# calc word problem

• January 6th 2009, 01:07 PM
mathh18
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(Speechless)
• January 6th 2009, 01:43 PM
galactus
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
• January 6th 2009, 03:20 PM
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?
• January 6th 2009, 04:05 PM
galactus
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(Sleepy). You can porobably find its derivation in a calc book or some place.
• January 6th 2009, 04:11 PM
Chris L T521
Quote:

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?