# Calculus one and several variables 10E Salas solutions manual ch15

July 30, 2017 | Author: 高章琛 | Category: Ellipse, Differential Topology, Geometric Shapes, Differential Geometry, Manifold

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Calculus one and several variables 10E Salas solutions manual...

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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 all points 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; 5. dom (f ) = the entire plane;

range (f ) = (0, ∞)

range (f ) = (−1, 1) since ex − ey ex + ey − 2ey 2 = = 1 − x−y x y x y e +e e +e e +1

and 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 plane above the parabola y = x2 ; range (f ) = (0, ∞) 10. dom (f ) = the range (f ) = [0, 3]

set

of

all

points

(x, y)

with

−3 ≤ x ≤ 3,

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

−1 ≤ y ≤ 1

(a rectangle);

(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 = y 2 —that is, all points of space except for those which 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 + y 2 < 9 —in other words, the set of all points of the plane inside the circle x2 + y 2 = 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 in space 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 + y 2 + z 2 ≤ 4 — in other words, the set of all points inside and on the sphere x2 + y 2 + z 2 = 4; range (f ) = [1, e2 ] 21. dom (f ) = all of space;

range (f ) = (0, 1]

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SECTION 15.1 22. dom (f ) = the set of all points (x, y, z) with −1 ≤ x ≤ 1, −2 ≤ y ≤ 2, −3 ≤ z ≤ 3 solid); range (f ) = [0, 3]

759

(a rectangular

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, dom (g) = all of space, 25.

range (f ) = [−1, 1]

range (g) = [−1, 1]

f (x + h, y) − f (x, y) 2(x + h)2 − y − (2x2 − y) 4xh + 2h2 = lim = lim = 4x h→0 h→0 h→0 h h h lim

f (x, y + h) − f (x, y) 2x2 − (y + h) − (2x2 − y) = lim = −1 h→0 h→0 h h lim

f (x + h, y) − f (x, y) xy + hy + 2y − (xy + 2y) = lim = lim y = y. h→0 h→0 h h f (x, y + h) − f (x, y) xy + xh + 2y + 2h − (xy + 2y) lim = lim = lim (x + 2) = x + 2 h→0 h→0 h→0 h h

26.

27.

lim

h→0

f (x + h, y) − f (x, y) 3(x + h) − (x + h)y + 2y 2 − (3x − xy + 2y 2 ) 3h − hy = lim = lim =3−y h→0 h→0 h→0 h h h lim

f (x, y + h) − f (x, y) 3x − x(y + h) + 2(y + h)2 − (3x − xy + 2y 2 ) = lim h→0 h→0 h h lim

−xh + 4yh + 2h2 = −x + 4y h→0 h

= lim

28.

lim

f (x + h, y) − f (x, y) x sin y + h sin y − x sin y = lim = lim sin y = sin y. h→0 h→0 h h

lim

f (x, y + h) − f (x, y) x sin(y + h) − x sin y sin(y + h) − sin y = lim = x lim = x cos y h→0 h→0 h h h

h→0

h→0

29.

lim

h→0

f (x + h, y) − f (x, y) cos[(x + h)y] − cos[xy] = lim h→0 h h cos[xy] cos[hy] − sin[xy] sin[hy] − cos[xy] = lim h→0 h   cos[hy] − 1 sin hy = cos[xy] lim − sin[xy] lim h→0 h→0 h h   cos[hy] − 1 sin hy = y cos[xy] lim − y sin[xy] lim h→0 h→0 hy hy = −y sin[xy]

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SECTION 15.2 and lim

h→0

f (x, y + h) − f (x, y) cos[x(y + h)] − cos[xy] = lim h→0 h h cos[xy] cos[hx] − sin[xy] sin[hx] − cos[xy] h   cos[hx] − 1 sin hx = cos[xy] lim − sin[xy] lim h→0 h→0 h h   cos[hx] − 1 sin hx = x cos[xy] lim − x sin[xy] lim h→0 h→0 hx hx

= lim

h→0

= −x sin[xy] 30.

f (x + h, y) − f (x, y) (x2 + 2xh + h2 )ey − x2 ey = lim = lim (2x + h)ey = 2xey . h→0 h→0 h→0 h h lim

f (x, y + h) − f (x, y) x2 ey+h − x2 ey ey+h − ey = lim = x2 lim = x2 ey . h→0 h→0 h→0 h h h lim

(b) f (x, y) = |2 i × (x i + y j)| = 2|y|

(b) f (x, y) = πx2 y

31. (a) f (x, y) =Ay

32. (a) f (x, y, z) = xy + 2xz + 2yz (b) f (x, y, z) = cos−1

(i + j) · (x i + y j + z k) x+y = cos−1 √  i + jx i + y j + z k 2 x2 + y 2 + z 2

(c) f (x, y, z) = [i × (i + j)] · (x i + y j + z k) = z 33. Surface area: S = 2lw + 2lh + 2hw = 20 =⇒ w = Volume: V = lwh = 34. wlh = 12

=⇒

35. V = πr2 h + 36. A =

lh(10 − lh) l+h

h=

12 ; wl

20 − 2lh 10 − lh = 2l + 2h l+h

C = 4wl + 2(2wh + 2lh) = 4wl +

48 48 + l w

4 3 πr 3

1 [2(12 − 2x) + 2x cos θ] · x sin θ = (12 − 2x + x cos θ) · x sin θ 2

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

11. a hyperbolic paraboloid

10.

a hyperbolic cylinder

12.

an elliptic paraboloid

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SECTION 15.2 13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

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SECTION 15.2

25. elliptic paraboloid xy-trace: the origin xz-trace: the parabola x2 = 4z yz-trace: the parabola y 2 = 9z surface has the form of Figure 15.2.5

26. ellipsoid xy-trace: the ellipse 9x2 + 4y 2 = 36 xz-trace: the ellipse x2 + 4z 2 = 4 yz-trace: the ellipse y 2 + 9z 2 = 9 surface has the form of Figure 15.2.1

27. elliptic cone xy-trace: the origin xz-trace: the lines x = ±2z yz-trace: the lines y = ±3z surface has the form of Figure 15.2.4

28. hyperboloid of one sheet xy-trace: the ellipse 9x2 + 4y 2 = 36 xz-trace: the hyperbola x2 − 4z 2 = 4 yz-trace: the hyperbola y 2 − 9z 2 = 9 surface has the form of Figure 15.2.2

29. hyperboloid of two sheets xy-trace: none xz-trace: the hyperbola 4z 2 − x2 = 4 yz-trace: the hyperbola 9z 2 − y 2 = 9 surface has the form of Figure 15.2.3

30. hyperbolic paraboloid xy-trace: the lines y = ± 32 x xz-trace: the parabola x2 = 4z yz-trace: the parabola y 2 = −9z surface has the form Figure 15.2.6

31. hyperboloid of two sheets xy-trace: the hyperbola 9x2 − 4y 2 = 36 xz-trace: the hyperbola x2 − 4z 2 = 4 yz-trace: none see Figure 15.2.3

32.

