Vector Calculus Study Guide & Solutions Manual

CONTENTS

How to Use This Book

IV

Acknowledgments

IV

CHAPTER 1

The Geometry of Euclidean Space

CHAPTER 2

Differentiation

21

CHAPTER 3

Higher-Order Derivatives; Maxima and Minima

43

CHAPTER 4

Vector-Valued Functions

63

CHAPTER 5

Double and Triple Integrals

77

CHAPTER 6

The Change of Variables Formula and Applications of Integration

97

CHAPTER 7

Integrals over Paths and Surfaces

17

CHAPTER 8

The Integral Theorems of Vector Analysis

143

CHAPTER 9

Sample Exams

161

APPENDIX

Answers to Chapter Te ts and Sample Exams

167

1

1

THE GEOMETRY OF EUCLIDEAN SPACE

1.1: VECTORS IN TWO- A ND T H REE-D IM ENSIO N A L SP ACE GOALS 1. Be able to perform the following operations on vectors: addition, subt raction , scalar multipli­ cation. 2. Given a vector and a point, be able to write the equation of the line passing through the point in the direction of the vector. 3. Given two points, be able to write the equation of the line passing through them.

STUDY HINT S 1. Space notation. T he symbol ]R or ]R I refers to all points on the real number line or a one­ dimensional space. ]R 2 refers to all ordered pairs (X , y) which lie in the plane, a two-dimensional space. ]R 3 refers to all ordered triples (x, y, z) which lie in three-dimensional space. In general, the "exponent" in ]R71 tells you how many components there are in each vector.

2. Vectors and scalars. A vector has both length (magnitude) and direction. Scalars are just ~u mbers . Sc~lars do not have direction. Two vectors are equal If and only If they both have the same length and the same direction . Pictorially, they do not need to originate from the same starting point. The vectors shown here are equal.

~~

3. Vector notation. Vectors are often denoted by boldface let ters , underlined letters, arrows over letters, or by an n-tuple (Xl, X2 , ... , x 71 ) . Each Xi of the n-tuple is called the lth component. BEWARE that the n-tuple may represent either a point or a vector. T he vector (0,0, ... , 0) is denoted O. Your instructor or other textbooks may use other notations such as a squiggly line underneath a letter. A circumflex over a letter is sometimes used to represent a unit vector. 4. Vector addition. Vectors may be added componentwise , e.g., in

;;;?J £?:j v

U

I I

_ _ _ _ _ _ __ 1

U

+

V

(XI , yr)

]R2

+ (X2, Y2) = (Xl +YI,X2+Y2).

Pictorially, two vectors may be thought of as the sides of a parallelogram. Star ting from the vertex formed by the two vectors, we fo rm a new vector which ends at the opposite corner of the parallelogram. This new vector is the sum of the other two. Alternatively, one could simply translate v so that the tail of v meets the head of u . The vector joining the tail of u to the head of v is u + v .

CHAPTER 1

2

5 . Vector subtmction . Just as with addition, vectors may be su btracted componentwise. Think

a

hb

b

~ a-

~

of this as adding a negative vector . Pictorially, the vectors a , b and a - b form a triangle. To determine the correct direction , you should Ibe able to add a - b and b to get a. Thus a ­ b go," f 0, then we should look at the level curves for constant k, i.e., analyze c - 9k 2 = 4x 2 + y2 . We recognize that if 9k 2 < c, then the level curves are ellipses which get smaller as Ikl approaches v'C/ 3. Similarly, we see that level sections parallel to the yz plane have the equation c - 4k 2 = y2 + 9z 2 , which are ellipses with decreasing "radii" as Ikl approaches .;c/2. Also , the level sect ions parallel to the xz plane have the equation c - k 2 = 4x 2 + 9z 2 , which are again ellipses which get smaller as Ikl approaches .;c. The level surfaces are ellipsoids if c is positive.

17. If c < 0, the level curve is empty. If c = 0, t he level curve is the x-axis. If c > 0, it is the pair of parallel lines Iyl = c; that is, the lines y = c and y = -c. In the yz plane, we sketch the graph of z = Iyl. Since x does not appear in the equation , we get a "cy,)inder" and this sketch is shifted along the x axis to obtain the graph in 1R 3 .

y

x 22. The equation can be written as (x 2 - 2x + 1) + y2 = 1 or (x - 1)2 + y'2 = 1. In the xy plane, this is a circle with r adius 1, centered at (1, 0). Since z is not in the equation, z may take on any value, so the circle is sh ifted up and down the z axis.

CH APTER 2

24

y - .---:;,.- 0 such that 0 < x < 6 implies that 1/1xl > N. Let 0 < 6 < l i N , then Ixl < liN implies that 1/1xl > N. T his is not true if t he absolute values are omitted , i.e., limx~o(llx) may be +ex> or -ex> depending on which side of 0 we are approaching from (see the figure below.) y

x x

y=

l/lxl

y

= l/x

27. (a) By the triangle inequality, 1a. 3 + 3a 2 + al < la 2 1+ 31a 2 1+ lal, since lal < 1 (we assume it is small, since 6 ,is small), this is less than (or equal to) 514 Choose 6 < 1/500 , then for lal < 6, la 3 + 3a 2 + al ::; 1/100 (note that this is a very rough estimate; a bigger 6 would probably work if we work harder to improve the inequality.)

2.3: DIFFE RENTIATION GOALS 1. Be able to state the definition of partial derivat ives.

2. Be able to compute a partial derivative or a matrix of partial derivatives. 3. Be able to compute a gradient . 4. Be able to compute a tangent plane.

STUDY HIN TS 1. Nota-tion. Class

en means that the nth derivative is continuous.

2. Partial derivatives . Know the definition

of = l In1 ' f( Xl, "" Xi + h, ... , x n) - OXi h~ O h

f(Xl' ... , Xi, ... , Xn)

=---..:........:....:.---''---'---'----'-~~-'--'----=-'--'--'-'--'----'-'-'-

To compute ofI OXi in examples, consider all variables except Xi to be constant and differentiate by one-variable methods . Differen iation is performed with respect to the variable Xi.

CHAPTER 2

28 3. No tation lor partial deriv atives. In many texts,

Ix

is used for 8f/8x. If we wish to evaluate

at a given point, we write

81

ax

1(xo,Yo)

'

fx l(xo ,yo) '

~: I(:Co ,Yo)

or



if

z

= f (x , y).

4. Tangent plane. The tangent plane to the graph of a function f (x , y) at (x o, Yo, f(xo, Yo) ) is given by

z= f(x o,yo)+ 8f ax l

81 1 (x -xo) + {)

(Y -Yo).

y (Xo ,Yo)

(Xo,Yo )

T his equation is also used to compute linear approxi m ation . Compare this equation to the equation of a tangent line and the linear approximation in the one-variable case. See section 2.6 for a gener lization.

5. Differ ntiability. Equation (2) in the text tells you that f : R 2 -+ is differentiable if t he tangent plane approaches f (xo,yo ) as (x,y) approaches (x o,yo ). Now, if f : U C ]Rn -+ ]Rffl, then · Il f(x ) - I(xo ) - T(x - x o) 11 0 I1m (2) X--+Xo Il x - xoll - , where T : ]Rn -+ ]Rm is the derivative. You should be able to get equat ion (2) from this definition . T his definition of differe ntiab.ility is most important for theortical work .

6. Gradient. T he gradient is a vector whose components are the partial derivatives of f , with 8f/8x; in the ,t h position. Here, fis a real-valued function. Thi operation is denoted by the sy mbol 'V . Someti, es , it is denoted "grad." For a function I : ]R3 -+ JR. , you should remember the form ula

7. Derivative of vector-valued fu.nctions. Con ider a fu nction I : R n -+ ]R m. T he derivative is an m x n matrix of partial derivativ s. The range consists of vectors with m components. Think of t he components as real-valued vector fu nctions; th n each row of the derivative matrix is a gradient. The derivative m atrix of f, evaluated at Xo, is denoted Df(xo) . 8. Important facts . Diffe rentiability implies continuity of the fun ct.ion, but continuity does not imply differen tiability. The existence of continuous partial derivatives im plies differentiability but the converse is not true. If a funct ion is differentiable, then its partial derivatives exist , but the converse is, again , not true.

SOLUTIONS TO SELECTED EXERCISES 1. (b) Holding y constant and differentiating with respect to x, we get 81/8x = ye xy . By symmetry, 8 f / 8y = xe XY • In this problem, all we did to compute fJ!I fJy was to switch x and y. T his is what we mean by "symmetry" ; it only works for functions that are unchanged when x and yare swapped.

2. (b) Hold y con taut and use the chain rule to differentiate wi th respect to x. We get f}z --

1

1

.-

1

ox - J1+ xy 2y'1 + xy

At (1,2), 8z/8x

'y -

y

.

- 2{1 + xy) '

. '1 1 ar y,

SImI

8z 8y

= 1/3 and 8z/ oy = 1/6. At (0,0), az/ax = az/ 8y = o.

3. (b) Holding y constant and using the quotient rule, we calculate:

8w _ 2x (x 2 - y2 ) - 2x (x + 2 + y2) 8x (x 2 _ y2 )2

x

2(1

+ xy) .

DIFFERENTIATION

29

Holding x constant and differentiating with the quotient rule , we get:

ow _ 2y(x 2 - y2) - (-2y)(x ay (x 2 - y2)2

+ 2 + y2)

_ 4x 2y 2 - (x - y2)2 .

4. (b) We must show that the partials are continuous in the domain : of/ax = l/y - y/x 2 = (x 2 - y2)/ x 2y, which is continuous for x f. 0 and y f. OJ a f/ ay = _X/y2 + l/x :;;:: (y2 - x 2 )/xy2, which is continuous for x f. O. T hus, f(x) is C l since its partial derivatives are continuous.

6. (b) The equation of the tangent plane is given by Z

= Zo + [fx(xo, yo)](x -

xo)

+ [fy(x o, yo)](y -

Yo).

Using .the result of exercise l(b) , we compute:

afl - l' ax (0 ,1) - , Therefore, the tangent plane is z

afl =0', f( 0,1 ) = 1. oy (0,1)

= 1 + l(x -

0) - O(y - 1) or z - 1 + x.

7. (b) The first row contains the partial derivatives of xe Y + cosy. The second roW contains those of x, and the third row con tains those of x + eY • The fi rst column contains the partial derivatives with resp ect to x, and t he second colu mn contains those wi th respect to y. Thus, the matrix of partial derivatives is

eY xeY - sin y 1 0

[ 1

1.

eY

8. (b) The function f is a m apping from JR3 to JR 2, so the matrix of the partials is 2 x 3. Let It (x, y, z) x - y, the first component of f. Similarly, let Iz(x, y, z) y + z. Then

=

=

af l Df(x, y , z)

=[

ay

a;: aiz

alz

ax

ay

11. Using t.he result of exercise 1(b), x(of/ax )

%1 ] _ [1

olz

-

-1 ()]

0

1

1

.

az

= xye xy

= y(af/ay) .

12. (b) Use the linear approximation form ula, which is the same as the equation of the tangent plane: z = Zo + [fx(xo, yo)](x - xo) + [f y(xo , yo)](y - Yo). Let z = f(x, y) = x 3 + y3 - 6xy, IO = 1, I = 0.99, Yo = 2 and y = 2. 01. We com pute:

8flax

= 3x 2 -

6y,

a f /ay=3 y2_6x, Therefore, our linear approxim ation is z value is -2.8485 .

~

so

fx(1 ,2) = -9.

so

/ y(1, 2) =6.

- 3 + (-9)( -0.01)

+ 6(0.01) = -2.85.

13. (c) The gradient is defined as the vector (of/ox , of/oy , a f/az) . Thus,

14. (c) The tangent plane is defined by \If(xo) . (x - xo)

+ 2e (z -

Exercise 13 (c) gives us

= ei + 2ek,

1) = 0 or x + 2z = 3.

\l f (1, 0,1)

so the tangent plane is e(x - 1)

= O.

The actual

CHAPTER 2

30 17. We compute 'V/ (0,0 , 1)

= (2:1:, 2y , - 2z )1(0 ,0,1) = (0 , 0, - 2) = - 2k.

20. We want to find T in equation (4). By linearity, / (x) - f( x o) by h , we want to find T so that ·

I hl:To

= f{ x -

xo ). Denoting x - Xo

°

II/ (h) - Thll Ilhll -.

=

If we choose T f, the numerator vanishes for all h, so this T sati ties the condition; that is, t he deriv tive of a linear map is the map itself. For exam ple, in one variable, consider f (x) = ax . From one-variable calculus, T = f ' (x o) = a fo r all Xo.

2.4: INTROD U CTIO N T O P ATHS AND CURVES G OALS 1. Given a path, be abl to compute t he velocity vector. 2. Be able to find a tangent line for a given path.

ST U DY HIN TS 1. Paths . A path is a "formula" that describes a curve in space . The picture of the pat h, which we can draw on paper, is called the image of the path or the curve of the pa th . 2. Path images. Often, it is convenient to express a path in terms of x and y when you want to know the image of a path. This is done by elim inating the parameter. For example (x, y) = (t2,t4) means t so y = t 4 = (-J x)4 = x 2. Caution: In this example, x = t 2, so x is always non-negative.

=..;x,

3. Circular functions. If a path is parametr ized by an expression involving cos t nd sin t , the parameter can usually be eli minated by squaring and adding. Use the identity cos 2 t +sin 2 t = 1 and 0 her trigonometric identities. 4. Velocity. T he velocity vector 's components are first deri vat ives of the components of the path. The velocity vector is tangent to the path. 5. Tangent lines. It is easy to find a tangent line if you rem ember that a line can be described as x + tv. The vector x is chosen to be a point on the path at to and v is the velocity vector c' (t o). Thus, the tangent line to a path i

l(t ) = c(to ) + (t - to )c'(to). SOLUTIONS TO SELECT ED EXERC ISES 1. From y = 4 cost , we get y/4 = cost. Use the fact that

and substitution to obtain 1 Since 0 ~ t ::; 211", th curve is an ellipse with y intercepts at ± 4 and x intercepts at ± l .

-4

x

31

DIF FE RENTIATION 3. As in example 1, c(t) has the for m (xo, Yo , zo) + tv,

where (xo, Yo , zo) = (- 1,2,0) and v = (2, 1,1). T hus,

c(t) is a line in R3. Specifically, it is the line t hrough (- 1, 2,0) with direction (2, 1,1 ).

z 2

(-1,2,0)

x 7. A path's velocity vector is found by differentiating the individual components. In this case,

d 2 d 3 d ) ( dt (cos t ), dt (3t - t ), dt (t)

1" ( t)

(- 2costsint,3 - 3t 2 , 1). 12. T he tangent vector to a curve c(t) tangent vector is (2t, 0).

= (x(t), y(t»

is c' (t )

= (x'(t),lI(t».

15. T he equation for the tangent line is l(t) = c(to) + (t - to) c'(t o). Here , to (3 cos 3t, - 3 sin 3t, 5t 3 / 2 ). T herefore , the tangent line is

l(t) = c(l)

+ (t - l) c'(I)

(sin 3, co 3, 2)

In this case, the

= 1 and

c'(t) =

+ (t - 1)(3 cos 3, -3 sin 3, 5) + t(3 cos 3, - 3 sin 3, 5).

(sin 3 - 3 cos 3, cos 3 + 3 sin 3, -3)

17. F irst , we need to find the tangent lin at to. We compute c (2) = (4,0 ,0) and c'(t) (2t , 3t2 - 4,0) , so c' (2) (4,8, 0). T hus, the tangent line is l(t) (4,0,0) + (t - 2)(4 ,8, 0). T he position of the particle a t t l = 3 is 1(3) = (4,0,0) + (3 - 2)(4 ,8, 0) = (8 , 8, 0).

=

=

2.5: PRO PERTIES OF THE DERIVATIVE

GOALS 1. Be able to state the ch ain rule. 2. Be able to compute a partial derivative by using the chain rule .

STUDY HINT S 1. Chain rule. Suppose

f is

a funct ion of Yl, Y2, ... , Yn and each Yi is a function of x. Then

df dY2 ' =of - .dYl - +of - . -+ ... +of- . ­dYn

dx

aYI

dx

aY2

dx

aYn

dx '

ay,

Notice how each term appears to be dJl dx wit h a.nd dYi "cane ling." However, bewar that the "su m" on t he right-hand side is df / dx, not n times df / dx. Also note the different d's: "a" is for a fun ction of many variables, while "d' is for a function of one variable. 2. Multivariable chain rule. T he multivar iable chain rule st ates th at

D(f 0 g)(xo ) =::; Df(yo) Dg (x o) , where Yo = g(xo) . T his is he product of two deriv tive m atrices , so any desired partial m y be obtained by multi plica ion .

32

CHAPTER 2

3. Chain rule, gradient relationship. Know that if fis r ai-valued and h (t )

== f( c (t)) , then

dh

dt = \7f(c(t)) . c' (t ).

SOLUTIONS TO SELECTED EXERCISES

f is differentiable by the sum rule. Its derivative is

2. (b) The function

af af] [ az' ay

= [1 , 1] .

(f) The function is differentiable by the chain rule. We know that z2 and y2 are differentiable by he product rule and that 1- Z2 - y2 is differentiable by the sum rule. Also, the square root function is differentiable (where its argument is positive), so th entire function is differentiable . Its derivative is

3. (b) This is a spe ial case of the first special case of the chain rule: dh af dx dx = ax· dz + af al az +

au .

5. (b) First, we compute I(c (t))

af du af dv au . dx + av . dx du af dv dx + av . dx·

= exp(3t 2 • t 3 ) = exp(3t 5 ), so f' (t)

= 15t4 exp(3t 5 ).

Next, by the chai n ruie, a 1 dx dy -dl = - . - + -aayf .-dx ax dt dt

Substitute x

= ye '" Y . 6t + :te'" Y • 3t 2 .

= 3t 2 and y = t 3 to get dl j dt

= =

t 3 exp (3t 5 ) • 6t + 3t 2 exp(3t 5 ) · 3t 2 15t4 exp(3t6 ),

which is the same as we got from a direct computation. 6. (b) Take th derivatives of each component to get c'(t) = (6t,3t 2 ) . 9. Substitute u = e"'- Y and v

=x -

y to get

By the chain rule, D(f 0 9)( X, y) = D/(u, v )D9(X, y). First, we calculat

Df(u, v) =

[ ~~~~~: ~~~~~~] = [ sec\uu-

When (x, y) = (1 , I ), we have g(l , 1)

= (e 1 - 1 , 1 -

D/(I, O) =

1)

1)

= (1, 0) . Hence,

[12 -1° ].

