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Arclength in polar coordinates?

  1. Oct 28, 2004 #1
    You know this should be simple but it's just not. A friend asked me this earlier and I was unable to disprove him. We're all aware of how one derives the area of a polar equation.. it's [tex] \pi r^2 \frac {\theta}{2 \pi}[/tex] and make theta infinitely small and integrate. Why can't a similar process be performed to find the arclength? IE [tex] 2 \pi r \frac{\theta}{2 \pi}[/tex] and make theta infinitely small and integrate. Obviously I can't derive this from rectangular coordinates because it only works for constant r but I just can't seem to disprove it.
     
    Last edited by a moderator: Oct 28, 2004
  2. jcsd
  3. Oct 28, 2004 #2
    Nevermind I think I solved it. The arclength formula he proposed to be when interpreted graphically had gaps between the lengths of r(d(theta)) and it was losing arclength that way. (edit here's my rendition, in case someone is stumped like I was). The figure is supposed to be a small section of curve with r varying.
     

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    Last edited by a moderator: Oct 29, 2004
  4. Oct 28, 2004 #3

    Tide

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    I'm not sure what you are trying to do but in the case of area you are changing both r and [itex]\theta[/itex] so that [itex]dA = dr \times r d\theta[/itex] whereas for path length you change r and [itex]\theta[/itex] according to Pythagoras being applied to infinitesimal displacements so that [itex]ds = \sqrt {dr^2 + r^2 d\theta^2}[/itex].
     
  5. Oct 29, 2004 #4
    Yeah I guess what threw me off at first was that r = sqrt(x^2+y^2) in rectangular coordinates so I figured the pathagorean theorem was already being applied (which it wasn't). Thanks for your input.
     
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