Answer: \(\displaystyle \frac{dr}{dt}=\frac{1}{6\pi}\text{ cm s}^{-1}\).

Differentiate first. Stationary points occur where the derivative is zero, while tangent and normal problems use the gradient at the given point.

Differentiate

\(V=\frac43\pi r^3\)

with respect to \(r\):

\(\frac{dV}{dr}=4\pi r^2.\)

At \(r=6\),

\(\frac{dV}{dr}=4\pi(6)^2=144\pi.\)

Using the chain rule,

\(\frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}.\)

Since \(\frac{dV}{dt}=24\),

\(24=144\pi\frac{dr}{dt}.\)

Hence

\(\frac{dr}{dt}=\frac{24}{144\pi}=\frac{1}{6\pi}.\)