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Math Help - Length between curve

  1. #1
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    Length between curve

    I need to find the length of the curve y=(1/2)x^2 - (1/4)lnx between x=1 and x=8.

    I derived it and squared it and added one to it so now I'm left with the integral from 1 to 8 of sqrt (x^2 + (1/2) + 1/16x^2) dx. How do I integrate this?
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  2. #2
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    L = \int_1^8 \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ dx

    Now, as you found: \frac{dy}{dx} = x - \frac{1}{4x}

    So:
    \begin{aligned} L & = \int_1^8 \sqrt{1 + \left(x - \frac{1}{4x}\right)^2} \ dx \\ & = \int_1^8 \sqrt{{\color{red}\left(\frac{x^2}{x^2}\right)}\le  ft(x^2 + \frac{1}{16x^2} + \frac{1}{2}\right)} \ dx \\ & =  \int_1^8 \sqrt{\frac{x^4 +\frac{1}{2}x^2 + \frac{1}{16}}{x^2}} \ dx \\ & = \int_1^8 \sqrt{\frac{\left(x + \frac{1}{4}\right)^2}{x^2}}  \end{aligned}

    So, taking the square root and our final integral becomes: L = \int_1^8 \frac{x + \frac{1}{4}}{x} \ dx \ = \ \int_1^8 \left(1 + \frac{1}{4x}\right) dx
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  3. #3
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    Hello, elitespart!

    Find the length of the curve: . y\:=\:\tfrac{1}{2}x^2 - \tfrac{1}{4}\ln x between x=1 and x=8.

    I derived it and squared it and added one to it.

    So now I'm left with: . L \;=\;\int^8_1\sqrt{x^2 + \tfrac{1}{2} + \tfrac{1}{16x^2}}\,dx
    How do I integrate this?
    Believe it or not, there is a perfect square under the radical . . .

    . . \sqrt{x^2 + \frac{1}{2} + \frac{1}{16x^2} }\;=\;\sqrt{\left(x + \frac{1}{4x}\right)^2} \;=\;x + \frac{1}{4x} \quad\hdots see?

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  4. #4
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    Awesome! Thanks a lot guys.
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  5. #5
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    It's actually \sqrt{\left( x+\frac{1}{4} \right)^{2}\cdot \frac{1}{x^{2}}}=\left| \left( x+\frac{1}{4} \right)\cdot \frac{1}{x} \right|, but note that \left( x+\frac{1}{4} \right)\cdot \frac{1}{x}>0 for 1\le x\le8, hence \left| \left( x+\frac{1}{4} \right)\cdot \frac{1}{x} \right|=\left( x+\frac{1}{4} \right)\cdot \frac{1}{x}.
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