How fast is the height of the pile increasing when the pile is 22 feet high?

Question

Gravel is being dumped from a conveyor belt at a rate of 50 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 22 feet high?

Answer 1

Since the base diameter and height of the cone are the same, we know that d = 2r = h and so r = h/2.

From geometry we know that the volume of a cone is given by V = 1/3 π r²h. Since r = h/2 for our specific cone, we have that V = 1/3 π (h/2)²h = 1/12 π h³.

Assuming that V and h change with respect to time t, we can use implicit differentiation with respect to t to find that

dV/dt = 1/12 π * 3h² * dh/dt = 1/4 π h² * dh/dt

Solving for dh/dt, we get dh/dt = (4 dV/dt) / (π h²) .

Substituting the values dV/dt = 50 ft³/min and h = 22 ft, we find that

dh/dt = (4*50 ft³/min) / (π (22 ft)²) = 50/(121 π) ft/min ≈ 0.13 ft/min.