Say I use the arc length $s$ for the parametrization of a curve ${\bf r}(s)$. The normalized tangent vector $\hat{\bf t}$ at $s$ is given by $\hat{\bf t}=\frac{d {\bf r}(s)}{ds}$ with $\left|\frac{d {\bf r}(s)}{ds}\right|=1$. Now I could (think of an ellipse) parametrize the curve using $\varphi$ from polar coordinates. This results in \begin{align} \frac{d {\bf r}(s)}{ds}= \frac{\partial {\bf r}}{\partial\varphi}\frac{d \varphi}{ds}\;. \end{align} In polar coordinates I know that ${\bf e}_\varphi = \frac{1}{r}\frac{\partial {\bf r}}{\partial\varphi}$ and I also know that the curvature radius $\rho$ is given by $ds=\rho d\varphi$. This misleads to say
\begin{align} \frac{d {\bf r}(s)}{ds}= \underbrace{\frac{\partial {\bf r}}{\partial\varphi}}_{r{\bf e}_\varphi}\underbrace{\frac{d \varphi}{ds}}_{1/\rho}\;. \end{align} Since $|\hat{\bf t}|=1$ and $|{\bf e}_\varphi|=1$ this would mean that $r=\rho$ which is obviously not true in general. Where am I doing the stupid mistake?
I think I know where I'm doing the mistake: Saying $ds=\rho d\varphi$ implies that the angle $\varphi$ is the one of the osculating circle and NOT the angle from the polar coordinates. So going into polar coordinates mixes here two different quantities.