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Arc length of a polar curve

  1. Sep 25, 2012 #1
    If we divide the polar curve into infinitely thin sectors, the arc length of a single sector can be approximated by [itex] ds = \frac{dθ}{2π}2πr = rdθ[/itex]. So why can't we model the arc length of the curve as [itex] \int^{β}_{α} rdθ[/itex]

    It turns out that the correct formula is actually
    [itex]\int^{β}_{α}\sqrt{r^{2}+(\frac{dr}{dθ})^{2}} \ dθ [/itex]
    I know how the correct formula is derived, I just can't figure out why the reasoning for the first formula is incorrect.

    BiP
     
  2. jcsd
  3. Sep 25, 2012 #2

    haruspex

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    By Pythagoras, ds2 = (rdθ)2+dr2. You can't go ignoring the dr term in general.
     
  4. Sep 25, 2012 #3
    My question was, why can't you use the arc length of the sector?

    BiP
     
  5. Sep 26, 2012 #4

    haruspex

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    Because in general the radial movement is of comparable magnitude to the tangential movement. In the extreme case, part of the curve might move directly away from the origin, so there dθ would be zero and you would only have dr.
     
  6. Sep 26, 2012 #5

    arildno

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    A straight line radiating from the origin must also have an arc length formular in polar coordinates, even though the angular coordinate remains a constant along the whole curve.
     
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