=;:] .

DIFFERENTIATION

33

Next , we calculate

Dg(I, 1)

so Therefore

D(f 0 g)(I,

1) == [~

Alternatively, we may calcul ate D (f

0

~1] [~ =~]

=[

1 1

-1 ] -1 .

[~ ~2]'

=

g)(x, y) directly from (f

g)(x, V) .

0

13. (a) By the chain rule,

dT = V'T(c(t)) . c'(t), dt

where c( t)

= (cos t, sin t).

Substituting x

Differentiate:

= cos t, Y = sin t gives

(2 cos te sin t

V'T(c(t)) c' (t)

-

sin 3 t, cos 2 te sin t

-

3 cos t sin 2 t),

( - sin t, cos t) .

Thus,

(b) Plug in x

' 2 ')

-dT = -2sintcoste Sm. / +sin 4 t+cos 3 te smt -3cos tsin~t . dt = cos t, Y = sin t into the expression for T and get

T(t) = cos 2 te sin t

-

cost sin 3 t.

Using techniques from one-variable calculus, we have

. t cos team . t + cos 3 teS int + -dT = -2 S1l1 ~

· 4

S1l1

' ? t - 3 cos 2 t S1l1~ t,

which is the same as the answer obtained by the chain rule in part (a). 17. (a) If y(x) and G are differentiable, then by the chain rule, and the fact that G(x, y(x)) is constant, dC = aC . dx + aC . dy = aC + oG . dy = 0 dx ax dx ay dx ox oy dx . Solve for dy/dx: If oC/ay

¥ 0, then dy dx

oC/ox - oC/oy'

(b) As in part (a), we differentiate C I and G 2 by the ch ain r ule:

dCI dx

= aC I

+ aC I . ciYI + oC I . dY2 ox aYI dx 8Y2 dx

=0

and

dC 2 dx

= aC 2 + ox

aC 2 . dYI + oC 2 . dY2 = O.

OYI dx oY2 dx

Assuming YI (x), Y2 (x) and G are differentiable and

I

aCt/aYI oCdOYI

aC I /aY2 8C2/ aY2

11=0

for all x ,

then we can solve for dyt/dx and dY2/dx. Rewrite the two equations as

oC I dYI . - + oG _I .dY2 -

oYI dx oG 2 dYI -. OYI dx

oY2 dx oG 2 dY2 +_.- aY2 dx

oC I

ox OC 2 ax

(1)

(2)

34 Mul tiply (1) by OG 2 / OYl and (2) by - 8GI/OYl . Add t h two together to get :

8G 2 8G 2 8G l -8G -1 . + _ .­ dY2 8x 8Yl 8x 8Yl dx = - 8" G=l ---; oo u exp(-u) = O. By I'H6pital's rule, lim ue - u u--+ 00

= u--+oo lim (.!!:..) = lim e tI --+ U

00

( .-!..) U

e

= o.

In general,

= x--+li mo+ /

f(n)(o)

f( n-l)(o)

(n - 1)(X) _

.

X

I"( x ) = .-!..2 exp ( -1) , f(3 )( x ) = (-32 + .-!..) exp ( - 1) , x

X

X

:1;4

X

so if we can show t hat lim.. --+ oo une - u = 0 for all n, then we can conclude that fis Coo at x :::: 0 as well. Again, use I'Hopital's rule n times: .

lI m une u--+oo

_

. un . n! " = u--+ hm - = ... = lim - = O. oo etl u--+ oo e"

Thus, f is C oo , with all derivatives equaling

f (O) + Hence,

f is not analytic.

f

I

°at x = 0. Now, 1 (0 + h ) = exp( -l/h) > 0, but

f(k)(O) k (O)h + ... + - k !- h + .. . =

o.

HI GHER-ORDER DERIVATIVES ; MAXIMA AND MINIM A EXTREMA OF REAL-VALUED FUNCTIONS GOALS 1. Be able to fin d th critical points of a real-valued two-variable fu n tion. 2. Be able to use the Hessian to classify the critical points of a functi on as maxima, minima, or

saddles .

STUDY HINTS 1. Defini tions. (a) A local or relative extremum is a point Xo where ! (x o) is largest or smallest in a sm all neigh borhood of Xo . (b) An absolute extremu m is a point Xo where !(xo) is largest or smalles on the entire doma in under consideration . (c) Critica l points occu.r where all first partial derivatives are zero. (d) A saddle point. is a critical poin L which is not a local extremum . 2. Criti cal po in t-extrem um relatzonshlP. All extrem a occur a t critical points, but not all critical points are extrema. C ritical points m ay also be saddle points. 3. Real-valued f unctions. Note t hat we are comparing function values for re I-valued fu nctions, not vector-val ued functions .

If a! / ox = 0 and {}!/ {}y = 0 have more than one solut ion , each com binat ion of (x, y) which satisfies t hese conditions must be considered. You must he complete in your analysis. See example 8.

4. F in ding extrema.

5. Hessian . T his is denoted by H! (x o)(h ) and it is equal to the second term of Taylor 's formu la, . . 1

wh ich

IS

"2

L n

; ,j= 1

[p! hi hj ~ ( xo). Xt

xJ

To determine definiteness in genet I, we need to know the determinants of the diagonal submatrices of the Hessian matrix. Starting from the upper left-hand corner , i.e., all , compute the deter­ minants . If they are all positive, t hen the Hessian is positive defini te . If a11 < 0 and t he signs alternate, then the Hessian is negative defi­ nite. ote that this test includes t heorem 6.

6. Deter-mining definiteness .

7. Usefulne ss of H essian. At a critical point, the first partial derivatives are all zero , so Taylor 's fo rmul a reduces to f (xo + h ) ! (x o) + Hessian + remainder,

=

where the remainder is small compared to the Hessian. Thus , if H ! (xo) is positive definite , then !(Xo + h ) = ! (xo ) + Hessian + remainder > ! (xo), so ! (xo) is a rel ative minimum . Similarly, if H ! (x o) is negative definite, then ! (xo

+ h ) = !(xo) + Hessian + remainder < !(xo) ,

and f(xo) is a relative m aximum .

8. Classifyi ng a critical poin t Xo. (a) If H! (xo ) is positive definite, then Xo is a local m inimum. (b) If H ! (x o) is negative definite, then Xo is a local maximu m. (c) If Hf (x o) does not satisfy (a) or (b) and not aU of the submatrices are zero, t.hen Xo is a saddle point.

CHAPTE R 3

48

9. Second deriv ative test. T heorem 6 is a special case of how the Hessian is used and is often

used for problem solving. To use this test , one must compute the discriminant D

°

= ((J2 f) (()2f) _(fl) 2 2 ()X

()y2

() X()y

°

If ()2 f I ()X 2 > and D > at a critical point , then we have a local minimum. If ()2 /1 ()x 2 and D > 0 at a critical p oint, then we have a local m aximu m. And if ()2 /1 ()x 2 > and D at a crit.ical point, then we have a saddle point.

°

., is the righ thing to do.

3. Solving equati ons. In general , we are not interested in the value of

4. Cautions. Remember , all of your equations must be solved simultaneously. Solving one equa­

tion alone does not complete the job of ,finding an ext remum. 5. Generalization. If there is more than one constraint, then "Y f (xo )

... + >'n "Ygn(XO)'

= >'1 "Y g1 (xo) + >'2 "Y g2(XO) ­

The right-hand sides of equations (2) will have the form

t

>'i

g;i, in plact'

i=1

J

of >.. : :' and there will be extra const raint equat ions . If you prefer to use equations (3), Ie J

h(x , >.) = f (x) + I: Ajgj (X) and solve 8h/ 8xj = 0 and 8h/ 8>., = 0 simultaneously. 6. B ordet'ed Hessian. This is a Hessian with a "border ," which are the additional top row and

the additional 1 ft colu mn. T he entries from left to right or top to bottom are 0, - 8g/ axl -8g/ 8X 2, ... , - 8g /8 xn , where 9 is t he const raint . Note that all entries except for the border are second partial derivatives. Xo . (a) If the k X k subm atrices of the bordered Hessian are aL negative for k ~ 3, then Xo is a local minimum. (b) If the signs alternate: positive, negative, positive, .. . , starting with the 3 x 3 matrix, then X o is a local maximum. (c) If t he pattern does not satisfy (a) or (b) , and the submatrices are not all zero, then Xo is at a saddle point .

7 . Classifying critical points

53

HIGHER-ORD ER DERIVATIVES; MAXIMA AND MINIMA

8. Extrema on a region. T he method of Lagrange multipliers is only good for locating extrema on a boundary. Don't forget to analyze the critical points inside the region.

9. Geometry. As in exam ple 8, section 3.3, we can analyze the square of the d istance rather th an the distance itself. 10. Economics. Isoquants are curves showing all possible combinations of capital and labor which

produce the sam output. In t hese examples, ). tells you how much more can be produced with one extra unit of capital or labor . Note that). has significance only a t the optim al point.

SOLUTIONS TO SELECTED EXERCISES 1. Use the method of Lagrange m ultipliers. We have \If(x, y, z) = (1 , - 1, 1), and the const raint is g(x ,y,z) = x 2 + y2 + z2 - 2, so )'\lg(x , y, z ) = ).( 2x , 2y , 2z ). Thus, Vf = >'Vg gives us 1 = )'2x , -1 = )'2y a nd 1 = )'2z. So we have x = z = - y = 1/2),. Su bstitute this into the constr aint: {1/ 2>.)2 + (-1/2).)2 + (1/2).)2 = 3/ 4).2 = 2, or ). = ±(1/2) J3/2. For ). = + (1/2)J3/2, we have (x,y, z) = (J2/3, - J2/3,J2/3), and for)' = -(1/2)J3/2, we have (x, y , z) = (- J2/3, .j2f3, - J2/3). These are the two extreme points and the maximum is v'6, while the minimum value is - v'6.

3. We want to find t he extrema of f(x , y) = x subj ect to x 2 + 2y2 = 3. Use the method of Lagrange multipliers. From the constraint , let g(x, y) = x 2 + 2y2 - 3, so V f = (1,0) and Vg = (2x ,4y). We want t o sim ultaneously solve Vf = ).Vg and the constrai nt equation:

>.2x My 3.

(1) (2) (3)

From (2), we get y = O. From (1), x = 1/ 2), . Substituting for x a nd y in (3) gives us (1/ 2).)2 = 3, so 1/2>. = ± J3; therefore, x = ± v'3. At (v'3, 0), f( x, y) = v'3 and at (-v'3, 0), f (x , y) = - -/3. We concl ude that the m aximum occurs at (v'3,0) and the minimum is at

(- y'3, 0). 8. On S, y is restricted to be cos x , so f (x, y ) = x 2 _ y2 = x 2 - cos 2 x. Applying one-variable methods , we cal­ culate f'( x ) = 2x + 2cos xsin:!.' . The derivative van­ ishes when x = - cos x sin x = - (1/ 2) sin 2x. This is a transcendental equation and can be solved by graphi­ sin 2x / 2 cal m ethods. The graphs of y x and y only intersect at the origin , so (0 , 0) is an extremum. Since x 2 ~ 0 and 0 ~ cos 2 X ~ 1, we conclude th at (0,0) is a minimum.

=

=-

13. We want to m inimize the surf ce area of th cylinder subject to t he constraint of the volume. That is, we want to minimize S(r , h) = 2rrl' h + 2rrr2 subject to rrr 2 h = 1000 cm 3 . Use the met hod of Lagrange multipliers. From the constraint, we get g(r, h) = rrr 2 h - 1000. We compute th following first partial derivatives: 8S/ 8r = 2rrh + 4rrr , 8g /8r = 2rrrh, 8S/ 8h = 2rrr ) 8g/ 8h = rrr2 . Now we want to solve the fo llowing system of equations: 2rrh

+ 4rrr 2rrr rr7· 2h

1000.

(1) (2) (3)

CHAPTE R 3

54

Factor out rrr from (2) to get 2 = Ar or A = 2/r . Substitution into (1) and factoring out 2rr gives = (2/ r )rh = 2h , or h = 2r. Substitution into (3) gives rrr 2 h = rr7,2 . 2r = 2rrr 3 = 1000, so r = 10/(2rr)1/3 and h = 2r = 20/(2rr)1/3. To check that our result sat.isfies the const.raint we calculate 2 100 20 rrr h = Tt (2rr)2/ 3 (2rr)l/3 = 1000.

h + 2r

Therefore, the desired cylinder has height 20/(2rr)1/3 cm and b ase radius 10/(2rr)1/3 cm. 18. Use the method of Lagrange multipliers. Let

= x + 2y sec B + A( XY + y2 tan B-

f(x , y, A)

A).

Then of/ax = 1 + AY; of/oy = 2 sec B+ A(X+2y tan B) ; a nd OflaA = xy+ y2 tan B- A. From af/ox 0, we get A = -1/y; whereas from of/oy 0, we get

=

=

A=

- 2sec B x + 2y tan B

Hence , 2y = (x + 2y tan B)/(sec B) , so 2y(1 - sin B) = x cos B. Thus, x = 2y(sec B - tan B). Substitute this into a fI OA = 0 to get 2y2(sec B-tan B)+y2 tan B = A. T hen y2 (2 sec B-tan B) A , so

=

y

2

= 2 sec BA-

Acos () 2 - sin B.

tan B

25 . (a) Use the method of Lagrange multipliers on t he auxiliary fu nction h(x, y, A) A(2x2 + y2 -1). We compute the following partial derivative : hx

hy

= = 0 or A = 1. h).,

= x + y2 ­ (1) (2) (3)

1 - 4XA == 0 2y - 2YA = 0 _(2x2 + y2 - 1) = O.

From (2) , we get either y For the case where A = 1, we get x = 1/4 from (1) and then y = ±V778 from (3) . W hen y = 0, we get x = ±1/V2 fro m (3) . Thus , the critical point.s for the constrained function a re located at (1/4, ±V778) and (±1/V2, 0) . (b) We let the const raint be g( x , y) = 2x2 + y2 - 1, so h(x, y, A) = f(x , y) - Ag(X, V). Then the bordered Hessian (T heorem 10) is

og ax

og oy

og ax

02h ox 2

a2 h oyox

og ay

02h oxoy

02h oy2

0

/H/=

At (x, y, A)

- 4x

-2y

- 4A 0

0 2 - 2A

= (1/4, ..ft78, I), we find /H/=

(01: , y) = (1/4 , we have

SO

0

-4x - 2y

0 -1

-Jr72

-1 -4 0

-j7fi 0 0

= 14> 0,

V778) is a relative m axi mum point. At t he point (x, y, A) = (1/4, -V778,1). I 0 -1 Vf72 IHI = I 0~2

~4

~

= 14 > 0,

55

HIGHER-ORDER DERIVATIVES; MAXIMA A ND MINIMA

so (x, y) = (1/4, -ftl8) is also a relative maximum point. When (x, y) = (1/V2, 0) , equation (1) tells us that>. so at the point (x, y, >.) (1/V2, 0, we see t hat

= -Ji78,

°

IH,I:o: -J8 I

- 078),

=

° °

2 + V2

-J8 V2

° °

= (2 + \1'2)( -8) < 0.

Thus, (1/V2,0) is a relative minimum. Similarly, when (x,y) at the point (x,y,>.) = (-1/V2, 0, Ji78), we get

°

IHI = J8

° 2 - °V2

J8

-V2

° °

= (2 -

= (-1/V2,0), >. = +Jl78, so

V2)(-8) < 0 .

Thus, (-1/V2,0) is also a relative minimum. Evaluating the function f(x,y) at the critical points tells us that (1/4,±ftl8) are absolute maxima, (-1/V2,0) is an absolute minimum, and (1/V2, 0) is only a relative minimum. 26. With a hyperbola, there is only a minimum distance from a point. With a parabola, there is also only a minimum. There can be no maximum distance from these geometric figures because both figures extend to infinity. 31. Let the price of labor be p and let the price of capital be q. We want to optimize Q given the constraint S::;;: pL + qK == B. We compute the partials of Q and S: {)Q/{)K = AaK,,-l L1-", {)S/ {)K q, {)Q/ {)L A(l - a)L -" K", {)S/ {)L = p. Use the method of Lagrange multipliers to get the following system of equations:

=

=

AaK,,-l L 1-"

>.q

A(l- a)K" L-"

>.p

pL+ qK

B.

From (1 ), we get q = (A/>')aK"-lL 1-,,, and from (2), we get p Substitution into (3) gives us (A/ >.) (l-a)K" L 1-" + (A/ >.)aK" L1-" B or >. = (A/ B)K" L 1-". Substitute for >. in (1) and (2): AaK,,-l L 1-" A(l - a)K" L-"

From (4), we get K

=

= (a:

= (A/>.)(l - a)K"L - ". = B or (A/ >.)K" L 1- " =

(A/ B)K" L i -"q (A/ B)K" L 1-"p.

= aB/q, and from (5), we get (K, L)

(1) (2) (3)

(4) (5)

L = (1 - a)B/p. Thus, the point

, (1 -pa)B)

optimizes the profit .

3.5: THE IMPLICIT FUNCTION THEOREM GOALS 1. Be able to determine if an inverse function exists near a point x. 2. If an inverse exists, be able to find a derivative by implicit methods.

STUDY HINTS 1. Advanced material. The theorems presented in this section are usually proved in more advanced courses . You should be most concerned with understanding the statements of the theorems .

- -- - - - - -

-

-

-

CHAPTER J

56

2. Notation. DxF is used in th is section. It is just another notation for " F with respect to x although as in Chap ter 2, DxF m ay be a m atrix if F is vector-valued . 3. Local Theorems. T he theorems introduced in this section may not apply if th range or domain

is too large. 4. Special implicit fu nction theorem. If o f / oz =F 0, then z can be wri tten in terms ofx at a given point (xo, zo), and the derivative of z = g( x) is

_ - DxF(x, z) D 9 (x ) of . a.;- (x, z) Note that we can differentiate z even though we don 't have a formula for z. Don't forget the minus sign in the derivat ive. 5 . Commonly used formu la . When z is a fu nction of x and y, we get t he formul a dy dx

oz/ox - oz/oy'

It looks almost like division of fr actions, except fo r the minus sign . Again , don 't forget the minus sign . 6. General implicit function theorem. In general , z m ay be a vector. We form a matrix with the top row being the parti al of Fl wi th res pect to Z j, j 1, .. . , m (alm ost like grad ient) I, ...m . If the determ inant of thi Similarly, the 0 her rows consist of th part ials of Fk, k

= =

m x m matrix is non-zero, then z is a function of x and a derivative exists.

