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Math Help - Curve Lengths

  1. #1
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    Curve Lengths

    Don't have any real idea how to go about this one, even thought its probably really easy... thnanks for the help

    Find the length of the curve:
    x=[3y^(4/3)]-(3/32)y^(2/3)

    -27<y<27 (both those are less than or equal to)

    thanks
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  2. #2
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    Hello, N736RA!

    Find the length of the curve: .  <br />
x\:=\:3y^{\frac{4}{3}} - \frac{3}{32}y^{\frac{2}{3}} \qquad -27 \:\leq\: y\:\leq \:27
    We will use this formula: . L \;=\;\int^b_a\sqrt{1 + \left(\frac{dx}{dy}\right)^2}\,dy


    We have: . . . . .  x \;=\;3y^{\frac{4}{3}} - \frac{3}{32}y^{\frac{2}{3}}

    Differentiate: . \frac{dx}{dy} \;=\;4y^{\frac{1}{3}} - \frac{1}{16}y^{-\frac{1}{3}}

    . . . . . . . . \left(\frac{dx}{dy}\right)^2 \;=\;16y^{\frac{2}{3}} - \frac{1}{2} + \frac{1}{256}y^{-\frac{2}{3}}

    . . . . . . 1 + \left(\frac{dx}{dy}\right)^2 \;=\;1 + \left(16y^{\frac{2}{3}} - \frac{1}{2} + \frac{1}{256}y^{-\frac{2}{3}}\right) \;=\;16y^{\frac{2}{3}} + \frac{1}{2} + \frac{1}{256}y^{-\frac{2}{3}} \;=\;\left(4y^{\frac{1}{3}} + \frac{1}{16}y^{-\frac{1}{3}}\right)^2<br />

    . . . . \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \;=\;4y^{\frac{1}{3}} + \frac{1}{16}y^{-\frac{1}{3}}


    You must integrate and evaluate: . L \;=\;\int^{27}_{\text{-}27}\left(4y^{\frac{1}{3}} + \frac{1}{16}y^{-\frac{1}{3}}\right)\,dy

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  3. #3
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    thanks for the reply! Although, when I do that integral I get 0 (which makes sense), yet the program I'm using tells me that 0 isnt the right answer, any ideas? Edit: nevermind, I figured it out, thanks for your help!
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