Question:
A balloon, which always remains spherical, has a variable diameter $\frac{3}{2}(2 x+1)$. Find the rate of change of its volume with respect to $x$.
Solution:
The volume of a sphere (V) with radius (r) is given by,
$V=\frac{4}{3} \pi r^{3}$
It is given that:
Diameter $=\frac{3}{2}(2 x+1)$
$\Rightarrow r=\frac{3}{4}(2 x+1)$
$\therefore V=\frac{4}{3} \pi\left(\frac{3}{4}\right)^{3}(2 x+1)^{3}=\frac{9}{16} \pi(2 x+1)^{3}$
Hence, the rate of change of volume with respect to x is as
$\frac{d V}{d x}=\frac{9}{16} \pi \frac{d}{d x}(2 x+1)^{3}=\frac{9}{16} \pi \times 3(2 x+1)^{2} \times 2=\frac{27}{8} \pi(2 x+1)^{2}$