# Problem with the logistic function

1. Oct 24, 2012

### la6ki

1. The problem statement, all variables and given/known data

The native Hawaiians lived for centuries in isolation from other peoples. When foreigners finally came to the islands they brought with them diseases such as measles, whooping cough, and smallpox, which decimated the population. Suppose such an island has a native population of 5000, and a sailor from a visiting ship introduces measles, which has an infection rate of 0.00005. Also suppose that the model for spread of an epidemic described in Example 3* applies.

a. Write an equation for the number of natives who remain uninfected. Let t represent time in days.

2. Relevant equations

*The model for spread of an epidemic described in Example 3: $\frac{dy}{dt}$=k(1-y/N)y

Where in modeling population growth y is the population, k is a growth constant, and N is the carrying capacity.

3. The attempt at a solution

Okay, so the general solution to this differential equation is:

y=$\frac{N}{1+be^{-kt}}$

where b=$\frac{N-y_{0}}{y_{0}}$

(y$_{0}$ is the initial population size).

If we apply this model to the current problem, y could represent the number of people affected. Therefore, the number of people unaffected is x=N-y. Solving for x:

N-$\frac{N}{1+be^{-kt}}$

Setting y$_{0}$=1 (since there is one affected individual initially), the solution of the above equation is:

x=$\frac{24955000}{4999+e^{0.00005t}}$

Now, I thought I had solved the problem, but when I looked at the correct answer, it was:

x=$\frac{24955000}{4999+e^{0.25t}}$

I checked several times if I made a mistake anywhere in my calculations, but I couldn't find it (not that it's still not possible I keep missing something obvious; hopefully you guys will help me spot it if so).

I went back to the chapter where the logistic function was introduced for the first time and there it said (right after the general solution was given):

"This function was introduced in Section 4.4 on Derivatives of Exponential Functions in the form:

G(t)=$\frac{m}{1+(m/G_{0})e^{-kmt}}$

where m is the limiting value of the population, G$_{0}$ is the initial number present, and k is a positive constant."

First of all, I can't see how this equation follows from N-$\frac{N}{1+be^{-kt}}$ (or how the latter could be rewritten in this form. But second, I tried using it and still couldn't get the correct answer. I decided to go back to the chapter where this second form was first introduced and I found the following:

G(t)=$\frac{m}{1+(m/G_{0}-1)e^{-kmt}}$

Which is different from the one above it (there is an extra negative one in the denominator). This is all very confusing. First of all, is this a just a typo (that is, they just missed writing the negative in the first equation I wrote above) or am I missing something? Second, when I now plug in m and G$_{0}$, I do get the correct response. And the reason is, there is an extra m at the exponent of the e term. Where did it come from? I keep solving and resolving the differential equation $\frac{dy}{dt}$=k(1-y/N)y and simply can't get to this general solution.

I am very confused and have been stuck on this problem. Any input will be appreciated.

2. Oct 24, 2012

### HallsofIvy

Staff Emeritus
You are assuming that the "infection rate", 0.00005, is the same as the "growth constant", k. Is that true? Also, I believe, in the equation you give, y is the number of infected persons while your question asks you ton "Write an equation for the number of natives who remain uninfected".

3. Oct 24, 2012

### la6ki

No, I did solve for the uninfected, as I subtracted the number of infected from 5000 (which is the total population).

It is true that I'm assuming infection rate to be the same as the growth constant. Is this a mistake? Perhaps this is where the problem's coming from? Could you explain?

4. Oct 25, 2012

### la6ki

Sorry to bump this but could anybody explain the relationship between the infection rate and the growth rate (in this the growth of the number of infected people)? Are we supposed to multiply the carrying capacity with the infection rate to get the growth rate? This doesn't make any immediate sense to me, but it would lead to the correct answer. If that's indeed what must be done, can anybody explain the logic behind it?

5. Oct 25, 2012

### Ray Vickson

I would venture the guess that since YOU are taking the course in which these concepts appear, you will know better than any of us what is meant by the terms. Is it in your textbook? Is it in your course notes? If not, can you ask your instructor for clarification? Then, there is always Google: a search under 'epidemic models' turns up lots of hits.

RGV

6. Oct 27, 2012

### la6ki

I'm not taking any course, I'm just studying on my own. And I was stuck on this problem for a few hours (reading the chapters of the book very carefully) but couldn't find the answer to the question.

7. Oct 27, 2012

### Ray Vickson

Well, the ".25" in the exponent given as the answer is 5000 times the 0.00005 that you used. As I see it (not being expert in epidemiology), the 0.00005 is a kind of "transmission" probability per contact, so if the population is 5000 there are 5000 potential contacts, hence a 5000 times larger probability of infecting somebody.

RGV

8. Oct 27, 2012

### la6ki

Yeah, I guess it could be that, it does make sense. The strange thing is that it doesn't seem obvious, yet there is no word on this in the book where this problem was for.

Thanks for the responses though!