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Math Help - Finding the equation using Implict differentiation

  1. #1
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    Finding the equation using Implict differentiation

    Hi
    I need help on the following question;

    Find the equation to the normal to the curve x^3 + y^3 -2xy + 4x = 8 at the point (0,2)

    I have found the eqaution but i don't understand why the answer is x=0.

    3x^2 + 3y^2 \frac{dy}{dx} - (2x \frac{dy}{dx} + 2y) +4 = 0

    \frac{dy}{dx} 3y^2 \frac{dy}{dx} +2y -3x^2

    \frac{dy}{dx} = \frac{-4-2y-3x^2}{3y^2-2x}

    to find the equation i sub the values (0,2)

    which gets: \frac{-2}{3}

    so equation i get is y=\frac{3}{2}x + 2

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    I need help on the following question;

    Find the equation to the normal to the curve x^3 + y^3 -2xy + 4x = 8 at the point (0,2)

    I have found the eqaution but i don't understand why the answer is x=0.

    3x^2 + 3y^2 \frac{dy}{dx} - (2x \frac{dy}{dx} + 2y) +4 = 0

    \frac{dy}{dx} 3y^2 \frac{dy}{dx} +2y -3x^2

    \frac{dy}{dx} = \frac{-4-2y-3x^2}{3y^2-2x}

    to find the equation i sub the values (0,2)

    which gets: \frac{-2}{3}

    so equation i get is y=\frac{3}{2}x + 2

    P.S
    sign error ...

    \frac{dy}{dx} = \frac{-4+2y-3x^2}{3y^2-2x}

    have a look see at the graph ...
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  3. #3
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    that means when i sub (0,2) into the equation \frac{dy}{dx} = 0, i don't understand how the book's answer is x=0,
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  4. #4
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    Quote Originally Posted by Paymemoney View Post
    that means when i sub (0,2) into the equation \frac{dy}{dx} = 0, i don't understand how the book's answer is x=0,
    the tangent line at (0,2) is y = 2

    the line perpendicular to a horizontal line is a vertical line ... the vertical line passing through the point (0,2) is x = 0 ... the y-axis.
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