Calculus one and several variables 10E Salas solutions manual ch16

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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 + x2 j] 4. ∇f =

(x2

1 [(y 2 − x2 + 2xy)i + (y 2 − x2 − 2xy)j] + y 2 )2

  5. ∇f = 2y 2 sin(x2 + 1) + 4x2 y 2 cos(x2 + 1) i + 4xy sin(x2 + 1) j 6. ∇f =

2x 2y i+ 2 j 2 +y x + y2

x2

7. ∇f = (ex−y + ey−x ) i + (−ex−y − ey−x ) j = (ex−y + ey−x )(i − j) 8. ∇f =

AD − BC [ yi − xj ] (Cx + Dy)2

9. ∇f = (z 2 + 2xy) i + (x2 + 2yz) j + (y 2 + 2zx) k x

10. ∇f = 

x2 + y 2 + z 2

y

i+ 

x2 + y 2 + z 2

j+ 

z x2 + y 2 + z 2

k

11. ∇f = e−z (2xy i + x2 j − x2 y k)    xyz xyz 12. ∇f = + yz ln(x + y + z) i + + xz ln(x + y + z) j x+y+z x+y+z   xyz + + xy ln(x + y + z) k x+y+z 

      13. ∇f = ex+2y cos z 2 + 1 i + 2ex+2y cos z 2 + 1 j − 2zex+2y sin z 2 + 1 k 14. ∇f = eyz

2

/x3

−

3yz 2 z2 2yz i + 3j + 3 k 4 x x x



  2 1 15. ∇f = 2y cos(2xy) + i + 2x cos(2xy) j + k x z 16. ∇f =

2xy − 3z 4 z



x2 j− i+ z

17. ∇f = (4x − 3y) i + (8y − 3x) j;



x2 y + 12xz 3 z2

k

at (2, 3), ∇f = −i + 18j

18. ∇f =

1 (−2yi + 2xj), (x − y)2

1 3 ∇f (3, 1) = − i + j 2 2

19. ∇f =

2y 2x i+ 2 j; x2 + y 2 x + y2

at (2, 1), ∇f =

4 2 i+ j 5 5

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SECTION 16.1 20. ∇f =

xy tan (y/x) − 2 x + y2 −1



i+

x2 x2 + y 2

21. ∇f = (sin xy + xy cos xy) i + x2 cos xy j; 22. ∇f = e−(x

2

+y 2 )



∇f (1, 1) =

j,

π 1 − 4 2

i+

1 j 2

at (1, π/2), ∇f = i ∇f (1, −1) = e−2 (i − j)

[(y − 2x2 y)i + (x − 2xy 2 )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 = cos πzi − cos πzj − π(x − y) sin πzk, 25. ∇f = i − 

y y2

+

z2

j− 

z y2

+

z2

k;

∇f

1, 0,

at (2, −3, 4),

26. ∇f = − sin(xyz )(yz i + xz j + 2xyzk), 2

27. (a) ∇f (0, 2) = 4 i

2

2

(b) ∇f

1

1 4 π, 6 π



∇f

1 2

= −πk

∇f = i +

1 π, , −1 4



3 4 j− k 5 5

√

π 2 1 =− i+πj− k 2 4 2

 =

 √ √ −1 + 3 1 −1 + 3 √ √ −1 − i+ − − j 2 2 2 2

(c) ∇f (1, e) = (1 − 2e) i − 2 j 1 1 27 28. (a) ∇f (1, 2, −3) = √ i + √ j − √ k 8 2 2 2 8 2 √ 3 π (c) ∇f (1, e2 , π/6) = i+ j+k 2 12e2

(b) ∇f (1, −2, 3) = −

1 1 5 i+ j+ k 18 9 18

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) + 3h21 − h1 h2 = [(6x − y) i + (1 − x) j] · h + 3h21 − h1 h2 The remainder g(h) = 3h21 − h1 h2 = (3h1 i − h1 j) · (h1 i + h2 j) ,

and

|g(h)| 3h1 i − h1 j  · h  · cos θ = ≤ 3h1 i − h1 j  h  h  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] + 12 h21 + 2h1 h2 + h22 ; g(h) = 12 h21 + 2h1 h2 + h22 is o(h).

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SECTION 16.1

31. For the function f (x, y, z) = x2 y + y 2 z + z 2 x, 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 ) − x2 y + y 2 z + z 2 x       = 2xy + z 2 h1 + 2yz + x2 h2 + 2xz + y 2 h3 + (2xh2 + yh1 + h1 h2 ) h1 + (2yh3 + zh2 + h2 h3 ) h2 + (2zh1 + xh3 + h1 h3 ) h3        = 2xy + z 2 i + 2yz + x2 j + 2xz + y 2 k · h + g(h) · h, where

g(h) = (2xh2 + yh1 + h1 h2 ) i + (2yh3 + zh2 + h2 h3 ) j + (2zh1 + xh3 + h1 h3 ) k

Since

|g(h)| → 0 as h → 0 h 

it follows that

      ∇f = 2xy + z 2 i + 2yz + x2 j + 2xz + y 2 k

 32. f (x + h) − f (x) = (2xy + 2h2 x + h1 y) i + 2x2 j + g(h) = h21 h2 is o(h) and ∇f = 4xy i = 2x2 j +   33. ∇f = F(x, y) = 2xy i + 1 + x2 j

⇒

∂f = x2 + g  (y) = 1 + x2 ∂y Thus, f (x, y) = x2 y + y + C

k.

∂f = 2xy ∂x g  (y) = 1

⇒

Now,

1 z2

 1 k · (h1 i + h2 j + h3 k) + h21 ; z(z + h3 )

⇒ ⇒

f (x, y) = x2 y + g(y) for some function g. g(y) = y + C, C a constant.

1 34. ∇f = (2xy + x)i + (x2 + y)j =⇒ fx = 2xy + x =⇒ f (x, y) = x2 y + x2 + g(y) 2 Now, fy = x2 + g  (y) = x2 + y =⇒ g  (y) = y =⇒ g(y) = 12 y 2 + C Thus,

f (x, y) = x2 y + 12 x2 + 12 y 2 + C

35. ∇f = F(x, y) = (x + sin y) i + (x cos y − 2y) j ⇒ for some function g. ∂f Now, = x cos y + g  (y) = x cos y − 2y ∂y Thus, f (x, y) = 12 x2 + x sin y − y 2 + C.

⇒

∂f = x + sin y ⇒ f (x, y) = ∂x g  (y) = −2y

⇒

1 2

x2 + x sin y + g(y)

g(y) = −y 2 + C, C a constant.

36. ∇f = yzi + (xz + 2yz)j + (xy + y 2 )k =⇒ fx = yz =⇒ f (x, y, z) = xyz + g(y, z). fy = xz + gy = xz + 2yz =⇒ gy = 2yz =⇒ g(y, z) = y 2 z + h(z) =⇒ f (x, y, z) = xyz + y 2 z + h(z). fx = xy + y 2 + h (z) = xy + y 2 =⇒ h (z) = 0 =⇒ h(z) = C. Thus, f (x, y, z) = xyz + y 2 z + C. 37. With r = (x2 + y 2 + z 2 )1/2 we have ∂r x = , ∂x r (a)

∇(ln r) =

∂r y = , ∂y r

∂r z = . ∂z r

∂ ∂ ∂ (ln r) i + (ln r) j + (ln r)k ∂x ∂y ∂z

1 ∂r 1 ∂r 1 ∂r i+ j+ k r ∂x r ∂y r ∂z x y z r = 2 i+ 2 j+ 2 k= 2 r r r r =

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SECTION 16.1 (b)

∇(sin r) =

791

∂ ∂ ∂ (sin r) i + (sin r) j + (sin r)k ∂x ∂y ∂z

∂r ∂r ∂r i + cos r j + cos r k ∂x ∂y ∂z x y z = (cos r) i + (cos r) j + (cos r) k r r r

cos r  = r r

= cos r

(c) ∇er =

er r

r

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

38. With rn = (x2 + y 2 + z 2 )n/2 we have ∂rn n = (x2 + y 2 + z 2 )(n/2)−1 (2x) = n(x2 + y 2 + z 2 )(n−2)/2 x = nrn−2 x. ∂x 2 Similarly ∂rn = nrn−2 y ∂y

and

∂rn = nrn−2 z. ∂z

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

39. (a) ∇f = 2x i + 2y j = 0

=⇒

x = y = 0;

∇f = 0 at (0, 0).

(b)

40. (a) (c)

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

∇f = 

−1 4 − x2 − y 2

(xi + yj) = 0 at (0, 0)

f has a maximum at (0, 0)

(b)

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SECTION 16.2

41. (a) Let c = c1 i + c2 j + c3 k. First, we take h = hi. Since c · h is o(h), c·h c1 h = lim = c1 . h→0 h h

0 = lim

h→0

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.

lim g(h) = lim

h→0

h→0

g(h) h h





=

lim h

h→0

g(h) lim h→0 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 2 2y 3 = 4 = ∂x y y

∂f 2y(y 2 − x2 ) . As (x, y) tends to (0, 0) along the positive y-axis, = ∂x (x2 + y 2 )2

tends to ∞.

SECTION 16.2 1. ∇f = 2xi + 6yj,

∇f (1, 1) = 2i + 6j,

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

u=

1 2

√

∇f (0, 0) = 2i + j,

3. ∇f = (ey − yex ) i + (xey − ex ) j, ∇f (1, 0) = i + (1 − e)j, 1 fu (1, 0) = ∇f (1, 0) · u = (7 − 4e) 5 1 (−2yi + 2xj), ∇f (1, 0) = 2j, (x − y)2 √ fu (1, 0) = ∇f (1, 0) · u = − 3

4. ∇f =

u=

1 u = √ (2i + j), 5

u=

1 (3i + 4j), 5

√ 1 (i − 3j), 2

(a − b)y (b − a)x a−b (i − j), i+ j, ∇f (1, 1) = (x + y)2 (x + y)2 4 1√ fu (1, 1) = ∇f (1, 1) · u = 2 (a − b) 4

5. ∇f =

√ fu (1, 1) = ∇f (1, 1) · u = −2 2

2 (i − j),

u=

1√ 2 (i − j), 2

1 d−c [(d − c)yi + (c − d)xj] , ∇f (1, 1) = (i − j), 2 (cx + dy) (c + d)2 d−c √ fu (1, 1) = ∇f (1, 1) · u = (c + d) c2 + d2

6. ∇f =

2x 2y i+ 2 j, ∇f (0, 1) = 2 j, 2 +y x + y2 2 fu (0, 1) = ∇f (0, 1) · u = √ 65

7. ∇f =

x2

1 u = √ (8 i + j), 65

u= √

c2

1 (ci − dj), + d2

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SECTION 16.2 π π 8. ∇f = 2xyi + (x2 + sec2 y)j, ∇f (−1, ) = − i + 3j, 4 2

 π π 1 π  fu −1, = ∇f −1, · u = −√ +6 4 4 5 2

1 u = √ (i − 2j) 5

9. ∇f = (y + z)i + (x + z)j + (y + x)k, ∇f (1, −1, 1) = 2j, 2√ fu (1, −1, 1) = ∇f (1, −1, 1) · u = 6 3 10. ∇f = (z 2 + 2xy)i + (x2 + 2yz)j + (y 2 + 2zx)k, √ 10  fu (1, 0, 1) = ∇f (1, 0, 1) · u = 10

u=

1 6

√

6 (i + 2j + k),

∇f (1, 0, 1) = i + j + 2k,

   11. ∇f = 2 x + y 2 + z 3 i + 2yj + 3z 2 k , ∇f (1, −1, 1) = 6(i − 2j + 3k), √ fu (1, −1, 1) = ∇f (1, −1, 1) · u = −3 2 12. ∇f = (2Ax + Byz)i + (Bxz + 2Cy)j + Bxyk, u= √

1 (Ai + Bj + Ck); 2 A + B2 + C 2

13. ∇f = tan−1 (y + z) i + 1 u = √ (i + j − k), 3

793

1 u = √ (3j − k) 10

u=

1 2

√

2 (i + j),

∇f (1, 2, 1) = 2(A + B)i + (B + 4C)j + 2Bk

fu (1, 2, 1) = ∇f (1, 2, 1) · u =

2A2 + B 2 + 2AB + 6BC √ A2 + B 2 + C 2

π 1 1 ∇f (1, 0, 1) = i + j + k, 4 2 2 √ π 3 fu (1, 0, 1) = ∇f (1, 0, 1) · u = √ = π 12 4 3

x x j+ k, 1 + (y + z)2 1 + (y + z)2

14. ∇f = (y 2 cos z − 2πyz 2 cos πx + 6zx)i + (2xy cos z − 2z 2 sin πx)j + (−xy 2 sin z − 4yz sin πx + 3x2 )k ∇f (0, −1, π) = (2π 3 − 1)i; 15. ∇f =

x2

x y i+ 2 j, 2 +y x + y2

u=

1 (2i − j + 2k), 3 1

u= 

x2

+

y2

(−xi − yj) ,

17. ∇f = (2Ax + 2By) i + (2Bx + 2Cy) j, ∇f (a, b) = (2aA + 2bB)i + (2aB + 2bC) j √ √ (a) u = 12 2 (−i + j), fu (a, b) = ∇f (a, b) · u = 2 [a(B − A) + b(C − B)] √ √ (b) u = 12 2 (i − j), fu (a, b) = ∇f (a, b) · u = 2 [a(A − B) + b(B − C)] z z i − j + ln x y 1 u = √ (i + j − k); 3



x k, y

∇f (1, 1, 2) = 2i − 2j

fu (1, 1, 2) = ∇f (1, 1, 2) · u = 0

1

fu (x, y) = ∇f · u = − 

  16. ∇f = exy (y 2 + xy 3 − y 3 )i + (x − 1)(2y + xy 2 )j , ∇f (0, 1) = −2j 1 4√ u = √ (−i + 2j), fu (0, 1) = ∇f (0, 1) · u = − 5 5 5

18. ∇f =

2 (2π 3 − 1). 3

fu (0, −1, π) = ∇f (0, −1, π) · u =

x2

+ y2

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SECTION 16.2

19. ∇f = ey

2

−z 2

∇f (1, 2, −2) = i + 4j + 4k, r (t) = i − 2 sin (t − 1) j − 2et−1 k, √ 7√ r (1) = i − 2k, u = 15 5 (i − 2k), fu (1, 2, −2) = ∇f (1, 2, −2) · u = − 5 5

(i + 2xyj − 2xzk),

at (1, 2, −2) t = 1, 20. ∇f = 2xi + zj + yk,

∇f (1, −3, 2) = 2i + 2j − 3k 1 1√ Direction: r (−1) = −2i + 3j − 3k, u = √ (−2i + 3j − 3k), fu (1, −3, 2) = ∇f (1, −3, 2) · u = 22 2 22   21. ∇f = (2x + 2yz) i + 2xz − z 2 j + (2xy − 2yz) k, ∇f (1, 1, 2) = 6 i − 2 k

1 The vectors v = ±(2 i + j − 3 k) are direction vectors for the given line; u = ± √ [2 i + j − 3 k] 14 18  √ are corresponding unit vectors; fu (1, 1, 2) = ∇f (1, 1, 2) · (±u) = ± 14 22. ∇f = ex (cos πyzi − πz sin πyzj − πy sin πyzk),

∇f (0, 1, 12 ) = −

π j−πk 2



The vectors v = ±(2 i + 3 j + 5 k) are direction vectors for the line; u = ± 13π fu (0, 1, 12 ) = ∇f (0, 1, 12 ) · (±u) = ∓ √ 2 38

are corresponding unit vectors;

1 √ [2 i + 3 j + 5 k] 38

√ ∇f  = 2 2,

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

23. ∇f = 2y 2 e2x i + 2ye2x j,

∇f (0, 1) = 2 i + 2 j,

24. ∇f = [1 + cos(x + 2y)]i + 2 cos(x + 2y)j, ∇f (0, 0) = 2i + 2j √ 1 Fastest increase in direction u = √ (i + j), rate of change ∇f (0, 0) = 2 2 2 √ 1 Fastest decrease in direction v = − √ (i + j), rate of change −2 2 2 x

25. ∇f = 

x2

+

y2

z2

y

i+ 

x2

y2

z2

j+ 

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

z x2

+ y2 + z2

k,

1 f increases most rapidly in the direction u = √ (i − 2 j + k); the rate of change is 1. 6 1 f decreases most rapidly in the direction v = − √ (i − 2 j + k); the rate of change is −1. 6 26. ∇f = (2xzey + z 2 )i + x2 zey j + (x2 ey + 2xz)k, ∇f (1, ln 2, 2) = 12i + 4j + 6k 1 Fastest increase in direction u = (6i + 2j + 3k), rate of change ∇f (1, ln 2, 2) = 14 7 1 Fastest decrease in direction v = − (6i + 2j + 3k), rate of change −14 7 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.

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SECTION 16.2 28. 0;

the vector c =

795

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

points along the level curve of f at (x0 , y0 ). 29. (a) (b) 30. (a) (b)

√ f (h, 0) − f (0, 0) h2 |h| lim = lim = lim does not exist h→0 h→0 h→0 h h h no; by Theorem 16.2.5 f cannot be differentiable at (0, 0) g(x + h)o(h) o(h) = g(x + h) → g(x)(0) = 0 h h |[g(x + h) − g(x)]∇f (x) · h| [g(x + h) − g(x)]∇f (x)h ≤ h h

by Schwarz’s inequality = |g(x + h) − g(x)| · ∇f (x) → 0 31. ∇λ(x, y) = − 83 xi − 6yj 8 (a) ∇λ(1, −1) = − i = 6j, 3

8 i − 6j −∇λ(1, −1) 2√ = 32 √ , λu (1, −1) = ∇λ(1, −1) · u = − 97 ∇λ(1, −1) 3 3 97   (b) u = i, λu (1, 2) = ∇λ(1, 2) · u = − 83 i − 12j · i = − 83 √ √   1√  26 (c) u = 12 2 (i + j), λu (2, 2) = ∇λ(2, 2) · u = − 16 2 3 i − 12 j · 2 2 (i + j) = − 3

u=

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 = −2y 2 ; the particle will follow the parabolic path x = −2y 2 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),

x(t) = e−2t

and

y(t) = e−6t .

y(0) = 1.

