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An edge of a variable cube is increasing at the rate of $3 \mathrm{~cm} / \mathrm{s}$. How fast is the volume of the cube increasing when the edge is $10 \mathrm{~cm}$ long?
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Let the edge of the cube $=\mathrm{x} \mathrm{cm} \therefore=\frac{\mathrm{dx}}{\mathrm{dt}}=3$
$\therefore \quad \frac{d v}{d t}=\frac{d v}{d x} \times \frac{d x}{d t}=3 x^2 \times 3=9 x^2 \mathrm{~m}^3 / \mathrm{sec} \text {. }$
When $x=10, \frac{d v}{d t}=9 \times(10)^2=900 \mathrm{~cm}^2 / \mathrm{sec}$.
$\therefore \quad \frac{d v}{d t}=\frac{d v}{d x} \times \frac{d x}{d t}=3 x^2 \times 3=9 x^2 \mathrm{~m}^3 / \mathrm{sec} \text {. }$
When $x=10, \frac{d v}{d t}=9 \times(10)^2=900 \mathrm{~cm}^2 / \mathrm{sec}$.
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