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Math Help - transformation - sorry really don't know how to describe the question!

  1. #1
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    transformation - sorry really don't know how to describe the question!

    For φ in [0, 2pi) and (x1, x2) in R2 we define a transformation.

    R(
    φ):R2 ---> R2
    x ---> R(
    φ)x=y=(y1,y2)
    where
    y1=x1cos
    φ+x2sinφ
    y2=-x1sinφ+x2cosφ

    Prove ||y||=||R(
    φ)x||=||x||
    ie. R(
    φ) leaves the euclidean norm invariant.

    Now geometry isn't my strong point to start with but I usually grasp the topics enough to do the homeworks well, but this may as well be in dutch for as much as I can figure out what to do. We haven't done anything even remotely similar in class and I'm at a loss as to where to even start. I would be really really grateful if someone could help me out a little. Thanks
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  2. #2
    GJA
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    Re: transformation - sorry really don't know how to describe the question!

    Hi carla1985,

    I think the notation of this problem may be making things more confusing than anything else. Remember that \|x\|=\sqrt{x_{1}^{2}+x_{2}^{2}} and \|y\|=\sqrt{y_{1}^{2}+y_{2}^{2}}. Try computing y_{1}^{2} and y_{2}^{2} individually, then add them together and see if you can get x_{1}^{2}+x_{2}^{2}. The trig. identity \cos^{2}\varphi+\sin^{2}\varphi=1 will be useful when doing this.

    Give it another shot. If you're still stuck let me know. Good luck!
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    Re: transformation - sorry really don't know how to describe the question!

    I agree, the notation is confusing, its still all qute new and is a lot to take in lol.
    I'l have another go at it now, using your suggestions and see how it goes. Thank you
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  4. #4
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    Re: transformation - sorry really don't know how to describe the question!

    here is what is happening:

    R(φ) takes a point x = (x1,x2) to another point y = (y1,y2).

    we want to show if |x| = a, that |R(φ)(x)| = a, as well.

    what do we mean by |x|? we mean: √[(x1)2 + (x2)2]

    (this is just the pythagorean theorem in disguise).

    so now we have to calculate |R(φ)(x)| = |y|.

    this is going to be √[(y1)2 + (y2)2].

    the algebra is going to get messy, here is where we start:

    |R(φ)(x)| = √[(y1)2 + (y2)2]

    = √[(x1cosφ + x2sinφ)2 + (-x1sinφ + x2cosφ)2]

    = √[(x1)2cos2φ + 2x1x2sinφcosφ + (x2)2sin2φ + (x1)2sin2φ - 2x1x2sinφcosφ + (x2)2cos2φ]

    i've done the hard part, you just have to simplify.

    (something which doesn't help with the calculation but makes what R(φ) "make sense" is this: R(φ) represents what a rotation through the angle φ around the origin does to the point x. |x| represents "how far it is from x to the origin". so the upshot of all of this is that you are proving: "rotations about the origin preserve distances from the origin", something that is intuitively obvious).
    Last edited by Deveno; November 19th 2012 at 02:33 PM.
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