7. Jacobian. T his is the de erminant of the m x m m atrix descri bed in item 6 above. It is denot d

8. Inverse functio n theorem. As stated in item 6 above , if J f (xo) =F 0, then z can be written in terms of x. It may not be easy to express z in terms of x , but it is possible in principle. g. Example 3. T his is a t ypical problem . St udy it carefully. Note that when more than one function is given , you will n ed to solve a system of simultaneous equations to fin d a partial

deri vati ve.

SOLUTION S TO SELECTED EXERCISES

=

2. Let F( x, y , z) xy + z + 3xz 5 - 4. Since we want to know if we can solve for z as a funct ion of (x, y), we need to know t hat o f /oz does not vanish near the desired point , so o f/oz 1 + 15x z4. ear (1,0, I), Fz 16 =F 0, so F 0 is solvable for z as a function of (x, y) .

=

=

=

Therefore, oz ax

- Fx Fz

+ 3z 5 1 + 15x z 4 y

and

Oz _ - Fy _ oy Fz

x 1 + 15xz 4

.

= (1 , 0), z = 1, so oz/ox = - 3/ 16 and oz/oy = -1 /16. 7. Let Fl = y + x + uv and F2 = uxy + v. Then we want At (x, y)

~

/ OU = I OFI o Fdou

oFI/ ov o F /ov 2

The entries of the determinant are a FI/ou We see that at (0,0,0 , 0),

I =F 0 = v,

at (x, y , u, v) aFi/ ov

= (0, 0,0, 0).

= u, o F 2 / o u = xy and oF2 /8v = 1.

HIGHER-ORDER DERI VATI VES; MAX IM A AND MINIMA

so we may not b able to sol ve for tI, v in terms of x , y near (x, y , u , v ) = (0,0 , 0,0) . To check directly, the firs t equation gives us ltv = - (y + x ), so v = -( y + x)/u . Combining this with the second equation , we get ux y = (x + y)/u, or 1.1 2 = (x + y)/ xy. For (x , y) near (0,0) , either there is no solution for u small, or t here are 2 solutions for u. 10. (a) Using the definition of a(x , y)/ o(r , 0) and computing the part ial derivatives , we get

a( x, y) _l ax/or 8(r , O) - ay/or

ox/ ao oy/oO

I_I cosO - sinO

-r sinS [_

rcos O

- r.

At ( 1'0 , ( 0 ) , r = 1'0 . (b) By the inverse function t heorem, we can form a smooth inverse fu nction (r( x, y) , O(x,y)) as long as r I- 0. As a direct check, solve fo r r and 0 in terms of x and y: x 2 + y2 = r and 0 = arctan(y/x) . Since we have written r and () in terms of x and y, the result above is confirmed . Note t hat if x 0, then 0 7r/ 2 or 37r/ 2, depending on the sign of y. If, in addition, y 0, then r 0, a.nd B can have any value, so we cannot find an inverse, as we did above.

J

=

=

=

=

12. Let F l = xy2+XZU +yv 2 -3 and F2 = u 3 yz+ 2x v- u 2 V 2 - 2. T hen oFt/f)u = xz , oFt/ov = 2yv, oF2/ol.l = 3u 2 yz - 2uv 2, oF2/ov = 2x - 2u 2 v. At (x,y, z ) = (1,1,1 ) and (1.1, v) = (1, 1),

Since L\. I- 0, it is possible to solve for ov/ ay, we use the chain rule:

tJ, tJ

in terms of x, y, z near the given point. To compute

and

1 3 OV -oF2 = 3u201. - yz + u z + 2x - oy oy oy

AU 2 av 2 2u - v - 2v - u oy oy

= O.

At (x , y, z) = (1,1, 1) and (u , v) = (1, I) , those equa tions become 2+ ou/ay + 1 +2(ov/oy) = 0 and 3(8u /oy) + 1 + 2(ov/oy) - 2(01l./oy) - 2(av/oy) = 0, or au/oy + 2(ov/oy) - 3 and ou/oy - 1, so ov/8y -1.

=

=

=

SOLUT IONS TO SELECTED R EV IEW EXERCISES FOR CHAPTER 3 2. (c) Compute the partials of l and set them equal to 0:

°

2x + 1 i = 2xy + 4 y 3 = 0.

(1)

(2)

°

From (I ), y2 = - 2x (so x ~ 0). Substitute into (2) to get y(_y2 ) + 411 = y3 = or y = 0. T hus, x = also and (0,0) is the critical point. T he reader should verify that the discriminant Dis 0, so this test is inconclusive. However, l(x, y) x 2 + 2x y2 + y4. - x y2 = (x + y2 )2 - x y2, so 1 (1: , y) > 0 for all x < 0, and for x positive, (z + y2 )2 < xy2 implies x 2 + y4 < _ z y2 but this is impossible since Z2 + y4. is always positive and _ x y2 is always negative. We can conclude that the minimum value of l( x, y) is 0 at (0, 0).

°

4. The Taylor expansion is

=

CH APTEP

58

where all of the partial derivatives are evaluated at (3': 0, Yo). We compute th partial derivati and evaluate at (xo , yo) = (0,0) as follow:

f (x, y) f ,, (x, y) fy(x , y) f" x(x, y) f"y (x, y) fy y(x , y)

e"Ycosx; f (O , O) = 1, ye"y cos x - eXY sin x ; fx (O,O) = 0, xe xy cos z; fy (0,0) = 0, 2 xy y e cos x - 2ye"Y sin z - eXY cos z; Ix;c(O, 0) = - 1, e"Y cos x + zye xy cos x - xe xy sin z; f;c y (0, 0) = 1, x 2 exy cosx; f yy (O , O) = 0.

Thus, the second-order Taylor expansion is Z2

f (x, y) =1-

2

+ x y.

7. Th critical points are tho e where the first parti al derivatives vanish . T hus, we need to so

of

y1T cos (11'z )

ax

of oy

sin (11'x) =

= 0,

o.

The equ ation si n(11'x) = 0 implies that x is an integer. Since the sine and cosine functions a never 0 at the same angle, y11' cos (11'x ) = 0 implies that y = O. Thus, the critical points a (n , 0) , where n is an integ r. Next, we calculate the second parti 1 derivatives. They are 02f

ox 2 82 f

11' cos (11'z ),

ozoy f)2 f

oy2

=

O.

=

At the cri tical points (n ,O), we have fxx (n, 0) = 0, f;cy (n, O) = (- 1)" 11' and fyy (n ,O ) Using T heorem 6 from Section 3.3 , we compute D = [I,,;c(n, O)][ly y{n , 0)] - (fxy(n , O)P o- (( _1) n11')2 = _ 11'2 . Since D < 0, we conclude that the points (n, 0) are saddle points. 11. Use the method of Lagrange multipliers with g(x,y) = x 2 + y2 constraint eq uation, we get

og og

= 1.

From

Vf = >'Vg and

=

t

8x>' = 2x>., 8y>'

= 2y>. ,

l. If x and yare both nonzero, t hen the first two equations tell us that - sin (x 2 - y2 ) ar sin (x 2 - y2) must both equal >., so >. must be 0; therefore, x 2 - y2 must be 0 also. Notice th x 2 - y2 cannot be a multiple of 11' because x 2 + y2 1 means both IxI and Iyl must be less th or equal to 1. From x 2 - y2 0, Z2 + y2 1, Ixl ~ 1 and Iyl ~ 1, we get four critical poin t

=

=

=

If x = 0 and y # 0, then x 2 + y2 = 1 tells us that y = ± 1 and consequently, >. = ± sin (- l

Thus, (0, ±l) are criti al points . Simila.rly, when y = 0, z = ± 1 and>' = ± sin (l ). Therefor

-

-------------------------------------------

59

HIGH ER-OR DER DER IVATIV ES; MAXIMA AND MINIMA

(±1,0) are also critical points. At the points (±,ff, ± Vt), f has the value 1. At (±1,0) and (0 , ± l) , f has the value cos(I). Thus, f has a maximum value of 1 at four points and a minimum value of cost 1) at four other points. 13. By the method of Lagrange multipliers, we get he foll owing system of equations:

x

>. >.

x+y

1.

y

Substituti ng into the last quation gives us 2>' = 1, or >. = 1/2. Thus, x = 1/2 and y = 1/ 2. However, in this case , it would have been much easier to solve for x and substitute to get z = x(1 - x ) = x - x 2 . By one-variable calculus methods, x = 1/2 is the critical point, and the maximu m value of z subject to x + y = 1 is 1/ 4. 16. (b) By the implicit function theorem ,

of/ 8x - of/ 8y '

dy

dx In this case , F( x, y)

= xS -

sin y + y4 - 4 = O. Thus,

18. (c) T he rectangul ar parallelpiped (box) is symmetri ,so if x is a coordinate of a point (x , y, z) on the corner of the box, then the di mension corresponding to that coordinate has to be 2x. W want to maximize V (x , y, z) = (2x )( 2y)(2z ) = 8xyz subject to x 2/ a2 + y2 /b 2 + z2/ e2 = 1. Use the method of Lagrange multipliers:

oV/ 8x 8V/o y 8V/ 8z 2

x /a

2

= 8yz = 8xz

2x>./ a2 2y>"/ b2 2z>. /e 2

= 8xy

+ y2/h + z2 / eZ 2

1.

Solve (1) for x: x = 4a 2 yz/>.. Substitute into (2): 8(4a Zyz / >.. )z = 2y>'/b 2, or z2 Substitute fo r x in (3): 8( 4a 2 yz / >.) y = 2z>.. / c2 , or y2 = >..2 / 16a2 e2 . Then

P lug these results into (4) to get

>.2 16a2 b2c2

i.e ., xyz

>.2 >.2 + 16a:lb2c2 + 16a 2 b2 c2 = 1.

= abe/ 3v'3. T herefore, the maximum volume is V = 8xyz

abe

8abe

= 8 . 3V3 = 3V3 '

(1)

(2) (3) (4)

= >.2/ 16a2bz .

CHAPTE R

60 22 . First check for extrem a on the interior of the circle ofradius J2. We have f( x , y)

= xy-y+ x -

_

so set its partial derivatives equal to O. We have

-of ax = y + 1

of = x - I , By

and

-

so the point (1 , - 1) is an extremum. Since (1) 2+(_1)2 = 2, this point is within the constrain To find ot her extrema (if any), we use the method of Lagrange multipliers: y

+1

= =

x-I

x

2

2x"\ 2y..\ 2.

+ y2

First consider the case when ..\ = O. Equations (1) and (2) gi ve us (x, y) = (1, -1), which had considered earlier. If ..\ :j; 0, then (1) and (2) added together yields x + y = 2"\(x + y) or (2,,\ - 1)(x + y) T hus , either ..\ = 1/2 or x = -yo For x = - y, equation (3) gives us the points (1, -1) ~_ (- 1, 1). If >. = 1/2, then equation (1) reduces to x = Y + 1. Substit ution into (3) yiel (y + 1) 2 + y2 = 2 or 2y2 + 2y - 1 = 0 or y = (-1 ± -/3) /2 . Therefore , two other possib extrema occur at ((1 + V3)/2, (-1 + V3)/2) and ((1 - -/3)/2 , (-1 - -/3)/2). Evaluating at of these possible extrema, we get

=

f

f

e

+2v'3, - 1; v'3 )

(

-

'

2'

1 ­ -/3 -1 ­ v'3) 2 ' 2

2'

1

f( l , - 1)

0;

f(-I , I)

-4 .

Thus, the maximum points are ((1 + v'3) /2 , (-1 + V3)/2) and ((1- -/3)/2, (-1- -/3)/2) wi maximum vlaue of 1/2. T he minimum point i (-1,1) with minimum value of -4. 25. We use the implicit function theorem. We need to show that the determinant

of

of

au ov oe oc au av is not 0 near (x, y,

1.1,

v) = (2, -1 , 2, 1). T he determinant is

I

- 31.12 2v -4u 121.1 3

I = I --812

2 12

I= - 144 + 16:j; O. ou/ox, "

Since the determinant is not 0, 1.1 and v exist as functions of x and y. To compute implicitly differentiate the given equations with respect to x. Keep in mind that 1.1 and v functions of x and y. We get

201.1 2x - 31.1 2y -

OV

ox + 2v -ax aU 3 0V 4u ax + 12v ax =

0 O.

To make the calculation si m pler, we can plug in (x, y, 1.1, v)

au

av

= (2, -1, 2,1).

4 - 12 ax + 2 ax

0

au + 12- ov ox ax

O.

-2 - 8-

Then

~

61

HIGHER-ORDER DERIVATIVES; MAXIMA AND MIN IM A

Solve this simple system of two equations by your favorite method. You should get 13/32.

au/ox =

30. (a) Using the given formula, we write

s ;: f(m, b)

= =

+ (3 - 2m - b)2 + (3 - 4m - b)2 (46m + 16b) + ((m + b)2 + (2m + b)2 + (4m + b)2).

(1 - 1 . m - b)2

19 -

The problem is to find m and b which minimize f(m, b), so we take derivatives:

of

am

_ -46+2(m+b)+4(2m+b)+8(4m+b)

=

-46 + 42m + 14b,

and

of

-16 + 2(m

ob

-

+ b) + 2(2m + b) + 2(4m + b)

-16+14m+6b.

Next, we set them equal to 0:

0 O.

-46+42m+14b -16+14m+6b

Solving this system of 2 equations, we get b = 1/2 and m = 13/14. The reader is encouraged to verify that this is indeed a minimum point. T herefore, the best-fitting straight line to the points (I, I), (2,3), (4,3) is Y = 13x/14 + 1/2, as shown in the graph below.

x

33. If y

= mx + b is the best-fitting straight line, we must have, in particular,

Performing this differentiation, we get

"

-22)Yi - mXi - b)

.=1

= 0,

which implies that the summation has to be 0, or the positive and negative deviations cancel.

TEST FOR CHAPTER 3 1. True of false. If false, explain why.

(a) If y is a differentiable function of x, then

z is a differentiable function of x and Y, and oz/oy i= 0, dy oz/ox dx - oz/oy'

CHAPTER 3

62

= w + xyz + y2 has no critical p oints. (c) For any f(x,y,z), we have o4 f loxo y 2 oz = o4flozoyoxoy. (d) A fourt h -order T aylor series for f (x ) y, z ) w) = x 3 + y 4 - z2 + w (b) The function g(w, x, y, z)

as

f

W

z 2 is exactly the same

itself.

(e) Suppose D is a closed region on the xy plane. If f (x y) is cont inuou and has a minimllI!' on D, t hen f (x , y) also has a m aximum on D. 2. Use Taylor 's theorem to calculate a fi rst-order and a second-order approxim ation for xy2z 3 a

th e point (1. 1, 2.03,0.98).

3. Consider the surface described by z cross-section in the plane y = 4.

= 5x 3 y -

4. F ind all of the critical poin ts of h(u, v, w , r) 5. Let

f

be a function of x and yand of lax

2x y2. Discuss the concavity of the surface '~

= v 3 + w 3 + u 2 + 7"2 - 3v - 12w + 4r - 8.

= 3x 2 yex . Wh ich of t h following, if any, can Ix

aflay? (a) x 3 -2y+cosy

(b) 3x 2 +y-8 (c) 3-e Y +x3 6. F ind the minimum and m aximum values of x

+ yon

7. F ind the m in im um and m axim u m values of x 2

-

x

the cur ve where 2x 2 + y2

+ y2 -

= l.

Y on the followi ng sq uares and thei!

interiors: (a) The square with vertices a t (0,0), (1,0), (1, 1) a nd (0,1), but not including the x or axes. (b) T he square with ver tices at (0, 0), (2,0), (2,2) and (0,2), bu t only t he border on the ~ a nd y axes is included in the region .

8. Let a particle move in a potential field in R 2 given by V( x, y) Find the critical points of Vand classify t hem.

= x 2 + 4x y + y2 -

2x + 4y + E

9. Consider the following system of equ ations :

x 2u

+ yv + uv 2 u + x v + yu

o k,

where k is a constant. If .T and y may be wri tten in terms of u a nd v, com pute o x l o v at th given point and for the given k.

(a) (u, v, x, y) = (1 , -2, 0, 1); k (b) (u, v, x, y) = (2 , 1,0, -2); k

= 2. = -6.

10. Farmer Jones owns a golden goose farm and he wants to maximize p rod uc tion of gol den eggs He knows that more geese will lay m ore eggs, but t oo m any geese will inhibit egg layin g as t h geese instinctively will not overpopu late. On the other hand , too m any coyotes will prey upo most of the golden geese and too fe w will cause the geese to overpopulate . Farmer Jones h& determined that th e golden egg production is propor tional t o

E(g , c)

= 10gexp( -O.lg) -

2cexp ( -0 .2c),

where 9 is the goose popul ation and c is t he coyote popu lation. (a) F ind the critical poi nts of E. (b) Classify the critical points of E.

63

4

VECTOR-VALUED FUNCTIONS

~.l:

ACCELERATION AND NEWTON'S SECOND LAW GOALS 1. Given a path, be able to compute the velocity vector , the acceleration vector and the speed at a given point . 2. Be able to use Newton's second law and Kepler 's law.

STU DY HINTS 1. Velocity, acceleration and speed. Reca.ll that the velocity vector's components are the first derivatives of the components of the path with respect to time. Speed is the length of the velocity vector. The second derivatives make up the acceleration. Beware that t he derivative of speed is not the accelerat ion . Note t hat speed is a scalar, while velocity and acceleration are vectors. 2. Differentiation rules. No tice how the differentiation rules for paths are comparable to the rules you learned in your one-variable calculus course. 3. Ne wton }~ second law. T his states that F should remember this.

4. R egular path. If c'{t)

1= 0, t hen

= rna, where F is force and a is the acceleration . You

c{t) is a regular path and the image curve looks smooth.

