The radius of a circular oil spill is increasing at the rate of $2$ cm s$^{-1}$. Find the rate at which the area is increasing when the radius is $5$ cm.

$A = \pi r^2$, so $\dfrac{dA}{dt} = \dfrac{dA}{dr} \cdot \dfrac{dr}{dt} = 2\pi r \cdot \dfrac{dr}{dt}$. With $r = 5$ and $\dfrac{dr}{dt} = 2$: $\dfrac{dA}{dt} = 2\pi(5)(2) = 20\pi$. Using $\pi r^2$ directly gives $25\pi$, the area rather than its rate of change.