Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 15

The equation $s = 4 \pi r^2$ represents the surface area of a spherical balloon that is being inflated. Find the rate of increase with respect to the radius $r$ when $r$ is (a) 1 ft, (b) 2 ft, and (c) 3ft. What conclusion can you make?


$
\begin{equation}
\begin{aligned}
\frac{ds}{dr} &= 4 \pi \frac{d}{dr}(r^2)\\
\\
\frac{ds}{dr} &= 4 \pi (2r)\\
\\
\frac{ds}{dr} &= 8 \pi r
\end{aligned}
\end{equation}
$

a.) when $r = 1$ft,
$\displaystyle \frac{ds}{dt} = 8 \pi (1) = 8\pi \frac{\text{area}}{\text{ft}}$

b.) when $r = 2$ft,
$\displaystyle \frac{ds}{dt} = 16 \pi (2) = 16\pi \frac{\text{area}}{\text{ft}}$

c.) when $r =3$ft,
$\displaystyle \frac{ds}{dt} = 8 \pi (3) = 24\pi \frac{\text{area}}{\text{ft}}$

We can conclude that the rate of the surface area increases as the radius increases.