The volume of a cube increases at a constant rate.

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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.

Solution:

Let’s assume x to be the length of the cube.

So, the volume of the cube V = x3 …. (1)

Given that, dV/dt = K

Now, on differentiating the equation (1) w.r.t. t, we get

dV/dt = 3x2. dx/dt = K (constant)

So, dx/dt = K/3x2

Now,

Surface area of the cube, S = 6x2

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

$\frac{\overline{d s}}{d t}=6 \cdot 2 \cdot x \cdot \frac{d x}{d t}=12 x \cdot \frac{\mathrm{K}}{3 x^{2}}$

$\frac{d s}{d t}=\frac{4 \mathrm{~K}}{x} \Rightarrow \frac{d s}{d t} \propto \frac{1}{x}$ $(4 \mathrm{~K}=$ constant $)$

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

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