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Thread: equation of normal problem

  1. #1
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    equation of normal problem

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


    a)Find the value of $\displaystyle k$

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

    $\displaystyle 5y + x = 7$

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


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


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

    Substituting into the derivative gives

    $\displaystyle k(1)^2 - 1 = 5$

    $\displaystyle k - 1 = 5$

    $\displaystyle k = 6$.


    Now that you know $\displaystyle k$, you know

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

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

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


    Since $\displaystyle (1, -2)$ lies on this curve

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

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

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

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


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