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Let $\alpha:I\rightarrow \mathbb{R}^3$ a parametrized curve. What is the definition of strong (weak) tangent of $\alpha$ at the point $t_0$?

Thanks!

Ivo Terek
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EQJ
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1 Answers1

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(Weak tangent) $\alpha: I \to \Bbb R^3$ has a weak tangent at $t_0 \in I$, if the line determined by $\alpha(t_0 + h)$ and $\alpha(t_0)$ has a limit position when $h \to 0$.

(Strong tangent) $\alpha: I \to \Bbb R^3$ has a strong tangent at $t_0 \in I$, if the line determined by $\alpha(t_0 + h)$ and $\alpha(t_0 + k)$ has a limit position when $h \to 0$ and $k \to 0$.

Ivo Terek
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  • What does it mean "a limit position"? – EQJ Sep 10 '14 at 01:55
  • That exists the limit: $$\lim_{h \to 0} \alpha(t_0 + h) - \alpha(t_0)$$ for example. – Ivo Terek Sep 10 '14 at 01:59
  • If $\alpha$ is continuous at $t_0$ then that limit is always $0$. It does not help :( – EQJ Sep 10 '14 at 02:01
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    I took that definition from do Carmo's Differential Geometry of Curves and Surfaces. Perhaps I could have translated it better. Here's a link so you can look yourself: http://gen.lib.rus.ec/search.php?req=do+carmo&open=0&view=simple&column=def It is in one of the exercices of chapter 1 or 2, not far in the book. – Ivo Terek Sep 10 '14 at 02:05
  • I know the book, and I found the same definition as you. This notion is intuitive clear but it is not very useful to calculate.Thanks! :) – EQJ Sep 10 '14 at 02:17
  • I found this thread because I had the same question. @IvoTerek that definition does not suffice, as YTS points out. If you normalize that difference vector and assert that the limit is not the zero vector (i.e. it determines a line, not a single point), then I believe we get the definition we're looking for. I'm a bit mad at do Carmo's lack of a definition for "limit position" – terrygarcia Mar 13 '19 at 20:08
  • @terrygarcia I have the same frustration too. I have just attempted to propose one in here: https://math.stackexchange.com/a/4250801/179461 You are welcome to take a look and join to improve the proposal. – Hopf eccentric Sep 15 '21 at 11:29