Question

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

Answer 1

Let x be the length of the cube

∴ Volume of the cube V = x³   ......(1)

Given that ${\displaystyle \frac{\text{dV}}{\text{dt}}}$ = K

Differentiating Equation (1) w.r.t. t, we get

${\displaystyle \frac{\text{dV}}{\text{dt}}=3x^2\cdot \frac{\text{dx}}{\text{dt}}}$ = K  .....(constant)

∴ ${\displaystyle \frac{\text{dx}}{\text{dt}}=\frac{\text{K}}{3x^2}}$

Now surface area of the cube, S = 6x²

Differentiating both sides w.r.t. t, we get

${\displaystyle \frac{\text{ds}}{\text{dt}}=6\cdot 2\cdot x\cdot \frac{\text{dx}}{\text{dt}}}$

= ${\displaystyle 12x\cdot \frac{\text{K}}{3x^2}}$

⇒ ${\displaystyle \frac{\text{ds}}{\text{dt}}=\frac{4\text{K}}{x}}$

⇒ ${\displaystyle \frac{\text{ds}}{\text{dt}}\infty \frac{1}{x}}$   .....(4K = constant)

Hence, the surface area of the cube varies inversely as the length of the side.