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Math Help - Slope of the tangent line to the curve

  1. #1
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    Slope of the tangent line to the curve



    at

    Please help. I'm so confused
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  2. #2
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    Quote Originally Posted by Bilbo94 View Post


    at

    Please help. I'm so confused
    f(x) = 3x^2 - 4xy + 3y^3 = -1.3 e + 3
    f'(x) = 6x - 4y - 4x \frac{dy}{dx} + 9 y^2 \frac{dy}{dy} = -1.3e
    \implies 6x - 4y +1.3e = 4x \frac{dy}{dx} - 9y^2 \frac{dy}{dy}
     6x - 4y +1.3e = \frac{dy}{dx} (4x - 9y^2)
    \frac{dy}{dx} = \frac{6x - 4y + 1.3e}{4x - 9y^2}

    Gradient found when (5 , -8) substituted into \frac{dy}{dx}.

    Gradient = -0.118 to 3sf

    Equation of tangent:
    y - (-8) = -0.118 (x - 5)
    y =  -0.118x -7.41
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  3. #3
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    I followed all your steps, but the homework module still says it's wrong.
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  4. #4
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    Quote Originally Posted by Bilbo94 View Post


    at

    Please help. I'm so confused
    Even assuming that Air's calculations are correct, there is still something funny about how this is written.
    2x^2 - 4xy +3y^3 = -1.3e + 03

    Could this possibly be
    2x^2 - 4sy + 3y^3 = -1.3 \times 10^3?

    -Dan
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