1. ## Cone Word Problem

Gravel is being dumped from a conveyor belt at a rate of 40 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 20 feet high?
volume of cone:
V=(1/3)(pi)(r^2)(h)

2. Originally Posted by biermann33
Gravel is being dumped from a conveyor belt at a rate of 40 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 20 feet high?
volume of cone:
V=(1/3)(pi)(r^2)(h)
not possible if the cone is increasing in size ... recheck the problem statement again.

3. Originally Posted by biermann33
Gravel is being dumped from a conveyor belt at a rate of 40 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 20 feet high?
volume of cone:
V=(1/3)(pi)(r^2)(h)
Did you mean that the diameter & the height would always be equal as the pile rises?
It is unlikely that you'd ever get any material to achieve that angle of repose (approx. 63 degrees).

If that is the case then the volume is

V = $\displaystyle \dfrac{2 \pi r^3}{3}$

You could think of this as as inverted cone, in which the diameter of the cone is equal to the depth. If the cone were being filled with water (with axis of the cone being plumb), what is the area of the surface when the depth is 20 feet. ($\displaystyle \pi 10^2 = 314.16$ Sq.Ft.)
You know that 40 cu.ft are being deposited each minute.

$\displaystyle \dfrac{314.16}{40}=0.13$
That gives you a rough idea of the value you are seeking.

A little differentiation will give you the exact value.