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Vector Function?

  1. Jul 10, 2010 #1
    1. The problem statement, all variables and given/known data

    Sketch a function V= -yx'+xy' ?

    2. Relevant equations



    3. The attempt at a solution

    i have compared it with r= xi'+yj'. and putting different values of y and x to sketch it on y -axis and x axis. is it correct. how should i do it.
     
  2. jcsd
  3. Jul 10, 2010 #2
    You have here a vector function of x and y, that means that with every point on the xy plane, there is associated a vector, with magnitude and direction.
    I suggest you find both of these separately and then sketch your diagram using something like this method:

    http://tutorial.math.lamar.edu/Classes/CalcIII/VectorFields.aspx

    Each little arrow starts at the point it describes, and gives both the direction of the vector field there, and its relative magnitude (Longer arrows for greater values of the magnitude)

    Can you find the magnitude and direction of a general point of the vector field?

    For the magnitude:
    [tex]|V| = \sqrt{V_x^2 + V_y^2}[/tex]

    For the direction:
    [tex]\hat V = \frac{\vec V}{|V|}[/tex]

    After you get the direction you may find it useful to look at it using polar coordinates. That should make the orientation of the direction very clear when you compare it with the simple radial direction: [tex]\hat r = \cos{\theta} \hat x + \sin{\theta} \hat y[/tex]
    Try and rearrange what you get into the form [tex]\hat V = \cos{(\theta + \phi)} \hat x + \sin{(\theta + \phi)}\hat y[/tex] and then you'll see by how much to rotate the radial unit vector to get the direction at the point.

    If you know how to take the curl (rotor) of a function, you may find that tool helpful here for a qualitative feel of what this field is like.

    For a complete explanation on the subject (With your particular example in use, even!) :
    http://math.etsu.edu/Multicalc/Chap5/Chap5-1/part4.htm [Broken]
     
    Last edited by a moderator: May 4, 2017
  4. Jul 10, 2010 #3
    Thanks .. i got it.. really good explanation
     
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