Question

An edge of a variable cube is increasing at the rate of 3cm/s. How fast the volume of the cube increasing when the edge is 10 cm long?

Answer 1

Let x be the length of an edge of cube and V be the volume of the cube. Then, $V=x^3$
$\therefore$ Rate of change of volume w.r.t time
$\frac{dV}{dt}=\frac{d}{dt}\left(x^3\right)=3x^2\frac{dx}{dt}$ (By chain rule)
It is given that edge of the cube is increasing at the rate of 3 cm/s, so $\frac{dx}{dt}=3cm/s\therefore \frac{dV}{dt}=3x^2\left(3\right)=9x^{2}c{m}^3/s$
Thus, when x=10 cm, $\frac{dV}{dt}=9(10{)}^2=900c{m}^3/s$
Hence, the volume of the cube is increasing at the rate of 900 $c{m}^3/s$ when the edge is 10 cm long.