# Parametrically Defined Curves: $f'$ and $g'$ Are Not Simultaneously Zero

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.

• 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 '18 at 14:12
• @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 '18 at 14:14
• @DavidC.Ullrich: I meant locally, it was more of a comment than a complete answer. Jul 20 '18 at 14:21
• @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 '18 at 14:27
• @copper.hat In fact I'm pretty sure that any rectifiable curve has a $C^1$ parametrization... Jul 20 '18 at 14:39

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 $f'(0)=g'(0)=0$.)

• 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 '18 at 14:25
• 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 '18 at 14:26
• @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 '18 at 14:30
• @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 '18 at 14:31
• @ThePointer Right. Jul 20 '18 at 15:30

Because when $f'=g'=0$, the direction of the curve is no more defined. At such a point, the curve could change direction abruptly.

• Singular point: mathworld.wolfram.com/SingularPoint.html Jul 20 '18 at 14:44
• Thanks for the response. So the reason for this condition is that we want the parameterised curve to be unidirectional? Jul 20 '18 at 14:45
• @ThePointer: what do you mean by unidirectional ?
– user65203
Jul 20 '18 at 14:47
• @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.
– user65203
Jul 20 '18 at 14:50
• @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. Jul 20 '18 at 15:11