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Math Help - Differential Eq stuck again

  1. #1
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    Post Differential Eq stuck again

    A radioactive substance decays at a rate directly proportional to the amount present. If the substance is present in the amount of Qo g initially (t = 0), find the amount present at any time t.

    I started off using the restricted growth model of

    dQ/dt = kQ(C-Q)

    That is about as far as I made it.
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by vodka View Post
    A radioactive substance decays at a rate directly proportional to the amount present. If the substance is present in the amount of Qo g initially (t = 0), find the amount present at any time t.

    I started off using the restricted growth model of

    dQ/dt = kQ(C-Q)

    That is about as far as I made it.
    Well I can help you integrate that:

    \frac{dQ}{dt} = kQ(C-Q) \Rightarrow \int_{Q_0}^{Q} \frac{dQ}{Q(C-Q)} = \int_0^t k\, dt

    Now \frac1{Q(C-Q)} = \frac1{C}\left(\frac1{Q} + \frac1{C-Q}\right),

    \int_{Q_0}^{Q} \frac{dQ}{Q(C-Q)} = \int_0^t k\, dt \Rightarrow  \frac1{C}\left(\int_{Q_0}^{Q} \frac{dQ}{Q} + \int_{Q_0}^{Q} \frac{dQ}{C-Q}\right) = kt

    \frac1{C}\left(\ln\left(\frac{Q}{Q_0}\right) - \ln\left(\frac{C - Q}{C - Q_0}\right) \right) = kt

    \left(\frac{Q(C - Q_0)}{Q_0(C - Q)}\right) = e^{Ckt}

    Q = \frac{CQ_0}{Q_0 + (C- Q_0)e^{-Ckt}}
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  3. #3
    Senior Member Peritus's Avatar
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    After making the following substitution:

    <br />
z = \frac{1}<br />
{Q}

    we get the following simple ODE:

    \frac{{dz}}<br />
{{dt}} + kcz - k = 0

    I'm sure you'll be able to take it from here...
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by vodka View Post
    A radioactive substance decays at a rate directly proportional to the amount present. If the substance is present in the amount of Qo g initially (t = 0), find the amount present at any time t.

    I started off using the restricted growth model of

    dQ/dt = kQ(C-Q)

    That is about as far as I made it.
    You apparently have the wrong equation. If Q' is directly proportional to Q then
    Q' = -kQ
    (negative because the amount present is getting smaller.)

    I'm guessing you will have no trouble solving this equation, but if you do simply say so.

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    You apparently have the wrong equation. If Q' is directly proportional to Q then
    Q' = -kQ
    (negative because the amount present is getting smaller.)

    I'm guessing you will have no trouble solving this equation, but if you do simply say so.

    -Dan

    Cheers again thats what I was looking for,

    dQ/dt =-kQ

    Q(0) = Qo
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  6. #6
    Lord of certain Rings
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    Quote Originally Posted by topsquark View Post
    You apparently have the wrong equation. If Q' is directly proportional to Q then
    Q' = -kQ
    (negative because the amount present is getting smaller.)
    -Dan
    I thought so too... But I thought probably the laws have changed over the past 4 years when I did not touch physics*

    *Actually I just dont want to admit that I have got rusty
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