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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.

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Given: a cube with volume increasing at a constant rate

To prove: the relation between the increase int eh surface area with the length of the side is inversely proportional.

Explanation: Let ‘a’ the length of the side of the cube.

Take the volume of the cube ‘V’

Then  V=a^3........(i)

As mentioned in the question, the rate of volume increase is constant, then

\frac{\mathrm{d} \mathrm{V}}{\mathrm{dt}}=\mathrm{k}
After substituting in the above equation, the values from equation (i) we get

Differentiating the equation with respect to t,

\\ \Rightarrow 3 a^{2} \times \frac{d(a)}{d t}=k$\\ $\Rightarrow \frac{\mathrm{d}(\mathrm{a})}{\mathrm{dt}}=\frac{\mathrm{k}}{3 \mathrm{a}^{2}} \ldots \ldots(\mathrm{ii})

Take S as the surface area of the cube, then \mathrm{S}=6 \mathrm{a}^{2}
After differentiating the surface area with respect to t, we get

\frac{\mathrm{dS}}{\mathrm{dt}}=\frac{\mathrm{d}\left(6 \mathrm{a}^{2}\right)}{\mathrm{dt}}
Applying the derivatives, we get

\Rightarrow \frac{\mathrm{dS}}{\mathrm{dt}}=6 \times 2 \mathrm{a} \times \frac{\mathrm{d}(\mathrm{a})}{\mathrm{dt}}

After substituting the value from equation (ii) in the given equation we get

\Rightarrow \frac{\mathrm{dS}}{\mathrm{dt}}=12 \mathrm{a}\left(\frac{\mathrm{k}}{3 \mathrm{a}^{2}}\right)

After taking out the like terms,

\Rightarrow \frac{\mathrm{dS}}{\mathrm{dt}}=4\left(\frac{\mathrm{k}}{\mathrm{a}}\right)

Now converting it to proportional, we get

\Rightarrow \frac{\mathrm{dS}}{\mathrm{dt}} \propto \frac{1}{\mathrm{a}}

Therefore, the relation between the length and the side of the cube is inversely proportional in the given condition.

Hence Proved.

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