Question

Sand falls from a conveyor belt onto a conical pile at a rate of 6ft/mi^3. The radius of the base is always equal to two-thirds of the pile’s height. At what rate is the height of the pile changing when the radius of the base of the pile is 2 ft?

I honestly have no idea how to do this. Please help.

Answer 1

sand falls at the rate of 6 ft/mi^3???

Do you mean 6 ft^3/min ?

r = 2h/3, so h = 3r/2
so, when r=2, h=3

v = 1/3 πr^2 h
= 1/3 π * (2/3 h)^2 * h
= 4/27 πh^3

dv/dt = 4/9 πh^2 dh/dt

You know that dv/dt = 6, so

4/9 π * 9 dh/dt = 6
dh/dt = 6/(4π) = 3/(2π) ft/min

Answer 2

Sweet! I managed to get as far as dv/dt = 4/9 πh^2 dh/dt , but couldn't figure it out from there, thanks!