Example 4.2.1. Understanding related rates.
The radius of a circle is growing at a rate of 5 in⁄h. At what rate is the circumference growing?
Solution.
The circumference and radius of a circle are related by \(C = 2\pi r\text{.}\) We are given information about how the length of \(r\) changes with respect to time; that is, we are told \(\lz{r}{t}\) is 5 in⁄h. We want to know how the length of \(C\) changes with respect to time, i.e., we want to know \(\lz{C}{t}\text{.}\)
Implicitly differentiate both sides of \(C = 2\pi r\) with respect to \(t\text{:}\)
\begin{align*}
C \amp = 2\pi r\\
\lzoo{t}{C} \amp = \lzoo{t}{2\pi r}\\
\lz{C}{t} \amp =2\pi \lz{r}{t}\text{.}
\end{align*}
As we know \(\lz{r}{t}\) is 5 in⁄h, we know
\begin{equation*}
\lz{C}{t} = 2\pi 5 = 10\pi \approx 31.4\text{ in/hr }\text{.}
\end{equation*}