Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 6 REVIEW ¤ 289
29. (a) The parabola has equation y = ax 2 with vertex at the origin and passing through
(4, 4). 4=a · 4 2 ⇒ a = 1 4
⇒ y = 1 4 x2 ⇒ x 2 =4y ⇒
x =2 y. Each circular disk has radius 2 y and is moved 4 − y ft.
W = 4
2 π 2 4
y
0 62.5(4 − y) dy = 250π y(4 − y) dy
0
= 250π 2y 2 − 1 y3 4
=250π
3
32 − 64
0 3 =
8000π
3
≈ 8378 ft-lb
(b) In part (a) we knew the final water level (0) but not the amount of work done. Here
we use the same equation, except with the work fixed, and the lower limit of
integration (that is, the final water level — call it h) unknown: W = 4000
250π 2y 2 − 1 y3 4
16
3
=4000 ⇔ =
h π
32 − 64 3 − 2h 2 − 1 h3 3
⇔
⇔
h 3 − 6h 2 +32− 48 =0. We graph the function f(h) π =h3 − 6h 2 +32− 48
π
on the interval [0, 4] to see where it is 0. From the graph, f(h) =0for h ≈ 2.1.
So the depth of water remaining is about 2.1 ft.
31. lim
h→0
f ave =lim
h→0
1
(x + h) − x
x+h
x
F (x + h) − F (x)
f(t) dt =lim
,whereF (x) = x
f(t) dt. But we recognize this
h→0 h
a
limit as being F 0 (x) by the definition of a derivative. Therefore, lim
h→0
f ave = F 0 (x) =f(x) by FTC1.