I have to prove that for any regular curve in $\mathbb{R^3}$ it's curvature $k(t)$ is the same as the curvature of its projection on the osculator plane at $t$
Basically what i did was writting on the canonical local form, and projecting on the First two coordinates, deriving twice and taking the norm, but it lead me nowhere I get $\sqrt{({ks^2})^2+(k+sk')^2}$ if i dont carry the aproximation errors.
If you use Taylor expansion at point $t_0$, you will get something like that
$$ \mathbf{r}(t) = \mathbf{r}(t_0) + \mathbf{r'}(t_0)\Delta t + \mathbf{r''}(t_0)\frac{\Delta t^2}2 + \mathbf{O}(\Delta t^3) $$
Then you show that vectors $\mathbf{r'}(t_0)$ and $\mathbf{r''}(t_0)$ lie on the osculating plane. That leads to the expansion of the projection curve $\mathbf{\tilde r}(t)$:
$$ \mathbf{\tilde r}(t) = \mathbf{r}(t_0) + \mathbf{r'}(t_0)\Delta t + \mathbf{r''}(t_0)\frac{\Delta t^2}2 + \mathbf{O}(\Delta t^3) $$
So $\mathbf{r}(t)$ and $\mathbf{\tilde r}(t)$ share first three terms of expansions. Finally, you know that curvature depends only on first and second derivatives.