Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 27

At what rate is the height of the pile increasing when the pile is 10ft high?
"Please refer to the image in the textbook."
Recall that the volume of the cone is $\displaystyle V = \frac{1}{3} \pi r^2 h$
It is known that $d = h = 2r $ so we have $\displaystyle r = \frac{h}{2}$ then,

$
\begin{equation}
\begin{aligned}
V & = \frac{1}{3} \pi \left( \frac{h}{2} \right)^2 h\\
\\
V &= \frac{\pi}{12} h^3
\end{aligned}
\end{equation}
$

Taking the derivative with respect to time we get,

$
\begin{equation}
\begin{aligned}
\frac{dV}{dt} &= \frac{\pi}{12} \frac{d}{dh} (h^3) \frac{dh}{dt}\\
\\
\frac{dV}{dt} &= \frac{\pi}{12} (3h^2) \frac{dh}{dt}\\
\\
\frac{dV}{dt} &= \frac{\pi}{4} h^2 \frac{dh}{dt}\\
\\
\frac{dh}{dt} &= \frac{4 \frac{dV}{dt}}{\pi h^2}\\
\\
\frac{dh}{dt} &= \frac{4(30)}{\pi(10)^2}\\
\\
\frac{dh}{dt} &= 0.3820 \frac{\text{ft}}{\text{min}}
\end{aligned}
\end{equation}
$