Timeline for Parametrically Defined Curves: $f'$ and $g'$ Are Not Simultaneously Zero
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Jul 20, 2018 at 15:31 | vote | accept | The Pointer | ||
| Jul 20, 2018 at 15:30 | comment | added | David C. Ullrich | @ThePointer Right. | |
| Jul 20, 2018 at 15:21 | comment | added | The Pointer | Makes total sense now! If we didn't have this condition, then we would classify non-differentiable curves as differentiable just by parameterising them in a certain way, right? And that would compromise our definition of what it means for a curve to be or not be differentiable? | |
| Jul 20, 2018 at 15:19 | comment | added | The Pointer | Your answer, combined with your comment in Yves Daoust's answer, clarifies it for me: "This is undesirable behavior because if we don't disallow this behavior we find the the curve $y = |x|$ is "differentiable", as I showed in my answer. We certainly don't want that - if $y = |x|$." is a "differentiable" curve then differentiable curves are nothing like what we think they should be." | |
| Jul 20, 2018 at 15:18 | comment | added | The Pointer | @DavidC.Ullrich Ahh, yes, because, as you said, "we want to say a curve is smooth, for example $C^1$ (continuously differentiable), if it has a smooth parametrization." | |
| Jul 20, 2018 at 15:14 | comment | added | David C. Ullrich | @ThePointer Yes, more or less. I'm not quite saying what you said about proving things about curves that are not smooth. Before we can prove anything like that we need to define what a smooth curve is... If we make the definition the way we do then what you say about non-smooth curves is true, just by definition. | |
| Jul 20, 2018 at 15:10 | comment | added | The Pointer | I've studied your answer and I think I understand now. So are you saying that it can be proven that all curves $C$ that aren't smooth will have some point $c \in [a, b]$ in their parameterisations $x = f(t)$, $y = g(t)$ such that $f'(c) = g'(c) = 0$? And, since we want to avoid curves that aren't smooth, we therefore also want to avoid curves that have a parameterisation where the derivatives of the coordinates are simultaneously zero? | |
| Jul 20, 2018 at 14:52 | comment | added | David C. Ullrich | @ThePointer I hadn't noticed the notation you were using. Answer revised to make the notation more consistent with the notation in the question. | |
| Jul 20, 2018 at 14:51 | history | edited | David C. Ullrich | CC BY-SA 4.0 |
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| Jul 20, 2018 at 14:49 | comment | added | David C. Ullrich | @Rahul Thanks for that! I hadn't noticed his "$f$" and "$g$"... | |
| Jul 20, 2018 at 14:47 | comment | added | user856 | @ThePointer: The functions $x(t)$ and $y(t)$ in this answer are what you call $f(t)$ and $g(t)$. They are both differentiable and have derivatives equal to $0$ at $t=0$. | |
| Jul 20, 2018 at 14:35 | comment | added | David C. Ullrich | @copper.hat As I'm sure you understand, just to prevent any possible confuusion: The parametrization is differentiable. The curve is not differentiable - that's the point to the technicality in the definition of "differentiable curve". | |
| Jul 20, 2018 at 14:33 | history | edited | David C. Ullrich | CC BY-SA 4.0 |
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| Jul 20, 2018 at 14:31 | comment | added | copper.hat | @ThePointer: The point here is that the curve is continuously differentiable everywhere, but it has a kink when you just look at the range of the curve. Our natural inclination is to consider such curves as not being smooth even though they are. | |
| Jul 20, 2018 at 14:30 | comment | added | David C. Ullrich | @ThePointer The point is that although the curve $y=|x|$ is not differentiable, the paramteriization $(x(t),y(t))$ that I gave is differentiable! So we need to not allow that parametrization. If add the condition about non-vanishing derivatives we rule out that parametrization, since $x'(0)=y'(0)=0$. | |
| Jul 20, 2018 at 14:26 | comment | added | The Pointer | Again, I appreciate the answer, but this was written by someone who understands the concept, for someone who understands the concept. I cannot derive anything useful from this, despite putting in the effort to understand it. | |
| Jul 20, 2018 at 14:25 | comment | added | The Pointer | Thanks for the response. So what is the connection between this and the condition that the derivatives of the points of the curve cannot simultaneously be zero? For instance, you used the curve $y=|x|$ as an example, but this doesn't have derivatives equal to $0$ at any point; rather, it is only undefined at $x = 0$, since this is a corner. Furthermore, I don't see any reference at all to the concept that my question referenced, which is the points being simultaneously being zero. [...] | |
| Jul 20, 2018 at 14:11 | history | answered | David C. Ullrich | CC BY-SA 4.0 |