Answer:
[tex]\frac{2}{9\pi}ft^3/min[/tex]
Step-by-step explanation:
Given
[tex]\frac{dV}{dt}=2ft^3/min\\ \\\frac{dh}{dt}=?\\ \\V_{cone}=\frac{1}{3}\pi r^2h\\\\d=2r=3h\\\\\frac{d}{2}=r=\frac{3}{2}h\\\\h=2[/tex]
Calculation
[tex]V=\frac{1}{3}\pi r^2h\\ \\V=\frac{1}{3}\pi(\frac{3}{2}h) ^2h\\\\V=\frac{1}{3}\pi(\frac{9}{4}h^2)h\\ \\V=\frac{9}{12}\pi h^3\\ \\V=\frac{3\pi}{4}h^3\\ \\\frac{dV}{dt}=\frac{9\pi}{4}h^2\frac{dh}{dt}\\ \\2=\frac{9\pi}{4}(2)^2\frac{dh}{dt}\\ \\ 2=9\pi\frac{dh}{dt}\\ \\ \frac{2}{9\pi} =\frac{dh}{dt}[/tex]
Therefore, the rate that the height of the pile is changing when the pile is 2 feet high is [tex]\frac{2}{9\pi}ft^3/min[/tex].
