Here we go. Carbon 14 dating assumes that carbon dioxide on earth today has the same
radioactive content it did centuries ago. If this is true then the amount of carbon
14 absorbed by a tree that grew hundreds of years ago should be the same as the amount
of carbon 14 absorbed by a tree growing today. A piece of charcoal has about 19% as much
carbon-14 as a new piece of charcoal. Knowing that the half-life of carbon-14 is 5715 years,
how long ago was an ancient tree burned to create the charcoal?
Now this is not a science class, it is a calculus class, and our solving of these separable
differential equations, led us into our work here with exponential growth and decay models.
So there is a probably a nice, easy formula for doing at a time that streamlines the process,
but for now because we are not in a science class, we just hopefully recognize we are
talking about here exponential decay, so going to that model, a= c times e to the kt power,
and I have to come up with a formula so I can finish the problem. So we have an initial
amount, a final amount, we have e which is of course approximately 2.718 and I don’t
know my constants, that’s what we need to figure out right now using the fact that half-life
is 5715 years. Half-life, you lose 50%. When you talk about percentage of change, if you
don’t care about the amount of change but the percent of change, the initial amount
is arbitrary. So, let’s just say that we start off with 2 units of this carbon 14 and
we end with 1. I am saying units, because again I am not a science teacher and I don't
want to say the wrong unit and have someone mad at me. So we start off with 2 of something
and end up with 1. We have that times e to the kt power, we actually know what t is,
it is of course 5715. Divide both sides by 2 and you will see that 50% come up. I could
have just said that the initial amount was 1 and our final amount was 0.5. So dividing
both sided by 2, you get 0.5 is equal to e to the 5715 times k power. I am trying to
keep my board from rattling there, in case you were wondering what I was doing. You can’t
solve for variables when it is exponent, so we take the natural of of both sides, that’s
the easiest with the base of e here. Of course we are using the natural log or log base e
because that will cancel out with the base of e on this exponential function. And we
get the natural log of ½ =5715k. And finishing this up by dividing both sides of the equation
by 5715, k= 1 over 5715 times the natural log of ½. You can make that a decimal right
now, I am going to keep this in this form for a little bit longer. I would caution you
with rounding off very much at all is going to cause our final answer to be off by A LOT.
If you are going to take that to decimal approximation, leave it off to 4-5 decimals, but I am going
to leave it off in exact format to minimize the round off errors by as much as possible.
Okay so now... I am putting this in the way of my problem, so let me put this down here:)
My advanced picture in picture. Now that we know the constant, we can write the equation.
Our final amount is going to be our initial amount times e to the1/5715 times the
natural log of ½ t. I got a lot of information here right? I know that I am looking for t,
how old this piece of charcoal, our final amount is 19% of what we started with so we
can let our initial amount be a 100 and let A be 19 and that would add an extra step of
dividing both sides by 100; or I could start with an initial amount of 1 and say that the
final amount is 0.19 units of this carbon-14. 1 e to the 1/5715 times the natural log of
½ t. Again, you can’t solve for variables when it is exponent, so to get rid of that
base e we take the natural of both sides. That’s going to cancel out. So we get ln
(0.19) = 1/5715 times the natural log of ½ t. We are going to solve for t, so we are
going to multiply both sides of the equation by 5715, we are going to divide both sides
by the natural log of ½ to undo the multiplication and now with our scientific calculators, and
rounding it off to 2 decimal places, t comes out to be approximately 13,692.73 years. That’s
one old clump of charcoal! That’s also the end of my last example. I am Mr. Tarrou. BAMM!
Go do your homework