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Math Help - Bacteria growth

  1. #1
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    Bacteria growth

    A type of bacteria increases continuously at a rate proportional to the number present. If there are 1000 present at a given time and 2000 present two hours later, how many hours (from the initial time) will it take for the number of bacteria to reach 5000?

    Could someone nudge me in the right direction?
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  2. #2
    MHF Contributor matheagle's Avatar
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    I haven't done one of these in years, so I had to solved it from scratch.
    Assuming that dX/dt=kX, we have the separable DE...
    {dX\over dt}=kX or {dX\over X}=kdt
    integrating we have \ln X=kt+c.
    Solving for X(t) we have X(t)=ae^{kt}
    where a=e^c=X(0).
    From what you said, I believe that X(0)=1000
    so X(t)=1000e^{kt}. Next we solve for k.
    But the trick is not to do it completely.
    X(2)=2000=1000e^{2k}, so 2=e^{2k}.
    There is no need to try to approximate k.
    The base is here 2, not e. Instead, insert this into our function...
    X(t)=1000e^{kt}=1000(e^{2k})^{t/2}=1000(2)^{t/2}.
    Now check to see if this makes sense. Look at X(0) and X(2).
    Finally, set 1000(2)^{t/2}=5000, which yields (2)^{t/2}=5
    or (t/2)\ln 2=\ln 5, thus t=2\ln 5/\ln2.
    Last edited by matheagle; March 11th 2009 at 10:23 PM.
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  3. #3
    Member Mentia's Avatar
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    So what you have here is a differential equation:

    \frac {dx}{dt}=cx

    where c is a constant of proportionality and x indicates the number of bacteria present. Basically this says the rate of change of bacteria is equal to some number, c, times the number of bacteria present.

    Ok so lets solve this differential equation. Using the usual method:
    \frac {dx}{dt}=cx ---> \frac{dx}{x} = cdt

    Now integrate both sides:

    \int_{}^{} \frac{dx}{x} = \int_{}^{} cdt

    Then,

    ln(x) = ct + {k}_{1}

    where k1 is the constant of integration.

    Then,

    x(t) = {k}_{2}e^{ct}

    Now we apply our given conditions, lets call "a given time" t = 0:


    x(0) = 1000 = {k}_{2}e^{0}, x(2) = 2000 = {k}_{2}e^{c(2)}


    The first equation immediately gives: {k}_{2} = 1000

    Then the second equation gives: 2000 = 1000e^{2c} --> c = ln(\sqrt[]{2})

    Thus:

    x(t) = 1000e^{ln(\sqrt[]{2})t} = 1000 (\, \sqrt[]{2} \, )^t

    Then we solve:

    5000 = 1000 (\, \sqrt[]{2} \, )^t

    which gives,

    t = \frac {ln(5)}{ln( \, \sqrt[]{2} \, )} \approx 4.64 hours.

    Note that Matheagle and my answers look different because we wrote our x(t)'s slightly differently but in the end they are exactly the same.
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  4. #4
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    Quote Originally Posted by matheagle View Post
    I haven't done one of these in years, so I had to solved it from scratch.
    Assuming that dX/dt=kX, we have the separable DE...
    {dX\over dt}=kX or {dX\over X}=kdt
    integrating we have \ln X=kt+c.
    Solving for X(t) we have X(t)=ae^{kt}
    where a=e^c=X(0).
    From what you said, I believe that X(0)=1000
    so X(t)=1000e^{kt}. Next we solve for k.
    But the trick is not to do it completely.
    X(2)=2000=1000e^{2k}, so 2=e^{2k}.
    There is no need to try to approximate k.
    The base is here 2, not e. Instead, insert this into our function...
    X(t)=1000e^{kt}=1000(e^{2k})^{t/2}=1000(2)^{t/2}.
    Now check to see if this makes sense. Look at X(0) and X(2).
    Finally, set 1000(2)^{t/2}=5000, which yields
    t=2\ln 5/\ln2.
    Why do you not have to find k?
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  5. #5
    MHF Contributor matheagle's Avatar
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    If you can avoid solving for k, you're better off.
    You're never asked to solve for it anyhow.
    Most books solve for it and get an annoying roundoff.
    The problem is that these books are forcing you to have a base of e.
    In this problem the correct base is 2 or likewise \sqrt{2}.
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