Rate of change of circular oil spill area

Start with the relationship between area and radius for a circle: $A = \pi r^2$. Since both $A$ and $r$ are functions of time $t$, differentiate both sides implicitly with respect to $t$ using the chain rule: $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}.$ This equation relates the rate of change of the area to the rate of change of the radius.

Now substitute the known values $r = 20$ m and $\frac{dr}{dt} = 3$ m/min into the differentiated equation: $\frac{dA}{dt} = 2\pi (20)(3) = 120\pi.$ So the area is increasing at $120\pi \approx 377$ square meters per minute at that instant.