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Math Help - population decrease problem

  1. #1
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    population decrease problem

    After treatment, a bacteria population decreases at a rate of 20% an hour.

    A. Whan t=0 there are 150,000 bacteria. What is an equation that satisfies the equation?

    B. How many remain after 6 hr?

    C. After how long do half remain?

    I have detirmined that dN/dt = -0.20N.
    Other than that, I am stumped.
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  2. #2
    MHF Contributor
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    Good start, now separate the variables...

    \frac{1}{N} \frac{\mathrm{d}N}{\mathrm{d}t}\ =\ -\frac{1}{5}

    ... and integrate. Just in case a picture helps...



    ... where



    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to t, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

    Expressing the constant as ln A rather than c can help the subsequent steps.

    Spoiler:



    _________________________________________
    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
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  3. #3
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    Im positive I do not follow you.
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  4. #4
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    You say you arrived at \frac{dN}{dt}= -.20N. "Separating variables", referred to by Tom, means rewriting the equation as \frac{dN}{N}= -.20dt ( = -\frac{1}{5}dt, of course) and then integrating both sides: ln(N)= -.20t+ C.

    Taking e to the power of both sides gives e^{ln(N)}= e^{-.20t+ C} or N= e^{-.20t}e^{C} which we can write as N= C'e^{-.20t} with C'= e^C.

    Now you are told "When t=0 there are 150,000 bacteria" so that 15000= C'e^{-.20(0)} and that tells you what C'[ is.

    For (B): "How many remain after 6 hr?" just evaluate that at t= 6.

    For(C) "After how long do half remain?" set N equal to (1/2)(150000)= 75000 and solve for t.
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