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Math Help - can u solve the problem???

  1. #1
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    can u solve the problem???

    radium decomposes at a rate propotional to the amount present.if p percent of the original amount disappears in l years how much will remain at the end of 2l years?
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  2. #2
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    Let m be the percentage mass of radium present at time t.

    Let m_0 be the mass present at time t=0.

    \frac{dm}{dt} \propto m

    \frac{dm}{dt}=km where k is a real constant (k is actually negative since this is "exponential decay". You can write k=-b or something to show that you know it is negative. I just left it because it doesn't change the overall answer at the end.)

    Integrating gives:

    \int \frac{1}{m} dm=\int k dt

    ln m=kt + C

    m=e^{kt+C}

    m=Ae^{kt} where A=e^C

    using the intial conditions:

    m=m_0 e^{kt}

    when t=I years:

    m_I=m=m_0e^{Ik}

    At t=2I years:

    m=m_0e^{2Ik}

    m=m_0(e^{Ik})^2

    m=m_0(e^{Ik})(e^{Ik})

    m=m_I(e^{Ik})

    and I think we're done! =D

    P.S: I did change notation. I used m instead of p. If you like you can work it through again with m=p or just write this at the end.
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  3. #3
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    Hello, padmini priya!

    Radium decomposes at a rate propotional to the amount present.
    If p percent of the original amount disappears in L years.
    how much will remain at the end of 2L years?
    Let Q = quantity of radium at time t.

    We have: . \frac{dQ}{dt} \:=\:-kQ \quad\Rightarrow\quad \frac{dQ}{Q} \:=\:-kt

    Integrate: . \ln(Q) \:=\:-kt + C \quad\Rightarrow\quad Q \:=\:e^{-kt+c} \quad\Rightarrow\quad Q \:=\:Ce^{-kt}


    When t = 0,\;Q = Q_o (initial amount)
    We have: . Q_o \:=\:Ce^0 \quad\Rightarrow\quad C \:=\:Q_o

    The function (so far) is: . Q \;=\;Q_o\,e^{-kt}


    When t = L,\;Q \:=\:(1-p)Q_o

    So we have: . (1-p)Q_o \;=\;Q_o\,e^{-kL} \quad\Rightarrow\quad e^{-kL} \:=\:1-p \quad\Rightarrow\quad -kL \:=\:\ln(1-p)

    . . Hence: . k \:=\:-\frac{\ln(1-p)}{L}

    The function is: . Q \;=\;Q_o\,e^{\text{-}\left[\text{-}\frac{ln(1-p)}{L}\right]t}  \quad\Rightarrow\quad Q \;=\;Q_o\,e^{\frac{\ln(1-p)}{L}\,t}


    When t = 2L\!:\;\;Q \;=\;Q_o\,e^{\frac{\ln(1-p)}{L}\cdot2L} \;=\;Q_o\,e^{2\ln(1-p)} \;=\;Q_o\,e^{\ln(1-p)^2}

    We have: . Q \;=\;Q_o(1-p)^2


    Therefore: . \frac{Q}{Q_o} \;=\;(1-p)^2\quad _{\Leftarrow\;\;\text{Percent of radium after 2L years}}

    . . \overbrace{_{\text{Fraction of radium after 2L years}}}^{\Uparrow}

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