Question 12

# The radius of an air bubble is increasing at the rate of 1/2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

The air bubble is in the shape of a sphere.

Now, the volume of an air bubble (V) with radius (r) is given by,

\begin{align} V = \frac{4}{3}\pi r^3 \end{align}

The rate of change of volume (V) with respect to time (t) is given by,

\begin{align} \frac{dV}{dt} = \frac{4}{3}\pi \frac{d}{dr}(r^3).\frac{dr}{dt} \;\;\;[By\; Chain\; Rule] \end{align}

\begin{align} = \frac{4}{3}\pi (3r^2).\frac{dr}{dt} \end{align}

\begin{align} = \frac{4}{3}\pi r^2.\frac{dr}{dt} \end{align}

It is given that

\begin{align} \frac{dr}{dt}=\frac{1}{2} cm/s .\end{align}

Therefore, when r = 1 cm,

\begin{align} \frac{dV}{dt}=4\pi(1)^2.(\frac{1}{2})=2\pi\; cm^3/s \end{align}

Hence, the rate at which the volume of the bubble increases is 2π cm3/s.