Calculus one and several variables 10E Salas solutions manual ch15

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

JWDD027-15 JWDD027-Salas-v1 December 5, 2006 16:40

758 SECTION 15.1

CHAPTER 15

SECTION 15.1

1. dom (f) = the first and third quadrants, including the axes; range (f) = [0,∞)

2. dom (f) = the set of all points (x, y) with xy ≤ 1; the two branches of the hyperbola xy = 1 and allpoints in between; range (f) = [0,∞)

3. dom (f) = the set of all points (x, y) except those on the line y = −x; range (f) = (−∞, 0) ∪ (0,∞)

4. dom (f) = the set of all points (x, y) other than the origin; range (f) = (0,∞)

5. dom (f) = the entire plane; range (f) = (−1, 1) since

ex − ey

ex + ey=

ex + ey − 2ey

ex + ey= 1 − 2

ex−y + 1and the last quotient takes on all values between 0 and 2.

6. dom (f) = the set of all points (x, y) other than the origin; range (f) = [0, 1]

7. dom (f) = the first and third quadrants, excluding the axes; range (f) = (−∞,∞)

8. dom (f) =the set of all points (x, y) between the branches of the hyperbola xy = 1;range (f) = (−∞,∞)

9. dom (f) = the set of all points (x, y) with x2 < y —in other words, the set of all points of the planeabove the parabola y = x2; range (f) = (0,∞)

10. dom (f) = the set of all points (x, y) with −3 ≤ x ≤ 3, −1 ≤ y ≤ 1 (a rectangle);range (f) = [0, 3]

11. dom (f) = the set of all points (x, y) with −3 ≤ x ≤ 3, −2 ≤ y ≤ 2 (a rectangle);

range (f) = [−2, 3]

12. dom (f) = all of space; range (f) = [−3, 3]

13. dom (f) = the set of all points (x, y, z) not on the plane x + y + z = 0; range (f) = {−1, 1}

14. dom (f) = the set of all points (x, y, z) with x2 �= y2 —that is, all points of space except for thosewhich lie on the plane x− y = 0 or on the plane x + y = 0; range (f) = (−∞,∞)

15. dom (f) = the set of all points (x, y, z) with |y| < |x|; range (f) = (−∞, 0 ]

16. dom (f) =the set of all points (x, y, z) not on the plane x− y = 0; range (f) = (−∞,∞)

17. dom (f) = the set of all points (x, y) with x2 + y2 < 9 —in other words, the set of all points of theplane inside the circle x2 + y2 = 9; range (f) = [ 2/3,∞)

18. dom (f) = all of space; range (f) = [0,∞)

19. dom (f) = the set of all points (x, y, z) with x + 2y + 3z > 0 — in other words, the set of all points inspace that lie on the same side of the plane x + 2y + 3z = 0 as the point (1, 1, 1); range (f) = (−∞,∞)

20. dom (f) = the set of all points (x, y, z) with x2 + y2 + z2 ≤ 4 — in other words, the set of all pointsinside and on the sphere x2 + y2 + z2 = 4; range (f) = [1, e2]

21. dom (f) = all of space; range (f) = (0, 1]



SECTION 15.1 759

22. dom (f) = the set of all points (x, y, z) with −1 ≤ x ≤ 1, −2 ≤ y ≤ 2, −3 ≤ z ≤ 3 (a rectangularsolid); range (f) = [0, 3]

23. dom (f) = {x : x ≥ 0}; range (f) = [0, ∞)

dom (g) = {(x, y) : x ≥ 0, y real}; range (g) = [0, ∞)

dom (h) = {(x, y, z) : x ≥ 0, y, z real}; range (h) = [0, ∞)

24. dom (f) = the entire plane, range (f) = [−1, 1]

dom (g) = all of space, range (g) = [−1, 1]

25. limh→0

f(x + h, y) − f(x, y)h

= limh→0

2(x + h)2 − y − (2x2 − y)h

= limh→0

4xh + 2h2

h= 4x

limh→0

f(x, y + h) − f(x, y)h

= limh→0

2x2 − (y + h) − (2x2 − y)h

= −1

26. limh→0

f(x + h, y) − f(x, y)h

= limh→0

xy + hy + 2y − (xy + 2y)h

= limh→0

y = y.

limh→0

f(x, y + h) − f(x, y)h

= limh→0

xy + xh + 2y + 2h− (xy + 2y)h

= limh→0

(x + 2) = x + 2

27. limh→0

f(x + h, y) − f(x, y)h

= limh→0

3(x + h) − (x + h)y + 2y2 − (3x− xy + 2y2)h

= limh→0

3h− hy

h= 3 − y

limh→0

f(x, y + h) − f(x, y)h

= limh→0

3x− x(y + h) + 2(y + h)2 − (3x− xy + 2y2)h

= limh→0

−xh + 4yh + 2h2

h= −x + 4y

28. limh→0

f(x + h, y) − f(x, y)h

= limh→0

x sin y + h sin y − x sin y

h= lim

h→0sin y = sin y.

limh→0

f(x, y + h) − f(x, y)h

= limh→0

x sin(y + h) − x sin y

h= x lim

h→0

sin(y + h) − sin y

h= x cos y

29. limh→0

f(x + h, y) − f(x, y)h

= limh→0

cos[(x + h)y] − cos[xy]h

= limh→0

cos[xy] cos[hy] − sin[xy] sin[hy] − cos[xy]h

= cos[xy](

limh→0

cos[hy] − 1h

)− sin[xy] lim

h→0

sinhy

h

= y cos[xy](

limh→0

cos[hy] − 1hy

)− y sin[xy] lim

h→0

sinhy

hy

= −y sin[xy]



760 SECTION 15.2

and

limh→0

f(x, y + h) − f(x, y)h

= limh→0

cos[x(y + h)] − cos[xy]h

= limh→0

cos[xy] cos[hx] − sin[xy] sin[hx] − cos[xy]h

= cos[xy](

limh→0

cos[hx] − 1h

)− sin[xy] lim

h→0

sinhx

h

= x cos[xy](

limh→0

cos[hx] − 1hx

)− x sin[xy] lim

h→0

sinhx

hx

= −x sin[xy]

30. limh→0

f(x + h, y) − f(x, y)h

= limh→0

(x2 + 2xh + h2)ey − x2ey

h= lim

h→0(2x + h)ey = 2xey.

limh→0

f(x, y + h) − f(x, y)h

= limh→0

x2ey+h − x2ey

h= x2 lim

h→0

ey+h − ey

h= x2ey.

31. (a) f(x, y) =Ay (b) f(x, y) = πx2y (b) f(x, y) = |2 i × (x i + y j)| = 2|y|

32. (a) f(x, y, z) = xy + 2xz + 2yz

(b) f(x, y, z) = cos−1 (i + j) · (x i + y j + z k)‖i + j‖‖x i + y j + z k‖ = cos−1 x + y√

2√x2 + y2 + z2

(c) f(x, y, z) = [i × (i + j)] · (x i + y j + z k) = z

33. Surface area: S = 2lw + 2lh + 2hw = 20 =⇒ w =20 − 2lh2l + 2h

=10 − lh

l + h

Volume: V = lwh =lh(10 − lh)

l + h

34. wlh = 12 =⇒ h =12wl

; C = 4wl + 2(2wh + 2lh) = 4wl +48l

+48w

35. V = πr2h +43πr3

36. A =12[2(12 − 2x) + 2x cos θ] · x sin θ = (12 − 2x + x cos θ) · x sin θ

SECTION 15.2

1. an elliptic cone 2. an ellipsoid

3. a parabolic cylinder 4. a hyperbolic paraboloid

5. a hyperboloid of one sheet 6. an elliptic cylinder

7. a sphere 8. a hyperboloid of two sheets

9. an elliptic paraboloid 10. a hyperbolic cylinder

11. a hyperbolic paraboloid 12. an elliptic paraboloid



SECTION 15.2 761

13. 14. 15.

