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 $x,y:\Bbb R\to\Bbb R$ by $y(t)=t^2$ and $$x(t)=\begin{cases} t^2,&(t\ge0), \\-t^2,&(t<0).\end{cases}$$Then $x$ and $y$ are both $C^1$, and $(x(t),y(t))$ 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 $x'(0)=y'(0)=0$.)

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

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 $x,y:\Bbb R\to\Bbb R$ by $y(t)=t^2$ and $$x(t)=\begin{cases} t^2,&(t\ge0), \\-t^2,&(t<0).\end{cases}$$Then $x$ and $y$ are both $C^1$, and $(x(t),y(t))$ is a parametrization of the curve $y=|x|$. Without the condition on non-vanishing derivatives this would make $y=|x|$ a $C^1$ curve.

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 $x,y:\Bbb R\to\Bbb R$ by $y(t)=t^2$ and $$x(t)=\begin{cases} t^2,&(t\ge0), \\-t^2,&(t<0).\end{cases}$$Then $x$ and $y$ are both $C^1$, and $(x(t),y(t))$ 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 $x'(0)=y'(0)=0$.)