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I am revising line integrals. Thomas's Calculus, 14th Edition, says the following:

enter image description here

As I understand it, a smooth curve is a curve that is defined by a smooth function, and a smooth function is defined as a function that has derivatives of all orders defined everywhere in its domain.

What I'm confused about is where $\mathbf{v} = \dfrac{ d\mathbf{r} }{ dt }$ never being zero came from? I don't see how $C$ being a smooth curve demands this?

I would greatly appreciate it if people could please take the time to clarify this.

EDIT:

enter image description here

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  • $\begingroup$ there seems to be some part of the text missing (first paragraph after the box, ends with 'We'). $\endgroup$ – Thomas Jul 1 '18 at 9:41
  • $\begingroup$ @Thomas The rest is irrelevant to the point in question, so I left it out. I have edited the main post with the next part, just in case. $\endgroup$ – The Pointer Jul 1 '18 at 9:55
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    $\begingroup$ Definition of smoothness usually does not include the requirement that the derivate does not vanish. If it is defined that way in that book then this is quite uncommon. As pointed out in my answer the requirement that the derivative does not vanish has the consequence that the curve is geometrically looking like one would expect (it has not cusps). After your edit I would add it also implies that the curve admits a reparametrization by arclenght (which is, in general, not necessarily true). $\endgroup$ – Thomas Jul 1 '18 at 13:46
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If $v$ is zero at some point, then this means that the 'velocity' of the curve is $=0$ In such a point the image of the parametrization may have all kinds of singularities, which -- in a geometric context -- is not desired.

As a simple example you may consider the curve

$$c(t) = (t^5, |t|^{5/2})$$

which is (quite obviously) $C^1$, but which is a parametrization of the graph of $y= \sqrt{|x|}$, which has a cusp at $t=0$. Of course this kind of example can be given in any dimension.

With the help functions like $\exp(-\frac{1}{|x|^2})$ it's possible to create parametriziations of such curves which are actually smooth.

The requirement $v\neq 0$ prevents the curve from having geometric singularities, for this reason one calls a curve with this property regular.

Edit (in response to a question in a comment): if you want to define line integrals then it is not really necessary to assume that the curves are regular. With this assumption the definition is, however, much easier and straightforward. Just assuming that there is a smooth parametrization, is, for example, not sufficient.

The example I gave is still a 'nice' curve and would allow do define a line integral along the curve without much of a problem. One can write down more nasty ones so that the parametrization is still smooth, but the curve no longer has finite length or has other nasty features like fractal behavior. This does not automatically mean that the definitions of length and line integral don't make any sense anymore, it may just be more difficult to give a consistent defintion.

The case of corners is often the first generalization which is introduced as admissible, since that class of curves is often more convenient in applications.

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  • $\begingroup$ Thanks for the response. From en.wikipedia.org/wiki/Curve#Differential_geometry: "A differentiable curve is said to be regular if its derivative never vanishes. (In words, a regular curve never slows to a stop or backtracks on itself.)" $\endgroup$ – The Pointer Jul 1 '18 at 10:03
  • $\begingroup$ Just to clarify: So reason they have the requirement that $\mathbf{v} = \dfrac{ d\mathbf{r} }{ dt } \not= 0$ is to ensure "good" geometry? Can you please elaborate on what this means and why it is desired (in this context)? Thanks again. $\endgroup$ – The Pointer Jul 1 '18 at 14:28
  • $\begingroup$ @ThePointer it means, informally, that the curve has no cusps or corners. I wrote down an example explicitly, to give you an idea what this means. I suggest you take that example (in a neighbourhood of $t=0$) and draw a picture of that curve. $\endgroup$ – Thomas Jul 1 '18 at 14:35
  • $\begingroup$ I read your example. What I mean is, why is this necessary/important for line integrals? Let's say there's a cusp or corner in the curve -- so what? For instance, the boundary of a rectangle has corners, so what would be the problem with calculating the line integral for that? $\endgroup$ – The Pointer Jul 1 '18 at 15:11
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    $\begingroup$ @ThePointer I made an edit to my answer to address your question. $\endgroup$ – Thomas Jul 1 '18 at 15:39

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