Consider a curve in 2D $\vec{p}(s)$ parameterized by arclength $s$ and the usual local coordinate system on the curve (Frenet Frame with unit vectors $\vec{n}(s)\perp\vec{t}(s)$, no torsion, curvature non-zero). Further, consider a point $\vec{r}(t)$ and it's projection onto the curve $\vec{p}(s)$ at $\vec{r}_0=\vec{p}(s_0)$ as below:
I'm interested in the time derivative of the distance of $\vec{r}(t)$ from $\vec{p}(s)$ with respect to time, i.e. the derivative of $\delta(t) = \vec{n}^T(\vec{r}-\vec{r}_0)$ and here is what I've come up with: $$ \begin{aligned} \frac{d}{dt}\delta = & \frac{d}{dt}\vec{n}^T(\vec{r}-\vec{r}_0)\\ = & \left(\frac{ds}{dt}\frac{d}{ds}\vec{n}^T\right)(\vec{r}-\vec{r}_0) + \vec{n}^T(\frac{d}{dt}\vec{r}-\frac{ds}{dt}\frac{d}{ds}\vec{r}_0)\\ = & \underbrace{\frac{ds}{dt}\kappa\vec{t}(\vec{r}-\vec{r}_0)}_{\equiv 0,~\vec{t}\perp (\vec{r}-\vec{r}_0)} + \vec{n}^T\frac{d}{dt}\vec{r} - \frac{ds}{dt}\underbrace{\vec{n}^T\frac{d}{ds}\vec{r}_0}_{\equiv 0, \frac{d}{ds}\vec{r}_0\propto \vec{t}, \vec{t}\perp\vec{n}}\\ = & \vec{n}^T\frac{d}{dt}\vec{r} \end{aligned} $$
where I've used Frenet Sorret, i.e. $d/ds\vec{n}=\kappa\vec{t}$ and that $d/ds\vec{r}_0$ must always be proportional to $\vec{t}$ by construction of the projection.
If the above is correct, then the velocity of change of distance to the curve is only the projection of the object's velocity onto the curve's normal. That seems totally unintuitive to me as I would expect the curvature of $\vec{p}(s)$ to enter here as well.
Where am I wrong and what is the correct formula? Thanks for any help or reference suggestions.