hello,

the volume of a sphere is given by $\displaystyle V =\frac{4}{3}\pi r^3$. an elastic balloon is being blown up so that the radius is increasing at the rate of $\displaystyle 1cm s^{-s}$ calculate the rate at which the volume of the baloon is increasing when the radius is 5 cm

this is what i did

$\displaystyle \frac {dV}{dt}$ = $\displaystyle 1cm s^{-s} $

find$\displaystyle \frac {dr}{dt}$

V =$\displaystyle \frac{4}{3}\pi r^3$ therefore $\displaystyle \frac {dV}{dr}$= $\displaystyle 4\pi r^2$

aplying chain rule $\displaystyle \frac {dV}{dt}$= $\displaystyle \frac {dV}{dr}$ . $\displaystyle

\frac {dr}{dt}

$

1= $\displaystyle 4\pi r^2$ x $\displaystyle \frac {dr}{dt}$

$\displaystyle \frac {dr}{dt}= $$\displaystyle \frac {1}{4\pi r^2}$

when r = 5

$\displaystyle \frac {dr}{dt}= $$\displaystyle \frac {1}{4\pi 5^2}$

=0.0032 cm/sec