If the radius of a circular blot of oil is in | vedclass

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(B) Given that the radius $r$ of the circular blot is increasing at a rate of $\frac{dr}{dt} = 2 	ext{ cm/min}$.We need to find the rate of change of

Vedclass · Accepted Answer

$12 \pi 	ext{ cm}^2/	ext{min}$

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(B) Given that the radius $r$ of the circular blot is increasing at a rate of $\frac{dr}{dt} = 2 	ext{ cm/min}$.We need to find the rate of change of the area $A$ when $r = 3 	ext{ cm}$.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 $r = 3 	ext{ cm}$ and $\frac{dr}{dt} = 2 	ext{ cm/min}$:$\frac{dA}{dt} = 2 \pi (3)(2) = 12 \pi 	ext{ cm}^2/	ext{min}$.Thus,the rate of change of the area is $12 \pi 	ext{ cm}^2/	ext{min}$.

If the radius of a circular blot of oil is increasing at the rate of $2 \text{ cm/min}$,then the rate of change of its area when its radius is $3 \text{ cm}$ is:

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