related rates.

[maeveoneill]

Helium is blown into a spherical weather balloon that increases steadily in radius at the rate of 5cm/s.

a) find the rate of increase of the volume of the balloon when the radius is 30cm.
b) Find the rate of increase of volume of the balloon when the volum is pi/6 m^3.

i have solved part a) and got 180,000 pi cm^3/s which is correct. any help with part b??? i just need help setting up teh equation.. you dont need to show me all the steps. thankssssss...

 

[stapel]

You have V = (4/3)(pi)(r³), and you are given that dr/dt = 5.

Plug the given volume in for "V", and solve for the radius "r".

Then differentiate the volume formula with respect to time "t". You know dr/dt, you were given V, and you found r. Plug these values in, and solve for dV/dt, the only remaining variable.

Eliz.

 

[Daniel_Feldman]

For part a, the answer should be 18,000 I think.

 

[soroban]

Hello, maeveoneil!

]Helium is blown into a spherical weather balloon that increases steadily in radius at the rate of 5cm/s.

a) Find the rate of increase of the volume of the balloon when the radius is 30cm.
b) Find the rate of increase of volume of the balloon when the volume is pi/6 m^3.

i have solved part a) and got 180,000 pi cm^3/s which is correct. any help with part b???
i just need help setting up the equation. . . . . really? It's the same equation!

We have: .\(\displaystyle V\;=\;\frac{4}{3}\pi R^3\) . [1]

Hence: .\(\displaystyle \frac{dV}{dt}\;=\;4\pi R^2\left(\frac{dR}{dt}\right)\) . [2]

We already know that: .\(\displaystyle \frac{dR}{dt}\,=\,5\) cm/s.
. . We need \(\displaystyle R\) at that instant.

At that instant, \(\displaystyle V\,=\,\frac{\pi}{6}\) cm\(\displaystyle ^3\).
. . From [1]: .\(\displaystyle \frac{\pi}{6}\,=\,\frac{4}{3}\pi R^3\;\;\Rightarrow\;\;R\,=\,\frac{1}{2}\)

Substitute into [2]: .\(\displaystyle \frac{dV}{dt}\:=\:4\pi\left(\frac{1}{2}\right)^2(5)\:=\:5\pi\) cm\(\displaystyle ^3\)/s

 

[maeveoneill]

yup it should be 18,000 i typed tha in wrong.
sorry could you actually work out the problem for me.. im not getting it. thank you/

 

[Daniel_Feldman]

At that instant, \(\displaystyle V\,=\,\frac{\pi}{6}\) cm\(\displaystyle ^3\).

Remember, though, volume given was in meters, so you need to convert to cm.


Unless, of course, the given was a typo, and the volume is, in fact, in cm^3.