The radius of a circular plate is increasing at the rate of $0.01 \mathrm{~cm} / \mathrm{s}$ when the radius…
The radius of a circular plate is increasing at the rate of $0.01 \mathrm{~cm} / \mathrm{s}$ when the radius is $12 \mathrm{~cm}$. Then, the rate at which the area increases, is
$0.24 \pi \mathrm{sq} \mathrm{cm} / \mathrm{s}$
$60 \pi \mathrm{sq} \mathrm{cm} / \mathrm{s}$
$24 \pi \mathrm{sq} \mathrm{cm} / \mathrm{s}$
$1.2 \pi \mathrm{sq} \mathrm{cm} / \mathrm{s}$
Solution
The area of circular plate is
$A=\pi r^2$
On differentiating w.r.t. $t$, we get
$\begin{aligned}
& \frac{d A}{d t}=2 \pi r \frac{d r}{d t} \\
& \frac{d A}{d t}=2 \pi(12)(0.01) \\
& \Rightarrow \quad {\left[\because \text { given } \frac{d r}{d t}=0.01 \Rightarrow r=12\right] } \\
& \frac{d A}{d t}= 0.24 \pi \mathrm{sq} \mathrm{cm} / \mathrm{sec}
\end{aligned}$