16. 17. 18.

19. 20. 21.

22. 23. 24.



762 SECTION 15.2

25. elliptic paraboloidxy-trace: the originxz-trace: the parabola x2 = 4zyz-trace: the parabola y2 = 9zsurface has the form of Figure 15.2.5

26. ellipsoidxy-trace: the ellipse 9x2 + 4y2 = 36xz-trace: the ellipse x2 + 4z2 = 4yz-trace: the ellipse y2 + 9z2 = 9surface has the form of Figure 15.2.1

27. elliptic conexy-trace: the originxz-trace: the lines x = ±2zyz-trace: the lines y = ±3zsurface has the form of Figure 15.2.4

28. hyperboloid of one sheetxy-trace: the ellipse 9x2 + 4y2 = 36xz-trace: the hyperbola x2 − 4z2 = 4yz-trace: the hyperbola y2 − 9z2 = 9surface has the form of Figure 15.2.2

29. hyperboloid of two sheetsxy-trace: nonexz-trace: the hyperbola 4z2 − x2 = 4yz-trace: the hyperbola 9z2 − y2 = 9surface has the form of Figure 15.2.3

30. hyperbolic paraboloidxy-trace: the lines y = ± 3

2x

xz-trace: the parabola x2 = 4zyz-trace: the parabola y2 = −9zsurface has the form Figure 15.2.6

31. hyperboloid of two sheetsxy-trace: the hyperbola 9x2 − 4y2 = 36xz-trace: the hyperbola x2 − 4z2 = 4yz-trace: nonesee Figure 15.2.3

32. hyperboloid of one sheetxy-trace: the hyperbola x2 − 9y2 = 9xz-trace: the circle x2 + y2 = 9yz-trace: the hyperbola z2 − 9y2 = 9surface has the form of Figure 15.2.2,rotated 90◦ about the x-axis.

33. elliptic paraboloidxy-trace: the parabola x2 = 9yxz-trace: the originyz-trace: the parabola z2 = 4ysurface has the form of Figure 15.2.5

34. elliptic conexy-trace: the lines x = ±2yxz-trace: the originyz-trace: the lines z = ±3ysurface has the form of Figure 15.2.4,rotated 90◦ about the x-axis.

35. hyperboloid of two sheetsxy-trace: the hyperbola 9y2 − 4x2 = 36xz-trace: noneyz-trace: the hyperbola y2 − 4z2 = 4see Figure 15.2.3

36. elliptic paraboloidxy-trace: the parabola y2 = 4xxz-trace: the parabola z2 = 9xyz-trace: the originsurface has the form of Figure 15.2.5,but opening along the positive x-axis.

37. paraboloid of revolutionxy-trace: the originxz-trace: the parabola x2 = 4zyz-trace: the parabola y2 = 4zsurface has the form of Figure 15.2.5

38. ellipsoidxy-trace: the ellipse 4x2 + y2 = 4xz-trace: the ellipse 9x2 + z2 = 9yz-trace: the ellipse 9y2 + 4z2 = 36the surface has the form of Figure 15.2.1,rotated 90◦ about the x-axis.

39. (a) an elliptic paraboloid (vertex down if A and B are both positive, vertex up if A and B are bothnegative)

(b) a hyperbolic paraboloid

(c) the xy-plane if A and B are both zero; otherwise a parabolic cylinder

40. The xz-plane and all planes parallel to the xy-plane.



SECTION 15.2 763

41. x2 + y2 − 4z = 0 (paraboloid of revolution)

42. c2x2 + c2y2 − b2z2 = b2c2 (hyperboloid of revolution, one sheet)

43. (a) a circle

(b) (i)√x2 + y2 = −3z (ii)

√x2 + z2 = 1

3y

44. (a) the ellipse b2x2 + y2 = b2

(b) ellipse approaches parallel lines x = ±1 in the plane z = 1

(c) paraboloid approaches parabolic cylinder z = x2

45. x + 2y + 3(x + y − 6

2

)= 6 or 5x + 7y = 30, a line

46. 3x + y − 2(4 − x + 2y) = 1, or 5x− 3y = 9, a line

47.x2 + y2 + (z − 1)2 = 3

2

x2 + y2 − z2 = 1

}(z2 + 1) + (z − 1)2 =

32; (2z − 1)2 = 0, z =

12

so that x2 + y2 =54

48. z2 + (z − 2)2 = 2 =⇒ 2(z − 1)2 = 0 =⇒ z = 1 =⇒ x2 + y2 = 1, a circle.

49. x2 + y2 +(x2 + 3y2

)= 4 or x2 + 2y2 = 2, an ellipse

50. y2 + (x2 + 3y2) = 4 =⇒ x2 + 4y2 = 4, an ellipse.

51. x2 + y2 = (2 − y)2 or x2 = −4(y − 1), a parabola

52. x2 + y2 =(

2 − y

2

)2

=⇒ 4x2 + 3y2 + 4y = 4, an ellipse.

53. (a) Set x = a cosu cos v, y = b cosu sin v, z = c sinu. Then:x2

a2+

y2

b2+

z2

c2= 1.

(b)

-2

0

2

-4

-2

0

2

4

-2

-1

0

1

2

-2

0

2



764 SECTION 15.3

54. (a) Set x = a secu cos v,

y = b secu sin v,

z = c tanu

(b) -10-5

0510

-10

0

10

-20

0

20

10

0

10

55. (a) Set x = av cosu,

y = bv sinu,

z = cv

-2-1

01

2

-4

-2

0

2

4

0

2

4

6

1

SECTION 15.3

1. lines of slope 1: y = x− c 2. lines of slope 2 : y = 2x− c

3. parabolas: y = x2 − c 4. parabolas: x− y2 =1c



SECTION 15.3 765

5. the y-axis and the lines y =(

1 − c

c

)x with

the origin omitted throughout

6. the x-axis and the parabolas y = cx2 with the

origin omitted throughout

7. the cubics y = x3 − c 8. the coordinate axes and the hyperbolas xy = ln c

9. the lines y = ±x and the hyperbolas

x2 − y2 = c

10. pairs of vertical lines x = ±√c and the y-axis

11. pairs of horizontal lines y = ±√c and the

x-axis

12. the lines x = 0, y = 1 and the hyperbolas

y =c

x+ 1



766 SECTION 15.3

13. the circles x2 + y2 = ec, c real 14. the parabolas y = ecx2 with the origin omitted

throughout

15. the curves y = ecx2

with the point (0, 1)

omitted

16. the coordinate axes and pairs of hyperbolas

xy = ±√c

17. the coordinate axes and pairs of lines

y = ±√

1 − c√c

x, with the origin omitted

throughout

18. the curves y = ecx with the point (0, 1) omitted

19. x + 2y + 3z = 0, plane through the origin

20. circular cylinder x2 + y2 = 4 (Figure 15.2.8)

21. z =√x2 + y2, the upper nappe of the circular cone z2 = x2 + y2 (Figure 15.2.4)

22. ellipsoidx2

4+

y2

6+

z2

9= 1 (Figure 15.2.1)

23. the elliptic paraboloidx2

18+

y2

8= z (Figure 15.2.5)



SECTION 15.3 767

24. hyperboloid of two sheetsx2

(1/6)2+

y2

(1/3)2− z2 = −1 (Figure 15.2.3)

25. (i) hyperboloid of two sheets (Figure 15.2.3)

(ii) circular cone (Figure 15.2.4)

(iii) hyperboloid of one sheet (Figure 15.2.2)

26. (i) hyperboloid of two sheets

(ii) elliptic cone

(iii) hyperboloid of one sheet

27. The level curves of f are: 1 − 4x2 − y2 = c. Substituting P (0, 1) into this equation, we have

1 − 4(0)2 − (1)2 = c =⇒ c = 0

The level curve that contains P is: 1 − 4x2 − y2 = 0, or 4x2 + y2 = 1.

