Differential Geometry/Curvature and Osculating Circle

Consider a curve $$C$$ of class of at least 2, parametrized by the arc length parameter, $$f(s)$$.

The magnitude of $$f''(s)$$ is called the curvature of the curve $$C$$ at the point $$f(s)$$. The multiplicative inverse of the curvature is called the radius of curvature.

The curvature is 0 at every point if and only if the curve is a straight line. Suppose that the curvature is always 0. Then $$f''(s)$$ is always 0, which proves that it is a straight line through elementary integrations.

We can also consider the normal vector $$f''(s)$$ to be the curvature vector.

The point that is away from $$f(s)$$ by a distance of the radius of curvature in the direction of the principal normal unit vector is called the center of curvature of the point $$f(s)$$ and the circle with the center on the center of curvature and with the radius as the radius of curvature is called the osculating circle at the point $$f(s)$$. It is very obvious that the unit tangent vector at the point $$f(s)$$ is tangent to the osculating circle at $$f(s)$$.