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Math Help - Calculating Arc length of a curve.

  1. #1
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    Calculating Arc length of a curve.

    The curve: y=\frac{x^2}{2} - \frac{lnx}{4}, 2\leq x \leq4

    my work so far:
    x(t)=\frac{x^2}{2} - \frac{ln}{4}

    x'(t)=x - \frac{1}{4x}

    y(t)=t

    y'(t)=1

    \int^4_2 \sqrt{1^2 + (x - \frac{1}{4x})^2}dx

    \int^4_2 \sqrt{1+\frac{9}{16}x^2}dx

    anyways after that mess I am really not sure how to integrate this with u-substitution. If I made a mistake please let me know or some direction if I am right so far would be very helpful!! Thanks a lot!
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  2. #2
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    Quote Originally Posted by jfinkbei View Post
    \int^4_2 \sqrt{1^2 + (x - \frac{1}{4x})^2}dx

    \int^4_2 \sqrt{1+\frac{9}{16}x^2}dx
    How did you make this step? What is \left(x-\frac{1}{4x}\right)^2?

    \left(x-\frac{1}{4x}\right)\left(x-\frac{1}{4x}\right)

    I'm getting something different under the radical.

    By the way, the 1 doesn't need to be squared. The formula for arclength has a 1+[f'(x)]^2 under the radical always.
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  3. #3
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    for the integrand I'm getting \sqrt{\frac{16x^4+8x^2+1}{16x^2}}

    Both the top and bottom are perfect squares, so factor and simplify and let u=x^2

    EDIT: It might not be that simple actually, I haven't done the whole problem. You may need to make an additional substitution. Let me know if you get stuck again and I'll try to work it out.

    EDIT2: you shouldn't need to sub at all for this integral. Sorry, I'm all mixed up right now.
    Last edited by Shananay; February 5th 2010 at 10:21 PM.
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  4. #4
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    \left(x-\frac{1}{4x}\right)^2= x^2- \frac{1}{2}+ \frac{1}{16x^2} so

    1+ \left(x-\frac{1}{4x}\right)^2= x^2+ \frac{1}{2}+ \frac{1}{16x^2}

    which is just the first formula with " -\frac{1}{2}" changed to " +\frac{1}{2}". It is
    \left(x+ \frac{1}{4x}\right)^2 so the square root just gives

    \int x+ \frac{1}{4x} dx
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    \left(x-\frac{1}{4x}\right)^2= x^2- \frac{1}{2}+ \frac{1}{16x^2} so

    1+ \left(x-\frac{1}{4x}\right)^2= x^2+ \frac{1}{2}+ \frac{1}{16x^2}

    which is just the first formula with " -\frac{1}{2}" changed to " +\frac{1}{2}". It is
    \left(x+ \frac{1}{4x}\right)^2 so the square root just gives

    \int x+ \frac{1}{4x} dx
    Faster and easier than what I was doing. This is the way to go.
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