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Math Help - Equation of the tangent to the curve.

  1. #1
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    Equation of the tangent to the curve. [Solved]

    Find the Equation of the tangent to the curve of e^2y + xy= 4x + 1,

    What I've done so far,

    e^2y + xy= 4x + 1

    e^2y + xy - 4x - 1 = 0

    dy/dx = 2ye^2y + 1 - 4 = 0

    dy/dx = 2ye^2y - 3 = 0

    I'm not sure if I've done it right because there is no x value, does this mean I sub y=0 into dy/dx instead.

    Any help would be appreciated.
    Last edited by driverfan2008; April 22nd 2009 at 07:56 PM.
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  2. #2
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    Quote Originally Posted by driverfan2008 View Post
    Find the Equation of the tangent to the curve of e^2y + xy= 4x + 1,

    What I've done so far,

    e^2y + xy= 4x + 1

    e^2y + xy - 4x - 1 = 0

    dy/dx = 2ye^2y + 1 - 4 = 0

    dy/dx = 2ye^2y - 3 = 0

    I'm not sure if I've done it right because there is no x value, does this mean I sub y=0 into dy/dx instead.

    Any help would be appreciated.
    This should be in the calculus section, but never mind...

    Since this equation can be solved for x, use the formula \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}.


    e^{2y} + xy = 4x + 1

    xy - 4x = 1 - e^{2y}

    x(y - 4) = 1 - e^{2y}

    x = \frac{1 - e^{2y}}{y - 4}

    \frac{dx}{dy} = \frac{(y - 4)\frac{d}{dy}(1 - e^{2y}) - (1 - e^{2y})\frac{d}{dy}(y - 4)}{(y - 4)^2}

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

    \frac{dy}{dx} = \frac{(y - 4)^2}{e^{2y}(4 - y) + e^{2y} - 1}.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    This should be in the calculus section, but never mind...

    Since this equation can be solved for x, use the formula \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}.


    e^{2y} + xy = 4x + 1

    xy - 4x = 1 - e^{2y}

    x(y - 4) = 1 - e^{2y}

    x = \frac{1 - e^{2y}}{y - 4}

    \frac{dx}{dy} = \frac{(y - 4)\frac{d}{dy}(1 - e^{2y}) - (1 - e^{2y})\frac{d}{dy}(y - 4)}{(y - 4)^2}

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

    \frac{dy}{dx} = \frac{(y - 4)^2}{e^{2y}(4 - y) + e^{2y} - 1}.
    Oh sorry for posting in the wrong area. Many thanks for your help its very appreciated =)
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