric surface

We now make the assumption th a t xu and x v are non-collinear a t each point of S. This guarantees th a t the two coordinate curves through each point of S are not tangent, bu t cross a t a non-zero angle. In addition, it guarantees the existence of a well-defined tangent plane a t each point of the surface. For the tangent plane is the plane determined by the non-collinear tangent vectors xu and x„. Further evidence: the image of a curve u = u ( t ) , v = v{t) in D is the curve x{t) = x[ u ( t ), Its velocity vector (tangent to the surface) is

dx du dv dt Xu dt * v d t ’

a vector which lies in the plane of xu and x v.Now the vector xu X x v plays an im portant role. I t is not zero because xM

and x„ are non-collinear; it is perpendicular to the surface in the sense th a t it is perpendicular to xu and to x„ hence to the tangent plane. Therefore

X u X X,n =

|xtt X x„|

is one of the two unit normals to the surface. The way it points defines for us the top of the parametric surface.

Let us return to our project of finding a reasonable definition for surface area. A small rectangle in D with sides du and dv maps to a small region on the surface. According to the formula dx — xu du + x„ dv, this region is closely approximated by the parallelogram (Fig. 6.2) in the tangent plane with sides xM du and x v dv. Its area is

(*) dA = | (xu du) X (x v dv)\ = |xtt X xv| dudv.

V


dv

du

F ig . 6.2 approximation to area by a small parallelogram

Recall th a ti j k

y u z u Zu Xu Xu y u

Xu X X x> — Xu Vu Zu = i + i +

y v z v Zv X v Xv y v

Xv y v Zv

d(y, g ) . d(g, x) . a(x, y) ^ d(w , «>) 3(m , v ) * d(u, v)

Consequently

|X v x I2 = + ^ (2’ z )Y 4- ( d(x’ y ) \\ d ( u , v ) / \ d ( u , v ) / \ d ( u , v ) ) '

We substitute this into (*), then add up the small pieces of area by an integral. Thus we arrive a t the following definition:

Surface Area Let x = x(u, v) define a param etric surface with domain D. Its surface area is

A = J J \xu X xv\ du dvD

g) V , ( d( z9 x ) \* ( d(x, y) Y , , d (u , v ) J \ d ( u , v ) ) \ d ( u , v ) ) U V'

There is one possible flaw in the definition. When we look a t a surface, we see a definite area, quite independent of how we have described the surface. If the


same geometric surface has two different parametrizations, how do we know th a t we get the same area from the two corresponding integrals?

This means th a t we have two plane domains D and E, each defining the surface S by differentiable functions x(u, v) and y (s, t) respectively (Fig. 6.3).

F ig . 6.3 a surface described by two different parametrizations

Since both D and E are in one-one correspondence with S, they are in one- one correspondence with each other. Thus there is a transform ation

0:

from E onto D such th a t

u = u(s, t)

V = v(s, t)

y (s, t) = x[>($, t), v(s, 0 1By the Chain Rule,

henceys — xuzis ~I- Xd s, yt — “I- x®t

d(u , v)ys X ]ft (usvt VsW't)xw X Xfl= f ~ X xv

d(s, t)

We also assume th a t there is a definite top side of the surface, the same whether

> 0.

computed by means of x(u, v) or y (s, t ) . This means th a t xu X xv and ys X yt point in the same direction, th a t is,

d(u, v) d(s, t)

Hence 0 is & proper transformation, and

|y* x y<l = lx« x x»l-d(s, t)

Now we can compare the D-area and the E-area of the surface, using the change of variable formula:

A 0 = J J \xu X x„| du dv

d(u , v)- / / 1*. X a (s , t )

ds dt - j f l». x »,l ds dt = A e.E E

The areas are the same, computed either way. The definition is OK!

EXAMPLE 6.3

Find the area of the spiral ram p x = (u cos v, u sin v, bv) corresponding to the rectangle D: 0 < w < a, 0 < ^ < c .

Solution: Although not necessary, it is nice to sketch the surface (Fig. 6.4). Since

xu = (cos v, sin v, 0) and x„ = ( — u sin v, u cos v, b),

the element of area is


dA =/ sin v 0 2 0 COS V 2 cos v sin v 2

1 / + + du dvu cos v b b — u sin v — u sin v u cos v

V ^ 2sin 2 v + b2 COS2 V + II

rs y / b 2 + u2 du dv.

As (u , v) ranges over the rectangle, the point x(u, v) runs over the spiral ramp. Hence

A = j j \ / b 2 + u2 du dv = \ / b 2 + u2 dv j .

Answer: -2

A non-parametric surface, the graph of 2 = /(z , ?/), is a special case of a parametric surface, provided we think of x and y as parameters. The variable point on the surface is x = (x, y , f (x , y ) ) . Then

~ = (1>0 ,/* ), J - = (o, l , /„ ) . dx dy

Consequently

^ X T 5 = (1, 0,/*) X (0, 1, A) = ( - /* , 1), dz dy

and the resulting formula for the element of area is

dA = V 1 + f* + fv dx dy ,

a most useful expression.The formula has a geometric interpretation. The unit normal to the surface

is

N =1

v i + /x 2 + / » 2

Its third component (direction cosine) is

( - /* , - f v , 1).

V i + /*2 + /*2 ’

where 7 is the angle between the normal and the 2-axis. Thus

(cos 7 ) dA = dx dy ,

which means th a t the small piece of surface of area dA projects onto a small portion of the x, y-plane of area dx dy.

EXAMPLE 6.4

Find the area of the portion of the hyperbolic paraboloid (saddle surface) z = — x2 + y2 defined on the domain x2 + y2 < a2.


Solution: First sketch the surface, then the portion corresponding to the range x2 + y2 < a2. See Fig. 6.5. By the formula,

dA = y / 1 + f x2 + fy1 dx dy = V 1 + ( “ 2x ) 2+ (% /)2 dx dy

= \ / l + 4x2 + 4y2 dx dy.


Use polar coordinates:

A - J f V T + 4 r2 r dr dd

f a ________ r ** 1= / ry/ 1 + 4r2 dr dd = — [ ( 1 + 4a2) 3/2 —

c/o y o

Answer: - [(1 + 4a2) 3/2 — 1].

Green's Theorem

There is a useful connection between certain line integrals and double integrals. Suppose D is a domain in the x, ?/-plane whose boundary consists of one or several nice closed curves, curves consisting of arcs with continuously turning tangent vectors. The counterclockwise sense of rotation in the plane im parts a direction to each of the boundary curves (Fig. 6 .6 ). We think of walking around the boundary curves so th a t D is always on our left.

We use the symbol dD to denote the whole boundary of D, with this direction imposed. Then we know the meaning of a line integral

/OD

P dx + Q dy.

The following theorem says th a t this line integral over dD is equal to a certain double integral over D.


Green's Theorem Suppose P ( x , y ) and Q ( x ,y ) are continuously differentiable functions on the domain D. Then

We shall not give a complete proof of the theorem but only prove one special case. The idea in general is to trea t the theorem as two separate formulas,

J P dx = — J J P y and J Q dy = J J Qx. For the first, cut the domain by

vertical lines into pieces, each of which is the region between the graphs of two functions. (This is the case we shall prove in a moment.) Then add up the results. The double integrals add up to the double integral over the whole domain. The line integrals add up to the line integral over the boundary, because the contributions from the vertical division lines cancel in pairs.

We shall prove

dD D

dPdy

dx dy

for the domain D of Fig. 6.7.

The boundary dD consists of four pieces, and accordingly the line integral decomposes into four summands. On the two vertical sides x is constant, hence

dx = 0 , no contribution. Therefore

J P dx = J P\_x, h ( x ) ~ \ d x J P[_x, g(x)~\dxd o 6 °

= I {-P[_x,h(x)~] +P[ x , g ( x ) ~] } dx.J a

- - m s * ) * - - V S * *This completes the proof.

Area Formula If D is a plane domain with a smooth boundary, then

|D| = ^ J — y dx + x dy.dD

Proof: Apply Green’s Theorem with P = —y and Q = x. Then Qx — P y = 2 , so

j —y dx + x dy = J J 2 dx dy = 2 |D|.dD D

R e m a r k : By similar applications of Green’s Theorem, we also have

|D| = J x d y = — J y dx.do o d

The boxed formula is often more convenient than either of these because it has a certain amount of symmetry.


EXAMPLE 6.5

Find the area enclosed by the ellipse

™ 4- t = 1 a2 b2

Solution: Let D denote the domain bounded by the ellipse. Param eterize the ellipse as usual by

x = a cos 0, y = b sin 6, 0 < 0 < 2t .Then ^

lDl = - / - y d x + x d ydD

1 f 2*— - / — (b sin 9) ( — a sin 6 dd) + (a cos 0) (b cos 6 dd)

2 Jo

(ab sin2 d + ab cos2 d) dd = irab.

Answer: rob.

EXERCISES

Evaluate F(t) for t > 0 and then compute F' (t). Now compute F' (() by the Leibniz Rule and compare your answers:

I . F(t) = J e~tx dx 2. F(t) = J ^ a rc ta n ^ d z , t > -

3. F(t) = J (t + x )n dx 4. F(t) = f x* dx, t > — 1.

5*. Prove under suitable hypotheses:d A< 0 A<o- f (x , t )d x = h'(t)f[h(t)y «] — g'(t)f[g(t), t ] + / f t (x,t)dx. dt Jo(t) 7^(0

Use the Chain Rule.]

Parametrize and set up an integral for surface area. Evaluate when you can:6. sphere of radius a7. lateral surface of a right circular cylinder of radius a and height h8. lateral surface of a right circular cone of radius a and lateral height L9. right circular torus obtained by revolving a circle of radius a about an axis in

its plane at distance A from its center (where a < A )

10. ellipsoid ^ + g + 5 = la1 b1 cz

[Hint: Use spherical coordinates and x = (a sin 0 cos d, b sin <t> sin 6, c cos 0 ).]II. (cont.) Reduce the double integral to a simple integral in the special case a = b.

Set up the area for the given non-parametric surface. Evaluate when you can:12. z = xy; — 1 < x < 1, — 1 < ?/ < 1 (do not evaluate)13. z = ax + by\ (x, y) in a domain D

14. z = x2 + y2) x2 + y2 < 115. z = V l — x2 — y2', x2 + y2 < 1.16. Find the area of the triangle with vertices (a, 0, 0), (0, 6, 0), and (0, 0, c).

17. Prove that / / N dA = 0 for a closed surface.

[Hint: Show that i • J J N dA = 0, etc.]

18. From each point of the space curve x = x(s) draw a segment of length 1 in the direction of the unit tangent. These segments sweep out a surface. Show that its

area is \ Jk(s ) ds, where k (s) is the curvature and the integral is taken over the

length of the curve.


19. (cont.) Interpret for a circle. Can you show that J,k(s) ds = 2ir for any closed

oval (convex curve) in the plane? If so, interpret Ex. 18 for ovals.20. Let x = x (s) be a curve of length L on the unit sphere x = 1. Connect each point

of the curve to the origin, forming a surface. Show that the area of this surface is \ L .

Let P and Q be continuously differentiable on a rectangle I.21. Suppose (P, Q) — grad0 . Prove

/ P dx -f- Q dy = 0

on every closed curve in I.22*. (cont.) Prove the converse.

23. Evaluate J s y dx 4x dy over the boundary of the rectangle with vertices at

(0, 1), (6, 1), (6, 3), (0, 3) taken counterclockwise.

24. Evaluate J y 2ex dx + 2yex dy over the circle x2 + y2 = 1.

25. Prove Green’s formula under suitable hypotheses:

/ (— uvy dx + uvx dy) — I {—vuy dx + vux dy) do do

= j 'f \jll (VXx “f~ Vyy ) i^xx “1“ Uyy ) ] dx dy.D

26. (cont.) Test this formula when D is the unit disk, u — 1, and v = In r. Explain the result.

27*. Suppose P and Q are continuously differentiable functions on a rectangle I, and dP/dy = dQ/dx. Prove there is a function f(x, y) on I such that df/dx = P and df/dy = Q.

28. (cont.) Let a and b be points of I. Prove that the line integral / P dx + Q dy a

is the same for all paths in I from a to b.

7. IMPROPER INTEGRALS [optional]

A multiple integral is improper if either the domain of integration is unbounded or the integrand has singularities or both. There are always two questions: Does the integral converge? If so, w hat is its value?

The Whole Plane

First we define

R-

y ) dx dy.

Once we have this, it is easy to define the integral over any unbounded domain D. The integrand will be a real-valued function f (x, y) defined on the whole plane R2. We can’t expect to define the integral for all functions, so we shall restrict attention to functions having the following reasonable property:

A function / on R2. is locally integrable if / is integrable on each bounded R-measurable set D.

We shall define the integral by means of limits of the type

/ dx dy ,lim//'n->oo J J

where {Dn} is an expanding sequence of bounded R-measurable sets, Di CI D2 Q D3 Q th a t in some sense fill out the plane.

Approximating Sequence A sequence {Dn} of bounded R-measurable subsets of R2 is an approximating sequence provided

(1) Di C D2 C D3 C(2) if B > 0, then there exists an n such th a t the disk of radius B and

center 0 lies in Dn, th a t is,

{x I |x| < B ) C Dn.

Thus each disk centered a t 0 , no m atter how large, is eventually covered by the sets of any approximating sequence. In fact, any bounded set is eventually covered, because a bounded set is contained in some disk centered a t 0.

Now we have enough preliminary definitions to define the integral.

Integral on R2 Let / be a locally integrable function on R2. Then / is called integrable on R2 provided:

(1) For each approximating sequence {Dn}, the limit

limn-* oo

exists.

(2) These limits all have the same value, which is then called

J J f d x d y .R2

/ / f dx dy

In every other case, we say

J J f dx dyR!


diverges. One common case of divergence occurs when

for each approximating sequence {Dn}. Then we write

J J f d x d y = + oo.

The preceding definition has the drawback th a t it is impossible to test every approximating sequence. However, when the integrand has constant sign, it is enough to test any one approximating sequence.

Theorem 7.1 If / > 0 on R2 and

exists (finite or + 00) for one single approximating sequence {Dn}, then / is integrable on R2.

Proof: Take any other approximating sequence {Em}. For each m, the set Em is bounded. Hence there exists an n such th a t Em C Dn. Since f > 0 , we conclude

Now we can apply the same argument with the D's and E’s interchanged. The result is

so equality follows. Since {Em} is an arbitrary approximating sequence, the proof is complete.

From this, we conclude th a t the increasing sequence

converges, possibly to + oo, and

There is a related result about absolute convergence. We shall state it, but save the (somewhat technical) proof for the exercises.

Theorem 7.2 Suppose

( 1) / i s locally integrable.(2) | / | is integrable on R2.

Then / is integrable on R2.

The point is th a t the test of Theorem 7.1 can be applied to | / |.

EXAMPLE 7.1

Prove th a t the integral

J J e-x2-y2 dx dyR2

is convergent and evaluate it.

Solution: The integrand is positive so we can use any convenient approximating sequence. We choose disks,

Dn = {x | |x| < n},

and switch to polar coordinates:

J J e~x2~y2 dx dy = J J e~r2r dr dd = ^ J r e~‘* ^ )Dn Dn

= (2tt) ^ (1 — e-"2) -------- > 7T.

Answer: ir.

This result has a surprising application.

EXAMPLE 7.2

Evaluate

f°° e d x .

Solution: We solve Example 7.1 again, this time using a sequence of squares:

En = {(x, y) | \x\ < n, \y\ < n) .Now

J J e - x 2- y2 ^ = ^ J e - x 2 d x 'j^ J e~y2 dy = (^J er*2 dx'j .

We know from Theorem 7.1 and Example 7.1 th a t


I f ' * ' dx d y -------- » 7r.

We conclude

( £ « - ■ * )Ansiver: y / 1r.

R e m a r k : In probability theory, the function

is im portant; it is the density function of the normal distribution. The simple change of variable x = u /y / 2 in Example 7.2 yields

/ "./ —(X</>(x) = 1.

The graph of </> is the familiar bell-shaped curve (Fig. 7.1); the area under the curve is 1.

1 "

y = <t>(x) = — = e~x*12 y /2 v

F ig . 7.1 normal distribution

Arbitrary Unbounded Domains

Now let D be an arbitrary unbounded subset of R2. We wish to define

/ / / < * , y ) dx dy

To do so, we extend / to the whole plane by making it 0 outside of D. Thus

we define

Then we set

F ( x , y ) =f ( x , y), (*, y) € D

0, (x, y) $ D.

J J f (x , y) d x d y = J J F{x, y) dx dyD R2

if the right-hand side exists. For the domains th a t occur in practice, things go routinely from this point.

R e m a r k : Actually, it is reasonable to restrict attention to domains D for which the characteristic function kD (one on D, zero off) is locally integrable. This means th a t the intersection of D with any bounded R-measurable set is itself a (bounded) R-measurable set.

Before reading the next example, it is advisable to review Theorem 8 .2, p. 110.

EXAMPLE 7.3

Evaluate

f 00 e—ax — e~~bx1 dx, 0 < a < b.

Jo *

Solution: The starting point is the observation

e~ax — e~-bx /*&

— = e~xJ a

dy.

Therefore it is natural to consider

- I I - xy d x d y , D = { (x, y) \ x > 0, a < y < b}.

To ev a lu a te I, ta k e th e rectan g le

= { (x, y) | 0 < x < n, a < y < b}.

On th e on e hand ,

J J e~xv d x d y = J ^ J e~xy dy'j dxDn 0 o

f n e~ax — e~bx f= / ------ -------d x --------- > I

J O X J 0

J J e~xy d x d y = Jso

-ax _ p—bxdx.

On the other hand,

f f e~xy d x d y = [ ( [* d x ) dy - I ----- — dy.J J J a V o / J a y

We infer from this th a t

/ ■*/odx = lim f -

oo ./ ady.

The trick now is to take the limit under the integral sign on the right. T hat is where Theorem 8 .2 , Chapter 3 comes in. We observe th a t

1 - e~ny

uniformly on [a, 6] because

1 — e~ny 1

y y

Q~ny e~na

y ~ a0 as n ■ -> 00 .

Therefore by Theorem 8.2,

1 g—ex _ g—bx

'0/ ' X

(b i — 6- ^ 7 [ b dy b= / l im -----------dy = / — = In - .

J a n-oo V J a V a

Answer: In (6/a ) .

Singularities

Suppose there is a single point in the (bounded) domain of / a t which / becomes infinite, or is indefined in some other way. We try to squeeze down on this isolated singularity of / by a growing sequence of subdomains, in the same way th a t we expanded to infinity in the case of the whole plane. If we limit ourselves to functions of constant sign, a result similar to Theorem 7.1 says th a t one Such sequence suffices. The following example should illustrate the idea. Recall the integral

n d x

J o * p

which is convergent for p < 1 and divergent for p > 1.

EXAMPLE 7.4

Find all p for which

/ / ; , < & %r<a

is convergent.

( r2 = x2 + y2)

Solution: The integrand is positive, except for a singularity a t r = 0 when p > 0. In polar coordinates,

// h ixdy- !“ // 7. dxdy- f de /' -21 f! £ •■


€ < r < l

Answer: For p < 2 , it converges; for p > 2 it diverges.

Iteration

The main point about iteration of improper integrals is th a t it often doesn’t work. Be careful; there are few useful results, and generally each case must be treated on its own merits. I t is easy to get incorrect results.

EXAMPLE 7.5

Apply iteration blindly, both ways, to

o<sx<i (x2 + y2) 20<y<,\

dx dy.

Solution: The formula

r 2

/ dx(x2 + a2) 2 x2 + a2

is valid, as is easily seen by differentiating the right-hand side. Therefore

/ = r ( r ~ x2 ~ y2 d i ) dy = r ( ~ x ) | 1 %J o \ Jo ( x2 + y ) ) J o v 2 + y ) 11=0

Now look carefully a t the integrand. If x and y are interchanged, it changes sign. Therefore integration first on y , then on x, leads to I = + j 7r, an impossible situation.

W hat is wrong is th a t the improper double integral does not exist. This example is worth careful study.

To end on a happier note, we give an example where the result is correct, bu t it is beyond our technique to justify the formal steps. (We shall use integral 41 inside the back cover of the book.) Set

■ / /0 < x0 < y

e~xy sin x dx dy.

First,

Second,

c °° / r °° \ c °°_p~xv x~ °°dy /■” / / ■* \ f K —e~xyI — I I e~xy sin x d x ) dy = / -—-— - ( y sin x + cos x)

J o V o / Jo 1 + 2/2

' I" i

x=0

+ y2 2

Therefore

f - s i r

J 0

This result is correct, bu t this derivation is full of holes.

sina: 7 7r dx = - .

x 2

EXERCISES

Prove for integrals over R5>2.

f f dxdy _ o f f _ dxdy _ _2' J J i + x* + y> ^ J J l + ^ + y2 + x V r '

3. Suppose 0 < g(x,y) < f(x, y) on R2 and j j f < 00 • Prove that ^ 0 < 00

4*. Prove Theorem 7.2. [Hint: Use a Cauchy Condition.]5*. Prove divergent: j j _dx dy

6. Prove convergent:

f i r+ x V '

dx dyR- - + X4 + y*'

Prove convergent and evaluate:

7. JJ (In r ) d x d y 8. JJ (1 - r)» dx dy, p > - 1r<> 1 r<a

9. j j j p8 dx dy dz, s > —3 10. j j j (In p) dx dy dzP< 1

///< -p — 1.p< l

12. Prove convergent for p > 1 and evaluate:

8. Numerical Integration 511

dx dy r2p

r> 1

13*. (cont.) Prove1

I f '

I < 00(m2 + n2 ) p

for p > 1. The sum is taken over all integers m, n except (0, 0).

Compute:

14. j ertxl dix, t > 0 15*. J xLe~x2 dx.

8. NUMERICAL INTEGRATION [optional]

In this section we discuss one method for approximating double integrals. I t is an extension of Simpson’s Rule.

Let us recall Simpson’s Rule. To approximate an integral

f f (x ) dx,J a

we divide the interval a < x < b into 2m equal parts of length h :

b — aa = x0 < Xi < x2 < • • • < x2m = b, h = — — ,

2m

and use the formula2m

f b h \ \J f ( x ) d x t t ~ 2 ^ B i f ( x i ) ,

where the coefficients are 1, 4, 2, 4, 2, 4, 2, • • •, 2, 4, 1.We extend Simpson’s Rule to double integrals in the following way. To

approximate

JJ f (x , y ) dx dy,I

where I denotes the rectangle a < x < b and c < y < d, we divide the z-interval into 2m parts as before and also divide the ^/-interval into 2n equal parts of length k :

d cC = 2/0 < yi < V2 < • • • < yin = d, k = .


We obtain (2m + 1) (2n + 1) points of the rectangle I, shown in Fig. 8.1. The Rule is

/ /2m 2 n

hk V Vf ix , y) d x d y ^ y^ ’9

i =0 j = 0

where the coefficients A are certain products of the coefficients in the ordinary Simpson’s Rule. Precisely,

A ij BiCj,

|84 21

n

1!16

X q

1

X i Xi X 3 x4

2'

xb

^ordinary Simpson’s Rule coefficients

F ig . 8.2

where B 0, B h • • •, B^m are the coefficients in the ordinary Simpson’s Rule2m

J p ( x ) dx « ^ B i p ( x i ) , a 1=0

and Co, Ci, • • •, C2n are the coefficients in the ordinary Simpson’s Rule2 n

J q(y) d y t t ^ ^ Cjq{yj).° j= 0

In Fig. 8 .2 , several of these products are formed. Since Bi and Cy take values1, 2 , and 4, the coefficients A ,7 take values 1, 2 , 4, 8 , and 16. The A i j can be written in a matrix corresponding to the points (xi, yj) as in Fig. 8 .2 . For example, if m = 3 and n = 2, the matrix is

"1 4 2 4 2 4 1“

4 16 8 16 8 16 4

2 8 4 8 4 8 2 .

4 16 8 16 8 16 4

1 4 2 4 2 4 1

EXAMPLE 8.1

Estimate / / (x + y )3 dx dy by Simpson’s Rule with m = n = 1.

Compare the result with the exact answer.

Solution: Here h = k = i. The coefficient matrix is

[ A i j ] —

1 4 1

4 16 4

-1 4 !.Write the value (Xi + i/y) 3 in a matrix:

[(*< + 2/j)3] =

(1 + 0)3( - 0*

(1 + 1)*

6 + ° ) G + 0 0 + i )(0 + 0)3 H ) ‘ (o + 1)3

1 ¥ s '

1 278 8

0 - 1

Now estimate the integral by

9 1 1 Ai j {x i + y iyi) *•i =0 j =0

To evaluate this sum, multiply corresponding terms of the two matrices and add the nine products:

M l 27 1 27 1 1-1 + 4-— + 1-8 + 4 -- + 16-1 + 4-— + 1-0 + 4 -- + 1 1 1

1 T 27 1 27 1 ] 54 3 = 36L1 + Y + 8 + 2 + 1 6 + Y + 2 + 1 J = 3 6 " 2 '

The exact value is

(x + y Y dx dy = I'm :if (X + y Y d x ) dy0 < x , y ^ l

)4 Jo

l ( y + l ) 4 - y4] dy = — l ( y + l ) 6 - 2/5]0 ^

1 _ 30 _ 3o “ 20 ~ 2

Answer; Simpson’s Rule gives the estimate §, which is exact.

R e m a r k 1: Because Simpson’s Rule is exact for cubics, the double integral rule is exact for cubics in two variables. (See exercises below.)

R e m a r k 2: The matrix of values [/(«», yi) ] is arranged to conform to the layout of points (xi, yj) in the plane (Fig. 8.1).

The next example is an integral that cannot be evaluated exactly, only approximated.

EXAMPLE 8.2

c cEstimate / / sin(zt/) dx dy, using m = n = 1.

0 £x,y tt/2

Solution: Here h = k = 1r/4, and the coefficient matrix is

"1 4 1"

z a <h = 4 16 4

1 4 1

The matrix of values of sin(xy) is

• ^ r T sin 0 sin — sin — 8 4

[sin xiyf] = • • T . TTsin 0 sin — sin — 16 8

sin 0 sin 0 sin 0

The estimate is

sin (#2/) dx dy7r__

144 ( . TC . TT16 sin — + 8 sin —0 < x,y <7r/2

Error Estimate

The error estimate for Simpson’s Rule in two variables is analogous to that in one variable:

where

a4/ < M andd4f

dx4 dy4

We omit the proof.

EXAMPLE 8.3

Estimate the error in Example 8.2.

Solution:

d4 . . a4—- (sin xy) = y4 sin xy , —- (sin xy) = x4 sin x y . dxr dy4

But |sin xy\ < 1. Hence in the square 0 < x, y < ir/2,

Apply the error estimate, with m = n — 1, h = k = 7r/4, and M = N = (tt/2)4:

|error| <(6 - a) ( d - c)

180[h4M + fcW],

EXERCISES

1. Use tables to complete Example 8.2.2. Do Example 8.2 using m = n = 2, and compare your estimate to that of Ex. 1.

Also estimate the error.3. Suppose f ( x , y ) = p(x)q(y) . Show that the double integral Simpson’s Rule

estimate is just the product of the Simpson’s Rule estimate for J p (x) dx by that

for / q(y) dy.

4. (cont.) Conclude that the rule is exact for polynomials involving only xsys, xsy2, x2yz, x*y, x2y2, xy3, and lower degree terms.

5. The analogue of the Trapezoidal Rule is

/ / f ( x , y ) d x d y ~ ^ [ / ( 0 , 0 ) + / ( 0 , l ) + / ( l , l ) + / ( l , 0 ) ] .0 < x , y < 1

Show this rule is exact for polynomials/^, y ) = A + Bx + Cy + Dxy.6. (cont.) Find the corresponding rule for a rectangle a < x < b , c < x < d , divided

into rectangles of size h by k with h = (b — a ) /m and k = (d — c)/n.7. (cont.) Let I denote the unit square 0 < x, y < 1. Suppose f(x, y ) = 0 at its

four vertices. Prove that

I J f ( x , y ) d x d y = — ^ J J y ( l — ' i i

- - i f ' . * 1-1 -

y)fw( .x,y) dxdy

x ) [ f * x ( x , 1) + f x x ( x , 0)] dx.

8. (cont.) Suppose also that \fxx\ < M and \fm\ < jV on I. Prove that

Iff-9. (cont.) Conclude that for any function/ (x,y) , the error in the trapezoidal estimate

(Ex. 5) is at most (M + N ) / 12.[Hint: Use the result of Ex. 5 and interpolation.]

10. (cont.) Suppose f ( x , y ) is defined on 0 < x < h, 0 < y < k, and / = 0 at the vertices of this rectangle I. Suppose also \fxx\ ^ M and \fyy\ < N. Deduce that

Ifff(x, y ) dx dyhh< ~ (h*M + k2N).iZ

11. Extend Simpson’s Rule for double integrals to triple integrals.12. (cont.) For which polynomials in 3 variables is the extended Simpson’s Rule

exact?13. (cont.) Estimate the integral of sin(xyz) over the cube 0 < x, y, z < 1, dividing

the cube into 8 equal parts. Carry your work to 3 places.

13. Differential Equations

1. INTRODUCTION

We are familiar with differential equations of the type

dx u \ d i ~ m -

Each solution is an antiderivative of f ( x ) . We are also familiar with the differential equation

dy

Tx = y '

each solution of which is a function y = cex.These differential equations are instances of the most general first order

differential equation,

y = q( x , y ), dx

where q(x, y) is a function of two variables. “First order” refers to the presence of the first derivative only, not the second or third, etc.

A differential equation d y / d x = q(x, y) defines a direction field. At each point (x, y) of the domain of q(x, y ) , imagine a short line segment of slope q(x, y ) , as in Fig. 1 .1 . A solution of the differential equation is a functionV = y(%) whose graph has slope matching that of the direction field (Fig. 1.2).

The most important problem in this subject is the initial-value problem:

Initial-Value Problem Find the solution y — y( x ) of the differential equation

dy (— = q(x, y)dx

whose graph passes through a given point (a, b), that is, which satisfies the initial condition

y ( a ) = b.

518 13. DIFFERENTIAL EQUATIONS

It is shown in advanced courses that if q(x, y ) is a reasonably behaved function, then the initial-value problem has a unique solution locally. That is, there is a neighborhood of a on which there exists one and only one differentiable function y{ x ) satisfying y ' {x ) = x, y{x)~\ and y ( a ) — b. This is true, for instance, if q(x, y ) has continuous partial derivatives in a neighborhood of (a, b ) . In Section 7, we shall discuss one method of proving such results, successive approximation.

\ \ \ \\ \ \ \ \ ' , \ \

\ \ \ \ \ \ ' . \ \\ \ \ \ \ \ \ ' . \ \\ \ \ \ \ \ \ '\ \ \ \ \ \ \ ' S.

\ \ \ \ \ \ N 'V -V -V -^7-

\ \ \ \ \ ^ —

\ \ \ \ -------\ \ v. - - — - - ^\ \ — ------ ---- /

^ y y / /

\ \ \ \ \

- - — - - / /------ - / / /

" ^ / / /' y / / /7* X / / /

/ / / / / / / / / / / / / /

/ / / / / I I/ / / / I I/ / / / I

/ / I I

F ig . 1.1

F ig . 1.2

\\A \ \ > \ \ \

• \ \ V \ \ N< \ \ \ —

\ \ \ A \ \ V \ \ — —/ / y\ \ \ \ \ \ >s. \ — — X

\ \ \ \ \ \ V >V \ — — — V /

\ \ \ \ \ \ \ c- — / / /

\ \ \ \ \ \ —— / / /\ V / / / fc\ \ \ \ / / r X

\ \ \ —— / / / / / /

\ \ \ —— X- / / / / / / /

\ \ s — /✓/ / / / / / /

\ s — — / // / / / / / /—— / /f / / / / /

/ / / 'f / / / /

y = y(*)

1. Introduction 519

Figure 1.3 shows two solutions of

dydx = Q(x, y ),

one satisfying the initial condition y ( 0) = 0, the other satisfying the initial condition 2/(1) = — 1.

F ig . 1.3

Usually we first find the general solution to the differential equation, ignoring the initial condition. Somewhere along the line we are forced to integrate, thus introducing a constant of integration. We then select the constant so that the initial condition is satisfied.

EXAMPLE 1.1

Find a solution of the initial-value problem

<MIISal H *8

13

U ( - 3 ) = 1.Solut ion: Integrate:

y(x) = - x3 + c.


This is the general solution of the differential equation. Now select the constant by substituting the initial condition:

l = i ( - 3 ) 3 + c, c = 10.

Answer: y{x) = - x z + 10.o

EXERCISES

Solve the initial-value problem:

i

dx x

7 . 1 - , , ,(1) — 10

9. ^ = a: + sin y(0) = 0

1 1 .$ =dX y / t f + 1

dy

y(0) = o

2 . ^ = x* + x - 5 , 2/(0) = - 2

A. — cos 2x, y( r) = 0 ax

6. ^ = ze**, 2/(0) = 1

8 . f x = y, 2/(1) = 0

dy

13. -g = (1 + x2)(l + x3), 2/(0) = 0.

14. Sketch the direction field of dy/dx = (0, 1); through (0, —1). Rewrite the

15. Solve xy' + y = 2x.[Hi'ntf: Compute (xy)'.]

17. Solve x + yy' = z2 + y2- [Hint: Compute the derivative of In Or2 + y2).]

= i/2. Draw the solutions passing through equation as dx/dy = y~2 and solve.16. Solve y' + y = e~x sin x.

[Hint: Compute (exy)'.]18. Solve xy' — y = x2 sin x.

[Find the trick yourself this time.]

2. SEPARATION OF VARIABLES

The general first order differential equation is

2. Separation of Variables 521

In this section, we study the special case in which q(x, y) = f (x ) g (y ) , the product of a function of x alone by a function of y alone. Examples are

dy 2 3 dv dv v— = x 2y 3, — = ex cos y, — = — ..... -ax ax ax _|_ x2

Not included in this special case are

dy dy . , K dy /--------------—^ ~ V l + *■ + »*,

since in each of these equations the right-hand side is not of the form f (x)g (y ) .

Solve

^ = f (x )g(v)

by first “separating the variables,” putting all the “y ” on one side and all the ltx ” on the other. Write

= /(*) dx.g(y)

Then integrate both sides (if you can):

G(y) = F(x) + c,

where G(y) is an antiderivative of 1 / g ( y ) and F(x) is an antiderivative of f ix ) . If this equation can be solved for y in terms of x, the resulting function is a solution of the differential equation. (This technique will be justified after several examples.)

Note that in separating the variables you must assume that g(y) 0. If g(y) = 0 for 2/ = 2/o, it is simple to check that the constant function y(x) = ?/o is a solution of the equation.

EXAMPLE 2.1

a i dy Solve — = xy.dx

Solut ion: Obviously y(x) = 0 is a solution. Now assume y(x) ^ 0 and separate variables:

i y . x i x , ( * - ( . * ,y

In \y\ = i x2 + c,

\y\ = kex2/2 (k = ec)j

where the constant k = ec is positive. If y > 0, then \y\ = y , if y < 0, then |2/| = — y. Hence

f kex2,i if y > 0y j —jfce-V* if y < 0 .


This is equivalent to

where a is any non-zero constant.

y = aex2/2,

Answer: y — aex2/2, a any constant. (Note that the solution y(x) = 0 is included in this answer.)

Check:

— = — (aex2/2) = axex2/2 = xy. dx dx

EXAMPLE 2.2

dy 2x(y - 1)Solve* ' “^ + r '

Solution: Obviously y(x) = 1 is a solution. Now assume y{x) 1 and separate variables:

dy 2x dx: C dy f 2x dx

r J y - l = J x 2 + ly - 1 x 2 +

In \y — 1| = ln(x2 + 1) + c,

12/ — 1| = e°(x2 + 1) = k ( x2 + 1).

Thereforey - 1 = ± k ( x 2 + 1),

where k is a positive constant. The answer may be written

y = 1 + a(z2 + 1),

where a is an arbitrary constant. This includes the special solution y{x) = 1.

An sw e r: y = 1 + a(z2 + 1).

Check:dV o o a(*2 + x) 2x(2/ - 1)_ = 2ax = 2x — = — 2 - , 1 - *dx x 2 + 1 x 2 + 1

EXAMPLE 2.3


Solution:

— = exe~y, ey dy = ex dx , ey = ex + c.dy

dx

Answer: y = ln(e* + c), ex + c > 0.

Check :

ex + cex— = ex~v. ey

These examples show that the me.thod works. Why does it work? Suppose y = y(x) satisfies

Let G{y) be any antiderivative of 1 / g ( y ) . Note that G(y) is indirectly a function of x. By the Chain Rule,

Hence G(y) is an antiderivative of f (x ) . Therefore, if F(x) is any antiderivative of J(x),

If you can solve this equation for y as a function of x, you have the general solution. If not, then at least you have a relation between x and y, often adequate for applications.

EXAMPLE 2.4

(5 y 4 — 1) dy = 2x dx> y h — y = x 2 + c.

We cannot solve this fifth degree equation for y ; it is hopeless. Still we can substitute the initial data:

0 = 1 + c, c — — lj y b — y = x 2 — 1.

This relation between x and y is as far as we shall get with the problem. But it is adequate for a graph and can be used to calculate y (given x) to any degree of accuracy.

t x " f ( x ) g (y ) ■

dG _ dG dy _ _ 1 _

dx dy dx g(y)

G(y) = F(x) + c.

Solve the initial-value problem

Solution:

Answer: y & — y = x 2 — 1.


EXAMPLE 2.5

dySolve the initial-value problem — = x 2y 2, y {0) = b.

Solution: Obviously y(x) = 0 is the solution if b = 0. Assume y ^ 0 and separate variables:

- l - Z + c.v y 3

Substitute the initial data:

— i = o + c = c. b

Hence1 x 3 1 bx3 — 3y 3 b 3b

3bAnswer: y — - ------—-• (The answer

3 — bx3includes the solution y(x) = 0.)

EXAMPLE 2.6

Find all curves with the following geometric property: the slope of the curve at each point P is twice the slope of the line through P and the origin.

Solution: Translate the geometrical property into an equation. Suppose the graph of y{x) is such a curve and P = (x, y) a point on it. The slope at P is dy / dx . The slope of the line through P and (0, 0) is y / x . Hence, the geometrical property is expressed by the differential equation

dy = 2 vdx x

Obviously y(x) = 0 is a solution. Now assume y(x) ^ 0 and separate variables :

— = 2 — ) In \y\ = 2 In \x\ + c = In x 2 + c. y x

Take exponentials, setting k = e°\

\y\ = k x \

As in Example 2.1, it follows that y = ax 2, where a is any constant.

Answer: The curves are the parabolas y = a x2 and the line y = 0. See Fig. 2.1.


F ig . 2.1

EXAMPLE 2.7

Find all curves whose slope at each point P is the reciprocal of the slope of the line through P and the origin.

Solution: The problem calls for all functions y{x) satisfying

dy x

dx ySeparate variables:

y dy = x dx, 1 2 1 2 .2 V 2 + C '

Answer: The curves are the rectangular hyperbolas y 2 — x 2 = c. See Fig. 2.2.

Note the special case c = 0, y = ± x .


Solve:

EXERCISES

x dy = & 2 — = —dx y dx xy

- dy5. -y- = xe2' dx

dy = ( 1 + y\ 2 dx y i + x j

( ? ) -

Show that the substitution y = ux changes the differential equation into an equation for u = u(x) whose variables separate, and solve:

dy = x2 + 2/2 8 — = + V •dx 2x2 dx x — y


9- S - - S ' »(1>- 3 10.g-2„co., ,- D ( » - 2) , ( 4 ) , 0

cfc a:

12. ^ = y3 sin x, y(0) = ■

Solve the given differential equation; interpret the equation and its solution geometrically:

13. dJ * = - y- 1 4 . ? = - - dx x dx y

1 5 . ^ - * - dx x

3. LINEAR DIFFERENTIAL EQUATIONS

A differential equation of the form

t - + p(x)y = q(x)dx

is called a first order linear differential equation.The term “linear” can be described in the following way. Imagine a

“black box” or “processor” that converts a function y(x) into a function w(x) , as in Fig. 3.1.

input output

F ig . 3.1

3. Linear Differential Equations 527

The black box is called linear if from a linear combination* of inputs, it produces the same linear combination of outputs (Fig. 3.2):

---------> Wi(x)y i (x )2/2 (x)

ciyi(x) + c2y2(x)-> w 2 (x )- » C iW i(x ) + C2W2 ( x ) .

