Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 9 PROBLEMS PLUS ¤ 409
(b) The floor area of the silo is F = π · 200 2 =40,000π ft 2 , and the area of the base of the pile is
A = πr 2 = π 3
2 h2 = 9π 4 h2 .Sotheareaofthefloor which is not covered when h =60is
F − A =40,000π − 8100π =31,900π ≈ 100,217 ft 2 .NowA = 9π 4 h2 ⇒ dA/dt = 9π 4
· 2h (dh/dt),
and from () in part (a) we know that when h =60, dh/dt = 80,000 = 200
3(60) 2 27
ft/h. Therefore,
dA/dt = 9π (2)(60)
200
4 27 =2000π ≈ 6283 ft 2 /h.
(c) At h =90ft, dV/dt =60,000π − 20,000π =40,000π ft 3 /h. From () inpart(a),
dh
dt = 4(dV/dt) = 4(40,000π) = 160,000 ⇒ 9h 2 dh = 160,000 dt ⇒ 3h 3 = 160,000t + C. Whent =0,
9πh 2 9πh 2 9h 2
h =90; therefore, C =3· 729,000 = 2,187,000. So3h 3 =160,000t +2,187,000. At the top, h = 100
⇒
3(100) 3 =160,000t +2,187,000 ⇒ t = 813,000 ≈ 5.1. The pile reaches the top after about 5.1 h.
160,000
13. Let P (a, b) be any point on the curve. If m is the slope of the tangent line at P ,thenm = y 0 (a), and an equation of the
normal line at P is y − b = − 1 m (x − a), or equivalently, y = − 1 m x + b + a .They-intercept is always 6,so
m
b + a m =6 ⇒ a m =6− b ⇒ m = a
dy
. We will solve the equivalent differential equation
6 − b dx = x
6 − y
(6 − y) dy = xdx ⇒ (6 − y) dy = xdx ⇒ 6y − 1 2 y2 = 1 2 x2 + C ⇒ 12y − y 2 = x 2 + K.
⇒
Since (3, 2) is on the curve, 12(2) − 2 2 =3 2 + K ⇒ K =11.Sothecurveisgivenby12y − y 2 = x 2 +11 ⇒
x 2 + y 2 − 12y +36=−11 + 36 ⇒ x 2 +(y − 6) 2 =25, a circle with center (0, 6) and radius 5.