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 $f(s)$.