# CSM_Chapters13.pdf

September 14, 2017 | Author: Clau Amaiia | Category: Trigonometric Functions, Sine, Sphere, Mathematical Objects, Mathematical Analysis

#### Description

Chapter 13

Partial Derivatives 13.1

Functions of Several Variables

1. {(x, y)|(x, y) 6= (0, 0)} 2. {(x, y)|x 6= x ± 3y} 3. {(t, Y )|y 6= x2 } 4. {(x, y)|y ≥ −4} 5. {(s, t)|s, t any real numbers} S 6. {(u, b)|(u, v(6= (0, 0)} {(u, v)|u2 + v 2 6= 1} 7. {(r, s)| |s| ≥ 1} 8. {(θ, φ) | tan θ tan φ 6= 1} kπ, k an integer}

T

{(θ, φ)

θ 6= π/2 + kπ, k an integer}

T

{(θ, φ) | φ 6= π/2 +

9. (u, v, w)|u2 + v 2 + q 2 ≥ 16 10. {(x, y, z)|x2 + y 2 + < 25 and z 6= 5} √ √ 11. (c); The domain of f (x, y) = x+ y − x is {(x, y)|x ≥ 0, y−x ≥ 0} = {(x, y)|x ≥ 0, y ≥ x} √ 12. (e); The domain of f (x, y) = xy is {(x, y)|xy ≥ 0} = {(x, y)|x ≥ 0, y ≥ 0 or x ≤ 0, y ≤ 0}   13. (b); The domain of f (x, y) = ln(x − y 2 ) is (x, y)|x − y 2 > 0 = (x, y)|x > y 2 p x2 + y 2 − 1  14. (h); The domain of f (x, y) = is (x, y)|x2 + y 2 − 1 ≥ 0, y 6= x = y−x  (x, y)|x2 + y 2 ≥ 1, y 6= x n o n o q 15. (d); The domain of f (x, y) = xy − 1 is (x, y)| xy − 1 ≥ 0 = (x, y)| xy ≥ 1 77

78

CHAPTER 13. PARTIAL DERIVATIVES

16. (g); The domain of f (x, y) =

x4 + y 4 is {(x, y)|xy 6= 0} = {(x, y)|x 6= 0, y 6= 0} xy

17. (f ); The domain of f (x, y) = sin−1 (xy) is {(x, y)||xy| ≤ 1} p  18. (a); The domain of f (x, y) = y − x2 is {(x, y)|y − x2 ≥ 0} = ∗x, y)y ≥ x2 19. {(x, y)|x ≥ 0 and y ≥ 0}

y

x

20. {(x, y)| x2 ≤ 1 and y 2 ≥ 4}

\

{(x, y)| x2 ≥ 1 and y 2 ≤ 4} [ {(x, y)| |x| ≤ 1 and |y| ≥ 2} {(x, y)| |x| ≥ 1 and |y| ≤ 2}

y 2

1 x

21. {(x, y)|y − x ≥ 0} y

22. {(x, y)|xy ≥ −1} y

x

23. {z | z ≥ 10}

25. {w ||; −1 ≤ w ≤ 1} R4 27. f (2, 3) = 2 (2t − 1)dt = (t2 − t)|42 = 12 − 2 = 10 R1 f (−1, 1) = −1 (2t − 1)dt = (t2 − t)|1−1 = 0 − 2 = −2

x

24. all real numbers

26. {x | w < 7}

13.1. FUNCTIONS OF SEVERAL VARIABLES f (5, −5) = ln

28. f (3, 0) = ln 9/9 = ln 1 = 0; 29. f (−1, 1, −1) = (−2)2 = 4;

79

25 1 = ln = − ln 2 25 + 25 2

f (2, 3, −2) = 22 = 4

√ √ √ 30. f ( 3, 3, 6) = 1/3 + 1/2 + 1/6 = 1;

f (1/4, 1/5, 1/3) = 16 + 25 + 9 = 50

31. A plane through the origin perpendicular to the xz-plane 32. A parabolic cylinder perpendicular to the yz-plane 33. The upper half of a cone lying above the xy-plane with axis along the positive z-axis 34. The upper half of a hyperboloid of two sheets with axis lying along the positive z-axis 35. The upper half of an ellipsoid

36. A hemisphere lying below the yy-plane

37. y = − 21 x + C

38. x = y 2 − c y

y

x

x

39. x2 − y 2 = 1 + c2

40. 4x2 + 9y 2 = 36 − c2 , −6 ≤ c ≤ 6 y

y

x x

80

CHAPTER 13. PARTIAL DERIVATIVES

41. y = x2 + ln c, c > 0

42. y = x + tan c, −π/2 < c < π/2 y

y

x

x

43. x2 /9 + z 2 /4 = c; elliptical cylinder

44. Setting f (x, y, z) equal to a constant√c, we have (x − 1)2 + (y − 2)2 + (z − 2)2 = c which is the equation of a sphere of radius c centered at (1, 2, 3). Therefore, the level curves are concentric spheres centered at (1, 2, 3).

45. x2 + 3y 2 + 6z 2 = c; ellipsoid

46. 4y − 2z + 1 = c; plane

47.

c=0

c0 z

z

y x

y

x

y

48. Setting x = −4, y = 2, and z = −3 in x2 /16 + y 2 /4 + z 2 /9 = c we obtain c = 3. The equation 2 2 2 of the the x-intercepts are √ surface is x /16 + y /4 + z /9 =√3. Setting y = z = 0 we find √ ±4 3. Similarly, the y-intercepts are ±2 3 and the z-intercepts are ±3 3.

13.1. FUNCTIONS OF SEVERAL VARIABLES

81

49. P

v

50. From V = s2 h we obtain h = V /s2 . 51. C(r, h) = πr2 (1.8) + πr2 (1) + 2πh(2.3) = 2.8πr2 + 4.6πrh 250 − xy . Thus, x+y

52. Let the height of the box be h. Then 2xy + 2xh + 2yh = 500 and h = 250xy + x2 y 2 . x+y  2 53. V + πr2 g + 13 πr2 32 h = 11 9 πr h V = xyh =

54. From the figure, we see that t = x tan θ = x

z p

y2 − z2 xz

=p y2 − z2

! x θ t y θ

My2-z2

z

55. X = 2(156)(50) = 15, 600 sq cm √ √ 56. h(20, −6.67) + (10 20 − 20 + 10.5)(33 + 6.67) = (20 5 − 9.5)(39.67) ≈ 1397 kcal/m2 h 57. (a) The distance the water falls in time t is s(t) = 12 gt2 + vt where vis the velocity of the water at the top level (t = 0). The velocity of the water at time t is v(t) = gt + v. If t1 is the time it takes a cross-section of water to fall from the top level to the bottom level, then V = gt1 + v and t1 = (V − v)/g. The distance traveled in time t1 is  2   1 1 V −v V −v h = gt21 + vt1 = g +v 2 2 g g Simplifying the equation we obtain 2gh = V 2 − v 2 . Now the rates at the top and bottom levels are Z = vπr2 and Q = V πr2 (recall that the flow rate is constant). Solving for 2 2 2 2 2 2 v and V and substituting into 2gh √ = V − v we obtain 2gh = (Q/πr ) − (Q/πR ) . πr2 R2 2gh Solving for Q we find Q = √ . R4 − r 4 (b) When r = 0.2 cm, R = 1 cm, and h = 10, Q ≈ 7.61 cm3 /s.

82

CHAPTER 13. PARTIAL DERIVATIVES

13.2 1.

Limits and Continuity (x2 + y 2 ) = 25 + 1 = 26

lim

(x,y)→(5,−1)

2.

x2 − y 4−1 = =3 2−1 (x,y)→(2,1) x − y lim

5x2 + y 2 5x2 = lim = 5. (x,y)→(0,0) x2 + y 2 (x,y)→(0,0) x2 y2 5x2 + y 2 = lim = 1. The limit does not exist. On x = 0, lim 2 2 (x,y)→(0,0) y 2 (x,y)→(0,0) x + y

3. On y = 0,

lim

4.

4x2 + y 2 4+4 1 = = 4 4 16 + 16 4 (x,y)→(1,2) 16x + y

5.

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

lim

lim

−y 2x2 − y = lim = ∞. (x,y→(0,0) 2y 2 (x,y→(0,0) x2 + 2y 2 2x2 − y 2x2 On y = 0, lim = lim = 2.. The limit does not exist. (x,y→(0,0) x2 + 2y 2 (x,y→(0,0) x2

6. On x = 0,

lim

x2 y x3 x = lim = lim = 0. 2 4 +y (x,y→(0,0) x + x2 (x,y→(0,0) x2 + 1 1 x4 x2 y = lim = . The limit does not exist. On y = x2 , lim 4 2 4 4 2 (x,y→(0,0) x + x (x,y→(0,0) x + y

7. On y = x,

lim

(x,y→(0,0) x4

6xy 2 6x3 6x = lim = lim = 0. (x,y→(0,0) x2 + y 4 (x,y→(0,0) x2 + x4 (x,y→(0,0) 1 + x2 2 4 x y 6y On x = y 2 , lim = lim = 3. The limit does not exist. 4 2 4 (x,y→(0,0) x + y (x,y→(0,0) y + y 4

8. On y = x,

9.

lim

lim

x3 y 2 (x + y)3 = 1(4)(27) = 108

(x,y)→(1,2)

10.

lim (x,y)→(2,3)

11.

6 xy 6 = =− x2 − y 2 4−9 5

exy 1 = =1 1 (x,y)→(0,0) x + y + 1 lim

sin xy sin mx2 = lim 2 2 (x,y)→(0,0) x + y (x,y)→(0,0) (1 + m2 )x2 m sin mx2 m = lim = . 1 + m2 (x,y)→(0,0) 1 + m2 mx2 The limit does not exist.

12. On y = mx,

lim

13.2. LIMITS AND CONTINUITY 13.

lim (x,y)→(2,2)

14.

83

xy 4 1 = = x3 + y 2 8+4 3 √ cos(3x + y) = cos(3π + π/4) = cos 13π/4 = − 2/2

lim (x,y)→(π,π/4)

15.

x2 − 3y + 1 1 =− 3 (x,y)→(0,0) x + 5y − 3 lim

16. On y = mx,

x2 m2 x2 m2 x2 y 2 = lim = . 4 4 4 4 + 5y 1 + 5m4 (x,y)→(0,0) x + 5m x

lim

(x,y)→(0,0) x4

The limit does not exist. 17.

lim

xy 2

(x,y)→(4,3)

18.

19.

x + 2y 4+6 = 4(9) = 360 x−y 4−3

0 x2 y = =0 1+0 (x,y)→(1,0) x+ y 3 lim

xy − x − y + 1 (x − 1)(y − 1) = lim 2 + y − 2x − 2y + 2 (x,y)→(1,1) (x − 1)2 + m2 (x − 1)2 On y − x = m(x − 1), (x − 1)(y − 1) (x − 1)m(x − 1) m lim = lim = . 2 2 2 2 2 2 1 + m2 (x,y)→(1,1) (x − 1) + m (x − 1) (x,y)→(1,1) (x − 1) + m (x − 1) The limit does not exist. lim

(x,y)→(1,1) x2

20. On x = 0,

lim (x,y)→(0,3)

21.

−3y xy − 3y = lim . The limit does not exist. x2 + y 2 − 6y + 9 (x,y)→(0,3) (y − 3)2

x3 y + xy 3 − 3x2 − 3y 2 xy(x2 + y 2 ) − 3(x2 + y 2 = lim x2 + y 2 x2 + y 2 (x,y)→(0,0) (x,y)→(0,0) lim

=

lim

(xy − 3) = −3

(x,y)→(0,0)

22. 23.

4 y 3 + 2x3 8 − 16 = = −2 − 40 21 (x,y)→(−2,2) x + 5xy 2 lim

lim

ln(2x2 − y 2 ) = ln(2 − 1) = 0

(x,y)→(1,1)

24.

sin−1 (x/y) sin−1 (1/2) π/6 1 = = = −1 (x − y cos−1 (−1) π 6 (x,y)→(1,2) cos In Problems 25-30 let x = r cos θ and y = r sin θ. Then x2 + y 2 = r2 and (x, y) → (0, 0) if lim

and only if r → 0. We also use the facts that | cos θ| ≤ 1 and | sin θ| ≤ 1 for all θ. 25.

(x2 − y 2 )2 (r2 cos2 θ − r2 sin2 θ)2 r4 (cos2 θ − sin2 θ)2 = lim = lim r→0 r→0 r2 r2 (x,y)→(0,0) x2 + y 2 lim

= lim r2 cos2 2θ = 0 r→0

84 26.

CHAPTER 13. PARTIAL DERIVATIVES sin 3r2 sin(3x2 + 3y 2 ) = lim Use L’Hˆopital’s Rule r→0 x2 + y 2 r2 (x,y)→(0,0) 6r cos 3r2 = lim = lim 3 cos 3r2 = 3 r→0 r→0 2r lim

27.

6r2 cos θ sin θ 6xy p √ = lim 3|r| sin 2θ = 0 = lim r→0 r→0 (x,y)→(0,0) r2 x2 + y 2

28.

x2 − y 2 r2 cos2 θ − r2 sin2 θ p √ = lim |r| cos 2θ = 0 = lim r→0 r→0 (x,y)→(0,0) r2 x2 + y 2

29.

x3 r3 cos3 θ = lim = lim r cos3 θ = 0 r→0 r→0 r2 (x,y)→(0,0) x2 + y 2

30.

x3 + y 3 r3 cos3 θ + r3 sin3 θ = lim = lim r(cos3 θ + sin3 θ) = 0 2 2 r→0 r→0 r2 (x,y)→(0,0) x + y

lim

lim

lim

lim

31. {(x, y) | x ≥ 0 and y ≥ −x} 32. {(x, y) | x 6= 0 and y 6= 0} 33. {(x, y) | y 6= 0 and x/y 6= π/2 + kπ, k and integer} 34. {(x, y) | x and y are real} 35. (a) For x2 + y 2 < 1, f (x, y) = 0 is continuous (b) For x ≥ 0, f (x, y) is not continuous since it is discontinuous at (2, 0). (c) For y > x, f (x, y) is not continuous since it is discontinuous at (2, 3). 36. (a) For y ≥ 3, f (x, y) is not continuous since it is not defined at (0, 3). (b) For |x| + |y| < 1, f (x.y) is discontinuous since it is not defined at (0, 0). (c) For (x − 2)2 + y 2 < 1, f (x, y) is discontinuous since it is not defined at (2, 0). 37. Since lim (x,y)→(0,0)

f (x, y) =

6x2 y 3 6r5 cos2 θ sin3 θ = lim = lim 6r cos2 θ sin3 θ = 0 = f (0, 0) r→0 r→0 r4 (x,y)→(0,0) (x2 + y 2 )2 lim

the function is continuous at (0, 0). 38. Since f (x, 0) = 0 for all x and f (0, y) = 0 for all y, f (x, 0) and f (0, y) are continuous at x = 0 and y = 0, respectively. On y = x, lim

f (x, y) =

(x,y)→0,0)

so f (x, y) is not continuous at (0, 0).

x2 1 = , 4 (x,y)→(0,0) 2x2 + 2x2 lim

13.3. PARTIAL DERIVATIVES

85

39. Choose  > 0. Using x = r cos θ and y = r sin θ we have 3t cos θr2 sin2 θ 3 3xy 2 = r cos θ sin2 θ. 2x2 + 2y 2 2r2 2 p Let δ = 2 x2 + y 2 < δ, we have 3 . Now, whenever r =   3 3xy 2 3 3 3 2 2 | = |r cos θ sin θ| ≤ |r| < δ = | 2 = . 2x + 2y 2 2 2 2 2 3 Thus

lim

3xy 2 = 0. + y2

(x,y)→(0,0) x2

40. Choose  > 0. Using x = r cos θ and y = r sin θ we have x2 y 2 r2 cos2 θr2 sin2 θ = = r2 cos2 θ sin2 θ. x2 + y 2 r2 p √ √ x2 y 2 2 2 = r2 cos2 θ sin2 θ ≤ r2 ≤ . Thus, Now, whenever r = x + y <  (for δ = ), 2 x + y2 x2 y 2 lim = 0. 2 (x,y)→(0,0) x + y 2 41. Where y 6= x, we have f (x, y) =

x3 − y 3 (x − y)(x2 + xy + y 2 ) = = x2 + xy + y 2 . x−y x−y

When y = x, we have x2 + xy + y 2 = x2 + x2 + x2 = 3x2 = f (x, y). Therefore, f (x, y) = x2 + xy + y 2 throughout the entire plane. Since x2 + xy + y 2 is a polynomial, f must be continuous throughout the plane and thus has no discontinuities. p 42. Choose  > 0. Then for δ = , whenever 0 < (x − a)2 + (y − b)2 < δ, we have p |f (x, y) − b| = |y − b| ≤ (x − a)2 + (y − b)2 < δ = . Thus,

lim

y = b.

(x,y)→(a,b)

13.3 1.

Partial Derivatives

∂z 7(x = 4x) + 8y 2 − 7x − 8y 2 74x = lim = lim =7 4→0 4x ∂x 4→0 4x ∂z 7x + 8(y + 4y)2 − 7x − 8y 2 16y 4 y + 8(4y)2 = lim = lim 4y→0 ∂y 4y→0 4y 4y = lim (16y + 8 4 y) = 16y 4y→0

86

CHAPTER 13. PARTIAL DERIVATIVES 2.

3.

∂z (x + 4x)y − xy y4x = lim = lim = y; 4x→0 4x→0 ∂x 4x 4x ∂z x(y + 4y) − xy x4y = lim = lim =x 4y→0 4y→0 ∂y 4y 4y 3(x + 4x)2 y + 4x + 4x)y 2 − 3x2 y − 4xy 2 ∂z = lim ∂x 4x→0 4x 3x2 y + 6x(4x)y + 3(4x)2 y + 4xy 2 + 4(4x)y 2 − 3x2 y − 4xy 2 = lim 4x→0 4x 6x(4x)y + 3(4x)2 y + 4(4x)y 2 = lim = lim (6xy + 3(4x)y + 4y 2 ) = 6xy + 4y 2 4x→0 4x→0 4x 3x2 (y + 4y) + 4x(y + 4y)2 − 3x2 y − 4xy 2 ∂z = lim ∂y 4y→0 4y 3x2 y + 3x2 4 y + 4xy 2 + 8xy 4 y + 4x(4y)2 − 3x2 y − 4xy 2 = lim 4y→0 4y 3x2 4 y + 8xy 4 y + 4x(4y)2 = lim = lim (3x2 + 8xy + 4x 4 y) = 3x2 + 8xy 4y→0 4y→0 4y

x+4 x − ∂z x2 + x 4 x + xy + (4x)y − x2 − x 4 x − xy x+4+y x+y 4. = lim = lim 4x→0 ∂x 4x→0 4x (x + 4x + y)(x + y) 4 x y (4x)y = = lim 4x→0 (x + 4x + y)(x + y) 4 x (x + y)2 x x − x2 + xy − x2 − xy − x 4 y ∂z x + y + 4y x + y = lim = lim 4y→0 (x + y + 4y)(x + y) 4 y ∂y 4y→0 4y −x 4 y x = lim =− 4y→0 (x + y + 4y)(x + y) 4 y (x + y)2 5. zx = 2x − y 2 ; zy = −2xy + 20y 4 6. zx = −3x2 + 12xy 3 ; zy = 18x2 y 2 + 10y 7. zx = 20x3 y 3 − 2xy 6 + 30x4 ; zy = 15x4 y 2 − 6x2 y 5 − 4 8. zx = 3x2 y 2 sec2 (x3 y 2 ); zy = 2x3 sec2 (x3 y 2 ) √ 2 24y x 9. zx = √ ; z = − y (3y 2 + 1)2 x(3y 2 + 1) 10. zx = 12x2 − 10x + 8; zy = 0 11. zx = −(x3 − y 2 )−2 (3x2 ) = −3x2 (x3 − y 2 )−2 ; zy = −(x3 − y 2 )−2 (−2y) = 2y(x3 − y 2 )−2 12. zx = 6(−x4 + 7y 2 + 3y)5 (−4x) = −24x( − x4 + 7y 2 − 3y)5 ; zy = 6(−x4 + 7y 2 + 3y)5 (14y + 3) 13. zx = 2(cos 5x)(− sin 5x)(5) = −10 sin 5x cos 5x; zy = 2(sin 5y)(cos 5y)(5) = 10 sin 5y cos 5y

13.3. PARTIAL DERIVATIVES 14. zx = (2x tan−1 y 2 )ex 3

2

tan−1 y 2

87 ; zy =

3

2x2 y x2 tan−1 y2 e 1 + y4

3

15. fx = x(3x2 yex y + ex y ; fy = x4 ex y       θ θ 1 θ θ θ θ 2 2 16. fθ = φ cos ; fφ = φ cos − 2 + 2φ sin = −θ cos + 2φ sin φ φ φ φ φ φ φ 7y (x + 2y)3 − (3x − y) = ; (x + 2y)2 (x + 2y)2   (x2 − y 2 )2 y − xy 2(x2 − y 2 )2x 18. fx = = (x2 −y 2 )4  (x2 − y 2 )x − xy 2(x2 − y 2 )(−2y) fy = (x2 − y 2 )4

17. fx =

fy =

(x + 2y)(−1) − (3x − y)(2) −7x = (x + 2y)2 (x + 2y)2

−3x2 y − y 3 ; (x2 − y 2 )3 3xy 2 + x3 = 2 (x − y 2 )3

8u 15v 2 ; gv = 2 3 − 5v 4u + 5v 3 √ √ 1 s r 1 20. hr = √ + 2 ; hx = − 2 − √ r s 2s r 2s r     y  y2 y √ √ y 1 y/z 21. wx = √ ; wy = 2 x − y e =2 x− + 1 ey/z ; wz = −yey/z − 2 = 2 ey/z z z z z x   1 xy 22. wx = xy + (ln xz)y = y + y ln xz; wy = x ln xz; wz = x z 19. gu =

4u2

23. Fu = 2uw2 − v 3 − vwt2 sin(ut2 ); Fv = −3uv 2 + w cos(ut2 ); Fx = 3(2x2 t)3 (4xt) = 16xt(2x2 t)3 = 128x7 t4 ; Ft = −2uvwt sin(ut2 ) + 64x8 t3 4 5

24. Gp = 2pq 3 e2r s 4 5 Gq = 3p2 q 2 e2r s 4 5 4 5 Gr = p2 q 3 (8r3 s5 )e2r s = 8p2 q 3 r3 s5 e2r s 4 5 4 5 Gs = p2 q 3 (10r4 s4 )e2r s = 10p1 q 3 r4 s4 e2r s 25. zy = 16x3 y 3 , zy (1, −1) = −16 26. zx = 12x2 y 4 , zx (1, −1) = 12 18x2 (x + y)18x − 18xy = , fy (−1, 4) = 2. An equation of the tangent line is given (x + y)2 (x + y)2 by x = −1 and z + 24 = 2(y − 4). Parametric equations of the line are x = −1, y = 4 + t, z = −24 + 2t.

27. fy =

(x + y)18y − 18xy 18y 2 = , fx (−1, 4) = 32. An equation of the tangent line is given (x + y)2 (x + y)2 z + 24 by y = 4 and z + 24 = 32(x + 1). Symmetric equations of the line are x + 1 = , y = 4. 32

28. fx =

88

CHAPTER 13. PARTIAL DERIVATIVES

−x 29. zx = p , zx (2, 2) = −2 9 − x− y 2 √ √ √ 3 , zy ( 2, 3) = − 30. zy = p 2 2 2 9−x −y −y

31.

∂z ∂2z = y 2 exy = yexy ; ∂x ∂x2

32.

