# 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 15, 2018

The rate of increase of Radius $\left(r\right)$ at time $t = 9 s$ is $\frac{\mathrm{dr}}{\mathrm{dt}} = \frac{1}{9} \left(\frac{c m}{s}\right)$ $=$ The rate of expansion of the balloon.

#### Explanation:

At $t = 9 s$ the radius of the balloon is $r = 0.03 m = 3 c m$

Let the Instantaneous Volume of the balloon be $V$

So, here $V = \frac{4 \pi {r}^{3}}{3}$ , where $r$ is the instantaneous radius of the spherical balloon.

Given that the rate of change of volume is $\frac{4 \pi c {m}^{3}}{s}$

$\Rightarrow \frac{\mathrm{dV}}{\mathrm{dt}} = \frac{4 \pi c {m}^{3}}{s}$

$\Rightarrow \frac{d \left(\frac{4 \pi {r}^{3}}{3}\right)}{\mathrm{dt}} = \frac{4 \pi c {m}^{3}}{s}$

$\Rightarrow \left\{\frac{\cancel{4 \pi}}{3}\right\} \frac{d \left({r}^{3}\right)}{\mathrm{dt}} = \frac{\cancel{4 \pi} c {m}^{3}}{s}$

$\Rightarrow \left\{\frac{1}{\cancel{3}}\right\} \cancel{3} {r}^{2} \frac{\mathrm{dr}}{\mathrm{dt}} = \frac{1 c {m}^{3}}{s}$

$\Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}} = \left(\frac{1}{r} ^ 2\right) \frac{c {m}^{3}}{s}$

Now at $t = 9 s$, $r = 3 c m$ :-

$\Rightarrow \frac{\mathrm{dr}}{\mathrm{dt}} = \left(\frac{1}{{3}^{2} c {m}^{2}}\right) \frac{c {m}^{3}}{s}$

$\therefore \frac{\mathrm{dr}}{\mathrm{dt}} = \frac{1}{9} \left(\frac{c m}{s}\right)$