- #1

- 25

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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?

- Thread starter Brandon_R
- Start date

- #1

- 25

- 1

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

HallsofIvy

Science Advisor

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No, it is not. Had you shown your work, I could have helped you more.

- #3

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

dv/dt = -2 in

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

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]r

dr/dt = -1/(18[tex]\pi[/tex])

- #4

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Could anyone spare the time to shed some light on this problem please. Thanks :)

- #5

Dick

Science Advisor

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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).Could anyone spare the time to shed some light on this problem please. Thanks :)

- #6

HallsofIvy

Science Advisor

Homework Helper

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Ah, thanks, Dick!

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