∂3z ∂z ∂2z 4 −4 = 6x y ; = −24x4 y −5 = −2x4 y −3 ; ∂y ∂y 2 ∂y 3

33. fx = 10xy 2 − 2y 3 ; fxy = 20xy − 6y 2 34. f (p, q) = ln(p + q) − 2 ln q, fq =

2 1 1 − , fqp = − p+q q (p + q)2

35. wt = 3u2 v 3 t2 , wtu 6uv 3 t2 ; wtuv 18uv 2 t2 36. wv = −

u2 sin(u2 v) u4 cos(u2 x) 3u4 cos(u2 v) ; wvv − ; wvvt = 3 3 t t t4 2

2

2

2

2

37. Fr = 2rer cos θ; Frθ − 2rer sin θ; Frθr − 2r(2rer ) sin θ − 2er sin θ = −2er (2r2 + 1) sin θ 2s 4s (s − t) − (s + t)(−1) = ; Htt = ; 2 2 (s − t) (s − t) (s − t)3 4 2 −8s − 4t (s − t) − 4x(3)(s − t) = = 6 (x − t) (s − t)4

38. Ht = Htts 39. 40.

∂2z ∂z ∂2z ∂z = −5x4 y 2 +8xy; = −60x3 y 2 +8y; = 6x5 −20x3 y 3 +4y 2 ; = −60x3 y 2 +8y ∂y ∂x∂y ∂x ∂y∂x ∂z 2x (1 + 4x2 y 2 )2 − 2x(8xy 2 ) 2y ∂z 2 − 8x2 y 2 ∂z = = = ; = ; 2 2 2 2 2 2 2 2 ∂y 1 + 4x y ∂x∂y (1 + 4x y ) (1 + 4x y ) ∂x 1 + 4x2 y 2 2 2 2 2 2 (1 + 4x y )2 − 2y(8x y) 2 − 8x y ∂z = = ∂y∂x (1 + 4x2 y 2 )2 (1 + 4x2 y 2 )2

41. wu = 3u2 v 4 − 8uv 2 t3 , wuv = 12u2 v 3 − 16uvt3 , wuvt = −48uvt2 ; wt = −12u2 v 2 t2 + v 2 , wtv = −24u2 vt2 + 2v, wtvu = −48uvt2 ; wv = 4u3 v 3 − 8u2 vt3 + 2vt, wvu = 12u2 v 3 − 16uvt3 , wvut = −48uvt2 42. Fη = 6η 2 (η 3 +ξ 2 +τ ) = 6η 5 +6η 2 ξ 2 +6η 2 τ, Fηξ = 12η 2 ξ, Fηξη = 24ηξ; Fξ = 4ξ(η 3 +ξ 3 +τ ) = 4η 3 ξ + 4xi3 + 4xiτ, Fξη = 12η 2 ξ, Fξηη 24ηξ; Fηη = 30η 4 + 12ηξ 2 , Fηηξ = 24ηξ 43. 2x + 2zzx = 0, zx = −x/z; 2y + 2zzy = 0, zy = −y/z 2x ; 2z − y 2 2yz 2zzy = y 2 zy + 2yz =⇒ (2z − y 2 )zy = 2yz =⇒ zy = 2z − y 2

44. 2zzx = 2x + y 2 zx =⇒ (2z − y 2 )zx = 2x =⇒ zx =

13.3. PARTIAL DERIVATIVES

89

vz − 2uv 3 ; 2z − uv 2 2 uz − 3u v 2zzv + 3u2 v 2 − uvzv − uz = 0 =⇒ (2z − uv)zv = uz − 3u2 v 2 =⇒ zv = 2z − uv

45. 2zzu + 2uv 3 − uvzu − vz = 0 =⇒ (2z − uv)zu = vz − 2uv 3 =⇒ zu =

46. sez zs + ez − test + 12s2 t = zs =⇒ (sez − 1)zs = tes − ez − 12s2 t =⇒ zs = sez zt − sest + 4s3 = zt =⇒ (sez − 1)zt = sest − 4s3 =⇒ zt =

sest − 4s3 sez − 1

test − ez − 12s2 t ; sez − 1

47. ax = y sin θ, Ay = x sin θ, Aθ = xy cos θ 48. Vh = (π/3)(r2 + rR + R2 ), Vr = (π/3)h(2r + R), VR = (π/3)h(r + 2R) 49.

50.

51.

52.

53.

∂2u ∂u = −4π 2 (cosh 2πy + sinh 2πy) sin 2πx; = 2π(cosh 2πy + sinh 2πy) cos 2πx; ∂x ∂x2 ∂u ∂2u = (2π sinh 2πy + 2π cosh 2πy) sin 2πx; = (4π 2 cosh 2πy + 4π 2 sinh 2πy) sin 2πx; ∂y ∂y 2 ∂2u ∂2u + 2 = −4π 2 (cosh 2πy + sinh 2πy) sin 2πx + 4π 2 (cosh 2πy + sinh 2πy) sin 2πx = 0 ∂x2 ∂y  nπ  ∂ 2 u  nπ  ∂u nπ n2 π 2 −(nπx/L) = − e−(nπx/L sin = e sin ; y; ∂x L L ∂x2 L2 L    2 2 2 nπ −(nπx/L) nπ ∂ u n π nπ ∂u = e cos y; = − 2 e−(nπx/L) sin y; ∂y L L ∂y 2 L L  nπ  n2 π 2  nπ  n2 π 2 −(nπx/L) ∂2u ∂2u + 2 = e sin − e−(nπx/L) sin =0 2 2 2 ∂x ∂y L L L L ∂z 2x ∂2z (x2 + y 2 )2 − 2x(2x) 2y 2 − 2x2 ∂z 2y = 2 , = = ; = 2 , ∂x x + y 2 ∂x2 (x2 + y 2 )2 (x2 + y 2 ) ∂y x + y2 2 2 2 2 2 2 2 2 2 2 ∂ z (x + y )2 − 2y(2y) 2x − 2y ∂ z ∂ z 2y − 2x + 2x − 2y 2 = = ; + = =0 ∂y 2 (x2 + y 2 )2 (x2 + y 2 )2 ∂x2 ∂y 2 (x2 + y 2 )2 ∂z 2 2 2 2 2 2 = 2yex −y sin 2xy + 2xex −y cos 2xy = 2ex −y (x cos 2xy − y sin 2xy), ∂x ∂2z 2 2 2 2 = 2ex −y (−2xy sin 2xy + cos 2xy − 2y 2 cos 2xy) + 4xex −y (x cos 2xy − y sin 2xy); ∂x2 ∂z − 2 − 2 − 2 = −2xex y sin 2xy − 2yex y cos 2xy = −2ex y (x sin 2xy + y cos 2xy), ∂y ∂2z − 2 − 2 = −2ex y (2x2 cos 2xy − 2xy sin 2xy + cos 2xy) + 4yex y (x sin 2xy + y cos 2xy); ∂y 2 − 2 ∂2z ∂2z + 2 =2ex y (−2xy sin 2xy + cos 2xy − 2y 2 cos 2xy + 2x2 cos 2xy − 2xy sin 2xy 2 ∂x ∂y − 2x2 cos 2xy + 2xy sin 2xy − cos 2xy + 2xy sin 2xy − 2y 2 cos 2xy) = 0 ∂u x =− 2 ; ∂x (x + y 2 + z 2 )3/2 ∂2u 2x2 − y 2 − z 2 = ; ∂x2 (x2 + y 2 + z 2 )5/2

∂u y ∂u z =− 2 ; =− 2 ; ∂y (x + y 2 + z 2 )3/2 ∂z (x + y 2 + z 2 )3/2 ∂2u −x2 + 2y 2 − z 2 ∂2u −x2 − y 2 + 2z 2 = ; = ; ∂y 2 (x2 + y 2 + z 2 )5/2 ∂z 2 (x2 + y 2 + z 2 )5/2

90

CHAPTER 13. PARTIAL DERIVATIVES 2x2 − y 2 − z 2 − x2 + 2y 2 − z 2 − x2 − y 2 + 2z 2 ∂2u ∂2u ∂2u + + = =0 ∂x2 ∂y 2 ∂z 2 (x2 + y 2 + z 2 )5/2

54.

√ √ ∂u √ 2 ∂2u 2 2 2 2 m2 +n2 x = (m + n )e = m + n2 e m +n x cos my sin nz; cos my sin nz; ∂x ∂x2 2 √ √ ∂u ∂ u 2 2 2 2 = −m2 e m +n x cos my sin nz; = −me m +n x sin my sin nz; ∂y ∂y 2 √ √ ∂u ∂2u 2 2 2 2 = −n2 e m +n x cos my sin nz; = ne m +n x cos my cos nz; 2 ∂z ∂z √ √ ∂2u ∂2u ∂2u 2 2 2 2 + 2 + 2 = (m2 + n2 )e m +n x cos my sin nz − m2 e m +n x cos my sin nz 2 ∂x ∂y ∂z

− n2 e 55.

56.

57.

m2 +n2 x

cos my sin nz = 0

∂u ∂u ∂2u ∂2u = − cos at sin x; = −a1 cos at sin x; = cos at cos x, = −a sin at sin x, ∂x ∂x2 ∂t ∂t2 ∂2u ∂2u a2 2 = a2 (− cos at sin x) = 2 ∂x ∂t ∂2u ∂u = − sin(x + at) + cos(x − at), = − cos(x + at) − sin(x − at); ∂x ∂x2 ∂u ∂2u ∂2u = −a sin(x + at) − a cos(x − at), = −a2 cos(x + at) − a2 sin(x − at); a2 2 = 2 ∂t ∂t ∂x ∂2u 2 2 −a cos(x + at) − a sin(x − at) = 2 ∂t 2x ∂2C 4x2 −1/2 −x2 /kt 2 ∂C / 2 = − t−/12 e−x kt , = t e − t−1/2 e−x /kt ; 2 2 2 ∂x kt ∂x k t kt 2 ∂C t−3/2 −x2 /kt k ∂ 2 C x2 −1/2 −x2 /kt t−1/2 −x2 /kt ∂C −1/2 x −x2 /kt =t e − e ; = 2t e − e = 2 2 ∂t kt 2 4 ∂x kt 2t ∂t

58. (a) Pv = −k(T /V 2 ) (b) P V = kt, P VT = k, VT = k/P (c) P V = kT, V = kTp , Tp = V /k  ∂u −gx/z, 0 ≤ x ≤ at 59. (a) = −gt, x > at ∂t For x > at, the motion is that of a freely falling body. ∂u (b) For x > at, = 0. For x > at, the string is horizontal. ∂x 60.

∂S = 0.0790975w0.425 h−0.275 ; Sh (60, 36) + 0.0790975(60)0.425 (36)−0.275 ≈ 0.1682 ∂h The approximate increase in skin-area as h increases from 36 to 37 inches is 0.1682 sq ft. ∂2z fx (x + ∆x, y) − fx (x, y) = lim 2 ∆x→0 ∂x ∆x ∂2z fy (x, y + ∆y) − fx (x, y) (b) = lim ∆y→0 ∂y 2 ∆y

61. (a)

13.3. PARTIAL DERIVATIVES

91

∂2z fy (x + ∆x, y) − fy (x, y) = lim ∂x∂y ∆x→0 ∆x

(c)

62. Integrating zx = 2xy 3 + 2y + 1/x with respect to x, we obtain z = x2 y 3 + 2xy + ln x + φ(y). Then 3x2 y 2 + 2x + 1 = zy = 3x2 y 2 + 2x + φ0 (y). Since φ0 (y) = 1, φ(y) = y + C, and z = x2 y 3 + 2xy + ln x + y + C. 63. Consider the mixed partials: ∂ ∂2z = ∂y∂x ∂y



∂z ∂x

 = 2y

and

∂2z ∂ = ∂x∂y ∂x



∂z ∂y

 = 2x.

∂z ∂z ∂2z ∂2z , , , and are all continuous on an open set, we should have ∂x ∂y ∂y∂x ∂x∂y ∂2z ∂2z = on that set. But the mixed partials are equal only on the line y = x, which ∂y∂x ∂x∂y contains no open set in the plane. Therefore, such a function cannot exist. Since

64. (a) There are 10 different third-order partial derivatives: Fxxx , Fxxy , Fxxz , Fxyy , Fxyz , Fxzz , Fyyy , Fyyz , Fyzz , Fzzz (b) Since the mixed partials are equal, the order in which differentiation occurs is irrelevant. The nth order partial derivatives are given by ∂nz ∂nz ∂nz ∂nz ∂nz , , , . . . , , . ∂xn ∂xn−1 ∂y ∂xn−2 ∂y 2 ∂x∂y n−2 ∂y n Hence, there are n + 1 different nth order partial derivatives. 65. (a) There slopes of the surface in the x and y directions are zero everywhere. This implies that the surface must have constant height everywhere. Therefore f must have the form f (x, y) = c. (b) Since the mixed partials are both zero, we have     ∂ ∂z ∂z = 0 and df rac∂∂y =0 ∂x ∂y ∂x ∂z ∂z is a function of y alone and is a function of x alone. Therefore, z ∂y ∂x has no term that depends on both x and y. Hence z is of the form z = g(x) + h(y) + c where g and h are twice continuously differentiable functions of a single variable. which implies

66. The level curves suggest that the surface height is decreasing as we move slightly to the right ∂z of the point, and increasing as we move slightly up from the point. This implies < 0 and ∂x ∂z > 0. ∂y ∂z f (0 + ∆x, 0) − f (0, 0) 0/2(∆x)2 67. = lim = lim = 0; ∆x→0 ∂x (0,0) ∆x→0 ∆x ∆x f (0, 0 + ∆y) − f (0, 0) 0/2(∆y)2 ∂z = lim = lim =0 ∆y→0 ∂y (0,0) ∆y→0 ∆y ∆y

92

CHAPTER 13. PARTIAL DERIVATIVES

∂z y 5 − 4x2 y 3 − x4 ∂z −x5 + 4x3 y 2 + xy 4 ∂z ∂z ; = y; ; = −x = = 68. (a) ∂x (x2 + y 2 )2 ∂x (0,y) ∂y x2 + y 2 )2 ∂y (x,0) (b)

13.4 1.

2.

3.

4.

5.

∂2z ∂2z ∂2z ∂2z = 1; = −1 =⇒ 6= ∂y∂x ∂x∂y ∂y∂x ∂x∂y

Linearization and Differentials

∂f ∂f = 4y 2 − 6x2 y so (1, 1) = −2 ∂x ∂x ∂f ∂f = 8zy − 2x3 so (1, 1) = 6 ∂y ∂y f (1, 1) = 2 The linearization is L(x, y) = 2 − 2(x − 1) + 6(y − 1) = −2x + 6y − 2 ∂f ∂f 3x2 y so = p (2, 2) = 3 3 ∂x ∂x 2 x y ∂f ∂f x3 so = p (2, 2) = 1 3 ∂y ∂y 2 x y f (2, 2) = 4 The linearization is L(x, y) = 4 + 3(x − 2) + (y − 2) = 3x + y − 4 p ∂f x2 ∂f 353 = x2 + y 2 + p so (8, 15) = 2 2 ∂x ∂x 17 x +y ∂f xy ∂f 120 =p so (8, 15) = ∂y ∂y 17 x2 + y 2 120 f (8, 15) = 136 The linearization is L(x, y) = 136+ 353 17 (x−8)+ 17 (y −15) = ∂f π 3π  −3 ∂f = 3 cos x cos y so , = ∂x ∂x 4 4 2 ∂f ∂f π 3π  −3 = 3 sin x sin y so = , ∂y ∂y 4 4 2  −3 π 3π f 4, 4 = The linearization is L(x, y) = 2 3 2 (π − 1)

−3 2

3 2

x−

π 4



3 2

y−

3π 4

353 120 17 x+ 17 y −136



=

−3 2 x

− 32 y +

2x ∂f ∂f = 2 so (−1, 1) = −1 ∂x x + y3 ∂x ∂f 3y 2 ∂f 3 = 2 so (−1, 1) = ∂y x + y3 ∂y 2

3 f (−1, 1) = ln(2) The linearization is L(x, y) = ln(2) − (x + 1) + 32 (y − 1) = −x + y − 52 + ln(2) 2  ∂f ∂f 2π 6. = 3e−2y cos 3x so 0, π3 = 3e− 3 ∂x ∂x  ∂f ∂f = −2e−2y sin 3x so 0, π3 = 0 ∂y ∂y  2π 2π f 0, π3 = 0 The linearization is L(x, y) = 3e− 3 (x − 0) = 3xe− 3 √ √ 7. Note that we are trying to approximate f (102, 80) where f (x, y) = x + 4 y. Since (102, 80) is reasonably close to (100, 81), we can use the linearization of f at (100, 81) to approximate

13.4. LINEARIZATION AND DIFFERENTIALS

93

the value at (102, 80). To do this, we compute ∂f ∂f ∂f 1 1 ∂f 1 1 1 , = √ , (100, 81) = , = (100, 81) = = , and 3/2 ∂x ∂x 20 ∂y ∂y 4(27) 108 2 x 4y f (100, 81) = 13 1 1 (x − 100) + 108 (y − 81). For the approximation, we have The linearization is L(x, y) = 13 + 20 1 1 1 1 − 108 = 7069 L9102, 80) = 13 + 20 (102 − 100) + 108 (80 − 81) = 13 + 10 540 ≈ 13.0907 √ x 8. We are trying to approximate f (36, 63) where f (x, y) = √ . Use the linearization of f at the y point (36, 64). To do this, compute √ ∂f ∂f 1 1 ∂f ∂f 3 3 − x = √ √ , (36, 64) = , = 2y (35, 64) = − , and f (36, 64) = . 3/2 ; ∂x 96 ∂y ∂y 512 4 2 x y ∂x 3 1 3 The linearization if L(x, y) = + (x − 36) − (y − 64). For the approximation, we have 4 96 512 3 1 3 387 L(36, 63) = + (36 − 36) − (63 − 64) = ≈ .7559. 4 96 512 512 9. First, linearize f at (2, 2). To do this, compute ∂f ∂f ∂f ∂f = 2(x2 + y 2 )2x, (2, 2) = 64, = 2(x2 + y 2 )2y, (2, 2) = 64, and f (2, 2, ) = 64. ∂x ∂x ∂y ∂y The linearizationis L(x, y) = 64 + 64)x − 2) + 64)y − 2). For the approximation, we have L(1.95, 2.01) = 64 + 64(−0.05) + 64(0.01) ≈ 61.44.  10. First, linearize f at 12 , 3 . To do this, compute  ∂f 1  ∂f 1  ∂f ∂f π = −πy sin(πxy), = −πx sin(πxy), , 3 = 3π, , 3 = , and f 21 , 3 = 2 2 ∂x ∂x ∂y ∂y 2 0.   π 1 + (y − 3). For the approximation, we have The linearization is L(x, y) = 3π x − 2 2 π L(0.52, 2.96) = 3π(0.02) + (−0.04) ≈ 0.1257. 2 11. dz = 2x sin 4ydx + 4x2 cos 4ydy h i 2 2 2 2 2 2 2 2 2 2 12. dz = x(2xex −y ) + ex −y dx − 2yxex −y dy = (2x2 + 1)ex −y dx − 2xyex −y dy 13. dz = p

2x 2x2 − 4y 3

dx − p

6y 2 2x2 − 4y 3

dy

14. dz = 45x2 y(5x3 y + 4y 5 )2 dx + (15x3 + 60y 4 )(5x3 y + 4y 5 )2 dy 15. df =

(s + 3t)2 − (2s − t) (s + t)(−1) − (2s − t)3 7t 7s ds + dt = − dt (s + 3t)2 (s + 3t)2 (s + 3t)2 (s + 3t)2

16. dg = 2r cos 3θdr − 3r2 sin 3θdθ 17. dw = 2xy 4 z −5 dx + 4x2 y 3 z −5 dy − 5x2 y 4 z −6 dz 2

2

2

18. dw = −2xe−z sin(x2 + y 4 )dx − 4y 3 e−z sin(x2 + y 4 )dy − 2ze−z cos(x2 + y 4 )dz 19. dF = 3r2 dr − 2s−3 ds − 2t−1/2 dt

94

CHAPTER 13. PARTIAL DERIVATIVES

20. dG = sin φ cos θdρ − ρ sin φ sin θdθ + ρ cos φ cos θdφ 21. w = ln u + ln v − ln s − ln t; dw =

22. dw = √

du dv ds dt + − − u v s t

u v st2 s2 t du − √ dv + √ ds + √ dt u2 + s2 t2 − v 2 u2 + s2 t2 − v 2 u2 + s2 t2 − v 2 u2 + s2 t2 − v 2

23. ∆z = z(2.2, 3.9) − z(2, 4) = (6.6 + 15.6 + 8) − (6 + 16 + 8) = 0.1; dz = 3dx + 4dy When x = 2, y = 4, dx = 0.2, and dy = −0.1, dz = 3(0.2) + 4(−0.1) = 0.2 24. ∆z = z(0.2, −0.1) − z(0, 0) = 2(0.2)2 (−0.1) + 5(−0.1) − 0 = −0.508; dz = 4xydx + (2x+ 5)dy When x = y = 0, dx = 0.2, and dy = −0.1, dz = 5(−0.1) = −0.5. 25. ∆z = z(3.1, 0.8) − z(3, 1) = (3.1 + 0.8)2 − (3 + 1)2 = 15.21 − 16 = −0.79; dz = 2(x + y)dx + 2(x + y)dy. When x = 3, y = 1, dx = 0.1, and dy = −0.2, 2(3 + 1)(0.1) + 2(3 + 1)(0.2) = 0.8 − 1.6 = −0.8

dz =

26. ∆z = z(0.9, 1.1) − z(1, 1) = [(0.9)2 + (0.9)2 (1.1)2 + 2] − [1 + 1 + 2] = 3.7901 − 4 = −0.2099; dz = (2x + 2xy 2 )dx + 2xy dy. When x = y = 1, dx = −0.1, and dy = 0.1, dz = 4(−0.1) + 2(0.1) = −0.2. 27. ∆z = 5(x + ∆x)2 + 3(y + ∆y) − (x + ∆x)(y + ∆y) − (5x2 + 3y − xy) = 10x∆x + 5(∆x)2 + 3∆y − x∆y − y∆x − ∆x∆y = (10x − y)∆x + (3 − z)∆y + (5∆x)∆x − (∆x)∆y 1 = 5∆x, 2 = −∆x 28. ∆z = 10(y + ∆y)2 + 3(x + ∆x) − (x + ∆x) − (10y 2 + 3x − x2 ) = 20y∆y + 10(∆y)2 + 3∆x − 2x∆x − (∆x)2 = (3 − 2x)∆x + 20y∆y − (∆x)∆x + (10∆y)∆y 1 = −∆x, 2 = 10∆y 29. ∆x = (x + ∆x)2 (y + ∆y)2 − x2 y 2 = [x2 + 2x∆x + (∆x)2 ][y 2 + 2y∆y + (∆y)2 ] − x2 y 2 = 2x2 y∆y + x2 (∆y)2 + 2xy 2 ∆x + 4xy(∆x)∆y + 2x(∆x)(∆y)2 + y 2 (∆x)2 + 2y(∆x)2 ∆y + (∆x)2 (∆y)2 = 2xy 2 ∆x + 2x2 y∆y + [4xy∆y + 2x(∆y)2 + y 2 x]∆x + [x2 ∆y + 2y(∆x)2 + (∆x)2 ∆y]∆y 1 = 4xy∆y + 2x(∆y)2 + y 2 x, 2 = x2 ∆y + 2y(∆x)2 + (∆x)2 ∆y (Several other choices of 1 and 2 are possible.) 30. ∆z = (x + ∆x)3 − (y + ∆y)3 − (x3 − y 3 ) = 3x2 ∆x + 3x(∆x)2 + (∆x)3 − 3y 2 ∆y − 3y(∆y)2 − (∆y)3 = 3x2 ∆x − 3y 2 ∆y + [3x∆x + (∆x)2 ] − [3y∆y + (∆y)2 ]∆y 1 = 3x∆x + (∆x)2 , 2 = −3y∆y − (∆y)2 31. R =

R1 R2 R3 ; ∆R1 = ±0.009R1 , ∆R! = ±0.009R2 , ∆R3 = ±0.0009R3 R2 R3 + R1 R3 + R1 R2

13.4. LINEARIZATION AND DIFFERENTIALS

95

R12 R32 R22 R32 (±0.009R1 ) + (±0.009R2 ) |∆R| ≈ |dR| ≤ 2 2 (R2 R3 + R1 R3 + R1 R2 ) (R2 R3 + R1 R3 + R1 R2 ) 2 2 R R 1 2 + (±0.009R3 ) (R2 R3 + R1 R3 + R1 R2 )2   R2 R3 + R1 R3 + R1 R2 = 0.009R = 0.009R R2 R3 + R1 R3 + R1 R2 The maximum percentage error is approximately 0.9%. 32. We are given ∆T = ±0.006T and ∆V = ±0.008V. Then

kT k kT kT |∆P | ≈ |dP | = (±0.006T ) − 2 (±0.008V ) ≤ (0.006) + (0.008) = P (0.014). V V V V Thus, the approximate maximum percentage error in P is 1.4%. (2r2 + R2 − R(2R) −R(4r) 2r2 − R2 4rR dR + mg dr = mg dR − mg dr 2 2 2 2 2 2 + 2 2 (2r + R ) 2r + R ) (2r R ) (2r+ R2 )2 When R = 4, r = 0.8, dR = 0.1, and dr = 0.1,

33. dT = mg

2(0.8)2 − 42 4(0.8)4 ∆T ≈ dT = mg (0.1) − (0.1) [2(0.8)2 + 42 ]2 [2(0.8)2 + 42 ]2   −1.472 − 1.28 = mg ≈ −0.009 mg. 298.598 



The tension decreases. 34. V = πr2 h, dV = 2πrhdr + πr2 dh. When r = 5, h = 10, dr = 0.3, and dh = 0.5, ∆V ≈ dV = 2π(5)(1−)(0.3) + π(52 )(0.5) = 42.5π cm3 . Since V (5, 10) = 250π cm3 , V (5.3, 10.5) = V (5, 10) + ∆V ≈ V (5, 10) + dV = 250π + 42.5π = 292.5π cm3 . 35. V = lwh, dV = whdl + lhdq + lwdh. With dl = ±0.02l, dw = ±0.05w, and dh = ±0.08h, |∆V | ≈ |dV | = |wh(±0.02l) + lh(±0.05w) + lw(±0.08h)| ≤ lwh(0.02 + 0.5 + 0.8) = 0.15V. The approximate percentage increase in volume is 15%. 36. S = 2lw + 2lh + 2wh, dS = (2w + 2h)dl + (2l + 2h)dw + (2l + 2w)dh With l = 3, w = 1, h = 2, sl = 0.06, dw = 0.05, and dh = 0.16, ∆S ≈ dS = (2 + 4)(0.06) + (6 + 4)(0.05) + (6 + 2)(0.16) = 2.14 ft2 . Since S(3, 1, 2) = 22 ft2 , the new surface area is approximately S(3, 1, 2) + dS = 24.14 ft2 .

