Example 5.2.
Suppose the volume of a balloon increases as a function of time. The radius, being a function of the volume, will therefore increase at a different rate. Writing \(R = f(V)\) and \(V = g(t)\text{,}\) we have \(R = f(g(t))\text{.}\) Therefore by the chain rule,
\begin{equation*}
\frac{dR}{dt} = \frac{dR}{dV} \frac{dV}{dt} \, .
\end{equation*}
This notation hides where each derivative is evaluated but the meaning is clear. Letting primes denote time derivatives, \(R' = V' \cdot dR/dV\text{.}\)
The rate of increase of radius and the rate of increase of volume are therefore called related rates. Knowing one always gives you the other, provided you know the present volume and can compute \(dR/dV\text{.}\) For a spherical balloon, \(V = (4\pi/3) R^3\text{,}\) therefore \(R = \sqrt[3]{3V/(4\pi)} = (3/(4\pi))^{1/3} V^{1/3}\) and we can compute \(dR/dV = (1/3) \cdot (3/(4\pi))^{1/3} V^{-2/3}\text{.}\) In other words, if the present volume is \(V\text{,}\) then the rate the radius is growing in, say, cm/sec, is equal to \(\sqrt[3]{3/(4\pi)} / 3\) times the rate the volume is growing in cm\(^3\)/sec divided by the two thirds power of the volume.
