Answer
$\frac{5}{8 \pi}$ $ft $/min
Work Step by Step
Let volume of the conical pile=V
Radius of the conical pile=r
Height of the conical pile=h
$V=\frac{1}{3} \pi r^2 h$ ...................... eq (1)
According to the given condition
$r=\frac{h}{2}$
Putting in equation (1)
$V=\frac{1}{3} \pi (\frac{h}{2})^2 h=\frac{ \pi h^3}{12}$
Taking derivative with respect to t
$\frac{dV}{dt}=\frac{d( \frac{ \pi h^3}{12} )}{dt} =\frac{\pi}{12} \frac{d(h^3)}{dt}=\frac{\pi}{12} \frac{d(h^3)}{dh} \frac{dh}{dt} $
$\frac{dV}{dt}=\frac{\pi}{12} (3h^2)\frac{dh}{dt}= \frac{\pi h^2}{4} \frac{dh}{dt}$
$\frac{dV}{dt}= \frac{\pi h^2}{4} \frac{dh}{dt}$ ...................... eq (2)
Putting $\frac{dV}{dt}=10 ft^3/min$, $ h=8$ ft in equation (2)
$ 10 =\frac{\pi 8^2}{4} \frac{dh}{dt}= 16\pi \frac{dh}{dt} $
$\frac{dh}{dt}= \frac{10}{16\pi}=\frac{5}{8\pi}$ ft/min
