Timeline for Parametrically Defined Curves: $f'$ and $g'$ Are Not Simultaneously Zero
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 20, 2018 at 16:43 | comment | added | Michael Hoppe | The curve may intersect itself, hence it is not necessarily injective. | |
| Jul 20, 2018 at 16:04 | comment | added | copper.hat | @DavidC.Ullrich: To be frank, I have been stuck on this part for a few decades now, starting with an early course in complex analysis where the instructor did not distinguish parameterised curves from their ranges. I dream of being a real mathematician... | |
| Jul 20, 2018 at 15:31 | vote | accept | The Pointer | ||
| Jul 20, 2018 at 15:09 | comment | added | copper.hat | @DavidC.Ullrich: That is where I get stuck... | |
| Jul 20, 2018 at 15:04 | comment | added | David C. Ullrich | @copper.hat Thinking about it I'm really not so sure. The idea, which may or may not work, is to slow down to zero velocity at every point where the curve is not differentiable, as in the answer I gave. Say $\gamma(t)$ is Lipschitz and $E$ is the set of $t$ where $\gamma$ is not differentiable. I believe I can do this if $E$ happens to be compact. In fact $E$ is an $F_\sigma$... | |
| Jul 20, 2018 at 14:52 | comment | added | copper.hat | @DavidC.Ullrich: Do you have a pointer for that please? I can see Lipschitz, but am curious how the reparametrisation would be done on the null set. | |
| Jul 20, 2018 at 14:39 | answer | added | user65203 | timeline score: 0 | |
| Jul 20, 2018 at 14:39 | comment | added | David C. Ullrich | @copper.hat In fact I'm pretty sure that any rectifiable curve has a $C^1$ parametrization... | |
| Jul 20, 2018 at 14:27 | comment | added | copper.hat | @DavidC.Ullrich: I guess I would settle for rectifiable, but as your answer nicely demonstrates, this allows a lot of curves that don't fit with our intuition. | |
| Jul 20, 2018 at 14:21 | comment | added | copper.hat | @DavidC.Ullrich: I meant locally, it was more of a comment than a complete answer. | |
| Jul 20, 2018 at 14:14 | comment | added | David C. Ullrich | @copper.hat So for example $t\mapsto(\cos(t),\sin(t))$ is injective? Ok, you meant locally injective. But I really don't think this is exactly the point - the "bad" parametrization of the curve $y=|x|$ in my answer is injective... | |
| Jul 20, 2018 at 14:12 | comment | added | copper.hat | It means the curve has a non zero speed for each value of the parameter. This means that the curve in an injective map so that any point on the curve defines a unique value of the parameter. | |
| Jul 20, 2018 at 14:11 | answer | added | David C. Ullrich | timeline score: 3 | |
| Jul 20, 2018 at 13:36 | history | asked | The Pointer | CC BY-SA 4.0 |