section 15.4 answers

658 CHAPTER 14 I DIFFERENTIATION IN SEVERAL VARIABLES (LT CHAPTER 15)

5. Assuming the hypotheses of Clairaut's Theorem are satisfied, which of the following partial fxxy?

(a) fxyx (b) /yyx (c) fxyy

SOLUTION fxxy involves two differentiations with respect to x and one differentiation with f satisfies the assumptions of Clairaut's Theorem, then

fxxy = fxyx = /yxx

Exercises 1. Use the limit definition of the partial derivative to verify the formulas

a -xv2 = y2 ax - ,

a -xi =2xy ay

SOLUTION Using the limit definition of the partial derivative, we have

a 2 . (x+h)y 2 -xi -xy = hm -'-----__:_::_ _ ____::_ ax h-+0 h

. xi +hi -xy2 hm ___:__ _ ___.::_____ _ __::__

h-+0 h I. hy2 ]' 2 2 1m-= 1my =y

h-+0 h h-+0

a 2 . x(y + k)2 - xy2 . x(y2 + 2yk + k2) - xy2 . xy2 + 2xyk -xy = hm = hm = hm _:_ __ ..:...____. ay k-+O k k-+o k k-+o k

k(2xy + xk) = lim = lim (2xy + k) = 2xy + 0 = 2xy

k-+0 k k-+0

a 2. Use the Product Rule to compute -(x2 + y)(x + y4 ).

ay

SOLUTION Using the Product Rule we obtain

~(x2 + y)(x + il = (x2 + y)~(x + il + (x + y4)~(x2 + y) ~ ~ ~

= (x 2 + y) ·4y3 + (x + y4) ·I =4x2y3 +5l+x

. a v 3. Use the Quotient Rule to compute - ---.

ay x + y

SOLUTION Using the Quotient Rule we obtain

~_Y_ (x + y)~(y)- y~(x + y) (x + y) · I - y · 1 x

(x + y)2 = (x + y)2 ay X+ y (X+ y)2

a 4. Use the Chain Rule to compute - ln(u2 + uv).

au

SOLUTION By the Chain Rule fu In w = ~~.Applying this with w = u2 + uv gives

a 2 1 a 2 2u + v -ln(u +uv) = ----(u +uv) = --au u2 + uv au u2 + uv

5. Calculate fz(2, 3, 1), where f(x, y, z) = xyz.

SOLUTION We first find the partial derivative fz(x, y, z):

a fz(X, y, Z) = az (xyz) = XY

Substituting the given point we get

fz(2,3,1)=2·3=6

6. ~ Explain the relation between the following two formulas (c is a constant).

d . - sm(cx) = ccos(cx), dx .-

a . - sm(xy) = ycos(xy) ax

• SECT I 0 N 14.3 I Partial Derivatives (LT SECTION 15.3) 659

SOLUTION fx sin(cx) is the derivative of the single-variable function sin(cx ), where c is a constant. lx sin(xy) is the partial derivative of the two-variable function sin(xy) with respect to x. While differentiating, the variable y is considered constant, hence it resembles the first differentiation, and the results are the same where c is replaced by y.

7. The plane y = 1 intersects the surface z = x 4 + 6xy - y4 in a certain curve. Find the slope of the tangent line to

this curve at the point P = (1, l, 6). sOLUTION The slope of the tangent line to the curve z = z(x, l) = x 4 + 6x - l, obtained by intersecting the surface

z = x4 + 6xy- y4 with the plane y = 1, is the partial derivative #O. 1).

az = ~ (x4 + 6xy - i) = 4x 3 + 6y ax ax

m= az(l,l)=4·1 3 +6·1=10 ax

8. Determine whether the partial derivatives af/ax and af/ay are positive or negative at the point P on the graph in

Figure 7.

X

y

FIGURE 7

SOLUTION The graph shows that f is increasing in the direction of growing x and f is decreasing in the direction of

growing y. Therefore, ¥x I p > 0 and %f I p < 0.

In Exercises 9-12, refer to Figure 8.

-4 -2 0 4

FIGURE 8 Contour map of f(x, y).

9. Estimate fx and /y at point A. SOLUTION To estimate fx we move horizontally to the next level curve in the direction of growing x, to a point A'. The change in f from A to A' is the contour interval, fl./ = 40 - 30 = 10. The distance between A and A' is approximately

l::l.x ~ 1.0. Hence,

l::l.f 10 fx(A) ~ l::l.x = 1.0 = 10

To estimate /y we move vertically from A to a point A" on the next level curve in the direction of growing y. The change in f from A to A" is l::l.f = 20 - 30 = -10. The distance between A and A" is l::l.y ~ 0.5. Hence,

l::l.f -10 /y(A) ~ - = - ~ -20.

l::l.y 0.5

10. Is fx positive or negative at B? SOLUTION To estimate fx at B, we move horizontally to the next level curve in the direction of growing x, to a point B1• The change in f from B to B' is the contour interval fl./ = 10 - 20 = -10 while the distance between B and B' is

approximately l::l.x ~ l. Hence

l::l.f -10 fx(B) ~-=- = -10 < 0

l::l.x 1

Therefore fx(B) is negative.

SECT I 0 N 14.3 I Partial Derivatives (LT SECTION 15.3) 665

( 3) 2 2 2 -5/2 3xy

= x. -- (x + y + z ) · 2y =- 512 2 (x2 + y2 + z2)

aw a 1 a 2 2 2 -3/2 - = X- = X- (X + y + Z ) az az (x2 + y2 + z2)3/2 az

( 3) 2 2 2 -5/2 3xz = x. -- (x + y + z ) · 2z =- 512 2 (x2+y2+z2)

In Exercises 41-44, compute the given partial derivatives.

SOLUTION Differentiating with respect to x gives

Evaluating at (1, 2) gives

fx(l, 2) = 6 · 1 · 2 + 12-12 · 22 -7 · 25 = -164.

42. f(x, y) = sin(x2 - y), /y(O, :rr)

SOLUTION We differentiate with respect toy, using the Chain Rule. This gives

a /y(x, y) = cos(x2 - y)-(x2 - y) = cos(x2 - y) · ( -1) =- cos(x2 - y)

ay

Evaluating at (0, :rr) we obtain

/y(O,:rr) = -cos(02 -:rr) = -cos(-:rr) = -cos:rr = 1.

43. g(u, v) = u ln(u + v), gu (1, 2)

SOLUTION Using the Product Rule and the Chain Rule we get

a 1 u gu(u, v) = -(uln(u + v)) = l·ln(u + v) +u · -- = ln(u + v) + --

au u + v u + v

At the point (1, 2) we have

1 1 gu(l, 2) = ln(l + 2) + -- = ln3 + -.

1+2 3

2 3 44. h(x, z) = exz-x z , h2 (3, 0)

SOLUTION We obtain the following partial:

Substituting x = 3, z = 0 we obtain the partial derivative at the point (3, 0):

hz(3, 0) = (3- O)e0- 0 = 3.

Exercises 45 and 46 refer to Example 5.

45. Calculate N for L = 0.4, R = 0.12, and d = 10, and use the linear approximation to estimate !:iN if dis increased from 10 to 10.4.

SOLUTION From the example in the text we have

N = c2~R) 1.9

Calculating N for L = 0.4, R = 0.12, and d = 10 we have

(2200 ° 0.12) 1.

9 N = 0.4 . 10 :::::: 2865.058


SOLUTION

(a) Let us compute I when T = 95 and H = 50:

I (95, 50) = 45.33 + 0.6845(95) + 5.758(50) - 0.00365(95)2 - 0.1565(50)(95) + 0.001 (50)(95)

2

= 73.19125

(b) The partial derivative we are looking for here is a/ jaT:

:~ = 0.6845 - 0.00730T - 0.1565H + 0.002HT

and evaluating we have:

a1 ar (95, 50) = o.6845 - o.oo730(95) - 0.1565(50) + o.002(50)(95) = 1.666

48. The wind-chill temperature W measures how cold people feel (based on the rate of heat loss from exposed skin) when the outside temperature is T°C (with T :::0: 10) and wind velocity is v mls (with v 2:: 2):

W = 13.1267 + 0.6215T- 13.947v0·16 + 0.486Tv

0·16

Calculate aw ;av at (T, v) = ( -10, 15) and use this value to estimate b. W if b.v = 2.

SOLUTION Computing the partial derivative we get:

aw = ~ (13.1267 + 0.6215T- 13.947v0·16 + 0.486Tv

0·16

) av av

= -13.947(0.16)v-0·84 + 0.486(0.16)Tv-0·84

aw (-10, 15) = -13.947(0.16)(15)-0·84 + 0.486(0.16)(-10)(15)-0·84 ~ -0.30940

av

Now using this information we would like to estimate b. W if b.v = 2:

aw b.W = -b.v ~ -0.30940 · 2 ~ -0.6188

av

49. The volume of a right-circular cone of radius rand height his V = )-r2h. Suppose that r = h = 12 em. What leads to a greater increase in V, a 1-cm increase in r or a 1-cm increase in h? Argue using partial derivatives.

SOLUTION We obtain the following derivatives:

av = ~ (:!.r2h) = rrh ~r2 = lrh . 2r = 2rrhr ar ar 3 3 ar 3 3

~~ = :h (~r2h) = ~r2 An increase b.r = 1 em in r leads to an increase of¥, (12, 12) · 1 in the volume, and an increase b.h = 1 em in h leads

to an increase of ~ ( 12, 12) · 1 in V. We compute these values, using the partials computed. This gives

av 2 2

_ 2rrhr I _ 2rr . 12. 12 _ 30 6 (1 , 1 ) - - - 1. ar 3 (12,12) 3

av rr 2 a,;-02. 12) = 3 . 12 = 150.8

We conclude that an increase of 1 em in r leads to a greater increase in V than an increase of 1 em in h.

50. Use the linear approximation to estimate the percentage change in volume of a right-circular cone of radius r = 40

em if the height is increased from 40 to 41 em. SOLUTION First, the volume of a right-circular cone is V = ~rrr2 h. We obtain the following partial derivative:

av 1 2 - = -rrr ah 3

Then an increase b.h = 1 em in h leads to an increase of a v 1 a h . 1 in v. To compute the percent change in volume of the right-circular cone we consider:

_b._V ~ av ;ah. M = -=-~-rr_r2_b._h = _b._h = _!_ = O.Q25 V V ~rrr2h h 40

Therefore, the percent change is about 2.5%.

668 CHAPTER 14 I DIFFERENTIATION IN SEVERAL VARIABLES (LTCHAPTER 15)

51. Calculate aWiaE and aWiaT, where W = e-EjkT, where k is a constant.

SOLUTION We use the Chain Rule

d u u du -e =e- and dE dE

d u u du -e =e-dT dT

with u = - fr, to obtain

aW = e-EjkT ~ (-_!__) = e-EjkT (-~) = -~e-EfkT aE aE kT kT kT

aw = e-EjkT ~ (-_!__) = e-EjkT. (-~) ~ (_!_) = e-EjkT (-~) (-~) = _!_e-EfkT aT aT kT k aT T k T2 kT2

52. Calculate apIa T and apIa v, where pressure p, volume v, and temperature T are related by the ideal gas P V = nRT (Rand n are constants).

SOLUTION We differentiate the two sides of the equation P V = n RT with respect to V (treating T as a constant). the Product Rule we obtain

Hence,

a aP av aP -PV = V- + P- = V- + P· av av av av '

aP V-+P=O

av

a -nRT =0 av

We substitute P = n~T and solve for U- This gives

ap nRT ap nRT V-+-- =0 =*

av v av v2

We now differentiate PV = nRT with respect toT, treating Vas a constant:

Hence,

a aP -PV=V-; aT aT

a -nRT =nR aT

ap aP nR V-=nR =}

aT aT v

53. ~ Use the contour map of f(x, y) in Figure 9 to explain the following statements.

(a) /y is larger at P than at Q, and fx is smaller (more negative) at P than at Q.

(b) fx (x, y) is decreasing as a function of y; that is, for any fixed value x = a, fx (a, y) is decreasing in y.

y

X

FIGURE 9 Contour interval 2.

SOLUTION

(a) A vertical segment through P meet more level curves than a vertical segment of the same size increasing more rapidly in they at P than at Q. Therefore, /yare both larger at P than at Q.

Similarly, a horizontal segment through P meet more level curves at P than at Q, but f is nPt'rPill<in;

x-direction, so f is decreasing more rapidly in the x-direction at P than at Q. Therefore, fx is more at Q.

(b) For any fixed value x = a, a horizontal segment meets fewer level curves as we move it indicates that fx(a, y) in a decreasing function of y.


(b) The derivative ¥f lc is estimated by 1,~. Since x varies in the horizontal direction, we move horizontally · to a point on the next level curve in the direction of increasing x (leftwards). Since the value of D on this level greater than on the level curve of C, 6.D = l. Also 6.x > 0, hence

aD' 6.D 1 a:; C ~ 6.x = 6.x > 0.

The statement is correct. (c) Moving from C vertically upward (in the direction of increasing y ), we come to a point on a level curve value of D. Therefore, 6.D = -1 and 6.y > 0, so we obtain

Hence, the statement is false.

56. Refer to Table 1 .

(a) Estimate ap;ar and ap;as at the points (S, T) = (34, 2) and (35, 10) by computing the average of right-hand difference quotients.

(b) ~ For fixed salinity S = 33, is p concave up or concave down as a function ofT? Hint: Determine

quotients 6.p/ 6.T are increasing or decreasing. What can you conclude about the sign of a2 pjaT2?

SOLUTION

TABLE 1 Seawater Density p as a Function of Temperature T and Salinity s

~ 30 31 32 33 34 35 36

12 22.75 23.51 24.27 25.07 25.82 26.6 27.36

10 23.07 23.85 24.62 25.42 26.17 26.99 27.73

8 23.36 24.15 24.93 25.73 26.5 27.28 29.09

6 23.62 24.44 25.22 26 26.77 27.55 28.35

4 23.85 24.62 25.42 26.23 27 27.8 28.61

2 24 24.78 25.61 26.38 27.18 28.01 28.78

0 24.11 24.92 25.72 26.5 27.34 28.12 28.91

(a) We estimate lf at the given points using the values in Table 1 and the following approximation:

ap p(34, 2 + 2) - p(34, 2) p(34, 4) - p(34, 2) 27- 27.18 -(34 2) ~ = = = -0.09 ar • 2 2 2

ap p(35, 10 + 2) - p(35, 10) p(35, 12) - p(35, 10) 26.6- 26.99 -(35, 10) ~ = = = -0.195 ar 2 2 2

Therefore, the average of the left-hand and right-hand difference quotients is:

~ (:;(34, 2) + :;(35, 10)) ~ ~(-0.09- 0.195) = -0.1425

We estimate the partial derivative ~ at the given points:

ap (34, 2) ~ p(34 + 1. 2) - p(34, 2) = p(35, 2) - p(34, 2) = 28.01 - 27.18 = 0.83 as 1 1

ap (35, 10) ~ p(35 + 1, 10)- p(35, 10) = p(36, 10)- p(35, 10) = 27.73- 26.99 = 0.74 as 1 1

Therefore, the average of the left-hand and right-hand difference quotients is:

~ (:~ (34, 2) + :~ (35, 10)) ~ ~(0.85 + 0.74) = 0.795

(b) The function p(33, T) is concave up (concave down) if -3f (33, T) is an increasing (decreasing) function ofT. We

use Table 1 to estimate whether the function lf<33, T) is increasing or decreasing. We compute the following values:

ap 3 2 p(33, 4)- p(33, 2) 26.23- 26.38 ar <3 • > ~ 2 = 2 = -om5


ap 3

~ p(33, 6)- p(33, 4) _ 26- 26.23 _ aT(3,4)~

2 - 2 --0.115

ap p(33, 8)- p(33, 6) 25.73- 26 aT (33, 6) "'"

2 = 2 = -0.135

ap p(33, 10)- p(33, 8) 25.42- 25.73 aT (33, 8) "'"

2 = 2 = -0.155

ap p(33, 12)- p(33, 10) 25.07-25.42 aT (33, 10)"'"

2 = 2 = -0.175

These values indicate that gj. (33, T) is a decreasing function ofT, which means that the second derivative is negative,

i.e., ~~ (33, T) < 0 and the graph of p(33, T) is concave down.

