Calculus one and several variables 10E Salas solutions manual ch16

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU

JWDD027-16 JWDD027-Salas-v1 December 7, 2006 16:36

788 SECTION 16.1

CHAPTER 16

SECTION 16.1

1. ∇f = (6x− y) i + (1 − x) j 2. ∇f = (2Ax + By)i + (Bx + 2Cy)j

3. ∇f = exy[ (xy + 1) i + x2j]

4. ∇f =1

(x2 + y2)2[(y2 − x2 + 2xy)i + (y2 − x2 − 2xy)j]

5. ∇f =[2y2 sin(x2 + 1) + 4x2y2 cos(x2 + 1)

]i + 4xy sin(x2 + 1) j

6. ∇f =2x

x2 + y2i +

2yx2 + y2

j

7. ∇f = (ex−y + ey−x) i + (−ex−y − ey−x) j = (ex−y + ey−x)(i − j)

8. ∇f =AD −BC

(Cx + Dy)2[ yi − xj ]

9. ∇f = (z2 + 2xy) i + (x2 + 2yz) j + (y2 + 2zx)k

10. ∇f =x√

x2 + y2 + z2i +

y√x2 + y2 + z2

j +z√

x2 + y2 + z2k

11. ∇f = e−z(2xy i + x2 j − x2y k)

12. ∇f =[

xyz

x + y + z+ yz ln(x + y + z)

]i +

[xyz

x + y + z+ xz ln(x + y + z)

]j

+[

xyz

x + y + z+ xy ln(x + y + z)

]k

13. ∇f = ex+2y cos(z2 + 1

)i + 2ex+2y cos

(z2 + 1

)j − 2zex+2y sin

(z2 + 1

)k

14. ∇f = eyz2/x3

(−3yz2

x4i +

z2

x3j +

2yzx3

k)

15. ∇f =[2y cos(2xy) +

2x

]i + 2x cos(2xy) j +

1z

k

16. ∇f =(

2xyz

− 3z4

)i +

x2

zj −

(x2y

z2+ 12xz3

)k

17. ∇f = (4x− 3y) i + (8y − 3x) j; at (2, 3), ∇f = −i + 18j

18. ∇f =1

(x− y)2(−2yi + 2xj), ∇f(3, 1) = −1

2i +

32j

19. ∇f =2x

x2 + y2i +

2yx2 + y2

j; at (2, 1), ∇f =45

i +25

j



SECTION 16.1 789

20. ∇f =(

tan−1(y/x) − xy

x2 + y2

)i +

(x2

x2 + y2

)j, ∇f(1, 1) =

(π

4− 1

2

)i +

12

j

21. ∇f = (sinxy + xy cosxy) i + x2 cosxy j; at (1, π/2), ∇f = i

22. ∇f = e−(x2+y2)[(y − 2x2y)i + (x− 2xy2)j], ∇f(1,−1) = e−2(i − j)

23. ∇f = −e−x sin (z + 2y) i + 2e−x cos (z + 2y) j + e−x cos (z + 2y)k;

at (0, π/4, π/4), ∇f = − 12

√2 (i + 2j + k)

24. ∇f = cosπzi − cosπzj − π(x− y) sinπzk, ∇f

(1, 0,

12

)= −πk

25. ∇f = i − y√y2 + z2

j − z√y2 + z2

k; at (2,−3, 4), ∇f = i +35

j − 45

k

26. ∇f = − sin(xyz2)(yz2i + xz2j + 2xyzk), ∇f

(π,

14,−1

)= −

√2

2

(14

i + π j − π

2k)

27. (a) ∇f(0, 2) = 4 i (b) ∇f(

14π,

16π

)=

(−1 − −1 +

√3

2√

2

)i +

(−1

2− −1 +

√3√

2

)j

(c) ∇f(1, e) = (1 − 2e) i − 2 j

28. (a) ∇f(1, 2,−3) =1

8√

2i +

12√

2j − 27

8√

2k (b) ∇f(1,−2, 3) = − 5

18i +

19

j +118

k

(c) ∇f(1, e2, π/6) =√

32

i +π

12e2j + k

29. For the function f(x, y) = 3x2 − xy + y, we have

f(x + h) − f(x) = f(x + h1, y + h2) − f(x, y)

= 3(x + h1)2 − (x + h1)(y + h2) + (y + h2) −[3x2 − xy + y

]= [(6x− y) i + (1 − x) j] · (h1 i + h2 j) + 3h2

1 − h1h2

= [(6x− y) i + (1 − x) j] · h + 3h21 − h1h2

The remainder g(h) = 3h21 − h1h2 = (3h1 i − h1 j) · (h1 i + h2 j) , and

|g(h)|‖h ‖ =

‖3h1 i − h1 j ‖ · ‖h ‖ · cos θ‖h ‖ ≤ ‖3h1 i − h1 j ‖

Since ‖3h1 i − h1 j ‖ → 0 as h → 0 it follows that

∇f = (6x− y) i + (1 − x) j

30. f(x + h) − f(x) = [(x + 2y) i + (2x + 2y) j] · [h1 i + h2 j] + 12h

21 + 2h1h2 + h2

2;

g(h) = 12h

21 + 2h1h2 + h2

2 is o(h).



790 SECTION 16.1

31. For the function f(x, y, z) = x2y + y2z + z2x, we have

f(x + h) − f(x) = f(x + h1, y + h2, z + h3) − f(x, y, z)

= (x + h1)2 (y + h2) + (y + h2)

2 (z + h3) + (z + h3)2 (x + h1) −

(x2y + y2z + z2x

)=

(2xy + z2

)h1 +

(2yz + x2

)h2 +

(2xz + y2

)h3 + (2xh2 + yh1 + h1h2)h1+

(2yh3 + zh2 + h2h3)h2 + (2zh1 + xh3 + h1h3)h3

=[(

2xy + z2)

i +(2yz + x2

)j +

(2xz + y2

)k]

· h + g(h) · h,

where g(h) = (2xh2 + yh1 + h1h2) i + (2yh3 + zh2 + h2h3) j + (2zh1 + xh3 + h1h3) k

Since|g(h)|‖h ‖ → 0 as h → 0 it follows that

∇f =(2xy + z2

)i +

(2yz + x2

)j +

(2xz + y2

)k

32. f(x + h) − f(x) =[(2xy + 2h2x + h1y) i + 2x2 j +

1z(z + h3)

k]· (h1 i + h2 j + h3 k) + h2

1;

g(h) = h21h2 is o(h) and ∇f = 4xy i = 2x2 j + 1

z2 k.

33. ∇f = F(x, y) = 2xy i +(1 + x2

)j ⇒ ∂f

∂x= 2xy ⇒ f(x, y) = x2y + g(y) for some function g.

Now,∂f

∂y= x2 + g′(y) = 1 + x2 ⇒ g′(y) = 1 ⇒ g(y) = y + C, C a constant.

Thus, f(x, y) = x2y + y + C

34. ∇f = (2xy + x)i + (x2 + y)j =⇒ fx = 2xy + x =⇒ f(x, y) = x2y +12x2 + g(y)

Now, fy = x2 + g′(y) = x2 + y =⇒ g′(y) = y =⇒ g(y) = 12y

2 + C

Thus, f(x, y) = x2y + 12x

2 + 12y

2 + C

35. ∇f = F(x, y) = (x + sin y) i + (x cos y − 2y) j ⇒ ∂f

∂x= x + sin y ⇒ f(x, y) = 1

2 x2 + x sin y + g(y)

for some function g.

Now,∂f

∂y= x cos y + g′(y) = x cos y − 2y ⇒ g′(y) = −2y ⇒ g(y) = −y2 + C, C a constant.

Thus, f(x, y) = 12 x

2 + x sin y − y2 + C.

36. ∇f = yzi + (xz + 2yz)j + (xy + y2)k =⇒ fx = yz =⇒ f(x, y, z) = xyz + g(y, z).

fy = xz + gy = xz + 2yz =⇒ gy = 2yz =⇒ g(y, z) = y2z + h(z) =⇒ f(x, y, z) = xyz + y2z + h(z).

fx = xy + y2 + h′(z) = xy + y2 =⇒ h′(z) = 0 =⇒ h(z) = C.

Thus, f(x, y, z) = xyz + y2z + C.

37. With r = (x2 + y2 + z2)1/2 we have

∂r

∂x=

x

r,

∂r

∂y=

y

r,

∂r

∂z=

z

r.

(a)∇(ln r) =

∂

∂x(ln r) i +

∂

∂y(ln r) j +

∂

∂z(ln r)k

=1r

∂r

∂xi +

1r

∂r

∂yj +

1r

∂r

∂zk

=x

r2i +

y

r2j +

z

r2k =

rr2



SECTION 16.1 791

(b)∇(sin r) =

∂

∂x(sin r) i +

∂

∂y(sin r) j +

∂

∂z(sin r)k

= cos r∂r

∂xi + cos r

∂r

∂yj + cos r

∂r

∂zk

= (cos r)x

ri + (cos r)

y

rj + (cos r)

z

rk

=(cos r

r

)r

(c) ∇er =(er

r

)r [ same method as in (a) and (b) ]

38. With rn = (x2 + y2 + z2)n/2 we have

∂rn

∂x=

n

2(x2 + y2 + z2)(n/2)−1(2x) = n(x2 + y2 + z2)(n−2)/2x = nrn−2x.

Similarly∂rn

∂y= nrn−2y and

∂rn

∂z= nrn−2z.

Therefore

∇rn = nrn−2xi + nrn−2yj + nrn−2zk = nrn−2(xi + yj + zk) = nrn−2r

39. (a) ∇f = 2x i + 2y j = 0 =⇒ x = y = 0; ∇f = 0 at (0, 0).

(b) (c) f has an absolute minimum at (0, 0)

40. (a) ∇f =−1√

4 − x2 − y2(xi + yj) = 0 at (0, 0) (b)

(c) f has a maximum at (0, 0)



792 SECTION 16.2

41. (a) Let c = c1i + c2j + c3k. First, we take h = hi. Since c · h is o(h),

0 = limh→0

c · h‖h‖ = lim

h→0

c1h

h= c1.

Similarly, c2 = 0 and c3 = 0.

(b) (y − z) · h = [f(x + h) − f(x) − z · h] + [y · h − f(x + h) + f(x) ] = o(h) + o(h) = o(h),

so that, by part (a), y − z = 0.

42. limh→0

g(h) = limh→0

(‖h‖g(h)

‖h‖

)=

(limh→0

‖h‖) (

limh→0

g(h)‖h‖

)= (0)(0) = (0).

43. (a) In Section 15.6 we showed that f was not continuous at (0, 0). It is therefore not differentiable

at (0, 0).

(b) For (x, y) �= (0, 0),∂f

∂x=

2y(y2 − x2)(x2 + y2)2

. As (x, y) tends to (0, 0) along the positive y-axis,

∂f

∂x=

2y3

y4=

2y

tends to ∞.

SECTION 16.2

1. ∇f = 2xi + 6yj, ∇f(1, 1) = 2i + 6j, u = 12

√2 (i − j), f ′

u(1, 1) = ∇f(1, 1) · u = −2√

2

2. ∇f = [1 + cos(x + y)]i + cos(x + y)j, ∇f(0, 0) = 2i + j, u =1√5(2i + j),

f ′u(0, 0) = ∇f(0, 0) · u =

√5

3. ∇f = (ey − yex) i + (xey − ex) j, ∇f(1, 0) = i + (1 − e)j, u =15(3i + 4j),

f ′u(1, 0) = ∇f(1, 0) · u =

15(7 − 4e)

4. ∇f =1

(x− y)2(−2yi + 2xj), ∇f(1, 0) = 2j, u =

12(i −

√3j),

f ′u(1, 0) = ∇f(1, 0) · u = −

√3

5. ∇f =(a− b)y(x + y)2

i +(b− a)x(x + y)2

j, ∇f(1, 1) =a− b

4(i − j), u =

12

√2 (i − j),

f ′u(1, 1) = ∇f(1, 1) · u =

14

√2 (a− b)

6. ∇f =1

(cx + dy)2[(d− c)yi + (c− d)xj] , ∇f(1, 1) =

d− c

(c + d)2(i − j), u =

1√c2 + d2

(ci − dj),

f ′u(1, 1) = ∇f(1, 1) · u =

d− c

(c + d)√c2 + d2

7. ∇f =2x

x2 + y2i +

2yx2 + y2

j, ∇f(0, 1) = 2 j, u =1√65

(8 i + j),

f ′u(0, 1) = ∇f(0, 1) · u =

2√65



SECTION 16.2 793

8. ∇f = 2xyi + (x2 + sec2 y)j, ∇f(−1,π

4) = −π

2i + 3j, u =

1√5(i − 2j)

f ′u

(−1,

π

4

)= ∇f

(−1,

π

4

)· u = − 1√

5

(π

2+ 6

)

9. ∇f = (y + z)i + (x + z)j + (y + x)k, ∇f(1,−1, 1) = 2j, u = 16

√6 (i + 2j + k),

f ′u (1,−1, 1) = ∇f(1,−1, 1) · u =

23

√6

10. ∇f = (z2 + 2xy)i + (x2 + 2yz)j + (y2 + 2zx)k, ∇f(1, 0, 1) = i + j + 2k, u =1√10

(3j − k)

f ′u(1, 0, 1) = ∇f(1, 0, 1) · u =

√10

10

11. ∇f = 2(x + y2 + z3

) (i + 2yj + 3z2k

), ∇f(1,−1, 1) = 6(i − 2j + 3k), u = 1

2

√2 (i + j),

f ′u(1,−1, 1) = ∇f(1,−1, 1) · u = −3

√2

12. ∇f = (2Ax + Byz)i + (Bxz + 2Cy)j + Bxyk, ∇f(1, 2, 1) = 2(A + B)i + (B + 4C)j + 2Bk

u =1√

A2 + B2 + C2(Ai + Bj + Ck); f ′

u(1, 2, 1) = ∇f(1, 2, 1) · u =2A2 + B2 + 2AB + 6BC√

A2 + B2 + C2

13. ∇f = tan−1(y + z) i +x

1 + (y + z)2j +

x

1 + (y + z)2k, ∇f(1, 0, 1) =

π

4i +

12

j +12

k,

u =1√3(i + j − k), f ′

u(1, 0, 1) = ∇f(1, 0, 1) · u =π

4√

3=

√3

12π

14. ∇f = (y2 cos z − 2πyz2 cosπx + 6zx)i + (2xy cos z − 2z2 sinπx)j + (−xy2 sin z − 4yz sinπx + 3x2)k

∇f(0,−1, π) = (2π3 − 1)i; u =13(2i − j + 2k), f ′

u(0,−1, π) = ∇f(0,−1, π) · u =23(2π3 − 1).

15. ∇f =x

x2 + y2i +

y

x2 + y2j, u =

1√x2 + y2

(−xi − yj) , f ′u(x, y) = ∇f · u = − 1√

x2 + y2

16. ∇f = exy[(y2 + xy3 − y3)i + (x− 1)(2y + xy2)j

], ∇f(0, 1) = −2j

u =1√5(−i + 2j), f ′

u(0, 1) = ∇f(0, 1) · u = −45

√5

17. ∇f = (2Ax + 2By) i + (2Bx + 2Cy) j, ∇f(a, b) = (2aA + 2bB)i + (2aB + 2bC) j

(a) u = 12

√2 (−i + j), f ′

u(a, b) = ∇f(a, b) · u =√

2 [a(B −A) + b(C −B)]

(b) u = 12

√2 (i − j), f ′

u(a, b) = ∇f(a, b) · u =√

2 [a(A−B) + b(B − C)]

18. ∇f =z

xi − z

yj + ln

(x

y

)k, ∇f(1, 1, 2) = 2i − 2j

u =1√3(i + j − k); f ′

u(1, 1, 2) = ∇f(1, 1, 2) · u = 0



794 SECTION 16.2

19. ∇f = ey2−z2

(i + 2xyj − 2xzk), ∇f(1, 2,−2) = i + 4j + 4k, r′(t) = i − 2 sin (t− 1) j − 2et−1k,

at (1, 2,−2) t = 1, r′(1) = i − 2k, u = 15

√5 (i − 2k), f ′

u(1, 2,−2) = ∇f(1, 2,−2) · u = −75

√5

20. ∇f = 2xi + zj + yk, ∇f(1,−3, 2) = 2i + 2j − 3k

Direction: r′(−1) = −2i + 3j − 3k, u =1√22

(−2i + 3j − 3k), f ′u(1,−3, 2) = ∇f(1,−3, 2) · u =

12

√22

21. ∇f = (2x + 2yz) i +(2xz − z2

)j + (2xy − 2yz)k, ∇f(1, 1, 2) = 6 i − 2k

The vectors v = ±(2 i + j − 3k) are direction vectors for the given line; u = ±(

1√14

[2 i + j − 3k])

are corresponding unit vectors; f ′u(1, 1, 2) = ∇f(1, 1, 2) · (±u) = ± 18√

14

22. ∇f = ex(cosπyzi − πz sinπyzj − πy sinπyzk), ∇f(0, 1, 12 ) = −π

2j − π k

The vectors v = ±(2 i + 3 j + 5k) are direction vectors for the line; u = ±(

1√38

[2 i + 3 j + 5k])

are corresponding unit vectors; f ′u(0, 1, 1

2 ) = ∇f(0, 1, 12 ) · (±u) = ∓ 13π

2√

38

23. ∇f = 2y2e2x i + 2ye2x j, ∇f(0, 1) = 2 i + 2 j, ‖∇f ‖ = 2√

2,∇f

‖∇f ‖ =1√2(i + j)

f increases most rapidly in the direction u =1√2(i + j); the rate of change is 2

√2.

f decreases most rapidly in the direction v = − 1√2(i + j); the rate of change is −2

√2.

24. ∇f = [1 + cos(x + 2y)]i + 2 cos(x + 2y)j, ∇f(0, 0) = 2i + 2j

Fastest increase in direction u =1√2(i + j), rate of change ‖∇f(0, 0)‖ = 2

√2

Fastest decrease in direction v = − 1√2(i + j), rate of change −2

√2

25. ∇f =x√

x2 + y2 + z2i +

y√x2 + y2 + z2

j +z√

x2 + y2 + z2k,

∇f(1,−2, 1) =1√6(i − 2 j + k), ‖∇f ‖ = 1

f increases most rapidly in the direction u =1√6(i − 2 j + k); the rate of change is 1.

f decreases most rapidly in the direction v = − 1√6(i − 2 j + k); the rate of change is −1.

26. ∇f = (2xzey + z2)i + x2zeyj + (x2ey + 2xz)k, ∇f(1, ln 2, 2) = 12i + 4j + 6k

Fastest increase in direction u =17(6i + 2j + 3k), rate of change ‖∇f(1, ln 2, 2)‖ = 14

Fastest decrease in direction v = −17(6i + 2j + 3k), rate of change −14

27. ∇f = f ′ (x0) i. If f ′ (x0) �= 0, the gradient points in the direction in which f increases: to the right

if f ′(x0) > 0, to the left if f ′(x0) < 0.



SECTION 16.2 795

28. 0; the vector c =∂f

∂y(x0, y0)i −

∂f

∂x(x0, y0)j is perpendicular to the gradient ∇f(x0, y0) and

points along the level curve of f at (x0, y0).

29. (a) limh→0

f(h, 0) − f(0, 0)h

= limh→0

√h2

h= lim

h→0

|h|h

does not exist

(b) no; by Theorem 16.2.5 f cannot be differentiable at (0, 0)

30. (a)g(x + h)o(h)

‖h‖ = g(x + h)o(h)‖h‖ → g(x)(0) = 0

(b)|[g(x + h) − g(x)]∇f(x) · h|

‖h‖ ≤ ‖[g(x + h) − g(x)]∇f(x)‖‖h‖‖h‖

by Schwarz’s inequality

= |g(x + h) − g(x)| · ‖∇f(x)‖ → 0

31. ∇λ(x, y) = − 83xi − 6yj

(a) ∇λ(1,−1) = −83i = 6j, u =

−∇λ(1,−1)‖∇λ(1,−1)‖ =

83 i − 6j23

√97

, λ′u(1,−1) = ∇λ(1,−1) · u = −2

3

√97

(b) u = i, λ′u(1, 2) = ∇λ(1, 2) · u =

(− 8

3 i − 12j)

· i = − 83

(c) u = 12

√2 (i + j), λ′

u(2, 2) = ∇λ(2, 2) · u =(− 16

3 i − 12 j)

·[12

√2 (i + j)

]= − 26

3

√2

32. ∇I = −4xi − 2yj. We want the curve r(t) = x(t)i + y(t)j which begins at (−2, 1) and has tangent

vector r′(t) in the direction ∇I. We can satisfy these conditions by setting

x′(t) = −4x(t), x(0) = −2; y′(t) = −2y(t), y(0) = 1.

These equations imply that

x(t) = −2e−4t, y(t) = e−2t.

Eliminating the parameter, we get x = −2y2; the particle will follow the parabolic path x = −2y2

toward the origin.

33. (a) The projection of the path onto the xy-plane is the curve

C : r(t) = x(t)i + y(t)j

which begins at (1, 1) and at each point has its tangent vector in the direction of −∇f. Since

∇f = 2xi + 6yj,

we have the initial-value problems

x′(t) = −2x(t), x(0) = 1 and y′(t) = −6y(t), y(0) = 1.

From Theorem 7.6.1 we find that

x(t) = e−2t and y(t) = e−6t.

Eliminating the parameter t, we find that C is the curve y = x3 from (1, 1) to (0, 0).



796 SECTION 16.2

(b) Here

x′(t) = −2x(t), x(0) = 1 and y′(t) = −6y(t), y(0) = −2

so that

x(t) = e−2t and y(t) = −2e−6t.

Eliminating the parameter t, we find that the projection of the path onto the xy-plane is the curve

y = −2x3 from (1,−2) to (0, 0).

34. z = f(x, y) =12x2 − y2; ∇f = xi − 2yj, so we choose the projection r(t) = x(t)i + y(t)j of the path

onto the xy-plane such that x′(t) = x(t), y′(t) = −2y(t)

(a) With initial point (−1, 1,− 12 ), we get x(t) = −et, y(t) = e−2t, or y =

1x2

from (−1, 1), in the direction of decreasing x.

(b) With initial point (1, 0, 12 ), we get x(t) = et, y(t) = 0, or the x-axis

from (1, 0), in the direction of increasing x.

35. The projection of the path onto the xy-plane is the curve

C : r(t) = x(t)i + y(t)j

which begins at (a, b) and at each point has its tangent vector in the direction of

−∇f = −(2a2xi + 2b2yj

). We can satisfy these conditions by setting

x′(t) = −2a2x(t), x(0) = a2 and y′(t) = −2b2y(t), y(0) = b

so that

x(t) = ae−2a2t and y(t) = be−2b2t.

Since [xa

]b2=

(e−2a2t

)b2

=[yb

]a2

,

C is the curve (b)a2xb2 = (a)b

2ya

2from (a, b) to (0, 0).

36. The particle must go in he direction −∇T = −ey cosxi − ey sinxj, so we set x′(t) = −ey(t) cosx(t),

y′(t) = −ey(t) sinx(t). Dividing, we havey′(t)x′(t)

=sinx(t)cosx(t)

, ordy

dx= tanx. With initial point

(0, 0), we get y = ln | secx|, in the direction of decreasing x (since x′(0) < 0).

37. We want the curve

C : r(t) = x(t)i + y(t)j

which begins at (π/4, 0) and at each point has its tangent vector in the direction of

∇T = −√

2 e−y sinx i −√

2 e−y cosx j.

From

x′(t) = −√

2 e−y sinx and y′(t) = −√

2 e−y cosx



SECTION 16.2 797

we obtaindy

dx=

y′(t)x′(t)

= cotx

so that

y = ln | sinx| + C.

Since y = 0 when x = π/4, we get C = ln√

2 and y = ln |√

2 sinx|. As ∇T (π/4, 0) = −i − j, the curve

y = ln |√

2 sinx| is followed in the direction of decreasing x.

38. ∇z = (1 − 2x)i + (2 − 6y)j, so the projection of the path onto the xy-plane satisfies x′(t) =

1 − 2x(t), y′(t) = 2 − 6y(t), ordy

dx=

2 − 6y1 − 2x

. With initial point (0, 0), this gives the curve

3y = (2x− 1)3 + 1, in the direction of increasing x.

