If the rate of increase of the radius of a ci | vedclass

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(B) Let $r$ be the radius and $A$ be the area of the circle. Given that the rate of change of the radius is $\frac{dr}{dt} = 5 	ext{ cm/sec}$. The ar

Vedclass · Accepted Answer

$200 \pi 	ext{ cm}^2/	ext{sec}$

Vedclass · Answer

(B) Let $r$ be the radius and $A$ be the area of the circle. Given that the rate of change of the radius is $\frac{dr}{dt} = 5 	ext{ cm/sec}$. The area of a circle is given by $A = \pi r^2$. Differentiating both sides with respect to time $t$, we get: $\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2 \pi r \frac{dr}{dt}$. Substituting the given values $r = 20 	ext{ cm}$ and $\frac{dr}{dt} = 5 	ext{ cm/sec}$: $\frac{dA}{dt} = 2 \pi (20)(5) = 200 \pi 	ext{ cm}^2/	ext{sec}$. Thus, the rate of increase of the area is $200 \pi 	ext{ cm}^2/	ext{sec}$.

If the rate of increase of the radius of a circle is $5 \text{ cm/sec}$, then the rate of increase of its area, when the radius is $20 \text{ cm}$, will be

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