Question

The volume of a cube increases at a constant rate. Prove that the increase in its surface area varies inversely with the length of the side.

Answer 1

Let the side of a cube be x unit.
$\therefore$ Volume of cube (V) $=x^3$
On differentiating both side w.r.t..t, we get
$\frac{dV}{dt}=3x^2\frac{dx}{dt}=k\Rightarrow \frac{dx}{dt}=\frac{k}{3x^2}$
Also, surface area of cube, $S=6x^2$
On differentiating w.r.t.t, we get
$\frac{dS}{dt}=12x.\frac{dx}{dt}\Rightarrow \frac{dS}{dt}=12x.\frac{k}{3x^2}\Rightarrow \frac{dS}{dt}=\frac{12k}{3x}=4\left(\frac{k}{x}\right)\Rightarrow \frac{dS}{dt}isproportionalto\frac{1}{x}$
Hence, the surface area of the cube varies inverely as the length of the side.