Cant figure out how to solve need little help

Question

Gravel is being dumped from a conveyor belt at a rate of 40 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 19 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 the rate of change of the volume, dV/dt of 40 cubic feet per minute and we want to find the rate of change for the height, dh/dt.

We know that diameter is equal to height at any given moment (d=h) in the same manner since the relationship between the radius and diameter is 2r = d, we have 2r = h or r=(h/2).

We can use our equation given for the volume to relate v to h

V = (pi/3) *(r^2)*h = (pi/3)*((h/2)^2)*h = (pi/3) * ((h^2)/4) * h = (pi/3)*((h^3)/4)= (pi*(h^3))/12

Now we need to take the derivative of both sides with respect to t, remember since the height is changing with time we have the form dh/dt in the derivative

dV/dt=((pi*3*(h^2))/12)*dh/dt=((pi*(h^2))/4)*dh/dt, substitute 40 for dV/dt and 19 for h and solve for dh/dt and we get

40=(pi/4)*(19^2)*dh/dt --> 40 = (pi/4)*361 * (dh/dt)

multiply both sides by 4 --> 160 = (361*pi) *(dh/dt)

divide both sides by 361*pi --> 160/(361*pi) = dh/dt

therefore dh/dt= 160/(361*pi) ft/min,

the height of the pile is therefore increasing at a rate of 160/(361*pi) ft/min at the height of 19 ft.