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INSTRUCTOR’S
SOLUTIONS MANUAL
SUSAN JANE COLLEY
DANIEL H. STEINBERG
VECTOR CALCULUS
THIRD EDITION
Upper Saddle River, New Jersey 07458
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Contents
Preface
v
1
Vectors
1.1 Vectors in Two and Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 More About Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5 Equations for Planes; Distance Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6 Some n-dimensional Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7 New Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.8 True/False Exercises for Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.9 Miscellaneous Exercises for Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
6
12
16
25
31
42
51
51
2
Differentiation in Several Variables
2.1 Functions of Several Variables; Graphing Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Properties; Higher-order Partial Derivatives; Newton’s Method . . . . . . . . . . . . . . . . . . . . . . .
2.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6 Directional Derivatives and the Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7 True/False Exercises for Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8 Miscellaneous Exercises for Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
70
70
79
86
94
105
115
124
125
3
Vector-Valued Functions
3.1 Parametrized Curves and Kepler’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Arclength and Differential Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Vector Fields: An Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 Gradient, Divergence, Curl, and the Del Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 True/False Exercises for Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.6 Miscellaneous Exercises for Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
137
137
145
154
163
170
171
4
Maxima and Minima in Several Variables
4.1 Differentials and Taylor’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Extrema of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.3 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Some Applications of Extrema . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5 True/False Exercises for Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6 Miscellaneous Exercises for Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
181
181
188
198
208
212
213
5
Multiple Integration
5.1 Introduction: Areas and Volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3 Changing the Order of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.5 Change of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.6 Applications of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.7 True/False Exercises for Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.8 Miscellaneous Exercises for Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
223
223
225
234
243
252
260
272
273
6
Line Integrals
6.1 Scalar and Vector Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.3 Conservative Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
289
289
295
302
iii
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iv
Contents
6.4
6.5
True/False Exercises for Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306
Miscellaneous Exercises for Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307
7
Surface Integrals and Vector Analysis
7.1 Parametrized Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.3 Stokes’s and Gauss’s Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.4 Further Vector Analysis; Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.5 True/False Exercises for Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7.6 Miscellaneous Exercises for Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
321
321
330
338
354
362
363
8
Vector Analysis in Higher Dimensions
8.1 An Introduction to Differential Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 Manifolds and Integrals of k -forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.3 The Generalized Stokes’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.4 True/False Exercises for Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.5 Miscellaneous Exercises for Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
382
382
386
391
396
396
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Preface
This manual accompanies the third edition of Susan Colley’s Vector Calculus. Solutions to all of the text’s main exercises
are provided. Brief answers to each chapter’s true/false exercises are also included. Most of the solutions were prepared
by Daniel H. Steinberg; the remainder were written by Susan Colley.
In addition to this manual, a Student Solutions manual is available. It provides solutions to many of the odd-numbered
exercises and answers to all of the odd-numbered true/false questions.
v
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C H A P T E R
1
Vectors
1.1 VECTORS IN TWO AND THREE DIMENSIONS
1. Here we just connect the point (0, 0) to the points indicated :
y
3
b
2.5
2
c
1.5
a
1
0.5
-1
1
2
3
x
2. Although more difficult for students to represent this on paper, the figures should look something like the following.
Note that the origin is not at a corner of the frame box but is at the tails of the three vectors.
3
z
a
2
b
1
c
2
0
x
0
-2
-2
0
2
y
In problems 3 and 4, we supply more detail than is necessary to stress to students what properties are being used :
3. (a) 3, 1 C −1, 7 D 3 C [−1], 1 C 7 D 2, 8.
(b) −28, 12 D −2 · 8, −2 · 12 D −16, −24.
(c) 8, 9 C 3−1, 2 D 8 C 3−1, 9 C 32 D 5, 15.
(d) 1, 1 C 52, 6 − 310, 2 D 1 C 5 · 2 − 3 · 10, 1 C 5 · 6 − 3 · 2 D −19, 25.
(e) 8, 10 C 38, −2 − 24, 5 D 8 C 38 − 2 · 4, 10 C 3−2 − 2 · 5 D 8, −26.
4. (a) 2, 1, 2 C −3, 9, 7 D 2 − 3, 1 C 9, 2 C 7 D −1, 10, 9.
(b)
1
8, 4, 1 C 2 5, −7, 1 D 4, 2, 1 C 10, −14, 1 D 14, −12, 1.
2
4
2
2
(c) −2 2, 0, 1 − 6 1 , −4, 1 D −22, 0, 1 − 3, −24, 6 D −2−1, 24, −5 D 2, −48, 10.
2
5. We start with the two vectors a and b. We can complete the parallelogram as in the figure on the left. The vector
from the origin to this new vertex is the vector a C b. In the figure on the right we have translated vector b so
that its tail is the head of vector a. The sum a C b is the directed third side of this triangle.
 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
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2
Chapter 1
Vectors
y
y
7
7
6
a+b
6
a+b
5 b translated
5
4
b
4
b
3
3
2
2
a
1
1
-2
-1.5
-1
a
-0.5
0.5
1
6. a D 3, 2 b D −1, 1
a − b D 3 − −1, 2 − 1 D 4, 1
x
-2
-1.5
1
a D 3 , 1
2
2
-1
-0.5
0.5
1
x
a C 2b D 1, 4
y
4
a+2b
3
a
2
b
-2
a-b
1
(1/2)a
-1
1
2
5
4
3
x
-1
−
'
−
'
−
'
7. (a) AB D −3 − 1, 3 − 0, 1 − 2 D −4, 3, −1 BA D −AB D 4, −3, 1
−
'
(b) AC D 2 − 1, 1 − 0, 5 − 2 D 1, 1, 3
−
'
BC D 2 − −3, 1 − 3, 5 − 1 D 5, −2, 4
−
'
−
'
AC C CB D 1, 1, 3 − 5, −2, 4 D −4, 3, −1
(c) This result is true in general:
B
A
C
Head-to-tail addition demonstrates this.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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Section 1.1
Vectors in Two and Three Dimensions
3
8. The vectors a D 1, 2, 1, b D 0, −2, 3 and a C b D 1, 2, 1 C 0, −2, 3 D 1, 0, 4 are graphed below.
Again note that the origin is at the tails of the vectors in the figure.
Also, −11, 2, 1 D −1, −2, −1. This would be pictured by drawing the vector (1, 2, 1) in the opposite
direction. Finally, 41, 2, 1 D 4, 8, 4 which is four times vector a and so is vector a stretched four times as
long in the same direction.
4
b
a+b
z
2
a
0
-2
1 0
x
0
2
y
9. Since the sum on the left must equal the vector on the right componentwise:
−12 C x D 2, 9 C 7 D y, and z C −3 D 5. Therefore, x D 14, y D 16, and z D 8.
10. If we drop a perpendicular
from
√
√ (3, 1) to the x-axis we see that by the Pythagorean Theorem the length of the
vector 3, 1 D 32 C 12 D 10.
y
1
0.8
0.6
0.4
0.2
x
0.5
1
1.5
2
2.5
3
11. Notice that b (represented by the dotted line) D 5a (represented by the solid line).
y
10
b
8
6
4
2
a
x
1
2
3
4
5
 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:11 — page 4 — #4
4
Chapter 1
Vectors
12. Here the picture has been projected into two dimensions so that you can more clearly see that a (represented by
the solid line) D −2b (represented by the dotted line).
8
a
6
4
2
-4
-6
-8
-2
4
2
-2
b
-4
13. The natural extension to higher dimensions is that we still add componentwise and that multiplying a scalar by a
vector means that we multiply each component of the vector by the scalar. In symbols this means that:
a C b D a1 , a2 , ... , an C b1 , b2 , ... , bn D a1 C b1 , a2 C b2 , ... , an C bn and ka D ka1 , ka2 , ... , kan .
In our particular examples, 1, 2, 3, 4 C 5, −1, 2, 0 D 6, 1, 5, 4,
and 27, 6, −3, 1 D 14, 12, −6, 2.
14. The diagrams for parts (a), (b) and (c) are similar to Figure 1.12 from the text. The displacement vectors are:
(a) (1, 1, 5)
(b) −1, −2, 3
(c) 1, 2, −3
(d) −1, −2
Note: The displacement vectors for (b) and (c) are the same but in opposite directions (i.e., one is the negative
of the other). The displacement vector in the diagram for (d) is represented by the solid line in the figure below:
y
1
0.75
P1
0.5
0.25
x
0.5
1
1.5
2
2.5
3
-0.25
-0.5
P2
-0.75
-1
15. In general, we would define the displacement vector from a1 , a2 , ... , an to b1 , b2 , ... , bn to be b1 − a1 , b2 −
a2 , ... , bn − an .
In this specific problem the displacement vector from P1 to P2 is 1, −4, −1, 1.
−'
16. Let B have coordinates (x, y, z). Then AB D x − 2, y − 5, z C 6 D 12, −3, 7 so x D 14, y D 2, z D 1
so B has coordinates (14, 2, 1).
17. If a is your displacement vector from the Empire State Building and b your friend’s, then the displacement vector
from you to your friend is b − a.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
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Section 1.1
Vectors in Two and Three Dimensions
5
you
a
b-a
Empire State Bldg
b
friend
18. Property 2 follows immediately from the associative property of the reals:
a C b C c D a1 , a2 , a3 C b1 , b2 , b3 C c1 , c2 , c3 D a1 C b1 , a2 C b2 , a3 C b3 C c1 , c2 , c3 D a1 C b1 C c1 , a2 C b2 C c2 , a3 C b3 C c3 D a1 C b1 C c1 , a2 C b2 C c2 , a3 C b3 C c3 D a1 , a2 , a3 C b1 C c1 , b2 C c2 , b3 C c3 D a C b C c.
Property 3 also follows from the corresponding componentwise observation:
a C 0 D a1 C 0, a2 C 0, a3 C 0 D a1 , a2 , a3 D a.
19. We provide the proofs for R3 :
1 k C la D k C la1 , a2 , a3 D k C la1 , k C la2 , k C la3 D ka1 C la1 , ka2 C la2 , ka3 C la3 D ka C la.
2 ka C b D ka1 , a2 , a3 C b1 , b2 , b3 D ka1 C b1 , a2 C b2 , a3 C b3 D ka1 C b1 , ka2 C b2 , ka3 C b3 D ka1 C kb1 , ka2 C kb2 , ka3 C kb3 D ka1 , ka2 , ka3 C kb1 , kb2 , kb3 D ka C kb.
3 kla D kla1 , a2 , a3 D kla1 , la2 , la3 D kla1 , kla2 , kla3 D lka1 , lka2 , lka3 D lka1 , ka2 , ka3 D lka.
20. (a) 0a is the zero vector. For example, in R3 :
0a D 0a1 , a2 , a3 D 0 · a1 , 0 · a2 , 0 · a3 D 0, 0, 0.
(b) 1a D a. Again in R3 :
1a D 1a1 , a2 , a3 D 1 · a1 , 1 · a2 , 1 · a3 D a1 , a2 , a3 D a.
21. (a) The head of the vector sa is on the x-axis between 0 and 2. Similarly the head of the vector tb lies somewhere
on the vector b. Using the head-to-tail method, sa C tb is the result of translating the vector tb, in this case,
to the right by 2s (represented in the figure by tb*). The result is clearly inside the parallelogram determined
by a and b (and is only on the boundary of the parallelogram if either t or s is 0 or 1.
b
x
tb
tb*
sa
a
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6
Chapter 1
Vectors
(b) Again the vectors a and b will determine a parallelogram (with vertices at the origin, and at the heads of a, b,
and a C b. The vectors sa C tb will be the position vectors for all points in that parallelogram determined
by (2, 2, 1) and (0, 3, 2).
−
−'
22. Here we are translating the situation in Exercise 21 by the vector OP0 . The vectors will all be of the form
−
−'
OP0 C sa C tb for 0 … s, t … 1.
√
23. (a) The speed of the flea is the length of the velocity vector D −12 C −22 D 5 units per minute.
(b) After 3 minutes the flea is at 3, 2 C 3−1, −2 D 0, −4.
(c) We solve 3, 2 C t−1, −2 D −4, −12 for t and get that t D 7 minutes. Note that both 3 − 7 D −4
and 2 − 14 D −12.
(d) We can see this algebraically or geometrically: Solving the x part of 3, 2 C t−1, −2 D −13, −27
we get that t D 16. But when t D 16, y D −29 not −27. Also in the figure below we see the path taken by
the flea will miss the point −13, −27.
y
3,2
-15 -12.5 -10
-7.5
-5
-2.5
2.5
5
x
-5
-10
-15
-20
-13,-27
-25
-30
24. (a) The plane is climbing at a rate of 4 miles per hour.
(b) To make sure that the axes are oriented so that the plane passes over the building, the positive x direction
is east and the positive y direction is north. Then we are heading east at a rate of 50 miles per hour at the
same time we’re heading north at a rate of 100 miles per hour. We are directly over the skyscraper in 1/10
of an hour or 6 minutes.
(c) Using our answer in (b), we have traveled for 1/10 of an hour and so we’ve climbed 4/10 of a mile or
2112 feet. The plane is 2112 − 1250 or 862 feet about the skyscraper.
25. (a) Adding we get: F1 C F2 D 2, 7, −1 C 3, −2, 5 D 5, 5, 4.
(b) You need a force of the same magnitude in the opposite direction, so F3 D −5, 5, 4 D −5, −5, −4.
26. (a) Measuring the force in pounds we get 0, 0, −50.
(b) The z components of the two vectors along the ropes must be equal and their sum must be opposite of the z
component in part (a). Their y components must also be opposite each other. Since the vector points in the
direction (0, ;2, 1), the y component will be twice the z component. Together this means that the vector in
the direction of 0, −2, 1 is 0, −50, 25 and the vector in the direction (0, 2, 1) is (0, 50, 25).
1.2 MORE ABOUT VECTORS
It may be useful to point out that the answers to Exercises 1 and 5 are the “same”, but that in Exercise 1 i D 1, 0 and
in Exercise 5 i D 1, 0, 0. This comes up when going the other direction in Exercises 9 and 10. In other words, it’s not
always clear whether the exercise “lives” in R2 or R3 .
1. 2, 4 D 21, 0 C 40, 1 D 2i C 4j.
2. 9, −6 D 91, 0 − 60, 1 D 9i − 6j.
3. 3, π, −7 D 31, 0, 0 C π0, 1, 0 − 70, 0, 1 D 3i C πj − 7k.
4. −1, 2, 5 D −11, 0, 0 C 20, 1, 0 C 50, 0, 1 D −i C 2j C 5k.
5. 2, 4, 0 D 21, 0, 0 C 40, 1, 0 D 2i C 4j.
6. i C j − 3k√
D 1, 0, 0 C 0, 1, 0 − 30, 0, √
1 D 1, 1, −3.
√
7. 9i − 2j C 2k D 91, 0, 0 − 20, 1, 0 C 20, 0, 1 D 9, −2, 2.
8. −32i − 7k D −6i C 21k D −61, 0, 0 C 210, 0, 1 D −6, 0, 21.
9. πi − j D π1, 0 − 0, 1 D π, −1.
10. πi − j D π1, 0, 0 − 0, 1, 0 D π, −1, 0.
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“runall” — 2005/8/1 — 12:11 — page 7 — #7
Section 1.2
More about Vectors
7
Note: You may want to assign both Exercises 11 and 12 together so that the students may see the difference. You should
stress that the reason the results are different has nothing to do with the fact that Exercise 11 is a question about R2 while
Exercise 12 is a question about R3 .
c C c2 D 3, and
11. (a) 3, 1 D c1 1, 1 C c2 1, −1 D c1 C c2 , c1 − c2 , so 1
c1 − c2 D 1.
Solving simultaneously (for instance by adding the two equations), we find that 2c1 D 4, so c1 D 2 and
c2 D 1. So b D 2a1 C a2 .
(b) Here c1 C c2 D 3 and c1 − c2 D −5, so c1 D −1 and c2 D 4. So b D −a1 C 4a2 .
c C c2 D b1 , and
(c) More generally, b1 , b2 D c1 C c2 , c1 − c2 , so 1
c1 − c2 D b2 .
b1 C b2
b − b2
Again solving simultaneously, c1 D
and c2 D 1
. So
2
2
b C b2
b1 − b2
bD 1
a1 C a2 .
2
2
12. Note that a3 D a1 C a2 , so really we are only working with two (linearly independent) vectors.
(a) 5, 6, −5 D c1 1, 0, −1 C c2 0, 1, 0 C c3 1, 1, −1; this gives us the equations:


 5 D c1 C c3
6 D c2 C c3

 −5 D −c1 − c3 .
The first and last equations contain the same information and so we have infinitely many solutions. You will
quickly see one by letting c3 D 0. Then c1 D 5 and c2 D 6. So we could write b D 5a1 C 6a2 . More
generally, you can choose any value for c1 and then let c2 D c1 C 1 and c3 D 5 − c1 .
(b) We cannot write (2, 3, 4) as a linear combination of a1 , a2 , and a3 . Here we get the equations:


 c1 C c3 D 2
c2 C c3 D 3

 −c1 − c3 D 4.
The first and last equations are inconsistent and so the system cannot be solved.
(c) As we saw in part (b), not all vectors in R3 can be written in terms of a1 , a2 , and a3 . In fact, only vectors
of the form a, b, −a can be written in terms of a1 , a2 , and a3 . For your students who have had linear
algebra, this is because the vectors a1 , a2 , and a3 are not linearly independent.
Note: As pointed out in the text, the answers for 13–21 are not unique.


xD2 C t
13. rt D 2, −1, 5 C t1, 3, −6 so y D −1 C 3t

 z D 5 − 6t.


 x D 12 C 5t
14. rt D 12, −2, 0 C t5, −12, 1 so y D −2 − 12t

 z D t.
15. rt D 2, −1 C t1, −7 so xD2 C t
y D −1 − 7t.


xD2 C t
16. rt D 2, 1, 2 C t3 − 2, −1 − 1, 5 − 2 so y D 1 − 2t

 z D 2 C 3t.


xD1 C t
17. rt D 1, 4, 5 C t2 − 1, 4 − 4, −1 − 5 so y D 4

 z D 5 − 6t.
18. rt D 8, 5 C t1 − 8, 7 − 5 so x D 8 − 7t
y D 5 C 2t.
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“runall” — 2005/8/1 — 12:11 — page 8 — #8
8
Chapter 1
Vectors
Note: In higher dimensions we switch our notation
to xi .


x
D
1 − 2t

1

 x D 2 C 5t
2
19. rt D 1, 2, 0, 4 C t−2, 5, 3, 7 so

x3 D 3t


 x4 D 4 C 7t.


x1 D 9 − 10t




− πt
 x2 D π C 1 √
√
20. rt D 9, π, −1, 5, 2 C t−1 − 9, 1 − π, 2 C 1, 7 − 5, 1 − 2 so x3 D −1 C 2 C 1t



x4 D 5 C 2t


 x5 D 2 − t.


 x D −1 C 2t
21. (a) rt D −1, 7, 3 C t2, −1, 5 so y D 7 − t

 z D 3 C 5t.


 x D 5 − 5t
(b) rt D 5, −3, 4 C t0 − 5, 1 C 3, 9 − 4 so y D −3 C 4t

 z D 4 C 5t.
(c) Of course, there are infinitely many solutions. For our variation on the answer to (a) we note that a line
parallel to the vector 2i − j C 5k is also parallel to the vector −2i − j C 5k so another set of equations
for part (a) is:


 x D −1 − 2t
yD7 C t

 z D 3 − 5t.
For our variation on the answer to (b) we note that the line passes through both points so we can set up the
equation with respect to the other point:


 x D −5t
y D 1 C 4t

 z D 9 C 5t.
(d) The symmetric forms are:
x C 1
z − 3
D7 − yD
2
5
y C 3
5 − x
z − 4
D
D
5
4
5
z − 3
x C 1
Dy − 7D
−2
−5
y
−
1
x
z − 9
D
D
−5
4
5
for (a)
for (b)
for the variation of (a)
for the variation of (b)
22. Set each piece of the equation equal to t and solve:
x − 2
D t * x − 2 D 5t * x D 2 C 5t
5
y − 3
D t * y − 3 D −2t * y D 3 − 2t
−2
z C 1
D t * z C 1 D 4t * z D −1 C 4t.
4
Note: In Exercises 23–25, we could say for certain that two lines are not the same if the vectors were not multiples of
each other. In other words, it takes two pieces of information to specify a line. You either need two points, or a point and
a direction (or in the case of R2 , equivalently, a slope).
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:11 — page 9 — #9
Section 1.2
More about Vectors
9
23. If we multiply each of the pieces in the second symmetric form by −2, we are effectively just traversing the same
path at a different speed and with the opposite orientation. So the second set of equations becomes:
y C 6
x C 1
z C 5
D
D
.
3
7
5
This looks a lot more like the first set of equations. If we now subtract one from each piece of the second set
of equations (as suggested in the text), we are effectively just changing our initial point but we are still on the
same line:
y C 6
x C 1
3
7
z C 5
5
− D
− D
− .
3
3
7
7
5
5
We have transformed the second set of equations into the first and therefore see that they both represent the same
line in R3 .
24. If you first write the equation of the two lines in vector form we can see immediately that their direction vectors
are the same so either they are parallel or they are the same line:
r1 t D −5, 2, 1 C t2, 3, −6
r2 t D 1, 11, −17 − t2, 3, −6.
The first line contains the point −5, 2, 1. If the second line contains −5, 2, 1 then the equations represent
the same line. Solve just the x component to get that −5 D 1 − 2t * t D 3. Checking we see that r2 3 D
1, 11, −17 − 32, 3, −6 D −5, 2, 1 so the lines are the same.
25. Here again the vector forms of the two lines can be written so that we see their headings are the same:
r1 t D 2, −7, 1 C t3, 1, 5
r2 t D −1, −8, −3 C 2t3, 1, 5.
The point 2, −7, 1 is on line one, so we will check to see if it is also on line two. As in Exercise 24
we check the equation for the x component and see that −1 C 6t D 2 * t D 1/2. Checking we see that
r2 1/2 D −1, −8, −3 C 1/223, 1, 5 D 2, −7, 2 Z 2, −7, 1 so the equations do not represent
the same lines.
Note: It is a good idea to assign both Exercises 26 and 27 together. Although they look similar, there is a difference
that students might miss.


 x D 3u C 7,
26. If you make the substitution u D t 3 , the equations become: y D −u C 2, and

 z D 5u C 1.
The map u D t 3 is a bijection. The important fact is that u takes on exactly the same values that t does, just at
different times. Since u takes on all reals, the parametric equations do determine a line (it’s just that the speed
along the line is not constant).

 x D 5u − 1,

27. This time if you make the substitution u D t 2 , the equations become: y D 2u C 3, and

 z D −u C 1.
The problem is that u cannot take on negative values so these parametric equations are for a ray with endpoint
−1, 3, 1 and heading 5, 2, −1.
28. (a) The vector form of the equations is: rt D 7, −2, 1 C t2, 1, −3. The initial point is then r0 D
7, −2, 1, and after 3 minutes the bird is at r3 D 7, −2, 1 C 32, 1, −3 D 13, 1, −8.
(b) 2, 1, −3
(c) We only need to check one component (say the x): 7 C 2t D 34/3 * t D 13/6. Checking we see that
r 13 D 7, −2, 1 C 13 2, 1, −3 D 34 , 1 , − 11 .
6
6
3 6
2
(d) As in part (c), we’ll check the x component and see that 7 C 2t D 17 when t D 5. We then check to
see that r5 D 7, −2, 1 C 52, 1, −3 D 17, 3, −14 Z 17, 4, −14 so, no, the bird doesn’t reach
17, 4, −14.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:11 — page 10 — #10
10
Chapter 1
Vectors
29. We can substitute the parametric forms of x, y, and z into the equation for the plane and solve for t. So 3t −
5 C 32 − t − 6t D 19 which gives us t D −3. Substituting back in the parametric equations, we find
that the point of intersection is −14, 5, −18.
30. Using the same technique as in Exercise 29, 51 − 4t − 2t − 3/2 C 2t C 1 D 1 which simplifies to
t D 2/5. This means the point of intersection is −3/5, −11/10, 9/5.
31. We will set each of the coordinate equations equal to zero in turn and substitute that value of t into the other two
equations.
x D 2t − 3 D 0 * t D 3/2. When t D 3/2, y D 13/2 and z D 7/2.
y D 3t C 2 D 0 * t D −2/3, so x D −13/3 and z D 17/3.
z D 5 − t D 0 * t D 5, so x D 7 and y D 17.
The points are (0, 13/2, 7/2), −13/3, 0, 17/3, and (7, 17, 0).
32. We could show that two points on the line are also in the plane or that for points on the line:
2x − y C 4z D 25 − t − 2t − 7 C 4t − 3 D 5, so they are in the plane.
33. For points on the line we see that x − 3y C z D 5 − t − 32t − 3 C 7t C 1 D 15, so the line does
not intersect the plane.
34. As explained in the text, we can’t just set the two sets of equations equal to each other and solve. If the two lines
intersect at a point, we may get to that point at two different times. Let’s call these times t1 and t2 and solve the
equations



 2t1 C 3 D 15 − 7t2 ,
3t1 C 3 D t2 − 2, and


 2t1 C 1 D 3t2 − 7.
Eliminate t1 by subtracting the third equation from the first to get t2 D 2. Substitute back into any of the equations
to get t1 D −1. Using either set of equations, you’ll find that the point of intersection is 1, 0, −1.
35. The way the problem is phrased tips us off that something is going on. Let’s handle this the same way we did in
Exercise 34.



 2t1 C 1 D 3t2 C 1,
−3t1 D t2 C 5, and


 t1 − 1 D 7 − t2 .
Adding the last two equations eliminates t2 and gives us t1 D 13/2. This corresponds to the point (14, −39/2,
11/2). Substituting this value of t1 into the third equation gives us t2 D 3/2, while substituting this into the
first equation gives us t2 D 13/3. This inconsistency tells us that the second line doesn’t pass through the point
14, −39/2, 11/2.
√
36. (a) The distance is 3t − 5 C 22 C 1 − t − 12 C 4t C 7 − 52 D 26t 2 − 2t C 13.
(b) Using a standard first year calculus trick, the distance is minimized when the square of the distance is
minimized. So we find D D 26t 2 − 2t C 13 is minimized (at the vertex of
the parabola) when t D 1/26.
Substitute back into our answer for (a) to find that the minimal distance is 337/26.
37. (a) As in Example 2, this is the equation of a circle of radius 2 centered at the origin. The difference is that you
are traveling around it three times as fast. This means that if t varied between 0 and 2π that the circle would
be traced three times.
(b) This is just like part (a) except the radius of the circle is 5.
(c) This is just like part (b) except the x and y coordinates have been switched. This is the same as reflecting
the circle about the line y D x and so this is also a circle of radius 5. If you care, the circle in (b) was drawn
starting at the point (5, 0) counterclockwise while this circle is drawn starting at (0, 5) clockwise.
(d) This is an ellipse with major axis along the x-axis intersecting it at ;5, 0 and minor axis along the y-axis
2
2
intersecting it at 0, ;3 : x C y D 1.
25
9
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“runall” — 2005/8/1 — 12:11 — page 11 — #11
Section 1.2
More about Vectors
11
y
4
2
x
-4
-2
2
4
-2
-4
38. The discussion in the text of the cycloid looked at the path traced by a point on the circumference of a circle of
radius a as it is rolled without slipping on the x-axis. The vector from the origin to our point P was split into
−
'
−
'
two pieces: OA (the vector from the origin to the center of the circle) and AP (the vector from the center of the
circle to P). This split remains the same in our problem.
The center of the circle is always a above the x-axis, and after the wheel has rolled through a central angle of
−
'
t radians the x coordinate is just at. So OA D at, a. This does not change in our problem.
−
'
The vector AP was calculated to be −a sin t, −a cos t. The direction of the vector is still correct but the
−
'
length is not. If we are b units from the center then AP D −bsin t, cos t.
We conclude then that the parametric equations are x D at − b sin t, y D a − b cos t. When a D b this is the
case of the cycloid described in the text; when a > b we have the curtate cycloid; and when a < b we have the
prolate cycloid.
For a picture of how to generate one consider the diagram:
Here the inner circle is rolling along the ground and the prolate cycloid is the path traced by a point on the
outer circle. There is a classic toy with a plastic wheel that runs along a handheld track, but your students are too
young for that. You could pretend that the big circle is the end of a round roast and the little circle is the end of a
skewer. In a regular rotisserie the roast would just rotate on the skewer, but we could imagine rolling the skewer
along the edges of the grill. The motion of a point on the outside of the roast would be a prolate cycloid.
39. You are to picture that the circular dispenser stays still so Egbert has to unwind around the dispenser. The direction
is cos θ, sin θ. The length is the radius of the circle a, plus the amount of tape that’s been unwound. The tape
that’s been unwound is the distance around the circumference of the circle. This is aθ where θ is again in radians.
The equation is therefore x, y D a1 C θcos θ, sin θ.
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“runall” — 2005/8/1 — 12:11 — page 12 — #12
12
Chapter 1
Vectors
1.3 THE DOT PRODUCT
Exercises 1–12 are just straightforward calculations. For 1–6 use Definition 3.1 and formula (1). For 7–9 use formula (4).
For 10–12 use formula (5).
1. 1, 5 · −2,3 D 1−2 C 53
√ D 13,
−2, 3 D −22 C 32 D 13.
1, 5 D
√
12 C 52 D
√
26,
√
4, −1 D 42 C −12 D 17
2. 4, −1 · 1/2,
2 D 41/2 −√12 D 0,
1/2, 2 D 1/22 C 22 D 17/2.
3. −1, 0, 7 · 2, 4, −6 D −12 C 04 C 7−6 D −44, −1, 0, 7 D −12 C 02 C 72 D
√
√
√
√
50 D 5 2, and 2, 4, −6 D 22 C 42 C −62 D 56 D 2 14.
√
√
4. 2, 1, 0 · 1, −2, 3 D√
21 C 1−2 C 03 D 0, 2, 1, 0 D 22 C 1 D 5, and 1, −2, 3 D
12 C −22 C 32 D 14.
√
√
5. 4i − 3j C k · i C√j C k D 41 √
C −31 C 11 D 2, 4i − 3j C k D 42 C 32 C 12 D 26,
and i C j C k D 1 C 1 C 1 D 3.
√
6. i C 2j − k · −3j
− 12 D −8, i C 2j − k D 12 C 22 C −12 D 6, and
C 2k D 2−3
√
− 3j C 2k D −32 C 22 D 13.
√
√
3i C j · − 3i C j
−3 C 1
−1
2π
√
√
7. θ D cos−1
D cos−1
D cos−1 .
D
3i C j − 3i C j
22
2
3
8. θ D cos−1
i C j − k · −i C 2j C 2k
i C j − k − i C 2j C 2k
9. θ D cos−1
1, −2, 3 · 3, −6, −5
1, −2, 3 3, −6, −5
D cos−1
D cos−1
−1 C 2 − 2
√
√
3 3
3 C 12 − 15
√ √
14 70
−1
D cos−1 √ .
3 3
D cos−1 0 D
π
.
2
Note: The answers to 10 and 11 are the same. You may want to assign both exercises and ask your students why √
this
should be true. You might then want to ask what would happen if vector a was the same but vector b was divided by 2.
i C j · 2i C 3j − k
5 5
2 C 3
i C j D
1, 1, 0 D , , 0.
i C j · i C j
1 C 1
2 2


iCj
1
√ 2 C 3
√ 2 · 2i C 3j − k 


1, 1, 0
5 5

iCj
2




√
D , , 0 .
11. proj iCj 2i C 3j − k D 


 √2 D
1C1
iCj
iCj
√
2 2
2
√ · √ 2
10. projiCj 2i C 3j − k D
2
12. projiCjC2k 2i − 4j C k D
2
2
i C j C 2k · 2i − 4j C k
2 − 4 C 2
i C j C 2k D
1, 1, 2 D 0.
i C j C 2k · i C j C 2k
1 C 1 C 4
2i − j C k
1
D √ 2, −1, 1.
||2i − j C k||
6
1
i − 2k
D √ 1, 0, −2.
14. Here we take the negative of the vector divided by its length:
i − 2k
5
√
3i C j − k
3
15. Same idea as Exercise 13, but multiply by 3:
D √ 1, 1, −1 D 31, 1, −1.
i C j − k
3
16. There are a whole plane full of perpendicular vectors. The easiest three to find are when we set the coefficients
of the coordinate vectors equal to zero in turn: i C j, j C k, and −i C k.
17. We have two cases to consider.
If either of the projections is zero: proja b D 0 3 a · b D 0 3 projb a D 0.
If neither of the projections is zero, then the directions must be the same. This means that a must be a multiple
of b. Let a D cb, then on the one hand
13. We just divide the vector by its length:
proja b D projcb b D
cb · b
cb D b.
cb · cb
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“runall” — 2005/8/1 — 12:11 — page 13 — #13
Section 1.3
On the other hand
projb a D projb cb D
The Dot Product
13
b · cb
b D cb.
b · b
These are equal only when c D 1.
In other words, proja b D projb a when a · b D 0 or when a D b.
18. Property 2: a · b D a1 , a2 , a3 · b1 , b2 , b3 D a1 b1 C a2 b2 C a3 b3 D b1 a1 C b2 a2 C b3 a3 D b · a.
Property 3: a · b C c D a1 , a2 , a3 · b1 , b2 , b3 C c1 , c2 , c3 D a1 , a2 , a3 · b1 C c1 , b2 C c2 , b3 C
c3 D a1 b1 C c1 C a2 b2 C c2 C a3 b3 C c3 D a1 b1 C a2 b2 C a3 b3 C a1 c1 C a2 c2 C a3 c3 D
a · b C a · c.
Property 4: ka · b D ka1 , a2 , a3 · b1 , b2 , b3 D ka1 , ka2 , ka3 · b1 , b2 , b3 D ka1 b1 C ka2 b2 C
ka3 b3 for the
1st equality D ka1 b1 C a2 b2 C a3 b3 D ka · b. for the 2nd equality D a1 kb1 C a2 kb2 C a3 kb3 D
a1 , a2 , a3 · kb1 , kb2 , kb3 D a · kb.
√ √
√
19. We have ka D ka · ka D k 2 a · a D k 2 a · a D |k| a.
20. The following diagrams might be helpful:
y
2
y
F1
1.5
0.5
1
1.5
2
x
a
1
-0.5
0.5
x
0.5
1
1.5
2
2.5
3
3.5
4
F2
-1
-0.5
F
-1
-1.5
-1.5
F
-2
-2
To find F1 , the component of F in the direction of a, we project F onto a:
F1 D proja F D
i − 2j · 4i C j
2
i − 2j D
4, 1.
i − 2j · i − 2j
17
To find F2 , the component of F in the direction perpendicular to a, we can just subtract F1 from F:
F2 D 1, −2 −
2
9 −36
9
4, 1 D ,
1, −4.
D
17
17 17
17
Note that F1 is a multiple of a so that F1 does point in the direction of a and that F2 · a D 0 so F2 is perpendicular
to a.
−
'
21. (a) Work done by the force is given to be the product of the length of the displacement PQ and the
−' F or in the case pictured in the text,
component of force in the direction of the displacement (;proj−
PQ
F cos θ). That is,
−
'
−
'
Work D PQ F cos θ D F · PQ
using Theorem 3.3.
(b) The amount of work is
−
'
Work D PQ F| cos θ| D 40|500 cos30◦ | D 40250 D 10000 ft-lbs.
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“runall” — 2005/8/1 — 12:11 — page 14 — #14
14
Chapter 1
Vectors
22. You could either use the three right triangles determined by the vector a and the three coordinate axes, or you could
a · i
a1
a2
use Theorem 3.3. By that theorem, cos α D
D . Similarly, cos β D 2
2
2
2
a i
a1 C a2 C a3
a1 C a22 C a23
a3
and cos γ D .
a21 C a22 C a23
23. Consider the figure:
A
P2
P1
C
B
−
'
−
'
If P1 is the point on AB located r times the distance from A to B, then the vector AP1 D rAB. Similarly, since
−
'
−
'
P2 is the point on AC located r times the distance from A to C, then the vector AP2 D rAC. So now we can
look at the line segment P1 P2 using vectors.
'
−
'
−
'
−
'
−
'
−
'
−
'
−
−−' −
P1 P2 D AP2 − AP1 D rAC − rAB D rAC − AB D rBC.
−
'
−
−−'
The two conclusions now follow. Because P1 P2 is a scalar multiple of BC, they are parallel. Also the positive
−
'
−
'
−
−−'
scalar r pulls out of the norm so P1 P2 D rBC D rBC.
24. This now follows immediately from Exercise 23 or Example 6 from the text. Consider first the triangle ABC.
A
M1
M4
B
M2
C
D
M3
If M1 is the midpoint of AB and M2 is the midpoint of BC, we’ve just shown that M1 M2 is parallel to AC and
has half its length. Similarly, consider triangle DAC where M3 is the midpoint of CD and M4 is the midpoint of
DA. We see that M3 M4 is parallel to AC and has half its length. The first conclusion is that M1 M2 and M3 M4
have the same length and are parallel. Repeat this process for triangles ABD and CBD to conclude that M1 M4 and
M2 M3 have the same length and are parallel. We conclude that M1 M2 M3 M4 is a parallelogram. For kicks—have
your students draw the figure for ABCD a non-convex quadrilateral. The argument and the conclusion still hold
even though one of the “diagonals” is not inside of the quadrilateral.
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“runall” — 2005/8/1 — 12:11 — page 15 — #15
Section 1.3
The Dot Product
15
25. In the diagram in the text, the diagonal running from the bottom left to the top right is a C b and the diagonal
running from the bottom right to the top left is b − a.
a C b D −a C b
−a C b · −a C b
√
a · a C b · b C 2a · b D −12 a · a C b · b − 2a · b
a · b D 0
a C b · a C b D
3
3
3
Since neither a nor b is zero, they must be orthogonal.
26. Using the same set up as that in Exercise 25, we note first that
a C b · −a C b D a · −a C b · −a C a · b C b · b D −a2 C b2 .
It follows immediately that
a C b · −a C b D 0 3 a D b,
in other words that the diagonals of the parallelogram are perpendicular if and only if the parallelogram is
a rhombus.
27. (a) Let’s start with the two circles with centers at W1 and W2 . Assume that in addition to their intersection at
point O that they also intersect at point C as shown below.
C
W1
W2
O
The polygon OW1 CW2 is a parallelogram. In fact, because all sides are equal, it is a rhombus. We can,
−
−
'
−
'
−
−
'
therefore, write the vector c D OC D OW1 C OW2 D w1 C w2 . Similarly, we can write b D w1 C w3
and a D w2 C w3 .
(b) Let’s use the results of part (a) together with the hint. We need to show that the distance from each of the
−
'
points A, B, and C to P is r. Let’s show, for example, that CP is r:
−
'
−
'
−
'
CP D OP − OC D w1 C w2 C w3 − w1 C w2 D w3 D r.
The arguments for the other two points are analogous.
(c) What we really need to show is that each of the lines passing through O and one of the points A, B,
−
'
or C is perpendicular to the line containing the two other points. Using vectors we will show that OA ⬜
−
' −
'
−
'
−
'
−
'
BC, OB ⬜ AC, and OC ⬜ AB by showing their dot products are 0. It’s enough to show this for one
−
' −
'
of them: OA · BC D w2 C w3 · w1 C w2 − w1 C w3 D w2 C w3 · w2 − w3 D
w2 · w2 C w3 · w2 − w2 · w3 − w3 · w3 D r2 − r2 D 0.
28. (a) This follows immediately from Exercise 26 if you notice that the vectors are the diagonals of the rhombus
with two sides ba and ab.
Or we can proceed with the calculation: ba C a · ba − ab. The only bit of good news
here is that the cross terms clearly cancel each other out and we’re left with: b2 a · a − a2 b · b D
b2 a2 − a2 b2 D 0.
(b) As in (a) the slicker way is to recall (or reprove geometrically) that the diagonals of a rhombus bisect the
vertex angles. Then note that ba C ab is the diagonal of the rhombus with sides ba and ab and
so bisects the angle between them which is the same as the angle between a and b.
Another way is to let θ1 be the angle between a and ba C ab, and let θ2 be the angle between b and
ba C ab.
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“runall” — 2005/8/1 — 12:11 — page 16 — #16
16
Chapter 1
Vectors
Then
cos−1 θ1 D
a · ba C ab
a2 b C aa · b
a b C a · b
D
D
.
aba C ab
aba C ab
ba C ab
cos−1 θ2 D
b · ba C ab
bb · a C b2 a
b · a C a b
D
D
.
bba C ab
bba C ab
ba C ab
Also
So ba C ab bisects the angle between the vectors a and b.
1.4 THE CROSS PRODUCT
For Exercises 1–4 use Definition 4.2.
1. 23 − 41 D 2.
2. 06 − 5−1 D 5.
3. 123 C 37−1 C 500 − 52−1 − 170 − 303 D −5.
4. −262 C 0−14 C 1/23−8 − 1/264 − −2−1−8 − 032 D −32.
Note: In Exercises 5–7, the difference between using (2) and (3) really amounts to changing the coefficient of j from
a3 b1 − a1 b3 in formula (2) to −a1 b3 − a3 b1 in formula (3). The details are only provided in Exercise 5.
5. First we’ll use formula (2):
1, 3, −2 * −1, 5, 7 D [37 − −25]i C [−2−1 − 17]j C [15 − 3−1]k
D 31i − 5j C 8k D 31, −5, 8.
If instead we use formula (3), we get:
i j
k 1, 3, −2 * −1, 5, 7 D 1 3 −2 −1 5
7 3 −2 1
D
i − 5
7 −1
1
−2 j C 7 −1
3 k
5 D 31i − 5j C 8k D 31, −5, 8.
6. Just using formula (3):
i
3i − 2j C k * i C j C k D 3
1
j
−2
1
k −2
1 D 1
1 3
1 i
−
1
1
3
1 j
C
1
1
−2 k
1 D −3i − 2j C 5k D −3, −2, 5.
7. Note that these two vectors form a basis for the xy-plane so the cross product will be a vector parallel to (0, 0,
1). Again, just using formula (3):
i j k 1 0 1 0 j C 1 1 k D 5k D 0, 0, 5.
i C j * −3i C 2j D 1 1 0 D i
−
−3 0 −3 2 2
0
−3 2 0 8. By (1) a C b * c D −c * a C b.
By (2), this D −c * a C −c * b.
By (1), this D a * c C b * c.
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“runall” — 2005/8/1 — 12:11 — page 17 — #17
Section 1.4
The Cross Product
17
9. a C b * a − b D a * a C b * a − a * b − b * b. The cross product of a vector with itself
is 0 and also b * a D −a * b, so
a C b * a − b D −2a * b.
You may wish to have your students consider what this means about the relationship between the cross product of
the sides of a parallelogram and the cross product of its diagonals. In any case, we are given that a * b D 3, −7,
−2, so a C b * a − b D −6, 14, 4.
10. If you plot the points you’ll see that they are given in a counterclockwise order of the vertices of a parallelogram.
To find the area we will view the sides from (1, 1) to (3, 2) and from (1, 1) to −1, 2 as vectors by calculating
the displacement vectors: 3, 2 − 1, 1 and −1, 2 − 1, 1. We then embed the problem in R3 and take a
cross product. The length of this cross product is the area of the parallelogram.
y
3
2.5
2
1.5
1
0.5
x
-1
-0.5
-0.5
0.5
1
1.5
2
2.5
3
-1
3 − 1, 2 − 1, 0 * −1 − 1, 2 − 1, 0 D 2, 1, 0 * −2, 1, 0 D 4k D 0, 0, 4.
So the area is 0, 0, 4 D 4.
11. This is tricky, as the points are not given in order. The figure on the left shows the sides connected in the order
that the points are given.
2
2
y
y
0
0
-2
-2
2
2
z
z
0
0
-2
-2
-2
-2
0
0
x
2
x
4
2
4
As the figure on the right shows, if you take the first side to be the side that joins the points (1, 2, 3) and 4, −2, 1
then the next side is the side that joins 4, −2, 1 and 0, −3, −2. We will again calculate the length of the
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“runall” — 2005/8/1 — 12:11 — page 18 — #18
18
Chapter 1
Vectors
cross product of the displacement vectors. So the area of the parallelogram will be the length of
0 − 4, −3 − −2, −2 − 1 * 1 − 4, 2 − −2, 3 − 1 D −4, −1, −3 * −3, 4, 2 D 10, 17, −19.
√
√
The length of 10, 17, −19 is 102 C 272 C −192 D 750 D 5 30.
12. The cross product will give us the right direction; if we then divide this result by its length we will get a unit
vector:
1, −5, −1
2, 1, −3 * 1, 0, 1
1
D
D √ 1, −5, −1.
2, 1, −3 * 1, 0, 1
1, −5, −1
27
13. For a * b · c to be zero either
• One or more of the three vectors is 0,
• a * b D 0 which would happen if a D kb for some real k, or
• c is in the plane determined by a and b.
For Exercises 14–17 we’ll just take half of the length of the cross product. Unlike Exercises 10 and 11, in Exercises 16
and 17 we don’t have to worry about the ordering of the points. In a triangle, whichever order we choose we are traveling
either clockwise or counterclockwise. Just choose any of the vertices as the base for the cross product. Our choices may
differ, but the solution won’t.
14. 1/21, 1, 0 * 2, −1, 0 D 1/20, 0, −3 D 3/2.
√
15. 1/21, −2, 6 * 4, 3, −1 D 1/2−16, 25, 11 D 1002/2.
16. 1/2−1 − 1, 2 − 1, 0 * −2 − 1, −1 − 1, 0 D 1/2−2, 1, 0 * −3, −2, 0 D
1/20, 0, 7 D 7/2.
17. 1/20 − 1, 2, 3 − 1 *√ −1 − 1,√
5, −2 − 1 D 1/2−1, 2, 2 * −2, 5, −3 D
1/2−16, −7, −1 D 306/2 D 3 34/2.
The triple scalar product is used in Exercises 18 and 19 and the equivalent determinant form mentioned in the text is
proved in Exercise 20.
Some people write this product as a · b * c instead of a * b · c. Exercise 28 shows that these are equivalent.
18. Here we are given the vectors so we can just use the triple scalar product:
3 −1 0 0 1 a * b · c D 3i − j * −2i C k · i − 2j C 4k D −2
1 −2 4 0 1 0 − −1 −2 1 C 0 −2
D 3 1 4 1 −2 D 32 C −9 D −3.
−2 4 Volume D |a * b · c| D 3.
19. You need to figure out a useful ordering of the vertices. You can either plot them by hand or use a computer
package to help or you can make some observations about them. First look at the z coordinates. Two points have
z D −1 and two have z D 0. These form your bottom face. Of the remaining points two have z D 5—these will
match up with the bottom points with z D −1, and two have z D 6—these will match up with the bottom points
with z D 0. The parallelepiped is shown below.
We’ll use the highlighted edges as our three vectors a, b, and c. You could have based the calculation at any
vertex. I have chosen 4, 2, −1. The three vectors are:
a D 0, 3, 0 − 4, 2, −1 D −4, 1, 1
b D 4, 3, 5 − 4, 2, −1 D 0, 1, 6
c D 3, 0, −1 − 4, 2, −1 D −1, −2, 0
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“runall” — 2005/8/1 — 12:11 — page 19 — #19
Section 1.4
The Cross Product
19
6
z
4
2
0
0 0
2
x
1
2
y
3
4
4
We can now calculate
−4
1 1 1 6 a * b · c D −4, 1, 1 * 0, 1, 6 · −1, −2, 0 D 0
−1 −2 0 1 6 1 − 1 0 6 C 1 0
D −4 −1 0 −1 −2 D −412 − 6 C 1 D −53.
−2 0 Finally, Volume D |a * b · c| D 53.
Note: The proofs of Exercises 20 and 28 are easier if you remember that if matrix A is just matrix B with any two rows
interchanged then the determinant of A is the negative of the determinant of B. If you don’t use this fact (which is explored
in exercises later in this chapter), you can prove this with a long computation. That is why the author of the text suggests
that a computer algebra system could be helpful—and this would be a great place to use it in a class demonstration.
20. This is not as bad as it might first appear.


i
j
k 





a * b · c D 
 · c1 , c2 , c3  a1 a2 a3 
b1 b2 b3 a1 a3 a3 C k a1 a2 · c1 , c2 , c3 −
j
b1 b3 b1 b2 b3 a a3 − c2 a1 a3 C c3 a1 a2 D c1 2
b1 b3 b1 b2 b2 b3 c1 c2 c3 a1 a2 a3 a1 a2 a3 D a1 a2 a3 D − c1 c2 c3 D b1 b2 b3 b1 b2 b3 b1 b2 b3 c1 c2 c3 b1
a3 b3 by Exercise 20. Similarly, a · b * c D b * c · a D c1
a1
c3 a
D i 2
b2
a1 a2
21. a * b · c D b1 b2
c1 c2
Exercise 20.
Expand these determinants to see that they are equal.
a1 a2 a3 b1 b2 b3 D a1 b2 c3 − b3 c2 − a2 b1 c3 − b3 c1 C a3 b1 c2 − b2 c1 c1 c2 c3 b1 b2 b3 c1 c2 c3 D b1 a3 c2 − a2 c3 − b2 a3 c1 − a1 c3 C b3 a2 c1 − a1 c2 a1 a2 a3 b2
c2
a2
b3 c3 by
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“runall” — 2005/8/1 — 12:11 — page 20 — #20
20
Chapter 1
Vectors
22. The value of |a * b · c| is the volume of the parallelepiped determined by the vectors a, b, c. But so is
|b · a * c|, so the quantities must be equal.
23. (a) We have
−
−−' −
−−'
Area D 1 P1 P2 * P1 P3 2
D 1 x2 − x1 , y2 − y1 , 0 * x3 − x1 , y3 − y1 , 0
2
i
j
k −
−−' −
−−' Now P1 P2 * P1 P3 D x2 − x1 y2 − y1 0 x3 − x1 y3 − y1 0 D [x2 − x1 y3 − y1 − x3 − x1 y2 − y1 ]k
y
P3
P2
P1
x
Hence the area is 1 |x2 − x1 y3 − y1 − x3 − x1 y2 − y1 |. On the other hand
2
1
1 x1
2
y1
1
x2
y2
1 x3 D 1
2
y3 x2
y2
x
x3 − 1
y3 y1
x
x3 C 1
y3 y1
x2 .
y2 Expanding and taking absolute value, we obtain
1
|x2 y3 − x3 y2 C x3 y1 − x1 y3 C x1 y2 − x2 y1 |.
2
From here, its easy to see that this agrees
with theformula above.
1 1
1 1 (b) We compute the absolute value of 1 2 −4 D 1 −8 − 8 C 3 − 4 C 12 C 4 D 1 −1 D − 1 .
2
2
2
2
2 3 −4 Thus the area is − 1 D 1 .
2
2
24. Surface area D 1 a * b C b * c C a * c C b − a * c − a
2
b
c
a
25. We assume that a, b, and c are non-zero vectors in R3 .
(a) The cross product a * b is orthogonal to both a and b.
(b) Scale the cross product to a unit vector by dividing by the length and then multiply by 2 to get 2
(c) proja b D a * b
.
a * b
a · b
a.
a · a
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“runall” — 2005/8/1 — 12:11 — page 21 — #21
Section 1.4
(d) Here we divide vector a by its length and multiply it by the length of b to get
The Cross Product
21
b
a.
a
(e) The cross product of two vectors is orthogonal to each: a * b * c.
(f) A vector perpendicular to a * b will be back in the plane determined by a and b, so our answer is a * b * c.
26. I love this problem—students tend to go ahead and calculate without thinking through what they’re doing first.
This would make a great quiz at the beginning of class.
(a) Vector: The cross product of the vectors a and b is a vector so you can take its cross product with vector c.
(b) Nonsense: The dot product of the vectors a and b is a scalar so you can’t dot it with a vector.
(c) Nonsense: The dot products result in scalars and you can’t find the cross product of two scalars.
(d) Scalar: The cross product of the vectors a and b is a vector so you can take its dot product with vector c.
(e) Nonsense: The cross product of the vectors a and b is a vector so you can take its cross product with vector
that is the result of the cross product of c and d.
(f) Vector: The dot product results in a scalar that is then multiplied by vector d. We can evaluate the cross
product of vector a with this result.
(g) Scalar: We are taking the dot product of two vectors.
(h) Vector: You are subtracting two vectors.
Note: You can have your students use a computer algebra system for these as suggested in the text. I’ve included worked
out solutions for those as old fashioned as I.
27. Exercise 25(f) shows us that a * b * c is in the plane determined by a and b and so we expect the solution
to be of the form k1 a C k2 b for scalars k1 and k2 .
Using formula (3) from the text for a * b:
i
k
j
a2 a3 a1 a3 a1 a2 −
a * b * c D b1 b3 b1 b2 b2 b3
c1
c2
c3
a1 a3 a1 a2 a1 a2 a2 a3 c j
D −
c − c i − c − b1 b3 3
b1 b2 2
b2 b3 3
b1 b2 1
a2 a3 a1 a3 c k
C c C b2 b3 2
b1 b3 1
Look first at the coefficient of i: −a1 b3 c3 C a3 b1 c3 − a1 b2 c2 C a2 b1 c2 . If we add and subtract a1 b1 c1 and
regroup we have: b1 a1 c1 C a2 c2 C a3 c3 − a1 b1 c1 C b2 c2 C b3 c3 D b1 a · c − a1 b · c. Similarly
for the coefficient of j. Expand then add and subtract a2 b2 b3 and regroup to get b2 a · c − a2 b · c. Finally
for the coefficient of k, expand then add and subtract a3 b3 c3 and regroup to obtain b3 a · c − a3 b · c. This
shows that a * b * c D a · cb − b · ca.
Now here’s a version of Exercise 27 worked on Mathematica. First you enter the following to define the vectors
a, b, and c.
a D {a1, a2, a3}
b D {b1, b2, b3}
c D {c1, c2, c3}
The reply from Mathematica is an echo of your input for c. Let’s begin by calculating the cross product. You
can either select the cross product operator from the typesetting palette or you can type the escape key followed
by “cross” followed by the escape key. Mathematica should reformat this key sequence as * and you should be
able to enter
a * b * c.
Mathematica will respond with the calculated cross product
{a2b1c2 − a1b2c2 C a3b1c3 − a1b3c3,
−a2b1c1 C a1b2c1 C a3b2c3 − a2b3c3,
−a3b1c1 C a1b3c1 − a3b2c2 C a2b3c2}.
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“runall” — 2005/8/1 — 12:11 — page 22 — #22
22
Chapter 1
Vectors
Now you can check the other expression. Use a period for the dot in the dot product.
a.cb − b.c a
Mathematica will immediately respond
{b1a1c1 C a2c2 C a3c3 − a1b1c1 C b2c2 C b3c3,
b2a1c1 C a2c2 C a3c3 − a2b1c1 C b2c2 C b3c3,
b3a1c1 C a2c2 C a3c3 − a3b1c1 C b2c2 C b3c3}
This certainly looks different from the previous expression. Before giving up hope, note that this one has been
factored and the earlier one has not. You can expand this by using the command
Expand[a.cb − b.ca]
or use Mathematica’s command % to refer to the previous entry and just type
Expand[%].
This still might not look familiar. So take a look at
a * b * c − [a.cb − b.ca].
If this still isn’t what you are looking for, simplify it with the command
Simplify[%]
and Mathematica will respond
{0, 0, 0}.
28. The exercise asks us to show that six quantities are equal.
The most important pair is a · b * c D c · a * b. Because of the commutative property of the dot product
c · a * b D a * b · c and so we are showing that a · b * c D a * b · c.
a1 a2 a3 c · a * b D a * b · c D b1 b2 b3 c1 c2 c3 b1 b2 b3 D c1 c2 c3 a1 a2 a3 D b * c · a D a · b * c.
The determinants of the 3 by 3 matrices above are equal because we had to interchange two rows twice to get
from one to the other. This fact has not yet been presented in the text. This would be an excellent time to use a
computer algebra system to show the two determinants are equal. Of course, you could use Mathematica or some
other such system to do the entire problem.
To show that a · b * c D b · c * a we use a similar approach:
b 1 b2 b3 a · b * c D b * c · a D c1 c2 c3 a1 a2 a3 c1 c2 c3 D a1 a2 a3 b1 b2 b3 D c * a · b D b · c * a.
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“runall” — 2005/8/1 — 12:11 — page 23 — #23
Section 1.4
The Cross Product
23
So we’ve established that the first three triple scalars are equal.
We get the rest almost for free by noticing that three pairs of equations are trivial:
a · b * c D −a · c * b,
b · c * a D −b · a * c, and
c · a * b D −c · b * a.
Each of the above pairs are equal by the anticommutativity property of the cross product. If you prefer the matrix
approach, this also follows from the fact that interchanging two rows changes the sign of the determinant.
29. By Exercise 28, a * b · c * d D c · d * a * b.
By anticommutativity, c · d * a * b D −c · a * b * d.
By
Exercise 27, −c
· a * b * d D −c · a · db − b · da D a · cb · d − a · db · c D
a · c a · d b · c b · d .
30. Apply the results of Exercise 27 to each of the three components:
a * b * c C b * c * a C c * a * b D [a · cb − b · ca] C [b · ac − c · ab]
C [c · ba − a · bc] D 0.
(For example, the a · cb cancels with the c · ab because of the commutative property for the dot product.)
31. If your students are using a computer algebra system, they may not notice that this is exactly the same problem
as Exercise 27. Just replace c with c * d on both sides of the equation in Exercise 27 to obtain the result here.
32. First apply Exercise 29 to the dot product to get
a * b · b * c * c * a D [a · b * c][b · c * a] − [a · c * a][b · b * c].
You can either observe that two of these quantities must be 0, or you can apply Exercise 28 to see a · c * a D
c · a * a D 0. Exercise 28 also shows that b · c * a D a · b * c. The result follows.
33. We did this above in Exercise 29.
34. The amount of torque is the product of the length of the “wrench” and the component of the force perpendicular
to the “wrench”. In this case, the wrench is the door—so the length is four feet. The 20 lb force is applied
◦
perpendicular to the plane√of the doorway
√ and the door is open 45 . So from the text, the amount of torque D
aF sin θ D 420 2/2 D 40 2 ft-lb.
35. (a) Here the length of a is 1 foot, the force F D 40 pounds and angle θ D 120 degrees. So
Torque D 140 sin 120◦ D 40
√
3
2
√
D 20 3 foot-pounds.
(b) Here all that has changed is that a is 1.5 feet, so
√
◦
Torque D 3/240 sin 120 D 60
3
2
√
D 30 3 foot-pounds.
36. a D 2 in but torque is measured in foot-pounds so a D 1/6 ft.
5
1
Torque D a * F D , 0, 0 * 0, 15, 0 D 0, 0, .
6
2
So Egbert is using 5/2 foot-pounds straight up.
37. From the figure
sin θ D
1
1.5
D
3
2
* θ D π/6.
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“runall” — 2005/8/1 — 12:11 — page 24 — #24
24
Chapter 1
Vectors
This is the angle the seesaw makes with horizontal. The angle we want is
π/6 C π/2 D 2π/3.
Since r D 3 and F D 50, the amount of torque is
T D r * F D r F sin
√
D 3 · 50 ·
2π
3
√
3
D 75 3 ft-lb
2
r
3⬘
1.5⬘
␪
2␲/3
F
F (translated)
38. (a) The linear velocity is v D ω * r so that
v D ω r sin θ.
We have that the angular speed is 2π radians D π radians/hr (this is ω.) Also r D 3960, so at 45◦
24 hrs
12
√
North latitude, v D π · 3960 · sin 45◦ D 330π
L 733.08 mph.
12
2
(b) Here the only change is that θ D 90◦ . Thus v D π · 3960 · sin 90◦ D 330π L 1036.73 mph.
2
39. Archie’s actual experience isn’t important in solving this problem—he could have ridden closer to the center. Since
we are only interested in comparing Archie’s experience with Annie’s, it turns out that their difference would be
the same so long as the difference in their distance from the center remained at 2 inches. The difference in speed
is 331/32π6 − 331/32π4 D 331/32π2 D 4π331/3 D 1331/3π D 400π/3 in/ min.
40. (a) v D ω * r D 0, 0, 12 * 2, −1, 3 D 12, 24, 0 D 12i C 24j.
(b) The height of the point doesn’t change so we can view this as if it were a problem in R2 . When x D 2
and y D −1, we can find the central angle by taking tan−1 −1/2. In one second the angle has moved 12
radians so the new point is
√
√
5 costan−1 −1/2 C 12, 5 sintan−1 −1/2 C 12, 3 L 1.15, −1.92, 3.
41. Consider the rotations of a sphere about each of the two parallel axes pictured below.
W2
W1
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“runall” — 2005/8/1 — 12:11 — page 25 — #25
Section 1.5
Equations for Planes; Distance Problems
25
Assume the two corresponding angular velocity vectors ω1 and ω2 (denoted w1 and w2 in the diagram) are “parallel”
and even have the same magnitude. Let them both point straight up (parallel to (0, 0, 1)) with magnitude 2π radians per
second. The idea is that as “free vectors” ω1 and ω2 are both equal to 0, 0, 2π. but that the corresponding rotational
motions are very different.
In the case of ω1 , each second every point on the sphere has made a complete orbit around the axis. The corresponding
motion is that the sphere is rotating about this axis. (More concretely, take your Vector Calculus book and stand it up on
its end. Imagine an axis anywhere and spin it around that axis at a constant speed.)
In the case of ω2 , each second every point on the sphere has made a complete orbit around the axis. In this case that
means that the corresponding motion is that the sphere is orbiting about this axis. (Hold your Vector Calculus book at
arms length and you spin around your axis.)
1.5 EQUATIONS FOR PLANES; DISTANCE PROBLEMS
1. This is a straightforward application of formulas (1) and (2):
1x − 3 − y C 1 C 2z − 2 D 0
)*
x − y C 2z D 8.
2. Again we apply formula (2):
x − 9 − 2z C 1 D 0
x − 2z D 11.
)*
So what happened to the y term? The equation is independent of y. In the x − z plane draw the line x − 2z D 11
and then the plane is generated by “dragging” the line either way in the y direction.
3. We first need to find a vector normal to the plane, so we take the cross product of two displacement vectors:
3 − 2, −1 − 0, 2 − 5 * 1 − 2, −2 − 0, 4 − 5 D 1, −1, −3 * −1, −2, −1 D −5, 4, −3.
Now we can apply formula (2) using any of the three points:
−5x − 3 C 4y C 1 − 3z − 2 D 0
−5x C 4y − 3z D −25.
)*
4. We’ll again find the cross product of two displacement vectors:
A, −B, 0 * 0, −B, C D −BC, −AC, −AB.
Now we apply formula (2):
−BCx − A − ACy − ABz D 0
BCx C ACy C ABz D ABC.
)*
5. If the planes are parallel, then a vector normal to one is normal to the other. In this case the normal vector is
n D 5, −4, 1. So using formula (2) we get:
5x − 2 − 4y C 1 C z C 2 D 0
5x − 4y C z D 12.
)*
6. The plane contains the line rt D −1, 4, 7 C 2, 3, −1t and the point (2, 5, 0). Choose two points on the
line, for example −1, 4, 7 and (13, 25, 0) and proceed as in Exercises 3 and 4.
−1 − 2, 4 − 5, 7 − 0 * 13 − 2, 25 − 5, 0 D −3, −1, 7 * 11, 20, 0 D −140, 77, −49
D 7−20, 11, −7.
We are just looking for the plane perpendicular to this vector so we can ignore the scalar 7.
−20x − 2 C 11y − 5 − 7z D 0
)*
−20x C 11y − 7z D 15.
7. The only relevant information contained in the equation of the line rt D −5, 4, 7 C 3, −2, −1t is the
vector coefficient of t. This is the normal vector n D 3, −2, −1.
3x − 1 − 2y C 1 − z − 2 D 0
)*
3x − 2y − z D 3.
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“runall” — 2005/8/1 — 12:11 — page 26 — #26
26
Chapter 1
Vectors
8. We have two lines given by the vector equations:
r1 t D 2, −5, 1 C 1, 3, 5t
r2 t D 5, −10, 9 C −1, 3, −2t
The vector 1, 3, 5 * −1, 3, −2 D 21, 3, −6 D 37, 1, −2 is orthogonal to both lines. So the equation
of the plane containing both lines is:
7x − 2 C y C 5 − 2z − 1 D 0
7x C y − 2z D 7.
)*
9. The line shared by two planes will be orthogonal each of their normal vectors. First, calculate: 1, 2, −3 *
5, 5, −1 D 13, −14, −5. Now find a point on the line by setting z D 0 and solving the two equations
x C 2y D 5
5x C 5y D 1
to get x D −23/5 and y D 24/5. The equation of the line is rt D −23/5, 24/5, 0 C 13, −14, −5t, or
in parametric form:


23


 x D 13t − 5
y D −14t C 24

5


 z D −5t.
10. The normal to the plane is n D 2, −3, 5 and the line passes through the point P D 5, 0, 6. The equation of
the line
rt D P C nt D 5, 0, 6 C 2, −3, 5 t.
In parametric form this is:


 x D 2t C 5
y D −3t

 z D 5t C 6.
11. The easiest way to solve this is to check that the vector from the coefficients of the first equation 8, −6, 9A is
a multiple of the coefficients of the second equation (A, 1, 2). In this case the first is −6 times the second. This
means that 8 D −6A or A D −4/3. Checking we see this is confirmed by 9A D −62.
12. For perpendicular planes we check that 0 D A, −1, 1 · 3A, A, −2. This yields the quadratic 0 D 3A2 −
A − 2 D 3A C 2A − 1. The two solutions are A D −2/3 and A D 1.
13. This is a direct application of formula (10):
xs, t D sa C tb C c D s2, −3, 1 C t1, 0, −5 C −1, 2, 7.
In parametric form this is:


 x D 2s C t − 1
y D −3s C 2

 z D s − 5t C 7
14. Again this follows from formula (10):
xs, t D s−8, 2, 5 C t3, −4, −2 C 2, 9, −4
or


 x D −8s C 3t C 2
y D 2s − 4t C 9

 z D 5s − 2t − 4
15. The plane contains the lines given by the equations:
r1 t D 5, −6, 10 C t2, −3, 4, and
r2 t D −1, 3, −2 C t5, 10, 7.
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“runall” — 2005/8/1 — 12:11 — page 27 — #27
Section 1.5
Equations for Planes; Distance Problems
27
So we use formula (10) with the vectors 2, −3, 4 and (5, 10, 7) and either of the two points to get:
xs, t D t2, −3, 4 C s5, 10, 7 C −1, 3, −2
or


 x D 2t C 5s − 1
y D −3t C 10s C 3

 z D 4t C 7s − 2.
16. We need to find two out of the three displacement vectors and use any of the three points:
a D 0, 2, 1 − 7, −1, 5 D −7, 1, −4
and b D 0, 2, 1 − −1, 3, 0 D 1, −2, 1


 x D −7s C t
y D s − 2t C 2
xs, t D s−7, 1, −4 C t1, −2, 1 C 0, 2, 1 or

 −4s C t C 1.
so
17. The equation of the line rt D −5, 10, 9 C t3, −3, 2 immediately gives us one of the two vectors
a D 3, −3, 2. The displacement vector from a point on the line to our given point gives us the vector b D
−5, 10, 9 − −2, 4, 7 D −3, 6, 2. So our equations are:
xs, t D s3, −3, 2 C t−3, 6, 2 C −5, 10, 9
or


 x D 3s − 3t − 5
y D −3s C 6t C 10

 z D 2s C 2t C 9.
18. To convert to the parametric form we will need two vectors orthogonal to the normal direction n D 2, −3, 5
and a point on the plane. The easiest way to find an orthogonal vector is to let one coordinate be zero and find
the other two. For example if the x component is zero then 2, −3, 5 · 0, y, z D −3y C 5z is solved when
y D 5k and z D 3k for any scalar k. In other words, the vectors a D 0, 5, 3 and b D 3, 2, 0 are orthogonal
to n. For a point in the plane 2x − 3y C 5z D 30, set any two of x, y, and z to zero. For example (0, 0, 6) is
in the plane. Our parametric equations are:


 x D 3t
y D 5s C 2t
xs, t D s0, 5, 3 C t3, 2, 0 C 0, 0, 6 or

 z D 3s C 6.
19. We combine the parametric equations into the single equation:
xs, t D s3, 4, 1 C t−1, 1, 5 C 2, 0, 3.
Use the cross product to find the normal vector to the plane:
n D 3, 4, 1 * −1, 1, 5 D 19, −16, 7.
So the equation of the plane is:
19x − 2 − 16y C 7z − 3 D 0
or
19x − 16y C 7z D 59.
−
−'
20. Using method 1 of Example 7, choose a point B on the line, say B D −5, 3, 4. Then BP0 D −5, 3, 4 −
1, −2, 3 D −6, 5, 1, and a D 2, −1, 0. So


−
−'
−
−'  a · BP0 
 a D 2, −1, 0 · −6, 5, 1 2, −1, 0 D −17 2, −1, 0.
proja BP0 D 
a · a
2, −1, 0 · 2, −1, 0
5
The distance is
√
−
−'
−
−'
−17
BP0 − proja BP0 D −6, 5, 1 −
2, −1, 0
D 1/54, 8, 5 D 105/5.
5
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:11 — page 28 — #28
28
Chapter 1
Vectors
21. This time we’ll use method 2 of Example 7. Again choose a point B on the line and a vector a parallel to the
line. The distance is then
DD
−
−'
3, 5, 0 * 7 − 2, −3 C 1, 0
a * BP0 31
D
D √ .
a
3, 5, 0
34
−
−'
For a method 3, you could have solved for an arbitrary point on the line B such that BP0 · a D 0 and then found
−
'
the length of BP0 . In R2 , the calculation is not too bad.
−
−'
22. Using method 1 of Example 7, choose a point B on the line, say B D 5, 3, 8. Then BP0 D −11, 10, 20 −
5, 3, 8 D −16, 7, 12, and a D −1, 0, 7. So


−
−'
−
−'  a · BP0 
 a D −1, 0, 7 · −16, 7, 12 −1, 0, 7 D −2, 0, 14.
proja BP0 D 
a · a
−1, 0, 7 · −1, 0, 7
The distance is
√
−
−'
−
−'
BP0 − proja BP0 D −16, 7, 12 − −2, 0, 14 D −14, 7, −2 D 249.
−
−−'
23. Use Example 9 and for two points B1 D −1, 3, 5 on l1 and B2 D 0, 3, 4 on l2 calculate B1 B2 D 1, 0, −1.
To find the vector n, calculate the cross product n D 8, −1, 0 * 0, 3, 1 D −1, −8, 24.
−
−−'
−
−−'
−1, −8, 24 · 1, 0, −1
n · B1 B2
projn B1 B2 D
nD
−1, −8, 24
n · n
−1, −8, 24 · −1, −8, 24
D−
25
−1, −8, 24.
641
25
25
Finally −
−1, −8, 24
D √ 641 .
641
−
−−'
24. Again, use Example 9 and for two points B1 D −7, 1, 3 on l1 and B2 D 0, 2, 1 on l2 calculate B1 B2 D
7, 1, −2. To find the vector n, calculate the cross product n D 1, 5, −2 * 4, −1, 8 D 38, −16, −21.
−
−−'
−
−−'
38, −16, −21 · 7, 1, −2
n · B1 B2
projn B1 B2 D
nD
38, −16, −21
n · n
38, −16, −21 · 38, −16, −21
292
38, −16, −21.
2141
292
292
38, −16, −21
D √
.
Finally 2141
2141
25. (a) Again, use Example 9 with the two points B1 D 4, 0, 2, and B2 D 2, 1, 3 and normal vector n D
−
−−'
−
−−'
3, 1, 2 * 1, 2, 3 D −1, −7, 5. The displacement vector is B1 B2 D −2, 1, 1. Note that B1 B2 is
−
−−'
orthogonal to n and so the projection projn B1 B2 D 0 (if you’d like, you can go ahead and calculate this)
and so the lines are distance 0 apart.
(b) This means that the lines must have a point in common (that they intersect at least once). The lines are not
parallel so they have exactly one point in common (i.e., they aren’t the same line).
26. (a) The shortest distance between a point P0 and a line l is a straight line that meets P0 orthogonally. If we
have two nonparallel lines then we can use the cross product to find the one direction n that is orthogonal to
each. The shortest segment between two lines will meet each orthogonally, for two skew lines l1 and l2 the
line that joins them at these closest points will be parallel to n.
If instead l1 is parallel to l2 we get a whole plane’s worth of orthogonal directions. We have no way of
choosing a unique vector n that is used in the calculation.
O.K., that’s why we can’t use the method of Example 9. What can we do instead?
D
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“runall” — 2005/8/1 — 12:11 — page 29 — #29
Section 1.5
Equations for Planes; Distance Problems
29
(b) Fix a point on l1 , say P1 D 2, 0, −4. Then as we saw in an earlier exercise, the distance from P1 to an
arbitrary point P2 D 1 C t, 3 − t, −5 C 5t on l2 is
−
−−'
P1 P2 D t − 12 C 3 − t2 C −1 C 5t2 D 27t 2 − 18t C 11.
−
−−'
−
−−'
P1 P2 is minimized when P1 P2 2 is minimized. This is at the vertex of the parabola, when 54t − 18 D 0
or t D 1/3. At this point the distance is
√
√
√
271/32 − 181/3 C 11 D 3 − 6 C 11 D 8 D 2 2.
−
−−'
Note: In Exercises 27–29 we could just cut to the end of Example 8 and realize that the length of projn P1 P2 D
−
−−'
|n · P1 P2 |
. Instead we will stay true to the spirit of the examples and follow the argument through.
n
27. These planes are parallel so we can use Example 8. The point P1 D 1, 0, 0 is on plane one and the point
−
−−'
P2 D 8, 0, 0 is on plane two. We project the displacement vector P1 P2 D 7, 0, 0 onto the normal direction
n D 1, −3, 2:
−
−−'
projn P1 P2 D
7, 0, 0 · 1, −3, 2
7
1
1, −3, 2 D
1, −3, 2 D 1, −3, 2.
1, −3, 2 · 1, −3, 2
14
2
√
−
−−'
So the distance is projn P1 P2 D 14/2.
28. These planes are also parallel. We choose point P1 D 0, 0, 6 on plane one and P2 D 0, 0, −2 on plane two.
−
−−'
The displacement vector is therefore P1 P2 D 0, 0, −8, and the normal vector is n D 5, −2, 2. So
−
−−'
0, 0, −8 · 5, −2, 2
−16
5, −2, 2 D
5, −2, 2.
projn P1 P2 D
5, −2, 2 · 5, −2, 2
33
−
−−'
The distance is projn P1 P2 D √16 .
33
29. As in Exercises 23 and 24 we’ll choose a point P1 D D1 /A, 0, 0 on plane one and P2 D D2 /A, 0, 0 on
plane two. The displacement vector is
−
−−'
D − D1
P1 P2 D 2
, 0, 0 .
A
A vector normal to the plane is n D A, B, C.


D2 −D1
,
0,
0
·
A,
B,
C

−
−−' 
D2 − D1


A



A, B, C.
projn P1 P2 D 


 A, B, C D 2

A, B, C · A, B, C
A C B2 C C 2
The distance between the two planes is:
−
−−'
projn P1 P2 D
|D2 − D1 |
|D2 − D1 |
A, B, C D √
.
A2 C B2 C C2
A2 C B2 C C2
30. (a) Plane one is normal to n1 D 9, −5, 9 * 3, −2, 3 D 3, 0, −3 while plane two is normal to n2 D
−9, 2, −9 * −4, 7, −4 D 55, 0, −55. So n1 D 3/55n2 , i.e. they normal vectors are parallel so
the planes are parallel.
−
−−'
(b) We’ll use the two points in the given equations to get the displacement vector P1 P2 D 8, −4, 12, and the
normal vector n D 3, 0, −3. So
−
−−'
8, −4, 12 · 3, 0, −3
−12
projn P1 P2 D
3, 0, −3 D
3, 0, −3.
3, 0, −3 · 3, 0, −3
18
−
−−'
The distance is projn P1 P2 D √12 D
18
√
12
4
√ D √ D 2 2.
3 2
2
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:11 — page 30 — #30
30
Chapter 1
Vectors
31. This exercise follows immediately from Exercise 29 (and can be very difficult without it). Here A D 1, B D
3, C D −5 and D1 D 2. The equation in 29 becomes:
3D So
|2 − D2 |
|2 − D |
D √ 2 .
2
2
2
35
1 C 3 C −5
√
3 35 D |2 − D2 | or
√
2 − D2 D ;3 35.
√
So D2 D 2 ; 3 35 and the equations of the two planes are:
√
x C 3y − 5z D 2 ; 3 35.
32. The lines are parallel, so the distance between them is the same as the distance between any point on one of the
lines and the other line. Thus take b2 —the position vector of a point on the second line—and use Example 7.
a * b2 − b1 Then D D
.
a
a
b2
a
b1
−'
33. We have AB D b − a.
D D projn b − a D
D
|n · b − a|
n
n2
|n · b − a|
.
n
B
n
A
(As for the motivation, consider Example 8 with A as P1 , B as P2 .)
34. The parallel planes have equations n · x − x1 D 0 and n · x − x2 D 0. The desired distance is given by
−
−−'
−
−−'
projn P1 P2 where Pi is the point whose position vector is xi . Thus P1 P2 D x2 − x1 so
−
−−'
|n · x2 − x1 |
n
projn P1 P2 D
n2
D
|n · x2 − x1 |
.
n
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“runall” — 2005/8/1 — 12:11 — page 31 — #31
Section 1.6
Some n-dimensional Geometry
31
35. By letting t D 0 in each vector parametric equation, we obtain b1 , b2 as position vectors of points B1 , B2 on the
−
−−'
respective lines. Hence B1 B2 D b2 − b1 . A vector n perpendicular to both lines is given by n D a1 * a2 . Thus
−
−−'
−
−−'
−
−−'
|n · B1 B2 |
|n · B1 B2 |
n D
D D projn B1 B2 D
n2
n
D
|a1 * a2 · b2 − b1 |
.
a1 * a2 1.6 SOME n-DIMENSIONAL GEOMETRY
1. (a) 1, 2, 3, ... , n D 1, 0, 0, ... , 0 C 20, 1, 0, 0, ... , 0 C · · · C n0, 0, 0, ... , 0, 1 D e1 C 2e2 C 3e3 C
· · · C nen .
(b) 1, 0, −1, 1, 0, −1, ... , 1, 0, −1 D e1 − e3 C e4 − e6 C e7 − e9 C · · · C en−2 − en .
2. e1 C e2 C · · · C en D 1, 1, 1, ... , 1.
3. e1 − 2e2 C 3e3 − 4e4 C · · · C −1nC1 nen D 1, −2, 3, −4, ... , −1nC1 n.
4. e1 C en D 1, 0, 0, ... , 0, 1.
5. (a) a C b D 1 C 2, 3 − 4, 5 C 6, 7 − 8, ... , 2n − 1 C −1nC1 2n D 3, −1, 11, −1, 19, −1, ... , 2n −
4n − 1 if n is odd, and
1 C −1nC1 2n. The nth term is −1
if n is even.
(b) a − b D 1 − 2, 3 C 4, 5 − 6, 7 C 8, ... , 2n − 1 − −1nC1 2n D −1, 7, −1, 15, −1, ... , 2n −
4n − 1 if n is even, and
1 − −1nC1 2n. The nth term is −1
if n is odd.
(c) −31, 3, 5, 7, ... , 2n
− 1 D −3, −9, −15, −21, ... , −6n C 3.
√
(d) a D a · a D 12 C 32 C 52 C · · · C 2n − 12 .
(e) a · b D 12 C 3−4 C 56 C · · · C 2n − 1−1nC1 2n D 2 − 12 C 30 − 56 C · · · C
−1nC1 2n2n − 1.
6. We want to show that a C b … a C b. Here n is even and a and b are vectors in Rn ,
a D 1, 0, 1, 0, ... , 0
b D 0, 1, 0, 1, ... , 1, and
a C b D 1, 1, 1, 1, ... , 1.
√
√
a C b D 12 C 12 C · · · C 12 D n D 2 n/4 … 2 n/2 D 2 12 C 12 C · · · C 12 D a C b.
√
n times
n/2 times
7. First we calculate
a D 12 C 22 C 32 C · · · C n2 D
b D
√
12 C 12 C · · · C 12 D
√
nn C 12n C 1
6
n, and
n times
|a · b| D 1 C 2 C 3 C · · · C n D




√
nn
C
12n
C
1
2n
C
12n
C
1






 n D n 



a b D 


.

6
2
3
So
For n D 1,
For n D 2,
nn C 1
2
2nC12nC1
D 2 D n C 1.
3
√
2nC12nC1
D 10 Ú 3 D n C 1.
3
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“runall” — 2005/8/1 — 12:11 — page 32 — #32
32
Chapter 1
Vectors
For n Ú 3,


2n
C
12n
C
1
n 2n C 1

n
n 



Ú n C 1 D |a · b|.

 Ú √

2
3
2
3
2
8. As always,
proja b D D
a · b
2 − 5 C 27 − 2
22
aD
a
a D
a · a
1 C 1 C 49 C 9 C 4
64
11 −11 77 11
11
1, −1, 7, 3, 2 D ,
, , .
32
32 32 32 16
9. This is just the triangle inequality:
a − b D a − c C c − b … a − c C c − b.
10. We are given that c D a C b so
c2 D a C b2 D a C b · a C b D a · a C a · b C b · a C b · b.
But a · b D 0 D b · a, so
a · a C a · b C b · a C b · b D a · a C b · b D a2 C b2 .
This is analogous to the Pythagorean Theorem. Here a and b are playing the role of the legs. They are orthogonal
vectors. The third side of the triangle is a C b D c. The theorem in this case says that the sum of the squares of
the lengths of the “legs” is the square of the length of the “hypotenuse”.
11. We have
a C b D a − b * a C b2 D a − b2
* a C b · a C b D a − b · a − b.
Expand to find
a · a C 2a · b C b · b D a · a − 2a · b C b · b
* 4a · b D 0,
so a and b are orthogonal.
12. As above, if a − b > a C b, then −2a · b > 2a · b so −4a · b > 0 3 a · b < 0. Thus cos θ D
Hence π < 0 … π.
2
13. The equation could also be written in the more suggestive form:
a · b
< 0.
a b
2, 3, −7, 1, −5 · [x1 , x2 , x3 , x4 , x5 − 1, −2, 0, 4, −1] D 0.
These are the points in R5 so that x1 , x2 , x3 , x4 , x5 − 1, −2, 0, 4, −1 is orthogonal to the vector
2, 3, −7, 1, −5. This is the four dimensional hyperplane in R5 orthogonal to 2, 3, −7, 1, −5 containing
the point 1, −2, 0, 4, −1.
14. Half of each type of your inventory gives T-shirts in quantities of 10, 15, 12, 10 (in order of lowest to highest
selling price). Half of each type of your friend’s inventory gives 15, 8, 10, 14 baseball caps. The value of your
half of the inventory is
8, 10, 12, 15 · 10, 15, 12, 10 D $524.
The value of your friend’s inventory is
8, 10, 12, 15 · 15, 8, 10, 14 D $530.
Thus your friend might be reluctant to accept your offer, unless he’s quite a good friend.
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“runall” — 2005/8/1 — 12:11 — page 33 — #33
Section 1.6
Some n-dimensional Geometry
33
15. (a) We have
p D 200, 250, 300, 375, 450, 500
Total cost D p · x D 200x1 C 250x2 C 300x3 C 375x4 C 450x5 C 500x6
(b) With p as in part (a), the customer can afford commodity bundles x in the set
{x ∈ R6 |p · x … 100,000}.
The budget hyperplane is p · x D 100,000 or 200x1 C 250x2 C 300x3 C 375x4 C 450x5 C 500x6 D 100,000.
16.
3A − 2B D 3 1
−2
3
−4
− 2
1 0
2
0
9
−8
− 3 0
D
3
−6
6
0
D
11
−6
−12
−6
18
6
9
3
5
0 10
0 −1
3 17.
1
AC D −2
2
0

1
3 
2
1
0

0
7 
−2
−1
0
3
D
11 C 22 C 30
−21 C 02 C 10
D
5
−2
8
5
1−1 C 20 C 33
−2−1 C 00 C 13
10 C 27 C 3−2
−20 C 07 C 1−2 8
.
−2 18.
DB D 1
2
0
−4
−3 0
D
1−4 C 00
2−4 − 30
D
−4
−8
9
9
9
3
5
0 19 C 03
29 − 33
15 C 00
25 − 30 5
.
10 19.


−4 0
1
0
BT D D  9 3  2 −3 5 0


−41 C 02 −40 C 0−3
90 C 3−3 
D  91 C 32
51 C 02
50 C 0−3


−4
0
D  15 −9 
5
0
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“runall” — 2005/8/1 — 12:11 — page 34 — #34
34
Chapter 1
Vectors
20. (a)

I2 D 1
0
1
0
, I3 D  0
1
0
0
1
0

0
0 ,
1

1
 0
and I4 D 
 0
0
0
1
0
0
0
0
1
0

0
0 
.
0 
1
(b) The ijth entry of the product of matrices A and B is the product of the ith row of A and the jth column of
B. So in case i. we have:
AIn ij D [ai1
In case ii. we have:
ai2
ai3
...
ain ]ej T D ai1 , ai2 , ai3 , ... , ain · ej D aij .


a1j





a2j 











a


3j 









D ei · a1j , a2j , a3j , ... , anj D aij .
In Aij D ei 






#








# 







 # 
anj
In both cases we’ve shown that the ij th component of the product is the ij th component of matrix A, so
AIn D A D In A.
21. We’ll expand on the first row:
7
0 −1
0 0 1
2
3 0
3 2
0
1
3 2 − 1 −3
2 D 7 −3 0
1 −3
0
2 5 1 −2 0
5 −2 0
5
1 −2 −3
−3 0 −3 2 1 −3 2 D 7 −1 C 3
− 2
C 3
5 −2 5 1 5 −2 0
5 D 7−1−4 C 3−3 − 2−4 C 35 D −42.
Note: Exercises 22 and 23 are good exploration problems for students before they’ve done Exercise 25.
22. Note that if we expand along the first row only one term survives. If at each step we expand along the first row,
the pattern continues. What we are left with is the product of the elements along the diagonal.
8 0
0 0 1
0 0 15 1
0
0
−7 6 −1 0 D 8 6 −1 0 1
9
7
8 1
9 7 −1 0 D 81 9 7 D 81−17 D −56.
23. This is similar to Exercise 22. Either we could expand along the last row of each matrix at each step or we could
expand along the first column at each step. It is easier to keep track of signs if we choose this second approach.
We again find that the determinant is the product of the diagonal elements.
5 −1 0
8 11 2 1
9
7 0
2 1
9
7 0 4 −3
5 0
0 4 −3
5 D 5 2
1 0
0 0
0 0
2
1 0 0
0
−3
0
0 0
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“runall” — 2005/8/1 — 12:11 — page 35 — #35
Section 1.6
Some n-dimensional Geometry
35
4
D 52 0
0
−3
5 2
1 0 −3 2
1 D 524 0 −3 D 5242−3 D −240.
24. There really isn’t anything to show. Using the convenient fact provided after Example 8:
• If row i consists of all zeros (i.e., aij D 0 for 1 … j … n) then expand along row i. Using the cofactor
notation:
|A| D −1iC1 ai1 |Ai1 | C −1iC2 ai2 |Ai2 | C · · · C −1iCn ain |Ain |
D −1iC1 0|Ai1 | C −1iC2 0|Ai2 | C · · · C −1iCn 0|Ain | D 0.
• If column j consists of all zeros (i.e., aij D 0 for all 1 … i … n) then expand along column j. As above
we get
|A| D −11Cj a1j |A1j | C −12Cj a2j |A2j | C · · · C −1nCj anj |Anj |
D −11Cj 0|A1j | C −12Cj 0|A2j | C · · · C −1nCj 0|Anj | D 0.
25. (a) A lower triangular matrix is an n * n matrix whose entries above the main diagonal are all zero. For
example the matrix in Exercise 22 is lower triangular.
(b) If we expand the determinant of an upper triangular matrix along its first column we get:
|A| D −11C1 a11 |A11 | C −12C1 a21 |A21 | C · · · C −1nC1 an1 |An1 |
D −11C1 a11 |A11 | C −12C1 0|A2j | C · · · C −1nC1 0|Anj | D a11 |A11 |.
Looking back on what we have found: The determinant of an upper triangular matrix is equal to the term in the
upper left position multiplied by the determinant of the matrix that’s left when the top most row and left most
column are removed. Each time we remove the top row and left column we are left with an upper triangular
matrix of one dimension lower. Repeat the process n times and it is clear that
|A| D a11 |A11 | D a11 a22 |A11 11 | D · · · D a11 a22 a33 · · · ann .
26. (a) Type I Rule: If matrix B results from matrix A by exchanging rows i and j then |A| D −|B|.
As one example,
1 0 0 0 1 0 0 1 0 D 1, while 1 0 0 D −1.
0 0 1 0 0 1 A more important example is
a1 a2 D a1 b2 − a2 b1 D −b1 a2 − b2 a1 D b1
a1
b1 b2 b2 .
a2 The reason this second example is more important is that you can always expand the determinants of A and
B so that you are left with a sum of scalars times the determinants of 2 by 2 matrices involving only the two
rows being switched. Since the scalars will be the same in both cases, this second example shows that the
effect of switching rows i and j is to switch the sign of every component in the sum and so |A| D −|B|.
(b) Type III Rule: If matrix B results from matrix A by adding a multiple of row i to row j and leaving row i
unchanged then |A| D |B|.
As one example,
1 0 0 1 1 0 0 1 0 D 1, and also 0 1 0 D 1.
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“runall” — 2005/8/1 — 12:11 — page 36 — #36
36
Chapter 1
Vectors
To see what’s going on, let’s look at the example
a1 C nb1 a2 C nb2 D a1 C nb1 b2 − a2 C nb2 b1 D a1 b2 − a2 b1 C nb1 b2 − b2 b1 b1
b2
D a1 b2 − a2 b1 .
Another way to look at the example above is to see that the determinant splits into two pieces:
b b a a a a 1
2
1
2
1
2
a1 b2 − a2 b1 C nb1 b2 − b2 b1 D b b C n b b D b b .
1
2
1
2
1
2
Note: A more general case of this rule will be proved in Exercise 28.
(c) Type II Rule: If matrix B results from matrix A by multiplying the entries in the ith row of A by the scalar
c then |B| D c|A|.
We will prove this by expanding the determinant for B along the ith row. Because row i is the only one
changed, the cofactors Bij are the same as the cofactors Aij .
|B| D −1iC1 bi1 |Bi1 | C −1iC2 bi2 |Bi2 | C · · · C −1iCn bin |Bin |
D −1iC1 cai1 |Ai1 | C −1iC2 cai2 |Ai2 | C · · · C −1iCn cain |Ain |
D c−1iC1 ai1 |Ai1 | C −1iC2 ai2 |Ai2 | C · · · C −1iCn ain |Ain | D c|A|.
27. Here we go: at each step we’ll specify what we’ve done.
2 1 −2
2 1 −2
7
8 7
8 1 0
−1 1
1 −2
4 2
3 −5 switched rows 2
−1 1
D −1 1 0
2
3
−5
1
−2
4 and 3
0 2
0 2
3
1
7
3
1
7
−3 2 −1
−3 2 −1
0
1 0
1 2 1 −2
7
8 0 1
3
1 −1 ( row 2 C row 3
1 −2
4 D −1 1 0
0 2
3
1
7 0 2
2 −6 13 ( row 5 C 3row 3
2 1 −2
7
8 0 1
3
1 −1 −1 2
0
2
−4
8 ( 2row 3
D
2 0 2
3
1
7 0 2
2 −6 13 2
1 −2
7
8 0
1
3
1 −1 −1 4 −11
0 ( row 3 − row 1
D
0 −1
2 0
2
3
1
7
0
2
2
−6 13 2 1 −2
7
8 0 1
3
1 −1 −1 7 −10 −1 ( row 3 C row 2
D
0 0
2 0 0 −3
−1
9 ( row 4 − 2row 2
0 0 −4
−8 15 ( row 5 − 2row 2
2 1
−2
7
8 0 1
3
1 −1 −1
0 0
7
−10
−1 D
277 0 0 −21
−7
63 ( 7row 4
0 0 −28 −56 105 ( 7row 5
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“runall” — 2005/8/1 — 12:11 — page 37 — #37
Section 1.6
2
0
−1
0
D
277 0
0
Some n-dimensional Geometry
37
8 −1 −1 60 101 1 −2
7
1
3
1
0
7 −10
0
0 −37
( row 4 C 3row 4
0
0 −96
( row 5 C 4row 3
2 1 −2
7
8 0 1
3
1
−1 −1
0 0
7
−10
−1
D
277−37 0 0
0
−37
60
0 0
0 3552 −3737 ( −37row 5
2 1 −2
7
8 0 1
3
1
−1 −1
0 0
7
−10
−1
D
277−37 0 0
0
−37
60
0 0
0
0 2023 ( row 5 C 96row 4
D
−1
2023
217−372023 D
D −289.
277−37
7
1 0
0 0
and B D then 1 D detA C B but detA D detB D 0. So in
0 0 0 1 general detA C B Z detA C detB.
1
1
1
2
7
2 7 2 7 1 5 C 2 −1 1 D −32 − 26 D −58. It
(b) 3 C 2 1 − 1 5 C 1 D −58, while 3
0 −2 0 0 −2 0 0
−2
0
makes sense that these should be equal; if you imagine expanding on the second row we see that
1
2
7
3 C 2 1 − 1 5 C 1 D 3 C 2 2 7 C 1 − 1 1 7 C 5 C 1 1 7 −2 0 0 0 0 0 0
−2
0
1 7 2 7 1 7 1 7 2 7 1 7 − C 1
C 2
C C 5
D 3
0 0 0 0 0 0 −2 0 −2 0 0 0 1
1
2 7 2 7 1 5 C 2 −1 1 .
D 3
0 −2 0 0 −2 0 28. (a) If you let A D 1
(c) 0
−1
3
4
0
1
2 C 3 −1 C 5 D 0, while 0
−1
0 − 2 3
4
0
1
2 −1 C 0
−1
0 3
4
0
3 5 D −11 C 11 D 0.
−2 (d) We might characterize the rules for rows as follows:
Let A, B and C be three matrices whose elements are the same except for those in row i where cij D aij C bij
for 1 … j … n. Then detC D detA C detB. We prove this by expanding the determinant along row i
noting that in that case the cofactors for all three matrices are equal (i.e., Aij D Bij D Cij for 1 … j … n):
|C| D −1iC1 ci1 |Ci1 | C −1iC2 ci2 |Ci2 | C · · · −1iCn cin |Cin |
D −1iC1 ai1 C bi1 |Ci1 | C −1iC2 ai2 C bi2 |Ci2 | C · · · −1iCn ain C bin |Cin |
D −1iC1 ai1 |Ci1 | C −1iC2 ai2 |Ci2 | C · · · −1iCn ain |Cin |
C −1iC1 bi1 |Ci1 | C −1iC2 bi2 |Ci2 | C · · · −1iCn bin |Cin |
D −1iC1 ai1 |Ai1 | C −1iC2 ai2 |Ai2 | C · · · −1iCn ain |Ain |
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“runall” — 2005/8/1 — 12:11 — page 38 — #38
38
Chapter 1
Vectors
C −1iC1 bi1 |Bi1 | C −1iC2 bi2 |Bi2 | C · · · −1iCn bin |Bin |
D |A| C |B|
The rule for columns is exactly the same:
Let A, B and C be three matrices whose elements are the same except for those in column j where cij D
aij C bij for 1 … i … n. Then detC D detA C detB. We could prove this by expanding the determinant
along column j just as above. Instead note that AT , BT , and CT satisfy the above rule for rows and that the
determinant of a matrix is equal to the determinant of its transpose. Our proof is then:
|C| D |CT | D |AT | C |BT | D |A| C |B|.
29. This is a pretty cool fact. If AB and BA both exist, these two matrices may not be equal. It doesn’t matter. They
still have the same determinant. The proof is straightforward: detAB D det AdetB D det Bdet A D
det BA.
30. (a) Check the products in both directions ...
1
1
0
1
1 −1
0
1 C 0
D
1 1 − 1
D
1 C 0
−1 C 1
0 C 0
1
D
0 C 1 0
0
1 0 C 0
1
D
0 C 1 −1
0
1
1 1
0
.
1 (b) Again, the products in both directions yield the identity matrix:


 

1 2 3
−40 16 9
−40 C 26 C 15 16 − 10 − 6 9 − 6 − 3
 2 5 3   13 −5 −3  D  −80 C 65 C 15 32 − 25 − 6 18 − 15 − 3 
1 0 8
5 −2 −1
−40 C 0 C 40 16 C 0 − 16 9 C 0 − 8



 
−40 C 32 C 9 −80 C 80 C 0 −120 C 48 C 72
−40 16 9
1 2 3
39 − 15 − 24  .
D  13 −5 −3   2 5 3  D  13 − 10 − 3 26 − 25 C 0
1 0 8
5 − 4 − 1
10 − 10 C 0
15 − 6 − 8
5 −2 −1
31. Say the given matrix is A. Then the top left entry in the inverse must be 1/2 because 1 is the top left entry of the
product of A−1 A and it is twice the top left entry in the inverse matrix.
Looking at the second row of A, in the product AA−1 it “picks out” the element in the second row. This means
that the second row of A−1 is (0, 1, 0). Similarly, the third row of A picks out the opposite of the element in the
third row in the product AA−1 so the third row of A−1 is 0, 0, −1.
The third column of A tells us that the first and third elements of the top row of A−1 must be the same. The
final element to solve for is the middle element of the top row of A−1 . It must be the opposite of the middle
element of the third row of A−1 . Putting this information together, we have that


1/2 −1 1/2
1
0 
A−1 D  0
0
0
−1
32. Since the first column is 0, the determinant is 0. This means that the matrix could not have an inverse. We’ll
actually show this in Exercise 35 below. Say, for a minute that you don’t accept the results of Exercise 35 and
you think you have found an inverse matrix A−1 for the given matrix A. Then look at the product A−1 A. It
should be the identity matrix I3 but the first column of the product will be all 0’s. For this reason, no inverse for
A could exist.
33. Using the hint, assume that A has two inverses B and C. Then
B D BI D BAC D BAC D IC D C.
34. We just verify that B−1 A−1 behaves as an inverse:
B−1 A−1 AB D B−1 A−1 AB D B−1 In B D B−1 B D In
ABB−1 A−1 D ABB−1 A−1 D AIn A−1 D AA−1 D In
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:11 — page 39 — #39
Section 1.6
Some n-dimensional Geometry
39
35. (a) If A is invertible, consider the product AA−1 D I. By the formula in Exercise 29, det Adet A−1 D
detAA−1 D det I D 1. From this we see that det A Z 0. In fact, we see more − the results of part
(b) follow immediately.
(b) See part (a).
36. (a)
d −b
ad − bc
0
a b
1
1
D I2
D
a c d 0
ad − bc ad − bc −c
ad − bc
a
c
(b)
2
−1

2
37. If A D  0
1

2
38. If A D  1
3
1
2
0
−1
2
0
b
d
1
·
d ad − bc −c
−1
4
2 
D
−b
ad − bc
1
D
a 0
ad − bc
2
1
2 · 2 − 4−1 1
−4
1 2
D 2 8 1
0
D I2
ad − bc −4
1/4
D
2 1/8
−1/2
1/4 1
4 , then det A D 12 C 4 − 2 D 14, so the formula gives
3


2 4 1 1 − 1 1 
0 3 0 3 2 4 




0
4
2
1
2
1

1
−

A−1 D
 − 
1
3
1
3
0
4


14 

0 2 2 1 2 1 

1 0 − 1 0 0 2 

3
1


−3

7
14
7 
6 −3
2

1 
4 
5
2

−
4
5 −8  D 
D

7
14
7 
14 −2


1
4
1
2
−1
7

14
7
3
−2 , then det A D 4 C 6 − 18 C 1 D −7, so the formula gives
1
 
2 −2 −1 3 −1
3 −
 0
0 1 2 −2 
1 


2 3 3 
1

1 −2 − 2
A−1 D − 
−
3
1 −2 
7
 1
3 1 
2 −1 2 −1 
 1 2 − 3 0 1
3
0 2 

4


−2 −1
7
7
7 
2
1 −4



1
1
1 −1 
7 D
D −  −7 −7


7 −6 −3
5
3
6
−5
7
7
7
39. We’ll transform the cross product into a determinant. To make the determinant easier to calculate we’ll replace
the fourth row with the sum of the fourth row and five times the second row. Finally we’ll expand along the
first column.
e1 e2 · · · en 1 2 −1 3 1, 2, −1, 3 * 0, 2, −3, 1 * −5, 1, 6, 0 D 0 2 −3 1 −5 1
6 0  2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:11 — page 40 — #40
40
Chapter 1
Vectors
e1
1
D 0
0
e2
2
2
11
···
−1
−3
1
en 3 1 15 ( row 4 C 5row 2
2
D e1 2
11
−1
−3
1
e2
3 1 − 2
11
15 e3
−3
1
e4 1 15 D 32e1 C 46e2 C 19e3 − 35e4 D 32, 46, 19, −35.
40. (a) We use the matrix form to write the cross product as a determinant. We then switch row i C 1 (the row
consisting of ai1 , ai2 , ... , ain ) with row j C 1 (the row consisting of aj1 , aj2 , ... , ajn ) which multiplies the
determinant by −1:
e2
···
en
e1
a
a12
···
ain
11
#
#
#
#
#
#
#
#
#
#
#
#
a
a
·
·
·
a
i1
i2
in
a1 * · · · * ai * · · · * aj * · · · * an−1 D #
#
#
#
#
#
#
#
#
#
#
#
aj1
aj2
···
ajn
#
#
#
#
#
#
ô
#
#
#
an−11 an−12 · · · an−1n e2
···
en
e1
a
a12
···
a1n
11
#
#
#
#
#
#
#
#
#
#
#
#
( Switch this row i C 1
a
a
·
·
·
a
i1
i2
in
D −1 #
#
#
#
#
#
#
#
#
#
#
#
( with this row j C 1
aj1
a
·
·
·
a
j2
jn
#
#
#
#
#
#
ô
#
#
#
an−11 an−12 · · · an−1n D −a1 * · · · * aj * · · · * ai * · · · * an−1 (b) Again we will change to the matrix form and then use the rule for the row operation of type II to pull the
scalar k out and then rewrite as a cross product.
e2
...
en
e1
a11
a12
...
a1n
#
#
#
#
#
#
#
#
#
#
#
#
a1 * · · · * kai * · · · * an−1 D kai2
...
kain kai1
#
#
#
#
#
#
ô
#
#
#
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:11 — page 41 — #41
Section 1.6
e1
a11
#
#
#
D k a
i1
#
#
#
an−11
e2
a12
...
...
#
#
#
ai2
#
#
#
...
#
#
#
ô
an−12
...
en
a1n
#
#
#
ain
#
#
#
an−1n Some n-dimensional Geometry
41
( row i C 1 divided by k
D ka1 * · · · * ai * · · · * an−1 (c) Once again, we will change to the matrix form. This time we will use the rule we developed in Exercise 28
to write this as two determinants. Finally we will convert each back to the cross product form.
e2
...
en
e1
a11
a12
...
a1n
#
#
#
#
#
#
#
#
#
#
#
#
a1 * · · · * ai C b * · · · * an−1 D a
C
b
a
C
b
...
a
C
b
i1
1
i2
2
in
n #
#
#
#
#
#
ô
#
#
#
an−11
an−12
...
an−1n e2
...
en
e2
...
en
e1
le1
a11
a
...
a
a
a
...
a
12
1n
11
12
1n
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
#
C D ai1
b1
ai2
...
ain
b2
...
bn
#
#
#
#
#
#
#
#
#
#
#
#
ô
ô
#
#
#
#
#
#
an−11 an−12 ... an−1n an−11 an−12 ... an−1n D a1 * · · · * ai * · · · * an−1 C a1 * · · · * b * · · · * an−1 (d) Expand the determinant along the first row; we’ll refer to the cross product matrix as C:
e2
...
en
e1
a11
a
...
a
12
1n
b · |C| D b · a1 * · · · * ai * · · · * an−1 D b · #
#
#
#
#
#
ô
#
#
#
an−11 an−12 ... an−1n D b · e1 |C11 | − e2 |C12 | C · · · C −11Cn en |C1n |
b1
a11
D b1 |C11 | − b2 |C12 | C · · · C −11Cn bn |C1n | D #
#
#
an−11
b2
a12
···
···
#
#
#
ô
an−12
···
#
#
#
an−1n bn
a1n
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“runall” — 2005/8/1 — 12:11 — page 42 — #42
42
Chapter 1
Vectors
41. This follows immediately from part (d) of Exercise 28. For 1 … i … n − 1,
ai2
···
ai1
a11
a
···
12
ai · a1 * · · · * ai * · · · * an−1 D #
#
#
#
ô
#
#
an−11 an−12 · · ·
.
#
#
#
an−1n ain
a1n
Replace the first row with the difference between row 1 and row i C 1 and you will get (by Exercise 26) a matrix
with the same determinant, namely:
0
0
···
0
a11
a12
···
a1n D 0.
#
#
#
#
#
#
ô
#
#
#
an−11 an−12 · · · an−1n Therefore b is orthogonal to ai for 1 … i … n − 1.
42. To find the normal direction n we’ll take the cross product of the displacement vectors:
−
−−'
P0 P1 D 2, −1, 0, 0, 5 − 1, 0, 3, 0, 4 D 1, −1, −3, 0, 1
−
−−'
P0 P2 D 7, 0, 0, 2, 0 − 1, 0, 3, 0, 4 D 6, 0, −3, 2, −4
−
−−'
P0 P3 D 2, 0, 3, 0, 4 − 1, 0, 3, 0, 4 D 1, 0, 0, 0, 0
−
−−'
P0 P4 D 1, −1, 3, 0, 4 − 1, 0, 3, 0, 4 D 0, −1, 0, 0, 0
We take the cross product which is the determinant (expand along the fourth row, and then along the last row):
e1
e2
e3 e4
e5 e3 e4
e5 e2
1 −1 −3 0
1
e5 e3 e4
−1 −3 0
1 6
0 −3 2 −4 D 1 D −3 0
0 −3 2 −4 1
0
0 0
0 −3 2 −4 −1
0 0
0 0 −1
0 0
0 D −2e3 − 15e4 − 6e5 D −0, 0, 2, 15, 6.
We can choose any of the points, say P0 to find the equation of the hyperplane:
2x3 − 3 C 15x4 C 6x5 − 4 D 0
or
2x3 C 15x4 C 6x5 D 30.
1.7 NEW COORDINATE SYSTEMS
In Exercises√
1–3 use equations
x D r cos θ and y D r√
sin θ.
√ (1)√
1. x D √ 2 cos π/4 D √2 2/2
D
1,
and
y
D
2 sin π/4
coordinates
√
√ D 1. The rectangular
√
√ are (1, 1).
2. x D 3 cos 5π/6 D √3− 3/2 D −3/2, and y D 3 sin 5π/6 D 31/2 D 3/2. The rectangular
coordinates are −3/2, 3/2.
3. x D 3 cos 0 D 31 D 3, and y D 3 sin 0 D 0. The rectangular coordinates are (3, 0).
In Exercises 4–6 use equations (2) r2 D x2 C y2 , and tan θ D y/x.
√
√
√
4. r2 D 2 32 C 22 D 16, so r D 4. Also, tan θ D 2/2 3 D 1/2/ 3/2. Since we are in the first quadrant
the polar coordinates are 4, π/6. √
5. r2 D −22 C 22 D 8,√so r D 2 2. Also, tan θ D 2/−2 D −1. Since we are in the second quadrant the
polar coordinates are 2 2, 3π/4. √
6. r2 D −12 C −22 D 5, so r D 5. Also, tan θ D −2/−1 D 2. If the point were in√the first quadrant,
then the angle would be tan−1 2. Since we are in the third quadrant the polar coordinates are 5, π C tan−1 2.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:11 — page 43 — #43
Section 1.7
New Coordinate Systems
43
Exercises 7–9 involve exactly the same idea as Exercises 1–3. Since the z coordinates are the same again we use
equations (1) or (3).
7. Here there’s nothing to do; the rectangular coordinates are (2 cos 2, 2 sin 2, 2).
8. x D π cos π/2 D π0, y D π sin π/2 D π1, and
√ z D 1. The rectangular coordinates are 0, π, 1.
3/2, and z D −2. The rectangular coordinates are
9. x D 1√
cos 2π/3 D −1/2, y D 1 sin 2π/3 D
−1/2, 3/2, −2.
In Exercises 10–13 use equations (7) x D ρ sin ϕ cos θ, y D ρ sin ϕ sin θ, and z D ρ cos ϕ.
√
√
10. x D 4sin π/2cos π/3 D 411/2 D 2, y D 4sin π/2sin
π/3 D 41 3/2 D 2 3, and z D
√
4 cos π/2 D 40 D 0. So the rectangular
coordinates are 2, 2 3, 0.
√
√
√
D 3 3/21 D 3 3/2, and
11. x D 3sin π/3cos π/2 D 3 3/20 D 0, y D 3sin π/3sin π/2
√
z D 3 cos π/3 D 31/2 D 3/2. √
So the rectangular coordinates
are 0, 3 3/2, 3/2.
√
√
√
3/2 D
12. x√ D sin 3π/4cos 2π/3 D √2/2−1/2 D − 2/4, y D sin 3π/4sin√2π/3√D 2/2
√
6/4, and z D cos 3π/4 D − 2/2. So the rectangular coordinates are − 2/4, 6/4, − 2/2. I gave
the answer in this form because most students have been told throughout high school that you can never
leave √
a square
√ √root in √the denominator. They should, of course, feel comfortable leaving the answer as
−1/ 8, 3/ 8, −1/ 2, but√most won’t.
√
13. x D 2sin πcos π/4 D 20 2/2 D 0, y D 2sin πsin π/4 D 20 2/2 D 0, and z D 2 cos π D
2−1 D −2. So the rectangular coordinates are 0, 0, −2.
Exercises 14–16 are basically the same as Exercises 4–6 since the z coordinates are the same in both coordinate
systems. Use equations (2) or (4).
14. r2 D −12 C 02 D 1, so r D 1. Also, tan θ D 0/−1 D 0, so θ D π. The cylindrical coordinates are
1, π, 2.
√
√
√
15. r2 D −12 C 32 , so r D 2. Also, tan θ D 3/−1 D 3/2/−1/2, so θ D 2π/3. The cylindrical
coordinates are 2, 2π/3, 13.
√
√
16. r2 D 52 C 62 , so r D 61. Also tan θ D 6/5, so θ D tan−1 6/5. The cylindrical coordinates are ( 61,
tan−1 6/5, 3.
In Exercises 17 and 18 use equations (7) ρ2 D x2 C y2 C z2 , tan ϕ D x2 C y2 /z, and tan θ D y/x.
√
√ √
√
√
√
2 C −12 C 62 D 8, so ρ D
2 C −12 / 6 D
8
D
2
2.
Also,
tan
ϕ
D
1
2/ 6 D
17. ρ2 D 1
√
1/2/ 3/2, so ϕ D π/6. Finally, tan θ D −1/1 D −1, so θ D 7π/4 (since
√ the point, when projected onto
the xy-plane is in the fourth quadrant). In spherical coordinates
the point is 2 2, π/6, 7π/4.
√
√
√
18. ρ2 D 02 C 32 C 12 D 4, so ρ D 2. Also tan ϕ D 02 C 32 /1 D 3, so ϕ D π/3. Finally, when we
project the point onto
√ the xy-plane we see that the point is on the positive y-axis so θ D π/2. Or, just using the
equation tan θ D 3/0, so θ D π/2. In spherical coordinates the point is 2, π/3, π/2.
The figures in Exercises 19–21 are a progression. To complete the continuum, the next in line following Exercise 21
would be a sphere.
19. As in Example 5, θ does not appear so the surface will be circularly symmetric about the z-axis. Once we have
our answer to part (a), we can just rotate it about the z-axis to generate the answer to part (b).
(a) We are slicing in the direction π/2 which puts us in the yz-plane for positive y. This means that r − 22 C
z2 D 1 becomes y − 22 C z2 D 1. This is a circle of radius 1 centered at (0, 2, 0).
z
2
1.5
1
0.5
y
0.5 1
1.5 2 2.5
3 3.5 4
-0.5
-1
-1.5
-2
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“runall” — 2005/8/1 — 12:11 — page 44 — #44
44
Chapter 1
Vectors
(b) As we start to rotate this about the z-axis, we get a feel for the shape being generated (see below left). In
the figure above we see the result of the condition that r Ú 0. Without that restriction we would see two
circles, each sweeping out a trail like that above. We would end up tracing our surface twice. Rotating this
circle (with the restriction on r) about the z-axis, we will end up with a torus (see below right).
20. (a) As in Example 2, we could reason that our result is a circle that is traced twice (in the figures a is taken to
be 1):
y
2
1.5
1
0.5
-1
-0.5
0.5
1
x
(b) When we move to spherical coordinates ϕ takes on the role of θ from part (a). Note θ does not explicitly
appear in this spherical equation. As in the case for cylindrical equations, this means that the surface will
be circularly symmetric about the z-axis. As we start to revolve about the z-axis we get the figure on
the left.
1
0.5
z
0
2
-0.5
1.5
-1
0
1
y
0.5
0.5
1
x
1.5
2
0
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:11 — page 45 — #45
Section 1.7
New Coordinate Systems
45
Again, the completed figure is a torus (see above right), but this time the “hole” closes off at the
origin.
Note: You might want to assign both Exercises 21 and 22. They look so similar and yet the results are very different.
21. As noted above, surface will be circularly symmetric about the z-axis (the equation does not involve θ). In this
case we are rotating a piece of the cardioid 1 − cos ϕ shown below left:
z
0.2 0.4 0.6 0.8 1 1.2
x
-0.5
-1
-1.5
-2
As we start to rotate it we see a “flattened” circle sweeping out the figure pictured above right. The completed
figure is like a “dimpled” sphere:
0
-0.5
5
z
-1
1
-1.5
.5
-1
-2
-1
0
x
0
y
1
1
22. Once again, the surface will be circularly symmetric about the z-axis (the equation does not involve θ). In this
case we are rotating a piece of the cardioid 1 − sin ϕ shown below left:
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:11 — page 46 — #46
46
Chapter 1
Vectors
z
1
0.5
x
0.05 0.15 0.25
-0.5
-1
As we start to rotate it we see a “double hump” sweeping out the figure pictured above right. The completed
figure is shown below:
23. The equation: ρ sin ϕ sin θ D 2 is clearly a spherical equation (it involves all three of the spherical coordinates).
• Use equation (7) to convert it to cartesian coordinates: y D ρ sin ϕ sin θ so the cartesian form is simply
y D 2.
This is a vertical plane parallel to the xz -plane.
 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:11 — page 47 — #47
Section 1.7
New Coordinate Systems
47
• Use equation (6) to convert to cylindrical coordinates. sin θ stays sin θ and ρ sin ϕ D r. So the cylindrical
form is
r sin θ D 2.
24. The equation
z2 D 2x2 C 2y2
is clearly a cartesian equation (it involves all three of the cartesian coordinates).
• Use equation (4) to convert it to cylindrical coordinates: z2 D 2x2 C y2 D 2r2 so the cylindrical form is
simply
z2 D 2r2 .
This is a cone which is symmetric about the z-axis, whose vertex is at the origin, one nappe above and one
below the xy-plane.
• Use equation (7) to convert to spherical coordinates. z2 D 2x2 C 2y2 , so 0 D 2x2 C y2 C z2 − 3z2 .
So the cylindrical form is
0 D 2ρ2 − 3ρ cos ϕ2
cos ϕ D ; 2/3.
or
In this final form it is again clear that the surface is a cone.
25. r D 0 is an equation in cylindrical coordinates. If r D 0 then it doesn’t matter what θ is and z is free to take on
any value. This is the z-axis. In cartesian coordinates this is
x D y D 0,
and in spherical coordinates ρ and θ are not constrained but
ϕD0
or ϕ D π.
26. You are slicing a wedge out of a cylinder. The result looks like a quarter of a wheel of cheese.
x
0
3
1
y
2
2
1
3
2
0
2
1
1
z
z
0
0
-1
-1
0
1
0
1
2
y
x
3
2
3
27. Here you are taking the triangular region above the ray z D r and below the ray z D 5 in a plane for which θ is
fixed (say θ D 0) and rotating it through half a rotation to get half of a cone. The figure is below left.
 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:11 — page 48 — #48
48
Chapter 1
Vectors
y
-1 -0.5 0
-5
x
-2.5
0
0.5 1
2.5
5
4
4
z
z
2
2
0
-1
-0.5
0
0.5 x
0
0
2
4
1
y
28. Again we are rotating a triangular region—but this time it is above the line z D 2r and below the line z D 5 − 3r.
This gives us an image that looks like a diamond spun on a diagonal. The figure is above right.
29. This solid is bound by two paraboloids.
4
z 2
0
-1
0
-1
0
y
1
x
1
Note: For Exercises 30–32 no sketch is included. I’ve just roughly described the figure.
30. This is a hollow sphere. The sphere of radius 2 is missing a spherical hole of radius 1.
31. This is the top half of the unit sphere.
32. This is a quarter of the unit sphere sitting over (and under) the first quadrant.
33. This looks like an ice cream cone:
z
y
x
34. This may look complicated, but it is the cone without the ice cream from the previous problem. The equation
ρ D 2/ cos ϕ looks worse than it is. Remember that z D ρ cos ϕ so this is equivalent to z D 2. So we get a flat
topped cone with height 2 and tip on the origin.
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“runall” — 2005/8/1 — 12:11 — page 49 — #49
Section 1.7
New Coordinate Systems
49
35. This is a sphere of radius 3 centered at the origin from which we’ve removed a sphere of radius 1 centered at
(0, 0, 1).
z
y
x
36. (a) Look for where x, y D r, θ. We know also that x D r cos θ, so r cos θ D r. This implies that cos θ D 1
so θ D 0. Also y D r sin θ, but sin θ D 0 so y D 0. So points of the form (a, 0) are the same in both cartesian
and polar coordinates.
(b) The only difference between this and part (a) is that a z coordinate has been added to each. So points of the
form (a, 0, b) are the same in both rectangular and cylindrical coordinates.
(c) Here (x, y, z) must equal ρ, ϕ, θ, where x D ρ sin ϕ cos θ, y D ρ sin ϕ sin θ, and z D ρ cos ϕ.
By the first equation ρ sin ϕ cos θ D ρ. This implies that sin ϕ D 1 which in turn implies that cos ϕ D 0,
cos θ D 1, and sin θ D 0. But then z D ρ cos ϕ D 0, and y D ρ sin ϕ sin θ D ρ10 D 0. It looks as if
we’re headed to solutions on the x-axis again. But wait a minute, if y D 0, then ϕ D 0, but if sin ϕ D 1 then
ϕ can’t be zero. The only point satisfying all of the conditions is the origin (0, 0, 0).
37. (a) Picture drawing the graph of the polar equation r D f θ by standing at the origin and turning to angle θ
and then walking radially out to f θ. You can see that if instead you walked radially out to −f θ you
would be heading the same distance in the opposite direction. This tells you that the graph r D −f θ is
just the graph r D f θ reflected through the origin.
(b) Although we now have an additional degree of freedom the idea is the same. For each direction specified by
ϕ and θ we would be heading the same distance in the opposite direction. Again this tells you that the graph
ρ D −f ϕ, θ is just the graph ρ D f ϕ, θ reflected through the origin.
(c) We’re back to the situation in part (a). This time you head in the same direction, you just walk three times as
far. So r D 3f θ is as if we expanded the graph r D f θ to three times its original size without changing
its shape or orientation.
(d) Analogously, ρ D 3f ϕ, θ is as if we expanded the graph ρ D f ϕ, θ to three times its original size
without changing its shape or orientation.
38. Because there is no dependence on θ it means that for each r and the corresponding z D f r you have a solution
set that corresponds to rotating the point (r, f r) about the z-axis.
39. (a) We need to take six dot products. Each vector dotted with itself must be 1 and each vector dotted with any
other must be 0.
er · er D cos θ, sin θ, 0 · cos θ, sin θ, 0 D cos2 θ C sin2 θ D 1.
eθ · eθ D − sin θ, cos θ, 0 · − sin θ, cos θ, 0 D sin2 θ C cos2 θ D 1.
ez · ez D 0, 0, 1 · 0, 0, 1 D 1.
er · eθ D cos θ, sin θ, 0 · − sin θ, cos θ, 0 D − cos θ sin θ C sin θ cos θ D 0.
er · ez D cos θ, sin θ, 0 · 0, 0, 1 D 0.
eθ · ez D − sin θ, cos θ, 0 · 0, 0, 1 D 0.
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“runall” — 2005/8/1 — 12:11 — page 50 — #50
50
Chapter 1
Vectors
(b) We now do the same for the spherical basis vectors.
eρ · eρ D sin ϕ cos θ, sin ϕ sin θ, cos ϕ · sin ϕ cos θ, sin ϕ sin θ, cos ϕ D sin2 ϕ cos2 θ C sin2 ϕ sin2 θ C cos2 ϕ
D sin2 ϕ C cos2 ϕ D 1.
eϕ · eϕ D cos ϕ cos θ, cos ϕ sin θ, − sin ϕ · cos ϕ cos θ, cos ϕ sin θ, − sin ϕ D cos2 ϕ cos2 θ
C cos2 ϕ sin2 θ C sin2 ϕ D cos2 ϕ C sin2 ϕ D 1.
eθ · eθ D − sin θ, cos θ, 0 · − sin θ, cos θ, 0 D sin2 θ C cos2 θ D 1.
eρ · eϕ D sin ϕ cos θ, sin ϕ sin θ, cos ϕ · cos ϕ cos θ, cos ϕ sin θ, − sin ϕ D sin ϕ cos ϕ cos2 θ C sin ϕ cos ϕ cos2 θ
− sin ϕ cos ϕ D sin ϕ cos ϕ − sin ϕ cos ϕ D 0.
eρ · eθ D sin ϕ cos θ, sin ϕ sin θ, cos ϕ · − sin θ, cos θ, 0 D − sin ϕ cos θ sin θ C sin ϕ sin θ cos θ D 0.
eϕ · eθ D cos ϕ cos θ, cos ϕ sin θ, − sin ϕ · − sin θ, cos θ, 0 D − cos ϕ cos θ sin θ C cos ϕ sin θ cos θ D 0.
40. Begin with
er D cos θ i C sin θ j
eθ D − sin θ i C cos θ j.
Then
sin θ er C cos θ eθ D sin θ cos θ i C sin2 θ j C − cos θ sin θ i C cos2 θ j
D j.
Similarly, cos θ er − sin θ eθ D i. Thus, all together
i D cos θ er − sin θ eθ
j D sin θ er C cos θ eθ
k D ez .
41. First note that, from (9),
sin ϕ eρ C cos ϕ eϕ D sin2 ϕ cos θ i C sin2 ϕ sin θ j C cos2 ϕ cos θ i C cos2 ϕ sin θ j
D cos θ i C sin θ j.
Hence
cos θ sin ϕ eρ C cos ϕ eϕ − sin θ eθ D cos2 θ i C cos θ sin θ j C sin2 θ i − sin θ cos θ j
D i.
and, similarly, sin θ sin ϕ eρ C cos ϕ eϕ C cos θ eθ D j.
Finally, verify that cos ϕ eρ − sin ϕ eϕ D k.
So our results are
i D sin ϕ cos θ eρ C cos ϕ cos θ eϕ − sin θ eθ
j D sin ϕ sin θ eρ C cos ϕ sin θ eϕ C cos θ eθ
k D cos ϕ eρ − sin ϕ eϕ .
42. The exercise is more naturally set up for spherical coordinates.
√
(a) Here we are inside the portion of the sphere ρ D 3 for | tan ϕ| … 1/ 8.
√
Ice cream cone D {ρ, ϕ, θ|0 … ρ … 3, 0 … ϕ … tan−1 1/ 8, and 0 … θ < 2π}.
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“runall” — 2005/8/1 — 12:11 — page 51 — #51
Section 1.8
True/False Exercises for Chapter 1
51
√
(b) Here√z’s lower limit is the cone portion so z Ú 8r. The upper limit is the portion of the sphere so
z … 32 − r2 . The variable r is free to be anything between 0 and 1 and θ is free to take on values between
0 and 2π. The cylindrical description is:
√
{r, θ, z| 8r … z … 9 − r2 , 0 … r … 1, and 0 … θ … 2π}.
1.8 TRUE/FALSE EXERCISES FOR CHAPTER 1
1. False. (The corresponding components must be equal.)
2. True. (Apply two kinds of distributive laws.)
3. False. −4, −3, −3 is the displacement vector from P2 to P1 .
4. True.
5. False. (Velocity is a vector, but speed is a scalar.)
6. False. (Distance is a scalar, but displacement is a vector.)
7. False. The particle will be at 2, −1 C 21, 3 D 4, 5.
8. True.
9. False. From the parametric equations, we may read a vector parallel to the line to be −2, 4, 0. This vector is
not parallel to −2, 4, 7.
10. True. Note that a vector parallel to the line is 1, 2, 3 − 4, 3, 2 D −3, −1, 1.)
11. False. The line has symmetric form x−2 D y − 1 D zC3 .
−3
2
12. True. Check that the points −1, 2, 5 and 2, 1, 7 lie on both lines.
13. False. (The parametric equations describe a semicircle because of the restriction on t.)
14. False. (The dot product is the cosine of the angle between the vectors.)
15. False. ka D |k| a.
16. True.
17. False. Let a D b D i, and c D j.
18. True.
19. True.
20. True.
21. True. (Check that each point satisfies the equation.)
22. False. (No values of s and t give the point (1, 2, 1).)
23. False. (The product BA is not defined.)
24. False. (The expression gives the opposite of the determinant.)
25. False. det2A D 2n det A.
26. True.
27. False. (The surface with equation ρ D 4 cos ϕ is a sphere.)
28. True. (It’s the plane x D 3.)
29. True.
30. False. The spherical equation should be ϕ D tan−1 1 .
2
1.9 MISCELLANEOUS EXERCISES FOR CHAPTER 1
−'
1. Solution 1. We add the vectors head-to-tail by parallel translating OP 2 so its tail is at the vertex P1 , translating
−'
−'
−'
OP 3 so that its tail is at the head of the translated OP 2 , etc. Since each vector OP i has the same length and,
−'
−'
for i D 2, ... , n, the vector OP i is rotated 2π/n from OP i−1 , the translated vectors will form a closed (regular)
n-gon, as the figure below in the case n D 5 demonstrates.
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“runall” — 2005/8/1 — 12:11 — page 52 — #52
52
Chapter 1
Vectors
P1
P5
P2
O
P4
P3
−'
Thus, using head-to-tail addition with the closed n-gon, we see that #niD1 OP i D 0.
−'
Solution 2. Suppose that #niD1 OP i D a Z 0. Imagine rotating the entire configuration through an angle of
2π/n about the center O of the polygon. The vector a will have rotated to a different nonzero vector b. However,
the original polygon will have rotated to an identical polygon (except for the vertex labels), so the new vector
−'
sum #niD1 OP i must be unchanged. Hence a D b, which
 is a contradiction. Thus a D 0.

