Question

The volume of a cube is increasing at the rate of $9c{m}^3/sec$. How fast is its surface area increasing when the length of an edge is $10cm$?

Answer 1

Let $x$ be the side of the cube, $V$ be the volume and $S$ be the surface area of the cube.

Then, $V$ = $x^3$ and S =$6x^2$

$\therefore \frac{dV}{dt}=3x^2\frac{dx}{dt}$, where $x$ is a function of time $\left(t\right)$

Given, $\frac{dV}{dt}=9c{m}^3/sec$

$\Rightarrow 9=3x^2\frac{dx}{dt}$

$\Rightarrow \frac{dx}{dt}=\frac{9}{3x^2}...\left(i\right)$

Now, $\frac{dS}{dt}=\frac{d}{dt}\left(6x^2\right)$

$=12x\frac{dx}{dt}$

$=12x\left(\frac{9}{3x^2}\right)\left[\text{using}\right(i\left)\right]$

=$\frac{36}{x}c{m}^2/sec$

$\text{When length of edge}x=10cm,$

$\frac{dS}{dt}=\frac{36}{x}=\frac{36}{10}=\frac{18}{5}=3\frac{3}{5}c{m}^2/sec$.