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

Homework 12

We are given the function:

1/(1+e^-t)

In order to change the height of the horizontal asymptote on the right, we must add a variable 'c' to the equation:

(1/(1+e^-t))+ c

Here 'c' is the factor which vertically translates the graph and this therefore changes the horizontal asymptote from its original position by a value of 'c'. The horizontal asymptote of the original function is a y=1. Therefore we can say that K = c + 1.

If we substitute in a value of 3 for 'c' the right asymptote changes to y=4, which can also be found by using the equation:

K = c + 1

K= 3+1 =4

We can see the following pattern occur by changing the value that that function is vertically translated by the value of 'c' using this formula.

To change the y-intercept to any number between 0 and K (which we chose to be 4 in this case) we can do this by adding another variable 'b' into the equation:

(1/(1+(b)e^-t))+3 , where 'b' cannot be equal to zero

Here 'b' provides us with a horizontal stretch with a dilation factor of 'b'.

The values of the original function are plotted as follows:

x=-1, y=3.26

x=0, y=3.5

x=1, y=3.73

x=2, y=3.88

x=3, y=3.95

x=4, y=3.982

x=5, y=3.9933

x=6, y=3.9975

x=7, y=3.9991

x=8, y=3.9997

x=9, y=3.9999

x=10, y=4

If we let b=7, then (1/(1+(7)e^-t))+3, then the values are shown as:

x=-1, y=3.04

x=0, y=3.125

x=1, y=3.2797

x=2, y=3.51

x=3, y=3.74

x=4, y=3.8864

x=5, y=3.955

x=6, y=3.9829

x=7, y=3.9937

x=8, y=3.9977

x=9, y=3.9991

x=10, y=3.9997

x=11, y=3.9999

x=12, y=4

Here we can see that by placing 'b' into the equation we can alter the y-intercept but still maintain the same horizontal asymptote for both funtions!

BONUS

In order to change the slope of the curved part so that the slope can go from very close to zero to almost vertical we must add yet another variable which we can call 'a' to the function:

a(1/(1+e^-)), where 'a' cannot be equal to zero.

If we make 'a' a very large number such as 1,000,000 the slope becomes almost vertical. If we make 'a' a very small number such as 1/10000000, the slope gets close to zero. This is an example of a vertical strech of the function by a dilation factor of 'a'.

Application

We can determine the numerator (a) to represent the carrying capacity of a population growth.

A village is formed in the mountain valley of Trella with 150 settlers. The area of the village has only enough resources to accommodate 250 settlers. Assuming the reproduction rate of the villagers is constant the maximum capacity will be reached eventually. In order to find the max capacity of the population we can use the equation:

P(t)= (x)/ 1+e^-t

Then we can plug 250 in for x and solve for t P(t)= (250)/ 1+e^-t In order to dictate 90% of our maximum capacity we can evaluate the number by: 250 x 0.90= 225 We can then plug in 45 as p(t) and find t

225= 250 / 1+e^-t

225(1+e^-t) =250

225+225e^-t= 250

225e^-t= 250-225

225e^-t/45= 25/225

e^-t= 0.11

lne^-t= ln(0.11)

-t= ln(0.11)

-t= -2.21

t=2.21 years

This means that within 2.21 years the village will reach 90% of its maximum capacity.

If we wanted to change the rate at which the population increases we can add a numerical value (b) in front of the "t" value. This will change the rate at which it will reach the carrying capacity.

Once you've played with the function enough, try to find an application of the graph to model something. It can be anything which starts at a value and then goes to another one (think for a population, it goes from 0 to it's carrying capacity). Explain what you are modelling and how you decide to attribute a numerical value to each of the 2 or 3 parameters that you researched just above. Then use the model to make a prediction. For example, if your model is suppose to describe a population for which you have its initial population and carrying capacity (potentially its rate of increase if you solved the bonus part), then use that data to make a prediction for the population in 20 years, or use the model to predict when will the population reach 95% of its carrying capacity). When doing this last part, explain well where you're taking your data from (real data or imagined data), what it is that you're modelling and how you are doing the math to answer a predictive question.