Solving a Related Rate Problem

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 \(6.283~{\rm m}^{2}/{\rm s}\).