Question

An edge of a variable cube is increasing at the rate of $5cm/s$. How fast is the volume of the cube increasing when the edge is $10cm$ long?

Answer 1

Let $x$ be the length of a side and $V$ be the volume of the cube.
Then, $V=x^3$
$\frac{dV}{dt}=3x^2\cdot \frac{dx}{dt},$ (By chain rule)
It is given that,
$\frac{dx}{dt}=3cm/s$
$\therefore \frac{dV}{dt}=3x^2\left(3\right)=9x^2$
Thus, when $x=10cm$,
$\therefore \frac{dV}{dt}=9(10{)}^2=900c{m}^3/s$