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(C) Let $V$ be the volume and $S$ be the surface area of the sphere with radius $r$.Given,$\frac{dV}{dt} = 12 	ext{ cm}^3/	ext{sec}$.The volume of a

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$4$

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(C) Let $V$ be the volume and $S$ be the surface area of the sphere with radius $r$.Given,$\frac{dV}{dt} = 12 	ext{ cm}^3/	ext{sec}$.The volume of a sphere is $V = \frac{4}{3} \pi r^3$.Differentiating with respect to $t$,we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Given diameter $d = 12 	ext{ cm}$,so radius $r = 6 	ext{ cm}$.Substituting the values: $12 = 4 \pi (6)^2 \frac{dr}{dt} \implies 12 = 144 \pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{12}{144 \pi} = \frac{1}{12 \pi} 	ext{ cm/sec}$.The surface area of a sphere is $S = 4 \pi r^2$.Differentiating with respect to $t$,we get $\frac{dS}{dt} = 8 \pi r \frac{dr}{dt}$.Substituting $r = 6$ and $\frac{dr}{dt} = \frac{1}{12 \pi}$:$\frac{dS}{dt} = 8 \pi (6) \left( \frac{1}{12 \pi} ight) = 48 \pi 	imes \frac{1}{12 \pi} = 4 	ext{ cm}^2/	ext{sec}$.

If the volume of a sphere is increasing at the rate of $12 \text{ cm}^3/\text{sec}$,then the rate (in $\text{cm}^2/\text{sec}$) at which its surface area is increasing,when the diameter of the sphere is $12 \text{ cm}$,is:

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