Question

If surface area of a cube is changing at a rate of $6\sqrt{3}{m}^2/s$ , find the rate (in $\text{m/s}$) at which length of the body diagonal changes, at a moment when side length is $1m$.

Answer 1

Surface area of a cube is
$S=6a^2$ (where $a$ = side of cube)

Body diagonal length $l$ is
$l=\sqrt{3}a$

$\therefore S=2l^2$

Differentiating $S$ w.r.t. time,
$\frac{dS}{dt}=2\left(2l\right)\frac{dl}{dt}$
$\Rightarrow \frac{dl}{dt}=\frac{1}{4\left(\sqrt{3}a\right)}\frac{dS}{dt}(\because \frac{dS}{dt}=6\sqrt{3}\frac{m^2}{s},a=1m)$

$\Rightarrow \frac{dl}{dt}=\frac{6\sqrt{3}}{4\sqrt{3}}=\frac{3}{2}\text{m/s}$