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Question about changing line element.

  • Thread starter ozone
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  • #1
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Homework Statement


Say we have a function such that
[itex] x = uv , y = (u^2 - v^2) /2 [/itex]
Hence our line element in Cartesian coordinates is.
[itex] ds^2 = dx^2 + dy^2 [/itex]

Now I have two questions. I like to work on math problems algebraically if possible so I thought to convert our line element I could take the implicit derivative of both x and y.

Hence we have

[itex] dx = (du*v) + (u*dv), dy = (2u du - 2v dv) [/itex]
However if i were to square these terms out I would then have some mixed components such as [itex] 4vu du dv[/itex]

I tested this technique with polar coordinates and using [itex] x = r \cos \theta, y = r \sin\theta [/itex] and I managed to come out to the proper line element.

Can someone please point me in the right direction as to what I need to do to get the proper line element?

Also I had one more question. If i wanted to find the equation of the unit circle centered at the origin in our new coordinate system can I just take the cartesian equation [itex] x^2 + y^2 = 1 [/itex] , plug in our substitutions, and then solve?
 
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Answers and Replies

  • #2
vela
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You forgot the factor of 1/2 when you calculated dy.
 
  • #3
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thats true I forgot to write that down when I was making the post.. so this is the correct way to set up the line element?
 
  • #4
vela
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Yup, and you should find with the correction the cross-term vanishes.

And to answer your other question, yes, you can just substitute for x and y into the equation for a circle and get the equivalent equation in terms of u and v. It works the way you think it should.
 

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