F ig . 3.2

In particular, the black box that converts y into d y / d x + p ( x ) y is linear: when the input is ciyi + C22/2, the output is

~ (c<yi + C22/2) + p( x) (cxyi + C22/2) dx

= ci + p ( x ) y 1 + C2 + p ( x ) y = ci(outputi) + C2(output2).

This is why the differential equation (d y /d x) + p ( x ) y = q(x) is called linear.

R e m a r k : The black boxes discussed here are nothing else than the linear functions of linear algebra, in an appropriate setting. (See Ch. 7, Sec. 1.) For example, let D be the set of all differentiable functions y ( x ) . Then the “black box” defined by L y = y' + p ( x ) y is a linear function defined on D since

L (a y i + by2) = a L y x + bLy2.

Not every black box is linear, however. For instance, consider the black box in Fig. 3.3.

yblackbox

dydx

F ig . 3.3This box is not linear since the output of a sum is not the sum of the outputs (Fig. 3.4).

2/1-

2/2*

2/i + 2/2 ’

dmdx

fdy2dx

F ig . 3.4

* An expression aX + bY + cZ + • • •, where a, b, c, • • • are numbers, is called a linear combination of X, Y, Z, • • •.


Figure 3.5 shows another non-linear process.

2 / i -

2/2

+ 2/2 '

, 2 3 - + x y ?dx

+ x y 22

T (y 1 + 2/2) + x (y 1 + 2/2)2 dx

>4 Property of Linear Processes

An extremely useful property of linear black boxes is shown in Fig. 3.6.

F ig . 3.6

If the output of 2/2 is 0, then the output of 2/1 + 2/2 is the output of 2/1 • Hence output is unchanged when input is augmented by a function which is converted to zero.

This simple fact is the basis for a systematic study of linear differential equations. We return to the process that really interests us, the one that converts y(x) into d y / d x + p(x)y . For brevity, we write

— + p ( x )y = Ly, dx

indicating that L y is the output corresponding to the input y. The differential equation in question is

~ + p ( x ) y = q(x), dx

or L y = q.

A solution is an input y(x) that results in the given output q(x).Suppose we can find a solution 2/1 . That means L y i = q. If z is any

function for which L z = 0, then L ( y i + z) = q. Thus 2/1 + z is also a solution. Furthermore, every solution must be of the form 2/1 + z. To see why, assume 2/1 and 2/2 are solutions:

L y i = q, L y 2 = q.

Consider the difference y i — y 2 . Because L is a l inear process,

L ( y i - y 2) = Lyx - L y 2 = q - q = 0.

Therefore, when the input is y i — y 2 , the output is 0; this says the difference of two solutions is one of the functions z.

3. Linear Differential Equations 529

Any two solutions of the differential equation

~-x + p ( x ) y = q(x)

differ by a solution of the homogeneous equation

^ + p f r ) y = o.

Therefore, the general solution of

~ + p ( x )y = q{x)

isu(x) + z(x),

where u(x) is any solution of the equation and z{x) is the general solution of the associated homogeneous equation.

Because of this analysis, the differential equation (often called the non- homogeneous equation)

+ p ( x )y = q(x)

is solved in three steps:

(1) Find the general solution of the associated homogeneous equation

% + - °-(2) Find any solution of

j - + P(x)y = q{x). ax

(3) Add the results.

In the next two sections we shall treat steps (1) and (2) separately.

EXERCISES

Suppose the input to a black box is y(x) and the output is one of the following. Determine in each case whether the black box is linear or non-linear:

1. J ^ y(x) dx 2. y(x) + x

3- S + 6 s + » 4' 3»w

5. In \y(x)\ 6. y(x + 1).7. What is the general solution of the differential equation dy/dx — x2? Is your

answer a particular solution plus the general solution of the associated homogeneous equation?

8. A big black box consists of two linear black boxes in series:


Is the big box linear?

4. HOMOGENEOUS EQUATIONS

Let us solve the homogeneous equation

£ + - o.

Obviously y(x) = 0 is a solution. Assume y{x) 0 and separate variables:

— = - p ( x ) dx, In \y\ = - P ( x ) + c,y

where P(x) is any antiderivative of p(x) . Now exponentiate both sides, setting k = ec:

\y(x)\ = ke~p(>x).

Hence,

/ N ( ke~p{x) if y > 0 y{ ) ~ ) - h e r ™ if y < 0.

These facts are summarized in the following statement.

4. Homogeneous Equations 531

The general solution of

dyTx + v(-x)v ~ 0

isy(x) = ae~p( x\

where P (x) is any antiderivative of p(x) and a is a constant.

EXAMPLE 4.1

dyFind the general solution of - — b (sin x ) y — 0.

dx

Solution: The equation is of the form

~ + p ( x ) y = 0, dx

where p(x) = sin x. An antiderivative of p(x) is P (x ) = — cos x. By the formula, the general solution is

y(x) = ae~p(x) = aeCOBX.

A ns we r : y(x) = aecoax.

EXAMPLE 4.2

Solve the initial-value problem y' + 3y = 0, i/(l) = 4.

Solution: In this case p(x) = 3. An antiderivative is P {x) = 3x and the general solution is

y(x) = ae~'Zx.

To choose the correct constant, substitute x = 1, y = 4:

4 = ae~3j a = 4e3.

A n sw e r : y(x) = 463<1~*}.

Solve the initial-value problem ~ + - y = 0, y ( 1) = 2.dx x

Solution: Since p(x) = 1/a: is not defined at x = 0, the solution may behave badly near x = 0. Therefore consider only x > 0.

An antiderivative of 1 / x is In x; the general solution is

y(x) = ae~Xnx = - x

(which----> oo as x ---->0). To choose the correct constant, substitutex = 1, y = 2:

532 13. DIFFERENTIAL EQUATIONS1

Answer: y(x) = x > 0. x

EXERCISES

Solve:

l . ? + 3^ = 0 ax x

3. + V tan 0 = 0

5. x + (1 + x)y = 0.


6. + xy = 0, 2/(0) = - 1

8. ^ + J/ V x = 0, y(0) = 2

10. Solve y" — y' = 0 by substituting = y'.

dx2. L — + Ri = 0, L, R constants

at1 dy y

*2x dx x2 + 1 = 0

7. (1 - x2) + xy - 0, y(0) = 3

9 - 1 + 1 - 0 .

5. NON-HOMOGENEOUS EQUATIONS

We now consider the non-homogeneous equation

^ + P<»2/ = qix).

By our analysis of the problem, we need only one particular solution of this equation, to be added to the general solution of the associated homogeneous equation. We shall discuss two methods for finding a particular solution.METHOD 1. GUESSING.

This works particularly well when(a) p(x) is a constant, and(b) q{x) and all of its derivatives can be expressed in terms of a few

functions. For example, a quadratic polynomial and its derivatives can be

expressed in terms of 1, x, and x 2. The function sin x and its derivatives can be expressed in terms of sin x and cos x. Each of the functions

ex, xex, x 2exj x cos x, ex sin x

has a similar property; it and all its derivatives involve only a few functions. However, 1 /x is not of this type; each of its derivatives involves a different function.

EXAMPLE 5.1

dyFind a solution of ------y = x 2.

dx

Solution: The right side of the equation is a quadratic polynomial. Guess a quadratic polynomial

y(x) = A x 2 + B x + C,

and try to determine suitable coefficients A, B, C. Substitute y(x) into the differential equation:

(2A x + B) - ( A x 2 + B x + C) = x 2,

- A x 2 + (2A - B ) x + (B - C) = x2.

Equate coefficients:

— A = 1, 2A - B = 0, B - C = 0,from which

A = —1, B = - 2 , C = - 2 .

i n s w ; $/(x) = —(a;2 + 2x + 2) is a solution.

EXAMPLE 5.2

dyFind a solution of ------y — e3x.

dx

Solution: Try y = A e 3x. The equation becomes

3 A e 3x — A e 3x = e3x.Hence,

2 A - l , A - 1- -

5. Non-Homogeneous Equations 533

Answer: y(x) = e3x is a solution. A

EXAMPLE 5.3

Find a solution of — y = xe3x. dx

Solution: Notice that xe3x and all its derivatives involve only the functions xe3x and e3x. Try

y(x) = A x e Sx + B e 3x.

The differential equation becomes

( A e3x + 3 A x e 3x + 3 B e 3x) — (Axe 3x + B e 3x) = xe3x,

2 A x e 3x + (A + 2 B ) e 3x = xe3x.

Thus, 2A = 1, A + 2B = 0, from which A = B = — £•

Answer: y(x) = - xe3x — - e3x

is a solution.

EXAMPLE 5.4

Find a solution of — y = sin x. dx

Solution: All derivatives of sin x involve only sin x and cos x. Try

y(x) — A cos x + B sin x.

The differential equation becomes

{ — A sin x + B cos x) — (.A cos x + B sin x) = sin x ,

— (A + B) sin x + (B — A ) cos x = sin x.

Therefore— (A + B) = 1, B - A = 0.

Thus A = B = —

Answer; ~ cos £ — ~ sin xA A

is a solution.

There is a special situation in which the method must be modified. Consider for instance the differential equation

dy— - y = e* dx

Try 2/(x) = A e x:

-j- ( A e x) — Aex =


Ae* - Aex = ex, 0 =


which is impossible. The guess fails because ex is a solution of the homogeneous equation

Consequently, substituting y = A e x makes the left side zero; there is no hope of equating it to the right side. The function xex is a solution, however, as is easily checked. This is an example of a general situation:

If q{x) is a solution of the homogeneous equation

This fact soon will seem more natural when we discuss Method 2. (See Example 5.8.) It is easily verified:

Solution: Notice that y(x) = x is a solution of the homogeneous equation

But x occurs also on the right-hand side of the differential equation. Therefore, according to the rule, a solution is x • x — x 2. Check it!

— + p ( x ) y = 0,

then y(x) = xq(x) is a solution of the non-homogeneous equation

t - + p ( * ) v = q ( x ) .

= q{x) + x • 0 = q{x).

EXAMPLE 5.5

Find a solution of — - y = x.dx x

Answer: y(x) = x 2 is a solution.

Summary

Here is a list of standard guesses for solving


q (x) Guess

a, constant A , constantax + b A x + Bax2 + bx + c A x 2 + B x + Cex A e xsin x ) cos x A cos x + B sin xex(ax + b) ex(A x + B)ex(ax2 + bx + c) ex( A x 2 + B x + C)(ax + b) sin x (Ax + B) cos x + (Cx + D) sin xex sin x , ex cos x ex(A cos x + B sin x)

METHOD 2.

Begin with any non-zero solution u(x) of the homogeneous equation

du , N— + p(x )u = 0. dx

Look for a solution y(x) of the non-homogeneous equation in the form

y(x) = u(x) • v(x).

The unknown function is now v(x). Substitute y — uv and y' = uv' + vu' into the differential equation y' + p y = q:

uv' + vu' + puv = q,

uv' + v(u' + pu) = q.

But the expression in parentheses is 0. Hence,

uv' = q.

Since u is not 0,

dv _ q(x)

dx u(x)

Antidifferentiate to find v(x). (This may be difficult or impossible, a disadvantage of the method.)

If V(x) is an antiderivative, then V{x)u{x) is a solution of the differential equation. If you keep the constant of integration, then you get [V(x) + c]u(x)} which is the general solution of the differential equation. (Why?)

EXAMPLE 5.6

dy yFind the general solution of -------- = x z + 1.

dx x


Solution: The associated homogeneous equation

du u _dx x

has a solution u(x) = x. Set y = xv and substitute:

dv xvX — + V ------- = X3 + 1,

dx x

dv 1 . 1 dv 2 . 1 x — = x 3 + 1, *— = x 2 H----ax dx x

Antidifferentiate:

v = - x z + In \x\ + c. o

The general solution is xv(x).

Answer: y(x) = - x4 + x In \x\ + ex.o

EXAMPLE 5.7

duSolve — + 2xy = x.

dx

Solution: Solve the associated homogeneous equation

du- — h 2 xu = 0: dx

du— = —2x dx, In Iu\ = —x 2 + c. u

One solution is

Now setu(x) = e~x\

y = e~x2v,

soy' = —2xe~x\ + e~xV.

Substitute into the differential equation:

— 2xe~xlv + e~xV + 2xe~x*v = x y

v' = xex\Antidifferentiate:

Hence the general solution is

y = e~x*v = - + ce~x\A

Hindsight: We should have guessed the particular solution y = \ .

Alternate Solution: Separate variables:

s - * 1- * * r— \ In |1 — 2y\ = \ x 2 — a,

2 2 2

In 11 — 2 /| = a — x2,

|1 — 2i/| = 2ke~x* (2 k = e°).


It follows that

1 - 2y = ± 2 k e ~ x\

y = + ce- *2, c any constant. a

Next we derive an assertion made in the discussion of Method 1 (rather than just pulling it out of a hat).

EXAMPLE 5.8

dyFind a solution of — + p ( x ) y = q{x) if q(x) is a solution of the

dxassociated homogeneous equation.

Solution: Try a solution y(x) = v(x) • g(x).

dy d dv dg

dv / dg \ dv , ^- s * + " f e + f ’3/) * ? s + o '

Hence, the differential equation becomes

dy dvq T x - q’ * - 1 '

Answer: 2/(x) = xg(x) is a solution.

EXERCISES

Find a particular solution:

z . d£ + i , . , .

5. ^ + y = x2e* 6. ~ + 2y = cos xax dx

7. 2 ^ = 3 sin x 8. 6 ^ + y = e3* — x — 1ax ax

9. ^ + 2y = cos x 10. L ^ + ifo* = e~*‘ax a£

11. + Ri = E cos 2 12. ~ + ay = cos 20dt dd

13. J - y = x* 14. ^ + 2y = e"2*ax ax

15. — — y = ex dx

Use Method 2 of the text to find a particular solution:

16. y - + xy = x 17. ^ = x2 + 1 dx dx x

is = I iq 2xydx x x2 dx x2 + 1

Find the general solution:

20 . ^ + 32/ = 0 21 . ~ + 32/ = 1

22. ^ - 22/ = 3x2 + 2x + 1 23. ^ + y = xe2* + 1

24. ? - 22/ = x2e- 25. L ^ = # cosax dt

26. L ^ = Eekt 27. ^ + 2xy = 2x<« ax

28- S + « ^ ? I - * » . g +dv30. Show that the Bernoulli Equation — + p(x)y = q(x)yn can be reduced to a

linear equation by the substitution z = y l~n.

31. (cont.) Solve y' + xy = \ / y ; find a formula for the answer, but do not try to evaluate it.

32* Suppose q(x) satisfies \q(x)\ < M and that y(x) is a solution of the initial-valueproblem^ + y = q(x), y(0) = 0. Show that \y(x)\ < M for x > 0.

ax33. Sketch the direction field of y' = x + y. By inspection of the field find a par

ticular solution of y' — y = x.


6. APPLICATIONS

We now consider some applications of the preceding material. In each case we must first interpret the geometrical or physical data in terms of a differential equation.

EXAMPLE 6.1

Find all curves y = y(x) with the property that the tangent line at each point (x, y) on the curve meets the x-axis at (x + 3, 0).

Solution: Sketch the data (Fig. 6.1). The slope of the tangent line is d y / d x on the one hand and [y — 0] /[x — (x + 3)] = —y / 3 on the other.

Hence the differential equation is

dy 1 dx ~ S y '

Answer: y = ke~xlz.

EXAMPLE 6.2

Find all curves that intersect each of the rectangular hyperbolas xy — c at right angles.

6. Applications 541

Solution: The family of hyperbolas is shown in Fig. 6.2. First construct the direction field which at each point (x, y) is tangent to the hyperbola xy — c through the point.

F ig . 6.2

Differentiate:

y + xy' = 0.

Hence

The direction field perpendicular to the hyperbola through (x, y) is determined by the negative reciprocal of y ' . In other words, the differential equation of the direction field is

dy xdx y

Separate variables:

y dy — x dxy y 2 = x2 — k.

A n sw er : The perpendicular curves (Fig. 6.3) are the rectangular hyperbolas x2 — y 2 = k.

EXAMPLE 6.3

A spherical mothball evaporates at a rate proportional to its surface area. If half of it evaporates in 3 weeks, when will the mothball disappear?

Solution: Let r = r(t) denote the radius at time t in weeks, with r(0) = r 0 . The volume at time t is V(t) = r 3 = |7r[r( )]3, and the surface area is A( t) = 4ttr2 = 4ir[r(t)]2.

The data are

<ii) V(3) - I VIO) - i ■ t2 3

By (i),

that is,

from which

Hence

d_dt (rr‘) - -

4/c7rr2,

4t r 2 — = — 4kTr2, dt

dr— = —k. dt

r = —kt + c.

Evaluate the constant c by substituting the initial condition, r = r 0 when t = 0:

T0 = C.

Hencer = r0 — kt.

To find k } use condition (ii):

ir[r(3)]3 = ^ | irr03, hence r(3) = ■

But r(3) = r 0 — 3k, consequently

i k = r° ~ 3k’ fc = 3°(1 _ ^ i ) ‘Therefore, the formula for r is

From this it follows that the mothball disappears (r = 0) when

' - R T “

6. Applications 543

Answer: The mothball disappears after approximately 14.5 weeks.

EXAMPLE 6.4

Let T = T(t) denote the average temperature of a small piece of hot metal in a cooling bath of fixed temperature 50°. According to Newton's Law of Cooling, the metal cools at a rate proportional to the difference T — 50. Suppose the metal cools from 250° to 150° in 30 sec. How long would it take to cool from 450° to 60° ?

Solution: In mathematical terms, the Law of Cooling is a linear differential equation:

d T— = - k ( T - 50), dt

that is,d T— + k T = 50/c. dt

The associated homogeneous equation

has the general solutionT = ce~kt.

It is easy to guess the particular solution T = 50. Conclusion: the general solution is

T = 50 + ce~kt.

Setting t = 0, we find

T ( 0) = 50 + c, c = T (0) - 50.

HenceT = 50 + [T(0) - 50]e~kt.

We must determine the constant k. By hypothesis, if T (0) = 250, then T(30) = 150:

T(30) = 150 = 50 + [250 - 50]e~S0k,

100 = 200e-30*, k = - ^ ln i 111 2.


Therefore,

T = 50 + [T(0) — 50]e~kt, where k = ln 2.oU

In this formula we substitute T = 60 and T(0) = 450, then solve for t :

60 = 50 + [450 - 50]e“*‘, = er**,400

1, 1 1 „ 30 ln 40t = - - ln — = - ln 40 = —— — •

k 40 k ln 2

30 In 40Answer: —-—-— « 160 sec.

In 2EXAMPLE 6.5

Water flows from an open tank through a small hole of area A ft2 at the base. When the depth in the tank is x ft, the rate of flow is (0.60) A y / 2gxy where g — 32.17 ft/sec2. How long does it take to empty a full spherical tank of diameter 10 ft through a circular hole of diameter 2 in. at the bottom? (Assume a small hole at the top admits air.)

Solution: Let V denote the volume when the depth is x. Then_ _

— = —k y / x j where k = (0.60) A \ / 2 g. dt

By the Chain Rule,d V dx _ r~T x d i - ~ k V x ■

Finding d V / d x is a problem in volume by slicing. Draw a cross section of the tank (Fig. 6.4). The change in volume is d V « 7xy2 dx. Hence

d V— = iry2 = 7r[52 — (x — 5)2] = 7r(10x — x 2). ax

6. Applications 545

x = 0

The differential equation is

7r(10x — x 2) ^ = — k y / x ,

(10x1'2 - x*'*) ~ = — > dt 7r

Separate variables and integrate:

s(0) = 10.

20 , 2 , k— x3 /2 -----x5/2 = ------£ + c.3 5 7r

Substitute the initial condition, x = 10 when t = 0:

Solve for t when x = 0 :

k 7tc 0 = -----t c, t = — y

7T k

where k = (0.60) A \ / 2 g , and A is the area of a circle of 2 in. diameter,


Thus

k = (0.60) A V Y g =

The answer to the problem is

80tt V lOttc 3 15

t = — = --------y = — = 6400 \ / — sec.* 7T V 2 g x 9

240

An sw er : 6400 \ / - « 2520 sec.

EXAMPLE 6.6

When a uniform steel rod of length L and cross-sectional area 1 is subjected to a tension T, it stretches to length L + D, where

j - = kT, k a constant.JL

(This is Hooke's Law, valid if T does not exceed a certain bound.)Suppose the rod is suspended vertically by one end. How much

does it stretch? Assume the weight of the rod is W.

Solution: First imagine the rod lying on a horizontal surface. Calibrate the rod by measuring distances from one end. In this way each point of the rod is identified with a number x between 0 and L.

Now suppose the rod is hung by the “zero” end. Let y(x) denote the distance from the top to the point designated by x. Since the rod is stretched due to its own weight, y{x) > x for x > 0. The amount the rod stretches is y(L) - L.

Consider a short portion of the rod from x to x + h. See Fig. 6.5. It is under tension T(x) , approximately the weight of the portion of the rod between the points marked x and L, i.e.,

L — x T(x) « W — —

J j

How much is this portion stretched by the tension T(x) ? Its original length is h. When stretched, its ends are at y(x) and y (x + h), so its new length is y{x + h) — y(x) . The increase in length is y (x + h) — y(x) — h. By Hooke's Law,

y (x + h) - y(x) - In, L — x--------------------------- , m — ■

6. Applications 547

y(x)-

y(x + h ) -

y{L)-

//

L — x

Let h ----» 0:

F ig . 6.5

dy t L — x------1 = —------dx L

This is a differential equation for the position y(x) of the point x. Solve it subject to the initial condition y (0) = 0. First integrate both sides:

k W (L - x Y y - x = — -------- -------- h c.

By the initial condition,

hence

Set x = L :

k W Lo = - — + C,

1 , 1 k W /ry - x = - k W L — - (L — x ) \

y {L) — L = - kW L .A

Answer: The hanging rod stretches

- k W L units of length.


EXERCISES

1. Find all curves that intersect each of the ellipses x2 + 3 y2 — c at right angles. Sketch the curves. (See Example 6.2.)

2. Find all curves that intersect each of the curves y = ce~*2 at right angles. Sketch them.

3. The half-life of a radioactive substance is the length of time in which a quantity of the substance is reduced to half its original mass. The rate of decay is proportional to the undecayed mass. If the half-life of a substance is 100 years, how long will it take a quantity of the substance to decay to yo its original mass?

4. (cont.) If -j- of a radioactive substance decays in 2 hr, what is its half-life?5. The population of a city is now 110,000. Ten years ago it was 100,000. If the

rate of growth is proportional to the population, predict the population 20 years from now.

6. When quenched in oil at 50°C, a piece of hot metal cools from 250°C to 150°C in 30 sec. At what constant oil temperature will the metal cool from 250°C to 150°C in 45 sec? (See Example 6.4.)

7. (cont.) The same piece of metal is quenched in oil whose temperature is 50°C, but rising 0.5°C/sec. How long will it take for the metal to cool from 250°C to 150°C?

8. When a cool object is placed in hot gas it heats up at a rate proportional to the difference in temperature between the gas and the object. A cold metal bar at 0°C is placed in a tank of gas at constant temperature 200°C. In 40 sec its temperature rises to 50°C. How long would it take to heat up from 20°C to 100°C?

9. (cont.) How hot should the gas be so that the metal bar heats up from 0°C to 50°C in 20 sec?

10. Two substances, U and V, combine chemically to form a substance X ; one gram of U combines with 2 gm of V to form 3 gm of X. Suppose 50 gm of U are allowed to react with 100 gm of V. According to the Law of Mass Action, the chemicals combine at a rate proportional to the product of the untransformed masses. Denote by x(t) the mass of X at time t. (The reaction starts at t = 0.) Show that x(t) satisfies the initial-value problem

If 75 gm of X are formed in the first second, find a formula for x(t).11. (cont.) Suppose that 50 gm of XJ are allowed to react with 150 gm of V. (Not all

of V will be transformed.) Set up an initial-value problem for x(t) and solve. Verify that x ( t )----» 150 as t ----> oo.

12. A cylindrical tank of diameter 10 ft contains water to a depth of 4 ft. How long will it take the tank to empty through a 2-in. hole in the bottom? (gee Example 6.5.)

13. (cont.) Suppose the tank in Ex. 12 is a paraboloid of revolution opening upward. It contains water to a depth of 4 ft, and the diameter of the tank at the 4-ft level is 10 ft. How long will it take the tank to empty through a 2-in. hole in the bottom?

14. A falling body of mass m is subject to a downward force mg, where g is the gravitational constant. Due to air resistance, it is subject also to a retarding (up-

7. Approximate Solutions 549

ward) force proportional to its velocity. Verify that Newton’s Law, F = ma, asserts in this case that

Solve, and show that the velocity does not increase indefinitely but approaches a limiting value. (Assume an initial velocity v0.)

15*A thermometer is plunged into a hot liquid. A few seconds later it records T 0°C. Five seconds later it records T ° C . Five seconds later still it records T 2°C.Show that the temperature of the liquid is

(Assume the reading T changes at a rate proportional to the difference between T and the actual temperature.)

16. In Example 6.6, is the amount of stretching of the top half of the rod half the total stretching?

17*Suppose the rod of Example 6.6 is a long right circular cone. Suspend it vertically by its base. How much does it stretch?

An initial-value problem may be converted into an integral equation, an equation in which the unknown function occurs under the integral sign.

TV - T 0T 2- T 0 + 2 T l - T2

7. APPROXIMATE SOLUTIONS

Successive Approximations

Solving the initial-value problem

= f L x , y ( x ) J , y ( a ) = b,

is equivalent to solving the integral equation

It is easy to verify this equivalence. On the one hand, if y ( x ) satisfies the integral equation, then y ( a ) = b and

On the other hand, if y { x ) satisfies the initial-value problem, then

y ( x ) - b = y { x ) - y ( a ) = J ^ dt = J f i t , y ( t ) ] dt.

One way to attack the integral equation is to compute a sequence

yo(x) , y i ( x ) , y2(x ) , •••

of approximate solutions. The following is an iteration procedure that generally produces better and better approximate solutions.

S tep (0 ) : Choose an approximate solution yo{x) . (The constant function yo(x) = b is a reasonable choice.)Set


Step ( 1 ) :

Step (2 ) :

y i (x ) = b + If *fZt,J a

2/o (<)] dt.

Usually, y i (x ) is a better approximation than yo(x) . Set

y2(x) = b + I f i t , 2/i(<)] dt,J a

a better approximation still.

Step ( n) : Set

y n(x) = b + I Un-\J a

( 0 ] dt.

EXAMPLE 7.1

Construct y ^ x ) for the initial-value problem

dydx

V 2/(0) « 1.

Start with yo(x) « 1.

Solution: The integral equation is

y ( x ) = 1 + I y(0 dt. Jo

Apply the iteration procedure starting with 2/0(2;) = 1:

y x(x) = 1 + 1 ?/o(0 dt = 1 + 1-dt = 1 + x,J o J 0

y t (x ) = 1 + / (1 + t) dt = 1 + x + - x2,


fJ o

yz(x) = 1 + j ( 1 + t + - t2) dt = 1 + x + ~ x2 + - x3,

nl 1 \ i l l

1 + t + - t2 + - t*J dt = 1 + x + - x2 + - x* + — x4.2 6 24

Answer: 1 + x + ~ x 2 + \ x * + x*.2 6 24

R e m a r k : The answer is the 4-th degree Taylor polynomial of ex.

EXAMPLE 7.2

Start with the approximation yo(x) = 2 and compute y*(x) for the initial-value problem

y' = x2y - 1 2/(0 ) = 2.

Solution: The corresponding integral equation is

y ( x ) = 2 + 1 [ t2y ( t ) - 1] dt.

The successive approximations are

/ ■Jo

Joy i (x ) = 2 + [<22/0(0 - 1] dt

= 2 + 1 ( - 1 + 2t2) dt = 2 - z +Jo

Ih(x) = 2 + 1 [<22/i(0 — l]d< Jo


Answer:

* « ■ > - * - * + ! * - i » , + 5 » - s ' + a *R e m a r k : The pattern of the iteration should be clear. If we want y\(x)>

we repeat each term in yz(x) and add the two new terms:

I t2 ( - — t1 + — i9) dt = — x10 + — x12.J o \ 28 81 / 280 972

This gives us y ( l ) ~ y*( 1) ^ 1.50187. Similarly the next two terms aref * / - l 1 \ - 1 1/ t2 ( — t10 H------ t12) dt = --------- x13 H------------x15,

J 0 \280 972 / 13-280 ^ 15-972 ’

which gives us y ( l ) y b( 1) ^ 1.50166.This method of successive approximations, also called the Picard Method,

is often useful in finding approximate solutions. It is of great theoretical importance because it can be used to prove what we have taken for granted, that each initial-value problem for a sufficiently smooth direction field has a solution.

Taylor Polynomial Approximations

Consider an initial value problem

y - = f£z , y ( z ) J , y ( a ) = b, dx

where f (x , y) is a polynomial, rational function, or more generally, a function that can be differentiated as often as we please. If we assume that y ( x ) is differentiable (once) and d y / d x = f\_x, y{x)~], then

= y W D = + f « l x> v W \ y ' (*)

exists by the Chain Rule. Hence y ( x ) is differentiable twice. This implies the existence in turn of

- { fxlx, 2/(z)]} + ^ { f y[x, y (x )~ \ }y ' (x ) + /„ [x , y{x)~\y"{x)

= fxx + 2f xvy' + fyy(y ' )2 + f yy".

By continuing this bootstrap operation, we find that y (4\ y (b\ • • • all exist.What is more, we can calculate the values of these derivatives at x = a

in terms of an assumed initial value y ( a ) = b of y ( x ) :

y' (a) = /(a, 6), y" (a) = f x(a, b) + / y(o, b ) y ' ( a ) ,

y' ' '(a) = fxx (a, b) + 2fxy(a, b ) y ' (a ) + f yy (a, b ) y ' ( a ) 2 + / y(a, b)y"(a),

etc. Hence we can compute the n-th degree Taylor polynomial of y ( x ) at x = a:

/ \ , \ , 2/'(a) , \ , v " (a ) , N, , y (n)(a) ,Pn(x) = y ( a ) + — 7 7 - (a; - a) H-----— ( z - a ) 2 + • • • H------r r ~ ( x ~ a ) •1! 2! n!

It is proved in advanced courses that p n(x) is a good approximation to the solution y ( x ) for n sufficiently large and x near a.

Let us try this technique on the same examples we used for successive approximations.

EXAMPLE 7.1

Compute the 5-th degree Taylor polynomial at x = 0 of the function y ( x ) that satisfies the initial-value problem

dydx y 2/(0 ) = 1.

Solution:2/' = y 2/'(0) =

y " = y' = y 2/"(0 ) =

y">= 2/' = y 2/" '(0 ) =

yW = y j/«)(0) =

2/<w = y 2/«)(0) =

Thus

X2 Xs X? X 5

x2 xz x* Answ er: 1 + x + — + — + •— + — <

2! 3! 4! 5!

R e m a r k : This simple problem has an explicit solution, y = ex. The answer to the example is its Taylor polynomial p*>(x) at x = 0. Observe that

* '(* ) = 1 + x + 2 \ + V + I \ = M X ) ~ 5! •

Thus Pb(x) satisfies the initial condition and, near x = 0, nearly satisfies the differential equation, y' = y.

EXAMPLE 7.2

Approximate the solution y ( x ) of the initial-value problem

dydx

x2y - 1 y ( 0 ) = 2

by its 9-th degree Taylor polynomial at x ~ 0.

Solution: The value y ( 0) = 2 is given; the values y ' ( 0), y " ( 0),2/(9)(0) must be computed. Now

y' = x2y - 1, y ' ( 0) = - 1 .

Differentiate the expression for y ':

y" = 2 xy + x W , 2/"( 0) = 0.Differentiate again:

y'" = 2 y + ±xy' + x*y".

Substitute x = 0 and 2/(0 ) = 2 to obtain

2/'"(0) = 4.Continue in this way:

yW = 6y' -j- 6xy" + x2y"

2/(5) = 12?/' + 8xy"' + x2i/(4)

2/(6) = 20 i/'" + 10;n/(4) + £22/(!

?/(7) = 302/(4) + 12^2/(5) + x22/((

(8) = 42?/(5) + 14 2/(6) + x2y (‘

2/(9) = 562/(6) + 16x2/(7) + x22/°

The coefficients of the Taylor approximation are

4480 1


y w (0) = 6^(0) = - 61 2/®(0) = 12?/" (0) = 0r (5 ) t/(6)( 0) = 80i(6) 2/<7) (0) = -1 8 0r(7) ?/(8)(0) = 0,(8) y»)(0) = 4480.

2, - 1 ,4

°- « -2

: 3 ’- 64!

80 1 -1 8 0 - 10,6! “ 9 ’ 7! 28 ’ 9! 81

Answer:

y ( x ) t t 2 - * + ? * » -

EXERCISES

Compute 2/3(x) by successive approximations for the initial-value problem; choose Vo(x) = 2/(0 ):1. 1f = xy + x; 2/(0) = 0 2. 2/' = xy + y; 2/(0) = 13. y' = xy + 1; 2/(0) = 2 4. 2/' = y + sin x\ 2/(0) = 15. y' ii * i 2/(0) = 1 6. 2/' = 1 + xy2-, 2/(0) = 07. i/' = x + y3; 2/(0) = 0 8. 2/' = —x2 + y2; <si —s o II 0

9. y' = x2 + y2; 2/(0) = 0 10. 2/' = 2/(0) = 1.


Without solving, compute the 5-th degree Taylor polynomial at x = 0 of the function satisfying the initial-value problem:11. y' = xy + x; y ( 0) = 0 12. yf = xy + y; y ( 0) = 113. y' = x y + 1; 2/(0) = 2 14. y' = y + sins; 2/(0) = 115. y' = z3t/ — s; 2/(0) = 1 16. 2/' = 1 + xy2; 2/(0) = 0I7 . y ' = x + y*; 2/(0) = 0 18. 2/' = — x2 + y2; 2/(0) = 0.Compute the 4-th degree Taylor polynomial at x = 0 for the initial-value problem: 19. y' = y sin(z2); 2/(0) = 1 20. y' = 1 + x2e~v; 2/(0) = 121. 2/' = e* + xy2\ 2/(0 ) = —1 22. y' = cos(:n/)-- 1; 2/(0 ) = —123. 2/ = x sinh y ; y(0) = 0 24. y’ = (1 + z2)(l + 2/2); 2/(0) = 2.25. The height x(l) of a certain balloon satisfies tx = 0.5 ( — z)2. When 2 = 10 sec,

x = 10 ft. Estimate x(\2) to 0.1 ft accuracy.

14. Second Order Equations and Systems

1. LINEAR EQUATIONS

We shall study second order linear differential equations with constant coefficients:

d 2x dx ..

where p and q are constants. Equations of this type have many physical applications, particularly to elastic or electric phenomena.

First order linear equations are solved by finding one particular solution and adding it to the general solution of the associated homogeneous equation. The same method applies to second order linear equations for exactly the same reason. Regard the left-hand side of the equation as a “black box” ; an input x(t) leads to an output x + p x + qx. The sum x + y of two inputs leads to the sum of the corresponding outputs:

(x + y )*' + p ( x + y)' + q(x + y) = (x + p x + qx) + (y + p y + qy)-

A constant multiple ax of an input leads to the same constant multiple of the output:

(ax)" + p(ax)' + q(ax) = a(x + p x + qx).

Thus the “black box” is linear.

The general solution of

isx(t) + z(t),

where x(t) is any solution of the differential equation and z(t) is the general solution of the homogeneous equation

d 2x dx"77 + P T + qx = 0. dt2 y dt *


Therefore, as in the last chapter, we first treat homogeneous equations, then look for particular solutions of non-homogeneous equations.

2. HOMOGENEOUS EQUATIONS

In this section we study homogeneous equations

x + p x + qx = 0,

where p and q are constants. There are three important special cases, each with p = 0, namely q = 0, q = — k 2, and q = k 2.

C a s e 1 : x = 0. This is easy. The solution is

x = at + b.

C a s e 2: x — k 2x = 0. Here, the second derivative of x is proportional to x itself, which reminds us of the exponential function. It is not hard to guess two solutions: x = ekt and x = e~kt. By linearity,

x = aekt + berki

is also a solution. In Section 7 we shall prove this is the general solution.

C a se 3: x + k2x = 0. Again it is easy to guess two solutions: x = cos kt and x = sin kt. In Section 7 we shall prove the general solution is

x = a cos kt + b sin kt.

Summary

Differential equation General solution

x = 0 x = at + bx — k 2x = 0 x = aekt + be~~ktx + k 2x = 0 x = a cos kt + b sin kt

These special cases are important because each homogeneous equation

x + p x + qx = 0

can be reduced to one of them. The trick is analogous to reducing a quadratic equation X 2 + p X + q = 0 to the form Y 2 ± k 2 = 0 by completing the square. Set

x(t) = ehty ( t ),

where y(t) is a new unknown function and h is a constant. Compute derivatives by the Product Rule:

x = hehty + ehty

= eht(y + hy) ;

x = heht(y + hy) — eht(y + hy)

= eht(y + 2 hy + h2y).

Substitute these expressions into the differential equation:

eht(y + 2 hy + h2y) + peht(y + hy) + qehty = 0.

Cancel eht and group terms:

y + (2h + p ) y + (h2 + ph + q)y = 0.

Now choose h = —p / 2 to make the term in y drop out. The coefficient of y is then

p 2 p 2 —p 2 + 4 qh* + ph + q = | - | + q = q~ ,

and the differential equation becomes

p 2 — 4g y ---------^— y = 0.

This is one of the three special cases, depending on whether p 2 — 4q is zero, positive, or negative.

EXAMPLE 2.1

Solve x + 6x + 9x = 0.

Solution: Here

v o p 2 - np = 6, q = 9, 2 = ~ ----4---- =

The substitution x = e~zty transforms the differential equation into

V = 0,from which

y = at + b.

Answer: x = e~zt(at + 6), where a and b are constants.

558 14. SECOND ORDER EQUATIONS AND SYSTEMS

EXAMPLE 2.2

Solve x — 3x + 2x = 0.

Solution: Here


p = - 3, q = 2, h = - - =p 3 p2 — 4g = H ) *2 2 4

The substitution x = eu/2y transforms the differential equation into

y - ( O ’ * - 0-

one of the standard forms. Conclusion:

y = aet/2 + be~tl2.

Answer: x = e w (aet]2 + be~tl2) = ae2t + beK

EXAMPLE 2.3

Solve x + 2x + 5x = 0.

Solution: In this example

I. v , p 2 - 4qP = 2, q = 5, ft = - - = - 1 , ----- ---- = - 4 = —22.

Set x = e- *?/. Theny + 2 2y = 0,

y = a cos 21 + b sin 2£.

Answer; x = e-<(a cos 2Z + & sin 2t).

Formula for Solving x + px + qx = 0

When you setx = e~(p,2)ty ,

the differential equationa; + px + g = 0

is transformed into

Ay — —y — 0, where A = p 2 — 4g.

Its solution depends on the sign of A:

C a s e 1: A = 0. Then

x{t) = e~~(pl2)t(at + b).

C a s e 2: A > 0. Write A = a 2. Then

x(t) = g-(p/2)*(ag(W2)* -j- be~^,2)t)

= a6^—p+<r)/2ii +

C a s e 3: A < 0. Write A = — a2. Then

x(t) = e~(pl2)t cos ^ ^ + b sin ^ j .

Now compare these three cases with the three cases describing the roots of the quadratic equation

X 2 + p X + q = 0.

By the quadratic formula, the roots are

. . - l + i v z . 0 - - H V I

C a s e 1: A = 0. The roots are real and equal:

V

C a s e 2: A > 0. Write A = a 2. The roots are real and distinct:

- p + a - p - aa = ------------ > p = ---------------

2 2

C a s e 3: A < 0. Write A = —a 2. The roots are complex:

— p + ai —p — ai . /— 7a = „ -> j8 = — = V - 1 .


