Amount Remaining = 18.517508 grams
Now let’s suppose that the original amount gets constantly increased at a linear rate - let’s say 5 grams of substance X gets added every 30 days. This is a continual process, which is to say the rate is 5 grams per 30 days or 1 gram per 6 days or one sixth of a gram per day. (The amount does not get increased by 5 gram “chunks” at precise 30 day intervals.)
So finally, the question:
Getting back to the original problem, you start with 40 grams of substance X, half-life of 45 days, and you have 18.517508 grams after 50 days.
How much would you have after 50 days if the original amount is 40 grams but is increasing by 5 grams per 30 days (or one sixth gram per day, etc)?
halflife = 45;
days = 50;
grams(1)= 40;
gramsf = grams(1) * 1/2^(days/halflife)
gramsadd = 1/6;
for x = 2:51
grams(x)= (gramsadd + grams (x-1)) * 1/2^(1/halflife);
end
grams(51)
gramsf is grams final when you aren’t adding anything.
for x = 2:51 means that x varies from 2 to 51 in increments in one (which is required when using (x) to name the index. I had to start at 2 because the initial, given value of grams(1) was already taken (and you can’t use zero).
This is a pretty simple problem to solve one you know about differential equations; you’re asking for a solution to the equation dN/dt = - a N + b, where a is the decay rate (ln 2 divided by the half-life) and b is the rate at which you’re adding material. The solution turns out to be
N(t) = C * 2[sup]- t / t[sub]1/2[/sub][/sup] + b * t[sub]1/2[/sub] / ln 2
where C is a constant determined by how much of the stuff you have initially. In your case, you will have
N(t) = ( 29.180 g * 2[sup]- t / t[sub]1/2[/sub][/sup] ) + 10.820 g
which gives 24.328 g after 50 days. Santo Rugger’s answer is slightly different from mine, I believe, because he used discrete one-day steps to calculate the final answer rather than approximating it as a continuous process.
If you want an intuitive way of thinking about this, try this on for size: it’s not too hard to see that there’s a given amount of material where the decay exactly balances the rate at which you’re adding material. If you have too much material, then the decay rate will be greater than the addition rate; if you have too little, the decay rate will be less than the addition rate. This amount happens to be b/a in my notation; in yours, it would be (addition rate)*(half-life)/(ln 2). So imagine dividing your initial material into two piles: one with this special amount, and one with the rest. If you only add material to the first pile, and ignore the second one, the first pile will stay constant while the second pile will decay normally — just with a smaller amount of material to start with.
This is what makes the solution a striaght forward differential equation as** Santo Rugger** explains. If the additions were instead discreet, the problem becomes a bit messier, requiring a different function to discribe the decay after each addition.
Of note in MikeS’s explaination is that the amount of material will always converge to the condition where the supply rate just balances the decay rate: The “leftover” pile will always converge to zero. Even if the supply rate starts out higher than the decay rate, material will accumulate to balance the two rates.
Just to see if I understand this correctly, after 100 days, the amount would equal 17.07376 grams, correct?
Now that I think about it, maybe the problem could be simplified even further if we just concentrated on the amount being added? In this way, we wouldn’t have to do that calculation for the special amount of material where decay rate = rate of material being added.
So, how about an equation just for solving the problem with no beginning amount?
That is to say, substance X has a half-life of 45 days. You start out with 0 grams and the amount increases (on a continual basis) of .166666 grams per day. How much of it do you have after 50 days?
Sorry for bumping this thread but the math on this is rather complex.
Basically, as I stated previously, how would the equation be set up so that no beginning amount is present?
I didn’t study differential equations in college and it seems something that would be well worth it. In college, I took 2 semesters of calculus but I’ve never found any practical application for it. I guess you have to get past all that introductory “delta x” material and “chain rules” before you get to things you can actually use.
Sorry for the late reply — didn’t see that you had asked a couple more questions.
Correct.
Well, technically speaking you don’t have to do that calculation; it’s just a handy way of thinking about how the decay works. Besides, I don’t quite see what you mean about “concentrating on the amount being added”; both “the amount being added” and “the original amount” are going to decay, and to accurately predict the total you’re going to have to take them both into account.
You would just change the coefficient in front of the exponentially decaying part; in general, it will be N(0) - b * t[sub]1/2[/sub] / ln 2, as I mentioned in post #7. This means that if N(0) = 0 (i.e., you start with no material), your solution will be
N(t) = -10.82 g * 2[sup]-t/t[sub]1/2[/sub][/sup] + 10.82 g
If you plot this function, you’ll find that it starts out at zero, increases somewhat quickly to start with, and then increases more and more slowly as the material accumulates and the decay processes become more important. The amount of material will, as noted by Kevbo, asymptotically approach 10.82 g — just from below this time.
MikeS, I think when (s)he says “concentrate the amount being added”, they mean you can just use a discrete algebraic solution.
FWIW, after 100 days, my MatLab code shows 17.01 grams, so they’re still slightly different, but they will always be less than 1/6 (.167) off after a short time.
