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.

  • $\begingroup$ 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. $\endgroup$ Apr 1 '13 at 8:49
  • $\begingroup$ @leena adam Some of your questions might be answered here. $\endgroup$
    – Git Gud
    Apr 1 '13 at 8:50
  • $\begingroup$ I think the OP is using "equivalent" to mean "iff." $\endgroup$ Apr 1 '13 at 9:32
  • $\begingroup$ this i mean equivalent <=> $\endgroup$
    – leena adam
    Apr 1 '13 at 9:44
  • 1
    $\begingroup$ Why is this question tagged algebraic-geometry? This is not algebraic geometry, it is differential geometry. $\endgroup$ 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

$$\begin{align*} \kappa_n = 0 & \iff c'' \perp n \\ & \iff \kappa N \perp n \\ & \iff N \perp n \\ & \iff B \parallel n. \end{align*}$$

In answer to your questions:

  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 $N$.

  • $\begingroup$ thank you very much great explain $\endgroup$
    – leena adam
    Apr 2 '13 at 10:05

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