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View Full Version : Proving that an osculating circle is unique


Eponym
Feb18-08, 07:58 PM
1. The problem statement, all variables and given/known data

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

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

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

\gamma (0) = \beta (0), \gamma' (0) = \beta' (0), \gamma'' (0) = \beta'' (0)

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

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

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?

EnumaElish
Feb18-08, 09:16 PM
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?My guess is, it is the Kronecker delta (http://en.wikipedia.org/wiki/Kronecker_delta).