96

CHAPTER 13. PARTIAL DERIVATIVES

37. dS = 0.1091(0.425)w−0.575 h0.725 dw + 0.1091(0.725)w0.425 h−0.275 dh With dw = ±0.03w and dh = ±0.05h, |∆S| ≈ |dS| = 0.1091|0.425w−0.575 h0.725 (±0.03w) + 0.725w0.425 h−0.275 (±0.05h)| ≤ 0.1091[0.425w0.425 h0.725 (0.03)] + 0.1091[0.725w0.425 h0.725 (0.05)] = 0.1091w0.425 h0.725 (0.013 + 0.036) = 0.049S. The approximate maximum percentage error is 4.9%. "

 38. Z = R + 100L − 2

1 1000c

2 #1/2 ;

"  2 #−1/2    1 1 1 2 r + 100L − 2RdR + 2 100L − (100)dL dZ = 2 1000c 1000c     1 1 2 100L − dC 1000c 1000c2   X 2 2 −1/2 = (R + X ) RdR + 1000XdL + dC 1000c2 With R = 4000, L + 0.4, C = 10−5 , dR = 25, d := 0.05, and dC = 1.1 × 10−5 − 10−5 = 10−6 , we have X = 300 and   300 2 2 −1/2 −6 dZ = (400 + 300 ) 400(25+ 100(300)(0.05) + 10 1000(10)−10 1 = (10000 + 15000 + 3000) = 56 ohms. 500 The new impedance is approximately Z(400, 0.4, 10−5 ) + dZ = 500 + 56 = 556 ohms. 39. (a) If a function w = f (x, y, z) is differentiable at a point (x0 , y0 , z0 ), then the function L(x, y, z) = f (x0 , y0 , z0 )+fx (x0 , y0 , z0 )(x−x0 )+fy (x0 , y0 , z0 )(y−y0 )+fz (x0 , y0 , z0 )(z−z0 ) is a linearization of f at (x0 , y0 , z0 ). p (b) Let f (x, y, z) = x2 + y 2 + z 2 . Then we wish to approximate f (9.1, 11.75, 19.98). To do this, linearize f at (9, 12, 20). Compute x ∂f 9 ∂f =p , (9, 12, 20) = 2 2 2 ∂x ∂x 25 x +y +z ∂f y ∂f 12 =p (9, 12, 20) = , 2 2 2 ∂y ∂y 25 x +y +z ∂f z ∂f 4 =p , (9, 12, 20) = and f (9, 12, 20) = 25 ∂z 5 x2 + y 2 + z 2 ∂z

13.4. LINEARIZATION AND DIFFERENTIALS

97

9 12 4 (x − 9) + (y − 12) + (z − 20). For the 25 25 5 12 4 9 approximation, we have L(9.1, 11.75, 19.98) = 25 + (0.1) + (−0.25) + (−0.02) = 25 25 5 24.9 The linearization is L(x, y, z) = 25 +

40. According to Theorem 13.4.3, if f were differentiable at (0, 0), then f would have to be continuous at (0, 0). However, as shown in Problem 38 in Exercises 13.2, f is not continuous at (0, 0). Therefore, f cannot be differentiable at (0, 0).

41. (a) The graph of z = f (x, y) is an inverted cone with vertex at the origin. Since the graph comes to a sharp ”point” at the origin, there is no possible increment formula for ∆z that will work in every direction there.

(b) We show that the partial derivative fx does not exist at (0, 0). If h > 0,

√ √ h2 + 02 − 02 + 02 f (0 + h) − f (0, 0) = h h √ 2 h |h| = = =1 h h

f (0 + h) − f (0, 0) = −1. h f (0 + h) − f (0, 0) Therefore, lim does not exist. But this means fx does not exist at h→0 h (0, 0) and thus f is not differentiable at (0, 0).

But if h < 0, then

98

CHAPTER 13. PARTIAL DERIVATIVES

42.

Δz z

z x

y

x

y

Δy

Δx ∆V

y

x

z z

x

y

dV

43.

∆V − dV

(a) From the figure we see that α = π − (θ + φ). Then xh = L cos θ + l cos α = L cos θ + l cos(π − θ − φ) = L cos θ − l cos(θ + φ) and yh = L sin θ − l sin α = L sin θ − l sin(π − θ − φ) = L sin θ − l sin(θ + φ).

L

θ

θ

φ

l

α

(x0,y0)

13.5. CHAIN RULE

99

(b) Using l sin(θ + φ) = ye − yh and l cos(θ + φ) = xe − xh , we have dxh = (−L sin θ + l sin(θ + φ))dθ + l sin(θ + φ)dφ = −yh dθ + (ye − yh )dφ dyh = (L cos θ − l cos(θ + φ))dθ − l cos(θ + φ)dφ = xh − (xe − xh )dφ. (c) One position has the lower arm reaching straight up, with the elbow on the x-axis, so that θ = 0 and φ = 270◦ . The other position has the lower arm reaching straight across, with the elbow on the y-axis, so that θ = φ = 90◦ . In both cases, (xh , yh ) = (L, L). In the first case, (xe , ye ) = (L, 0), and in the second case (xe , ye ) = (0, L). In general, the approximate maximum error in xh is |dxh | = | − hh dθ + (ye yh )dφ| ≤ L|dθ| + |ye − L||dφ| = (L + |ye − L|)

π . 180

Thus, in the first case the approximate maximum error is 2πL/180, while in the second case it is only πL/180. 44. (a) The horizontal and vertical components of velocity are v cos θ and v sin θ, respectively. The projectile strikes the cliff wall at time t = D/v cos θ. At this time its height is  2 1 D 1 1 D2 H = tv sin θ − gt2 = D tan θ − g = D tan θ − g 2 sec2 θ. 2 2 v cos θ 2 v   ∂H D2 d2 ∂H 2 2 2 dv + dθ = g 3 sec θdv + D sec θ − g 2 sec θ tan θ dθ (b) dH = ∂v ∂θ v v (c) When D = 100, g = 32, v = 100, and θ = 45◦ , we have H = 68 ft. (d) Taking |dv|≤ 1 and  |dθ| ≤ π/180  we find 1002 1002 π |dH| ≤ 32 sec2 sec2 4 (1) + 100 sec2 π4 − 32 3 100 1002 3.01 ft.

π 4

tan π4



π 180



=

34π 16 + ≈ 25 45

(e) We have dH =

∂H ∂H ∂H + + ∂v ∂θ ∂D

=g

    D2 D2 D 2 2 2 2 sec θdv + D sec θ − g sec θ tan θ + tan θ − g sec θ dD. v3 v2 v2

With |dD| ≤ 2, we obtain dH ≤

13.5 1.

2.

16 34π 18 34 34π + + = + ≈ 3.73 ft. 25 45 25 25 45

Chain Rule

∂z dx ∂z dy 2x dz 2y (2t) + 2 (−2t−3 ) = + = 2 dt ∂x dt ∂y dt x + y2 x + y2 4xt − 4yt−3 = x2 + y 2 dz ∂z dx ∂z dy = + dt ∂x dt ∂y dt = (3x2 y − y 4 )(5e5t ) + (x3 − 4xy 3 ) (5 sec(t) tan(t))

100

CHAPTER 13. PARTIAL DERIVATIVES

dz ∂z dx ∂z dy = + = −3 sin(3x + 4y)(2) − 4 sin(3x + 4y)(−1) dt ∂x dt ∂y dt 5π and y = −5π At t = π, x = 4 2     15π dz 15π = −6 sin so − 5π + 4 sin − 5π = −6 + 4 = −2 dt t=π 2 2   dz ∂z dx ∂z dy −8 xy 4. + xexy (3) = + = ye dt ∂x dt ∂y dt (2t + 1)2 At t = 0, x = 4 and y = 5 dz so = −40e20 + 12e20 = −28e20 dt t=0     2 r 1 r 1 2r 4r dp 2u √ − 3 − − √ = (2u)− + 5. = du 2s + t (2s + t)2 u (2s + t)2 2 u 2s + t u3 (2s + t)2 2 u(2s + t)2

3.

y2 2xy 3xy 2 y 2 sin s 2xy cos s 3xy 2 sec2 s dr = 3 (− sin s) + 3 (cos s) − 4 (sec2 s) = − + − ds z z z z3 z3 z4 2 2 ∂z ∂x ∂z ∂y + = y 2 exy (3u2 ) + 2xyexy (1) 7. zu = ∂x ∂u ∂y ∂u

6.

2

2

= 3u2 y 2 exy + 2xyexy 2 2 ∂z ∂x ∂z ∂y + = y 2 exy (0) + 2xyexy (−2v) zv = ∂x ∂v ∂y ∂v = −4vxyexy

2

∂z ∂x ∂z ∂y + = 2x cos 4y(2uv 3 ) − 4x2 sin 4y(3u2 ) ∂x ∂u ∂y ∂u = 4uv 3 x cos 4y − 12u2 x2 sin 4y ∂z ∂x ∂z ∂y + = 2x cos 4y(3u2 v 2 ) − 4x2 sin 4y(3v 2 ) zv = ∂x ∂v ∂y ∂v = 6u2 v 2 x cos 4y − 12v 2 x2 sin 4y

8. zu =

9. zu = 4(4u3 ) − 10y[2(2u − v)(2)] = 16u3 − 40(2u − v)y zv = 4(−24v 2 ) − 10y[2(2u − v)(−1)] = −96v 2 + 20(2u − v)y  2 2y 1 2xv 2 −2x v 2y 10. zu = + 2 + − 2 = 2 2 2 (x + y) v (x + y) u v(x + y) u (x + y)2    u 2y −2x 2yu 4xv 2v zv = − 2 + =− 2 − 2 (x + y) v (x + y)2 u v (x + y)2 u(x + y)2 3 3 2 (u + v 2 )1/2 (2u)(−e−t sin θ) + (u2 + v 2 )1/2 (2v)(−e−t cos θ) 2 2 = −3u(u2 + v 2 )1/2 e−t sin θ − 3v(u2 + v 2 )1/2 e−t cos θ 3 3 wθ = (u2 + v 2 )1/2 (2u)e−t cos θ + (u2 + v 2 )1/2 (2v)(−e−t sin θ) 2 2 = 3u(u2 + v 2 )1/2 e−t cos θ − 3v(u2 + v 2 )1/2 e−t sin θ

11. wt =

13.5. CHAIN RULE

101

√ √ u/2 uv rs2 u v/2 uv rv (2r) + (2rs2 ) = √ +√ 12. wr = 1 + uv 1 + uv uv(1 + uv) uv(1 + uv) √ √ v/2 uv u/2 uv −sv r2 su ws = (−2s) + (2r2 s) = √ +√ 1 + uv 1 + uv uv(1 + uv) uv(1 + uv) 2

2

13. Ru = s2 t4 (ev ) + 2rst4 (−2uve−u ) + 4rs2 t3 (2uv 2 eu 2

2

2 2

v

2

2

) = s2 t4 ev − 4uvrst4 e−u + 8uv 2 rs2 t3 eu

2 2

2

2

16. sφ = 2pe3θ + 2q[− sin(φ + θ)] − 2rθ2 + 4(2) = 2pe3θ − 2q sin(φ + θ) − 2rθ2 + 8 sθ = 2p(3e3θ ) + 2q[− sin(φ + θ)] − 2r(2φθ) + 4(8) = 6pφe3θ − 2q sin(φ + θ) − 4rφθ + 32 17. (a) 3x2 − 2x2 (2yy 0 ) − 4xy 2 + y 0 = 0 =⇒ (1 − 4x2 y)y 0 = 4xy 2 − 3x2 =⇒ y 0 = (b) fx = 3x2 − 4xy 2 , fy = −4x2 y + 1; y 0 = −

4xy 2 − 3x2 1 − 4x2 y

3x2 − 4xy 2 4xy 2 − 3x2 = −4x2 y + 1 1 − 4x2 y

18. (a) 1 + 4yy 0 = ey y 0 =⇒ 1 = (ey − 4y)y 0 =⇒ y 0 =

ey

1 − 4y

1 1 = y 4y − ey e − 4y y cos xy 19. (a) y 0 = (cos xy)(xy 0 + y) =⇒ (1 − x cos xy)y 0 = y cos xy =⇒ y 0 = 1 − x cos xy (b) f (x, y) = y − sin xy; fx = −y cos xy, fy = 1 − x cos xy; −y cos xy y cos xy y0 = − = 1 − x cos xy 1 − x cos xy (b) f (x, y) = x + 2y 2 − ey ; fx = 1, fy = 4y − ey ; y 0 = −

2 (x + y)−1/3 (1 + y 0 ) = xy 0 + y =⇒ 2(x + y)−1/3 + 2(x + y)−1/3 y 0 = 3xy 0 + 3y 3   3y − 2(x + y)−1/3 =⇒ 2(x + y)−1/3 − 3x y 0 = 3y − 2(x + y)−1/3 =⇒ y 0 = 2(x + y)−1/3 − 3x 2 2 (b) f (x, y) = (x + y)2/3 − xy; fx = (x + y)−1/3 − y, fy = (x + y)−1/3 − x; 3 3 2 −1/3 −1/3 (x + y) − y 3y − 2(x + y) 3 y0 = − 2 = −1/3 − x 2(x + y)−1/3 − 3x 3 (x + y)

20. (a)

v

2 2

Rv = s2 t4 (2uvev ) + 2rst4 (e−u ) + 4rs2 t3 (2u2 veu v ) = 2s2 t4 uvev + 2rst4 e−u + 8rs2 t3 u2 veu       1 1 1 1 t2 t 1 t2 1/t √ √ + + 2+ + = 14. Qx = 2 2 2 2 2 P q t r 1 + (x/t) qt r(t + x2 ) 1−x p 1−x     1 2t sin−1 x x 1 1 2x −x/t2 2x = Qt = (2t sin−1 x) + − 3 + − 3− 2 P q t r 1 + (x/t) p qt r(t2 + x2 ) 2x u 2y cosh rs xu y cosh rs 15. wt = p + p =p + p 2 2 2 2 2 2 u 2 x + y rs + tu 2 x + y x + y (rs + tu) u x2 + y 2 s st sinh rs yst sinh rs 2x 2y xs − p wr = p + p =p 2 2 2 2 2 2 rs + tu u 2 x +y 2 x +y x + y (rs + tu) u x2 + y 2 2x yt cosh rs t 2y −t cosh rs xt wu = p − p + p =p 2 2 2 2 2 2 2 2 u 2 x + y rs + tu 2 x + y x + y (rs + tu) u x2 + y 2

2 2

v

102

CHAPTER 13. PARTIAL DERIVATIVES

21. Fx = 2x, Fy = 2y, Fz = −2z;

∂z 2x x ∂z 2y y =− = ; =− = ∂x −2z z ∂y −2z z

z 1/3 2 2 ∂z (2/3)x−1/3 2 −1/3 = − ; x , Fy = y −1/3 , Fz = z −1/3 ; =− 3 3 3 ∂x (2/3)z −1/3 x1/3 ∂z z 1/3 (2/3)y −1/3 = − 1/3 =− −1/3 ∂y (2/3)z y

22. Fx =

23. F (x, y, z) = xy 2 z 3 + x2 − y 2 − 5z 2 , Fx = y 2 z 3 + 2x, Fy = 2xyz 3 − 2y, Fz = 3xy 2 z 2 − 10z ∂z ∂z y 2 z 3 + 2x y 2 z 3 + 2x 2xyz 3 − 2y 2xyz 3 − 2y ; =− = = − = ∂x 3xy 2 z 2 − 10z 10z − 3xy 2 z 2 ∂y 3xy 2 z 2 − 10z 10z − 3xy 2 z 2 1 1 1 ∂z −1/x z 24. F (x, y, z) = z − ln(xyz); Fx = − , Fy = − , Fz = 1 − ; =− = ; x y z ∂x 1 − 1/z xz − x ∂z −1/y z =− = ∂y 1 − 1/z yz − y 25. Let y = x + at and z = x − at. Then u(x, t) = F (y) + G(z) and dF ∂y dG ∂z dF dG ∂u = + = + ; ∂x dy ∂x dz ∂x dy dz dF ∂y dG ∂z dF dG ∂u = + =a −a ; ∂t dy ∂t dz ∂t dy dz Thus, a2

∂2u d2 F ∂y d2 G ∂z d2 F d2 G = + = + ; ∂x2 dy 2 ∂x dz 2 ∂x dy 2 dz 2 ∂2u d2 F ∂y d2 G ∂z d2 F d2 G =a 2 −a 2 = a2 2 + a2 2 . 2 ∂xt dy ∂t dz ∂t dy dz

2 2 ∂2u ∂2u 2d F 2d G = a + a = . ∂x2 dy 2 dz 2 ∂t2

26. Solving η = x + at and ξ = x − at for x and t, we obtain x = (η + ξ)/2 and t = (η − ξ)/2a. Then ∂u ∂x ∂u ∂t 1 ∂u 1 ∂u ∂u = + = − ∂ξ ∂x ∂ξ ∂t ∂ξ 2 ∂x 2a ∂t and

Setting

∂2u 1 ∂ 2 u ∂x 1 ∂ 2 u ∂t 1 ∂2u 1 ∂2u = − = − ∂η∂ξ 2 ∂x2 ∂η 2a ∂t2 ∂η 4 ∂x2 4a2 ∂t2 1 ∂2u 1 ∂2u ∂2u ∂2u ∂2u = 0, we have − 2 2 = 0 or a2 2 = 2 . 2 ∂η∂ξ 4 ∂x 4a ∂t ∂x ∂t

27. With x = r cos θ and y = r sin θ ∂u ∂x ∂u ∂y ∂u ∂u ∂u = + = cos θ + sin θ ∂r ∂x ∂r ∂y ∂r ∂x ∂y ∂2u ∂ 2 u ∂x ∂ 2 u ∂y ∂2u ∂2u 2 = cos θ + sin θ = cos θ + sin2 θ ∂r2 ∂x2 ∂r ∂y 2 ∂r ∂x2 ∂y 2 ∂u ∂u ∂x ∂u ∂y ∂u ∂u ∂ 2 u ∂y = + = (r sin θ) + (−r sin θ) + 2 (r cos θ) ∂θ ∂x ∂θ ∂y ∂θ ∂y ∂y ∂y ∂θ ∂u ∂2u ∂u ∂2u = −r cos θ + r2 2 sin2 θ − r sin θ − r2 2 cos2 θ. ∂x ∂x ∂y ∂y

13.5. CHAIN RULE

103

∂2u ∂2u + 2 = 0, we have ∂x2 ∂y   ∂ 2 u 1 ∂u ∂2u ∂2u 1 ∂u 1 ∂2u ∂u 2 2 + cos θ + 2 sin θ + + 2 2 = cos θ + sin θ ∂r2 r ∂r r ∂θ ∂x2 ∂y r ∂x ∂y   ∂u ∂u 1 ∂2u ∂2u + 2 −r + r2 2 sin2 θ − r sin θ + r2 2 cos2 θ r ∂x ∂x ∂y ∂y   2 ∂2u ∂ u ∂u 1 1 2 2 2 2 = (cos θ + sin θ) + (cos θ + sin θ) + cos θ − cos θ ∂x2 ∂y 2 ∂x r r   ∂u 1 1 + sin θ − sin θ ∂y r r ∂2u ∂2u + 2 =0 = ∂x2 ∂y

Using

28.