 x D 1 C 3t
2. The line will be rt D 1, 0, −2 C t3, −7, 1, or y D −7t

 z D −2 C t.
3. The displacement vector 3t0 C 1, 5 − 7t0 , t0 C 12 − 1, 0, −2 D 3t0 , 5 − 7t0 , t0 C 14 is orthogonal
to 3, −7, 1. This means that
0 D 3t0 , 5 − 7t0 , t0 C 14 · 3, −7, 1 D 9t0 − 35 C 49t0 C t0 C 14 D 59t0 − 21.
So t0 D 21/59. The displacement vector gives us the direction of the line:
3t0 , 5 − 7t0 , t0 C 14 D 1/5963, 148, 847.
So the equation of the line is
rt D 1, 0, −2 C t63, 148, 847,
or


 x D 1 C 63t
y D 148t

 z D −2 C 847t.
−
'
−
'
−
'
−
'
−
−'
−
−'
4. (a) If rt D OP0 C t P0 P 1 , then r0 D OP0 and r1 D OP0 C P0 P 1 D OP1 .
(b) Part (a) set us up for part (b). We know that r(0) and r(1) give us the end points of the line segment so
−
'
−
−'
rt D OP0 C t P0 P 1 , for 0 … t … 1 is the equation of the line segment.
(c) We can just plug into our equation in part (b) to get rt D 0, 1, 3 C t2, 4, −10 for 0 … t … 1. In
parametric form this is


 x D 2t
y D 1 C 4t
for 0 … t … 1.

 z D 3 − 10t
5. (a) The desired line must pass through the midpoint of P1 P 2 , which has coordinates −1C5 , 3−7 D 2, −2.
2
2
−
−−'
−
−−'
The line must also be perpendicular to P1 P2 . The vector P1 P2 is 5 C 1, −7 − 3 D 6, −10. A vector
perpendicular to this must satisfy 6, −10 · a1 , a2 D 0 so 3a1 − 5a2 D 0 Hence a D 5, 3 will serve.
A vector parametric equation for the line is lt D 2, −2 C t5, 3, yielding
x D 5t C 2
y D 3t − 2.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:11 — page 53 — #53
Section 1.9
Miscellaneous Exercises for Chapter 1
53
P1
P2
−
−'
(b) We generalize part (a). Midpoint of P1 P 2 is a1 Cb1 , a2 Cb2 . Vector P1 P 2 is b1 − a1 , b2 − a2 . A
2
2
−
−'
vector v perpendicular to P1 P 2 satisfies b1 − a1 , b2 − a2 · v D 0 We may therefore take v to be
v D b2 − a2 , a1 − b1 so lt D a1 Cb1 , a2 Cb2 C tb2 − a2 , a1 − b1 yielding
2
2
a1 C b1
2
a C b2
.
y D a1 − b1 t C 2
2
x D b2 − a2 t C
P1
P2
−
−'
6. (a) Desired plane passes through midpoint M 1, 2, −1 and has P1 P 2 D −10, −2, 2 as normal vector. So
the equation is
−10x − 1 − 2y − 2 C 2z C 1 D 0 )* 5x C y − z D 8.
−
−'
(b) M is a1 Cb1 , a2 Cb2 , a3 Cb3 I P1 P 2 D b1 − a1 , b2 − a2 , b3 − a3 .
2
2
2
Equation for plane is
b1 − a1 x −
a1 C b1
a2 C b2
a3 C b3
C b2 − a2 y −
C b3 − a3 z −
D0
2
2
2
or
b1 − a1 x C b2 − a2 y C b3 − a3 z D
1 2
b1 C b22 C b32 − a21 − a22 − a23 .
2
−
−−'
7. (a) Midpoint of segment is 1−3 , 6−2 , 0C4 , 3C1 −2C0 D −1, 2, 2, 2, −1. Normal to hyperplane is P1 P2 D
2
2
2
2
2
−4, −8, 4, −2, 2 so the equation of the hyperplane is −4x1 C 1 − 8x2 − 2 C 4x3 − 2 −
2x4 − 2 C 2x5 C 1 D 0 or 2x1 C 4x2 − 2x3 C x4 − x5 D 5.
(b) Very similar to 6(b). Equation for plane is
b1 − a1 x1 −
a1 C b1
an C bn
C · · · C bn − an xn −
D0
2
2
or
b1 − a1 x1 C · · · C bn − an xn D
1 2
b1 C · · · C bn2 − a21 − · · · − a2n .
2
8. We have
a * b D a b sin θ D sin θ,
a · b D a b cos θ D cos θ,
since a, b are unit vectors. Thus a * b2 C a · b2 D sin2 θ C cos2 θ D 1.
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“runall” — 2005/8/1 — 12:11 — page 54 — #54
54
Chapter 1
Vectors
9. (a) No. a · b D a · c just means that the angle between vectors a and b and the angle between vectors a and b
have the same cosine. If you would prefer, rewrite the equation as a · b − c D 0 and you can see that
what this says is that one of the following is true: vector a is orthogonal to the vector b − c or a D 0 or
b − c D 0.
(b) No. Use the distributive property of cross products to rewrite the equation as a * b − c D 0. This could
be true if a is parallel to b − c or if a D 0 or if b − c D 0.
10. The lines are r1 t D −3, 1, 5 C t1, −2, 2 and r2 t D 4, 3, 6 C t−2, 4, −4. The direction vector for
line 2 is −2 times the direction vector for line 1 so either they are parallel or they are the same line. Look at the
displacement vector from a point on line 1 to a point on line 2, for example 4, 3, 6 − −3, 1, 5 D 7, 2, 1.
This is not a multiple of the direction vector so they are not the same line. Now, to find the normal direction
we’ll take
7, 2, 1 * 1, −2, 2 D 6, −13, −16.
The equation of the plane is therefore
6x C 3 − 13y − 1 − 16z − 5 D 0
or
6x − 13y − 16z D −111.
11. (a) The angle between the two planes will be the same as the angle between the normal vectors. The normal to
x C y D 1 is n1 D 1, 1, 0, and the normal to y C z D 1 is n2 D 0, 1, 1.
The angle is then
1, 1, 0 · 0, 1, 1
1
π
cos−1
D cos−1 D .
1, 1, 0 0, 1, 1
2
3
(b) The line common to both planes must be orthogonal to both n1 and n2 . We use the cross product to find:
n1 * n2 D 1, 1, 0 * 0, 1, 1 D 1, −1, 1.
The line must also pass through the point (0, 1, 0). Of course this isn’t the only point you could have come
up with, but it is the easiest to see. So parametric equations for the line are:


xDt
yD1 − t

 z D t.
12. Set up a cube so that one vertex is at the origin and the rest of the bottom face has vertices at (1, 0, 0), (0, 1, 0),
and (1, 1, 0). Then the top face will have vertices at (0, 0, 1), (1, 0, 1), (0, 1, 1), and (1, 1, 1).
(a) The angle between the diagonal and one of the edges is
cos−1
1, 1, 1 · 1, 0, 0
1, 1, 1 1, 0, 0
1
D cos−1 √ .
3
(b) You might be tempted to think that the angle between the diagonal of the cube and the diagonal of one of
its faces is (by inspection) half of a right angle. The triangle with√the diagonal
of the cube as its hypotenuse
√
and the diagonal of one of the faces as one of the legs is a 1 : 2 :√ 3√right triangle. The cosine of the
angle between the diagonal of the cube and the diagonal of a side is 2/ 3. Using the formula above we
also see
√
√
1, 1, 1 · 1, 1, 0
2
2
6
cos−1
D cos−1 √ D cos−1 √
D cos−1
.
1, 1, 1 1, 1, 0
6
3
3
13. The dot product of your two vectors indicates how much you agree with your friend on these five questions. When
you both agree or both disagree with an item, the contribution to your dot product is 1. When one of you agrees
and the other disagrees the contribution is −1. Your dot product will be an odd number between −5 and 5.
−'
−
'
−
'
−'
−
'
14. (a) Following the instructions, we can write BM1 D AM1 − AB D 1 AC − AB because M1 is the midpoint of
2
−
'
−
'
−'
AC. Similarly, CM2 D 1 AB − AC.
2
−
'
−'
−
'
−'
(b) P is on BM1 so we can write BP as some multiple of BM1 . For definiteness, let’s say that BP D k BM1
−
'
−'
where 0 < k < 1. Similarly, CP D lCM2 where 0 < l < 1. Putting this together with our results from
−
'
−
'
−
'
−
'
−
'
−
'
part (a), BP D k 1 AC − AB and CP D l 1 AB − AC.
2
2
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:11 — page 55 — #55
Section 1.9
Miscellaneous Exercises for Chapter 1
55
−
'
−
'
−
'
−
'
−
'
−
'
−
'
−
'
−
'
(c) First, CB D CP C PB D CP − BP. From part (b), this is l 1 AB − AC − k 1 AC − AB D l C
2
2
2
−
'
−
'
−
'
−
'
−
'
−
'
−
'
kAB − l C k AC. But, CB also equals CA C AB D AB − AC. Equating the coefficients gives us the
2
simultaneous equations l C k D 1 and l C k D 1. This easily gives us l D k D 2/3.
2
2
−'
(d) Repeat steps (a) through (c) with AM3 and either of the other median vectors. You will again get a point of
intersection, say Q. You will show that Q is 2/3 of the way down each median and so must be the same
point as P.
15. We are assuming that the plane  contains the vectors a, b, c, and d. The vector n1 D a * b is orthogonal to ,
and the vector n2 D c * d is orthogonal to . So the vectors n1 and n2 are parallel. This means that n1 * n2 D 0.
16. The first two ways that may come to mind to your students each depends on prior knowledge:
Method One: Recall that the area of a triangle is 1/2a b sin C, where C is the angle between a and b.
So the area is
1
1
a b sin C D a2 b2 sin2 C
2
2
1 D a2 b2 1 − cos2 C
2
1 D a2 b2 − a2 b2 cos2 C
2
1 D a2 b2 − a · b2 .
2
Method Two: The area of a triangle is 1/2 the area of the parallelogram determined by the same two vectors.
The area of the parallelogram is the length of the cross product. So the area is
1
1 a * b D a * b · a * b
2
2
1 by Section 1.4, Exercise 29 D a · ab · b − a · bb · a
2
1 D a2 b2 − a · b2 .
2
17. (a) The vertices are given so that if they are connected in order ABDC we will sketch a parallelogram. From
Exercise 16 we could say that
−
'
−
'
−
' −
'
Area D AB2 AC2 − AB · AC2
D [4 − 1, −1 − 3, 3 C 12 2 − 1, 5 − 3, 2 C 12
− 4 − 1, −1 − 3, 3 C 1 · 2 − 1, 5 − 3, 2 C 12 ]1/2
D 3, −4, 42 1, 2, 32 − 3, −4, 4 · 1, 2, 32
√
√
D 4114 − 72 D 525 D 5 21.
(b) When we project the parallelogram in the xy-plane we get the same points with the z coordinate equal to 0.
We do the same calculation as in part a with the new vectors:
−
'
−
'
−
' −
'
Area D AB2 AC2 − AB · AC2
D 4 − 1, −1 − 3, 02 2 − 1, 5 − 3, 02 − 4 − 1, −1 − 3, 0 · 2 − 1, 5 − 3, 02
D 3, −4, 02 1, 2, 02 − 3, −4, 0 · 1, 2, 02
√
D 255 − 52 D 100 D 10.
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“runall” — 2005/8/1 — 12:11 — page 56 — #56
56
Chapter 1
Vectors
18. (a) Students raised on the slope-intercept form of a line may be more comfortable once you point out that the
slope of the line ax C by D d is y D −a . Now the direction that the vector points is clear: v D b, −a.
x
b
(b) A vector n normal to the line l must be orthogonal to the vector v you found in part (a). We are also told
that the first component of n is a. This means that
0 D n · v D a, ? · b, −a D ab−? a.
So n D a, b.
(c) Choose a point P1 on the line ax C by D d. For example, if P1 has x component zero then y D d/b. In
other words, choose P1 D 0, d/b. It doesn’t matter. We are going to project the displacement vector from
the point P1 to the point P0 D x0 , y0 onto n.
−
−−' −
−−'
−
−−'
|a, b · x0 , y0 − d/b|
|ax0 C by0 − d|
|n · P0 P1 |
n · P0 P1
√
n D
D
D
.
projn P0 P1 D n · n
n
a, b
a2 C b 2
(d) We plug into our brand new formula:
Distance from (3, 5) to l : 3x − 5y D 2 is
|83 − 55 − 2|
3
√
D √ .
89
82 C 52
19. (a) As should be expected, this is similar to the calculation in Exercise 18. We choose any point P1 in the plane
 : Ax C By C Cz D D. For example, let P1 D 0, 0, D/C and P0 D x0 , y0 , z0 . The normal vector
n D A, B, C. Again the distance from P0 to  is
−
−−' −
−−'
n · P0 P1
n
projn P0 P1 D n · n
D
−
−−'
|n · P0 P1 |
n
D
|A, B, C · x0 , y0 , z0 − D/C|
A, B, C
D
|Ax0 C By0 C Cz0 − D|
√
.
A2 C B2 C C2
(b) We plug into our formula from part (a):
Distance from 1, 5, −3 to : x − 2y C 2z C 12 D 0 is
|11 − 25 C 2−3 C 12|
3
D √ D 1.
2
2
2
9
1 C −2 C 2
20. We have the equation that if α is the angle between vectors x and the vector k D 0, 0, 1, then
cos α D
x · k
x · k
D
.
x k
x
√
Since we are given that this last quantity D 1/ 2, x makes an angle of 45 degrees with the positive z-axis. So
the points P satisfying the condition of this exercise sweep out the top nappe of the cone making an angle of 45
degrees with the positive z-axis minus the origin.
21. The equation a * x D b tells us that x points in the direction of b * a. Now we have to determine the length of
x. We can choose any vector in the direction of x. For convenience, let y be the unit vector in direction of x:
yD
b * a
.
b * a
The angle between a and x is the same as that between a and y so
a · y
a · x
c
D
D
.
a
a x
a x
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:11 — page 57 — #57
Section 1.9
Miscellaneous Exercises for Chapter 1
57
So if c Z 0,
x D
c
,
a · y
and,
xD
c
y.
a · y
If c D 0 then a is orthogonal to x (and y). Use the fact that
b D a * x D a x sin θ D a x sin π/2 D a x.
So when c D 0,
x D
b
a
b
y.
a
and x D
22. (a) In the figure below left, the cross product a * b is a vector outwardly normal to the face containing edges
a and b with length equal to twice the area of the face. To keep the diagram uncluttered, it has been split
into two:
c
e
a
b
d
So the sum of the four vectors v1 , v2 , v3 , and v4 asked for in the exercise can be expressed as
1/2[a * b C b * c C c * a C e * d].
But d D b − a and e D c − a so
e * d D c − a * b − a
D c * b − a * b − c * a C a * a
D −a * b − b * c − c * a.
We put this together with the above to conclude:
v1 C v2 C v3 C v4 D 1/2[a * b C b * c C c * a C e * d]
D 1/2[a * b C b * c C c * a − a * b − b * c − c * a]
D 0.
(b) Denote the vectors associated with the first tetrahedron as v1 , v2 , v3 , and v4 and the vectors associated with
the second tetrahedron as v1 , v2 , v3 , and v4 . Let the vectors associated with the sides being glued together
be v1 and v1 .
By construction v1 and v1 have equal lengths and point in opposite directions so v1 C v1 D 0. From part
(a) we know that
v1 D −v2 C v3 C v4 and v1 D −v2 C v3 C v4 .
This means that
v2 C v3 C v4 C v2 C v3 C v4 D 0.
(c) Just as we can break any polygon into triangles, we can break any polyhedron into tetrahedra. The key to
part (b) was that when we glue two tetrahedra together, the vector of the face being hidden is equal to the
sum of the three vectors being introduced. In symbols,
v1 D v2 C v3 C v4 .
From part (a) we know that for any tetrahedron v1 C v2 C v3 C v4 D 0. So as we build up our polyhedron
by gluing tetrahedra together, at each stage (by parts (a) and (b)) the sum of the outward normals with length
equal to the area of the face will be zero.
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“runall” — 2005/8/1 — 12:11 — page 58 — #58
58
Chapter 1
Vectors
23. (a) Remember, if the adjacent sides of a parallelogram are a and b, then the diagonals are a C b and a − b.
So the sum of the squares of the lengths of the diagonals are
a C b2 C a − b2 D a C b · a C b C a − b · a − b
D a · a C 2a · b C b · b C a · a − 2a · b C b · b D 2a2 C 2b2
which is the sum of the squares of the lengths of the four sides (opposite sides have equal lengths).
(b) a C b2 C a − b2 D 2a2 C b2 .
24. The last line of the proof of the Cauchy-Schwarz inequality in Section 1.6 is
a2 b2 Ú a · b2 .
Now we only need to notice that
2
n
a · b D $ % ai bi &
2
iD1
n
a2 D % a2i
iD1
n
b2 D % bi2
iD1
and the result follows immediately:
n
n
n
iD1
iD1
iD1
2
2
2
$ % ai & $ % bi & Ú $ % ai bi & .
25. (a)
AD
1
0
1
1
, A2 D 1 0
2
1
, A3 D 1 0
3
1
, A4 D 1 0
4
1 (b) It seems reasonable to guess that
An D 1
0
n
.
1 (c) We need only show the inductive step:
AnC1 D AAn D 1
0
1
1
1 0
n
1
D
1 0
n C 1
.
1
26. (a) There’s nothing much to show. A2 D 0.
(b) You shouldn’t need a calculator or computer for this. The diagonal of 1’s keeps moving to the left so that




0 0 0 0 0
0 0 0 0 0
 0 0 0 0 0 
 0 0 0 0 0 




2
3



A D  1 0 0 0 0 , A D 
 0 0 0 0 0 ,
 0 1 0 0 0 
 1 0 0 0 0 
0 0 1 0 0
0 1 0 0 0


0 0 0 0 0
 0 0 0 0 0 


5

A4 D 
 0 0 0 0 0  , and A D 0.
 0 0 0 0 0 
1 0 0 0 0
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“runall” — 2005/8/1 — 12:11 — page 59 — #59
Section 1.9
Miscellaneous Exercises for Chapter 1
59
27. (a) The determinants are:
1
1
1
, |H3 | D
, |H4 | D
,
12
2160
6048000
1
1
|H5 | D
, and |H6 | D
266716800000
186313420339200000
|H2 | D
The determinants are going to 0 as n gets larger. As for writing out the matrices, note that H2 is the upper
left two by two matrix in H10 in part (b). Similarly, H3 is the upper left three by three ... H6 is the upper left
six by six matrix in H10 . I would consider deducting points from any student who actually writes these out.
They can use a computer algebra system to accomplish this. For Mathematica the command for generating
H10 would be
Table[1/i C j − 1, {i, 10}, {j, 10}]//MatrixForm.
The command for calculating the determinant would be
Det[Table[1/i C j − 1, {i, 10}, {j, 10}]].
(b) Using the Mathematica commands described in part (a), the determinant
|H10 | D 1/46206893947914691316295628839036278726983680000000000
L 2.16 * 10−53 .
The matrix is














H10 D 












1
1
2
1
3
1
4
1
5
1
6
1
7
1
8
1
9
1
10
1
2
1
3
1
4
1
5
1
6
1
7
1
8
1
9
1
10
1
11
1
3
1
4
1
5
1
6
1
7
1
8
1
9
1
10
1
11
1
12
1
4
1
5
1
6
1
7
1
8
1
9
1
10
1
11
1
12
1
13
1
5
1
6
1
7
1
8
1
9
1
10
1
11
1
12
1
13
1
14
1
6
1
7
1
8
1
9
1
10
1
11
1
12
1
13
1
14
1
15
1
7
1
8
1
9
1
10
1
11
1
12
1
13
1
14
1
15
1
16
1
8
1
9
1
10
1
11
1
12
1
13
1
14
1
15
1
16
1
17
1
9
1
10
1
11
1
12
1
13
1
14
1
15
1
16
1
17
1
18
1
10
1
11
1
12
1
13
1
14
1
15
1
16
1
17
1
18
1
19



























(c) Again, the code examples will be from Mathematica. Let’s first calculate a numerical approximation A of
H10 with the command
A D N[Table[1/i C j − 1, {i, 10}, {j, 10}]].
We can then calculate the inverse B and A with the command
B D Inverse[A].
You can display these as matrices by appending “//MatrixForm” to the command. Now generate AB and BA
with the commands
A.B//MatrixForm and B.A//MatrixForm.
You should note that these aren’t equal and neither is the 10 * 10 identity matrix I10 .
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“runall” — 2005/8/1 — 12:11 — page 60 — #60
60
Chapter 1
Vectors
28. The center of the moving circle is at a − bcos t, sin t. Notice that as the moving circle rolls so that its
center moves counterclockwise it is turning clockwise relative to its center. When the small circle has traveled
completely around the large circle it has rolled over a length of 2πa. Its circumference is 2πb so if it were
rolling along a straight line it would have revolved a/b times. The problem is that it is rolling around in a circle
and so it has lost a rotation each time the center has traveled completely around. In other words the smaller wheel
is turning at a rate of
a/b − 1t D a − bt/b.
The position of P relative to the center of the moving circle is
b cos −
a − bt
a − bt
a − bt
a − bt
, − sin
, sin −
D b cos
.
b
b
b
b
Putting this together, the position of P is the sum of the vector from the origin to the center of the moving circle
and the vector from the center of the moving circle to P. This is
a − bcos t, sin t C b cos a − bt
a − bt
, − sin .
b
b
29. Not much changes here. The center of the moving circle is now at a C bcos t, sin t. Now the moving circle
gains one revolution each time around the fixed circle and so turns at a rate of a/b C 1t D a C bt/b. Since
we are starting P at (a, 0), the initial angle from the center of the moving circle to P is π so the position of P
relative to the center of the moving circle is b cos π C
aCbt
a−bt
aCbt
aCbt
, sin b D −b cos b , sin b .
b
As in Exercise 28 we sum the same two vectors to get the expression:
a C bcos t, sin t − b cos a C bt
a C bt
, sin .
b
b
30. (a) Let’s look at diagrams of hypocycloid (below on the left) and an epicycloid (below on the right) with a D 6
and b D 5:
y
y
-6
-4
-2
15
4
10
2
5
2
4
6
x
-15
-10
-5
5
10
15
x
-5
-2
-10
-4
-15
What are the roles of a and b? You can see in the figure on the left that there are 6 cusps. This is also
true, but harder to see, in the figure on the right. Let’s look at what portion of these curves correspond to
0 … t … 2π.
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“runall” — 2005/8/1 — 12:11 — page 61 — #61
Section 1.9
Miscellaneous Exercises for Chapter 1
61
y
x
2.5
4
3.5
4.5
5
5.5
6
y
15
-1
10
-2
5
-3
-15
-5
-10
5
-4
-5
-5
-10
10
x
The figure on the left shows that 6/5 of the hypocycloid is covered for 0 … t … 2π. The figure on the right
is the corresponding portion of the epicycloid. Usually what we call the hypocycloid is what we draw until
the ends close up. In this case, the hypocycloid is complete when t D 52π. Again, although it is harder to
see, this epicycloid will close up when t D 52π.
If a and b have no common divisors and are both rational, then the hypocycloid or epicycloid will have
a cusps and will close up after t D b2π. If a and b have common divisors then write a/b in lowest
terms. The hypocycloid or epicycloid will have as many cusps as the numerator. The same answer holds for
epicycloids.
(b) We noted in part (a) that if a/b is rational in lowest terms, the hypocycloid or epicycloid closes up when
t D b2π. In the case of the hypocycloid, this is because then a − bcos b2π, sin b2π C
b cos a−b b2π, − sin a−b b2π D a − bcos 0, sin 0 C b cos a−b 0, − sin a−b 0. In
b
b
b
b
words, its because the angle is a rational multiple of 2π.
A picture of part of an epicycloid for which a/b is irrational is:
-4
4
4
2
2
-2
2
4
-4
-2
2
-2
-2
-4
-4
4
If a/b is irrational then the curve will never close up. It can’t. At no time when the center of the moving
circle comes back to its original position will P be back in its original position.
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“runall” — 2005/8/1 — 12:11 — page 62 — #62
62
Chapter 1
Vectors
A picture of part of a hypocycloid for which a/b is irrational is:
y
-1
y
1
1
0.5
0.5
-0.5
0.5
1
x
-1
-0.5
0.5
-0.5
-0.5
-1
-1
1
x
In each case, the figure on the left shows several periods. For the figure on the right we let t get larger. If
we let t get arbitrarily large the curve is dense.
31. Look at the second part of the answers in Exercises 28 and 29. The only difference is that we are changing the
distance from the center of the moving wheel to P from b to c. The formula for a hypotrochoid is:
a − bcos t, sin t C c cos a − bt
a − bt
, − sin .
b
b
In parametric form, the formulas for a hypotrochoid are:
x D a − b cos t C c cos a − bt
a − bt
, y D a − b sin t − c sin .
b
b
The formula for an epitrochoid is:
a C bcos t, sin t − c cos a C bt
a C bt
, sin .
b
b
In parametric form, the formulas for an epitrochoid are:
x D a C b cos t − c cos a C bt
a C bt
, y D a C b sin t − c sin .
b
b
32. (a) Here (below left) we get the four leaf rose:
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“runall” — 2005/8/1 — 12:11 — page 63 — #63
Section 1.9
Miscellaneous Exercises for Chapter 1
63
y
1
0.5
-1
-0.5
0.5
1
x
-0.5
-1
(b) We just erect a cylinder on that base and get the above right image.
(c) There is no θ explicitly in the equation, so the rose is being rotated about the z-axis (we show both the
completed figure and a partial to see how it is formed):
(d) Here we show half of the figure and then the completed figure. From the outside, the figure looks as if the
rose has been first rotated about the x-axis and then about the y-axis.
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“runall” — 2005/8/1 — 12:11 — page 64 — #64
64
Chapter 1
Vectors
33. Parts (a), (b), and (d) are pictured below top, left, and right. They look very similar to the graphs from the previous
exercise.
y
0.75
0.5
0.25
-0.75 -0.5 -0.25
x
0.25 0.5 0.75
-0.25
-0.5
-0.75
(c) This looks very different from its counterpart for Exercise 32. It looks like a dented sphere.
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“runall” — 2005/8/1 — 12:11 — page 65 — #65
Section 1.9
Miscellaneous Exercises for Chapter 1
65
34. (a) We begin with a three leaf rose (the path is traced twice) shown below left.
y
0.75
0.5
0.25
-0.5 -0.25
0.25 0.5 0.75
1
x
-0.25
-0.5
-0.75
(b) The cylindrical equation again adds nothing. A cylinder is built over the rose. It is shown above right.
(c) This interesting and different image is shown below left.
(d) This three leaf version of what we saw in Exercises 32 and 33 is shown above right.
35. The polar plot, cylinder and part (d) are similar to the corresponding solutions for Exercise 34. They are shown
below.
y
0.5
0.25
x
-0.75 -0.5 -0.25
0.25 0.5 0.75
-0.25
-0.5
-0.75
-1
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“runall” — 2005/8/1 — 12:11 — page 66 — #66
66
Chapter 1
Vectors
(c) Here in the figure shown below you see a difference in the solid generated by using sine instead of cosine.
36. (a) The nephroid is shown below left.
y
1.5
1
0.5
x
-2 -1.5
-1 -0.5
0.5
1
-0.5
-1
-1.5
(b) The cylinder based on it is shown above right.
(c) The first spherical graph is a dimpled sphere.
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“runall” — 2005/8/1 — 12:11 — page 67 — #67
Section 1.9
Miscellaneous Exercises for Chapter 1
67
(d) The second spherical graph has a lot of complexity so I have included a partial graph and the completed
graph.
37. (a) The curve is a spiral and is pictured below left.
y
z
7.5
5
2.5
-10
-5
5
10
x
-2.5
y
-5
-7.5
-10
x
(b) The cylinder based on the spiral in part (a) is shown above right.
(c) Because only part of the spiral is used, the resulting surface is a dimpled ball.
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“runall” — 2005/8/1 — 12:11 — page 68 — #68
68
Chapter 1
Vectors
z
y
x
(d) Finally, we see a lovely and intricate shell-like surface.
z
y
x
38. (a) In spherical coordinates the flat top of the hemisphere is the xy-plane with spherical equation ϕ D π/2. The
hemispherical bottom has equation ρ D 5, but only with π/2 … ϕ … π. Thus we may describe the object as
{ρ, ϕ, θ|0 … ρ … 5,
π/2 … ϕ … π,
0 … θ < 2π}.
(b) Now the flat top is described in√cylindrical coordinates as z D 0 and the bottom hemisphere as z2 C r2 D 25
with z … 0, that is, as z D − 25 − r2 . Bearing this in mind, the solid object is the set of points
{r, θ, z| − 25 − r2 … z … 0, 0 … r … 5, 0 … θ < 2π}.
39. Position the cylinder so that the center of the bottom disk is at the origin and the z-axis is the axis of the cylinder.
(a) In cylindrical coordinates θ is free to take on any values between 0 and 2π. The z coordinate is bounded by
0 and 3, and 0 … r … 3. To sum up:
{r, θ, z|0 … r … 3, 0 … z … 3, 0 … θ < 2π}.
(b) The key to solving this problem using spherical coordinates is to remember that the equation of a vertical
line in the plane is x D c where c is constant. In polar coordinates this is r cos θ D c or r D c sec θ. Similarly
the polar equation for a horizontal line is r D c csc θ.
Since the exercise is circularly symmetric with respect to the z-axis, 0 … θ < 2π. We are spinning the
rectangle with vertices (in cartesian coordinates) (0, 0, 0), (3, 0, 0), (0, 0, 3), and (3, 0, 3) about the z-axis.
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“runall” — 2005/8/1 — 12:11 — page 69 — #69
Section 1.9
Miscellaneous Exercises for Chapter 1
69
z
3.5
3
2.5
2
1.5
1
0.5
-0.5
0.5
1
1.5
2
2.5
3
x
3.5
-0.5
For ϕ between π/4 and π/2, ρ is bounded by the vertical line, so 0 … ρ … 3 csc ϕ. For ϕ between 0 and
π/4, ρ is bounded by the horizontal line, so 0 … ρ … 3 sec ϕ. To sum up:
{ρ, ϕ, θ|0 … ρ … 3 sec ϕ, 0 … ϕ … π/4, 0 … θ < 2π}
∪{ρ, ϕ, θ|0 … ρ … 3 csc ϕ, π/4 … ϕ … π/2, 0 … θ < 2π}.
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“runall” — 2005/8/1 — 12:14 — page 70 — #1
C H A P T E R
2
Differentiation in Several Variables
2.1 FUNCTIONS OF SEVERAL VARIABLES; GRAPHING SURFACES
1. f : R ' R : x & 2x2 C 1
(a) Domain f D {x ∈ R}, Range f D {y ∈ R|y Ú 1}.
(b) No. For instance f 1 D 3 D f −1.
(c) No. For instance if y D 0, there is no x such that f x D 0.
2. f : R2 ' R : x, y & 2x2 C 3y2 − 7
(a) Domain g D {x, y ∈ R2 }, Range g D {z ∈ R|z Ú −7}.
(b) Let Domain g D {x, x ∈ R2 |x Ú 0}.
(c) Let Codomain g D Range g.
3. Domain f D {x, y ∈ R2 |y Z 0}, Range f D R.
4. Domain f D {x, y ∈ R2 |x C y > 0}, Range f D R.
5. Domain g D R3 , Range g D {w ∈ R|w Ú 0}.
6. Domain g D {x ∈ R3 | ||x|| < 2}, Range g D {y ∈ R|y Ú 1/2}.
7. Domain f D {x, y ∈ R2 |y Z 1}, Range f D {x, y, z ∈ R3 |y Z 0, y2 z D xy − y − 12 C y C 12 }.
8. If x D x1 , x2 , x3 , then f1 x D x1 , f2 x D x2 C 3, and f3 x D x3 .
9. (a) fx D −2x/||x||.
(b) The component functions are
f1 x, y, z D −2x
x 2 C y 2 C z2
, f2 x, y, z D −2y
x 2 C y 2 C z2
,
f3 x, y, z D and
−2z
x 2 C y 2 C z2
.
10. Here there is nothing to show. Everything is at level 3. This surface is a plane parallel to the xy-plane 3 units
above it so the level set is the entire xy-plane if c D 3 and is the empty√set if c Z 3.
11. For c > 0 the level sets are circles centered at the origin of radius c. For c D 0 the level set is just the
origin. There are no values corresponding to c < 0. Note that the curves get closer together, indicating that
we are climbing faster as we head out radially from the origin. The second figure below shows the plot of the
level curves shaded to indicate the height of the level set (lighter is higher). The surface is therefore a paraboloid
symmetric about the z-axis. We show it with and without the surface filled in.
z
z
7.5
5
2.5
0
-2.5
-5
y
y
x
x
-7.5
-7.5 -5 -2.5 0 2.5 5 7.5
70
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“runall” — 2005/8/1 — 12:14 — page 71 — #2
Section 2.1
Functions of Several Variables; Graphing Surfaces
71
12. This is exactly the same as Exercise 11 except that the paraboloid has been shifted down 9 units so the level
curves begin in the center at c D −9, not c D 0.
13. Again this time for c > 0 the level curves are circles. This time, however, the circles corresponding to the level
sets at height c are of radius c. In other words, they are evenly spaced. We are climbing at a constant rate as we
head out radially, so the surface is a cone.
z
y
x
14. This time the level curves are ellipses. The sections as we cut in the direction x is constant or y is constant are
still parabolas.
z
x
y
15. The graphs xy D c are hyperbolas (unless c D 0 in which case it is the union of the two axes). When x and y
are both positive the height of the level curves are positive and so the hyperboloid is increasing as we head away
from the origin radially in either the first or third quadrant. When x and y are of different signs, the heights of
the level curves are negative and so the hyperboloid is decreasing as we head out radially in either the second or
fourth quadrant.
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“runall” — 2005/8/1 — 12:14 — page 72 — #3
72
Chapter 2
Differentiation in Several Variables
7.5
5
2.5
z
0
-2.5
x
-5
-7.5
-7.5
-5
-2.5
0
2.5
5
y
7.5
16. This is exactly the same as Exercise 17 except that the image has been reflected about the plane y D x.
17. We have a problem when y D 0. When k < 0, the section by x D k looks like the hyperbola in the figure on the
left, when k > 0, the section looks like the hyperbola in the figure on the right:
You can see that as y ' 0 from either side, along a line where x is constant and not 0, the z values won’t
match up. We are going to get a tear down the line y D 0. The level sets look like:
4
2
0
-2
-4
-10
-5
0
5
10
Notice that you can see that tear on the right center part of the above graph. The solid black and solid white
areas which are on either side of the x-axis point to the behavior around the tear.
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“runall” — 2005/8/1 — 12:14 — page 73 — #4
Section 2.1
Functions of Several Variables; Graphing Surfaces
73
Graph each side of the x-axis and you will see the following piece of the surface:
200
100
0
z
-100
-200
-1
0.08
0.06
0.04
y
-0.5
0.02
0
x
0.5
1 0
Our final surface is what you get when you try to glue two of those together:
18. The surface is a plane. Level sets for which f x, y D c are lines c D 3 − 2x − y or y D −2x C 3 − c.
Level sets are pictured below on the left. The surface is pictured below on the right.
7.5
5
2.5
0
20
-2.5
z 0
-5
-20
5
0y
-5
-7.5
-7.5 -5 -2.5 0
2.5
5
0
x
7.5
-5
5
19. Here we are looking at the graph of z D |x|. For c > 0, level sets for z D c will be the lines x D ; c. For c D 0
the level set is the y-axis. The graph is like a folded plane.
7.5
5
2.5
z
0
-2.5
-5
x
-7.5
-7.5 -5 -2.5
0
2.5
5
7.5
y
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“runall” — 2005/8/1 — 12:14 — page 74 — #5
74
Chapter 2
Differentiation in Several Variables
Note: In Problems 20–23 the level curves are shown along with the contour shading so you get an idea at what height
to hang the curves. You should be able to figure out the orientation of the surface from the contour plot.
20. Figures below:
7.5
5
2.5
0
500
-2.5
4
z 0
2
-500
-2
-5
0
0
y
-2
-7.5
2
-7.5
-5
-2.5
0
2.5
5
x
7.5
-4
4
21. Figures below:
7.5
5
2.5
0
-2.5
z
-5
x
-7.5
-7.5 -5 -2.5
0
2.5
5
y
7.5
22. Figures below: (note only a portion of the surface has been sketched so that you get a better idea of what’s
going on)
3
2
1
0
3
2
z
1
0
-3
-1
-2
2
0 y
-2
x
-3
-3
-2
-1
0
1
2
3
-2
-1
0
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“runall” — 2005/8/1 — 12:14 — page 75 — #6
Section 2.1
Functions of Several Variables; Graphing Surfaces
75
23. Figures below:
3
2
1
0
1
0.5
-1
z
2
0
-0.5
-2
0 y
-2
0
-3
-2
x
-3
-2
-1
0
1
2
2
3
24. (a) We solve the equation PV D kT for T , obtaining T D f P, V D 1/kPV. This is the same as we
considered in Exercise 15. See the figures for Exercise 15 for the general shape of the level curves.
(b) Here V D gP, T D kT . This is the same as the cases we considered in Exercises 16 and 17. We will get a
P
“torn” surface similar to the one shown in Exercise 17. The level curve V D c is the line through the origin:
P D k/cT .
25. (a) The surface z D x2 is graphed below left and z D y2 below right.
z
z
y
y
x
x
(b) Consider first the surface z D f x by considering the curve in the uv -plane given by v D f u. The
intersection of the surface with planes of the form y D c will look the same as the curve in the uv -plane
for any value of y. This helps us see that if we “drag” this curve in each direction along the y-axis, the trail
will trace out the surface. Similarly, but along the x-axis for surfaces of the form z D f y. The lack of
dependence on x is our clue.
(c) The graph of the surface y D x2 is shown below. It’s what we would expect from parts (b) and (a).
z
y
x
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“runall” — 2005/8/1 — 12:14 — page 76 — #7
76
Chapter 2
Differentiation in Several Variables
26. See the solution to Exercise 17 and the note in Exercise 16.
27. They can’t intersect—even though they may sometimes appear to. Say that two different level curves f x, y D c1
and f x, y D c2 where c1 Z c2 intersect at some point (a, b). Then f a, b would have assigned to it two
non-equal values. This can’t happen for a function (it’s our vertical line test). On the other hand, if the limit as
you approach (a, b) along different paths is different, those level curves may appear to intersect at (a, b) no matter
how good the resolution on your contour plot.
28. The level surfaces are planes x − 2y C 3z D c.
29. The level surfaces at level w D c are elliptic paraboloids.
√
30. The level surfaces at level w D c are nested spheres of radius c centered at the origin.
31. The level surfaces at level w D c are nested ellipsoids.
32. The level surfaces are of the form yx − z D c. If c D 0 we get the union of the xz -plane and the plane x D z.
If c Z 0 we get the hyperbola in the xy-plane y D c/x; this generates the solution surfaces when translated by
m1, 0, −1.
33. (a) These are cylinders
√ with the z-axis being the axis of the cylinder. For the surface at level w D c, the radius
of the cylinder is c.
(b) This is related to Exercise 25. A level surface at w D c will be the surface generated by building a cylinder
on the curve hx, y D c in the z D 0 plane. You are dragging the curve both directions along the z-axis so
that all cross sections for z D c1 look identical.
(c) Same thing in the y direction.
(d) If you said “same thing in the x direction,” read the problem again. You are solving equations that look like
hx D c. For each xi that solves this equation, you have no dependency on y or z so the level set looks
like a plane in R3 parallel to the yz -plane of the form x D xi .
34. (a) F is, of course, not uniquely determined. But if we let F x, y, z D x2 C xy − xz − 2, then the surface
is the level set F x, y, z D 0.
x2 C xy − 2
(b) x2 C xy − xz D 2 is equivalent to z D
D f x, y.
x
35. The ellipsoid is pictured below left. To see why you couldn’t express the surface as one function z D f x, y,
look for example at the intersection of the ellipsoid and the plane y D 0 pictured below on the right.
z
1
0.75
0.5
0.25
z
-2 -1.5 -1 -0.5
-0.25
0.5
1
1.5
2
x
-0.5
y
x
-0.75
-1
You can see that for −2
< x < 2 there correspond two values of z. We could express
the top portion of the
ellipsoid as f x, y D 1 − x2 /4 C y2 /9 and the bottom portion as gx, y D − 1 − x2 /4 C y2 /9.
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“runall” — 2005/8/1 — 12:14 — page 77 — #8
Section 2.1
Functions of Several Variables; Graphing Surfaces
77
36. The figure is a hyperbolic paraboloid shown below left.
z
y
x
37. The only difference here is that z is squared. Here we get a cone with axis of symmetry the x-axis. The figure is
shown above right.
38. This is Exercise 36 with the roles of x, y and z permuted and a change in the constants. The figure is shown
below left.
2
1
y
0
-1
z
-2
2
y
1
z
0
x
-1
-2
0
0.5
x
1
39. This is “cone” where the cross sections are ellipses, not circles. The figure is shown above right.
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“runall” — 2005/8/1 — 12:14 — page 78 — #9
78
Chapter 2
Differentiation in Several Variables
40. We see the figure is a hyperboloid. It is shown below left.
20
z
y
0
-20
10
y
z
0
x
-10
-10
0
x
10
41. This is a hyperboloid of two sheets. It is shown above right.
42. Here we have the parabola z D y2 C 2 translated arbitrarily in the x direction. This is what we call a cylinder
over the parabola z D y2 C 2.
3
2.75
z 2.5
2.25
2
-1
1
0.5
0
-0.5
y
-0.5
0
x
0.5
1
-1
Note: Except that your students have to complete the square first, these are similar to Exercises 36–42 above. You may
want them to be more explicit in reporting the translation as that’s sometimes hard to pick up from a diagram.
43. This is the equation of an elliptic cone with vertex at 1, −1, −3. The graph is shown below left.
z
z
y
x
y
x
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“runall” — 2005/8/1 — 12:14 — page 79 — #10
Section 2.2
Limits
79
44. Here we have an elliptic paraboloid. The graph is shown above right.
45. This is the equation of an ellipsoid 4x C 12 C y2 C z2 D 4. The graph is shown below left.
6
4
y
2
0
z
-2
2
1
z 0
-1
x
y
-2
-3
-2
-1
x
0
1
46. This is the equation of a hyperboloid of one sheet 4x C 12 C y − 22 − 4z2 D 4. The graph is shown
above right.
47. This is similar to Exercise 44. The equation is equivalent to z − 1 D x − 32 C 2y2 .
48. Here we get 9x2 C 4y − 12 − 36z C 42 D 684 which is similar to Exercise 46.
2.2 LIMITS
Note: In Exercises 1–6, the rule of thumb is that a set is closed if it contains all of its boundary points.
1. This is an annulus which doesn’t include its inner or outer boundary and so is open.
2. This is an annulus which includes all of its boundary points and so is closed.
3. This is an annulus which includes its inner boundary but not its outer boundary and so it is neither open nor
closed.
4. This is a hollowed out sphere which includes its boundary points and so is closed.
5. This may be a bit harder to see. This is the union of an infinite open strip in the plane −1 < x < 1 and a
closed line in the plane x D 2 and so is neither open nor closed.
6. This is the open infinite cylinder in R3 and so is open. You could follow up on this by asking about {x, y, z ∈
R3 |1 … x2 C y2 … 4}.
Note: As pointed out in the text, the most common and convincing way to prove that a limit of a function with domain in
R2 doesn’t exist is to show that you get two different answers when you follow two different paths. After doing Exercises
7–18 students may get in the habit of thinking that it is sufficient to check a few straight paths. Exercise 23 should make
them think twice.
7. There’s no trick to taking this limit. Just let x, y, z ' 0, 0, 0 and x2 C 2xy C yz C z3 C 2 ' 2.
√ |y|
8. We can see that
lim
doesn’t exist by looking at the limit along the paths x D 0 and y D 0. On
2
2
x Cy
x,y'0,0
the one hand
lim
0,y'0,0
|y|
x2 C y 2
|y|
D D1
y2
while
lim
x,0'0,0
|y|
x2 C y 2
0
D √ D 0.
x2
9. Again, the limit does not exist.
x C y2
x2 C 2xy C y2
2xy
D
lim
D1 C
lim
.
2
2
x,y'0,0 x C y
x,y'0,0
x,y'0,0 x2 C y2
x2 C y 2
lim
When x D y,
1 C
lim
2xy
x,y'0,0 x2 C y2
D 1 C lim
2x2
x'0 x2 C x2
D 1 C 1 D 2.
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“runall” — 2005/8/1 — 12:14 — page 80 — #11
80
Chapter 2
Differentiation in Several Variables
When x D 0,
1 C
2xy
lim
0,y'0,0 x2 C y2
D 1 C lim
0
y'0 y2
D 1.
10. Here nothing goes wrong so we can evaluate the limit by substituting in the expression.
e0e0
1
exey
D
D .
x,y'0,0 x C y C 2
0 C 0 C 2
2
lim
11. No limit exists.
2x2 C y2
x2
D1 C
lim
.
x,y'0,0 x2 C y2
x,y'0,0 x2 C y2
lim
We reason, as above, that if x D y then the limit is 3/2, but if y D 0 the limit is 2.
12. Here we can evaluate the function at the limit point and find that
2x2 C y2
6
D .
x,y'−1,2 x2 C y2
5
lim
13. Just as with limits in first semester Calculus, this is begging to be simplified.
x C y2
x2 C 2xy C y2
D
lim
D
lim x C y D 0.
x,y'0,0
x,y'0,0 x C y
x,y'0,0
x C y
lim
14. This is the same as the limit in Exercise 9 (once we simplified it). The limit does not exist.
15. This, too, is begging to be simplified.
x2 − y2 x2 C y2 x4 − y 4
D
lim
D
lim x2 − y2 D 0.
x,y'0,0 x2 C y2
x,y'0,0
x,y'0,0
x2 C y 2
lim
16. This is the same as the limit in Exercise 11 (once we simplified it). The limit does not exist.
17. This is another standard trick from first year Calculus.
√
√
√
√
x x C y x − y
xx − y
x2 − xy
lim
D
lim
D
lim
√
√
√
√
√
√
x,y'0,0,xZy
x,y'0,0,xZy
x,y'0,0,xZy
x − y
x − y
x − y
√
√
x x C y D 0.
D
lim
x,y'0,0,xZy
18. You can see that you would get different values depending on the path you took to x, y D 2, 0. If you
followed the path 2, y ' 2, 0 the limit would be −1. If you followed the path x, 0 ' 2, 0 the limit
would be 1. So the limit doesn’t exist.
√
19. The function is continuous so the limit is f 0, π, 1 D e0 cos π − 0 D −1.
20. As in Exercise 18, you get different values depending on the path you choose. Look, for example, at paths along
the three axes. Along x, 0, 0 ' 0, 0, 0 the limit is 2, along 0, y, 0 ' 0, 0, 0 the limit is 3 and along
0, 0, z ' 0, 0, 0 the limit is 1. We can see that no limit can exist.
21. Again the limit doesn’t exist because the value would differ on different paths. If you followed a path t, t, t '
0, 0, 0 the limit would be 1/3. If you followed the path x, 0, 0 ' 0, 0, 0 the limit would be 0.
22. (a) We know from single-variable calculus (either using l’Hôpital’s rule or the direct geometric argument) that
lim
θ'0
(b)
sin θ
D 1.
θ
sinxCy
D lim sin θ D 1.
xCy
θ'0 θ
sinxy
D lim sin θ D 1.
lim
θ'0 θ
x,y'0,0 xy
lim
x,y'0,0
(c)
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“runall” — 2005/8/1 — 12:14 — page 81 — #12
Section 2.2
Limits
81
Note: Exercise 23 is a classic and cool problem. You may wish to set it up in class before assigning it. Write the function
on the board and ask the students to evaluate the limit or explain why the limit fails to exist. For those who get it right,
this is wonderful. For those who get it wrong, they are now in a position to appreciate the subtlety of the problem.
x4 y 4
23. Our goal is to evaluate limx,y'0,0
or explain why the limit fails to exist. We divide the answer
x2 C y4 3
into parts to make it easier to follow—there are no corresponding parts (a)–(d) in the text.
(a) If you evaluate the limit along the lines x D 0 and y D 0 the limit is 0. We might be tempted to guess that
limx,y'0,0 f x, y D 0 but as we saw in Exercise 14, we could get a limit of 0 along the paths x D 0
and y D 0 but perhaps not along x D y.
(b) So now let’s follow the line y D mx into the origin and see where f heads off to.
x4 y 4
lim
x,y'0,0,yDmx x2 C y4 3
D lim
x4 mx4
x'0 x2 C mx4 3
m4 x8
x'0 x2 1 C m4 x2 3
D lim
x8
D m4 lim
x'0 x6 1 C m4 x2 3
D m4 lim
x2
x'0 1 C m4 x2 3
D 0.
This means then if we head into the origin along any straight line the limit of f is 0. Here is the point of this
problem: If we head into the origin in any constant direction, the limit of f is 0 and yet limx,y'0,0 f x, y
does not exist!
(c) For the limit to exist f must approach the same number no matter what path we choose to take to the origin.
So let’s approach along the parabola x D y2 .
y2 4 y4
x4 y 4
D lim
y'0 y2 2 C y4 3
x,y'0,0,xDy2 x2 C y4 3
lim
y12
y'0 2y4 3
D lim
y12
1
D .
y'0 8y12
8
D lim
(d) So we get different answers for limx,y'0,0
So the limit does not exist.
x4 y 4
depending on what path we follow into the origin.
x2 C y4 3
Note—In Exercises 24–27 your students may find better visual information by using a contour plot than a threedimensional plot.
24. Below see two graphs of the function. The three dimensional plot makes it seem as if there are mountains and
valleys quite close to the origin. The contour plot helps you see from the diagonal lines that meet at the origin
that the limit doesn’t exist.
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“runall” — 2005/8/1 — 12:14 — page 82 — #13
82
Chapter 2
Differentiation in Several Variables
0.4
0.2
0
-0.2
-0.4
-0.4
-0.2
0
0.2
0.4
Analytically, f is equivalent to 1 C x2 C 2xy/3x2 C 5y2 . Head in toward the origin on a path where
x D y and the limit is 13/8. Head in toward the origin on a path where x D 0 and the limit is 1. Head in toward
the origin on a path where y D 0 and the limit is 11/3. So the limit doesn’t exist.
25. Below see two graphs of the function. You actually get most of the picture from the three dimensional graph—
except that it looks as if things are joined smoothly. The contour plot shows the dramatic problems near the origin.
Particularly if you look along the vertical line x D 0 you’ll see that the limit does not exist at the origin.
0.4
0.2
0
-0.2
-0.4
-0.4 -0.2
0
0.2
0.4
Analytically, look at the path x D 0. Here we’re looking at the graph of z D −1/y. The limits as we approach
from positive and negative y values is ;q so no limit exists.
26. In the three-dimensional graph below you can see that the extreme behavior calms down near the origin. This is
confirmed in the contour plot. From the graphs it appears that the limit exists at the origin.
0.4
0.2
0
-0.2
-0.4
-0.4 -0.2
0
0.2
0.4
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“runall” — 2005/8/1 — 12:14 — page 83 — #14
Section 2.2
Limits
83
Before exploring this one analytically, consider the graph gx, t D xt/x2 C t 2 Its contour plot is shown
below.
0.4
0.2
0
-0.2
-0.4
-0.4 -0.2
0
0.2
0.4
So really the problem we are considering is the same with t D y5 . We’re not looking along a path that shows
us enough. Let’s look at the limit for our original function f as we approach the origin. Along a path where
x D 0 or y D 0 the limit is 0. Along a path where x D y5 the limit is 1/2. This is a good place to encourage your
students to be careful drawing conclusions from even very good graphs.
27. You’d think we would have learned our lesson from Exercise 26. On the other hand, it sure looks as if things are
calming down near the origin. Sure sin 1/y oscillates madly between −1 and 1 but x seems to dampen it. We’ll
boldly assert that the limit exists at the origin.
0.4
0.2
0
-0.2
-0.4
-0.4 -0.2
0
0.2
0.4
Actually, the discussion above leads us to the truth. The product of a bounded function and one going to 0
goes to 0. The limit exists and is 0.
28.
x2 y
lim
x,y'0,0 x2 C y2
29.
lim
r2 cos2 θ · r sin θ
D lim r sin θ cos2 θ D 0
r'0
r'0
r2
D lim
x2
x,y'0,0 x2 C y2
D lim
r'0
r2 cos2 θ
D lim cos2 θ D cos2 θ
r'0
r2
Limit does not exist as the result depends on θ.
30.
r2 C r cos θr sin θ
x2 C xy C y2
D lim
D lim 1 C cos θ sin θ D 1 C cos θ sin θ.
r'0
r'0
x,y'0,0
x2 C y 2
r2
lim
Thus the limit does not exist.
31.
lim
xyz
x,y,z'0,0,0 x2 C y2 C z2
D lim
ρ'0
ρ sin ϕ cos θρ sin ϕ sin θρ cos ϕ
ρ2
D lim ρ sin2 ϕ cos ϕ cos θ sin θ D 0
ρ'0
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“runall” — 2005/8/1 — 12:14 — page 84 — #15
84
Chapter 2
Differentiation in Several Variables
32.
lim
x,y,z'0,0,0
33.
x2 C y 2
x 2 C y 2 C z2
ρ2 sin2 ϕ cos2 θ C ρ2 sin2 ϕ sin2 θ
D lim ρ sin2 ϕ D 0
ρ'0
ρ'0
ρ
D lim
ρ2 sin ϕ cos ϕ cos θ
xz
D lim
D lim sin ϕ cos ϕ cos θ D sin ϕ cos ϕ cos θ
ρ'0
ρ'0
x,y,z'0,0,0 x2 C y2 C z2
ρ2
lim
The Limit does not exist.
In Exercises 34–41: as the rules on continuity show, if the components are continuous and we put the functions together
by adding, subtracting, multiplying, or composing, then the result is continuous. It should be clear to the students what
points need checking.
34. This is a polynomial and is continuous everywhere.
35. This too is a polynomial and is continuous everywhere.
To make the point about composition, you may want to assign Exercises 36 and 37 together.
36. The only place we could get into trouble is where the denominator is 0, but x2 C 1 Z 0 so g is always continuous.
37. Here we are composing a continuous function (cos) with the continuous function g from Exercise 22, so the
composition is continuous.
38. You can even rewrite the function as cos x2 − 2sin xy2 so that it is clear that this is just the composition of
continuous functions.
39. The only place we need to check is the origin. We need to show that the limit of f as we approach (0, 0) is 0. If
we add and subtract y2 to the numerator we find that:
x2 − y 2
y2
D1 − 2
lim
.
2
2
2
x,y'0,0 x C y
x,y'0,0 x C y2
lim
In Exercise 16 we showed that this limit doesn’t exist (in this case you get two different answers if you follow
the paths y D 0 and y D x) and so f is not continuous at (0, 0).
40. As in Exercise 39, the only point we need to check is the origin.
x2 C y2 x C 1
x3 C x2 C xy2 C y2
D
lim
D
lim x C 1 D 1.
x,y'0,0
x,y'0,0
x,y'0,0
x2 C y 2
x2 C y 2
lim
The good news is that the limit exists, the bad news is that
lim
x,y'0,0
gx, y D 1 Z 2 D g0, 0,
so g is not continuous at the origin.
41. A vector-valued function is continuous if each of its component functions is continuous. Each clearly is, so F is
continuous.
42. Notice that when x, y Z 0,
x2 C y2 x C 2
x3 C xy2 C 2x2 C 2y2
D
D x C 2.
x2 C y 2
x2 C y 2
So c D 2 and the function gx, y is seen to be equivalent to x C 2.
43. Here you can view f as being a function R3 ' R; then f x1 , x2 , x3 D 2x1 − 3x2 C x3 which is linear in
x1 , x2 , and x3 and therefore continuous.
44. This is equivalent to fx, y, z D −5y, 5x − 6z, 6y. Since each of the component functions from R3 ' R is
continuous, so is f.
We make students do at least a few of the following because “it’s good for them.” Exercise 45 is a review of how they
looked at limits in first semester Calculus—it prepares them for Exercise 46. Exercise 47 is a generalization of Exercise 46.
45. Here f x D 2x − 3.
(a) If |x − 5| < δ, then |f x − 7| D |2x − 3 − 7| D |2x − 10| D 2|x − 5| < 2δ.
(b) For any > 0, if 0 < |x − 5| < /2, then |f x − 7| < . This means that limx'5 f x D 7.
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“runall” — 2005/8/1 — 12:14 — page 85 — #16
Section 2.2
Limits
85
46. Now the function is f x, y D 2x − 10y C 3.
(a) Really we’re just arguing that the hypotenuse of a right triangle is at least as long as either leg.
δ > ||x, y − 5, 1|| D x − 52 C y − 12 Ú x − 52 D |x − 5|.
And
δ > ||x, y − 5, 1|| D x − 52 C y − 12 Ú y − 12 D |y − 1|.
(b) First:
|f x, y − 3| D |2x − 10y C 3 − 3| D |2x − 10y| D |2x − 5 − 10y − 1|.
(c) By the triangle inequality
|2x − 5 − 10y − 1| … |2x − 5| C |10y − 1| D 2|x − 5| C 10|y − 1|.
But we are assuming that ||x, y − 5, 1|| < δ and from part (a) we know that this implies that |x − 5| < δ
and |y − 1| < δ, so
2|x − 5| C 10|y − 1| < 2δ C 10δ D 12δ.
(d) We put these together to obtain: For any ε > 0, if 0 < ||x, y − 5, 1|| < ε/12, then |f x, y − 3| < ε.
In other words,
f x, y D 3.
lim
x,y'5,1
47. This is just a generalization of Exercise 46. We can use the same steps outlined there:
(a)
δ > ||x, y − x0 , y0 || D x − x0 2 C y − y0 2 Ú x − x0 2 D |x − x0 |.
And
δ > ||x, y − x0 , y0 || D x − x0 2 C y − y0 2 Ú y − y0 2 D |y − y0 |.
(b) Assume that ||x, y − x0 , y0 || < δ, then follow the steps in part (b) of Exercise 46:
|f x, y − Ax0 C By0 C C| D |Ax C By C C − Ax0 C By0 C C|
D |Ax − x0 C By − y0 | … |Ax − x0 | C |By − y0 |
D |A||x − x0 | C |B||y − y0 | < |A|δ C |B|δ D |A| C |B|δ.
(c) Now we’re ready to put this together: For any ε > 0, if 0 < ||x, y − x0 , y0 || < ε/|A| C |B|, then
|f x, y − Ax0 C By0 C C| < ε. In other words,
lim
x,y'x0 ,y0 f x, y D Ax0 C By0 C C.
48. (a) This is really what we just showed in Exercise 47 with x0 D 0 and y0 D 0.
√
||x, y|| D x2 C y2 Ú x2 D |x|.
And
||x, y|| D x2 C y2 Ú y2 D |y|.
3
3
3
3
3
3
(b) We
follow the hint given in the text: |x C y | … |x | C |y | D |x| C |y| . But by part (a), |x| … ||x, y|| D
2
2
2
2
x C y , and |y| … ||x, y|| D x C y . Therefore,
|x3 C y3 | … |x|3 C |y|3 … 2 x2 C y2 3 D 2x2 C y2 3/2 .
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“runall” — 2005/8/1 — 12:14 — page 86 — #17
86
Chapter 2
Differentiation in Several Variables
(c) If 0 < ||x, y|| < δ then by part (b),
3
x C y3 2x2 C y2 3/2 … D 2 x2 C y2 D 2||x, y|| < 2δ.
2
x C y2 x2 C y 2
x3 C y 3
can be made to be arbitrarily close to 0 by choosing (x, y) close
x2 C y 2
enough to the origin. This means that the limit is 0.
Assemble the pieces: For any ε > 0, if 0 < ||x, y|| < ε/2, then |f x, y| < ε. This shows that
(d) First we know by part (c) that
x3 C y 3
D 0.
x,y'0,0 x2 C y2
lim
49. (a) 0 … a C b2 D a2 C 2ab C b2 , so −2ab … a2 C b2 . Also 0 … a − b2 D a2 − 2ab C b2 , so
2ab … a2 C b2 . We combine these two results to get: 2|ab| … a2 C b2 .
(b) If ||x, y|| < δ, then we’ll use part (a) to rewrite |xy| in the following calculation:
1/2x2 C y2 |x2 − y2 |
|xy||x2 − y2 |
x2 − y 2 1
…
D |x2 − y2 |.
D
xy 2
x C y2 x2 C y 2
x2 C y 2
2
We can apply part (a) again with a D x C y and b D x − y so that
|x C yx − y| …
x C y2 C x − y2
D x2 C y 2 .
2
Noting that x2 C y2 D ||x, y||2 D δ2 , we have:
x2 − y 2 δ2
1
xy 2
… |x2 − y2 | D .
2
x C y 2
2
(c) As in Exercise 48, the limit has to be 0 because we can make f as small as we want by choosing (x, y) close
enough to the origin.
√
We summarize the above as: For any ε > 0, if 0 < ||x, y|| < 2ε, then |f x, y| < ε. This shows
that
x2 − y 2 lim
xy 2
D 0.
x,y'0,0 x C y2 2.3 THE DERIVATIVE
The general strategy for Exercises 1–11 is to treat all variables except for the one with respect to which we are differentiating as constants.
1. f x, y D xy2 C x2 y, so ⭸f /⭸x D y2 C 2xy, and ⭸f /⭸y D 2xy C x2 .
2
2
2
2
2
2
2. f x, y D ex Cy , so ⭸f /⭸x D 2xex Cy , and ⭸f /⭸y D 2yex Cy .
3. f x, y D sin xy C cos xy, so ⭸f /⭸x D y cos xy − y sin xy, and ⭸f /⭸y D x cos xy − x sin xy.
x3 − y 2
, so
4. f x, y D
1 C x2 C 3y4
1 C x2 C 3y4 3x2 − x2 − y2 2x
⭸f
D
⭸x
1 C x2 C 3y4 2
and
1 C x2 C 3y4 −2y − x2 − y2 12y3 ⭸f
.
D
⭸y
1 C x2 C 3y4 2
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“runall” — 2005/8/1 — 12:14 — page 87 — #18
Section 2.3
The Derivative
87
x2 C y2 2x − x2 − y2 2x
4xy2
⭸f
x2 − y 2
D
, so
D
2
2
2
2
2
2
x C y
⭸x
x C y x C y2 2
2
2
2
2
2
x C y −2y − x − y 2y
⭸f
−4x y
D
and
D
.
⭸y
x2 C y2 2
x2 C y2 2
⭸f
⭸f
2y
2x
1
6. f x, y D lnx2 C y2 , so
2x D
and
.
D
D
⭸x
x2 C y 2
x2 C y 2
⭸y
x2 C y 2
⭸f
⭸f
D − sin x3 y3yx2 D −3x2 y sin x3 y and
D −x3 sin x3 y.
7. f x, y D cos x3 y, so
⭸x
⭸y
8. F x, y, z D xyz, so ⭸F /⭸x D yz, ⭸F /⭸y D xz, and ⭸F /⭸z D xy.
9. F x, y, z D x2 C y2 C z2 D x2 C y2 C z2 1/2 . The partials derivatives are:
5. f x, y D
2x
x
⭸F
D D ,
2
2
2
2
⭸x
2 x C y C z
x C y 2 C z2
y
⭸F
D ⭸y
x 2 C y 2 C z2
and,
z
⭸F
D .
2
⭸z
x C y 2 C z2
10. F x, y, z D eax cos by C eaz sin bx so
⭸F
D aeax cos by C beaz cos bx,
⭸x
⭸F
D −beax sin by, and
⭸y
⭸F
D aeaz sin bx.
⭸z
11. F x, y, z D
x C y C z
1 C x2 C y2 C z2 3/2
Fx x, y, z D
D
1 C x2 C y2 C z2 3/2 − x C y C z3/21 C x2 C y2 C z2 1/2 2x
1 C x2 C y2 C z2 3
1 − 2x2 C y2 C z2 − 3xy − 3xz
1 C x2 C y2 C z2 5/2
Fy x, y, z D
1 C x2 − 2y2 C z2 − 3xy − 3yz
, and
1 C x2 C y2 C z2 5/2
Fz x, y, z D
1 C x2 C y2 − 2z2 − 3xz − 3yz
.
1 C x2 C y2 C z2 5/2
12. F x, y, z D sin x2 y3 z4 so this is similar to Exercise 7 above. Fx x, y, z D 2xy3 z4 cos x2 y3 z4 , Fy x, y, z D
3x2 y2 z4 cos x2 y3 z4 and Fz x, y, z D 4x2 y3 z3 cos x2 y3 z4 .
x3 C yz
13. F x, y, z D
We’ve seen this form a couple of times by now.
x2 C z2 C 1
Fx x, y, z D
x2 C z2 C 13x2 − x3 C yz2x
x4 C 3x2 z2 C 3x2 − 2xyz
D
2
2
2
x C z C 1
x2 C z2 C 12
Fy x, y, z D
x2 C z2 C 1z − x3 C yz0
z
D
2
2
2
2
x C z C 1
x C z2 C 1
Fz x, y, z D
x2 C z2 C 1y − x3 C yz2z
x2 y − yz2 C y − 2x3 z
D
x2 C z2 C 12
x2 C z2 C 12
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“runall” — 2005/8/1 — 12:14 — page 88 — #19
88
Chapter 2
Differentiation in Several Variables
The gradient of f is the function fx x, y, z, fy x, y, z, fz x, y, z. In Exercises 14–19 we are evaluating the
gradient at a given point.
14. f x, y D x2 y C ey/x , so §f x, y D 2xy C −y/x2 ey/x , x2 C 1/xey/x . This means that §f 1, 0 D
0, 2.
x − y
15. f x, y D
, so
x2 C y 2 C 1
§f x, y D D
x2 C y2 C 11 − x − y2x x2 C y2 C 1−1 − x − y2y
,
x2 C y2 C 12
x2 C y2 C 12
−x2 C y2 C 1 C 2xy −x2 C y2 − 1 − 2xy
,
.
x2 C y2 C 12
x2 C y2 C 12
So
§f 2, −1 D −
6 0
1
, D − , 0 .
36 36
6
16. f x, y, z D sin xyz, so §f x, y, z D cos xyzyz, xz, xy. This means that
§f π, 0, π/2 D cos 00, π2 /2, 0 D 0, π2 /2, 0.
17. f x, y, z D xy C y cos z − x sin yz, so §f x, y, z D y − sin yz, x C cos z − xz cos yz, −y sin z −
xy cos yz. So,
§f 2, −1, π D −1 − sin−π, 2 C cosπ − 2π cos−π, sinπ C 2 cos−π
D −1, 1 C 2π, −2.
18. f x, y D exy C lnx − y, so §f x, y D yexy C 1/x − y, xexy − 1/x − y. This means that
§f 2, 1 D e2 C 1, 2e2 − 1.
19. f x, y, z D x C ye−z , so §f x, y, z D e−z , e−z , −x C ye−z . So, §f 3, −1, 0 D 1, 1, −2.
The nth row of the derivative matrix is the gradient of the nth component function.
20. f x, y D x , Dfx, y D 1 , −x
. So Df3, 2 D [1/2, −3/4].
y
y y2
21. fx, y, z D xyz, x2 C y2 C z2 , so

yz
xz

Dfx, y, z D 
y/ x2 C y2 C z2
x/ x2 C y2 C z2
xy
z/ x2 C y2 C z2


.
This means,
Df1, 0, −2 D 0√
1/ 5
−2
0
0√
.
−2/ 5
22. ft D t, cos 2t, sin 5t, so

1
Dft D  −2 sin 2t 
5 cos 5t

1
Df0 D  0  .
5


and so
23. fx, y, z, w D 3x − 7y C z, 5x C 2z − 8w, y − 17z C 3w so


3 −7
1
0
0
2 −8  .
Dfx, y, z, w D  5
0
1 −17
3
Since all of the entries are constant, the matrix doesn’t depend on a.
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“runall” — 2005/8/1 — 12:14 — page 89 — #20
Section 2.3
The Derivative
89
24. fx, y D x2 y, x C y2 , cos πxy, so

2xy
Dfx, y D 
1
−πy sin πxy
This means,


4
−2  .
0


0
s .
2t


0
−1  .
2
−4
Df2, −1 D  1
0
25. fs, t D s2 , st, t 2 , so
2s
Dfs, t D  t
0
This means,

x2
.
2y
−πx sin πxy
−2
Df−1, 1 D  1
0
We will appeal to Theorem 3.5 for Exercises 26–28.
26. f x, y D xy − 7x8 y2 C cos x is differentiable because the two partials fx x, y D y − 56x7 y2 − sin x and
fy x, y D x − 14x8 y are continuous.
x C y C z
27. f x, y, z D
is differentiable because the three partials
2
x C y 2 C z2
fx x, y, z D
−x2 C y2 C z2 − 2xy − 2xz
x2 C y2 C z2 2
fy x, y, z D
x2 − y2 C z2 − 2xy − 2yz
x2 C y2 C z2 2
fz x, y, z D
x2 C y2 − z2 − 2xz − 2yz
x2 C y2 C z2 2
are all continuous.
xy2
y
x
28. fx, y D , C is differentiable because the partials in the matrix
2
4
x C y y
x

y 6 − x2 y 2

x2 C y4 2
Dfx, y D 
 1
y
−
y
x2

2x3 y − 2xy5

x2 C y4 2 
−x
1 
C
y2
x
are continuous in the domain of f.
29. (a) The graph of z D x3 − 7xy C ey has continuous partial derivatives at −1, 0, 0.
(b) By Theorem 3.3, the equation for the tangent plane is: z D f −1, 0 C fx −1, 0x − −1 C
fy −1, 0y − 0. In this case fx x, y D 3x2 − 7y so fx −1, 0 D 3. Also fy x, y D −7x C ey and
so fy −1, 0 D 8. The equation of the plane is z D 3x C 1 C 8y.
30. Again using Theorem 3.3, the equation for the tangent plane is: z D f π/3, 1 C fx π/3, 1x − π/3 C
fy π/3, 1y −
√ 1. Here z D 4 cos xy,√so fx x, y D −4y sin xy and fy x, y D −4x sin xy. Plugging in we
get z D 2 − 2 3x − π/3 − 2π/ 3y − 1.
31. Again using Theorem 3.3, the equation for the tangent plane is: z D f 0, 1 C fx 0, 1x C fy 0, 1y − 1.
Here z D exCy cos xy, so fx x, y D exCy cos xy − y sin xy and fy x, y D exCy cos xy − x sin xy. Plugging
in we get z D e − ex C ey − 1 or z D −ex C ey.
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“runall” — 2005/8/1 — 12:14 — page 90 — #21
90
Chapter 2
Differentiation in Several Variables
32. First find the two partials fx x, y D 2x − 6 and fy x, y D 3y2 . Then putting the tangent plane equation into the
same form as the plane 4x − 12y C z D 7 gives us z − 2a − 6x − a − 3b2 y − b D a2 − 6a C b3
or z − 2a − 6x − 3b2 y D −a2 − 2b3 . So 2a − 6 D −4 so a D 1 and 3b2 D 12 so b D ;2. This
gives two tangent planes. The equation for one is 4x − 12y C z D −17 and the equation for the other is
4x − 12y C z D 15.
33. For f x1 , ... , x4 D 10 − x12 C 3x22 C 2x32 C x42 , we have
§f D −2x1 , −6x2 , −4x3 , −2x4 so
§f 2, −1, 1, 3 D −4, 12, −4, −6.
Formula (8) gives that the hyperplane has equation
x5 D −8 C −4, 12, −4, −6x1 − 2, x2 C 1, x3 − 1, x4 − 3
D −8 − 4x1 − 2 C 12x2 C 1 − 4x3 − 1 − 6x4 − 3
or
x5 D −4x1 C 12x2 − 4x3 − 6x4 C 34.
34. (a)
f 1.98, 3 − f 2, 3
12.1 − 12
.1
D
D
D −5
1.98 − 2
−.02
−.02
f 2, 3.01 − f 2, 3
12.2 − 12
.2
fy 2, 3 L
D
D
D 20
3.01 − 3
0.01
.01
fx 2, 3 L
Thus, formula (4) of §2.3 would give an approximate equation for the tangent plane as
z D f 2, 3 C fx 2, 3x − 2 C fy 2, 3y − 3 L 12 − 5x − 2 C 20y − 3
or
z D −5x C 20y − 8.
(b)
f 1.98, 2.98 L 12 − 51.98 − 2 C 202.98 − 3 D 12 − 5−0.02 C 20−0.02
D 11.7
Exercises 35–37 have the student investigate the linear approximation h of f near a given point a. We use the formula
in Definition 3.8:
hx D fa C Dfax − a.
35. Here f x, y D exCy so the partials are fx x, y D exCy D fy x, y.
(a) h.1, −.1 D f 0, 0 C e0 , e0 · .1, −.1 D 1.
(b) f .1, −.1 D e0 D 1. So the approximation is exact.
36. Here f x, y D 3 C cos πxy so the partials are fx x, y D −πy sin πxy and fy x, y D −πx sin πxy.
(a) h.98, .51 D 3 C cos π1.5 − π.5 sin[π1.5], π1 sin[π1.5] · −.02, .01 D 3 −
π.5, 1 · −.02, .01 D 3.
(b) f .98, .51 D 3 C cos π.98.51 L 3.00062832.
37. f x, y, z D x2 C xyz C y3 z, so the partials are fx x, y, z D 2x C yz, fy x, y, z D xz C 3y2 z, and
fz x, y, z D xy C y3 .
(a) h1.01, 1.95, 2.2 D f 1, 2, 2 C fx 1, 2, 2, fy 1, 2, 2, fz 1, 2, 2 · .01, −.05, .2 D 21 C
6, 26, 10 · .01, −.05, .2 D 21.76.
(b) f 1.01, 1.95, 2.2 D 21.665725.
38.
x C x2 C · · · C xn
f x1 , x2 , ... , xn D 1
, so
x12 C x22 C · · · C xn2
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“runall” — 2005/8/1 — 12:14 — page 91 — #22
Section 2.3
fxi x1 , x2 , ... , xn D
D
The Derivative
91
x12 C x22 C · · · C xn2 − xi x1 C x2 C · · · C xn x12 C x22 C · · · C xn2 −1/2
x12 C x22 C · · · C xn2
x12 C x22 C · · · C xn2 − xi x1 C x2 C · · · C xn x12 C x22 C · · · C xn2 3/2
39. (a) For x, y Z 0, 0 we can find a neighborhood that misses the origin. In this neighborhood
f x, y D
xy2 − x2 y C 3x3 − y3
2x3
Dx − y C
.
2
2
2
x C y
x C y2
We can then easily compute the partials as
fx x, y D 1 C
2x4 C 6x2 y2
x2 C y2 2
and
fy x, y D −1 −
(b) Using Definition 3.2 of the partial derivative, if



2x3
 x − y C
2
f x, y D
x C y2


 0
then
and
4x3 y
x2 C y2 2
if x, y Z 0, 0
.
,
if x, y D 0, 0
f h, 0 − f 0, 0
⭸f
3h
0, 0 D lim
D lim
D 3,
h'0
h'0
⭸x
h
h
f 0, h − f 0, 0
⭸f
−h
0, 0 D lim
D lim
D −1.
h'0
h'0
⭸y
h
h
Note: Exercises 40–43 are review exercises for single-variable calculus. The idea is to see that near a point, the tangent
line approximates the curve. This idea will then be extended to a tangent plane and a surface in Exercises 45–49. For
Exercises 40–43 use either the point-slope equation y − f a D f ax − a or solve for y to get y D f ax C
f a − f aa.
40. For the tangent line to F x D x3 − 2x C 3 at a D 1 F x D 3x2 − 2 so F 1 D 1. The tangent line is
y D x C 1. The graph of F and the tangent line near x D 1 (in this case for .8 … x … 1.2) is shown below left.
√
41. For the tangent line to F x
D π/4 F x D√1 C cos x so F π/4 D 1 C 2/2. The
√ D x C sin x at a √
tangent line is y D 1 C 2/2x C π/4 C 2/2 − 1 C 2/2π/4. The graph of F and the tangent
line near x D π/4 is shown above right.
42. For the tangent line rewrite F x D x − 3 C 3/x2 C 1. F x D 1 − 6x/x2 C 12 so F 0 D 1 and
F 0 D 0. The tangent line is y D x. We can see that by looking at our rewritten version of F . The graph of F
and the tangent line near x D 0 is shown below left.
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“runall” — 2005/8/1 — 12:14 — page 92 — #23
92
Chapter 2
Differentiation in Several Variables
43. For the tangent line to F x D lnx2 C 1 at a D −1F x D 2x/x2 C 1 so F −1 D −1. The tangent
line is y D −x C ln 2 − 1. The graph of F and the tangent line near x D −1 is shown above right.
44. Looking at the graph below, we can see that there is a cusp at x D 2 (trust me, that’s where the cusp is). You
can also see that the limit of the derivative using points to the left of 2 would not be the same as the derivative
using points to the right of 2 as one set is negative and the other is positive. Finally, the tangent line looks to be
a vertical line. This has no slope and so the derivative wouldn’t exist.
45. (a) For the function f x, y D x3 − xy C y2 , fx x, y D 3x2 − y and fy x, y D −x C 2y. So at the point
(2, 1) these become f 2, 1 D 7, fx 2, 1 D 11, and fy 2, 1 D 0. The equation of the tangent plane is
z D 7 C 11x − 2.
(b)
(c) The partials are continuous so by Theorem 3.5, f is differentiable.
46. (a) To find the partial derivatives fx 1, 0 and fy 1, 0, we must look at appropriate partial functions of
f x, y D x − 1y2/3 :
f x, 0 K 0 * fx 1, 0 D 0
f 1, y K 0 * fy 1, 0 D 0
Since f 1, 0 D 0, the candidate tangent plane has equation z D 0 C 0x − 1 C 0y − 0 or z D 0.
(b) A computer graph looks as follows.
0.1
0.075
0.05
0.025
0
0.8
0.2
0.1
0
0.9
-0.1
1
1.1
-0.2
Zooming in closer to the point (1, 0, 0) doesn’t make things appear very different, so it’s tempting to conclude
that f must not be differentiable at (1, 0).
(c) From our calculations in part (a), the linear function hx, y D f 1, 0 C fx 1, 0x − 1 C fy 1, 0y −
0 D 0. Thus, for x, y Z 1, 0 we have
0 …
|f x, y − hx, y|
|f x, y|
D .
||x, y − 1, 0||
x − 12 C y2
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“runall” — 2005/8/1 — 12:14 — page 93 — #24
Section 2.3
The Derivative
93
Now
|f x, y| D |x − 1|2/3 |y|2/3 … x − 12 C y2 1/3 x − 12 C y2 1/3
D x − 12 C y2 2/3 .
Thus
|f x, y|
x − 12 C y2
…
x − 12 C y2 2/3
D x − 12 C y2 1/6 .
x − 12 C y2 1/2
Since this last expression approaches zero as x, y ' 1, 0, we see that f must be differentiable at (1, 0)
by Definition 3.4.
xy
−x2 y C y3 C y
47. (a) For the function f x, y D
, fx x, y D
and fy x, y D
x2 C y 2 C 1
x2 C y2 C 12
x3 − xy2 C y
. So at the point (0, 0) these become f 0, 0 D 0, fx 0, 0 D 0, and fy 0, 0 D 0. The
x2 C y2 C 12
equation of the tangent plane is z D 0.
(b) The surface is shown below left. It is shown with the tangent plane below right.
x
-2
-1
0
1
0.4
0.2
z 0
-0.2
-0.4
-2
2
1
0 y
-1
-1
0
x
2
0.4
0.2
z
0
-0.2
-0.4
-2
-1
1
2 -2
0
y
1
2
(c) This is the plane that best approximates the surface at that point. But we can see that it’s not a very good
approximation as you move way in any direction other than the two axes lines. Analytically, the reason is
that the partials are continuous in a neighborhood of (0, 0).
48. (a) For the function f x, y D sin x cos y, fx x, y D
√ cos x cos y and fy x, y D√− sin x sin y. So at the point
2/4,
π/6,
3π/4
these
become
f
π/6,
3π/4
D
−
√
√ fx π/6,√3π/4 D − 6/4, and
√ fy π/6, 3π/4 D
− 2/4. The equation of the tangent plane is z D − 2/4 − 6/4x − π/6 − 2/4y − 3π/4.
(b)
1
0.5
z 0
-0.5
-1
-0.5
3
0
2
0.5
x
1
2.5
y
1.5
1.5
(c) Again the partials are continuous in a neighborhood of π/6, 3π/4 so by Theorem 3.5, f is differentiable
at the point.
49. (a) For the function f x, y D x2 sin y C y2 cos x, fx x, y D 2x sin y −√y2 sin x and fy x, y D x2 cos y C
2
2
2y√cos x. So at √
the point π/3, π/4 these become
√ f π/3, π/4 D π 2/18 C π /32, fx π/3, π/4 D
2
2
2/3 − π 3/32, and fy√
π/3, π/4 D
of the tangent plane is z D
π √
√
√ π 2/18 C π/4. The equation
π2 2/18 C π2 /32 C π 2/3 − π2 3/32x − π/3 C π2 2/18 C π/4y − π/4.
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“runall” — 2005/8/1 — 12:14 — page 94 — #25
94
Chapter 2
Differentiation in Several Variables
(b)
4
2
z2
0
1.5
1 y
0
0.5
1
x
0.5
1.5
2 0
(c)
50. (a)
(b)
(c)
The partials are continuous near π/3, π/4 so by Theorem 3.5, f is differentiable there.
Yes gx, y D xy1/3 is continuous at (0, 0).
⭸g/⭸x D 1/3x−2/3 y1/3 , and ⭸g/⭸y D 1/3x1/3 y−2/3 .
Unfortunately we can’t just substitute the point (0, 0) in our answers to (b), but using Definition 3.2 of partial
derivatives, we see that the two partials must be 0. In other words we define gx 0, 0 D 0, and gy 0, 0 D 0.
(d) No (choose a path that crosses the x- and y-axes).
(e) You can see this answer if you look along the line y D x. There gx, x D x2/3 which has a corner at (0,
0). So there can’t be a tangent plane.
(f) No g isn’t differentiable at (0, 0).
51. If fx D Ax D nkD1 a1k xk , nkD1 a2k xk , ... , nkD1 amk xk . Let’s look at the entry in row i column j of Df(x).
This will be
n
⭸fi
⭸
D
aik xk D aij .
⭸xj
⭸xj kD1
So Dfx D A.
2.4 PROPERTIES; HIGHER-ORDER PARTIAL DERIVATIVES; NEWTON’S METHOD
In Exercises 1–4 there isn’t much to show ... the students just need to verify that the sum of the derivative is the derivative
of the sum (Proposition 4.1).
1. f x, y D xy C cos x, and gx, y D sinxy C y3 , so Df D [y − sin x, x], Dg D [y cos xy, x cos xy C 3y2 ],
and Df C g D [y − sin x C y cos xy, x C x cos xy C 3y2 ].
2. fx, y D exCy , xey , and gx, y D lnxy, yex , so
Df D e xCy
ey
and
Df + g D y
e xCy
, Dg D xy x
xey ye
y
xy
e y C yex
e xCy C
x
xy
ex
x
xy
xey C e x
e xCy C
.
Note the use of the product rule in Exercise 3 when calculating g1 x .
3. fx, y, z D x sin y C z, yex − 3x2 and gx, y, z D x3 cos x, xyz, so
Df D Df + g D sin y
−6x
x cos y
ez
1
3x2 cos x − x3 sin x
, Dg D z
ye
yz
sin y C 3x2 cos x − x3 sin x
−6x C yz
x cos y
e z C xz
0
xz
0
xy and
1
.
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“runall” — 2005/8/1 — 12:14 — page 95 — #26
Section 2.4
Properties; Higher-order Partial Derivatives; Newton’s Method
4. fx, y, z D xyz2 , xe−y , y sin xz and gx, y, z D x − y, x2 C y2 C z2 , lnxz C 2, so




1
−1
0
yz2
xz2
2xyz

 , Dg D 
2x
2y
2z
Df D  e −y
−xe−y
0
z/xz C 2 0 x/xz C 2
zy cos xz sin xz xy cos xz


1 C yz2
−1 C xz2
2xyz
.
Df + g D 
e −y C 2x
−xe−y C 2y
2z
zy cos xz C z/xz C 2
sin xz
xy cos xz C x/xz C 2
95
and
Exercises 5–8 are again mainly calculations to convince the students of the formulas given in Proposition 4.2; we hope
that they remember to apply them when confronted with a product or quotient. In Exercises 6 and 7 we notice that we just
get the quotient rule in each component which factors into the quotient rule given in the proposition (and we drop the
argument when convenient and clear).
f x, y
5. f x, y D x2 y C y3 , gx, y D x/y, f x, ygx, y D x3 C xy2 , and
D xy2 C y4 /x.
gx, y
So Df D [2xy, x2 C 3y2 ],
and
Dg D [1/y, −x/y2 ],
Dfg D [3x2 C y2 , 2xy]
D x2 y C y3 [1/y, −x/y2 ] C x/y[2xy, x2 C 3y2 ]
D fDg C gDf , and
f
D D [y2 − y4 /x2 , 2xy C 4y3 /x]
g
D y/x[2xy, x2 C 3y2 ] − y2 /x2 x2 y C y3 [1/y, −x/y2 ]
D
gDf − fDg
.
g2
6. f x, y D exy , gx, y D x sin 2y, f x, ygx, y D xexy sin 2y, and
So Df D [yexy , xexy ],
f x, y
exy
D
.
gx, y
x sin 2y
and Dg D [sin 2y, 2x cos 2y],
Dfg D [sin 2ye xy C xy exy , xxexy sin 2y C 2exy cos 2y]
D e xy [sin 2y, 2x cos 2y] C x sin 2y[yexy , xexy ]
D fDg C gDf , and
xyexy sin 2y − e xy sin 2y x2 e xy sin 2y − 2xexy cos 2y
f
D D ,
g
x2 sin2 2y
x2 sin2 2y
D
x sin 2y[yexy , xexy ] − e xy [sin 2y, 2x cos 2y]
x2 sin2 2y
D
gDf − fDg
.
g2
7. f x, y D 3xy C y5 , gx, y D x3 − 2xy2 , f x, ygx, y D 3x4 y C x3 y5 − 6x2 y3 − 2xy7 , and
f x, y
D
gx, y
3xy C y5
. So
x3 − 2xy2
Df D [3y, 3x C 5y4 ],
and Dg D [3x2 − 2y2 , −4xy],
Dfg D [12x3 y C 3x2 y5 − 12xy3 − 2y7 , 3x4 C 5x3 y4 − 18x2 y2 − 14xy6 ]
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“runall” — 2005/8/1 — 12:14 — page 96 — #27
96
Chapter 2
Differentiation in Several Variables
D 3xy C y5 [3x2 − 2y2 , −4xy] C x3 − 2xy2 [3y, 3x C 5y4 ]
D fDg C gDf , and
gx, yfx x, y − f x, ygx x, y gx, yfy x, y − f x, ygy x, y
f
D D ,
g
[gx, y]2
[gx, y]2
D
gDf − fDg
.
g2
8. f x, y, z D x cosyz, gx, y, z D x2 C x9 y2 C y2 z3 C 2, f x, ygx, y D x3 cosyz C x10 y2 cosyz C
f x, y
x cosyz
xy2 z3 cosyz C 2x cosyz, and
D
.
gx, y
x 2 C x 9 y 2 C y 2 z3 C 2
Df D [cosyz, −xz sinyz, −xy sinyz], and Dg D [2x C 9x8 y2 , 2x9 y C 2yz3 , 3y2 z2 ],
T

3x2 cos yz C 10x9 y2 cos yz C y2 z3 cos yz C 2 cos yz


Dfg D  −x3 z sin yz C 2x10 y cos yz − x10 y2 z sin yz C 2xyz3 cos yz − xy2 z4 sin yz − 2xz sin yz 
−x3 y sin yz − x10 y3 sin yz C 3xy2 z2 cos xy − xy3 z3 sin yz − 2xy sin yz
T
T


2x C 9x8 y2
cos yz
D x cos yz  2x9 y C 2yz3  C x2 C x9 y2 C y2 z3 C 2  −xz sin yz 
−xy sin yz
3y2 z2
So
D fDg C gDf, and
gf − fgx gfy − fgy gfz − fgz
f
D D x
,
,
g
g2
g2
g2
D
gDf − fDg
.
g2
In Exercises 9–17, students should verify that fxy D fyx . The fact that in these problems the derivative with respect
to y of fx is equal to the derivative with respect to x of fy is not trivial. Problem 18 explicitly asks them to examine the
mixed partials.
9. f x, y D x3 y7 C 3xy2 − 7xy so fx x, y D 3x2 y7 C 3y2 − 7y and fy x, y D 7x3 y6 C 6xy − 7x. The
second order partials are:
fxx x, y D 6xy7 ,
fxy x, y D fyx x, y D 21x2 y6 C 6y − 7, and
fyy x, y D 42x3 y5 C 6x.
10. f x, y D cosxy so fx x, y D −y sinxy and fy x, y D −x sinxy. The second order partials are:
fxx x, y D −y2 cos xy,
fxy x, y D fyx x, y D −xy cos xy − sin xy, and
fyy x, y D −x2 cos xy.
−y y/x
1
11. f x, y D ey/x − ye−x so fx x, y D
e
C ye−x and fy x, y D ey/x − e−x . The second order partials
x2
x
are:
fxx x, y D
2y y/x
y2 y/x
e
C
e
− ye−x ,
3
x
x4
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:14 — page 97 — #28
Section 2.4
Properties; Higher-order Partial Derivatives; Newton’s Method
97
y y/x
−1 y/x
e
−
e
C e −x , and
x2
x3
1
fyy x, y D e y/x .
x2
fxy x, y D fyx x, y D
12. f x, y D sin x2 C y2 so
x cos x2 C y2
fx x, y D x2 C y 2
y cos x2 C y2
fy x, y D .
x2 C y 2
and
The second order partials are:
fxx x, y D
√
−x sin x2 Cy2
x2 C y2 cos x2 C y2 C x √ 2 2 − x cos x2 C y2 √ 2x
x Cy
x2 C y 2
y2 cos x2 C y2 − x2 x2 C y2 sin x2 C y2
D
fyy x, y D
x Cy2
x2 C y2 3/2
x2 cos x2 C y2 − y2 x2 C y2 sin x2 C y2
, and by symmetry
, and
x2 C y2 3/2
−xy x2 C y2 sin x2 C y2 − xy cos x2 C y2
.
fxy x, y D fyx x, y D
x2 C y2 3/2
13. f x, y D
1
so
sin2 x C 2ey
fx x, y D
−2 sin x cos x
− sin 2x
D
sin2 x C 2ey 2
sin2 x C 2ey 2
and
fy x, y D
−2ey
.
sin x C 2ey 2
2
The second order partials are:
fxx x, y D
D
fxy x, y D fyx x, y D
fyy x, y D
sin2 x C 2ey 2 −2 cos 2x C sin 2x · 2sin2 x C 2ey sin 2x
sin2 x C 2ey 4
sin2 x C 2ey −2 cos 2x C 2 sin2 2x
,
sin2 x C 2ey 3
4ey sin 2x
, and
sin2 x C 2ey 3
2ey 2ey − sin2 x
.
sin2 x C 2ey 3
14. f x, y D ex Cy so fx x, y D 2xex Cy and fy x, y D 2yex Cy . The second order partials are:
2
2
2
2
2
2
fxx x, y D 2ex Cy C 2x · 2xex Cy
2
2
2
2
D e x Cy 2 C 4x2 ,
2
2
fxy x, y D fyx x, y D 4xyex Cy , and
2
2
fyy x, y D e x Cy 2 C 4y2 .
2
2
15. f x, y, z D x2 yz C xy2 z C xyz2 so fx x, y, z D 2xyz C y2 z C yz2 , fy x, y, z D x2 z C 2xyz C xz2 , and
fz x, y, z D x2 y C xy2 C 2xyz. The second order partials are:
fxx x, y, z D 2yz
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“runall” — 2005/8/1 — 12:14 — page 98 — #29
98
Chapter 2
Differentiation in Several Variables
fyy x, y, z D 2xz
fzz x, y, z D 2xy
fxy x, y, z D fyx x, y, z D 2xz C 2yz C z2
fxz x, y, z D fzx x, y, z D 2xy C y2 C 2yz
fyz x, y, z D fzy x, y, z D x2 C 2xy C 2xz
16. f x, y, z D exyz so fx x, y, z D yzexyz , fy x, y, z D xzexyz , and fz x, y, z D xyexyz . The second order
partials are:
fxx x, y, z D y2 z2 e xyz
fyy x, y, z D x2 z2 e xyz
fzz x, y, z D x2 y2 e xyz
fxy x, y, z D fyx x, y, z D zexyz 1 C xyz
fxz x, y, z D fzx x, y, z D yexyz 1 C xyz
fyz x, y, z D fzy x, y, z D xexyz 1 C xyz
17. f x, y, z D eax sin y C ebx cos z so fx x, y, z D aeax sin y C bebx cos z, fy x, y, z D eax cos y, and
fz x, y, z D −ebx sin z. The second order partials are:
fxx x, y, z D a2 e ax sin y C b2 e bx cos z
fyy x, y, z D −e ax sin y
fzz x, y, z D −e bx cos z
fxy x, y, z D fyx x, y, z D aeax cos y
fxz x, y, z D fzx x, y, z D −bebx sin z
fyz x, y, z D fzy x, y, z D 0
18. F x, y, z D 2x3 y C xz2 C y3 z5 − 7xyz so Fx x, y, z D 6x2 y C z2 − 7yz, Fy x, y, z D 2x3 C 3y2 z5 − 7xz,
and Fz x, y, z D 2xz C 5y3 z4 − 7xy.
(a) Fxx x, y, z D 12xy, Fyy x, y, z D 6yz5 , and Fzz x, y, z D 20y3 z3 C 2x.
(b) Fxy x, y, z D 6x2 − 7z D Fyx x, y, z, Fxz x, y, z D 2z − 7y D Fzx x, y, z, and Fyz x, y, z D
15y2 z4 − 7x D Fzy x, y, z.
(c) Fxyx x, y, z D 12x D Fxxy x, y, z. We knew that these would be equal because they are the mixed partials
of Fx (i.e., Fx yx D Fx xy ).
(d) Fxyz x, y, z D −7 D Fyzx x, y, z.
19. We will denote the degree of f by deg(f ) in this solution.
(a) degpx D 16, degpy D 16, degpxx D 15, degpyy D 15, and degpyx D 15.
(b) degpx D 3, degpy D 3, degpxx D 2, degpyy is undefined, and degpyx D 2.
(c) This is difficult because the term of highest degree can switch during the process of taking a derivative. For
example consider f x, y D xy2 C x3 y. Take the derivative with respect to y and the degree has decreased
by one as we would expect: fy x, y D 2xy C x3 so degfy D 3. Now take another derivative with respect
to y: fyy x, y D 2x and so the degree is now one.
For a polynomial f x1 , x2 , ... , xn which has degree d D d1 C d2 C · · · C dn because of a term
cx1d1 xxd2 ... xndn , ⭸k f /⭸xi1 ... ⭸xik has degree d –k if xj occurs at most dj times in the partial derivative—
otherwise we must look for the highest degree of any other surviving terms. If no terms survive, (i.e.,
⭸k f /⭸xi1 ... ⭸xik D 0) then the degree is undefined.
Exercises 20 and 21 have the students verify that certain functions are solutions to the given differential equations.
When the students studied exponential equations in first semester calculus they may have seen that f x D cekx solves
the differential equation y D ky. Here is a nice way to introduce the idea of a partial differential equation.
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“runall” — 2005/8/1 — 12:14 — page 99 — #30
Section 2.4
Properties; Higher-order Partial Derivatives; Newton’s Method
99
20. (a) For the first function, f x, y, z D x2 C y2 − 2z2 , fx x, y, z D 2x, fy x, y, z D 2y, and fx x, y, z D
−4z.
This means that fxx x, y, z D 2, fyy x, y, z D 2, and fzz x, y, z D −4. We see that fxx C fyy C fzz D
0 and conclude that f is harmonic.
For the second function, f x, y, z D x2 − y2 C z2 , fx x, y, z D 2x, fy x, y, z D −2y, and
fz x, y, z D 2z.
This means that fxx x, y, z D 2, fyy x, y, z D −2, and fzz x, y, z D 2. We see that fxx C fyy C fzz Z 0
and conclude that f is not harmonic.
(b) One possible example
is f x1 , x2 , ... , xn D x12 − x22 C 3x3 C 4x4 C 5x5 C · · · C nxn .


2
if i D 1,

Here fxi xi D −2 if i D 2, and we see that niD1 fxi xi D 0 so f is harmonic.

 0 if i > 2.
21. (a) To show that T x, t D e−kt cos x satisfies the differential equation kTxx D Tt we calculate the derivatives:
Tx x, t D −e −kt sin x
Txx x, t D −e −kt cos x
Tt x, t D −ke−kt cos x
so kTxx D Tt .
For t0 D 0 and t0 D 1 the graphs are:
z
1
z
1
0.75
0.5
0.25
0.75
0.5
0.25
x
-6
-4
-2
-0.25
-0.5
-0.75
-1
2
4
6
-4
-6
-2
-0.25
2
4
6
x
-0.5
-0.75
-1
For t0 D 10 the graph is further damped. The graph of the surface z D T x, t is:
1
5
T 0
-1
0
0 x
1
-5
t
10
(b) To show that T x, y, t D e−kt cos x C cos y satisfies the differential equation kTxx C Tyy D Tt we
calculate the derivatives:
Tx x, y, t D −e −kt sin x
Txx x, y, t D −e −kt cos x
Ty x, y, t D −e −kt sin y
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“runall” — 2005/8/1 — 12:14 — page 100 — #31
100
Chapter 2
Differentiation in Several Variables
Tyy x, y, t D −e −kt cos y
Tt x, y, t D −ke−kt cos x C cos y
so kTxx C Tyy D Tt .
The graphs of the surfaces given by z D T x, y, t0 for t0 D 0, 1, and 10 are:
z0
5
z 0
0
-5
5
0 y
-5
y
0
0
x
x
-5
-5
5
5
z0
5
0
-5
y
0
x
-5
5
(c) Finally, to show that T x, y, z, t D e−kt cos x C cos y C cos z satisfies the differential equation kTxx C
Tyy C Txx D Tt we calculate the derivatives:
Tx x, y, z, t D −e −kt sin x
Txx x, y, z, t D −e −kt cos x
Ty x, y, z, t D −e −kt sin y
Tyy x, y, z, t D −e −kt cos y
Tz x, y, z, t D −e −kt sin z
Tzz x, y, z, t D −e −kt cos z
Tt x, y, z, t D −ke−kt cos x C cos y C cos z
so kTxx C Tyy C Tzz D Tt .
22. (a) For x, y Z 0, 0, compute the partial derivatives:
fx x, y D y [x2 C y2 ]2x − [x2 − y2 ]2x
x2 − y 2
C xy 2
2
x C y
x2 C y2 2
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“runall” — 2005/8/1 — 12:14 — page 101 — #32
Section 2.4
Properties; Higher-order Partial Derivatives; Newton’s Method
D
yx2 − y2 x2 C y2 C xy4xy2 x2 C y2 2
D
yx4 C 4x2 y2 − y4 and similarly
x2 C y2 2
fy x, y D
101
xx4 − 4x2 y2 − y4 x2 C y2 2
(b) We use part (a):
fx 0, y D
y−y4 y2 2
D −y for y Z 0, and
fy x, 0 D x for x Z 0.
(c) From part (b), fxy 0, y D −1 while fyx x, 0 D 1 and fx 0, y and fy x, 0 are continuous at the origin
so you can conclude that fxy 0, 0 D −1 while fyx 0, 0 D 1. Why aren’t the mixed partials equal? The
answer is that the second partials are not continuous at the origin. We can see this by calculating
fxy x, y D
x6 C 9x4 y2 − 9x2 y4 − y6
.
x2 C y2 3
Therefore fxy x, 0 D 1 and
fxy 0, y D −1.
Hence
lim
x,y'0,0
fxy x, y does not exist.
In other words, fxy is not continuous at the origin.
23. An equation of a plane in the form z D f x, y is z D Ax C By C C. Here zx D A, zy D B and the second
derivatives are all 0. The partial differential equation for minimal surfaces is therefore trivially satisfied and a
plane is seen to be a minimal surface.
24. (a) Here’s an image of Scherk’s surface.
(b) In this case z D lncos x/ cos y. So zx D − tan x, zy D tan y, zxy D 0, zxx D − sec2 x, and zyy D sec2 y. So
1 C z2y zxx C 1 C z2x zyy D 1 C tan2 y− sec2 x C 1 C tan2 xsec2 y
D − sec2 x sec2 y C sec2 x sec2 y D 0.
This agrees with the right side of the equation as zxy D 0.
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“runall” — 2005/8/1 — 12:14 — page 102 — #33
102
Chapter 2
Differentiation in Several Variables
25. (a) Here’s an image of the helicoid:
2
0 z
-2
-2
-1
x
0
1
2-2
-1
0
y
1
2
(b) There’s no reason not to think of this surface as z D x tan y. Then zx D tan y, zy D x sec2 y, zxx D 0, zxy D
sec2 y, and zyy D 2 tan y sec2 y. So
1 C z2y zxx C 1 C z2x zyy D 1 C x2 sec4 y0 C 1 C tan2 y2 tan y sec2 y
D sec2 y2 tan y sec2 y D 2tan yx sec2 ysec2 y
D 2zx zy zxy
26. Let L denote limk'q xk . Then limk'q xk−1 D L and taking limits in (6), we have
L D L − [DfL]−1 fL.
Hence [DfL]−1 fL D 0. Now multiply by DfL on the left to obtain DfL[DfL]−1 fL D
DfL0 D 0 3 In fL D 0 3 fL D 0.
27. (a)
k
xk
yk
0 −1
1
1 −1.3
1.7
2 −1.2653846 1.55588235
3 −1.2649112 1.54920772
4 −1.2649111 1.54919334
5 −1.2649111 1.54919334
This table suggests that xk ' −1.2649111, 1.54919334 L − 8/5, 12/5.
(b)
k
xk
yk
0 1
−1
1 1.3
−1.7
2 1.26538462 −1.558824
−1.5492077
3 1.2649115
4 1.26491106 −1.5491933
5 1.26491106 −1.5491933
Here xk ' 8/5, − 12/5 it seems.
xk
yk
k
0 −1
1 −1.3
2 −1.2653846
3 −1.2649112
4 −1.2649111
5 −1.2649111
Here xk ' − 8/5, − 12/5.
−1
−1.7
−1.555824
−1.5492077
−1.5491933
−1.5491933
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“runall” — 2005/8/1 — 12:14 — page 103 — #34
Section 2.4
Properties; Higher-order Partial Derivatives; Newton’s Method
103
(c) The results don’t seem too strange; each initial vector is in a different quadrant and the limit is an intersection
point in the same quadrant.
28. (a)
k
xk
yk
(b)
0
1
2
3
4
5
6
7
8
9
10
k
1.4
54.7
28.0832917
14.8412307
8.35050251
5.34861164
4.2264792
4.00886454
4.0001468
4
4
xk
10
−317.452
−75583.381
−9364.2812
−1128.3294
−119.96986
−4.8602841
4.73583325
4.99959722
5
5
yk
0 1.3
10
1 −105.35
641.7815
2 −52.041661
606283.635
75622.9747
3 −25.420779
4 −12.17662
9372.7823
1132.95037
5 −5.6848239
6 −2.6823677
124.108919
7 −1.5599306
8.94078154
8 −1.3422068 −0.6614827
9 −1.333348
−0.9255223
10 −1.3333333 −0.9259259
11 −1.3333333 −0.9259259
(c) (1.3, 10) is a good deal closer to (4, 5) than it is to (−1.3333333, −0.9259259).
(d) It seems surprising that, beginning with x0 D 1.3, 10, we found the limit we did, especially when x0 D
1.4, 10 causes things to converge to (4, 5). This suggests that, when there are multiple solutions, it can be
difficult to know to which solution the initial vector will converge.
29. Formula (6) says xkC1 D xk − [Dfxk ]−1 fxk . But if xk solves (2) exactly, then fxk D 0. Thus xkC1 D
xk − [Dfxk ]−1 0 D xk . By the same argument xk D xkC2 D xkC3 D · · ·
f fy
gy −fy
1
30. Dfx, y D x
. By Exercise 36 of §1.6, [Dfx, y]−1 D
. If we evaluate at
−g
fx gy −fy gx
gx gy fx x
xk−1 , yk−1 and calculate, we find that formula (6) tells us that
xk
xk−1
y D y
−
k
k−1
gy
1
fx gy − fy gx −gx
−fy
f
.
fx g all evaluated at xk−1 ,yk−1 Expanding and taking entries we obtain the desired formulas.
31. DFx, y D [4y cosxy C 3x2 , 4x cosxy C 3y2 ], so we want to solve 4y cos xy C 3x2 D 0
. Using the
4x cos xy C 3y2 D 0
result of Exercise 30, we have
3
3 sinx
2
6yk−1
cosxk−1 yk−1 C xk−1 6xk−1
C 3yk−1
k−1 yk−1 −
xk−1 yk−1 9 C 8 sin 2xk−1 yk−1 xk D
3
3 sinx
22 − 9xk−1 yk−1 C 2 cos2xk−1 yk−1 C 6xk−1
C yk−1
k−1 yk−1 −
4xk−1 yk−1 sin2xk−1 yk−1 3
3
2
cosxk−1 yk−1 C yk−1 63xk−1
C yk−1
sinxk−1 yk−1 −
6xk−1
xk−1 yk−1 9 C 8 sin2xk−1 yk−1 yk D
3
3 sinx
22 − 9xk−1 yk−1 C 2 cos2xk−1 yk−1 C 6xk−1
C yk−1
k−1 yk−1 −4xk−1 yk−1 sin2xk−1 yk−1  2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
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“runall” — 2005/8/1 — 12:14 — page 104 — #35
104
Chapter 2
Differentiation in Several Variables
(This was obtained using Mathematica to simplify.)
Using initial vector x0 , y0 D −1, −1 and iterating the formulas above we find
k
0
1
2
3
4
5
xk
yk
−1
−0.9206484
−0.9073724
−0.9070156
−0.9070154
−0.9070154
−1
−0.9206484
−0.9073724
−0.9070156
−0.9070154
−0.9070154
(
Here’s the approximate root.


