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Differential Geometry: angle between a line to a curve and a vector

  1. Feb 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Let α(t) be a regular, parametrized curve in the xy plane viewed as a subset of ℝ^3. Let p be a fixed point not on the curve. Let u be a fixed vector. Let θ(t) be the angle that α(t)-p makes with the direction u. Prove that:

    θ'(t)=||α'(t) X (α(t)-p)||/(||(α(t)-p)||)^2

    2. Relevant equations



    3. The attempt at a solution
    I'm not really sure how to approach this problem. I know what it is asking though. I have tried to extend the tangent line to the point of intersection and meeting it with u to make a triangle and applying the law of cosines but that didn't get me anywhere. I suspect this problem will ask me to use the angle definition of cross product: aXb=absin(θ) but I really don't know.
     
  2. jcsd
  3. Feb 27, 2013 #2

    BruceW

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    Homework Helper

    Re: Differential Geometry: angle between a line to a curve and a vecto

    are you sure that you have written the question down correctly? The equation you are trying to prove doesn't contain u, but does contain theta. Which makes me think that it might not be right.

    EDIT: Ah, wait, I'm just being stupid. The equation is correct. The curve lies in a 2d space. So to begin with, imagine p was not in the equation, and imagine that u is a fixed vector in 2d space and so a is just a vector in 2d, and the equation only involves 2d space. Do you recognise the equation then? Then later, you can show that adding p and making u 3d doesn't change the equation.
     
    Last edited: Feb 27, 2013
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