33. elliptic paraboloid xy-trace: the parabola x2 = 9y xz-trace: the origin yz-trace: the parabola z 2 = 4y surface has the form of Figure 15.2.5

34. elliptic cone xy-trace: the lines x = ±2y xz-trace: the origin yz-trace: the lines z = ±3y surface has the form of Figure 15.2.4, rotated 90◦ about the x-axis.

35. hyperboloid of two sheets xy-trace: the hyperbola 9y 2 − 4x2 = 36 xz-trace: none yz-trace: the hyperbola y 2 − 4z 2 = 4 see Figure 15.2.3

36.

37. paraboloid of revolution xy-trace: the origin xz-trace: the parabola x2 = 4z yz-trace: the parabola y 2 = 4z surface has the form of Figure 15.2.5

38. ellipsoid xy-trace: the ellipse 4x2 + y 2 = 4 xz-trace: the ellipse 9x2 + z 2 = 9 yz-trace: the ellipse 9y 2 + 4z 2 = 36 the surface has the form of Figure 15.2.1, rotated 90◦ about the x-axis.

hyperboloid of one sheet xy-trace: the hyperbola x2 − 9y 2 = 9 xz-trace: the circle x2 + y 2 = 9 yz-trace: the hyperbola z 2 − 9y 2 = 9 surface has the form of Figure 15.2.2, rotated 90◦ about the x-axis.

elliptic paraboloid xy-trace: the parabola y 2 = 4x xz-trace: the parabola z 2 = 9x yz-trace: the origin surface has the form of Figure 15.2.5, but opening along the positive x-axis.

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

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SECTION 15.2 41. x2 + y 2 − 4z = 0

(paraboloid of revolution)

42. c2 x2 + c2 y 2 − b2 z 2 = b2 c2

(hyperboloid of revolution, one sheet)

43. (a) a circle  (b) (i) x2 + y 2 = −3z

(ii)

x2 + z 2 = 13 y

44. (a) the ellipse b2 x2 + y 2 = 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

46. 3x + y − 2(4 − x + 2y) = 1, 47.

x2 + y 2 + (z − 1)2 =

or or

5x + 7y = 30, 5x − 3y = 9,



3 2

(z 2 + 1) + (z − 1)2 =

x2 + y 2 − z 2 = 1

3 ; 2

a line a line (2z − 1)2 = 0,

z=

48. z 2 + (z − 2)2 = 2 =⇒ 2(z − 1)2 = 0 =⇒ z = 1 =⇒ x2 + y 2 = 1,   49. x2 + y 2 + x2 + 3y 2 = 4

or

x2 + 2y 2 = 2,

50. y 2 + (x2 + 3y 2 ) = 4 =⇒ x2 + 4y 2 = 4, 51. x2 + y 2 = (2 − y)2  2

2

52. x + y =

2−y 2

or

x2 = −4(y − 1),

a circle.

an ellipse

an ellipse. a parabola

2 =⇒ 4x2 + 3y 2 + 4y = 4,

an ellipse.

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

4 2 0 -2 -4 2 1 0 -1 -2 -2

1 5 so that x2 + y 2 = 2 4

0 2

x2 y2 z2 + + = 1. a2 b2 c2

763

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December 5, 2006

SECTION 15.3 (a) Set x = a sec u cos v,

55.

(a) Set x = av cos u,

y = b sec u sin v,

y = bv sin u,

z = c tan u

z = cv

(b)

10 0

-10-5 0 5 10

2

4

0

-10 10

-2

20

-4 6

0

4 2

-20

0 -2

-1

0

1

2

SECTION 15.3 1. lines of slope 1: y = x − c

3.

parabolas: y = x2 − c

2. lines of slope 2 : y = 2x − c

4.

parabolas: x − y 2 =

1 c

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SECTION 15.3  5. the y-axis and the lines y =

1−c c

 x with

the origin omitted throughout

7. the cubics y = x3 − c

9. the lines y = ±x and the hyperbolas

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

8. the coordinate axes and the hyperbolas xy = ln c

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

x2 − y 2 = c

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

765

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

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SECTION 15.3

13. the circles x2 + y 2 = ec , c real

14. the parabolas y = ec x2 with the origin omitted throughout

15.

2

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

17. the coordinate axes and pairs of lines √ 1−c y=± √ x, with the origin omitted c throughout

19. x + 2y + 3z = 0,

 x2 + y 2 ,

22. ellipsoid

18.

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

plane through the origin

20. circular cylinder x2 + y 2 = 4 21. z =

16. the coordinate axes and pairs of hyperbolas √ xy = ± c

(Figure 15.2.8)

the upper nappe of the circular cone z 2 = x2 + y 2

x2 y2 z2 + + =1 4 6 9

23. the elliptic paraboloid

(Figure 15.2.1)

x2 y2 + =z 18 8

(Figure 15.2.5)

(Figure 15.2.4)

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SECTION 15.3 x2 y2 + − z 2 = −1 (1/6)2 (1/3)2

24. hyperboloid of two sheets 25. (i) hyperboloid of two sheets (ii) circular cone

767

(Figure 15.2.3)

(Figure 15.2.3)

(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 − y 2 = 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 − y 2 = 0, or 4x2 + y 2 = 1. 28. (x2 + y 2 )exy = 1 29. The level curves of f are: y 2 arctan x = c. Substituting P (1, 2) into this equation, we have 4 arctan 1 = c

=⇒

c=π −1

2

The level curve that contains P is: y tan

x = π.

30. (x2 + y) ln(2 − x + ey ) = 5 31. The level surfaces of f are: x2 + 2y 2 − 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 + 2y 2 − 2xyz = 13. 32.

 x2 + y 2 − ln z = 4

33. (a)

4

2

50 0 -50 -4

4

0

2 -2

0 0

2

-2

-2 4 -4

-4 -4

-2

0

2

4

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SECTION 15.3 (b) 4 2 0 15 10 5 0

5

-2

0

-5 0

-4

-5

5

-4

34. (a) the level surfaces are planes

35. (a)

-2

0

2

4

(b)

3x + 2y + 1 3 = 4x2 + 9 5

(b) x2 + 2y 2 − z 2 = 21

36. (a) 2

(b)

1

0

-1

-2 -2

37.

