As sand falls from the conveyor belt, several features of the sand pile will change: the volume of the pile will grow, the height will increase, and the radius will get bigger, too. All of these quantities are related to one another, and the rate at which each is changing is related to the rate at which sand falls from the conveyor.
We begin by identifying which variables are changing and how they are related. In this situation, we observe that the radius and height of the pile are related to its volume by the standard equation for the volume of a cone,
\begin{equation*}
V = \frac{1}{3} \pi r^2 h\text{.}
\end{equation*}
Viewing each of \(V\text{,}\) \(r\text{,}\) and \(h\) as functions of \(t\text{,}\) we differentiate implicitly to arrive at an equation that relates their respective rates of change. Taking the derivative of each side of the equation with respect to \(t\text{,}\) we find
\begin{equation*}
\frac{d}{dt}[V] = \frac{d}{dt}\left[\frac{1}{3} \pi r^2 h\right]\text{.}
\end{equation*}
On the left, \(\frac{d}{dt}[V]\) is simply \(\frac{dV}{dt}\text{.}\) On the right, the situation is more complicated, as both \(r\) and \(h\) are implicit functions of \(t\text{.}\) Hence we need the product and chain rules. We find that
\begin{align*}
\frac{dV}{dt} \amp= \frac{d}{dt}\left[\frac{1}{3} \pi r^2 h\right]\\
\amp= \frac{1}{3} \pi r^2 \frac{d}{dt}[h] + \frac{1}{3} \pi h \frac{d}{dt}[r^2]\\
\amp= \frac{1}{3} \pi r^2 \frac{dh}{dt} + \frac{1}{3} \pi h 2r \frac{dr}{dt}
\end{align*}
(Note particularly how we are using ideas from
Section 2.7 on implicit differentiation. There we found that when
\(y\) is an implicit function of
\(x\text{,}\) \(\frac{d}{dx}[y^2] = 2y \frac{dy}{dx}\text{.}\) The same ideas are applied here when we compute
\(\frac{d}{dt}[r^2] = 2r \frac{dr}{dt}\text{.}\))
The equation
\begin{equation*}
\frac{dV}{dt} = \frac{1}{3} \pi r^2 \frac{dh}{dt} + \frac{2}{3} \pi rh \frac{dr}{dt}\text{,}
\end{equation*}
relates the rates of change of \(V\text{,}\) \(h\text{,}\) and \(r\text{.}\)
