Question

The radius of an air bubble is increasing at the rate of
$\frac{1}{2}\text{cm/s}$
At what rate is the volume of the bubble increasing when the radius is $1\text{cm}$?

Answer 1

Let $r$ be the radius of bubble
Given, Radius of the bubble is increasing at the rate of

$\frac{1}{2}\text{cm/s}$

so

$\frac{dr}{dt}=\frac{1}{2}\text{cm/s}$

We know that Volume of sphere is,

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

Now,

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

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

$\Rightarrow \frac{dV}{dt}=\frac{4}{3}\pi .3r^2.\frac{dr}{dt}$

$\Rightarrow \frac{dV}{dt}=4\pi r^2\times \frac{1}{2}[\because \frac{dr}{dt}=\frac{1}{2}]$

$\Rightarrow \frac{dV}{dt}=2\pi r^2$

When $r=1\text{cm}$

$\frac{dV}{dt}=2\pi {\text{cm}}^3/\text{s}$

Hence, the volume of the bubble is increasing at the rate of $2\pi {\text{cm}}^3/\text{s}$ when radius is $1\text{cm}$.