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Infinitesimal arc-length square

  1. Nov 30, 2009 #1

    I am trying to understand the "infinitesimal arc-length square." So (ds)^2=(dx)^2+(dy)^2+(dz)^2. What does this means?

    And then what does (ds)^2=(dx)^2+(1+x^2)(dy)^2 -2x(dy) +(dz)^2 mean? And how does this apply to a space?
  2. jcsd
  3. Nov 30, 2009 #2


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    If you have a parameterized curve

    [tex]\vec r(t) = \langle x(t), y(t), z(t) \rangle,\, t \in [a,b][/tex]

    you calculate the arc length by taking the limit of polygonal approcimations. If

    [tex]a = t_0 < t_1 < ...< t_n = b[/tex]

    the polygonal approximation to the arc length is

    [tex]P_n = \sum_{i=1}^n |\vec r(t_i) - \vec r(t_{i-1})|[/tex]

    Here the absolute value signs represent Euclidean distance so

    [tex] |\vec r(t_i) - \vec r(t_{i-1})| = \sqrt{(x(t_i)-x(t_{i-1}))^2+(y(t_i)-y(t_{i-1}))^2+(z(t_i)-z(t_{i-1}))^2}[/tex]

    This is sometimes written the the delta notation:

    [tex]\Delta s_i \approx \Delta P_i =\sqrt{(\Delta x_i)^2 + (\Delta y_i)^2 + (\Delta z_i})^2[/tex]

    Passing to the limit suggests the notation [itex]ds^2 = dx^2+dy^2+dz^2[/itex]
  4. Nov 30, 2009 #3
    Thanks, that's great.

    I'm not really sure if this question would be well defined, but how would I then apply this method to the second metric [tex]
    (ds)^2=(dx)^2+(1+x^2)(dy)^2 -2x(dy) +(dz)^2
  5. Nov 30, 2009 #4


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    I can't say I have seen that kind of expression. The type of line integral you usually see in calculus is one of these two:

    [tex]\int_C \vec F \cdot\, d\vec R\ or\ \int_C \delta(x,y,z)\ ds[/tex]

    The first might represent work in a force field and the second the mass of a wire. Perhaps you are thinking of some other setting or application.
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