28. (x2 + y2)exy = 1

29. The level curves of f are: y2 arctanx = c. Substituting P (1, 2) into this equation, we have

4 arctan 1 = c =⇒ c = π

The level curve that contains P is: y2 tan−1 x = π.

30. (x2 + y) ln(2 − x + ey) = 5

31. The level surfaces of f are: x2 + 2y2 − 2xyz = c. Substituting P (−1, 2, 1) into this equation, we

have

(−1)2 + 2(2)2 − 2(−1)(2)(1) = c =⇒ c = 13

The level surface that contains P is: x2 + 2y2 − 2xyz = 13.

32.√x2 + y2 − ln z = 4

33. (a)

-4

-2

0

2

4-4

-2

0

2

4

-50

0

50

-4 -2 0 2 4

-4

-2

0

2

4



768 SECTION 15.3

(b)

-5

0

5-5

0

5

0

5

10

15

-4 -2 0 2 4

-4

-2

0

2

4

34. (a) the level surfaces are planes (b)

35. (a)3x + 2y + 1

4x2 + 9=

35

(b) x2 + 2y2 − z2 = 21

36. (a)

-2 -1 0 1 2

-2

-1

0

1

2 (b)

37.GmM

x2 + y2 + z2= c =⇒ x2 + y2 + z2 =

GmM

c; the surfaces of constant gravitational force are

concentric spheres.

38. Circular cylinders around the positive y-axis: x2 + z2 =k2

c2



SECTION 15.4 769

39. (a) T (x, y, z) =k√

x2 + y2 + z2, where k is a constant.

(b)k√

x2 + y2 + z2= c =⇒ x2 + y2 + z2 =

k2

c2; the level surfaces are concentric spheres.

(c) T (1, 2, 1) =k√

12 + 22 + 12= 50 =⇒ k = 50

√6 =⇒ T (x, y, z) =

50√

6√x2 + y2 + z2

Now, T (4, 0, 3) =50√

6√32 + 42

= 10√

6 degrees

40. x2 + y2 = r2 − k2

c2; circles about the origin, for |c| > k

r.

41. f(x, y) = y2 − y3; F 42. D. 43. f(x, y) = cos√

x2 + y2; A

44. B. 45. f(x, y) = xye−(x2+y2)/2; E 46. C.

PROJECT 15.3

1. 2.

3. 4.

SECTION 15.4

1.∂f

∂x= 6x− y,

∂f

∂y= 1 − x 2.

∂g

∂x= 2xe−y,

∂g

∂y= −x2e−y

3.∂ρ

∂φ= cosφ cos θ,

∂ρ

∂θ= − sinφ sin θ



770 SECTION 15.4

4.∂ρ

∂θ= 2 sin(θ − φ) cos(θ − φ),

∂ρ

∂φ= −2 sin(θ − φ) cos(θ − φ)

5.∂f

∂x= ex−y + ey−x,

∂f

∂y= −ex−y − ey−x 6.

∂z

∂x=

x√x2 − 3y

,∂z

∂y=

−3

2√x2 − 3y

7.∂g

∂x=

(AD −BC)y(Cx + Dy)2

,∂g

∂y=

(BC −AD)x(Cx + Dy)2

8.∂u

∂x=

−ez

x2y2,

∂u

∂y=

−2ez

xy3,

∂u

∂z=

ez

xy2

9.∂u

∂x= y + z,

∂u

∂y= x + z,

∂u

∂z= x + y 10.

∂z

∂x= 2Ax + By,

∂z

∂y= Bx + 2Cy

11.∂f

∂x= z cos (x− y),

∂f

∂y= −z cos (x− y),

∂f

∂z= sin (x− y)

12.∂g

∂u=

2uu2 + vw − w2

,∂g

∂v=

w

u2 + vw − w2,

∂g

∂w=

v − 2wu2 + vw − w2

13.∂ρ

∂θ= eθ+φ [cos (θ − φ) − sin (θ − φ)],

∂ρ

∂φ= eθ+φ [cos (θ − φ) + sin (θ − φ)]

14.∂f

∂x= (x + y) cos(x− y) + sin(x− y),

∂f

∂y= −(x + y) cos(x− y) + sin(x− y)

15.∂f

∂x= 2xy secxy + x2y(secxy)(tanxy)y = 2xy secxy + x2y2 secxy tanxy

∂f

∂y= x2 secxy + x2y(secxy)(tanxy)x = x2 secxy + x3y secxy tanxy

16.∂g

∂x=

21 + (2x + y)2

,∂g

∂y=

11 + (2x + y)2

17.∂h

∂x=

x2 + y2 − x(2x)(x2 + y2)2

=y2 − x2

(x2 + y2)2∂h

∂y=

−2xy(x2 + y2)2

18.∂z

∂x=

x

x2 + y2,

∂z

∂y=

y

x2 + y2

19.∂f

∂x=

(y cosx) sin y − (x sin y)(−y sinx)(y cosx)2

=sin y(cosx + x sinx)

y cos2 x

∂f

∂y=

(y cosx)(x cos y) − (x sin y) cosx(y cosx)2

=x(y cos y − sin y)

y2 cosx

20.∂f

∂x= exy(y sinxz + z cosxz),

∂f

∂y= xexy sinxz,

∂f

∂z= xexy cosxz

21.∂h

∂x= 2f(x)f ′(x)g(y),

∂h

∂y= [f(x)]2g′(y)



SECTION 15.4 771

22.∂h

∂x= f ′(x)g(y)ef(x)g(x),

∂h

∂y= f(x)g′(y)ef(x)g(y)

23.∂f

∂x= (y2 ln z)zxy

2,

∂f

∂y= (2xy ln z)zxy

2,

∂f

∂z= xy2zxy

2−1

24.∂h

∂x= 2[f(x, y)]3g(x, z)

∂g

∂x+ 3[f(x, y)]2[g(x, z)]2

∂f

∂x

∂h

∂y= 3[f(x, y)]2[g(x, z)]2

∂f

∂y,

∂h

∂z= 2[f(x, y)]3g(x, y)

∂g

∂z

25.∂h

∂r= 2re2t cos(θ − t)

∂h

∂θ= −r2e2t sin(θ − t)

∂h

∂t= 2r2e2t cos(θ − t) + r2e2t sin(θ − t) = r2e2t[2 cos(θ − t) + sin(θ − t)]

26.∂u

∂x=

1x− yzexz,

∂u

∂y= −1

y− exz,

∂u

∂z= −xyexz

27.∂f

∂x= z

11 + (y/x)2

(−y

x2

)= − yz

x2 + y2

∂f

∂y= z

11 + (y/x)2

(1x

)=

xz

x2 + y2

∂f

∂x= arctan (y/x)