39. (a)limh→0

f(2 + h, (2 + h)2

)− f(2, 4)

h= lim

h→0

3(2 + h)2 + (2 + h)2 − 16h

= limh→0

4[4h + h2

h

]= lim

h→04(4 + h) = 16

(b) limh→0

f

(h + 8

4, 4 + h

)− f(2, 4)

h= lim

h→0

3(h + 8

4

)2

+ (4 + h) − 16

h

= limh→0

316h

2 + 3h + 12 + 4 + h− 16h

= limh→0

(316h + 4

)= 4

(c) u = 117

√17 (i + 4j), ∇f(2, 4) = 12i + j; f ′

u(2, 4) = ∇f(2, 4) · u = 1617

√17

(d) The limits computed in (a) and (b) are not directional derivatives. In (a) and (b) we have, in

essence, computed ∇f(2, 4) · r0 taking r0 = i + 4j in (a) and r0 = 14 i + j in (b). In neither case

is r0 a unit vector.

40. ∇f =−GMm

(x2 + y2 + z2)3/2(xi + yj + zk) =

−GMm

‖r‖3r

41. (a) u = cos θ i + sin θ j, ∇f(x, y) =∂f

∂xi +

∂f

∂yj;

f ′u(x, y) = ∇f · u =

(∂f

∂xi +

∂f

∂yj)

· (cos θ i + sin θ j) =∂f

∂xcos θ +

∂f

∂ysin θ

(b) ∇f =(3x2 + 2y − y2

)i + (2x− 2xy) j, ∇f(−1, 2) = 3 i + 2 j

f ′u(−1, 2) = 3 cos(2π/3) + 2 sin(2π/3) =

2√

3 − 32



798 SECTION 16.3

42. f ′u(x, y) =

∂f

∂xcos

5π4

+∂f

∂ysin

5π4

= 2xe2y

(−√

22

)+ 2x2e2y

(−√

22

)= −

√2xe2y(1 + x)

f ′u(2, ln 2) = −

√2 · 2 · e2 ln 2(1 + 2) = −24

√2

43. ∇(fg) =∂(fg)∂x

i +∂(fg)∂y

j +∂(fg)∂z

k =(f∂g

∂x+ g

∂f

∂x

)i +

(f∂g

∂y+ g

∂f

∂y

)j +

(f∂g

∂z+ g

∂f

∂z

)k

= f

(∂g

∂xi +

∂g

∂yj +

∂g

∂zk)

+ g

(∂f

∂xi +

∂f

∂yj +

∂f

∂zk)

= f ∇g + g∇f

44. ∇(f

g

)=

∂

∂x

(f

g

)i +

∂

∂y

(f

g

)j +

∂

∂z

(f

g

)k

=

∂f

∂xg − f

∂g

∂xg2

i +

∂f

∂yg − f

∂g

∂y

g2j +

∂f

∂zg − f

∂g

∂zg2

k

=g(x)∇f(x) − f(x)∇g(x)

g2(x)

45. ∇fn =∂fn

∂xi +

∂fn

∂yj +

∂fn

∂zk = nfn−1 ∂f

∂xi + nfn−1 ∂f

∂yj + nfn−1 ∂f

∂zk = nfn−1 ∇f

SECTION 16.3

1. f(b) = f(1, 3) = −2; f(a) = f(0, 1) = 0; f(b) − f(a) = −2

∇f =(3x2 − y

)i − x j; b − a = i + 2 j and ∇f · (b − a) = 3x2 − y − 2x

The line segment joining a and b is parametrized by

x = t, y = 1 + 2t, 0 ≤ t ≤ 1

Thus, we need to solve the equation

3t2 − (1 + 2t) − 2t = −2, which is the same as 3t2 − 4t + 1 = 0, 0 ≤ t ≤ 1

The solutions are: t = 13 , t = 1. Thus, c = ( 1

3 ,53 ) satisfies the equation.

Note that the endpoint b also satisfies the equation.

2. ∇f = 4zi − 2yj + (4x + 2z)k, f(a) = f(0, 1, 1) = 0, f(b) = f(1, 3, 2) = 3

b − a = i + 2j + k, so we want (x, y, z) such that

∇f · (b − a) = 4z − 4y + 4x + 2z = 6z − 4y + 4x = f(b) − f(a) = 3

Parametrizing the line segment from a to b by x(t) = t, y(t) = 1 + 2t, z(t) = 1 + t,

we get t = 12 , or c = ( 1

2 , 2,32 )

3. (a) f(x, y, z) = a1x + a2y + a3z + C (b) f(x, y, z) = g(x, y, z) + a1x + a2y + a3z + C

4. Using the mean-value theorem 16.3.1, there exists c such that ∇f(c) · (b − a) = 0



SECTION 16.3 799

5. (a) U is not connected

(b) (i) g(x) = f(x) − 1 (ii) g(x) = −f(x)

6. By the mean-value theorem

f(x1) − f(x2) = ∇f(c) · (x1 − x2)

for some point c on the line segment x1x2. Since Ω is convex, c is in Ω. Thus

|f(x1) − f(x2)| = |∇f(c) · (x1 − x2)| ≤ ‖∇f(c)‖‖x1 − x2‖ ≤ M‖x1 − x2‖.

by Schwarz’s inequality∧

7. ∇f = 2xyi + x2j;

∇f(r(t)) · r′(t) =(2i + e2tj

)· (eti − e−tj) = et

8. ∇f = i − j; ∇f(r(t)) · r′(t) = (i − j) · (ai − ab sin atj) = a(1 + b sin at)

9. ∇f =−2x

1 + (y2 − x2)2i +

2y1 + (y2 − x2)2

j, ∇f(r(t)) =−2 sin t

1 + cos2 2ti +

2 cos t1 + cos2 2t

j

∇f(r(t)) · r′(t) =( −2 sin t

1 + cos2 2ti +

2 cos t1 + cos2 2t

j)

· (cos t i − sin t j) =−4 sin t cos t1 + cos2 2t

=−2 sin 2t

1 + cos2 2t

10. ∇f =1

2x2 + y3(4xi + 3y2j)

∇f(r(t)) · r′(t) =1

2e4t + t(4e2ti + 3t2/3j) · (2e2ti +

13t−2/3j) =

8e4t + 12e4t + t

11. ∇f = (ey − ye−x) i + (xey + e−x) j; ∇f(r(t)) = (tt − ln t) i +(tt ln t +

1t

)j

∇f(r(t)) · r′(t) =(

(tt − ln t) i +(tt ln t +

1t

)j)

·(

1ti + [1 + ln t] j

)= tt

(1t

+ ln t + [ln t]2)

+1t

12. ∇f =2

x2 + y2 + z2(xi + yj + zk)

∇f(r(t)) · r′(t) =2

1 + e4t(sin ti + cos tj + e2tk) · (cos ti − sin tj + 2e2tk) =

4e4t

1 + e4t

13. ∇f = yi + (x− z)j − yk;

∇f(r(t)) · r′(t) =(t2i +

(t− t3

)j − t2k

)·

(i + 2tj + 3t2k

)= 3t2 − 5t4

14. ∇f = 2x i + 2y j

∇f(r(t)) · r′(t) = (2a cosωti + 2b sinωtj) · (−ωa sinωti + ωb cosωtj + bωk) = 2ω(b2 − a2) sinωt cosωt

15. ∇f = 2xi + 2yj + k;

∇f(r(t)) · r′(t) = (2a cos ωt i + 2b sin ωt j + k) · (−aω sin ωt i + bω cos ωt j + bωk)

= 2ω(b2 − a2

)sin ωt cos ωt + bω



800 SECTION 16.3

16. ∇f = y2 cos(x + z)i + 2y sin(x + z)j + y2 cos(x + z)k

∇f(r(t)) · r′(t)

= [cos2 t cos(2t + t3)i + 2 cos t sin(2t + t3)j + cos2 t cos(2t + t3)k] · (2i − sin tj + 3t2k)

= cos t[(2 + 3t2) cos t cos(2t + t3) − 2 sin t sin(2t + t3)]

17.du

dt=

∂u

∂x

dx

dt+

∂u

∂y

dy

dt= (2x− 3y)(− sin t) + (4y − 3x)(cos t)

= 2 cos t sin t + 3 sin2 t− 3 cos2 t = sin 2t− 3 cos 2t

18. du

dt=

∂u

∂x· dxdt

+∂u

∂y· dydt

=(

1 + 2√

y

x

)3t2 +

(2√

x

y− 3

) (− 1t2

)

=(

1 +2t2

)3t2 + (2t2 − 3)

(− 1t2

)= 3t2 + 4 +

3t2

19.du

dt=

∂u

∂x

dx

dt+

∂u

∂y

dy

dt

= (ex sin y + ey cosx)(

12

)+ (ex cos y + ey sinx) (2)

= et/2(

12 sin 2t + 2 cos 2t

)+ e2t

(12 cos 1

2 t + 2 sin 12 t

)

20.du

dt=

∂u

∂x· dxdt

+∂u

∂y· dydt

= (4x− y)(−2 sin 2t) + (2y − x) cos t

= 2 sin 2t(sin t− 4 cos 2t) + cos t(2 sin t− cos 2t)

21.du

dt=

∂u

∂x

dx

dt+

∂u

∂y

dy

dt= (ex sin y) (2t) + (ex cos y) (π)

= et2[2t sin(πt) + π cos(πt)]

22.du

dt=

∂u

∂x· dxdt

+∂u

∂y· dydt

+∂u

∂z· dzdt

= − z

x2t +

z

y

12√t

+ ln(y

x

)et(1 + t)

= − 2t2et

t2 + 1+

et

2+ ln

( √t

t2 + 1

)et(1 + t)

23.du

dt=

∂u

∂x

dx

dt+

∂u

∂y

dy

dt+

∂u

∂z

dz

dt

= (y + z)(2t) + (x + z)(1 − 2t) + (y + x)(2t− 2)

= (1 − t)(2t) + (2t2 − 2t + 1)(1 − 2t) + t(2t− 2)

= 1 − 4t + 6t2 − 4t3

24.du

dt=

∂u

∂x· dxdt

+∂u

∂y· dydt

+∂u

∂z· dzdt

= (sinπy + πz sinπx)2t + πx cosπy(−1) − cosπx(−2t)

= 2t[sin[π(1 − t)] + π(1 − t2) sin(πt2)

]− πt2 cos[π(1 − t)] + 2t cos(πt2)

25. V =13πr2h,

dV

dt=

∂V

∂r

dr

dt+

∂V

∂h

dh

dt=

(23πrh

)dr

dt+

(13πr2

)dh

dt.



SECTION 16.3 801

At the given instant,dV

dt=

23π(280)(3) +

13π(196)(−2) =

12883

π.

The volume is increasing at the rate of1288

3π in.3/ sec .

26. v = πr2h,dv

dt=

∂v

∂r· drdt

+∂v

∂h· dhdt

= 2πrhdr

dt+ πr2 dh

dtdr

dt= −2,

dh

dt= 3, r = 13, h = 18 =⇒ dv

dt= −429π : decreasing at the rate of

429π cm3/sec.

27. A = 12 xy sin θ;

dA

dt=

∂A

∂x

dx

dt+

∂A

∂y

dy

dt+

∂A

∂θ

dθ

dt= 1

2

[(y sin θ)

dx

dt+ (x sin θ)

dy

dt+ (xy cos θ)

dθ

dt

].

At the given instantdA

dt=

12

[(2 sin 1) (0.25) + (1.5 sin 1) (0.25) − (2(1.5) cos 1) (0.1)] ∼= 0.2871 ft2/s ∼= 41.34 in2/s

28.dz

dt= 2x

dx

dt+

y

2dy

dt. But x2 + y2 = 13 =⇒ 2x

dx

dt+ 2y

dy

dt= 0 =⇒ dy

dt= −x

y

dx

dt

=⇒ dz

dt= 2x

dx

dt+

y

2

(−x

y

dx

dt

)=

3x2

dx

dt= 15. z is increasing 15 centimeters per second

29.∂u

∂s=

∂u

∂x

∂x

∂s+

∂u

∂y

∂y

∂s= (2x− y)(cos t) + (−x)(t cos s)

= 2s cos2 t− t sin s cos t− st cos s cos t∂u

∂t=

∂u

∂x

∂x

∂t+

∂u

∂y

∂y

∂t= (2x− y)(−s sin t) + (−x)(sin s)

= −2s2 cos t sin t + st sin s sin t− s cos t sin s

30.∂u

∂s=

∂u

∂x· ∂x∂s

+∂u

∂y· ∂y∂s

= [cos(x− y) − sin(x + y)]t + [− cos(x− y) − sin(x + y)]2s

= (t− 2s) cos(st− s2 + t2) − (t + 2s) sin(st + s2 − t2)

∂u

∂t=

∂u

∂x· ∂x∂t

+∂u

∂y· ∂y∂t

= [cos(x− y) − sin(x + y)]s + [− cos(x− y) − sin(x + y)](−2t)

= (s + 2t) cos(st− s2 + t2) − (s− 2t) sin(st + s2 − t2)

31.∂u

∂s=

∂u

∂x

∂x

∂s+

∂u

∂y

∂y

∂s= (2x tan y)(2st) +

(x2 sec2 y

)(1)

= 4s3t2 tan(s + t2

)+ s4t2 sec2

(s + t2

)∂u

∂t=

∂u

∂x

∂x

∂t+

∂u

∂y

∂y

∂t= (2x tan y)

(s2

)+

(x2 sec2 y

)(2t)

= 2s4t tan(s + t2

)+ 2s4t3 sec2

(s + t2

)



802 SECTION 16.3

32.∂u

∂s=

∂u

∂x· ∂x∂s

+∂u

∂y· ∂y∂s

+∂u

∂z· ∂z∂s

= z2y secxy tanxy(2t) + z2x secxy tanxy + 2z secxy(2st)

= sec[2st(s− t2)](2s4t3(s− t2) tan[2st(s− t2)] + 2s3t2 tan[2st(s− t2)] + 4s3t2

)∂u

∂t= z2y secxy tanxy(2s) + z2x secxy tanxy(−2t) + 2z secxy(s2)

= sec[2st(s− t2)](2s5t2(s− t2) tan[2st(s− t2)] − 4s5t4 tan[2st(s− t2)] + 2s4t

)33.

∂u

∂s=

∂u

∂x

∂x

∂s+

∂u

∂y

∂y

∂s+

∂u

∂z

∂z

∂s

= (2x− y)(cos t) + (−x)(− cos (t− s)) + 2z(t cos s)

= 2s cos2 t− sin (t− s) cos t + s cos t cos (t− s) + 2t2 sin s cos s

∂u

∂t=

∂u

∂x

∂x

∂t+

∂u

∂y

∂y

∂t+

∂u

∂z

∂z

∂t

= (2x− y)(−s sin t) + (−x)(cos (t− s)) + 2z(sin s)

= −2s2 cos t sin t + s sin (t− s) sin t− s cos t cos (t− s) + 2t sin2 s

34.∂u

∂s=

∂u

∂x· ∂x∂s

+∂u

∂y· ∂y∂s

+∂u

∂z· ∂z∂s

= eyz2 1s

+ xz2eyz2 · 0 + 2xyzeyz

22s

=1set

3(s2+t2)2 + 4st3(s2 + t2) ln(st)et3(s2+t2)2

∂u

∂t=

∂u

∂x· ∂x∂t

+∂u

∂y· ∂y∂t

+∂u

∂z· ∂z∂t

= eyz2 1t

+ xz2eyz23t2 + 2xyzeyz

22t

=1tet

3(s2+t2)2 + t2(s2 + t2)(3s2 + 7t2) ln(st)et3(s2+t2)2

35.d

dt[f(r(t) ) ] =

[∇f(r(t) ) · r′(t)

‖ r′(t) ‖

]‖ r′(t) ‖

= f ′u(t)(r(t)) ‖ r′(t) ‖ where u(t) =

r′(t)‖ r′(t) ‖

36.∂

∂x[f(r)] =

d

dr[f(r)]

∂r

∂x= f ′(r)

∂r

∂x= f ′(r)

x

r; similarly

∂

∂y[f(r)] = f ′(r)

y

rand

∂

∂z[f(r)] = f ′(r)

z

r.

Therefore ∇f(r) = f ′(r)x

ri + f ′(r)

y

rj + f ′(r)

z

rk = f ′(r)

rr.

37. (a) (cos r)rr

(b) (r cos r + sin r)rr

38. (a) ∇(r ln r) = (1 + ln r)rr

(b) ∇(e1−r2) = −2re1−r2 r

r = −2e1−r2r



SECTION 16.3 803

39. (a) (r cos r − sin r)rr3

(b)(

sin r − r cos rsin2 r

)rr

40. (a) (b)du

dt=

∂u

∂x

dx

ds

ds

dt+

∂u

∂y

dy

ds

ds

dt

41. (a)

(b)∂u

∂r=

∂u

∂x

(∂x

∂w

∂w

∂r+

∂x

∂t

∂t

∂r

)+

∂u

∂y

(∂y

∂w

∂w

∂r+

∂y

∂t

∂t

∂r

)+

∂u

∂z

(∂z

∂w

∂w

∂r+

∂z

∂t

∂t

∂r

).

To obtain ∂u/∂s, replace each r by s.

42. (a)

(b)∂u

∂r=

∂u

∂x

∂x

∂r+

∂u

∂z

∂z

∂r+

∂u

∂w

∂w

∂r,

∂u

∂v=

∂u

∂y

∂y

∂v+

∂u

∂w

∂w

∂v



804 SECTION 16.3

43.du

dt=

∂u

∂x

dx

dt+

∂u

∂y

dy

dt

d2u

dt2=

∂u

∂x

d2x

dt2+

dx

dt

[∂2u

∂x2

dx

dt+

∂2u

∂y∂x

dy

dt

]+

∂u

∂y

d2y

dt2+

dy

dt

[∂2u

∂x ∂y

dx

dt+

∂2u

∂y2

dy

dt

]and the result follows.

44.∂u

∂s=

∂u

∂x

∂x

∂s+

∂u

∂y

∂y

∂s

∂2u

∂s2=

∂u

∂x

∂2x

∂s2+

∂x

∂s

(∂2u

∂x2

∂x

∂s+

∂2u

∂y∂x

∂y

∂s

)+

∂u

∂y

∂2y

∂s2+

∂y

∂s

(∂2u

∂x∂y

∂x

∂s+

∂2u

∂y2

∂y

∂s

)

=∂2u

∂x2

(∂x

∂s

)2

+ 2∂2u

∂x∂y

∂x

∂s

∂y

∂s+

∂2u

∂y2

(∂y

∂s

)2

+∂u

∂x

∂2x

∂s2+

∂u

∂y

∂2y

∂s2

45. (a) ∂u

∂r=

∂u

∂x

∂x

∂r+

∂u

∂y

∂y

∂r=

∂u

∂xcos θ +

∂u

∂ysin θ

∂u

∂θ=

∂u

∂x

∂x

∂θ+

∂u

∂y

∂y

∂θ=

∂u

∂x(−r sin θ) +

∂u

∂y(r cos θ)

(b)(∂u

∂r

)2

=(∂u

∂x

)2

cos2 θ + 2∂u

∂x

∂u

∂ycos θ sin θ +

(∂u

∂y

)2

sin2 θ,

1r2

(∂u

∂θ

)2

=(∂u

∂x

)2

sin2 θ − 2∂u

∂x

∂u

∂ycos θ sin θ +

(∂u

∂y

)2

cos2 θ,

(∂u

∂r

)2

+1r2

(∂u

∂θ

)2

=(∂u

∂x

)2 (cos2 θ + sin2 θ

)+

(∂u

∂y

)2 (sin2 θ + cos2 θ

)=

(∂u

∂x

)2

+(∂u

∂y

)2

46. (a) By Exercise 45 (a)

∂w

∂r=

∂w

∂xcos θ +

∂w

∂ysin θ,

∂w

∂θ= −∂w

∂xr sin θ +

∂w

∂yr cos θ.

Solve these equations simultaneously for∂w

∂xand

∂w

∂y.

(b) To obtain the first pair of equations set w = r;

to obtain the second pair of equations set w = θ.

(c) θ is not independent of x; r =√x2 + y2 gives

∂r

∂x=

x√x2 + y2

=r cos θ

r= cos θ

47. Solve the equations in Exercise 45 (a) for∂u

∂xand

∂u

∂y:

∂u

∂x=

∂u

∂rcos θ − 1

r

∂u

∂θsin θ,

∂u

∂y=

∂u

∂rsin θ +

1r

∂u

∂θcos θ

Then ∇u =∂u

∂xi +

∂u

∂yj =

∂u

∂r(cos θ i + sin θ j) +

1r

∂u

∂θ(− sin θ i + cos θ j)

48. u(r, θ) = r2 =⇒ ∇u = 2rer



SECTION 16.3 805

49. u(x, y) = x2 − xy + y2 = r2 − r2 cos θ sin θ = r2(1 − 1

2 sin 2θ)

∂u

∂r= r(2 − sin 2θ),

∂u

∂θ= −r2 cos 2θ

∇u =∂u

∂rer +

1r

∂u

∂θeθ = r(2 − sin 2θ)er − r cos 2θ eθ

50.∂u

∂θ=

∂u

∂x

∂x

∂θ+

∂u

∂y

∂y

∂θ= −r sin θ

∂u

∂x+ r cos θ

∂u

∂y

∂2u

∂r∂θ= − sin θ

∂u

∂x− r sin θ

(∂2u

∂x2

∂x

∂r+

∂2u

∂y∂x

∂y

∂r

)+ cos θ

∂u

∂y+ r cos θ

(∂2u

∂x∂y

∂x

∂r+

∂2u

∂y2

∂y

∂r

)

= − sin θ∂u

∂x+ cos θ

∂u

∂y+ r sin θ cos θ

(∂2u

∂y2− ∂2u

∂x2

)+ r(cos2 θ − sin2 θ)

∂2u

∂x∂y

51. From Exercise 45 (a),

∂2u

∂r2=

∂2u

∂x2cos2 θ + 2

∂2u

∂y ∂xsin θ cos θ +

∂2u

∂y2sin2 θ

∂2u

∂θ2=

∂2u

∂x2r2 sin2 θ − 2

∂2u

∂y ∂xr2 sin θ cos θ +

∂2u

∂y2r2 cos2 θ − r

(∂u

∂xcos θ +

∂u

∂ysin θ

).

The term in parentheses is∂u

∂r. Now divide the second equation by r2 and add the two equations.

The result follows.

52. u(x, y) = x2 − 2xy + y4 − 4,∂u

∂x= 2x− 2y,

∂u

∂y= −2x + 4y3

dy

dx= −

∂u

∂x∂u

∂y

=2y − 2x4y3 − 2x

=y − x

2y3 − x

53. Set u = xey + yex − 2x2y. Then∂u

∂x= ey + yex − 4xy,

∂u

∂y= xey + ex − 2x2

dy

dx= − ∂u/∂x

∂u/∂y= − ey + yex − 4xy

xey + ex − 2x2.

54. u(x, y) = x2/3 + y2/3,∂u

∂x=

23x−1/3,

∂u

∂y=

23y−1/3

=⇒ dy

dx= −

∂u

∂x∂u

∂y

= −(y

x

)1/3

55. Set u = x cosxy + y cosx− 2. Then

∂u

∂x= cosxy − xy sinxy − y sinx,

∂u

∂y= −x2 sinxy + cosx

dy

dx= − ∂u/∂x

∂u/∂y=

cosxy − xy sinxy − y sinx

x2 sinxy − cosx.



806 SECTION 16.4

56. Set u(x, y, z) = z4 + x2z3 + y2 + xy − 2. Then∂u

∂x= 2xz3 + y,

∂u

∂y= 2y + x,

∂u

∂z= 4z3 + 3x2z2

∂z

∂x= −

∂u

∂x∂u

∂z

= − 2xz3 + y

4z3 + 3x2z2,

∂z

∂y= −

∂u

∂y∂u

∂z

= − 2y + x

4z3 + 3x2z2

57. Set u = cosxyz + ln(x2 + y2 + z2

). Then

∂u

∂x= −yz sinxyz +

2xx2 + y2 + z2

,∂u

∂y= −xz sinxyz +

2yx2 + y2 + z2

, and

∂u

∂z= −xy sinxyz +

2zx2 + y2 + z2

.

∂z

∂x= − ∂u/∂x

∂u/∂z= − 2x− yz

(x2 + y2 + z2

)sinxyz

2z − xy (x2 + y2 + z2) sinxyz,

∂z

∂y= − ∂u/∂y

∂u/∂z= − 2y − xz

(x2 + y2 + z2

)sinxyz

2z − xy (x2 + y2 + z2) sinxyz.

