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Arc Length Problem

  1. Oct 28, 2007 #1
    [SOLVED] Arc Length Problem


    So you plug it into the formula for arc length. (integral of the sqrt of 1+y'^2)

    And it yields [tex]\int \sqrt{1+(\frac{3x^{2}}{2\sqrt{x^{3}}})^{2}dx[/tex]

    From there you would use trig substitution, 1+tan^2theta = sec^2theta. But converting the dx to dtheta is a complete pain. And from what I can tell, it looks like it gets ugly.

    So the ugliness makes me think I am wrong. Can anybody check this up to this point?

  2. jcsd
  3. Oct 28, 2007 #2


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    Homework Helper


    [tex]\int \sqrt{1+(\frac{3x^{2}}{2\sqrt{x^{3}}})^{2}dx[/tex]

    can be simplified even more to give

    [tex]\frac{1}{2}\int \sqrt{4+9x} dx[/tex]

    and from there it should become much easier
  4. Oct 28, 2007 #3
    Solved. Thanks =P
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