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Length of Curve (Calc 2) question

  1. Sep 25, 2005 #1
    So the length of a curve given by a vector r is:

    L = [tex] \int |r'(t)|dt[/tex]

    If the vector is given by r(t) = < t, t^2/srt2, t^3/3> on an interval from 0 to 1, find the length

    I am incredibly confused here, as most of the examples we've used have been using the cos^2+sin^2 = 1 identity.

    I took the derivative of r(t), and got

    r'(t) = 1, 2t/sqrt2, t^2>

    then i set up the integral using the magnitude of that equation, and end up with L = [tex] \int (1^2 + (2t/sqrt2)^2 + (t^2)^2)dt [/tex]

    I get this feeling like I need to set the t value equal to something in order to make the magnitude work out for an integral that would work out to integrating something like dt itself.

    Any help would be much appreciated
    Thank you
    Twiztidmxcn
     
  2. jcsd
  3. Sep 25, 2005 #2
    You forgot the squareroot and the limits, but the rest looks fine. You just have to calculate

    [tex] L = \int \limits_a^b | r'(t) |\mathrm{d} t = \int \limits_0^1 \sqrt{1^2 + (\sqrt{2}t)^2 + (t^2)^2 } \mathrm{d} t [/tex]
     
  4. Sep 25, 2005 #3

    arildno

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    Just to help you a bit further:
    Note that:
    [tex]1^{2}+(\sqrt{2}t)^{2}+(t^{2})^{2}=1+2t^{2}+t^{4}=(1+t^{2})^{2}[/tex]
     
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