# Showing a parametric equation is smooth?

• April 22nd 2008, 09:01 AM
WalkingInMud
Showing a parametric equation is smoothly parametrized?
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 ...?
• April 22nd 2008, 12:20 PM
Opalg
Quote:

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 $\textstyle\frac d{dt}(\tanh t) = \text{sech}^2t$ and $\textstyle\frac d{dt}(\text{sech}\, t) = -\text{sech}\, t\tanh t$, which gives $r'(t) = (1-\text{sech}^2t)\mathbf{i} - \text{sech}\, t\tanh t\mathbf{j}$. 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 $1-\text{sech}^2t = 0$ and $\text{sech}\, t\tanh t = 0$?