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Math Help - Worded Differential equation

  1. #1
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    Worded Differential equation

    I have this differential which I have no idea what to do with! I think it's because there's lots of pointless stuff in here that isn't needed.

    If you visit the medieval Great Hall, at the top of the high street in Winchester, just 20 minutes by car/train north of Southampton, you will see a circular table, 18 foot in diameter, that is claimed to be King Arthur's Round table. If Arthur really did exist, it is most likely he was a local chieftan in the dark ages with poor written records immediately following the Roman withdrawal in the 5th century AD. The table was recorded as having been in Winchester in 1484, but it was not known how old it really was. In 1976 the Council had it carbon dated. The decay rate of Carbon=14 atoms to Carbon-12 is 1.245 x 10^-4 per year. The number og radioactive Carbon-14 atoms decaying was measued to be

    dN/dt = -6.08 per minute per gramme. In living wood the rate of decay is 6.68 particles per minute per gramme. Work out how old the table is and hence deduce whether this could be Arthur's round table?



    Could someone help?

    I recognised that dN/dt = -kN

    So seperating variables:

    ln N = -kt + C

    I'm stuck after this.
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  2. #2
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    \ln N = -kt + C

     N =e^{ -kt + C}

     N =e^{ -kt}\times e^{ C}

    As   e^{ C} is constant then we can say

     N =N_0e^{ -kt} where  N_0 = e^{ C} and is the initial amount of carbon before decay.
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  3. #3
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    Quote Originally Posted by pickslides View Post
    \ln N = -kt + C

     N =e^{ -kt + C}

     N =e^{ -kt}\times e^{ C}

    As   e^{ C} is constant then we can say

     N =N_0e^{ -kt} where  N_0 = e^{ C} and is the initial amount of carbon before decay.
    Yeh I forgot to post the further bit.

    I got ln N = -kt + c
    N = e^-kt+c
    N = Ae^-kt

    t=0, N = 6.68

    So A = 6.68

    N = 6.68e^-kt

    I'm stuck after this...

    Failed to get any further
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