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Math Help - equation of normal problem

  1. #1
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    equation of normal problem

    The gradient function of a curve is kx^2-x, where k is constant.The equation of the normal to the curve at point (1,-2) is 5y+x=7..


    a)Find the value of k

    b)the equation of the curve.
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  2. #2
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    The normal is

    5y + x = 7

    y = -\frac{1}{5}x + \frac{7}{5}.


    Therefore, the gradient of the tangent will be 5, since the tangent and normal are perpendicular.


    So the gradient function at the point (1, -2) has a value of 5.

    Substituting into the derivative gives

    k(1)^2 - 1 = 5

    k - 1 = 5

    k = 6.


    Now that you know k, you know

    \frac{dy}{dx} = 6x^2 - x

    y = \int{6x^2 - x\,dx}

    y = 2x^3 - \frac{x^2}{2} + C.


    Since (1, -2) lies on this curve

    -2 = 2(1)^3 - \frac{1^2}{2} + C

    -2 = 2 - \frac{1}{2} + C

    -\frac{4}{2} = \frac{3}{2} + C

    -\frac{7}{2} = C.


    Thus the equation of the curve is y = 2x^3 - \frac{x^2}{2} - \frac{7}{2}.
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