∂z dz ∂u ∂z dz ∂u = , = ∂x du ∂x ∂y du ∂y

29. Letting u = y/x in Problem 28, we have ∂z dz ∂u dz ∂u dz  y  ∂z dz +y =x +y =x x − 2 +y ∂x ∂y du ∂x du ∂y du x du

    1 dz −y y = + = 0. x du x x

30. We first compute ∂u ∂u ∂r ∂u x p = = ∂x ∂r ∂x ∂r x2 + y 2 x ∂ 2 u x2 ∂u y2 ∂2u ∂u y2 ∂2u +p + 2 2 = = 2 2 2 2 3/2 2 2 3/2 ∂x ∂r (x + y ) ∂r (x + y ) ∂r x + y 2 x2 + y 2 ∂r ∂u x2 ∂ 2 u y2 ∂2u = . ∂y 2 ∂r (x2 + y 2 )3/2 ∂r2 x2 + y 2 Then

∂u x2 + y 2 ∂ 2 u 1 ∂u ∂2u ∂2u ∂ 2 u x2 + y 2 + 2 = = + . + 2 2 2 2 2 2 3/2 ∂x ∂y ∂r (x + y ) ∂r x + y ∂r2 r ∂r

31. We first compute ∂u ∂ = B erf ∂x ∂x



x √ 4kt



∂ =B ∂x

2 π

Z

√ x/ 4kt

! e

0

−v 2

dv

2 1 −x2 /4kt = B√ √ e π 4kt

2 2 1  x  −x2 /4kt x ∂2u √ √ = B − e = −B √ √ e−x /4kt 3 ∂x2 2kt π 4kt 2k π kt !      Z x/√4kt ∂ x ∂u ∂ 2 2 x 1 −3/2 −v 2 = B erf √ =B e dv = B √ − √ − t e−x/4kt ∂t ∂t ∂t π 0 2 π 4kt 2 k 2 x = −B √ √ e−x /4kt 3 2 π kt

Then k

∂2u ∂u = . ∂x2 ∂t

104

CHAPTER 13. PARTIAL DERIVATIVES dI ∂I dE ∂I dR 1 E = + = (2) − 2 (−1), dt ∂E dt ∂R dt R R 1 dI 2 60 3/5 8 = and when E = 60 and R = 50, = + + = amp/min. dt 50 502 25 25 125

32. We are given dE/dt = 2 and dR/dt = −1. Then

33. Since the height of the triangle is x sin θ, the area is given by A = 12 xy sin θ. Then dA ∂A dx ∂A dy ∂A dθ 1 dx 1 dy 1 dθ = + + = y sin θ + x sin θ + xy cos θ . dt ∂x dt ∂y dt ∂θ dt 2 dt 2 dt 2 dt When x = 10, y = 8, θ = π/6, dz/dt = 0.3, dy/dt = 0.5, and dθ/dt = 0.1,     1 1 1 1 (0.3) + (10) (0.5) + (10)(8) 2 2 2 2 √ √ = 0.6 + 1.25 + 2 3 = 1.85 + 2 2 ≈ 5.31 cm2 /s.

dA 1 = (8) dt 2

34.

35.

√ ! 3 (0.1) 2

dP (V − 0.0427)(0.08)dT /dt 0.08T (dV /dt) 3.6 dV = − + 3 dt (V − 0.0427)2 (V − 0.0427)2 V dt   0.08 3.6 dT 0.08T dV = + − V − 0.0427 dt V3 (V − 0.0427)2 dt   dS dw dh = 0.1091 0.425w−0.575 h0.725 + 0.725w0.425 h−0.275 dt dt dt When w = 25, h = 29, dw/dt = 4.2, and dh/dt = 2, dS = 0.1091[0.425(25)−0.575 (29(0.725 (4.2) + 0.725(25)0.425 (29)−0.275 (2)] ≈ 0.5976in2 /yr. dt

36.

dw ∂w dx ∂w dy ∂w dz xdx/dt + ydy/dt + zdz/dt −4x sin t + 4y cos t + 5z p = + + = =p 2 2 2 dt ∂x dt ∂y dt ∂z dt x +y +z 16 cos2 t + 16 sin2 t + 25t2 −16 sin t cos t + 16 sin t cos t + 25t 25t √ = =√ 2 16 + 25t 16 + 25t2 dw 125π 125π/2 =√ ≈ 4.9743 =p dt t=5π/2 64 + 625π 2 16 + 625π 2 /4

37. Since dT /dT = 1 and 0 = FT =

∂P = 0, ∂T

∂F ∂P ∂F ∂V ∂F ∂T ∂V ∂F/∂T 1 + + =⇒ =− =− . ∂P ∂T ∂V ∂T ∂T ∂T ∂T ∂F/∂V ∂T /∂V

38. (a) From the law of sines,

r 500 500 sin φ = so r = . sin φ sin(π − θ − φ) sin(θ + φ)

(b) r = 500 sin 75◦ / sin 137◦ ≈ 708 yds

13.5. CHAIN RULE

105

(c) Using the chain rule, we obtain dr ∂r dθ ∂r dφ = + dt ∂θ dt ∂φ dt 500 sin φ cos(θ + φ) dθ sin(θ + φ) cos φ − sin φ cos(θ + φ) dφ =− + 500 dt dt sin2 (θ + φ) sin2 (θ + φ) 500 sin φ cos(θ + φ) dθ =− dt sin2 (θ + φ) [sin θ cos φ + cos θ sin φ] cos φ − sin φ[cos θ cos φ − sin θ sin φ] dφ + 500 dt sin2 (θ + φ) 500 sin φ cos(θ + φ) dθ 500 sin θ dφ =− + . 2 dt sin (θ + φ) sin2 (θ + φ) dt When dθ/dt = 5◦ = 5π/180 and dφ/dt = −10π/180, we have     500 sin 75◦ cos 137◦ 5π 500 sin 62◦ 10π dr =− + − ≈ −99.4yd/min. dt 180 180 sin2 137◦ sin2 137◦ The distance from C to A is decreasing. 39. (a) Using f = π, l = 6, V = 100, and c = 330, 000 we obtain f ≈ 380.04 cycles per second. ! r  c A −1/2 A 1 ∂f c A 1 = − 2 =− =− f and (b) We first note that ∂V 4π lV lv 4π lV V 2V ∂f 1 = − f. ∂l 2l   df ∂f dV ∂f dl 1 dV 1 dl f 1 dV 1 dl Then = + =− f − f =− + . dt ∂V dt ∂l dt 2V dt 2l dt 2 V dt l dt Using dV /dt = −10, dl/dt = 1, V = 100, and l = 6 we find     df f 1 1 f 1 1 =− (−10) + (1) = − − < 0. dt 2 100 6 2 6 10 The frequency is decreasing. 40. (a)

w w x x

w y

w z

y

z

y x

z x x

x

106

CHAPTER 13. PARTIAL DERIVATIVES dw ∂w ∂w y ∂w dz = + + dx ∂x ∂y x ∂z dx (b) Using the formula from Part (a), we have dw = (y 2 + 1) + (2xy − 2z) dx

  1 + (−2y)(ex ) x

41. z

z u

z v

u u t1 u t2 t1

t2

z w

w

v v t1

u u t4 t3 t3

t4

t1

v t2 t2

v v t4 t3 t3

w t1 w t2 t4

t1

w w t4 t3

t2

t3

∂z ∂z ∂u ∂z ∂v ∂z ∂w = + + ∂t2 ∂u ∂t2 ∂v ∂t2 ∂w ∂t2 ∂z ∂z ∂u ∂z ∂v ∂z ∂w = + + ∂t4 ∂u ∂t4 ∂v ∂t4 ∂w ∂t4 42. Since w = F (x, y, z, u) = 0, ∂w/∂x = 0. Also dx/dx = 1, ∂y/∂x = 0, and ∂z/∂x = 0. Then dx ∂u ∂z ∂u ∂w = Fx (x, y, z, u) + Fu (x, y, z, u) + FZ (x, y, z, u) + Fu (x, y, z, u) ∂x dx ∂x ∂x ∂x implies ∂u/∂x = −Fx (x, y, z, u)/Fu (x, y, z, u). Similarly, ∂u/∂y = −FY (x, y, z, u)/Fu (x, y, z, u) and ∂u/∂z = −FZ (x, y, z, u)/Fu (x, y, z, u). 43. Letting F (x, y, z, u) = −xyz + x2 yu + 2xy 3 u − u4 − 8 we find FZ = −yz + 2xyu + 2y 3 u, Fy = −xz + x2 u + 6xy 2 u, Fz = −xy, and Fu = x2 y + 2xy 3 − 4u3 . Then ∂u −yz + 2xyu + 2y 3 u =− 2 , ∂x x y + 2xy 3 − 4u3

∂u −xz + x2 y + 6xy 2 u =− 2 , ∂y x y + 2xy 3 − 4u3

∂u xy = 2 . ∂z x y + 2xy 3 − 4u3

44. (a) Let u = λx and v = λy. Then f (u, v) = λn f (x, y), and differentiating both sides with respect to λ, we have ∂f ∂u ∂f ∂v + = nλn−1 f (x, y) or xfu (u, v) + yfu (u, v) = nλn−1 f (x, y). ∂u ∂λ ∂v ∂λ Letting λ = 1, we have u = x and y = v, so xfx (x, y) + yfy (x, y) = nf (x, y). (b) f (λx, λy) = 4(λx)2 (λy 3 ) − 3(λx)(λy)4 + (λx)5 = λ5 f (x, y)

t4

13.6. DIRECTIONAL DERIVATIVE

107

(c) xfx + yfy = x(8xy 3 − 3y 4 + 5x4 ) + y(12x2 y 2 − 12xy 3 ) = 8x2 y 3 − 3xy 4 + 5x5 + 12x2 y 3 − 12xy 4 = 20x2 − 15xy 4 + 5x5 = 5(4x2 y 3 − 3xy 4 + x5 ) = 5f (x, y)   y y λy = λ0 f , we see that z = f (d) By observing that f =f λx x x of degree zero.

13.6

y x



is homogeneous

Directional Derivative

1. ∇f = (2x − 3x2 y 2 )i + (4y 3 − 2x3 y)j 2

2

2. ∇f = 4xye−2x y i + (1 + 2x2 e−2x y )j 3. ∇F =

y2 2xy 3xy 2 i + j − k z3 z3 z4

4. ∇F = y cos yzi + (x cos yz − xyz sin yz)j − xy 2 sin yzk 5. ∇f = 2xi − 8yj; ∇f (2, 4) = 4i − 32j x3 − 4y 3 27 3x2 5 6. ∇f = p i+ p j; ∇f (3, 2) = √ i − √ j 3 4 3 4 38 2 38 2 x y−y 2 x y−y 7. ∇F = 2xz 2 sin 4yi + 4x2 z 2 cos 4yj + 2x2 z sin 4yk √ √ 4π 4π 4π i + 16 cos j + 8 sin k = 2 3i − 8j − 4 3k ∇F (−2, π/3, 1) = −4 sin 3 3 3 2x 2y 2z 4 3 1 i+ 2 j+ 2 k; ∇F (−4, 3, 5) = − i + j + k x2 + y 2 + z 2 x + y2 + z2 x + y2 + z2 25 25 25 √ √ f (x + h 3/2, y + h/2) − f (x, y) (x + h 3/2)2 + (y + h/2)2 )2x − y 2 9. Du f (x, y) = lim = lim h→0 h→0 h h √ 2 2 √ √ h 3x + 3h /4 = hy + h /4 = lim = lim ( 3x + 3h/4 + y + h/4) = 3x + y h→0 h→0 h √ √ f (x + h 2/2, y + h 2/2) − f (x, y) 10. Du f (x, y) = lim h→0 h √ √ 3x + 34 2/2 − (y + h 2/2)2 − 3x + y 2 = lim h→0 h √ √ √ √ √ √ 3h 2/2 − h 2yh2 /2 = lim = lim (3 2/2 − 2y − h/2) = 3 2/2 − 2y h→0 h→0 h √ √ 3 1 15 3 2 6 3 5 11. u = i j; ∇f = 15x y i + 30x y j; ∇f (−1, 1) = 15i − 30j; Du f (−1, 1) = − 15 = 2 2 2 √ 15 ( 3 − 2) 2 8. ∇F =

108

CHAPTER 13. PARTIAL DERIVATIVES √

√ 2 2 12. u = i j; ∇f = (4 + y 2 )i(2xy − 5)j; ∇f (3, −1) = 5i − 11j; 2 2 √ √ √ 5 2 11 2 − = −3 2 Du f (3, −1) = 2 2 √ √ x 1 3 10 −y 1 10 13. u = i+ 2 j; ∇f (2, −2) = i + j; i− j; ∇f = 2 2 2 10 10 y x +y 4 4 √ √ x +√ 10 3 10 10 Du f (2, −2) = − =− 40 40 20 x2 6 8 y2 i + j; ∇f (2, −1) = i + 4j; i + j; ∇f = 10 10 (x + y)2 (x + y)2 3 16 19 Du f (2, −1) = + = 5 5 5 √ 15. u = (2i + j)/ 5; ∇f = 2y(xy + 1)i + 2x(xy + 1)j; ∇f (3, 2) = 28i + 42j; 2(28) 42 98 Du f (3, 2) = √ + √ = √ 5 5 5 √ √ 16. u = −i; ∇F = 2x tan yi + x2 sec2 yj; ∇f (1/2, π/3) = 3i + j; Du f (1/2, π/3) = − 3

14. u =

1 1 17. u = √ j + √ k; ∇f = 2xy 2 (2z + 1)2 i2x2 y(2z + 1)2 j + 4x2 y 2 (2z + 1)k; ∇f (1, −1, 1) = 2 2 √ 18 12 6 18i − 18j + 12k; Du f (1, −1, 1) = − √ + √ = − √ = −3 2 2 2 2 2 1 2x 1 2y 2y 2 − 2x2 18. u = √ i − √ j + √ k; ∇f = 2 i − 2 j + k; ∇f (2, 4, −1) = 4i − 8j − 24k; z z z3 6 6 6 √ 4 16 24 Du f (2, 4, −1) = √ − √ − √ = −6 6 6 6 6 x2 + 4z y2 i+ p j+ p k; x2 y + 2y 2 z 2 x2 y + 2y 2 z x2 y + 2y 2 z ∇f (−2, 2, 1) = −i + j + k; Du f (−2, 2, 1) = −1

19. u = −k; ∇f = p

xy

√ 2 1 2 20. u = −(4i−4j+2k)/ 36 = − i+ j− k; ∇f = 2i−2yj+2zk; ∇f (4, −4, 2) = 2i+8j+4k; 3 3 3 4 16 4 8 Du f (4, −4, 2) = − + − = 3 3 3 3 √ 21. u = (−4i − j/ 17; ∇f = 2(x − y)i − 2(x − y)j; ∇f (4, 2) = 4i − 4j; 16 4 12 Du f (4, 2) = − √ + √ = − √ 17 17 17 √ 22. u = (−2i + 5j/ 29; ∇f = (3x2 − 5y)i − (5x − 2y)j; ∇f (1, 1) = −2i − 3j; 4 15 11 Du f (1, 1) = √ − √ = − √ 29 29 29 √ √ 2 2x 2x 23. ∇f = 2e sin yi + e cos yj; ∇f (0, π/4) = 2i + j 2 p √ √ √  √ 2 1/2 The maximum Du is ( 2) + ( 2/2)2 = 5/2 in the direction 2i + ( 2/2)j.

13.6. DIRECTIONAL DERIVATIVE

109

24. ∇f = (xyex−y + yex−y i + (−xyex−y + xex−y j; ∇f (5, 5) = 30i − 20j √  1/2 The maximum Du is 302 + (−20)2 = 10 13 in the direction 30i − 20j. 25. ∇f = (2x + 4z)i + 2z 2 j + (4x + 4yz)k; ∇f (1, 2, −1) = −2i + 2j − 4k √  1/2 The maximum Du is (−2)2 + (2)2 + (−4)2 = 2 6 in the direction −2i + 2j − 4k. 26. ∇f = yzi + xzj + xyk; ∇f (3, 1, −5) = −5i − 15j + 3k  1/2 √ The maximum Du is (−5)2 + (−15)2 + (3)2 = 259 in the direction −5i − 15j + 3k. 2 2 2 2 2 2 27. ∇f = p2x secp(x + y )ip+ 2y sec2 (x + y )j; p ∇f ( π/6, π/6) = 2 p π/6 sec (π/3)(i + j) = 8p π/6(i + j) The minimum Du is −8 π/6(12 + 12 )1/2 = −8 π/3 in the direction −(i + j).

28. ∇f = 3x2 i − 3y 2 j; ∇f (2, −2) = 12i − 12j = 12(i − j) √  1/2 The minimum Du is −12 12 + (−1)2 = −12 2 in the direction −(i − j) = −i + j. √ y √ √ ze x 3 2 y √ 29. ∇f = i + xze j + √ k; ∇f (16, 0, 9) = i + 12j + k. The minimum Du is 8 3 2 x 2 z √   3 2 2 2 2 1/2 − (3/8) + 12 + (2/3) = − 83281/24 in the direction − i − 12j − k. 8 3 1 1 1 i + j − k; ∇f (1/2, 1/6, 1/3) = 2i + 6j − 3k x y z  1/2 The minimum Du is − 22 + 62 (−3)2 = −7 in the direction −2i − 6j + 3k.

30. ∇f =

31. Using implicit differentiation on 2x2 + y 2 = 9 we find y 0 =√−2x/y. At (2, 1) the slope of the tangent line is −2(2)/1 = −4. √Thus, u√= ±(i − 4j)/ √ 17. Now, ∇f = i + 2yj and ∇f (3, 4) = i + 8j. Thus, Du = ±(1/ 17 − 32 17) = ±31/ 17. 2x + y − 1 x + 2y 3x + 3y − 1 √ √ 32. ∇f = (2x + y − 1)i + (x + 2y)j; Du f (x, y) = + √ = Solving 2 2 2 √ (3x + 3y − 1)/ 2 = 0 we see that Du is 0 for all points on the line 3x + 3y = 1.  33. (a) Vectors perpendicular to 4i + 3j are ±(3i − 4j). Take u = ±

 3 4 i− j . 5 5

√ 4 3 (b) u = (4i + 3j)/ 16 + 9 = i + j 5 5 4 3 (c) u = − i − j 5 5 34. D−u f (a, b) = ∇f (a, b) · (−u) = −∇f (a, b) · u = −Du f (a, b) = −6 35. (a) ∇f = (3x2 − 6xy 2 )i + (−6x2 y + 3y 2 )j 3(3x2 − 6xy 2 ) −6x2 y + 3y 2 9x2 − 18xy 2 − 6x2 y + 3y 2 √ √ √ Du f (x, y) = + = 10 10 10

110

CHAPTER 13. PARTIAL DERIVATIVES 3 3 (b) F (x, y) = √ (3x2 − 3xy 2 − 2x2 y + y 2 ); ∇F = √ [(6x − 6y 2 − 4xy)i + (−12xy − 10 10 2x2 + 2y)j]       3 3 3 1 2 √ √ (6x − 6y − 4xy) + √ (−12xy − 2x2 + 2y) Du F (x, y) = √ 10 10 10 10 9 3 1 = (3x − 3y 2 − 2xy) + (−6xy − x2 + y) = (27x − 27y 2 − 36xy − 3x2 + 3y) 5 5 5

12 5 α − β = 7 and Dv f (a, b) = 36. Let ∇f (a, b) = αi + βj. Then Du f (a, b) = ∇f (a, b) · u = 13 13 5 12 ∇f (a, b) · v = α − β = 3. Solving for α and β, we obtain α = 13 and β = −13/6. Thus, 13 13 ∇f (a, b) = 13i − (13/6)j. 37. 38. ∇f = h2x, −5yi, |∇f | =

p

10x2 + 25y 2 = 10, 4x2 + 25y 2 = 100,

x2 y2 + =1 25 4

y

x

39. ∇T = 4xi + 2yj; ∇T (4, 2) = 16i + 4j. The minimum change in temperature (that is, the maximum decrease in temperature) is in the direction −∇T (4, 3) = −16i − 4j. 40. Let x(t)i + y(t)j be the vector equation of the path. At (x, y) on this curve, the direction of a tangent vector is x0 (t)i + y 0 (t)j. Since we want the direction of motion to be −∇T (x, y), we have x0 (t)i + y 0 (t)j = −∇T (x, y) = 4xi + 2yj. Separating variables in dx/dt = 4x, we obtain dx/x = 4dt, ln x = 4t + c1 , and x = C1 e4t . Separating variables in dy/dt = 2y, we obtain dy/y = 2dt, ln y = 2t + c2 , and y = C2 e2t . Since x(0) = 4 and y(0) = 2, we have x = 4e4t and y = 2e2t . The equation of the path is 4e4t i + 2e2t j for t ≥ 0, or eliminating the parameter, x = y 2 , y ≥ 0. 41. Let x(t)i + y(t)j be the vector equation of the path. At (x, y) on this curve, the direction of a tangent vector is x0 (t)i + y 0 (t)j. Since we want the direction of motion to be ∇T (x, y), we have x0 (t)i + y 0 (t)j = ∇T (x, y) = −4xi − 2yj. Separating variables in dx/dt = −4x, we obtain dx/x = −4dt, ln x = −4t + c1 , and x = C1 e−4t . Separating variables in dy/dt = −2y, we obtain dy/y = −2dt, ln y = −2t + c2 , and y = C2 e−2t . Since x(0) = 3 and y(0) = 4, we have x = 3e−4t and y = 4e−2t . The equation of the path is 3e−4t i + 4e−2t j, or eliminating the parameter, 16x = 3y 2 , y ≥ 0.

13.6. DIRECTIONAL DERIVATIVE

111

42. Substituing x = 0, y = 0, z = 1, and T = 500 into t = and T (x, y, z) =

k we see that k = 500 x2 + y 2 + z 2

500 . x2 + y 2 + z 2

1 1 2 2 h1, −2, −2i = i − j − k 3 3 3 3 1000y 1000z 1000x i− 2 j− 2 k ∇T = − 2 2 2 2 2 2 2 (x + y + z ) (x + y + z ) (x + y 2 + z 2 )2 500 750 750 ∇T (2, 3, 3) = − i− j− k 121 121 121       1 500 2 750 2 750 2500 Du T (2, 3, 3) = − − − − − = 3 121 3 121 3 121 363

(a) u =

(b) The direction of maximum increase is ∇T (2, 3, 3) = −

500 750 750 252 i− j− k= (−2i − 3j − 3k). 121 121 121 121

(c) The maximum rate of change of T is |∇T (2, 3, 3)| =

√ 250 √ 250 22 4+9+9= . 121 121

Gmx Gmy Gm i+ 2 j= 2 (xi + yj) 2 3/2 2 3/2 +y ) (x + y ) (x + y 2 )3/2 The maximum and minimum values of Du U (x, y) are obtained when u is in the directions ∇U and −∇U , respectively. Thus, at a point (x,y), not (0,0), the directions of maximum and minimum increase in U are xi+yj and −xi−yj, respectively. A vector at (x, y) in the direction ±(xi+yj) lies on a line through the origin.

43. ∇U =

(x2

44. Since ∇f = fx (x, y)i + fy (x, y)j, we have ∂f /∂x = 3x2 + y 3 + yexy . Integrating, we obtain f (x, y) = x3 + xy 3 + exy + g(y). Then fy = 3xy 2 + xexy + g 0 (y) = −2y 2 + 3xy 2 + xexy . Thus, g 0 (y) = −2y 2 , g(y) = − 32 y 3 + c, and f (x, y) = x3 + xy 3 + exy − 23 + C. 45. ∇(cf ) =

∂ ∂ (cf )i + j = cfx i + cfy j = c(fx i + fy j) = c∇f ∂x ∂y

46. ∇(f + g) = (fx + gx )i + (fy + gy )j = (fx i + fy j) + (gx i + gy j) = ∇f + ∇g 47. ∇(f g) = (f gx + fx g)i + (f gy + fy g)j = f (gx i + gy j) + g(fx i + fy j) = f ∇g + g∇f     48. ∇(f /g) = (gfx − f gx )/g 2 i + (gfy − f gy )/g 2 j = g(fx i + fy j)/g 2 − f (gx i + gy j)/g 2 = g∇f /g 2 − f ∇g/g 2 = (g∇f − f ∇g)/g 2 p ∂r x x ∂r y y x2 + y 2 so =p = and =p = 2 2 2 2 ∂x r ∂y r x +y x +y Dx yE 1 r This gives ∇r = , = hx, yi = r r r r

49. r(x, y) =

112 50.

