# 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

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?