Question

The radius of a circle is increasing at a constant rate of $0.2$ meters per second. what is the rate increase in the area of the circle at the instant when the circumference of the circle is $20\pi$ meters?

Answer 1

Explanation:

Given,

radius of the circle is increasing at a constant rate of $0.2\frac{m}{sec}$

circumference of the circle at an instant is $20\pi$ meters

STEP-1:

Area of the circle $=\mathrm{\pi }\times {\mathrm{r}}^2$

Now differentiating the equation with respect to time $t$.

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

Given $\frac{dr}{dt}=0.2\frac{m}{sec}$

then,

$\frac{dA}{dt}=0.4\times \mathrm{\pi }\times \mathrm{r}$

STEP-2:

Circumference of the circle is C$=2\mathrm{\pi r}$

At $C$$=20\mathrm{\pi }$ we get,

$$\begin{gathered} 2\mathrm{\pi r}=20\mathrm{\pi } \\ \Rightarrow \mathrm{r}=10\mathrm{m} \end{gathered}$$

STEP-3:

Now if we put the value of $r$ in the above equation of $\frac{dA}{dt}$

$$\begin{gathered} \Rightarrow \frac{dA}{dt}=0.4\times \mathrm{\pi }\times 10 \\ \Rightarrow \frac{d\mathrm{A}}{d\mathrm{t}}=12.56\frac{{\mathrm{m}}^2}{\sec } \end{gathered}$$

Therefore, the rate of increase in the area of the circle is $12.56\frac{m^2}{sec}$