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Inverse mapping theorem & local inverses

  1. Nov 29, 2007 #1
    [Related concepts: Inverse mapping theorem, transformations and coordinate systems]

    1) For each of the following transformations (u,v) = f(x,y), (i) compute det Df, (ii) find formulas for the local inverses of f when they exist.
    a) u=x^2, v=y/x
    b) u=(e^x) cos y, v=(e^x) sin y



    I got part (i) easily, but I don't understand part (iii) at all.

    Here are the answers from the solutions manual:
    For 1a)(iii), the answer is f^-1 (u,v) = (sqrt u, v (sqrt u) )
    For 1b)(iii), the answer is
    x= ln(u^2 + v^2) / 2
    y is given up to multiples of 2pi by arctan(v/u) when u>0, pi/2 - arctan(u/v) when v>0, pi + arctan(v/u) when u<0, 3pi/2 - arctan(u/v) when v<0

    But they didn't show any of the steps, nor do they show me how they arrive at the answers. Can someone please explain this part? How exactly can I find the formula for the local inverse? I am terribly confused...

    Thanks a million!
     
  2. jcsd
  3. Nov 29, 2007 #2

    Hurkyl

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    You have two equations. Solve for x and y.
     
  4. Nov 30, 2007 #3
    The first one should be fairly easy. In the second one what you need to do is eliminate x so that you can solve for y in terms of u and v only and then eliminate y so that you can solve for x in terms of u and v only.

    If you're stuck on solving for x and y, then just consider u^2 + v^2 and u/v.
     
  5. Nov 30, 2007 #4

    HallsofIvy

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    The crucial point is what Hurkyl said: solve u= x2, v= y/x for x and y in terms of u and v. Try it. If you have difficulty get back to us.
     
  6. Nov 30, 2007 #5
    1a)(iii) Solving for x and y, I get x=+/-sqrt u, y=+/-v (sqrt u), but the answer says that f^-1 (u,v) = (sqrt u, v (sqrt u) ), they only take the positive signs, this is what I don't understand...

    How can I do so?

    Thanks!
     
    Last edited: Nov 30, 2007
  7. Nov 30, 2007 #6

    HallsofIvy

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    A function does not have two values! Since you say you got part i easily, I presume you showed that there is a local inverse as long as x is not 0. You can guarantee a local inverse by restricting x to be positive or negative. If you restrict x to be positive, the "local" inverse function is [itex]x= \sqrt{u}[/itex], [itex]y= v\sqrt{u}[/itex]. If you restrict x to be negative, the "local" inverse function is [itex]x= -\sqrt{u}[/itex], [itex]y= -v\sqrt{u}[/itex]. Each part has a unique inverse function- you can't write them together.

     
  8. Nov 30, 2007 #7
    Yes, I showed that there is a local inverse as long as x is not 0


    So now there are two answers to (iii) [itex]x= \sqrt{u}[/itex], [itex]y= v\sqrt{u}[/itex] and [itex]x= -\sqrt{u}[/itex], [itex]y= -v\sqrt{u}[/itex].
    Now how can I decide which one to take? If I just take one of them, would it still be a complete answer? Which one is the correct answer that the question is actually looking for?
     
    Last edited: Nov 30, 2007
  9. Dec 1, 2007 #8
    Can someone please explain this part? I really want to understand it...Thanks!
     
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