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Find the center of the circle of curvature

  1. Sep 15, 2016 #1
    1. The problem statement, all variables and given/known data
    For the curve with equation [itex]y={ x }^{ 2 }[/itex] at the point (1, 1) find the curvature, the radius of curvature, the equation of the normal line, the center of the circle of curvature, and the circle of curvature.

    2. Relevant equations


    3. The attempt at a solution
    [itex]\kappa \left( 1 \right) =\frac { 2\sqrt { 5 } }{ 25 } \\ r=\frac { 1 }{ \kappa } =\frac { 25 }{ 2\sqrt { 5 } } \\ y-1=\frac { -1 }{ 2 } \left( x-1 \right) [/itex] I have no idea how to find the center, I've tried the distance formula, and even a vector approach
     
  2. jcsd
  3. Sep 15, 2016 #2

    andrewkirk

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    The centre of the radius of curvature must be on the line that is perpendicular to the tangent to the curve at (1,1), the equation of which you can calculate. Having done that, what info do you already have that tells you how far along that line the centre of curvature is?
     
  4. Sep 15, 2016 #3
    Well, I have the line which goes straight through the center, and I also have the radius (how far I should go), but need an x and a y value
     
  5. Sep 15, 2016 #4

    andrewkirk

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    What is the equation of the line?
    Does it go through (1,1)?
    What are the coordinates of the point on that line that is distance r from (1,1), in the direction towards the inside of the curve?
     
  6. Sep 15, 2016 #5
    The equation of the line is [itex]y-1=-0.5(x-1)[/itex], and it goes through the point (1,1). The distance formula says [itex]r=\frac { 25 }{ 2\sqrt { 5 } } =\sqrt { { (1-{ x }_{ 0 }) }^{ 2 }+{ (1-y_{ 0 }) }^{ 2 } } [/itex], but there's an x and a y.
     
  7. Sep 15, 2016 #6

    andrewkirk

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    You have two equations, and two unknowns ##x_0## and ##y_0##.
     
  8. Sep 15, 2016 #7
    Got it, it's (x+4)^2 + (y-7/2)^2 = 125/4.
    Thanks
     
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