-1

GmM = c =⇒ + y2 + z2 concentric spheres. x2

0

1

2

x2 + y 2 + z 2 =

GmM ; the surfaces of constant gravitational force are c

38. Circular cylinders around the positive y-axis:

x2 + z 2 =

k2 c2

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SECTION 15.4 k

39. (a) T (x, y, z) = 

x2 + y 2 + z 2

(b) 

k x2 + y 2 + z 2

=⇒

, where k is a constant. k2 ; the level surfaces are concentric spheres. c2 √ √ 50 6 k = 50 6 =⇒ T (x, y, z) =  x2 + y 2 + z 2

x2 + y 2 + z 2 =

k = 50 =⇒ + 22 + 12 √ √ 50 6 T (4, 0, 3) = √ = 10 6 degrees 2 2 3 +4

(c) T (1, 2, 1) = √ Now,

=c

40. x2 + y 2 = r2 −

12

k2 ; c2

circles about the origin, for |c| >

41.

f (x, y) = y 2 − y 3 ; F

42.

D.

44.

B.

45.

f (x, y) = xye−(x

2

k . r

+y 2 )/2

; E

43.

f (x, y) = cos

46.

C.

PROJECT 15.3 1.

2.

3.

4.

SECTION 15.4 1.

3.

769

∂f = 6x − y, ∂x ∂ρ = cos φ cos θ, ∂φ

∂f =1−x ∂y ∂ρ = − sin φ sin θ ∂θ

2.

∂g = 2xe−y , ∂x

∂g = −x2 e−y ∂y



x2 + y 2 ; A

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SECTION 15.4 ∂ρ = −2 sin(θ − φ) cos(θ − φ) ∂φ

∂ρ = 2 sin(θ − φ) cos(θ − φ), ∂θ

5.

∂f = ex−y + ey−x , ∂x

7.

∂g (AD − BC)y , = ∂x (Cx + Dy)2

9.

∂u = y + z, ∂x

∂f = −ex−y − ey−x ∂y ∂g (BC − AD)x = ∂y (Cx + Dy)2

∂u = x + z, ∂y

∂u =x+y ∂z

6.

∂z x , = ∂x x2 − 3y

8.

∂u −ez = 2 2, ∂x x y

10.

∂z −3 =  ∂y 2 x2 − 3y

∂u −2ez , = ∂y xy 3

∂z = 2Ax + By, ∂x

∂z = Bx + 2Cy ∂y

11.

∂f = z cos (x − y), ∂x

12.

∂g 2u = 2 , ∂u u + vw − w2

13.

∂ρ = eθ+φ [cos (θ − φ) − sin (θ − φ)], ∂θ

∂ρ = eθ+φ [cos (θ − φ) + sin (θ − φ)] ∂φ

14.

∂f = (x + y) cos(x − y) + sin(x − y), ∂x

∂f = −(x + y) cos(x − y) + sin(x − y) ∂y

15.

∂f = 2xy sec xy + x2 y(sec xy)(tan xy)y = 2xy sec xy + x2 y 2 sec xy tan xy ∂x

∂f = −z cos (x − y), ∂y w ∂g = 2 , ∂v u + vw − w2

∂f = sin (x − y) ∂z v − 2w ∂g = 2 ∂w u + vw − w2

∂f = x2 sec xy + x2 y(sec xy)(tan xy)x = x2 sec xy + x3 y sec xy tan xy ∂y 16.

∂g 2 = , ∂x 1 + (2x + y)2

17.

x2 + y 2 − x(2x) ∂h y 2 − x2 = = 2 ∂x (x2 + y 2 ) (x2 + y 2 )2

18.

∂z x , = 2 ∂x x + y2

19.

(y cos x) sin y − (x sin y)(−y sin x) ∂f sin y(cos x + x sin x) = = ∂x (y cos x)2 y cos2 x

1 ∂g = ∂y 1 + (2x + y)2 ∂h −2xy = ∂y (x2 + y 2 )2

∂z y = 2 ∂y x + y2

∂f (y cos x)(x cos y) − (x sin y) cos x x(y cos y − sin y) = = ∂y (y cos x)2 y 2 cos x 20.

∂f = exy (y sin xz + z cos xz), ∂x

21.

∂h = 2f (x)f (x)g(y), ∂x

∂f = xexy sin xz, ∂y

∂h = [f (x)]2 g (y) ∂y

∂f = xexy cos xz ∂z

∂u ez = 2 ∂z xy

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SECTION 15.4 22.

∂h = f (x)g(y)ef (x)g(x) , ∂x

23.

∂f 2 = (y 2 ln z)z xy , ∂x

24.

25.

∂h = f (x)g (y)ef (x)g(y) ∂y

∂f 2 = (2xy ln z)z xy , ∂y

∂f 2 = xy 2 z xy −1 ∂z

∂h ∂g ∂f = 2[f (x, y)]3 g(x, z) + 3[f (x, y)]2 [g(x, z)]2 ∂x ∂x ∂x ∂h ∂h ∂f ∂g = 3[f (x, y)]2 [g(x, z)]2 , = 2[f (x, y)]3 g(x, y) ∂y ∂y ∂z ∂z ∂h = 2re2t cos(θ − t) ∂r

∂h = −r2 e2t sin(θ − t) ∂θ

∂h = 2r2 e2t cos(θ − t) + r2 e2t sin(θ − t) = r2 e2t [2 cos(θ − t) + sin(θ − t)] ∂t 26.

27.

28.

∂u 1 = − yzexz , ∂x x 1 ∂f =z ∂x 1 + (y/x)2 ∂f = arctan (y/x) ∂x

∂u 1 = − − exz , ∂y y 

−y x2

∂w = y sin z − yz cos x, ∂x

29. fx (x, y) = ex ln y,



∂u = −xyexz ∂z

∂w = x sin z − z sin x, ∂y

fx (0, e) = 1;

fy (x, y) =

30. gx = e−x [− sin(x + 2y) + cos(x + 2y)], gy = 2e−x cos(x + 2y), 31. fx (x, y) =

y , (x + y)2

32. gx (x, y) =

y2 , (x + y 2 )2

33. fx (x, y) = lim

h→0

h→0

34. fx (x, y) = 0,

  xz 1 = 2 x x + y2

∂w = xy cos z − y sin x ∂z

1 x e , y

fy (0, e) = e−1

gx (0, 14 π) = −1

gy (0, 14 π) = 0 fx (1, 2) =

2 ; 9

gx (1, 2) =

fy (x, y) =

4 25

(x + h)2 y − x2 y = lim y h→0 h

−x , (x + y)2

gy (x, y) = 

2xh + h2 h

fy (1, 2) = −

−2xy , (x + y 2 )2

1 9

gy (1, 2) = −

 = y lim (2x + h) = 2xy h→0

x (y + h) − x y x h = lim = lim x2 = x2 h→0 h h→0 h 2

fx (x, y) = lim

1 ∂f =z ∂y 1 + (y/x)2

yz =− 2 x + y2

2

2

fy (x, y) = 2y

  ln y(x + h)2 − ln x2 y ln y + 2 ln(x + h) − 2 ln x − ln y 35. fx (x, y) = lim = lim h→0 h→0 h h ln(x + h) − ln x d 2 = 2 lim = 2 (ln x) = h→0 h dx x