5. K epler 's law. T his law relates an orbiting body's period to its radius. Vector alculus is used to derive the equation. It is probably not necessary to remember the equation, but ask your instructor to be sure.

SOLUTIONS T O SELECTED EXERCISES 2. T ake the first derivative of each component to get the veloci ty vector; that is, c'{t) == v{t) = (t cos t + sin t , - t sin t + cos t, via). The acceleration is composed of the second derivatives, so c"{t) = a(t ) = (- tsin t + 2cost , - tcost - 2sint,0 ). Therefore, v (O) = (O ,I,vIa), a (O) = (2 ,0,0) and the tangent line is 1(>..) = c(O) + >"v(O) = 0 + >"(0, 1, via) = >"(0 , 1, via). 5. By the sum rule, we may add vector fu nctions first and th n differentiate , or we m ay d ifferen­ tiate first and then add the derivatives. By adding first,

! [Cl (t) + C2(t)]

~ (e t + e- t ,sint + cost , t 3 (e t

- e - t, cos t - sin t,

2t 3 )

- 3t 2 ) .

On the other hand, d dt [Cl (t)

+ C2(t)]

d

dtc dt)

d

+ dt C2 (t)

(e t , cos t, 3t 2 ) + (_e- t , - sin t, - 6t 2 ) (e t - e- t , cost - sin t, -3t 2 ), which is what we got by adding first and then differentiating.

CHAPTER

64

10. We wan F(O) = ma(O) , where m = 1. T he accelerat ion vector is given by a(t ) (- cos t , - 4 sin 2t ), so a (O) = (-1 , 0) . Thus, F( O) = - i g-cm / s 2 = - O.OOli newton.

= r l/(t )

13. To show that the speed is constant, we differentiate it with respect to time and show that deriva.tive is zero. Since speed is IIvll = .;v:v, it is more convenient to work with the squ of the speed. (If th square of t he speed is constant, then the speed must be a constant, t We calculate d d dv

dt II v ll2

= dt (v· v) = 2v·

dt

= 2v . a,

but the acceleration a is perpendicular to the velocity Vi therefore v . a (d/dt)ll v (t )W = 0, so Il v (t)1I is constant.

=

O.

Th

17. Integrating each component of c'(t) = (t,e t ,t2) gives us c(t ) = (t 2/2+ A,e t + B,t 3/ 3+ where A, B and C are constants. When t = 0, we have e(O) = (A, 1 + B, C) = (0, - 5, 1), so A = 0, B = -6 and C = 1. Therefore, the path is c(t ) = (t 2 / 2, et - 6 t 3 / 3 + 1). 19. (b) Let X = 2x, so X 2 + y2 = 1. Recogn izing this as a circle, we let X = cost and y = sint, so x = X / 2 = (cost )/ 2. OUf

path e(t ) is described by (x,y) = ((cost )/2,sin t) . From the

original equation , we recogn ize 4x 2 + y 2 = 1 as an ellipse with

x-intercepts at ±1/2 and y-intercepts at ± 1.

y

1/2

x

20. By the cross product rule in the box preceding Example 1 of the text ,

d [mc (t) x v(t)] = m dt d [c(t) x v(t )] = m [~ dt dt x v (t ) + e(t) x ~ dt ] .

But dc /dt

= v(t), dv / dt = a(t) and v (t) x v (t) = 0, so the m[O + c(t ) x a(t )].

xpression reduces to

If k is a constant,

k (u x w)

= ku x w = u x kw,

so

m [c(t ) x a (t)] = c(t ) x matt ) = c(t) x F(c(t )). If F(e(t )) and c(t) are parallel, their ross product is 0, so the angul ar momentum i con This is t he case of planetary motion; since

F(c (t))

=

GmMc (t) Il c(t)113 '

we see that F(c(t )) is a multiple of c(t), so it is parallel to c(t) .

4.2: ARC LENG TH G OALS 1. Be abl to compute the arc length of a given segment of a path.

STUDY HINTS 1. No tation. Often

5

is used to denote a path in space, rather than c.

2. A rc length. T his is j ust t he length of a curv . Lengths may be added together, so we may pute t he ar I ngth of curves which are not differentiable at finjtely many points by sum:

the lengths of the pieces .

­

65

VECTOR-VALUED FUNCTIONS

3. A rc length formula . You may find it easier to remember that arc length is the integral of speed (n ot velocity !) . T his m akes sense because arc length gives the distance traveled . In any case , you should know that the for mula is

L(c)

=

I

tl

Ile'(t) 11 dt =

to

Itl

J (X'(t))2

+ (y'(t))2 + (z'(t))2 dt.

to

4. Integration tricks. (a) Due to t.he nature of the arc length formula , you m ay want to memorize the for mula

T his formula may be derived by trigonometric substitution if you do not wi h to memorize it . Ask your instructor if this and similar form ulas will be provided on an exam . (b) Look for perfect squares. Radicals can be elim inated from the integrand if a p r£ ct square occurs. 5. Positive lengths only. If you compute a negati ve or a zero arc lengt h, then you m ade a mistake. Arc length is always positive. 6. R iemann sum derivati on . If the derivation does not make sense now, you should return to this discussion after R iemann sums are explained more thoroughly in C hapter 5. Understand ing the derivation of fo rmulas gives much more insight into the t h ory. 7. Parametrization warning. If you need to parametrize a curve, be sure the curve is traversed once and only once. T he orien ' at ion is also importan , and this will b come especially signif­ icant in chapter 7. T he bad consequences of an incorrect parametrization are illustrated in example 1.

S OLUTIONS TO SELECTED EXERCISES 3. We compute e'(t)

= (3cos3t, - 3sin 3t,3Vt), so

lie' (t) II = [(3 cos 3t) 2 + (-3 sin 3t )2 + (3vt) 2P/ 2 = ';9 + 9t = 3v'l+t. T hus, the arc length is

1

1

3.Jf+t dt

o

= 3 · -2 . (1 + t )3/2 11 = 2(23/2 3

1)

= 4h -

2.

0

6. Given e(t) = (t , t sin t , t cos t) , we co mpute e' (t) = (1 , sin t +t cos t, cos t-t sin t). T hen lie' (t ) II = [1 + (sin t + t cos t) 2 + (cos t - t sin t)2j1/2 ';2 + t 2. We want t.he arc length on the interval [0 , 1t'], so we need to compu te:

=

fo re J2+t2 dt. This m ay be integrated by he method of trigonometric substitut ion: One would let tan B = An alternative is to use the formula given in the integration tables (See study h int 4 above .):

V2t , aq,d t hen perform an integr ation involvingsec 3 8. This is left as an exercise.

i J2+t2 o

re

dt

=

'12 [tJi2+2 + 2ln(t + Ji2+2)] Ire0 ~ [ 1t' J 1t'2 + 2+ 2 In(1i + ~) - 21nh] .

(HAPTE

66

= (2t , t 2 , logt), we compute c' (t) = (2, 2t, l/t) . Then Il c'(t)11 = (4 + 4t 2 + l/t 2)1/2 = [(4t 2 + 4t4 + 1)/t2F/2 = (2t 2 + 1)/t = 2t + l/t. Since c(l ) = (2,1,0) and c(2) = (4, 4, log2), we want the arc length on [1,2]' so we n

9. Given e(t)

ed

compute

1 2 ( 2t +

~) dt = (t

2

+ 10gt{ = 3 + log2 .

=

=

12. (a) Since T(t) . T(t) = II T (t)112 1, we can differenti ate both sides of T( t ) . T(t) l. By product rule fo r dot products (see the box preceding example 1 in section 4.1), (d/dt)( T T(t)) (d/dt )( l ), i.e., 2T (t) . T'(t) 0, which implies that T (t) . T ' (t ) = O. (b) Beginning with T(t) = c'(t)/lle'(t)ll, we differentiate with respect to t, using the qUOl rule:

=

=

'( ) = !!.. ( ~) = c"(t)ll c' (t)1I -

T t

dt

c'(t)(d/dt)lle'(t)11

lie' (t) 112

Ilc'(t) II

.

Recall that Ilc'(t )112 = c' (t) . c'(t), so d -lle'(t)11

dt

= -d y'c'(t) . c'(t) = -1 (e' (t ) . e'(t))-1/2(2e' (t ) . e"(t)) = e'(t) . c"(t) . dt

2

Ile'(t)11

Su bsti tution yields

, T (t)

e" ( t ) ,

= lIe'(t )1I21Ie (i)11 -

c' (t) . e" (t) , lI e'(t) 11 3 e (t)

=

e" (t) II e' (t) 112 - (c' (t) . e" (t) )e' (t) lIe'(tlIl 3

17. (a) From the definitions of k and N in exercises 13(b) and 14,

dT II' ( ) II T' ( s) ds = T '() s = T s IIT'(s)11 = kN . T he vectors T, N and B have un it length and form a right-handed system of mutually ort onal vectors, so we have T x N

= B,

N x B

=T

and

B x T

= N.

Differentiate , using the product rule fo r cross products , to get: dN

d

ds = ds (B

dB

x T)

= ds

x T

+B

dT

x

dB

ds = ds

dT

xT -

ds

x B.

Using the fact that dB /ds = -TN (Exercise 15) along with the results derived earher i exercise and then factoring out a constant from t he cross prod ucts, we get -T( N x T) - k( N x B )

= -T( - B ) -

kT

= -kT + TB.

Finally,

d

dB ds

dT

dN

= ds (T x N) = ds x N + T x ds = kN x N + T = k N x N - kT x T + TT x B = -TN = Tx T = o.

since N x N (b) Let w = WI T and W3. We have

+ W2N + W3 B.

dT /ds

= =

w

x (- kT

+ TB)

We shall use the results from part (a) to find scalars

x T

= wt{ T

x T) +w2(N x T) +w3(B x T)

0 - w2B +W3aN

= kN .

67

VECTOR-VALUED FUNCTIONS

Here, we have used the facts that (1) B = T x Nand (2) T, Nand B form a right-handed system of mutually orthogonal vectors, and so there exists scalars a and b such that B x T = aN and N x B = bT. Thus, W2 0 and aW3 k.

=

=

Similarly,

dN/ds

= wdT x N) + w2(N x N) + w3(B x N) w1B + 0 - w3bT = - k T + rB. w xN

This gives us

Wl

= r,

bW3:; k

and

a

= b.

Finally,

dB/ds

w x

B=

(T x B) + w2(N x B) + w3(B x + 0 + 0 = -TaN = -rN,

Wl

wl(T x B)

which gives a ::: 1, and this implies that b == 1 and W3

= k.

B)

Therefore,

w == rT+ kB.

4.3: VECTOR FIELDS

GOALS 1. Be able to sketch simple vector fields. 2. Understand the relationship between flow lines and a vector field.

STUDY HINTS 1. Vector field. This is a mapping from ]Rn to ]Rn . Note that the dimensions of the spaces are equal. Each point x in the domain is assigned a vector. To depict a vector field, We draw the assigned vector originating from the point x. 2. Scalar field. A scalar field differs from a vector field in that each point x in the domain is assigned a scalar, not a vector. An example is the annual rainfall at each point on the earth's surface. The wind velocity at any instant of time is an example of a vector field. 3. Gradient vs. vector field. All gradient fields are vector fields, but not all vector fields are gradient fields. For example, the vector field i + xj is not a gradient field because you cannot find an f such that of/ox = 1 and of/ oy = x. If a vector field Pi + Qj is a gradient field then oP/ 8y = 8Q / ox. This statement comes from the fact that {j2 f / oxoy = 0 2 f / oyox for a well-behaved function f. 4. Flow lines. This is the path a par ticle would take if it was free to move along the vectors in the field. T hinking of the vector field as velocity, the flow lines would show displacement . A formula description of a flow line can be obtained by integrating each component of a vector field (or solving a system of differential equations). Flow lines c(t) must satisfy the equation

c'(t) = F(c(t)).

CHAPTER

68 SOLUTIO N S TO SELECTED EXE RC ISES 1. At each point (x, y) , we sketch the vector (2, 2) oriO"i­

nating from (x, y). Alternatively, we may draw a small multiple of the vector field. T he orientation of the vectors must be maintained at 45 0 fro m the positive x axis , and all of t he vectors must have the same ma.g­ nitude.

/' /' /'

/' /' /'

/'

/'

/'

/' /'

/'

/'

/' /' /'

/'

5. At each point (x , y), we sketch the vector (2y, x ) orig­

inating from (x , y). T he magnitude of the vectors in­ creases as (x, y) moves away from the origin .

/'

/'

/'

/'

/'

y

10 . At each point (x , y), draw a li ttle arrow in the direction (x , - y ), then connect the little arr to ge t flow lines. Al tern atively, one can solve the system of differential equations dxldt = dYI dt -y and write y in terms of x, then plot th flow lines y C I x fo r various consta C. Our compu ter-generated sket.ch is shown below.

=

=

x

13. If c(t) is a flow line of F , then c' (t) = F(c (t)) . The left-hand side is c' (t ) = (2e 2t , l i t , -lil­ t f::. O. On the right-hand side , we have F( c(t)) = (2e2t , lit, - 1It 2 ) since x = e2t and z = We got the same result for bo th sides of t he equation, so c(t ) is a flow line of F .

69

VECTO R-VALU ED FUNCTIO NS

18. From the ch ain rule, we have dV(c(t)) dt

= \7V(c(t)) . c'(t).

We are given that c(t) is a flow line of F = -\7V, so c'(t) = F == -\7V and

dV~~(t)) = -c'(t)

. c'(t) =

-llc/(t)112 :::; 0.

Thus, the derivative of V :::; 0, which shows that V is a decreasing function of t. 20 . First , we compute

At each point (x,y), we draw the vector -\7V. Notice that far from the origin, the denom ina­ tors of - \7V, (x 2+y2) 2, will be much larger than t he numerators . Thus , the magnitude of - \7V is very sm all at points far from the origin . A few computations reveal that - \7V is very large in m agnitude near the origin . For example, - \7 V(0.5, 0) == (4, -4) and -\7V(O. l , 0.1) = (50,50). A sketch of - \7V is shown below at the left. To sketch the equipotential surface V( x , y) = (x + y)/(x 2 + y2 ) 1, we rearrange and (x + (y Therefore, complete the squares to get (x 2 - x + + (y2 - Y +

i) =

t)2

=

t)2 = t. the equ ipotential surface V(x , y) = 1 is a circle of radius V"f, centered a t (~, t). This is shown

i)

below at the right .

y

,

,

2

/

/ -2

-1

2

0

x

-2

-

x

,

4.4: DIVERGENC E AN D CURL GOALS 1. Given any vector field , be able to compute its divergence.

2. Given any vector field in 1R 3 , be able to compute its curl. 3. Be able to explain the physical significance of the divergence and the curl. 4. Be able to manipulate expressions involving the cross product, the dot product and the del operator .

CHAPTER 4

70

STU DY HINTS 1. T he operation \1 . This operator , called "del," tells you to assem ble the vector of partial deri vat ives: (%x, %y, % z). 2. Divergence. Note th at the divergence is a scalar , not a vector. Know the t wo notations: \1 . F = di v F . You should know t hat the divergence is a rate of expansion or, if it is negative, compression. T he term incompressible means that div F = O.

3. Curl. Note that the curl is a vector , not a scalar. Know the two not ations: \1 x F ::: curl F. You should know t hat the curl is associated with rotations and that the term irrotational means that curl F = O. 4. Valid space f or curl. Note t hat curl is a property of ]R3. We do not attempt to take the curl of vectors in dimensions higher t han t hree. T he two-dimension al vector xi + xyj is taken to mean xi + xyj + Ok if a curl is desired .

5 . Laplacian. The expression \1 2 f means \1 . (\1 f) , which is a scalar.

=

=

6. Theorem s. T he facts t hat \1 x (\1 f ) 0 and V . (\1 x F) 0 are useful ; it is nice to commit t hese to mem ory. If you need to know one of t hese facts and you forget t hem, you can always do t he comp utation . 7. Formulas in JR3. Note that t he cross product or curl occurs in some of the basic identities of vector analysis in the t able preceding exam ple 15; therefore, it is assumed that the formulas are used in ]R 3 and not in higher dimensions . 8. Basic iden tities of vector analysis. T he for mulas in the table preced ing Example 15 are useful for developing the theory of vectors; however, you should not memorize the t able. You can refer back to the table as needed . It will become obvious which formulas are most important as you refer back t o them frequently.

g. Exercise 30. These fo rmulas are referred to quite frequently in the examples. You should do this exercise even if it is not assigned .

SOLUTIONS TO SELECTED EXERCISES 2. T he divergence is \1 . V

= tx (yz ) + :y (x z ) + : )x y) = O.

7. Recall t hat flow lin es are defined by

c' (t)

= F(c(t)) .

In this case , we have c' (t) = (dx / dt,dy/ dt) = (y, O) .

ince dy / dt = 0 , we know y is some const ant, k l .

Then dx/dt = y = k l , so X = kI t + k 2 , where kl and k2 are constants. Thus, the flow lines have the form (kIt + k2 , kJ). If kl is positive, flu id flo ws from

left to right as t increases , and if kl is negative , fluid

flows from right to left, as shown in t he sketch. We com pute the divergence: \1 . F = 8(y)/8x +8(O) /8 y =

O. The sketch shows that flu id appears to be neither expandin g nor contracting , which is consistent with our calculations.

t

• •... .=r=; y

1

....

­

....

.... ...

11. The divergence is

\1. F

= :x (sin (xy)) -

: y (cos( x 2 y))

= ycos(xy) + x 2 sin(x 2 y).

--..

x

71

VECTOR-VALUED FUNCTIONS 14. The curl is i

j

k

xz

0 oz XY

() a ax oy

\7x F

yz

= i (~(Xy) oy

~(xz)) oz

- j

(~(Xy) - ~(yz)) ax oz

+ k (:x (xz) - :y (Yz) )

+ (z -

(x - x)i - (y - y)j

z )k

= O.

17. The scalar curl is the coefficient of k when \7 x F is computed for a vector field F in lR 2. Here,

we have

\7 x F =

i

j

a ox

a oy

Sill X

cosx

k

a

=

oz

[:x (cos x)- ; y (sinx) ] k = (-sinx)k.

o

Therefore, the scalar curl is - sin x. 22. First, we compute that \7 J = (y + z)i

i

\7 x \7J

+ (x + z)j + (y + x)k.

o

o

J

k

ax y+z

oy x +z

az y +x

[ ~(y oy + x) (1 - 1) i - (1 -

Then

o

~(x + Z)] i - [~(y + x) - ~(y + z)] j oz ox . oz + [:x (x+ z)- ;y (y + z)] k l)j + (1 - 1) k = o.

