(C)
Step-by-step explanation:
The volume of the conical pile is given by
[tex]V = \dfrac{\pi}{3}r^2h[/tex]
Taking the derivative of V with respect to time, we get
[tex]\dfrac{dV}{dt} = \dfrac{\pi}{3}\dfrac{d}{dt}(r^2h)[/tex]
[tex]\:\:\:\:\:\:\:= \dfrac{\pi}{3}\left(2rh\dfrac{dr}{dt} + r^2\dfrac{dh}{dt}\right)[/tex]
Since r is always equal to h, we can set
[tex]\dfrac{dr}{dt} = \dfrac{dh}{dt}[/tex]
so that our expression for dV/dt becomes
[tex]\dfrac{dV}{dt} = \dfrac{\pi}{3}\left(3r^2\dfrac{dh}{dt}\right)[/tex]
[tex]\:\:\:\:\:\:\:= \pi r^2\dfrac{dh}{dt}[/tex]
Solving for dh/dt, we get
[tex]\dfrac{dh}{dt} = \dfrac{1}{\pi r^2}\dfrac{dV}{dt}[/tex]
[tex]\:\:\:\:\:\:\:= \dfrac{1}{9\pi\:\text{m}^2}(36\:\text{m}^3\text{/s})[/tex]
[tex]\:\:\:\:\:\:\:= \dfrac{4}{\pi}\:\text{m/s}[/tex]
