Showing that a parametric equation is smoothly parametrized?

[WalkingInMud]

A parametric equation, say r(t), is smoothly parametrized if:

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 lets say we have the tractrix:

r(t) = (t-tanht)i + sechtj,

then r'(t) = [ 1-1/(1+t^2) ]i + [ tantsect ]j, right?


without reverting to MatLab or Maple to view the graph, how do we mathematically explain that r'(t) is continuous (or not)?
:?
Do I just state that is is/isn't -by inspection, or ...?

 

[galactus]

A vector-valued function is continuous at the point given by t=a if the limit of r(t) exists as t-->a and

\(\displaystyle \lim_{t\to{a}}r(t)=r(a)\)

A vector-valued function is continuous on an interval if it is continuous at every point in that interval.

If you want to check the continuity of r'(t), then the derivative is

\(\displaystyle r'(t)=(\frac{-4}{e^{2t}+1}+\frac{4}{(e^{2t}+1)^{2}}+1)i-\frac{2e^{t}(e^{2t}-1)}{(e^{2t}+1)^{2}}j\)

Let the limit as t-->0 and we get 0, so r(0)=0 and we can conclude that r is continuous at t=0. We can also conclude by siialr reasoning that r is continuous at all real numbered values of t. Seems to me to be the way to go.