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At the beginning of the Gold Rush, the population of Coyote Gulch,Arizona was $365$.From then on ,the population would have grown by a factor of $e$ each year,except for the high rate of "accidental" death, amounting to one victim per day among every 100 citizens.By solving an appropriate differential equation determine, as functions of time:
(a) the actual population of Coyote Gulch $t$ years from the day the Gold Rush began, and
(b) the cumulative number of fatalities.

The question is from Apostol's Calculus I. In other questions,Apostol uses the statement "... increases at a rate proportional to the amount present. ..."But this one says "grown by a factor of $e$", I have difficulty in understanding the meaning of "a factor of $e$".Does it mean $y=be^{kt}$ or $y'=ey$ or what?

So I decide to denote "a factor of $e$" by $f_e$ and focus on the ' "accidental" death '.So let $y$ denote the population at present year, each day one victim dies among every 100 citizens .So the population remains in that year is $y(1 -\frac{1}{100})^{365}$. And the number of fatalities in that year is $$y-y(1 -\frac{1}{100})^{365} = y(1- (\frac{99}{100})^{365})$$ . And as far as I can get $$y'=f_e -y(1- (\frac{99}{100})^{365})$$

The answer of the question is given
a) $365e^{-2.65t}$
b) $365(1-e^{-2.65t})$

Any help is appreciated.

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  • $\begingroup$ Grown by a factor of $e$ probably if the population this year is $100$ next year it will be $271.8\ldots$(well not fractional population, but cut me some slack) $\endgroup$ – Guy Apr 6 '14 at 10:36
  • $\begingroup$ @Sabyasachi I tried your suggestion and substitute $f_e$ by $ey$ but I couldn't get the right answer :( $\endgroup$ – Detective King Apr 6 '14 at 10:55
  • $\begingroup$ I mean $f_e=e$, not $ey$. Does it work? $\endgroup$ – Guy Apr 6 '14 at 11:00
  • $\begingroup$ @Sabyasachi so $y'=e - y(1-(99/100)^{365})$ and I get $y=362e^{-0.974t}+0.974$. $\endgroup$ – Detective King Apr 6 '14 at 11:12
  • $\begingroup$ Oh. can't help then. Did you try googling for the question text $\endgroup$ – Guy Apr 6 '14 at 11:13
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Try to write it out in words.

The rate of change in population is the population we have minus the loss ratio of that population (of course, we could have other factors, but that is what we are working with here), so we have:

$$\dfrac{dP}{dt} = P - \alpha P = P(1 - \alpha)$$

Now how we can we find the loss ratio $\alpha$ of the population per year? The problem tells us that we lose "$1$ victim per day among every $100$ citizens", so for the year, we lose $365 \times \dfrac{1}{100}$ of the population $P$. This gives us $\alpha = \dfrac{365}{100}$.

So we have:

$$\dfrac{dP}{dt} = P\left(1 - \dfrac{365}{100}\right), P(0) = 365$$

Note: $P(0)$ is the initial population which is given as $365$.

This DEQ is separable and gives us:

$$20 \int \dfrac{1}{P}~dP = -53 ~\int dt, P(0) = 365$$

Thus,

$$\large P(t) = 365 ~e^{-\frac{53 t}{20}} = 365~ e^{-2.65~ t}$$

If we want to find the cumulative number of fatalities over time, we write that out in words as: we start out with a population of $365$ and we lose them at a rate of $P(t)$ (which we just calculated), so this is:

$$CP(t) = 365 - P(t) = 365 - 365 ~e^{-\frac{53 t}{20}} = 365\left(1 - e^{-\frac{53 t}{20}}\right) = 365\left(1 - e^{-2.65 ~t}\right)$$

We can plot these curves to verify that they are inverses of each other as:

enter image description here

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  • $\begingroup$ This was really nicely (and thoroughly) explained, Amzoti. Nice job. $\endgroup$ – Namaste Apr 6 '14 at 13:27
  • $\begingroup$ @Amzoti Your explaination is really nice!At the same time I think the question is quite "unrealistic"(it feels like "statistic") to me . $\endgroup$ – Detective King Apr 6 '14 at 14:01
  • $\begingroup$ @DetectiveKing: I think the point of these is just trying to set them up. It is very difficult to take the leap from knowing how to solve things, to actually producing models of the physical reality. By the way, it was not unreasonable for really bad things to happen to small populations as the result of a single issue, including having them totally wiped out. Regards $\endgroup$ – Amzoti Apr 6 '14 at 14:09

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