Course:MATH110/Archive/2010-2011/003/Teams/Solothurn/Homework 12

Homework 12: Team Problem

$P(t)={\frac {1}{1+e^{-t}}}$

For this function, the height of the horizontal asymptote on the right is equal to the value of the numerator. If the whole function is multiplied by a value, this value will be where the horizontal asymptote on the right hand side is located.

eg. $P(t)={\frac {K}{1+e^{-t}}}$ $K=$ height of right side horizontal asymptote.

The Y-intercept will always be a value between the left side horizontal asymptote and the right side horizontal asymptote because these are the floor and the ceiling of the function. Since the left side horizontal asymptote is 0 and the right side horizontal asymptote is K, it is a matter of choosing which value between 0 and K is desired for the Y-intercept. The Y-intercept can be changed by shifting the function to the left and to the right, therefore, it will take different Y-values at t=0. Shifting the function left and right is done by adding or subtracting directly to the t.

eg. $P(t)={\frac {1}{1+e^{-t+j}}}$

In order to select the specific number for the Y-intercept, you can use the following formula in place of j:

$ln({\frac {K}{J}}-1)$

Where K= the horizontal asymptote on the right, and where J= the Y-value you have chosen for your Y-intercept. So when you combine these manipulations, you get:

$P(t)={\frac {K}{1+e^{-t+ln({\frac {K}{J}}-1)}}}$

Bonus If you want to alter the slope of the function you need to add a coefficient to t. If you multiply t by an integer then the function will get from the horizontal asymptote on the left to the horizontal asymptote on the right sooner. If you divide t then the function will move from it's floor to it's ceiling more slowly.

eg. $P(t)={\frac {K}{1+e^{-Rt+ln({\frac {K}{J}}-1)}}}$

Here the value you plug in for R will control the slope of the function as it spans from one asymptote to the other.

In summary:

           K = horizontal asymptote on the right
           J = value of Y-intercept
           R > 1 = steeper slope between asymptotes
           0 < R < 1 = more gradual slope between asymptotes

Application of the graph to model squirrels population in Canada

$P(t)={\frac {1}{3+e^{-15t+6}}}$

A projection of the population in 20 years is P(20)=0.3 periodic.

$P(t)={\frac {1}{3+e^{-15{20}+6}}}$ = 0.3 periodic.

We compute the derivative P'(t):

In 20 years the instantaneous rate of growth P'(20) of the squirrel population will be 9.2 X 10^-8. This number tells us that the population is nearing carrying capacity. As time progresses and resources and space are abundant the growth of the squirrel population greatly increases; however, when space and resources become more limited the population of squirrels stops increasing and levels off at carrying capacity.