In Exercises 57--62, compute the derivatives indicated.

57. f(x, y) = 3x2y- 6xi, azf azf -andax2 ay2

SOLUTION We first compute the partial derivatives ~~ and ~{.:

af 4 ax = 6xy- 6y ;

af - = 3x 2 - 6x · 4v3 = 3x2 - 24xy3

ay .

We now differentiate ~~ with respect to x and ~{. with respect toy. We get

a2 J a ---! -6v· ax2 -ax X- •'

a2 J a 2 2 -

2 =-!y=-24x·3y =-72xy.

ay ay

xy a2g 58. g(x, y) = --,

X- y ax ay

SOLUTION By definition we have

Thus, we must find ~;:

ag =x~(-y-)=xl·(x-y)-y·(-1) = x2 ay ay X- y (X- y)2 (X- y)2

Differentiating* with respect to x, using the Quotient Rule, we obtain

u 59. h(u, v) = --, hvv(u, v)

u +4v

SOLUTION We first note

so thus

60. h(x, y) = ln(x3 + y3), hxy(x, y)

SOLUTION We first note that

2x(x- y)2 - x 2 · 2(x- y)

(x- y)4

ah -4u

au (u + 4u)2

ah2 a ( -4u ) 32u av2 = av (u + 4v)2 = (u + 4v)3

2xy

(x- y)3


so thus

61. f(x, y) = x In(y2), /yy(2, 3)

SOLUTION We find /y using the Chain Rule:

a 2 a 2 I 2x /y = -(x!ny) =x-Iny =x-z ·2y =-

ay ay y y

We now differentiate /y with respect toy, obtaining

a a ( 1) -2x /yy(x,y)=ay/y=2xay Y =?.

The derivative at (2, 3) is thus

-2·2 4 /yy(2, 3) = ~ = -9.

62. g(x,y)=xe-xY, gxy(-3,2)

SOLUTION We first compute:

so thus:

and

ag =X. e-xy. (-y) + e-xy = e-xy(I- xy) ax

a2 a _g_ = -(e-XY(J- xy)) = e-xy(-x) + (1- xy)e-xy · (-x) = -xe-XY(2- xy) ayax ay

6 6 gxy( -3, 2) = 3e (2 + 6) = 24e

63. Compute fxyxzy for

f(x, y, z) = y sin(xz) sin(x + z) + (x + z2) tan y + x tan ( z + z~~) y-y

Hint: Use a well-chosen order of differentiation on each term.

SOLUTION At the points where the derivatives are continuous, the partial derivative fxyxzy may be performed in order. To simplify the computation we first consider f (x, y, z) as the sum of the following terms:

so that

F(x, y, z) = y sin(xz) sin(x + z), G(x, y, z) = (x + z2) tan y, H(x, y, z) = x tan ( z + z~~) y-y

f(x, y, z) = F(x, y, z) + G(x, y, z) + H(x, y, z)

We can differentiate each in any order. First, let us work with F(x, y, z) = y sin(xz) sin(x + z):

then

hence,

a Fy(x, y, z) = -(y sin(xz) sin(x + z)) = sin(xz) sin(x + z)

ay

a Fyy(X, y, z) = ay (Fy(X, y, z)) = 0

Fyyxxz(X, y, z) = 0

Next, let us work with G(x, y, z) = (x + z2 ) tan y:

a Gx(x, y, z) = -((x + z2) tan y) =tan y

ax


Finally, we differentiate gxy with respect to z, obtaining

a 2 2 2 -3/2 ( 3) 2 2 2 -5/2 3xyz gnz=-xy-;-(x +y +z) =-xy· -- (x +y +z) ·2z= 52 · az 2 (x2 + y2 + 22) I

72. u(x, tl = sech2(x- t), Uxxx

SOLUTION Using the Chain Rule we have

a 2 a 2 Ux = -- sech (x- t) = 2 sech(x- t) · (- sech(x- t) tanh(x- t)) · - (x- t) = -2 sech (x- t) tanh(x- t)

ax ax

We now use the Product Rule and the Chain Rule to differentiate ux with respect to x:

Uxx = -2[2 sech(x- t) · (- sech(x- t) tanh(x- t)) tanh(x- t) + sech2(x- t). sech2(x- t)]

= 4 sech2(x- t) tanh2(x- t)- 2 sech4 (x- t) = 2 sech2(x- t)(2 tanh2(x- t)- sech2(x- t))

We find Uxxx, using the Product Rule and the Chain Rule:

ltxxx = 4 sech(x- t)(- sech(x- t) tanh(x- t) )(2 tanh2(x- t)- sech2(x- t))

+ 2 sech2 (x - n[ 4 tanh(x - t) . sech2 (x - t) - 2 sech(x - t)(- sech(x - t) tanh(x - t))]

= -8 sech2 (x - t) tanh3 (x - t) + 4 sech4 (x - t) tanh(x - t) + 12 sech4 (x - t) tanh(x - t)

= 16sech4 (x- t)tanh(x- t)- 8sech2(x- t)tanh\x- t)

. . of of 2 73. Fmd a functiOn such that- = 2xy and- = x .

ax ay

SOLUTION af af The function f(x, y) = x 2y satisfies ay = x 2 and ax = 2xy.

~I . . of of 2 . 74. ~--· Prove that there does not ex1st any functiOn f(x, y) such that-= xy and-= x . Hznt: Show that f

ax ay cannot satisfy Clairaut's Theorem.

SOLUTION Suppose that there exists a function f(x, y) such that~~ = xy and* = x2

. Hence,

fxy = ~ (af) = ~xy =x ay ax ay

hx = ;___ (af) = ~x2 = 2x ax ay ax

The mixed partials fxy and f.vx are continuous everywhere, but fxy =F jyx for x =F 0. This contradicts Clairaut's Theorem on Equality of Mixed Partials. We conclude that there does not exist any function f (x, y) with the given partials.

75. Assume that fry and jyx are continuous and that fyxx exists. Show that fxyx also exists and that hxx = fxyx.

SOLUTION Since f,y and fvx are continuous, Clairaut's Theorem implies that

fry= iyx

We are given that hxx exists. Using (I) we get

a a a a fyxx = -- !y = - hx = - fxv = fxyx

ax ax ax ax .

Therefore, fxyx also exists and f_vxx = fx_vx.

2 76. Show that u(x. t) = sin(nx) e-n 1 satisfies the heat equation for any constant n:

SOLUTION We compute ~7 using the Chain Rule:

We now find Ux:

au a 2 2 a 2 2 2 - = sin(nx)-e-n 1 = sin(nx)e-n 1 -(-n t) = -n sin(nx)e-n 1

at ar ar

2 a 2 2 Ux =e-n 1 ax sin(nx) =e-n 1 cos(nx) · n = n · cos(nx)e-n

1

(1)


Differentiating Ux with respect to x gives

Uxx = ne-n 1- cos(nx) = ne-n 1

- sin(nx)-(nx) = ne-n 1 (- sin(nx)) · n = -n e-n 1 sin(nx) 2o 2( a) 2 2 2 ax ax

We see that u1 = Uxx, therefore u satisfies the heat equation.

77. Find all values of A and B such that f(x, t) = eAx+BI satisfies Eq. (3).

SOLUTION We compute the following partials, using the Chain Rule:

of = ~(eAx+BI) = eAx+BI ~(Ax+ Bt) = BeAx+BI at at at

of a a _ = -(eAx+BI) = eAx+BI -(Ax+ Bt) = AeAx+BI ax ax ax

a2f a a a _ = -(AeAx+BI) = A-(eAx+BI) = AeAx+BI -(Ax+ Bt) = A2eAx+BI ax2 ax ax ax

Substituting these partials in the differential equation (3), we get

We divide by the nonzero eAx+Bt to obtain

We conclude that f(x, t) = eAx+BI satisfies equation (5) if and only if B = A2, where A is arbitrary.

78. The function

describes the temperature profile along a metal rod at timet > 0 when a burst of heat is applied at the origin (see 11 ). A small bug sitting on the rod at distance x from the origin feels the temperature rise and fall as heat diffuses the bar. Show that the bug feels the maximum temperature at timet = ~x 2 .

SOLUTION From the example in the text we see that:

of l -3/2 -x2j41 l 2 -5/2 -x2/41 - = ---t e + --x t e at 4Jlf 8Jlf

We take this expression, in order to find the maximum, and set it equal to 0 and solve fort:

__ 1_t-312e-x2j41 + _1_x2t-512e-x2j41 = 0 4J]r 8J]r

e-x2 /41 ( _ 2t-3!2 + x2t-512) = 0

t-512e-x2 /41 ( -2t + x2) = 0

Then since the exponential factor is never equal to 0 and the t-512 is not either, we only consider when

-2t+x2 =0 => I

t = -x2

2

Since we are told that the bug experiences the rise and then the fall of the temperature, we are assured that t = 1 the point in time when the bug experiences the maximum temperature.

In Exercises 79-82, the Laplace operator~ is defined by ~f = fxx + /yy. A function u(x, y) satisfying the equation ~u = 0 is called harmonic.

79. Show that the following functions are harmonic:

(a) u(x, y) = x

(c) u(x, y) = tan- 1 ~ X

(b) u(x, y) =ex cosy

(d) u(x, y) = ln(x2 + y 2)


SOLUTION

(a) We compute Uxx and Uyy for u(x, y) = x:

a Ux = ax(X) = 1;

a uv = -(x) = 0;

. ay

a Uxx = -(1) =0

ax

a Uyy· = -(0) = 0

. ay

Since Uxx + Uyy = 0, u is harmonic.

(b) We compute the partial derivatives of u(x, y) = ex cosy:

a a Ux = -(ex COS V) =COS V-ex =(COS y)ex

ax . . ax

a a u y = - (ex cosy) = ex - cosy = -ex sin y

ay ay

a a Uxx = ax ((COS y)ex) =COSy ax ex =(COS y)ex

a( t · ) Xa · X uyy=- -e· smy =-e -smy=-e cosy · ay ay

Thus,

!lxx + Uyy = (cos y)ex -ex cosy= 0

Hence u (x, y) = ex cosy is harmonic.

(c) We compute the partial derivatives of u(x, y) = tan-l ~using the Chain Rule and the formula X

We have

d _ 1 I -tan t=-dt 1 + r2

a Y ----;c

] + (yjx)2 ax X

I (-y) y l+(yjx)2 ~ =-x2+y2

a -t Y Ux =-tan -

ax X

a -t Y Uy =-tan -

· ay x

1 a y 1 + (y/x)2 ay ~

I (I) x I+ (yjx)2 ~ = x2 + y2

a ( y ) 2xy Uxx = ax - x2 + y2 = (x2 + y2)2

a X 2xy

Uyy = ay x2 + y2 =- (x2 + y2)2

Therefore Uxx + uxx = 0. This shows that u(x, y) = tan-l ~is harmonic. X

(d) We compute the partial derivatives of u(x, y) = ln(x 2 + y2) using the Chain Rule:

a 2 2 1 2x Ux = - in(x + y ) = - 2--2 · 2x = - 2--2 ax X + y X + y

a 2 2 1 2y uy = - in(x + y ) = -2--2 . 2y = -2--2

ay X + y X + y

We now find Uxx and Uyy using the Quotient Rule:

Thus,

2(y2- x2)

(x2 + y2)2

2(x 2 - y2) (x2 + y2)2

Therefore, u(x, y) = ln(x2 + y2) is harmonic.

SECT I 0 N 14.4 I Differentiability and Tangent Planes (LT SECTION 15.4) 681

4. Estimatef(2,3.1).

SOLUTION We use the linear approximation

f(a + h, b + k) "'=' f(a, b)+ fx(a, b)h + !y(a, b)k

We let (a, b)= (2, 3), h = 0, k = 3.1-3 = 0.1. Then,

f(2, 3.1) ~ f(2, 3) + fx(2, 3) · 0 + !y(2, 3) · 0.1 = 8 + 0 + 7 · 0.1 = 8.7

We get the estimation f(2, 3.1) ~ 8.7.

5. Estimate b.f at (2, 3) if t.x = -0.3 and b.y = 0.2.

SOLUTION The change in f can be estimated by the linear approximation as follows:

b.f ~ f,(a, b)t.x + !y(a, b)b.y

b.f ~ fx (2, 3) · ( -0.3) + !y (2, 3) · 0.2

or

b.f ~ 5 . ( -0.3) + 7 . 0.2 = -0.1

The estimated change is b.f ~ -0.1.

6. Which theorem allows us to conclude that f (x, y) = x 3 y8 is differentiable?

SOLUTION The function f (x, y) = x 3 y8 is a polynomial, hence fx (x, y) and !y (x, y) exist and are continuous.

Therefore the Criterion for Differentiability implies that f is differentiable everywhere.

Exercises 1. Use Eq. (2) to find an equation of the tangent plane to the graph off (x, y) = 2x

2 - 4xy

2 at (-I, 2).

SOLUTION The equation of the tangent plane at the point (-I, 2, 18) is

z = j(-1, 2) + fx(-1, 2)(x +I)+ !y(-1, 2)(y- 2) (I)

We compute the function and its partial derivatives at the point ( -1, 2):

f(x, y) = 2x2 - 4xy2 j(-1, 2) = 18

f,(x, y) = 4x- 4y2 =} fx(-1. 2) = -20

jy(x, y) = -8xy !y(-1, 2) = 16

Substituting in ( 1) we obtain the following equation of the tangent plane:

z = 18- 20(x +I)+ 16(y- 2) = -34- 20x + 16y

That is,

z = -34- 20x + 16y

2. Find the equation of the plane in Figure 9, which is tangent to the graph at (x, y) = ( 1, 0.8).

'

.)":-··· ··-·-·-:.~~- "

,,. \\: ." . .

~~·,/' FIGURE 9 Graph of f(x, y) = 0.2x

4 + y6- xy.

SOLUTION We know that the equation of the tangent plane at the point (1, 0.8) is:

z = f (I, 0.8) + fx (1, 0.8) (x - 1) + !y (l, 0.8)(y - 0.8)


We compute the function and its partial derivatives at the point (I, 0.8):

f(x,y)=0.2x 4 +i-xy => /(1,0.8)=-0.34

fx(X, y) = 0.8x3 - y

/y(X, y) = 6y5 - X

=> fx(l, 0.8) = 0

=> /y(l, 0.8) = 0.96608

Substituting in the equation of the tangent plane we obtain the following equation:

z = -0.34 + O(x- I)+ 0.96608(y- 0.8)

That is,

z = 0.96608y- 1.112864

In Exercises 3-10, find an equation of the tangent plane at the given point.

3. f(x,y) =x2y+xy3, (2, 1)

SOLUTION The equation of the tangent plane at (2, I) is

z = /(2, I)+ fx(2, I)(x- 2) + /y(2, 1)(y- I)

We compute the values off and its partial derivatives at (2, 1):

f(x,y)=x 2y+xy3 /(2,1)=6

fx(x, y) = 2xy + y 3 => fx(2, 1) = 5

/y(x, y) = x 2 + 3xi /y(2, I)= 10

We now substitute these values in (I) to obtain the following equation of the tangent plane:

z = 6 + 5(x - 2) + 10(y - 1) = 5x + lOy- 14

That is,

X 4. f(x, y) = -, (4, 4)

.jY

z = 5x +lOy- 14.

