Question

The surface area of a cube increases at the rate of $12$ $c{m}^2/sec$. Find the rate at which its volume increases, when its edge has length $5$cm.

Answer 1

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

$V$ (Vol. of cube) $=a^3$

Now,

$\frac{ds}{dt}=12a\times \frac{da}{dt}=12⟹a\frac{da}{dt}=1$

$⟹\frac{dv}{dt}=3a^2\frac{da}{dt}=3a^2\times \frac{1}{a}=3a$