If the surface area of a cube is increasing at a rate of $3.6 \mathrm{~cm}^{2} / \mathrm{sec}$, reta

Question

If the surface area of a cube is increasing at a rate of $3.6 \mathrm{~cm}^{2} / \mathrm{sec}$, retaining its shape; then the rate of change of its volume (in $\mathrm{cm}^{3} / \mathrm{sec}$.), when the length of a side of the cube is $10 \mathrm{~cm}$, is\

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(d) : Let $a$ be the side of a cube. Then, we have\ $S=6 a^{2} \Rightarrow \frac{d S}{d t}=12 a \frac{d a}{d t}$\ $\Rightarrow 3.6=12 \cdot 10 \frac{d a}{d t} \Rightarrow \frac{d a}{d t}=\frac{3}{100}$\ Now, $V=a^{3} \Rightarrow \frac{d V}{d t}=3 a^{2} \cdot \frac{d a}{d t}$ \ $ =3 \cdot(10)^{2} \cdot \frac{3}{100}=9 \mathrm{~cm}^{3} / \mathrm{sec} . $