Proving that an osculating circle is unique

[Eponym]

Homework Statement

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

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

Homework 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

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]

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.