1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    Ah, thanks, Dick!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Balloon Problem
Loading...