58. (a) Usedudt

=du1

dti +

du2

dtj and apply the chain rule to u1, u2.

(b) (i)dudt

= t(ex cos yi + ex sin yj) + π(−ex sin yi + ex cos yj)

= tet2/2(cosπti + sinπtj) + πet

2/2(− sinπti + cosπtj)

(ii) u(t) = et2/2 cosπti + et

2/2 sinπtj

dudt

= (−πet2/2 sinπt + tet

2/2 cosπt)i + (πet2/2 cosπt + tet

2/2 sinπt)j

59.∂u∂s

=∂u∂x

∂x

∂s+

∂u∂y

∂y

∂s,

∂u∂t

=∂u∂x

∂x

∂t+

∂u∂y

∂y

∂t

60.dudt

=∂u∂x

dx

dt+

∂u∂y

dy

dt+

∂u∂z

dz

dtwhere

∂u∂x

=∂u1

∂xi +

∂u2

∂xj +

∂u3

∂xk,

∂u∂y

=∂u1

∂yi +

∂u2

∂yj +

∂u3

∂yk,

∂u∂z

=∂u1

∂zi +

∂u2

∂zj +

∂u3

∂zk.

SECTION 16.4

1. Set f(x, y) = x2 + xy + y2. Then,

∇f = (2x + y)i + (x + 2y)j, ∇f(−1,−1) = −3i − 3j.

normal vector i + j; tangent vector i − j

tangent line x + y + 2 = 0; normal line x− y = 0



SECTION 16.4 807

2. Set f(x, y) = (y − x)2 − 2x, ∇f = −2(y − x + 1)i + 2(y − x)j, ∇f(2, 4) = −6i + 4j

normal vector −3i + 2j; tangent vector 2i + 3j

tangent line 3x− 2y + 2 = 0; normal line 2x + 3y − 16 = 0

3. Set f(x, y) =(x2 + y2

)2 − 9(x2 − y2

). Then,

∇f = [4x(x2 + y2) − 18x]i +[4y

(x2 + y2

)+ 18y

]j, ∇f

(√2, 1

)= −6

√2 i + 30j.

normal vector√

2 i − 5 j; tangent vector 5i +√

2 j

tangent line√

2x− 5y + 3 = 0; normal line 5x +√

2 y − 6√

2 = 0

4. Set f(x, y) = x3 + y3, ∇f = 3x2i + 3y2j, ∇f(1, 2) = 3i + 12j

normal vector i + 4j; tangent vector 4i − j

tangent line x + 4y − 9 = 0; normal line 4x− y − 2 = 0

5. Set f(x, y) = xy2 − 2x2 + y + 5x. Then,

∇f = (y2 − 4x + 5) i + (2xy + 1) j, ∇f(4, 2) = −7i + 17j.

normal vector 7i − 17j; tangent vector 17i + 7j

tangent line 7x− 17y + 6 = 0; normal line 17x + 7y − 82 = 0

6. Set f(x, y) = x5 + y5 − 2x3. ∇f = (5x4 − 6x2)i + 5y4j, ∇f(1, 1) = −i + 5j

normal vector i − 5j; tangent vector 5i + j

tangent line x− 5y + 4 = 0; normal line 5x + y − 6 = 0

7. Set f(x, y) = 2x3 − x2y2 − 3x + y. Then,

∇f = (6x2 − 2xy2 − 3) i + (−2x2y + 1) j, ∇f(1,−2) = −5i + 5j.

normal vector i − j; tangent vector i + j

tangent line x− y − 3 = 0; normal line x + y + 1 = 0

8. Set f(x, y) = x3 + y2 + 2x. ∇f = (3x2 + 2)i + 2yj, ∇f(−1, 3) = 5i + 6j

normal vector 5i + 6j; tangent vector 6i − 5j

tangent line 5x + 6y − 13 = 0; normal line 6x− 5y + 21 = 0

9. Set f(x, y) = x2y + a2y. By (15.4.4)

m = −∂f/∂x

∂f/∂y= − 2xy

x2 + a2.

At (0, a) the slope is 0.

10. Set f(x, y, z) = (x2 + y2)2 − z. ∇f = 4x(x2 + y2)i + 4y(x2 + y2)j − k, ∇f(1, 1, 4) = 8i + 8j − k

Tangent plane: 8x + 8y − z − 12 = 0

Normal: x = 1 + 8t, y = 1 + 8t, z = 4 − t



808 SECTION 16.4

11. Set f(x, y, z) = x3 + y3 − 3xyz. Then,

∇f = (3x2 − 3yz) i + (3y2 − 3xz) j − 3xyk, ∇f(1, 2, 3

2

)= −6i + 15

2 j − 6k;

tangent plane at(1, 2, 3

2

): −6(x− 1) + 15

2 (y − 2) − 6(z − 3

2

)= 0, which reduces to 4x− 5y + 4z = 0.

Normal: x = 1 + 4t, y = 2 − 5t, z = 32 + 4t

12. Set f(x, y, z) = xy2 + 2z2. ∇f = y2i + 2xyj + 4zk, ∇f(1, 2, 2) = 4i + 4j + 8k

Tangent plane: x + y + 2z − 7 = 0

Normal: x = 1 + t, y = 2 + t, z = 2 + 2t

13. Set z = g(x, y) = axy. Then, ∇g = ayi + axj, ∇g

(1,

1a

)= i + aj.

tangent plane at(

1,1a, 1

): z − 1 = 1(x− 1) + a

(y − 1

a

), which reduces to x + ay − z − 1 = 0

Normal: x = 1 + t, y = 1a + at, z = 1 − t

14. Set f(x, y, z) =√x +

√y +

√z. ∇f =

12√xi +

12√yj +

12√zk, ∇f(1, 4, 1) =

12i +

14j +

12k

Tangent plane: 2x + y + 2z − 8 = 0

Normal: x = 1 + 2t, y = 4 + t, z = 1 + 2t

15. Set z = g(x, y) = sinx + sin y + sin (x + y). Then,

∇g = [cosx + cos (x + y)] i + [cos y + cos (x + y)] j, ∇g(0, 0) = 2i + 2j;

tangent plane at (0, 0, 0) : z − 0 = 2(x− 0) + 2(y − 0), 2x + 2y − z = 0.

Normal: x = 2t, y = 2t, z = −t

16. Set f(x, y, z) = x2 + xy + y2 − 6x + 2 − z. ∇f = (2x + y − 6)i + (x + 2y)j − k, ∇f(4,−2,−10) =

−k

Tangent plane: z = −10

Normal: x = 4, y = −2, z = −10 + t

17. Set f(x, y, z) = b2c2x2 − a2c2y2 − a2b2z2. Then,

∇f (x0, y0, z0) = 2b2c2x0i − 2a2c2y0j − 2a2b2z0k;

tangent plane at (x0, y0, z0) :

2b2c2x0 (x− x0) − 2a2c2y0 (y − y0) − 2a2b2z0 (z − z0) = 0,

which can be rewritten as follows:

b2c2x0x− a2c2y0y − a2b2z0z = b2c2x02 − a2c2y0

2 − a2b2z02

= f (x0, y0, z0) = a2b2c2.

Normal: x = x0 + 2b2c2x0t, y = y0 − 2a2c2y0t, z = z0 − 2a2b2z0t



SECTION 16.4 809

18. Set f(x, y, z) = sin(x cos y) − z. ∇f = cos y cos(x cos y)i − x sin y cos(x cos y)j − k,

∇f(1, π2 , 0) = j − k

Tangent plane: y + z =π

2

Normal: x = 1, y =π

2+ t, z = t

19. Set z = g(x, y) = xy + a3x−1 + b3y−1.

∇g =(y − a3x−2

)i +

(x− b3y−2

)j, ∇g = 0 =⇒ y = a3x−2 and x = b3y−2.

Thus,

y = a3b−6y4, y3 = b6a−3, y = b2/a, x = b3y−2 = a2/b and g(a2/b, b2/a

)= 3ab.

The tangent plane is horizontal at(a2/b, b2/a, 3ab

).

20. z = g(x, y) = 4x + 2y − x2 + xy − y2. ∇g = (4 − 2x + y)i + (2 + x− 2y)j

∇g = 0 =⇒ 4 − 2x + y = 0, 2 + x− 2y = 0 =⇒ x =103, y =

83

The tangent plane is horizontal at (103 , 8

3 ,283 ).

21. Set z = g(x, y) = xy. Then, ∇g = yi + xj.

∇g = 0 =⇒ x = y = 0.

The tangent plane is horizontal at (0, 0, 0).

22. z = g(x, y) = x2 + y2 − x− y − xy. ∇g = (2x− 1 − y)i + (2y − 1 − x)j

∇g = 0 =⇒ 2x− 1 − y = 0 = 2y − 1 − x = 0 =⇒ x = 1, y = 1

The tangent plane is horizontal at (1, 1,−1).

23. Set z = g(x, y) = 2x2 + 2xy − y2 − 5x + 3y − 2. Then,

∇g = (4x + 2y − 5) i + (2x− 2y + 3) j.

∇g = 0 =⇒ 4x + 2y − 5 = 0 = 2x− 2y + 3 =⇒ x = 13 , y = 11

6 .

The tangent plane is horizontal at(

13 ,

116 , − 1

12

).

24. (a) Set f(x, y, z) = xy − z. ∇f = yi + xj − k, ∇f(1, 1, 1) = i + j − k

upper unit normal =√

33

(−i − j + k)

(b) Set f(x, y, z) =1x− 1

y− z. ∇f = − 1

x2i +

1y2

j − k, ∇f(1, 1, 0) = −i + j − k

lower unit normal: =√

33

(−i + j − k)

25.x− x0

(∂f/∂x)(x0, y0, z0)=

y − y0

(∂f/∂y)(x0, y0, z0)=

z − z0

(∂f/∂z)(x0, y0, z0)



810 SECTION 16.4

26. All the tangent planes pass through the origin. To see this, write the equation of the surface as

xf(x/y) − z = 0. The tangent plane at (x0, y0, z0) has equation

(x− x0)[x0

y0f ′

(x0

y0

)+ f

(x0

y0

)]− (y − y0)

[x0

2

y02f ′

(x0

y0

)]− (z − z0) = 0.

The plane passes through the origin:

−x02

y0f ′

(x0

y0

)− x0f

(x0

y0

)+

x02

y0f ′

(x0

y0

)+ z0 = z0 − x0f

(x0

y0

)= 0.

27. Since the tangent planes meet at right angles, the normals ∇F and ∇G meet at right angles:∂F

∂x

∂G

∂x+

∂F

∂y

∂G

∂y+

∂F

∂z

∂G

∂z= 0.

28. The sum of the intercepts is a. To see this, note that the equation of the tangent plane at

(x0, y0, z0) can be writtenx− x0√

x0+

y − y0√y0

+z − z0√

z0= 0.

Setting y = z = 0 we havex− x0√

x0=

√y0 +

√z0.

Therefore the x-intercept is given by

x = x0 +√x0(

√y0 +

√z0) = x0 +

√x0(

√a−√

x0) =√x0

√a.

Similarly the y-intercept is√y0√a and the z-intercept is

√z0√a. The sum of the intercepts is

(√x0 +

√y0 +

√z0)

√a =

√a√a = a.

29. The tangent plane at an arbitrary point (x0, y0, z0) has equation

y0z0 (x− x0) + x0z0 (y − y0) + x0y0 (z − z0) = 0,

which simplifies to

y0z0x + x0z0y + x0y0z = 3x0y0z0 and thus tox

3x0+

y

3y0+

z

3z0= 1.

The volume of the pyramid is

V =13Bh =

13

[(3x0) (3y0)

2

](3z0) =

92x0y0z0 =

92a3.

30. The equation of the tangent plane at (x0, y0, z0) can be written

x0−1/3(x− x0) + y0

−1/3(y − y0) + z0−1/3(z − z0) = 0

Setting y = z = 0, we get the x-intercept x = x0 + x01/3(y0

2/3 + z02/3) = x0 + x0

1/3(a2/3 − x02/3)

=⇒ x = x01/3a2/3

Similarly, the y-intercept is y01/3a2/3 and the z-intercept is z0

1/3a2/3.

The sum of the squares of the intercepts is

(x02/3 + y0

2/3 + z02/3)a4/3 = a2/3a4/3 = a2.



SECTION 16.4 811

31. The point (2, 3,−2) is the tip of r(1).

Since r′(t) = 2i − 3t2

j − 4tk, we have r′(1) = 2i − 3j − 4k.

Now set f(x, y, z) = x2 + y2 + 3z2 − 25. The function has gradient 2xi + 2yj + 6zk.

At the point (2, 3,−2),

∇f = 2(2i + 3j − 6k).

The angle θ between r′(1) and the gradient gives

cos θ =(2i − 3j − 4k)√

29· (2i + 3j − 6k)

7=

197√

29∼= 0.504.

Therefore θ ∼= 1.043 radians. The angle between the curve and the plane isπ

2− θ ∼= 1.571 − 1.043 ∼= 0.528 radians.

32. The curve passes through the point (3, 2, 1) at t = 1, and its tangent vector is r′(1) = 3i + 4j + 3k.

For the ellipsoid, set f(x, y, z) = x2 + 2y2 + 3z2. ∇f = 2xi + 4yj + 6zk,

∇f(3, 2, 1) = 6i + 8j + 6k, which is parallel to r′(1).

33. Set f(x, y, z) = x2y2 + 2x + z3. Then,

∇f = (2xy2 + 2) i + 2x2yj + 3z2k, ∇f(2, 1, 2) = 6i + 8j + 12k.

The plane tangent to f(x, y, z) = 16 at (2, 1, 2) has equation

6(x− 2) + 8(y − 1) + 12(z − 2) = 0, or 3x + 4y + 6z = 22.

Next, set g(x, y, z) = 3x2 + y2 − 2z. Then,

∇g = 6xi + 2yj − 2k, ∇g(2, 1, 2) = 12i + 2j − 2k.

The plane tangent to g(x, y, z) = 9 at (2, 1, 2) is

12(x− 2) + 2(y − 1) − 2(z − 2) = 0, or 6x + y − z = 11.

34. Sphere: f(x, y, z) = x2 + y2 + z2 − 8x− 8y − 6z + 24, ∇f = (2x− 8)i + (2y − 8)j + (2z − 6)k

∇f(2, 1, 1) = −4i − 6j − 4k

Ellipsoid: g(x, y, z) = x2 + 3y2 + 2z2, ∇g = 2xi + 6yj + 4zk

∇g(2, 1, 1) = 4i + 6j + 4k

Since their normal vectors are parallel, the surfaces are tangent.

35. A normal vector to the sphere at (1, 1, 2) is

2xi + (2y − 4) j + (2z − 2)k = 2i − 2j + 2k.

A normal vector to the paraboloid at (1, 1, 2) is

6xi + 4yj − 2k = 6i + 4j − 2k.



812 SECTION 16.4

Since

(2i − 2j + 2k) · (6i + 4j − 2k) = 0,

the surfaces intersect at right angles.

36. Surface A: Set f(x, y, z) = xy − az2, ∇f = yi + xj − 2azk

Surface B: Set g(x, y, z) = x2 + y2 + z2, ∇g = 2xi + 2yj + 2zk

Surface C: Set h(x, y, z) = z2 + 2x2 − c(z2 + 2y2). ∇h = 4xi − 4cyj + 2(1 − c)zk

Where surface A and surface B intersect, ∇f · ∇g = 4(xy − az2) = 0

Where surface A and surface C intersect, ∇f · ∇h = 4(1 − c)(xy − az2) = 0

Where surface B and surface C intersect, ∇g · ∇h = 4[2x2 − 2cy2 + (1 − c)z2] = 0

37. (a) 3x + 4y + 6 = 0 since plane p is vertical.

(b) y = − 14 (3x + 6) = − 1

4 [3(4t− 2) + 6] = −3t

z = x2 + 3y2 + 2 = (4t− 2)2 + 3(−3t)2 + 2 = 43t2 − 16t + 6

r(t) = (4t− 2)i − 3tj + (43t2 − 16t + 6)k

(c) From part (b) the tip of r(1) is (2,−3, 33). We take

r′(1) = 4i − 3j + 70j as d to write

R(s) = (2i − 3j + 33k) + s(4i − 3j + 70k).

(d) Set g(x, y) = x2 + 3y2 + 2. Then,

∇g = 2xi + 6yj and ∇g(2,−3) = 4i − 18j.

An equation for the plane tangent to z = g(x, y) at (2,−3, 33) is

z − 33 = 4(x− 2) − 18(y + 3) which reduces to 4x− 18y − z = 29.

(e) Substituting t for x in the equations for p and p1, we obtain

3t + 4y + 6 = 0 and 4t− 18y − z = 29.

From the first equation

y = − 34 (t + 2)

and then from the second equation

z = 4t− 18[− 3

4 (t + 2)]− 29 = 35

2 t− 2.

Thus,

(∗) r(t) = ti − ( 34 t + 3

2 )j +(

352 t− 2

)k.

Lines l and l′ are the same. To see this, consider how l and l′ are formed; to assure yourself, replace

t in (∗) by 4s + 2 to obtain R(s) found in part (c).



SECTION 16.4 813

38. (a) normal vector: 1225 i − 14

25 j; normal line: x = 2 + 1225 t, y = 1 − 14

25 t

(b) tangent line: x = 2 + 1425 t, y = 1 + 12

25 t

(c)

1 2 3x

1

2y

39. (a) normal vector: 2 i + 2 j + 4k; normal line: x = 1 + 2t, y = 2 + 2t, z = 2 + 4t

(b) tangent plane: 2(x− 1) + 2(y − 2) + 4(z − 2) = 0 or x + y + 2z − 7 = 0

(c)

40. (a)

-1

0

1-1

0

10

2

-1

0

1

(b)

-1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

(c) ∇f = 0 at (±1, 0),(0,±

√3/2

)



814 SECTION 16.5

41. (a)

-1

0

1

-1

0

1

0

1

2

3

-1

0

1

(b)

-1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

(c) ∇f =(4x3 − 4x

)i −

(4y3 − 4y

)j;

∇f = 0 : 4x3 − 4x = 0 =⇒ x = 0,±1; 4y3 − 4y = 0 =⇒ y = 0,±1

∇f = 0 at (0, 0), (±1, 0), (0,±1), (±1,±1)

42. (a)

-2-1

0

1

2-2

-1

0

1

2

-1

0

1

-2-1

0

1

(b)

-2 -1 0 1 2-2

-1

0

1

2

(c) ∇f = 0 at (0, 0),(±√

2/2, ±√

2/2)

SECTION 16.5

1. ∇f = (2 − 2x) i − 2y j = 0 only at (1, 0).

The difference

f(1 + h, k) − f(1, 0) =[2(1 + h) − (1 + h)2 − k2

]− 1 = −h2 − k2 ≤ 0

for all small h and k; there is a local maximum of 1 at (1, 0).

2. ∇f = (2 − 2x) i + (2 + 2y) j = 0 only at (1,−1).

The difference

f(1 + h,−1 + k) − f(1,−1)

= [2(1 + h) + 2(−1 + k) − (1 + h)2 + (−1 + k)2 + 5] − 5 = −h2 + k2

does not keep a constant sign for small h and k; (1,−1) is a saddle point.



SECTION 16.5 815

3. ∇f = (2x + y + 3) i + (x + 2y) j = 0 only at (−2, 1).

The difference

f(−2 + h, 1 + k) − f(−2, 1)

= [(−2 + h)2 + (−2 + h)(1 + k) + (1 + k)2 + 3(−2 + h) + 1] − (−2) = h2 + hk + k2

is nonnegative for all small h and k. To see this, note that

h2 + hk + k2 ≥ h2 − 2|h||k| + k2 = (|h| − |k|)2 ≥ 0;

there is a local minimum of −2 at (−2, 1).

4. ∇f = (3x2 − 3) i + j is never 0 ; there are no stationary points and no local extreme values.

5. ∇f = (2x + y − 6) i + (x + 2y) j = 0 only at (4,−2).

fxx = 2, fxy = 1, fyy = 2.

At (4,−2), D = 3 > 0 and A = 2 > 0 so we have a local min; the value is −10.

6. ∇f = (2x + 2y + 2) i + (2x + 6y + 10) j = 0 only at (1,−2).

∂2f

∂x2= 2,

∂2f

∂y∂x= 2,

∂2f

∂y2= 6; D = 2 · 6 − 22 > 0, A = 2 =⇒ local min; the value is −8.

7. ∇f = (3x2 − 6y) i +(3y2 − 6x

)j = 0 at (2, 2) and (0, 0).

fxx = 6x, fxy = −6, fyy = 6y, D = 36xy − 36.

At (2, 2), D = 108 > 0 and A = 12 > 0 so we have a local min; the value is −8.

At (0, 0), D = −36 < 0 so we have a saddle point.

8. ∇f = (6x + y + 5) i + (x− 2y − 5) j = 0 at(− 5

13,−35

13

).

∂2f

∂x2= 6,

∂2f

∂y∂x= 1,

∂2f

∂y2= −2; D = 6 · (−2) − 12 < 0; (−5/13, −35/13) is a saddle point.

9. ∇f = (3x2 − 6y + 6) i + (2y − 6x + 3) j = 0 at (5, 272 ) and (1, 3

2 ).

fxx = 6x, fxy = −6, fyy = 2, D = 12x− 36.

At (5, 272 ), D = 24 > 0 and A = 30 > 0 so we have a local min; the value is − 117

4 .

At (1, 32 ), D = −24 < 0 so we have a saddle point.

10. ∇f = (2x− 2y − 3) i + (−2x + 4y + 5) j = 0 at ( 12 ,−1).

∂2f

∂x2= 2,

∂2f

∂y∂x= −2,

∂2f

∂y2= 4; D = 2 · 4 − (−2)2 > 0, A = 2 =⇒ local minimum;

the value is − 134 .

11. ∇f = sin y i + x cos y j = 0 at (0, nπ) for all integral n.

fxx = 0, fxy = cos y, fyy = −x sin y.

Since D = − cos2 nπ = −1 < 0, each stationary point is a saddle point.



816 SECTION 16.5

12. ∇f = sin y i + (1 + x cos y) j = 0 at (−1, 2nπ) and (1, (2n + 1)π) for all integral n.

∂2f

∂x2= 0,

∂2f

∂y∂x= cos y,

∂2f

∂y2= −x sin y; D = 0 · (−x sin y) − cos2 y < 0 at the above points

so they are all saddle points

13. ∇f = (2xy + 1 + y2) i +(x2 + 2xy + 1

)j = 0 at (1,−1) and (−1, 1).

fxx = 2y, fxy = 2x + 2y, fyy = 2x, D = 4xy − 4(x + y)2.

At both (1,−1) and (−1, 1) we have saddle points since D = −4 < 0.

14. ∇f =(

1y

+y

x2

)i +

(− x

y2− 1

x

)j =

x2 + y2

x2yi − x2 + y2

xy2j is never 0;

no stationary points, no local extreme values.

15. ∇f = (y − x−2) i +(x− 8y−2

)j = 0 only at

(12 , 4

).

fxx = 2x−3, fxy = 1, fyy = 16y−3, D = 32x−3y−3 − 1.

At(

12 , 4

), D = 3 > 0 and A = 16 > 0 so we have a local min; the value is 6.

16. ∇f = (2x− 2y) i + (−2x− 2y) j = 0 only at (0, 0)

∂2f

∂x2= 2,

∂2f

∂y∂x= −2,

∂2f

∂y2= −2; D = 2 · (−2) − (−2)2 < 0; (0, 0) is a saddle point.

17. ∇f = (y − x−2) i +(x− y−2

)j = 0 only at (1, 1).

fxx = 2x−3, fxy = 1, fyy = 2y−3, D = 4x−3y−3 − 1.

At (1, 1), D = 3 > 0 and A = 2 > 0 so we have a local min; the value is 3.

18. ∇f = (2xy − y2 − 1) i + (x2 − 2xy + 1) j = 0 at (1, 1), (−1,−1)

∂2f

∂x2= 2y,

∂2f

∂y∂x= 2(x− y),

∂2f

∂y2= −2x; D = −4xy − 4(x− y)2 < 0 at the above points;

(1, 1) and (−1,−1) are saddle points.

