Question

Gravel is being dumped from a conveyor belt at a rate of 30 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter is alwaysequal to its height. How fast is the height of the pile increasing when the pile is 18 feet high? Recall that the volume of a right circular cone with height h andradius of the base r is given by V=1/3*pi*r^2*h

What isr requested rate of increase of height of pile per feet/minute?

Answer 1

Start of with realizing what is given in the problem
Given:
1. rate of 30 cubic ft/m
2. d = h

30 ft/m = dV/dt (meaning that the volume is changing with respect to time)

Since h=d, and we know that D=2r, we can substitute this equation in for r in the volume formula

V=1/3π(h/2)²h

V=1/3π(h³/4)

V=π/12(h³)

Differentiate:

dV/dt = (3π/12)(h²)dh/dt

30=(π/4)(10²)dh/dt

30=25π(dh/dt)

dh/dt = 30/25π = 6/5π ft/min

Final answer: (6/5π) ft³/minute