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

Given, the diameter of the balloon $=\frac{3}{2}(2x+1)$
Therefore Radius of the balloon $=\frac{Diameter}{2}=\frac{1}{2}\left[\frac{3}{2}\right(2x+1\left)\right]=\frac{3}{4}(2x+1)$
For the volume V, the balloon is given by
$V=\frac{4}{3}\pi (radius{)}^3=\frac{4}{3}\pi {\left[\frac{3}{4}\right(2x+1\left)\right]}^3=\frac{9\pi }{16}(2x+1{)}^3$
For the rate of change of volume, differentiate w.r.t to x, we get
$\frac{dV}{dx}=\frac{9\pi }{16}\times 3(2x+1{)}^2\times 2=\frac{27\pi }{8}(2x+1{)}^2$
Thus, rate of change of volume is $\frac{27\pi }{8}(2x+1{)}^2$.