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Proving that an osculating circle is unique

  1. Feb 18, 2008 #1
    1. The problem statement, all variables and given/known data

    e1, e2 are vector parameters such that dot(e1, e2) = [tex] \delta_{ij} [/tex]
    c is another vector parameter
    r, s are real valued parameters
    [tex]\beta[/tex] is a unit speed curve in E3

    [tex]\gamma[/tex] is a parametrization of a circle
    [tex]\gamma (s) = \overbar{c} + rcos (\frac{s}{r})\overbar{e_1} + rsin (\frac{s}{r})\overbar{e_2}[/tex]

    Prove there is one and only one circle [tex]\gamma[/tex] that approximates [tex]\beta[/tex] near [tex]\beta (0)[/tex] such that

    [tex] \gamma (0) = \beta (0), \gamma' (0) = \beta' (0), \gamma'' (0) = \beta'' (0) [/tex]

    2. Relevant equations
    Given a curve in R3 [tex]\beta[/tex] with unit speed,
    [tex] \beta' = T [/tex], where T is the tangent vector
    [tex] \beta'' = \kappa N [/tex] where N is the normal vector
    [tex] B = T \times N [/tex] where B is the binormal vector

    3. The attempt at a solution

    I found the first and second derivatives of gamma and evaluated them at 0:

    [tex]\gamma (0) = \overbar{c} + rcos (0)\overbar{e_1} + rsin (0)\overbar{e_2} = \overbar{c} + r \overbar{e_1}[/tex]
    [tex]\gamma' (0) = -sin (\frac{0}{r})\overbar{e_1} + cos (\frac{0}{r})\overbar{e_2} = -sin (0)\overbar{e_1} + cos (0)\overbar{e_2} = \overbar{e_2}[/tex]
    [tex]\gamma'' (0) = -\frac{1}{r} cos (0)\overbar{e_1} - \frac{1}{r} sin (0)\overbar{e_2} = -\frac{1}{r} \overbar{e_1} [/tex]

    In addition, I have the relationship from the beginning of the problem between e1 and e2. I actually don't understand this - what does delta mean?

    How do I prove that I will have a unique solution if I fix s?
    Last edited: Feb 19, 2008
  2. jcsd
  3. Feb 18, 2008 #2


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    Science Advisor
    Homework Helper

    My guess is, it is the Kronecker delta.
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