# Proving that an osculating circle is unique

• Eponym
In summary, the conversation discusses vector parameters e1 and e2, another vector parameter c, real valued parameters r and s, and a unit speed curve beta in E3. It introduces a parametrization of a circle gamma, and the problem is to prove the existence of a unique circle gamma that approximates beta near beta(0) and satisfies certain conditions. The solution attempts to find the first and second derivatives of gamma at 0 and uses the relationship between e1 and e2 to prove uniqueness.
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?

Last edited:
Eponym said:
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.

## 1. What is an osculating circle?

An osculating circle is a circle that is tangent to a curve at a specific point and has the same curvature as the curve at that point. In other words, it is the circle that best approximates the curve at that particular point.

## 2. Why is it important to prove that an osculating circle is unique?

Proving that an osculating circle is unique allows us to accurately analyze and understand the behavior of a curve at a specific point. It also helps us to make predictions and solve problems in various fields, such as physics, engineering, and mathematics.

## 3. How do you prove that an osculating circle is unique?

To prove that an osculating circle is unique, we use a mathematical concept called the radius of curvature. This concept involves finding the radius of the circle that is tangent to the curve at a specific point and has the same curvature as the curve at that point. If there is only one circle that satisfies these conditions, then it is the unique osculating circle for that point.

## 4. Can an osculating circle be unique at every point on a curve?

Yes, it is possible for an osculating circle to be unique at every point on a curve. This occurs when the curve has a constant curvature, meaning that the curvature at every point is the same. In this case, the osculating circle will have the same radius and will be unique at every point.

## 5. How is the uniqueness of an osculating circle related to the smoothness of a curve?

The uniqueness of an osculating circle is closely related to the smoothness of a curve. A curve that is smooth, meaning it has no abrupt changes in direction, will have a unique osculating circle at every point. However, if a curve has a sharp corner or a discontinuity, the osculating circle at that point will not be unique.

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