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Homework Help: DE modeling practice

  1. Jul 8, 2011 #1
    This is not a homework question or anything assigned, but I would like some practice / assistance with modeling as we have not covered that in my DE class.

    1. The problem statement, all variables and given/known data

    The rhinoceros is now extremely rare. Suppose enough game preserve land is set aside so that there is sufficient room for many more rhinoceros territories than there are rhinoceroses. Consequently there will be no danger of overcrowding. However if the population is too small, the fertile adults will have difficulty finding each other when it is time to mate. Write a differential equation that models the rhinoceros population based on these assumptions. (Note that there is more than one reasonable model that fits these assumptions.)

    2. Relevant equations



    3. The attempt at a solution
    Ok so if I were to model this--I would say that the rate of change in the rhinoceros population with respect to time is proportional to the size of the population times the parameter which would be the growth coefficient...and the growth coefficient would be affected if the population is small. There are no limiting resources or predators...

    Independent variable = time (t)
    Dependent variable = population of rhinoceroses (P)
    Parameter = (k)

    I've never used LaTex before -_-
    so it would be basic unlimited population growth model to start with
    dP/dt = kP
    but I think there would need to be another parameter (g) the smallest possible population for which the adults can still meet. So if P < g then dP/dt < 0....and I'm not sure how to express that in the model :biggrin:.
     
  2. jcsd
  3. Jul 8, 2011 #2

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    Good! You defined your variables and constants.
    And you set up a proper reasoning how the population expansion would be proportional to the population size.

    I think you just did! :smile:

    Aren't you saying something like:
    [tex]\frac {dP} {dt} =
    \left\{ \begin{array}{ll}
    < 0, & \mbox{ if } P < g \\
    k P, & \mbox{ otherwise}
    \end{array} \right.[/tex]

    Perhaps you might consider graphing a function that describes what you need.
    What kind of graph would match the function just defined?
     
    Last edited: Jul 8, 2011
  4. Jul 8, 2011 #3
    If rhino's being to spread apart affects your model, try relating your parameter to a density function. (rhinos/area) good luck!
     
  5. Jul 8, 2011 #4
    Yayyy! My programming teacher done taught me well :biggrin:


    Yes this is what I am saying. I will think about the graph and mess around in Maplesoft a bit and come back with my findings :) Thanks I Like Serena!!
    This is very interesting, I did not think of this at all...I will see how I can implement this. I have to go, but when I come back I will post what I've come up with. Thanks elegysix.
     
  6. Jul 9, 2011 #5

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    Well, here's a link to WolframAlpha, and the corresponding picture.
    Somehow, I don't think this is what you had in mind! :wink:

    I have set g=2, k=1/2, and the negative part to -1.
    http://www.wolframalpha.com/input/?i=Piecewise[{{-1,+P+<+2}},+(1/2)+P

    attachment.php?attachmentid=37048&stc=1&d=1310242802.gif

    Can you pinpoint what is wrong with this graph for being a model to your problem?
     

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  7. Jul 9, 2011 #6
    Sorry I like Serena...I'm studying for a chemistry exam I have on Monday. I have a slope field that resembles what I have in mind
     

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    Last edited by a moderator: Apr 26, 2017
  8. Jul 10, 2011 #7

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    No hurry, finish your chemistry exam first.
    I have to admit that I was wondering if you had already lost interest, but for now I'll assume you just had pressing business elsewhere. :)

    Now, let's take a look at your attempt.
    attachment.php?attachmentid=37057&d=1310260188.png

    It seems you've tried to set up the model, calculate, and show the results all in one go.
    (My graph was only of the growth rate dP/dt versus population P.)
    I presume your field is population P versus time t?

    Assuming it is, there are a few things wrong with it.
    I'll mention one: apparently the population growth is near infinite at any given time when there is a positive population!
    That can't be right.
    Can you point out a few other flaws?

