# Suppose you are blowing a spherical bubble, filling it in with air at a uniform rate of 400mm^3/s. How fast is the radius of the bubble increasing by the time it is already 20mm long?

Apr 4, 2018

$\frac{1}{4 \pi}$ $\text{mm/s}$

#### Explanation:

$V = \frac{4}{3} \pi {r}^{3}$ (volume of sphere)

if $r = 20 m m$, $V = \frac{4}{3} \pi {\left(20 m m\right)}^{3} = \frac{32000}{3} \pi {\left(m m\right)}^{3}$

differentiate both sides with respect to time:

$\frac{\mathrm{dV}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left(\frac{4}{3} \pi {r}^{3}\right)$
$\frac{\mathrm{dV}}{\mathrm{dt}} = 4 \pi {r}^{2} \cdot \frac{\mathrm{dr}}{\mathrm{dt}}$

volume of air increasing at uniform rate of 400mm^3/s = $\frac{\mathrm{dV}}{\mathrm{dt}}$

substitute: $400 {\left(m m\right)}^{3} / s = 4 \pi {r}^{2} \cdot \frac{\mathrm{dr}}{\mathrm{dt}}$

$\frac{\mathrm{dr}}{\mathrm{dt}} = \frac{400}{4 \pi {r}^{2}} {\left(m m\right)}^{3} / s$
$\frac{\mathrm{dr}}{\mathrm{dt}} = \frac{400}{4 \pi {\left(20 m m\right)}^{2}} {\left(m m\right)}^{3} / s$ ($r = 20 m m$)
$\frac{\mathrm{dr}}{\mathrm{dt}} = \frac{1}{4 \pi} \frac{m m}{s}$

$\frac{\mathrm{dr}}{\mathrm{dt}}$ is rate radius is increasing