From Theorem 7.6.1 we find that

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

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SECTION 16.2 (b) Here x (t) = −2x(t),

x(0) = 1

y  (t) = −6y(t),

and

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). 1 2 x − y 2 ; ∇f = xi − 2yj, so we choose the projection r(t) = x(t)i + y(t)j of the path 2 onto the xy-plane such that x (t) = x(t), y  (t) = −2y(t) 1 (a) With initial point (−1, 1, − 12 ), we get x(t) = −et , y(t) = e−2t , or y = 2 x from (−1, 1), in the direction of decreasing x.

34. z = f (x, y) =

(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 = − 2a2 xi + 2b2 yj . We can satisfy these conditions by setting x (t) = −2a2 x(t),

x(0) = a2

and

y  (t) = −2b2 y(t),

y(0) = b

so that x(t) = ae−2a

2

Since

 x b 2 a a2

C is the curve (b)

b2

x

b2

= (a) y

a2

t

and

y(t) = be−2b t . 2

b2  y a2 2 = e−2a t = , b

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

36. The particle must go in he direction −∇T = −ey cos xi − ey sin xj, so we set x (t) = −ey(t) cos x(t), y  (t) = −ey(t) sin x(t). Dividing, we have

sin x(t) dy y  (t) = , or = tan x. x (t) cos x(t) dx

With initial point

(0, 0), we get y = ln | sec x|, 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 sin x i − 2 e−y cos x j. From √ x (t) = − 2 e−y sin x

and

√ y  (t) = − 2 e−y cos x

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SECTION 16.2

797

we obtain dy y  (t) =  = cot x dx x (t) so that y = ln | sin x| + C. √ √ Since y = 0 when x = π/4, we get C = ln 2 and y = ln | 2 sin x|. As ∇T (π/4, 0) = −i − j, the curve √ y = ln | 2 sin x| 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

dy 2 − 6y = . With initial point dx 1 − 2x 3y = (2x − 1)3 + 1, in the direction of increasing x. 1 − 2x(t), y  (t) = 2 − 6y(t),

39.

or

(0, 0),

x (t) =

this gives the curve

  f 2 + h, (2 + h)2 − f (2, 4) 3(2 + h)2 + (2 + h)2 − 16 lim = lim h→0 h→0 h h   2 4h + h = lim 4 = lim 4(4 + h) = 16 h→0 h→0 h

2 h+8 h+8 f 3 + (4 + h) − 16 , 4 + h − f (2, 4) 4 4 lim = lim h→0 h→0 h h

(a)

(b)

+ 3h + 12 + 4 + h − 16 h→0 h 3  = lim 16 h + 4 = 4 = lim

3 2 16 h

h→0

(c) u =

1 17

√

17 (i + 4j),

∇f (2, 4) = 12i + j;

fu (2, 4) = ∇f (2, 4) · u =

16 17

√

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 =

−GM m −GM m (xi + yj + zk) = r r3 (x2 + y 2 + z 2 )3/2

∂f ∂f i+ j; ∂x ∂y

∂f ∂f ∂f ∂f  fu (x, y) = ∇f · u = i+ j · (cos θ i + sin θ j) = cos θ + sin θ ∂x ∂y ∂x ∂y   (b) ∇f = 3x2 + 2y − y 2 i + (2x − 2xy) j, ∇f (−1, 2) = 3 i + 2 j √ 2 3−3 fu (−1, 2) = 3 cos(2π/3) + 2 sin(2π/3) = 2

41. (a) u = cos θ i + sin θ j,

∇f (x, y) =

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SECTION 16.3

 √  √ √ ∂f 2 2 5π ∂f 5π 2y 2 2y − − 42. = cos + sin = 2xe + 2x e = − 2xe2y (1 + x) ∂x 4 ∂y 4 2 2 √ √ fu (2, ln 2) = − 2 · 2 · e2 ln 2 (1 + 2) = −24 2





∂(f g) ∂(f g) ∂(f g) ∂g ∂f ∂g ∂f ∂g ∂f i+ j+ k= f +g i+ f +g j+ f +g k 43. ∇(f g) = ∂x ∂y ∂z ∂x ∂x ∂y ∂y ∂z ∂z



∂g ∂g ∂g ∂f ∂f ∂f =f i+ j+ k +g i+ j+ k ∂x ∂y ∂z ∂x ∂y ∂z fu (x, y)

= f ∇g + g ∇f







f ∂ f ∂ f ∂ f ∇ = i+ j+ k g ∂x g ∂y g ∂z g

44.

∂f ∂g ∂f ∂f ∂g ∂g g−f g−f g−f ∂y ∂y ∂x ∂x ∂z ∂z k = i+ j+ g2 g2 g2 = 45. ∇f n =

g(x)∇f (x) − f (x)∇g(x) g 2 (x)

∂f ∂f ∂f ∂f n ∂f n ∂f n i+ j+ k = nf n−1 i + nf n−1 j + nf n−1 k = nf n−1 ∇f ∂x ∂y ∂z ∂x ∂y ∂z

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

The solutions are: t = 13 , t = 1. Thus, c = ( 13 , 53 )

3t2 − 4t + 1 = 0, 0 ≤ t ≤ 1

satisfies the equation.

Note that the endpoint b also satisfies the equation. 2. ∇f = 4zi − 2yj + (4x + 2z)k, b − a = i + 2j + k,

f (a) = f (0, 1, 1) = 0,

f (b) = f (1, 3, 2) = 3

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 = ( 12 , 2, 32 ) 3. (a) f (x, y, z) = a1 x + a2 y + a3 z + C

(b)

f (x, y, z) = g(x, y, z) + a1 x + a2 y + a3 z + C

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

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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 x1 x2 .

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 + x2 j;

  ∇f (r(t)) · r (t) = 2i + e2t j · (et i − e−t j) = et

8. ∇f = i − j;

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

−2x 2y −2 sin t 2 cos t ∇f (r(t)) = i+ j 2 i+ 2 j, 2 2t 2 2 2 2 1 + cos 1 + cos2 2t 1 + (y − x ) 1 + (y − x )

−2 sin t 2 cos t −4 sin t cos t −2 sin 2t  ∇f (r(t)) · r (t) = i+ j · (cos t i − sin t j) = = 1 + cos2 2t 1 + cos2 2t 1 + cos2 2t 1 + cos2 2t

9. ∇f =

10. ∇f =

1 (4xi + 3y 2 j) 2x2 + y 3

1 −2/3 8e4t + 1 2t 2/3 2t (4e t i + 3t j) · (2e i + j) = 2e4t + t 3 2e4t + t

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





1 1 1 1 ∇f (r(t)) · r (t) = (tt − ln t) i + tt ln t + j · i + [1 + ln t] j = tt + ln t + [ln t]2 + t t t t 1

∇f (r(t)) · r (t) =

12. ∇f =

2 (xi + yj + zk) x2 + y 2 + z 2

∇f (r(t)) · r (t) =

2 4e4t (sin ti + cos tj + e2t k) · (cos ti − sin tj + 2e2t k) = 4t 1+e 1 + e4t

13. ∇f = yi + (x − z)j − yk;       ∇f (r(t)) · r (t) = t2 i + t − t3 j − t2 k · i + 2tj + 3t2 k = 3t2 − 5t4 14. ∇f = 2x i + 2y j ∇f (r(t)) · r (t) = (2a cos ωti + 2b sin ωtj) · (−ωa sin ωti + ωb cos ωtj + bωk) = 2ω(b2 − a2 ) sin ωt cos ωt 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

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SECTION 16.3

16. ∇f = y 2 cos(x + z)i + 2y sin(x + z)j + y 2 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 + 3t2 k) = cos t[(2 + 3t2 ) cos t cos(2t + t3 ) − 2 sin t sin(2t + t3 )] 17.

du ∂u dx ∂u dy = + = (2x − 3y)(− sin t) + (4y − 3x)(cos t) dt ∂x dt ∂y dt = 2 cos t sin t + 3 sin2 t − 3 cos2 t = sin 2t − 3 cos 2t 



du y x ∂u dx ∂u dy 1 2 = · + · = 1+2 3t + 2 −3 − 2 dt ∂x dt ∂y dt x y t



2 1 3 = 1 + 2 3t2 + (2t2 − 3) − 2 = 3t2 + 4 + 2 t t t

18.

19.

du ∂u dx ∂u dy = + dt ∂x dt ∂y dt

  = (ex sin y + ey cos x) 12 + (ex cos y + ey sin x) (2)     = et/2 12 sin 2t + 2 cos 2t + e2t 12 cos 12 t + 2 sin 12 t

20.

21.

du ∂u dx ∂u dy = · + · = (4x − y)(−2 sin 2t) + (2y − x) cos t dt ∂x dt ∂y dt = 2 sin 2t(sin t − 4 cos 2t) + cos t(2 sin t − cos 2t) du ∂u dx ∂u dy = + = (ex sin y) (2t) + (ex cos y) (π) dt ∂x dt ∂y dt 2

= et [2t sin(πt) + π cos(πt)] 22.

23.

y du ∂u dx ∂u dy ∂u dz z z 1 √ + ln = · + · + · = − 2t + et (1 + t) dt ∂x dt ∂y dt ∂z dt x y2 t x √

2t2 et et t =− 2 + + ln 2 et (1 + t) t +1 2 t +1 du ∂u dx ∂u dy ∂u dz = + + dt ∂x dt ∂y dt ∂z 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 ∂u dx ∂u dy ∂u dz = · + · + · = (sin πy + πz sin πx)2t + πx cos πy(−1) − cos πx(−2t) dt ∂x dt ∂y dt ∂z dt   = 2t sin[π(1 − t)] + π(1 − t2 ) sin(πt2 ) − πt2 cos[π(1 − t)] + 2t cos(πt2 )

25. V =

1 2 πr h, 3

dV ∂V dr ∂V dh = + = dt ∂r dt ∂h dt



2 πrh 3



dr + dt



1 2 πr 3



dh . dt

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SECTION 16.3

801

At the given instant, dV 2 1 1288 = π(280)(3) + π(196)(−2) = π. dt 3 3 3 1288 The volume is increasing at the rate of π in.3 / sec . 3 26. v = πr2 h, dr = −2, dt

dv dh ∂v dr ∂v dh dr = · + · = 2πrh + πr2 dt ∂r dt ∂h dt dt dt dh dv = 3, r = 13, h = 18 =⇒ = −429π : dt dt

decreasing at the rate of

429π cm3 /sec. 27. A =

1 2

xy sin θ;

dA ∂A dx ∂A dy ∂A dθ = + + = dt ∂x dt ∂y dt ∂θ dt

1 2

  dx dy dθ (y sin θ) + (x sin θ) + (xy cos θ) . dt dt dt

At the given instant dA 1 = [(2 sin 1) (0.25) + (1.5 sin 1) (0.25) − (2(1.5) cos 1) (0.1)] ∼ = 41.34 in2 /s = 0.2871 ft2 /s ∼ dt 2 dz dx dx y dy dy dy x dx = 2x + . But x2 + y 2 = 13 =⇒ 2x + 2y = 0 =⇒ =− dt dt 2 dt dt dt dt y dt

dz dx y x dx 3x dx =⇒ = 2x + − = = 15. z is increasing 15 centimeters per second dt dt 2 y dt 2 dt ∂u ∂u ∂x ∂u ∂y 29. = + = (2x − y)(cos t) + (−x)(t cos s) ∂s ∂x ∂s ∂y ∂s 28.

= 2s cos2 t − t sin s cos t − st cos s cos t ∂u ∂u ∂x ∂u ∂y = + = (2x − y)(−s sin t) + (−x)(sin s) ∂t ∂x ∂t ∂y ∂t = −2s2 cos t sin t + st sin s sin t − s cos t sin s 30.

∂u ∂u ∂x ∂u ∂y = · + · ∂s ∂x ∂s ∂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 ∂u ∂x ∂u ∂y = · + · ∂t ∂x ∂t ∂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 ∂u ∂x ∂u ∂y = + = (2x tan y)(2st) + x2 sec2 y (1) ∂s ∂x ∂s ∂y ∂s     3 2 = 4s t tan s + t2 + s4 t2 sec2 s + t2     ∂u ∂u ∂x ∂u ∂y = + = (2x tan y) s2 + x2 sec2 y (2t) ∂t ∂x ∂t ∂y ∂t     4 = 2s t tan s + t2 + 2s4 t3 sec2 s + t2

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SECTION 16.3 ∂u ∂x ∂u ∂y ∂u ∂z ∂u = · + · + · ∂s ∂x ∂s ∂y ∂s ∂z ∂s = z 2 y sec xy tan xy(2t) + z 2 x sec xy tan xy + 2z sec xy(2st)   = sec[2st(s − t2 )] 2s4 t3 (s − t2 ) tan[2st(s − t2 )] + 2s3 t2 tan[2st(s − t2 )] + 4s3 t2 ∂u = z 2 y sec xy tan xy(2s) + z 2 x sec xy tan xy(−2t) + 2z sec xy(s2 ) ∂t   = sec[2st(s − t2 )] 2s5 t2 (s − t2 ) tan[2st(s − t2 )] − 4s5 t4 tan[2st(s − t2 )] + 2s4 t

33.

∂u ∂u ∂x ∂u ∂y ∂u ∂z = + + ∂s ∂x ∂s ∂y ∂s ∂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 ∂u ∂x ∂u ∂y ∂u ∂z = + + ∂t ∂x ∂t ∂y ∂t ∂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 ∂u ∂x ∂u ∂y ∂u ∂z = · + · + · ∂s ∂x ∂s ∂y ∂s ∂z ∂s = eyz

2

1 2 2 + xz 2 eyz · 0 + 2xyzeyz 2s s

1 t3 (s2 +t2 )2 3 2 2 2 + 4st3 (s2 + t2 ) ln(st)et (s +t ) e s ∂u ∂u ∂x ∂u ∂y ∂u ∂z = · + · + · ∂t ∂x ∂t ∂y ∂t ∂z ∂t =

= eyz

2

1 2 2 + xz 2 eyz 3t2 + 2xyzeyz 2t t

1 t3 (s2 +t2 )2 3 2 2 2 + t2 (s2 + t2 )(3s2 + 7t2 ) ln(st)et (s +t ) e t   d r (t) [f (r(t) ) ] = ∇f (r(t) ) ·  r (t)  dt  r (t)  =

35.

 = fu(t) (r(t))  r (t) 

36.

where

u(t) =

∂ ∂r x d ∂r [f (r)] = [f (r)] = f  (r) = f  (r) ; ∂x dr ∂x ∂x r ∂ y [f (r)] = f  (r) ∂y r

similarly

z ∂ [f (r)] = f  (r) . ∂z r x y z r ∇f (r) = f  (r) i + f  (r) j + f  (r) k = f  (r) . r r r r

Therefore 37. (a) (cos r)

r (t)  r (t) 

and

r r

38. (a) ∇(r ln r) = (1 + ln r)

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

2

r r

(b) ∇(e1−r ) = −2re1−r

2

r r

2

= −2e1−r r

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SECTION 16.3

r 39. (a) (r cos r − sin r) 3 r

(b)

40. (a)

(b)

sin r − r cos r sin2 r



r r

du ∂u dx ds ∂u dy ds = + dt ∂x ds dt ∂y ds dt

41. (a)

∂u ∂u = (b) ∂r ∂x



∂x ∂w ∂x ∂t + ∂w ∂r ∂t ∂r



∂u + ∂y



∂y ∂w ∂y ∂t + ∂w ∂r ∂t ∂r



To obtain ∂u/∂s, replace each r by s. 42. (a)

(b)

∂u ∂x ∂u ∂z ∂u ∂w ∂u = + + , ∂r ∂x ∂r ∂z ∂r ∂w ∂r

∂u ∂u ∂y ∂u ∂w = + ∂v ∂y ∂v ∂w ∂v

∂u + ∂z



∂z ∂w ∂z ∂t + ∂w ∂r ∂t ∂r

.

803

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SECTION 16.3 ∂u dx ∂u dy du = + dt ∂x dt ∂y dt     2 2 d u ∂u d x dx ∂ 2 u dx ∂ 2 u dy ∂u d2 y dy ∂ 2 u dx ∂ 2 u dy = + + + + + 2 dt2 ∂x dt2 dt ∂x2 dt ∂y∂x dt ∂y dt2 dt ∂x ∂y dt ∂y dt and the result follows.

44.

