Question

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

Answer 1

The air bubble is in the shape of a sphere.
Let r be the radius of bubble and V be the volume of bubble at any time t.
Then, rate of change of radius $\frac{dr}{dt}=\frac{1}{2}cm/s$ and r=1 cm
Now, volume of the bubble $V=\frac{4}{3}\pi r^3$
On differentiating w.r.t. t, we get
Rate of volume increasing $\frac{dV}{dt}=\left(\frac{4}{3}\pi \right)\left(3r^2\frac{dr}{dt}\right)$
$=4\pi r^2\frac{dr}{dt}=4\pi (l{)}^2\frac{1}{2}(onputtingr=1and\frac{dr}{dt}=\frac{1}{2})$
$=2\pi c{m}^3/s$
Hence, the rate at which the volume of the bubble increases is $2\pi c{m}^3/s$.