4 25

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SECTION 15.4   ln x2 (y + h) − ln x2 y ln x2 + ln(y + h) − ln x2 − ln y = lim fy (x, y) = lim h→0 h→0 h h = lim

h→0

36. fx (x, y) = −(x + 4y)−2 ,

ln(y + h) − ln y d 1 = (ln y) = h dy y

fy (x, y) = −4(x + 4y)−2

 1 1 1 −h − = lim h→0 h (x + h) − y x − y (x + h − y) (x − y) −1 −1 = lim = h→0 (x + h − y) (x − y) (x − y)2   1 1 h 1 1 − = lim fy (x, y) = lim h→0 h h→0 h x − (y + h) x − y (x − y − h) (x − y) 1 1 = lim = h→0 (x − y − h) (x − y) (x − y)2

1 37. fy (x, y) = lim h→0 h



38. fx (x, y) = 2e2x+3y , 39. fx (x, y, z) = lim

h→0

fy (x, y, z) = lim

h→0

fy (x, y) = 3e2x+3y

(x + h)y 2 z − xy 2 z = lim y 2 z = y 2 z h→0 h x(y + h)2 z − xy 2 z xz(2yh + h2 ) = lim h→0 h h

= lim xz(2y + h) = 2xyz h→0

fz (x, y, z) = lim

h→0

40. fx (x, y, z) =

xy 2 (z + h) − xy 2 z = lim xy 2 = xy 2 h→0 h

2xy , z

fy (x, y, z) =

x2 , z

fz (x, y, z) = −

x2 y 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 , 42. fx (x, y) = 2x,

fx (1, 3) = 2,

z − z0 = fy (x0 , y0 )(y − y0 )

equations for the tangent line are:

43. Let z = f (x, y) = x2 + y 2 . Then f (2, 1) = 5, equations for the tangent line are:

x = 2,

fy (x, y) = 2y

y = 3,

z − 10 = 2(x − 1).

and fy (2, 1) = 2;

z − 5 = 2(y − 1)

x2 −x2 2y , fy (x, y) = 2 −3 (y − 3)2 Tangent line: x = x0 , z − z0 = fy (x0 , y0 )(y − y0 ) =⇒ x = 3, z − 9 = −36(y − 2)

44. f (x, y) =

y2

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SECTION 15.4 x2 2x . Then f (3, 2) = 9, fx (x, y) = 2 y2 − 3 y −3 equations for the tangent line are: y = 2, z − 9 = 6(x − 3)

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

46. f (x, y) = (4 − x2 − y 2 )1/2 , fx (x, y) = −x(4 − x2 − y 2 )−1/2 , √ √ √ 2 2 (a) fy (1, 1) = − =⇒ x = 1, z − 2 = − (y − 1) 2 2 √ √ √ 2 2 (b) fx (1, 1) = − =⇒ y = 1, z − 2 = − (x − 1) 2 2 √

and fx (3, 2) = 6;

fy (x, y) = −y(4 − x2 − y 2 )−1/2

(c) l1 and l2 have direction vectors j − 22 k, i − 22 k respectively. The normal to the plane is

√ √ √ 2 2 2 2 j− k × i− k =− i− j − k, so the tangent plane is 2 2 2 2 √ √ √ √ √ 2 2 − (x − 1) − (y − 1) − (z − 2) = 0, or (x − 1) + (y − 1) + 2(z − 2) = 0 2 2 47. (a) mx = −6;

tangent line:

y = 2,

z = −6x + 13

(b) my = 18;

tangent line:

x = 1,

z = 18y − 29

tangent line:

y = 2,

z=

tangent line:

x = 1,

z=

48. (a) mx = (b) my =

7 25 ; 1 − 25 ;

uy (x, y) = −2y = −vx (x, y)

49. ux (x, y) = 2x = vy (x, y); 50. ux = ex cos y, 51. ux (x, y) =

1 25 (7x − 12) 1 − 25 (y + 3)

uy = −ex sin y,

vx = ex sin y = −uy ,

1 x 1 2x = 2 ; 2 x2 + y 2 x + y2

vy = ex cos y = ux

vy (x, y) =

1 1 + (y/x)2

vx (x, y) =

1 1 + (y/x)2

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

1 y 1 2y = 2 ; 2 x2 + y 2 x + y2

  1 x = 2 x x + y2 

−y x2

 =

−y x2 + y 2

Thus, uy (x, y) = −vx (x, y). 52. ux =

y 2 − x2 , (x2 + y 2 )2

uy =

53. (a) f depends only on y.

−2xy , 2 (x + y 2 )2

vx =

2xy = −uy , 2 (x + y 2 )2

vy =

y 2 − x2 = ux (x2 + y 2 )2

(b) f depends only on x.

√ 1/2  54. (a) a0 = b20 + c20 − 2b0 c0 cos θ0 =5 7 √   ∂a 1 7 2 2 −1/2 (b) = (2b − 2c cos θ) (b + c − 2bc cos θ) = ∂b 2 14 √ √ √ √ 7 7 7 ∼ (c) a = a0 + (b − b0 ) = 5 7 + · (−1) decreases by about inches. 14 14 14 15 √ ∂a 1 21 (d) = 2bc sin θ( )(b2 + c2 − 2bc cos θ)−1/2 = ∂θ 2 7

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SECTION 15.4 (e) Diﬀerentiate implicitly:

0 = 2c

∂c bc sin θ 15 √ = =− 3 ∂θ b cos θ − c 2 √ 75 3 55. (a) in.2 2 ∂A 1 (b) = c sin θ; ∂b 2

at time t0 ,

∂c ∂c − 2b cos θ + 2bc sin θ ∂θ ∂θ

√ ∂A 15 3 = ∂b 4

∂A 1 75 ∂A = bc cos θ; at time t0 , = ∂θ 2 ∂θ 2 π ∂A π 75 5π (d) with h = , A (b, c, θ + h) − A (b, c, θ) ∼ = = in.2 =h 180 ∂θ 180 2 24   1 ∂c ∂c −c 3 (e) 0 = sin θ b + c ; at time t0 , = =− 2 ∂b ∂b b 2 (c)

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 + y0 j + f (x, y0 )k

∂f (x0 , y0 )k tangent line: R(t) = [x0 i + y0 j + f (x0 , y0 )k ] + t i + ∂x

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

∂f tangent line: R(t) = [x0 i + y0 j + f (x0 , y0 ) k] + t j + (x0 , y0 )k ∂y (c) For (x, y, z) in the plane     ∂f ∂f [(x − x0 )i + (y − y0 )j + (z − f (x0 , y0 ))k ]. i + (x0 , y0 )k × j + (x0 , y0 )k = 0. ∂x ∂y From this it follows that ∂f ∂f z − f (x0 , y0 ) = (x − x0 ) (x0 , y0 ) + (y − y0 ) (x0 , y0 ). ∂x ∂y 58. Fix y and set F (x) = f (x, y). Then, for that value of y, h(x, y) = g(F (x)) and thus d hx (x, y) = [g(F (x))] = g (F (x))F (x) = g (f (x, y))fx (x, y). dx The second formula can be derived in the same manner. 59. (a) Set u = ax + by.

