I am trying to figure out a question but am having trouble solving it. The question reads:
Let $\alpha:I\rightarrow\mathbb{R}^3$ be a regular curve without critical points of order $1$, and let $t_0\in I$. Let $\beta(t)$ be the curve obtained by projecting $\alpha(t)$ onto its osculating plane at $t=t_0$. Show that $\alpha(t)$ and $\beta(t)$ have the same curvature at $t=t_0$.
Now I know I'm supposed to use the Local Canonical Form for this, so the projection of $\alpha(t)$ onto the osculating plane would give:
$$\beta(t)=(t-\frac{\kappa_0}{6}t^3+R_x, \frac{\kappa_0}{2}t^2+\frac{\kappa_0'}{6}t^3+r_y)$$ Where the term $\kappa_0$ denotes the curvature at $\kappa(t_0)$. Am I approaching this right? Would I now compute the curvature of $\beta(t)$ to determine the curvature?
\begin{align*} \kappa &= \frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{(\dot{x}^2+\dot{y}^2)^{3/2}} \\ &= \frac{\left( 1-\dfrac{\kappa_0 t^2}{2} \right)(\kappa_0+\kappa_0't)- (-\kappa_0 t)\left( \kappa_0t+\dfrac{\kappa_0't^2}{2} \right)} {\left[ \left( 1-\dfrac{\kappa_0 t^2}{2} \right)^2+ \left( \kappa_0t+\dfrac{\kappa_0't^2}{2} \right)^2 \right]^{3/2}} \\ \kappa(0) &= \kappa_0 \end{align*}