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Homework Help: Balloon Problem

  1. Jun 13, 2010 #1
    A punctured balloon, in the shape of a sphere, is losing air at the rate of
    2 in.3/sec. At the moment that the balloon has volume 36π cubic inches,
    how is the radius changing?



    I got

    dr/dt = -1/18π in/sec.

    Is that correct?
     
  2. jcsd
  3. Jun 13, 2010 #2

    HallsofIvy

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    No, it is not. Had you shown your work, I could have helped you more.
     
  4. Jun 13, 2010 #3
    Sorry about that Sir, here is what i did.

    Volume = (4/3)[tex]\pi[/tex]r3

    dv/dt = -2 in3/sec

    dv/dt = 4[tex]\pi[/tex]r2dr/dt

    I substituted 36[tex]\pi[/tex] into the volume equation to get the radius which is equal to 3

    dr/dt = -2/(4[tex]\pi[/tex]r2)

    dr/dt = -1/(18[tex]\pi[/tex])
     
  5. Jun 16, 2010 #4
    Could anyone spare the time to shed some light on this problem please. Thanks :)
     
  6. Jun 16, 2010 #5

    Dick

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    That looks fine to me. Halls may have called it wrong since you wrote -1/18 pi instead of -1/(18 pi).
     
  7. Jun 16, 2010 #6

    HallsofIvy

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    Ah, thanks, Dick!
     
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