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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}

6. {(u, b)|(u, v( 6= (0, 0)}⋃{(u, v)|u2 + v2 6= 1}

7. {(r, s)| |s| ≥ 1}

8. {(θ, φ) | tan θ tanφ 6= 1}⋂{(θ, φ) θ 6= π/2 + kπ, k an integer}

⋂{(θ, φ) | φ 6= π/2 +

kπ, k an integer}

9. (u, v, w)|u2 + v2 + q2 ≥ 16

10. {(x, y, z)|x2 + y2+ < 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− y2) is{

(x, y)|x− y2 > 0}

={

(x, y)|x > y2}

14. (h); The domain of f(x, y) =

√x2 + y2 − 1y − x

is{

(x, y)|x2 + y2 − 1 ≥ 0, y 6= x}

={(x, y)|x2 + y2 ≥ 1, y 6= x

}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 + y4

xyis {(x, y)|xy 6= 0} = {(x, y)|x 6= 0, y 6= 0}

17. (f); The domain of f(x, y) = sin−1(xy) is {(x, y)||xy| ≤ 1}

18. (a); The domain of f(x, y) =√y − x2 is {(x, y)|y − x2 ≥ 0} =

{∗x, y)y ≥ x2

}y

x

19. {(x, y)|x ≥ 0 and y ≥ 0}

y

x

2 1

20. {(x, y)| x2 ≤ 1 and y2 ≥ 4}⋂{(x, y)| x2 ≥ 1 and y2 ≤ 4}

{(x, y)| |x| ≤ 1 and |y| ≥ 2}⋃{(x, y)| |x| ≥ 1 and |y| ≤ 2}

21. {(x, y)|y − x ≥ 0}

y

x

22. {(x, y)|xy ≥ −1}

y

x

23. {z | z ≥ 10} 24. all real numbers

25. {w ||;−1 ≤ w ≤ 1} 26. {x | w < 7}

27. f(2, 3) =∫ 4

2(2t− 1)dt = (t2 − t)|42 = 12− 2 = 10

f(−1, 1) =∫ 1

−1(2t− 1)dt = (t2 − t)|1−1 = 0− 2 = −2

13.1. FUNCTIONS OF SEVERAL VARIABLES 79

28. f(3, 0) = ln 9/9 = ln 1 = 0; f(5,−5) = ln25

25 + 25= ln

12

= − ln 2

29. f(−1, 1,−1) = (−2)2 = 4; 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 = − 12x+ C

y

x

38. x = y2 − c

y

x

39. x2 − y2 = 1 + c2

y

x

40. 4x2 + 9y2 = 36− c2, −6 ≤ c ≤ 6

y

x


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

y

x

42. y = x+ tan c, −π/2 < c < π/2

y

x

43. x2/9 + z2/4 = 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 whichis 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 + 3y2 + 6z2 = c; ellipsoid

46. 4y − 2z + 1 = c; plane

47. c = 0 c < 0 c > 0

z

xy y

z

xy

z

x

48. Setting x = −4, y = 2, and z = −3 in x2/16+y2/4+z2/9 = c we obtain c = 3. The equationof the surface is x2/16 + y2/4 + z2/9 = 3. Setting y = z = 0 we find the x-intercepts are±4√

3. Similarly, the y-intercepts are ±2√

3 and the z-intercepts are ±3√

3.

13.1. FUNCTIONS OF SEVERAL VARIABLES 81

49.

v

P

50. From V = s2h we obtain h = V/s2.

51. C(r, h) = πr2(1.8) + πr2(1) + 2πh(2.3) = 2.8πr2 + 4.6πrh

52. Let the height of the box be h. Then 2xy + 2xh + 2yh = 500 and h =250− xyx+ y

. Thus,

V = xyh =250xy + x2y2

x+ y.

53. V + πr2g + 13πr

2(

23h)

= 119 πr

2h

y

x

z

θ

θ

t

y2-z2

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

(z√

y2 − z2

)=

xz√y2 − z2

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/m2h

57. (a) The distance the water falls in time t is s(t) = 12gt

2 + vt where vis the velocity of thewater at the top level (t = 0). The velocity of the water at time t is v(t) = gt+ v. If t1is 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

h =12gt21 + vt1 =

12g

(V − vg

)2

+ v

(V − vg

)Simplifying the equation we obtain 2gh = V 2−v2. Now the rates at the top and bottomlevels are Z = vπr2 and Q = V πr2 (recall that the flow rate is constant). Solving forv and V and substituting into 2gh = V 2 − v2 we obtain 2gh = (Q/πr2)2 − (Q/πR2)2.

Solving for Q we find Q =πr2R2

√2gh√

R4 − r4.

(b) When r = 0.2 cm, R = 1 cm, and h = 10, Q ≈ 7.61 cm3/s.


13.2 Limits and Continuity

1. lim(x,y)→(5,−1)

(x2 + y2) = 25 + 1 = 26

2. lim(x,y)→(2,1)

x2 − yx− y

=4− 12− 1

= 3

3. On y = 0, lim(x,y)→(0,0)

5x2 + y2

x2 + y2= lim

(x,y)→(0,0)

5x2

x2= 5.

On x = 0, lim(x,y)→(0,0)

5x2 + y2

x2 + y2= lim

(x,y)→(0,0)

y2

y2= 1. The limit does not exist.

4. lim(x,y)→(1,2)

4x2 + y2

16x4 + y4=

4 + 416 + 16

=14

5. lim(x,y)→(1,1)

4− x2 − y2

x2 + y2=

4− 1− 11 + 1

= 1

6. On x = 0, lim(x,y→(0,0)

2x2 − yx2 + 2y2

= lim(x,y→(0,0)

−y2y2

=∞.

On y = 0, lim(x,y→(0,0)

2x2 − yx2 + 2y2

= lim(x,y→(0,0)

2x2

x2= 2.. The limit does not exist.

7. On y = x, lim(x,y→(0,0)

x2y

x4 + y2= lim

(x,y→(0,0)

x3

x4 + x2= lim

(x,y→(0,0)

x

x2 + 1= 0.

On y = x2, lim(x,y→(0,0)

x2y

x4 + y2= lim

(x,y→(0,0)

x4

x4 + x4=

12. The limit does not exist.

8. On y = x, lim(x,y→(0,0)

6xy2

x2 + y4= lim

(x,y→(0,0)

6x3

x2 + x4= lim

(x,y→(0,0)

6x1 + x2

= 0.

On x = y2, lim(x,y→(0,0)

x2y

x4 + y2= lim

(x,y→(0,0)

6y4

y4 + y4= 3. The limit does not exist.

9. lim(x,y)→(1,2)

x3y2(x+ y)3 = 1(4)(27) = 108

10. lim(x,y)→(2,3)

xy

x2 − y2=

64− 9

= −65

11. lim(x,y)→(0,0)

exy

x+ y + 1=

11

= 1

12. On y = mx, lim(x,y)→(0,0)

sinxyx2 + y2

= lim(x,y)→(0,0)

sinmx2

(1 +m2)x2

= lim(x,y)→(0,0)

m

1 +m2

sinmx2

mx2=

m

1 +m2.

The limit does not exist.

13.2. LIMITS AND CONTINUITY 83

13. lim(x,y)→(2,2)

xy

x3 + y2=

48 + 4

=13

14. lim(x,y)→(π,π/4)

cos(3x+ y) = cos(3π + π/4) = cos 13π/4 = −√

2/2

15. lim(x,y)→(0,0)

x2 − 3y + 1x+ 5y − 3

= −13

16. On y = mx, lim(x,y)→(0,0)

x2y2

x4 + 5y4= lim

(x,y)→(0,0)

x2m2x2

x4 + 5m4x4=

m2

1 + 5m4.


17. lim(x,y)→(4,3)

xy2x+ 2yx− y

= 4(9)4 + 64− 3

= 360

18. lim(x,y)→(1,0)

x2y

x+y3=

01 + 0

= 0

19. lim(x,y)→(1,1)

xy − x− y + 1x2 + y2 − 2x− 2y + 2

= lim(x,y)→(1,1)

(x− 1)(y − 1)(x− 1)2 +m2(x− 1)2

On y − x = m(x− 1),

lim(x,y)→(1,1)

(x− 1)(y − 1)(x− 1)2 +m2(x− 1)2

= lim(x,y)→(1,1)

(x− 1)m(x− 1)(x− 1)2 +m2(x− 1)2

=m

1 +m2.


20. On x = 0, lim(x,y)→(0,3)

xy − 3yx2 + y2 − 6y + 9

= lim(x,y)→(0,3)

−3y(y − 3)2

. The limit does not exist.

21. lim(x,y)→(0,0)

x3y + xy3 − 3x2 − 3y2

x2 + y2= lim

(x,y)→(0,0)

xy(x2 + y2)− 3(x2 + y2

x2 + y2

= lim(x,y)→(0,0)

(xy − 3) = −3

22. lim(x,y)→(−2,2)

y3 + 2x3

x+ 5xy2=

8− 16−2− 40

=421

23. lim(x,y)→(1,1)

ln(2x2 − y2) = ln(2− 1) = 0

24. lim(x,y)→(1,2)

sin−1(x/y)cos−1(x− y

=sin−1(1/2)cos−1(−1)

=π/6π

=16

In Problems 25-30 let x = r cos θ and y = r sin θ. Then x2 + y2 = r2 and (x, y) → (0, 0) if

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

25. lim(x,y)→(0,0)

(x2 − y2)2

x2 + y2= limr→0

(r2 cos2 θ − r2 sin2 θ)2

r2= limr→0

r4(cos2 θ − sin2 θ)2

r2

= limr→0

r2 cos2 2θ = 0


26. lim(x,y)→(0,0)

sin(3x2 + 3y2)x2 + y2

= limr→0

sin 3r2

r2Use L’Hopital’s Rule

= limr→0

6r cos 3r2

2r= limr→0

3 cos 3r2 = 3

27. lim(x,y)→(0,0)

6xy√x2 + y2

= limr→0

6r2 cos θ sin θ√r2

= limr→0

3|r| sin 2θ = 0

28. lim(x,y)→(0,0)

x2 − y2√x2 + y2

= limr→0

r2 cos2 θ − r2 sin2 θ√r2

= limr→0|r| cos 2θ = 0

29. lim(x,y)→(0,0)

x3

x2 + y2= limr→0

r3 cos3 θ

r2= limr→0

r cos3 θ = 0

30. lim(x,y)→(0,0)

x3 + y3

x2 + y2= limr→0

r3 cos3 θ + r3 sin3 θ

r2= limr→0

r(cos3 θ + sin3 θ) = 0

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 + y2 < 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 + y2 < 1, f(x, y) is discontinuous since it is not defined at (2, 0).

37. Since

lim(x,y)→(0,0)

f(x, y) = lim(x,y)→(0,0)

6x2y3

(x2 + y2)2= limr→0

6r5 cos2 θ sin3 θ

r4= limr→0

6r cos2 θ sin3 θ = 0 = f(0, 0)

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 = 0and y = 0, respectively. On y = x,

lim(x,y)→0,0)

f(x, y) = lim(x,y)→(0,0)

x2

2x2 + 2x2=

14,

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

13.3. PARTIAL DERIVATIVES 85

39. Choose ε > 0. Using x = r cos θ and y = r sin θ we have

3xy2

2x2 + 2y2=

3t cos θr2 sin2 θ

2r2

32r cos θ sin2 θ.

Let δ = 2ε3 . Now, whenever r =

√x2 + y2 < δ, we have

| 3xy2

2x2 + 2y2| = 3

2|r cos θ sin2 θ| ≤ 3

2|r| < 3

2δ =

32

(2ε3

)= ε.

Thus lim(x,y)→(0,0)

3xy2

x2 + y2= 0.

40. Choose ε > 0. Using x = r cos θ and y = r sin θ we have

x2y2

x2 + y2=r2 cos2 θr2 sin2 θ

r2= r2 cos2 θ sin2 θ.

Now, whenever r =√x2 + y2 <

√ε (for δ =

√ε),∣∣∣∣ x2y2

x2 + y2

∣∣∣∣ = r2 cos2 θ sin2 θ ≤ r2 ≤ ε. Thus,

lim(x,y)→(0,0)

x2y2

x2 + y2= 0.

41. Where y 6= x, we have

f(x, y) =x3 − y3

x− y=

(x− y)(x2 + xy + y2)x− y

= x2 + xy + y2.

When y = x, we have

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

Therefore, f(x, y) = x2 + xy + y2 throughout the entire plane. Since x2 + xy + y2 is apolynomial, f must be continuous throughout the plane and thus has no discontinuities.

42. Choose ε > 0. Then for δ = ε, whenever 0 <√

(x− a)2 + (y − b)2 < δ, we have

|f(x, y)− b| = |y − b| ≤√

(x− a)2 + (y − b)2 < δ = ε.

Thus, lim(x,y)→(a,b)

y = b.

13.3 Partial Derivatives

1.∂z

∂x= lim4→0

7(x = 4x) + 8y2 − 7x− 8y2

4x= lim4→0

74 x

4x= 7

∂z

∂y= lim4y→0

7x+ 8(y +4y)2 − 7x− 8y2

4y= lim4y→0

16y4 y + 8(4y)2

4y= lim4y→0

(16y + 84 y) = 16y


2.∂z

∂x= lim4x→0

(x+4x)y − xy4x

= lim4x→0

y4 x

4x= y;

∂z

∂y= lim4y→0

x(y +4y)− xy4y

= lim4y→0

x4 y

4y= x

3.∂z

∂x= lim4x→0

3(x+4x)2y + 4x+4x)y2 − 3x2y − 4xy2

4x

= lim4x→0

3x2y + 6x(4x)y + 3(4x)2y + 4xy2 + 4(4x)y2 − 3x2y − 4xy2

4x

= lim4x→0

6x(4x)y + 3(4x)2y + 4(4x)y2

4x= lim4x→0

(6xy + 3(4x)y + 4y2) = 6xy + 4y2

∂z

∂y= lim4y→0

3x2(y +4y) + 4x(y +4y)2 − 3x2y − 4xy2

4y

= lim4y→0

3x2y + 3x2 4 y + 4xy2 + 8xy4 y + 4x(4y)2 − 3x2y − 4xy2

4y

= lim4y→0

3x2 4 y + 8xy4 y + 4x(4y)2

4y= lim4y→0

(3x2 + 8xy + 4x4 y) = 3x2 + 8xy

4.∂z

∂x= lim4x→0

x+4x+4+ y

− x

x+ y

4x= lim4x→0

x2 + x4 x+ xy + (4x)y − x2 − x4 x− xy(x+4x+ y)(x+ y)4 x

= lim4x→0

(4x)y(x+4x+ y)(x+ y)4 x

=y

(x+ y)2

∂z

∂y= lim4y→0

x

x+ y +4y− x

x+ y

4y= lim4y→0

x2 + xy − x2 − xy − x4 y

(x+ y +4y)(x+ y)4 y

= lim4y→0

−x4 y

(x+ y +4y)(x+ y)4 y= − x

(x+ y)2

5. zx = 2x− y2; zy = −2xy + 20y4

6. zx = −3x2 + 12xy3; zy = 18x2y2 + 10y

7. zx = 20x3y3 − 2xy6 + 30x4; zy = 15x4y2 − 6x2y5 − 4

8. zx = 3x2y2 sec2(x3y2); zy = 2x3 sec2(x3y2)

9. zx =2√

x(3y2 + 1); zy = − 24y

√x

(3y2 + 1)2

10. zx = 12x2 − 10x+ 8; zy = 0

11. zx = −(x3 − y2)−2(3x2) = −3x2(x3 − y2)−2; zy = −(x3 − y2)−2(−2y) = 2y(x3 − y2)−2

12. zx = 6(−x4 + 7y2 + 3y)5(−4x) = −24x(− x4 + 7y2− 3y)5; zy = 6(−x4 + 7y2 + 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


14. zx = (2x tan−1 y2)ex2 tan−1 y2

; zy =2x2y

1 + y4ex

2 tan−1 y2

15. fx = x(3x2yex3y + ex

3y; fy = x4ex3y

16. fθ = φ2

(cos

θ

φ

)(1φ

); fφ = φ2

(cos

θ

φ

)(− θ

φ2

)+ 2φ sin

θ

φ= −θ cos

θ

φ+ 2φ sin

θ

φ

17. fx =(x+ 2y)3− (3x− y)

(x+ 2y)2=

7y(x+ 2y)2

; fy =(x+ 2y)(−1)− (3x− y)(2)

(x+ 2y)2=

−7x(x+ 2y)2

18. fx =(x2 − y2)2y − xy

[2(x2 − y2)2x

](x2 − y2)4

=−3x2y − y3

(x2 − y2)3;

fy =(x2 − y2)x− xy

[2(x2 − y2)(−2y)

](x2 − y2)4

=3xy2 + x3

(x2 − y2)3

19. gu =8u

4u2 − 5v3; gv =

15v2

4u2 + 5v3

20. hr =1

2s√r

+√s

r2; hx = −

√r

s2− 1

2s√r

21. wx =y√x

; wy = 2√x− y

(1zey/z

)= 2√x−

(yz

+ 1)ey/z; wz = −yey/z

(− y

z2

)=y2

z2ey/z

22. wx = xy

(1x

)+ (lnxz)y = y + y lnxz; wy = x lnxz; wz =

xy

z

23. Fu = 2uw2 − v3 − vwt2 sin(ut2); Fv = −3uv2 + w cos(ut2);Fx = 3(2x2t)3(4xt) = 16xt(2x2t)3 = 128x7t4; Ft = −2uvwt sin(ut2) + 64x8t3

24. Gp = 2pq3e2r4s5

Gq = 3p2q2e2r4s5

Gr = p2q3(8r3s5)e2r4s5 = 8p2q3r3s5e2r4s5

Gs = p2q3(10r4s4)e2r4s5 = 10p1q3r4s4e2r4s5

25. zy = 16x3y3, zy(1,−1) = −16

26. zx = 12x2y4, zx(1,−1) = 12

27. fy =(x+ y)18x− 18xy

(x+ y)2=

18x2

(x+ y)2, fy(−1, 4) = 2. An equation of the tangent line is given

by x = −1 and z+24 = 2(y−4). Parametric equations of the line are x = −1, y = 4+t, z =−24 + 2t.

