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Two Species Population Models

  1. Jan 5, 2013 #1
    The academics and the technicians of a university faculty are planning a paintball battle and the mathematicians are trying to predict their chances of winning with a continuous model. They are representing the numbers in the two teams at any time t as and respectively and have decided that the system to model the encounter should be dA/dT= - k2T and dT/dt= - K1A, where k1 and k2 are positive constants.
    Based on the initial conditions A(0)=100, T(0)=80, The system of ODEs can of course be solved analytically using eigenvalue techniques of the populations over a 20 minutes battle.

    = By using the separation of variables, then integrating bothe sides of dA/dT=-k2t
    i got A=-k2Tt + c1, where c1 is constant.
    At initial , A=100, A=c1= 100,
    therefore i got A=-k2Tt + 100.
    Similary for dT/dt= - K1A,
    I got T=-k1At + 80
    Can some one help after that to solve the systems of ODEs.

    ADDITIONAL QS FOR THE ABOVE SOLUTION. YOU CAN IGNORE MAPLE BIT FROM b) but does it have something to do with the above solution ans part c). TO solve part c, do i have to use part b) qs part. THE ABOVE SOLUTION I HAVE DONE IS FOR PART C BUT ITS STILL INCOMPLETE AND NOT SURE WITH ANSWER TOO PLEASE HELP.

    CONTINUE QS
    b)The muscles and eyesight of the academics have of course suffered from too much ‘book learning’ over the years, but at least they are aware of their limitations as good soldiers. They estimate that although both sides can fire paintballs at the same rate as each other, denoted by f_A =f_T =2 shots per minute, the technicians have a probability of 0.035 of hitting their target with a single paintball shot whilst the academics have only a 0.01 probability of hitting theirs. Based on the initial conditions A(0)=100, T(0)=80, use Maple to produce a numerical approximation and graph of the populations over a 20 minutes battle then comment on the outcome after that time.

    (c)The system of ODEs can of course be solved analytically using eigenvalue techniques. Do this and compare it with predicted team numbers remaining standing after 20 minutes from your Maple simulation to ensure correctness.
     
    Last edited: Jan 5, 2013
  2. jcsd
  3. Jan 5, 2013 #2

    haruspex

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    dA/dt, surely
    Doesn't look right to me.
    You can use one equation to substitute for A or T in the other and get a 2nd order ODE in one dependent variable only.
     
  4. Jan 5, 2013 #3
    From your idea, I got
    dT/dA= k1A/k2T
    Integrate using separation of variables
    K2 T^2= K1 A^2 + C, where C is constant
    K2 T^2-K1 A^2=C

    What happen do constant K1 and K2. Does it disappear while integrating or will it stay?

    After that I got : K2 T= C1 e^t + C2 e^-t,
    Im not sure with what K1 A=?

    After that i really don't know what to do?
     
  5. Jan 5, 2013 #4
    From part b) f_A= f_T= 2 shots per minute? What is this? And
    technicians have a probability of 0.035 of hitting their target and academic have a probability of 0.035 of hitting their target. whats is this all abou? Is there any concern to work part c) which i solving it analytically for system of ODEs using eigenvalue techniques.

    Are probability and f_A and f_t values require to solve part c)?
     
  6. Jan 5, 2013 #5

    haruspex

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    k1 and k2 should feature in the answers.
    No, you've lost a constant somewhere. Try K2 T= C1 eλt + C2 e-λt.
    Substitute this back in the differential equations to find A as a function of t, then substitute for that in the other ODE to get an equation that will tell you what λ is. Then plug in the initial conditions to calculate C1 and C2 .
     
  7. Jan 5, 2013 #6
    K2 T= C1 eλt + C2 e-λt where do i substitute this, i mean which differentiaal equation to get A?

    I try to substitute in dA/dT=-k2t, but -k2t is not same as k2T.

    i also don't know equation for K2 A?
     
  8. Jan 5, 2013 #7
    subs K2 T= C1 eλt + C2 e-λt, in dT/dA= k1A/k2T, we get
    dT/dA= k1A/C1 eλt + C2 e-λt
    ∫(C1 eλt + C2 e-λt) dT = ∫k1A dA
    (C1 eλt + C2 e-λt )T = (k1 A^2)/2 + D
    2T(C1 eλt + C2 e-λt) -D= k1 A^2
    √{2T(C1 eλt + C2 e-λt)- D}= k1 A
    From there how do calculate λ?

    if i i did wrong, Would please explain me more deeply, please..........
     
  9. Jan 5, 2013 #8

    haruspex

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    No, I mean in the original equations, the ones with dA/dt and dT/dt.
     