∂u ∂u ∂x ∂u ∂y = + ∂s ∂x ∂s ∂y ∂s ∂2u ∂u ∂ 2 x ∂x = + ∂s2 ∂x ∂s2 ∂s =

∂2u ∂x2



∂x ∂s



∂ 2 u ∂x ∂ 2 u ∂y + ∂x2 ∂s ∂y∂x ∂s

2 +2



∂ 2 u ∂x ∂y ∂ 2 u + 2 ∂x∂y ∂s ∂s ∂y

∂u ∂ 2 y ∂y + + ∂y ∂s2 ∂s



∂y ∂s

2 +



∂ 2 u ∂x ∂ 2 u ∂y + 2 ∂x∂y ∂s ∂y ∂s



∂u ∂ 2 x ∂u ∂ 2 y + ∂x ∂s2 ∂y ∂s2

∂u ∂x ∂u ∂y ∂u ∂u ∂u = + = cos θ + sin θ ∂r ∂x ∂r ∂y ∂r ∂x ∂y

45. (a)

∂u ∂u ∂x ∂u ∂y ∂u ∂u = + = (−r sin θ) + (r cos θ) ∂θ ∂x ∂θ ∂y ∂θ ∂x ∂y 2 2 2 ∂u ∂u ∂u ∂u ∂u 2 (b) cos θ sin θ + = cos θ + 2 sin2 θ, ∂r ∂x ∂x ∂y ∂y 2 2 2 1 ∂u ∂u ∂u ∂u ∂u 2 = sin θ − 2 cos2 θ, cos θ sin θ + 2 r ∂θ ∂x ∂x ∂y ∂y 2 2 2 2 2 2   ∂u 1 ∂u ∂u  2 ∂u  2 ∂u ∂u + 2 = + cos θ + sin2 θ + sin θ + cos2 θ = ∂r r ∂θ ∂x ∂y ∂x ∂y

46. (a) By Exercise 45 (a) ∂w ∂w ∂w = cos θ + sin θ, ∂r ∂x ∂y ∂w ∂x (b) To obtain the first pair of equations set w = r; Solve these equations simultaneously for

∂w ∂w ∂w =− r sin θ + r cos θ. ∂θ ∂x ∂y and

∂w . ∂y

to obtain the second pair of equations set w = θ.  (c) θ is not independent of x; r = x2 + y 2 gives ∂r r cos θ x = = = cos θ 2 2 ∂x r x +y 47. Solve the equations in Exercise 45 (a) for

∂u ∂u and : ∂x ∂y

∂u ∂u 1 ∂u = cos θ − sin θ, ∂x ∂r r ∂θ Then

∇u =

∂u ∂u 1 ∂u = sin θ + cos θ ∂y ∂r r ∂θ

∂u ∂u 1 ∂u ∂u i+ j= (cos θ i + sin θ j) + (− sin θ i + cos θ j) ∂x ∂y ∂r r ∂θ

48. u(r, θ) = r2 =⇒ ∇u = 2rer

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SECTION 16.3  49. u(x, y) = x2 − xy + y 2 = r2 − r2 cos θ sin θ = r2 1 −

1 2

∂u = r(2 − sin 2θ), ∂r

sin 2θ

805



∂u = −r2 cos 2θ ∂θ

∂u 1 ∂u er + eθ = r(2 − sin 2θ)er − r cos 2θ eθ ∂r r ∂θ ∂u ∂u ∂x ∂u ∂y ∂u ∂u = + = −r sin θ + r cos θ ∂θ ∂x ∂θ ∂y ∂θ ∂x ∂y

∇u = 50.

∂u ∂2u = − sin θ − r sin θ ∂r∂θ ∂x = − sin θ



∂ 2 u ∂x ∂ 2 u ∂y + ∂x2 ∂r ∂y∂x ∂r

∂u ∂u + cos θ + r sin θ cos θ ∂x ∂y





∂u + cos θ + r cos θ ∂y

∂2u ∂2u − ∂y 2 ∂x2



∂ 2 u ∂x ∂ 2 u ∂y + 2 ∂x∂y ∂r ∂y ∂r

+ r(cos2 θ − sin2 θ)



∂2u ∂x∂y

51. From Exercise 45 (a), ∂2u ∂2u ∂2u ∂2u sin θ cos θ + 2 sin2 θ = cos2 θ + 2 2 2 ∂r ∂x ∂y ∂x ∂y ∂2u ∂2u 2 2 ∂2u 2 ∂2u 2 = r sin θ − 2 sin θ cos θ + r cos2 θ − r r ∂θ2 ∂x2 ∂y ∂x ∂y 2 The term in parentheses is The result follows.

∂u = 2x − 2y, ∂x

∂u = −2x + 4y 3 ∂y

∂u y−x dy 2y − 2x = − ∂x = 3 = 3 ∂u dx 4y − 2x 2y − x ∂y 53. Set u = xey + yex − 2x2 y.

Then ∂u = xey + ex − 2x2 ∂y

dy ∂u/∂x ey + yex − 4xy =− =− y . dx ∂u/∂y xe + ex − 2x2 54. u(x, y) = x2/3 + y 2/3 ,

∂u ∂u cos θ + sin θ . ∂x ∂y

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

52. u(x, y) = x2 − 2xy + y 4 − 4,

∂u = ey + yex − 4xy, ∂x



2 ∂u = x−1/3 , ∂x 3

2 ∂u = y −1/3 ∂y 3

∂u

y 1/3 dy = − ∂x = − =⇒ ∂u dx x ∂y 55. Set u = x cos xy + y cos x − 2. ∂u = cos xy − xy sin xy − y sin x, ∂x

Then ∂u = − x2 sin xy + cos x ∂y

dy ∂u/∂x cos xy − xy sin xy − y sin x =− = . dx ∂u/∂y x2 sin xy − cos x

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SECTION 16.4

56. Set u(x, y, z) = z 4 + x2 z 3 + y 2 + xy − 2. Then

∂u = 2xz 3 + y, ∂x

∂u = 2y + x, ∂y

∂u ∂z 2y + x ∂y =− =− 3 ∂u ∂y 4z + 3x2 z 2 ∂z

∂u ∂z 2xz 3 + y = − ∂x = − 3 , ∂u ∂x 4z + 3x2 z 2 ∂z

  57. Set u = cos xyz + ln x2 + y 2 + z 2 . ∂u 2x , = −yz sin xyz + 2 ∂x x + y2 + z2

Then ∂u 2y , = −xz sin xyz + 2 ∂y x + y2 + z2

and

∂u 2z . = −xy sin xyz + 2 ∂z x + y2 + z2

  2x − yz x2 + y 2 + z 2 sin xyz ∂z ∂u/∂x =− =− , ∂x ∂u/∂z 2z − xy (x2 + y 2 + z 2 ) sin xyz   2y − xz x2 + y 2 + z 2 sin xyz ∂z ∂u/∂y =− =− . ∂y ∂u/∂z 2z − xy (x2 + y 2 + z 2 ) sin xyz

58. (a)

Use

(b) (i)

du du1 du2 = i+ j and apply the chain rule to u1 , u2 . dt dt dt du = t(ex cos yi + ex sin yj) + π(−ex sin yi + ex cos yj) dt 2

= tet

/2

2

(ii) u(t) = et

2

(cos πti + sin πtj) + πet

/2

2

cos πti + et

/2

/2

(− sin πti + cos πtj)

sin πtj

du 2 2 2 2 = (−πet /2 sin πt + tet /2 cos πt)i + (πet /2 cos πt + tet /2 sin πt)j dt

59.

∂u ∂u ∂x ∂u ∂y = + , ∂s ∂x ∂s ∂y ∂s

60.

du ∂u dx ∂u dy ∂u dz = + + dt ∂x dt ∂y dt ∂z dt

∂u ∂u ∂x ∂u ∂y = + ∂t ∂x ∂t ∂y ∂t

∂u ∂u1 ∂u2 ∂u3 = i+ j+ k, ∂y ∂y ∂y ∂y

where

∂u ∂u1 ∂u2 ∂u3 = i+ j+ k, ∂x ∂x ∂x ∂x

∂u ∂u1 ∂u2 ∂u3 = i+ j+ k. ∂z ∂z ∂z ∂z

SECTION 16.4 1. Set f (x, y) = x2 + xy + y 2 .

Then,

∇f = (2x + y)i + (x + 2y)j, normal vector i + j; tangent vector i − j tangent line x + y + 2 = 0; normal line x − y = 0

∇f (−1, −1) = −3i − 3j.

∂u = 4z 3 + 3x2 z 2 ∂z

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SECTION 16.4 2. Set f (x, y) = (y − x)2 − 2x,

∇f = −2(y − x + 1)i + 2(y − x)j,

807

∇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   2  3. Set f (x, y) = x2 + y 2 − 9 x2 − y 2 . Then,

√  √     ∇f = [4x(x2 + y 2 ) − 18x]i + 4y x2 + y 2 + 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 ∇f = 3x2 i + 3y 2 j,

4. Set f (x, y) = x3 + y 3 ,

∇f (1, 2) = 3i + 12j

tangent vector 4i − j

normal vector i + 4j;

tangent line x + 4y − 9 = 0;

normal line 4x − y − 2 = 0

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

Then,

∇f = (y 2 − 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 + y 5 − 2x3 . ∇f = (5x4 − 6x2 )i + 5y 4 j, normal vector i − 5j;

∇f (1, 1) = −i + 5j

tangent vector 5i + j

tangent line x − 5y + 4 = 0;

normal line 5x + y − 6 = 0

7. Set f (x, y) = 2x3 − x2 y 2 − 3x + y.

Then,

∇f = (6x2 − 2xy 2 − 3) i + (−2x2 y + 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 + y 2 + 2x. ∇f = (3x2 + 2)i + 2yj, normal vector 5i + 6j;

tangent vector 6i − 5j

tangent line 5x + 6y − 13 = 0; 9. Set f (x, y) = x2 y + a2 y.

∇f (−1, 3) = 5i + 6j

normal line 6x − 5y + 21 = 0

By (15.4.4) m=−

∂f /∂x 2xy =− 2 . ∂f /∂y x + a2

At (0, a) the slope is 0. 10. Set f (x, y, z) = (x2 + y 2 )2 − z. ∇f = 4x(x2 + y 2 )i + 4y(x2 + y 2 )j − k, Tangent plane: Normal:

8x + 8y − z − 12 = 0

x = 1 + 8t,

y = 1 + 8t,

z =4−t

∇f (1, 1, 4) = 8i + 8j − k

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SECTION 16.4

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

Then,

  ∇f = (3x2 − 3yz) i + (3y 2 − 3xz) j − 3xyk, ∇f 1, 2, 32 = −6i + 15 2 j − 6k;     3 tangent plane at 1, 2, 32 : −6(x − 1) + 15 2 (y − 2) − 6 z − 2 = 0, which reduces to 4x − 5y + 4z = 0. Normal:

y = 2 − 5t,

x = 1 + 4t,

z=

3 2

+ 4t

12. Set f (x, y, z) = xy 2 + 2z 2 . ∇f = y 2 i + 2xyj + 4zk, x + y + 2z − 7 = 0

Tangent plane: Normal:

∇f (1, 2, 2) = 4i + 4j + 8k

x = 1 + t,

y = 2 + t,

z = 2 + 2t

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

1 tangent plane at 1, , 1 : z − 1 = 1(x − 1) + a y − a Normal: x = 1 + t, y = a1 + at, z = 1 − t 14. Set f (x, y, z) =

√

Tangent plane: Normal:

√

1 1, = i + aj.

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

√

1 1 1 z. ∇f = √ i + √ j + √ k, 2 y 2 x 2 z 2x + y + 2z − 8 = 0

x+

x = 1 + 2t,

y+

y = 4 + t,

Then,

∇g = [cos x + cos (x + y)] i + [cos y + cos (x + y)] j, tangent plane at (0, 0, 0) : z − 0 = 2(x − 0) + 2(y − 0), x = 2t,

y = 2t,

1 1 1 i+ j+ k 2 4 2

z = 1 + 2t

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

Normal:

∇f (1, 4, 1) =

∇g(0, 0) = 2i + 2j;

2x + 2y − z = 0.

z = −t

16. Set f (x, y, z) = x2 + xy + y 2 − 6x + 2 − z. ∇f = (2x + y − 6)i + (x + 2y)j − k, −k Tangent plane: Normal:

x = 4,

z = −10 y = −2,

z = −10 + t

17. Set f (x, y, z) = b2 c2 x2 − a2 c2 y 2 − a2 b2 z 2 .

Then,

∇f (x0 , y0 , z0 ) = 2b2 c2 x0 i − 2a2 c2 y0 j − 2a2 b2 z0 k; tangent plane at (x0 , y0 , z0 ) : 2b2 c2 x0 (x − x0 ) − 2a2 c2 y0 (y − y0 ) − 2a2 b2 z0 (z − z0 ) = 0, which can be rewritten as follows: b2 c2 x0 x − a2 c2 y0 y − a2 b2 z0 z = b2 c2 x0 2 − a2 c2 y0 2 − a2 b2 z0 2 = f (x0 , y0 , z0 ) = a2 b2 c2 . Normal: x = x0 + 2b2 c2 x0 t,

y = y0 − 2a2 c2 y0 t,

z = z0 − 2a2 b2 z0 t

∇f (4, −2, −10) =

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SECTION 16.4 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: Normal:

x = 1,

y+z = y=

π 2

π + t, 2

z=t

19. Set z = g(x, y) = xy + a3 x−1 + b3 y −1 .     ∇g = y − a3 x−2 i + x − b3 y −2 j,

∇g = 0 =⇒ y = a3 x−2 and x = b3 y −2 .

Thus, y = a3 b−6 y 4 , y 3 = b6 a−3 , y = b2 /a, x = b3 y −2 = a2 /b   The tangent plane is horizontal at a2 /b, b2 /a, 3ab .

  and g a2 /b, b2 /a = 3ab.

20. z = g(x, y) = 4x + 2y − x2 + xy − y 2 . ∇g = (4 − 2x + y)i + (2 + x − 2y)j 10 8 ∇g = 0 =⇒ 4 − 2x + y = 0, 2 + x − 2y = 0 =⇒ x = , y= 3 3 8 28 The tangent plane is horizontal at ( 10 , , ). 3 3 3 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 + y 2 − 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 − y 2 − 5x + 3y − 2.

Then,

∇g = (4x + 2y − 5) i + (2x − 2y + 3) j. ∇g = 0

4x + 2y − 5 = 0 = 2x − 2y + 3   1 The tangent plane is horizontal at 13 , 11 6 , − 12 . =⇒

=⇒

x = 13 ,

y=

24. (a) Set f (x, y, z) = xy − z.√ ∇f = yi + xj − k, ∇f (1, 1, 1) = i + j − k 3 upper unit normal = (−i − j + k) 3 1 1 1 1 (b) Set f (x, y, z) = − − z. ∇f = − 2 i + 2 j − k, ∇f (1, 1, 0) = −i + j − k x y√ x y 3 lower unit normal: = (−i + j − k) 3 25.

x − x0 y − y0 z − z0 = = (∂f /∂x)(x0 , y0 , z0 ) (∂f /∂y)(x0 , y0 , z0 ) (∂f /∂z)(x0 , y0 , z0 )

11 6 .

809

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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

   2  x0 x0  x0 x0  x0 +f − (y − y0 ) − (z − z0 ) = 0. (x − x0 ) f f y0 y0 y0 y0 2 y0

The plane passes through the origin:







x0 2  x0 x0 2  x0 x0 x0 − − x0 f + + z0 = z0 − x0 f = 0. f f y0 y0 y0 y0 y0 y0 27. Since the tangent planes meet at right angles, the normals ∇F and ∇G meet at right angles: ∂F ∂G ∂F ∂G ∂F ∂G + + = 0. ∂x ∂x ∂y ∂y ∂z ∂z 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 written x − x0 y − y0 z − z0 + √ + √ = 0. √ x0 y0 z0 Setting y = z = 0 we have √ x − x0 √ = y 0 + z0 . √ x0 Therefore the x-intercept is given by √ √ √ √ √ √ √ 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. x = x0 +

√

29. The tangent plane at an arbitrary point (x0 , y0 , z0 ) has equation y0 z0 (x − x0 ) + x0 z0 (y − y0 ) + x0 y0 (z − z0 ) = 0, which simplifies to y0 z0 x + x0 z0 y + x0 y0 z = 3x0 y0 z0

and thus to

The volume of the pyramid is

x y z + + = 1. 3x0 3y0 3z0

  1 1 (3x0 ) (3y0 ) 9 9 V = Bh = (3z0 ) = x0 y0 z0 = a3 . 3 3 2 2 2

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 + x0 1/3 (y0 2/3 + z0 2/3 ) = x0 + x0 1/3 (a2/3 − x0 2/3 ) =⇒ x = x0 1/3 a2/3 Similarly, the y-intercept is y0 1/3 a2/3 and the z-intercept is z0 1/3 a2/3 . The sum of the squares of the intercepts is (x0 2/3 + y0 2/3 + z0 2/3 )a4/3 = a2/3 a4/3 = a2 .

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SECTION 16.4

811

31. The point (2, 3, −2) is the tip of r(1). 3 Since r (t) = 2i − 2 j − 4tk, we have r (1) = 2i − 3j − 4k. t Now set f (x, y, z) = x2 + y 2 + 3z 2 − 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 (2i − 3j − 4k) (2i + 3j − 6k) 19 √ · = √ ∼ = 0.504. 7 29 7 29 Therefore θ ∼ = 1.043 radians. The angle between the curve and the plane is π −θ ∼ = 1.571 − 1.043 ∼ = 0.528 radians. 2 cos θ =

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 + 2y 2 + 3z 2 . ∇f = 2xi + 4yj + 6zk, ∇f (3, 2, 1) = 6i + 8j + 6k, 33. Set

which is parallel to r (1).

f (x, y, z) = x2 y 2 + 2x + z 3 .

Then,

∇f = (2xy 2 + 2) i + 2x2 yj + 3z 2 k,

∇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 + y 2 − 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 + y 2 + z 2 − 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 + 3y 2 + 2z 2 ,

∇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.

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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 − az 2 ,

∇f = yi + xj − 2azk

Surface B: Set g(x, y, z) = x2 + y 2 + z 2 ,

∇g = 2xi + 2yj + 2zk

Surface C: Set h(x, y, z) = z 2 + 2x2 − c(z 2 + 2y 2 ). ∇h = 4xi − 4cyj + 2(1 − c)zk Where surface A and surface B intersect,

∇f · ∇g = 4(xy − az 2 ) = 0

Where surface A and surface C intersect,

∇f · ∇h = 4(1 − c)(xy − az 2 ) = 0

Where surface B and surface C intersect,

∇g · ∇h = 4[2x2 − 2cy 2 + (1 − c)z 2 ] = 0

37. (a) 3x + 4y + 6 = 0 since plane p is vertical. (b) y = − 14 (3x + 6) = − 14 [3(4t − 2) + 6] = −3t z = x2 + 3y 2 + 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 + 3y 2 + 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 − 34 (t + 2) − 29 =

35 2 t

− 2.

Thus, (∗)

r(t) = ti − ( 34 t + 32 )j +

 35 2

 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).