Then b

∂w ∂w −a = b(a g (u)) − a(b g (u)) = 0. ∂x ∂y

(b) Set u = xm y n . Then nx

    ∂w ∂w − my = nx mxm−1 y n g (u) − my nxm y n−1 g (u) = 0. ∂x ∂y

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SECTION 15.5 ∂x ∂y ∂x ∂y − = (cos θ) (r cos θ) − (−r sin θ) (sin θ) = r ∂r ∂θ ∂θ ∂r     ∂P T kT ∂P ∂P T k 61. V = V − 2 = −k = −P ; V +T = −k + T =0 ∂V V V ∂V ∂T V V

60.

R1 R 2 R 3 62. R = ; R1 R2 + R1 R3 + R2 R3

∂R = ∂R1



R2 R3 R1 R2 + R1 R3 + R2 R3

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 without 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.

2

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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 = the faces of the cube; set is neither open nor closed (upper face of cube is omitted)

9. interior = { (x, y, z) : x2 + y 2 < 1, 0 < z ≤ 4} (the inside of the cylinder), boundary = the total surface of the cylinder (the curved part, the top, and the 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};

15. interior = the entire set, boundary = {1}; 16. interior ={x : x < −1},

boundary = {−1};

set is neither open nor closed.

set is open. set is closed.

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SECTION 15.6 17. interior = {x : |x| > 1}, boundary = {1, −1}; 18. interior = φ,

boundary = the entire set;

777

set is neither open nor closed.

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 = 2A, ∂x2

∂2f = 2C, ∂y 2

2.

∂2f = 6Ax + 2By, ∂x2

3.

∂2f = Cy 2 exy , ∂x2

4.

∂2f = 2 cos y − y 2 sin x, ∂x2

5.

∂2f = 2, ∂x2

∂2f = 2Cx, ∂y 2

∂2f = Cx2 exy , ∂y 2

∂2f 1 =− , ∂x2 4(x + y 2 )3/2

∂2f ∂2f = = 2Bx + 2Cy ∂y∂x ∂x∂y ∂2f ∂2f = = Cexy (xy + 1) ∂y∂x ∂x∂y

∂2f = 2 sin x − x2 cos y, ∂y 2

∂2f = 4(x + 3y 2 + z 3 ), ∂y 2

∂2f ∂2f = = 4y, ∂x∂y ∂y∂x 6.

∂2f ∂2f = = 2B ∂y∂x ∂x∂y

∂2f ∂2f = = 2(y cos x − x sin y) ∂y∂x ∂x∂y

∂2f = 6z(2x + 2y 2 + 5z 3 ) ∂z 2

∂2f ∂2f = = 6z 2 , ∂z∂x ∂x∂z ∂2f x = , ∂y 2 (x + y 2 )3/2

∂2f ∂2f = = 12yz 2 ∂z∂y ∂y∂z ∂2f ∂2f y = =− ∂x∂y ∂y∂x 2(x + y 2 )3/2

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SECTION 15.6

7.

∂2f 1 1 = − 2, ∂x2 (x + y)2 x

∂2f 1 = , ∂y 2 (x + y)2

8.

∂2f 2C(AD − BC)y =− , 2 ∂x (Cx + Dy)3

9.

∂2f = 2(y + z), ∂x2

∂2f ∂2f 1 = = ∂y∂x ∂x∂y (x + y)2

∂2f 2D(AD − BC)x = , 2 ∂y (Cx + Dy)3

∂2f = 2(x + z), ∂y 2

∂2f (AD − BC)(Cx − Dy) ∂2f = = ∂y∂x ∂x∂y (Cx + Dy)3

∂2f = 2(x + y) ∂z 2

all the second mixed partials are 2(x + y + z) 10.

∂2f 2xy 3 z 3 =− , 2 ∂x (1 + x2 y 2 z 2 )2

∂2f 2yx3 z 3 =− 2 ∂y (1 + x2 y 2 z 2 )2

∂2f 2zx3 y 3 = − , ∂z 2 (1 + x2 y 2 z 2 )2

∂2f ∂2f z(1 − x2 y 2 z 2 ) = = ∂y∂x ∂x∂y (1 + x2 y 2 z 2 )2

∂2f ∂2f y(1 − x2 y 2 z 2 ) , = = ∂z∂x ∂x∂z (1 + x2 y 2 z 2 )2

∂2f ∂2f x(1 − x2 y 2 z 2 ) = = ∂y∂z ∂z∂y (1 + x2 y 2 z 2 )2

11.

∂2f = y(y − 1)xy−2 , ∂x2

∂2f = (ln x)2 xy , ∂y 2

∂2f ∂2f = = xy−1 (1 + y ln x) ∂y∂x ∂x∂y

12.

∂2f = − sin(x + z y ), ∂x2

∂2f = z y (ln z)2 [cos(x + z y ) − z y sin(x + z y )] ∂y 2

∂2f = y(y − 1)z y−2 cos(x + z y ) − y 2 z 2y−2 sin(x + z y ) ∂z 2 ∂2f ∂2f = = −z y ln z sin(x + z y ), ∂y∂x ∂x∂y

∂2f ∂2f = = −yz y−1 sin(x + z y ) ∂z∂x ∂x∂z

∂2f ∂2f = = z y−1 (1 + y ln z) cos(x + z y ) − yz 2y−1 (ln z) sin(x + z y ) ∂z∂y ∂y∂z 13.

∂2f = yex , ∂x2

∂2f = xey , ∂y 2

∂2f ∂2f = = ey + ex ∂y∂x ∂x∂y

14.

∂2f 2xy = 2 , 2 ∂x (x + y 2 )2

∂2f −2xy = 2 , 2 ∂y (x + y 2 )2

∂2f ∂2f y 2 − x2 = = 2 ∂y∂x ∂x∂y (x + y 2 )2

15.

∂2f y 2 − x2 = , 2 ∂x (x2 + y 2 )2

∂2f x2 − y 2 = , 2 ∂y (x2 + y 2 )2

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

16.

