Results 1 to 2 of 2

Thread: Phase Line & Limits

  1. #1
    Junior Member
    May 2011

    Phase Line & Limits

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


    ...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 $\displaystyle x(0)=-4$ is defined on the interval $\displaystyle a<t<b $, find
    $\displaystyle lim_{t\to a+} x(t)= $
    $\displaystyle 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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Nov 2010
    Madrid, Spain
    The function $\displaystyle v:\mathbb{R}\to \mathbb{R},\;v(x)=5x^2(x+5)(x-3)^3$ represents a velocity field. That is, the equation $\displaystyle x'(t)=v(x)$ is equivalent to find the movement $\displaystyle x=x(t)$ of a particle in the real line such that its velocity in any instant of time $\displaystyle t$ is exactly $\displaystyle v(x)$ . The arrows represent the movement sense. Hope this helps.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Phase Space/ State Space/ Phase Portrait
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: Oct 12th 2010, 01:40 PM
  2. How to do: Phase Shift of tan
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Oct 17th 2009, 10:54 AM
  3. Replies: 1
    Last Post: May 31st 2009, 12:31 PM
  4. Line integrals - how to determine limits
    Posted in the Calculus Forum
    Replies: 13
    Last Post: Jan 6th 2008, 02:16 PM
  5. Out of phase
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Nov 30th 2007, 02:59 PM

Search Tags

/mathhelpforum @mathhelpforum