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SECTION 16.4 38. (a) normal vector: (b) tangent line: (c) 2

12 25

i−

x=2+

14 25 14 25

j;

normal line:

t, y = 1 +

12 25

x=2+

12 25

t, y = 1 −

14 25

t

t

y

1

1

39. (a) normal vector: (b) tangent plane:

2

3

x

2 i + 2 j + 4 k;

normal line:

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

2(x − 1) + 2(y − 2) + 4(z − 2) = 0

or x + y + 2z − 7 = 0

(c)

40. (a)

(b)

1.5 1 0.5

2 1

0 0

-1

-0.5 -1

0

-1

1

(c) ∇f = 0 at (±1, 0),

0

-1.5

0, ±



 3/2

-1.5 -1

-0.5

0

0.5

1

1.5

813

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SECTION 16.5

41. (a)

(b)

1.5 1 0.5 0 -0.5

1

-1

3 2

-1.5 -1.5 -1 -0.5

0

0

0.5

1

1.5

1 0 -1

-1 0 1

    (c) ∇f = 4x3 − 4x i − 4y 3 − 4y j; ∇f = 0 :

4x3 − 4x = 0

x = 0, ±1;

=⇒

4y 3 − 4y = 0

y = 0, ±1

=⇒

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

42. (a)

(b)

2

1

1

2

0

1

-1 -2

0

0

-1

-1

0 1

2 -2

-1

-2 -2

-1

0

1

√  √  (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 − k 2 − 1 = −h2 − k 2 ≤ 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 + k 2 does not keep a constant sign for small h and k; (1, −1) is a saddle point.

2

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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 + k 2 is nonnegative for all small h and k. To see this, note that h2 + hk + k 2 ≥ h2 − 2|h||k| + k 2 = (|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 ∂2f = 2, ∂x2

∂2f = 2, ∂y∂x

∂2f = 6; ∂y 2

only at (1, −2).

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

  7. ∇f = (3x2 − 6y) i + 3y 2 − 6x j = 0 at (2, 2) and (0, 0). fxx = 6x,

fxy = −6,

D = 36xy − 36.

fyy = 6y,

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 ∂2f = 6, ∂x2

∂2f = 1, ∂y∂x

∂2f = −2; ∂y 2

−

5 35 ,− . 13 13

D = 6 · (−2) − 12 < 0;

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

fxy = −6,

fyy = 2,

27 2 )

(−5/13, −35/13) is a saddle point.

and (1, 32 ).

D = 12x − 36.

117 At (5, 27 2 ), D = 24 > 0 and A = 30 > 0 so we have a local min; the value is − 4 .

At (1, 32 ), D = −24 < 0 so we have a saddle point. 10. ∇f = (2x − 2y − 3) i + (−2x + 4y + 5) j = 0 ∂2f ∂2f = 2, = −2, ∂x2 ∂y∂x the value is − 13 4 .

∂2f = 4; ∂y 2

11. ∇f = sin y i + x cos y j = 0 fxx = 0,

fxy = cos y,

at

( 12 , −1).

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

at (0, nπ) for all integral n.

fyy = −x sin y.

Since D = − cos nπ = −1 < 0, each stationary point is a saddle point. 2

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SECTION 16.5

12. ∇f = sin y i + (1 + x cos y) j = 0 ∂2f = 0, ∂x2

∂2f = cos y, ∂y∂x

at (−1, 2nπ) and (1, (2n + 1)π) for all integral n.

∂2f = −x sin y; ∂y 2

D = 0 · (−x sin y) − cos2 y < 0

at the above points

so they are all saddle points   13. ∇f = (2xy + 1 + y 2 ) i + x2 + 2xy + 1 j = 0 at fxx = 2y,

fxy = 2x + 2y,

(1, −1) and (−1, 1).

D = 4xy − 4(x + y)2 .

fyy = 2x,

At both (1, −1) and (−1, 1) we have saddle points since D = −4 < 0. 14. ∇f =

1 y + 2 y x





x 1 i+ − 2 − y x

j=

x2 + y 2 x2 + y 2 j i − x2 y xy 2

is never 0;

no stationary points, no local extreme values.   1  15. ∇f = (y − x−2 ) i + x − 8y −2 j = 0 only at 2, 4 . fxx = 2x−3 , fxy = 1, fyy = 16y −3 , D = 32x−3 y −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 = 2, ∂x2

∂2f = −2, ∂y∂x

∂2f = −2; ∂y 2

D = 2 · (−2) − (−2)2 < 0;

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

fxy = 1,

fyy = 2y −3 ,

(0, 0) is a saddle point.

(1, 1).

D = 4x−3 y −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 − y 2 − 1) i + (x2 − 2xy + 1) j = 0 at (1, 1), (−1, −1) ∂2f = 2y, ∂x2

∂2f = 2(x − y), ∂y∂x

∂2f = −2x; ∂y 2

D = −4xy − 4(x − y)2 < 0

at the above points;

(1, 1) and (−1, −1) are saddle points. 19. ∇f =

fxx =

  2 x2 − y 2 − 1 (x2

+

y2

2

+ 1)

i+

(x2

−4x3 + 12xy 2 + 12x , (x2 + y 2 + 1)3

f (1, 0) = −1;

4xy j=0 + y 2 + 1)2 fxy =

at

(1, 0) and (−1, 0).

4y 3 + 4y − 12x2 y , (x2 + y 2 + 1)3

fyy =

4x3 + 4x − 12xy 2 . (x2 + y 2 + 1)3

point

A

B

C

D

(1, 0)

1

0

1

1

loc. min.

(−1, 0)

−1

0

−1

1

loc. max.

f (−1, 0) = 1

result

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SECTION 16.5 20. ∇f =

3 ln xy + 1 − x

∂2f 1 3 = + 2, 2 ∂x x x



x−3 j=0 y

i+

∂2f 1 = , ∂y∂x y

∂2f 2 = , ∂x2 3

At (3, 1/3),

3−x ∂2f = 2 ∂y y2

∂2f = 3, ∂y∂x

  21. ∇f = 4x3 − 4x i + 2y j = 0 fxx = 12x2 − 4,

at (3, 1/3)

fxy = 0,

at

∂2f =0 ∂y 2

and D = −9 < 0 =⇒ saddle point.

(0, 0), (1, 0), and (−1, 0).

fyy = 2. point

A

B

C

D

result

(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. 2

22. ∇f = 2xex

−y 2

2

(1 + x2 + y 2 ) i + 2yex

−y 2

(1 − x2 − y 2 ) j = 0

at (0, 0), (0, 1), (0, −1)

2

A=

∂ f 2 2 2 2 = 2xex −y (2x + 2x3 + 2xy 2 ) + ex −y (2 + 6x2 + 2y 2 ) ∂x2

B=

∂2f 2 2 2 2 = −2yex −y (2x + 2x3 + 2xy 2 ) + ex −y (4xy) ∂y∂x

C=

∂2f 2 2 2 2 = −2yex −y (2y − 2yx2 − 2y 3 ) + ex −y (2 − 2x2 − 6y 2 ) ∂y 2

At (0, 0), At (0, ±1),

AC − B 2 = (2)(2) = 4 > 0, A > 0

local minimum; the value is 0.

AC − B 2 = (4e−1 )(−4e−1 ) = −8e−2 > 0, 1

23. ∇f = cos x sin y i + sin x cos y j = 0 at fxx = − sin x sin y,

1 2 π, 2 π

point  1 2 π, 2 π 1 3  2 π, 2 π

 1 3      , 2 π, 2 π , (π, π), 32 π, 12 π , 32 π, 32 π .

fyy = − sin x sin y

fxy = cos x cos y, 1

saddle points

A

B

C

D

−1

0

−1

1

loc. max.

1

0

1

1

loc. min.

(π, π) 0 1 0 −1  1 1 0 1 1 2 π, 2 π 3 3  −1 0 −1 1 2 π, 2 π 1 1  3 3  1 3  3 1  f 2 π, 2 π = f 2 π, 2 π = 1; f 2 π, 2 π = f 2 π, 2 π = −1 3

result

saddle loc. min. loc. max.

24. ∇f = − sin x cosh y i + cos x sinh y j = 0 at (−π, 0) , (0, 0) , (π, 0). fxx = − cos x cosh y,

fxy = − sin x sinh y,

D = − cos2 x cosh y − sin2 x sinh y < 0; 2

2

fyy = cos x cosh y

(−π, 0) , (0, 0) , (π, 0) are saddle points.

817

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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 − k 2 . Thus, D < 0 for |k| > 2 and (0, 0) is a saddle point

(c) D = 4 − k 2 > 0 for |k| < 2. Since A = fxx = 2 > 0, (0, 0) is a local minimum. (d) The test is inconclusive when D = 4 − k 2 = 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 (b)

∂2f = 2, ∂x2

∂2f = k, ∂y∂x

∂2f = 8; ∂y 2

at (0, 0). we want

16 − k 2 < 0, or |k| > 4

(c) We want 16 − k 2 > 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) = 2

2



x2 + y 2 + z 2 .

2

However, it is sufficient to minimize the square of the distance: F (x, y, z) = x + y + z . 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 + z 2 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

32 16 , from which it follows that y = − . 9 9   16 32 The point in the plane that is closest to the origin is P 32 , − , . 9 9 9

The only solution to this pair of equations is: x = z =

The distance from the origin to the plane is: F (P ) = Check using (13.6.5): d(P, 0) =

16 3 .

|2 · 0 − 0 + 2 · 0 − 16| 16  = . 2 2 2 3 2 + (−1) + 2

28. We want to minimize (x + 1)2 + (y − 2)2 + (z − 4)2 on the plane. Since z = −16 − 32 x + 2y, we need to minimize f (x, y) = (x + 1)2 + (y − 2)2 + (−20 − 32 x + 2y)2 ;   ∇f = 13 2 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 + z 2 = (x − 1)2 + (y − 2)2 + x2 + 2y 2

 since z =



x2 + 2y 2



1 2 , y= . 2 3 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 = 16 114 ∇f = [2(x − 1) + 2x] i + [2(y − 2) + 4y] j = 0

=⇒

x=

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SECTION 16.5 30. V = 8xyz,

x2 + y 2 + z 2 = a2 =⇒ V (x, y) = 8xy



a2 − x2 − y 2 , x > 0, y > 0, x2 + y 2 < a2

8y(a2 − x2 − y 2 ) − 8x2 y 8x(a2 − x2 − y 2 ) − 8xy 2   i+ j=0 a2 − x2 − y 2 a2 − x2 − y 2 2a 2a 2a 8 3√ dimensions: √ × √ × √ , maximum volume: a 3 9 3 3 3

∇V =

31. (a)

-2

-1

2 4

0

0.5 0 -0.5

1

-1

-0.5

1

1

2

-2

√ √  a/ 3, a/ 3

(b) 0

3

0

at

1

-1

0 2

-1

0.5

1

(c) ∇f = (4y − 4x3 ) i + (4x − 4y 3 ) j = 0 at (0, 0), (1, 1), (−1, −1). fxx = −12x2 ,

fxy = 4,

fyy = −12y 2 ,

D = 144x2 y 2 − 16

point

A

B

C

D

result

(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)

(b)

1.5 1 0.5

3 2

0

1

1 0

-1 0

-0.5

-1

1

-1 -1.5 -1.5

-1

-0.5

(c) ∇f = (4x3 − 4x) i + 2y j = 0 at (0, 0), (1, 0), (−1, 0). fxx = 12x2 − 4,

fx,y = 0,

f (1, 0) = f (−1, 0) = 0

fyy = 2,

D = 24x2 − 8

point

A

B

C

D

result

(0, 0)

−8

0

2

−8

saddle

(1, 0)

8

0

2

16

loc. min.

(−1, 0)

8

0

2

16

loc. min.

0

0.5

1

1.5

819

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SECTION 16.5

33. (a)

(b)

4

1.5 1 0.5

2

0 -0.5 -1

0 -2

0

-1.5

2

-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)

(b)

2

1 0.6 0.4 0.2 0 -2

2 1

0

0 -1

-1

0

1

2

-1

-2

-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) 0

2

-2

0

(b)

2

2

-2 1

1

0

0

-1

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

-1

-2 -2

-1

0

1

2

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SECTION 16.6 36. (a)

(b)

821

10

8

3 2 1 0 -1 0

6

10 8

4

6 2

4

4 6

8

2

2 100

0

0

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 y 4

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: C1 : r1 (t) = t i,

1

2

x

t ∈ [ 0, 2 ],

C2 : r2 (t) = 2 i + t j, C3 : r3 (t) = t i + 2t j,

t ∈ [ 0, 4 ], 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) + 1,

t ∈ [ 0, 4 ];

critical number: t = 1,

f3 (t) = f (r3 (t)) = 6t2 − 8t + 2,

t ∈ [ 0, 2 ];

critical number: t = 23 .

2

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; f2 (1) = f (2, 1) = 1;

f1 (1) = f (1, 0) = 0;

f2 (4) = f3 (2) = f (2, 4) = 10;

f1 (2) = f2 (0) = f (2, 0) = 2; 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: t ∈ [ 0, 4 ],  3  C2 : r2 (t) = t i + − 2 t + 6 j, C1 : r1 (t) = t i,

C3 : r3 (t) = t j,

t ∈ [ 0, 6 ],

t ∈ [ 0, 4 ],

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December 7, 2006

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, f3 (t) = f (r3 (t)) = 2 + 2t,

t ∈ [ 0, 4 ];

t ∈ [ 0, 6 ];

no critical numbers, no critical numbers.

Evaluating these functions at the endpoints of their domains, we find that: f1 (4) = f2 (4) = f (4, 0) = −10;

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

f2 (0) = f3 (6) = f (0, 6) = 14;

f takes on its absolute maximum of 14 at (0, 6) and its absolute minimum of −10 at (4, 0). 3. ∇f = (2x + y − 6) i + (x + 2y) j = 0 at (4, −2) in D;

y 1

f (4, −2) = −13

2

3

4

5

x

-1

Next we consider the boundary of D. We -2

parametrize each side of the rectangle: C1 : r1 (t) = −t j,

-3

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 =

9 2

f3 (t) = f (r3 (t)) = t2 − 5t − 6,

t ∈ [ 0, 3 ];

critical number: t =

5 2

f4 (t) = f (r4 (t)) = t2 − 6t − 1,

t ∈ [ 0, 5 ];

critical number: t = 3

Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f1 (0) = f4 (0) = f (0, 0) = −1;

f1 (−3) = f2 (0) = f (0, −3) = 8;

f2 (5) = f3 (3) = f (5, −3) = −12;

f3 (5/2) = f (5, −5/2) = − 49 4 ;

f2 (9/2) = f (9/2, −3) = − 49 4 ; 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; Next we consider the boundary of D. 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,

f (0, 0) = 0

We parametrize each side of the square:

t ∈ [ −2, 2 ],

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SECTION 16.6

823

and evaluate f : f1 (t) = f (r1 (t)) = t2 − 4t + 12,

t ∈ [ −2, 2 ];

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 .

no critical numbers,

Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f1 (−2) = f4 (−2) = f (−2, −2) = 24; f2 (2) = f3 (2) = f (2, 2) = 24;

f1 (2) = f2 (−2) = f (2, −2) = 8;

f3 (−2) = f4 (2) = f (−2, 2) = 8;

f2 (−2/3) = f (2, −2/3) = 83 ;

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 + 3y 2 = (x + y)2 + 2y 2 . The results follow immediately from this observation. 5. ∇f = (2x + 3y) i + (2y + 3x) j = 0 at (0, 0) in D; Next we consider the boundary of D. C : r(t) = 2 cos t i + 2 sin t j,

f (0, 0) = 2

We parametrize the circle by:

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, F  (t) = 12 cos2 t − 12 sin2 t :

F  (t) = 0

=⇒

t ∈ [ 0, 2π ]

cos t = ± sin t

=⇒

t = 14 π,

3 5 7 4 π, 4 π, 4 π

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

√ √ 

√ √      F (0) = F (2π) = f (2, 0) = 6; F 14 π = F 54 π = f 2, 2 = f ( − 2, − 2 = 12;

√ √ 

√ √  3  7  2, − 2 = 0 4 π = f − 2, 2 = F 4 π = f √  √ √   √ f takes on its absolute maximum of 12 at 2, 2 and at − 2, − 2 ; f takes on its absolute √  √  √ √  2, − 2 . minimum of 0 at − 2, 2 and at F

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 df = 9(2 cos2 t − cos t − 1); = 0 =⇒ cos t = 1, − 12 which yields the points A (3, 0), dt dt √ √ √ √ B (− 32 , 3 2 3 ), C (− 32 , − 3 2 3 ) : f (A) = 0, f (B) = − 274 3 min, f (C) = 274 3 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.

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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. C1 : r1 (t) = t j + t2 j, C2 : r2 (t) = t i + 4 j,

We parametrize the boundary by:

t ∈ [ −2, 2 ], t ∈ [ −2, 2 ],

and evaluate f : f1 (t) = f (r1 (t)) = t3 − t2 + t + 3, f2 (t) = f (r2 (t)) = 5t − 1,

t ∈ [ −2, 2 ];

t ∈ [ −2, 2 ];

no critical numbers,

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 − 2y 2 − 2 4xy i+ 2 j = 0 at (1, 0) and (−1, 0) in D; f (1, 0) = −1, (x2 + y 2 + 1)2 (x + y 2 + 1)2

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, 4 , +5 −2t , f2 (t) = f (r2 (t)) = 2 t +5 −4 f3 (t) = f (r3 (t)) = 2 , t +5 −2t f4 (t) = f (r4 (t)) = 2 , t +5

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

t2

t ∈ [ −2, 2 ]; critical number: t = 0 t ∈ [ −2, 2 ]; no critical numbers t ∈ [ −2, 2 ]; critical number: t = 0 t ∈ [ −2, 2 ]; no critical numbers

Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: f1 (−2) = f4 (−2) = f (−2, −2) = 49 ; f4 (2) = f3 (−2) = f (2, −2) = − 49 ;

f1 (0) = f (−2, 0) = 45 ; f3 (0) = f (2, 0) = − 45 ;

f1 (2) = f2 (−2) = f (−2, 2) = 49 ; 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 − 2y 2 − 2 4xy i+ 2 j = 0 at (1, 0) in D; f (1, 0) = −1. (x2 + y 2 + 1)2 (x + y 2 + 1)2

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 ],

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SECTION 16.6

825

and evaluate f : √ −2t , t ∈ [ 0, 2 ]; critical number: t = 1/ 2, 2 2t + 1 −4 f2 (t) = f (r2 (t)) = 2 , t ∈ [ −2, 2 ]; critical number: t = 0 t +5 √ −2t f3 (t) = f (r3 (t)) = 2 , t ∈ [ 0, 2 ]; critical number: t = 1/ 2. 2t + 1

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

Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: √ √ √ √ 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) = − 45 ; √ √ √ √ f3 (1/ 2) = f (1/ 2, 1/ 2) = −1/ 2.

f2 (2) = f3 (2) = f (2, 2) = − 49 ;

f takes on its absolute maximum of 0 at (0, 0) and its absolute minimum of −1 at (1, 0). 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 ∈ − 14 π, 14 π C2 : r2 (t) = t i − 14 π j, C3 : r3 (t) = 2 i + t j,

t ∈ [ 0, 2 ]   t ∈ − 14 π, 14 π t ∈ [ 0, 2 ]

C4 : r4 (t) = t i + 14 π j, Now, f1 (t) = f (r1 (t)) = 0; √

2 (4t − 2t2 ), 2 f3 (t) = f (r3 (t)) = 0; √ 2 f4 (t) = f (r4 (t)) = (4t − 2t2 ), 2 f2 (t) = f (r2 (t)) =

t ∈ [ 0, 2 ];

t ∈ [ 0, 2 ];

f at the vertices of the rectangle has the value 0;

critical number: t = 1;

critical number: t = 1;     √ f2 (1) = f4 (1) = f 1, − 14 π = f 1, 14 π = 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. df Boundary: On y = x2 , f = (x − 3)2 + x4 , = 2(x − 3) + 4x3 = 0 at x = 1 =⇒ (1, 1) dx 3 3 12 df On y = 4x, f = (x − 3)2 + 16x2 , = 2(x − 3) + 32x = 0 at x = =⇒ ( , ). dx 17 17 17 So the maximum and minimum occur at one or more of the following points: (0, 0), (4, 16), (1, 1), (

3 12 , ). 17 17

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 .

