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Arc Length Question

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data

    y = (2/3) * (x^2 - 1) ^ (3/2) 1 <= x <= 3
    Length = ?


    2. Relevant equations

    L = [tex]\int\sqrt{1 + (dy/dx)^2} dx[/tex]


    3. The attempt at a solution

    dy/dx y = (2/3) * (x^2 - 1) ^ (3/2) =
    2x * sqrt(x - 1)

    Any ideas for a proper substitution? The answer on wolfram seems ridiculous. :bugeye:
     
    Last edited: Nov 24, 2008
  2. jcsd
  3. Nov 24, 2008 #2
    Hmm... it seems wolfram is being ridiculous on this problem. But you don't need a substitution here. Observe that the derivative is 2x√(x² - 1), not 2x√(x - 1). So once you go back and do the arithmetic correctly, you will find that 1 + (dy/dx)² is the square of a polynomial (specifically, it is (2x² - 1)²). After that, this problem should be a piece of cake.
     
  4. Nov 24, 2008 #3
    I actually knew it was x2, just a typing error. I actually got (2x2 - 1)2
    the first time I calculated it, but I didn't get the right answer for the definite integral at 1 through 3 (15.333).

    My solution:

    (2x)^2 * (x^2 - 1) + 1 = 4x^4 - 4x^2 + 1 = (2x^2 - 1)^2

    = (2/3) * x3 -x + C ] (1-3) = 15.333 :rolleyes: Nevermind . . .

    Thanks for your help :]
     
    Last edited: Nov 24, 2008
  5. Nov 24, 2008 #4
    Glad you figured it out. It looks like you have the theory right, and were just getting caught up with typos.
     
  6. Nov 25, 2008 #5
    Story of my mathematical life.
     
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