28. fx =(x+ y)18y − 18xy

(x+ y)2=

18y2

(x+ y)2, fx(−1, 4) = 32. An equation of the tangent line is given

by y = 4 and z+ 24 = 32(x+ 1). Symmetric equations of the line are x+ 1 =z + 24

32, y = 4.


29. zx =−x√

9− x−y2, zx(2, 2) = −2

30. zy =−y√

9− x2 − y2, zy(

√2,√

3) = −√

32

31.∂z

∂x= yexy;

∂2z

∂x2= y2exy

32.∂z

∂y= −2x4y−3;

∂2z

∂y2= 6x4y−4;

∂3z

∂y3= −24x4y−5

33. fx = 10xy2 − 2y3; fxy = 20xy − 6y2

34. f(p, q) = ln(p+ q)− 2 ln q, fq =1

p+ q− 2q, fqp = − 1

(p+ q)2

35. wt = 3u2v3t2, wtu6uv3t2; wtuv18uv2t2

36. wv = −u2 sin(u2v)

t3; wvv −

u4 cos(u2x)t3

; wvvt =3u4 cos(u2v)

t4

37. Fr = 2rer2

cos θ; Frθ − 2rer2

sin θ; Frθr − 2r(2rer2) sin θ − 2er

2sin θ = −2er

2(2r2 + 1) sin θ

38. Ht =(s− t)− (s+ t)(−1)

(s− t)2=

2s(s− t)2

; Htt =4s

(s− t)3;

Htts =(s− t)4 − 4x(3)(s− t)2

(x− t)6=−8s− 4t(s− t)4

39.∂z

∂y= −5x4y2+8xy;

∂2z

∂x∂y= −60x3y2+8y;

∂z

∂x= 6x5−20x3y3+4y2;

∂2z

∂y∂x= −60x3y2+8y

40.∂z

∂y=

2x1 + 4x2y2

;∂z

∂x∂y=

(1 + 4x2y2)2− 2x(8xy2)(1 + 4x2y2)2

=2− 8x2y2

(1 + 4x2y2)2;∂z

∂x=

2y1 + 4x2y2

∂z

∂y∂x=

(1 + 4x2y2)2− 2y(8x2y)(1 + 4x2y2)2

=2− 8x2y2

(1 + 4x2y2)2

41. wu = 3u2v4 − 8uv2t3, wuv = 12u2v3 − 16uvt3, wuvt = −48uvt2; wt = −12u2v2t2 + v2,wtv = −24u2vt2 +2v, wtvu = −48uvt2; wv = 4u3v3−8u2vt3 +2vt, wvu = 12u2v3−16uvt3,wvut = −48uvt2

42. Fη = 6η2(η3+ξ2+τ) = 6η5+6η2ξ2+6η2τ, Fηξ = 12η2ξ, Fηξη = 24ηξ; Fξ = 4ξ(η3+ξ3+τ) =4η3ξ + 4xi3 + 4xiτ, Fξη = 12η2ξ, Fξηη24ηξ; Fηη = 30η4 + 12ηξ2, Fηηξ = 24ηξ

43. 2x+ 2zzx = 0, zx = −x/z; 2y + 2zzy = 0, zy = −y/z

44. 2zzx = 2x+ y2zx =⇒ (2z − y2)zx = 2x =⇒ zx =2x

2z − y2;

2zzy = y2zy + 2yz =⇒ (2z − y2)zy = 2yz =⇒ zy =2yz

2z − y2


45. 2zzu + 2uv3 − uvzu − vz = 0 =⇒ (2z − uv)zu = vz − 2uv3 =⇒ zu =vz − 2uv3

2z − uv;

2zzv + 3u2v2 − uvzv − uz = 0 =⇒ (2z − uv)zv = uz − 3u2v2 =⇒ zv =uz − 3u2v2

2z − uv

46. sezzs + ez − test + 12s2t = zs =⇒ (sez − 1)zs = tes − ez − 12s2t =⇒ zs =test − ez − 12s2t

sez − 1;

sezzt − sest + 4s3 = zt =⇒ (sez − 1)zt = sest − 4s3 =⇒ zt =sest − 4s3

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.∂u

∂x= 2π(cosh 2πy + sinh 2πy) cos 2πx;

∂2u

∂x2= −4π2(cosh 2πy + sinh 2πy) sin 2πx;

∂u

∂y= (2π sinh 2πy + 2π cosh 2πy) sin 2πx;

∂2u

∂y2= (4π2 cosh 2πy + 4π2 sinh 2πy) sin 2πx;

∂2u

∂x2+∂2u

∂y2= −4π2(cosh 2πy + sinh 2πy) sin 2πx+ 4π2(cosh 2πy + sinh 2πy) sin 2πx = 0

50.∂u

∂x= −nπ

Le−(nπx/L sin

(nπL

);∂2u

∂x2=n2π2

L2e−(nπx/L) sin

(nπL

)y;

∂u

∂y=nπ

Le−(nπx/L) cos

(nπL

)y;

∂2u

∂y2= −n

2π2

L2e−(nπx/L) sin

(nπL

)y;

∂2u

∂x2+∂2u

∂y2=n2π2

L2e−(nπx/L) sin

(nπL

)− n2π2

L2e−(nπx/L) sin

(nπL

)= 0

51.∂z

∂x=

2xx2 + y2

,∂2z

∂x2=

(x2 + y2)2− 2x(2x)(x2 + y2)2

=2y2 − 2x2

(x2 + y2);∂z

∂y=

2yx2 + y2

,

∂2z

∂y2=

(x2 + y2)2− 2y(2y)(x2 + y2)2

=2x2 − 2y2

(x2 + y2)2;∂2z

∂x2+∂2z

∂y2=

2y2 − 2x2 + 2x2 − 2y2

(x2 + y2)2= 0

52.∂z

∂x= 2yex

2−y2sin 2xy + 2xex

2−y2cos 2xy = 2ex

2−y2(x cos 2xy − y sin 2xy),

∂2z

∂x2= 2ex

2−y2(−2xy sin 2xy + cos 2xy − 2y2 cos 2xy) + 4xex

2−y2(x cos 2xy − y sin 2xy);

∂z

∂y= −2xex

−y2sin 2xy − 2yex

−y2cos 2xy = −2ex

−y2(x sin 2xy + y cos 2xy),

∂2z

∂y2= −2ex

−y2(2x2 cos 2xy − 2xy sin 2xy + cos 2xy) + 4yex

−y2(x sin 2xy + y cos 2xy);

∂2z

∂x2+∂2z

∂y2=2ex

−y2(−2xy sin 2xy + cos 2xy − 2y2 cos 2xy + 2x2 cos 2xy − 2xy sin 2xy

− 2x2 cos 2xy + 2xy sin 2xy − cos 2xy + 2xy sin 2xy − 2y2 cos 2xy) = 0

53.∂u

∂x= − x

(x2 + y2 + z2)3/2;∂u

∂y= − y

(x2 + y2 + z2)3/2;∂u

∂z= − z

(x2 + y2 + z2)3/2;

∂2u

∂x2=

2x2 − y2 − z2

(x2 + y2 + z2)5/2;∂2u

∂y2=−x2 + 2y2 − z2

(x2 + y2 + z2)5/2;∂2u

∂z2=−x2 − y2 + 2z2

(x2 + y2 + z2)5/2;


∂2u

∂x2+∂2u

∂y2+∂2u

∂z2=

2x2 − y2 − z2 − x2 + 2y2 − z2 − x2 − y2 + 2z2

(x2 + y2 + z2)5/2= 0

54.∂u

∂x=√m2 + n2e

√m2+n2x cosmy sinnz;

∂2u

∂x2= (m2 + n2)e


∂u

∂y= −me

√m2+n2x sinmy sinnz;

∂2u

∂y2= −m2e


∂u

∂z= ne

√m2+n2x cosmy cosnz;

∂2u

∂z2= −n2e


∂2u

∂x2+∂2u

∂y2+∂2u

∂z2= (m2 + n2)e

√m2+n2x cosmy sinnz −m2e

√m2+n2x cosmy sinnz

− n2e√m2+n2x cosmy sinnz = 0

55.∂u

∂x= cos at cosx,

∂2u

∂x2= − cos at sinx;

∂u

∂t= −a sin at sinx,

∂2u

∂t2= −a1 cos at sinx;

a2 ∂2u

∂x2= a2(− cos at sinx) =

∂2u

∂t2

56.∂u

∂x= − sin(x+ at) + cos(x− at), ∂2u

∂x2= − cos(x+ at)− sin(x− at);

∂u

∂t= −a sin(x + at) − a cos(x − at),

∂2u

∂t2= −a2 cos(x + at) − a2 sin(x − at); a2 ∂

2u

∂x2=

−a2 cos(x+ at)− a2 sin(x− at) =∂2u

∂t2

57.∂C

∂x= −2x

ktt−/12e−x

/kt,∂2C

∂x2=

4x2

k2t2t−1/2e−x

2/kt − 2ktt−1/2e−x

2/kt;

∂C

∂t= t−1/2 x

2

kt2e−x

2/kt − t−3/2

2e−x

2/kt;k

4∂2C

∂x2=

x2

kt2t−1/2e−x

2/kt − t−1/2

2te−x

2/kt =∂C

∂t

58. (a) Pv = −k(T/V 2)

(b) PV = kt, PVT = k, VT = k/P

(c) PV = kT, V = kTp, Tp = V/k

59. (a)∂u

∂t={−gx/z, 0 ≤ x ≤ at−gt, x > at

For x > at, the motion is that of a freely falling body.

(b) For x > at,∂u

∂x= 0. For x > at, the string is horizontal.

60.∂S

∂h= 0.0790975w0.425h−0.275; Sh(60, 36) + 0.0790975(60)0.425(36)−0.275 ≈ 0.1682

The approximate increase in skin-area as h increases from 36 to 37 inches is 0.1682 sq ft.

61. (a)∂2z

∂x2= lim

∆x→0

fx(x+ ∆x, y)− fx(x, y)∆x

(b)∂2z

∂y2= lim

∆y→0

fy(x, y + ∆y)− fx(x, y)∆y


(c)∂2z

∂x∂y= lim

∆x→0

fy(x+ ∆x, y)− fy(x, y)∆x

62. Integrating zx = 2xy3 + 2y + 1/x with respect to x, we obtain z = x2y3 + 2xy + lnx+ φ(y).Then 3x2y2 + 2x + 1 = zy = 3x2y2 + 2x + φ′(y). Since φ′(y) = 1, φ(y) = y + C, andz = x2y3 + 2xy + lnx+ y + C.

63. Consider the mixed partials:

∂2z

∂y∂x=

∂

∂y

(∂z

∂x

)= 2y and

∂2z

∂x∂y=

∂

∂x

(∂z

∂y

)= 2x.

Since∂z

∂x,∂z

∂y,

∂2z

∂y∂x, and

∂2z

∂x∂yare all continuous on an open set, we should have

∂2z

∂y∂x=

∂2z

∂x∂yon that set. But the mixed partials are equal only on the line y = x, which

contains no open set in the plane. Therefore, such a function cannot exist.

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

∂xn,

∂nz

∂xn−1∂y,

∂nz

∂xn−2∂y2, . . . ,

∂nz

∂x∂yn−2,∂nz

∂yn.

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 impliesthat the surface must have constant height everywhere. Therefore f must have the formf(x, y) = c.

(b) Since the mixed partials are both zero, we have

∂

∂x

(∂z

∂y

)= 0 and dfrac∂∂y

(∂z

∂x

)= 0

which implies∂z

∂yis a function of y alone and

∂z

∂xis a function of x alone. Therefore, z

has no term that depends on both x and y. Hence z is of the form z = g(x) +h(y) + cwhere g and h are twice continuously differentiable functions of a single variable.

66. The level curves suggest that the surface height is decreasing as we move slightly to the right

of the point, and increasing as we move slightly up from the point. This implies∂z

∂x< 0 and

∂z

∂y> 0.

67.∂z

∂x

∣∣∣∣(0,0)

= lim∆x→0

f(0 + ∆x, 0)− f(0, 0)∆x

= lim∆x→0

0/2(∆x)2

∆x= 0;

∂z

∂y

∣∣∣∣(0,0)

= lim∆y→0

f(0, 0 + ∆y)− f(0, 0)∆y

= lim∆y→0

0/2(∆y)2

∆y= 0


68. (a)∂z

∂x=y5 − 4x2y3 − x4

(x2 + y2)2;∂z

∂x

∣∣∣∣(0,y)

= y;∂z

∂y=−x5 + 4x3y2 + xy4

x2 + y2)2;∂z

∂y

∣∣∣∣(x,0)

= −x

(b)∂2z

∂y∂x= 1;

∂2z

∂x∂y= −1 =⇒ ∂2z

∂y∂x6= ∂2z

∂x∂y

13.4 Linearization and Differentials

1.∂f

∂x= 4y2 − 6x2y so

∂f

∂x(1, 1) = −2

∂f

∂y= 8zy − 2x3 so

∂f

∂y(1, 1) = 6

f(1, 1) = 2 The linearization is L(x, y) = 2− 2(x− 1) + 6(y − 1) = −2x+ 6y − 2

2.∂f

∂x=

3x2y

2√x3y

so∂f

∂x(2, 2) = 3

∂f

∂y=

x3

2√x3y

so∂f

∂y(2, 2) = 1

f(2, 2) = 4 The linearization is L(x, y) = 4 + 3(x− 2) + (y − 2) = 3x+ y − 4

3.∂f

∂x=√x2 + y2 +

x2√x2 + y2

so∂f

∂x(8, 15) =

35317

∂f

∂y=

xy√x2 + y2

so∂f

∂y(8, 15) =

12017

f(8, 15) = 136 The linearization is L(x, y) = 136+ 35317 (x−8)+ 120

17 (y−15) = 35317 x+ 120

17 y−136

4.∂f

∂x= 3 cosx cos y so

∂f

∂x

(π4 ,

3π4

)=−32

∂f

∂y= 3 sinx sin y so

∂f

∂y

(π4 ,

3π4

)=−32

f(π4 ,

3π4

)=−32

The linearization is L(x, y) = −32 −

32

(x− π

4

)− 3

2

(y − 3π

4

)= −3

2 x −32y +

32 (π − 1)

5.∂f

∂x=

2xx2 + y3

so∂f

∂x(−1, 1) = −1

∂f

∂y=

3y2

x2 + y3so

∂f

∂y(−1, 1) =

32

f(−1, 1) = ln(2) The linearization is L(x, y) = ln(2)− (x+1)+ 32 (y−1) = −x+

32y− 5

2 +ln(2)

6.∂f

∂x= 3e−2y cos 3x so

∂f

∂x

(0, π3

)= 3e−

2π3

∂f

∂y= −2e−2y sin 3x so

∂f

∂y

(0, π3

)= 0

f(0, π3

)= 0 The linearization is L(x, y) = 3e−

2π3 (x− 0) = 3xe−

2π3

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

∂x=

12√x,

∂f

∂x(100, 81) =

120,

∂f

∂y=

14y3/2

,∂f

∂y(100, 81) =

14(27)

=1

108, and

f(100, 81) = 13The linearization is L(x, y) = 13 + 1

20 (x− 100) + 1108 (y− 81). For the approximation, we have

L9102, 80) = 13 + 120 (102− 100) + 1

108 (80− 81) = 13 + 110 −

1108 = 7069

540 ≈ 13.0907

8. We are trying to approximate f(36, 63) where f(x, y) =√x√y. Use the linearization of f at the

point (36, 64). To do this, compute∂f

∂x=

12√x√y,∂f

∂x(36, 64) =

196,∂f

∂y= −

√x

2y3/2 ;∂f

∂y(35, 64) = − 3

512, and f(36, 64) =

34.

The linearization if L(x, y) =34

+196

(x− 36)− 3512

(y− 64). For the approximation, we have

L(36, 63) =34

+196

(36− 36)− 3512

(63− 64) =387512≈ .7559.