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SECTION 16.6

13. ∇f = (3x2 − 3y) i + (−3x − 3y 2 ) j = 0 at (−1, 1) in D;

y 2

f (−1, 1) = 1 Next we consider the boundary of D. We -2

parametrize each side of the triangle: C1 : r1 (t) = −2 i + t j,

-1

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 :

x

-2

t ∈ [ −2, 2 ];

f3 (t) = f (r3 (t)) = t3 − 6t − 8,

2

√ t ∈ [ −2, 2 ]; critical numbers: t = ± 2,

f1 (t) = f (r1 (t)) = −8 + 6t − t3 , f2 (t) = f (r2 (t)) = −3t2 ,

1

critical number: t = 0,

√ critical numbers: t = ± 2.

t ∈ [ −2, 2 ];

Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: √ √ √ 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 ∼ f1 (2) = f3 (−2) = f (−2, 2) = −4; = −2.34; 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 + t3 j,

t ∈ [ 0, 2 ].

Note that f1 (t) = f (r1 (t)) = 17t2 − 8t + 16, f2 (t) = f (r2 (t)) = (t − 4)2 + t6 . Next f1 (t) = 34t − 8 = 0

=⇒

t = 4/17

and gives x = 4/17, y = 16/17

and f2 (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:  4 16  f (0, 0) = 16, f (2, 8) = 68, f 17 , 17 =

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.

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SECTION 16.6 15. ∇f =

827

4xy 2y 2 − 2x2 − 2 i + j = 0 at (0, 1) and (0, −1) in D; (x2 + y 2 + 1)2 (x2 + y 2 + 1)2

f (0, 1) = −1,

f (0, −1) = 1

Next we consider the boundary of D. C : r(t) = 2 cos t i + 2 sin t j,

We parametrize the circle by:

t ∈ [ 0, 2π ]

The values of f on the boundary are given by the function F (t) = f (r(t)) = − 45 sin t, F  (t) = − 45 cos t :

F  (t) = 0

=⇒

t ∈ [ 0, 2π ]

cos t = 0

Evaluating F at the endpoints and critical numbers, we have:   F (0) = F (2π) = f (2, 0) = 0; F 12 π = f (0, 2) = − 45 ;

=⇒

F

t = 12 π,

3 2 π.

3  2 π = 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 − 12 , 14 in D; Next we consider the boundary of D. C : r(t) = 2 cos t i + sin t j,

  f − 12 , 14 = − 12

We parametrize the ellipse by:

t ∈ [ 0, 2π ]

The values of f on the boundary are given by the function F (t) = f (r(t)) = 4 cos2 t + 4 sin2 t + 2 cos t − 2 sin t = 4 + 2 cos t − 2 sin t, F  (t) = −2 sin t − 2 cos t :

F  (t) = 0

=⇒

cos t = − sin t

=⇒

t ∈ [ 0, 2π ]

t = 34 π, or 74 π

Evaluating F at the endpoints and critical numbers, we have: F (0) = F (2π) = f (2, 0) = 6;

√ √ 

√ √ √  √ 3    F 4 π = f − 2, 2/2 = 4 − 2 2; F 74 π = 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 − 12 , 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. Next we consider the boundary of D. We 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 ]

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SECTION 16.6 and observe from f (r1 (t)) = t2 , 2

f (r2 (t)) = t ,

t ∈ [ 0, 12 ] 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 + y 2 )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 − x2 y − xy 2 in the triangular region. The gradient of f is:     ∇D = 18y − 2xy − y 2 i + 18x − x2 − 2xy j The gradient is 0 when 18y − 2xy − y 2 = 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) = 30yz − y z − yz , 2

2 2

3

∇f = (30z − 2yz − z )j + (60yz − 2y z − 3yz )k = 0 at 2



2

15 , 15 2

3



2

2

154 . 4

(other points are not in the interior);

f

On the line y + z = 30, f (y, z) = 0

so the maximum of xyz 2 occurs at x = y =

21. f (x, y) = xy(1 − x − y),

0 ≤ x ≤ 1,

=

15 , 15 2

15 , 2

z = 15

0 ≤ y ≤ 1 − x.

[ dom (f ) is the triangle with vertices (0, 0), (1, 0), (0, 1).]   ∇f = (y − 2xy − y 2 )i + x − 2xy − x2 j = 0 =⇒ x = y = 0, x = 1, y = 0, x = 0, y = 1, x = y = 13 . (Note that [ 0, 0 ] is not an interior point of the domain of f .) fxx = −2y, fxy = 1 − 2x − 2y, fyy = −2x. 1 1 1 At 3 , 3 , D = 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.

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SECTION 16.6

829

x y z x y x y + + = 1 =⇒ V (x, y) = xyc 1 − − , x > 0, y > 0, + 0. 

V (x, y) = xy

S − 2xy 2(x + y)



and

    S − 2xy 2(x + y)(−2y) − (S − 2xy)(2) ∇V = y + xy i 2(x + y) 4(x + y)2     S − 2xy 2(x + y)(−2x) − (S − 2xy)(2) + x + xy j 2(x + y) 4(x + y)2 ∂V ∂V Setting = = 0 and simplifying, we get the pair of equations ∂x ∂y 2S − 4x2 − 8xy = 0 2S − 4y 2 − 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.

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SECTION 16.6

26. V = xyz,

S = xy + 2xz + 2yz

=⇒

V (x, y) = xy

(S − xy) , x > 0, y > 0, xy < S. 2(x + y)

y 2 (S − x2 − 2xy) x2 (S − y 2 − 2xy) i + j 2(x + y)2 2(x + y)2   S S ∇V = 0 =⇒ x = , y= ; dimensions for maximum volume: 3 3 ∇V =

27.

f (x, y) =

3  

(x − xi )2 + (y − yi )2



S × 3



S 1 × 3 2



S 3



i=1

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

x1 + x2 + x3 y1 + y2 + y3 ∇f = 0 only at , = (x0 , y0 ) . 3 3 The difference

f (x0 + h, y0 + k) − f (x0 , y0 ) 3    = (x0 + h − xi )2 + (y0 + k − yi )2 − (x0 − xi )2 − (y0 − yi )2 i=1

=

3    2h (x0 − xi ) + h2 + 2k (y0 − yi ) + k 2 i=1

= 2h (3x0 − x1 − x2 − x3 ) + 2k (3y0 − y1 − y2 − y3 ) + 3h2 + 3k 2 = 3h2 + 3k 2 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  1 x tan θ , A = xy + x 2 2

x  P = x + 2y + 2 sec θ , 2 1 P 0 < θ < π, 0 < x < . 2 1 + sec θ

29.

A(x, θ) = 12 x(P − x − x sec θ) + 14 x2 tan θ,

2

P x2 x x 2 ∇A = − x − x sec θ + tan θ i + sec θ − sec θ tan θ j, 2 2 4 2 (Here j is the unit vector in the direction of increasing θ.) ∇A =

1 x2 [P + x(tan θ − 2 sec θ − 2)] i + sec θ (sec θ − 2 tan θ) j. 2 4

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SECTION 16.6 From

831

∂A ∂A = 0 we get θ = 16 π and then from = 0 we get ∂θ ∂x √ √  √  P + x 13 3 − 43 3 − 2 = 0 so that x = (2 − 3)P.

Next, Axx = 12 (tan θ − 2 sec θ − 2), x sec θ (sec θ − 2 tan θ), 2   x2 = sec θ sec θ tan θ − sec2 θ − tan2 θ . 2

Axθ = Aθθ Apply the second-partials test:

√ √ C = − 13 P 2 3 (2 − 3 )2 , D < 0. √ √ Since, D > 0 and A < 0, the area is a maximum when θ = 16 π, x = (2 − 3 ) P and y = 16 (3 − 3 )P. A = − 12 (2 +

√

3 ),

B = 0,

30. (a) ∇f = (2ax + by)i + (bx + 2cy)j ∂2f = 2a, ∂x2

∂2f = b, ∂y∂x

∂2f = 2c; ∂y 2

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     |a|x − |c|y = 0 (ii) if b > 0, f (x, y) = −( |a|x − |c|y)2 ; every point on the line 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 = 12 y = 13 z = t

and

x=y−2=z =s

(t, 2t, 3t)

and

(s, 2 + s, s)

we take

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; ftt = 28,

fts = −12,

fss = 6,

∇f = 0

=⇒

t = 0, s = −2/3.

D = 6(28) − (−12)2 = −24 < 0.

By the second-partials test, the distance is a minimum when t = 0, s = −2/3;  √ problem tells us the minimum is absolute. The distance is f (0, −2/3) = 23 6.

the nature of the

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SECTION 16.6

32. We want to minimize S = 4πr2 + 2πrh given that V = 43 πr3 + πr2 h = 10, 000.

4 V 4 2V 2 S(r) = 4πr + 2πr − r = πr2 + πr2 3 3 r  V 8πr3 − 6V 4 3 6V S  (r) = , h= 2 − r=0 = 0 =⇒ r = 3r2 8π πr 3  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 ∂V = = 0, we get the pair of equations ∂x ∂y ∂V = 108y − 4xy − 2y 2 = 0 ∂x ∂V = 108x − 4xy − 2x2 = 0 ∂y

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 r2 l,

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 − 2π 2 r3 . Now dV = 216π r − 6π 2 r2 dr Setting

dV = 0, we get dr 216π r − 6π 2 r2 = 0

=⇒

r=

36 π

=⇒

Now, at r = 36/π, we have d2 V 36 = − 216π < 0 = 216π − 12π 2 2 dr π Thus, V is a maximum when r = 36/π inches and l = 36 inches.

l = 36

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SECTION 16.6 34. Let (x, y, z) be on the ellipsoid, x > 0, y > 0, z > 0.

833

Then

V = 2x · 2y · 2z = 8xyz. Note that V achieves its maximum ⇐⇒ Let s = x2 y 2 z 2 ,

=⇒

then

2x2 y2 + = 1, 9 4

x2 y 2 z 2 achieves its maximum.



x2 y2 s = x2 y 2 1 − − , 9 4

2x2 ∂s y2 = 2xy 2 1 − − =0 ∂x 9 4

∂s x2 2y 2 2 = 2x y 1 − − =0 ∂y 9 4 x2 2y 2 + =1 9 4

Thus, Vmax

=⇒

3 x= √ , 3

2 y=√ , 3

1 z=√ 3

√ 3 2 1 16 3 = 8xyz = 8 · √ · √ · √ = . 3 3 3 3

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

1 1 (12 − 2x + 12 − 2x + 2x cos θ) x sin θ = 12x sin θ − 2x2 sin θ + x2 sin 2θ, 2 2

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 ∂S = = 0, we get the pair of equations ∂x ∂θ 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 3√ 3, B = Sxθ = 12 cos θ − 4x cos θ + 2x cos 2θ = −6, 2 √ = −12x sin θ + 2x2 sin θ − 2x2 sin 2θ = −24 3 and D = 108 − 36 > 0.

A = Sxx = −4 sin θ + sin 2θ = − C = Sθθ

Thus, S is a maximum when x = 4 inches and θ = π/3. 36.

96 = xyz, C = 30xy + 10(2xz + 2yz) 96 = 30xy + 20(x + y) . xy

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SECTION 16.7   64 64 + , C(x, y) = 30 xy + x y ∇C = 30(y − 64x−2 )i + 30(x − 64y −2 ) j = 0

=⇒

Cxx = 128x−3 ,

x = y = 4. Cyy = 128y −3 .

Cxy = 1,

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, fmm = 10,

fb = 6m + 6b − 2;

fmb = 6,

fbb = 6,

Answer: the line y = x −

fm = fb = 0

D = 24 > 0

=⇒

m = 1, b = − 23 .

=⇒

a minimum.

2 3.

(b) f (α, β) = [2 − β]2 + [−5 − (α + β)]2 + [4 − (4α + β)]2 . fα = 34α + 10β − 22, fαα = 34,

fβ = 10α + 6β − 2;

⎧ ⎨α =

fα = fβ = 0

fαβ = 10,

fββ = 6, D = 104 > 0   1 Answer: the parabola y = 13 14x2 − 19 .

=⇒

=⇒

⎩

14 13

β = − 19 13

⎤ ⎦.

a minimun.

38. (a) f (m, b) = [2 − (−m + b)]2 + [−1 − b]2 + [1 − (m + b)]2 fm = 4m + 2, fmm = 4,

fb = 6b − 4,

2 1 m=− , b= 2 3 =⇒ minimum

fm = fb = 0

=⇒

fmb = 0,

fbb = 6, D = 24 > 0 1 2 Answer: the line y = − x + 2 3 (b) f (α, β) = [2 − (α + β)]2 + [−1 − β]2 + [1 − (α + β)]2 fα = 4α + 4β − 6, fαα = 4,

fβ = 4α + 6β − 4;

fα = fβ = 0

fαβ = 4,

fββ = 6, D = 8 > 0 5 Answer: the parabola y = x2 − 1 2

=⇒

=⇒

α=

5 , β = −1 2

minimum

SECTION 16.7 g(x, y) = xy − 1

f (x, y) = x2 + y 2 ,

1.

∇f = 2xi + 2yj, ∇f = λ∇g

=⇒

∇g = yi + xj.

2x = λy and 2y = λx.

Multiplying the first equation by x and the second equation by y, we get 2x2 = λxy = 2y 2 . 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.

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SECTION 16.7 2. f (x, y) = xy,

835

g(x, y) = b2 x2 + a2 y 2 − a2 b2 ∇f = yi + xj,

∇g = 2b2 xi + 2a2 yj

∇f = λ∇g =⇒ y = 2λb2 x,

x = 2λa2 y =⇒ a2 y 2 = b2 x2 a b From g(x, y) = 0 we get 2b2 x2 = a2 b2 =⇒ x = ± √ , y = ± √ 2 2 √ √ √ √ 1 The maximum value of xy is 2 ab, achieved at (a/ 2, b 2) and at (−a/ 2, −b 2). g(x, y) = b2 x2 + a2 y 2 − a2 b2

f (x, y) = xy,

3.

∇f = yi + xj, ∇f = λ∇g

∇g = 2b2 xi + 2a2 yj. =⇒

y = 2λb2 x and x = 2λa2 y.

Multiplying the first equation by a2 y and the second equation by b2 x, we get a2 y 2 = 2λa2 b2 xy = b2 x2 . √ √ Thus, ay = ±bx. From g(x, y) = 0 we conclude that x = ± 12 a 2 and y = ± 12 b 2. Since f is continuous and the ellipse is closed and bounded, the minimum exists. It occurs at √  1 √  1 √ 1 √  1 1 2 a 2, − 2 b 2 and − 2 a 2, 2 b 2 ; the minimum is − 2 ab. 4. f (x, y) = xy 2 ,

g(x, y) = x2 + y 2 − 1

∇f = y 2 i + 2xyj,

∇g = 2xi + 2yj

∇f = λ∇g =⇒ y 2 = 2λx,

2xy = 2λy =⇒ y = 0

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

or y 2 = 2x2

f (±1, 0) = 0

√ 2 y = ±√ 3

1 y = 2x : From g(x, y) = 0 we get 3x = 1, =⇒ x = ± √ , 3 √ √ √ √ 2 The Minimum of xy 2 is: − 3 at (−1/ 3, ± 2 3) 9 2

2

2

5. Since f is continuous and the ellipse is closed and bounded, the maximum exists. g(x, y) = b2 x2 + a2 y 2 − a2 b2

f (x, y) = xy 2 , ∇f = y 2 i + 2xyj, ∇f = λ∇g

=⇒

∇g = 2b2 xi + 2a2 yj.

y 2 = 2λb2 x and

2xy = 2λa2 y.

Multiplying the first equation by a2 y and the second equation by b2 x, we get a2 y 3 = 2λa2 b2 xy = 2b2 x2 y. We can exclude y = 0; it clearly cannot produce the maximum. Thus, a2 y 2 = 2b2 x2 and, from g(x, y) = 0, 3b2 x2 = a2 b2 . √ √ √ √ This gives us x = ± 13 3 a and y = ± 13 6 b. The maximum occurs at x = 13 3 a, y = ± 13 6 b; the √ value there is 29 3 ab2 .