    Basically that's what modelling is about. You try to find a simple model to fit the facts.
    Work it out. See if your model fits. And if it's not complete, add something to your model.
    Until you cannot solve it anymore. Then you simplify the model again...... :smile:
     
  9. Jul 10, 2011 #8
    I never lose interest in math :) And modeling is something I have not figured out so I am FAR from losing interest in it :biggrin:. I will examine these things and will get back to you after my exam. We have to turn in our homework on our exam date and my prof gives 100+ problems for every chapter and they either require a lot of steps or are so simple and repetitious that it becomes a pain doing them -_-


    Yes the field was dP/dt but I have to confess I did it very quickly just to give an impression of what I was idealising and I was trying to relate it to the figures you were using lol.
    I definitely do not want an infinite growth at all times, I was idealising two points of equilibrium, exponential growth above the highest point and a curve declining between the highest and lowest equilibrium points and ending at the lowest point. This leads me to believe there has to be a fraction in the model with at least one variable (numerically speaking...it could be a fraction of a parameter over a variable) in the numerator and the denominator--and I am considering that along with elegysix's hint about relating the parameter to a density function, but I have not given this the attention I want to yet.
     
  10. Jul 10, 2011 #9

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    So we're agreed that we'll first try to find a function that properly describes the relation between current population P and growth rate dP/dt?
    (Note that this is a function and not a vector field?)

    I'll leave you to it for now, and I'll look forward to your response in which you may show some incredible insight! :wink:
    (Or just show you want to learn more... whatever.)

    Btw, will you let us know how your chem test went?
     
  11. Jul 10, 2011 #10
    omg lol yes I know this is a function we are trying to define. I really should not have responded until I have given it proper thought. And I will definitely show you that I need to learn more lol this is my second attempt at modeling ever :) Will certainly let you know about the chem exam also.

    And I should note I am notorious for making the simple complicated and doing things backwards lol
     
    Last edited: Jul 10, 2011
  12. Jul 10, 2011 #11
    I wonder now if there is something fundamentally wrong with my reasoning. Isn't a vector field defined by a function? Like the divergence of a vector field is a function? Sometimes it is easier for me to visualise the vector field if I am not sure of the model for the function or sure how the equation would look precisely in space...the vector field helps me to visualise this. I have not learned about vector fields in calculus yet, but "slope fields" in differential equations which I understand to be the same thing.
     
  13. Jul 10, 2011 #12

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    In a literal sense you are right. :)

    A vector field is indeed a function. A special one, where a vector is assigned to each point in the plane.
    I was talking more of a regular simple function, where a scalar value (population growth) is assigned to a scalar value (population size).

    So we have a simple function f(P) that describes the population growth as a (simple) function of population.

    I guess you can also visualize this as a vector field, based on population P and time t.
    That is, a vector field that for each time t and population P shows what the growth rate of P versus time is.
     
  14. Jul 10, 2011 #13
    -_- The "S" word again lol

    Yes this is what I was doing. I will come back to this on Tuesday or Monday evening :) Thanks for all of your help so far I like Serena :biggrin:! And thanks for the explanation.
     
  15. Jul 11, 2011 #14
    dP/dt = kP((P/g)-1)?
    Because of this "However if the population is too small, the fertile adults will have difficulty finding each other when it is time to mate"
    so P/g provides a ratio of the current population to the smallest possible population for which the adults can still meet. P < g then dP/dt < 0 if kP is multiplied by (P/g)-1 because if the population is greater than the smallest possible population for which adults can still meet then P/g should be > 1; if g>P then P/g<1 and if we subtract 1 from P/g when g>p then (P/g) - 1 < 0 and dP/dt<0.

    I hope this makes sense I am severely lacking sleep and I feel I may be over looking something.
     
  16. Jul 11, 2011 #15
    It seems like it could be a possible model...
     
  17. Jul 12, 2011 #16

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    Looks good.
    Could you make a graph?
    Since I like pictures. :smile:

    And can you solve the DE?
     
  18. Jul 12, 2011 #17
    Really? lol Yayyyyy :biggrin:
    Will do as soon as I finish writing up this lab :)
     
  19. Jul 14, 2011 #18

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    And? Did you finish your lab? :)

    Btw, you may have found it's rather hard to solve the DE.
     
