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Help With Arc Length Integral

  1. Mar 26, 2014 #1
    ∫sqrt(x^4/4 + 1/(x^4) + 1/2) dx from x = 1 to 4

    Could someone help me solve this? I can't seem to find a substitution that works, or find the square root of (x^4/4 + 1/(x^4). Any help would be very appreciated.

    Thanks in advance!
     
  2. jcsd
  3. Mar 26, 2014 #2

    Mark44

    Staff: Mentor

    What's inside the radical is almost a perfect square, so it's possible that you have made an error in the work leading up to the integral. Please show the function whose arc length you're trying to find, and what you did to get your integral.
     
  4. Mar 26, 2014 #3
    Here is the work that i have done so far:

    1) Find the length of the curve. y = x^3/6 + 1/(2x) from x = 1 to x = 4

    L = ∫sqrt(1 + (d(x^3/6 + 1/(2x)/dx))^2) dx from x = 1 to 4
    L = ∫sqrt(1 + (x^2/2-1/(2 x^2))^2) dx from x = 1 to 4
    L = ∫sqrt(1 + x^4/4 + 1/(x^4) - 1/2) dx from x = 1 to 4
    L = ∫sqrt(x^4/4 + 1/(x^4) + 1/2) dx from x = 1 to 4
     
  5. Mar 26, 2014 #4

    Mark44

    Staff: Mentor

    Your mistake is in the next line. Notice that you have a sign error and an error in the coefficient.
    If y = (1/6)x3 + 1/(2x), y' = (1/2)x2 - 1/(2x2)
    Also, I hope this isn't how you're doing your work. Instead of starting each line with L = ... and dragging the integral along at each step, it's much easier, and easier to read, if you write down what you need for the step at hand. In other words, find the derivative, square it, add 1, and take the square root. That will be your integrand. Each step from then on until you evaluate the integral will show the integral symbol, but you don't need it in the steps leading up to getting the integrand.

    I see that you didn't use the homework template, which has a section for the problem statement. Without knowing what the original problem was, I was not able to tell where you went wrong. This is one reason why we ask that you include this information. Please include the problem statement in any future posts.
     
  6. Mar 26, 2014 #5
    Thanks a lot for your help, Mark44! I will be sure to in the future.
     
  7. Mar 27, 2014 #6
    Now how would I go about factoring the expression that I get when I correct my mistakes? What I have now is

    ∫sqrt(x^6/36 + x^2/6 + 1/(4x^2) + 1) dx from x = 1 to 4
     
  8. Mar 27, 2014 #7

    Mark44

    Staff: Mentor

    You still have mistakes. It's possible you're skipping steps.

    Starting from y = x3/6 + 1/(2x), show me what you get for the following:
    1. y'
    2. (y')2
    3. 1 + (y')2

    If you do the three steps above correctly, what you get is actually fairly easy to factor inside the radical. Only then should you worry about the radical and the integral.
     
  9. Mar 28, 2014 #8
    Here is the work that i have done so far, corrected for the mistake that you pointed out earlier.


    1) Find the length of the curve. y = x^3/6 + 1/(2x) from x = 1 to x = 4

    L = ∫sqrt(1 + (d(x^3/6 + 1/(2x)/dx))^2) dx from x = 1 to 4

    (d(x^3/6 + 1/(2x)/dx) = (x^3/6 + 1/(2x)
    ((x^3/6 + 1/(2x))^2 = x^6/36 + x^2/6 + 1/(4x^2)

    Putting it together,

    ∫sqrt(1 + (d(x^3/6 + 1/(2x)/dx))^2) dx from x = 1 to 4 = ∫sqrt(x^6/36 + x^2/6 + 1/(4x^2) + 1) dx from x = 1 to 4
     
  10. Mar 28, 2014 #9

    Mark44

    Staff: Mentor

    No.

    Please do as I suggested in my previous post.
    Your first line should NOT start with "L = ∫<something>".
     
    Last edited: Mar 28, 2014
  11. Mar 28, 2014 #10

    Mark44

    Staff: Mentor

    Your differentiation is incorrect here.
     
  12. Mar 28, 2014 #11

    Mark44

    Staff: Mentor

    Also, it's probably easier to read if you write the left side like so:
    (d/dx)(x3/6 + 1/(2x))
     
  13. Mar 29, 2014 #12
    Hint: You can rewrite the stuff inside the square root as a three term polynomial over 4x^4.
     
  14. Mar 31, 2014 #13
    How's this?

    1. (d/dx)((x^3/6 + 1/(2x)) = x^2/2 - 1/(2x^2)
    2. (dy/dx)^2 = (x^4 -1)^2/4x^4 = ((x - 1)^2 (x+1)^2 (x^2+1)^2) / 4x^2
    3. (dy/dx)^2 + 1 = (x^4 + 1)^2 / (4x^4)
    4. sqrt((dy/dx)^2 + 1) = (x^4+1) / 4x^2 = x^2 / 4 + 1 / (4x^2)
    5.∫ x^2 / 4 + 1 / (4x^2) dx from x = 1 to 4 = x^3/12 - 1/ (4x) from x = 1 to 4 = (4^3/12 - 1/ (4(4))) - (1^3/12 - 1/ (4)) = 87 / 16

    Thank you for being so patient with me. I'm 15 and have very little clue what I'm doing. I skipped Algebra II and Precalculus (testing out of them by the skin of my teeth.)
     
    Last edited: Mar 31, 2014
  15. Mar 31, 2014 #14

    Mark44

    Staff: Mentor

    It's helpful to simplify the above by factoring out 1/2. This gives you (1/2)(x2 - 1/x2), which will make the following step easier.
    I can't tell what you did on the right side of the first =.
    [(1/2)(x2 - 1/x2)]2 = (1/4)(x4 - 2 + 1/x4)
    When you add 1 to the above (= 4/4), you get a perfect square trinomial, which is easy to factor.
    That was probably not a good idea. In any case, for the areas in which you're weak, it would be very helpful for you to review. You could pick up a precalculus textbook from Amazon or look at the algebra and precalc topic on khanacademy.org. Some time spent on a regular basis would be a very good investment.
     
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