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Math Help - The Shroud of Turin

  1. #1
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    The Shroud of Turin

    The mathematical law is that the rate of decay of C14 is proportional to the quanitity of C14. That is, C14' = -kC14.

    Let t=0 be the year that the cotton for the shroud was harvested. Let "age" be the age of the shroud in 1989 when the analysis of the shroud was conducted. The facts that are given are these:

    1. The amount of C14 in the shroud as measure by scientists in 1989 (C14(age)), was about 92% of the amount in living plants today.

    2. The amount of C14 in living plants today is assumed to be the same as the amount of C14 in living plants when the shroud was made. (The shroud was made from plant material, thus, C14(age)= 0.92 times C14(0).

    3. The function, C14(t), is a solution to C14' = -kC14, and so that function must be C14(t) = Ae^-kt, for some constant A.

    4. A serviceable value for k is 0.0001245.

    *During your calculations you'll be faced with an algebraic expression of the form p = e^qt, which you'll have to solve for t. That means getting t out of the exponent, and that means using ln.

    The question is to find the age of the Shroud of Turin as well as to find the half-life of Carbon 14 (aka C14).

    I'm at a loss as to how I should go about solving this, so please, any help will be much appreciated!

    Thank you.

    -Maggie
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  2. #2
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    Off Topic: A long time ago I saw something really funny on the History channel about the Shroud. Since the face on the Shourd is not so well seen it is hard to identity who the person it, however there are some scientists who believe the face is Leonardo. And this Shroud is kept in a Church, were it is treated as a holy object. Now imagine that the scientists are right; that would mean all those Church members are praying to Leonardo.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by blurain View Post
    The mathematical law is that the rate of decay of C14 is proportional to the quanitity of C14. That is, C14' = -kC14.

    Let t=0 be the year that the cotton for the shroud was harvested. Let "age" be the age of the shroud in 1989 when the analysis of the shroud was conducted. The facts that are given are these:

    1. The amount of C14 in the shroud as measure by scientists in 1989 (C14(age)), was about 92% of the amount in living plants today.

    2. The amount of C14 in living plants today is assumed to be the same as the amount of C14 in living plants when the shroud was made. (The shroud was made from plant material, thus, C14(age)= 0.92 times C14(0).

    3. The function, C14(t), is a solution to C14' = -kC14, and so that function must be C14(t) = Ae^-kt, for some constant A.

    4. A serviceable value for k is 0.0001245.

    *During your calculations you'll be faced with an algebraic expression of the form p = e^qt, which you'll have to solve for t. That means getting t out of the exponent, and that means using ln.

    The question is to find the age of the Shroud of Turin as well as to find the half-life of Carbon 14 (aka C14).

    I'm at a loss as to how I should go about solving this, so please, any help will be much appreciated!

    Thank you.

    -Maggie
    the half-life, t_h, is given by:

    t_h = \frac {\ln 2}k



    Let C be C14. if we want to solve C = Ae^{-kt}, we can solve for t as follows:

    C = Ae^{-kt} ...................divide both sides by A

    \Rightarrow \frac CA = e^{-kt} .............take the log of both sides

    \Rightarrow \ln \frac CA = \ln e^{-kt} .......simplify

    \Rightarrow \ln \frac CA = -kt .............solve for t

    \Rightarrow t = - \frac {\ln \frac CA}k = \frac {\ln A - \ln C}k

    can you continue?
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    the half-life, t_h, is given by:

    t_h = \frac {\ln 2}k



    Let C be C14. if we want to solve C = Ae^{-kt}, we can solve for t as follows:

    C = Ae^{-kt} ...................divide both sides by A

    \Rightarrow \frac CA = e^{-kt} .............take the log of both sides

    \Rightarrow \ln \frac CA = \ln e^{-kt} .......simplify

    \Rightarrow \ln \frac CA = -kt .............solve for t

    \Rightarrow t = - \frac {\ln \frac CA}k = \frac {\ln A - \ln C}k

    can you continue?
    I'm sorry--I'm not quite sure how to continue this...is it possible for you to elaborate?
    Last edited by blurain; November 13th 2007 at 12:55 PM.
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