9. First, linearize f at (2, 2). To do this, compute∂f

∂x= 2(x2 + y2)2x,

∂f

∂x(2, 2) = 64,

∂f

∂y= 2(x2 + y2)2y,

∂f

∂y(2, 2) = 64, and f(2, 2, ) = 64.

The linearizationis L(x, y) = 64 + 64)x − 2) + 64)y − 2). For the approximation, we haveL(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

∂x= −πy sin(πxy),

∂f

∂x

(12 , 3)

= 3π,∂f

∂y= −πx sin(πxy),

∂f

∂y

(12 , 3)

=π

2, and f

(12 , 3)

=

0.

The linearization is L(x, y) = 3π(x− 1

2

)+π

2(y − 3). For the approximation, we have

L(0.52, 2.96) = 3π(0.02) +π

2(−0.04) ≈ 0.1257.

11. dz = 2x sin 4ydx+ 4x2 cos 4ydy

12. dz =[x(2xex

2−y2) + ex

2−y2]dx− 2yxex

2−y2dy = (2x2 + 1)ex

2−y2dx− 2xyex

2−y2dy

13. dz =2x√

2x2 − 4y3dx− 6y2√

2x2 − 4y3dy

14. dz = 45x2y(5x3y + 4y5)2dx+ (15x3 + 60y4)(5x3y + 4y5)2dy

15. df =(s+ 3t)2− (2s− t)

(s+ 3t)2ds+

(s+ t)(−1)− (2s− t)3(s+ 3t)2

dt =7t

(s+ 3t)2− 7s

(s+ 3t)2dt

16. dg = 2r cos 3θdr − 3r2 sin 3θdθ

17. dw = 2xy4z−5dx+ 4x2y3z−5dy − 5x2y4z−6dz

18. dw = −2xe−z2

sin(x2 + y4)dx− 4y3e−z2

sin(x2 + y4)dy − 2ze−z2

cos(x2 + y4)dz

19. dF = 3r2dr − 2s−3ds− 2t−1/2dt


20. dG = sinφ cos θdρ− ρ sinφ sin θdθ + ρ cosφ cos θdφ

21. w = lnu+ ln v − ln s− ln t; dw =du

u+dv

v− ds

s− dt

t

22. dw =u√

u2 + s2t2 − v2du− v√

u2 + s2t2 − v2dv +

st2√u2 + s2t2 − v2

ds+s2t√

u2 + s2t2 − v2dt

23. ∆z = z(2.2, 3.9)− z(2, 4) = (6.6 + 15.6 + 8)− (6 + 16 + 8) = 0.1; dz = 3dx+ 4dyWhen 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)dyWhen 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, dz =2(3 + 1)(0.1) + 2(3 + 1)(0.2) = 0.8− 1.6 = −0.8

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 + 2xy2)dx + 2xydy. 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

ε1 = 5∆x, ε2 = −∆x

28. ∆z = 10(y + ∆y)2 + 3(x+ ∆x)− (x+ ∆x)− (10y2 + 3x− x2)

= 20y∆y + 10(∆y)2 + 3∆x− 2x∆x− (∆x)2

= (3− 2x)∆x+ 20y∆y − (∆x)∆x+ (10∆y)∆yε1 = −∆x, ε2 = 10∆y

29. ∆x = (x+ ∆x)2(y + ∆y)2 − x2y2 = [x2 + 2x∆x+ (∆x)2][y2 + 2y∆y + (∆y)2]− x2y2

= 2x2y∆y + x2(∆y)2 + 2xy2∆x+ 4xy(∆x)∆y + 2x(∆x)(∆y)2 + y2(∆x)2 + 2y(∆x)2∆y + (∆x)2(∆y)2

= 2xy2∆x+ 2x2y∆y + [4xy∆y + 2x(∆y)2 + y2x]∆x+ [x2∆y + 2y(∆x)2 + (∆x)2∆y]∆yε1 = 4xy∆y + 2x(∆y)2 + y2x, ε2 = x2∆y + 2y(∆x)2 + (∆x)2∆y (Several other choices of ε1and ε2 are possible.)

30. ∆z = (x+ ∆x)3 − (y + ∆y)3 − (x3 − y3) = 3x2∆x+ 3x(∆x)2 + (∆x)3 − 3y2∆y − 3y(∆y)2 − (∆y)3

= 3x2∆x− 3y2∆y + [3x∆x+ (∆x)2]− [3y∆y + (∆y)2]∆yε1 = 3x∆x+ (∆x)2, ε2 = −3y∆y − (∆y)2

31. R =R1R2R3

R2R3 +R1R3 +R1R2; ∆R1 = ±0.009R1, ∆R! = ±0.009R2, ∆R3 = ±0.0009R3


|∆R| ≈ |dR| ≤∣∣∣∣ R2

2R23

(R2R3 +R1R3 +R1R2)2(±0.009R1)

∣∣∣∣+∣∣∣∣ R2

1R23

(R2R3 +R1R3 +R1R2)2(±0.009R2)

∣∣∣∣+∣∣∣∣ R2

1R22

(R2R3 +R1R3 +R1R2)2(±0.009R3)

∣∣∣∣= 0.009R

(R2R3 +R1R3 +R1R2

R2R3 +R1R3 +R1R2

)= 0.009R

The maximum percentage error is approximately 0.9%.

32. We are given ∆T = ±0.006T and ∆V = ±0.008V. Then

|∆P | ≈ |dP | =∣∣∣∣ kV (±0.006T )− kT

V 2(±0.008V )

∣∣∣∣ ≤ kT

V(0.006) +

kT

V(0.008) = P (0.014).

Thus, the approximate maximum percentage error in P is 1.4%.

33. dT = mg(2r2 +R2 −R(2R)

(2r2 +R2)2dR+mg

−R(4r)2r2 +R2)2

dr = mg2r2 −R2

(2r+R2)2dR−mg 4rR

(2r+R2)2dr

When R = 4, r = 0.8, dR = 0.1, and dr = 0.1,

∆T ≈ dT = mg

[2(0.8)2 − 42

[2(0.8)2 + 42]2(0.1)− 4(0.8)4

[2(0.8)2 + 42]2(0.1)

]= mg

(−1.472− 1.28

298.598

)≈ −0.009 mg.

The tension decreases.

34. V = πr2h, dV = 2πrhdr + πr2dh. 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)dhWith 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.


37. dS = 0.1091(0.425)w−0.575h0.725dw + 0.1091(0.725)w0.425h−0.275dhWith dw = ±0.03w and dh = ±0.05h,

|∆S| ≈ |dS| = 0.1091|0.425w−0.575h0.725(±0.03w) + 0.725w0.425h−0.275(±0.05h)|≤ 0.1091[0.425w0.425h0.725(0.03)] + 0.1091[0.725w0.425h0.725(0.05)]

= 0.1091w0.425h0.725(0.013 + 0.036) = 0.049S.

The approximate maximum percentage error is 4.9%.

38. Z =

[R2 +

(100L− 1

1000c

)2]1/2

;

dZ =12

[r2 +

(100L− 1

1000c

)2]−1/2 [

2RdR+ 2(

100L− 11000c

)(100)dL

2(

100L− 11000c

) (1

1000c2

)dC

]= (R2 +X2)−1/2

(RdR+ 1000XdL+

X

1000c2dC

)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

dZ = (4002 + 3002)−1/2

[400(25+100(300)(0.05) +

3001000(10)−10

10−6

]=

1500

(10000 + 15000 + 3000) = 56 ohms.

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).

(b) Let f(x, y, z) =√x2 + y2 + z2. Then we wish to approximate f(9.1, 11.75, 19.98). To do

this, linearize f at (9, 12, 20). Compute

∂f

∂x=

x√x2 + y2 + z2

,∂f

∂x(9, 12, 20) =

925

∂f

∂y=

y√x2 + y2 + z2

,∂f

∂y(9, 12, 20) =

1225

∂f

∂z=

z√x2 + y2 + z2

,∂f

∂z(9, 12, 20) =

45

and f(9, 12, 20) = 25


The linearization is L(x, y, z) = 25 +925

(x − 9) +1225

(y − 12) +45

(z − 20). For the

approximation, we have L(9.1, 11.75, 19.98) = 25 +925

(0.1) +1225

(−0.25) +45

(−0.02) =24.9

40. According to Theorem 13.4.3, if f were differentiable at (0, 0), then f would have to becontinuous at (0, 0). However, as shown in Problem 38 in Exercises 13.2, f is not continuousat (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 graphcomes to a sharp ”point” at the origin, there is no possible increment formula for ∆zthat will work in every direction there.

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

f(0 + h)− f(0, 0)h

=√h2 + 02 −

√02 + 02

h

=

√h2

h=|h|h

= 1

But if h < 0, thenf(0 + h)− f(0, 0)

h= −1.

Therefore, limh→0

f(0 + h)− f(0, 0)h

does not exist. But this means fx does not exist at

(0, 0) and thus f is not differentiable at (0, 0).


42.

xy

z

Δy Δx

Δz

xy

z

∆V

xy

z

xy

z

dV ∆V − dV

L lθ α

θ

φ

(x0,y0)

43. (a) From the figure we see that α = π − (θ + φ). Thenxh = L cos θ + l cosα = L cos θ + l cos(π − θ − φ)

= L cos θ − l cos(θ + φ)and yh = L sin θ − l sinα = L sin θ − l sin(π − θ − φ)

= L sin θ − l sin(θ + φ).

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 = −yhdθ + (ye − yh)dφdyh = (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 thatθ = 0 and φ = 270◦. The other position has the lower arm reaching straight across, withthe elbow on the y-axis, so that θ = φ = 90◦. In both cases, (xh, yh) = (L,L). In the firstcase, (xe, ye) = (L, 0), and in the second case (xe, ye) = (0, L). In general, the approximatemaximum error in xh is

|dxh| = | − hhdθ + (yeyh)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 caseit 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

H = tv sin θ − 12gt2 = D tan θ − 1

2g

(D

v cos θ

)2

= D tan θ − 12gD2

v2sec2 θ.

(b) dH =∂H

∂vdv +

∂H

∂θdθ = g

D2

v3sec2 θdv +

(D sec2 θ − g d

2

v2sec2 θ tan θ

)dθ

(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

|dH| ≤ 32(

1002

1003

)sec2 π

4 (1) +[100 sec2 π

4 − 321002

1002sec2 π

4 tan π4

] (π

180

)=

1625

+34π45≈

3.01 ft.(e) We have

dH =∂H

∂v+∂H

∂θ+∂H

∂D= g

D2

v3sec2 θdv +

(D sec2 θ − gD

2

v2sec2 θ tan θ

)+(

tan θ − g Dv2

sec2 θ

)dD.

With |dD| ≤ 2, we obtain dH ≤ 1625

+34π45

+1825

=3425

+34π45≈ 3.73 ft.

13.5 Chain Rule

1.dz

dt=∂z

∂x

dx

dt+∂z

∂y

dy

dt=

2xx2 + y2

(2t) +2y

x2 + y2(−2t−3)

=4xt− 4yt−3

x2 + y2

2.dz

dt=∂z

∂x

dx

dt+∂z

∂y

dy

dt

= (3x2y − y4)(5e5t) + (x3 − 4xy3) (5 sec(t) tan(t))


3.dz

dt=∂z

∂x

dx

dt+∂z

∂y

dy

dt= −3 sin(3x+ 4y)(2)− 4 sin(3x+ 4y)(−1)

At t = π, x =5π2

and y = −5π4

sodz

dt

∣∣∣∣t=π

= −6 sin(

15π2− 5π

)+ 4 sin

(15π2− 5π

)= −6 + 4 = −2

4.dz

dt=∂z

∂x

dx

dt+∂z

∂y

dy

dt= yexy

(−8

(2t+ 1)2

)+ xexy(3)

At t = 0, x = 4 and y = 5

sodz

dt

∣∣∣∣t=0

= −40e20 + 12e20 = −28e20

5.dp

du=

12s+ t

(2u)− 2r(2s+ t)2

(− 2u3

)− r

(2s+ t)2

(1

2√u

)=

2u2s+ t

+4r

u3(2s+ t)2− r

2√u(2s+ t)2

6.dr

ds=y2

z3(− sin s) +

2xyz3

(cos s)− 3xy2

z4(sec2 s) = −y

2 sin sz3

+2xy cos s

z3− 3xy2 sec2 s

z4

7. zu =∂z

∂x

∂x

∂u+∂z

∂y

∂y

∂u= y2exy

2(3u2) + 2xyexy

2(1)

= 3u2y2exy2

+ 2xyexy2

zv =∂z

∂x

∂x

∂v+∂z

∂y

∂y

∂v= y2exy

2(0) + 2xyexy

2(−2v)

= −4vxyexy2

8. zu =∂z

∂x

∂x

∂u+∂z

∂y

∂y

∂u= 2x cos 4y(2uv3)− 4x2 sin 4y(3u2)

= 4uv3x cos 4y − 12u2x2 sin 4y

zv =∂z

∂x

∂x

∂v+∂z

∂y

∂y

∂v= 2x cos 4y(3u2v2)− 4x2 sin 4y(3v2)

= 6u2v2x cos 4y − 12v2x2 sin 4y

9. zu = 4(4u3)− 10y[2(2u− v)(2)] = 16u3 − 40(2u− v)y

zv = 4(−24v2)− 10y[2(2u− v)(−1)] = −96v2 + 20(2u− v)y

10. zu =2y

(x+ y)2

1v

+−2x

(x+ y)2

(− v

2

u2

)=

2yv(x+ y)2

+2xv2

u2(x+ y)2

zv =2y

(x+ y)2

(− u

v2

)+−2x

(x+ y)2

(2vu

)= − 2yu

v2(x+ y)2− 4xvu(x+ y)2

11. wt =32

(u2 + v2)1/2(2u)(−e−t sin θ) +32

(u2 + v2)1/2(2v)(−e−t cos θ)

= −3u(u2 + v2)1/2e−t sin θ − 3v(u2 + v2)1/2e−t cos θ

wθ =32

(u2 + v2)1/2(2u)e−t cos θ +32

(u2 + v2)1/2(2v)(−e−t sin θ)

= 3u(u2 + v2)1/2e−t cos θ − 3v(u2 + v2)1/2e−t sin θ

13.5. CHAIN RULE 101

12. wr =v/2√uv

1 + uv(2r) +

u/2√uv

1 + uv(2rs2) =

rv√uv(1 + uv)

+rs2u√

uv(1 + uv)

ws =v/2√uv

1 + uv(−2s) +

u/2√uv

1 + uv(2r2s) =

−sv√uv(1 + uv)

+r2su√

uv(1 + uv)

13. Ru = s2t4(ev2) + 2rst4(−2uve−u

2) + 4rs2t3(2uv2eu

2v2) = s2t4ev2− 4uvrst4e−u

2+ 8uv2rs2t3eu

2v2

Rv = s2t4(2uvev2) + 2rst4(e−u

2) + 4rs2t3(2u2veu

2v2) = 2s2t4uvev2

+ 2rst4e−u2

+ 8rs2t3u2veu2v2

14. Qx =1P

(t2√

1− x2

)+

1q

(1t2

)+

1r

(1/t

1 + (x/t)2

)=

t2

p√

1− x2+

1qt2

+t

r(t2 + x2)

Qt =1P

(2t sin−1 x) +1q

(−2xt3

)+

1r

(−x/t2

1 + (x/t)2

)=

2t sin−1 x

p− 2xqt3− x

r(t2 + x2)

15. wt =2x

2√x2 + y2

u

rs+ tu+

2y

2√x2 + y2

cosh rsu

=xu√

x2 + y2(rs+ tu)+

y cosh rs

u√x2 + y2

wr =2x

2√x2 + y2

s

rs+ tu+

2y

2√x2 + y2

st sinh rsu

=xs√

x2 + y2(rs+ tu)− yst sinh rs

u√x2 + y2

wu =2x

2√x2 + y2

t

rs+ tu+

2y

2√x2 + y2

−t cosh rsu2

=xt√

x2 + y2(rs+ tu)− yt cosh rs

u2√x2 + y2

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′)− 4xy2 + y′ = 0 =⇒ (1− 4x2y)y′ = 4xy2 − 3x2 =⇒ y′ =4xy2 − 3x2

1− 4x2y

(b) fx = 3x2 − 4xy2, fy = −4x2y + 1; y′ = −3x2 − 4xy2

−4x2y + 1=

4xy2 − 3x2

1− 4x2y

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

ey − 4y

(b) f(x, y) = x+ 2y2 − ey; fx = 1, fy = 4y − ey; y′ = − 14y − ey

=1

ey − 4y

19. (a) y′ = (cosxy)(xy′ + y) =⇒ (1− x cosxy)y′ = y cosxy =⇒ y′ =y cosxy

1− x cosxy(b) f(x, y) = y − sinxy; fx = −y cosxy, fy = 1− x cosxy;

y′ = − −y cosxy1− x cosxy

=y cosxy

1− x cosxy

20. (a)23

(x+ y)−1/3(1 + y′) = xy′ + y =⇒ 2(x+ y)−1/3 + 2(x+ y)−1/3y′ = 3xy′ + 3y

=⇒[2(x+ y)−1/3 − 3x

]y′ = 3y − 2(x+ y)−1/3 =⇒ y′ =

3y − 2(x+ y)−1/3

2(x+ y)−1/3 − 3x

(b) f(x, y) = (x+ y)2/3 − xy; fx =23

(x+ y)−1/3 − y, fy =23

(x+ y)−1/3 − x;

y′ = −23 (x+ y)−1/3 − y23 (x+ y)−1/3 − x

=3y − 2(x+ y)−1/3

2(x+ y)−1/3 − 3x


21. Fx = 2x, Fy = 2y, Fz = −2z;∂z

∂x= − 2x−2z

=x

z;∂z

∂y= − 2y−2z

=y

z

22. Fx =23x−1/3, Fy =

23y−1/3, Fz =

23z−1/3;

∂z

∂x= − (2/3)x−1/3

(2/3)z−1/3= − z

1/3

x1/3;

∂z

∂y= − (2/3)y−1/3

(2/3)z−1/3= −z

1/3

y1/3

23. F (x, y, z) = xy2z3 + x2 − y2 − 5z2, Fx = y2z3 + 2x, Fy = 2xyz3 − 2y, Fz = 3xy2z2 − 10z∂z

∂x= − y2z3 + 2x

3xy2z2 − 10z=

y2z3 + 2x10z − 3xy2z2

;∂z

∂y= − 2xyz3 − 2y

3xy2z2 − 10z=

2xyz3 − 2y10z − 3xy2z2

24. F (x, y, z) = z − ln(xyz); Fx = − 1x, Fy = −1

y, Fz = 1− 1

z;∂z

∂x= − −1/x

1− 1/z=

z

xz − x;

∂z

∂y= − −1/y

1− 1/z=

z

yz − y

25. Let y = x+ at and z = x− at. Then u(x, t) = F (y) +G(z) and

∂u

∂x=dF

dy

∂y

∂x+dG

dz

∂z

∂x=dF

dy+dG

dz;

∂2u

∂x2=d2F

dy2

∂y

∂x+d2G

dz2

∂z

∂x=d2F

dy2+d2G

dz2;

∂u

∂t=dF

dy

∂y

∂t+dG

dz

∂z

∂t= a

dF

dy− adG

dz;

∂2u

∂xt2= a

d2F

dy2

∂y

∂t− ad

2G

dz2

∂z

∂t= a2 d

2F

dy2+ a2 d

2G

dz2.

