# Logistics Model

• Jan 19th 2009, 10:12 PM
banshee.beat
Logistics Model
I have no clue how to even begin this problem..

Okay, so the function:

$\displaystyle P(t) = \frac{1}{1+Ae^{-kt}}$

satisfies this logistic model:

$\displaystyle \frac{dP}{dt}=kP(1-P)$

For which value of t is P'(t) maximum? What is the value of P for this t?

For this problem, this function is used to model the spread of a disease where P is the proportion of people infected. Each person infects, on average, k other people.

Now, I understand that the first function is the integral of the second. But I'm so confused when it comes to understanding what exactly I need to derive. Can someone please help me walk through this?
• Jan 20th 2009, 12:11 AM
DeMath
Quote:

Originally Posted by banshee.beat
I have no clue how to even begin this problem..

Okay, so the function:

$\displaystyle P(t) = \frac{1}{1+Ae^{-kt}}$

satisfies this logistic model:

$\displaystyle \frac{dP}{dt}=kP(1-P)$

For which value of t is P'(t) maximum? What is the value of P for this t?

For this problem, this function is used to model the spread of a disease where P is the proportion of people infected. Each person infects, on average, k other people.

Now, I understand that the first function is the integral of the second. But I'm so confused when it comes to understanding what exactly I need to derive. Can someone please help me walk through this?

Find the second derivative of the function $\displaystyle P(t) = \frac{1}{1+Ae^{-kt}}$ and find $\displaystyle t$ of $\displaystyle P''\left( t \right) = 0$.

Because $\displaystyle \frac{{dP}}{{dt}} = P'\left( t \right) \to \max$ if $\displaystyle \frac{{{d^2}P}}{{d{t^2}}} = P''\left( t \right) \to 0$

The first

$\displaystyle P'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{1}{{1 + A{e^{ - kt}}}}} \right] = - \frac{{{{\left( {1 + A{e^{ - kt}}} \right)}^\prime }}} {{{{\left( {1 + A{e^{ - kt}}} \right)}^2}}} = \frac{{Ak{e^{ - kt}}}} {{{{\left( {1 + A{e^{ - kt}}} \right)}^2}}}.$

The second

$\displaystyle P''\left( t \right) = \frac{{{d^2}}}{{d{t^2}}}P\left( t \right) = \frac{d}{{dt}}\left[ {P'\left( t \right)} \right] = \frac{d}{{dt}}\left[ {\frac{{Ak{e^{ - kt}}}}{{{{\left( {1 + A{e^{ - kt}}} \right)}^2}}}} \right] =$

$\displaystyle = \frac{{{{\left( {Ak{e^{ - kt}}} \right)}^\prime }{{\left( {1 + A{e^{ - kt}}} \right)}^2} - Ak{e^{ - kt}}{{\left[ {{{\left( {1 + A{e^{ - kt}}} \right)}^2}} \right]}^\prime }}}{{{{\left( {1 + A{e^{ - kt}}} \right)}^4}}} =$

$\displaystyle = - \frac{{A{k^2}{e^{ - kt}}{{\left( {1 + A{e^{ - kt}}} \right)}^2} + 2{A^2}{k^2}{e^{ - 2kt}}\left( {1 + A{e^{ - kt}}} \right)}} {{{{\left( {1 + A{e^{ - kt}}} \right)}^4}}} =$

$\displaystyle = \frac{{2{A^2}{k^2}{e^{ - 2kt}} - A{k^2}{e^{ - kt}}\left( {1 + A{e^{ - kt}}} \right)}}{{{{\left( {1 + A{e^{ - kt}}} \right)}^3}}} =$

$\displaystyle = \frac{{2{A^2}{k^2}{e^{ - 2kt}} - A{k^2}{e^{ - kt}} - {A^2}{k^2}{e^{ - 2kt}}}}{{{{\left( {1 + A{e^{ - kt}}} \right)}^3}}} =$

$\displaystyle = \frac{{{A^2}{k^2}{e^{ - 2kt}} - A{k^2}{e^{ - kt}}}}{{{{\left( {1 + A{e^{ - kt}}} \right)}^3}}} = \frac{{A{k^2}{e^{ - kt}}\left( {A{e^{ - kt}} - 1} \right)}}{{{{\left( {1 + A{e^{ - kt}}} \right)}^3}}}.$

So, we let $\displaystyle P''\left( t \right) = 0$ to find $\displaystyle t$
(which value is $\displaystyle P'\left( t \right) = \frac{{Ak{e^{ - kt}}}}{{{{\left( {1 + A{e^{ - kt}}} \right)}^2}}}$ maximum).

