Scalar fields

  • Thread starter Nibbler
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  • #1
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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?!
 

Answers and Replies

  • #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.
 
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  • #3
Dick
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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.
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.
 
  • #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.
 
  • #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
 
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  • #6
Dick
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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=-1,0.5,1[/tex]

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

ZM
I'm really impressed! You never considered that the circles might not be centered on the origin. Did you?
 
  • #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
 
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  • #8
5
0
Thanks guys! I'll have Complete The Square burned into my memory from now on.
 

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