Thus, a2 ∂2u

∂x2= a2 d

2F

dy2+ a2 d

2G

dz2=∂2u

∂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

∂ξ=∂u

∂x

∂x

∂ξ+∂u

∂t

∂t

∂ξ=

12∂u

∂x− 1

2a∂u

∂t

and∂2u

∂η∂ξ=

12∂2u

∂x2

∂x

∂η− 1

2a∂2u

∂t2∂t

∂η=

14∂2u

∂x2− 1

4a2

∂2u

∂t2

Setting∂2u

∂η∂ξ= 0, we have

14∂2u

∂x2− 1

4a2

∂2u

∂t2= 0 or a2 ∂

2u

∂x2=∂2u

∂t2.

27. With x = r cos θ and y = r sin θ

∂u

∂r=∂u

∂x

∂x

∂r+∂u

∂y

∂y

∂r=∂u

∂xcos θ +

∂u

∂ysin θ

∂2u

∂r2=∂2u

∂x2

∂x

∂rcos θ +

∂2u

∂y2

∂y

∂rsin θ =

∂2u

∂x2cos2 θ +

∂2u

∂y2sin2 θ

∂u

∂θ=∂u

∂x

∂x

∂θ+∂u

∂y

∂y

∂θ=∂u

∂y(r sin θ) +

∂u

∂y(−r sin θ) +

∂2u

∂y2

∂y

∂θ(r cos θ)

= −r ∂u∂x

cos θ + r2 ∂2u

∂x2sin2 θ − r ∂u

∂ysin θ − r2 ∂

2u

∂y2cos2 θ.


Using∂2u

∂x2+∂2u

∂y2= 0, we have

∂2u

∂r2+

1r

∂u

∂r+

1r2

∂2u

∂θ2=∂2u

∂x2cos2 θ +

∂2u

∂y2sin2 θ +

1r

(∂u

∂xcos θ +

∂u

∂ysin θ

)+

1r2

(−r ∂u

∂x+ r2 ∂

2u

∂x2sin2 θ − r ∂u

∂ysin θ + r2 ∂

2u

∂y2cos2 θ

)=∂2u

∂x2(cos2 θ + sin2 θ) +

∂2u

∂y2(cos2 θ + sin2 θ) +

∂u

∂x

(1r

cos θ − 1r

cos θ)

+∂u

∂y

(1r

sin θ − 1r

sin θ)

=∂2u

∂x2+∂2u

∂y2= 0

28.∂z

∂x=dz

du

∂u

∂x,∂z

∂y=dz

du

∂u

∂y

29. Letting u = y/x in Problem 28, we have

x∂z

∂x+ y

∂z

∂y= x

dz

du

∂u

∂x+ y

dz

du

∂u

∂y= x

dz

du

(− y

x2

)+ y

dz

du

(1x

)=dz

du

(−yx

+y

x

)= 0.

30. We first compute∂u

∂x=∂u

∂r

∂r

∂x=∂u

∂r

x√x2 + y2

∂2u

∂x2=∂u

∂r

y2

(x2 + y2)3/2+

x√x2 + y2

∂2u

∂r2=∂u

∂r

y2

(x2 + y2)3/2+∂2u

∂r2

x2

x2 + y2

∂2u

∂y2=∂u

∂r

x2

(x2 + y2)3/2

∂2u

∂r2

y2

x2 + y2.

Then∂2u

∂x2+∂2u

∂y2=∂u

∂r

x2 + y2

(x2 + y2)3/2+∂2u

∂r2

x2 + y2

x2 + y2=∂2u

∂r2+

1r

∂u

∂r.

31. We first compute

∂u

∂x= B

∂

∂xerf

(x√4kt

)= B

∂

∂x

(2π

∫ x/√

4kt

0

e−v2dv

)= B

2√π

1√4kt

e−x2/4kt

∂2u

∂x2= B

2√π

1√4kt

(− x

2kt

)e−x

2/4kt = −B x

2k√π√kt3

e−x2/4kt

∂u

∂t= B

∂

∂terf

(x√4kt

)= B

∂

∂t

(2π

∫ x/√

4kt

0

e−v2dv

)= B

2√π

[− x

2√k

(−1

2t−3/2

)]e−x/4kt

= −B x

2√π√kt3

e−x2/4kt

Then k∂2u

∂x2=∂u

∂t.


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

dt=

∂I

∂E

dE

dt+∂I

∂R

dR

dt=

1R

(2)− E

R2(−1),

and when E = 60 and R = 50,dI

dt=

250

+60502

=125

+3/525

=8

125amp/min.

33. Since the height of the triangle is x sin θ, the area is given by A = 12xy sin θ. Then

dA

dt=∂A

∂x

dx

dt+∂A

∂y

dy

dt+∂A

∂θ

dθ

dt=

12y sin θ

dx

dt+

12x sin θ

dy

dt+

12xy cos θ

dθ

dt.

When x = 10, y = 8, θ = π/6, dz/dt = 0.3, dy/dt = 0.5, and dθ/dt = 0.1,

dA

dt=

12

(8)(

12

)(0.3) +

12

(10)(

12

)(0.5) +

12

(10)(8)

(√3

2

)(0.1)

= 0.6 + 1.25 + 2√

3 = 1.85 + 2√

2 ≈ 5.31 cm2/s.

34.dP

dt=

(V − 0.0427)(0.08)dT/dt(V − 0.0427)2

− 0.08T (dV/dt)(V − 0.0427)2

+3.6V 3

dV

dt

=0.08

V − 0.0427dT

dt+(

3.6V 3− 0.08T

(V − 0.0427)2

)dV

dt

35.dS

dt= 0.1091

(0.425w−0.575h0.725 dw

dt+ 0.725w0.425h−0.275 dh

dt

)When w = 25, h = 29, dw/dt = 4.2, and dh/dt = 2,

dS

dt= 0.1091[0.425(25)−0.575(29(0.725(4.2) + 0.725(25)0.425(29)−0.275(2)] ≈ 0.5976in2/yr.

36.dw

dt=∂w

∂x

dx

dt+∂w

∂y

dy

dt+∂w

∂z

dz

dt=xdx/dt+ ydy/dt+ zdz/dt√

x2 + y2 + z2=

−4x sin t+ 4y cos t+ 5z√16 cos2 t+ 16 sin2 t+ 25t2

=−16 sin t cos t+ 16 sin t cos t+ 25t√

16 + 25t2=

25t√16 + 25t2

dw

dt

∣∣∣∣t=5π/2

=125π/2√

16 + 625π2/4=

125π√64 + 625π2

≈ 4.9743

37. Since dT/dT = 1 and∂P

∂T= 0,

0 = FT =∂F

∂P

∂P

∂T+∂F

∂V

∂V

∂T+∂F

∂T

∂T

∂T=⇒ ∂V

∂T= − ∂F/∂T

∂F/∂V= − 1

∂T/∂V.

38. (a) From the law of sines,r

sinφ=

500sin(π − θ − φ)

so r =500 sinφ

sin(θ + φ).

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


(c) Using the chain rule, we obtain

dr

dt=∂r

∂θ

dθ

dt+∂r

∂φ

dφ

dt

= −500 sinφ cos(θ + φ)sin2(θ + φ)

dθ

dt+ 500

sin(θ + φ) cosφ− sinφ cos(θ + φ)sin2(θ + φ)

dφ

dt


dθ

dt

+ 500[sin θ cosφ+ cos θ sinφ] cosφ− sinφ[cos θ cosφ− sin θ sinφ]

sin2(θ + φ)dφ

dt


dθ

dt+

500 sin θsin2(θ + φ)

dφ

dt.

When dθ/dt = 5◦ = 5π/180 and dφ/dt = −10π/180, we have

dr

dt= −500 sin 75◦ cos 137◦

sin2 137◦

(5π180

)+

500 sin 62◦

sin2 137◦

(−10π

180

)≈ −99.4yd/min.

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.

(b) We first note that∂f

∂V=

c

4π

(A

lV

−1/2)(− A

lv2

)= − c

4π

√A

lV

1V

= − 12V

f and

∂f

∂l= − 1

2lf.

Thendf

dt=∂f

∂V

dV

dt+∂f

∂l

dl

dt= − 1

2VfdV

dt− 1

2lfdl

dt= −f

2

(1V

dV

dt+

1l

dl

dt

).

Using dV/dt = −10, dl/dt = 1, V = 100, and l = 6 we find

df

dt= −f

2

[1

100(−10) +

16

(1)]

= −f2

(16− 1

10

)< 0.

The frequency is decreasing.

40. (a)

�w�x

�w�y

�w�z

�y�x

�z�x

w

x y z

x x


dw

dx=∂w

∂x+∂w

∂y

y

x+∂w

∂z

dz

dx

(b) Using the formula from Part (a), we have

dw

dx= (y2 + 1) + (2xy − 2z)

(1x

)+ (−2y)(ex)

41.

�u�t1 �u

�t2�u�t3

�u�t4

�v�t1 �v

�t2�v�t3

�v�t4

�w�t1 �w

�t2�w�t3

�w�t4

�z�u

�z�v

�z�w

z

u v w

t1 t1 t1t2 t2 t2t3 t3 t3t4 t4 t4

∂z

∂t2=∂z

∂u

∂u

∂t2+∂z

∂v

∂v

∂t2+∂z

∂w

∂w

∂t2∂z

∂t4=∂z

∂u

∂u

∂t4+∂z

∂v

∂v

∂t4+∂z

∂w

∂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

∂w

∂x= Fx(x, y, z, u)

dx

dx+ Fu(x, y, z, u)

∂u

∂x+ FZ(x, y, z, u)

∂z

∂x+ Fu(x, y, z, u)

∂u

∂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 + x2yu+ 2xy3u− u4 − 8 we find FZ = −yz + 2xyu+ 2y3u,Fy = −xz + x2u+ 6xy2u, Fz = −xy, and Fu = x2y + 2xy3 − 4u3. Then

∂u

∂x= −−yz + 2xyu+ 2y3u

x2y + 2xy3 − 4u3,

∂u

∂y= −−xz + x2y + 6xy2u

x2y + 2xy3 − 4u3,

∂u

∂z=

xy

x2y + 2xy3 − 4u3.

44. (a) Let u = λx and v = λy. Then f(u, v) = λnf(x, y), and differentiating both sides withrespect to λ, we have

∂f

∂u

∂u

∂λ+∂f

∂v

∂v

∂λ= nλn−1f(x, y) or xfu(u, v) + yfu(u, v) = nλn−1f(x, y).

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(λy3)− 3(λx)(λy)4 + (λx)5 = λ5f(x, y)

13.6. DIRECTIONAL DERIVATIVE 107

(c) xfx + yfy = x(8xy3 − 3y4 + 5x4) + y(12x2y2 − 12xy3)

= 8x2y3 − 3xy4 + 5x5 + 12x2y3 − 12xy4 = 20x2 − 15xy4 + 5x5

= 5(4x2y3 − 3xy4 + x5) = 5f(x, y)

(d) By observing that f(λy

λx

)= f

(yx

)= λ0f

(yx

), we see that z = f

(yx

)is homogeneous

of degree zero.

13.6 Directional Derivative

1. ∇f = (2x− 3x2y2)i + (4y3 − 2x3y)j

2. ∇f = 4xye−2x2yi + (1 + 2x2e−2x2y)j

3. ∇F =y2

z3i +

2xyz3

j− 3xy2

z4k

4. ∇F = y cos yzi + (x cos yz − xyz sin yz)j− xy2 sin yzk

5. ∇f = 2xi− 8yj; ∇f(2, 4) = 4i− 32j

6. ∇f =3x2

2√x3y − y4

i +x3 − 4y3

2√x3y − y4

j; ∇f(3, 2) =27√38

i− 52√

38j

7. ∇F = 2xz2 sin 4yi + 4x2z2 cos 4yj + 2x2z sin 4yk

∇F (−2, π/3, 1) = −4 sin4π3

i + 16 cos4π3

j + 8 sin4π3

k = 2√

3i− 8j− 4√

3k

8. ∇F =2x

x2 + y2 + z2i +

2yx2 + y2 + z2

j +2z

x2 + y2 + z2k; ∇F (−4, 3, 5) = − 4

25i +

325

j +125

k

9. Duf(x, y) = limh→0

f(x+ h√

3/2, y + h/2)− f(x, y)h

= limh→0

(x+ h√

3/2)2 + (y + h/2)2)2x − y2

h

= limh→0

h√

3x+ 3h2/4 = hy + h2/4h

= limh→0

(√

3x+ 3h/4 + y + h/4) =√

3x+ y

10. Duf(x, y) = limh→0

f(x+ h√

2/2, y + h√

2/2)− f(x, y)h

= limh→0

3x+ 34√

2/2− (y + h√

2/2)2 − 3x+ y2

h

= limh→0

3h√

2/2− h√

2yh2/2h

= limh→0

(3√

2/2−√

2y − h/2) = 3√

2/2−√

2y

11. u =√

32

i12j; ∇f = 15x2y6i + 30x3y5j; ∇f(−1, 1) = 15i − 30j; Duf(−1, 1) =

15√

32− 15 =

152

(√

3− 2)


12. u =√

22

i

√2

2j; ∇f = (4 + y2)i(2xy − 5)j; ∇f(3,−1) = 5i− 11j;

Duf(3,−1) =5√

22− 11

√2

2= −3

√2

13. u =√

1010

i− 3√

1010

j; ∇f =−y

x2 + y2i +

x

x2 + y2j; ∇f(2,−2) =

14i +

14j;

Duf(2,−2) =√

1040− 3√

1040

= −√

1020

14. u =610

i +810

j; ∇f =y2

(x+ y)2i +

x2

(x+ y)2j; ∇f(2,−1) = i + 4j;

Duf(2,−1) =35

+165

=195

15. u = (2i + j)/√

5; ∇f = 2y(xy + 1)i + 2x(xy + 1)j; ∇f(3, 2) = 28i + 42j;

Duf(3, 2) =2(28)√

5+

42√5

=98√

5

16. u = −i; ∇F = 2x tan yi + x2 sec2 yj; ∇f(1/2, π/3) =√

3i + j; Duf(1/2, π/3) = −√

3

17. u =1√2j +

1√2k; ∇f = 2xy2(2z + 1)2i2x2y(2z + 1)2j + 4x2y2(2z + 1)k; ∇f(1,−1, 1) =

18i− 18j + 12k; Duf(1,−1, 1) = − 18√2

+12√

2= − 6√

2= −3

√2

18. u =1√6i− 2√

6j +

1√6k; ∇f =

2xz2

i− 2yz2

j +2y2 − 2x2

z3k; ∇f(2, 4,−1) = 4i− 8j− 24k;

Duf(2, 4,−1) =4√6− 16√

6− 24√

6= −6

√6

19. u = −k; ∇f =xy√

x2y + 2y2zi +

x2 + 4z

2√x2y + 2y2z

j +y2√

x2y + 2y2zk;

∇f(−2, 2, 1) = −i + j + k; Duf(−2, 2, 1) = −1

20. u = −(4i−4j+2k)/√

36 = −23i+

23j− 1

3k; ∇f = 2i−2yj+2zk; ∇f(4,−4, 2) = 2i+8j+4k;

Duf(4,−4, 2) = −43

+163− 4

3=

83

21. u = (−4i− j/√

17; ∇f = 2(x− y)i− 2(x− y)j; ∇f(4, 2) = 4i− 4j;

Duf(4, 2) = − 16√17

+4√17

= − 12√17

22. u = (−2i + 5j/√

29; ∇f = (3x2 − 5y)i− (5x− 2y)j; ∇f(1, 1) = −2i− 3j;

Duf(1, 1) =4√29− 15√

29= − 11√

29

23. ∇f = 2e2x sin yi + e2x cos yj; ∇f(0, π/4) =√

2i +√

22

j

The maximum Du is[(√

2)2 + (√

2/2)2]1/2

=√

5/2 in the direction√

2i + (√

2/2)j.


