Thread: Showing a parametric equation is smooth?

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?


FIRSTLY, do I have the derivative correct? --and

SECONDLY, without reverting to MatLab or Maple to view the graph, how do we deduce that r'(t) is continuous (or not)?

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

Originally Posted by 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?


FIRSTLY, do I have the derivative correct? --and

SECONDLY, without reverting to MatLab or Maple to view the graph, how do we deduce that r'(t) is continuous (or not)?

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

The derivatives of the hyperbolic functions are and , which gives . Those hyperbolic functions are continuous, so there's no problem about r'(t) being continuous. You just have to check whether it's ever zero. In other words, are there any values of t for which both and ?