A curve with positive curvature is asymptotic if and only if its binormal is parallel to the unit normal of the surface I want some one explain to me; this part is not clear to me.

Q: Show that curve $c$ with positive curvature is asymptotic if and only if if its binormal $B$ is parallel to the unit normal of $S$ at all points of $c$ A: Since $K_n=0 \iff$ $c''$ is perpendicular to $N$ $\iff$ $N$ is perpendicular to $n$ $\iff$ $N$ is parallel to $B$ (since $N$ is perpendicular to $T$).

Here, $c''$ is the second derivative, $B$ binormal vector, $K_n$ normal curvature.
My question is

*

*I know from Frenet–Serret formulas that $N$ is perpendicular to $B$ and $N$ is perpendicular to $T$ and $B$ is perpendicular to $T$. In the question it changes, it said that $N$ is parallel to $B$. Why did this change? Can any one explain this part please?


*Why is $N$ perpendicular to $n$? Where this come from?
These are my questions, I hope someone can help me to understand please. If possible, draw a figure. Thank you.
A: I'm assuming you have the following setup: The curve $c$ lies in the surface $S$, which has unit normal vector $n$.  We let $\{T, N, B\}$ denote the Frenet frame of $c$.  We also let $\kappa$ denote the curvature of $c$, and $\kappa_n$ the normal curvature of $c$.  We can assume that $c$ is parametrized by arc length.
By the Frenet formulas, $c'' = T' = \kappa N$.  Since $\kappa_n = c'' \cdot n$, we have
$$\begin{align*}
\kappa_n = 0 & \iff c'' \perp n \\
& \iff \kappa N \perp n \\
& \iff N \perp n \\
& \iff B \parallel n.
\end{align*}$$
In answer to your questions:


*

*There is a typo in what you wrote.  It should be that $B$ is parallel to $n$ (not $N$).  But why is this true?  Well, $n$ always lies in the plane spanned by $N$ and $B$ (because $n$ is perpendicular to $T$), so if $n$ is perpendicular to $N$, then $n$ must be parallel to $B$.

*Because $c'' = \kappa N$ is parallel to $N$.
