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Scalar fields

  1. Mar 24, 2007 #1
    Ok, I've been trying to work this out for a couple of hours now and I'm completely stumped. Not even Google was helping much.

    The question is:
    Find and Sketch the level curves of the scalar field T(x,y) = (x +y)/(x2 + y2) for T = -1, -0.5, 0, 0.5, 1

    I know that I should equate the equation to the values of T given and I suspect that I'll be getting 4 circles but it doesn't seem to work.

    Here's what I've done:

    (x +y)/(x2 + y2) = -1
    x + y = -(x2 + y2)

    Where do I go from here?!
     
  2. jcsd
  3. Mar 24, 2007 #2
    0 is easy. 1 and .5 should be easy too. The negative values are a bit tricky. A brute force way would be to put the x's on one side and y's on the other and then set the [tex] x+x^2 [/tex] to an arbitary value and then try to solve the quadratic equation for [tex] -y-y^2 = value [/tex]. The only trick is that your teacher probably doesn't want you to wander into imaginary (i) numbers.
     
    Last edited: Mar 24, 2007
  4. Mar 24, 2007 #3

    Dick

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    The correct way is to simply complete the squares to read off the properties of the circles. This has nothing to do with imaginary numbers.
     
  5. Mar 24, 2007 #4

    HallsofIvy

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    You have
    (x +y)/(x2 + y2) = -1
    x + y = -(x2 + y[/sup]2[/sup])
    Don't stop there!
    x+ y= -x2- y2
    x2+ x+ y2+ y= 0

    Now, as Dick said, complete the square to get the equation of the circle.
     
  6. Mar 24, 2007 #5
    Not circles

    I'm surprised everyone thinks this field's level curves are circles. They can't be.

    Switch to a polar coordinate system:

    [tex]x=r\cos\theta[/tex]
    [tex]y=r\sin\theta[/tex]
    [tex]T(r,\theta)=(\cos\theta+\sin\theta)/r[/tex]

    For [tex]T=0[/tex], you get [tex]\cos\theta+\sin\theta=0[/tex] which occurs at [tex]\theta=-\pi/4[/tex]

    For the other values, you have

    [tex]\cos\theta+\sin\theta=Tr[/tex] for [tex]T=-1,-0.5,0.5,1[/tex]

    and the only way these could be circles is if you could eliminate the [tex]\theta[/tex] dependence.

    ZM
     
    Last edited: Mar 24, 2007
  7. Mar 24, 2007 #6

    Dick

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    I'm really impressed! You never considered that the circles might not be centered on the origin. Did you?
     
  8. Mar 24, 2007 #7
    Doh!

    It's early in the morning. :)

    I just gave the polar equation for a circle for [tex]T\ne0[/tex], didn't I? :-)

    ZM---going back to bed for a few hours
     
    Last edited: Mar 24, 2007
  9. Mar 25, 2007 #8
    Thanks guys! I'll have Complete The Square burned into my memory from now on.
     
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