24. ∇f = (xyex−y + yex−yi + (−xyex−y + xex−yj; ∇f(5, 5) = 30i− 20jThe maximum Du is

[302 + (−20)2

]1/2 = 10√

13 in the direction 30i− 20j.

25. ∇f = (2x+ 4z)i + 2z2j + (4x+ 4yz)k; ∇f(1, 2,−1) = −2i + 2j− 4kThe maximum Du is

[(−2)2 + (2)2 + (−4)2

]1/2 = 2√

6 in the direction −2i + 2j− 4k.

26. ∇f = yzi + xzj + xyk; ∇f(3, 1,−5) = −5i− 15j + 3kThe maximum Du is

[(−5)2 + (−15)2 + (3)2

]1/2 =√

259 in the direction −5i− 15j + 3k.

27. ∇f = 2x sec2(x2 + y2)i + 2y sec2(x2 + y2)j;∇f(

√π/6,

√π/6) = 2

√π/6 sec2(π/3)(i + j) = 8

√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 = 3x2i− 3y2j; ∇f(2,−2) = 12i− 12j = 12(i− j)The minimum Du is −12

[12 + (−1)2

]1/2 = −12√

2 in the direction −(i− j) = −i + j.

29. ∇f =√zey

2√x

i +√xzeyj +

√x

2√zk; ∇f(16, 0, 9) =

38i + 12j +

23k. The minimum Du is

−[(3/8)2 + 122 + (2/3)2

]1/2 = −√

83281/24 in the direction −38i− 12j− 2

3k.

30. ∇f =1x

i +1y

j− 1zk; ∇f(1/2, 1/6, 1/3) = 2i + 6j− 3k

The minimum Du is −[22 + 62(−3)2

]1/2 = −7 in the direction −2i− 6j + 3k.

31. Using implicit differentiation on 2x2 + y2 = 9 we find y′ = −2x/y. At (2, 1) the slope ofthe 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.

32. ∇f = (2x + y − 1)i + (x + 2y)j; Duf(x, y) =2x+ y − 1√

2+x+ 2y√

2=

3x+ 3y − 1√2

Solving

(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 = ±(

35i− 4

5j).

(b) u = (4i + 3j)/√

16 + 9 =45i +

35j

(c) u = −45i− 3

5j

34. D−uf(a, b) = ∇f(a, b) · (−u) = −∇f(a, b) · u = −Duf(a, b) = −6

35. (a) ∇f = (3x2 − 6xy2)i + (−6x2y + 3y2)j

Duf(x, y) =3(3x2 − 6xy2)√

10+−6x2y + 3y2

√10

=9x2 − 18xy2 − 6x2y + 3y2

√10


(b) F (x, y) =3√10

(3x2 − 3xy2 − 2x2y + y2); ∇F =3√10

[(6x − 6y2 − 4xy)i + (−12xy −

2x2 + 2y)j]

DuF (x, y) =(

3√10

)(3√10

)(6x− 6y2 − 4xy) +

(1√10

)(3√10

)(−12xy − 2x2 + 2y)

=95

(3x− 3y2 − 2xy) +35

(−6xy − x2 + y) =15

(27x− 27y2 − 36xy − 3x2 + 3y)

36. Let ∇f(a, b) = αi + βj. Then Duf(a, b) = ∇f(a, b) · u =513α− 12

13β = 7 and Dvf(a, b) =

∇f(a, b) · v =513α− 12

13β = 3. Solving for α and β, we obtain α = 13 and β = −13/6. Thus,

∇f(a, b) = 13i− (13/6)j.

37.

38. ∇f = 〈2x,−5y〉, |∇f | =√

10x2 + 25y2 = 10, 4x2 + 25y2 = 100,x2

25+y2

4= 1

y

x

39. ∇T = 4xi + 2yj; ∇T (4, 2) = 16i + 4j. The minimum change in temperature (that is, themaximum 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 ofa tangent vector is x′(t)i + y′(t)j. Since we want the direction of motion to be −∇T (x, y), wehave x′(t)i + y′(t)j = −∇T (x, y) = 4xi + 2yj. Separating variables in dx/dt = 4x, we obtaindx/x = 4dt, lnx = 4t + c1, and x = C1e

4t. Separating variables in dy/dt = 2y, we obtaindy/y = 2dt, ln y = 2t + c2, and y = C2e

2t. Since x(0) = 4 and y(0) = 2, we have x = 4e4t

and y = 2e2t. The equation of the path is 4e4ti+2e2tj for t ≥ 0, or eliminating the parameter,x = y2, 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 ofa tangent vector is x′(t)i + y′(t)j. Since we want the direction of motion to be ∇T (x, y), wehave x′(t)i + y′(t)j = ∇T (x, y) = −4xi− 2yj. Separating variables in dx/dt = −4x, we obtaindx/x = −4dt, lnx = −4t + c1, and x = C1e

−4t. Separating variables in dy/dt = −2y, weobtain dy/y = −2dt, ln y = −2t+ c2, and y = C2e

−2t. Since x(0) = 3 and y(0) = 4, we havex = 3e−4t and y = 4e−2t. The equation of the path is 3e−4ti + 4e−2tj, or eliminating theparameter, 16x = 3y2, y ≥ 0.


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

x2 + y2 + z2we see that k = 500

and T (x, y, z) =500

x2 + y2 + z2.

(a) u =13〈1,−2,−2〉 =

13i− 2

3j− 2

3k

∇T = − 1000x(x2 + y2 + z2)2

i− 1000y(x2 + y2 + z2)2

j− 1000z(x2 + y2 + z2)2

k

∇T (2, 3, 3) = −500121

i− 750121

j− 750121

k

DuT (2, 3, 3) =13

(−500

121

)− 2

3

(−750

121

)− 2

3

(−750

121

)=

2500363

(b) The direction of maximum increase is

∇T (2, 3, 3) = −500121

i− 750121

j− 750121

k =252121

(−2i− 3j− 3k).

(c) The maximum rate of change of T is |∇T (2, 3, 3)| = 250121√

4 + 9 + 9 =250√

22121

.

43. ∇U =Gmx

(x2 + y2)3/2i +

Gmy

(x2 + y2)3/2j =

Gm

(x2 + y2)3/2(xi + yj)

The maximum and minimum values of DuU(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 minimumincrease 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.

44. Since ∇f = fx(x, y)i + fy(x, y)j, we have ∂f/∂x = 3x2 + y3 + yexy. Integrating, we obtainf(x, y) = x3 + xy3 + exy + g(y). Then fy = 3xy2 + xexy + g′(y) = −2y2 + 3xy2 + xexy. Thus,g′(y) = −2y2, g(y) = − 2

3y3 + c, and f(x, y) = x3 + xy3 + exy − 2

3 + C.

45. ∇(cf) =∂

∂x(cf)i +

∂

∂yj = cfxi + cfyj = c(fxi + fyj) = c∇f

46. ∇(f + g) = (fx + gx)i + (fy + gy)j = (fxi + fyj) + (gxi + gyj) = ∇f +∇g

47. ∇(fg) = (fgx + fxg)i + (fgy + fyg)j = f(gxi + gyj) + g(fxi + fyj) = f∇g + g∇f

48. ∇(f/g) =[(gfx − fgx)/g2

]i +[(gfy − fgy)/g2

]j = g(fxi + fyj)/g2 − f(gxi + gyj)/g2

= g∇f/g2 − f∇g/g2 = (g∇f − f∇g)/g2

49. r(x, y) =√x2 + y2 so

∂r

∂x=

x√x2 + y2

=x

rand

∂r

∂y=

y√x2 + y2

=y

r

This gives ∇r =⟨xr,y

r

⟩=

1r〈x, y〉 =

rr


50.∂ (f(r))∂x

=df

dr

∂r

∂xand

∂ (f(r))∂y

=df

dr

∂r

∂yso that ∇f(r) = 〈∂ (f(r))

∂x,∂ (f(r))∂y

〉

= 〈dfdr

∂r

∂x,df

dr

∂r

∂y〉 =

df

dr〈 ∂r∂x,∂r

∂y〉 = f ′(r)∇r = f ′(r)r/r

51. Let u = u1i + u2j and v = v1i + v2j.Dvf = (fxi + fyj) · v = v1fx + v2fy

DuDvf =[∂

∂x(v1fx + v2fy)i +

∂

∂y(v1fx + v2fy)j

]· u = [(v1fxx + v2fyz)i + (v1fxy + v2fyy)j] · u

= u1v1fxx + u1v2fyx + u2v1fxy + u2v2fyy

D − uf = (fxi + fyj) · u = u1fx + u2fy

DvDuf =[∂

∂x(u1fx + u2fy)i +

∂

∂y(u1fx + u2fy)j

]· v = [(u1fxx + u2fyx)i + (u1fxy + u2fyy)j] · v

= u1v1fxx + u2v1fyx + u1v2fxy + u2v2fyySince the second partial derivatives are continuous, fxy = fyx and DuDvf = DvDuf. [Notethat this result is a generalization fxy = fyx since DiDjf = fyx and DjDif = fxy]

52. ∇× F =

∣∣∣∣∣∣∣i j k∂

∂x

∂

∂y

∂

∂zf1 f2 f3

∣∣∣∣∣∣∣=(∂f3

∂y− ∂f2

∂z

)i−(∂f3

∂x− ∂f1

∂z

)j +(∂f2

∂x− ∂f1

∂y

)k

13.7 Tangent Planes and Normal Lines

y

x

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.

∇f = − y

x2i +

1x

j; ∇f(1, 3) = −3i + j

13.7. TANGENT PLANES AND NORMAL LINES 113

y

x

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 + y2 = 10.∇f = 2xi + 2yj; ∇f(−1, 3) = −2i + 6j

y

x

5. Since f(−2,−3) = 2, the level curve is x2/4 + y2/0 = 2 or

x2/8 + y2/18 = 1. ∇f =x

2i +

2y9

j; ∇f(−2,−3) = −i− 23j

y

x

6. Since f(2, 2) = 2, the level curve is y2 = 2x, x 6= 0.

∇f = −y2

x2i +

2yx

j; ∇f(2, 2) = −i + 2j


y

x

7. Since f(1, 1) = −1, the level curve is (x − 1)2 − y2 = −1 ory2 − (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 = sinx or

y = 1 + sinx, sinx 6= 0. ∇f =−(y − 1) cosx

sin2 xi +

1sinx

j;

∇f(π/6, 3/2) = −√

3i + 2j

z

y

x

2

2

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

yx

z10. Since f(1, 1, 3) = −1, the level curve is x2 + y2 − z = −1 or

z = 1 + x2 + y2. ∇f = 2xi + 2yj− k; ∇f(1, 1, 3) = 2i + 2j− k

z

yx

5

5

11. Since F (3, 4, 0) = 5, the level curve is x2 + y2 + z2 = 25.∇F =

x√x2 + y2 + z2

i +y√

x2 + y2 + z2j +

z√x2 + y2 + z2

k;

∇F (3, 4, 0) =34i +

45j


z

y

x

12. Since F (0,−1, 1) = 0, the level curve is x2 − y2 + z = 0 or z = y2 − x2.∇F = 2xi− 2yj + k; ∇F (0,−1, 1) = 2i + k

13. F (x, y, z) = x2 + y2 − z; ∇F = 2xi + 2yj − k. We want ∇F = c(4i + j + 1

2k)

or 2x =4c, 2y = c, −1 = c/2. From the third equation c = −2. Thus, x = −4 and y = −1. Sincez = x2 + y2 = 16 + 1 = 17, the point on the surface is (−4,−1,−17).

14. F (x, y, z) = x3 + y3 + z; ∇F = 3x2i + 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 − y2 =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 +y2 + z2; ∇F = 2xi+ 2yj+ 2zk. ∇F (−2, 2, 1) = −4i+ 4j+ 2k. The equationof 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−y2 +4z2; ∇F = 10xi−2yj+8zk; ∇F (2, 4, 1) = 20i−8j+8k.The equationof 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− y2− 3z2;∇F = 2xi− 2yj− 6zk; ∇F (6, 2, 3) = 12i− 4j− 18k. The equationof 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 + y2 + z; ∇F = 2xi + 2yj + k; ∇F (3,−4, 0) = 6i− 8j + k. The equation ofthe 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 is3(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) =√

2i+√

22

j−k. The equation of the tangent plane is√

2(x− π

2

)+√

22

(y − π

4

)−(z +

1√2

)=

0, 2(x− π

2

)+(y − π

4

)−√

2(z +

1√2

)= 0, or 2x+ y −

√2z =

5π4

+ 1.

22. F (x, y, z) = x2y3 +6z; ∇F = 2xy3i+3x2y2j+6k; ∇F (2, 1, 1) = 4i+12j+6k. The equationof the tangent plane is 4(x− 2) + 12(y − 1) + 6(z − 1) = 0 or 2x+ 6y + 3z = 13.

23. F (x, y, z) = ln(x2 + y2) − z; ∇F =2x

x2 + y2i +

2yx2 + y2

j − k; ∇F (1/√

2, 1/√

2, 0) =√

2i +

√2j − k. The equation of the tangent plane is

√2(x− 1√

2

)+√

2(y − 1√

2

)− (z − 0) =

0, 2(x− 1√

2

)+ 2

(y − 1√

2

)−√

2z = 0, or 2x+ 2y −√

2z = 2√

2.


24. F (x, y, z) = 8e−2y sin 4x − z; ∇F = 32e−2y cos 4xi − 16e−2y sin 4xj − k; ∇F (π/24, 0, 4) =16√

3i− 8j− k. The equation of the tangent plane is

16√

3(x− π/24)− 8(y − 0)− (z − 4) = 0 or 16√

3x− 8y − z =2√

3π3− 4.

25. The gradient of F (x, y, z) = x2 + y2 + z2 is ∇F = 2xi + 2yj + 2zk, so the normal vector tothe surface at (x0, y0, z0) is 2x0i + 2y0j + 2z0k. A normal vector to the plane 2x+ 4y+ 6z = 1is 2i + 4j + 6k. Since we want the tangent plane to be parallel to the given plane, we find cso that 2x0 = 2c, 2y0 = 4c, 2z0 = 6c or x0 = c, y0 = 2c, z0 = 3c. Now, (x0, y0, z0) is onthe surface, so c2 + (2c)2 + (3c)2 = 14c2 = 7 and c = ±1/

√2. Thus, the points on the surface

are (√

2/2,√

2, 3√

2/2) and −√

2/2,−√

2,−3√

2/2).

26. The gradient of F (x, y, z) = x2−2y2−3z2 is ∇F (x, y, z) = 2xi−4yj−6zk, so a normal vectorto the surface at (x0, y0, z0) is ∇F (x0, y0, z0) = 2x0i − 4y0j − 6z0k. A normal vector to theplane 8x+4y+6z = 5 is 8i+4j+6k. Since we want the tangent plane to be parallel to the givenplane, we find c so that 2x0 = 8c, −4y0 = 4c, −6z0 = 6c or x0 = 4c, y0 = −c, z0 = −c.Now (x0, y0, z0) is on the surface, so (4c)2−2(−c)2−3(−c)2 = 11c2 = 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+y2+z2−2z is∇F = (2x+4)i+2yj+(2z−2)k, so a normalto the surface at (x0, y0, z0) is (2x0 +4)i+2y0j+(2z0−2)k. A horizontal plane has normal ckfor 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 andc = ±4. The points on the surface are (−2, 0, 5) and (−2, 0,−3).

28. The gradient of F (x, y, z) = x2 + 3y2 + 4z2 − 2xy is ∇F = (2x− 2y)i + (6y − 2x)j + 8zk, soa normal to the surface at (x0, y0, z0) is 2(x0 − y0)i + 2(3y0 − x0)j + 8z0k.

