Results 1 to 2 of 2

Math Help - Showing a parametric equation is smooth?

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    4

    Question 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 ...?
    Last edited by WalkingInMud; April 22nd 2008 at 10:23 AM. Reason: corrections
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by WalkingInMud View Post
    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?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: March 26th 2011, 09:10 AM
  2. Showing Analytic Using Derivative Equation
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: March 10th 2011, 09:10 AM
  3. Replies: 1
    Last Post: October 7th 2010, 08:54 PM
  4. Showing a piecewise function is smooth
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: March 16th 2010, 09:12 AM
  5. Showing a root is of an equation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 23rd 2009, 02:10 AM

Search Tags


/mathhelpforum @mathhelpforum