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Arc length and surfaces

  1. Feb 9, 2009 #1
    latex2png.2.php?z=100&eq=y%3D%20x%5E3%2F12%2B1%2Fx%20%2CA(1%2C13%2F2)%2C%20B(2%2C7%2F6).jpg

    latex2png.2.php?z=100&eq=y'%3D(x%5E4-4)%2F4x%5E2.jpg

    latex2png.2.php?z=100&eq=L%3D%5Csqrt%7B1%2B(x%5E4-4)%2F4x%5E2)%5E2%7D.jpg

    latex2png.2.php?z=100&eq=%3D(1%2F4x%5E2)%5Csqrt%7B8x%5E4%2Bx%5E8%2B16%7D.jpg

    latex2png.2.php?z=100&eq=u%3D8x%5E4%2Bx%5E8%2B16.jpg

    latex2png.2.php?z=100&eq=du%3D8x%5E7%2B32x%5E3.jpg

    HELP
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 9, 2009 #2

    Mark44

    Staff: Mentor

    What, are we supposed to guess what the problem is from the title on the thread? Presumably you want to find the arc length along the curve between points A and B.

    What does this have to do with surfaces, though?

    For the arc length, the integrand is sqrt(1 + (y')^2), which can be written as
    [tex]\sqrt{1 + (\frac{x^2}{4} - \frac{1}{x^2})^2}[/tex]
    [tex]=\sqrt{1 + \frac{x^4}{16} -1/2 + \frac{1}{x^4}}[/tex]

    The last three terms under the radical are a perfect square. When you add the first term, you'll still have a perfect square, which makes it easy to take the square root, which means you'll have an easy function to integrate.
     
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