Question

The circumference of the circle is increasing at a rate of $0.5\mathrm{meters}\mathrm{per}\mathrm{minute}$. What's the rate of change of the area of the circle when the radius is $4\mathrm{meters}$?

• A. $3\mathrm{meters}\mathrm{per}\mathrm{minute}$
• B. $4\mathrm{meters}\mathrm{squared}\mathrm{per}\mathrm{minute}$
• C. $4\mathrm{meters}\mathrm{per}\mathrm{minute}$
• D. $2\mathrm{meters}\mathrm{squared}\mathrm{per}\mathrm{minute}$
• E. $7\mathrm{meters}\mathrm{per}\mathrm{minute}$

Answer 1

The correct option is D

$2\mathrm{meters}\mathrm{squared}\mathrm{per}\mathrm{minute}$

Step 1: Compute the rate of change of the radius of the circle per minute.

The formula of the circumference of the circle is $C=2\mathrm{\pi r}$, where, $r$ is the radius of the circle.

Differentiate both sides of $C=2\mathrm{\pi r}$ with respect to $t$ i.e., time,

we get,

$\frac{dC}{dt}=2\mathrm{\pi }\frac{\mathrm{dr}}{\mathrm{dt}}$

Since the circumference of the circle is increasing at a rate of $0.5\mathrm{meters}\mathrm{per}\mathrm{minute}$, $\frac{dC}{dt}=0.5$.

$\therefore 2\mathrm{\pi }\frac{dr}{dt}=0.5$

$\Rightarrow$ $\frac{dr}{dt}=\frac{0.5}{2\mathrm{\pi }}$

Step 2: Compute the rate of change of the area of the circle per minute.

The formula of the area of the circle is $A={\mathrm{\pi r}}^2$, where, $r$ is the radius of the circle.

Differentiate both sides of $A={\mathrm{\pi r}}^2$ with respect to $t$ i.e., time,

we get,

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

Substitute $\frac{0.5}{2\mathrm{\pi }}$for $\frac{dr}{dt}$ and $4$ for $r$ in $\frac{dA}{dt}=2\mathrm{\pi r}\frac{\mathrm{dr}}{\mathrm{dt}}$, we get,

$\frac{dA}{dt}=2\mathrm{\pi r}\times \frac{0.5}{2\mathrm{\pi }}$

$\Rightarrow$ $\frac{dA}{dt}=2\pi \times 4\times \frac{0.5}{2\mathrm{\pi }}$

$\therefore$ $\frac{dA}{dt}=2$

Hence, the rate of change in the area of the circle is $2\mathrm{meters}\mathrm{squared}\mathrm{per}\mathrm{minute}$.

Hence, option (D) is the correct option.