Student Solution Manual to accompany the 5th edition of Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach

315 Pages·5·5.765 MB·English

by  John H. Hubbard and Barbara Burke Hubbard

Student Solution Manual
to accompany the 5th edition of
Vector Calculus, Linear Algebra, and
Differential Forms: A Unified Approach
John Hamal Hubbard Barbara Burke Hubbard
Cornell University
Universite´ Aix-Marseille
Matrix Editions Ithaca, NY 14850 MatrixEditions.com
Copyright2015byMatrixEditions
214UniversityAve. Ithaca,NY14850
www.MatrixEditions.com
All rights reserved. This book may not be translated, copied, or reproduced, in
whole or in part, in any form or by any means, without prior written permission
fromthepublisher,exceptforbriefexcerptsinconnectionwithreviewsorscholarly
analysis.
Printed in the UnitedStatesof America
10 9 8 7 6 5 4 3 2
ISBN: 978-0-9715766-9-8
Contents
Note to students iv
Chapter 0: Preliminaries 1
Chapter 1: Vectors, matrices, and derivatives 9
Chapter 1 review exercises 31
Chapter 2: Solving equations 40
Chapter 2 review exercises 74
Chapter 3: Higher partial derivatives, quadratic forms,
and manifolds 85
Chapter 3 review exercises 122
Chapter 4: Integration 133
Chapter 4 review exercises 173
Chapter 5: Volumes of manifolds 182
Note on parametrizations 183
Chapter 5 review exercises 204
Chapter 6: Forms and vector calculus 207
Chapter 6 review exercises 266
Appendix A: Some harder proofs 277
Index 305
Note to Students
This manual gives solutions to all odd-numbered exercises in the fifth edi-
tion of Vector Calculus, Linear Algebra, and Di↵erential Forms: A Unified
Approach.
You should not attempt to use Pleasethinkabouttheexercisesbeforelookingatthesolutions! Youwill
this manual with a previous edi- get more out of them that way. Some exercises are straightforward; their
tion of the textbook, as some sec- solutions are correspondingly straightforward. Others are “entertaining”.
tions of the textbook have been
You may be tempted go straight to the solution manual to look for
renumbered, new sections have
problems similar to those assigned as your homework, without reading the
been added, some exercises have
text. This is a mistake. You may learn some of the simpler material that
been moved from one section to
another, some earlier exercises way,butifyoudon’treadthetext,soonyouwillfindyoucannotunderstand
have been discarded, and some the solutions. The textbook contains a lot of challenging material; you
new ones have been added. really have to come to terms with the underlying ideas and definitions.
Inthismanual,wedonotnum- Errata for this manual and for the textbook will be posted at
ber every equation. When we do,
www.matrixeditions.com/errata.html
weuse(1),(2),... ,startingagain
with (1) in a new solution.
If you think you have found errors either in the text or in this manual,
please email us at
[email protected]
or
[email protected]
so that we can make the errata lists as complete as possible.
John H. Hubbard and Barbara Burke Hubbard
iv
Solutions for Chapter 0
0.2.1 a. The negation of the statement is: There exists a prime number
such that if you divide it by 4 you have a remainder of 1, but which is not
the sum of two squares.
Solution0.2.1: Thenegationin b. The negation is: There exist x R and ✏>0 such that for all � >0
part a is false; the original state- there exists y R with y x <� an2d y2 x2 ✏.
mentistrue: 5=4+1,13=9+4, 2 | � | | � |�
c. Thenegationis: Thereexists✏>0suchthatforall� >0,thereexist
17 = 16 + 1, 29 = 25 + 4,...,
97=81+16,.... x,y R with x y <� and y2 x2 ✏.
2 | � | | � |�
If you divide a whole number
(prime or not) by 4 and get a 0.3.1 a. (A B) (A B) b. (A A) (B B) c. A A
⇤ ⇤ ⇤ ⇤ ⇤ ⇤ ⇤
remainderof3,thenitisneverthe
sum of two squares: 3 is not, 7 0.4.1 a. No: Many people have more than one aunt, and some have none.
is not, 11 is not, etc. You may b. No, 1 is not defined.
0
well be able to prove this; it isn’t
c. No.
all that hard. But the original
statement about primes is pretty
0.4.3 Here are some examples:
tricky.
