My Math Forum Half-Life

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 January 8th, 2011, 11:09 PM #1 Newbie   Joined: Jan 2011 Posts: 6 Thanks: 0 Half-Life I have a very interesting half-life problem that I cannot figure out for the life of me. I am starting with a container of arbitrary size and has 100L of water in it. It is at a constant high temperature, so the half-life (due to evaporation) of the water is 8 hours. But, every hour, 2L of water of the same temperature is poured in. It takes 2 minutes to pour in the extra water. This is not necessarily a realistic question but it applies in theory. I want to graph the amount of water in the container. Any thoughts? This is driving me crazy!
 January 9th, 2011, 05:52 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,948 Thanks: 1139 Math Focus: Elementary mathematics and beyond Re: Half-Life Some thoughts: You start with 100L of water. Every hour you lose 50/8 liters of water due to evaporation, but add 2 liters. The formula 100 - 4.25h, where h is time in hours, should give how much water is in the tank after h hours.
 January 9th, 2011, 09:29 AM #3 Newbie   Joined: Jan 2011 Posts: 6 Thanks: 0 Re: Half-Life I don't know if this is starting to get into calculus or not, but I would like to be able to know how much water is in the take at any given time, not just at hour intervals. The relationship will not be linear in this case, since the more water that is in the tank the faster it is evaporating. Is that right?
 January 9th, 2011, 09:33 AM #4 Newbie   Joined: Jan 2011 Posts: 22 Thanks: 0 Re: Half-Life Starting with 100 L, with t = time in minutes, I think it might be this: 100 * 0.5^(t/480) + 1/60t actually, that doesn't account for the water added every two minutes because the amount to halve would no longer be 100 so it would be (100 + 1/60t) * 0.5^(t/480) Is that right?
January 9th, 2011, 09:42 AM   #5
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Re: Half-Life

Quote:
 Originally Posted by jerakahol . . . since the more water that is in the tank the faster it is evaporating. Is that right?
I believe the rate of evaporation will be constant. You may need a piecewise function to accurately determine the amount of water in the tank at any given time.

January 9th, 2011, 09:45 AM   #6
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Re: Half-Life

Quote:
Originally Posted by greg1313
Quote:
 Originally Posted by jerakahol . . . since the more water that is in the tank the faster it is evaporating. Is that right?
I believe the rate of evaporation will be constant.
I think that's true in real life, but I think he's suggesting the water has a half-life of 8 hours, which means the more water in it, the more it will lose. Maybe its some kind of alien radioactive water?

 January 9th, 2011, 06:44 PM #7 Newbie   Joined: Jan 2011 Posts: 6 Thanks: 0 Re: Half-Life Ha, yes. It is radioactive alien water. I realize the problem is flawed, but in theory the water evaporates according to a half-life. What I ultimately want to do with this problem is discover the balancing point of the water. At what point does the decay balance with the addition of water? First off, the water will turn from 100L to 50L in the first eight hours, while you only add 16L of water. So now (roughly, not actually) there are only 66L of water after 8 hours. After another 8 hours, 66L turns to 33L but we add another 16L, leaving us with 49L. Eventually we reach a point where the decay balance the addition, and we wind up bouncing back and forth between two amounts when its at its highest and lowest.
 January 10th, 2011, 12:25 AM #8 Newbie   Joined: Jan 2011 Posts: 22 Thanks: 0 Re: Half-Life I can't think how to work in the 'it takes 2 mins to pour in' but, if it gets poured in at a steady rate across the hour, the answer would be\; $(100 + \frac{1}{60}t) * 0.5^{\frac{t}{480}}$
 January 10th, 2011, 01:33 PM #9 Newbie   Joined: Jan 2011 Posts: 6 Thanks: 0 Re: Half-Life I like it. However, if we assume that we start with exactly 100L at t=0 and add 2L of water over the course of dt=1/30hr, even 2 minutes later by the time we have poured the water in, we do not have 102L, we have less. Of course, because the 100L is "evaporating" slightly for 2 minutes and also the water is technically evaporating -- albeit slowly -- while we are pouring it in, once it hits the container.
 January 10th, 2011, 01:43 PM #10 Newbie   Joined: Jan 2011 Posts: 6 Thanks: 0 Re: Half-Life At the moment I'm liking the piecewise suggestion. Since I'm concerned about where the function balances, we could say that there would be a starting point, an ending point, and then something in between. Then, at the end, we would start back at the beginning. So what I need to figure out is the initial value at t=0 that would cause the final to value at t=8hrs to equal it exactly. I'll give it a shot and get back to you. Thanks to everyone for your help so far!!!

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