To solve the differential equation

x + p x + qx = 0,

find the roots a and /3 of the quadratic equation

X 2 + p X + q = 0.

1. If a = 13, then the solution is

x = eat(at + b).

2. If a and 13 are real and distinct, then the solution is

x = aeat + befit.

3. If a and fi are complex,

a = h + hi, (3 = h — k i ,

then the solution is

x = eht(a cos kt + b sin kt).


Let us test these formulas on some numerical examples.

EXAMPLE 2.4

Solve each equation by the preceding rule:*

(a) x + 6x + 9a; = 0,

(b) x — Sx + 2x = 0,

(c) x + 2x + 5x = 0.

Solution:

(a) X2 + 6X + 9 = 0,

(X + 3)2 = 0, a = fi = - 3 ,

x = e~Zi(at + b).

(b) X 2 - 3X + 2 = 0,

( X - 2 ) ( X - 1 ) = 0 , a = 2, f i = 1,

x = ae2t + be1.

(c) X 2 + 2X + 5 = 0,

a = - ( — 2 + 16) = — 1 + 2z,

/3 = | ( — 2 — = “ I “ 2t,

x = e- *(a cos 2t b sin 2 t ) .

Answer:

(a) x = e-~u (at + b),

(b ) x = ae2* + be1,

(c) x = e~*(a cos 2£ + 5 sin 2t).

EXERCISESFind the general solution:

1. x — 6a; + 5a; = 0 2. x + 7a; + 10a; = 0

6. x — Sx + 16a: = 0

7 . 1 + U - O 8' 4 S + 4 s f + !' - 0

11. — 4 - + 4ax = 0 12. x + x + x = 0.a3 a

Find the most general solution, subject to the indicated condition:

13. g + 9y = 0, 2/(0) = 0 14. g + 4j/ = 0, j,'(0) = 0

15. x — 4x + 4x = 0, x(0) = 1 16. x + 6x + 8 = 0, y(0) = 0

17- S - 2 ^ + 2r = 0, rW = l 18. g - 2 % + 2r = 0, r'fcr) = 1


d 0 v / r f 0 2 d d

d2r dr d f 2 ~ dd19. S + r - 0 , r Q = 0 20. S “ 5 ^ + 6r = 0, r'(0) = 3.

d^y dy21. If a and b are positive constants, show that each solution of — + a - — [- by = 0(IX CLX

tends to zero as x ---- > <x>.22. Let u(t) and v(t) be solutions of x + P(t)x + Q(t)x = 0. Set w = uv — uv. Show

that w + Pw = 0. Solve for w. (w is called the Wronskian of u and v.)

3. PARTICULAR SOLUTIONS

We return to the equation

x + px + qx = r(t).

The problem is to find any one solution. There are several ingenious methods for doing this, some quite complicated. We consider only the easiest one: guessing. This works just as in the last chapter.

EXAMPLE 3.1

Find a solution of x + 3x — x = t 2 — 1.

Solution: Try x = A + Bt + Ct2. Then x = B + 2Ct and x = 2C, so substitution in the equation yields

2C + 3(B + 2Ct) - (A + B t + Ct2) = t2 - 1.

Equate coefficients:

t2: - C = 1,

t : + 6C = 0,

1: —A + SB + 2C = - 1 .

Therefore C = — 1, B = —6, A = —19.

Answer: x = —19 — 6£ — is a solution.

3. Particular Solutions 563

EXAMPLE 3.2

Find a solution of x + 3x — x = e~*‘.

Solut ion: Try x = Ae ~ 4t. Then x = —4Ae ~ 4t and x = 16Ae ~ 4t. Substitute, and cancel e~4t:

16A - 12A - A = 1, 3A = 1.

Answer: x = ~ e~~4t is a solution,o

Find a solution of x + Sx — x = cos t.

Solution: Try x = A cos Z + B sin Then x = — A sin t + B cos £ and x = —A cos t — B sin t. Substitute:

( — A cos t — B sin t) + 3( — A sin t + B cos t) — (A cos t + B sin t) = cos t.

Equate coefficients of cos t and sin t :

/ - 2 A + SB = 1

I —3A - 2B = 0.

The solution of this system is A = — and B =

Answer: a: = ( —2 cos t + 3 sin t)l o

m i ' is a solution.

EXAMPLE 3.4

Find the general solution of x — x = t.

Solution: The quadratic equation X 2 — 1 = 0 has roots 1 and — 1. Hence the solution of the homogeneous equation x — x = 0 is x = ael + be~l. We guess the particular solution x = —t.

Answer: x(t) = ael + be~l — t} where a and b are constants.

Initial-Value Problems

The general solution of the differential equation

x + p x + qx = r(t)

involves two arbitrary constants. By choosing them suitably you can usually find a solution satisfying two additional conditions. Most important is the case of initial conditions:

Find a solution x = x(t) of the differential equation

x + p x + qx = r(t)such that

x( t0) = a , x( t0) = b.


EXAMPLE 3.5

Solve the initial-value problem x — x = t, x(0) = 0, i:(0) = 1.

Solution: By the last example, the general solution of the differential equation is

x = ael + be~l — t.Hence

x = ael — be~l — 1.

Substitute the initial conditions:

j a + b = 0

\ a — b — 1 = 1.

This system has the solution a = 1 and b = — 1.

Answer: x = el

EXERCISES

Find a particular solution:1. x + Sx = 1 2. 2x + Sx = 53. x — 4x = 2t + 1 4. x + x = t25. x + Sx — x = t2 + 4t + 6 6. x + x + x = t*7. x + x + 2x = 2ezt 8. x + x = cosh 2

9- 3 S + l _2/ = e“2l + X 10. 2 g - 3 » - 8 i n *

1L S + I " 32/ = 4sin2x 12‘ S + 2| + 2/= 1+X + C0S3Xdv d2i/13. —j r + -— (-2y = ex cos x 14. 2 -r-r — y = x cos zrfx2 ax dx2

15' l x 2 + y = e* + *** + e3x + eU + e5x

< P y _ d y _ Sxlb- dx2 dx y xe ■

Find the general solution:17. x + x = t2 18. x - 7x + lOx = 3e‘

4. Applications 565

19. x + x — 6x = te 1 20. x + x — 5x = 2t + 3

21. x + 3i = cosh 2t 22. 3 + y = x2 + 5

(i2i di d2v23. 3 — + 4 — + 2i = 10 cos t 24. — - 2r = sin 0 -f cos 0.dt2 cm dv1

Solve the initial-value problem:25. x + x = 2t - 5, z(0) = 0, x(0) = 126. x + 3x = e‘, x(0) = 0, x(0) = 127. a: — 4x = sin 2t, x(0) = 1, x(0) = 028. x - 2x - 15x = 1, x(l) = 0, x(l) = 029. 4x - 7x + Sx = e2t, x(0) = -1 , x(0) = 230. x + 2x + x = t2, x(0) = 0, x(0) = 5.31. The usual method for finding a particular solution of x — 4x = e2t fails. Why?

Try x = Ate2t.32. (cont.) Find a solution of x + x = sin t by guessing.33. Let z be a solution of the homogeneous equation x + px + qx = 0. Substitute

x(0 = z(t)u(t) in the equation x + px + qx = r(t). Show that the resulting equation for u can be reduced to a first order equation by setting w = u.

34. (cont.) Apply this technique to find a particular solution of x + x = sin t.35. Show that the equation x + k2x = r(t) has the particular solution

1 f tx (f) = 1 / sin k{t — s)r(s) ds.

k Jo

You may presuppose the formula

4. APPLICATIONS

In applications, the expression a cos kt + b sin kt occurs frequently. There is a useful equivalent expression for it. Assume a and b are not both zero. If c denotes the positive square root of a2 + b2, then

Thus the point (b / c , —a/c) lies on the unit circle (Fig. 4.1). Hence there is an angle kto, determined up to an integer multiple of 2n, such that

sin kto -------- , cos kt0 = - .c c

By the addition law for the sine,

a cos kt + b sin kt = c ( - cos kt + ~ sin k t )\ c c /

= c( — sin kt0 cos kt + cos kt0 sin kt)

= c sin k( t — t0) .

Thus the expression a cos kt + b sin kt is nothing but a sine function (sinusoidal wave) in disguise.

[Alternatively, t0 may be defined by a / c = cos kt0, b/c = sin kt0, from which a cos kt + b sin kt = c cos k( t — t0) .]


Simple Harmonic Motion

Imagine a point Q moving counterclockwise about the circle x2 + y 1 = c2. See Fig. 4.2. Starting at Q0 when t = 0, the point Q moves with constant speed, making k revolutions in sec. Its projection P on the x-axis has coordinate

x(t ) = c cos k( t — to).

As Q traverses its circle, P oscillates between (c, 0) and ( —c, 0), making one complete oscillation in 2ir/k sec. The motion of P is called simple harmonic motion with amplitude c, period 2 r / k , and time lag t0 (phase angle —kt0).

Note that the speed of P, unlike that of Q, is not constant. Why? When is it greatest?

R e m a r k : Since cos a = sin(7r/2 — a), harmonic motion may also be described in terms of a sine:

Conclusion:

7rx( t) = c sin k( t — ti), where = to — — .

2 k

4. Applications 567

EXAMPLE 4.1 (Simple Harmonic Motion)

A particle moves along the z-axis. The only force acting on it is a force directed towards the origin and proportional to the displacement of the particle from the origin (a spring). Describe the motion. (Assume the spring has negligible length when unloaded.)

Solution: Call the mass m. The force is x times a negative constant,—mk2, with k > 0. According to Newton’s Law of Motion,

mass X acceleration = force:

mx = —mk2x, x + k2x = 0.The solution is

x = a cos kt + b sin kt = c cos k( t — t0),

where the constants c and to depend on the initial position and velocity and c > 0. Hence the motion is simple harmonic.

Answer: x = c cos k ( t — to).

EXAMPLE 4.2

Solve the same problem

(a) with initial conditions z(0) ==0, £(0) = v0 > 0;

(b) with initial conditions z(0) = x* > 0, £(0) = 0.

Solution: By Example 4.1, the general solution is x = c cos k( t — to), where c and k are positive. Now determine c and £o.

(a) Since x = — ck sin k ( t — to), the initial conditions are:

0 = ccos( — kto), Vo = — ck sin( — kto),that is,

Vocos kto = 0, sin kt0 = — .

ck

From the first equation, kt0 = tt/ 2 or kto = — tt/2; from the second, sin kto > 0. Hence the only possibility is kto = ir/2, that is, to = x /2k . It follows that Vo = ck, so c = Vo/k. Therefore the desired solution is


From the second equation, kto = 0 or r ; from the first, cos kto > 0. Hence kt0 = 0. It follows that c = x0, and the desired solution is x = x0 cos kt.

R e m a r k : As this example shows, it is simpler to describe harmonic motion by the sine form when the initial position is x0 = 0 and by the cosine form when the initial velocity is v0 = 0.

EXAMPLE 4.3

A 164b weight hangs at rest on the end of an 8-ft spring attached to a ceiling. When the spring is stretched, it exerts a restoring force proportional to displacement from equilibrium position. Suppose the weight is pulled down 6 in. and released. Describe its motion.

Solution: Let x denote the distance from the ceiling in feet (Fig. 4.3). The force due to the spring is F( x) = — ex. When the spring is at rest, F = —16 and x = 8. Hence c = 2 and F ( x ) = —2x lb.

Answer: (a) x = ~ sin kt;K

(b) x = Xo cos kt.

equilibrium position

8 ft

spring stretchedF ig . 4.3

4. Applications 569

When the weight is at position x, the forces acting on it are its weight, 16 lb, and the spring, —2x lb. Since the units of force and length are pounds and feet, mass must be in slugs: m = 1 Q/g = slugs. Thus mass X acceleration = force,

16— x = 16 — 2x, x + \ x = 32.oZ

Initial conditions: £(0) = 8.5, x(0) = 0.

The associated homogeneous equation has solution

x = c cos 2 (t — to),

and it is easy to guess the particular solution, x = 8. Consequently,

x = c cos 2 (t — to) + 8

is the general solution. It describes simple harmonic motion centered about x = 8 rather than x = 0. That is reasonable since you expect the weight to oscillate about its equilibrium point at x = 8.

The initial conditions are

s(0) = 8.5, x(0) = 0.

From these you can solve for c and to. But good scientists are lazy; they refuse to do the same work twice. Use the answer to part (b), of Example 4.2. It describes harmonic motion with 0 initial velocity by x = x0 cos kt, where Xo is initial displacement. In this problem, initial displacement is Hence, you can write down the answer.

Answer: x = cos 21 + 8.

Other Applications

EXAMPLE 4.4

A particle of mass 1 attached to a spring slides along a straight surface. It is subject to a restoring force proportional to displacement and a retarding force (due to friction) assumed proportional to velocity. Describe the motion of the particle assuming it starts at the origin (equilibrium point) with initial velocity v0.

Solution: The forces are — k2x (spring) and —px (friction), where k and p are positive constants. The equation of motion is

mass X acceleration = force,

x = —px — k2x, x + px + k2x = 0.

Initial conditions: x(0) = 0, x(0) = v0.

Before solving, think about the problem. If there is no friction, that is if p is zero, then the motion is simple harmonic. If the friction is small, then the motion should be nearly simple harmonic motion, except for a gradual slowing down due to friction. If the friction is large, the motion should be considerably inhibited, in fact there might not be oscillations.

From this physical reasoning, it seems clear that the relative size of the constants k and p is crucial.

To solve the differential equation, examine the roots of the corresponding quadratic equation:

X 2 + p X + k2 = 0.They are

a = - | ^ \ / A , P = V A , w h ere A = p2 — 4fc2.

The nature of the solutions depends on the sign of A. In terms of k and p, the crucial question is whether p > 2k, p < 2k, or p = 2k.

C a s e 1: p < 2k (underdamped case: friction small compared to spring force). In this case A < 0; set A = — 4q2. Then the general solution of the differential equation is

x = e~(pl2)t(a cos qt + b sin q t ) .

Use the first initial condition, z(0) = 0; it implies a = 0. Hence

x = be~(pl2)t sin qt.

Now use the second initial condition, £(0) = v0:

x ( t ) = b ^ e~(pl2)t sin qt + qe~(pl2)t cos q t j ,

v0Vo = x (0 ) = bq, b = — .

Q

Therefore, the solution of the initial-value problem is:

v0x = — e~(pl2)t sin qt.

Q

This is a damped oscillatory motion (Fig. 4.4). The particle oscillates withconstant period 2ir/q, but the amplitude--------> 0 as t ----------> oo.

C a s e 2: p > 2k (overdamped case: friction large compared to spring force). Now A > 0. The roots of the quadratic are

t

« == ~ { - V + ^ ( - P ~ a / A ) -


Both roots are negative ( \ /A = y / p 2 — 4k2 < p ) . Let us call the roots — r

4. Applications 571

and — s, where r > s > 0. The general solution is

x = aerTt + be~8t.Since

x = —are~rt — bse~8t,

the initial conditions become

0 = a + by Vo = —ar — bs.It follows that

Therefore, the solution of the initial-value problem is

x = c(e~8t — e~rt) , where c = — ~— > 0.r — s

The graph of x as a function of t is shown in Fig. 4.5. The particle moves away from the origin at first, then reverses direction and approaches the origin, never quite reaching it again.

C a s e 3: p = 2k (borderline case, critical damping). Now A = 0, and the quadratic equation has two equal roots, a = fi = —p / 2 . The general solution is

x = e~(p,2)t(a + bt).

Thus z(0) = a and £(0) = b — \ pa , so the initial conditions reduce to

0 = a, v0 = b.

Therefore the solution of the initial-value problem is

x = vote~(p,2)t = v0te~kt.

By solving x = 0, it is seen that x reaches its maximum value 2v0/ p e = v0/k e when t = 2 / p . See Fig. 4.6. The particle moves 2v0/ p e units from the origin, reverses direction and approaches the origin again as t --------> oo. As is physically plausible, the maximum distance from the origin is proportional to the initial velocity, inversely proportional to the coefficient of friction.

F ig . 4.6

Answer: Let A = p 2 ~ 4k2.

A X a...............

If A < 0, then x = - e~W2)< sm qt, where q = - y / — A.(£ A

If A > 0, then x = —.. (e~st — e~rt) , wherer — s

r = i ( p + v / A), s = ^ ( p - V A ) -

If A = 0, then x = v0te - (pl2)t = v0te~k‘.

4. Applications 573

The next example provides a simple mechanical model for a variety of physical phenomena.

EXAMPLE 4.5

Suppose in Example 4.4 an external force F = A sin cot is imposed. What is the nature of the motion when t is large? (Describe the solution x(t) as t --------> co. ) Assume small friction.

Solution: The differential equation is

x + px + k2x = A sin cot.

Guess a particular solution of the form

x = a cos cot + b sin cot.

Substitute; after some computation the differential equation becomes

l ( k 2 — co2) a + pcob~] cos cct + [(/c2 — co2) b — pcoa] sin cct = A sin cot.

Equation coefficients of cos cot and sin cot:

ra + sb = 0

Solve:

where r = k2 — co2 and s = pco.— sa + rb = A

—As Ar a = —----- : , b =

r2 + s2 ’ r2 + s2

Hence, a particular solution isA

x = —------- ( — s cos cot + r sin cot) .r2 + s2

By the usual method, convert it to the form

x = /-o.., - - - sinco(< - to),\ / r 2 + s2

where

sin coto = —> _ , 0 and cos co0 =

is

y / r 2 + s2 y / r 2 + s2 '

If friction is small, the general solution of the homogeneous equation

x + px + k2x = 0

x = ce~(pl2)t sin q(t — t\), q = - \ / Ak2 — p2.2

(See the answer to Example 4.4, first case.)


Combine results: the general solution of

x + px + k2x = A sin cotis

ce -(p/2)* sin q(t — t\) +y / (k2 — co2) 2 + ( p c o )2

sin co (t — to) .

The constants c and t\ can be determined from initial conditions, but they arenot needed to predict the behavior of x for large values of t. As t --------» oo, theterm ce~(pl2)t sin q(t — t i ) --------> 0. (It is called a transient.) Thus aftersufficient time has elapsed,

x(t) C sin co(t — t0) ;

all that is visible is simple harmonic motion. The amplitude of this motion is

AC =

y/(k? - u 2) 2 + (poo)2

In case p is small and co k, the denominator is small, hence the amplitude C is large. This is the phenomenon of resonance. It occurs when there is a periodic force with frequency near the “natural frequency” of the system. Destructive vibrations in machinery and vibrations caused by soldiers marching in step on a bridge are results of resonance.

The approximationx(t) « C sin co(t — £0) ,

valid for large values of t, is called the steady state solution. Notice that it is independent of the initial conditions; these affect only the constants in the transient term.

Answer,

C

For large t, x(t)

A

C sin co(t

A

to), where

\ / (k2 — co2) 2 + (poo)2 y / r 2 +

and

sin co to =\ / r 2 + s2 k2

poo.COS ooto

y / r 2 + s2

The standard electric model for this situation is a simple circuit with resistance R, inductance L, capacitance C, current 7, and with voltage E = —A cos cot. The equation of the circuit is

AcoI ( t ) ^ ------ sin co( t — to) .

4. Applications 575

The steady state solution is

w g m ’ + g -)'Resonance (tuned circuit) occurs when w2 = 1/LC. Then sin uto = 1 and cos uto = 0 . Hence ut0 = t / 2 and

EXERCISES

Express as c cos k ( t — t0) and as c sin k (t — fo). Use trigonometric tables if necessary: t . cos 3t + sin 3t 2. sin t — cos t

(§). \ / 3 cos 21 — sin 2£ 4. cos t + y / 3 sin t. t , £

5. 4 sin t — 3 cos Z 6. sin - + 5 cos - . z

The next four problems concern a particle in simple harmonic motion about the origin. Its position at time t is x ( t ) and its velocity is v ( t ) .

7. Find x ( t ) if the period is 2 sec, the amplitude is 6, and x(0) = 3.8. Find x { t ) if z(0) = 0, v(0) = 5, and the period is 2 sec.

Find the amplitude if x(0) = 0, v(0) = 10, and v ( 3 ) = 5.10. Estimate the amplitude to 3 significant digits if x(0) = 0, v(O) = 10, and

x ( l ) = 6.(Q) A 16-lb weight hangs at rest on the end of an 8-ft spring attached to a ceiling.

(See Example 4.3.) The weight is pulled down k in. and released. As it passes through its original (equilibrium) position, its speed is 1.5 ft/sec. Find k.

12. (cont.) Suppose the weight slides along a wall. Due to friction there is a retarding force proportional to the velocity. The motion is a damped vibration with period 2 t / \ / 3 sec. After how long will the amplitude of the oscillation be reduced to half its original magnitude?

13. Suppose the weight in Example 4.3 is w lb, not 16 lb. What is the period of its motion?

14. A pendulum is made of a small weight at the end of a long wire. Its motion is described by the differential equation

d?0 , g . - _~dt2 ~LSm

where 6 is the angle between the wire and the vertical. If the pendulum swings only through a small arc, then sin 6 can be approximated by 6, thus simplifying the differential equation. Do so and find the approximate period of the pendulum.

15. A cylindrical buoy floats vertically in the water. Its weight is 100 lb and its diameter is 2 ft. When depressed slightly and released, it oscillates with simple harmonic motion. Find the period of the oscillation.[H in t: This is just a spring problem in disguise. Use Archimedes’ Law: A body

in water is subjected to an upward buoyant force equal to the weight of the water displaced. Take the density of water to be 62.4 lb/ft3.]

16. In the overdamped motion of Fig. 4.5, find the maximum value of x(t). Show that x (t) > 0 for t > 0 and x (t) < 0 for t < 0.

17. Consider the underdamped motion of Fig. 4.4. Show that the graph crosses the £-axis infinitely often for t > 0.

18. A rocket sled is subjected to 6<? acceleration for 5 sec. After the engine shuts down, the sled undergoes a deceleration (ft/sec2) equal to 0.05 times its velocity (ft/sec). What is the speed of the sled 10 sec after engine shutdown? How far has it traveled?

The external forces acting on a projectile are gravity and air resistance. At low altitude and low speed, it may be assumed that air resistance is proportional to speed.

If a projectile is shot straight up, it rises to its maximum height and then falls to the ground. Whether it takes longer to rise or to fall, or equal times, is not obvious. However, as we shall see, in falling there is a terminal speed. Hence a projectile shot up with initial speed faster than the terminal speed necessarily takes longer in falling than in rising. The next five exercises show this is always so.

19. A projectile is shot straight up with initial velocity Show that its height satisfies the initial-value problem y + ky = — g, y (0) = 0, y (0) = Vq. Derive the solution

y(t) = A (1 — e~k t) — 1 1, where A = ~t2 ° •K K20. (cont.) Show that v approaches a terminal velocity as t ---------> oo . Find it.21. (cont.) Show that the projectile reaches its maximum height at time

Show that the projectile returns to ground at time t2 > 0, where

(1 - = k t2.

22. (cont.) Show that t \ ---------> vo/g as k ---------> 0 and interpret.[Hint: Express the derivative of ln (1 + cx) at x = 0 as a limit; alternatively use the first order Taylor Approximation.]Guess what t2 approaches as k ---------> 0.

23*. (cont.) Prove (t2 — h) > t\. Begin by showing that y (t\ 4- r ) + y {t\ — r) > 0 for r > 0. Interpret physically. Then integrate the inequality over the interval0 < r < h to obtain y {t\ + r ) > y (h — r ) for r > 0. Deduce that 2t\ < t2.

24*. (cont.) Now try a concrete problem. Suppose the initial velocity is 500 ft/sec and the maximum height is 3000 ft. Assume g = 32.2 ft/sec2. Estimate k to 3 significant digits. Then estimate h and t2 to 2 significant digits.

5. POWER SERIES SOLUTIONS

In the theory of differential equations, there is an important result concerning initial-value problems of the type

T = f ( x , V ) , Via) = b, ax

5. Power Series Solutions 577

where f(x, y) has a convergent power series expansion near (a, b). I t says that the solution y = y(x) has a convergent power series expansion near x = a. Therefore we may assume a solution of the form

oo

y(x ) = ^ an{x — a )nn = 0

and then try to find suitable coefficients a0, ah a , • • •. This technique often leads to an exact power series solution, rather than to a Taylor polynomial approximation as in the last chapter.

Similar remarks apply to initial-value problems of the form

y (a) = b0, y' (a) = h.

EXAMPLE 5.1

Find a power series solution to

y ' * xy, 2/(0 ) = oo.

Solution: Try a solution of the form

y(x) = a0 + aix + «2X2 + • • •,

where the coefficients must be determined. Substitute this power series into the differential equation:

d— (oo + axx + (hx2 + • • •) = x(do + aix + a x2 + • • •), dx

d\ -f- 2oqX -|- Sa3x2 -f- • • • -f- (fc -|- l)dk+ixk • • •

= a0x + ai^2 + • • • + ak-ixk + • • •.Equate coefficients:

a,\ = 0, 2a2 = do, 3a3 = ai, • • •, (fc + 1)^+1 =

Hence

at+i _ *; + 1 ■

This recurrence relation expresses each coefficient in terms of the next to last coefficient. From it, all coefficients can be computed successively. For instance, all odd coefficients are zero; since ai = 0 , it follows that a3 = 0 , from which it follows that a5 = 0, and so on. The even coefficients can be expressed in terms of a0. Apply the recurrence relation with k = 1, 3, 5:

d2y ( dyd ^ ~ f V ’ y’ Tx

The pattern is clear:


1 1 1 1 12ft 2n — 2 2ft — 4 6 4 2

Therefore the desired power series is

do = 2n • ft! do•

y O ) = O o [ l + |■2 x4 a:6

+ +

oo

V a:2”J 00 2n-nl

n = 022 * 2 ! 23-3! 2 »-n!

When written in a slightly different form, this series is a familiar one:

y(x ) = a° X ^ ( f ) = a o e l2 /2 ‘n =0

Answer: y = aoe*/2.

R e m a r k : This differential equation can be solved also by separation of variables. The next example, however, is not as simple.

EXAMPLE 5.2

Obtain a power series solution of the initial-value problem

*y" + xy' + y = 0

2/(0) = 0 , y ' m = 3.

Solution: Try a power series

y(x) = a0 + aix + d2x2 + • • • + anxn + • • •.

From the initial conditions, a0 = 0 and ai = 3. Now substitute the series into the differential equation:

y" ”f" xy' -f* y = (2ct2 + 3 • 2asx -f- 4 «3ci4X2 + • • • -f- ft (ft — l ) a nxn~2 -f- • • •)

+ x(ai + 2 a2x + 3 a3x2 + • • • + nanxn~l + • • •)

+ (oo + dix + a2x2 + a3x3 + • • • + dnxn + • • • ) = 0 .

Collect powers of x:

(do -f- 2o^) -|- (2a i 3 '2 ds )x 4- (3g&2 4*3(14) x2 -j- • • •

+ [a* + nan + (ft + 2 ) (ft + l ) a n+2]xn + • • • = 0 .

All coefficients on the left-hand side are zero, in particular the coefficient of xn. I t follows that

° n+2 ~ ( n + 2) a”'

5. Power Series Solutions 579

From this recurrence relation, the even coefficients can be expressed as multiples of a0 and the odd coefficients as multiples of ax. Since do = 0 , all even coefficients equal 0. Apply the recurrence relation with n = 1, then n = 3, then n = 5, and so on:

1dz — — - ax, a$

* ■ ~ \ a' = ~ z h at-

In general,

«2n+l — ( “ l ) n q C 7 To i~ i \ 0/11 -3 . 5 . 7 . . . (2n + 1)

. 2*4«6***(2n) 2 n*n!1 • 2 • 3 • • • (2n -f- 1) (2n -J- 1)!

Since ax = 3, all coefficients are now determined.

EXERCISES

Solve by a power series at x = 0; check your answer by solving the differential equationexactly:

2‘ t +3y = gX[Hint: Expand ex in a power series.]

I 3 + * - &■!<§

*1 II *

Obtain a power series solution at x = 0 :, cPy b- XM = V

(Exercise 6 is a special case of Bessel's Equation, which has important applications in fluid flow, electric fields, aerodynamics, and other physical problems.)

n 2 1 / 2 1 \ dv o &y i dy7- x M + ( x + x ) T x - y = 0 8- M + x T x - y = 0 -

For the initial-value problem, find a power series solution at x = 0 up to and including the term in x4:

1L % = 1 - y + xiyi ’ 2/(0 ) = — 1 12. ^ + y = ; 2/(0 ) = 1.

Find a power series solution at x = 2 up to and including the term in (x — 2 )4:

i3. x ^ = t / 2; 2/(2 ) = 1 14. ^ + (a; — 2) ^ + y = a:; 2/(2) = 1.

Write x = (x — 2 ) + 2.]

6. MATRIX POWER SERIES

Power series whose variable takes matrix values are a useful tool in the study of linear systems of differential equations (next section). Given a power series

/ (Z) — Co + C\Z + c2z 2 + • • • + cnz n + • • •and a matrix

dll &12A =

021 &22we boldly substitute A for 2 :

f (A ) = c07 + C\A + c2A 2 + • • • + cnA n + • • •.

(Note that the constant term is replaced by c0 times the identity matrix I.) Is this substitution meaningful? Each power A n can be written

II1

©

1----------a2\ n(n) 022 _

where (n) is a superscript, not an exponent. The sum /(A ) is a matrix computed component-wise:

f(A ) = Co

Co

Co

"1 0“+ y cn

r„(n)an

1

_0 1 n =1n(n)a21 n(n)a2 2 _

+ ^ Cnin = 1

OO

i c»

n(n)Gil

a f t co

X cna®n = 1

00

+ ^ Cni n = 1

„ (n)a22

If the series for f(z) converges for \z\ < R, it can be shown that the four series in f (A ) converge, provided the characteristic roots X and m of A satisfy

6. Matrix Power Series 581

|X| < R and \n\ < R. In particular, if R = oo, then /(A ) converges for all matrices A. We shall sketch a proof of this fact for the case of real characteristic roots. The complex case is proved similarly, once the proper groundwork is laid (next chapter and an extension of Chapter 7, Section 8). We begin with two important examples.

EXAMPLE 6.1

L et/(^) = c0 + Ciz + Cxz2 + * • • converge for \z\ < R and let

"x 0”A = , where |X| < R> |m| < R •

0 fJL

Prove that the power series fo r/(A ) converges and find its sum.

Solution:

A n =

hence

f U ) =

Co

Xn 0 '

0 Mn_

071=1

Con =1

/(X) o '

. 0 /(m)_

Answer:m o

. o / ( m) .

EXAMPLE 6.2

Let f(z) = Co + Ciz + C2Z2 + • • • converge for \z\ < R, and let

A -"x 1"

0 X_, where |X| < R.

Prove that the power series for f (A ) converges and find its sum.

Solution:

where

N =

A = XI + N,

"0 1"

0 0N2 = 0.

Hence by the Binomial Theorem,

A n = (X/ + N ) n - \ nI + nXn_W.

(Higher powers A2, N 3, • • • are all 0.) ThusCO 00

f (A ) = CqI + ^ CnA n = Col + ^ cn( \ nI + n \n-W)n = 1

= (c0 + 2 C»X”) 7 + ( i AT = /(X )/ + r' n = 1 ' n = 1 '

We know that /(X) converges because |X| < R. We also know that/'(X ) converges by Theorem 8.8(3), p. 114.

Answer:/(X) f(X )

0 /(X)

In general, if A has real characteristic roots, then there is a non-singular matrix P such that either

‘ x o' "\ 1'

II *0

_op-x or

QsII

.0 x_(See Theorem 8.1, p. 272 and Theorem 8 .2 , p. 273.) An easy computation as in Examples 6,1 and 6.2 shows that

V 0 ' V n \n~lA n = P

.0 Mn_

p - 1 or

0*II8

0 Xn

hence either

7 (x ) 0 ' 7 0 0 f ' Wf U ) = P p-1 or f (A ) = P

0 /(/*)_ 0 f W .

This establishes the convergence of f(A ) provided the characteristic roots satisfy |X| < R and |ju| < R, where R is the radius of convergence of f(z) .

The Exponential Function

The matrix exponential is particularly useful in the study of linear systems. I t is given by

eA = I + V — A n,Z-v n!n =1

6. Matrix Power Series 583

a formula valid for all A since ez = 1 + J2 zn/n \ converges for all z. Let us evaluate eA in two important instances.

EXAMPLE 6.3

Find eA, where A =

Solution: Set

C =

0 t

- t 0

0 1

-1 0so A = tC.

Note that

C2 =

1T—10 1_______

1r—101 'orH11

1ot-H1

_______1

- 1 0 1 0 1 1____

___

= - I .

I t follows readily that successive powers of C are

I, c, -I, -c, I, c, -I, -c,repeating in groups of four (compare to the powers of i). Since A n = tHCn,

t «2 t3 t* <5 <6 t1

/ <2 t4 \ T / / t3 P C \ „= V1 " 2! + 4! " 6! + V ! “ 3! + 51 “ 7! + " 7 °

= (cos t ) I + (sini)C\

Answer:cos t

— sin t

sin t

cos t

EXAMPLE 6.4

Find eA, where A0 t

t 0

Solution: Proceed as in the last example:

0 fA = tC, C =

Successive powers of C are

1 0, C2 = I.

I, C, I, c,

hence

e i - I + T i c + T i , + T i c + -


( i + —! + —! + • • I + C = (cosh t ) I + (sinh t)C.

cosh t sinh tAnswer:

sinh t cosh t*

We close with two properties of the matrix exponential, whose proofs will be given in Section 8 :

(1) = eA+B if AB = BA.(2) (eA)~l = erA.

EXERCISES

Set A~ -3 91

. Find:_ - l 3 j

1. eA3. (I — A )_1

rx o'Set A =

2. sin A 4. cos A .

. Find:.1 x_

5. f (A ) for/(z) = co + ciz + Compute eA:

6. e,tA

7

9. Set

■CI] and A

- □

- t : : ]Verify that ePAP~l = PeAP~\

10. Set B = ln(/ — A). Test the formula eB = I — A in two cases (where |X| <1 and |ju| < 1):

11. IfLo mJ Lo \

7. Systems 585

show that

12. Compute eA for A =

eA = ea

’a b

b a

cos 6 sin b~

— sin b cos b

7. SYSTEMS

We consider first order systems of the form

x = f(t, x, y)

V = 0 (*> y ) •Here /(£,£, 2/) and </(£, X, 2/) are given continuously differentiable functions of three variables. In the corresponding initial-value problem, an initial point (to, Xo, yo) is given and we seek two functions, x(t) and y(t) , satisfying

dx(t)dt

dy(t)dt

= f i t , x(t), 2/(0 ]

= g[t, x(t), i/(0 ]

and

x(to) = x0

y(to) = yo.

Such systems include the most general second order equation

x = f(t, x, x)

in one unknown x(t). For simply set y = x and a system results:

x = y

V = f(t, x ,y ) .

A linear system has the form

x = an (t)x + a12(t)y + bx(t)

y = d2i(t)x + 022(02/ + &*(0 -

I t can be written in matrix form as

x = Ax + b,where

X X O il Oi2 ' b l lX =

.y .X =

J/_

IIII

O21 O22

O" II XT cs* II

1

(We do not use our usual notation x' for column vectors because of a possible confusion with derivatives.)

We shall emphasize linear systems with constant coefficient matrix A.

Homogeneous Systems

Consider a homogeneous linear system with constant coefficients


X a n ai2, A =

JJ . _021 a22_

The solution of the initial-value problem

x = Ax, x(0 ) = x0is

x = etAx0.

We shall prove here only that x = etAx0 is a solution, not that it is the only solution. This fact is harder and will be proved in the next section. We have

CO

Xtn — A nxo,

n =1

hence x(0 ) = x0 and

* - t T ^ W \ A t Ax-n = 1 n = 1

R e m a r k : The solution is formally the same as that of the initial-value problem for one differential equation

x = ax, x(0 ) = x0.Here the constant a can be thought of as a 1 by 1 matrix, and the solution is

x = x0eat.EXAMPLE 7.1

Solve the initial-value problemx = \x $itSH

y *• m 2/(0 ) = yo.

Solution: In matrix form, the system is x = Ax and x(0) = x0, whereX "X 0“ x 0

X = , A = , Xo =

JJ. .0 .yo .

7. Systems 587

By Example 6.1,

0 e^hence the solution is

x = etAx0 =eu 0 x0 x0eu

_ 0 0"\ .Vo_ _y0e _

x = XtfUAnswer: •

y = Voe .

R e m a r k : The system really consists of two entirely unrelated equations. Such a system is called uncoupled. The next example is not so simple.

EXAMPLE 7.2

Solve the initial-value problem

x = \x + y x(0) = x0

se ll M m .2/(0 ) = 2/o.

Solution: In matrix form the system is x = Ax, where

X "x f Xo

x = , A = , Xo =

JJ. .0 \ JJo_By the calculation in Example 6 .2 ,

( tA )n = tnA n =\ ntn n \n~ltn

0 \ ntnhence

Therefore the solution in matrix form is

e\t

'1 0" 00 ~\ntn n \n~Hn~ eu teuelA = + y =

.0 1_ O 3 3 0 ex‘_

x =0

Answer: x = (x0 + yot)eKt

y = yoeu .


Second Order Equations

We already noted that the general second order equation x = f(t, x, x) can be reduced to a first order system. One technique is to set y = x and then the system is x = y , y = f(t, x, y ) . But there are other techniques; for instance set y = x + tx. Then y = x + x + t x = f ( t , x, y) + x + t(y — tx), so the system is

x = - t x + y, y = (1 - t2)x + ty + f(t, x, y).

This particular reduction is not especially recommended, but it indicates the variety possible. Let us try one easy example.

EXAMPLE 7.3

Find the general solution of x + k2x = 0, k > 0.

Solution: Setx = ky.

Thenky + k2x = x + k2x = 0 ,

y = —kx.

Thus the system of first order equations

x = ky

V = —lex,

is equivalent to the given second order equation. The system may be written

0 kx = Ax, where A =

- k 0By Example 6.3,

cos kt sin kt

— sin kt cos kt

hence the solution to the system is

cos kt sin kt

— sin kt cos kt

X

JJ_

Xo

. y o .

Answer: x — Xq cos kt + yo sin kt, Xo and y0 constants.

Suppose we wish to solve a non-homogeneous system with constant coefficients :

7. Systems 589

Non-homogeneous Systems

an aJ2 M t yx = Ax + b (/), A = , b =

021 CL22 M t ) .

Here is a direct and effective method. Define y(t) by x = etAy. Then

(etAy)' = AetA y + b,

hence the given differential equation is equivalent to

AetA y + etAy = AetA y + b, etAy = b, y = e~tAb.

Now integrate:

y(0 = Yo + j e~sAb (s) ds,J o

where the integral of a vector function is taken componentwise. Therefore we have the solution formula

x(t) = etA ^x0 + e~sAb(s) ds'j .

EXAMPLE 7.4

Solve the system

dx— = x + y + t2 dt< X(0) = y (0 ) = 0.

dyT t = y + 1’

Solution: Write the system as x = Ax + b, where

x{t) I 1X =

.2/(0 ., A =

.0 1., b«) =

t_

In Example 7.2 it was shown that

e* e~s — se~s

ct> £ II 1 II

0 1lo 0 1 Co 1—

Hence

[ e~sAb(s) ds = f J o J o


e~8 — se~8

0 er>

s2 n " 0 "ds = /

s J 0 _se~8_

f o <J 0

/ v . ./ 0

ds -e ~ ' ( t + 1) + 1

The initial condition is x(0) = 0, so the solution formula yields

x (0 = e‘A ^0 + J e~°Ab(s) d s j

iSIdi_______

i01tct — t(t -f- 1)

iIdO

_______I

1 1 1

+

1—I 1 1.....

......

..

1 cs* + 1-

Answer: x = tel — t2 — t, y = e* — t — I.

EXERCISES

1. Show that the substitution x = ky changes x — k2x = 0 to the system

Ix = ky

y = kx}and solve.

2 . Show that the substitution y = x — \x changes x — 2\x + \ 2x = 0 to the systemx = \x + y

[y = Xy,and solve.

3. Let x be a solution of x = Ax. Show that y = P~lx is a solution of y = (P~lA P )y. (Here P is a non-singular constant matrix.)

4. Solve

ix = ky + cos kt

y = — kx + sin kt.5. Solve

i = Xz + y +

.2/ = Xy + eu .