CHAPTER 13. PARTIAL DERIVATIVES ∂ (f (r)) df ∂r ∂ (f (r)) df ∂r ∂ (f (r)) ∂ (f (r)) = and = so that ∇f (r) = h , i ∂x dr ∂x ∂y dr ∂y ∂x ∂y df ∂r df ∂r df ∂r ∂r =h , i= h , i = f 0 (r)∇r = f 0 (r)r/r dr ∂x dr ∂y dr ∂x ∂y

51. Let u = u1 i + u2 j and v = v1 i + v2 j. Dv f = (fx i+ fy j) · v = v1 fx + v2 fy  ∂ ∂ Du Dv f = (v1 fx + v2 fy )i + (v1 fx + v2 fy )j · u = [(v1 fxx + v2 fyz )i + (v1 fxy + v2 fyy )j] · u ∂x ∂y = u1 v1 fxx + u1 v2 fyx + u2 v1 fxy + u2 v2 fyy D − uf = (fx i + fy j) · u = u1 fx + u2 fy   ∂ ∂ (u1 fx + u2 fy )i + (u1 fx + u2 fy )j · v = [(u1 fxx + u2 fyx )i + (u1 fxy + u2 fyy )j] · v Dv Du f = ∂x ∂y = u1 v1 fxx + u2 v1 fyx + u1 v2 fxy + u2 v2 fyy Since the second partial derivatives are continuous, fxy = fyx and Du Dv f = Dv Du f. [Note that this result is a generalization fxy = fyx since Di Dj f = fyx and Dj Di f = fxy ] i j k ∂ ∂ ∂ 52. ∇ × F = ∂x ∂y ∂z f f2 f3 1       ∂f3 ∂f2 ∂f3 ∂f1 ∂f2 ∂f1 = − i− − j+ − k ∂y ∂z ∂x ∂z ∂x ∂y

13.7

Tangent Planes and Normal Lines

1. Since f (6, 1) = 4, the level curve is x − 2y = 4. ∇f = i − 2j; ∇f (6, 1) = i − 2j

y

x

2. Since f (1, 3) = 5, the level curve is y+2x = 5x or y = 3x, x 6= 0. y 1 ∇f = − 2 i + j; ∇f (1, 3) = −3i + j x x

y

x

13.7. TANGENT PLANES AND NORMAL LINES

113

3. Since f (2, 5) = 1, the level curve is y = x2 + 1. ∇f = −2xi + j; ∇f (2, 5) = −4i + j

y

x

4. Since f (−1, 3) = 10, the level curve is x2 + y 2 = 10. ∇f = 2xi + 2yj; ∇f (−1, 3) = −2i + 6j

y

x

5. Since f (−2, −3) = 2, the level curve is x2 /4 + y 2 /0 = 2 x 2y x2 /8 + y 2 /18 = 1. ∇f = i + j; ∇f (−2, −3) = −i − 2 9

y

or 2 j 3

x

6. Since f (2, 2) = 2, the level curve is y 2 = 2x, x 6= 0. 2y y2 ∇f = − 2 i + j; ∇f (2, 2) = −i + 2j x x

y

x

114

CHAPTER 13. PARTIAL DERIVATIVES

7. Since f (1, 1) = −1, the level curve is (x − 1)2 − y 2 = −1 or y 2 − (x − 1)2 = 1. ∇f = 2(x − 1)i − 2yj; ∇f (1, 1) = −2j

y

x

8. Since f (π/6, 3/2) = 1, the level curve is y − 1 = sin x or −(y − 1) cos x 1 y = 1 + sin x, sin x 6= 0. ∇f = i+ j; 2 sin x sin x √ ∇f (π/6, 3/2) = − 3i + 2j

y

x

9. Since f (3, 1, 1) = 2, the level curve is y + z = 2 ∇f = j + k; ∇f (3, 1, 1) = j + k

z

2

2

y

x

10. Since f (1, 1, 3) = −1, the level curve is x2 + y 2 − z = −1 or z = 1 + x2 + y 2 . ∇f = 2xi + 2yj − k; ∇f (1, 1, 3) = 2i + 2j − k

z

y

x 11. Since F (3, 4, 0) = 5, the level curve is x2 + y 2 + z 2 = 25. x y z ∇F = p i+ p j+ p k; 2 2 2 2 2 2 2 x +y +z x +y +z x + y2 + z2 3 4 ∇F (3, 4, 0) = i + j 4 5

z 5 5 y x

13.7. TANGENT PLANES AND NORMAL LINES

115

12. Since F (0, −1, 1) = 0, the level curve is x2 − y 2 + z = 0 or z = y 2 − x2 . ∇F = 2xi − 2yj + k; ∇F (0, −1, 1) = 2i + k

z

y x

 13. F (x, y, z) = x2 + y 2 − z; ∇F = 2xi + 2yj − k. We want ∇F = c 4i + j + 21 k or 2x = 4c, 2y = c, −1 = c/2. From the third equation c = −2. Thus, x = −4 and y = −1. Since z = x2 + y 2 = 16 + 1 = 17, the point on the surface is (−4, −1, −17). 14. F (x, y, z) = x3 + y 3 + z; ∇F = 3x2 i + 2yj + k. We want ∇F = c(27i + 8j + k) or 3x2 = 27c, 2y = 8c, 1 = c. From c = 1 we obtain x = ±3 and y = 4. Since z = 15 − x3 − y 2 = 15 − (±3)3 − 16 = −1 ∓ 27, the points on the surface are (3, 4, −28) and (−3, 4, 26). 15. F (x, y, z0 = x2 + y 2 + z 2 ; ∇F = 2xi + 2yj + 2zk. ∇F (−2, 2, 1) = −4i + 4j + 2k. The equation of the tangent plane is −4(x + 2) + 4(y − 2) + 2(z − 1) = 0 or −2x + 2y + z = 9. 16. F (x, y, z) = 5x2 − y 2 + 4z 2 ; ∇F = 10xi − 2yj + 8zk; ∇F (2, 4, 1) = 20i − 8j + 8k.The equation of the tangent plane is 20(x − 2) − 8(y − 4) + 8(z − 1) = 0 or 5x − 2y + 2z = 4. 17. F (x, y, z) = x2 − y 2 − 3z 2 ; ∇F = 2xi − 2yj − 6zk; ∇F (6, 2, 3) = 12i − 4j − 18k. The equation of the tangent plane is 12(x − 6) − 4(y − 2) − 18(z − 3) = 0 or 6x − 2y − 9z = 5. 18. F (x, y, z) = xy + yz + zx; ∇F = (y + z)i + (x + z)j + (y + x)k; ∇F (1, −3, −5) = −8i − 4j − 2k. The equation of the tangent plane is −8(x − 1) − 4(y + 3) − 2(z + 5) = 0 or 4x + 2y + z = −7. 19. F (x, y, z) = x2 + y 2 + z; ∇F = 2xi + 2yj + k; ∇F (3, −4, 0) = 6i − 8j + k. The equation of the tangent plane is 6(x − 3) − 8(y + 4) + z = 0 or 6x − 8y + z = 50. 20. F (x, y, z) = xz; ∇F = zi + xk; ∇F (2, 0, 3) = 3i + 2k. The equation of the tangent plane is 3(x − 2) + 2(z − 3) = 0 or 3x + 2z = 12. √ 21. F (x, y,√z) = cos(2x+y)−z; ∇F = −2 sin(2x+y)i−sin(2x+y)j−k; ∇F (π/2, π/4, −1 2) = √   √ √  2 π 2 π 1 2i+ j−k. The equation of the tangent plane is 2 x − + y− − z+√ = 2 2 2 4 2    √ π  π √ 1 5π 0, 2 x − + y− − 2 z+√ = 0, or 2x + y − 2z = + 1. 2 4 4 2 22. F (x, y, z) = x2 y 3 + 6z; ∇F = 2xy 3 i + 3x2 y 2 j + 6k; ∇F (2, 1, 1) = 4i + 12j + 6k. The equation of the tangent plane is 4(x − 2) + 12(y − 1) + 6(z − 1) = 0 or 2x + 6y + 3z = 13. √ √ √ 2x 2y 23. F (x, y, z) = ln(x2 + y 2 ) − z; ∇F = 2 i+ 2 j − k; ∇F (1/ 2, 1/ 2, 0) = 2i + 2 2 x +y x  +y    √ √ √ 1 1 √ √ 2j − k. The equation of the tangent plane is 2 x − + 2 y− − (z − 0) = 2 2     √ √ √ 1 1 +2 y− √ − 2z = 0, or 2x + 2y − 2z = 2 2. 0, 2 x − √ 2 2

116

CHAPTER 13. PARTIAL DERIVATIVES

−2y 24. F (x, sin 4x − z; ∇F = 32e−2y cos 4xi − 16e−2y sin 4xj − k; ∇F (π/24, 0, 4) = √ y, z) = 8e 16 3i − 8j − k. The equation of the tangent plane is √ √ √ 2 3π − 4. 16 3(x − π/24) − 8(y − 0) − (z − 4) = 0 or 16 3x − 8y − z = 3

25. The gradient of F (x, y, z) = x2 + y 2 + z 2 is ∇F = 2xi + 2yj + 2zk, so the normal vector to the surface at (x0 , y0 , z0 ) is 2x0 i + 2y0 j + 2z0 k. A normal vector to the plane 2x + 4y + 6z = 1 is 2i + 4j + 6k. Since we want the tangent plane to be parallel to the given plane, we find c so that 2x0 = 2c, 2y0 = 4c, 2z0 = 6c or x0 = c, y0 = √ 2c, z0 = 3c. Now, (x0 , y0 , z0 ) is on 2 2 2 2 2. Thus, the points on the surface the surface, so c + (2c) + (3c) = 14c = 7 and c = ±1/ √ √ √ √ √ √ are ( 2/2, 2, 3 2/2) and − 2/2, − 2, −3 2/2). 26. The gradient of F (x, y, z) = x2 − 2y 2 − 3z 2 is ∇F (x, y, z) = 2xi − 4yj − 6zk, so a normal vector to the surface at (x0 , y0 , z0 ) is ∇F (x0 , y0 , z0 ) = 2x0 i − 4y0 j − 6z0 k. A normal vector to the plane 8x+4y+6z = 5 is 8i+4j+6k. Since we want the tangent plane to be parallel to the given plane, we find c so that 2x0 = 8c, −4y0 = 4c, −6z0 = 6c or x0 = 4c, y0 = −c, √z0 = −c. 2 2 2 Now (x0 , y0 , z0 ) is on the surface, 2(−c)2 − 3(−c) √so (4c) √ −√ √ = √ 11c √ = 33 and c = ± 3. Thus, the points on the surface are (4 3, − 3, − 3) and (−4 3, 3, 3). 27. The gradient of F (x, y, z) = x2 +4x+y 2 +z 2 −2z is ∇F = (2x+4)i+2yj+(2z−2)k, so a normal to the surface at (x0 , y0 , z0 ) is (2x0 + 4)i + 2y0 j + (2z0 − 2)k. A horizontal plane has normal ck for c 6= 0. Thus, we want 2x0 + 4 = 0, 2y0 = 0, 2z0 − 2 = c or x0 = −2, y0 = 0, z0 = c + 1. Since (x0 , y0 , z0 ) is on the surface, (−2)2 + 4(−2) + (c + 1)2 − 2(c + 1) = c2 − 5 = 11 and c = ±4. The points on the surface are (−2, 0, 5) and (−2, 0, −3). 28. The gradient of F (x, y, z) = x2 + 3y 2 + 4z 2 − 2xy is ∇F = (2x − 2y)i + (6y − 2x)j + 8zk, so a normal to the surface at (x0 , y0 , z0 ) is 2(x0 − y0 )i + 2(3y0 − x0 )j + 8z0 k. (a) A normal to the xz plane is cj for c 6= 0. Thus, we want 2(x0 − y0 ) = 0, 2(3y0 − x0 ) = c, 8z0 = 0 or x0 = y0 , 3y0 −x0 = c/2, z0 = 0. Solving the first two equations, we obtain 2 x0 = y0 = c/4. Since (x0 , y0√ , z0 ) is on the surface, (c/4)2 +3(c/4)2 +4(0)√ −2(c/4)(c/4) = √ 2 2c /16 = 16 and c = ±16/ 2. Thus, the points on the surface are (4/ 2, 4/ 2, 0) and √ √ (−4 2, −4 2, 0). (b) A normal to the yz-plane is ci for c 6= 0. Thus, we want 2(x0 − y0 ) = c, 2(3y0 − x0 ) = 0, 8z0 = 0 or x0 − y0 = c/2, x0 = 3y0 , z0 = 0. Solving the first two equations, we obtain x0 = 3c/4 and y0 = c/4. Since (x0 , y0 , z0 )√is on the surface, (3c/4)2 + 3(c/4)2 + 4(0)2√− 2(3c/4)(c/4) = 6c2 /16 √ √ = 16 and √ c = ±16 6. Thus, the points on the surface are (12/ 6, 4/ 6, 0) and (−12/ 6, −4/ 6, 0). (c) A normal to the xy-plane is ckfor c 6= 0. Thus, we want 2(x0 − y0 ) = 0, 2(3y0 − x0 ) = 0, 8z0 = c or x0 = y0 , 3y0 −x0 = 0, z0 = c/8. Solving the first two equations, we obtain x0 = y0 = 0. Since (x0 , y0 , z0 ) is on the surface, 02 +3(0)2 +4(c/8)2 −2(0)(0) = c2 /16 = 16 and c = ±16. Thus, the points on the surface are (0, 0, 2) and (0, 0, −2).

29. If (x0 , y0 , z0 ) is on x2 /a2 + y 2 /b2 + z 2 /c2 = 1, then x20 /a2 + y02 /b2 + z02 /c2 = 1 and x0 , y0 , z0 ) is on the plane xx0 /a2 + yy0 /b2 + zz0 /c2 = 1. A normal to the surface at (x0 , y0 , z0 ) is

13.7. TANGENT PLANES AND NORMAL LINES

117

∇F (x0 , y0 , z0 ) = (2x − 0/a2 )i + (2y0 /b2 )j + (2z0 /c2 )k. A normal to the plane is (x0 /a2 )i + (y0 /b2 )j + (z0 /c2 )k. Since the normal to the surface is a multiple of the normal to the plane, the normal vectors are parallel and the plane is tangent to the surface. 30. If (x0 , y0 , z0 ) is on x2 /a2 − y 2 /b2 + z 2 /c2 = 1, then x20 /b2 − y02 /b2 + z02 /c2 = 1 and (x0 , y0 , z0 ) is on the plane xx0 /a2 − yy0 /b2 + zz0 /c2 = 1. A normal to the surface at (x0 , y0 , z0 ) is ∇F (x0 , y0 , z0 ) = (2x0 /a2 )i − (2y0 /b2 )j + (2z0 /c2 )k. A normal to the plane is (x0 /a2 )i − (y0 /b2 )j + (z0 /c2 )k. Since the normal to the surface is a multiple of the normal to the plane, the normal vectors are parallel, and the plane is tangent to the surface. 31. F (x, y, z) = x2 + 2y 2 + z 2 ; ∇F = 2xi + 4yj + 2zk; ∇F (1, −1, 1) = 2i − 4j + 2k. Parametric equations of the line are x = 1 + 2t, y = −1 − 4t, z = 1 + 2t. 32. F (x, y, z) = 2x2 − 4y 2 − z; ∇F = 4xi − 8yj − k; ∇F (3, −2, 2) = 12i + 16j − k. Parametric equations of the line are x = 3 + 12t; y = −2 + 16t, z = 2 − t. 33. F (x, y, z) = 4x2 + 9y 2 − z; ∇F = 8xi + 18yj − k; ∇F (1/2, 1/3, 3) = 4i + 6j − k. Symmetric y − 1/3 z−3 x − 1/2 = = . equations of the line are 4 6 −1 34. F (x, y, z) = x2 + y 2 − z 2 ; ∇F = 2xi + 2yj − 2zk; ∇F (3, 4, 5) = 6i + 8j − 10k. Symmetric x−3 y−4 z−5 equations of the line are = = . 6 8 −10 35. Let F (x, y, z) = x2 + y 2 − z 2 . Then ∇F = 2xi + 2yj − 2zk and a normal to the surface at (x0 , y0 , z0 ) is x0 i + y0 j − z0 k. An equation of the tangent plane at (x0 , y0 , z0 ) is x0 (x − x0 ) + y0 (y − y0 ) − z0 (z − z0 ) = 0 or x0 x + y0 y − z0 z = x20 + y02 − z02 . Since (x0 , y0 , z0 ) is on the surface, z02 = x20 + y02 and x20 + y02 − z02 = 0. Thus, the equation of the tangent plane is x0 x + y0 y − z0 z = 0, which passes through the origin. 1 1 1 √ √ x+ y + z. Then ∇F = √ i+ √ j+ √ k and a normal to the surface 2 x 2 y 2 z 1 1 1 at (x0 , y0 , z0 ) is √ i + √ j + √ k. An equation of the tangent plane at (x0 , y0 , z0 ) is 2 x0 2 y0 2 z0 1 1 1 1 1 1 √ √ √ √ (x−x0 )+ √ (y−y0 )+ √ (z−z0 ) = 0 or √ x+ √ y+ √ z = x0 + y0 + z0 = 2 x0 2 y0 2 z0 x0 y0 z0 √ √ √ √ √ √ √ √ √ √ √ √a. The √ sum of the intercepts is x0 a + y0 a + z0 a = ( x0 + y0 + z0 ) a = a · a = a.

36. Let F (x, y, z) =

37. A normal to the surface at (x0 , y0 , z0 ) is ∇F (x0 , y0 , z0 ) = 2x0 i + 2y0 j + 2z0 k. Parametric equations of the normal line are x = x0 + 2x0 t, y = y0 + 2y0 t, z = z0 + 2z0 t. Letting t = −1/2, we see that the normal line passes through the origin. 38. The normal lines to F (x, y, z) = 0 and G(x, y, z) = 0 are Fx i+Fy j+Fz k and Gx i+Gy j+Gz k, respectively. These vectors are orthogonal if and only if their dot product is 0. Thus, the surfaces are orthogonal at P if and only if Fx Gx + Fy Gy + Fz Gz = 0. 39. We have F (x, y, z) = x2 + y 2 + z 2 and G(x, y, z) = x2 + y 2 − z 2 . ∇F = h2x, 2y, 2zi = 6 0 except at the origin

118

CHAPTER 13. PARTIAL DERIVATIVES ∇G = h2x, 2y, −2zi = 6 0 except at the origin Therefore, the gradient vectors are nonzero at each of the intersection points. Now Fx Gx + Fy Gy + Fx Gz = (2x)(2x) + (2y)(2y) + (2z)(−2z) = 4x2 + 4y 2 − 4z 2 = 4(x2 + y 2 + z 2 ) = 4(0) = 0 The second to last equality follows from the fact that the intersection points lie on both surfaces and hence satisfy the second equation x2 + y 2 − z 2 = 0.

40. Let F (x, y, z) = x2 − y 2 + z 2 − 4 and G(x, y, z) = 1/xy 2 − z. Then Fx Gx + Fy Gy + Fz Gz = (2x)(−1/x2 y 2 ) + (−2y)(−2/xy 3 ) + (2z)(−1) = −2/xy 2 + 4/xy 2 − 2z = 2(1/xy 2 − z). For (x, y, z) on both surfaces, F (x, y, z) = G(x, y, z) = 0. Thus, Fx Gx + Fy Gy + Fz Gz = 2(0) and the surfaces are orthogonal at points of intersection.

13.8

Extrema of Multivariable Functions

1. fx = 2x; fxx = 2; fxy = 0; fy = 2y; fyy = 2; D = 4. Solving fx = 0 and fy = 0, we obtain the critical point (0, 0). Since D(0, 0) = 4 > 0 and fxx (0, 0) = 2 > 0, f (0, 0) = 5 is a relative minimum. 2. fx = 8x; fxx = 8; fxy = 0; fy = 16y; fyy = 16; D = 128. Solving fx = 0 and fy = 0, we obtain the critical point (0, 0). Since D(0, 0) = 128 > 0 and fxx (0, 0) = 8 > 0, f (0, 0) = 0 is a relative minimum. 3. fx = −2x + 8; fxx = −2; f xy = 0; fy = −2y + 6; fyy = −2; D = 4. Solving fx = 0 and fy = 0 we obtain the critical point (4, 3). Since D(4, 3) = 4 > 0 and fxx (4, 3) = −2 < 0, f (4, 3, ) = 25 is a relative maximum. 4. fx = 6x − 6; f xx = 6; f xy = 0; fy = 4y + 8; fyy = 4; D = 24. Solving fx = 0 and fy = 0, we obtain the critical point (1, −2). Since D(1, −2) = 24 > 0 and fxx (1, −2) = 6 > 0, f (1, −2) = −11 is a relative minimum. 5. fx = 10x + 20; fxx = 10; fxy = 0; fy = 10y − 10; fyy = 10; D = 100. Solving fx = 0 and fy = 0, we obtain the critical point (−2, 1). Since D(−2, 1) = 100 > 0 and fxx (−2, 1) = 10 > 0, f (−2, 1) = 15 is a relative minimum. 6. fx = −8x − 8; fxx = −8; fxy = 0; fy = −4y + 12; fyy = −4; D = 32. Solving fx = 0 and fy = 0, we obtain the critical point (−1, 3). Since D(−1, 3) = 32 > 0 and fxx (−1, 3) = −8 < 0, f (−1, 3) = 27 is a relative maximum. 7. fx = 12x2 − 12; fxx = 24x; fxy = 0; fy = 3y 2 − 3; fyy = 6y; D = 144xy. Solving fx = 0 and fy = 0, we obtain the critical points (−1, −1), (−1, 1), (1, −1), and (1, 1). Since D(−1, 1) = −144 < 0 and D(1, −1) = −144 < 0, these points do not give relative extrema. Since D(−1, −1) = 144 > 0 and fxx (−1, −1) = −24 < 0, f (−1, −1) = 10 is a

13.8. EXTREMA OF MULTIVARIABLE FUNCTIONS

119

relative maximum. Since D(1, 1) = 144 > 0 and fxx (1, 1) = 24 > 0, relative minimum.

f (1, 1) = −10 is a

8. fx = −3x2 + 27; fxx = −6x; fxy = 0; fy = 6y 2 − 24; fyy = 12y; D = −72xy. Solving fx = 0, fy = 0, we obtain the critical points (−3, −2), (−3, 2), (3, −2), and (3, 2). Since D(−3, −2) = −432 < 0 and D(3, 2) = −432 < 0, these points do no give relative extrema. Since D(−3, 2) = 432 > 0 and fxx (−3, 2) = 18 > 0, f (−3, 2) = 432 > 0 and fxx (3, −2) = −18 < 0, f (3, −2) = 89 is a relative maximum. 9. fx = 4x − 2y − 10; fxx = 4; fxy = −2; fy = 8y − 2x − 2; fyy = 8; D = 32 − (−2)2 = 28. Setting fx = 0 and fy = 0, we obtain 4x − 2y = 10 and 8y − 2x = 2 or 2x − y = 5 and 4y − x = 1. Solving, we obtain the critical point (3, 1). Since D(3, 1) = 28 > 0 and fxx (3, 1) = 4 > 0, f (3, 1) = −14 is a relative minimum. 10. fx = 10x + 5y − 10; fxx = 10; fxy = 5; fy = 10y + 5x − 5; fyy = 10; D = 100 − (5)2 = 75. Setting fx = 0 and fy = 0, we obtain 10x + 5y = 10 and 10y + 5x = 5 or 2x + y = 2 and 2y + x = 1. Solving, we obtain the critical point (1, 0). Since D(1, 0) = 75 > 0 and fxx (1, 0) = 10 > 0, f (1, 0) = 13 is a relative minimum. 11. fx = 2t − 8; fxx = 0; fxy = 2; fy = 2x − 5; fyy = 0; D(x, y) = −4 < 0 for all (x, y), there are no relative extrema.

