Question

Oil spilled from a ruptured tanker spread in a circle whose area increases at a constant rate of $7$$\raisebox{1ex}{$m{i}^2$}\!\left/ \!\raisebox{-1ex}{$h$}\right.$.

How fast is the radius of the spill increasing when the area is $1m{i}^2$ $?$

$\frac{\mathrm{dr}}{\mathrm{dt}}$ when $\text{A}=1$

(answer should be in $\raisebox{1ex}{$mi$}\!\left/ \!\raisebox{-1ex}{$h$}\right.$)

Round your answer to two decimal places.

Answer 1

Find the rate of increase in radius of the spill:

It is given that area of a circle increases at a constant rate of $7$$\raisebox{1ex}{$m{i}^2$}\!\left/ \!\raisebox{-1ex}{$h$}\right.$ .

To find $\frac{\mathrm{dr}}{\mathrm{dt}}$ when $\text{A}=1$.

Let $\text{r}$ be the radius of the circle and $\text{A}$ be area of the circle formed. $\left[A={\mathrm{\pi r}}^2\right]$

Find $\text{r}$ when $\text{A}=1$, by substituting $1$ for $\text{A}$ in the formula :

$\begin{array}{rcl}A& =& {\mathrm{\pi r}}^2\\ & \Rightarrow & 1={\mathrm{\pi r}}^2\\ & \Rightarrow & \mathrm{r}=\frac{1}{\sqrt{\mathrm{\pi }}}\end{array}$

(first equation)

As the area of the circle increases at a constant rate of $7$$\raisebox{1ex}{$m{i}^2$}\!\left/ \!\raisebox{-1ex}{$h$}\right.$:

So, $\frac{dA}{dt}=7$

(second equation)

Now, differentiate the second equation with respect to $\text{t}$:

$\frac{\text{dA}}{\text{dt}}=2\mathrm{\pi r}\frac{\mathrm{dr}}{\mathrm{dt}}$

Substitute $7$ for $\frac{\mathrm{dA}}{\mathrm{dt}}$ from the second equation:

$$\begin{gathered} 7=2\mathrm{\pi r}\frac{\mathrm{dr}}{\mathrm{dt}} \\ \Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}}=\frac{7}{2\mathrm{\pi r}} \end{gathered}$$

Then, substitute the value of $\text{r}$ from the first equation:

$\begin{array}{rcl}\frac{dr}{dt}& =& \frac{7}{2\mathrm{\pi }{\displaystyle \frac{1}{\sqrt{\mathrm{\pi }}}}}\\ & =& \frac{7}{2\sqrt{\mathrm{\pi }}}\\ & \approx & 1.98\end{array}$

Hence, the radius of the spill is increasing at the rate of $\mathbf{1}\mathbf{.}\mathbf{98}\mathbf{}\raisebox{1ex}{$\mathbf{m}\mathbf{i}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{h}\mathbf{o}\mathbf{u}\mathbf{r}\mathbf{}$}\right.$.