The negation in part b is false a. “DNAof”frompeople(excludingclonesandidenticaltwins)toDNA
also, and the original statement patterns.
is true: it is the definition of the
b. f(x)=x.
mapping x x2 being continu-
7!
ous. Butinpartc,thenegationis
0.4.5 The following are well defined: g f : A C, h k : C C,
true,andtheoriginalstatementis � ! � !
k g : B A, k h : A A, and f k : C B. The others are not,
false. Indeed, if you take ✏ = 1 � ! � ! � !
unless some of the sets are subsets of the others. For example, f g is
and any � > 0 and set x = 1, �
y= 1 + �, then � not because the codomain of g is C, which is not the domain of f, unless
� 2
C A.
y2 x2 = y x y+x ⇢
| � | | � || |
� 2 � 0.4.7 a. f g(h(3)) =f g( 1) =f( 3)=8
= + >1. � �
2 � 2 b. f g(h(1)) =f g( 2) =f( 5)=25
✓ ◆ � � �� � �
The original statement says that
the function x x2 is uniformly 0.4.9I�fthearg�umen�tofthe�squarerootisnonnegative,thesquarerootcan
7! be evaluated, so the open first and the third quadrants are in the natural
continuous, and it isn’t.
domain. The x-axis is not (since y = 0 there), but the y-axis with the
origin removed is in the natural domain, since x/y =0 there.
0.4.11 The function is defined for x R 1<x<0, or 0<x . It is
{ 2 | � }
also defined for the negative odd integers.
0.5.1 Without loss of generality we may assume that the polynomial is of
odd degree d and that the coe�cient of xd is 1. Write the polynomial
xd+a xd 1+ +a .
d 1 � 0
� ···
Let A= a + + a +1. Then
0 d 1
| | ··· | � |
a Ad 1+a Ad 2+ +a (A 1)Ad 1,
d 1 � d 2 � 0 �
� � ···  �
� �
� �
� 1 �
2 Solutionsfor Chapter0
and similarly,
a ( A)d 1+a ( A)d 2+ +a (A 1)Ad 1.
d 1 � d 2 � 0 �
� � � � ···  �
� �
Therefor�e, �
� �
p(A) Ad (A 1)Ad 1 >0 and p( A) ( A)d+(A 1)Ad 1 <0.
� �
� � � �  � �
By the intermediate value theorem, there must exist c with c < A such
| |
that p(c)=0.
0.5.3 Suppose f : [a,b] [a,b] is continuous. Then the function
!
g(x) = x f(x) satisfies the hypotheses of the intermediate value theo-
�
rem (Theorem 0.5.9): g(a) 0 and g(b) 0. So there must exist x [a,b]
 � 2
such that g(x)=0, i.e., f(x)=x.
0.5.5 First solution
Let 1k=1bk be a rearrangement of 1k=1ak, which means that there
exists aPbijective map ↵:N!N such thPat b↵(k) =ak.
We give two solutions to Exer- Since the series 1k=1ak is absolutely convergent, we know it is conver-
cise 0.5.5. gent. Let A= 1k=P1ak. We must show that
P m
( ✏>0)( M) m M = b A <✏ .
k
8 9 � ) � � � !
�kX=1 �
� �
First choose N such that 1N+1|ak|<✏��/2; in that��case,
n P
1 1
(n>N) = a A a a <✏/2.
k k k
) � � � | | | |
�kX=1 � kX=n k=XN+1
� �
Now set � �
� �
M =max ↵(1),...,↵(N) ,
{ }
i.e., M is so large that all of a ,...,a appear among b ,...,b . Then
1 N 1 M
m N
1
m>M = b A a A + a <✏.
k k k
) � � �� � � | |
�kX=1 � �kX=1 � k=XN+1
� � � �
Second solution � � � �
� � � �
The series 1k=1ak+|ak| is a convergent series of positive numbers, so
its sum is the sup of all finite sums of terms.
