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I am studying for the first actuary exam, and I came across this problem in the very first section (reviewing calculus, algebra, set theory, etc.) I have many questions of my own, so please bear with me. The actual question is:

"A model for world population assumes a population of 6 billion at reference time 0, with population increasing to a limiting population of 30 billion. The model assumes that the rate of population growth at time $t \geq 0$ is $\frac{Ae^t}{(.02A+e^t)^2}$ billion per year, where $t$ is regarded as a continuous variable. According to this model, at what time will the population reach 10 billion (nearest .1)?"

They actually post steps to the solution, but they do so without many words. There are a couple steps that I am unsure on. Hopefully someone can help me out. Here are the steps to the solution:

"We define $F(t)$ to be the population at time $t$. Then $F(0) = 6$, $\lim_{t\to\infty} F(t)=30$, and $F'(t)= \frac{Ae^t}{(.02A+e^t)^2}$. Then (using substitution $u = .02A + e^t$), we have $F(s) - F(0) = \int_{0}^{s}\frac{Ae^t}{(.02A+e^t)^2}dt = -\frac{A}{.02A+e^t}|_{t=0}^{t=s} = \frac{A}{.02A+1} - \frac{A}{.02A + e^s}$, so that $F(s) = 6 + \frac{A}{.02A+1} - \frac{A}{.02A+e^s}$."

Okay, to this point I generally comprehend what is going on. $F(0) = 6$ is our starting population of 6 billion, and the population will keep growing infinitely until it reaches a cap of 30 billions. The growth rate given in the question is the rate of change, thus being $F'(t)$. As I understand it, $s$ is the time constant that we are trying to solve for, the time constant that will give us the desired population amount when plugged into our function F. How did they come up with the trick to use $F(s)-F(0)$? I understand how it works. Is this used frequently in population growth problems, something that I should memorize? Lastly, I understand using $u=.02A+e^t$. This gives $du=e^t dt$.Thus the integral becomes $\int \frac{du}{u^2}$ (from to $s$). But what about the $A$ in the numerator? What happened to it? Other than those things, I'm good thus far. Now for the next part:

"Then $\lim_{s\to\infty}F(s) = 6 + \frac{A}{.02A+1} = 30$, $\frac{A}{.02A+1}=24$, $A=46.15$. Therefore, $F(s) = 30 - \frac{46.15}{.923 + e^s}$.

Right here I am struggling a little bit. Plugging in the value for A into in the equation for $F(s)$ gives $6 + 23.998 - \frac{46.15}{.923 +e^s}$. Is it okay to go ahead and round that to 30 since our answer has to be to the nearest tenth?

In order to have $F(t) = 10$, we have $30 - \frac{46.15}{.923+e^s}=10$, which gives $s=.325$.

There at the end is where I'm confused. Just to double check, $F(s) = F(t)$, correct? I think it makes sense to me if that is the case. Going into the problem, I thought that I would solve for $t$ by finding the time it would take to go from 6 to 10 billion. The end at first glance appears to go from 30 down to 10 billion, which threw me off. I guess that we simply found a function/equation ($F(s)=F(t)=30-\frac{46.15}{.923+e^s}$) that would give us $t$ for what ever population size is plugged in on the other side? Last but not least, will there be any population growth problems on exam P? Sorry for such a long post.

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Good luck on Exam P!

The problem is that the solver rounded, writing $A=46.15$ when he should have written $A=600/13$. The actual solution then is $$F(s) = 30 -\frac{600}{12+13e^s}$$ which you should work out for yourself to confirm.

Then you just need to solve $F(s)=10$, getting $$s=\log\left(\frac{\frac{600}{30-10}-12}{13}\right)=\log\frac{18}{13}\approx 0.32542$$

Also, the whole $s$ vs $t$ business is just because the solver used $t$ as the dummy variable in the integration, so had to come up with another variable name for the argument of $F$. The name of that variable isn't important, it's just a placeholder.

And at the end, there is indeed a typo. He should have said "In order to have $F(s)=10$..." since he then solved for $s$.

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  • $\begingroup$ That makes sense now. Thanks so much! $\endgroup$ – Josh Jan 15 '16 at 4:26

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