  20. Jul 14, 2011 #19
    Patience is a virtue ILS :biggrin: Actually I did not finish the lab lol the Prof extended the due date because we have 4 lab reports due at the same time, so I have been working on all of them plus an extra credit assignment :biggrin: I got my DE lab back and got 100% though :) I have not even attempted to solve this equation, but have pondered it and I cannot think of how to solve it readily. Please do not give me any hints or anything yet because I have not seriously considered solving it yet. Sorry that I am taking long to get back to you with this model, but my chemistry class is coming to an end and the prof postponed some things because of the holidays, so now she is trying to cram everything in before the class ends.
     
  21. Jul 14, 2011 #20

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    Whaaaat? I think you have it backward lol.
    Patience is not a virtue, impatience is!
    See for instance here: http://patrickg.net/the-3-virtues-of-a-programmer/ [Broken]

    (Weren't you interested in learning how to program too? :wink:)
     
    Last edited by a moderator: May 5, 2017
  22. Jul 14, 2011 #21
    :rofl:

    I promise I am not being lazy this time >_>

    I am trying to get everything finished by Saturday and then I will be completely devoted to DE :biggrin: Although.... I have been trying to sneak some of my DE skills into my chem lab reports :tongue2:
     
  23. Jul 15, 2011 #22

    Redbelly98

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    I'm just now joining in this interesting thread.
    I was going to chime in with a somewhat lengthy explanation of a suitable equation, then realized that it was equivalent in form to this one. I'll just add that I would take g=1, the reasoning is that if there are even 2 rhinos then there is a nonzero probability that they could produce offspring together.

    I'll just mention, be prepared that you may have to solve this numerically.

    EDIT: ... or go to wolfram alpha to see the solution. (The solution surprised me, but I won't spoil it for HeLiXe.)
     
    Last edited: Jul 15, 2011
  24. Jul 15, 2011 #23

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    Hey RB, welcome to this thread! :smile:
    (The more the merrier.)

    Yes, but since the chance that they would meet is supposedly low, wouldn't the expectancy for the number of rhino's in the next generation be a decreasing number?
     
  25. Jul 15, 2011 #24
    Yes I'll take this into consideration. All lengthy explanations are welcome :biggrin:

    Thank you Redbelly98!! Please no one tell me -_- i will avoid Wolfram Alpha like the plague! Thanks for putting up with me taking long to respond :biggrin: I am going through my Chem work at a good pace and should be finished tomorrow. As soon as I am done, I will be immersed in DE until Wednesday of next week =)
     
  26. Jul 15, 2011 #25

    Redbelly98

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    Thanks!

    I had the impression that we were ignoring rhinos dying, in which case the expectancy is still a slightly larger number. If we want to include dying off in the model, I am thinking we would add a term [itex]-k_2 \ p[/itex] to the differential equation:

    [tex]\frac{dp}{dt} = k_1 p (p/g - 1) -k_2 \ p[/tex]

    Oh, hey, wait a minute...

    [tex]\begin{equation*}
    \begin{split}
    & = k_1 \ p \ [p/g - (1+k_2/k_1)] \\
    & = k_1 \ (1+k_2/k_1) \ p \ [\frac{p}{g (1+k_2/k_1)} - 1] \\
    & = k \ p \ [\frac{p}{g'} - 1]
    \end{split} \end{equation*}
    [/tex]
    So never mind -- we get the same equation again. Mortality can be accounted for by choosing a larger value for g.

    Okay, here is what I was thinking:

    1. The probability that a single rhino meets another rhino and reproduces (in some time interval) is proportional to the density of rhinos (or to the number of rhinos, if we take the land area to be a fixed parameter).

    2. The rate of population growth is proportional to the number of rhinos, and to the probability that a single rhino reproduces, i.e. [itex]\frac{dp}{dt}=kp^2[/itex]

    3. If we modify the probability that a single rhino reproduces to include a threshold, or include a mortality rate, we get [itex]\frac{dp}{dt}=kp(p/g - 1)[/itex] instead.
     
    Last edited: Jul 15, 2011
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