Some Advanced Math

This is a problem that might interest deer hunters. I'm just taking a break from my math homework and would like to share one of the types of problems that I am working on.

A certain area or region can support a maximum of 5000 deer. In 1990, the population was estimated to be 3,000 (wow that is a lot of deer!). Five years later, the population was 3,500.
According to the information you received, estimate the population for today (year 2000).

What's the Answer?
3919 deer

How did you get that?

The fun part about story problems is that you have to use math to figure them out. Let's start explaining this darn thing, ok?

You need to know little about population growth and modeling. Let P=Population and t= time variables

The rate of change of population is proportional to the population itself

dP
-- = kP (equation #1)
dt

whereas
dP
-- is a derivative or rate of change or something
dt

k is just a constant of the proportion

Let's define an initial or beginning population
at t=0 (starting time) P=Po Thus Po represents our starting population

We have to manipulate the rate of change equation #1 and then integrate it. First, place like variables on one side of the equation

dP
-- = kdt (the k is just a constant)
P

Next, integrate both sides of the equation

ln P = kt + C

Raise both sides to the e
ln P kt + C
e = e

We can simplify the righthand side of this equation by using the rules of exponents. Remember that e^c is just the arbitrary constant and we can rewrite it as ec


kt+C kt C kt
e = e * e = Ce


Thus
kt
P = Ce

Let's find the initial population Po
k*0
Po = Ce

Po = C

We now have our UNRESTRICTED population growth model
kt
P(t) = Po*e

This equation works just peachy if the deer have unlimited amounts of food and don't have any diseases or other factors.

Now what about taking into account of these other factors? We need a better equation!

Consider
dP
-- = kP(1-P/L)
dt

L is defined as the carrying capicity or limit of the number of species. for example in our problem, the deer has a limiting capacity of 5,000. Now what does this equation mean?
Ignore everything else for now and just look at (1-L/P)
Now if the population is greater that the carrying capacity, the expression will be negative and the rate will be negative (deer will start to die). This makes sense because if the deer exceed
their population capacity then they are going to die off. If P is less than the capacity, then the deer will continue to increase. For very small valus of P we notice a very high exponential growth but as the the population increases we notice a limit of the population.

I tried to make this graph look decent!

this line is the limit or capactity of deer
|___________________________
| * * *
| *
| *
| *
| *
| * <-- exponential growth
|*
|____________________________
|

Now I'm going to skip some complicated calculus and give you a final equation that relates all the variables

P(t) = L*Po
---------------
-kt
Po + (L-Po)*e

This is called a Logistic Equation

Now it's your turn to prove the answer. Just plug in the values given in the problem to find "k", and then rewrite the equation
with k and solve for the year 2000.

Did you have fun?

SMILE =)

The text got screwed up when I submited the equations and I don't know if I can fix it!

You know it's not easy being a princess.
I have to perform royal duties ordained only to me, the Princess. The Princess must keep herself
looking beautiful all the time, even when in sleep (incase she gets awakened with a kiss by the Prince). She must stay informed and keep a somewhat concerned view for the affairs of her kingdom. At dances, anyone caught wearing the same gown as the Princess will be punished severely. She may dance and court whomeever she pleases because she's the Princess. Anyone disagreeing with the princess will have a trail in her royal court with her royalness the Princess being the judge. The word of the Princess if final. In otherwords, the Princess RULES!