19. ∇f =2

(x2 − y2 − 1

)(x2 + y2 + 1)2

i +4xy

(x2 + y2 + 1)2j = 0 at (1, 0) and (−1, 0).

fxx =−4x3 + 12xy2 + 12x

(x2 + y2 + 1)3, fxy =

4y3 + 4y − 12x2y

(x2 + y2 + 1)3, fyy =

4x3 + 4x− 12xy2

(x2 + y2 + 1)3.

point A B C D result

(1, 0) 1 0 1 1 loc. min.

(−1, 0) −1 0 −1 1 loc. max.

f(1, 0) = −1; f(−1, 0) = 1



SECTION 16.5 817

20. ∇f =(

lnxy + 1 − 3x

)i +

x− 3y

j = 0 at (3, 1/3)

∂2f

∂x2=

1x

+3x2

,∂2f

∂y∂x=

1y,

∂2f

∂y2=

3 − x

y2

At (3, 1/3),∂2f

∂x2=

23,

∂2f

∂y∂x= 3,

∂2f

∂y2= 0 and D = −9 < 0 =⇒ saddle point.

21. ∇f =(4x3 − 4x

)i + 2y j = 0 at (0, 0), (1, 0), and (−1, 0).

fxx = 12x2 − 4, fxy = 0, fyy = 2.


(0, 0) −4 0 2 −8 saddle

(1, 0) 8 0 2 16 loc. min.

(−1, 0) 8 0 2 16 loc. min.

f(±1, 0) = −3.

22. ∇f = 2xex2−y2

(1 + x2 + y2) i + 2yex2−y2

(1 − x2 − y2) j = 0 at (0, 0), (0, 1), (0,−1)

A =∂2f

∂x2= 2xex

2−y2(2x + 2x3 + 2xy2) + ex

2−y2(2 + 6x2 + 2y2)

B =∂2f

∂y∂x= −2yex

2−y2(2x + 2x3 + 2xy2) + ex

2−y2(4xy)

C =∂2f

∂y2= −2yex

2−y2(2y − 2yx2 − 2y3) + ex

2−y2(2 − 2x2 − 6y2)

At (0, 0), AC −B2 = (2)(2) = 4 > 0, A > 0 local minimum; the value is 0.

At (0,±1), AC −B2 = (4e−1)(−4e−1) = −8e−2 > 0, saddle points

23. ∇f = cosx sin y i + sinx cos y j = 0 at(

12π,

12π

),

(12π,

32π

), (π, π),

(32π,

12π

),

(32π,

32π

).

fxx = − sinx sin y, fxy = cosx cos y, fyy = − sinx sin y

point A B C D result(12π,

12π

)−1 0 −1 1 loc. max.(

12π,

32π

)1 0 1 1 loc. min.

(π, π) 0 1 0 −1 saddle(32π,

12π

)1 0 1 1 loc. min.(

32π,

32π

)−1 0 −1 1 loc. max.

f(

12π,

12π

)= f

(32π,

32π

)= 1; f

(12π,

32π

)= f

(32π,

12π

)= −1

24. ∇f = − sinx cosh y i + cosx sinh y j = 0 at (−π, 0) , (0, 0) , (π, 0).

fxx = − cosx cosh y, fxy = − sinx sinh y, fyy = cosx cosh y

D = − cos2 x cosh2 y − sin2 x sinh2 y < 0; (−π, 0) , (0, 0) , (π, 0) are saddle points.



818 SECTION 16.5

25. (a) ∇f = (2x + ky) i + (2y + kx) j and ∇f(0, 0) = 0 independent of the value of k.

(b) fxx = 2, fxy = k, fyy = 2, D = 4 − k2. Thus, D < 0 for |k| > 2 and (0, 0) is a saddle point

(c) D = 4 − k2 > 0 for |k| < 2. Since A = fxx = 2 > 0, (0, 0) is a local minimum.

(d) The test is inconclusive when D = 4 − k2 = 0 i.e., for k = ±2. (If k = ±2, f(x, y) = (x± y)2 and

(0, 0) is a minimum.)

26. (a) ∇f = (2x + ky) i + (kx + 8y) j = 0 at (0, 0).

(b)∂2f

∂x2= 2,

∂2f

∂y∂x= k,

∂2f

∂y2= 8; we want 16 − k2 < 0, or |k| > 4

(c) We want 16 − k2 > 0, or |k| < 4

(d) k = ±4. (If k = ±4, f(x, y) = (x± 2y)2 and (0, 0) is a minimum.)

27. Let P (x, y, z) be a point in the plane. We want to find the minimum of f(x, y, z) =√

x2 + y2 + z2.

However, it is sufficient to minimize the square of the distance: F (x, y, z) = x2 + y2 + z2. It is clear

that F has a minimum value, but no maximum value. Since P lies in the plane, 2x− y + 2z = 16

which implies y = 2x + 2z − 16 = 2(x + z − 8). Thus, we want to find the minimum value of

F (x, z) = x2 + 4(x + z − 8)2 + z2

Now,

∇F = [2x + 8(x + z − 8)] i + [8(x + z − 8) + 2z]k

The gradient is 0 when

2x + 8(x + z − 8) = 0 and 8(x + z − 8) + 2z = 0

The only solution to this pair of equations is: x = z =329, from which it follows that y = −16

9.

The point in the plane that is closest to the origin is P(

329 , − 16

9 , 329

).

The distance from the origin to the plane is: F (P ) = 163 .

Check using (13.6.5): d(P, 0) =|2 · 0 − 0 + 2 · 0 − 16|√

22 + (−1)2 + 22=

163.

28. We want to minimize (x + 1)2 + (y − 2)2 + (z − 4)2 on the plane. Since z = −16 − 32x + 2y,

we need to minimize f(x, y) = (x + 1)2 + (y − 2)2 + (−20 − 32x + 2y)2;

∇f =(

132 x− 6y + 62

)i + (−84 − 6x + 10y) j = 0 at (−4, 6)

Closest point (−4, 6, 2), distance=√

(−1 − (−4))2 + (2 − 6)2 + (4 − 2)2 =√

29

29. f(x, y) = (x− 1)2 + (y − 2)2 + z2 = (x− 1)2 + (y − 2)2 + x2 + 2y2[since z =

√x2 + 2y2

]∇f = [2(x− 1) + 2x] i + [2(y − 2) + 4y] j = 0 =⇒ x =

12, y =

23.

fxx = 4 > 0, fxy = 0, fyy = 6, D = 24 > 0. Thus, f has a local minimum at (1/2, 2/3).

The shortest distance from (1, 2, 0) to the cone is√f

(12 ,

23

)= 1

6

√114



SECTION 16.5 819

30. V = 8xyz, x2 + y2 + z2 = a2 =⇒ V (x, y) = 8xy√a2 − x2 − y2, x > 0, y > 0, x2 + y2 < a2

∇V =8y(a2 − x2 − y2) − 8x2y√

a2 − x2 − y2i +

8x(a2 − x2 − y2) − 8xy2√a2 − x2 − y2

j = 0 at(a/

√3, a/

√3

)

dimensions:2a√

3× 2a√

3× 2a√

3, maximum volume:

89a3√

3

31. (a) -2-1

01

2

-2-1

01

20

1

2

3

4

0

1

2

3

(b)

-1 -0.5 0 0.5 1

-1

-0.5

0

0.5

1

(c) ∇f = (4y − 4x3) i + (4x− 4y3) j = 0 at (0, 0), (1, 1), (−1,−1).

fxx = −12x2, fxy = 4, fyy = −12y2, D = 144x2y2 − 16


(0, 0) 0 4 0 −16 saddle

(1, 1) −12 4 −12 128 loc. max.

(−1,−1) −12 4 −12 128 loc. max.

f(1, 1) = f(−1,−1) = 3

32. (a)

-1

0

1-1

0

11

2

3

-1

0

1

(b)

-1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

(c) ∇f = (4x3 − 4x) i + 2y j = 0 at (0, 0), (1, 0), (−1, 0).

fxx = 12x2 − 4, fx,y = 0, fyy = 2, D = 24x2 − 8


(0, 0) −8 0 2 −8 saddle

(1, 0) 8 0 2 16 loc. min.

(−1, 0) 8 0 2 16 loc. min.

f(1, 0) = f(−1, 0) = 0



820 SECTION 16.5

33. (a)

-2 0 2

0

2

4 (b)

-1.5 -1 -0.5 0 0.5 1 1.5

-1.5

-1

-0.5

0

0.5

1

1.5

f(1, 1) = 3 is a local max.; f has a saddle at (0, 0).

34. (a)

-2-1

01

2-2

-1

0

1

2

00.20.4

0.6

-2-1

01

2

(b)

-2 -1 0 1 2-2

-1

0

1

2

f(0, 0) = 0 is a local min.; f(0, 1) = f(0,−1) = 2e−1 are local maxima; f has a saddle at (±1, 0).

35. (a) -20

2

-2

02

-1

0

1

(b)

-2 -1 0 1 2-2

-1

0

1

2

f(1, 0) = −1 is a local min.; f(−1, 0) = 1 is a loc. max.



SECTION 16.6 821

36. (a)

02

46

8100

2

4

6

8

10

-10123

02

46

8

(b)

0 2 4 6 8 100

2

4

6

8

10

f(π/2, π/2) = f(π/2, 5π/2) = f(5π/2, π/2) = f(5π/2, 5π/2) = 3 are local maxima;

(π/2, 3π/2), (3π/2, π/2), (3π/2, 3π/2), (3π/2, 5π/2), (5π/2, 3π/2) are saddle points of f ;

f(7π/6, 7π/6) = f(11π/6, 11π/6) = − 32 are local minima.

SECTION 16.6

1. ∇f = (4x− 4) i + (2y − 2) j = 0 at (1, 1) in D;

f(1, 1) = −1

Next we consider the boundary of D. We

parametrize each side of the triangle:1 2

x

4

y

C1 : r1(t) = t i, t ∈ [ 0, 2 ],

C2 : r2(t) = 2 i + t j, t ∈ [ 0, 4 ],

C3 : r3(t) = t i + 2t j, t ∈ [ 0, 2 ],

Now,

f1(t) = f(r1(t)) = 2(t− 1)2, t ∈ [ 0, 2 ]; critical number: t = 1,

f2(t) = f(r2(t)) = (t− 1)2 + 1, t ∈ [ 0, 4 ]; critical number: t = 1,

f3(t) = f(r3(t)) = 6t2 − 8t + 2, t ∈ [ 0, 2 ]; critical number: t = 23 .

Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that:

f1(0) = f3(0) = f(0, 0) = 2; f1(1) = f(1, 0) = 0; f1(2) = f2(0) = f(2, 0) = 2;

f2(1) = f(2, 1) = 1; f2(4) = f3(2) = f(2, 4) = 10; f3(2/3) = f(2/3, 4/3) = − 23 .

f takes on its absolute maximum of 10 at (2, 4) and its absolute minimum of −1 at (1, 1).

2. ∇f = −3 i + 2 j �= 0; no stationary points in D;

Next we consider the boundary of D. We parametrize each side of the triangle:

C1 : r1(t) = t i, t ∈ [ 0, 4 ],

C2 : r2(t) = t i +(− 3

2 t + 6)

j, t ∈ [ 0, 4 ],

C3 : r3(t) = t j, t ∈ [ 0, 6 ],



822 SECTION 16.6

and evaluate f :

f1(t) = f(r1(t)) = 2 − 3t, t ∈ [ 0, 4 ]; no critical numbers,

f2(t) = f(r2(t)) = −6t + 14, t ∈ [ 0, 4 ]; no critical numbers,

f3(t) = f(r3(t)) = 2 + 2t, t ∈ [ 0, 6 ]; no critical numbers.

Evaluating these functions at the endpoints of their domains, we find that:

f1(0) = f3(0) = f(0, 0) = 2; f1(4) = f2(4) = f(4, 0) = −10; f2(0) = f3(6) = f(0, 6) = 14;


3. ∇f = (2x + y − 6) i + (x + 2y) j = 0 at (4,−2) in

D; f(4,−2) = −13


parametrize each side of the rectangle:

1 2 3 4 5x

-3

-2

-1

y

C1 : r1(t) = −t j, t ∈ [ 0, 3 ]

C2 : r2(t) = t i − 3 j, t ∈ [ 0, 5 ]

C3 : r3(t) = 5 i − t j, t ∈ [ 0, 3 ]

C4 : r4(t) = t i, t ∈ [ 0, 5 ]

Now,

f1(t) = f(r1(t)) = t2 − 1, t ∈ [ 0, 3 ]; no critical numbers

f2(t) = f(r2(t)) = t2 − 9t + 8, t ∈ [ 0, 5 ]; critical number: t = 92

f3(t) = f(r3(t)) = t2 − 5t− 6, t ∈ [ 0, 3 ]; critical number: t = 52

f4(t) = f(r4(t)) = t2 − 6t− 1, t ∈ [ 0, 5 ]; critical number: t = 3


f1(0) = f4(0) = f(0, 0) = −1; f1(−3) = f2(0) = f(0,−3) = 8; f2(9/2) = f(9/2,−3) = − 494 ;

f2(5) = f3(3) = f(5,−3) = −12; f3(5/2) = f(5,−5/2) = − 494 ; f3(0) = f4(5) = f(5, 0) = −6.

f4(3) = f(3, 0) = −10

f takes on its absolute maximum of 8 at (0,−3) and its absolute minimum of −13 at (4,−2).

4. ∇f = (2x + 2y) i + (2x + 6y) j = 0 at (0, 0) in D; f(0, 0) = 0

Next we consider the boundary of D. We parametrize each side of the square:

C1 : r1(t) = t i − 2 j, t ∈ [−2, 2 ],

C2 : r2(t) = 2 i + t j, t ∈ [−2, 2 ],

C3 : r3(t) = t i + 2 j, t ∈ [−2, 2 ],

C4 : r4(t) = −2 i + t j, t ∈ [−2, 2 ],



SECTION 16.6 823

and evaluate f :

f1(t) = f(r1(t)) = t2 − 4t + 12, t ∈ [−2, 2 ]; no critical numbers,

f2(t) = f(r2(t)) = 4 + 4t + 3t2, t ∈ [−2, 2 ]; critical number: t = − 23 ,

f3(t) = f(r3(t)) = t2 + 4t + 12, t ∈ [−2, 2 ]; no critical numbers,

f4(t) = f(r4(t)) = 4 − 4t + 3t2, t ∈ [−2, 2 ]; critical number: t = 23 .


f1(−2) = f4(−2) = f(−2,−2) = 24; f1(2) = f2(−2) = f(2,−2) = 8; f2(−2/3) = f(2,−2/3) = 83 ;

f2(2) = f3(2) = f(2, 2) = 24; f3(−2) = f4(2) = f(−2, 2) = 8; f4(2/3) = f(−2, 2/3) = 83 .

f takes on its absolute maximum of 24 at (−2,−2) and (2, 2) and its absolute minimum of 0 at (0, 0).

Note that x2 + 2xy + 3y2 = (x + y)2 + 2y2. The results follow immediately from this observation.

5. ∇f = (2x + 3y) i + (2y + 3x) j = 0 at (0, 0) in D; f(0, 0) = 2

Next we consider the boundary of D. We parametrize the circle by:

C : r(t) = 2 cos t i + 2 sin t j, t ∈ [ 0, 2π ]

The values of f on the boundary are given by the function

F (t) = f(r(t)) = 6 + 12 sin t cos t, t ∈ [ 0, 2π ]

F ′(t) = 12 cos2 t− 12 sin2 t : F ′(t) = 0 =⇒ cos t = ± sin t =⇒ t = 14π,

34π,

54π,

74π

Evaluating F at the endpoints and critical numbers, we have:

F (0) = F (2π) = f(2, 0) = 6; F(

14π

)= F

(54π

)= f

(√2,√

2)

= f((−√

2,−√

2)

= 12;

F(

34π

)= f

(−√

2,√

2)

= F(

74π

)= f

(√2,−

√2)

= 0

f takes on its absolute maximum of 12 at(√

2,√

2)

and at(−√

2,−√

2); f takes on its absolute

minimum of 0 at(−√

2,√

2)

and at(√

2,−√

2).

6. ∇f = yi + (x− 3)j = 0 at (3, 0), which is not in the interior of D. The boundary is

r(t) = 3 cos t i + 3 sin t j. f(r(t)) = 3 sin t(3 cos t− 3) = 9 sin t(cos t− 1), t ∈ [0, 2π].

df

dt= 9(2 cos2 t− cos t− 1);

df

dt= 0 =⇒ cos t = 1, − 1

2 which yields the points A (3, 0),

B (− 32 ,

3√

32 ), C (− 3

2 , − 3√

32 ) : f(A) = 0, f(B) = − 27

√3

4 min, f(C) = 27√

34 max

7. ∇f = 2(x− 1)i + 2(y − 1) j = 0 only at (1, 1) in D. As the sum of two squares, f(x, y) ≥ 0.

Thus, f(1, 1) = 0 is a minimum. To examine the behavior of f on the boundary of D, we note that

f represents the square of the distance between (x, y) and (1, 1). Thus, f is maximal at the point

of the boundary furthest from (1, 1). This is the point(−√

2, −√

2); the maximum value of f is

f(−√

2, −√

2)

= 6 + 4√

2.



824 SECTION 16.6

8. ∇f = (y + 1) i + (x− 1) j = 0 at (1,−1) which is not in the interior of D.

Next we consider the boundary of D. We parametrize the boundary by:

C1 : r1(t) = t j + t2 j, t ∈ [−2, 2 ],

C2 : r2(t) = t i + 4 j, t ∈ [−2, 2 ],

and evaluate f :

f1(t) = f(r1(t)) = t3 − t2 + t + 3, t ∈ [−2, 2 ]; no critical numbers,

f2(t) = f(r2(t)) = 5t− 1, t ∈ [−2, 2 ]; no critical numbers.

Evaluating these functions at the endpoints of their domains, we find that:

f1(−2) = f2(−2) = f(−2, 4) = −11; f1(2) = f2(2) = f(2, 4) = 9.

f takes on its absolute maximum of 9 at (2, 4) and its absolute minimum of −11 at (−2, 4).

9. ∇f =2x2 − 2y2 − 2(x2 + y2 + 1)2

i +4xy

(x2 + y2 + 1)2j = 0 at (1, 0) and (−1, 0) in D; f(1, 0) = −1, f(−1, 0) = 1.

Next we consider the boundary of D. We parametrize each side of the squre:

C1 : r1(t) = −2 i + t j, t ∈ [−2, 2 ]

C2 : r2(t) = t i + 2 j, t ∈ [−2, 2 ]

C3 : r3(t) = 2 i + t j, t ∈ [−2, 2 ]

C4 : r4(t) = t i, t ∈ [−2, 2 ]

Now,

f1(t) = f(r1(t)) =4

t2 + 5, t ∈ [−2, 2 ]; critical number: t = 0

f2(t) = f(r2(t)) =−2tt2 + 5

, t ∈ [−2, 2 ]; no critical numbers

f3(t) = f(r3(t)) =−4


f4(t) = f(r4(t)) =−2tt2 + 5

, t ∈ [−2, 2 ]; no critical numbers


f1(−2) = f4(−2) = f(−2,−2) = 49 ; f1(0) = f(−2, 0) = 4

5 ; f1(2) = f2(−2) = f(−2, 2) = 49 ;

f4(2) = f3(−2) = f(2,−2) = − 49 ; f3(0) = f(2, 0) = − 4

5 ; f2(2) = f3(2) = f(2, 2) = − 49 .

f takes on its absolute maximum of 1 at (−1, 0) and its absolute minimum of −1 at (1, 0).

10. ∇f =2x2 − 2y2 − 2(x2 + y2 + 1)2

i +4xy

(x2 + y2 + 1)2j = 0 at (1, 0) in D; f(1, 0) = −1.

Next we consider the boundary of D. We parametrize each side of the triangle:

C1 : r1(t) = t i − t j, t ∈ [ 0, 2 ]

C2 : r2(t) = 2 i + t j, t ∈ [−2, 2 ]

C3 : r3(t) = t i + t j, t ∈ [ 0, 2 ],



SECTION 16.6 825

and evaluate f :

f1(t) = f(r1(t)) =−2t

2t2 + 1, t ∈ [ 0, 2 ]; critical number: t = 1/

√2,

f2(t) = f(r2(t)) =−4


f3(t) = f(r3(t)) =−2t

2t2 + 1, t ∈ [ 0, 2 ]; critical number: t = 1/

√2.


f1(0) = f3(0) = f(0, 0) = 0; f1(1/√

2) = f(1/√

2,−1/√

2) = −1/√

2;

f1(2) = f2(−2) = f(2,−2) = − 49 ; f2(0) = f(2, 0) = − 4

5 ; f2(2) = f3(2) = f(2, 2) = − 49 ;

f3(1/√

2) = f(1/√

2, 1/√

2) = −1/√

2.


11. ∇f = (4 − 4x) cos y i − (4x− 2x2) sin y j = 0 at (1, 0) in D: f(1, 0) = 2

Next we consider the boundary of D. We parametrize each side of the rectangle:

C1 : r1(t) = t j, t ∈[− 1

4π,14π

]C2 : r2(t) = t i − 1

4π j, t ∈ [ 0, 2 ]

C3 : r3(t) = 2 i + t j, t ∈[− 1

4π,14π

]C4 : r4(t) = t i + 1

4π j, t ∈ [ 0, 2 ]

Now,

f1(t) = f(r1(t)) = 0;

f2(t) = f(r2(t)) =√

22

(4t− 2t2), t ∈ [ 0, 2 ]; critical number: t = 1;

f3(t) = f(r3(t)) = 0;

f4(t) = f(r4(t)) =√

22

(4t− 2t2), t ∈ [ 0, 2 ]; critical number: t = 1;

f at the vertices of the rectangle has the value 0; f2(1) = f4(1) = f(1,− 1

4π)

= f(1, 1

4π)

=√

2.

f takes on its absolute maximum of 2 at (1, 0) and its absolute minimum of 0 along the lines x = 0

and x = 2.

12. ∇f = 2(x− 3)i + 2yj = 0 at (3, 0) which is not in the interior of D.

Boundary: On y = x2, f = (x− 3)2 + x4,df

dx= 2(x− 3) + 4x3 = 0 at x = 1 =⇒ (1, 1)

On y = 4x, f = (x− 3)2 + 16x2,df

dx= 2(x− 3) + 32x = 0 at x =

317

=⇒ (317

,1217

).

So the maximum and minimum occur at one or more of the following points:

(0, 0), (4, 16), (1, 1), (317

,1217

).

Evaluating f at these points, we find that f(1, 1) = 5 is the absolute minimum of f ; f(4, 16) = 257

is the absolute maximum of f .



826 SECTION 16.6

13. ∇f = (3x2 − 3y) i + (−3x− 3y2) j = 0 at (−1, 1) in D;

f(−1, 1) = 1


parametrize each side of the triangle:-2 -1 1 2

x

-2

2

y

C1 : r1(t) = −2 i + t j, t ∈ [−2, 2 ],

C2 : r2(t) = t i + t j, t ∈ [−2, 2 ],

C3 : r3(t) = t i + 2 j, t ∈ [−2, 2 ],

and evaluate f :

f1(t) = f(r1(t)) = −8 + 6t− t3, t ∈ [−2, 2 ]; critical numbers: t = ±√

2,

f2(t) = f(r2(t)) = −3t2, t ∈ [−2, 2 ]; critical number: t = 0,

f3(t) = f(r3(t)) = t3 − 6t− 8, t ∈ [−2, 2 ]; critical numbers: t = ±√

2.


f1(−2) = f2(−2) = f(−2,−2) = −12; f1(−√

2) = f(−2,−√

2) = −8 − 4√

2 ∼= −13.66;

f1(√

2) = f(−2,√

2) = −8 + 4√

2 ∼= −2.34; f1(2) = f3(−2) = f(−2, 2) = −4;

f2(0) = f(0, 0) = 0; f2(2) = f3(2) = f(2, 2) = −12;

f3(−√

2) = f(−√

2, 2) = −8 + 4√

2; f3(√

2) = f(√

2, 2) = −8 − 4√

2

f takes on its absolute maximum of 1 at (−1, 1) and its absolute minimum of −8 − 4√

2 at (√

2, 2)

and (−2,−√

2).

14. ∇f = 2(x− 4)i + 2yj = 0 at (4, 0) which is not in the

interior of D. Next we examine f on the boundary

of D:

C1 : r1(t) = ti + 4tj, t ∈ [ 0, 2, ],

C2 : r2(t) = ti + t3j, t ∈ [ 0, 2 ].