∂2f = 6xy 2 cos(x3 y 2 ) − 9x4 y 4 sin(x3 y 2 ), ∂x2

∂2f = 2x3 cos(x3 y 2 ) − 4x6 y 2 sin(x3 y 2 ) ∂y 2

∂2f ∂2f = = 6x2 y cos(x3 y 2 ) − 6x5 y 3 sin(x3 y 2 ) ∂y∂x ∂x∂y 17.

∂2f = −2 y 2 cos 2xy, ∂x2

∂2f = −2 x2 cos 2xy, ∂y 2

∂2f ∂2f = = −[sin 2xy + 2xy cos 2xy] ∂y∂x ∂x∂y

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SECTION 15.6 18.

∂2f 2 = y 4 exy , ∂x2

19.

∂2f = 0, ∂x2

∂2f 2 = exy (2x + 4x2 y 2 ), ∂y 2

∂2f = xz sin y, ∂y 2

∂2f = zex , ∂x2

∂2f = xey , ∂y 2

∂2f ∂2f = = ex , ∂z∂x ∂x∂z 21. x2

∂2f ∂2f 2 = = exy (2y + 2xy 3 ) ∂y∂x ∂x∂y

∂2f = − xy sin z, ∂z 2

∂2f ∂2f = = sin z − z cos y, ∂y∂x ∂x∂y 20.

779

∂2f ∂2f = = y cos z − sin y, ∂x∂z ∂z∂x

∂2f = yez , ∂z 2

∂2f ∂2f = = x cos z − x cos y ∂y∂z ∂z∂y

∂2f ∂2f = = ey , ∂y∂x ∂x∂y

∂2f ∂2f = = ez ∂z∂y ∂y∂z

∂2u ∂2u ∂2u + y 2 2 = x2 + 2xy 2 ∂x ∂x∂y ∂y



−2y 2 (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 xm y n are mnxm−1 y n−1 23. (a) no, since

24.

∂2f ∂2f = ∂y∂x ∂x∂y

∂h = g (x + y) + g (x − y), ∂x ∂2h = g (x + y) + g (x − y), ∂x2

25.

(b)

no, since

∂2f ∂2f = for x = y ∂y∂x ∂x∂y

∂h = g (x + y) − g (x − y) ∂y ∂2h ∂2h = g (x + y) + g (x − y) = 2 ∂y ∂x2

 2   2        ∂ ∂ f ∂ ∂ f ∂2 ∂f ∂2 ∂f ∂ ∂2f ∂3f ∂3f = = = = = = ∂x2 ∂y ∂x ∂x∂y ∂x ∂y∂x ∂x∂y ∂x ∂y∂x ∂x ∂y ∂x2 ∂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) = 27. (a) lim

x→0

(c) lim

x→0

(x)(0) = lim 0 = 0 x2 + 0 x→0

lim

y→0

(0)(y) = lim 0 = 0 y→0 0 + y2

(x)(mx) m m = lim = x→0 1 + m2 x2 + (mx)2 1 + m2

(d) lim+ θ→0

(b)

(θ cos θ)(θ sin θ) = lim+ cos θ sin θ = 0 (θ cos θ)2 + (θ sin θ)2 θ→0

1 2

tends to 12 ;

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SECTION 15.6 (e) By L’Hospital’s rule

lim

x→0

f (x) = lim f (x) = f (0). Thus x→0 x

xf (x) f (x)/x f (0) = lim = . x→0 x2 + [ f (x) ]2 x→0 1 + [ f (x)/x ]2 1 + [ f (0) ]2 lim

(f)

lim

θ→(π/3)−

=

(cos θ sin 3θ)(sin θ sin 3θ) 1√ = lim − cos θ sin θ = 3 2 2 (cos θ sin 3θ) + (sin θ sin 3θ) 4 θ→(π/3)

(1/t)(sin t)/t sin t  2  = lim ; 2 2 t→∞ 1/t + sin t /t t→∞ 1 + sin2 t

(g) lim

28. (a) lim

x→0 (x2

(c) lim

x→0 (x2

(d) lim

θ→0+

x(0)2 = lim 0 = 0 x→0 + 02 )3/2

(b)

does not exist

0 · y2 = lim 0 = 0 y→0 (0 + y 2 )3/2 y→0 lim

xm2 x2 m2 x3 m2 x = lim 3 = lim ; 2 2 3/2 2 3/2 x→0 |x| (1 + m ) x→0 |x| (1 + m2 )3/2 +m x )

does not exist

(θ cos θ)(θ sin θ)2 = lim+ cos θ sin2 θ = 0 θ→0 [(θ cos θ)2 + (θ sin θ)2 ]3/2

x[f (x)]2 [f (x)/x]2 x3 [f (0)]2 = lim = lim ; x→0 (x2 + [f (x)]2 )3/2 x→0 (1 + [f (x)/x]2 )3/2 x→0 |x|3 (1 + [f (0)]2 )3/2

(e) lim

(f) lim π− θ→ 3

(cos θ sin 3θ)(sin θ sin 3θ)2 3 = lim cos θ sin2 θ = 2 2 3/2 − π 8 [(cos θ sin 3θ) + (sin θ sin 3θ) ] θ→ 3

(1/t) (sin t/t)2 sin2 t (g) lim  = lim   3/2 ; 3/2 t→∞ t→∞ (1/t2 ) + (sin2 t/t2 ) 1 + sin2 t 29. (a)

does not exist

does not exist

∂g g(h, 0) − g(0, 0) (0, 0) = lim = lim 0 = 0, h→0 h→0 ∂x h ∂g g(0, h) − g(0, 0) (0, 0) = lim = lim 0 = 0 h→0 h→0 ∂y h

(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) =

1 2

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) =

31. For y = 0,

Since

x−1 1 = 2 x3 − 1 x +x+1

tends to

1 3

∂f f (h, y) − f (0, y) y(y 2 − h2 ) (0, y) = lim = lim = y. h→0 h→0 h2 + y 2 ∂x h ∂f f (h, 0) − f (0, 0) (0, 0) = lim = lim 0 = 0, h→0 h→0 ∂x h

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SECTION 15.6 we have For x = 0,

781

∂f (0, y) = y for all y. ∂x ∂f f (x, h) − f (x, 0) x(h2 − x2 ) = −x. (x, 0) = lim = lim h→0 h→0 x2 + h2 ∂y h ∂f f (0, h) − f (0, 0) (0, 0) = lim = lim 0 = 0, h→0 h→0 ∂y h

Since we have Therefore In particular

∂f (x, 0) = −x for all x. ∂y ∂2f (0, y) = 1 for all y and ∂y∂x ∂2f (0, 0) = 1 ∂y∂x

∂2f (x, 0) = −1 for all x. ∂x∂y

∂2f (0, 0) = −1. ∂x∂y

while

  f (x0 + h, y0 ) − f (x0 , y0 ) lim [f (x0 + h, y0 ) − f (x0 , y0 )] = lim (h) h→0 h→0 h    f (x0 + h, y0 ) − f (x0 , y0 ) = lim h lim h→0 h→0 h

32.

