The radius of a circle is increasing at a rat | vedclass

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(B) Let the radius of the circle be $r 	ext{ cm}$ and its area be $A 	ext{ cm}^2$. Given that the rate of change of the radius is $\frac{dr}{dt} = 0

Vedclass · Accepted Answer

$\pi 	ext{ cm}^2 	ext{ s}^{-1}$

Vedclass · Answer

(B) Let the radius of the circle be $r 	ext{ cm}$ and its area be $A 	ext{ cm}^2$. Given that the rate of change of the radius is $\frac{dr}{dt} = 0.1 	ext{ cm s}^{-1}$. The area of a circle is given by $A = \pi r^2$. Differentiating both sides with respect to time $t$,we get $\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$. Substituting the given values $r = 5 	ext{ cm}$ and $\frac{dr}{dt} = 0.1 	ext{ cm s}^{-1}$,we have: $\frac{dA}{dt} = 2 \pi (5) (0.1) = 10 \pi (0.1) = \pi 	ext{ cm}^2 	ext{ s}^{-1}$. Thus,the rate of change of the area is $\pi 	ext{ cm}^2 	ext{ s}^{-1}$.

The radius of a circle is increasing at a rate of $0.1 \text{ cm s}^{-1}$. Then the rate of change of area,when its radius is $5 \text{ cm}$,is .........

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