$\displaystyle P''\left( t \right) = \frac{{A{k^2}{e^{ - kt}}\left( {A{e^{ - kt}} - 1} \right)}}{{{{\left( {1 + A{e^{ - kt}}} \right)}^3}}} = 0 \Rightarrow$

$\displaystyle \Rightarrow A{k^2}{e^{ - kt}}\left( {A{e^{ - kt}} - 1} \right) = 0 \Rightarrow A{e^{ - kt}} - 1 = 0 \Rightarrow$

$\displaystyle \Rightarrow A{e^{ - kt}} = 1 \Rightarrow {e^{ - kt}} = \frac{1}{A} \Rightarrow \frac{1}{{{e^{kt}}}} = \frac{1}{A} \Rightarrow$

$\displaystyle \Rightarrow {e^{kt}} = A \Rightarrow kt = \ln A \Rightarrow \boxed{t = \frac{{\ln A}}{k}}.$

So, we have $\displaystyle \color{red}P'{\left( t \right)_{\max }}$ if $\displaystyle \color{red}t = \frac{{\ln A}}{k}$.

$\displaystyle P'{\left( t \right)_{\max }} = P'\left( {\frac{{\ln A}}{k}} \right) = \frac{{Ak{e^{ - \ln A}}}}{{{{\left( {1 + A{e^{ - \ln A}}} \right)}^2}}} = \frac{k}{{{{\left( {1 + 1} \right)}^2}}} = \frac{k}{4}.$

Quote:

Originally Posted by banshee.beat
What is the value of P for this t?

$\displaystyle P\left( {\frac{{\ln A}}{k}} \right) = \frac{1}{{1 + A{e^{ - \ln A}}}} = \frac{1}{{1 + 1}} = \frac{1}{2}.$
• Jan 20th 2009, 05:36 PM
banshee.beat
Thank you so much! Could you explain to me what happens to the negative sign from the $\displaystyle -lnA$in the second derivative?

Could someone help me with graphing P(t) and P'(t)?

If I take A = 50 and k = 2, would my domain be [0, infinity) for P(t) and P'(t)?

=======P(t)========
For intercepts I got none because for
$\displaystyle \frac{1}{1+50e^{-2t}} = 0$
it would be $\displaystyle t = \frac{ln(-50)}{2}$ is that correct?

Also, I realize that eventually all values of t go to P(t) = 1, but I don't know how to express that because it's not quite an asymptote, is it?

For the intervals of increase/decrease, I set
$\displaystyle P'(t) = \frac{Ake^{-kt}}{(1+Ae^{-kt})^2} = \frac{100e^{-2t}}{(1+50e^{-2t})^2} = 0$
but I'm not able to calculate any actual values. I know it's increasing. I can't really find the max/min either...

For concavity and inflection, the value I got for the concave down interval, (0, 1.96), does not match up with the graph on my calculator.

=========P'(t)========

Domain: [0, infinity]
Intercepts: none
Asymptotes: as clueless as I am for P(t)
Intervals Inc/Dec: decrease (0, difficulties calculating)
Max/min: min(??,0) max(.5,1)
Concavity/inflection: do I need to calculate P'''(t) to find this one?

• Jan 20th 2009, 07:49 PM
DeMath
Quote:

Originally Posted by banshee.beat
Thank you so much! Could you explain to me what happens to the negative sign from the $\displaystyle -lnA$in the second derivative?

I used there rules:

1. $\displaystyle \alpha \cdot \ln \left( b \right) = \ln \left( {{b^a}} \right)$

in your case $\displaystyle - \ln \left( A \right) = - 1 \cdot \ln \left( A \right) = \ln \left( {{A^{ - 1}}} \right) = \ln \left( {\frac{1}{A}} \right)$

2. $\displaystyle {e^{\ln \left( b \right)}} = b$

in your case $\displaystyle {e^{ - \ln \left( A \right)}} = {e^{\ln \left( {{1 \mathord{\left/{\vphantom {1 A}} \right.\kern-\nulldelimiterspace} A}} \right)}} = \frac{1}{A}$
• Jan 20th 2009, 08:14 PM
DeMath
Quote:

Could someone help me with graphing P(t) and P'(t)?
This is a graph of the function $\displaystyle P'\left( t \right) = \frac{{Ak{e^{ - kt}}}}{{{{\left( {1 + A{e^{ - kt}}} \right)}^2}}}$
for $\displaystyle k=2$ and $\displaystyle A=50$

Do you need a graph of the function $\displaystyle P(t) = \frac{1}{1+Ae^{-kt}}$
For which $\displaystyle k$ and $\displaystyle A$?
• Jan 20th 2009, 09:30 PM
banshee.beat
I got the graph for P(t) off my calculator.. but I have to show my professor that I calculated everything out by hand for both, and that's where I'm running in to many of my problems. I have to show the exact values for:

domain
intercepts
symmetry
asymptotes
intervals of incr/decreasing
local max/min
concavity/inflection

I know how to do these with most functions, but I'm having a hard time calculating them for P(t) and P'(t).

For example, if the function can achieve the value of P(t) = 1, then 1 can't be an asymptote, can it?

I am also having trouble with the calculations of incr/decr, max/min, conc/infl pts.

Oh, and the same values for k and A. k = 2, A = 50.