Sand is falling from a conveyor belt at a rate of 12ft^3/min into a conical pile. Diameter of the pile is 4x the height. At what rate is the height of the pile changing when the pile is 10 feet high?

Question

Sand is falling from a conveyor belt at a rate of 12ft^3/min into a conical pile. The diameter of the pile is four times the height. The volume of the cone is V=1/3pir^2h. At what rate is the height of the pile changing when the pile is 10 ft high?

Answer 1

First write down the formula for the volume of a cone V=1/3pir²h. Note that it is a given that D=4h (r=2h).

Second substitute 2h for r in yielding V=4/3pih³

Third take the derivative of V to get dV/dt=4pih²dh/dt note that dV/dt is given at 12f³/min

Fourth: Rearrange to get dh/dt=(1/4ph²)dV/dt

Fifth substitute h=10 and dV/dt=12 to get dh/dt= 3/100pi