1. ## calc word problem

in 1950, a research team digging near folsom, new mexico, found charred bison bones along with some leaf-shaped projectile points that had been made by a paleo-indian hunting culture. it was clear from the evidence that the bison had been coked and eaten by the makers of the points, so that carbon-14 dating of the bones made it possible for the researchers to determine when the hunters roamed North America. tests showed that the bones contained between 27% and 30% of their original carbon-14. use this information to show that the hunters lived roughly between 9000 BC and 8000 BC.

i know to use the equation y(t)= C*e^(kt)
where y represents the population at a given time, C represents the population, k represents the rate with respect to time, and t represents time
i also know that 1950, 27% and 30%, and 9000 and 8000 BC are important but i don't know what to do with them...especially since they are ranges and not exact numbers

2. The decay constant is $k=\frac{-1}{T}ln(2)$

Since the half life of Carbon 14 is about T=5750 years, then we have:

$k=\frac{-1}{5750}ln(2)=-.00012$

Now, let A be the original amount.

27% remaining:

$.27A=Ae^{-.00012t}$

$t\approx 10911$ years.

30% remaining:

$.30A=Ae^{-.00012t}$

$t\approx 10033$ years.

1950-10911=8961 BC

1950-10033=8083 BC

3. Thanks Galactus. The only part I didn't follow was how did you get k= (-1/5750) ln(2)
I know that 5750 is the half life but where did ln2 come from?

4. That is derived from DE's. Just use it as an identity, so to speak. That is how to find k if you know half life. No need to derive it. We can do it, but I would just as soon not right now. You can porobably find its derivation in a calc book or some place.

5. Originally Posted by mathh18
Thanks Galactus. The only part I didn't follow was how did you get k= (-1/5750) ln(2)
I know that 5750 is the half life but where did ln2 come from?
The equation for decay is $y=y_0 e^{-kt}$.

You want to find the time it takes for half of the substance to decay, or when $y=\tfrac{1}{2}y_0$.

Substituting this into the equation, we have $\tfrac{1}{2}y_0=y_0e^{-kt}\implies\tfrac{1}{2}=e^{-kt}$

Taking the natural log of both sides, we end up with $\ln\left(\tfrac{1}{2}\right)=-kt\implies t=-\frac{\ln\left(\tfrac{1}{2}\right)}{k}=\frac{\ln\l eft(2\right)}{k}$.

This value $t$ is the half life of the substance. It is usually denoted by $T_{\frac{1}{2}}$ or $\lambda$.

As a result, the decay constant is $k=\frac{\ln(2)}{T_{\frac{1}{2}}}$.

Does this make sense?