2
2
2

 x C y C z D4
2
2
32. (a) Here we’re trying to solve the system
Hence we define fx, y, z D x2 C y2 C
x C y D1


 4y2 C z2 D 4.
z2 − 4, x2 C y2 − 1, 4y2 C z2 −
 4.
2x 2y 2z
0 . It follows (see Exercise 37 of §1.6) that
Thus Dfx, y, z D  2x 2y
0 8y 2z


1
1
3
−

8x
8x
8x 



1
1
1 
.
−
[Dfx, y, z]−1 D 

8y
8y
8y 




1
1
−
0
2z
2z


 


2
2
xk−1
C yk−1
C z2k−1 − 4
xk
xk−1
 2

2
Thus  yk  D  yk−1  − [Dfxk−1 , yk−1 , zk−1 ]−1  xk−1
C yk−1
− 1
.
2
2
zk
zk−1
4yk−1 C zk−1 − 4
This simplifies to give
3
xk−1
C
2
8xk−1
y
1
yk D k−1 C
2
8yk−1
xk D
zk D
3
zk−1
C
2
2zk−1
Newton’s method with x0 D 1, 1, 1 gives the following set of results
k
xk
yk
zk
0
1
0.875
1
2
0.86607143
3
0.86602541
0.8660254
4
5
0.8660254
0.8660254
6
With x0 D 1, −1, 1, we find
k
xk
0
1
2
3
4
5
1
0.875
0.86607143
0.86602541
0.8660254
0.8660254
1
0.625
0.5125
0.50015244
0.50000002
0.5
0.5
1
2
1.75
1.73214286
1.73205081
1.73205081
1.73205081
yk
zk
−1
−0.625
−0.5125
−0.5001524
−0.5
−0.5
1
2
1.75
1.73214286
1.73205081
1.73205081
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“runall” — 2005/8/1 — 12:14 — page 105 — #36
Section 2.5
The Chain Rule
105


2
2
2

 x C y C z D4
2
2
by hand. First insert the second equation into the first: 1 C z2 D 4 3 z D
(b) We solve
x C y D1


2
2
 4y C z D 4
√
; 3. Use this in the third equation 4y2 C 3 D 4 3 y D ; 1 .
2√
Now use this in the second equation: x2 C 1 D 1 3 x D ; 3 .
4
2
So we have 8 solutions:
√
√
√
√
3 1 √
3 1 √
3
3
1 √
1 √
, , 3 , −
, , 3 , −
, − , 3 , , − , 3
2 2
2 2
2
2
2
2
√
√
√
√
√
√
√
√
3 1
3 1
3
3
1
1
, , − 3 , −
, , − 3 , −
, − , − 3 , , − , − 3
2 2
2 2
2
2
2
2
We found two of them above.
2.5 THE CHAIN RULE
In Exercises 1–3 students see that if you have a composite function you can take the derivative either by substituting or
by using the chain rule.
1. f x, y, z D x2 − y3 C xyz, x D 6t C 7, y D sin 2t, and z D t 2 .
Substitution:
f xt, yt, zt D 6t C 72 − sin 2t3 C 6t C 7sin 2tt 2 D 6t C 72 − sin 2t3 C 6t 3 C 7t 2 sin 2t and so
df
D 26t C 76 − 3sin 2t2 2 cos 2t C 18t 2 C 14t sin 2t C 6t 3 C 7t 2 2 cos 2t
dt
Chain Rule:
df
⭸f dx
⭸f dy
⭸f dz
D
C
C
dt
⭸x dt
⭸y dt
⭸z dt
D 2x C yz6 C −3y2 C xz2 cos 2t C xy2t
D [26t C 7 C sin 2tt 2 ]6 C [−3 sin2 2t C 6t C 7t 2 ]2 cos 2t C [6t C 7 sin 2t]2t
2. f x, y D sinxy, x D s C t, and y D s2 C t 2 .
(a) f xt, yt D sinxtyt D sin[s C ts2 C t 2 ].
⭸f
D cos[s C ts2 C t 2 ][s2 C t 2 C s C t2s]
⭸s
⭸f
D cos[s C ts2 C t 2 ][s2 C t 2 C s C t2t]
⭸t
(b)
⭸f
⭸f ⭸x
⭸f ⭸y
D
C
⭸s
⭸x ⭸s
⭸y ⭸s
D y cosxy C x cosxy2s
D cos[s C ts2 C t 2 ][s2 C t 2 C s C t2s] and
⭸f
⭸f ⭸x
⭸f ⭸y
D
C
⭸t
⭸x ⭸t
⭸y ⭸t
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“runall” — 2005/8/1 — 12:14 — page 106 — #37
106
Chapter 2
Differentiation in Several Variables
D y cosxy C x cosxy2t
D cos[s C ts2 C t 2 ][s2 C t 2 C s C t2t]
3. (a) We want
⭸P dx
⭸P dy
⭸P dz
dP
D
C
C
dt
⭸x dt
⭸y dt
⭸z dt
D
6x2
12xz
6x2 z
2 cos t C
−2 sin t −
3
2
y
y
y
D
64 cos2 t3t
64 cos2 t
122 cos t3t
−2 sin t −
2
cos
t
C
3
2 sin t
4 sin2 t
2 sin t
D −72t cos t − 36t
cos3 t
36 cos2 t
C
.
sin2 t
sin t
Therefore,
36 − 27π
dP √
D
.
dt tDπ/4
2
2
2
(b) Pxt, yt, zt D 62 cos t 3t D 36t cos t , so
2 sin t
sin t
sin t36 cos2 t − 36t · 2 cos t sin t − 36t cos2 tcos t
dP
D
.
dt
sin2 t
Therefore,
36 − 27π
dP √
D
.
dt tDπ/4
2
(c) Using differentials,
P L 36 − 27π
dP √
dt D .01 L −.34523.
dt tDπ/4
2
So (writing P as a function of t),
9π
Pπ/4 C .01 L Pπ/4 C P L √ − .34523 L 19.6477.
2
4. We are thinking of z D zs, t D [xs, t]2 C [ys, t]3 . So
⭸y ⭸z
⭸z ⭸x ⭸z 2, 1 D
·
C
·
D 2x|2,1 · s|2,1 C 0 D 8.
⭸t
⭸x 2,1 ⭸t 2,1
⭸y 2,1 ⭸t 2,1
5. Here V D LWH, so
⭸V dL
⭸V dW
⭸V dH
dV
D
C
C
dt
⭸L dt
⭸W dt
⭸H dt
D WH dW
dH
dL
C LH C LW dt
dt
dt
D 5 · 4.75 C 7 · 4.5 C 7 · 5−1
D −6 in3 /min.
Since
dV
< 0, the volume of the dough is decreasing at this instant.
dt
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“runall” — 2005/8/1 — 12:14 — page 107 — #38
Section 2.5
The Chain Rule
107
6. Let the length of the butter be y and the length of an edge of the cross section be x. Then the volume V D x2 y.
The rate at which the volume is changing is
dy
dV
dx
D 2xy
C x2
D 21.56−.125 C 1.52 −.25 D −2.8125 in3 /min.
dt
dt
dt
7. Note that in 6 months:
x D 1 C .6 − cos π D 2.6
y D 200 C 12 sin π D 200
The chain rule gives
dy dP ⭸P dx ⭸P D
C
dt tD6
⭸x xD2.6 dt tD6
⭸y xD2.6 dt tD6
yD200
yD200
πt π
sin 6
6 tD6
2
πt
2πt
πt −
− 4y 3 |yD200 2 sin
C
cos 6
6
6
D 100.1x C 10
−1
2
0.1|xD2.6 0.1 −
tD6
D 10.26
−1
2
0.1 − 4200
−2
3
−2π
D 0.031219527 C 0.734885812 D 0.766105339 units/month (demand in rising slightly).
8. (a) The chain rule gives
⭸BMI dw
⭸BMI dh
d
BMI D
C
dt
⭸w dt
⭸h dt
10,000 dw
20,000 w dh
D
−
h2 dt
h3
dt
On the child’s 10th birthday: w D 33 kg, h D 140 cm,
dw
D 0.4,
dt
dh
D 0.6.
dt
So
dBMI
10,000
20,000 · 33
D
0.4 −
0.6
2
dt
140
1403
L 0.0598 points/month.
(b) The rate we found in part (a) is greater than the typical rate by about 49%. I’d monitor the situation monthly
so that it doesn’t persist for too long, but I wouldn’t be very concerned, since the current BMI is roughly
16.84, which is quite low.
9. Since x D er cos θ and y D er sin θ we can write
⭸y
⭸z
⭸z
⭸x
⭸z
⭸z
⭸z
D C D e r cos θ C e r sin θ.
⭸r
⭸x
⭸r
⭸y
⭸r
⭸x
⭸y
Similarly,
⭸y
⭸z
⭸z
⭸x
⭸z
⭸z
⭸z
D C D −e r sin θ C e r cos θ.
⭸θ
⭸x
⭸θ
⭸y
⭸θ
⭸x
⭸y
Therefore,
2
2
2
⭸z
⭸z
⭸z
2r
2
2
C D e cos θ C sin θ ⭸r
⭸θ
⭸x
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:14 — page 108 — #39
108
Chapter 2
Differentiation in Several Variables

2
⭸
z
⭸
z
⭸
z
Ccos2 θ C sin2 θ C 2 cos θ sin θ − 2 cos θ sin θ 
⭸y
⭸x ⭸y


2
2
⭸z
⭸z
D e 2r  C  .
⭸x
⭸y
The result follows.
Exercises 10–13 are fun exercises. You may want to stress that we are showing that the partial differential equations
are true without even knowing the “outside” function.
10. We’ll start by calculating the components on the left side:
⭸z
⭸z ⭸u
⭸z ⭸v
D
C
⭸x
⭸u ⭸x
⭸v ⭸x
⭸z
⭸z
1 C
1
D
⭸u
⭸v
⭸z
⭸z
D
C
and
⭸u
⭸v
⭸z
⭸z ⭸u
⭸z ⭸v
D
C
⭸y
⭸u ⭸y
⭸v ⭸y
⭸z
⭸z
D
1 C
−1
⭸u
⭸v
⭸z
⭸z
−
so
D
⭸u
⭸v
⭸z ⭸z
⭸z
⭸z
⭸z
⭸z
D
C
−
⭸x ⭸y
⭸u
⭸v
⭸u
⭸v
2
D
2
⭸z
⭸z
− .
⭸u
⭸v
11. First calculate:
yy2 − x2 ⭸u
D
and
⭸x
x2 C y2 2
xx2 − y2 ⭸u
D
⭸y
x2 C y2 2
Now
x
⭸w
⭸w
⭸w ⭸u
⭸w ⭸u
C y
Dx
C y
⭸x
⭸y
⭸u ⭸x
⭸u ⭸y
D
⭸w
⭸u
⭸u
C y x
⭸u
⭸x
⭸y
D
yy2 − x2 xx2 − y2 ⭸w
C y
x 2
2
2
⭸u
x C y x2 C y2 2
D 0.
12. First calculate:
4xy2
⭸u
D
and
⭸x
x2 C y2 2
−4x2 y
⭸u
D
⭸y
x2 C y2 2
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:14 — page 109 — #40
Section 2.5
The Chain Rule
109
Now
x
⭸w
⭸w
⭸w ⭸u
⭸w ⭸u
C y
Dx
C y
⭸x
⭸y
⭸u ⭸x
⭸u ⭸y
D
⭸w
⭸u
⭸u
C y x
⭸u
⭸x
⭸y
D
4xy2
−4x2 y
⭸w
C y
x 2
2
2
2
⭸u
x C y x C y2 2
D 0.
13.
⭸u
⭸u
⭸v
⭸v
−1 ⭸u
1
−1 ⭸v
1
D
D
D 0. Also
D
D 0, and
D . Now it is just a matter of using
,
, and
,
⭸x
x2 ⭸y
y2
⭸z
⭸x
x2 ⭸y
⭸z
z2
the chain rule and plugging in:
x2
⭸w
⭸w
⭸w
⭸w ⭸u
⭸w ⭸v
⭸w ⭸v
⭸w ⭸v
2 ⭸w ⭸u
2 ⭸w ⭸u
C y2
C z2
D x2 C
C
C
C y C z ⭸x
⭸y
⭸z
⭸u ⭸x
⭸v ⭸x
⭸u ⭸y
⭸v ⭸y
⭸u ⭸z
⭸v ⭸z
⭸w 2 ⭸u
⭸u
⭸u
⭸w 2 ⭸v
⭸v
⭸v
C y2
C z2 C
C y2
C z2 x
x
⭸u
⭸x
⭸y
⭸z
⭸v
⭸x
⭸y
⭸z


1
⭸w  2 −1
⭸w 2 −1
2 1
2
D
x x 2 C 0 C z 2 C y 2 C 0 C
⭸u
x2
y
⭸v
x
z
D
D 0.
14.
⭸u
⭸u
⭸v
⭸v
⭸v
1 ⭸u
−x
−z
1
, and
, and
D ,
D
D 0. Also
D 0,
D
D . Again, it is just a matter of using
⭸x
y ⭸y
y2
⭸z
⭸x
⭸y
y2
⭸z
y
the chain rule and plugging in:
x
⭸w
⭸w
⭸w
⭸w ⭸u
⭸w ⭸v
⭸w ⭸u
⭸w ⭸v
⭸w ⭸u
⭸w ⭸v
C y
C z
D x
C
C
C
C y
C z
⭸x
⭸y
⭸z
⭸u ⭸x
⭸v ⭸x
⭸u ⭸y
⭸v ⭸y
⭸u ⭸z
⭸v ⭸z
⭸w ⭸u
⭸u
⭸u
⭸w ⭸v
⭸v
⭸v
C y
C z C
C y
C z x
x
⭸u ⭸x
⭸y
⭸z
⭸v ⭸x
⭸y
⭸z




−x
1
⭸w  1
⭸w 
−z
D
x C y
0 C y
C 0 C
C z 
⭸u
y
y2
⭸v
y2
y
D
D 0.
15. (a) f ◦ g D 3s − 7t5 , e2s−14t so
15s − 7t4
2e2s−14t
−105s − 7t4
−14e2x−14t
15s − 7t4
15x4
D 2s−14t
2x
2e
2e
and Dg D 1
Df ◦ g D (b)
Df D −7 We can easily see that Df Dg D Df ◦ g.
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“runall” — 2005/8/1 — 12:14 — page 110 — #41
110
Chapter 2
Differentiation in Several Variables
16. (a) f ◦ g D st2 − 3s C t 2 2 D s2 t 2 − 3s2 − 6st 2 − 3t 4 , so
Df ◦ g D [2st 2 − 6s − 6t 2
(b) Df D 2x
−6y D
Df Dg D
2st
17. (a) f ◦ g D s/ts2 t −
−6s − 6t 2 !, and Dg D 2st
−6s − 6t 2 ! t
1
2s2 t − 12st − 12t 3 ].
t
1
s
, so
2t s
D [2st 2 − 6s − 6t 2
2t 2s2 t − 12st − 12t 3 ].
s2 t s/t
1
,
C s6 t 3 D s3 − st2 ,
C s6 t 3 , so
s/t s2 t
st 2
Df ◦ g D 3s2 − t 2
−1/s2 t 2 C 6s5 t 3
−2st
−2/st 3 C 3s6 t 2
(b)

Df D 
and
Dg D y
x2
y C
1
y
1
t
2st
−x
y2
x
C 3y2

D
s2 t C
1
s2 t
s2 t
s2 /t 2
s
− t
t
s
−s/t
4t2
C
3s
s4 t 2


D
s2 t C t 3
1
s2 t
s2 −t 2
st
− 31 3 C 3s4 t 2
s t


− s2
t so
s2

Df Dg D 

x − 1
s2 t C t 3
s2 −t 2
st
− 31 3
s t
1
s2 t
C 3s4 t 2


1
t
2st
3s2 − t 2
− s2
t D −1
C 6s5 t 3
s2
s2 t 2
−2
st 3
−2st
.
C 3s6 t 2 3
18. (a) f ◦ g D t − 22 3t C 7 C 3t C 72 t 3 , t − 23t C 7t 3 , et so


45t 4 C 168t 3 C 156t 2 − 10t − 16


15t 4 C 4t 3 − 42t 2
Df ◦ g D 
.
3
3t 2 e t
(b)

2xy x2 C 2yz y2
Df D  yz
xz
xy 
0
0
ez


2t − 23t C 7 t − 22 C 23t C 7t 3
3t C 72


3t C 7t 3
t − 2t 3
t − 23t C 7 
D
3
0
0
et




45t 4 C 168t 3 C 156t 2 − 10t − 16
1


15t 4 C 4t 3 − 42t 2
and Dg D  3  so DfDg D 
.
3
3t 2
3t 2 e t

2
19. (a) f ◦ g D st C tu C su, s3 t 3 − estu so
Df ◦ g D t C u
2
3s2 t 3 − tu2 e stu
s C u
2
3s3 t 2 − su2 e stu
s C t
2 .
−2stuestu
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:14 — page 111 — #42
Section 2.5
The Chain Rule
111
(b)
Df D 1
3x2

t
and Dg D  0
u
1
−zeyz

s 0
u t 
0 s
1
1
D
−yeyz 3s2 t 2
so
DfDg D 1
2
−suestu
1
2 −tuestu
t C u
2
3s2 t 3 − tu2 e stu
s C u
2
3s3 t 2 − su2 e stu
s C t
2 .
−2stuestu
20. This is a matter of seeing what we have to determine and which formula to use. We calculate Df ◦ g1, −1, 3
as Dfg1, −1, 3Dg1, −1, 3. The second piece is given in the exercise. For the first we calculate
2y 2x 10 4
2y 2x D D
.
Dfg1, −1, 3 D 3 −1 2,5 3 −1 3 −1 g1,−1,3
Then we can multiply the matrices to get the result
Df ◦ g1, −1, 3 D 10
3
4
1
−1 4
−1
0
−10
−3
0
26
D
7 −1
28
.
−7 21. (a) This is similar to Exercise 20.
Df ◦ g1, 2 D Dfg1, 2Dg1, 2 D Df3, 5Dg1, 2
D
1
3
1
2
5 5
3
7
D
7 31
10
44 (b)
Dg ◦ f4, 1 D Dgf4, 1Df4, 1 D Dg1, 2Df4, 1
D
3
−1
7 1
2
5
2
1
D
3 2
13
31 22. We’ll start with the right hand side of the equation because we can easily calculate the partials of x and y with
respect to r and θ.
2
2
2
2
1 ⭸z
1 ⭸z ⭸x
⭸z
⭸z ⭸x
⭸z ⭸y
⭸z ⭸y
C
C
C 2
C 2 D
⭸r
r
⭸θ
⭸x ⭸r
⭸y ⭸r
r ⭸x ⭸θ
⭸y ⭸θ
2
D
2
2
2
2
2
⭸y
1 ⭸x
1 ⭸y
⭸z
⭸x
⭸z
C 2 C C 2 ⭸x
⭸r
r
⭸θ
⭸y
⭸r
r
⭸θ
C 2
1 ⭸x ⭸y
⭸z ⭸z ⭸y ⭸x
C
⭸x ⭸y ⭸r ⭸r
r2 ⭸θ ⭸θ
2
2
1 2 2
1 2
⭸z
⭸z
D cos2 θ C
r sin θ C sin2 θ C
r cos2 θ
2
⭸x
r
⭸y
r2
2
C 2
2
1
⭸z ⭸z
⭸z
⭸z
sin θ cos θ C 2 −r sin θr cos θ D C ⭸x ⭸y
r
⭸x
⭸y
23. (a) From formula (10) in Section 2.5, we have
sin θ ⭸
⭸
⭸
D cos θ
−
and
⭸x
⭸r
r ⭸θ
cos θ ⭸
⭸
⭸
D sin θ
C
⭸y
⭸r
r ⭸θ
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:14 — page 112 — #43
112
Chapter 2
Differentiation in Several Variables
Hence if z D f x, y, then
sin θ ⭸ ⭸z
⭸2 z
⭸ ⭸z
⭸ ⭸z
D
D cos θ −
⭸x2
⭸x ⭸x
⭸r ⭸x
r ⭸θ ⭸x
D cos θ
⭸
⭸z
⭸z
sin θ ⭸z
sin θ ⭸
sin θ ⭸z
−
−
cos θ
−
cos θ
⭸r
⭸r
r ⭸θ
r ⭸θ
⭸r
r ⭸θ
Now use the product rule:
sin θ ⭸2 z
sin θ ⭸z
⭸2 z
⭸2 z
−
D cos θ cos θ
C
⭸x2
⭸r2
r2 ⭸θ
r ⭸r⭸θ
⭸z
⭸2 z
sin θ
cos θ ⭸z
sin θ ⭸2 z
C cos θ
−
−
− sin θ
r
⭸r
⭸θ⭸r
r ⭸θ
r ⭸θ2
−
D cos2 θ
2 sin θ cos θ ⭸z
2 sin θ cos θ ⭸2 z
sin2 θ ⭸z
sin2 θ ⭸2 z
⭸2 z
C
.
−
C
C
2
2
⭸r
r
⭸θ
r
⭸r⭸θ
r ⭸r
r2 ⭸θ2
Follow the same steps to calculate
cos θ ⭸ ⭸z
⭸2 z
⭸ ⭸z
⭸ ⭸z
D
D sin θ C
⭸y2
⭸y ⭸y
⭸r ⭸y
r ⭸θ ⭸y
D sin θ
⭸
⭸z
⭸z
cos θ ⭸z
cos θ ⭸
cos θ ⭸z
C
C
sin θ
C
sin θ
⭸r
⭸r
r ⭸θ
r ⭸θ
⭸r
r ⭸θ
D sin2 θ
2 sin θ cos θ ⭸2 z
cos2 θ ⭸z
cos2 θ ⭸2 z
2 sin θ cos θ ⭸z
⭸2 z
C
C
C
−
.
2
2
⭸r
r
⭸θ
r
⭸θ⭸r
r ⭸r
r2 ⭸θ2
(b) Adding the two equations above we easily see that
⭸2
⭸2
⭸2
1 ⭸2
1 ⭸
C
C
D
C
.
⭸x2
⭸y2
⭸r2
r ⭸r
r2 ⭸θ2
⭸2
⭸2
⭸2
⭸ ⭸2
1 ⭸
C
C
D
C
. Since the z-coordinate means
⭸x2
⭸y2
⭸r2
r ⭸r
r2 ⭸θ2
the same thing in both Cartesian and cylindrical coordinates, the result follows.
⭸w
⭸w ⭸r
⭸w ⭸θ
⭸w ⭸z
25. (a) The chain rule gives
D
C
C
for any appropriately differentiable function w.
⭸ρ
⭸r ⭸ρ
⭸θ ⭸ρ
⭸z ⭸ρ
Now (6) of §1.7 gives z D ρ cos ϕ, r D ρ sin ϕ. Hence
24. Given Exercise 23, this is easy: We know
⭸w
⭸w
⭸w
⭸w
⭸w
D sin ϕ
C 0 C cos ϕ
D sin ϕ
C cos ϕ
.
⭸ρ
⭸r
⭸z
⭸r
⭸z
Also
⭸w
⭸w
⭸w
D ρ cos ϕ
− ρ sin ϕ
⭸ϕ
⭸r
⭸z
from a similar chain rule computation.
From this, we have
ρ sin ϕ
Thus
⭸w
⭸w
⭸w
⭸w
⭸w
2 ⭸w
C cos ϕ
D ρ sin2 ϕ
C ρ sin ϕ cos ϕ
− ρ cos ϕ sin ϕ
C ρ cos ϕ
⭸ρ
⭸ϕ
⭸r
⭸z
⭸r
⭸z
⭸w
Dρ
.
⭸r
cos ϕ ⭸w
⭸w
⭸w
D sin ϕ
C
⭸r
⭸ρ
ρ ⭸ϕ
or
cos ϕ ⭸
⭸
⭸
D sin ϕ
C
.
⭸r
⭸ρ
ρ ⭸ϕ
(Alternatively, consider formula (10) in this section with x D z, y D r, θ replaced by ϕ, and r replaced
by ρ.)
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
“runall” — 2005/8/1 — 12:14 — page 113 — #44
Section 2.5
The Chain Rule
113
⭸2
⭸2
1 ⭸2
1 ⭸
. From z D ρ cos ϕ, r D ρ sin ϕ, we may treat
C
C
C
2
2
⭸r
⭸z
r2 ⭸θ2
r ⭸r
z and r as if they are Cartesian coordinates, so that
(b) The cylindrical Laplacian is
⭸2
⭸2
⭸2
1 ⭸2
1 ⭸
C
D
C
C
2
2
2
⭸r
⭸z
⭸ρ
ρ2 ⭸ϕ2
ρ ⭸ρ
Now we know
Cartesian/cylindrical
⭸
from part (a). So, with r D ρ sin ϕ, we have
⭸r
1 ⭸
1 ⭸2
1 ⭸
1 ⭸2
⭸2
⭸2
⭸2
D
C
C
C
C 2 2 C
2
2
2
⭸r
⭸z
r ⭸θ
r ⭸r
⭸ρ
ρ2 ⭸ϕ2
ρ ⭸ρ
C
D
1
⭸2
ρ2 sin2 ϕ ⭸θ2
C
cos ϕ ⭸
1
⭸
C
sin ϕ
ρ sin ϕ
⭸ρ
ρ ⭸ϕ
cot ϕ ⭸
⭸2
⭸2
1
1 ⭸2
2 ⭸
C
C
C
C
2
2
2
2
2
2
⭸ρ
ρ ⭸ϕ
ρ ⭸ρ
ρ sin ϕ ⭸θ
ρ2 ⭸ϕ
as desired.
Exercises 26–28 puts the implicit differentiation techniques which the students learned in a previous course in the
context of the current discussion. This is one of those problems where it would be immediately clear if we were able to talk
to each other. The problem is explaining to you which derivative with respect to x is being considered. One solution is to
introduce another variable. You might want to use this as an example of why the author introduces the notation she does for
Exercises 31–35. One other note is that the results hold also for F x, y or F x, y, z being constant (not necessarily 0).
26. (a) View x and y as functions of t, where x D xt D t and y D yt. Since F x, y D 0 we know that
Ft x, y D 0. This means that we know:
0D
But
dy
dx
dF
D Fx x, y
C Fy x, y .
dt
dt
dt
F x, y
dy
dy
dy
dx
D 1 and
D
so
D− x
.
dt
dt
dx
dx
Fy x, y
(b-i) If F x, y D x3 − y2 then Fx x, y D 3x2 and Fy x, y D −2y so
(b-ii) y2 D x3 so y D x3/2 so
dy
3x2
3x2
D−
D
.
dx
−2y
2y
dy
3
D x1/2 . Multiply numerator and denominator by x3/2 to get the answer in
dx
2
(b-i).
27. Here we’ll just use the formula from Exercise 26(a) where here F x, y D sinxy − x2 y7 C ey .
y cosxy − 2xy7
dy
D−
.
dx
x cosxy − 7x2 y6 C e y
The results of Exercise 28 are used in Exercises 33 and 35 in a nice way. None of them is very time consuming—it is
worth assigning all three.
28. (a) We have the same problem here with ambiguity about what is meant by the derivative with respect to x and
y. Let x D xs, t D s, y D ys, t D t, and z D zs, t. Then
0D
⭸y
⭸F
⭸x
⭸z
⭸z
D Fx x, y, z
C Fy x, y, z
C Fz x, y, z D Fx x, y, z C Fz x, y, z .
⭸s
⭸s
⭸s
⭸s
⭸x
Solving we get
An analogous calculation gives
F x, y, z
⭸z
D− x
.
⭸x
Fz x, y, z
Fy x, y, z
⭸z
D−
.
⭸y
Fz x, y, z
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“runall” — 2005/8/1 — 12:14 — page 114 — #45
114
Chapter 2
Differentiation in Several Variables
(b-i) F x, y, z D xyz − 2 so by part (a):
(b-ii) z D 2/xy so
yz
z
⭸z
D− D−
⭸x
xy
x
and
xz
z
⭸z
D− D− .
⭸y
xy
y
⭸z
−2
D
⭸x
x2 y
and
⭸z
−2
.
D
⭸y
xy2
29. Use the equations from Exercise 28(a) for F x, y, z D x3 z C y cos z C sin y/z D 0:
−3x2 z
−3x2 z3
⭸z
D
D
⭸x
x3 − y sin z − sin y/z2
x3 z2 − yz2 sin z − sin y
and
− cos z − cos y/z
−z2 cos z − z cos y
⭸z
D
.
D
3
2
3
⭸y
x − y sin z − sin y/z
x z2 − yz2 sin z − sin y
Exercise 30 is a good example of why you can not just blindly apply formulas such as the chain rule without first
checking that all of the hypotheses are met.
30. (a) By definition
f h, 0 − f 0, 0
0
D lim D 0, and
h'0 h
h
f 0, h − f 0, 0
0
fy 0, 0 D lim
D lim D 0.
h'0
h'0 h
h
fx 0, 0 D lim
h'0
(b)
f ◦ xD



at
1
C
a2

 0
if t Z 0
if t D 0
at
a
and so Df ◦ x0 D
.
1 C a2
1 C a2
(c) By definition, Df 0, 0 D [fx 0, 0, fy 0, 0]. We calculated these in part (a) to be 0 so
therefore f ◦ x D
Df0, 0Dx0 D 0
0 1
D 0.
a The function f is not differentiable at the origin and so not all of the assumptions of the chain rule are met.
⭸w
⭸w
⭸w
⭸w
⭸w
D −10, D 1, D 7, D 1 − 102x D 1 − 20x, and D
⭸x y,z
⭸y x,z
⭸z x,y
⭸x y
⭸y x
7 − 102y D 7 − 20y.
⭸y
⭸y
⭸w
⭸w
⭸x
⭸w
⭸w
⭸z
⭸x
⭸w
(b) D 1 and
D 0 so D
C C
. But
D
⭸x y
⭸x y,z ⭸x
⭸y x,z ⭸x
⭸z x,y ⭸x
⭸x
⭸x
⭸x y
⭸w
⭸w
⭸z
C .
⭸x y,z
⭸z x,y ⭸x
31. (a) 32. ⭸w
⭸w
⭸w
⭸w
D 3x2 , D 3y2 , D 3z2 , D 3x2 C 3z2 2 D 3x2 C 62x − 3y2 , and
⭸x y,z
⭸y x,z
⭸z x,y
⭸x y
⭸w
D 3y2 C 3z2 −3 D 3y2 − 92x − 3y2 .
⭸y x
33. ⭸y
⭸s
⭸s
⭸s
⭸s
D xw − 2z, so D C .
⭸z x,y,w
⭸z x,w
⭸z x,y,w
⭸y x,z,w ⭸z x,w
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“runall” — 2005/8/1 — 12:14 — page 115 — #46
Section 2.6
To calculate Directional Derivatives and the Gradient
115
⭸y
we can use the results of Exercise 28 with F x, y, z, w D xyw − y3 z C xz:
⭸z x,w
F x, y, z, w
⭸y
−y3 C x
D−
.
D− z
⭸z x,w
Fy x, y, z, w
xw − 3y2 z
y3 − x
⭸s
D xw − 2z C x2 .
⭸z x,w
xw − 3y2
34. U D F P, V , T and PV D kT.
k
⭸U
⭸U
⭸U
⭸V
C (a) D
D FT P, V , T C FV P, V , T .
⭸T P
⭸T P,V
⭸V P,T ⭸T
P
k
⭸U
⭸U
⭸U
⭸P
(b) C D
D FT P, V , T C FP P, V , T .
⭸T V
⭸T P,V
⭸P V ,T ⭸T
V
V
⭸U
⭸U
⭸U
⭸T
C (c) D
D FP P, V , T C FT P, V , T .
⭸P V
⭸P V ,T
⭸T P,V ⭸P
k
So 35. Fy x, y, z
Fz x, y, z
Fx x, y, z
⭸y
⭸x
⭸z
D −
−
−
D −1.
⭸y z ⭸z x ⭸x y
Fx x, y, z
Fy x, y, z
Fz x, y, z
36. In this case P D kT/V so ⭸P/⭸T V D k/V . Similarly, V D kT/P so ⭸V /⭸PT D −kT/P 2 and T D PV/k
so ⭸T /⭸V P D P/k. So
⭸y
k
−kt
P
⭸x
⭸z
−kT
−kTP
D
D −1.
D 2 D
2
⭸y z ⭸z x ⭸x y
V
P
k
VP
VP
The last equality holds since PV D kT.
37. It is easiest to use implicit differentiation and solve. For example, for the equation ax2 C by2 C cz2 − d D 0,
hold z constant and take the derivative with respect to y. You get 2ax⭸x/⭸yz C 2by D 0. Solve this and get
⭸x/⭸yz D −by/ax. Similarly we get that ⭸y/⭸zx D −cz/by and ⭸z/⭸xy D −ax/cz. So
−by
⭸y
−cz
−ax
⭸x
⭸z
D
D −1.
⭸y z ⭸z x ⭸x y
ax
by
cz
2.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT
1. (a) §f x, y, z · −k is the directional derivative of f x, y, z in the direction −k (i.e., the negative
z direction).
(b) §f x, y, z · −k D ⭸f , ⭸f , ⭸f · 0, 0, −1 D − ⭸f .
⭸x
⭸y
⭸z
⭸z
In Exercises 2–8, the students should notice that the given vector u is not always a unit vector and that they may have
to normalize it first.
√
2. §f x, y D ey cos x, ey sin x so §f π/3, 0 D 1/2, 3/2.
√
√
3 − 3
√
Du f π/3, 0 D §f π/3, 0 · 3, −1/ 10 D
.
2 10
3. §f x, y D 2x − 6x2 y, −2x3 C 6y2 , so §f 2, −1 D 28, −10 and
Du f 2, −1 D 28, −10 ·
1, 2
8
√
D √ .
5
5
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116
Chapter 2
Differentiation in Several Variables
4. As noted above, here we have to normalize u so Du f a D §f a ·
§f x, y D u
.
||u||
−2y
−2x
,
so §f 3, −2 D 1/169−6, 4 and
x2 C y2 2 x2 C y2 2
Du f a D 1, −1
−10
−6 4
√
√ .
D
,
·
169 169
2
169 2
5. §f x, y D ex − 2x, 0 so §f 1, 2 D e − 2, 0 and
2, 1
2e − 4
√
D √
.
5
5
Du f a D e − 2, 0 ·
6. §f x, y, z D yz, xz, xy so §f −1, 0, 2 D 0, −2, 0 and
−1, 0, 2
√
D 0.
5
Du f a D 0, −2, 0 ·
7. §f x, y, z D −e−x Cy Cz 2x, 2y, 2z so §f 1, 2, 3 D −e−14 2, 4, 6 and
2
2
2
Du f a D −e −14 2, 4, 6 ·
8. §f x, y, z D √
1, 1, 1
√
D −4 3e −14 .
3
y
y
ey
, xe , −6xe z so §f 2, −1, 0 D e−1 , 2e−1 , 0 and
3z2 C1 3z2 C1 3z2 C12
Du f a D e −1 , 2e−1 , 0 ·
1, −2, 3
−3
√
D √ .
14
e 14
9. (a)
f h, 0 − f 0, 0
0 − 0
D lim
D 0.
h'0
h'0
h
h
f 0, h − f 0, 0
0 − 0
fy 0, 0 D lim
D lim
D 0.
h'0
h'0
h
h
fx 0, 0 D lim
(b)
Du,v f 0, 0 D lim
h'0
f hu, hv − 0
hu|hv|
D lim √
h'0 h h2 u2 C h2 v2
h
But (u, v ) is a unit vector so this
D lim
hu|h||v|
h'0 h|h|1
D u|v|
for all unit vectors (u, v ).
(c) The graph is shown below.
x 0.5 1
0
-0.5
-1
0.5
z
0
-0.5
1
0.5
0
y
-0.5
-1
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“runall” — 2005/8/1 — 12:14 — page 117 — #48
Section 2.6
10. (a)
Directional Derivatives and the Gradient
117
f h, 0 − f 0, 0
0 − 0
D lim
D 0.
h'0
h
h
f 0, h − f 0, 0
0 − 0
fy 0, 0 D lim
D lim
D 0.
h'0
h'0
h
h
fx 0, 0 D lim
h'0
(b)
Du,v f 0, 0 D lim
h'0
f hu, hv − 0
huhv
D lim √
h'0 h h2 u2 C h2 v2
h
But (u, v ) is a unit vector so this
h2 uv
D uvsgnh
h'0 h|h|
D lim
for all unit vectors (u, v ) where sgn(h) is 1 for h Ú 0 and −1 for h < 0. Unless u or v are zero, this limit
doesn’t exist.
(c) The graph is shown below.
0.5
1
z
0
0.5
-0.5
-1
0
-0.5
0
x
y
-0.5
0.5
1 -1
11. The gradient direction for the function h is §h D −6xy2 , −6x2 y.
(a) Head in the direction §h1, −2 D −24, 12. If you prefer your directions given by a unit vector, we
normalize to obtain:
§h1, −2
−24, 12
−2, 1
D √
.
D √
2
2
||§h1, −2||
5
24 C 12
1,2
5
(b) Head in a direction orthogonal to your answer for part (a): ; √ .
12. fx 3, 7 D 3 and fy 3, 7 D −2 so the gradient is §f 3, 7 D 3, −2.
√
(a) To warm up we head in the direction of the gradient; this is the unit vector 3,√−2/ 13.
(b) To cool off we head in the opposite direction; this is the unit vector −3, 2/ 13.
√
(c) To maintain temperature we head in a direction orthogonal to the gradient, namely ;2, 3/ 13.
13. We begin by heading east and keep heading towards lower levels while intersecting each level curve orthogonally.
See the solution given in the text.
14. We’re looking at the top half of this ellipsoid. The equation is f x, y D z D 4 − x2 − y2 /4. For the path of
steepest descent, we look at the negative gradient
−§f x, y D 1/24 − x2 − y2 /4−1/2 2x, y/2.
This means that
y/2
dy
y
D
D
.
dx
2x
4x
This is the separable differential equation 4/y dy D 1/x dx or 4 ln y D ln x C c. Work the usual magic and
get y4 D kx. So the raindrops will follow curves of that form where z is constrained by the surface of the ellipsoid.
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“runall” — 2005/8/1 — 12:14 — page 118 — #49
118
Chapter 2
Differentiation in Several Variables
15. We want to head in the direction of the negative gradient. Since Mx, y D 3x2 C y2 C 5000, the negative
gradient is −§Mx, y D −6x, −2y. This means that
−2y
y
dy
D
D
.
dx
−6x
3x
This is the separable differential equation 3/y dy D 1/x dx or 3 ln y D ln x C c. Work the usual magic and
get y3 D kx. Substitute in the point (8, 6) to solve for k to end up with the path y3 D 27x.
For Exercises 16–22 we can use equations (5) and (6) from Section 2.6 in the text.
16. f x, y, z D x3 C y3 C z3 D 7 so §f x, y, z D 3x2 , 3y2 , 3z2 and §f 0, −1, 2 D 0, 3, 12. So the
equation of the tangent plane is:
0 D 0, 3, 12 · x − 0, y C 1, z − 2
or y C 4z D 7.
17. f x, y, z D zey cos x D 1 so §f x, y, z D −zey sin x, zey cos x, ey cos x and §f π, 0, −1 D 0, 1, −1.
So the equation of the tangent plane is:
0 D 0, 1, −1 · x − π, y, z C 1 or y − z D 1.
18. f x, y, z D 2xz C yz − x2 y C 10 D 0 so §f x, y, z D 2x − 2xy, z − x2 , 2x C y and §f 1, −5, 5 D
20, 4, −3. So the equation of the tangent plane is:
0 D 20, 4, −3 · x − 1, y C 5, z − 5 or 20x C 4y − 3z D −15.
19. f x, y, z D 2xy2 − 2z2 C xyz so §f x, y, z D 2y2 C yz, 4xy C xz, xy − 4z and §f 2, −3, 3 D
9, −18, −18. So the equation of the tangent plane is:
0 D 9, −18, −18 · x − 2, y C 3, z − 3 or x − 2y − 2z D 2.
20. (a) First we use the formula (4) from Section 2.3 in the text: z D f a, b C fx a, bx − a C fy a, by −
b. If x2 − 2y2 C 5xz D 7 then z D 7C2y −x D f x, y. Calculate the two partial derivatives:
2
2
5x
−7 − 2y2 − x2
−8
so fx −1, 0 D
5x2
5
4y
fy x, y D
so fy −1, 0 D 0.
5x
fx x, y D
and
At −1, 0, −6/5 formula (4) gives the equation of the tangent plane as
zD
−8
−6
C
x C 1.
5
5
(b) Now we’ll use formula (6) from this section and calculate the gradient of f x, y, z D x2 − 2y2 C 5xz
as §f x, y, z D 2x C 5z, −4y, 5x so §f −1, 0, −6/5 D −8, 0, −5 and so the equation for the
plane is
0 D −8, 0, −5 · x C 1, y, z C 6/5 or − 8x − 5z D 14.
This agrees with the answer we found in part (a).
21. (a) First we use the formula (4) from Section 2.3 in the text: x D f a, b C fy a, by − a C fz a, bz −
yz
b. If x sin y C xz2 D 2eyz then x D 2e 2 D f y, z. Calculate the two partial derivatives:
sin yCz
fy y, z D 2eyz
z sin y C z3 − cos z
sin y C z2 2
so
fy π/2, 0 D 0
and fz y, z D 2eyz
y sin y C yz2 − 2z
sin y C z2 2
so
fz π/2, 0 D π.
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“runall” — 2005/8/1 — 12:14 — page 119 — #50
Section 2.6
Directional Derivatives and the Gradient
119
At 2, π/2, 0 formula (4) gives the equation of the tangent plane as
x D 2 C πz.
(b) Now we’ll use formula (6) from this section and calculate the gradient of f x, y, z D x sin y C xz2 − 2eyz
as §f x, y, z D sin y C z2 , −x cos y − 2zeyz , 2xz − 2yeyz so §f 2, π/2, 0 D 1, 0, −π and so the
equation for the plane is
0 D 1, 0, −π · x − 2, y − π/2, z or
x − 2 − πz D 0.
This agrees with the answer we found in part (a).
22. Using formula (6) we get that the gradient of f x, y, z D x3 − 2y2 C z2 at x0 , y0 , z0 is §f
√ x0 , y0 , z0 D
perpendicular to the given line, 3x02 , −4y0 , 2z0 D k3, 2, − 2. This means
3x02 , −4y0 , 2z0 . For this to be √
that x02 D −2y0 and z0 D − 2/2x02 . Substituting this back into the equation
of the surface, we get that
√
x03 − 2x04 /4 C x04 /2 D 27 or x0 D 3. Our point is, therefore 3, −9/2, −9 2/2.
23. The tangent plane to the surface at a point x0 , y0 , z0 is
0 D 18x0 x − x0 − 90y0 y − y0 C 10z0 z − z0 .
For this to be parallel to x C 5y − 2z D 7, the vector
18x0 , −90y0 , 10z0 D k1, 5, −2.
This means that y0 D −x0 and z0 D −18/5x0 . Substitute these back into the equation of the hyperboloid:
9x2 − 45y2 C 5z2 D 45 to get:
45 D 9x02 − 45x02 C 5182 /52 x02
therefore x0 D ;5/4.
This means that the points are 5/4, −5/4, −9/2 and −5/4, 5/4, 9/2.
24. First note that 2, 1, −1 lies on both surfaces: 7 · 22 − 12 · 2 − 5 · 1 D −1, 2 · 1−12 D 2. The normal
to the first surface at 2, 1, −1 is given by fx 2, 1, fy 2, 1, −1 where f x, y D 7x2 − 12x − 5y2 . This
is 14x − 12|2,1 , −10y|2,1 , −1 D 16, −10, −1. The normal to the second surface at 2, 1, −1 is
§F 2, 1, −1 where F x, y, z D xyz2 . This is yz2 , xz2 , 2xyz|2,1,−1 D 1, 2, −4. We have
16, −10, −1 · 1, 2, −4 D 16 − 20 C 4 D 0.
Since the normals are orthogonal, the tangent planes must be so as well.
25. The two surfaces are tangent at x0 , y0 , z0 3 the tangent planes at x0 , y0 , z0 are the same 3 normal vectors
at x0 , y0 , z0 are parallel (since the surfaces intersect at x0 , y0 , z0 3 §F x0 , y0 , z0 * §Gx0 , y0 , z0 D 0.
26. (a) S is the level set at height 0 of f x, y, z D x2 C 4y2 − z2 so §f D 2x, 8y, −2z *
§f 3, −2, −5 D 6, −16, 10. Thus formula (6) gives the equation of the tangent plane as 6x − 3 −
16y C 2 C 10z C 5 D 0 or 3x − 8y C 5z D 0.
(b) §f 0, 0, 0 D 0, 0, 0 so formula (6) cannot be used. Note that there’s no tangent plane at the origin,
which is the vertex of the cone (i.e., the surface is not “locally flat” there).
27. (a) For f x, y, z D x3 − x2 y2 C z2 , §f x, y, z D 3x2 − 2xy2 , −2x2 y, 2z so §f 2, −3/2, 1 D
3, 12, 2. Thus the equation of the tangent plane is
3x − 2 C 12y C 3/2 C 2z − 1 D 0 or 3x C 12y C 2z C 10 D 0.
(b) §f0, 0, 0 D 0, 0, 0 so the gradientcannot be used as a normal vector. If we solve z D ; y2 x2 − x3 D
;x y2 − x, we see that gx, y D x y2 − x fails to be differentiable at (0, 0)—so there is no tangent
plane there.
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“runall” — 2005/8/1 — 12:14 — page 120 — #51
120
Chapter 2
28. (a) 2x C 2y
Differentiation in Several Variables
dy
D 0 so
dx
√
dy −x − 2
√
D
D
D 1.
dx −√ 2,√ 2
y −√ 2,√ 2
− 2
√
√
The equation of the line is y − 2 D x C 2.
2
2
(b) The equation of the tangent line√is 0 √
D §f x0 , y0√
· x
√ − x0 , y − y0 . Here f x, y D x C y D 4 so
§f x, y D 2x, 2y or §f − 2, 2 D −2 2, 2 2. The equation of the tangent line is
√
√
√
√
√
0 D −2 2, 2 2 · x C 2, y − 2 or x − y D −2 2.
dy
dy 5
D
. The equation of the tangent line is
D 2x C 3x2 so
29. (a) 3y2
2/3
3
dx
dx 1, √
32
2
y −
√
3
2D
5
x − 1.
322/3
√
(b) f x, y D y3 − x2 − x3 so §f x, y D −2x − 3x2 , 3y2 so §f 1, 3 2 D −5, 322/3 . The equation
of the tangent line is
√
3
0 D −5, 322/3 · x − 1, y − 2 or − 5x C 322/3 y D 1.
dy
dy
dy −76
−19
D
C 3y2
D 0 so
D
. The equation of the tangent line is
30. (a) 5x4 C 2y C 2x
dx
dx
dx 2,−2
16
4
y C 2D
−19
x − 2.
4
(b) f x, y D y3 − x2 − x3 so §f x, y D 5x4 C 2y, 2x C 3y2 so §f 2, −2 D 76, 16. The equation
of the tangent line is
0 D 76, 16 · x − 2, y C 2 or 19x C 4y D 30.
31. If f x, y D x2 − y2 then §f 5, −4 D 10, 8 so the equations of the normal line are
xt D 10t C 5
and yt D 8t − 4 or 8x − 10y D 80.
√
√
32. If f x, y D x2 − x3 − y2 then §f −1, 2 D 5, 2 2 so the equations of the normal line are
√
√
√
√
xt D 5t − 1 and yt D 2 2t − 2 or 2 2x − 5y D −7 2.
33. If f x, y D x3 − 2xy C y5 then §f 2, −1 D 14, 1 so the equations of the normal line are
xt D 14t C 2
and yt D t − 1 or
x − 14y D 16.
34. If f x, y, z D x3 z C x2 y2 C sinyz then
§f x, y, z D 3x2 C 2xy2 , 2x2 y C z cosyz, x3 C y cosyz.
(a) The plane is given by 0 D §f −1, 0, 3 · x C 1, y, z − 3 D 9x C 1 C 3y − z − 3 or
9x C 3y − z D −12.
(b) The normal line to the surface at −1, 0, 3 is given by


 x D 9t − 1
y D 3t

 z D −t C 3.
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“runall” — 2005/8/1 — 12:14 — page 121 — #52
Section 2.6
Directional Derivatives and the Gradient
121
35. Using the method above for f x, y, z D exy C ezx − 2eyz , we find that §f D yexy C zexz , xexy − 2zeyz , xexz −
2yeyz so §f −1, −1, −1 D e−2, 1, 1. So




 x D −2et − 1
 x D −2t − 1
y D et − 1
yDt − 1
or, factoring out e,


 z D et − 1
 z D t − 1.
36. Remember in the equation of a plane 0 D v · x − x0 , y − y0 , z − z0 that v is a vector orthogonal to the plane.
We saw in this section that we can use §f x0 , y0 , z0 for v. This means that the equation of the line normal to
a surface given by the equation F x, y, z D 0 at a given point x0 , y0 , z0 is
x, y, z D §F x0 , y0 , z0 t C x0 , y0 , z0 .
37. The hyper surface is the level set at height −1 of the function f x1 , ... , x5 D sin x1 C cos x2 C sin x3 C
3π
cos x4 C sin x5 . We find §f π, π,
, 2π, 2π D −1, 0, 0, 0, 1. Hence the tangent hyperplane has equation
2
−1x1 − π C 1x5 − 2π D 0
or x5 − x1 D π.
nn C 1
of the function f x1 , ... , xn D x12 C 2x22 C · · · C nxn2 . We
2
have §f D 2x1 , 4x2 , 6x3 , ... , 2nxn * §f −1, ... , −1 D −21, 2, 3, ... , n. An equation for the tangent
hyperplane is thus
38. The surface is the level set at height
1x1 C 1 C 2x2 C 1 C 3x3 C 1 C · · · C nxn C 1 D 0
or
x1 C 2x2 C 3x3 C · · · C nxn C
nn C 1
D0
2
39. Here f x1 , x2 , ... , xn D x12 C x22 C · · · C xn2 so §f x1 , x2 , ... , xn D 2x1 , 2x2 , ... , 2xn . Using the techniques
√ section,√ the tangent
√ hyperplane to the n − 1-dimensional sphere f x1 , x2 , ... , xn D 1 at
√ of this
1/ n, 1/ n, ... , 1/ n, −1/ n is
√
√
√
√
√
√
√
0 D §f 1/ n, ... , 1/ n, −1/ n · x1 − 1/ n, x2 − 1/ n, ... , xn−1 − 1/ n, xn C 1/ n
√
1
1
1
1
2
2
2
−2
D √ x1 − √ C √ x2 − √ C · · · C √ xn−1 − √ C √ xn C √ n
n
n
n
n
n
n
n
√
√
√
√
0 D x1 − 1/ n C x2 − 1/ n C · · · C xn−1 − 1/ n − xn C 1/ n so
or
n D x1 C x2 C · · · C xn−1 − xn .
40. F x, y, z D z2 y3 C x2 y D 2.
(a) We can write z D f x, y when Fz Z 0. Fz x, y, z D 2zy3 is not 0 when both z Z 0 and y Z 0.
(b) We can write x D f y, z when Fx Z 0. Fx x, y, z D 2xy is not 0 when both x Z 0 and y Z 0.
(c) We can write y D f x, z when Fy Z 0. Fy x, y, z D 3z2 y2 C x2 is not 0 everywhere but on the y- or
z-axis (i.e., except when x D 0 at the same time that either y D 0 or z D 0).
⭸F
D xexz . This is non-zero whenever x Z 0. There we can solve for z to get
41. (a)
⭸z
zD
ln1 − sin xy − x3 y
.
x
(b) Looking only at points in S we only need to stay away from points in yz-plane (i.e., where x D 0).
(c) You shouldn’t then make the leap from your answer to part (b) that you can graph z D
ln1 − sin xy − x3 y
for any values of x and y just so x Z 0. Your other restriction is that 1 − sin xy −
x
3
x y > 0 as it is the argument of the natural logarithm. A sketch that gives you an idea of the surface is:
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“runall” — 2005/8/1 — 12:14 — page 122 — #53
122
Chapter 2
Differentiation in Several Variables
4
2
z 0
-2
-4
-1
1
0.5
0
y
-0.5
-0.5
0
x
0.5
1
-1
Now the actual surface S includes the plane x D 0 since x D 0 satisfies the original equation: sin xy C
exz C x3 y D 1. S will actually look a bit like:
y
0.5
0
-0.5
-1
4
x
1-2
-1
0
1
2
z 0
-2
-4
42. The point of this problem is that since F x, y D c defines a curve C in R2 such that either fx x0 , y0 Z 0 or
fy x0 , y0 Z 0 then by the implicit function theorem we can represent the curve near x0 , y0 as either the graph
of a function x D gy or a function y D gx.
Exercise 43 poses a bit of a puzzle. Here we can write the equation of C as y D f x even though Fy is zero at the
origin. Why doesn’t this contradict the implicit function theorem? What “goes bad” in Exercise 43 is that we have a corner
at the origin. You may also want to assign the students the same problem for the function F x, y D x − y3 .
43. (a) F 0, 0 D 0 so the origin lies on the curve C. Fy x, y D 3y2 and so Fy 0, 0 D 0.
(b) We can write C as the graph of y D x2/3 . The graph of C is
y
1
0.8
0.6
0.4
0.2
-1
-0.5
0.5
1
x
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“runall” — 2005/8/1 — 12:14 — page 123 — #54
Section 2.6
Directional Derivatives and the Gradient
123
(c) So here we are with Fy 0, 0 D 0 but we can express the graph of C everywhere as y D x2/3 . On second
look we see that C is not a C1 function—it has a corner at the origin—and so the implicit function theorem
doesn’t apply.
44. (a) F x, y D xy C 1 so Fy x, y D x and so we cannot solve F x, y D c for y when x D 0 or when
c D 0y C 1 D 1. In other words, level sets are unions of smooth curves in R2 except for c D 1.
(b) Here the function is F x, y, z D xyz C 1. Using a similar argument to that in part (a), Fz x, y, z D xy
and this is only 0 when xy D 0. This means that we cannot solve F x, y, z D c for z when xy D 0 or when
c D z0 C 1 D 1. So level sets of this family are unions of smooth surfaces in R3 except for level c D 1.
45. (a) G−1, 1, 1 D F −1 − 2 C 1, −1 − 1 C 3 D F −2, 1 D 0.
(b) To invoke the implicit function theorem, we need to show that Gz −1, 1, 1 Z 0.
⭸x3 − 2y2 C z5 ⭸xy − x2 z C 3 Gz −1, 1, 1 D Fu −2, 1
C Fv −2, 1
⭸z
⭸z
−1,1,1
−1,1,1
D 75 C 51 D 40 Z 0.
46. Let F1 D x2 y2 − x1 cos y1 D 5 and F2 D x2 sin y1 C x1 y2 D 2. Solving for y in terms of x means that we have
to look at the determinant


⭸F1 ⭸F1
 ⭸ y1 ⭸ y2 
x1 sin y1 x2
2
2

det 
 ⭸F2 ⭸F2  D det x2 cos y1 x1 D x1 sin y1 C x2 cos y1 .
⭸ y1 ⭸ y2
To see that you can solve for y1 and y2 in terms of x1 and x2 near x1 , x2 , y1 , y2 D 2, 3, π, 1, evaluate the
determinant at that point. We get −9. This is not 0 so you can, at least in theory, solve for the y’s in terms of
the x’s.
To see that you can solve for y1 and y2 as functions of x1 and x2 near x1 , x2 , y1 , y2 D 0, 2, π/2, 5/2, evaluate
the determinant at that point. We get 0. We can not solve for the y’s in terms of the x’s.
47. (a) Let F1 D x12 y22 − 2x2 y3 D 1, F2 D x1 y15 C x2 y2 − 4y2 y3 D −9, and F3 D x2 y1 C 3x1 y32 D 12. Solving
for y’s in terms of x’s means that we have to look at the determinant


⭸F1 ⭸F1 ⭸F1




 ⭸y1 ⭸y2 ⭸y3 


0
2x12 y2
−2x2
 ⭸F2 ⭸F2 ⭸F2 


4

det 
 ⭸y1 ⭸y2 ⭸y3  D det  5x1 y1 x2 − 4y3 −4y2 


x2
0
6x1 y3


 ⭸F3 ⭸F3 ⭸F3 
⭸y1 ⭸y2 ⭸y3
D −60x14 y14 y2 y3 − 8x12 x2 y22 C 2x23 − 8x22 y3 .
Evaluating this at the point x1 , x2 , y1 , y2 , y3 D 1, 0, −1, 1, 2 results in −120 Z 0. This means that we
can solve for y1 , y2 , and y3 in terms of x1 and x2 .
(b) Take the partials of the three equations with respect to x1 to get


⭸y
⭸y


y22 C 2x1 y2 2 − 2x2 3 D 0



⭸
x
⭸x1
1



⭸
y
⭸
y
⭸y
⭸y
y15 C 5x1 y14 1 C x2 2 − 4y3 2 − 4y2 3 D 0

⭸x1
⭸x1
⭸x1
⭸x1




⭸
y
⭸
y

1
3
2
 x2
C 3y3 C 6x1 y3
D 0.


⭸x1
⭸x1
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“runall” — 2005/8/1 — 12:14 — page 124 — #55
124
Chapter 2
Differentiation in Several Variables
At the point 1, 0, −1, 1, 2 this system of equations becomes:



 1 C 2 ⭸y2 D 0



⭸x1



⭸y
⭸y
⭸y
−1 C 5 1 − 8 2 − 4 3 D 0

⭸
x
⭸
x
⭸x1

1
1



⭸
y

3


 12 C 12 ⭸x D 0.
1
Solving, we find that
⭸y1
−7 ⭸y2
1
,
D
D− ,
⭸x1
5 ⭸x1
2
and
⭸y3
D −1.
⭸x1
48. (a) We need to consider where the following determinant is non-zero.
⭸F /⭸r ⭸F /⭸θ ⭸F /⭸z cos θ −r sin θ 0 1
1
1
⭸F2 /⭸r ⭸F2 /⭸θ ⭸F2 /⭸z D sin θ r cos θ 0 D r cos2 θ C r sin2 θ D r.
⭸F3 /⭸r ⭸F3 /⭸θ ⭸F3 /⭸z 0
0
1 In other words, for any points for which r Z 0.
(b) This makes complete sense. When the radius is 0 then r and z completely determine the point. You get no
extra information from the θ component. Without the z coordinate, this is the standard problem when using
polar coordinates in the plane.
49. (a) As with Exercise 48, we need to consider where the same determinant is non-zero. In this case the determinant is
sin ϕ cos θ ρ cos ϕ cos θ −ρ sin ϕ sin θ sin ϕ sin θ ρ cos ϕ sin θ ρ sin ϕ cos θ D ρ2 sin ϕ cos2 ϕ C ρ2 sin3 ϕ D ρ2 sin ϕ.
cos ϕ
−ρ sin ϕ
0
In other words, for any points for which ρ Z 0 and for which sin ϕ Z 0.
(b) Again, this makes complete sense. When the radius is 0, then ρ completely determines the point as being the
origin. When sin ϕ D 0 you are on the z-axis so θ no longer contributes any information.
2.7 TRUE/FALSE EXERCISES FOR CHAPTER 2
1. False.
2. True.
3. False. (The range also requires v Z 0.)
4. False. (Note that fi D fj.)
5. True.
6. False. (It’s a paraboloid.)
7. False. (The graph of x2 C y2 C z2 D 0 is a single point.)
8. True.
9. False.
10. False. (The limit does not exist.)
11. False. limx,y'0,0 f x, y D 0 Z 2.
12. False.
13. False.
14. True.
15. False. §f x, y, z D 0, cos y, 0.
16. False. (It’s a 4 * 3 matrix.)
17. True.
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“runall” — 2005/8/1 — 12:14 — page 125 — #56
Section 2.8
Miscellaneous Exercises for Chapter 2
125
18. False.
19. False. (The partial derivatives must be continuous.)
20. True.
21. False. fxy Z fyx .
22. False. (f must be of class C2 .)
23. True. (Write the chain rule for this situation.)
24. True.
25. False. (The correct equation is x C y C 2z D 2.)
26. False. (The plane is normal to the given vector.)
27. True.
28. False. (The directional derivative equals −⭸f /⭸z.)
29. False.
30. True.
2.8 MISCELLANEOUS EXERCISES FOR CHAPTER 2
i
1
x1
1. (a) Calculate the determinant
j
0
x2
k 1 D −x2 , x1 − x3 , x2 .
x3 More explicitly, the component functions are f1 x1 , x2 , x3 D −x2 , f2 x1 , x2 , x3 D x1 − x3 , and
f3 x1 , x2 , x3 D x2 .
(b) The domain is all of R3 while the range restricts the first component to be the opposite of the last component.
In other words the range is the set of all vectors a, b, −a.
2. (a) It might help to see f explicitly first as
3, −2, 1 · x, y, z
3x − 2y C z
3, −2, 1.
3, −2, 1 D
3, −2, 1 · 3, −2, −1
14
(b) The domain is all of R3 and the range are vectors of the form 3a, −2a, a.
3. (a) The domain of f is {x, y|x Ú 0 and y Ú 0} ∪ {x, y|x … 0 and y … 0}. The range is all real numbers
greater than or equal to 0.
(b) The domain is closed. The quarter planes are closed on two sides because they include the axes.
4. (a) The domain of f is {x, y|x Ú 0 and y > 0} ∪ {x, y|x … 0 and y < 0}. The range is all real numbers
greater than or equal to 0.
(b) The domain is neither open nor closed. The quarter planes are closed on one side because they include the
y-axis but they don’t include the x-axis and so aren’t closed.
5.
f (x, y)
Graph Level curves
1/x2 C y2 C 1
sin x2 C y2
D
d
B
e
3y2 − 2x2 e
A
b
y3 − 3x2 y
E
c
−x2 −y2
F
a
C
f
−x2 −2y2
x2 y 2 e
−x2 −y2
ye
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“runall” — 2005/8/1 — 12:14 — page 126 — #57
126
Chapter 2
Differentiation in Several Variables
6. (a) See below left.
4
2
0
5
4
z 3
2
-2
4
2
-4
0
-2
x
-4
-4
-2
0
2
4 -4
4
2
y
-2
0
(b) See above right.
7. First we’ll substitute x D r cos θ and y D r sin θ while noting that x, y ' 0, 0 is equivalent to r ' 0.
r sin θr2 cos2 θ − r3 sin3 θ
yx2 − y3
D lim
r'0
x,y'0,0 x2 C y2
r2 cos2 θ C r2 sin2 θ
lim
r3 cos2 θ − sin2 θ sin θ
r'0
r2
D lim r cos 2θ sin θ D 0
D lim
r'0
8. (a)
2xy
x2 C y 2
D
2r2 cos θ sin θ
D 2 cos θ sin θ D sin 2θ. So
r2
f x, y D "
sin 2θ
0
if r Z 0
.
if r D 0
(b) We’re looking for (x, y) such that f x, y D c. For −1 < c < 1 the level sets are pairs of radial lines
symmetric about θ D π/4.
For example, if c D 1/2 then we are looking for θ such that sin 2θ D 1/2. In this case, θ D π/12, 5π/12,
13π/12, and 17π/12. So the level sets are the lines θ D π/12 and θ D 5π/12. These could also be written
as θ D π/4 ; π/6.
For c D 1 the level set is the line θ D π/4, for c D −1 the level set is the line θ D 3π/4 and for |c| > 1
the level set is the empty set.
(c) f is constant along radial lines, so the figure below just shows a ribbon corresponding to .4 < r < 1.
1
y
0.5
0
-0.5
-1
1
0.5
z
0
-0.5
-1
-1
-0.5
0
x
0.5
1
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“runall” — 2005/8/1 — 12:14 — page 127 — #58
Section 2.8
Miscellaneous Exercises for Chapter 2
127
(d) limx,y'0,0 f x, y D limr'0 sin 2θ which doesn’t exist.
(e) Since the limit doesn’t exist at the origin, f couldn’t be continuous there. Also, f takes on every value
between 1 and −1 in every open neighborhood of the origin.
Before assigning Exercise 9 you may want to ask the students if it is true that if a function F (x, y) is continuous in each
variable separately it is continuous. The calculations in Exercise 9 are fairly routine but the conclusion is very important.
9. gx D F x, 0 K 0 and so is continuous at x D 0 and hy D F 0, y K 0 and so is continuous at y D 0.
Consider px D F x, x D 1 when x Z 0 and F 0, 0 D 0. Clearly, p(x) is not continuous at 0 so F (x, y) is
not continuous at (0,0).
10. (a) You can see as x √
gets closer and closer to 0 that 1/x2 gets larger and larger. More formally, for any N > 0,
2
if 0 < |x| < 1/ N then 1/x
> N.
(b) Here
||x, y − 1, 3|| D x − 12 C y − 32 so for any N > 0, if 0 < ||x, y − 1, 3|| <
2/N, then
2
2
2
D N.
D
>
x − 12 C y − 32
||x, y − 1, 3||2
2/N
(c) The definition is analogous to that for above: limx'a f x D −q means that given any N < 0 there is
some δ > 0 such that if 0 < ||x − a|| < δ then f a < N.
(d) We are considering limx,y'0,0 so let’s restrict our attention to |x| < 1 and |y| < 1. For |x| < 1 we have
−1
1 − x
D
.
xy4 − y4 C x3 − x2
y 4 C x2
For |y| < 1 we have y4 < y2 so
−1
−1
−1
<
D
.
y 4 C x2
y 2 C x2
||x, y||2
√
So for any N < 0 if 0 < ||x, y|| < min{1, 1/ −N} then
1 − x
< N.
xy4 − y4 C x3 − x2
11. We read right from the table in the text:
(a) 15◦ F.
(b) 5◦ F.
12. (a) If the temperature of the air is 10◦ F we read off the chart that when the windspeed is 10 mph the windchill is
−4; when the windspeed is 15 mph the windchill is −7. Since we are looking to estimate when the windchill
is −5 you might be tempted to stop here and just conclude that the answer is between 10 mph and 15 mph
(and you’d be correct) but we want to say more. Our first estimate will just use linear interpolation (similar
triangles) to get x D 5 or the distance from 15 is x D 10/3. We would then conclude that, to the nearest
2
3
degree, the windspeed is 15 − 10/3 L 12 mph.
(b) Before you feel too good about your answer to part (a) you should notice further that when the windspeed is 20
the windchill is −9 and when the windspeed is 25 the windchill is −11. In other words, the rate at which the
windchill is dropping is slowing slightly. In calculus terms, for the function f s D W s mph, 30◦ , f s
seems to be positive so the curve is concave up. The line used to estimate in part (a) then probably lies above
the curve and our guess of 12 mph is, most likely, too high.
13. For the function W s mph, t ◦ , we want to estimate
W 30, 35 C h − W 30, 35
⭸W .
D lim
⭸t 30 mph,35◦ h'0
h
We will use the slopes of the two secant lines:
W 30, 40 − W 30, 35
28 − 22
D
D 1.2
5
5
W 30, 30 − W 30, 35
15 − 22
D
D 1.4
−5
−5
We average them to get an estimate of 1.3.
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128
Chapter 2
Differentiation in Several Variables
14. We will use the same technique as in Exercise 13 and estimate the derivative with respect to windspeed by
averaging the slopes of the two secant lines.
W 20, 25 − W 15, 25
11 − 13
D
D −0.4
5
5
W 10, 25 − W 15, 25
15 − 13
D
D −0.4
−5
−5
so we average them to get an estimate of −0.4.
15. (a) Comparison with Exercise 11: With an air temperature of 25◦ F, windspeed of 10 mph,
√
W 10, 25 D 91.4 C 25 − 91.40.474 C 0.304 10 − 0.203
L 9.573
or 10◦ F
(as compared to 15◦ F in 11(a)).
If s D 20 mph, then W D −15◦ F if
√
91.4 C t − 91.40.474 C 0.304 20 − 0.406 D −15.
15 C 91.4
√
L 16.866 or 17◦ F (as compared to 5◦ F in 11(b)).
0.474 C 0.304 20 − 0.406
√
Comparison with Exercise 12: With W s, t D 91.4 C t − 91.4.474 C .304 s − .0203s, we must
solve
√
20 D 91.4 C 30 − 91.4.474 C .304 s − .0203s
Hence t D 91.4 −
or
√
20 − 91.4
D .474 C .304 s − .0203s so that
30 − 91.4
√
1.16287 L .474 C .304 s − .0203s or
√
0 L .0203s − .304 s C .688866
√
√
.304 ; .0364801
Now solve the quadratic: s L
. The two solutions are 7.74682 and 148.646.
.0406
(b) The windchill effect of windspeed appears to be greater in the Siple formula than that which may be inferred
from the table.
(c) For temperatures greater than 91.4 the model has the wind actually making the apparent temperature warmer
than air temperature. Physically, the model probably falls apart because between 91.4 and 106 you are too
close to body temperature for the wind to have much effect and if you are in temperatures much greater than
106 a breeze won’t replace a frosty beverage. For winds below 4 mph, the effect is negligible and won’t be
reflected in the model.
16. Comparison with Exercise 13: We want to calculate Wt 30, 35. ⭸W /⭸t D 0.621 C 0.4275s0.16 , so Wt 30, 35 D
0.621 C 0.4275300.16 L 1.358 (this is close).
Comparison with Exercise 14: We want Ws 15, 25.
⭸W /⭸s D −35.750.16s−0.84 C 0.42750.16ts−0.84
D −5.72 C 0.0684ts−0.84
Ws 15, 25 D −5.72 C 0.0684 · 2515−0.84 L −0.412 (again close).
17. (a) Pictured (left) are the pairs W1 s, 40 and W2 s, 40 and, on the right, the pairs W1 s, 5 and W2 s, 5.
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“runall” — 2005/8/1 — 12:14 — page 129 — #60
Section 2.8
W(s,40)
Miscellaneous Exercises for Chapter 2
129
W(s,5)
10
40
35
20
30
40
60
80
100
120
s
-10
W2
25
W2
-20
20
-30
15
W1
10
20
40
60
80
100
120
-40
s
W1
From these graphs, we see that windspeed depresses apparent temperature in the Siple formula much more
than in the National Weather Service Formula.
(b) Pictured (left) are the pairs W1 10, t and W2 10, t and, on the right, the pairs W1 30, t; and W2 30, t.
Again we see that the Siple formula results in lower apparent temperatures predicted, only the effect appears
to be more of a constant difference.
W(10,t)
W(30,t)
20
20
-40
-40
-20
20
40
-20
t
40
t
-40
-20
-60
W2
-40
W2
20
-20
-80
-100
-60
W1
W1
-120
(c) The surfaces z D W1 s, t and z D W2 s, t are pictured. Note that the Siple surface determined by W1 is
more curved, demonstrating a more nonlinear effect of windspeed.
0
-50
-100
40
W1
20
0
20
-20
s
40
0
-50
-100
40
W2
20
0
t
20
s
40
-40
-40
60
t
-20
60
18. The equation of the sphere is F x, y, z D x2 C y2 C z2 D 9 so §F D 2x, 2y, 2z and the plane tangent to
the sphere at (1, 2, 2) is 0 D 2, 4, 4 · x − 1, y − 2, z − 2 or x C 2y C 2z D 9. This intersects the x-axis
when y D 0 and z D 0 so x D 9.
19. Without loss of generality we can locate the center of the sphere at the origin and so the equation of the sphere is
F x, y, z D x2 C y2 C z2 D r2 so §F D 2x, 2y, 2z and the equation of the plane tangent to the sphere at
P D x0 , y0 , z0 is 0 D 2x0 , 2y0 , 2z0 · x − x0 , y − y0 , z − z0 or x0 x C y0 y C z0 z D x02 C y02 C z20 .
This is orthogonal to the vector x0 , y0 , z0 which is the vector from the center of the sphere to P.
20. Because we’re looking at a curve in the plane 2x − y D 1 we know the x and y components of the parametric
equations. What is left to determine is z. Substitute in 2x − 1 for y in the equation of the surface to get
z D 3x2 C x3 /6 − x4 /8 − 42x − 12 D −5x2 C x3 /6 − x4 /8 C 4. We can now calculate the derivative
⭸z/⭸x D −10x C x2 /2 − x3 /2 and evaluate it at the point 1, 1, −23/24 to get −10. Because the value of z
is −23/24 when x D 1, this component of the tangent line is derived by looking at z C 23/24 D −10x − 1.
So the parametric equations for the tangent line are t, 2t − 1, −10t C 192/24.
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130
Chapter 2
Differentiation in Several Variables
21. (a) For the function f x, y, z D x2 C y2 − z2 D 0 we consider §f x0 · x − x0 D 0. Here we get
23, 2−4, −25 · x − 3, y C 4, z − 5 D 0 or the equation is 6x − 3 − 8y C 4 − 10z −
5 D 0.
(b) In general we get 2a, 2b, −2c · x − a, y − b, z − c D 0. This amounts to 2ax − a C
2by − b − 2cz − c D 0.
(c) Note that (0, 0, 0) is a solution so the plane passes through the origin.
22. Show that the two surfaces
S1 : z D xy and S2 : z D
3 2
x − y2
4
intersect perpendicularly at the point (2, 1, 2). First we see that 2 D 12 and 2 D 3/44 − 1 so (2, 1, 2) is
a point on both surfaces. Rewrite the surfaces so that they are level sets of functions:
F1 x, y, z D xy − z and F2 x, y, z D z C y2 −
3 2
x .
4
The gradients are normal to the tangent planes (see Section 2.6, Exercise 36), so we calculate the two gradients
at the given point: §F1 2, 1, 2 D 1, 2, −1 and §F2 2, 1, 2 D −3, 2, 1 so
§F1 2, 1, 2 · §F2 2, 1, 2 D 0.
So the two surfaces intersect perpendicularly at (2, 1, 2).
23. (a) As we have done before we find the plane tangent to the surface given by F x, y, z D z − x2 − 4y2 D 0
by formula (6):
0 D §F 1, −1, 5 · x − 1, y C 1, z − 5 D −2, 8, 1 · x − 1, y C 1, z − 5
or − 2x C 8y C z D −5.
(b) The line is parallel to a vector which is orthogonal to §F 1, −1, 5 D −2, 8, 1 and with no component
in the x direction. So it is of the form (0, a, b) with 0, a, b 
· −2, 8, 1 D 0 so the line has the direction

 xD1
yDt − 1
0, 1, −8 and passes through 1, −1, 5. The equations are

 z D −8t C 5.
24. We are assuming that the collar is fairly rigid so that it is maintaining a cylindrical shape throughout this process.
dr
⭸V
⭸V
dh
We want
at t D t0 . Since V D πr2 h,
D 2πrh
C πr2 . We are given that the rate of change of the
⭸t
⭸t
dt
dt
circumference at t D t0 is −.2 in/min. This means
⭸2πr dr ⭸C −.2 D
D
D 2π .
⭸t t0
⭸t t0
dt t0
We also know that at t D t0 , 2πr D 18, h D 3, and
dh
D .1. Substituting into the equation above, we get:
dt
2
18
8.1
2.7
−.2
−5.4
⭸V C
π
C
D
.
D
183
.1 D
⭸t t0
2π
2π
π
π
π
So the volume is increasing at t D t0 .
25. First note that 0.2◦ C/day D 0.2 · 24 D 4.8◦ C/month. Then, with time measured in months, the chain rule
tells us
⭸P dS
⭸P dT
dP
D
C
.
dt
⭸S dt
⭸T dt
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“runall” — 2005/8/1 — 12:14 — page 131 — #62
Section 2.8
Here
Miscellaneous Exercises for Chapter 2
131
dT
dS
D −2,
D 4.8. With PS, T D 330S 2/3 T 4/5 , we have
dt
dt
dP D 220S −1/3 T 4/5 |75,15 −2 C 264S 2/3 T −1/5 |75,15 4.8
dt SD75,T D15
D 22075−1/3 154/5 −2 C 264752/3 15−1/5 4.8
D 12,201.4 units/month
or 508.392 units/day
26. We want to know
tells us
dy
du
dx
(t in weeks) when x D 80, y D 240, given that
D 5 and
D −15. The chain rule
dt
dt
dt
⭸u dx
⭸u dy
du
D
C
dt
⭸x dt
⭸y dt
Thus
D 0.002xe −0.001x −0.00005y 2
2
2
2 dy
dx
C 0.0001ye −0.001x −0.00005y dt
dt
2
2
du D e −0.00180 −0.00005240 [0.00280 · 5 − 0.0001240 · 15]
dt xD80,yD240
L 0.000041.
So the utility function is increasing ever so slightly
27.
w D x 2 C y 2 C z2 ,
x D ρ cos θ sin ϕ,
y D ρ sin θ sin ϕ
and
z D ρ cos ϕ
(a)
⭸w
⭸w ⭸x
⭸w ⭸y
⭸w ⭸z
D
C
C
⭸ρ
⭸x ⭸ρ
⭸y ⭸ρ
⭸z ⭸ρ
D 2x cos θ sin ϕ C 2y sin θ sin ϕ C 2z cos ϕ
D 2ρ cos2 θ sin2 ϕ C 2ρ sin2 θ sin2 ϕ C 2ρ cos2 ϕ
D 2ρ,
⭸w
⭸w ⭸x
⭸w ⭸y
⭸w ⭸z
D
C
C
⭸ϕ
⭸x ⭸ϕ
⭸y ⭸ϕ
⭸z ⭸ϕ
D 2xρ cos θ cos ϕ C 2yρ sin θ cos ϕ − 2zρ sin ϕ
D 2ρ2 cos2 θ cos ϕ sin ϕ C 2ρ2 sin2 θ cos ϕ sin ϕ − 2ρ2 cos ϕ sin ϕ
D 0,
and
⭸w
⭸w ⭸x
⭸w ⭸y
⭸w ⭸z
D
C
C
⭸θ
⭸x ⭸θ
⭸y ⭸θ
⭸z ⭸θ
D −2xρ sin θ sin ϕ C 2yρ cos θ sin ϕ
D −2ρ2 cos θ sin θ sin2 ϕ C 2ρ2 cos θ sin θ sin2 ϕ
D 0.
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132
Chapter 2
Differentiation in Several Variables
(b) First substitute: w D x2 C y2 C z2 D ρ cos θ sin ϕ2 C ρ sin θ sin ϕ2 C ρ cos ϕ2 D ρ2 . Now taking
the derivatives from part (a) is trivial: wρ D 2ρ, wϕ D 0, and wθ D 0.
28. If w D f xCy , let u D xCy . So
xy
xy
x2
⭸w
⭸w
⭸w ⭸u
⭸w ⭸u
− y2
D x2
− y2
⭸x
⭸y
⭸u ⭸x
⭸u ⭸y
D x2
2
⭸w −y2
2 ⭸w −x
2 2 − y
2 2
⭸u x y
⭸u x y
D 0.
29. (a) First use the chain rule to find
⭸z
⭸z
and
:
⭸r
⭸θ
⭸z
⭸z ⭸x
⭸z ⭸y
D
C
⭸r
⭸x ⭸r
⭸y ⭸r
⭸z r
⭸z r
D
e cos θ C
e sin θ, and
⭸x
⭸y
⭸z
⭸z ⭸x
⭸z ⭸y
D
C
⭸θ
⭸x ⭸θ
⭸y ⭸θ
⭸z
⭸z r
D
−e r sin θ C
e cos θ.
⭸x
⭸y
Now solve for
⭸z
⭸z
and
:
⭸x
⭸y
⭸z
⭸z
⭸z
D e −r cos θ
− e −r sin θ ,
⭸x
⭸r
⭸θ
⭸z
⭸z
⭸z
−r
−r
D e sin θ
C e cos θ .
⭸y
⭸r
⭸θ
(b) Given the results for
and
⭸z
⭸z
and
in part (a), we compute:
⭸x
⭸y
⭸2 z
⭸ ⭸z
⭸ ⭸z
⭸ ⭸z
−r
−r
D
D e cos θ − e sin θ ⭸x2
⭸x ⭸x
⭸r ⭸x
⭸θ ⭸x
D e −r cos θ
⭸
⭸z
⭸z
⭸
⭸z
⭸z
−r
− e −r sin θ − e −r sin θ e −r cos θ
− e −r sin θ e cos θ
⭸r
⭸r
⭸θ
⭸θ
⭸r
⭸θ
D e −r cos θ −e −r cos θ
⭸z
⭸2 z
⭸z
⭸2 z
C e −r cos θ
− e −r sin θ
C e −r sin θ
⭸r
⭸r2
⭸θ
⭸r⭸θ
− e −r sin θ −e −r sin θ
D e −2r sin2 θ − cos2 θ
⭸z
⭸2 z
⭸z
⭸2 z
C e −r cos θ
− e −r cos θ
− e −r sin θ
⭸r
⭸θ⭸r
⭸θ
⭸θ2
⭸z
⭸2 z
⭸z
⭸2 z
⭸2 z
C cos2 θ
− 2 sin θ cos θ
C sin2 θ
C 2 sin θ cos θ
2
⭸r
⭸r
⭸θ
⭸r⭸θ
⭸θ2
A similar calculation gives:
⭸2 z
⭸ ⭸z
⭸ ⭸z
⭸ ⭸z
−r
−r
D
D e sin θ C e cos θ ⭸y2
⭸y ⭸y
⭸r ⭸y
⭸θ ⭸y
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Section 2.8
D e −r sin θ
Miscellaneous Exercises for Chapter 2
133
⭸
⭸z
⭸z
⭸
⭸z
⭸z
−r
C e −r cos θ C e −r cos θ e −r sin θ
C e −r cos θ e sin θ
⭸r
⭸r
⭸θ
⭸θ
⭸r
⭸θ
D e −2r cos2 θ − sin2 θ
⭸z
⭸2 z
⭸z
⭸2 z
⭸2 z
− 2 sin θ cos θ
C sin2 θ
C 2 sin θ cos θ
C cos2 θ
.
2
⭸r
⭸r
⭸θ
⭸r⭸θ
⭸θ2
Now add these to get:
⭸2 z
⭸2 z
C
D e −2r [cos2 θ C sin2 θzθθ C cos2 θ C sin2 θzrr ] D e −2r [zθθ C zrr ].
⭸x2
⭸y2
30. (a) Consider w D f x, y D xy D ey ln x . Then d uu can be calculated by taking the derivative and evaluating
du
at the point (u, u).
dw
⭸w dx
⭸w dy
D
C
D yxy−1 C ln xey ln x D u · uu−1 C ln u uu D uu 1 C ln u.
du
⭸x du
⭸y du
(b) Here x D sin t and y D cos t. So
⭸w dx
⭸w dy
dw
D
C
D yxy−1 cos t C ln x ey ln x − sin t D cos2 tsin tcos t−1 − sin t lnsin t sin t cos t .
dt
⭸x dt
⭸y dt
z
31. This is an extension of the preceding exercise. This time w D f x, y, z D xy . If x D u, y D u, and z D u we
again calculate
z
z
z ln y
dw
⭸w dx
⭸w dy
⭸w dz
D
C
C
D yz xy −1 C e y ln x z ln xyz−1 C e e ln x ln x e z ln y ln y
du
⭸x du
⭸y du
⭸z du
D uu uu −1 C uu u ln uuu−1 C uu ln u2 uu D uu uu 1/u C ln u C ln u2 .
u
32. With
u
u
#
r D ||x|| D x12 C · · · C xn2 ,
u
xi
x
⭸r
D D i.
⭸xi
r
x12 C · · · C xn2
⭸f
dg ⭸r
x
D
D g r i
⭸xi
dr ⭸xi
r
By the product and chain rules:
The chain rule gives
g r
⭸2 f
⭸
d g r ⭸r
xi
D
r
g
D
C
x
i
⭸xi2
⭸xi
r
r
dr
r
⭸xi
D
rg r − g r xi
1 g r C xi r
r2
r
D
g r
g r
1 −
g r C xi2 .
r
r2
r3
Add these to find
g r
g r
⭸2 f
n
2
2
D g r C −
x1 C · · · C xn 2
2
r
r
r3
iD1 ⭸xi
n
§2 f D Dr 2
g r
n g r C g r −
r
r
1
D n − 1g r C g r.
r
D
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134
Chapter 2
Differentiation in Several Variables
33. (a)
§2 §2 f x, y D
D
⭸2 f
⭸2 f
⭸2 ⭸2 f
⭸2 ⭸2 f
2 C
C
2 C
2
2
2
⭸x ⭸x
⭸y
⭸y ⭸x
⭸y2
⭸4 f
C
⭸x4
⭸4 f
⭸4 f
C
2
2
⭸x ⭸y
⭸y2 ⭸x2
C
⭸4 f
⭸y 4
these are equal−f is of class C4
D desired expression.
(b) Similar:




n
n
2f
2f
2
⭸
⭸
⭸2 
⭸



 C ··· C


§ § f D 2  ⭸x1 jD1 ⭸xj2
⭸xn2 jD1
⭸xj


n
n
n
2f
⭸
⭸4 f
⭸2 

D

.
D
2
2 2
2
iD1 ⭸xi
jD1 ⭸xj
i,jD1 ⭸xi ⭸xj
2
2
34. Livinia is at (0, 0, 1) and T x, y, z D 10xe−y C ze−x √
(a) The unit vector in the direction from (0, 0, 1) to (2, 3, 1) is u D 2, 3, 0/ 13.
√
√
Du T D §T 0, 0, 1 · u D 101, 0, 1 · 2, 3, 0/ 13 D 20/ 13 deg/cm.
√
(b) She should head in the direction√of the negative
gradient: −1, 0, −1/ 2.
√
(c) 3101, 0, 1 · −1, 0, −1/ 2 D −30 2 deg/sec.
35. z D r cos 3θ
(a) z D r[cos θ cos 2θ − sin θ sin 2θ] D r[cos θcos2 θ − sin2 θ − sin θ2 sin θ cos θ] so
2
zD
2
x3 − 3xy2
r3 [cos3 θ − cos θ sin2 θ − 2 sin2 θ cos θ]
D
.
2
r
x2 C y 2
(b) Note that limr'0 r cos 3θ D 0 which is the value of the function at the origin. So yes, f x, y D z is
continuous at the origin.
x2 C y2 3x2 − 3y2 − x3 − 3xy2 2x
x4 − 3y4 C 6x2 y2
(c) (i) fx D
D
.
x2 C y2 2
x2 C y2 2
x2 C y2 −6xy − x3 − 3xy2 2y
−8x3 y
D
.
x2 C y2 2
x2 C y2 2
f h, 0 − f 0, 0
h − 0
(iii) fx 0, 0 D limh'0
D lim
D 1.
h'0
h
h
f 0, h − f 0, 0
0 − 0
D lim
D 0.
(iv) fy 0, 0 D limh'0
h'0
h
h
(d) gr, θ D r cos 3θ so gr r, θ D cos 3θ. This is the directional derivative Du f .
−8x3 y
. In particular, when y D x, fy D −2. From part (c) fy 0, 0 D
(e) When x, y Z 0, 0, fy x, y D
x2 C y 2
0 so fy is not continuous at the origin.
(ii) fy D
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“runall” — 2005/8/1 — 12:14 — page 135 — #66
Section 2.8
Miscellaneous Exercises for Chapter 2
135
(f) Below are two sketches; the one on the left just shows a ribbon of the surface:
1
0.5
y
0
1
y
-0.5
0.5
0
-1
1
-0.5
-1
1
0.5
0.5
z
z
0
0
-0.5
-0.5
-1
-1
-1
-1
-0.5
-0.5
0
0
x
x
0.5
1
0.5
1
36. (a) u D cosx − t C sinx C t − 2ezCt − y − t3 so
• ux D − sinx − t C cosx C t and uxx D − cosx − t − sinx C t.
• uy D −3y − t2 and uyy D −6y − t.
• uz D −2ezCt and uzz D −2ezCt .
• ut D sinx − t C cosx C t − 2ezCt C 3y − t2 and utt D − cosx − t − sinx C t −
2ezCt − 6y − t.
We have, therefore, the result: uxx C uyy C uzz D utt .
(b) ux, y, z, t D f1 x − t C f2 x C t C g1 y − t C g2 y C t C h1 z − t C h2 z C t so
⭸x − t
⭸x C t
C f2 xCt
D f1 x−t C f2 xCt so
⭸x
⭸x
2
2
2
⭸ f1
⭸ f2
⭸ u
D
C
•
2
2
⭸x
⭸x − t
⭸x C t2
⭸y − t
⭸y C t
• uy D g1 y−t
C g2 yCt
D g1 y−t C g2 yCt so
⭸y
⭸y
⭸2 g1
⭸2 g2
⭸2 u
•
D
C
.
⭸y2
⭸y − t2
⭸y C t2
⭸z − t
⭸z C t
C h2 zCt
D h1 z−t C h2 zCt so
• uz D h1 z−t
⭸z
⭸z
⭸2 u
⭸2 h1
⭸2 h2
D
C
.
•
2
2
⭸z
⭸z − t
⭸z C t2
⭸x − t
⭸x C t
⭸y − t
⭸y C t
C f2 xCt
C g1 y−t
C g2 yCt
C
• ut D f1 x−t
⭸t
⭸t
⭸t
⭸t
⭸z − t
⭸z C t
h1 z−t
C h2 zCt
so ut D −f1 x−t C f2 xCt − g1 y−t C g2 yCt −
⭸t
⭸t
⭸z − t
⭸z C t
C h2 zCt
and
h1 z−t
⭸t
⭸t
• utt D uxx C uyy C uzz .
• ux D f1 x−t
37. F tx, ty D t 3 x3 C t 3 xy2 − 6t 3 y3 D t 3 F x, y so F is homogeneous of degree 3.
38. F tx, ty, tz D t 3 x3 y − t 4 x2 z2 C t 8 z8 so, no, F is not homogeneous.
39. F tx, ty, tz D t 3 zy2 − t 3 x3 C t 3 x2 z D t 3 F x, y, z so yes F is homogeneous of degree 3.
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“runall” — 2005/8/1 — 12:14 — page 136 — #67
136
Chapter 2
Differentiation in Several Variables
40. F tx, ty D ety/tx D ey/x D F x, y so F is homogeneous of degree 0.
t 3 x3 C t 3 x2 y − t 3 yz2
D F x, y, z so F is homogeneous of degree 0.
41. F tx, ty, tz D
t 3 xyz C 7t 3 xz2
42. Make sure that the students realize (as in Exercises 40 and 41) that a function can be homogeneous and not be a
polynomial. In the special case that F is a polynomial, F is homogeneous when all of the terms are of the same
degree.
43. F tx1 , tx2 , ... , txn D t d F x1 , x2 , ... , xn so that, by differentiating both sides with respect to t:
x1
⭸F
⭸F
tx1 , ... , txn C · · · C xn
tx1 , ... , txn D dt d−1 F x1 , ... , xn .
⭸x1
⭸xn
Now let t D 1 and we get the result:
x1
44. The conjecture is:
n
i1 ,...,ik D1
⭸F
⭸F
C · · · C xn
D dF .
⭸x1
⭸xn
D xi1 xi2 · · · xik Fxi1 xi2 ···xik D
d!
F.
d − k!
Although not asked in the text, a good exercise is to ask the students to establish the formula given in this exercise.
Show that
n
⭸
⭸2 F
⭸F
[dF ] D xj
C
.
⭸xi
⭸xi ⭸xj
⭸xi
jD1
Then you can show
n
d2 F D d xi
iD1
n
⭸F
⭸2 F
D xi xj
C dF .
⭸xi
⭸xi ⭸xj
i,jD1
You can finish from there.
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“runall” — 2005/8/1 — 15:09 — page 137 — #1
C H A P T E R
3
Vector-Valued Functions
3.1 PARAMETRIZED CURVES AND KEPLER’S LAWS
1. The graph is a line segment with slope −1/2 and y intercept 3:
y
4
3
2
1
x
-3
-2
-1
1
2. In this case y D 1/x and both x and y are positive:
y
4
3
2
1
1
2
x
4
3
3. This is the spiral r D θ (note x D r cos θ and y D r sin θ):
y
15
10
5
-15
-10
-5
5
10
15
x
-5
-10
-15
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“runall” — 2005/8/1 — 15:09 — page 138 — #2
138
Chapter 3
Vector-Valued Functions
4. This is a lemniscate beginning and ending at the point (3, 0):
y
3
2
1
-3
-2
-1
1
2
3
x
-1
-2
-3
5. Although this is a curve in R3 , because z K 0 the curve lives in the xy-plane. It is the parabola y D 3x2 C 1:
y
4
3
2
1
-1.5
-1
-0.5
0.5
1
x
1.5
6. It’s hard to see what this curve looks like in R3 (below left):
y
1
0.75
0.5
0.25
-1 -0.75 -0.5 -0.25
-0.25
0.25 0.5 0.75
1
x
-0.5
-0.75
-1
so I have also projected it onto the three coordinate planes:
z
z
1
1
0.75
0.75
0.5
0.5
0.25
0.25
x
-1
-0.75 -0.5 -0.25
-0.25
0.25
0.5
0.75
1
y
-1
-0.75 -0.5 -0.25
-0.25
-0.5
-0.5
-0.75
-0.75
-1
-1
0.25
0.5
0.75
1
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“runall” — 2005/8/1 — 15:09 — page 139 — #3
Section 3.1
Parametrized Curves and Kepler’s Laws
139
For Exercises 7–10, the velocity is the derivative of position, the speed is the length of the velocity vector and
the acceleration is the derivative of the velocity vector.
√
√
7. xt D 3t − 5, 2t C 7 so velocity D vt D x t D 3, 2 and speed D vt D 32 C 22 D 13.
Finally, acceleration D at D x t D 0, 0.
8. xt D 5 cos t, 3 sin t so velocity
D vt D x t D −5 sin t, 3 cos t and speed D vt D
−5 sin t2 C 3 cos t2 D 9 C 16 sin2 t. Acceleration D at D x t D −5 cos t, −3 sin t D −xt.
9. xt D tsin t, t cos t, t 2 so velocity D vt D x t D sin t C t cos t, cos t − t sin t, 2t and speed D
√
vt D sin2 t C 2t sin t cos t C t 2 cos2 t C cos2 t − 2t sin t cos t C t 2 sin2 t C 4t 2 D 1 C 5t 2 . Finally,
acceleration D at D x t D 2 cos t − t sin t, −2 sin t − t cos t, 2.
√
t 2t
t
t
2t
t
2t
4t
10. xt
√ D e , e , 2e so velocity D vt D x t D e , 2e , 2e and speed D vt D 5e C 4e D
t
t
2t
t
2t
e 5 C 4e . Finally, acceleration D at D x t D e , 4e , 2e .
11. (a)
(b) To verify that the curve lies on the surface check that
y2
9 cos2 πt
16 sin2 πt
x2
C
D
C
D cos2 πt C sin2 πt D 1.
9
16
9
16
The z component just determines the speed traveling up the cylinder.
12. (a)
(b) To verify that the curve lies on the surface check that
x2 C y2 D t 2 cos2 t C t 2 sin2 t D t 2 cos2 t C sin2 t D t 2 D z2 .
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140
Chapter 3
Vector-Valued Functions
13. (a)
(b) To verify that the curve lies on the surface check that
x2 C y2 D t 2 sin2 2t C t 2 cos2 2t D t 2 sin2 2t C cos2 2t D t 2 D z.
14. (a)
(b) To verify that the curve lies on the surface check that
x2 C y2 D 4 cos2 t C 4 sin2 t D 4cos2 t C sin2 t D 4.
In Exercises 15–18 use formulas (2) and (3) from the text. In each case we will need to calculate the position
and velocity at the given time.
15. xt D te−t , e3t so x0 D 0, 1 and x t D e−t − te−t , 3e3t so x 0 D 1, 3. The equation of the
tangent line at t D 0 is lt D 0, 1 C 1, 3t D t, 1 C 3t.
√
16. xt D 4 cos t, −3
sin t, 5t so xπ/3 D 2, −3 3/2, 5π/3 and x t D −4 sin t, −3 cos t, 5 so
√
x π/3 D −2 3, −3/2, 5. The equation of the tangent line at t D π/3 is
√
√
lt D 2, −3 3/2, 5π/3 C −2 3, −3/2, 5t − π/3.
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“runall” — 2005/8/1 — 15:09 — page 141 — #5
Section 3.1
Parametrized Curves and Kepler’s Laws
141
17. xt D t 2 , t 3 , t 5 so x2 D 4, 8, 32 and x t D 2t, 3t 2 , 5t 4 so x 2 D 4, 12, 80. The equation of the
tangent line at t D 2 is
lt D 4, 8, 32 C 4, 12, 80t − 2 D 4t − 4, 12t − 16, 80t − 128.
18. xt D coset , 3 − t 2 , t so x1 D cos e, 2, 1 and x t D −et sinet , −2t, 1. Therefore, x 1 D
−e sin e, −2, 1. The equation of the tangent line at t D 1 is
lt D cos e, 2, 1 C −e sin e, −2, 1t − 1 D cos e C e sin e − e sin et, 4 − 2t, t.
19. (a) The sketch of xt D t, t 3 − 2t C 1 is:
y
20
15
10
5
x
-3
-2
-1
1
2
3
(b) x2 D 2, 5 and since x t D 1, 3t 2 − 2 we get x 2 D 1, 10. The equation of the line is then
lt D 2, 5 C 1, 10t − 2 D t, 10t − 15.
y
20
15
10
5
-3
-2
-1
1
2
3
x
(c) Since x D t we see that y D f x D x3 − 2x C 1.
(d) So the equation of the tangent line at x D 2 is y − f 2 D f 2x − 2, where f is as in part (c).
Substituting, we get y − 5 D 10x − 2 or y D 10x − 15. This is consistent with our answer for part (b).
20. From the first equation t D x/v0 cos θ. Substitute this into the second equation to get
yD
v0 sin θx
g
1
x2
D tan θx −
x2 .
− g
2
v0 cos θ
2 v0 cos θ
2v0 cos θ2
This is of the form y D ax2 C bx and the graph is a parabola.
21. We know from the text that Roger is on the ground at t D 0 and t D 2v0 sin θ/g. By symmetry,√Roger is at
his maximum height at t D v0 sin θ/g. For this exercise this is at time t D 100 sin 60◦ /32 D 25 3/16. The
maximum height is found by substituting into the equation for y:
y D v0 sin θ √
√
√ 2
√
6253
6253
25 3
25 3
1
25 3
D
− 32 D 50 3 −
16
2
16
16
16
16
Roger’s maximum height is 117.1875 feet.
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“runall” — 2005/8/1 — 15:09 — page 142 — #6
142
Chapter 3
Vector-Valued Functions
22. By forumula (5) from the text, x D v20 sin 2θ/g. In this case we can say that 2640 D v20 sin 120◦ /32. Solve this for
v0 D
√
264032
√
D 32 55 3 L 312.329.
3/2
23. We use the same formula as in Exercise 22 but now solve for θ. So, x D v20 sin 2θ/g becomes 1500 D
2502 sin 2θ/32 or
150032
96
sin 2θ D
D
.
62500
125
So
θ D 1/2 sin−1 96/125 L .875712 L 50.176◦ .
24. This is similar to Example 6 from the text. We have the equation:
xt D −1/2gt 2 j C tv0 C x0 j.
√
√
(a) Here the angle is given as 45◦ and the initial speed of the water is 7 m/s, therefore, v0 D 7 2/2, 2/2.
Also x0 is the initial height of 1 m and gravity is about −9.8 m/s2 . This means that
√
√
√
√
7 2
7 2
xt D −4.9t 2 j C 7 2/2, 2/2t C j D t, −4.9t 2 C
t C 1 .
2
2
√
√
We want to know the height when the x distance is 5 so first solve 7 2 t D 5 for t to get t D 10/7 2.
2
Substitute this into our vertical equation to find that the height would be 1 so the answer is yes, Egbert
gets wet.
(b) Here the idea is the same
√ as in part (a). The initial speed of the water is 8 m/s and we don’t know the
direction so v0 D 8u, 1 − u2 for some u between 0 and 1. So
xt D −4.9t 2 j C 8u,
√
1 − u2 t C j D 8ut, −4.9t 2 C
√
1 − u2 t C 1.
We want the height when
the horizontal distance is 5 or when t D 5/8u. In that case, the height is
√
−4.95/8u2 C 8 1 − u2 5/8u C 1. For what values of u is this between 0 and 1.6? Consider
the figure:
height
2
1.5
1
0.5
0.4
0.5
0.6
0.7
0.8
u
0.9
-0.5
Explore with Mathematica or a graphing calculator and you will find the u values in the two intervals which
correspond to the correct heights. This gives the two approximate ranges for α as between 11.2◦ and 34.2◦ and
as between 62.6◦ and 67.5◦ .
25. We have x2 D e4 , 8, 3 and x 2 D 2e4 , 10, 5 . If the rocket’s engines cease when t D 2, then the rocket
2
4
will follow the tangent line path
lt D x2 C t − 2x 2 D e 4 2t − 3, 10t − 12, 5 t − 1 .
4
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“runall” — 2005/8/1 — 15:09 — page 143 — #7
Section 3.1
Parametrized Curves and Kepler’s Laws
143
For this path to reach the space station, we must have
5
4
4
e 2t − 3, 10t − 12, t − 1 D 7e , 35, 5.
4
Thus, in particular
e 4 2t − 3 D 7e4 3 2t − 3 D 7 3 t D 5.
However l5 D 7e4 , 38, 21 Z 7e4 , 35, 5. Hence the rocket does not reach the repair station.
4
26. (a) We set xt D yt and solve for t:
2
t − 2,
t2
− 1 D t, 5 − t 2 .
2
Comparing first components, we have t 2 − 2 D t 3 t 2 − t − 2 D 0 3 t D −1, 2. Now x−1 D
1
−1, − 2 and y−1 D −1, 4, so this is not a collision point. However, x2 D 2, 1 D y2. So the
balls collide when t D 2 at the point (2, 1).
(b) We have x 2 D 4, 2, y 2 D 1, −4. The angle between the paths is the angle between these tangent
vectors, which is
cos−1
x 2 · y 2
−4
−2
D cos−1 √ √
D cos−1 √ √ .
x 2y 2
20 17
5 17
27. The calculation is fairly straightforward:
d
d
x · y D x1 ty1 t C x2 ty2 t C · · · C xn tyn t
dt
dt
D x1 ty1 t C x1 ty1 t C x2 ty2 t C x2 ty2 t C · · · C xn tyn t C xn tyn t
D [x1 ty1 t C x2 ty2 t C · · · C xn tyn t] C [x1 ty1 t C x2 ty2 t C · · · C xn tyn t]
Dy ·
dy
dx
C x ·
.
dt
dt
28. This is similar to Exercise 27:
d
d
x * y D [x2 y3 − x3 y2 i − x1 y3 − x3 y1 j C x1 y2 − x2 y1 k]
dt
dt
D x2 y3 − x3 y2 C x2 y3 − x3 y2 i − x1 y3 − x3 y1 C x1 y3 − x3 y1 j
C x1 y2 − x2 y1 C x1 y2 − x2 y1 k
D [x2 y3 − x3 y2 ] C [x2 y3 − x3 y2 ]i − [x1 y3 − x3 y1 ] C [x1 y3 − x3 y1 ]j
C [x1 y2 − x2 y1 ] C [x1 y2 − x2 y1 ]k
D
dy
dx
* y C x *
.
dt
dt
29. You’re asked to show that if xt is constant, then x is perpendicular to dx/dt. If xt is constant, then
d
xt K 0. So
dt
0D
1
d
d√
dx
x · xD √
xt D
· x .
2
dt
dt
2 x · x
dt
This means that dx · x D 0.
dt
30. (a) xt2 D cos2 t C cos2 t sin2 t C sin4 t D cos2 t C sin2 tcos2 t C sin2 t D 1.
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“runall” — 2005/8/1 — 15:09 — page 144 — #8
144
Chapter 3
Vector-Valued Functions
(b) This follows from Proposition 1.7 since xt K 1. The exercise really wants you to calculate the velocity
vector: v D − sin t, − sin2 t C cos2 t, 2 sin t cos t. Then v · x D 0.
(c) If xt is a path on the unit sphere, xt K 1 so by Proposition 1.7 the position vector is perpendicular to
its velocity vector.
31. (a) Computer graphs are shown for (i), (ii), (iii). The constant a affects the size (radius) of the rings; the constant
b affects the size (radius) of the coils; the constant ω affects the number of coils going around the ring.
(b) If x D a C b cos ωt cos t, y D a C b cos ωt sin t, then x2 C y2 D a C b cos ωt2 , so that x2 C y2 −
a2 D a C b cos ωt − a2 D b2 cos2 ωt. (Note a > b > 0.) Hence
x2 C y2 − a2 C z2 D b2 cos2 ωt C b2 sin2 ωt D b2 .
4
1
z
0
2
-1
-4
0
y
-2
-2
0
x
2
4
5
z 1
0
-1
0
-5
y
-2.5
0
2.5
x
-5
5
5
z 1
0
-1
0
-5
y
-2.5
0
x
2.5
-5
5
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“runall” — 2005/8/1 — 15:09 — page 145 — #9
Section 3.2
Arclength and Differential Geometry
145
32. The distance between a point on the image and the origin is xt and this is minimized when t D t0 . Thus
the function
f t D xt2 D xt · xt
is also minimized when t D t0 . Hence
0 D f t0 D
d
xt · xt
D xt0 · x t0 C x t0 · xt0 dt
tDt0
D 2xt0 · x t0 .
Thus xt0 and x t0 are orthogonal.
3.2 ARCLENGTH AND DIFFERENTIAL GEOMETRY
In Exercises 1–6 we are using Definition 2.1 to calculate the length of the given paths.
1. xt D 2t C 1, 7 − 3t so x t D 2, −3. The length of the path is then
Lx D
2
−1
x dt D
2 −1
22 C −32 dt D
2 √
−1
13 dt D
√
√
2
13t −1 D 3 13.
2. xt D t 2 , 2/32t C 13/2 so x t D 2t, 22t C 11/2 . The length of the path is then
4
Lx D
4
2t2 C 42t C 1 dt D
0
0
4
D2
0
4
4t 2 C 8t C 4 dt D 2
|t C 1| dt
0
4
t C 1 dt D 2t 2 /2 C t 0 D 24.
3. xt D cos 3t, sin 3t, 2t 3/2 so x t D −3 sin 3t, 3 cos 3t, 3t 1/2 . The length of the path is then
2
Lx D
2√
9 sin2 3t C 9 cos2 3t C 9t dt D 3
0
0
4. xt D 7, t, t 2 so x t D 0, 1, 2t. The length of the path is then
3√
Lx D
1
3
1 C 4t 2 dt D 2
1
√
3
u du D 2u3/2 1 D 6 3 − 2.
3
1/4 C t 2 dt D t 1/4 C t 2 C 1/4 lnt C 1/4 C t 2 1
1
3√
1 C t dt D 3
1
5
C ln3 C 37/4 − ln1 C 5/4
4
4


√
√
√
6 C 37 
3 37 − 5
1
√ L 8.2681459.
D
C ln 2
4
2 C 5
D3
5. xt D ln t, t 2 /2,
37
−
4
√
2t so x t D 1/t, t,
4
Lx D
√
2. The length of the path is then
4
1/t 2 C t 2 C 2 dt D
1
4
1/t C t2 dt D
1
1/t C t dt
1
4
D [ln t C t 2 /2] 1 D ln 4 C 8 − 1/2 D ln 4 C 15 .
2
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“runall” — 2005/8/1 — 15:09 — page 146 — #10
146
Chapter 3
Vector-Valued Functions
√
√
6. xt D 2t cos t, 2t sin t, 2 2t 2 so x t D 2 cos t − 2t sin t, 2 sin t C 2t cos t, 4 2t. The length of the path
is then
3
Lx D
4 cos2 t − 8t cos t sin t C 4t 2 sin2 t C 4 sin2 t C 8t sin t cos t C 4t 2 cos2 t C 32t 2 dt
0
3
D
3
4 C 4t 2 C 32t 2 dt D
0
4 C 36t 2 dt
0
√
3
D [t 1 C 9t 2 C sinh−1 3t/3] 0 D 3 82 C sinh−1 9/3.
7. A sketch of the curve xt D a cos3 t, a sin3 t for 0 … t … 2π is:
1
0.5
-1
-0.5
0.5
1
-0.5
-1
Because of the obvious symmetries we will compute the length of the portion of the curve in the first quadrant
and multiply it by 4:
π/2 π/2
−3a cos2 t sin t, 3a sin2 t cos t dt D 4
Lx D 4
0
9a2 cos4 t sin2 t C sin4 t cos2 t dt
0
π/2 π/2
9a2 sin2 t cos2 tcos2 t C sin2 t dt D 4
D4
0
0
π/2
3a sin t cos t dt D 6a sin2 t 0
D 6a.
8. If f is a continuously differentiable function then we can calculate the length of the curve y D f x between
(a, f a) and (b, f b) by viewing the curve as the path yx D x, f x so y x D 1, f x, and so by
Definition 2.1 the length is
b
Ly D
1 C [f x]2 dx.
a
9. Here f x D mx C b and f x D m so by Exercise 8, the length of the curve is
LD
x1 √
√
1 C m2 dt D x1 − x0 1 C m2 .
x0
A quick look at a sketch shows why this should be the case:
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“runall” — 2005/8/1 — 15:09 — page 147 — #11
Section 3.2
Arclength and Differential Geometry
147
mx1-mx0
x1-x0
If x1 > x0 then the horizontal distance is |x1 − x0 | D x1 − x0 and the vertical distance is |mx1 − mx0 | D
|m|x1 − x0 . By the Pythagorean theorem, the length of the hypotenuse is
√
x1 − x0 2 C |m|x1 − x0 2 D x1 − x0 2 m2 C 1 D x1 − x0 m2 C 1.
10. (a) xt D a1 t C b, a2 t C b so x t D a1 , a2 so
Lx D
t1 t0
a21 C a22 dt D t1 − t0 a21 C a22 .
(b) The equation of the line in Exercise√9 could be given as xt D t, mt C b in which case part (a) would
tell us that the length is x1 − x0 1 C m2 .
(c) xt D at C b so x t D a. Then
Lx D
t1
a dt D at1 − t0 .
t0
This, of course, is the same as our answer in part (a).
11. (a) A sketch of x D |t − 1|i C |t|j, −2 … t … 2 is:
2
1.5
1
0.5
0.5
1
1.5
2
2.5
3
(b) Except for two points, the path is smooth (more than C1 ). In fact, the path is comprised of three line segments
joined end to end. In other words, on the open intervals −2 … t < 0, 0 < t < 1, and 1 < t … 2, the path
x is C1 . We say that the path x is piecewise C1 .
(c) We could figure out the length of each piece and add them together. In the process we will find that we’re
working too hard.



 1 − ti − tj
1 − ti C tj
xt D


 t − 1i C tj
−2 … t … 0
0 < t … 1
1 < t … 2
so



 −1, −1
−1, 1
x t D


 1, 1
−2 … t … 0
0 < t … 1 .
1 < t … 2
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“runall” — 2005/8/1 — 15:09 — page 148 — #12
148
Chapter 3
Vector-Valued Functions
√
So we see that x t K 2. This means that to calculate the length of the curve we don’t have to break
up the integral into three pieces:
2 √
√
Lx D
2 dt D 4 2.
−2
t
12. (a) We’ll use the equation: st D 0 x τ dτ. For xτ D eaτ cos bτi C eaτ sin bτj C eaτ k the derivative is
x τ D aeaτ cos bτ, sin bτ,√
1 C eaτ −b sin bτ, b cos bτ, 0. Therefore the speed is given by x τ D
aτ
2
2aτ
2
2aτ
a e 2 C b e D e
2a2 C b2 . This means that
t
st D
√
e aτ 2a2 C b2 dτ D
0
√
t
2a2 C b2 aτ
e
D
a
0
√
2a2 C b2 at
e − 1.
a
(b) Just solve the above for t:
t D ln √
as
C 1 a.
2a2 C b2
For Problems 13–16, we’ll use
TD
Also
ds
D x t,
dt
x t
,
x t
ND
dT/ dt
,
dT/ dt
dT dT/ dt
D
κt D
ds ,
ds/ dt
and B D T * N.
and
dB/ dt
dB
D
D −τN.
ds
ds/ dt
You may want to ask your students to make a guess about τ before they do Exercises 14 and 16. The curves are
planar—what might that suggest about τ? Also see Section 3.6, Exercise 28.
√
√
13. xt D 5 cos 3t, 6t, 5 sin 3t so x t D −15 sin 3t, 6, 15 cos 3t and x t D 225 C 36 D 261.
√
T D 1/ 261−15 sin 3t, 6, 15 cos 3t
√
D 1/ 29−5 sin 3t, 2, 5 cos 3t.
√
dT/dt D 1/ 29−15 cos 3t, 0, −15 sin 3t so
N D − cos 3t, 0, − sin 3t,
√
B D T * N D 1/ 29−2 sin 3t, −5, 2 cos 3t, and
√
−5 cos 3t, 0, −5 sin 3t
5
1/ 29−15 cos 3t, 0, −15 sin 3t √
κD
D
.
D
261
29
29
Finally,
√
−2 cos 3t, 0, −2 cos 3t
1/ 29−6 cos 3t, 0, −6 sin 3t
√
D
τN D
261
29
2
τD− .
29
so
14. xt D sin t − t cos t, cos t C t sin t, 2 with t Ú 0. So x t D t sin t, t cos t, 0 and x t D |t| D t.
t sin t, t cos t, 0
D sin t, cos t, 0, and
t
N D cos t, − sin t, 0, B D 0, 0, −1, and
TD
κD
cos t, − sin t, 0
1
D .
t
t
Finally, dB/dt D 0 so τ D 0.
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“runall” — 2005/8/1 — 15:09 — page 149 — #13
Section 3.2
Arclength and Differential Geometry
149
15. xt D t,1/3t C 13/2 , 1/31 − t3/2 so x t D 1, 1/2t C 11/2 , −1/21 − t1/2 , and
x t D 3/2.
TD
ND
√
√
2
1
1
1, 2 t C 1, − 2 1 − t , and
3
2/30, 1/4t C 1−1/2 , 1/41 − t−1/2 2/31/16 1 C 1 1−t
tC1
√
√
1
D √ 0, 1 − t, t C 1, and
2
√
√
√
√
1 1
1
1
BD
2 t C 1 C 2 1 − t, − t C 1, 1 − t D √ 1, − t C 1, 1 − t.
3
3
Also,
2/30, 1/4t C 1−1/2 , 1/41 − t−1/2 1
.
D κD
3/2
3 21 − t 2 Finally,
dB
1
1
1
D √ 0, − √
,− √
so
dt
3
2 t C 1
2 1 − t
1
1
1
dB
D √ 0, − √
,− √
3/2 D −τN.
ds
3
2 t C 1
2 1 − t
Solving,
1
.
τD 3 1 − t 2 16. xt D e2t sin t, e2t cos t, 1 so x t D e2t 2 sin t C cos t, 2 cos t − sin t, 0, and x t D e2t
TD
2 sin t C cos t, 2 cos t − sin t, 0
√
,
5
ND
2 cos t − sin t, −2 sin t − cos t, 0
√
, and
5
√
5.
B D 0, 0, −1.
Also,
κD
2 cos t − sin t, −2 sin t − cos t, 0
1
√
√ .
D
2t
2t
e
5
e
5
Finally, again we see that dB/dt D 0 so dB/ds D 0 and hence τ D 0.
17. (a) By formula (17): κ D x * x3 . Let y D f x and view the problem as sitting inside of R3 . Then x D
x x, f x, 0, x D 1, f x, 0, and x D 0, f x, 0. We calculate the cross product x * x D
0, 0, f x so
|f x|
0, 0, f x
D
.
κD
3
1, f x, 0
[1 C f x2 ]3/2
(b) If y D lnsin x, then y D cos x/ sin x and y D −1/ sin2 x. By our results for part (a),
κD
| − 1/ sin2 x|
D | sin x|.
[1 C cos2 x/ sin2 x]3/2
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“runall” — 2005/8/1 — 15:09 — page 150 — #14
150
Chapter 3
Vector-Valued Functions
18. (a) Formula (17) requires the use of the cross product, so we view this problem as sitting inside of R3 . Let
x D xs, ys, 0. Then x D x s, y s, 0, and x D x s, y s, 0. By formula (17):
κD
0, 0, x y − x y x * x D
.
x 3
x s, y s, 03
But the curve is parametrized by arclength so x s, y s, 0√
D 1 so κ D |x y − x y |.
2
−1
(b) √
Here xs D 1/21 − s and ys D 1/2cos s − s 1 − s2 so x s D −s and y s D
1 − s2 so x s2 C y s2 D 1. So the curve is parametrized by arclength. We can then compute
its curvature using the formula from part (a):
κ D |x y − x y | D −s √
19. (a) The curvature is calculated to be
√
−s
1
.
− −1 1 − s2 D √
1 − s2
1 − s2
2
.
cos2 t C 4 sin2 t3/2
(b) The path is pictured below left while the corresponding curvature is plotted below right.
␬
y
-2
1
2
0.5
1.5
-1
1
2
1
x
0.5
-0.5
1
-1
2
3
4
5
t
6
20. (a) The curvature is calculated to be (with some simplification)
31 C cos t
.
16 cos3 t/2
(b) The path is pictured below left while the corresponding curvature is plotted below right.
y
2
1
␬
1
2
3
4
x
25
20
-1
15
10
-2
5
1
2
3
4
5
6
t
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“runall” — 2005/8/1 — 15:09 — page 151 — #15
Section 3.2
Arclength and Differential Geometry
151
21. (a) The curvature is the same as in Exercise 20.
(b) The path is pictured below left while the corresponding curvature is plotted below right.
y
2
1
25
-1
1
2
3
␬
x
20
15
-1
10
5
-2
1
22. (a) The curvature is calculated to be
2
5
4
3
t
6
√
2|7 sin t C sin 7t|
.
35 C cos 6t C 4 cos 8t3/2
(b) The path is pictured below left while the corresponding curvature is plotted below right.
y
␬
2
3
1
-3
-2
-1
1
-1
-2
2
3
x
2
1
1
2
3
4
5
6
t
For Exercises 23–28, calculate the tangential component s̈ and then subtract it from the length of the acceleration
to obtain the normal component.
√
2
23. xt D t 2 , t so x t D 2t, 1 and x t D
√ 2, 0. The speed is then x t D 1 C 4t and so the
2
2
2
tangential component of acceleration is √
s̈ D 4t/ 1 C 4t . Since a D 2, a − s̈ D 4/1 C 4t 2 , so the
normal component of acceleration is 2/ 1 C 4t 2 .
√
2t
4t
24. xt D 2t, e2t so x t D 2, 2e2t and x t D
√0, 4e . The speed is then x t D 2 1 C e and so the
2
4t
4t
2
4t
4t
tangential component of acceleration is s̈ D 4e /√ 1 C e . Since a D 16e , a − s̈ D 16e /1 C e4t ,
so the normal component of acceleration is 4e2t / 1 C e4t .
25. xt D et cos 2t, et sin 2t so x t D et cos 2t − 2 sin 2t, et sin√2t C 2 cos 2t and x t D et −3 cos 2t −
4 sin 2t, et 4 cos√2t − 3 sin 2t. The speed is then x t D et 5 and so the tangential component of acceleration
is s̈ D et 5. Since a D 5et , a2 − s̈2 D 25e2t − 5e2t , so the normal component of acceleration is
√
2 5et .
26. xt D 4 cos 5t, 5 sin 4t, 3t so x t D −20
sin 5t, 20 cos 4t, 3 and we also have that x t D −100 cos 5t,
−80 sin 4t, 0. The speed is then x t D 400 sin2 5t C 400 cos2 4t C 9 and so the tangential component of
acceleration is
−3200 cos 4t sin 4t C 4000 cos 5t sin 5t
s̈ D
.
400 sin2 5t C 400 cos2 4t C 9
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“runall” — 2005/8/1 — 15:09 — page 152 — #16
152
Chapter 3
Vector-Valued Functions
Since a D 20 25 cos2 5t C 16 sin2 4t,
a2 − s̈2 D 10000 cos2 5t C 6400 sin2 4t −
3200 cos 4t sin 4t C 4000 cos 5t sin 5t2
,
4400 sin2 5t C 400 cos2 4t C 9
so the normal component of acceleration is the square root of this last quantity.
√
27. xt D t, t, t 2 so x t D 1, 1, 2t and x t D√0, 0, 2. The speed is then x t D 2 C 4t 2 and so
the tangential component of acceleration is √
s̈ D 4t/ 2 C 4t 2 . Since a D 2, a2 − s̈2 D 4/1 C 2t 2 , so
the normal component of acceleration is 2/ 1 C 2t 2 .
28. xt D 3/51 − cos t, sin t, 4/5 cos t so x t D 3/5 sin t, cos t, −4/5 sin t and x t D
3/5 cos t, − sin t, −4/5 cos t. The speed is then x t D 1 and so the tangential component of acceleration is s̈ D 0. Since a D 1, a2 − s̈2 D 1, so the normal component of acceleration is 1.
29. (a) Tangential component:
s̈ D
√
dx d x · x
1
x · x
dṡ
D
D
D √
.
2x · x D
dt
dt
dt
2 x · x
x Normal component (using formula (17)):
κṡ2 D v * a
v * a
x * x 2
D
.
v D
3
v
v
x component
(b) xt D t C 2, t 2 , 3t
√
√ so x t D 1, 2t, 3 and x t D 0, 2, 0. So by part (a),√the tangential
2
of acceleration is 4t/ 10 C 4t , and the normal component of acceleration is 2 10/ 10 C 4t 2 .
30. Here x D x, f x, 0, x D 1, f x, 0, and x D 0, f x, 0. Further, you need to calculate x D
1 C [f x]2 , x · x D f xf x, and x * x D 0, 0, f x D |f x|. Substituting into the
formulas from Exercise 29 gives us:
atang D f xf x
1 C [f x]2
,
and anorm D |f x|
1 C [f x]2
.
31. To establish the formula, first note that v * a D κṡ3 B (see, for example, the calculation leading up to formula
(17)) and v * a D κṡ3 D κv3 . So
v * a · a
κṡ3 B · a
B · a
D
D
.
6
2
2
v * a
κ ṡ
κṡ3
Now, at D s̈T C κṡ2 N and by the Frenet equations N s D −κT C τB. Since we are calculating the dot
product of a with B, the only piece that will survive is the coefficient of B, so
a t D something without B C κṡ2 N s
ds
dt
D something else without B C κṡ3 τB
and so, putting it all together,
v * a · a
κṡ3 B · a
B · a
B · κṡ3 τB
D
D
D
D τ.
6
3
2
2
v * a
κ ṡ
κṡ
κṡ3
32. By equations (11) and (13) we have T D κN and B D −τN
Hence
−T · B D −κN · −τN D κτ N · N D κτ,
since N is a unit vector.
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“runall” — 2005/8/1 — 15:09 — page 153 — #17
Section 3.2
33. From formula (17) κ D
By Exercise 31,
Arclength and Differential Geometry
153
v * a
D x * x since x must be a unit speed path as it is parametrized by arclength.
v3
τD
v * a · a
x * x · x
D
2
v * a
x * x 2
Thus
κ2 τ D x * x 2 · x * x · x
D x * x · x .
x * x 2
34. (a) Really, there’s nothing much to show in this part—but it really helps you solve part (b). B is T * N so it
is perpendicular to the plane determined by them. In this case, we interpret that as B is perpendicular to the
osculating plane. Make the analagous observations for the other two cases.
(b) Example 9 gives us the formulas for T, N, and B. Using the result of part (a) we can use the perpendicular
vector to write down the equation of the plane. First, B is perpendicular to the osculating plane. So at t D t0
the osculating plane must be of the form b sin t0 x − a cos t0 − b cos t0 y − a sin t0 C az − bt0 D 0.
Similarly the rectifying plane can be obtained from N as − cos t0 x − a cos t0 − sin t0 y − a sin t0 D 0.
Finally, the normal plane is obtained from T as −a sin t0 x − a cos t0 a cos t0 y − a sin t0 C bz − bt0 D
0.
35. We have x − x0 2 D x − x0 · x − x0 D a2 . Thus x − x0 D a, so xt lies on a sphere of radius a.
36. The normal plane to x at any point xt is the plane passing through xt and perpendicular to Tt. Thus the
plane has equation x − xt · Tt D 0 (Here xt and Tt are used as “constant” vectors.) Thus, using the
product rule,
d
xt − x0 · xt − x0 D 2xt − x0 · x t.
dt
Hence
0 D x0 − xt · T D −xt − x0 ·
x t
1
D−
xt − x0 · x t
x t
x t
Thus xt − x0 · x t D 0 for all t. Hence xt − x0 · xt − x0 D constant, which implies that we
have a sphere curve.
−2 sin 2t, −2 cos 2t, −2 sin t
37. We have Tt D
.
4 C 4 sin2 t
Now we check that xt − 1, 0, 0 · Tt D 0. This equation is
cos 2t − 1, − sin 2t, 2 cos t ·
D 1
4 C 4 sin2 t
−2 sin 2t, −2 cos 2t, −2 sin t
4 C 4 sin2 t
−2 cos 2t sin 2t C 2 sin 2t C 2 sin 2t cos 2t C 4 cos t sin t
D 0.
38. By Exercise 27 of §1.4: a * b * c D a · cb − b · ca, so
T * B D T * T * N D −T * N * T D −[T · TN − N · TT] D −N
N * B D N * T * N D −T * N * N D −[T · NN − N · NT] D T
39.
w2 D w · w D τT C κB · τT C κB D τ2 T · T C κτB · T C κτT · B C κ2 B · B D τ2 C κ2
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“runall” — 2005/8/1 — 15:09 — page 154 — #18
154
Chapter 3
40. (a)
Vector-Valued Functions
w * T D τT C κB * T D τT * T C κB * T
D κB * T D κN by Exercise 38
D T by Frenet–Serret
w * N D τT C κB * N D τT * N C κB * N
D τB − κT by Exercise 38
D N by Frenet–Serret
w * B D τT C κB * B D τT * B D −τN by Exercise 38
D B by Frenet–Serret
(b) T D w * T D τT C κB * T D κN by manipulations and Exercise 38. The other equations are similar.
41. w is a constant vector 3 w s D 0. So
0 D w s D τ T C τT C κ B C κB
D τ T C κ B C τκN − κτN using Frenet–Serret
D τ T C κ B.
T and B are always perpendicular—hence we can never have T D c B (or vice versa). Thus τ D κ D 0 so
τ, κ are constant and nonzero because x * x Z 0. Thus by Theorem 2.5 the path must be a helix. Conversely,
having a helix implies constant τ, κ so w K 0. Thus w must be constant.
3.3 VECTOR FIELDS: AN INTRODUCTION
The figures can be generated using Mathematica or Maple. The axes are in the ‘usual’ positions with the origin at the
center. The relative length of the shaft of the arrows corresponds to the length of the vectors. The students should then
compare the results in Exercises 1–3 and Exercises 4–6. The differences between the equations for the vector fields should
be compared to the differences in the resulting sketches.
1. F D yi − xj D y, −x is shown below left.
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“runall” — 2005/8/1 — 15:09 — page 155 — #19
Section 3.3
Vector Fields: An Introduction
155
2. F D xi − yj D x, −y is shown above right.
3. F D −x, y is shown below left.
4. F D x, x2 is shown above right.
5. F D x2 , x is shown below left.
6. F D y2 , y is shown above right.
Now we are looking at sketches of vector fields in R3 . These are harder to see. In most cases, I have also included
a sketch of a slice.
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“runall” — 2005/8/1 — 15:09 — page 156 — #20
156
Chapter 3
Vector-Valued Functions
7. F D 3i C 2j C k D 3, 2, 1 is constant. The figure on the right shows the slice in the xy-plane:
1
0.5
y
0
-0.5
-1
1
0.5
z
0
-0.5
-1
-1
-0.5
0
0.5
x
1
8. F D y, −x, 0. The figure on the right shows the slice in the xy-plane—compare this to Exercise 1:
1
y
0
-1
1
0.5
z
0
-0.5
-1
-1
0
x
1
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“runall” — 2005/8/1 — 15:09 — page 157 — #21
Section 3.3
Vector Fields: An Introduction
157
9. F D 0, z, −y; compare this to Exercise 8:
1
y
0
-1
1
z
0
-1
-1
-0.5
0
x
0.5
1
10. F D y, −x, 2. The figure on the right shows the slice in the xy-plane—compare this to Exercise 8:
1
y
0.5
0
-0.5
-1
1
0.5
z
0
-0.5
-1
-1
-0.5
0
x
0.5
1
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“runall” — 2005/8/1 — 15:09 — page 158 — #22
158
Chapter 3
Vector-Valued Functions
11. F D y, −x, z. The figure on the right shows the slices in the z D 1 and z D −1 planes—compare this to
Exercises 8 and 10:
1
y
0
-1
1
z
0
-1
-1
0
x
1
12. F D y, −x, z/ x2 C y2 C z2 except at the origin. The figure on the right shows the slices in the z D 1 and
z D −1 planes—compare this to Exercise 11 (they are the same except the vectors in this problem are all unit
vectors—they may not look like unit vectors because of the vertical components):
1
y
0
-1
1
z
0
-1
-1
0
x
1
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“runall” — 2005/8/1 — 15:09 — page 159 — #23
Section 3.3
Vector Fields: An Introduction
159
13. The figure is below left.
14. The figure is above right.
15. The figure is below left.
16. The figure is above right.
In Exercises 17–19 we will show that x is a flow line of F using Definition 3.2, by showing x t D Fxt.
17. xt D x, y, z D sin t, cos t, 0 so x t D cos t, − sin t, 0 D y, −x, 0 D Fxt. We can see below
how the path, in bold, is a flow line for the vector field we saw above in Exercise 8. The figure on the right is
the xy-plane slice of the figure on the left.
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“runall” — 2005/8/1 — 15:09 — page 160 — #24
160
Chapter 3
Vector-Valued Functions
1
0.5
z
0
-0.5
1
-1
0
y
-1
0
x
-1
1
18. xt D x, y, z D sin t, cos t, 2t so x t D cos t, − sin t, 2 D y, −x, 2 D Fxt. Below we see the
view from almost directly above one “period” of the path.
The path, below in bold, is a flow line of the vector field we saw above in Exercise 10.
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“runall” — 2005/8/1 — 15:09 — page 161 — #25
Section 3.3
Vector Fields: An Introduction
161
x
-1
y
0
1
1
0
-1
5
z
0
-5
19. xt D x, y, z D sin t, cos t, e2t so
x t D cos t, − sin t, 2e2t D y, −x, 2z D Fxt.
The projection of this path onto the xy-plane is the same as that of the path in Exercise 18. The difference is that
the rate at which the path climbs is changing:
z
1
0.75
0.5
0.25
0
1
0
-1
0
x
y
-1
1
20. If xt D x, y then x t D Fx, y D −x, y. Consider x for a moment. This says that dx/dt D −x. The
solution to this is x D ce−t . Our initial condition is that x0 D 2 so c D 2. Similarly, dy/dt D y so y D ket .
The initial condition y0 D 1 tells us that k D 1. The equation of the flow line is xt D 2e−t , et .
21. If xt D x, y then x t D Fx, y D x2 , y. As in Exercise 20, we know that y D ket and y1 D e tells
us that k D 1. As for x, dx/dt D x2 . This is a separable differential equation dx/x2 D dt. Integrating and solving
for x gives us x D −1/t C c. From the initial condition x1 D 1 we find that 1 D −1/1 C c or c D −2.
The equation, therefore, of the flow line is xt D 1/2 − t, et .
22. If xt D x, y, z then x t D Fx, y, z D 2, −3y, z3 . We see immediately that the x coordinate function
must be linear and of the form 2t C c. From the initial condition, this constant is 3 so x D 2t C 3. As in
Exercise 20, we know that y D ke−3t and y0 D 5 tells us that k D 5. As for z, dz/dt D z3 . This is a separable
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“runall” — 2005/8/1 — 15:09 — page 162 — #26
162
Chapter 3
Vector-Valued Functions
differential equation dz/z3 D dt. Integrating and solving for z gives us z D 1/ −2t C c. From the initial
condition z0 D 7 we find that 7 D 1/ −2c or c D −1/98. The equation, therefore, of the flow line is
√
xt D 2t C 3, 5e−3t , 7/ 1 − 98t.
23. (a) For the function f x, y, z D 3x − 2y C z, §f D F so F is a gradient field.
(b) The equipotiential surfaces are those for which f (x, y, z) is constant. 3x − 2y C z D c. These are planes
with normal vector 3, −2, 1.
24. (a) For the function f x, y, z D x2 C y2 − 3z, §f D F so F is a gradient field.
(b) The equipotiential surfaces are those for which f (x, y, z) is constant. x2 C y2 − 3z D c is equivalent to
z D 1/3x2 C y2 − c. These are paraboloids with z intercept 0, 0, −c/3. A typical surface is:
25. Let x be a flow line of a gradient vector field F D §f and let Gt D f xt. We will show that G is an
increasing function of t by showing G t Ú 0. First, G t D §f xt · x t D Fxt · x t since
F D §f .
Now we use the fact that x is a flow line of F:
Fxt · x t D x t · x t D x t2 Ú 0.
For Exercises 26–28, verify that ⭸ φx, t D Fφx, t and φx, 0 D x.
⭸t
26. First we see that φx, y, 0 D xCy e0 C x−y e0 , xCy e0 C y−x e0 D x, y. Next,
2
2
2
2
x − y −t x C y t
y − x −t
x C y t
⭸
φx, y, t D e −
e ,
e −
e ⭸t
2
2
2
2
D φy, x, t D Fφx, y, t.
27. First we see that φx, y, 0 D y sin 0 C x cos 0, y cos 0 − x sin 0 D x, y. Next,
⭸
φx, y, t D y cos t − x sin t, −y sin t − x cos t
⭸t
D φy, −x, t D Fφx, y, t.
28. First we see that φx, y, z, 0 D x cos 0 − y sin 0, y cos 0 C x sin 0, ze0 D x, y, z. Next,
⭸
φx, y, z, t D −2x sin 2t − 2y cos 2t, −2y sin 2t C 2x cos 2t, −ze−t ⭸t
D φ−2y, 2x, −z, t D Fφx, y, z, t.
29. We are assuming that φ is a flow of F and that xt D φx0 , t. Then
x t D
⭸
φx0 , t D Fφx0 , t D Fxt.
⭸t
The middle equality holds because φ is a flow of F.
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“runall” — 2005/8/1 — 15:09 — page 163 — #27
Section 3.4
Gradient, Divergence, Curl, and The Del Operator
163
30. Using the hint, we can apply the results of Exercise 29. If φ is a flow of the vector field F then for any fixed
point x0 in X, the map xt D φx0 , t is a flow line of F.
So φx0 , s C t is where we are if we flow for t C s seconds while φφx0 , t, s is where we are if we first
flow for t seconds and then we flow for s seconds. It should be clear that we end up the same place in either case.
It is worth checking that your students understand the idea behind the problem—the author of the text has taken
great care to make sure that these symbols make some physical sense to them.
31. We know that
⭸
φx, t D Fφx, t. So
⭸t
⭸
⭸
Dx φx, t D Dx φx, t D Dx Fφx, t.
⭸t
⭸t
Now by the chain rule (Theorem 5.3):
Dx Fφx, t D DFφx, tDx φx, t.
3.4 GRADIENT, DIVERGENCE, CURL, AND THE DEL OPERATOR
For Exercises 1–6 calculate the divergence of F: div F D § · F.
⭸F1
⭸F2
C
D 2x C 2y.
⭸x
⭸y
⭸F1
⭸F2
2. F D y2 , x2 , so div F D § · F D
C
D 0 C 0 D 0.
⭸x
⭸y
⭸F1
⭸F2
⭸F3
3. F D x C y, y C z, x C z, so div F D § · F D
C
C
D 1 C 1 C 1 D 3.
⭸x
⭸y
⭸z
√
2
⭸F1
⭸F2
⭸F3
C
C
D 0 C 0 C 0 D 0.
4. F D z cosey , x z2 C 1, e2y sin 3x, so div F D § · F D
⭸x
⭸y
⭸z
⭸F1
⭸F2
⭸Fn
5. F D x12 , 2x22 , ... , nxn2 , so div F D § · F D
C
C ··· C
D 2x1 C 4x2 C · · · C 2nxn .
⭸ x1
⭸x2
⭸ xn
⭸F1
⭸F2
⭸Fn
C
C ··· C
D 1 C 0 C · · · C 0 D 1.
6. F D x1 , 2x1 , ... , nx1 , so div F D § · F D
⭸x1
⭸ x2
⭸xn
For Exercises 7–11 calculate the curl of F: curl F D § * F.
i
j
k
1. F D x2 , y2 , so div F D § · F D
7. curl F D
⭸
⭸
⭸
⭸x
⭸y
⭸z
−xey
x2
8. curl F D
2xyz
i
j
⭸
⭸
⭸
⭸x
⭸y
⭸z
x
y
z
9. curl F D
D 2xz, −2yz, −ey .
k
D 0, 0, 0.
i
j
⭸
⭸
⭸
⭸x
⭸y
⭸z
x C yz
10. curl F D
k
y C xz
D x − x, −y C y, z − z D 0, 0, 0.
z C xy
i
j
⭸
⭸
k
⭸
⭸x
⭸y
⭸z
cos yz − x cos xz − y cos xy − z
D xsin xz − sin xy, ysin xy − sin yz, zsin yz − sin xz.
11. curl F D
i
j
⭸
⭸
⭸
⭸x
⭸y
⭸z
y2 z
e xyz
k
D x2 − xyexyz , y2 − 2xy, yzexyz − 2yz.
x2 y
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“runall” — 2005/8/1 — 15:09 — page 164 — #28
164
Chapter 3
Vector-Valued Functions
12. (a) We denote the vector field from Exercise 8 by F8 and sketch it below on the left. The figure on the right
represents any planar slice through the origin. Every point is being pushed outwards. If you imagine a twig
caught in this body of water and you think in terms of spherical coordinates, the change in position is an
increase in ρ with no change to ϕ or θ.
(b) Note that F D √xiCyjCzk
D √ 2 F8 2
2
2
2
x Cy Cz
x Cy Cz2
. At each point the direction of F is the same as that of F8 but F
is made up of unit vectors. As in part (a) we would argue that the motion of each point is in the direction of
increasing ρ and so the curl is again 0.
(c) In Exercise 24 below we’ll show that § * f F D f § * F C §f * F. If you don’t like citing a future
problem, you can follow through the steps in Exercise 24 for this exercise. Note that we know from Exercise
8 that § * F8 D 0, 0, 0.


