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The volume of a cube is increasing at the rate of $8 \mathrm{~cm}^3 / \mathrm{s}$. How fast is the surface area increasing when the length of an edge is $12 \mathrm{~cm}$ ?
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Let $x$ be the length of the cube volume $V=x^3$, $\mathrm{S}=6 \mathrm{x}^2 \Rightarrow \frac{d s}{d x}=12 x \& \mathrm{~V}=\mathrm{x}^3$
$\begin{aligned}
\therefore \frac{\mathrm{dV}}{\mathrm{dt}} &=\frac{\mathrm{dV}}{\mathrm{dx}} \times \frac{\mathrm{dx}}{\mathrm{dt}} \text { We have } \frac{\mathrm{dV}}{\mathrm{dt}}=8 \mathrm{~cm}^3 / \mathrm{sec} \\
\frac{\mathrm{dx}}{\mathrm{dt}} &=\frac{8}{3 \mathrm{x}^2} \therefore \frac{\mathrm{dS}}{\mathrm{dt}}=\frac{\mathrm{dS}}{\mathrm{dx}} \times \frac{\mathrm{dx}}{\mathrm{dt}}=2 \frac{2}{3} \mathrm{~cm}^2 / \mathrm{sec}
\end{aligned}$
$\begin{aligned}
\therefore \frac{\mathrm{dV}}{\mathrm{dt}} &=\frac{\mathrm{dV}}{\mathrm{dx}} \times \frac{\mathrm{dx}}{\mathrm{dt}} \text { We have } \frac{\mathrm{dV}}{\mathrm{dt}}=8 \mathrm{~cm}^3 / \mathrm{sec} \\
\frac{\mathrm{dx}}{\mathrm{dt}} &=\frac{8}{3 \mathrm{x}^2} \therefore \frac{\mathrm{dS}}{\mathrm{dt}}=\frac{\mathrm{dS}}{\mathrm{dx}} \times \frac{\mathrm{dx}}{\mathrm{dt}}=2 \frac{2}{3} \mathrm{~cm}^2 / \mathrm{sec}
\end{aligned}$
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