SOLUTION The equation of the tangent plane at ( 4, 4) is

z = /(4, 4) + fx(4, 4)(x- 4) + /y(4, 4)(y- 4)


X f(x, y) =

.jY /(4, 4) = 2

I fx(x,y)= .jY 1

=> fx(4, 4) = 2

/y(x, y) = x~y-112 = x. (-~) y-3/2 = __ x_ ay 2 2y3/2

Substituting these values in (I) gives

1 I 1 I z = 2 + - (x - 4) - - (y - 4) = - x - - y + I.

2 4 2 4

5. f(x, y) = x 2 + y-2 , (4, 1)

SOLUTION The equation of the tangent plane at (4, I) is

z = /(4, I)+ fx(4, I)(x- 4) + /y(4, l)(y- 1)


f(x, y) = x 2 + y-2 /(4, 1) = 17

fx(X, y) = 2x => fx(4, I)= 8

/y(x, y) = -2y-3 /y(4, 1) = -2

SECTION 14.4 I Differentiability and Tangent Planes (LT SECTION 15.4) 683

Substituting in (I) we obtain the following equation of the tangent plane:

z = 17 + 8(x- 4)- 2(y- 1) = Sx- 2y- 13.

6. G(u, w) = sin(uw), (~. 1)

SOLUTION The equation of the tangent plane at ( ~, 1) is

z = f (~. 1) + fu (~. 1) (u- ~) + fw (~. 1) (w- 1)

We compute the following values:

f(u, w) = sin(uw)

fu(u,w)=wcos(uw) =}

fw(u, w) = u cos(uw)

f (~. 1)

fu (~. 1)

fw (~. 1)

. lf 1 =Sill-=-

6 2

lf ,J3 =!·COS-=-

6 2

rr rr ..f3rr =-cos-=--

6 6 12

Substituting in ( 1) gives the following equation of the tangent plane:

1 ,J3 ( rr ) ..f3rr z =- +- u-- + --(w- 1) 2 2 6 12

That is,

,J3 ..f3rr I .J3rr z=-u+--w+----.

2 12 2 6

SOLUTION The equation of the tangent plane at (2, 1) is

Z = j(2, 1) + fr(2, l)(r- 2) + f 5 (2, l)(s- 1)

We compute f and its partial derivatives at (2, 1):

f(r, s) = r2s- 112 + s-3 j(2, 1) = 5

fr(r, s) = 2rs- 112 =} fr(2, I)= 4

I f5

(r, s) = --r2s-312 - 3s-4 f 5 (2, 1) = -5 2

We substitute these values in ( 1) to obtain the following equation of the tangent plane:

z = 5 + 4(r- 2)- S(s- 1) = 4r- 5s + 2.

8. g(x, y) = exfy, (2, 1)

SOLUTION The equation of the tangent plane at (2, I) is:

z = g(2, I)+ gx(2, l)(x- 2) + gy(2, l)(y- 1)

We compute g and its partial derivatives at (2, I):

g(x, y) = exfy g(2, I)= e2

gx(X, y) = ~exfy, y

gv(X, v) =- x2exfy, . . y

2 gy(2, 1) = -2e

We substitute these values in the tangent plane equation to obtain the following equation of the tangent plane:

I

(I)

(I)


9. f(x, y) = sech(x- y), (ln4, ln2)

SOLUTION The equation of the tangent plane at (In 4, In 2) is:

We compute f and its partial derivatives at (In 4, In 2):

j(x, y) = sech(x- y), 4

f (In 4, In 2) = sech(ln 2) = 5

fx(x, y) =- tanh(x- y) sech(x- y),

!y(x, y) = tanh(x- y) sech(x- y),

12 fx (In 4, In 2) = -tanh (In 2) sech(ln 2) = -

25

12 !y (In 4, In 2) = tanh (In 2) sech(ln 2) =

25

We substitute these values in the tangent plane equation to obtain:

4 12 12 4 z =-- -(x -ln4) + -(x -In 2) = --(3x- 3y- 5 -In 8)

5 25 25 25

10. j(x, y) = ln(4x2 - y 2 ), (1, 1)

SOLUTION The equation of the tangent plane at (1, 1) is

Z = f(l, 1) + fx(l, l)(x- 1) + !y(l, 1)(y- 1)


f(x, y) = ln(4x2 - i), j(1, 1) =In 3

8x fx (x, y) = 4 2 2,

X - y

-2y !y(x, y) = 4 2 2'

X - y

8 fx(l, 1) = 3

2 !y(l, 1) = -3

Substituting these values into the equation for the tangent plane we obtain:

8 2 8 2 z =In 3 + 3(x- 1)- 3(y- I)= 3x- 3y +In 3-2

11. Find the points on the graph of z = 3x2 - 4y2 at which the vector n = (3, 2, 2) is normal to the tangent plane.

SOLUTION The equation of the tangent plane at the point (a, b, f(a, b)) on the graph of z = j(x, y) is

z = f(a, b)+ fx(a, b)(x- a)+ jy(a, b)(y- b)

or

fx(a, b)(x- a)+ !y(a, b)(y- b)- z + f(a, b)= 0

Therefore, the following vector is normal to the plane:

v = (fx(a, b), !y(a, b), -I)

We compute the partial derivatives of the function f(x, y) = 3x2 - 4y2:

fx(X, y) = 6x ::::} fx(a, b)= 6a

!y(x, y) = -8y ::::} !y(a, b) = -8b

Therefore, the vector v = (6a, -8b, -1) is normal to the tangent plane at (a, b). Since we want n = (3, 2, 2) to be to the plane, the vectors v and n must be parallel. That is, the following must hold:

6a -8b

3 2 2

which implies that a = -! and b = ~. We compute the z-coordinate of the point:

z=3·(-~r-4Gr =~ The point on the graph at which the vector n = (3, 2, 2) is normal to the tangent plane is (-!, ~, ~).

'------------r-~--.---


12. Find the points on the graph of z = x y 3 + 8 y -i where the tangent plane is parallel to 2x + 7 y + 2z = 0.

SOLUTION The equation of the tangent plane at the point (a, b, f (a, b)) on the graph of z = f (x, y) is

z = f(a, b)+ fx(a, b)(x- a)+ !y(a, b)(y- b)

or

fx(a, b)(x- a)+ !y(a, b)(y- b)- z + f(a, b)= 0

Therefore, the following vector is normal to the plane:

v = (fx(a, b), !y(a, b), -1)

We compute the partial derivatives of the function z = xy3 + 8y- 1:

fx(X, y) = y3 , fx(a, b)= b3

!y(x, y) = 3xi- 8y-2, !y(a, b)= 3ab2 - 8b-2

Therefore, the vector v = {b3 , 3ab2 - 8b-2 , -I) is normal to the tangent plane at (a, b). For two planes to be parallel,

the vectors v and n must be parallel. The corresponding normal vector here is n = (2, 7. 2). The following must hold:

which implies that b = -1 and a = 3/2. We compute the z-coordinate of the point:

3 3 -] 19 z = -(-1) + 8(-1) = --2 2

The point on the graph at which the tangent plane is parallel to 2x + 7 y + 2z = 0 is ( ~, -1, -1;).

13. Find the linearization L(x, y) of f(x, y) = x 2y 3 at (a, b) = (2, 1). Use it to estimate f(2.01, 1.02) and f( 1.97, 1.01) and compare with values obtained using a calculator.

SOLUTION

(a) We compute the value of the function and its partial derivatives at (a, b) = (2, 1):

f(x, y) = x 2y 3 f(2, I)= 4

fx(x, y) = 2xy3 =} fx(2, I)= 4

!y(x, y) = 3x2i !y(2, I)= 12

The linear approximation is therefore

L(x, y) = f(2, 1) + fx(2, l)(x- 2) + !y(2, I)(y- 1)

L(x, y) = 4+4(x- 2) + 12(y- I)= -16+4x + 12y

(b) For h = x- 2 and k = y- 1 we have the following form of the linear approximation at (a, b)= (2, 1):

L(x, y) = f(2, I)+ fx(2, l)h + fy(2, l)k = 4 + 4h + 12k

To approximate f(2.01, 1.02) we seth = 2.01 - 2 = 0.01, k = 1.02- 1 = 0.02 to obtain

L(2.01, 1.02) = 4 + 4 · 0.01 + 12 · 0.02 = 4.28

The actual value is

f(2.01, 1.02) = 2.01 2 . 1.023 = 4.2874

To approximate f(l.97, 1.01) we seth = 1.97- 2 = -0.03, k = 1.01 - I = 0.01 to obtain

L(l.97, 1.01) = 4 + 4 · (-0.03) + 12 · O.Ql = 4.

The actual value is

f(l.97, 1.01) = 1.972 . 1.01 3 = 3.998 .

•


20. Find the linearization to f(x, y, z) = xyjz at the point (2, 1, 2). Use it to estimate /(2.05, 0.9, 2.01) and with the value obtained from a calculator.

SOLUTION The linear approximation to f at the point (2, 1, 2) is:

f(x, y, z) ~ /(2, 1, 2) + fx(2, 1, 2)(x- 2) + /y(2, 1, 2)(y- 1) + /2

(2, 1, 2)(z- 2)

We compute the values off and its partial derivatives at (2, 1, 2):

xy f(x,y,z)=

z

y fx(x, y, z) =

z

X /y(x, y, z) = -

z

xy fz(X, y, z) = -2

z

/(2, 1, 2) = 1

1 => fx(2, 1, 2) = 2

/y(2, 1, 2) = 1

1 / 2 (2, 1, 2) = -2

We substitute these values in (I) to obtain the following linear approximation:

xy I I ~~I+ 2(x- 2) +I· (y- I)- 2(z- 2)

xy I I - ~ -x + y- -z z 2 2

To estimate /(2.05, 0.9, 2.01) we will have:

I 1 /(2.05, 0.9, 2.01) ~ 2(2.05) + 0.9- 2(2.01) = 0.92

Comparing this with the calculator value we get:

2.05. 0.9 /(2.05, 0.9, 2.01) = ~ 0.9179

2.01

21. Estimate f (2.1, 3.8) assuming that

/(2, 4) = 5, fx(2, 4) = 0.3, /y(2, 4) = -0.2

SOLUTION We use the linear approximation off at the point (2, 4), which is

/(2 + h, 4 + k) ~ /(2, 4) + fx(2, 4)h + /y(2, 4)k

Substituting the given values and h = 0.1, k = -0.2 we obtain the following approximation:

/(2.1, 3.8) ~ 5 + 0.3. 0.1 + 0.2. 0.2 = 5.o7.

22. Estimate /(1.02, O.oJ, -0.03) assuming that

f(I, 0, 0) = -3,

/y(l, 0, 0) = 4,

fx (1, 0, 0) = -2,

/ 2 (1, 0, 0) = 2

SOLUTION The linear approximation at (1, 0, 0) is

f(I + h, k, l) ~ j(I, 0, 0) + fxO. 0, O)h + /y(l, 0, O)k + /2

(1, 0, 0)/

We substitute h = 0.02, k = 0.01, l = -0.03 and the given values to obtain the following estimation:

/(1.02, 0.01' -0.03) ~ -3 + ( -2) . 0.02 + 4. 0.01 + 2( -0.03) = -3.06

That is,

/(1.02, O.oJ, -0.03) ~ -3.06.

--


8.01 28 •

./(1.99)(2.01)

SOLUTION We use the linear approximation of the function f(x, y, z) = jyz at the point (8, 2, 2), which is

/(8 + h, 2 + k, 2 +I)~ /(8, 2, 2) + fx(8, 2, 2)h + /y(8, 2, 2)k + fz(8, 2, 2)1

We compute the values of the function and its partial derivatives at (8, 2, 2). This gives

X f(x,y,z)= ""'

vYZ

1 fx(x,y,z) = ""'

vYZ

a 1 1 /y(x, y, z) =X ay (yz)-1/2 = -2x(yz)-3/2z = -2xy-3f2z-l/2

a I I fz(X, y, z) =X az (yz)-1/2 = -2x(yz)-3f2y = -2xy-lf2z-3j2

/(8, 2, 2) = 4

1 =} fx(8, 2, 2) = 2

/y(8, 2, 2) = -1

fz(8, 2, 2) =-I

Substituting these values and h = 0.01, k = -0.01, I= 0.01 in (1) gives the following approximation:

8.01 1 -._,~r:;o;::::;.9~9:;::::;)(""'2 .""01~) = 4 + 2 . 0.01- 1. (-O.OI)- 1. 0.01 = 4.005

The value given by a calculator is 4.00505. The error is 0.00005 and the percentage error is at most

0.00005 . 100 Percentage error~ ~ 0.00125%

4.00505 29. Find an equation of the tangent plane to z = f (x, y) at P = (I, 2, I 0) assuming that

f(i, 2) = 10, /(1.1, 2.01) = 10.3, /(1.04, 2.1) = 9.7

SOLUTION The equation of the tangent plane at the point (I, 2) is

Z = f(i, 2) + fxO. 2)(x- I)+ /y(i. 2)(y- 2)

Z = 10 + fx(i, 2)(x- I)+ fvO. 2)(y- 2)

Since the values of the partial derivatives at (I, 2) are not given, we approximate them as follows:

2 /(1.1, 2)- f(i, 2) /(1.1, 2.01)- f(i, 2)

fxO. ) ~ ~ = 3 0.1 0.1

f(l, 2.1)- f(i, 2) /(1.04, 2.1)- j(l, 2) /y(l,2)~ ~ =-3

0.1 0.1

Substituting in (1) gives the following approximation to the equation of the tangent plane:

z = 10 + 3(x- I)- 3(y- 2)

That is. z = 3x- 3y + 13.

(I)

(I)

30. Suppose that the plane tangentto z = f (x, y) at (-2, 3, 4) has equation 4x + 2 y + z = 2. Estimate f (-2. I, 3 .I).

SOLUTION The tangent plane z = 2 - 4x - 2y is also a linear approximation for f near (-2, 3), so we can thus calculate the following:

f(-2.1, 3.1) ~ 2- 4(-2.1)- 2(3.1) = 4.2

In Exercises 31-34, let I = W j H 2 denote the BMI described in Example 5.

31. A boy has weight W = 34 kg and height H = 1.3 m. Use the linear approximation to estimate the change in I if (W, H) changes to (36, 1.32).

SOLUTION Let 1'1I = I (36, 1.32) - 1 (34, 1.3) denote the change in 1. Using the linear approximation of I at the point (34, 1.3) we have

ai ai I(34 + h, 1.3 + k)- I(34, 1.3) ~ -(34, 1.3)h + -(34, 1.3)k

aw aH

For h = 2, k = 0.02 we obtain

ai ai 1'1I ~ -(34, 1.3). 2 + -(34, 1.3). 0.02

aw aH (1)

•

sEcT I 0 N 14.5 I The Gradient and Directional Derivatives (LT SECTION 15.5) 701

(b) Using the Chain Rule gives

d d ( 1 2 6) d ( 1 8) 7 - f (c(t)) = - -t · t = - -t = 4t dt dt 2 dt 2

Substituting x = !t2 andy = t 3, we obtain

d 6 12 3 2 7 dt f (c(t)) = t · t + 2 · :zt · 3 . t . t = 4t

At the point t = 1 and t = - 1, we get

!!._ (f (c(t))) I = 4 · 17 = 4, dt t=J

d I 7 - (f (c(t))) = 4 · (-1) = -4. dt t=-1

2. Let f(x, y) = exy and c(t) = (t\ I+ t).

(a) Calculate V f and c' (t).

(b) Use the Chain Rule for Paths to calculate!!._ f(c(t)). dt

(c) Write out the composite f(c(t)) as a function oft and differentiate. Check that the result agrees with part (b).