28.∂w

∂x= y sin z − yz cosx,

∂w

∂y= x sin z − z sinx,

∂w

∂z= xy cos z − y sinx

29. fx(x, y) = ex ln y, fx(0, e) = 1; fy(x, y) =1yex, fy(0, e) = e−1

30. gx = e−x[− sin(x + 2y) + cos(x + 2y)], gx(0, 14π) = −1

gy = 2e−x cos(x + 2y), gy(0, 14π) = 0

31. fx(x, y) =y

(x + y)2, fx(1, 2) =

29; fy(x, y) =

−x

(x + y)2, fy(1, 2) = −1

9

32. gx(x, y) =y2

(x + y2)2, gx(1, 2) =

425

gy(x, y) =−2xy

(x + y2)2, gy(1, 2) = − 4

25

33. fx(x, y) = limh→0

(x + h)2 y − x2y

h= lim

h→0y

(2xh + h2

h

)= y lim

h→0(2x + h) = 2xy

fx(x, y) = limh→0

x2(y + h) − x2y

h= lim

h→0

x2h

h= lim

h→0x2 = x2

34. fx(x, y) = 0, fy(x, y) = 2y

35. fx(x, y) = limh→0

ln(y(x + h)2

)− lnx2y

h= lim

h→0

ln y + 2 ln(x + h) − 2 lnx− ln y

h

= 2 limh→0

ln(x + h) − lnx

h= 2

d

dx(lnx) =

2x



772 SECTION 15.4

fy(x, y) = limh→0

ln(x2(y + h)

)− lnx2y

h= lim

h→0

lnx2 + ln(y + h) − lnx2 − ln y

h

= limh→0

ln(y + h) − ln y

h=

d

dy(ln y) =

1y

36. fx(x, y) = −(x + 4y)−2, fy(x, y) = −4(x + 4y)−2

37. fy(x, y) = limh→0

1h

{1

(x + h) − y− 1

x− y

}= lim

h→0

1h

{ −h

(x + h− y) (x− y)

}

= limh→0

−1(x + h− y) (x− y)

=−1

(x− y)2

fy(x, y) = limh→0

1h

{1

x− (y + h)− 1

x− y

}= lim

h→0

1h

{h

(x− y − h) (x− y)

}

= limh→0

1(x− y − h) (x− y)

=1

(x− y)2

38. fx(x, y) = 2e2x+3y, fy(x, y) = 3e2x+3y

39. fx(x, y, z) = limh→0

(x + h)y2z − xy2z

h= lim

h→0y2z = y2z

fy(x, y, z) = limh→0

x(y + h)2z − xy2z

h= lim

h→0

xz(2yh + h2)h

= limh→0

xz(2y + h) = 2xyz

fz(x, y, z) = limh→0

xy2 (z + h) − xy2z

h= lim

h→0xy2 = xy2

40. fx(x, y, z) =2xyz

, fy(x, y, z) =x2

z, fz(x, y, z) = −x2y

z2

41. (b) The slope of the tangent line to C at the point P (x0, y0, f(x0, y0)) is fy(x0, y0)

Thus, equations for the tangent line are:

x = x0, z − z0 = fy(x0, y0)(y − y0)

42. fx(x, y) = 2x, fx(1, 3) = 2, equations for the tangent line are: y = 3, z − 10 = 2(x− 1).

43. Let z = f(x, y) = x2 + y2. Then f(2, 1) = 5, fy(x, y) = 2y and fy(2, 1) = 2;

equations for the tangent line are: x = 2, z − 5 = 2(y − 1)

44. f(x, y) =x2

y2 − 3, fy(x, y) =

−x22y(y2 − 3)2

Tangent line: x = x0, z − z0 = fy(x0, y0)(y − y0) =⇒ x = 3, z − 9 = −36(y − 2)



SECTION 15.4 773

45. Let z = f(x, y) =x2

y2 − 3. Then f(3, 2) = 9, fx(x, y) =

2xy2 − 3

and fx(3, 2) = 6;

equations for the tangent line are: y = 2, z − 9 = 6(x− 3)

46. f(x, y) = (4 − x2 − y2)1/2, fx(x, y) = −x(4 − x2 − y2)−1/2, fy(x, y) = −y(4 − x2 − y2)−1/2

(a) fy(1, 1) = −√

22

=⇒ x = 1, z −√

2 = −√

22

(y − 1)

(b) fx(1, 1) = −√

22

=⇒ y = 1, z −√

2 = −√

22

(x− 1)

(c) l1 and l2 have direction vectors j −√

22 k, i −

√2

2 k respectively. The normal to the plane is(j −

√2

2k

)×

(i −

√2

2k

)= −

√2

2i −

√2

2j − k, so the tangent plane is

−√

22

(x− 1) −√

22

(y − 1) − (z −√

2) = 0, or (x− 1) + (y − 1) +√

2(z −√

2) = 0

47. (a) mx = −6; tangent line: y = 2, z = −6x + 13

(b) my = 18; tangent line: x = 1, z = 18y − 29

48. (a) mx = 725 ; tangent line: y = 2, z = 1

25 (7x− 12)

(b) my = − 125 ; tangent line: x = 1, z = − 1

25 (y + 3)

49. ux(x, y) = 2x = vy(x, y); uy(x, y) = −2y = −vx(x, y)

50. ux = ex cos y, uy = −ex sin y, vx = ex sin y = −uy, vy = ex cos y = ux

51. ux(x, y) =12

1x2 + y2

2x =x

x2 + y2; vy(x, y) =

11 + (y/x)2

(1x

)=

x

x2 + y2

Thus, ux(x, y) = vy(x, y).

uy(x, y) =12

1x2 + y2

2y =y

x2 + y2; vx(x, y) =

11 + (y/x)2

(−y

x2

)=

−y

x2 + y2

Thus, uy(x, y) = −vx(x, y).

52. ux =y2 − x2

(x2 + y2)2, uy =

−2xy(x2 + y2)2

, vx =2xy

(x2 + y2)2= −uy, vy =

y2 − x2

(x2 + y2)2= ux

53. (a) f depends only on y. (b) f depends only on x.

54. (a) a0 =(b20 + c20 − 2b0c0 cos θ0

)1/2 = 5√

7

(b)∂a

∂b= (2b− 2c cos θ)

(12

)(b2 + c2 − 2bc cos θ)−1/2 =

√7

14

(c) a ∼= a0 +√

714

(b− b0) = 5√

7 +√

714

· (−1) decreases by about√

714

inches.

(d)∂a

∂θ= 2bc sin θ(

12)(b2 + c2 − 2bc cos θ)−1/2 =

157

√21



774 SECTION 15.4

(e) Differentiate implicitly: 0 = 2c∂c

∂θ− 2b

∂c

∂θcos θ + 2bc sin θ

∂c

∂θ=

bc sin θ

b cos θ − c= −15

2

√3

55. (a)75√

32

in.2

(b)∂A

∂b=

12c sin θ; at time t0,

∂A

∂b=

15√

34

(c)∂A

∂θ=

12bc cos θ; at time t0,

∂A

∂θ=

752

(d) with h =π

180, A (b, c, θ + h) −A (b, c, θ) ∼= h

∂A

∂θ=

π

180752

=5π24

in.2

(e) 0 =12

sin θ

(b∂c

∂b+ c

); at time t0,

∂c

∂b=

−c

b= −3

2

56. By theorem 7.6.1, f(x, y) = Cekx where C is independent of x. Since C may depend on y, we write

C = g(y).

57. (a) y0-section: r(x) = xi + y0j + f(x, y0)k

tangent line: R(t) = [x0i + y0j + f(x0, y0)k ] + t

[i +

∂f

∂x(x0, y0)k

]

(b) x0-section: r(y) = x0i + yj + f(x0, y)k

tangent line: R(t) = [x0i + y0j + f(x0, y0)k] + t

[j +

∂f

∂y(x0, y0)k

](c) For (x, y, z) in the plane

[(x− x0)i + (y − y0)j + (z − f(x0, y0))k ].[(

i +∂f

∂x(x0, y0)k

)×

(j +

∂f

∂y(x0, y0)k

)]= 0.

From this it follows that

z − f(x0, y0) = (x− x0)∂f

∂x(x0, y0) + (y − y0)

∂f

∂y(x0, y0).