(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 obtainx0 = y0 = c/4. Since (x0, y0, z0) is on the surface, (c/4)2 +3(c/4)2 +4(0)2−2(c/4)(c/4) =2c2/16 = 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, weobtain 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 obtainx0 = y0 = 0. Since (x0, y0, z0) is on the surface, 02+3(0)2+4(c/8)2−2(0)(0) = c2/16 = 16and 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 + y2/b2 + z2/c2 = 1, then x20/a

2 + y20/b

2 + z20/c

2 = 1 and x0, y0, z0)is on the plane xx0/a

2 + yy0/b2 + zz0/c

2 = 1. A normal to the surface at (x0, y0, z0) is


∇F (x0, y0, z0) = (2x − 0/a2)i + (2y0/b2)j + (2z0/c

2)k. A normal to the plane is (x0/a2)i +

(y0/b2)j + (z0/c

2)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 − y2/b2 + z2/c2 = 1, then x20/b

2 − y20/b

2 + z20/c

2 = 1 and (x0, y0, z0)is on the plane xx0/a

2 − yy0/b2 + zz0/c

2 = 1. A normal to the surface at (x0, y0, z0) is∇F (x0, y0, z0) = (2x0/a

2)i − (2y0/b2)j + (2z0/c

2)k. A normal to the plane is (x0/a2)i −

(y0/b2)j + (z0/c

2)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 + 2y2 + z2; ∇F = 2xi + 4yj + 2zk; ∇F (1,−1, 1) = 2i− 4j + 2k. Parametricequations of the line are x = 1 + 2t, y = −1− 4t, z = 1 + 2t.

32. F (x, y, z) = 2x2 − 4y2 − z; ∇F = 4xi− 8yj− k; ∇F (3,−2, 2) = 12i + 16j− k. Parametricequations of the line are x = 3 + 12t; y = −2 + 16t, z = 2− t.

33. F (x, y, z) = 4x2 + 9y2 − z; ∇F = 8xi + 18yj− k; ∇F (1/2, 1/3, 3) = 4i + 6j− k. Symmetric

equations of the line arex− 1/2

4=y − 1/3

6=z − 3−1

.

34. F (x, y, z) = x2 + y2 − z2; ∇F = 2xi + 2yj − 2zk;∇F (3, 4, 5) = 6i + 8j − 10k. Symmetric

equations of the line arex− 3

6=y − 4

8=z − 5−10

.

35. Let F (x, y, z) = x2 + y2 − z2. Then ∇F = 2xi + 2yj − 2zk and a normal to the surfaceat (x0, y0, z0) is x0i + y0j − z0k. An equation of the tangent plane at (x0, y0, z0) is x0(x −x0) + y0(y − y0) − z0(z − z0) = 0 or x0x + y0y − z0z = x2

0 + y20 − z2

0 . Since (x0, y0, z0) is onthe surface, z2

0 = x20 + y2

0 and x20 + y2

0 − z20 = 0. Thus, the equation of the tangent plane is

x0x+ y0y − z0z = 0, which passes through the origin.

36. Let F (x, y, z) =√x+√y+√z. Then ∇F =

12√x

i+1

2√y

j+1

2√zk and a normal to the surface

at (x0, y0, z0) is1

2√x0

i +1

2√y0

j +1

2√z0

k. An equation of the tangent plane at (x0, y0, z0) is

12√x0

(x−x0)+1

2√y0

(y−y0)+1

2√z0

(z−z0) = 0 or1√x0x+

1√y0y+

1√z0z =√x0+√y0+√z0 =

√a. The sum of the intercepts is

√x0√a +√y0√a +√z0√a = (

√x0 +

√y0 +

√z0)√a =√

a ·√a = a.

37. A normal to the surface at (x0, y0, z0) is ∇F (x0, y0, z0) = 2x0i + 2y0j + 2z0k. Parametricequations of the normal line are x = x0 + 2x0t, y = y0 + 2y0t, z = z0 + 2z0t. Lettingt = −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 Fxi+Fyj+Fzk and Gxi+Gyj+Gzk,respectively. These vectors are orthogonal if and only if their dot product is 0. Thus, thesurfaces are orthogonal at P if and only if FxGx + FyGy + FzGz = 0.

39. We have F (x, y, z) = x2 + y2 + z2 and G(x, y, z) = x2 + y2 − z2.∇F = 〈2x, 2y, 2z〉 6= 0 except at the origin


∇G = 〈2x, 2y,−2z〉 6= 0 except at the originTherefore, the gradient vectors are nonzero at each of the intersection points. Now

FxGx + FyGy + FxGz = (2x)(2x) + (2y)(2y) + (2z)(−2z)

= 4x2 + 4y2 − 4z2

= 4(x2 + y2 + z2) = 4(0) = 0

The second to last equality follows from the fact that the intersection points lie on bothsurfaces and hence satisfy the second equation x2 + y2 − z2 = 0.

40. Let F (x, y, z) = x2 − y2 + z2 − 4 and G(x, y, z) = 1/xy2 − z. Then

FxGx + FyGy + FzGz = (2x)(−1/x2y2) + (−2y)(−2/xy3) + (2z)(−1)

= −2/xy2 + 4/xy2 − 2z = 2(1/xy2 − z).

For (x, y, z) on both surfaces, F (x, y, z) = G(x, y, z) = 0. Thus, FxGx + FyGy + FzGz = 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, weobtain the critical point (0, 0). Since D(0, 0) = 4 > 0 and fxx(0, 0) = 2 > 0, f(0, 0) = 5 is arelative minimum.

2. fx = 8x; fxx = 8; fxy = 0; fy = 16y; fyy = 16; D = 128. Solving fx = 0 and fy = 0, weobtain the critical point (0, 0). Since D(0, 0) = 128 > 0 and fxx(0, 0) = 8 > 0, f(0, 0) = 0 isa relative minimum.

3. fx = −2x+ 8; fxx = −2; fxy = 0; fy = −2y + 6; fyy = −2; D = 4. Solving fx = 0 andfy = 0 we obtain the critical point (4, 3). Since D(4, 3) = 4 > 0 andfxx(4, 3) = −2 < 0, f(4, 3, ) = 25 is a relative maximum.

4. fx = 6x − 6; fxx = 6; fxy = 0; fy = 4y + 8; fyy = 4; D = 24. Solving fx = 0 andfy = 0, we obtain the critical point (1,−2). Since D(1,−2) = 24 > 0 andfxx(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. Solvingfx = 0 and fy = 0, we obtain the critical point (−2, 1). Since D(−2, 1) = 100 > 0 andfxx(−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. Solvingfx = 0 and fy = 0, we obtain the critical point (−1, 3). Since D(−1, 3) = 32 > 0 andfxx(−1, 3) = −8 < 0, f(−1, 3) = 27 is a relative maximum.

7. fx = 12x2 − 12; fxx = 24x; fxy = 0; fy = 3y2 − 3; fyy = 6y; D = 144xy. Solvingfx = 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 relativeextrema. 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, f(1, 1) = −10 is arelative minimum.

8. fx = −3x2 + 27; fxx = −6x; fxy = 0; fy = 6y2 − 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 relativeextrema. Since D(−3, 2) = 432 > 0 and fxx(−3, 2) = 18 > 0, f(−3, 2) = 432 > 0 andfxx(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 = 5and 4y − x = 1. Solving, we obtain the critical point (3, 1). Since D(3, 1) = 28 > 0 andfxx(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 = 2and 2y + x = 1. Solving, we obtain the critical point (1, 0). Since D(1, 0) = 75 > 0 andfxx(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 = 0 − 22 = −4. SinceD(x, y) = −4 < 0 for all (x, y), there are no relative extrema.

12. fx = 2y + 6; fxx = 0; fxy = 2; fy = 2x + 10; fyy = 0; D = 0 − 22 = −4. SinceD(x, y) = −4 < 0 for all (x, y), there are no relative extrema.

13. fx = −6x2 + 6y; fxx = −12x; fxy = 6; fy = −6y2 + 6x; fyy = −12y; D = 144xy − 36.Setting fx = 0 and fy = 0, we obtain −6x2 + 6y = 0 and −6y2 + 6x = 0 or y = x2 andx = y2. Substituting x = y2 into y = x2, we obtain y = y4 or y(y3 − 1) = 0. Thus, y = 0 andy = 1. The critical points are (0, 0) and (1, 1). Since D(0, 0) = −36 < 0, (0, 0) does not givea relative extremum. Since D(1, 1) = 108 > 0 and fxx(1, 1) = −12 < 0, f(1, 1) = 12 is arelative maximum.

14. fx = 3x2 − 6y; fxx = 6x; fxy = −6; fy = 3y2 − 6x; fyy = 6y; D = 36xy − 36.Setting fx = 0 and fy = 0, we obtain 3x2 − 6y = 0 and 3y2 − 6x = 0 or x2 = 2y andy2 = 2x. Substituting y = x2/2 into y2 = 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) isnot an extremum. Since D(2, 2) = 108 > 0 and fxx(2, 2) = 12 > 0, f(2, 2) = 19 is a relativeminimum.

15. fx = y + 2/x2; fxx = −4/x3; fxy = 1; fy = x + 4/y2; fyy = −8/y3;D = 32/x3Y 3 − 1.Setting fx = 0 and fy = 0 we obtain y+ 2/x2 = 0 and x+ 4/y2 = 0. Substituting y = −2/x2

into x + 4/y2 = 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 andfxx(−1,−2) = 4 > 0, f(−1,−2) = 14 is a relative minimum.

16. fx = −6xy − 3y2 + 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 = 0we obtain −6xy − 3y2 + 36y = 0 and −3x2 − 6xy + 36x = 0 or −3y(2x + y − 12) = 0and −3x(x + 2y − 12) = 0. Letting y = 0, the first equation is satisfied and the second


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 = 12we obtain the critical point (4, 4). Since D(0, 0) = −362 < 0, D(0, 12) = −362 < 0, andD(12, 0) = −362 < 0, none of these points give relative extrema. Since D(4, 4) = 432 > 0 andfxx(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) = 0and fy(x, y) = 0 we obtain (xex + ex) sin y = 0 and xex cos y = 0. Since ex > 0 for allx, we have (x + 1) sin y = 0 and x cos y = 0. When x = −1, we must have cos y = 0or y = π/2 + kπ, k an integer. When x = 0, we must have sin y = 0 or y = kπ, kan integer. Thus, the critical points are (0, kπ) and (−1, π/2 + kπ), k an integer. SinceD(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 fork even, and f(−1, π/2 + kπ) = e−1 are relative maxima for k odd.

18. fx = (2x+ 4)ey2−3y+x2+4x; fxx = [(2x+ 4)2 + 2]ey

2−3y+x2+4x;fxy = (2x+ 4)(2y − 3)ey

2−3y+x2+4x; fy = (2y − 3)ey2−3y+x2+4x;

fyy = [(2y− 3)2 + 2]ey2−3y+x2+4x;D = [(2x+ 4)2 + 2][(2y− 3)2 + 2] · e2(y2−3y+x2+4x)− [(2x+

4)(2y− 3)]2e2(y2−3y+x2+4x). Setting fx = 0 and fy = 0 and using the fact that an exponentialfunction is always positive, we obtain 2x + 4 = 0 and 2y − 3 = 0. Thus, a critical pointis (−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 = cosx; fxx = − sinx; fxy = 0; fy = cos y; fyy = − sin y; D = sinx sin y. Solvingfx = 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 mand 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 fxx > 0 and f(π/2 + mπ, π/2 + nπ) = −2 are relativeminima.

20. fx = y cosxy; fxx = −y2 sinxy; fxy = −xy sinxy + cosxy; fy = x cosxy;fyy = −x2 sinxy; D = x2y2 sin2 xy − (−xy sinxy + cosxy)2 = 2xy sinxy cosxy − cos2 xy.Setting fx = 0 and fy = 0 we see that (0, 0) is a critical point. Also, solving cosxy = 0 weobtain 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 = 0and no conclusion can be drawn from the second partials test. Since −1 ≤ sinxy ≤ 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 fork 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 − x2y − xy2. Now Px = 21y − 2xy − y2; 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 thecritical points (0, 0), (0, 21), and (21, 0). Each of these results in P = 0 which is clearly nota maximum. Solving 21− 2x− y = 0 and 21− x− 2y = 0, we obtain the critical point (7, 7).


Since D(7, 7) = 147 > 0 and Pxx(7, 7) = −14 < 0, P (7, 7) = 343 is a maximum. The threenumbers 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, itsheight is 1/xy and S = 2xy + 2x(1/xy) + 2y(1/xy) = 2xy + 2/y + 2/x, x > 0, y > 0. NowSx = 2y − 2/x2; Sxx = 4/x3; Sxy = 2; Sy = 2x − 2/y2; Syy = 4/y3; D = 16/x3y3 − 4.Setting Sx = 0 and Sy = 0 we obtain y = 1/x2 and x = 1/y2. 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 footon 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 + y2 + (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 obtain2x−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 aminimum. 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 squareof 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 + 2y2 − 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 andx + 2y = 3. Thus, (1/3, 4/3) is a critical point. Since D = 12 > 0 and fxx = 4 > 0 for all(x, y), f(1/3, 4/3) = 25/3 is a minimum. Thus, the least distance between the point and theplane 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 distanceto the origin or f(x, y) = x2 + y2 + 64/x2y2. Now fx = 2x − 128/x3y2; fxx = 2 + 384/x4y2;fxy = 256/x3y3; fy = 2y − 128/x2y3; fyy = 2 + 384/x2y4; D = (2 + 384/x4y2)(2 +384/x2y4) − (256)2/x6y6. Setting fx = 0 and fy = 0 we obtain 2x − 128/x3y2 = 0 and2y − 128/x2y3 = 0 or x4y2 = 64 and x2y4 = 64; x 6= 0, y 6= 0. This gives x4y2 =x2y4, x2y2(x2 − y2) = 0 or x2 = y2. Thus, x6 = 64 and x = ±2. Similarly, y = ±2 andthe critical points are (−2,−2), (−2, 2), (2,−2), and (2, 2). Since D(±2,±2) = 48 > 0 andfxx(±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; fss24; fst = 8; ft = 12t+8s−12; ftt = 12; D = 24(12)−64 = 224. Solving24s+8t−8 = 0 and 12t+8s−12 = 0 we obtain the critical point (0, 1). Since D(0, 1) = 224 > 0and fss(0, 1) = 24 > 0, we see that f(0, 1) is a minimum. The corresponding points on thelines are (3, 6, 8) on L2 and (1, 2, 2) on L1. The minimum distance is

√f(0, 1) =

√56 = 2

√14.


27. We will maximize the square of the volume of the box in the first octant,

V (x, y) = x2Y 2z2 = x2y2(c2 − c2x2/a2 − c2y2/b2).

Vx = 2c2xy2−4c2x3y2/a2−2c2xy4/b2; Vxx = 2c2y2−12c2x2y2/a2−2c2y4/b2; Vxy = 4c2xy−8c2x3y/a2 − 8c2xy3/b2; Vy = 2c2x2y − 2c2x4/a2 − 4c2x2y3/b2; Vyy = 2c2x2 − 2c2x4/a2 −12c2x2y2/b2; D = VxxVyy − V 2

xy. Setting Vx = 0 and Vy = 0 we obtain xy2 − 2x3y2/a2 −xy4/b2 = 0, x2y−x4y/a2−2x2y3/b2 = 0, or, assuming x > 0 and y > 0, 2b2x2+a2y2 = a2b2.Solving, we obtain x2 = a2/3 and y2 = b2/3. Thus, (a/

√3, b/√

3) is a critical point. Since

D(a/√

3, b/√

3) = (−149b2c2)(−14

9a2c2)− (−4

9abc2)2 =

209a2b2c4 > 0

and vxx = −149b2c2 < 0, V (a/

√3, b/√

3) = a2b2c2/27. The maximum volume is

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− a2b− ab2)/3.

Va =4π3

(kb− 2ab− b2); Vaa = −8π3

; Vab =4π3

(k − 2a− 2b); Vb =4π3

(ka− a2 − 2ab);

Vbb = −8π3a; D =

64x2

9ab − 16x2

9(k − 2a − 2b)2. Setting Va = 0 and Vb = 0 we obtain

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) = 16π2k2/27 > 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 theellipsoid is a sphere.

x x

x tanθ

θx secθ

y

29. The perimeter is given by P = 2x + 2y + 2x sec θ and thearea is a = 2xy + x2 tan θ. Solving P for 2y and substitut-ing in A, we obtain A = Px − 2x2(1 + sec θ) + x2 tan θ. NowAx(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 thatx > 0 and 0 ≤ θ ≤ π/2.

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 amaximum. 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◦


xθθ

24-2x x cosθ

x sinθ

30. We want to maximize A(x, θ) = (x sin θ)(24 −2x + x cos θ) = 24x sin θ − 2x2 sin θ +

12x2 sin 2θ.

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.

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 and2x cos2 θ − 2x cos θ + 2 cos θ − x = 0. Solving the first equation for cos θ and substituting into thesecond equation, we obtain 2x(2− 12/x)2 − 2x(12− 12/x) + 24(2− 12/x)− x = 0. Simplifying, wefind 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.

