Question

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

Answer 1

As volume of cube, V=$a^3$ $\Rightarrow \frac{dV}{dt}=3a^2\frac{da}{dt}=9\frac{c{m}^3}{s}\Rightarrow \frac{da}{dt}=\frac{3}{a^2}\frac{cm}{s}$

Now Surface area of cube,S=$6a^2\Rightarrow \frac{dS}{dt}=12a\frac{da}{dt}=12a\times \frac{3}{a^2}=\frac{36}{a}c{m}^2{s}^{-1}$

${\begin{array}{c}\therefore \frac{dS}{dt}\end{array}]}_{\text{at a=10 cm}}=\frac{36}{10}=\frac{18}{5}c{m}^2{s}^{-1}or3.6c{m}^2{s}^{-1}.$