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Arc Length Problem

  1. Feb 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the arc length of the curve:

    [tex]y=\frac{x^5}{6}+\frac{1}{10x^3}[/tex]

    1[tex]\leq[/tex]x[tex]\leq[/tex]2


    2. Relevant equations

    [tex]ds=\sqrt{dx^2+dy^2}[/tex]

    [tex]ds=\sqrt{1+\frac{dy}{dx}^2}dx[/tex]

    3. The attempt at a solution

    [tex]\frac{dy}{dx}=\frac{5}{6}x^4-\frac{3}{10x^4}[/tex]

    [tex]ds=\sqrt{1+(\frac{5}{6}x^4-\frac{3}{10x^4})^2}dx[/tex]

    If I use Trig. Sub, I have [tex]tan \theta=(\frac{5}{6}x^4-\frac{3}{10x^4})[/tex].

    However, I don't know how to solve this for X. If this is the right path, can anyone help with the next steps, and is there a better way to do this problem?

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



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 22, 2010 #2

    Mark44

    Staff: Mentor

    OK, hold it right there. Just carry out the multiplication, and add the 1. You'll see that you still have a perfect square so you can take its square root.

    This is a perfect example of a problem that has been laboriously cooked-up so that it's doable. Most of the arc length problems come out with difficult integrands, so some effort has to go into one to make it not too difficult to integrate.
     
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