Question about changing line element.

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The discussion centers on converting the line element from Cartesian to a new coordinate system defined by the functions x = uv and y = (u^2 - v^2)/2. The user attempts to derive the line element algebraically by taking implicit derivatives but encounters mixed components in the squared terms. It is clarified that correcting for a missing factor of 1/2 in dy resolves the issue, allowing the cross-term to vanish. Additionally, substituting the Cartesian equation of the unit circle into the new coordinate system is confirmed to yield the correct equivalent equation. The overall focus is on ensuring accurate transformations and derivatives in the context of line elements and coordinate substitutions.
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Homework Statement


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

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

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

I tested this technique with polar coordinates and using x = r \cos \theta, y = r \sin\theta 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 x^2 + y^2 = 1 , plug in our substitutions, and then solve?
 
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You forgot the factor of 1/2 when you calculated dy.
 
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?
 
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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