Volume of balloon - Derivative Question

1. May 5, 2013

kingstar

1. The problem statement, all variables and given/known data
The volume of a spherical balloon of radius 'r' is Vcm^3, where V =4/3pir^3
The volume of the balloon increases with time 't' seconds according to the formula

dV/dt = 1000/(2t+1)^2, t>0

i) Find an expression in terms of 'r' and 't' for dr/dt
ii) Given that V = 0 and t = 0, solve the differential equation

dV/dt = 1000/(2t+1)^2, to obtain V in terms of t

iii) Find the radius of the balloon at time t =5

iv) Find the rate of increases of the radius of the balloon

2. Relevant equations

3. The attempt at a solution
i) Well we know dV/dt = (dv/dr) x (dr/dt)
Which means that I found dr/dt to be - (250/ (pir^2(2t+1)^2))

For the rest of them, i have no clue what to do. :S

Thanks in advance

2. May 5, 2013

voko

What methods of solving differential equations are you familiar with?

3. May 5, 2013

kingstar

I'm not exactly sure what you mean? :S

Usually I just differentiate lol :S

4. May 5, 2013

voko

So you do not understand what a "differential equation" is?

5. May 5, 2013

kingstar

I do, but I'm not sure what kind of "methods" you are referring to

6. May 5, 2013

voko

Do you know how to solve diff. eq.'s in any way?

7. May 5, 2013

kingstar

Yeah, I know how to solve differential equations. As in product rule etc? And using the 2nd/3rd derivative? :S

8. May 5, 2013

voko

So why can't you solve the differential equation in ii)?

9. May 5, 2013

kingstar

I'm not sure what to do?

Am i just meant to differentiate it?

I thought you would have to integrate it so you would get the original equation?

I got V = 1000t/(2t+1)^2 + C

Is that it? :S

10. May 5, 2013

voko

You are given $\frac {dV} {dt}$. You need to find $V(t)$. Your answer to this is wrong, which you can easily check by differentiating your "solution".

11. May 5, 2013

kingstar

Oh ok,

So you integrate dv/dt to get V(t) which is -500/(2t+1)?

12. May 5, 2013

voko

Very well.

13. May 5, 2013

kingstar

How do i find the radius at t=5?

I know what dr/dt is but i cant integrate it because it has two variables :S

14. May 5, 2013

voko

Now that you have integrated the equation, you have volume as a function of time. Volume is directly related to the radius. Eh?

15. May 5, 2013

Alcubierre

By methods, voko meant if you can use integrating factor, Laplace transform, etc, I believe.

As for the dr/dt, have you tried separating the variables so you have the 'r' on one side and 't' on the other?

Last edited: May 5, 2013
16. May 5, 2013

kingstar

OH, so because we now have V = -500/(2t+1)

We can let V = 4/3pir^3 and let t=5!

Ahaha, thanks for being so patient with me Voko! i truly appreciate it! :D

Ugh, So i got the radius as 2.21 from plugging the above in.

iv)
However when I sub this radius and time into dr/dt = (250/ (pir^2(2t+1)^2))
I'm not getting the right answer
:'(

What am i doing wrong?

Last edited: May 5, 2013
17. May 5, 2013

voko

I hope you enjoyed your "aha!" moment :)

18. May 5, 2013

kingstar

Went from happiness to sadness within seconds. This world is too cruel.

19. May 5, 2013

voko

This is a physical forum, not philosophical. So you have to render your sadness in a way we can deal with - if we are supposed to.

20. May 5, 2013

kingstar

Lol, maybe you could help me with the last part? :( I did it but the answer book says im wrong

So i got the radius as 2.21 from plugging V = 4/3pir^3 and t = 5 into V = -500/(2t+1)

iv)
However when I sub this radius and time into dr/dt = (250/ (pir^2(2t+1)^2))
I'm not getting the right answer
:'(

What am i doing wrong?

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