Derivative of a Vector-Valued Function I was completing my Webassign earlier and a particular problem came up.

Find the open interval(s) on which the curve given by the vector-valued function is smooth. (Enter your answer using interval notation.)
$r(t) = 9t^2i + 7t^3j$

The answer turned out to be $(-\infty, 0) \cup (0, \infty)$.
But the issue I have with the problem is the intuition, and thus leading to my question:
Why can't the derivative of the vector-valued function equal $0$ and be smooth?

There was a similar question here. Basically, for parametric curves (and this is parametric — the parameter is $t$), the definition of smooth means that the derivative is continuous and not equal to the zero vector at any point.
Intuitively, if the derivative of a "normal" (non-parametric) function is zero, then that tells us we have a max or min. In a parametric curve, it could mean that the curve reaches a point, stops, and heads sharply off in another direction, creating a three dimensional cusp.
Clearly that doesn't happen here, but the non-zero requirement has still worked its way into the definition.