P
If 1k=1bk is a rearrangement of the same series, then 1k=1bk+|bk| is
also aPrearrangement of 1k=1ak+|ak|. P
But the finite sums oPf terms of 1k=1ak +|ak| and 1k=1bk +|bk| are
exactly the same set of numbers, so these two series converge to the same
P P
limit. The same argument says that
1 1
a = b .
k k
| | | |
k=1 k=1
X X
Solutionsfor Chapter0 3
Thus
The first equality is justified
becauseboth series 1 1 1 1 1 1
b = (b + b ) b = (a + a ) a = a .
k k k k k k k k
| | � | | | | � | |
1 1 k=1 k=1 k=1 k=1 k=1 k=1
(b + b ) and b X X X X X X
k k k
| | | |
k=1 k=1
X X
0.6.1 a. Begin by listing the integers between 1 and 1, then list the
converge, so their di↵erence does
�
also. numbersbetween 2and2thatcanbewrittenwithdenominators 2and
� 
whichhaven’talreadybeenlisted,thenlisttherationalsbetween 3and3
�
Solution 0.6.1, part a: There thatcanbewrittenwithdenominators 3andhaven’talreadybeenlisted,

areinfinitelymanywaysofsolving etc. This will eventually list all rationals. Here is the beginning of the list:
thisproblem,andmanyseemjust 3 1 1 3 8 5 7 5 4 2 1 1 2
as naturalas the one we propose. 1,0,1, 2, , , , ,2, 3, , , , , , , , , ,...
� � �2 �2 2 2 � �3 �2 �3 �3 �3 �3 �3 3 3
b. Justasbefore,listfirstthefinitedecimalsin[ 1,1]withatmostone
�
digitafterthedecimalpoint(thereare21ofthem),thentheonesin[ 2,2]
�
with at most two digits after the decimal, and which haven’t been listed
earlier (there are 380 of these), etc.
0.6.3 It is easy to write a bijective map ( 1,1) R. For instance,
� !
Solution0.6.3: Thesameargu- x ⇡x
f(x)= or g(x)=tan .
mentholdsifoneusesthefunction 1 x2 2
�
g, with derivative
The derivative of the first mapping is
⇡
g0(x)= 2(1+cos2 ⇡x). (1 x2)+2x2 1+x2
2 f0(x)= � = >0
(1 x2)2 (1 x2)2
� �
so the mappingis monotoneincreasingon ( 1,1), henceinjective. Sinceit
�
is continuous on ( 1,1) and
�
lim f(x)= and lim f(x)=+ ,
x 1 �1 x +1 1
&� %
it is surjective by the intermediate value theorem.
0.6.5 a. To the right, if a A is an element of a chain, then it can be
2
continued:
a,f(a),g f(a) ,f g(f(a)) ,...,
and this is clearly the only pos�sible�con�tinuation�.
Similarly, if b is an element of a chain, then the chain can be continued
b,g(b),f g(b) ,g f(g(b)) ,...,
and again this is the only poss�ible c�ont�inuation�.
Let us set A = g(B) and B = f(A). Let f : A B be the unique
1 1 1 1
!
map such that f (g(b)) = b for all b B, and let g : B A be the
1 1 1
2 !
unique map such that g (f(a)) = a for all a A. To the left, if a A is
1 1
2 2
an element of a chain, we can extend to f (a),a. Then if f(a ) B , we
1 1 1
2
can extend one further, to g (f (a)), and if g (f (a)) A , we can extend
1 1 1 1 1
2
further to
f g (f (a)) , g f (a) , f (a), a.
1 1 1 1 1 1
� � � �
4 Solutionsfor Chapter0
Eitherthiswillcontinueforeveroratsomepointwewillrunintoanelement
of A that is not in A or into an element of B that is not in B ; the chain
1 1
necessarily ends there.
b. As we saw, any element of A or B is part of a unique infinite chain
to the right and of a unique finite or infinite chain to the left.
Part c: c. Since every element of A and of B is an element of a unique chain,
(1) can be uniquely continued and since this chain satisfies either (1), (2), or (3) and these are exclusive,
forever the mapping h is well defined.
(2) can be continued to an el- If h(a ) = h(a ) and a belongs to a chain of type 1 or 2, then so does
1 2 1
ement of A that is not in f(a ),hencesodoesf(a ),hencesodoesa ,andthenh(a )=h(a )implies
1 2 2 1 2
the imageof g
a =a , since f is injective.
