Question

If the radius of a circular plate is increasing at the rate of $0.01\mathrm{cm}{\mathrm{s}}^{-1}$, when the radius is $12\mathrm{cm}$. Then, the rate at which the area increases is

• A. $0.24\mathrm{\pi }\mathrm{sq}\mathrm{cm}{\mathrm{s}}^{-1}$
• B. $60\mathrm{\pi }\mathrm{sq}\mathrm{cm}{\mathrm{s}}^{-1}$
• C. $24\mathrm{\pi }\mathrm{sq}\mathrm{cm}{\mathrm{s}}^{-1}$
• D. $1.2\mathrm{\pi }\mathrm{sq}\mathrm{cm}{\mathrm{s}}^{-1}$

Answer 1

The correct option is A

$0.24\mathrm{\pi }\mathrm{sq}\mathrm{cm}{\mathrm{s}}^{-1}$

Explanation for Correct answer:

Option (C) $0.24\mathrm{\pi }\mathrm{sq}\mathrm{cm}{\mathrm{s}}^{-1}$

Step 1 Given data

The radius($\mathrm{r}$) of a circular plate is increasing at the rate of $0.01\mathrm{cm}{\mathrm{s}}^{-1}$

That is, $\frac{\mathrm{dr}}{\mathrm{dt}}=0.01\mathrm{cm}{\mathrm{s}}^{-1}$

The radius of the circular plate is $12\mathrm{cm}$ that is, $1200\mathrm{m}$

Step 2 Formula

The area of the circle is $\mathrm{A}={\mathrm{\pi r}}^2$

Step 3 The rate at which the area increases

The radius ($\mathrm{r}$) of a circular plate is increasing at the rate $\frac{\mathrm{dr}}{\mathrm{dt}}=0.01\mathrm{cm}{\mathrm{s}}^{-1}$

The radius of the circular plate is $\mathrm{r}=12\mathrm{cm}$

The area of the circular plate is $\mathrm{A}={\mathrm{\pi r}}^2$

The rate at which the area is,

$$\begin{gathered} \frac{\mathrm{dA}}{\mathrm{dt}}=\frac{\mathrm{d}\left({\mathrm{\pi r}}^2\right)}{\mathrm{dt}} \\ \frac{\mathrm{dA}}{\mathrm{dt}}=2\mathrm{\pi r}\frac{\mathrm{dr}}{\mathrm{dt}}.....\left(1\right) \end{gathered}$$

Substituting values in the equation (1)

$$\begin{gathered} \frac{\mathrm{dA}}{\mathrm{dt}}=2\times \mathrm{\pi }\times 12\times 0.01 \\ \frac{\mathrm{dA}}{\mathrm{dt}}=0.24\mathrm{\pi }\mathrm{sq}\mathrm{cm}{\mathrm{s}}^{-1} \end{gathered}$$

Therefore, the rate at which the area increases is $0.24\mathrm{\pi }\mathrm{sq}\mathrm{cm}{\mathrm{s}}^{-1}$.