I have the circle $\gamma(t) = (\cos t, \sin t, 0)$ in the plane $z=0$.
Now I understand that normal curvature is related to the second fundamental form, and an expression for it is $\kappa_n=L\dot{u}^2 + 2M\dot{u}\dot{v} + N\dot{v}^2$, where, if there is a surface patch $\sigma(u,v)$, then $L=\sigma_{uu}\cdot \vec{N}, M=\sigma_{uv}\cdot \vec{N}, N=\sigma_{vv} \cdot \vec{N}$ where $\vec{N}$ is the normal vector to the surface.
I assume that the surface patch for a plane $z=0$ is $\sigma(u,v)=(u,v,0)$. So $\sigma_{uu}=\sigma_{uv}=\sigma_{vv}=0$, and so $L=M=N=0$ thus $\kappa_n=0$
I am looking for verification that my answer is correct.
Also, apparently if the normal curvature is equal to zero, the curve is called asymptotic. Can I then say that this curve $\gamma$ on $\sigma$ is indeed asymptotic?
Thank you in advance!