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Find the radius of the osculating circle

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data

    Let C be the curve of intersection of the cylinder x^2-2x+y^2 =3 and the plane z=y+1. Find the radius of the osculating circle of the curve C at point (1,2,3)


    3. The attempt at a solution

    I am not really sure how to start it...
    i tried finding the point of intersection by setting y=0 and got z=1, x=3, but doesnt really make sense to me to do so. I also tried setting the equation to each other, but it doesnt seem to make sense too..

    it would be appreciated if anyone can tell me how to begin and guide me through it

    Thanks in advance
     
  2. jcsd
  3. Nov 8, 2009 #2

    Dick

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    Your first job is to find a parametric expression for the curve r(t)=(x(t),y(t),z(t)). Write the first equation as (x-1)^2+y^2=2. That's the equation of a circle with center (1,0) and radius sqrt(2), right? Can you write expressions for x and y in terms of cos(t) and sin(t) where t is the angle the point makes with the circle center? Now can you use z=y+1 to write z in terms of t? Now I'm guessing you probably know how to find the radius from the parametric expression using the curvature.
     
  4. Nov 9, 2009 #3
    I think i get it, but here is what i got

    re parametrize the intersection, and i got c=(x=2cost+1,y=2sint, z= 2sint+1) and point (1,2,3) is t=pi/2
    and to find radius is R=1/curvature at t=pi/2?
     
  5. Nov 9, 2009 #4

    Dick

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    That looks ok to me.
     
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