Curse

FengShuiDove · Feb 13, 2008, 08:32 AM // 08:32

Hi, call me Dove. Help me on my Diff. Eq.

K, so she's a word problem, dealing with the half-life decay of carbon-14. It gives us:

Q(t) is the amount of carbon-14 at time t and Qnot is the original amount, so Q(t)/Qnot should be able to be determined.

a. Assuming that Q satisfies the differential equation Q' = -rQ, determine the decay constant r for carbon-14.

b. Find an expression for Q(t) at any time t, if Q(t) = Qnot

c. Suppose that certain remains are discovered in which the current residual amount of carbon-14 is 20% of the original amount. Determine the age of these remains.

Any help or a tutorial of some sort would be vastly appreciated.

Father of The Son of God · Feb 14, 2008, 07:46 PM // 19:46

We'll take this in 4 parts. I think the first two parts are the hardest (it's always hardest to get started...), but the last two parts use
basic calculations that you've probably been doing for years. Don't look at math as a bunch of numbers. We'll be using logic along
the way to infer some equations that are not stated explicitly.

First, the differential equation Q' = -rQ. Note that Q (the amount of carbon-14) is a function of t (time), e.g. at t=0 there could be
40 grams of carbon-14, then at t=100 there could be 20 grams of carbon-14. The amount of Q depends only on t, not on anything
else (such as temperature, location, etc.)

So Q' is shorthand for Q'(t), i.e. dQ/dt. To solve the equation, think of it like you're treating dQ/dt as a fraction.
dQ/dt = -rQ ; now get all the Q's on one side, and all the t's on the other side, using simple algebraic manipulation
dQ/Q = -r dt ; now anti-differentiate
ln Q = -rt + C ; where C is a constant of integration

Did you follow all of that above?

To finish up solving, you take the anti-log (i.e. "exponentiate"), and you'll get Q(t) = ke^(-rt), where k is some constant (C got absorbed into k).
Do NOT worry about carrying along all your constants when doing differential equations. You won't be given enough information to find every
constant.

The second part will be to determine r (so that we can answer (a) in your question). Note that there are 3 unknowns/variables aside from r:
Q, k, t. We will actually be able to find r using only 2 equations, and we will get these two equations using an easily looked-up fact (this is where we're going to do some thinking):
"The half-life of carbon-14 is 5730 years."
This implies a first equation: Q(0) = ke^(-r * 0) = k
and a second equation: Q(5730) = ke^(-r * 5730); moreover, Q(5730) = (1/2) * Q(0) -- in words: the amount of carbon-14 at t=5730 years is equal
to 1/2 of the amount of carbon-14 at t=0 years.

Did you follow all of this?

So by substitution, that second equation becomes (1/2) * Q(0) = ke^(-r * 5730)
Now we take the ratio of equation 1 / equation 2, to get: 2 = e^(+r * 5730), which is easily solved, to yield r = (1/5730) ln 2.

The third part that you do with differential equations is, if given the info, determine k.
It looks to me like you got a typo in your info. I believe it should say: Q(0) = Qnot.
Working with this alteration, you're supposed to take your Q(t) = ke^(-rt), and evaluate at t=0, because the given info is Q(t=0) = Qnot.
You get from your equation: Q(0) = ke^(-r * 0) = k, but on the other hand you are given that Q(0) = Qnot. By transitivity of equality,
k = Qnot, so we can specify our equation: Q(t) = Qnot * e^(-rt), where r = (1/5730) ln 2
That finishes part (b) of your question.

The fourth part will be to calculate an age (i.e. value for t) given that Q(t) / Qnot = 0.20.
From our expresssion, we know that Q(t) / Qnot = e^(-rt). Thus we can calculate t from: e^(-rt) = 0.20, where r = (1/5730) ln 2.
You should get t = 13,300 years

Splitisoda · Feb 14, 2008, 09:01 PM // 21:01

What grade is this? This looks way complicated.

Flem · Feb 14, 2008, 09:02 PM // 21:02

Tldr
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I pwnd U · Feb 14, 2008, 09:03 PM // 21:03

DiffyQ is what the subject is I do believe and that is college level stuff. My gf is in it right now.

FengShuiDove · Feb 15, 2008, 12:11 AM // 00:11

Thanks much. I did a little of this myself and cheated a bit by looking up the general exponential decay formula, namely Q(t) = Qnot(e^rt). Once I got the equation integrated, I did a few different things, though you're probably correct =P.

I set r = .5/5730; so decay/time. Plugging that in and solving, I calculated 11,383 years. Admittedly, you're probably more correct as I'm new at this =P.

At any rate, thanks for the help and advice.

/close

doinchi · Feb 15, 2008, 04:26 AM // 04:26

And I thought O Level Physics was tough.