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  1. #1
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    Exclamation 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
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  2. #2
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    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
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  3. #3
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    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?
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  4. #4
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    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.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mathh18 View Post
    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?
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