1


§ * F D § * 
F8 
x 2 C y 2 C z2
 




D
1
x 2 C y 2 C z2
D 0, 0, 0 −
D−

 § * F8 C § 

1
x 2 C y 2 C z2

 * F8
x, y, z
* F8
x2 C y2 C z2 3/2
x, y, z
x2 C y2 C z2 3/2
* x, y, z D 0, 0, 0.
13. (a) At each point “more is moving away than towards” so div F > 0 on all R2 .
(b) At each point “more is moving towards than away” so div F < 0 on all R2 .
(c) Here we have a mixed bag. At each point to the left of the y-axis “more is moving towards than away”
and at each point to the right of the y-axis “more is moving away than towards” so div F < 0 for x < 0,
div F > 0 for x > 0, and div F D 0 for x D 0.
(d) Again we have a mixed bag. At each point the above the x-axis “more is moving towards than away” and
at each point below the x-axis “more is moving away than towards” so div F < 0 for y > 0, div F > 0
for y < 0, and div F D 0 for y D 0.
In Exercises 14 and 15, the student is asked to work examples of the results of Theorems 4.3 and 4.4. Exercise
16 has the student prove Theorem 4.4.
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“runall” — 2005/8/1 — 15:09 — page 165 — #29
Section 3.4
Gradient, Divergence, Curl, and The Del Operator
165
14. f x, y, z D x2 sin y C y2 cos z so §f D 2x sin y, x2 cos y C 2y cos z, −y2 sin z.
§ * §f D
i
j
⭸
⭸
k
⭸
⭸x
⭸y
⭸z
x2 cos y C 2y cos z
2x sin y
−y2 sin z
D −2y sin z C 2y sin z, 0 − 0, 2x cos y − 2x cos y D 0, 0, 0.
15. Fx, y, z D xyzi C ez cos xj C xy2 z3 k so
§ * FD
i
j
⭸
⭸
k
⭸
⭸x
⭸y
⭸z
e z cos x
xyz
D 2xyz3 C e z cos x, −y2 z3 C xy, e z sin x − xz.
xy2 z3
Finally we calculate
§ · § * F D
⭸
⭸
⭸ z
2xyz3 C e z cos x C
−y2 z3 C xy C
e sin x − xz
⭸x
⭸y
⭸z
D 2yz3 − e z sin x − 2yz3 C x C e z sin x − x D 0.
16. We want to show that § · § * F D 0.
§ · § * F D
D
⭸ ⭸F3
⭸F2
⭸ ⭸F1
⭸ F3
⭸ ⭸F2
⭸F1
−
−
−
C
C
⭸x ⭸y
⭸x
⭸y ⭸z
⭸x
⭸z ⭸x
⭸y
⭸2 F3
⭸2 F2
⭸2 F1
⭸2 F3
⭸2 F2
⭸2 F1
−
C
−
C
−
⭸x⭸y
⭸x⭸z
⭸y⭸z
⭸y⭸x
⭸z⭸x
⭸z⭸y
Finally, because F is of class C2 , the mixed partials are equal and so this last quantity is 0.
17. This is a good warm-up.
§rn D i
n
⭸
⭸
⭸
C j
C k x2 C y2 C z2 n/2 D x2 C y2 C z2 n−2/2 2x, 2y, 2z
⭸x
⭸y
⭸z
2
D nrn−2 x, y, z D nrn−2 r.
18. This is similar to Exercise 17 as most of the derivative of the ln pulls out.
§ln r D §lnx2 C y2 C z2 1/2 D 1 §lnx2 C y2 C z2 2
D 1
2
1
1
r
2x, 2y, 2z D 2 x, y, z D 2
x 2 C y 2 C z2
r
r
19. In this exercise and the next we’ll need to know that rn r D x2 C y2 C z2 n/2 x, y, z.
§ · rn r D
⭸
⭸
⭸
[xx2 C y2 C z2 n/2 ] C
[yx2 C y2 C z2 n/2 ] C
[zx2 C y2 C z2 n/2 ]
⭸x
⭸y
⭸z
n
n
D rn C x 2xx2 C y2 C z2 n−2/2 C rn C y 2yx2 C y2 C z2 n−2/2
2
2
n
C rn C z 2zx2 C y2 C z2 n−2/2
2
D [rn C nx2 rn−2 ] C [rn C ny2 rn−2 ] C [rn C nz2 rn−2 ]
D 3rn C nx2 C y2 C z2 rn−2 D 3rn C nr2 rn−2 D 3rn C nrn D n C 3rn
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“runall” — 2005/8/1 — 15:09 — page 166 — #30
166
Chapter 3
Vector-Valued Functions
20. Here
§ * r r D
i
j
⭸
⭸
⭸
⭸x
⭸y
⭸z
n
xx2 C y2 C z2 n/2
k
yx2 C y2 C z2 n/2
zx2 C y2 C z2 n/2
Let’s begin by calculating the coefficient of i:
⭸
⭸
[zx2 C y2 C z2 n/2 ] −
[yx2 C y2 C z2 n/2 ]
⭸y
⭸z
D zx2 C y2 C z2 n−2/2 2y − yx2 C y2 C z2 n−2/2 2z D 0.
The calculation is the same for the coefficients of j and k.
Exercises 21 and 22 follow quickly from properties we explored in Chapter 1. The § seems to distribute over the sum
because the derivative of a sum is the sum of the derivatives. Exercises 23–25 are product rules.
21.
n
n
n
⭸
⭸
⭸
Fi C Gi D $
Fi C $
Gi D § · F C § · G.
iD1 ⭸xi
iD1 ⭸xi
iD1 ⭸xi
§ · F C G D $
22. You can expand the first matrix below and see the result pretty quickly. On the other hand, you can use the result
of Exercise 28 from Section 1.6 and the fact that d F C G D d F C d G.
dxi
i
j
⭸
⭸
⭸
⭸x
⭸y
⭸z
F1 C G1
F2 C G2
F3 C G3
§ * F C G D
D
dxi
k
i
j
k
i
j
⭸
⭸
⭸
⭸
⭸
⭸
⭸x
⭸y
⭸z
⭸x
⭸y
⭸z
F1
F2
F3
G1
G2
G3
C
dxi
k
D § * F C § * G.
23.
n
⭸f
⭸
⭸F
f Fi D $ Fi C f i ⭸ xi
iD1 ⭸xi
iD1 ⭸xi
n
§ · f F D $
n
D $
iD1
n
⭸f
⭸F
Fi C $ f i D §f · F C f § · F
⭸xi
⭸ xi
iD1
D f § · F C F · §f .
24.
i
j
⭸
⭸
⭸
⭸x
⭸y
⭸z
f F1
f F2
f F3
§ * f F D
k
D
⭸
⭸
⭸
⭸
⭸
⭸
f F3 −
f F2 i − f F3 −
f F1 j C
f F2 −
f F1 k
⭸y
⭸z
⭸x
⭸z
⭸x
⭸y
D
⭸f
⭸f
⭸f
⭸f
⭸F3
⭸F2
⭸F3
⭸F1
F3 −
F2 C
f −
f i − F3 −
F1 C
f −
fj
⭸y
⭸z
⭸y
⭸z
⭸x
⭸z
⭸x
⭸z
C
⭸f
⭸f
⭸F2
⭸F1
F2 −
F1 C
f −
f k D f § * F C §f * F.
⭸x
⭸y
⭸x
⭸y
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“runall” — 2005/8/1 — 15:09 — page 167 — #31
Section 3.4
Gradient, Divergence, Curl, and The Del Operator
167
25.
§ · F * G D D
⭸ ⭸ ⭸
, , · F2 G3 − F3 G2 , F3 G1 − F1 G3 , F1 G2 − F2 G1 ⭸x ⭸y ⭸z
⭸F2
⭸G3
⭸F3
⭸G2
⭸F3
⭸G1
⭸F1
⭸G3
G3 C F2
−
G2 − F3
C
G1 C F3
−
G3 − F1
⭸x
⭸x
⭸x
⭸x
⭸y
⭸y
⭸y
⭸y
⭸F1
⭸G2
⭸F2
⭸G1
C
G2 C F1
−
G1 − F2
⭸z
⭸z
⭸z
⭸z
D G1 ⭸F3
⭸F2
⭸F3
⭸F1
⭸F2
⭸F1
⭸G3
⭸G2
−
−
−
−
− G2 C G3 − F1 ⭸y
⭸z
⭸x
⭸z
⭸x
⭸y
⭸y
⭸z
C F2 ⭸G3
⭸G1
⭸G2
⭸G1
−
−
− F3 ⭸x
⭸z
⭸x
⭸y
D G · § * F − F · § * G.
26. We will use formulas (6) and (7) from the text. First we establish formula (3):
§f D
⭸f
⭸f
⭸f
i C
j C
k
⭸x
⭸y
⭸z
D cos θ
⭸f
⭸f
⭸f
sin θ ⭸f
cos θ ⭸f
−
C
k
i C sin θ
j C
⭸r
r ⭸θ
⭸r
r ⭸θ
⭸z
D
⭸f
⭸f
1 ⭸f
cos θi C sin θj C − sin θi C cos θj C
k
⭸r
r ⭸θ
⭸z
D
⭸f
⭸f
1 ⭸f
er C eθ C
ez .
⭸r
r ⭸θ
⭸z
Now we establish formula (5). Again we need formulas (6) and (7). First use (6) to obtain: Fr er C Fθ eθ D
Fr cos θ − Fθ sin θi C Fr sin θ C Fθ cos θj.
i
j
k
⭸
⭸
⭸
⭸x
⭸y
⭸z
Fx
Fy
Fz
§ * FD
D sin θ
i
D
⭸
sin θ ⭸
cos θ ⭸r − r ⭸θ Fr cos θ − Fθ sin θ
j
k
cos θ ⭸
sin θ ⭸r C r ⭸θ Fr sin θ C Fθ cos θ
⭸z
⭸
Fz
cos θ ⭸
⭸
⭸
C
Fr sin θ C Fθ cos θ i
Fz −
⭸r
r ⭸θ
⭸z
− cos θ
⭸
⭸
sin θ ⭸
−
Fr cos θ − Fθ sin θ j
Fz −
⭸r
r ⭸θ
⭸z
C cos θ
⭸
sin θ ⭸
−
Fr sin θ C Fθ cos θ
⭸r
r ⭸θ
− sin θ
D
⭸
cos θ ⭸
⭸
C
Fr cos θ − Fθ sin θ k
⭸r
r ⭸θ
1 ⭸
⭸
⭸
⭸
Fz −
Fθ cos θi C sin θj − Fz −
Fr − sin θi C cos θj
r ⭸θ
⭸z
⭸r
⭸z
C
1 ⭸
⭸
Fθ − Fr k
⭸r
r ⭸θ
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168
Chapter 3
Vector-Valued Functions
D
D
1 ⭸
⭸
⭸
⭸
Fz −
Fθ er − Fz −
Fr eθ C
r ⭸θ
⭸z
⭸r
⭸z
er
eθ
ez
⭸
1 ⭸
r ⭸θ
Fθ
⭸
⭸r
Fr
⭸z
Fz
1
D
r
ez
er
reθ
⭸
⭸
⭸
⭸r
⭸θ
⭸z
Fr
rFθ
1 ⭸
⭸
Fθ − Fr ez
⭸r
r ⭸θ
.
Fz
27. We will need formula (9) from Section 1.7:



 eρ D sin ϕ cos θi C sin ϕ sin θj C cos ϕk
eϕ D cos ϕ cos θi C cos ϕ sin θj − sin ϕk


 eθ D − sin θi C cos θj.
From the chain rule, we have the following relations between rectangular and spherical differential operators:

⭸
⭸
⭸
⭸


D sin ϕ cos θ
C sin ϕ sin θ
C cos ϕ



⭸ρ
⭸x
⭸y
⭸z


 ⭸
⭸
⭸
⭸
D ρ cos ϕ cos θ
C ρ cos ϕ sin θ
− ρ sin ϕ

⭸ϕ
⭸x
⭸y
⭸z




⭸
⭸
⭸


D −ρ sin ϕ sin θ
C ρ sin ϕ cos θ .

⭸θ
⭸x
⭸y
Solving for ⭸/⭸x, ⭸/⭸y, and ⭸/⭸z:


cos ϕ cos θ ⭸
sin θ ⭸
⭸
⭸


D sin ϕ cos θ
C
−



⭸x
⭸ρ
ρ
⭸ϕ
ρ sin ϕ ⭸θ


 ⭸
cos ϕ sin θ ⭸
⭸
cos θ ⭸
D sin ϕ sin θ
C
C

⭸y
⭸ρ
ρ
⭸ϕ
ρ sin ϕ ⭸θ




sin ϕ ⭸
⭸
⭸


D cos ϕ
−
.

 ⭸z
⭸ρ
ρ ⭸ϕ
Now we calculate the gradient:
§f D
⭸f
⭸f
⭸f
i C
j C
k
⭸x
⭸y
⭸z
D sin ϕ cos θ
⭸f
cos ϕ cos θ ⭸f
sin θ ⭸f
C
−
i
⭸ρ
ρ
⭸ϕ
ρ sin ϕ ⭸θ
C sin ϕ sin θ
D
⭸f
⭸f
cos ϕ sin θ ⭸f
sin ϕ ⭸f
cos θ ⭸f
C
C
−
j C cos ϕ
k
⭸ρ
ρ
⭸ϕ
ρ sin ϕ ⭸θ
⭸ρ
ρ ⭸ϕ
⭸f
1 ⭸f
sin ϕ cos θi C sin ϕ sin θj C cos ϕk C cos ϕ cos θi C cos ϕ sin θj − sin ϕk
⭸ρ
ρ ⭸ϕ
C ⭸f
⭸f
⭸f
1
1 ⭸f
1
− sin θi C cos θj D
er C eϕ C eθ .
ρ sin ϕ ⭸θ
⭸ρ
ρ ⭸ϕ
ρ sin ϕ ⭸θ
⭸2
⭸2
⭸2
C
C
.
2
2
⭸x
⭸y
⭸z2
⭸
⭸
⭸
⭸
⭸
⭸
⭸ ⭸
⭸ ⭸
⭸ ⭸
⭸2
⭸2
⭸2
(a) § · § D C
C
D
C
C
C
C
C
C
D
·
D
⭸x
⭸y
⭸z
⭸x
⭸y
⭸z
⭸x ⭸x
⭸y ⭸y
⭸z ⭸z
⭸x2
⭸y2
⭸z2
§2 .
28. §2 D
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“runall” — 2005/8/1 — 15:09 — page 169 — #33
Section 3.4
Gradient, Divergence, Curl, and The Del Operator
169
(b) We use ideas from Exercise 23:
§2 f g D § · §f g D § · §f g C f §g D §2 f g C §f · §g C §f · §g C f §2 g
D f §2 g C g§2 f C 2§f · §g.
(c) Again we use Exercise 23:
§ · f §g − g§f D §f · §g C f §2 g − §g · §f − g§2 f D f §2 g − g§2 f .
29. f §f D f ⭸f , f ⭸f , f ⭸f hence
⭸x
⭸y
⭸z
§ · f §f D
⭸f
⭸f
⭸f
⭸
⭸
⭸
f
C
f
C
f
⭸x
⭸x
⭸y
⭸y
⭸z
⭸z
2
2
2
⭸f
⭸2 f
⭸f
⭸2 f
⭸f
⭸2 f
D C f
C C f
C C f
2
2
⭸x
⭸x
⭸y
⭸y
⭸z
⭸z2
2
2
2
⭸f
⭸f
⭸f
⭸2 f
⭸2 f
⭸2 f
C
C C C f 2 C
2
⭸x
⭸y
⭸z
⭸x
⭸y
⭸z2
D
D §f 2 C f §2 f .
30. Write F D Mi C Nj C Pk. Then
§ * FD
i
j
⭸
⭸
k
⭸
⭸x
⭸y
⭸z
M
N
P
D Py − Nz i C Mz − Px j C Nx − My k
and thus
§ * § * F D
i
j
⭸
⭸
⭸
⭸x
⭸y
⭸z
Py − Nz
Mz − P x
k
Nx − My
D Nxy − Myy − Mzz C Pxz i C Pyz − Nzz − Nxx C Myx j
C Mzx − Pxx − Pyy C Nzy k.
On the other hand,
§§ · F D §Mx C Ny C Pz D Mxx C Nyx C Pzx i C Mxy C Nyy C Pzy j C Mxz C Nyz C Pzz k
and
§2 F D § · §F D ⭸2
⭸2
⭸2
C
C
F
2
2
⭸x
⭸y
⭸z2
D Mxx C Myy C Mzz i C Nxx C Nyy C Nzz j C Pxx C Pyy C Pzz k
Hence,
§§ · F − §2 F D Nyx C Pzx − Myy − Mzz i C Mxy C Pzy − Nxx − Nzz j C Mxz C Nyz − Pxx − Pyy k
By assumption, F is of class C2 so Mxy D Myx , etc.
Thus we have shown: § * § * F D §§ · F − §2 F, as desired.
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170
Chapter 3
Vector-Valued Functions
31. (a) Let Gt D Fa C tv. Then
1
1
Dv Fa D lim Fa C tv − Fa D lim Gt − G0
t'0 t
t'0 t
D G 0.
Thus
Dv Fa D
d
Fa C tv
.
dt
tD0
(b) d Fa C tv D DFa C tv d a C tv D DFa C tvv. Now evaluate at t D 0 to get Dv Fa D DFav.
dt
dt
32. By definition,
Dv Fa D lim
1
D lim
1
h'0 h
h'0 h
Fa C hv − Fa
F1 a C hv − F1 a, ... , Fn a C hv − Fn a
D lim F a C hv − Fn a
F1 a C hv − F1 a
, ... , n
h
h
D lim
F1 a C hv − F1 a
F a C hv − Fn a
, ... , lim n
D Dv F1 a, ... , Dv Fn a
h'0
h
h
h'0
h'0
using Definition 6.1 of Chapter 2.
33. We use part (b) of Exercise 31 since F is evidently differentiable.




0 z y
0 1 2
DFx, y, z D  z 0 x  so DF3, 2, 1 D  1 0 3 
y x 0
2 3 0






√
√
1/ √3
1/√ 3
0 1 2

 

Di−jCk/√ 3 F3, 2, 1 D  1 0 3   −1/√ 3  D  4/ √3 
2 3 0
1/ 3
−1/ 3

34.
1
Dv Fa D DFav D  0
0
0
1
0
 


0
v1
v1
0   v2  D  v2  D v.
v3
v3
1
In general if F D x1 , ... , xn , then DFa D In (n * n identity matrix), so Dv Fa D DFav D In v D v.
3.5 TRUE/FALSE EXERCISES FOR CHAPTER 3
1. True.
2. False. (The path has unit speed.)
3. True.
4. False.
5. False. (There should be a negative sign in the second term on the right.)
6. False. (κ D dT/ds, where s is arclength.)
7. True.
8. True.
9. False.
10. False. (dT/ds must be normalized to give N.)
11. True.
12. True.
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“runall” — 2005/8/1 — 15:09 — page 171 — #35
Section 3.6
Miscellaneous Exercises for Chapter 3
171
13. True.
14. False. (It’s a vector field.)
15. False. (It’s a scalar field.)
16. True.
17. True.
18. False. (It’s a scalar field.)
19. False. (It’s a meaningless expression.)
20. False. Fxt Z x t.
21. True. (Check that Fxt D x t.)
22. True. (Verify that § · F D 0.)
23. False. § * F Z 0.)
24. False. (f must be of class C2 .)
25. False. (Consider F D yi C xj.)
26. False. (The first term on the right needs a negative sign.)
27. True.
28. False. § * F Z 0.
29. False. § · § * F Z 0.
30. True.
3.6 MISCELLANEOUS EXERCISES FOR CHAPTER 3
1. Here are the answers: (a) D
(d) B
(b) F
(c) A
(e) C
(f) E
Here’s some explanation: The formulas in (a) and in (f) are the only ones that keep x and y bounded (between
−1 and 1), so they must correspond to D and E. Note that in (a) x0 D 0, 0, but the graph in E does not
pass through the origin. Note that in (c) x Ú 1 and the only graph with that property is A. In (b) we see that
x−t D −t − sin 5t, t 2 C cos 6t D −xt, yt. This means that the graph will be symmetric about the
y-axis and the only plot that remains with this property is F. What remains is to match the formulas in (d) and
(e) with the graphs in B and C. This is easy: in (d) large positive values of t give points in the first quadrant. The
graph in C has no points in the first quadrant.
2. Answers: (a) E (b) F (c) C (d) B (e) A (f) D
Explanation: (a) Must have z between −1 and 1, but y can be arbitrarily large and positive.
(b) All three coordinates should be bounded (making the only choices B or F). The projection of the curve into
the xy-plane should be an astroid—giving choice F.
(c) This is an elliptical helix—so choice C.
(d) All three coordinates are between −1 and 1, so graph B.
(e) Note that x2 C y2 D 4t 2 D 1 z2 —thus the graph lies on a cone (so A).
4
(f) Only D remains, but note that we must have x Ú 1.
3. First note that d x t D d x t · x t D x t · x t/x t. So x has constant (non-zero) speed
dt
dt
if and only if d x t D 0 if and only if x t · x t D 0 (i.e., its velocity and acceleration vectors are
dt
perpendicular).
4. (a) If we forget about gravity, the glasses travel along the tangent line to x at t D 90. We need the position
along this tangent line two seconds after we lose our glasses:
1t D x90 C 2x 90 D −e 3/2 , 0, 80 C 2−e 3/2 /60, −πe 3/2 /30, 0
D −31e 3/2 /30, −πe 3/2 /15, 80.
(b) The only component that changes when we factor in gravity is the height ht of the glasses at time t. We
know that gravity is h t D −32 f t/sec2 . The initial vertical velocity is zero so h t D −32t. We know
that when the glasses fall off they are 80 feet off the ground, so ht D −16t 2 C 80 so h2 D 16 and the
position of the glasses two seconds after they fall off is −31e3/2 /30, −πe3/2 /15, 16.
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172
Chapter 3
Vector-Valued Functions
5. The velocity is x t D − sint − 1, 3t 2 , − 12 so x 1 D 0, 3, −1. At t D 1 the position is x1 D
t
1, 0, −1. If we define a surface by the equation f x, y, z D x3 C y3 C z3 − xyz D 0 then §f x, y, z D
3x2 − yz, 3y2 − xz, 3z2 − xy so §f 1, 0, −1 D 3, 1, 3. In general this vector is normal to the tangent
plane at 1, 0, −1 and by observation it is also perpendicular to x 1 so the curve is tangent to the surface
when t D 1.
6. (a) We’ll convert distance to inches and then r D 240 − 3t while θ D 4πt.
(b) x D r cos θ and y D r sin θ so xt D 240 − 3t cos 4πt and yt D 240 − 3t sin 4πt.
(c) It takes Gregor 80 minutes to reach the center so
80 Distance D
[x t]2 C [y t]2 dt
0
80 D
[−3 cos 4πt − 4π240 − 3t sin 4πt]2 C [−3 sin 4πt C 4π240 − 3t cos 4πt]2 dt
0
80 D
9 C 16π2 240 − 3t2 dt D 120638 inches L 1.90401 miles.
0
7. For w D 0 we just get the line segment joining the points x1 and x2 . As w increases the curve becomes more
bent in the direction of the control point.
y
1
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1
1
2
3
4
x
8. We see the same pattern as in Exercise 7.
y
3
2
1
x
-1
-1
9. (a) There’s nothing much to show. In the equations given in (1), at t D 0 all but the first terms in the numerator
and denominator disappear and you get x0, y0 D x1 , y1 . At t D 1 all but the last terms in the
numerator and denominator disappear and you get x1, y1 D x3 , y3 .
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Section 3.6
Miscellaneous Exercises for Chapter 3
173
(b) Here we get
x /4 C wx2 /2 C x3 /4 y1 /4 C wy2 /2 C y3 /4
,
x1/2 D 1
1 C w/2
1 C w/2
D
y C y3
1
x1 C x3
C wx2 , 1
C wy2 1 C w
2
2
D
x1 C x3 y1 C y3
1
,
C wx2 , wy2 1 C w
2
2
D
1
x1 C x3 y1 C y3
w
,
x2 , y2 .
C
1 C w
2
2
1 C w
Note that for w Ú 0, both 1/1 C w and w/1 C w are between 0 and 1 and sum to 1. This tells us that
the point x(1/2) lies on the line segment joining x2 , y2 to the midpoint of the line segment joining x1 , y1 to
x3 , y3 .
10. If you’re doing this by hand, first simplify the denominator in the expression for xt and yt by realizing the
sum of the first and last terms is 1. In other words, 1 − t2 C 2wt1 − t C t 2 D 2wt1 − t. Crank it
through and find that x 0 D 2 wx2 − x1 , y2 − y1 and x 1 D 2 wx3 − x2 , y3 − y2 .
Let l0 be the tangent line to the curve at x(0). Then
l0 t D x0 C tx 0 D x1 , y1 C 2wtx2 − x1 , y2 − y1 .
This cries out for us to check out t D 1/2 w. We see that
l0 1/2 w D x1 , y1 C x2 − x1 , y2 − y1 D x2 , y2 .
Similarly, let l1 be the tangent line to the curve at x(1). Then
l1 t D x1 C tx 1 D x3 , y3 C 2wtx3 − x2 , y3 − y2 .
At t D −1/2 w we see that
l1 −1/2 w D x3 , y3 − x3 − x2 , y3 − y2 D x2 , y2 .
In other words, the point x2 , y2 is on both of the tangent lines.
11. (a) Use part (b) of Exercise 9.
1
y1 C y3
x1 C x3
a D x1/2 − x2 , y2 D C wx2 ,
C wy2 − x2 , y2 1 C w
2
2
1
D x1 − 2x2 C x3 , y1 − 2y2 C y3 21 C w
D
1
x1 − 2x2 C x3 2 C y1 − 2y2 C y3 2
21 C w
(b) This is a similar calculation.
x1 C x3 y1 C y3 ,
b D x1/2 − 2
2
1
y1 C y3
x1 C x3 y1 C y3 x1 C x3
,
−
C
wx
C
wy
,
D
2
2
1 C w
2
2
2
2
w
D −x1 C 2x2 − x3 , −y1 C 2y2 − y3 21 C w
D
w
x1 − 2x2 C x3 2 C y1 − 2y2 C y3 2
21 C w
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174
Chapter 3
Vector-Valued Functions
(c) It’s kind of amazing, but
b
D
a
w
21Cw x1 − 2x2 C x3 2 C y1 − 2y2 C y3 2
1
21Cw D w.
x1 − 2x2 C x3 2 C y1 − 2y2 C y3 2
12. (a) Start with y D 2x. So y −2 D −4 and y 2 D 4. The two tangent lines y − 4 D −4x C 2 and
y − 4 D 4x − 2 can be rewritten as y D −4x − 4 and y D 4x − 4. The point of intersection is
0, −4 and so, by Exercise 10, this is the third control point.
(b) The deal here is that we are actually going to end up with y D x2 between x D −2 and x D 2. Because
x(1/2) must be on the line segment connecting the control point we found in (a) to the midpoint of the the
line segment connecting the other two control points, it must be on the y-axis. The only point on the parabola
that satisfies this is the origin. The constant w is the ratio of the distance between (0, 0) and (0, 4) and the
distance between (0, 0) and 0, −4. In this case, w D 1.
(c) The Bézier parametrization is
'
xt D −21 − t2 C 2t 2 D 4t − 2
yt D 41 − t2 − 42t1 − t C 4t 2 D 4t − 22 .
13. (a) We have x t D cos t, − sin t C 1 cot t sec2 t . Using the double angle formula, we have
2
2
2
1
t
t
1
cot sec2 D
2
2
2
2 sin t cos t
2
2
D
1
.
sin t
Hence x t D cos t, 1 − sin t . Thus, x t D 0, 0 if and only if t D π/2.
sin t
(b) The tangent line to the tractrix at the point xt0 is given by ls D xt0 C sx t0 . This line crosses the
y-axis when x D 0 and if we explicitly compute the first component of ls, we see that the condition for
crossing is
sin t0 C s cos t0 D 0 3 s D − tan t0 .
The length of the segment we seek is given by
l− tan t0 − xt0 D xt0 − tan t0 x t0 − xt0 D | tan t0 |x t0 .
Using the work from part (a), the length of the segment is
| tan t0 | cos2 t0 C csc t0 − sin t0 2 D | tan t0 | cos2 t0 C csc2 t0 − 2 sin2 t0
D | tan t0 | csc2 t0 − 1 D | tan t0 || cot t0 | D 1.
√
14. (a) Now we have y r D er , 1 − e2r by the fundamental theorem of calculus. So the tangent line at the
point yr0 is ms D yr0 C sy r0 . This line crosses the y-axis when x D 0 3 er0 C ser0 D 0 3 s D
−1.
√ As in Exercise 13, we compute m−1 − yr0 D yr0 − y r0 − yr0 D y r0 D
2r
2r
e 0 C 1 − e 0 D 1.
√
r √
(b) Note that, for ρ < 0, the integrand 1 − e2ρ is positive. Hence for r < 0, the integral 0 1 − e2ρ dρ
is negative. Since the exponential er varies between 0 and 1 as r varies from −q to 0, we see that y covers
just the bottom half of the tractrix.
15. If r D f θ, then we may write xθ D f θ
Hence v D x θ D f θ cos θ −
cos θ, f θ sin θ. √
f θ sin θ, f θ sin θ C f θ cos θ and v D f θ2 C f θ2 D r2 C r2 . Also
a D x θ D f θ cos θ − 2f θ sin θ − f θ cos θ, f θ sin θ C 2f θ cos θ − f θ sin θ.
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“runall” — 2005/8/1 — 15:09 — page 175 — #39
Section 3.6
Miscellaneous Exercises for Chapter 3
175
If we calculate v * a, we find (after same algebra)
v * a D −f θf θ C 2f θ2 C f θ2 k D r2 − rr C 2r2 k.
Hence, using formula (17), we have
κD
v * a
|r2 − rr C 2r2 |
D
.
3
v
r2 C r2 3/2
16. For the lemniscate r2 D cos 2θ, so that, differentiating with respect to θ, we have 2rr D −2 sin 2θ. Hence
1
r D − sin 2θ
r
so r2 D
Thus
r2 C r2 D cos 2θ C
1
sin2 2θ
.
sin2 2θ D
r2
cos 2θ
1
sin2 2θ
D
.
cos 2θ
cos 2θ
Now, if we differentiate the equation rr D − sin 2θ, we obtain
rr C r2 D −2 cos 2θ.
From Exercise 15, we must compute κ D
we have
|r2 − rr C 2r2 |
. The denominator is easy; for the numerator,
r2 C r2 3/2
r2 − rr C 2r 2 D r2 − rr C r 2 C 3r 2
D cos 2θ − −2 cos 2θ C
D 3 cos 2θ C
3 sin2 2θ
cos 2θ
3 sin2 2θ
3
D
.
cos 2θ
cos 2θ
√
|3/ cos 2θ|
D 3 cos 2θ.
3/2
1/ cos 2θ
17. (a) The involute of xt D a cos t, a sin t is yt D a cos t, a sin t − at− sin t, cos t.
(b) The circle and involute are shown below.
Hence κθ D
y
3
2
1
-4
-3
-2
-1
1
x
-1
-2
-3
-4
18. We actually can afford to be a bit sloppy. Look first at y t D x t − s tTt − stT t. By the fundamental
theorem of calculus, s t D x t so s tTt D x t. So we can now say that y t D −stT t. But
by the Frenet–Serret formulas, T t D s tκN. This means that y t D −sts tκN. In other words, the
tangent vector to the involute is in the opposite direction to the normal vector to the curve, so the unit tangent
vector to the involute at t is the opposite of the unit normal vector Nt to the original path x.
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“runall” — 2005/8/1 — 15:09 — page 176 — #40
176
Chapter 3
Vector-Valued Functions
19. (a) This first conclusion is pretty much by definition. Analytically,
yt − xt D xt − stTt − xt D stTt D |st|Tt D |st| D st.
This last fact follows because st Ú 0. Finally we note that st is the distance traveled from xt0 to xt
along the underlying curve of x.
(b) We calculated the distance from x to y in part (a). We should also observe that this is the distance along
the tangent line to x at time t as it included the point xt and was in the direction Tt. The conclusion
follows—it is as if you are unwinding a taut string from around x: at each point yt is at a point in the
direction of the tangent to xt of distance equal to the distance already traveled along x. In other words,
the distance is equal to the string already unraveled.
√
20. (a) The tangent vector is T D x t/x t D 1, 2t/ 1 C 4t 2 . The normal √
vector is in the xy-plane
perpendicular to T, pointing in the direction that T is changing: N D −2t, 1/ 1 C 4t 2 . We’ll use the
formula (from Section 3.2):
κD
1, 2t, 0 * 0, 2, 0
0, 0, 2
x * x 2
D
D
D
.
3/2
3
3
2
x 1, 2t, 0
1 C 4t 1 C 4t 2 3/2
(b) The formula for the evolute is:
yt D xt C
1
Nt D t C 1 C 4t 2 −t, t 2 C 1 C 4t 2 /2 D −4t 3 , 3t 2 C 1/2.
κ
(c) In the figure below, the evolute increases in the opposite direction as the parabola. We can see that because
the parabola is “straightening out” so the curvature is decreasing so 1/κ is increasing. The evolute is made
up of points distance 1/κ from the parabola in the normal direction.
y
12
10
8
6
4
2
-8
-6
-4
-2
2
x
21. The curvature of a circle of radius a is κ D 1/a. Recall from high school geometry that a tangent to a circle at a
given point is perpendicular to a radial line at that point. If the normal vector is oriented inward, then the evolute
consists of points of distance equal to the radius of the circle in the direction of the center of the circle. In other
words, the evolute of a circle is the center of that circle. Analytically, this is et D xt C aNt.
22. (a) Using a computer we get that the evolute of the ellipse (a cos t, b sin t) is


2
2
2
2
2
2
2
2

.
cos t a − bb cos t C a sin t , cos t b − ab cos t C a sin t 
|ab|
|ab|
(b) An example when a D 3 and b D 4 is shown below. As a gets close to b the ellipse approaches a circle and
the evolute shrinks to a point.
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“runall” — 2005/8/1 — 15:09 — page 177 — #41
Section 3.6
Miscellaneous Exercises for Chapter 3
177
y
3
2
1
-4
-2
2
4
x
-1
-2
-3
23. You may initially get an ugly looking expression. After some coaxing and explicit help, your computer algebra
system should be able to reduce the formula for your evolute of a cycloid to at − 3a sin t, 3a − 3a cos t. This
is another cycloid.
24. Punch this into Mathematica and you will get
2a cos t1 C a cos t −
2a3 1 C a2 C 2a cos tcos t C a cos 2t
,
|a2 C 2a4 C 3a3 cos t|
2a sin t1 C a cos t −
a2 1 C 2a cos t1 C a2 C 2a cos t
.
|a2 C 2a4 C 3a3 cos t|
25. Assume that x is a unit speed curve. To get the direction of the tangent, consider e t D x t C
−1/κ2 κ Nt C 1/κN t. By the Frenet–Serret equations, N t D −κT since x is a planar curve.
So we see what remains is e t D −1/κ2 κ Nt. This tells us that the unit tangent vector to the evolute is
the parallel to the unit normal vector to the original path.
26. First, κ D v * a/v3 . We know that v is a unit vector so v3 D 1. This means that κ D v * a and also
that the tangential component of acceleration is d v D 0 so v is perpendicular to a. Finally, this means that
dt
v * a D va D 1.
27. (a) [x s]2 C [y s]2 D [cos gs]2 C [sin gs]2 D 1.
(b) vs D cos gs, sin gs and as D −g s sin gs, g s cos gs so
κD
v * a
D 0, 0, g s D |g s|.
v3
(c) We use the defining equations with g s D κs.
(d) There is more than one solution. For s Ú 0 we have κ D s therefore, gs D s2 /2 so
s
xs D
s
cost 2 /2 dt
and
ys D
0
sint 2 /2 dt.
0
For s < 0, one solution corresponds to gs D −s2 /2 because for s < 0, g s D −s D |s|. By formula
(8) in Section 3.2, κ will always be non-negative, so we can also take gs D s2 /2 for s < 0. Because
cosine is an even function and sine is an odd function, our two solutions are
s
xs D
s
cost 2 /2 dt
0
and
ys D ;
sint 2 /2 dt.
0
(e) The graph of the clothoid is shown below.
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“runall” — 2005/8/1 — 15:09 — page 178 — #42
178
Chapter 3
Vector-Valued Functions
y
1
0.5
x
-1
-0.5
0.5
1
-0.5
-1
28. (a) −τN D dB/ds so if τ K 0 then B is constant.
(b) The velocity v(0) is in the tangent direction T, so T lies in the xy-plane. The acceleration a(0) has components
in the direction of T and N and since a(0) and T are in the xy-plane, so is N. The binormal vector B must
be length one and perpendicular to the plane containing T and N, so B D ;k.
(c) Combining the results from parts (a) and (b), we know that B K k. It is always true that v · B D 0 and a · B D
0. In this case that is equivalent to v · j D 0 and a · k D 0. But vt · k D x t, y t, z t · 0, 0, 1 D
z t. We conclude that z t is always zero so zt is constant. Since we assumed that z0 D 0, zt K 0
and the path remains in the xy-plane.
(d) Look at the plane determined by v(0) and a(0). By part (b), B will be perpendicular to that plane. By part (a),
B will be constant. Part (c) shows that motion will always be orthogonal to the direction of B. It is harder to
see in this case, but we can translate the problem so that x(0) is the origin and rotate so that a(0) and v(0)
are in the xy-plane, make our conclusions then translate and rotate the solution curve back.
29. Note that we may write xs D xs, ys, 0, where s is the arclength parameter.
T s
1
(a) T D x s, y s, 0, so N D
D x s, y s, 0. (Hence κ D x2 C y2 .) Thus
T s
x2 C y2
B D T * N D 0, 0,
x y − x y
D 0, 0, ;1
κ
because B must be a unit vector. Now B must vary continuously, so either B D 0, 0, 1 or 0, 0, −1—but
in either case, it must be constant.
(b) B s D −τN. B is constant (by part (a)), so B s D 0. Thus, since N is never zero, we may conclude that
τ K 0.
30. We have 0 D κ D dT . Thus dT/ds D 0 so T must be a constant vector. Since x is parametrized by arclength,
ds
x s D T is constant. We may integrate to find:
s
xs D
x σ dσ C xs0 D
s0
s
T dσ C xs0 D s − s0 T C xs0 s0
D sT C xs0 − s0 T,
which is of the form sa C b * straight line.
31. (a) The plan is to find the curvature of the strake by finding the curvature of the helix along the pipe. The radius
r will be the reciprocal of this curvature (since the curvature of a circle of radius r is 1/r). The path is
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“runall” — 2005/8/1 — 15:09 — page 179 — #43
Section 3.6
Miscellaneous Exercises for Chapter 3
179
xt D a cos t, a sin t, ht/2π so x t D −a sin t, a cos t, h/2π and x t D −a cos t, −a sin t, 0.
ah/2π sin t, −ah/2π cos t, a2 −a sin t, a cos t, h/2π * −a cos t, −a sin t, 0
D
3
−a sin t, a cos t, h/2π
a2 C [h/2π]2 3/2
a h2 /4π2 C a2
a2 C h2 /4π2 a
D
D
so r D
.
a2 C [h/2π]2 3/2
a2 C h2 /4π2
a
κD
(b) If a D 3 and h D 25 then r D 9 C 625/4π2 /3 L 8.2771.
32. Since xt D a cos t, a sin t, bt is not parametrized by arclength, we may rewrite it as xs D a cos √ 2s 2 ,
a Cb
√
2 C b2 t is arclength (see Example 3 in §3.2). Following Example 9,
a sin √ 2s 2 , √ bs
a
where
s
D
a Cb
a2 Cb2
we have
a
s
a
s
b
Ts D − √
sin √
,√
cos √
,√
.
a2 C b 2
a2 C b 2
a2 C b 2
a2 C b 2
a2 C b 2
√
√
So the tangent spherical image is a circle of radius a/ a2 C b2 in the plane z D b/ a2 C b2 .
Ns D − cos √ 2s 2 , − sin √ 2s 2 , 0; normal spherical image is a unit circle in the xy-plane. Bs D
a Cb
a Cb
b
s
b
s
√
√
√
√
sin
,
−
cos
, √ 2a 2 . Thus the binormal spherical image is a circle of radius
a2 Cb2
a2 Cb2
a2 Cb2
a Cb
√ a2 Cb2
√
b/ a2 C b2 in the plane z D a/ a2 C b2 .
33. By Example 7 of §3.2 and Exercise 30: x is a straight-line path 3 κ D 0 D dT 3 T is constant.
ds
34. By Exercises 28 and 29, x is a plane curve 3 B is constant.
35. N D −κT C τB by the Frenet–Serret formula. Now for N to be defined κ Z 0, so if τ D 0, then N D −κT Z 0
(hence N is not constant). If τ Z 0, then for N to be 0, T and B would have to be parallel, which they aren’t.
36. (a) xt D rt cos θti C rt sin θtj C ztk D rtcos θti C sin θtj C ztk D rter C ztez .
(b) We prepare for part (c) by calculating:
der
D −θ t sin θti C θ t cos θtj D θ teθ ,
dt
deθ
D −θ t cos θti − θ t sin θtj D −θ ter , and
dt
dez
D0
dt
(c) We use the results of parts (a) and (b) to calculate:
d
d
xt D [rter C ztez ]
dt
dt
de
de
D r ter C rt r C z tez C zt z
dt
dt
vt D
D r ter C rtθ teθ C z tez , and
at D
d [r ter C rtθ teθ C z tez ]
dt
D r ter C r tθ teθ C r tθ teθ C rtθ teθ − rt[θ t]2 er C z tez
D r t − rt[θ t]2 er C 2r tθ t C rtθ teθ C z tez .
√
37. xt D sin 2t, 2 cos 2t, sin 2t − 2
(a) The loop first closes up when t D π so the length of the loop is
π
Length D
0
√
[2 cos 2t]2 C [−2 2 sin 2t]2 C [2 cos 2t]2 dt D
π
√
√
2 2 dt D 2π 2.
0
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“runall” — 2005/8/1 — 15:09 — page 180 — #44
180
Chapter 3
Vector-Valued Functions
(b) By Definition 3.2, the path is a flow line if x t D Fxt. Here
√
√
√
x t D 2 cos 2t, −2 2 sin 2t, 2 cos 2t and Fxt D 2 cos 2t, −2 sin 2t, 2 cos 2t.
√
√
√
√
So x is a flow line of the vector field 2Fx, y, z D 2yi − 2 2xj C 2yk.
38. Poor Livinia, she’s been caught in an oven back in Chapter 2, and now here in Chapter 3 she’s still looking to
get warm.
(a) We saw in Section 2.6 that the gradient is the direction of quickest increase. Livinia should head in a direction
parallel to the gradient. In other words, at each point she should travel in the direction k§T so x t D k§T
so x is a path of k§T for k Ú 0.
(b) If T x, y, z D x2 − 2y2 C 3z2 then §T D 2x, −4y, 6z. We also know that the initial position is
2, 3, −1. This means that x D 2x and x0 D 2 so xt D 2e2kt . Similarly, yt D 3e−4kt and
zt D −e6kt . So the equation of the path is xt D 2e2kt , 3e−4kt , −e6kt .
39. F D ux, yi − vx, yj is an incompressible, irrotational vector field and so § · F D 0 and § * F D 0.
(a) The Cauchy–Riemann equations follow immediately from the assumptions:
0D§ · FD
⭸u
⭸v
−
, so
⭸x
⭸y
⭸u
⭸v
D
, and
⭸x
⭸y
0 D § * F · k D −
⭸v
⭸u
−
, so
⭸x
⭸y
⭸u
⭸v
D−
⭸y
⭸x
(b) Take the partial derivative with respect to x of both sides of the equation:
⭸u
⭸v
D
:
⭸x
⭸y
⭸2 u
⭸ ⭸v
⭸ ⭸v
⭸ ⭸u
⭸2 u
D
D−
D−
D
.
2
⭸x
⭸x ⭸y
⭸y ⭸x
⭸y ⭸y
⭸y2
An analogous calculation shows the result for v.
40. F is a gradient field so F D §f xt. Also F D ma. From Section 3.3 we know that if x is a path on an
equipotential surface of F then f (xt) is constant so d f xt D 0. So
dt
0D
d
d
f xt D §f ·
xt D ma · v.
dt
dt
From Section 3.2, formulas (14) and (16), we see that
ma · v D m ṡ s̈ D
1 d 2
m ṡ .
2 dt
So, since the derivative of ṡ2 D 0, we conclude that ṡ2 is constant and hence ṡ is constant.
41. Using Section 3.1, Exercise 28:
d
dx
dv
dl
D x * mv D
* mv C x * m
dt
dt
dt
dt
D v * mv C x * ma
D 0 C x * ma D M.
42. If F is a central force, then F is always parallel to x. Hence M D x * F D 0, By Exercise 41, M D dl so l must
dt
be constant.
43. Notice that § * F D 0, 2e−x cos z, 0 Z 0. If F were a gradient field §f of class C2 , then, by Theorem 4.3,
§ * F D § * §f D 0.
44. Note that § · F D y2 C 1 C ez C x2 ez > 0 for all x, y, z ∈ R3 . But if F D § * G, then § · F D § · § * G K 0
for any vector field G of class C2 . Thus F Z § * G.
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“runall” — 2005/8/1 — 15:11 — page 181 — #1
C H A P T E R
4
Maxima and Minima in Several Variables
4.1 DIFFERENTIALS AND TAYLOR’S THEOREM
In Exercises 1–7 we will first calculate f x, f x, ... , f k x and f a, f a, ... , f k a. Then we’ll plug into the
formula for Taylor’s theorem in one variable (Theorem 1.1 in the text):
pk x D f a C f ax − a C · · · C
1. Here a D 0 and k D 4:
f k a
x − ak .
k!
f 0 D 1
f n 0 D 2n
f x D e2x
f n x D 2n e2x
so
4 2
8
16 4
x C x3 C
x
2
6
24
4
2
D 1 C 2x C 2x2 C x3 C x4 .
3
3
p4 x D 1 C 2x C
2. Here a D 0 and k D 3:
f x D ln1 C x
f 0 D 0
1
f x D
1 C x
1
f x D −
1 C x2
f 0 D 1
f x D −2 f 0 D −1
−1
1 C x3
f 0 D 2,
so
1 2
2
x C x3
2
6
1 2
1 3
Dx − x C x .
2
3
p3 x D 0 C x −
3. Here a D 1 and k D 4:
f x D
1
x2
f x D −
f x D
f 1 D 1
2
x3
6
x4
24
x5
120
f x D
x6
f x D −
f 1 D −2
f 1 D 6
f 1 D −24
f 1 D 120,
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“runall” — 2005/8/1 — 15:11 — page 182 — #2
182
Chapter 4
Maxima and Minima in Several Variables
so
6
24
120
x − 12 −
x − 13 C
x − 14
2
6
24
p4 x D 1 − 2x − 1 C
D 1 − 2x − 1 C 3x − 12 − 4x − 13 C 5x − 14 .
Students sometimes forget that the Taylor polynomial depends on the choice of a. Some texts include the parameter a
in the notation to stress this fact. A nice way to remind your students of this dependence on a is to either assign Exercises
4 and 5 or 6 and 7 together.
We’ll do the scratch work for both Exercises 4 and 5 together:
√
f x D x
f 1 D 1
1
f x D √
2 x
−1
f x D
4x3/2
3
f x D
8x5/2
f 1 D
1
2
f 1 D −
f 1 D
f 9 D 3
f 9 D
1
4
3
8
1
6
1
108
1
f 9 D
.
648
f 9 D −
4. Here a D 1 and k D 3 so, using the work above:
p3 x D 1 C
1
1
1
x − 1 − x − 12 C
x − 13 .
2
8
16
5. Here a D 9 and k D 3 so, using the work above:
p3 x D 3 C
1
1
1
x − 9 −
x − 92 C
x − 93 .
6
216
3888
We’ll do the scratch work for both Exercises 6 and 7 together:
f x D sin x
f x D cos x
f x D − sin x
f x D − cos x
f x D sin x
f x D cos x
f 0 D 0
f 0 D 1
f 0 D 0
f 0 D −1
f 0 D 0
f 0 D 1
f π/2 D 1
f π/2 D 0
f π/2 D −1
f π/2 D 0
f π/2 D 1
f π/2 D 0.
6. Here a D 0 and k D 5 so, using the work above:
p5 x D x −
x3
x5
C
.
6
120
7. Here a D π/2 and k D 5 so, using the work above:
p5 x D 1 −
x − π/24
x − π/22
C
.
2
24
Three notes:
• It makes sense to assign Exercises 8, 9, 14 and 17 together as they explore the same function. Exercise 12 is
a higher-dimensional analogue.
• In Exercises 8–13, we again do the preliminary calculations and then substitute into the formulas given in
Theorem 1.3
p1 x D f a C Dfax − a
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“runall” — 2005/8/1 — 15:11 — page 183 — #3
Section 4.1
Differentials and Taylor’s Theorem
183
and Theorem 1.5
n
1 n
fxi xj axi − ai xj − aj 2 i,jD1
p2 x D f a C fxi axi − ai C
iD1
D p1 x C
1 n
fxi xj axi − ai xj − aj .
2 i,jD1
• Just as in the one-variable versions of Taylor’s theorem, note the lower degree polynomials are contained in
the expressions for the higher degree ones.
We’ll do the scratch work for both Exercises 8 and 9 together:
1
x2 C y 2 C 1
−2x
fx x, y D
x2 C y2 C 12
−2y
fy x, y D
2
x C y2 C 12
f x, y D
fxx x, y D
6x2 − 2y2 − 2
x2 C y2 C 13
6y2 − 2x2 − 2
x2 C y2 C 13
8xy
fxy x, y D
x2 C y2 C 13
fyy x, y D
f 0, 0 D 1
f 1, −1 D 1/3
fx 0, 0 D 0
fx 1, −1 D −2/9
fy 0, 0 D 0
fy 1, −1 D 2/9
fxx 0, 0 D −2
fxx 1, −1 D 2/27
fyy 0, 0 D −2
fyy 1, −1 D 2/27
fxy 0, 0 D 0
fxy 1, −1 D −8/27
8. a D 0, 0 so, using the work above:
p1 x D f 0, 0 C Df0, 0x D 1 and
p2 x D p1 x C
1
fxx 0, 0x2 C 2fxy 0, 0xy C fyy 0, 0y2 2
D 1 − x2 − y 2 .
9. a D 1, −1 so, using the work above:
p1 x D f 1, −1 C Df1, −1x − 1, −1 D
2
1
C −
3
9
x − 1
2
y C 1 9
2x − 1
2y C 1
1
−
C
and
3
9
9
1
p2 x D p1 x C fxx 1, −1x − 12 C 2fxy 1, −1x − 1y C 1 C fyy 1, −1y C 12 2
D
D
2x − 1
2y C 1
x − 12
8x − 1y C 1
y C 12
1
−
C
C
−
C
.
3
9
9
27
27
27
10. Here a D 0, 0 and
f x, y D e2xCy
f 0, 0 D 1
fx x, y D 2e2xCy
fx 0, 0 D 2
fy x, y D e2xCy
fxx x, y D 4e2xCy
fyy x, y D e2xCy
fxy x, y D 2e2xCy
fy 0, 0 D 1
fxx 0, 0 D 4
fyy 0, 0 D 1
fxy 0, 0 D 2,
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“runall” — 2005/8/1 — 15:11 — page 184 — #4
184
Chapter 4
Maxima and Minima in Several Variables
so
p1 x D f 0, 0 C Df0, 0x D 1 C 2x C y
1
4x2 C 22xy C y2 2
p2 x D 1 C 2x C y C
D 1 C 2x C y C 2x2 C 2xy C
11. Here a D 0, π and
and
y2
2
f x, y D e2x cos 3y
f 0, π D −1
fx x, y D 2e2x cos 3y
fy x, y D −3e2x sin 3y
fxx x, y D 4e2x cos 3y
fyy x, y D −9e2x cos 3y
fxy x, y D −6e2x sin 3y
fx 0, π D −2
fy 0, π D 0
fxx 0, π D −4
fyy 0, π D 9
fxy 0, π D 0,
so
p1 x D −1 − 2x
and
1
−4x2 C 9y − π2 2
9
D −1 − 2x − 2x2 C y − π2 .
2
p2 x D −1 − 2x C
12. Here a D 0, 0, 0 and there is quite a bit of symmetry so we’ll only calculate:
1
x 2 C y 2 C z2 C 1
−2x
fx x, y, z D
2
2
x C y C z2 C 12
6x2 − 2y2 − 2z2 − 2
fxx x, y, z D
x2 C y2 C z2 C 13
8xy
fxy x, y D
2
2
x C y C z2 C 13
f x, y, z D
f 0, 0, 0 D 1
fx 0, 0, 0 D 0 D fy 0, 0, 0 D fz 0, 0, 0
fxx 0, 0, 0 D −2 D fyy 0, 0, 0 D fzz 0, 0, 0
fxy 0, 0, 0 D 0 D fxz 0, 0, 0 D fyz 0, 0, 0
so
p1 x D 1 and
p2 x D 1 C
1
−2x2 − 2y2 − 2z2 2
D 1 − x 2 − y 2 − z2 .
13. Again a D 0, 0, 0 and there is quite a bit of symmetry so we’ll only calculate:
f x, y, z D sin xyz
f 0, 0, 0 D 0
fx x, y, z D yz cos xyz
fx 0, 0, 0 D 0 D fy 0, 0, 0 D fz 0, 0, 0
fxx x, y, z D −y2 z2 sin xyz
fxx 0, 0, 0 D 0 D fyy 0, 0, 0 D fzz 0, 0, 0
fxy
x, y D z cos xyz − xyz2 sin xyz
so
p1 x D 0
fxy 0, 0, 0 D 0 D fxz 0, 0, 0 D fyz 0, 0, 0
and p2 x D 0.
14. From Exercise 8 we can read off that the Hessian Hf0, 0 D −2
0
0
.
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“runall” — 2005/8/1 — 15:11 — page 185 — #5
Section 4.1
Differentials and Taylor’s Theorem
185
15. f x, y, z D x3 C x2 y − yz2 C 2z3
fx x, y, z D 3x2 C 2xy
fy x, y, z D x2 − z2
fz x, y, z D −2yz C 6z2
fxx x, y, z D 6x C 2y
fyx x, y, z D 2x
fzx x, y, z D 0
fxy x, y, z D 2x
fyy x, y, z D 0
fzy x, y, z D −2z
fxz x, y, z D 0
fyz x, y, z D −2z
fzz x, y, z D −2y C 12z

so
6
Hf1, 0, 1 D  2
0

0
−2  .
12
2
0
−2
16. f x, y, z D e2x−3y sin 5z
fx x, y, z D 2e2x−3y sin 5z
fy x, y, z D −3e2x−3y sin 5z
fz x, y, z D 5e2x−3y cos 5z
fxx x, y, z D 4e2x−3y sin 5z
fyx x, y, z D −6e2x−3y sin 5z
fzx x, y, z D 10e2x−3y cos 5z
fxy x, y, z D −6e2x−3y sin 5z
fyy x, y, z D 9e2x−3y sin 5z
fzy x, y, z D −15e2x−3y cos 5z
fxz x, y, z D 10e2x−3y cos 5z
fyz x, y, z D −15e2x−3y cos 5z
fzz x, y, z D −25e2x−3y sin 5z

so
0
Hf0, 0, 0 D  0
10
0
0
−15

10
−15  .
0
For Exercises 17–19 you’ll need formula (10): p2 x D f a C Dfah C 1/2hT Hfah where h D x − a.
17. Use the work from Exercises 8 and 14:
p2 x D f 0, 0 C Df0, 0x C
D1 C
1
2
x
y
−2
0
1 T −2
x 0
2
0
x
−2 0
x
.
−2 y 18. Use the work from Exercise 15:
−2
0
1
x − 1, 0, 1
x − 1, 0, 1T 0 −2 2



6
2
0
x − 1
.
x − 1 y z − 1  2
0 −2  
y
0 −2 12
z − 1
p2 x D f 1, 0, 1 C Df1, 0, 1x − 1, 0, 1 C

D3 C
3
0
6

x − 1

 C 1
y
2
z − 1
Exercises 19 and 20 are related and could be assigned together. To make it a cohesive single problem, you may want
to tell the students to use the function from Exercise 20 in place of the function given in Exercise 19.
19. The function is f x1 , x2 , ... , xn D ex1 C2x2 C···Cnxn .
(a) Df x1 , x2 , ... , xn D ex1 C2x2 C···Cnxn 1 2 · · · n , and therefore Df 0, 0, ... , 0 D
1 2 · · · n . Taking second derivatives and evaluating at the origin results in:


1 2
3 ··· n
 2 4
6 · · · 2n 


 3 6
9 · · · 3n 


.
Hf0, 0, ... , 0 D 
 #
#
#
# 
..


#
#
# 
 #
.
 #
#
#
# 
n 2n 3n · · · n2
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“runall” — 2005/8/1 — 15:11 — page 186 — #6
186
Chapter 4
Maxima and Minima in Several Variables
(c) Since (c) follows immediately from (a) we will skip (b) for a moment.
1 T
x Hf0, 0, ... , 0x
2

p2 x D f 0, 0, ... , 0 C Df0, 0, ... , 0x C

D1 C
1
2
···
n

x1
 x2 




 #  C 1


2
 # 
 # 
xn
x1
x2
···
1
 2

 3


 #

 #
 #
n
xn
2
4
6
3
6
9
···
···
···
#
#
#
2n
#
#
#
3n
..



n
x1

2n  

 x2 
3n 


 # .


#  # 

#  # 
# 
xn
n2
.
···
(b) Now we can read the answer to (b) right off of our answer to (c).
p1 x D 1 C x1 C 2x2 C · · · C nxn
and
p2 x D 1 C x1 C 2x2 C · · · C nxn C
1 n
ijxi xj .
2 i,jD1
20. This is an extension of a special case of Exercise 19. Note that fxi xj xk 0, 0, 0 D ijk so
p3 x D 1 C x C 2y C 3z C
C
1 2
x C 4y2 C 9z2 C 4xy C 6xz C 12yz
2
1 3
x C 8y3 C 27z3 C 6x2 y C 9x2 z C 12xy2 C 36y2 z C 27xz2 C 54yz2 C 36xyz.
6
21. Dfx, y, z D 4x3 C 3x2 y − z2 C 2xy C 3y
x3 C 6y2 C x2 C 3x

12x2 C 6xy C 2y

Hfx, y, z D
3x2 C 2x C 3
−2z
−2xz − 1 and
3x2 C 2x C 3
12y
0

−2z
0 .
−2x
The only non-zero third derivatives are
fxxx x, y, z D 24x C 6y
fxxy x, y, z D 6x C 2
fxzz x, y, z D −2
fyyy x, y, z D 12
and their permutations.

(a) Here a D 0, 0, 0 so f 0, 0, 0 D 2, Df0, 0, 0 D
p3 x D 2 − z C 3xy C
0
0
0
−1 , and Hf0, 0, 0 D  3
0
3
0
0

0
0 .
0
1
6x2 y − 6xz2 C 12y3 6
D 2 − z C 3xy C x2 y − xz2 C 2y3 .

(b) Here f 1, −1, 0 D −4, Df1, −1, 0 D
−4
11
4
−1 , and Hf1, −1, 0 D  8
0
8
−12
0

0
0 .
−2
p3 x D −4 − 4x − 1 C 11y C 1 − z
1
[4x − 12 C 16x − 1y C 1 − 12y C 12 − 2z2 ]
2
1
C [18x − 13 C 38x − 12 y C 1 − 32x − 1z2 C 12y C 13 ]
6
C
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“runall” — 2005/8/1 — 15:11 — page 187 — #7
Section 4.1
Differentials and Taylor’s Theorem
187
D −4 − 4x − 1 C 11y C 1 − z C 2x − 12 C 8x − 1y C 1 − 6y C 12 − z2
C 3x − 13 C 4x − 12 y C 1 − x − 1z2 C 2y C 13 .
Exercises 22 and 26 are used in Exercise 27 (a) and (b). From Definition 1.4, the total differential of f is
n
⭸f
dfa, h D a dxi .
iD1 ⭸xi
22. f x, y D x2 y3 so dfx, y, h D 2xy3 dx C 3x2 y2 dy.
23. f x, y, z D x2 C 3y2 − 2z3 so dfx, y, z, h D 2x dx C 6y dy − 6z2 dz.
24. f x, y, z D cosxyz so dfx, y, z, h D −yz sinxyz dx − xz sinxyz dy − xy sinxyz dz.
25. f x, y, z D ex cos y C ey sin z so dfx, y, z, h D ex cos y dx C −ex sin y C ey sin z dy C ey cos z dz.
√
26. f x, y, z D 1/ xyz so dfx, y, z, h D − 1 xyz−3/2 yz dx C xz dy C xy dz.
2
27. (a) Use the function from Exercise 22: f x, y D x2 y3 with x D 7, y D 2, dx D .07, and dy D −.02. So
7.072 1.983 L 72 23 C df 7, 2, .07, −.02 D 2723 .07 C 372 22 −.02
D −3.92 D 388.08.
√
(b) Use the function from Exercise 26: f x, y, z D 1/ xyz with x D 4, y D 2, z D 2, dx D .1, dy D −.04,
and dz D .05. So
1
1
1
L − 16−3/2 4.1 C 8−.04 C 8.05
2
4.11.962.05
422
D
1
1
−
.48 D .24625.
4
128
(c) Here the function is f x, y, z D x cosyz with x D 1, y D π, z D 0, dx D .1, dy D −.03, and
dz D .12. So
1.1 cosπ − 0.030.12 L 1 C cos 0.1 − π sin 0.12 D 1.1.
28. dgx, y, z, h D 3x2 − 2y C 2xz dx C −2x dy C x2 C 7 dz, so dg1, −2, 1, h D 9 dx − 2 dy C 8 dz.
This means that changes in x have the most effect.
29. Although students will probably solve this more formally, they should see that, intuitively, changes in the upper
left entry are multiplied by the largest number so that is the entry for which the value of the determinant is most
sensitive.
30. r D 2, dr D .1, h D 3, and dh D .05.
(a) V D πr2 h, so dV D 2πrh dr C πr2 dh D 2π23.1 C π22 .05 D 1.4π.
(b) S D 2πrh C 2πr2 , so dS D 2πh C 4πr dr C 2πr dh D 2π3 C 4π2.1 C 2π2.05 D 1.6π.
31. We are told that dr D dh and know that V D 1/3πr2 h. So dV D 2/3πrh dr C 1/3πr2 dh D 28π/3 dr.
Now we want |dV | to be at most .2 so |dV | D 28π/3|dr| … .2 or |dr| … .3/14π L .0068209.
32. V D xyz where x D 3, y D 4, z D 2 and we assume that dx D dy D dz. So dV D 42 dx C 32 dy C
34 dz D 26dx. We want |dV | … .2 so |dx| … .2/26 L .00769. This is a percentage error of .2/24 D .8333%.
33. (a) We do the preliminary calculations:
f x, y D cos x sin y
f 0, π/2 D 1
fx x, y D − sin x sin y
fx 0, π/2 D 0
fy x, y D cos x cos y
fy 0, π/2 D 0
fxx x, y D − cos x sin y
fxx 0, π/2 D −1
fyy x, y D − cos x sin y
fyy 0, π/2 D −1
fxy x, y D − sin x cos y
fxy 0, π/2 D 0
So p2 x D 1 − x2 /2 − y − π/22 /2.
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“runall” — 2005/8/1 — 15:11 — page 188 — #8
188
Chapter 4
Maxima and Minima in Several Variables
(b) We’ll just follow the estimate in Example 12 in the text: “since all partial derivatives of f will be the product
of sines and cosines and hence no larger than 1 in magnitude” and |h1 | and |h2 | are each no more than .3,
|R2 0, π/2, h1 , h2 | …
1
1
|h1 |3 C 3h21 |h2 | C 3|h1 |h22 C |h2 |3 … 8 · 0.33 D .036.
6
6
34. (a) We do the preliminary calculations:
f x, y D exC2y
f 0, 0 D 1
fx x, y D exC2y
fx 0, 0 D 1
fy x, y D 2exC2y
fy 0, 0 D 2
xC2y
fxx x, y D e
fxx 0, 0 D 1
fyy x, y D 4exC2y
fyy 0, 0 D 4
fxy x, y D 2exC2y
fxy 0, 0 D 2.
So p2 x D 1 C x C 2y C x2 /2 C 2xy C 2y2 .
(b) This time each third derivative has a factor of exC2y in it. Each derivative with respect to y brings out an
additional factor of two. Here |h1 | and |h2 | are no more than .1 and on our set exC2y … e.3 < 2. So
1
1
|R2 0, 0, h1 , h2 | … 2 |h1 |3 C 6h21 |h2 | C 12|h1 |h22 C 8|h2 |3 … 27 · 0.13 D .009.
6
3
4.2 EXTREMA OF FUNCTIONS
1. f x, y D 4x C 6y − 12 − x2 − y2 so fx x, y D 4 − 2x, fy x, y D 6 − 2y, fxx x, y D −2, fxy x, y D 0,
and fyy x, y D −2.
(a) To find the critical point we will set each of the first partial derivatives equal to 0 and solve: fx x, y D 0
when 4 − 2x D 0 or when x D 2 and fy x, y D 0 when 6 − 2y D 0 or when y D 3. So f has a unique
critical point at (2, 3).
(b) The increment
f D f 2 C x, 3 C y − f 2, 3
D 42 C x C 63 C y − 12 − 2 C x2 − 3 C y2
− 42 C 63 − 12 − 22 − 32 D −x2 − y2 .
This tells us that little changes in x and/or y result in a decrease in the value of f . This means that f must
have a local maximum at (2, 3).
−2
0
(c) The Hessian is Hf2, 3 D so d1 D −2 and d2 D 4 so by the second derivative test, f has a
0 −2 local maximum at (2, 3).
2. gx, y D x2 − 2y2 C 2x C 3 so gx x, y D 2x C 2, gy x, y D −4y, gxx x, y D 2, gxy x, y D 0, and
gyy x, y D −4.
(a) To find the critical point we will set each of the first partial derivatives equal to 0 and solve: gx x, y D 0
when 2x C 2 D 0 or when x D −1 and gy x, y D 0 when −4y D 0 or when y D 0. So g has a unique
critical point at −1, 0.
(b) The increment
g D g−1 C x, y − g−1, 0
D −1 C x2 − 2y2 C 2−1 C x C 3 − −12 C 2−1 C 3
D x2 − 2y2 .
This tells us that any changes in x result in an increase in the value of g and little changes in y result in a
decrease in the value of g. This means that f must have a saddle at −1, 0.
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“runall” — 2005/8/1 — 15:11 — page 189 — #9
Section 4.2
(c) The Hessian is Hg−1, 0 D 2
0
Extrema of Functions
189
0
so d1 D 2 and d2 D −8, so by the second derivative test, g has
−4 a saddle at −1, 0.
In Exercises 3–20, most of the mistakes will be algebra mistakes made in solving for the critical points. For Exercises
3–14, you are using the familiar rule for the second derivative test at a point a D a, b where fx a D 0 D fy a. The
determinant of the Hessian is often referred to as the discriminant:
Da, b D |Hfa, b| D fxx a, bfyy a, b − [fxy a, b]2 .
The second derivative test (see Example 4) is then
• if Da, b > 0 and
— if fxx a, b > 0 then f has a local minimum at (a, b)
— if fxx a, b < 0 then f has a local maximum at (a, b)
• if Da, b < 0 then f has a saddle at (a, b).
• Otherwise the test tells us nothing.
In many calculus classes students never see the extension of this test to higher dimensions. In Exercises 15–20, the students
will need to use the R3 version of the second derivative test.
3. f x, y D 2xy − 2x2 − 5y2 C 4y − 3, so fx x, y D 2y − 4x and fy x, y D 2x − 10y C 4. At a critical
point 2y − 4x D 0 so y D 2x. Also 4 D 10y − 2x D 10y − y D 9y so y D 4/9 and x D 2/9. So f has a
critical point at (2/9, 4/9).
−4
2
We easily calculate the Hessian Hf D so d1 D −4 and d2 D 36. So f has a local maximum at
2 −10 (2/9, 4/9).
2y
2x
and fy x, y D
. The only critical
4. f x, y D lnx2 C y2 C 1, so fx x, y D
2
2
2
x C y C 1
x C y2 C 1
point of f is at the origin.
−2x2 C 2y2 C 2
2x2 − 2y2 C 2
, fyy x, y D
, and also
The second derivatives are fxx x, y D
2
2
2
x C y C 1
x2 C y2 C 12
4xy
2 0
fxy x, y D
. At the origin, the Hessian Hf0, 0 D so d1 D 2 and d2 D 4. So f
0 2 x2 C y2 C 12
has a local minimum at (0, 0).
5. f x, y D x2 C y3 − 6xy C 3x C 6y, so fx x, y D 2x − 6y C 3 and fy x, y D 3y2 − 6x C 6.
At a critical point for f , 2x D 6y − 3 and 0 D 3y2 − 6x C 6 so 0 D y2 − 2x C 2. Substituting,
0 D y2 − 6y C 5 D y − 1y − 5. We have critical points at (3/2, 1) and (27/2, 5).
The second derivatives are fxx x, y D 2, fyy x, y D 6y, and fxy x, y D −6. d1 D 2 and d2 D 12y − 36.
In other words, d1 is always positive and d2 is positive when y D 5 and negative when y D 1 so by the second
derivative test f has a saddle point at (3/2, 1) and f has a local minimum at (27/2, 5).
6. f x, y D y4 − 2xy2 C x3 − x, so fx x, y D −2y2 C 3x2 − 1√
and fy x, y D 4y3 − 4xy D 4yy2 − x.
2
At a critical point for f , y D 0 or y D x. If y D 0 then x D ;1/ 3. If y2 D x then 0 D 3x2 − 2x − 1 D
3x C 1x − 1.
√ This gives us that x D 1 or x D −1/3 but x can’t be negative. So there are four critical
points for f : ;1/ 3, 0, and 1, ;1.
The second derivatives are fxx x, y D 6x, fyy x, y D 12y2 − 4x, and fxy x, y D −4y. d1 D 6x and
d2 D 89xy2 − 2y2 − 3x2 . We’ll calculate di at each critical point to classify them:
Critical Point
d1
d2
Classification
√
√
1/ √
3, 0
6/√ 3 −8 saddle
−1/ 3, 0 −6/ 3 −8 saddle
1, −1
6
32 local minimum
6
32 local minimum
(1, 1)
1
8
1
1
8
C , so fx x, y D y −
and fy x, y D x −
. At a critical point for f , x D
7. f x, y D xy C
2
2
x
y
x
y
y2
8
4
3
and y D
D 8y so 0 D y8y − 1 so either y D 0 or y D 1/2. Since y D 0 is not in the domain of f , the
x2
only critical point of f is at (4, 1/2).
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“runall” — 2005/8/1 — 15:11 — page 190 — #10
190
Chapter 4
Maxima and Minima in Several Variables
16
2
16
32
, fyy x, y D , and fxy x, y D 1. d1 D
and d2 D
− 1.
x3
y3
x3
x3 y 3
At our critical point both d1 and d2 are positive so (4, 1/2) is a local minimum.
8. f x, y D ex sin y so fx x, y D ex sin y and fy x, y D ex cos y. There are no values of x and y for which both
first partials are 0 so there are no critical points.
9. f x, y D e−y x2 − y2 , so fx x, y D 2xe−y and fy x, y D −e−y x2 − y2 C 2y. At a critical point for
f , x D 0 and 0 D −y2 C 2y D −yy − 2 so the critical points of f are at (0, 0) and (0, 2).
The second derivatives are fxx x, y D 2e−y , fyy x, y D e−y x2 − y2 C 4y − 2, and fxy x, y D −2xe−y .
d1 > 0 and d2 0, y D −2e−2y y2 − 4y C 2. In other words, d1 is always positive and d2 is negative when
y D 0 and positive when y D 2 so by the second derivative test f has a saddle point at (0, 0) and f has a local
minimum at (0, 2).
10. f x, y D x C y − x2 y − xy2 , so fx x, y D 1 − 2xy − y2 and fy x, y D 1 − 2xy − x2 . At a critical
√
point for f , x2 D y2 so x D ;y. If x D y, then 0 D 1 − 2xy − y2 D 1 − 3x2 so x D y D ;1/ 3. If
x D −y,√then 0√
D 1 − 2xy − y2 D 1 C y2 for which there are no real solutions. So the critical points for f
are ;1/ 3, 1/ 3.
The second derivatives are fxx x, y D −2y, fyy x, y D −2x, and fxy x, y D −2x − 2y. d1 D −2y and
2 − 4xy − 4y2 . At the critical points d is negative and d is non-zero so f has a saddle point at
d2 D −4x√
2
1
√
both ;1/ 3, 1/ 3.
11. f x, y D x2 − y3 − x2 y C y, so fx x, y D 2x − 2xy D 2x1 − y√and fy x, y D −3y2 − x2 C 1. At
a critical point for f , either x D 0 or y√
D 1. When x D 0, y must be ;1/ 3. No solution corresponds to y D 1,
So the critical points for f are 0, ;1/ 3.
− 2y, fyy x, y D −6y, and fxy x, y D −2x. d1 D 2 − 2y and
The second derivatives are fxx x, y D 2 √
2 − 4x2 . At 0, −1/ 3, d is positive and d is positive so f has a local minimum at
C
12y
d2 D −12y
1
2
√
√
√
0, −1/ 3. At 0, 1/ 3, d1 is positive and d2 is negative so f has a saddle point at 0, 1/ 3.
12. f x, y D e−x x2 C 3y2 , so fx x, y D 2x − x2 − 3y2 e−x and fy x, y D 6ye−x . From fy we see that
at a critical point for f , we must have y D 0. Plugging back into fx we conclude that there are critical points at
(0, 0) and at (2, 0).
The second derivatives are fxx x, y D 2 − 4x C x2 C 3y2 e−x , fyy x, y D −6e−x , and fxy x, y D
−6ye−x . d1 D 2 − 4x C x2 C 3y2 e−x and d2 D 6e−2x 1 − 4x C x2 − 3y2 . At (0, 0), d1 and d2 are
positive so f has a local minimum at (0, 0). At (2, 0), d1 and d2 are negative so f has a saddle point at (2, 0).
13. f x, y D 2x − 3y C ln xy, so fx x, y D 2 C 1/x and fy x, y D −3 C 1/y. The critical point is
−1/2, 1/3.
The second derivatives are fxx x, y D −1/x2 , fyy x, y D −1/y2 , and fxy x, y D 0. d1 D −1/x2 and
d2 D 1/x2 y2 . At −1/2, 1/3, d1 is negative and d2 is positive so f has a local max at −1/2, 1/3.
14. f x, y D cos x sin y, so fx x, y D − sin x sin y and fy x, y D cos x cos y. The critical points are of the form
nπ, π/2 C mπ and π/2 C nπ, mπ where m and n are integers.
The second derivatives are fxx x, y D − cos x sin y, fyy x, y D − cos x sin y, and fxy x, y D − sin x cos y.
d1 D − cos x sin y and d2 D cos2 x sin2 y − sin2 x cos2 y. At points of the form nπ, π/2 C mπ, d1 alternates
between negative and positive values while d2 is positive so f has an alternating string of local maxs and mins
at such points. At the point 0, π/2, for example, f has a local max. At points of the form π/2 C nπ, mπ,
d1 D 0 and d2 is negative so such points are saddle points.
15. f x, y, z D x2 − xy C z2 − 2xz C 6z, so fx x, y, z D 2x − y − 2z, fy x, y, z D −x and fz x, y, z D
2z − 2x C 6. From the second equation, x D 0. From the third, then, z D −3 and from the first it follows that
y D 6.
The second derivatives are fxx x, y, z D 2, fyy x, y, z D 0, fzz x, y, z D 2, fxy x, y, z D −1,
fxz x, y, z D −2 and fyz x, y, z D 0. d1 D 2, d2 D −1 and d3 D −2 so f has a saddle point at 0, 6, −3.
16. f x, y, z D x2 C 2y2 C 1 cos z, so fx x, y, z D 2x cos z, fy x, y, z D 4y cos z and fz x, y, z D
−x2 C 2y2 C 1 sin z. From the third equation, z D nπ. The other two equations imply that x and y both
are 0. So the critical points are of the form 0, 0, nπ.
The second derivatives are fxx x, y, z D 2 cos z, fyy x, y, z D 4 cos z, fzz x, y, z D −x2 C 2y2 C 1 cos z,
fxy x, y, z D 0, fxz x, y, z D −2x sin z and fyz x, y, z D −4y sin z. d1 D 2 cos z and d2 D 8 cos2 z. It is
easier to calculate d3 at our critical point. In this case d3 0, 0, nπ D <8 while d1 0, 0, nπ D ;2, d2 D 8. So
f has saddle points at 0, 0, nπ.
17. f x, y, z D x2 C y2 C 2z2 C xz so fx x, y, z D 2x C z, fy x, y, z D 2y, and fz x, y, z D 4z C x. It
is easy to see that the only critical point is at the origin.
The second derivatives are fxx x, y D
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“runall” — 2005/8/1 — 15:11 — page 191 — #11
Section 4.2
Extrema of Functions
191

2 0 1
The Hessian is Hf D  0 2 0  so d1 D 2, d2 D 4, and d3 D 14. By the second derivative test, f has a local
1 0 4
minimum at (0, 0, 0).
18. f x, y, z D x3 C xz2 − 3x2 C y2 C 2z2 so fx x, y, z D 3x2 C z2 − 6x, fy x, y, z D 2y, and
fz x, y, z D 2xz C 4z D 2zx C 2. We see immediately that at a critical point of f , y D 0 and either
z D 0 or x D −2. If z D 0 then 0 D 3x2 − 6x D 3xx − 2 so x D 0 or x D 2. If x D −2 then z2 D −24
for which there are no real solutions. We conclude
0, 0).
 that f has critical points at (0, 0, 0) and (2, 
6x − 6 0
0
6x − 6 0
2z
. This makes it
 so Hfx, 0, 0 D 
0
2
0
0
2
0
The Hessian is Hf D 
0
0 2x C 4
2z
0 2x C 4
easier to calculate d1 x, 0, 0 D 6x − 6, d2 x, 0, 0 D 2d1 x, 0, 0, and d3 x, 0, 0 D 2x C 4 d2 . At (0, 0,
0) all three di ’s are negative and at (2, 0, 0) all three are positive. By the second derivative test, f has a saddle
point at (0, 0, 0) and a local minimum at (2, 0, 0).
1
1
19. f x, y, z D xy C xz C 2yz C so fx x, y, z D y C z − , fy x, y, z D x C 2z, and fz x, y, z D x C 2y.
x
x2
1
1
We see immediately that at a critical point of f , y D z so both 2z D −x and 2z D
so −x D
so x D −1.
x2
x2
Therefore, f has a critical
point at −1,

 1/2, 1/2.
2/x3 1 1
The Hessian is Hf D  1
0 2  so d1 −1, 1/2, 1/2 D −2, d2 −1, 1/2, 1/2 D −1, and
1
2 0
d3 −1, 1/2, 1/2 D 12. This is the case of the second derivative test where the conditions are valid but neither
of the first two cases holds so f has a saddle point at −1, 1/2, 1/2.
20. f x, y, z D ex x2 − y2 − 2z2 so fx x, y, z D ex x2 C 2x − y2 − 2z2 , fy x, y, z D −2yex , and
fz x, y, z D −4zex . We see immediately that at a critical point of f , y D z D 0 and therefore 0 D x2 C 2x D
xx C 2. The two critical points of f are (0, 0, 0) and −2, 0, 0.


ex x2 C 4x C 2 − y2 − 2z2 −2yex −4zex
 so
The Hessian is Hf D 
−2yex
−2ex
0
0
−4ex
−4zex


ex x2 C 4x C 2
0
0
Hfx, 0, 0 D 
0
−2ex
0 .
0
0
−4ex

For (0, 0, 0), d1 > 0, d2 < 0, and d3 < 0 so f has a saddle at (0, 0, 0). For −2, 0, 0, d1 < 0, d2 > 0, and
d3 < 0 so f has a local maximum at −2, 0, 0.
6x2y3 − 3y2 − 36y C 2
2y3 − 3y2 − 36y C 2
and fy x, y D
so fx x, y D
21. (a) f x, y D
2
1 C 3x
1 C 3x2 6y2 − y − 6
6y − 3y C 2
D
. From fy we see that either y D 3 or y D −2. Neither of these
1 C 3x2
1 C 3x2
values makes fx D 0 so x D 0. The critical points for f are 0, −2 and (0, 3).
(b)


63x − 13x C 12y3 − 3y2 − 36y C 2
36xy − 3y C 2
−


3x2 C 13
3x2 C 12


Hf D 
 and


36xy − 3y C 2
62y − 1
−
3x2 C 12
3x2 C 1
Hf0, y D −62y3 − 3y2 − 36y C 2
0
0
62y − 1
.
At 0, −2 we find that d1 < 0 and d2 > 0 so f has a local maximum at 0, −2. At (0, 3) we find that
d1 > 0 and d2 > 0 so f has a local minimum at (0, 3).
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192
Chapter 4
Maxima and Minima in Several Variables
22. (a) f x, y D kx2 − 2xy C ky2 so fx x, y D 2kx − 2y and fy x, y D −2x C 2ky. We see that the origin
2k −2
is a critical point for any value of k. The Hessian is so d1 D 2k and d2 D 4k 2 − 4. For f to
−2 2k have a non-degenerate local maximum or minimum d2 > 0 so k 2 − 1 > 0 so either k > 1 or k < −1.
If k > 1, then d1 > 0 and the origin is a non-degenerate local minimum. If k < −1, then d1 < 0 and the
origin is a non-degenerate local maximum.
(b) gx, y, z D kx2 C kxz − 2yz − y2 C kz2 /2 so gx x, y, z D2kx C kz, gy x, y, z D −2z − 2y, and
2k
0
k
gz x, y, z D kx − 2y C kz. The Hessian is  0 −2 −2 . First note that d1 D 2k and d2 D −4k.
k −2
k
These are of opposite signs so a non-degenerate local minimum is not possible. For a non-degenerate local
maximum we need d1 < 0 and d2 > 0 so k < 0. We also need d3 D 2k−k − 4 < 0 so k < −4. So
we have a non-degenerate local maximum when k < −4.
23. If you think of this problem geometrically it should be reasonably straightforward. The slices through the origin
where only one variable is allowed to change are parabolas. They open up if the coefficient of the term containing
that variable is positive and down if it is negative. This tells you that if all of the coefficients are positive then
we have a local minimum, if all of the coefficients are negative then we have a local maximum, and if some are
positive and some are negative then we have a saddle point.
(a) f x, y D ax2 C by2 so fx x, y D 2ax and fy x, y D 2by. Since neither a nor b is 0, the critical point
2a 0
must be the origin. The Hessian is Hf D . The first condition is that d2 > 0 so 4ab > 0 so a
0 2b and b are the same sign. Also, d1 D 2a so when a and b are negative the origin is a local maximum and
when a and b are positive the origin is a local minimum.
(b) f x, y D ax2 C by2 C cz2 so fx x, y, z D 2ax, fy x, y, z D 2by and fzx, y, z D 2cz.Since none
2a 0
0
of a, b and c is 0, the critical point must be the origin. The Hessian is Hf D  0 2b 0 . Again, in
0
0 2c
either case d2 > 0 so 4ab > 0 so a and b are the same sign. Also, d1 D 2a and d3 D 8abc. In either case
d1 and d3 must be the same sign. When a, b and c are negative the origin is a local maximum and when a,
b and c are positive the origin is a local minimum.
(c) Really the analysis is no harder, it is just harder to write down. The function is now f x1 , x2 , ... , xn D
a1 x12 C a2 x22 C · · · C an xn2 . The first derivatives are fxi x1 , x2 , ... , xn D 2ai xi . Because none of the ai
is zero and all of the first derivatives are 0, we conclude that the only critical point is at the origin. The
Hessian is an n * n matrix with zeros everywhere off of the main diagonal and the entry in position (i, i)
is 2ai . We easily calculate di D 2i a1 a2 ... ai . As above, d2 must be positive so both a1 and a2 are of the
same sign. We could continue to argue that d4 D 4a3 a4 d2 so a3 and a4 must be of the same sign. In fact,
we can continue that reasoning to say for k odd, ak and akC1 must be of the same sign. For f to have a
local maximum d1 < 0 so a1 and a2 are both negative. Also, dk D 2ak dk−1 and for k odd dk < 0 so we
can move up through the entries and argue that all of the ai ’s must be negative. Similarly, for f to have a
local minimum all of the ai ’s must be positive.
Note: In Exercises 24–27 we have used a computer algebra system. In fact, I’ve used Mathematica. In Exercise 24,
I’ve included a list of the relevant commands. These were adapted for each of the exercises.
24. We’ll use the following sequence of commands:
• f [x− , y− ] D y4 − 2xy2 C x3 − x
• Solve [{D[f [x, y], x] DD 0, D[f [x, y], y] DD 0}]
• H D {{⭸x,x f [x, y], ⭸x,y f [x, y]}, {⭸y,x f [x, y], ⭸y,y f [x, y]}}
• MatrixForm [H/.{x ' 1, y ' −1}] (since 1, −1 is the critical point found in the second step)
This is how you define the function, solve §f D 0, create the Hessian and display it √
at the critical points. √
In this case we get the following solutions to the simultaneous equations: (−1/3, ;i/ 3), (1, ;1), and (;1/ 3,
0). Let’s examine the real-valued solutions.
6 −4
At (1, 1) the Hessian is . This means that d1 > 0 and d2 > 0 so (1, 1) is a local minimum.
−4
8 6 4
At 1, −1 the Hessian is . This means that d1 > 0 and d2 > 0 so 1, −1 is a local minimum.
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“runall” — 2005/8/1 — 15:11 — page 193 — #13
Section 4.2
Extrema of Functions
193
√
√
0√
−2 3
At −1/ 3, 0 the Hessian is . This means that d1 < 0 and d2 < 0 so −1/ 3, 0 is a
0
4/ 3
saddle point.
√
√
√
0√
2 3
At 1/ 3, 0 the Hessian is . This means that d1 < 0 and d2 < 0 so 1/ 3, 0 is a saddle
0
−4/ 3
point.
25. The commands are the same as those outlined in Exercise 24. The critical points are (0, 0), ; 3/2, 0, and
√
√
;1/ 2, −1/ 2.
0 −3
At (0, 0) the Hessian is . This means that d1 D 0 and d2 < 0 so (0, 0) is a saddle point.
−3 −2 0
6
. Again, d1 D 0 and d2 < 0 so both 3/2, 0 and − 3/2, 0 are
At ; 3/2, 0 the Hessian is 6 −2 saddle points.
√
√
√
√
−6
0
At ;1/ 2, −1/ 2 the Hessian is . This means that d1 < 0 and d2 > 0 so both 1/ 2, −1/ 2
0 −2 √
√
and −1/ 2, 1/ 2 are local maxima.
√
26. We need to slightly alter the commands from the previous two exercises. The command to find the roots specified
by the three first partials is now:
Solve [{D[f [x, y, z], x] DD 0, D[f [x, y, z], y] DD 0, D[f [x, y, z], z] DD 0}].
We also need to change the specification of the Hessian to:
H D {{⭸x,x f [x, y, z], ⭸x,y f [x, y, z], ⭸x,z f [x, y, z]},
{⭸y,x f [x, y, z], ⭸y,y f [x, y, z], ⭸y,z f [x, y, z]},
{⭸z,x f [x, y, z], ⭸z,y f [x, y, z], ⭸z,z f [x, y, z]}}
Finally, it will be helpful to use the computer to calculate the determinant. For Mathematica you type Det[M]
where M is the matrix for which you wish to calculate the determinant.
√
√
√
√ √
√
The critical points
are at 1 − 2 2, − 24 − 2, − 4 − 2, 1
− 2 2, 24 − 2, 4 − 2,
√
√
√
√
√
√
C 2 2, 24 C 2, − 4 C 2, and (0, 0, 0).
1 C 2 2, − 24 C 2,
 4 C 2, 1 
−2
0
0
1 . So d1 < 0, d2 > 0 and d3 < 0 so (0, 0, 0) is a local max.
At (0, 0, 0) the Hessian is  0 −2
0
1 −4


√
√
4 − 2
24 − 2
−2

 √
√
√
√
√
.
At 1 − 2 2, − 24 − 2, − 4 − 2 the Hessian is 
−2
2 2

 4 − 2
√
√
24 − 2
2 2
−4
√
√
√
√
√
So, d1 D −2 < 0 and d2 D 2 > 0 and d3 D 64 − 16 2 > 0 so 1 − 2 2, − 24 − 2, − 4 − 2
is a saddle point.


√
√
−2
− 4 − 2 − 24 − 2


√ √
√
√
√
.
At 1 − 2 2, 24 − 2, 4 − 2 the Hessian is 
4 − 2
−2
2 2

 −
√
√
− 24 − 2
2 2
−4
√
√
√ √
√
So, d1 D −2 < 0 and d2 D 2 > 0 and d3 D 64 − 16 2 > 0 so 1 − 2 2, 24 − 2, 4 − 2 is
a saddle point.


√
√
24 C 2
−2
− 4 C 2


√
√
√
√
√
.
At 1 C 2 2, − 24 C 2, 4 C 2 the Hessian is 
−2
−2 2

 − 4 C 2
√
√
24 C 2
−2 2
−4
√
√
√
√
√
So, d1 D −2 < 0 and d2 D − 2 < 0 and d3 D 64 C 16 2 > 0 so 1 C 2 2, − 24 C 2, 4 C 2
is a saddle point.
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“runall” — 2005/8/1 — 15:11 — page 194 — #14
194
Chapter 4
Maxima and Minima in Several Variables

√
− 24 C 2

√
.
−2
−2 2

√
−2 2
−4
√
√
√ √
√
So, d1 D −2 < 0 and d2 D − 2 < 0 and d3 D 64 C 16 2 > 0 so 1 C 2 2, 24 C 2, − 4 C 2
is a saddle point.
√
√
√
27. The√commands
√
√are extended
√ as they
√ were in
√ Exercise 26. The
√ critical
√ points
√ are (0, 0, 0, 0), − 2, − 2, 2 2, 1,
− 2, 2, 2 2, −1, 2,− 2, −2 2, −1,and 2, 2, −2 2, 1.
−2 0
0 0
 0 0
0 1 

. So d1 D −2 < 0, d2 D 0, d3 D 0, and d4 D −8 < 0, so
At (0, 0, 0, 0) the Hessian is 
0 0 −4 0 
0 1
0 2
(0, 0, 0, 0) is a saddle point.


√
−2 −1 −2√ 2 0


√
√
√

2 1 
−1 √ 0
√
At − 2, − 2, 2 2, 1 the Hessian is 
. So d1 D −2 < 0, d2 D −1 < 0,
 −2 2
2
−4 0 
0
1
0 2
√
√
√
d3 D 16 > 0, and d4 D 32 > 0, so − 2, − 2, 2 2, 1 is a saddle point.