D = 0 − 22 = −4. Since

12. fx = 2y + 6; fxx = 0; fxy = 2; fy = 2x + 10; fyy = 0; D(x, y) = −4 < 0 for all (x, y), there are no relative extrema.

D = 0 − 22 = −4. Since

13. fx = −6x2 + 6y; fxx = −12x; fxy = 6; fy = −6y 2 + 6x; fyy = −12y; D = 144xy − 36. Setting fx = 0 and fy = 0, we obtain −6x2 + 6y = 0 and −6y 2 + 6x = 0 or y = x2 and x = y 2 . Substituting x = y 2 into y = x2 , we obtain y = y 4 or y(y 3 − 1) = 0. Thus, y = 0 and y = 1. The critical points are (0, 0) and (1, 1). Since D(0, 0) = −36 < 0, (0, 0) does not give a relative extremum. Since D(1, 1) = 108 > 0 and fxx (1, 1) = −12 < 0, f (1, 1) = 12 is a relative maximum. 14. fx = 3x2 − 6y; fxx = 6x; fx y = −6; fy = 3y 2 − 6x; fyy = 6y; D = 36xy − 36. Setting fx = 0 and fy = 0, we obtain 3x2 − 6y = 0 and 3y 2 − 6x = 0 or x2 = 2y and y 2 = 2x. Substituting y = x2 /2 into y 2 = 2x we obtain x4 = 8x or x(x3 − 8) = 0. Thus, x = 0 and x = 2. The critical points (0, 0) and (2, 2). Since D(0, 0) = −36 < 0, f (0, 0) is not an extremum. Since D(2, 2) = 108 > 0 and fxx (2, 2) = 12 > 0, f (2, 2) = 19 is a relative minimum. 15. fx = y + 2/x2 ; fxx = −4/x3 ; fxy = 1; fy = x + 4/y 2 ; fyy = −8/y 3 ; D = 32/x3 Y 3 − 1. Setting fx = 0 and fy = 0 we obtain y + 2/x2 = 0 and x + 4/y 2 = 0. Substituting y = −2/x2 into x + 4/y 2 = 0 we obtain x + x4 = x(1 + x3 ) = 0. Since x = 0 is not in the domain of f, the only critical point is (−1, −2). Since D(−1, −2) = 3 > 0 and fxx (−1, −2) = 4 > 0, f (−1, −2) = 14 is a relative minimum. 16. fx = −6xy − 3y 2 + 36y; fxx = −6y; fxy = −6x − 6y + 36 = 6(6 − x − y); fy = −3x2 − 6xy + 36x; fyy = −6x; D = 36xy − 36(6 − x − y)2 . Setting fx = 0 and fy = 0 we obtain −6xy − 3y 2 + 36y = 0 and −3x2 − 6xy + 36x = 0 or −3y(2x + y − 12) = 0 and −3x(x + 2y − 12) = 0. Letting y = 0, the first equation is satisfied and the second

120

CHAPTER 13. PARTIAL DERIVATIVES equation becomes −3x(x − 12) = 0. Thus, (0, 0) and (12, 0) are critical points. Similarly, letting x = 0 we obtain the critical point (0, 12). Finally solving 2x + y = 12 and x + 2y = 12 we obtain the critical point (4, 4). Since D(0, 0) = −362 < 0, D(0, 12) = −362 < 0, and D(12, 0) = −362 < 0, none of these points give relative extrema. Since D(4, 4) = 432 > 0 and fxx (4, 4) = −24 < 0, f (4, 4) = 192 is a relative maximum.

17. fx = (xex + ex ) sin y; fxx = (xex + 2ex ) sin y; fxy = (xex + ex ) cos y; fy = xex cos y; fyy = −xex sin y; D = −xe2x (x + 2) sin2 y − e2x (x + 1)2 cos2 y. Setting fx (x, y) = 0 and fy (x, y) = 0 we obtain (xex + ex ) sin y = 0 and xex cos y = 0. Since ex > 0 for all x, we have (x + 1) sin y = 0 and x cos y = 0. When x = −1, we must have cos y = 0 or y = π/2 + kπ, k an integer. When x = 0, we must have sin y = 0 or y = kπ, k an integer. Thus, the critical points are (0, kπ) and (−1, π/2 + kπ), k an integer. Since D(0, kπ) = 0−cos2 kπ < 0, (0, kπ) does not give a relative extrema. Now, D(−1, π/2+kπ) = e−2 sin2 (π/2+kπ)−0 > 0 and fxx (−1, π/2+kπ) = e−1 sin(π/2+kπ). Since fxx (−1, π/2+kπ) is positive for k even and negative for k odd, f (−1, π/2 + kπ) = −e−1 are relative minima for k even, and f (−1, π/2 + kπ) = e−1 are relative maxima for k odd. 2

2

2

2

18. fx = (2x + 4)ey −3y+x +4x ; fxx = [(2x + 4)2 + 2]ey −3y+x +4x ; 2 2 2 2 fxy = (2x + 4)(2y − 3)ey −3y+x +4x ; fy = (2y − 3)ey −3y+x +4x ; 2 2 2 2 fyy = [(2y − 3)2 + 2]ey −3y+x +4x ; D = [(2x + 4)2 + 2][(2y − 3)2 + 2] · e2(y −3y+x +4x) − [(2x + 2 2 4)(2y − 3)]2 e2(y −3y+x +4x) . Setting fx = 0 and fy = 0 and using the fact that an exponential function is always positive, we obtain 2x + 4 = 0 and 2y − 3 = 0. Thus, a critical point is (−2, 3/2). Since D(−2, 3/2) = 4e2(9/4−9/2+4−8) > 0 and fxx (−2, 3/2) = 2e9/4−9/2+4−8 > 0, f (−2, 3/2) = e9/4−9/2+4−8 = e−25/4 is a relative minimum. 19. fx = cos x; fxx = − sin x; fxy = 0; fy = cos y; fyy = − sin y; D = sin x sin y. Solving fx = 0 and fy = 0, we obtain the critical points (π/2 + mπ, π/2 + nπ) for m and n integers. For m even and n odd or m odd and n even, D < 0 and no relative extrema result. For m and n both even, D > 0 and fxx < 0 and f (π/2 + mπ, π/2 + nπ) = 2 are relative maxima. For m and n both odd, D > 0 and f xx > 0 and f (π/2 + mπ, π/2 + nπ) = −2 are relative minima. 20. fx = y cos xy; fxx = −y 2 sin xy; fxy = −xy sin xy + cos xy; fy = x cos xy; fyy = −x2 sin xy; D = x2 y 2 sin2 xy − (−xy sin xy + cos xy)2 = 2xy sin xy cos xy − cos2 xy. Setting fx = 0 and fy = 0 we see that (0, 0) is a critical point. Also, solving cos xy = 0 we obtain xy = π/2 + kπ or y = π(1 + 2k)/2x for k an integer. Since D(0, 0) = −1 < 0, (0, 0) does not give a relative extrema. For any of the critical points (x, π(1 + 2k)/2x), D = 0 and no conclusion can be drawn from the second partials test. Since −1 ≤ sin xy ≤ 1 for all (x, y)f (x, π(1 + 2k)/2x) = −1 for k odd are relative minima and f (x, π(1 + 2k)/2x) = 1 for k even are relative maxima. 21. Let the numbers be x, y, and 21 − x − y. We want to maximize P (x, y) = xy(21 − x − y) = 21xy − x2 y − xy 2 . Now Px = 21y − 2xy − y 2 ; Pxx = −2y; Pxy = 21 − 2x − 2y; Py = 21x − x2 − 2xy; Pyy = −2x; D = 4xy − (21 − 2x − 2y)2 . Setting Px = 0 and Py = 0, we obtain y(21 − 2x − y) = 0 and x(21 − x − 2y) = 0. Letting x = 0 and y = 0, we obtain the critical points (0, 0), (0, 21), and (21, 0). Each of these results in P = 0 which is clearly not a maximum. Solving 21 − 2x − y = 0 and 21 − x − 2y = 0, we obtain the critical point (7, 7).

13.8. EXTREMA OF MULTIVARIABLE FUNCTIONS

121

Since D(7, 7) = 147 > 0 and Pxx (7, 7) = −14 < 0, P (7, 7) = 343 is a maximum. The three numbers are 7,7, and 7. 22. Let the sides of the base of the box by x and y. Then, since the volume of the box is 1, its height is 1/xy and S = 2xy + 2x(1/xy) + 2y(1/xy) = 2xy + 2/y + 2/x, x > 0, y > 0. Now Sx = 2y − 2/x2 ; Sxx = 4/x3 ; Sxy = 2; Sy = 2x − 2/y 2 ; Syy = 4/y 3 ; D = 16/x3 y 3 − 4. Setting Sx = 0 and Sy = 0 we obtain y = 1/x2 and x = 1/y 2 . The critical point is (1, 1). Since D(1, 1) = 12 > 0 and Sxx (1, 1) = 4 > 0, S(1, 1) = 6 is a minimum. The box is 1 foot on each side. 23. Let (x, y, 1 − x − 2y) be a point on the plane x + 2y + z = 1. We want to minimize f (x, y) = x2 + y 2 + (1 − x − 2y)2 . Now fx = 2x − 2(1 − x − 2y); fxx = 4; fxy = 4; fy = 2y − 4(1 − x − 2y); fyy = 10; D = 40 − 42 = 24. Setting fx = 0 and fy = 0 we obtain 2x−2(1−x−2y) = 0 and 2y−4(1−x−2y) = 0 or 2x+2y = 1 and 2x+5y = 2. Thus, (1/6, 1/3) is a critical point. Since D = 24 > 0 and fxx = 4 > 0 for all (x, y), f (1/6, 1/3) = 1/6 is a minimum. Thus, the point on the plane closest to the origin is (1/6, 1/3, 1/6). 24. Let (x, y, 1 − x − y) be a point on the plane x + y + z = 1. We want to minimize the square of the distance between the point and the plane. This is given by f (x, y) = (x − 2)2 + (y − 3)2 + (−x − y)2 = 2x2 + 2y 2 − 4x − 6y + 2xy + 13. fx = 4x − 4 + 2y; fxx = 4; fxy = 2; fy = 4y − 6 + 2x; fyy = 4; D = 16 − 22 = 12. Setting fx = 0 and fy = 0 we obtain 4x − 4 + 2y = 0 and 4y − 6 + 2x = 0 or 2x + y = 2 and x + 2y = 3. Thus, (1/3, 4/3) is a critical point. Since D = 12 > 0 and fxx = 4 > 0 for all (x, y), fp (1/3, 4/3) =√25/3 is a minimum. Thus, the least distance between the point and the plane is 25/3 = 5/ 3. 25. Let (x, y, 8/xy) be a point on the surface. We want to minimize the square of the distance to the origin or f (x, y) = x2 + y 2 + 64/x2 y 2 . Now fx = 2x − 128/x3 y 2 ; fxx = 2 + 384/x4 y 2 ; fxy = 256/x3 y 3 ; fy = 2y − 128/x2 y 3 ; fyy = 2 + 384/x2 y 4 ; D = (2 + 384/x4 y 2 )(2 + 384/x2 y 4 ) − (256)2 /x6 y 6 . Setting fx = 0 and fy = 0 we obtain 2x − 128/x3 y 2 = 0 and 2y − 128/x2 y 3 = 0 or x4 y 2 = 64 and x2 y 4 = 64; x 6= 0, y 6= 0. This gives x4 y 2 = x2 y 4 , x2 y 2 (x2 − y 2 ) = 0 or x2 = y 2 . Thus, x6 = 64 and x = ±2. Similarly, y = ±2 and the critical points are (−2, −2), (−2, 2), (2, −2), and (2, 2). Since D(±2, ±2) = 48 > 0 and fxx (±2, ±2) = 8 > 0, f (±2, ±2) = 12 are minima. The points closest to√the origin √ are (−2, −2, 2), (−2, 2, −2), (2, −2, −2), and (2, 2, 2). The minimum distance is 12 = 2 3. 26. We will minimize the square of the distance between the lines. This is given by f (s, t) = [(3 + 2s) − t]2 + [(6 + 2s) − (4 − 2t)]2 + [(8 − 2s) − (1 + t)]2 = (2s − t + 3)2 + (2s + 2t + 2)2 + (−2s − t − 7)2 = 12s2 + 6t2 + 8st − 8s + 12t + 62. fs = 24s+8t−8; fss 24; fst = 8; ft = 12t+8s−12; ftt = 12; D = 24(12)−64 = 224. Solving 24s+8t−8 = 0 and 12t+8s−12 = 0 we obtain the critical point (0, 1). Since D(0, 1) = 224 > 0 and fss (0, 1) = 24 > 0, we see that f (0, 1) is a minimum. The corresponding points on√the p √ lines are (3, 6, 8) on L2 and (1, 2, 2) on L1 . The minimum distance is f (0, 1) = 56 = 2 14.

122

CHAPTER 13. PARTIAL DERIVATIVES

27. We will maximize the square of the volume of the box in the first octant, V (x, y) = x2 Y 2 z 2 = x2 y 2 (c2 − c2 x2 /a2 − c2 y 2 /b2 ). Vx = 2c2 xy 2 −4c2 x3 y 2 /a2 −2c2 xy 4 /b2 ; Vxx = 2c2 y 2 −12c2 x2 y 2 /a2 −2c2 y 4 /b2 ; Vxy = 4c2 xy− 8c2 x3 y/a2 − 8c2 xy 3 /b2 ; Vy = 2c2 x2 y − 2c2 x4 /a2 − 4c2 x2 y 3 /b2 ; Vyy = 2c2 x2 − 2c2 x4 /a2 − 2 12c2 x2 y 2 /b2 ; D = Vxx Vyy − Vxy . Setting Vx = 0 and Vy = 0 we obtain xy 2 − 2x3 y 2 /a2 − 4 2 2 4 2 2 3 2 2 2 2 2 2 2 xy /b = 0, x y−x y/a −2x y /b = 0, or, assuming x √ y > 0, 2b x +a y = a b . √> 0 and 2 2 2 2 Solving, we obtain x = a /3 and y = b /3. Thus, (a/ 3, b/ 3) is a critical point. Since √ √ 14 4 20 2 2 4 14 D(a/ 3, b/ 3) = (− b2 c2 )(− a2 c2 ) − (− abc2 )2 = a b c >0 9 9 9 9 and vxx = −

√ √ 14 2 2 b c < 0, V (a/ 3, b/ 3) = a2 b2 c2 /27. The maximum volume is 9 q √ √ √ 8 V (a/ 3, b/ 3) = 8 3abc/9.

28. Let a + b + c = k. Then c = k − a − b and we want to maximize V (a, b) = 4πab(k − a − b)/3 = 4π(kab − a2 b − ab2 )/3. 4π 8π 4π 4π (kb − 2ab − b2 ); Vaa = − ; Vab = (k − 2a − 2b); Vb = (ka − a2 − 2ab); 3 3 3 3 8π 64x2 16x2 Vbb = − a; D = ab − (k − 2a − 2b)2 . Setting Va = 0 and Vb = 0 we obtain 3 9 9 kb − 2ab − b2 = 0 or ka − a2 − 2ab = 0, a 6= 0, b 6= 0, or 2a + b = k and a + 2b = k. Solving, we get a = b = k/3. Since D(k/3, k/3) = 16π 2 k 2 /27 > 0 and Vaa (k/3, k/3) = −8πk/9 < 0, the volume is maximized when a = b = k/3. Since c = k − a − b = k/3, a = b = c and the ellipsoid is a sphere.

Va =

29. The perimeter is given by P = 2x + 2y + 2x sec θ and the area is a = 2xy + x2 tan θ. Solving P for 2y and substituting in A, we obtain A = P x − 2x2 (1 + sec θ) + x2 tan θ. Now Ax (x, θ) = P −4x(1+sec θ)2x tan θ; Axx (xθ) = −4(1+sec θ)+2 tan θ; Axθ (x, θ) = −4x sec θ tan θ + 2x sec2 θ; Aθ (xθ) = x2 sec θ(sec θ − 2 tan θ); Aθθ (x, θ) = 2x2 sec θ(tan θ − 2 sec2 θ + 1). We assume that x > 0 and 0 ≤ θ ≤ π/2.

x tanθ

x secθ

θ x

x

y

Setting Ax = 0 and Aθ = 0, we obtain P −4x(1+sec θ)+2x tan θ = 0 and x2 sec θ(sec θ−2 tan θ) = 0. We note from the second equation and the fact that sec θ 6= 0 for all θ that sec θ − 2√ tan θ = 0. Solving for θ, we obtain θ = 30◦ and solving Ax = 0 for x, we obtain x) = P/(4 + 2 3). Since √ √ √ √ D(x0 , 30◦ ) = (−2 3 + 2)(4x20 ( √3 − 5)/3 3) − 02 > 9 and Axx = 2 − 2 3 < 0, A(x0 , 30◦ ) √ is a maximum. Letting x = P/(4+2 3) and θ = 30√◦ in P = 2x+2y+2x sec θ, we obtain P = 2y+P ? 3. √ √ Thus, the area is maximized for x = P/(4 + 2 3), y = P ( 3 − 1)/2 3, and θ = 30◦

13.8. EXTREMA OF MULTIVARIABLE FUNCTIONS (x sin θ)(24 − 1 2x + x cos θ) = 24x sin θ − 2x sin θ + x2 sin 2θ. 2 Now Ax = 24 sin θ − 4x sin θ + x sin 2θ; Axx = −4 sin θ + sin 2θ; Axθ = 24 cos θ − 4x cos θ + 2x cos 2θ; Aθ = 24x cos θ − 2x2 cos θ + x2 cos 2θ; Aθθ = −24x sin θ + 2x2 sin θ − 2x2 sin 2θ.

30. We want to maximize A(x, θ)

123

=

2

x θ 24-2x

x sinθ

θ x cosθ

We assume 0 < x < 12 and =< θ < π/2. Setting Ax = 0 and Aθ = 0 we obtain 24 sin θ − 4x sin θ + 2x sin θ cos θ = 0 and 24x cos θ − 2x2 cos θ + x2 (2 cos2 θ − 1) = 0 or 12 − 2x + x cos θ = 0 and 2x cos2 θ − 2x cos θ + 2 cos θ − x = 0. Solving the first equation for cos θ and substituting into the second equation, we obtain 2x(2 − 12/x)2 − 2x(12 − 12/x) + 24(2 − x = 0. Simplifying, we √ − 12/x)√ find x = 8 and√cos θ = 1/2 or θ = 60◦ .√Since D(8, 60◦ ) = (−3 3/2)(−96 3) − (−12)2 = 288 > 0 and Axx = −3 3/2 < 0, A(8, 60◦ = 48 3) square inches is the maximum area. 2 2 31. fx = − x−1/3 , fy = − y −1/3 . Since fx = 0 and fy = 0 have no solutions, f (x, y) has no 3 3 critical points and Theorem 13.8.2 does not apply. However, for all (x, y), f (0, 0) = 16 ≥ 16 − (x1/3 )2 − (y 1/3 )2 = f (x, y), and f (0, 0) = 16 is an absolute maximum. 32. fx = −4x3 y 2 ; fxx = −12x2 y 2 ; fxy = −8x3 y; fy = −2x4 y; fyy = −2x4 ; D = 24x6 y 2 − 64x6 y 2 = −40x6 y 2 . Setting fx = 0 and fy = 0 we see that (0, y) and (x, 0) are critical points for any x and y. Since, for any critical point, D = 0, Theorem 13.8.2 does not apply. However, for all (x, y), f (0, 0) = 1 ≥ 1 − (x2 y)2 = f (x, y), and f (0, 0) = 1 is an absolute maximum. 33. fx = 10x; fxx = 10; fxy = 0; fy = 4y 3 ; fyy = 12y 2 ; D = 120y 2 . Solving fx = 0 and fy = 0 we obtain the critical point (0, 0). Since D(0, 0) = 0, Theorem 13.8.2 does not apply. However, for any (x, y), f (0, 0) = −8 ≤ 5x2 +y 4 −8 = f (x, y) and f (0, 0) = −8 is an absolute minimum. x y2 xy y 34. fx = p ; fxx = 2 ; ; fxy = − 2 ; fy = p (x + y 2 )3/2 (x + y 2 )3/2 x2 + y 2 x2 + y 2 x2 −xy x2 y 2 fyy = −( 2 ; D = )2 = 0. Since D = 0 for all (x, y), 2 2 3 2 2 3/2 (x + y ) (x + y ) (x + y 2 )3/2 p Theorem 13.8.2 does not apply. However, for all (x, y), f (0, 0) = 0 ≤ x2 + y 2 = f (x, y), so f (0, 0) = 0 is an absolute minimum.

In Problems 35-38 we parameterize the boundary of R by letting x = cos t and y = sin t; 0 ≤ t ≤ 2π. Then, for (x(t), y(t)) on the boundary, we maximize or minimize F (t) = f (cos t, sin t) on [0, 2π]. √ 35. fx = 1; fy = 3. There √ are no critical points on the interior √ of R. On the boundary we consider F (t) = cos t + 3 sin t. Solving F 0 (t) = − sin t + 3 cos t = 0, we obtain critical points at t = π/3 2, and F (4π/3) = −2, we √ and t = 4π/3. Comparing F (0) = 1, F (π/3) = √ see that f (1/2, 3/2) = 2 is an absolute maximum and f (−1/2, − 3/2) = −2 is an absolute minimum.

