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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?

Not around so much now that school's started

"Live a good life.

If there are gods and they are just,

then they will not care how devout you have been,

but will welcome you based on the virtues you have lived by.

If there are gods, but unjust, then you should not want to worship them.

If there are no gods, then you will be gone, but will have lived a noble life

that will live on in the memories of your loved ones."

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.

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.

Not around so much now that school's started

"Live a good life.

If there are gods and they are just,

then they will not care how devout you have been,

but will welcome you based on the virtues you have lived by.

If there are gods, but unjust, then you should not want to worship them.

If there are no gods, then you will be gone, but will have lived a noble life

that will live on in the memories of your loved ones."