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(A) Let the side length of the cube be $a$. The surface area $S$ of the cube is given by $S = 6a^2$.Given that $\frac{dS}{dt} = 3.6 	ext{ cm}^2/	ext

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$9$

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(A) Let the side length of the cube be $a$. The surface area $S$ of the cube is given by $S = 6a^2$.Given that $\frac{dS}{dt} = 3.6 	ext{ cm}^2/	ext{sec}$.Differentiating $S$ with respect to $t$,we get $\frac{dS}{dt} = 12a \frac{da}{dt}$.Substituting the given values: $3.6 = 12(10) \frac{da}{dt} \Rightarrow 3.6 = 120 \frac{da}{dt} \Rightarrow \frac{da}{dt} = \frac{3.6}{120} = 0.03 	ext{ cm}/	ext{sec}$.The volume $V$ of the cube is $V = a^3$.Differentiating $V$ with respect to $t$,we get $\frac{dV}{dt} = 3a^2 \frac{da}{dt}$.Substituting $a = 10$ and $\frac{da}{dt} = 0.03$: $\frac{dV}{dt} = 3(10)^2(0.03) = 3(100)(0.03) = 300 	imes 0.03 = 9 	ext{ cm}^3/	ext{sec}$.

If the surface area of a cube is increasing at a rate of $3.6 \text{ cm}^2/\text{sec}$,while maintaining its shape,then the rate of change of its volume (in $\text{cm}^3/\text{sec}$),when the length of a side of the cube is $10 \text{ cm}$,is:

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