25. We know that the curl of a gradient is the zero vector. Thus, it suffices to show that \7 x F We have

i j k 000 ax oy oz y cos x x sin y 0

\7 x F =

= (0 - O)i + (0 - O)j + (sin y - cos x)k

i=

O.

i= o.

28. We refer to the table of basic identities of vector analysis, which precedes Example 15. (a) From identity (5), div (F + G) = div F + di v G = 'V . F + \7 . G, which sums to zero, according to the given hypot hesis . Therefore F + G does have zero divergence. (b) From identity (8), div (F x G) = G· curl F - F·curl G, which , in general, does not equal O. As a simple example, let F = xyi - z yk and G = i. We have \7 . F = \7 . G = 0 and div (F x G) = z. 30. (a) We have \7(l/r) = \7(1/ Jx 2 + y2 + z2) . Begin by findi ng (8/ox)(1/r) or

a ( ox

Jx2

1 )

+ y2 + z2

-2x -x-x = 2( x 2 + y2 + Z2)3/2 = (x 2 + y2 + z2)3/2 = -;:J'

By symmetry, (%y)(l/r) = _ y/ r 3 and ({)j oz)(l /r) = -z/r 3 • Putting these together, V'{l/r) = -(xi + yj + zk)/r3 = - r / r 3 . In general, \7 (rn )

= \7((x 2 + y2 + z2)n/2)

CHAPTER 4

72

and

(O /O X)( X2

+ y2 + Z2r/2) = (n/ 2)(x 2 + y2 + Z2)n/2-1)2x = nx1'n - 2,

so by symmetry,

\7 (1'n) = nrn - 2(xi + yj

+ zk ) =

n1' n - 2r.

Finally,

= \7(log( J

\7(logr) and

a

-(log( Jx 2 + y2 ox

+ z2 )) =

so by sym metry, \7(logr) = (xi + yj (b) Using the results of part (a),

\7 2 (1/1')

x2

+ y2 + z2)) ,

a

1

J x 2+y2 +z2

. - ( J x2 ox

+ y2 + z2) =

x 2' l'

+ zk)/1' 2 = r/1'2 .

\7. \7 (l / r ) = \7. (-r / 1'3 )

=

= -

[: x C JX 2 +:2+ Z2 )3 ) +:y

CJX2+~2+ Z2)3)

C + ~2 +

+ :z

J x2

z2) 3) ].

The first partial derivative is

x

(x 2 + y2

+ z2)3/2 _ ~Jx 2 + y2 + z2 (x 2 + y2 + z2 )3 2 x + y2 + z2 _ 3x 2 (x2 + y2 + z2)5/2 .

:x Cx2 + y2 + z2).3/2) By symmetry,

a( oy

and

a( OZ

Then

\7.

y (x 2 + y2

(x 2 + y2

)

x 2 + y2 + z2 _ 3y2

= (x2 + y2+ z2)5 / 2

)

x 2 + y2 + z2 - 3z 2

= (x2 + y2 + z2)5/2 .

+ z2)3/2

z

+ z2 )3/2

.2x

(-r) = 3(x 2 + y2 + z2) - 3(x 2 + y2 + z2 ) = O.

1'3 (x2 + y2 + z 2)5/2

Simil arly, part (a) tells us that in the general case,

\7 2 (1'n) = \7 . \7(1'n)

= \7 . (nr n - 2r) .

We compute

: x « x 2 + y2

+ z2) n/ 2-1

. x ) = n [( x 2 + y2

+ z 2 t/ 2 - 1 + 2x2 (~

_

1) (x 2 + y2 + z2) n/ 2-2] .

Again, by sy mmetry, \7 . (n1' n - 2r)

n [3(x2 + y2 + z 2t/ 2- 1 + (x 2 + y2 + z2)(n _ 2)(x 2 + y2 n1'n- z(3 + n - 2) = n(n + 1)1'n-2.

+ z2r/ 2-2j

(c) The identity \7 . (r/r 3) = 0 follows immediately from part (b) since n = -1 in that case. For the general case , use part (b) again to compute \7 . (1'''r). Note that we only need to divide

73

VECTOR-VALUED FUNCTIONS the general result by n and ch ange n - 2 to k (Why?) . T hen \7 . (rkr ) = (k + 3)rk = (n (d ) By a dired computation, we get

i

0 8x x

'\7 x r =

J

0 8y y

+ 3)rn.

k

8 az z

= 0.

Using the fact that the urI of a gradient is the zero vector, the calculation j ust completed, and t he general case in part (a), we get

\l x (rnr ) = rn ('\7 x r ) + \l(rn ) x r = 0

+ m· n - 2 (r x r ) = o.

32 . (a) 1

j 8

3x y

x3 + y3

o ox2

curl F = \l x F

k(3x 2

-

k

oy

o oz o

= o.

3x 2 )

(b) Note that if F = \l I, then 3x 2 y = (%x) /(x, y) and x 3 each equation:

f( x, y) =

+ 11

= (%

y)/(x, y). Integrate

J

3x 2 y dx = x 3 y+g(y),

where 9 is a fu nction of y only, and

where h is a function of x only. In bo th cases, I (x , y) must be the same , so compare both sides. If we let g(y) = y4 / 4 and h(x) be an arbitrary constant , then f(x,y) = x 3 y + y4 / 4 + C satisfies \l f = F.

33. (c) Let F( x, y) = e'" cos yi - eXsin yj , where the j component is the real part of ex - iy and th j component is the imaginary part . We calculate div F = (0/ ox)( eX cos y) - (%y)( eXsin y) = eX cos y - eX cos y = 0 and i

curl F

j

o ox

o oy

k

a

8z

eX cos y -e'" sin y 0 k( - ex sin y + e" sin y) = o. Thus, F is both incompressi bl and irrota tional.

SOLUTIONS TO SELECTED REVIEW E XERCISE S FOR CHAPTER 4 2. The velocity vector is

vet)

= e'(t) = 2ti + (- 2t sin(t

))j + 4t 3 k. The acceleration vector is 4t cos(t )lj + 12t k. When t = ."fi, we have v( y'ir) = 2

+ [-2sin(t ) 2y'1ri + 47r."fik and a(y'ir) = 2i + 47rj + 127rk. The speed is the length of the velocity vector. When t = ."fi, the speed is [47r + 1611"3]1/ 2 = 2-/11" + 4~ . The equation of the tangent lin IS given by l(t) e(y'ir) + tv (y'ir) = (r. - 1, - 1, r. 2 ) + t(2."fi, 0, 4r.y'ir). aCt) = e" (t ) = 2i

2

2

2

2

=

7. We use t he equation F = rna = me". Here, e" = (2, - sin t, - cos t) . At t = 0, a (O) (2,0, -1), so F(O) = rn(2, 0, - 1).

_• ....a. _____ .

CHAPTER 4

74

9_ From x 2 = .; = Zll , we get y = X 2 / 3 and z = X 2 / 5 . Let x = t, so the path of the curve is c(t) = (t, t 2/ 3, t 2/ 5). Here, we have 1 ~ t ~ 4 and c' (t ) = (1 , it-1/3, i t- 3/S). The arc length is

t VI + i9t

1

4I1C'(t)11 dt =

11

1

-2/3

+ i. t -6/ 5 dt. 25

13. We set the given equation equal to t and solve for x, y and z: x- I = 2y + 1 = 3z + 2 = t. From x -I = t, we get r = t + 1. Similarly, 2y + 1 = t yields y = (t - 1)/2 and 3z + 2 = t yields z (t - 2)/3. Therefore, we get (.7:, y, z) (t + 1, (t - 1)/2, (t - 2)/3) .

=

=

15. We want to show that c'(t) = F (c(t) ). Here, x = 1/(1 - t) , Y = 0 and z = et /(1 - t). We ~et dx / dt = 1/(1- t) 2, dy / dt = 0 and dz/ dt = (2e t - t et )/(l- t)2. T hus, dx / dt = .:z:2 , dy/ dt = 0 and dz /dt = z (l + x) = let /( 1- t )][l + 1/(1- t )l = (2e t - tet) / (l - t )2.

+ /y (y2 ) + f. (z2 ) = 2x + 2y + 2z .

18. We compute div F = \l . F = fx( .:z:2) i

\l x F = I f}.:z: .:z:2

­

8u y2

8z z2

i

8 ay z

{}x

y

8x

f}z

8x

= f}y/ f}x + az/ 8y + 8x / 8z = O. k 8

J

a

[8 - (z 2) - -8 (x 2]. ) J + [-{} (y2) - -8(r 2] ) k = o.

1-

Z ) -

21. The divergence is \l - F

=I

-

[-aya( 2 -8za( y2)] '

=

\l x F

k

888

=

curl F

j

Also,

= ({}X_ {}Z ) 8y

8z

{}z

ay

The curl is

i + ( f}y _ {}x ) j + (8Z _ 8Y ) k = -i-j -k. {}Z ax (}x 8y

x

25. We compute

\lJ

= ( ~~ ) i + ( ~~ ) j + ( ~~ ) k = [2.:z: exp(z2 ) + y2 sin (zy2)]i + [2xy sin(xy2 )li + Ok.

Next, we compute I

\l x \lJ

=

8x 2x exp(x 2 ) + y2 sin(xy2 )

[:Y (0) -

k

J

{}

:z (2xYSin( X

8 ay 2xysin(zy2)

y2))] i +

[:z(2x

8 8z 0

exp (z2 ) + y2 sin(xy2 )) - : z (0)] j

+ [:x(2x y sin (x y2)) - : y (2xex p (x 2 ) + y2Sin(Xy2 ))] k

=

(0 - O)i

=

o.

T hus, for J (x, y)

+ (0 -

= exp (x

2

) -

O)j + [2y sin(xy2 ) + 2x y3 cos (xy2 ) - 2y sin (xy2) - 2xy3 cos(xy2 )] k

Cos(xy2 ), we have \l x \l J

= O.

J exists, then we know that F = \l1 = ({}f/{}x )i + (al/ ay )j + (8f/8z)k . Thue , we have {} 1/ 8x = 2xye Z , 01/ oy = eZ x 2 and 81/ oz = x2ye~ + z2. Integrating aI/ox = 2xye Z with respect to x gives us I(x, y, z ) x2ye~ +A(y, z ), where A is some function involving yand Z only. (You can check this result by computing (}f/o x. ) Similarly, integrating (}f/ ay with respect to y gives us l (x, !I , z) = z2ye~ + B (x, z ), where B is some function involving x and z only. Next, in tegrating oj/ 8z with respect to z gives us I( x , y, z ) = z2ye z + z3/ 3 + D(x , y), where D is some function involving x and y only. Comparing the three resulte for I(z,y , z ), we see that A(y, z ) = z3 / 3 + C, B(x , z ) = z3 / 3 + C and D(x , y) = C, where C is an arbitrary constant. Therefore, f (x, !I, z ) = z2ye~ + z3 / 3 + C.

28 . (b) If such

=

75

VE CTOR-VALUED FUNCTIO NS

32. (a) At the intersection, we know that y = 1, so t he intersecting curve has the equation x 2+( 1)2+z2 = ;3, or x 2+z 2 = 2. Eq uivalently, we have x 2/2+z 2 /2 = 1, or (x/V2)2+(z/V2)2 = 1. Since we know that cos 2 t+sin 2 t = 1, we get x/V2 = cost and z/V2 = sin t, or x = V2cost and z = V2 sin t. To get one revolution of the intersecting curve, we let t vary from 0 to 2r.. Thus, a parametrization of the in t ersecting curve is x = V2 cost, Y 1, z V2 sint,

=

o < t < 2r..

=

(b) From Sect ion 2.4, recall that an equation for a tangent line is l(t ) = c(to) + (t - to)c'(to). Here, c(to) = (1, 1,1), where c(t) is the param etriza tion given in part (a). We want (V2 cos to, 1, V2 sin to) = (1, 1, 1), so t = r./ 4. Differentiating each component of the parametrizat ion, we get cl(t) = (-V2 sin t,O,V2 cost) . T hus, the tangent line is

l(t)

= (1,1,1) +

(c) In par t (b), we computed c'(t), so Illc'(t)11

f 2rr

io

(t - r./4)( - 1, 0,1).

= V2. Thus,

1

the arc length is

2rr

=

Ilc'(t)11 dt

0

h dt

= 2hr..

As expected, this is the circumference of a circle of radius V2. 36. (a) T he direction of the rotation vector w is the same as the axis of rotation, in this case, the positive z axis, or k direction. vVe're given the magnitude as 4, so w = 4k. ,( b) When r = !5v'2 ~ i ­ j), the velocity is

v

=w

x r

=

0 5V2

j 0 -5V2

k 4 0

= 20hi ­

20hj .

(c) The vector r is the vector from the axis to the point , so r is the vector from (0,0,5) to (0, 5V3, 5), or r = .5V3j. For the point (0, .5V3, .5), we get J

v

=w x r =

0

o

0 .5V3

k 4 0

= -20V3i.

T EST FOR CHAPTER 4 1. True or false. If false, explain why.

(a) The curl of any vector field in

]R3

is another vector field .

(b) One revolution of the path c(t) = (cos2t,sin2t) has arc lengt h (c) If band c are vector-valued func t ions of t, then d(b x c)/dt

f;rr Ilc'(t)11 dt.

= db/dt x c +

dc/ dt x b.

(d) Two particles which have t he same acceleration must travel at the same speed. (e) Suppose G(x, y) is the velocity field of a gas. If G(x, y) = xi, then the gas is expanding and if G(x, y) = yi, then the gas is not expanding . 2. Let c(t) = (2t, t - 3, t 2 + 1) be a possible flow line for a velocity vector field . W hich of the following, if any, could be such a velocity vector field?

(a) F(x, y, z)

= (2, x -

2y + 7, x).

(b) F(x,y,z)=(yiz -l-y+5,1, 2y -6).

3. Suppose the velocity vector fields represent the flow of a fl uid. At what poi nts in does the flui d lack rotation?

]R 3,

if any,

-'---­

--

-

76

CHAPTER 4

(a) F( x, y, z)

= 4x yi + 2x 2j + k.

(b) F(x , y, z ) = xyi

+ zj + xk.

4. A particle travels counterclockwise along the ellipse 9x 2

+ 4y2 = 36.

(a) Express the distance traveled going from (2,0) to (0, -3) as an integral of the form J + b cos 2 t dt for constants a and b.

va

(b) Suppose t he part icle had flown off on a tangent line at (-l ,3 V3/2) . Where will the particle hi t the x-axis? 5. For a given function , g(t), let a particle's posi tion be described by 2g(t)i

+ [g(t)j2j -

k.

(a) What force is acting on the particle if its mass is 3 units? (b) Wh at conditions must be imposed on g(t) for the force to be O? 6. Let F(x , y, z) = xi + yj gas expanding?

+ (z2 + y) k

be the velocity vector field for a gas. Where , in ~3, is the

7. In R 4 , a path is described by y = 2x, z = 4x 2 - 6 and w as an integral in y for 0 ~ x ~ 2. Do not evaluate.

=e

Z

•

Find the arc length of the path

8. Which of the following, if any, is the curl of some vector field G such that F

= curl G?

(a) F(x, y, z ) = 2i + 3j - 2k.

(b) F (x, y,z ) =xi + 2j - zk.

(c) F (x , y,z )= yi + yj + x 2 k. 9. (a) Let v(t) = (sin 2 t)i respect t o x?

+ (t 4 - t 3 )j + 3k.

What is t he deri vative of v( x 3

(b) Let c(t) be a path in ]R 3 and let p(t) be a scalar function. acceleration of p(t )c(t) .

-

x2

+ 3x -

4) with

Find a formula for the

10. At the annual camel races, camel A 's position is given by (cos 5t, 2 sin 5t) and camel B's position is given by (cos 4t , 2 sin 4t). (a) Show that the camels are running on the same path . (b) Suppose camel A 's mass is 500 . What force is being generated by camel A at t = 7r/5? (c) Camel B saw an oasis at t = 7r/ 5. It ran off on a tangent line. What is the equation of the tangent line? (d) Suppose camel B had continued to run on the track. How far behind (in distance) was camel B when camel A had com pleted one lap? Leave your answer in the form of an integral.

77

5

DOUBLE AND TRIPLE INTEGRALS

5.1: INTRODUCTION GOALS

•

1. Be able to calculate double integrals over rectangles.

2. Be able to use Cavalieri' principle.

STUDY HINTS 1. Some notation. (a) The cartesian product of two intervals in lR 2 is a rectangle. If a ~ x ~ b

and c ~ y ~ d, then it is denoted [a, b] x [c, d]. (b) Sometimes dx dy or dy dx is abbreviated dA, the "differential of area."

2. Review. Before continuing, you should re iew integration techniques for one variable. It is essential that you remember how to integrate by parts and by subst itution .

lID

3. Geometric interpretation. If f(x, y) ~ 0, then the double int gral f( x, y) dA is the volume under the surface defined by the graph of z f (x, y). Recall that with one variable, f(x) dx is the area under the curve y = f(x).

=

I

4. Computing a double integral. Just as with partial d rivatives , all but one variable is held

constant in each step. We integra te from the inside and work our way to the outside. For example,

i

bi df(x, y)dy dx = Ja,b [F(x ,d)-F(x,c)]dx,

a

c

where Fis an antiderivative of fwhen x is held constant. Now , we com pute the second x-integral using one-variable methods. 5. Cavalieri's principle. This principle is used in most one-variable calculus courses to derive volume formulas. These fo rmul as are often referred to by names: disk or slice method.

2:7:=0 f(Ci)( Xi +! - Xi ), ci can be chosen anywhere in [Xi , xi+d . In a Riemann sum, we take the limi t as n -+ 00, so in most cases, ! (c;) is almost independent of Ci because X i +! and Xi eventually get close together.

6. Riemann sums. Be aware that in the sum m ation

SOLUTIO NS TO SELECTED EXERCISES 1. (b) First hold x constant and integr ate with respect to y, then integrate with respect to x:

t (ycos x +2)dy dx ior/ Jo 2

r/ (12' io 2

cos

)

x+ 2

dx

= ( 2'1 sin x + 2x ) 17f/2 0 = '12 + 'fr.

