- #1

WalkingInMud

- 5

- 0

**, is smoothly parametrized if:**

*r(t)*1. its derivative is continuous, and

2. its derivative does not equal zero for all t in the domain of r.

Now that sounds simple enough. Now let's say we have the

*tractrix*:

**, ...**

*r(t) = (t-tanht)i + sechtj*then

**, right?**

*r'(t) = [ 1/(1+x^2) ]i + [ tantsect ]j*Now, without reverting to MatLab or Maple to view the graph, how do we mathematically explain that

**is continuous (or not)?**

*r'(t)*Do I just state that is is/isn't -by inspection, or ...?