A spherical balloon is inflated at the rate of $10~{\rm{m}}^3/{\rm min}$. Find the rate at which the surface area of the balloon is increasing when the radius of the sphere is $3 \,{\rm m}$.

Figure 3

Solution We follow the steps for solving a related rate problem.

Step 1 Figure 3 shows a sketch of the balloon with its radius labeled.

Step 2 Identify the variables of the problem: $$\begin{eqnarray*} t &=& \hbox{the time (in minutes) measured from the moment the balloon begins} \\[-3pt] && \hbox{inflating} \\ R &=& \hbox{the radius (in meters) of the balloon at time } t \\ V &=& \hbox{the volume (in meters cubed) of the balloon at time } t \\ S &=& \hbox{the surface area (in meters squared) of the balloon at time } t \end{eqnarray*}$$

Identify the rates of change: $$\begin{eqnarray*} \dfrac{\textit{dR}}{\textit{dt}} &=& \hbox{the rate of change of the radius of the balloon (in meters per minute)} \\[6pt] \dfrac{\textit{dV}}{\textit{dt}} &=& \hbox{the rate of change of the volume of the balloon (in meters cubed per}\\[-5pt] && \hbox{minute)} \\[6pt] \dfrac{\textit{dS}}{\textit{dt}} &=& \hbox{the rate of change of the surface area of the balloon (in meters squared}\\[-5pt] && \hbox{per minute)} \end{eqnarray*}$$

We are given $\dfrac{\textit{dV}}{\textit{dt}}=10~{\rm{m}}^3 /{\rm min}$, and we seek $\dfrac{\textit{dS}}{\textit{dt}}$ when $R=3 \,{\rm m}$.

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Step 3 Since both the volume $V$ of the balloon (a sphere) and its surface area $S$ can be expressed in terms of the radius $R,$ we use two equations to relate the variables. $$V=\frac{4}{3}\pi\! R^{3}\quad\hbox{ and }\quad S=4\pi\! R^{2} \qquad \hbox{ where } V, S,\hbox{ and }R\hbox{ are functions of }t$$

Step 4 Differentiate both sides of the equations with respect to time $t.$ $$\frac{\textit{dV}}{\textit{dt}}=4\pi\! R^{2} \,\frac{\textit{dR}}{\textit{dt}}\qquad \hbox{ and }\qquad \frac{ \textit{dS}}{ \textit{dt}}=8 \pi\! R \,\frac{\textit{dR}}{\textit{dt}}$$

Combine the equations by solving for $\dfrac{\textit{dR}}{\textit{dt}}$ in the equation on the left and substituting the result into the equation for $\dfrac{\textit{dS}}{\textit{dt}}$ on the right. Then $$\frac{ \textit{dS}}{ \textit{dt}}=8\pi\! R\!\left(\!\dfrac{\dfrac{\textit{dV}}{\textit{dt}}}{{4\pi\! R^{2}}}\!\right) = \frac{2}{R}\frac{\textit{dV}}{\textit{dt}}$$

Step 5 Substitute $R=3 \,{\rm m}$ and $\dfrac{\textit{dV}}{\textit{dt}}=10 \,{\rm{m}}^3 /{\rm min}$. $$\frac{ \textit{dS}}{ \textit{dt}}=\left( \frac{2}{3}\right) (10)\approx 6.667$$

When the radius of the balloon is $3 \,{\rm m}$, its surface area is increasing at the rate of about $6.667 \,{\rm{m}}^2/{\rm min}$.