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

Homework 12

Starting with the function

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

Your goal is to modify the function so that we can use it to model a real-life problem. We want to be able to control the following things:

• Change the height of the horizontal asymptote on the right, we'll denote it by K.
• Change the y-intercept to any number between 0 and K

As we can see from the graph above, the function has a horizontal asymptote at y=1 and a y-intercept at (0,5)

If we manipulate the function by changing the numerator, we can also change the horizontal asymptote. K= the value of the horizontal asymptote

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

For example, If we change the K to 12 then the the function will become

$P(t)=\frac{12}{1+e^{-}t}$

The horizontal asymptote will be at y=12 and the y-intercept will be at (0,6)

This shows that the y-intercept is always K/2. This means that the y-intercept will always be between 0 and K.

Applied Model

An invasive species of frog was introduced to a small pond ecosystem. They are reproducing fast and an ecologist wants to know when 95% of the carrying capacity of 200 frogs will be reached. $95\mathrm{\%}of200frogs=0.95x200=190$

$Time=190=\frac{200}{1+e^{-}t}$

$190+190e^{-}t=200$

$190e^{-}t=10$

$e^{-}t=\frac{10}{/}190$

$ln(e^{-}t)=ln(\frac{10}{190})$

$-tlne=ln\frac{10}{190}$

$-t=ln(\frac{10}{190})$

$-t=-2.94$

$t=2.94$

95% of the carrying capacity will be reached in around 3 years.