Note that

f1(t) = f (r1(t)) = 17t2 − 8t + 16,

f2(t) = f (r2(t)) = (t− 4)2 + t6.

Next

f ′1(t) = 34t− 8 = 0 =⇒ t = 4/17 and gives x = 4/17, y = 16/17

and

f ′2(t) = 6t5 + 2t− 8 = 0 =⇒ t = 1 and gives x = 1, y = 1.

The extreme values of f can be culled from the following list:

f(0, 0) = 16, f(2, 8) = 68, f(

417 ,

1617

)= 256

17 , f(1, 1) = 10.

We see that f(1, 1) = 10 is the absolute minimum and f(2, 8) = 68 is the absolute maximum.



SECTION 16.6 827

15. ∇f =4xy

(x2 + y2 + 1)2i +

2y2 − 2x2 − 2(x2 + y2 + 1)2

j = 0 at (0, 1) and (0,−1) in D;

f(0, 1) = −1, f(0,−1) = 1




F (t) = f(r(t)) = − 45 sin t, t ∈ [ 0, 2π ]

F ′(t) = − 45 cos t : F ′(t) = 0 =⇒ cos t = 0 =⇒ t = 1

2π,32π.


F (0) = F (2π) = f(2, 0) = 0; F(

12π

)= f(0, 2) = − 4

5 ; F(

32π

)= f(0,−2) = 4

5

f takes on its absolute maximum of 1 at (0,−1) and its absolute minimum of −1 at (0, 1).

16. ∇f = (2x + 1) i + (8y − 2) j = 0 at(− 1

2 ,14

)in D; f

(− 1

2 ,14

)= − 1

2

Next we consider the boundary of D. We parametrize the ellipse by:

C : r(t) = 2 cos t i + sin t j, t ∈ [ 0, 2π ]


F (t) = f(r(t)) = 4 cos2 t + 4 sin2 t + 2 cos t− 2 sin t = 4 + 2 cos t− 2 sin t, t ∈ [ 0, 2π ]

F ′(t) = −2 sin t− 2 cos t : F ′(t) = 0 =⇒ cos t = − sin t =⇒ t = 34π, or 7

4π


F (0) = F (2π) = f(2, 0) = 6;

F(

34π

)= f

(−√

2,√

2/2)

= 4 − 2√

2; F(

74π

)= f

(√2,−

√2/

)= 4 + 2

√2

f takes on its absolute maximum of 4 + 2√

2 at(√

2,−√

2/2); f takes on its absolute minimum of

− 12 at

(− 1

2 ,14

).

17. ∇f = 2(x− y)i − 2(x− y) j = 0 at each point of the

line segment y = x from (0, 0) to (4, 4). Since

f(x, x) = 0 and f(x, y) ≥ 0, f takes on its minimum

of 0 at each of these points.


parametrize each side of the triangle:

C1 : r1(t) = tj, t ∈ [ 0, 12 ]

C2 : r2(t) = ti, t ∈ [ 0, 6 ]

C3 : r3(t) = ti + (12 − 2t) j, t ∈ [ 0, 6 ]



828 SECTION 16.6

and observe from

f(r1(t)) = t2, t ∈ [ 0, 12 ]

f(r2(t)) = t2, t ∈ [ 0, 6 ]

f(r3(t)) = (3t− 12)2, t ∈ [ 0, 6 ]

that f takes on its maximum of 144 at the point (0, 12).

18. ∇f =1

(x2 + y2)3/2(−xi − yj) �= 0 in D. Note that f(x, y) is the reciprocal of the distance of (x, y)

from the origin. The point of D closest to the origin (draw a figure) is (1, 1). Therefore f(1, 1) = 1/√

2

is the maximum value of f. The point of D furthest from the origin is (3, 4). Therefore f(3, 4) = 1/5

is the least value taken on by f .

19. Using the hint, we want to find the maximum value of f(x, y) = 18xy − x2y − xy2 in the triangular

region. The gradient of f is:

∇D =(18y − 2xy − y2

)i +

(18x− x2 − 2xy

)j

The gradient is 0 when

18y − 2xy − y2 = 0 and 18x− x2 − 2xy = 0

The solution set of this pair of equations is: (0, 0), (18, 0), (0, 18), (6, 6).

It is easy to verify that f is a maximum when x = y = 6. The three numbers that satisfy x + y + z = 18

and maximize the product xyz are: x = 6, y = 6, z = 6.

20. f(y, z) = 30yz2 − y2z2 − yz3, ∇f = (30z2 − 2yz2 − z3)j + (60yz − 2y2z − 3yz2)k = 0 at

(152, 15

)

(other points are not in the interior); f

(152, 15

)=

154

4.

On the line y + z = 30, f(y, z) = 0 so the maximum of xyz2 occurs at x = y =152, z = 15

21. f(x, y) = xy(1 − x− y), 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x.

[ dom (f) is the triangle with vertices (0, 0), (1, 0), (0, 1).]

∇f = (y − 2xy − y2)i +(x− 2xy − x2

)j = 0 =⇒ x = y = 0, x = 1, y = 0, x = 0, y = 1, x=y= 1

3 .

(Note that [ 0, 0 ] is not an interior point of the domain of f .)

fxx = −2y, fxy = 1 − 2x− 2y, fyy = −2x.

At(

13 ,

13

), D = 1

3 > 0 and A < 0 so we have a local max; the value is 1/27.

Since f(x, y) = 0 at each point on the boundary of the domain, the local max of 1/27 is also the

absolute max.



SECTION 16.6 829

22. V = xyz,x

a+

y

b+

z

c= 1 =⇒ V (x, y) = xyc

(1 − x

a− y

b

), x > 0, y > 0,

x

a+

y

b< 1

∇V = yc

(1 − 2x

a− y

b

)i + xc

(1 − x

a− 2y

b

)j = 0 at

(a

3,b

3

)

Maximum volume V =a

3· b3· c3

=abc

27

23. (a) ∇f = 12x i − 2

9y j = 0 only at (0, 0).

(b) The difference

f(h, k) − f(0, 0) = 14h

2 − 19k

2

does not keep a constant sign for all small h and k; (0, 0) is a saddle point. The function has no

local extreme values.

(c) Being the difference of two squares, f can be maximized by maximizing 14x

2 and minimizing19y

2; (1, 0) and (−1, 0) give absolute maximum value 14 . Similarly, (0, 1) and (0,−1) give ab-

solute minimum value − 19 .

24. (a) ∇f = anxn−1i + cnyn−1j = 0 at (0, 0)

(b)∂2f

∂x2= an(n− 1)xn−2,

∂2f

∂y∂x= 0,

∂2f

∂y2= cn(n− 1)yn−2; at (0, 0), D = 0.

(c) (i) (0, 0) gives absolute min of 0 if n is even; no extreme value if n is odd.

(ii) (0, 0) gives absolute max of 0 if n is even; no extreme value if n is odd.

(iii) no extreme values.

25. Let x, y and z be the length, width and height of the box. The surface area is given by

S = 2xy + 2xz + 2yz, so z =S − 2xy2(x + y)

, where S is a constant, and x, y, z > 0.

Now, the volume V = xyz is given by:

V (x, y) = xy

[S − 2xy2(x + y)

]

and

∇V ={y

[S − 2xy2(x + y)

]+ xy

2(x + y)(−2y) − (S − 2xy)(2)4(x + y)2

}i

+{x

[S − 2xy2(x + y)

]+ xy

2(x + y)(−2x) − (S − 2xy)(2)4(x + y)2

}j

Setting∂V

∂x=

∂V

∂y= 0 and simplifying, we get the pair of equations

2S − 4x2 − 8xy = 0

2S − 4y2 − 8xy = 0

from which it follows that x = y =√S/6. From practical considerations, we conclude that V has a

maximum value at (√S/6,

√S/6). Substituting these values into the equation for z, we get z =

√S/6

and so the box of maximum volume is a cube.



830 SECTION 16.6

26. V = xyz, S = xy + 2xz + 2yz =⇒ V (x, y) = xy(S − xy)2(x + y)

, x > 0, y > 0, xy < S.

∇V =y2(S − x2 − 2xy)

2(x + y)2i +

x2(S − y2 − 2xy)2(x + y)2

j

∇V = 0 =⇒ x =

√S

3, y =

√S

3; dimensions for maximum volume:

√S

3×

√S

3× 1

2

√S

3

27. f(x, y) =3∑

i=1

[(x− xi)

2 + (y − yi)2]

∇f(x, y) = 2 [(3x− x1 − x2 − x3) i + (3y − y1 − y2 − y3) j]

∇f = 0 only at(x1 + x2 + x3

3,y1 + y2 + y3

3

)= (x0, y0) .

The difference f(x0 + h, y0 + k) − f (x0, y0)

=3∑

i=1

[(x0 + h− xi)

2 + (y0 + k − yi)2 − (x0 − xi)

2 − (y0 − yi)2]

=3∑

i=1

[2h (x0 − xi) + h2 + 2k (y0 − yi) + k2

]= 2h (3x0 − x1 − x2 − x3) + 2k (3y0 − y1 − y2 − y3) + 3h2 + 3k2

= 3h2 + 3k2

is nonnegative for all h and k. Thus, f has its absolute minimum at (x0, y0) .

28. Profit P (x, y) = N1(x− 50) + N2(y − 60) = 250(y − x)(x− 50) + [32, 000 + 250(x− 2y)](y − 60)

∇P = 250(2y − 2x− 10)i + [32, 000 + 250(2x + 70 − 4y)]j = 0

=⇒ x = 89, y = 94

29. A = xy +12x

(x

2tan θ

),

P = x + 2y + 2(x

2sec θ

),

0 < θ <12π, 0 < x <

P

1 + sec θ.

A(x, θ) = 12x(P − x− x sec θ) + 1

4x2 tan θ,

∇A =(P

2− x− x sec θ +

x

2tan θ

)i +

(x2

4sec2 θ − x2

2sec θ tan θ

)j,

(Here j is the unit vector in the direction of increasing θ.)

∇A =12[P + x(tan θ − 2 sec θ − 2)] i +

x2

4sec θ (sec θ − 2 tan θ) j.



SECTION 16.6 831

From∂A

∂θ= 0 we get θ = 1

6π and then from∂A

∂x= 0 we get

P + x(

13

√3 − 4

3

√3 − 2

)= 0 so that x = (2 −

√3)P.

Next,

Axx = 12 (tan θ − 2 sec θ − 2),

Axθ =x

2sec θ (sec θ − 2 tan θ),

Aθθ =x2

2sec θ

(sec θ tan θ − sec2 θ − tan2 θ

).

Apply the second-partials test:

A = − 12 (2 +

√3 ), B = 0, C = − 1

3P2√

3 (2 −√

3 )2, D < 0.

Since, D > 0 and A < 0, the area is a maximum when θ = 16π, x = (2 −

√3 )P and y = 1

6 (3 −√

3 )P.

30. (a) ∇f = (2ax + by)i + (bx + 2cy)j

∂2f

∂x2= 2a,

∂2f

∂y∂x= b,

∂2f

∂y2= 2c; D = 4ac− b2.

(b) The point (0, 0) is the only stationary point. If D < 0, (0, 0) is a saddle point; if D > 0,

(0, 0) is a local minimum if a > 0 and a local maximum if a < 0.

(c) (i) if b > 0, f(x, y) = (√ax +

√cy)2; every point on the line

√ax +

√cy = 0

is a stationary point and at each such point f takes on a local and absolute min of 0

if b < 0, f(x, y) = (√ax−√

cy)2; every point on the line√ax−√

cy = 0

is a stationary point and at each such point f takes on a local and absolute min of 0

(ii) if b > 0, f(x, y) = −(√|a|x−

√|c|y)2; every point on the line

√|a|x−

√|c|y = 0

is a stationary point and at each such point f takes on a local and absolute max of 0

if b < 0, f(x, y) = −(√|a|x +

√|c|y)2; every point on the line

√|a|x +

√|c|y = 0

is a stationary point and at each such point f takes on a local and absolute max of 0

31. From x = 12y = 1

3z = t and x = y − 2 = z = s

we take (t, 2t, 3t) and (s, 2 + s, s)

as arbitrary points on the lines. It suffices to minimize the square of the distance between these

points:

f(t, s) = (t− s)2 + (2t− 2 − s)2 + (3t− s)2

= 14t2 − 12ts + 3s2 − 8t + 4s + 4, t, s real.

Let i and k be the unit vectors in the direction of increasing t and s, respectively.

∇f = (28t− 12s− 8)i + (−12t + 6s + 4) j; ∇f = 0 =⇒ t = 0, s = −2/3.

ftt = 28, fts = −12, fss = 6, D = 6(28) − (−12)2 = −24 < 0.

By the second-partials test, the distance is a minimum when t = 0, s = −2/3; the nature of the

problem tells us the minimum is absolute. The distance is√f(0,−2/3) = 2

3

√6.



832 SECTION 16.6

32. We want to minimize S = 4πr2 + 2πrh given that V = 43πr

3 + πr2h = 10, 000.

S(r) = 4πr2 + 2πr(

V

πr2− 4

3r

)=

43πr2 +

2Vr

S′(r) =8πr3 − 6V

3r2= 0 =⇒ r = 3

√6V8π

, h =V

πr2− 4

3r = 0

The optimal container is a sphere of radius r = 3√

7500/π meters.

33. (a) Let x and y be the cross-sectional measurements of the box, and let l be its length.

Then

V = xyl, where 2x + 2y + l ≤ 108, x, y > 0

To maximize V we will obviously take 2x+2y+l=108. Therefore, V (x, y)=xy(108−2x−2y) and

∇V = [y(108 − 2x− 2y) − 2xy] i + [x(108 − 2x− 2y) − 2xy] j

Setting∂V

∂x=

∂V

∂y= 0, we get the pair of equations

∂V

∂x= 108y − 4xy − 2y2 = 0

∂V

∂y= 108x− 4xy − 2x2 = 0

from which it follows that x = y = 18 =⇒ l = 36.

Now, at (18, 18), we have

A = Vxx = −4y = −72 < 0, B = Vxy = 108 − 4x− 4y = −36,

C = Vyy = −4x = −72, and D = (36)2 − (72)2 < 0.

Thus, V is a maximum when x = y = 18 inches and l = 36 inches.

(b) Let r be the radius of the tube and let l be its length.

Then

V = π r2l, where 2π r + l ≤ 108, r > 0

To maximize V we take 2π r + l = 108. Then V (r) = π r2(108 − 2π r) = 108π r2 − 2π2r3. Now

dV

dr= 216π r − 6π2r2

SettingdV

dr= 0, we get

216π r − 6π2r2 = 0 =⇒ r =36π

=⇒ l = 36

Now, at r = 36/π, we have

d2V

dr2= 216π − 12π2 36

π= − 216π < 0

Thus, V is a maximum when r = 36/π inches and l = 36 inches.



SECTION 16.6 833

34. Let (x, y, z) be on the ellipsoid, x > 0, y > 0, z > 0. Then

V = 2x · 2y · 2z = 8xyz.

Note that V achieves its maximum ⇐⇒ x2y2z2 achieves its maximum.

Let s = x2y2z2, then

s = x2y2

(1 − x2

9− y2

4

),

∂s

∂x= 2xy2

(1 − 2x2

9− y2

4

)= 0

∂s

∂y= 2x2y

(1 − x2

9− 2y2

4

)= 0

=⇒ 2x2

9+

y2

4= 1,

x2

9+

2y2

4= 1 =⇒ x =

3√3, y =

2√3, z =

1√3

Thus,

Vmax = 8xyz = 8 · 3√3· 2√

3· 1√

3=

16√

33

.

35. Let S denote the cross-sectional area. Then

S =12

(12 − 2x + 12 − 2x + 2x cos θ)x sin θ = 12x sin θ − 2x2 sin θ +12x2 sin 2θ,

where 0 < x < 6, 0 < θ < π/2

Now, with j in the direction of increasing θ,

∇S = (12 sin θ − 4x sin θ + x sin 2θ) i + (12x cos θ − 2x2 cos θ + x2 cos 2θ) j

Setting∂S

∂x=

∂S

∂θ= 0, we get the pair of equations

12 sin θ − 4x sin θ + x sin 2θ = 0

12x cos θ − 2x2 cos θ + x2 cos 2θ = 0

from which it follows that x = 4, θ = π/3.

Now, at (4, π/3), we have

A = Sxx = −4 sin θ + sin 2θ = − 32

√3, B = Sxθ = 12 cos θ − 4x cos θ + 2x cos 2θ = −6,

C = Sθθ = −12x sin θ + 2x2 sin θ − 2x2 sin 2θ = −24√

3 and D = 108 − 36 > 0.

Thus, S is a maximum when x = 4 inches and θ = π/3.

36. 96 = xyz,

C = 30xy + 10(2xz + 2yz)

= 30xy + 20(x + y)96xy

.



834 SECTION 16.7

C(x, y) = 30[xy +

64x

+64y

],

∇C = 30(y − 64x−2)i + 30(x− 64y−2) j = 0 =⇒ x = y = 4.

Cxx = 128x−3, Cxy = 1, Cyy = 128y−3.

When x = y = 4, we have D = 3 > 0 and A = 2 > 0 so the cost is minimized by making the dimensions

of the crate 4 × 4 × 6 meters.

37. (a) f(m, b) = [2 − b]2 + [−5 − (m + b)]2 + [4 − (2m + b)]2.

fm = 10m + 6b− 6, fb = 6m + 6b− 2; fm = fb = 0 =⇒ m = 1, b = − 23 .

fmm = 10, fmb = 6, fbb = 6, D = 24 > 0 =⇒ a minimum.

Answer: the line y = x− 23 .

(b) f(α, β) = [2 − β]2 + [−5 − (α + β)]2 + [4 − (4α + β)]2.

fα = 34α + 10β − 22, fβ = 10α + 6β − 2; fα = fβ = 0 =⇒

⎧⎨⎩

α = 1413

β = − 1913

⎤⎦ .

fαα = 34, fαβ = 10, fββ = 6, D = 104 > 0 =⇒ a minimun.

Answer: the parabola y = 113

(14x2 − 19

).

38. (a) f(m, b) = [2 − (−m + b)]2 + [−1 − b]2 + [1 − (m + b)]2

fm = 4m + 2, fb = 6b− 4, fm = fb = 0 =⇒ m = −12, b =

23

fmm = 4, fmb = 0, fbb = 6, D = 24 > 0 =⇒ minimum

Answer: the line y = −12x +

23

(b) f(α, β) = [2 − (α + β)]2 + [−1 − β]2 + [1 − (α + β)]2

fα = 4α + 4β − 6, fβ = 4α + 6β − 4; fα = fβ = 0 =⇒ α =52, β = −1

fαα = 4, fαβ = 4, fββ = 6, D = 8 > 0 =⇒ minimum

Answer: the parabola y =52x2 − 1

SECTION 16.7

1. f(x, y) = x2 + y2, g(x, y) = xy − 1

∇f = 2xi + 2yj, ∇g = yi + xj.

∇f = λ∇g =⇒ 2x = λy and 2y = λx.

Multiplying the first equation by x and the second equation by y, we get

2x2 = λxy = 2y2.

Thus, x = ±y. From g(x, y) = 0 we conclude that x = y = ±1. The points (1, 1) and (−1,−1) clearly

give a minimum, since f represents the square of the distance of a point on the hyperbola from the

origin. The minimum is 2.



SECTION 16.7 835

2. f(x, y) = xy, g(x, y) = b2x2 + a2y2 − a2b2

∇f = yi + xj, ∇g = 2b2xi + 2a2yj

∇f = λ∇g =⇒ y = 2λb2x, x = 2λa2y =⇒ a2y2 = b2x2

From g(x, y) = 0 we get 2b2x2 = a2b2 =⇒ x = ± a√2, y = ± b√

2

The maximum value of xy is 12ab, achieved at (a/

√2, b

√2) and at (−a/

√2,−b

√2).

3. f(x, y) = xy, g(x, y) = b2x2 + a2y2 − a2b2

∇f = yi + xj, ∇g = 2b2xi + 2a2yj.

∇f = λ∇g =⇒ y = 2λb2x and x = 2λa2y.

Multiplying the first equation by a2y and the second equation by b2x, we get

a2y2 = 2λa2b2xy = b2x2.

Thus, ay = ±bx. From g(x, y) = 0 we conclude that x = ± 12a

√2 and y = ± 1

2b√

2.

Since f is continuous and the ellipse is closed and bounded, the minimum exists. It occurs at(12a

√2,− 1

2b√

2)

and(− 1

2a√

2, 12b√

2); the minimum is − 1

2ab.

4. f(x, y) = xy2, g(x, y) = x2 + y2 − 1

∇f = y2i + 2xyj, ∇g = 2xi + 2yj

∇f = λ∇g =⇒ y2 = 2λx, 2xy = 2λy =⇒ y = 0 or y2 = 2x2

y = 0: From g(x, y) = 0 we get x = ±1; f(±1, 0) = 0

y2 = 2x2: From g(x, y) = 0 we get 3x2 = 1, =⇒ x = ± 1√3, y = ±

√2√3

The Minimum of xy2 is: −29

√3 at (−1/

√3,±

√2√

3)

5. Since f is continuous and the ellipse is closed and bounded, the maximum exists.

f(x, y) = xy2, g(x, y) = b2x2 + a2y2 − a2b2

∇f = y2i + 2xyj, ∇g = 2b2xi + 2a2yj.

∇f = λ∇g =⇒ y2 = 2λb2x and 2xy = 2λa2y.

Multiplying the first equation by a2y and the second equation by b2x, we get

a2y3 = 2λa2b2xy = 2b2x2y.

We can exclude y = 0; it clearly cannot produce the maximum. Thus,

a2y2 = 2b2x2 and, from g(x, y) = 0, 3b2x2 = a2b2.

This gives us x = ± 13

√3 a and y = ± 1

3

√6 b. The maximum occurs at x = 1

3

√3 a, y = ± 1

3

√6 b; the

value there is 29

√3 ab2.



836 SECTION 16.7

6. f(x, y) = x + y, g(x, y) = x4 + y4 − 1

∇f = i + j, ∇g = 4x3i + 4y3j

∇f = λ∇g =⇒ 1 = 4λx3, 1 = 4λy3 =⇒ x = y

From g(x, y) = 0 we get 2x4 = 1 =⇒ x = y = ±2−1/4

The maximum of x + y is: 2 · 2−1/4 = 23/4, achieved at (2−1/4, 2−1/4).

7. The given curve is closed and bounded. Since x2 + y2 represents the square of the distance from points

on this curve to the origin, the maximum exists.

f(x, y) = x2 + y2, g(x, y) = x4 + 7x2y2 + y4 − 1

∇f = 2xi + 2yj, ∇g =(4x3 + 14xy2

)i +

(4y3 + 14x2y

)j.

We use the cross-product equation (16.7.4) :

2x(4y3 + 14x2y) − 2y(4x3 + 14xy2) = 0,

20x3y − 20xy3 = 0,

xy(x2 − y2) = 0.

Thus, x = 0, y = 0, or x = ±y. From g(x, y) = 0 we conclude that the points to examine are

(0, ±1), (±1, 0),(± 1

3

√3, ± 1

3

√3

).

The value of f at each of the first four points is 1; the value at the last four points is 2/3. The

maximum is 1.

8. f(x, y, z) = xyz, g(x, y, z) = x2 + y2 + z2 − 1

∇f = yzi + xzj + xyk, ∇g = 2xi + 2yj + 2zk

∇f = λ∇g =⇒ yz = 2λx, xz = 2λy, xy = 2λz =⇒ x2 = y2 = z2 or λ = 0.

λ = 0: In this case, at least two of x, y, z are 0 and f = 0.

x2 = y2 = z2 From g(x, y, z) = 0 we get 3x2 = 1 =⇒ x = ± 1√3, y = ± 1√

3, z = ± 1√

3

The minimum of xyz is: −19

√3 at (−1/

√3,−1/

√3,−1/

√3), (−1/

√3, 1/

√3, 1/

√3),

(1/√

3,−1/√

3, 1/√

3), (1/√

3, 1/√

3,−1/√

3).

9. The maximum exists since xyz is continuous and the ellipsoid is closed and bounded.

f(x, y, z) = xyz, g(x, y, z) =x2

a2+

y2

b2+

z2

c2− 1

∇f = yzi + xzj + xyk, ∇g =2xa2

i +2yb2

j +2zc2

k.

