Application of Derivatives — Class 12 Maths Solution

Let at any instant of time $t$, the edge of the cube be $x$ and its volume be $V$
then $V = {x^3}$ …(i)

Differentiating (i) wr.t. $t$, we get
$\Rightarrow \cfrac{{dV}}{{dt}} = 3{x^2}\cfrac{{dx}}{{dt}} = 3{{\rm{(10cm)}}^{\rm{2}}}\left( {{\rm{3cm/sec}}} \right)$

$= 900{\rm{c}}{{\rm{m}}^{\rm{3}}}{\rm{/sec}}$
Therefore, rate of increase of volume of the cube is $900{\rm{c}}{{\rm{m}}^{\rm{3}}}{\rm{/sec}}$.