Solução_Calculo_Stewart_6e

F.
TX.10
31. By similar triangles, r 5 = h 16
⇒
CHAPTER 3 PROBLEMS PLUS ¤ 145
r = 5h . The volume of the cone is
16
2 5h
V = 1 3 πr2 h = 1 π h = 25π
3
16 768 h3 ,so dV
dt = 25π dh
256 h2 dt .Nowtherateof
change of the volume is also equal to the difference of what is being added
(2 cm 3 /min) and what is oozing out (kπrl,whereπrl is the area of the cone and k
Equating the two expressions for dV
dt
is a proportionality constant). Thus, dV
dt
and substituting h =10,
dh
dt
=2− kπrl.
= −0.3, r =
5(10)
16
= 25 8 ,and l
√
281
= 10
16
⇔
l = 5 8
√ 25π
281,weget
256 (10)2 (−0.3) = 2 − kπ 25
8 · 5 √ 125kπ √ 281
281 ⇔
8
64
=2+ 750π . Solving for k gives us
256
256 + 375π
k =
250π √ . To maintain a certain height, the rate of oozing, kπrl, must equal the rate of the liquid being poured in;
281
that is, dV =0. Thus, the rate at which we should pour the liquid into the container is
dt
kπrl =
256 + 375π
250π √ 281 · π · 25 8 · 5 √ 281 256 + 375π
= ≈ 11.204 cm 3 /min
8 128