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Sketching a graph from its differential equation Watch

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    I have the DE of the Verhulst Model:

     \frac{dN}{dt} = rN\left(1 - \frac{N}{K} \right)

    where 'r' and 'K' are constants.

    I have solved it and got

     N(t) = \frac{K}{\left( \frac{K}{N_0} - 1 \right) e^{-rt} + 1}

    where  N(0) = N_0 .

    My question then says:

    "Sketch the graph of the solution of N(t) against 't' for  t \geq 0 , at fixed values of parameters 'r', 'K' and for two fixed values of  N_0 , one in the range  0 < N_0 <K and other in the range  N_0 > k . Determine the behaviour of the solution for large times, i.e in the limit  t \rightarrow + \infty for a given $N_0 \geq 0 [/tex]."

    I am stuck on how to sketch this graph. I now that there are two stationary points at N = 0 and N= K, but I'm not sure what to do now. K is the carrying capacity for the population, so I'm guessing that is always positive. So that means, when N > K, I get  1 - \frac{N}{K} > 0 and so  \frac{dN}{dt} < 0  \implies \mathrm{for} \, 0 < N < K, \frac{dN}{dt} > 0.. I think this is all correct.

    The bit I don't get is how I bring the  N_0 into it. If we say that if our starting point goes from  N_0 is slightly greater than 0 and we look at the range from when  N_0 is slightly less than K, does that mean that as we get closer to K, the rate at which N(t) is increasing slowly decreases? And so if [/tex] N_0 > K [/tex], then N(t) will constantly decrease?

    So what does this tell us about the graph as  t \rightarrow \infty ? All I can think is that if  N_0 > K , then N(t) constantly decreases and so will die out quickly. If  0 < N_0 < K then the closer it gets to K, the longer the population will go on for as the lower the rate of increase and so the lower the rate of decrease?

    Is that right? Does that make sense?
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    If N_0>k then dN/dt will initially be <0 and N will decrease until N=k and dn/dt=0 and the population stops changing.

    If N_0<k then dN/dt will be positive and N will increase until n=k and ...
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    (Original post by BabyMaths)
    If N_0>k then dN/dt will initially be <0 and N will decrease until N=k and dn/dt=0 and the population stops changing.

    If N_0<k then dN/dt will be positive and N will increase until n=k and ...
    Ok, but after it gets to n=k, why will the population stay constant? Once its passed that point, would it increase or decrease again? Or IS that not really a pure maths question and more to do with how the model works?

    This was posted from The Student Room's Android App on my HTC Desire
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    Actually I should have said N-> k and dN/dt ->0.

    Reality is another matter.
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    (Original post by claret_n_blue)
    Ok, but after it gets to n=k, why will the population stay constant? Once its passed that point, would it increase or decrease again? Or IS that not really a pure maths question and more to do with how the model works?

    This was posted from The Student Room's Android App on my HTC Desire
    When N=k, dN/dt = 0, which means...
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    (Original post by BabyMaths)
    Actually I should have said N-> k and dN/dt ->0.

    Reality is another matter.

    (Original post by around)
    When N=k, dN/dt = 0, which means...
    Yes, but why does this show an asymptote? What happens when N passes point K, then according to dN/dt, the gradient starts changing again doesn't it?
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    (Original post by claret_n_blue)
    Yes, but why does this show an asymptote? What happens when N passes point K, then according to dN/dt, the gradient starts changing again doesn't it?
    The point is that that we have an asymptote at N=k, so the graph never gets past N=k.
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    How can

    \dfrac{dN}{dt} = 0

    imply an asymptote at N=k?
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    (Original post by Indeterminate)
    How can

    \dfrac{dN}{dt} = 0

    imply an asymptote at N=k?
    The only thing I can think of is that if we look at the function of N(t), it isn't defined at N = K, and so there is an asymptote there. I'm not sure why its a horizontal asymptote though?
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    (Original post by claret_n_blue)
    The only thing I can think of is that if we look at the function of N(t), it isn't defined at N = K, and so there is an asymptote there. I'm not sure why its a horizontal asymptote though?


    N(t) = K \Rightarrow N_0 = K

    I've never heard of a function being stationary at a point that's meant to be an asymptote to it.
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    (Original post by Indeterminate)
    How can

    \dfrac{dN}{dt} = 0

    imply an asymptote at N=k?
    Because, if you look at the differential equation, you see that as N approaches k, dN/dt approaches 0 from above.
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    (Original post by around)
    Because, if you look at the differential equation, you see that as N approaches k, dN/dt approaches 0 from above.
    Is it possible for a function to jump across its asymptote? And so once this has happened, something else happens further down the line after the point N = K?
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    (Original post by claret_n_blue)
    Is it possible for a function to jump across its asymptote? And so once this has happened, something else happens further down the line after the point N = K?
    No it's not: if N crosses the line y=K, then it would have to have a non-negative gradient at this crossing point, but the differential equation tells us that WHENEVER N=K dN/dt=0.
 
 
 
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