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Math Help - Logistics Model

  1. #1
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    Logistics Model

    I have no clue how to even begin this problem..

    Okay, so the function:

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

    satisfies this logistic model:

    \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?
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by banshee.beat View Post
    I have no clue how to even begin this problem..

    Okay, so the function:

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

    satisfies this logistic model:

    \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 P(t) = \frac{1}{1+Ae^{-kt}} and find t of P''\left( t \right) = 0.

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

    The first

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

    The second

    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] =

    = \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}}} =

    =  - \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)}}<br />
{{{{\left( {1 + A{e^{ - kt}}} \right)}^4}}} =

    = \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}}} =

    = \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}}} =

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

    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<br />

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

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

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

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

    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 View Post
    What is the value of P for this t?
    P\left( {\frac{{\ln A}}{k}} \right) = \frac{1}{{1 + A{e^{ - \ln A}}}} = \frac{1}{{1 + 1}} = \frac{1}{2}.
    Last edited by DeMath; January 20th 2009 at 10:41 AM.
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  3. #3
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    Thank you so much! Could you explain to me what happens to the negative sign from the -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
    \frac{1}{1+50e^{-2t}} = 0
    it would be 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
    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?

    ...ahh. help please.
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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by banshee.beat View Post
    Thank you so much! Could you explain to me what happens to the negative sign from the -lnA in the second derivative?
    I used there rules:

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

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

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

    in your case {e^{ - \ln \left( A \right)}} = {e^{\ln \left( {{1 \mathord{\left/{\vphantom {1 A}} \right.\kern-\nulldelimiterspace} A}} \right)}} = \frac{1}{A}
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  5. #5
    Senior Member DeMath's Avatar
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    Could someone help me with graphing P(t) and P'(t)?
    This is a graph of the function P'\left( t \right) = \frac{{Ak{e^{ - kt}}}}{{{{\left( {1 + A{e^{ - kt}}} \right)}^2}}}
    for k=2 and A=50



    Do you need a graph of the function P(t) = \frac{1}{1+Ae^{-kt}}<br />
    For which k and A?
    Last edited by DeMath; January 20th 2009 at 09:27 PM.
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  6. #6
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    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.
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