A spherical balloon is inflated so that its volume grows at $36\pi$ cm$^3$/s. When the radius is 3 cm, how fast is the balloon's surface area increasing?

Core Concept

From $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$, $36\pi=4\pi(9)\dfrac{dr}{dt}$, so $\dfrac{dr}{dt}=1$. Then $\dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}=8\pi(3)(1)=24\pi$ cm$^2$/s.