# Thread: World Population Problem Using Integrals

1. ## 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.

2. Originally Posted by Kristen
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