SOLUTION

(a) We first find the partial derivatives of f(x, y) = exy:

The gradient vector is thus

of xy - =ye ' ax

of xy - =xe . ay

Differentiating c(t) = (t3 , 1 + t) componentwise, we obtain

c'(t) = (Ct3{ (1 + t)') = (3t2

, 1)

(b) We find f, f (c(t)) using the Chain Rule and the results of part (a). This gives

!!._ f (c(t)) = of dx + of dy = (yeXY) . 3t2 + (xeXY) . 1 dt ax dt ay dt

To write the answer in terms oft only, we substitute x = t3 and y = I + t. This gives

d 34 2 334 2 334 - f (c(t)) = (1 + t)e1 +t · 3t + (t )e1 +t = (3t + 4t )e1

+t dt

(c) We substitute x = t 3 , y = 1 + t in f (x, y) = exy to obtain the composite function f (c(t)):

3 4 f (c(t)) = et +t

We now differentiate the composite function to obtain

d d34 2 334 - f (c(t)) = -(e1 +t ) = (3t + 4t )e1

+t dt dt

This result agrees with the result obtained in part (a).

3. Figure 14 shows the level curves of a function f(x, y) and a path c(t), traversed in the direction indicated. State

d whether the derivative - f(c(t)) is positive, negative, or zero at points A-D.

dt

y

4

0

-4

~-20

~ A 10

20 30

L-~--~---+--~-x

-4 0 4

FIGURE 14


SOLUTION At points A and D, the path is (temporarily) tangent to one of the contour lines, which means that along

path c(t) the function j(x, y) is (temporarily) constant, and so the derivative ft j(c(t)) is zero. At point B, the moving from a higher contour (of -10) to a lower one (of -20), so the derivative is negative. At the point C, where path moves from the contour of -I 0 towards the contour of value 0, the derivative is positive.

4. Let f(x, y) = x 2 + y2 and c(t) =(cost, sin t). d

(a) Find- j(c(t)) without making any calculations. Explain. dt

(b) Verify your answer to (a) using the Chain Rule.

SOLUTION

(a) The level curves of f (x, y) are the circles x 2 + y 2 = c2. Since c(t) is a parametrization of the unit circle, f constant value I on c. That is, f (c(t)) = I, which implies that ft f (c(t)) = 0.

(b) We now find ft f ( c(t)) using the Chain Rule:

d aj dx aj dy - j(c(t)) = -- + -dt ax dt ay dt

We compute the derivatives involved in ( 1):

af a - = - (x 2 + i) = 2x, ax ax

aj = ~ (x 2 + i) = 2y ay ay

dx d - = - (cost) = - sin t, dt dt

dy d - = -(sint) =cost dt dt

Substituting the derivatives in (I) gives

d . - f ( c (t)) = 2x (- sm t) + 2 y cos t dt

Finally, we substitute x = cost and y = sin t to obtain

d - f (c(t)) = -2 cost sin t + 2 sin t cost = 0. dt

In Exercises 5-8, calculate the gradient.

5. j(x, y) = cos(x2 + y)

SOLUTION We find the partial derivatives using the Chain Rule:

:~ =-sin (x2 + y) aax (x

2 + y) = -2x sin (x2 + y)

:~ = -sin ( x2 + y) aay ( x

2 + y) = -sin ( x2 + y)


V' f = ( :~, :~) = { -2x sin ( x 2 + y) , -sin ( x 2 + y)) = -sin ( x 2 + y) (2x, 1}

X 6. g(x, y) = -2--2

X + y

SOLUTION We compute the partial derivatives. We first find * using the Quotient Rule:

1 · (x 2 + y2)- x · 2x

(x2 + y2)2

We compute ~;, using the Chain Rule:

ag a 1 -I -2xy -=x----=x·-------=-2 ·2y=

2 ay ay x2 + y2 (x2 + y2) (x2 + y2)


•

sECT 1 oN 14.5 I The Gradient and Directional Derivatives (LT SECTION 15.5) 707

20. g(x,y,z,w)=x+2y+3z+5w, c(t)=(t2,t3 ,t,t-2), t=1

soLUTION We compute the gradient and c' ( t):

'Vg= -,-,-,- =(1,2,3,5) (ag ag ag ag) ax ay az aw

c' (t) = { 2t, 3t2, 1, 1)

At the point t = 1 we have (notice that the gradient is a constant vector)

'Vgc(l) = (1, 2, 3, 5)

c'(l) = (2, 3, 1, I)

We now use the Chain Rule for Paths to obtain the following derivative:

-g(c(t)) = 'Vgc(l) ·C (1) = (1, 2,3,5) · (2,3, 1, 1) = 2+6+ 3 +5 = 16 d I I

dt t=l

In Exercises 21-30, calculate the directional derivative in the direction of vat the given point. Remember to normalize

the direction vector or use Eq. ( 4 ).

21. f(x, y) = x 2 + y3, v = (4, 3), P = (1, 2)

SOLUTION We first normalize the direction vector v:

v (4, 3) (4 3) u = w = J 42 + 32 = 5 · 5

We compute the gradient of f(x, y) = x 2 + y 3 at the given point:

'V f = ( !~, !~) = {2x, 3i) =:} 'V /(1,2) = (2, 12)

Using the Theorem on Evaluating Directional Derivatives, we get

(4 3) 8 36 44

Duf(1, 2) = 'V !(1,2) · u = (2, 12) · 5' 5 = 5 + 5 = 5 = 8.8

22. f(x,y)=x 2y3, v=i+j, P=(-2,1)

SOLUTION We normalize v to obtain a unit vector u in the direction of v:

v 1 . . 1 . 1 . u = - = - (1 + J) = -I+ -J

IIVI\ ./2 ./2 ./2

We compute the gradient of f (x, y) = x 2 y3 at the point P:

'V f = (!~, !~) = {2xy3, 3x2l) =:} 'V f(-2,1) = (-4, 12} = -4i + 12j

The directional derivative in the direction of vis therefore

( 1 1 ) 4 12 8

Duf(-2,1)='Vf(-2,l)"u=(-4i+12j)· .Jii+ .Jij =- ./2+ ./2= ./2=4J2

23. f(x,y)=x 2 y3 , v=i+j, P=(i,3)

SOLUTION We normalize v to obtain a unit vector u in the direction of v:

v I . 1 . 1 . u = - = - (1 + j) = -I+ -J

1\VI\ ./2 ./2 ./2

We compute the gradient of f(x, y) = x 2y 3 at the point P:

( 1 3 1 2) ( 3 ) . 3 . '\1 f(I ) = 2 ·- · 3 , 3 · 2 · 3 = 9,- = 91 + -j

6•3 6 6 4 4

The directional derivative in the direction v is thus


24. f(x, y) = sin(x- y), v = (1, 1), P = (}-, '6-) SOLUTION We normalize v to obtain a unit vector u in the direction v:

v I u = w = V2 (I, 1)

We compute the gradient of f(x, y) = sin(x- y) at the point P:

(af aJ) 'Vf= -,- =(cos(x-y),-cos(x-y)) ax ay \1 f(]f ]f)= (cos~. -cos~)=(~.-~)

2• 6 3 3 2 2


Duf(P) = \1 f(!!. !!.) · u = (~. -~) · ~ (1, 1) = 0 2'6 2 2 v2

25. f(x,y)=tan- 1(xy), v=(l,l), P=(3,4)

SOLUTION We first normalize v to obtain a unit vector u in the direction v:

v I u = M = V2 (1, I)

We compute the gradient of f(x, y) = tan- 1 (xy) at the point P = (3, 4):

(af a1) ( y x ) 1 \1 f = -,- = , = (y, x) ax ay 1 + (xy) 2 1 + (xy)2 1 + x2y2

1 I \1 !(34) = 2 2 (4, 3) = - (4, 3)

, 1 + 3 . 4 145

Therefore, the directional derivative in the direction v is

1 I 1 7 7v0, Duf(3,4) = \lf(34) ·U= 145

(4,3) · r, (1, 1) = ~(4+3) = ~ = 290 ' v2 145-v2 145-v2

26. f(x, y) = exy-y2

, v = (12, -5), P = (2, 2)

SOLUTION We first normalize v to obtain a unit vector u in the direction v:

v (12, -5) 1 u = - = = - (12, -5)

llvll J122 + (-5)2 13

2 We compute the gradient of f(x, y) = exy-y at the point P = (2, 2):

\1 f = ( :~, :~) = {.vexy-y2

, (x- 2y)exy-y2

) = exy-y2 (y, x- 2y)

\1 1(2,2) = e0 (2, -2) = (2, -2)

Therefore, the directional derivative in the direction v is thus

1 34 Duf(2, 2) = \1 1(2,2) · u = (2, -2) ·13 (12, -5) = 13

27. f(x,y)=ln(x 2 +y2), V=3i-2j, P=(l,O)

SOLUTION We normalize v to obtain a unit vector u in the direction v:

v 1

3' 2' 1 3' 2' U = - = ( I - J) = 1M ( I - J) llvll j 32+(-2)2 v13

We compute the gradient off (x, y) = In ( x 2 + y 2 ) at the point P = (1, 0):

(aj aj) ( 2x 2y ) 2 'Vf=-- = --,-- =--(x,y) ax ' ay x2 + y2 x2 + y2 x2 + y2

\1 f(l.O) = ]Z ~ 02 (1, 0) = (2, 0) = 2i

•

sECTIoN 14.5 I The Gradient and Directional Derivatives (LT SECTION 15.5) 709

The directional derivative in the direction vis thus

Duf(J, 0) = 'V fo o) · u = 2i · ~ (3i- 2j) = ~ , v 13 v 13

28. g(x, y, z) = z2 - xy2, v = (-1, 2, 2), P = (2, 1, 3)

SOLUTION We normalize v to obtain a unit vector u in the direction v:

v (-1,2,2) 1 u=-= =-(-1,2,2)

llvll J<-1)2+22+22 3

We compute the gradient of f(x, y, z) = z2 - xi at the point P = (2, 1, 3):

'Vf=(:~. :~. iz)=(-i,-2xy,2z} =* 'Vf(2,1,3)=(-1,-4,6)

The directional derivative in the direction v is thus 1 1 5

D0

f(2, 1, 3) = 'V f,(2 1 3) · u = (-1, -4, 6) ·- (-1, 2, 2) = -(1- 8 + 12) = -3 ' , 3 3

29. g(x, y, z) = xe-YZ, v = (1, 1, 1), P = (1, 2, 0)

SOLUTION We first compute a unit vector u in the direction v:

We find the gradient of f(x, y, z) = xe-yz at the point P = (1, 2, 0):

'V f = (:~, :~, :~) = (e-yz, -xze-yz, -xye-yz) = e-yz (1, -xz, -xy)

'V !(1.2,0) = e0 (1, 0, -2) = (1, 0, -2)

The directional derivative in the direction vis thus 1 1 1

Duf(J, 2, 0) = 'V !(1,2,0) · u = (1, 0, -2) · v'3 (1, 1, 1) = v'3(1 + 0- 2) =- v'3

30. g(x, y, z) = x Jn(y + z), v = 2i- j + k, P = (2, e, e)

SOLUTION We first find a unit vector u in the direction v:

v 2i- j + k 1 • • u = - = = - (21 - J + k)

llvll J22+(- 1)2+12 ./6

We compute the gradient of f(x, y, z) = x ln(y + z) at the point P = (2, e, e):

'V f =(at, aj, at)= (ln(y + z), _x_, _x_) ax ay az y + z y + z

'V f(2,e,e) =(In 2e, ;e, ;e)= (1n 2e, e-1

, e-1} =(In 2e)i + e-

1j + e-

1k


31. Find the directional derivative of f(x, y) = x 2 + 4yl at P = (3, 2) in the direction pointing to the origin.

-> SOLUTION The direction vector is v =PO= (-3, -2). A unit vector u in the direction vis obtained by normalizing v.

That is,

u = _.!._ = (-3· - 2) =-=..!.._ (3, 2) II vii .)32 + 22 v'l3

I


Hence,

Substituting in (I) we get

38. Let T(x, y) be the temperature at location (x, y). Assume that VT = (y- 4, x + 2y). Let c(t) = (t 2 , t) be a path the plane. Find the values oft such that

d - T(c(t)) = 0 dt

SOLUTION By the Chain Rule for Paths we have

d 1 -T (c(t)) = VTc(tJ · c (t) dt

We compute the gradient vector VT for x = t 2 andy= t:

VT = (r - 4, t 2 + 2t)

Also c1(t) = (2t, 1). Substituting in (I) gives

:t T (c(t)) = (r- 4, t2 + 2t) · (2t, I)= (t- 4) · 2t + (r 2 + 2t). I= 3t2 - 6t

We are asked to find the values oft such that

We solve to obtain

d -T (c(t)) = 3t2 - 6t = 0 dt

3t2

- 6t = 3t (t - 2) = o =? t 1 = 0, t2 = 2

39. Find a vector normal to the surface x2 + y 2 - z2 = 6 at P = (3, I, 2).

SOLUTION The gradient V f p is normal to the level curve f (x, y, z) = x 2 + y2 - z2 = 6 at P. We compute this

fx(x, y, z) = 2x

/y(x, y, z) = 2y =? V fp = V 1(3.1,2) = (6, 2, -4)

fz(x, y, z) = -2z

The vector (6, 2, -4) is normal to the surface x2 + y 2 - z2 = 6 at P.

40. Find a vector normal to the surface 3z 3 + x 2 y - y 2 x = I at P = (I, -I, I).

SOLUTION The gradient is normal to the level surfaces, that is V fp is normal to the level surface f(x, y, z) = 3z3 x

2y- y

2x = I. We compute the gradient vector at P = (I, -I, 1):

(at at at) ( 2 2 2) Vf = - - - = 2xy- }' x - 2vx 9z ax' ay' az ' . , Vfp = (-3,3,9)

41. Find the two points on the ellipsoid

where the tangent plane is normal to v = (I, 1, - 2).

x2 y2 SOLUTION The gradient V fp is normal to the level surface f(x, y, z) =

4 + g + z2 = 1. Ifv =(I, 1, -2) is

normal, then V f p and v are parallel, that is, V f p = kv for some constant k. This yields the equation

x 2v Vfp = (2, g-.2z) =k(l, 1,-2)

sEcT 1 0 N 14.5 I The Gradient and Directional Derivatives (LT SECTION 15.5) 713

Thus x = 2k, y = 9kj2, and z = -k. To determine k, substitute in the equation of the ellipsoid:

x2 y2 2 (2k )2 (9k /2)2 2 4 + 9 + z = -4- + -9- + (-k) = l

This yields k2 + ~k2 + k2 = I or k = ±2/ v?i. The two points are

9 ( 4 9 2 ) (x,y,z)=(2k,-k,-k)=± 11";' 11";'- 11"; 2 vl7 vl7 vl7

In Exercises 42-45, find an equation of the tangent plane to the surface at the given point.

