The radius of a circular plate is increasing | vedclass

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(A) The area $A$ of a circular plate with radius $r$ is given by $A = \pi r^2$. To find the rate of change of the area with respect to time $t$,we dif

Vedclass · Accepted Answer

$0.24 \pi 	ext{ cm}^2/	ext{s}$

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(A) The area $A$ of a circular plate with radius $r$ is given by $A = \pi r^2$. To find the rate of change of the area with respect to time $t$,we differentiate both sides with respect to $t$: $\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt}$. Given that $\frac{dr}{dt} = 0.01 	ext{ cm/s}$ and $r = 12 	ext{ cm}$,we substitute these values into the equation: $\frac{dA}{dt} = 2 \pi (12) (0.01) = 0.24 \pi 	ext{ cm}^2/	ext{s}$. Thus,the area increases at the rate of $0.24 \pi 	ext{ cm}^2/	ext{s}$.

The radius of a circular plate is increasing at the rate of $0.01 \text{ cm/s}$ when the radius is $12 \text{ cm}$. Then,the rate at which the area increases,is

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