Course:MATH110/Archive/2010-2011/003/Teams/Basel/homework 12

Homework 12

Team Problem

Starting with the function:

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

• Change the height of the horizontal asymptote on the right and denote it by K.
• Change the y-intercept to any number between 0 and K.
• BONUS - Change the slope of the curved part. Find a way so that the slope can go from very close to zero to almost vertical.

Solution

If we change our numerator to K, this will change the height of our hoirzontal asymptote. As K increases, the height of horizontal asymptote (y-value) increases.

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

• the value of the y-intercept will always be half the value of K.
• example: if we want our y-intercept to be 8, we make K=16.

Model Example

A large kennel in a pet shop contains 20 rabbits. The kennel has enough space and food to only feed 40 rabbits. If these rabbits are able to mate and reproduce, the carrying capacity will be reached assuming that the reproduction rate of the rabbits is constant; the number of rabbits in the kennel can be modelled by the equation:

$P(t)={\frac {40}{1+e^{-t}}}\quad$

• where P is the population of rabbits
• where t is number of weeks.

95% of our maximum carrying capacity for the rabbits kennel is 38 rabbits.

$40*0.95=38\quad$

• we use 38 as our P(t)

$38={\frac {40}{1+e^{-t}}}\quad$

• solve for t

$38+38e^{-t}=40\quad$

$38e^{-t}=4\quad$

${\frac {1}{e^{-t}}}={\frac {4}{38}}\quad$

$lne^{-t}=ln{\frac {38}{4}}\quad$

$t=ln{\frac {38}{4}}\quad$

• Answer

2.25