Solution Manual - Calculus 8th Edition Varberg, Purcell, Rigdon Ch12 PDF

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Summary

CHAPTER 12 Derivatives for Functions of Two or More Variables 12 Concepts Review 4. a. 6 1. function of two real variables b. 12 2. level contour map c. 2 3. concentric circles d. (3cos 2 1 3 4. parallel lines Problem Set 12 1. a. 5 b. 0 c. 6 d. a6 a 2 e. 2 x2 , x 0 f. Undefined 5. F (t cos t , sec2...

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12

CHAPTER

Derivatives for Functions of Two or More Variables

12.1 Concepts Review

4. a.

6

1. real-valued function of two real variables

b. 12

2. level curve; contour map

c.

2

3. concentric circles

d.

(3cos 6)1/ 2 + 1.44 ≈ 3.1372

4. parallel lines

5. F (t cos t , sec2 t ) = t 2 cos 2 t sec2 t = t 2 , cos t ≠ 0

Problem Set 12.1 1. a.

6. F ( f (t ), g (t)) = F (ln t 2, e t / 2) = exp(ln t 2 ) + ( et / 2) 2 = t 2 + et , t ≠ 0

5

b. 0

7. z = 6 is a plane.

c.

6

d.

a6 + a 2

e.

2x 2 , x ≠ 0

f.

Undefined

The natural domain is the set of all (x, y) such that y is nonnegative. 2. a.

4

8. x + z = 6 is a plane.

b. 17 c.

17 16

d.

1 + a2 , a ≠ 0

e.

x3 + x, x ≠ 0

f.

Undefined

The natural domain is the set of all (x, y) such that x is nonzero. 3. a.

744

9. x + 2y + z = 6 is a plane.

sin(2 π ) = 0

b.

⎛π⎞ 4sin ⎜ ⎟ = 2 ⎝6⎠

c.

⎛ π⎞ 16sin ⎜ ⎟ = 16 ⎝2⎠

d.

π sin(π ) ≈ –4.2469

2

2

Section 12.1

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10. z = 6 – x 2 is a parabolic cylinder.

11. x2 + y2 + z2 = 16 , z ≥ 0 is a hemisphere.

12.

x2 y2 z2 + + = 1 , z ≥ 0 is a hemi-ellipsoid. 4 16 16

13. z = 3 – x2 – y2 is a paraboloid.

14. z = 2 – x – y 2

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15. z = exp[–( x 2 + y 2 )]

16. z =

x2 y

,y>0

17. x2 + y2 = 2 z; x2 + y2 = 2 k

18. x = zy, y ≠ 0 ; x = ky, y ≠ 0

19. x 2 = zy, y ≠ 0; x2 = ky, y ≠ 0

Section 12.1

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20. x 2 = –( y – z ); x 2 = –( y – k )

21. z =

x 2 +1

24. ( x – 2)2 + ( y + 3)2 =

25. a.

San Francisco and St. Louis had a temperature between 70 and 80 degrees Fahrenheit.

b.

Drive northwest to get to cooler temperatures, and drive southeast to get warmer temperatures.

c.

Since the level curve for 70 runs southwest to northeast, you could drive southwest or northeast and stay at about the same temperature.

26. a.

The lowest barometric pressure, 1000 millibars and under, occurred in the region of the Great Lakes, specifically near Wisconsin. The highest barometric pressure, 1025 millibars and over, occurred on the east coast, from Massachusetts to South Carolina.

b.

Driving northwest would take you to lower barometric pressure, and driving southeast would take you to higher barometric pressure.

c.

Since near St. Louis the level curves run southwest to northeast, you could drive southwest or northeast and stay at about the same barometric pressure.