124

CHAPTER 13. PARTIAL DERIVATIVES

36. fx = y; fy = x. Solving fx = 0 and fy = 0 we obtain the critical point (0, 0) with corre1 sponding function value f (0, 0) = 0. On the boundary we consider F (t) = cos t sin t = sin 2t. 2 Solving F 0 (t) = cos 2t = 0, we obtain critical points at π/4, 3π/4, 5π/4, and 7π/4. Comparing f (0, 0, ) = 0, F (0) = 0, F√ (π/4) √= 1/2, F (3π/4) =√−1/2, F (5π/4) = 1/2, √ 2/2, 2/2) = f (− 2/2, − 2/2) = 1/2 are absolute and F (7π/4) = −1/2, we see that f ( √ √ √ √ maxima and f (− 2/2, 2/2) = f ( 2/2, − 2/2) = −1/2 are absolute minima. 37. fx = 2x + y; fy = x + 2y. Solving fx = 0 and fy = 0 we obtain the critical point (0, 0) with corresponding function value f (0, 0) = 0. On the boundary we consider F (t) = 1 cos2 t + cos t sin t + sin2 t = 1 + sin 2t. Solving F 0 (t) = cos 2t = 0 we obtain critical points at 2 π/4, 3π/4, 5π/4, and 7π/4. Comparing f (0, 0) = 0, F (0) = 1, F (π/4) = 3/2; √ √ F (3π/4) =√ 1/2, F (5π/4) = 3/2, and F (7π/4) = 1/2, we see that f ( 2/2, w/2) = √ f (− 2/2, − 2/2) = 3/2 are absolute maxima and f (0, 0) = 0 is an absolute minimum. 38. fx = −2x; fy = −6y + 4. Solving fx = 0 and fy = 0, we obtain the critical point (0, 2/3), which is inside R, with corresponding function value f (0, 2/3) = 5. On the boundary we consider F (t) = − cos2 t − 3 sin2 t + 4 sin t + 1. Solving F 0 (t) = 2 cos t sin t − 6 sin t cos t + 4 cos t = 4 cos t − 4 sin t cos t = 0, we obtain critical points at π/2 and 3π/2. Comparing f (0, 2/3) = 5, F (0) = 0, F (π/2) = 2, and F (3π/2) = −6, we see that f (0, −1) = −6 is an absolute minimum and f (0, 2/3) = 5 is an absolute maximum. 39. fx = 4; fy = −6. There are no critical points over the region R, so absolute extrema must occur on the boundary. We parameterize the boundary by x = 2 cos t and y = sin t for 0 ≤ t ≤ 2π. Considering F (t) = 8 cos t − 6 sin t we obtain F 0 (t) = −8 sin t − 6 cos t. Solving F 0 (t) = 0 we find tan t = −3/4. Using 1 + tan2 t = sec2 t we see that sec2 t = 25/16 and cos t = −4/5, t is in the second quadrant and sin t = 3/5. The corresponding points on the boundary of R are (8/5, −3/5) and (−8/5, 3/5). Comparing f (0) = F (2π) = f (2, 0) = 8, f (8/5, −3/5) = 10, and f (−8/5, 3/5) = −10 we see that the absolute minimum is f (−8/5, 3/5) = −10 and the absolute maximum is f (8/5, −3/5) = 10. 40. fx = y − 2; fy = x − 1. Solving fx = 0 and fy = 0 we obtain the critical point (1, 2) in the region. On x = 0, F (y) = f (0, y) = −y + 6, which has no critical points for 0 ≤ y ≤ 8. The endpoints of the interval are (0, 0) and (0, 8). On y = 0, G(x) = f (x, 0) = −2x + 6, which has no critical points for 0 ≤ x ≤ 4. The endpoints of the interval are (0, 0) and (4, 0). On y = −2x + 8, H(x) = f (x, −2x + 8) = x(−2x + 8) = x(−2x + 8) − 2x − (−2x + 8) + 6 = −2x2 + 8x − 2. Solving H 0 (x) = −4x + 8 = 0 we obtain x = 2. The corresponding point on the triangle is (2, 4). Comparing f (0, 0) = 6, f (0, 8) = −2; f (4, 0) = −2; f (2, 4) = 6, and f (1, 2) = 4 we see that absolute maxima are f (0, 0) = f (2, 4) = 6 and absolute minima are f (0, 8) = f (4, 0) = −2. 41. (a) fx = y cos xy; fy = x cos xy. Setting fx = 0 and fy = 0 we obtain y cos xy = 0 and x cos xy = 0. If y = 0 from the first equation, then necessarily x = 0 from the second equation. Thus, (0, 0) is a critical point. For x 6= 0 and y 6= 0 we have cos xy = 0 or xy = π/2. Thus, all points (x, π/2x) for 0 ≤ x ≤ π are also critical points. (b) Since 0 ≤ sin xy ≤ 1 for 0 ≤ x ≤ π and 0 ≤ y ≤ 1, f (x, y) = sin xy has absolute minima at any points for which sin xy = 0 and absolute maxima at any points for which

13.9. METHOD OF LEAST SQUARES

125

sin xy = 1. Thus, f (x, y) has absolute minima when xy = 0 or xy = π, that is, at the points (0, y), (x, 0), and (π, 1) which are in the region. Absolute maxima occur when xy = π/2 or along the curve y = π/2x in the region (c) z

1

y

x

42. We want to maximize P (x, y) = R(x, y) − C(x, y) = 108x − 8x2 + 192y − 6y 2 − 4xy − 20. Now Px = 108 − 16x − 4y; Pxx = −16; Pxy = −4; Py = 192 − 4x; Pyy = −12; D = 192 − 16 = 176. Setting Px = 0 and Py = 0 we obtain 108 − 16x − 4y = 0 and 192 − 12y − 4x = 0 or 4x + y = 27 and x + 3y = 48. Solving, we see that (3, 15) is a critical point. Since D(3, 15) = 175 > 0 and Pxx (3, 15) = −16 < 0, P (3, 15) = 1582 is the maximum profit 43. Since the volume of the box is 60, the height is 60/xy. Then C(x, y) = 10xy + 20xy + 2[2x60/xy + 2y60/xy] = 30xy + 240/y + 240/x. Cx = 30y − 240/x2 ; Cxx = 480/x3 , Cxy = 30; Cy = 30x − 240/y 2 ; cyy = 480/y 3 ; D = 4802 /x3 y 3 −900. Setting Cx = 0 and Cy = 0 we obtain 30y −240/x2 = 0 and 30x−240/y 2 = 0 or y = 8/x2 and x = 8/y 2 . Substituting the first equation into the second, we have x = x4 /8 or x(x3 − 8) = 0. Thus, (2, 2) is a critical point. Since D(2, 2) = 2700 > 0 and Cxx (2, 2) = 60 > 0, C(2, 2) is a minimum. Thus, the cost is minimized when the base of the box is 2 feet square and the height is 15 feet.

13.9 1.

4 X

Method of Least Squares xi = 14,

i=1

4 X

yi = 8,

i=1

4 X

xi yi = 30,

i=1

4 X

x2i = 54, m =

4(30) − 14(8) = 0.4, 4(54) − (14)2

x2i = 14, m =

4(34) − 6(14) = 2.6, 4(14) − (6)2

i=1

54(8) − 30(14) b= = 0.6, y = 0.4x + 0.6 4(54) − (14)2 2.

4 X i=1

xi = 6,

4 X i=1

yi = 14,

4 X i=1

xi yi = 34,

4 X i=1

14(14) − 34(6) b= = −0.4, y = 2.6x − 0.4 4(14) − (6)2

126

3.

CHAPTER 13. PARTIAL DERIVATIVES 5 X

5 X

xi = 15,

i=1

yi = 15,

5 X

i=1

5 X

xi yi = 56,

i=1

x2i = 55, m =

5(56) − 15(15) = 1.1, 5(55) − (15)2

x2i = 54, m =

5(55) − 14(14) ≈ 1.06757, 5(54) − (14)2

i=1

55(15) − 56(15) b= = −0.3, y = 1.1x − 0.3 5(55) − (15)2 4.

5 X

4 X

xi = 14,

i=1

yi = 14,

5 X

i=5

5 X

xi yi = 55,

i=1

i=1

54(14) − 55(14) ≈ −0.189189, y ≈ 1.06757x − 0.189189 b= 5(54) − (14)2 5.

7 X

7 X

xi = 21,

i=1

yi = 42,

7 X

i=1

xi yi = 164,

i=1

7 X

x2i = 91, m =

i=1

7(164) − 21(42) ≈ 1.35714, 7(91) − (21)2

91(42) − 164(21) ≈ 1.92857, y ≈ 1.35714x + 1.92857 b= 7(91) − (21)2 6.

7 X

7 X

xi = 28,

i=1

yi = 17.2,

7 X

i=1

xi yi = 80.2,

i=1

7 X

x2i = 140, m =

i=1

7(80.2) − 28(17.2) ≈ 0.407143, 7(140) − (28)2

140(17.2) − 80.2(28) ≈ 0.828571, y ≈ 0.407143x + 0.828571 b= 7(140) − (28)2 7.

6 X

Ti = 420,

6 X

vi = 1055,

Ti vi = 68, 000,

i=1

i=1

i=1

6 X

6 X

Ti2 = 36, 400, m =

i=1

6(68, 000) − 420(1055) ≈ 6(36, 400) − (420)2

36, 400(1055) − 68, 000(420) ≈ 234.333, v ≈ −0.835714T + 234.333. −0.835714, b = 6(36, 400) − (420)2 8.

6 X i=1

Ti = 3150,

6 X

Ri = 29.57,

6 X i=1

i=1

Ti Ri = 17, 878,

6 X

Ti2 = 1, 697, 500, m =

i=1

6(17, 878) − 3150(29.57) ≈ 6(1, 697, 500) − (3150)2

1, 697, 500(29.57) − 17, 878(3150) 0.05, b = ≈ −23.32, R ≈ 0.05T − 23.32 − 23.32. When 6(1, 697, 500) − (3150)2 T = 700, R ≈ 14.34. 9. (a) least-squares line: y = 0.5966x + 4.3665 least-squares quadratic: y = −0.0232x2 + 0.5618x + 4.5942 least-squares cubic: y = 0.00079x3 − 0.0212x2 + 0.5498x + 4.5840 (b)

least-squares line

least-squares cubic

y

y

x

y

x

x

13.10. LAGRANGE MULTIPLIERS

127

10. The least-squares line is given by y = 2.0533x − 3837.115. Plugging 2020 in for x, we predict that the population will be 310.551 million.

13.10

Lagrange Multipliers

1. f has constrained extrema where the level lines intersect the circle.

2. f has constrained extrema where the level curve intersect the line.

y

y

2

1 x

1

x

3. fx = 1; fy = 3; gx = 2x; gy = 2y. We need to solve 1 = 2λx, 3 = 2λy, x2 + y 2 − 1 = 0. Dividing the second equation by the first, we obtain √ 3 = y/x or y =√3x. Substituting √ into 2 2 the third equation, we have x + 9x = 1 or x = ±1/ 10. For x = 1/ 10, y = 3/ √ √ √ √ √ 10 and for x = −1/ 10, y = −3/ 10. A constrained maximum is f (1/ 10, 3/ 10) + 10 and a √ √ √ constrained minimum is f (−1/ 10, −3/ 10) = − 10. 4. fx = y; fy = x; gx = 1/2; gy = 1. We need to solve y = λ/2, x = λ, x/2 + y − 1 = 0. From the first two equations y = x/2. Substituting into the second equation, we have x = 1. Thus, f (1, 1/2) = 1/2 is a constrained extremum. Since (0, 1) satisfies the constraint and f (0, 1) = 0 < 1/2, f (1, 1/2) = 1/2 is a constrained maximum. 5. fx = y; fy = x; gx = 2x; gy = 2y. We need to solve y = 2λx, x = 2λy, x2 + y 2 − 2 = 0. Substituting the second equation into the first, we obtain y = 4λ2 y or y(4λ2 − 1) = 0. If y = 0, then from the second equation x = 0. Since g(0, 0) = −2 6= 0, (0, 0) does not satisfy the constraint. Thus, λ = ±1/2 and y = ±x. Substituting into the third equation, we have 2x2 = 2 or x = ±1. Solutions of the system are x = 1, y = 1, λ = 1/2, x = −1, y = −1; λ = 1/2, x = 1, y = −1, λ = −1/2, and x = −1, y = 1, λ = −1/2. Thus, f (1, 1) = f (−1, −1) = 1 are constrained maxima and f (1, −1) = f (−1, 1) = −1 are constrained minima. 6. fx = 2x; fy = 2y; gx = 2; gy = 1. We need to solve 2x = 2λ, 2y = λ, 2x + y − 5 = 0. Substituting the second equation into the first, we find x = 2y. Substituting into the third equation, we have 4y + y − 5 = 0 or y = 1. A constrained extremum is f (2, 1) = 5. Since (0, 5) satisfies the constraint and f (0, 5) = 25 > 5, f (2, 1) = 5 is a constrained minimum. 7. fx = 6x; fy = 6y; gx = 1; gy = −1. We need to solve 6x = λ, 6y = −λ; x − y − 1 = 0. From the first two equations, we obtain x + y = 0. Solving this with the third equation, we

128

CHAPTER 13. PARTIAL DERIVATIVES obtain x = 1/2, y = −1/2. Thus, f (1/2, −1/2) = 13/2 is a constrained extremum. Since (1, 0) satisfies the constraint and f (1, 0) = 8 > 13/2, f (1/2 − 1/2) = 13/2 is a constrained minimum.

8. fx = 8x; fy = 4y; gx = 8x; gy = 2y. We need to solve 8x = 8λx, 4y = 2λy, 4x2 + y 2 − 4 = 0 or x(λ − 1) = 0, y(λ − 2) = 0, 4x2 + y 2 = 4. If x = 0, then from the third equation y = ±2. If y = 0, then from the third equation x = ±1. The cases λ = 1 and λ = 2 lead also to y = 0 and x = 0, respectively. Thus, f (0, 2) = f (0, −2) = 18 are constrained maxima and f (1, 0) = f (−1, 0) = 14 are constrained minima. 9. fx = 2x; fy = 2y; gx = 4x3 ; gy = 4y 3 . We need to solve 2x = 4λx3 , 2y = 4λy 3 , x4 + y 4 − 1 = 0 or x(2λx2 − 1) = 0, y(2λy 2 − 1) = 0, x4 + y 4 = 1. If x = 0, the from the 2 third equation y = ±1. If y = 0, then x = ±1. From 2λx = 1 = 2λy 2 we have x2 = y 2 . √ √ 4 4 Substituting into the third equation, we obtain x = ±1/ 2 and y = ±1/ 2. Solutions of √ 4 the system are (0, ±1), (±1, 0), and (±1/ 2, ±1/[4]2). Thus, f (0, ±1) = f (±1, 0) = 1 are √ √ √ constrained minima and f (±1/ 4 2, ±1/ 4 2) = 2 are constrained maxima. 10. fx = 16x−8y; fy = 4y −8x; gx = 2x; gy = 2y. We need to solve 16x−8y = 2λx, 4y −8x = 2λy, x2 + y 2 − 10 = 0 or 8 − 47/x = λ, x2 + y 2 = 10. From the first two equations, we obtain 6 − 4y/x = −4x/y, 6(y/x) − 4(y/x)2 = −4, and 2(y/x)2 − 3(y/x) − 2 = 0. Factoring, we have (2y/x + 1)(y/x − 2) = 0. Then y = −x/2 and y = 2xs. √ Substituting y = −x/2 into the third equation, we have x2 + x2 /4 = 10 and x =√±2 2. Substituting 2 2 y = 2x into the we have of the √ third √ equation, √ √ √ x √+ 4x = 10√ and x√ = ± 2. Solutions √ √ system (− 2, √ are √ (2 2, − 2), (−2 2, 2), ( 2, 2 2), and √ √−2 2). Thus, √ f (2√ 2, − 2) = f (−2 2, 2) = 100 are constrained maxima and f ( 2, 2 2) = f (− 2, −2 2) = 0 are constrained minima. √ √ √ 11. fx = 3x2 y; fy = x3 ; gx = 1/2 x; gy = 1/2 y. We need to solve 3x2 y = λ/2 x, x3 = √ √ √ √ √ λ/2 y, x + y − 1 = 0 or 6x5/2 y = λ, 2x3 y 1/2 = λ, x + y = 1. From the first two √ √ equations, we obtain 3x5/2 y = x3 y 1/2 and 3 y = x. Substituting into the third equation, √ √ √ we have 3 y + y = 4 y = 1. Then, y = 1/16 and x = 9/16. Since (1/4, 1/4) satisfies the constraint and f (1/4, 1/4) = 1/256, f (9/16, 1/16) + 729/65, 536 is a constrained maximum. We also consider x = 0, which requires y = 1; and y = 0, which requires x = √ 1. Since x ≥ 0 √ and y ≥ 0, f (0, 1) = f (1, 0) = 0 ≤ x3 y = f (x, y) for all (x, y) which satisfy x + y = 1. Thus, f (0, 1) = 0 and f (1, 0) = 0 are constrained minima. 12. fx = y 2 ; fy = 2xy; gx = 2x; gy = 2y. We need to solve y 2 = 2λx, 2xy = 2λy, x2 + 2 2 y 2 − 27 = 0 or y 2 = 2λx, y(x √ − λ) = 0, x + y = 27. When y = 0 in the third equation, 2 we obtain x = 27 or x = ±3 3, and λ = 0. When x = λ in the second equation, we obtain 2 2 2 y 2 = 2x2 from the first √ equation and √ x + 2x = 3x = 27 from the√third equation.√This gives x = ±3 and y = ±3 2. Since f (±3√ 3, 0) = 0, we√see that f (−3, 3 2) = f (−3, −3 2) = −54 are constrained minima and f (3, 3 2) = f (3, −3 2) = 54 are constrained maxima. 13. Fx = 1; Fy = 2; Fz = 1; gx = 2x; gy = 2y; gz = 2z. We need to solve 1 = 2λx, 2 = 2λy; 1 = 2λz, x2 + y 2 + z 2 − 30 = 0. From the first and second equations, we obtain y = 2x. From the first and third equations, we obtain z √ = x. Substituting into √ the fourth √ 2 2 2 2 equation, we have x + 4x + x = 6x = 30. Thus, x = ± 5, y = ±2 5, z = ± 5.√Then, √ √ √ √ √ √ √ F ( 5, 2 5, 5) = 6 5 is a constrained maximum and F (− 5, −2 5, − 5) = −6 5 is a constrained minimum.

13.10. LAGRANGE MULTIPLIERS

129

14. Fx = 2x; Fy = 2y; Fx = 2z; gx = 1; fy = 2; gz = 3. We need to solve 2x = λ, 2y = 2λ, 2z = 3λ, x + 2y + 3z − 4 = 0. From the first and second equations, y = 2x. From the first and third equations, z = 3x. Substituting into the fourth equation, we have x+4x+9x = 14x = 4. Thus, x = 2/7, y = 4/7, andz = 6/7. Then, F (2/7, 4/7, 6/7) = 56/49 is a constrained extremum. Since (4, 0, 0) satisfies the constraint and F (4, 0, 0) = 16 > 56/49, F (2/7, 4/7, 6/7) = 56/49 is a constrained minimum. 15. Fx = yz; Fy = xz; Fz = xy; gx = 2x; gy = y/2; gz = 2z/9. We need to solve yz = 2λx, xz = λy/2, xy = 2λz/9 or xyz/2 = λx2 , xyz/2 = λy 2 /4, xyz/2 = λz 2 /9 along with x2 + y 2 /4 + z 2 /9 − 1 = 0 or x2 + Y 2 /4 + z 2 /9 = 1 for x > 0, y > 0, z > 0. From the first three equations and the fact that λ 6= 0, x2 = y 2 /4 = z 2 /9. Substituting into the 2 2 2 2 2 2 2 third equation, we √ obtain √ x +√x + x = 3x = 1, so x = 1/3, y =√4/3, and z = 3. √ Since 23/6, 1, 1) Thus, F ( 3/3, 2 3/3,√ 3) = 2 3/3 is√a constrained √ extremum. √ √ ( √ √ satisfies the constraint and F ( 23/6, 1, 1) = 23/6 < 2 3/3, F ( 3/3, 2 3/3, 3) = 2 3/3 is a constrained maximum. 16. Fx = yz; Fy = xz; Fz = xy; gx = 3x2 ; gy = 3y 2 ; gz = 3z 2 . We need to solve yz + 3λx2 , xz = 3λy 2 , xy = 3λz 2 or xyz = 3λx3 , syz = 3λy 3 , xyz = 3λz 3 along √ √ with 3 3 3 3 3 3 3 x + y + z√ − 24 = 0 or x + y + z = 24. Taking λ = 0 we see that ( 24, 0, 0), (0, 3 24, 0), and (0, 0, 3 24) satisfy the system. If λ 6= 0, the the first three equations imply x3 = y 3 = z 3 . Substituting into the fourth√ equation,√we obtain x3 +x3 +x3 = 3x√3 = 24 or x = 2. Then √ √ (2, 2, 2) 3 3 3 satisfies the system. Since 24 = 2 3, F (2 3, 0, 0) = F (0, 2 3 3, 0) = F (0, 0, 2 3 3) = 5 is a constrained minimum and F (2, 2, 2) = 13 is a constrained maximum. 17. Fx = 3x2 ; Fy = 3y 2 ; Fz = 3z 2 ; gx = 1; gy = 1; gz = 1. We need to solve 3x2 = λ, 3y 2 = λ, 3z 2 = λ, x + y + z − 1 = 0 for x > 0, y > 0, z > 0, and hence λ > 0. From the first three equations x2 = y 2 = z 2 , and since x, y, and z are positive, x = y = z. Then, from the fourth equation, x = y = z = 1/3 and F (1/3, 1/3, 1/3) = 1/9 is a constrained extremum. Since (1/2, 1/4, 1/4) satisfies the constraint and F (1/2, 1/4, 1/4) = 5/32 > 1/9, F (1/3, 1/3, 1/3) = 1/9 is a constrained minimum. 18. Fx = 8xy 2 z 2 ; Fy = 8x2 yz 2 ; Fz = 8x2 Y 2 z; gx = 2x; gy = 2y; gz = 2z. We need to solve 8xy 2 z 2 = 2λx, 8x2 yz 2 = 2λy, 8x2 y 2 z = 2λz or 4x2 y 2 z 2 = λx2 , 4x2 y 2 z 2 = λy 2 , 4x2 y 2 z 2 = λz 2 along with x2 + y 2 + z 2 − 9 = 0 or x2 + y 2 + z 2 = 9 for x > 0, y > 0, z > 0, and hence lambda > 0. From the first three equations, we see x2 = y 2 = z 2 . Substituting into 2 2 2 2 the third equation, we obtain √ √ √ x √+ x + x = 3x = 9. Thus, since x, y, and z are positive, x = y = z = 3 and F ( 3, 3, 3) = 108 is √ a constrained extremum. Since (1, 2, 2) satisfies √ √ the constraint and F (1, 2, 2) = 64 < 108, F ( 3, 3, 3) = 108 is a constrained maximum. 19. Fx = 2x; Fy = 2y; Fz = 2z; gx = 2; gy = 1; gz = 1; hx = −1; hy = 2; hz = −3. We need to solve 2x = 2λ−µ, 2y = λ+2µ, 2z = λ−3µ subject to 2x+y+z = 1, −x+2y−3z = 4. Solving the first three equations for x, y, and z, respectively, and substituting into the constraint equations, we obtain 2λ−µ+λ/2−3µ/2 = 1, −λ+µ/2 +λ+ 2µ−3λ/2 +9µ/2 = 4 or 6λ − 3µ = 2, −3λ + 14µ = 8. From this, we obtain λ = 52/75 and µ = 54/75. Then x = 1/3, y = 16/15, and z = −11/15. Thus, F (1/3, 16/15, −11/15) = 134/75 is a constrained minimum. 20. Fx = 2x; Fy = 2y; Fz = 2z; gx = 4; gy = 0; gz = 1; hx = 2x; hy = 2y; hz = −2z. We need to solve 2x = 4λ + 2xµ, 2y = 2yµ, 2z = λ − 2zµ subject to 4x + z = 7, z 2 = x2 + y 2 .

130

CHAPTER 13. PARTIAL DERIVATIVES Consider the second equation. If y = 0, then the constraint equations become 4x + z = 7 and z 2 = x2 . The solutions of these equations are x = z = 7/5 and x = −z = 7/3. In either case, the first and third equations can be solved for λ and µ. Thus, (7/5, 0, 7/5) and (7/3, 0, −7/3) are candidates for constrained extrema. Now, if y 6= 0, then from the second equations µ = 0. In this case, 2x = 4λ and 2z = λ or x = 4z. Then the first constraint equation becomes 16z + z = 17z = 7, so z = 7/17. Then, x = 28/17 and y 2 = z 2 − x2 < 0. Hence, the system has no solution when y 6= 0. Thus, F (7/5, 0, 7/5) = 98/25 is a constrained minimum and F (7/3, 0, −7/3) = 98/9 is a constrained maximum.

