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Length of a curve

  1. Dec 22, 2008 #1
    is there a way to find the length of a curve between two x values?
    if so, what is it.
  2. jcsd
  3. Dec 22, 2008 #2
    Yes since the element of length is (assuming two dimensions)


    If you integrate you get the length of the curve [itex]s[/itex] from [itex][a,b][/itex]

    [tex]s=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt=\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}dx[/tex]
  4. Dec 22, 2008 #3


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    In three dimensions, write x, y, z as functions of parameter t and do the same:
    [tex]\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2+ \left(\frac{dz}{dt}\right)^2}dt[/tex]
  5. Dec 23, 2008 #4
    I am sorry for my broken english.

    How can one find the length of a curve, if the coordinate system is not rectangular (for instance, it is spherical)?

    Please if not inconvenient to you, point out my mistakes in my english.
  6. Dec 23, 2008 #5


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    By using exactly the same formulas but converting from the other coordinate system.

    For example, suppose a path is given in spherical coordinates by [itex]\rho= 1[/itex], [itex]\phi= \pi/3[/itex], [itex]\theta= t[/itex], with parameter t. In Cartesian coordinates that is [itex]x= \rho cos(\theta) sin(\phi)= (\sqrt{3}/2)cos(t)[/itex], [itex]y= \rho sin(\theta)sin(\phi)= (\sqrt{3}/2)sin(t)[/itex], [itex]z= \rho cos(t)= cos(t)[/itex].

    And my only criticism is that you should stop apologizing for your English. It is excellent. Far better than my (put whatever language you wish in here!).
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