8. Uniqueness of Solutions 591

8. UNIQUENESS OF SOLUTIONS [optional]

We differentiate vector functions and matrix functions componentwise. Let us note a few useful formulas:

If x = x (0 , B = B (t) , and C = C(t), then

(Bx)B = B-x + Bx, (BCY = B-C + BC',

(B~lY = —B~lB'B~l if B is non-singular.

The first two formulas are easily proved. For instance,

bijCjk) = ^ (bijCjk “l- bijdjk)' J

implies the second. Note the order of multiplication in the products. The third formula follows from the second:

BB- 1 = I, B 'B -1 + B(B~iy = 7* = 0, B-'B-B~l + (B~lY = 0.

Homogeneous Systems

We claimed (p. 557) that the general solution of x — k2x = 0 is

x = aekt + be~kt,

assuming k ^ 0. An equivalent form of this solution is

x = a cosh kt + b sinh kt.

We also claimed that the general solution of x + k2x = 0 is

x = a cos kt + b sin kt.

Both are reasonable from the system point of view. If we set x = ky and

xx =

Xthe two systems are x = i x , where either

0 kor A =

0 k

* 0_ — k 0

We saw earlier that for these matrices either

cosh kt sinh kt cos kt sin ktelA = or e‘A -

sinh kt cosh kt — sin kt cos kt

The claims about x =b k2x = 0 follow from the assertion that x = eiAx0. We shall prove this in general for any homogeneous system.

A Uniqueness Theorem


Theorem 8.1 The general solution of

x = Ax,

where A is any constant matrix, is

x = etAx0, x0 = x(0).

Proof: Suppose x is any solution of x = Ax. Set B{t) = etA. Then

B = J ( i + V - A") = V - 1 A n = AB. d t \ n\ J Z-/ (n — 1)!\ n = 1 / n = 1

Therefore

(B~lxY = (B~l)-x + B~lx = - B ~ 1(A B )B -1x + B~l {Ax)

= —B~lAx + B~lAx = 0.

I t follows that B~lx = x0, a constant vector, hence x — Bx . Obviously x(0) = x0 since # (0) = I. This completes the proof.

Another Uniqueness Proof

The theorem we just proved shows that the initial value problem x = Ax, x(0) = x0 has only one solution, namely x = eMx0. There is another way to prove there is at most one solution without actually knowing there is any solution. The method involves estimating successive approximations, and it is important because it applies with suitable modifications to many systems besides homogeneous ones with constant coefficients.

Suppose Xi(t) and x2(t) are two solutions of the initial value problem x = Ax, x(0) = x0. Their difference x(t) = Xi(t) — x2(t) is a solution of the initial-value problem x = Ax, x(0) = 0. If we can conclude that x(t) = 0, then Xi = x2.

Theorem 8.2 Let x(t) be a solution of the differential equation x = Ax with x(0) = 0. Then x(t) = 0.

Proof: We may write

Let us work on a fixed interval — b < t < b . We know (p. 240) that there is a constant c such that \Av\ < c |v| for all v. Also, since x(s) is continuous on [ —6, 6], there is a constant M such that |x(s) | < M for — b < s < b. Now we are set up for a remarkable bootstrap operation. We start:

< c|x(0 | < c \ I x(s) dsI J o

Next step:

|x(0 | < c \ j |x(s)| dsI J o

Once again:

K O I < C I f |x(s)| ds1 J 0

Continuing in this way, we prove

/ |x(s)| ds = c / Jo I Jo

M d s = Me |<|.

< c I/ 'I J 0

- I B

Me Isl ds

Me2 Isl2 ds

= - Me2 | |2.

for all n. Hence

As n -

|x(0l < — Mcn I n\

| x (0 | < — il/c-fe- =n! n!

-> 0. Hence x(t) = 0 for |/| < b.—> oo, we have (cb)n/ ( n !) - But 6 is arbitrary, so x(£) = 0 for all t. End of the proof.

Here is a simple, but useful, corollary of the uniqueness theorem.

Theorem 8.3 The only matrix solution of

X' = AX, X (0 ) = Xois

X = e‘AX 0.

Proof: Write X = [x, y] in columns. Then

x = Ax x(0) = x0

y = Ay y(0) = y0,

so the uniqueness theorem for vector systems implies

x = etAx0, y = etAy0.

Hence X = etAX 0. (See Ex. 1 for another proof.)

In Section 6 , we stated without proof the following two properties of the

matrix exponential:

(1) if AB = BA.

(2) (eA)~l = e"A.

Property (1) can be proved by multiplying power series, but there is a tricky convergence problem. A better way is to use differential equations. We know that etU+B) is the only solution of X 9 = (A + B )X , X (0 ) = I.

Set Y(t) = etAetB. From AB = BA we deduce etAB = BetA:


Therefore

Ym = AetAeiB + etABetB = AetAetB + BetAetB = (A + B) Y.

Since F (0 ) = 7, uniqueness implies Y(t) = et(A+B). Set t = 1 to complete the proof of (1).

Certainly A ( —A) = ( — A) A. Hence

I = e0 = eA+ ~A) = eAe~A, (eA) -1 = e~A.

This proves (2 ).

EXERCISES

1. Prove the Theorem 8.3 by considering e tAX.2*. Show that Theorem 8.2 is true when A = A (t) is a matrix of continuous functions.3. Set

Compute eA+B and eAeB. Explain the discrepancy.4. Interpret as trigonometric identities (exp A is another notation for eA):

exp0 a + i

L - (« + 0 ) 0The trace of a 2 X 2 matrix A is tr A = an + a22.5. Show that |2?(0|* = tr(5* cof B).6. (cont.) Set g(t) = \etA\. Prove g = (tr A)g.7. (cont.) Prove |e | = etrA.

n r 0 ai r ° /n= exp exp

J OJ L -0 Oj


8. (cont.) Let X(t) satisfy X* = AX. The Wronskian of X is w(t) = \X (0|. Show that wit) = w(0)e(tT A)t.

9. Show that

x (f) = / (f ~ s )d(s ) ds

is the solution of x = g, x(0) = x(0) = 0.10. Suppose x is the solution of x = Ax + b (t) with x (0) = 0 and y is the solution of

y = Ay + c(0 with y(0) = 0. Show that x + y is the solution of z = Az + (b + c) with z (0) = 0. (This is the principle of superposition.)

15. Complex Analysis

1. INTRODUCTION

The simple equationx2 + 1 = 0

has no solution in terms of real numbers, the numbers of ordinary experience; neither do the equations

x2 + 3 = 0, x2 + x + 1 = 0, x2 — 4x -f- 10 = 0.

Yet such equations arise in scientific computations. For this reason, the real number system is enlarged by introducing a new number i satisfying

i 2 = - 1.

The result is the system of complex numbers. I t consists of all expressions a + bi, where a and b are real numbers. These expressions are treated by the usual rules of algebra, except that i 2 is replaced by —1. Thus

(a ~b bi) + (c + di) = (a + c) -f- (b -f- d)i,

(a + bi) (c + di) = ac + bdi2 + adi + bci

= (ac — bd) + (ad + bc)i.

Two complex numbers a + bi and c + di are equal if and only if a = c and b = d. A real number a is considered as a special type of complex number: a — a + 0 • i.

R e m a r k s o n N o t a t i o n : Sometimes a + ib is the preferred notation. For example, —1 + i \ / 3 looks better than — 1 + a/3~i because of the possible confusion with — 1 + V Si. Similarly cos 6 + i sin 6 is better than cos 6 + sin 6 i. In engineering, j is often used instead of i.

1. Introduction 597

In terms of complex numbers, the quadratic equation

x2 + I = 0

has two roots, ± i . Furthermore, every quadratic equation

ax2 + bx + c = 0

has complex roots. By the quadratic formula, the roots are

_ _ b_ V~D2 a ~ 2 a

where D = b2 — 4ac. If D is non-negative, then \ / D is a real number, so the roots are real. If D is negative, write

V d = V ( - 1) ( - D ) = V ^ l V ^ - D = i V ^ D ;

then the equation has complex roots

- b V ^ - D .x = —— ± —----- 1 .

2 a 2 a

EXAMPLE 1.1

Find the roots of x2 + x + 1 0 and of x2 — 4x + 10 = 0.

Solution: Apply the quadratic formula. For the first equation a = b — c = 1, D = —3. For the second equation a = 1, b = —4, c = 10, D = —24.

Answer: — - J— — i ; 2 4- i a / 6 .2 ~ 2

All quadratic equations with real coefficients can be solved in terms of complex numbers. W hat about cubic equations

axz + bx2 + cx + d = 0 ?

If a new quantity i is needed to solve quadratics, is another new quantity needed to solve cubics, still another to solve quartics, and so on?

This question was answered in 1799 by C. F. Gauss in the Fundamental Theorem of Algebra. I t asserts that each polynomial equation of any degree with real coefficients, has a complex root. Thus, the complex number system is rich enough to contain solutions for all polynomial equations. Once the number i is adjoined to the real number system, that is enough.

R e m a r k : Later we shall learn that each complex number has two complex square roots. Hence, the quadratic formula provides complex roots for any quadratic equation with complex coefficients. The Fundamental Theorem of Algebra also includes the statement tha t each polynomial equation with complex coefficients has complex roots.

598 15. COMPLEX ANALYSIS

EXERCISES

Solve and check your answers by direct substitution:1. x2 — Sx + 25 = 03. x2 + x + 2 = 05. 3x2 — 2x + 3 = 07. 225a;2 + 15x + 61 = 0

Find all roots:9. x3 — 1 = 0

11. x3 — x2 + x — 1 = 013. x4 - 1 = 0

2. z2 + 25 = 04. x2 — 6x + 9 = 06. 2x2 + x + 2 = 08. 5x2 — 4x + 1 = 0.

10. x3 + 1 = 012. x4 - 2x2 + 1 = 014. x4 + 5x2 + 4 = 0.

15. Compute (1 + i)2. Use the result to solve the equation x2 = i.16. Compute [( — 1 + i \/3 ) /2 ]3.17. Compute i + t’2 + i z -f * * * + i1492.

2. COMPLEX ARITHMETIC

Complex numbers can be pictured as vectors in a plane. Think of the number 1 as a unit horizontal vector and the number i as a unit vertical vector (Fig. 2.1a). Then a complex number a + bi is a linear combination of these two vectors (Fig. 2.1b).

F ig . 2.1a

The horizontal component of a + bi is called its real part, written R e(a + bi); the vertical component is called its imaginary part, written Im (a + bi). Thus

Re(a + bi) = a, Im(a + bi) = b.

Note that Im(a + bi) is the real number 6, not bi.

2. Complex Arithmetic 599

Associated with each complex number a + bi is the number a — bi called its (complex) conjugate (Fig. 2.2). Conjugates are denoted by bars:

a + bi = a — bi.

a — bi

F ig . 2 .2

Set z = a + bi; the following basic relations hold:

z + z = 2a = 2 Re(z), z — z — 2bi = 2i Im(z),

z = z, zz = a2 + b2.

By the quadratic formula, if the roots of a quadratic equation with real coefficients are not real, then they are conjugate complex numbers.

The modulus or absolute value of a complex number a + bi is

| a + bi\ = \ / a2 + b2.

I t is the length of the vector a + bi. Notice that

\z\2 = zz.

If z 0, then \z\ > 0.The absolute value of a complex number is a measure of its size. Since

complex numbers fill the plane, you cannot say that one complex number is greater than or less than another; the statements

i < 2, 2 -j- i < 4 — 3i

make no sense. Yet you can compare absolute values; the statements

\i\ < |2 |, |2 + i\ < |4 - 3t|

do make sense. The latter, for example, says that the length of the vector2 + i is less than the length of the vector 4 — Si. See Fig. 2.3.

All complex numbers of the same absolute value determine a circle (Fig. 2.4).

F ig . 2.4

EXAMPLE 2.1

1 — iExpress - .—— in the form a + bi.

o i” Ax

2. Complex Arithmetic 601

Solution: Here is a standard trick. Multiply numerator and denominator by 3 — 2i, the conjugate of the denominator:

1 - i 1 - t 3 - 2 i (3 - 2) - (3 + 2) i 1 - 5i3 + 2i " .3 + 2i ’ 3 - 2i ~ 32 + 22 13

in s w r : - 4 ---- i. 13------------------- 13

EXAMPLE 2.2

If |z| = 1, show that - = z. z

Solution: Let z = a + bi. Then by the trick of the last example,

1 1 1 a — bi a — bi z a + bi a + bi a — bi a2 + b2

But a2 + b2 — \z\2 = 1. Hence,

- = a — bi = z. z

Shorter Solution:

zz = \z\2 = 1, z = - • z

EXERCISES

Express z in the form a + bi:

!. z = | - £i 2. z = 0.4 + 1.7i3. z = (2 - i) 4. z = (2 + t*)-15. z = (— 1 — i ) ( —2 + 3i) 6. 2 = (1 + t) (2 -f~ i) (3 -b i)_ i 1 — i 7. z = ------. 8. Z = ---;—1 —l I

2 - i — 1 + i11. Find |z| in Ex. 1-10. 12. Show thatzw = zw.13. Show that \zw\ = \z\ |tu|. 14. Show that |z| = |z|.15. Show that the equation z = z is satisfied only by real numbers.16. Let f(x) = clqx* + aix2 + a2x + a3, where the coefficients are real numbers.

Let z be a complex zero of f(x). Prove that z is also a zero. (Hint: 0 = 0.)


17. Do the same for an n-th degree polynomial with real coefficients.18. Show that if (a + bi)A = 1, then a + bi is one of the numbers Jrl, ±i.

3, POLAR FORM

When a complex number z is written z = a + bi, it is said to be in rectangular form. If z 9 0, it is sometimes convenient to express z in polar form :

z = r(cos 0 + i sin 0), r > 0 .

As is seen in Fig. 3.1,

r = \z\ = V a 2 + 62; cos 0 = -> sin 0 = - •r r

iV

z = a -f- bi

S '/ I

—---

-

^----------*----------------1--------- »a x

F ig . 3.1

The complex number cos 0 + i sin 0 is a unit vector making angle 0 with the positive z-axis. Hence, the polar form z — r(cos 0 + i sin 0) expresses the vector z as a unit vector in the same direction stretched by a factor \z\. The angle 0 is called the argument of z and is written arg z. I t is determined only up to a multiple of 2t. For example, arg(l + i) can be taken to be 7t/4 or 97t / 4 or — 157t/4, etc. If z = r(cos 0 + i sin 6), then

z = r[cos( — 6) + i sin( — 6)] = r(cos 6 — i sin 6).See Fig. 3.2.

Polar form is particularly useful in situations involving multiplication (or division) of complex numbers; it makes multiplication easy. Suppose

Z\ = ri(cos 0i + i sin 0i), z2 = r2(cos 02 + i sin 02).

ThenZiZ2 = rir2[(cos 0i cos 02 — sin 0i sin 02) + i’(sin 0i cos 02 + cos 0i sin 02)]

= rir2[cos(0i 02) -f- i sin(0i -f- 02)].

3. Polar Form 603

y z

X

z

F ig . 3 .2

The product is again in polar form; its modulus is rir2 and its argument is + 02 .

To multiply two complex numbers, multiply their absolute values and add their arguments:

Similarly, to divide two complex numbers, divide their absolute values and subtract their arguments:

From this rule follows De Moivre’s Theorem, a formula for powers of a complex number.

De Moivre’s Theorem For each positive integer n,

[r(cos 0 + i sin 0)]n = rn(cos nO + i sin nd).

EXAMPLE 3.1

Describe geometrically the effect of multiplying a complex number by i , by 1 — i.

Solution: Write i and 1 — i in polar form:

\ziz2\ = |zi| |z2|, argziz2 = argzi + argz2.

Z i \zi\ Z2 | 21

Ziarg — = arg zi — arg z2.

z2


According to the rule for multiplication,

7r\iz\ = \i\ \z\ = \z\y arg iz = arg z + arg i = arg z -\-----

2

Therefore multiplying the vector z by i simply rotates the vector 90° counterclockwise (Fig. 3.3).

According to the rule,

|(1 - i)z\ = V 2 \z\, arg(l - i)z = arg 2 -

Therefore multiplying the vector z by 1 — i rotates the vector 45° clockwise and stretches it by a factor y / 2 . See Fig. 3.3.

Answer: rotation by 90°;rotation by —45° and stretching by a factor y / 2 .

EXAMPLE 3.2

Compute (1 + i)8.

Solution: By the Binomial Theorem,

(1 + i)* = 1 + Si + 2 Si2 + 56z3 + 70 i* + 56 i 5 + 2 8z6 + Si7 + z8

= 1 + Si - 28 - 56i + 70 + 56z - 28 - Si + 1 = 16.

Alternate Solution: Write 1 + i in polar form, then use De Moivre's Theorem:

I + i = y / 2 [ cos 7 + i sin ~ ) >V 4 4 /

(1 + i ) s = ( y / 2) 8 ^cos + i sin = 16(1 + 0 • i) = 16.

Answer: 16.

The following example illustrates a powerful technique for deriving trigonometric identities.

EXAMPLE 3.3

Derive the identities

cos 30 = cos3 0 — 3 cos 0 sin2 0,

sin 30 = 3 cos2 0 sin 0 — sin3 0.

Solution: Compute (cos 0 + i sin 0)3 two ways, by De Moivre’s Theorem and by the Binomial Theorem. The results must be equal:

cos 30 + i sin 30 = cos3 0 + 3 cos2 0 (i sin 0) + 3 cos 0 (i sin 0)2 + (i sin 0)3

= (cos3 0 — 3 cos 0 sin2 0) + i(3 cos2 0 sin 0 — sin3 0).

Now equate real and imaginary parts on both sides of this equation.

3. Polar Form 605

Roots of UnityThe equation

z4 = 1

has four complex roots, ± 1 and ± i . Thus the number 1 has four 4-th roots, which are complex numbers equally spaced around the circle \z\ = 1.

The situation for n-th roots is similar. Write the equation zn = 1 in polar form, setting z = r(cos 0 + i sin 0) :

fr(cos 0 + i sin 0)]n = l(cos 0 + i sin 0), r > 0 .

By De Moivre’s Theorem,

rn(cos nO + i sin nO) = l(cos 0 + i sin 0 ).Consequently,

rn = 1, cos nO = 1, sin nO = 0.I t follows that

2wkr = 1, nO = 2irk, 0 = ----> k an integer.

nThus

2irk . . 2irk z — cos-------b z s in ---- )

n n

where k is an integer. This formula yields exactly n distinct values: for k = 0,1, 2 , • • • , n — l. (Why?) For k — 1, call the root

2tt . . 2tco = cos-----[- i s in ----

n n

Since (De Moivre again)

2irk . 2irk cok = cos------ h t s in ---- >

n n

the other roots are the powers of co, namely 1 = co°, co, co2, • • • , con - 1 .

The equationzn = 1

has exactly n complex roots 1, co, co2, • • • , con~l, where

2t . . 2irco = cos-----1- i s in ----

n n

These numbers are called n-th roots of unity. They lie equally spaced around the circle \z\ = 1. See Fig. 3.4.


EXAMPLE 3.4

Find the cube roots of 1.

areSolution: Use the formulas given above with n = 3. The cube roots

co° = 1 ,

2tt . . 2tco = cos----- \- i sin — >

3 3

4tt . 4ttco2 = cos-----b i sin — •

3 3

3. Polar Form 607

Ansiver: 1, — - H— — i ,1 , V 3 . 1 V 3 .X H 7T~ h

As a check, cube the complex numbers in the answer by the Binomial Theorem.

Not only the number 1, but any (non-zero) complex number a has complex n-th roots, exactly n of them.

Each non-zero complex number a has exactly n complex n-th roots. If

I t follows that the n-th roots of a are equally spaced points on the circle of radius |a |1/n centered at the origin.

The assertion is easily verified. By De Moivre's Theorem

Hence 0 is an n-th root of a. Furthermore,

(0 c o * )n = pn * cofcw = a • 1,

so j8 , 0co, jSco2, • • • , jftcow_1 are n-th roots of a. There are no others; for if y is an n-th root of a , then

Hence y/fi is an n-th root of unity; y/fi = co* for some k. Therefore y = /3co*.In practice, we compute n-th roots from the above formula for /?co&, not

by actually multiplying 13 and co*.

EXAMPLE 3.5

fin — r (COs 6 + i sin 0) = a.

Find the cube roots of — (1 + i \ / 3) •£J

Solution: The polar form of this number is

27(cos 60° + i sin 60°).

Apply the above formula for n-th roots with n = 3 and 6 = 60°. The three cube roots are

/3 = 3 (cos 20° + i sin 20°),

0CO = 3[cos(20 + 120)° + i sin(20 + 120)°],

0co2 = 3[cos(20 + 240)° + i sin(20 + 240)°].

Answer: 3(cos 20° + i sin 20°),

3 (cos 140° + i sin 140°),

3 (cos 260° + i sin 260°).

See Fig. 3.5.


F ig . 3.5

EXERCISES

Express in polar form:1. - 1 + i 2. - 1 - i3. 1 + i y /S 4. 3 - 2i5 . - 4 6. - 3 i7. 1 + 4i 8. 4 - Si.

Multiply and express the answer in rectangular form:

3. Polar Form 609

10. (cos 17° + i sin 17°) (cos 208° + i sin 208°)11. (cos 135° + i sin 135°) (1 + i)12. (cos 17° — i sin 17°) (cos 197° + t sin 197°). Express in polar form:

13. i + i 14. 1 "1 - * — 1 + * - \ / 3

15. - .— 16. -------------------— 1 — t V 3 ir . . . ITv cos - 1 sin —

5 5

Compute by De Moivre’s Theorem:

17. (1 — i)4 18. ( -1 + i y

19. (a /3 - 1y 20. [ 1 + 2 V /3 ] 11

2 L o h - “ ' [ v j j

23. Verify De Moivre’s Theorem for n = 2, 3, 4, 5.24. Find the 5-th roots of unity.25. Find the 8-th roots of unity.26. Find the 5-th roots of (1 + i).27. Find the 6-th roots of —1.28. Find the 7-th roots of — i.29. Find the 10-th roots of i.30. Examine the 8-th roots of unity, 1, co, co2, • • • . One root has the value 1,

another (which?) is a square root of unity, and two others (which?) are 4-th roots of unity. The remaining four (which?) are called primitive 8-th roots of unity.

31. (cont.) Similarly analyze 6-th roots, 9-th roots, and 10-th roots of unity. Which are the primitive ones?

32. Show that each 5-th root of unity other than 1 satisfies x4 + x3 + x2 + x + 1 = 0. (Hint: Factor x5 — 1.)

33. Factor x6 — 1. Show that each primitive 6-th root of unity satisfies x2 — x + 1 = 0.

34. Set a = ——-~ and co = a ~ X/*------ * Show that co5 = 1.A Z(Hint: Use Ex. 32.)

35. Verify that De Moivre’s Theorem holds also for exponents which are negative integers.

36. Let 1, co, co2, • • • , con_1 be the n-th roots of unity. Show that 1 + co + co2 + • • • + con_1 = 0.

+ <11


37. (cont.) Show that2tt , 47r 2(n — l)7r

1 -f- cos — + cos — + • • • + cos = 0,n n n. 27T . 4ir . 2 (n — l)7rsin — + sin — + * * * + sin = 0 .n n n

38. Prove that cos 30 = 4 cos3 0 — 3 cos 0. (Hint: See Example 3.3.)39. Express cos 40 in terms of cos 0.

4. COMPLEX EXPONENTIALS

Let us try to give a meaning to ez for complex numbers z. The exponential series

x2 xn

converges for all real values of x. We now boldty substitute iy for x:

i , (w )2 , (iy)3 ,ew = I _L iy J---------- -----------L . . .2! 3!

, . . 2/2 *2/3 , y i . w 6— 1 -j- 'I'll — -- — --- —I— — —I— --- — . . .y 2 ! 3! 4! 5!

_ L y * y * y* , \ , - f y 3 , y 6 \\ 2l 4] 67 / \ 3! 5! J'

The quantities in parentheses are familiar; they are the series for cos y and sin y. These observations suggest the following definition:

For any complex number of the form iy with y real, define

eiv = cos y + i sin y .

For a general complex number z = x + i y , define ez — ex . eiv — e * ( c o s y i sin y).

If this is to be a reasonable definition of ez, the usual rules of exponents should hold. Let us verify that ez1 • ez* = ez*+z*. Suppose Z\ — X\ + iyi and z2 = x2 + iy2 • Then

e*i • e*t = e*i(cos y 1 + i sin y 1) • e*a(cos y 2 + i sin y 2)

= • ex*(cos yi + i sin y 1) (cos y 2 + i sin y 2)

= e*i+x2[Cos(?/i + 2/ 2) + i sin(?/i + y 2)]

— gaJI+X2+*(l/i+J/2) = g(*l+*l/i)+(*2+t’i'2) = eZl+Z2;

which completes the verification.

We observe that

e- iV _ cos( _ ^ ) i sin( — y) = cos y — i sin y.

Hence e~iv is both the reciprocal and the conjugate of eiy. This is not surprising since \eiv\ = 1, and, as shown in Example 2 .2 , if \z\ = l , t h e n l / z = z.

4. Complex Exponentials 611

EXAMPLE 4.1

Find the value of:

(1 ) e<, (2) e « , (3) eri, (4) eln2+(”74), (5) e2Hln.

Solution: In each case, use the definition

ex+iv — ex(cos y + i sin y).

(1) ei = cos 1 + i sin 1 (1 rad).

(2) e2vi = cos 2t + i sin 2t = 1.

(3) eri = cos 7r + i sin w = — 1.

(4) eln 2+ir»/4 = eln 2 ^cos ^ + i sin ^ = 2 + i = V^2 (1 + ii).

/r\ o •/ 2tt , . . 2tt(5) e2irtln = cos-----1- sin —n n

R e m a r k : From the answer to (5) we recognize that e2iriln is an n-th root of unity. This agrees with the answer to (2) since

g2iri/n n _ g2iri — }

Trigonometric Functions

From the basic relations

eid = cos 6 + i sin 0,

e~ie = cos 0 — i sin 0,

follow two important formulas:

g i O g — id

cos 0 = ----- ------ = Re(e^),

sin 0 =eie — e~ie

2 i= Im (e1’9).

These formulas are extremely useful and are worth memorizing. They

convert problems about trigonometric functions into problems about exponentials which are often simpler to handle.

EXAMPLE 4.2

Derive the identity

sin ( n + i J 61 -f“ 2 cos 6 -j~ 2 cos 26 -f- • * • -f- 2 cos n6 = *—


sin - 62

Solution: Let Cn denote the left-hand side of the identity,

Cn — 1 “h 2 cos 6 -\- 2 cos 26 • • • -|- 2 cos n6

= 1 + (e* + e~ie) + (e2ie + e~2ie) + • • • + (enid + e~nie).

Rearrange:

Cn = e~nie + e~(n~1)ie + • • • + e~i9 + 1 + + • • • + e"*'*

= e~nie( l + eie + e2ie + • • • + e2nie).

The expression in parentheses is a finite geometric series whose sum is known:

_ g(2n4-l)i0 g—niO _ g(n+l)*— o—niQ . _ _ _ _ _ _ _ _ _ _ _ _ — _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

1 - eie 1 - eie

Here is an important trick. Remembering that eia — e~ia = 2i sin a, multiply numerator and denominator by e~idl2:

g —(n + l)iO __ g ( n + i )Cn =

e~iei2 - ei012 6 6— 2i sin - s i n -

2 2

Hyperbolic Functions

There is a close connection between trigonometric and hyperbolic functions. Recall the definitions

ex + e~x . . ex — e~xcosh x = ---------- j sinh x -- -------------

2 2

Formally replacing x by iy , we find

cosh iy = cos y , sinh iy — i sin y .

In order to give meaning to these identities, we define sin 2, cos z, sinh z, and cosh z for complex numbers z.

4. Complex Exponentials 613

If z is a complex number,

eiz + e~izcos zcos z = ----- ------ )

2sin z

2 i

Similarly,

ez + e~*cosh z = ---- -----

2sinh z —

2

Thus cos z and cosh z are closely related functions, as are sin z and sinh z. Directly from the definitions, one obtains the basic relations:

valid for all complex numbers z. Equivalent relations are obtained on replacing z by iw :

valid for all complex numbers w.I t is proved in more advanced courses that each identity involving sines

and cosines, valid for all real values of the variables, is valid also for all complex values of the variables. (For example, the identity

holds for all complex values of z and w as well as for all real values.) In particular, in any identity involving sines and cosines of variables z, w, • • • , the variables may be replaced by iz, iw, • • • . I t follows that each identity involving sines and cosines has a counterpart involving hyperbolic functions.

EXAMPLE 4.3

Show that cosh(2 + w) = cosh z cosh w + sinh z sinh w .

Solution:

cosh(2 + w) = cos i(z + w) = cos (iz + iw)

cos z = cosh iz, sin z = - sinh iz = —i sinh iz.i

cosh w = cos iw, i sinh w = sin iw,

cos(z + w) = cos z cos w — sin z sin w

cos iz cos iw — sin iz sin iwcosh z cosh w — (i sinh z) (i sinh w)

cosh z cosh w + sinh z sinh w.

EXAMPLE 4.4

Find a formula for 1 + 2 cosh z + 2 cosh 2z + • * • + 2 cosh nz.

Solution: Use the corresponding formula for cosines found in Example 4.2:

1 + 2 cosh z + • • • + 2 cosh nz = 1 + 2 cos iz + • • • + 2 cos niz


B ut

sin i ( n + ^

. iz sin —2

sin i z i sinh + i )

IZ zsin — i sinh -

2 2

Answer:

1 + 2 cosh z + 2 cosh 2z + • • • + 2 cosh nz

sinhinh(n + i )s inh i

1. Evaluate(a) e*il3 (b) el~{

in (i+ 2i)

EXERCISES

(c) gi/C1-2*) (d) sin

(e) cos 3i (f) cosh tti.2. Show that

sin(z + iy) = sin x cosh y + i sinh y cos x,

cos(x + iy) = cos x cosh y — i sin x sinh y.

3. Find all complex numbers z for which(a) sin z is real (b) cos z = 0.

4. Show that for all complex numbers z,(a) et+2iri = ez (b) sin(z + 2t) = sin z(c) cos (z + 2ir) = cos z (d) sinh (z + 2ti) = sinh z(e) cosh(z + 27ri) = cosh z.

5. Is |cos z\ < 1 for all complex numbers z ?

6. Derive the trigonometric identities:

(a) cos4 x — — [cos 4x + 4 cos 2x + 6 + 4 cos( —2x) + cos( —4x)]jL

(b) cos6 x = ^ [cos 6x + 6 cos 4x + 15 cos 2x + 20+ 15 cos( — 2x) + 6 cos(—4x) + cos( — 6x)].

(Hint: Write cos x = ^(eix + e~ix) and use the Binomial Theorem.) Guess a general formula for cos2n x.

7. Derive the identity

sinh f n + i j z1 + 2 cosh 2 + 2 cosh 2z + • • • + 2 cosh nz = -------------- —

sinh|

by converting the left-hand side into a sum of exponentials, sinh z

8. Define tanh z = — -— > sech z = — -— • Find a relation between tanh2 zcosh z cosh zand sech2 z.

9. Express cosh 4z in terms of cosh z.(Hint: Express cos 4z in terms of cos z.)

10. Derive the hyperbolic identities:(a) sinh 2z = 2 sinh z cosh z (b) cosh 2z = 2 cosh2 z — 1(c) sinh Sz = 4 sinh3 2 + 3 sinh z.

5. Integration and Differentiation 615

5. INTEGRATION AND DIFFERENTIATION

In this section we deal with functions having complex values, for example,

f(x) = eix = cos x + i sin x.

Such a function can be written as

f(x) = u(x) + i v(x),

where u(x) and v(x) are real-valued functions.

If f(x) = u(x) + i v(x), then the derivative of f(x) is defined by

f'(x) = u'(x) + i vf(x). Similarly,

J f(x) dx = J u(x) dx + i J v(x) dx.

Many formulas for differentiation and integration extend to complexvalued functions. We shall consider just one case, complex exponentials.


If a is a complex number, then

d / X— (eax) = aeadx

and

/ eax dx = - eax + C, a 5* 0 . a

Let us verify the first formula; the second formula follows from it. Suppose a = a + bi. Then

~~ (eax) = -f- [e(a+6i)z] = -f- [e°*(cos bx + i sin bx)] dx dx dx

= -7- (eax cos bx) + i (eax sin bx). dx dx

By ordinary differentiation,

d— (eax) = (aeax cos bx — beax sin bx) + i{aeax sin bx + beax cos bx) dx

= aeax(cos bx + i sin bx) + ibeax(cos bx + i sin bx)

= (a + bi)eax(cos bx + i sin bx) = aeax.

EXAMPLE 5.1

Evaluate J eax cos bx dx.

Solution: This can be done using integration by parts twice. I t is easier, however, using complex exponentials. From

e (a+bi)x e ax cog f a _|_ i e ax sJn f a

follows

J e(a+bi)x dx = Je (a+bi)x f a _ / e ax cog f a ^ _|_ I / e ax gj n f a f a

■ '/Therefore the desired integral is the real part of the integral on the left. By the preceding rule,

g(a+ b i)xe (a+bi)x f a = ---------- _(_ Q

a -j- bi/To find the real part, write

e (a+ bi)x eax(cos bx + i sin bx) a — bi a + bi a + bi a — bi

eax[(a cos bx + b sin bx) + i(a sin bx — b cos bx)] a2 + b2

5. Integration and Differentiation 617

eax(a cos bx + b sm bx) „Answer: ----------------- —---------- - + C.

a2 + b2

R e m a r k : By c o m p a r in g im a g in a r y p a r ts , w e g e t fr e e o f c h a r g e t h e fo r m u la

/ . , , eax(a sin bx — b cos bx) „eax sm bx dx = — -------------- ----------- - + C.

a2 + b2

EXAMPLE 5.2

/ .Compute / einx dx, where n is a non-zero integer.

Solution:r 2ir r2 ir r2 ir

/ elnx dx = / cos nx dx + i / sin nx dx = 7 o yo yo

sm nxn

. cos nx— i --------

o n

2ir= 0 .

Alternate Solution: An antiderivative of einx is einx/{in). Hence

/ ; einx dx =m

2x 1 1= — (62 - 6°) = — (1 - 1) = 0 .

o in tn

(Remember, e2ri = 1 . )

Answer: 0.

EXAMPLE 5.3f2x

Compute / sin kx cos nx dx, Jo

fc, n positive integers.

Solution: Write

/'2x /’27r ( e ikx — e~ikx\ (e inx + e~inx\/ sin kx cos nx dx = / I ------—----- ) I ------- -------J dx

Jo Jo \ 2i J \ 2 /

1 r f2ir f2"= — / ei(k+n)x dx — / e~i(k+n)x dx4* I. Jo Jo

+ J ei{Jc~n)x dx — J e ^ k~n)x dx j .Now use the result of the last example. If k 5* n, then all four integrals on the right are 0. If k = n, the first two integrals are 0 , the third and fourth cancel.

Answer: 0.


EXAMPLE 5.4

Evaluate J x cos3 x dx.

Solution: Write

(gtx g—i x \ 3 x

-----2-----/ = 8 3e2ixe~ix + 3eixe~2ix + e~3ix)

= - (e*ix + 3eix + 3e~ix + e~*ix).8

Hence,

J x cos3 x dx

= - xezix dx + 3 J xeix dx + 3 J xe~ix dx + J xe~zix dx'j

- (Iz + 3 /i + 3/_i + /_ 3),

where

I n = j xeinxdx.

To evaluate 7n , integrate by parts:

r r ( einx\ xeinx [ einx ,I n = / xetnx dx = / x d [ — J = —------ / — dx =

J J \ w / m J m

\ n 2 m /

xe,tnx p in x

in (in)2qXTIX

= ( - 9 + -f ) = — (1 - inx).

Notice that I n and 7_n are conjugates. Hence

’ (cos nx + i sin nx) (1 — inx)I n + I-n — 2 R e(/n) = 2 Re

2= — (cos nx + nx sin nx).

n2Now the answer follows easily:

J x cos3 x dx = - ( / 3 + /_ 3) + - (11 + /_ 1) + C

1 2 3= - • — (cos 3a; + 3a; sin 3a;) + - • 2 (cos x + x sin x) + C.

8 0 8

Answer:

EXERCISES

6. Applications to Differential Equations 619

1. Compute the 12-th derivative of exV3/2 ^ cos ?

Evaluate the integral using complex exponentials:

2. J cos x cosh x dx f C°S4 % 4. / cos2n x dx 5. sin kx sin nx dx,

Jo Jok, n positive integers

6. J cos kx cos nxdx} 7. J x sin3 x dx

k} n positive integers.

5. J x cos2 x dx

■ i :'2ir

9. Compute I (a0 + ai cos x + a2 cos 2x + • • • + an cos nx)2 dx.

(Hint: Use Ex. 6.)

6. APPLICATIONS TO DIFFERENTIAL EQUATIONS

In Chapter 14 we developed a systematic method for solving the linear differential equation

ay" + by' + cy = 0 .

Now we present another approach, via complex exponentials. The basic idea is this:

If y(x) = u(x) + iv(x) is a complex-valued function that satisfies the differential equation (with real coefficients)

ay" + by' + cy = 0 ,

then u(x) and v(x) also satisfy the equation. In other words, the real and the imaginary parts of a solution are also solutions.

Let us verify this statement. We are given

ay" + by' + cy = 0 .Hence

a(u" + iv") + b(u' + iv') + c(u + iv) = 0 ,that is,

(au" + bu' + cu) + i(av" + bv' + cv) = 0 .


Since the left-hand side equals 0 , so do its real and imaginary parts:

au" + bu' + cu = 0 and av" + bv' + cv = 0 .

Thus u{x) and v{x) are solutions.

EXAMPLE 6.1

Solve y" + 9y = 0.

Solution: Try y = epx, where p is allowed to be complex:

y" + 9 y = p 2epx + 9epx = 0, p 2 + 9 = 0, p = ± 3 i.

Thus e3ix is a complex solution. Therefore, its real and imaginary parts, cos Sx and sin 3x, are also solutions. Hence, the general solution, which involves two arbitrary constants, is C\ cos Sx + c2 sin Sx. Note that e~zix is another complex solution, but its real and imaginary parts are cos Sx and— sin Sx, nothing new. (The general solution could be written aezix + be~3ix.)

Answer: y = c\ cos Sx + c2 sin Sx.

EXAMPLE 6.2

Solve y" — 5y f + Qy = 0.

Solution: Try y = epx:

y" — by' + 6 y = p 2epx — hpepx + 6epx = 0 ,

p 2 — 5p + 6 = 0, p = 2 , 3.

Hence e2x and e3x are solutions.

Answer: y — C\e2x + c2eZx.

EXAMPLE 6.3

Solve y" + 4y' + 13y = 0 .

Solution: Try y = epx:

y" + 4 y f + 13 y = (;p2 + 4 p + lS)epx = 0,

p 2 + 4p + 13 = 0, p = —2 ± Si.

Hence y = e ~2+zi)x is a complex solution. Its real and imaginary parts are e~2x cos Sx and e~2x sin Sx, also solutions.

Answer: y = e~2x(c\ cos Sx + c2 sin 3a;).

Particular Solutions

Complex exponentials can be applied also to non-homogeneous equations

ay" + by' + cy = f(x),

5. Applications to Differential Equations 621

where the function f(x) can be expressed in terms of complex exponentials.

EXAMPLE 6.4

Find a particular solution of the differential equation y" + 9y = cos 2x.

Solution: Since cos 2x = Re(e2lx), find a complex solution of the differential equation

and take its real part. Try y = ae2ix:

y" + 9 y = (2i)2ae2ix + 9 ae2ix = - 4 ae2ix + 9 ae2ix = e2tx,

5a = 1.

Thus \ e 2ix is a complex solution.

y" + 9 y = e2ix,

Answer: Re = - cos 2x.5

EXAMPLE 6.5

Find a particular solution of y" + 4y' + 13y = cos x.

