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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.
    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
    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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    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.

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