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