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Thread: An application in growth

  1. #1
    No one in Particular VonNemo19's Avatar
    Apr 2009
    Detroit, MI

    An application in growth

    The population of a bacterial culture grows at a rate proportional to the number of bacteria present at any time. After 3 hours it is observed that there are 400 bacteria present. After 10 hours there are 2000 present. What was the initial number of bacteria?

    My attempt:

    It is implied by context that a model for the rate of growth is

    $\displaystyle \frac{dP}{dt}=kP$ which implies that $\displaystyle P(t)=ce^{kt}$.

    But the fact that $\displaystyle P(3)=200$, and $\displaystyle P(10)=2000$ we see that

    $\displaystyle 400=ce^{3k}$ and $\displaystyle 2000=ce^{10k}$

    Solving for $\displaystyle c$ in the first equation gives

    $\displaystyle c=400e^{-3k}$. Substituting into the second...

    $\displaystyle 2000=400e^{-3k}e^{10k}$ and we see that $\displaystyle k=\frac{\ln5}{7}$

    Now, again using the fact that $\displaystyle P(3)=400$ we can conclude that $\displaystyle c=400e^{(-3\ln5)/7}$

    So that $\displaystyle P(t)=400e^{\frac{\ln5}{7}(t-3)}$

    and $\displaystyle P(0)=400e^{\frac{\ln5}{7}(0-3)}$

    Have I done this right. If so, was there a faster way?
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  2. #2
    A Plied Mathematician
    Jun 2010
    CT, USA
    Everything looks good to me. The only thing that might have speeded things up a hair, would have been if you had divided one equation by the other. That would have gotten rid of the c a little quicker. There's not a lot of fluff in your solution, though.
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