"a rate of 10 cubic feet per minute" ---> dV/dt = 10 ft^3/min
base diameter and height are always equal ---> 2r = h
r = h/2
V = (1/3)πr^2 h
= (1/3)π (h^2/4)(h) = (1/12)π h^3
dV/dt = (1/4)π h^2 dh/dt
when h = 16
10 = (1/4)π(256) dh/dt
dh/dt = 40/(256π) = 5π/32 ft/min
Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 16 feet high? Recall that the volume of a right circular cone with height h and base radius r is given by V=1/3ðr^2h.
ft/min
Note: See number 25 on pg 268 for a picture of this.
1 answer