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I can't find a clear, comprehensive explanation, on this site or elsewhere, for why parametrically defined curves frequently have the condition that the the derivatives of their points $x = f(t)$ and $y = g(t)$ cannot simultaneously be zero on the interval $[a, b]$.

Most of the explanations use language that assumes that the reader already understands the concept they're explaining, or the explanations make the meaningless claim that the curve must be "nice".

I would appreciate it if people could please take the time to explain, comprehensively (not rigorously), what is meant by this condition. If you're going to use words that are likely to be unfamiliar to someone who doesn't understand this concept, like "regular", then please take the time to define what it means.

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  • $\begingroup$ 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. $\endgroup$
    – copper.hat
    Commented Jul 20, 2018 at 14:12
  • $\begingroup$ @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... $\endgroup$
    – David C. Ullrich
    Commented Jul 20, 2018 at 14:14
  • $\begingroup$ @DavidC.Ullrich: I meant locally, it was more of a comment than a complete answer. $\endgroup$
    – copper.hat
    Commented Jul 20, 2018 at 14:21
  • $\begingroup$ @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. $\endgroup$
    – copper.hat
    Commented Jul 20, 2018 at 14:27
  • $\begingroup$ @copper.hat In fact I'm pretty sure that any rectifiable curve has a $C^1$ parametrization... $\endgroup$
    – David C. Ullrich
    Commented Jul 20, 2018 at 14:39

2 Answers 2

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The point is we want to say a curve is smooth, for example $C^1$ (continuously differentiable), if it has a smooth parametrization. And without that condition on non-vanishing derivatives that would admit "smooth" curves that we really don't want to call "smooth".

Example: Consider the curve in the plane defined by $y=|x|$. That has a corner at the origin - our definition of "$C^1$ curve" had better not include this curve, right?

But define $f,g:\Bbb R\to\Bbb R$ by $g(t)=t^2$ and $$f(t)=\begin{cases} t^2,&(t\ge0), \\-t^2,&(t<0).\end{cases}$$Then $f$ and $g$ are both $C^1$, and $x=f(t)$, $y=g(t)$ is a parametrization of the curve $y=|x|$. Without the condition on non-vanishing derivatives this would make $y=|x|$ a $C^1$ curve. (With the condition on non-vanishing derivatives everything's fine - this parametization is disallowed because $f'(0)=g'(0)=0$.)

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  • $\begingroup$ 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. [...] $\endgroup$
    – The Pointer
    Commented Jul 20, 2018 at 14:25
  • $\begingroup$ 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. $\endgroup$
    – The Pointer
    Commented Jul 20, 2018 at 14:26
  • $\begingroup$ @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 $x'(0)=y'(0)=0$. $\endgroup$
    – David C. Ullrich
    Commented Jul 20, 2018 at 14:30
  • $\begingroup$ @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. $\endgroup$
    – copper.hat
    Commented Jul 20, 2018 at 14:31
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    $\begingroup$ @ThePointer Right. $\endgroup$
    – David C. Ullrich
    Commented Jul 20, 2018 at 15:30
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Because when $f'=g'=0$, the direction of the curve is no more defined. At such a point, the curve could change direction abruptly.

Example.

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  • $\begingroup$ Singular point: mathworld.wolfram.com/SingularPoint.html $\endgroup$
    – The Pointer
    Commented Jul 20, 2018 at 14:44
  • $\begingroup$ Thanks for the response. So the reason for this condition is that we want the parameterised curve to be unidirectional? $\endgroup$
    – The Pointer
    Commented Jul 20, 2018 at 14:45
  • $\begingroup$ @ThePointer: what do you mean by unidirectional ? $\endgroup$
    – user65203
    Commented Jul 20, 2018 at 14:47
  • $\begingroup$ @ThePointer: "in one direction" means following a straight line. I doubt this is what you have in mind. But I have completely rephrased my answer. $\endgroup$
    – user65203
    Commented Jul 20, 2018 at 14:50
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    $\begingroup$ @ThePointer 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. $\endgroup$
    – David C. Ullrich
    Commented Jul 20, 2018 at 15:11

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