31. fx = −23x−1/3, fy = −2

3y−1/3. Since fx = 0 and fy = 0 have no solutions, f(x, y) has no

critical points and Theorem 13.8.2 does not apply. However, for all (x, y),f(0, 0) = 16 ≥ 16− (x1/3)2 − (y1/3)2 = f(x, y), and f(0, 0) = 16 is an absolute maximum.

32. fx = −4x3y2; fxx = −12x2y2; fxy = −8x3y; fy = −2x4y; fyy = −2x4;D = 24x6y2 − 64x6y2 = −40x6y2. 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 doesnot apply. However, for all (x, y), f(0, 0) = 1 ≥ 1 − (x2y)2 = f(x, y), and f(0, 0) = 1 is anabsolute maximum.

33. fx = 10x; fxx = 10; fxy = 0; fy = 4y3; fyy = 12y2; D = 120y2. Solving fx = 0 andfy = 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 +y4−8 = f(x, y) and f(0, 0) = −8 is an absoluteminimum.

34. fx =x√

x2 + y2; fxx =

y2

(x2 + y2)3/2; fxy = − xy

(x2 + y2)3/2; fy =

y√x2 + y2

;

fyy =x2

(x2 + y2)3/2; D =

x2y2

(x2 + y2)3− (

−xy(x2 + y2)3/2

)2 = 0. Since D = 0 for all (x, y),

Theorem 13.8.2 does not apply. However, for all (x, y), f(0, 0) = 0 ≤√x2 + y2 = 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 weconsider F (t) = cos t +

√3 sin t. Solving F ′(t) = − sin t +

√3 cos t = 0, we obtain critical

points at t = π/3 and t = 4π/3. Comparing F (0) = 1, F (π/3) = 2, and F (4π/3) = −2, wesee that f(1/2,

√3/2) = 2 is an absolute maximum and f(−1/2,−

√3/2) = −2 is an absolute

minimum.


36. fx = y; fy = x. Solving fx = 0 and fy = 0 we obtain the critical point (0, 0) with corre-

sponding function value f(0, 0) = 0. On the boundary we consider F (t) = cos t sin t =12

sin 2t.

Solving F ′(t) = cos 2t = 0, we obtain critical points at π/4, 3π/4, 5π/4, and 7π/4. Com-paring f(0, 0, ) = 0, F (0) = 0, F (π/4) = 1/2, F (3π/4) = −1/2, F (5π/4) = 1/2,and F (7π/4) = −1/2, we see that f(

√2/2,√

2/2) = f(−√

2/2,−√

2/2) = 1/2 are absolutemaxima 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) =

cos2 t+ cos t sin t+ sin2 t = 1 +12

sin 2t. Solving F ′(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) = 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 weconsider F (t) = − cos2 t − 3 sin2 t + 4 sin t + 1. Solving F ′(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. Comparingf(0, 2/3) = 5, F (0) = 0, F (π/2) = 2, and F (3π/2) = −6, we see that f(0,−1) = −6 is anabsolute 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 mustoccur 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 ′(t) = −8 sin t− 6 cos t. Solving F ′(t) = 0we find tan t = −3/4. Using 1 + tan2 t = sec2 t we see that sec2 t = 25/16 and cos t = −4/5, tis 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 theabsolute 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 theregion. On x = 0, F (y) = f(0, y) = −y + 6, which has no critical points for 0 ≤ y ≤ 8. Theendpoints of the interval are (0, 0) and (0, 8). On y = 0, G(x) = f(x, 0) = −2x + 6, whichhas no critical points for 0 ≤ x ≤ 4. The endpoints of the interval are (0, 0) and (4, 0). Ony = −2x + 8, H(x) = f(x,−2x + 8) = x(−2x + 8) = x(−2x + 8) − 2x − (−2x + 8) + 6 =−2x2 + 8x − 2. Solving H ′(x) = −4x + 8 = 0 we obtain x = 2. The corresponding point onthe triangle is (2, 4). Comparing f(0, 0) = 6, f(0, 8) = −2; f(4, 0) = −2; f(2, 4) = 6, andf(1, 2) = 4 we see that absolute maxima are f(0, 0) = f(2, 4) = 6 and absolute minima aref(0, 8) = f(4, 0) = −2.

41. (a) fx = y cosxy; fy = x cosxy. Setting fx = 0 and fy = 0 we obtain y cosxy = 0 andx cosxy = 0. If y = 0 from the first equation, then necessarily x = 0 from the secondequation. Thus, (0, 0) is a critical point. For x 6= 0 and y 6= 0 we have cosxy = 0 orxy = π/2. Thus, all points (x, π/2x) for 0 ≤ x ≤ π are also critical points.

(b) Since 0 ≤ sinxy ≤ 1 for 0 ≤ x ≤ π and 0 ≤ y ≤ 1, f(x, y) = sinxy has absoluteminima at any points for which sinxy = 0 and absolute maxima at any points for which

13.9. METHOD OF LEAST SQUARES 125

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

(c)

z

y

x

1

42. We want to maximize P (x, y) = R(x, y)−C(x, y) = 108x− 8x2 + 192y− 6y2− 4xy− 20. NowPx = 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 = 0or 4x + y = 27 and x + 3y = 48. Solving, we see that (3, 15) is a critical point. SinceD(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/y2; cyy = 480/y3; D =4802/x3y3−900. Setting Cx = 0 and Cy = 0 we obtain 30y−240/x2 = 0 and 30x−240/y2 = 0or y = 8/x2 and x = 8/y2. Substituting the first equation into the second, we have x = x4/8or x(x3 − 8) = 0. Thus, (2, 2) is a critical point. Since D(2, 2) = 2700 > 0 andCxx(2, 2) = 60 > 0, C(2, 2) is a minimum. Thus, the cost is minimized when the base of thebox is 2 feet square and the height is 15 feet.

13.9 Method of Least Squares

1.4∑i=1

xi = 14,4∑i=1

yi = 8,4∑i=1

xiyi = 30,4∑i=1

x2i = 54, m =

4(30)− 14(8)4(54)− (14)2

= 0.4,

b =54(8)− 30(14)4(54)− (14)2

= 0.6, y = 0.4x+ 0.6

2.4∑i=1

xi = 6,4∑i=1

yi = 14,4∑i=1

xiyi = 34,4∑i=1

x2i = 14, m =

4(34)− 6(14)4(14)− (6)2

= 2.6,

b =14(14)− 34(6)4(14)− (6)2

= −0.4, y = 2.6x− 0.4


3.5∑i=1

xi = 15,5∑i=1

yi = 15,5∑i=1

xiyi = 56,5∑i=1

x2i = 55, m =

5(56)− 15(15)5(55)− (15)2

= 1.1,

b =55(15)− 56(15)5(55)− (15)2

= −0.3, y = 1.1x− 0.3

4.5∑i=1

xi = 14,4∑i=5

yi = 14,5∑i=1

xiyi = 55,5∑i=1

x2i = 54, m =

5(55)− 14(14)5(54)− (14)2

≈ 1.06757,

b =54(14)− 55(14)5(54)− (14)2

≈ −0.189189, y ≈ 1.06757x− 0.189189

5.7∑i=1

xi = 21,7∑i=1

yi = 42,7∑i=1

xiyi = 164,7∑i=1

x2i = 91, m =

7(164)− 21(42)7(91)− (21)2

≈ 1.35714,

b =91(42)− 164(21)

7(91)− (21)2≈ 1.92857, y ≈ 1.35714x+ 1.92857

6.7∑i=1

xi = 28,7∑i=1

yi = 17.2,7∑i=1

xiyi = 80.2,7∑i=1

x2i = 140, m =

7(80.2)− 28(17.2)7(140)− (28)2

≈ 0.407143,

b =140(17.2)− 80.2(28)

7(140)− (28)2≈ 0.828571, y ≈ 0.407143x+ 0.828571

7.6∑i=1

Ti = 420,6∑i=1

vi = 1055,6∑i=1

Tivi = 68, 000,6∑i=1

T 2i = 36, 400, m =

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

≈

−0.835714, b =36, 400(1055)− 68, 000(420)

6(36, 400)− (420)2≈ 234.333, v ≈ −0.835714T + 234.333.

8.6∑i=1

Ti = 3150,6∑i=1

Ri = 29.57,6∑i=1

TiRi = 17, 878,6∑i=1

T 2i = 1, 697, 500, m =

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

≈

0.05, b =1, 697, 500(29.57)− 17, 878(3150)

6(1, 697, 500)− (3150)2≈ −23.32, R ≈ 0.05T − 23.32 − 23.32. When

T = 700, R ≈ 14.34.

9. (a) least-squares line: y = 0.5966x+ 4.3665least-squares quadratic: y = −0.0232x2 + 0.5618x+ 4.5942least-squares cubic: y = 0.00079x3 − 0.0212x2 + 0.5498x+ 4.5840

(b) least-squares line least-squares quadratic least-squares cubic

y

x x

y

x

y

13.10. LAGRANGE MULTIPLIERS 127

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

13.10 Lagrange Multipliers

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

x

y

2

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

y

x

1

1

3. fx = 1; fy = 3; gx = 2x; gy = 2y. We need to solve 1 = 2λx, 3 = 2λy, x2 + y2 − 1 = 0.Dividing the second equation by the first, we obtain 3 = y/x or y = 3x. Substituting intothe third equation, we have x2 + 9x2 = 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 aconstrained 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 andf(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 + y2 − 2 = 0.Substituting the second equation into the first, we obtain y = 4λ2y or y(4λ2 − 1) = 0. Ify = 0, then from the second equation x = 0. Since g(0, 0) = −2 6= 0, (0, 0) does notsatisfy 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 areconstrained 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 thirdequation, 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


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 constrainedminimum.

8. fx = 8x; fy = 4y; gx = 8x; gy = 2y. We need to solve 8x = 8λx, 4y = 2λy, 4x2 +y2−4 =0 or x(λ − 1) = 0, y(λ − 2) = 0, 4x2 + y2 = 4. If x = 0, then from the third equationy = ±2. If y = 0, then from the third equation x = ±1. The cases λ = 1 and λ = 2 lead alsoto y = 0 and x = 0, respectively. Thus, f(0, 2) = f(0,−2) = 18 are constrained maxima andf(1, 0) = f(−1, 0) = 14 are constrained minima.

9. fx = 2x; fy = 2y; gx = 4x3; gy = 4y3. We need to solve 2x = 4λx3, 2y = 4λy3, x4 +y4 − 1 = 0 or x(2λx2 − 1) = 0, y(2λy2 − 1) = 0, x4 + y4 = 1. If x = 0, the from thethird equation y = ±1. If y = 0, then x = ±1. From 2λx2 = 1 = 2λy2 we have x2 = y2.Substituting into the third equation, we obtain x = ±1/ 4

√2 and y = ±1/ 4

√2. Solutions of

the system are (0,±1), (±1, 0), and (±1/ 4√

2,±1/[4]2). Thus, f(0,±1) = f(±1, 0) = 1 areconstrained 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 + y2 − 10 = 0 or 8 − 47/x = λ, x2 + y2 = 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. Substitutingy = −x/2 into the third equation, we have x2 + x2/4 = 10 and x = ±2

√2. Substituting

y = 2x into the third equation, we have x2 + 4x2 = 10 and x = ±√

2. Solutions of thesystem are (2

√2,−√

2), (−2√

2,√

2), (√

2, 2√

2), and (−√

2,−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 areconstrained minima.

11. fx = 3x2y; fy = x3; gx = 1/2√x; gy = 1/2

√y. We need to solve 3x2y = λ/2

√x, x3 =

λ/2√y,√x +√y − 1 = 0 or 6x5/2y = λ, 2x3y1/2 = λ,

√x +√y = 1. From the first two

equations, we obtain 3x5/2y = x3y1/2 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 ≥ 0and y ≥ 0, f(0, 1) = f(1, 0) = 0 ≤ x3y = 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 = y2; fy = 2xy; gx = 2x; gy = 2y. We need to solve y2 = 2λx, 2xy = 2λy, x2 +y2 − 27 = 0 or y2 = 2λx, y(x − λ) = 0, x2 + y2 = 27. When y = 0 in the third equation,we obtain x2 = 27 or x = ±3

√3, and λ = 0. When x = λ in the second equation, we obtain

y2 = 2x2 from the first equation and x2 + 2x2 = 3x2 = 27 from the third equation. This givesx = ±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 + y2 + z2 − 30 = 0. From the first and second equations, we obtainy = 2x. From the first and third equations, we obtain z = x. Substituting into the fourthequation, we have x2 + 4x2 + x2 = 6x2 = 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 aconstrained minimum.


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 havex+4x+9x = 14x = 4. Thus, x = 2/7, y = 4/7, andz = 6/7. Then, F (2/7, 4/7, 6/7) = 56/49is 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 solveyz = 2λx, xz = λy/2, xy = 2λz/9 or xyz/2 = λx2, xyz/2 = λy2/4, xyz/2 = λz2/9 alongwith x2 + y2/4 + z2/9 − 1 = 0 or x2 + Y 2/4 + z2/9 = 1 for x > 0, y > 0, z > 0. Fromthe first three equations and the fact that λ 6= 0, x2 = y2/4 = z2/9. Substituting into thethird equation, we obtain x2 + x2 + x2 = 3x2 = 1, so x2 = 1/3, y2 = 4/3, and z2 = 3.Thus, F (

√3/3, 2

√3/3,√

3) = 2√

3/3 is a constrained extremum. Since (√

23/6, 1, 1) satisfiesthe constraint and F (

√23/6, 1, 1) =

√23/6 < 2

√3/3, F (

√3/3, 2

√3/3,√

3) = 2√

3/3 is aconstrained maximum.

16. Fx = yz; Fy = xz; Fz = xy; gx = 3x2; gy = 3y2; gz = 3z2. We need to solveyz + 3λx2, xz = 3λy2, xy = 3λz2 or xyz = 3λx3, syz = 3λy3, xyz = 3λz3 along withx3 + y3 + z3− 24 = 0 or x3 + y3 + z3 = 24. Taking λ = 0 we see that ( 3

√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 = y3 = z3.Substituting into the fourth equation, we obtain x3+x3+x3 = 3x3 = 24 or x = 2. Then (2, 2, 2)satisfies the system. Since 3

√24 = 2 3

√3, F (2 3

√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 = 3y2; Fz = 3z2; gx = 1; gy = 1; gz = 1. We need to solve 3x2 = λ, 3y2 =λ, 3z2 = λ, x+ y + z − 1 = 0 for x > 0, y > 0, z > 0, and hence λ > 0. From the first threeequations x2 = y2 = z2, and since x, y, and z are positive, x = y = z. Then, from the fourthequation, 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 = 8xy2z2; Fy = 8x2yz2; Fz = 8x2Y 2z; gx = 2x; gy = 2y; gz = 2z. We need to solve8xy2z2 = 2λx, 8x2yz2 = 2λy, 8x2y2z = 2λz or 4x2y2z2 = λx2, 4x2y2z2 = λy2, 4x2y2z2 =λz2 along with x2 + y2 + z2 − 9 = 0 or x2 + y2 + z2 = 9 for x > 0, y > 0, z > 0, andhence lambda > 0. From the first three equations, we see x2 = y2 = z2. Substituting intothe third equation, we obtain x2 + x2 + x2 = 3x2 = 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) satisfiesthe 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. Weneed 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 theconstraint equations, we obtain 2λ−µ+λ/2−3µ/2 = 1, −λ+µ/2+λ+2µ−3λ/2+9µ/2 = 4or 6λ − 3µ = 2, −3λ + 14µ = 8. From this, we obtain λ = 52/75 and µ = 54/75. Thenx = 1/3, y = 16/15, and z = −11/15. Thus, F (1/3, 16/15,−11/15) = 134/75 is a constrainedminimum.

20. Fx = 2x; Fy = 2y; Fz = 2z; gx = 4; gy = 0; gz = 1; hx = 2x; hy = 2y; hz = −2z. Weneed to solve 2x = 4λ+ 2xµ, 2y = 2yµ, 2z = λ− 2zµ subject to 4x+ z = 7, z2 = x2 + y2.