√
−2
1 −2√ 2 0


√
√ √

− 2 1 
√0
√1
At − 2, 2, 2 2, −1 the Hessian is 
. So d1 D −2 < 0, d2 D −1 < 0,
 −2 2 − 2
−4 0 
0
1
0 2
√ √
√
d3 D 16 > 0, and d4 D 32 > 0, so − 2,  2, 2 2, −1 is a saddlepoint.
√
−2
1 2√ 2 0


√
√
√

1 √0
2 1 
At 2, − 2, −2 2, −1 the Hessian is  √
. So d1 D −2 < 0, d2 D −1 < 0,
 2 2
2
−4 0 
0
1
0 2
√
√
√
d3 D 16 > 0, and d4 D 32 > 0, so 2,
is a saddle
 point.
 − 2, −2 2, −1 √
−2 −1 −2√ 2 0


√ √
√

2 1 
−1 √ 0
√
At 2, 2, −2 2, 1 the Hessian is 
. So d1 D −2 < 0, d2 D −1 < 0,
 −2 2
2
−4 0 
0
1
0 2
√ √
√
d3 D 16 > 0, and d4 D 32 > 0, so 2, 2, −2 2, 1 is a saddle point.
28. We want to maximize V D xyz subject to the constraint 2xy C 2xz C 2yz D c. Solve the second equation for
c − 2xy
and substitute to get
zD
2x C 2y
cxy − 2x2 y2
V x, y D
.
2x C 2y


√ √
√
At 1 C 2 2, 24 C 2, − 4 C 2 the Hessian is 

−2
√
4 C 2
√
− 24 C 2
4 C
√
2
y2 2x2 C 4xy − c
x2 2y2 C 4xy − c
and Vy D −
. Since neither x nor
2x C y2
2x C y2
y could be zero (we wouldn’t have a box), a critical point of f occurs when both 2x2 C 4xy − c D 0
and 2y2 C 4xy − c D 0. Solving these together we find that x2 D y2 and since x and y are positive we
conclude that x D y. Substituting back in, 0 D 2x2 C 4xy − c D 2x2 C 4x2 − c D 6x2 − c so x D
c − c/3
c − 2xy
y D c/6. z D
D c/6. So our only critical point is when the box is a cube.
D
2x C 2y
4 c/6
y2 c C 2y2 To conclude that this is a local maximum we see that d1 D −
< 0 and at our critical point
x C y3
2x2 y2 2x2 C 8xy C 2y2 − 3c
2x4 12x2 − 3c
2c − 3c
> 0. So the largest rectangular
D−
D−
d2 D −
x C y4
2x4
8
box with fixed surface area is a cube.
The derivatives are Vx D −
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“runall” — 2005/8/1 — 15:11 — page 195 — #15
Section 4.2
Extrema of Functions
195
29. We will actually minimize the square of the distance (i.e., the sum of the squares of the differences in each
direction): Dx, y D x2 C y2 C 3x − 4y − 242 so Dx x, y D 20x − 24y − 144 and Dy x, y D
34y − 24x C 192. Set these equal to 0 and solve to get that the point on the plane closest to the origin is
36/13, −48/13, −12/13.
Exercises 30–33 force us to check values on the border of our region.
30. f x, y D x2 C xy C y2 − 6y so fx x, y D 2x C y and fy x, y D x C 2y − 6. At a critical point for
f , y D −2x so 6 D x C 2y D −3x. Our only critical point is −2, 4. We need to check the value of f at the
critical point and along the boundary of the region −3 … x … 3, 0 … y … 5.
• f −2, 4 D −12,
• f −3, y D 9 − 9y C y2 has a minimum of −11.25 at y D 4.5 and a maximum of 9 at y D 0,
• f 3, y D 9 − 3y C y2 has a minimum of 27/4 at y D 3/2 and a maximum of 19 at y D 5,
• f x, 0 D x2 which has a minimum of 0 at x D 0 and a maximum of 9 at x D ;3,
• f x, 5 D x2 C 5x − 5 has a minimum of −11.25 at x D −5/2 and a maximum of 19 at x D 3.
The absolute maximum is, therefore, 19 at (3, 5) and the absolute minimum is −12 at −2, 4.
31. f x, y, z D x2 C xz − y2 C 2z2 C xy C 5x so fx x, y, z D 2x C y C z C 5, fy x, y, z D x − 2y, and
fz x, y, z D x C 4z. At a critical point for f , x D 2y D −4z so −5 D 2x C y C z D −8z − 2z C z D −9z.
Our only critical point is −20/9, −10/9, 5/9 which is not within our region. We need to check the value of
f along the boundary of the region −5 … x … 0, 0 … y … 3, 0 … z … 2. This consists of six two-dimensional
faces, twelve one-dimensional edges and eight vertices.
• f x, 0, 0 D x2 C 5x has a minimum of −6.25 at x D −5/2 and a maximum of 0 at x D −5 or 0,
• f x, 0, 2 D x2 C 7x C 8 has a minimum of −4.25 at x D −7/2 and a maximum of 8 at x D 0,
• f x, 3, 0 D x2 C 8x − 9 has a minimum of 25 at x D −4 and a maximum of −9 at x D 0,
• f x, 3, 2 D x2 C 10x − 1 has a minimum of −26 at x D −5 and a maximum of −1 at x D 0,
• f −5, y, 0 D −y2 − 5y has a minimum of −24 at y D 3 and a maximum of 0 at y D 0,
• f 0, y, 0 D −y2 has a minimum of −9 at y D 3 and a maximum of 0 at y D 0,
• f −5, y, 2 D −y2 − 5y − 2 has a minimum of −26 at y D 3 and a maximum of −2 at y D 0,
• f 0, y, 2 D 8 − y2 has a minimum of −1 at y D 3 and a maximum of 8 at y D 0,
• f −5, 0, z D 2z2 − 5z has a minimum of −25/8 at z D 5/4 and a maximum of 0 at z D 0,
• f 0, 0, z D 2z2 has a minimum of 0 at z D 0 and a maximum of 8 at z D 2,
• f −5, 3, z D 2z2 − 5z − 24 has a minimum of −217/8 at z D 5/4 and a maximum of −24 at z D 0,
• f 0, 3, z D 2z2 − 9 has a minimum of −9 at z D 0 and a maximum of −1 at z D 2.
You also must check for extrema on each face and at each vertex. When you do you find: The absolute maximum
is 8 at (0, 0, 2) and the absolute minimum is −191/7 at −32/7, 3, 8/7.
32. In a fit of compassion, the author of the text has not forced Livinia the housefly to walk around the metal plate
in search of the hottest and coldest points. The temperature is T x, y D 2x2 C y2 − y − 3 so Tx x, y D 4x
and Ty x, y D 2y − 1. We have a critical point for T at (0, 1/2) and T 0, 1/2 D 2.75. To check the
temperature of the boundary we note that it is a unit disk and so x D cos θ and y D sin θ. We can rewrite
T θ D 2 cos2 θ C sin2 θ − sin θ C 3 D cos2 θ − sin θ C 4. Then Tθ θ D −2 cos θ sin θ − cos θ D
− cos θ2 sin θ C 1. We, therefore, have critical points on the boundary when cos θ D 0 (so θ D π/2 or 3π/2)
and when sin θ D −1/2 (so θ D 7π/6 or 11π/6). Checking the values we see that T π/2 D 3, T 3π/2 D 5
and T 7π/6 D T 11π/6 D 21/4. We conclude that
√ the coldest spot on the plate is at (0, 1/2) where the
temperature is 11/4 and the two hottest spots are at ; 3/2, −1/2 where the temperature is 21/4.
33. Because the function is “separable”, we can analyze it without calculus. The maximum value for f is 1 and the
minimum value for f is −1. The absolute maximum is achieved at π/2, 0, π/2, 2π, and 3π/2, π. The
absolute minimum is achieved at 3π/2, 0, 3π/2, 2π, and π/2, π.
34.
⭸f
D −2 sin x
⭸x
⭸f
D 3 cos y
⭸y
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“runall” — 2005/8/1 — 15:11 — page 196 — #16
196
Chapter 4
Maxima and Minima in Several Variables
So “ordinary” critical points on {x, y|0 … x … 4, 0 … y … 3} are at 0, π , π, π . (In fact, π, π is the
2
2
2
only critical point that’s actually in the interior of the rectangle.)
Now we look at the boundary of the rectangle:
f1 x D f x, 0 D 2 cos x
f1 x D −2 sin x so critical points at 0, 0, π, 0I
f2 x D f x, 3 D 2 cos x C 3 sin 3
f2 x D −2 sin x so critical points at 0, 3, π, 3I
f3 y D f 0, y D 2 C 3 sin y
f3 y D 3 cos y so critical point at 0, π I
f4 y D f 4, y D 2 cos 4 C 3 sin y
f4 y D 3 cos y so critical point at 4, π .
2
2
Now we compare values:
(x, y)
f x, y D 2 cos x C 3 sin y
π
0, 2 5
π
π, 2 1
2
−2
2 C 3 sin 3 L 2.423
−2 C 3 sin 3 L −1.577
(0, 0)
π, 0
(0, 3)
π, 3
π
4, 2 (4, 0)
(4, 3)
2 cos 4 C 3 L 1.693
2 cos 4 L −1.307
2 cos 4 C 3 sin 3 L −0.884
Thus the absolute minimum occurs at π, 0 and is −2. The absolute maximum occurs at 0, π and is 5.
2
35. The boundary of the closed ball is given by x2 C y2 − 2y C z2 C 4z D 0. Completing the square, we
find x2 C y2 − 2y C 1 C z2 C 4z C 4 D 5 or x2 C y − 12 C z C 22 D 5. (Note also that
x2 C y2 − 2y C z2 C 4z D x2 C y − 12 C z C 22 − 5.)
2
2
2
The function f x, y, z D e1−x −y C2y−z −4z has
fx x, y, z D −2xe1−x −y C2y−z −4z D 0
2
2
2
fy x, y, z D −2y C 2e1−x −y C2y−z −4z D 0
2
2
2
fz x, y, z D −2z − 4e1−x −y C2y−z −4z D 0
2
2
2
when x D 0
when y D 1
when z D −2
So 0, 1, −2 is an interior critical point (the only one). Note that on the boundary x2 C y2 − 2y C z2 C 4z D 0,
we have
f x, y, z D e1−0 D e
f 0, 1, −2 D e1−−5 D e6 ( so absolute max is at 0, 1, −2.
The absolute minimum of e occurs at all points of the boundary. If we set w D x2 C y2 − 2y C z2 C 4z,
then f x, y, z D e1−w , so that it’s clear that the minimum must occur when w D 0 (since w … 0 defines the
domain we are to consider). Likewise, the maximum must occur at the center of the ball.
It’s good to take a step back and see that sometimes we can tell what type of critical point we have without using the
tools we’ve developed. In single-variable calculus, when the second derivative test failed to tell us anything we returned
either to the first derivative test or to an analysis of the function.
In Exercises 36–41, the exponents are all at least two so (see, for example, Section 2.4, Exercise 19) when the Hessian
is evaluated at the origin, all of the entries will be 0. The fact that Hf0 D 0 means that the Hessian doesn’t provide us
with any information about the nature of the critical point at the origin. This is part (a) for Exercises 36–41. By a deleted
neighborhood of the origin, we will mean a neighborhood of the origin with the origin removed.
36. f x, y D x2 y2 : in every deleted neighborhood of the origin f x, y > 0 so f 0, 0 < f x, y for every point
(x, y) near but not equal to (0, 0) so f has a local minimum at the origin.
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“runall” — 2005/8/1 — 15:11 — page 197 — #17
Section 4.2
Extrema of Functions
197
37. f x, y D 4 − 3x2 y2 : in every deleted neighborhood of the origin x2 y2 > 0 so −3x2 y2 < 0 so f x, y < 4
so f 0, 0 > f x, y for every point (x, y) near but not equal to (0, 0) so f has a local maximum at the origin.
38. f x, y D x3 y3 : in every deleted neighborhood of the origin in quadrants I and III f x, y > 0 and in quadrants
II and IV f x, y < 0 so f has neither a minimum nor a maximum at the origin.
39. f x, y, z D x2 y3 z4 : in every deleted neighborhood of the origin where y > 0, f x, y, z > 0; when y < 0,
f x, y, z < 0 so f has neither a minimum nor a maximum at the origin.
40. f x, y, z D x2 y2 z4 : in every deleted neighborhood of the origin f x, y, z > 0 so f 0, 0, 0 < f x, y, z for
every point (x, y, z) near but not equal to (0, 0, 0) so f has a local minimum at the origin.
41. f x, y, z D 2 − x4 y4 − z4 : in every deleted neighborhood of the origin x4 y4 C z4 > 0 so f x, y, z < 2
so f 0, 0, 0 > f x, y, z for every point (x, y, z) near but not equal to (0, 0, 0) so f has a local maximum at
the origin.
2
2
42. f x, y D ex C5y . Notice that eu is a monotone increasing function of u and x2 C 5y2 has a unique minimum
at (0, 0). So f has a local minimum at (0, 0) so f 0, 0 D 1 is a global minimum.
2
2
4
43. f x, y, z D e2−x −2y −3x . Notice that eu is a monotone increasing function of u and 2 − x2 − 2y2 − 3x4
has a unique maximum of 2 at (0, 0, 0). So f has a local maximum at (0, 0, 0), so f 0, 0, 0 D e2 is a global
maximum.
44. f x, y D x3 C y3 − 3xy C 7.
(a) The first partial derivatives are fx x, y D 3x2 − 3y and fy x, y D 3y2 − 3x so we have critical points
at (0, 0) and (1, 1). At the origin we have a saddle point. For the behavior at (1, 1), d1 1, 1 D 6 and
d2 1, 1 D 36 − 9 D 27. By the second derivative test we have a local minimum.
(b) We know there are no global extrema. Look along the x-axis. The function is f x, 0 D x3 C 7. As x ' q
f increases without bound and as x ' −q f decreases without bound.
45. There can’t be a global maximum because, for example, for fixed y, as x ' 0C the function grows without
bound. fx x, y D y − 1/x and fy x, y D x C 2 − 2/y so f has a critical point at (2, 1/2). From the Hessian
1/4 1
we see that there is a local minimum at (2, 1/2) of 2 C ln 2. Note that fx 2, y D y − 1/2.
1
8 We would like to now conclude that f has a unique critical point at (2, 1/2) which is a local minimum and hence
it is a global minimum—such a conclusion seems reasonable, but, as Exercise 46 will demonstrate, is not correct.
Consider fx x, 1/2 D 1/2 − 1/x. For x > 2 this is positive and so f is increasing along this line. Now look
at fy x, y D x C 2 − 2/y for x Ú 2. When y > 1/2 this is positive and when 0 < y < 1/2 this is negative.
So as we move vertically away from the line y D 1/2 for x Ú 2 we see that f is increasing. A similar analysis
for the remaining regions shows that f has a global minimum at (2, 1/2).
46. (a) Say that f has a local maximum at x0 and no other critical points. Assume that f x0 is not the global
maximum. Then there exists a point x1 such that f x1 > f x0 . By the extreme value theorem, on the
closed interval with endpoints x0 and x1 there must be a global maximum and a global minimum somewhere
on that closed interval. The global minimum could not be at x0 since it is a local maximum. It could not
be at x1 , since f x1 > f x0 . The global minimum must be somewhere on the open interval and it must
be at a critical point. This contradicts the assumption that there were no other critical points. If instead the
unique critical point of f were a local minimum, then just modify the argument appropriately.
(b) f x, y D 3yex − e3x − y3 so fx x, y D 3yex − 3e3x and fy x, y D 3ex − 3y2 . Solving, y D 0 or
y D 1, but y can’t be 0 since ex D y2 . The only critical point for f is at (0, 1) and f 0, 1 D 1. Also,
d1 0, 1 D fxx 0, 1 D −6 and d2 0, 1 D 27 so at (0, 1) f has a local maximum. Along the y-axis,
f 0, y D 3y − 1 − y3 , so as y ' −q we see that f increases without bound.
47. (a) Let the local maxima occur at a < b. Consider f on [a, b]. By the extreme value theorem, f must attain both
a maximum and minimum somewhere on [a, b]. The minimum cannot occur at a or b since local maxima
occur there. Hence there must be some c is the open interval (a, b) that gives an absolute minimum on [a,
b]—hence it must be at least a local minimum on R.
(b)
fx x, y D −2xy2 − y − 1y2
fy x, y D −2xy2 − y − 12xy − 1 − 4y2 − 1y
For fx D 0, either xy2 − y − 1 D 0 or y2 D 0 (so y D 0). If y D 0, then the fy D 0 equation
becomes −2−1−1 D 0, which is false. Thus xy2 − y − 1 D 0 and the fy D 0 equation becomes
−4yy2 − 1 D 0. Since y Z 0, we must have y2 − 1 D 0 or y D ;1. With y D 1 in the fx D 0 equation,
we have −2x − 2 D 0 * x D 2. With y D −1 in the fx D 0 equation, we have −2x C 1 − 1 D 0
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198
Chapter 4
Maxima and Minima in Several Variables
so x D 0. So we have two critical points: (2, 1) and 0, −1. The Hessian matrix is
Hfx, y D −2y4
−24xy3 − 3y2 − 2y
so
Hf2, 1 D −2
−6
Hf0, −1 D −2
2
−24xy3 − 3y2 − 2y
−26x2 y2 − 6xy − 2x C 1 − 43y2 − 1
−6
sequence of minors is −2, 16 * local maxI
−26 2
sequence of minors is −2, 16 * local max .
−10 (c) Best left to a computer. Stay close to the critical points to see the surface details well.
2
0
z
-2
0
0.5
x
1
0.5
1
0
1.5
-0.5
2
-1
y
4.3 LAGRANGE MULTIPLIERS
1. The plane is given by 2x − 3y − z D 4. There will be only one critical point in each case. Geometrically, it
cannot be a local maximum because there will always be points nearby which are farther away. There is at least
one point on the plane closest to the origin so the single critical point will be at this point. You can also perform
the second derivative test.
(a) We’ll minimize the square of the distance: Dx, y D x2 C y2 C 2x − 3y − 42 . The partials are
Dx x, y D 10x − 12y − 16 and Dy x, y D 20y − 12x C 24. Set these equal to zero and solve
simultaneously to find the critical point 4/7, −6/7, −2/7.
(b) Minimize f x, y, z D x2 C y2 C z2 subject to the constraint gx, y, z D 2x − 3y − z D 4. We solve
the system


2x D 2λ


 2y D −3λ

2z D −λ


 2x − 3y − z D 4.
We see that x D λ so y D −3/2x and z D −x/2. Substituting into the last equation: 2x C 9x/2 C x/2 D 4
so x D 4/7 and our critical point is 4/7, −6/7, −2/7.
2. The function is f x, y D y subject to the constraint gx, y D 2x2 C y2 D 4. We solve the system


 0 D 4λx
1 D 2λy

 2x2 C y2 D 4.
From the first equation, λx D 0, but λ Z 0. since 2λy Z 0. Hence we must have x D 0, so y2 D 4, therefore the
critical points are 0, ;2.
3. The function is f x, y D 5x C 2y subject to the constraint gx, y D 5x2 C 2y2 D 14. We solve the system


 5 D 10λx
2 D 4λy

 5x2 C 2y2 D 14.
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“runall” — 2005/8/1 — 15:11 — page 199 — #19
Section 4.3
Lagrange Multipliers
199
By either of the first two equations we
√ see
√ that λ Z 0. Together, the first two equations imply that x D y so
7x2 D 14 so the critical points are ; 2, 2.
4. The function is f x, y D xy subject to the constraint gx, y D 2x − 3y D 6. We solve the system


 y D 2λ
x D −3λ

 2x − 3y D 6.
If λ were 0, then both x and y would be 0 which would contradict the third equation. In short, λ Z 0. In that
case, the first two equations imply that x D −3/2y so −3y − 3y D 6 or y D −1. The critical point is at
3/2, −1.
5. The function is f x, y, z D xyz subject to the constraint gx, y, z D 2x C 3y C z D 6. We solve the system


yz D 2λ


 xz D 3λ

xy D λ


 2x C 3y C z D 6.
One possibility is that two of x, y, and z are zero. In this case the three possible critical points are (3, 0, 0),
(0, 2, 0), and (0, 0, 6). If none of x, y, and z is zero then the first two equations imply that x D 3/2y, and the
second and third equations together imply that 3y D z. Hence, 3y C 3y C 3y D 6, so the final critical point is
(1, 2/3, 2).
6. The function is f x, y, z D x2 C y2 C z2 subject to the constraint gx, y, z D x C y − z D 1. We solve
the system


2x D λ


 2y D λ

2z D −λ


 x C y − z D 1.
We see immediately that x D y D −z, which implies that x C x C x D 1. Therefore, the critical point is
1/3, 1/3, −1/3.
7. The function is f x, y, z D 2x C y2 − z2 subject to the two constraints g1 x, y, z D x − 2y D 0 and
g2 x, y, z D x C z D 0. We solve the system


2Dλ C µ




 2y D −2λ
−2z D µ



x D 2y


 x D −z.
Solving, we see that 2 D λ C µ D −y − 2z D −x/2 C 2x D 3x/2. So the critical point is 4/3, 2/3, −4/3.
8. The function is f x, y, z D x C y C z subject to the two constraints g1 x, y, z D y2 − x2 D 1 and
g2 x, y, z D x C 2z D 1. We solve the system


1 D −2λx C µ




 1 D 2λy
1 D 2µ



y 2 − x2 D 1


 x C 2z D 1.
Solving, we see that µ D 1/2 and 2λ D 1/y √
so 1/2√D −2λx √
D −x/y or y D√−2x. This
√ means that
√ 1 D
y2 − x2 D 3x2 , so the critical points are −1/ 3, 2/ 3, 3 C 3/6 and 1/ 3, −2/ 3, 3 − 3/6.
9. (a) The function is f x, y D x2 C y subject to the constraint gx, y D x2 C 2y2 D 1. We solve the system


 2x D 2xλ
1 D 4yλ

 x2 C 2y2 D 1.
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200
Chapter 4
Maxima and Minima in Several Variables
From the first equation, we see that either x D 0 or λ D 1. If λ D 1, then y D 1/4, so x D ; 7/8. If x D 0,
then y D ; 1/2. In short, the critical points are ; 7/8, 1/4 and 0, ; 1/2.
(b) LλI x, y D x2 C y − λx2 C 2y2 − 1 so


0
−2x −4y
0 .
HλI x, y D  −2x 2 − 2λ
−4y
0 −4λ
So−d3 D −16y[x2 C 1/2 − 2y]. Substitute
the critical points to find that there are local maxima at
; 7/8, 1/4 and local minima at 0, ; 1/2.
10. (a) The function is f x, y, z, w D x2 C y2 C z2 C w2 subject to the three constraints g1 x, y, z, w D
2x C y C z D 1, g2 x, y, z, w D x − 2z − w D −2 and g3 x, y, z, w D 3x C y C 2w D −1. We
solve the system



2x D 2λ C µ C 3ν



2y D λ C ν




 2z D λ − 2µ
2w D −µ C 2ν



2x C y C z D 1



x − 2z − w D −2



 3x C y C 2w D −1.
1
−11, 15, 75, −25.
68

−2 −1 −1
0
−1
0
2
1 

−3 −1
0 −2 

2
0
0
0 
.
0
2
0
0 

0
0
2
0 
0
0
0
2
After a great deal of fussing we find that there is a critical point at
(b)

0
 0

 0

HLλ, µ, ν, x, y, z, w D 
 −2
 −1

 −1
0
0
0
0
−1
0
2
1
0
0
0
−3
−1
0
−2
We calculate −d7 D 628 and conclude that f has a local minimum at the critical point.
Note: For Exercises 11–15 the Mathematica code would be similar to that in Exercise 11.
11. Input the following three lines into Mathematica (or the equivalent into your favorite computer algebra system)
f D 3xy − 4z
g D 3x C y − 2xz
Solve [{D[f , x] DD λD[g, x], D[f , y] DD λD[g, y], D[f , z] DD λD[g, z], 3x C y − 2x z DD 1}]
The solutions are
√
√
• λ D 6, x, y, z D 2/3, 1/2, 12 − 6/8 and
√
√
• λ D − 6, x, y, z D − 2/3, 1/2, 12 C 6/8.
12. Use the same basic code you used in Exercise 11, allowing for two Lagrange multipliers. The solution is λ1 D
482/121, λ2 D −107/121, x, y, z D 31, 29, 5/11.
13. Many solutions are returned by Mathematica. They are
• 0, −1, 0 for λ D −3/2
• (0, 1, 0) for λ D 3/2
• −2/3, −2/3, −1/3 and 2/3, −2/3, 1/3 for λ D −4/3
• −1, 0, 0 and (1, 0, 0) for λ D −1
• 0, 0, −1 and (0, 0, 1) for λ D 0 and
• 11/2/8, −3/8, −3 11/2/8 and − 11/2/8, −3/8, 3 11/2/8 for λ D 1/8.
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“runall” — 2005/8/1 — 15:11 — page 201 — #21
Section 4.3
Lagrange Multipliers
201
14. The solutions given are
• 1, −1/2, ; 3/2 for λ D −1
√
√
√
√
• −1 − 5/2, −3 − 5/4, ;i51/4 / 2 for λ D 1 C 5/2.
√
√
√
√
• 1 − 5/2, −3 C 5/4, ;51/4 / 2 for λ D 1 − 5/2.
• −i, i, 0 and i, −i, 0 for λ D −2, and
• −1, −1, 0 and (1, 1, 0) for λ D 2.
Note that several of the solutions are complex and, for the purposes of this discussion, can be discarded.
15. Here there are two solutions:
√
√
√
√
√
√
• w, x, y, z D 1 − 2/2, 1/ 2, 1/ 2, 1 − 2/2 for λ1 D 2 − 1/ 2, λ2 D 1 − 2, and
λ3 D 0, and
√
√
√
√
√
√
• w, x, y, z D 1 C 2/2, −1/ 2, −1/ 2, 1 C 2/2 for λ1 D 2 C 1/ 2, λ2 D 1 C 2, and
λ3 D 0.
16. (a) We need to solve


 3x2 D λy
6y D λx

 xy D −4
Substitute y D −4/x into the second equation to get λ D −24/x2 . Substitute both of these into the right
5
side of the first equation to get
 x D −32 or xD 2. So y D −2 and λ D −6.
0
2 −2
6 . Following the rule for the second derivative test for con(b) The Hessian in this case is  2 12
−2
6
6
strained local extrema, note that n D 2 and k D 1 so the only relevant term in sequence (1) is
−11 d3 D −1[−224 − 236] D 120 > 0.
We conclude that there is a constrained local minimum at the point 2, −2.
(c) You can see from the figure below that there is a constrained local minimum at 2, −2 on the curve. This
will be the point at which the constraint curve is tangent to one of the level curves.
0
-1
-2
-3
-4
0
1
2
3
4
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“runall” — 2005/8/1 — 15:11 — page 202 — #22
202
Chapter 4
Maxima and Minima in Several Variables
17. The symmetry of the problem suggests the answer, but we are maximizing f x, y, z D xyz subject to the
constraint gx, y, z D x C y C z D 18. We solve the system


yz D λ


 xz D λ

xy D λ


 x C y C z D 18.
None of the solutions that corresponds to one of x, y, and z being zero is a maximum. The solution we get is
x D y D z, so 3x D 18, so the maximum product occurs at the point (6, 6, 6).
18. First, a sphere is a compact surface so, by the extreme value theorem, we know that there is both a minimum and
a maximum attained. We find the extrema of f x, y, z D x C y − z with the constraint x2 C y2 C z2 D 81.
We solve the system


1 D 2λx


 1 D 2λy

−1 D 2λz


 x2 C y2 C z2 D 81.
√
√
√
√
√
We see
D −z, so there is a maximum of 9 3 at 3 3, 3 3, −3 3 and a minimum of −9 3 at
√ that x√D y √
−3 3, −3 3, 3 3.
19. This is a nice problem to assign because by this point some students are only checking boundary values. We are
looking for the maximum and minimum values of f x, y D x2 C xy C y2 constrained to be inside the closed
disk gx, y D x2 C y2 … 4. First we find the critical points without paying attention to the constraint. The
partial derivatives are fx x, y D 2x C y and fy x, y D x C 2y so we have a critical point at the origin, and
f 0, 0 D 0. Next we look for extrema of the function on the boundary of the disk by solving the system


 2x C y D 2λx
x C 2y D 2λy

 x2 C y2 D 4.
√
From the first two equations we see that x2 D y2 so√x D
√ ;y and x D ; 2. Substituting, we find that the
minimum is 0 at the origin and the maximum is 6 at 2, 2.
20. We are maximizing V x, y, z D xyz subject to the constraint gx, y, z D 2x C 2y C z … 108. In this case, the
maximum must occur on the boundary because the only unconstrained critical point requires two of the coordinates
to be zero—these points are on the boundary and give the (degenerate) minimum solution of 0. We solve the
system


yz D 2λ


 xz D 2λ

xy D λ


 2x C 2y C z D 108.
Since none of x, y, or z can be zero, we find that x D y D z/2, so 3z D 108 and the critical point is (18, 18,
36). So the dimensions are 18” by 18” by 36”.
21. We are maximizing f r, h D πr2 h subject to the constraint that gr, h D 2πrh C 2πr2 D c. We solve the
system


 2πrh D λ2πh C 4πr
πr2 D 2λπr

 2πrh C 2πr2 D c.
Since r Z 0 the second equation implies that r D 2λ, so, substituting this into the first equation, we see that
h D 2r. Hence, the height should equal the diameter.
22. We are minimizing the cost which is Cr, h D πr2 C 22πrh C 52πr2 D 11πr2 C 4πrh subject to the
constraint gr, h D πr2 h C 2/3πr3 D 900π. We solve the system


2

 22πr C 4πh D πλ2rh C 2r 2
4πr D λπr


 πr2 h C 2/3πr3 D 900π.
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“runall” — 2005/8/1 — 15:11 — page 203 — #23
Section 4.3
Lagrange Multipliers
203
As above, we see that 4 D λr so 22πr C 4πh D 4π/r2rh C 2r2 or 14r D 4h. Substituting, 900 D
7/2r3 C 2/3r3 D 25/6r3 so the radius is 6 feet and the height is 21 feet.
23. We wish to minimize Mx, y, z D xz − y2 C 3x C 3 subject to the constraint gx, y, z D x2 C y2 C z2 D 9.
We solve the system


z C 3 D 2λx


 −2y D 2λy

x D 2λz


 x2 C y2 C z2 D 9.
√
Either y D 0 or λ D −1. If y D 0, then z D −3 or 3/2 so we get 0, 0, −3 and ;3 3/2, 0, 3/2 as critical
points. If λ D −1, we find the critical points are −2, 2, 1 and −2, −2, 1. Comparing values of M, the
minimum of −9 is attained at either −2, 2, 1 or −2, −2, 1.
24. It’s easier to maximize the square of the area f x, y, z D ss − xs − ys − z subject to x C y C z D
2s D P, a constant.
Thus §f D λ§g (where gx, y, z D x C y C z) gives us the system:


−ss − ys − z D λ


 −ss − xs − z D λ

−ss − xs − y D λ



x C y C z D 2s
0 < x, y, z … s
Hence −ss − ys − z D −ss − xs − z D −ss − xs − y. The first equality implies z D s or
x D y. Note that z D s means f D 0—so there’s zero area which cannot possibly be maximum. Thus x D y.
2
From −ss − xs − z D −ss − xs − y we similarly conclude that y D z. Hence x D y D z D s
3
gives us our critical point and corresponds to having an equilateral triangle. Our constraint looks like a portion of
a plane. The dark triangle in the figure below is the part to be considered—it’s where f is Ú 0. Therefore, the
2 2 2
point s, s, s yields the maximum.
3 3 3
z
(0,0,2s)
(2s,0,0)
(0,2s,0)
x
y
25. A sphere centered at the origin has equation x2 C y2 C z2 D r2 . Thus we want to maximize f x, y, z D
x2 C y2 C z2 subject to the constraint gx, y, z D 3x2 C 2y2 C z2 D 6. We can solve this using Lagrange
multipliers, but we must make sure we find an inscribed sphere. We consider the system


2x D 6λx
1st equation gives x D 0 or λ D 1/3




2nd equation gives y D 0 or λ D 1/2
 2y D 4λy
2z D 2λz
3rd equation gives z D 0 or λ D 1


2
2
2

 3x C 2y C z D 6 Note that we can’t have x D y D z D 0


and still satisfy the constraint.
√
√
Thus if λ D 1/3, y D z D 0 and the √
constraint implies x D ; 2. If λ D 1/2, x D z D 0 and y D ; 3. Finally,
if λ D 1, then x D y D 0 and z D ; 6. Comparing values, we have
√
√
√
f ; 2, 0, 0 D 2, f 0, ; 3, 0 D 3, f 0, 0, ; 6 D 6,
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“runall” — 2005/8/1 — 15:11 — page 204 — #24
204
Chapter 4
Maxima and Minima in Several Variables
√
so that it’s tempting to say that the largest sphere has a radius of 6. However, such a sphere is not actually
√
inscribed in the ellipsoid. The largest sphere that actually remains inscribed in the ellipsoid has a radius of 2.
26. This is just Exercise 1 with two constraints. We are minimizing f x, y, z D x2 C y2 C z2 with the constraints
g1 x, y, z D 2x C y C 3z D 9 and g2 x, y, z D 3x C 2y C z D 6. We solve the system


2x D 2λ C 3µ




2y
D λ C 2µ


2z D 3λ C µ



2x
C y C 3z D 9



 3x C 2y C z D 6.
Eliminate λ and µ and then solve to get a critical point at (1, 2/5, 11/5).
27. This is just Exercise 18 translated by 2, 5, −1. We are minimizing f x, y, z D x − 22 C y − 52 C
z C 12 with the constraints g1 x, y, z D x − 2y C 3z D 8 and g2 x, y, z D 2z − y D 3. We solve the
system


2x − 2 D λ





 2y − 5 D −2λ − µ
2z C 1 D 3λ C 2µ



x
− 2y C 3z D 8



 2z − y D 3.
Eliminate µ by combining the second and third equations and then substitute 2x − 2 for λ. Solve to get a
critical point at (9/2, 2, 5/2).
28. We want to minimize and maximize x2 C y2 C z2 but note that this is equivalent to finding the extrema of the
square of the distance f x, y, z D x2 C y2 C z2 subject to the constraints of the two surfaces g1 x, y, z D
x C y C z − 4 D 0 and g2 x, y, z D x2 C y2 − z D 0. The function f is continuous and an ellipse is
compact so the extreme value theorem guarantees that f will have a global minimum and a global maximum on
the ellipse. Either by hand, or using a computer algebra system as in Exercises 11–15, we find that the nearest
point to the origin is at (1, 1, 2) and the farthest is at −2, −2, 8.
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“runall” — 2005/8/1 — 15:11 — page 205 — #25
Section 4.3
Lagrange Multipliers
205
29. This is the same as Exercise 28 except that we are trying to find extrema for f x, y, z D z and the plane has
the equation g1 x, y, z D x C y C 2z D 2. Again, using a computer algebra system we find that the lowest
point is at (1/2, 1/2, 1/2) and the highest is at −1, −1, 2.
30. Minimize f x, y, u, v D x − u2 C y − v2 subject to the two constraints: g1 x, y, u, v D x2 C 2y2 D 1
and g2 x, y, u, v D u C v D 4. We solve the system



2x − u D 2λx




−2x − u D µ



 2y − v D 2λy

−2y − v D µ





x2 C 2y2 D 1



 u C v D 4.
√
√
Solving you get two critical points x, y, u, v D 2/3, 1/6, 2 C 6/12, 2 − 6/12 for which the square
√
√
√
of the distance is 35/4 − 2 6 L 3.85 and x, y, u, v D − 2/3, − 1/6, 2 − 6/12, 2 C 6/12 for which
√
√
the square of the distance is 35/4 C 2 6 L 13.65. The minimum distance is 35/4 − 2 6 L 1.96.
31. (a) f x, y D x C y with the constraint xy D 6 so we solve the system



 1 D 2λy
1 D 2λx


 xy D 6.
√ √
So x D y and the critical points are at ; 6, 6.
√
√
(b) The constraint curve is not
√ connected. There are two distinct components. Although − 6, − 6 produces
at any point on the other component
a local maximum of −2
√ 6√on its component, the value of the function
√
is greater. Similarly, 6, 6 produces a local minimum of 2 6 on its component, but the value of the
function at any point on the other component is less.
32. This is a non-linear version of Exercise 26. Minimize f x, y, z D x2 C y2 C z2 subject to the constraints
g1 x, y, z D x2 C y2 D 4 and g2 x, y, z D 2x C 2y C z D 2. We solve the system


2x D 2λx C 2µ





 2y D 2λy C 2µ
2z D µ



x2 C y 2 D 4



 2x C 2y C z D 2.
√
√
Solving
√ we√see that either
√ x D y or λ D 1. If x D y then
√ x D y D ; 2 and
√ z D 2 < 4 2. The farthest point is
− 2, − 2, 2 C 4 2. If λ D 1 then x D 1 ; 7/2, y D 1 < 7/2, and z D 0—these last two are
the closest points.
33. (a) The system is

3
2


 1 D λ16x − 12x 0 D 2λy


 y2 − 4x3 C 4x4 D 0.
The second equation implies that either y D 0 or λ D 0. But λ D 0 cannot satisfy the first equation, so
y D 0. The last equation implies 4x3 1 − x D 0; thus x D 0 or 1. But x D 0 cannot satisfy the first
equation. Thus the only solution to the system is (1, 0).
(b) The graph of the curve (known as the piriform) is shown in the figure below. From it, it’s clear that the
maximum value of f x, y D x occurs at (1, 0) and the minimum value at (0, 0).
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“runall” — 2005/8/1 — 15:11 — page 206 — #26
206
Chapter 4
Maxima and Minima in Several Variables
y
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1
x
-0.2
-0.4
-0.6
(c) Note that §gx, y D 16x3 − 12x2 , 2y D 0, 0 at (0, 0) (and at (3/4, 0)). (0, 0) is a point on the curve
((3/4, 0) is not). It’s the singular point of the piriform and, although not a solution to the Lagrange multiplier
system in part (a), it must be considered as a possible site for extrema.
34. (a) The relevant Lagrange multiplier system to solve is


2x D 0


 2y D 0

0Dλ


 zDc
The obvious unique solution is 0, 0, c with λ D 0.
(b) LlI x, y, z D x2 C y2 − lz − c. With c as a constant and x1 D x, x2 D y, x3 D z, we have

0
 0

HLlI x, y, z D 
0
−1
0
2
0
0
0
0
2
0

−1
0 
 D HL0I 0, 0, c.
0 
0
The second derivative test asks us to calculate −11 d3 and −11 d4 or

0
−d3 D −det  0
0
0
2
0

0
0  D 0I
2

0
 0

−d4 D −det 
0
−1
0
2
0
0
0
0
2
0

−1
0 
 D −−4 D 4.
0 
0
Thus the second derivative test seems to suggest that we’ve found a saddle point.
(c) Now we let x1 D z, x2 D y, x3 D x and look at

0
 −1

HLlI z, y, x D 
0
0
−1
0
0
0
0
0
2
0

0
0 
.
0 
2
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“runall” — 2005/8/1 — 15:11 — page 207 — #27
Section 4.3
Lagrange Multipliers
207
In this case we find

0
−d3 D −det  −1
0

0
 −1

−d4 D −det 
0
0

0
0  D −−2 D 2 and
2

0 0
0 0 
 D −−4 D 4.
2 0 
0 2
−1
0
0
−1
0
0
0
This time the sound derivative test suggests a local minimum.
(d) Indeed, inspection tells us that the expression x2 C y2 attains a global minimum at x D y D 0. So to satisfy
the constraint z D c, we see that (0, 0, c) yields a global minimum. The difference between the results of
(b) and (c) can be explained by looking at ⭸g/⭸x vs. ⭸g/⭸z: ⭸g/⭸x D 0, but ⭸g/⭸z D 1 Z 0.
In part (b), we did not satisfy the hypothesis of the second derivative test that the variables he ordered so that

⭸g1
a
 ⭸x1




#
det 
#


#

 ⭸gk
a
⭸ x1
...
..
.
...

⭸g1
a

⭸xk




#
 Z 0.
#


#


⭸gk
a
⭸xk
(The determinant in this situation is just ⭸g/⭸x.) In part (c), we did satisfy the hypothesis, since ⭸g/⭸z Z 0.
35. (a) In order for λ, a to be a solution of the constrained problem, λ, a must solve the system



 f a D
k
λj gj xi a
jD1


 gj a D cj
xi
for 1 … i … n
for 1 … j … k.
On the other hand, an unconstrained critical point for L must be where all first partials are zero. In other
words, we must have
Llj D 0, 1 … j … k
and Lxj D 0, 1 … j … n.
Upon explicit calculation of the partials these equations are:
flj a − gj a − cj D 0
fxj a − kiD1 λi gi xj a D 0
for 1 … j … k, and
for 1 … j … n.
This is the same system as that for the constrained case.
(b) Calculate the Hessian in four blocks. All of the entries in the upper left k * k block are 0. This is because
the entry in position (i, j) is Lli lj and the highest power of any li appearing in L is 1. The top right block
with k rows and n columns gives back the negative first partials of the constraint conditions because the
entry in position k C i, j is Lxi lj D −gj − cj xi D −gj xi . The lower left block of n rows and k
columns is just the transpose of this last block. The lower right n * n block is such that the entry in position
k C i, k C j D Lxi xj D f − kqD1 lq gq xi xj . When λ and a are substituted for l and x, the desired
matrix is obtained.
36. We find extreme values of f x1 , ... , xn , y1 , ... , yn D niD1 xi yi subject to the two constraints g1 x1 , ... , xn ,
y1 , ..., yn D x1 2 C · · · C xn 2 D 1 and g2 x1 , ... , xn , y1 , ... , yn D y1 2 C · · · C yn 2 D 1. Thus we look at
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“runall” — 2005/8/1 — 15:11 — page 208 — #28
208
Chapter 4
Maxima and Minima in Several Variables
§f D λ1 §g1 C λ2 §g2 together with the constraints to solve




y1 D 2λ1 x1 





The first n equations (and the last) imply 1 D yi 2 D 4λ1 2 xi 2






#
D 4λ1 2 · 1


#


1




#
so λ1 D ; .





2


yn D 2λ1 xn








 x1 D 2λ2 y1 






The next n equations (and the next-to-last) imply 1 D xi 2 D 4λ2 2 yi 2 D 4λ2 2


#

1



so λ2 D ; .
 #



#

2


 x D 2λ y 


n
2
n







2



xi D 1




2

yi D 1
Putting all the information together, we find that x D y (when λ1 D λ2 D 1 ) and x D −y (when λ1 D λ2 D − 1 ).
2
2
When x D y, f x, y D xi 2 D 1. When x D −y, f x, −x D −xi 2 D −1. Though it takes a little bit of
argumentation, the hypersphere in Rn is compact—hence so is the product of hyperspheres in R2n D Rn * Rn .
Thus we find maximum and minimum values of C1 and −1, respectively.
37. (a)
n
x1 2
x2 2
xn 2
xi 2
2
2
2
ui D u1 C · · · C un D 2 2 C 2 2 C · · · C 2 2 D x 2 D 1.
i
xi xi xi iD1
So u is an the unit hypersphere. The case for v is identical.
(b) By Exercise 36, we have −1 … niD1 ui vi … 1. Hence
−1 … i
3 −
2
yi
xi
2 j xj
2 … 1
j yj
2
j xj j yj … i xi yi …
2
2
j xj j yi
3 − xy … x · y … xy
|x · y| … xy.
3
4.4 SOME APPLICATIONS OF EXTREMA
1. This problem can be done using calculators or the following table to help with Proposition 4.1:
xi
yi
xi 2
xi yi
0
1
2
3
4
5
6
2
3
5
3
2
7
7
0
1
4
9
16
25
36
0
3
10
9
8
35
42
21
29
91
107
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“runall” — 2005/8/1 — 15:11 — page 209 — #29
Section 4.4
Some Applications of Extrema
So m0 D
7107 − 2129
140
35
D
D
L .71428
2
791 − 21
196
49
and b0 D
9129 − 21107
392
98
D
D
D 2.
791 − 212
196
49
209
The equation of the least squares line is y D 35/49x C 2.
2. Again, using Proposition 4.1,
mD
bD
2x1 y1 C x2 y2 − x1 C x2 y1 C y2 x − x2 y1 − y2 y − y2
D 1
D 1
.
2x12 C x22 − x1 C x2 2
x1 − x2 2
x1 − x2
x12 C x22 y1 C y2 − x1 C x2 x1 y1 C x2 y2 2x12 C x22 − x1 C x2 2
D
x1 y2 − x2 y1 x1 − x2 x y − x2 y1
D 1 2
.
x1 − x2 2
x1 − x2
You can check that x1 , y1 and x2 , y2 are both on the line
y − y2
x1 y2 − x2 y1
.
yD 1
x C
x1 − x2
x1 − x2
3. (a) As in the text, the function D(a, b) will be the sum of the squares of the differences between the observed
y values and the y values on the curve y D a/x C b. This means that
n
Da, b D yi − a/xi C b2 .
iD1
(b) Make the substitution Xi D 1/xi and then fit the line y D aX C b to this transformed data using Proposition 4.1. We get
aD
n X2i − ! Xi "
X i ! y i " − ! X i " ! X i y i "
2
n Xi yi − ! Xi " ! yi "
and b D
2
n X2i − ! Xi "
2
.
Transform the data back, replacing Xi with 1/xi , then the curve of the form y D a/x C b that best fits the
data has
aD
n yi /xi − ! 1/xi " ! yi "
n 1/xi2 − ! 1/xi "
2
1/xi ! yi " − ! 1/xi " ! yi /xi "
2
and
bD
n 1/xi2 − ! 1/xi "
2
.
4. We’ll use the results of Exercise 3 and organize our sums with the following table:
1/xi2
yi /xi
0
−1
1
−1/2
1
1/4
4
1/9
0
−1/2
2
−1/6
−1/2
193/36
8/6
1/xi
yi
1
1/2
2
1/3
23/6
So a D
and
bD
48/6 − 23/6−1/2
29
261
D
D
4193/36 − 23/62
243
27
193/36−1/2 − 23/68/6
187
561
D−
.
D−
4193/36 − 23/62
486
162
The equation of the least squares curve of the desired form is y D 29/27x − 187/162.
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“runall” — 2005/8/1 — 15:11 — page 210 — #30
210
Chapter 4
Maxima and Minima in Several Variables
5. Again the function D(a, b, c) will be the sum of the squares of the differences between the observed y values
and the y values on the curve y D ax2 C bx C c. This means that
n
Da, b, c D yi − axi2 C bxi C c2
iD1
n
n
n
n
n
n
iD1
iD1
iD1
D yi2 C a2 xi4 C b2 xi2 C nc2 − 2a xi2 yi − 2b xi yi − 2c yi
iD1
iD1
iD1
n
n
n
iD1
iD1
iD1
C 2ab xi3 C 2ac xi2 C 2bc xi
so
n
n
n
n
iD1
iD1
iD1
iD1
n
n
n
n
iD1
iD1
iD1
iD1
Da a, b, c D 2a xi4 − 2 xi2 yi C 2b xi3 C 2c xi2 ,
Db a, b, c D 2b xi2 − 2 xi yi C 2a xi3 C 2c xi ,
n
n
n
iD1
iD1
iD1
and
Dc a, b, c D 2cn − 2 yi C 2a xi2 C 2b xi .
Set each of the partial derivatives equal to zero, move the term with coefficient −2 to the other side, and divide
by 2 to get the desired equations.
6. You may want to point out to the students that the independent variable x corresponds to hours of sleep because
that is what (in theory) Egbert can control.
(a) To get a line y D ax C b we’ll need
xi
yi
xi2
xi yi
8
8.5
9
7
4
8.5
7.5
6
85
72
95
68
52
75
90
65
64
72.25
81
49
16
72.25
56.25
36
680
612
855
476
208
637.5
675
390
58.5
602
446.75
4533.5
Using the formulas in Proposition 4.1 you’ll find that the least squares line is
y D 4204/607x C 14935/607 L 6.93x C 24.6.
(b) We will need some additional data:
xi3
xi4
xi2 yi
8
8.5
9
7
4
8.5
7.5
6
512
614.125
729
343
64
614.125
421.875
216
4096
5220.0625
6561
2401
256
5220.0625
3164.0625
1296
5440
5202
7695
3332
832
5418.75
5062.5
2340
58.5
3514.125
28214.1875
35322.25
xi
Use the formulas given in Exercise 5 to obtain the system


 28214.1875a C 3514.125b C 446.75c D 35322.25
3514.125a C 446.75b C 58.5c D 4533.5

 446.75a C 58.5b C 8c D 602.
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“runall” — 2005/8/1 — 15:11 — page 211 — #31
Section 4.4
Some Applications of Extrema
211
Solve this system to get the following (approximate) quadratic:
y D −.192044054x2 C 9.42923983x C 17.02314387.
(c) Plugging 6.8 into the linear model predicts that Egbert will get 71.7, plugging 6.8 into the quadratic model
predicts that Egbert will get 72.26.
7. (a) We are required to show that F is a gradient (conservative) vector field. Clearly if V x, y D x2 C 2xy C
3y2 C x C 2y then −§V D −2x − 2y − 1i C −2x − 6y − 2j D F.
(b) We find equilibrium points of F when F D 0. Solve the system of equations
−2x − 2y D 1
# −2x − 6y D 2
and find one solution at −1/4, −1/4. The Hessian is
HV D 2
2
2
6 so both d1 and d2 are positive so the equilibrium is stable.
8. Here V x, y D 2x2 − 8xy − y2 C 12x − 8y C 12 so §V D −F D 4x − 8y C 12, −8x − 2y − 8.
This is 0 at −11/9, 8/9. The Hessian is
HV D 4
−8
−8
.
−2 Note that d1 > 0 and d2 < 0 so the equilibrium at −11/9, 8/9 is not stable.
9. Here V x, y, z D 3x2 C 2xy C z2 − 2yz C 3x C 5y − 10 so §V D −F D 6x C 2y C 3, 2x − 2z C
5, −2y C 2z. This is 0 at −1, 3/2, 3/2. The Hessian is


6
2
0
0 −2  .
HV D  2
0 −2
2
Note that d1 > 0, d2 < 0, and d3 > 0 so the equilibrium at −1, 3/2, 3/2 is not stable.
10. (a) Here we are looking for constrained equilibria (as in Example 3 in the text). Our equation is F − §V D λ§g
where gx, y, z D 2x2 C 3y2 C z2 D 1, F D −mgk, and V x, y, z D 2x. So our system of equations is


−2 D 4λx


 0 D 6λy

−mg D 2λz


 2x2 C 3y2 C z2 D 1.
Note from the first equation that λ Z 0 so by the second equation y D 0. From the third equation
2λ D −mg/z
2 C m2 g2 x2 D 1 so x D ;1/ 2 C m2 g2 . So
so z D mgx. Substituting into the equation of the ellipsoid,
2x
our two equilibria are at ;1/ 2 C m2 g2 , 0, mg/ 2 C m2 g2 .
(b) Note the direction of the force is −2, 0, −mg so −1/ 2 C m2 g2 , 0, mg/ 2 C m2 g2 is a stable
equilibrium.
11. Maximize Rx, y, z D xyz2 − 25000x − 25000y − 25000z subject to the constraint x C y C z D 200000.
Our system of equations is


yz2 − 25000 D λ


 xz2 − 25000 D λ

2xyz − 25000 D λ


 x C y C z D 200000.
The hidden condition is that all of the variables are non-negative. This means that we are finding a maximum on
the triangular portion of the plane that lies in the first octant. The maximum revenue will occur at a boundary
point or at a critical point. Along the boundary at least one of the variables is 0 and the revenue is at most 0
when at least one of x, y and z is 0. We will see the value of R at the critical point is greater and therefore that
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“runall” — 2005/8/1 — 15:11 — page 212 — #32
212
Chapter 4
Maxima and Minima in Several Variables
it is our global maximum. Assume none of the variables is zero. Then, from the first two equations, since z Z 0
then x D y. From the third equation paired with either of the first two we see that z D 2x D 2y. Finally, since
their sum is 200000 we find the solution (50000, 50000, 100000) is where the maximum revenue occurs.
12. This is similar to Example 4 from the text. We are maximizing Ux1 , x2 , x3 D x1 x2 C 2x1 x3 C x1 x2 x3 subject
to the constraint gx1 , x2 , x3 D x1 C 4x2 C 2x3 D 90. Our system of equations is


x2 C 2x3 C x2 x3 D λ


 x C x x D 4λ
1
1 3

2x1 C x1 x2 D 2λ


 x1 C 4x2 C 2x3 D 90.
The only solution of this system with all three of the xi ’s non-negative is (33.0149, 6.37314, 15.7463). You
can only order integer amounts, so experiment with the different ways of rounding to obtain a maximum at
(28, 7, 17).
13. (a) This is an example of the Cobb-Douglas production function with p D w D 1 (see Example 5 from the
text). The only critical point will be K, L D 1/3360000, 2/3360000 D 120000, 240000.
(b) ⭸Q/⭸K D 20L/K2/3 and so at (120000, 240000), ⭸Q/⭸K D 2022/3 . On the other hand, ⭸Q/⭸L D
40K/L1/3 and so at (120000, 240000), ⭸Q/⭸L D 401/21/3 . These quantities are equal at the critical
point.
14. This time we are minimizing pK C wL D M subject to the constraint QK, L D c. Our system of equations is


⭸Q


 p D λ ⭸K

⭸Q


.
 wDλ
⭸L
Since none of p, q, and λ is 0, we can divide the top equation by pλ, divide the bottom equation by qλ and the
result is immediate.
4.5 TRUE/FALSE EXERCISES FOR CHAPTER 4
1. True.
2. False. (The increment measures the change in the function.)
3. True.
4. True.
5. True.
6. False. p2 x, y D 1 − 3x C y C 3x2 C 2xy.
7. False. (f is most sensitive to changes in y.)
8. False. (The result is true if f is of class C2 .)
9. False.
10. True.
11. True.
12. False. (The set is not bounded.)
13. False. (Consider the function f x, y D x2 C y2 .)
14. True. (This ball is compact.)
15. True.
16. False. (The point a might not be a critical point.)
17. False. (The point is not a critical point of the function.)
18. False. (The point (0, 0, 0) gives a local minimum.)
19. True.
20. True.
21. False. (The critical point is a saddle point.)
22. False. (A local extremum can occur where a partial derivative fails to exist.)
23. False. (Extrema may also occur at points where g D c and §g D 0.)
24. False. (Solutions to the system only give critical points.)
25. False. (You will have to solve a system of 7 equations in 7 unknowns.)
26. True.
27. True.
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“runall” — 2005/8/1 — 15:11 — page 213 — #33
Section 4.6
Miscellaneous Exercises for Chapter 4
213
28. True.
29. False. (The equilibrium points are the critical points of the potential function.)
30. False. (This is only true at values of labor and capital that maximize the output.)
4.6 MISCELLANEOUS EXERCISES FOR CHAPTER 4
1. If V D πr2 h then dV D 2πrh dr C πr2 dh, so in order for V to be equally sensitive to small changes in r and
h, we must be at a point r0 , h0 where 2πr0 h0 L πr02 so r0 D 2h0 .
2
2
2
2
2
2
2. (a) If f x1 , x2 , ... , xn D e−x1 −x2 −···−xn , then fxi x1 , x2 , ... , xn D −2xi e−x1 −x2 −···−xn and is 0 only when
xi D 0. So the only critical point is at the origin.
2
2
2
(b) If i Z j, then fxi xj x1 , x2 , ... , xn D 4xi xj e−x1 −x2 −···−xn , so fxi xj 0, 0, ... , 0 D 0. Also fxi xi x1 , x2 , ... ,
2
2
2
xn D −2 C 4xi2 e−x1 −x2 −···−xn , so fxi xi 0, 0, ... , 0 D −2. The Hessian is an n * n diagonal matrix
with −2’s on the main diagonal and 0’s everywhere else. It is easy to calculate di 0, 0, ... , 0 D −2i and
so by the second derivative test, f has a local maximum at the origin.
3. We are asked to maximize the profit Px, y D x − 280 − 100x C 40y C y − 420 C 60x − 35y D
−100x2 C 40x − 35y2 C 80y C 100xy − 240. The partial derivatives are Px x, y D −200x C 100y C 40
and Py x, y D 100x − 70y C 80. These are both zero at (27/10, 5). You can read the Hessian right off the
first derivatives and you see that d1 D −200 < 0 and d2 D 4000 > 0 so profit is maximized when you charge
$2.70 for Mocha and $5 for Kona.
4. (a) Revenue is Rx, y, z D 1000x4 − .02x C 1000y4.5 − .05y C 1000z5 − .1z D −20x2 C
4000x − 50y2 C 4500y − 100z2 C 5000z.
(b) The prices are $3.88, $4.25, and $4.60 at x, y, z D 6, 5, 4 and the prices are $3.98, $4.35, and $4.70 at
x, y, z D 1, 3, 3. The difference is R1, 3, 3 − R6, 5, 4 D 31130 − 62930 D −31800. The revenue
will decline by $31800 if the prices are raised.
(c) The partials are Rx x, y, z D 4000 − 40x, Ry x, y, z D 4500 − 100y, and Rz x, y, z D 5000 − 200z
so the critical point is (100, 45, 25) so the selling prices should be $2 for X, $2.25 for Y , and $2.50 for Z.
5. We note that there must be a maximum and a minimum value because we are constrained to lie on the surface of
a sphere, which is compact.
√
(a) Maximize f x, y, z D x − 3y subject to gx, y, z D x2 C y2 C z2 D 4. Using the Lagrange multiplier
method, we solve


1 D 2λx


 −√ 3 D 2λy

0 D 2λz


 x2 C y2 C z2 D 4.
√
From the first equation λ Z 0 so from the third equation
z D 0. From the first two equations, y D − 3x
√
2
2
;1, − 3, 0. We check to find we have a maximum of 4 at
so since
√ x C 3x D 4 our critical points are √
1, − 3, 0 and a minimum of −4 at −1, 3, 0.
√
(b) Now
√ we are looking at the function f θ, ϕ D 2 sin√ϕ cos θ − 2 3 sin ϕ sin θ so fθ θ, ϕ D −2 sin ϕ sin θ −
2 3 sin ϕ cos θ and fϕ θ, ϕ D 2 cos ϕ cos θ − 2 3 cos ϕ sin θ so
√
0 D −2 sin ϕsin θ C√ 3 cos θ
0 D 2 cos ϕcos θ − 3 sin θ.
√
√
Either ϕ D 0 or π and cos θ D 3 sin θ, or ϕ D π/2 and sin θ D − 3 cos θ. When you check these six
points you get the same results as in part (a).
6. (a) Here we are maximizing T x, y, z D 200xyz2 subject to the constraint gx, y, z D x2 C y2 C z2 D 1.
Using the Lagrange multiplier method, we solve


200yz2 D 2λx


 200xz2 D 2λy

400xyz D 2λz


 x2 C y2 C z2 D 1.
that
From the third equation, λ Z 0 so 2λ D 400xy so 2x2 D 2y2 D z2 . From the last equation we see √
4x2 D 1 so our critical points are the eight possible combinations of x D ;1/2, y D ;1/2 and z D ;1/ 2.
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“runall” — 2005/8/1 — 15:11 — page 214 — #34
214
Chapter 4
7. (a)
(b)
(c)
(d)
(e)
Maxima and Minima in Several Variables
The temperature√is a maximum of 25 when the sign of x and y are the same. This is at the four points
;1/2, 1/2, ;1/ 2.
fx x, y D 2x−3y C 4x2 while fy x, y D 2y − 3x2 . From fx we see that either x D 0 or y D 4/3x2 .
But from the second equation y D 3/2x2 . So we conclude that the only solution is at (0, 0).
0 0
fxx x, y D 6−y C 4x2 , fxy x, y D −6x, and fyy x, y D 2. At the origin, the Hessian is 0 2 and so the determinant is 0 and the critical point is degenerate.
If y D mx then the original equation becomes F x D m2 x2 − 3mx3 C 2x4 . We calculate F x D
2m2 x − 9mx2 C 8x3 D 2xm2 − 9mx/2 C 4x2 . From the second derivative we see that F x D
2m2 − 18mx C 24x2 . This is positive at x D 0 for all m Z 0 so there is a minimum for x D 0 along any
line other than the two axes. When m D 0, F x D 8x3 and so the first derivative test implies that there is
a minimum at x D 0 when m D 0. Finally, consider Gy D f 0, y D y2 . This clearly has a minimum at
y D 0. We’ve shown that along any line through the origin, f has a minimum at (0, 0).
Consider gx D f x, 3x2 /2 D −x2 /2x2 /2 D −x4 /4. From the derivative g x D −x3 we see
that g has a maximum at x D 0 and hence f has a maximum at the origin when constrained to the given
parabola. This means that the origin is actually a saddle point for f .
A portion of the surface is shown below.
z
y
x
8. (a) Here we are finding the critical points of f x, y D xy subject to the constraint gx, y D x2 C y2 − 1 D 0.
So taking the partials of f x, y D λgx, y along with the constraint we get the following system of
equations.


 y D 2λx
x D 2λy

 1 D x2 C y 2 .
√
√
√
√
The√solutions√correspond to λ
√ D ;1/2√and are the four critical points 1/ 2, 1/ 2, −1/ 2, 1/ 2,
1/ 2, −1/ 2, and −1/ 2, −1/ 2.
(b) Here is a contour plot of f x, y D xy along with the constraint curve x2 C y2 D 1 and the four critical
points.
1.5
1
0.5
0
-0.5
-1
-1.5
-1.5
-1
-0.5
0
0.5
1
1.5
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“runall” — 2005/8/1 — 15:11 — page 215 — #35
Section 4.6
Miscellaneous Exercises for Chapter 4
215
(c) You can see from the figure that f is at its highest value along the constraint curve at two
√ of the
√ critical
points and at its lowest at two√
of the others.
In
particular,
f
has
a
constrained
max
at
;1/
2,
1/
2 and
√
has a constrained min at ;1/ 2, −1/ 2.
9. (a) Here we are finding the critical points of f x, y, z D xy subject to the constraint gx, y, z D x2 C y2 C
z2 − 1 D 0. So taking the partials of f x, y, z D λgx, y, z along with the constraint we get the following
system of equations.


y D 2λx



 x D 2λy

0 D 2λz



 1 D x 2 C y 2 C z2
This problem is very similar to Exercise 8 and
it is no surprise
we again get
points
√ so √
√ that √
√ four critical
√
corresponding
to
λ
D
;1/2.
They
are
1/
2,
1/
2,
0,
−1/
2,
1/
2,
0,
1/
2,
−1/
2,
0,
and
√
√
−1/ 2, −1/ 2, 0. We also get critical points at the two poles corresponding to λ D 0. These are
at 0, 0, ;1.
(b) Of course, it is harder to represent this situation than its lower-dimensional counterpart. Here are some level
sets, the unit sphere and the critical points.
1
0.5
z
2
0
1
-0.5
-1
-2
0
-1
y
-1
0
x
1
-2
2
√
√
√
(c) The √
arguments that f has a constrained max at ;1/ 2, 1/ 2, 0 and has a constrained min at ;1/ 2,
−1/ 2, 0 are the same as in Exercise 8. The two poles must be saddle points. If you travel in a direction
where y D x, f x, y is increasing while if you travel in a direction where y D −x, f x, y is decreasing.
So there are saddle points at 0, 0, ;1.
10. From the diagram you can see that we are minimizing f x, y D x C yy subject to the constraint that
x2 C y2 D 1. Because this is a physical problem, we can assume that x > 0 and y > 0. A look at the contour
plot for f along with the constraint curve lets us see that this solution will be a max.
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“runall” — 2005/8/1 — 15:11 — page 216 — #36
216
Chapter 4
Maxima and Minima in Several Variables
1.5
1
0.5
0
-0.5
-1
-1.5
-1.5
Our system of equations is
-1
-0.5


 y
x C 2y

 1
0
0.5
1
1.5
D 2λx
D 2λy
D x2 C y 2
√
√
Solving
gives
us
one
solution
for
which
x
and
y
are
positive,
namely
x
D
2 C 2/2 2 − 1 and
√
√
y D 2 C 2/2. The area of the rectangle is, therefore, 2 C 1/2.
11. Minimize f x1 , x2 , ... , xn D x12 C x22 C · · · C xn2 subject to the constraint gx1 , x2 , ... , xn D a1 x1 C a2 x2 C
· · · C an xn D 1 where not all the ai ’s are zero. We solve
for 1 … i … n
2xi D ai λ
# a x C a x C · · · C a x D 1.
1 1
2 2
n n
This means that our constrained critical point is at xi D ai λ/2 and 2/a21 C a22 C · · · C a2n D λ so xi D
ai /a21 C a22 C · · · C a2n . So our minimum is
a1
a2
an
f x1 , x2 , ... , xn D f 2
,
, ... , 2
a1 C a22 C · · · C a2n a21 C a22 C · · · C a2n
a1 C a22 C · · · C a2n
2
2
2
a2
an
a1
D 2
C 2
C ··· C 2
a1 C a22 C · · · C a2n
a1 C a22 C · · · C a2n
a1 C a22 C · · · C a2n
D
1
a21 C a22 C · · · C a2n
.
12. Minimize the function f x1 , x2 , ... , xn D a1 x1 C a2 x2 C · · · C an xn 2 subject to the constraint gx1 , x2 , ... ,
xn D x12 C x22 C · · · C xn2 D 1 where not all the ai ’s are zero. We solve
2ai a1 x1 C a2 x2 C · · · C an xn D 2λxi for 1 … i … n, and
x2 C x2 C · · · C x2 D 1.
1
2
n
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“runall” — 2005/8/1 — 15:11 — page 217 — #37
Section 4.6
Miscellaneous Exercises for Chapter 4
217
From the first equation, xi2 D ai xi /λa1 x1 C a2 x2 C · · · C an xn , so
λ D a1 x1 a1 x1 C a2 x2 C · · · C an xn C a2 x2 a1 x1 C a2 x2 C · · · C an xn C · · ·
C an xn a1 x1 C a2 x2 C · · · C an xn D a1 x1 C a2 x2 C · · · C an xn 2 so
ai
and finally,
xi D
a1 x1 C a2 x2 C · · · C an xn
n
2
ai
iD1
2
xi D a x C a x C · · · C a x 2 D 1.
1 1
2 2
n n
iD1
n
Now we can substitute back into the original equation:
f x1 , x2 , ... , xn D f D
D
a1
an
, ... ,
a1 x1 C a2 x2 C · · · C an xn
a1 x1 C a2 x2 C · · · C an xn
a21 C a22 C · · · C a2n
a1 x1 C a2 x2 C · · · C an xn
2
a21 C a22 C · · · C a2n
a1 x1 C a2 x2 C · · · C an xn 2
a1 C a2 C · · · C an 2
2
2
D a21 C a22 C · · · C a2n .
13. Since the faces are parallel to the coordinate planes, we can reduce the problem to maximizing Mx, y, z D xyz
subject to the constraint gx, y, z D x2 C 2y2 C 4z2 D 12, where x, y, and z are all positive. Here, by the
symmetry of the problem, we are maximizing the volume of one eighth of the box and therefore we will have the
dimensions of the box itself by doubling x, y, and z. We solve


yz D 2λx



 xz D 4λy

xy D 8λz



 x2 C 2y2 C 4z2 D 12.
√
So x2 D 2y2 D 4z2 and 12z2 D 12
√ so a critical point is 2, 2, 1. The dimensions of the box are twice these
values so the largest box is 4 * 2 2 * 2.
14. We are minimizing the cost of producing a sphere and a cylinder of equal radii with the given constraints. We also
need to convert 8000 gallons to 8000/7.480520 L 1069.444 cubic feet. So minimize V r, h D 2πrh C 8πr2
subject to gr, h D πr2 h C 4/3πr3 D 1069.444. We solve

2


 2πh C 16πr D λ2πrh C 4πr 2
2πr D λπr 

 πr2 h C 4/3πr3 D 1069.444.
Physically, r cannot be zero, so by the second equation λ D 2/r and then by the first h √
D 4r and so by the third
1069.444 D 4πr3 C 4/3πr3 D 16π/3r3 . Therefore, the best dimensions are r L 3 63.8277 L 3.9964 feet
and h L 15.9856 feet.
15. Minimize Mx, y, z D x2 C y2 C z2 subject to x2 − y − z2 D 1. We solve


2x D 2λx



 2y D −2λy − z

2z D 2λy − z



 x2 − y − z2 D 1.
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“runall” — 2005/8/1 — 15:11 — page 218 — #38
218
Chapter 4
Maxima and Minima in Several Variables
Since the last equation implies that x Z 0, the first equation gives us that λ D 1, so y D z D 0 and thus x D ;1.
The minimum distance is, therefore, 1.
16. Place the vertex of the cone at the North Pole (0, 0, a), with the axis of symmetry of the cone coinciding with
the z-axis. The height of the cone is h and the radius is r. We are maximizing V r, h D 1/3πr2 h subject to
the constraint h − a2 C r2 D a2 or gh, a D h2 − 2ha C r2 D 0. We solve



 2/3πrh D 2λr
1/3πr2 D 2λh − a


 h2 − 2ha C r2 D 0.
From the first equation we know λ Z 0 and πh/3 D λ. So substitute this into the second
equation to find that
√
r2 D 2h2 − 2ah. Solve this with the third equation to find that h D 4a/3 and r D 2 2a/3.
17. We want to maximize V D xyz subject to bcx C acy C abz D abc.
z
corner (x,y,z)
on plane
x
y
or



 yz D λbc


 xz D λac
xy D λab



y
x
z


C
C D 1.

a
b
c
yD
bc
b
x D x.
ac
a
Thus we solve
§V D λ§bcx C acy C abz
# bcx C acy C abz D abc
Hence λ D
yz
xy
xz
D
D
.
bc
ac
ab
yz
xz
D
3 z D 0 or
bc
ac
Now z D 0 makes V D 0, so this cannot possibly maximize. Thus y D b/ax. Now
or z D
bc
c
x or z D x. Hence the constraint becomes
ab
a
x
x
x
C
C D1
a
a
a
so
xy
yz
D
3 y D 0 (reject)
bc
ab
x D a/3 * y D b/3 z D c/3.
2
x
π
18. We have V x, y D π y D x2 y with πx C y … 108.
2
4
(a) We maximize V subject to gx, y D πx C y D 108. Thus, with a Lagrange multiplier we solve


πxy




D
πλ





xy
y
2
πx2
πx
πx

* λD
D
so x D 0 (reject) or
D
, so y D
.

πx2

2
4
2
4
2

Dλ 



4


 πx C y D 108
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“runall” — 2005/8/1 — 15:11 — page 219 — #39
Section 4.6
Miscellaneous Exercises for Chapter 4
219
πx
2 · 108
72
D 108 so x D
D
. Hence the maximizing dimensions
2
3π
π
72
are
diameter, 36 length. (That these dimensions really do maximize volume may be seen from the
π
following picture.)
In the constraint we must have πx C
constraint line
100
80
increasing
V
60
40
20
0
0
5
10
15
20
25
30
35
(b) Perhaps this is an easier method: πx C y D 108 3 y D 108 − πx so vx D V x, 108 − πx D
108
π
72
πx2
2
108 − πx defined on 0,
.
. Thus v x D 216x − 3πx so critical points are x D 0,
4
π
4
π
108
72
Compare values: v0 D 0, v > 0, v D 0, so x D 72/π must give the absolute maximum.
π
π
2
19. The two equations are x D y/2 − 1 and x D y . We will minimize the square of the distance between a point
x1 , y1 on the line and a point x2 , y2 on the parabola. Maximize f y1 , y2 D y1 /2 − 1 − y22 2 C y2 − y1 2 .
Take the first partials:
fy1 y1 , y2 D
5
y1 − 1 − y22 − 2y2
2
and
fy2 y1 , y2 D 4y22 − 2y1 y2 C 6y2 − 2y1 .
Set these equal
√ to zero and solve to find the critical point at y1 , y2 D 5/8, 1/4. The minimal distance is
therefore 3 5/8.
y
3
2
1
-1
1
2
3
4
x
-1
-2
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“runall” — 2005/8/1 — 15:11 — page 220 — #40
220
Chapter 4
Maxima and Minima in Several Variables
20. (a) For each section the time is the distance divided by the rate and the hypotenuse is the altitude divided by
the cosine of the angle that is formed by the altitude and the hypotenuse. So
T θ1 , θ2 D
a
b
C
.
v1 cos θ1
v2 cos θ2
(b) We are minimizing time subject to the constraint that the horizontal separation is constant: a tan θ1 C
b tan θ2 D c. We solve


a sin θ1
λa


D

2θ
2θ

v
cos
cos

1
1
 1
λb
b sin θ2
D



v2 cos2 θ2
cos2 θ2



 a tan θ1 C b tan θ2 D c.
sin θ1
v
D 1.
sin θ2
v2
21. We are minimizing the square of the distance f x, y D x − x0 2 C y − y0 2 subject to the constraint
ax C by D d. We solve



 2x − x0 D aλ
2y − y0 D bλ


 ax C by D d.
The first two equations immediately give the result:
Solving, we see that x D aλ C 2x0 /2 and y D bλ C 2y0 /2 so substituting for x and y in the third equation
a2 C b2 λ D 2d − ax0 − by0 . Also substituting for x and y in f we see that
2
f x, y D D
2
aλ
bλ
a2 C b 2
a2 C b2 2d − ax0 − by0 2
C D
λ D
2
2
4
4
a2 C b 2
2
d − ax0 − by0 2
a2 C b 2
|ax0 C by0 − d|
√
.
a2 C b 2
22. This is very similar to Exercise 21. Minimize the square of the distance f x, y, z D x − x0 2 C y − y0 2 C
z − z0 2 subject to the constraint ax C by C cz D d. We solve



 2x − x0 D aλ

 2y − y0 D bλ
 2z − z0 D cλ



 ax C by C cz D d.
so the distance D is the square root of this: D D
Solving, we see that x D aλ C 2x0 /2, y D bλ C 2y0 /2 and z D cλ C 2z0 /2 so substituting for x, y
and z in the fourth equation a2 C b2 C c2 λ D 2d − ax0 − by0 − cz0 . Also substituting for x, y and z
in f we see that
d − ax0 − by0 − cz0 2
a2 C b 2 C c 2
2
f x, y, z D λ D
4
a2 C b 2
so the distance D is the square root of this: D D
23. (a) We solve
|ax0 C by0 C cz0 − d|
√
.
a2 C b 2 C c 2



2xy2 z2 D 2λx



 2x2 yz2 D 2λy
 2x2 y2 z D 2λz




 x 2 C y 2 C z 2 D a2 .
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“runall” — 2005/8/1 — 15:11 — page 221 — #41
Section 4.6
Miscellaneous Exercises for Chapter 4
221
If λ D 0, then at least one of x, y and z is 0 and this corresponds to a minimum. If λ Z 0, we see that, at a
critical point, x2 D y2 D z2 , so 3x2 D a2 or x2 D a2 /3. Therefore, at a critical point,
3
f x, y, z D a6
a2
.
D
3
27
(b) In part (a) we showed x2 y2 z2 … a2 /33 D [x2 C y2 C z2 /3]3 and so this result follows immediately.
(c) We make the appropriate adjustments to parts (a) and (b) and maximize f x1 , x2 , ... , xn D x12 x22 · · · xn2
subject to x12 C x22 C · · · C xn2 D a2 . Because, as in part (a), the case λ D 0 corresponds to a minimum,
we see that at a maximum no xi is 0. So we solve
2x12 x22 · · · xn2 /xi D 2λxi
x12 C x22 C · · · C xn2 D a2 .
for 1 … i … n
At a maximum, x12 D x22 D · · · D xn2 , so xi2 D a2 /n. Therefore, the maximum of f is a2 /nn . So we
conclude that x12 x22 · · · xn2 … a2 /nn D [x12 C x22 C · · · C xn2 /n]n . The result follows immediately.
(d) We found that f was maximized when x12 D x22 D · · · D xn2 so, since here we are assuming xi > 0 for all
i, the equality holds when x1 D x2 D · · · D xn .
24. (a)
n
n
n
⭸f
D akj xj C aik xi D ajk C akj xj
⭸xk
jD1
iD1
jD1
⭸g
D 2xk .
⭸xk
Thus the Lagrange multiplier system is




aj1 C a1j xj D 2λx1


j






 #
#
#




 ajn C anj xj D 2λxn



j



 x12 C · · · C xn2 D 1.
(b) Because A is symmetric, ajk D akj so the system becomes




2a1j xj D 2λx1


j





 #

#
#





 2anj xj D 2λxn


j



 x2 C · · · C x2 D 1.
1
n
The first n equations come from §f D λ§g and simplify to




a1j xj D λx1


j




 #
#


#





anj xj D λxn .


 j
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“runall” — 2005/8/1 — 15:11 — page 222 — #42
222
Chapter 4
Maxima and Minima in Several Variables
Note that akj xj is the dot product of the kth row of A with x. So the n equations, taken together, express
j
Ax D λx.
(c)
f x1 , ... , xn D xT Ax D xT λx x is an eigenvector
D λxT x D λx · x
D λx2 D λ · 1,
since x is assumed to be a unit vector.
25. (a) To set things up using Lagrange multipliers, we solve






 2ax C 2by D 2λx
 a − λx C by D 0
2bx C 2cy D 2λy 3
bx C c − λy D 0




 x2 C y 2 D 1
 x2 C y2 D 1.
In the last system, multiply the first equation by λ − c and the second by b, then add to obtain:
a − λλ − c C b2 x D 0.
Now multiply the first equation by b and the second by λ − a, then add to get:
b2 C λ − ac − λy D 0.
Since x2 C y2 D 1, we cannot have both x and y equal to 0. Thus
b2 C λ − ac − λ D 0 3 λ2 − a C cλ C ac − b2 D 0.
Hence
λ1 , λ2 D
a C c ; a C c2 − 4ac − b2 a C c ; a − c2 C 4b2
2
.
. a − c2 C 4b2 Ú 0 so the eigenvalues are always real.
2
26. (a) λ1 D λ2 3 a − c2 C 4b2 D 0 3 a D c, b D 0 so f x, y D ax2 C y2 .
(b) The eigenvalues are the max and min values of f on the circle. If both are positive, then f has a positive
minimum on the circle; hence f must be positive on the entire circle.
(c) If both eigenvalues are negative, then f has a negative maximum on the circle—so f must be negative on
the entire circle.
27. (a)
(b) Rewriting, λ1 , λ2 D
n
n
i,jD1
i,jD1
f kx1 , ... , kxn D aij kxi kxj D k 2 aij xi xj
(b) Let u D x/x when x Z 0. Then u is a point on the unit hypersphere. If f has a positive minimum on the
hypersphere, then f must be positive on the entire hypersphere. Hence, for x Z 0:
f x D f ku D k 2 f u > 0
k D x.
The case where f has a negative maximum on the hypersphere is similar.
(c) Clearly the converses of the results of part (b) hold (i.e., if f x > 0 for all x Z 0, then f is positive on
the hypersphere ...). From Exercise 24, the minimum value of f is the smallest eigenvalue of A. Thus the
quadratic form is positive definite 3f x > 0 for all x Z 0 3 f is positive on the hypersphere 3 the
smallest eigenvalue of A is positive 3 all eigenvalues are positive. (The negative definite result is similar.)
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“runall” — 2005/8/1 — 15:13 — page 223 — #1
C H A P T E R
5
Multiple Integration
5.1 INTRODUCTION: AREAS AND VOLUMES
1.
2 3
0
x2 C y dy dx D
1
2
0
D
π 2
0
2
2x2 C 4 dx D 2x3 /3 C 4x D 40/3.
0
yD2
π
y2
1
sin
x
dx
D
2 sin x − sin x dx
2
2
0
0
yD1
π
π
3
3
3
3
D
sin x dx D − cos x D C D 3.
2 0
2
2
2
0
π
y sin x dy dx D
1
3.
4 1
xe dy dx D
4
y
−2 0
yD1
4
xe dx D
xe − 1 dx
y
−2
π/2 1
4
x2
D e − 1
2
π/2
e cos y dx dy D
x
0
0
−2
yD0
−2
4.
0
yD1
2
0
2.
yD3
2
x2 y C y2 /2
dx D
3x2 C 9/2 − x2 C 1/2 dx
D 8 − 2e − 1 D 6e − 1.
xD1
π/2
e cos y
dy D
e − 1 cos y dy
x
0
xD0
0
π/2
D e − 1.
D e − 1 sin y
0
5.
2 1
1
e xCy C x2 C ln y dx dy D
0
2 1
1
e x e y C x2 C ln y dx dy D
0
2
1
x y
e e C
xD1
x3
dy
C x ln y 3
xD0
2
y
1
y
C ln y dy D e − 1e C
C y ln y − y D
e − 1e C
3
3
1
1
2
y
D e − 1e 2 − e C
1
2
C 2 ln 2 − 1 D e 3 − 2e2 C e −
C 2 ln 2.
3
3
6.
9 e
1
1
9 e
√
ln x
ln x
1
dx dy D
dx dy treat y as a constant—use substitution
xy
2 1 1 xy
xDe
9
9
9
ln x2 1
1
ln 9
1
1
D
dy D
.
dy D ln y D
2 1
2y xD1
2 1 2y
4
4
1
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“runall” — 2005/8/1 — 15:13 — page 224 — #2
224
Chapter 5
Multiple Integration
7. (a) Here we are fixing x and finding the area of the slices:
2
y3
C 2y D 2x2 C 20/3.
x C y C 2 dy D x y C
Ax D
3
0
0
2
2
2
2
Now we “add up the areas of these slices”:
V D
2
−1
Ax dx D
2
40
2
20
2
20 16
x D C
2x2 C 20/3 dx D x3 C
− − −
D 26.
3
3
3
3
3
3
−1
−1
2
(b) Now we fix y and find the area of the slices:
Ay D
2
−1
D
x2 C y2 C 2 dx D 2
x3
C y2 x C 2x 3
−1
8
1
C 2y2 C 4 − − − y2 − 2 D 9 C 3y2 .
3
3
Adding up the area of these slices:
V D
2
Ay dy D
2
0
0
2
9 C 3y2 dy D 9y C y3 D 26.
0
8. Here we are calculating:
2 3
1
3
2
x2
9
C 3yx C x dy D
C 9y C 3 dy
2
2
1
1
0
2
2
15
15
9 2 D
C 9y dy D y C y 2
2
2
1
1
x C 3y C 1 dx dy D
0
2
D 15 C 18 − 15/2 C 9/2 D 21.
9. Here we are calculating
2 1
−1 0
1
2
y4
2y4
2 3
2
−
dy
D
x
cos
πx
C
dy
π
π
−1 3
−1 3
0
2
2y5 2
4
64
2
2
66
D y C
.
D C
− − −
D2 C
3
5π −1
3
5π
3
5π
5π
2x2 C y4 sin πx dx dy D
2
10. This is the volume of the “rectangular box” bounded by the plane z D 2, the xy-plane, and the planes x D 1,
x D 3, y D 0, and y D 2. Here we could just calculate the volume of this 2 * 2 * 2 box as 8 without
integrating—or
2
2 3
2
2 3
2 dx dy D
2x dy D
4 dy D 4y D 8.
V D
0
1
0
0
1
0
11. This is the volume of the region bounded by the paraboloid z D 16 − x2 − z2 , the xy-plane, and the planes
x D 1, x D 3, y D −2, and y D 2. The volume is
2
3
y3 16
2
V D
16 − x − y dy dx D
16y − x y −
64 − 4x −
dx D
dx
3 −2
3
1 −2
1
1
3
16 4 3
x D 192 − 36 − 16 − 64 − 4/3 − 16/3 D 248/3.
D 64x − x −
3
3
1
3 2
3
2
2
2
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“runall” — 2005/8/1 — 15:13 — page 225 — #3
Section 5.2
Double Integrals
225
12. This is the volume of the region bounded by z D sin x cos y, the xy-plane, and the planes x D 0, x D π, y D
−π/2, and y D π/2. The volume is
π/2
π/2 π
V D
−π/2 0
π/2
D2
−π/2
sin x cos y dx dy D
π/2
cos y dy D 2 sin y
−π/2
−π/2
π
− cos x cos y dy
0
D 4.
13. This is the volume of the region bounded by z D 4 − x2 , the xy-plane, and the planes x D −2, x D 2, y D 0,
and y D 5. The volume is
V D
5 2
0
D
5
0
−2
4 − x2 dx dy D
5
0
2
5
4x − x3 /3 dy D
8 − 8/3 − −8 C 8/3 dy
−2
5
32
32 dy D
y D 160/3.
3
3 0
0
14. This is the volume of the region bounded by z D |x| sin πy, the xy-plane, and the planes x D −2, x D 3, y D 0,
and y D 1. The volume is
V D
3 1
−2 0
|x| sin πy dy dx D
3
−2
−
1
3
2|x|
|x|
cos πy dx D
dx.
π
−2 π
0
At this point we use the definition of absolute value to split this into two quantities:
3
0
2
− x dx C
V D
π
−2
0
0
3
x2 4
x2 9
13
2
C
D
x dx D − C
D
.
π
π −2
π 0
π
π
π
15.
5 2
−5 −1
D
5 − |y| dx dy D
5
−5
5
−5
2
5 − |y|x
5 − |y| · 3 dy D 150 − 3
D 150 − 3
D 150 C
0
−5
−y dy − 3
dy
xD−1
5
5
−5
|y| dy
y dy
0
0
5
75
3 75
3 2 y −
D 75.
− y2 D 150 −
2 −5
2 0
2
2
The iterated integral gives the volume of the region bounded by the graph of z D 5 − |y|, the xy-plane, and the
planes x D −1, x D 2, y D −5, y D 5. (The solid so described is a rectangular prism.)
b d
16. We have V D
f x, y dy dx. Since 0 … f x, y … M, the solid bounded by y D f x, y, the xy-plane,
a
c
and the planes x D a, x D b, y D c, y D d sits inside the rectangular block of height M and base bounded by
x D a, x D b, y D c, y D d. Hence V … Mb − ad − c
5.2 DOUBLE INTEGRALS
Note: you may want to discuss Exercise 1 (b) before assigning it, to get your students in the habit of looking critically at
problems before working on them.
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“runall” — 2005/8/1 — 15:13 — page 226 — #4
226
Chapter 5
Multiple Integration
1. (a) We are computing
2 4−x2
−2 0
x3 dy dx D
2
−2
4−x2
2
2
x3 y
dx D
4x3 − x5 dx D x4 − x6 /6
−2
0
−2
D 0.
(b) The integrand is an odd function depending only on x and the region is symmetric about the y-axis. The
a
students encounter this situation when they looked at −a x3 dx in first year calculus.
1
1 x3
1
1 x3
3 3
3 dy dx D
3y dx D
3x3 dx D x4 D . The region over which we are integrating is:
2.
4
4
0 0
0
0
0
0
y
1
0.8
0.6
0.4
0.2
0.2
2 y2
3.
0
y dx dy D
0
0.4
0.6
0.8
x
1
y2
2
2
y4 xy dy D
y3 dy D D 4. The region over which we are integrating is:
4 0
0
0
0
2
y
2
1.5
1
0.5
1
2 x2
4.
0
0
y dy dx D
2
0
2
3
4
x
x2
2
2 4
y2 32
x5 16
x
D
dx D
D
. The region over which we are integrating is:
dx D
2 0
10 0
10
5
0 2
y
4
3
2
1
x
0.5
1
1.5
2
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“runall” — 2005/8/1 — 15:13 — page 227 — #5
Section 5.2
Double Integrals
2xC1
3
xy2 1
1 3
4
x2
5.
xy dy dx D
dx D
3x3 C 4x2 C x dx D x4 C x3 C
2 −1
2 4
3
2
−1 x
−1 2 x
The region over which we are integrating is:
3 2xC1
3
3
−1
D
227
152
.
3
y
6
4
2
x
-1
1
2
3
sin x
π
π 2
y
1
y cos x dy dx D
dx D
sin2 x cos x dx D using the substitution u D sin x
cos x
2
2
0 0
0
0
0
π
xDπ
1
sin3 x u2 du D
D 0. The region over which we are integrating is:
2 xD0
6 0
π sin x
6.
y
1
0.8
0.6
0.4
0.2
0.5
1
1.5
2
2.5
3
x
Note: After you assign Exercises 7 and 8, together you can probe to see whether students see that they are the
same. This is a nice set-up for Section 5.3 where they will learn about interchanging the order of integration.
√ 1−x2
1 √
7.
3 dy dx D
3y √
dx D
6 1 − x2 dx D using the substitution x D sin t D 3π/2.
√
0 − 1−x2
0
0
− 1−x2
You can also see that the region over which we are integrating is a half-circle of radius 1 so we have found the
volume of the cylinder over this region of height 3. This figure is:
1 √ 1−x2
1
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“runall” — 2005/8/1 — 15:13 — page 228 — #6
228
Chapter 5
Multiple Integration
y
1
0.5
x
0.2
0.6
1
-0.5
-1
8. This is the same as Exercise 7 with the limits of integration reversed. The solution is again 3π/2.
1 ex
1 4 ex
1
y 3
9.
y dy dx D
dx D
0 dx D 0. The region over which we are integrating is:
0 −ex
0 4 x
0
−e
y
3
2
1
0.2
0.4
0.6
0.8
1
x
-1
-2
-3
10. For each square in the domain we need to estimate the height of the square and multiply it by the length times
the width. For our estimate we will choose the value of the height f cij in the lower right corner of the square
in row i column j as our height for the square. The heights are then:
4
4
4
4
4
5
5
5
5
5
6
6
6
6
6
7
7
7
7
7
8
8
8
8
7
9
9
9
8
8
9
10
10
9
8
10
11
10
9
8
9
10
10
9
8
9
9
9
9
8
Each box has a base of area 25 so the sum of the products of 25 times the heights is 92500. Of course, this answer
depends on what point in each box we chose for our estimate—your mileage may vary.
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“runall” — 2005/8/1 — 15:13 — page 229 — #7
Section 5.2
Double Integrals
229
11. A quick sketch of the region over which we are integrating helps us set up our double integral.
y
2
1.5
1
0.5
x
0.5
2 2−x
0
0
1
1.5
2
2−x
2
xy2 1 − xy dy dx D
dx D
2 − 3x C 2x2 − x3 /2 dx
y −
2 0
0
0
2
2 3
x4 3 2
D 2x − x C x −
D 4 − 6 C 16/3 − 2 D 4/3.
2
3
8 2
0
12. Again a sketch
√of the region over which we are integrating helps us set up our double integral. The top bounding
curve is y D x and the bottom curve is y D 32x3 .
y
0.5
0.4
0.3
0.2
0.1
x
-0.05
0.1
0.15
0.2
0.25
√ x
1/4
3 2
7
3xy dy dx D
dx D
x − 1536x dx
2 32x3
2
32x3
0
0
1/4
1
3
5
3
−
D
.
D x2 − 192x8 D
2
128
1024
1024
0
1/4 √ x
0
0.05
1/4
3xy2
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“runall” — 2005/8/1 — 15:13 — page 230 — #8
230
Chapter 5
Multiple Integration
13. We can easily determine the limits of integration from the sketch and/or by solving for where x C y D 2 intersects
the parabola y2 − 2y − x D 0.
y
2
1.5
1
0.5
-1
1
2
x
3
-0.5
-1
2−y
2
y2
y4
x2
C xy
C y3 −
C 2 dy
x C y dx dy D
dy D
−
2
2
2
2
y2 −2y
−1
−1
y −2y
2 2−y
−1
2
2
y5
y4
y3
99
D −
C
−
C 2y D
.
10
4
6
20
−1
14. We see from the sketch that we need to divide the
√ integral into two pieces. For 0 … x … 1/9 we see that x … y … 3
and for 1/9 … x … 1 we see that x … y … 1/ x.
y
3
2.5
2
1.5
1
0.5
0.2
6D
3y dA D
0.4
1/9 3
0
x
0.6
3y dy dx C
1
0.8
x
1 1/√ x
3y dy dx
1/9 x
3
1/√ x
1
3 2
3 2
D
y dx C
y dx
2 x
0
1/9 2 x
1/9
1
3
3
27
3
D
− x2 dx C
− x2 dx
2
2
2
0
1/9 2x
1/9
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“runall” — 2005/8/1 — 15:13 — page 231 — #9
Section 5.2
Double Integrals
231
1/9
1
27
3
1 3 1 3 D x − x C ln x − x 2
2
2
2
0
1/9
1
3
3
−
− ln1/9 D 1 C ln 27.
2
2
2
D
15. From the sketch below we see that this is a fairly straightforward integral.
y
6
4
2
x
-2
-1
1
2
-2
6D
x − 2y dA D
D
2 x2 C2
−2 2x2 −2
2
x − 2y dy dx
x2 C2
2
xy − y dx D
3x4 − x3 − 12x2 C 4x dx
2
2
−2
−2
2x −2
2
D 3x /5 − x /4 − 4x C 2x 5
4
3
2
−2
D 192/5 − 64 D −128/5
16. From the sketch below we see that this integral needs to be done in two pieces.
y
3
2.5
2
1.5
1
0.5
0.25
6D
x C y dA D
2
1 3x
2
1
0
D
1
0
0.75
1
x C y dy dx C
2
0
D
0.5
1.25
1.5
√ 3 3/x
2
x
3x
x2 y C y3 /3 dx C
x
32/3x3 dx C
√3
1
√
1
x
1.75
x2 C y2 dy dx
x
3
3/x
x2 y C y3 /3 dx
x
9/x3 C 3x − 4x3 /3 dx D 8/3 C 10/3 D 6
1
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“runall” — 2005/8/1 — 15:13 — page 232 — #10
232
Chapter 5
Multiple Integration
17. As in the proof of property 1 in the text, we note that the Riemann sum whose limit is
n
6R
18.
6R
g dA D
6R
n
cf cij Aij D c
cf dA is
i,jD1
f cij Aij ' c
i,jD1
f C [g − f ] dA which, by property 1, equals
6R
f dA C
6R
6R
f dA.
[g − f ] dA. But g − f Ú 0
[g − f ] dA Ú 0 and so
g dA Ú
f dA.
6R
6R
6R
19. Define f C D maxf , 0 and f − D max−f , 0. Note that both f C and f − have only non-negative values.
Then f D f C − f − and |f | D f C C f − . Since f ; … |f | D f C C f − we can see that |f | is Riemann
integrable. Also we can use property 2 to conclude that
C
−
C
−
f dA D f − f dA D f dA −
f dA
6
6R
6R
6R
R
so
…
6R
f C dA C
6R
f − dA D
6R
|f | dA.
20. (a) Intuitively, the volume of a figure with constant height should be the area of the base times the height. In
this case that is just the area of the base. More formally, by Definition 2.3,
6D
1 dA D
n
lim
all xi ,yj '0 i,jD1
xi yj .
We are assuming that D is an elementary region; let’s consider the case of a Type 1 region, then we can
rewrite the above sum as
b
n
δci − γci xi D
δx − γx dx D the area of D.
lim
all xi '0 iD1
a
The proof is not much different for
the other elementary regions.
a √ a2 −x2
a √
1 dA D
dy dx D 2
a2 − x2 dx. We’ve seen this above in Exercises
(b) We integrate
√
−a − a2 −x2
−a
6D
7 and 8. Let x D a sin t and integrate to get the desired result.
21. Using Exercise 20, the area is
6A
1 dA D
1 x2
0
x3
1 dy dx D
1
x2 − x3 dx
0
1
D x3 /3 − x4 /4 D 1/3 − 1/4 D 1/12.
0
22. Again using Exercise 20, the area is
6A
1 dA D
√ 5−2 1−2x−x2
0
1 dy dx D
2x
√ 5−2
1 − 4x − x2 dx
0
√ 5−2
√
D x − 2x − x /3
D 1/310 5 − 22.
2
3
0
a
23. We integrate
√
−a −
b2 −b2 x2 /a2
√ 2
b −b2 x2 /a2
a
dy dx D 2
−a
b2 −
a √
b2 x2
2b
b
dx D
a2 − x2 dx D πa2 D
a2
a
a
−a
πab.
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“runall” — 2005/8/1 — 15:13 — page 233 — #11
Section 5.2
Double Integrals
233
24. (a) For x Ú 0 the curve x3 − x lies below the curve y D
ax2 between 0 and their positive point of intersection
√
aC√ a2 C4/2 ax3
a C a2 C 4
. So the area is given by
dy dx.
xD
2
x3 −x
0
(b) The graph of area against a is:
Area
2
1.5
1
0.5
a
0.2
0.4
0.6
1
0.8
1.2
1.4
The area is 1 at a L .995.
25. The graphs of y D x2 − 10 and y D 31 − x − 12 intersect at x D −4 and x D 5 with the graph of
y D x2 − 10 lying below the graph of y D 31 − x − 12 on this interval.
31−x−12
5
5 31−x−12
4x C 2y C 25 dy dx D
4xy C y2 C 25y
dx
−4 x2 −10
−4
D
5
−4
x2 −10
−12x3 − 78x2 C 330x C 1800 dx
5
D −3x4 − 26x3 C 165x2 C 1800x
−4
D 11664.
26. (a) This is√a special case of the region over which we integrated in Exercise 20 (b). The integral is
2 4−x2
x2 − y2 C 5 dy dx.
√
−2 − 4−x2
(b) You can use your favorite computer algebra system. Using Mathematica, enter the command:
Integrate[Integrate[x2 − y2 C 5, {y, −Sqrt[4 − x2 ], Sqrt[4 − x2 ]}], {x, −2, 2}] or
Integrate[x2 − y2 C 5, {x, −2, 2}, {y, −Sqrt[4 − x2 ], Sqrt[4 − x2 ]}] and get the answer 20π.
27. By symmetry we see that the volume is four times the volume of the piece over the first quadrant x, y Ú 0. In
this region |x| D x and |y| D y so the volume is
2−x
2 2−x
2
2
4
2 − x − y dy dx D 4
2y − xy − y2 /2
dx D 4
2 − 2x C x2 /2 dx
0
0
0
0
2
D 42x − x2 C x3 /6 D 16/3.
0
0
The results demonstrated in Exercises 28 and 29 are arrived at easily but worth seeing. In Exercise 28 we have
the dream situation where the double integral of a product can be split into the product of integrals. We quickly
see that this only works in a very special case. In Exercise 29 we examine a function where
6
f dy dx exists but
f dA does not.
6
28. (a) The function hx, y D f xgy satisfies the conditions of Theorem 2.6 (Fubini’s theorem) on [a, b] *
[c, d]. So:
6R
b
f xgy dA D
d
f xgy dy dx.
a
c
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“runall” — 2005/8/1 — 15:13 — page 234 — #12
234
Chapter 5
Multiple Integration
For emphasis, we rewrite this last integral with parentheses and, since f x does not depend on y, we have:
b
d
b d
f xgy dy dx D
f x gy dy dx.
a
c
a
c
d
But
gy dy is constant so we can pull it out of this last integral to get the result:
c
b
a
d
d
b
f x gy dy dx D gy dy f x dx .
c
c
a
(b) If D is an elementary region we can perform the first step above, if D is not an elementary region, there’s
not much we can do. For example, if D is a Type 1 region, D D {x, y|γx … y … δx, a … x … b} then
δx
b δx
b
f xgy dA D
f xgy dy dx D
f x gy dy dx.
a
γx
a
γx
6
R
2
2
2
1
29. (a) If x is rational, then
f x, y dy D
1 dy D 2. If x is irrational, then
f x, y dy D
0 dy C
0
0
0
0
2
2 dy D 2.
1
1 2
1
(b) Using our answer from part (a),
f x, y dy dx D
2 dx D 2.
0
0
0
(c) If cij has a rational x coordinate, then f cij D 1 and so the Riemann sum will converge to the area of the
region, which is 2.
(d) In this case f cij D 1 for our points in the region [0, 1] * [0, 1] and f cij D 2 for our points in the
region [0, 1] * [1, 2]. In short, the Riemann sums will converge to 11 C 21 D 3.
(e) As we saw in parts (c) and (d), the Riemann sum does not have a well defined limit and so f fails to be
integrable on R, even though in part (b) we actually computed the iterated integral.
5.3 CHANGING THE ORDER OF INTEGRATION
This is a good section in which to encourage students to explore with a computer system.
1. (a)
2 2x
0
x2
2x C 1 dy dx D
2
2x C 12x − x2 dx D
0
2
x4
3
2 D −
C x C x D 4.
2
0
2
−2x3 C 3x2 C 2x dx
0
(b) The region of integration is bounded above by y D 2x and below by y D x2 :
y
4
3
2
1
x
0.5
1
1.5
2
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“runall” — 2005/8/1 — 15:13 — page 235 — #13
Section 5.3
Changing The Order of Integration
235
(c)
4 √y
0
√ y
4
√
y
y2
C
C y dy
x2 C x dy D
−
4
2
0
0
y/2
4
y3
y2
2y3/2 D −
C
C
D 4.
12
4
3
0
2x C 1 dx dy D
y/2
4
Note: In Exercises 2–9, most students will find the biggest challenge in reversing the order of integration (the topic
of this section). You may want to suggest that they reverse the order of integration in all of the exercises, but that they
evaluate both iterated integrals only in Exercises 7–9.
2. The region of integration is:
y
1
0.8
0.6
0.4
0.2
0.2
1 x
0
0.6
2 − x − y dy dx D
0
1 1
0
0.4
1
0.8
x
1
2x − 3x2 /2 dx D 1/2
and
0
2 − x − y dx dy D
1
0
y
3 2
y − 2y C 1 dy D 1/2.
2
3. The region of integration is:
y
4
3
2
1
0.5
1
1.5
2
x
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“runall” — 2005/8/1 — 15:13 — page 236 — #14
236
Chapter 5
Multiple Integration
2 4−2x
0
0
4 2−y/2
0
y dy dx D
2
2x2 − 8x C 8 dx D 16/3 and
0
y dx dy D
4
0
−y2 /2 C 2y dy D 16/3.
0
4. The region of integration is:
y
2
1.5
1
0.5
1
2 4−y2
2
x dx dy D
2
0
0
4 √ 4−x
4
0
0
x dy dx D
0
4
3
x
4 − y2 2
dy D 128/15 and
2
√
x 4 − x dx D 128/15.
0
5. The region of integration is:
y
8
6
4
2
x
0.5
1
9 3
0
x C y dx dy D
√
1.5
9
y
0
3 x2
9
0
0
x C y dy dx D
2
2.5
3
1
−2y3/2 C 5y C 9 dy D 891/20 and
2
x4 /2 C x3 dx D 891/20.
0
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“runall” — 2005/8/1 — 15:13 — page 237 — #15
Section 5.3
Changing The Order of Integration
237
6. The region of integration is:
y
20
15
10
5
x
0.5
1
3 ex
0
1
1
ln y
1.5
2 dy dx D
3
2
2.5
2ex − 2 dx D 2e3 − 8
3
and
0
e3 3
2 dx dy D
3
6 − 2 ln y dy D 2e3 − 8.
0
7. The region of integration is:
y
1
0.8
0.6
0.4
0.2
x
0.5
1 2y
1 x
0
x/2
0
e x dy dx C
x/2
1
1.5
e 2y − e y dy D
0
y
2 1
1
e x dx dy D
1
e x dy dx D
1
0
xex /2 dy C
2
1
2
1 2
e − 2e C 1 and
2
e x − xex /2 dy D
1
e2
C
− e.
2
2
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“runall” — 2005/8/1 — 15:13 — page 238 — #16
238
Chapter 5
Multiple Integration
8. The region of integration is:
y
1
0.8
0.6
0.4
0.2
x
0.25
π/2 cos x
0
0.5
0.75
0
1.25
1.5
π/2
sin x dy dx D
0
1 cos−1 y
1
cos x sin x dx D 1/2 and
0
sin x dx dy D
1
0
1 − y dy D 1/2.
0
9. The region of integration is:
y
2
1.5
1
0.5
x
-2
-1
2 √ 4−y2
−
0
√
4−y2
2 √ 4−x2
−2 0
1
y dx dy D
2
2
2y 4 − y2 dy D 16/3
and
0
y dy dx D
2
−2
−x2 /2 C 2 dx D 16/3.
10. The limits of integration describe a region D bounded on the top by the line y D −x and on the bottom by the
parabola y D x2 − 2, as shown in the figure.
y
2
1
-2
-1.5
-1
-0.5
0.5
1
x
-1
-2
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“runall” — 2005/8/1 — 15:13 — page 239 — #17
Section 5.3
Changing The Order of Integration
239
To reverse the order of integration we must divide D into two regions by the line y D −1. Then the original
integral is equivalent to the sum
−1 √ yC2
2 −y
x − y dx dy C
x − y dx dy
√
√
−2
− yC2
The first of these integrals is
−1 √ yC2
−2
√
− yC2
D
1
−1 − yC2
x − y dx dy D
−1
−2
−2y y C 2 dy
√
−2u − 2 u du D −2
1
0
u3/2 − 2u1/2 du
0
1
2
2
4
28
4
.
D −2 u5/2 − u3/2 D −2 − D
5
3
5
3
15
0
The second integral is
2 −y
2
1 2
1
2
y − y C 2 C y − y y C 2 dy
2
−1 − yC2
−1 2
2
2
4
√
1 3
1 2
3
D y − y − y −
y y C 2 dy D −
u − 2 u du
2
4
4
−1
1
−1
4
2
3
64
32
2
4
4
3
D − u5/2 − u3/2 D −
C
C
−
4
5
3
4
5
3
5
3
1
√
D−
x − y dx dy D
139
.
60
28
139
9
−
D− .
15
60
20
11. The limits of integration describe a region D bounded on the left by x D y − 4 and on the right by the parabola
x D 4y − y2 .
Thus the final answer is
y
4
3
2
1
-4
-2
2
4
x
-1
To reverse the order of integration, divide D into two regions by the line x D 0 (the y-axis). The original integral
is equivalent to
0 xC4
4 2C√ 4−x
y C 1 dy dx C
y C 1 dy dx
√
√
−5 2− 4−x
D
0
0
2− 4−x
√
√
1
1
2
2
x C 4 C x C 4 − 2 − 4 − x − 2 C 4 − x dx
2
−5 2
4
√
√
√
√
1
1
2
2
C
2 C 4 − x C 2 C 4 − x − 2 − 4 − x − 2 − 4 − x dx
2
2
0
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“runall” — 2005/8/1 — 15:13 — page 240 — #18
240
Chapter 5
Multiple Integration
0
√
11 2
1
x C x2 dx C
6 C 3 4 − x C
2
2
−5
625
241
C 32 D
.
D
12
12
D
4
√
6 4 − x dx
0
12. The limits of integration of the first integral describe the triangular region D1 bounded on top by y D x:
y
1
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
x
1
The limits of integration of the second integral describe the triangular region D2 bounded by y D 2 − x:
y
1
0.8
0.6
0.4
0.2
0.25 0.5 0.75
1
1.25 1.5 1.75
2
x
Taken together, we obtain the triangular region D below
y
1
0.8
0.6
0.4
0.2
x
0.25 0.5 0.75
1
1.25 1.5 1.75
2
Reversing the order of integration, we find that the sum of the integrals equals
1 2−y
1
sin x dx dy D
− cos2 − y C cos y dy
0
y
0
1
D sin2 − y C sin y D sin 1 C sin 1 − sin 2
0
D 2 sin 1 − sin 2.
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“runall” — 2005/8/1 — 15:13 — page 241 — #19
Section 5.3
Changing The Order of Integration
241
13. The limits of integration of the first integral describe the region D1 bounded on the left by the x-axis, on the right
by x D y/3 (or, equivalently, by y D 3x2 ) and on top by x D 8.
y
8
6
4
2
0.25
0.5
0.75
1
1.25
x
1.5
The limits of integration of the second integral describe the region D2 bounded on the bottom by y D 8, on the
√
left by x D y − 8 (which is equivalent to y D x2 C 8), and on the right by x D −y/3.
y
12
10
8
6
4
2
0.25 0.5 0.75
1
1.25 1.5 1.75
2
x
Together, D1 and D2 give the full region D of integration.
y
12
10
8
6
4
2
0.25 0.5 0.75
1
1.25 1.5 1.75
2
x
When we reverse the order of integration, the sum of integrals is equal to
2 x2 C8
0
3x2
y dy dx D
2
0
D
1
2
1
x2 C 82 − 9x4 dx
2
2
−8x4 C 16x2 C 64 dx
0
1
256
128
896
D −
C
C 128 D
.
2
5
3
15
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“runall” — 2005/8/1 — 15:13 — page 242 — #20
242
Chapter 5
Multiple Integration
14. We reverse the order of integration:
1 3
0
cos x dx dy D
2
3y
3 x/3
0
cos x dy dx D
2
3
0
0
x/3
y cos x dx
2
3
1
sin x2 sin 9
D
.
x cos x2 dx D
D
3 0
6 0
6
3
0
15. We reverse the order of integration:
1 1
0
x2 sin xy dx dy D
1 x
0
y
x2 sin xy dy dx D
0
1
0
x
−x cos xy dx
0
1
1 2
1
2 x − x cos x dx D x − sin x D 1 − sin 1.
D
2
2
0
0
1
2
16. We reverse the order of integration:
π π
0
y
sin x
dx dy D
x
π x
0
0
sin x
dy dx D
x
π
0
x
π
π
y sin x dx D
sin x dx D − cos x D 2.
x 0
0
0
17. We reverse the order of integration:
3 9−x2
0
0
√ 9−y
9
xe3y
x2 e 3y dy
dx dy D
9 − y
0 0
0 29 − y 0
9
9
e 27 − 1
3y
3y
D
.
e /2 dx D e /6 D
6
0
0
xe3y
dy dx D
9 − y
9 √ 9−y
18. We reverse the order of integration:
2 1
0
e −x dy dx D
2
1 2x
0
y/2
D
1
0
e −x dy dx D
2
0
1
0
2x
2 e −x y dx
0
1
1
−x2
−x2 2xe dx D −e
D 1 − .
e
0
Note: It’s kind of interesting to see, in Exercises 19–21, that order of integration matters to us and to computer algebra
systems.
19. (a) After churning for a while the program returned a sum of terms that included Bessel functions, Gamma
functions and other non-trivial and non-elightening results.
(b) You would use integration by parts twice and then substitute back in to eliminate the integral.
1 2y
(c) In a blink of an eye you get
y2 cos xy dx dy D 1/41 − cos 2.
0
0
20. (a) Again, the program thought for a while and warned that inverse functions were being used and that values
could be lost for multivalued inverses. This time, however, it did come up with the correct answer of
1/41 − cos 81. √
9 y
(b) The calculation
x sin y2 dx dy resulted in the same answer, but the solution came much more quickly.
0
0
21. (a) The software did nothing more than typeset the integral and leave it unevaluated.
π/2 sin x
(b) This time Mathematica quickly calculated the integral
ecos x dy dx D e − 1.
0
0
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“runall” — 2005/8/1 — 15:13 — page 243 — #21
Section 5.4
Triple Integrals
243
5.4 TRIPLE INTEGRALS
In Exercises 1–3, use Theorem 4.5, Fubini’s Theorem, to integrate in the most convenient order. Exercise 4 asks the
students to reconsider what happened in Exercise 1. Exercise 3 is a nice opportunity to look back at a result from Section 5.2.
1. If we integrate with respect to x first, the integral simplifies:
l
[−1,1] * [0,2] * [1,3]
xyz dV D
D
3 2 1
1
0
1
0
3 2
−1
xyz dx dy dz D
3 2
1
0
1
x2 yz dy dz
2 −1
0 dy dz D 0.
2. Here order doesn’t matter.
l
[0,1] * [0,2] * [0,3]
x2 C y2 C z2 dV D
1 2 3
0
D
1 2
0
D
1
1
x2 C y2 C z2 dz dy dx
0
2
2
x z C y z C
3
z3 dy dx
3 0
3x2 C 3y2 C 9 dy dx
0
2
3x y C y C 9y dx
2
0
D
0
1 2
0
D
0
3
0
6x2 C 26 dx
0
1
D 2x3 C 26x D 28.
0
3. You could work this out as Exercise 2, or suggest to your students that they could extend the result they established
in Exercise 24 of Section 5.2:
e
e
e
1
1
1
1
dx dy dz
dV D xyz
x
y
1
1
1 z
l
[1,e] * [1,e] * [1,e]
e 3
3
e
1
D
dx D ln x D 13 D 1.
1 x
1
4. This works for the same reason that Exercise 1 simplified.
We are integrating an odd function of z on an interval
3
that is symmetric in the z coordinate and so, since
5.
2 z2 yCz
3yz dx dy dz D
2 z2
2
−1 1
0
−1 1
−3
z dz D 0, the triple integral will also be 0.
yCz
2
3xyz 0
dy dz D 3
2 z2
−1 1
y2 z2 C yz3 dy dz
z2
2
y 3 z2
y2 z3 z8
z7
z3
z2
C
C
−
−
dz D 3
dz
3
2
2
2
3
−1
−1 3
1
2
z9
1539
z8
z4
z3 D 3
D
C
−
−
.
27
16
8
9 −1
16
D3
2
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“runall” — 2005/8/1 — 15:13 — page 244 — #22
244
Chapter 5
6.
Multiple Integration
3 z xz
1
0
x C 2y C z dy dx dz D
1
D
3 z
1
0
1
0
3 z
xz
xy C y2 C zy dx dz
1
x2 z C x2 z2 C xz2 − x − z − 1 dx dz
z
x3 z
x 3 z2
x 2 z2
x2
D
C
C
−
− xz − x dz
3
3
2
2
1
0
3 5
z
5z4
3z2
574
D
C
−
− z dz D
.
3
6
2
9
1
1 2y yCz
1 2y yCz
z dx dz dy D
xz
dz dy
3
7.
0
1Cy
0
z
D
1Cy
1 2y
0
z
yz dz dy D
1Cy
1
0
2y
yz2 /2
dy
1Cy
1
y
3y3
5
− y2 − dy D − .
2
2
24
0
8. (a) This is a higher-dimensional analogue of Exercise 20 from Section 5.2. Again the idea would be that if we
were in four-dimensional space that a figure of constant height would have volume equal to the volume of
the base multiplied by the height. In this case that would be just the volume of the base. Somehow this is a
lot less physically appealing or intuitive. By Definition 4.3,
D
lW
1 dA D
n
lim
all xi ,yj ,zk'0 i,j,kD1
xi yj zk .
The intuition follows from examining the formula above on the right. This converges to the volume of W .
More formally, we are assuming that W is an elementary region; let’s consider the case of a Type 1 region,
then we can rewrite the sum above as
δci n
n
xi yj ψcij − ϕcij D lim
xi ψy − ϕy dy
lim
all xi ,yj '0 i,jD1
all xi '0 iD1
D
b δci a
γci ψy − ϕy dy dx D volume of W .
γci The proof is not much different for the other elementary regions.
(b) Work out that the equation of the circle where the two paraboloids intersect is x2 C y2 D 9/2 so
√ √
2
2
2 Volume D
D
3/ 2
9/2−x
√
√
−3/ 2 − 9/2−x2
3/√ 2 √ 9/2−x2
9−x −y
x2 Cy2
1 dz dy dx
9 − 2x2 − 2y2 dy dx
√
√
−3/ 2 − 9/2−x2
3/√ 2
8 2
9
D
12
−
x
− x2 dx
√
3
2
−3/ 2

3
 9 − x2 15x − 2x
D
2
2
3
D
 √
√
3/ 2
81
2x 

C
arcsin √
4
3
−3/ 2
81π
.
4
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“runall” — 2005/8/1 — 15:13 — page 245 — #23
Section 5.4
Triple Integrals
245
9. Of course there are other ways to calculate the volume of the sphere.
Volume D
a √ a2 −x2 √ a2 −x2 −y2
√
−a − a2 −x2
−
√
a2 −x2 −y2
1 dz dy dx D
a √ a2 −x2
2 a
√
−a − a2 −x2
a
D
−a
y a2 − x2 − y2 − a2 − x2 arcsin
a
Dπ
−a
a
a2 − x2 dx D πa2 x − x3 /3
−a
D
2 − x2 − y2 dy dx
√ a2 −x2
y
√
dx
√
2
2
a − x
− a2 −x2
4πa3
.
3
10. The students have seen this as the volume of a solid of revolution. We’ll orient the cone so that the vertex is
down at the origin and the axis is along the z-axis. Then the horizontal cross sections are circles of radius rz/h.
This simplifies the following computation:
Volume D
h rz/h √ rz/h2 −x2
0
−rz/h −
√
rz/h2 −x2
h
dy dx dz D
π
0
r2 2
1
z dz D πr2 h.
2
h
3
11.
1 2 y2
0
−2 0
y2
z2 2x − y C z dz dy dx D
2xz − yz C
dy dx
2 0
0 −2
1 2
2xy2 − y3 C y4 /2 dy dx
D
1 2
0
−2
2
y4
y5 2xy3
D
dx
−
C
3
4
10 −2
0
1
64
32x
C
D
dx
3
10
0
1
16x2
176
32 D
C
x D
.
3
5
15
0
1
12.
1 √ 1−x2 2−x−z
√
−1 − 1−x2
y dy dz dx D
0
D
D
1 √ 1−x2
−1
2−x−z
y2 dz dx
√
2 0
− 1−x2
1 √ 1−x2
−1 −
1
−1
√
1−x2
2 − x − z2 /2 dz dx
1√
1 − x2 2x2 − 12x C 13
3


√
3 − 16x2 C 25x C 16
1
9
2x


D  1 − x2 C arcsin x
12
4
−1
D
3 9 9−y
13. Here
−3 x2
9π
.
4
8 xyz dz dy dx D 0, because we are integrating an odd function in x over an interval that is
0
symmetric in x (see Exercises 1 and 4).
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“runall” — 2005/8/1 — 15:13 — page 246 — #24
246
Chapter 5
Multiple Integration
3 3 √ 9−y2
14.
0
x
z dz dy dx D
0
D
√ 9−y2
dy dx
2
3 3
0
x
0
x
3 3
z2 0
9 − y2 /2 dy dx
3
−y3 C 27y D
dx
6
0
x
3 3
81
x − 27x C 54
D
.
dx D
6
8
0
3
15. Here we are again integrating a polynomial. The only difficulty is in the set up:
1 2−2x 3−3x−3y/2
0
0
1 − z2 dz dy dx D
0
1
.
10
16. Again, the set up and solution are:
2 √ 4−x2 4
0
17.
3 3−x √ 3−x2 /3
0
−
0
√
3−x2 /3
x2 Cy2
0
x C y dz dy dx D
3 3−x
0
D
3x dz dy dx D
64
.
5
2x C y 3 − x2 /3 dy dx
0
3
9 − x2 3 − x2 /3 dx
0


√
3
3
45x
−
2x
3
81

2

D  3 − x /3 arcsinx/3
C
8
8
0
√
81 3π
D
.
16
18.
2 √ 1−x2 /4 xC2
−2 −
√
1−x2 /4 0
z dz dy dx D
D
2 √ 1−x2 /4
−2 −
2
−2
D
√
1−x2 /4
x C 22 /2 dy dx
x C 22 1 − x2 /4 dx
2
3x3 C 16x2 C 18x − 64
1 − x2 /4 C 5 arcsinx/2 12
−2
D 5π.
19.
1 √ 1−y2 /2 2−y2
−1 −
√
1−y2 /2 4x2 Cy2
dz dx dy D
1 √ 1−y2 /2
−1 −
√
1−y2 /2
2 − 4x2 − 2y2 dx dy
√
4 2 2
y − 1 1 − y2 dy
D
3
−1
1
π
D √ .
2
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“runall” — 2005/8/1 — 15:13 — page 247 — #25
Section 5.4
Triple Integrals
247
20.
a √ a2 −x2 √ a2 −x2
√
−a − a2 −x2
√
− a2 −x2
dz dy dx D
a √ a2 −x2
√
−a − a2 −x2
√
2 a2 − x2 dy dx
a
D
D
−a
4a2 − x2 dx
16a3
.
3
21. The region looks like a wedge of cheese:
The five other forms are:
1 1 1−x
−1 y2
f x, y, z dz dx dy D
0
1 √ x 1−x
√
− x 0
f x, y, z dz dy dx
√
− x
1 1−z √ x
f x, y, z dy dz dx
√
− x
f x, y, z dy dx dz
0
D
D
D
D
1 1−x √ x
0
0
0
0
1 1−y2 1−z
f x, y, z dx dz dy
−1 0
y2
1 √ 1−z 1−z
0
√
− 1−z y2
f x, y, z dx dy dz.
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“runall” — 2005/8/1 — 15:13 — page 248 — #26
248
Chapter 5
Multiple Integration
22. The five other forms are:
1 1 x2
0
0
f x, y, z dz dx dy D
0
D
D
1 1 x2
f x, y, z dz dy dx
0
0
0
0
1 1 1
0
f x, y, z dx dz dy
z
√
f x, y, z dx dy dz
z
1 x2 1
f x, y, z dy dz dx
0
0
D
√
1 1 1
0
D
0
0
1 1 1
√
0
f x, y, z dy dx dz.
z 0
23. The five other forms are:
2 x y
0
0
f x, y, z dz dy dx D
0
D
D
D
D
2 2 y
f x, y, z dz dx dy
0
y
0
0
0
z
0
z
z
0
0
y
0
z
y
2 x x
f x, y, z dy dz dx
2 2 x
f x, y, z dy dx dz
2 y 2
f x, y, z dx dz dy
2 2 2
f x, y, z dx dy dz.
24. (a) The solid W is bounded below by the surface z D 5x2 , above by the paraboloid z D 36 − 4x2 − 4y2 , on
the left by the xz -plane (i.e., y D 0), and in back by the yz -plane (i.e., x D 0). The solid is shown below.
y
0
1
2
3
30
z
20
10
0
0.5
2 1.5
x
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“runall” — 2005/8/1 — 15:13 — page 249 — #27
Section 5.4
Triple Integrals
249
(b) The shadow of the solid in the xy-plane is a quarter of the ellipse 9x2 C 4y2 D 36 (obtained by finding the
intersection curve of z D 5x2 and z D 36 − 4x2 − 4y2 .) The shadow looks like:
y
3
2.5
2
1.5
1
0.5
0.5
1
1.5
x
2
Using the shadow region to reverse the order of integration between x and y, we find that the original integral
is equivalent to
3 1 √ 36−4y2 36−4x2 −4y2
3
2 dz dx dy.
5x2
0
0
(c) In this case, we need to consider the shadow of W in the xz -plane.
z
35
30
25
20
15
10
5
0.25 0.5 0.75
1
1.25 1.5 1.75
2
x
This region is bounded on the left by x D 0, on the bottom by z D 5x2 , and on top by z D 36 − 4x2 (the
1
36 − 4x2 − z
section by y D 0). Now the full solid W is bounded in the y-direction by y D 0 and y D
2
(the latter is just the paraboloid surface). Hence the desired iterated integral is
2 36−4x2 1 √ 36−4x2 −z
2
0
5x2
2 dy dz dx.
0
(d) Here we use the same shadow in the xz -plane as in part (c), only to integrate with respect to x before integrating
with respect to z will require dividing the shadow into two regions by the line z D 20. (Equivalently, we are
dividing the solid W into two solids by the plane z D 20.) This is why we need a sum of integrals. They are
√ 1√
1√
1√
2
2
20
0
z/5
0
2
0
36−4x −z
2 dy dx dz C
36
20
2
0
36−z
2
36−4x −z
2 dy dx dz.
0
(e) To integrate with respect to x first, we need to divide W in a very different manner. The shadow in the yz plane shows a region bounded on the left by y D 0 and above by z D 36 − 4y2 (the section of the paraboloid
by x D 0). However, the curve of intersection of the surfaces z D 5x2 and z D 36 − 4x2 − 4y2 with x
20y2
. It is along this curve that we must divide the yz -shadow and
eliminated yields the equation z D 20 −
9
thus the integrals. (Note: This curve is just the shadow of the intersection curve of the two surfaces projected
into the yz -plane.)
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“runall” — 2005/8/1 — 15:13 — page 250 — #28
250
Chapter 5
Multiple Integration
z
35
30
25
20
15
10
5
0.5
1
1.5
Thus the desired sum of integrals is
√
2
3
0
20−20y /9
0
z/5
2 dx dz dy C
0
2
2.5
1 √ 36−4y2 −z
3 36−4y2
2
20−20y2 /9
0
y
3
2 dx dz dy
0
25. (a) The solid W is bounded below by the paraboloid z D x2 C 3y2 , above by the surface z D 4 − y2 and in
back by the plane y D 0. The solid is shown below.
2
x
0
1
-1
-2
4
3
2
z
1
0
0
0.25
0.5
0.75 y
1
(b) The shadow of W in the xy-plane is half of the region inside the ellipse x2 C 4y2 D 4 (the half with y Ú 0).
It may be obtained by finding the intersection curve of z D x2 C 3y2 and z D 4 − y2 and eliminating z.
The shadow looks like
y
1
0.8
0.6
0.4
0.2
-2
-1
1
2
x
Using the shadow to reverse the order of integration between x and y, we find that the original integral is
equivalent to
1 2√ 1−y2 4−y2
x3 C y3 dz dx dy.
√
0
−2
1−y2
x2 C3y2
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“runall” — 2005/8/1 — 15:13 — page 251 — #29
Section 5.4
Triple Integrals
251
(c) We need to consider the shadow of W in the yz -plane.
z
4
3
2
1
0.2
0.4
0.6
0.8
1
y
The region is bounded on the left by y D 0, on the bottom by z D 3y2 (the section by x D 0) and on the
top by z D 4 − y2 . The full solid W is bounded in the x-direction by the paraboloid z D x2 C 3y2 , which
must be expressed in terms of x as x D ; z − 3y2 . Putting all this information together, we find the desired
iterated integral is
1 4−y2 √ z−3y2
x3 C y3 dx dz dy.
√
0
−
3y2
z−3y2
(d) Here we use the same shadow in the yz -plane as in part (c), only to integrate with respect to y before
integrating with respect to z requires dividing the shadow into two regions by the line z D 3. (Equivalently,
we are dividing the solid W by the plane z D 3.) This is why we need a sum of integrals. They are
3 √ z/3 √ z−3y2
0
0
−
√
z−3y2
x3 C y3 dx dy dz C
4 √ 4−z √ z−3y2
3
0
−
√
z−3y2
x3 C y3 dx dy dz.
(e) To integrate with respect to y first, we need to divide W in a different manner. The shadow in the xz -plane
shows a region bounded by z D x2 (the section of the paraboloid by y D 0) and z D 4. However, the
curve of intersection of the surfaces z D x2 C 3y2 and z D 4 − y2 with y eliminated yields the equation
2
z D x C 3. It is along this curve that we must divide the xz -shadow and thus the integrals. (Note: This
4
curve is just the shadow of the intersection curve of the two surfaces projected into the xz -plane.)
z
4
3
2
1
x
-2
-1
1
2
Thus the desired sum of integrals is
2 x2 /4C3 √ z−x2 /3
x C y dy dz dx C
3
−2 x2
0
2 4
√ 4−z
3
−2 x2 /4C3 0
x3 C y3 dy dz dx.
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“runall” — 2005/8/1 — 15:13 — page 252 — #30
252
Chapter 5
Multiple Integration
5.5 CHANGE OF VARIABLES
1. (a)
3
0
Tu, v D
0
−1 u
.
v (b) In this case we can see by inspection that the transformation stretches by 3 in the horizontal direction and
reflects without a stretch in the vertical direction. Therefore the image D D TD∗ where D ∗ is the unit
square is the rectangle [0, 3] * [−1, 0].
√
2. (a) This is similar to the map in Example 4 with a scaling factor of 1/ 2. We can also rewrite

1
 √2
Tu, v D 
 1
√
2

1
−√ 
2 

1
√
2
u
.
v This is a rotation matrix (the determinant is 1 so there is no stretching) which rotates the unit square
counterclockwise by 45◦ leaving the vertex at the origin in place.
(b) We rewrite


