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Distance Traveled

  1. Jan 3, 2015 #1
    If we wish to find the distance traveled (not the distance from the origin) by a particle along a path ##C## defined by ##y = f(t)## and ##x = g(t)## we would use this integral:

    $$L = \int_C ds = \int_{t_1}^{t_2} \sqrt{({\frac{dy}{dt}})^2 + ({\frac{dx}{dt}})^2} dt$$

    My question is, does this give a nonzero answer if the particle were at the same position at ##t = t_1## and at ##t = t_2##?

    Also, is this a line integral? I don't know much about line integrals and vector calculus, all I know is work done is a good example of a line integral.
    Last edited: Jan 3, 2015
  2. jcsd
  3. Jan 3, 2015 #2


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    You can simply understand it yourself by calculating this integral for [itex] x=\cos t \ , \ y=\sin t [/itex] from [itex] t=0 \ to \ 2\pi [/itex].
  4. Jan 3, 2015 #3


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    As long as the particle is moving, the integrand is always positive even if the particle ends up where it started.
  5. Jan 5, 2015 #4


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    Yes, that is a line integral and gives the length of the line. If you go around a circle with circumference 100 meters, you are right back where you started but have walked 100 meters, not 0!
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