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Homework Help: Length of the curve

  1. Oct 31, 2005 #1
    I need to find length of the curve x=cos(2t), y=3sin(2t), t[0,p]

    I'm getting a length of this curve to be equal zero.
    I think it is because this curve is closed and start point = end point.
    Is it normal?
    How can I get a length of this curve?
     
  2. jcsd
  3. Oct 31, 2005 #2
    No you should still be getting a length, what procedure are you using? Show the integral you set up.
     
  4. Oct 31, 2005 #3
    length of a arc is given by

    s = # * r

    # is the internal angle in rads

    r is radias
     
  5. Oct 31, 2005 #4
    I got

    s(t)=Integral (2sin(2t)-6cos(2t))dt

    integral from 0 to pi
     
  6. Oct 31, 2005 #5
    I remember the arc length integral being more complex than that.. can you explain how you set it up/
     
  7. Oct 31, 2005 #6

    TD

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    Homework Helper

    The length of a curve f given in parametric form can be calculated with

    [tex]\int_a^b {\left\| {\frac{{d\vec f}}{{dt}}} \right\|dt} [/tex]

    Which is, written out in 2 variables:

    [tex]\int_a^b {\sqrt {x'\left( t \right)^2 + y'\left( t \right)^2 } dt} [/tex]
     
  8. Nov 1, 2005 #7
    A tip: [tex]\sqrt {x^2 } = \left| x \right| \ne x,\left| x \right| = x \Leftrightarrow x \ge 0[/tex]
     
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