Question

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

Answer 1

Let the diameter and radius of the balloon be $d$ and $r$ respectively
Given:
$d=\frac{3}{2}(2x+1)$

$\Rightarrow r=\frac{d}{2}=\frac{3}{4}(2x+1)$

$\Rightarrow \frac{dr}{dx}=\frac{3}{4}.\frac{d}{dx}(2x+1)$

$\Rightarrow \frac{dr}{dx}=\frac{3}{4}\times 2=\frac{3}{2}\cdots \left(i\right)$

Volume of the balloon

$V=\frac{4}{3}\pi r^3$

Now,

$\frac{dV}{dx}=\frac{d}{dx}\left(\frac{4}{3}\pi r^3\right)$

$=\frac{4\pi }{3}\times \frac{d}{dx}\left(r^3\right)$

$=\frac{4\pi }{3}\times 3r^2\times \frac{dr}{dx}$

$=4\pi r^2\times \frac{3}{2}$

[From $\left(i\right)$]

$=6\pi r^2$

$=6\pi {\left[\frac{3}{4}\right(2x+1\left)\right]}^2$

$=6\pi \times \frac{9}{16}(2x+1{)}^2$

$\therefore \frac{dV}{dx}=\frac{27\pi }{8}(2x+1{)}^2$

Hence, rate of change of volume with respect to $x$ is

$\frac{27\pi }{8}(2x+1{)}^2$