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

1. 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?

2. 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.

• When you write the word "equivalent", do you mean the symbol $\equiv$ ? Please note that you can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. Apr 1 '13 at 8:49
• I think the OP is using "equivalent" to mean "iff." Apr 1 '13 at 9:32
• this i mean equivalent <=> Apr 1 '13 at 9:44
• Why is this question tagged algebraic-geometry? This is not algebraic geometry, it is differential geometry. Apr 1 '13 at 9:58

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*}

1. 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 $B$.
2. Because $c'' = \kappa N$ is parallel to $N$.