, k = 1, 2, 4 x2 + y2

k = 1: y 2 = 1 or y = ±1 ; two parallel lines k = 2: 2 x2 + 2 y2 = x2 + 1 x2 y2 + = 1 ; ellipse 1 1 2

k = 4: 4 x2 + 4 y2 = x2 + 1 x2 1 3

+

y2 1 4

= 1; ellipse

16 V2

22. y = sin x + z; y = sin x + k

23. x = 0, if T = 0: ⎛ 1 ⎞ y 2 = ⎜ –1 ⎟ x2 , if y ≠ 0 . ⎝T ⎠

27. x 2 + y 2 + z2 ≥ 16 ; the set of all points on and outside the sphere of radius 4 that is centered at the origin 28. The set of all points inside (the part containing the z-axis) and on the hyperboloid of one sheet; x2 y2 z2 – + = 1. 9 9 9 29.

746

Section 12.1

x2 y2 z2 + + ≤ 1 ; points inside and on the 9 16 1 ellipsoid

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30. Points inside (the part containing the z-axis) or on x2 y2 z2 the hyperboloid of one sheet, + = 1, – 9 9 16 excluding points on the coordinate planes 31. Since the argument to the natural logarithm function must be positive, we must have x 2 + y 2 + z 2 > 0 . This is true for all ( x , y , z )

except ( x, y, z ) = ( 0, 0, 0 ) . The domain consists all points in  3 except the origin. 32. Since the argument to the natural logarithm function must be positive, we must have xy > 0 .

This occurs when the ordered pair ( x , y ) is in the first quadrant or the third quadrant of the xy-plane. There is no restriction on z. Thus, the domain consists of all points (x , y, z ) such that x and y are both positive or both negative. 33. x 2 + y2 + z2 = k, k > 0; set of all spheres centered at the origin

37. 4x 2 – 9y 2 = k , k in R;

k 100

+

y2 k 16

+

z2 k 25

= 1; set of all ellipsoids centered

k 4

–

y2 k 9

= 1, if k ≠ 0;

2x (for k = 0) and all hyperbolic 3 cylinders parallel to the z-axis such that the ratio ⎛ 1 ⎞ ⎛ 1⎞ a:b is ⎜ ⎟ : ⎜ ⎟ or 3:2 (where a is associated ⎝ 2 ⎠ ⎝ 3⎠ with the x-term)

planes y = ±

38. ex

2 y2 z 2 + +

= k, k > 0

x2 + y2 + z2 = ln k concentric circles centered at the origin.

39. a.

All ( w, x, y, z ) except ( 0, 0, 0, 0 ) , which would cause division by 0.

b.

All ( x 1, x 2, …, xn ) in n-space.

c.

All ( x 1, x 2, …, xn ) that satisfy x12 + x 22 +  + xn2 ≤ 1 ; other values of

( x1 , x2 ,… , xn )

would lead to the square root of a negative number.

34. 100x 2 +16 y2 + 25 z2 = k, k > 0; x2

x2

40. If z = 0, then x = 0 or x = ± 3 y .

at origin such that their axes have ratio ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1⎞ or 2:5:4. ⎜ 10 ⎟ : ⎜ 4 ⎟ :⎜ 5⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 35.

x2 1 16 2

+

y2 1 4 2

–

z2 = k; the elliptic cone 1

x y z2 + = and all hyperboloids (one and two 9 9 16 sheets) with z-axis for axis such that a:b:c is ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1⎞ ⎜ 4 ⎟ :⎜ 4 ⎟ :⎜ 3 ⎟ or 3:3:4. ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

36.

x2 1 9 2

–

y2 1 4 2

–

z2 = k ; the elliptical cone 1 2

41. a.

b.

AC is the least steep path and BC is the most steep path between A and C since the level curves are farthest apart along AC and closest together along BC. AC ≈ (5750) 2 + (3000) 2 ≈ 6490 ft BC ≈ (580) 2 + (3000) 2 ≈ 3060 ft

y z x + = and all hyperboloids (one and two 9 36 4 sheets) with x-axis for axis such that a:b:c is ⎛ 1⎞ ⎛ 1⎞ ⎜ 3 ⎟ : ⎜ 2 ⎟ :1 or 2:3:6 ⎝ ⎠ ⎝ ⎠

42. Completing the squares on x and y yields the equivalent equation f (x , y ) + 25.25 = ( x – 0.5)2 + 3( y + 2) 2 , an elliptic paraboloid.