58. Fix y and set F (x) = f(x, y). Then, for that value of y, h(x, y) = g(F (x)) and thus

hx(x, y) =d

dx[g(F (x))] = g′(F (x))F ′(x) = g′(f(x, y))fx(x, y).

The second formula can be derived in the same manner.

59. (a) Set u = ax + by. Then

b∂w

∂x− a

∂w

∂y= b(a g′(u)) − a(b g′(u)) = 0.

(b) Set u = xmyn. Then

nx∂w

∂x−my

∂w

∂y= nx

[mxm−1yng′(u)

]−my

[nxmyn−1g′(u)

]= 0.



SECTION 15.5 775

60.∂x

∂r

∂y

∂θ− ∂x

∂θ

∂y

∂r= (cos θ) (r cos θ) − (−r sin θ) (sin θ) = r

61. V∂P

∂V= V

(−kT

V 2

)= −k

T

V= −P ; V

∂P

∂V+ T

∂P

∂T= −k

T

V+ T

(k

V

)= 0

62. R =R1R2R3

R1R2 + R1R3 + R2R3;

∂R

∂R1=

(R2R3

R1R2 + R1R3 + R2R3

)2

SECTION 15.5

1. interior = {(x, y) : 2 < x < 4, 1 < y < 3} (the

inside of the rectangle), boundary = the union of

the four boundary line segments; set is closed.

2. same interior and same boundary as in Exercise 1;

set is open

3. interior = the entire set (region between the two

concentric circles), boundary = the two circles, one

of radius 1, the other of radius 2; set is open.

4. interior = {(x, y) : 1 < x2 < 4} = {(x, y) : −2 < x <

− 1} ∪ {(x, y) : 1 < x < 2} (two vertical stripes with-

out the boundary lines), boundary = {(x, y) : x = −2,

x = −1, x = 1, or x = 2} (four vertical lines); set is closed.

5. interior = {(x, y) : 1 < x2 < 4} =

{(x, y) : −2 < x < −1} ∪ {(x, y) : 1 < x < 2}(two vertical strips without the boundary lines),

boundary = {(x, y) : x = −2, x = −1, x = 1,

or x = 2} (four vertical lines); set is neither open

nor closed.



776 SECTION 15.5

6. interior = the entire set (region below the parabola

y = x2), boundary = the parabola y = x2; the set

is open.

7. interior = region below the parabola y = x2,

boundary = the parabola y = x2; the set is closed.

8. interior = the inside of the cube; boundary = thefaces of the cube; set is neither open nor closed (upperface of cube is omitted)

9. interior = { (x, y, z) : x2 + y2 < 1, 0 < z ≤ 4}(the inside of the cylinder), boundary = the totalsurface of the cylinder (the curved part, the top, andthe bottom); the set is closed.

10. interior = the entire set (the inside of the ball of radius 12 , centered at (1,1,1)),

boundary = the spherical surface; set is open.

11. (a) φ (b) S (c) closed

12. interior = the entire set, boundary = {1, 3}; set is open.

13. interior = {x : 1 < x < 3}, boundary = {1, 3}; set is closed.

14. interior = {x : 1 < x < 3}, boundary = {1, 3}; set is neither open nor closed.

15. interior = the entire set, boundary = {1}; set is open.

16. interior ={x : x < −1}, boundary = {−1}; set is closed.



SECTION 15.6 777

17. interior = {x : |x| > 1}, boundary = {1,−1}; set is neither open nor closed.

18. interior = φ, boundary = the entire set; set is closed.

19. interior = φ, boundary = {the entire set} ∪ {0}; the set is neither open nor closed.

20. (a) φ is open because it contains no boundary points,

φ is closed because it contains its boundary (the boundary is empty).

(b) X is open because it contains a neighborhood of each of its points,

X is closed because it contains its boundary (the boundary is empty).

(c) Suppose that U is open. Let x be a boundary point of X − U . Then every neighborhood of

x contains points from X − U . The point x can not be in U because U contains a

neighborhood of each of its points. Thus x ∈ X − U . This shows that X − U contains its

boundary and is therefore closed.

Suppose now that X − U is closed. Let x be a point of U . If no neighborhood of x lies entirely

in U , then every neighborhood of x contains points from X − U . This makes x a boundary

point of X − U and, since X − U is closed, places x in X − U . This contradiction shows that

some neighborhood of x lies entirely in U . Thus U contains a neighborhood of each of its points

and is therefore open.

(d) Set U = X − F and note that F = X − U . By (c)

F = X − U is closed iff X − F = U is open.

SECTION 15.6

1.∂2f

∂x2= 2A,

∂2f

∂y2= 2C,

∂2f

∂y∂x=

∂2f

∂x∂y= 2B

2.∂2f

∂x2= 6Ax + 2By,

∂2f

∂y2= 2Cx,

∂2f

∂y∂x=

∂2f

∂x∂y= 2Bx + 2Cy

3.∂2f

∂x2= Cy2exy,

∂2f

∂y2= Cx2exy,

∂2f

∂y∂x=

∂2f

∂x∂y= Cexy(xy + 1)

4.∂2f

∂x2= 2 cos y − y2 sinx,

∂2f

∂y2= 2 sinx− x2 cos y,

∂2f

∂y∂x=

∂2f

∂x∂y= 2(y cosx− x sin y)

5.∂2f

∂x2= 2,

∂2f

∂y2= 4(x + 3y2 + z3),

∂2f

∂z2= 6z(2x + 2y2 + 5z3)

∂2f

∂x∂y=

∂2f

∂y∂x= 4y,

∂2f

∂z∂x=

∂2f

∂x∂z= 6z2,

∂2f

∂z∂y=

∂2f

∂y∂z= 12yz2

6.∂2f

∂x2= − 1

4(x + y2)3/2,

∂2f

∂y2=

x

(x + y2)3/2,

∂2f

∂x∂y=

∂2f

∂y∂x= − y

2(x + y2)3/2



778 SECTION 15.6

7.∂2f

∂x2=

1(x + y)2

− 1x2

,∂2f

∂y2=

1(x + y)2

,∂2f

∂y∂x=

∂2f

∂x∂y=

1(x + y)2

8.∂2f

∂x2= −2C(AD −BC)y

(Cx + Dy)3,

∂2f

∂y2=

2D(AD −BC)x(Cx + Dy)3

,∂2f

∂y∂x=

∂2f

∂x∂y=

(AD −BC)(Cx−Dy)(Cx + Dy)3

9.∂2f

∂x2= 2(y + z),

∂2f

∂y2= 2(x + z),

∂2f

∂z2= 2(x + y)

all the second mixed partials are 2(x + y + z)

10.∂2f

∂x2= − 2xy3z3

(1 + x2y2z2)2,

∂2f

∂y2= − 2yx3z3

(1 + x2y2z2)2

∂2f

∂z2= − 2zx3y3

(1 + x2y2z2)2,

∂2f

∂y∂x=

∂2f

∂x∂y=

z(1 − x2y2z2)(1 + x2y2z2)2

∂2f

∂z∂x=

∂2f

∂x∂z=

y(1 − x2y2z2)(1 + x2y2z2)2

,∂2f

∂y∂z=

∂2f

∂z∂y=

x(1 − x2y2z2)(1 + x2y2z2)2

11.∂2f

∂x2= y(y − 1)xy−2,

∂2f

∂y2= (lnx)2xy,

∂2f

∂y∂x=

∂2f

∂x∂y= xy−1(1 + y lnx)

12.∂2f

∂x2= − sin(x + zy),

∂2f

∂y2= zy(ln z)2[cos(x + zy) − zy sin(x + zy)]

∂2f

∂z2= y(y − 1)zy−2 cos(x + zy) − y2z2y−2 sin(x + zy)