21. We want p to maximize A(x, yx y/2 subject to P (x, y) = Mx2+y2 x + y + x2p+ y 2 − 4 = 0. Ax =p y/2; Ay = x/2; y ; Py = 1 + y/ x2 + y 2 .pWe need Px = 1 + x/ x2 + y 2p to solve p y/2 = λ + λx/ x2 + y 2 , x/2 = λ + λy/ x2 + y 2 , x x + y + x2 + y 2 − 4 = 0 for x > 0, y > 0, and hence p x2 + y 2 or λ > 0. Subtracting the second equation from the first, we have (y − x)/2 = λ(x − y)/ p p 2 2 2 2 (y − x) = (y − x)(−2λ/ x + y ). Since −2λ/ x + y is negative, 1, and hence √ it cannot equal √ 2 = (2 + y − x = 0 or y √ = x. Substituting in the third equation gives 2x + 2x 2)x √ √ √ = 4. Thus, x = y = 4/(2 + 2) and this maximum area is A(4/(2 + 2), 4/(2 + 2)) = 4/(3 + 2 2). 22. Let the base of the box have dimensions x and y and let the height be z. We want to maximize V (x, y, z) = xyz subject to S(x, y, z) = xy + 2yz + 2xz − 75 = 0. Now Vx = yz; Vy = xz; Vz = xy; Sx = y + 2z; Sy = x + 2z; Sz = 2y + 2x. We need to solve yz = λ(y + 2z), xz = λ(x + 2z), xy = λ(2y + 2x), xy + 2yz + 2xz − 75 = 0 or xyz = λ(xy + 2xz), xyz = λ(xy + 2yz), xyz = λ(2yz + 2xz), xy + 2yz + 2xz = 75, for x > 0, y > 0, z > 0, and thus λ > 0. From the first three equations, we have xy + 2xz = xy + 2yz, which gives xz = yz or x = y; and xy + 2yz = 2yz + 2xz, which gives xy = 2xz or y = 2z. Substituting x = y = 2z into the fourth equation, we obtain 4z 2 + 4z 2 + 4z 2 = 12z 2 = 75. Thus, z = 5/2cm and x = y = 5cm. When the box is closed, S(x, y, z) = 2(xy + yz + xz) − 75, Sx = 2(y + z), Sy = 2(x + z), Sz = 2(x + y), and we need to solve xyz = 2λ(xy + xz), xyz = 2λ(xy + yz), xyz = 2λ(yz + xz), 2(xy + yz + xz) = 75 for x > 0, y > 0, z > 0, and thus λ > 0. From the first three equations, we have xy+xz = xy+yz, which gives xz = yz or x = y; and xy+yz = yz +xz, which gives xy = xz or 2 2 2 2 y = z. Substituting x √ = y = z into the fourth equation, we obtain √ 2(x + x + x ) = 6x = 75. Thus, x = y = z = 5/ 2. The box is a cube with each side 5/ 2cm. p 9 + y 2 − 81π. 23. We want to maximize V (x, y) = 9πx + 3πy subject to S(x, y) = 9π + 6πx + 3π p 2 Now Vp 6π; Sy = 3πy/ 9 + y . We need to solve 9π = 6πλ, 3π = x = 9π; Vy = 3π; Sx = p 3πλy/ 9 + y 2 , 9π + 6πx + 3 9 + y 2 − 81π = 0 for x > 0 and y > √ 0. From the first equation, λ = 3/2. Usingpλ = 3/2 in the second equation gives y = 6/ 5. From the third √ equation, we √ have 6x + 3 9 +√36/5 = 72 or x = 12 − 9/2 5. The volume is maximum when x = 12 − 9/2 5m and y = 6/ 5m. 1 2 1 24. Ux = x−2/3 y 2/3 ; Uy = x1/3 y −1/3 ; gx = 1; gy = 6. We need to solve x−2/3 y 2/3 = 3 3 3 2 1/3 1 λ, x y−1/3 = 6λ, x + 6y − 18 = 0 or y = 3λx2/3 y 1/3 , x = 3λx2/3 y 1/3 , x + 3 3 6y = 18. From the first two equations, y = x/3. Substituting into the third equation, we

13.10. LAGRANGE MULTIPLIERS

131

have x + 2x = 3x = 18. Thus, x = 6 and y = 2. Since (12, 1) satisfies the constraint and U (12, 1) = 121/3 , U (6, 2) = 61/3 22/3 = 241/3 is a constrained maximum. 25. We want to maximize z(x, y) = P − x − y subject to z 2 /xy 3 = k or (P − x − y)2 − kxy 3 = 0. Now zx = −1; zy = −1; gx = −2(P − x − y) − ky 3 ; gy = −2(P − x − y) − 3kxy 2 . We need to solve −1 = −2λ(P − x − y) − λky 3 , −1 = −2λ(P − x − y) − 3λkxy 2 , (P − x − y)2 = kxy 3 for x > 0, y > 0, and z > 0. From the first two equations, √ we have y = 3x. Substituting into the third equation, we obtain (P − 4x)2 = 27kx4 or 27kx2 = P − 4x. (Since z > 0, z = P − x − y = P − 4x > 0.) Using the quadratic formula and the fact that x > 0, we find p √ √ −2 + 4 + P 27k 16 + 4P 27k √ = x= . 27k 2 27k p √ √ Then the maximum value of z is P − 4x = P + 4(2 − 4 + P 27k/ 27k. −4 +

p

26. (a) See part (b). (b) Maximizing 1/(x21 · · · x2n ) is equivalent to minimizing the denominator F (x1 , . . . xn ) = x21 + · · · + x2n . The constraint is still x1 + · · · xn = 1, which we can write as g(x1 , . . . xn ) = x1 + · · · + xn − 1 = 0. Since ∂F/∂xi = 2xi and ∂g/∂xi = 1, we can get the equations 2x1 = λ,, 2x2 = λ, , . . . , 2xn = λ, x1 + · · · xn = 1. The solution is x1 = . . . = 1/2 ( and λ = 1/2n). 4 27. f (x, y) is the square of the distance from a point on the graph of x√ + y 4 =√1 to the origin. 4 The points (0, ±1) and (±1, 0) are closest to the origin, while (±1/ 2, ±1/ 4 2) are farthest from the origin.

28. F (x, y, z) is the square of the distance from a point on the plane x + 2y + 3z = 4 to the origin. The point (2/7, 4/7, 6/7) is closest to the origin. 29. F is the square of the distance of points on the intersection of the planes 2x + y + z = 1 and −x + 2y − 3z = 4 from the origin. The point (1/3, 16/15, −11/15) is closest to the origin. 30. F is the square of the distance of points on the intersection of the plane 4x + z = 7 and the circular cone z 2 = x2 = y 2 . The point (7/5, 0, 7/5) is closest to the origin and the point (7/3, 0, −7/3) is farthest from the origin. 31. We want to minimize f (x, ) = x2 + y 2 subject to xy 2 = 1. Now fx = 2x; fy = 2y; gx = y 2 ; gy = 2xy. We need to solve 2x = λy 2 , 2y = 2λxy, xy 2 − 1 = 0 or 2xy = λy 3 , 2xy = 2λx2 y, xy 2 = 1 for x > 0, y > 0, and hence λ > 0. From the first two equations, we have y 3 = 2x2 y or y 2 = 2x2 . Substituting into the third equation gives 2x3 = 1 or x = 2−1/3 . Again, from the third equation we have y = 1/(2−1/3 )1/2 = 21/6 . Thus, the point closest to the origin is (2−1/3 , 21/6 ). Since the surface is F (x, y, z) = xy 2 − 1 = 0, ∇F = y 2 i + 2xyj is normal to the surface at (x, y, z). Thus, a normal to the surface at (2−1/3 , 21/6 , 0) is ∇F (2−1/3 , 21/6 , 0) is ∇F (2−1/3 , 21/6, 0) = 21/3 i + 2(2−1/3 )(21/6 )j = 21/3 i + 25/6 j = 22/3 (2−1/3 i + 21/6 j). Since ∇F (2−1/3 , 21/6 , 0) is a multiple of the vector from the origin to P (2−1/3 , 21/6 , 0), this vector is perpendicular to the surface.

132

CHAPTER 13. PARTIAL DERIVATIVES 1 −2/3 1/3 1/3 1 1 x y z ; Fy = x1/3 y −2/3 z 1/3 ; Fz = x1/3 y 1/3 z −2/3 ; gx = 1; gy = 3 3 3 1 1/3 −2/3 1/3 1 1/3 1/3 −2/3 1 x y z = λ, x y z = 1; gz = 1. We need to solve x−2/3 y 1/3 z 1/3 = λ, 3 3 3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 λ, x + y + z − k = 0 or x y Z1/3 = 3λx, x y z = 3λy, x y z = λz, x + y + z = k. From the first three equations, x = y = z. Substituting into the fourth equation, x + x + x = 3x = k. Thus, x = y = z = k/3 and F (k/3, k/3, k/3) = k/3 is a constrained maximum.

32. Fx =

33. For any x + y + z = k, by Problem 32,

√ 3

xyz ≤

k x+y+z = . 3 3

34. Distance from the xz-plane is measured by |y|. Alternatively, we will find the extreme values of F (x, y, z) = y 2 subject to g(x, y, z) = x2 + z 2 − 1 = 0 and h(x, y, z) = x + y + 2z − 4 = 0. Now Fx = 0; Fy = 2y; Fz = 0; gx = 2x; gy = 0; gz = 2z; hx = 1; hy = 1; hz = 2. We need to solve 0 = 2λx + µ, 2y = µ, 0 = 2λz + 2µ, x2 + z 2 = 1, x + y + 2z = 4. By inspection we see that if λ = 0, then µ = 0 and y = 0. Similarly, if µ = 0, then λ = 0 and y = 0. Substituting y = 0 into the fourth and fifth equations, we obtain the system x2 + z 2 = 1, x + 2z = 4, which is inconsistent. Thus, µ 6= 0 and λ 6= 0. Now, solving the first three equations for x, y, and z and substituting into the fourth and fifth equations, we obtain the system

µ2 µ µ 2µ µ2 + = 1, − + − = 4 or 5µ2 = 4λ2 , (λ − 5)µ = 8λ. 4λ2 λ2 2λ 2 λ 8λ 2 Solving the second equation for µ and substituting into the first, we obtain 5( ) = 4λ2 λ −5 √ or 80 = (λ − 5)2 . Thus, √ λ = 5 ± 4 5. From µ = 8λ/(λ − 5) we find that corresponding 2 values of µ are 8 ± 2 5. Since √ 2y = µ we see that the objective √ function F (x, y, z) = y is minimized when µ = 8 − 2 5 and maximized when µ = 8 + 2 5. Corresponding values of √ √ µ 8±2 5 1 √ = ∓ √ ≈ |mp0.45, y − µ/2 = 4 ± 5 ≈ 6.24 and x, y, and z are x = − =− 2λ 10 ± 8 5 5 √ √ √ √ 1.76, z = −µ/λ = 2x = ∓2/ 5 ≈ ∓0.89. The closest point is (−1/ 5, 4 − 5, −2 5) or about (−4.5, 6.24, −0.89).

Chapter 13 in Review A. True/False 1. False; see Example 3 in Section 13.2 in the text. 2. False; (0,4.1) is in the domain of g but not in the domain of f . 3. True 4. True 5. False; consider z = y 2 .

CHAPTER 13 IN REVIEW

133

6. False; consider f (x, y) = xy at (0, 0). 7. False; ∇f is perpendicular to the level curve f (x, y) = c. 8. True 9. True 10. False; at a saddle point fx = fy = 0, but there is no extremum.

B. Fill in the Blanks 1.

3x2 + xy 2 − 3xy − 2y 3 3+1−3−2 1 = =− 2 2 5x − y 5−1 4 (x,y)→(1,1) lim

2. where x − y + 1 = 0 3. 3x2 + y 2 = 3(2)2 + (−4)2 = 28 4.

∂ ∂T ∂p ∂T ∂q T (p, q) = + = Tp gξ + Tq hξ ∂ξ ∂p ∂ξ ∂q ∂ξ

5.

∂F dr ∂F ds d F (r, s) = + = Fr g 0 (w) + Fs h0 (w) dw ∂r dw ∂s dw

6. dg = gs ∆s + gt ∆t =

4s 2 ∆s − 3 ∆t t2 t

7. fyyzx 8.

∂3f ∂y 2 ∂x

 ∂f ∂ R y 9. Using the Fundamental Theorem of Calculus, we have (x, y) = F (t)dt = F (y) ∂y ∂y x i i  ∂ hR x ∂f ∂ R y ∂ h Rx − y F (t)dt = − F (t)dt = −F (x) (x, y) = F (t)dt = x y ∂x ∂x ∂x ∂x 10. ∇F (x0 , y0 , z0 ) = i + j + k ∂2 ∂ fx (x, y)g(y)h(z) = [fx (x, y)g 0 (y)h(z) + fxy (x, y)g(y)h(z)] ∂z∂y ∂z = fx (x, y)g 0 (y)h0 (z) + fxy (x, y)g(y)h0 (z)

11. Fx,y,z =

12. The distinct fourth-order partial derivatives are fxxxx , fxxxy , fxxyy , fxyyy , and fyyyy .

134

CHAPTER 13. PARTIAL DERIVATIVES

C. Exercises 3

1. zy = −x3 ye−x

2. zu = −

3. fr =

y

+ e−x

3

y

v sin uv = −v tan uv cos uv

3 3 2 3 r (r + θ2 )−1/2 ; frθ = − r2 θ(r3 + θ2 )−3/2 2 2

4.

∂f ∂2f = 2(2 + y 2 )2 = 2(2x + xy 2 )(2 + y 2 ) = 2(2 + y 2 )2 x; ∂x ∂x2

5.

∂z ∂2z = 3x2 y 2 sinh x2 y 3 ; = 9x4 y 4 cosh x2 y 3 + 6x2 y sinh x2 y 3 ∂y ∂y 2

6.

∂z ∂2z ∂3z 2 2 2 2 2 = −4y(ex + e−y ); = −8xyex ; 2 = −16x2 yex − 8yex ∂y ∂x∂y ∂ x∂y

7. Fs = 3s2 t5 v −4 ; Fst = 15s2 t4 v −4 ; Fstv = −60s2 t4 v −5

8.

xy x y ∂2w x x 1 ∂3w 2x ∂4w 2 ∂w = 2 + + ; =− 2 − 2 + ; 2 = 3; = 3 2 ∂z z y x ∂y∂z z y x ∂ y∂z y ∂x∂ y∂z y

9. ∇f = −

10. ∇F =

y 1 1 1 y x 1 1 i+ j=− 2 i+ 2 j; ∇f (1, −1) = i + j 2 2 2 2 2 2 2 x 1 + y /x x 1 + y /x x +y x +y 2 2

2x 9y 2 4(x2 − 3y 3 ) i− 4 j− k; ∇F (1, 2, 1) − 2i − 36j + 92k 4 z z z5

6 1 2 11. ∇f = (2xy − y 2 )i + (x2 − 2xy)j; u = √ i + √ j = √ (i + 3j); 40 40 10 1 1 2 2 2 Du f = √ (2xy − y + 3x − 6xy) = √ (3x − 4xy − y 2 ) 10 10 2x 2y 2z 2 1 2 i+ 2 j+ 2 k; u = − i + j + k; 2 2 2 2 2 2 +y +z x +y +z x +y +z 3 3 3 −4x + 2y + 4z Du F = 3(x2 + y 2 + z 2 )

12. ∇F =

x2

CHAPTER 13 IN REVIEW 13. {(x, y)|(x + y)2 ≤ {(x, y)| |x + y| ≤ 1}

135 1}

=

14. {(x, y)|y > x, y 6= x + 1} y

y

x x

15. ∆z = 2(x + ∆x)(y + ∆y) − (y + ∆y)2 − (2xy − y 2 ) = 2x∆y + 2y∆x + 2∆x∆y − 2y∆y − (∆y)2 16. ∆z = (x + ∆x)2 − 4(y + ∆y)2 + 7(x + ∆x) − 9(y + ∆y) + 10 − (x2 − 4y 2 + 7x − 9y + 10) = 2∆x + (∆x)2 − 8y∆y − 4(∆y)2 + 7∆x − 9∆y 4x + 3y − (x − 2y)4 11y (4x + 3y)(−2) − (x − 2y)3 −11x = ; zy = = ; (4x + 3y)2 (4x + 3y)2 (4x + 3y)2 (4x + 3y)2 11y 11x dz = dx − dy (4x + 3y)2 (4x + 3y)2

17. zx =

18. Ax = 2y + 2z; Ay = 2x + 2z; Az = 2y + 2x; dA = 2(y + z)dx + 2(x + z)dy + 2(x + y)dz p √ √ √ 19. zy = 4y/ x2 + 4y 2 , zy (− 5, 1) = 4/3, z(− 5, 1) = 3. The line is given by x = − 5 and √ z−3 y−1 4 = . z − 3 = (y − 1). Symmetric equations of the line are x = − 5, 3 4 3 −−→ 1 1 20. The direction vector is P Q = 2i + 2j. ∇z = (y + 2x)i + xj. u = √ i + √ j; Du = ∇z · u = 2 2 √ √ √ √ (y + 2x + x)/ 2 = (y + 3x)/ 2; Du (2, 3) = 9/ 2. The slope of the tangent line is 9/ 2. 21. fx = 2xy 4 , fy = 4x2 y 3 . (a) u = i, Du (1, 1) = fx (1, 1) = 2 √ √ √ (b) u = (i − j/ 2, Du (1, 1) = (2 − 4)/ 2 = −2/ 2 (c) u = j, Du (1, 1) = fy (1, 1) = 4 22. (a)

dw ∂w dx ∂w dxy ∂w dz = + + dt ∂x dt ∂y dt ∂z dt x y z =p 6 cos 2t + p (−8 sin 2t) + p 15t2 2 2 2 2 2 2 2 x +y +z x +y +z x + y2 + z2 (6x cos 2t − 8y sin 2t + 15zt2 ) p = x2 + y 2 + z 2

136

CHAPTER 13. PARTIAL DERIVATIVES (b)

dw ∂w dx ∂w dxy ∂w dz = + + dt ∂x dt ∂y dt ∂z dt   6 8r 2r 2t y z x sin 15t2 r3 cos + p +p =p 2 2 2 2 2 2 2 2 2 2 r r t t x +y +z x +y +z x +y +z   2r 6x 2t 8yr cos + 2 sin + 15zt2 r3 r r t t p = x2 + y 2 + z 2

√ π 1 23. F (x, y, z) = sin xy − z; ∇F = y cos xyi + x cos xyj − k; ∇F (1/2, 2π/3, 3/2) = i + j − k. 3 4 √ π 1 1 2π 3 The equation of the tangent plane is (x− )+ (y − )−(z − ) = 0 or 4πx+3y −12z = 3 2 4 3 2 √ 4π − 6 3. 24. We want to find a normal to the surface that is parallel to k. ∇F = (y −2)i+(x−2y)j+2zk. We need y − 2 = 0 and x − 2y = 0. The tangent√plane is parallel√to z = 2 when y = 2 and x = 4. In this case z 2 = 5. The points are (4, 2, 5) and (4, 2, − 5). 25. ∇F = 2xi + 2yj; The equation of the tangent plane is 6(x − 3) + 8(y − 4) = 0 or 3x + 4y = 25. 26. We want to minimize 1 3 Du f = u · ∇f = √ (i + j) · [(3x2 + 3y − 6x)i + (3x + 3y 2 )j] = √ (x2 + y − 2x + x + y 2 ) 2 2 or equivalently, we want to minimize F (x, y) = x2 − x + y 2 + y. Now Fx = 2x − 1; Fxx = 2; Fxy = 0; Fy = 2y + 1; Fyy = 2; D = 4. Solving Fx = 0 and Fy = 0 we obtain x = 1/2 and y = −1/2. Since D = 4 > 0 and Fxx = 2 > 0, F, and hence Du f, has a minimum at (1/2, −1/2). 27. We want to maximize v(x, y, z) = xyz subject to x + 2y + z = 6. Now Vx = yz; Vy = xz; VZ = xy; gx = 1; gy = 2; gz = 1. We need to solve yz = λ, xz = 2λ, xy = λ or xyz = λx, xyz = 2λy, xyz = λz along with x + 2y + z − 6 = 0 or x + 2y + z = 6. From the first three equations, we have x = 2y = z. Substituting into the fourth equation gives x + x + x = 3x = 6 or x = 2. Then y = 1 and z = 2 and V (2, 1, 2) = 4 is the maximum volume. c2 Dθ2 G c2 (b) dM = (θ2 dD + 2Dθdθ) G   dM c2 θ2 2Dθ θ2 2Dθ dD dθ (c) We have = dD + dθ = dD + dθ = + 2 , so M G M M Dθ2 Dθ2 D θ dM dD dθ dθ dD = ≤ 0.10 + 2(0.02) = 0.14 = 14%. + 2 ≤ + 2 M D θ θ D

28. (a) M =

CHAPTER 13 IN REVIEW

137

√ 29. We√are given v = 14 5ry −1/2 , dr = −1, dy = 1, r = 20, and y = 25. Now, dv = √ 14 5y −1/2 dr − 7 5ry −3/2 dy and the approximate change in volume is √ √ √ ∆v ≈ 14 5(25)−1/2 (−1) − 7 5(20)(25)−3/2 (1) = −98 5/25 ≈ −8.77cm/s. 30. ∆f = 2xi + 2yj, ∆f (3, 4) = 6i = 8j √ √ − 2j)/ 5; (a) ∆f (1, −2)2i −√4j; u =√(2i − 4j) √20 = (i √ Du f (3, 4) = 6 5 − 16 5 = −10 5 = −2 5 √ (b) v = (6i + 8j)/ 100 = (3i + 4j)/5; Dv f (3, 4) = 18/5 + 32/5 = 10

R2 − r 2 . Then, after a straightforward but lengthy comR2 − 2rR cos(θ − φ) + r2 putaiton, we find

31. Let g(r, θ) =

gr = grr = gθθ =

(2r2 R + 2R3 ) cos(θ − φ) − 4rR2 (R2 − 2rR cos(θ − φ) + r2 )2 ,

8R4 cos2 (θ − φ) + (−12rR3 − 4r3 R) cos(θ − φ) − 4R4 + 12r2 R2 , [R2 − 2rR cos(θ − φ) + r2 ]3

(4r4 R2 − 4r2 R4 ) cos2 (θ − φ) + (2r5 R − 2rR5 ) cos(θ − φ) − 8r4 R2 + 8r2 R4 (R2 − 2rR cos θ + r2 )3

and r2 grr + rgr + gθθ . Then Z Z π Z π r2 π r 1 2 r Urr + rUr + Uθθ = g(r, θ)f (φ)dφ + g(r, θ)f (φ)dφ + g(r, θ)f (φ)dφ 2π π 2π π 2π π Z π 1 = (r2 grr + rgr + gθθ )dφ = 0. 2π π αz βAxα y β βy αAxα y β = ; fy = Aβxα y β−1 = = ; x x y y xαzx − αz xα(αz/x) − αz α2 z − αz α(α − 1) fxx = = = = ; x2 x2 x2 x2 2 yβzy − βz yβ(βz − βz) β z − βz β(β − 1)z fyy = = = = ; y2 y2 y2 y2 α(βz)/y αβz fxy = fyz = = x xy

32. fx = Aαxα−1 y β =

33. Since D = 4(6) − 52 = −1 < 0, f (a, b) is not a relative extremum. 34. Since D = 2(7) − 02 = 14 > 0 and fxx = 2 > 9, f (a, b) is a relative minimum. 35. Since D = (−5)(−9) − 62 = 9 > 0 and fxx = −5 < 0, f (a, b) is a relative maximum. 36. Since D = (−2)(−8) − 42 = 0, no determination is possible.

138

CHAPTER 13. PARTIAL DERIVATIVES

37. Since x = L cos θ and y = L sin θ, 1 1 1 A = xy = L2 sin θ cos θ = l2 sin 2θ. 2 2 4

L

y

θ x

38. Substituting x = h cot φ into tan θ = h=

tan θ . 1 − tan θ cos φ

h and solving, we obtain 1+x

h φ

θ 1

x

39. A = xy − (y − 2z)(x − 2z) − z 2 = 2(x + y)z − 5z 2 40. We are given V (x, y, z) = xyz, x = 30, y = 40, z = 25, and dx = dy = −1 and dz = −1/2. Then dV = yzdx + dzdy + dydz, so the approximate volume of plastic is |dV | = 40(25)(1) + 30(25)(1) + 30(40)(1/2) = 2350cm3 .   p p 41. V = (2x)(2y)z = 4xy 4 − x2 − y 2 = 16xy − 4xy x2 + y 2 42. C(x, y, z) = 1.5(2xy + 2xz + 2yz + xz + 5yz) =

3 (2xy + 3xz + 7yz) 2