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Thread: 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


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

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

    $\displaystyle Gradient = -0.118$ to $\displaystyle 3sf$

    Equation of tangent:
    $\displaystyle y - (-8) = -0.118 (x - 5)$
    $\displaystyle y = -0.118x -7.41$
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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


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

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

    -Dan
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