Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 3.9 RELATED RATES ¤ 121
23. If C = the rate at which water is pumped in, then dV
dt = C − 10,000,where
V = 1 3 πr2 h is the volume at time t. By similar triangles, r 2 = h 6
⇒ r = 1 3 h ⇒
V = 1 π 1
3 3 h2 h = π 27 h3 ⇒ dV
dt = π dh
9 h2 .Whenh =200cm,
dt
dh
dt =20cm/min,soC − 10,000 = π 9 (200)2 (20) ⇒ C =10,000 + 800,000 π ≈ 289,253 cm 3 /min.
9
25. The figure is labeled in meters. The area A of a trapezoid is
1
(base1 + base2)(height), and the volume V of the 10-meter-long trough is 10A.
2
Thus, the volume of the trapezoid with height h is V =(10) 1 2 [0.3+(0.3+2a)]h.
By similar triangles, a h = 0.25
0.5 = 1 2 ,so2a = h ⇒ V =5(0.6+h)h =3h +5h2 .
Now dV
dt = dV dh
dh dt
⇒
0.2 =(3+10h) dh
dt
dh
dt = 0.2
3 + 10(0.3) = 0.2
6 m/min = 1 10
m/min or
30 3 cm/min.
⇒
dh
dt = 0.2 .Whenh =0.3,
3+10h
27. We are given that dV
dt =30ft3 /min. V = 1 3 πr2 h = 1 2 h
3 π h = πh3
2 12
⇒
dV
dt = dV dh
dh dt
⇒
30 = πh2
4
dh
dt
⇒
dh
dt = 120
πh . 2
When h =10ft, dh
dt = 120
10 2 π = 6 ≈ 0.38 ft/min.
5π
29. A = 1 2 bh,butb =5mandsin θ = h 4
⇒
h =4sinθ,soA = 1 (5)(4 sin θ) =10sinθ.
2
We are given dθ
dA
=0.06 rad/s, so
dt
When θ = π 3 , dA
dt =0.6 cos π 3
dθ
=(10cosθ)(0.06) = 0.6cosθ.
dt
dt = dA
dθ
=(0.6)
1
2 =0.3 m 2 /s.
31. Differentiating both sides of PV = C with respect to t and using the Product Rule gives us P dV
dt + V dP dt =0
dV
dt = −V P
dP
dP
.WhenV = 600, P = 150 and
dt dt
decreasing at a rate of 80 cm 3 /min.
33. With R 1 =80and R 2 = 100,
dV
=20,sowehave
dt = − 600 (20) = −80. Thus, the volume is
150
1
R = 1 + 1 = 1 R 1 R 2 80 + 1
100 = 180
8000 = 9 400
,soR =
400
with respect to t,wehave− 1 dR
R 2 dt = − 1 dR 1
− 1 dR 2
R1
2 dt R2
2 dt
R 2 = 100, dR
dt = 4002 1
9 2 80 (0.3) + 1
2 100 (0.2) = 107 ≈ 0.132 Ω/s.
2 810
⇒
dR 1
dt = dR 1
R2 + 1 R1
2 dt R2
2
dR 2
dt
⇒
9 . Differentiating 1 R = 1 + 1 R 1 R 2
.WhenR 1 =80and