Question

Gravel is being dumped from a conveyor belt at a rate of 50 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 21 feet high? Recall that the volume of a right circular cone with height h and radius of the base r is given by V=1/3πr^2h.

Answer 1

Given
\[\frac{dV}{dt} = 50\]
\[h = 2r \rightarrow r = h/2\]
Find volume in terms of h
\[V = \frac{1}{3}\pi (h/2)^{2} h = \frac{1}{12}\pi h^{3}\]
Recall that:
\[\frac{dh}{dt} = \frac{dV}{dt}*\frac{dh}{dV}\]
Find dh/dV using implicit differentiation
\[\rightarrow 1 = \frac{1}{12}\pi (3h^{2})\frac{dh}{dV}\]
\[\frac{dh}{dV} = \frac{4}{\pi h^{2}}\]
Now we can find dh/dt
\[\frac{dh}{dt} = 50*\frac{4}{\pi h^{2}} = \frac{200}{\pi h^{2}}\]
Plug in 21 for h to find the rate height is changing at that moment