1 2
(3) can be continued to an el-
Now suppose that a belongs to a chain of type 3. Then a is not the
ement of B that is not in 1 1
first element of the list, so a A , and h(a ) = f (a ) is well defined.
the imageof f 1 2 1 1 1 1
The element h(a ) is a part of the same chain, hence also of type 3, and
2
h(a ) = f (a ). But then a = g(f (a )) = g(f (a )) = a . So h is
2 1 2 1 1 1 1 2 2
injective. Nowanyelementb B belongstoamaximalchain. Ifthischain
2
is of type 1 or 2, then b is not the first element of the chain, so b B and
1
2
h(g (b)) = f(g (b)) = b, so b is in the image. If b is in a chain of type 3,
1 1
then b=f (g(b))=h(g(b)), so again b is in the image of h, proving that h
1
is surjective.
d. This is the only entertaining part of the problem. In this case, there
is only one chain of type 1, the chain of all 0’s. There are only two chains
of type 2, which are
f 1 g 1 f 1 g 1 f 1
+1 ....
! 2 ! 2 ! 4 ! 4 ! 8
f 1 g 1 f 1 g 1 f 1
1 ....
� !�2 !�2 !�4 !�4 !�8
Allotherchainsareoftype3,andendtotheleftwithanumberin(1/2,1)
orin( 1, 1/2), whicharethepointsinB f(A). Suchasequencemight
� � �
be
3 g 3 f 3 g 3 f 3
....
4 ! 4 ! 8 ! 8 ! 16
Theexistenceofthisfunctionh Following our definition of h, we see that
is Bernstein’s theorem.
0 if x=0
Solution0.6.7: TwosetsAand h(x)= x/2 if x= 1/2k for some k 0
8
B have the same cardinality if ± �
>< x if x= 1/2k for some k 0.
thereexistsaninvertiblemapping
6 ± �
A ! B. Here, finding such an >:
invertible map would be di�cult, 0.6.7 a. There is an obvious injective map [0,1) [0,1) [0,1), given
! ⇥
so instead we use Bernstein’s the- simply by g :x (x,0).
orem, which says that if there is Weneedtoco7!nstructaninjectivemapintheoppositedirection;thatisa
an injective map A B and an
! lotharder. Takeapointin(x,y) [0,1) [0,1),andwritebothcoordinates
injective map B A, then there 2 ⇥
! as decimals:
is a invertiblemap A B.
!
x=.a a a ... and y =.b b b ...;
1 2 3 1 2 3
Solutionsfor Chapter0 5
ifeithernumbercanbewrittenintwoways,oneendingin0’sandtheother
in 9’s, use the one ending in 0’s.
Remember that Now consider the number f(x,y) = .a1b1a2b2a3b3.... The mapping f
injective = one to one is injective: using the even and the odd digits of f(x,y) allows you to
surjective = onto reconstruct x and y. The only problem you might have is if f(x,y) could
bijective = invertible. be written in two di↵erent ways, but as constructed f(x,y) will never end
in all 9’s, so this doesn’t happen.
Bernstein’s theorem is the ob- Note that this mapping is not surjective; for instance, .191919... is not
ject of Exercise0.6.5. intheimage. ButBernstein’stheoremguaranteesthatsincethereareinjec-
tive maps both ways, there is a bijection between [0,1) and
[0,1) [0,1).
⇥
b. The proof of part a works also to construct a bijective mapping
(0,1) (0,1) (0,1). But it is easy to construct a bijective mapping
! ⇥
(0,1) R, for instance x cot(⇡x). If we compose these mappings, we
! 7!
find bijective maps
R (0,1) (0,1) (0,1) R R.
! ! ⇥ ! ⇥
c. We can use part b repeatedly to construct bijective maps
Part c: The map
R f R R(f,id)(R R) R=R R2 (f,id)(R R) R2
(f,id):R R R ! ⇥ ! ⇥ ⇥ ⇥ ! ⇥ ⇥
mapsthefirstRt⇥oR⇥!Randthe =R⇥R3 (f!,id)(R⇥R)⇥R3 =R⇥R4··· .
secondR to itself. Continuing this way, it is easy to get a bijective map R Rn.
!
0.6.9 It is not possible. For suppose it were, and consider as in equation
0.6.3 the decimal made up of the entries on the diagonal. If this number
is rational, then the digits are eventually periodic, and if you do anything
systematic to these digits, such as changing all digits that are not 7 to
7’s, and changing 7’s to 5’s, the sequence of digits obtained will still be
Solution 0.7.3, part b: If you eventually periodic, so it will represent a rational number. This rational
hadtroublewiththis,notethatit numbermustappearsomeplaceinthesequence,butitdoesn’t,sinceithas
followsfromProposition0.7.5(ge- a di↵erent kth digit than the kth number for every k.
ometrical representation of mul- The only weakness in this argument is that it might be a number that
tiplication of complex numbers)
can be written in two di↵erent ways, but it isn’t, since it has only 5’s and
that
7’s as digits.
