# A spherical balloon inflated at a rate of 4pi cm^3/sec. initial volume of balloon 0, how fast is radius expanding after 9 sec (r=0.03m)?

Mar 14, 2018

$\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right) {|}_{r = 3} = \frac{1}{9}$ (in units of $\frac{c m}{s}$)

#### Explanation:

Denoting volume by V, the volume of the sphere is

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

where $r$ (and hence $V$) is a function of time.

Rearranging,

$r = {\left(\frac{3 V}{4 \pi}\right)}^{\frac{1}{3}}$

Noting

$\frac{\mathrm{dr}}{\mathrm{dt}} = \frac{\mathrm{dr}}{\mathrm{dV}} \frac{\mathrm{dV}}{\mathrm{dt}}$ (chain rule)

and further noting

$\frac{\mathrm{dr}}{\mathrm{dV}} = \left(\frac{1}{3}\right) {\left(\frac{3 V}{4 \pi}\right)}^{- \frac{2}{3}} \left(\frac{3}{4 \pi}\right)$

it is given that

$\frac{\mathrm{dV}}{\mathrm{dt}} = 4 \pi$ (noting units for later!)

So

$\frac{\mathrm{dr}}{\mathrm{dt}} = \frac{\mathrm{dr}}{\mathrm{dV}} \frac{\mathrm{dV}}{\mathrm{dt}} = \left(\frac{1}{3}\right) {\left(\frac{3 V}{4 \pi}\right)}^{- \frac{2}{3}} \left(\frac{3}{4 \pi}\right) \left(4 \pi\right)$

$= {\left(\frac{3 V}{4 \pi}\right)}^{- \frac{2}{3}}$

$= {\left({r}^{3}\right)}^{- \frac{2}{3}}$

$= \frac{1}{r} ^ 2$

So (with $r$ in $c m$)

$\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right) {|}_{r = 3} = \frac{1}{3} ^ 2 = \frac{1}{9}$ (in units of $\frac{c m}{s}$)