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Length of a 3D parametric function

  1. Jan 1, 2014 #1
    1. The problem statement, all variables and given/known data

    find the length of a circular helix expressed in parametric form x= cos(t), y=sin(t) and z = t
    from t = 0 to t =2pi


    2. Relevant equations

    L = integrate ds

    (ds)^2 = (dx)^2+(dy)^2+(dz)^2

    3. The attempt at a solution

    I got to ds = (1 + (dt)^2)^0.5

    but I can't integrate as the dt is squared and inside the square root.

    :(
     
  2. jcsd
  3. Jan 1, 2014 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    It shouldn't be! You have x= cos(t) so dx= -sin(t)dt, y= sin(t) so dy= cos(t)dt, and z= t so dz= dt. Looks to me like you forgot the "dt" in both dx and dy.
    [tex]ds= \sqrt{dx^2+ dy^2+ dz^2}= \sqrt{sin^2(t)dt^2+ cos^2(t)dt^2+ dt^2}= \sqrt{2}dt[/tex].
     
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