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43.

(2x – y 2 ) exp(– x 2 – y 2 )

sin 2x 2 + y 2

46.

44.

sin x sin y (1 + x 2 + y 2 )

sin(x 2 + y2 ) x2 + y2

12.2 Concepts Review [( f ( x0 + h, y 0 ) – f ( x0, y0 )] ; partial h derivative of f with respect to x

1. lim

h→ 0

2. 5; 1 3. 45.

∂2f ∂ y∂ x

4. 0

Problem Set 12.2 1. f x (x , y ) = 8(2x – y ) 3; f y( x , y ) = –4(2 x – y ) 3 2. f x (x , y ) = 6(4 x – y 2 )1/ 2 ; f y ( x, y) = –3 y(4 x – y2 )1/ 2

748

Section 12.2

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3.

fx ( x, y) =

fy ( x, y) =

( xy)(2 x) – ( x2 – y2 )( y) (xy ) 2

x2 + y2

14.

x2y

fs ( s, t ) = –2 set

2 – s2

; fs ( s, t) = 2 tet

2 –s2

15. Fx ( x, y) = 2 cos x cos y; Fy ( x, y) = –2sin x sin y

( xy)( −2 y) −( x 2 − y 2 )( x) ( xy ) 2

=−

4.

=

(x 2 + y 2 ) xy 2

16.

fr (r , θ ) = 9 r 2 cos 2θ ; fθ (r , θ ) = –6 r3 sin 2θ

17.

fx ( x, y) = 4 xy3 – 3 x2 y5 ; fxy ( x, y) =12 xy2 –15 x2 y4

fx ( x , y ) = e x cos y; f y ( x, y ) = – e x sin y

fy ( x, y ) = 6 x 2 y 2 – 5 x 3 y 4;

5.

fx ( x , y ) = e y cos x ; f y ( x, y) = e y sin x

6.

⎛ 1⎞ fx ( x, y ) = ⎜ – ⎟ (3 x 2 + y 2 ) –4 / 3 (6 x) ⎝ 3⎠

fyx ( x, y ) = 12 xy2 – 15 x2 y4

18.

fxy ( x, y) =60 x2( x3 + y2) 3(2 y)

= – 2 x (3x 2 + y 2 ) –4 / 3 ;

= 120x 2 y ( x3 + y2 )3

⎛ 1⎞ fy ( x, y) = ⎜ – ⎟ (3 x2 + y2) –4 / 3(2 y) ⎝ 3⎠ ⎛ 2y ⎞ = ⎜ – ⎟ (3x 2 + y 2 ) –4 / 3 ⎝ 3 ⎠

7.

fx ( x, y) =

x( x2

−

fy ( x, y ) = 5( x3 + y2 )4 (2 y); fyx ( x, y ) = 40 y( x3 + y2 )3 (3 x2 ) = 120x 2 y ( x3 + y2 )3

y2 ) −1/ 2;

19.

fy ( x, y) = – y( x 2 – y 2) –1/ 2

8.

uv

fu (u, v) = ve ; fv ( u, v) = ue

9. g x ( x, y ) = – ye 10.

– xy

20.

