The parametric equations of a circle of radius R are obviously of the form: $$x=R \cos\left(\frac{s}{R}\right) , y= R \sin \left(\frac{s}{R}\right)$$

where $s$ is the natural parameter (length of arc).

we have $$\ddot{x}=-\frac{1}{R}\cos\left(\frac{s}{R}\right)\;\;\;,\;\ddot{y}=\frac{1}{R}\sin\left(\frac{s}{R}\right)$$ We have:$$k(s)=\frac{1}{R}\;\;\; $$for all $s$

therefore, the curvature of the circle is constant and equal to the inverse of its radius.

Along any curve, the curvature is a determined function of the length of the arc:$$\frac{1}{\rho}=f(s) \;\;\tag{$1$}$$ one shows reciprocally, that to any equation equation of the form (1) corresponds a determined curve. Indeed, we choose any direction for the X axis and let $\varphi$ be the angle formed by the tengent to the curve with this axis. we know that: $\;\frac{1}{\rho}=\pm\frac{d\varphi}{ds}$ and equation (1) becomes:$$\pm\frac{d\varphi}{ds}=f(s)$$ from where:$$\varphi=\pm\int_{0}^{s}f(s)ds+C$$ We can consider that the direction of the X axis corresponds to the direction of the tengent for $s=0$, so that we can consider $C=0$ in the last formula, i.e. we have for the angle $\varphi$:$$\varphi=\pm F(s) \;\;{where} \;\;F(s)=\int_{0}^{s}f(s)ds$$ we know that $$\frac{dx}{ds}=\cos(\varphi)\;\;,\;\;\frac{dy}{ds}=\sin(\varphi)$$ whence, given the previous equality $$x=\int_{0}^{s}\cos[F(s)]ds+C_{1}\;\;,\;\;y=\int_{0}^{s}\sin[F(s)]ds+C_{2}$$

If we take the origin of the coordinates at the point of the curve where $s=0$, we must take $C_{1}=C_{2}=0$ and we obtain a well-defined curve:$$x=\int_{0}^{s}\cos[F(s)]ds\;\;,\;\;y=\int_{0}^{s}\sin[F(s)]ds$$ Equation (1) is the natural equation of the curve in the sense that this curve is in no way linked to the arbitrary choice of the coordinate axes and that it corresponds to a well-determined curve 'up to a symmetry'.

The parametric equations of a circle of radius R are obviously of the form: $$x=R \cos\left(\frac{s}{R}\right) , y= R \sin \left(\frac{s}{R}\right)$$

where $s$ is the natural parameter (length of arc).

we have $$\ddot{x}=-\frac{1}{R}\cos\left(\frac{s}{R}\right)\;\;\;,\;\ddot{y}=-\frac{1}{R}\sin\left(\frac{s}{R}\right)$$ We have:$$k(s)=\frac{1}{R}\;\;\; $$for all $s$

therefore, the curvature of the circle is constant and equal to the inverse of its radius.

Along any curve, the curvature is a determined function of the length of the arc:$$\frac{1}{\rho}=f(s) \;\;\tag{$1$}$$ one shows reciprocally, that to any equation equation of the form (1) corresponds a determined curve. Indeed, we choose any direction for the X axis and let $\varphi$ be the angle formed by the tengent to the curve with this axis. we know that: $\;\frac{1}{\rho}=\pm\frac{d\varphi}{ds}$ and equation (1) becomes:$$\pm\frac{d\varphi}{ds}=f(s)$$ from where:$$\varphi=\pm\int_{0}^{s}f(s)ds+C$$ We can consider that the direction of the X axis corresponds to the direction of the tengent for $s=0$, so that we can consider $C=0$ in the last formula, i.e. we have for the angle $\varphi$:$$\varphi=\pm F(s) \;\;{where} \;\;F(s)=\int_{0}^{s}f(s)ds$$ we know that $$\frac{dx}{ds}=\cos(\varphi)\;\;,\;\;\frac{dy}{ds}=\sin(\varphi)$$ whence, given the previous equality $$x=\int_{0}^{s}\cos[F(s)]ds+C_{1}\;\;,\;\;y=\int_{0}^{s}\sin[F(s)]ds+C_{2}$$

If we take the origin of the coordinates at the point of the curve where $s=0$, we must take $C_{1}=C_{2}=0$ and we obtain a well-defined curve:$$x=\int_{0}^{s}\cos[F(s)]ds\;\;,\;\;y=\int_{0}^{s}\sin[F(s)]ds$$ Equation (1) is the natural equation of the curve in the sense that this curve is in no way linked to the arbitrary choice of the coordinate axes and that it corresponds to a well-determined curve 'up to a symmetry'.