Example 4.2.2. 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 1.
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*}
This problem was relatively straightforward, owing to the linear relationship between radius and circumference. The video in Figure 4.2.3 explores what would happen if we had instead been asked for the rate at which the area is changing.