1
1
√
 √2
2 
 u .
Tu, v D 
 1
1  v √
−√
2
2
This is a rotation followed by a reflection. You can apply Proposition 5.1 and see where T maps each of the
vertices to completely determine the image
square. You will √
see that the
√ unit √
√ vertices (0, 0), (1, 0), (1,
√ of the
1), and (1, 0) are mapped to 0, 0, 1/ 2, 1/ 2, 2, 0, and 1/ 2, −1/ 2.
3. Again, since T has non-zero determinant, we can apply Proposition 5.1 and see where T maps each of the vertices.
We conclude that T maps D ∗ to the parallelogram whose vertices are: (0, 0), (11, 2), (4, 3), and (15, 5).
4. We are trying to determine the entries a, b, c, and d in the expression:
Tu, v D
a
c
b
d u
.
v Since T0, 0 D 0, 0 we know that the motion is not a translation. Also T0, 5 D 4, 1 so b D 4/5
and d D 1/5. Now T1, 2 D 1, −1 so a D −3/5 and c D −7/5. We check with the remaining vertex:
T−1, 3 D 3, 2.
Tu, v D
−3/5
−7/5
4/5
1/5 u
.
v 5. As noted in the text, we have a result for R3 that is analogous to Proposition 5.1, so as in Exercises 3 and 2
(b) we can just compute the images of the vertices of W ∗ . We conclude that W ∗ maps to the parallelepiped with
vertices: (0, 0, 0), (3, 1, 5), −1, −1, 3, 0, 2, −1, (2, 0, 8), (3, 3, 4), −1, 1, 2, and (2, 2, 7).
6. You can see that Tu, v D u, uv is not one-one on D ∗ by observing that all points of the form (0, v) get
mapped to the origin under T. In fact, you can imagine the map by picturing the left vertical side of the unit
square being shrunk down to a point at the origin. The image is the triangle:
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“runall” — 2005/8/1 — 15:13 — page 253 — #31
Section 5.5
Change of Variables
253
y
1
0.8
0.6
0.4
0.2
x
0.2
0.4
0.6
1
0.8
7. This map should be a happy memory for the students:
x, y, z D Tρ, ϕ, θ D ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ
is familiar from their work with spherical coordinates.
(a) This is the unit ball: D D {x, y, z|x2 C y2 C z2 … 1}.
(b) This is the portion of the unit ball in the first octant: D D {x, y, z|x2 C y2 C z2 … 1, x, y, z Ú 0}.
(c) You can think of this as the region from part (b) with the portion corresponding to 0 … ρ < 1/2 removed. It
is the portion in the first octant of the shell 1/2 unit thick around a sphere of radius 1/2: D D {x, y, z|1/2 …
x2 C y2 C z2 … 1, x, y, z Ú 0}.
y/2C2
1 y/2C2
1
1
8. (a)
2x − y dx dy D
x2 − xy
dy D
4 dy D 4. A sketch of D is shown below.
0
0
y/2
0
y/2
(b) We again can apply Proposition 5.1 and see that the vertices are mapped: 0, 0 ' 0, 0, 2, 0 '
4, 0, 1/2, 1 ' 0, 1, and 5/2, 1 ' 4, 1 so D ∗ is [0, 4] * [0, 1].
(c) First note that
⭸u, v
⭸x, y
2 −1
1
D det
D .
D 2 so
0
1
⭸x, y
⭸u, v
2
Then, using the change of variables theorem,
1 y/2C2
0
2x − y dx dy D
1 4
0
y/2
u1/2 du dv D
0
1
0
4
1
u2 4 dv D 4.
du D
4 0
0
y
1
0.8
0.6
0.4
0.2
0.5
9. First,
⭸u, v
D det
⭸x, y
1
1
−1
1.5
0
D 2 so
2 2
2.5
x
⭸x, y
1
D .
⭸u, v
2
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“runall” — 2005/8/1 — 15:13 — page 254 — #32
254
Chapter 5
Multiple Integration
2
2
Also we can rewrite x5 2y − xe2y−x D u5 vev , and the transformed region is [0, 2] * [0, 2] so
2 x/2C1
0
2
x5 2x − ye 2x−y dy dx D
2 2
0
x/2
2
u5 vev 1/2 du dv D
0
16
3
2
2
vev dv D
0
8 4
e − 1.
3
10. The original region D is sketched below left. The transformation u D x C y and v D x − 2y maps D to the
region D ∗ sketched below right. ⭸x, y/⭸u, v is easily calculated to be −1/3.
v
y
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
u
x
0.2
0.4
0.6
0.8
0.2
1
0.4
0.6
0.8
1
So, using the change of variables theorem, our integral becomes
1 u
0
0
1 u1/2
dv du D
3 v1/2
1
0
u
1
1
2 1/2 1/2 2
u2 1
u v du D
u du D
D .
3
3
3
3
0
0
0
11. Here the problem cries out to you to let u D 2x C y and v D x − y. Once you’ve made that move you can
easily figure that ⭸x, y/⭸u, v D −1/3 and that the new region is [1, 4] * [−1, 1]. So the integral is
4 1
1
u e 1/3 dv du D
3
−1
4
2 v
1
1
1
2 v
u e −1
du D e − e
−1
4
u3 D 7e − e −1 .
9 1
12. If we sketch the region we get the square:
y
1
0.5
x
0.5
1
1.5
2
2.5
3
-0.5
-1
-1.5
-2
The transformation we use is u D 2x C y − 3 and v D 2y − x C 6 so ⭸x, y/⭸u, v D 1/5 and the
transformed region is the square:
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“runall” — 2005/8/1 — 15:13 — page 255 — #33
Section 5.5
Change of Variables
255
v
6
5
4
3
2
1
-3
-2
-1
1
2
u
Our integral is
6 2
6
1
1
1
u2
du dv D
2
15
−3 5v
2
6
6
1
7
7
35
u3 .
dv D
v−2 dv D − D
2
v −3
3 1
3
v 1
18
Note: In Exercises 13–17 the Jacobian for the change of variables is r. Assign Exercise 16 so that your students
appreciate the role of the extra r.
1 √ 1−x2
2π 1
2π
3
13.
dθ D 3π.
3 dy dx D
3r dr dθ D
√
2
2
−1 − 1−x
0
0
0
2 √ 4−x2
π/2 2
π/2
14.
dy dx D
r dr dθ D
2 dθ D π.
0
0
0
2π 3
15.
0
a
r4 dr dθ D
0
√
16.
0
a2 −y2
e
3 x
dy dx
x2 C y 2
√
ln1 C 2.
0
0
0
0
3
2π
r5 243
486π
dθ D
.
dθ D
5
5
5
0
0
a
π/2
2
1 r2 1 a2
e dθ D
e − 1dθ D πea − 1/2.
dx dy D
re dr dθ D
−π/2 0
−π/2 2
−π/2 2
0
π/4 3 sec θ
π/4
√
π/4
D
dr dθ D
3 sec θ dθ D 3 lnsec θ C tan θ|0 D ln1 C 2 − ln 1 D
x2 Cy2
−a 0
17.
2π
0
0
π/2 a
π/2
r2
0
18. This is a job for polar coordinates. The given disk has boundary circle with equation x2 C y − 12 D 1. In
polar coordinates this equation becomes
r2 cos2 θ C r sin θ − 12 D 1 3 r2 cos2 θ C r2 sin2 θ − 2r sin θ C 1 D 1
3 r2 D 2r sin θ.
Factoring out r, the boundary circle has equation r D 2 sin θ. In fact, this circle is completely traced by letting θ
vary from 0 to π. Thus the region D inside the disk is given by
D D {r, θ|r … 2 sin θ,
0 … θ … π}.
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“runall” — 2005/8/1 — 15:13 — page 256 — #34
256
Chapter 5
Multiple Integration
Hence
1
6D 4 − x2 − y2
π 2 sin θ
1
r dr dθ
4 − r2
0
0
2 sin θ
π
π
1 √
D
− 2 4 − r2 dθ D −
4 − 4 sin2 θ − 2 dθ
2
0
0
0
π
π
D −2
cos2 θ dθ C
2 dθ
dA D
√
0
0
π/2
D −2 π
cos θ dθ C
0
π/2
D −2 sin
− cos θ dθ C 2π
π
π
− sin 0 − sin π C sin C 2π
2
2
D −4 C 2π D 2π − 4.
19. The region in question looks like
y
(-1,1)
(1,1)
x
(-1,-1)
We find
6D
y2 dA D
6square
6square
6disk
y2 dA −
y2 dA D
y2 dA D
6
disk
y2 dA
1 1
−1 −1
2π 1
0
D
(1,-1)
1
8
y2 dx dy D
1
−1
2y2 dy D
r2 sin2 θ · r dr dθ D
0
2π
0
1 − cos 2θ dθ D
0
Thus
6D
2π
y2 dA D
1
2 3 4
y D
3 −1
3
1 2
sin θ dθ
4
2π
1
1
π
θ
−
sin
2θ
D .
8
2
4
0
4
π
16 − 3π
−
D
.
3
4
12
20. A sketch of the rose is shown below. One leaf means that 0 … θ … π/2. The area of one leaf is
π/2 sin 2θ
0
0
1
r dr dθ D
2
π/2
π/2
1
π
sin 2θ dθ D
D .
4θ − sin 4θ
16
8
0
2
0
The total area is then four times this, or π/2.
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“runall” — 2005/8/1 — 15:13 — page 257 — #35
Section 5.5
Change of Variables
257
y
0.75
0.5
0.25
-0.75 -0.5 -0.25
0.25
0.5
0.75
x
-0.25
-0.5
-0.75
21. We sketch the graphs of the cardioid r D 1 − cos θ and the circle r D 1:
y
1
0.5
-2
-1.5
-1
-0.5
0.5
1
x
-0.5
-1
The two curves intersect when 1 − cos θ D 1 which is when cos θ D 0 so the two points of intersection are
r, θ D 1, π/2 and 1, 3π/2. The region between the two graphs is where π/2 … θ … 3π/2. The area is
3π/2 1−cos θ
π/2
3π/2
cos2 θ
− cos θ dθ
2
π/2
3π/2
π
1
D2 C .
D 2θ − 8 sin θ C sin 2θ
8
4
π/2
r dr dθ D
1
22. We want the area “inside” the spiral shown below. The area is
2π 3θ
0
0
r dr dθ D
2π
0
2π
3 3
9 2
θ dθ D θ D 12π3 .
2
2 0
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“runall” — 2005/8/1 — 15:13 — page 258 — #36
258
Chapter 5
Multiple Integration
y
5
-10
-5
5
10
15
x
-5
-10
23. The integral is
π
1
π
sin r dθ D sin 1
D sin 1.
2
3
π/3
π/3
1
r cos r dr dθ D
2
0
2
π/3
1
2
π
π 1
0
24. Since B is a ball we will use spherical coordinates:
dV
l
B x 2 C y 2 C z2 C 3
D
2π π 2
0
D
0
0
D
0
ρ2 C 3
0
2π π
2π
ρ2 sin ϕ
√
7 −
dρ dϕ dθ
√
3
arcsinh2/ 3 sin ϕ dϕ dθ
2
√
√
2 7 − 3 arcsinh2/ 3 dθ
0
√
D 4 7π − 6π arcsinh2/ 3 which is the same as the text’s solution
√
√
D 4 7 − 6 ln2 C 7 C 3 ln 3π.
√
25. Here we will use cylindrical coordinates:
l
W
x C y C 2z dV D
2
2
2 2π 2
2
D
D
−1 0
2 2π
−1 0
2
−1
rr2 C 2z2 dr dθ dz
0
4z2 C 4 dθ dz
8πz2 C 8π dz D 48π.
26. It is natural to use spherical coordinates.
dV
l
W x 2 C y 2 C z2
D
2π π b
ρ sin ϕ dρ dϕ dθ
0
0
a
2π π
1
b2 − a2 sin ϕ dϕ dθ
2 0
0
2π
b2 − a2 dθ D 2πb2 − a2 .
D
D
0
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“runall” — 2005/8/1 — 15:13 — page 259 — #37
Section 5.5
Change of Variables
259
27. Once again we use spherical coordinates.
x2 C y2 C z2 e x Cy Cz dV D
2
lW
2
2
2π π b
2
ρ3 e ρ sin ϕ dρ dϕ dθ
0
0
a
2π π
2
2
1
[1 − a2 e a C b2 − 1e b ] sin ϕ dϕ dθ
2 0
0
2π
2
2
1 − a2 e a C b2 − 1e b dθ
D
D
0
2
2
D 2π1 − a2 e a C b2 − 1e b .
28. We are integrating over a cone with vertex at the origin and base the disk at height 6 with radius 3. We will use
cylindrical coordinates.
l
W
2 C x2 C y2 dV D
D
D
3 2π 6
0
0
0
0
3 2π
3
r2 C r dz dθ dr
2r
−2r3 C 2r2 C 12r dθ dr
2π−2r3 C 2r2 C 12r dr D 63π.
0
You should assign one of Exercises 29 or 30 so that your students see the benefits of using another coordinate system
even when it is not explicitly called for. You might want to stress that the symmetries of the problem are what lead
you, in this case, to choose cylindrical coordinates. Exercise 31 is fun because students will be tempted to use spherical
coordinates—life is much easier if they use cylindrical coordinates.
29. We will use cylindrical coordinates.
l
W
dV D
D
D
1 2π √ 10−2r2
0
0
0
0
1 2π
1
r dz dθ dr
√
− 10−2r 2
2r 10 − 2r2 dθ dr
4πr 10 − 2r2 dr
0
D
√
4π √
5 10 − 8 2.
3
30. We will again use cylindrical coordinates.
lW
dV D
D
D
2 2π 9−r2
r dz dθ dr
0
0
0
0
2 2π
2
0
9r − r3 dθ dr
18πr − 2πr3 dr
0
D 28π.
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“runall” — 2005/8/1 — 15:13 — page 260 — #38
260
Chapter 5
Multiple Integration
31. We will again use cylindrical coordinates.
lW
2 C x C y dV D
2
5 2π √ 25−z2
2
3
2 C r2 r dr dθ dz
0
0
5 2π
z4
27z2
725
−
C
dθ dz
4
2
4
3 0

5
4
2
z
27z
725


 dz D 656π .
2π −
C
D
4
2
4
5
3
D
32. You can draw a million pictures, but the easiest way to visualize this is by taking an apple corer and a potato and
cutting in the three orthogonal directions. This will provide you with a model that the students can hold and pass
around to aid their discussion. They can easily identify symmetries and cut the model along the coordinate planes
to set up the integral.
If you do this and look in the first octant, you will see a seam along the line y D x. If we split the integral along
this line we will have 1/16 of the desired volume. Using cylindrical coordinates this means that 0 … θ … π/4
and the cylinder with axis of symmetry the z-axis gives us that 0 … r … a. The hard one to see is z, but because
we are only looking at the wedge on one side of θ D π/4 we need only worry about one other cylinder so
0 … z … a2 − r2 cos2 θ.
So the volume is
π/4 a √ a2 −r2 cos2 θ
√
V D 16
r dz dr dθ D 8a3 2 − 2.
0
0
0
5.6 APPLICATIONS OF INTEGRATION
Exercises 1–9 concern average value.
1. (a) Let’s assume a 30-day month.
[f ]avg D
1
30
30
Ix dx D
0
1
30
30
0
75 cos
πx
C 80 dx
15
30
πx
1 1125
sin
C 80x D 2400/30 D 80 cases.
D
30
π
15
0
(b) Here the 2 cents will be a constant that pulls through the integral so the average holding cost is just 2 cents
times the average daily inventory, or $1.60.
2. We will divide the integral by the area:
4π
2π 4π
2π
1
1
sin2 x
sin2 x cos2 y dy dx D
sin 2y C 2y dx
2π4π 0
8π2 0
4
0
0
2π
2π
1
2π2
1
1
π
2π sin2 x dx D
D .
D
2x − sin 2x D
2
2
2
8π 0
8π
2
8π
4
0
[f ]avg D
3. Again we will divide the integral by the area:
1−x
1 1−x
1
1
e 2xCy dy dx D 2
e 2xCy dx
1/2 0 0
0
0
1
1
xC1
2x
1Cx
2x e
− e dx D 2e
− e D e 2 − 2e C 1.
D2
[f ]avg D
0
0
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“runall” — 2005/8/1 — 15:13 — page 261 — #39
Section 5.6
Applications of Integration
261
4. Here we are finding the average over a ball of volume 4π/3. We’ll integrate using cylindrical coordinates because
z appears explicitly in the integrand.
1 2π √ 1−z2
1 2π
1
3
ez
2
[f ]avg D
rez dr dθ dz D
1 − z dθ dz
4π/3 −1 0
4π
2
0
−1 0
1
3
3
3 4π
D
D .
πe z 1 − z2 dz D
4π −1
4π e
e
5. (a) We are told that in the 2 * 2 * 2 cube centered at the origin, T x, y, z D cx2 C y2 C z2 . The average
temperature of the cube is
[T ]avg D
c
8
c
D
8
1 1 1
−1 −1 −1
1
−1
x2 C y2 C z2 dx dy dz D
4z2 C 8/3 dz D
c
8
1 1
−1 −1
2z2 C 2y2 C 2/3 dy dz
c
8 D c.
8
(b) T x, y, z D c when x2 C y2 C z2 D 1 so the temperature is equal to the average temperature on the
surface of the unit sphere.
6. The region looks like
y
(-1,1)
(1,1)
x
(-1,-1)
(1,-1)
and the area of it is 22 − π D 4 − π. Hence the average value is
1
1
x2 C y2 dA D
x2 C y2 dA −
x2 C y2 dA ,
4 − π 6D
4 − π 6D1
6D2
where D1 denotes the square and D2 the disk.
1
1 3
2 x C y dA D
x C y dx dy D
dy
x C y x
−1 −1
−1 3
6
D1
xD−1
1
1
4
2
2
2 3 4
8
2
D
dy
D
C
2y
y
C
y
D C D
3
3
3
3
3
3
−1
−1
2π 1
2π
1
π
dθ D
x2 C y2 dA D
r2 · r dr dθ D
4
2
0
0
0
6
D2
1 1
2
2
1
2
2
Therefore the average value is
8
π
16 − 3π
3π − 16
1
D
L 1.2766.
− D
4 − π 3
2
24 − 6π
6π − 24
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“runall” — 2005/8/1 — 15:13 — page 262 — #40
262
Chapter 5
Multiple Integration
7. The volume of W is 8 −
4
24 − 4π
xD
. Thus the average value is
3
3
3
3
x2 C y2 C z2 dV D
x2 C y2 C z2 dV
24 − 4π lW
24 − 4π l
W1
−
x2 C y2 C z2 dV l
W2
where W1 denotes the cube and W2 the ball. Using Cartesian coordinates to integrate over W1 and spherical
coordinates to integrate over W2 , this may be calculated as
1 1 1
2π π 1
3
x2 C y2 C z2 dz dy dx −
ρ4 sin ϕ dρ dϕ dθ
24 − 4π −1 −1 −1
0
0
0
D
3π − 30
.
5π − 30
1
0.5
y
0
-0.5
-1
1
0.5
z
0
-0.5
-1
-1
-0.5
0
x
0.5
1
8. We are looking for the average value of the minimum of x and y in the 6 * 6 box. This is 1/36 times the sum of
the average value for x in the region where x … y and the average value for y in the region where y … x. Because
of the symmetry, the average value can be calculated by doubling the result for either region and dividing by 36:
6 x
6 2
x
2
1
dx
[Time]avg D
y dy dx D
36 0 0
18 0 2
6
1 x3 D
D 2.
18 6 0
9. This is an extension of Exercise 8. The domain is [0, 6] * [0, 6] * [0, 6]. This time there is six-fold symmetry
so we will calculate the average value for z in the region where z … y … x and multiply by 6 and then divide by
63 which is the volume of the domain.
6 x y
6 x 2
y
6
1
z dz dy dx D
dy dz
[Time]avg D
216 0 0 0
36 0 0 2
6
6 3
1 x4 x
1
dx D
D
D 3/2.
36 0 6
36 24 0
So with three train lines the average wait is 90 seconds.
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“runall” — 2005/8/1 — 15:13 — page 263 — #41
Section 5.6
Applications of Integration
263
Exercises 10–24 concern centers of mass. We use the formula:
b
xδx dx
Center of mass D b
a
δx dx
a
and its variants.
10. (a) The curve y D 8 − 2x2 intersects the x-axis at ;2. So
2
2 8−2x2
2
c dy dx D c
8 − 2x2 dx D c 8x − 2x3 /3
−2 0
My D
Mx D
2 8−2x2
−2 0
2 8−2x2
−2 0
−2
cx dy dx D c
2
−2
−2
2
8x − 2x3 dx D c 4x2 − x4 /2
cy dy dx D c/2
D 64c/3
−2
D 0 and
2
2
16 3
8 − 2x2 2 dx D c x5 −
x C 32x D 1024c/15
5
3
−2
−2
2
1024c/15
D 16/5.
64c/3
(b) Again, we see the symmetry with respect to x so x D 0. The following integrals are straightforward so we
leave out the details, but
2 8−2x2
3cy2 dy dx
32768c/35
D
D 32/7.
y D −2 0
2
8−2x2
1024c/5
3cy dy dx
So x D 0 and y D
−2 0
11. We assume that the plate has uniform density and place it so that the center of the straight border is at the origin
and the semicircle is symmetric with respect to the y-axis. Once again this means that x D 0.
a √ a2 −x2
cy dy dx
−a 0
yD
D
πa2 c/2
12. First calculate
MD
My D
Mx D
2 2x
0
x2
0
x2
0
x2
2 2x
2 2x
2a3 c/3
4a
D
.
πa2 c/2
3π
1 C x C y dy dx D
24
5
[x1 C x C y] dy dx D
28
5
[y1 C x C y] dy dx D
328
35
Thus,
xD
13. Again we first calculate
MD
9 √x
0
My D
0
9 √x
0
Mx D
28/5
D 7/6
24/5
xy dy dx D
0
328/35
D 41/21.
24/5
243
2
x2 y dy dx D
6561
8
xy2 dy dx D
1458
7
0
9 √x
0
and y D
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“runall” — 2005/8/1 — 15:13 — page 264 — #42
264
Chapter 5
Multiple Integration
so
xD
6561/8
D 27/4 and
243/2
yD
1458/7
D 12/7.
243/2
14. We’ll take δ to be 1. A look at the figure below tells us again that x D 0. We’ll use polar integrals to calculate y.
y
-1
-0.5
0.5
x
1
-0.5
-1
-1.5
-2
We first calculate M D
2π 1−sin θ
0
r dr dθ D
0
−5π/4
D −5/6.
3π/2
15. We first calculate
3π
and Mx D
y dA D
2
6D
2π 1−sin θ
0
0
r2 sin θ dr dθ D
−5π
,
4
so y D
MD
π/3 4 cos θ
0
My D
Mx D
6D
6D
0
x dA D
y dA D
r dr dθ D
√
π/3 4 cos θ
0
0
0
0
π/3 4 cos θ
3 C
4π
3
7
8π
r2 cos θ dr dθ D √ C
3
3
and
r2 sin θ dr dθ D 5,
√
7 3 C 8π
15
√
and y D √
.
so x D
3 3 C 4π
3 3 C 4π
16. The region is a slice of pie:
y
2
1.5
1
0.5
x
0.5
1
1.5
2
2.5
3
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“runall” — 2005/8/1 — 15:13 — page 265 — #43
Section 5.6
Total mass M D
δ dA D
π/4 3
6D
π/4
D
0
My D
xδ dA D
6D
63
D √
4 2
0
Mx D
yδ dA D
2
2r −
0
π/4 3
9π
4
4r − r cos θ dr dθ D
3
0
π/4 3
0
6D
√
63
2 − 2
D
8
0
4r2 − r3 sin θ drdθ D
π/4
0
0
265
3
1 3 r 3
rD0
π/4
2
0
Applications of Integration
π/4
4 − rr dr dθ D
18 − 9 dθ D
0
36 −
81
cos θ dθ
4
36 −
81
sin θ dθ
4
√
63
7 2
4
D √ ·
D
M
2π
4 2 9π
√
√
632 − 2 4
72 − 2
M
yD x D
D
M
8
9π
2π
Thus
xD
My
17.
y
1
0.5
0.5
1
1.5
2
x
-0.5
-1
Total mass M D
D
δ dA D
6D
2π
0
2π
2π 1Ccos θ
r2 dr dθ
0
0
1
1 C cos θ3 dθ
3
1
5π
1 C 3 cos θ C 3 cos2 θ C cos3 θ dθ D
3
3
2π 1Ccos θ
2π
1
My D
1 C cos θ4 cos θ dθ
xδ dA D
r3 cos θ dr dθ D
4
0
0
0
6D
7π
after some effort!
D
4
D
0
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“runall” — 2005/8/1 — 15:13 — page 266 — #44
266
Chapter 5
Multiple Integration
Mx D
D
yδ dA D
6D
2π
0
2π 1Ccos θ
0
r3 sin θ dr dθ
0
1
1
1 C cos θ4 sin θ dθ D −
4
4
2
u4 du D 0
2
(It’s also possible to see this from symmetry.) Thus
xD
7π
3
21
·
D
,
4
5π
20
y D 0.
18. Because the volume of the tetrahedron is 1, we can find the centroid by calculating:
xD
1 2−2x 3−3y/2−3x
0
yD
0
zD
0
y dz dy dx D
1
2
z dz dy dx D
3
.
4
0
1 2−2x 3−3y/2−3x
0
1
4
0
1 2−2x 3−3y/2−3x
0
x dz dy dx D
0
0
and,
19. (a) First calculate:
MD
Myz D
Mxz D
Mxy D
2 1 3
−1 −1 3y2
2 1 3
−1 −1 3y2
2 1 3
−1 −1 3y2
2 1 3
−1 −1
3y2
dz dy dx D 12
x dz dy dx D 6
y dz dy dx D 0
z dz dy dx D
108
.
5
This means that x, y, z D 1/2, 0, 9/5.
(b) Next we calculate the center of mass with the given density function.
MD
Myz D
Mxz D
Mxy D
2 1 3
−1 −1 3y2
2 1 3
−1 −1 3y2
2 1 3
−1 −1
3y2
−1 −1
3y2
2 1 3
z C x2 dz dy dx D
168
5
xz C x2 dz dy dx D
129
5
yz C x2 dz dy dx D 0
zz C x2 dz dy dx D
2376
.
35
This means that x, y, z D 43/56, 0, 99/49.
Note that in Exercises 20–22, the symmetry with respect to the z-axis implies that x D 0 and y D 0. We only explicitly
set up all of the integrals in the solution of Exercise 20.
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“runall” — 2005/8/1 — 15:13 — page 267 — #45
Section 5.6
20. First calculate:
MD
3 2π √ 18−r2
r 2 /3
3 2π √ 18−r2
0
Myz D
r 2 /3
3 2π √ 18−r2
0
Mxz D
0
267
√
63π
r dz dθ dr D 36 2π −
2
r cos θ dz dθ dr D 0
0
r 2 /3
3 2π √ 18−r2
0
Mxy D
0
Applications of Integration
r sin θ dz dθ dr D 0
0
0
r 2 /3
rz dz dθ dr D
189π
.
4
21
√
.
28 2 − 7
21. As noted above, x D 0 and y D 0. First calculate:
5/2 2π 9−r2
625π
rdz dθ dr D
MD
2
8
3r −16
0
0
5/2 2π 9−r2
10625π
Mxy D
rz dz dθ dr D −
2
96
3r −16
0
0
This means that x, y, z D 0, 0,
17
.
12
22. As noted above, x D 0 and y D 0. First calculate:
√ 5 2π √ 25−r2
√
π
r dz dθ dr D 250 − 100 5
MD
3
2r
0
0
√ 5 2π √ 25−r2
125π
Mxy D
rz dz dθ dr D
4
2r
0
0
This means that x, y, z D 0, 0, −
This means that x, y, z D 0, 0,
15
√ .
85 − 2 5
23. By symmetry x D y D z. z is easiest to find. Volume of W is
1 4 3
πa3
.
πa D
8 3
6
x
0.25 0
0.5
0.75
1
1
0.75
z
0.5
0.25
0
0
0.25
0.5
y
0.75
1
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“runall” — 2005/8/1 — 15:13 — page 268 — #46
268
Chapter 5
Thus
Multiple Integration
π/2 π/2 a
6
6
z dV D
ρ cos ϕ · ρ2 sin ϕ dρ dϕ dθ
πa3 l
πa3 0
0
0
W
π/2
π/2 π/2 4
a
1
6
3a
D
cos ϕ sin ϕ dϕ dθ D
dθ
πa3 0
4
2π 0
2
0
3a
.
D
8
zD
24. If we put the bottom of the cylinder in the xy-plane, then δx, y, z D h − z2 .
z
y
x
Therefore the total mass is
2π a h
MD
δ dV D
h − z2 r dz dr dθ
0
0 0
l
W
h
2π a
1
3
− h − z r dr dθ
D
3
0
0
D
2π a
0
0
zD0
πh3 a2
h3
r dr dθ D
.
3
3
x D y D 0 by symmetry, so we compute
2π a h
Mxy D
zδ dV D
zh − z2 r dz dr dθ
0
0 0
l
W
2π a h
2π a
1 4
πh4 a2
D
h2 z − 2hz2 C z3 r dz dr dθ D
r ·
h dr dθ D
.
12
12
0
0 0
0
0
Thus
zD
πh4 a2
h
3
D .
·
12
πh3 a2
4
25. (a) By symmetry we can see that the moment of inertia about each of the coordinate axes is the same.
1 1−x 1−x−y
1
Ix D Iy D Iz D
.
y2 C z2 dz dy dx D
30
0 0
0
(b) Again, by symmetry we see that the radius of gyration about each of the coordinate axes is the same. We
calculate
1 1−x 1−x−y
1
MD
dz dy dx D .
6
0 0
0
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“runall” — 2005/8/1 — 15:13 — page 269 — #47
Section 5.6
Applications of Integration
269
√
1/30
D 1/ 5.
1/6
26. By symmetry we can see that the moment of inertia about each of the coordinate axes is the same
2 2 2
y2 C z2 x C y C z C 1 dz dy dx D 96.
Ix D Iy D Iz D
Then rx D ry D rz D
0
0
0
Again, by symmetry we see that the radius of gyration about each of the coordinate axes is the same. We calculate
2 2 2
x C y C z C 1 dz dy dx D 32.
MD
0
0
0
√
96
D 3.
32
27. (a) The problem cries out to be solved using cylindrical coordinates. For Iz this means that the x2 C y2 in the
integrand is r2 so
3 2π 9
6561π
Iz D
2zr3 dz dθ dr D
and
4
0 0
r2
3 2π 9
2zr dz dθ dr D 486π, so
MD
Then rx D ry D rz D
0
r2
0
√
3 3
rz D √ .
2 2
(b) This time
Iz D
MD
3 2π 9
0
0
r2
0
0
r2
3 2π 9
√
3 3
rz D √ .
7
r4 dz dθ dr D
8748π
35
and
r2 dz dθ dr D
324π
,
5
so
28. Although it may be tempting to move to spherical coordinates, it is nice to have a z-coordinate so we will stay
with cylindrical coordinates.
(a)
a 2π √ a2 −r2
8πca5
and
r3 c dz dθ dr D
Iz D
√
15
− a2 −r 2
0 0
a 2π √ a2 −r2
4πca3
MD
rc dz dθ dr D
so
√
3
− a2 −r 2
0 0
2
.
rz D a
5
(b)
Iz D
a 2π √ a2 −r2
MD
0
rz D a
0
0
and
4πa5
5
so
r3 r2 C z2 dz dθ dr D
√
rr2 C z2 dz dθ dr D
− a2 −r 2
a 2π √ a2 −r2
0
8πa7
21
√
− a2 −r 2
10
.
21
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“runall” — 2005/8/1 — 15:13 — page 270 — #48
270
Chapter 5
Multiple Integration
(c)
Iz D
a 2π √ a2 −r2
r5 dz dθ dr D
32πa7
105
and
√
r3 dz dθ dr D
8πδa5
15
so
− a2 −r 2
a 2π √ a2 −r2
0
MD
√
0
0
0
− a2 −r 2
2a
rz D √ .
7
29.
Ix D
6D
1
y2 δ dA D
1 3
−1 x2 C2
y2 x2 C 1 dy dx
1 2
3
2
D
9 − x C 2 x C 1 dx D
3
−1
D
1
1
19 C 7x2 − 18x4 − 7x6 − x8 dx
−1 3
1
14
36
2
1496
−
− 2 − D
38 C
3
3
5
9
135
y
3
2.5
2
1.5
1
0.5
x
-1
-0.5
0.5
1
30.
Iz D
2 1
x C y δ dA D
x2 C y2 1 C y dy dx
0 0
6[0,2] * [0,1]
2
2 1
1
1
1
2
x2 C y2 C x2 y C y3 dy dx D
D
C x2 C dx
x C
3
2
4
0 0
0
2
4
1
31
8
C
C D
D C
3
3
3
2
6
2
2
31. The only adjustment in the formula for Ix is because we are using the square of the distance from the line y D 3
and not the formula given in text which squares the distance from the x-axis. This is a straightforward application
of formula (8).
2 √ 4−x2
116π
IyD3 D
.
x2 3 − y2 dy dx D
√
3
−2 − 4−x2
What follows is preliminary work for Exercises 32–34. You should probably assign all three together.
We will be calculating
Gmδx, y, z dV
.
V 0, 0, r D −
lW x2 C y2 C z − r2
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“runall” — 2005/8/1 — 15:13 — page 271 — #49
Section 5.6
Applications of Integration
271
In this special case, W is the shell bound by spheres centered at the origin of radii a and b where a < b. The
volume of W is therefore 4πb3 − a3 /3. The density is assumed to be constant and so the density is mass
divided by volume, so
M
3M
δD
D
.
3
3
[4πb − a /3]
4πb3 − a3 So
V 0, 0, r D −Gmδ
dV
lW x2 C y2 C z − r2
3GmM
dV
3
3
2
2
4πb − a lW x C y C z − r2
2π b π
ρ2 sin ϕ
3GmM
dϕ dρ dθ
D−
3
3
4πb − a 0
a 0
ρ2 C r2 − 2rρ cos ϕ
π
2π b
ρ
3GmM
2
2
D−
dρ dθ
ρ C r − 2rρ cos ϕ 3
3
4πb − a 0
r
a
ϕD0
2π b
ρ
3GmM
D−
|ρ C r| − |ρ − r| dρ dθ.
4πb3 − a3 0
r
a
D−
Our final note before proceeding to Exercises 32–34 is that
2ρ
ρ
|ρ C r| − |ρ − r| D 2ρ2 /r
r
if ρ Ú r, and
if ρ < r.
32. See preliminary work above. When r Ú b, then in the range a … ρ … b, we have that ρ … r, so
2π b
2ρ2
dρ dθ
r
0
a
2π 3 b
2ρ 3GmM
D−
dθ
3
3
4πb − a 0
3r a
2π
2b3 − a3 3GmM
dθ
D−
3
3
4πb − a 0
3r
2π
GmM
GmM
D−
dθ D −
.
2πr 0
r
V 0, 0, r D −
3GmM
4πb3 − a3 33. See preliminary work above. When r … a, then in the range a … ρ … b, we have that ρ Ú r, so
V 0, 0, r D −
D−
D
3GmM
4πb3 − a3 3GmM
4πb3 − a3 2π b
2ρ dρ dθ
0
2π
a
b2 − a2 dθ
0
−3GmMb2 − a2 .
2b3 − a3 What is striking about this result is that V (0, 0, r) is independent of r. Therefore, since F D −§V , we see that
there is no gravitational force.
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“runall” — 2005/8/1 — 15:13 — page 272 — #50
272
Chapter 5
Multiple Integration
34. (b) Students might consider the connection before they explicitly find it in part (a). If a < r < b, then we have
a combination of the two cases dealt with in Exercises 32 and 33. For a … ρ … r, we are in a case similar
to Exercise 32, and for r … ρ … b we are in a case similar to Exercise 33.
(a) We must break the integral at ρ D r:
V 0, 0, r D −
D−
3GmM
4πb3 − a3 3GmM
4πb3 − a3 2π r
2π b
0
2ρ2
3GmM
dρ dθ −
r
4πb3 − a3 a
0
2r3 − a3 3GmM
dθ −
3r
4πb3 − a3 2π
2ρ dρ dθ
0
r
2π
b2 − r2 dθ
0
D−
2r3 − a3 3GmM
3GmM
b2 − r2 −
2b3 − a3 3r
2b3 − a3 D−
GmM
3b2 r − 2a3 − r3 .
2rb3 − a3 5.7 TRUE/FALSE EXERCISES FOR CHAPTER 5
1. False. (Not all rectangles must have sides parallel to the coordinate axes.)
2. True.
3. True.
4. True.
5. False. (Let f x, y D x, for example.)
6. True.
7. False. (The integral on the right isn’t even a number!)
8. True.
9. True.
10. False. (It’s a type 1 region.)
11. True.
12. True.
13. False. (The value of the integral is 3.)
14. True. (Use symmetry.)
15. True.
16. False. (The y part of the integrand gives a nonzero value.)
17. True. (The inner integral with respect to z is zero because of symmetry.)
18. False. (The triple integral of y is zero because of symmetry, but not the triple integral of x.)
19. True.
20. False. (The area is 30 square units.)
21. False. (The integrals are opposites of one another.)
22. True.
23. False. (A factor of r should appear in the integrand.)
24. True.
25. False. (A factor of ρ is missing in the integrand.)
26. True.
27. True.
1
28. False. (The centroid is at 0, 0, h.)
4
29. True.
30. True.
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“runall” — 2005/8/1 — 15:13 — page 273 — #51
Section 5.8
Miscellaneous Exercises for Chapter 5
273
5.8 MISCELLANEOUS EXERCISES FOR CHAPTER 5
1. First let’s split the integrand:
l
B
z3 C 2 dV D
l
B
z3 dV C
l
B
2 dV D
l
B
z3 dV C 2
l
B
dV .
Here B is the ball of radius 3 centered at the origin. The integral of an odd function of z over a region which is
symmetric with respect to z is 0. The other integral is twice the volume of a sphere of radius 3 so
l
B
4
z3 C 2 dV D 2 π33 D 72π.
3
2. As in Exercise 1 we see that our integrand is the sum of odd functions in x and y and a constant which we are
integrating over a region which is symmetric with respect to x and y. Our answer will be −3 times the volume
of the hemisphere of radius 2. In symbols,
V D
l
W
x3 C y − 3 dV D −3
4
1
dV D −3 π23 D −16π.
2
3
l
W
3. (a) We’ll use the bounds given for z in both integrals and just reverse the order of integration for x and y. We
are integrating over the ellipse:
y
3
2
1
x
-2 -1.5 -1 -0.5
0.5 1 1.5 2
-1
-2
-3
l
W
3 dV D
D
3 √ 3−y2 /3 9−x2
3 dz dx dy
9−x2
√
− 9−3x2
2x2 Cy2
0
and
2x2 Cy2
−3 0
√ 3 √ 9−3x2
3 dz dy dx.
√
(b) Using Mathematica, the result was 81 3π/4 in either order.
4. First follow the hint (noting that x and y are just dummy variables) and write
x
y
F x, y D
gx , y dx where gx , y D
f x , y dy .
a
c
By the fundamental theorem of calculus,
⭸F
⭸
D
⭸x
⭸x
x
gx , y dx D gx, y.
a
Also, again by the fundamental theorem,
⭸2 F
⭸
⭸
D
[gx, y] D
⭸y⭸x
⭸y
⭸y
y
f x, y dy D f x, y.
c
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“runall” — 2005/8/1 — 15:13 — page 274 — #52
274
Chapter 5
Multiple Integration
By Fubini’s theorem,
x y
a
As above, write
F x, y D
f x , y dy dx D
y x
b
y
c
hx, y dy
f x , y dx dy .
a
where hx, y D
c
x
f x , y dx .
a
Proceeding as above we see that
⭸F
D hx, y and
⭸y
⭸2 F
D f x, y.
⭸x⭸y
5. I think the given form is the easiest to integrate:
2π 1 √ 9−r2
0
r dz dr dθ D
0
0
2π 1
0
r 9 − r2 dr dθ
0
√
16 2
dθ
3
0
√
16 2
D 2π 9 −
.
3
D
2π
9 −
(a) In Cartesian coordinates, z doesn’t really change and for the outer two limits, we are integrating over a unit
circle so our answer is
1 √ 1−x2 √ 9−x2 −y2
dz dy dx.
√
−1 − 1−x2
0
(b) The solid is the intersection of the top half of a sphere of radius 3 centered at the origin and a cylinder
of radius 1 with axis of symmetry the z-axis. In spherical coordinates this means that we have to split the
integral into two pieces: one that corresponds to the spherical cap and one that corresponds to the straight
sides. The “cone” of intersection is when ϕ D sin−1 1/3. For the integral that corresponds to the “straight
sides”, 0 … r … 1. In spherical coordinates that is 0 … ρ sin ϕ … 1 or 0 … ρ … csc ϕ. The integrals are, therefore,
2π sin−1 1/3 3
0
0
ρ2 sin ϕ dρ dϕ dθ C
0
2π π/2
0
csc ϕ
sin−1 1/3 0
ρ2 sin ϕ dρ dϕ dθ.
6. (a) This solid is similar to that in Exercise 5. It is the intersection of a cylinder over the circle of radius 2 with
center 0, 2 (i.e., x2 D 4y − y2 ) and the plane x D 0 with caps on either end that are portions of the
sphere of radius 4 centered at
; 16 − x2 − y2 .
√ the origin z D √
2
(b) In cylindrical coordinates, − 16 − r … z … 16 − r2 and we are above the first quadrant so 0 … θ … π/2.
Since x2 C y2 D 4y, r2 D 4r sin θ so in the first quadrant, r D 4 sin θ. The volume is therefore
V D
π/2 4 sin θ √ 16−r2
0
D
0
√
− 16−r 2
r dz dr dθ D
π/2 4 sin θ
0
2r 16 − r2 dr dθ
0
π/2
128
64
1 − cos3 θ dθ D
3π − 4.
3
9
0
Exercises 7 and 8 are a good lesson in the advantage of choosing the right coordinate system in which to work. This
simple problem in Cartesian coordinates is a pain using either cylindrical or spherical coordinates.
7. Orient the cube so that a vertex is at the origin and the edges that meet at that vertex lie along the x-, y- and
z-axes so that the cube is in the first octant. We’ll double the volume of half of the cube. In this case 0 … z … a,
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“runall” — 2005/8/1 — 15:13 — page 275 — #53
Section 5.8
Miscellaneous Exercises for Chapter 5
275
0 … θ … π/4 and the only difficulty is with r. The radius varies from 0 to the line x D a. In cylindrical coordinates
x D r cos θ so r D a sec θ and our limits for r are 0 … r … a sec θ. The volume is
a π/4 a sec θ
a π/4
a
V D2
r dr dθ dz D
a2 sec2 θ dθ dz D
a2 dz D a3 .
0
0
0
0
0
0
The above calculation wouldn’t change much if you followed the hint in the text and placed the center of the
cube at the origin. In this case you would have 1/8 of the figure in the first octant and you would be calculating
the volume of a cube with sides a/2.
8. We again orient the cube so that a vertex is at the origin and the edges that meet at that vertex lie along the x-, yand z-axes so that the cube is in the first octant. As in Exercise 7 we will double the volume of half of the cube
corresponding to 0 … θ … π/4. We will have to split the integral into two pieces: the piece in which ρ is bounded
by the top of the cube (z D a or ρ D a sec ϕ) and the piece in which ρ is bounded by the side of the cube (x D a
or ρ D a csc ϕ sec θ). The boundary value of ϕ depends on θ. Set a D ρ cos ϕ equal to a D ρ sin ϕ cos θ and solve
to obtain ϕ D cot−1 cos θ. So the volume is
V D2
π/4 cot−1 cos θ a sec ϕ
0
D 2
π/4 π/2
ρ sin ϕ dρ dϕ dθ C 2
a csc ϕ sec θ
2
0
0
ρ2 sin ϕ dρ dϕ dθ
cot−1 cos θ 0
0
a3
a3
3
C
Da .
6
3
Exercises 9–17 are examples where a change of variables helps. Exercise 14 depends on Exercise 11 and together they
are much less difficult than they may first appear.
9. Here we will let u D x − 2y and v D x C y. We calculate
⭸u, v 1 −2 ⭸x, y
D
D 3 so
D 1/3.
1
1 ⭸x, y
⭸u, v
The three boundary lines x C y D 1, x D 0, and y D 0 correspond to v D 1, 2v D −u, and u D v. We have all
of the pieces to assemble our integral:
v
1 v
1
x − 2y
u
u 1
v
cos du dv D
sin cos du
dA D
x C y
v
v −2v
0 −2v 3
0 3
6D
1
1
v2
1
v
D
sin 1 − sin−2 dv D
sin 1 C sin 2 D sin 1 C sin 2.
6
6
0 3
0
10. Let u D y3 and v D x C 2y. Then 0 … u … 216, 0 … v … 1, and
⭸u, v 0 3y2 D
D −3y2 D −3u2/3 so
1
2 ⭸x, y
⭸x, y
1
D−
.
⭸u, v
3u2/3
Then
6 1−2y
0
−2y
3
y3 x C 2y2 exC2y dx dy D
216 1
0
0
u 2 v3
v e dv du
3u2/3
216 1
1 1/3 2 v3
u v e dv du
3
0
0
1
216
1 1/3 v3 du
D
u e 9
0
D
D
216
0
vD0
1 1/3
u e − 1 du
9
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“runall” — 2005/8/1 — 15:13 — page 276 — #54
276
Chapter 5
Multiple Integration
216
1 3 4/3
D u e − 1
9 4
0
D 108e − 1.
a b√ 1−x2 /a2
11. (a) As we’ve seen before, we can write the integral as
−a −b
√
dy dx.
1−x2 /a2
(b) When we scale by letting x D ax and y D by, the ellipse is transformed into the unit circle E ∗ in the xyplane. To√rewrite the integral we also quickly calculate that ⭸x, y/⭸x, y D ab. The transformed integral
1 1−x2
ab dy dx.
is
√
−1 − 1−x2
(c) Because we are integrating over a unit circle, we transform to polar coordinates (o.k., really we do it because
the text tells us to):
1 √ 1−x2
−1 −
√
1−x2
ab dy dx D
2π 1
0
abr dr dθ D
0
2π
0
1
1
ab dθ D ab2π D π ab.
2
2
u C v
2v − u
which implies y D
. Substituting
3
3
these expressions into the equation for the ellipse, we obtain
12. (a) With u D 2x − y, v D x C y, we see u C v D 3x so x D
2
13 2
u C v
u C v
2v − u
2v − u
C 14 C 10 D 9.
3
3
3
3
Expanding and simplifying, we find
u2
C v2 D 1.
9
⭸x, y 1 dA D
1 dx dy D
(b) Area D
du dv where E ∗ denotes the corresponding ellipse in the
6E
6E∗ ⭸u, v 6E
⭸x, y
1/3 1/3
1
D det
uv -plane given above. Now
D − so
2/3 −1/3 ⭸u, v
3
1
1
1
∗
Area D
du dv D area of E D π · 3 · 1 D π using the result of part (c) of Exercise 11.
3
3
6E∗ 3
u C v
v − u
13. With u D x − y, v D x C y we find that x D
,yD
. Substituting these expressions into the
2
2
equation for E, we find
2
5
2
v − u
v − u
u C v
u C v
C 6
C 5
D 4,
2
2
2
2
which simplifies to
u2
C v2 D 1.
4
The area of ellipse E ∗ in the uv -plane is 2π. The area of the original ellipse E is
1/2
⭸x, y det
du
dv
D
1 dA D
dx dy D
−1/2
∗
∗
⭸
u,
v
6E
6E
6E
6E
1
du dv D 1 area of E∗ D 1 2π D π.
D
∗
2
2
6E 2
1/2 du dv
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“runall” — 2005/8/1 — 15:13 — page 277 — #55
Section 5.8
Miscellaneous Exercises for Chapter 5
277
14. We follow the steps in Exercise 11, inserting the same letters for ease in locating the corresponding parts.
(a) First we write the integral in Cartesian coordinates as
a b√ 1−x2 /a2 c√ 1−x2 /a2 −y2 /b2
−a −b
√
1−x2 /a2
−c
√
dz dy dx.
1−x2 /a2 −y2 /b2
(b) We now scale the variables using x D ax, y D by, and z D cz. Note that the ellipsoid E is transformed into
the unit sphere E ∗ and that ⭸x, y, z/⭸x, y, z D abc. The transformed integral is:
1 √ 1−x2 √ 1−x2 −y2
√
−1 − 1−x2
−
√
abc dz dy dx.
1−x2 −y2
(c) Because we are integrating over a unit sphere, we will transform to spherical coordinates:
1 √ 1−x2 √ 1−x2 −y2
−1 −
√
1−x2
−
√
1−x2 −y2
abc dz dy dx D
2π 1 π
abc ρ2 sin ϕ dϕ dρ dθ
0
D
D
0
2π 1
0
π
2π 1
− cos ϕabcρ dρ dθ D
2abc ρ2 dρ dθ
2
0
0
0
4
2
abc dθ D πabc.
3
3
2π
0
0
0
15. If you didn’t first sketch the region you may be tempted to use the numerator and denominator of the integrand
as your new variables. The diamond-like shape is bounded on two sides by the hyperbolas x2 − y2 D 1 and
x2 − y2 D 4 and on the other two sides by the ellipses x2 /4 C y2 D 1 and x2 /4 C y2 D 4.
2
1.75
1.5
1.25
1
0.75
0.5
0.25
0.5
1
1.5
2
2.5
3
4
3.5
We, therefore, make the change of variables u D x2 − y2 and v D x2 /4 C y2 . Then
⭸u, v 2x −2y D
D 5xy so
x/2
2y ⭸x, y
⭸x, y
D 1/5xy.
⭸u, v
The integral greatly simplifies:
xy
dA D
2 − x2
y
6
D
4 4
1
1
D−
5
1
xy 1
1
du dv D −
−u 5xy
5
4 4
1
1
1
du dv
u
4
3
ln 4 dv D − ln 4.
5
1
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278
Chapter 5
Multiple Integration
16. The region D is bounded on the left and right by x2 − y2 D 1 and x2 − y2 D 9 and on the bottom and top by
xy D 1 and xy D 4. It looks
y
3
2.5
2
1.5
1
0.5
0.5
1
1.5
2
2.5
3
3.5
4
x
This suggests we try the change of variables u D x2 − y2 , v D xy. Then
⭸u, v
D det
⭸x, y
so that
2x
y
−2y
D 2x2 C y2 x ⭸x, y
1
D
.
2
⭸u, v
2x C y2 Moreover, the region D* in the uv -plane corresponding to D is
v
4
1
1
9
u
Thus, using the change of variables theorem, we have
2
6D
4 9
1 u
1 u
e du dv D
e du dv
∗
2
1 1 2
6D
4
3
1 9
D
e − e dV D e9 − e.
2
1 2
x2 C y2 ex −y dA D
2
17. The region D is bounded on the top and bottom by y D 1 and y D 2 and on the left and right by xy D 1 and
xy D 4; the region looks like the following figure.
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“runall” — 2005/8/1 — 15:13 — page 279 — #57
Section 5.8
Miscellaneous Exercises for Chapter 5
279
y
3
2.5
2
1.5
1
0.5
1
2
3
x
4
With this in mind, we try the change of variables u D xy, v D y. Then
⭸u, v
D det
⭸x, y
y
0
x
D y D v.
1 Moreover, the region D∗ in the uv -plane corresponding to D is the rectangle [1, 4] * [1, 2]:
v
2
1
u
1
4
The change of variables theorem tells us that
4 2
1
1
v
dA D
· v du dv D
dv du
2
2
2
2
∗
1 1 u C 1
6D u C 1
6D x y C 1
4
4
3/2
3
−1 du D tan u
D
2
2
1 u C 1
1
D
3
π
−1
tan 4 − .
2
4
18. (a) Follow the same steps as in defining the double and triple integrals.
• Define a partition of B D [a, b] * [c, d] * [p, q] * [r, s] of order n to be four collections of partition
points that break up B into a union of n4 subboxes. See Definition 4.1 and add that r D w0 < w1 <
· · · < wn D s and wl D wl − wl−1 .
• Define a Riemann sum. For a function f defined on B, partition B as above and let cijkl be any point
in the subbox
Bijkl D [xi−1 , xi ] * [yj−1 , yj ] * [zk−1 , zk ] * [wl−1 , wl ].
The Riemann sum of f on B corresponding to the partition is
SD
n
i,j,k,lD1
f cijkl xi yj zk wl D
n
f cijkl Vijkl .
i,j,k,lD1
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“runall” — 2005/8/1 — 15:13 — page 280 — #58
280
Chapter 5
Multiple Integration
• Define the quadruple integral of f on B, written
oB
f x, y, z, w dV D
oB
f x, y, z, w dx dy dz dw
to be
o
B
f x, y, z, w dV D
n
lim
all xi ,yj ,zk ,wl '0 i,j,k,lD1
f cijkl xi yj zk wl .
• We extend the definition to compact non-box regions W by defining the function f ext which is f
everywhere in W and is 0 everywhere else. Then if B is a box containing W we can define
o
W
f dV D
oB
f ext dV .
• As in the cases of the double and triple integrals, Fubini’s theorem allows us to evaluate the integral as
an iterated integral.
(b) We calculate:
o
W
x C 2y C 3z − 4w dV D
2 3 4 2
−1 0
0
D4
D4
D4
2 3 4
0
−1 0
0
−1
2 3
2
−2
x C 2y C 3z − 4w dw dz dy dx
x C 2y C 3z dz dy dx
4x C 8y C 24 dy dx
16x C 32 C 96 dx D 64
0
2
x C 8 dx
0
D 642 C 16 D 1152.
19. (a) We are just generalizing what we’ve done to set up the area of a circle or the volume of a sphere (more
recently see Exercises 11 and 14 from this section). Here our integral is:
√
√
√
a2 −x2
a
√
−a − a2 −x2
a2 −x2 −y2
−
√
a2 −x2 −y2
a2 −x2 −y2 −z2
−
√
dw dz dy dx.
a2 −x2 −y2 −z2
(b) You should get π2 a2 /4.
(c) For n D 5 you should get 8π2 a5 /15, and for n D 6 you should get π3 a6 /6. If you include the cases for
n D 2 and n D 3 you may begin to see a pattern for the even exponents. If n is even, the volume of the
n-sphere of radius a is πn/2 an /n/2!. Fitting in the odd terms looks really hard and the pattern shouldn’t
occur to any of your students. In fact, the general formula depends on the Gamma function which is beyond
what we would expect the students to know at this point. For kicks, the volume of the n-sphere of radius a is
πn/2 an
.
n/2 C 1
Note that the volume of an n-sphere of radius a decreases to 0 as n increases.
20. (a) To obtain the mass we compute the following integral (which is straightforward so the details are omitted):
MD
2π π 4
0
0
.12ρ2 ρ2 sin ϕ dρ dϕ dθ D 74.976π L 235.5440508g.
3
(b) Because the shell is sealed, the volume is V D 4/3π43 D 256π/3 cm3 , so the mass of that volume of
water is greater, and so the shell would float.
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“runall” — 2005/8/1 — 15:13 — page 281 — #59
Section 5.8
Miscellaneous Exercises for Chapter 5
281
(c) If the core of the shell fills with water then the volume that the shell has to displace is V D 4/3π43 −
33 D 4/3π37. The water for that volume would have mass of about 155 grams so the shell would sink.
21. When you average the height of the hemisphere of radius a, first you integrate
2π a
0
zr dr dθ D
0
2π a √
2
r a2 − r2 dr dθ D πa3 .
3
0
0
For the average height, we divide this by the area of the region over which we are integrating:
2/3πa3
2
D a.
πa2
3
√
We now solve to see √
which values of r correspond to this height: 2/3a D√ a2 − r2 when 4/9a2 D a2 − r2 ,
which is when r D 5a/3. Therefore, the pole can be installed at most 5a/3 from the center of the floor of
the dome.
22. (a) By the fundamental theorem of calculus
d
dy
y
y b
d
dy
Gy dy D Gy so
c
c
fy x, y dx dy D
a
b
fy x, y dx.
a
(b) On the other hand, by Fubini’s theorem,
d
dy
y b
c
fy x, y dx dy D
a
D
d
dy
d
dy
b y
a
b
fy x, y dy dx
c
f x, y − f x, c dx D
a
d
dy
b
f x, y dx.
a
Combine parts (a) and (b) to obtain the desired results.
23. (a)
I , δ D
1− 1−δ
δ
√
D 4 1 −
(b)
lim
,δ'0,0
24. For 0 <
<
1−
√
1
2 √
√ 1 − δ − δ dx
√ dy dx D
xy
x
√
√
√
−
1 − δ − δ
I , δ D 4 · 1 · 1 D 4
1
1
1
, 0 < δ < we consider I , δ D
dA where D ,δ D [ , 1 − ] * [δ, 1 − δ]. Then
2
2
6D ,δ x C y
I , δ D
1− 1−δ
δ
D
1−
1
dy dx D
x C y
1−
1−δ
lnx C y
dx
yDδ
lnx C 1 − δ − lnx C δ dx.
Using integration by parts, we find that
ln u du D u ln u − u C C so that
I , δ D [x C 1 − δ lnx C 1 − δ − x C 1 − δ − x C δ lnx C δ C x C δ|1−
xD
D 2 −
− δ ln2 −
C 1 −
− δ − 2 −
− δ − 1 −
C δ ln1 −
C δ
C δ − C 1 − δ ln C 1 − δ C C 1 − δ
C C δ ln C δ − C δ.
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282
Chapter 5
To evaluate
Multiple Integration
lim
,δ'0C ,0C I , δ we first note that, by l’Hôpital’s rule,
ln u
lim u ln u D lim
u'0C
u'0C 1/u
1/u
D lim
u'0C −1/u2
D − lim u D 0.
u'0C
Thus C δ ln C δ ' 0 as , δ ' 0C , 0C . The other terms in the expression have evident limits so
I , δ D 2 ln 2 − 2 − ln 1 C 1 − ln 1 C 1 C 0 − 0 D 2 ln 2.
that
lim
,δ'0C ,0C 25. For 0 < < 1 , 0 < δ < 1 , let D ,δ D [ , 1 −
2
2
I , δ D
D
x
dA D
6D ,δ y
Note that
lim
,δ'0,0
1−δ 1−
y
I , δ D −q since 1 −
2
x
dx dy
y
δ
1−δ 1 − 2
δ
] * [δ, 1 − δ] and consider
dy D 1
− ln1 − δ − ln δ.
2
and ln1 − δ remain finite, but ln δ ' −q. Thus the improper
integral does not converge.
In Exercises 26 and 27 the students will need integration by parts and l’Hôpital’s rule.
26.
2π 1
ln x2 C y2 dA D lim
ln x2 C y2 dA D lim
r ln r dr dθ
'0 6
'0 0
6D
D
1
2π 2
2π
2
r2 r
1
D lim
dθ D lim
ln
ln r −
− −
'0 0
'0 0
2
4
4
2
D lim 2π −
'0
2
1
−
ln
4
2
2
C
4
C
2
4
dθ
D −π/2.
27. Define B D {x, y, z| … x2 C y2 C z2 … 1}. Then
l
B
ln x2 C y2 C z2 dV D lim
'0 l
B
D lim
ln x2 C y2 C z2 dV
2π 1 π
'0 0
D lim
2π 1
'0 0
D lim
2π
'0 0
2ρ2 ln ρ dρ dθ
1
2/3ρ3 ln ρ − 2ρ3 /9 dθ
D lim 2π −
'0
28. (a)
a b
Ia, b D
1
a
D
1
(b) As a, b ' q, Ia, b '
1
ln ρρ2 sin ϕ dϕ dρ dθ
0
2
2 3
−
ln
9
3
1
dy dx D
x2 y 3
C
2 3
D −4π/9.
9
b
1 −
dx
2y2 x2 yD1
1
a
1
1
1
1
1
1
−
2 dx D −
1 − 2
2
2
2b x
2
2b
a
1
.
2
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“runall” — 2005/8/1 — 15:13 — page 283 — #61
Section 5.8
Miscellaneous Exercises for Chapter 5
283
29. Let Da,b D [1, a] * [1, b] and consider, for p, q Z 1,
Ia, b D
1
dA D
p yq
x
6Da,b
b
D
1
D
b a
1
1
a
1
q
p−1
1 − py x
1
dx dy
xp y q
dy D
xD1
b
1
1
1
dy
p−1 − 1
q
1 − p a
1 y
1
1
1
− 1 − 1 .
1 − p1 − q ap−1
bq−1
1
1
D
, so the integral converges in this case.
1−p1−q
p−1q−1
p−1
p−1
' q. Similarly if q < 1, 1/b
' q.
If p < 1, then 1/a
If p > 1, q > 1 then as a, b ' q, Ia, b '
If p D 1, q Z 1, then
b a
1
1
1
1
1 xyq dx dy D ln a 1−q bq−1
− 1. This becomes infinite as a, b ' q. Similarly, if
q D 1, p Z 1, Ia, b becomes infinite as a, b ' q. If p D q D 1, then Ia, b D ln a ln b ' q as a, b ' q.
To summarize: the integral converges if and only if p > 1 and q > 1—in which case the value of the integral
is 1/p − 1q − 1.
30. (a) We use polar coordinates to make the evaluation. Let
Ia D
6Da
1 C x2 C y2 −2 dA D
a
1
2 −1 D 2π −1 C r 2
rD0
D π 1 −
2π a
0
1 C r2 −2 r dr dθ
0
D −π 1
− 1
1 C a2
1
.
1 C a2
lima'q Ia D π. Thus the integral converges.
2π a
1 C x2 C y2 p dA D
1 C r2 p r dr dθ. If p Z −1, then
(b) Let Ia D
0
0
6Da
Ia D
π
1 C a2 pC1 − 1.
p C 1
Now lima'q 1 C a2 pC1 is finite (and equals 0) just in case p C 1 < 0 or p < −1. In such case, the integral
2π a
r
π
dr dθ D π ln1 C a2 ' q
converges and its value is −
. If p D −1, then I−1 D
2
p C 1
1
C
r
0
0
as a ' q. So the integral converges if and only if p < −1.
31. Consider
2π π a
ρ2 sin ϕ
1
dV D
dρ dϕ dθ
2
2
2 3/2
2 3/2
0
0
0 1 C ρ l
Ba 1 C x C y C z a 2π π
a
ρ2
ρ2
D
sin ϕ dϕ dθ dρ D
4π ·
dρ
2 3/2
1 C ρ2 3/2
0 0
0 1 C ρ 0
Ia D
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284
Chapter 5
Multiple Integration
Now let ρ D tan u so dρ D sec2 u du. Then
tan−1 a
tan−1 a
tan2 u · sec2 u du
sin2 u
D 4π
du
3
sec u
cos u
0
0
tan−1 a
tan−1 a
1 − cos2 u
D 4π
du D 4π
sec u − cos u du
cos u
0
0
tan−1 a
√
D 4πln | sec u C tan u| − sin u
D 4π ln a2 C 1 C a −
Ia D 4π
0
a
.
a2 C 1
Since lim Ia D q the integral does not converge.
a'q
32. Consider
Ia D
D
e−
√
x2 Cy2 Cz2
lBa
a 2π π
0
0
dV D
2π π a
0
0
e−ρ · ρ2 sin ϕ dρ dϕ dθ
0
e−ρ ρ2 sin ϕ dϕ dθ dρ D 4π
a
0
ρ2 e−ρ dρ
0
Now use integration by parts twice: First let u D ρ2 and dv D e−ρ dρ. Then
a
2 −ρ a
0
0
Ia D 4π −ρ e
C 2
−ρ
ρe
2 −a
dρ D −4πa e
a
C 8π
ρe−ρ dρ.
0
Now let u D ρ and dv D e−ρ dρ so that
Ia D −4πa2 e−a C 8π −ρe−ρ |a0 C
a
0
e−ρ dρ
D −4πa2 e−a − 8πae−a − 8πe−a C 8π.
lim Ia D 8π, so the integral converges and has value 8π.
a'q
The importance of Exercise 33 can not be overemphasized. The students have come from a course where they learned
one technique of integration after another. They also learned some numerical methods (at least a brief introduction to
Riemann sums, the trapezoid rule and Simpson’s rule). In a way Exercise 33 is the payoff—it is a chance to mention:
q
2
• Until now they couldn’t calculate −q e−x dx. The fact that you need the tools of multivariable calculus (or complex
analysis) is pretty cool.
b
2
• They still can’t calculate a e−x dx. There is a need for numerical methods to calculate a function as common
as the bell curve (with a constant that stretches in the vertical direction and another constant that stretches in the
horizontal direction, this is the normal curve). Many will encounter this function in a course on statistics and use
the tables; they should know that this is because we can’t find the definite integral over a general finite interval.
• The technique is pretty and unexpected and is one of the tricks that they should see some time in their mathematical
training. The problem is surprisingly straightforward once someone shows you the trick.
1
2
2
2
33. (a)
e−x dx is finite since e−x is bounded on [−1, 1]. Since 0 … e−x … 1/x2 on both [1, q and
−1
q
−1
q
2
2
2
−q, −1] and the improper integrals
1/x dx and
1/x dx converge, we see that
e−x dx and
1
−1
q1
−1 −q
1
q
−x2
−x2
−x2
−x2
−x2
e dx both converge. Hence
e dx D
e dx C
e dx C
e dx converges.
−q
−q
−q
−1
1
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“runall” — 2005/8/1 — 15:13 — page 285 — #63
Section 5.8
Miscellaneous Exercises for Chapter 5
285
(b) We have
q
q
q
q
2
2
2
2
e−x dx e−x dx D e−x dx e−y dy
I2 D −q
−q
q q
D
e−x e−y dx dy D
2
−q −q
−q
2
e−x −y dA.
2
6R2
−q
2
(c) We’ll use polar coordinates.
6Da
−x2 −y2
e
dA D
2π a
1
· r dr dθ D −
2
−r 2
e
0
0
−a2
D π1 − e
2π
e−a − 1 dθ
2
0
e−x −y dA D
2
(d) Note that, as a ' q, the disk Da fills out more and more of R2 . Thus lim
a'q 6
Da
2
e−x −y dA D I 2 .
2
2
6
(e) Putting everything together:
R2
I 2 D lim π1 − e−a D π.
2
a'q
q
√
2
Thus I D −q e−x dx D π.
34. (a) First, clearly f x Ú 0. And second, by symmetry,
q
−2|x|
e
−q
q
dx D 2
−2x
e
q
D 1.
−2x dx D −e
0
0
(b) We will reduce the calculations in Egbert’s problem by recentering.
350
350
1 −|x−300|
1 −x−300
e
e
dx D 2
dx
250 2
300 2
50
50
−x
−x e dx D −e D 1 − e−50 .
D
P250 … x … 350 D
0
35. (a) Since
0
2x C y
Ú 0 on [0, 5] * [0, 4], f x, y Ú 0 for all (x, y). Now
140
6R2
f x, y D
4 5
0
0
2x C y
1
dx dy D
140
140
4
1
5 2 25 C 5y dy D
25y C y 140
2
0
0
4
1
D
100 C 40 D 1.
140
(b) Since f x, y D 0 for x < 0 or y < 0,
Probx … 1, y … 1 D Probx, y ∈ [0, 1] * [0, 1] D
1
D
140
1
0
1 1
0
1 C y dy D
0
2x C y
dx dy
140
1
1
3
L 0.0107.
1 C D
140
2
280
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“runall” — 2005/8/1 — 15:13 — page 286 — #64
286
Chapter 5
Multiple Integration
36. (a) Since ye−x−y Ú 0 for y Ú 0 (note the exponential term is strictly positive), we have that f x, y Ú 0 for
all (x, y). Now we check
q q
ye−x−y dA D
ye−y e−x dx dy
0
0
6R2
q
D
ye−y dy
0
q
D −ye−y − e−y D 1.
0
(b) Probx C y … 2 D Probx, y ∈ D where D is the triangular region bounded by x D 0, y D 0 and
x C y D 2.
y
2
1.5
1
0.5
0.5
1
1.5
2
x
Thus the desired probability is
2
2 2−x
−x
ye−y e−x dy dx D
−ye−y − e−y |2−x
dx
0 e
0
0
D
2
0
x − 2ex−2 − ex−2 C 1e−x dx D
0
2
x − 2e−2 − e−2 C e−x dx
0
D 1 − 5e−2 .
37. First, we know that C Ú 0. Second,
q q
q q
1D
Ce−a|x|−b|y| dx dy D 4
Ce−ax−by dx dy
−q −q
q
D 4C −ax
e
0
q
dx 0
−by
e
0
0
q
q
1 −ax 1 −by 4C
dy D 4C − e − e
D ab .
a
b
0
0
So C D ab/4.
38. Note that if C Ú 0, then f x, y Ú 0 for all x since a and b are nonnegative. Thus we calculate
1 1
1
1
f x, y dA D
Cax C by dx dy D C
a C by dy
2
2
0 0
0
6R
1
1
a C b
D C a C b D C .
2
2
2
For this to equal 1, we must have C D
2
.
a C b
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“runall” — 2005/8/1 — 15:13 — page 287 — #65
Section 5.8
Miscellaneous Exercises for Chapter 5
287
39. (a) If C Ú 0, then f x, y Ú 0 for all (x, y). Thus we calculate
b a
b 2
a
y dy
f x, y dA D
C xy dx dy D C
2
0 0
0 2
6R
DC
a2 b 2
.
4
4
.
a2 b 2
b
(b) Probbx − ay Ú 0 is the probability that a point (x, y) falls below the line y D x. Since f is zero
a
outside the rectangle [0, a] * [0, b], we see that the desired probability is the same as the probability
Probx, y ∈ D where D is the triangular region shown below.
For this to equal 1 we must have C D
y
b
a
x
This last probability is calculated as
a b/ax
6D
f dA D
0
a
4
a2 b 2
0
xy dy dx
2
4
1 b
x · x dx D
D
2 b2
a
2 a
0
a
1
1 4 x D .
D
2a4 2
a
0
2 3
x dx
a4
0
40. We are integrating over the triangle where 0 … x C y … 60. The integral is fairly straightforward so the details
are omitted:
60 60−x
60
1
1
e−x/50 e−y/5 dy dx D −
e−x/50 [e−12Cx/5 − 1] dx
250 0
50 0
0
10 −6/5
1
D1 −
C e−12 L .665340.
e
9
9
41. We use polar coordinates:
1/2 √ 1/4−x2
−1/2
1 −x2 −y2
dy dx D
e
√
2
π
− 1/4−x
D
2π 1/2
0
0
2π
0
1 −r2
re dr dθ
π
1 1
−1/4
dθ
1 − e
π 2
D 1 − e−1/4 L .22199.
42. The joint density function of the components is
F x, y D f x · f y D



1
e−xCy/2000
2
2000

 0
if x Ú 0, y Ú 0
otherwise.
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“runall” — 2005/8/1 — 15:13 — page 288 — #66
288
Chapter 5
Multiple Integration
2000 2000
1
e−x/2000 e−y/2000 dx dy
2
2000
0
0
2000
2000
−x/2000 −2000e
e−y/2000 dy
So we want Probx … 2000, y … 2000 D
1
20002 0
0
2000
2
1
1
1 −y/2000
D
dy D 1 − .
1 − e
2000 0
e
e
D
43. Formula (8) in Section 5.6 is I D
6W
d 2 δx, y, z dV . If we choose the coordinates so that the center of mass is
xδx, y, z dV D 0 and
yδx, y, z dV D 0. We can also choose the coordinates so
l
l
W
W
that L is the z-axis. L is a line parallel to the z-axis distance h away, so L is the line corresponding to x D a
and y D b where a2 C b2 D h2 . Then
at the origin, then
IL D
l
W
x2 C y2 δx, y, z dV
and
IL D
lW
x2 C y2 C h2 − 2ax − 2byδx, y, z dV D
l
W
x2 C y2 C h2 δx, y, z dV
so
IL − IL D
l
W
h2 δx, y, z dV D h2
l
W
δx, y, z dV D h2 M.
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“runall” — 2005/8/1 — 15:23 — page 289 — #1
C H A P T E R
6
Line Integrals
6.1 SCALAR AND VECTOR LINE INTEGRALS
1. (a) x t D −3, 4 and x t D 5 so by Definition 1.1,
2
25 2 f ds D
x C 2y5 dt D 5
[2 − 3t C 8t − 2] dt D 5
5t dt D
t D 50.
2 0
x
0
0
0
2
2
2
(b) x t D − sin t, cos t and x t D 1 so by Definition 1.1,
π
f ds D
x
π
x C 2y1 dt D
0
0
[cos t C 2 sin t] dt D [sin t − 2 cos t]|π
0 D 4.
For Exercises 2–5 we will use Definition 1.1. For each calculate x , x , and f x.
2.
f ds D
2
x
[t2t3t 12 C 22 C 32 ] dt D 6
√
0
2
√
√
6 14 4 14
t dt D
t D 24 14.
4
0
0
2
3
3.
3
3/2 3
t C t 3/2
8
9
9
9
f ds D
1 C 1 C t dt D
2 C t dt D
2 C t t C t 3/2
4
4
27
4
x
1
1
1
√
√
D 35 35 − 17 17/27.
3
4.
f ds D
x
√
16 C 9
2π
3 cos 4t C cos 4t sin 4t C 27t 3 dt D 5
0
2π
0
2π
1
27 4 3
cos 8t C
t D 540π4 .
D 5 sin 4t −
4
16
4
1
sin 8t C 27t 3 dt
2
3 cos 4t C
0
5.
f ds D
x
1
3
√
[2t − t 1 C 4t 2 ] dt C
2 − 1 C 2t 2 − 4t C 2 dt
0
D 5
1
3/2
− 1/12 C 22/3 D 5
3/2
C 87/12.
For Exercises 6–10 we will use Definition 1.2.
6.
F · ds D
x
1
0
2t C 1, t, 3t − 1 · 2, 1, 3 dt D
1
1
14t − 1 dt D 7t − t D 6.
2
0
0
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“runall” — 2005/8/1 — 15:23 — page 290 — #2
290
Chapter 6
Line Integrals
7.
π/2
F · ds D
π/2
2 − cos t, sin t · cos t, sin t dt D
2 cos t − cos2 t C sin2 t dt
0
x
0
π/2
D 2 sin t − sin 2t/2
D 2.
0
8.
F · ds D
1
x
9.
F · ds D
4π
x
−1
t, 3t 2 , −2t 3 · 1, 6t, 6t 2 dt D
t, sin2 t, 2t · − sin t, cos t, 1/3 dt D
0
F · ds D
1
−1
4π
3
t2
D
−t sin t C sin2 t cos t C 2t/3 dt
sin t
C
3
3
4π
2
D 12π C 16π .
3
0
t 2 cos t 3 , t sin t 3 , t 3 sin t 6 · 1, 2t, 3t 2 dt D
0
x
t C 18t 3 − 12t 5 dt D 0.
0
D t cos t − sin t C
10.
1
1
t 2 cos t 3 C 2t 2 sin t 3 C 3t 5 sin t 6 dt
0
1
2 cos t 3
cos t 6 7 − 7 cos 1 C 2 sin 1
sin t 3
−
−
.
D
3
3
2
6
0
Assign at least one of Exercises 11 and 13 so that the students are exposed to the notation before they encounter Green’s
theorem in the next section.
π
π
11.
x dy − y dx D
[3cos 3t2 C 3sin 3t2 ] dt D 3
dt D 3π.
0
0
x
2
2
12.
F · ds D
2t, 1, 0 · 1, 6t, 0 dt D
8t dt D 16.
x
0
0
13. The good news is: there is a ton of cancellation.
2π 2t
e cos 3t2e2t cos 3t − 3e2t sin 3t C e2t sin 3t2e2t sin 3t C 3e2t cos 3t
x dx C y dx
D
dt
2
2 3/2
e4t cos2 3t C e4t sin2 3t3/2
x x C y 0
2π
2π
−2t
−2t D
2e
dt D −e D 1 − e−4π .
0
0
√
14. Note that x D t, 2 t and x D 1, t −1/2 , so
9
√
√
1
3y ds D
6t 1/2 1 C dt D 40 10 − 8 2.
t
C
1
15. (a)
F · ds D
x
1
2t , t − t · 1, 2t dt D
2
0
F · ds D
y
1/2
1
2
2t 3 dt D
0
1
.
2
2 − 8t C 8t 2 , 4t 2 − 2t · −2, 8t − 4 dt
0
D
1/2
0
1
32t 3 − 48t 2 C 24t − 4 dt D − .
2
(b) The path y is an orientation-reversing reparametrization of x.
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“runall” — 2005/8/1 — 15:23 — page 291 — #3
Section 6.1
Scalar and Vector Line Integrals
291
16. We write the path as xt D t C 1, −4t C 1, 2t C 1, 0 … t … 1. This means that x t D 1, −4, 2, therefore
Work D
F · ds D
1
0
C
D
1 C t2 1 − 4t − 41 C 2t C 221 C t − 1 − 4t dt
1
10
−4t 3 − 7t 2 C 2t − 1 D − .
3
0
0
1
17. First we organize the information we need for each of the four paths (each is for 0 … t … 1).
i
xi
xi t
Fxi t · xi t
1
1 − 2t, 1, 3
−2, 0, 0
−2486 − 31 − 2t
2
−1, 1 − 2t, 3
0, −2, 0
−2−1
3
−1 C 2t, −1, 3
(2, 0, 0)
2486 − 31 − 2t
4
1, −1 C 2t, 3
(0, 2, 0)
2−1
So
F · ds D
Work D
1
iD1 xi
C
D
4
Fxi t · xi t dt
−2486 − 31 − 2t dt C
1
0
2 dt C
0
1
2486 − 31 − 2t dt C
0
1
−2 dt
0
D 0.
18. The path is xt D 2t C 1, 4t C 1, 0 … t … 1. The integral is
x − y dx C x − y dy D
2
2
1
[4t 2 2 C −16t 2 − 6t4] dt
0
C
D
1
−56t 2 − 24t dt D −
0
92
.
3
19. The curve C looks like
y
1
0.8
0.6
0.4
0.2
x
0.5
x2 y dx − x C y dy D
Then
C
1
C
C1
1.5
C
C2
2
2.5
3
C
C3
C4
— C1 is the segment from (0, 0) to (3, 0), given as xt D t, 0, 0 … t … 3 * x t D 1, 0.
3
Then
D
0 dx − t · 0 D 0.
C1
0
C2
0
— C2 is the segment from (3, 0) to (3, 1), given by xt D 3, t, 0 … t … 1 * x t D 0, 1.
1
1
1
Then
D
0 − 3 C t dt D −3t − t 2 D −7/2.
2
0
— C3 is the segment from (3, 1) to (1, 1), given by xt D 3 − t, 1, 0 … t … 2 * x t D −1, 0.
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“runall” — 2005/8/1 — 15:23 — page 292 — #4
292
Chapter 6
Line Integrals
2
1
1
26
3 − t3 D 1 − 27 D − .
3
3
3
C3
0
0
— C4 is the segment from (1, 1) to (0, 0), given by xt D 1 − t, 1 − t, 0 … t … 1 * x t D −1, −1.
1
1
1
D
[1 − t3 −1 C 2 − 2t] dt D 1 − t4 C 2t − t 2 4
C4
0
0
D
Then
2
3 − t2 −1 dt − 4 − t · 0 D
D2 − 1 −
3
1
D .
4
4
26
3
137
7
−
C D−
.
2
3
4
12
C
2
3
20. Parametrize C as xt D t , t , −1 … t … 1, so that x t D 2t, 3t 2 . Then
D−
So
x2 y dx − xy dy D
C
1
−1
t 7 2t − t 5 3t 2 dt D
1
2
3 4
2t 8 − 3t 7 dt D t 9 − t 8 D .
9
8
9
−1
−1
1
21. The path is xt D 4t C 1, 2t C 1, −t C 2, 0 … t … 1. The integral is
1
yz dx − xz dy C xy dz D
[4−2t 2 C 3t C 2 − 2−4t 2 C 7t C 2 − 8t 2 C 6t C 1] dt
0
C
D
1
[−8t 2 − 8t C 3] dt D −
0
11
.
3
22. We must parametrize C. Along the cylinder we may take x D 2 cos t, y D 2 sin t. Then z D x2 so we have
z D 4 cos2 t. The curve is traced once as t varies from 0 to 2π, so we have
2π
z dx C x dy C y dz D
[4 cos2 t−2 sin t C 2 cos t2 cos t C 2 sin t8 cos t− sin t] dt
0
C
D
2π
8 cos2 t− sin t C 21 C cos 2t − 16 sin2 t cos t dt
0
2π
8
16 3 3
D cos t C 2t C sin 2t −
sin t D 4π.
3
3
0
23. Using formula (3) in §6.1, we have
T · ds D T · T ds D 1 ds D length of x.
x
x
x
24. Of course it’s left to Becky Thatcher to figure out that the path is xt D 5 cos t, 5 sin t, 0 … t … π/2, so the
area of one side of the fence is
π/2
10 − x − y ds D
510 − 5 cos t − 5 sin t dt D 25[π − 2] L 28.54 ft2 .
0
C
25. (a) The force that Sisyphus is applying is 50x t/x t. The
√ path is given as xt D 5 cos 3t, 5 sin 3t, 10t
and so x t D −15 sin 3t, 15 cos 3t, 10 and x t D 325. The total work done is
10
0
50x t
· x t dt D
x t
10
0
50x t dt D
10
√
√
50 325 dt D 2500 13 ft-lb.
0
(b) This time the 75 pounds is applied straight down. The total work done is
10
10
0, 0, 75 · −15 sin 3t, 15 cos 3t, 10 dt D
750 dt D 7500 ft-lb.
0
0
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“runall” — 2005/8/1 — 15:23 — page 293 — #5
Section 6.1
Scalar and Vector Line Integrals
293
26. The force is applied in the direction of 24, 32 − 14t. Force is applied in the opposite direction to the tension.
The total work done is
1
1
24, 32 − 14t
32 − 14t
√
25 · 0, −14 dt D −725
dt D −250 ft-lb.
2
2
400 − 224t C 49t 2
24 C 32 − 14t
0
0
27. The path is xt D t, f t, a … t … b, and x t D 1, f t. Since F D yi,
b
F · ds D
C
f t, 0 · 1, f t dt D
b
f t dt.
a
a
28. We take the sphere to be of radius c, so that x2 C y2 C z2 D c2 . Begin with the hint and take the derivative with
respect to t of [xt]2 C [yt]2 C [zt]2 D c2 . Divide the result by 2 to obtain: xtx t C yty t C
ztz t D 0. Now we are ready to calculate the integral.
b
F · ds D
x
xt, yt, zt · x t, y t, z t dt
a
b
D
xtx t C yty t C ztz t dt
a
b
D
0 dt D 0.
a
29. If x is a parametrization of C, then formula (3) of §6.1 gives
§f · ds D
§f · T ds, where T D
C
x
x t/x t. But T is tangent to C and §f is perpendicular to level sets of f (including C), so §f
· T D 0,
and thus the integral must be zero.
20
ds
ds
ds
ds
30. (a) We have
D vs, so dt D
. Hence the total time for the trip is dt D
D2
dt
vs
vs
0 2s C 20
20
(where l’ve used symmetry) D ln2s C 200 D ln 60 − ln 20 D ln 3 L 1.0986 hours or 65.92 min.
(b) On a semicircular path you can travel at a maximum constant speed of 60 mph. You must do so for 20π
20π
miles, so the trip will take
D π/3 L 1.047 hrs or 62.83 min.
60
(c) Traveling through the center of Cleveland (as in part (a)) will take
20
2
0
π/4
20
ds
16 ds
20 sec2 θ dθ
D 32
D2
2
2
s2 /16 C 25
202 sec2 θ
0 s C 20
0
π/4
2π
8 π
32
D
L 1.2566 hrs or 75.40 min.
dθ D ·
D
20 0
5
4
5
20π
2π
D
—same time!
50
5
31. (a) Newton’s second law gives ma D qE C v * B. Take the dot product with v: ma · v D qE · v C v *
B · v D qE · v since v * B ⬜ v.
(b)
b
b
Work D E · ds D
Ext · x t dt D
Ext · vt dt
Going around Cleveland will take
x
b
D
a
a
mat · vt dt by part (a).
a
If the path has constant speed, then vt is constant. Hence v · v is constant so that
0 3 2a · v D 0 3 a · v D 0. Therefore the integrand of the work integral is zero.
d
v · v D
dt
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“runall” — 2005/8/1 — 15:23 — page 294 — #6
294
Chapter 6
Line Integrals
32. (a) In this case all xk D x D
1
1
3
5
7
, while y1 D
, y2 D
, y3 D
, y4 D
. Then
4
16
16
16
16
3
T4 D 03 C 2 3
3
1
1
9
3 1/4
C 2 C 2 C 1 16
4
16
2
2
1
C −02 − 4
2
1/16
1
1
C − − 2
4
2
2
2
2
3/16
3
1
C − − 2
2
4
5/16
2
2
7/16
3
2675
C − − 12
D−
L −0.326538.
4
2
8192
(b) With y D x2 we have dy D 2x dx so that
1
1
3
2
2 3
2
y dx − x dy D
x dx − x 2x dx D
x6 − 2x3 dx
0
C
0
1
1
1
D x7 − x4 D −5/14 D −0.357143.
7
2
0
1 1 3
1
3 3 9
3
33. (a) We have x0 D 0, 0, 0, x1 D , , , x2 D , 1, , x3 D , , , x4 D 1, 2, 3. Then all
4 2 4
2
2
4 2 4
1
1
3
xk D , yk D , zk D . Then
4
2
4
T4 D 0 C 2 ·
1/4
1/2
3
3
27
C 2 ·
C 2 ·
C 6
C 0 C 2 · 1 C 2 · 2 C 2 · 3 C 4
8
2
8
2
2
C 0 C 2 ·
3/4
1
27
245
1
C 2 ·
C 2 ·
C 2
D
D 7.65625.
32
4
32
2
32


 xDt
y D 2t,
(b) Parametrize C as

 z D 3t
0 … t … 1. Then
yz dx C x C z dy C x y dz D
1
2
C
6t 2 C 4t · 2 C 2t 3 · 3 dt
0
1
3 4 15
D 7.5.
D 2t C 4t C t D
2
2
0
3
2
34. (a) We have: x1 D 1, x2 D 0, x3 D 1, x4 D x5 D x6 D x7 D 0, x8 D −1;
y1 D 3, y2 D 1, y3 D y4 D 0, y5 D −1, y6 D 0, y7 D −1, y8 D 0;
z1 D 0, z2 D 1, z3 D z4 D 1, z5 D 0, z6 D z7 D −1, z8 D 0. Then
x1
x
x
x
x
C 0 C 1 2 C 1 C 2 3 C 2 C 2 4 C 2 C 2 5
2
2
2
2
2
y
x
x
x
C 2 C 3 6 C 3 C 4 7 C 4 C 4 8 C 0 C 1 1
2
2
2
2
y2
y3
y4
y5
C 1 C 1
C 1 C 1
C 1 C 2
C 2 C 3
2
2
2
2
y6
y7
y8
C 3 C 3
C 3 C 3
C 3 C 3
2
2
2
z
z
z
z
z
C 1 C 2 1 C 2 C 2 2 C 2 C 2 3 C 2 C 2 4 C 2 C 3 5
2
2
2
2
2
T8 D 0 C 0
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“runall” — 2005/8/1 — 15:23 — page 295 — #7
Section 6.2
C 3 C 3
D
Green’s Theorem
295
z
z
z6
C 3 C 3 7 C 3 C 4 8
2
2
2
3
5
11
3
C −4 C
C 1 C − C −3 C 2 C 2 C 2 C −3 C −3 D − .
2
2
2
2
(b) Now
x1 D x2 − x0 D 1
x2 D x4 − x2 D 1
x3 D x6 − x4 D 0
x4 D x8 − x6 D −1
y1 D 4
y2 D 0
y3 D −1
y4 D −1
z1 D 1
z2 D 2
z3 D −1
z4 D −1
Then
x
x
x
x1
C 1 C 2 2 C 2 C 3 3 C 3 C 4 4
2
2
2
2
y2
y3
y
y1
C 1 C 2
C 2 C 3
C 3 C 3 4
C 0 C 1
2
2
2
2
z
z
z
z
C 1 C 2 1 C 2 C 2 2 C 2 C 3 3 C 3 C 4 4
2
2
2
2
T4 D 0 C 1
3
7
5
3
5
7
1
C
C − C 2 C − C −3 C
C 4 C − C − 2
2
2
2
2
2
2
11
D− .
2
D
6.2 GREEN’S THEOREM
1. Mx, y D −x2 y and Nx, y D xy2 .
• For the line integral the path is xt D 2 cos t, 2 sin t, 0 … t … 2π.
2π
M dx C N dy D
−8 cos2 t sin t, 8 cos t sin2 t · −2 sin t, 2 cos t dt
0
C⭸D
2π
D 32
sin2 t cos2 t dt
0
2π
D 4t − sin 4t D 8π.
0
• For the area calculation, we use polar coordinates:
6D
Nx − My dA D
D
y2 C x2 dA D
6D
2π
2π 2
r3 dr dθ
0
0
4 dθ D 8π.
0
2. Mx, y D x2 − y and Nx, y D x C y2 .
• For the line integral, the path is split into four pieces, in each case 0 … t … 1 : x1 t D 2t, 0, x2 t D 2, t,
x3 t D 2 − 2t, 1, and x4 t D 0, 1 − t.
1
M dx C N dy D
[24t 2 C 2 C t 2 − 24t 2 − 8t C 3 − t 2 − 2t C 1] dt
0
C⭸D
1
[18t − 5] dt D 4.
D
0
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“runall” — 2005/8/1 — 15:23 — page 296 — #8
296
Chapter 6
Line Integrals
• The area calculation is straightforward:
6D
Nx − My dA D
6D
2 dA D
1 2
0
2 dx dy D 4.
0
3. Mx, y D y and Nx, y D x2 .
• For the line integral, the path is again split into four pieces, in each case 0 … t … 1 : x1 t D 1 − 2t, 1,
x2 t D −1, 1 − 2t, x3 t D −1 C 2t, −1, and x4 t D 1, −1 C 2t.
1
M dx C N dy D
[−21 C −21 C 2−1 C 21] dt
0
C⭸D
1
−4 dt D −4.
D
0
• The area calculation is again straightforward:
6D
Nx − My dA D
6D
2x − 1 dA D
1 1
−1 −1
1
2x − 1 dx dy D
−1
−2 dy D −4.
4. Mx, y D 2y and Nx, y D x.
• For the line integral, the path is split into two pieces: x1 t D a cos t, a sin t, 0 … t … π, and x2 t D
−a C 2at, 0, 0 … t … 1.
π
1
M dx C N dy D
2a sin t, a cos t · −a sin t, a cos t dt C
a0 dt
0
0
C⭸D
π
π
πa2
2
2
2
2
−2 sin t C cos t dt D a
−2 C 3 cos2 t dt D −
.
Da
2
0
0
• We’ll use polar coordinates for the area calculation:
6D
Nx − My dA D
6D
π a
1 − 2 dA D
π
−r dr dθ D
0
0
−
0
πa2
a2
dθ D −
.
2
2
5. (a) By Green’s theorem:
CC
y2 dx C x2 dy D
6D
2x − 2y dA D
1 1
0
0
2x − 2y dx dy D
1
1 − 2y dy D 0.
0
(b) Our path is made up of four pieces with 0 … t … 1 on each: x1 t D t, 0, x2 t D 1, t, x3 t D 1 − t, 1,
and x4 t D 0, 1 − t. So
1
y2 dx C x2 dy D
[10 C 11 − 11 − 10] dt D 0.
0
CC
6. Mx, y D 3xy and Nx, y C 2x2 .
• For the line integral, the path is split into four pieces: x1 t D 0, −2t, x2 t D 2t, −2, and x3 t D
2, −2 C 2t, with 0 … t … 1, and x4 t D cos t C 1, sin t with 0 … t … π. So
1
π
F · ds D
[−20 C 2−12t C 28] dt C
[−3 sin2 tcos t C 1 C 2 cos tcos t C 12 ] dt
0
0
CC
1
π
D
[−24t C 16] dt C
[2 cos t C 4 cos2 t C 2 cos3 t − 3 sin2 t − 3 cos t sin2 t] dt
0
0
D 4 C π/2.
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“runall” — 2005/8/1 — 15:23 — page 297 — #9
Section 6.2
Green’s Theorem
297
• If D is the region bounded by C, then
CC
F · ds D
4x − 3x dA D
x dA
6D
6D
π 1
0 2
x dx dy C
rr cos θ C 1 dr dθ
D
−2 0
D
0
−2
π
2 dy C
0
0
0
1
1
cos θ C dθ D 4 C π/2.
3
2
7. Note that the curve is oriented clockwise so the square lies on the right side of the curve.
CC
x − y dx C x C y dy D −
2
2
2
1 1
2
0
8. As we saw in Section 6.1, Work D
Green’s theorem
CC
CC
2x C 2y dy dx D −
0
1
2x C 1 dx D −2.
0
F · ds. If D is the ellipse x2 C 4y2 D 4 and its boundary is C, then by
4y − 3x, x − 4y D
D
6D
2
−2
1 − 4 dA D
2 √ 1−x2 /4
−2 −
√
1−x2 /4
−3 dy dx
−6 1 − x2 /4 dx D −6π.
For Exercises 9, 10, 12, 13, 14 we will calculate the area of a region using formula (1):
area of D D
1
− y dx C x dy.
2 C⭸D
1
− y dx C x dy. We can work this out, as in the case of Exercises 2 and 3, by
2 C⭸R
enumerating the paths along the four sides and calculating the integral. We can, however, eliminate a lot of work
by first noting that dy D 0 along both horizontal parts of the path and that x D 0 along the left vertical portion
of the path. Also, dx D 0 along both vertical parts of the path and y D 0 along the bottom portion. So
9. By formula (1), the area is
a
b
1
1
− y dx C x dy D b dx C
a dy D ab.
2 C⭸R
2 0
0
10. One arch of the cycloid is produced from t D 0 to t D 2π.
y
2
1
pa
2pa
x
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“runall” — 2005/8/1 — 15:23 — page 298 — #10
298
Chapter 6
Line Integrals
Because of the orientation shown,
1
1
y dx − x dy D
Area D
2 CC
2
1
2
1
D
2
C1
2π
a2
D
2
a2
D
2
2π
a2
1 − cos t − t sin t C sin t dt D
2
2
2π
0
1
C
2
C1
.
C2
a1 − cos t · a1 − cos t − at − sin t · a sin t dt
0
0
2π
2
1 − 2 cos t C cos2 t − t sin t C sin2 t dt
0
2π
a2
a2
2 − 2 cos t − t sin t dt D 2t − 2 sin t C t cos t − sin t D 4π C 2π D 3πa2 .
2
2
0
11. By Green’s theorem:
CC
x4 y5 − 2y dx C 3x C x5 y4 dy D −
3 C 5x4 y4 − 5x4 y4 − 2 dA
6D
5 dA D −5 · area of D D −52 C 3 C 4 D −45.
D−
6D
(Note the minus sign because of the orientation of the curve.)
12. A sketch of a hypocycloid with a D 1 is:
y
1
0.5
-1
-0.5
0.5
1
x
-0.5
-1
Let D be the interior of the hypocycloid and let C be the bounding curve traced by the path
xt D a cos3 t, a sin3 t. Then by Green’s theorem,
6D
2π
1
− y dx C x dy D
[a sin3 t3a cos3 t sin t C a cos3 t3a sin2 t cos t] dt
2 C⭸D
0
2π
3πa2
3a2
.
cos2 t sin4 t C cos4 t sin2 t dt D
D
2 0
8
dy dx D
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“runall” — 2005/8/1 — 15:23 — page 299 — #11
Section 6.2
Green’s Theorem
299
13. (a) Shown below are three views of the curve xt D 1 − t 2 , t 3 − t.
y
1.5
1
0.5
-1.5
-1
-0.5
0.5
1
y
x
0.4
0.2
-0.5
0.2
0.4
0.6
0.8
1
x
-1
-0.2
-1.5
-0.4
y
0.3
0.2
0.1
0.2
0.4
0.6
0.8
1
x
The figure on the top left is for −3 … t … 3, the top right figure is for −1 … t … 1, and the figure on the
bottom is for −1 … t … 0. The first gives a feel for the curve, the second isolates the closed portion of the
curve and the third gives us the orientation: that as t increases from −1 to 1 the path moves clockwise.
(b) We again must make an adjustment because the path moves clockwise. The area is
1
1
− y dx C x dy D −
[t 3 − t2t C 1 − t 2 3t 2 − 1] dt
2 C⭸D
−1
1
8
t 4 − 2t 2 C 1 dt D
D
.
15
−1
14. In this exercise, we are finding the area of the region D that is outside the ellipse and inside the circle.
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“runall” — 2005/8/1 — 15:23 — page 300 — #12
300
Chapter 6
Line Integrals
y
4
2
-4
-2
4
2
x
-2
-4
We need to orient the boundary curve C so that the path travels counterclockwise around the circle and clockwise
around the ellipse. We split this path into two pieces, each with 0 … t … 2π, x1 t D 5 cos t, 5 sin t and
x2 t D 3 cos t, −2 sin t. By Green’s theorem,
6D
1
− y dx C x dy
2 C⭸D
2π
1
[−5 sin t−5 sin t C 5 cos t5 cos t] dt
D
2 0
2π
1
[2 sin t−3 sin t C 3 cos t−2 cos t] dt
C
2 0
2π
2π
1
1
25 dt C
−6 dt D 19π.
D
2 0
2 0
dA D
15. It is easier if we work from the line integral to the double integral. By Green’s theorem,
C⭸D
x dy D
6D
⭸x
dA D Area of D.
dA D
⭸x
6D
Similarly, also by Green’s theorem,
⭸y
−
y dx D −
−
⭸y
C⭸D
6D
dA D
6D
dA D Area of D.
Note: Assign Exercises 15 and 16 together to point out that the students cannot mix the two line integrals given in
Exercise 15. The quadrilateral in Exercise 16 has one vertical side and one horizontal side so there is a temptation to use
the integral with a dx in it along the vertical side and the integral with a dy in it along the horizontal side so that they
disappear. You must choose one or the other for the entire problem.
16. We’ll use the results of Exercise 15. If we use the formula
x dy, then for the side connecting (1, 1) to −1, 1,
C⭸D
since there is no change in y, this integral is 0. Therefore,
1
1
3
Area of D D x dy D
2 − t2 C 1−1 C 0 C −1 C 3t−1 dt D
4 − 5t dt D .
2
0
0
CC
17. (a) By the divergence theorem:
CC
2yi − 3xj · n ds D
6D
[2yx C −3xy ] dA D
6D
0 dA D 0.
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“runall” — 2005/8/1 — 15:23 — page 301 — #13
Section 6.2
Green’s Theorem
301
(b) For direct computation, n D cos θ, sin θ and x D cos θ and y D sin θ. Therefore,
1
F · n D 2y, −3x · cos θ, sin θ D 2 cos θ sin θ − 3 cos θ sin θ D − cos θ sin θ D − sin 2θ.
2
Then
1
F · n ds D −
2
CC
2π
sin 2θ dθ D 0.
0
18. Similar to what was done in the proof of the divergence theorem, we will calculate the line integral
along a C1 segment of ⭸D. Recall that Tt D x t/x t.
b
Fxt · Tt x t dt D
a
b
D
F · T ds
Fxt · x t dt
a
b
C⭸D
Mxt, yt x t C Nxt, yt y t dt D
M dx C N dy.
x
a
We extend this result to the entire curve and apply Green’s theorem.
C⭸D
F · T ds D
C⭸D
M dx C N dy D
6D
Nx − My dA.
19. By Green’s Theorem, if D is the region bounded by C,
CC
3x2 y dx C x3 dy D
6D
3x2 − 3x2 dA D 0.
(Note that in this case the orientation of C is not important as −3x2 − 3x2 D 3x2 − 3x2 .)
20. If C is oriented as required and D is the region bounded by C, then by Green’s Theorem,
CC
− y3 dx C x3 C 2x C y dy D
6D
3x2 C 2 C 3y2 dA > 0.
21. Let δ D 1 if C is oriented counterclockwise and δ D −1 if C is oriented clockwise. Let D be the region bounded
by C. Then by Green’s Theorem,
CC
x2 y3 − 3y dx C x3 y2 dy D δ
6D
3x2 y2 − 3x2 y2 C 3 dA D 3δ (the area of the rectangle).
22.
b
Flux D
D
r · n ds D
CC
b
a
D
D
x
6D
6D
y i − x j 2
xi C yj · x C y2 dt
a
x2 C y2
dy
dx
− y dx C x dy
− y dt D
dt
dt
6C
1 − −1 dA
by Green’s theorem
2 dA D 2 · area inside C.
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“runall” — 2005/8/1 — 15:23 — page 302 — #14
302
Chapter 6
Line Integrals
23. We have u §v D u
CC
⭸v ⭸v
,u
⭸x ⭸y
so that
⭸v
⭸v
u dx C u dy
⭸y
CC ⭸x


⭸
⭸
v
⭸
⭸
v

 dA

D
u
−
u 
⭸y
⭸x
6D ⭸x ⭸y
u §v · ds D
by Green’s theorem
D
⭸u ⭸v
⭸2 v
⭸u ⭸v
⭸2 v
C u
−
− u
⭸x ⭸y
⭸y ⭸x
⭸y ⭸x
6D ⭸x ⭸y
D
⭸u ⭸v
⭸u ⭸v
−
⭸y ⭸x
6D ⭸x ⭸y
D
⭸u, v
6D ⭸x, y
dA
dA since v is of class C2
dA.
24. Let D be the region bounded by C. By Green’s theorem,
⭸f
⭸f
⭸2 f
⭸2 f
C
dx −
dx D
⭸x
⭸x2
CC ⭸y
6D ⭸y2
25. First,
⭸f
⭸f
⭸f
ds D
i C
j
§f · n ds D
⭸y
C⭸D ⭸n
C⭸D
C⭸D ⭸x
dA D
6D
0 dA D 0.
· n ds. You can continue the calculation or note that
this is the same computation done in the proof of the divergence theorem with F D Mi C Nj D
Therefore, applying Green’s theorem,
⭸f
⭸f
⭸f
⭸2 f
⭸2 f
ds D
dx C
dy D
−
C
2
⭸x
⭸y2
C⭸D ⭸n
C⭸D ⭸y
6D ⭸x
dA D
6D
⭸f
⭸f
i C
j.
⭸x
⭸y
§2 f dA.
6.3 CONSERVATIVE VECTOR FIELDS
1. (a) Let C be the path parametrized by xt D t, t, t with 0 … t … 1. Then
1
1
5
z2 dx C 2y dy C xz dz D
t 2 C 2t C t 2 dt D 2
t 2 C t dt D .
3
0
0
C
(b) Let xt D t, t 2 , t 3 with 0 … t … 1. Then
1
1
11
2
6
2
4
2
z dx C 2y dy C xz dz D
t C 2t 2t C t 3t dt D 4
t 6 C t 3 dt D
.
7
0
0
C
(c) Parts (a) and (b) show that line integrals are not path-independent. By Theorem 3.3, therefore, F is not
conservative.
2. (a) Let C be the path parameterized by xt D t 2 , t 3 , t 5 with 0 … t … 1. Then
1
1
F · ds D
2t 5 , t 4 C t 10 , 2t 8 · 2t, 3t 2 , 5t 4 dt D
7t 6 C 13t 12 dt D 2.
C
0
0
(b) Let C be comprised of the two paths: x1 t D t, 0, 0 and x2 t D 1, t, t each with 0 … t … 1. The
integral along x1 is easily seen to be zero (y, z, and dx are all identically zero along x1 ). We have that
1
F · ds D
2t, 1 C t 2 , 2t · 0, 1, 1 dt D 2.
C
0
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“runall” — 2005/8/1 — 15:23 — page 303 — #15
Section 6.3
Conservative Vector Fields
303
(c) Obviously the fact that our answers to part (a) and (b) are the same is not enough to convince us that F is
conservative. We can, however, easily see that F D § x2 y C yz2 so F is conservative.
In Exercises 3–7, we will check to see whether F D Mi C Nj is conservative by checking to see whether ⭸N/⭸x D
⭸M/⭸y (formula (1)).
⭸N
⭸M
3.
D yexy Z exCy D
, so F is not conservative.
⭸x
⭸y
⭸N
⭸M
4.
D 2x cos y D
, so F is conservative. We want to find f where F D §f x, y. We find that the indefinite
⭸x
⭸y
integral of 2x sin y with respect to x is x2 sin y. To see whether any adjustments need to be made, we check to
⭸ 2
x sin y D x2 cos y. It does, so we conclude that f x, y D §x2 sin y.
make certain that
⭸y
1 − x2 y 2
1 − x2 y 2
⭸N
⭸M
5.
D 3x2 sin y C
, so F is not conservative.
Z −3x2 sin y C
D
2
2
2
⭸x
1 C x y 1 C x2 y2 2
⭸y
2xy
x2 y 2
⭸N
⭸M
D
D
, so F is conservative. F D §
.
6.
2
2
⭸x
1 C x ⭸y
21 C x2 ⭸ −y
⭸
− y sinxy D −e−y − sin xy − xy cos xy D
7. Note that
e
−xe−y − x sin xy. Since the domain
⭸y
⭸x
⭸f
D e−y − y sin xy * f x, y D
of F is all of R2 , the vector field is conservative. Thus F D §f , so
⭸x
⭸f
D −xe−y − x sin xy C g y D −xe−y − x sin xy so
xe−y C cos xy C gy for some g. Hence
⭸y
g y D 0. Thus f x, y D xe−y C cos xy C C is a potential for any C.
In Exercises 8–14, we will check to see whether F D Mi C Nj C Pk is conservative by checking whether § * F D 0.
This amounts to checking whether ⭸N/⭸x D ⭸M/⭸y, ⭸P/⭸x D ⭸M/⭸z, and ⭸P/⭸y D ⭸N/⭸z. We also need to check
that the domain of F is simply-connected. This last condition is only an issue in Exercise 12.
8. § * F D 2xexyz C 2x2 yzexyz i − 2zexyz C 2xyz2 exyz k Z 0. Hence F is not conservative.
9. We see that ⭸N/⭸x D 1 D ⭸M/⭸y, ⭸P/⭸x D 0 D ⭸M/⭸z, and ⭸P/⭸y D cos yz − yz sin yz D ⭸N/⭸z. So F is
conservative. F D §x2 C xy C sin yz.
10. Here, ⭸N/⭸x D 0 Z 1 D ⭸M/⭸y, so F is not conservative.
11. We see that ⭸N/⭸x D ex cos y D ⭸M/⭸y, ⭸P/⭸x D 0 D ⭸M/⭸z, and ⭸P/⭸y D 0 D ⭸N/⭸z. So F is conservative.
F D §ex sin y C z3 C 2z.
12. We see that ⭸N/⭸x D 0 D ⭸M/⭸y, ⭸P/⭸x D 0 D ⭸M/⭸z, and ⭸P/⭸y D 2z/y D ⭸N/⭸z. So F is conservative
in each of the simply-connected regions on which it is defined: {x, y, z|y > 0} and {x, y, z|y < 0}. On
each, F D §x3 C z2 ln|y|.
13. We see that ⭸N/⭸x D ze−yz C exyz xyz2 C z D −⭸M/⭸y, so F is not conservative.
14. We see that, for G, ⭸N/⭸x D 2x D ⭸M/⭸y, ⭸P/⭸x D 0 D ⭸M/⭸z, and ⭸P/⭸y D 2y D ⭸N/⭸z. So G D
2xy, x2 C 2yz, y2 is conservative and G D §x2 y C y2 z. We know, therefore, that F is not conservative
because the wording of the problem assured us that exactly one of F and G was conservative. It may be more
satisfying to verify that for F, ⭸M/⭸y D 2xyz3 while ⭸N/⭸x D 4xy. These are different so F is not conservative.
1
⭸N
⭸
D
ye2x C 3x2 ey D e2x C 3x2 ey . Thus Nx, y D e2x C
15. For F to be conservative, we must have
⭸x
⭸y
2
x3 ey C uy where u is any C2 function of y.
16. For F to be conservative, we must have § * F D 0. Thus we impose
i
j
k
⭸
⭸
⭸
0D§ * FD
⭸x
⭸y
⭸z
2
2
2
3x C 3y z sin xz ay cos xz C bz 3xy sin xz C 5y D 6xy sin xz C 5 C axy sin xz − bi
C 3y2 sin xz C 3xy2 z cos xz − 3y2 sin xz − 3xy2 z cos xzj
C −ayz sin xz − 6yz sin xzk.
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“runall” — 2005/8/1 — 15:23 — page 304 — #16
304
Chapter 6
Line Integrals
From this, it is easy to see that only the choices a D −6, b D 5 will work. Moreover, the resulting vector field
is clearly defined on all of R3 (a simply-connected region), so the vanishing of the curl is enough to guarantee
that F is conservative.
17. (a) As above we check that ⭸N/⭸x D 0 D ⭸M/⭸y, ⭸P/⭸x D 0 D ⭸M/⭸z, and ⭸P/⭸y D cos y cos z D ⭸N/⭸z.
So F is conservative. F D §x3 /3 C sin y sin z.
(b) By Theorem 3.3,
F · ds D f x1 − f x0 D f 2, e, e2 − f 1, 1, 1 D 7/3 C sin e sine2 − sin2 1.
x
18. Since 7y − 5xx D −5 D 3x − 5yy , F D 3x − 5yi C 7y − 5xj is conservative and the integral is
path independent. We’ll integrate along the path xt D 4t C 1, −t C 3 for 0 … t … 1.
3x − 5y dx C 7y − 5x dy D
1
[434t C 1 − 5−t C 3 − 7−t C 3 − 54t C 1] dt
0
C
D
1
0
95t − 64 dt D −
33
.
2
Using Theorem 3.3,
3x − 5y dx C 7y − 5x dy D f 5, 2 − f 1, 3,
where F D §f .
C
In this case, f x, y D 3x2 /2 − 5xy C 7y2 /2. Therefore,
7
3
7
33
3
3x − 5y dx C 7y − 5x dy D 25 − 552 C 4 − 1 − 513 C 9 D − .
2
2
2
2
2
C
19. Here




xy
y
⭸ 
⭸
x



D−
.


D
⭸x
x2 C y2 3/2
⭸y
x2 C y 2
x2 C y 2
So F is conservative so long as we restrict the domain. Our domain must be simply-connected and must contain
the upper half of the circle of radius 2 centered at the origin. Our domain also must not contain the origin as F is
not defined at the origin. We can choose, for example, the upper half disk of radius 3 centered at the origin minus
the upper half disk of radius one centered at the origin. This “semi-annular” region meets all of our conditions.
Therefore, the given integral is path independent. We’ll integrate along the path xt D 2 cos t, 2 sin t, 0 … t … π.
The integral
π
π
π
x dy C y dx
sin 2t 4 cos2 t − 4 sin2 t
dt
D
D
cos
2t
dt
D
D 0.
2
2
2 0
C
0 4 cos t C 4 sin t
0
x2 C y 2
Using Theorem 3.3, and the fact that F D §f where f x, y D x2 C y2 ,
x dy C y dx
D f −2, 0 − f 2, 0 D −22 C 0 − 22 C 0 D 0.
C
x2 C y 2
20. This time we check that three pairs of partial derivatives are equal:
⭸
⭸
2x C z D 2 D
2y − 3z
⭸x
⭸y
⭸
⭸
y − 3x D −3 D
2y − 3z
⭸x
⭸z
⭸
⭸
y − 3z D 1 D 2x C z.
⭸y
⭸z
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“runall” — 2005/8/1 — 15:23 — page 305 — #17
Section 6.3
Conservative Vector Fields
305
We conclude that F is conservative, because the domain of F is all of R3 . The given integral, therefore, is path
independent. We’ll integrate along the paths x1 t D 0, t, t, 0 … t … 1, and x2 t D t, t C 1, 2t C 1,
0 … t … 1. The integral
2y − 3z dx C 2x C z dy C y − 3x dz
C
D
1
0−t C 1t C 1t dt C
0
D
1
2t dt C
0
1
1
−4t − 1 C 4t C 1 C 2−2t C 1 dt
0
−4t C 2 dt D 1 C 0 D 0.
0
Using Theorem 3.3, and the fact that F D §f where f x, y, z D 2xy − 3xz C yz, we obtain
2y − 3z dx C 2x C z dy C y − 3x dz D f 1, 2, 3 − f 0, 0, 0 D 1.
C
21. (a) We’ll check to see where Nx D My .
⭸ x2 C 1
⭸ x C xy2
2x
D
D
,
⭸x
y3
y3
⭸y
y2
therefore, F is conservative on each of the two simply-connected sets on which it is defined. More precisely,
F is conservative on {x, y|y > 0} and on {x, y|y < 0}.
x2 C x2 y 2 C 1
.
(b) The scalar potential is f x, y D
2y2
(c) As the particle moves from (0, 1) to (1, 1) along the parabola y D 1 C x − x2 we note that y > 0 and
so the path lies entirely in one of the simply-connected regions. We can, therefore, apply Theorem 3.3 and
calculate the work done as f 1, 1 − f 0, 1 D 3/2 − 1/2 D 1.
22. (a) We need to check that three pairs of partial derivatives are equal:
⭸
⭸
x C gy C z D 1 D
f x C y C z
⭸x
⭸y
⭸
⭸
x C y C hz D 1 D f x C y C z
⭸x
⭸z
⭸
⭸
x C y C hz D 1 D x C gy C z.
⭸y
⭸z
(b) F D §φx, y, z where, for constants a, b, and c,
x
φx, y, z D xy C xz C yz C
y
f t dt C
a
z
gt dt C
b
ht dt.
c
(c) Using Theorem 3.3,
F · ds D φx1 , y1 , z1 − φx0 , y0 , z0 C
D x1 y1 − x0 y0 C x1 z 1 − x0 z 0 C y1 z 1 − y0 z 0 C
x1
x0
f t dt C
y1
gt dt C
y0
z1
ht dt.
z0
23. (a) F is conservative since F D §f where f x, y, z D sinx2 C xz C cosy C yz.
(b) Since we have a potential function,
F · ds D §f · ds D f x1 − f x0
x
x
D f 1, 1, π − 1 − f 0, 0, 0 D −2.
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“runall” — 2005/8/1 — 15:23 — page 306 — #18
306
Chapter 6
Line Integrals
24. (a) G D F C xj, where F is given in Exercise 23. Now § * G D § * F C § * xj D k Z 0, so G is not
conservative.
(b) Here we have that
G · ds D F C xj · ds D F · ds C
xj · ds.
x
x
x
x
F · ds D −2, so
From Exercise 23, we have that
x
G · ds D −2 C
x
1
0
D −2 C
0, t 3 , 0 · 3t 2 , 2t, π −
1
0
π
πt
cos dt
2
2
8
2t 4 dt D −2 C 2/5 D − .
5
25. You could check that F is conservative by confirming that § * F D 0 on any simply-connected region that misses
the origin. It is, however, easy enough to find the scalar potential for F is f x, y, z D GMmx2 C y2 C z2 −1/2 .
So the work done by F as a particle of mass m moves from x0 to x1 is
f x1 , y1 , z1 − f x0 , y0 , z0 D GMm
x12 C y12 C z21
− GMm
x02 C y02 C z20
D GMm
1
1
−
.
x1 x0 6.4 TRUE/FALSE EXERCISES FOR CHAPTER 6
1. True.
2. False. (The value is 2.)
3. False. (The integral is negative.)
4. True.
5. False. (The integral is 0.)
6. True.
7. False. (There is equality only up to sign.)
8. False.
9. True.
10. True.
11. True.
12. False. (§f is everywhere normal to C.)
13. True.
14. False. (The work is at most 3 times the length of C.)
15. False. (The line integral could be ; C F ds, depending on whether F points in the same or the opposite direction
as C.)
16. True. (Just use Green’s theorem.)
17. False. (Let F D yi − xj and consider Green’s theorem.)
18. True. (Use the divergence theorem in the plane.)
19. False. (Under appropriate conditions, the integral is f B − f A.)
20. True.
21. True.
22. False. (There’s a negative sign missing.)
23. False. (For the vector field to be conservative, the line integral must be zero for all closed curves, not just a
particular one.)
24. True.
25. False. (The vector field ex cos y sin z, ex sin y sin z, ex cos y cos z is not conservative.)
26. False. (F must be of class C1 on a simply-connected region.)
27. False. (The domain is not simply-connected.)
28. True.
29. False. (f is only defined up to a constant.)
30. True.
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“runall” — 2005/8/1 — 15:23 — page 307 — #19
Section 6.5
Miscellaneous Exercises for Chapter 6
307
6.5 MISCELLANEOUS EXERCISES FOR CHAPTER 6
1. Partition the curve into n pieces each of length sk D length of C/n. The right side of the given formula is
just our calculation of arclength for a rectifiable curve:
f ds
C f ds
D C
.
length of C
C ds
Now, if on the kth sub-interval we choose any ck , then on the interval f x L f ck . Therefore,
n
limn'q nkD1 f ck sk
f ck 1 n
C f ds
D
D
.
f ck D lim
lim
n'q n
n'q
n
length
of
C
length
of C
kD1
kD1
For each value of n we are calculating an average of n values of f at points on the curve. As n grows large, if
this limit exists, it is reasonable to define it as the average value of f along C.
√
2. Here f xt D 2 C 2t 2 , and x t D 2. Therefore,
√
3π
3π
2t 2 C 2 2 dt
f ds
2
2
D 0
9π3 C 3π D 6π2 C 2.
t 2 C 1 dt D
D
[f ]avg D C
3π √
ds
3π
3π
2 dt
0
C
0
3. We may parametrize the semicircle as xt D a cos t, a sin t, where 0 … t … π. Therefore, x t D a. The
length of the semicircle is πa and so the average y-coordinate may be found by calculating
π
π
a
2a
1
a sin t · a dt D
sin t dt D
.
πa 0
π 0
π
z ds
C
. The total length of C is just the sum of the lengths of four straight segments:
4. Calculate [z]avg as
length of C
√
√
√
√
2 C 1 C 4 C 0 C 1 C 1 C 1 C 1 D 3 C 5 C 3.
Now C z ds D C1 z ds C · · · C C4 z ds, but z is clearly zero on two of the four segments.
The segment C3 joining (2, 1, 0) and (0, 1, 1) may be parametrized as
xt D 1 − t2, 1, 0 C t0, 1, 1,
0 … t … 1
D 2 − 2t, 1, t.
√
Thus x t D −2, 0, 1 and x t D 5. Therefore,
√
1
√
5
z ds D
t · 5 dt D
.
2
0
C3
The segment C4 joining (0, 1, 1) and (1, 0, 2) may be parametrized as
xt D 1 − t0, 1, 1 C t1, 0, 2,
0 … t … 1
D t, 1 − t, t C 1.
√
Thus x t D 1, −1, 1 and x t D 3. Hence
z ds D
C4
1
√
√
√ t2
3 3
C t D
.
t C 1 3 dt D 3
2
2
0
0
1
Putting all this together, we find
√
[z]avg D
√
√
√
5/2 C 3 3/2
5 C 3 3
√
√ D
√
√
L 0.5333.
3 C 5 C 3
23 C 5 C 3
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“runall” — 2005/8/1 — 15:23 — page 308 — #20
308
Chapter 6
Line Integrals
√
5. The curve may be parametrized as xt D 5 cos t, sin t, 2 sin t, 0 … t … 2π. Then x t
√
√
√
2π √
D − 5 sin t2 C cos t2 C 2 cos t2 D 5 so the length of C is 0
5 dt D 2π 5. Now
f ds D
2π
0
C
D
√
4 sin2 t C
2π
5
√
√
5 cos t · esin t 5 dt
21 − cos 2t C
0
D
√
52t − sin 2t C
√
sin t
5e
√
5 esin t · cos t dt
2π
√
√
√
D 54π C 5 − 0 C 0 − 5
0
√
D 4π 5.
√
4 5π
Hence [f ]avg D √
D 2.
2 5π
6. (a) For the total mass we integrate the density along the curve:
π
3 − y ds D
23 − 2 sin t dt D 6π − 8.
0
C
(b) The density depends only on y and the wire is symmetric with respect to x so x D 0 (if you write out the
formula you’ll see that the numerator is an integral of an odd function of x over a curve that is symmetric
with respect to x). Also, since z K 0, we quickly conclude that z D 0. What remains is to calculate
π
2 sin t3 − 2 sin t2 dt
24 − 4π
12 − 2π
C yδx, y, z ds
yD D 0
D
D
.
δx,
y,
z
ds
6π
−
8
6π
−
8
3π − 4
C
7. Locate the wire in the first quadrant of the xy-plane. Then the center is at √a , √a and δx, y, z D x−
2
2
a 2
a 2
√ C y − √ . From symmetry considerations, we must have x D y. Now parametrize the quarter circle as
2
2
x D a cos t
y D a sin t,
0 … t … π/2.
Then x t D a. We have
π/2
δ ds D
MD
0
C
2
a
a
a cos t − √ C a sin t − √ 2
2
π/2
Da
3
cos t −
2
0
π/2
D a3
2 −
0
√
π/2
√
2 cos t C
2 cos t −
√
2
a dt
√
1
1
C sin2 t − 2 sin t C dt
2
2
√
2 sin t dt D π − 2 2a3
2
2
a
a
a cos t a cos t − √ C a sin t − √ · a dt
2
2
C
0
√
π/2
π/2
√
√
√
2
1 C cos 2t − 2 sin t cos t
D a4
cos t2 − 2 cos t − 2 sin t dt D a4
2 cos t −
2
0
0
π/2
√
√
√
2
2
2 2 D a4 2 sin t −
t −
sin 2t −
sin t 2
4
2
0
Myz D
xδ ds D
dt
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“runall” — 2005/8/1 — 15:23 — page 309 — #21
Section 6.5
Miscellaneous Exercises for Chapter 6
309
√
2π
2
− 0 −
− 0
Da 2 −
4
2
√
√
8 − 2π − 2 2 4
D
a .
4
√
4
Hence
xDyD
8 −
√
√
√
√
√
2π − 2 2a4
8 − 2π − 2 2
3
√
π
−
2
2a
D
a.
4
4π − 2 2
8. (a) By symmetry
x D y D 0 and z D 8π.
√
(b) x D 9 C 16 D 5; δx, y, z D x2 C y2 C z2 . Hence the mass of the wire is
MD
δ ds D
x
xD
1
M
1
yD
M
zD
1
M
4π
0
xδ ds D
9 C 16t 2 · 5 dt D
x
1
M
1
yδ ds D
M
x
zδ ds D
x
1
M
4π
20π
27 C 256π2 3
3 cos t9 C 16t 2 · 5 dt D
1
· 1920π
M
3 sin t9 C 16t 2 · 5 dt D
1
−3840π2 M
0
4π
0
4π
4t9 C 16t 2 · 5 dt D
0
1
160π2 9 C 128π2 .
M
Thus
xD
3 · 1920π
288
L 0.112781
D
20π27 C 256π2 27 C 256π2
yD
3−3840π2 576π
D−
L −0.708625
2
20π27 C 256π 27 C 256π2
zD
3 · 160π2 9 C 128π2 24π9 C 128π2 L 37.5662.
D
20π27 C 256π2 27 C 256π2
9. (a) Parametrize the wire as xt D 2 cos t, 2 sin t, 0 … t … π. Then x D 2 and
π
2
x δ ds D
4 cos2 t3 − 2 sin t · 2 dt
Iy D
0
C
π
D
π
24 cos2 t − 16 cos2 t sin t dt D
0
D 12π C
121 C cos 2t C 16 cos2 t− sin t dt
0
16
32
36π − 32
16
−1 −
1 D 12π −
D
.
3
3
3
3
(b) The square of the distance between a point on the wire and the z-axis is x2 C y2 . Thus Iz D
y2 δx, y, z ds. Using the given information,
C x
2
C
π
Iz D
4 · 3 − 2 sin t2 dt D 83π − 4 D 24π − 32.
0
The total mass was found in Exercise 6 to be 6π − 8. Hence the radius of gyration is
rz D
24π − 32
D 2.
6π − 8
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“runall” — 2005/8/1 — 15:23 — page 310 — #22
310
Chapter 6
Line Integrals
√
t
C 2 , −2 … t … 2. Then x D 5 and
2
2
√
2
2
2
3
√
1
t
t
t
5
Ix D
y2 δ ds D
5
C
2
C
2
·
dt
D
C 2 · dt
2
2
2
2
2
−2
−2
C
3
√
3
√
√
5 4 u D 20 5
u3 du D
D 5
4
1
1
2
√
√
2
√
t
5
5 1 2
MD
δ ds D
dt D
C 2
t C 2t D 4 5
2
2
4
−2 2
C
−2
10. Parametrize the curve as xt D t,
Hence
√
√
20 5
√ D 5.
4 5
√
11. We use x as parameter so xt D t, t 2 , 0 … t … 2, and x D 1 C 4t 2 . Then
2
2 √
√
Ix D
y2 δ ds D
t 4 · t 1 C 4t 2 dt D
t 5 1 C 4t 2 dt.
rx D Ix /M D
0
C
0
1
sec2 θ dθ. Then
2
tan−1 4
1
1
tan5 θ sec θ sec2 θ dθ
Ix D
32
2
0
tan−1 4
1
D
tan4 θ sec2 θsec θ tan θ dθ
64 0
tan−1 4
1
D
sec2 θ − 12 sec2 θsec θ tan θ dθ
64 0
tan−1 4
1
D
sec6 θ − 2 sec4 θ C sec2 θ dsec θ
64 0
tan−1 4
1 1
2
1
7
5
3
D
sec θ − sec θ C sec θ 64 7
5
3
0
Now let 2t D tan θ so dt D
2
1
1 1 7/2
2 5/2
1 3/2
1
17
17
C
− −
C
−
17
64 7
5
3
7
5
3
√
7769 17 − 1
D
L 38.1326.
840
D
We also have
δ ds D
MD
C
D
2
2 √
1 2
t 1 C 4t 2 dt D · 1 C 4t 2 3/2 8 3
0
0
1
173/2 − 1 L 5.75773.
12
Hence
rx D
Ix
D
M
√
12
7769 17 − 1
√
·
D
840
17 17 − 1
√
7769 17 − 1
√
1190 17 − 70
L 2.57349.
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“runall” — 2005/8/1 — 15:23 — page 311 — #23
Section 6.5
Miscellaneous Exercises for Chapter 6
311
12. (a) Ix D C y2 C z2 δx, y, z ds, Iy D C x2 C z2 δx, y, z ds, Iz D C x2 C y2 δx, y, z ds
√
(b) For the given parametrization, x D 9 C 16 D 5.
4π
4π
9
2
9 sin2 t C 16t 2 dt D 5δ
Ix D 5δ
1 − cos 2t C 16t dt
2
0
0
10π27 C 512π2 δ
3
4π
4π
9
2
Iy D 5δ
9 cos2 t C 16t 2 dt D 5δ
1 C cos 2t C 16t dt
2
0
0
D 5δ 18π C
1024π3
3
D
10π27 C 512π2 δ
3
4π
9 dt D 180πδ
Iz D 5δ
D
0
δ ds D
Now M D
C
4π
5δ dt D 20δ. Thus
0
π27 C 512π2 ,
6
rx D ry D
rz D
√
180πδ
D 3 π.
20δ
13. We may parametrize the segment
as xt D√1 − t−1, 1, 2 C t2, 2, 3, 0 … t … 1, or xt D 3t − 1, t C
√
1, t C 2. Then x D 9 C 1 C 1 D 11.
1
√
[3t − 12 C t C 12 ][1 C t C 22 ] · 11 dt
Iz D x2 C y2 δ ds D
C
D
D
√
√
1
11
0
10t 4 C 36t 3 C 36t 2 − 12t C 10 dt D
0
√
1
112t 5 C 9t 4 C 12t 3 − 6t 2 C 10t
0
√
112 C 9 C 12 − 6 C 10 D 27 11
1
1
√
√
1
MD
δ ds D
1 C t C 22 11 dt D 11 t C t C 23 3
C
0
0
√
√
22 11
8
.
D 11 1 C 9 − D
3
3
√
9 22
81
Hence rz D Iz /M D
D
.
22
22
Exercises 14, 15, and 23 explore polar versions of results we’ve seen in this Chapter.
14. (a) The path is
xθ D f θ cos θ, f θ sin θ.
Using the product rule, we find that
x θ D f θ cos θ − f θ sin θ, f θ sin θ C f θ cos θ.
The length of x θ is a straightforward calculation:
x t D f θ cos θ − f θ sin θ2 C f θ sin θ C f θ cos θ2 D f θ2 C f θ2 .
We conclude that the arclength of the curve between f a, a and f b, b is
b
ds D
x θ dθ D
f θ2 C f θ2 dθ.
C
xθ
a
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“runall” — 2005/8/1 — 15:23 — page 312 — #24
312
Chapter 6
Line Integrals
(b) The sketch of r D sin2 θ/2 is
y
0.6
0.4
0.2
-1
-0.8
-0.6
-0.4
x
-0.2
-0.2
-0.4
-0.6
The length is
2π !
sin θ/2 C sin θ/2 cos2 θ/2 dθ D
4
2π
2
0
15. (a)
sinθ/2 dθ D −2−1 − 1 D 4.
0
gxθx θ dθ D
gx, y ds D
xθ
CC
(b) We’ll use the formula from part (a).
g ds D
2π
b
gf θ cos θ, f θ sin θ f θ2 C f θ2 dθ.
a
[e3θ cos θ2 C e3θ sin θ2 − 2e3θ cos θ] e6θ C 9e6θ dθ
0
C
D
2π
√
[e6θ − 2e3θ cos θ] 10e3θ dθ
0
√
D
10
37e18π − 108e12π C 71.
333
In this text κ is always non-negative.
In cases where the curvature is signed, differential geometers are often interested
κ2 ds.
in the total squared curvature:
C
16. In Section 3.2 it was shown that
κD
v * a
.
v3
Here v D x and a D x . So
KD
b
κ ds D
C
a
v * a x dt D
v3
b
a
v * a
v2
dt.
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“runall” — 2005/8/1 — 15:23 — page 313 — #25
Section 6.5
Miscellaneous Exercises for Chapter 6
313
17. We use the results of Exercise 16:
10π
−3 sin t, 3 cos t, 4 * −3 cos t, −3 sin t, 0
dt
−3 sin t, 3 cos t, 42
10π
10π
−12 sin t, −12 cos t, 9
3
dt D
dt D 6π.
D
25
5
0
0
KD
0
18. We use the results of Exercise 16 with xt D t, At 2 , 0:
b
b
1, 2At, 0 * 0, 2A, 0
0, 0, 2A
dt D
dt
2
2 2
1, 2At, 0
a
a 1 C 4A t
b
b
2A
−1
D
dt D tan 2At D tan−1 2Ab − tan−1 2Aa.
2
2
a 1 C 4A t
a
KD
19. We parameterize the ellipse by the path xt D a cos t, b sin t, 0 for 0 … t … 2π. Then, using Exercise 16,
KD
D
2π
0
−a sin t, b cos t, 0 * −a cos t, −b sin t, 0
dt D
−a sin t, b cos t, 02
0
ab
dt D 2π.
a2 sin2 t C b2 cos2 t
2π
2π
0, 0, ab
a2 sin2 t C b2 cos2 t
0
dt
This verifies Fenchel’s Theorem for the given ellipse. This final integral was calculated using Mathematica. With
work it can also be done by hand.
20. (a) By Fenchel’s theorem (see Exercise 19), we know that for C (a simple, closed C1 curve in R3 ), K Ú 2π, so
KD
κ ds Ú 2π.
C
But 0 … κ … 1/a, therefore
KD
1
1
ds D
a
C a
κ ds …
C
ds D
C
L
.
a
Putting these two inequalities together we see that
L
Ú K Ú 2π so
a
L Ú 2πa.
(b) To conclude that L D 2πa we would need both of the preliminary inequalities to be equalities. As we saw
in Exercise 19, we have K D 2π when C is also a plane convex curve. Also, as we saw above, K D L/a
when κ D 1/a. Together, these two conditions imply that C is a circle of radius a.
21. The work done is
F · ds D
x
1
sint , cos−t , t · 3t , −2t, 1 dt D
3
2
4
2
0
1
3t 2 sint 3 − 2t cos−t 2 C t 4 dt
0
1
D − cost 3 C sin−t 2 C t 5 /5 D 6/5 − cos 1 − sin 1.
0
22. The first thing to note is that we are traversing the path in the wrong direction to apply Green’s theorem. If C1
is the triangular path described in the problem from the origin, to (0, 1), to (1, 0), back to the origin, then let C2
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“runall” — 2005/8/1 — 15:23 — page 314 — #26
314
Chapter 6
Line Integrals
be the path traversed in the opposite direction and let D be the region bounded by C1 and C2 . Then
CC1
x2 y dx C x C yy dy D − x2 y dx C x C yy dy D −
y − x2 dA
CC2
6D
1−x
1 1−x
1
2
2
2 D−
y − x dy dx D −
y /2 − x y
0
D−
0
1
0
0
x2
1
− x −
C x3
2
2
0
1
x4 x3
1
1
2
C
dx D − x − x −
D− .
2
3
2 0
12
23. In Section 6.2 we saw that Green’s theorem implied the formula
Area D
1
− y dx C x dy.
2 C⭸D
In general the boundary ⭸D of the region D consists of the curve r D f θ, which may be parametrized by
xθ D xθ, yθ D f θ cos θ, f θ sin θ, and possibly straight line segments along θ D a and θ D b.
The line θ D a may be parametrized by yr D xr, yr D r cos a, r sin a and the line θ D b may be
parametrized similarly. Note that, along the straight segment C1 given by θ D a, we have
1
2
−y dx C x dy D
C1
1
2
f a
−r sin a cos a C r cos a sin a dr D 0.
0
An identical result holds for the straight segment C2 given by θ D b. Therefore, the area of D may be evaluated
by computing the line integral over the path x described above:
1
Area D
2
D
D
1
2
1
2
−y dx C x dy
x
b
−f θ sin θf θ cos θ − f θ sin θ
a
C f θ cos θf θ sin θ C f θ cos θ dθ
b
f θ2 dθ.
a
24. By Green’s theorem, if D is the region with C D ⭸D,
CC
f x dx C gy dy D
⭸
6D ⭸x
gy −
⭸
0 dx dy D 0.
f x dx dy D
⭸y
6D
25. We begin by applying Green’s theorem (here D has constant density δ):
x dx dy
1
1
6
2
D
x dy D
2x dx dy D
2 · area of D C⭸D
2 · area of D 6D
dx dy
6D
D 6D
x δ dx dy
6D
D x.
δ dx dy
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“runall” — 2005/8/1 — 15:23 — page 315 — #27
Section 6.5
Miscellaneous Exercises for Chapter 6
315
Similarly,
y dx dy
1
1
6
D
xy dy D
y dx dy D
area of D C⭸D
area of D 6D
dx dy
6D
D 6D
y δ dx dy
6D
D y.
δ dx dy
For the second pair of formulas, we proceed in an entirely similar manner with Green’s theorem.
−
1
1
xy dx D −
− x dx dy
area of D C⭸D
area of D 6D
D
x dx dy
1
x dx dy D 6D
area of D 6D
dx dy
6D
D 6D
x δ dx dy
6D
D x.
δ dx dy
Also,
−
1
1
y2 dx D −
− 2y dx dy
2 · area of D C⭸D
2 · area of D 6D
D
y δ dx dy
1
D
y dx dy D 6
D y.
area of D 6D
δ dx dy
6D
26. Along the bottom of the triangle, dy is zero and along the left side x is zero, so the first pair of line integrals
in Exercise 25 must be zero except along the side connecting (1, 0) to (0, 2). Parametrize this side by xt D
1 − t, 2t with 0 … t … 1. Note also that the area of the triangle is 1. Using the results of Exercise 25, we have
1
1
1
1
xD
x2 dy D
1 − t2 2 dt D
t 2 − 2t C 1 dt
2 C⭸D
2 0
0
D
1
1
− 1 C 1D
3
3
and
yD
C⭸D
xy dy D
1
0
1 − t2t2 dt D 2
1
−2t 2 C 2t dt
0
D 2 −
2
2
C 1 D .
3
3
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“runall” — 2005/8/1 — 15:23 — page 316 — #28
316
Chapter 6
Line Integrals
27. The region in question looks like:
y
x
The area of this region is
36π − π − π D 34π.
Using the result of Exercise 25, we calculate
xD
1
x2 dy
2 · area of D C⭸D
and y D −
1
y2 dx
2 · area of D C⭸D
(other computations are possible).
We may parametrize the outer boundary of the region by
xt D 6 cos t, 6 sin t,
0 … t … 2π
and the inner two circles by
yt D 4 C sin t, cos t, 0 … t … 2π and
zt D sin t − 2, cos t C 2, 0 … t … 2π.
Hence
1
x2 dy
68π C⭸D
2π
2π
2π
1
D
36 cos2 t · 6 cos t dt C
sin t C 42 − sin t dt C
sin t − 22 − sin t dt 68π 0
0
0
2π
2π
1
2
D
2161 − sin t cos t dt −
sin3 t C 8 sin2 t C 16 sin t dt
68π 0
0
2π
−
sin3 t − 4 sin2 t C 4 sin t dt xD
0
D
2π
1
68π 0
2161 − sin2 t cos t dt −
2π
0
2 sin3 t C 4 sin2 t C 20 sin t dt 2π
2π
1
3 D
21 − cos2 t sin t C 21 − cos 2t C 20 sin t dt 216 sin t − 72 sin t −
68π
0
0
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“runall” — 2005/8/1 — 15:23 — page 317 — #29
Section 6.5
Miscellaneous Exercises for Chapter 6
317
2π
1
2
3
D
0 C 2 cos t − cos t − 2t C sin 2t C 20 cos t 68π
3
0
D
1
1
−4π D −
68π
34
and
2π
2π
2π
1
1
y2 dx D −
36 sin2 t · −6 sin t dt C
cos2 t · cos t dt C
cos t C 22 cos t dt 68π C⭸D
68π 0
0
0
2π
2π
2π
1
D−
2161 − cos2 t− sin t dt C
cos3 t dt C
cos3 t C 4 cos2 t C 4 cos t dt 68π 0
0
0
2π
2π
1
3 D−
21 − sin2 t cos t C 21 C cos 2t C 4 cos t dt 216 cos t − 72 cos t C
68π
0
0
2π
1
2 3
D−
0
C
2
sin
t
−
sin
t
C
2t
C
sin
2t
C
4
sin
t
68π
3
0
yD−
D−
1
1
[4π] D − .
68π
34
28. We can write f §g as f ⭸g/⭸x, f ⭸g/⭸y. Now apply the divergence theorem and collect the appropriate terms.