∇f = λ∇g =⇒ yz =2xa2

λ, xz =2yb2

λ, xy =2zc2

λ.



SECTION 16.7 837

We can assume x, y, z are non-zero, for otherwise f(x, y, z) = 0, which is clearly not a maximum. Then

from the first two equations

yza2

x= 2λ =

xzb2

yso that a2y2 = b2x2 or

x2

a2=

y2

b2.

Similarly from the second and third equations we get

b2z2 = c2y2 ory2

b2=

z2

c2.

From g(x, y, z) = 0, we get3x2

a2= 1 =⇒ x± a√

3, from which it follows that y = ± b√

3,

z = ± c√3. The maximum value is 1

9

√3 abc.

10. f(x, y, z) = x + 2y + 4z, g(x, y, z) = x2 + y2 + z2 − 7

∇f = i + 2j + 4k, ∇g = 2xi + 2yj + 2zk

∇f = λ∇g =⇒ 1 = 2λx, 2 = 2λy, 4 = 2λz =⇒ y = 2x, z = 4x

From g(x, y, z) = 0 we get 21x2 = 7 =⇒ x = ± 1√3

Minimum of x + 2y + 4z is: −7√

3, achieved at (−1/√

3,−2/√

3,−4/√

3).

11. Since the sphere is closed and bounded and 2x + 3y + 5z is continuous, the maximum exists.

f(x, y, z) = 2x + 3y + 5z, g(x, y, z) = x2 + y2 + z2 − 19

∇f = 2i + 3j + 5k, ∇g = 2xi + 2yj + 2zk.

∇f = λ∇g =⇒ 2 = 2λx, 3 = 2λy, 5 = 2λz.

Since λ �= 0 here, we solve the equations for x, y and z:

x =1λ, y =

32λ

, z =52λ

,

and substitute these results in g(x, y, z) = 0 to obtain1λ2

+9

4λ2+

254λ2

− 19 = 0,384λ2

− 19 = 0, λ = ±12

√2.

The positive value of λ will produce positive values for x, y, z and thus the maximum for f. We get

x =√

2, y = 32

√2, z = 5

2

√2, and 2x + 3y + 5z = 19

√2.

12. f(x, y, z) = x4 + y4 + z4, g(x, y, z) = x + y + z − 1

∇f = 4x3i + 4y3j + 4z3k, ∇g = i + j + k

∇f = λ∇g =⇒ 4x3 = λ, 4y3 = λ, 4z3 = λ =⇒ x = y = z

From g(x, y, z) = 0 we get 3x = 1, =⇒ x = 13 = y = z

Minimum is:127

13. f(x, y, z) = xyz, g(x, y, z) =x

a+

y

b+

z

c− 1

∇f = yzi + xzj + xyk, ∇g =1a

i +1bj +

1ck.

∇f = λ∇g =⇒ yz =λ

a, xz =

λ

b, xy =

λ

c.



838 SECTION 16.7

Multiplying these equations by x, y, z respectively, we obtain

xyz =λx

a, xyz =

λy

b, xyz =

λz

c.

Adding these equations and using the fact that g(x, y, z) = 0, we have

3xyz = λ(x

a+

y

b+

z

c

)= λ.

Since x, y, z are non-zero,

yz =λ

a=

3xyza

, 1 =3xa, x =

a

3.

Similarly, y =b

3and z =

c

3. The maximum is

127

abc.

14. Maximize area A = xy given that the perimeter P = 2x + 2y

f(x, y) = xy, g(x, y) = 2x + 2y − P

∇f = yi + xj, ∇g = 2i + 2j; ∇f = λ∇g =⇒ y = 2λ, x = 2λ =⇒ x = y.

The rectangle of maximum area is a square.

15. It suffices to minimize the square of the distance from (0, 1) to a point on the parabola. Clearly, the

minimum exists.

f(x, y) = x2 + (y − 1)2, g(x, y) = x2 − 4y

∇f = 2xi + 2(y − 1)j, ∇g = 2xi − 4j.

We use the cross-product equation (16.7.4):

2x(−4) − 2x(2y − 2) = 0, 4x + 4xy = 0, x(y + 1) = 0.

Since y ≥ 0, we have x = 0 and thus y = 0. The minimum is 1.

16. Minimize f(x, y) = (x− p)2 + (y − 4p)2 subject to g(x, y) = 2px− y2 = 0

∇f = 2(x− p)i + 2(y − 4p)j, ∇g = 2pi − 2yj

∇f = λ∇g =⇒ 2(x− p) = 2λp, 2(y − 4p) = −2λy =⇒ x =4p2

y

From g(x, y) = 0 we get8p3

y= y2 =⇒ y = 2p, x = 2p

Distance to parabola is:√f(x, y) =

√5p

17. It suffices to maximize and minimize the square of the distance from (2, 1, 2) to a point on the sphere.

Clearly, these extreme values exist.

f(x, y, z) = (x− 2)2 + (y − 1)2 + (z − 2)2, g(x, y, z) = x2 + y2 + z2 − 1

∇f = 2(x− 2) i + 2(y − 1) j + 2(z − 2)k, ∇g = 2x i + 2y j + 2z k.

∇f = λ∇g =⇒ 2(x− 2) = 2xλ, 2(y − 1) = 2yλ, 2(z − 2) = 2zλ



SECTION 16.7 839

Thus,

x =2

1 − λ, y =

11 − λ

, z =2

1 − λ.

Using the fact that x2 + y2 + z2 = 1, we have(

21 − λ

)2

+(

11 − λ

)2

+(

21 − λ

)2

= 1 =⇒ λ = −2, 4

At λ = −2, (x, y, z) = (2/3, 1/3, 2/3) and f(2/3, 1/3, 2/3) = 4

At λ = 4, (x, y, z) = (−2/3, −1/3, −2/3) and f(−2/3,−1/3,−2/3) = 16

Thus, (2/3, 1/3, 2/3) is the closest point and (−2/3, −1/3, −2/3) is the furthest point.

18. f(x, y, z) = sinx sin y sin z, g(x, y, z) = x + y + z − π

∇f = cosx sin y sin zi + sinx cos y sin zj + sinx sin y cos zk, ∇g = i + j + k

∇f = λ∇g =⇒ cosx sin y sin z = λ = sinx cos y sin z = sin z sin y cos z =⇒ cosx = cos y = cos z

=⇒ x = y = z =π

3

Maximum of sinx sin y sin z is:3√

38

19. f(x, y, z) = 3x− 2y + z, g(x, y, z) = x2 + y2 + z2 − 14

∇f = 3 i − 2 j + k, ∇g = 2x i + 2y j + 2z k.

∇f = λ∇g =⇒ 3 = 2xλ, −2 = 2yλ, 1 = 2zλ.

Thus,

x =32λ

, y = − 1λ, z =

12λ

.

Using the fact that x2 + y2 + z2 = 14, we have(

32λ

)2

+(− 1

λ

)2

+(

12λ

)2

= 14 =⇒ λ = ± 12.

At λ =12, (x, y, z) = (3,−2, 1) and f(3,−2, 1) = 14

At λ = −12, (x, y, z) = (−3, 2,−1) and f(−3, 2,−1) = −14

Thus, the maximum value of f on the sphere is 14.

20. f(x, y, z) = xyz, g(x, y, z) = x2 + y2 + z − 4

∇f = yzi + xzj + xyk, ∇g = 2xi + 2yj + k

∇f = λ∇g =⇒ yz = 2λx, xz = 2λy, xy = λ =⇒ x2 = y2 =z

2From g(x, y, z) = 0 we get 4x2 = 4 =⇒ x = 1, y = 1, z = 2.

Maximum volume is 2



840 SECTION 16.7

21. It’s easier to work with the square of the distance; the minimum certainly exists.

f(x, y, z) = x2 + y2 + z2, g(x, y, z) = Ax + By + Cz + D

∇f = 2xi + 2yj + 2zk, ∇g = Ai + Bj + Ck.

∇f = λ∇g =⇒ 2x = Aλ, 2y = Bλ, 2z = Cλ.

Substituting these equations in g(x, y, z) = 0, we have

12λ

(A2 + B2 + C2

)+ D = 0, λ =

−2DA2 + B2 + C2

.

Thus, in turn,

x =−DA

A2 + B2 + C2, y =

−DB

A2 + B2 + C2, z =

−DC

A2 + B2 + C2

so the minimum value of√

x2 + y2 + z2 is |D|(A2 + B2 + C2

)−1/2.

22. f(x, y, z) = xyz, g(x, y, z) = 2xy + 2xz + 2yz − 6a2

∇f = yzi + xzj + xyk, ∇g = 2(y + z)i + 2(x + z)j + 2(x + y)k

∇f = λ∇g =⇒ yz = 2λ(y + z), xz = 2λ(x + z), xy = 2λ(x + y) =⇒ x = y = z

From g(x, y, z) = 0 we get 6x2 = 6a2 =⇒ x = y = z = a

Maximum volume is a3.

23. area A = 12ax + 1

2by + 12cz.

The geometry suggests that

x2 + y2 + z2

has a minimum.

f(x, y, z) = x2 + y2 + z2, g(x, y, z) = ax + by + cz − 2A

∇f = 2xi + 2yj + 2zk, ∇g = ai + bj + ck.

∇f = λ∇g =⇒ 2x = aλ, 2y = bλ, 2z = cλ.

Solving these equations for x, y, z and substituting the results in g(x, y, z) = 0, we have

a2λ

2+

b2λ

2+

c2λ

2− 2A = 0, λ =

4Aa2 + b2 + c2

and thus

x =2aA

a2 + b2 + c2, y =

2bAa2 + b2 + c2

, z =2cA

a2 + b2 + c2.

The minimum is 4A2(a2 + b2 + c2)−1.



SECTION 16.7 841

24. Use figure 16.7.6 and write the side condition as x + y + z = 2π.

For (a) maximize f(x, y, z) = 8R3 sin 12x sin 1

2y sin 12z.

For (b) maximize f(x, y, z) = 4R2(sin2 12x + sin2 1

2y + sin2 12z).

Each maximum occurs with x = y = z =2π3

. This gives an equilateral triangle.

25. Since the curve is asymptotic to the line y = x as x → −∞ and as x → ∞, the maximum exists. The

distance between the point (x, y) and the line y − x = 0 is given by

|y − x|√1 + 1

=12

√2 |y − x|. (see Section 1.4)

Since the points on the curve are below the line y = x, we can replace |y − x| by x− y. To simplify

the work we drop the constant factor 12

√2.

f(x, y) = x− y, g(x, y) = x3 − y3 − 1

∇f = i − j, ∇g = 3x2i − 3y2j.

We use the cross-product equation (16.7.4):

1(−3y2

)−

(3x2

)(−1) = 0, 3x2 − 3y2 = 0, x = −y (x �= y).

Now g(x, y) = 0 gives us

x3 − (−x)3 − 1 = 0, 2x3 = 1, x = 2−1/3.

The point is(2−1/3,−2−1/3

).

26. Let r, s, t be the intercepts. We wish to minimize the volume

V =16rst [volume of pyramid =

13base × height]

subject to the side conditiona

r+

b

s+

c

t= 1. The minimum occurs when all the intercepts are:

r = 3a, s = 3b, t = 3c

27. It suffices to show that the square of the area is a maximum when a = b = c.

f(a, b, c) = s(s− a)(s− b)(s− c), g(a, b, c) = a + b + c− 2s

∇f = −s(s− b)(s− c)i − s(s− a)(s− c) j − s(s− a)(s− b)k, ∇g = i + j + k.

(Here i, j, k are the unit vectors in the directions of increasing a, b, c.)

∇f = λ∇g =⇒ −s(s− b)(s− c) = −s(s− a)(s− c) = −s(s− a)(s− b) = λ.

Thus, s− b = s− a = s− c so that a = b = c. This gives us the maximum, as no minimum exists. [The

area can be made arbitrarily small by taking a close to s.]

28. f(x, y, z) = 8xyz, g(x, y, z) = a2 − x2 − y2 − z2, x > 0, y > 0, z > 0.

∇f = 8yzi + 8xzj + 8xyk, ∇g = −2x i − 2y j − 2z k

∇f = λ∇g =⇒ 8yz = −2λx, 8xz = −2λy, 8xy = −2λz =⇒ x = y = z

The rectangular box of maximum volume inscribed in the sphere is a cube.



842 SECTION 16.7

29. (a) f(x, y) = (xy)1/2, g(x, y) = x + y − k, (x, y ≥ 0, k a nonnegative constant)

∇f =y1/2

2x1/2i +

x1/2

2y1/2j, ∇g = i + j.

∇f = λ∇g =⇒ y1/2

2x1/2= λ =

x1/2

2y1/2=⇒ x = y =

k

2.

Thus, the maximum value of f is: f(k/2, k/2) =k

2.

(b) For all x, y (x, y ≥ 0) we have

(xy)1/2 = f(x, y) ≤ f(k/2, k/2) =k

2=

x + y

2.

30. (a) The maximum occurs when x = y = z =k

3, where (xyz)1/3 =

k

3.

(b) If x + y + z = k, then, by (a), (xyz)1/3 ≤ k

3=

x + y + z

3.

31. Simply extend the arguments used in Exercises 29 and 30.

32. T (x, y, z) = xy2z, g(x, y, z) = x2 + y2 + z2 − 1

∇T = y2zi + 2xyzj + xy2k, ∇g = 2xi + 2yj + 2zk

∇T = λ∇g =⇒ y2z = 2λx, 2xyz = 2λy, xy2 = 2λz =⇒ x2 =y2

2= z2

From g(x, y, z) = 0 we get 4x2 = 1 =⇒ x = ± 12 , y = ± 1√

2, z = ± 1

2

Maximum 18 at ( 1

2 ,± 1√2, 1

2 ), (− 12 ,± 1√

2,− 1

2 )

Minimum − 18 at ( 1

2 ,± 1√2,− 1

2 ), (− 12 ,± 1√

2, 1

2 )

33. S(r, h) = 2πr2 + 2πrh, g(r, h) = πr2h− V, (V constant)

∇S = (4πr + 2πh) i + 2πr j, ∇g = 2πrh i + πr2 j.

∇S = λ∇g =⇒ 4πr + 2πh = 2πrhλ, 2πr = πr2λ =⇒ r =2λ, h =

4λ.

Now πr2h = V, =⇒ λ = 3

√16πV

=⇒ r = 3

√V

2π, h = 3

√4Vπ

.

To minimize the surface area, take r = 3

√V

2π, and h = 3

√4Vπ

.

34. f(x, y, z) = xyz, g(x, y, z) = x + y + z − 18

∇f = yzi + xzj + xyk, ∇g = i + j + k

∇f = λ∇g =⇒ yz = xz = xy =⇒ x = y = z =⇒ x = y = z = 6

35. Same as Exercise 13.



SECTION 16.7 843

36. f(x, y, z) = xyz2, g(x, y, z) = x + y + z − 30

∇f = yz2i + xz2j + 2xyzk, ∇g = i + j + k

∇f = λ∇g =⇒ yz2 = λ = xz2 = 2xyz =⇒ x = y =z

2=⇒ x = y =

152, z = 15

37. Let x, y, z denote the length, width and height of the box. We want to maximize the volume V of the

box given that the surface area S is constant. That is:

maximize V (x, y, z) = xyz subject to S(x, y, z) = 2xy + 2xz + 2yz = S constant

Let g(x, y, z) = 2xy + 2xz + 2yz − S. Then

∇V = yz i + xz j + xy k, ∇g = (2y + 2z) i + (2x + 2z) j + (2x + 2y)k

∇V = λ∇g and the side condition yield the system of equations:

yz = λ (2y + 2z)

xz = λ (2x + 2z)

xy = λ (2x + 2y)

xy + 2xz + 2yz = S.

Multiply the first equation by x, the second by y and subtract. This gives

0 = 2λ z(x− y) =⇒ x = y since z = 0 =⇒ V = 0.

Multiply the second equation by y, the third by z and subtract. This gives

0 = 2λx(y − z) =⇒ y = z since x = 0 =⇒ V = 0.

Thus the closed rectangular box of maximum volume is a cube. The cube has side length x =√

S/6.

38. Let x, y, z denote the length, width and height of the box. We want to maximize the volume V of the

box given that the surface area S is constant. That is:

maximize V (x, y, z) = xyz subject to S(x, y, z) = 2xy + 2xz + 2yz = S constant

Let g(x, y, z) = xy + 2xz + 2yz − S. Then

∇V = yz i + xz j + xy k, ∇g = (y + 2z) i + (x + 2z) j + (2x + 2y)k

∇V = λ∇g and the side condition yield the system of equations:

yz = λ (y + 2z)

xz = λ (x + 2z)

xy = λ (2x + 2y)

xy + 2xz + 2yz = S.

Multiplying the first equation by x, the second by y and subtracting, we get

0 = 2λ z(x− y) =⇒ x = y since z = 0 =⇒ V = 0.



844 SECTION 16.7

Now put y = x in the third equation. This gives

x2 = 4λx =⇒ x(x− 4λ) = 0 =⇒ x = 4λ since x = 0 =⇒ V = 0.

Thus, x = y = 4λ. Substituting x = 4λ in the second equation gives z = 2λ.

Finally, substituting these values for x, y, z in the fourth equation, we get

48λ2 = S =⇒ λ2 =S

48=⇒ λ =

14

√S

3

To maximize the volume, take x = y =

√S

3and z =

12

√S

3.

39. S(r, h) = 4πr2 + 2πrh, g(r, h) =43πr3 + πr2h− 10, 000

∇S = (8πr + 2πh)i + 2πrj, ∇g = (4πr2 + 2πrh)i + πr2j

(Here i, j are the unit vectors in the directions of increasing r and h.)

∇S = λ∇g =⇒ 2π(4r + h) = 2πrλ(2r + h), 2πr = λπr2 =⇒ h = 0

Maximum volume for sphere of radius r = 3√

7500/π meters.

40. (a) f(x, y, l) = xyl, g(x, y, l) = 2x + 2y + l − 108,

∇f = yl i + xl j + xy k, ∇g = 2 i + 2 j + k.

∇f = λ∇g =⇒ yl = 2λ, xl = 2λ, xy = λ =⇒ y = x and l = 2x.

Now 2x + 2y + l = 108, =⇒ x = 18 and l = 36.

To maximize the volume, take x = y = 18 in. and l = 36 in.

(b) f(r, l) = πr2l, g(r, l) = 2πr + l − 108,

∇f = 2πrl i + πr2 j, ∇g = 2π i + j.

∇f = λ∇g =⇒ 2πrl = 2πλ, πr2 = λ, l = πr.

Now 2πr + l = 108, =⇒ r =36π

and l = 36.

To maximize the volume, take r = 36/π in. and l = 36 in.

41. f(x, y, z) = 8xyz, g(x, y, z) = 4x2 + 9y2 + 36z2 − 36.

∇f(x, y, z) = 8yzi + 8xzj + 8xyk, ∇g(x, y, z) = 8xi + 18yj + 72zk.

∇f = λ∇g gives

yz = λx, 4xz = 9λy, xy = 9λz.

4xyz

λ= 4x2, 4

xyz

λ= 9y2, 4

xyz

λ= 36z2.

Also notice

4x2 + 9y2 + 36z2 − 36 = 0

We have

12xyz

λ= 36 =⇒ x =

√3, y =

2√3, z =

1√3.



SECTION 16.7 845

Thus,

V = 8xyz = 8 ·√

3 · 2√3· 1√

3=

16√3.

42. Let x, y, z denote the length, width and height of the crate, and let C be the cost. Then

C(x, y, z) = 30xy + 20xz + 20yz subject to g(x, y, z) = xyz − 96 = 0

∇C = (30y + 20z) i + (30x + 20z) j + (20x + 20y)k, ∇g = yz i + xz j + xy k

∇C = λ∇g implies

30y + 20z = λ yz

30x + 20z = λxz

20x + 20y = λxy

xyz = 96

Multiplying the first equation by x, the second by y and subtracting, we get

20z(x− y) = 0 =⇒ x = y since z �= 0

Now put y = x in the third equation. This gives

40x = λx2 =⇒ x(λx− 40) = 0 =⇒ x =40λ

since x �= 0

Thus, x = y = 40/λ. Substituting x = 40/λ in the second equation gives z = 60/λ.

Finally, substituting these values for x, y, z in the fourth equation, we get40λ

40λ

60λ

= 96 =⇒ 96λ3 = 96, 000 =⇒ λ3 = 1000 =⇒ λ = 10

To minimize the cost, take x = y = 4 meters and z = 6 meters.

43. To simplify notation we set x = Q1, y = Q2, z = Q3.

f(x, y, z) = 2x + 8y + 24z, g(x, y, z) = x2 + 2y2 + 4z2 − 4, 500, 000, 000

∇f = 2i + 8j + 24k, ∇g = 2xi + 4yj + 8zk.

∇f = λ∇g =⇒ 2 = 2λx, 8 = 4λy, 24 = 8λz.

Since λ �= 0 here, we solve the equations for x, y, z:

x =1λ, y =

2λ, z =

3λ,

and substitute these results in g(x, y, z) = 0 to obtain

1λ2

+ 2(

4λ2

)+ 4

(9λ2

)− 45 × 108 = 0,

45λ2

= 45 × 108, λ = ±10−4.

Since x, y, z are non-negative, λ = 10−4 and

x = 104 = Q1, y = 2 × 104 = Q2, z = 3 × 104 = Q3.



846 SECTION 16.7

44. f(x, y, z) = 8xyz, g(x, y, z) =x2

a2+

y2

b2+

z2

c2− 1. We take a, b, c, x, y, z > 0

∇f(x, y, z) = 8yzi + 8xzj + 8xyk; ∇g(x, y, z) =2xa2

i +2yb2

j +2zc2

k.

∇f = λ∇g and the side condition yield the system of equations:

8yz =2xλa2

8xz =2yλb2

8xy =2zλc2

x2

a2+

y2

b2+

z2

c2= 1

Multiply the first equation by x, the second by y and subtract. This gives

0 =2λx2

a2− 2λy2

b2=⇒ y =

b

ax since λ = 0 =⇒ V = 0.

Multiply the second equation by y, the third by z and subtract. This gives

0 =2λy2

b2− 2λz2

c2=⇒ z =

c

by =

c

ax.

Substituting these results into the side condition, we get:

3x2

a2= 1 =⇒ x =

a√3

=⇒ y =b√3

and z =c√3.

The volume of the largest rectangular box is: V = 8(

a√3

) (b√3

) (c√3

)=

8√

39

abc.

PROJECT 16.7

1. f(x, y, z) = xy + z2, g(x, y, z) = x2 + y2 + z2 − 4, h(x, y, z) = y − x

∇f = y i + x j + 2z k, ∇g = 2x i + 2y j + 2z k, ∇h = −i + j.

∇f = λ∇g + μ∇h =⇒ y = 2λx− μ, x = 2λy − μ, 2z = 2λz

2z = 2λz =⇒ λ = 0 or z = 1.

λ = 0 =⇒ y = −x which contradicts y = x.

z = 1 =⇒ x2 + y2 = 3, which, with y = x implies x = ±√

3/2;(±

√3/2,±

√3/2

)



SECTION 16.7 847

Adding the first two equations gives

x + y = 2λ(x + y) =⇒ (x + y)[2λ− 1] = 0 =⇒ λ =1c

or x = y = 0.

x = y = 0 =⇒ z = ±2; (0, 0,±2).

λ =12

=⇒ z = 0 and y = x =⇒ 2x4 = 4; x = ±√

2; (±√

2,±√

2, 0).

f(±

√3/2,±

√3/2, 1

)=

52; f(0, 0,±2) = 4; f(±

√2,±

√2, 0) = 2.

The maximum value of f is 4; the minimum value is 2.

2. D(x, y, z) = x2 + y2 + z2, g(x, y, z) = x + 2y + 3z, h(x, y, z) = 2x + 3y + z − 4

∇D = 2xi + 2yj + 2zk, ∇g = i + 2j + 3k, ∇h = 2i + 3j + k

∇D = λ∇g + μ∇h =⇒ 2x = λ + 2μ, 2y = 2λ + 3μ, 2z = 3λ + μ =⇒ z = 5y − 7x

Then g(x, y, z) = 0 and h(x, y, z) = 0 give x =6875

, y =1615

, z = −7675

Closest point(

6875

,1615

,−7675

)

3. f(x, y, z) = x2 + y2 + z2, g(x, y, z) = x + y − z + 1, h(x, y, z) = x2 + y2 − z2

∇f = 2x i + 2y j + 2z k, ∇g = i + j − k, ∇h = 2x i + 2y j − 2z k.

