## Calculus (3rd Edition)

$9600\pi\,cm^{2}/min$
We know that the surface area of a sphere is given by: $S=4\pi r^2$ We can find the rate of change of the surface area with respect to time as: $\frac{dS}{dt}=\frac{dS}{dr}\times\frac{dr}{dt}=\frac{d}{dr}(4\pi r^{2})\times30\,cm/min$ $=8\pi r\times 30\,cm/min$ Thus, when $r=40\,cm$, $\frac{dS}{dt}=8\pi(40\,cm)\times30\,cm/min$ $=9600\pi\,cm^{2}/min$