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Plotting a circular vector field

  1. Jun 24, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm supposed to sketch the vector field and verify that all the vectors of the following equation have the same length.


    2. Relevant equations

    [tex]G(x,y) = \frac{-iy + jx}{\sqrt{x^2+y^2}}[/tex]


    3. The attempt at a solution

    If I start plugging in numbers, for example the point (1,1) into [tex]\frac{-iy}{\sqrt{x^2+y^2}}[/tex] and into [tex]\frac{jy}{\sqrt{x^2+y^2}}[/tex] I move minus 1 unit along the x axis, and up one unit along the Y axis. So that vector has a length of 1 unit. But if I put in (2,2) to the same equation I move [tex]\frac{-2}{\sqrt{8}}[/tex] in the negative X direction and the same in the positive Y direction. I don't see ho..Oh. Now that I'm typing it out in LaTeX I see it. If I take the magnitude of the new i,j vector I get from evaluating the equation, I'm going to get 1 aren't I? And no matter what values I plug in to the original equation, the magnitude of the resulting vector is always going to work out to 1. Does that sound correct?
     
    Last edited: Jun 24, 2009
  2. jcsd
  3. Jun 24, 2009 #2

    lanedance

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    Homework Helper

    hey bitrex, sounds like you're heading in teh right direction

    try taking the magintude of an arbitrary vector (in terms of x,y) and see if it simplifies to one, then you've shown it for every x,y, though you may have to be careful at the origin.

    Then plotting will only involve the direction of the vector as the magnitude is constant.
     
  4. Jun 25, 2009 #3

    HallsofIvy

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    What is
    [tex]\sqrt{\left(\frac{-y}{\sqrt{x^2+ y^2}}\right)^2+ \left(\frac{x}{\sqrt{x^2+ y^2}}\right)^2[/tex]
    It's that easy.
     
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