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SECTION 16.7

6. f (x, y) = x + y, ∇f = i + j,

g(x, y) = x4 + y 4 − 1

∇g = 4x3 i + 4y 3 j

∇f = λ∇g =⇒ 1 = 4λx3 ,

1 = 4λy 3 =⇒ 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 + y 2 represents the square of the distance from points on this curve to the origin, the maximum exists. g(x, y) = x4 + 7x2 y 2 + y 4 − 1     ∇g = 4x3 + 14xy 2 i + 4y 3 + 14x2 y j.

f (x, y) = x2 + y 2 , ∇f = 2xi + 2yj,

We use the cross-product equation (16.7.4) : 2x(4y 3 + 14x2 y) − 2y(4x3 + 14xy 2 ) = 0, 20x3 y − 20xy 3 = 0, xy(x2 − y 2 ) = 0. Thus, x = 0, y = 0, or x = ±y. From g(x, y) = 0 we conclude that the points to examine are √   1√ (0, ±1), (±1, 0), ± 3 3, ± 13 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 + y 2 + z 2 − 1

∇f = yzi + xzj + xyk,

∇g = 2xi + 2yj + 2zk

∇f = λ∇g =⇒ yz = 2λx,

xz = 2λy,

xy = 2λz =⇒ x2 = y 2 = z 2 or λ = 0.

λ = 0: In this case, at least two of x, y, z are 0 and f = 0. x2 = y 2 = z 2 From g(x, y, z) = 0 we get 3x2 = 1 =⇒ x = ± √13 , y = ± √13 , z = ± √13 √ √ √ √ √ √ 1√ 3 at (−1/ 3, −1/ 3, −1/ 3), (−1/ 3, 1/ 3, 1/ 3), 9 √ √ √ √ √ √ (1/ 3, −1/ 3, 1/ 3), (1/ 3, 1/ 3, −1/ 3).

The minimum of xyz is:

−

9. The maximum exists since xyz is continuous and the ellipsoid is closed and bounded. x2 y2 z2 + + −1 a2 b2 c2 2x 2y 2z ∇f = yzi + xzj + xyk, ∇g = 2 i + 2 j + 2 k. a b c 2x 2y 2z ∇f = λ∇g =⇒ yz = 2 λ, xz = 2 λ, xy = 2 λ. a b c

f (x, y, z) = xyz,

g(x, y, z) =

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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 xzb2 = 2λ = x y

so that a2 y 2 = b2 x2

or

x2 y2 = . a2 b2

Similarly from the second and third equations we get b2 z 2 = c2 y 2

y2 z2 = 2. 2 b c

or

3x2 a b From g(x, y, z) = 0, we get 2 = 1 =⇒ x ± √ , from which it follows that y = ± √ , a 3 3 √ c 1 z = ± √ . The maximum value is 9 3 abc. 3 g(x, y, z) = x2 + y 2 + z 2 − 7

10. f (x, y, z) = x + 2y + 4z, ∇f = i + 2j + 4k,

∇g = 2xi + 2yj + 2zk

∇f = λ∇g =⇒ 1 = 2λx, 2 = 2λy, 4 = 2λz =⇒ y = 2x, z = 4x 1 From g(x, y, z) = 0 we get 21x2 = 7 =⇒ x = ± √ 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. g(x, y, z) = x2 + y 2 + z 2 − 19

f (x, y, z) = 2x + 3y + 5z, ∇f = 2i + 3j + 5k, ∇f = λ∇g

∇g = 2xi + 2yj + 2zk.

=⇒

2 = 2λx, 3 = 2λy, 5 = 2λz.

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

1 , λ

y=

3 , 2λ

z=

5 , 2λ

and substitute these results in g(x, y, z) = 0 to obtain 1 9 25 + 2 + 2 − 19 = 0, 2 λ 4λ 4λ

38 − 19 = 0, 4λ2

λ=±

1√ 2. 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 = 52 2, and 2x + 3y + 5z = 19 2. g(x, y, z) = x + y + z − 1

12. f (x, y, z) = x4 + y 4 + z 4 , ∇f = 4x i + 4y j + 4z k, 3

3

3

∇g = i + j + k

∇f = λ∇g =⇒ 4x = λ, 4y 3 = λ, 4z 3 = λ =⇒ x = y = z 3

From g(x, y, z) = 0 we get 3x = 1, =⇒ x = 1 Minimum is: 27 13.

1 3

=y=z

x y z + + −1 a b c 1 1 1 ∇f = yzi + xzj + xyk, ∇g = i + j + k. a b c λ λ λ ∇f = λ∇g =⇒ yz = , xz = , xy = . a b c f (x, y, z) = xyz,

g(x, y, z) =

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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

x y z  3xyz = λ + + = λ. a b c Since x, y, z are non-zero, yz = Similarly, y =

λ 3xyz = , a a

1=

3x , a

x=

a . 3

b c 1 and z = . The maximum is abc. 3 3 27

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 ∇f = 2(x − p)i + 2(y − 4p)j,

subject to

g(x, y) = 2px − y 2 = 0

∇g = 2pi − 2yj

∇f = λ∇g =⇒ 2(x − p) = 2λp,

2(y − 4p) = −2λy =⇒ x =

8p3 = y 2 =⇒ y = 2p, x = 2p y  √ Distance to parabola is: f (x, y) = 5p

4p2 y

From g(x, y) = 0 we get

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

=⇒

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

2(x − 2) = 2xλ, 2(y − 1) = 2yλ, 2(z − 2) = 2zλ

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SECTION 16.7

839

Thus, x=

2 , 1−λ

y=

1 , 1−λ

2 . 1−λ

z=

Using the fact that x2 + y 2 + z 2 = 1, we have

At λ = −2, At λ = 4,

2 1−λ



2 +

1 1−λ

(x, y, z) = (2/3, 1/3, 2/3)



2 +

2 1−λ

2 =1

=⇒

λ = −2, 4

and f (2/3, 1/3, 2/3) = 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) = sin x sin y sin z,

g(x, y, z) = x + y + z − π

∇f = cos x sin y sin zi + sin x cos y sin zj + sin x sin y cos zk,

∇g = i + j + k

∇f = λ∇g =⇒ cos x sin y sin z = λ = sin x cos y sin z = sin z sin y cos z =⇒ cos x = cos y = cos z π =⇒ x = y = z = 3 √ 3 3 Maximum of sin x sin y sin z is: 8 g(x, y, z) = x2 + y 2 + z 2 − 14

f (x, y, z) = 3x − 2y + z,

19.

∇f = 3 i − 2 j + k, ∇f = λ∇g

∇g = 2x i + 2y j + 2z k.

=⇒

3 = 2xλ,

3 , 2λ

y=−

−2 = 2yλ,

1 = 2zλ.

Thus, x=

1 , λ

z=

1 . 2λ

Using the fact that x2 + y 2 + z 2 = 14, we have

At λ =

1 , 2

3 2λ

2



2 2 1 1 + − + = 14 λ 2λ

=⇒

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

1 At λ = − , (x, y, z) = (−3, 2, −1) and f (−3, 2, −1) = −14 2 Thus, the maximum value of f on the sphere is 14. 20. f (x, y, z) = xyz,

g(x, y, z) = x2 + y 2 + z − 4

∇f = yzi + xzj + xyk,

∇g = 2xi + 2yj + k

∇f = λ∇g =⇒ yz = 2λx, xz = 2λy, xy = λ =⇒ x2 = y 2 = From g(x, y, z) = 0 we get 4x2 = 4 =⇒ x = 1, y = 1, z = 2. Maximum volume is 2

z 2

1 λ=± . 2

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SECTION 16.7

21. It’s easier to work with the square of the distance; the minimum certainly exists. f (x, y, z) = x2 + y 2 + z 2 ,

g(x, y, z) = Ax + By + Cz + D

∇f = 2xi + 2yj + 2zk, ∇f = λ∇g

=⇒

∇g = Ai + Bj + Ck.

2x = Aλ, 2y = Bλ, 2z = Cλ.

Substituting these equations in g(x, y, z) = 0, we have  1  2 −2D . λ A + B 2 + C 2 + D = 0, λ = 2 2 A + B2 + C 2 Thus, in turn, −DA −DB −DC , y= 2 , z= 2 A2 + B 2 + C 2 A + B2 + C 2 A + B2 + C 2   −1/2 so the minimum value of x2 + y 2 + z 2 is |D| A2 + B 2 + C 2 . x=

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 = 12 ax + 12 by + 12 cz. The geometry suggests that x2 + y 2 + z 2 has a minimum.

g(x, y, z) = ax + by + cz − 2A

f (x, y, z) = x2 + y 2 + z 2 , ∇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 a2 λ b2 λ c2 λ + + − 2A = 0, 2 2 2

λ=

a2

4A + b2 + c2

and thus x=

a2

2aA , + b2 + c2

The minimum is 4A2 (a2 + b2 + c2 )−1 .

y=

a2

2bA , + b2 + c2

z=

a2

2cA . + b2 + c2

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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 12 x sin 12 y sin 12 z. For (b) maximize f (x, y, z) = 4R2 (sin2 12 x + sin2 12 y + sin2 12 z). 2π Each maximum occurs with x = y = z = . This gives an equilateral triangle. 3 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√ √ 2 |y − x|. (see Section 1.4) = 2 1+1 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, ∇f = i − j,

g(x, y) = x3 − y 3 − 1 ∇g = 3x2 i − 3y 2 j.

We use the cross-product equation (16.7.4):     1 −3y 2 − 3x2 (−1) = 0,

3x2 − 3y 2 = 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 1 1 V = rst [volume of pyramid = base × height] 6 3 a b c subject to the side condition + + = 1. The minimum occurs when all the intercepts are: r s t 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 − y 2 − z 2 , 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.

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SECTION 16.7 g(x, y) = x + y − k, (x, y ≥ 0, k a nonnegative constant)

29. (a) f (x, y) = (xy)1/2 , ∇f =

y 1/2 x1/2 i + j, 2x1/2 2y 1/2

∇g = i + j.

∇f = λ∇g

=⇒

y 1/2 x1/2 = λ = 2x1/2 2y 1/2

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

=⇒

x=y=

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 x+y = . 2 2

k k , where (xyz)1/3 = . 3 3 x+y+z k 1/3 (xyz) . ≤ = 3 3

30. (a) The maximum occurs when x = y = z = (b) If x + y + z = k,

then, by (a),

31. Simply extend the arguments used in Exercises 29 and 30. 32. T (x, y, z) = xy 2 z,

g(x, y, z) = x2 + y 2 + z 2 − 1

∇T = y 2 zi + 2xyzj + xy 2 k,

∇g = 2xi + 2yj + 2zk

∇T = λ∇g =⇒ y 2 z = 2λx,

2xyz = 2λy, xy 2 = 2λz =⇒ x2 =

y2 = z2 2

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

1 8

at ( 12 , ± √12 , 12 ),

(− 12 , ± √12 , − 12 )

Minimum − 18 at ( 12 , ± √12 , − 12 ),

(− 12 , ± √12 , 12 ) g(r, h) = πr2 h − V, (V constant)

S(r, h) = 2πr2 + 2πrh,

33.

∇S = (4πr + 2πh) i + 2πr j, ∇S = λ∇g

∇g = 2πrh i + πr2 j.

4πr + 2πh = 2πrhλ, 2πr = πr2 λ =⇒    4V 3 V 3 16π 2 Now πr h = V, =⇒ λ = =⇒ r = , h= 3 . V 2π π   4V 3 V To minimize the surface area, take r = , and h = 3 . 2π π 34. f (x, y, z) = xyz,

r=

g(x, y, z) = x + y + z − 18

∇f = yzi + xzj + xyk, ∇f = λ∇g

=⇒

=⇒

35. Same as Exercise 13.

∇g = i + j + k

yz = xz = xy

=⇒

x=y=z

=⇒

x=y=z=6

2 , λ

h=

4 . λ

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SECTION 16.7 36. f (x, y, z) = xyz 2 ,

g(x, y, z) = x + y + z − 30

∇f = yz 2 i + xz 2 j + 2xyzk, ∇f = λ∇g

=⇒

843

∇g = i + j + k

yz 2 = λ = xz 2 = 2xyz

=⇒

x=y=

z 2

=⇒

x=y=

15 , 2

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 = 2λ x(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.

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SECTION 16.7 Now put y = x in the third equation. This gives x2 = 4λ x

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  S 1 S 2 2 48λ = S =⇒ λ = =⇒ λ = 48 4 3   S 1 S To maximize the volume, take x = y = and z = . 3 2 3 39. S(r, h) = 4πr2 + 2πrh,

4 3 πr + πr2 h − 10, 000 3 ∇g = (4πr2 + 2πrh)i + πr2 j

g(r, h) =

∇S = (8πr + 2πh)i + 2πrj,

(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  Maximum volume for sphere of radius r = 3 7500/π meters.

40. (a)

=⇒

=⇒

h=0

g(x, y, l) = 2x + 2y + l − 108,

f (x, y, l) = xyl, ∇f = yl i + xl j + xy k, ∇f = λ∇g

∇g = 2 i + 2 j + k. =⇒

Now 2x + 2y + l = 108,

yl = 2λ,

=⇒

xl = 2λ,

xy = λ

=⇒

y = x and l = 2x.

x = 18 and l = 36.

To maximize the volume, take x = y = 18 in. and l = 36 in. g(r, l) = 2πr + l − 108,

f (r, l) = πr2 l,

(b)

∇g = 2π i + j.

∇f = 2πrl i + πr j, 2

∇f = λ∇g

2πrl = 2πλ, πr2 = λ, l = πr. 36 and l = 36. Now 2πr + l = 108, =⇒ r = π To maximize the volume, take r = 36/π in. and l = 36 in. 41. f (x, y, z) = 8xyz,

=⇒

g(x, y, z) = 4x2 + 9y 2 + 36z 2 − 36.

∇f (x, y, z) = 8yzi + 8xzj + 8xyk, ∇f = λ∇g

∇g(x, y, z) = 8xi + 18yj + 72zk.

gives yz = λx, 4

xyz = 4x2 , λ

4xz = 9λy, 4

xyz = 9y 2 , λ

xy = 9λz. 4

xyz = 36z 2 . λ

Also notice 4x2 + 9y 2 + 36z 2 − 36 = 0 We have 12

xyz = 36 λ

=⇒

x=

√

3,

2 y=√ , 3

1 z=√ . 3

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SECTION 16.7 Thus, V = 8xyz = 8 ·

√

2 1 16 3· √ · √ = √ . 3 3 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

g(x, y, z) = xyz − 96 = 0

subject to

∇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 = λ xz 20x + 20y = λ xy 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 = λ x2

=⇒

x(λ x − 40) = 0

=⇒

x=

40 λ

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 get 40 40 60 = 96 λ λ λ

=⇒

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 , f (x, y, z) = 2x + 8y + 24z,

y = Q2 ,

g(x, y, z) = x2 + 2y 2 + 4z 2 − 4, 500, 000, 000

∇f = 2i + 8j + 24k, ∇f = λ∇g

z = Q3 .

∇g = 2xi + 4yj + 8zk. =⇒

2 = 2λx,

8 = 4λy,

24 = 8λz.

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

1 , λ

y=

2 , λ

z=

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



4 1 9 + 2 + 4 − 45 × 108 = 0, λ2 λ2 λ2

3 , λ

45 = 45 × 108 , λ2

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 .

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SECTION 16.7

44. f (x, y, z) = 8xyz,

g(x, y, z) =

x2 y2 z2 + + − 1. We take a, b, c, x, y, z > 0 a2 b2 c2

∇f (x, y, z) = 8yzi + 8xzj + 8xyk;

∇g(x, y, z) =

2x 2y 2z i + 2 j + 2 k. 2 a b c

∇f = λ ∇g and the side condition yield the system of equations: 8yz =

2xλ a2

8xz =

2yλ b2

8xy =

2zλ c2

x2 y2 z2 + + =1 a2 b2 c2 Multiply the first equation by x, the second by y and subtract. This gives 0=

2λx2 2λy 2 − 2 2 a b

=⇒

y=

b x since λ = 0 a

=⇒

V = 0.

Multiply the second equation by y, the third by z and subtract. This gives 0=

2λy 2 2λz 2 − 2 2 b c

c c z = y = x. b a

=⇒

Substituting these results into the side condition, we get: 3x2 =1 a2

a x= √ 3

c and z = √ . 3 √





a b c 8 3 √ √ The volume of the largest rectangular box is: V = 8 √ abc. = 9 3 3 3 =⇒

=⇒

b y=√ 3

PROJECT 16.7 1. f (x, y, z) = xy + z 2 ,

g(x, y, z) = x2 + y 2 + z 2 − 4,

∇f = y i + x j + 2z k,

∇g = 2x i + 2y j + 2z k,

∇f = λ∇g + μ∇h

=⇒

2z = 2λz

=⇒

λ=0

y = 2λx − μ,

h(x, y, z) = y − x

∇h = −i + j.

x = 2λy − μ,

2z = 2λz

or z = 1.

λ=0

=⇒

y = −x which contradicts y = x.

z=1

=⇒

 x2 + y 2 = 3, which, with y = x implies x = ± 3/2;

   ± 3/2, ± 3/2

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SECTION 16.7 Adding the first two equations gives x + y = 2λ(x + y) x=y=0

z = ±2;

=⇒

(x + y)[2λ − 1] = 0

=⇒

=⇒

λ=

1 c

or x = y = 0.

(0, 0, ±2).

1 =⇒ z = 0 and y = x =⇒ 2

  5  f ± 3/2, ± 3/2, 1 = ; f (0, 0, ±2) = 4; 2

2x4 = 4;

λ=

√ x = ± 2;

√ √ (± 2, ± 2, 0).

√ √ f (± 2, ± 2, 0) = 2.

The maximum value of f is 4; the minimum value is 2.

2. D(x, y, z) = x2 + y 2 + z 2 , ∇D = 2xi + 2yj + 2zk, ∇D = λ∇g + μ∇h

g(x, y, z) = x + 2y + 3z, ∇g = i + 2j + 3k,

=⇒

∇h = 2i + 3j + k

2x = λ + 2μ,

2y = 2λ + 3μ,

Then g(x, y, z) = 0 and h(x, y, z) = 0 give x = Closest point

68 16 76 , ,− 75 15 75



3. f (x, y, z) = x2 + y 2 + z 2 , ∇f = 2x i + 2y j + 2z k, ∇f = λ∇g + μ∇h

h(x, y, z) = 2x + 3y + z − 4

68 , 75

y=

g(x, y, z) = x + y − z + 1, ∇g = i + j − k, =⇒

2z = 3λ + μ 16 , 15

z=−

=⇒

z = 5y − 7x

76 75

h(x, y, z) = x2 + y 2 − z 2

∇h = 2x i + 2y j − 2z k.

