Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute.

Question

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 25 feet high? Recall that the volume of a right circular cone with height h and radius of the base r is given by

Answer 1

We are given dV/dt=10 cubic ft/min and we want to find dh/dt. Therefore, first we need to write a formula relating v and h.

V=pi*r^2*h/3 since d=h at any moment, we know that r=h/2 at any moment and if we substitute h/2 for r, we get

V=pi(h/2)^2*h/3 and if we simplify this, we get

V=pi*(h^3)/12

Now we need to take the derivative of both sides with respect to t

dV/dt=pi*(3*h^2)*/12 *dh/dt=pi*h^2/4*dh/dt, substitute 10 for dV/dt and 25 for h and solve for dh/dt and we get

10=pi^625/4*dh/dt ==> dh/dt=10*4/(625*pi)=8/(125*pi) ft/min,

that is, the height of the pile is increasing 8/(pi*125) ft/min when the height is 25 ft.