Example 2.6.1. Related Rate: Volume and Radius.
Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 \(\text{cm}^3/\text{s}.\) How fast is the radius of the balloon increasing when the diameter is 50cm?
Solution.
First, we need to look up a formula for the volume of a sphere. It is \(V=\frac{4}{3}\pi r^3.\) Next note that we are told the change in volume is 100. Change in volume means we know the derivative of \(V\) with respect to time. We also note that time is not an explicit variable (i.e., there is no variable \(t\)). Thus to use this formula we need to differentiate the it.
\begin{equation*}
V^\prime=\frac{4}{3}\pi 3r^2 r^\prime.
\end{equation*}
Note we need the \(r^\prime\) at the end because the radius is also a function of time; thus we use implicit differentiation. Finally we can plug in the values we were given. Note \(r=25\) cm because the diameter is 50 cm.
\begin{align*}
V^\prime = \amp \frac{4}{3}\pi 3r^2 r^\prime.\\
100 = \amp \frac{4}{3}\pi 3(25)^2 r^\prime.\\
\frac{100 \cdot 3}{4\pi 25^2} = \amp r^\prime.
\end{align*}