⭸
g
⭸
g
⭸
⭸
 dA
 f C


f §g · n ds D
f
⭸x
⭸y
⭸y
6D ⭸x
CC
⭸f ⭸g
⭸2 g
⭸f ⭸g
⭸2 g
C
dA
C f
C f
⭸x2
⭸y ⭸y
⭸y 2
6D ⭸x ⭸x


2g
2g
⭸
f
⭸
g
⭸
f
⭸
g
⭸
⭸

 dA

C
D
C f 2 C f 2 
⭸
x
⭸
x
⭸
y
⭸
y
⭸x
⭸y
6D
D
D
6D
§f · §g C f §2 g dA.
29. Apply the results of Exercise 28 to both parts of the line integral.
CC
f §g − g§f · n ds D
D
D
CC
f §g · n ds −
6D
6D
CC
g§f · n ds
f §2 g C §f · §g dA −
6D
g§2 f C §g · §f dA
f §2 g − g§2 f dA
30. With f x, y K 1 in Green’s first identity, we have §f K 0, so
6D
f §2 g C §f · §g dA D
But if g is harmonic, §2 g D 0, so
6D
§2 g dA D
C⭸D
§g · n ds D
⭸g
ds D 0.
C⭸D ⭸n
31. Now use Green’s first identity with f D g and f harmonic to obtain
Since C D ⭸D and §f · n D
⭸g
ds.
C⭸D ⭸n
⭸f
, the desired result follows.
⭸n
6D
§f · §f dA D
CC
f §f · n ds.
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“runall” — 2005/8/1 — 15:23 — page 318 — #30
318
Chapter 6
Line Integrals
⭸f
ds D
f
§f · §f dA. But
6D
C⭸D ⭸n
§f · §f D §f 2 Ú 0. Thus the right integral is of a nonnegative, continuous integrand. For it to be zero, the
integrand must be identically zero. That is, §f · §f vanishes an D. We conclude that §f is zero on D and so
f must be constant. Since f x, y D 0 on ⭸D and f is constant on D, we must have f K 0 on D.
33. Let f D f1 − f2 . Then since f1 D f2 on ⭸D, f D 0 on ⭸D. Also f is harmonic if f1 and f2 are. Hence, by
Exercise 32, f K 0 on D so f1 D f2 on D.
34. (a) Exercise 25 from Section 6.3 is a particularly nice example of a nontrivial radially symmetric vector field
because there is a compelling physical reason for the field F to be radially symmetric. There F is the
gravitational force field of a mass M on a particle of mass m.
32. If f is zero on the boundary of D, then Exercise 31 implies that 0 D
FD−
xi C yj C zk
GMm
GMm
GMm
D−
xi C yj C zk D −
eρ .
x2 C y2 C z2 3/2
x2 C y2 C z2 xi C yj C zk
ρ2
(b) Apply the formula for the curl in spherical coordinates found in Theorem 4.6 in Chapter 3.
eρ
1
§ * FD
⭸/⭸ρ
ρ2 sin ϕ f ρ
ρ eϕ
⭸/⭸ϕ
0
ρ sin ϕ eθ 
⭸/⭸ρ
1

⭸/⭸θ D 0, 0, eθ f ρ
ρ
0

⭸/⭸ϕ 
 D 0, 0, 0.
0 When students get to complex analysis and learn to integrate around poles, texts often refer to their experience with
Green’s theorem in multivariable calculus. At least one of Exercises 35 and 36 should be assigned so that this reference
might ring a bell.
35. (a) The boundary is in two pieces which we separately parametrize as x1 θ D cos θ, sin θ and x2 θ D
a cos θ, −a sin θ, each for 0 … θ … 2π. The line integral is then
−y
x
dx C
dy D
2
2
2
x C y2
CC x C y
D
2π
sin θ C cos θ dθ C
2
2π
2
0
2π
−
0
a2 cos2 θ
a2 sin2 θ
−
a2
a2
dθ
1 − 1 dθ D 0.
0
The double integral is

x
⭸
2
6D ⭸x x C y2

−y2 C x2
−x2 C y2
−y
⭸
 dx dy D
C
−
2
2 2
⭸y x2 C y2
x2 C y2 2
6
D x C y D
6
D
dx dy
0 dx dy D 0.
Thus the conclusion of Green’s theorem holds for F in the given annular region.
(b) This time the line integral is only taken over the outer boundary and so
−y
x
dx C
dy D 2π.
2 C y2
2 C y2
x
x
CC
The same cancellation takes place in the double integral as in part (a), so


x
⭸
2 C y2
⭸
x
x
6D
−
−y
⭸
 dx dy D 0.
⭸y x2 C y2
The problem is that F is not defined at the origin.
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“runall” — 2005/8/1 — 15:23 — page 319 — #31
Section 6.5
Miscellaneous Exercises for Chapter 6
319
(c) Let D be the region so that ⭸D consists of the given curve C oriented counterclockwise and also the curve
Ca , the circle of radius a centered at the origin oriented clockwise. Then F is defined everywhere in the
region D. Green’s theorem holds so
CC
−y
dx C
x2 C y 2
x
dy C
x2 C y 2
CCa

x

⭸
D
6D ⭸x x2 C y2
−y
CCa x2 C y2
dx C
−y
x2 C y 2
x
x2 C y 2
dy D
CC∪Ca

−y
⭸

 dx dy D 0,
−
⭸y x2 C y2
x
dy D −2π.
x2 C y 2
36. (a)
dx C
Therefore,
i
⭸
/
⭸x
§ * F D y
−
2
x C y2
−y
CC x2 C y2
j
⭸/⭸y
x
2
x C y2
dx C
−y
x2 C y 2
dx C
x
x2 C y 2
dy
but
x
dy D 2π.
x2 C y 2
k ⭸/⭸z D 0, 0, 0.
0 (b) Here the path is xθ D cos θ, sin θ for 0 … θ … 2π. Then
CC
2π
F · ds D
− sin θ− sin θ C cos θcos θ dθ D
0
2π
dθ D 2π.
0
(c) We saw in part (b) that the line integral around a closed path is not zero, so F cannot be conservative on its
domain.
(d) The conditions are not met for the theorem as the domain of F is not a simply-connected region.
37. (a) By the divergence theorem, the flux
1 5
F · n ds D
0
C
0
⭸ y
⭸ 4
[e ] C
[x ] dy dx D 0.
⭸x
⭸y
(b) Again, by the divergence theorem, the flux
F · n ds D
C
⭸
6D ⭸x
[f y] C
⭸
[f x] dy dx D 0.
⭸y
38. Over the path xt,
F · ds D
x
ma · ds D
b
x
mx t · x t dt D m
a
b
a
b
1
1
1
D mx t2 D m[vb]2 − m[va]2 .
2
2
2
1 d [x t · x t] dt
2 dt
a
39. We’ll first replace F with −§V and apply Theorem 3.3.
F · ds D
x
−§V · ds D −V B C V A,
x
where A D xa and B D xb. However, in Exercise 38 we showed that
F · ds D
x
1
1
m[vb]2 − m[va]2 .
2
2
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“runall” — 2005/8/1 — 15:23 — page 320 — #32
320
Chapter 6
Line Integrals
Therefore,
1
1
m[vb]2 − m[va]2 , or
2
2
1
1
2
V A C m[va] D V B C m[vb]2 .
2
2
−V B C V A D
We see, therefore, that the sum of the potential and kinetic energies of the particle remains constant.
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“runall” — 2005/8/1 — 15:18 — page 321 — #1
C H A P T E R
7
Surface Integrals and Vector Analysis
7.1 PARAMETRIZED SURFACES
1. (a) To find a normal vector we calculate
Ts s, t D 2s, 1, 2s
so
Ts 2, −1 D 4, 1, 4
Tt s, t D −2t, 1, 3
so
Tt 2, −1 D 2, 1, 3.
Then a normal vector is
N2, −1 D Ts 2, −1 * Tt 2, −1 D −1, −4, 2.
(b) We find an equation for the tangent plane using
0 D N2, −1 · x − 3, 1, 1 D −1, −4, 2 · x − 3, 1, 1 D −x C 3 − 4y C 4 C 2z − 2.
This is equivalent to x C 4y − 2z D 5.
2. First we figure that since
√ 2 sin t D 1, either t D π/6 or 5π/6. Since 2 cos t < 0 we know that t D 5π/6. Then we
can see that sin s D 2/2 so s D π/4. Next, find a normal vector to the surface at the given point by calculating
Ts s, t D −5 C 2 cos t sin s, 5 C 2 cos t cos s, 0
Tt s, t D −2 sin t cos s, −2 sin t sin s, 2 cos t
and
so
Ns, t D Ts s, t * Tt s, t D 25 C 2 cos tcos s cos t, sin s cos t, sin t. Therefore,
√
√
3 − 5 √ √
√
Nπ/4, 5π/6 D
3, 3, − 2.
2
We calculate an equation for the tangent plane by writing N · x − x0 , y0 , z0 D 0 or, equivalently in this case,
√
0 D 3,
√


√
√
√
5
−
5
−
3
3
x − √

3, − 2 · 
, √
, 1
2
2
or
√
3x C
√
3y −
√
√
√
2z D 5 6 − 4 2.
3. Since x D es at x D 1, we know that s D 0. Also since z D 2e−s C t, when z D 0 and s D 0, we have t D −2.
As above we calculate,
Ts s, t D e s , 2t 2 e 2s , −2e−s and
Tt s, t D 0, 2te 2s , 1.
Thus, N0, −2 D Ts 0, −2 * Tt 0, −2 D 1, 8, −2 * 0, −4, 1 D 0, −1, −4. Then an equation of
the tangent plane is 0 D N0, −2 · x − 1, 4, 0 D 0, −1, −4 · x − 1, 4, 0. We can simplify this
to y C 4z D 4.
4. (a) Ts s, t D 2s cos t, 2s sin t, 1 so Ts −1, 0 D −2, 0, 1. Also, Tt s, t D −s2 sin t, s2 cos t, 0 so
Tt −1, 0 D 0, 4, 0. Therefore, N−1, 0 D −2, 0, 1 * 0, 4, 0 D −4, 0, −8.
(b) An equation of the tangent plane is −4, 0, −8 · x − 1, 0, −1 D 0. This simplifies to x C 2z D −1.
(c) Note that the x-component of X is s2 cos t and the y-component is s2 sin t and the z-component is a function
of s. We can eliminate the t by looking at x2 C y2 . So without much work we have found that an equation
for the image of X is x2 C y2 − z4 D 0.
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“runall” — 2005/8/1 — 15:18 — page 322 — #2
322
Chapter 7
Surface Integrals and Vector Analysis
5. (a) Using Mathematica and the command:
ParametricPlot3D[{s, sG 2 C t, tG 2}, {s, −2, 2}, {t, −2, 2}, AxesLabel ' {x,y,z}],
we obtain the image
6
4
y
2
0
-2
4
3
z 2
1
0
-2
-1
0
x
1
2
(b) To determine whether the surface is smooth we need to calculate N. First, Ts s, t D 1, 2s, 0, and
Tt s, t D 0, 1, 2t so N D Ts * Tt D 4st, −2t, 1. We conclude that N Z 0 for any (s, t) so N is
smooth.
(c) If s, s2 C t, t 2 D 1, 0, 1, then s D 1 and t D −1. So N1, −1 D −4, 2, 1 and an equation of the
tangent plane at this point is −4, 2, 1 · x − 1, 0, 1 D 0 or more simply, 4x − 2y − z D 3.
6. In Exercise 1, x D s2 − t 2 , y D s C t, and so if we note that x D s C ts − t D ys − t, then
x/y D s − t. This allows us to solve for s and t separately: 2s D y C x/y and 2t D y − x/y. This means that
z D s2 C 3t can be written as z D y C x/y2 /4 C 3y − x/y.
7. (a) For the surface, Xs, t D s cos t, s sin t, s2 , where s Ú 0 and 0 … t … 2π. This means that Ts s, t D
cos t, sin t, 2s and Tt s, t D −2 sin t, s cos t, 0. Then a normal vector is given by N D Ts * Tt D
−2s2 cos t, 2s2 sin t, s. This means that the surface is smooth except when s D 0. In other words, S is
smooth except at the origin.
(b) See the figure below and note that z D x2 C y2 so we see that S is a paraboloid.
y
0
2
-2
x
0
2
-2
8
6
z
4
2
0
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“runall” — 2005/8/1 — 15:18 — page 323 — #3
Section 7.1
Parametrized Surfaces
323
(c) Again, z D x2 C y2 .
(d) Part (a) above takes care of every point except the origin. At the origin N D 0, but we easily see that
the tangent plane there is the horizontal plane z D 0. Thus smoothness in the sense defined in Section 7.1
depends on the parametrization as well as the geometry of the underlying surface.
8. Really there’s not much to show. You know that if the image of the parametrized surface is to be an ellipsoid,
you need a2 sin s cos t2 C b3 sin s sin t2 C ccos s2 D 1. So a D 1/4, b D 1/9, and c D 1. Therefore the
y2
x2
C
C z2 D 1.
image satisfies
4
9
9. For t D t0 , Xs, t0 D a C b cos t0 cos s, a C b cos t0 sin s, b sin t0 . So z is constant and x2 C y2 D
a C b cos t0 2 . This is a circle of radius a C b cos t0 centered
√
√ at 0, 0, b sin t0 .
10. (a) When θ D π/3, the r-coordinate curve is given by r/2, r 3/2, π/3 where r Ú 0. This is the ray y D 3x
where x Ú 0 and z D π/3. In general, the r-coordinate curve when θ D θ0 is a ray in the z D θ0 plane.
The solution is simpler than the following four cases make it seem. If cos θ0 Z 0 then y D tan θ0 x where
x Ú 0 if cos θ0 > 0 and x … 0 if cos θ0 < 0. If cos θ0 D 0, then the ray is x D 0 with y Ú 0 if sin θ0 > 0
and y … 0 if sin θ0 < 0.
(b) When r D 1 the θ-coordinate curve is the helix cos θ, sin θ, θ. In general, when r D r0 the θ-coordinate
curve is the helix r0 cos θ, r0 sin θ, θ.
(c) You can see that the helicoids are made up of the helices that are the θ-coordinate curves.
11. (a) First we consider the sphere as the graph of the function f x, y D 4 − x − 22 − y C 12 . The
partial derivatives are
fx D
−x − 2
4 − x − 22 − y C 12
fy D
−y C 1
4 − x − 22 − y C 12
.
√
√
√
√
3.3 √
of Chapter 2, z D f a,
So fx 1, 0, 2 D 1/ 2 and fy 1, 0, 2 D −1/ 2. By Theorem
√
√ b C
fx a, bx − a C fy a,√by − b. In this case, this is z D 2 C 1/ 2x − 1 − 1/ 2y, or
equivalently, −x C y C 2z D 1.
(b) Now we look at the sphere as a level surface of F x, y, z D x√− 22 C y C √12 C z2 . The gradient
§F x, y, z D 2x − 2, y C 1, z and, therefore, §F 1, 0, 2 D −2, 2, 2 2. By formula (5) of
Section 2.6, the tangent plane is given by
√
√
√
√
0 D §F 1, 0, 2 · x − 1, 0, 2 D −2, 2, 2 2 · x − 1, 0, 2.
√
This too is equivalent to −x C y C 2z D 1.
√
(c) Now
√ we’ll use the results of this section. Considering the z-component, we see 2 cos s D 2 so cos s D
2/2. Considering the y- and x-components, 2 sin s sin t D 1 and sin s cos t D −1. Thus we have that
s D π/4 and t D 3π/4. Also Ts s, t D 2 cos s cos t, 2 cos s sin t, −2 sin s and Tt s, t D −2 sin s sin t,
2 sin s cos t, 0. A normal vector to the sphere at the specified point is
√
√ √
Nπ/4, 3π/4 D Ts π/4, 3π/4 * Tt π/4, 3π4 D −1, 1, − 2 * −1, −1, 0 D − 2, 2, 2.
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“runall” — 2005/8/1 — 15:18 — page 324 — #4
324
Chapter 7
Surface Integrals and Vector Analysis
√ √
√
The tangent
plane is given by − 2, 2, 2 · x − 1, 0, 2 D 0 which is also equivalent to −x C
√
y C 2z D 1.
12. (a) X1, −1 D 1, −1, −1 and we have Ts D 3s2 , 0, t, Tt D 0, 3t 2 , s. Hence the normal at (1, −1,
−1)—which is when s D 1, t D −1—is N1, −1 D Ts 1, −1 * Tt 1, −1 D 3, 0, −1 * 0, 3, 1 D
3, −3, 9. So an equation for the tangent plane is
3x − 1 − 3y C 1 C 9z C 1 D 0 or x − y C 3z D −1.
(b) In general we have that the standard normal is given by
i
3s2
0
Ns, t D Ts * Tt D
j
0
3t 2
k
t
s
D −3t 3 , −3s3 , 9s2 t 2 .
Note that N D 0 when s D t D 0, i.e., at (0, 0, 0). So the surface fails to be smooth there.
(c) A computer graph is shown.
1
0.5
y
0
-0.5
-1
1
0.5
z
0
-0.5
-1
-1
-0.5
0
x
√
3
0.5
1
√
3
√
(d) With x D s3 , y D t 3 , z D st, xy D s3 t 3 D st D z. Sometimes a computer will graph z D 3 xy for
points only where x and y are nonnegative (or sometimes where xy Ú 0).
13. (a) y2 z D t 2 · s2 D st2 D x2
(b) The standard normal is
Ns, t D Ts * Tt D t, 0, 2s * s, 1, 0
D
i
t
s
j
0
1
k
2s
0
D −2s, 2s2 , t.
So N D 0 when s D t D 0, i.e., at (0, 0, 0). At this point X fails to be smooth.
(c) A computer graph is shown.
4
3
2 z
1
0
2
-4
-2
0
x
2
1
0 y
-1
4 -2
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“runall” — 2005/8/1 — 15:18 — page 325 — #5
Section 7.1
Parametrized Surfaces
325
(d) Xs1 , t1 D Xs2 , t2 when s1 t1 D s2 t2 , t1 D t2 , s12 D s22 . Thus if t1 D t2 D 0 and s1 D ;s2 we get the
same image—i.e., Xs, 0 D X−s, 0 D 0, 0, s2 . Thus the positive z-axis (which lies on the image of
X) is not uniquely determined.
(e) Note that 2, 1, 4 D X2, 1. From work in part (b), N2, 1 D −4, 8, 1 so an equation for the tangent
plane is −4x − 2 C 8y − 1 C 1z − 4 D 0 or −4x C 8y C z D 4.
(f) 0, 0, 1 D X−1, 0 D X1, 0.
N−1, 0 D 2, 2, 0
N1, 0 D −2, 2, 0
So the corresponding tangent planes have equations x C y D 0 and x − y D 0 respectively.
(If you look at the graph in part (c), you can see two parts of the surface intersecting, so this makes sense.)
14. Here we generalize the results of parts (a) and (c) of Exercise 11. If we view S as the graph of a function f (x,
y) then we can apply formula (4) of Section 2.3:
z D f a, b C fx a, bx − a C fy a, by − b.
We can rewrite this equation as 0 D fx a, b, fy a, b, −1 · x − a, b, f a, b, where a D xs0 , t0 , b D
ys0 , t0 , and f a, b D zs0 , t0 . In other words, we are also considering S to be a surface that is parametrized
by Xs, t D xs, t, ys, t, zs, t and so, using the Chain Rule,
Ts s, t D xs s, t, ys s, t, fx x, yxs s, t C fy x, yys s, t and
Tt s, t D xt s, t, yt s, t, fx x, yxt s, t C fy x, yyt s, t.
We calculate the normal vector N by taking the cross product Ts * Tt and simplifying to obtain
Ns, t D [xt s, tys s, t − xs s, tyt s, t]fx x, y, fy x, y, −1.
So an equation of the tangent plane at s0 , t0 is N · x − Xs0 , t0 D 0 which in this case is
[xt s0 , t0 ys s0 , t0 − xs s0 , t0 yt s0 , t0 ]fx a, b, fy a, b, −1 · x − a, b, f a, b D 0
or
fx a, b, fy a, b, −1 · x − a, b, f a, b D 0.
So we see that in this case the results of the two methods agree.
15. (a) To find an equation for the tangent plane to a surface described by the equation y D gx, z at the point (a,
g(a, c), c) we basically permute the case detailed in the text and in Exercise 14 to obtain either
gx a, c, −1, gz a, c · x − a, ga, c, c D 0
or
gx a, cx − a − y − ga, c C gz a, cz − c D 0.
(b) Similarly, an equation for the tangent plane to a surface described by the equation x D hy, z at the point
(h(b, c), b, c) is either
−1, hy b, c, hz b, c · x − hb, c, b, c D 0
or
−x − hb, c C hy b, cy − b C hz b, cz − c D 0.
16. We have X : D ' R3 and by Definition 3.8 of Chapter 2, the linear approximation is given by
x D Xs0 , t0 C DXs0 , t0 s − s0
t − t0
.
Here DXs0 , t0 is the matrix

xs s0 , t0 DXs0 , t0 D  ys s0 , t0 zs s0 , t0 
xt s0 , t0 yt s0 , t0  D [Ts s0 , t0 T
zt s0 , t0 Tt s0 , t0 T ].
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“runall” — 2005/8/1 — 15:18 — page 326 — #6
326
Chapter 7
Surface Integrals and Vector Analysis
Thus the tangent plane to the surface is given by
x, y, z D Xs0 , t0 C [Ts s0 , t0 T
Tt s0 , t0 T ]
s − s0
t − t0
D Xs0 , t0 C Ts s0 , t0 s − s0 C Tt s0 , t0 t − t0 .
17. By Exercise 16,

1
x, y, z D 1, 0, 1 C  2
0

0
1 
−2
s − 1
t C 1
x, y, z D s, 2s C t − 1, −2t − 1.
We check this against our result for Exercise 5(c):
4x − 2y − z D 4s − 22s C t − 1 − −2t − 1 D 3.
18. In Exercise 3 we parametrized a cylinder of radius a and height h by Xs, t D a cos s, a sin s, t for 0 … t … h
and 0 … s < 2π. Then Ts s, t D −a sin s, a cos s, 0, Tt s, t D 0, 0, 1, and Ts s, t * Tt s, t D
a cos s, a sin s, 0. Then, by formula (6), the surface area of S is
2π h
2π h
||Ts s, t * Tt s, t|| dt ds D
a dt ds D 2πah.
0
0
0
0
19. As in Exercise 18 √
we need to calculate
√ ||Ts s, t * Tt s, t||. We have that Xs, t D s C t, s − t, s for
−1 … s … 1 and − 1 − s2 … t … 1 − s2 . Therefore,
√ Ts s, t D 1, 1, 1, Tt s,
√ t D 1, −1, 0, Ts s, t *
Tt s, t D 1, 1, −2 and ||Ts s, t * Tt s, t|| D 6. So we are integrating 6 over the unit disk in the
√
√
6 dt ds D 6π.
st-plane. Therefore, the surface area of XD D
6D
20. For the parametrization of the helicoid, Xr, θ D r cos θ, r sin θ, θ so Tr r, θ D cos θ, sin θ,√0, Tθ r, θ D
−r sin θ, r cos θ, 1, Tr r, θ * Tθ r, θ D sin θ, − cos θ, r and ||Tr r, θ * Tθ r, θ|| D 1 C r2 . Then
the surface area of n “turns” of the helicoid is
2πn 1 √
2πn
2πn
√
1 √
1 √
1 C r2 dr dθ D
[ 2 C sinh−1 1] dθ D
[ 2 C ln 2 C 1] dθ
2
2
0
0
0
0
√
√
D [ 2 C ln 2 C 1]πn.
21. A quick look at the figure below shows a cutaway of a quarter of the xz
√-plane intersection of the cylindrical hole
of radius b bored in a sphere of radius a. The height of the hole is 2 a2 − b2 . The top half of the ring is the
region swept out by the portion of the diagram containing the letter ‘h’.
z
b
a
h
x
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“runall” — 2005/8/1 — 15:18 — page 327 — #7
Section 7.1
Parametrized Surfaces
327
If Xs, t D a sin s cos t, a sin s sin t, a cos s, then Ts s, t D a cos s cos t, a cos s sin t, −a sin s, Tt s, t D
−a sin s sin t, a sin s cos t, 0, Ts s, t * Tt s, t D a2 sin ssin s cos t, sin s sin t, cos s and ||Ts s, t *
2
−1
Tt s, t||
√ D a sin s. Notice that the angle s made with the z-axis has lower limit cos h/a D
−1
2
2
cos a − b /a and upper limit π/2. So the surface area is
2π π/2
2
√
cos−1 a2 −b2 /a
0
a sin s ds dt D 2
√
2π
2
2
0
a √
a2 − b 2
dt D 4πa a2 − b2 .
a
22. The parametrization of the paraboloid is Xs, t D s cos t, s sin t, 9 − s2 where 0 … t … 2π and 0 … s … 3. So
2
2
Ts s, t D cos t, sin t, −2s,
√ Tt s, t D −s sin t, s cos t, 0, Ts s, t * Tt s, t D 2s cos t, 2s sin t, s, and
2
||Ts s, t * Tt s, t|| D s 4s C 1. The surface area is then
2π 3 √
2π
1
π
3
s 4s2 C 1 ds dt D
1 C 4s2 3/2 0 dt D 373/2 − 1.
12 0
6
0
0
23. We’ll parametrize the surface by Xs, t D s cos t, s sin t, 2s2 for 0 … t … 2π and 1 … s … 2. So Ts s, t D
2
2
cos t, sin t, 4s,
√ Tt s, t D −s sin t, s cos t, 0, Ts s, t * Tt s, t D 4s cos t, 4s sin t, s, and ||Ts s, t *
Tt s, t|| D s 16s2 C 1. So the surface area is
2π 2
0
s 16s2 C 1 ds dt D
1
1
48
2π
653/2 − 173/2 dt D
0
π
653/2 − 173/2 .
24
24. (a) First we use the parametrization Xs, t D s, t, a − s − t and calculate T√s s, t D 1, 0, −1, Tt s, t D
0, 1, −1, Ts s, t * Tt s, t D 1, 1, 1 and ||Ts s, t * Tt s, t|| D 3. The surface area is then the
√
√
√
integral of 3 over the disk of radius a, which is
3 ds dt D 3πa2 .
6D
(b) To use formula (9), we view the surface as z D f x, y D a − x − y, so fx x, y D −1 and fy x, y D −1.
Therefore, formula (9) gives the surface area as
6D
−12 C −12 C 1 dx dy D
√
6D
3 dx dy D
√
3πa2 .
25. We have z D f x, y and fx2 C fy2 D a so, by formula (9), the surface area is
6D
fx2 C fy2 C 1 dx dy D
√
6D
a C 1 dx dy D
√
a C 1area of D.
26. (a) Here is a sketch of the surface for z Ú 1.
y
z
x
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328
Chapter 7
Surface Integrals and Vector Analysis
(b) We can calculate the volume under the infinite funnel by disks:
q
1
b
π
π
dz D lim −
D π.
b'q
z2
z 1
1
(c) To calculate the surface area, we’ll parametrize the funnel as Xs, t D s cos t, s sin t, , where 0 < s … 1
s
1
and 0 … t < 2π. Then Ts s, t D cos t, sin t, − , Tt s, t D −s sin t, s cos t, 0, Ts s, t * Tt s, t D
2
s
1
1
1
C s2 . Therefore, using tables or a computer algebra
cos t, sin t, s and ||Ts s, t * Tt s, t|| D
s
s
s2
system, we see that the surface area is given by
2π 1 2π 1 1
1
2
C s ds dt D lim
C s2 ds dt
a'0C 0
s2
s2
0
0
0
√
√
√
√
π lim [ 2 − ln 2 C 1 − a4 C 1 − ln1 C 1 C a4 C lna2 ].
a'0C
Each term in this last expression possesses a finite limit except lna2 . Since lima'0C lna2 D −q, we
see that the surface area is infinite.
27. The first octant portion of the intersection is shown below.
x
a
a
z
y
a
Note that 1/16 of the total surface area is that of the graph of z D
bounded by y D x, x D a, and y D 0.
√
a2 − x2 lying over the triangular region
y
a
a
x
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“runall” — 2005/8/1 — 15:18 — page 329 — #9
Section 7.1
For z D
√
Parametrized Surfaces
329
x
⭸z
⭸z
D −√
D 0. Hence
and
2
2
⭸x
⭸
y
a − x
a a x
x2
x 2 C a2 − x 2
C 0 C 1 dy dx D 16
x
dx
Surface area D 16
2
2
a − x
a2 − x 2
0 0
0
a
ax
√
D 16
dx.
2
a − x2
0
a2 − x 2 ,
Let u D a2 − x2
du D −2x dx. Then
so
Surface area D −8a
0
a2
a2
du
−1/2
1/2
u
du
D
8a
·
2u
D
8a
√
u
a2
0
0
D 16a2 .


 x D r cos θ
y D r sin θ
28. We have

 z D f r, θ
r, θ ∈ D . Therefore,
Tr D cos θ, sin θ,
⭸f
,
⭸r
Tθ D −r sin θ, r cos θ,
⭸f
.
⭸θ
So
Nr, θ D Tr * Tθ D sin θ
⭸f
⭸f
⭸f
⭸f
− r cos θ , −r sin θ
− cos θ , r cos2 θ C r sin2 θ .
⭸θ
⭸r
⭸r
⭸θ
Hence
2
||N||2 D sin2 θ 2
⭸f
⭸f ⭸f
⭸f
C r2 cos2 θ − 2r sin θ cos θ
⭸θ
⭸r ⭸θ
⭸r
2
C cos2 θ 2
⭸f
⭸f ⭸f
⭸f
C r2 sin2 θ C r2
C 2r sin θ cos θ
⭸θ
⭸r ⭸θ
⭸r
2
D
2
⭸f
⭸f
2
C r C 1 .
⭸θ
⭸r
Thus we have
2
2
||N|| D r
⭸f
1 ⭸f
C C 1
r2 ⭸θ
⭸r
2
2
and
⭸f
1 ⭸f
Surface area D
r
C C 1 dr dθ, using formula (6) in §7.1.
2
r
⭸
θ
⭸r
6
D


x
D
f
ϕ,
θ
sin
ϕ
cos
θ

y D f ϕ, θ sin ϕ sin θ from spherical/Cartesian conversions. From this,
29. We have

 z D f ϕ, θ cos ϕ
Tϕ D fϕ sin ϕ cos θ C f cos ϕ cos θ, fϕ sin ϕ sin θ C f cos ϕ sin θ, fϕ cos ϕ − f sin ϕ
Tθ D fθ sin ϕ cos θ − f sin ϕ sin θ, fθ sin ϕ sin θ C f sin ϕ cos θ, fθ cos ϕ.
After some careful computation and using cos2 α C sin2 α D 1, we find Nϕ, θ D Tϕ * Tθ D f fθ sin θ −
f fϕ sin ϕ cos ϕ cos θ C f 2 sin2 ϕ cos θ, f 2 sin2 ϕ sin θ − f fϕ sin ϕ cos ϕ sin θ − f fθ cos θ, f 2 sin ϕ cos ϕ C
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“runall” — 2005/8/1 — 15:18 — page 330 — #10
330
Chapter 7
Surface Integrals and Vector Analysis
f fϕ sin2 ϕ. After still more computation, one finds ||N||2 D f fθ 2 C f fϕ 2 sin2 ϕ C f 4 sin2 ϕ, so that
using formula (6) in §7.1,
Surface area D
f fθ 2 C f fϕ 2 sin2 ϕ C f 4 sin2 ϕ dϕ dθ
6D
D
f ϕ, θ fθ2 C sin2 ϕfϕ C f 2 dϕ dθ.
6D
7.2 SURFACE INTEGRALS
Many of your students will apply the formulas and techniques introduced in this section by first finding a parametrization
for the given surface. In many cases, as was shown in the text, if they examine the geometry of the surface, an easier
solution might present itself. In several of the solutions below, each approach is outlined.
f dS D
f Xs, t||Ts * Tt || ds dt. Here Xs, t D
6
6D
X
s, s C t, t,√Ts s, t D 1, 1, 0, Tt s, t D 0, 1, 1, Ns, t D Ts s, t * Tt s, t D 1, −1, 1, and
||Ns, t|| D 3. Also, f Xs, t D s2 C s C t2 C t 2 D 2s2 C st C t 2 . So
1. We will use Definition 2.1 to calculate the integral:
6X
√
x C y C z dS D 2 3
2
2
2
2 1
0
√
D 2 3
√
s C st C t ds dt D 2 3
2
2
0
2
0
1
t
C
C t 2 dt
3
2
2
8
26
C 1 C D √ .
3
3
3
2. (a) Since Xs, t D s C t, s − t, st, we can calculate Ts s, t√D 1, 1, t, Tt s, t D 1, −1, s, Ns, t D
Ts s, t * Tt s, t D s C t, t − s, −2, and ||Ns, t|| D 2s2 C 2t 2 C 4. Using polar coordinates in
the double integral, we obtain
6X
4 dS D
4
√
2s2 C 2t 2 C 4 ds dt D
π/2 1
0
0
6D
π/2
√
√
2
π
D
[6 6 − 8] dθ D [6 6 − 8].
3 0
3
√
4r 2r2 C 4 dr dθ
F · dS D
FXs, t · Ns, t ds dt. Here FXs, t D s C t, s − t, st and
6X
6D
so, from part (a), we know that Ns, t D s C t, t − s, −2. This means that F · N D s C t2 − s −
t2 − 2st D 2st. Therefore,
(b) By Definition 2.2,
6X
F · dS D
1 √ 1−t 2
2st ds dt D
2st ds dt
0 0
6D
1
1
√
1
t4
t2
1
1−t 2
2
D
s t 0
dt D
t − t 3 dt D − D .
2
4 0
4
0
0
F · dS. The surface is given by a level set of f x, y, z D 2x − 2y C z. Since
6X
1
§f D 2, −2, 1, the upward-pointing unit normal is 2, −2, 1. So, since 2x − 2y C z D 2,
3
3. We need to calculate
6X
F · dS D
D
1
1
x, y, z · 2, −2, 1 dS D
2x − 2y C z dS
3 6S
3 6S
1
2
2
2dS D
dS D ||N||area of D D 2area of D.
3 6S
3 6S
3
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“runall” — 2005/8/1 — 15:18 — page 331 — #11
Section 7.2
Surface Integrals
331
Here D is the “shadow” of S in the xy-plane. D is a right triangle in the xy-plane with legs each of length
1. Hence
6X
F · dS D 2area of D D 21/211 D 1.
4. (a) You can easily verify that both X and Y parametrize the surface z D 3x2 C 3y2 for 0 … x2 C y2 … 4. The
major difference is that X covers the surface once while Y covers the surface twice.
(b) For X, the standard normal N is
cos t, sin t, 6s * −s sin t, s cos t, 0 D −6s2 cos t, −6s2 sin t, s
so
6X
yi − xj C z2 k · dS D
2π 2
0
D
s sin t, −s cos t, 9s4 · −6s2 cos t, −6s2 sin t, s ds dt
0
2π 2
0
9s5 ds dt D
2π
0
0
2
9s6
dt D
6 0
2π
96 dt D 192π.
0
For Y, the standard normal N is
2 cos t, 2 sin t, 24s * −2s sin t, 2s cos t, 0 D −48s2 cos t, −48s2 sin t, 4s
so
6Y
yi − xj C z2 k · dS D
4π 1
0
D
2s sin t, −2s cos t, 144s4 · −48s2 cos t, −48s2 sin t, 4s ds dt
0
4π 1
0
576s5 ds dt D
0
4π
0
1
576s6
dt D
6 0
4π
96 dt D 384π.
0
As noted in part (a), the integral over Y should be twice the integral over X since they both parametrize the
same space but Y covers the space twice.
5. We will parametrize the six faces of the cube as follows (in each case −2 … s, t … 2:
i
Xs, t for Si
face
1
2
3
4
5
6
(s, t, 2)
s, t, −2
(s, 2, t)
s, −2, t
(2, s, t)
−2, s, t
top
bottom
right
left
front
back
[xs, t]2 ||Ns, t|| ds dt D
[xs, t]2 ds dt for 1 … i … 6.
Note that in each case ||Ns, t|| D 1, so
6
6
Si
Si
2 2
2 2
Also,
[xs, t]2 ds dt D
s2 ds dt for i D 1, 2, 3, 4 and
[xs, t]2 ds dt D
4 ds dt for
−2 −2
−2 −2
6Si
6Si
i D 5, 6. Then
6
6S
x2 dS D Si
iD1 6
D4
[xs, t]2 ||Ns, t|| ds dt
2 2
−2 −2
2
s2 ds dt C 2
2 2
−2 −2
4 ds dt
2
2
s3
2
dt C 8
s −2 dt
−2 3 −2
−2
2
2
640
256
16
D4
dt C 8
C 128 D
.
4 dt D
3
3
−2 3
−2
D4
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“runall” — 2005/8/1 — 15:18 — page 332 — #12
332
Chapter 7
Surface Integrals and Vector Analysis
6. We parametrize the lateral surface of the cylinder by Xs, t D a cos s, a sin s, t where 0 … s … 2π and
0 … t … h. So we have Ts s, t D −a sin s, a cos s, 0, Tt s, t D 0, 0, 1, Ns, t D Ts s, t * Tt s, t D
a cos s, a sin s, 0, and ||Ns, t|| D a. So
6S
x2 C y2 dS D
2π h
0
a2 cos2 s C a2 sin2 sa dt ds D
2π h
0
0
a3 dt ds D 2πha3 .
0
A quicker approach is to note that on the cylinder x2 C y2 D a2 , so
6S
x2 C y2 dS D
6S
a2 dS D a2 · area of S D a2 2πah D 2πha3 .
7. (a) Because x2 C y2 C z2 D a2 on the surface,
6S
x2 C y2 C z2 dS D a2 surface area of S D 4πa4 .
(b) Here we note that by part (a)
6S
x2 dS C
6S
y2 dS C
6S
z2 dS D 4πa4
and by the symmetries of the sphere
6S
x2 dS D
6S
y2 dS D
6S
z2 dS. So
6S
y2 dS D 4πa4 /3.
x dS D 0 as for each small piece of the sphere
6S
with coordinate x > 0 (and x … a), there is a corresponding piece with coordinate x < 0. Hence contributions
in an appropriate Riemann sum will cancel.
xi C yj C zk
(b) For x2 C y2 C z2 D a2 the outward unit normal is given by n D
. Thus
a
8. (a) The sphere is symmetric about the plane x D 0. Hence
6S
F · dS D
6S
D
F · n dS D
1
x C y C z dS
6S a
1
x dS C
y dS C
z dS D 0
a 6S
6S
6S
since each surface integral is zero
 via reasoning as in part (a).

 x D 2 cos t
y D 2 sin t
9. (a) We parametrize the cylinder as
0 … t < 2π, −2 … s … 2.

 zDs
Then
||Ts * Tt || D ||0, 0, 1 * −2 sin t, 2 cos t, 0|| D ||−2 cos t, −2 sin t, 0||
D 2.
Hence
6S
z − x − y dS D
2
2π 2
2
0
D
2π
−2
s − 4 · 2 ds dt D
2π
0
2
s2 − 8s
dt
sD−2
−32 dt D −64π.
0
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“runall” — 2005/8/1 — 15:18 — page 333 — #13
Section 7.2
(b)
6S
z − x2 − y2 dS D
x2 C y2 D 4 on S. Hence
−44π · 4 D −64π.
6S
6S
z dS −
6S
Surface Integrals
333
x2 C y2 dS. S is symmetric about the z D 0 plane and
z dS D 0 and −
x2 C y2 dS D −
4 dS D −4 · surface area of S D
6S
6S
The following calculations are useful for Exercises 10–18. Let’s parametrize the surface of the cylinder in three pieces:
• S1 D the lateral surface, Xs, t D 3 cos s, 3 sin s, t for 0 … s … 2π and 0 … t … 4.
• S2 D the bottom surface, Xs, t D t cos s, t sin s, 0 for 0 … s … 2π and 0 … t … 3.
• S3 D the top surface, Xs, t D t cos s, t sin s, 4 for 0 … s … 2π and 0 … t … 3.
For S1 , Ts s, t D −3 sin s, 3 cos s, 0, Tt s, t D 0, 0, 1, Ts s, t * Tt s, t D 3 cos s, 3 sin s, 0, and ||Ts s, t *
Tt s, t|| D 3. For both S2 and S3 , Ts s, t D −t sin s, t cos s, 0, Tt s, t D cos s, sin s, 0, Ts s, t * Tt s, t D
0, 0, −t, and ||Ts s, t * Tt s, t|| D t. Because we are orienting with outward normals, Ns, t D 0, 0, −t on S2
and Ns, t D 0, 0, t on S3 .
In Exercises 10–13 we use Definition 2.1:
6X
f dS D
6D
f Xs, t||Ns, t|| ds dt. And we’ll break down the
C
C
.
6
6S2
6S3
S1
2π 4
2π 3
2π 3
z dS D
3t dt ds C
0 dt ds C
4t dt ds D 48π C 36π D 84π.
10.
0
0
0
0
6S
0 4 2π
0 3 2π
11.
y dS D
9 cos s ds dt C 2
t 2 cos s ds dt D 0 C 0 D 0. Alternatively, you could notice that we
0 0
0 0
6S
are integrating an odd function of y over a region that is symmetric with respect to y.
4 2π
3 2π
3 2π
12.
xyz dS D
27t cos s sin s ds dt C
0 ds dt C
4t 3 cos s sin s ds dt D 0. Use the substitu0 0
0 0
0 0
6S
tion u D sin s. Again, alternatively, you could use a symmetry argument. We are again integrating an odd function
of y over a region that is symmetric with respect to y.
13.
4 2π
3 2π
2
2
x dS D
27 cos s ds dt C 2
t 3 cos2 s ds dt
0 0
0 0
6S
2π
2π
4
3
4
3
1
1
s
s
D 27
C sin 2s
C sin 2s
dt C 2
t3
dt D 27
π dt C 2
πt 3 dt
2
9
2
4
0
0
0
0
0
0
integral as
6S
D
D 108π C
81π
297π
D
.
2
2
F · dS D
F · n dS D
FXs, t · Ns, t ds dt. For another
6X
6X
6D
way of solving these exercises, recall from Section 2.6, that if S is a surface in R3 defined by an equation of the form
f x, y, z D c, then if x0 ∈ X, the gradient vector §f x0 is a vector normal to the plane tangent to S at x0 . Therefore
the unit normal to S1 (a surface given by x2 C y2 D 9) is n D xi C yj/3, while the unit normal to S2 is −k and the
unit normal to S3 is k.
For Exercises 14–18, we use Definition 2.2:
14.
6S
xi C yj · dS D
C
3 2π
0
D
4 2π
0
4 2π
0
3 cos s, 3 sin s, 0 · 3 cos s, 3 sin s, 0 ds dt
0
t cos s, t sin s, 0 · 0, 0, −t ds dt C
0
3 2π
0
t cos s, t sin s, 0 · 0, 0, t ds dt
0
9 ds dt D 72π.
0
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334
Chapter 7
Surface Integrals and Vector Analysis
A different approach would be to observe that as the unit normals for S2 and S3 are ;k then F · n D 0 on S2
and S3 . On S1 the unit normal is xi C yj/3 So F · n D x2 C y2 /3 D 9/3 D 3. Therefore we obtain,
15.
6S1
F · n dS D 3area of S1 D 32π34 D 72π.
6S
zk · dS D
4 2π
0
0, 0, t · 3 cos s, 3 sin s, 0 ds dt C
0
3 2π
C
0
0, 0, 4 · 0, 0, t ds dt D
0
3 2π
3 2π
0
0
0, 0, 0 · 0, 0, −t ds dt
0
4t ds dt D
0
3
8πt dt D 36π.
0
A different approach would have been to notice that, since the unit normal vector to the lateral surface S1 has no
k component,
6S
6S1
zk · dS D 0. Also, z D 0 on S2 so
zk · dS D
6S3
16.
6S
y3 i · dS D
C
3 2π
0
D 81
zk · dS D
4 2π
0
6S3
6S2
4k · k dS D
zk · dS D 0. Finally, z D 4 on S3 and therefore
6S3
4 dS D 4 · area of S3 D 4π32 D 36π.
27 sin3 s, 0, 0 · 3 cos s, 3 sin s, 0 ds dt
0
t 3 sin3 s, 0, 0 · 0, 0, −t ds dt C
0
0
4 2π
81
sin s cos s ds dt D
4
4
3
0
3 2π
0
t 3 sin3 s, 0, 0 · 0, 0, t ds dt
0
2π
dt D 0.
4
sin s
0
0
Again, a careful student should have noticed that there is no k component and so the integrals over S2 and S3 are
each 0.
17.
6S
−yi C xj · dS D
C
3 2π
0
D 81
4 2π
0
27 sin3 s, 0, 0 · 3 cos s, 3 sin s, 0 ds dt
0
t 3 sin3 s, 0, 0 · 0, 0, −t ds dt C
0
0
4 2π
0
3 2π
81
4
sin3 s cos s ds dt D
0
4
t 3 sin3 s, 0, 0 · 0, 0, t ds dt
0
2π
dt D 0.
sin4 s
0
0
Again, a careful student should have noticed that there is no k component and so the integrals over S2 and S3 are
each 0. Therefore, a different approach would be to calculate
6S
−yi C xj · dS D
18.
6S
x2 i · dS D
C
3 2π
0
t cos s, 0, 0 · 0, 0, −t ds dt C
0
3 2π
2
cos s ds dt D 27
4 2π
3
0
6S1
0 dS D 0.
9 cos2 s, 0, 0 · 3 cos s, 3 sin s, 0 ds dt
0
4 2π
−yi C xj · xi C yj/3 dS D
0
2
0
D 27
4 2π
6S1
0
t 2 cos2 s, 0, 0 · 0, 0, t ds dt
0
1 − sin s cos s ds dt D 27
4
2
0
0
0
2π
[sin s − sin3 s/3]
dt D 0.
0
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“runall” — 2005/8/1 — 15:18 — page 335 — #15
Section 7.2
Surface Integrals
335
Again, a careful student should have noticed that there is no k component and so the integrals over S2 and S3 are
each 0.
We calculate the flux from
6S
F · n dS D
FXs, t · Ns, t ds dt. For Exercises 19–22 we have that
6S
Xϕ, θ D a sin ϕ cos θ, a sin ϕ sin θ, a cos ϕ
for 0 … θ < 2π and 0 … ϕ … π,
Tϕ ϕ, θ D a cos ϕ cos θ, a cos ϕ sin θ, −a sin ϕ,
Tθ ϕ, θ D −a sin ϕ sin θ, a sin ϕ cos θ, 0,
and
Nϕ, θ D Tϕ ϕ, θ * Tθ ϕ, θ D a sin ϕ cos θ, a2 sin2 ϕ sin θ, a2 cos ϕ sin ϕ
2
2
D a2 sin ϕsin ϕ cos θ, sin ϕ sin θ, cos ϕ.
19.
6S
yj · n dS D
2π π/2
0
D
2π π/2
a sin ϕ sin θ dϕ dθ D a
3
0
D a3
3
2
0
2π
0
D
0, a sin ϕ sin θ, 0 · a2 sin2 ϕ cos θ, a2 sin2 ϕ sin θ, a2 cos ϕ sin ϕ dϕ dθ
0
2a
3
0
cos3 ϕ
3
0
1 − cos2 ϕ sin ϕ sin2 θ dϕ dθ
0
π/2
− cos ϕ C
2π
3
0
2π π/2
3
sin2 θ dϕ dθ D
1 − cos 2θ
2a3 θ
sin 2θ
dθ D
−
2
3 2
4
2a3
3
2π
0
2π
D
0
sin2 θ dθ
2πa3
.
3
20.
6S
yi − xj · n dS
D
2π π/2
0
Da
a sin ϕ sin θ, −a sin ϕ cos θ, 0 · a2 sin2 ϕ cos θ, a2 sin2 ϕ sin θ, a2 cos ϕ sin ϕ dϕ dθ
0
2π π/2
3
0
sin3 ϕ sin θ cos θ − sin3 ϕ cos θ sin θ dϕ dθ D 0.
0
Actually it is simpler not to resort to the parametrization. Since n D xi C yj C zk/a for the sphere we see
that yi − xj · n D 0 and so
21.
6S
6S
yi − xj · n dS D 0.
−yi C xj − k · n dS D −
k · n dS −
yi − xj · n dS
6S
6S
D−
k · n dS since, by Exercise 20,
yi − xj · n dS D 0
6S
6S
2π π/2
0, 0, 1 · a2 sin2 ϕ cos θ, a2 sin2 ϕ sin θ, a2 cos ϕ sin ϕ dϕ dθ
D−
0
D −a
0
2π π/2
2
0
a2
D−
2
2π
cos ϕ sin ϕ dϕ dθ D −a
0
2π
2
0
π/2
sin2 ϕ
2 0
dθ
dθ D −πa2 .
0
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336
Chapter 7
Surface Integrals and Vector Analysis
22.
6S
x2 i C xyj C xzk · n dS
D
π/2 2π
[a2 sin2 ϕ cos2 θ, a2 sin2 ϕ cos θ sin θ, a2 cos ϕ sin ϕ cos θ
0
0
· a2 sin2 ϕ cos θ, a2 sin2 ϕ sin θ, a2 cos ϕ sin ϕ] dθ dϕ
D a4
D a4
D a4
π/2 2π
0
0
0
0
π/2 2π
π/2 2π
0
sin4 ϕ cos3 θ C sin4 ϕ cos θ sin2 θ C cos2 ϕ sin2 ϕ cos θ dθ dϕ
sin4 ϕ cos θ C sin2 ϕ cos2 ϕ cos θ dθ dϕ
π/2
sin2 ϕ cos θ dθ dϕ D a4
0
2π
dθ D 0.
[sin2 ϕ sin θ]
0
0
A different approach would be to see that
6S
x2 i C xyj C xzk · n dS D
D
6S
xxi C yj C zk ·
xi C yj C zk
dS
a
a2
x dS.
dS D a
6S a
6S
x
The integrand is an odd function of x which is being integrated over a region which is symmetric with respect to
x2 i C xyj C xzk · n dS D 0.
6S
23. (a) Below left is just the portion of S for 0 … z … 2 so that you can more clearly see the funnel shape. Below
right is a sketch of S.
x; therefore
(b) For the cylindrical portion of S, Xs, t D cos s, sin s, t for 0 … s < 2π and 0 … t … 1. In that case
Ts s, t D − sin s, cos s, 0, Tt s, t D 0, 0, 1, Ts s, t * Tt s, t D cos s, sin s, 0 and so the outward
pointing unit normal for this portion is n D cos s, sin s, 0 D xi C yj.
For the conical portion of S, Xs, t D t cos s, t sin s, t for 0 … s < 2π and 1 … t … 9. In that case Ts s, t D
Ts s, t * Tt s, t D t cos s,√t sin s, −t and so the outward
−t sin s, t cos s, 0, Tt s, t D cos s, sin s, 1, √
pointing unit normal for this portion is n D 1/ 2cos s, sin s, −1 D 1/ 2x/zi C y/zj − k.
(c)
2π 9
F · dS D
−yi C xj C zk · dS D
−t sin s, t cos s, t · t cos s, t sin s, −t dt ds
0
1
6S
6S
2π 1
2π 9
2π 1
2
C
− sin s, cos s, t · cos s, sin s, 0 dt ds D
−t dt ds C
0 dt ds
D
2π
0
0
−
0
9
t3
3 1
ds D
2π
0
0
−
1
0
0
1
1456π
729
C
ds D −
.
3
3
3
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“runall” — 2005/8/1 — 15:18 — page 337 — #17
Section 7.2
Surface Integrals
337
24. We know that the heat flux density H D −k§T D −k2x, 2y, 6z − 12. On the ground k D 3 and Xs, t D
t cos s, t sin s, 0 for 0 … t … 2 and 0 … s … 2π. Also, Ts s, t D −t sin s, t cos s, 0, Tt s, t D cos s, sin s, 0
and so Ns, t D 0, 0, −t. Along the glass we have k D 1 and Xs, t D t cos s, t sin s, 8 − 2t 2 for
0 … t … 2 and 0 … s … 2π. Also, Ts s, t D −t sin s, t cos s, 0, Tt s, t D cos s, sin s, −4t and therefore
Ns, t D −4t 2 cos s, −4t 2 sin s, −t. The outward normal must be −Ns, t D 4t 2 cos s, 4t 2 sin s, t.
6S
H · dS D
H · dS C
H · dS
6S1
6S2
2π 2
D
−32t cos s, 2t sin s, −12 · 0, 0, −t dt ds
0
0
−
2π 2
0
D
2π 2
0
D
0
D
−36t dt ds C
0
2π 2
2π
2t cos s, 2t sin s, 36 − 12t 2 · 4t 2 cos s, 4t 2 sin s, t dt ds
0
2π 2
0
4t 3 − 72t dt ds D
−8t 3 − 36t C 12t 3 dt ds
0
2π
0
2
[t 4 − 36t 2 ] ds
0
0
[16 − 144] ds D −256π.
0
25. (a) A sketch of the surface for a D 2 using Mathematica is:
(b) At t D 0 we have that sin t D sin 2t D 0 and so the s-coordinate curve is given by x, y, z D
a cos s, a sin s, 0. This is a circle of radius a in the xy-plane.
(c) A computer algebra system would help the following calculation. It is not difficult; it is just very easy to
drop a term here or there.
1
s
1
s
s
s
Ts s, t D cos s − sin sin t − cos sin 2t − sin s a C cos sin t − sin sin 2t ,
2
2
2
2
2
2
s
1
s
s
s
1
sin s − sin sin t − cos sin 2t C cos s a C cos sin t − sin sin 2t ,
2
2
2
2
2
2
s
1
s
1
cos sin t − sin sin 2t ,
2
2
2
2
so
Ts s, 0 D −a sin s, a cos s, 0.
Tt s, t D cos s cos
s
s
s
s
cos t − 2 cos 2t sin
, sin s cos cos t − 2 cos 2t sin
,
2
2
2
2
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“runall” — 2005/8/1 — 15:18 — page 338 — #18
338
Chapter 7
Surface Integrals and Vector Analysis
2 cos
s
s
cos 2t C cos t sin 2
2
Tt s, 0 D cos s cos
so
s
s
s
s
s
s
− 2 sin
, sin s cos − 2 sin
, 2 cos C sin .
2
2
2
2
2
2
Calculate the cross product Ts s, 0 * Tt s, 0 to obtain
Ns, 0 D a cos s 2 cos
s
s
s
s
s
s
C sin
, a sin s 2 cos C sin
, 2a sin − a cos .
2
2
2
2
2
2
We note that
X0, 0 D a, 0, 0 D X2π, 0
but
N0, 0 D 2a, 0, −a
while N2π, 0 D −2a, 0, a.
When you travel around the s-coordinate curve at t D 0 once, you find that the normal vector is now pointing
in the opposite direction. The conclusion is that the Klein bottle cannot be orientable.
7.3 STOKES’S AND GAUSS’S THEOREMS
Exercises 1–4 are similar to Example 1 from the text. Recall from Section 2.6 that if S is a surface in R3 defined by an
equation of the form f x, y, z D c, then if x0 ∈ X, the gradient vector §f x0 is a vector normal to the plane tangent
to S at x0 .
1. Calculate
i
j
k
§ * F D ⭸/⭸x ⭸/⭸y
D 2y − yi C −2x C xj D yi − xj.
⭸/⭸z
xz
yz
x2 C y 2
By symmetry we can see that the integral will be zero; however, let’s follow the instructions. View the surface
as a level set at height 1 of f x, y, z D x2 C y2 C 5z Then N D §f D 2xi C 2yj C 5k. So,
6S
§ * F · dS D
D
6D
6D
yi − xj · 2xi C 2yj C 5k dx dy
2xy − 2xy dx dy D 0.
On the other hand, ⭸S consists of C D {x, y, z|x2 C y2 D 1 and z D 0} which we parametrize by xt D
cos t, sin t, 0. Then,
⭸S
F · ds D
2π
Fxt · x t dt D
0
2π
0, 0, 1 · − sin t, cos t, 0 dt D 0.
0
These two answers agree.
2. S is a helicoid. We begin by calculating
§ * FD
i
⭸/⭸x
z
j
⭸/⭸y
x
k
⭸/⭸z
y
D i C j C k.
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“runall” — 2005/8/1 — 15:18 — page 339 — #19
Section 7.3
Stokes’s and Gauss’s Theorems
339
We calculated a normal vector in Exercise 20 of Section 7.1: N D sin t, − cos t, s. So,
6S
§ * F · dS D
i C j C k · sin ti − cos tj C sk dt ds
6D
1
1 π/2
π
s ds
sin t − cos t C s dt ds D
D
0 0
0 2
1
D
π
π 2
s D .
4 0
4
On the other hand, ⭸S consists of four pieces which we parametrize by x1 s D s, 0, 0 for 0 … s … 1, x2 t D
cos t, sin t, t for 0 … t … π/2, x3 s D 0, 1 − s, π/2 for 0 … s … 1, and x4 t D 0, 0, π/2 − t for
0 … t … π/2. Then,
F · ds D
⭸S
1
0
1
C
D
π/2
0, s, 0 · 1, 0, 0 ds C
1
π/2
π/2, 0, 1 − s · 0, −1, 0 ds C
0
0 ds C
π/2
0
D
t, cos t, sin t · − sin t, cos t, 1 dt
0
π/2 − t, 0, 0 · 0, 0, −1 dt
0
−t sin t C cos2 t C sin t dt C
0
1
0
0 ds C
π/2
0 dt
0
π
.
4
These two answers agree.
3. We see that
§ * FD
i
⭸/⭸x
x
j
⭸/⭸y
y
k
⭸/⭸z
z
D0
so
6S
§ * F · dS D 0.
On the other hand, ⭸S consists of C D {x, y, z|y2 C z2 D 16 and x D 0} which we parametrize by
xt D 0, 4 cos t, 4 sin t for 0 … t … 2π. Then,
⭸S
F · ds D
2π
Fxt · x t dt D
0
2π
0, 4 cos t, 4 sin t · 0, −4 sin t, 4 cos t dt D 0.
0
These two answers agree.
4. For S,
§ * FD
i
j
⭸/⭸x
⭸/⭸y
2y − z x C y2 − z
k
⭸/⭸z
4y − 3x
D 4 − −1i C 3 − 1j C 1 − 2k D 5i C 2j − k.
If we parametrize S by Xs, t D 2 cos s sin t, 2 sin s sin t, 2 cos t, a downward normal vector is given by
N D 4 cos s sin2 t, 4 sin s sin2 t, 4 sin t cos t. So,
6S
§ * F · dS D
5i C 2j − k · 4 cos s sin2 ti C 4 sin s sin2 tj C 4 sin t cos tk ds dt
6D
π 2π
D
20 cos s sin2 ti C 8 sin s sin2 tj − 4 sin t cos tk ds dt
π/2 0
π
D
4π sin2t dt D 4π.
π/2
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“runall” — 2005/8/1 — 15:18 — page 340 — #20
340
Chapter 7
Surface Integrals and Vector Analysis
On the other hand, ⭸S consists of C D {x, y, z|y2 C z2 D 4 and z D 0} which we parametrize by xt D
2 cos t, −2 sin t, 0. Then,
2π
F · ds D
Fxt · x t dt
⭸S
0
D
2π
−4 sin t, 2 cos t C 4 sin2 t, −8 sin t − 6 cos t · −2 sin t, −2 cos t, 0 dt
0
D
2π
8 sin2 t − 4 cos2 t − 8 sin2 t cos t dt
0
D
2π
8 − 61 C cos 2t − 8 sin2 t cos t dt
0
D
2π
2 − 6 cos 2t − 8 sin2 t cos t dt D 4π.
0
These two answers agree.
5. Stokes’s Theorem implies that we don’t need to be concerned that S is defined as the union of S1 and S2 if
we choose the calculation along the boundary. Then ⭸S is parametrized by xt D 3 cos t, 3 sin t, 0 where
0 … t … 2π, and so
2π
F · ds D
Fxt · x t dt
⭸S
0
D
2π
0
D −34
27 cos3 ti C 37 sin7 tj · −3 sin t, 3 cos t, 0 dt
2π
cos3 t sin t dt C 38
0
2π
sin7 t cos t dt D 0.
0
6. Note that § · F D 3 so
l
D
§ · F dV D 3
D3
l
D
3
dV D 3
3 √ 9−x2
9 − x
√
−3 − 9−x2
2
− y2 dy dx
4
243π
9 − x2 3/2 dx D
.
3
2
−3
On the other hand, the boundary of D is in two pieces: S1 D the disk at height z D 0 and S2 D the portion
of the paraboloid about the xy-plane. Parametrize S1 by X1 s, t D t cos s, t sin s, 0 for 0 … s … 2π and
0 … t … 3. Then N1 s, t D 0, 0, −t. Also parametrize S2 by X2 s, t D t cos s, t sin s, 9 − t 2 . Then
N2 s, t D 2t 2 cos s, 2t 2 sin s, t. So
#S
F · dS D
F · dS C
F · dS
6S1
6S2
2π 3
D
t cos s, t sin s, 0 · 0, 0, −t dt ds
0
C
0
2π 3
0
D
2π 3
0
D
2π
0
t cos s, t sin s, 9 − t 2 · 2t 2 cos s, 2t 2 sin s, t dt ds
0
9t C t 3 dt ds D
0
2π
0
3
9t 2
t4
C ds
2
4 0
243π
243
ds D
.
4
2
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“runall” — 2005/8/1 — 15:18 — page 341 — #21
Section 7.3
Stokes’s and Gauss’s Theorems
341
These two answers agree.
7. Here § · F D 0 so
§ · F dV D 0. As for the integral over the surface, because the normal vectors of each
l
D
of the three opposite pairs of sides are equal and opposite, everything will cancel. So
#S
F · dS
D
6top
C
C
D
6front
6right
6top
C
y − x, y − 1, x − y · 0, 0, 1 dS C
y − 1, y − z, 1 − y · 1, 0, 0 dS C
1 − x, 1 − z, x − 1 · 0, 1, 0 dS C
x − y dS C
6right
6bottom
1 dS C
6bottom
6left
y − x dS C
6front
y − x, y − −1, x − y · 0, 0, −1 dS
6back
6left
y − −1, y − z, −1 − y · −1, 0, 0 dS
−1 − x, −1 − z, x − −1 · 0, −1, 0 dS
−1 dS C
6back
−1 dS
1 dS D 0.
These two answers agree.
8. Note that § · F D 2x C 2 so
l
D
§ · F dV D 2
D2
x C 1 dV
l
D
2 √ 4−x2 5
D2
8
D
3
√
−2 − 4−x2
√
2 4−x2
x2 Cy2 C1
x C 1 dz dy dx
[x C 14 − x
√
−2 − 4−x2
2
−2
2
− y2 ] dy dx
[x C 14 − x2 3/2 ] dx D
8
6π D 16π.
3
On the other hand, the boundary of D can be split into two pieces: the flat top piece S1 and the surface of
the paraboloid S2 . A parametrization of S1 is X1 s, t D t cos s, t sin s, 5 for 0 … s … 2π and 0 … t … 2.
Then a normal vector is N1 s, t D 0, 0, t. A parametrization of S2 is X2 s, t D t cos s, t sin s, t 2 C 1 for
0 … s … 2π and 0 … t … 2. Then a normal vector is N2 s, t D 2t 2 cos s, 2t 2 sin s, −t. So,
#S
F · dS D
F · dS C
F · dS
6S1
6S2
2π 2
D
t 2 cos2 s, t sin s, 5 · 0, 0, t dt ds
0
C
0
2π 2
0
D
2π 2
t 2 cos2 s, t sin s, t 2 C 1 · 2t 2 cos s, 2t 2 sin s, −t dt ds
0
2t cos s C 2t sin s − t C 4t D
4
0
0
3
3
2
2π
3
0
8 C
64 cos3 s
− 4 cos 2s ds D 16π.
5
These two answers agree.
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“runall” — 2005/8/1 — 15:18 — page 342 — #22
342
Chapter 7
Surface Integrals and Vector Analysis


y 2 C z2
⭸ 
x


D
9. Since
we see that
⭸x
x2 C y2 C z2 3/2
x 2 C y 2 C z2
§ · FD
lD
2x2 C y2 C z2 2
D
,
3/2
2
2
2
2
x C y C z x C y 2 C z2
2
and
2π π b
2 2
ρ sin ϕ dρ dϕ dθ
0
0
a ρ
l
D x 2 C y 2 C z2
2π π
2π π b
b
2ρ sin ϕ dρ dϕ dθ D
[ρ2 sin ϕ] dϕ dθ
D
§ · F dV D
0
0
a
D b2 − a2 dV D
2π π
0
0
0
sin ϕ dϕ dθ D b2 − a2 0
2π
a
2 dθ D 4πb2 − a2 .
0
On the other hand the boundary consists of two pieces: S1 is the sphere of radius a and S2 is the sphere of
radius b. Parametrize S1 by X1 s, t D a sin s cos t, a sin s sin t, a cos s for 0 … s … π and 0 … t … 2π.
Then a normal vector is N1 s, t D −a2 sin ssin s cos t, sin s sin t, cos s. A similar calculation for S2 yields
N2 s, t D b2 sin ssin s cos t, sin s sin t, cos s. Note that N1 is oriented pointing inward and N2 is oriented
pointing outward. Then,
#S
F · dS D
F · dS C
F · dS
6S1
6S2
2π π
1
D
a sin s cos t, a sin s sin t, a cos s · [−a2 sin ssin s cos t, sin s sin t, cos s] ds dt
0
0 a
2π π
1
C
b sin s cos t, b sin s sin t, b cos s · [b2 sin ssin s cos t, sin s sin t, cos s] ds dt
0
0 b
2π π
2π π
2
[−a sin s] ds dt C
[b2 sin s] ds dt
D
0
D
2π
0
[−2a2 ] dt C
0
2π
0
0
[2b2 ] dt D 4πb2 − a2 .
0
These two answers agree.
10. For Stokes’s theorem we assume that S is a bounded, piecewise smooth, oriented surface in R3 . To specialize
to Green’s theorem we must further assume that S is in the xy-plane. In each case we assume that the boundary
C D ⭸S consists of finitely many simple, closed curves which are oriented so that S is on the left as you traverse
C. In each case, F is a vector field of class C1 whose domain includes S. In general, this would mean that
Fx, y, z D mx, y, zi C nx, y, zj C px, y, zk but because S is planar we assume that F is independent of
z and that its k-component is identically zero. In other words, we take Fx, y, z D Mx, yi C Nx, yj. Then
6S
⭸N
⭸M
§ * F · dS.
−
dx dy D
⭸x
⭸y
6S
But by Stokes’s theorem,
6S
§ * F · dS D
⭸S
F · ds.
Then, by the formula for the differential form of the line integral given in Section 6.1,
F · ds D
M dx C N dy.
⭸S
C
And so we get Green’s theorem from Stokes’s theorem.
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“runall” — 2005/8/1 — 15:18 — page 343 — #23
Section 7.3
Stokes’s and Gauss’s Theorems
343
11. Begin by calculating
§ * FD
i
⭸/⭸x
2xyz C 5z
j
k
⭸/⭸y
⭸/⭸z
e x cosyz x2 y
D x2 C e x y sinyzi C 5j C e x cosyz − 2xzk.
As in Example 2, we see that this looks difficult, but that Stokes’s theorem implies that
6S
§ * F · dS D
6S1
§ * F · dS
where S and S1 have the same boundary. So let S1 be the disk in the y D 1 plane bounded by the circle
x2 C z2 D 9. The rightward pointing unit normal to S1 is (0, 1, 0) and so
6S
§ * F · dS D
D
D
6S1
6S1
6S1
§ * F · dS
x2 C e x y sinyzi C 5j C e x cosyz − 2xzk · 0, 1, 0 dS
5 dS D 5area of S1 D 5π32 D 45π.
12. The boundary of S is the ellipse 4x2 C y2 D 4 in the z D 0 plane. By Stokes’s theorem
6S
§ * F · dS D
⭸S
F · dS D
6S § * F · dS
where S is any piecewise smooth, orientable surface with ⭸S D ⭸S (subject to appropriate orientation). One
computes that
i
j
k
§ * F D ⭸/⭸x ⭸/⭸y ⭸/⭸z D xzexy i − yzexy j.
2
ey
zexy
x3
This has no k-component. So let us take for S the portion of the z D 0 plane inside the ellipse. Hence n D k
so that
6S § * F · dS D
D
6S 6S xzexy i − yzexy j · k dS
0 dS D 0.
13. (a) By the double angle formula we have z D sin 2t D 2 sin t cos t D 2xy.
y3 C cos x dx C sin y C z2 dy C x dz D
(b)
C
F · ds where F D y3 C cos xi C sin y C z2 j C xk.
C
§ * F · dS where S is the
6S
portion of z D 2xy bounded by C. (Note that S lies over the unit disk in the xy-plane.) Now § * F D
−4xyi − j − 3y2 k D −2zi − j − 3y2 k on S. Note that the orientation of C is compatible with an upward
orientation of S. So we may take for normal
By Stokes’s theorem we may calculate the line integral by evaluating
nD
Hence
6S
§ * F · dS D
6D
−2yi − 2xj C k
4x2 C 4y2 C 1
unit normal of S.
8xy2 C 2x − 3y2 dx dy D D unit disk in xy-plane.
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“runall” — 2005/8/1 — 15:18 — page 344 — #24
344
Chapter 7
Surface Integrals and Vector Analysis
Now use polar coordinates, so that the integral becomes
2π 1
8r3 sin2 θ cos θ C 2r cos θ − 3r2 sin2 θr dr dθ
D
0
0
0
8 2
2
3 1
sin θ cos θ C cos θ − 1 − cos 2θ dθ
5
3
4 2
2π
2π
D
3
3
2
3π
8
sin3 θ C sin θ − θ C
sin 2θ
D− .
15
3
8
16
4
0
14. S is the portion of the “bell” surface for which z D e1−x −y and z Ú 1. Take S2 to be the disk in the plane
z D 1 bounded by the circle x2 C y2 D 1. Then S ! S2 is the boundary of a solid V . S is oriented with an
upward pointing normal and S2 is oriented with a downward pointing normal.
2
§ · FD0
so
l
V
2
§ · F dV D 0.
Also,
6S2
But along S2 , z D 1, so
F · dS D
6S2
6S2
x, y, 2 − 2z · 0, 0, −1 dS D
2z − 2 dS D
6S
F · dS D
l
V
6S2
6S2
2z − 2 dS.
2 − 2 dS D 0. So
§ · F dV −
6S2
F · dS D 0 − 0 D 0.
15. Let X : D ' R3 , Xu, v D xu, v, yu, v, zu, v parametrize S and (ut, vt), a … t … b parametrize ⭸D
so that X(ut, vt) parametrizes ⭸S. (Note the assumption that ⭸D can be parametrized by a single path—this
is not a problem.) Write F as Mi C Nj C Pk. We need to show that
∗
⭸S
Mi C Nj C Pk · ds D
6S
⭸P
⭸N ⭸M
⭸P ⭸N
⭸M
−
,
−
,
−
· dS.
⭸y
⭸z ⭸z
⭸x ⭸x
⭸y
Consider the line integral in (∗). We may write it in differential form as
⭸S
M dx C N dy C Pdz. Consider, for
⭸x du
⭸x dv
dx
D
C
. Hence,
dt
⭸
u
dt
⭸v dt
⭸S
b
⭸x du
⭸x dv
⭸x
⭸x
C
du C M ◦ X dv.
M dx D
MXut, vt M ◦ X
dt D
⭸u dt
⭸v dt
⭸u
⭸v
⭸S
⭸D
a
the moment, just the piece
M dx. By the chain rule,
The last line integral in just an integral in the uv -plane and so we may apply Green’s theorem to find
⭸S
M dx D
⭸
⭸x
⭸
⭸x
M ◦ X du dv.
M ◦ X −
⭸v
⭸v
⭸u
6
D ⭸u
We need to apply the chain rule again, along with the product rule:
⭸
⭸x
⭸M ⭸x
⭸M ⭸y
⭸M ⭸z ⭸x
⭸2 x
C
C
C M ◦ X
M ◦ X D ⭸u
⭸v
⭸x ⭸u
⭸y ⭸u
⭸z ⭸u ⭸v
⭸u ⭸v
⭸
⭸x
⭸M ⭸x
⭸M ⭸y
⭸M ⭸z ⭸x
⭸2 x
C
C
C M ◦ X
.
M ◦ X D ⭸v
⭸u
⭸x ⭸v
⭸y ⭸v
⭸z ⭸v ⭸u
⭸v ⭸u
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“runall” — 2005/8/1 — 15:18 — page 345 — #25
Section 7.3
Stokes’s and Gauss’s Theorems
Since the exercise allows us to assume that X is of class C2 , the mixed partials are equal:
Therefore, our double integral becomes, after cancellation,
345
⭸2 x
⭸2 x
D
.
⭸u⭸v
⭸v⭸u
⭸M ⭸x ⭸y
⭸x ⭸y
⭸M ⭸x ⭸z
⭸x ⭸z
−
−
C
du dv.
⭸u ⭸v
⭸z ⭸v ⭸u
⭸u ⭸v
6D ⭸y ⭸v ⭸u
∗∗
Now consider the surface integral in (∗). Using the parametrization X, and calculating the normal, we have that
it is equal to
6D
Py − Nz , Mz − Px , Nx − My · yu zv − yv zu , zu xv − zv xu , xu yv − xv yu du dv.
Next, calculate the dot product and isolate just those terms that contain M. Then the piece of the surface integral
in (∗) that involves just M is
⭸M ⭸z ⭸x
⭸z ⭸x
⭸M ⭸x ⭸y
⭸x ⭸y
−
−
−
du dv.
⭸
z
⭸
u
⭸
v
⭸
v
⭸
u
⭸
y
⭸
u
⭸
v
⭸v ⭸u
6D
This is the same as the double integral in (∗∗).
"
"
In an entirely analogous way, we may show that ⭸S N dy and ⭸S Pdz are equal to the remaining pieces of the
surface integral in (∗), completing the proof.
§ · F dV for the closed cube and then subtract
F · dS where S2 is the bottom. Orient
#S2
l
V
all of the faces of the cube with an outward pointing normal. In particular this means that the normal to S2 is
downward pointing.
1 1 1
2
§ · F dV D
2xze x C 3 − 7yz6 dx dy dz
0 0 0
lV
1 1
1 1
xD1
2
[zex C 3x − 7xyz6 ]
dy dz D
[ez − z C 3 − 7yz6 ] dy dz
D
16. We will calculate
0
D
1
0
0
xD0
ezy − zy C 3y −
0
7 2 6
y z
2
yD1
dz D
1
0
ez − z C 3 −
0
yD0
7 6
z dz
2
1
D
Also,
#S2
F · dS D
z2
1
e
z2
e −
C 3z − z7 D C 2.
2
2
2
2
0
1 1
0
0, 3y, 2 · 0, 0, −1 dy dx D
0
1 1
0
6S
F · dS D −2 dy dx D −2. Therefore,
0
e
e
C 2 − −2 D C 4.
2
2
17. The proof is outlined in the proof of Theorem 3.5 of Chapter 6. One direction has already been proved in
Theorem 4.3 of Chapter 3. There it was established that if F D §f , then § * F D 0. Now suppose that
§ * F D 0. We show that then
F · ds D 0 where C is any piecewise C1 , simple closed curve in R 8 R3 . The
C
idea is to “fill in C”, that is, to find a surface S 8 R whose boundary is C. Since R is simply-connected, this is
possible. If we orient S consistently with C, then we may apply Stokes’s theorem to conclude
F · ds D
C
6S
§ * F · dS D
6S
0 · dS D 0.
This shows, among other things, that F has path-independent integrals over curves in R. Therefore, by Theorem 3.3
of Chapter 6, F D §f for some function f on R.
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“runall” — 2005/8/1 — 15:18 — page 346 — #26
346
Chapter 7
Surface Integrals and Vector Analysis
18. (a) Note that the boundary of D is made up of two components:
• S5 D the sphere centered at the origin of radius 5 oriented with outward pointing normal and
• S7 D the sphere centered at the origin of radius 7 oriented with outward pointing normal.
Then by Gauss’s theorem
l
D
§ · F dV D
#S7
F · dS −
#S5
F · dS D 7a C b − 5a C b D 2a.
(b) By Theorem 4.4 of Section 3.4, § · § * G D 0. So if D is the solid sphere centered at the origin with
radius r then, since F D § * G,
ar C b D
D
#Sr
F · dS next apply Gauss’s theorem
l
D
§ · F dV D
l
D
§ · § * G dV D 0.
Therefore ar C b D 0 for all values of r. We conclude that a D b D 0.
2xi C 2yj C 2zk
2xi C 2yj C 2zk
19. (a) If f x, y, z D lnx2 C y2 C z2 then §f x, y, z D
D
on S.
x 2 C y 2 C z2
a2
2xi C 2yj C 2zk
D
Also, the unit normal to the sphere that points away from the origin is nx, y, z D
4x2 C y2 C z2 2xi C 2yj C 2zk
xi C yj C zk
D
. So,
2a
a
⭸f
dS D
§f · n dS
6S
6S ⭸n
(b) First calculate that § · §f D
l
D
D
2xi C 2yj C 2zk xi C yj C zk
·
dS
a
6S x2 C y2 C z2
D
2
2
dS D surface area of S
a
6S a
D
2 4πa2
D πa.
a
8
2
. We’ll use spherical coordinates to integrate.
x 2 C y 2 C z2
π/2 π/2 a
2
2
2
dV D
2 ρ sin ϕ dρ dϕ dθ
2 C y 2 C z2
x
ρ
0
0
0
l
D
π/2 π/2 a
π/2 π/2
D2
sin ϕ dρ dϕ dθ D 2a
sin ϕ dϕ dθ
§ · §f dV D
0
π/2
D 2a
0
0
0
0
dθ D πa.
0
§ · §f dV D
§f · dS. The boundary of D consists of four pieces: S, the
#⭸D
l
D
surface from part (a); Sx , the intersection of D and the plane x D 0I Sy , the intersection of D and the plane
2yj C 2zk
y D 0; and Sz , the intersection of D and the plane z D 0. On Sx we know that §f 0, y, z D
y 2 C z2
and n D −1, 0, 0 so
(c) By Gauss’s theorem,
6Sx
§f · dS D
6Sx
§f · n dS D
6Sx
0 dS D 0.
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“runall” — 2005/8/1 — 15:18 — page 347 — #27
Section 7.3
A similar analysis gives us
l
D
20. By Gauss’s Theorem,
l
D
6Sy
§f · dS D 0 and
§ · §f dV D
§ · §f dV D
#⭸D
#⭸D
6Sz
Stokes’s and Gauss’s Theorems
347
§f · dS D 0. Therefore,
§f · dS D
6S
§f · dS D
⭸f
dS.
6S ⭸n
§f · dS. Here the boundary of D consists of finitely many
§ · §f dV D 0.
lD
⭸2 f
⭸2 f
⭸2 f
C
C
.
This is true for any solid D, so § · §f D 0. As we saw earlier in the text § · §f D
⭸x2
⭸y2
⭸z2
So f is harmonic.
21. For Sa , the outward unit normal is given by n D x, y, z/a. Let Da be the solid ball whose boundary is Sa .
Then by the integral formula from Proposition 3.4,
piecewise smooth, closed orientable surfaces Si . By assumption,
3
div FP D lim
a'0 4πa3 #
S
3
F · dS D lim
a'0 4πa3 l
D
a
3
D lim
a'0 4πa3 l
D
#⭸Si
a
§f · dS D 0 and so
§ · F dV
by Gauss’s theorem
a
⭸F1 x, y, z
⭸F2 x, y, z
⭸F3 x, y, z
C
C
dV .
⭸x
⭸y
⭸z
⭸F1 x, y, z
⭸F2 x, y, z
⭸F3 x, y, z
C
C
. By the mean value theorem for triple integrals,
⭸x
⭸y
⭸z
there is a point Qa ∈ Da such that
Let Gx, y, z D
l
Da
Gx, y, z dV D GQa · volume of Da D
4 3
πa GQa .
3
Therefore,
3
div FP D lim
a'0 4πa3 #
S
F · dS D lim
a
D GP D 3
a'0 4πa3
4 3
πa GQa D lim GQa a'0
3
⭸F1
⭸F2
⭸F3
C
C
.
⭸x
⭸y
⭸z x,y,zDP
22. We will shrink the region D specified in the problem down to a point P. The volume decreases monotonically as
we shrink the solid. Let DV be the shrunken version of D which is the solid of volume V and let SV D ⭸DV for
0 … V … the volume of D. Then, by Gauss’s theorem,
#SV
F · dS D
l
DV
§ · F dV .
By the mean value for triple integrals, there exists a QV ∈ DV so that
lDV
§ · F dV D
lDV
§ · FQV dV D § · FQV volume of D.
So
lim
1
V '0 V #
S
V
F · dS D lim § · FQV D § · FP D div FP.
V '0
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“runall” — 2005/8/1 — 15:18 — page 348 — #28
348
Chapter 7
Surface Integrals and Vector Analysis
23. (a) F D Fr er C Fθ eθ C Fz ez . The area of the top face is
θ/2π[πr C r/22 − πr − r/22 ] D θ/22rr D rθr.
Therefore,
6top
F · dS D
6
top
F · n dS D
6top
F · ez dS D
6top
Fz dS
L Fz r, θ, z C z/2area of top D Fz r, θ, z C z/2rθr.
The calculation for the bottom face is similar. The differences are that the normal vector points down and
Fz is evaluated at a different point. The result is that
6bottom
F · dS L −Fz r, θ, z − z/2rθr.
The area of the outer face is
zθ/2π[2πr C r/2] D θzr C r/2.
Therefore,
6outer
F · dS D
6outer
F · n dS D
6outer
F · er dS D
6outer
Fr dS
L Fr r C r/2, θ, zarea of outer D Fr r C r/2, θ, zr C r/2θz.
The calculation for the inner face is similar. The differences are that the normal vector points inward, Fr is
evaluated at a different point, and the area of the face is slightly different. The result is that
6inner
F · dS L −Fr r − r/2, θ, zr − r/2θz.
The area of either the left or right face is just rz. Therefore, the integral along the left face (looking from
the origin out at the solid) is
6left
F · dS D
6left
F · n dS D
6
left
F · eθ dS D
6left
Fθ dS
L Fθ r, θ C θ/2, zarea of left D Fθ r, θ C θ/2, zrz.
The calculation for the right face is similar. The differences are that the normal vector points the opposite
direction and Fθ is evaluated at a different point. The result is that
6right
F · dS L −Fθ r, θ − θ/2, zrz.
We sum these to obtain
#S
F · dS L Fz r, θ, z C z/2rθr − Fz r, θ, z − z/2rθr
C Fr r C r/2, θ, zr C r/2θz − Fr r − r/2, θ, zr − r/2θz
C Fθ r, θ C θ/2, zrz − Fθ r, θ − θ/2, zrz.
(b) To calculate the divergence using the results of Exercise 22 we will divide the answer to part (a) by V L
rθrz and take the limit as V ' 0. Two notes before the calculation: 1) We can replace L with D because
in the limit our approximation assumptions are true and 2) in evaluating each of the limits we use the note
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“runall” — 2005/8/1 — 15:18 — page 349 — #29
Section 7.3
Stokes’s and Gauss’s Theorems
349
given in the text (although you may want to break the argument of the middle limit down further to see what
is going on).
1
div FP D lim
V '0 V #
S
D lim
V '0
Fz r, θ, z C z/2rθr − Fz r, θ, z − z/2rθr
rθrz
C lim
Fr r C r/2, θ, zr C r/2θz − Fr r − r/2, θ, zr − r/2θz
rθrz
C lim
Fθ r, θ C θ/2, zrz − Fθ r, θ − θ/2, zrz
rθrz
V '0
V '0
D lim
z'0
Fz r, θ, z C z/2 − Fz r, θ, z − z/2
z
C lim
Fr r C r/2, θ, zr C r/2 − Fr r − r/2, θ, zr − r/2
rr
C lim
Fθ r, θ C θ/2, z − Fθ r, θ − θ/2, z
rθ
r'0
θ'0
D
F · dS
⭸Fz
1 ⭸Fθ
1 ⭸
C
rFr C
⭸z
r ⭸r
r ⭸θ
.
P
24. Follow the steps from Exercise 23. This time F D Fρ eρ C Fθ eθ C Fϕ eϕ . Again, for each face,
F · dS is
6S
approximately the product of the component of F in the normal direction evaluated at the center point of the face
and the area of that face. So summing up we have that
#S
F · dS L Fϕ ρ, θ, ϕ C ϕ/2ρ sinϕ C ϕ/2θρ − Fϕ ρ, θ, ϕ − ϕ/2ρ sinϕ − ϕ/2θρ
C Fρ ρ C ρ/2, θ, ϕρ C ρ/22 sin ϕθϕ − Fρ ρ − ρ/2, θ, ϕρ − ρ/22 sin ϕθϕ
C Fθ ρ, θ C θ/2, ϕρρϕ − Fθ ρ, θ − θ/2, ϕρρϕ.
Divide through by V L ρ2 sin ϕρθϕ and simplify to obtain
Fϕ ρ, θ, ϕ C ϕ/2 sinϕ C ϕ/2 − Fϕ ρ, θ, ϕ − ϕ/2 sinϕ − ϕ/2
1
F · dS L V #S
ρ sin ϕϕ
C C Fρ ρ C ρ/2, θ, ϕρ C ρ/22 − Fρ ρ − ρ/2, θ, ϕρ − ρ/22
ρ2 ρ
Fθ ρ, θ C θ/2, ϕ − Fθ ρ, θ − θ/2, ϕ
.
ρ sin ϕθ
Take the limit as V ' 0 to conclude
div FP D lim
1
V '0 V #
S
D lim F · dS
Fϕ ρ, θ, ϕ C ϕ/2 sinϕ C ϕ/2 − Fϕ ρ, θ, ϕ − ϕ/2 sinϕ − ϕ/2
ϕ'0
C lim ρ'0
ρ sin ϕϕ
Fρ ρ C ρ/2, θ, ϕρ C ρ/22 − Fρ ρ − ρ/2, θ, ϕρ − ρ/22
ρ2 ρ
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“runall” — 2005/8/1 — 15:18 — page 350 — #30
350
Chapter 7
Surface Integrals and Vector Analysis
Fθ ρ, θ C θ/2, ϕ − Fθ ρ, θ − θ/2, ϕ
ρ sin ϕθ
C lim θ'0
D
1
⭸
1 ⭸ 2
1 ⭸Fθ
sin ϕFϕ C
ρ Fρ C
.
ρ sin ϕ ⭸ϕ
ρ2 ⭸ρ
ρ sin ϕ ⭸θ P
25. Let F, P, n, S and C be as described in the text. As in Exercise 22, we will assume that C shrinks down to the
point P so that the area of the surface bounded decreases monotonically. We will then refer to SA and CA as the
surface and bounding curve that corresponds to area A. Then by Stokes’s theorem,
F · ds D
CA
6SA
§ * F · dS D
6SA
§ * F · n dS.
By the mean value theorem for surface integrals, there is some point QA ∈ SA such that
6SA
§ * F · n dS D § * FQA · narea of SA D § * FQA · n A.
Therefore,
lim
1
A'0 A
F · ds D lim
1
A'0 A
CA
[§ * FQA · nA] D lim § * FQA · n
A'0
D n · § * FP D n · curl FP.
1
A
26. (a) By Exercise 25, ez · curl FP D limA'0
CA
F · ds L −Fr r, θ C
F · ds. Here A L rrθ.
CA
r
θ
r
, z r − Fθ r −
, θ, z r −
θ
2
2
2
C Fr r, θ −
θ
r
r
, z r C Fθ r C
, θ, z r C
θ.
2
2
2
Therefore,
ez · curl FP D lim
1
A'0 A

F · ds
CA

θ
θ
,
z
−
F
,
z
r,
θ
C
r,
θ
−
F
%
%
r$
r$

2
2
D lim −

θ'0
rθ


r
r
r
r
F
,
θ,
z
r
C
,
θ,
z
r
−
r
C
−
F
r
−
%
$
%
$
$
%
$
%
θ
 θ
2
2
2
2 
C lim 

r'0
rr
D−
1 ⭸Fr
1 ⭸
C
rFθ .
r ⭸θ
r ⭸r
(b) Again by Exercise 25, er · curl FP D limA'0
CA
F · ds L Fz r, θ C
1
A
F · ds. Here A L r z θ.
CA
θ
z
, z z − Fθ r, θ, z C
r θ
2
2
− Fz r, θ −
θ
z
, z z C Fθ r, θ, z −
r θ.
2
2
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“runall” — 2005/8/1 — 15:18 — page 351 — #31
Section 7.3
Stokes’s and Gauss’s Theorems
351
Therefore,
1
er · curl FP D lim
A'0 A

F · ds
CA

θ
θ
,
z
−
F
,
z
r,
θ
C
r,
θ
−
F
%
%
z$
 z$
2
2
D lim 

θ'0
rθ


z
z
F
r,
θ,
z
C
−
F
r,
θ,
z
−
%
%
θ$
θ$

2
2 
C lim −

z'0
z
D
1 ⭸Fz
⭸Fθ
−
.
r ⭸θ
⭸z
(c) Again by Exercise 25, eθ · curl FP D limA'0
CA
F · ds L Fz r −
1
A
F · ds. Here A D rz.
CA
r
z
, θ, z z C Fr r, θ, z C
r
2
2
− Fz r C
r
z
, θ, z z − Fr r, θ, z −
r.
2
2
Therefore,
eθ · curl FP D lim
1
A'0 A

F · ds
CA

r
r
,
θ,
z
−
F
,
θ,
z
F
r
C
r
−
%
%
$
$
z
z


2
2
D lim −

r'0
r


z
z
F
r,
θ,
z
C
−
F
r,
θ,
z
−
$
%
$
%
r
r

2
2 
C lim 

z'0
z
D−
⭸Fz
⭸Fr
C
.
⭸r
⭸z
The final conclusion is just a matter of putting the three pieces together and checking that the sum agrees
with the determinant given.
1
27. This is similar to Exercise 26. By Exercise 25, eρ · curl FP D limA'0
F · ds. Here A L ρ2 sin ϕϕθ.
A CA
CA
F · ds D −Fϕ ρ, θ C
C Fϕ ρ, θ −
ϕ
ϕ
θ
, ϕ ρϕ − Fθ ρ, θ, ϕ −
ρ sin ϕ −
θ
2
2
2
ϕ
ϕ
θ
, ϕ ρϕ C Fθ ρ, θ, ϕ C
ρ sin ϕ C
θ.
2
2
2
Therefore,
eρ · curl FP D lim
1
A'0 A
F · ds
CA
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“runall” — 2005/8/1 — 15:18 — page 352 — #32
352
Chapter 7
Surface Integrals and Vector Analysis


θ
θ
,
ϕ
−
F
,
ϕ
F
ρ,
θ
C
ρ,
θ
−
%
%
$
$
ϕ
ϕ


2
2
D lim −

θ'0
ρ sin ϕθ

ϕ
ϕ
ϕ
ϕ
F
ρ,
θ,
ϕ
C
sin
ϕ
C
−
F
ρ,
θ,
ϕ
−
sin
ϕ
−
$
%
$
%
$
%
$
%
θ
θ

2
2
2
2 
C lim 

ϕ'0
ρ sin ϕϕ

D
⭸Fϕ
1
⭸
C
sin ϕFθ .
−
ρ sin ϕ
⭸θ
⭸ϕ
Again, by Exercise 25, eθ · curl FP D limA'0
CA
1
A
F · ds. Here A L ρϕρ.
CA
ρ
ρ
ϕ
, θ, ϕ ρ C
ϕ − Fρ ρ, θ, ϕ C
ρ
2
2
2
F · ds D Fϕ ρ C
− Fϕ ρ −
ρ
ρ
ϕ
, θ, ϕ ρ −
ϕ C Fρ ρ, θ, ϕ −
ρ.
2
2
2
Therefore,
eθ · curl FP D lim
1
F · ds
A'0 A

CA

ρ
ρ
ρ
ρ
,
θ,
ϕ
ρ
C
,
θ,
ϕ
ρ
−
ρ
C
−
F
ρ
−
F
%
$
%
$
$
%
$
%
ϕ
ϕ

2
2
2
2 
D lim 

ρ'0
ρρ


ϕ
ϕ
ρ,
θ,
ϕ
C
−
F
ρ,
θ,
ϕ
−
F
$
%
$
%
ρ
ρ

2
2 
C lim −

ϕ'0
ρϕ
D
⭸Fρ
1 ⭸
ρFϕ −
.
ρ ⭸ρ
⭸ϕ
Again, by Exercise 25, eϕ · curl FP D limA'0
CA
F · ds D Fθ ρ −
1
A
F · ds. Here A L ρ sin ϕρθ.
CA
ρ
ρ
θ
, θ, ϕ ρ −
, ϕ ρ
sin ϕθ C Fρ ρ, θ C
2
2
2
− Fθ ρ C
ρ
ρ
θ
, θ, ϕ ρ C
, ϕ ρ.
sin ϕθ − Fρ ρ, θ −
2
2
2
Therefore,
eϕ · curl FP D lim
1
A'0 A

F · ds
CA

ρ
ρ
ρ
ρ
−F
,
θ,
ϕ
ρ
C
,
θ,
ϕ
ρ
−
ρ
C
C
F
ρ
−
%
$
%
$
$
%
$
%
θ
θ

2
2
2
2 
D lim −

ρ'0
ρρ
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“runall” — 2005/8/1 — 15:18 — page 353 — #33
Section 7.3
Stokes’s and Gauss’s Theorems
353


