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Thread: Another arc length

  1. #1
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    Another arc length

    Find the length of the arc formed by

    $\displaystyle y= \frac{1}{8}(-4x^{2}+2ln(x))$ from x=2 to x=6.
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  2. #2
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    Arc lenght formula states

    $\displaystyle \int_a^b\sqrt{1+(y')^2}\,dx$

    Apply it.
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  3. #3
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    I did, but then I have a crazy integral to deal with.
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  4. #4
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    Hello, tttcomrader!

    Watch your algebra . . . The integral comes out rather neat . . .


    Find the length of the arc formed by

    $\displaystyle y\:= \:\frac{1}{8}\left[-4x^{2}+2\ln(x))\right]$ .from $\displaystyle x=2$ to $\displaystyle x=6.$

    $\displaystyle y' \:=\:\frac{1}{8}\left(-8x + \frac{2}{x}\right) \;=\;\frac{1}{4x} - x$

    $\displaystyle (y')^2 \;=\;\left[\frac{1}{4x} - x\right]^2 \;=\;\frac{1}{16x^2} - \frac{1}{2} + x^2$

    $\displaystyle 1 + (y')^2\;=\;1 + \left(\frac{1}{16x^2} - \frac{1}{2} + x^2\right) \;=\;\frac{1}{16x^2} + \frac{1}{2} + x^2 \;=\;\left(\frac{1}{4x} + x\right)^2$

    . . Hence: .$\displaystyle \sqrt{1 + (y')^2} \;=\;\sqrt{\left(\frac{1}{4x} + x\right)^2} \;=\;\frac{1}{4x} + x $


    Therefore: .$\displaystyle L \;=\;\int^6_2\left(\frac{1}{4x} + x\right)\,dx $

    I assume you can finish it . . .

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