The diameter of a circle is increasing at the rate of 1 cm/sec.

Question:

The diameter of a circle is increasing at the rate of 1 cm/sec. When its radius is π, the rate of increase of its area is

(a) $\pi \mathrm{cm}^{2} / \mathrm{Sec}$

(b) $2 \pi \mathrm{cm}^{2} / \mathrm{sec}$

(c) $\pi^{2} \mathrm{~cm}^{2} / \mathrm{sec}$

 

(d) $2 \pi^{2} \mathrm{~cm}^{2} / \mathrm{sec}^{2}$

Solution:

(c) $\pi^{2} \mathrm{~cm}^{2} / \mathrm{sec}$

Let $D$ be the diameter and $A$ be the area of the circle at any time $t .$ Then,

$A=\pi r^{2}$ (where $r$ is the radius of the cicle)

$\Rightarrow A=\pi \frac{D^{2}}{4}$               $\left[\because r=\frac{D}{2}\right]$

$\Rightarrow \frac{d A}{d t}=2 \pi \frac{D}{4} \frac{d D}{d t}$

$\Rightarrow \frac{d A}{d t}=\frac{\pi}{2} \times 2 \pi \times 1$        $\left[\because \frac{d D}{d t}=1 \mathrm{~cm} / \mathrm{sec}\right]$

$\Rightarrow \frac{d A}{d t}=\pi^{2} \mathrm{~cm}^{2} / \mathrm{sec}$

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