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I was given this word problem by a friend, and it's stumped me on how to set it up. Hopefully you guys can help. Here it is:

"Exactly one person is an isolated population of 10,000 people comes down with a certain disease on a certain day. Suppose the rate at which this disease spreads is proportional to the product of the number of people who hace the disease and the number of people who do not have it yet. If 50 People have the disease after 5 days, how many have it after ten days?"

So I guess what I am asking is not really what the answer is (although that would be cool too), but rather how to set up and do this problem.

apparently it involves integrals somehow.

Thanks in advanced.

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This is a typical differential equation problem. It leads to what is known as a "logistics model".

Let $N(t)$ denote the number of infected people at time $t$ after the first infection, with $t$ measured in days. The rate at which the disease spreads is the rate of change of $N(t)$, which is the derivative $N'(t)$. We are told that the rate of change is proportional to $N(t)$ (the number of people who already have the disease) times $10000-N(t)$ (the number of people who don't have the disease). That is: $$N'(t) = kN(t)(10000-N(t))$$ for some constant $k$.

We are also told that $N(0) = 1$ (one person gets sick) and $N(5)=50$. We are asked for the value of $N(10)$.

Note that we cannot have $N(t)=0$ for all $t$, nor can we have $N(t)=10000$ for all $t$.

This differential equation is of a kind called "separable". We can solve it as follows: $$\begin{align*} \frac{dN}{dt} &= kN(10000-N)\\ \frac{dN}{N(10000-N)} &= k\,dt\\ \int\frac{dN}{N(10000-N)} &= \int k\,dt\\ \int\frac{dN}{N(10000-N)} &= kt + C. \end{align*}$$ for some constant $C$. To solve the integral on the left, we can use Partial Fractions: $$\frac{1}{N(10000-N)} = \frac{\quad\frac{1}{10000}\quad}{N} + \frac{\quad\frac{1}{10000}\quad}{10000-N}$$ so $$\begin{align*} \int\frac{dN}{N(10000-N)} &= \frac{1}{10000}\int\left(\frac{1}{N} + \frac{1}{10000-N}\right)\,dN\\ &= \frac{1}{10000}\left(\ln|N| - \ln|10000-N|\right) + D. \end{align*}$$ Putting it all together, we get $$\frac{1}{10000}\left(\ln |N| - \ln|10000-N|\right) = kt + E,$$ where $E$ is some constant. Rewriting, we have: $$\ln\left|\frac{N}{10000-N}\right| = Kt + F,$$ where $K$ and $F$ are constants. Since $N$ is always between $1$ and $10000$, we can drop the absolute values.

Plugging in $t=0$, we know that $N=1$, so we have $\ln\frac{1}{9999} = F$. Plugging in $N=5$, we know that $N(t)=50$, so we have $$\ln\frac{50}{9950} = 50K + \ln\frac{1}{9999}.$$

From here, we can plug in $t=10$ and solve for $N$.

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It is a logistic function. this means that it will first cause and exponential increase, which will slow down and eventually level off. I would look something like this: enter image description here

The equation of this is $P(t)=\frac{KP_0 e^{rt}}{K+P_0(e^{rt}-1)}$

$P_0 =$ population at time $t = 0$

$K$ = final population after some (long) time, also called the “carrying capacity”, which limits growth

$r$ = initial growth rate

Initial: $1$ person Total amount of people: $10,000$ $50$ people after $5$ days $x$ people after $10$ days

$t$ is measured in days $$ $P(t)=\frac{KP_0 e^{rt}}{K+P_0(e^{rt}-1)}$

$50=\frac{10000*1*e^{5r}}{10,000+1(e^{5r}-1)}$

$50=\frac{10000e^r5}{9,999+e^{5r}}$

$50(9999)+50(e^{5r})=10,000e^{5r}$

$499950+50e^{5r}=10,000e^{5r}$

$499950=10,000e^{5r}-50e^{5r}$

$499950=9950e^{5r}$

$499950=9950e^{5r}$

$50.2462=e^{5r}$

$\ln(50.2462)=\ln(e^{5r})$

$\ln(50.2462)=5r$

$\frac{\ln(50.2462)}{5}=r$

This then needs to be added to the initial equation

$P(t)=\frac{KP_0 e^{\frac{\ln(50.2462)}{5} t}}{K+P_0 (e^((ln(50.2462))/5 t)-1)}$

${t = 10}$

$P(10)=\frac{10,000*1e^((\ln(50.2462))/5(10))}{10,000+1(e^((ln(50.2462))/5(10))-1)}$

$P(10)=\frac{10,000*e^((\ln(50.2462))/5(10))}{9,999+e^((ln(50.2462))/5(10))}$

$P(10)=\frac{25246837.4700}{12523.6838}$

$P(10) = 2015.9274$

It will take after $10$ days, $2016$ people will be infected.

I got my information (the equation) from this website: http://www.intmath.com/blog/h1n1-and-the-logistic-equation/3498

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