The volume of a cube is increasing at the rate of 9 cm³/s. How fast is its surface area increasing when the length of an edge is 10 cm?
 

Let the length of the sides of the cube be x, 
t is time in seconds 
We know that, Volume of cube V = x³
Given, Volume of cube is increasing at rate of 9cm³/s, so
$\begin{array}{l}\frac{\mathrm{dv}}{\mathrm{dt}}=9\\ \frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{x}}^3\right)=9\\ 3{\mathrm{x}}^2\frac{\mathrm{dx}}{\mathrm{dt}}=9\\ \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{3}{{\mathrm{x}}^2}\dots \dots \end{array}$
Surface area of cube S = 6x²
For finding $\frac{\mathrm{ds}}{\mathrm{dt}}$
$\begin{array}{l}\frac{\mathrm{ds}}{\mathrm{dt}}=\frac{\mathrm{d}\left(6{\mathrm{x}}^2\right)}{\mathrm{dt}}\\ =12\mathrm{x}\frac{\mathrm{dx}}{\mathrm{dt}}\end{array}$
From (i) equation, we get
$12\mathrm{x}\times \frac{3}{{\mathrm{x}}^2}\Rightarrow \frac{36}{\mathrm{x}}$
When length of edge, $\mathrm{x}=10\mathrm{cm}$
$\frac{\mathrm{ds}}{\mathrm{dt}}=\frac{36}{10}\Rightarrow 3.6{\mathrm{cm}}^2/\mathrm{s}$
Hence, rate of increasing in surface area when edge length is 10cm is 3.6cm²/s.