1. checking rate of change answer

hello,

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

this is what i did

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

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

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

$

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

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

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

=0.0032 cm/sec

2. Is $\frac{\mathrm{d}V}{\mathrm{d}t} = 1$ cm/s or is $\frac{\mathrm{d}r}{\mathrm{d}t}$? I think you have mixed up the radius and the volume.....

3. seems like i have, can you show me the right way to work this problem.

4. Originally Posted by sigma1
seems like i have, can you show me the right way to work this problem.
Just substitute the correct things into the correct parts of your expression for dV/dt !