– xy

fx ( x, y) = x(1 + x 2 y 2) –1;

fs ( s, t ) = 2 s( s2 – t 2 ) –1 ;

fxy ( x, y) = (1 − x 2 y 2)(1 + x 2 y 2) − 2

fx ( x, y) = 4[1 + (4 x – 7 y) 2 ] –1;

21. Fx ( x, y) =

fy ( x, y) = –7[1 +(4 x – 7 y) 2] –1

1

12. Fw ( w, z) = w 1– w z

= 1–

( wz)

2

( wz )

1 1–

( )

w 2 z

Fy ( x, y) =

2

⎛ w⎞ ⎜ – 2⎟ = ⎝ z ⎠

( wz ) 2 1 – ( zw ) 2

–

2

2

2

fy ( x, y) = –2 y sin( x + y ) +cos( x + y )

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( xy )

2

y2

=

2 2

x y

=

1 x2

;

1 9

( xy )(–1) – (2 x – y)( x) (x y)

Fy (3, − 2) = −

fx ( x, y ) = –2 xy sin( x 2 + y 2 ); 2

( xy )(2) – (2 x – y)( y)

Fx (3, − 2) =

⎛ 1⎞ –1 ⎛ w ⎞ ⎜ z ⎟ +sin ⎜ z ⎟ ⎝ ⎠ ⎝ ⎠

⎛w⎞ + sin –1 ⎜ ⎟ ; ⎝z⎠

Fz = (w, z ) = w

13.

2

fx ( x, y) = y(1 + x2 y2 ) –1; fxy ( x, y) = (1 − x 2 y 2)(1 + x 2 y 2) − 2

ft ( s, t ) = −2t ( s2 − t 2 ) −1

11.

fx ( x, y) = 6 e2 x cos y; fxy ( x, y) = –6 e2 x sin y fy ( x, y) = –3e 2x sin y; fyx ( x, y) = –6 e 2x sin y

uv

; g y ( x, y) = – xe

fx ( x, y) = 5( x3 + y2 ) 4 (3 x2 );

2

=

–2 x2 x2 y2

=–

2 x2

;

1 2

22. Fx ( x, y) = (2 x + y)( x2 + xy + y2 ) –1; Fx (–1, 4) =

2 ≈ 0.1538 13

Fy ( x, y) = ( x +2 y)( x2 + xy + y2 ) –1;

Fy (–1, 4) =

7 ≈ 0.5385 13

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23. f x (x , y ) = – y 2 ( x 2 + y 4 ) –1; 4 f x 5, – 2 = – ≈ –0.1905 21 f y ( x, y) = 2 xy( x 2 + y 4) –1;

(

fy

24.

(

)

)

4 5 5, – 2 = – ≈ –0.4259 21

f x ( x, y ) = e y sinh x; f x (–1, 1) = e sinh(–1) ≈ –3.1945 f y ( x, y) = ey cosh x;

kT V k PT (V , T ) = ; V

31. P (V , T ) =

PT (100, 300) =

k lb/in.2 per degree 100

32. V [PV (V , T )] +T [PT (V , T )] =V (–kTV

–2

) +T (kV

–1

) =0

⎛ kT ⎞ ⎛ k ⎞⎛ V PV VT TP = ⎜ – ⎟ ⎜ ⎟⎜ ⎝ V 2 ⎠ ⎝ P ⎠⎝ k

kT PV ⎞ =– = –1 ⎟= – PV PV ⎠

f y (–1, 1) = e cosh(–1) ≈ 4.1945 x 2 y2 . 25. Let z = f (x , y ) = + 9 4 y f y ( x, y ) = 2 The slope is fy (3, 2) = 1.

26. Let z = f ( x , y ) = (1/ 3)(36 – 9 x 2 – 4 y 2) 1/ 2. ⎛ 4⎞ f y ( x, y) = ⎜ − ⎟ y(36 −9 x 2 − y 2) − 1/ 2 ⎝ 3⎠ 8 ≈ 0.8040. The slope is fy (1, – 2) = 3 11 ⎛1⎞ 27. z = f ( x, y) = ⎜ ⎟ (9 x 2 +9 y 2 −36) 1/ 2 ⎝2⎠ 9x f x (x , y ) = 2(9 x 2 +9 y 2 – 36) 1/ 2 f x (2, 1) = 3 ⎛5⎞ 28. z = f (x , y ) = ⎜ ⎟ (16 – x 2 )1/ 2 . ⎝4⎠ ⎛ 5⎞ f x (x , y ) = ⎜ – ⎟ x(16 – x 2) –1/ 2 ⎝ 4⎠ f x (2, 3) = –

