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Homework Help: Volume of balloon - Derivative Question

  1. May 5, 2013 #1
    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. jcsd
  3. May 5, 2013 #2
    What methods of solving differential equations are you familiar with?
  4. May 5, 2013 #3
    I'm not exactly sure what you mean? :S

    Usually I just differentiate lol :S
  5. May 5, 2013 #4
    So you do not understand what a "differential equation" is?
  6. May 5, 2013 #5
    I do, but I'm not sure what kind of "methods" you are referring to
  7. May 5, 2013 #6
    Do you know how to solve diff. eq.'s in any way?
  8. May 5, 2013 #7
    Yeah, I know how to solve differential equations. As in product rule etc? And using the 2nd/3rd derivative? :S
  9. May 5, 2013 #8
    So why can't you solve the differential equation in ii)?
  10. May 5, 2013 #9
    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
  11. May 5, 2013 #10
    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".
  12. May 5, 2013 #11
    Oh ok,

    So you integrate dv/dt to get V(t) which is -500/(2t+1)?
  13. May 5, 2013 #12
    Very well.
  14. May 5, 2013 #13
    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
  15. May 5, 2013 #14
    Now that you have integrated the equation, you have volume as a function of time. Volume is directly related to the radius. Eh?
  16. May 5, 2013 #15
    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
  17. May 5, 2013 #16
    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.

    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
  18. May 5, 2013 #17
    I hope you enjoyed your "aha!" moment :)
  19. May 5, 2013 #18
    Went from happiness to sadness within seconds. This world is too cruel.
  20. May 5, 2013 #19
    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.
  21. May 5, 2013 #20
    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)

    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?
  22. May 5, 2013 #21
    iv) does not mention any time, so I suspect you have to find the function. Again, you know the volume as a function of time, so you should be able get the radius as a function of time, which I think you did in iii). If then you need to know its value at some particular time, just sub the time.
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