Question

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

Answer 1

Let $x$ be the edge length

Given, Edge of cube is increasing at the rate of $3\text{cm/s}$

$\therefore \frac{dx}{dt}=3\text{cm/s}$

Volume of cube, $V=x^3$
Differentiating w.r.t time,

$\frac{dV}{dt}=\frac{d\left(x^3\right)}{dt}$

$\Rightarrow \frac{dV}{dt}=3x^2.\frac{dx}{dt}=9x^2[\because \frac{dx}{dt}=3]$

Putting $x=10$

$\frac{dV}{dt}=9\times {10}^2=900{\text{cm}}^3/\text{s}$

Hence, volume of a cube is increasing at the rate of $900{\text{cm}}^3/\text{s}$ when edge length is
$10\text{cm}.$