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Topic: Wine and Water (Read 1630 times) 

Pietro K.C.
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Wine and Water
« on: Sep 18^{th}, 2002, 8:04am » 
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This is a very cool one I solved some time ago. Suppose a 1liter bottle of wine is hanging from the ceiling, and from it hangs a 1liter bottle of water. On the bottom of both bottles there is a small hole, through which a constant amount of fluid pours out (neglecting pressure differences). So the wine bottle is becoming empty and the water bottle remains full, though the concentration of wine is increasing. Assuming that wine and water mix instantly and completely, find out the concentration of wine in the water bottle after the top one is empty. Give two different methods of solution.

« Last Edit: Nov 7^{th}, 2002, 8:51am by Pietro K.C. » 
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James Fingas
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Re: NEW PROBLEM: WINE AND WATER
« Reply #1 on: Sep 18^{th}, 2002, 11:52am » 
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This is a differential equations problem. All you're saying is that for c, the concentration of wine in the water bottle, dc/dt = k  k*c (k is the constant rate of drippage. The k term is the wine drippage into the water bottle, and the k*c term is the wine drippage out of the water bottle) c(0) = 0 (we start with pure water) c(1/k) = ? (we want c when the wine bottle is empty) The solution is the exponential function heading towards 1. It is pretty simple to find the numerical value of the concentration. The thing that bothers me about this question is that the rates of drippage won't be constant. As the wine bottle gets emptier, it will drip a lot slower.


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Pietro K.C.
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Re: NEW PROBLEM: WINE AND WATER
« Reply #2 on: Sep 18^{th}, 2002, 1:01pm » 
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Yeah, I assumed the drip rate was constant for simplicity, but also for another reason. I was afraid someone would resort to integrals or differential equations and do away with the challenge. That's why I said "give two different solutions". It is possible to solve this one without any of this, and it's a lot prettier. Give it a shot! If you want a more complicated problem, that WILL probably require differential equations, you could calculate the drip rate as a function of pressure (Bernoulli's law), and hence of time. Assuming wine and water are equally dense,the drip rate in the bottom bottle will be constant. I haven't tackled this one, though.


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James Fingas
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Re: NEW PROBLEM: WINE AND WATER
« Reply #3 on: Sep 18^{th}, 2002, 1:20pm » 
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If I were a real sicko, I would use a series to find the solution. I'd start by mixing in two steps: 1) Pour all the wine into the water bottle 2) Pour half the result out Then I would try 1) Pour half the wine into the water bottle 2) Pour a third of the result out 3) Pour the rest of the wine into the water bottle 4) Pour a third of the result out Then I take the limit as the number of steps goes to infinity.


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Pietro K.C.
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Re: NEW PROBLEM: WINE AND WATER
« Reply #4 on: Sep 19^{th}, 2002, 12:07pm » 
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Yes, you truly WOULD have to be a sicko to take the mixing as the limit of THAT process. Think simpler.


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James Fingas
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Re: NEW PROBLEM: WINE AND WATER
« Reply #5 on: Sep 19^{th}, 2002, 12:42pm » 
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If I were even more of a sicko, I'd consider an infinite number of bottles of water hanging one below the other, with the wine bottle at the top. I'd solve for the amount of wine in the nth bottle after the wine bottle is empty, sum all the results except for the top bottle, then whatever wine is not in the other bottles is in the first one. QED. I suppose if I were going to think simpler, I'd try this alternate process: 1) Dump the water out of the water bottle into a bucket, and dump the wine bottle into the water bottle. I get zero wine in the bucket. Next case: 1) Dump half the water out of the water bottle into the bucket, then dump half the wine bottle into the water bottle. 2) Dump half the result into the bucket, then dump the rest of the wine into the bucket. Now take the limit as the number of steps goes to infinity. This is probably a little more direct than my last process.


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Pietro K.C.
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Re: NEW PROBLEM: WINE AND WATER
« Reply #6 on: Sep 19^{th}, 2002, 1:39pm » 
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Now you're talking! I wasn't quite able to see how the limit of the last process could be taken easily  maybe I'm just dense. To solve this one, I considered the water concentration instead  it isn't too hard to see that, if we divide the water and wine into n equal parts, and do like you did for n = 2, the concentration of water decreases by a factor of (11/n) at each step. So the water concentration is (11/n)^{n} in the end, which is a limit we all know pretty well. I think part of the reason I like this problem is because, when I first tried it, I didn't have a CLUE how to begin (come on, I was just 16 ), and after a while just decided to do this to see where it was going. I never really expected to get anywhere, and then bam!  the answer right in front of me. It was one of my first tastes of "experimental" mathematics.


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