42. x2 + 3y2 + 4z2 = 20, P = (2, 2, l)

SOLUTION The equation of the tangent plane is

'V fp · (x- 2, y- 2, z- l) = 0

We compute the gradient of f(x, y, z) = x 2 + 3y2 + 4z2 at P = (2, 2, 1):

At the point P we have

'Vf=jat, at, at)=(2x,6y,8z) \ax ay az

'Vfp = (2·2,6·2,8·1) = (4, 12,8)

Substituting in (1) we obtain the following equation of the tangent plane:

or

(4, 12, 8) · (x- 2, y- 2, z- I) = 0

4(x- 2) + l2(y- 2) + 8(z- l) = 0

x - 2 + 3(y - 2) + 2(z - 1) = 0

x + 3y + 2z = 10

SOLUTION The equation of the tangent plane at P is

'V f p · (x - 2, y - l, z - l) = 0

We compute the gradient of f(x, y, z) = xz + 2x2y + iz3 at the point P = (2, I, 1):

'Vf= -,-,- = z+4xy,2x2 +2yz3,x+3iz2 (

at at at) ( ) ax ay az

At the point P we have

'V fp = (9, 10, 5)

Substituting in ( l) we obtain the following equation of the tangent plane:

or

(9, 10, 5) · (x- 2, y- I, z- 1) = 0

9(x- 2) + IO(y- I)+ 5(z- 1) = 0

9x +lOy+ 5z = 33

44. x2 + z2ey-x = 13, P = ( 2, 3, ~)

SOLUTION We compute the gradient of f(x, y, z) = x 2 + z2ey-x at the point P = ( 2, 3, }e):

'Vf = - - - = 2x- z2eY x z eY x 2zeY x (at at at) ( _ 2 - - ) ax' ay' az ' '

•

(I)

(1)

1

714 C H A PH R 14 I DIFFERENTIATION IN SEVERAL VARIABLES (LT CHAPTER 15)

At the point P = ( 2, 3, Je) we have

( 9 9 3 ) v fp = 4-;. e,;. e, 2. ve. e = (-5, 9, 6v'e)

The equation of the tangent plane at P is

'\1 j p · (X - 2, y - 3, Z- Je) = 0

That is,

-5(x- 2) + 9(y- 3) + 6ve (z- Je) = o or

-5x + 9y + 6vez = 35

45. In[! +4x2 + 9y4]- O.lz2 = 0, P = (3, I, 6.1876)

SOLUTION The equation of the tangent plane at P is

V fp · (x- 3, y- 1, z- 6.1876) = 0

We compute the gradient of f(x, y, z) = In(! + 4x 2 + 9y4 ) - O.lz2 at the point P:

v f =(a! aj aj) _ ( 8x 36y3 _

0 2z)

ax' ay' az - 1 + 4x2 + 9y4' 1 + 4x2 + 9y4' .

At the point P = (3, 1, 6.1876) we have

( 24 36 ) v fp = '-, -1.2375 = (0.5217, 0.7826, -1.2375) I+ 36 + 9 46

We substitute in (I) to obtain the following equation of the tangent plane:

0.5217(x - 3) + 0.7826(y - I) - 1.2375(z - 6.1876) = 0

or

0.5217x + 0.7826y- 1.2375z = -5.309

46. Verify what is clear from Figure 15: Every tangent plane to the cone x 2 + y 2 - z2 = 0 passes through the origin.

/

y

X

SOLUTION The equation of the tangent plane to the surface f (x. y, z) = x 2 + y 2 - z2 = 0 at the point P = (xo, Yo, on the surface is

V fp · (x- xo, y- Yo. z- zo)

We compute the gradient of f(x, y, z) = x 2 + y 2 - z2 at P:

(af aJ aJ) Vf = - - - = (2x 2v -2z) ax' ay' az ' -'

Hence,

v fp = (2xo, 2yo, -2zo) .... _____ -.--- ---------

sEcT 1 oN 14.5 I The Gradient and Directional Derivatives (LT SECTION 15.5) 715

Substituting in (I) we obtain the following equation of the tangent plane:

(2xo, 2yo, -2zo) · (x - xo, y -Yo. z - zo) = 0

xo(x - xo) + Yo(Y -Yo) - zo(z - zo) = 0 2 2 2

x0x + YOY - ZQZ = x0 + Yo - z0

Since P = (xo, Yo, zo) is on the surface, we have x5 + Y5 - z6 = 0. The equation of the tangent plane is thus

XQX + YOY - zoz = 0

This plane passes through the origin.

47. CAS Use a computer algebra system to produce a contour plot of f(x, y) = x 2 - 3xy + y- y2 together with its gradient vector field on the domain [ -4, 4] x [ -4, 4].

SOLUTION

48. Find a function f(x, y, z) such that V' f is the constant vector (I, 3, 1).

SOLUTION The gradient of j(x, y, z) must satisfy the equality

Equating corresponding components gives

(af af af)

V' f = ax ' ay ' az = ( 1' 3' 1)

af = 1 ax

af = 3 ay

af = 1 az

One of the functions that satisfies these equalities is

f(x,y,z)=x+3y+z

49. Find a function j(x, y, z) such that V' f = (2x, 1, 2).

SOLUTION The following equality must hold:

Equating corresponding components gives

V'f=(af. af. aj)=(2x,l,2) ax ay az

aj = 2x ax

af =I ay

af = 2 az

One of the functions that satisfies these equalities is f (x, y, z) = x 2 + y + 2z.


50. Find a function f(x, y, z) such that \7 f = (x, y 2 , z3 ).

SOLUTION The following equality must hold:

That is,

\7-----x· (at at at) ( 2 3)

f - ax' ay' az - 'y 'z

at -=x ax

at 2 -=y ay

at 3 -=z az

One of the functions that satisfies these equalities is

I I I f(x Y z) = -x2 + -)'3 + -74

' ' 2 3 4 ~

51. Find a function f(x, y, z) such that \7 f = {z, 2y, x).

SOLUTION f(x, y, z) = xz + y 2 is a good choice.

52. Find a function f(x, y) such that \7 f = {y, x).

SOLUTION We must find a function f (x, y) such that

(at at) 'Vf = -,- = {y.x) ax ay

That is,

at ax= y,

at -=x ay

We integrate the first equation with respect to x. Since y is treated as a constant, the constant of integration is a function ofy. We get

f(x, y) = j ydx = yx + g(y)

We differentiate f with respect toy and substitute in the second equation. This gives

at a , - =- (yx + g(y)) = x + g (y) ay ay

Hence,

x+g'(y)=x =} g1(y)=O =} g(y)=C

Substituting in (I) gives

f(x, y) = yx + C

One of the solutions is f(x, y) = yx (obtained for C = 0).

53. Show that there does not exist a function f (x, y) such that \7 f = (y2 , x ). Hint: Use Clairaut's Theorem fxy = /yx·

SOLUTION Suppose that for some differentiable function f (x, y ),

That is, fx = y2 and /y = x. Therefore,

a a 2 fxv = -fx = -y =2y and · ay ay

a a /yx = ~ fv = ~X = I

ax . ax

Since fxy and /yx are both continuous, they must be equal by Clairaut's Theorem. Since fxy f= /yx we conclude such a function f does not exist.

sEcT 1 oN 14.5 I The Gradient and Directional Derivatives (LT SECTION 15.5) 719

(c) Note that f is not defined for y = 0. For x = 0, the level curve of fis they-axis, and the gradient vector is ( t, 0), which is perpendicular to the y-axis. For y i= 0 and x i= 0, the level curves off are the curves where f (x, y) is con.stant. That is,

tan- 1 ~ =k y

X -=tank y

I y = tankx

(fork i= 0)

We conclude that the lines y = mx, m i= 0, are level curves for f. (d) By part (c), the level curve through P = (xo, yo) is the line y = ..l:'Qx. This line has a direction vector (1, ..l:'Q). The

XQ XQ

gradient at Pis, by part (a), V' fp = h (yo, -xo). We verify that the two vectors are orthogonal: xa+Yo

( YO) ( YO) I I ( XOYO) 1,- ·'Vfp= 1,- ·--(yo.-xo)=-- yo-- =0 xo xo x 2 + y 2 x 2 + y2 xo 0 0 0 0

Since the dot products is zero, the two vectors are orthogonal as expected (Theorem 6).

59. ~ Suppose that the intersection of two surfaces F (x, y, z) = 0 and G (x, y, z) = 0 is a curve C, and let P be a point on C. Explain why the vector v = V' F p x V' G p is a direction vector for the tangent line to C at P.

SOLUTION The gradient V' Fp is orthogonal to all the curves in the level surface F(x, y, z) = 0 passing through P. Similar! y, V' G p is orthogonal to all the curves in the level surface G (x, y, z) = 0 passing through P. Therefore, both V' F p and V' G p are orthogonal to the intersection curve C at P, hence the cross product V' F p x V' G p is parallel to the tangent line to C at P.

60. Let C be the curve of intersection of the spheres x 2 + y 2 + z2 = 3 and (x - 2)2 + (y - 2)2 + z2 = 3. Use the result of Exercise 59 to find parametric equations of the tangent line to C at P = ( l, 1, l).

SOLUTION The parametric equations of the tangent line to C at P = ( 1, 1, 1) are

x=1+at, y=l+bt, z=l+ct

where v = (a, b, c) is a direction vector for the line. By Exercise 59 v may be chosen as the following cross product:

v = 'V'Fp X 'V'Gp

where F(x, y, z) = x 2 + y2 + z2 and G(x, y, z) = (x- 2)2 + (y- 2)2 + z2. We compute V' Fp and V'G p:

Hence,

Fx(x, y, z) = 2x

Fy(x,y,z)=2y =} V'Fp=(2·1,2·1,2·1)=(2,2,2)

Fz(X, y, z) = 2z

Gx(x, y, z) = 2(x- 2)

Gy(x, y, z) = 2(y- 2) =} V'G p = (2(1- 2), 2(1- 2), 2 · 1) = (-2, -2, 2)

Gz(x, y, z) = 2z

j k v=(2,2,2)x(-2,-2,2)= 2 2 2 =(4+4)i-(4+4)j+(-4+4)k=8i-8j=(8,-8,0)

-2 -2 2

(l)

(2)

Therefore, v = (a, b, c) = (8, -8, 0), yielding a = 8, b = -8, c = 0. Substituting in (l) gives the following equations of the tangent line: x = I+ 8t, y = 1- 8t, z = 1.

61. Let C be the curve obtained by intersecting the two surfaces x 3 + 2xy + yz = 7 and 3x2 - yz = 1. Find the parametric equations of the tangent line to Cat P = (l, 2, 1).

SOLUTION The parametric equations of the tangent line to C at P = ( 1, 2, 1) are

x = 1 +at, y = 2 + bt, z = l + ct (1)

where v = (a, b, c) is a direction vector for the line. By Exercise 59, v may be chosen as the cross product:

v='V'Fpx'V'Gp (2)


where F(x, y, z) = x 3 + 2xy + yz and G(x, y, z) = 3x2 - yz. We compute the gradient vectors:

Hence,

Fx(x, y, z) = 3x2 + 2y

Fy(x,y,z)=2x+z

Fz(X, y, z) = y

Gx(x, y, z) = 6x

Fx(l. 2, I)= 7

=} Fy(1,2, I)= 3 =} VFp = (7,3,2)

Fz(l, 2, 1) = 2

Gx(l. 2, I)= 6

Gy(x, y, z) = -z =} Gy(l. 2, I)= -I =} VGp = (6, -I, -2)

Gz(x, y, z) = -y Gz(l, 2, I)= -2

j k V = (7, 3, 2) X (6, -], -2) = 7 3 2 = -4i + 26j- 25k = (-4, 26, -25)

6 -1 -2

Therefore, v = (a, b, c) = (-4, 26, -25), so we obtain

a = -4, b = 26, c = -25.

Substituting in (1) gives the following parametric equations of the tangent line:

x=l-4t, y=2+26t, z=l-25t.

62. Verify the linearity relations for gradients:

(a) V(f+g)=Vt+Vg

(b) V(ct) = cV t SOLUTION

(a) We use the linearity relations for partial derivative to write

V(f +g)= (U + g)x, (f + g)y, (f + g)z) = (fx + gx, ty + gy. tz + gz)

= (tx. /y, tz) + (gx. gy. gz) = V t + Vg

(b) We use the linearity properties of partial derivatives to write

63. Prove the Chain Rule for Gradients (Theorem 1).

SOLUTION We must show that if F (t) is a differentiable function oft and t (x, y, z) is differentiable, then

V F (f(x, y, z)) = F' (f(x, y, z)) V t

Using the Chain Rule for partial derivatives we get

( a a a ) V F (f(x, y, z)) = ax F (f(x, y, z)), ay F (f(x, y, z)), az F (f(x, y, z))

=(dF. at. dF. at. dF. at)= dF(at. at. at)=F'(f(x,y,z))VF dt ax dt ay dt az dt ax ay az

64. Prove the Product Rule for Gradients (Theorem I).

SOLUTION We must show that if t(x, y, z) and g(x, y, z) are differentiable, then

V(fg) = tVg + gV t

Using the Product Rule for partial derivatives we get

V(fg) = ((fg)x, (fg)y, (fg)z) = (/xg + tgx. tyg + tgy. tzg + tgz)

= (/xg, tyg, tzg) + (tgx. tgy, tgz) = (/x. ty. tz)g + t (gx. gy. gz) = gV t + tVg


SOLUTION For a function f(x, y) where x = x(r, s) andy = y(r, s), the Chain Rule states that the partial ¥, is as given in (b). That is,

ajax aj ay --+-ax ar ay ar

5. Suppose that x, y, z are functions of the independent variables u, v, w. Which of the following tenns Chain Rule expression for af/aw?

(a) ajax av av

(b) aj aw aw ax

SOLUTION By the Chain Rule, the derivative Sf is

Therefore (c) is the only correct answer.

aJ aj ax af ay aj az -=--+--+-aw axaw ayaw azaw

aJ az (c)-

az aw

6. With notation as in the previous question, does axjav appear in the Chain Rule expression for af/au?

SOLUTION The Chain Rule expression for M; is

aj aj ax aj ay aj az -=--+--+-au ax au ay au az au

The derivative ~~ does not appear in differentiating f with respect to the independent variable u.

Exercises 1. Let f(x, y, z) = x 2y 3 + z4 and x = s2, y = st2, and z = s2t.

. . . af af af (a) Calculate the pnmary denvat1ves -, -, -.

ax ay az ax ay az

(b) Calculate-,-,-. as as as

(c) Compute aj using the Chain Rule: as

aj af ax aj ay aJ az -=--+--+-as ax as ay as az as

Express the answer in terms of the independent variables s, t.

SOLUTION

(a) The primary derivatives of f (x, y, z) = x 2 y3 + z4 are

af 3 - = 2xy, ax

aj = 3x2y2 ay ,

(b) The partial derivatives of x, y, and z with respect to s are

ax -=2s as '

ay 2 - =t as '

aJ 3 - =4z az

az - = 2st as

(c) We use the Chain Rule and the partial derivatives computed in parts (a) and (b) to find the following derivative:

aj aJ ax aj ay aj az 3 2 2 2 3 3 2 2 2 3 - = -- + -- + -- = 2xy · 2s + 3x y t + 4z · 2st = 4xy s + 3x y t + 8z st as ax as ay as az as

To express the answer in terms of the independent variables s, t we substitute x = s2, y = s t 2, z = s2 t. This gives

aj = 4s2

(st 2)3 s + 3(s2

)2

(st 2/ t 2 + 8(s2t)3 st = 4s6t6 + 3s6t6 + 8s 7 t4 = 7s6t6 + 8s 7 t4 .

as

2. Let f(x, y) = x cos(y) and x = u2 + v2 andy= u- v. . . . af aJ

(a) Calculate the pnmary denvatJves -, -. ax ay

(b) Use the Chain Rule to calculate aflav. Leave the answer in terms of both the dependent and the independent (c) Determine (x, y) for (u, v) = (2, l) and evaluate af/av at (u, v) = (2, l).

SOLUTION

(a) The primary derivatives of f(x, y) = x cos(y) are

af - = cos(y), ax

af . - = -xsm(y). ay

SECT I 0 N 14.6 I The Chain Rule (LT SECTION 15.6) 727

(b) By the Chain Rule, we have

We compute the partial derivatives ~~ and *: at at ax at ay -=--+-av ax av ay av

ax - =2v, av

ay - = -1. av

Substituting these derivatives and the primary derivatives computed in part (a) in the Chain Rule (1) gives

at = cos(y). 2v- x sin(y) · (-I)= 2v cos(y) + x sin(y) av

(c) We substitute u = 2, v =I in x = u2 + v2 andy= u- v, and determine (x, y) for (u, v) = (2, 1). This gives

x=22 +1 2 =5, y=2-l=l.

