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(C) Let the radius of the sphere be $r$ and its surface area be $S$. The surface area of a sphere is given by $S = 4\pi r^2$. We are given that the ra

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$r$

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(C) Let the radius of the sphere be $r$ and its surface area be $S$. The surface area of a sphere is given by $S = 4\pi r^2$. We are given that the rate of change of the radius is $\frac{dr}{dt} = 2 	ext{ cm/s}$. Differentiating $S$ 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 $r$.

If the radius $r$ of a sphere is increasing at a rate of $2 \text{ cm/s}$,then the rate of change of its surface area is proportional to which of the following?

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