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Rotation of cartesian coordinate system

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Please see the rotation formula in the attachment.

    2. Relevant equations



    3. The attempt at a solution
    I understand this formula rotates x,y into x',y' by some angle theta. Problem is, how is this formula derived? I cannot for the life of me visualize the cosine and sine transformation physically. Can someone explain to me how you get this formula. Thank you very much.
     

    Attached Files:

  2. jcsd
  3. Sep 10, 2012 #2

    gneill

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    Staff: Mentor

    Consider the vector that extends from the origin to the point (x,y) in the base coordinate system. It has some magnitude R and angle β with respect to the x-axis of the coordinate system. In fact, x = Rcos(β) and y = Rsin(β).

    Rotating that point around the origin by some angle θ is equivalent to rotating the vector by θ, so what would the coordinates of its endpoint be?
     
  4. Sep 11, 2012 #3
    so the end points would be x=Rcos(β+θ), y=Rsin(β+θ). Then what? I still can't see how this relates to the formula, espcially how in the formula for x' and y' individually there are x and y terms together.
     
  5. Sep 11, 2012 #4
    What you just found are the [itex]x'[/itex] and [itex]y'[/itex] coordinates. Expand the sines and cosines using angle sum formulas and put any sines or cosines of [itex]\beta[/itex] in terms of the origina [itex]x,y[/itex].
     
  6. Sep 11, 2012 #5
    x' = R(cosβcosθ-sinβsinθ)
    y' = R(sinβcosθ+sinβcosθ)

    x' = R[(x/R)cosθ-(y/R)sinθ]
    y' = R[(y/R)cosθ+(y/R)cosθ]

    arrrrgh almost there. First term in y' is wrong. I get y' = ycosθ .... instead of y' = xsinθ ...


    Can someone point out myt mistake? Thanks a lot for your help!
     
  7. Sep 11, 2012 #6
    In your second line, you forgot to switch beta and theta in the second term.
     
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