Question
A balloon which always remains spherical, has a variable diameter $\frac{3}{2} (2\text{x + 1)}.$ Find the rate of change of its volume with respect to x.

Answer

Given: Diameter of the balloon $=\frac{3}{2}(2\text{x}+1)$
$\therefore $ Radius of the balloon $=\frac{3}{4}(2\text{x}+1)$
$\therefore$ Volume of the balloon $=\frac{4}{3}\pi\Big(\frac{3}{4}\pi2\text{x}+1\Big)=\frac{9\pi}{16}(2\text{x}+1)^3$ cu. units
$\therefore$ Rate of change of volume $\text{w.r.t.x}\ = \frac{\text{dV}}{\text{dx}}= \frac{9\pi}{16}.3(2\text{x}+1)^2 .\frac {\text{d}}{\text{dx}}(2\text{x}+\text{x})$
$=\frac{ 27\pi}{16}(2\text{x}+1 )^{2}.2=\frac{27\pi}{8}(2\text{x} + 1)^{2}$

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