The rate of change of the surface area of a s | vedclass

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(C) Let the surface area of the sphere be $S = 4\pi r^2$. Given that the rate of change of the radius is $\frac{dr}{dt} = 2 	ext{ cm/sec}$. Different

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$r$

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(C) Let the surface area of the sphere be $S = 4\pi r^2$. Given that the rate of change of the radius is $\frac{dr}{dt} = 2 	ext{ cm/sec}$. Differentiating the surface area with respect to time $t$,we get: $\frac{dS}{dt} = \frac{d}{dt}(4\pi r^2) = 4\pi 	imes 2r 	imes \frac{dr}{dt}$. Substituting the given value of $\frac{dr}{dt}$: $\frac{dS}{dt} = 8\pi r 	imes 2 = 16\pi r$. Since $16\pi$ is a constant,we have $\frac{dS}{dt} \propto r$. Therefore,the rate of change of the surface area is proportional to $r$.

The rate of change of the surface area of a sphere of radius $r$ when the radius is increasing at the rate of $2 \text{ cm/sec}$ is proportional to

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