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Finding The Coordinates of The Center Of Curvature

  1. Jul 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Let C be a curve given by y = f(x). Let K be the curvature ([itex]K \ne 0[/itex]) and let [itex]z = \frac{1+ f'(x_0)^2}{f''(x_0)}[/itex]. Show that the coordinates [itex]( \alpha , \beta )[/itex] of the center of curvature at P are [itex]( \alpha , \beta ) = (x_0 -f'(x_0)z , y_0 + z)[/itex]


    2. Relevant equations



    3. The attempt at a solution

    I attached a picture of the solution. The portion of the solution with a half-box I would appreciate someone helping me with.
     

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    Last edited: Jul 3, 2013
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  3. Jul 3, 2013 #2

    LCKurtz

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  4. Jul 3, 2013 #3
    No, it is no longer that day in which I learn Linear Algebra. Today is devoted to reviewing multivariable calculus and electricity and magnetism. Tomorrow is when I resume Linear Algebra.
     
  5. Jul 3, 2013 #4

    LCKurtz

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    With respect to your question in this thread, if f'' > 0 the curve is concave up, so the circle is above the curve. At the point shown, f'' < 0 so the curve is concave down and the circle is below the curve. That determines whether ##\beta## is less or greater than ##y_0##.
     
  6. Jul 3, 2013 #5
    Ooh, you are very right. Thank you much for your help.
     
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