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Thread: Need help with setup for arc length of x^(2/3) + y^(2/3) = 4

  1. #1
    Junior Member
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    Need help with setup for arc length of x^(2/3) + y^(2/3) = 4

    Find the arc length in the second quadrant of the curve

    x^(2/3) + y^(2/3) = 4 from x = -8 to x = -1.

    Ok what I'm thinking i should do is:

    y^(2/3) = 4 - x^(2/3)

    But here's where I get stuck.

    I'm thinking I should ^(1/6) both sides, but that gets really ugly. It seems to me there should be an easier way to set this up.

  2. #2
    Elite Member
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    This thing is an astroid, I believe. There's symmetry, so you can change your limits of integration to positive values, 1 and 8.

    Solve for y, then use the arc length formula integral.

    [tex]\L\\y=(4-x^{\frac{2}{3}})^{\frac{3}{2}}[/tex]

    The derivative is:

    [tex]\L\\\frac{-\sqrt{4-x^{\frac{2}{3}}}}{x^{\frac{1}{3}}}[/tex]

    Square it and get:

    [tex]\L\\\frac{4-x^{\frac{2}{3}}}{x^{\frac{2}{3}}}[/tex]

    [tex]1+\frac{4-x^{\frac{2}{3}}}{x^{\frac{2}{3}}}=\frac{4}{x^{\fra c{2}{3}}}[/tex]

    [tex]\L\\\sqrt{\frac{4}{x^{\frac{2}{3}}}}=\frac{2}{x^{\ frac{1}{3}}}[/tex]

    Now, integrate:

    [tex]\L\\\int_{1}^{8}\frac{2}{x^{\frac{1}{3}}}dx=3x^{\f rac{2}{3}}[/tex]

    [tex]\L\\3(8)^{\frac{2}{3}}-3(1)^{\frac{2}{3}}=12-3=9[/tex]

  3. #3
    arc length =[1+[dy/dx]^2]^1/2 dx integrated from -8 to -1

    x^2/3 + y^2/3 =4
    let us determine dy /dx
    2/3 x^(-1/3)dx +2/3 y^(-1/3)dy=0
    dy/y^1/3 =- dx/x^1/3
    dy/dx = - [y/x]^1/3

    arc length = [1+ [-[y/x]^1/3]^2 ]^1/2 dx integrated from -8 to -1
    arc length = [1+y^2/3/x^2/3]^1/2 dx int. from -8 to -1

    but y^2/3 = 4-x^2/3
    arc length = [1 + [4-x^2/3]/x^2/3 ] ^1/2 dx int from -8 to -1
    arc length = [x^2/3 +4-x^2/3]/x^2/3 ]^1/2 dx int fro -8 to -1
    arc length = [4/x^2/3]^1/2 dx int. from -8 to -1
    arc length = 2/x^1/3 dx int. from -8 to -1
    arc length = 2 int [-8,-1]x^-1/3 dx
    arc length =2 x^2/3 [3/2] evaluated at -1,-8
    arc length =2[1] - 2[4]
    arc length = -6 answer

    I assume I lost a minus sign, but the technique is right
    Arthur

  4. #4
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    Re: Need help with setup...

    Hello, Hank!

    Find the arc length in the second quadrant of the curve:
    . . [tex]x^{\frac{2}{3}}\,+\,y^{\frac{2}{3}} \:=\:4[/tex] from [tex]x\,=\,-8[/tex] to [tex]x\,=\,-1[/tex]

    [tex]\text{Solve for }y:\;\;y^{\frac{2}{3}}\:=\:4\,-\,x^{\frac{2}{3}}\;\;\Rightarrow\;\;y\;=\;\left(4\ ,-\,x^{\frac{2}{3}}\right)^{\frac{3}{2}}[/tex]

    Hence: [tex]\L\:\frac{dy}{dx} \:=\:\frac{3}{2}\left(4\,-\,x^{\frac{2}{3}}\right)^{\frac{1}{2}}\left(-\frac{2}{3}x^{-\frac{1}{3}}\right) \;=\;-\frac{\sqrt{4\,-\,x^{\frac{2}{3}}}}{x^{\frac{1}{3}}}[/tex]

    Then: [tex]\L\:1\,+\,\left(\frac{dy}{dx}\right)^2 \:=\:1\,+\,\frac{4\,-\,x^{\frac{2}{3}}}{x^{\frac{2}{3}}} \;=\;\frac{4}{x^{\frac{2}{3}}}[/tex]

    Hence: [tex]\L\:ds\;=\;\sqrt{\frac{4}{x^{\frac{2}{3}}}} \;=\;\frac{2}{x^{\frac{1}{3}}} \;=\;2x^{-\frac{1}{3}}[/tex]

    Can you finish it now?

    I'm the other of the two guys who "do" homework.

  5. #5
    I found my error
    from arc length = [1+[4-x^2/3]/x^2/3]^1/2 dx int. from -8 to -1
    arc length = [4/x^2/3]^1/2 dx int from -8 to -1
    arc length = +/-2 x^(-1/3) dx int from -8 to -1
    arc length = +/-2 x^2/3 / 2/3 evaluated at -1,-8
    arc length = +/-3 x^2/3 evaluated at -1,-8
    arc length = +/- 3 [1-4]
    arc length= +/- 9
    arc length =9 answer, -9 is a extraneous answer, arc length must be positive

    Arthur

  6. #6
    Junior Member
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    Thanks tons, everyone.

    What screwed me up was simply solving for y.

    Shoulda been 3/2 instead of 1/6. Thx.

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