Parallel derivatives of two curves with a common end point

Question

Let $\xi, \nu$ two smooth ($C^{1}$) parametric curves in $\mathbb{R}^{k}$ defined on $[a,b]$ and $[c,d]$ with $k>2$. Let us assume that the two curves intersect in a sequence of points $x_{n}$ such that $x_{n}$ converges to $\xi(b)=\nu(d)$. Under which generic conditions can we say that the vectors $\dot{\xi}(b),\dot{\nu}(d)$ are parallel in $\mathbb{R}^{k}$? An answer for $k=3$ would be quite significant.

Answer 1

I think Ted Shifrin is right, but let's be clear about the hypotheses:

Thm: Assume $\xi,\nu$ are $C^1$ maps of $[a,b],[c,d]$ respectively into $\mathbb R^k,$ with $\xi(b) =\nu(d).$ Also assume both maps are injective, and that $\xi',\nu'$ never vanish. Suppose $x_1,x_2, \cdots$ are distinct points in $\xi ([a,b])\cap \nu([c,d])$ such that $x_n \to \xi(b) =\nu(d).$ Then $\xi'(b)$ is a positive scalar multiple of $\nu'(d).$

Lemma 1: Suppose $\gamma: [a,b]\to \mathbb R^k$ is continuous and injective. Then $\gamma$ is a homeomorphism of $[a,b]$ onto $\gamma([a,b]).$

Lemma 2: Suppose $\gamma: [a,b]\to \mathbb R^k$ is $C^1$ and injective, with $\gamma'$ never vanishing. Then there exist $0<c<C$ such that

$$c|t-s| \le |\gamma(t) - \gamma(s)| \le C|t-s|$$

for all $s,t \in [a,b].$

The lemmas are pretty standard results, so I'll leave them unproved here.

Proof of the theorem: Because $\xi$ is a homeomorphism (Lemma 1), $s_n= \xi^{-1}(x_n)$ converges to $b.$ Similarly, $t_n= \nu^{-1}(x_n)$ converges to $d.$ We then have

$$\tag 1 \frac{\xi(s_n) - \xi(b)}{s_n-b} = \frac{\nu(t_n) - \nu(d)}{s_n-b} = \frac{t_n-d}{s_n - b}\frac{\nu(t_n) - \nu(d)}{t_n-d}.$$

Let $f_n = (t_n-d)/(s_n - b).$ Then $f_n >0$ for each $n.$ Furthermore, $|s_n-b|, |t_n-d|$ are each comparable to $|\xi(s_n)-\xi(b)| = |\nu(t_n)-\nu(d)|$ by Lemma 2. It follows that the fractions $f_n$ lie in some $[\alpha,\beta]$ with $0<\alpha.$ Passing to a subsequence $f_{n_j},$ we can assume $f_{n_j} \to f\in [\alpha,\beta].$

Now take the limit of $(1)$ through the subsequence $n_j$ to see

$$\xi'(b)= f\cdot \nu'(d),$$

which proves the theorem.