Solution: Since cos x = Re(eix), find a complex solution ot

y " + 42/' + 13 y = e*

and take its real part. Try y = aeix:

y" + 4 y' + 13 y = - a e ix + 4 aieix + 13 aeix = eix.Hence

— a -f- 4ai + 13a = 1, a1 12 — 4i 3 - i

12 + 4i 122 + 42 40

Therefore a complex solution is

Its real part is

R Q(y) = ^ (3 cos a; + sin x).

Answer: — (3 cos x + sin x).

EXAMPLE 6.6

Find a particular solution of y" + 2y' + Sy = ez sin 2x.

Solution: Since ex sin 2x = Im(e(1+2*)ir), find a complex solution of

y" + 2 y' + 3 y = e<1+2*>*,

and take its imaginary part. Try y = ae(1+2i)l:

y" + 2 y' + 3 y = a[(l + 2*)2 + 2(1 + 2 i) + 3]e<1+2f>- = e<1+2*>1,

a[(l + 41 + 4i2) + (2 + 4») + 3] = 1,

1 2 - 8 i 1 - 4* “ “ 2 + 8t “ 22 + 82 “ 34

Thus a complex solution is

y = (1 — 4i)e(1+2<)* = ^ (1 — 4i)e*(cos 2z + i sin 2z),

and its imaginary part is


Im (2/) = ~~ e*(sin 2x — 4 cos 2x).

Answer; e*(sin 2a: — 4 cos 2z).

EXERCISES

Using complex exponentials, find the general real solution:1. y'" - y = 0 2. y'" + y" + y ' + y = 03. i/<4> - y" - 6y = 0 4. */(4> - y = 0.5. Find the general solution of y" — 6y' + 10?/ = cos 2x by the method of this

section.Find a particular solution by the method of this section:

6. y" — a2y = cos bx 7. y" + a2y = x cos bx8. y" + a2y = e~bx cos ax 9. y" + y' — Qy = 3 sin x

, d2/ d/ 1 d2r dr10. L — + # — + — I = COS cot 11. -rrr--- — = a S in 0.d*2 dt C dd2 dd

7. APPLICATIONS TO POWER SERIES [optional]

Here is a puzzling fact about power series. We know the geometric series 1 + x + x2 + • • • diverges for \x\ > 1. This is not surprising because it represents 1/(1 — x), a function which “blows up” at x = 1. But the power series for 1/(1 + x2) also diverges for \x\ > 1 even though 1/(1 + x2) is perfectly well-behaved for all x. W hat goes wrong?

The answer to this question requires a broader view of power series. When investigating a real series

a0 + di(x — x0) + a2(x — x0)2 + * * • + an{x — x0)n + • • • ,

it often helps to regard the series as a special case of the complex series

a0 + ax(z — x0) + a2{z - x0)2 + * • • + an(z — x0)n + • * • .

(The coefficients are the same, but z is allowed to be complex.) For example, the real series

= i - *2 + - *6 + • • •

is a special case of the complex series

— = 1 - Z2 + 2" - Z6 + • • • .1 + z2

The complex function 1/(1 + z2) is well-behaved for all real values of z but “blows up” at 2 = i and z = —i. Hence, its power series cannot be expected to converge for all z. In fact, the series converges only in the disk \z\ < 1. The same is true of

— = 1 + Z + Z2 + Z3 + • • • ,1 — z

whose convergence is limited to the disk \z\ < 1 because of the “bad point” at z = 1. In both examples, the series fail to converge for \z\ > 1 due to troublesome points on the circle \z\ = 1.

In general, the complex power series

do + ai(z — z0) + a2(z — z0)2 + • • •

converges to the function/(z) in a region S of the complex plane if

|f(z) - [a0 + ai(z - z0) + • • • an(z - z0)n] \ ----> 0

as n ----* oo for each point z of S. Formally, this definition is the same as thecorresponding definition for real power series.

The basic fact about convergence of complex power series is th is:

Given a power series

Oo + Oi(« — So) + Ci2(z — z 0) 2 + • • • + an(z — z 0) n + • • * ,

precisely one of three cases holds:(i) The series converges only for z = z0.

(ii) The series converges for all values of z.(iii) There is a positive number R such that the series converges for

each z satisfying |z — z0\ < R and diverges for each z satisfying \z — z0\ > R.

7. Applications to Power Series 623

The proof of the corresponding fact for real power series (p. 75) applies almost without change.


In case (iii), the series converges in the circle (Fig. 7.1) with center at z0 and radius R. The number R is called the radius of convergence.

The real power series

a0 + ai(x — Xo) + • • • + an(x — x0)n + • • •

is a special case of the complex power series

a0 + ai(z — x0) + • • • + an(z - x0)n + • • • .

The latter converges in a circle of radius R centered at x0 on the real axis (Fig. 7.2). The real series converges on that part of the real axis contained in this circle, namely, an interval (diameter of the circle of convergence). Half of this interval is a radius of the circle; hence the term “radius of convergence” as applied to real series.

F ig . 7 .2


I t is shown in more advanced courses that the power series for f{z) at z = z0 converges inside the largest circle centered at z0 in which/(z) and all its derivatives are well-behaved. For example, take/(z) = 1/(1 — z), which has one bad point, at z = 1. Its power series at z = i converges in the circle shown in Fig. 7.3.

F ig . 7.3

I t is known that the ratio test applies to complex power series as well as to real ones.

Partial Fractions

Complex power series have applications to problems involving real functions. An example is the power series expansion by partial fractions, where the denominator has complex zeros.

EXAMPLE 7.1

Find the power series for —------------ — at x = 0 . W hat is its rare2 - 6z + 10

dius of convergence?

Solution: First, factor the denominator. Since the equation

x2 — 6x + 10 = 0

has roots 3 + i and 3 — i,

x2 - 6x + 10 = [x - (3 + i)] [x - (3 - i ) l

Now use partial fractions. (Use abbreviations a = 3 + i and 0 = 3 — i. Note that afi = 10.)

G26 15. COMPLEX ANALYSIS

— i _____ = ______ 1______ = +— 6x + 10 (x — a) (x — P) a — 0 \ x — /3 x — a )

= _ j _ / V — - V — 1= — V P ______M ra - fi \ L / Pn+1 L ( an+1) a - P L / \P n+1 an+1J

n = 0 n = 0 n = 0

= _ J _ V ^ "+1 - gn+1>\ X" = - L - V A"+1 - £"+1\a — /3 Zv \ (a:/3)n+1 / a — fi Zy \ 10n+1 /

n =0 n = 0

This is the answer. I t can be written in a slightly different form by observing that

a — /3 2 i

and that /3n+1 is the conjugate of an+1 (since 0 is the conjugate of a). Thus

an+l — (3n+1 1--------------- = — • 2i Im (an+1) = Im (an+1).

a — 2 i

The series converges inside the largest circle centered at the origin in which l / ( x 2 — 6x + 10) is well-behaved. This circle passes through 3 + i and 3 — i, the two points where the function “blows up.” Its radius is |3 + i\ = |3 - i\ = V lO .

oo

1 V IAnswer: —------— —— = > — — Im (an+1)xn,x2 - 6a; + 10 L ( 10n+1

n — 0

a = 3 + i; R = V lO .

R e m a r k : L et i s w r ite o u t a few term s o f th e p reced in g series:

a = 3 + i Im(a) = 1

a 2 = (3 + i )2 = 32 + 6 i + i2 Im (a2) = 6

a 3 = (3 + i )3 = 33 + 3 • 32z + 3 • 3*'2 + i 3 Im (a3) = 3 • 32 - 1 = 26

a 4 = (3 + i y = 34 + 4 • 3H + 6 • 32i2 + 4 • Si3 + i4 Im (a4) = 4 • 33 - 4 • 3 = 96.

H en ce

As a quick check, let us try x = 1 (permissible since the series converges for \x\ < y / 10). The left side is then ^ = 0.2000; the sum of the first four terms on the right side is

0.1000 + 0.0600 + 0.0260 + 0.0096 = 0.1956.

EXERCISES

1. Expand 1/(1 — z) in powers of z — i.[Hint: Write (1 — z) = (1 — i) — (z — t).]

Find the power series at z = 0 and write out the first four terms explicitly. What is the radius of convergence?

2. 3. 1z2 + z + 1 z2 + 4z + 133 + 1 3z + 5z24. —------- —r 52 — 2 + 1 1 - 2z + 5z2

6. Find the sum of the series 1 + r cos 0 + r2 cos 20 + • • • + rn cos nO + • • • , - 1 < r < 1.[Hint: rn cos nd — Re(zn), where z = reie.]

d d7. If z = x + iy, prove — (zn) = nzn~l, — (zn) = nizn~l.ox dy8. Write zn = u(x, y) + i v(x, y), in real and imaginary parts. Compute u and

v for n = ±1, ±2, and 3.g\ / ■ \ cii_ T-. xi x du dv du dv9. (cont.) bhow, using Ex. 7, that -r— = -r— > —— = — ——•dx dy dy dx

10. (cont.) Show that g + g = 0, g + g = 0.

11. (cont.) Let/(z) = a0 + ai2 + a2z2 + * * * be the sum of a complex power series. Separate real and imaginary parts: f(z) = w(x, 2/) + i v( , y).Show that

TABLE 1 Trigonometric Functions

Degrees sin cos tan cot

0° .0000 1.000 .0000 ' — CO o o

1° .0175 .9998 .0175 57.29 89°2° .0349 .9994 .0349 28.64 88°3° .0523 .9986 .0524 19.08 87°4° .0698 .9976 .0699 14.30 86°5° .0872 .9962 .0875 11.43 85°6° .1045 .9945 .1051 9.514 84°7° .1219 .9925 .1228 8.144 83°8° .1392 .9903 .1405 7.115 82°9° .1564 .9877 .1584 6.314 81°

10° .1736 .9848 .1763 5.671 00 o o

11° .1908 .9816 .1944 5.145 79°12° .2079 .9781 .2126 4.705 78°13° .2250 .9744 .2309 4.331 77°14° .2419 .9703 .2493 4.011 76°15° .2588 .9659 .2679 3.732 75°16° .2756 .9613 .2867 3.487 74°17° .2924 .9563 .3057 3.271 73°18° .3090 .9511 .3249 3.078 72°19° .3256 .9455 .3443 2.904 71°20° .3420 .9397 .3640 2.747

0O

21° .3584 .9336 .3839 2.605 69°22° .3746 .9272 .4040 2.475 68°23° .3907 .9205 .4245 2.356 67°24° .4067 .9135 .4452 2.246 66°25° .4226 .9063 .4663 2.145 65°26° .4384 .8988 .4877 2.050 64°27° .4540 .8910 .5095 1.963 63°28° .4695 .8829 .5317 1.881 62°29° .4848 .8746 .5543 1.804 61°

00 o o .5000 .8660 .5774 1.732 60°

31° .5150 .8572 .6009 1.664 59°32° .5299 .8480 .6249 1.600 58°33° .5446 .8387 .6494 1.540 57°34° .5592 .8290 .6745 1.483 56°

oCO .5736 .8192 .7002 1.428 55°36° .5878 .8090 .7265 1.376 54°37° .6018 .7986 .7536 1.327 53°38° .6157 .7880 .7813 1.280 52°39° .6293 .7771 .8098 1.235 51°

oO

.6428 .7660 .8391 1.192 50°41° .6561 .7547 .8693 1.150 49°42° .6691 .7431 .9004 1.111 48°43° .6820 .7314 .9325 1.072 47°44° .6947 .7193 .9657 1.036 46°45° .7071 .7071 1.000 1.000 45°

cos sin ctn tan Degrees

TABLE 2 Trigonometric Functions for Angles in Radians*

Rad. Sin Tan Cot Cos Rad Sin Tan Cot Co8

.00 .00000 .00000 oo 1.00000 .50 .47943 .54630 1.8305 .8775801 .01000 .01000 99.997 0.99995 .51 .48818 .55936 1.7878 .87274

.02 .02000 .02000 49.993 .99980 .52 .49688 .57256 1.7465 .86782

.03 .03000 .03001 33.323 .99955 .53 .50553 .58592 1.7067 .86281

.04 .03999 .04002 24.987 .99920 . 54 .51414 .59943 1.6683 .85771

.05 .04998 .05004 19.983 .99875 .55 .52269 .61311 1.6310 .85252

.06 .05996 .06007 16.647 .99820 .56 .53119 .62695 1.5950 .84726

.07 .06994 .07011 14.262 .99755 .57 .53963 .64097 1.5601 .84190

.08 .07991 .08017 12.473 .99680 .58 .54802 .65517 1.5263 .83646

.09 .08988 .09024 11.081 .99595 .59 .55636 .66956 1.4935 .83094

.10 .09983 .10033 9.9666 .99500 .60 .56464 .68414 1.4617 .82534

.11 .10978 .11045 9.0542 .99396 .61 .57287 .69892 1.4308 .81965

.12 .11971 .12058 8.2933 .99281 .62 .58104 .71391 1.4007 .81388

.13 .12963 .13074 7.6489 .99156 .63 .58914 .72911 1.3715 .80803

.14 .13954 .14092 7.0961 .99022 .64 .59720 .74454 1.3431 .80210

.15 .14944 .15114 6.6166 .98877 .65 .60519 .76020 1.3154 .79608

.16 .15932 .16138 6.1966 .98723 .66 .61312 .77610 1.2885 .78999

.17 .16918 .17166 5.8256 .98558 .67 .62099 !79225 1.2622 .78382

.18 .17903 .18197 5.4954 .98384 .68 .62879 .80866 1.2366 .77757

.19 .18886 .19232 5.1997 .98200 .69 .63654 .82534 1.2116 .77125

.20 .19867 .20271 4.9332 .98007 .70 .64422 .84229 1.1872 .76484

.21 .20846 .21314 4.6917 .97803 .71 .65183 .85953 1.1634 .75836

.22 .21823 .22362 4.4719 .97590 .72 .65938 .87707 1.1402 .75181

.23 .22798 .23414 4.2709 .97367 .73 .66687 .89492 1.1174 .74517

.24 .23770 .24472 4.0864 .97134 .74 .67429 .91309 1.0952 .73847

.25 .24740 .25534 3.9163 .96891 .75 .68164 .93160 1.0734 .73169

.26 .25708 .26602 3.7591 .96639 .76 .68892 .95045 1.0521 .72484

.27 .26673 .27676 3.6133 .96377 .77 .69614 .96967 1.0313 .71791

.28 .27636 .28755 3.4776 .96106 .78 .70328 .98926 1.0109 .71091

.29 .28595 .29841 3.3511 .95824 .79 .71035 1.0092 .99084 .70385

.30 .29552 .30934 3.2327 .95534 .80 .71736 1.0296 .97121 .69671

.31 .30506 .32033 3.1218 .95233 .81 .72429 1.0505 .95197 .68950

.32 .31457 .33139 3.0176 .94924 .82 .73115 1.0717 .93309 .68222

.33 .32404 .34252 2.9195 .94604 .83 .73793 1.0934 .91455 .67488

.34 .33349 .35374 2.8270 .94275 .84 .74464 1.1156 .89635 .66746

.35 .34290 .36503 2.7395 .93937 .85 .75128 1.1383 .87848 .65998

.36 .35227 .37640 2.6567 .93590 .86 .75784 1.1616 .86091 .65244

.37 .36162 .38786 2.5782 .93233 .87 .76433 1.1853 .84365 .64483

.38 .37092 .39941 2.5037 .92866 .88 .77074 1.2097 .82668 .63715

.39 .38019 .41105 2.4328 .92491 .89 .77707 1.2346 .80998 .62941

.40 .38942 .42279 2.3652 .92106 .90 .78333 1.2602 .79355 .62161

.41 .39861 .43463 2.3008 .91712 .91 .78950 1.2864 .77738 .61375

.42 .40776 .44657 2.2393 .91309 .92 .79560 1.3133 .76146 .60582

.43 .41687 .45862 2.1804 .90897 .93 .80162 1.3409 .74578 .59783

.44 .42594 .47078 2.1241 .90475 .94 .80756 1.3692 .73034 .58979

.45 .43497 .48306 2.0702 .90045 .95 .81342 1.3984 .71511 .58168

.46 .44395 .49545 2.0184 .89605 .96 .81919 1.4284 .70010 .57352

.47 .45289 .50797 1.9686 .89157 .97 .82489 1.4592 .68531 .56530

.48 .46178 .52061 1.9208 .88699 .98 .83050 1.4910 .67071 .55702

.49 .47063 .53339 1.8748 .88233 .99 .83603 1.5237 .65631 .54869

.50 .47943 .54630 1.8305 .87758 1.00 .84147 1.5574 .64209 .54030

Rad. Sin Tan Cot Cos Rad. Sin Tan Cot Cos

* Tables 2-5 are reproduced from “Handbook of Tables for Mathematics” (R. C. Weast and S. M. Selby, eds.), 3rd ed., Chemical Rubber Co., Cleveland, Ohio, and used by permission.

TABLE 2 Trigonometric Functions for Angles in Radians (Continued)

Rad. Sin Tan Cot Cos Rad. Sin | Tan Cot Cos

1.00 .84147 1.5574 .64209 .54030 1 .50 ! .99749 14.101 07091 .070741 01 .84683 1.5922 .62806 .53186 1 51 1 .99815 16.428 .06087 .060761.02 .85211 1.6281 61420 .52337 1 52 1 99871 19 670 .05084 .050771.03 .85730 1.6652 .60051 .51482 i 1.53 .99917 24.498 .04082 .040791.04 .86240 1.7036 .58699 .50622 i 1 .54 .99953 32.461 .03081 .03079

1.05 .86742 1.7433 .57362 .49757 1.55 .99978 48.078 .02080 .020791 .06 .87236 1 7844 .56040 .48887 ! 1.56 | 99994 92 621 .01080 .010801.07 .87720 1.8270 .54734 .48012 1.57 I 1 00000 1255 8 .00080 .000801.08 .88196 1.8712 53441 47133 1 .58 .99996 - 1 0 8 .6 5 - .00920 - .009201.09 .88663 1.9171 .52162 46249 1 .59 .99982 - 5 2 067 - .01921 -.01920

1.10 89121 1.9648 .50897 .45360 1.60 99957 -34.233 - .02921 - .029201.11 .89570 2.0143 .49644 44466 1.61 99923 -25.495 - .03922 - .039191 12 .90010 2.0660 .48404 .43568 1.62 .99879 -2 0 307 - .04924 -.049181.13 .90441 2.1198 .47175 .42666 1 63 99825 - 1 6 .8 7 1 - .05927 - .059171,14 .90863 2.1759 .45959 41759 1.64 .99761 -14.427 - .06931 - .06915

1.15 91276 2 2345 44753 .40849 1.65 .99687 -1 2 599 - .07937 - .079121.16 .91680 2.2958 .43558 39934 1 66 99602 -11.181 - .08944 - .089091 17 .92075 2 3600 .42373 39015 1 67 .99508 -10.047 - .09953 - .099041.18 .92461 2 4273 41199 .38092 1.68 .99404 - 9 .1208 - .10964 - .108991.19 92837 2.4979 .40034 37166 1.69 .99290 - 8.3492 -.11977 -.11892

1.20 93204 2.5722 .38878 36236 1.70 99166 - 7.6966 - .12993 - .128841.21 .93562 2.6503 .37731 35302 1.71 .99033 - 7.1373 -.14011 - .1 3 8 7 51.22 .93910 2.7328 36593 .34365 1.72 .98889 - 6.6524 - .15032 -.148651.23 .94249 2.8198 35463 .33424 1.73 .98735 - 6.2281 -.16056 -.158531 24 .94578 2.9119 .34341 .32480 1 74 .98572 - 5.8535 - .17084 - .1 6 8 4 0

1.25 .94898 3.0096 .33227 31532 1.75 .98399 - 5 5204 -.18115 - .1 7 8 2 51.26 .95209 3 1133 32121 .30582 1.76 .98215 - 5 2221 - .19149 - .188081.27 .95510 3.2236 .31021 29628 1.77 .98022 - 4 9534 - .20188 - .197891.28 .95802 3.3413 .29928 .28672 1 78 .97820 - 4 7101 - .21231 - .2 0 7 6 81.29 .96084 3.4672 .28842 27712 1.79 .97607 - 4 .4887 - .22278 -.21745

1.30 .96356 3.6021 27762 26750 1.80 .97385 - 4.2863 -.23330 - .2 2 7 2 01.31 96618 3.7471 26687 25785 1.81 .97153 - 4 1005 - .24387 - .236931 32 .96872 3.9033 25619 .24818 1.82 .96911 - 3 9294 - .25449 - .246631.33 .97115 4 0723 .24556 .23848 1 83 96659 - 3.7712 - 26517 -.256311.34 .97348 4.2556 .23498 .22875 1 84 .96398 - 3.6245 - .27590 - .26596

1.35 .97572 4 4552 22446 .21901 1.85 .96128 - 3.4881 - .28669 -.275591 36 .97786 4 6734 21398 .20924 1.86 .95847 - 3 .3608 - .29755 - .285191 37 .97991 4 9131 20354 .19945 1.87 95557 - 2 2419 - 30846 - .294761.38 .98185 5.1774 19315 .18964 1.88 .95258 - 3 1304 - .31945 - .304301.39 .98370 5.4707 18279 .17981 1.89 .94949 - 3 .0257 - .33051 -.31381

1.40 .98545 5.7979 .17248 16997 1.90 94630 - 2 9271 - .34164 - .323291.41 .98710 6.1654 16220 .16010 1.91 94302 - 2 8341 - 35284 - 332741.42 .98865 6 5811 .15195 15023 1 92 93965 - 2 7463 - .36413 - .342151.43 .99010 7.0555 14173 .14033 1 93 93618 - 2 6632 - .37549 -.351531.44 .99146 7 6018 .13155 13042 1.94 93262 - 2.5843 - .38695 - .36087

1.45 .99271 8.2381 .12139 .12050 1 95 .92896 - 2.5095 - .39849 - .3 7 0 1 81.46 .99387 8.9886 .11125 .11057 1 96 92521 - 2 4383 - 41012 - 379451.47 .99492 9.8874 .10114 10063 1.97 92137 - 2 3705 - 42185 - .388681.48 .99588 10 983 .09105 .09067 1 98 91744 - 2 3058 - .43368 - .397881.49 99674 12.350 .08097 .08071 1 99 91341 - 2 2441 - .44562 - .40703

1.50 .99749 14.101 .07091 .07074 2 .00 .90930 - 2 .1850 - .45766 - .41615

Rad. Sin Tan Cot Cos Rad. Sin Tan Cot Cos

631

TABLE 3 Four-Place Mantissas for Common Logarithms

N 0 1 2 3 4 5 6 7 8 9 1 2Proportional Parts

3 4 5 6 7 8 9

10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 *4 8 12 17 21 25 29 33 3711 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 4 8 11 15 19 23 26 30 3412 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 3 7 10 14 17 21 24 28 3113 1139 f 173 1206 1239 1271 1303 1335 1367 1399 1430 3 6 10 13 16 19 23 26 2914 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 3 6 9 12 15 18 21 24 27

15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 *3 6 8 11 14 17 20 22 2516 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 3 5 8 11 13 16 18 21 2417 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 2 5 7 10 12 15 17 20 2218 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 2 5 7 9 12 14 16 19 2119 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 2 4 7 9 11 13 16 18 20

20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 2 4 6 8 11 13 15 17 1921 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 2 4 6 8 10 12 14 16 1822 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 2 4 6 8 10 12 14 15 1723 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 2 4 6 7 9 11 13 15 1724 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 2 4 5 7 9 11 12 14 16

25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 2 3 5 7 9 10 12 14 1526 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 2 3 5 7 8 10 11 13 1527 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 2 3 5 6 8 9 11 13 1428 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 2 3 5 6 8 9 11 12 1429 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 1 3 4 6 7 9 10 12 13

30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 1 3 4 6 7 9 10 11 1331 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 1 3 4 6 7 8 10 11 1232 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 1 3 4 5 7 8 9 11 1233 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 1 3 4 5 6 8 9 10 1234 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 1 3 4 5 6 8 9 10 11

35 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 1 2 4 5 6 7 9 10 1136 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 1 2 4 5 6 7 8 10 1137 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 1 2 3 5 6 7 8 9 1038 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 1 2 3 5 6 7 8 9 1039 5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 1 2 3 4 5 7 8 9 10

40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 1 2 3 4 5 6 8 9 1041 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 1 2 3 4 5 6 7 8 942 6232 6243 6253 6203 6274 6284 6294 6304 6314 6325 1 2 3 4 5 6 7 8 943 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 1 2 3 4 5 6 7 8 944 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 1 2 3 4 5 6 7 8 9

45 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 1 2 3 4 5 6 7 8 946 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 1 2 3 4 5 6 7 7 847 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 1 2 3 4 5 5 6 7 848 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 1 2 3 4 4 5 6 7 849 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 1 2 3 4 4 5 6 7 8

50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 1 2 3 3 4 5 6 7 851 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 1 2 3 3 4 5 6 7 852 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 1 2 2 3 4 5 6 7 753 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 1 2 2 3 4 5 6 6 754 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 1 2 2 3 4 5 6 6 7

N 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

* Interpolation in this section of the table is inaccurate.

632

TABLE 3 Four-Place Mantissas for Common Logarithms (Continued)

N 0 1 2 3 4 5 6 7 8 9 1 2Proportional Parts

3 4 5 6 7 8 9

55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 1 2 2 3 4 5 5 6 756 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 1 2 2 3 4 5 5 6 757 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 1 2 2 3 4 5 5 6 758 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 1 1 2 3 4 4 5 6 759 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 1 1 2 3 4 4 5 6 7

60 7782 7789 7796 7808 7810 7818 7825 7832 7839 7846 1 1 2 3 4 4 5 6 661 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 1 1 2 3 4 4 5 6 662 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 1 1 2 3 3 4 5 6 663 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 1 1 2 3 3 4 5 5 664 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 1 1 2 3 3 4 5 5 6

65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 1 1 2 3 3 4 5 5 666 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 1 1 2 3 3 4 5 5 667 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 1 1 2 3 3 4 5 5 668 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 1 1 2 3 3 4 4 5 669 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 1 1 2 2 3 4 4 5 6

70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 1 1 2 2 3 4 4 5 671 8513 8519 8525 8531 8537 8543 8549 8555 8561 8567 1 1 2 2 3 4 4 5 572 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 1 1 2 2 3 4 4 5 573 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 1 1 2 2 3 4 4 5 574 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 1 1 2 2 3 4 4 5 5

75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 1 1 2 2 3 3 4 5 576 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 1 1 2 2 3 3 4 5 577 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 1 1 2 2 3 3 4 4 578 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 1 1 2 2 3 3 4 4 579 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 1 1 2 2 3 3 4 4 5

80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 1 1 2 2 3 3 4 4 581 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 1 1 2 2 3 3 4 4 582 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 1 1 2 2 3 3 4 4 583 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 1 1 2 2 3 3 4 4 584 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 1 1 2 2 3 3 4 4 5

85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 1 1 2 2 3 3 4 4 586 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 1 1 2 2 3 3 4 4 587 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 0 1 1 2 2 3 3 4 488 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 0 1 1 2 2 3 3 4 489 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 0 1 1 2 2 3 3 4 4

90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 0 1 1 2 2 3 3 4 491 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 0 1 1 2 2 3 3 4 492 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 0 X 1 2 2 3 3 4 493 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 0 1 1 2 2 3 3 4 494 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 0 1 1 2 2 3 3 4 4

95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 0 1 1 2 2 3 3 4 496 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 0 1 1 2 2 3 3 4 497 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 0 1 1 2 2 3 3 4 498 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 0 1 1 2 2 3 3 4 499 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 0 1 1 2 2 3 3 3 4

N 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

9

22222

22233

33333

33333

33344

44444

44445

55555

55666

66666

9

TABLE 4 Antilogarithms

0 1 2 3 4 5 6 7 8 9 1 2Proportional Parts

3 4 5 6 7 8

1000 1002 1005 1007 1009 1012 1014 1016 1019 1021 0 0 1 1 1 1 2 21023 1026 1028 1030 1033 1035 1038 1040 1042 1045 0 0 1 1 1 1 2 21047 1050 1052 1054 1057 1059 1062 1064 1067 1069 0 0 1 1 1 1 2 21072 1074 1076 1079 1081 1084 1086 1089 1091 1094 0 0 1 1 1 1 2 21096 1099 1102 1104 1107 1109 1112 1114 1117 1119 0 1 1 1 1 2 2 2

1122 1125 1127 1130 1132 1135 1138 1140 1143 1146 a 1 1 1 1 2 2 21148 1151 1153 1156 1159 1161 1164 1167 1169 1172 9 1 1 1 1 2 2 21175 1178 1180 1183 1186 1189 1191 1194 1197 1199 0 1 1 1 1 2 2 21202 1205 1208 1211 1213 1216 1219 1222 1225 1227 0 1 1 1 1 2 2 21230 1233 1236 1239 1242 1245 1247 1250 1253 1256 0 1 1 1 1 2 2 2

1259 1262 1265 1268 1271 1274 1276 1279 1282 1285 0 1 1 1 1 2 2 21288 1291 1294 1297 1300 1303 1306 1309 1312 1315 0 1 1 1 2 2 2 31318 1321 1324 1327 1330 1334 1337 1340 1343 1346 0 1 1 1 2 2 2 21349 1352 1355 1358 1361 1365 1368 1371 1374 1377 0 1 1 1 2 2 2 31380 1384 1387 1390 1393 1396 1400 1403 1406 1409 0 1 1 1 2 2 2 3

1413 1416 1419 1422 1426 1429 1432 1435 1439 1442 0 1 1 1 2 2 2 31445 1449 1452 1455 1459 1462 1466 1469 1472 1476 0 1 1 1 2 2 2 31479 1483 1486 1489 1493 1496 1500 1503 1507 1510 0 1 1 1 2 2 2 31514 1517 1521 1524 1528 1531 1535 1538 1542 1545 0 1 1 1 2 2 2 31549 1552 1556 1560 1563 1567 1570 1574 1578 1581 0 1 1 1 2 2 3 3

1585 1589 1592 1596 1600 1603 1607 1611 1614 1618 0 1 1 1 2 2 3 31622 1626 1629 1633 1637 1641 1644 1648 1652 1656 0 1 1 2 2 2 3 31660 1663 1667 1671 1675 1679 1683 1687 1690 1694 0 1 1 2 2 2 3 31698 1702 1706 1710 1714 1718 1722 1726 1730 1734 0 1 1 2 2 2 3 31738 1742 1746 1750 1754 1758 1762 1766 1770 1774 0 1 1 2 2 2 3 3

1778 1782 1786 1791 1795 1799 1803 1807 1811 1816 0 1 1 2 2 2 3 31820 1824 1828 1832 1837 1841 1845 1849 1854 1858 0 1 1 2 2 3 3 31862 1866 1871 1875 1879 1884 1888 1892 1897 1901 0 1 1 2 2 3 3 31905 1910 1914 1919 1923 1928 1932 1936 1941 1945 0 1 1 2 2 3 3 41950 1954 1959 1963 1968 1972 1977 1982 1986 1991 0 1 1 2 2 3 3 4

1995 2000 2004 2009 2014 2018 2023 2028 2032 2037 0 1 1 2 2 3 3 42042 2046 2051 2056 2061 2065 2070 2075 2080 2084 0 1 1 2 2 3 3 42089 2094 2099 2104 2109 2113 2118 2123 2128 2133 0 1 1 2 2 3 3 42138 2143 2148 2153 2158 2163 2168 2173 2178 2183 0 1 1 2 2 3 3 42188 2193 2198 2203 2208 2213 2218 2223 2228 2234 1 1 2 2 3 3 4 4

2239 2244 2249 2254 2259 2265 2270 2275 2280 2286 1 1 2 2 3 3 4 42291 2296 2301 2307 2312 2317 2323 2328 2333 2339 1 1 2 2 3 3 4 42344 2350 2355 2360 2366 2371 2377 2382 2388 2393 1 1 2 2 3 3 4 42399 2404 2410 2415 2421 2427 2432 2438 2443 2449 1 1 2 2 3 3 4 42455 2460 2466 2472 2477 2483 2489 2495 2500 2506 1 1 2 2 3 3 4 5

2512 2518 2523 2529 2535 2541 2547 2553 2559 2564 1 1 2 2 3 4 4 52570 2576 2582 2588 2594 2600 2606 2612 2618 2624 1 1 2 2 3 4 4 52630 2636 2642 2649 2655 2661 2667 2673 2679 2685 1 1 2 2 3 4 4 52692 2698 2704 2710 2716 2723 2729 2735 2742 2748 1 1 2 3 3 4 4 52754 2761 2767 2773 2780 2786 2793 2799 2805 2812 1 1 2 3 3 4 4 5

2818 2825 2831 2838 2844 2851 2858 2864 2871 2877 1 1 2 3 3 4 5 52884 2891 2897 2904 2911 2917 2924 2931 2938 2944 1 1 2 3 3 4 5 52951 2958 2965 2972 2979 2985 2992 2999 3006 3013 1 1 2 3 3 4 5 53020 3027 3034 3041 3048 3055 3062 3069 3076 3083 1 1 2 3 4 4 5 63090 3097 3105 3112 3119 3126 3133 3141 3148 3155 1 1 2 3 4 4 5 6

0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8

634

TABLE 4 Antilogarithms (Continued)

0 1 2 3>4 5 6 7 8 9 1 2

Proportional Parts 3 4 5 6 7 8 9

.50 3162 3170 3177 3184 3192 3199 3206 3214 3221 3228 1 1 2 3 4 4 5 6 7

.51 3236 3243 3251 3258 3266 3273 3281 3289 3296 3304 1 2 2 3 4 5 5 6 7

.52 ] 3311 3319 3327 3334 3342 3350 3357 3365 3373 3381 1 2 2 3 4 5 5 6 7

.53 | 3388 3396 3404 3412 3420 3428 3436 3443 3451 3459 1 2 2 3 4 5 6 6 7

.54 3467 3475 3483 3491 3499 3508 3516 3524 3532 3540 1 2 2 3 4 5 6 6 7

.55 3548 3556 3565 3573 3581 3589 3597 3606 3614 3622 1 2 2 3 4 5 6 7 7

.56 3631 3639 3648 3656 3664 3673 3681 3690 3698 3707 1 2 3 3 4 5 6 7 8

.57 3715 3724 3733 3741 3750 3758 3767 3776 3784 3793 1 2 3 3 4 5 6 7 8

.58 i 3802 3811 3819 3828 3837 3846 3855 3864 3873 3882 1 2 3 4 4 5 6 7 8

.59 3890 3899 3908 3917 3926 3936 3945 3954 3963 3972 1 2 3 4 5 5 6 7 8

.60 3981 3990 3999 4009 4018 4027 4036 4046 4055 4064 1 2 3 4 5 6 6 7 8

.61 4074 4083 4093 4102 4111 4121 4130 4140 4150 4159 1 2 3 4 5 6 7 8 9

.62 4169 4178 4188 4198 4207 4217 4227 4236 4246 4256 1 2 3 4 5 6 7 8 9

.63 4266 4276 4285 4295 4305 4315 4325 4335 4345 4355 1 2 3 4 5 6 7 8 9

.64 4365 4375 4385 4395 4406 4416 4426 4436 4446 4457 1 2 3 4 5 6 7 8 9

.65 4467 4477 4487 4498 4508 4519 4529 4539 4550 4560 1 2 3 4 5 6 7 8 9

.66 4571 4581 4592 4603 4613 4624 4634 4645 4656 4667 1 2 3 4 5 6 7 9 10

. 67 4677 4688 4699 4710 4721 4732 4742 4753 4764 4775 1 2 3 4 5 7 8 9 10

.68 4786 4797 4808 4819 4831 4842 4853 4864 4875 4887 1 2 3 4 6 7 8 9 10

.69 4898 4909 4920 4932 4943 4955 4966 4977 4989 5000 1 2 3 5 6 7 8 9 10

.70 5012 5023 5035 5047 5058 5070 5082 5093 5105 5117 1 2 4 5 6 7 8 9 11

.71 5129 5140 5152 5164 5176 5188 5200 5212 5224 5236 1 2 4 5 6 7 8 10 11

.72 5248 5260 5272 5284 5297 5309 5321 5333 5346 5358 1 2 4 5 6 7 9 10 11

.73 5370 5383 5395 5408 5420 5433 5445 5458 5470 5483 1 3 4 5 6 8 9 10 1174 5495 5508 5521 5534 5546 5559 5572 5585 5598 5610 1 3 4 5 6 8 9 10 12

.75 5623 5636 5649 5662 5675 5689 5702 5715 5728 5741 1 3 4 5 7 8 9 10 12

.76 5754 5768 5781 5794 580 > 5821 5834 5848 5861 5875 1 3 4 5 7 8 9 11 12

.77 5888 5902 5916 5929 5943 5957 5970 5984 5998 6012 1 3 4 5 7 8 10 11 12

.78 6026 6039 6053 6067 6081 6095 6109 6124 6138 6152 1 3 4 6 7 8 10 11 13

.79 6166 6180 6194 6209 6223 6237 6252 6266 6281 6295 1 3 4 6 7 9 10 11 13

.80 6310 6324 6339 6353 6368 6383 6397 6412 6427 644? 1 3 4 6 7 9 10 12 13

.81 6457 6471 6486 6501 6516 6531 6546 6561 6577 6592 2 3 5 6 8 9 11 12 14

.82 6607 6622 6637 6653 6668 6683 6699 6714 6730 6745 2 3 5 6 8 9 11 12 14

.83 6761 6776 6792 6808 6823 6839 6855 6871 6887 6902 2 3 5 6 8 9 11 13 14

.84 6918 6934 6950 6966 6982 6998 7015 7031 7047 7063 2 3 5 6 8 10 11 13 15

.85 7079 7096 7112 7129 7145 7161 7178 7194 7211 7228 2 3 5 7 8 10 12 13 15

.86 7244 7261 7278 7295 7311 7328 7345 7362 7379 7396 2 3 5 7 8 10 12 13 15

.87 7413 7430 7447 7464 7482 7499 7516 7534 7551 7568 2 3 5 7 9 10 12 14 16

.88 7586 7603 7621 7638 7656 7674 7691 7709 7727 7745 2 4 5 7 9 11 12 14 16

.89 7762 7780 7798 7816 7834 7852 7870 7889 7907 7925 2 4 5 7 9 11 13 14 16

.90 ! 7943 7962 7980 7998 8017 8035 8054 8072 8091 8110 2 4 6 7 9 11 13 15 1791 8128 8147 SI 66 8185 8204 8222 8241 8260 8279 8299 2 4 6 8 9 11 13 15 17

.92 8318 8337 8356 8375 8395 8414 8433 8453 8472 8492 2 4 6 8 10 12 14 15 17

.93 8511 8531 8551 8570 8590 8610 8630 8650 8670 8690 2 4 6 8 10 12 14 16 18

.94 8710 8730 8750 8770 8790 8810 8831 8851 8872 8892 2 4 6 8 10 12 14 16 18

.95 8913 8933 8954 8974 8995 9016 9036 9057 9078 9099 2 4 6 8 10 12 15 17 19

.96 9120 9141 9162 9183 9204 9226 9247 9268 9290 9311 2 4 6 8 11 13 15 17 19

.97 9333 9354 9376 9397 9419 9441 9462 9484 9506 9528 2 4 7 9 11 13 15 17 20

.98 9550 9572 9594 9616 9638 9661 9683 9705 9727 9750 2 4 7 9 11 13 16 18 20

.99 9772 9795 9817 9840 9863 9886 9908 9931 9954 9977 2 5 7 9 11 14 16 18 20

0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

635

TABLE 5 Exponential Functions

X e x Log,0(e*) e~x

0.00 1.0000 0.00000 1.0000000.01 1.0101 .00434 0.9900500.02 1.0202 .00869 .9801990.03 1 0305 .01303 .9704460.04 1.0408 .01737 .960789

0.05 1.0513 0.02171 0.9512290.06 1.0618 .02606 .9417650.07 1 0725 .03040 .9323940 08 1 0833 .03474 .9231160.09 1 0942 .03909 .913931

0.10 1 1052 0.04343 0.904837O i l 1 1163 .04777 .8958340.12 1.1275 05212 .8869200.13 1.1388 .05646 .8780950.14 1.1503 .06080 .869358

0.15 1 1618 0.06514 0.8607080.16 1.1735 .06949 .8521440.17 1.1853 .07383 .8436650.18 1.1972 .07817 .8352700.19 1 2092 .08252 826959