78

CHAPTER 5

2. (b) We can change t he order of integration . Thus , we hold y constant and integrate with respect to x firs t:

r/ (y cosx + 2) dx dy Jor Jo i

2

Jot

1"'/2 ]

[(y sin x + 2x ) x=O dy

1\y + dy= (y; + 7r I: = ~ + Y)

7r)

7r.

As expected, we get the sa me answer regardless of the order of integration.

3. By Cavalieri 's principle , the volume of a solid is b

V

=

1 a

A(x ) dx, •

where A (x ) is the cross-sectional area cut out by a plane. Note that at the same height, a cross-section of the left-hand side is a circle of radius 7' and a cross-section of the right-hand side is also a circle of radius r . Since A (x ) is equal in both cases, the volumes must be equal.

5. With the setup in figure 5. 1.12, we slice W vertically by planes to produce triangles Rx of area A(x ) in the figure. T he base b of the triangle is b = .Jr2 - x 2 and its height is given by h = b tan (j = .Jr2 - x 2 tan (j, Thus A(x ) = ~bh = ~ (r2 - x 2) t an (j . Hence, the volume is

j

r _rA(x) dx

=

jT 21 (r

2

21 t an (j

( 2r 3

_r

32r

-

8. Note that when y is in the interval [-1 ,0], then

Jl (Iyl

cos

~x )

r

3

2 1 ( r 2 x -'3 x ) -x )tan(jdx= 2(tan(j) 3 )

= 32r

I

-r

3

tan (j.

Iyl = -y,

and so

2

dydx

= Jor jO-1 (-YCOS 7r4X) dy dx O 2 r [(_y2 7rX)I Jo -2- cos 4" y=-l

1 2

o

1dx (1 7rx) dx = -2. 7rX - cos 2 4

sm -

I

7r

12

2

0

7r

4

10. Since f( x, y) ~ 0 for all points in R, ffR f( x, y) dy dx is the desired volume. It is

1 2

1 \1+ 2x + 3Y)dY dx =

12

[(Y+ 2X Y +

12 (2X+~)

3f )I~=J dx

dx=

5. 2: THE D OUBLE INTEG RAL OVER A R ECTA N GLE G OALS

1. Be able to com pute a double integral over a rectangular region.

2. Understand Fubini's theorem .

(x2+ 52X)

I:

11 2

DOUBLE AND TRIPLE INTEG RALS

79

STU DY HINTS 1. Definition. The definition of the integral is more importa nt fo r theoretical rather than com­ putational work . It is defined to be a limit of Riemann sums:

J1

! (x, y) dA

= nl~n;,

R

n -1

L

f ( Cjk )

~x ~y.

j ,k=O

Although it is not required, it is usually convenient to use a regular partition, i.e., a partit ion with ual spacings. 2. Properties. Many of the properties that hold for single integrals also hold for dou ble integrals. Some of these include the integrabili ty of any continuous or even piecewise conti nuous fu nction .

3. Warning. As in one-variable integration, [fJRfdA] [fJR g dA]

=1= JJR!gdA, in general. In other words, the product of two integrals does not usually equal the integral of the product.

4. New terminology. T h nam s of the proper ties are known as lineari ty, homogeneity, mono­

tonici ty and additivity. You should know the statements even if you don 't know the names.

5. Fubini's theorem . Th is tells you tha.t, for most reasonable functions, you can int grate one variable at a t ime and the or der of integration does not matter. 6. Double integrals and volumes. Recall that if !(x , y) 2: 0, t he double integral is simply the volume of the region between the graph of !(x, y) and the xy plane. If f (x, y) < 0, then we subtra t the correspond ing volume between the graph of !(x, y) and the xy plane.

SOLUTIONS TO SELECTED EXERCISES 1. (b) We compute t he do uble integral as an iterated in tegral:

Note that we chose to integrate in x first be ause that integral is simple (integrat ing with respect to y first would requ ire integration by parts). 2. (b) T his is the integrated integral

fa 1[ (a~2 + (by + C)x) [=al dy fa (~ + by + c) dy = [b~2 + (~ + c) yJI~ 1

a

b

"2 +"2 + c. 5. We will use t h fact t hat J cf(x ) d:c = c J f( x ) d:t for any cons tant c. If we integrate in x first, then g(y) is held constant in the fir t st p, and so

J

k[J(IX )g(y)]dx dy

=

ldib ld lb

f (x)g(y) dx dy

[g( y)

b = ld [ i

1

! (x ) dx dy ,

1

f(x )g(y) d:c dy

CHAPT ER 5

80 where

J: I( z ) dz is a constant in

y. Factoring out the constant integral gives us

[i d

[ib I(X)dXl

n, so t he double integral represents

7. The fu nction x 2 + y is positive over

111\x2 +y) dydx

g{Y)dyl .

11[

=

(

x3

3

+ 3X ) 11 2

J

J J J

=~ + ~ =~. 3

0

2

6

x 2 - Z2

=

tan 2 () 2

x' sec 411u (xsec 0dO)

y

,>

2

x3(1 - tan B) dB X4 sec 2 B

=

J

2

2 cos 0 - sin (J dB z

cos 2B dO = sin 2B + C. x 2x

From the triangle, we have sin e = y/ ';'2 z-::--+-y2 ~ and CO! 0 = z / sin 2B 2x

dx

x sec 2 OdO and

11. Begin by substituting y danO, so dy J x 2 + y2 = X sec e. We get

y . .. d (x~" + y 2)"

dx

y; )

1 16 .

Thus, the volu me is

x2 - y2

[=J = 11(x2 + ~)

+

2

(YX

the desired volu me.

+C =

sin e cos e x

+C =

~ . __ y x Vx2+ y2

x

J z2 + y2. Then

x J x2 + y2

+C=

Y +C X2 + y2 '

Evaluating, we get

[I x 2 Then

111x2 -

1 o

_ y2

y2

(2

2)2

x +y

0

y

(x 2 + y2 )2 dy =

10

dydx

=

Xl

11

+ yl

y2 (x2 + y2 )2 z2 -

J

1

0

SO

dx

2

J -J11 (1 y2

1

y=0

-2~-dz x +1

For the second integral, we substitute z = ytan , This gi ves -:-;:--~:-;::- dx

11

= x2 + l '

l

=rr/4

= tan- 1 (2), we get the follo wing:

1 )

r

vis + Vi.1 o

/2

dO

3rr ( 1

=2

1 )

Vi - v'5 .

z

2

y

1

6.3: APPLICATIONS GOALS 1. Be able to use double and triple integrals to compute averages, centers of mass . moments of inerti a and gravitation al potentials.

STUDY HINTS 1. Averages. In general, an average is defined to be the integral of fd ivided by the integral of 1. Hence, the two average formulas given in this section are

IffII' fdV fffll'dV

and

ffD f dA ffD dA '

2. Cente1' of mass. This is a weighted average. In general, the center of mass of a region in JR n is (Xl,X2,X3, ... ,X n ), where Xj is fR x j c5 dV

fR c5 dV and dV = dXl dX2 ... dX n and R is the region. Since c5 is the density, fR c5 dV is the mass. If the region is in JR2 use double integrals, and if the region is in IRs use t riple integrals.

THE CHANGE OF VARIABLES FORMULA AND APPLICATIONS OF INTEGRATION

105

3 . Moment of intertta. A good way to remember Ix = IIIw (y2 + z2) 8 dV is to notice that the integrand lacks an x term. Similarly, Iy and l z lack a y and a z term , respectively.

4. Gmvitational potential. Its formula is V = - OM m/ R. Although you do not ne d to under­ stand the physic 1 t heory behind the discussion the mathematics shou ld be clear to you . You probably will not need to reproduce the discussion for an exam . 5. Geometry. Recall that if the integrand is 1, then the triple integral IIIw dV gives the volume of W. Sim il arly, the double integral lID dA gives the area of D.

SOLUTIONS TO SELECTED EXERCISES 3. The fo rmulas used to compute t he center of mass (x, y)

for a region in the plane are

_ IIDxtS(;z;,y)dxdy = lIDtS(x, y) dx dy

x

_ lID ycS(x, y) dx dy y­ - lID 8(x , y) dx dy .

and

In this case, we sketch the region D and see that it can

be described by

o ~ x ~ 1

x 2 ~ y ~ x.

and

x

T he numerator of x is

[11" x(x +y) dydx

Jo

rl

T he deno minator is

11[(Xy + y; )[=".] 11(3;2 _ ~4 )

( l "(X + y) dydx

Jo

r'

x3 _

Therefore,

dx

dx =

X3 _ (

X4 _

2

X

5

)1

10

4

1

0

18 120

65 x = i! . We leave it to you to calculate y = 126'

6. Th mass of the plate is lID 8 dA, which is

[21
2(X)

l

f(x, y) dy dx.

¢>,(X)

a

(b) The region D is like the one described in part (a) . Since the limits of integration are con­ stant, we can integrate each variable separately and then multiply the results. The integrand is factora ble as fo llows:

xy ex p(-x 2

y2)

_

= [xexp(-x2)][yexp(-y2)],

so

JI f

[1

1

(x,Y)dA

[/00 xex p (-x 2 )dX]

yex p (-y2)d Y]

= (. ~)

b

11)

2 p lim (ex (-x ) I ) b-t oo -2 1

exp(-y2) ( -2 a

4

(! _~) . e2

e

7. Fubini's theorem tells us t hat multiple integration can be performed in any order as long as

the integrand is positive. In this case,

e- xy

> 0, so we have

00

/2 1

e-XY

dx dy

=

100 /2 e-

xy

dydx .

For the left-hand side, we compute

rJo[00

e-xy

dx dy

J1

1

2 (

1

2 ( / ,1

-xy Ib ) dy lim ~ b-too y x=O

1) dy= 12 dy

by

lim b-too

- = logy 12

_e­ -+-

y

y

1

Y

~log2.

1

On the right-hand side, we compute

1 1 00

a

2

1

= 100 -_e--

XY

e- XY dy dx

o

:Z:

12 y=1

dx

100

=

e-

X ­

e-

2x

dx .

0

Since Fubini's theorem tells us t he two integrals are equal, we have the desired result.

THE CHANGE OF VARI ABLES FORMU LA AND APP LI CATION S OF INTEGRAT IOI\l

109

9. The integrand sim plifies to 1/(x+y). T he integral is improper whenever y == -x, wh'ich occurs in our region at t he point (0,0). We compu te

1111+

_1_ dy dx

o

0

x

y

=

11[

(In Ix +

yl)1

0

which we interpret as

[In(1 + x) -In(x)] dx,

dx

0

y-O

11

lim

~ --+o+

1_] =11

[In (1

~

+ x) -In(x)] dx .

Integrating by parts, we get

E~+ { [((x + 1) In(x + 1) -

x) - (x Inx - x)]I: } ·

Using I'Hopital 's r ule, we know hat

· I1m

E--+O+

In f = I'1m In/f = 0 ,

f

[ --+0+

1

f

and so the integral is 2 1n 2. 11. Since x 2 + y2 + z2 appears in the integrand, try using spherical coordinates. In the first octant, we have o ~ () ~ 7r/2 and 0 ~ cp ~ 7r/2. Since x 2 + y2 + z 2 ~ a 2 , we also have 0 ~ p ~ a. Recall that the Jacobian for spherical coordinates is p2 sin e/; . Subs itute p2 = x 2 + y2 + z2 and p cos cp = z to get

l

1f/2 1 ff /2 1 Q

o

0

Substitute u

pl/2. ----r====p2 sm cp dp de/; d() p cos cp + p4

J

0

=l

1f/2 1 1f/2 1 Q 0

0

0

p2 sin cp

J cos cp + p3

dp dcp d().

= cos cp + p3:

3

1f /2 1 1f /2 j COS ¢+a d u SI·n '+' '+' dcp d() o 0 cos¢

l

3y'U

r/1 2

1f

2

/

Jo 0 1f/2 o Jo

~ sin e/; ( y'Ul cos ¢+a 3 2

3 )

dcpd()

u= c os¢

r/2"3 sin CPJcos cp + a3 dcp d()

1r/ r/ 2

- Jo

r/ Jo

2

Jo

2

3" sin CP Jcos cp dcp d() 2

2

[_i(coscp+a 3 )3/zl1f/ + i(cosCP)3/2'11f/2] d() 9 ¢=o 9 ¢=o

27r [(1 + a 3 )3/2 9 14. W hen 0

~

g(:I:, y)

~

_

a9/ 2

_

1]

.

f(x, y), integration over D should give us

.lin

0 dA

~

.IL

g(x , y) dA

~

.IL

f(x, y) dA.

However , since the integral of 9 is sm aller than the integral of f and exist , we m ust conclude that fiDf (x , y) dA also does not exist.

lID g(x, y) dA does not

CHAPTER 6

110

16. Since x 2 + y2 + z2 appears in the integral , we will try to use spherical coordinates. The region D can be described by

I:=;p

and

0 :=; O:=;27r

In spherical coordinates, the integrand is fore, we get

27f 1 7f

1

1/ p4

and

0 :=;¢:=; 7r.

and recall that the J acobian is p2 sin r/!. There­

Joo p2 sin ¢ dp d¢ dO. '----;-4-'-

o 0 1 P Since all of the limits of integration are constant , we may integrate each variable separately and multiply the results:

{27f

f"Joo ~ sinriJ dp d 1d¢ dO

- 1 " /2

o

[1

27r

0

r/ SintPdl,b] io 2

- 'If/2

(- sin¢J ) d¢J +

27r [ cOS¢J IO

- 'If/ 2

+ (.- cos 4» 1'If/ 2] = 47r. 0

T he answer is 47r because when 4> varies from -7["/ 2 to 7r/ 2, t he parametrization covers the upper hemisphere twice, so the surface area is the same as one compl te sphere . If we vary I/; from 0 to 27r we should get 87r , since the sphere is parametrized twice:

1 1 211" 2

1[

Isin ¢ I d¢J dB

=

27r

[111" sin tP d¢ + 12'1f(- sin ¢J) d¢]

27r[ (- cos¢J)I : + costP C ] = 27r(2 +2) = 87r. If we had fo rgo ten the absolute values and simply integrated sin 1/;, we would have computed a surface area of 0 in both cases, but tills is not possib le. Surface areas have to be positi ve .

lID II Ttl

5. The ar a of of/ (D) is

8(x, y) = 8(u, v)

I

1 -1

1 1= 2; 1

x Tv II du dv. Given of/(u, v ) = (u - v, u + v, u'v) , we get

8(y, z) = 1- 1 1 1= - u _ v; 8(x, z) = 11 1 1= u - v. 8(u, v) v u 8(u, v) v u

Squaring t he de terminants, adding them and then taking the squar root gives us the integrand II Ttl x Tv ll = V2 2 + (- u - V)2 + (u - v)2 = V4 + 21.12 + 2v 2 = V2Vu 2 + v 2 + 2. Since we are integrating over the unit disk, use polar coordina tes. Let 1.1 r sin () , 0 ~ r ~ I , 0 ~ B ~ 27r. Then

JLh Vu

2

+

v2 +

=

2dA

111211" h

Vr2 + 2rdBdr = 2V27r 2

2V27r' i (r + 2)3/ 2

8. We choose the parametrization x = ~, y = t,

Z

1: =

11v,z

2V;7r (33 / 2 -

= z, 0 :S Z

= r cos B and v =

2 3/2 )

+ 2r dl' =

i (6V6 -

8).

~ 1, - 1 ~ t :S 1. To compute

the sur face area integrand, we first compute

1

8(x,y~ 1= 0,

8 (y,Z) I- l 18(x,Z )I 8(t, z) - , 8(t, z) 1

8(t, z }

then, II T t x Tz II =

2

J

t l dt dz 02 + 12 + --Zt +

Thus , the surface area is

2+ 1111 [ffit a

- 2-

-1

t

-

1 dt dz -_

+1

=

t

Jt2+T'

[ffi t2 + t +

1 dt dz. 1

- 2- -

11 [ffi t2+ - 2-

-1

t

-

1 dt.

+1

We do not a ttem pt to simplify further. An alternative parametrization is to use the hyperbolic functions sinh t and cosh t; the integral only gets nastier.

(H APTER 7

126 10. Let x

= ucos v, Y = f(u ), z = usin v, a :::; u:::; b, 0:::; v:::; 2r. . The reader should verify that 8 (X,y) 1= - U!, (U) sin v, 18(u, v)

8(y, z) 1= u!,(u ) cos v, and I8(u,

v)

8 (X,Z) 8(u, v)

1= u.

1

Thus, the surface area is

=

A(S)

JIn

.ju2 + u 2(f'(u)) 2 du dv.

Since the integrand does not depend on v, the v integral can be performed, and we get the desired formula:

A(S) = 211"

lb lul.jl +

(f'(u))2 duo

We are rotating a. curve about the y axis, so consider the distance from the y-axis to the curve as t he "height," which is IxI. Thus, a cross-sectional circumference of the surface at a fixed Yo is 211" 1xl· Next, describe the cur ve y = f (x ), a < x < b as a path c (t) = (t, f (t )) . T hen an infin itesimal arc length can be expressed as J1 + (f'(t ))2 dt or simply ds . T he surface area is obtained by integrating the cross-sectional circumferences along the path c and the above formula reduce to A (S ) = 2r.lxl ds.

Ie

13. We are interested in the area of the surface z (x, y) = f(x, y) = 1 - x - y, inside x 2 + 2y2 :::; l. First, compute

VI + r; + fJ

dx dy

= v'3 dx dy .

To compute the surface area, we need to parametrize the disc z = 0, x 2 + 2y2 :::; 1 using polar coordinates: x = rcosB , y = (r / V2) sinB, 0 :::; r:::; 1, 0 :::; e :::; h, and the Jacobian is r/V2. Our integral then becomes

2" fl

1 o

Jo

~r. v'3 dr dB =

r.V6 2 .

