Question

Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of $1\mathrm{m}/\mathrm{s}$, how fast is the area of the spill increasing when the radius is $37\mathrm{m}$?

Answer 1

Find the rate at which the area of spill increases.

Area of circle with radius $r$ is given by $A={\mathrm{\pi r}}^2$.

Differentiate both sides of $A={\mathrm{\pi r}}^2$ w.r.t. $t$.

$$\begin{gathered} \frac{dA}{dt}=\frac{d}{dt}\left({\mathrm{\pi r}}^2\right) \\ \Rightarrow \frac{dA}{dt}=\mathrm{\pi }\frac{d}{dt}\left(r^2\right) \\ \Rightarrow \frac{d\mathrm{A}}{d\mathrm{t}}=\mathrm{\pi }\left(2\mathrm{r}\right)\frac{d\mathrm{r}}{d\mathrm{t}}\left[\because \frac{d}{d\mathrm{x}}\left({\mathrm{x}}^{\mathrm{n}}\right)={\mathrm{nx}}^{\mathrm{n}-1}\right] \\ \therefore \frac{d\mathrm{A}}{d\mathrm{t}}=2\mathrm{\pi r}\frac{d\mathrm{r}}{d\mathrm{t}}\dots \left(1\right) \end{gathered}$$

It is given that $\frac{dr}{dt}=1\mathrm{m}/\mathrm{s}$ and $r=37\mathrm{m}$.

Substituting $\frac{dr}{dt}=1$ and $r=37$ in equation $\left(1\right)$, we get

$$\begin{gathered} \Rightarrow \frac{d\mathrm{A}}{d\mathrm{t}}=2\mathrm{\pi }\left(37\right)\left(1\right) \\ \Rightarrow \frac{d\mathrm{A}}{d\mathrm{t}}=74\mathrm{\pi } \\ \therefore \frac{d\mathrm{A}}{d\mathrm{t}}=232.47 \end{gathered}$$

Hence, the rate at which the area of spill increases is $232.47{\mathrm{m}}^2/\mathrm{s}$.