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. √ =⇒ z = ± 2 x.

=⇒

Therefore, we must have y = x √ Substituting y = x, z = 2 x into the equation x + y − z + 1 = 0, we get √ √ √ 2 2 x = −1 − =⇒ y = −1 − , z = −1 − 2 2 2 √ Substituting y = x, z = − 2 x into the equation x + y − z + 1 = 0, we get √ √ √ 2 2 x = −1 + =⇒ y = −1 + , z = −1 + 2 2 2

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SECTION 16.8 Since 

√ √ √ 2 2 −1 − , −1 − , −1 − 2 = 6 + 4 2 and 2 2

f

√



√ √ √ 2 2 −1 + , −1 + , −1 + 2 = 6 − 4 2, 2 2

f

√

 it follows that 

√ √ 2 2 −1 + , −1 + , −1 + 2 is closest to the origin and 2 2 √

√ √ 2 2 −1 − is furthest from the origin. , −1 − , −1 − 2 2 2 √

SECTION 16.8     1. df = 3x2 y − 2xy 2 Δx + x3 − 2x2 y Δy 2. df =

∂f ∂f ∂f Δx + Δy + Δz = (y + z)Δx + (x + z)Δy + (x + y)Δz ∂x ∂y ∂z

3. df = (cos y + y sin x) Δx − (x sin y + cos x) Δy 4. df = 2xye2z Δx + x2 e2z Δy + 2x2 ye2z Δz   5. df = Δx − (tan z) Δy − y sec2 z Δz 

   x−y x−y 6. df = + ln(x + y) Δx + − ln(x + y) Δy x+y x+y 7. df =

y(y 2 + z 2 − x2 ) x(x2 + z 2 − y 2 ) 2xyz Δx + Δz 2 2 Δy − 2 2 2 2 2 2 2 (x + y + z ) (x + y + z ) (x + y 2 + z 2 )2 

8. df =

   2x 2y xy 2 xy Δy + e (1 + xy) Δx + + x e x2 + y 2 x2 + y 2

9. df = [cos(x + y) + cos(x − y)] Δx + [cos(x + y) − cos(x − y)] Δy 10. df = ln

1+y 1−y

Δx +

2x Δy 1 − y2

  x 11. df = y 2 zexz + ln z Δx + 2yexz Δy + xy 2 exz + Δz z 12. df = y(1 − 2x2 )e−(x

2

+y 2 )

Δx + x(1 − 2y 2 )e−(x

2

+y 2 )

Δy

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SECTION 16.8     Δu = (x + Δx)2 − 3(x + Δx)(y + Δy) + 2(y + Δy)2 − x2 − 3xy + 2y 2     = (1.7)2 − 3(1.7)(−2.8) + 2(−2.8)2 − 22 − 3(2)(−3) + 2(−3)2

13.

= (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)   − x2 z − 2yz 2 + 3xyz     = (2.1)2 (2.8) − 2(1.3)(2.8)2 + 3(2.1)(1.3)(2.8) − (2)2 3 − 2(1)(3)2 + 3(2)(1)(3) = 2.896     du = (2xz + 3yz) Δx + −2z 2 + 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 =

y 3 + yz 2 x3 + xz 2 xyz 77 Δx + Δy − 2 Δz = 2 2 2 3/2 2 2 2 3/2 2 2 3/2 (x + y + z ) (x + y + z ) (x + y + z ) 4(14)3/2

17. f (x, y) = x1/2 y 1/4 ;

x = 121, y = 16, Δx = 4, Δy = 1

f (x + Δx, y + Δy) ∼ = f (x, y) + df = x1/2 y 1/4 + 12 x−1/2 y 1/4 Δx + 14 x1/2 y −3/4 Δy √ √ √ √ 4 4 125 17 ∼ = 121 16 + 12 (121)−1/2 (16)1/4 (4) + 14 (121)1/2 (16)−3/4 (1)   1 (2)(4) + 14 (11) 18 = 11(2) + 12 11 = 22 +

4 11

+

11 32

∼ = 22 249 352 = 22.71

√ √ 18. f (x, y) = (1 − x)(1 + y), x = 9, y = 25, Δx = 1, Δy = −1 √ √ 1+ y 1− x 4 df = − √ Δx + √ Δy = − 2 y 5 2 x 4 4 f (10, 24) ∼ = f (9, 25) − = −12 5 5 π π π f (x, y) = sin x cos y; x = π, y = , Δx = − , Δy = − 19. 4 7 20 df = cos x cos y Δx − sin x sin y Δy f (x + Δx, y + Δy) ∼ = f (x, y) + df 6 1 π π π π π  sin π cos π ∼ − − sin π sin − = sin π cos + cos π cos 7 5 4 4 7 4 20 √

 1√ π π 2∼ = 0+ 2 +0= = 0.32 2 7 14

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SECTION 16.8 √

1 π , Δx = −1, Δy = π 4 16 √ 1 1 3π df = √ tan yΔx + x sec2 yΔy = − + 6 8 2 x



 π 1 3π 17 3 ∼ 5 − + = + π = 4.01 f 8, π ∼ = f 9, 16 4 6 8 6 8

20. f (x, y) =

x tan y,

x = 9, y =

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 + x2 yexy Δx + x3 exy Δy = 2(3)e0 + (3)2 (0)e0 (−0.1) + 33 e0 (0.01) = − 0.33 Thus, f (2.9, .01) ∼ = 32 e0 − 0.33 = 8.67. 22. x = 2, y = 3, z = 3, Δx = 0.12, Δy = −0.08, Δz = 0.02 df = 2xy cos πzΔx + x2 cos πzΔy − πx2 y sin πzΔz = −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 xz xy π Δy + Δz = (−0.06) + (1.5)(0.1) + (1.5)(−0.08) ∼ = −0.0171 2 2 2 2 1+y z 1+y z 4 Thus, f (2.94, 1.1, 0.92) ∼ = 34 π − 0.0171 ∼ = 2.3391 df = tan−1 yz Δx +

24. x = 3, y = 4, Δx = 0.06, Δy = −0.12 x y 3 4 df =  Δx +  Δy = (0.06) + (−0.12) = −0.06 2 2 2 2 5 5 x +y x +y ∼ f (3.06, 3.88) = f (3, 4) − 0.06 = 4.94 ∂z 2x ∂z 2y Δx − Δy Δx + Δy = 2 ∂x ∂y (x + y) (x + y)2 With x = 4, y = 2, Δx = 0.1, Δy = 0.1, we get

25. df =

4 8 1 df = 36 (0.1) − 36 (0.1) = − 90 . 4.1 − 2.1 4 − 2 2 1 1 The exact change is − = − =− . 4.1 + 2.1 4 + 2 6.2 3 93

26. V (r, h) = πr2 h,

r = 8,

h = 12,

Δr = −0.3,

Δh = 0.2

dV = 2πrhΔr + πr Δh = 192π(−0.3) + 64π(0.2) = −44.8π 2

decreases by approximately 44.8π cubic inches. 27. S = 2πr2 + 2πrh;

r = 8, h = 12, Δr = −0.3, Δh = 0.2 dS =

∂S ∂S Δr + Δh = (4πr + 2πh) Δr + (2πr) Δh ∂r ∂h = 56π(−0.3) + 16π(0.2) = −13.6π.

The area decreases about 13.6π in.2 .

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SECTION 16.8 28. dT = 2x cos πzΔx − 2y sin πzΔy − (πx2 sin πz + πy 2 cos πz)Δz 2 4 = 4(0.1) − 4π(0.2) = − π 5 5 4 2 T decreases by about π − ∼ = 2.11 5 5 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. πr2 h , r = 7, h = 10, Δr = 0.2, Δh = 0.15 3 2πrh πr2 140 49 π df = Δr + Δh = π(0.2) + π(0.15) = (35.35) 3 3 3 3 3 π π ∼ ∼ f (7.2, 10.15) = f (7, 10) + (35.35) = 525.35 = 550.15 3 3

30. f (r, h) =

31. (a) dV = yz Δx + xz Δy + xy Δz = (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 =

2 16

2 Δf ∼ ) = 468 = df = yzΔx + xzΔy + xyΔz = 3774( 16

in.

The amount of paint is approximately 468 cubic inches. 35. (a) πr2 h = π(r + Δr)2 (h + Δh)

=⇒

Δh =

df = (2πrh) Δr + πr2 Δh,

r2 h (2r + Δr)h −h=− Δr. 2 (r + Δr) (r + Δr)2 df = 0

=⇒

Δh =

−2h Δr. r

(b) 2πr2 + 2πrh = 2π(r + Δr)2 + 2π(r + Δr)(h + Δh). Solving for Δh, r2 + rh − (r + Δr)2 2r + h + Δr −h=− Δr. r + Δr r + Δr

2r + h df = (4πr + 2πh) Δr + 2πr Δh, df = 0 =⇒ Δh = − Δr. r Δh =

851

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SECTION 16.9

36. The area is given by A =

1 2

x2 tan θ.

(a) The change in area is approximated by: dA = x tan θ Δx +

1 2

x2 sec2 θ Δθ = 3Δx +

25 2 Δθ.

(b) The actual change in area is 1 1 1 (x + Δx)2 tan(θ + Δθ) − x2 tan θ = [4 + Δx]2 tan[arctan (3/4) + Δθ] − 6. 2 2 2 (c) The area is more sensitive to a change in θ. 1 2 x2 x sin θ; ΔA ∼ cos θΔθ = dA = x sin θΔx + 2 2 (b) The area is more sensitive to changes in θ if x > 2 tan θ, otherwise it is more sensitive to changes

37. (a) A =

in x. 38. (a) dV ∼ = yzΔx + xzΔy + xyΔz,

x = 60 in, y = 36 in, z = 42 in

1 Maximum possible error = 6192( 12 ) = 516 cubic inches.

(b) dS ∼ = 2(y + z)Δx + 2(x + z)Δy + 2(x + y)Δz 1 Maximum possible error = 552( 12 ) = 46 square inches

39. s =

A ; A = 9, W = 5, ΔA = ±0.01, ΔW = ±0.02 A−W A ∂s ∂s −W ΔA + ΔW ds = ΔA + ΔW = 2 ∂A ∂W (A − W ) (A − W )2 =−

5 9 (±0.01) + (±0.02) ∼ = ±0.014 16 16

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 = xy 2 , ∂x Thus,

2. 3.

4.

∂f = x, ∂x

f (x, y) = 12 x2 y 2 + φ(y),

φ (y) = 0, φ(y) = C,

and

∂f = x2 y + φ (y) = x2 y. ∂y f (x, y) = 12 x2 y 2 + C.

∂f 1 = y =⇒ f (x, y) = (x2 + y 2 ) + C ∂y 2

∂f ∂f = y, f (x, y) = xy + φ(y), = x + φ (y) = x. ∂x ∂y Thus, φ (y) = 0, φ(y) = C, and f (x, y) = xy + C. ∂f x3 = x2 + y =⇒ f (x, y) = + xy + φ(y); ∂x 3

∂f 1 1 = x + φ (y) = y 3 + x =⇒ f (x, y) = x3 + y 4 + xy + C ∂y 3 4     ∂ ∂ 5. No; y 3 + x = 3y 2 whereas x2 + y = 2x. ∂y ∂x

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SECTION 16.9 6.

7.

8.

9.

∂f = y 2 ex − y =⇒ f (x, y) = y 2 ex − xy + φ(y); ∂x ∂f = 2yex − x + φ (y) = 2yex − x =⇒ f (x, y) = y 2 ex − xy + C ∂y ∂f ∂f = cos x − y sin x, f (x, y) = sin x + y cos x + φ(y), = cos x + φ (y) = cos x. ∂x ∂y Thus, φ (y) = 0, φ(y) = C, and f (x, y) = sin x + y cos x + C. ∂f = 1 + ey =⇒ f (x, y) = x + xey + φ(y); ∂x ∂f y3 = xey + φ (y) = xey + y 2 =⇒ f (x, y) = x + xey + +C ∂y 3 ∂f = ex cos y 2 , ∂x Thus,

10.

11.

12.

15.

17.

φ (y) = 0, φ(y) = C,

and

∂f = −2yex sin y 2 + φ (y) = −2yex sin y 2 . ∂y

f (x, y) = ex cos y 2 + C.

∂2f ∂2f = ex sin y = ; ∂x∂y ∂y∂x

not a gradient.

∂f ∂f = xex − e−y , f (x, y) = xyex + e−y + φ(x), = yex + xyex + φ (x) = yex (1 + x). ∂y ∂x Thus, φ (x) = 0, φ(x) = C, and f (x, y) = xyex + e−y + C. ∂f = ex + 2xy =⇒ f (x, y) = ex + x2 y + φ(y); ∂x =⇒ f (x, y) = ex + x2 y − cos y + C  ∂  xy xe + x2 = x2 exy ∂y

∂f = x2 + φ (y) = x2 + sin y ∂y

∂ (yexy − 2y) = y 2 exy ∂x

whereas

∂f = x sin x + 2y + 1 =⇒ f (x, y) = xy sin x + y 2 + y + φ(x) ∂y ∂f = y sin x + xy cos x + φ (x) = y sin x + xy cos x =⇒ f (x, y) = xy sin x + y 2 + y + C ∂x ∂f = 1 + y 2 + xy 2 , f (x, y) = x + xy 2 + ∂x Thus,

16.

f (x, y) = ex cos y 2 + φ(y),

∂2f = −ex sin y, ∂y∂x

13. No;

14.

853

φ (y) = y + 1,

φ(y) =

1 2

1 2

x2 y 2 + φ(y),

y2 + y + C

∂f = 2xy + x2 y + φ (y) = x2 y + y + 2xy + 1. ∂y

and f (x, y) = x + xy 2 +

∂f 1 = 2 ln 3y + =⇒ f (x, y) = 2x ln 3y + ln |x| + φ(y); ∂x x y3 +C f (x, y) = 2x ln 3y + ln |x| + 3 ∂f x , = 2 ∂x x + y2 Thus,

f (x, y) =

φ (y) = 0, φ(y) = C,



x2 y 2 +

1 2

y 2 + y + C.

2x ∂f 2x = + φ (y) = + y2 ∂y y y

∂f y y + φ (y) =  . = 2 2 2 ∂y x +y x + y2  f (x, y) = x2 + y 2 + C.

x2 + y 2 + φ(y), and

1 2

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SECTION 16.9 ∂f x2 = x tan y + sec2 x =⇒ f (x, y) = tan y + tan x + φ(y); ∂x 2 ∂f x2 π x2 x2 = sec2 y + φ (y) = sec2 y + πy; =⇒ f (x, y) = tan y + tan x + y 2 + C. ∂y 2 2 2 2 ∂f = x2 sin−1 y, ∂x Thus,

f (x, y) = 13 x3 sin−1 y + φ(y),

φ (y) = − ln y,

=⇒

∂f x3 x3 + φ (y) =  − ln y. =  ∂y 3 1 − y2 3 1 − y2

φ(y) = y − y ln y + C,

and

1 f (x, y) = x3 sin−1 y + y − y ln y + C. 3 20.

x x2 ∂f tan−1 y + =√ =⇒ f = sin−1 x tan−1 y + + φ(y); ∂x y 2y 1 − x2 ∂f x2 sin−1 x x2 x2 sin−1 x  − + φ (y) = − 2 + 1 =⇒ f (x, y) = sin−1 x tan−1 y + = + y + C. 2 2 2 ∂y 1+y 2y 1+y 2y 2y

21. (a) Yes

(b) Yes

22. (a) f (x, y) = (x − y)e−x

2

y

(c) No

+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 = f (x, y), ∂x Thus,

24.

∂f /∂x = 1, f (x, y)

ln f (x, y) = x + φ(y),

φ (y) = 1, φ(y) = y + K,

and

∂f /∂y = 0 + φ (y), f (x, y)

∂f = f (x, y). ∂y

f (x, y) = ex+y+K = Cex+y .

∂f = eg(x,y) gx (x, y) =⇒ f (x, y) = eg(x,y) + φ(y); ∂x ∂f = eg(x,y) gy (x, y) + φ (y) = eg(x,y) gy (x, y) =⇒ f (x, y) = eg(x,y) + C. ∂y

25. (a) P = 2x, Q = z, R = y;

∂P ∂Q =0= , ∂y ∂x

∂P ∂R =0= , ∂z ∂x

∂Q ∂R =1= ∂z ∂y

(b), (c), and (d) ∂f = 2x, f (x, y, z) = x2 + g(y, z). ∂x ∂f ∂g ∂f ∂g =0+ with = z =⇒ = z. ∂y ∂y ∂y ∂y Then, g(y, z) = yz + h(z)

=⇒

∂f = 0 + y + h (z) and ∂z Thus, h(z) = C 26.

and

f (x, y, z) = x2 + yz + h(z), ∂f =y ∂z

=⇒

h (z) = 0.

f (x, y, z) = x2 + yz + C.

∂f ∂f ∂g = yz =⇒ f (x, y, z) = xyz + g(y, z); = xz + = xz =⇒ f = xyz + h(z) ∂x ∂y ∂y ∂f = xy + h (z) = xy =⇒ f (x, y, z) = xyz + C ∂z

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SECTION 16.9 27. The function is a gradient by the test stated before Exercise 25. Take

P = 2x + y,

R = y − 2z.

Q = 2y + x + z, ∂P ∂Q =1= , ∂y ∂x

Then

∂P ∂R =0= , ∂z ∂x

∂Q ∂R =1= . ∂z ∂y

∇f = P i + Qj + Rk.

Next, we find f where

∂f = 2x + y ∂x

=⇒

∂f ∂g = x+ ∂y ∂y

with

f (x, y, z) = x2 + xy + g(y, z). ∂f = 2y + x + z ∂y

=⇒

∂g = 2y + z. ∂y

Then, g(y, z) = y 2 + yz + h(z), f (x, y, z) = x2 + xy + y 2 + yz + h(z). ∂f = y + h (z) = y − 2z ∂z Thus, 28.

h(z) = −z 2 + C

h (z) = −2z.

