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I was reading through Paul's Online Math Notes and I bumped into the following definition for a smooth curve here:

A smooth curve is any curve for which $\dot{\vec{r}}(t)$ is continuous and  $\dot{\vec{r}}(t)\neq 0$ for any $t$ except possibly at the endpoints. 

This looks to me very different from the usual definition of a smooth curve, i.e. the function is "sufficiently" differentiable and continuous for our purposes. (For example check out Walfram Alpha, Wikipedia and a previous question on MSE.

Are these definitions equivalent then? Is it maybe a mistake from Paul's notes?

I was thinking maybe he is referring to a different object, although I am pretty sure he is introducing smooth curves in that section of Calculus 2 in order to be able to talk about line integrals and in general the Integral Theorems of vector calculus (Stoke's, Green's, Divergence, etc) further on, so this should be the same object as in the links above.

I tried to think about how they could be equivalent, however I could not see why a smooth curve should have non-vanishing derivative except at the end points, I am pretty sure I saw loads of exercises where the curve was smooth and it certainly attained a maximum or a minimum not in the boundaries and therefore the derivative was zero there..

Can you help me clarify this?

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Euler_Salter
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  • When it says $\dot{r}(t)$ is continuous, what exactly does it mean? – user160738 Dec 23 '16 at 13:13
  • I guess the standard $\epsilon-\delta$ definition – Euler_Salter Dec 23 '16 at 13:23
  • Doesn't it give some kind of examples of this "smooth" function? So in those example he varifies $\dot{r}(t):\mathbb{R}\to\mathbb{R}^n$ is continous directly using $\epsilon-\delta$? – user160738 Dec 23 '16 at 13:26
  • "A helix is a smooth curve, for example." this is one example he gives. No I don't think he gives any $\epsilon-\delta$ proof in this section(as it would rather be Calculus 1 stuff). However it normally deals with standard definitions – Euler_Salter Dec 23 '16 at 13:28
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    The definitions aren't equivalent. Requiring that the derivative vanishes nowhere ensures a) that the curve has a tangent everywhere [well, locally, the curve may intersect itself, then you have no tangent as in tangent space of a submanifold at (transversal) intersection points, but if you consider the curve restricted to a small interval around one of the pertinent parameter values, it has], and b) you can reparametrise the curve by arclength (and still get a smooth curve). For some applications these things are good to have, for others it's not important. – Daniel Fischer Dec 23 '16 at 13:35
  • @DanielFischer so is this the "smoothness" we want also for the Integral Theorems? Furthermore, how can someone define different things with the same name? I thought mathematicians tried to be rigorous – Euler_Salter Dec 23 '16 at 13:41
  • It's quite common to use the same name for different things ("normal" is a word used for many different things, for example). Even such fundamental things like the set of natural numbers have different definitions. One definition includes $0$, the other doesn't. One definition is more convenient than the other in circumstances $\alpha,\beta,\gamma,\dotsc$, the other is more convenient in circumstances $x,y,z,\dotsc$. No problem with that, as long as one says upfront which definition is in use [or it doesn't matter]. – Daniel Fischer Dec 23 '16 at 13:46
  • Which integral theorems are you thinking of? Some things are certainly more convenient if you demand the derivative non-vanishing, but off the top of my head, I don't know what would break (in connection with line integrals and such) if that isn't required. – Daniel Fischer Dec 23 '16 at 13:48
  • For example Stoke's Theorem, Green's Theorem (i.e. Stoke's in $x-y$ plane) and Gauss' Theorem (or Divergence Theorem), for example Stokes says "Let $S$ be a piecewise continuous, smooth, orientable surface in $\mathbb{R}^3$, whose boundary $\partial S$ consists of a finite number of piecewise $C^1$ simple closed curves. Let $\underline{F}$ be a $C^1$ vector field, whose domain contains $S$, then $$\iint_S (\nabla \times \underline{F})\cdot d\underline{S} = \oint_{\partial S}\underline{F}\cdot d\underline{r}$$ – Euler_Salter Dec 23 '16 at 13:57
  • Here it is a smooth surface, but I guess it is the same, as it is a parametrised curve, just with one more dimension. – Euler_Salter Dec 23 '16 at 13:58
  • Things get way more fussy if you don't require that the Jacobi determinant of the parametrisation doesn't vanish. I think it remains true if you allow it, but it's not worth the effort, usually. – Daniel Fischer Dec 23 '16 at 14:06
  • @DanielFischer , hold on, you mean requiring it in the Stoke's Theorem or when you define a smooth curve? – Euler_Salter Dec 23 '16 at 14:09
  • Stokes. If you allow the $C^1$ pieces of the parametrisation of the surface and/or the boundary to have vanishing Jacobian determinant/derivative, it's way more complicated to prove the theorem. Unless you need the greater generality that allows bad surfaces and boundaries and dive into geometric measure theory, it's not worth the fuss. – Daniel Fischer Dec 23 '16 at 14:27
  • Okay this makes sense. If you have time to read that page of Paul's Notes, can you please tell me why you think here that definition was more suitable? – Euler_Salter Dec 23 '16 at 15:10

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