0.20 1.2214 0.08686 0.8187310.21 1.2337 .09120 .8105840.22 1.2461 .09554 .8025190.23 1.2586 .09989 .7945340.24 1 2712 .10423 .786628

0.25 1.2840 0.10857 0.7788010.26 1.2969 .11292 .7710520.27 1.3100 .11726 .7633790.28 1.3231 .12160 .7557840.29 1 3364 12595 .748264

0.30 1 3499 0.13029 0.7408180.31 1.3634 .13463 7334470.32 1.3771 .13897 .7261490.33 1 3910 14332 .7189240.34 1 4049 .14766 .711770

0.35 1.4191 0.15200 0.7046880.36 1 4333 15635 6976760.37 1 4477 .16069 J .6907340.38 1 4623 .16503 .6838610.39 1.4770 16937 677057

0.40 1.4918 0.17372 0.6703200.41 1.5068 17806 .6636500.42 1.5220 .18240 .6570470.43 1.5373 18675 .6505090.44 1.5527 19109 .644036

0.45 1 5683 0.19543 0.6376280.46 1.5841 .19978 .6312840.47 1.6000 .20412 .6250020.48 1.6161 .20846 .6187830.49 1 6323 .21280 612626

0.50 1.6487 0.21715 0.606531

X e x Logi0(ex) e~z

0.50 1.6487 0.21715 0.6065310.51 1 6653 .22149 .6004960.52 1.6820 22583 5945210.53 1.6989 .23018 5886050 54 1 7160 23452 .582748

0.55 1 7333 0 23886 0.5769500.56 1.7507 .24320 .5712090.57 1 7683 .24755 .5655250.58 1 7860 .25189 .5598980.59 1 8040 25623 .554327

0.60 1.8221 0.26058 0.5488120.61 1 8404 .26492 .5433510 62 1 8589 .26926 .5379440.63 1.8776 .27361 .5325920 64 1.8965 .27795 .527292

0.65 1 9155 0.28229 0 5220460 66 1 9348 .28663 .5168510.67 1.9542 .29098 .5117090.68 1.9739 .29532 5066170.69 1.9937 .29966 .501576

0.70 2.0138 0 30401 0.4965850.71 2.0340 .30835 .4916440.72 2.0544 .31269 4867520.73 2.0751 .31703 .4819090.74 2.0959 .32138 .477114

0.75 2.1170 0.32572 0 4723670.76 2 1383 .33006 4676660.77 2.1598 33441 4630130.78 2.1815 33875 4584060.79 2.2034 34309 453845

0.80 2.2255 0 34744 0 4493290.81 2.2479 .35178 .4448580.82 2.2705 35612 4404320.83 2.2933 .36046 4360490.84 2.3164 .36481 .431711

0.85 2.3396 0.36915 0 4274150.86 2.3632 .37349 .4231620.87 2.3869 .37784 .4189520.88 2.4109 38218 .4147830.89 2 4351 .38652 .410656

0.90 2.4596 0 39087 0 4065700.91 2.4843 .39521 .4025240.92 2.5093 .39955 .3985190.93 2.5345 40389 3945540 94 2,5600 .40824 390628

0.95 2.5857 0 41258 0 3867410 96 I 2.6117 41692 .3828930.97 2 6379 42127 .3790830.98 2.6645 42561 .3753110.99 2.6912 42995 .371577

1.00 2.7183 0.43429 0.3678791

636

TABLE 5 Exponential Functions (Continued)

X e x Log10(ex; e 1 X e x Log,0(ex) e ~ z

1.00 2 7183 0.43429 0.367879 1.50 4.4817 0.65144 0.2231301.01 2 7456 43864 .364219 1.51 4 5267 .65578 .2209101.02 2.7732 44298 .360595 1.52 4 5722 .66013 .218712

, 1.03 2.8011 44732 .357007 1.53 4.6182 .66447 .2165361.04 2.8292 .45167 .353455 1.54 4.6646 .66881 .214381

1.05 2.8577 0.45601 0.349938 1.55 4 7115 0.67316 0.2122481.06 2.8864 .46035 .346456 1.56 4.7588 .67750 .2101361.07 2.9154 .46470 .343009 1.57 4 8066 .68184 .2080451.08 2 9447 .46904 .339596 1.58 4.8550 .68619 .2059751.09 2.9743 47338 .336216 1.59 4.9037 .69053 .203926

1.10 3.0042 0.47772 0.332871 1.60 4.9530 0.69487 0.2018971.11 3 0344 .48207 .329559 1.61 5.0028 .69921 .1998881.12 3.0649 .48641 .326280 1.62 5.0531 .70356 .1978991.13 3.0957 .49075 .323033 1.63 5.1039 .70790 .1959301.14 3.1268 .49510 .319819 1.64 5.1552 .71224 .193980

1.15 3.1582 0.49944 0.316637 1.65 5.2070 0.71659 0.1920501.16 3.1899 .50378 .313486 1.66 5.2593 .72093 .1901391.17 3.2220 .50812 .310367 1.67 5.3122 .72527 .1882471.18 3.2544 .51247 .307279 1.68 5.3656 .72961 .1863741.19 3.2871 .51681 .304221 1.69 5.4195 .73396 .184520

1.20 3.3201 0.52115 0.301194 1.70 5.4739 0.73830 0.1826841.21 3 3535 .52550 .298197 1.71 5.5290 .74264 .1808661.22 3.<872 .52984 .295230 1.72 5.5845 .74699 .1790661.23 3 4212 .53418 .292293 1.73 5.6407 .75133 .1772841.24 3.4556 .53853 .289384 1.74 5.6973 .75567 .175520

1.25 3 4903 0.54287 0.286505 1.75 5.7546 0.76002 0.1737741.26 3.5254 .54721 .283654 1.76 5.8124 .76436 .1720451.27 3.5609 .55155 .280832 1.77 5.8709 .76870 .1703331.28 3.5966 .55590 .278037 1.78 5.9299 .77304 .1686381.29 3.6328 .56024 .275271 1.79 5.9895 .77739 .166960

1.30 3.6693 0.56458 0.272532 1.80 6.0496 0.78173 0.1652991.31 3 7062 .56893 .269820 1.81 6.1104 .78607 .1636541.32 3.7434 57327 .267135 1.82 6.1719 .79042 .1620261 33 3.7810 .57761 .264477 1.83 6.2339 .79476 .1604141 34 3 8190 .58195 .261846 1.84 6.2965 .79910 .158817

1.35 3.8574 0.58630 0.259240 1.85 6 3598 0.80344 0 1572371.36 3.8962 .59064 256661 1.86 6 4237 .80779 .1556731.37 3.9354 .59498 254107 1.87 6.4883 .81213 .1541241.38 3 9749 .59933 .251579 1.88 6 5535 .81647 .1525901.39 4 0149 .60367 .249075 1.89 6.6194 .82082 .151072

1.40 4 0552 0 60801 0 246597 1.90 6.6859 0.82516 0.1495691.41 4 0960 61236 244143 1.91 6 7531 .82950 .1480801 42 4 1371 61670 241714 1 92 6.8210 .83385 .1466071.43 4.1787 .62104 .239309 1.93 6.8895 .83819 1451481 44 4.2207 62538 .236928 1.94 6.9588 .84253 .143704

1.45 4 2631 0 62973 0 234570 1.95 7.0287 0.84687 0.1422741 46 4 3060 .63407 232236 1 96 7.0993 .85122 .1408581 47 4.3492 .63841 .229925 1.97 7.1707 85556 .1394571.48 4 3929 64276 227638 1.98 7 2427 .85990 .1380691.49 4.4371 64710 .225373 1 99 7 3155 .86425 .136695

1.50 4 4817 0 65144 0 223130 2.00 7.3891 0 86859 0.135335

637


X e x Logi0(ez) e ~ z

2.00 7.3891 0.86859 0.1353352.01 7.4633 .87293 .1339892.02 7.5383 .87727 .1326552.03 7.6141 .88162 .1313362.04 7.6906 .88596 .130029

2.05 7.7679 0.89030 0.1287352.06 7.8460 .89465 .1274542.07 7.9248 .89899 .1261862.08 8.0045 .90333 .1249302.09 8.0849 .90768 .123687

2.10 8.1662 0.91202 0.1224562.11 8.2482 .91636 .1212382.12 8.3311 .92070 .1200322.13 8.4149 .92505 .1188372.14 8.4994 .92939 .117655

2.15 8.5849 0.93373 0.1164842.16 8.6711 .93808 .1153252.17 8.7583 .94242 .1141782.18 8.8463 .94676 .1130422.19 8.9352 .95110 .111917

2.20 9.0250 0.95545 0.1108032.21 9.1157 .95979 .1097012.22 9.2073 .96413 .1086092.23 9.2999 .96848 .1075282.24 9.3933 .97282 .106459

2.25 9.4877 0.97716 0.1053992.26 9.5831 .98151 .1043502.27 9.6794 .98585 .1033122.28 9.7767 .99019 .1022842.29 9.8749 .99453 .101266

2.30 9.9742 0.99888 0.1002592.31 10.074 1.00322 .0992612.32 10.176 1.00756 .0982742.33 10.278 1.01191 .0972962.34 10.381 1.01625 .096328

2.35 10.486 1.02059 0.0953692.36 10.591 1.02493 .0944202.37 10.697 1.02928 .0934812.38 10.805 1.03362 .0925512 39 10.913 1.03796 .091630

2.40 11.023 1.04231 0.0907182.41 11.134 1.04665 .0898152.42 11.246 1.05099 .0889222.43 11.359 1.05534 .0880372.44 11 473 1.05968 .087161

2.45 11.588 1.06402 0.0862942.46 11.705 1.06836 .0854352.47 11.822 1.07271 .0845852.48 11.941 1.07705 .0837432.49 12.061 1.08139 082910

2.50 12.182 1.08574 0.082085

X e* Log10(e*) e~*

2.50 12.182 1.08574 0.0820852.51 12.305 1.09008 .0812682.52 12.429 1.09442 .0804602.53 12 554 1.09877 .0796592.54 12.680 1.10311 .078866

2.55 12.807 1.10745 0.0780822.56 12 936 1.11179 .0773052.57 13.066 1.11614 .0765362.58 13.197 1.12048 .0757742.59 13.330 1.12482 .075020

2.60 13.464 1.12917 0.0742742.61 13.599 1.13351 .0735352.62 13.736 1.13785 .0728032.63 13.874 1.14219 .0720782.64 14.013 1.14654 .071361

2.65 14.154 1.15088 0.0706512.66 14.296 1.15522 .0699482.67 14.440 1.15957 .0692522.68 14.585 1.16391 .0685632.69 14.732 1.16825 .067881

2.70 14.880 1.17260 0.0672062.71 15.029 1.17694 .0665372.72 15.180 1.18128 .0658752.73 15.333 1.18562 .0652192.74 15.487 1.18997 .064570

2.75 15.643 1.19431 0.0639282.76 15.800 1.19865 .0632922.77 15.959 1.20300 .0626622.78 16.119 1.20734 .0620392.79 16.281 1.21168 .061421

2.80 16 445 1.21602 0.0608102.81 16.610 1.22037 .0602052.82 16 777 1.22471 .0596062.83 16.945 1 22905 .0590132.84 17.116 1 23340 .058426

2.85 17.288 1.23774 0.0578442 86 17.462 1.24208 .0572692.87 17.637 1.24643 .0566992.88 17.814 1.25077 .0561352.89 17.993 1.25511 .055576

2.90 18.174 1.25945 0.0550232.91 18.357 1.26380 .0544762.92 18.541 1.26814 .0539342.93 18.728 1 27248 .0533972.94 18.916 1.27683 .052866

2.95 19.106 1 28117 0.0523402.96 19.298 1.28551 .0518192 97 19.492 1.28985 .0513032.98 19.688 1.29420 .0507932.99 19.886 1.29854 .05Q287

3.00 20.086 1.30288 0.043787

638


X e* Logio(e*) e *

3.00 20 086 1.30288 0.0497873.01 20.287 1 30723 .0492923.02 20.491 1 31157 .0488013.03 20 697 1 31591 .0483163.04 20 905 1.32026 .047835

3.05 21.115 1 32460 0.0473593.06 21.328 1.32894 .0468883.07 21 542 1 33328 .0464213.08 21.758 1.33763 .0459593.09 21.977 1.34197 .045502

3.10 22.198 1.34631 0.0450493.11 22.421 1.35066 .0446013.12 22 646 1.35500 .0441573.13 22.874 1.35934 .0437183.14 23.104 1.36368 .043283

3.15 23.336 1.36803 0.0428523.16 23.571 1.37237 .0424263.17 23 807 1.37671 .0420043.18 24 047 1.38106 .0415863.19 24.288 1.38540 .041172

3.20 24.533 1.38974 0.0407623.21 24.779 1.39409 .0403573.22 25.028 1.39843 .0399553.23 25.280 1.40277 .0395573.24 25.534 1.40711 .039164

3.25 25.790 1.41146 0.0387743.26 26.050 1.41580 .0383883.27 26.311 1.42014 .0380063.28 26.576 1 42449 .0376283.29 26 843 1.42883 .037254

3.30 27.113 1.43317 0.0368833.31 27.385 1.43751 .0365163.32 27.660 1.44186 .0361533.33 27.938 1.44620 .0357933.34 28.219 1.45054 .035437

3.35 28.503 1.45489 0.0350843.36 28.789 1.45923 .0347353.37 29.079 1.46357 .0343903.38 29.371 1.46792 .0340473.39 29.666 1 47226 .033709

3.40 29.964 1.47660 0.0333733.41 30.265 1.48094 .0330413.42 30.569 1.48529 .0327123.43 30.877 1.48963 .0323873 44 31.187 1.49397 .032065

3.45 31.500 1.49832 0.0317463 46 31.817 1.50266 .0314303.47 32 137 1.50700 .0311173.48 32 460 1.51134 .0308073.49 32.786 1 51569 .030501

3.50 33.115 1.52003 0.030197

X e* Logio(e*) r *

3.50 33 115 1.52003 0.0301973.51 33.448 1.52437 .0298973 52 33.784 1.52872 .0295993.53 34 124 1.53306 .0293053 54 34.467 1.53740 029013

3.55 34.813 1 54175 0 0287253 56 35 163 1.54609 .0284393 57 35.517 1.55043 .0281563.58 35.874 1.55477 .0278763.59 36 234 1.55912 .027598

3.60 36.598 1.56346 0.0273243 61 36 966 1.56780 .0270523 62 37.338 1.57215 .0267833.63 37.713 1.57649 .0265163.64 38.092 1.58083 .026252

3.65 38 475 1.58517 0.0259913.66 38.861 1.58952 .0257333.67 39.252 1.59386 .0254763.68 39 646 1 59820 .0252233.69 40 045 1.60255 .024972

3.70 40.447 1.60689 0.0247243.71 40.854 1.61123 .0244783.72 41.264 1.61558 .0242343.73 41.679 1.61992 .0239933.74 42.098 1.62426 .023754

3.75 42.521 1.62860 0.0235183.76 42.948 1.63295 .0232843.77 43.380 1.63729 .0230523.78 43.816 1.64163 .0228233.79 44 256 1.64598 .022596

3.80 44.701 1.65032 0.0223713.81 45.150 1.65466 .0221483.82 45.604 1.65900 .0219283.83 46.063 1.66335 .0217103.84 46.525 1.66769 .021494

3.85 46.993 1.67203 0.0212803 86 47.465 1.67638 .0210683.87 47.942 1.68072 .0208583.88 48.424 1.68506 .0206513 89 48 911 1 68941 .020445

3.90 49.402 1.69375 0.0202423.91 49.899 1.69809 .0200413.92 50.400 1.70243 .0198413.93 50.907 1.70678 .0196443 94 51.419 1.71112 .019448

3.95 51.935 1.71546 0.0192553.96 52.457 1.71981 .0190633.97 52.985 1.72415 .0188733.98 53.517 1.72849 .0186863.99 54.055 1.73283 .018500

4.00 54.598 1.73718 0.018316

639


X eX Logi0(ex) r *

4.50 90.017 1.95433 0.0111094.51 90.922 1.95867 .0109984.52 91 836 1.96301 .0108894.53 92. 759 1.96735 .0107814.54 93 691 1.97170 .010673

4.55 94 632 1.97604 0.0105674.56 95 583 1.98038 .0104624.57 96. 544 1.98473 .0103584.58 97. 514 1.98907 .0102554.59 98. 494 1.99341 .010153

4.60 99. 484 1.99775 0.0100524 61 100.48 2 00210 .0099524 62 101 .49 2 .00644 .0098534.63 102 .51 2 .01078 .0097554 64 103 54 2 .01513 .009658

4.65 104 .58 2 .01947 0.0095624 66 105 64 2.02381 .0094664 67 106 .70 2.02816 .0093724.68 107 .77 2 03250 .0092794 69 108 .85 2 03684 .009187

4.70 109 .95 2 04118 0.0090954.71 111 .05 2 04553 .0090054.72 112 .17 2 .04987 .0089154 73 113 .30 2 05421 .0088264.74 114 43 2 05856 .008739

4.75 115.58 2 06290 0.0086524 76 116 75 2 06724 .0085664 77 117 .92 2 07158 .0084804.78 119 .10 2.07593 .0083964.79 120 .30 2 .08027 .008312

4.80 121 .51 2 08461 0.0082304.81 122 73 2 .08896 .0081484.82 123 .97 2 09330 .0080674.83 125 .21 2 09764 .0079874.84 126.47 2 10199 .007907

4.85 127 .74 2. 10633 0 0078284.86 129 .02 2. 11067 .0077504.87 130 .32 2. 11501 .0076734.88 131 .63 2 11936 .0075974.89 132 95 2 12370 007521

4.90 134 .29 2 12804 0.0074474 91 135 64 2. 13239 .0073724.92 137 .00 2 13673 .0072994 93 138 .38 2. 14107 .0072274.94 139 .77 2 14541 .007155

4.95 141 17 2 14976 0.0070834.96 142 59 2 15410 .0070134.97 144 .03 2 15844 .0069434.98 145 47 2. 16279 .0068744.99 146 .94 2. 16713 .006806

5.00 148 41 2 17147 0.006738

X es Logi0(ex) e~x

4.00 54.598 1.73718 0.0183164.01 55.147 1.74152 .0181334.02 55.701 1.74586 .0179534.03 56.261 1.75021 .0177744.04 56.826 1.75455 .017597

4.05 57.397 1.75889 0.0174224.06 57.974 1.76324 .0172494.07 58.557 1.76758 .0170774.08 59.145 1.77192 .0169074.09 59.740 1.77626 .016739

4.10 60.340 1.78061 0.0165734.11 60 947 1.78495 .0164084.12 61.559 1.78929 .0162454.13 62.178 1.79364 .0160834.14 62.803 1.79798 .015923

4.15 63.434 1.80232 0.0157644.16 64 072 1.80667 .0156084.17 64.715 1.81101 .0154524.18 65 366 1.81535 .0152994.19 66.023 1.81669 .015146

4.20 66.686 1.82404 0.0149964.21 67.357 1.82838 .0148464.22 68.033 1.83272 .0146994.23 68.717 1.83707 .0145524.24 69.408 1.84141 .014408

4.25 70.105 1.84575 0.0142644.26 70 810 1.85009 .0141224.27 71.522 1.85444 .0139824.28 72.240 1.85878 .0138434.29 72 966 1.86312 .013705

4.30 73.700 1.86747 0 0135694.31 74.440 1.87181 .0134344.32 75.189 1.87615 .0133004.33 75 944 1.88050 .0131684.34 76.708 1.88484 .013037

4.35 77.478 1.88918 0.0129074.36 78.257 1.89352 .0127784.37 79 044 1.89787 .0126514 38 79.838 1.90221 .0125254 39 80.640 1.90655 .012401

4.40 81 451 1.91090 0.0122774.41 82.269 1 91524 .0121554 42 83.096 1.91958 .0120344 43 83 931 1.92392 .0119144 44 84.775 1.92827 .011796

4.45 85.627 1 93261 0 0116794.46 86.488 1.93695 .0115624 47 87.357 1.94130 .0114474.48 88.235 1.94564 .0113334.49 89.121 1 9499.8 .011221

4.50 90.017 1.95433 0 011109

640


X e x Log,0(e*) e *

5.00 148.41 2.17147 0.0067385.01 149.90 2.17582 .0066715.02 151.41 2.18016 .0066055.03 152 93 2.18450 .0065395.04 154.47 2.18884 .006474

5.05 156 02 2.19319 0.0064095.06 157 59 2.19753 .0063465.07 159 17 2.20187 0062825.08 160 77 2.20622 0062205.09 162.39 2 21056 .006158

5.10 164.02 2.21490 0.0060975.11 165.67 2.21924 .0060365.12 167.34 2 22359 .0059765.13 169.02 2 22793 .0059175 14 170 72 2.23227 .005858

5.15 172.43 2 23662 0.0057995.16 174 16 2.24096 .0057425.17 175.91 2.24530 .0056855.18 177.68 2.24965 .0056285.19 179.47 2.25399 .005572

5.20 181.27 2.25833 0.0055175.21 183.09 2.26267 .0054625.22 184.93 2.26702 .0054075.23 186.79 2.27136 0053545.24 188.67 2.27570 .005300

5.25 190.57 2.28005 0.0052485.26 192.48 2.28439 .0051955.27 194.42 2.28873 .0051445.28 196.37 2.29307 .0050925.29 198.34 2.29742 .005042

5.30 200.34 2.30176 0.0049925.31 202.35 2 30610 .0049425.32 204.38 2.31045 .0048935 33 206.44 2.31479 .0048445.34 208.51 2.31913 .004796

5.35 210.61 2.32348 0 0047485.36 212.72 2.32782 .0047015.37 214 86 2.33216 .0046545.38 217.02 2.33650 .0046085.39 219 20 2.34085 .004562

5.40 221.41 2.34519 0.0045175.41 223.63 2 34953 .0044725 42 225.88 2.35388 .0044275.43 228.15 2.35822 .0043835 44 230.44 2.36256 .004339

5.45 232 76 2.36690 0.0042965.46 235 10 2.37125 .0042545.47 237 46 2.37559 .0042115.48 239.85 2.37993 .0041695.49 242 26 2.38428 .004128

5.50 244.69 2.38862 0.004087

X e x Log10(e*) e ~ x

5.50 244 69 2.38862 0.00408685.55 257.24 2.41033 .00388755.60 270.43 2.43205 .00369795.65 284.29 2 45376 .00351755.70 298.87 2.47548 .0033460

5.75 314.19 2 49719 0.00318285.80 330.30 2.51891 .00302765.85 347.23 2.54062 .00287995.90 365.04 2.56234 .00273945.95 383.75 2.58405 .0026058

6.00 403.43 2.60577 0.00247886 05 424.11 2.62748 .00235796.10 445.86 2 64920 .00224296.15 468.72 2.67091 .00213356.20 492.75 2 69263 .0020294

6.25 518.01 2.71434 0.00193056.30 544.57 2.73606 .00183636.35 572.49 2.75777 .00174676.40 601.85 2.77948 .00166166.45 632.70 2.80120 .0015805

6.50 665.14 2.82291 0.00150346.55 699.24 2 84463 .00143016.60 735.10 2.86634 .00136046.65 772.78 2.88806 .00129406.70 812.41 2.90977 .0012309

6.75 854.06 2.93149 0 00117096.80 897.85 2.95320 .00111386.85 943.88 2.97492 .00105956.90 992.27 2.99663 .00100786.95 1043.1 3.01835 .0009586

7.00 1096.6 3.04006 0.00091197.05 1152 9 3.06178 .00086747.10 1212.0 3.08349 .00082517.15 1274.1 3.10521 .00078497.20 1339.4 3.12692 .0007466

7.25 1408.1 3 14863 0.00071027.30 1480.3 3.17035 .00067557.35 1556.2 3.19206 .00064267.40 1636.0 3.21378 .00061137.45 1719.9 3.23549 .0005814

7.50 1808.0 3.25721 0 00055317.55 1900.7 3.27892 .00052617.60 1998.2 3.30064 .00050057.65 2100.6 3.32235 .00047607.70 2208.3 3.34407 .0004528

7.75 2321.6 3.36578 0.00043077.80 2440.6 3.38750 .00040977.85 2565.7 3.40921 .00038987.90 2697.3 3.43093 .00037077.95 2835.6 3.45264 .0003527

8.00 2981.0 3.47436 0.0003355

641


X e* Logi0(e*) e~*

8.00 2981.0 3.47436 0.00033558.05 3133.8 3.49607 .00031918.10 3294.5 3.51779 .00030358.15 3463.4 3.53950 .00028878.20 3641.0 3.56121 .0002747

8.25 3827.6 3.58293 0.00026138.30 4023.9 3 60464 .00024858.35 4230.2 3.62636 .00023648.40 4447.1 3.64807 .00022498.45 4675.1 3.66979 .0002139

8.50 4914.8 3.69150 0.00020358.55 5166.8 3.71322 .00019358.60 5431.7 3.73493 .00018418.65 5710.1 3.75665 .00017518.70 6002.9 3.77836 .0001666

8.75 6310.7 3.80008 0.00015858.80 6634.2 3.82179 .00015078.85 6974.4 3.84351 .00014348.90 7332.0 3.86522 .00013648.95 7707.9 3.88694 .0001297

9.00 8103.1 3.90865 0.00012349.05 8518.5 3.93037 .00011749.10 8955.3 3.95208 .00011179.15 9414.4 3.97379 .00010629.20 9897.1 3.99551 .0001010

9.25 10405 4.01722 0.00009619.30 10938 4.03894 .00009149.35 11499 4.06065 .00008709.40 12088 4.08237 .00008279.45 12708 4.10408 .0000787

9.50 13360 4.12580 0.00007499.55 14045 4.14751 .00007129.60 14765 4.16923 .00006779.65 15522 4.19094 .00006449.70 16318 4.21266 .0000613

9.75 17154 4.23437 0.00005839.80 18034 4.25609 .00005559.85 18958 4.27780 .00005279.90 19930 4.29952 .00005029.95 20952 4.32123 0.0000477

10.00 22026 4.34294 0.0000454

642

Answers To Selected Exercises

CHAPTER 1

Section 1, page 2

1. § (1 - 3~10) 3. 255/256 5. 3(z"+l - 3n+1) / x n(x - 3) 7. ( V r + r ) ( l - r * ) / ( l - r) 9. f 11. 2"9

13. 1/(1 + x2) 15. 9 ft17. 60 miles (one hour at 60 mph) 19. ^21. 43/99 23. -J- -f- ••• -f- l/2?i =

2 (1 + h + * • • + lA O --------* . 25. 1600 because

101 + “ * + 1600 “ ( lo l + " ' + 200) + (201 + " ■ + 400)

+ ( — + • • • + — ^ + ( — + • • • - + ---- — \\401 800/ \801 1600/

1 1 > 2 + * " + 2 = 2-

Section 2, page 8

1. C 3. D 5. C 7. C 9. C11. sn < 1 + 1 / (p — 1) for p = 2 .13. E 5 - i (aj + W = (Z 3 -i «i) + (E y -i W --------» A + B.15. Yj ^ converges, so an--------> 0. Therefore 0 < an < 1 for n > N, so

0 < an2 < an. Then an2 converges by comparison with

Section 3, page 7 7

1. C 3. D 5. C 7. C 9. C 11. C13. All real x. Given x, sfct an = x2n/n\. Then an+i/an = x2/ ( n + 1 ) --------> 0,

so the series converges by the ratio test.15. <C x <C ^

644 ANSWERS TO SELECTED EXERCISES

17. an < rn and r < 1; convergence by comparison with the geometric series. 19. Take an = 1 /n and bn = 1 /n2. There is no contradiction; the test in the

text requires bn/ a n--------> L > 0 ,

Section 4, page 15

I . C, not abs 3. D 5. C abs 7. C, not abs 9. C abs 11. D

13' ' - i + i r r s T i + I e b - s ^ 0'60651

15. $2n — $2n—2 + («2n-l + 02n) > $2n-2 and$2n = di + C(o2 + 03) + (a4 + a5) ' ++ («2n-2 + «2n-i) + ((kn) ] < since each term in parentheses is negative. Similarly s2n+i < s2n-i and s2n+1 > 0 .

Section 5, page 22

1. | 3. 1 5. tt/4 7. 2 9. i ln (3 + V IO ) I I . tt/4 13. 2/ s3 15. l/(« - a) 17. s/(«2 - 1)19. infinite 21. infinite 23. ln 2 ^ 0.6931525. ln 100 « 4.60517

Section 6, page 29

1. D 3. C 5. D 7. C 9. C 11. D

/oo r 00dx/x2 = t , A > / cta/z = 00. The word “paint” implies

a layer of uniform thickness. 17. s > 0 19. s > %

Section 7, page 34

j ;1. D 3. C 5. D 7. C 9. Z r i/™ 2< 1 + / dx/x2 = 2

11. about e1000 « 1.97 X 10434 13. C 15. D 17. The substitution x2 = u reduces it to Ex. 16.19. By the Mean Value Theorem,

„. . arc tan bx — arc tan ax bx — ax 1f{x) = -----------------------------= ----------- — — ,

X X 1 + 02

where ax < 9 < bx. H ence/(z) < (6 — o ) / ( l + a2x2), etc.

Section 8, page 39

1. C 3. D 5. D 7. C 9. C 11. D 13. C 15. C

r * rTT/2 r 7T /* ir/21. / = / = / + / = / z ln s in z

»/0 0 ^jr/2 0

n r/2 Ttt/2+ / + i 71*) In cos x = j 7r ln cos x

^0 y 0

rW2 /*t/2+ / x(ln sin z + ln cos x ). Now / ln cos x

*/o . /0

= —^7t In 2 by Example 9.1, andrw 2 r t /2/ x(ln sin x + ln cos z) = / z ln(J sin 2x)

./0 ^ 0

/•»/ 2 r »/2 r ir= / x ln sin 2x — (ln 2) / x = J / z ln sin z

J0 ./o «/0— Jir2 In 2. Hence / = — | 7r2 ln 2 + j / — ftf2 ln 2, f / = — §7r2 ln 2.

3. Set cos x — t. Then 1 = 1 In \ / l — <2 dtJ 0

= 7) f ln (l - P) dt = \ [ ln (l + <) dt + ^ [ ln (l — t) d< J 0 * J 0 J 0

i [2 i n i n 2= - In u du + - / ln w dw = - I ln u du = %(u\n u — u)

J 1 ^ ./0 ^ ./0 0

= ln 2 - 1.Z*1 1 Cl

nx(ln x)n_1(dx/x)

Chapter 1 645

Section 9, page 43

ri i r i5. By parts, I n = I (In x)n dx = x(ln x)n — /

0 0 ^ 0-T il

J .

= —n l n- 1, etc.

7. 1 = 1 sech xdx = 2 arc tan

9. Set

= 2( 7T — Jr) = J7T.0

.A# -j- 5 ^ (7x -j- Z)x4 + 1 x2 + x / 2* + 1 X2 - V 2^ + 1 ’

clear of fractions, and equate powers of x.

Section 7 0, poge 5 7

1. 100! = \/2007T(100/e) 10°. By 4-place tables, log(100!) ^ 158.0. Answer: 158. 3. In 2 5. In3

7. From a figure like 10.1, with y = l /V ^ ;


n — 1

X I f n dx _, v t ’ i, v i + ”■ ' 2V“ ~ 2 +

where ax < < a3 < • • • < 1 . Hence an--------» L,n

2 V i r 2V S - 7 ; - 2 + “• — , i _ 2 '1

CHAPTER 2

Section 2, page 55

1. (x — l ) 2 -f- 7(3 — 1) —|— 8 3. 2(3 -|- l )3 — (3 4- l ) 2 + 9(3 + 1) 5. 2(3 + 2)4 - 11(3 + 2)3 + 18(3 + 2)27. 5(3 + l ) 5 - 21(3 + l ) 4 + 31(3 + l ) 3 - 19(3 + l ) 2 + 5(3 + 1)9. x4 - 7xs + 5z2 + 3x - 6 11. 62.23 13. -157.2

Section 3, page 62

8 25 22* " 1 *• 2X - 3 ! * ’ + 5 !X‘ ' ' ' ' + <' 1) ’“ (2n - l ) ! x!" ‘’

x 3 3 4 /»*n3. x + x2 + - + - + . . . +

2! 3! (» - 1 ) ! ’

|rn(3)| < n y — xn+l if x > 0, \rn(x)\ < \x\n+l/n\ if x < 0(n + 1)!

5. (x - 1) + 3(3 - l ) 2/2! + 2(x - l ) 3/3! - 2(3 - l ) 4/4! +4(a: - l ) 5/5! + ••• + ( - l ) n2(x - 1 Y / n { n - 1) (n - 2); if n > 4: |rn(x) | < 2 (n — 2) \(x — 1 ) n+1/(w + 1)! for 3 > 1,

Ir»(a0i < 2(n — 2) !(1 — 3)n+1/ ( n + 1) * n_1 for 0 < 3 < 1 7. 32 — 3 3 + J 34 — ^ 35 + • • • + ( — l ) n3 n/ ( n — 2 ) !;

|rn(3) | < 3n+1/ ( n — 1)! for 3 > 0,\vn(3) j < [_x2 — 2 (w + 1)3 + n(n + 1)] |3 |n+1/ex(n + 1)! for 3 < 0

9. 32 - 34/ 3 ! + 3 6/ 5 ! + ••• + ( — l ) n_13 2n/ (2n - 1)!;M * ) | < [|*| + 2n + 2] |3|2-+2/(2 n + 2)!

11. 1 + 3 — \ x 2 — ^33 + 34/4! + 35/5! + • • • + <rnxn/ n \ , where0-4* = 04*+l = 1, (TAk+2 = CT4it+3 = — lj |^n(3)| < |3|n+1/ ( n + 1) !

13. 2x + 2aty5! + 2x9/9! + • • • + 2x4n+1/ (4 n + 1)!;|**4»+i(aj) | < (1 + cosh x) |x|4n+3/ (4n + 3)!

15. |error| < Tf ? < 0.024

Section 4, page 66

1. 1 - x2 + ix 4 - fx6 + x8/4! - x10/ 5 ! + xI2/ 6 !3. 1/(11 X 2n) < 5 X 10- 55. p4(x) = x2 — ^x4, |error| < 5 (2x)6/ 6 ! < 5 X 10-87. |sin x — p9(x)| < (7t / 2) u/ 11! < 3.6 X 10-69. sm(5x/8) « 1 - ( | 7t)2/ 2 ! + ( | 7r)4/ 4 ! - ( j r ) 6/ 6 ! » 0.92388,

| error | < 2 X 10-8

Section 5, page 68

1. Z r ( - l ) <- 132*-1x2* - V ( 2 t - 1)! 3. 22” ( — 1) <-122<_1x2’/ (2i) ! 5. Z o ( ~ l ) {2<x</i\ 7. Eo3*(x + f)*A-! 9. E o3 ‘x*

11. Zo ( - ! )< (* - 2)</2*«

Chapter 3

CHAPTER 3

Section 1, page 7 7

1. 1 / ( 4 - x ) , l 3 . l / ( l + x2) , l 5 . e*+l, a7 . 1 / ( 1 - 125x*), i 9 . e_(l_1)4, » 11 . 1 / ( 1 - «*), x < 0

13 . e_8in*, a ll x 15 . 1 / ( 1 - 2 y / x ) , 0 < x < \17 . 1 / ( 1 — a 2 — x2) , |x| < (1 — a2) 1/2 if |a| < 1, n o x if |a| > 1 19 . cos x 1/3, a ll x

Section 2, page 81

1 . 1 3 . 1 5 . 2 7 . 1 9 . 1 /e 11 . i 1 3 . 1 15 . 5 17 . 1 19 .

Section 3, page 89

1. E ? 5 » x ” 3 . Z o ^ / n l 5 . Z o x V ( 2 n ) !7 . - 1 + E ? ( » - l ) x n/w ! 9 . Z o x n+2

11. ( — 1) n+122n~ 1x2n/ ( 2 n ) !13 . 1 + 2x + 7X2 + 14x3 + 37x* + 74 x 6 + 175x«15. 1 + x2 + x 3 + f x 4 + V -x6 + f f x 6 17 . x 3 - | x 519. 8! 2 1 . 8 ! /4 ! = 1680

Section 4, page 95

1 . Z ? ( 3 /2 " + l - 4 / 3 n+1) x ”, |x | < 2 3. § Z ? ( l /2 n - 1 - 1 /Z”+1) x", |x| < 1

5. ZS>( - l ) " ( z SB + z3a+1), W < 1 7. (4 - 3 x ) /( l - x)2, \x\ < 1 9. —} ln ( l — x*), |x| < 1 11. (2 — x ) / ( l — x)2, \x\ < 1

Section 5, page 10]

1- Z o 5 ( - l ) n(rc+ l)(w + 2)x" 3. 1 + z ? ( n + l)4"x2» 5. 1 + ix3 + Z ” ( —1)"_1[3 -7 -11* • • (4n - 5)]x3"/2n-n! 7. + i ( * - 1) + E 2 ( - l ) n[ 3 -5 -7 - - - ( 2n - 3 )](x - l )" /4 n-n!} 9. 1 + §x2 + ix3 + fx4 11. \ / 3 ( 2;e + h*2 ~ H*3 ~ snre*4)

13. 1 - ix - fx2 + ff* 3 - txz4 15. 3(1 + s t) 1/4 ~ 3 + Tws ~ 2-6 X 3-6

3.00000 + 0.00926 - 0.00004 « 3.0092

Y (2n ) ! x2"+1 ‘ L ) 22"(n!)22n + 1

0

Section 6, page 104

1 . The alternating series 1 — | + J — • • • has 1/ n --------» 0 .3. ( i ) 5/5! < 4 X 10-6, hence eru* « 1 - * + J ( i )2 ~ M i) 3 +

^4 ( i )4 ^ 0.81874 5. 4, including the constant

Section 7, page 108

1. Z ? ( - l ) nz2n+1/(2 n + l ) ( 2 n + 1)1 3. £ £ ( - l ) ^ " + 2/(4 n + 2)5. 0.10003 7. 0.25049 9. k = t9t , length = (2tt) (41) (1 - S),

S « (J)2(t9t )2 + ( M ) 2( i) (A ) 4 « 0.01216; length « 254.5

Section 8, page 115

1. Let e > 0. Then there exists N such that ( i^ )n < e for all n > N since(To)n--------> 0 . Therefore | f n(x) — 0 | < (A )n < e for n > iV and all xsuch that 0 < x < so f n --------♦ 0 uniformly on [0, A ] . Supposef n --------» F uniformly on [0, 1]. Then F would be continuous on [0, 1]by Theorem 8.1 since each/n is continuous. But F(x) = 0 for 0 < x < 1 and F( 1) = 1, so F is not continuous.

3. By elementary calculus, 0 < xe~nx < 1/ne for 0 < x < oo. Therefore xe~nx--------> 0 uniformly on [0, °o).

5. | (sin nx)/n2\ < 1/n2; use the M-test.7. Let 0 < a. Then \e~nx sin nx | < e~na for a < x < oo. By the M-test, the

series converges uniformly on [a, <*>). By Theorem 8.1, the sum is continuous there. This is true for each a > 0, hence the sum is continuous for 0 < x < oo.

9. Both series converge uniformly by the M-test. Apply Theorem 8.4 (3).11. |an sin nx| < |an|. Apply the M-test with M n = \an\.

Chapter 4 649

CHAPTER 4

9. yes 11. no 13. (5, 2,4)15. (3, - 2 , 10) 17. (1 ,-1 6 ,9 )19. ( — 1, 10, 4) 21. ui + Vi = Vi + Ui, etc.23. (a + b)vi = avi + bvi, etc.25. The diagonals of a parallelogram bisect each other. But i ( v + w) is the

midpoint of the diagonal from 0 to v + w. Hence it is the midpoint of the diagonal from v to w. Alternate solution: The vector w — v is parallel to the segment from v to w, and has the same length. Hence the midpoint is i(w - v) + v = J(v + w).

27. \(u + v + w) = | [ i ( u + v )] + §w. Hence %(u + v + w) is on the segment joining w to the midpoint of uv (a median). Likewise it is on the other two medians, so all three medians intersect.

Section 2, page 129

1. 29 3 . - 1 5. y / n 7. 19. arc cos( — 12/25) 11.

