A golfer hits a ball into a pond of still water, causing a circular ripple as shown in Figure 2. If the radius of the circle increases at the constant rate of $0.5 \,{\rm m}/{\rm s}$, how fast is the area of the circle increasing when the radius of the ripple is $2 \,{\rm m}$?

Solution The quantities that are changing, that is, the variables of the problem, are $$\begin{eqnarray*} t &=& \hbox{the time (in seconds) elapsed from the time when the ball hits the water} \\ r &=& \hbox{the radius (in meters) of the ripple after } t \hbox{ seconds} \\ A &=& \hbox{the area (in square meters) of the circle formed by the ripple after } t \hbox{ seconds} \end{eqnarray*}$$

The rates of change with respect to time are $$\begin{eqnarray*} \dfrac{\textit{dr}}{\textit{dt}} &=& \hbox{the rate (in meters per second) at which the radius of the ripple is increasing} \\[5pt] \dfrac{\textit{dA}}{\textit{dt}} &=& \hbox{the rate (in meters squared per second) at which the area of the circle is}\\[-5pt] && \hbox{increasing} \end{eqnarray*}$$

It is given that $\dfrac{\textit{dr}}{\textit{dt}}=0.5 \,{\rm m}/{\rm s}$. We seek $\dfrac{\textit{dA}}{\textit{dt}}$ when $r=2 \,{\rm m}$.

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The relationship between $A$ and $r$ is given by the formula for the area of a circle: $$A=\pi r^{2}$$

Since $A$ and $r$ are functions of $t$, we differentiate with respect to $t$ to obtain $$\frac{\textit{dA}}{\textit{dt}}=2\pi r\frac{\textit{dr}}{\textit{dt}}$$

Since $\dfrac{\textit{dr}}{\textit{dt}}=0.5 \,{\rm m}/{\rm s}$, $$\dfrac{ \textit{dA}}{\textit{dt}}=2\pi r( 0.5) =\pi r$$

When $r=2$m, $$\frac{\textit{dA}}{\textit{dt}}=\pi (2)=2 \pi$$

The area of the circle is increasing at a rate of about $6.283~{\rm m}^{2}/{\rm s}$.