=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.

35.

0.5 0 -0.5

2

-2

2

0.5 0 -2

0 0

0 0

2 -2

2 -2

36.

0.5

2

0 -0.5 -2

0 0 2 -2

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SECTION 15.6

PROJECT 15.6 1. (a)

∂u x2 y 2 + 2xy 3 = , ∂x (x + y)2 x

(b)

x2 y 2 + 2x3 y ∂u = ∂y (x + y)2

∂u ∂u 3x2 y 2 (x + y) = 3u +y = ∂x ∂y (x + y)2

∂u ∂u ∂u = 2xy + z 2 , = 2yz + x2 , = 2xz + y 2 ∂x ∂y ∂z ∂u ∂u ∂u + + = 2xy + z 2 + 2yz + x2 + 2xz + y 2 = (x + y + z)2 ∂x ∂y ∂z

2. (a) (i)

∂2f ∂2f + 2 = 6x − 6x = 0 2 ∂x ∂y

(ii)

∂2f ∂2f + 2 = (− cos x sinh y − sin x cosh y) + (cos x sinh y + sin x cosh y) = 0 2 ∂x ∂y

(iii)

∂2f y 2 − x2 = , ∂x2 (x2 + y 2 )2

∂2f x2 − y 2 = ∂y 2 (x2 + y 2 )2

∂2f ∂2f y 2 − x2 x2 − y 2 + = + =0 ∂x2 ∂y 2 (x2 + y 2 )2 (x2 + y 2 )2 (b) (i) (ii)

3. (i) (ii)

∂2f ∂2f ∂2f 2x2 − y 2 − z 2 2y 2 − x2 − z 2 2z 2 − x2 − y 2 + 2 + 2 = 2 + 2 + 2 =0 2 2 2 5/2 2 2 5/2 ∂x ∂y ∂z (x + y + z ) (x + y + z ) (x + y 2 + z 2 )5/2 √  ∂2f x+y = e cos 2z , ∂x2

√  √  ∂2f ∂2f x+y x+y = e cos 2 z , = − 2e cos 2z ∂y 2 ∂z 2 √  √   √  ∂2f ∂2f ∂2f x+y x+y x+y + + = e cos 2 z + e cos 2 z + − 2e cos 2z = 0 ∂x2 ∂y 2 ∂z 2

∂2f ∂2f ∂2f ∂2f = = 0 =⇒ − c2 2 = 0 2 2 2 ∂t ∂x ∂t ∂x ∂2f = − 5c2 sin(x + ct) cos(2x + 2ct) − 4c2 cos(x + ct) sin(2x + 2ct) ∂t2 ∂2f = − 5 sin(x + ct) cos(2x + 2ct) − 4 cos(x + ct) sin(2x + 2ct) ∂x2 It now follows that

(iii)

(iv)

4.

∂2f c2 = − , ∂t2 (x + ct)2

∂2f ∂2f − c2 2 = 0 2 ∂t ∂x

2 ∂2f 1 ∂2f 2∂ f = − =⇒ − c =0 ∂x2 (x + ct)2 ∂t2 ∂x2

 kx   ckt     ∂2f ∂2f 2 2 −kx −ckt Ae Ce , = c k + Be + De = k 2 Aekx + Be−kx Ceckt + De−ckt 2 2 ∂t ∂x 2 ∂2f 2∂ f It now follows that −c =0 ∂t2 ∂x2

  ∂2f ∂2f − c2 2 = c2 g (x + ct) + c2 h (x − ct)] − c2 [g (x + ct) + h (x − ct) = 0 2 ∂t ∂x

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REVIEW EXERCISES 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 + y 2 },

dange [0, +∞)

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

range R

1 5. (a) f (x, y) = πx2 y; 3 1 (b) f (x, y) = yx2 ; 2 x + 2y (c) θ = arccos √  5 x2 + y 2 6. Assume one  of the vertices is (x, y, z), x > 0, y > 0, z > 0. x2 y2 V = 8cxy 1 − 2 − 2 a b 7. ellipsoid xy−trace: ellipse 4x2 + 9y 2 = 36 xz−trace: ellipse 4x2 + 36z 2 = 36 yz−trace: ellipse 9y 2 + 36z 2 = 36 9. hyperbolic paraboloid xy−trace: lines x = ±y xz−trace: parabola z = −x2 yz−trace: parabola z = y 2

8. hyperboloid of two sheets xy−trace: none xz−trace: hyperbola 4z 2 − x2 = 4 yz−trace: hyperbola 4z 2 − y 2 = 4 10. elliptic paraboloid xy−trace: parabola 4x2 = y xz−trace: (0, 0) yz−trace: parabola 9z 2 = y

11. cone xy−trace: lines x = ±y xz−trace: lines x = ±z yz−trace: (0, 0)

12. hyperboloid of one sheet xy−trace: ellipse 9x2 + 4y 2 = 36 xz−trace: hyperbola z 2 = 9x2 − 36 yz−trace: hyperbola z 2 = 4y 2 − 36

13.

14.

4 2 2

0

-2 -2 0 2

783

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REVIEW EXERCISES

15.

16.

2 2

0 -2 2 -2

0

2

0 1

17. c = 0, =⇒ 0 = 2x2 + 3y 2 =⇒ (0, 0)

2

18. c = 0, =⇒ 0 = x2 + y 2 − 4, circle

c = 6, =⇒ 6 = 2x2 + 3y 2 , ellipse

c = 1, =⇒ 5 = x2 + y 2 , circle

c = 12, =⇒ 6 = 2x2 + 3y 2 , ellipse

c = 2, =⇒ 8 = x2 + y 2 , circle √ c = 5, =⇒ 9 = x2 + y 2 , circle

19. c = −4, =⇒ x = −4y 2 , parabola

20. c = 1, =⇒ x2 + y 2 = 1 circle

c = −1, =⇒ x = −y , parabola

c = 4, =⇒ x2 + y 2 = 4 circle

c = 1, =⇒ x = y 2 , parabola

c = 9, =⇒ x2 + y 2 = 9 circle

2

c = 4, =⇒ x = 4y 2 , parabola the origin is omitted 21.

c = 6, 2x + y + 3z = 6, plane

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

22.

c = 16, x2 + y 2 + 4z 2 = 16, ellipsoid

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

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

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

25.

f (x + h, y) − f (x, y) (x + h)2 + 2(x + h)y − x2 − 2xy = lim h→0 h→0 h h = lim (2x + h + 2y) = 2x + 2y

fx = lim

h→0

f (x, y + h) − f (x, y) x + 2x(y + h) − x2 − 2xy = lim = 2x h→0 h→0 h h 2

fy = lim 26.

f (x + h, y) − f (x, y) y 2 cos 2(x + h) − y 2 cos 2x = lim h→0 h→0 h h cos 2(x + h) − cos 2x = y 2 lim h→0 h = y 2 cos 2x = −2y 2 sin 2x

fx = lim

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REVIEW EXERCISES f (x, y + h) − f (x, y) (y + h)2 cos 2x − y 2 cos 2x = lim h→0 h→0 h h (y + h)2 − y 2 = cos 2x lim = 2y cos 2x h→0 h

fy = lim

27. fx = 2xy − 2y 3 ;

fy = x2 − 6xy 2

28. gx = (x2 + y 2 )−1/2 − x2 (x2 + y 2 )−3/2 ; 29.

