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rate of decay - June 4th 2009, 06:11 PM

I'm having a bit of trouble figuring out where to start with this question, and what it wants. Can anyone help me out?

A sample of radioactive material has decay constant .25, where time is measured in hours. How fast will the sample be disintegrating when the sample size is 8 grams? For what sample size will the sample size be decreasing at the rate of 2 grams per day?


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Re: rate of decay - June 4th 2009, 07:27 PM

For which class is this? Most of these types of problems ask for a certain amount of material at a specific time, or they might ask at what time does some amount decay to another amount, etc...

Since you are not given a specific time that you want to find the rate of change or the amount of substance, I assume you are seeking the values at the initial time (t=0), which makes this problem rather easy using differential equations.

Your material is decaying with a .25 proportionality constant to your amount of material at that time, so your equation is:

N' = -.25N

N' = rate of change of the material
N = amount of material

If N = 8 grams, then N' = -.25(8) = -2 grams/hour

If N' = -2 grams/day, convert to -1/12 grams/hour, then -1/12 = -.25N, and N = 1/3 grams

I'm actually pretty tired, so hopefully this is correct.

Here is the full derivation of the decay formula:

Your rate of change is proportional to your amount of material:

N' = sN (where your proportionality constant is .25 in the negative direction)

N' = -.25N, this is just a separable differential equation, after integrating you will get:

ln[N(t)] - ln[N(0)] = -.25t, and let's let N(0) = A (which is the initial value)

ln[N/A] = -.25t
N/A = e^(-.25t)
N(t) = Ae^(-.25t)

This equation gives the amount of material at time t, N(t), with initial material, A.

If we differentiate, we will get:

N'(t) = -.25Ae^(-.25t)

Where N'(t) gives the rate of change of the material at time t.

My assumption is that either this problem does not give enough information, because to find a rate of change or an initial amount (given the rate of change), you have to have a specific time as well (yes the rate of change is different for different values of t, this is an exp function).

And that is why I assumed that they want the values when t=0, which is easier to work with than t=pi or something.

But if t=0, then our entire exponential portion turns to 1 and we are again left with:

N' = -.25A, which is the same equation I used up top. :-)

Last edited by pbandjay; June 4th 2009 at 07:44 PM.
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Re: rate of decay - June 5th 2009, 12:27 AM

It's differential calculus with applications to social studies and economics, first year. Thank you for your help - I wasn't quite sure what to do without an initial amount.


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