To find %f at (u, v) = (2, I) we substitute u = 2, v = I, x = 5, andy= I in ~~ computed in part (b). We obtain

at I = 2 . I cos 1 + 5 sin 1 = 2 cos 1 + 5 sin 1. av (u. v)=(2, l)

(I)

In Exercises 3-10, use the Chain Rule to calculate the partial derivatives. Express the answer in terms otthe independent

variables.

at at 3. - -· t<x y z) = xy + z2 x = s2 y = 2rs z = r2

as' ar' ' ' ' ' ' SOLUTION We perform the following steps:

Step 1. Compute the primary derivatives. The primary derivatives oft (x, y, z) = xy + z2 are

at ax= y,

Step 2. Apply the Chain Rule. By the Chain Rule,

at -=X, ay

at= 2z az

at at ax at ay at az -=-·-+-·-+-·-as ax as ay as az as

at at ax at ay at az -=-·-+-·-+-·-ar ax ar ay ar az ar

We compute the partial derivatives of x, y, z with respect to sand r:

ax ay az - =2s - =2r --0 as , as , as- .

ax ay az = 2r. -=0 - =2s, ar , ar ar

Substituting these derivatives and the primary derivatives computed in step I in ( 1) and (2), we get

at - = y · 2s + x · 2r + 2z · 0 = 2ys + 2xr as

at - = y · 0 + x · 2s + 2z · 2r = 2xs + 4zr ar

(l)

(2)

Step 3. Express the answer in terms of r and s. We substitute x = s2, y = 2r s, and z = r2 in ~~ and ~~ in step 2, to obtain

at - = 2rs · 2s + s2 · 2r = 4rs 2 + 2rs2 = 6rs2

. as

at - = 2s2 · s + 4r2 · r = 2s3 + 4r3

. ar

ar at 4. -'--, -; t(x, y, z) = xy + z2 , x = r + s- 2t, y =3rt, z = s2

ar at SOLUTION We use the following steps:

Step 1. Compute the primary derivatives. The primary derivatives oft (x, y, z) = xy + z2 are

at ax= y,

at -=X ay .

• at= 2z az



at = at ax + at ay + at az = Y ax + x ay + 2z az ar ax ar ay ar az ar ar ar ar

at = at ax + at ay + at az = ax + x ay + 2

z az at ax at ay at az at Y at at at

We compute the partial derivatives of x, y with respect tor and t:

ax ay = 3t, az = 0 - = 1, ar ar ar ax ay az =O -=-2 - =3r, at • at at

Substituting in (1) and (2), we get

at - = y + 3tx + 2z · 0 = y + 3xt ar

at - = y · ( -2) + x · 3r + 2z · 0 = -2y + 3xr at

Step 3. Express the answer in terms of rand t. We substitute x = r + s - 2t, y = 3rt, and z = s 2 in ~ and¥, in step 2. This gives

aj = 3rt+ 3(r + s - 2t)t =3rt+ 3rt+ 3st - 6t 2 = 6rt + 3st - 6t2 ar

aj = -2 ·3rt+ 3(r + s- 2t)r = -6rt + 3r2 + 3sr- 6tr = -12rt + 3rs + 3r2 at

ag ag 5. -;-, -; g(x, y) = cos(x- y), x = 3u- 5v, y = -7u + 15v

au av

SOLUTION We use the following steps:

Step l. Compute the primary derivatives. The primary derivatives of g(x, y) = cos(x- y) are:

ag . - = -sm(x- y), ilx

ag . - = sm(x- y) ay


~ ~~ ~~ . ~ . ~ - = -- + -- = -sm(x- y)- +sm(x- y)au ax au ay au au au

ag ag ax ag ay . ax . ay - = -- + -- = -sm(x- y)- +sm(x -y)av ax av ay av av av

We compute the partial derivatives of x, y with respect to u and v:

substituting in the expressions above we have:

ax -=3 au '

ay -=-7 au ,

ax -=-5 av

ay = 15 av

ag =- sin(x- y)(3) + sin(x- y)( -7) = -10sin(x- y) au

ag = -sin(x- y)(-5) + sin(x- y)(l5) = 20sin(x- y) av

Step 3. Express the answer in terms of u and v. We substitute x = 3u - 5v andy = -7u + 15v in agjau and found in step 2. This gives:

ag . o - = -IOsm(lOu- 2 v) au

ag - = 20 sin(IOu - 20v) av


8. of; f(x, y) = X2 + y2, X = eu+v' y = U + V au


Step 1. Compute the primary derivatives. The primary derivatives of f(x, y) = x 2 + y 2 are

of - =2x, ox

of - =2y ay


We compute ax and £I: au au

Hence,

of of ax of ay ax ay - = --+-- =2x-+2you ax au oy au au au

ax u+v au= e '

ay = 1 au

of - = 2xeu+v + 2y au

Step 3. Express the answer in terms of u and v. We substitute x = eu+v andy = u + v in (1) to obtain

of - = 2eu+v eu+v + 2(u + v) = 2(e2(u+v) + u + v) au

9. !..!!__; h(x, y) = ::, x = t112, y = 1f12 o12 y


Step 1. Compute the primary derivatives. The primary derivatives of h(x, y) = ~ are

ah ah X

ax y


oh oh OX oh oy 1 OX X oy -=--+--=-----0(2 ax o12 ay o12 y o12 y2 a12

We compute the partial derivatives of x andy with respect to 12:

Hence,

ay 2 --1 o12 - I

ah 11 x 2 -----1 o12 - Y y2 I

Step 3. Express the answer in terms of 11 and 12. We substitute x = 1112, y = 1f12 in ~ computed in step 2, to

ah t1 11 t2 · 1[ 1 1 -=----=---=0 o12 1[12 Ct[ 12 )2 t1t2 1112

Remark: Notice that h(x(tl, 12), y(11, 12))= h(11, 12) = ~ = t· h(11, 12) is independent of t2, hence#,',= 0 t 1 t2 I 2

obtained in our computations).

10. :;; f(x, y, z) = xy- z2,x = rcos8, y = cos2 e, z = r


Step 1. Compute the primary derivatives. The primary derivatives off (x, y, z) = xy - z2 are

of of ax= y, -=X,

ay

of - =-2z oz

SECT I 0 N 14.6 I The Chain Rule (LT SECTION 15.6) 731


of of ax of ay of az ax ay az ae = ax ae + ay ae + "ih ae = Y ae + x ae - 2z ae

We compute the partial derivatives of x, y, and z with respect toe:

:; = -rsine, :~ = -2cosesine = -sin2e, oz = 0 ae

Step 3. Express the answer in terms of e and r. We substitute x = r cos e, y = cos2 e, and z = r in ( 1) to obtain

of = -r cos2 e sine - r cos e sin 2e = -r . ~cos e sin 2e - r cos e sin 2e = -~cos e sin 2e ae 2 2

In Exercises /1-16, use the Chain Rule to evaluate the partial derivative at the point specified.

11. ofjou and ofjov at (u, v) = (-1, -1), where f(x, y, z) = x 3 + yz2 , x = u2 + v, y = u + v2 , z = uv.

SOLUTION The primary derivatives of f(x, y, z) = x 3 + yz2 are

By the Chain Rule we have

of 2 ay = z ,

of - =2yz oz

of of ax of ay of az 2 ax 2 ay az - = --+ --+-- =3x- +z- +2yz-au ax au ay au {Jz au au au au

of of ax of ay of az 2 ax 2 ay az - = --+--+-- =3x -+z -+2yzav ax av oy av {Jz av av av av

We compute the partial derivatives of x, y, and z with respect to u and v:

Substituting in ( 1) and (2) we get

ax - =2u, au

ax -=1, av

ay - = 1 au

ay - =2v av .

oz - =v au

oz -=U av

of = 6x 2u + z2 + 2vzv au .

of = 3x 2 + 2vz2 + 2yzu av

We determine (x, y, z) for (u, v) = ( -1, -1):

x=(-1)2 -1=0, y=-1+(-1)2 =0, z=(-1)·(-1)=1.

Finally, we substitute (x, y, z) = (0, 0, 1) and (u, v) = ( -1, -1) in (3), ( 4) to obtain the following derivatives:

of I = 6. o2 . (-1) + 12 + 2. o. 1. (-1) = 1 au {u,v)={-1,-1)

of I = 3 . o2 + 2 . c -I) . 12 + 2. o. 1 . ( -1) = -2 ov (u,v)=(-1,-1)

12. ofjos at (r, s) = (1, 0), where f(X, y) = ln(xy), X= 3r + 2s, y = 5r + 3s.

SOLUTION The primary derivatives of f(x, y) = ln(xy) are


of y X

OX xy X oy xy y

of of ax of ay 1 ax 1 ay -=--+--=--+-as OX OS oy OS X OS y OS

(1)

(2)

(3)

(4)

(I)

SECTION 14.6 I TheChainRule (LTSECTION 15.6) 733

We now determine (x, y) for s = 4:

Substituting (x, y) = (17, -7) and s = 4 in (2) gives the following derivative:

ag I - =4·17·4-4·7=244 as s=4

15. agjau at (u, v) = (0, 1), where g(x, y) = x 2 - y2 , x = e" cos v, y = e" sin v.

SOLUTION The primary derivatives of g(x, y) = x 2 - y2 are


We find ax and ay · au au.

Substituting in ( 1) gives

ag - =2x ax '

ag - =-2y ay

ag ag ax ag ay ax ay - = -·-+-·- =2x- -2yau ax au ay au au au

ax - = eu COSV, au

ay - = eu sin v au

ag = 2xe" cos v- 2ye" sin v = 2eu(x cos v- y sin v) au

We determine (x, y) for (u, v) = (0, 1):

x = e0 cos 1 = cos 1, y = e0 sin 1 = sin 1

(1)

(2)

Finally, we substitute (u, v) = (0, 1) and (x, y) = (cos 1, sin 1) in (2) and use the identity cos2 a - sin2 a = cos 2a, to obtain the following derivative:

ag I = 2e0 (cos21- sin2 1) = 2 · cos2 · 1 = 2cos2 au (u,v)=(O,I)

ah 16. -at (q, r) = (3, 2), where h(u, v) = uev, u = q3, v = qr2 .

aq

SOLUTION We first find the primary derivatives of h (u, v) = uev:

ah v -=e' au

ah v - =ue av

By the Chain Rule, we have

ah = ah. au+ ah. av =evau +uevav =ev(au +uav) aq au aq av aq aq aq aq aq

We compute au and av · aq aq ·

Substituting in ( 1) gives

We now determine (u, v) for (q, r) = (3, 2):

au 2 - =3q, aq

av 2 -=r aq

u=33 =27, v=3·22 =12

Substituting in (2) gives the following derivative:

(I)

(2)

SECTION 14.6 I The Chain Rule (LTSECTION 15.6) 739

The partial derivatives ofF are


or esfr r2esfr -at 2r + E..esfr

r2 2r3 + stesfr

at 2r + E..esfr 2r3 + stesfr ; st r2 = 2re-s r +-or esfr r2esfr r2

aw 29. ay,

1 1 2 2 + 2 2 =1at(x,y,w)=(l,1,1)

w +x w + y

SOLUTION Using the formula obtained by implicit differentiation (Eq. (7)), we have

ow _ Fy

oy Fw

We find the partial derivatives of F(x, y, w) = ~ + ~- 1: w +x w +y

2w

We substitute in (1) to obtain

aw

ay -2w 2w (w2+x2)2 (w2+y2)2

y(w2 + x2/ -y(w2 + x2/

w(w2 + y2)2 + w(w2 + x2)2 w( (w2 + y2)2 + (w2 + x2)2)

30. oU joT and oT joU, (TV- V) 2 In(W- UV) = 1 at (T, U, V, W) = (1, 1, 2, 4)

SOLUTION Using the formulas obtained by implicit differentiation (Eq. (7)) we have,

au FT aT

aT - Fu • au

Fu

FT

We compute the partial derivatives of F(T, U, V, W) = (TU- V) 2 In(W- UV)- 1:

FT = 2U(TU- V) In(W- UV)

2 -V Fu = 2T(TU- V) In(W- UV) + (TU- V) · _W ___ U_V_

( V(TU-V))

= (TU- V) 2Tln(W- UV)- W _ UV

At the point (T, U, V, W) = (1, 1, 2, 4) we have

FT = 2(1- 2)ln(4- 2) = -2ln2

( 2(1-2)) Fu=(1-2) 2ln(4-2)-

4_

2 =(-2ln2-1)=-1-2ln2

Substituting in (1) we obtain

au I - = aT (1, 1,2,4)

2ln2

1 + 21n2' aT I - = au (1.1,2,4)

I +21n2

21n2

(1)

(1)

31. Let r = (x, y, z) and er = r /llrll. Show that if a function f (x, y, z) = F (r) depends only on the distance from the origin r = llrll = Jx 2 + y 2 + z2, then

Y' f = F' (r)er


36. For all x > 0, there is a unique value y = r(x) that solves the equation y3 + 4xy = 16.

(a) Show that dyldx = -4yl(3i + 4x).

(b) Let g(x) = f(x, r(x)), where f(x, y) is a function satisfying

fx(I. 2) = 8, fy(I, 2) = 10

Use the Chain Rule to calculate g1(!). Note that r(l) = 2 because (x, y) = (1, 2) satisfies y3 + 4xy = 16.

SOLUTION

(a) Using implicit differentiation we see:

2 dy dy 3y -+4x-+4y=0

dx dx

dy 2 -(3y + 4x) = -4y dx

dy -4y

dx 3y2+4x

(b) Note that r 1(1) =- ~(2) =-~Therefore, 3(2) + 4(1) 2

g'(I) = txci. 2) + ~yc1. 2). r'o) = 8 + 10 ( -D = 3

37. The pressure P, volume V, and temperature T of a van der Waals gas with n molecules (n constant) are related the equation

( an2) P + \f2 (V- nb) = nRT

where a. b. and R are constant. Calculate apIa T and a vI a p.

SOLUTION Let F be the following function:

By Eq. (7),

( an2) F(P, V, T) = P + \f2 (V- nb)- nRT

aP aT

We compute the partial derivatives ofF:


aF - = V -nb aP aF -=-nR aT

aF ( an2

) 2an3b an

2 - = -2an2v-3(V- nb) + P + - = P + -- - -av vz v3 vz

ap -nR nR aT=- V- nb V -nb

av

aP V -nb

38. When x, y, and z are related by an equation F(x, y, z) = 0, we sometimes write (azlax)y in place . indicate that in the differentiation, z is treated as a function of x withy held constant (and similarly for the

(a) Use Eq. (7) to prove the cyclic relation


We obtain the following equations for the critical points (a, b):

{a= 2b

a= b3

Equating the two equations, we get

2b =b3

b3 - 2b = b(b2 - 2) = 0 ::::} l :~ : ~

b3 = -../2

Since a= 2b, we have a, = 0, az = 2../2, a3 = -2../2. The critical points are thus

(b) Referring to Figure 14, we see that P1 = (0, 0) is a saddle point and Pz = (2../2, ..ti). P3 = ( -2../2, -JZ) local minima. The absolute minimum value off is -4.

2. Find the critical points of the functions

f(x, y) = x 2 + 2y2 - 4y + 6x, g(x, y) = x 2 - 12xy + y

Use the Second Derivative Test to determine the local minimum, local maximum, and saddle points. Match j(x, y) g(x, y) with their graphs in Figure 17.

X

X

(A) (B)

FIGURE 17

SOLUTION

Step 1. Find the critical points. We set the first partial derivatives equal to zero and solve:

The critical point is ( -3, 1).

