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Thread: Word Problem - Integration

  1. #1
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    Word Problem - Integration

    Hi,

    I don't think I know how to do all of this yet however:

    The evolution of a population of rhinoceros, R(t), in hundreds, time in years, in anAfrican National Park is given by the equation,

    dR/dt =1/5R(2 − R); R(0) = 0.5

    (a) Solve the system exactly for R(t).
    (b) What happens as the time t → ∞, i.e. what is the population a long time in the future?
    (c) Write an Euler scheme and compute until the population levels off (using Excel, Matlab,Octave, LibreCalc or similar).
    (d) Plot both of your solutions on the same set of axes and comment.

    (a) dR/dt = 1/5 R(2-R)
    Multiply R through brackets - dR/dt = 1/5 2R + R2 or dR/dt = 2R+R2/5
    Divide both sides by dR - dt = (2R+R2)/5

    If I am correct I can see dt = a number plus its derivative - does this mean I can use integration by substitution setting R2 as the number to be substituted? Or am I on the complete wrong planet?

    Also for(b) am I looking for the absolute maxima of the equation?

    Kind regards
    Beetle
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  2. #2
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    Re: Word Problem - Integration

    For part (a) - I meant to say - is that just the product rule backwards?
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    Re: Word Problem - Integration

    Just want to check the function. Is it \frac {dR}{dt} =\frac{1}{5}R(2-R) or is it \frac {dR}{dt} =\frac{1}{5}t(2-t)?
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  4. #4
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    Re: Word Problem - Integration

    Hi, it is the first one.
    Cheers
    Beetle
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    Re: Word Problem - Integration

    Quote Originally Posted by Beetle37 View Post
    For part (a) - I meant to say - is that just the product rule backwards?
    Integration by separation of variables, then partial fraction decomposition ...
    Last edited by skeeter; Apr 1st 2016 at 03:36 AM.
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    Re: Word Problem - Integration

    Again (I hate to say it), have a look at the example in:
    https://en.wikipedia.org/wiki/Separation_of_variables

    For (b) you need to consider the limit of the function R(t) as t ->infinity.
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  7. #7
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    Re: Word Problem - Integration

    dR/dt =1/5R(2 − R); R(0) = 0.5
    $\dfrac{dR}{R(2-R)} = \dfrac{1}{5} \, dt$

    partial fraction decomposition ...

    $\dfrac{1}{R(2-R)} = \dfrac{A}{R} + \dfrac{B}{2-R}$

    $1 = A(2-R) + BR$

    $R = 0 \implies A = \dfrac{1}{2}$ ; $R=2 \implies B = \dfrac{1}{2}$

    $\displaystyle \dfrac{1}{2} \int \dfrac{1}{R} + \dfrac{1}{2-R} \, dR = \dfrac{1}{5} \int \, dt$

    $\displaystyle \int \dfrac{1}{R} - \dfrac{-1}{2-R} \, dR = \dfrac{2}{5} \int \, dt$

    $\ln|R| - \ln|2-R| = \dfrac{2}{5} t + C$

    $\ln\left|\dfrac{R}{2-R}\right| = \dfrac{2}{5} t + C$

    $\dfrac{R}{2-R} = e^{2t/5} \cdot e^C$ ; let $e^C = A$

    $\dfrac{R}{2-R} = Ae^{2t/5}$

    $R = 2Ae^{2t/5} - RAe^{2t/5}$

    $R + RAe^{2t/5} = 2Ae^{2t/5}$

    $R(1 + Ae^{2t/5}) = 2Ae^{2t/5}$

    $R = \dfrac{2Ae^{2t/5}}{1 + Ae^{2t/5}}$

    $R(0) = \dfrac{1}{2} \implies A = \dfrac{1}{3}$

    $R = \dfrac{2e^{2t/5}}{3 + e^{2t/5}}$

    $R = \dfrac{2}{1 + 3e^{-2t/5}}$ ... a logistic curve


    $\displaystyle \lim_{t \to \infty} \dfrac{2}{1 + 3e^{-2t/5}} = 2$
    Attached Thumbnails Attached Thumbnails Word Problem - Integration-logistic.png  
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  8. #8
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    Re: Word Problem - Integration

    shortcut ...

    general model for logistic growth is

    $\dfrac{dy}{dt} = k\left(1 - \dfrac{y}{M}\right) \implies y = \dfrac{M}{1 + Be^{-kt}}$ where $M$ is the carrying capacity for $y$ and $B = \dfrac{M-y(0)}{y(0)}$

    transforming $\dfrac{dR}{dt} = \dfrac{1}{5}R(2-R)$ to the form shown above ...

    $\dfrac{dR}{dt} = \dfrac{2}{5}R\left(1-\dfrac{R}{2}\right)$

    $k = \dfrac{2}{5}$, $M = 2$, $B = \dfrac{2-0.5}{0.5} = 3$

    $y = \dfrac{M}{1 + Be^{-kt}} \implies R = \dfrac{2}{1+3e^{-2t/5}}$
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