The radius of a circular plate is increasing at the rate of 0.01 cm/sec.
Question:

The radius of a circular plate is increasing at the rate of $0.01 \mathrm{~cm} / \mathrm{sec}$. The rate of increase of its area when the radius is $12 \mathrm{~cm}$, is

(a) $144 \pi \mathrm{cm}^{2} / \mathrm{sec}$

(b) $2.4 \pi \mathrm{cm}^{2} / \mathrm{sec}$

(c) $0.24 \pi \mathrm{cm}^{2} / \mathrm{sec}$

(d) $0.024 \pi \mathrm{cm}^{2} / \mathrm{sec}$

Solution:

(c) $0.24 \pi \mathrm{cm}^{2} / \mathrm{sec}$

Let $r$ be the radius and $A$ be the area of the circular plate at any time $t .$ Then,

$A=\pi r^{2}$

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

$\Rightarrow \frac{d A}{d t}=2 \pi(12)(0.01)$

$\Rightarrow \frac{d A}{d t}=0.24 \pi \mathrm{cm}^{2} / \mathrm{sec}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.