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Hard arc-length integral

  1. Feb 24, 2013 #1
    Find the length of the curve
    y = x^2 − ln(x), 1 ≤ x ≤ 2.

    I have have to find the arc-length of the given equation. I have checked my work several times and i ultimately come up with the integral of ((4x^4 -3x^2 +1)/(x^2))^(1/2), which I just cannot seem to solve. Any help would be appreciated.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 24, 2013 #2

    SammyS

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    Hello rhothi. Welcome to PF !

    Please give details as to how you got that integral, so we can help you.
     
  4. Feb 24, 2013 #3
    I used the formula for calculating arc length. The formula is ((1 + (f'(x))^(2))^(1/2) from a to b. As a result, I got f'(x) to be 2x - 1/x. When I squared the function i received ((4x^4 -4x^2 +1). From there I plugged f'(x)^2 into the equation to get the integral that I mentioned.
     
  5. Feb 24, 2013 #4
    ##\displaystyle \left(2x - \frac{1}{x}\right)^2## is not ##4x^4 -4x^2 +1##.

    Note that ##(a+b)^2 = a^2 + 2ab + b^2##.

    Try it again.
     
  6. Feb 24, 2013 #5
    I meant that it is (4x^4−4x^2+1)/(x^2). After adding 1 underneath the radical I received ((4x^4 -3x^2 +1)/(x^2))^((1/2)).
     
  7. Feb 24, 2013 #6

    SammyS

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    It looks like you have a typo.

    Are you saying that [itex]\displaystyle \ \ 1+\left(2x-\frac{1}{x}\right)^2=\frac{4x^4 -3x^2 +1}{x^2}\ ?[/itex]

    That's correct.

    The result for the integral, [itex]\displaystyle \ \ \int\sqrt{\frac{4x^4 -3x^2 +1}{x^2}}\,dx\ \ [/itex] looks very messy.

    You can solve for x as a function of y for x > 0 . That result doesn't look much more promising.

    Off hand, I can't think of any parametrization which would give a nicer result.
     
  8. Feb 24, 2013 #7
    Yes that is what I meant. How would I solve the equation in terms of y? I have no idea how to compute the integral in terms of x.
     
  9. Feb 24, 2013 #8

    SammyS

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    Here is a parametrization which I thought looked promising, but not surprisingly, it gives a similarly messy result.

    Let x = et, then y = (et) - ln(et) = e2t - t .
     
  10. Feb 24, 2013 #9

    Dick

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    If you put the integral into Wolfram Alpha, it does give you an indefinite integral in terms of elementary functions. I didn't think it would. That probably means you could do it by completing a square, thinking of a clever substitution, maybe doing an integration by parts or two and voila. But the complexity of the answer makes it look like a really tough road for mere mortals. I'd guess if this was an exercise in computing arclength, you want to make it a lot easier than that. I'm going to guess there is probably a typo in the problem statement.
     
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