Balloon Problem

  • Thread starter Brandon_R
  • Start date
  • #1
25
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?
 

Answers and Replies

  • #2
HallsofIvy
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No, it is not. Had you shown your work, I could have helped you more.
 
  • #3
25
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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])
 
  • #4
25
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Could anyone spare the time to shed some light on this problem please. Thanks :)
 
  • #5
Dick
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Could anyone spare the time to shed some light on this problem please. Thanks :)
That looks fine to me. Halls may have called it wrong since you wrote -1/18 pi instead of -1/(18 pi).
 
  • #6
HallsofIvy
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Ah, thanks, Dick!
 

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