Question

The volume of a cube is increasing at a rate of $9$ cubic centimetres per second. How fast is the surface area increasing when the length of an edge is $10$ centimetres?

Answer 1

Let the length of the sides of the cube be $x$,

$t$ is time in seconds

We know that,

Volume of cube $V=x^3$

Given, Volume of cube is increasing at rate of $9{\text{cm}}^3/\text{s}$, so

$\frac{dV}{dt}=9$

$\Rightarrow \frac{d}{dt}\left(x^3\right)=9$

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

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

Surface area of cube $S=6x^2$

Finding $\frac{dS}{dt},$

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

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

$=12x\times \frac{3}{x^2}$

$=\frac{36}{x}$

[From equation $\left(i\right)$]

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

$\frac{dS}{dt}=\frac{36}{10}=3.6{\text{cm}}^2/\text{s}$

Hence, rate of increasing in surface area when edge length is $10\text{cm}$ is $3.6{\text{cm}}^2/\text{s}$