17. Completing squares, the equation x 2 + y2 = x becomes

(x 2 - x+ t )+ y2 = i.e. , (x - t )2+ y2 = ( ~ t Th is

equation represents a cylinder whose base circle is centered

at (~, 0) with radius ~, as shown . To find the surface area

of 5 ] , we need to consider where the cylinder "sticks out"

of the sphere. Consider the positive octant. The surface

area is + r; + f3dx dy , where D is half of the base

circle (shaded), and z = f (x, y) = J1 - x 2 - y2 is the sphere. Since we will be integrating over a circular region, we

t,

".. ,­

~,

~

x

IID ) 1

can use polar coordinates: x 2 + y2 = x is the same as r2 = r cos B or r = cos B. From the r :::; cos Band 0 :::; 8 :::; r. /2. Also , we com­ figure , one can see that D is described by 2 pute f x = - x /J 1 - x - y2 and by symmetry, fy = -y/Jl- x 2 - y2. So + + f~ =

°:: ;

)1 n

1/ J1 - x 2 - y2 , which becomes 1 /~ in polar coordi nates. Remembering that the J aco­ bian is r and that SJ consists of four equal surfaces, we get A(S t}

4

1f/21cos e 1o f7r/2

4Jo

0

r

v'1=T2 drdB 1 - r2

= 4 1 0 7r/ 2 (

7r/2 (l-sinB)dB = 4 (8+ cosB)lo

By high school geometry, we know that A (S:d (r. + 2) /(r. - 2).

= 411" -

- ~

= 21T' -

(211" - 4)

Icos e)

dB

r=O

4.

= 211" + 4, so A(52 )/ A(SI) =

127

INTEGRALS OVER PATHS AND SUR FACES

20. First, we are going to figure the volume of the material removed to make the hole in the sphere. Use cylindrical coordinates to describe the hole: 0::; r ::; 1, 0 ::; B ::; 21!' - ~ ::;

z ::;

~.

The limits for z were fou nd from the equ ations of the upper and lower hemispheres: z = J4 - x 2 - y2 and z = x 2 - y2 and substituting 1'2 = x 2 + y2. Remembering that the Jacobian is r, we get

- /4 -

Vhole

= ('" tl~ l ' dz drdB = t -~

Jo Jo

1

~ 1'2)3/2

2

"

(

-2{4

[=J

"t

Jo Jo

dB =

2r/4 - r 2 dr dB

~ (8 -

1 2

3\1'3)

"

dB = 4; (8 - 3\1'3).

Since the volume of the sphere is 3211" / 3, th volume of the coupler is 3211" /3 - (3211"/3 -4rrvS) = 4rrV3. For the surface area, it suffices to calcula e the surface area of one "cap" of the hole. In rect­ angular coordinates, the surface area of I (x, y ) = J 4 - x 2 ­ y2 over D, the circle of radi us 1, is ffD + Ii + I~ dx dy. We calc ulate Ix = - x/ J4 - x 2 - y2 and Iy = -y/ / 4 - x 2 - y2,

VI so V I+ Ii + n

4/J4 - x 2

:=

-

y2 ,

" [

-

Changing to polar coordinates, the surface rea of one

cap is

12"11 h o

4- r

0

1

2

drdB = w

0

4~llr=O ]

dB

= (8 -

1 2

4V3)

".

dB:= rr(16 - 8V3).

0

Si.nce the surface area of the sph re is 411"1'2, or 16rr , and the surface area of the two caps is 2rr(16 - 8V3), the outer surfa.ce area of t he coupler is 161!'(V3 -1). 22 . (b) We computel:r: = y + l/( y + 1) and Iy

VI + Ii + fJ

=

dA

=x -

x/(y + 1)2. Thus

1 + (y2 + y + 1)2 + (x(y + 1)2 - x )2 dA (y + l )2 (y+l)4 1

( )2 J( y + 1)4 + (0 + 2y2 y +l

+ 2y + 1)2 +

(x (y + 1)2 -

xF dA ,

and the surface area is

1412 1

1

(

1

\2

Y + I,

/ (y + 1)4 + (0 + 2y2 + 2y + 1)2 + (x (y + 1)2 - x)2 dx dy.

7.5: INTEGRALS OF SCALAR FUNCTIONS OVER SURFACES GOALS 1. Be able to compute the integral of a given scalar function over a given surface. 2. Understand why the integral is defined the way it is and to interpret it physically.

STUDY HINTS In t his section, we introduce dS, which stands for li T" x T v II du dv , which was discussed in section 7.4.

l. Notation.

2. Importance. This section and the next will be used extensively in chapter 8. In this section , we integrate scalar function as we did in section 7.1, and in the next section, we will integrate vector-valued functions .

128

CHAPTER 7

3. Computation. The form ula for the scalar surface integral of a scalar function

f

is

fisfdS = flf liTu XTtl ll dudV. Here, fis usually given as a func tion of (z, y, z ) and we rewrite it in terms of t he parameters tI and v by substituting x, y, z as functions of tI, v. 4. Physical interpretation .

(a ) If f

= 1, then

we get the surface area. T his may help you to remember t he formula.

(b) If fis the mass density per unit area at each point of t he surface, we get the m ass of the surface.

5. Scalar integral over a graph. If z

fl

= g(x, y), then the formula becomes

f (z , y,g (x'Y ))V 1 +

(:!r

You should remember this or be able to derive it from

+ (~:r dx dy.

lIDf li T.. x Tvll du dv.

O. Integra ting over a plane. If S is a plane, we can simplify the integrati on formula in equ ation (5) of the text :

fis f dS fLc~ =

=

() dx dy ,

where cos () D . k and D is the unit vector normal to the plane. (Review the geometry of the dot product, section 1.2.) The region D is the projection of S onto the xy plane .

SOLUTIONS TO SELECTED EXERCISES 3 . Since we're integating over a hemisphere, it is wise to use spherical coordinates. For th hemispherical surface, we have p = a, so x = a cos () sin , y = a sin () sin 4> and z = a cos fo r o ~ () ~ 271" and 0 ~ ~ 71"/2 . Thus

II (- a sin () sin i + a cos () sin j -

liTex T¢> Il

Ok)

x(a cos ()cos 4>i + asin () cos j - asio k)11 a 2 sin . Then

f1

'{J J

zdS

{2'1f

D

a cos II T e x T 4> 11 d4> d8 =

3

r /2 sin 4> cos ¢ d¢ = 21ra

3

271"a Jo

r a ¢)1"/2 = 7I"a

Jo Jo

l2

2 a cos . sin ¢ d4> dB

( . ry sm;

5. (a) The equation of the sphere is x 2 + y 2 + z2 = 2Rz, R > O.

Upon com pleting the squares, we see t hat it is equivalent to the equati on z 2 + y2 + (z - R)2 = R2 , wbich is a spbere of

radius R centered at (0, 0, R ). The tip of the cone z2 = x 2 + y2

intersects (by design) the sphere at the origin. To fin d the other

intersections, note that x 2 + y2 + (z - R )2 R2 and x 2+ y2 z2

implies R2 - (z - R)2 = z2 or z = R. Parametrize t he cone as

p cos B, y = p sin 8, z p with 0 ~ p ~ Rand

follows: x o ~ B ~ 271" . You should verify that IITp x Tell V2p. Thus ,

the area of the portion of the cone that is inside the sphere is

=

=

=

12"lR o

V2p dp dB

0

=

0

z

=

= 271"V2 -p2 1R = 7I" V2R2. 2

3

0

x

y

129

INTEGRALS OVER PATHS AND SURFACES

(b) By "area of the portion of the sphere inside the cone," the authors presumably mean the area of the piece of the sphere that is the "ice cream" part of this configuration. T his is si mply the area of the hemisphere of radius R, or 21l'R2 . 7. Parametrize S usin polar coordinates: x = r cosB , y = 1'sinB, 0 ~ r ~ 1, 0 ~ B ~ 21l'. Since the su rface is described by z = x 2 + y2 , we substitute in x and y, and get z = 1'2. Then, T r x T 8 = (cos B,sin B, 21') x (-1' sin B,rcosB,O) = (- 2r 2 cosB, -2r 2 sinB, 1'), and so

dS = IIT r x T elld1'dB= V41' 4 +r 2 d1'dB=1'V4r 2 +1d1'dB .

Calculat,jng

IIs ;; dS:

This integral can be done using integration by parts (or the tables) : let 1

u = r2, dv =rV4r 2 +1dr; du=2 1' d1', v= 12 (41'2+ 1)3/2 . Then t he integral becomes

8. First, integrate over the portion of the cube in the plane z = 1, which we will call 51 . We have -1 ~ x ~ 1 and -1 ~ y ~ 1, so T x x T y = i x j = k and II T x x T y II = l. Then

J1 11Jl

z211Tx x Tyll dx d y

-I

1·1· d x dy

= 4.

y

-1

The integral over t he portion of the cube in the plane z = -1, which we wIll call 52 , is done in the same way, so z2 d52 = 4.

J.{ is,

ow, for S3, which is in the plane x = l. Let D be the square -1 and no e that Ty x Tz = j x k = j and IITy x Tzil = l. Then

Similarly,

J.{J~

z2 dS4 =

J'J~{

z2 d55 =

J'hsf

z2 dS6 =

~ y ~

±,

1 and -1 ~ z ~ 1,

3 where S4 is the par t of the cube that is in the plane x = -1, S5 is in the plane y = 1 and 56 is in the plane y = -1. Therefore, z2 d5 is the sum of the integrals over the six surfaces, which is 4 + 4 + ~ + ~ + ~ + ~ = ~o.

IIs

CHAPTER 7

130

x 2 + y2 + z2 = R2. One can substit ute x for

y, y for z, and z fo r x, or any other perm utation , and still get the same equation. T hus, t he three integrals ought to be equal. (This is what "by symmetry" usuaUy means.) Geometrically, a sphere "looks the same" no matter how you look at it. (b) Using part (a) , we have

11. (a) The equation of the spbere is

Jis

(x

2

+ y2 + z 2) dS =

Substitute x 2 + y2

11 s

+ z2 = R2

x 2 dS =-1

3

Jis

2 x dS

+

Jis

y2 dS +

Jis

z2 dS

=3

Jis

2

x dS.

and recall that ffs dS is the surface area of 5 to get

fl

(x 2 + y2 + z2) dS = -1 s 3

fl

s

411" 4. R 2 dS =R2 - ·411"R 2 = -R 3 3

(c) Due to symmetry of the sphere, if we integrate x 2 + y2 over the entire sphere, we should get twice the mass desired in exercise 10. So the desired mass is

~

Jis

(x

+ y2) dS = ~2

2

Jis

2

x dS

= ~11" R4.

14. By exercise 12, the average z coordinate is

Jis

A tS)

z dS.

In this case, A (S), the surface area of a hemisphere ofradius r, is 2 11"1' 2. By exercise 3, ffs z dS is 11"1'3 , so Z = 1I" r 3/ 211"1'2 = r /2 . Since there is as much of the sphere on one side of the z axis as there is on the other (by symmetry ), x = y = O. 20. If z

= g( x,y), then we can use t he fo rmula stated in equation (4), namely,

Jis

!(x, y, z) dS =

JIn

f( x, y, g(x, y) )J1 + (8g/ ox )2

Since 5 is defined implicitly by F (x , y, z)

8z 8x Given that ! (x, y, z)

8g ox

+ (og/ oy)2 dx dy.

= 0, we can use implicit differentiation to get

-(oF/o x ) of/ oz

and

oz oy

og oy

-(oF/oy) of/oz .

= loF/ 8z l, we get

10F IdS f·r Js oZ

=

r::

JrJD 10F8z Iy

1 i

"f

f JD JIn

[- (OF/8 x)]2 8F/ 8z

+ [-(8 F/8 y )]2 dxdy 8F/8 z

18F I J (8F/8z )2 + (8F/ oX)2 + (8F/8yF dx dy OZ J(8 F/ 8zP

J(8F/8z)2

+ (oF/8 x )2 + (8F/8y)2 dx dy.

7.6: SURFACE INTEGRALS OF VECTOR FIELDS GOALS 1. Be able to compute a surface integral of a vector fu nction.

2. Understand its derivation and physical in terpretation.

131

INTEGRALS OVE R PATHS AND SURFACES STUDY HINTS

1. Notation. The symbol n( ~ (u o, vo)) is used to describe the un it vector which is normal to (a param trized surface) ~ at (uo, va ); note that ~(uo, va ) is the base poin t of n. 2. Orientation. As with line integrals, the orientation is important. T he sign of the integral changes with the opposite orientation. 3. ParametrizatlOn. As with line integrals , we can reparametrize a surface. orientation is retained, the value of the integral is unchanged .

As long as the

4. Definition . The surface integral of a vec or field F over a parametrized surface D is

~

ff~F . dS=

with domain

flF.(Tu X TtJ ) du dv .

The integral of a vector field is a scalar.

5. Generalizations. _'otice that scalar integrals (sections 7.1 and 7.5) do not depend on orienta­ tion. However the sign will change in the integration of a vector fie ld (sections 7.2 and 7.6) if the orientation has been reversed . Finally, note that scalar integrals involve the length of a vector and the integral of vector fields involve the dot product. 6. Reduction to calor integrals. If we know he unit vector to t he surface ~ , t hen the surface integral reduces to n) dS. Letting f = F . n, we get the scalar integral of section 7. 5. If he orientation switches, n changes sign and so does this f.

IIs(F .

7. Surface integral ouer a graph. If z = g(x, y) and F = (FI ' F2, F3 ), then a normal vector (- ag/ox , - og/oy,l) and the surface integral becomes

fin [Fl (- ~!) +

F2 ( -

1S

~:) + F3 ] dx dy.

8. Physical interpretation. If F represents the velocity field of a flu id , the surface integral of F

over

~

gives the rate of flow of the fluid across the surface

~.

T his is known as fl ux .

9. Un it normal to unit sphere. It is useful to know that for the un1t sphere, n = r = (x , y, z).

10. Good example. Example 6 shows three ways of computi ng the same integral.

11. Summary. The formulas following example 6 give a nice summary of sections 7.5 and 7. 6. An understanding of how to use these formulas is important. If any of t he formulas do not make sense to you . it is time to review .

SOLUT I O NS TO SELECTED EXERCISES

IIs

1. As in example 4, the heat flux across the surface S is - 'i1T · dS . Parametrize the surface 2 x + z2 = 2. Let x = V2 cos 8, z = v'2 sin 8 with 0 ~ 0 ~ 211". The surface can be parametri zed by S (8,y) = (V2cosO,y,V2sin8) , 0 ~ 8 ~ 211" and 0 ~ y ~ 2. Then the outward pointing normal to S is To x Ty = V2cosBi + v'2sinBk. Given T (x,y,z ) = 3x 2 + 3z 2 , we compute - 'ilT = - 6(.1:,0, z) = - 6( v'2cos 8,0, v'2sin B). Thus

f

l - 'ilT . dS

=

JL

- 'ilT· (T o x Ty ) dydB

-6 2""12 [ J2 cos 8) 2+ ( v'2 sin B) 2] dy dB

1 0

0

(

-1212""12 dydO = -4811". If the Ty x To cross product (hence the opposite orientation ) is used, the answer would be + 4811" .

(HAPTER 7

132

6. First, we compute k

j %y x 2 + y - 4 3xy

%x

\7xF =

= - 2zj + (3y - l) k .

0/8z 2xz + z2

Use spherical coordinates to parametrize S: x = 4 cos Bsin u(p ) to m aintain the average. Therefore, tI must be constant on some disk centered at p .

8. 2: STOKES ' TH EOREM GOALS 1. Be a ble to state and use Stokes ' theorem. 2. Be able to use Stokes ' theorem t o calculate a line integral on a closed curve or a surface integral over a surface wi th a closed curve as its boundary.

STUDY HINTS 1. R eview. Surface integrals are used in t his section . You should review section 7.6 if you have

forgott en how to calculate a surface integral.

2, R elation to Green's theorem. Like Green 's theorem , Stokes ' t heorem converts an integral in one dimension to an integral in two dimensions. Stokes ' theorem is a generalization of Green 's theorem . 3. Stokes ' th eorem for graphs. If z

= f(x , y) , then

Jis

curl F · dS

=

lasF .

ds.

This is a formula you should memorize. As with Green 's theorem , oS is oriented so that the surface is on your left as you walk upright around as, which must be a closed curve . 4. Genemlized surfaces. If the surface S is not the graph of a functio n, then Stokes ' theorem is still true if S can be descri bed using a one-to-one parametriz ation. T he boundary of D, oD, gets m apped to as, so oD should h ave t he correct orientation. For an example when the orient ation becomes imp ortant , see the solution of exercise 5 of section 8. 1.

THE IN TEGRAL TH EOREMS OF VECTOR ANALYSIS

147

5 , Application, According to Stokes' theorem, to evaluate ffs curl F ' dS , we can change the surface 5 to any other surface with the same boundary 85, In most cases, we will change to

a planar surface or another simple surface, Imagine a loop of wire with an elastic sheet S on it , The surface integral of the curl of a vector field over any surface formed by a deformation of the elastic sheet will be equ al to t he line integral over the wire (assu ming the wire itself is not deformed), 6, Circulation, You should know t hat curl V, n is the circulation of V per unit area of surface

perpendicular to n ,

SOLUTIONS TO SELEC TED EXERCISES 1. By Stokes' theorem , we only need to evaluate

r F . ds,

Jas

where 8S is the circle x 2 + y2 = 1, Z = 0, the boundary of OUI surface . Parame trize the circle using polar coordinates, i.e., x = cos B, y = sinB, Z = 0, 0 ~ B ~ 27r. T hen the integral becomes

1 1 F · ds =

&S

2~

[sin 8(- sin 8)- cos 8(cos B)] d8= -

0

12~

d8 =- 27r .

0

4. We want to show that the derivative of he magnetic flu x with respect to time is O. We begin

wit h

~

J'{

J'r-

J' {

H . dS = 8H . dS = 8H . dS .

8t Js Js at Js at Th first step is justified because S is not a function of time. By Faraday's law ,

-Jis

('v x E ) . dS .

Then by Stokes' theorem and the fact t.hat E· ds = 0, since E is perpendicular to t he boundary of 5, we get

- { E· ds = O.

Jas

5. The boundary of th surface is a closed curve, so we m ay take advan age of Stokes' theorem:

Jis T he bou ndary is the circle x 2

+ y2

(V' x F ) . dS:;:: la s F . ds .

= 1,

Z

{ F · ds

Jas

= O. The right-hand side becomes

={

Jas

(x, y) . (dx, dy).

Now use polar coordin ates: Let x = cos 8, dx = - sin 8 d8; y = sin 8, dy = cos 8 d8; 0 ~ 8 ~ 27r . Substitution into the last integral gives us

r (- cos 8 sin 8 + sin Bcos 8) dB = Jo 21r

O.

