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Thread: Arc length of an eqation

  1. #1
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    Arc length of an eqation



    from x=6 and x=9

    I need to find the arc length of this equation:

    What I know: The equation for arc length is [integral of sqrt(1 + (f'(x))^2)]

    What I'm stuck on: When I plug the equation into the arc length equation, I get lost. The hard part for me is after I find the number under the square root, then I have NO IDEA on how to do the integral under the square root!

    Help please, thanks
    Last edited by johnnycho; Sep 28th 2009 at 03:47 PM.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by johnnycho View Post


    I need to find the arc length of this equation:

    What I know: The equation for arc length is [integral of sqrt(1 + (f'(x))^2)]

    What I'm stuck on: When I plug the equation into the arc length equation, I get lost. The hard part for me is after I find the number under the square root, then I have NO IDEA on how to do the integral under the square root!

    Help please, thanks
    Recall the formula

    $\displaystyle s=\int_a^b\sqrt{1+[y']^2}dx$

    So, we need $\displaystyle y'$

    $\displaystyle
    y'=3\frac{u'}{u}=3\frac{\frac{2}{9}x}{\frac{x^2}{9 }-1}
    =\frac{6x}{x^2-9}$

    Now we have

    $\displaystyle s=\int_6^9\sqrt{1+\left(\frac{6x}{x^2-9}\right)^2}dx$

    Now simplifying

    $\displaystyle s=\int_6^9\frac{\sqrt{x^4+18x^2+81}}{\sqrt{x^4-18x^2+81}}dx$


    Can you see that these are perfect trinomial squares?

    Once you factor them, the radicals will dissapear...
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by johnnycho View Post


    from x=6 and x=9

    I need to find the arc length of this equation:

    What I know: The equation for arc length is [integral of sqrt(1 + (f'(x))^2)]

    What I'm stuck on: When I plug the equation into the arc length equation, I get lost. The hard part for me is after I find the number under the square root, then I have NO IDEA on how to do the integral under the square root!

    Help please, thanks

    I'm sorry, I'll edit it right away.
    It is : x=6 and x=9
    $\displaystyle y = 3\ln \left( {{{\left( {\frac{x}{3}} \right)}^2} - 1} \right) = 3\ln \left( {\frac{{{x^2}}}{9} - 1} \right).$

    $\displaystyle y' = \frac{3}{{\frac{{{x^2}}}
    {9} - 1}} \cdot {\left( {\frac{{{x^2}}}
    {9} - 1} \right)^\prime } = \frac{{27}}
    {{{x^2} - 9}} \cdot \frac{{2x}}
    {9} = \frac{{6x}}
    {{{x^2} - 9}}.$

    $\displaystyle \sqrt {1 + {{\left( {\frac{{6x}}
    {{{x^2} - 9}}} \right)}^2}} = \sqrt {1 + \frac{{36{x^2}}}
    {{{{\left( {{x^2} - 9} \right)}^2}}}} = \sqrt {\frac{{{{\left( {{x^2} - 9} \right)}^2} + 36{x^2}}}
    {{{{\left( {{x^2} - 9} \right)}^2}}}} =$

    $\displaystyle = \sqrt {\frac{{{x^4} - 18{x^2} + 81 + 36{x^2}}}
    {{{{\left( {{x^2} - 9} \right)}^2}}}} = \sqrt {\frac{{{x^4} + 18{x^2} + 81}}
    {{{{\left( {{x^2} - 9} \right)}^2}}}} =$

    $\displaystyle = \sqrt {\frac{{{{\left( {{x^2} + 9} \right)}^2}}}
    {{{{\left( {{x^2} - 9} \right)}^2}}}} = \left| {\frac{{{x^2} + 9}}
    {{{x^2} - 9}}} \right| = \left\{ \begin{gathered}
    \frac{{{x^2} + 9}}{{{x^2} - 9}}\qquad{\text{if}}\, x \in \left( { - \infty ; \, - 3} \right) \cup \left( {3; \,+ \infty } \right), \hfill \\
    - \frac{{{x^2} + 9}}
    {{{x^2} - 9}} \quad \, {\text{if}}\, x \in \left( { - 3;\,3} \right). \hfill \\
    \end{gathered} \right.$

    $\displaystyle {L_y} = \int\limits_{{x_1}}^{{x_2}} {\sqrt {1 + {{\left( {y'} \right)}^2}}\,dx} = \int\limits_6^9 {\frac{{{x^2} + 9}}
    {{{x^2} - 9}}\,dx} = \ldots$
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  4. #4
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    Thanks so much for your help.
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