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Math Help - Finding the Arclength

  1. #1
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    Finding the Arclength

    Use the arc length formula to find the length L of the curve below, 0 ≤ x2. Check your answer by noting that the curve is part of a circle.
    y=sqrt(8-x^2)

    Here is my work:
    ds = sqrt(1+(dy/dx)^2)

    dy/dx = x/sqrt(8-x^2)

    s=int(sqrt(1-(x^2/(x^2-8)))) from 0 to 2

    where ds is a smal portion of the arc and s is the total arclength

    My problem comes from where s is equal to the integral, I don't know how to integrate it. Thanks for any help!
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  2. #2
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    Quote Originally Posted by dillonmhudson View Post
    Use the arc length formula to find the length L of the curve below, 0 ≤ x2. Check your answer by noting that the curve is part of a circle.
    y=sqrt(8-x^2)

    Here is my work:
    ds = sqrt(1+(dy/dx)^2)

    dy/dx = x/sqrt(8-x^2)

    s=int(sqrt(1-(x^2/(x^2-8)))) from 0 to 2

    where ds is a smal portion of the arc and s is the total arclength

    My problem comes from where s is equal to the integral, I don't know how to integrate it. Thanks for any help!
    let r^2 = 8

    y = \sqrt{r^2-x^2}

    y' = \frac{-x}{\sqrt{r^2-x^2}}

    (y')^2 = \frac{x^2}{r^2-x^2}

    1+(y')^2 = \frac{r^2-x^2}{r^2-x^2} + \frac{x^2}{r^2-x^2} = \frac{r^2}{r^2-x^2} = \frac{1}{1 - \left(\frac{x}{r}\right)^2}<br />

    \sqrt{\frac{1}{1 - \left(\frac{x}{r}\right)^2}} = \frac{1}{\sqrt{1 - \left(\frac{x}{r}\right)^2}}


    \int_0^2 \frac{1}{\sqrt{1 - \left(\frac{x}{r}\right)^2}} \, dx

    r \int_0^2 \frac{\frac{1}{r}}{\sqrt{1 - \left(\frac{x}{r}\right)^2}} \, dx

    r \left[ \arcsin\left(\frac{x}{r}\right) \right]_0^2

    r \arcsin\left(\frac{2}{r}\right)

    2\sqrt{2} \arcsin\left(\frac{1}{\sqrt{2}}\right)

    2\sqrt{2} \cdot \frac{\pi}{4} = \frac{\pi \sqrt{2}}{2}
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