Why is a vector function not smooth if $r'=0$

Question

It is stated that a vector function is smooth if its derivative is continuous and nowhere zero, but I can’t find a proof or a geometric interpretation. Is this a definition or a theorem?

Answer 1

This is a definition. See for example page $910$ of Thomas' Calculus, which also gives a geometric interpretation of a smooth vector function.

Answer 2

I suspect the source of your statement is sloppy or misquoted; not knowing the source, I can't tell which one. Usually, a function is called smooth if it has a continuous derivative (of 1st order... or of all orders, in other people's usage). A curve is smooth if it's parametrized by a smooth function whose derivative is nowhere zero.

For example, $\vec r(t) = (t-\sin t, 1-\cos t)$ is a smooth function but it does not describe a smooth curve: the image is the cycloid.

The cusps become possible when $\vec r{}'(t)$ turns to zero; this is why the definition of a smooth curve prohibits $\vec r{}'(t)=0$.

Here is a sketch of the proof that $\vec r{}'(t)\ne 0$ implies smoothness in some geometric meaning. Given $t_0$, pick an index $j$ such that the $j$th component of the velocity vector is nonzero at $t_0$. Then it's also nonzero in a neighborhood of $t_0$. By the inverse function theorem the function $t\mapsto x_j(t)$ is invertible. Let its inverse be $t=f(x_j)$. Then $x_j\mapsto \vec r(f(x_j))$ is a non-parametric equation of the curve, which expresses coordinates $x_i$, $i\ne j$, as smooth functions of $x_j$. Such a curve has tangent line given by a familiar equation.