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Math Help - Phase Line & Limits

  1. #1
    Junior Member
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    Phase Line & Limits

    I understand how to create a horizontal phase line of a function...
    \frac{dx}{dt}=5x^2(x+5)(x-3)^3

    --->-----(-5)-----<------(0)------<------(3)----->----


    ...but I don't really understand what it means. What does the horizontal line represent and what do the arrows represent?

    And what procedure would I use for answering the question:

    The solution of the above equation with x(0)=-4 is defined on the interval  a<t<b , find
     lim_{t\to a+} x(t)=
     lim_{t\to b-} x(t)=

    Is there a way to find the limit analytically? I'm guessing not since we're using the phase line to do it.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    The function v:\mathbb{R}\to \mathbb{R},\;v(x)=5x^2(x+5)(x-3)^3 represents a velocity field. That is, the equation x'(t)=v(x) is equivalent to find the movement x=x(t) of a particle in the real line such that its velocity in any instant of time t is exactly v(x) . The arrows represent the movement sense. Hope this helps.
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