Question

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?

Answer 1

as you know the shape of air bubble is spherical. therefore, volume of spherical bubble is $\bf{V=\frac{4}{3}\pi r^3}$

so, $\bf{V=\frac{4}{3}\pi r^3}$
now differentiate V with respect to t,
$\bf{\frac{dV}{dt}=\frac{4}{3}\pi\frac{dr^3}{dt}}\\\\=\bf{\frac{4}{3}\pi.3r^{3-1}\frac{dr}{dt}}\\\\=\bf{4\pi r^2\frac{dr}{dt}}$

hence, rate of change of volume with respect to time is $\bf{\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}}$
here $\bf{\frac{dr}{dt}}$ is the rate of change of radius with respect to time.
given , $\bf{\frac{dr}{dt}}=\frac{1}{2}cm/s$
and radius , r = 1cm
so, $\bf{\frac{dV}{dt}=4\pi (1)^2 \frac{1}{2}}$
hence, $\bf{\frac{dv}{dt} =2π cm^3/s}$