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Math Help - differential equation problem(i think)

  1. #1
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    differential equation problem(i think)

    I didn't know what to title this post

    Suppose that the population of bacteria P(t) changes according to

    P(t) = 4000(1 + (2t/1+t^2)) where t is given in hours and t is eual to or greater than 0

    a) when does the population reach the maximum number?
    b) What is the maximum population?

    I been looking through this from a previous test and trying to figure out how I got the answer...According to my professor I did some things wrong and got marked down 3 out of 10 points, but I still ended up with the right answer.
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  2. #2
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    Quote Originally Posted by jeph View Post

    P(t) = 4000(1 + (2t/1+t^2)) where t is given in hours and t is eual to or greater than 0

    a) when does the population reach the maximum number?
    b) What is the maximum population?
    First, use paranthesis

    P=4000 \left(1+\frac{2t}{1+t^2} \right)
    P=4000+\frac{8000t}{1+t^2}
    \frac{dP}{dt}=\frac{(8000t)'(1+t^2)-(8000t)(1+t^2)'}{(1+t^2)^2}
    \frac{8000(1+t^2)-8000t(2t)}{(1+t^2)^2}=0
    8000(1+t^2)-8000t(2t)=0
    1+t^2-t(2t)=0
    -2t^2+t^2+1=0
    t^2=1 and t>0
    t=1
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  3. #3
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    what happened to the 2 8000s here?

    8000(1+t^2)-8000t(2t)=0
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  4. #4
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    Quote Originally Posted by jeph View Post
    what happened to the 2 8000s here?

    8000(1+t^2)-8000t(2t)=0
    I divided both sides by 8000.

    The reason why I skipped explanations is because since you are in a Calculus class you need to be able to do this stuff well already. Thus, I assumed an explanation was not necessary.
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  5. #5
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    Oh!

    LOL

    I didnt think of that....
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