Question

Inflating a Balloon A spherical balloon is being inflated.

Find the rate of change of the surface area ($S=4\pi r^2$) with respect to the radius r when r = 2 ft.

Short answer

The rate of change is $16\pi {\text{ft}}^{\text{2}}\text{/sec}$.

Step-by-step solution

Step 1.  Given

The radius is r = 2 ft.

Step 2.  To determine

Find the rate of change of the surface area ($S=4\pi r^2$) with respect to the radius r when r = 2 ft.

Step 3.  Calculation

We differentiate both sides of $S=4\pi r^2$with respect to r.

$\begin{array}{l}S=4\pi r^2\\ \frac{dS}{dr}=\frac{d}{dr}\left(4\pi r^2\right)\\ \frac{dS}{dr}=4\pi \frac{d}{dr}\left(r^2\right)\\ \frac{dS}{dr}=4\pi \left(2r\right)\\ \frac{dS}{dr}=8\pi r\end{array}$

Given that $r=2$.

So, $\frac{dS}{dr}=8\pi r=8\pi \left(2\right)=16\pi$

The rate of change is $16\pi {\text{ft}}^{\text{2}}\text{/sec}$.