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Arc Length integration always unusually difficult? Any special trick?

  1. Oct 30, 2012 #1

    B3NR4Y

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    Gold Member

    I don't understand why finding the arc length is always difficult. I understand the formula and know pretty much all the integration methods, but whenever I try to find the arc length of a function like 8x2 = 27y3 from 1 to 8 it's unusually difficult.
    I would start by solving for x
    \begin{equation}
    \begin{split}
    &\frac{2}{3} x^{\frac{2}{3}} = y\\
    &\frac{4}{9x^{\frac{1}{3}}} = y'\\
    \end{split}
    \end{equation}
    Now I add this into the integration formula [itex]\int_a^b \sqrt{1+(y')^{2}} \, dx[/itex]
    [tex]\begin{equation}
    \begin{split}
    L^{8}_{1} &= \int_1^8 \, \sqrt{1+ (\frac{4}{9x^{\frac{1}{3}}})^{2}} \, dx \\
    &= \int_1^8 \, \sqrt{1+ (\frac{16}{81x^{\frac{2}{3}}})} \, dx \\
    &= \int_1^8 \sqrt{\frac{81x^{\frac{2}{3}}+16}{81x^{\frac{2}{3}}}} \, dx \\
    &= \int_1^8 \frac{\sqrt{81x^{\frac{2}{3}}+16}}{9x^{\frac{1}{3}}} \, dx
    \end{split}
    \end{equation}[/tex]
    From here I would do a u-substitution of [itex]u=81x^{\frac{2}{3}}+16[/itex] and then take the derivative to find [itex]\frac{1}{54} du = \frac{1}{x^{\frac{1}{3}}} dx [/itex]
    My new limits would be 97 and 340
    \begin{equation}
    \begin{split}
    L^{8}_{1} &= \frac{1}{9*54} \int_{97}^{340} \sqrt{u} \, du \\
    &= \frac{1}{486} \Bigg(\frac{3}{2} u^{\frac{3}{2}} |^{97}_{340} \Bigg) \\
    &=7.28... \\
    \end{split}
    \end{equation}
    I got the question right, by going slowly and typing it all out, derp. Anyway, is there a special trick for these types of problems? Or is it to just go slow and remember rules.
     
    Last edited by a moderator: Oct 30, 2012
  2. jcsd
  3. Oct 31, 2012 #2

    lurflurf

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    Homework Helper

    Those are often messy because of the form of the integrals.
    Many calculus books choose y to make the integral simple.
    Even simple y live x^3, sin(x), and 1/x give non-elementary integrals.
     
  4. Oct 31, 2012 #3

    arildno

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    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Finding the arc length of the hyperbolic cosine curve is, however, fairly trivial.
    :smile:
     
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