I was doing an exercise in which I am supposed to conclude that because all of the normal curvatures at a given point $p$ are zero, $dN_p=0$. This is supposed to be trivial but I do not really understand why.
The idea of the exercise:
We are considering the rotation if the curve $z=y^4$ about the $z$ axis.
The goal is to show that, at $p=(0,0,0)$, $dN_p=0$.
To see this, we observed that the curvature of our curve at $p$ is zero.
Since the $xy$ plane is a tangent plane to the surface at $p$, the normal vector $N(p)$ is parallel to the $z$ axis.
Therefore, any normal section has curvature zero, since they are all obtained through rotations of our curve.
Now, because all normal curvatures are zero at $p$, we can conclude $dN_p=0$.
And this is the conlcusion I do not understand.
Thank you
Exercise in Manfredo do Carmo's book.
HINT: Compute $dN_p(v)$ by the definition. What is the surface normal along the normal slice of the surface in direction $v$? (Use the fact that this is a surface of revolution.)
EDIT: As I suggested both in the hint and in a further comment, you compute $dN_p(v)=(N\circ\alpha)'(0)$ where $\alpha(t)$ is a curve in the surface $S$ with $\alpha(0)=p$ and $\alpha'(0)=v\in T_pS$. Take $v$ to be a unit vector. The obvious curve $\alpha$ to take is the slice of $S$ by the vertical plane containing $v$. Along $\alpha$, we have $N=\pm n$, where $n$ is the principal normal of the plane curve $\alpha$. Now, $dN_p(v) = (N\circ\alpha)'(0) = \pm n'(0) = \mp k(0)t(0) = 0$.