1. ## checking rate of change answer

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

2. Is $\displaystyle \frac{\mathrm{d}V}{\mathrm{d}t} = 1$ cm/s or is $\displaystyle \frac{\mathrm{d}r}{\mathrm{d}t}$? 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 !