Question

Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of $6{\mathrm{m}\mathrm{i}}^2/\mathrm{h}$.

How fast is the radius of the spill increasing when the area is $9{\mathrm{m}\mathrm{i}}^2?$

Answer 1

Finding how fast is the radius of the spill increasing:

Let $A$ be the area and $r$ be the radius of spill

$\frac{dA}{dt}=6m{i}^2/h$

We have to find $\frac{dr}{dt}$ when $A=9m{i}^2$

Since $A=9m{i}^2$:

$$\begin{gathered} {\mathrm{\pi r}}^2=9{\mathrm{mi}}^2 \\ \Rightarrow {\mathrm{r}}^2=\frac{9}{\mathrm{\pi }}{\mathrm{mi}}^2 \\ \Rightarrow \mathrm{r}=\frac{3}{\sqrt{\mathrm{\pi }}}\mathrm{mi} \end{gathered}$$

Now:

$$\begin{gathered} A={\mathrm{\pi r}}^2 \\ \Rightarrow \frac{d\mathrm{A}}{d\mathrm{t}}=2\mathrm{\pi r}\frac{d\mathrm{r}}{d\mathrm{t}} \end{gathered}$$

Substituting values:

$$\begin{gathered} 6m{i}^2/h=2\mathrm{\pi }\left(\frac{3}{\sqrt{\mathrm{\pi }}}\mathrm{mi}\right)\frac{d\mathrm{r}}{d\mathrm{t}} \\ \Rightarrow \frac{d\mathrm{r}}{d\mathrm{t}}=\frac{1}{\sqrt{\mathrm{\pi }}}\mathrm{mi}/\mathrm{h} \end{gathered}$$

Hence, radius of the spill increasing at $\frac{1}{\sqrt{\mathrm{\pi }}}mi/h$ when the area is $9m{i}^2$.