13. arc cos( — l/- \ / l5 ) 15. -\/42 17. 5-\/219. \ y / 2 , 0, 21. y \/l4> — A \ / l4 :23. (1, 1, 0), (0, 2, 1) for instance25. |v w | = |v| • |w| |cos0| < | v| • |w|. Equality if v = 0, w = 0, or 6 = 0, tt.27. jv + w|2 — |v — w|2 = (v + w) • (v + w) — (v — w) • (v — w) =

(|v|2 + 2v*w + |w|2) — (|v|2 — 2v*w + |w|2) = 4v*w.29. u • v = |u| • |v| cos 6 = cos 6 , etc. When u = (cos a, sin a, 0) and

v = (cos 0 , sin p, 0) are plane vectors, the formula reduces to the addition law cos (a — fi) = cos a cos fi + sin a sin fi.

Section 3, page 132

1. x-n = i n = (J, - | , | )3. x-n = 3, n = ( - - f, i ± - * )5. x-n = «\/3, n = J \ /3 ( l , 1, 1)7. x = x0 + t(xi — x0) for instance9. The plane through x0 parallel to the given plane is x-n = x0*n. The

line x = tn, perpendicular to these parallel planes, meets them in pn and (x0-n)n respectively. Hence the distance between the planes is | pn — (x0-n)n| = | p — x0-n|.

11. They are parallel if and only if the direction of the line is perpendicular to the normal to the plane, that is, v-n = 0 .

13. Substitute x = x0 + £v into x-n = p and solve for t. The result is t = (p — x0-n )/(v -n ) . Now substitute this t into x = x0 + rv.

Section J, page 122

15. Let x = Xo + tube the required point. Then (x — y0) *u = 0, so t = (x x0) • u = (x - y0 + y0 - x0)*u = (y0 - x0)-u.

Therefore x = x0 + [(y 0 — x0) • u]u.

Section 4, page 140

1. ( - i f ) 3. ( - 1 ,1 ) 5. (1/19,7/19) 7. ( -8 /1 9 , 1/19, 7/19) 9. (0, 0, 0)

11. (6/28, —7/28, —3/28) 13. two parallel planes 15. The first and third planes are parallel. 17. (t, §t — |)19. (t, ~l ' t — —f t + %)21. Subtract a times the second equation from the third; then subtract a

times the first equation from the second. The results are (6 — a)y + (c — a) z = d2 and b(b — a)y + c(c — a)z = d3 . Subtract b times the first of these from the second: (c — b) (c — a)z = d3", etc.

Section 5, page 147

1. ( - 5 ,2 , -1 4 ) 3. ( - 4 ,8 , - 4 ) 5. (2, - 2 ,0 ) 7. (0 ,0 ,0) 9. ( - 7 ,8 ,2 ) 11. (0 ,-1 0 ,1 0 ) 13. ( 3 , - 1 ,4 )

15. A determinant changes sign when two rows are interchanged, hence u* (v X w) = — v* (u X w) = + v (w X u).

17. (av2)w3 — (av3)w2 = a(v2w3 — v3w2), etc.19. Express both sides in terms of components and compare.23. By the second formula, (a X b) X (a X c) = [(a X c) -a]b —

[(a X c) *b]a = - [ ( a X c)*b]a.

Section 6, page 155

1. 2 3. x = ( - 2, 1, 0) + t ( - 5 , 1, 1)5. x = (1, i, 1) + f(3, 1, - 5 )7. (3,1) 9. (1,4, - 7 ) 11. (6, - 5 , -1 9 )

13. s(5, 0, - 4 ) + *(0, 5, - 3 ) 15. x = 1 17. 2x — y — z = 3 19. x/a + y/b + z/c = 121. Up to sign, the determinant is the volume V of a parallelepiped with sides

a = (ai2 + Oo2 + a32) 1/2, b = (bi2 + b22 + b32) 112, c = (ci2 + c22 + c32) 1/2. The base area is A = be sin a < be. The height is h = a sin 7 , where y is the angle between e and the base. Hence V = Ah < abc.

23. F = (0 ,0 ,n ) , p = (E iP j ) / n25. (p + c) X F - (q + c) X F = (p - q ) X F

CHAPTER 5

Section 1, page 161

1. x = (e\ 2e2t, 3e30 3. (1, 3, 4) 5. v = (21, St2 + 4*3, 0), |v| = (412 + 9P + 2W> + 16*6) 1/2

Chapter 5 651

7. v = ( — Ai* sin cot, Aoj cos cat, B ) , |v| = (A2a>2 + B 2) 1129. d |x|2/d t = d(x ' x ) /d t = 2x*x = 0, hence |x|2 = const

11. (xi + yi)' - xi + i/i, etc. 13. — 2:32/2) ‘ = (*22/3 - x3y2) +(xii/3 — x3i/i), etc.

15. [x(<) — y(r0) ]* [x (0 — y(r0) ] is minimum at t = t,}. hence its derivative is 0 there: [x(<o) — y(To)]*x(tfo) = 0, etc.

Section 2, page 167

I. (ai2 + 022 + as2) 1'2 3. 2tt /2 5. / * (1 + a2n2x2n~2) dxJ xo

7. V"l7 — j I n ( \ / l7 — 4) 9. ( \ /2 /2 ) (1, —sin t, cos 011 . (ai2 + «22 + a32)_1/2(ai, 02, a3)

Section 3, page 173

1. 2V5/25 3. V l 9 / 7 \ / i 45. N-N = Jd(N-N)/ctt = 0; t -N + T-N = 0, t = /cN, hence N-T = - k .

Thus N = - k J + ON = —kT.7. (7t/2, 1) 9. k = |y" \ / ( • • • ) . But y" = 0 at an inflection.

11. da/dt = (da/ds) (ds/dt) = 3e2/ ( l + e 4)3/2 rad/sec13. & = (2/z3) / [ l + ( — l / z 2)2]3/2--------> 0 as 3 --------* 0 or x -------->00

Section 4, page 180

1. Use the solution of Example 4.3. The shell strikes the hill when y/x = tan 0 , that is, when t = (2 vQ/g) (sin a — tan 0 cos a).

3. v = (1, 21), a = (0, 2)5. v = T and a = d l /d s = /cN since ds/d£ = 1- Hence the tangential com

ponent of a is always 0 and the normal component is a itself. Since x = (x, sin x), T (ds/dx) = (1, cos x) and a (ds/dx) + T (d2s/dx2) = (0, —sin x). Hence, (ds/dx ) 2 = 1 + cos2 x, (ds/dx) (d2s/dx2) =— sin x cos x. Evaluate at x = 0: a = 0; at x = 7t / 2 : a = (0, — \ / 2/ 2).

7. v = ( — aco sin co£, aco cos co£, 6), a = — aco2(cos co£, sin co£, 0). Since these are perpendicular, the tangential component of a is 0; the normal component is a.

I I . 80007J-/24 ^ 1047 mph, 80007t(cos 40°) /24 ^ 802 mph, 0

Section 5, page 186

1. 3 3. 6tt2 5. I 7. 1 9. g 11. (§, 2, f )13. ( 0 , | , 0 )15. T ds = dx = (dx, dy). Rotate 7t/2 clockwise. Then T rotates into the

outward normal N0: N0 ds = (dy, —dx). The area of the small triangle formed by the two nearby radii and dx is Jx* N0 ds = \ ( x d y — y dx) .

17. The center is (ad, a) and x(0) = (moving point) — (center) =— a (sin 6 , cos 6 ); hence, the moving point is x = (ad, a) — a (sin 6 , cos 6 ).

19. 8a 23. (1 - 6—1, |( 1 - e~2), |( 1 - e~*))25. Since x and x are perpendicular unit vectors, x X x is a unit vector.

Section 6, page 192

1. (0, 1) 3. (V2, y / 2 ) 5. ( |V 3 . h) 7. {y /2 , 7tt/4}9. {2, 5tt/6} 11. iaC V 2 + ln (l + y/2)~\

fir 1613. 6a ( 8 sin2 36 + 1) 1/2 dd 15. 7ra2/4 17. 7ra2/2

J o19. 37ra2/2 21. a223. (262 + a2) (tt/2 — arc cos b/a) + 3by/a 2 — b2

CHAPTER 6

Section 2, page 204

1. Componentwise : xnj --------* a, and y nj ■* bj, henceXnj “t- Vnj * aj + &/•

3. Xn* Yn %niynl ~~Xn2?/n2 ~~I- Xn3?/n3 ' >X1J/1+ x2y 2 + x3y 3 = x* y.5. Xn \jnZ Xn3yn2 ------* x m X32/2, etc.7. Let (x„, y n) — * (*. y) and yn > x„2. Then y n — Xn2 > 0

Vn Xn y - X2, so y - x2 > 0, ;*/ > X2.

9. If (Xn, ?/n) -> (x, 2/) and x„2/ a 2 — 2/;n7&2 < 1, thenx2/a 2 — y2/b2 = lim(xn2/a 2 — y 2/b2) < 1.

11. If xn G R3 and xn--------» x, then x £ R3.13. Let {x0, 2/0} 6 S so 0 < x0 < 1. Set 8 = min{x0, 1 — xQ) > 0. Then

{x | |x — x0| < 8\ C S.15. Let x0 £ S n T. Since x0 £ S, there is 5i > 0 such that

{x I |x — x0| < di\ C S. Likewise there is 52 > 0 for T. Set 8 = min{5i, 82} > 0 . Then |x — x0| < 5 implies x 6 S and x £ T, hence x e s nT.

17. open, not closed 19. neither 21. closed, not open

Section 3, page 208

1. Let fi and / 2 be continuous at a. If e > 0, there exist 81 > 0 and 82 > 0 such that |x — a| < 8j implies | /y(x) — f j (a) | < |e. Set 5 = min{5i, 52}. Then |x — a | < 8 implies

|[ /i(x ) + / , ( x ) ] - [ / 1(a) + / , ( a ) ] |< l/i(x ) — /i(a ) | + | / 2(x) — / 2(a) | < §e + §e =

3. By the triangle inequality, ||x| — |a|j < |x — a|, etc.

Chapter 6 653

5. g ( x , y ) = x + y is continuous (sum of continuous functions), so = 1 /g(x>y) is continuous (quotient of continuous functions,

non-zero denominator). Now set xn = (1/n, 1/n ) . Then|xn+i — xn| --------> 0, but | /(x n+i) - /(x n) | = J —X > 0. Hence / is notuniformly continuous.

7. Let /(a ) > c. Set e = /(a ) — c > 0 . Then there exists 6 > 0 such that |x — a| < 8 implies | /(x ) — / (a ) | < e, in particular/(x) > / (a ) — e = c. Hence a is an interior point; the set is open.

9. A continuous function has a minimum on a bounded closed set.11. Every point on the surface of a sphere is closest to the center.13. Since/is continuous on a bounded closed set, the circle, / has a minimum

there. But f(x, y) = (x — 3y ) 2 + y2 > 0 on the circle, hence the minimum is positive.

15. 1, at 0 17. yes, composite of e* and t = —x2 — y2 19. x --------> (u, v) = (x, b) is continuous and (u, v ) --------> f(u, v) is con

tinuous. The composite is x --------» f(x, b).21. At each point of the set {(x, y ) | x ^ y ), the function / is continuous, a

quotient of continuous functions. Next, if x ^ y, the Mean Value Theorem says there is a z between x and y such that f(x, y ) = g'(z) . To p rove/is continuous at (a, a ) , let e > 0 . Choose 8 so \z — a\ < 8 implies \g'(z) — g'(a) | < e. If |(x, y) — (a, a)\ < 8 , then either (1) x = y and I fix, y) - f ia, a) I = \g'(x) - g'(a) | < e, or (2) x ^ y and f(x, y) = gf (z) where z is between x and y and hence \z — a\ < 8. ThenI f(x>y) — a) | = |g'(z) — g'(a)\ < e. Thus/ is continuous at (a, a).

Section 4, page 212

15. z = 2x2 + 3c2, y = c17. Let (xn, yn, zn) --------> (x, y, z) a n d /(x n, yn) = zn. Then

(xn, yn) --------> (x, y) (z D since D is closed, and zn z. ButZ n = f ( x n, y n) -------->f (x ,y ) by continuity. Hence f ( x , y ) = z , so(x, y, z) is also on the graph.

K Section 5, page 216

1. 1, 2 3. 3y, Sx 5. 4x / ( y + 1), - 2x2/ ( y + l )27. sin y, x cos y 9. 2 cos 2z, —3 sin 3y

11. 2y cos 2xy, 2x cos 2xy 13. l / y — y / x 2, —x/y2 + 1/x 15. ey, xey 17. yexy, xexy 19. 2e2x sin y, e2x cos y 21. 4x, 1 23. 32, 0, Sxh225. zx + Szy = 6(3z — y) + 3 [—2(3x — y)~\ = 027. zx2 — zv2 = (2x)2 — ( — 2 y )2 = 4(x2 — ?/2) = 4,2

Section 6, page 223

1. max = 4 at (0, 0) 3. min = 0 at (2, —3)5. min = 4 at (0, 0)

7. max = 2 \ /3 /9 at (\/3 /3> 0), neither max nor min at ( — -y/3/3, 0) 9. cube of side 2 ft 11. x = y = (2 F /3 )1'3, z = (9F /4 ) 1'3

CHAPTER 7

Section 1, page 226

I . Sx + 2y — 5z3. F(x) = p • x for some vector p t* 0. Hence F(x) = 0 defines the plane

through 0 perpendicular to p.5. F (x , y , z ) = xF( i) + yF( j) + z F (k ) . Given F(i) and F (j) the value

F (k } may still be any constant, so F is not determined.7. Let F and G be linear and suppose H = F + G. Then

H(au + 6v) = [F + (?](au + bv) = F{a u + bv) + G(a u + bv)= aF(u) + bF(v) + a(?(u) + 6G(v)= a[F (u) + (?(u)] + blF(y) + (?(v)]= alF + GQ(u) + b[_F + (?](v) = aH{ u) + bH(v) .

9. L(af + bg) = f [a/(«) + %(<)] dt = a J 0

= a L ( f ) + bL(g).I I . Expand the squares on the left side. The result is J^=i 1 a^xf, all

mixed terms canceling out. This is precisely the same as the right side.13. Let F(x) = P’X + d, where d 9 0. If a + b = 1, then

F(au + by) = p* (au + by) + d = ap*u + 6p* v + (a + b)d = a(p*u + d) + b( p*v + d) = aF( u) + bF(y).

15. I t is enough to show that G(x) = F(x) — F(0) is a homogeneou^linear function.G(au + by) = F(au + by) — F( 0)

= F(au + by + [1 — a — 6] 0) — F( 0)= aF(u) + bF(y) + (1 - a - b)F(0) - F(0)= a[_F(u) - F (0) ] + b[_F(y) - F (0)] = oG(u) + W7(v).

Section 2, page 231

1. According to the text, we may take c2 = |pi|2 + |p2|2 + IP3I2, where Pi, p2, P3 are the rows of A. Thus |pi|2 = an2 + a122 + ai32, etc.

3. F(bx + cy) = a X (bx + cy) = a X (bx) + a X (cy)= b( a X x) + c(a X y ) = bF(x) + cF( y).

5. F(x 1, 2/1, 2:1) + F (z2, y2, z2) = (xi + x2, yl + y2, 2)^ F(x 1 + j*2, 2/1 + y2) zi + z2)

"1 0 0"9.

"0 1 o"0 1 0 0 1 01 1 1 0 2 4

f f(t) dt + b f Jo Jo

git) dt

Chapter 7 655

11. Identity transformation; each vector is fixed13. Projection onto the xs, Xi-plane15. Each vector is transformed into its negative; reflection through 0.17. Reflection in the x2-plane19. Stretching by a factor of 2 in the Xi-direction, and by a factor of 3 in the

^-direction, followed by a reflection in the x1} x2-plane21. F(x) = F(xi + y\ + zk) = xF( i) + yF( j) + zF(k), again a vector in

the plane of / ' ’ ( j ) , ^ ( k ) .

23. Let <?(x) = a •Fix). Then G(bx + cy) = a-F(bx + cy) =a* [bF(x) + c F ( y ) ] = 6 [ a * F ( x ) ] + c [ a * F ( y ) ] = bG(x) + c(?(y) .

Section 3, page 238

1. 1

5.

3. 10

nL39J7.

~2~ ra 01 ”- 7 20 26~9. 3 11. 13.2 .9 0J — 1 3 2

2 - 2 4 5

- 6 - 6 - 6 a 1 a2 a315. 17.3 3 3 0 0 0

6 6 6 0 0 0

19.

23.

0 0 00 0 00 0 0

r° onLo oJ

21. x2 + 2 y2 + 3 z2

25. No. For example,a b 0" "o 0 0"

0 0 0 0 0 0 = 0.0 0 0 c d e

27. xA = (x, y, z)Cl\ (h ds bi b2 63

Ci c2 c3

= (aix + biy + ciz, a x + b2y + c2z, asx + b3y + c&z), hence

= A'x'.a ix + 6 iy + ci z a 1 bx Ci X

(xA)' = (hx + b2y + dz (h b2 C2 Va3x + b3y + c3z a3 bs C3_ z

29. xAy' = (xi, x2, x3)dn d\2 dis

(hi (hi CI2 3#31 a32 #33


&n2/i + #122/2 + &132/3 = (X\, X2, Xz) (hilJl + 022 /2 232/3

dZiyi + &322/2 + #332/3= £i(#n2/i + di2y2 + 0132/3) + £2(0212/1 + #222/2 + #232/3)

+ Xz(an y\ + #322/2 + ^332/3)

Xidijy j — yjdijXi — y idjiXj.i j i j

Similarly

31. (A + B ) r = [a*,- + b**y]/ = CaJ* "t“ ^ '0 = + [by/U — A' + Bf

Section 4, page 243

1. (98, —1, 11) ' 3. F 0 G has the matrix’5 1 O"

AB = 8 1 0O O O

#1 6l Ci X dx5.

02 b2 c2 y = d2

03 b3 c3_ z dz7. X = 26r + 11s -- 26£, y

d2Xz — dzX2

10 1 § $

1

Xi

9. a' X x' = dzXi — diXz = e10

X2

diX2 — O2X1 10e§1

1 Xz

11. a ' X (b' X x') = A (b ' X x') = A(Bx') = (AjB)x'. However,

AB =0 — o3 02

1O 1 O- CO b2

= o3 0 -#1 b3 0 - h

— fl2 #1 0 — bi bi 0

— d2)j2 — dzbz ajbi CS3&1

= d\b2

rCe11 a 3b%

d A

= — (d\bi + 0262 + dzbz)I +

#2 3 —#1 1 — #2 2

#A 02 1 O361

Ol?>2 #2 2 #3^2

d i b 3 d i b s #363

= - ( a ' - b ' ) i + b'a.

Chapter 7 657

13. By induction: suppose |Anx'| < cn |x'|. Then|An+1x'| = \A (A nx')\ = c \Anx'\ < c*cn |x'| = cn+1 |x'|, etc.00 m

15. / —- A n(bx' + cy') = lim / — A n(bx' + cy') nl t - l nl

n =0 n =0m m

" 1“ [b X ^ Jl”x' + C S »! ^*y']n = 0 n = 0m m

= b lim / — A nx' + c lim / —- A ny'm-+ oo . m-*oo * -4 ?! •

n =0 n = 000 00

= 6 / —7 A”x' + c / —, A"y'.Z_/ n! /_/ n!

21.

23.

25.27.

29.

31.

n =0

1. 4x2 — 2 xy + 3i/2

Section 5, page 252

3. x2 + 2y2 + 8?/£ + 3 y2

- K - 3 [ I i ]ab'

ab b2.

0 12 i a2 ab ac11. 13.12 1 12 ab b2 be

1_2 12 0 ac be c215. No 17. Yes 19. H A + .

Let /(x) = xAx' and g(x) = xBx' be positive definite. Then /(x) + g(x) > 0 + 0 = 0 with equality if and only if x' = 0. Since /(x ) + g(x) = x(A B)x', the matrix of the sum is A + B.Let B = H A + A'). Then B' = (*A + ±A'Y = ( W + (JA ') ' = \ A ' + \ A = A + A') = 5 .U # ) ' = (BAY = A 'fi' = AB.We seek the smallest positive k such that x2 + 2xy + 2?/2 — k(x2 + y2) = (1 — k)x2 + 2xy + (2 — k)y2 is not positive definite. I t is positive definite if 1 — k > 0, 2 — k > 0, and (1 — k) (2 — fc) — 1 > 0 . The zeros of (1 — k) (2 — k) — 1 = k2 — 3/c + 1 are k = J(3 ± \ / 5 ). The answer is fc = ^(3 — \ / 5 ) , which satisfies 0 < k < 1./(x ) is negative semi-definite if and only if — /(x ) is positive semi- definite if and only if —a > 0, — c > 0 , and ( — a) ( — c) — ( — b)2 > 0 if and only if a < 0 , c < 0, ac — b2 > 0 .There exists x0 ^ 0 such th a t/(x 0) = 0. By rotating coordinates, we may assume x0 = (0, 1). Then c = 0, so / = x(ax + 2by). If b = 0, then / = ax2 is not indefinite, hence b 5* 0 . The factors x, ax + 2by are not proportional, and remain that way when we rotate back.


Section 6, page 26 J

11. z = (x + l )2 + 2/2 — 1 ; paraboloid with axis x = — 1, y =

Chapter 7 659

17. 19.

21. 23.

25. f(x/z, y/z) = 0

1.

7.

27. f(y, y / & + ?) = 0

Section 7, page 268

1COCM11_______

s- L,lr* _ 1 iL 3 4J a2 Lo oj

, IP - 61a LO aJ 11. -- 6 - 3 - 2 3

2 0

3- 1

- 2

13. - -23 22 -30~

15- I”- 1 - 3 5~

- 1 1 - 1 0 12 - 4 - 7 10- 5 - 4 6

fJ- 3 - 9 10

17. Cf, 2, f j 19. C - f , - I H J 21. (AB) (B - ' A -1) = A(BB~l)A~l = AIA~1 = A 4 "1 = /, so AB has an

inverse and it is B~lA _123. A ' ( A - 1)' f _ (A--y.) / _ V = /, so A' has an inv

”o 1 o" CLl #2 o3 6 i 62 b 3"

25. PuA = 1 0 0 bi b2 bs = Oi (X2 O3

0 0 i Ci C2 cs Cl C2 c3


”l 0 c ai 02 a* ai + cdi d2 cdh a,z + cdz27. Rn(c)A = 0 1 0 b2 bs = b! b2 bz

0 0 1 di d2 dz d,l (k dz

29. We first apply to the rows of A the steps of the elimination method to change A to the form

bn b\2 biz

B = 0 b22 b2z = Q r Q r - V * ' Q i Q l A

0 bz2 bzz

Each Q in these steps is a Pij or Rij(c) by Exs. 25, 27. Also each of these steps either does not change \A\ or changes its sign only, hence \B\ = ± \A\ 0. Therefore b22 and b32 are not both 0. Further elimination steps result in

Cn Ci2 Ciz Cn 0 0 "

0 C22 c2z = Qs* • •QiA, then 0 c22 0

0 0 Czz 0 0 C33

Now multiply the first row by Cn-1, etc. By Ex. 26, these operations involve further^multiplications by Q’s, now of the form -Di(cn_1), etc. The final result is Qn • • • QiA = I.

31. Each row operation is applied simultaneously to A and to I :

' - [ . I a — c ■ ; ] — [ : n — i : n

The explicit sequence of steps is explained byA - 1 = R n ( S ) D 2( \ ) R l i ( - 2 ) D 1( - l ) P li .

33."1 1 1" "l 1 f "1 1 1"

1 - 1 2 0 - 2 1 --------> 0 1 —121 1 4 0 0 3 0 0 1

’ l 0 u>|co ^ ’ 1 0 0"

--------> 0 1 - i --------> 0 1 0 =0 0 1 0 0 1

Chapter 7 661

I =

1 0 0

0 1 0

0 0 1

1 0 0

1 1 0

1 0 1

■ 2

0

35.

A =2 2 6 1 3 - 21 2 - 9 - 1 2 - 91 3 - 2 2 2 6

~1 3 — 2--------»■ 0 1 - 1

0 - 4 10

/ =

0 o"12 0

0 13_

1 l”2 21 1■ 2 60 1

3 _

" l 3 -- 20 5 - 11

0 - 4 10

" l 0 f0 1 - 1

0 0 6

= A~l

1 0

0 1

0 0

"l 0 o’0 1 0>0 0 1

= I

"l 0 0~ ’o 0 f ’o 0 f0 1 0 0 1 0 0 1 10 0 1 1 0 0 1 0 - 2

0 0 1 - 3 - 3 4"-> 1 1 - 1 1 1 - 1

1 0 - 2 5 4 - 6

- 3 - 3 4~|

1 1 - f c J* A O ^¥ ¥ —Gj

- 2 3 - 2 2 3011 10 - 1 2

5 4 - 6

A~l

37. ( A + B y = [a,:, + fe y ]' = [ d y + fe y ] = [ d y ] + [ f e y ] = A' + B \

39. ( A B ) ' — [XI (Lijbik]' — dijbjk + Uijbjk] = [ S dijbjk~] + [ T . a,ijbjk~\ — A ' B + AB*

Section 8, page 278

1 . 2, aT 'O' T r 11a ; - 3 , 6 3. 3, a ; 2' 6 U_0_ . 1. .0.

1

L lJ

5. 7. 0 , a [ J ; 3 ,6

9. 1, « [ _ * ] ; 5 , 6 ^ ] 11. - 2 , a

13. 3,a[ j ;

17. A = (a - d)2 + 46c > 0 if 6c > 0.19. Equate coefficients in/(<) = (t — X) (t —

= t2 — (a + d)t + (ad — be).

21. [ 1 ‘ 1. - 2 2j(See 9.) 23• [ _ ! ; ] (See 7.)

T "o~ "o"ro n

25. 27. 3, a 0 ; i) 6 l ; 2, c 0.1 OJ

0 0 i

0 0 329. — 2, a 0 31. 0 , a 1 ; i,& 0% i 0 1

33. m = (t ■- X)(< - m) = *2 — (a + c)t + (ac — b2), hence(a — X) (a — n) = /(a ) = a2 — a(a + c) + (ac — b2) = —b2. Therefore (a — X) (a — /x) >< 0, so X < a < /x

35. Let A . By Theorem 8.5, there are real characteristic roots X

and /x. If X ^ /x, use Theorem 8.1. If X = /x, the discriminant of/(£) is 0, so A = (a — c) 2 + 462 = 0, a = c, 6 = 0. Hence A is already diagonal.

37. Equate coefficients of t2 in the expansion of f ( t) . Only the term (t — an )( t — 022) (t — a33) contributes to t2, and the contribution is- (an + a22 + a33) .

Chapter 8 \ 663 '

39. Try0 a b

0 0 c

0 0 0

CHAPTER 8

Section J, page 2 8 7

1. Take hx + ky = 0; | /(x ) /|x || = \xy2/\x\\ < |x|3/|x | = |x|2--------> 0.3. The rational function / is continuous where the denominator is non-zero,

that is, x 7 0. I t is also continuous at 0 because if x ^ 0, then

I flu'll ^ + y \ - lXl lXl (IXI + - - M|;W I - |xp - |x|- - 2|X |-

Hence / ( x ) --------> 0 = / ( 0) as x --------> 0.5. /(0) = fz(0) = /j,(0) = 0, so if / is differentiable at 0, then

/ ( x ) / |x |--------> 0 as x --------> 0. B u t/(x , x)/\ (x, x) | = ± J \ / 2 —^ —* 0.7. Let / = / i / 2, w here/,(c + x) = /*(c) + ki*x + ei(x) and

e<(x)/|x|--------> 0 as x --------> 0. Then / ( c + x) = /(c) + k*x + e(x),where k is constant and

e(x) = /i(c )e 2(x) + / 2(c)ei(x) + (ki*x)(k2*x)+ (ki* x)e2(x) + (k2*x)e!(x) + ei(x)e2(x).

I t follows, one term at a time, that e(x) / |x | --------* 0 as |x |--------> 0. Theonly questionable term is (k rx ) (k2*x). But |k,*x| < jk,|-|x|, so

|(k i-x )(k 2-x)| |ki| ♦ | k2[ • |x|2_____|x| - |x|

9- f /g = /• (1/sO ; use 7 and 8 .

Section 2, page 291 s

1. (9t2 + 2<)e‘2<3(+1) 3. 4<3cos(l/<) + i2sin(l/<) — 2t 5. 91*7. 4fe-<[(2 — <) sin 41 + 41 cos 41~\9. (s2 - 0 2(2s2 + t ) /2s3<2, - (s2 - «)2(2s2 + <)/4s2<3

11. (si4 + st2 + t) /z, (s + s~t + 2$H3)/z, where z2 = 1 + sH4 + (1 + st)213. 3.5x ft3/h r15. (d/du)F(u + v, u — v) + (d/dv)F(u + y, u — v) = (Fx + Fv) +

(Fx - Fy) = 2Fx17. 1 19. 2 2 1 . - 123. Apply d/dx to f(tx) = tnf(x) by the Chain Rule: tfx{tx) = tnfx(x).


Section 3, page 296

I. z = 0 3. 4x + 8y — 2 = 8 5. 4x — 13 y + 2 = —207. ( - 1 , - 1 , 1 ) /V 3 9. (0, - 3 , 1 ) /V I6 11. y = ± x , 2 = 0

Section 4, page 3 0 7

1. lines; grad 2 = (1, —2) 3. parabolas; grad 2 = (2x, 1)5. planes normal to (1, 1, 1)7. Consider xyz = 1, a typical level surface. Because of symmetry in the

coordinate axes, there are four sheets to the surface. The first octant sheet looks something like a paraboloid of revolution with axis (a, a, a ) , a > 0 , but it flattens out towards the coordinate planes as you move away from the origin.

9. (1, 1, 1), (2x, 8y, 182 ), {yz, zx, xy), (2x, 2y, - 2 2 )I I . — 2r_3(cos 30, sin 36)13. x — y = 0 15. 2x — 2 = 1 17. a

/ x \ 1 • / j* j* j* \ a (/u ) , , a(/w )19. d iv (/u ) = d iv (/u ,/y ,/w ) = — ------b — — H------—dx dy d2

( d f , , M , M , / d /

= ( £ M + ^ „ + l O + / ( ^ + ^\ax ay d z ) \ d x dy

df _ , f dw\ w + } 7 , )

dw\+ W

= (grad/) • u + /d iv u.21. curl(/u) = curl ( f u , f v , f u > )

= i ( f w ) y ~ ( f r ) , , ( /« ) * - ( / « ) . , ( M x ~ ( f u ) y)= (C f W y + fyW — }V2 — f zV~\, [ / « . + /,M— f w z — / XW],

[ f v x + fxV - f u y - f vu ] )- /• (wv — v „ uz — w x, vx - Uy)

+ ( fyW — fzV, f zU — f xw , f xV — f yu )= } curl u + (f x, f y , f z) X (u, v, w ) = / curl u + (grad/) X u.

23. By 21, cu rl[/(p )x ] = [g rad /(p )] X x + /( p ) curl x = [ ( / '( p ) /p ) x ] X x + 0 = 0.

25. curl(a X x) = curl(6e — cy, cx — az, ay — bx) = ((ay — bx)v —(cx — az)z, (bz — cy)z — (ay — bx)x, (cx — az)x — (bz — cy)y) =(2a, 26, 2c) = 2a

Section S, page 306

1. 1, 1, 1 3. 0, - 3 , 1 5. 1 7. 6

Section 6, page 309

1. —2, — i3. F = grad[f(x3 + j/3 + z3) + xt/z], K ®3 + 63 + c3) + a6c.

5. Take x = a cos 0, y = a sin 0, 0 < 0 < 2ir. Then

( ’ " S + X* ) / + * ■ ) - « ’/« ’ - 1. * > / - l d ) - 2 r .

7. (n — 2)_1a2_n

Section 7, page 315

1. (1 — sin 2/ ) / (sc cos y — 1) 3. y(exy — 3y) /x(6y — exy)5. (ex sin y + ey sin x ) / ( e y cos x — ex cos y) 7. — 2x3/9 2/59. l , i

11. -~~:2)/2 (z2 + y2 + z2) w/2 < xn + yn + zn < (x2 + y2 + z2) n/2

13. /'"[x, y , z(x, 2/) ] = 0. Compute d/dx by the Chain Rule: + Fzzx = 0, etc.

Section 8, page 318

1. — (z/x) dx — (z / y ) dy 3. (1/22) dx + (2/A) dy5 r2 = x2 _j_ y2 hence 2r dr = 2x dx + 2y dy. Also x dy — y dx =

(r cos 0) (dr sin 0 + r dd cos 0) — (r sin 0) (dr cos 6 — r dd sin 0) = r2(cos2 d + sin2 d) dd = r2 dd.

7. /*(£ dx + x dt) + fy(t dy + y dt) + fz(t dz + z dt) = ntn~lf dt + tn( fx dx + f y dy + f z dz) . Hence xfx(tx) + yfy(tx) + zfz(tx) = ntn~lf(x) and/*(Zx) = tn~lf x{x), fy(tx) = tn~lfy( x ) , f z(tx) = tn~lf z(x).

9. du = ( — sin 0 d0, cos 0 dd) = w dd, dw = ( —cos d dd, —sin d dd) =— u d0, dx = d(ru) = u dr + r du = u dr + r w dd.

11. Set 2 = (x 2 — 2/3) 1/2 so dz = J ( 2 x dx — 3y2 dy) ( x 2 — y*)~112. Setx = 6, y = 3, dx = —0.01, d?/ = 0 .02. Then dz = —0.11 so z + dz 3.00 — 0.11 = 2.89. (By 5-place tables: 2.8895.)

Chapter 9 665

CHAPTER 9

Section 1, page 328

1. 20x32/4 3. —2/y* 5. — sin(x + y) 7. —ex,v( x Jr y ) / y *9. mnxm~lyn~l 11. xy~l (1 + y ln x) 13. — 2(x + y ) / (x — 2/)3

15. 2 (2/ — x) / ( I + #2/)3 17. ( — 9x2 + 25xy — 18y2) / ( x — y ) z(x — 2y)z19. 2b 21. fxx = f ^ = g"(x + y) + h"(x - 2/)

Section 2, page 333

1. 18xy2, 18x2y 3. 8ys, 24xy25. —2y cos(xy) + xy2sin(xy), —2x cos(xy) + x2y sin(xy)7. 6xy(sin x) (2y + xy2 — x) + 2exy(cos x) (1 + X2/),

x2exy cos x + (x2y + 2x)exy sin x

9.11.13.17.

21.

23.

25.

(a:1,v/ x 2y i) l ( y ~ 1) (In x) + i f — 2y~\, (xllv/xyb) [2y2 + 4?/(ln x) + In2 x]8y(y2 — 5x2) / ( x 2 + y2)4, 8x(x2 — 5y2) / ( x 2 + y2)4f(x, y) = g(x') + xh(y) + k(y) 15. cubic polynomialsax + by + c 19. none

m(m — l ) x m~2ynzp mnxm~1yn~1zp

mnx"' 1 y n - l z p

mpxm~1ynzv~

n(n — 1 )xmyn~2zp

npxmyn~1zp~1

mpxm~lynzp~1

npxmyn~,zp~1

p(p — 1 )xmynzp~

w = x + 2y + dz

3.7.9.

11.13.15.17.

— sin it; — 2 sin w) — 3 sin w— 2 sin w — 4sinw — 6 sin w— 3 sin w — 6 sin w — 9 sin

10 27. g(x, y) + h(y, z) + k(z, x)

Section 3, page 338

Vi(x, y) = 1 + 2(x — 1) + 2 (y — 1) + (x - l ) 2 + 4(x — 1) (y — 1) + (y — l ) 2, pi(x, y) = 1 + 2 (x — 1) + 2 (y — 1), the linear part of Pi(x, y)Pi(x, y) = 0, p2(x, y) = xy 5. pi(x, y) = 1, p2(x, y) = 1 + (x - l ) y Pi(x, y) = p2(x, y) = - x - (y - ix )Vi(x, y) = (x - i ) + 2 (y - J),Pz(x, y) = pi(x, y) - | [ ( x - §)2 + 4(x - \ ) ( y - \ ) + M y - i ) 2]1.1200; exact: ( l . l ) 1-2^ 1.121173.9993; exact: 3.99929499It is the tangent plane at (a, b, f(a, b) ).Pi(x, y) = 1 + Jx + y at (0, 0 ). Also f xx = —1/4(1 + x + 2y )w , fxy = - l / 2 ( - - - ) 3/2, fvv = —! /( • • -)3/2- Hence | fxx\, \ f xy\, \ fm\ are bounded by

1 1 1 1(1 + x + 2y)m ~ (1 - 0.1 - 0 .2)3'2 (0.7)3/2

1(0.8)3 1

< (0.64)3'2

10.512

0.04.

< 2.

Take M 2 = 2 : \ri(x, y)\ < M 2 |(x, */)|2 < 2 (0 .02)

Section 4, page 344

min 3. neither 5. max 7. min 9. neither

Section 5, page 348

(1 ,1 ,1 ) 3. 1 , - 1P = ( E l xk/n, X i Vk/n), where P k = (xk, 2/*) 7- V 2 distance ^ 1.207 at x 0.7533, y 0.5674, z ^ 0.7527

Chapter 10 667

1. min 3. min 5. min 7. neither 9. min11. neither 13. max 0 15. min — f at ( — J, — 1, f)17. min 19. max 21. neither

Section 7, page 359

1. \ / l 3 , - V l 33. None; no level curve x — y = c is tangent to the hyperbola.5- b ~ i 7. square 9. height = \ / 2 (radius) 11. none

13. max 21_p/9, min 1 15. l \ / 2

Section 3, page 367

1. 8 abc/S\ /3 3. \ / l i , — y / l i 5. cube7. ^ 7, at (£, £) 9. max 31-p/g, min 1

11. Maximize xy + yz + zx subject to x2 + y2 + z2 = 1.13. max: f(27 + \ /3 ) ft3; min: f(27 — \ /3 ) ft3; sides: n, n, 6 — 2ju, where

m = 1(6=F \/3) ft15. The conditions for an extremum at x are 2xA — 2XxB = 0, xBx' = 1.

Hence x ^ 0 and x.l = XxB, that is, xAB~l = Xx. (Note that xAx' = X.)

Section 6, page 35 J

~2 0 o" "o 0 o"17. H, - \ H g = 0 2 0 - X 0 0 0

0 0 - 2 0 0 0is not positive definite.

CHAPTER 10

Section 2, page 376

1. ^ 3. i 5. ^ 7. 0 9. 0 11. | 13. 0 15. 0 17. tt/2 19. f ln 3 + 3 ln 2 -

Section 3, page 380

1. l n | 3. i 5. \ 7. 0 9. 1811. (3n+2 — 2n+3 + l ) / ( » + 1) (n + 2) 15. A = f f f(x, y) dx dy

Section 4, page 386

1. 16/3 3. 16/3 5. 46/3 7. 27/4 gm 9. 6.5 gm11. (7/15, 7/15) 13. (r - 1, (fx2 - x + l) /(§ * - 1)) 15. (0, 14/5)

Section 5, page 397

1. e — 1 3. 1 5. -£o 7. yV 9. §7r —11. - | V 3 13. f 15. | 17. | V 519. Fig. 5.17: 0 < x < 1, 0 < y < ex; Fig. 5.18: 0 < x < 1, 3? < y < x2;

Fig. 5.19: < x < §*\/3> 1 - (1 - x2) 1/2 < y < (1 - X2) 1'2;Fig. 5.20: 0 < y < 1, — y2 < x < y2; Fig. 5.21: J(1 — \ / 5 ) < z ^ V ^ 1 21. T -g 23. 25. 104

Section 6, page 407

1. 15tt/8 3. 15/16 5. 2tt 7. £tt|>3 - (a2 - 62)3'2]11. (fa , 0) 13. 3tt/640 15. 20/2717. 7t / 8j 7t/24, 7t/8 19. 0, 0, 0 by odd symmetry 21. 16/3

CHAPTER 11


1. 3. -£5 5. ■g'xir 7. /ca9/216 9. V"

r \/3 /2 / r V i ^

- V 3 / 2 \ J l - V l - * *

/ \/3 /2 / r V 1-X* \ f 1 / f v— arc sin x \

( / __ _f dy) dx 19. / ( / /d tfje fe- \ / 3 / 2 \ J l— \ / l — x2 ' J 0 arc sin x /

21. tetrahedron with vertices at (0, 0, 0), (0, 0, 1), (0, 1, 1), (1, 1, 1);

i' tr or' *) *] f [/.' if, p *) *] *"■

23. tetrahedron with vertices (0, 0, 0), (0, 0, 2), (0, 3, 2), (6, 3, 2);

f f f ( P “ p dx) del dyJ 0 LJ2J//3 w 0 / J

25. r r f 5 ( r x ^ = « •J4 Ly 10—x W n-x-y / j

Chapter 11 669

1. ( - V 2 , - y / 2 , - 3 ) , ( - 1 ,0 ,2 )3. By direct calculation, x2 + y2 + z2 = r2 + r202 + z2.5. dx = d(r cos 0, r sin 0, z)

= (dr cos 6 — r dO sin 0, dr sin 0 + r d0 cos 0, dz)= dr(cos 0, sin 0, 0) + r d0( —sin 0, cos 0, 0) + dz(0, 0, 1).

