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Arc length help

  1. Mar 12, 2008 #1

    rock.freak667

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    1.The problem statement, all variables and given/known data

    [tex]y=x^{\frac}{1}{2}}- \frac{1}{3}x^{\frac{3}{2}}+\lambda[/tex]

    For [itex]0 \leq x \leq 3[/itex]
    Show that the arc length,s=[itex]2\sqrt{3}[/itex]
    2. Relevant equations
    [tex]s=\int_{x_1} ^{x_2} \sqrt{1+ (\frac{dy}{dx})^2} dx[/tex]


    3. The attempt at a solution


    [tex]\frac{dy}{dx}=\frac{1}{2\sqrt{x}} - \frac{1}{2}x^{\frac{1}{2}}[/tex]

    [tex] (\frac{dy}{dx})^2=\frac{1}{4x}-\frac{1}{2}+\frac{x}{2}[/tex]

    [tex]s=\int_{0} ^{3} \sqrt{\frac{1}{4x}+\frac{x}{2}+\frac{1}{2}} dx[/tex]

    and now that integral seems a bit too odd to get the answer [itex]2\sqrt{3}[/itex]
     
  2. jcsd
  3. Mar 13, 2008 #2

    Gib Z

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    You made a tiny mistake when you squared the derivative, the coefficient of the linear term should be 1/4.

    You integral, when corrected, then becomes

    [tex]s= \int^3_0 \sqrt{ \frac{1}{4x} + \frac{x}{4} + \frac{1}{2} } dx[/tex]

    [tex]= \frac{1}{2} \int^3_0 2\sqrt{ \frac{1}{4x} + \frac{x}{4} + \frac{1}{2} } dx[/tex]

    [tex]= \frac{1}{2} \int^3_0 \sqrt{x + 2 + \frac{1}{x} } dx[/tex]

    I think I've already given out too much =]. Now just ignore that the bottom limit of the integral is x=0, or be rigorous and take limits instead.
     
  4. Mar 13, 2008 #3
    It might also be a good idea to think about completing the square ...
     
  5. Mar 13, 2008 #4

    rock.freak667

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    I was trying to do that but I kept on squaring it wrongly...even though I squared it out like 9 times...Seems I have to keep my eye out for those tiny mistakes...Thanks Gib Z & Pere Callahan!
     
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