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Math Help - Help on exponential question

  1. #1
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    Help on exponential question

    Thank you for your help,
    Last edited by kiranizzle; December 3rd 2009 at 09:26 AM.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    I'll give you the answer for the growing case in i and ii

    Note y = Ae^(kt)

    y(0) = A


    y = y(0)e^(kt) = y0 e^(kt)

    if y(t1) = y1

    Then y1 = y0 e^(kt1)

    1/kln(y1/y0) = t1

    Similarly for ii y(tn) = yn

    yields 1/k ln(yn/y0) = tn

    For your Lake problem You know your solution can't be correct

    If P(t)=100,000e^(-t/1000) then P(0) =100,00

    But if the concentration is initially .1% and there are 108 m^3

    P(0) = .108m^3
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  3. #3
    MHF Contributor Calculus26's Avatar
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    lake pblm revisited

    It occurred to me you mean 10^8 m^3 and 10^5 m^3/day

    otherwise if it were 108 and 105 you would have a mud puddle.

    so dP/dt = pollution in -pollution out

    dp/dt = 0 - %pollution*flow rate


    dP/dt = -(P(t)/10^8) (10^5 m^2/day)

    dP/dt = -P(t)/1000 P(0) = .001* 10^8 = 10^5



    dp/P = -1/1000dt

    ln(P) = -t/1000 +C

    e^(ln(P) = e^(-t/1000+C) = e^C e^(-t/1000)

    Since e^C is a constant we can just call it C


    P = C*e^(-t/1000)

    P(0) = 100,000 = C

    P = 100,000 e^(-t/1000)

    To find when it 1/10 original solve 10,000 = 100,000 e^(-t/1000)

    1/10 = e^(-t/1000)

    1/1000ln(10) = t
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