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Arc length problem calc II

  1. Oct 2, 2014 #1

    LBK

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    1. The problem statement, all variables and given/known data
    Find the exact length of the curve: y= 1/4 x2-1/2 ln(x) where 1<=x<=2

    2. Relevant equations
    Using the Length formula (Leibniz) given in my book, L=Int[a,b] sqrt(1+(dy/dx)2)
    I found derivative of f to be (x2-1)/2x does that look correct?

    3. The attempt at a solution
    I found f' to be (x2-1)/2x does that look correct? I'm thinking I must be off on something because I get a wrong answer. Either that or my integration. I am trying to use tables in the book for that, and not sure about that either. I am trying to sub u for that (dy/dx) part when integrating, for which I think I can use the form:
    int sqrt(u2+a2)=1/2 (u*sqrt(u2+a2)+a2ln|u+ sqrt(u2+a2)
    When I put my value back in for u, it turns in to a big mess. When I plug in my x values I got 15/32 +ln 2. This isn't right, the book has 3/4 + 1/2 ln 2 if that helps. Thank you.
     
  2. jcsd
  3. Oct 2, 2014 #2

    pasmith

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    Homework Helper

    Write [itex]y'[/itex] in the simpler form [itex]\frac12(x - \frac{1}{x})[/itex]. Take a closer look at [itex]1 + y'^2[/itex], and consider the identity [itex](a + b)^2 = (a - b)^2 + 4ab[/itex].
     
  4. Oct 2, 2014 #3

    Mark44

    Staff: Mentor

    Like many textbook problems on arc length, this one is cooked up so that the square root simplifies to something nice. As it turns out, 1 + y'2 is a perfect square, so the final integral doesn't involve a radical and is pretty simple. I get the same answer as the textbook.
     
  5. Oct 3, 2014 #4

    LBK

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    ah, I get it now. Thanks, you guys rock. I'll definitely be back lol.
     
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