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Proving Magnutude Of Vector Valued Function Is Constant

  • Thread starter Lancelot59
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  • #1
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Proving Magnitude Of Vector Valued Function Is Constant

I have a rather neat problem. I need to prove that the magnitude of this function:

[tex]F_{2}(x,y)=\frac{<-x,y>}{x^{2}+y^{2}}[/tex]

is constant along circles centred about the origin. Now while proving that the magnitude is inversely proportional I had to get the magnitude, and it wound up being:

[tex]\sqrt{\frac{1}{(x^{2}+y^{2})}}[/tex]

Which looks like the basic equation for a circle to me. That particular function will give elliptical level curves. I'm not sure how to go about this. Can I just say that because it's the inverse of a the form of an ellipse that if the level curve x^2+y^2 is equal to a constant c^2 then the magnitude will also be a constant because it has the same form?
 
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Answers and Replies

  • #2
Char. Limit
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that's not what I get for the magnitude. Can you show me how you got that answer?
 
  • #3
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This is how I did it:

First I multiplied in the scalar x^2+y^2.
[tex]=<\frac{-x}{x^{2}+y^{2}},\frac{y}{x^{2}+y^{2}}>[/tex]

Then put it in the usual form:
[tex]=\sqrt{(\frac{-x}{x^{2}+y^{2}})^{2}+(\frac{y}{x^{2}+y^{2}})^{2}}[/tex]

They have a common denominator, so I added them:
[tex]=\sqrt{\frac{x^{2}+y^{2}}{(x^{2}+y^{2})^{2}}[/tex]

Which then simplifies down to:
[tex]=\sqrt{\frac{1}{x^{2}+y^{2}}[/tex]
 
  • #4
Char. Limit
Gold Member
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Whoops, my bad.

Well, your magnitude is equal to [tex]\frac{1}{r}[/tex], isn't it? Seems pretty obvious to me.
 
  • #5
634
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Whoops, my bad.

Well, your magnitude is equal to [tex]\frac{1}{r}[/tex], isn't it? Seems pretty obvious to me.
I see what you did there. Thanks!
 
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