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 21 feet high?

dx/dt = 0,029 ft

Step-by-step explanation:

Let call "x " diameter of the base of the cone then h (height) is equal to x

V = (1/3) * A (b) * h

A (b) = π * (x/2) ²

Then

Volume of the cone is

V = (1/3) * π * (x/2) ²*x

V = (1/12) * π*x³

Then:

dV/dt = (1/12) * π * 3*x²*dx/dt ⇒ dV/dt = (1/4) * π*x²*dx/dt

dV/dt = (1/4) * π * (21) ²*dx/dt

As dV/dt = 10 ft³

10*4 = 441*π * dx/dt ⇒ dx/dt = 40 / 1384,74

dx/dt = 0,029 ft