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How to show a parametric equation is continuous?

  1. Apr 22, 2008 #1
    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 ...?
     
  2. jcsd
  3. Apr 22, 2008 #2

    HallsofIvy

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    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.


    A vector valued function is continuous if and only if its components are continuous. Where are sech t and tanh t continuous?
     
  4. Apr 22, 2008 #3
    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?
     
  5. Apr 22, 2008 #4

    HallsofIvy

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