9. Use Stokes' theorem . The boundary of S is the circle y2 + z2 = I , x = O. On the boundary, F becomes - i'j . Use polar coordinates : y = sinO, Z = cosB , 0 ~ f) ~ 27r. By Stokes ' theorem , we get

(HAPTER 8

148

12. The flow rate is JJs curl I) ·dS and by Stokes ' theorem, this is Ja s I) . ds. We have the boundary on the xy plane where z ~ O. Parametrize the boundary by

x = (R/4) cos O, Then the fl ux is

f2Tr R2

1

y

= (R/4) sinO . 11"R2

asl)·ds = io 16(sin B+cos B) dO=-8-·

14. By Stokes' theorem,

J1s

2

(V' x F) . dS

2

~ las F . ds.

J

Since F is perpendicular to the tangent to the boundary S, F·ds ~ 0. Hence, Js ('V' x F ) ·dS ~ O. If F is an electric field, this means the rate of change of magnetic flux is zero by Faraday's

law. See example 5 in the text . 18. The solu tion relies on the basic identities of vector analysis (see table in section 4.4) . Since C is a closed curve which is the boundary of a surface S, we have C = as in Stokes' theorem. T hus, Stokes' theorem tells us JJs (V' x F) . dS = Je F . ds . (a) Here, we use Stokes' theorem where F = IV' g. We need to show that \l x fV' 9 ~ V'I x V' g, or curl(JV'g ) = \llx\lg . Bybasicidentity10,we havecurl (lV'g) =/ curl(V'g )+\llx V'g, but by basic identity 11, curl( V' g) = 0 , and 80 we have curl (IV' g) = 1 (0) + V'I x V' 9 ~ V'I x V'g , as required. Al tern ati vely, we could have computed the curl of IV'g = I (ag/o x )i + I(og/oy)j + I (og/ oz) k and t hen used the equality of mixed partials since I and 9 are C 2 func tions. (b) Here, we use Stokes' theorem with F = I V' g+ gV' I. By basic identity 6, curl(fV' g+ gV' f) = curl(fV' g) + curl (g\l I ). Applying basic identity 10, we have

f cUrl(V'g) + V' I x \lg + I curl( V'g) + 9 curl (\lf)

curl (IV'9 + 9 V'f)

9 curl (V' f) + V'g x V'I

since V'I x V' 9 = -(V' 9 x V' f) by properties of the cross product. According to basic identity 11, curl(\ll) = cu rl (V'g) = 0, and so curl(l\lg + gV'f) = O. By Stokes' theorem ,

J JIs o·

[(IV'9 + gV' f) . ds

is V' x (IV' 9 + 9V'/) . dS dS

~ O.

2l. First , we wili calcuiate JJs (V' xF ) .dS . Weco m pute \lxF ~ (1,1, 1). Alsol)r = (cosO , sin O,O) and 1) 0 = (- r sin 0, "cos 0, 1), so I)r x I) e = (sin e,- cos 0, r ). Therefore , we get 1

Jis (1,1,1) . {Sin 8, - cOs8, r ) d7>dO = 1 1"/2 (sino -coso+ r ) dBdr =

11

Cir)

dr = ~.

On the other hand , the boundary, aS, is composed of four parts. First, when r = 1, we have 1)(1, B) = (cos e, sin fJ, B), so F = (e, cos B, sin e) and ds = dl) (1, 8) = (- sin e, cos B, 1) dB . T herefore, Tr / 2 F· ds = (0, cos 8,8in B) . (- sin 0, cos B, 1) dO.

1

l

BS,

0

-e sin B and the half-angle formula to integrate cos 2 8,

Using integration by parts to integrate we get

[(Bcos B When

e=

sine) +

. 20) - cosO]1"'0/2 ( "20+ SI:

11" 4

11"/2, orientation is maintained by letting r go from 1 to O. T hus, we have

f

iaS2

F . ds =

fO(~ ,o, r). (O,1, O)dr=O.

il

THE INTEGRAL THEOREMS OF VECTOR AN ALYSIS

149

W hen r = 0, B goes from 7r/2 to 0, so we get

[0 (B , 0,0)

[ F · ds =

las.

o.

. (0 , 0,1 ) dB =

l-rr /2

Si milarly, when B = 0, we get [ F · ds = [l(O,,·, O).(l , O, O) dr =O .

las.

lo

Adding all the parts together, we get verified.

Ias F . ds

= 7r/4 + 0 + 0+0 = 7r/4, so theorem 6 is

25. For a direct computation , parametrize the sUl'face as follows: Le x = T' cos B and y = rsinB, so z = t(x2 + y2 ) = r 2 /2. Also , we want ~ z ~ 2, so 0 ~ ,. 2 / 2 ~ 2, or ~ r ~ 2. In addition, we have 0 ~ B ~ 210 . W calculate T o = (- 1' sin B,rcosB, O) and Tr = (cosB , in B, r), so the outward normal is T 8 X T r = (1'2cos e, r2 sin B, -1'). Also, we calcu la te

°

i \7 x F =

Finally,

j

°

k

a/ax a/ay a/az 3y - xz _ yz 2

-(z 2 + x , 0 , -z - 3) --

(-1 - r

4

4

-1 )

+ 1'cosB ' 0 -1'2 -3 . , 2

IIs(\7 x F) . dS becomes

2 r2rr

1lo o

(\7 x F) . (T o x T r ) dBd1'

12

(2 7r1· 3 + 67rr ) dr = ( 7r;4 + 37r7'2)

I1s (\7

I: = 207r .

Ias

On the other hand, by Stokes' theorem, x F ) . dS = F . ds. T he boundary is as, which is the circle ofradi us 2 in the plane z = 2. It can be parametrized by (2 cos t, - 2sin t , 2) for ~ t ~ 27r. We use this orientation b cause the surface lies below the boundary, so we should traverse it in a clockwise orientation. We com pute ds = (- 2 sin t , - 2 cos t , 0) dt, so

°

[ F · ds

las

=

1

211'

(- 6 sint, - 4cost,8sint )· (- 2sint, -2 cost ,O) dt

r -rr (12 sin t + 8 cos t ) dt 2

lo

[12

2

2

C

(~ - Si: 2t ) + 8 (~ + Si:2t )]

= 207r.

If one choo es the other orientation, the answer should be - 201T.

8.3: CONSERVATIVE FIELDS GOALS 1. Understand that the line integral of a gradient field is path independent. 2. Be able to determine whether a field is conservative.

CHAPTER 8

150

3. Given a conservative vector field , be able to fi nd a scalar function whose gradient is equal to that vector field .

STUDY HINTS 1. Theorem 7. T his is a very important theorem . To summarize, it states that if F is a gradient, then \l x F = 0, the line integral depends only on the endpoints, and all lin e integrals around closed curves are O. Also, if one these con ditions hold, then F is a gradient. If the conditions of the t heorem hold , then we say that t he line integrals are "independent of path." Note that if a single line integral is 0, then F may not necessarily be a gradient. 2 . Example 1. In method 2, part (a), notice how after integrating in x, we add a "constant" h 1 (y, z). It is a "constant" because it only involves the other variables. Study example 1

carefully. You should know how to use at least one of the two methods .

3. Gra di ents in 1R 2. The corollary precedi ng example 3 tells you that if the integrand in Green's theorem, oQ/ Bx - BP/By , is 0, then F = Pi + Qj is a gradient . Be careful!! F must be C 1 on all of ]R. 2, unlike theorem 7, which allows some exceptional points in 1R3. Exercise 12 el aborates this point . If the origin is an exceptional point in IR 2, then the integral x of F over t.he path on the left (a closed path) is 0 because the path does not include the origin, while the integral of F over the path on the right is not.

= O.

4. Is F a curl? F is a curl of some vector field if div F for fin ding a G such t hat F = curl G.

Exercise 16 explains the procedure

SOLU TIONS TO SELECTED EXERCISES 2. (a) Si nce y = 2X2, we have dy = 4x dx, and so

[

t

[1

1

8

19

lc F· ds = 10 (x · 2X2 ) dx + (2X 2 ) 2. 4xdx = 10 (2x 3 + 16x 6 )dx = 2" + 3' = 6' (b) The answer is yes, since

\7 x F

=I

j

k

B/ox a/ ay o/Bz xy

y2

0

1= (0 , 0, -x) f- O.

Alternatively, one can pick a different path and show t hat the line integral is not equal to T his would mean that t he line integral of F is path dependent. 3. If F

= \l f , then t he x component of F

must be

li.

of /ax, i. e.,

of/ox = 2xyz + sin x.

(1)

Similarly, the y component of F must be oj/ oy and the z component of F must be oj/ oz, % .e. ,

of/ay Bj/oz

x2z 2

x y.

(2) (3)

Integrating (1) with respect to x, we get f = f( 2xyz + sin x) dx = z 2yz - cos x + h(y, z), where h is a fu nction of y and z only. When we integrate with respect to x we should restore all the terms with x. The terms without x are treated as a constant when differentiated , and

THE INTEGRAL TH EOREMS OF VECTOR ANALYSIS

151

integration with respect to x cannot restore them. Similarly, integrate (2) with respect to y to get 1= x 2z dy = x 2z y + g( X, z),

J

where g(x, z) is a function of x and z only. Integrating (3) with respect to z gives us I = x 2y dz = x 2yz + k (x, V), where k(x , y) is a fu nction of x and y only. Compare the three results: I(x, y, z) = x 2yz - cos x + h(y, z ) = x 2yz + g(x, z) = x 2yz + k(x, V) .

f

We conclude (by insp ction) t hat g(x , z) = k( x, y) = - cos x + C and h(y, z) a constant. Thus, I(x , y, z) = x 2 yz - cos X + C.

= C,

where Cis

6. (a) This fact should already be familiar (see Example 6 in Section 2.6) . Here are t he details. We use the chain rule to compute (alox )(1 /J x2 + y2 + z2 ) = -xl (x 2 + y2 + z2)3/2. Then, by sym metry, we get "V

(~)

1

= '\7 (

J x2

l'

+ y2 + Z2

)

=

= ~ = -r

- (xi + yj + zk) (x 2 + y2 + z2)3/2

IIrl13

r3 .

(b) Since F is the gradient of a function j, fcF . ds = I (c(a)) - I (c (b)) for all paths c(t) beginning at a and ending at b, unless the path passes t hrough the origin , where 1/r is not continuous. Thus , t he work done by F in moving a particle from ro "to 00" is 1 J x 2 + y2

+ z2

( ~)

_ lim

r

r-+ oo

_ 1

-..;x 2 + y2

+ z2 .

Notice th at this tells you it is more difficult to move a particle from a position near the origin. 9. T he reader should verify that '\7 x F = O. Hence, F is a gradient of some function I(x, y, z). Integrate the i component with respect to x, the j component with respect to y, and the k component wi h respect to z. Comparing the results, we see t hat I(x, y, z) = eX sin y + z 3 /3. Also, we compute c(O) = (0,0 , 1) and c(l ) = (1 , 1, e). Since F is a gradient, we have

1

c F· ds

= l(c (l))

e3

- l(c(O))

1

= esin 1 + 3 - 3'

=

13. (b) F is not the gradient of a scalar function f. If such j were to exist, then a I I ax xy and 2 2 of I 8y xv· By t h equality of mixed partials, we would expect that 8 I I 8x8y 8 I I 8y8x. But in this case, (818y)(8f!8x ) = x and (818x )(8118y) y.

=

=

=

15. (b) To show that F is conservative, we can show t hat it is a gradient . (We can also show that anyone of the four conditions in theorem 7 is met) . We will use the same method as in exercise 13. The partial derivatives are:

8 ( 2X ) 8y y2+1

=

(2 x )( - 2y) -4xy (y2 +1)2 = (y2 +1)2 ;

1))

8 (-2y(x 2 + 8x \ (y2 + 1)2

=

- 4x y

(y2 + 1)2 '

Since these part ials are equal, F is conservative. Finding I so that F = "V I makes the evaluation of the path integral easy. You should verify t hat if I(x, y) = (x 2 + 1) /(y2 + 1) , then F = '\7 f. Then substitute x = t 3 - 1 and y = t 6 - t to get I (t) = [(t 3 - 1)2 + 1]/[(t 6 - t)2 + 1], and so

[

lc F· ds =

[(t3 _1)2+ 1]1 (t 6 _ t)2

+1

1

o::::;;-l.

CH APTER 8

152 18 . First, we compute "V . F = (8/8x)xz + (8/oy)( -yz) + (%z)y = z - z + 0 exists a G such that F = "V x G . To find G , use the result of exercise 16:

G1

=

Hence, G answer:

l

z

o

-yt dt -

lY

t dt

_

y2 = -yz-2 - -, 2

0

2

= (_~)[(yz2 + y2)i + xz 2jJ.

"Vx G

21 I

G2

J a/oy yz2 + y2 xz 2 xzi - yzj + yk = F.

a/ax

=-

lz 0

= O.

Thus, there

2

xt dt

- xz = --, 2

and G 3

= O.

It is a good idea to compute curl G to verify your

k 8/8z

= ~l (- 2xzi + 2yzj + (z2 -

z2 - 2y) k)

0

Note that G is not unique; for example, arbitrary constants may be added to each component of G, or any gradient may be added to G, and "V x G would still be equal to F . 23. (a) Recall from chapter 4 that F is not irrotational when curl F # O. Thus, we com pute curl F : J k "V x F = I 0/ ox 0/ 8y 8/ OZ I = 2k # O.

x

- y

0

(b) Let c(t) = (x(t), y(t), z(t)) be the trajectory of the cork. By the definition of flow lines, c/(t ) = F(c(t)), and so c/(t) = (-y(t),x(t),O) . Equi valently, we have the following system of differential equations :

Xl (t) yl (t) Zl

(t)

(1) (2) (3)

-y(t) x(t)

o.

Equati on (3) can be solved easily. Its solution is z(t) = constant. T aking one more deri vative of (1) with respect to t yields X"(t) _yl(t), but (2) implies that x"(t) - x (t), or x"(t) + x (t ) = O. T his equation represents a harmonic oscillator. T he famous solution to the differential equation x "(t) + x(t) = 0 is

=

=

x(t) =Asint+B cos t,

where A and B are cons t ants. If t he reader is un familiar with the techniques of solving ord in ary differential equations , the reader may verify this is indeed the solution by substitution. Similarly, y(t) = Csin t + Dcos t , where C and D are constants. Since x"(t) = -y'(t), we differentiate x and y and compare terms to find that C = B and D = -A. T herefore, y(t) = B sin t - A cos t. Squaring x and y and addi ng them gives us

x2

+ y2

A 2 sin 2 t + 2AB sin t cos t + B 2 cos 2 t + B2 sin 2 t -2A B sin t cos t + A2 cos 2 t = A2 + B2.

Since A2 + B2 is a constant, we recognize t he equation x 2 + y2 = A 2 + B 2 as the equation of a circle of radius VA2 + B2 centered at (0,0) . Thus, the cork has a ci rcu lar trajectory about the z axis in a plane parallel to the ~'y pl ane. (c) As y increases, x decreases, since xl(t) = -yo We also kn ow t hat the cork is goin g in a circle. T hus t he cork is revolving counterclock wise. 24. (c) T he property of being rotational is a local property, t.h at is, the field is rotational at a poin t. In exercise 23, the cork whi rls while going around in a circle, but in exercise 24 , the cork does not. Single trajectories have little to do with the r otationalness of the fluid.

THE INTEGRAL TH EOR EMS OF VECTOR ANALYSIS

153

8.4: GAUSS' THEOREM GOALS 1. Be able to s ate and use Gauss' theorem.

2. Be able t o use Gauss ' t heorem to compute a double integral over a closed surface or a triple int egral over a volume enclosed by a surface.

STUDY HINTS 1. D efi nition. A clos ed surf ace is a surface wh ich, rough ly speaking, must be p unctured in order

to get into the region it encloses . The enclosed region is denoted Wand the closed surface is denoted oW. 2. Gauss} dw ergence theorem. If oW is a closed surface, then

fffw (diV F)

dV

= fhw(F.n ) dS .

T hus, a triple integral is reduced to a double integral , or vice versa. Compare th is to the divergence theorem in the plane (section 8.1) .

3. Physlcal interpretation. div F(P) is the net outward flow at the poi nt P per unit volume. If div F(P) > 0 material flo ws out from P, and if div F (P) < 0, material flows in towards P. If div F(P ) = 0, t he vector field is divergence free, that is, what goes in must come out .

SOLUTIONS TO SELECTED EXERCISES 4. (a) For the faces parallel to the yz plane, the outward pointing unit normal vectors are j and

- i, respectively. For those two faces , we have

fiaw F'dS

1

1

= 1 1 \ i + j + k) .idY dZ + 1 1 \i + j +k).( - i)dy dz

111\

i

+ j + k) . (i -

i) dy dz

= o.

T he outward pointing unit normal vectors n for any two parallel faces are the exact opposite, so the integrals over any two parallel faces of the cube cancel. Therefore, the int gral is O. Next , we calculate that div F = 0, so by the divergence theorem, t he desired integral is

fffwdiVFdV =O. 6. (b) First , we see th at div F

= 1 + 1 + 1 = 3, so

flaw

F . dS =

by the di vergence theorem ,

1ffw div F

dV = 3

fffw

dV,

which is t hree times the vol ume of W. T he problem now is to find th volume of the region of interest. Use "cylindrical" coordinates. We will let f) range fro m -1r / 2 t o 7r / 2 since x ~ O. In addition , we have :S r :S 1 and x 2 + y2 = 1'2 :S Z :S 1. Therefore,

°

fli

W

dV

=

J,"/2 1111r dz dr de = -1f' /

2 0

r 2

7r

11

r (l - 1'2) dr

0

Thus,

flaw F· dS = 3 (~) = 3:.

= 7r

(1 1) -

2

-

­

4

7r

4

CHAPTER 8

154

= (x 2 + y2)2, so

10 . Use the divergence theorem . We have div F

fhsF . dS = f fis (x2 + y2)2 dV. Using cylindrical coordinates, the cylinder S can be described by 0 ::;: 1. Since the Jacobian is r, we get

T ::;:

1, 0 ::;: B ::;: 21!', and

o::;: z ::;:

fhsF . dS = 121