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Geometrical interpretation of this coordinate transformation

  1. May 20, 2014 #1
    How can I geometrically interpret this coordinate transformation (from x,y space to [itex]\check{x}[/itex],[itex]\check{y}[/itex] space)?

    x = [itex]\check{x}[/itex]cos(β) - [itex]\check{y}[/itex]sin(β)
    y = [itex]\frac{1}{2}[/itex]([itex]\check{x}[/itex]2 -[itex]\check{y}[/itex]2)sin(2β) -[itex]\check{x}[/itex][itex]\check{y}[/itex]cos (2β)
  2. jcsd
  3. May 20, 2014 #2


    Staff: Mentor

    Did you mean to interpret the transformation from x'y' to xy space?

    This looks like a homework assignment so you need to show some work for us to help you. I would suggest trying to sketch it out and then interpret your sketch of coordinate lines.

    So to start if you pick beta to be zero radians then x=x' and y= - x'y'

    then sub in x for x' you get y = - x^2 which is what kind of curve?
  4. May 21, 2014 #3
    Its not a HW problem. I am trying to solve an integral where the integrand is a transcendental function.

    Using the coordinate transformation here, I came up with the above coordinate transformation for my case. What I need to do now is to draw the new [itex]\hat{x}[/itex][itex]\hat{y}[/itex] on top of x,y cartesian coordinate system to get the new limits of integration, before attempting to solve the integral.

    The integral is from (0 → 1) in x,y so I need to figure out only the upper limit of integration in the new coordinate system. Also, both sin (β) and cos (β) are known.

    So, I need some direction on how to proceed to draw the new coordinate system on x,y.

    Let me know if more information is required.
    Last edited: May 21, 2014
  5. May 21, 2014 #4


    Staff: Mentor

    I don't know how to help at this point. Can you show us the integral?

    Are you integrating from x=0 to x=1 over some curve in x and y ie a line integral? and now you have a new coordinate system x', y' where you need to find the x'(0,y(0)) and x'(1,y(1)) or something like that? and the beta is a constant or does it vary with x' and y'?
  6. May 21, 2014 #5
    Exactly. So, assuming I have done everything correctly - here are the details.

    I am trying to evaluate

    01y(x,y)dx, where

    y= (a+b)*x - b*x*exp(p*x+q*y) - c*(p*x2 + q*y)

    with a, b, c, p and q all known constants. Also, its known that y = 0 at x = 0 and y = 0 at x =1.

    Now using the coordinate transformation above, and sin(β) = p/h and cos (β) = q/h together with h2 = p2 + q2, I get (in the transformed coordinates)

    [itex]\check{y}[/itex] = [itex]\frac{h}{p}[/itex] [(a + b) - b*exp (h*[itex]\check{x}[/itex]) - (c*h + q/h)[itex]\check{x}[/itex]] which is only a function of [itex]\check{x}[/itex] in R.H.S.

    Now I need to evaluate UL [itex]\check{y}[/itex]d[itex]\check{x}[/itex]. U and L need to be determined using a geometric interpretation of the new transformed coordinates. U should 0 but I need some direction on how to determine L.

    Ideally I think I need to draw the new coordinate system on x,y which I am not good at. Any help is much appreciated.
  7. May 21, 2014 #6


    Staff: Mentor

    If this is a line integral can't you parameterize it and then using the parametrized equations to determine the upper and lower limits.

    If you're trying to plot the coordinate lines to get an understanding of the mapping then perhaps MATLAB (or Freemat) could do that given your equations and iterating them where you vary the beta for each plot line holding x and then y constant for each line.
    Last edited by a moderator: Sep 25, 2014
  8. May 22, 2014 #7
    Do you mean that I integrate in (x,y) but over the curve [itex]\check{y}[/itex]?
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