Answer
$\frac{8}{9\pi} $ $ft/min$
Work Step by Step
Let $r$ be the radius and $h$ be the height of the conical pile.
According to the given condition
$h=2r$ ................ eq(1).
Let $V$ be the volume of the conical pile. Then
$V= \frac {1}{3} \pi r^2 h$ .................... eq (2)
From equation (1)
$r=\frac{h}{2} $
Putting in equation (2)
$V=\frac {1}{3} \pi (\frac{h}{2}) ^2 h=\frac {\pi h^3}{12} $
Taking derivative with respect to t
$ \frac{dV}{dt}=\frac{d ( \frac{\pi h^3}{12})}{dt}= \frac{ \pi}{12} \frac{dh^3}{dt}=\frac{\pi}{12} \frac{dh^3}{dh} \frac{dh}{dt}=\frac{3\pi h^2}{12} \frac{dh}{dt}=\frac{\pi h^2}{4} \frac{dh}{dt}$ ............. eq(3)
Putting $h=6$ ft, $ \frac{dV}{dt}=8 $ ft/min in equation (3)
$ 8=\frac{\pi\times 6^2}{4} \frac{dh}{dt}=9\pi \frac{dh}{dt}$ Or
$\frac{dh}{dt}=\frac{8}{9\pi}$ $ft/min$
