# How to show a parametric equation is continuous?

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 let's 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 ...?

Homework Helper
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 let's say we have the tractrix:

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

then r'(t) = [ 1/(1+x^2) ]i + [ tantsect ]j, right?
No, not right. r'(t)= [1- sech t]i- tanh t sech t j/b]. I don't know why you have changed to "x" in the first component but you surely should not have tan and sec rather than tanh and sech in the second component.

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 ...?
A vector valued function is continuous if and only if its components are continuous. Where are sech t and tanh t continuous?

WalkingInMud
How does this look?

r(t)=(t-tanht)i+(secht)j ...so
r'(t)=(1-secht)i+(-tanhtsecht)j

Now, secht is continuous for all real t, and also
tanht is continuous for all real t, BUT...

since tanh(0)=0 i.e. (-tanhtsecht)=0 for t=0, and furthermore sech(0)=1 i.e. (1-secht)=0 for t=0,

...r'(0)=0 so r(t) is smoothly parametrized for all t NOT EQUAL TO 0

Is this OK?

Homework Helper
r(t)=(t-tanht)i+(secht)j ...so
r'(t)=(1-secht)i+(-tanhtsecht)j

Now, secht is continuous for all real t, and also
tanht is continuous for all real t, BUT...

since tanh(0)=0 i.e. (-tanhtsecht)=0 for t=0, and furthermore sech(0)=1 i.e. (1-secht)=0 for t=0,

...r'(0)=0 so r(t) is smoothly parametrized for all t NOT EQUAL TO 0

Is this OK?
Yes, that is correct. And if you look at the graph of the tractrix, you can see what happens at t= 0.