∇f = λ∇g + μ∇h =⇒ 2x = λ + 2xμ, 2y = λ + 2yμ, 2z = −λ− 2zμ

Multiplying the first equation by y, the second equation by x and subtracting, yields

λ(y − x) = 0.

Now λ = 0 =⇒ μ = 1 =⇒ x = y = z = 0. This is impossible since x + y − z = −1.

Therefore, we must have y = x =⇒ z = ±√

2x.

Substituting y = x, z =√

2x into the equation x + y − z + 1 = 0, we get

x = −1 −√

22

=⇒ y = −1 −√

22

, z = −1 −√

2

Substituting y = x, z = −√

2x into the equation x + y − z + 1 = 0, we get

x = −1 +√

22

=⇒ y = −1 +√

22

, z = −1 +√

2



848 SECTION 16.8

Since

f

(−1 −

√2

2, −1 −

√2

2, −1 −

√2

)= 6 + 4

√2 and

f

(−1 +

√2

2, −1 +

√2

2, −1 +

√2

)= 6 − 4

√2,

it follows that

(−1 +

√2

2, −1 +

√2

2, −1 +

√2

)is closest to the origin and

(−1 −

√2

2, −1 −

√2

2, −1 −

√2

)is furthest from the origin.

SECTION 16.8

1. df =(3x2y − 2xy2

)Δx +

(x3 − 2x2y

)Δy

2. df =∂f

∂xΔx +

∂f

∂yΔy +

∂f

∂zΔz = (y + z)Δx + (x + z)Δy + (x + y)Δz

3. df = (cos y + y sin x) Δx− (x sin y + cos x) Δy

4. df = 2xye2zΔx + x2e2zΔy + 2x2ye2zΔz

5. df = Δx− (tan z) Δy −(y sec2 z

)Δz

6. df =[x− y

x + y+ ln(x + y)

]Δx +

[x− y

x + y− ln(x + y)

]Δy

7. df =y(y2 + z2 − x2)(x2 + y2 + z2)2

Δx +x(x2 + z2 − y2)(x2 + y2 + z2)2

Δy − 2xyz(x2 + y2 + z2)2

Δz

8. df =[

2xx2 + y2

+ exy(1 + xy)]

Δx +[

2yx2 + y2

+ x2exy]

Δy

9. df = [cos(x + y) + cos(x− y)] Δx + [cos(x + y) − cos(x− y)] Δy

10. df = ln(

1 + y

1 − y

)Δx +

2x1 − y2

Δy

11. df =(y2zexz + ln z

)Δx + 2yexz Δy +

(xy2exz +

x

z

)Δz

12. df = y(1 − 2x2)e−(x2+y2)Δx + x(1 − 2y2)e−(x2+y2)Δy



SECTION 16.8 849

13. Δu =[(x + Δx)2 − 3(x + Δx)(y + Δy) + 2(y + Δy)2

]−

(x2 − 3xy + 2y2

)=

[(1.7)2 − 3(1.7)(−2.8) + 2(−2.8)2

]−

(22 − 3(2)(−3) + 2(−3)2

)= (2.89 + 14.28 + 15.68) − 40 = −7.15

du = (2x− 3y) Δx + (−3x + 4y) Δy

= (4 + 9)(−0.3) + (−6 − 12)(0.2) = −7.50

14. du =(√

x− y +x + y

2√x− y

)Δx +

(√x− y − x + y

2√x− y

)Δy = 1

15. Δu =[(x + Δx)2(z + Δz) − 2(y + Δy)(z + Δz)2 + 3(x + Δx)(y + Δy)(z + Δz)

]−

(x2z − 2yz2 + 3xyz

)=

[(2.1)2(2.8) − 2(1.3)(2.8)2 + 3(2.1)(1.3)(2.8)

]−

[(2)23 − 2(1)(3)2 + 3(2)(1)(3)

]= 2.896

du = (2xz + 3yz) Δx +(−2z2 + 3xz

)Δy +

(x2 − 4yz + 3xy

)Δz

= [2(2)(3) + 3(1)(3)](0.1) + [−2(3)2 + 3(2)(3)](0.3) + [22 − 4(1)(3) + 3(2)(1)](−0.2) = 2.5

16. du =y3 + yz2

(x2 + y2 + z2)3/2Δx +

x3 + xz2

(x2 + y2 + z2)3/2Δy − xyz

(x2 + y2 + z2)3/2Δz =

774(14)3/2

17. f(x, y) = x1/2y1/4; x = 121, y = 16, Δx = 4, Δy = 1

f(x + Δx, y + Δy) ∼= f(x, y) + df

= x1/2 y1/4 + 12x

−1/2 y1/4 Δx + 14x

1/2 y−3/4 Δy

√125 4

√17 ∼=

√121 4

√16 + 1

2 (121)−1/2 (16)1/4 (4) + 14 (121)1/2 (16)−3/4 (1)

= 11(2) + 12

(111

)(2)(4) + 1

4 (11)(

18

)= 22 + 4

11 + 1132 = 22 249

352∼= 22.71

18. f(x, y) = (1 −√x)(1 +

√y), x = 9, y = 25, Δx = 1, Δy = −1

df = −1 +√y

2√x

Δx +1 −√

x

2√y

Δy = −45

f(10, 24) ∼= f(9, 25) − 45

= −1245

19. f(x, y) = sinx cos y; x = π, y =π

4, Δx = −π

7, Δy = − π

20df = cosx cos yΔx− sinx sin yΔy

f(x + Δx, y + Δy) ∼= f(x, y) + df

sin67π cos

15π ∼= sinπ cos

π

4+

(cosπ cos

π

4

) (−π

7

)−

(sinπ sin

π

4

) (− π

20

)

= 0 +(

12

√2

) (π

7

)+ 0 =

π√

214

∼= 0.32



850 SECTION 16.8

20. f(x, y) =√x tan y, x = 9, y =

π

4, Δx = −1, Δy =

116

π

df =1

2√x

tan yΔx +√x sec2 yΔy = −1

6+

3π8

f

(8,

516

π

)∼= f

(9,

π

4

)− 1

6+

3π8

=176

+38π ∼= 4.01

21. f(2.9, 0.01) ∼= f(3, 0) + df, where df is to be evaluated at x = 3, y = 0, Δx = −0.1, Δy = 0.01.

df =(2xexy + x2yexy

)Δx + x3exy Δy =

[2(3)e0 + (3)2(0)e0

](−0.1) + 33e0(0.01) = − 0.33

Thus, f(2.9, .01) ∼= 32e0 − 0.33 = 8.67.

22. x = 2, y = 3, z = 3, Δx = 0.12, Δy = −0.08, Δz = 0.02

df = 2xy cosπzΔx + x2 cosπzΔy − πx2y sinπzΔz = −12(0.12) + 4(0.08) = −1.12

f(2.12, 2.92, 3.02) ∼= f(2, 3, 3) − 1.12 = −13.12

23. f(2.94, 1.1, 0.92) ∼= f(3, 1, 1) + df, where df is to be evaluated at x = 3, y = 1, z = 1,

Δx = −0.06, Δy = 0.1, Δz = −0.08

df = tan−1 yz Δx +xz

1 + y2z2Δy +

xy

1 + y2z2Δz =

π

4(−0.06) + (1.5)(0.1) + (1.5)(−0.08) ∼= −0.0171

Thus, f(2.94, 1.1, 0.92) ∼= 34π − 0.0171 ∼= 2.3391

24. x = 3, y = 4, Δx = 0.06, Δy = −0.12

df =x√

x2 + y2Δx +

y√x2 + y2

Δy =35(0.06) +

45(−0.12) = −0.06

f(3.06, 3.88) ∼= f(3, 4) − 0.06 = 4.94

25. df =∂z

∂xΔx +

∂z

∂yΔy =

2y(x + y)2

Δx− 2x(x + y)2

Δy

With x = 4, y = 2, Δx = 0.1, Δy = 0.1, we get

df = 436 (0.1) − 8

36 (0.1) = − 190 .

The exact change is4.1 − 2.14.1 + 2.1

− 4 − 24 + 2

=2

6.2− 1

3= − 1

93.

26. V (r, h) = πr2h, r = 8, h = 12, Δr = −0.3, Δh = 0.2

dV = 2πrhΔr + πr2Δh = 192π(−0.3) + 64π(0.2) = −44.8π

decreases by approximately 44.8π cubic inches.

27. S = 2πr2 + 2πrh; r = 8, h = 12, Δr = −0.3, Δh = 0.2

dS =∂S

∂rΔr +

∂S

∂hΔh = (4πr + 2πh) Δr + (2πr) Δh

= 56π(−0.3) + 16π(0.2) = −13.6π.

The area decreases about 13.6π in.2.



SECTION 16.8 851

28. dT = 2x cosπzΔx− 2y sinπzΔy − (πx2 sinπz + πy2 cosπz)Δz

= 4(0.1) − 4π(0.2) =25− 4

5π

T decreases by about45π − 2

5∼= 2.11

29. S(9.98, 5.88, 4.08) ∼= S(10, 6, 4) + dS = 248 + dS, where

dS = (2w + 2h) Δl + (2l + 2h) Δw + (2l + 2w) Δh = 20(−0.02) + 28(−0.12) + 32(0.08) = −1.20

Thus, S(9.98, 5.88, 4.08) ∼= 248 − 1.20 = 246.80.

30. f(r, h) =πr2h

3, r = 7, h = 10, Δr = 0.2, Δh = 0.15

df =2πrh

3Δr +

πr2

3Δh =

1403

π(0.2) +493π(0.15) =

π

3(35.35)

f(7.2, 10.15) ∼= f(7, 10) +π

3(35.35) = 525.35

π

3∼= 550.15

31. (a) dV = yz Δx + xz Δy + xyΔz = (8)(6)(0.02) + (12)(6)(−0.05) + (12)(8)(0.03) = 0.24

(b) ΔV = (12.02)(7.95)(6.03) − (12)(8)(6) = 0.22077

32. (a) S(x, y, z) = 2(xy + xz + yz), x = 12, y = 8, z = 6, Δx = 0.02, Δy = −0.05, Δz = 0.03

dS = 2(y + z)Δx + 2(x + z)Δy + 2(x + y)Δz = 28(0.02) + 36(−0.05) + 40(0.03) = −0.04

(b) ΔS = S(12.02, 7.95, 6.03) − S(12, 8, 6) = −0.0438

33. T (P ) − T (Q) ∼= dT = (−2x + 2yz) Δx + (−2y + 2xz) Δy + (−2z + 2xy) Δz

Letting x = 1, y = 3, z = 4, Δx = 0.15, Δy = −0.10, Δz = 0.10, we have

dT = (22)(0.15) + (2)(−0.10) + (−2)(0.10) = 2.9

34. Amount of paint is increase in volume. f(x, y, z) = xyz, x = 48 in, y = 24 in, z = 36 in,

Δx = Δy = Δz = 216 in. Δf ∼= df = yzΔx + xzΔy + xyΔz = 3774( 2

16 ) = 468

The amount of paint is approximately 468 cubic inches.

35. (a) πr2h = π(r + Δr)2(h + Δh) =⇒ Δh =r2h

(r + Δr)2− h = − (2r + Δr)h

(r + Δr)2Δr.

df = (2πrh) Δr + πr2 Δh, df = 0 =⇒ Δh =−2hr

Δr.

(b) 2πr2 + 2πrh = 2π(r + Δr)2 + 2π(r + Δr)(h + Δh).

Solving for Δh,

Δh =r2 + rh− (r + Δr)2

r + Δr− h = −2r + h + Δr

r + ΔrΔr.

df = (4πr + 2πh) Δr + 2πrΔh, df = 0 =⇒ Δh = −(

2r + h

r

)Δr.



852 SECTION 16.9

36. The area is given by A = 12 x

2 tan θ.

(a) The change in area is approximated by:

dA = x tan θΔx + 12 x

2 sec2 θΔθ = 3Δx + 252 Δθ.

(b) The actual change in area is12(x + Δx)2 tan(θ + Δθ) − 1

2x2 tan θ =

12[4 + Δx]2 tan[arctan (3/4) + Δθ] − 6.

(c) The area is more sensitive to a change in θ.

37. (a) A =12x2 sin θ; ΔA ∼= dA = x sin θΔx +

x2

2cos θΔθ

(b) The area is more sensitive to changes in θ if x > 2 tan θ, otherwise it is more sensitive to changes

in x.

38. (a) dV ∼= yzΔx + xzΔy + xyΔz, x = 60 in, y = 36 in, z = 42 in

Maximum possible error = 6192( 112 ) = 516 cubic inches.

(b) dS ∼= 2(y + z)Δx + 2(x + z)Δy + 2(x + y)Δz

Maximum possible error = 552( 112 ) = 46 square inches

39. s =A

A−W; A = 9, W = 5, ΔA = ±0.01, ΔW = ±0.02

ds =∂s

∂AΔA +

∂s

∂WΔW =

−W

(A−W )2ΔA +

A

(A−W )2ΔW

= − 516

(±0.01) +916

(±0.02) ∼= ±0.014

The maximum possible error in the value of s is 0.014 lbs; 2.23 ≤ s + Δs ≤ 2.27

40. Assuming A > W , s is more sensitive to change in A.

SECTION 16.9

1.∂f

∂x= xy2, f(x, y) = 1

2x2y2 + φ(y),

∂f

∂y= x2y + φ′(y) = x2y.

Thus, φ′(y) = 0, φ(y) = C, and f(x, y) = 12x

2y2 + C.

2.∂f

∂x= x,

∂f

∂y= y =⇒ f(x, y) =

12(x2 + y2) + C

3.∂f

∂x= y, f(x, y) = xy + φ(y),

∂f

∂y= x + φ′(y) = x.

Thus, φ′(y) = 0, φ(y) = C, and f(x, y) = xy + C.

4.∂f

∂x= x2 + y =⇒ f(x, y) =

x3

3+ xy + φ(y);

∂f

∂y= x + φ′(y) = y3 + x =⇒ f(x, y) =

13x3 +

14y4 + xy + C

5. No;∂

∂y

(y3 + x

)= 3y2 whereas

∂

∂x

(x2 + y

)= 2x.



SECTION 16.9 853

6.∂f

∂x= y2ex − y =⇒ f(x, y) = y2ex − xy + φ(y);

∂f

∂y= 2yex − x + φ′(y) = 2yex − x =⇒ f(x, y) = y2ex − xy + C

7.∂f

∂x= cosx− y sinx, f(x, y) = sinx + y cosx + φ(y),

∂f

∂y= cosx + φ′(y) = cosx.

Thus, φ′(y) = 0, φ(y) = C, and f(x, y) = sinx + y cosx + C.

8.∂f

∂x= 1 + ey =⇒ f(x, y) = x + xey + φ(y);

∂f

∂y= xey + φ′(y) = xey + y2 =⇒ f(x, y) = x + xey +

y3

3+ C

9.∂f

∂x= ex cos y2, f(x, y) = ex cos y2 + φ(y),

∂f

∂y= −2yex sin y2 + φ′(y) = −2yex sin y2.

Thus, φ′(y) = 0, φ(y) = C, and f(x, y) = ex cos y2 + C.

10.∂2f

∂y∂x= −ex sin y,

∂2f

∂x∂y= ex sin y �= ∂2f

∂y∂x; not a gradient.

11.∂f

∂y= xex − e−y, f(x, y) = xyex + e−y + φ(x),

∂f

∂x= yex + xyex + φ′(x) = yex(1 + x).

Thus, φ′(x) = 0, φ(x) = C, and f(x, y) = xyex + e−y + C.

12.∂f

∂x= ex + 2xy =⇒ f(x, y) = ex + x2y + φ(y);

∂f

∂y= x2 + φ′(y) = x2 + sin y

=⇒ f(x, y) = ex + x2y − cos y + C

13. No;∂

∂y

(xexy + x2

)= x2exy whereas

∂

∂x(yexy − 2y) = y2exy

14.∂f

∂y= x sinx + 2y + 1 =⇒ f(x, y) = xy sinx + y2 + y + φ(x)

∂f

∂x= y sinx + xy cosx + φ′(x) = y sinx + xy cosx =⇒ f(x, y) = xy sinx + y2 + y + C

15.∂f

∂x= 1 + y2 + xy2, f(x, y) = x + xy2 + 1

2 x2y2 + φ(y),

∂f

∂y= 2xy + x2y + φ′(y) = x2y + y + 2xy + 1.

Thus, φ′(y) = y + 1, φ(y) = 12 y

2 + y + C and f(x, y) = x + xy2 + 12 x

2y2 + 12 y

2 + y + C.

16.∂f

∂x= 2 ln 3y +

1x

=⇒ f(x, y) = 2x ln 3y + ln |x| + φ(y);∂f

∂y=

2xy

+ φ′(y) =2xy

+ y2

f(x, y) = 2x ln 3y + ln |x| + y3

3+ C

17.∂f

∂x=

x√x2 + y2

, f(x, y) =√x2 + y2 + φ(y),

∂f

∂y=

y√x2 + y2

+ φ′(y) =y√

x2 + y2.

Thus, φ′(y) = 0, φ(y) = C, and f(x, y) =√

x2 + y2 + C.



854 SECTION 16.9

18.∂f

∂x= x tan y + sec2 x =⇒ f(x, y) =

x2

2tan y + tanx + φ(y);

∂f

∂y=

x2

2sec2 y + φ′(y) =

x2

2sec2 y + πy; =⇒ f(x, y) =

x2

2tan y + tanx +

π

2y2 + C.

19.∂f

∂x= x2 sin−1 y, f(x, y) = 1

3x3 sin−1 y + φ(y),

∂f

∂y=

x3

3√

1 − y2+ φ′(y) =

x3

3√

1 − y2− ln y.

Thus, φ′(y) = − ln y, =⇒ φ(y) = y − y ln y + C, and

f(x, y) =13x3 sin−1 y + y − y ln y + C.

20.∂f

∂x=

tan−1 y√1 − x2

+x

y=⇒ f = sin−1 x tan−1 y +

x2

2y+ φ(y);

∂f

∂y=

sin−1 x

1 + y2− x2

2y2+ φ′(y) =

sin−1 x

1 + y2− x2

2y2+ 1 =⇒ f(x, y) = sin−1 x tan−1 y +

x2

2y+ y + C.

21. (a) Yes (b) Yes (c) No

22. (a) f(x, y) = (x− y)e−x2y + C

(b) f(x, y) = sin(x + y) − cos(x− y) + C; f(π/3, π/4) = 6 =⇒ C = 6

f(x, y) = sin(x + y) − cos(x− y) + 6.

23.∂f

∂x= f(x, y),

∂f/∂x

f(x, y)= 1, ln f(x, y) = x + φ(y),

∂f/∂y

f(x, y)= 0 + φ′(y),

∂f

∂y= f(x, y).

Thus, φ′(y) = 1, φ(y) = y + K, and f(x, y) = ex+y+K = Cex+y.

24.∂f

∂x= eg(x,y)gx(x, y) =⇒ f(x, y) = eg(x,y) + φ(y);

∂f

∂y= eg(x,y)gy(x, y) + φ′(y) = eg(x,y)gy(x, y) =⇒ f(x, y) = eg(x,y) + C.

25. (a) P = 2x, Q = z, R = y;∂P

∂y= 0 =

∂Q

∂x,

∂P

∂z= 0 =

∂R

∂x,

∂Q

∂z= 1 =

∂R

∂y(b), (c), and (d)

∂f

∂x= 2x, f(x, y, z) = x2 + g(y, z).

∂f

∂y= 0 +

∂g

∂ywith

∂f

∂y= z =⇒ ∂g

∂y= z.

Then,

g(y, z) = yz + h(z) =⇒ f(x, y, z) = x2 + yz + h(z),

∂f

∂z= 0 + y + h′(z) and

∂f

∂z= y =⇒ h′(z) = 0.

Thus, h(z) = C and f(x, y, z) = x2 + yz + C.

26.∂f

∂x= yz =⇒ f(x, y, z) = xyz + g(y, z);

∂f

∂y= xz +

∂g

∂y= xz =⇒ f = xyz + h(z)

∂f

∂z= xy + h′(z) = xy =⇒ f(x, y, z) = xyz + C



SECTION 16.9 855

27. The function is a gradient by the test stated before Exercise 25.

Take P = 2x + y, Q = 2y + x + z, R = y − 2z. Then∂P

∂y= 1 =

∂Q

∂x,

∂P

∂z= 0 =

∂R

∂x,

∂Q

∂z= 1 =

∂R

∂y.

Next, we find f where ∇f = P i + Qj + Rk.

∂f

∂x= 2x + y =⇒ f(x, y, z) = x2 + xy + g(y, z).

∂f

∂y= x +

∂g

∂ywith

∂f

∂y= 2y + x + z =⇒ ∂g

∂y= 2y + z.

Then,

g(y, z) = y2 + yz + h(z),

f(x, y, z) = x2 + xy + y2 + yz + h(z).

∂f

∂z= y + h′(z) = y − 2z =⇒ h′(z) = −2z.

Thus, h(z) = −z2 + C and f(x, y, z) = x2 + xy + y2 + yz − z2 + C.

28.∂f

∂x= 2x sin 2y cos z =⇒ f(x, y, z) = x2 sin 2y cos z + g(y, z);

∂f

∂y= 2x2 cos 2y cos z +

∂g

∂y= 2x2 cos 2y cos z =⇒ f(x, y, z) = x2 sin 2y cos z + h(z)

∂f

∂z= −x2 sin 2y sin z + h′(z) = −x2 sin 2y sin z =⇒ f(x, y, z) = x2 sin 2y cos z + C

29. The function is a gradient by the test stated before Exercise 25.

Take P = y2z3 + 1, Q = 2xyz3 + y, R = 3xy2z2 + 1. Then

∂P

∂y= 2yz3 =

∂Q

∂x,

∂P

∂z= 3y2z2 =

∂R

∂x,

∂Q

∂z= 6xyz2 =

∂R

∂y.

Next, we find f where ∇f = P i + Qj + Rk.

∂f

∂x= y2z3 + 1,

f(x, y, z) = xy2z3 + x + g(y, z).

∂f

∂y= 2xyz3 +

∂g

∂ywith

∂f

∂y= 2xyz3 + y =⇒ ∂g

∂y= y.

Then,

g(y, z) = 12 y

2 + h(z),

f(x, y, z) = xy2z3 + x + 12 y

2 + h(z).

∂f

∂z= 3xy2z2 + h′(z) = 3xy2z2 + 1 =⇒ h′(z) = 1.

Thus, h(z) = z + C and f(x, y, z) = xy2z3 + x + 12 y

2 + z + C.



856 REVIEW EXERCISES

30.∂f

∂x=

y

z− ez =⇒ f(x, y, z) =

xy

z− xez + g(y, z)

∂f

∂y=

x

z+

∂g

∂y=

x

z+ 1 =⇒ f(x, y, z) =

xy

z+ y − xez + h(z)

∂f

∂z= −xy

z2− xez + h′(z) = −xez − xy

z2=⇒ f(x, y, z) =

xy

z− xez + y + C

31. F(r) = ∇(GmM

r

)

32.h(r) =

⎧⎪⎪⎨⎪⎪⎩

∇(

k

n + 2rn+2

), n �= 2

∇ (k ln r) , n = −2.

REVIEW EXERCISES

1. ∇f(x, y) = (4x− 4y)i + (3y2 − 4x)j 2. ∇f(x, y) =y3 − x2y

(x2 + y2)2i +

x3 − xy2

(x2 + y2)2j

3. ∇f(x, y) = (yexy tan 2x + 2exy sec2 2x) i + xexy tan 2x j

4. ∇f =1

x2 + y2 + z2(x i + y j + z k)

5. ∇f(x, y) = 2xe−yz sec z i,−zx2e−yz sec zj − (x2ye−yz sec z − x2e−yz sec z tan z)k

6. ∇f(x, y) = ye−3z cosxy i + e−3z(x cosxy + sin y) j,−3e−3z(sinxy − cos y)k

7. ∇f(x, y) = (2x− 2y) i − 2x j, ∇f(1,−2) = 6 i − 2 j; ua =1√5

i +2√5

j;

f ′ua

(1,−2) = ∇f(1,−2) · ua =2√5.

