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Equation of the Osculating circle

  1. Nov 17, 2013 #1
    1. The problem statement, all variables and given/known data
    The ellipse has equations x = 2cos(t) and y = 3sin(t) where 0 <= t <= 2*pi
    The problem asked me to calculate the curvature at points (2,0) and (0,3). I did that, but now the problem asks what the equation of the osculating circle is at each of those points. I know the radius of curvature of each of those points since I already calculated the curvature. But I'm not sure how to figure out where the center of the osculating circle needs to be.


    2. Relevant equations
    I used this equation for calculating curvature of a parametric curve: http://math.info/image/58/curvature_parametric.gif


    3. The attempt at a solution
     
  2. jcsd
  3. Nov 17, 2013 #2

    LCKurtz

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    You know the radius of the osculating circle is equal to the radius of curvature. To find the center of the osculating circle, move from the curve a distance equal the radius along the direction of the principle normal (the direction of the normal component of acceleration).
     
  4. Nov 17, 2013 #3
    Thank you for your response. Before you replied I thought of something. Tell me if it's a valid argument. Since the graph of this parametric is symmetrical, and both my points are on either the y or x axis, wouldn't the center of the osculating circle be on the x or y axis? So I could just subtract the radius from 2 for the first one and 3 for the second one. Is that valid? For example: the point (2,0) had a radius of curvature of 27/6. I could subtract that from 2 and get -5/2. So the center would be -5/2. Is that correct?
     
  5. Nov 17, 2013 #4

    LCKurtz

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    That looks correct. Luckily, you have points where the direction of the normal is easy.
     
  6. Nov 17, 2013 #5
    I hope my instructor doesn't dock me points for taking a shortcut haha
     
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