Air is pumped into a spherical balloon such that the radius of the balloon increases at the rate of $\dfrac{1}{20}\pi$ cm/s when the radius is $8.5 \text{cm}$. Find the rate of change of the surface area of the balloon at this instant.
How do I do this? I still don't really understand how rate of change works here.
The surface area $S$ is related to the radius $r$ via
$S = 4\pi r^2; \tag 1$
thus, $\dot S$ is related to $\dot r$ via
$\dot S = 8\pi r \dot r; \tag 2$
when
$r = 8.5 \; \text{cm.} \tag 3$
and
$\dot r = \dfrac{1}{20 \pi} \; \text{cm./sec.}, \tag 4$
we have
$\dot S = 8 \pi (8.5) \dfrac{1}{20\pi} = \dfrac{68}{20} \; \text{cm.^2/sec.} = \dfrac{17}{5} \; \text{cm.^2/sec.}. \tag 5$
