Browse Questions

# 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.

Toolbox:
• If $y = f(x)$, then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
• $\bigg(\large \frac{dy}{dx} \bigg)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1
Let the surface area of the cube be
$S=6x^2$
and the volume of the cube be
$v=x^3$
where $x$ is the edge of the cube
$v=x^3$
differentiating w.r.t $x$ we get,
$\large\frac{dv}{dt}=3x^2\large\frac{dx}{dt}$
But it is given that $\large\frac{dv}{dt}=k$ ( constant)
$\therefore k = 3x^2\large\frac{dx}{dt} \Rightarrow \large\frac{dx}{dt}=\large\frac{k}{3x^2}$
$s=6x^2$
differentiating w.r.t $x$ we get,
$\large\frac{ds}{dt}=12x.\large\frac{dx}{dt}$
Now substituting for $\large\frac{dx}{dt}$ we get,
$\large\frac{ds}{dt}=\large\frac{k}{3x^2} \times 12x$
$=\large\frac{4k}{x}$
$\Rightarrow \large\frac{ds}{dt} \alpha \large\frac{1}{x}$
Hence the rate of increase in surface area varies inversely as the length of the edge of the cube.