∂2f

∂y∂x=

∂2f

∂x∂y= −zy ln z sin(x + zy),

∂2f

∂z∂x=

∂2f

∂x∂z= −yzy−1 sin(x + zy)

∂2f

∂z∂y=

∂2f

∂y∂z= zy−1(1 + y ln z) cos(x + zy) − yz2y−1(ln z) sin(x + zy)

13.∂2f

∂x2= yex,

∂2f

∂y2= xey,

∂2f

∂y∂x=

∂2f

∂x∂y= ey + ex

14.∂2f

∂x2=

2xy(x2 + y2)2

,∂2f

∂y2=

−2xy(x2 + y2)2

,∂2f

∂y∂x=

∂2f

∂x∂y=

y2 − x2

(x2 + y2)2

15.∂2f

∂x2=

y2 − x2

(x2 + y2)2,

∂2f

∂y2=

x2 − y2

(x2 + y2)2,

∂2f

∂y∂x=

∂2f

∂x∂y= − 2xy

(x2 + y2)2

16.∂2f

∂x2= 6xy2 cos(x3y2) − 9x4y4 sin(x3y2),

∂2f

∂y2= 2x3 cos(x3y2) − 4x6y2 sin(x3y2)

∂2f

∂y∂x=

∂2f

∂x∂y= 6x2y cos(x3y2) − 6x5y3 sin(x3y2)

17.∂2f

∂x2= −2 y2 cos 2xy,

∂2f

∂y2= −2x2 cos 2xy,

∂2f

∂y∂x=

∂2f

∂x∂y= −[sin 2xy + 2xy cos 2xy]



SECTION 15.6 779

18.∂2f

∂x2= y4exy

2,

∂2f

∂y2= exy

2(2x + 4x2y2),

∂2f

∂y∂x=

∂2f

∂x∂y= exy

2(2y + 2xy3)

19.∂2f

∂x2= 0,

∂2f

∂y2= xz sin y,

∂2f

∂z2= −xy sin z,

∂2f

∂y∂x=

∂2f

∂x∂y= sin z − z cos y,

∂2f

∂x∂z=

∂2f

∂z∂x= y cos z − sin y,

∂2f

∂y∂z=

∂2f

∂z∂y= x cos z − x cos y

20.∂2f

∂x2= zex,

∂2f

∂y2= xey,

∂2f

∂z2= yez,

∂2f

∂y∂x=

∂2f

∂x∂y= ey,

∂2f

∂z∂x=

∂2f

∂x∂z= ex,

∂2f

∂z∂y=

∂2f

∂y∂z= ez

21. x2 ∂2u

∂x2+ 2xy

∂2u

∂x∂y+ y2 ∂2u

∂y2= x2

( −2y2

(x + y)3

)+ 2xy

(2xy

(x + y)3

)+ y2

( −2x2

(x + y)3

)= 0

22. (a) mixed partials are 0

(b) mixed partials are g′(x)h′(y)

(c) by the hint mixed partials for each term xmyn are mnxm−1yn−1

23. (a) no, since∂2f

∂y∂x�= ∂2f

∂x∂y(b) no, since

∂2f

∂y∂x�= ∂2f

∂x∂yfor x �= y

24.∂h

∂x= g′(x + y) + g′(x− y),

∂h

∂y= g′(x + y) − g′(x− y)

∂2h

∂x2= g′′(x + y) + g′′(x− y),

∂2h

∂y2= g′′(x + y) + g′′(x− y) =

∂2h

∂x2

25.∂3f

∂x2∂y=

∂

∂x

(∂2f

∂x∂y

)=

∂

∂x

(∂2f

∂y∂x

)=

∂2

∂x∂y

(∂f

∂x

)=

∂2

∂y∂x

(∂f

∂x

)=

∂

∂y

(∂2f

∂x2

)=

∂3f

∂y∂x2

∧by def.

∧(15.6.5)

∧by def.

∧(15.6.5)

∧by def.

∧by def.

26. (a) as (x, y) tends to (0, 0) along the x-axis, f(x, y) = f(x, 0) = 1 tends to 1;

as (x, y) tends to (0, 0) along the line y = x, f(x, y) = f(x, x) = 0 tends to 0;

(b) as (x, y) tends to (0, 0) along the x-axis, f(x, y) = f(x, 0) = 0 tends to 0;

as (x, y) tends to (0, 0) along the line y = x, f(x, y) = f(x, x) = 12 tends to 1

2 ;

27. (a) limx→0

(x)(0)x2 + 0

= limx→0

0 = 0 (b) limy→0

(0)(y)0 + y2

= limy→0

0 = 0

(c) limx→0

(x)(mx)x2 + (mx)2

= limx→0

m

1 + m2=

m

1 + m2

(d) limθ→0+

(θ cos θ)(θ sin θ)(θ cos θ)2 + (θ sin θ)2

= limθ→0+

cos θ sin θ = 0



780 SECTION 15.6

(e) By L’Hospital’s rule limx→0

f(x)x

= limx→0

f ′(x) = f ′(0). Thus

limx→0

xf(x)x2 + [ f(x) ]2

= limx→0

f(x)/x1 + [ f(x)/x ]2

=f ′(0)

1 + [ f ′(0) ]2.

(f) limθ→(π/3)−

=(cos θ sin 3θ)(sin θ sin 3θ)

(cos θ sin 3θ)2 + (sin θ sin 3θ)2= lim

θ→(π/3)−cos θ sin θ =

14

√3

(g) limt→∞

(1/t)(sin t)/t1/t2 +

(sin2 t

)/t2

= limt→∞

sin t

1 + sin2 t; does not exist

28. (a) limx→0

x(0)2

(x2 + 02)3/2= lim

x→00 = 0 (b) lim

y→0

0 · y2

(0 + y2)3/2= lim

y→00 = 0

(c) limx→0

xm2x2

(x2 + m2x2)3/2= lim

x→0

m2x3

|x|3(1 + m2)3/2= lim

x→0

m2x

|x| (1 + m2)3/2; does not exist

(d) limθ→0+

(θ cos θ)(θ sin θ)2

[(θ cos θ)2 + (θ sin θ)2]3/2= lim

θ→0+cos θ sin2 θ = 0

(e) limx→0

x[f(x)]2

(x2 + [f(x)]2)3/2= lim

x→0

[f(x)/x]2

(1 + [f(x)/x]2)3/2= lim

x→0

x3[f ′(0)]2

|x|3(1 + [f ′(0)]2)3/2; does not exist

(f) limθ→π

3−

(cos θ sin 3θ)(sin θ sin 3θ)2

[(cos θ sin 3θ)2 + (sin θ sin 3θ)2]3/2= lim

θ→π3

−cos θ sin2 θ =

38

(g) limt→∞

(1/t) (sin t/t)2[(1/t2) + (sin2 t/t2)

]3/2 = limt→∞

sin2 t(1 + sin2 t

)3/2; does not exist

29. (a)∂g

∂x(0, 0) = lim

h→0

g(h, 0) − g(0, 0)h

= limh→0

0 = 0,

∂g

∂y(0, 0) = lim

h→0

g(0, h) − g(0, 0)h

= limh→0

0 = 0

(b) as (x, y) tends to (0, 0) along the x-axis, g(x, y) = g(x, 0) = 0 tends to 0;

as (x, y) tends to (0, 0) along the line y = x, g(x, y) = g(x, x) = 12 tends to 1

2

30. No; as (x, y) tends to (1, 1) along the line x = 1, f(x, y) = 1 tends to 1; as (x, y) tends to (1, 1)

along the line y = 1,

f(x, y) =x− 1x3 − 1

=1

x2 + x + 1tends to

13

31. For y �= 0,∂f

∂x(0, y) = lim

h→0

f(h, y) − f(0, y)h

= limh→0

y(y2 − h2)h2 + y2

= y.