Th .. I . . ( 1 1) e cntica pomt IS 12, 72 .

fx = 2x +6 = 0

/y = 4y- 4

gx = 2x- 12y = 0

gy=-12x+1=0

X= -3

y=1

1 y = 72

1 x=-

12

Step 2. Compute the Discriminant. We compute the second-order partial derivatives:

fxx = 2

/yy=4

fxy = 0

The discriminant is D(x, y) = fxx /yy- J}y = 2 · 4-02 = 8.

gxx = 2

gyy = 0

gxy=-12

The discriminant is D(x, y) = gxxgyy - g;Y = 2 · 0- 144 = -144.

sEcT I 0 N 14.7 I Optimization in Several Variables (LT SECTION 15. 7) 753

Step 3. Apply the Second Derivative Test.

For f, we have D > 0 and fxx > 0, therefore f( -3, 1) is a local minimum.

For g, we have D < 0, hence g ( f2 , jz) is a saddle point.

The graph in Figure 17(A) has a saddle point, therefore it is the graph of g(x, y ). The graph in Figure 17(B) corresponds to f(x, y), since it has a local minimum.

3. Find the critical points of

Use the contour map in Figure 18 to determine their nature (local minimum, local maximum, or saddle point).

y

I

-I

---=~~.,.----0.1

~iiiiiiiiiiiiWif--o.z

~------~----~-X -1 0

FIGURE 18 Contourmapoff(x,y)=8y4 +x2 +xy-3y2 -y3.

SOLUTION The critical points are the solutions of fx = 0 and /y = 0. That is,

fx(x, y) = 2x + y = 0

/y(x, y) = 32y3 + x- 6y- 3i = 0

The first equation gives y = -2x. We substitute in the second equation and solve for x. This gives

32( -2x)3 + x- 6( -2x)- 3( -2x)2 = 0

-256x3 + 13x - 12x2 = 0

-x(256x2 + 12x- 13) = 0

Hence x = 0 or 256x2 + 12x- 13 = 0. Solving the quadratic,

-12 ± J122- 4. 256. (-13) Xl, 2 = 512

-12 ± 116

512

13 x =

64 or

Substituting in y = - 2x gives the y-coordinates of the critical points. The critical points are thus

(0, 0), ( ~' - ~~) , (-~ ~) 4'2

4

We now use the contour map to determine the type of each critical point. The level curves through (0, 0) consist of two intersecting lines that divide the neighborhood near (0, 0) into four regions. The function is decreasing in the y direction

and increasing in the x-direction. Therefore, (0, 0) is a saddle point. The level curves near the critical points ( ~, -H) and (-!, t) are closed curves encircling the points, hence these are local minima or maxima. The graph shows that both

(13 13) d( I I) 1 1 · · 64• -TI an - 4 , 2 are oca m1mma.


SOLUTION

Step 1. Find the critical points. We set the first-order partial derivatives off (x, y) = x 3 - xy + y 3 equal to zero solve:

fx(x,y)=3x 2 -y=0

2 /y(x,y)=-x+3y =0

Equation (1) implies that y = 3x2 . Substituting in equation (2) and solving for x gives

-x + 3(3x2)2

= 0

1 -x+27x4 =x(-1 +27x3) =0 X =0, X=-

3

The y-coordinates are y = 3 · 02 = 0 and y = 3 · ( 1) 2 = 1. The critical points are thus (0, 0) and ( 1, 1).

Step 2. Compute the Discriminant. We find the second-order partials:

fxx(x, y) = 6x, /yy(x, y) = 6y, fxy(x, y) = -1

The discriminant is

2 2 D(x, y) = fxx/yy- fxy = 6x · 6y- (-1) = 36xy- I

Step 3. Apply the Second Derivative Test. We have

D(O, 0) = -I < 0

D G, D = 36 · ~ · ~ - I = 3 > 0, fxx G, D = 6 · ~ = 2 > 0

Thus, (0, 0) is a saddle point, whereas f ( 1, 1) is a local minimum.

9. f(x, y) = x 3 + 2xy- 2yl- lOx

SOLUTION

Step 1. Find the critical points. We set the first-order partial derivatives of f(x, y) = x 3 + 2xy- 2y2 - lOx equal zero and solve:

fx(x,y)=3x 2 +2y-10=0

_fy(x, y) = 2x- 4y = 0

Equation (2) implies that x = 2y. We substitute in (I) and solve for y. This gives

-!±JI-4·6·(-5) Yl.2 =

12 12 Yl =-I

We find the x-coordinates using x = 2y:

5 5 Xj = 2 · (-J) = -2, X2 = 2 · 6 = 3

The critical points are thus (-2, -1) and ( ~, t). Step 2. Compute the Discriminant. We find the second-order partials:

and

fxx(x, y) = 6x, .{yy(x, y) = -4, fxy(x, y) = 2

The discriminant is

2 2 2 D(x,y)=fxx/yy-fxy=6x·(-4)-2 =- 4x-4

•

SECT I 0 N 14.7 I Optimization in Several Variables (LT SECTION 15. 7l 757


D(-2, -1) = -24 · (-2)- 4 = 44 > 0,

fxx(-2, -1) = 6 · (-2) = -12 < 0

D (~ ~) = -24 · ~ - 4 = -44 < 0 3, 6 3

We conclude that f (-2, -1) is a local maximum and ( ~, t) is a saddle point.

10. f(x, y) = x 3y + 12x2 - 8y

SOLUTION

Step 1. Find the critical points. We set the first-order partial derivatives of f(x, y) = x 3y + 12x2 - 8y equal to zero and solve:

fx(x, y) = 3x2y + 24x = 3x(xy + 8) = 0

3 /y(x,y)=x -8=0

Equation (2) implies that x = 2. We substitute in equation (1) and solve for y to obtain

6(2y + 8) = 0 or y = -4

The critical point is (2, -4).

Step 2. Compute the Discriminant. We find the second-order partials:

fxx(x, y) = 6xy + 24, /yy = 0, fxy = 3x2

The discriminant is thus

2 4 D(x, y) = fxx/yy- fxy = -9x


Hence (2, -4) is a saddle point.

11. f(x, y) = 4x- 3x3 - 2xy2

SOLUTION

D(2, -4) = -9 · 24 < 0

(I)

(2)

Step 1. Find the critical points. We set the first-order derivatives off (x, y) = 4x - 3x3 - 2xy2 equal to zero and solve:

fx(x, y) = 4- 9x 2- 2i = 0

/y(x, y) = -4xy = 0

Equation (2) implies that x = 0 or y = 0. If x = 0, then equation (1) gives

4-2i = 0 =} i=2 =} y = h,

If y = 0, then equation (1) gives

4- 9x2 = 0 9x 2 = 4 2

=} =} X= 3'

The critical points are therefore

y = -h

2 X=--

3

(o.h). (o.-h), G·o). (-~,o) Step 2. Compute the discriminant. The second-order partials are

fxx(x, y) = -18x, /yy(x, y) = -4x, fxy = -4y

The discriminant is thus

2 2 2 2 D(x, y) = fxx/yy- fxy = -18x · (-4x)- (-4y) = 72x - 16y

•

(I)

(2)


18. f(x, y) = x In(x + y)

SOLUTION

Step 1. Find the critical points. We set the first-order partial derivatives off (x, y) = x In(x + y) equal to zero and solve:

I X fx(x, y) = In(x + y) +x · -- = ln(x + y) + -- = 0

x+y x+y

X /y(X, }') = -- = 0

x+y

The second equation implies x = 0. Substituting in the first equation gives

In y + 0 = 0 =} In y = 0 =} y = I.

We obtain the critical point (0, 1). fx and /y do not exist at the points where x + y = 0, but these points are not in the domain off, hence they are not critical points. The critical point is thus (0, 1).

Step 2. Compute the discriminant. We find the second-order derivatives:

iJ ( x ) I I · (x + y)- x · I I y fxx=-i) In(x+y)+-+ =-+ + 2 =x+y+(x+y)2

X X }' X }' (X+ y)

/yy = iJi)y C: Y) =- (x: y)2

fty - /yx - - -- - -----;;--_ _ iJ ( x ) _ I · (x + y)- x ·I y

iJx x + y (x + y)2 - (x + y)2

The discriminant is

2 x(x+2y) D(x, y) = fxxfvv- fxy =- 4

. . (x + y)


12 D(0,1)=0- 4 =-1<0

(0 + 1)

Therefore, (0, I) is a saddle point.

19. f(x, y) = Inx + 2Iny- x- 4y

SOLUTION

X +2y

(x + y)2

Step 1. Find the critical points. We set the first-order partials off (x, y) = In x + 2In y - x - 4 y equal to zero and solve:

I fx(x,y) =- -1 =0,

2 /y(x,y)=--4=0

y X

The first equation gives x = I, and the second equation gives y = ~. We obtain the critical point ( 1, ~). Notice that fx

and /y do not exist if x = 0 or y = 0, respectively, but these are not critical points since they are not in the domain off

The critical point is thus (I, ~).

Step 2. Compute the discriminant. We find the second-order partials:

The discriminant is

I fxx(x, y) = -2,

X

2 /yy(x, y) = -2,

y /xy(X, y) = 0

2 2 D(x, y) = fxx/yy- fxy = 22

X y


We conclude that f ( 1, ~) is a local maximum.


Step 2. Compute the discriminant. We compute the second-order partial derivatives:

fxx(x,y) = aa (1- _+I ) = I 2 X X y (X+ y)

/yy(X, y) = !.__ (-2y- -1-) = -2 + --1

-----,. ay X+ y (X + y)2

a ( 1 ) 1 fxy(X, y) =- 1--- = 2 ay x + y (x + y)

The discriminant is

D(x, y) = fxx/yy- fx = -2 + - = ----= 2 1 ( 1 ) 1 -2 Y (x+y)2 (x+y)2 (x+y)4 (x+y)2


DG, -D ~ (l ='l), ~ -2 <0

We conclude that ( ~, - ~) is a saddle point.

2 2 22. f(x, y) = (x- y)ex -y

SOLUTION Find the critical points. We set the partial derivatives of f(x, y) = (x- y)ex2-Y

2 equal to zero and solve:

fx(x, y) =ex -y + (x- y)ex -y · 2x =ex -y 2x - 2xy + 1 = 0 2 2 2 2 2 2( 2 )

/y(x, y) =-ex -y + (x- y)ex -y · (-2y) =ex -y 2y - 2xy- 1 = 0 2 2 2 2 2 2( 2 )

Since ex2-Y

2 =!= 0, we have the following equations:

2x2 - 2xy + 1 = 0

2i- 2xy- 1 = 0

We add and subtract the two equations to obtain the following equations:

2 (x2 + y 2)- 4xy = 0

2 (x2- i) + 2 = 0

The first equation can be rewritten as x2 - 2xy + y2 = 0 or (x- y)2 = 0, yielding x = y. Substituting in the second equation gives 2 = 0, we conclude that the two equations have no solutions, that is, there are no critical points (notice that fx and /y exist everywhere). Since local minima and local maxima can occur only at critical points, it follows that

f (x, y) = (x - y )ex2

- Y2

does not have local minima or local maxima. 2

23. f(x, y) = (x + 3y)ey-x

SOLUTION

Step 1. Find the critical points. We compute the partial derivatives of f(x, y) = (x + 3y)eY-X2, using the Product

and the Chain Rule:

fx(X, y) = I · ey-x2 + (x + 3y)ey-x

2 · ( -2x) = ey-x

2 (I - 2x 2 - 6xy)

2 2 2 /y(x, y) = 3ey-x + (x + 3y)ey-x · 1 = ey-x (3 + x + 3y)

We set the partial derivatives equal to zero and solve to find the critical points:

ey-x2

(I - 2x2 - 6xy) = 0

2 eY-X (3 +X+ 3y) = 0

Since eY-x2

=1= 0, we obtain the following equations:

I - 2x2 - 6xy = 0

3 +x + 3y = 0

SECT I 0 N 14.7 I Optimization in Several Variables (LT SECTION 15.7) 773

Step 3. Conclusions. We compare the values of f(x, y) at the points obtained in step (2), and determine the global extrema of f(x, y). This gives

f(O, 0) = 0, /(1, 0) = I, /(0, I)= 2, f G· D = ~

We conclude that the global minimum off in the given domain is f(O, 0) = 0 and the global maximum is /(0, 1) = 2.

SOLUTION We use the following steps.

Step 1. Examine the critical points in the interior of the domain. We set the partial derivatives equal to zero and solve:

fx(x, y) = 3x2 - 3y = 0

/y(x, y) = 3i- 3x = 0

The first equation gives y = x 2 . We substitute in the second equation and solve for x:

3(x2)

2- 3x = 0

3x4

- 3x = 3x ( x 3 - 1) = 0 =} X = 0, y = 02 = 0

or x=l, y=1 2 =1

The critical points (0, 0) and ( 1, 1) are not in the interior of the domain.

Step 2. Find the extreme values on the boundary. We consider each part of the boundary separately.

y

D(O. I)._ _____ C(l, I)

L------~-x A(O,O) B(I.O)

The edge AB: On this edge, y = 0, 0 :::: x :::: 1, and f (x, 0) = x 3. The maximum value is obtained at x = I and the minimum value is obtained at x = 0. The corresponding extreme points are (1, 0) and (0, 0).

The edge BC: On thisedgex = 1, 0:::: y :::: 1, and f(l, y) = y 3 - 3y + 1. The critical points are JY (y 3 - 3y + I) =

3y2 - 3 = 0, that is, y = ±1. The point in the given domain is y = 1. The candidates for extreme values are thus y = 1 andy= 0, giving the points (1, 1) and (1, 0). The edge DC: On this edge y = 1, 0:::: x :::: I, and f(x, I)= x 3 - 3x + 1. Replacing the values of x andy in the previous solutions we get the points (1, 1) and (0, 1). The edge AD: On this edge x = 0, 0 :::: y :::: 1, and f (0, y) = y 3. Replacing the values of x and y obtained for the edge AB, we get (0, 1) and (0, 0).

By Theorem 3, the extreme values occur either at a critical point in the interior of the square or at a point on the boundary of the square. Since there are no critical points in the interior of the square, the candidates for extreme values are the following points:

(0, 0), (1, 0), (1, 1), (0, 1)

We compute f (x, y) = x 3 + y 3 - 3x y at these points:

J (O, O) = o3 + o3 - 3 . o = o f(l, 0) = 13 + 03 -3 ·I· 0 =I

f(l, I)= 13 + 13 -3 ·I· I= -1

f (0, I) = 03 + 13 - 3 · 0 · I = 1

We conclude that in the given domain, the global maximum is f ( 1, 0) /(1, 1) = -1.

• /(0, 1) 1 and the global minimum is


42. f(x, y) = x 2 + y 2 - 2x- 4y, x 2: 0, 0 ~ y ~ 3, y 2: x


Step 1. Examine the critical points in the interior of the domain. We set the partial derivatives equal to zero and solve:

fx(x, y) = 2x- 2, /y(x, y) = 2y- 4

Setting each equal to zero and solving we get: x = I andy = 2. Evaluating at the point (1, 2) we see:

f(l, 2) = -5

Step 2. Find the extreme values on the boundary. We consider each part of the boundary separately. The region that is described is the triangle bounded by the lines x = 0, y = 3, andy = x with vertices (0, 0), (3, 3), (0, 3).

First consider the line x = 0:

f(O, y) = y 2 - 4y => !' = 2y- 4

Setting f' equal to zero and solving we get y = 2. So we must consider the point (0, 2):

/(0, 2) = -4

We must also consider the endpoints of this line segment, (0, 0) and (0, 3):

f(O, 0) = 0, f(O, 3) = -3

Next, consider the line y = 3:

f(x, 3) = x 2 + 9- 2x- 12 = x 2 - 2x- 3 => !' = 2x- 2

Setting f' equal to zero and solving we get x = I. So we must also consider the point (1, 3):

f(I, 3) = -4

We must also consider the endpoints of this line segment, (0, 3) and (3, 3):

/(0, 3) = -3, /(3, 3) = 0

Finally, consider the line y = x:

f(x, x) = x 2 + x 2- 2x- 4x = 2x 2 - 6x => !' = 4x- 6

Setting!' equal to zero and solving, we get x = 3/2. So we must also consider the point (3/2, 3/2):

/(3/2, 3/2) = -~

We have already examined the endpoints of this line segment in the steps above.

