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Math Help - World Population Problem Using Integrals

  1. #1
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    World Population Problem Using Integrals

    Here is the factoids from the original problem:

    F(0) = 6
    lim(x -> infinity) F(x) = 30
    rate of population growth = (A*e^t)/((.02*A+e^t)^2)
    where t is a continuous variable and t >= 0

    I need to figure out at what t will the population reach 10.

    I managed to figure out that A = 46.15, but after that I get lost.

    The book says I should be solving for t with the equation 30-((46.15)/(.923+e^t)) = 10, but I keep coming up with the equation 6+((46.15)/(.923+e^t)) = 10 and I don't know where I am going wrong.

    If anyone has some insight for me, that would be super. I've been trying to figure this out for two days now and I need some help.
    Thanks.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Kristen View Post
    Here is the factoids from the original problem:

    F(0) = 6
    lim(x -> infinity) F(x) = 30
    rate of population growth = (A*e^t)/((.02*A+e^t)^2)
    where t is a continuous variable and t >= 0

    I need to figure out at what t will the population reach 10.

    I managed to figure out that A = 46.15, but after that I get lost.

    The book says I should be solving for t with the equation 30-((46.15)/(.923+e^t)) = 10, but I keep coming up with the equation 6+((46.15)/(.923+e^t)) = 10 and I don't know where I am going wrong.

    If anyone has some insight for me, that would be super. I've been trying to figure this out for two days now and I need some help.
    Thanks.
    You are given:

    F'(t)=\dfrac{Ae^t}{(0.02A+e^t)^2}

    so now integrate this to find F(t) and use that together with the given data to solve the problem.

    F(t)=30-\dfrac{A}{0.02A+e^t}

    the 30 is there because that is the value of F(\infty) where the second term on the right is 0

    CB
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