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Math Help - Length of a curve

  1. #1
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    Length of a curve

    Find the length of a curve given by:

    y= (x^4)/4 + 1/8x^2 for 1<x<3

    I understand length of curves, but I am confused how to find the square root of:

    sqr(x^6 + 1/2 x^4 + 1/16 x^2 + 1)

    help? Am I writing this term wrong in the first place?

    thanks!
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  2. #2
    MHF Contributor matheagle's Avatar
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    I know how these problems are rigged to work.
    They are quite artifical and what happens is the +1 changes a -.5 to a +.5

    Hence MY guess is that the second power is -2 not +2

    y= (x^4)/4 + 1/8x^{-2} for 1<x<3

    I GUESS you meant to put 1/(8x^2), which I know works.
    But your math showed that you wanted (1/8)x^2.
    Last edited by matheagle; July 30th 2009 at 06:54 AM.
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  3. #3
    Junior Member enjam's Avatar
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    Quote Originally Posted by klooless View Post
    I understand length of curves, but I am confused how to find the square root of:

    sqr(x^6 + 1/2 x^4 + 1/16 x^2 + 1)

    help? Am I writing this term wrong in the first place?
    Hmm... how did you end up with the +1 at the end of the equation, I just get sqrt. (x^6 + 1/2 x^4 + 1/16x^2).
    Anyway, it'd equal:
    x^3 + sqrt. 1/2 x^2 + 1/4 x
    You would then integrate this to get:
    x^4 / 4 + x^3 / 3sqrt(1/2) + x^2 / 8
    Length between 3 and 1 would equal:
    (20.25 + 12.73 + 1.13) - (0.25 + 0.47 + 0.13)
    = 33.26
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  4. #4
    Senior Member DeMath's Avatar
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    Quote Originally Posted by klooless View Post
    Find the length of a curve given by:

    y= (x^4)/4 + 1/8x^2 for 1<x<3

    I understand length of curves, but I am confused how to find the square root of:

    sqr(x^6 + 1/2 x^4 + 1/16 x^2 + 1)

    help? Am I writing this term wrong in the first place?

    thanks!
    y = \frac{{{x^4}}}{4} + \frac{1}{{2{x^2}}},{\text{ }}1 \leqslant x \leqslant 3,{\text{ }}{L_y} = ?

    {L_y} = \int\limits_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } .

    \sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}}  = \sqrt {1 + {{\left( {\frac{d}{{dx}}\left( {\frac{{{x^4}}}{4} + \frac{1}{{8{x^2}}}} \right)} \right)}^2}}  = \sqrt {1 + {{\left( {{x^3} - \frac{1}{{4{x^3}}}} \right)}^2}}  =

    = \sqrt {{x^6} + \frac{1}{2} + \frac{1}{{16{x^6}}}}  = \sqrt {{{\left( {{x^3} + \frac{1}{{4{x^3}}}} \right)}^2}}  = {x^3} + \frac{1}<br />
{{4{x^3}}}{\text{ }}\left( {1 \leqslant x \leqslant 3} \right).

    So we have

    {L_y} = \int\limits_1^3 {\left( {{x^3} + \frac{1}{{4{x^3}}}} \right)dx}  = \left. {\left( {\frac{{{x^4}}}{4} - \frac{1}{{8{x^2}}}} \right)} \right|_1^3 =

    = \frac{{81}}{4} - \frac{1}{{72}} - \left( {\frac{1}{4} - \frac{1}{8}} \right) = \frac{{1457}}{{72}} - \frac{1}{8} = \frac{{1448}}{{72}} = \frac{{181}}{9}.

    See picture

    Last edited by DeMath; July 30th 2009 at 02:12 AM. Reason: picture
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