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Homework Help: Calculus II - Lengths of Curves - Hard

  1. Aug 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the arc length of the following curves on the given interval by integrating with respect to x

    y=x^4/4+ 1/(8x^2); [1,2]

    2. Relevant equations

    Let f have a continuous first derivative on the interval [a,b]. The length of the curve from (a,f(a)) to (b,f(b)) is

    L = integral[a,b] sqrt(1+f'(x)^2) dx

    3. The attempt at a solution

    y=x^4/4+ 1/(8x^2); [1,2]
    I took the derivative and got

    dy/dx = x^3 - 1/(4x^4)

    I than square this and got

    x^6 - 1/2 + 1/(16*x^6)

    plunging into the formula I get

    integral[1,2] sqrt(1/2 + x^6 + 1/(16*x^6)) dx

    I have no idea were to go from here. I tried making several substitutions and than realized that there was nothing in my techniques of integration to evaluate this in any way or form. I plugged the indefinite integral into wolfram alpha to see what it got for the antiderivative and apparently it doesn't know how to integrate this neither.
    http://www.wolframalpha.com/input/?i=integral+sqrt%281%2F2%2Bx^6%2B1%2F%2816x^6%29%29
    well apparently it came up with a antiderivative it just has no steps to prove it to be correct when you click on show steps "(step-by-step results unavailable)". So I conclude that there is something really simple that I'm not seeing or something really complicated. This came from my textbook in -Applications of Integration-Lengths of Curves

    Thanks for any help!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 11, 2011 #2
    Try factoring 1/(16x6) out of x6 - 1/2 + 1/(16*x6) first.
     
  4. Aug 11, 2011 #3
    x^6 - 1/2 + 1/(16*x6) = 1/(16*x^6)[16x^12-8x^6+1]
    plugging into L = integral[a,b] sqrt(1+f'(x)^2) dx
    yields
    integral[1,2] sqrt(1+1/(16*x^6)[16x^12-8x^6+1])dx

    wasn't sure what to do at this point so I wrote it all as the same fraction

    integral[1,2] sqrt((16x^6+16x^12-8x^6+1)/16x^6)dx

    simplified -> 1/sqrt(16x^6) = 1/(4x^3)

    1/4 integral[1,2] sqrt(16x^6+16x^12-8x^6+1)/x^3 dx

    I'm not sure how to proceed from here
     
  5. Aug 11, 2011 #4
    I copied down the wrong part :blushing: I meant to say factor 1/(16x6) from x6 - 1/2 + 1/(16*x6)
    You basically already did it though, so you would get sqrt(1/(16*x6)[16x12+8x6+1])

    What can you do with 16x12+8x6+1?
     
  6. Aug 11, 2011 #5
    thanks i got it factor it and then it's easy
     
  7. Aug 12, 2011 #6

    Ray Vickson

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    Science Advisor
    Homework Helper

    If your y is *exactly* as written [y = x^4/4 + 1/8/x^2 --- note that a/b/c means a/(b*c) ], then your derivative is wrong, so you have the
    wrong integral. The last term in dy/dx is -1/4/x^3, not your -1/4/x^4.

    RGV
     
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