Solving a Related Rate Problem
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 \(3 \,{\rm m}\).
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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Geometry formulas\(\!\!\!\!\!\!\!\!\) are discussed in Appendix A.2, p. A-15.
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 \(6.667 \,{\rm{m}}^2/{\rm min} \).