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Math Help - Length of the Curve

  1. #1
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    Length of the Curve

    Find the exact length of the curve analytically by antidifferentiation. You will need to simplify the integrand algebraically before finding an antiderivative.


    x= \frac{y^3}{3} + \frac{1}{4y}
    from y=1 to y= 3

    Hint: 1 + (\frac{dy}{dx})^2) is a perfect square.

    I don't see the perfect square.

    Thanks.
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  2. #2
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    Quote Originally Posted by Truthbetold View Post
    Find the exact length of the curve analytically by antidifferentiation. You will need to simplify the integrand algebraically before finding an antiderivative.


    x= \frac{y^3}{3} + \frac{1}{4y}
    from y=1 to y= 3

    Hint: 1 + (\frac{dy}{dx})^2) is a perfect square.

    I don't see the perfect square.

    Thanks.
    1 = \left ( y^2 - \frac{1}{4y^2} \right ) \frac{dy}{dx}

    \frac{dy}{dx} = \frac{4y^2}{4y^4 - 1}

    So
    1 + \left ( \frac{dy}{dx} \right )^2

    = 1 + \left ( \frac{4y^2}{4y^4 - 1} \right )^2

    = \frac{(4y^4 - 1)^2 + 16y^4}{(4y^4 - 1)^2}

    = \frac{16y^8 - 8y^4 + 1 + 16y^4}{(4y^4 - 1)^2}

    = \frac{16y^8 + 8y^4 + 1}{(4y^4 - 1)^2}

    = \frac{(4y^4 + 1)^2}{(4y^4 - 1)^2}

    -Dan
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