Since∂f

∂x(0, 0) = lim

h→0

f(h, 0) − f(0, 0)h

= limh→0

0 = 0,



SECTION 15.6 781

we have∂f

∂x(0, y) = y for all y.

For x �= 0,∂f

∂y(x, 0) = lim

h→0

f(x, h) − f(x, 0)h

= limh→0

x(h2 − x2)x2 + h2

= −x.

Since∂f

∂y(0, 0) = lim

h→0

f(0, h) − f(0, 0)h

= limh→0

0 = 0,

we have∂f

∂y(x, 0) = −x for all x.

Therefore∂2f

∂y∂x(0, y) = 1 for all y and

∂2f

∂x∂y(x, 0) = −1 for all x.

In particular∂2f

∂y∂x(0, 0) = 1 while

∂2f

∂x∂y(0, 0) = −1.

32. limh→0

[f(x0 + h, y0) − f(x0, y0)] = limh→0

[(h)

(f(x0 + h, y0) − f(x0, y0)

h

)]

=(

limh→0

h

) (limh→0

f(x0 + h, y0) − f(x0, y0)h

)

= 0 · ∂f∂x (x0, y0) = 0

33. Since fxy(x, y) = 0, fx(x, y) must be a function of x alone, and fy(x, y) must be a function of y

alone. Then f must be of the form

f(x, y) = g(x) + h(y).

34.

-2

0

2-2

0

2

-0.5

0

0.5

35.

-2

0

2-2

0

2

0

0.5

36.

-2

0

2-2

0

2

-0.5

0

0.5



782 SECTION 15.6

PROJECT 15.6

1. (a)∂u

∂x=

x2y2 + 2xy3

(x + y)2,

∂u

∂y=

x2y2 + 2x3y

(x + y)2

x∂u

∂x+ y

∂u

∂y=

3x2y2(x + y)(x + y)2

= 3u

(b)∂u

∂x= 2xy + z2,

∂u

∂y= 2yz + x2,

∂u

∂z= 2xz + y2

∂u

∂x+

∂u

∂y+

∂u

∂z= 2xy + z2 + 2yz + x2 + 2xz + y2 = (x + y + z)2

2. (a) (i)∂2f

∂x2+

∂2f

∂y2= 6x− 6x = 0

(ii)∂2f

∂x2+

∂2f

∂y2= (− cosx sinh y − sinx cosh y) + (cosx sinh y + sinx cosh y) = 0

(iii)∂2f

∂x2=

y2 − x2

(x2 + y2)2,

∂2f

∂y2=

x2 − y2

(x2 + y2)2

∂2f

∂x2+

∂2f

∂y2=

y2 − x2

(x2 + y2)2+

x2 − y2

(x2 + y2)2= 0

(b) (i)∂2f

∂x2+

∂2f

∂y2+

∂2f

∂z2=

2x2 − y2 − z2

(x2 + y2 + z2)5/2+

2y2 − x2 − z2

(x2 + y2 + z2)5/2+

2z2 − x2 − y2

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

(ii)∂2f

∂x2= ex+y cos

(√2 z

),

∂2f

∂y2= ex+y cos

(√2 z

),

∂2f

∂z2= − 2ex+y cos

(√2 z

)∂2f

∂x2+

∂2f

∂y2+

∂2f

∂z2= ex+y cos

(√2 z

)+ ex+y cos

(√2 z

)+

[− 2ex+y cos

(√2 z

)]= 0

3. (i)∂2f

∂t2=

∂2f

∂x2= 0 =⇒ ∂2f

∂t2− c2

∂2f

∂x2= 0

(ii)∂2f

∂t2= − 5c2 sin(x + ct) cos(2x + 2ct) − 4c2 cos(x + ct) sin(2x + 2ct)

∂2f

∂x2= − 5 sin(x + ct) cos(2x + 2ct) − 4 cos(x + ct) sin(2x + 2ct)

It now follows that∂2f

∂t2− c2

∂2f

∂x2= 0

(iii)∂2f

∂t2= − c2

(x + ct)2,

∂2f

∂x2= − 1

(x + ct)2=⇒ ∂2f

∂t2− c2

∂2f

∂x2= 0

(iv)∂2f

∂t2= c2k2

(Aekx + Be−kx

) (Ceckt + De−ckt

),

∂2f

∂x2= k2

(Aekx + Be−kx

) (Ceckt + De−ckt

)It now follows that

∂2f

∂t2− c2

∂2f

∂x2= 0

4.∂2f

∂t2− c2

∂2f

∂x2=

[c2g′′(x + ct) + c2h′′(x− ct)] − c2[g′′(x + ct) + h′′(x− ct)

]= 0



REVIEW EXERCISES 783

REVIEW EXERCISES

1. domain {(x, y) : y > x2}, range (0,∞)

2. domain {(x, y) : x, y ∈ R}, range (0,∞)

3. domain {(x, y, x) : z ≥ x2 + y2}, dange [0,+∞)

4. domain {(x, y, z) : x + 2y + z > 0}, range R

5. (a) f(x, y) =13πx2y;

(b) f(x, y) =12yx2;

(c) θ = arccosx + 2y√

5√x2 + y2

6. Assume one of the vertices is (x, y, z), x > 0, y > 0, z > 0.

V = 8cxy

√1 − x2

a2− y2

b2

7. ellipsoidxy−trace: ellipse 4x2 + 9y2 = 36xz−trace: ellipse 4x2 + 36z2 = 36yz−trace: ellipse 9y2 + 36z2 = 36

8. hyperboloid of two sheetsxy−trace: nonexz−trace: hyperbola 4z2 − x2 = 4yz−trace: hyperbola 4z2 − y2 = 4

9. hyperbolic paraboloidxy−trace: lines x = ±y

xz−trace: parabola z = −x2

yz−trace: parabola z = y2

10. elliptic paraboloidxy−trace: parabola 4x2 = y

xz−trace: (0, 0)yz−trace: parabola 9z2 = y

11. conexy−trace: lines x = ±y

xz−trace: lines x = ±z

yz−trace: (0, 0)

12. hyperboloid of one sheetxy−trace: ellipse 9x2 + 4y2 = 36xz−trace: hyperbola z2 = 9x2 − 36yz−trace: hyperbola z2 = 4y2 − 36

13. 14.

-2

0

2

2

4

-2

0

2

-2

0

2

4



784 REVIEW EXERCISES

15.

-2

0

2

-2

0

2

2

16.

