How to show 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+x^2) ]i + [ tantsect ]j, right?


Now, 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 ...?

[HallsofIvy]

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+x^2) ]i + [ tantsect ]j, right?


Now, without reverting to MatLab or Maple to view the graph, how do we mathematically explain that r'(t) is continuous (or not)?
Look at the components separately. However, you seem to have confused "tanh" with "arctan". The derivative of t- tanh(t) is 1+ sech(t), not 1/(1+ t^2) and the derivative of sech(t) is -sech(t)tanh(t). For what values of t are those continuous? For what values of t are the 0?