5 4 3

33. f x (x , y ) = 3x 2y – y 3; fxx ( x, y) = 6 xy; f y ( x, y) = x3 – 3 xy2 ; fyy ( x, y ) = –6 xy

Therefore, fxx ( x, y) + f yy ( x, y) = 0. 34. f x (x , y ) = 2 x ( x 2 + y 2 ) –1; f xx( x, y ) = –2( x2 – y2 )( x2 + y2 ) –1 f y ( x, y) = 2 y( x 2 + y 2) –1; − f yy( x, y) = 2( x 2 − y 2)( x 2 + y 2) 1

35. Fy ( x, y) =15 x 4 y 4 – 6 x 2 y 2; Fyy ( x, y ) = 60 x4 y3 −12 x2 y; Fyyy ( x, y ) = 180 x 4 y 2 – 12 x 2

36. f x (x , y ) = [– sin(2x 2 – y 2 )](4 x ) = – 4x sin(2 x 2 – y 2 )

f xx( x, y) = (–4 x)[cos(2 x 2 – y 2)](4 x) + [sin(2 x 2 – y 2 )](–4) f xxy ( x, y) = –16 x 2[– sin(2 x 2 – y 2 )](–2 y)

≈ –0.7217

29. V r (r , h ) = 2πrh ; V r (6, 10) = 120π ≈ 376.99 in.2

– 4[cos(2 x 2 – y 2 )](–2 y) = –32x 2 y sin(2 x 2 – y 2 ) + 8 y cos(2 x 2 – y 2 )

37. a.

2

30. Ty ( x, y) = 3 y ; T y (3, 2) = 12 degrees per ft b.

c.

750

Section 12.2

∂ 3f ∂ y3 ∂ 3y ∂ y∂ x 2 ∂ 4y ∂ y 3∂ x

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38. a.

fyxx

b.

fyyxx

c.

fyyxxx

39. a. b.

fx ( x, y, z) = 6 xy – yz fy ( x, y, z ) = 3 x2 – xz + 2 yz2 ; fy (0, 1, 2) = 8

c. 40. a. b.

45. Domain: (Case x < y) The lengths of the sides are then x, y – x, and 1 – y. The sum of the lengths of any two sides must be greater than the length of the remaining side, leading to three inequalities: 1 x + (y – x) > 1 – y ⇒ y > 2 1 (y – x) + (1 – y) > x ⇒ x < 2 1 x + (1 – y) > y – x ⇒ y < x + 2

Using the result in a, f xy ( x, y, z) = 6 x– z. 12x 2 ( x3 + y2 + z)3 fy ( x, y, z) = 8 y( x3 + y2 + z)3 ; fy (0, 1, 1) = 64

c.

fz ( x, y, z) = 4( x3 + y2 + z)3 ; 2

2

fzz ( x, y, z) =12( x + y + z)

2

41.

fx (x , y , x) = – yze –xyz – y ( xy – z 2 ) –1

42.

⎛ 1 ⎞⎛ xy ⎞ fx ( x, y, z) = ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ z ⎠

–1/ 2

⎛ 1 ⎞⎛ 1 ⎞ fx (–2, – 1, 8) = ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 4 ⎠

⎛ y⎞ ⎜ z⎟; ⎝ ⎠

–1/ 2

⎛ 1⎞ 1 ⎜– ⎟ = – 8 8 ⎝ ⎠

43. If f ( x, y) = x4 + xy3 + 12, fy ( x, y) = 3 xy2 ; fy (1, – 2) =12. Therefore, along the tangent line

Δy = 1 ⇒ Δz = 12, so 0, 1, 12 is a tangent

vector (since Δx = 0). Then parametric equations ⎧ x =1 ⎫ ⎪ ⎪ of the tangent line are ⎨ y = –2 + t ⎬ . Then the ⎪ z = 5 +12 t⎪ ⎩ ⎭ point of xy-plane at which the bee hits is (1, 0, 29) [since y = 0 ⇒ t = 2 ⇒ x = 1, z = 29]. 44. The largest rectangle that can be contained in the circle is a square of diameter length 20. The edge of such a square has length 10 2, so its area is 200. Therefore, the domain of A is {( x, y ) : 0 ≤ x 2 + y 2 < 400}, and the range is (0, 200].