1

2

0

2

0

17. c = 0,=⇒ 0 = 2x2 + 3y2 =⇒ (0, 0)

c = 6,=⇒ 6 = 2x2 + 3y2, ellipse

c = 12,=⇒ 6 = 2x2 + 3y2, ellipse

18. c = 0,=⇒ 0 = x2 + y2 − 4, circle

c = 1,=⇒ 5 = x2 + y2, circle

c = 2,=⇒ 8 = x2 + y2, circle

c =√

5,=⇒ 9 = x2 + y2, circle

19. c = −4,=⇒ x = −4y2, parabola

c = −1,=⇒ x = −y2, parabola

c = 1,=⇒ x = y2, parabola

c = 4,=⇒ x = 4y2, parabola

the origin is omitted

20. c = 1,=⇒ x2 + y2 = 1 circle

c = 4,=⇒ x2 + y2 = 4 circle

c = 9,=⇒ x2 + y2 = 9 circle

21. c = 6, 2x + y + 3z = 6, plane 22. c = 16, x2 + y2 + 4z2 = 16, ellipsoid

23. (a) f(0, 0) = 1, level curve: f(x, y) = 1

(b) f(ln 2, 1) = 4, level curve: f(x, y) = 4

(c) f(1,−1) = 2e, level curve: f(x, y) = 2e

24. (a) f(2, 0, 1) = 4, level surface: f(x, y, z) = 4

(b) f(1, π,−1) = −1, level surface:

f(x, y, z) = −1

(c) f(4, π, 1/2) = 0, level surface: f(x, y, z) = 0

25. fx = limh→0

f(x + h, y) − f(x, y)h

= limh→0

(x + h)2 + 2(x + h)y − x2 − 2xyh

= limh→0

(2x + h + 2y) = 2x + 2y

fy = limh→0

f(x, y + h) − f(x, y)h

= limh→0

x2 + 2x(y + h) − x2 − 2xyh

= 2x

26. fx = limh→0

f(x + h, y) − f(x, y)h

= limh→0

y2 cos 2(x + h) − y2 cos 2xh

= y2 limh→0

cos 2(x + h) − cos 2xh

= y2 cos′ 2x = −2y2 sin 2x




fy = limh→0

f(x, y + h) − f(x, y)h

= limh→0

(y + h)2 cos 2x− y2 cos 2xh

= cos 2x limh→0

(y + h)2 − y2

h= 2y cos 2x

27. fx = 2xy − 2y3; fy = x2 − 6xy2

28. gx = (x2 + y2)−1/2 − x2(x2 + y2)−3/2; gy = −xy(x2 + y2)−3/2

29.∂z

∂x= 2x sin(xy2) + x2y2 cos(xy2);

∂z

∂y= 2x3y cos(xy2).

30. fx = yexy ln(y/x) − 1xexy fy = xexy ln(y/x) + exy

y

31. hx = −e−x cos(2x− y) − 2e−x sin(2x− y) hy = e−x sin(2x− y)

32. ux = y2 secx tanx + 2x tan y uy = 2y secx + x2 sec2 y

33. fx =2y2 + 2yz

(x + y + z)2; fy =

2x2 + 2xz(x + y + z)2

; fz =−2xy

(x + y + z)2

34. wx = arctan(y − z) wy =x

1 + (y − z)2wz =

−x

1 + (y − z)2

35.∂g

∂x=

x

x2 + y2 + z2;

∂g

∂y=

y

x2 + y2 + z2;

∂g

∂z=

z

x2 + y2 + z2.

36. hu = veuv sinuw + weuv cosuw; hv = ueuv sinuw; hw = ueuv cosuw

37. fx = 3x2y2 − 4y3 + 2, fy = 2x3y − 12xy2 − 1;

fxx = 6xy2, fyy = 2x3 − 24xy, fyx = fxy = 6x2y − 12y2

38. gx = 2x ln(y − x) − x2

y − x, gxx = 2 ln(y − x) − 4x

y − x− x2

(y − x)2;

gy =x2

y − x, gyy = − x2

(y − x)2, gxy = gyx =

2xy − x

+x2

(y − x)2

39. gx = y sinxy + xy2 cosxy, gxx = 2y2 cosxy − xy3 sinxy;

gy = x sinxy + x2y cosxy, gyy = 2x2 cosxy − yx3 sinxy,

gxy = gyx = sinxy + 3xy cosxy − x2y2 sinxy

40. fx = 2xex/y +x2

yex/y, fxx = 2ex/y +

4xyex/y +

x2

y2ex/y;

fy = −x3

y2ex/y, fyy =

2x3

y3ex/y +

x4

y4ex/y, fxy = fyx = −3x2

y2ex/y − x3

y3ex/y



786 REVIEW EXERCISES

41. fx = 2xe2y cos(2z + 1), fy = 2x2e2y cos(2z + 1), fz = −2x2e2y sin(2z + 1);

fxx = 2e2y cos(2z + 1), fyy = 4x2e2y cos(2z + 1), fzz = −4x2e2y cos(2z + 1);

fxy = fyx = 4xe2y cos(2z + 1), fxz = fzx = −4xe2y sin(2z + 1), fyz = fzy = −4x2e2y sin(2z + 1)

42. gx = 4xyz3 + yzexyz, gy = 2x2z3 + xzexyz, gz = 6x2yz2 + xyexyz;

gxx = 4yz3 + y2z2exyz, gyy = x2z2exyz, gzz = 12x2yz + x2y2exyz;

gxy = gyx = 4xz3 + zexyz + xyz2exyz, gxz = gzx = 12xyz2 + yexyz + xy2zexyz

gyz = gzy = 6x2z2 + xexyz + x2yzexyz

43.∂z

∂x= 4x + 6|(1,2,8) = 10 44.

∂z

∂y= 3x|(2,−1,2) = 6

x = 1 + t, y = 2, z = 8 + 10t x = 2, y = −1 + t, z = 2 + 6t

45. (a) zy(2, 1) =−6y

2√

20 − 2x2 − 3y2(2, 1) = −1; the equation for l1 is:

x = 2; y = 1 − t; z = 3 + t

(b) zx(2, 1) =−4x

2√

20 − 2x2 − 3y2(2, 1) = −4

3; the equation for l2 is:

x = 2 − 34t; y = 1; z = 3 + t

(c) The normal vector for this plane is: −i − 34

j − 34

k or 4 i + 3 j + 3k;

an equation for the plane is: 4(x− 2) + 3(y − 1) + 3(z − 3) = 0.

46. Neither.

interior: {(x, y) : 0 < x < 3, 2 < y < 5}boundary: {(x, y) : x = 0 or x = 3, 2 ≤ y ≤ 5} ∪ {(x, y) : y = 2 or y = 5, 0 ≤ x ≤ 3}

47. Open.

interior: {(x, y) : 0 < x2 + y2 < 4}boundary: {(0, 0)} ∪ {(x, y) : x2 + y2 = 4}

48. Closed.

interior: {(x, y) : x + y > 4}boundary: {(x, y) : x + y = 4}

49. Closed.

interior: {(x, y, z) : 0 < x < 2, 0 < y, 0 < z, y2 + z2 < 4}boundary: the quarter disks x = 0, y2 + z2 ≤ 4; x = 2, y2 + z2 ≤ 4;

the squares z = 0, 0 ≤ x, y ≤ 2; y = 0, 0 ≤ x, z ≤ 2; and

the cylindrical surface y2 + z2 = 4, 0 ≤ x ≤ 2, y, z ≥ 0




50. Neither.

interior: {(x, y, z) : 0 < x2 + y2 < z < 4}boundary: the cone z = x2 + y2 and the disk x2 + y2 ≤ 4, z = 4

51. (a) fx = yg′(xy), fy = xg′(xy); xfx − yfy = xyg′ − xyg′ = 0

(b) fxx = y2g′′(xy), fyy = x2g′′(xy); x2fxx − y2fyy = x2y2g′′ − x2y2g′′ = 0

52. fx = − y

x2 + y2, fy =

x

x2 + y2; fxx =

2xy(x2 + y2)2

, fyy =−2xy

(x2 + y2)2

fxx + fyy = 0

53. No.∂2f

∂y∂x= x2exy �= y2exy =

∂2f

∂x∂y

54. (a) limx→0

f(x, 0) = limx→0

0 = 0 (b) limy→0

f(0, y) = limy→0

0 = 0

(c) limx→0

2x2mx

x4 + m2x2= lim

x→0

2mx

m2 + x2= 0 (d) lim

x→0

2x2ax2

x4 + a2x4=

2a1 + a2

lim(x,y)→(0,0) f(x, y) does not exist.

Calculus one and several variables 10E Salas solutions manual ch15

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Transcript of Calculus one and several variables 10E Salas solutions manual ch15