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Thread: Applications of Differentiation

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    Applications of Differentiation

    The parametric equations of a curve are x = cos 4A + 2cos 2A, y = sin 4A - 2sin2A where A is the parameter. Find (a ) dy/dx and (b) equation of the normal to the curve at the point where A = pi/4
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    Quote Originally Posted by puggie View Post
    The parametric equations of a curve are x = cos 4A + 2cos 2A, y = sin 4A - 2sin2A where A is the parameter. Find (a ) dy/dx and (b) equation of the normal to the curve at the point where A = pi/4
    For part a) you need to use the Chain Rule.

    Recall that $\displaystyle \frac{dy}{dx} = \frac{dy}{dA}\times\frac{dA}{dx}$.


    $\displaystyle \frac{dy}{dA} = 4\cos{(4A)} - 4\cos{(2A)}$


    $\displaystyle \frac{dA}{dx} = \frac{1}{\frac{dx}{dA}}$

    $\displaystyle = \frac{1}{-4\sin{(4A)} - 4\sin{(2A)}$.


    So $\displaystyle \frac{dy}{dx} = \frac{4\cos{(4A)} - 4\cos{(2A)}}{-4\sin{(4A)} - 4\sin{(2A)}}$

    $\displaystyle = -\frac{\cos{(4A)} - \cos{(2A)}}{\sin{(4A)} + \sin{(2A)}}$.


    For part b) , you need to remember that a normal is a LINE.

    So it is of the form $\displaystyle y - y_1 = m(x - x_1)$.


    At the point $\displaystyle A = \frac{\pi}{4}$...


    $\displaystyle x_1 = \cos{\pi} + 2\cos{\frac{\pi}{2}} = -1$


    $\displaystyle y_1 = \sin{\pi} - 2\sin{\frac{\pi}{2}} = -2$.



    The gradient of the curve at point A is $\displaystyle \frac{dy}{dx} = -\frac{\cos{\pi} - \cos{\frac{\pi}{2}}}{\sin{\pi} + \sin{\frac{\pi}{2}}}$

    $\displaystyle = -\frac{-1}{1}$

    $\displaystyle = 1$.


    The gradient of the normal is the negative reciprocal. So $\displaystyle m = -1$.


    Thus the normal's equation is given by

    $\displaystyle y - (-2) = -1[x - (-1)]$

    $\displaystyle y + 2 = -x - 1$

    $\displaystyle y = - x - 3$.



    Hope that helped.
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