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Math Help - arc length

  1. #1
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    arc length

    I have been working on this problem for a bit and obviously missing something simple. I am finding arc length

    Given the equation y=\frac{3{x^{2/3}}}{2}+4. I am coming up with L=\int_{1}^{27}\sqrt{1+x^{1/9}}~dx. Can I get some help?
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  2. #2
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    Quote Originally Posted by Possible actuary View Post
    I have been working on this problem for a bit and obviously missing something simple. I am finding arc length

    Given the equation y=\frac{3{x^{2/3}}}{2}+4. I am coming up with L=\int_{1}^{27}\sqrt{1+x^{1/9}}~dx. Can I get some help?
    Arc length is:

    <br />
L = \int_a^b \sqrt{1+[f'(x)]^2}\ dx<br />

    here:

    <br />
f(x)=\frac{3{x^{2/3}}}{2}+4<br />

    so:

    <br />
f'(x)=x^{-1/3}<br />

    so:

    <br />
L = \int_a^b \sqrt{1+x^{-2/3}}\ dx<br />

    RonL
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  3. #3
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    Re:

    Quote Originally Posted by Possible actuary View Post
    I have been working on this problem for a bit and obviously missing something simple. I am finding arc length

    Given the equation y=\frac{3{x^{2/3}}}{2}+4. I am coming up with L=\int_{1}^{27}\sqrt{1+x^{1/9}}~dx. Can I get some help?
    RE:
    Attached Thumbnails Attached Thumbnails arc length-45.jpg  
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  4. #4
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    Quote Originally Posted by qbkr21 View Post
    RE:
    Yes, but the exact value can be found using the u substitution.
    Attached Thumbnails Attached Thumbnails arc length-june29.gif  
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  5. #5
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    Quote Originally Posted by curvature View Post
    Yes, but the exact value can be found using the u substitution.
    Aside from having my (y`)^2 wrong, this is what I was having trouble with. Thanks! I have a graphing calculator, TI 85, but do not know if it can compute arc length. I am sure it can just have not messed with it to find out.
    Last edited by Possible actuary; June 27th 2007 at 05:01 PM. Reason: TI 85 not TI 83
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