  10. Jan 6, 2013 #9
    so what about K2 A? do you know the equation for that? how did you find the equation for K2 A?
    subs, K2 T= C1 e^λt + C2 e^-λt, in dA/dt= - k2T ,
    i got dA/dt= -C1 e^λt - C2 e^-λt
    ∫ dA = ∫(-C1 e^λt - C2 e^-λt) dt
    A= -λC1 e^λt + λC2 e^-λt + D

    after that i sub value of A in dT/dt= - K1A
    dT/dt= -k1 (-λC1 e^λt + λC2 e^-λt + D)
    dT/dt= (k1 λC1 e^λt - k1 λC2 e^-λt + D k1 )
    ∫dT= ∫(k1 λC1 e^λt - k1 λC2 e^-λt + D k1 ) dt
    T= k1 λ^2 C1 e^λt + k1 λ^2 C2 e^-λt + t D k1 +E
    Now, I have
    A= -λC1 e^λt + λC2 e^-λt + D
    T= k1 λ^2 C1 e^λt + k1 λ^2 C2 e^-λt + t D k1 +E
    What to do next? Could please check my my answer too for T and A.
     
    Last edited: Jan 6, 2013
  11. Jan 6, 2013 #10
    Given that :dA/dt= - k2T and dT/dt= - K1A, initial conditions A(0)=100, T(0)=80
    I got
    A= -λC1 e^λt + λC2 e^-λt + D
    T= k1 λ^2 C1 e^λt +k1 λ^2 C2 e^-λt + t D k1 +E

    how to calculate λ, C1, C2, D, E and k1
    There must be something wrong
    please help me.
     
    Last edited: Jan 6, 2013
  12. Jan 6, 2013 #11

    haruspex

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    Rather simpler if you substitute on the LHS of the ODE:
    dT/dt = -k1A = (1/k2) d(C1 eλt + C2 e-λt)/dt = (λ/k2)(C1 eλt - C2 e-λt)
    Using that to substitute for A in the other ODE will show you what λ is.
     
  13. Jan 6, 2013 #12
    what you have done here?
    dT/dt = -k1A = (1/k2) d(C1 eλt + C2 e-λt)/dt = (λ/k2)(C1 eλt - C2 e-λt)
    i know u sub T=( C1 e^λt + C2 e^-λt)/k2, in dT/dt= - k2T ,
    only in left hand side, but how did you get (λ/k2)(C1 eλt - C2 e-λt) of the right hand side?


    What is this (λ/k2)(C1 eλt - C2 e-λt)? where to subs this to get A?
     
  14. Jan 6, 2013 #13

    haruspex

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    Maybe it's less confusing if I write it like this:
    dT/dt = -k1A (1)
    T=( C1 e^λt + C2 e^-λt)/k2 (2)
    Differentiating (2)
    dT/dt = λ( C1 e^λt - C2 e^-λt)/k2 (3)
    Combining (1) and (3):
    -k1A = λ( C1 e^λt - C2 e^-λt)/k2 (4)
    Now use that to substitute for A in the dA/dt equation.
     
  15. Jan 6, 2013 #14
    dA/dt =- k2T
    A=λ( -C1 e^λt + C2 e^-λt)/(k1 k2) .........eqn i
    Differentiating (eqni)
    dA/dt= λ^2( -C1 e^λt - C2 e^-λt)/(k1 k2).......combining eqn i and eqn ii
    - k2T= λ^2( -C1 e^λt - C2 e^-λt)/(k1 k2)
    we have K2 T= C1 e^λt + C2 e^-λt
    (-C1 e^λt - C2 e^-λt)= λ^2( -C1 e^λt - C2 e^-λt)/(k1 k2)
    (k1 k2)= λ^2
    therefore λ= √(k1 k2)
    But we do not know what k1 and k2 are?
     
  16. Jan 6, 2013 #15

    haruspex

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    Those are given parameters. You don't care what they are. They will feature in the answer.
     
  17. Jan 7, 2013 #16
    can i assume k1 and k2 = e( 2.718) and work out C1 and C2 using initial condition.
    Because qs ask to find A and T after 20 minute, and if k1 and k2 feature in the answer then i won't able to get numerical answer.

    please suggest me what to do? As I getting closer with getting answer and if I didn't get the final answer it will be make me really sad :(
     
  18. Jan 7, 2013 #17
    if they feature in the answer, does it has somethign do with f_A =f_T =2 shots per minute, what do you mean by f_A =f_T =2?
    OR
    technicians probability = 0.035
    academics probability= 0.01
     
  19. Jan 7, 2013 #18

    haruspex

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    The additional information in part (b) allows you to determine k1 and k2. If each individual in the Academic team, say, is firing 2 shots/minute, and 0.01 of the shots are successful, what does that tell you about the rate of change of T?
     
  20. Jan 7, 2013 #19
    how? is it like f_A is same as DA/dt=2 and f_T is same DT/dt=2
    and what about the probabilty 0.01 and 0.035 tells about , where to subs these to get
    K1 and K2?
     
  21. Jan 7, 2013 #20

    haruspex

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    dT/dt is the rate of change (increase, but it will be negative) of the number of players in the Tech team. At a given moment, there are A members of the Academic team, all firing away at 2 rounds per minute. Of those rounds, 0.01 (1%) succeed in hitting and thereby taking out a member of the Tech team.
    How many hits per minute is one member of the Academic team scoring?
    How many hits per minute is the Academic team scoring as a whole?
     
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