Thomas Calculus 13th [Solutions]

188 Chapter 3 Derivatives
2 2 2 2 2 2 2
dy
18. (a) s x y z s x y z 2s ds 2x dx 2y 2z
dz
dt dt dt dt
ds x dx y dy z dz
dt 2 2 2 dt 2 2 2 dt 2 2 2
x y z x y z x y z
dt
(b) From part (a) with dx 0 ds y dy z dz
dt dt 2 2 2 dt 2 2 2
x y z x y z
dt
(c) From part (a) with ds 0 0 2 dx 2 dy
y dy
x y 2z
dz dx z dz 0
dt dt dt dt dt x dt x dt
19. (a) A 1 sin dA 1 cos d
2 ab dt 2
ab dt
(b) A 1 absin dA 1 ab cos d 1 b sin da
2 dt 2 dt 2 dt
(c) A 1 absin
dA 1 abcos d 1 bsin da 1 a sin db
2
dt 2 dt 2 dt 2 dt
20. Given
2
A r , dr 0.01 cm/sec, and r 50 cm. Since dA 2 r dr , then dA
dt
dt dt
r 50 2 (50)
dt
1
100
2
cm /min.
21. Given d 2 cm/sec, dw 2 cm/sec, 12 cm and w 5 cm.
dt
dt
(a) A w dA dw w d dA
dt dt dt dt
12(2) 5( 2) 2
14 cm /sec, increasing
(b) P 2 2w dP 2 d 2 dw 2( 2) 2(2) 0 cm/sec, constant
dt dt dt
2 2 2 2 1/2
1/2
(c) D w ( w ) dD 1 2 2
dt 2 w 2 w dw 2 d dD
dt dt dt
14 cm/sec, decreasing
13
w
dw d
dt dt
2 2
w
(5)(2) (12)( 2)
25 144
22. (a) V xyz dV dx dy dz
dt
yz dt xz dt xy dV
dt dt (4, 3, 2)
3
(3)(2)(1) (4)(2)( 2) (4)(3)(1) 2 m /sec
(b) 2 2 2 dS
dy
S xy xz yz (2y 2 z) dx (2x 2 z) (2x 2 y)
dz
dt
dt dt dt
dS
2
(10)(1) (12)( 2) (14)(1) 0 m /sec
dt (4, 3, 2)
2 2 2 2 2 2 1/2
(c) x y z ( x y z ) d x dx y dy z dz
dt 2 2 2
x y z
dt 2 2 2
x y z
dt 2 2 2
x y z
dt
d
4 3
2
dt (4, 3, 2) (1) ( 2) (1) 0 m/sec
29 29 29
23. Given: dx 5 ft/sec, the ladder is 13 ft long, and x 12, y 5 at the instant of time
dt
2 2
(a) Since x y dy
169 x dx 12 (5) 12 ft/sec,
dt y dt 5
the ladder is sliding down the wall
(b) The area of the triangle formed by the ladder and walls is 1 1 dy
A xy dA x y dx . The area is
2 dt 2 dt dt
changing at 1 2 [12( 12) 5(5)] 119
2
59.5 ft /sec.
(c)
cos
x
13
sin
d
dt
2
1 dx d 1 dx
13 dt dt 13sin dt
1
5
(5) 1 rad / sec
24.
2 2 2
s y x 2s ds 2x dx 2y dy ds 1 dy
x dx y ds 1 [5( 442) 12( 481)] 614 knots
dt dt dt dt s dt dt dt 169
25. Let s represent the distance between the girl and the kite and x represents the horizontal distance between the
2 2 2
girl and kite s (300) x ds x dx 400(25)
20 ft/sec.
dt s dt 500
26. When the diameter is 3.8 in., the radius is 1.9 in. and dr 1
2
in/min. Also V 6 r dV 12
dt 3000 dt
dV 12 (1.9) 1 0.0076 . The volume is changing at about 0.0239 in 3 /min.
dt
3000
r dr
dt
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