Why do we need tangent vector unequal to zero for smoothness of a vector function?

Question

]1

My textbook gives this definition of smoothness of a $\vec r(t)$ on an interval $I$ of $t$. Why do we need $\vec r'(t) \neq\vec0$ on $I$?

Answer 1

Let

$x(t)=\begin{cases}0&0<t\leq1\\(t-1)^2&1<t<2\end{cases}$ and $y(t)=\begin{cases}(1-t)^2&0<t\leq1\\0&1<t<2\end{cases}$

The path goes from $(1,0)$ to $(0,1)$, made up of two line segments with a right angle at the origin at $t=1$.

Now the vector derivative $\vec r'(1)=(0,0)$ at the origin (and is defined everywhere else of course), but we wouldn't want to call this motion "smooth"; it has a sharp corner at the origin.

Answer 2

Since $\mathbf{T}$ has $|\mathbf{r}'|$ in the denominator, the latter vanishing would leave the tangent vector undefined. A smooth curve should have a well-defined tangent everywhere.