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Air escaping from a balloon

by sonofjohn
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sonofjohn
#1
Jan12-09, 08:55 AM
P: 76
Air is escaping from a balloon at a rate of R(t) = 60/(1+t2) cubic feet per minute, where t is measured in minutes. How much air, in cubic feet, escapes during the first minute?

dr/dt = 60/(1 + t2)
if I plug the time in (1) for (t) I should get 60/2 or 30

This seems a tad bit too easy compared to the rest of the problems I have been doing.
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Dick
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Jan12-09, 09:46 AM
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That's because 30 cu ft/min is the rate that air is escaping at t=1. At t=0 it's 60 cu ft/min. Neither one answers the question "much air, in cubic feet, escapes during the first minute".
HallsofIvy
#3
Jan12-09, 09:54 AM
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Quote Quote by sonofjohn View Post
Air is escaping from a balloon at a rate of R(t) = 60/(1+t2) cubic feet per minute, where t is measured in minutes. How much air, in cubic feet, escapes during the first minute?

dr/dt = 60/(1 + t2)
if I plug the time in (1) for (t) I should get 60/2 or 30

This seems a tad bit too easy compared to the rest of the problems I have been doing.
R(1)= 30 cubic feet per minute is the rate at which air is escaping at t= 1 minute. It is NOT the amount of air that has escaped in that minute.

If the rate were a constant, then the amount of air escaping in "t" minutes would be that constant rate times t. Here, it is NOT a constant so the amount of air escaping between t= 0 and t= 1 (the first minute) is the integral of the rate function from t= 0 to 1.

Mark44
#4
Jan12-09, 03:39 PM
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Air escaping from a balloon

Quote Quote by sonofjohn View Post
Air is escaping from a balloon at a rate of R(t) = 60/(1+t2) cubic feet per minute, where t is measured in minutes. How much air, in cubic feet, escapes during the first minute?

dr/dt = 60/(1 + t2)
You have R(t) and dr/dt both equal to the same expression, which is almost certainly not what you meant. Per your description, R(t) is a rate, so is already the derivative of something (radius?) with respect to time.

dr/dt is the derivative of r (what is r? Is it the same as R?) with respect to time.

Try to use the same variable names consistently or you will increase the opportunities for confusion in both yourself and anyone reading your work.
Try to keep a function and its derivative separate. It's conceivable that you will soon be working with a function and two of its derivatives. If you don't keep them all separate, you'll get hopelessly confused.
Quote Quote by sonofjohn View Post
if I plug the time in (1) for (t) I should get 60/2 or 30

This seems a tad bit too easy compared to the rest of the problems I have been doing.
sonofjohn
#5
Jan13-09, 08:39 AM
P: 76
Ok, I see that 30t is the rate at which the air escapes the balloon. To find how much air leaves the balloon in a minute, I must take the integral from 0 to 1 of 30t.

Once completed, I got 15 cubic feet escaped the balloon in one minute. Is that even possible? If the air escapes at 30cubic feet a minute, wouldn't the answer be over 30 cubic feet a minute?
Dick
#6
Jan13-09, 08:43 AM
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Sure. It would have to be over 30 cu ft. (Be careful with your units). But WHY are you integrating 30t? They gave you dr/dt. Integrate that to get r(t).
sonofjohn
#7
Jan13-09, 09:00 AM
P: 76
Ahh, that does make sense that I don't integrate the amount of air that escapes the balloon within a minute but rather the rate at which the air escapes the balloon per minute. If I do that I should integrate 60/(1 + t^2) from 0 to 1. Once done I get 30 as an answer.
Dick
#8
Jan13-09, 09:03 AM
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Quote Quote by sonofjohn View Post
Ahh, that does make sense that I don't integrate the amount of air that escapes the balloon within a minute but rather the rate at which the air escapes the balloon per minute. If I do that I should integrate 60/(1 + t^2) from 0 to 1. Once done I get 30 as an answer.
No. You don't get 30 as an answer. The answer should be between 60 and 30. Since those were the beginning and ending rates.
sonofjohn
#9
Jan13-09, 09:11 AM
P: 76
I see what I did, I took the derivative instead of the integral. I need to use the arctan formula instead.

60(1/(1 + t^2)) when you take the integral with the arctan formula

formula:

integral(du/(a^2 + u^2)) = 1/a(arctan(u/a) + C)

so

integral 60(1/1(arctan(t/1)) from 0 to 1

Once solved I get pi/4
Mark44
#10
Jan13-09, 09:23 AM
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No you don't. Pi/4 is (roughly) 3/4. As Dick said, your answer should be between 30 and 60. Haven't you forgotten something?
Dick
#11
Jan13-09, 09:23 AM
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Yes. pi/4*60. Cubic feet.
sonofjohn
#12
Jan13-09, 09:25 AM
P: 76
Quote Quote by Mark44 View Post
No you don't. Pi/4 is (roughly) 3/4. As Dick said, your answer should be between 30 and 60. Haven't you forgotten something?
Yep I wrote it wrong I should actually get 15pi
Mark44
#13
Jan13-09, 09:35 AM
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Right. Also note that 15pi is about 45 (very rough approximate), so it fills the bill of being between 30 and 60.
sonofjohn
#14
Jan13-09, 09:38 AM
P: 76
I see. How are the bounds or how is the gap created?


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