1 1
= ,
z z 0.7.1 “Modulus of z,” “absolute value of z,” and z are synonyms. “Real
since zz�1 =���� 1����, wh|ic|h has length part of z” is the same as Rez = a. “Imaginary p|ar|t of z” is the same as
1. The modulus (absolute value) Imz =b. The “complex conjugate of z” is the same as z¯.
of 3+4i is p25=5, so the mod-
ulus of (3+4i)�1 is 1/5. It fol- 0.7.3 a. The absolute value of 2+4i is |2+4i| = 2p5. The argument
lowsfromthesecondstatementin (polarangle)of2+4iisarccos1/p5,whichyoucouldalsowriteasarctan2.
Proposition 0.7.5 that the polar b. The absolute value of (3+4i) 1 is 1/5. The argument (polar angle)
angleofz�1isminusthepolaran- is arccos(3/5). �
gle of z, since the two angles sum �
to 0, which is the polar angle of c. The absolute value of (1+i)5 is 4p2. The argument is 5⇡/4. (The
the product zz�1 = 1. The polar complex number 1 + i has absolute value p2 and polar angle ⇡/4. De
angleof 3+4i is arccos(3/5). Moivre’s formula says how to compute these for (1+i)5.)
6 Solutionsfor Chapter0
d. The absolute value of 1+4i is p17; the argument is arccos1/p17.
0.7.5 Parts 1–4 are immediate. For part 5, we find
(z z )z = (x x y y )+i(y x +x y ) (x +iy )
1 2 3 1 2 1 2 1 2 1 2 3 3
�
=�(x1x2x3 y1y2x3 y1x2y3 x1�y2y3)
� � �
+i(x x y y y y +y x x +x y x ),
1 2 3 1 2 3 1 2 3 1 2 3
�
which is equal to
z (z z )=(x +iy )((x x y y )+i(y x +x y ))
1 2 3 1 1 2 3 2 3 2 3 2 3
�
=(x x x x y y y y x y x y )
1 2 3 1 2 3 1 2 3 1 2 3
� � �
+i(y x x y y y +x y x +x x y ).
1 2 3 1 2 3 1 2 3 1 2 3
�
Parts 6 and 7 are immediate. For part 8, multiply out:
a b a2 b2 ab ab
(a+ib) i = + +i
a2+b2 � a2+b2 a2+b2 a2+b2 a2+b2 � a2+b2
✓ ◆ ✓ ◆
=1+i0=1.
Part 9 is also a matter of multiplying out:
z (z +z )=(x +iy ) (x +iy )+(x +iy )
1 2 3 1 1 2 2 3 3
=(x1+iy1)((x2+x3�)+i(y2+y3)) �
=x (x +x ) y (y +y )+i y (x +x )+x (y +y )
1 2 3 1 2 3 1 2 3 1 2 3
�
=x1x2 y1y2+i(y1x2+x1y2)�+x1x3 y1y3+i(y1x3+x�1y3)
� �
=z z +z z .
1 2 1 3
0.7.7 a. The equation z u + z v = c represents an ellipse with foci
| � | | � |
Solution0.7.7: Rememberthat at u and v, at least if c> u v . If c= u v it is the degenerate ellipse
| � | | � |
thesetofpointssuchthatthesum consistingofthesegment[u,v],andifc< u v itisempty,bythetriangle
of their distances to two points is inequality, which asserts that if there is a|z �sati|sfying the equality, then
constant is an ellipse, with foci at
thosepoints. c< u v u z + z v =c.
| � || � | | � |
b. Set z =x+iy; the inequality z <1 Rez becomes
| | �
x2+y2 <1 x,
�
corresponding to a region bopunded by the curve of equation
x2+y2 =1 x.
�
We should worry that squar- If we square this equation, wpe will get the curve of equation
ing the equation might have in-
1
troduced parasitic points, where x2+y2 =1 2x+x2, i.e., x= (1 y2),
x2+y2 < 1 x. This is not � 2 �
� �
the case, since 1 x is positive which is a parabola lying on its side. The original inequality corresponds
p �
throughoutthe region. to the inside of the parabola.
i p 1 8
0.7.9 a. The quadratic formula gives x= � ± � � , so the solutions
2
are x=i and x= 2i.
