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Line integral - finding the arc length

  1. Nov 16, 2008 #1
    a curve is given as 3 parameters of t:
    x=a(3t - t^3), y=3a(t^2), z=a(3t + t^3)

    i have to find the arc length measured from origin and curvature as functions of t.

    would i be correct in using the integral at the bottom of page 2 here: http://homepages.ius.edu/wclang/m311/fall2005/notes17.2.pdf
     
  2. jcsd
  3. Nov 16, 2008 #2

    arildno

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    Sure enough, with f=1.
    Then you get:
    [tex]\int3|a|\sqrt{(1-t^{2})^{2}+(2t)^{2}+(1+t^{2})^{2}}[/tex]
    If you are clever, you'll find a perfect square buried within the radicand.
     
  4. Nov 16, 2008 #3
    why is f=1?

    would my limits be 0 to t?
     
  5. Nov 16, 2008 #4

    arildno

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    Because you are to find [tex]\int{ds}=\int{1}*ds[/tex]
    Yes.
     
  6. Nov 16, 2008 #5
    ive ended up with integral 3a√2 (t^2 +1) dt
    so i take the 3a√2 out and integrate t^2 +1.

    but im confused since the limits are 0 to t, how would i replace the t with the limit t once i have integrated it?
     
  7. Nov 16, 2008 #6

    HallsofIvy

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    Don't worry about the limits of integration: you are finding the arclength as a function of t so find an anti-derivative and take the constant so that the integral is 0 when t= 0.
     
  8. Nov 16, 2008 #7
    then i get 3a√2 (t^3/3 + t). thats wrong though?
     
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