8. ∇f(x, y) = (exy + xyexy) i + x2exy j, ∇f(2, 0) = i + 4j; ua =12i +

√3

2j

f ′ua

(2, 0) = ∇f(2, 0) · ua =12

+ 2√

3.

9. ∇f(x, y, z) = (y2 + 6xz) i + (2xy + 2z) j + (2y + 3x2)k, ∇f(1,−2, 3) = 22 i + 2 j − k;

ua =13i − 2

3j +

23k; f ′

ua(1,−2, 3) = ∇f(1,−2, 3) · ua =

163

.

10. ∇f(x, y, z) =2x

x2 + y2 + z2i +

2yx2 + y2 + z2

j +2z

x2 + y2 + z2k, ∇f(1, 2, 3) =

17(i + 2 j + 3k);

ua =1√3

i − 1√3

j +1√3

k; f ′ua

(1, 2, 3) = ∇f(1, 2, 3) · ua =2

7√

3.



REVIEW EXERCISES 857

11. ∇f(x, y) = (6x− 2y2) i − 4xy j, ∇f(3,−2) = 10 i + 24 j;

a = (0, 0) − (3,−2) = (−3, 2) = −3 i + 2 j, ua =−3√13

i +2√13

j;

f ′ua

(3,−2) = ∇f(3,−2) · ua =18√13

.

12. ∇f(x, y, z) = (y2z − 3yz) i + (2xyz − 3xz) j + (xy2 − 3xy)k, ∇f(1,−1, 2) = 8 i − 10 j + 4k;

r′(t) = i − π sin πt j + 2et−1 k, a = r′(1) = i + 2k, ua =1√5

i +2√5

k;

f ′ua

(1,−1, 2) = ∇f(1,−1, 2) · ua =16√

5.

13. ∇f(x, y, z) =1√

x2 + y2 + z2(x i + y j + z k), ∇f(3,−1, 4) =

1√26

(3 i − j + 4k);

a = ±(4 i − 3 j + k), ua = ± 1√26

(4 i − 3 j + k); f ′ua

(3,−1, 4) = ∇f(3,−1, 4) · ua = ±1926

.

14. ∇f(x, y) = 2e2x(cos y − sin y) i − e2x(sin y + cos y) j, ∇f(

12 ,− 1

2π)

= 2e i + e j;

maximum directional derivative: ‖∇f(

12 ,− 1

2π)‖ = e

√5.

15. ∇f(x, y, z) = cosxyz(yz i + xz j + xy k), ∇f( 12 ,

13 , π) = π

√3

6 i + π√

34 j +

√3

12 k;

minimum directional derivative: f ′u = −‖∇f( 1

2 ,13 , π)‖ = −

√39π2+3

12

16. Let r(t) = x(t) i + y(t) j be the path of the particle. ∇I(x, y) = −2x i − 6y j. Then

x′(t) = −2x(t), y′(t) = −6y(t) =⇒ x(t) = C1e−2t, y(t) = C2e

−6t.

r(0) = (4, 3) =⇒ C1 = 4, C2 = 3.

Therefore the path of the particle is: r(t) = 4e−2t i + 3e−6t j, t ≥ 0, or, y = 364x

3, 0 < x ≤ 4

17. Let r(t) = x(t) i + y(t) j be the path of the particle. ∇T = −e−x cos y i − e−x sin y j. Then

x′(t) = −e−x(t) cos y(t), y′(t) = −e−x(t) sin y(t) =⇒ y′(t)x′(t)

= tan y(t) =⇒ dy

dx= tan y

The solution is sin y = Cex. Since r(0) = 0, C = 0 and y = 0. The particle moves to the right the

x-axis.

18. ∇z = 8x i + 2y j; r(t) = x(t) i + y(t) j.

x′(t) = −8x(t), y′(t) = −2y(t) =⇒ x(t) = C1e−8t, y(t) = C2e

−2t.

(a) r(0) = (1, 1) =⇒ C1 = 1, C2 = 1; x = e−8t, y = e−2t or x = y4.

(b) r(0) = (1,−2) =⇒ C1 = 1, C2 = −2; x = e−8t, y = −2e−2t or x = y4/16.




19. ∇f(x, y) = ex arctan y i + ex1

1 + y2j; ∇f(0, 1) =

π

4i +

12

j.

u =∇f(0, 1)

‖∇f(0, 1)‖ =1√

4 + π2(π i + 2 j); rate: ‖∇f(0, 1)‖ =

√π2 + 4

4

20. ∇f(x, y, z) =1

(y + z)2[(y + z) i + (z − x) j − (x + y)k]; ∇f(−1, 1, 3) =

14

i +14

j.

u =∇f(−1, 1, 3)

‖∇f(−1, 1, 3)‖ = 12

√2 i + 1

2

√2 j; rate: ‖∇f(−1, 1, 3)‖ = 1

4

√2

21. rate:df

dt= ∇f · r′ =

(4x i − 9y2 j

)·

(12t−1/2 i + 2e2t j

)= 2 − 18e6t

22. f(r(t) = sin t2 + cos t2, rate: f ′(r(t)) = 2t cos t2 − 2t sin t2

23. rate:df

dt= ∇f · r′ =

[(1y

+z

x2

)i − x

y2j − 1

xk]

· (cos t i − sin t j + sec2 tk) =1 − sin t

cos2 t

24.du

dt= ∇u · r′ =

11 + x2y2

(y i + x j) · (sec2 t i + 2e2t j) =e2t

1 + e4t tan2 t(sec2 t + 2 tan t)

25.du

dt= ∇u · r′ =

[(3y2 − 2x) i + 6xy j

]· [(2t + 2) i + 3 j] = 104t3 + 150t2 − 8t

26. u(r(t)) =1√

1 + t2,

du

dt=

−t

(1 + t2)3/2

27. area A = 12x(t)y(t) sin θ(t)

dA

dt= 0 = 1

2y(t)x′(t) sin θ(t) + 1

2x(t)y′(t) sin θ(t) + 12θ

′(t)x(t)y(t) cos θ(t) = 0

At x = 4, y = 5, θ = π/3,dx

dt=

dy

dt= 2, we have

5dθ

dt+ 2

√3 +

5√

32

= 0 =⇒ dθ

dt= −9

√3

10.

28. V = πr2h;dV

dt= 2πrh

dr

dt+ πr2 dh

dt

Measure in centimeters: at r = 12, h = 1000,dr

dt= 4,

dh

dt= 150,

dV

dt= 2π(12)(1000)(4) + π(144)(150) = 117, 600π cu.cm/yr ∼= 0.37 cu m/yr.

29.∂u

∂s=

∂u

∂x+

∂u

∂y;

∂u

∂t=

∂u

∂x− ∂u

∂y

∂u

∂s

∂u

∂t=

(∂u

∂x+

∂u

∂y

) (∂u

∂x− ∂u

∂y

)=

(∂u

∂x

)2

−(∂u

∂y

)2




30.∂u

∂s= uxe

s cos t + uyes sin t

∂2u

∂s2= uxxe

2s cos2 t + uxye2s sin t cos t + uxe

s cos t + uyes sin t + uyxe

2s cos t sin t + uyye2s sin2 t

∂u

∂t= −uxe

s sin t + uyes cos t

∂2u

∂t2= uxxe

2s sin2 t− uxye2s sin t cos t− uxe

s cos t− uyes sin t− uyxe

2s cos t sin t + uyye2s cos2 t

∂2u

∂s2+

∂2u

∂t2= e2s(uxx + uyy) =⇒ ∂2u

∂x2+

∂2u

∂y2= e−2s

[∂2u

∂s2+

∂2u

∂t2

]

31. ∇f(x, y) = (3x2 − 6xy) i + (−3x2 + 2y) j; ∇f(1,−1) = N = 9 i − 5 j

normal line: x = 1 + 9t, y = −1 − 5t; tangent line: x = 1 + 5t, y = −1 + 9t

32. ∇f(x, y) = −πy sin πxy i − πx sin πxy j; ∇f(1/3, 2) = −π√

3 i − π√

36

j, take N = 6i+ j;

normal line: x = 1/3 + 6t, y = 2 + t; tangent line: x = 1/3 + t, y = 2 − 6t

33. Set f(x, y, z) = x1/2 + y1/2 − z

∇f(x, y, z) =1

2√x

i +1

2√y

j − k; ∇f(1, 1, 2) = 12 i + 1

2 j − k. Take N = i + j − 2k.

tangent plane: (x− 1) + (y − 1) − 2(z − 2) = 0; normal line: x = 1 + t, y = 1 + t, z = 2 − 2t

34. Set f(x, y, z) = x2 + y2 + z2.

∇f(x, y, z) = 2x i + 2y j + 2z k; ∇f(1, 2,−2) = 2 i + 4 j − 4k. Take N = i + 2 j − 2k.

tangent plane: (x− 1) + 2(y − 2) − 2(z + 2) = 0; normal line: x = 1 + t, y = 2 + 2t, z = −2 − 2t

35. Set f(x, y, z) = z3 + xyz − 2.

∇f(x, y, z) = yz i + xz j + (3z2 + xy)k; ∇f(1, 1, 1) = i + j + 4k.

tangent plane: (x− 1) + (y − 1) + 4(z − 1) = 0; normal line: x = 1 + t; y = 1 + t; z = 1 + 4t

36. Set f(x, y, z) = e3x sin 3y − z.

∇f(x, y, z) = 3e3x sin 3y i + 3e3x cos 3y j − k; ∇f(0, π/6, 1) = 3i − k.

tangent plane: 3(x− 0) − (z − 1) = 0 or 3x− z + 1 = 0; normal line: x = 3t, y = π/6, z = 1 − t

37. The point (2, 2, 1) is on each hyperboloid. Set f(x, y, z) = x2 + 2y2 − 4z2, g(x, y, z) = 4x2 − y2 + 2z2.

∇f = 2x i + 4y j − 8z k, ∇f(2, 2, 1) = (4, 8,−8); ∇g = 8x i − 2y j + 4z k, ∇g(2, 2, 1) = (16,−4, 4).

Since ∇f(2, 2, 1) · ∇g(2, 2, 1) = 0, the hyperboloids are mutually perpendicular at (2, 2, 1).




38. Set f(x, y, z) = x2 + y2 + z2. At each point (x0, y0, z0), ∇f(x0, y0, z0) = 2x0 i + 2y0 j + 2z0 k.

normal line to the sphere: x = x0 + x0t, y = y0 + y0t, z = z0 + z0t. At t = −1, x = y = z = 0.

39. ∇f(x, y) = (2xy − 2y) i + (x2 − 2x + 4y − 15) j = 0 at (5, 0), (−3, 0), (1, 4).

fxx = 2y, fxy = 2x− 2, fyy = 4.


(5, 0) 0 8 4 −64 saddle

(−3, 0) 0 −8 4 −64 saddle

(1, 4) 8 0 4 32 loc. min.

f(1, 4) = −34

40. ∇f(x, y) = (6x− 3y2) i + (3y2 + 6y − 6xy) j = 0 at (0, 0), (2, 2), ( 12 ,−1).

fxx = 6, fxy = −6y, fyy = 6y − 6x + 6.


(0, 0) 6 0 6 36 loc. min.

(2, 2) 6 −12 6 −108 saddle

( 12 ,−1) 6 6 −3 −54 saddle

f(0, 0) = 0

41. ∇f(x, y) = (3x2 − 18y) i + (3y2 − 18x) j = 0 at (0, 0), (6, 6).

fxx = 6x, fxy = −18, fyy = 6y.


(0, 0) 0 −18 0 −182 saddle

(6, 6) 36 −18 36 > 0 loc. min.

f(6, 6) = −216

42. ∇f(x, y) = (3x2 − 12x) i + (2y + 1) j = 0 at (0,− 12 ), (4,− 1

2 ).

fxx = 6x− 12, fxy = 0, fyy = 2.


(0,− 12 ) −12 0 2 −24 saddle

(4,− 12 ) 12 0 2 24 loc. min.

f(4,− 12 ) = − 145

4




43. ∇f(x, y) = (1 − 2xy + y2) i + (−1 − x2 + 2xy) j = 0 at (1, 1), (−1,−1).

fxx = −2y, fxy = −2x + 2y, fyy = 2x.


(1, 1) −2 0 2 −4 saddle

(−1,−1) 2 0 −2 −4 saddle

44. ∇f(x, y) = e−(x2+y2)/2[(y2 − x2y2) i + (2xy − xy3) j

]= 0 at (±1,±

√2), (x, 0), x any real number.

fxx = e−(x2+y2)/2(−3xy2 + x3y2), fxy = e−(x2+y2)/2(2y − y3 − 2x2y + x2y3),

fyy = e−(x2+y2)/2(2x− 5xy2 + xy4).


(1,√

2) −4e−3/2 0 −4e−3/2 16e−3 loc. max

(1,−√

2) −4e−3/2 0 −4e−3/2 16e−3 loc. max

(−1,√

2) 4e−3/2 0 4e−3/2 16e−3 loc. min

(−1,−√

2) 4e−3/2 0 4e−3/2 16e−3 loc. min

local maxima: f(1,√

2) = f(1,−√

2) = 2e−3/2; local minima: f(−1,√

2) = f(−1,−√

2) = −2e−3/2.

At (x, 0), D = 0 and f(x, 0) ≡ 0. For x < 0, f(x, y) < f(x, 0) for all y �= 0; for x > 0,

f(x, y) > f(x, 0) for all y > 0; (0, 0) is a saddle point.

Here is a graph of the surface.

45. ∇f = (2x− 2) i + (2y + 2) j = 0 at (1,−1) in D; f(1,−1) = 0




F (t) = f(r(t)) = 6 − 4 cos t + 4 sin t, t ∈ [ 0, 2π ]

F ′(t) = 4 sin t + 4 cos t : F ′(t) = 0 =⇒ sin t = − cos t =⇒ t =34π,

74π





F (0) = F (2π) = f(2, 0) = 2; F(

34π

)= f

(−√

2,√

2)

= 6 + 4√

2;

F(

74π

)= f

(√2,−

√2)

= 6 − 4√

2.

f takes on its absolute maximum of 6 + 4√

2 at(−√

2,√

2); f takes on its absolute minimum of 0 at

(1,−1).

46. ∇f(x, y) = (4x− 4) i + (2y − 4) j = 0 at (1, 2) on the boundry of D; no critical points in D.


parametrize each side of the triangle:

C1 : r1(t) = t i + 2t j, t ∈ [ 0, 1 ]

C2 : r2(t) = (1 − t) i + 2 j, t ∈ [ 0, 1 ]

C3 : r3(t) = (2 − t) j, t ∈ [ 0, 2 ]

1 2 3x

1

2

3

y

Now,

f1(t) = f(r1(t)) = 6t2 − 8t + 3, t ∈ [ 0, 1 ]; critical number: t = 23

f2(t) = f(r2(t)) = 2t2 − 3+, t ∈ [ 0, 1 ]; critical number

f3(t) = f(r3(t)) = t2 − 1, t ∈ [ 0, 2 ]; critical number


f1(0) = f3(2) = f(0, 0) = 3; f1(2/3) = f(2/3, 4/3) = − 73 ; f1(1) = f2(0) = f(1, 2) = −3;

f2(1) = f3(0) = f(0, 2) = −1.


47. ∇f(x, y) = (8x− y) i + (−x + 2y + 1) j = 0 at (−1/15,−8/15) in D; f(−1/15,−8/15) = −4/15.

On the boundary of D : x = cos t, y = 2 sin t. Set

F (t) = f(cos t, 2 sin t) = 4 + 2 sin t− 2 sin t cos t, 0 ≤ t ≤ 2π.

Then

F ′(t) = 2 cos t− 4 cos2 t + 2 = −2(2 cos t + 1)(cos t− 1); F ′(t) = 0 =⇒ t =2π3,

4π3.

Evaluating F at the endpoints of the interval and at the critical points, we get

F (0) = F (2π) = f(1, 0) = 4, F (2π/3) = f(−1/2,√

3) = 4 +3√

32

,

F (4π/3) = f(−1/2,−√

3) = 4 − 3√

32

> − 415

f takes on its absolute maximum of 2 at (0, 1); f takes on its absolute minimum of −4/15 at

(−1/15,−8/15).




48. ∇f(x, y) = 4x3 i + 6y2 j = 0 at (0, 0) in D; f(0, 0) = 0.

On the boundary of D : x = cos t, y = sin t. Set

F (t) = f(cos t, sin t) = cos4 t + 2 sin3 t, 0 ≤ t ≤ 2π.

Then

F ′(t) = 4 cos3 t sin t + 6 sin2 t cos t = 2 sin t cos t(2 sin t− 1)(sin t + 2);

F ′(t) = 0 =⇒ t = π/6, π/2, 5π/6, π, 3π/2

Evaluating F at the endpoints of the interval and at the critical points, we get

F (0) = F (2π) = f(1, 0) = 1, F (π/6) = f(√

3/2, 1/2) = 13/16, F (π/2) = f(0, 1) = 2,

F (5π/6) = f(−√

3/2, 1/2) = 13/16, F (π) = f(−1, 0) = 1, F (3π/2) = f(0,−1) = −2.

f takes on its absolute maximum of 2 at (0, 1); f takes on its absolute minimum of −2 at (0,−1).

49. Set f(x, y, z) = D2 = (x− 1)2 + (y + 2)2 + (z − 3)2, g(x, y) = 3x + 2y − z − 5.

∇f = 2(x− 1) i + 2(y + 2) j + 2(z − 3)k, ∇g = 3 i + 2 j − k.

Set ∇f = λ∇g :

2(x− 1) = 3λ =⇒ x = 32λ + 1,

2(y + 2) = 2λ =⇒ y = λ− 2,

2(z − 3) = −λ =⇒ z = − 12λ + 3.

Substituting these values in 3x + 2y − z = 5 gives λ =97

=⇒ x =4114

, y = −57, z =

3314

.

The point on the plane that is closest to (1,−2, 3) is (41/14,−5/7, 33/14). The distance from the

point to the plane is9√14

.

50. Set f(x, y, z) = 3x− 2y + z, g(x, y, z) = x2 + y2 + z2 − 14,

∇f = 3 i − 2 j + k, ∇g = 2x i + 2y j + 2z k.

Set ∇f = λ∇g :

3 = 2λx =⇒ x = 3/2λ, −2 = 2λy =⇒ y = −1/λ, 1 = 2λz =⇒ z = 1/2λ.

Substituting these values in x2 + y2 + z2 = 14 gives λ = ± 12 =⇒ x = 3, y = −2, z = 1 or

x = −3, y = 2, z = −1. Evaluating f : f(3,−2, 1) = 14, f(−3, 2,−1) = −14. The maximum value of

f on the sphere is 14.

51. Set f(x, y, z) = x + y − z, g(x, y, z) = x2 + y2 + 4z2 − 4,

∇f = i + j − k, ∇g = 2x i + 2y j + 8z k.

Set ∇f = λ∇g :

1 = 2λx =⇒ x = 1/2λ, 1 = 2λy =⇒ y = 1/2λ, −1 = 8λz =⇒ z = −1/8λ.

Substituting these values in x2 + y2 + 4z2 = 4 gives λ = ± 38 =⇒ x = 4/3, y = 4/3, z = −1/3

or x = −4/3, y = −4/3, z = 1/3. Evaluating f : f( 43 ,

43 ,− 1

3 ) = 3, f(− 43 ,− 4

3 ,13 ) = −3. The maximum

value of f is 3, the minimum value is −3.




52. Let the length, width and height be x, y, z respectively. Then the total cost is

f(x, y, z) = 12 xy + 1

2 xz + 12 yz + 1

10 xy = 35 xy + 1

2 xz + 12 yz

with the condition

g(x, y, z) = xyz − 16 = 0.

Note first that xyz = 16 =⇒ x �= 0, y �= 0 z �= 0.

∇f =(

35 y + 1

2 z)

i +(

35 x + 1

2 z)

j +(

12 y + 1

2 x)

k, ∇g = yz i + xz j + xy k

∇f = λ∇g =⇒ 35 y + 1

2 z = λyz, 35 x + 1

2 z = λxz, 12 x + 1

2 y = λxy

Multiply the first equation by x, the second equation by y and subtract. This gives:

12 (xz − yz) = 0 =⇒ z(x− y) = 0 =⇒ y = x

Substituting y = x in the third equation yields x = λx2 =⇒ x =1λ

.

Substituting x = y =1λ

in the first equation yields z =65λ

.

Finally, substituting these values for x, y and z into the equation xyz = 16, we get λ =3√

32 3√

5.

Therefore, x = y =2 3√

53√

3=

103√

75∼= 2.37 and z =

65x =

123√

75∼= 2.85.

53. df = (9x2 − 10xy2 + 2) dx + (−10x2y − 1) dy

54. df = (2xy sec2 x2 − 2y2) dx + (tanx2 − 4xy) dy

55. df =y2z + z2y

(x + y + z)2dx +

xz2 + zx2

(x + y + z)2dy +

x2y + y2x

(x + y + z)2dz

56. df = − z

y2 + xzdx +

(zeyz − 2y

y2 + xz

)dy +

(yeyz − x

y2 + xz

)dz

57. Set f(x, y, z) = ex√y + z3. Then

df = ex√y + z3 Δx +

ex

21√

y + z3Δy +

ex

23z2√y + z3

Δz.

With x = 0, y = 15, z = 1, Δx = 0.02, Δy = 0.2, Δz = 0.01, df = 4 Δx + 18 Δy + 3

8 Δz ∼= 0.1088.

Therefore, e0.02√

15.2 + (1.01)3 ∼= e0√

15 + 1 + 0.1088 = 4.1088.

58. Set f(x, y) = x1/3 cos2 y. Then

df = 13 x

−2/3 cos2 yΔx− 2x1/3 cos y sin yΔy.

With x = 64, y = 30◦ = π/6, Δx = 0.5, Δy = −2◦ = − π

90, df =

164

Δx− 2√

3 Δy ∼= 0.1287.

Therefore, (64.5)1/3 cos2(28◦) ∼= 641/3 cos2(30◦) + 0.1287 = 3.1287.




59. V = πr2h; r = 5 ft., h = 22 ft., Δr = 0.01 in. = 11200 ft., Δh = 0.01 = 1

1200

dV = 2πrhΔr + πr2 Δh

Using the values given above,

dV = 2π(5)(22)1

1200+ π(25)

11200

∼= 0.6414 cu. ft. ∼= 1108.35 cu. in.;1108.35

231∼= 4.80.

Approximately 4.80 gallons will be needed.

60.∂P

∂y= 12x2y − 8x =

∂Q

∂x; the vector function is a gradient.

∂f

∂x= 6x2y2 − 8xy + 2x, f(x, y) = 2x3y2 − 4x2y + x2 + φ(y),

∂f

∂y= 4x3y − 4x2 + φ′(y) = 4x3y − 4x2 − 8.

Thus, φ′(y) = −8, φ(y) = −8y + C, and f(x, y) = 2x3y2 − 4x2y + x2 − 8y + C.

61.∂P

∂y= 2x− sin x =

∂Q

∂x; the vector function is a gradient.

∂f

∂x= 2xy + 3 − y sin x, f(x, y) = x2y + 3x + y cos x + φ(y),

∂f

∂y= x2 + cos x + φ′(y) = x2 + 2y + 1 + cos x.

Thus, φ′(y) = 2y + 1, φ(y) = y2 + y + C, and f(x, y) = x2y + 3x + y cos x + y2 + y + C.

62.∂P

∂y= 2xy + 4y;

∂Q

∂x= −2xy + 2;

∂P

∂y�= ∂Q

∂x; the vector function is not a gradient.

63.∂P

∂y= ey sin z =

∂Q

∂x,

∂P

∂z= ey cos z =

∂R

∂x,

∂Q

∂z= xey cos z =

∂R

∂y;

the vector function is a gradient.

f(x, y, z) =∫

(ey sin z + 2x) dx = xey sin z + x2 + φ(y, z),

fy = xey sin z +∂φ

∂y= xey sin z − y2 =⇒ ∂φ

∂y= −y2 =⇒ φ = −1

3y3 + ψ(z),

f(x, y, z) = xey sin z + x2 − 13y3 + ψ(z), fz = xey cos z + ψ′(z) = xey cos z =⇒ ψ′(x) = 0 =⇒

ψ(x) = C

Therefore f(x, y, z) = xey sin z + x2 − 13y

3 + C.

Calculus one and several variables 10E Salas solutions manual ch16

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Transcript of Calculus one and several variables 10E Salas solutions manual ch16