∂z = 2x sin(xy 2 ) + x2 y 2 cos(xy 2 ); ∂x

30. fx = yexy ln(y/x) −

1 xy e x

gy = −xy(x2 + y 2 )−3/2 ∂z = 2x3 y cos(xy 2 ). ∂y

fy = xexy ln(y/x) +

31. hx = −e−x cos(2x − y) − 2e−x sin(2x − y) 32. ux = y 2 sec x tan x + 2x tan y 33. fx =

2y 2 + 2yz ; (x + y + z)2

fy =

34. wx = arctan(y − z)

35.

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

gy =

x2 , y−x

−x 1 + (y − z)2

∂g z . = 2 ∂z x + y2 + z2

hv = ueuv sin uw;

hw = ueuv cos uw

fy = 2x3 y − 12xy 2 − 1;

fyy = 2x3 − 24xy, x2 , y−x

gyy = −

−2xy (x + y + z)2

wz =

∂g y ; = 2 ∂y x + y2 + z2

37. fx = 3x2 y 2 − 4y 3 + 2,

38. gx = 2x ln(y − x) −

fz =

x 1 + (y − z)2

36. hu = veuv sin uw + weuv cos uw;

fxx = 6xy 2 ,

hy = e−x sin(2x − y)

uy = 2y sec x + x2 sec2 y

wy =

∂g x ; = 2 ∂x x + y2 + z2

exy y

gxx = 2 ln(y − x) −

x2 , (y − x)2

39. gx = y sin xy + xy 2 cos xy, gy = x sin xy + x2 y cos xy,

fyx = fxy = 6x2 y − 12y 2

gxy = gyx =

4x x2 ; − y − x (y − x)2

2x x2 + y − x (y − x)2

gxx = 2y 2 cos xy − xy 3 sin xy; gyy = 2x2 cos xy − yx3 sin xy,

gxy = gyx = sin xy + 3xy cos xy − x2 y 2 sin xy 40. fx = 2xex/y + fy = −

x3 x/y e , y2

x2 x/y e , y fyy =

fxx = 2ex/y +

4x x/y x2 x/y + 2e ; e y y

2x3 x/y x4 x/y e + 4e , y3 y

fxy = fyx = −

3x2 x/y x3 x/y e − 3e y2 y

785

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REVIEW EXERCISES

41. fx = 2xe2y cos(2z + 1), fy = 2x2 e2y cos(2z + 1), fz = −2x2 e2y sin(2z + 1); fxx = 2e2y cos(2z + 1), fyy = 4x2 e2y cos(2z + 1), fzz = −4x2 e2y cos(2z + 1); fxy = fyx = 4xe2y cos(2z + 1), fxz = fzx = −4xe2y sin(2z + 1), fyz = fzy = −4x2 e2y sin(2z + 1) 42. gx = 4xyz 3 + yzexyz ,

gy = 2x2 z 3 + xzexyz ,

gxx = 4yz 3 + y 2 z 2 exyz ,

gyy = x2 z 2 exyz ,

gxy = gyx = 4xz 3 + zexyz + xyz 2 exyz ,

gz = 6x2 yz 2 + xyexyz ; gzz = 12x2 yz + x2 y 2 exyz ;

gxz = gzx = 12xyz 2 + yexyz + xy 2 zexyz

gyz = gzy = 6x2 z 2 + xexyz + x2 yzexyz 43.

∂z = 4x + 6|(1,2,8) = 10 ∂x x = 1 + t,

y = 2,

44.

∂z = 3x|(2,−1,2) = 6 ∂y x = 2, y = −1 + t, z = 2 + 6t

z = 8 + 10t

−6y 45. (a) zy (2, 1) =  (2, 1) = −1; the equation for l1 is: 2 20 − 2x2 − 3y 2 x = 2;

y = 1 − t;

z =3+t

−4x 4 (b) zx (2, 1) =  (2, 1) = − ; the equation for l2 is: 2 2 3 2 20 − 2x − 3y 3 x = 2 − t; 4

y = 1;

z =3+t 3 3 j − k or 4 i + 3 j + 3 k; 4 4 4(x − 2) + 3(y − 1) + 3(z − 3) = 0.

(c) The normal vector for this plane is: −i − an equation for the plane is: 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 + y 2 < 4} boundary: {(0, 0)} ∪ {(x, y) : x2 + y 2 = 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, y 2 + z 2 < 4} boundary: the quarter disks x = 0, y 2 + z 2 ≤ 4; the squares z = 0, 0 ≤ x, y ≤ 2;

x = 2, y 2 + z 2 ≤ 4;

y = 0, 0 ≤ x, z ≤ 2;

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

2

and

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REVIEW EXERCISES 50. Neither. interior: {(x, y, z) : 0 < x2 + y 2 < z < 4} boundary: the cone z = x2 + y 2 and the disk x2 + y 2 ≤ 4, z = 4 51. (a) fx = yg (xy),

fy = xg (xy);

(b) fxx = y 2 g (xy), 52. fx = −

y , x2 + y 2

xfx − yfy = xyg − xyg = 0

fyy = x2 g (xy);

fy =

x ; x2 + y 2

x2 fxx − y 2 fyy = x2 y 2 g − x2 y 2 g = 0

fxx =

2xy , (x2 + y 2 )2

fyy =

−2xy (x2 + y 2 )2

fxx + fyy = 0 ∂2f ∂2f = x2 exy = y 2 exy = ∂y∂x ∂x∂y

53. No.

54. (a) lim f (x, 0) = lim 0 = 0 x→0

x→0

2x2 mx 2mx (c) lim 4 = lim 2 =0 2 2 x→0 x + m x x→0 m + x2 lim(x,y)→(0,0) f (x, y) does not exist.

(b) lim f (0, y) = lim 0 = 0 y→0

y→0

2a 2x2 ax2 (d) lim 4 = 2 4 x→0 x + a x 1 + a2

787

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