Why is normal curvature well defined? I am reading Do Carmo's Differential Geometry and am confused by the definition of normal curvature (page 143). Here is the definition:
DEFINITION 3. Let $C$ be  a regular curve in $S$ passing through $p \in S, k$ the curvature of $C$ at $p$, and $\cos \theta = \langle n, N \rangle$, where $n$ is the normal vector to $C$ and $N$ is the normal vector to $S$ at $p$. The number $k_n = k \cos \theta$ is then called the normal curvature of $C \subset S$ at $p$
What is confusing is that curves with the same normal vector can have different curvatures for example the curves 
$$\alpha(t) = (t, t^2, 0), \beta(t) = (t, 2t^2, 0)$$
 on the xy-plane have the same tangent vectors at $p = (0,0,0)$ but different curvatures: $k_\alpha(0) = 2, k_\beta(0) = 4$. 
Why does this definition not depend on the choice of $C$? It seems to me that this is just assumed when Meusnier's Proposition is proven in the following page.
 A: The normal curvature is well-defined as a property of a given curve $\alpha(t)$ in a surface $S$, because clearly we can find $n$ (the normal to the curve), $N$ (the normal to the surface), and $k$ (the curvature of the curve) at some point $p=\alpha(0)$, and then find $k_n = \langle N,kn\rangle$. However, if $\alpha$ changes, then so do $n$ and $k$, and potentially therefore $k_n$ will change too, so a priori the normal curvature will depend on $\alpha'(0)$ and $\alpha''(0)$.
The calculation on the next page shows that $k_n$ only depends on $\alpha'(0)$, in particular that $II_p(\alpha'(0))=k_n$ for any curve $C$ with any $\alpha''(0)$. It is quite remarkable that if $\alpha'(0)$ stays the same but $\alpha''(0)$ changes, it nevertheless changes in such a way that $k_n$ stays the same.
Take the example of a unit sphere $S$, with $p$ the north pole. It is easy to see what $R$ (radius of curvature) is for various curves $C$ obtained by intersecting a plane $P$ through $p$ with $S$, because those curves are just circles. Then you can see the result just using simple plane geometry in the plane perpendicular to $P$ passing through the north pole. The result is just similar triangles, really, not even trigonometry in this case.
