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Math Help - Integrals and initial-value problems?

  1. #1
    Newbie oObutterfly-chaserOo's Avatar
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    Unhappy Integrals and initial-value problems?

    I thought I understood this problem, until I solved it and got an answer of -4000 bacteria. Here is the problem:


    A population of bacteria is changing at a rate of

    dP/dt = 3000 / ( 1+ 0.25t)

    where t is the time in days. The initial population is 1000. Write an equation that gives te population at any time t, and find the population when t = 3 days.


    From this information I know that P(0) = 1000 and I went to integrate the function.

    The equation I got was P = 3000 ln(1-0.25t) + k

    But I don't think this is right because, well, -4158.5 just doesn't seem like a plausable answer to me. What should I have done differently?
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  2. #2
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    Quote Originally Posted by oObutterfly-chaserOo View Post
    I thought I understood this problem, until I solved it and got an answer of -4000 bacteria. Here is the problem:


    A population of bacteria is changing at a rate of

    dP/dt = 3000 / ( 1+ 0.25t)

    where t is the time in days. The initial population is 1000. Write an equation that gives te population at any time t, and find the population when t = 3 days.


    From this information I know that P(0) = 1000 and I went to integrate the function.

    The equation I got was P = 3000 ln(1-0.25t) + k

    But I don't think this is right because, well, -4158.5 just doesn't seem like a plausable answer to me. What should I have done differently?
    You want to integrate like this:

    \int\frac{3,000}{1+0.25t}dt=3000\int\frac{1}{1+0.2  5t}dt

    You want this in the form \int\frac{f'(t)}{f(t)}dt, so you need a .25 in the numerator. So use the fact that 1=\frac{.25}{.25} to write the integral as \frac{3000}{.25}\int\frac{.25}{1+0.25t}dt=12,000\l  n(1+0.25t)+k

    Using your P(0)=1,000 gives:

    12,000\ln(1)+k=1,000.

    k=1,000

    So your function should be:

    P(t)=12,000\ln(1+0.25t)+1,000
    Last edited by adkinsjr; January 14th 2010 at 11:14 PM.
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