θ
θ
,
ϕ
−
F
,
ϕ
F
ρ,
θ
C
ρ,
θ
−
%
%
$
$
ρ
ρ


2
2
C lim 

θ'0
ρ sin ϕθ
D
⭸
1 ⭸Fρ
1
− ρFθ C
.
ρ
⭸ρ
sin ϕ ⭸θ
Again, the final conclusion is just a matter of assembling the pieces above and checking that the sum agrees with
the determinant.
28. We use the results of Exercises 22 and 25:
div FP D lim
1
V '0 V #
S
n · curl FP D lim
F · dS
1
A'0 A
F · ds.
C
The vector fields to be considered are planar, so the divergence results should actually be interpreted as
div FP D lim
1
A'0 A
F · n ds.
C
(See the discussion regarding two-dimensional flux in Section 6.2.) Here n is the outward unit normal to C that
lies in the plane. We need to find the four fields for which the divergence is identically zero. Intuitively, you can
see in figures (b) and (e) by looking at symmetric neighborhoods of the center point that at the center point the
divergence is not zero. We will be more precise than that. For the curl result, we need only take n to be the unit
vector pointing up out of the plane of the vector field. Using these results, we may categorize the vector fields by
drawing appropriate paths.
(a) Draw a rectangular path C with sides parallel to the x- and y-axes (see below left). Along such a path,
F · ds Z 0, since the path is tangent to the vector field along vertical segments and F has different
C
magnitudes along these segments. The integrals along the horizontal segments will be equal and opposite.
F · nds D 0 because F · n vanishes
This will be true in the limit, so curl F Z 0. On the other hand,
C
on vertical parts of C and has opposite sign on the two horizontal segments. Therefore, div F D 0.
C1
C2
C1
C2
(b) Draw a path contained in the upper right quarter of the diagram that is a “polar rectangle” (see above center).
In other words, we draw the path so that two of the sides are tangent to the vector field (one in the same
direction, one in the opposite direction) and the remaining two sides are sides each of whose distance to the
center of the figure is constant. Note that
once the path is oriented, the segments
labelled C1 and C2 will
F · n ds >
radial segments. Therefore, div F Z 0. On the other hand,
receive “opposite” orientations. Here
C1
C1
F · n ds and
F · n ds D 0 along the
F · ds D
F · ds D 0, since F is perpendicular
C2
C2
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“runall” — 2005/8/1 — 15:18 — page 354 — #34
354
Chapter 7
Surface Integrals and Vector Analysis
to C1 and C2 . However, F · T has opposite signs on the radial pieces so
F · ds D
C
F · Tds D 0.
C
Hence curl F D 0.
(c) Again our path will be a polar rectangle (see above right). This time orient the path clockwise and picture
the center of the coordinate system to be at the center of the right border of the figure. Denote the left-most,
“vertical” side C1 and the right-most, “vertical”
way. C1 and C2 will receive
side C2 . Orient the path either
“opposite” orientations. The idea here is that
F · ds is cancelled by
F · ds because the integral of
C1
C2
the smaller magnitude of F along the longer segment C1 is balanced by the integral of the larger magnitude
of F along the shorter segment C2 . Integrals along the other segments are 0 because F is perpendicular to
those segments. Hence, curl F D 0. The path is also arranged so
F · n ds D 0. It is zero along C1 and
C
C2 and cancels on the other segments. Hence, div F D 0.
(d) Again choose a polar rectangle for our path (see below left). This time picture the center of the coordinate
system to be at the center of the left border of the figure. What makes this different from the vector field in
F · ds Z 0 and, therefore, curl F Z 0. On the other hand,
(c) is that here ||F|| is constant. For this reason,
C
div F D 0 for the same reasons as in part (c).
(e) Let our path be an oriented circle centered at the center of the figure (see below center). It is clear that
F · nds Z 0 and therefore div F Z 0. Likewise,
F · ds Z 0, so curl F Z 0.
C
C2
C
C1
(f) Well, by elimination we must have div F D 0 and curl F D 0. For the divergence argument, choose a
rectangular path in the upper right quarter of the diagram with two sides parallel to and symmetric about the
diagonal from the lower left corner to the upper right corner of the diagonal. For the curl argument, use a
rectangular path with sides parallel to the coordinate axes (see above right).
7.4 FURTHER VECTOR ANALYSIS; MAXWELL’S EQUATIONS
1. Notice the similarities between this exercise and Exercise 24 in Section 6.5. By Gauss’s theorem (Theorem 3.3),
#⭸D
f §g · dS D
l
D
§ · f §g dV .
By the product rule,
lD
§ · f §g dV D
l
D
§f · §g C f §2 g dV D
l
D
§f · §g dV C
l
D
f §2 g dV .
2. Let f K 1 in Green’s first formula. Then §f D 0 so the first term in Green’s first formula is 0, so
l
D
§2 g dV D
#S
§g · dS.
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“runall” — 2005/8/1 — 15:18 — page 355 — #35
Section 7.4
Further Vector Analysis; Maxwell’s Equations
355
We assumed that g is harmonic so §2 g D 0. Also we know that S D ⭸D. Therefore, by the definition of the
normal derivative,
⭸g
0D
dS.
§g · dS D
§g · n dS D
#⭸D
#⭸D
#⭸D ⭸n
3. (a) Using Green’s first formula with f D g, we obtain
lD
§f · §f dV C
l
D
f §2 f dV D
#S
f §f · dS.
We are assuming that f is harmonic, so the second integral on the left side is 0. Therefore,
l
D
§f · §f dV D
#⭸D
f §f · dS D
#⭸D
f §f · n dS D
#
⭸D
f
⭸f
dS.
⭸n
(b) If f D 0 on the boundary of D, then part (a) implies that
0D
⭸f
dS D
f
§f · §f dV .
#⭸D ⭸n
l
D
But §f · §f D ||§f ||2 Ú 0. So the right-hand integral was of a non-negative, continuous integrand. For
this to be zero, the integrand must have been identically zero. In other words, §f · §f is zero on D. We
conclude that §f is zero on D and so f is constant on D. Since f x, y, z D 0 on ⭸D and f is constant on
D, we must have that f K 0 on D.
4. Use the hint and consider f D f1 − f2 . Then, since f1 D f2 on ⭸D, we have that f D 0 on ⭸D. Note that if f1
and f2 are harmonic on D, then f is harmonic on D. Therefore, by Exercise 3(b), f K 0 on all of D so f1 D f2
on D.
⭸ρ
5. (a) Using the hint we see that the rate of fluid flowing into W is
dV and the rate of fluid flowing out
l
W ⭸t
⭸ρ
ρF · dS. Hence we have
ρF · dS D −
of W is
dV . Also, by Gauss’s theorem, we have
#S
#S
l
W ⭸t
⭸ρ
§ · ρFdV D
ρF · dS; therefore
§ · ρFdV D −
dV . Finally, as in the arguments
#S
l
l
l
W
W
W ⭸t
in the text, we point out that the equation
⭸ρ
§ · ρF dV D −
dV
l
lR ⭸t
R
holds for any solid region R 8 W by the same argument. Thus, by shrinking R to a point, we can conclude
⭸ρ
§ · ρF D − .
⭸t
⭸ρ
(b) From (14) in the text, the current density field J is ρv. Therefore,
dV represents the current flowing
l
W ⭸t
J · dS represents the current flowing out of W (across S). Hence, the same argument as that
#S
⭸ρ
given in part (a) shows that § · J D − .
⭸t
⭸T
6. We are given that the total heat leaving D per unit time is −
σρ dV . This is equal to the flux
H · dS
⭸t
#S
lD
into W and
− k§T · dS. By Gauss’s theorem, we have
− k§T · dS D
#
#S
S
⭸T
− k§ · §T dV . Therefore, −
σρ dV D
− k§ · §T dV . Since D is arbitrary, shrink it to a
⭸t
lD
lD
l
D
⭸T
⭸T
point to conclude that −σρ
D −k§ · §T or σρ
D k§ · §T .
⭸t
⭸t
which, by the definition of H, is the same as
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“runall” — 2005/8/1 — 15:18 — page 356 — #36
356
Chapter 7
Surface Integrals and Vector Analysis
⭸T
7. We know from the argument in Exercise 6 that −
σρ dV D
− § · k§T dV . Use the product
⭸t
l
l
D
D
rule to conclude that this equals
l
D
− §k · §T C k§2 T dV . As before, shrink to a point to conclude
⭸T
D k§2 T C §k · §T .
⭸t
8. This is immediate from the heat equation since ⭸T /⭸t D 0 and σ, ρ, k are constants.
9. (a)
σρ
#⭸D
H · dS D
#⭸D
D −k
− k§T · dS D
lD
l
D
§2 T dV D 0
§ · −k§T dV
by Gauss’s theorem
by Exercise 8.
(b) By part (a), there can be no net inflow or outflow of heat. Thus, heat must be flowing into D from the inner
(hotter) sphere and out of D through the outer sphere at the same rate.
10. (a) Since w D T1 − T2 , §2 w D §2 T1 − T2 . But T1 and T2 each satisfy the heat equation given in the
exercise, so
⭸T
⭸T2
⭸
⭸w
§2 w D §2 T1 − T2 D 1 −
D T1 − T2 D
.
⭸t
⭸t
⭸t
⭸t
So w satisfies the heat equation. Now for x, y, z ∈ D we have
wx, y, z, 0 D T1 x, y, z, 0 − T2 x, y, z, 0 D αx, y, z − αx, y, z D 0.
So the first condition holds. Also for all x, y, z ∈ ⭸D and t Ú 0 we see
wx, y, z, t D T1 x, y, z, t − T2 x, y, z, t D φx, y, z, t − φx, y, z, t D 0.
So w satisfies the second condition.
(b) We take the derivative
E t D
d 1
1
⭸ 2
⭸w
w dV D
dV .
w2 dV D
w
dt 2 l
2
⭸
t
l
l
D
D
D ⭸t
From part (a) we know that w satisfies the heat equation so
⭸w
w§2 w dV .
dV D
l
l
D ⭸t
D
w
Using Green’s first formula with f D g D w, we have
lD
w§2 w dV D
#⭸D
w§w · dS −
l
D
§w · §w dV D −
§w · §w dV
l
D
||§w||2 dV … 0.
since we showed in part (a) that w K 0 on ⭸D. Thus, E t D −
lD
(c) In part (a) we showed that wx, y, z, 0 D 0 on D. Therefore,
E0 D
1
[wx, y, z, 0]2 dV D 0.
2l
D
Now Et is the integral of a non-negative integrand so Et Ú 0. On the other hand, from part (b) we
know that E is nonincreasing. Therefore, E is a nonincreasing, nonnegative function such that E0 D 0.
Hence Et D 0 for all t Ú 0.
w2 dV D 0 for all t Ú 0. Since w2 Ú 0, we must have [wx, y, z, t]2 D 0 for all x, y, z ∈
lD
D and t Ú 0. Therefore wx, y, z, t D 0 for all x, y, z ∈ D and t Ú 0. Hence T1 x, y, z, t D T2 x, y, z, t
for all x, y, z ∈ D and t Ú 0.
(d) By part (c),
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“runall” — 2005/8/1 — 15:18 — page 357 — #37
Section 7.4
11. From Ampère’s law we have J D
1
§ * B −
µ0
§ * JD
0
Further Vector Analysis; Maxwell’s Equations
D−
357
⭸E
. Therefore,
⭸t
1
§ · § * B −
µ0
D− 0
0§ ·
⭸E
⭸E
D − 0§ ·
⭸t
⭸t
⭸
⭸ ρ
§ · E D − 0 ⭸t
⭸t 0
by Gauss’s law,
⭸ρ
.
⭸t
12. We find where § · E D 0.
§ · ED
⭸ 3
⭸ 1 3
⭸ 1 3
x − x C
y C
z − 2z .
⭸x
⭸y 4
⭸z 9
2
2
So § · E D 3x2 C 3 y2 C 1 z2 − 3. This is zero for points on the ellipsoid x2 C y C z D 1.
4
3
4
9
13. First we check that § · F D 0 wherever F is defined (i.e., away from the origin):
§ · F D k
D k
y
x
z
⭸
⭸
⭸
C
C
3/2
3/2
2
2
2
2
2
2
2
2
⭸x x C y C z ⭸y x C y C z ⭸z x C y C z2 3/2
x2 C y2 C z2 3/2 − 3x2 x2 C y2 C z2 1/2
x2 C y2 C z2 3/2 − 3y2 x2 C y2 C z2 1/2
C
x2 C y2 C z2 3/2
x2 C y2 C z2 3/2
C
x2 C y2 C z2 3/2 − 3z2 x2 C y2 C z2 1/2
.
x2 C y2 C z2 3/2
Multiply numerator and denominator by x2 C y2 C z2 1/2 :
§ · FD
k
2
2
2 2
2 2
2
2
2 2
2
2
$3x C y C z − 3x x C y C z − 3y x C y C z x2 C y2 C z2 7/2
− 3z2 x2 C y2 C z2 D
kx2 C y2 C z2 3x2 C y2 C z2 − 3x2 − 3y2 − 3z2 K 0.
x2 C y2 C z2 7/2
F · dS D
§ · F dV D 0 if S D ⭸D and S does not enclose the origin. If S
#S
l
D
does enclose the origin, let D be the solid region between S and a small sphere Sb of radius b that encloses the
origin and is inside S (as in Figure 7.54). Then
Thus, by Gauss’s theorem,
0D
l
D
§ · F dV D
#S
where S and Sb are both oriented by outward normals.
1
kx
Hence
F · dS D
F · dS D
· x dS
3
b
#
#Sb
#
S
Sb ||x||
F · dS −
#Sb
F · dS
(outward normal n to Sb is
D
k||x||2
kb2
dS D
dS
3
#Sb b4
#Sb b||x||
D
k
k
· surface area of Sb D 4πb2 D 4πk.
b2
b2
1
x)
b
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“runall” — 2005/8/1 — 15:18 — page 358 — #38
358
Chapter 7
Surface Integrals and Vector Analysis
14. (a) We may write Ex D Eρ xeρ C Eϕ xeϕ C Eθ xeθ . The field of a point charge at the origin must
be symmetric about the origin. Thus Eϕ D Eθ D 0, so Ex D Eρ xeρ D Exeρ . Once again, by
symmetry, E(x) must be constant on any sphere centered at the origin, so E can only depend on ρ. Hence
Ex D Eρeρ .
(b) We have
#S
Eρeρ · dS D
D
#
S
E · dS from part (a),
l
D
§ · E dV
using Gauss’s theorem,
ρ
D
dV using Gauss’s law,
l
D 0
q
D
by definition of ρ and q.
0
(c) We have
#S
#S
Eρeρ · dS D
Eρeρ · dS D
q
#S
, we have
0
(d) By part (c), q/ 0 D
sphere q/ 0 D
#S
#S
Eρeρ · n dS D
#S
Eρ dS D
q
#S
Eρeρ · eρ dS D
#S
Eρ dS. Since, by part (b),
.
0
Eρ dS. But, obviously, ρ is constant on the sphere of radius a and so on that
Eρ dS D Ea · 4πa2 . Thus we see that Eρ D q/4π 0 ρ2 . Hence,
Ex D
q
4π 0
ρ2
eρ D
q
q
x
x
D
as desired.
2
4π 0 ||x|| ||x||
4π 0 ||x||3
15. (a) This is just a straightforward calculation. Write F D Mi C Nj C Pk. Then
§ * FD
i
⭸/⭸x
M
j
⭸/⭸y
N
k
⭸/⭸z
P
D Py − Nz i C Mz − Px j C Nx − My k
and
i
⭸/⭸x
Py − Nz
§ * § * F D
j
⭸/⭸y
Mz − P x
k
⭸/⭸z
Nx − My
D Nxy − Myy − Mzz C Pxz i C Pyz − Nzz − Nxx C Myx j
C Mzx − Pxx − Pyy C Nzy k.
On the other hand,
§§ · F D §Mx C Ny C Pz D Mxx C Nyx C Pzx i C Mxy C Nyy C Pzy j C Mxz C Nyz C Pzz k
and § · §F D ⭸2
⭸2
⭸2
C
C
F
⭸x2
⭸y2
⭸z2
D Mxx C Myy C Mzz i C Nxx C Nyy C Nzz j C Pxx C Pyy C Pzz k.
Hence, §§ · F − §2 F D Nyx C Pzx − Myy − Mzz i C Mxy C Pzy − Nxx − Nzz j
C Mxz C Nyz − Pxx − Pyy k.
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“runall” — 2005/8/1 — 15:18 — page 359 — #39
Section 7.4
Further Vector Analysis; Maxwell’s Equations
359
By assumption F is of class C2 and so the mixed partials are equal; thus we have the result:
§ * § * F D §§ · F − §2 F.
(b) First we show that E satisfies the wave equation.
§2 E D §§ · E − § * § * E from part (a),
D §
D
1
ρ
0
− § * −
§ρ C
⭸
§ * B
⭸t
§ρ C
⭸
µ0 J C
⭸t
0
D
1
⭸B
⭸t
0
D0 C
0 µ0
⭸2 E
⭸t 2
0 µ0
using Gauss’s and Faraday’s laws,
⭸E
⭸t
using Ampère’s law
since there are no charges or currents so ρ K 0 and J K 0.
⭸2 E
where k D 0 µ0 .
⭸t 2
Next we show that B satisfies the wave equation.
Thus §2 E D k
§2 B D §§ · B − § * § * B
D 0 − § * µ0 J C
D − 0 µ0 § *
⭸
§ * E
⭸t
D − 0 µ0
⭸
⭸B
− ⭸t
⭸t
So §2 B D k
(c) By part (a),
0 µ0
⭸E
⭸t
using Maxwell’s equations,
⭸E
since J K 0 no currents,
⭸t
D − 0 µ0
D
0 µ0
from part (a),
from Faraday’s law,
⭸2 B
.
⭸t 2
⭸2 B
where k D 0 µ0 .
⭸t 2
§§ · E − § · §E D § * § * E
D § * −
⭸B
⭸t
D−
⭸
§ * B
⭸t
D−
⭸
µ0 J C
⭸t
D −µ0
⭸
J C
⭸t
by Faraday’s law,
0 µ0
0
⭸E
⭸t
by Ampère’s law,
⭸E
.
⭸t
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“runall” — 2005/8/1 — 15:18 — page 360 — #40
360
Chapter 7
Surface Integrals and Vector Analysis
(d) Again from part (a),
§2 E D §§ · E − § * § * E
D 0 − § * § * E
D µ0
⭸J
C
⭸t
0 µ0
⭸2 E
⭸t 2
by Gauss’s law and the fact that ρ D 0,
from the argument in part (c).
16. Start with the non-static version of Ampère’s law.
⭸E
⭸E
D § · µ0 J C § · µ0 0
⭸t
⭸t
⭸ρ
⭸E
C µ0 0 § ·
from the continuity equation
D −µ0
⭸t
⭸t
⭸ρ
⭸ρ
C µ0
from Gauss’s law
D −µ0
⭸t
⭸t
D 0.
§ · § * B D § · µ0 J C µ0 0
17. (a) From Ampère’s law in the static case, § * B − µ0 J must be 0 when J does not depend on time. Otherwise,
the difference must depend on time. If F1 is a time-varying vector field then ⭸F1 /⭸t Z 0. If, on the other
hand, F1 does not depend on time, then ⭸F1 /⭸t D 0. Hence, if we take § * B − µ0 J D ⭸F1 /⭸t, then we
will have an equation that is valid in both the static and the non-static cases.
(b) This is similar to our calculation in Exercise 16.
⭸ρ
⭸F1
⭸F1
D −µ0
C § ·
from the continuity equation
⭸t
⭸t
⭸t
⭸E
⭸F1
D −µ0 0 § ·
C § ·
from Gauss’s law.
⭸t
⭸t
§ · § * B D § · µ0 J C § ·
So to have § · § * B D 0 we conclude that µ0 0 § ·
(c) If § ·
⭸F1
⭸E
D µ0 0 § , then by part (b),
⭸t
⭸t
⭸F1
⭸E
D µ0 0
C F2
⭸t
⭸t
⭸E
⭸F1
D§ ·
.
⭸t
⭸t
where § · F2 D 0.
Therefore the most general formulation is
§ * B D µ0 J C µ0 0
⭸E
C F2 .
⭸t
18. We first show that E satisfies the telegrapher’s equation. From Exercise 15(d) we know that
§2 E D µ0
⭸J
C
⭸t
0 µ0
⭸E
C
⭸t
0 µ0
but here J D σE, so
§2 E D µ0 σ
⭸2 E
,
⭸t 2
⭸2 E
.
⭸t 2
Next we show that B satisfies the telegrapher’s equation. Now,
§2 B D §§ · B − § * § * B from Exercise 15(a)
D 0 − § * µ0 J C
D −§ * µ0 σE C
0 µ0
0 µ0
⭸E
⭸t
by Maxwell’s equations,
⭸E
⭸t
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“runall” — 2005/8/1 — 15:18 — page 361 — #41
Section 7.4
Further Vector Analysis; Maxwell’s Equations
⭸
§ * E
⭸t
⭸B
⭸
⭸B
D −µ0 σ − − 0 µ0 − ⭸t
⭸t
⭸t
D −µ0 σ§ * E −
D µ0 σ
⭸B
C
⭸t
0 µ0
361
0 µ0
by Faraday’s law,
⭸2 B
.
⭸t 2
19. Since P D E * B,
#S
P · dS D
D
D
D
#S
E * B · dS
l
D
l
D
l
D
§ · E * B dV
by Gauss’s theorem,
B · § * E − E · § * B dV
−B ·
⭸B
− E · µ0 J C
⭸t
Since B and E are both assumed to be constant in time,
#S
P · dS D
0 µ0
⭸E
dV
⭸t
from Faraday and Ampère’s laws.
⭸B
⭸E
D
D 0. Therefore, we get the desired result:
⭸t
⭸t
lD
− µ0 E · J dV .
20. (a) If r D r1 , r2 , r3 and x D x, y, z,
Er D
D
r − x
1
ρx
dV
4π 0 lD
||r − x||3
r − y
r − x
r − z
1
ρx, y, z 1
dV ,
ρx, y, z 2
dV ,
ρx, y, z 3
dV .
3
3
4π 0 l
||r − x||
||r − x||
||r − x||3
lD
lD
D
(b) Look at the first component of E. (The arguments for the other two components are similar.) We have
|ρx, y, z| ||r − x||
ρx, y, z r1 − x
K
…
…
4π 0 ||r − x||3
4π 0
||r − x||3
||r − x||2
where K may be taken to be the maximum value of |ρ| on D divided by 4π 0 . Thus,
r − x
K
1
ρx, y, z 1
dV …
dV .
3
4π 0 l
||r
−
x||
||r
−
x||2
lD
D
(c) Use spherical coordinates with r as the origin so that the spherical coordinate ρ is ||r − x||. Then
K
2
l
D ||r − x||
dV D
K
2
l
Dρ
ρ2 sin ϕ dρ dϕ dθ D
l
D
K sin ϕ dρ dϕ dθ.
Note that K sin ϕ is a bounded, continuous integrand. Since D is a bounded region, this last integral must
converge. Hence, by the remarks in the exercise, the original triple integral must converge.
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“runall” — 2005/8/1 — 15:18 — page 362 — #42
362
Chapter 7
Surface Integrals and Vector Analysis
21. We are given Br D
i
J1
r1 − x
µ0
r − x
J *
dV . Now,
4π l
||r − x||3
D
j
J2
r2 − y
k
J3
r3 − z
D [r3 − zJ2 − r2 − yJ3 ] i C [r1 − xJ3 − r3 − zJ1 ] j
C [r2 − yJ1 − r1 − xJ2 ] k.
Hence the first component of the triple integral for B is
µ0
r − y
r3 − z
− J3 2
J2
dV .
4π lD
||r − x||3
||r − x||3
(The other components are of the same form.) Note that each term in the integrand is of the form described in
Exercise 20. Thus, using the arguments in Exercise 20, each component integral of B must converge.
7.5 TRUE/FALSE EXERCISES FOR CHAPTER 7
1. True.
2. False. (Note that the parametrization only gives y Ú 3.)
3. True. (Let u D s3 and v D tan t.)
4. False. (The standard normal vanishes when s or t is zero.)
5. False. (The limits of integration are not correct.)
6. True. (Use symmetry.)
7. False. (The value of the integral is 24.)
8. True. (Use symmetry.)
9. True.
10. True. F · n D 0.
11. False. (The integral has value 32π.)
12. True.
13. False. (The value is 3π.)
14. True.
15. False. (The surface must be connected.)
16. False. (Consider the Möbius strip.)
17. True. (The result follows from Stokes’s theorem.)
18. False. (The value is the same only up to sign.)
19. True. (Use Gauss’s theorem.)
20. True. (Apply Gauss’s theorem.)
21. False. (Gauss’s theorem implies that the integral is at most twice the surface area.)
22. False.
23. True.
24. True.
25. False. (Should be the flux of the curl of F.)
26. True. (This is what Gauss’s theorem says.)
27. True. (Apply Green’s first formula.)
28. False. (The negative sign is incorrect.)
29. False. (f is determined up to addition of a harmonic function.)
30. False. (Only if S doesn’t enclose the origin.)
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“runall” — 2005/8/1 — 15:18 — page 363 — #43
Section 7.6
Miscellaneous Exercises for Chapter 7
363
7.6 MISCELLANEOUS EXERCISES FOR CHAPTER 7
1. Here are the matches:
(a) C
(b) E (c) A
(d) D
(e) F (f) B
Brief reasons:
(c) The projection of X into the xy-plane, for fixed s, is a circle centered at the origin of radius 2 C cos s.
(b) Note that x2 C y2 D z2 , so we have a conical surface.
(a) Since y D s, the intersection of the surface with the plane y D 0 is the parameterized curve x D −t 3 , y D
0, z D −t 2 or z D −x2/3 , y D 0, which is a cuspidal curve.
(d) Let t D π/2. Then x D 0, y D s, z D sin s. So the intersection of the surface by the x D 0 plane is a
sinusoidal curve.
(f) For constant values of s we have a helix, so the surface should be a helicoid.
(e) By elimination, this must correspond to F.
2. (a) Consider all the lines through (0, 0, 1). Either such a line is tangent to the sphere, or else it passes through
another point of the sphere S. The lines tangent to S at (0, 0, 1) fill out the tangent plane z D 1. All the
other lines therefore have “slope vectors” with nonzero k-components. Hence they intersect the z D 0 plane
somewhere. Thus any line joining (0, 0, 1) and (s, t, 0) intersects S at a point other than (0, 0, 1).
(b) The line joining (0, 0, 1) and (s, t, 0) is given parametrically by
yu D 1 − u0, 0, 1 C us, t, 0 D us, ut, 1 − u.


 x D us
y D ut
To see where the line intersects the sphere, we insert the parametric equations

 zD1 − u
equation for S and solve for u. Thus:
into the
us2 C ut2 C 1 − u2 D 1 3 u2 s2 C t 2 C 1 − 2u C 1 D 1
3 us2 C t 2 C 1u − 2 D 0.
So either u D 0 (which corresponds to (0, 0, 1)) or u D
define Xs, t as
Xs, t D y D
2
s2 C t 2 C 1
2s
s2 C t 2 C 1
,
2
. For this second value of u, we may
s2 C t 2 C 1
2s
2t
2
,
,1 −
D 2
2
2
2
2
s C t C 1 s C t C 1
s C t2 C 1
2t
s2 C t 2 C 1
,
s2 C t 2 − 1
.
s2 C t 2 C 1
(c) Check that the coordinates of X(s, t) satisfy the equation for S, i.e., that
2
2
2t
s2 C t 2 − 1
2s
2
C 2
C 2
2
2
s C t C 1
s C t C 1
s C t2 C 1
2
D
4s2 C 4t 2 C s4 C t 4 C 2s2 t 2 − 2s2 − 2t 2 C 1
s2 C t 2 C 12
D
s2 C t 2 C 12
s4 C t 4 C 2s2 t 2 C 2s2 C 2t 2 C 1
D
K 1.
s2 C t 2 C 12
s2 C t 2 C 12
Note that there are no values for s and t so that Xs, t D 0, 0, 1. (To see thus, look at the first two
coordinates—we must have s D t D 0. But then X0, 0 D 0, 0, −1). Hence the parametrization misses
the north pole.
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“runall” — 2005/8/1 — 15:18 — page 364 — #44
364
Chapter 7
Surface Integrals and Vector Analysis
3. (a) If we use cylindrical coordinates x D r cos θ, √
y D r sin θ, z D z, then the equation x2 C y2 − z2 D 1
2 − z2 D 1 or, since r Ú 0, r D
z2 C 1. Hence the desired parametrization is Xz, θ D
becomes
r
√
√
z2 C 1 cos θ, z2 C 1 sin θ, z where z ∈ R and 0 … θ … 2π.
(b) Modify the cylindrical coordinate substitution by letting x D ar cos
√t, y D br sin t, z D cs. Substitution
2 − s2 D 1 so r D
into the equation
for
the
hyperboloid
yields
r
s2 C 1. Hence a parametrization is
√
√
Xs, t D a s2 C 1 cos t, b s2 C 1 sin t, cs, where s ∈ R and 0 … t … 2π.
(c) Substitute the parametric equations for l1 into the left side of the equation for the hyperboloid:
b2 x0 t C y0 2
a2 x0 − y0 t2
c2 t 2
C
−
D x02 − 2x0 y0 t C y02 t 2 C x02 t 2 C 2x0 y0 t C y02 − t 2
2
2
a
b
c2
D x02 C y02 C x02 C y02 t 2 − t 2
D 1 C t 2 − t 2 D 1,
since x02 C y02 D 1. Thus l1 lies on the hyperboloid. The calculation for l2 is similar.
(d) The plane tangent to the hyperboloid at the point ax0 , by0 , 0 is given by
§F ax0 , by0 , 0 · x − ax0 , y − by0 , 0 D 0 where F D
y2
x2
z2
C
−
.
2
2
a
b
c2
That is, the tangent plane is
∗
y
x0
x − ax0 C 0 y − by0 D 0.
a
b
If we substitute the parametric equations for l1 into the left side of (∗), we find
y
x0
ax0 − y0 t − ax0 C 0 bx0 t C y0 − by0 D −x0 y0 t C y0 x0 t D 0
a
b
can be√made for l2 .
for all t. Therefore, the line l1 lies in the plane. A similar calculation
√
2
4. Reconsider the parametrization from Exercise 3(a), Xz, θ D z C 1 cos θ, z2 C 1 sin θ, z where z ∈ R
and 0 … θ … 2π. Then we have,
Tz D √
z
z
cos θ, √
sin θ, 1 ,
z2 C 1
z2 C 1
Tθ D − z2 C 1 sin θ, z2 C 1 cos θ, 0.
Thus Tz * Tθ D − z2 C 1 cos θ, − z2 C 1 sin θ, z,
so that ||Tz * Tθ || D z2 C 1 C z2 D 2z2 C 1.
Therefore,
Surface area D
2π a
0
−a
√
√
√
√
2z2 C 1 dz dθ D π 2 ln 2a2 C 1 C 2a C 2a 2a2 C 1.
√
(Let tan u D 2z in the z-integral.)
5. (a) This is similar to Exercise 3(b). First, consider a variant of spherical coordinates: x D aρ cos θ sin ϕ, y D
bρ sin θ sin ϕ, and z D cρ cos ϕ. If we set ρ D 1, we get the desired parametrization: x D a sin ϕ cos θ, y D
b sin θ sin ϕ, and z D c cos ϕ where 0 … ϕ … π and 0 … θ … 2π.
(b) Here we have
Tϕ D a cos ϕ cos θ, b cos ϕ sin θ, −c sin ϕ and
Tθ D −a sin ϕ sin θ, b sin ϕ cos θ, 0.
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“runall” — 2005/8/1 — 15:18 — page 365 — #45
Section 7.6
Miscellaneous Exercises for Chapter 7
365
Therefore,
N D Tϕ * Tθ D bc sin2 ϕ cos θ, ac sin2 ϕ sin θ, ab cos ϕ sin ϕ
and
||N|| D b c sin ϕ cos θ C a c sin ϕ sin θ C a b cos ϕ sin ϕ.
2 2
4
2
2 2
4
2
2 2
2
2
Therefore,
Surface area D
2π π
0
b2 c2 sin4 ϕ cos2 θ C a2 c2 sin4 ϕ sin2 θ C a2 b2 cos2 ϕ sin2 ϕ dϕ dθ.
0
In the special case where a D b D c, we find that
2π π
a2 sin4 ϕ cos2 θ C sin4 ϕ sin2 θ C cos2 ϕ sin2 ϕ dϕ dθ
Surface area D
0
D a2
0
2π π
0
D a2
D a2
Da
2
sin4 ϕ C cos2 ϕ sin2 ϕ dϕ dθ
0
2π π
sin2 ϕ dϕ dθ
0
0
0
0
2π π
sin ϕ dϕ dθ
2π
π
− cos ϕ
0
Da
2
2π
dθ
0
2 dθ D 4πa2 .
0
6. (a) The t-coordinate curve is s0 , f s0 cos t, f s0 sin t, which is a circle of radius |f s0 | in the x D s0
plane. That is, the radius of this cross-sectional circle depends on f s0 .
(b) Ts D 1, f s cos t, f s sin t and Tt D 0, −f s sin t, f s cos t, so N D Ts * Tt D f sf s,
−f s cos t, −f s sin t. Thus ||N|| D [f s]2 [f s]2 C [f s]2 D |f s| [f s]2 C 1. So
2π b
Surface area D
|f x| [f x]2 C 1 dx dt
0
D
a
b 2π
a
D 2π
|f x| [f x]2 C 1 dt dx
0
b
|f x| [f x]2 C 1 dx.
a
7. (a) This should remind students of when they were using washer and shell methods for surfaces of revolution.
Of course, here we are finding a surface area and not volume. If you look at the specific value s D s0 then,
since we are revolving around the y-axis, we are sweeping out a circle of radius s0 in the plane y D f s0 .
Therefore, a parametrization is Xs, t D s cos t, f s, s sin t, a … s … b, 0 … t … 2π. Compare this with
Exercise 6(a).
(b) Using the parametrization in (a), Ts D cos t, f s, sin t and Tt D −s sin t, 0, s cos t. Therefore, N D
Ts * Tt D sf s cos t, −s, sf s sin t, so ||N|| D s2 [f s]2 C s2 . Hence
b 2π
s2 [f s]2 C s2 dt ds
Surface area D
a
0
b
D 2π
s [f s]2 C 1 ds
a
b
D 2π
x [f x]2 C 1 dx
a
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“runall” — 2005/8/1 — 15:18 — page 366 — #46
366
Chapter 7
Surface Integrals and Vector Analysis
by changing the variable of integration. Compare this result with that of Exercise 6(b).
8. (a) It would be helpful for you to first draw a picture. The surface is the curve z D f x in the xz -plane extended
so that the derivative in the y direction is identically zero (i.e., we’re dragging the curve in the y direction).
Then S1 , the portion of S lying over D, may be parametrized as Xx, y D x, y, f x, x, y ∈ D.
Then Tx D 1, 0, f x and Ty D 0, 1, 0, so that N D Tx * Ty D −f x, 0, 1, and so ||N|| D
1 C [f x]2 . Hence,
Surface area D
Since sx D
x
1 C [f x]2 dx dy D
6D
6
D
1 C [f x]2 dA.
1 C [f t]2 dt, then, by the fundamental theorem of calculus, we have that s x D
a
1 C [f x]2 . Thus the surface area is
6D
s x dA.
s x dA, which, by part (a), is the surface area.
6D
(c) Here we are working a specific example of what we worked out in part (a). The rectangle D is [1, 3] * [−2, 2].
sx dy D
(b) From Green’s theorem,
C
Using part (a), we compute the surface area as
f x D
s x dA. Now s x D 1 C [f x]2 where z D
6D
x3
1
1
C
, and so f x D x2 −
. Therefore,
3
4x
4x2
1 C [f x]2 D 1 C x4 −
2
1
1
1
1
1
4
C
C
D x 2 C
Dx C
.
2
16x4
2
16x4
4x2
Hence,
Surface area D
1 C [f x]2 dA D
6D
2 3
6D
2
1
dA
4x2
3
1
1 3
dy
x −
3
4x xD1
−2 1
−2
2
2
1
1
106
1
53
D
−
C dy D
dy D
.
9 −
12
3
4
3
−2
−2 6
D
2
x C
1
dx dy D
4x2
2
x2 C
9. (a) The surface integral
f dS, roughly speaking, represents the “sum” of all the values of f on S. The area
6S
of S is a measure of the size of S. So the quotient can be thought of as the “total” amount of f divided by
the size of the region being sampled.
(b) Parametrize the sphere as Xs, t D 7 cos s sin t, 7 sin s sin t, 7 cos t, 0 … s … 2π, 0 … t … π. Then,
following Example 11 in Section 7.1, ||Ts * Tt || D 49 sin t. Note that, on the surface S, the temperature
T x, y, z D x2 C y2 − 3z D 49 − z2 − 3z. As a result, we can calculate
2π π
T x, y, z dS D
49 − 49 cos2 t − 21 cos t49 sin t dt ds
0
0
6S
2π
π
21
49
D 49
cos3 t C
cos2 t ds
−49 cos t C
3
2
0
0
2π
492 42π
21
49
2 C
1 − 1 ds D
.
D 49
492 −
3
2
3
0
Now, since the surface area of a sphere of radius 7 is 4π49, we have
T x, y, z dS
492 42π
1
98
[T ]avg D 6S
D
D
.
surface area
3
4π49
3
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“runall” — 2005/8/1 — 15:18 — page 367 — #47
Section 7.6
Miscellaneous Exercises for Chapter 7
367
10. The surface area of the cylinder is 2π · 2 · 3 D 12π. If we parametrize the surface as


 x D 2 cos t
y D 2 sin t

 zDs
0 … t < 2π,
0 … s … 3
then ||Ts * Tt || D 2. Hence
3 2π
1
1
f dS D
4e s cos2 t − 4s sin2 t · 2 dt ds
12π 6S
12π 0 0
3 2π
1
D
[e s 1 C cos 2t − s1 − cos 2t] dt ds
3π 0 0
3
2π
1
1
1
s
ds
D
e t C sin 2t − s t − sin 2t 3π 0
2
2
tD0
[f ]avg D
D
1
3π
3
2πe s − 2πs ds D
0
2
3
3
e s − s ds
0
3
D
1 2
2 3
9
2 s
2e3 − 11
− 1 D
.
e − s D e −
3
2
3
2
3
0
11. The cone looks as follows.
y
2
0
-2
x
0
2
-2
6
4
z 2
0
-2
The upper nappe has a height of 6 and radius of 3; the lower nappe has a height of 2 and a radius of 1. Hence
the total surface area is
√
√
√
π · 1 · 5 C π · 3 · 3 5 D 10 5π.


 x D s cos t
y D s sin t
Next, parametrize the surface as

 z D 2s
with −1 … s … 3, 0 … t < 2π. Then
||N|| D ||Ts * Tt || D ||−2s cos t, −2s sin t, s|| D
√
5 |s|.
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“runall” — 2005/8/1 — 15:18 — page 368 — #48
368
Chapter 7
Surface Integrals and Vector Analysis
Therefore,
[f ]avg D
1
√
1
√
f dS D
2π 3
√
s2 − 3 5|s| ds dt
10 5π 6S
10 5π 0
−1
3 2π
1
D
s2 − 3|s| dt ds
10π −1 0
3
3
0
1
1
D
2πs2 − 3|s| ds D s2 − 3−s ds C
s2 − 3s ds
10π −1
5 −1
0
0
3
D
1
1
1 4
3 2
3
C s4 − s2 − s C s 5
4
2
4
2
−1
0
D
1 1
3
81
27
11
C
−
.
−
D
5 4
2
4
2
10
δ dS. For the helicoid, Ts D cos t, sin t, 0 and Tt D −s sin t, s cos t, 1. Then N D
6X
√
Ts * Tt D sin t, − cos t, s and ||N|| D 1 C s2 . Hence,
12. The total mass is
Total mass D
x2 C y2 dS
6X
4π 1
√
s cos t2 C s sin t2 1 C s2 ds dt
D
0
0
0
0
4π 1 √
D
s 1 C s2 ds dt
1 2
4π √
D
· 1 C s2 3/2 2 2 − 1.
2 3
3
sD0
1
D 4π 13. By the symmetry of the surface, we must have x D y D z. We compute z explicitly. Since δ is constant, it will
cancel from the center of mass integrals:
δ
z dS
z dS
6S
D 6S
D
.
surface area of S
δ dS
δ
dS
6S
6S
z D 6S
zδ dS
The surface area of the first octant portion of a sphere of radius a is
1 2
1
4πa2 D
πa . There8
2
2
z dS. We may parametrize the first octant portion of the sphere as Xϕ, θ D
πa2 6
S
a sin ϕ cos θ, a sin ϕ sin θ, a cos ϕ, 0 … ϕ … π/2, 0 … θ … π/2. Hence,
fore, z D
Tϕ D a cos ϕ cos θ, a cos ϕ sin θ, −a sin ϕ,
Tθ D −a sin ϕ sin θ, a sin ϕ cos θ, 0.
Therefore,
N D a2 sin2 ϕ cos θ, a2 sin2 ϕ sin θ, a2 sin ϕ cos θ and
Thus,
2
zD
πa2
2a3
D
πa2
||N|| D a2 sin ϕ.
π/2 π/2
a cos ϕa2 sin ϕ dϕ dθ
0
0
0
0
π/2 π/2
cos ϕ sin ϕ dϕ dθ
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“runall” — 2005/8/1 — 15:18 — page 369 — #49
Section 7.6
2a
D
π
D
2a
π
π/2
Miscellaneous Exercises for Chapter 7
369
π/2
0
1 2
dθ
sin ϕ
2
ϕD0
0
a
π
a
1
dθ D
·
D .
2
π
2
2
π/2
a
14. A quick sketch should convince you that, by symmetry, x D 0 and y D . The equation for the surface may be
2
√
⭸z
⭸z
x
written as z D a2 − x2 , so that
D −√
D 0. Then
and
⭸x
⭸y
a2 − x 2
'
(
(
dS D )1 C
2
2
⭸z
⭸z
C dx dy D
⭸x
⭸y
x2
1 C
dx dy D
2
a − x2
a2
a2 − x 2
dx dy.
Hence,
z D 6S
zδ ds
D 6S
z ds
δ ds
6S
dS
6S
z ds
,
D 6S
πa2
since the surface area of half a cylinder is πa2 . Now we calculate
a a √
1
1
a2
zD
z ds D
a2 − x 2
dx dy
πa2 6S
πa2 0 −a
a2 − x 2
a a
a
2a
1
.
a dx dy D
2a2 D
D
πa2 0 −a
πa2
π
S
15. By symmetry x D y D 0, so we only need to calculate z D 6
zδ dS
6
S
6S
δ ds D
6
S
z C a dS D
6S
. Now
δ dS
z dS C a
6S
1 dS
D 0 by symmetry C a · 4πa2 D 4πa3 .
zδ dS D
z2 C az dS D
6
S


 x D a cos s sin t
y D a sin s sin t
Now we parametrize the sphere:

 z D a cos t
Then ||N|| D a2 sin t (see Example 1 of §7.2). Thus
6S
6S
z2 ds D
π 2π
0
0
D −2πa
6S
z2 dS C a
6S
z dS D
6S
z2 dS.
0 … s < 2π
0 … t … π .
π
a cos t2 · a2 sin t ds dt D
π
π
cos2 t− sin t dt D
4
a4 cos2 t sin t · 2π dt
0
0
−2πa4
4πa4
cos3 t D
.
3
3
0
So
zD
4πa4 /3
a
D .
4πa3
3
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“runall” — 2005/8/1 — 15:18 — page 370 — #50
370
Chapter 7
Surface Integrals and Vector Analysis


 x D a cos s
yDt
0 … s < 2π, 0 … t … 2 .
16. Parametrize the cylinder as

 z D a sin s
Then N D Ts * Tt D −a sin s, 0, a cos s * 0, 1, 0 D −a cos s, 0, −a sin s so ||N|| D a. Hence
2 2π
δ ds D
a2 cos2 s C t · a ds dt
MD
0 0
6S
2 2π
2
a3
2
D
1
C
cos
s
ds
dt
D
2πat C πa3 dt
at
C
2
0 0
0
2
D πat 2 C πa3 t
D 4πa C 2πa3 D 2πaa2 C 2.
0
Symmetry implies z D 0, so we calculate
xD
1
2 2π
a cos s at C a2 cos2 s ds dt
2πaa2 C 2 0 0
2 2π
a
D
t cos s C a2 cos3 s ds dt
2πa2 C 2 0 0
2 2π
a
t cos s C a2 1 − sin2 s cos s ds dt
D
2πa2 C 2 0 0
2
2π
a2
a
D
t sin s C a2 sin s −
sin3 s dt D 0.
2πa2 C 2 0
3
0
(Actually, you can really see this from symmetry.)
1
yD
yx2 C y dS D
1
2 2π
t t C a2 cos2 s · a ds dt
2πaa2 C 2 6S
2πaa2 C 2 0 0
2 2π
2π
1
1
1
D
t 2 s C a2 t s C sin 2s
dt
2πa2 C 2 0 0
2
4
sD0
2
2
1
1
D
2πt 2 C πa2 t dt D
2t 2 C a2 t dt
2πa2 C 2 0
2a2 C 2 0
2
D
1
2 3
16
a2 2
1
t C
t D
C 2a2 2 C 2
2a2 C 2 3
2
2a
3
0
D
8
1
3a2 C 8
C a2 D
.
a2 C 2 3
3a2 C 6
3a2 C 8
, 0 .
3a2 C 6
17. (a) Parametrize the frustum z2 D 4x2 C 4y2 , 2 … z … 4, as Xr, θ D r cos θ, r sin θ, 2r, 0 … θ … 2π,
1 … r … 2. Then
So x, y, z D 0,
⭸x, y
D
⭸r, θ
cos θ
sin θ
−r sin θ
r cos θ
Dr
⭸x, z
D
⭸r, θ
cos θ
2
−r sin θ
0
D 2r sin θ
⭸y, z
D
⭸r, θ
sin θ
2
r cos θ
0
D −2r cos θ.
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“runall” — 2005/8/1 — 15:18 — page 371 — #51
Section 7.6
Miscellaneous Exercises for Chapter 7
371
Therefore,
Iz D
D
x2 C y2 dS D
6S
2π 2
0
2π 2
r2
0
√
5r dr dθ D
1
r2 C 4r2 dr dθ
2π √
3
1
√
0
√
5
15 5π
15 dθ D
.
4
2
Iz
. Assuming, as in part (a), that the density is 1, the total mass
M
is just the surface area of the frustum. This can be computed from the surface area of the cone without much
trouble. We view the frustum as a large cone (of height 4) with the tip (a similar cone of height 2) removed
and note that the surface area of a cone is π(radius)(slant height). Then
√
√
√
Surface area of frustum D π22 5 − π1 5 D 3 5π.
(b) The radius of gyration is given by rz D
Hence
rz D
Iz
D
M
√
15 5π 1
5
√
D
.
2
3 5π
2
2π 2
(Note: you can also compute the surface area as
0
√
5r dr dθ.)
1
(c) We recompute the integral for Iz with δ D kr. Thus
Iz D
x2 C y2 δ dS
6S
2π 2
D
r2 kr 5r2 dr dθ
0
D
0
D
1
2π 2
√
5kr4 dr dθ
1
2π √
√
5k 5
62 5πk
2 − 1 dθ D
.
5
5
0
The total mass of the frustum is
MD
D
δ dS D
6S
2π √
0
Hence
rz D
2π 2
0
√
kr 5r dr dθ
1
√
14 5πk
5 3
2 − 1 dθ D
.
3
3
Iz
D
M
√
3
62 5πk
93
√
D
.
5
14 5πk
35
18. (a)
Iz D
6S
x2 C y2 δ dS D δ
6S
a2 dS D δa2 · surface area
D δa2 · 2πa · 2b D 4πδa3 b
(b) M D
6S
δ dS D δ · 4πab, so rz D
Iz
D
M
4πδa3 b
D a.
4πδab
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“runall” — 2005/8/1 — 15:18 — page 372 — #52
372
Chapter 7
Surface Integrals and Vector Analysis


 x D a cos t
y D a sin t
19. (a) Ix D
y2 C z2 δ dS. If we parametrize S by

 zDs
6S
||Ts * Tt || D a and so
Ix D
b 2π
−b 0
b
a2 sin2 t C s2 δa dt ds D δa
D δa
Iy D
−b
πa2 C 2πs2 ds D πδa 2a2 b C
x2 C z2 δ dS D
b 2π
2π
−b
b
−b … s … b, 0 … t < 2π , then ||N|| D
a2
1
2
t − sin 2t C s t
ds
2
2
tD0
4 3
2πabδ
b D
3a2 C 2b2 3
3
a2 cos2 t C s2 δa dt ds
−b 0
6S
b
2π
2
1
4
a
2
D δa
ds D πδa 2a2 b C b3 t C sin 2t C s t 2
2
3
−b
tD0
as before.
(b) From Exercise 18, M D 4πabδ, so
rx D ry D
'
(
( πδa $2a2 b C 4 b3 %
)
3
D
4πabδ
D
'
( 2
) 2a C 4 b2
3
4
D
'
( 2
) a C 2 b2
3
2
3a2 C 2b2
.
6
20. (a) Let M be the maximum value of f on D and m the minimum value. (The numbers M and m must exist
since D is compact.) Then
m D 6D
mg dA
6D
… 6D
g dA
f g dA
6D
… 6D
g dA
Mg dA
6D
D M.
g dA
Hence by the intermediate value theorem, there must be some point P in D such that
D
f P D 6
f g dA
6D
which gives the result, provided
If
6D
6D
,
g dA
g dA Z 0.
g dA D 0 then we have
g dA D
mg dA …
f g dA …
Mg dA D M
g dA D 0,
0Dm
6D
6D
6D
6D
6D
so
6D
f g dA D 0 and any P in D gives the desired result.
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“runall” — 2005/8/1 — 15:18 — page 373 — #53
Section 7.6
Miscellaneous Exercises for Chapter 7
373
(b) Assume that S may be parametrized by a single function X. Then
6S
F · dS D
6S
F · n dS D
6D
FXs, t · ns, t||Ns, t|| ds dt
D FXs0 , t0 · ns0 , t0 6D
||Ns, t|| ds dt
by part (a),
D FP · nParea of S
where P D Xs0 , t0 .
21. (a) Let a D a1 , a2 , a3 and assume xt D xt, yt, zt parametrizes C. Then
b
a · ds D
C
a · x t dt
a
b
D
a1 x t C a2 y t C a3 z t dt
a
b
D a1 xt C a2 yt C a3 zt
a
b
D a · xt
D a · xb − a · xa
a
D0
since xa D xb because C is a closed curve.
(b) Let S be any smooth, orientable surface with boundary curve C. If we orient S appropriately and use Stokes’s
theorem, we have
a · ds D
C
6S
§ * a · dS D
6S
0 · dS D 0.
22. Note that C lies in the surface z D x2 − y2 . The line integral is
F · ds, where F D x2 C z2 i C yj C zk
C
Therefore, Stokes’s theorem implies that
F · ds D
C
6S
§ * F · dS,
where S is the portion of z D x2 − y2 bounded by C. Note that S lies over the unit disk in the xy-plane. We
may take for unit normal
nD
§ * FD
−2xi C 2yj C k
4x2 C 4y2 C 1
i
⭸/⭸x
x 2 C z2
j
⭸/⭸y
y
and
k
⭸/⭸z
z
D 2zj D 2x2 − y2 j on S.
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“runall” — 2005/8/1 — 15:18 — page 374 — #54
374
Chapter 7
Thus
6S
Surface Integrals and Vector Analysis
§ * F · dS D
2π 1
0
6D
4yx2 − y2 dA where D is the unit disk. This is
4r4 cos2 θ sin θ − sin3 θ dr dθ
0
2π
D
0
4
4
cos2 θ sin θ − sin3 θ dθ D
5
5
2π
cos2 θ sin θ − 1 − cos2 θ sin θ dθ
0
2π
4
2
3
D 0.
− cos θ C cos θ
5
3
0
D
23. By Stokes’s theorem
⭸S
f §g · ds D
D
6S
6S
§ * f §g · dS
§f * §g C f § * §g · dS D
6S
§f * §g · dS,
since § * §g D 0 (see §3.4).
24. Using the result of Exercise 23 (twice):
⭸S
f §g C g§f · ds D
6
S
§f * §g C §g * §f · dS D
6S
0 · dS D 0
because §f * §g D −§g * §f .
25.
⭸S
f §f · ds D
1
f §f C f §f · ds
⭸S 2
D 0 by Exercise 24.
26. (a) First apply Stokes’s theorem:
1
2
bz − cy dx C cx − az dy C ay − bx dz
C
i
⭸/⭸x
bz − cy
j
⭸/⭸y
cx − az
k
⭸/⭸z
ay − bx
D
1
2 6D
D
1
2a, 2b, 2c · dS D
a, b, c · dS
2 6D
6D
D
6D
n · n dS D
6D
dS
· dS
D is the region enclosed by C
since n is a unit vector,
D area enclosed by C.
(b) If C is contained in the xy-plane, then n D k, so a D b D 0 and c D 1 in the notation above. Hence the
result reduces to
1
−y dx C x dy D area enclosed by C.
2 C
27. By Faraday’s law
d
⭸B
§ * E · dS D −
B · dS.
· dS D −
dt 6S
6S
6S ⭸t
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“runall” — 2005/8/1 — 15:18 — page 375 — #55
Section 7.6
Miscellaneous Exercises for Chapter 7
375
On the other hand, using Stokes’s theorem,
6S
§ * E · dS D
⭸S
E · ds D
⭸S
E · T ds D 0,
d
B · dS D 0, so the magnetic flux does not vary with time.
dt 6S
28. For Gauss’s theorem to apply to the situation, S must be closed. Hence ⭸S is empty. But then there really is no
line integral
G · ds. If we try to apply Stokes’s theorem in general (i.e., to surfaces with nonempty boundary)
since E is everywhere perpendicular to ⭸S. Thus
⭸S
then we cannot also apply Gauss’s theorem.
Qa and the lateral surfaces L of ⭸W . With ⭸W oriented
29. Note that the boundary ⭸W of W consists of three parts: S, S
Q
by outward normal, and if we take S and Sa to be oriented in the same way,
x
x
x
x
· dS D ; · dS −
· dS C
· dS
3
3
3
||x||
||x||
||x||
Q
#⭸W ||x||3
6S
6Sa
6L
Qa are oriented with respect to the orientation of ⭸W .) Now L consists of a
(The ; sign depends on how S, S
collection of segments of the rays defining S, O. Thus L is tangent to x. Hence x · n D 0 where n is the
x
appropriate unit normal to L. Thus
· dS D 0. Thus
3
||x||
6L
x
x
x
· dS D ; · dS −
· dS .
#⭸W ||x||3
6S ||x||3
6SQa ||x||3
Gauss’s theorem implies
x
x
· dS D
0 dV .
§ ·
dV D
3
3
||x||
||x||
#⭸W
lW
l
W
Hence
x
6S ||x||3
· dS D
x
6SQa ||x||3
Qa , n D
· dS. On S
S, O D
D
x
3
6
S ||x||
x
, so
||x||
· dS D
x
6SQa ||x||3
·
x
dS
||x||
||x||2
1
dS D
dS.
4
Qa ||x||
6SQa ||x||2
6
S
Qa , ||x|| D a, so
But on S
S, O D
1
6SQa a2
dS D
1
Qa .
surface area of S
a2
x
· dS. Now x D xs, t, ys, t, zs, t, so
3
||x||
6S
that ||x|| D x2 C y2 C z2 . Moreover, the standard normal N D Ts * Tt is
30. From the definition of
ND
S, O, we calculate
S, O D
i
⭸x
⭸s
⭸x
⭸t
⭸y
⭸s
⭸y
⭸t
j
⭸y
⭸s
⭸y
⭸t
k
⭸z
⭸s
⭸z
⭸t
D
⭸z
⭸s
⭸z
⭸t
i −
⭸x
⭸s
⭸x
⭸t
⭸z
⭸s
⭸z
⭸t
j C
⭸x
⭸s
⭸x
⭸t
⭸y
⭸s
⭸y
⭸t
k.
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“runall” — 2005/8/1 — 15:18 — page 376 — #56
376
Chapter 7
Surface Integrals and Vector Analysis
Hence
x
1
· dS D
x · Ts * Tt ds dt
3
2
2
6D x C y C z2 3/2
6S ||x||
D
1
6D x2 C y2 C z2 3/2
x
y
z
⭸x/⭸s
⭸y/⭸s
⭸z/⭸s
⭸x/⭸t
⭸y/⭸t
⭸z/⭸t
ds dt
as desired.
31. First, if S does not enclose the origin then, by Gauss’s theorem
S, O D
x
6S ||x||3
x
· dS D ;
§ · 0 dV D 0.
dV D
||x||3
lW
lW
Here the ; sign depends on the orientation of S and W is the region enclosed by S.
Next, if S does enclose the origin, let Sa be the sphere of radius a centered at O and contained inside S. Let
D be the solid region in R3 between Sa and S.
S
Sa
O
x
throughout D since D doesn’t contain O. If Sa is oriented by inward normal (which points
||x||3
away from D), then, by Gauss’s theorem, we have:
Note that § · 0D
Hence
l
D
§ · x
x
x
x
· dS D ;
· dS C
· dS.
dV D
3
3
3
||x||
#⭸D ||x||
6S ||x||
6Sa ||x||3
1
x
x
D − so
· dS. On Sa , n D −
S, O D ;
3
||x||
||x||
a
6Sa
x
1
1
· − x dS D ;
− x · x dS
S, O D ;
a
a4
6Sa
6Sa a3
a2
1
D;
−
dS D ;
surface area of Sa 4
2
a
a
6Sa
D;
32. We may parametrize S as
1
4πa2 D ;4π.
a2


 x D s cos t
y D s sin t

 zD0
0 … s … a,
0 … t < 2π .
(0,0,z) O
x
S
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“runall” — 2005/8/1 — 15:18 — page 377 — #57
Section 7.6
Miscellaneous Exercises for Chapter 7
377
Then one way to orient S is with unit normal n D k. Also, we have the vector x from O to a point of S given by
x D s cos t, s sin t, −z * ||x|| D s2 C z2
Hence
2π a
x
−z
· n dS D
s ds dt
2
2 3/2
0
0 s C z 6S ||x||3
a 2π
s
D −z
dt ds
s2 C z2 3/2
0 0
a
a
2s
ds D −πzs2 C z2 −1/2 −2
D −πz
2
2 3/2
0 s C z 0
S, O D
1
1
1
−
D 2πz √ 2
|z|
|z|
a C z2
√
√
1
z 2 − a2 C z 2
1
D 2πz √
− √ D 2πz √ √
.
a2 C z 2
z2
z 2 a2 C z 2
D 2πz a2 C z2 −1/2 −
Now
C1
z
√ D*
−1
z2
if z > 0
if z < 0
z
−z
if z Ú 0
.
if z < 0
and
z2 D *
Thus

√


z − a2 C z 2


√

 2π a2 C z 2
S, O D
√


z C a2 C z 2

 2π √


a2 C z 2
if z > 0
if z < 0.
(z Z 0 because O should not
√
√
√ be a point of S.)
2
2
2
2
Note that if z > 0,√
z − a2 C z2 < 0 and |z
√ − a C z√| < a C z . Hence 0 > S, O > −2π.
2
2
2
2
2
2
If z < 0, then z C a C z > 0 and z C a C z < a C z . Hence 0 < S, O < 2π. Either
way −2π < S, O < 2π. Now as z ' 0C , S, O ' −2π and as z ' 0− , S, O ' 2π. Hence as O
passes through S, there is a jump of 4π.
33. We have
1
§ * GD§ *
tFtr * r dt where r D x, y, z,
D
1
0
§ * tFtr * r dt
0
D
1
t§ * Ftr * r dt
since t behaves as a constant with respect to §,
0
D
1
t{Ftr§ · r − r§ · Ftr C r · §Ftr − Ftr · §r} dt
by the first identity,
0
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“runall” — 2005/8/1 — 15:18 — page 378 — #58
378
Chapter 7
Surface Integrals and Vector Analysis
D
1
t{3Ftr − r§ · Ftr C r · §Ftr − Ftr} dt
0
D
1
t{2Ftr − r§ · Ftr C r · §Ftr} dt.
0
⭸
⭸F
Ftr D t
by the hint. This implies that § · Ftr D t§X,Y,Z FX, Y, Z
⭸x
⭸X
where §X,Y,Z signifies that all partials are to be taken with respect to X, Y, and Z where X D tx, Y D ty, and
Z D tz. Thus § · Ftr D 0 since F is assumed to be divergenceless. By the second identity given in the hint,
To compute § · Ftr, note that
r · §Ftr D
d
[tFtr] − Ftr.
dt
Hence,
§ * GD
1
0
D
1
0
D
1
0
t *2Ftr C
d
[tFtr] − Ftr+ dt
dt
t *Ftr C
d
[tFtr]+ dt
dt
d 2
[t Ftr] dt
dt
by the last identity in the hint,
1
D Fr.
D t 2 Ftr
tD0
34. Note § · F D 2 − 1 − 1 D 0 so, by the result of Exercise 33, a vector potential for F must exist. We can
compute it by
GD
1
t2xt, −yt, −zt * x, y, z dt D
0
D
1
1
t0, −3xzt, 3xyt dt
0
1
0, −3xzt 2 , 3xyt 2 dt D 0, −xzt 3 , xyt 3 0
tD0
D 0, −xz, xy.
35. § · F D 1 C 1 C 1 D 3 Z 0, so, by the result of Exercise 33, F has no vector potential.
36. § · F D 0 C 0 C 0 D 0, so, by the result of Exercise 33, a vector potential for F must exist. We compute it as
follows.
GD
1
t3yt, 2xzt 2 , −7x2 yt 3 * x, y, z dt
0
D
1
2xz2 t 3 C 7x2 y2 t 4 , −7x3 yt 4 − 3yzt 2 , 3y2 t 2 − 2x2 zt 3 dt
0
7
7
1
1
D xz2 C x2 y2 , − x3 y − yz, y2 − x2 z .
2
5
5
2
37. Since § * §φ D 0 for any C2 function, we have
§ * G C §φ D § * G C § * §φ D § * G C 0 D F.
Thus G C §φ is a vector potential for F.
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“runall” — 2005/8/1 — 15:18 — page 379 — #59
Section 7.6
38. (a) Write F D −
Miscellaneous Exercises for Chapter 7
379
GMm
xi C yj C zk. Then
x2 C y2 C z2 3/2
x2 C y2 C z2 3/2 − 3x2 x2 C y2 C z2 1/2
x
⭸F1
⭸
D
D
2
⭸x
⭸x x C y2 C z2 3/2
x2 C y2 C z2 3
x2 C y2 C z2 2 − 3x2 x2 C y2 C z2 .
x2 C y2 C z2 7/2
D
Similarly,
x2 C y2 C z2 2 − 3y2 x2 C y2 C z2 y
⭸F2
⭸
D
D
2
3/2
2
2
⭸y
⭸y x C y C z x2 C y2 C z2 7/2
x2 C y2 C z2 2 − 3z2 x2 C y2 C z2 z
⭸F3
⭸
D
.
D
2
⭸z
⭸z x C y2 C z2 3/2
x2 C y2 C z2 7/2
Thus
§ · FD
3x2 C y2 C z2 2 − 3x2 C 3y2 C 3z2 x2 C y2 C z2 ⭸F1
⭸F2
⭸F3
D 0.
C
C
D
⭸x
⭸y
⭸z
x2 C y2 C z2 7/2
(b) Let S be a sphere of radius a enclosing the origin. Consider S to be the union of hemispheres S1 and S2 ,
each oriented so that the normal vector points away from the center of the sphere. If F D § * G, then
6S
F · dS D
6S
D
⭸S1
§ * G · dS D
G · ds C
⭸S2
6S1
§ * G · dS C
6S2
§ * G · dS
G · ds by Stokes’s theorem
D 0,
since ⭸S1 and ⭸S2 inherit opposite orientations from S1 and S2 and are equal as unoriented curves. On the
other hand n D r/||r||, so
6S
F · dS D
6S
GMm
r
||r||2
r ·
dS
dS D −GMm
3
||r||
||r||
6S ||r||4
−
1
1
D −GMm
dS D −GMm
dS
2
||r||
a
6S
6S 2
D −GMm
since ||r|| D a on S,
4πa2
D −4πGMm Z 0.
a2
Hence, it cannot be that F D § * G.
(c) F is not of class C1 on R3 ; F is undefined at the origin. The C1 hypothesis is assumed in Exercise 33, so
there’s no contradiction.
39. We calculate the curl:
⭸A
⭸A
D§ * E C § *
⭸t
⭸t
⭸
D§ * E C
§ * A
⭸t
⭸B
D§ * E C
D0
⭸t
§ * E C
by Faraday’s law. Since E, B and thus A are all defined on a simply-connected region, we must have that
E C ⭸A/⭸t is conservative.
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“runall” — 2005/8/1 — 15:18 — page 380 — #60
380
Chapter 7
Surface Integrals and Vector Analysis
40. Substituting § * A for B in Ampère’s law, we have
§ * § * A D µ0 J C µ0 0
⭸E
.
⭸t
From the identity § * § * A D §§ · A − §2 A, we have
µ0 J D §§ · A − §2 A − µ0 0
Since E C ⭸A/⭸t is conservative, E D §f −
⭸E
.
⭸t
⭸A
, so that
⭸t
µ0 J D §§ · A − §2 A − µ0 0
⭸
⭸A
§f −
⭸t
⭸t
D §§ · A − §2 A − µ0 0 § Thus
µ0 J D § § · A − µ0 0
⭸f
⭸2 A
2
−
since f is of class C
⭸t
⭸t 2
⭸f
⭸2 A
2
− § A C µ0 0 2 ,
⭸t
⭸t
which is equivalent to the desired formula.
⭸A
⭸A
41. Again we have E D §f −
so that Gauss’s law becomes ρ/ 0 D § · E D § · §f −
D §2 f −
⭸t
⭸t
⭸
⭸
§ · A or §2 f D ρ/ 0 C
§ · A.
⭸t
⭸t
Q
42. (a) If A D A C §φ, then in order to have
Q
⭸A
⭸t
⭸φ
⭸A
DE C
C §
⭸t
⭸t
⭸φ
D §f C § ,
⭸t
§fQ D E C
⭸φ
⭸φ
.
we must have §fQ D § f C
. Thus, up to addition of a constant, fQ D f C
⭸t
⭸t
Q
Q D µ0 0 ⭸f is equivalent to
(b) The condition that § · A
⭸t
§ · A C §φ D µ0 0
⭸φ
⭸
f C
⭸t
⭸t
§ · A C §2 φ D µ0 0 §2 φ − µ0 0
⭸f
⭸2 φ
C
⭸t
⭸t 2
or
3
⭸2 φ
⭸f
.
D −§ · A C µ0 0
⭸t 2
⭸t
43. If the final equation in part (b) above can be solved for φ, then we may arrange things so that § · A D µ0 0
Then the equation in Exercise 40 is
,
⭸f
.
⭸t
0
-.
/
⭸
f
§2 A − µ0 0
D −µ0 J C § § · A − µ0 0
D −µ0 J
⭸t 2
⭸t
⭸2 A
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“runall” — 2005/8/1 — 15:18 — page 381 — #61
Section 7.6
Miscellaneous Exercises for Chapter 7
381
and the equation in Exercise 41 is
§2 f D
ρ
C µ0 0
0
⭸2 f
⭸t
or §2 f − µ0 0
⭸2 f
ρ
D .
2
⭸t
0
44. We check all the equations, given the assumptions.
§ · E D § · −
.
⭸f
⭸A
⭸
⭸
2
C §f D − § · A C §2 f D − µ0 0
C § f
⭸t
⭸t
⭸t
⭸t
/,
E
D −µ0 0
⭸2 f
⭸t 2
ρ
C §2 f D
0
from the second equation in Exercise 43.
§ * E D −§ * ⭸A
⭸A
⭸
⭸B
− §f D −§ *
D − § * A D −
⭸t
⭸t
⭸t
⭸t
§ · B D § · § * A D 0
§ * B D § * § * A D §§ · A − §2 A
identity
D §§ · A − µ0 0
⭸2 A
C µ0 J
⭸t 2
by the equation in part (b) of Exercise 42
D § µ0 0
using the condition § · A D µ0 0
⭸f
⭸
− µ0 0 −E C §f C µ0 J,
⭸t
⭸t
⭸f
⭸A
, and that
D §f − E
⭸t
⭸t
D µ0 0
⭸E
C µ0 J
⭸t
since we may assume f to be of class C2 .
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“runall” — 2005/8/1 — 15:24 — page 382 — #1
C H A P T E R
8
Vector Analysis in Higher Dimensions
8.1 AN INTRODUCTION TO DIFFERENTIAL FORMS
1. dx1 − 3 dx2 7, 3 D dx1 7, 3 − 3 dx2 7, 3 D 7 − 33 D −2.
2.
2 dx C 6 dy − 5 dz1, −1, −2
D 2 dx1, −1, 2 C 6 dy1, −1, −2 − 5 dz1, −1, −2
D 21 C 6−1 − 5−2 D 6.
3. 3 dx1 G dx2 4, −1, 2, 0 D 3 det dx1 4, −1
dx2 4, −1
4
dx1 2, 0
D 3 det dx2 2, 0
−1
2
D 32 D 6.
0
4.
4 dx G dy − 7 dy G dz0, 1, −1, 1, 3, 2
D 4 dx G dy0, 1, −1, 1, 3, 2 − 7 dy G dz0, 1, −1, 1, 3, 2
D 4 det dx0, 1, −1
dy0, 1, −1
D 4 det 0
1
dx1, 3, 2
dy0, 1, −1
− 7 det dy1, 3, 2
dz0, 1, −1
1
1
− 7 det 3
−1
dy1, 3, 2
dz1, 3, 2
3
D 4−1 − 75 D −39.
2
5.
2 dx1 G dx3 G dx4 C dx2 G dx3 G dx5 a, b, c



dx1 (a) dx1 (b) dx1 (c)
dx2 (a)



D 2 det  dx3 (a) dx3 (b) dx3 (c)  C det  dx3 (a)
dx4 (a) dx4 (b) dx4 (c)
dx5 (a)




1 0 5
0
0
0




0 
D 2 det  −1 9 0  C det  −1 9
4 1 0
2 −1 −2
dx2 (b)
dx3 (b)
dx5 (b)

dx2 (c)

dx3 (c) 
dx5 (c)
D 2−185 C 0 D −370.
6. ω3,−1,4 (a) D −9 dx C 4 dy C 192 dza1 , a2 , a3 D −9a1 C 4a2 C 192a3 .
7.
ω2,−1,−3,1 a, b D −6 dx1 G dx3 C dx2 G dx4 a, b
D −6 det dx1 a
dx3 a
dx2 a
dx1 b
C det dx3 b
dx4 a
dx2 b
dx4 b
D −6a1 b3 − a3 b1 C a2 b4 − a4 b2 .
382
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“runall” — 2005/8/1 — 15:24 — page 383 — #2
Section 8.1
An Introduction to Differential Forms
383
8.
ω0,−1,π/2 a, b D 1 dx G dy − 1 dy G dz C 4 dx G dza, b
D
dxa
dya
D
a1
a2
dxb
dyb
b1
b2
−
−
a2
a3
dya
dza
dyb
dzb
C 4
a1
a3
b2
b3
C 4
dxa
dza
dxb
dzb
b1
b3
D a1 b2 − a2 b1 − a2 b3 − a3 b2 C 4a1 b3 − a3 b1 9.
ωx,y,z 2, 0, −1, 1, 7, 5 D cos x
2
0
1
7
− sin z
0
−1
7
5
C y2 C 3
2
−1
1
5
D 14 cos x − 7 sin z C 11y2 C 3
10. From Definition 1.3 of exterior product,
3 dx C 2 dy − x dz G x2 dx − cos y dy C 7 dz
D 3x2 dx G dx C 2x2 dy G dx − x3 dz G dx − 3 cos y dx G dy − 2 cos y dy G dy C x cos y dz G dy
C 21 dx G dz C 14 dy G dz − 7x dz G dz
D 2x2 dy G dx − x3 dz G dx − 3 cos y dx G dy C x cos y dz G dy
C 21 dx G dz C 14 dy G dz
using 4,
D −2x2 C 3 cos y dx G dy C x3 C 21 dx G dz
C 14 − x cos y dy G dz
using 3.
11. Again from Definition 1.3 of exterior product,
y dx − x dy G z dx G dy C y dx G dz C x dy G dz
D yz dx G dx G dy − xz dy G dx G dy C y2 dx G dx G dz − xy dy G dx G dz
C xy dx G dy G dz − x2 dy G dy G dz
D 2xy dx G dy G dz
using 3 and 4.
12. Again from Definition 1.3 of exterior product,
2 dx1 G dx2 − x3 dx2 G dx4 G 2x4 dx1 G dx3 C x3 − x2 dx3 G dx4 D 4x4 dx1 G dx2 G dx1 G dx3 − 2x3 x4 dx2 G dx4 G dx1 G dx3
C 2x3 − x2 dx1 G dx2 G dx3 G dx4 − x3 x3 − x2 dx2 G dx4 G dx3 G dx4
D −2x3 x4 dx2 G dx4 G dx1 G dx3 C 2x3 − x2 dx1 G dx2 G dx3 G dx4
D 2x3 x4 C x3 − x2 dx1 G dx2 G dx3 G dx4
using 4,
using 3.
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“runall” — 2005/8/1 — 15:24 — page 384 — #3
384
Chapter 8
Vector Analysis in Higher Dimensions
13. Again from Definition 1.3 of exterior product,
x1 dx2 G dx3 − x2 x3 dx1 G dx5 G ex4 x5 dx1 G dx4 G dx5 − x1 cos x5 dx2 G dx3 G dx4 D x1 ex4 x5 dx2 G dx3 G dx1 G dx4 G dx5 − x2 x3 ex4 x5 dx1 G dx5 G dx1 G dx4 G dx5
− x12 cos x5 dx2 G dx3 G dx2 G dx3 G dx4 C x1 x2 x3 cos x5 dx1 G dx5 G dx2 G dx3 G dx4
D x1 ex4 x5 dx2 G dx3 G dx1 G dx4 G dx5 C x1 x2 x3 cos x5 dx1 G dx5 G dx2 G dx3 G dx4
D x1 ex4 x5 − x1 x2 x3 cos x5 dx1 G dx2 G dx3 G dx4 G dx5
using 4,
using 3.
14. Using Definition 1.1,
dxi1 G dxi2 G · · · G dxij G · · · G dxil G · · · G dxik a1 , a2 , ... , ak 

dxi1 a1 dxi1 a2 ... dxi1 ak 



#
#
#




#
#
#


#
#
#


 dxi a1 dxi a2 ... dxi ak 
j
j
j






#
#
#


D det 
#
#
#



#
#
#


 dxil a1 dxil a2 ... dxil ak 






#
#
#


#
#
#




#
#
#
dxik a1 dxik a2 ... dxik ak 

dxi1 a1 dxi1 a2 ... dxi1 ak 



#
#
#




#
#
#


#
#
#


 dxi a1 dxi a2 ... dxi ak 
l
l
l






#
#
#


D −det 
#
#
#



#
#
#


 dxij a1 dxij a2 ... dxij ak 






#
#
#


#
#
#




#
#
#
dxik a1 dxik a2 ... dxik ak (since switching rows l and j changes the sign of the determinant)
D −dxi1 G dxi2 G · · · G dxil G · · · G dxij G · · · G dxik a1 , a2 , ... , ak .
15. This is easier to show in person, but the point is that if you switch the two identical forms then, on the one hand,
nothing has changed and, on the other hand, formula (3) says that you now have the negative of what you started
with. So
dxi1 G dxi2 G · · · G dxij G · · · G dxij G · · · G dxik D −dxi1 G dxi2 G · · · G dxij G · · · G dxij G · · · G dxik
and therefore
dxi1 G dxi2 G · · · G dxij G · · · G dxij G · · · G dxik D 0.
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“runall” — 2005/8/1 — 15:24 — page 385 — #4
Section 8.1
An Introduction to Differential Forms
385
16. A k-form ω on Rn may be written as ω D
Fi1 ...ik dxi1 G · · · G dxik . For each summand, each of the k dxij ’s is
one of dx1 , dx2 , ... dxn . If k > n, then, by the pigeon hole principle, there must be at least one repeated term dxl
in dxi1 G · · · G dxik (i.e., it will look like dxi1 G · · · G dxl G · · · G dxl G · · · G dxik ). And so, by formula (4),
we have that dxi1 G · · · G dxik D 0. Hence every term of ω is zero.
Fi1 ...ik dxi1 G · · · G dxik , ω2 D
Gi1 ...ik dxi1 G · · · G dxik , and η D
Hj1 ...jl dxj1 G · · · G
17. Let ω1 D
dxjl . Then


Fi1 ...ik C Gi1 ...ik dxi1 G · · · G dxik  G
ω1 C ω2 G η D 
i1 ,...,ik
D
Hj1 ...jl dxj1 G · · · G dxjl
j1 ,...,jl
Fi1 ...ik C Gi1 ...ik Hj1 ...jl dxi1 G · · · G dxik G dxj1 G · · · G dxjl
i1 ,...,ik
j1 ,...,jl
D
Fi1 ...ik Hj1 ...jl dxi1 G · · · G dxik G dxj1 G · · · G dxjl
i1 ,...,ik
j1 ,...,jl
C
Gi1 ...ik Hj1 ...jl dxi1 G · · · G dxik G dxj1 G · · · G dxjl
i1 ,...,ik
j1 ,...,jl
D ω1 G η C ω2 G η.
18. Let ω D
Fi1 ...ik dxi1 G · · · G dxik , and η D
Gj1 ...jl dxj1 G · · · G dxjl . Then
Fi1 ...ik Gj1 ...jl dxi1 G · · · G dxik G dxj1 G · · · G dxjl .
ω G ηD
i1 ,...,ik
j1 ,...,jl
Now move dxj1 to the front by switching, in reverse order, with each of the dxip ’s. There are k switches so, by
formula (3), there are k sign changes and this last equation becomes
Fi1 ...ik Gj1 ...jl −1k dxj1 G dxi1 G · · · G dxik G dxj2 G · · · G dxjl .
ω G ηD
i1 ,...,ik
j1 ,...,jl
Similarly, we use k more interchanges to move dxj2 into the second position. We repeat this for each of the l
dxjq ’s. and our equation becomes
ω G ηD
i1 ,...,ik
j1 ,...,jl
Fi1 ...ik Gj1 ...jl −1k −1k · · · −1k dxj1 G · · · G dxjl G dxi1 G · · · G dxik D −1kl η G ω.
l times
19. Let ω D
Fi1 ...ik dxi1 G · · · G dxik , η D
dxum . Then
Gj1 ...jl dxj1 G · · · G dxjl , and τ D



ω G η G τ D 


Fi1 ...ik Gj1 ...jl dxi1 G · · · G dxik G dxj1 G · · · G dxjl 

i1 ,...,ik
j1 ,...,jl
Hu1 ...um dxu1 G · · · G
Hu1 ...um dxu1 G · · · G dxum
G
u1 ,...,um
D
Fi1 ...ik Gj1 ...jl Hu1 ...um dxi1 G · · · G dxik G dxj1 G · · · G dxjl G dxu1 G · · · G dxum .
i1 ,...,ik
j1 ,...,jl
u1 ,...,um
Similarly, calculate ω G η G τ and you will obtain the same result.
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“runall” — 2005/8/1 — 15:24 — page 386 — #5
386
Chapter 8
Vector Analysis in Higher Dimensions
20. Here ω D
Fi1 ...ik dxi1 G · · · G dxik , and η D
Gj1 ...jl dxj1 G · · · G dxjl and f is a function (or 0-form).
First we note that

 





f ω G η D 
fF
dx G · · · G dx  G 
G
dx G · · · G dx 
i1 ...ik
i1
ik
i1 ,...,ik
D
j1 ...jl
j1
jl
j1 ,...,jl
f Fi1 ...ik Gj1 ...jl dxi1 G · · · G dxik G dxj1 G · · · G dxjl
i1 ,...,ik
j1 ,...,jl
Df
Fi1 ...ik Gj1 ...jl dxi1 G · · · G dxik G dxj1 G · · · G dxjl
i1 ,...,ik
j1 ,...,jl
D f ω G η.
We will use this result to establish the second equality,
f ω G η D −1kl f η G ω by property 2 of Proposition 1.4,
D −1kl f η G ω
by the result established above,
D −1kl −1kl ω G f η by property 2 of Proposition 1.4,
D ω G f η.
Therefore, f ω G η D f ω G η D ω G f η.
8.2 MANIFOLDS AND INTEGRALS OF k-FORMS
1. Here the map is Xθ1 , θ2 , θ3 D 3 cos θ1 , 3 sin θ1 , 3 cos θ1 C 2 cos θ2 , 3 sin θ1 C 2 sin θ2 , 3 cos θ1 C 2 cos θ2 C
cos θ3 , 3 sin θ1 C 2 sin θ2 C sin θ3 .
Follow the lead of Example 2 from the text. Each component function is at least C1 so the mapping is at least
B B B
C1 . To see one-one, consider the equation Xθ1 , θ2 , θ3 D Xθ1 , θ2 , θ3 . The first two component equations
B
B
B
B
would then have cos θ1 D cos θ1 and sin θ1 D sin θ1 . Since 0 … θ1 , θ1 < 2π we see that θ1 D θ1 . Using this
B
information in the next two component functions, we make the same conclusion for θ2 and θ2 . Finally, use all of
B
this information in the last set of equations to see that θ3 D θ3 . So X is one to one and C1 . What is left to show
is that the tangent vectors Tθ1 , Tθ2 , and Tθ3 are linearly independent.
Tθ1 D −3 sin θ1 , 3 cos θ1 , −3 sin θ1 , 3 cos θ1 , −3 sin θ1 , 3 cos θ1 Tθ2 D 0, 0, −2 sin θ2 , 2 cos θ2 , −2 sin θ2 , 3 cos θ2 Tθ3 D 0, 0, 0, 0, sin θ3 , cos θ3 Because of the leading pair of zeros in Tθ2 and Tθ3 we can see that if c1 Tθ1 C c2 Tθ2 C c3 Tθ3 D 0, then c1 D 0.
Looking at the second pair of zeros in Tθ3 we can then see that c2 D 0. This would then force c3 D 0. So Tθ1 ,
Tθ2 , and Tθ3 are linearly independent. We have shown that the parametrized 3-manifold is a smooth parametrized
3-manifold.
2. As in Example 3, let’s begin by describing the location of the point x1 , y1 . It is anywhere in the annular region
described by l1 cos θ1 , l1 sin θ1 where 1 … l1 … 3 and 0 … θ1 < 2π. You can now describe x2 , y2 as being
this same annular region centered at x1 , y1 . Together this means that the locus of x2 , y2 is the interior of a
disk of radius 6. Using variables l2 and θ2 such that 1 … l2 … 3 and 0 … θ2 < 2π, the mapping is
Xl1 , θ1 , l2 , θ2 D l1 cos θ1 , l1 sin θ1 , l1 cos θ1 C l2 cos θ2 , l1 sin θ1 C l2 sin θ2 .
As before, the component functions are at least C1 so the mapping is at least C1 . As for one-one, consider
B B B B
Xl1 , θ1 , l2 , θ2 D Xl1 , θ1 , l2 , θ2 . From the first component functions we see that x1 , y1 lies on a circle
B
B
B
of radius l1 and xB 1 , yB 1 lies on a circle of radius l1 so l1 D l1 . Then, as in Exercise 1, cos θ1 D cos θ1 and
B
B
B
sin θ1 D sin θ1 . As 0 … θ1 , θ1 < 2π, we see that θ1 D θ1 . Now the rest of the argument follows in exactly the
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“runall” — 2005/8/1 — 15:24 — page 387 — #6
Section 8.2
Manifolds and Integrals of k-forms
387
same way since x2 , y2 is related to x1 , y1 in the same way that x1 , y1 is related to the origin. We now need
to show that the four tangent vectors are linearly independent.
Tl1 D cos θ1 , sin θ1 , cos θ1 , sin θ1 Tθ1 D −l1 sin θ1 , l1 cos θ1 , −l1 sin θ1 , l1 cos θ1 Tl2 D 0, 0, cos θ2 , sin θ2 Tθ2 D 0, 0, −l2 sin θ2 , l2 cos θ2 Look at the equation c1 Tl1 C c2 Tθ1 C c3 Tl2 C c4 Tθ2 D 0. Because of the leading pair of zeros in Tl2 and
Tθ2 we can see that c1 cos θ1 D c2 l1 sin θ1 and c1 sin θ1 D −c2 l1 cos θ1 . Solve for c1 in the first equation and
substitute into the second equation to get c2 l1 sin2 θ1 D −c2 l1 cos2 θ1 . Because l1 cannot be zero, this implies
that c2 D 0. This then implies that c1 D 0. Given that, we can make the same argument to show c3 D c4 D 0.
Therefore the four tangent vectors are linearly independent and we have described the states of the robot arm as
a smooth parametrized 4-manifold in R4 .
3. This is a combination of Example 3 and Exercise 2. Let’s begin by describing the location of the point x1 , y1 . It
is anywhere on a circle of radius 3 centered at the origin. So x1 , y1 D 3 cos θ1 , 3 sin θ1 where 0 … θ1 < 2π.
We can then describe x2 , y2 as being this same annular region centered at x1 , y1 . Together this means that
the locus of x2 , y2 is 3 cos θ1 C l2 cos θ2 , 3 sin θ1 C l2 sin θ2 where 1 … l2 … 2 and 0 … θ2 < 2π. Similarly
we describe x3 , y3 in terms of x2 , y2 using variables l3 and θ3 such that 1 … l3 … 2 and 0 … θ2 < 2π. The
mapping is
Xθ1 , l2 , θ2 , l3 , θ3 D 3 cos θ1 , 3 sin θ1 , 3 cos θ1 C l2 cos θ2 , 3 sin θ1 C l2 sin θ2 ,
3 cos θ1 C l2 cos θ2 C l3 cos θ3 , 3 sin θ1 C l2 sin θ2 C l3 sin θ3 .
As before, the component functions are at least C1 so the mapping is at least C1 . As for one-one, consider
B B B B B
B
Xθ1 , l2 , θ2 , l3 , θ3 D Xθ1 , l2 , θ2 , l3 , θ3 . From the first two component functions we see that cos θ1 D cos θ1
B
B
B
B
B
and sin θ1 D sin θ1 and 0 … θ1 , θ1 < 2π so θ1 D θ1 . Now, x2 , y2 lies on a circle of radius l2 and x2 , y2 B
B
lies on a circle of radius l2 with each circle centered at the same point x1 , y1 D xB 1 , yB 1 . So l2 D l2 . Then,
B
B
B
B
as above, cos θ2 D cos θ2 and sin θ2 D sin θ2 . As 0 … θ2 , θ2 < 2π, we see that θ2 D θ2 . Now the rest of the
argument follows in exactly the same way since x3 , y3 is related to x2 , y2 in the same way that x2 , y2 is
related to x1 , y1 .
We now need to show that the five tangent vectors are linearly independent.
Tθ1 D −3 sin θ1 , 3 cos θ1 , −3 sin θ1 , 3 cos θ1 , −3 sin θ1 , 3 cos θ1 Tl2 D 0, 0, cos θ2 , sin θ2 , cos θ2 , sin θ2 Tθ2 D 0, 0, −l2 sin θ2 , l2 cos θ2 , −l2 sin θ2 , l2 cos θ2 Tl3 D 0, 0, 0, 0, cos θ3 , sin θ3 Tθ3 D 0, 0, 0, 0, −l3 sin θ3 , l3 cos θ3 Look at the equation c1 Tθ1 C c2 Tl2 C c3 Tθ2 C c4 Tl3 C c5 Tθ3 D 0. Because of the leading pair of zeros in all but
the vector Tθ1 we conclude that c1 D 0. The remainder of the argument is exactly as in Exercise 2. Because the
first four components of Tl3 and Tθ3 are zero, we can see that c2 cos θ2 D c3 l2 sin θ2 and c2 sin θ2 D −c3 l2 cos θ2 .
Solve for c2 in the first equation and substitute into the second equation to get c3 l2 sin2 θ2 D −c3 l2 cos2 θ2 .
Because l2 cannot be zero, c3 D 0. This then implies that c2 D 0. Given that, we can make the same argument to
show c4 D c5 D 0. Therefore the five tangent vectors are linearly independent and we have described the states
of the robot arm as a smooth parametrized 5-manifold in R6 .
4. We can use spherical coordinates to describe the parametrized space. The point x1 , y1 , z1 can be written as
2 sin ϕ1 cos θ1 , 2 sin ϕ1 sin θ1 , 2 cos ϕ1 where 0 … ϕ1 … π and 0 … θ1 < 2π. We can then write x2 , y2 , z2 as x1 C sin ϕ2 cos θ2 , y1 C sin ϕ2 sin θ2 , z1 C cos ϕ2 where 0 … ϕ2 … π and 0 … θ2 < 2π. In other words,
our mapping is
Xθ1 , ϕ1 , θ2 , ϕ2 D 2 sin ϕ1 cos θ1 , 2 sin ϕ1 sin θ1 , 2 cos ϕ1 ,
2 sin ϕ1 cos θ1 C sin ϕ2 cos θ2 , 2 sin ϕ1 sin θ1 C sin ϕ2 sin θ2 , 2 cos ϕ1 C cos ϕ2 .
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“runall” — 2005/8/1 — 15:24 — page 388 — #7
388
Chapter 8
Vector Analysis in Higher Dimensions
As in the previous exercises, the fact that the component functions are at least C1 tells us that the mapping is at
least C1 . Checking one-one is a little more interesting than in the above exercises. Consider the implications of the
B B B B
B
equation Xθ1 , ϕ1 , θ2 , ϕ2 D Xθ1 , ϕ
1 , θ2 , ϕ2 . By the third component functions we see that cos ϕ1 D cos ϕ1 .
B
B
Because 0 … ϕ1 , ϕ1 … π we see that ϕ1 D ϕ1 . Substituting this into the sixth component function implies that
B . Now comparing the equations from the first two component functions we see that if ϕ D 0 or π then
ϕ2 D ϕ
2
1
B
θ1 need not be the same as θ1 . This is allowed—recall that the mapping might not be one-one on the boundary
B
B
B
of the domain. Other than on the boundary, cos θ1 D cos θ1 and sin θ1 D sin θ1 and so, as before θ1 D θ1 . Again,
substitute this into the equations that arise from the fourth and fifth component functions to conclude that, except
B
when ϕ2 is 0 or π, we must have θ2 D θ2 .
We now need to show that the four tangent vectors are linearly independent.
Tθ1 D −2 sin ϕ1 sin θ1 , 2 sin ϕ1 cos θ1 , 0, −2 sin ϕ1 sin θ1 , 2 sin ϕ1 cos θ1 , 0
Tϕ1 D 2 cos ϕ1 cos θ1 , 2 cos ϕ1 sin θ1 , −2 sin ϕ1 , 2 cos ϕ1 cos θ1 , 2 cos ϕ1 sin θ1 , −2 sin ϕ1 Tθ2 D 0, 0, 0, − sin ϕ2 sin θ2 , sin ϕ2 cos θ2 , 0
Tϕ2 D 0, 0, 0, cos ϕ2 cos θ2 , cos ϕ2 sin θ2 , − sin ϕ2 Look at the equation c1 Tθ1 C c2 Tϕ1 C c3 Tθ2 C c4 Tϕ2 D 0. There is a zero in the third component of all of
the tangent vectors except for Tϕ1 . This tells us that c2 D 0. If that is the case, then there is a zero in the sixth
component of all of the remaining tangent vectors except for Tϕ2 so c4 D 0. But then the leading trio of zeros in
Tθ2 implies that c1 D 0 which in turn would mean that c3 D 0. Therefore the four tangent vectors are linearly
independent and we have described the states of the robot arm as a smooth parametrized 4-manifold in R6 .
5. This is just an exercise in linear algebra. If x ∈ Rn is orthogonal to vi for i D 1, ... , k, then x · vi D 0 for
i D 1, ... , k. An arbitrary vector v in Span{v1 , ... , vk } is of the form v D c1 v1 C · · · C ck vk for scalars
c1 , ... , ck ∈ R. The calculation is straightforward:
x · v D x · c1 v1 C · · · C ck vk D c1 x · v1 C · · · C ck x · vk D c1 0 C · · · C ck 0 D 0.
In other words, x isorthogonal
to v.
π
ω D
6. By Definition 2.1,
x
ωxt x t dt. We have, x t D −a sin t, b cos t, c and also ω D b dx −
0
a dy C xy dz so that
π
ωD
[b−a sin t − ab cos t C ab cos t sin tc] dt
0
x
π
D ab
[− sin t − cos t C c sin t cos t] dt
0
π
D ab cos t − sin t C
c 2
sin t D −2ab.
2
0
7. Parametrize the unit circle C by xt D cos t, sin t, 0 … t … 2π. Then
2π
2π
ωD
ωxt − sin t, cos t dt D
sin t dx − cos t dy− sin t, cos t dt
0
C
D
2π
0
0
− sin2 t − cos2 t dt D
2π
−1 dt D −2π.
0
8. Parametrize the segment as xt D t, t, ... , t, 0 … t … 3. Then x t D 1, 1, ... , 1 and so
ωxt x t D t dx1 C t 2 dx2 C · · · C t n dxn 1, 1, ... , 1 D t C t 2 C · · · C t n .
Hence,
3
3
nC1 k
n
1
1
1
3
3kC1
t nC1 D
D
.
ωD
t C t 2 C · · · C t n dt D t 2 C t 3 C · · · C
2
3
n C 1
C
0
0
kD2 k
kD1 k C 1
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“runall” — 2005/8/1 — 15:24 — page 389 — #8
Section 8.2
Manifolds and Integrals of k-forms
389
ω D
ωXs,t Ts , Tt ds dt. For Xs, t D s cos t, s sin t, t, we have Ts D cos t,
6D
sin t, 0, and Tt D −s sin t, s cos t, 1. Then
9. By Definition 2.3,
S
ωXs,t Ts , Tt D t dx G dy C 3 dz G dx − s cos t dy G dzTs , Tt Dt
cos t
sin t
−s sin t
s cos t
0
cos t
C 3
1
−s sin t
− s cos t
sin t
0
s cos t
1
s
sin 2t.
2
D st − 3 cos t −
Thus
ωD
S
4π 1
0
0
s
sin 2t ds dt D
2
st − 3 cos t −
4π
0
1
1
t − 3 cos t − sin 2t dt
2
4
4π
1
1
D 4π2 .
D t 2 − 3 sin t C cos 2t 4
8
0
10. (a) First calculate the two tangent vectors for this parametrization of the helicoid. We have Tu1 D cos 3u2 ,
sin 3u2 , 0 and Tu2 D −3u1 sin 3u2 , 3u1 cos 3u2 , 5. Then

−5 sin 3u2

5 cos 3u2
T
,
T
D
det

u
u
Xu1 ,u2 1
2
−3u1
cos 3u2
sin 3u2
0

−3u1 sin 3u2

3u1 cos 3u2  D −9u21 − 25 < 0
5
for all u1 , u2 . Therefore this particular parametrization is incompatible with .
(b) There is more than one solution. One possible way to do this is to switch the ordering of the variables so that the resulting determinant is positive. Try the parametrization Yu1 , u2 D Xu2 , u1 D
u2 cos 3u1 , u2 sin 3u1 , 5u1 for 0 … u1 … 2π and 0 … u2 … 5. Then the tangent vectors are Tu1 D
−3u2 sin 3u1 , 3u2 cos 3u1 , 5 and Tu2 D cos 3u1 , sin 3u1 , 0. Then

−5 sin 3u1

T
,
T
D
det
 5 cos 3u1
u1
u2
Yu1 ,u2 −3u2
−3u2 sin 3u1
3u2 cos 3u1
5

cos 3u1

sin 3u1  D 9u21 C 25 > 0
0
for all u1 , u2 . Therefore this particular parametrization is now compatible with
(c) Since the goal is to change the sign of the resulting determinant, we can change


−5 sin 3u2 a1 b1


Xu1 ,u2 a, b D −det  5 cos 3u2 a2 b2  .
−3u1
a3 b3
.
to  where
(d) The discussion following Theorem 2.11 tells us what to do if the parametrization is compatible.
Since the
parametrization X is incompatible with
we make the following simple adjustment: S ω D − X ω. We
pause to calculate
ωXu1 ,u2 Tu1 , Tu2 D 5u2
cos 3u2
sin 3u2
−3u1 sin 3u2
3u1 cos 3u2
− u21 cos2 3u2 C u21 sin2 3u2 sin 3u2
0
3u1 cos 3u2
5
D 15u1 u2 − 5u21 sin 3u2 .
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“runall” — 2005/8/1 — 15:24 — page 390 — #9
390
Chapter 8
Vector Analysis in Higher Dimensions
Hence,
ωD−
5 2π
0
S
D5
0
5
0
15u1 u2 − 5u21 sin 3u2 du2 du1 D 5
2
u1 D −30π
2
5
5 2π
0
0
u21 sin 3u2 − 3u1 u2 du2 du1
5
3u1 u22 u2 D2π
− cos 3u2
du1 D 5
−6π2 u1 du1
−
3
2
0
u2 D0
5
u1 du1 D −15π2 u21 0 D −375π2 .
0
11. (a) For the parametrization given, we calculate the tangent vectors as Tu1 D cos u2 , sin u2 , 0,
Tu2 D −u1 sin u2 , u1 cos u2 , 0, and Tu3 D 0, 0, 1. Then

cos u2

T
,
T
,
T
D
det
 sin u2
u1
u2
u3
Xu
0
−u1 sin u2
u1 cos u2
0

0

0  D u1 .
1
√
As 0 … u1 … 5, this is positive when u1 Z 0. Note that when u1 D 0 the parametrization is not one-one
and also that Tu2 D 0 so Tu1 , Tu2 , and Tu3 are not linearly independent. In other words, the parametrization
is not smooth when u1 D 0. It is, however, smooth when u1 Z 0. You can easily see that the mapping
is one-one and at least C1 . To see that the tangent vectors are linearly independent, consider the equation
c1 Tu1 C c2 Tu2 C c3 Tu3 D 0. We see from the third components that c3 D 0. Look at the remaining
equations and we see that
cos u2 c1 − u1 sin u2 c2 D 0
sin u2 c1 C u1 cos u2 c2 D 0.
Multiply the first equation by − sin u2 and the second by cos u2 and add to obtain u1 c2 D 0. Because we
are assuming that u1 Z 0, this implies that c2 D 0 and therefore c1 D 0. This shows that the tangent vectors
are linearly independent and hence the parametrization is smooth when u1 Z 0. The conclusion is then that
the parametrization given is compatible with the orientation when it is smooth.
(b) We can read the boundary pieces right off of the original
√ parametrization: they are paraboloids that intersect
at z D −1 in a circle in the plane z D −1 of radius 5 centered at 0, 0, −1. The boundary is
⭸M D {x, y, z|z D x2 C y2 − 6, z … −1} ∪ {x, y, z|z D 4 − x2 − y2 , z Ú −1}.
We can easily adapt the parametrization to each of these pieces. For the bottom, use
Y1 : [0,
√
5] * [0, 2π ' R3 I
Y1 s1 , s2 D s1 cos s2 , s1 sin s2 , s12 − 6.
5] * [0, 2π ' R3 I
Y2 s1 , s2 D s1 cos s2 , s1 sin s2 , 4 − s12 .
For the top, use
Y2 : [0,
√
(c) On the bottom part of ⭸M the outward-pointing unit vector
V1 D 2x, 2y, −1
4x2 C 4y2 C 1
. In terms of Y1 , this is V1 D
2s1 cos s2 , 2s1 sin s2 , −1
.
4s12 C 1
On the top part of ⭸M the outward-pointing unit vector
V2 D 2x, 2y, 1
4x2 C 4y2 C 1
. In terms of Y2 , this is V2 D
2s1 cos s2 , 2s1 sin s2 , 1
.
4s12 C 1
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“runall” — 2005/8/1 — 15:24 — page 391 — #10
Section 8.3
The Generalized Stokes’s Theorem
391
12. The paraboloid can be parametrized as Xs, t D s, t, s2 C t 2 where 0 … s2 C t 2 … 4. Therefore, Ts D 1, 0, 2s
and Tt D 0, 1, 2t. Note that this parametrization is compatible with the orientation derived from the normal
N D −2x, −2y, 1 as


−2s 1 0


0 1  D 2s2 C 2t 2 C 1 > 0.
Xs,t Ts , Tt D det[N Ts Tt ] D  −2t
1
2s 2t
Therefore, we may compute ∫S ω as ∫X ω. So we begin by calculating
ωXs,t Ts , Tt D es Ct dx G dy C t dz G dx C s dy G dzTs , Tt 2
D es Ct
2
2
2
1
0
0
1
C t
2s
1
2t
0
C s
0
2s
1
2t
D es Ct − 2t 2 − 2s2 .
2
2
Use this in the calculation:
2
2
ωD
[es Ct − 2s2 C t 2 ] ds dt
S
60…s2 Ct 2 …4
2π 2
2
D
er − 2r2 r dr dθ using polar coordinates,
0
D
D
2π
0
2
0
1 r2
1 4
dθ
e − r 2
2
rD0
0
1 4
1
− 8 dθ D πe4 − 17.
e −
2
2
2π
13. We have, for the given parametrization, that Tu1 D 1, 0, 0, 42u1 − u3 , Tu2 D 0, 1, 0, 0, and Tu3 D
0, 0, 1, 2u3 − 2u1 . Thus,
ωXu1 ,u2 ,u3 Tu1 , Tu2 , Tu3 D u2 dx2 G dx3 G dx4 C 2u1 u3 dx1 G dx2 G dx3 Tu1 , Tu2 , Tu3 D u2
0
0
42u1 − u3 1
0
0
0
1
2u3 − 2u1 C 2u1 u3
1
0
0
0
1
0
0
0
1
D u2 8u1 − 4u3 C 2u1 u3 D 8u1 u2 − 4u2 u3 C 2u1 u3 .
Hence,
ωD
M
D
D
1 1 1
0
0
0
0
1 1
1
8u1 u2 − 4u2 u3 C 2u1 u3 du1 du2 du3
0
4u2 − 4u2 u3 C u3 du2 du3
2 − 2u3 C u3 du3 D 2 −
0
3
1
D .
2
2
8.3 THE GENERALIZED STOKES’S THEOREM
1. Using Definition 3.1,
dexyz D
⭸ xyz
⭸ xyz
⭸ xyz
e dx C
e dy C
e dz C exyz yz dx C xz dy C xy dz.
⭸x
⭸y
⭸z
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“runall” — 2005/8/1 — 15:24 — page 392 — #11
392
Chapter 8
Vector Analysis in Higher Dimensions
2. Using Definition 3.1,
dx3 y − 2xz2 C xy2 z D 3x2 y − 2z2 C y2 z dx C x3 C 2xyz dy C xy2 − 4xz dz.
3. Again, using Definition 3.1,
dx2 C y2 dx C xy dy D dx2 C y2 G dx C dxy G dy
D 2x dx C 2y dy G dx C y dx C x dy G dy
D 2y dy G dx C y dx G dy
D −y dx G dy
using (4) from Section 8.1,
using (3) from Section 8.1.
4. Again, using Definition 3.1,
dx1 dx2 − x2 dx1 C x3 x4 dx4 − x4 x5 dx5 D dx1 G dx2 − dx2 G dx1 C x4 dx3 C x3 dx4 G dx4 − x5 dx4 C x4 dx5 G dx5
D dx1 G dx2 − dx2 G dx1 C x4 dx3 G dx4 − x5 dx4 G dx5
D 2 dx1 G dx2 C x4 dx3 G dx4 − x5 dx4 G dx5
using 4 from Section 8.1,
using 3 from Section 8.1.
5. Again, using Definition 3.1,
dxz dx G dy − y2 z dx G dz D z dx C x dz G dx G dy − 2yz dy C y2 dz G dx G dz
D x dz G dx G dy − 2yz dy G dx G dz using 4 from Section 8.1,
D x C 2yz dx G dy G dz
using 3 from Section 8.1.
6. Again, using Definition 3.1,
dx1 x2 x3 dx2 G dx3 G dx4 C x2 x3 x4 dx1 G dx2 G dx3 D x2 x3 dx1 C x1 x3 dx2 C x1 x2 dx3 G dx2 G dx3 G dx4
C x3 x4 dx2 C x2 x4 dx3 C x2 x3 dx4 G dx1 G dx2 G dx3
D x2 x3 dx1 G dx2 G dx3 G dx4 C x2 x3 dx4 G dx1 G dx2 G dx3
D0
using 4 from Section 8.1,
using 3 from Section 8.1.
i means that the term dxi is omitted.
7. For this solution dx
dω D
n
i G · · · G dxn
dxi 2 G dx1 G · · · G dx
iD1
D
n
i G · · · G dxn
2xi dxi G dx1 G · · · G dx
iD1
D
n
−1i−1 2xi dx1 G · · · G dxn
using equation 3 of Section 8.1 repeatedly
iD1
D 2x1 − x2 C x3 − · · · C −1n−1 xn dx1 G · · · G dxn .
8. Let u D u1 , u2 , ... , un ; then
dfx0 u D fx1 x0 dx1 C fx2 x0 dx2 C · · · C fxn x0 dxn u
D fx1 x0 u1 C fx2 x0 u2 C · · · C fxn x0 un
D fx1 x0 , fx2 x0 , ... , fxn x0 · u
D §f x0 · u
D Du f x0 by Theorem 6.2 of Chapter 2.
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“runall” — 2005/8/1 — 15:24 — page 393 — #12
Section 8.3
The Generalized Stokes’s Theorem
393
9. For ω D F x, z dy C Gx, y dz, we have dω D Fx dx C Fz dz G dy C Gx dx C Gy dy G dz.
Expanding, this gives dω D Fx dx G dy C Gx dx G dz C Gy − Fz dy G dz. But we are told that
dω D z dx G dy C y dx G dz so
⭸F
⭸G
⭸G
⭸F
D z,
D y, and
−
D 0.
⭸x
⭸x
⭸y
⭸z
The first equation implies that F x, z D xz C f z for some differentiable function f of z alone. Similarly,
the second equation implies that Gx, y D xy C gy for some differentiable function g of y alone. Using
these results together with the third equation we see that x C g y D x C f z or g y D f z. This can
only be true if their common value is a constant C. So if g y D f z D C, then f z D Cz C D1 and
gy D Cy C D2 for arbitrary constants C, D1 , and D2 . We conclude that F x, z D xz C Cz C D1 and
Gx, y D xy C Cy C D2 .
10. If ω D 2x dy G dz − z dx G dy, then dω D 2 dx G dy G dz − dz G dx G dy D dx G dy G dz. From
Exercise 11 of Section 8.2, M is parametrized as X: D ' √
R3 I Xu1 , u2 , u3 D u1 cos u2 , u1 sin u2 , u3 where
2
2
D D {u1 , u2 , u3 |u1 − 6 … u3 … 4 − u1 , 0 … u1 … 5, 0 … u2 < 2π}. If we orient M by the 3-form
D dx G dy G dz, then

cos u2

X(u) Tu1 , Tu2 , Tu3 D det  sin u2
0
−u1 sin u2
u1 cos u2
0

0

0  D u1 Ú 0.
1
As before, this is strictly positive when the parametrization is smooth so the parametrization is compatible with
the orientation.
Therefore, using this orientation,
dω D
M
dω D
X
2π √ 5 4−u2
1
0
0
u21 −6
u1 du3 du1 du2 D 2π
√5
0
√
5
1
D 4π u21 − u41 2
4
0
5
D 4π u1 10 − 2u21 du1
25
25
−
D 25π.
2
4
On the other hand, ⭸M is parametrized on the bottom surface as
Y1 : [0,
√
5] * [0, 2π ' R3 I
Y1 s1 , s2 D s1 cos s2 , s1 sin s2 , s12 − 6
with tangent vector normal to ⭸M
V1 D
2s1 cos s2 , 2s1 sin s2 , −1
.
4s12 C 1
The boundary ⭸M is parametrized on the top surface as
Y2 : [0,
√
5] * [0, 2π ' R3 I
Y2 s1 , s2 D s1 cos s2 , s1 sin s2 , 4 − s12 with tangent vector normal to ⭸M
V2 D
2s1 cos s2 , 2s1 sin s2 , 1
.
4s12 C 1
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“runall” — 2005/8/1 — 15:24 — page 394 — #13
394
Chapter 8
Vector Analysis in Higher Dimensions
⭸M
Then we have that the induced orientation on ⭸M is given by
that on the bottom part of ⭸M
⭸M
Y1 s Ts1 , Ts2 D
a1 , a2 D
Xs1 ,s2 ,s12 −6 V1 , Ts1 , Ts2 
2s cos s2
1
 4s2 C 1

1

 2s1 sin s2
D det 
 4s2 C 1
1

 −1
4s12 C 1
cos s2
−s1 sin s2
sin s2
s1 cos s2
2s1
0
V, a1 , a2 . Therefore we see









4s3 C s1
… 0.
D − 1
4s12 C 1
The parametrization Y1 is incompatible with the induced orientation on ⭸M. Along the top part of ⭸M
⭸M
Y2 s Ts1 , Ts2 D
Xs1 ,s2 ,4−s12 V2 , Ts1 , Ts2 
2s cos s2
1
 4s2 C 1

1

 2s1 sin s2
D det 
 4s2 C 1
1

 1
4s12 C 1
cos s2
−s1 sin s2
sin s2
s1 cos s2
−2s1
0









4s3 C s1
D 1
Ú 0.
4s12 C 1
This parametrization is compatible with the induced orientation on ⭸M.
Therefore we set up our integral (changing signs in the first integrand because of the incompatibility of the
parametrization) to obtain the following.
⭸M
ωD−
ω C
Y1
D−
ω
Y2

2π √ 5 

2s1 cos s2



2π √ 5 

2s1 cos s2
C


0
0
0
D
2π √ 5
0
D
2π
0
D
0
2π
0
0
sin s2
2s1
s1 cos s2
0
− s12 − 6
cos s2
sin s2
sin s2
−2s1
s1 cos s2
0
− 4 − s12 cos s2
sin s2


−s1 sin s2 
ds1 ds2
s1 cos s2 



−s1 sin s2 
ds1 ds2
s1 cos s2 

[8s13 cos2 s2 C 2s12 − 10s1 ] ds1 ds2
4
#s1 cos 2s2 C
√
s D 5
1
3 4
s1 − 5s12 $
2
s1 D0
ds2
[25 cos 2s2 C 25/2] ds2 D [25/2 sin 2s2 C 25s2 /2]
2π
0
D 25π.
11. One integral is easy. Since ωD xy dz G dw and ⭸M D {x, y, z, w|x D 0, 8 − 2y2 − 2z2 − 2w2 D 0}, we
see that x D 0 along ⭸M so ⭸M ω D ⭸M 0 D 0.
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“runall” — 2005/8/1 — 15:24 — page 395 — #14
Section 8.3
The Generalized Stokes’s Theorem
395
Now dω D dxy G dz G dw D x dy G dz G dw C y dx G dz G dw. We can orient M any way we wish,
so we won’t worry about this—we’ll choose the orientation to be compatible with the parametrization.
Xu1 , u2 , u3 D 8 − 2u21 − 2u22 − 2u23 , u1 , u2 , u3 X : D ' R4 ,
where D D {u1 , u2 , u3 |u21 C u22 C u23 … 4} (i.e., the solid ball of radius 2). Then
dω D
dω D
dωXu Tu1 , Tu2 , Tu3 du1 du2 du3
M
X
lB



−4u1 −4u2 −4u3
1 0 0


0
1
0
8 − 2u21 − 2u22 − 2u23 0 1 0 C u1
D

lB 


0
0
1
0 0 1
D
lB









du1 du2 du3
8 − 2u21 C u22 C u23 − 4u21 du1 du2 du3 .
At this point it is helpful to switch to spherical coordinates. The previous quantity is then
2π π 2
D
8 − 2ρ2 − 4ρ2 sin2 ϕ cos2 θρ2 sin ϕ dρ dϕ dθ
0
0
0
D 8 · volume of B − 2
2π π 2
0
2π π
0
ρ4 sin ϕ C 2 sin3 ϕ cos2 θ dρ dϕ dθ
0
4
32
sin ϕ C 2 sin ϕ1 − cos2 ϕ cos2 θ dϕ dθ
D 8 · π23 − 2
3
0
0 5
2π
π
π
256π
64
{− cos ϕ C 2 cos2 θ− cos ϕ C cos3 ϕ/3
} dθ
D
−
3
5 0
0
ϕD0
2π
2π
256π
2
64
256π
128
4
2
2
D
−
−
2 C 2 cos θ · 2 − % dθ D
1 C cos θ dθ
3
5 0
3
3
5 0
3
2π
128
2
256π
5
−
D
C cos 2θ dθ using the half angle formula
3
5 0
3
3
2π
D
128 5
1
256π
256π
256π
−
−
D 0.
& 2π C sin 2θ ' D
3
5 3
3
3
3
0
12. (a) Using the generalized version of Stokes’s theorem (Theorem 3.2), we have
1
1
x dy G dz − y dx G dz C z dx G dy D
dx dy G dz − y dx G dz C z dx G dy
3 ⭸M
3 M
1
D
dx G dy G dz − dy G dx G dz C dz G dx G dy
3 M
1
3 dx G dy G dz using formula (3) of Section 8.1,
D
3 M
dx G dy G dz D
dx dy dz D volume of M.
D
M
l
M
(See Definition 2.6 and Example 6 of Section 8.2.)
(b) This generalizes the result demonstrated in part (a). Notice that the kth summand is −1k−1 xk multiplied
by the (n − 1)-form which is the wedge product of the dxi ’s in order with dxk missing. In other words, the
kth summand is
k G · · · G dxn
−1k−1 xk dx1 G · · · G dx
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“runall” — 2005/8/1 — 15:24 — page 396 — #15
396
Chapter 8
Vector Analysis in Higher Dimensions
k means that dxk is omitted. (Make the obvious adjustments to the expression if it is the first or last
where dx
term that is omitted.) Then
k G · · · G dxn .
dof the kth summand D −1k−1 dxk G dx1 G · · · G dx
Let ω denote the (n − 1)-form in the integrand. Then, using the generalized Stokes’s theorem,
1
1
ωD
dω
n ⭸M
n M
n
1
k−1
k G · · · G dxn ' .
D
dxk G dx1 G · · · G dx
& −1
n M kD1
Use formula (3) of Section 8.1 to “move” each dxk back into the slot from which it has been omitted and
collect terms to obtain
1
1
ωD
n dx1 G · · · G dxn D · · ·
dx1 · · · dxn .
n ⭸M
n M
M
It is entirely reasonable to take this last n-dimensional integral to represent the n-dimensional volume of M.
8.4 TRUE/FALSE EXERCISES FOR CHAPTER 8
1. True.
2. False. (There is a negative sign missing.)
3. True.
4. False.
5. True.
6. False. (A negative sign is missing.)
7. True.
8. False. (There should be no negative sign.)
9. True.
10. True.
11. False. X1, 1, −1 D X1, 1, 1, so X is not one-one on D.)
12. True. (Both manifolds are the same helicoid.)
13. False. (The agreement is only up to sign.)
14. True.
15. False. (This is only true if n is even.)
16. False. (A negative sign is missing.)
17. True.
18. False. dω D 0.
19. True. (dω would be an n C 1-form, and there are no nonzero ones on Rn .)
20. True. (This is the generalized Stokes’s theorem, since ⭸M D ∅.)
8.5 MISCELLANEOUS EXERCISES FOR CHAPTER 8
1. (a) First, by definition of the exterior product and derivative
df G g D
n
⭸f
⭸g
⭸
f g dxi D
g C f
&
' dxi
⭸
x
⭸
x
⭸
xi
i
i
iD1
iD1
n
Dg
by the product rule,
n
⭸f
⭸g
dxi C f
dxi
⭸
x
⭸
i
iD1
iD1 xi
n
D g G df C f G dg
D df G g C −10 f G dg.
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“runall” — 2005/8/1 — 15:24 — page 397 — #16
Section 8.5
Miscellaneous Exercises for Chapter 8
397
(b) If k D 0, then write ω D f so that
f Gj1 ...jl dxj1 G · · · G dxjl ) D
dω G η D df G η D d (
D
df Gj1 ...jl G dxj1 G · · · G dxjl
df G Gj1 ...jl C f G dGj1 ...jl G dxj1 G · · · G dxjl
D df G
Gj1 ...jl dxj1 G · · · G dxjl C f G
from (a),
dGj1 ...jl dxj1 G · · · G dxjl
D df G η C −1 f G dη.
0
(c) If l D 0, then write η D g so that
dω G η D dω G g D d (
gFi1 ...ik dxi1 G · · · G dxik )
D
dgFi1 ...ik G dxi1 G · · · G dxik
D
dg G Fi1 ...ik C g G dFi1 ...ik G dxi1 G · · · G dxik
D dg G ω C g G dω
D −1k ω G dg C dω G g
by part 2 of Proposition 1.4 (recall dg is a 1-form).
(d) In general,


dω G η D d 






D d



Fi1 ...ik dxi1 G · · · G dxik G
1…i1 <···<ik …n
1…i1 <···<ik …n
1…j1 <···<jl …n
Gj1 ...jl dxj1


G · · · G dx 
jl
1…j1 <···<jl …n






Fi1 ...ik Gj1 ...jl dxi1 G · · · G dxik G dxj1 G · · · G dxjl 



D
dFi1 ...ik Gj1 ...jl G dxi1 G · · · G dxik G dxj1 G · · · G dxjl
D
dFi1 ...ik G Gj1 ...jl C Fi1 ...ik G dGj1 ...jl G dxi1 G · · · G dxik G dxj1 G · · · G dxjl
D
dFi1 ...ik G Gj1 ...jl G dxi1 G · · · G dxik G dxj1 G · · · G dxjl
C
D
so by part (a),
Fi1 ...ik G dGj1 ...jl G dxi1 G · · · G dxik G dxj1 G · · · G dxjl
dFi1 ...ik G dxi1 G · · · G dxik G Gj1 ...jl G dxj1 G · · · G dxjl
C
Fi1 ...ik G −1k dxi1 G · · · G dxik G dGj1 ...jl G dxj1 G · · · G dxjl
since Gj1 ...jl is a 0-form and dGj1 ...jl is a 1-form,
D dω G η C −1k ω G dη.
2. (a) Define X : [0, 1] * [0, 1] * [0, 1] * [0, 1] ' R5 , Xu1 , u2 , u3 , u4 , u5 D u1 , u2 , u3 , u4 , u1 u2 u3 u4 .
Then Tu1 D 1, 0, 0, 0, u2 u3 u4 , Tu2 D 0, 1, 0, 0, u1 u3 u4 , Tu3 D 0, 0, 1, 0, u1 u2 u4 , and Tu4 D 0,
0, 0, 1, u1 u2 u3 . From this we see that

1

 0

Xu Tu1 , Tu2 , Tu3 , Tu4 D det 
 0
0
0
1
0
0
0
0
1
0

0

0 
 D 1.
0 

1
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“runall” — 2005/8/1 — 15:24 — page 398 — #17
398
Chapter 8
Vector Analysis in Higher Dimensions
(b) We can now calculate
dx1 G dx2 G dx3 G dx4
M

1

1 1 1 1

0
D
det 

0
0 0 0 0

u2 u3 u4
D
1 1 1 1
0
0
0
0
1
0
u1 u3 u4
0
0
1
u1 u2 u4
u1 u2 u3 du1 du2 du3 du4 D
0

0


0
 du1 du2 du3 du4

0

u1 u2 u3
1
.
8
3. (a) The curve C may be parametrized as xt D t, f t, a … t … b. Then x t D 1, f t and this is
compatible with the orientation of C. By Definition 2.1, we have
b
ωD
x
C
For ω D y dx this is
ωD
ωxt x t dt.
a
b
b
f t · 1 dt D
f t dt D area under the graph.
a
a
(b) Parametrize S by
X : [a, b] * [c, d] ' R3 I
Xu1 , u2 D u1 , u2 , f u1 , u2 .
The upward unit normal N is given by
−fx , −fy , 1
ND .
fx2 C fy2 C 1
The parametrization is compatible with the orientation since
Tu1 * Tu2 D
i
1
0
j
0
1
k
fu1
fu2
D −fu1 , −fu2 , 1
is parallel to N (when N is expressed in terms of the parametrization). Thus,
d b
ωD
ωD
X
S
c
a
ωXu1 ,u2 Tu1 , Tu2 du1 du2 .
For ω D z dx G dy, this is
d b
f u1 , u2 c
a
1
0
0
1
d b
du1 du2 D
f u1 , u2 du1 du2 D area under the graph.
c
a
(c) Parametrize M using
X : D ' Rn ,
Xu1 , ... , un−1 D u1 , ... , un−1 , f u1 , ... , un−1 .
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“runall” — 2005/8/1 — 15:24 — page 399 — #18
Section 8.5
Miscellaneous Exercises for Chapter 8
399
Then, depending on how M is oriented,
ω D ; ω D ; ···
ωXu Tu1 , ... , Tun−1 du1 · · · dun−1
X
M
D;
D
···
D
D;


f u1 , ... , un−1 det 

1
..

 du1 · · · dun−1
.
1
···
f u1 , ... , un−1 du1 · · · dun−1 D ;n-dimensional volume under the graph.
D
If you orient M with the unit normal
N D −1n fx1 , ... , fxn−1 , −1
fx1 2 C · · · C fxn−1 2 C 1
we can guarantee a C sign above.
4. (a) Define a parametrization
X : [0, 3] * [0, 2π ' R3 I
Xu1 , u2 D cos u2 , u1 , sin u2 .
Then we may define Xu a, b D det[N a b]. Note that X is compatible with this orientation as Tu1 D
0, 1, 0 and Tu2 D − sin u2 , 0, cos u2 so that

cos u2

0
T
,
T
D
det

u1
u2
Xu
sin u2
0
1
0

− sin u2

0
 D 1 > 0.
cos u2
(Note that the first column is the normal N in terms of the parametrization.)
(b) The boundary ⭸M consists of two disjoint pieces. The left piece is {x, 0, z|x2 C z2 D 1}, parametrized
by Y1 : [0, 2π ' R3 , Y1 t D cos t, 0, sin t. The right piece is {x, 3, z|x2 C z2 D 1}, parametrized
by Y2 : [0, 2π ' R3 , Y2 t D cos t, 3, sin t.
(c) We must first determine V, a unit vector tangent to M, normal to ⭸M, and pointing away from M. If you think
about the boundary pieces we looked at in part (b), a vector corresponding to the left side is V1 D 0, −1, 0
and corresponding to the right side is V2 D 0, 1, 0. Then, along the left circle of ⭸M,
and along the right circle of ⭸M,
⭸M
Y1 t a D
X0,t V1 , a
⭸M
Y2 t a D
X3,t V2 , a.

Note that
cos t

⭸M
0
Y1 t Tt D det 
sin t
So the parametrization Y1 is incompatible with
⭸M

⭸M

− sin t

0
 D −1.
cos t
. However,
cos t

⭸M
T
D
det
 0
t
Y2 t
sin t
So the parametrization Y2 is compatible with
0
−1
0
0
1
0

− sin t

0
 D 1.
cos t
.
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“runall” — 2005/8/1 — 15:24 — page 400 — #19
400
Chapter 8
Vector Analysis in Higher Dimensions
(d) If ω D z dx C x C y C z dy − x dz, we have
dω D dz G dx C dx C dy C dz G dy − dx G dz D dx G dy − dy G dz − 2 dx G dz.
Then using the orientation
and the parametrization X from part (a), we have

2π 3 
0
−
sin
u
1
0
0
−
sin
u
2
2



 du1 du2
−
− 2
dω D
dω D
0 cos u2
1
0
0 cos u2
M
X
0
0
2π 3
D
sin u2 − cos u2 du1 du2 D 3− cos u2 − sin u2 |2π
0 D 0.
0
0
On the other hand, using the parametrizations Y1 and Y2 for ⭸M in parts (b) and (c), we have (after reversing
the sign for the left piece because of the incompatibility with ⭸M )
⭸M
ωD−
ω C
Y1
D−
2π
ω
Y2
[sin t− sin t C cos t C sin t · 0 − cos tcos t] dt
0
C
2π
[sin t− sin t C cos t C 3 C sin t · 0 − cos tcos t] dt D 0.
0
5. If S 4 is the unit 4-sphere in R5 , then let B denote the 5-dimensional unit ball
B D {x1 , x2 , x3 , x4 , x5 |x12 C x22 C x32 C x42 C x52 … 1}.
Note that ⭸B D S 4 . Then using the generalized Stokes’s theorem, we have
S4
ωD
dω.
B
For ω D x3 dx1 G dx2 G dx4 G dx5 C x4 dx1 G dx2 G dx3 G dx5 we have dω D dx3 G dx1 G dx2 G dx4 G
dx5 C dx4 G dx1 G dx2 G dx3 G dx5 D dx1 G · · · G dx5 − dx1 G · · · G dx5 D 0. Hence S 4 ω D B 0 D 0.
⭸f
dxi and
6. (a) Let ω D f . Then df D
i ⭸xi
ddf D
i
D


2f
⭸f
⭸


 G dxi
d&
G dxj 
' G dxi D
⭸xi
i
j ⭸xj ⭸xi
⭸2 f
⭸2 f
dxj G dxi C
dxj G dxi ,
i<j ⭸xj ⭸xi
i>j ⭸xj ⭸xi
since the terms where i D j contain dxi G dxi D 0. By exchanging the roles of i and j in the second sum,
we find
ddf D
⭸2 f
⭸2 f
dxj G dxi C
dxi G dxj
i<j ⭸xj ⭸xi
i<j ⭸xi ⭸xj
D
&−
i<j
⭸2 f
⭸2 f
C
' dxi G dxj D 0
⭸xj ⭸xi
⭸xj ⭸xi
since the mixed partials are equal because f is of class C2 .
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“runall” — 2005/8/1 — 15:24 — page 401 — #20
Section 8.5
Miscellaneous Exercises for Chapter 8
401
(b) Now
 


d 
ddω D d 


D d
Fi1 ...ik dxi1




G · · · G dx 
ik
1…i1 <···<ik …n
dFi1 ...ik G dxi1


G · · · G dx 
ik
1…i1 <···<ik …n
D
[ddFi1 ...ik G dxi1 G · · · G dxik
1…i1 <···<ik …n
C −11 dFi1 ...ik G ddxi1 G · · · G dxik ] from Exercise 1,
dFi1 ...ik G ddxi1 G · · · G dxik D−
1…i1 <···<ik …n
since ddFi1 ...ik D 0 from part (a). But
ddxi1 G · · · G dxik D d1 dxi1 G · · · G dxik D d1 G dxi1 G · · · G dxik D 0.
Hence ddω D 0, as desired.
7. (a) If ω is a 0-form, write ω D f . Then, using the first row of the chart, the 1-form dω corresponds to the vector
field §f . Hence, from the second row of the chart, ddω is the 2-form that corresponds to § * §f . Thus
ddω D 0 “translates” to the statement § * §f D 0.
(b) If ω is a 1-form, it corresponds to the vector field F and, using the second row of the chart, dω is the 2-form
that corresponds to § * F, another vector field. Then, using the third row of the chart, ddω is the 3-form
that corresponds to § · § * F. Hence, ddω D 0 “translates” to the statement that § · § * F D 0.
8. (a) The outward unit normal N D x, y, z gives orientation form Xu a1 , a2 D det[N a1 a2 ] where X is a
parametrization of S. For a specific parametrization we can use
X : [0, π] * [0, 2π ' R3 I
Xu1 , u2 D sin u1 cos u2 , sin u1 sin u2 , cos u1 .
Then Tu1 D cos u1 cos u2 , cos u1 sin u2 , − sin u1 and Tu2 D − sin u1 sin u2 , sin u1 cos u2 , 0, so that

sin u1 cos u2

sin u1 sin u2
T
,
T
D
det

u1
u2
Xu
cos u1
cos u1 cos u2
cos u1 sin u2
− sin u1

− sin u1 sin u2

sin u1 cos u2  D sin u1 Ú 0.
0
In fact, this quantity is strictly greater than 0 when the parametrization is smooth and so the parametrization
is compatible with the orientation.
Next we note that on S we have ω D x dy G dz C y dz G dx C z dx G dy as the denominators in ω
are all 1 on S. Therefore,
ωD
ω
S
X

2π π 

D
sin u1 cos u2

0
0 
C sin u1 sin u2
C cos u1
cos u1 sin u2
− sin u1
sin u1 cos u2
0
− sin u1
cos u1 cos u2
cos u1 cos u2
cos u1 sin u2
0
− sin u1 sin u2


− sin u1 sin u2 
du1 du2
sin u1 cos u2 

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“runall” — 2005/8/1 — 15:24 — page 402 — #21
402
Chapter 8
Vector Analysis in Higher Dimensions
D
2π π
0
sin3 u1 C cos2 u1 sin u1 du1 du2 D
0
2π π
sin u1 du1 du2
0
0
D 2π− cos u1 |π
0 D 4π.
(b) For ω as given we calculate
d
d
d
x
y2 C z2 − 2x2 dx − 3xy dy − 3xz dz
x2 C y2 C z2 5/2
y
−3xy dx − x2 C z2 − 2y2 dy − 3yz dz
x2 C y2 C z2 5/2
D
x2 C y2 C z2 3/2
D
x2 C y2 C z2 3/2
z
x2 C y2 C z2 3/2
D
−3xz dx − 3yz dy − x2 C y2 − 2z2 dz
x2 C y2 C z2 5/2
Hence,
dω D
1
2
2
2
*y C z − 2x dx G dy G dz
x2 C y2 C z2 5/2
C x2 − 2y2 C z2 dy G dz G dx
Cx2 C y2 − 2z2 dz G dx G dy+
This is identically equal to 0 wherever it is defined.
(c) Since M does not include the origin, we have M dω D M 0 D 0 from part (b).
⭸M consists of two pieces. The outer piece S1 is the unit sphere x2 C y2 C z2 D 1, oriented by the
outward unit normal n1 D x, y, z. The inner piece is the sphere x2 C y2 C z2 D a2 of radius a, oriented
by inward unit normal n2 D −x, −y, −z/a. Then, using Proposition 2.4, we have
xi C yj C zk
ωD
F · dS where F D
.
x2 C y2 C z2 3/2
⭸M
6⭸M
In the following calculation we will use the fact that x2 C y2 C z2 is 1 on S1 and is a2 on S2 .
ωD
F · n1 dS C
F · n2 dS
⭸M
6S1
6S2
D
6S1
1 dS C
6S2
−
1
dS
a2
1
surface area of S2 a2
D 1surface area of S1 −
D 4π −
1
4πa2 D 0.
a2
This verifies Theorem 3.2.
(d) No—since ω is not defined at the origin, Theorem 3.2 does not apply.
(e) Let M be the 3-manifold bounded on the outside by S, oriented with the outward normal, and on the inside
/ M, so we have
by S , oriented by the inward normal. Then 0 ∈
0D
dω D
ωD ω C
ω D ω − 4π.
M
⭸M
S
S
S
The last equality follows from part (c). The conclusion is that ∫S ω D 4π.
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“runall” — 2005/8/1 — 15:24 — page 403 — #22
Section 8.5
Miscellaneous Exercises for Chapter 8
403
9. Because ⭸M D ∅, the note following Theorem 3.2 advises us to take ∫⭸M ω G η to be 0 in the equation
∫⭸M ω G η D ∫M dω G η. Now substitute the results of Exercise 1 to get
0D
dω G η D
dω G η C −1k ω G dη D
dω G η C −1k
ω G dη.
M
M
M
Pull this last piece to the other side to obtain the result
ω G dη D
−1kC1
M
M
dω G η.
M
10. By the generalized Stokes’s theorem,
fω D
df ω
⭸M
M
df G ω C f G dω by the result of Exercise 1,
D
M
D
df G ω C f dω.
M
Hence
f dω D
M
⭸M
fω −
df G ω.
M
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