The case for y < x yields similar inequalities (x and y interchanged). The graph of D A , the domain of A is given above. In set notation it is 1 1 1⎫ ⎧ D A = ⎨(x , y ) : x < , y > , y < x + ⎬ 2 2 2⎭ ⎩ 1 1 1⎫ ⎧ ∪ ⎨ ( x, y ) : x > , y < , x < y + ⎬ . 2 2 2⎭ ⎩ Range: The area is greater than zero but can be arbitrarily close to zero since one side can be arbitrarily small and the other two sides are bounded above. It seems that the area would be largest when the triangle is equilateral. An 1 equilateral triangle with sides equal to has 3 ⎛ 3⎤ 3 area . Hence the range of A is ⎜⎜ 0, ⎥. (In 36 ⎝ 36 ⎦ Sections 8 and 9 of this chapter methods will be presented which will make it easy to prove that the largest value of A will occur when the triangle is equilateral.) 46. a.

u = cos (x) cos (ct): u x = – sin(x )cos(ct ) ; ut = –c cos( x )sin(ct ) u xx = –cos( x) cos(ct ) utt = –c 2 cos(x ) cos(ct )

Therefore, c2 uxx = utt . u = e x cosh(ct ) : u x = e x cosh(ct ), u t = ce x sinh(ct ) u xx = e x cosh(ct ),u tt = c 2e x cosh(ct )

Therefore, c2 uxx = utt . Instructor’s Resource Manual

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b.

u = e –ct sin(x ) :

48. a.

sin( x + y 2 )

ux = e– ct cos x u xx = –e –ct sin x ut = –ce ct sin x Therefore, cu xx = u t . 2 u = t –1/ 2 e– x / 4ct : 2 / 4 ct ⎛

x ⎞ ⎜– ⎟ ⎝ 2 ct ⎠

ux = t –1/ 2 e– x

uxx = ut =

b.

D x (sin( x + y 2))

c.

D y (sin(x + y 2 ))

d.

D x ( D y (sin( x + y) 2))

( x 2 – 2ct ) (4 c2 t5 / 2 ex

2 / 4 ct

)

2

( x – 2ct ) 2 (4ct 5 / 2e x / 4ct )

Therefore, cu xx = u t . 47. a.

Moving parallel to the y-axis from the point (1, 1) to the nearest level curve and Δz approximating , we obtain Δy fy (1, 1) =

4–5 = –4. 1.25 –1

b. Moving parallel to the x-axis from the point (–4, 2) to the nearest level curve and Δz approximating , we obtain Δx 1– 0 2 fx (–4, 2) ≈ = . –2.5 – (–4) 3 c.

Moving parallel to the x-axis from the point (–5, –2) to the nearest level curve and Δz approximately , we obtain Δx 1– 0 2 fx (–4, – 5) ≈ = . – 2.5 – (–5) 5

d. Moving parallel to the y-axis from the point (0, –2) to the nearest level curve and Δz approximating , we obtain Δy fy (0, 2) ≈

fy ( x , y , z ) = lim

Δy →0

b.

Δz →0

c.

f ( x, y + Δy, z) − f ( x, y, z) Δy

fz ( x, y, z ) = lim

0 –1 –19 8

8 = . – (–2) 3

49. a.

f ( x, y, z + Δz) − f ( x, y, z) Δz

G x ( w, x, y, z ) G( w, x + Δx, y, z) − G( w, x, y, z) Δx →0 Δx