Step 3. Conclusions. The points that we have considered in this problem are

f(l, 2) = -5, /(0, 2) = -4, f(l, 3) = -4, /(3/2, 3/2) = -~

f(O, 0) = 0, /(0, 3) = -3, /(3, 3) = 0

Therefore the minimum value is -5 and occurs at the point (I, 2) and the maximum value is 0 and occurs in two at the points (0, 0) and (3, 3).

43. f(x, y) = (4y 2 - x 2)e-x2-Y

2, x 2 + y2 ~ 2

SOLUTION We use the following steps.

Step 1. Examine the critical points. We compute the partial derivatives of f(x, y) = equal to zero and solve. This gives

fx(x, y) = -2xe-x2-y

2 + ( 4i- x 2) e-x

2-y

2 · ( -2x) = -2xe-x2-y2 ( 1 + 4i- x 2) = 0

/y(x, y) = 8ye-x2-y2 + ( 4y2- x2) e-x2-y2 . ( -2y) = -2ye-x2-y2 ( -4 + 4y2- x2) = 0

Since e-x2

-Y2

f= 0, the first equation gives x = 0 or x 2 = 1 + 4y2. Substituting x = 0 in the second equation

SECT I 0 N 14.7 I Optimization in Several Variables (LT SECTION 15. 7) 775

. y2 0 Smce e-. ,P , we get

y(-1+y2)=y(y-1)(y+1)=0 ==> y=O, y=l, y=-1

We obtain the three points (0, 0), (0, -1), (0, 1). We now substitute x 2 = 1 + 4y2 in the second equation and solve for y:

-2ye-i-Sy2

( -4 + 4i- I - 4i) = 0

-2ye-i-Sy2

· (-5) = 0 ==> y = 0

The corresponding values of x are obtained from

x 2 = I+ 4. 02 = 1 ==> X= ±1

We obtain the solutions ( 1, 0) and ( -1, 0). We conclude that the critical points are

(0,0), (0,-1), (0,1), (1,0), and (-1,0).

All of these points are in the interior x 2 + y2 < 2 of the given disk.

Step 2. Check the boundary. The boundary is the circle x 2 + y 2 = 2. On this set y 2 = 2 - x 2 , hence the function f (x. y) takes the values

That is, g(x) = -5e- 2x 2 + 8e-2. We determine the interval of x. Since x 2 + y 2 = 2, we have 0 < x 2 < 2 or

-h ::5 X ::5 h.

We thus must find the extreme values of g(x) = -se-2 x 2 + 8e-2 on the interval -h ::5 x ::5 h. With the aid of the graph of g(x), we conclude that the maximum value is g(O) = 8e-2 and the minimum value is

We conclude that the points on the boundary with largest and smallest values off are

f ( 0, ±h) = 8e-2""" 1.083, f (±h. 0) = -2e-2 """ -0.271

Step 3. Conclusions. The extreme values either occur at the critical points or at the points on the boundary, found in step 2. We compare the values off at these points:

/(0,0)=0

/(0, -I)= 4e-i """1.472

f(O, I)= 4e-i """1.472

f(l, 0) = -e-i """-0.368

f(-1,0) = -e-i """-0.368

J (o, ±h)""" 1.083

f (±h. 0) """ -0.271

We conclude that the global minimum is f(l,O) = f(-1,0) = -0.368 and the global maximum is /(0, -1) f(O, 1) = 1.472.

sEcT I 0 N 14.8 I Lagrange Multipliers: Optimizing with a Constraint (LT SECTION 15.8) 791

(c) We substitute y = 2x in the constraint equation x 2 + y 2 - 5 = 0 and solve for x andy. This gives

x 2 + (2x) 2 - 5 = 0

5x2 = 5

x 2 = 1 =} X] = -1, X2 = 1

Since y = 2x, we have YI = 2xl = -2, Y2 = 2x2 = 2. The critical points are thus

(-1, -2) and (1, 2).

Extreme values can also occur at the points where \1 g = (2x, 2y) = (0, 0). However, (0, 0) is not on the constraint.

(d) We evaluate f(x, y) = 2x + 4y at the critical points, obtaining

/(-1, -2) = 2. (-1) + 4. (-2) = -10

/(1, 2) = 2. 1 + 4. 2 = 10

Since f is continuous and the graph of g = 0 is closed and bounded, global minimum and maximum points exist. So according to Theorem I, we conclude that the maximum of f (x, y) on the constraint is I 0 and the minimum is -I 0.

2. Find the extreme values of f(x, y) = x 2 + 2y2 subject to the constraint g(x, y) = 4x- 6y = 25.

(a) Show that the Lagrange equations yield 2x = 4A., 4y = -6A..

(b) Show that if x = 0 or y = 0, then the Lagrange equations give x = y = 0. Since (0, 0) does not satisfy the constraint, you may assume that x and y are nonzero.

(c) Use the Lagrange equations to show that y = - ~ x.

(d) Substitute in the constraint equation to show that there is a unique critical point P.

(e) Does P correspond to a minimum or maximum value off? Refer to Figure 11 to justify your answer. Hint: Do the values of f (x, y) increase or decrease as (x, y) moves away from P along the line g (x, y) = 0?

y

4

0

-4

-4 0 4

FIGURE 11 Level curves of f(x, y) = x 2 + 2y2 and graph of the constraint g(x, y) = 4x- 6y- 25 = 0.

SOLUTION

(a) The gradients \1 f and \1 g are

\1 f = (2x, 4y), \lg = (4, -6)

The Lagrange equations are thus

'llf=A.'Ilg

(2x, 4y) =A. (4, -6)

or

2x = 4A.

4y = -6A.

(b) If x = 0, the first equation gives 0 = 4A. or A. = 0. Substituting in the second equation gives 4y = 0 or y = 0. Similarly, if y = 0, the second equation implies that A. = 0, hence by the first equation also x = 0. That is, if x = 0, then y = 0 and if y = 0 also x = 0. The point (0, 0) does not satisfy the equation of the constraint, hence we may assume that x :;f 0 and y :;f 0.

(c) The first equation in part (a) gives A.= ~-Substituting in the second equation we get

X 4y = -6 · - = -3x

2 ==>

3 y = --x

4


(d) We substitute y = -~x in the constraint 4x- 6y = 25 and solve for x andy. This gives

4x- 6 ( -~x) = 25

9 4x + 2x = 25

17x =50 50

X=-17'

3 50 75 y = --.- = --

4 17 34

We conclude that there is a unique critical point, which is ( ~, - ~).

(e) We now refer to Figure 11. As (x, y) moves away from P along the line g (x, y) = 0, the values off (x, y) increase, hence P corresponds to a minimum value of f.

3. Apply the method of Lagrange multipliers to the function f (x, y) = (x 2 + I) y subject to the constraint x 2 + y2 = 5. Hint: First show that y f. 0; then treat the cases x = 0 and x f. 0 separately.

SOLUTION We first write out the Lagrange Equations. We have Y' f = {2xy, x 2 +I) and Y'g = (2x, 2y). Hence, the

Lagrange Condition for Y' g f. 0 is

Y'f=A.Y'g

{2xy, x 2 + 1) =A. (2x, 2y)

We obtain the following equations:

2xy = A.(2x)

x2 + I = A.(2y)

2x(y- A.)= 0

x2 + 1 = 2A.y

The second equation implies that y f. 0, since there is no real value of x such that x 2 + 1 = 0. Likewise, A. f. 0. The solutions of the first equation are x = 0 andy = A..

Case 1: x = 0. Substituting x = 0 in the second equation gives 2A.y = 1, or y = 21A.. We substitute x = 0, y = -ft

(recall that A. f. 0) in the constraint to obtain

4)...2 = ~ 5

1 1 A.=±-=±-

v'W 2vS The corresponding values of y are

We obtain the critical points:

I y= -- =J5 and 2· _I_

2v'5

I y = = -JS

2. (- 2~)

( 0, J5) and ( 0, -J5)

Case 2: x f. 0. Then the first equation in (l) implies y =A.. Substituting in the second equation gives

We now substitute y = A. and x 2 = 2A. 2 - 1 in the constraint x 2 + y2 = 5 to obtain

The solution (x, y) are thus


2A.2 - 1 + A. 2 = 5

3A.2 = 6

A.2 =2 => A.=±J2

A. = J2: y = J2, x = ±J2 · 2 - I = ±J3

)... = -J2: y = -J2, x = ±J2 · 2- 1 = ±J3

( J3, J2) , ( -J3, J2), ( J3, -J2), ( -J3, -J2)

SECT I 0 N 14.8 I Lagrange Multipliers: Optimizing with a Constraint (LT SECTION 15.8) 793

We conclude that the critical points are

( 0, Js) , ( 0, -v's) , ( v'3, h) , ( -v'3, h) , ( v'3, -h), ( -v'3, -h) .

We now calculate f (x, y) = ( x 2 + I) y at the critical points:

f ( 0, Js) = v's ~ 2.24

f ( 0, -Js) = -v's ~ -2.24

f ( v'3, h) = f ( -v'3, h) = 4h ~ 5.66

f ( v'3, -h) = f ( -v'3, -h) = -4h ~ -5.66

Since the constraint gives a closed and bounded curve, f achieves a minimum and a maximum under it. We conclude that the maximum of f(x, y) on the constraint is 4J2 and the minimum is -4J2.

In Exercises 4-13, find the minimum and maximum values of the function subject to the given constraint.

4. f(x, y) = 2x + 3y, x 2 + y 2 = 4

SOLUTION We find the extreme values of f(x, y) = 2x + 3y under the constraint g(x, y) = x 2 + y 2 - 4 = 0.

Step 1. Write the Lagrange Equations. We have 'V f = (2, 3) and V'g = (2x, 2y), hence the Lagrange Condition is

The corresponding equations are

'Vf=).'Vg

(2, 3) =).. (2x, 2y)

2 =A(2x)

3 = )..(2y)

Step 2. Solve for x and y using the constraint. The two equations imply that x -:1 0 and y -:1 0, hence

I A=

x

The two expressions for ).. must be equal, so we obtain

X

3 2y

3 and ).. =-

2y

3 y =-X

2

We now substitute y = ix in the constraint equation x 2 + y2 = 4 and solve for x andy:

x2 + Gxr =4

x 2 + ~x 2 = 4 4

i3x2 = 16

Since y = ix, the corresponding values of yare


4 4 XJ=--,

v'I3 X2 = ---

v'J3

Extreme points may occur also where 'V g = (2x, 2y) = (0, 0). However, the point (0, 0) is not on the constraint.


Step 3. Calculate f at the critical points. We evaluate f (x, y) = 2x + 3 y at the critical points:

( 4 6 ) 8 18 26

f .Jl3' v'13 = v'13 + v'13 = v'13 ~ 7.21

t(-~.-~)=-~- ;,=-;,~-7.21 We conclude that the maximum of f on the constraint is about 7.21 and the minimum is about -7 .21.

5. f(x, y) = x 2 + yZ, 2x + 3y = 6

SOLUTION We find the extreme values of f(x, y) = x 2 + y 2 under the constraint g(x, y) = 2x + 3y- 6 = 0.

Step 1. Write out the Lagrange Equations. The gradients off and g are V' f = (2x, 2 y) and V' g = (2, 3). The Lagrange Condition is

We obtain the following equations:

'Vf=J....'Vg

(2x, 2y) = J.... (2, 3)

2x = J.... ·2

2y =)... 3

Step 2. Solve for J.... in terms of x and y. Notice that if x = 0, then the first equation gives J.... = 0, therefore by the second equation also y = 0. The point (0, 0) does not satisfy the constraint. Similarly, if y = 0 also x = 0. We therefore may assume that x f. 0 and y f. 0 and obtain by the two equations:

A=X 2

and J.... = 3y.

Step 3. Solve for x and y using the constraint. Equating the two expressions for J.... gives

3 y = -x

2

We substitute y = ~x in the constraint 2x + 3y = 6 and solve for x andy:

3 2x + 3 · -x = 6

2

13x = 12

We obtain the critical point ( H, H).

12 X= 13'

3 12 18 y=2·13=13

Step 4. Calculate fat the critical point. We evaluate f(x, y) = x 2 + y 2 at the critical point:

( 12 18) ( 12)

2 ( 18)

2 468

f 13' 13 = 13 + 13 = 169 ~ 2"77

Rewriting the constraint as y = -~x + 2, we see that as lxl -+ +oo then so does lyl, and hence x 2 + yZ is without bound on the constraint as lxl-+ oo. We conclude that the value 468/169 is the minimum value off constraint, rather than the maximum value.

6. f(x, y) = 4x2 + 9y2 , xy = 4

SOLUTION We find the extreme values off (x, y) = 4x2 + 9y2 under the constraint g(x, y) = xy - 4 = 0.

Step 1. Write out the Lagrange Equations. The gradient vectors are V' f = (8x, 18 y) and V' g = (y, x), hence the condition is

or

'Vf=J....'Vg

(8x, 18y) = J.... (y, x)

8x = J....y

18y =Ax

r 796 CHAPTER 14 I DIFFERENTIATION IN SEVERAL VARIABLES (LTCHAPTER 15)

We find y by the relation y = ±jx:

2 4

4x2 4x2 +9 ·- = 32

9

8x 2 = 32 =} X = -2, X = 2

y = 3. (-2) = -3, 2 4 y=--·2=--

3 3

We obtain the following critical points:

( -2 -~) ' 3 ' ( -2 ~) '3 '

Extreme values can also occur at the point where 'V g = (8x, 18y) = (0, 0), that is, at the point (0, 0). However, the point doe~ not lie on the constraint.

Step 4. Calculate fat the critical points. We evaluate f(x, y) = xy at the critical points:

1 ( -2. -i) = 1 (2. D = ~

1 ( -2. D = 1 (2. -D = -~ Since f is continuous and the constraint is a closed and bounded set in R2 (an ellipse), f attains global extrema on the constraint. We conclude that ~ is the maximum value and - ~ is the minimum value.

8. f(x,y)=x 2y+x+y, xy=4

SOLUTION Under the constraint xy = 4, then f(x, y) = x(xy) + x + y = 4x + x + t· Therefore, as x ~ 0+, f(x. y) ~ +oo on the constraint, and as x ~ 0-, f(x, y) ~ -oo. Therefore there are no minimum and maximum values of f (x, y) under the constraint.

9. f(x,y)=x2+y2, x4+y4= I

SOLUTION We find the extreme values of f(x, y) = x 2 + y2 under the constraint g(x, y) = x4 + y4 - 1 = 0.

Step 1. Write out the Lagrange Equations. We have 'V f = (2x, 2y) and 'V g = (4x 3, 4y3), hence the Lagrange Condition

'V f = ). 'V g gives

or

2x =). ( 4x3 )

2y =). (4y3)

Step 2. Solve for). in terms of x andy. We first assume that x i= 0 andy i= 0. Then the Lagrange equations give

1 A=-

2x2 1

and ). = - 2 2y

Step 3. Solve for x and y using the constraint. Equating the two expressions for). gives

We now substitute y = ±x in the equation of the constraint x 4 + y4 = I and solve for x andy:

x 4 + (±x)4 = I

2x4 =I

4 1 X =-

2 1

X=2!f4'

The corresponding values of y are obtained by the relation y = ±x. The critical points are thus

We examine the case x = 0 or y = 0. Notice that the point (0, 0) does not satisfy the equation of the constraint, either x = 0 or y = 0 can hold, but not both at the same time.

section 15.4 answers

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