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Q.
If the radius of a circle be increasing at a uniform rate of $ 2 \,cm/s $ . The area of increasing of area of circle, at the instant when the radius is $ 20\, cm $ , is
Given $\frac{d r}{d t}=2 \,cm / s$, where $r$ be radius of circle and $t$ be the time.
Now, area of circle is giver. by $A=\pi r^{2}$.
On differentiating w.r.t. $t$, we get
$ \frac{d A}{d t}=2 \pi r \frac{d r}{d t}$
$\Rightarrow \,\frac{d A}{d t}=2 \pi \cdot 20 \cdot 2 $
$\Rightarrow \, \frac{d A}{d t}=80 \pi \,cm ^{2} / s$
Thus, the rate of change of area of circle with respect to time is $80 \,\pi\, cm ^{2} / s$.