7. d/ = (grad /) • dx, hence / r dr + f e dd + /* dz = (grad /) • u dr +(grad / ) • w r dd + (grad / ) • k dz. Therefore (grad / ) • u = fr, etc. by equating coefficients of dr, dd, dz, and the formula follows.

9. aAh2/ 16 11. 2a5/15 13. 7ra12/60

Section 3, page 436

1* ( i , \ y / 2 ) , ( § \ / 2> § \ / 2, 0)> (iv^3> 2> 1)3. [p, 7r — <t>, 0 + 7r], [3p, 0 , 0] 5. ^7ra sec a7. d \ = d(sin </> cos 0, sin </> sin 0, cos 0 ) =

d0 (cos 0 cos 0, cos <t> sin 0, —sin </>) + d0 ( — sin <t> sin 0, sin <£ cos 0, 0 ) = d<t> |x + sin </> d0 v, etc.

9. In spherical coordinates, [p, <£, 0] = p[l, </>, 0], hence /[p, <£, 0] =pn/ [ i , <t>, 0 ] = p " ^ , o)

11. 8w/(n + 1) (n + 2) (n + 3) 13. 2ir(6 - a) 15. 2 ■nah 17. 2wtAai19. Center the sphere at the origin and take its apex at (a csc a, 0, 0). Treat

like a volume of revolution problem, using polar coordinates in the x, y-plane:

7ra3 ( l — sin a )2

Section 2, page 426

/ ( t /2 ) - a / r a c s c ( e + a ) \ ,

y j (2ttr sin 0)r drj dd =-3 sin a

Check: if a --------» tc / 2 , then V --------» 0 ; if a --------> 0, thenV | 7ra3 csc a = i (?ra2) (a csc a ) , the volume of a cone of radius a and height a csc a.

23. Take the axis of the solid cylinder along the z-axis and bore along the ?/-axis. Section by planes perpendicular to the x-axis; the cross-section is a rectangle of dimensions 2\ / a 2 — x2, 2\ / b 2 — x2. This yields the first integral. Substitute x = a cos 0 for the second.

25. 2a2s arc sin(s/2a) + Js2-\/4a2 — s2s 2 f c + s r ______ 2 / x\

27. V - - J [ V x * - V + - a r c csc ( - j dx, where b = s/2 tan a

Section 4, page 447

1. (fa, fa, fa) 3. (0, 0, f a ( l + cos a))5. For the wedge 0 < r < a and — a < d < a, x = (f (sin a )o r la, 0). 7. For the wire r = a and 0 < 0 < t /2, x = (2a/ir, 2a/ it).9. Let the wire be x(0) = 100(cos 0, sin 0), where 0 < 0 < ir and


8(6) = 0.01 + O.240/7T. Then x = —4800/137r2 ^ —37.4 cm, y = 200/tt 63.7 cm

11. V = 2w2Aa2 for a circle of radius a revolved about an axis in its plane at distance A from the center of the circle. (See Ex. 17, p. 437)

13. Data is in Ex. 11, area = 47r2Aa. 15. (0, 0, Ja( 1 + cos a))19. ^(a + b + c + d)

Section 5, page 4 55

1. products: 0, I xx = Iyy = M(Sa2 + 4/i2)/12 gm-cm2, I sz = \ M a 2 gm-cm2 3. products: 0, Ixx = Iyy = I zz = fM a2 5. products: 0, I xx = \ M (b2 + c2), etc.7. M = 2w2Aa28. Let u = x — A. Then by slicing into cylindrical shells,

Izz = 8 (A + u)2[2 r (A + u) (2V «2 - u2) du] => \ M ( 4 A 2 + 3a2).J —a

9. I zz = \Mha 11. \M h(a + b)13. fM a2(10 - r ) / ( 6 - tr)15. v = o) X x, hence

x X v = x X ( c o X x ) = |x|2 co — (x*(o)x== (x2 “j- y2 “I- z2) (ccx, coy, o)z) (xccx “I- yojy zo)z) (x, 2/, z)

•)= (( |/2 + z2)cox - xyo)y - xza>2, y2 + z2 - x y

(w*, coy, wz) - 2/x —zx

Z 2 + X 2- z y

— xz

- y z

x2 + y2

Multiply by 8 and integrate.17. I xx = Iyy = \M h2 gm-cm2, I zz = 019. Ixx = Iyy = I zl = \M a 2 21. \ M ( 2 A 2 + 3a2)

CHAPTER 12

Section 2, Page 463

1. Si and s2 are constant on each rectangle of the common refinement; so is si + s2.

3. |si| is constant wherever si is.5. Take a partition with respect to which both Si and s2 are step functions.

Say Si(x, y) = Bjk and s2(x, y) = Cjk on the interior of I/*. Then

J J (S i + S2) = ^ (Bjk + Cjk) 11 /As |

= ^ Bjk |l/*| + ^ Cjk 11 /Ac | = j j S i + J J S2.

Chapter 12 671

Each B jk > 0, hence Bjk |lyjb| > 0.Given s, choose a defining partition for s that is finer than the given fixed partition of I. Then each I# is partitioned into rectangles Iy*,rg inside each of which s is constant, say s = Bjk,™- Hence

Given e > 0, choose s so | / — s\ < eh where ei = e /S \ \ \ . Then s — ei < f < s + ei, the functions s ± 6i are step functions, and JJ(s + ei) —

Suppose s < f < S. Let J be any rectangle on which s is constant. Since J contains points with irrational coordinates, s < 0 on J . Similarly, S > 1 on each rectangle where S is constant. Hence JJS — JJs > J J l — JJ0 = |l|, so JJ*S — JJs < e cannot be satisfied if e < |l|.From Ex. 4, Jff(y, x) = JJ/(x, y), hence 2JJ /O , y) = 0 .Assume/ (p ) > 0 . Given e > 0 , take a partition ly* so fine that each ly* has area less than e //(p ). Suppose p £ lrs. Define s(x) = 0; define aS(x) = / ( p) on lrs and S(x) = 0 otherwise. Then s < f < S and

Thus / is integrable and 0 < J J | / < e. Hence J J i / = 0 . A similar argument holds if /(p ) < 0 .Suppose / ( p) > 0 for some p in I. By continuity there exists a small square S centered at p on which/(x) > J/(p). Then JJi / > hf(P) *1 1 > 0 , a contradiction.

First step: the segment from x0 to Xi obviously can be enclosed in the rectangle with x0 and Xi as opposite vertices, sides parallel to the axes. Its area is \(x\ — x0) (yi — 2/0) | < i |xx — x0|2. Second step: divide the segment into n equal parts. The division points are Zy = x0 + ( j / n) (xi — x0) . By the first step, the segment from Zy to Zy_i is enclosed in a rectangle of area < J |zy — Zy_i| = |xi — x0|2/n 2. The whole segment is covered by these n rectangles; their total area is at most |xx — x0|2/n. This is anupper bound for the area of the segment. Let n --------» 00.The triangle with vertices 0, x2 — Xi, x3 — Xi is a translate of the given

Section 3, page 471

JJ(S — *i) = 2 JJei = fe < €.

j j (S — s) = I I S < /(p )[« //(p )] = e.

Section 4, page 482

Say | / | < M on I. If g(x) is defined, then

\f(x, y ) \ d y < M ( d - c ) .

triangle, so it has the same area. But by subtracting rows,xi yi 1 xi yi 1

1 1- x2 y* 1 = - x2 - xi 2/2 - 2/i 0

x3 2/3 1 & — xi 2/3 - 2/i 0

1

= - \ ( x 2 - X X) ( 2/3 - 2/ 1) - (^3 - X i ) ( 2/2 - 2 /i ) I ,

which is its area by Ex. 4.7 . Because <7 and h are continuous, the following three sets are closed:

On Si, /*(x, 2/) = /[z> <7(z) ] is continuous because it is a composite function of continuous functions. Similarly, /* is continuous on S3. Also /* is continuous on S2 because it equals/, a continuous function. The assertion follows directly from the following lemma. Lemma: If g is continuous on closed sets S and T, then g is continuous on S U T. For let x0 G S U T. To prove g is continuous at x0. Case 1: x0 6 S fl T. Then if e > 0, there is <$1 > 0 such that |gr(x) — g(*o)\ < e for all x such that x £ S and |x — x0| < 8h and there is 52 > 0 such that |gr(x) — gf(x0) | < e for all x such that x 6 T and |x — x0| < 52. Choose 8 = min{5i, 52}, etc. Case 2: x0 is in one set, not in the other. Say x0 £ S and x0 $ T. Since T is closed, there is 8 > 0 such that x £ T whenever |x — x0| < 8. The continuity of g at x0 in S U T now follows from the continuity of g at x0 in S.

1. 3; inverse: u = \x — \ y , v = — \ x + j y.3. 2(v — u ) ; inverse: u = \ x — i \ / 2 y — x2, v = %x + \ y / 2 y — x2 on the

domain y > \x 2.5. u2v ; inverse: u = x, v = y / x } w = z / y on x > 0, y > 0, z arbitrary.7 . - l / ( x 2 + 2/2)2- 9- The cube 0 0 .

Section 6, page 501

1. F = (1 - e"0 A, F ' = - 1A2 + <r*/t23. F = [(1 + 0 n+1 - n+1] / ( n + 1), F' = (1 + 0 W “5. Define

51 = {(x, y) I c < y < g(x), a < x < b}

52 = {(x, y) | gO) < 2/ < a < x < b}

53 = {(x, y) | h(x) < y < d, a < x < b } .

Section 5, page 489

G(u, v, t) = f(x, t) dx.

Chapter 12 673

By the Chain Rule,

^ G\_g(t)y h(t) , t] = Gug + Gvh + Gt

= - fZg( t ) , Qg + /[M O , t]h + I f t(x, t) dx.J g

'■ f ( f a dz'j dd = 2irah

9. x = ( (A + a cos a) cos 6, (A + a cos a) sin d, a sin a), 0 < d, a < 2ir

J J a(A + a cos a) dd da = 4:ir2Aa

11. 27ra / (c2 sin2 <t> + a2 cos2 0 ) 1/2 sin <£ d</>/ o

13. J J (1 + a2 + 62) 1/2 dx ^ = (1 + a2 + &*)»«*|D|

“ • f ( t e ) * - *17. Set A = J J N dA. It is enough to show that A*i = A*j = A*k = 0.

For example, A*k = / / N-k dA = / / cos y dA = J J dx dy. Each

point of the projection of the closed surface on the x, ?/-plane is covered an even number of times, half positively, half negatively, hence

If dx dy = 0 .

19. A = 7r = 7r[( v T + a 2)2 — a2]. For any oval, k ds = da, where a is the

angle between the tangent and the positive x-axis. Hence J k ds = 2t t .

Thus the area swept out by the unit tangent is r, regardless of the oval.

21. / P dx + Q dy = J J (Qx — P y) dx dyd D

- I IE ((fry) x (0*)i/] dx dy — 0 .

Alternate solution, not assuming second derivatives exist:

/ P dx + Q dy = / 4>x dx + <t>y dy = / d</> = <t>(*i) — 4>(x0). Xo 0 0

This is 0 if x0 = Xi (closed curve).23. 12

25. / ( — uvy + vu*) dx + (uvx — vux) dydD


- J JD

- / /

[(t^X “ ( — UVy + VUy)y] (fa dy

\u{vxx + Vyy) — v[uxx + Uyy)]dx dy.D

The other terms cancel each other. The proof requires u and v to have continuous second partials on D.

27. Let (a , b) be the lower left-hand corner of I and let (x, y) be any point of I. These points are opposite vertices of a sub-rectangle J , and

j P dx + Q dy = j j (Qx — P y) dx dy = 0 .dJ J

This implies

f P (s ,b ) d s + I Q(x, t) d t - / P(s, y) ds — / Q(a, t) dt = 0.J a J b J a J b

Hence we can define

f(x, y) = I P(s, b) ds + I Q(x, t) dt J a J b

= I P(s, y) ds + [ Q(a, J a J b

t) dt.

By the first form of /, we have df/dy = Q(x, y ) . By the second, df/dx = P ( x , y).

Section 7, page 510

f f f 2ir f a v dv1. J I dx d y / ( l + r2) = J dd J - = n ln (l + a2) --------♦ «

r <a ^as a --------> oo.

/ / • « / / - / / <

Chapter 12 675

Hence {J J Dn g\ is a bounded increasing sequence (since g > 0 ), so it converges.Let D be the domain 0 < y < 1/x, 1 < x < b. Then 1 + x2y2 < 2 on D, so

IfD

dxdy 1 1> - IDI = - ln b --------> oo.

1 + x2y2 2 2

Alternate solution:

f a f a dx f a arc tan u . f a arc tan u t duJ , dyJ, T + & - J , — ~ d u > j , — ^ V

= - ln a --------> oo .2

J J (ln r) dx dy = 2tt j ' r(ln r) dr

6”r_1= (27r) (— \ — je2 In e + ------> -Jttas e--------> 0.

J J J ps dx dy dz = (2t ) (2) J ps+2 dp, etc.€<p<1 6 Let 0 < a < 1. Then

J J J (1 — p)8 dxdydz = 47r J p2( 1 — p)s dpp < a °

= 4ir I p*( 1 — p)2 dpJ 1-0

= 47r f l ~ (1 ~ «),+1 _ 2 1 ~ (1 ~ a) ,+2 , 1 - (1 - a)*+3l _ 7 r L s + l s + 2 s + 3 J

r 1 2 1 i--------» 4tt-------------------- 1-------- .Ls + 1 s + 2 s -j- 3j

Drop the 4 points (± 1 , ± 1 ) and drop all terms (m, 0), (0, n) ; that part converges by the integral test. For each remaining point (m, n), let S m,n be the square of sides 1 for which (m, n) is the farthest point from the origin. Then these squares do not overlap and on S m,n,(m2 + n2)~p < (x2 + y2)~p, hence

+ ni) v ~ j j dx dy(m2 + n2) v “ J J (x2 + y2) p '

Sm, ,n

Now sum.

Section 8, page 516

1. 1.19463. i%h 2 BiPi) (ffc CiQi) = \hk £ BiCjPiqj = %hk £ A n f n 5. Both expressions are A + %B + %C + \ D 7. Integrate by parts twice:

[ 2/(1 - y)fn(.x, y) dy = f (2y - 1 )fv(x, y) dy J 0 j 0

= fix, 1) + f(x, 0) - 2 f f ix, y) dy.Jo

Now integrate on x:

JJ 2/(1 ~ y ) fm ix ,y ) dxdy = J [ f i x , 1) + f ix , 0 ) ]d x

: / / / < * , y) dx dy.

Integrate by parts twice:

[ x i l — * ) [ /„ (* , 1) + fxxix, 0 ) ] dxJ 0

= f {2x - 1 )[/*(x, 1) + fxix, 0)] dx = - 2 f [ f ix , 1) + fix, 0)] dx. Jo Jo

The last step uses the hypothesis /(0 , 0) = / ( 0, 1) = /(1 ,0 ) =/ ( I , 1) = 0 .

9. Define p(x, y) = A + Bx + Cy + Dxy by A = /(0 , 0), A + B = / ( l , 0), A + C = f(0, 1 ) , A + B + C + D = / ( l , l).T hen g(x, y) = f(x> y) — P(x> y) satisfies the hypotheses of Ex. 8, etc.

11. m u : y> z) dz'j dy^ dx

2 m 2 n 2 phjk V V V27 I I I A rBsCtF ( x r, y$) %t) >

r =0 s =0 t =0

b — a . d — c f — e h ~ ^ i J ^ i k ~ ,2m 2 n 2p

(Ao, A h A 2, • • •) = (1, 4, 2, 4, 2, • • •)> etc.

13. J J J f t J t ( J ) 3[64 s in | + 3(16) sin \ + 3(4) sin \ + sin 1] ft! 0.122

Chapter 13 677

CHAPTER 13

1. y = ix4 3. y = 1 + 3 ln x 5. y = 6 — e~x 7. y = 10ex_1 9. ?/ = Jx2 — cos x + 1 11. y = (x2 + 1)1/2 — 1

13. i/ = x + %x3 + ix4 + 15. y = x + c/x17. ln(x2 + y2) = 2 x + c

Section 2, page 526

1. y2 = %x* + c 3. y = ( V ^ + c)2 5._ y = ln[2/(c - z2)]7. y = x [l + 2 / (c — ln a )] 9. y = 3/%/®

11. y = (4 — x) / (2 — z) 13. (Slope at P) = — (slope O P ); xy = c;rectangular hyperbolas with axes y = ± x

15. (Slope at P) = (slope O P ); y = cx; straight lines through (0, 0)

Section 3, page 530

1. linear 3. linear 5. non-linear7. y = \x 3 + c; yes, y = \x? is a particular solution, and y — c is the

general solution of y ' = 0 .

Section 4, page 532

1. y = c/x3 3. y = c cos0 5. y = ce~x/x 7. y = 3(1 - x2) 1/2 9. y = fx~2

Section 5, page 539

1* V = \ 3. y = 5. y = \ex(2x2 — 2z + 1)7. ^V(15 sin x — 6 cos x) 9. y = Je_x(sin z + cos x)

11. i — E(R cos t + L sin t) / (R2 + L2) 13. i/ = - (r5 + 5X4 + 20x3 + 60x2 + 120z + 120)15. y = xex 17. y = z3 — x 19. ?/ = J(x2 + 1) ln(z2 + 1)21. y = ce~3x + 23. y = ce-x + \xe2x — \ c2x + 1 25. i = ce~RtlL + E(R cos cot + Leo sin a)t)/ (R2 + L2a>2) 27. ?/ = C6_x2 + 1 29. y = ce~3x + xe~3x33. The slope is constant c along each line x + y = c. But each such line

has slope — 1, so c = — 1 gives the solution x + y = — 1.

Section 6, page 548

1. y = cx3 3. 100(log 10)/(log 2) ^ 332 years5. I I3 X 102 ^ 133000 7. about 32.1 sec9. 100(2 + V 3 ) ~ 373.2°C 13. 8 0 0 0 / ~ 997 sec

Section 7, page 520

Section 7, page 554

1. + + 3. 2 + * + x2 + i x 3 + ix 4 + **» +5. 1 - ix 2 + Ix4 — iVx6 + iv *? - r h x 10 + m ^ 1!

7. ?x2 + -^x7 + a:12 + x17 +16-56 34*(56)2 22-(56)3

2 19. ix 3 + ^g-x7 H--------- x11 H--------------x15

33*63 15-(63)2

11. £x2 + fx4 13. 2 + * + x2 + **» + Ix4 + -^x5 15. 1 - §x2 + Jx4 17. |x 2 19. 1 + |x» 21. - 1 + x + x2 - §x3 - ^ x 4 23. 025. 10 + M t - 10)3 - M t - 10)4, x(12) f t 10.1 ft

CHAPTER 14

Section 2, page 561

1. x = ael + be5t 3. r = a cos 20 + b sin 265. y = ex/4[a cos(x \/7 /4 ) + b s in (x \/7 /4 )] 7. x = a + be-6*9. y = ae_x + be_4x 11. x = e2a2t(bt + c) 13. y = a sin 3x

15. x = e2‘(a£ + 1) 17. r = ee( — e-r cos 0 + 6 sin 0)19. r = a(cos 0 — sin 0)

Section 3, page 564

1. x = \ t 3. x = - i 5. x = - t 2 - l O t - 3 8 7. x = ye3* 9. y = \ e r 2x — x — 1

11. y = 5%( — 2 cos 2x — 7 sin 2x) 13. y = Jex(cos x + sin x)g5ic

15. y = -------- + + --------+ --------- + --------* P + i 22 + 1 32 + 1 42 + 1 52 + 1

17. x = a cos £ + 6 sin t + t2 — 219. x = ae2* + 6e_3t + — 6021. x = a + 6e-3* + rfr( —2 cosh 2£ + 3 sinh 2£)23. i = e_2‘/3(a cos t y / 2 / 3 + 6 sin t y /2 /3 ) + t t ( 4 sin t — cos t)25. x = 5 cos t — sin t + 2£ — 5 27. x = T£(9e2i + 7e~2t — 2 sin 2<)29. x = 10e* - ^ e 3*'4 + \ e 2i31. Because e2t is a solution of the homogeneous equation; x = \te2t33. zw + (2i + pz)w = r*

Section 4, page 575

1. \ / 2 cos 3 (t — T5"?r) = \ / 2 sin 3(£ + t r 71*)3. 2 cos 2(/ + t$t ) = 2 sin 2(£ + 7r)

Chapter 14 679

5. 5 cos(£ — a) = 5 sin(t — /3), where a = arc cos(—f) ^ 126° 52' and 13 = arc sin f ^ 36° 52'

7. x = 6 cos7r(£ =L J) 9. 90/7T 11. 9 13. 7r, assuming the equilibrium position is 8 ft from the ceiling; 2ir'\/L/g

if it is L ft from the ceiling 15. 27i-/[(62.4)7rg/100]1/2 ^ 0 .7 9 1 sec

Section 5, page 579

1. y = cex — x — 1 3. y = a cos 2x + b sin 2x5. y = ai]Li° nxn/ ( n \ ) 2 7. (2ai/x)(e~x + x — 1)9. y = 2 - 3x + 6x2 - W + -V-x4 + • • •

11. y = — 1 + 2x — x2 + §x3 + Jx4 + • • •13. y — 1 + J(x — 2) + f (x — 2)2 + -ri{x — 2 )3 + — 2)4 + • • •

-2 9 4.

Section 6, page 584

3. (7 - A ) - 1 = I + A =2 9

4.

11. A = a l +

r - 21. = / + A = I

r / ( X ) 0 1 r i e - 115. /(A ) = 7. I + (e - 1)4 =

L/'(X) /(X)J V Lo e J

— u i ] ' -[::]■ — c n ]r 0 bl r 0 6_|

, so eA = eaIe b 0 , etc.L - 6 OJ

Section 7, page 590

1. x = Xo cosh kt + y0 sinh kt 3. y = P -1x = P~lAx = P~]APy 5. x = (a + 6£)ex* + 22eXt, y = 6ex' -f- tex<

Section 8, page 594

1. (e -^ X )‘ = - e ~ tAA X + e~tAX = e~tA(X - AX) = 0, so e~'AX = X 0, etc.

3 . eA = I + A, eB = I + B,

\2 11 "cosh 1 sinh 11

<3> iu Ob to II eA+B =.1 l j _sinh 1 cosh l j

Since AB BA, there is no reason for equality of eAeB and eA+B.

5. Just compute:[a bl r d - b l

B- l J- J-t r ( 6 c o f B ) - t r ( [ “ % [ _ * “ ‘ ] )

[dd - b e - I = tr = (ad - be)’.

L ad — bej

7. Solve for g: g(t) = g(0) et(tT A) = et(tT A) since #(0) = |e°| = 1. Set t = 1.

9. Integrate twice and change the order of iteration:

£ (0 = / y(s) ds, z (0 = / x(u) du— / g ( s ) ds, x(t) = I x\Jo Jo

= J ( J g(s) ds) du = J ( J ' du ) g(s) ds

= [ (t - s)g(s) ds.J 0

CHAPTER 15

Section 1, page 598

1. 4 ± 3 i 3. §( —1 ± i y / 7 ) 5. \ ( l ± 2 i y / 2 ) 7. in)( — 1 i Qiy/Z) 9. 1, §■(— 1 ± i*s/3) 11* 1, dot

13. ± 1 , ± * 15. (1 + i ) 2 = 2i; d=(l + i ) / V 2 17. 0

Section 2, page 601

1. | + f i 3. 2 - i 5. 5 + i 7. - f - \ i 9. * - 11. [1 ] * V l3 , C3] V 5 , [5 ] V26, [7 ] i V 2 , [9 ] 1

Section 3, page 608

1. v ^ 0013 t^t + i sin f 7r) 3. 2 (cos J7r + i sin %t )5. 4 (cos 7r + i sin tt)7. \ / l 7 (cos 6 + i sin 6), where tan 6 = 4, 0 < 0 < 9. i

11. — \ / 2 13. cos \ ir + i sin15. i \ /3 (c o s 7tt/6 + i sin 7tt/6) 17. - 4 19. -64*v/3 + 64i21. — i — i i 25. ± 1 , dzi, (± 1 ± i ) / \^2 , all signs

Chapter 15 681

27. dz i, J ( dz y / 3 dz i ) , all signs29. cos 0 + i sin 0, 0 = -£qt dz ^7rk, 0 < k < 939. cos 40 = 8 cos4 0 — 8 cos2 0 + 1

Section 4, page 614

1. i ( l + i\/3 )» 6(cos 1 — i sin 1), e1/5(cos -f + i sin f ) ,(cosh 2 + i sinh 2) / \ /2 , cosh 3 , - 1

3. Set z = x + iy. By 2, sin z is real only if sinh y cos x = 0, i.e., y = 0 or x = ^7r dz h r, k = 0, 1, 2, • • •. Likewise cos z is real only if cos x cosh y = 0 and sin x sinh y = 0 , that is, cos x = 0 and y = 0, 2 = ^7r d= kn.

5. no; for example, cos iy = cosh 1/ --------» 00 as y --------> 009. cosh 4x = 8 cosh4 x — 8 cosh2 x + 1


1. exv/5/2(cos Jx + i sin Jx)3* i^C rr(cos 4x + 4 sin 4x) + i (cos 2x + 2 sin 2x) + 3] + c 5. 0 if k 5* n; 7r if k = n7. A ( — sin 3x + 3x cos 3x + 27 sin x — 27x cos x) + c 9. 7r(2ao2 + &i2 + &22 + * *' + &n2)

Section 6, page 622

1 . y = aex + e~x,2(b cos ^x\ / 3 + c sin ^ x \ / 3 )3. y = aeXN/3 + be~xV* + c cos x \ /2 + d sin x-\/2 5. y = e3x(a cos x + b sin x) + ^ 0 (cos 2x — 2 sin 2x)7. y = (a2 — 62)- 2[ (a 2 — fr2)x cos bx + 2b sin bx] if a ^ b;

y = £a_2(ax2 sin ax + x cos ax + a sin ax) if a = 5 9. y = 5%( — cos x — 7 sin x) 11. ^a(cos 0 — sin 0)

Section 7, page 627

1. Z o ( i ) n+1(l + i)"+1(z - i)"3. | Eo ( & ) n+1 Im ( - 2 + 3*) »+>*», = V l3 5. 1 - Eo R e(l + 2t)«+«a» 7? = V 5 /57. 2" = Zo ( D x ' - ' i i y y , d(z”) /dx = Z o _1 (n - fc) (?)x»-*-> (<»)■*. But

(» - k) © = » ( V ) , hence d(z")/dx = n Z S _1 =n (x + iy )n~l = 7izn_1; a similar argument for dzn/dy.

9. d(zn) / dy = i d(zn)/dx, uy + iv,, = i [ux + ivx~] = — vx + mx. Equate real and imaginary parts.

11. Apply Exs. 9 and 10 term-by-term.

Index

A

Absolute convergence, 14 value, 599

Acceleration, 174 polar form, 194

Addition law, for centers of gravity, 447 of matrices, 237 of vectors, 119

Additivity, 469, 481 Adjoint (cofactor), 263 Affine function, 227(Ex. 12)Alternating series, 12, 102 Amplitude, 566Angular momentum, 185, 456(Ex. 15)

speed, 180 velocity, 180

Approximating sequence, 503 Approximation

by Taylor polynomials, 56 to solutions of differential equations, 549

Arc length, 163, 189 Area, 191, 396, 475, 500 Argument, 602 Aristotle, 3 (Ex. 26)Associative law, 236, 242

B

Bernoulli’s equation, 539(Ex. 30)Bessel’s equation, 579(Ex. 6)Binomial series, 97 Bolzano-Weierstrass theorem, 205 Bounded set, 207

C

Cardioid, 193 (Ex. 19)Cauchy criterion, 5, 17, 205

test, 5Cauchy-Schwarz inequality, 129(Ex. 25)

Center of gravity, 383, 401, 440, 442 Central force, 194 Centripetal acceleration, 179 Centroid, 442 Chain rule, 159, 288 Change of variable, 482, 486 Characteristic function, 475

polynomial, 270 roots, 270 vector, 269

Cissoid, 193 (Ex. 20)Closed set, 200 Cofactor matrix, 263 Comparison of series and integrals, 31

test for series, 6 Completeness, 74 Complex exponential, 610

numbers, 596 Components of acceleration, 178

of a vector, 118 Composite functions, 206 Composition of linear functions, 241 Cone, 257 Conjugate, 599 Constrained maxima, 352 Continuity, 205 Contour line, 297 Convergence, 4, 199 Convex, 474 Coordinate curves, 491

functions, 227 planes, 116

Coordinates, space, 117 Couple, 156 (Ex. 25)Cramer’s rule, 140 Cross product, 141 Curl, 301 (Ex. 19)Curvature, 168 Curve length, 163 Cusp, 166Cycloid, 188 (Ex. 17)Cylinder, 260Cylindrical coordinates, 419

684 INDEX

D

Definite integrals, 105 De Moivre’s theorem, 603 Density, 382Derivative of a matrix, 269(Exs. 37-40), 591

of a vector, 158 Determinants, 139 Difference of vectors, 121 Differentiable function, 284, 286 Differential, 316

equation, 517 Direction cosines, 128

field, 517 Directional derivative, 302 Distance formula, 124 Divergence, 301 (Ex. 19)Divergent series, 4 Domain, 197 Dot product, 124 Double integral, 372

E

Eccentricity, 195 (Ex. 7)Electric current, 574 Elimination, 134 Ellipsoid, 255, 490 Elliptic integrals, 106

paraboloid, 259 Equilibrium, 154 Euler’s constant, 44

relation, 291 (Ex. 24)Even function, 86 Exponential function, 582

of a matrix, 244(Ex. 15), 582

F

First order differential equation, 517 system, 585

Folium, 188(Ex. 21)Functions of several variables, 197

G

Gauss, C. F., 597 Generating function, 96 (Ex. 17) Generator of a cone, 257

cylinder, 260 Geometric series, 1 Gradient, 297

Graph of a function, 209 Green’s formula, 502(Ex. 25)

theorem, 499

H

Half-life, 548(Ex. 3)Harmonic motion, 566

series, 2 Hessian matrix, 340Homogeneous differential equation, 529, 557

equations, 150 function, 291 (Exs. 16-26) linear function, 226 system, 586, 591

Hooke’s law, 546 Hyperbola, 187(Ex. 16)Hyperbolic functions, 612

paraboloid, 259 Hyperboloid, 256, 257

I

Identity matrix, 261 Imaginary part, 598 Implicit differentiation, 311

function, 310 Improper double integral, 502

integral, 17, 18, 35 Inconsistent system, 135 Indefinite form, 251 Inequalities, 414 Infinite series, 4 Infinitely differentiable, 68 Initial condition, 517

-value problem, 517, 563 Inner product, 124 Integrable function, 457, 463, 479

on R2, 503 Integral, 464

equation, 549 of step function, 457, 460 of vector function, 184

Interior point, 202 Intersection of planes, 137, 148

of sets, 198 Interval, 198

of convergence, 76 Inverse of a matrix, 262 Inversion, 489(Ex. 7)Iterated integral, 379

INDEX 685

Iteration, 409, 462, 473, 479 formula, 378, 393

J

Jacobi identity, 147(Ex. 24) Jacobian, 483

matrix, 321 Jordan form, 271

for symmetric matrices, 276

K

Kepler’s laws, 194-196 Kinetic energy, 184

L

Lagrange multiplier, 354 Laplace transform, 21 Leibniz rule, 489 Lemniscate, 193(Ex. 21)Length of a curve, 163

of a vector, 123 Level curve, 297

surface, 299 Limagon, 193(Ex. 23)Line integral, 182 Linear differential equation, 526

function, 224 system, 133, 266 transformation, 227

Locally integrable, 503

M

Mass, 382, 401 Matrix, 229

exponential, 582 of a linear transformation, 228 of inertia, 450 multiplication, 235

Maxima and minima, 207, 217 with constraints, 352

Mean value theorem, 285 Measurable, 475 Minor, 139Mixed moments of inertia, 450

partials, 327 Modulus, 599

Moment, 383, 440 of inertia, 449, 454

Momentum, 185 Mothball problem, 542 M - test, 112Multiplication by a scalar, 120 Multiplier, 354

N

Natural frame, 422, 428 Negative definite matrix, 248 Negative semi-definite, 249 Newton’s law of cooling, 543

of motion, 175 Non-homogeneous differential equation, 529

systems, 589 Non-parametric surface, 491 Non-singular matrix, 265 Normal distribution, 506

form, 131 to a surface, 295 vector, 172

Numerical integration, 511

O

Odd function, 86 Open set, 203 Orientation, 145 Orthogonal, 298

projection, 230

P

Pappus theorems, 444, 446 Paraboloids, 258 Parallel-axis theorem, 454 Parameters, 153 Parametric equation, 130

form of a plane, 152 surface, 491

Partial derivative, 213 fractions, 93, 625 sum, 4

Particular solution, 562 Partition, 457, 458, 481 Period, 566

of orbit, 196 Phase angle, 566 Picard method, 552

686 INDEX

Plane, 131 through three points, 154

Polar angle, 189coordinates, 188, 400

Positive definite, 246 semi-definite, 249

Potential, 308 Power series, 72 Principal minor, 250 Product of inertia, 450

of matrices, 235 Proper transformation, 484 p-series, 7 Pure partials, 326

Q

Quadratic form, 244, 366 Quadric surface, 253

R

Radius, 189 of convergence, 76, 624

Ratio test, 10, 78 Real part, 598 Recurrence relation, 577 Regular transformation, 484 Resonance, 574 Riemann integral, 465 Right-hand rule, 116 R-integrable, 465 R-measurable, 475 Root test, 12 (Ex. 17)Roots of unity, 605 Rose curves, 193(Exs. 12-17) Row-by-column multiplication, 232 Row-by-matrix-by-column, 234 Rulings on a quadric, 297(Exs. 11-13)

S

Saddle point, 344 Scalar multiple, 120 Second derivative test, 340, 367

order differential equation, 588 Semi-definite, 249 Separation of variables, 520 Set notation, 198

Simple harmonic motion, 566 integral, 369

Simpson’s rule, 512 Singular matrix, 265 Singularity, 35 Skew lines, 151 Solid angle, 435Solution of differential equation, 517 Speed, 158 Spherical area, 434

coordinates, 427 Spiral of Archimedes, 193 (Ex. 11) Steady state, 574 Step function, 457, 459 Steradian, 436 Stirling’s formula, 47 Strophoid, 193(Ex. 22)Successive approximations, 549 Sum of vectors, 119 Summation notation, 5 Superposition, 595 Surface area, 493 Symmetric matrix, 245 System determinant, 140 System of differential equations, 585

T

Tangent plane, 292 vector, 167

Taylor formula, 58 polynomials, 56, 334 series, 68

Term by term operations on series, 90 Tetrahedra, 416 Torque, 146, 156(Ex. 24) Transformation, linear, 227

of variables, 482 Transient, 574 Transpose, 236 Triangle inequality, 123 Trigonometric functions, 611 Triple cross product, 147(Ex. 21),

244(Ex. 12) integral, 409 scalar product, 145

U

Uncoupled system, 587 Under determined system, 136

INDEX 687

Uniform continuity, 208 convergence, 109

Union of sets, 197 Uniqueness of power series, 82

theorem, 592 Unit normal, 172

tangent, 167 vector, 127

Vector, 118 algebra, 122 field, 306

Velocity, 158, 194

Voltage, 309Volume of parallepiped, 147

W

Wallis’s product, 45 Weierstrass M-test, 112

-Bolzano theorem, 205 Witch of Agnesi, 188(Ex. 20) Wronskian, 562(Ex. 22), 595(Ex. 8)

Zeno’s paradoxes, 3 (Ex. 26)

5C 6 D 7 E 8 F 9 G 0 H 1I 2 J 3

i . o , x sin 2ax10. / sin2 ax dx ------------------2 4 a

i o 7 x sin 2ax11. / cos2 ax dx = - -\-----------' 2 4 a

12

13

ax dx = —na +

axdx =cosw_1 ax sin ax n —

na +

, , cos(a + b)x cos(a — b)x sin ax cos ox dx = ----- —— —77--------- —------ 77— ( a ^ ± 6)

). / s i ,

/. J sinn

. J cosn

, /

• / si

I

■ i i s i

■ I x

/ xn cos aa: n fxn sin ax dx -------------------- 1— / xn_1

a a J

/ xn sin ax n fxn cos ax dx = ------------------/ xn~l si

a a J

n - 1 f . ------- / sinn_

w J

1 f- / cosn~

ax dx

ax dx

15. sin ax sin 6x dx =

2 (a + 6) 2 (a - 6)

sin(a — b)x sin(a + b)x 2 (a — 6) 2(a + 6)

. sin(a - b)x sin(a + b)x16. / cos ax cos bx dx = — -------—---- ------------------

2 (a - b ) 2 (a + 6)

sin ax x cos ax

(a 5 ±b)

(a 7* ± b )

17. x sin ax dx =aL

. _ cos ax x sin ax 18. / x cos ax dx = — ----- 1------------

19 cos ax dx

20 sin ax dx

21. / tan ax dx = ---- In |cos ax|' a

/ ■

‘ /

* /

22. / cot ax dx = - In |sin ax| a

23. / tan2 ax dx = - tan ax — x a

24. I cot2 ax dx = -----cot ax — xa

/ tan n_1 ax ftann ax dx = --------—-----/ tann_2 ax dx (n > 1)

(n - l)a J

/ cotn_1 ax f ,cotn ax dx = — ------- —-----/ cotn~2 ax dx (n > 1)

(n — l)a y

27. / sec ax dx = - In |sec ax + tan ax\ J ci

28. I csc ax dx = - In |csc ax — cot ax a

29

30

• / ’

/ secn“2 ax tan ax n — 2 fsecn ax dx = ---- ;--------r-------- 1--------- / secn_2 ax dx

(n — 1 )a n — 1 J

/ cscn-2 ax cot ax n — 2 fcscn axdx = -------- ;------ —--------1-------- - / cscn~2 ax dx

(n — l)a n — 1 J

/ • x i • x /~«------ ;arc sin - dx — x arc sin — \- V a2 — x2 a a

/x x a

32. / arc tan - dx — x arc t a n ---- -- In (a2 + x2)1 a a 2

34

35. / In ax dx = x In ax — x

V T =

xw+1 dx+ X 2

/ xn+1 arc sin x 1 fxn arcsin x ax = -------------------- *— — ■ / -

n + 1 n + 1 J

/

;xn+1 arc tan x 1 f xnxn arc tan x dx — ------------------------------ / —

n + i n + l j l

■ f36. j (In ax)n dx = x(ln ax)n — n J (In ax)n_1 dx (n > 0)

37. J xn In ax dx = T 7 ", .,N0 (^5^ — 1)xn+1 In ax xn+1 n + 1 (n + l )2

38J x n + 1

/ g aX

X6ox dx — — (ax — 1) a2

/xneox dx = - xneax — — [ xn~leax dx a a J

40

eaxeax sin bx dx = —-----— (a sin bx — b cos bx)

a2 + b2

(n > 1)

(n > 1)

M ~ 1)

/ eaxeax cos bx dx = —---- — (a cos bx + b sin bx)

a2 + 62

Second Course in Calculus

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