Consider the second equation. If y = 0, then the constraint equations become 4x+ z = 7 andz2 = 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 becomes16z + z = 17z = 7, so z = 7/17. Then, x = 28/17 and y2 = z2 − x2 < 0. Hence, the systemhas no solution when y 6= 0. Thus, F (7/5, 0, 7/5) = 98/25 is a constrained minimum andF (7/3, 0,−7/3) = 98/9 is a constrained maximum.

x

yx2+y221. We want to maximize A(x, yxy/2 subject to P (x, y) =

x + y +√x2 + y2 − 4 = 0. Ax = y/2; Ay = x/2;

Px = 1 + x/√x2 + y2; Py = 1 + y/

√x2 + y2. We need

to solve y/2 = λ+ λx/√x2 + y2, x/2 = λ+ λy/

√x2 + y2,

x+ y +√x2 + y2 − 4 = 0 for x > 0, y > 0, and hence

λ > 0. Subtracting the second equation from the first, we have (y − x)/2 = λ(x− y)/√x2 + y2 or

(y − x) = (y − x)(−2λ/√x2 + y2). Since −2λ/

√x2 + y2 is negative, it cannot equal 1, and hence

y − x = 0 or y = x. Substituting in the third equation gives 2x +√

2x2 = (2 +√

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 tomaximize 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 tosolve yz = λ(y + 2z), xz = λ(x + 2z), xy = λ(2y + 2x), xy + 2yz + 2xz − 75 = 0 orxyz = 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 havexy + 2xz = xy + 2yz, which gives xz = yz or x = y; and xy + 2yz = 2yz + 2xz, whichgives xy = 2xz or y = 2z. Substituting x = y = 2z into the fourth equation, we obtain4z2 + 4z2 + 4z2 = 12z2 = 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 needto solve xyz = 2λ(xy + xz), xyz = 2λ(xy + yz), xyz = 2λ(yz + xz), 2(xy + yz + xz) = 75for x > 0, y > 0, z > 0, and thus λ > 0. From the first three equations, we havexy+xz = xy+yz, which gives xz = yz or x = y; and xy+yz = yz+xz, which gives xy = xz ory = z. Substituting x = y = z into the fourth equation, we obtain 2(x2 +x2 +x2) = 6x2 = 75.Thus, x = y = z = 5/

√2. The box is a cube with each side 5/

√2cm.

23. We want to maximize V (x, y) = 9πx+ 3πy subject to S(x, y) = 9π+ 6πx+ 3π√

9 + y2−81π.Now Vx = 9π; Vy = 3π; Sx = 6π; Sy = 3πy/

√9 + y2. We need to solve 9π = 6πλ, 3π =

3πλy/√

9 + y2, 9π + 6πx + 3√

9 + y2 − 81π = 0 for x > 0 and y > 0. From the firstequation, λ = 3/2. Using λ = 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 whenx = 12− 9/2

√5m and y = 6/

√5m.

24. Ux =13x−2/3y2/3; Uy =

23x1/3y−1/3; gx = 1; gy = 6. We need to solve

13x−2/3y2/3 =

λ,23x1/3y−1/3 = 6λ, x + 6y − 18 = 0 or y = 3λx2/3y1/3,

13x = 3λx2/3y1/3, x +

6y = 18. From the first two equations, y = x/3. Substituting into the third equation, we


have x + 2x = 3x = 18. Thus, x = 6 and y = 2. Since (12, 1) satisfies the constraint andU(12, 1) = 121/3, U(6, 2) = 61/322/3 = 241/3 is a constrained maximum.

25. We want to maximize z(x, y) = P − x− y subject to z2/xy3 = k or (P − x− y)2 − kxy3 = 0.Now zx = −1; zy = −1; gx = −2(P − x− y)− ky3; gy = −2(P − x− y)− 3kxy2. We needto solve −1 = −2λ(P −x− y)−λky3, −1 = −2λ(P −x− y)− 3λkxy2, (P −x− y)2 = kxy3

for x > 0, y > 0, and z > 0. From the first two equations, we have y = 3x. Substituting intothe 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

x =−4 +

√16 + 4P

√27k

2√

27k=−2 +

√4 + P

√27k

27k.

Then the maximum value of z is P − 4x = P + 4(2−√

4 + P√

27k/√

27k.

26. (a) See part (b).

(b) Maximizing 1/(x21 · · ·x2

n) is equivalent to minimizing the denominator F (x1, . . . xn) =x2

1 + · · ·+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 equations2x1 = λ,, 2x2 = λ, , . . . , 2xn = λ, x1 + · · ·xn = 1. The solution is x1 = . . . = 1/2 ( andλ = 1/2n).

27. f(x, y) is the square of the distance from a point on the graph of x4 + y4 = 1 to the origin.The points (0,±1) and (±1, 0) are closest to the origin, while (±1/ 4

√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 andthe circular cone z2 = x2 = y2. 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 + y2 subject to xy2 = 1. Now fx = 2x; fy = 2y; gx =y2; gy = 2xy. We need to solve 2x = λy2, 2y = 2λxy, xy2 − 1 = 0 or 2xy = λy3, 2xy =2λx2y, xy2 = 1 for x > 0, y > 0, and hence λ > 0. From the first two equations, we havey3 = 2x2y or y2 = 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 originis (2−1/3, 21/6). Since the surface is F (x, y, z) = xy2 − 1 = 0, ∇F = y2i + 2xyj is normal tothe 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) = 21/3i + 2(2−1/3)(21/6)j = 21/3i + 25/6j = 22/3(2−1/3i + 21/6j). 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 vectoris perpendicular to the surface.


32. Fx =13x−2/3y1/3z1/3; Fy =

13x1/3y−2/3z1/3; Fz =

13x1/3y1/3z−2/3; gx = 1; gy =

1; gz = 1. We need to solve13x−2/3y1/3z1/3 = λ,

13x1/3y−2/3z1/3 = λ,

13x1/3y1/3z−2/3 =

λ, x + y + z − k = 0 or x1/3y1/3Z1/3 = 3λx, x1/3y1/3z1/3 = 3λy, x1/3y1/3z1/3 =λz, x + y + z = k. From the first three equations, x = y = z. Substituting into the fourthequation, x + x + x = 3x = k. Thus, x = y = z = k/3 and F (k/3, k/3, k/3) = k/3 is aconstrained maximum.

33. For any x+ y + z = k, by Problem 32, 3√xyz ≤ k

3=x+ y + z

3.

34. Distance from the xz-plane is measured by |y|. Alternatively, we will find the extreme valuesof F (x, y, z) = y2 subject to g(x, y, z) = x2 + z2 − 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 + z2 = 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 = 0and y = 0. Substituting y = 0 into the fourth and fifth equations, we obtain the systemx2 + z2 = 1, x + 2z = 4, which is inconsistent. Thus, µ 6= 0 and λ 6= 0. Now, solving thefirst three equations for x, y, and z and substituting into the fourth and fifth equations, weobtain the system

µ2

4λ2+µ2

λ2= 1, − µ

2λ+µ

2− 2µ

λ= 4 or 5µ2 = 4λ2, (λ− 5)µ = 8λ.

Solving the second equation for µ and substituting into the first, we obtain 5(8λλ− 5

)2 = 4λ2

or 80 = (λ − 5)2. Thus, λ = 5 ± 4√

5. From µ = 8λ/(λ − 5) we find that correspondingvalues of µ are 8 ± 2

√5. Since 2y = µ we see that the objective function F (x, y, z) = y2 is

minimized when µ = 8 − 2√

5 and maximized when µ = 8 + 2√

5. Corresponding values of

x, y, and z are x = − µ

2λ= − 8± 2

√5

10± 8√

5= ∓ 1√

5≈ |mp0.45, y − µ/2 = 4 ±

√5 ≈ 6.24 and

1.76, z = −µ/λ = 2x = ∓2/√

5 ≈ ∓0.89. The closest point is (−1/√

5, 4 −√

5,−2√

5) orabout (−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 = y2.

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. lim(x,y)→(1,1)

3x2 + xy2 − 3xy − 2y3

5x2 − y2=

3 + 1− 3− 25− 1

= −14

2. where x− y + 1 = 0

3. 3x2 + y2 = 3(2)2 + (−4)2 = 28

4.∂

∂ξT (p, q) =

∂T

∂p

∂p

∂ξ+∂T

∂q

∂q

∂ξ= Tpgξ + Tqhξ

5.d

dwF (r, s) =

∂F

∂r

dr

dw+∂F

∂s

ds

dw= Frg

′(w) + Fsh′(w)

6. dg = gs∆s+ gt∆t =2t2

∆s− 4st3

∆t

7. fyyzx

8.∂3f

∂y2∂x

9. Using the Fundamental Theorem of Calculus, we have∂f

∂y(x, y) =

∂

∂y

[∫ yxF (t)dt

]= F (y)

∂f

∂x(x, y) =

∂

∂x

[∫ yxF (t)dt

]=

∂

∂x

[−∫ xyF (t)dt

]= − ∂

∂x

[∫ xyF (t)dt

]= −F (x)

10. ∇F (x0, y0, z0) = i + j + k

11. Fx,y,z =∂2

∂z∂yfx(x, y)g(y)h(z) =

∂

∂z[fx(x, y)g′(y)h(z) + fxy(x, y)g(y)h(z)]

= fx(x, y)g′(y)h′(z) + fxy(x, y)g(y)h′(z)

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


C. Exercises

1. zy = −x3ye−x3y + e−x

3y

2. zu = −v sinuvcosuv

= −v tanuv

3. fr =32r2(r3 + θ2)−1/2; frθ = −3

2r2θ(r3 + θ2)−3/2

4.∂f

∂x= 2(2x+ xy2)(2 + y2) = 2(2 + y2)2x;

∂2f

∂x2= 2(2 + y2)2

5.∂z

∂y= 3x2y2 sinhx2y3;

∂2z

∂y2= 9x4y4 coshx2y3 + 6x2y sinhx2y3

6.∂z

∂y= −4y(ex

2+ e−y

2);

∂2z

∂x∂y= −8xyex

2;

∂3z

∂2x∂y= −16x2yex

2 − 8yex2

7. Fs = 3s2t5v−4; Fst = 15s2t4v−4; Fstv = −60s2t4v−5

8.∂w

∂z=xy

z2+x

y+y

x;

∂2w

∂y∂z= − x

z2− x

y2+

1x

;∂3w

∂2y∂z=

2xy3

;∂4w

∂x∂2y∂z=

2y3

9. ∇f = − y

x2

11 + y2/x2

i +1x

11 + y2/x2

j = − y

x2 + y2i +

x

x2 + y2j; ∇f(1,−1) =

12i +

12j

10. ∇F =2xz4

i− 9y2

z4j− 4(x2 − 3y3)

z5k; ∇F (1, 2, 1)− 2i− 36j + 92k

11. ∇f = (2xy − y2)i + (x2 − 2xy)j; u =2√40

i +6√40

j =1√10

(i + 3j);

Duf =1√10

(2xy − y2 + 3x2 − 6xy) =1√10

(3x2 − 4xy − y2)

12. ∇F =2x

x2 + y2 + z2i +

2yx2 + y2 + z2

j +2z

x2 + y2 + z2k; u = −2

3i +

13j +

23k;

DuF =−4x+ 2y + 4z3(x2 + y2 + z2)


13. {(x, y)|(x + y)2 ≤ 1} ={(x, y)| |x+ y| ≤ 1}

y

x

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

x

y

15. ∆z = 2(x+ ∆x)(y+ ∆y)− (y+ ∆y)2− (2xy− y2) = 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 − 4y2 + 7x− 9y + 10)

= 2∆x+ (∆x)2 − 8y∆y − 4(∆y)2 + 7∆x− 9∆y

17. zx =4x+ 3y − (x− 2y)4

(4x+ 3y)2=

11y(4x+ 3y)2

; zy =(4x+ 3y)(−2)− (x− 2y)3

(4x+ 3y)2=

−11x(4x+ 3y)2

;

dz =11y

(4x+ 3y)2dx− 11x

(4x+ 3y)2dy

18. Ax = 2y + 2z; Ay = 2x+ 2z; Az = 2y + 2x; dA = 2(y + z)dx+ 2(x+ z)dy + 2(x+ y)dz

19. zy = 4y/√x2 + 4y2, zy(−

√5, 1) = 4/3, z(−

√5, 1) = 3. The line is given by x = −

√5 and

z − 3 =43

(y − 1). Symmetric equations of the line are x = −√

5,z − 3

4=y − 1

3.

20. The direction vector is−−→PQ = 2i + 2j. ∇z = (y+ 2x)i +xj. u =

1√2i +

1√2j; Du = ∇z ·u =

(y + 2x+ x)/√

2 = (y + 3x)/√

2; Du(2, 3) = 9/√

2. The slope of the tangent line is 9/√

2.

21. fx = 2xy4, fy = 4x2y3.

(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

dt=∂w

∂x

dx

dt+∂w

∂y

dxy

dt+∂w

∂z

dz

dt

=x√

x2 + y2 + z26 cos 2t+

y√x2 + y2 + z2

(−8 sin 2t) +z√

x2 + y2 + z215t2

=(6x cos 2t− 8y sin 2t+ 15zt2)√

x2 + y2 + z2


(b)dw

dt=∂w

∂x

dx

dt+∂w

∂y

dxy

dt+∂w

∂z

dz

dt

=x√

x2 + y2 + z2

6r

cos2tr

+y√

x2 + y2 + z2

(8rt2

sin2rt

)+

z√x2 + y2 + z2

15t2r3

=

(6xr

cos2tr

+8yrt2

sin2rt

+ 15zt2r3

)√x2 + y2 + z2

23. F (x, y, z) = sinxy−z; ∇F = y cosxyi+x cosxyj−k; ∇F (1/2, 2π/3,√

3/2) =π

3i+

14j−k.

The equation of the tangent plane isπ

3(x− 1

2)+

14

(y− 2π3

)−(z−√

32

) = 0 or 4πx+3y−12z =

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 andx = 4. In this case z2 = 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) = 0or 3x+ 4y = 25.

26. We want to minimize

Duf = u · ∇f =1√2

(i + j) · [(3x2 + 3y − 6x)i + (3x+ 3y2)j] =3√2

(x2 + y − 2x+ x+ y2)

or equivalently, we want to minimize F (x, y) = x2 − x + y2 + 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 Duf, 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 threeequations, we have x = 2y = z. Substituting into the fourth equation gives x+x+x = 3x = 6or x = 2. Then y = 1 and z = 2 and V (2, 1, 2) = 4 is the maximum volume.

28. (a) M =c2

GDθ2

(b) dM =c2

G(θ2dD + 2Dθdθ)

(c) We havedM

M=c2

G

(θ2

MdD +

2DθM

dθ

)=

θ2

Dθ2dD +

2DθDθ2

dθ =dD

D+ 2

dθ

θ, so

∣∣∣∣dMM∣∣∣∣ =

∣∣∣∣dDD + 2dθ

θ

∣∣∣∣ ≤ ∣∣∣∣dDD∣∣∣∣+ 2

∣∣∣∣dθθ∣∣∣∣ ≤ 0.10 + 2(0.02) = 0.14 = 14%.


29. We are given v = 14√

5ry−1/2, dr = −1, dy = 1, r = 20, and y = 25. Now, dv =14√

5y−1/2dr − 7√

5ry−3/2dy 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

(a) ∆f(1,−2)2i− 4j; u = (2i− 4j)√

20 = (i− 2j)/√

5;Duf(3, 4) = 6

√5− 16

√5 = −10

√5 = −2

√5

(b) v = (6i + 8j)/√

100 = (3i + 4j)/5; Dvf(3, 4) = 18/5 + 32/5 = 10

31. Let g(r, θ) =R2 − r2

R2 − 2rR cos(θ − φ) + r2. Then, after a straightforward but lengthy com-

putaiton, we find

gr =(2r2R+ 2R3) cos(θ − φ)− 4rR2

(R2 − 2rR cos(θ − φ) + r2)2,

grr =8R4 cos2(θ − φ) + (−12rR3 − 4r3R) cos(θ − φ)− 4R4 + 12r2R2

[R2 − 2rR cos(θ − φ) + r2]3,

gθθ =(4r4R2 − 4r2R4) cos2(θ − φ) + (2r5R− 2rR5) cos(θ − φ)− 8r4R2 + 8r2R4

(R2 − 2rR cos θ + r2)3

and r2grr + rgr + gθθ. Then

r2Urr + rUr + Uθθ =r2

2π

∫ π

π

g(r, θ)f(φ)dφ+r

2π

∫ π

π

g(r, θ)f(φ)dφ+1

2π

∫ π

π

g(r, θ)f(φ)dφ

=1

2π

∫ π

π

(r2grr + rgr + gθθ)dφ = 0.

32. fx = Aαxα−1yβ =αAxαyβ

x=αz

x; fy = Aβxαyβ−1 =

βAxαyβ

y=βy

y;

fxx =xαzx − αz

x2=xα(αz/x)− αz

x2=α2z − αz

x2=α(α− 1)

x2;

fyy =yβzy − βz

y2=yβ(βz − βz)

y2=β2z − βz

y2=β(β − 1)z

y2;

fxy = fyz =α(βz)/y

x=αβz

xy

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.


Ly

xθ

37. Since x = L cos θ and y = L sin θ,

A =12xy =

12L2 sin θ cos θ =

14l2 sin 2θ.

θx

φ

h

1

38. Substituting x = h cotφ into tan θ =h

1 + xand solving, we obtain

h =tan θ

1− tan θ cosφ.

39. A = xy − (y − 2z)(x− 2z)− z2 = 2(x+ y)z − 5z2

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

.

41. V = (2x)(2y)z = 4xy(

4−√x2 − y2

)= 16xy − 4xy

√x2 + y2

42. C(x, y, z) = 1.5(2xy + 2xz + 2yz + xz + 5yz) =32

(2xy + 3xz + 7yz)

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