Question

The radius of a sphere is decreasing at a constant rate of 3 centimeters per second. In terms of the surface area of the sphere S, what is the rate of change of the volume of the sphere, in cubic centimeters per second.

Answer 1

\[V = \frac{4}{3} \pi r^{3}\\A = 4\pi r^{2}\\ \frac{dr}{dt} = 3\]

Answer 2

\[\frac{dV}{dt} = 4 \pi (\frac{dr}{dt})^{2} \\ \frac{dV}{dt} = 4\pi (3)^{2} \\ \frac{dV}{dt} = 36\pi\]