=⇒

f (x, y, z) = x2 + xy + y 2 + yz − z 2 + C.

and

∂f = 2x sin 2y cos z =⇒ f (x, y, z) = x2 sin 2y cos z + g(y, z); ∂x ∂f ∂g = 2x2 cos 2y cos z + = 2x2 cos 2y cos z =⇒ f (x, y, z) = x2 sin 2y cos z + h(z) ∂y ∂y ∂f = −x2 sin 2y sin z + h (z) = −x2 sin 2y sin z =⇒ f (x, y, z) = x2 sin 2y cos z + C ∂z

29. The function is a gradient by the test stated before Exercise 25. Take

P = y 2 z 3 + 1,

Q = 2xyz 3 + y,

∂P ∂Q = 2yz 3 = , ∂y ∂x Next, we find f where

R = 3xy 2 z 2 + 1. ∂P ∂R = 3y 2 z 2 = , ∂z ∂x

Then ∂Q ∂R = 6xyz 2 = . ∂z ∂y

∇f = P i + Qj + Rk. ∂f = y 2 z 3 + 1, ∂x

f (x, y, z) = xy 2 z 3 + x + g(y, z). ∂f ∂g = 2xyz 3 + ∂y ∂y

with

∂f = 2xyz 3 + y ∂y

=⇒

∂g = y. ∂y

Then, g(y, z) =

1 2

y 2 + h(z),

f (x, y, z) = xy 2 z 3 + x +

1 2

y 2 + h(z).

∂f = 3xy 2 z 2 + h (z) = 3xy 2 z 2 + 1 ∂z Thus,

h(z) = z + C

and

f (x, y, z) = xy 2 z 3 + x +

1 2

=⇒

y 2 + z + C.

h (z) = 1.

855

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December 7, 2006

REVIEW EXERCISES y ∂f xy = − ez =⇒ f (x, y, z) = − xez + g(y, z) ∂x z z ∂f x ∂g x xy = + = + 1 =⇒ f (x, y, z) = + y − xez + h(z) ∂y z ∂y z z ∂f xy xy xy = − 2 − xez + h (z) = −xez − 2 =⇒ f (x, y, z) = − xez + y + C ∂z z z z

31. F(r) = ∇

GmM r



⎧ ⎪ ⎪ ⎨∇

32.

h(r) =

⎪ ⎪ ⎩

k n+2 , r n+2

n = 2

∇ (k ln r) ,

n = −2.

REVIEW EXERCISES 1.

∇f (x, y) = (4x − 4y)i + (3y 2 − 4x)j

2.

∇f (x, y) =

y 3 − x2 y x3 − xy 2 i + j (x2 + y 2 )2 (x2 + y 2 )2

3. ∇f (x, y) = (yexy tan 2x + 2exy sec2 2x) i + xexy tan 2x j 4. ∇f =

1 (x i + y j + z k) x2 + y 2 + z 2

5. ∇f (x, y) = 2xe−yz sec z i, −zx2 e−yz sec zj − (x2 ye−yz sec z − x2 e−yz sec z tan z)k 6. ∇f (x, y) = ye−3z cos xy i + e−3z (x cos xy + sin y) j, −3e−3z (sin xy − cos y) k 7. ∇f (x, y) = (2x − 2y) i − 2x j,

∇f (1, −2) = 6 i − 2 j;

1 2 ua = √ i + √ j; 5 5

2 fu a (1, −2) = ∇f (1, −2) · ua = √ . 5 8. ∇f (x, y) = (e

xy

xy

2 xy

+ xye ) i + x e

fu a (2, 0) = ∇f (2, 0) · ua =

j,

∇f (2, 0) = i + 4j;

√ 1 + 2 3. 2

9. ∇f (x, y, z) = (y 2 + 6xz) i + (2xy + 2z) j + (2y + 3x2 ) k, ua =

√ 3 1 ua = i + j 2 2

∇f (1, −2, 3) = 22 i + 2 j − k;

1 16 2 2 i − j + k; fu a (1, −2, 3) = ∇f (1, −2, 3) · ua = . 3 3 3 3

10. ∇f (x, y, z) =

x2

2x 2y 2z i+ 2 j+ 2 k, 2 2 2 2 +y +z x +y +z x + y2 + z2

1 1 1 ua = √ i − √ j + √ k; 3 3 3

∇f (1, 2, 3) =

2 fu a (1, 2, 3) = ∇f (1, 2, 3) · ua = √ . 7 3

1 (i + 2 j + 3 k); 7

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REVIEW EXERCISES 11. ∇f (x, y) = (6x − 2y 2 ) i − 4xy j,

857

∇f (3, −2) = 10 i + 24 j;

a = (0, 0) − (3, −2) = (−3, 2) = −3 i + 2 j,

2 −3 ua = √ i + √ j; 13 13

18 fu a (3, −2) = ∇f (3, −2) · ua = √ . 13 12. ∇f (x, y, z) = (y 2 z − 3yz) i + (2xyz − 3xz) j + (xy 2 − 3xy) k, r (t) = i − π sin πt j + 2et−1 k,

a = r (1) = i + 2k,

∇f (1, −1, 2) = 8 i − 10 j + 4 k;

1 2 ua = √ i + √ k; 5 5

16 fu a (1, −1, 2) = ∇f (1, −1, 2) · ua = √ . 5 13. ∇f (x, y, z) = 

1 x2 + y 2 + z 2

(x i + y j + z k),

1 ∇f (3, −1, 4) = √ (3 i − j + 4 k); 26

1 ua = ± √ (4 i − 3 j + k); 26

a = ±(4 i − 3 j + k),

fu a (3, −1, 4) = ∇f (3, −1, 4) · ua = ±

14. ∇f (x, y) = 2e2x (cos y − sin y) i − e2x (sin y + cos y) j, ∇f √   maximum directional derivative: ∇f 12 , − 12 π  = e 5. 15. ∇f (x, y, z) = cos xyz(yz i + xz j + xy k),

1

1 2, −2π



√ √ π 3 π 3 6 i+ 4 √ 2 +3 − 39π 12

∇f ( 12 , 13 , π) =

minimum directional derivative: fu = −∇f ( 12 , 13 , π) =

19 . 26

= 2e i + e j;

√

3 12

j+

k;

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) = C1 e−2t ,

r(0) = (4, 3)

=⇒

C1 = 4,

y(t) = C2 e−6t .

C2 = 3.

Therefore the path of the particle is: r(t) = 4e−2t i + 3e−6t j, t ≥ 0,

or,

y=

00

result saddle loc. min.

f (6, 6) = −216 42. ∇f (x, y) = (3x2 − 12x) i + (2y + 1) j = 0 at (0, − 12 ), (4, − 12 ). fxx = 6x − 12, fxy = 0, fyy = 2.

f (4, − 12 ) = − 145 4

point

A

B

C

D

result

(0, − 12 ) (4, − 12 )

−12

0

2

−24

saddle

12

0

2

24

loc. min.

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REVIEW EXERCISES

861

43. ∇f (x, y) = (1 − 2xy + y 2 ) i + (−1 − x2 + 2xy) j = 0 at (1, 1), (−1, −1). fxx = −2y, fxy = −2x + 2y, fyy = 2x.

44. ∇f (x, y) = e−(x

2

+y 2 )/2

fxx = e−(x

2

point

A

B

C

D

result

(1, 1)

−2

0

2

−4

saddle

(−1, −1)

2

0

−2

−4

saddle

√  2  (y − x2 y 2 ) i + (2xy − xy 3 ) j = 0 at (±1, ± 2), (x, 0), x any real number. +y 2 )/2

(−3xy 2 + x3 y 2 ), fyy = e−(x

2

fxy = e−(x

2

+y 2 )/2

point A B √ (1, 2) −4e−3/2 0 √ (1, − 2) −4e−3/2 0 √ (−1, 2) 4e−3/2 0 √ (−1, − 2) 4e−3/2 0 √ √ local maxima: f (1, 2) = f (1, − 2) = 2e−3/2 ;

+y 2 )/2

(2y − y 3 − 2x2 y + x2 y 3 ),

(2x − 5xy 2 + xy 4 ). C

D

result

−4e−3/2

16e−3

loc. max

−4e−3/2

16e−3

loc. max

4e−3/2

16e−3

loc. min

4e−3/2

16e−3

loc. min

√ √ 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; Next we consider the boundary of D. C : r(t) = 2 cos t i + 2 sin t j,

f (1, −1) = 0

We parametrize the circle by:

t ∈ [ 0, 2π ]

The values of f on the boundary are given by the function F (t) = f (r(t)) = 6 − 4 cos t + 4 sin t, F  (t) = 4 sin t + 4 cos t :

F  (t) = 0

=⇒

t ∈ [ 0, 2π ]

sin t = − cos t

=⇒

t=

7 3 π, π 4 4

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

√ √  √   F (0) = F (2π) = f (2, 0) = 2; F 34 π = f − 2, 2 = 6 + 4 2;

√ √  √ 7  π = f 2, − 2 = 6 − 4 2. 4 √  √ √  f takes on its absolute maximum of 6 + 4 2 at − 2, 2 ; f takes on its absolute minimum of 0 at F

(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. Next we consider the boundary of D. We

y 3

parametrize each side of the triangle: C1 : r1 (t) = t i + 2t j,

t ∈ [ 0, 1 ]

C2 : r2 (t) = (1 − t) i + 2 j, C3 : r3 (t) = (2 − t) j,

2

t ∈ [ 0, 1 ]

t ∈ [ 0, 2 ]

1

Now, f1 (t) = f (r1 (t)) = 6t2 − 8t + 3, f2 (t) = f (r2 (t)) = 2t2 − 3+, f3 (t) = f (r3 (t)) = t2 − 1,

t ∈ [ 0, 1 ];

t ∈ [ 0, 1 ];

t ∈ [ 0, 2 ];

critical number: t =

1

2 3

2

3

x

critical number critical number

Evaluating these functions at the endpoints of their domains and at the critical numbers, we find that: 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. f takes on its absolute maximum of 3 at (0, 0) and its absolute minimum of −3 at (1, 2). 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=

2π 4π , . 3 3

Evaluating F at the endpoints of the interval and at the critical points, we get √ √ 3 3 F (0) = F (2π) = f (1, 0) = 4, F (2π/3) = f (−1/2, 3) = 4 + , 2 √ √ 4 3 3 F (4π/3) = f (−1/2, − 3) = 4 − >− 2 15 f takes on its absolute maximum of 2 at (0, 1); (−1/15, −8/15).

f takes on its absolute minimum of −4/15 at

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REVIEW EXERCISES 48. ∇f (x, y) = 4x3 i + 6y 2 j = 0 at (0, 0) in D;

863

f (0, 0) = 0.

On the boundary of D : x = cos t, y = sin t. Set 0 ≤ t ≤ 2π.

F (t) = f (cos t, sin t) = cos4 t + 2 sin3 t, 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 , ∇f = 2(x − 1) i + 2(y + 2) j + 2(z − 3) k,

g(x, y) = 3x + 2y − z − 5.

∇g = 3 i + 2 j − k.

Set ∇f = λ∇g : 2(x − 1) = 3λ

=⇒

x = 32 λ + 1,

2(y + 2) = 2λ

=⇒

y = λ − 2,

2(z − 3) = −λ

=⇒

z = − 12 λ + 3.

9 41 5 33 =⇒ x = , y=− , z= . 7 14 7 14 The point on the plane that is closest to (1, −2, 3) is (41/14, −5/7, 33/14). The distance from the 9 point to the plane is √ . 14

Substituting these values in 3x + 2y − z = 5 gives λ =

50. Set f (x, y, z) = 3x − 2y + z, ∇f = 3 i − 2 j + k,

g(x, y, z) = x2 + y 2 + z 2 − 14,

∇g = 2x i + 2y j + 2z k.

Set ∇f = λ∇g : 3 = 2λx

=⇒

x = 3/2λ,

−2 = 2λy

=⇒

y = −1/λ,

Substituting these values in x2 + y 2 + z 2 = 14 gives λ = ± 12

1 = 2λz

=⇒

z = 1/2λ.

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, ∇f = i + j − k,

g(x, y, z) = x2 + y 2 + 4z 2 − 4,

∇g = 2x i + 2y j + 8z k.

Set ∇f = λ∇g : 1 = 2λx

=⇒

x = 1/2λ,

1 = 2λy

=⇒

y = 1/2λ,

−1 = 8λz

=⇒

Substituting these values in x2 + y 2 + 4z 2 = 4 gives λ = ± 38

=⇒

or x = −4/3, y = −4/3, z = 1/3. Evaluating f :

f (− 43 , − 43 , 13 )

value of f is 3, the minimum value is −3.

f ( 43 , 43 , − 13 )

= 3,

z = −1/8λ.

x = 4/3, y = 4/3, z = −1/3 = −3. The maximum

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REVIEW EXERCISES

52. Let the length, width and height be x, y, z respectively. Then the total cost is f (x, y, z) =

1 2

xy +

1 2

xz +

1 2

yz +

1 10

xy =

3 5

xy +

1 2

xz +

1 2

yz

with the condition g(x, y, z) = xyz − 16 = 0. xyz = 16 =⇒ x = 0, y = 0 z = 0.       ∇f = 35 y + 12 z i + 35 x + 12 z j + 12 y + 12 x k,

Note first that

∇f = λ ∇g

=⇒

3 5

y+

1 2

z = λyz,

3 5

x+

1 2

∇g = yz i + xz j + xy k

z = λxz,

1 2

x+

1 2

y = λxy

Multiply the first equation by x, the second equation by y and subtract. This gives: 1 2 (xz

− yz) = 0

=⇒

z(x − y) = 0

=⇒

Substituting y = x in the third equation yields x = λx2 =⇒ 1 6 Substituting x = y = in the first equation yields z = . λ 5λ

x=

y=x 1 . λ

√ 3 3 Finally, substituting these values for x, y and z into the equation xyz = 16, we get λ = √ . 3 2 5 √ 3 2 5 10 ∼ 6 12 ∼ Therefore, x = y = √ = √ = 2.37 and z = x = √ = 2.85. 3 3 3 5 3 75 75

53. df = (9x2 − 10xy 2 + 2) dx + (−10x2 y − 1) dy 54. df = (2xy sec2 x2 − 2y 2 ) dx + (tan x2 − 4xy) dy 55. df =

y2 z + z2 y xz 2 + zx2 x2 y + y 2 x dx + dy + dz (x + y + z)2 (x + y + z)2 (x + y + z)2





z 2y x yz yz 56. df = − 2 dx + ze − 2 dy + ye − 2 dz y + xz y + xz y + xz 57. Set f (x, y, z) = ex

 y + z 3 . Then df = ex

 ex ex 3z 2 1   y + z 3 Δx + Δy + Δz. 2 y + z3 2 y + z3

With x = 0, y = 15, z = 1, Δx = 0.02, Δy = 0.2, Δz = 0.01, df = 4 Δx +  √ Therefore, e0.02 15.2 + (1.01)3 ∼ = e0 15 + 1 + 0.1088 = 4.1088.

1 8

Δy +

3 8

Δz ∼ = 0.1088.

58. Set f (x, y) = x1/3 cos2 y. Then df =

1 3

x−2/3 cos2 y Δx − 2x1/3 cos y sin y Δy.

√ 1 π , df = Δx − 2 3 Δy ∼ = 0.1287. 90 64 Therefore, (64.5)1/3 cos2 (28◦ ) ∼ = 641/3 cos2 (30◦ ) + 0.1287 = 3.1287.

With x = 64, y = 30◦ = π/6, Δx = 0.5, Δy = −2◦ = −

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REVIEW EXERCISES 59. V = πr2 h;

r = 5 ft.,

h = 22 ft.,

Δr = 0.01 in. =

1 1200

ft.,

Δh = 0.01 =

865

1 1200

dV = 2πrh Δr + πr2 Δh Using the values given above, dV = 2π(5)(22)

1 1 ∼ + π(25) = 0.6414 cu. ft. ∼ = 1108.35 cu. in.; 1200 1200

1108.35 ∼ = 4.80. 231

Approximately 4.80 gallons will be needed. 60.

∂P ∂Q = 12x2 y − 8x = ; the vector function is a gradient. ∂y ∂x ∂f = 6x2 y 2 − 8xy + 2x, ∂x

f (x, y) = 2x3 y 2 − 4x2 y + x2 + φ(y),

∂f = 4x3 y − 4x2 + φ (y) = 4x3 y − 4x2 − 8. ∂y Thus, 61.

φ (y) = −8, φ(y) = −8y + C,

f (x, y) = 2x3 y 2 − 4x2 y + x2 − 8y + C.

and

∂P ∂Q = 2x − sin x = ; the vector function is a gradient. ∂y ∂x ∂f = 2xy + 3 − y sin x, ∂x

f (x, y) = x2 y + 3x + y cos x + φ(y),

∂f = x2 + cos x + φ (y) = x2 + 2y + 1 + cos x. ∂y Thus, 62.

63.

φ (y) = 2y + 1, φ(y) = y 2 + y + C,

∂P = 2xy + 4y; ∂y

∂Q = −2xy + 2; ∂x

and

∂P ∂Q = ; ∂y ∂x

f (x, y) = x2 y + 3x + y cos x + y 2 + y + C. the vector function is not a gradient.

∂P ∂P ∂Q ∂Q ∂R ∂R = ey sin z = , = ey cos z = , = xey cos z = ; ∂y ∂x ∂z ∂x ∂z ∂y the vector function is a gradient.  f (x, y, z) = (ey sin z + 2x) dx = xey sin z + x2 + φ(y, z), ∂φ ∂φ 1 = xey sin z − y 2 =⇒ = −y 2 =⇒ φ = − y 3 + ψ(z), ∂y ∂y 3 1 f (x, y, z) = xey sin z + x2 − y 3 + ψ(z), fz = xey cos z + ψ  (z) = xey cos z =⇒ ψ  (x) = 0 =⇒ 3 ψ(x) = C

fy = xey sin z +

Therefore f (x, y, z) = xey sin z + x2 − 13 y 3 + C.

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