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    Need help on this question please

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    dx/dt = 2sin2t
    dy/dt = 2cos2t

    gradient of the normal = -dx/dy = -tan2t
    at P : gradient = -tanpi/3
    y = sinpi/3
    x = 1- cospi/3

    y - sinpi/3 = -tanpi/3 (x-(1-cospi/3))
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    Substituting \large{ t = \frac{\pi}{6} } into the equations for x and y gives:

    \large{ y = \sin \frac{2 \pi}{6} = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \, \, \, x = 1 - \cos \frac{2 \pi}{6} = 1 - \cos \frac{\pi}{3} = 1 - \frac{1}{2} = \frac{1}{2} }

    You should know that \large{ \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}} (*)

    Thus, differentiating y and x with respect to t we get:

    \large{\frac{dy}{dt} = 2 \cos 2t \, \, \frac{dx}{dt} = 2 \sin 2t }

    Now using (*)

    \large{\frac{dy}{dx} = \frac{ 2 \cos 2t}{2 \sin 2t} = \cot 2t}

    Therefore, the gradient where \large{ t = \frac{\pi}{6} } is \large{ \cot \frac{2 \pi}{6} = \cot \frac{\pi}{3} = \frac{1}{\sqrt{3}}}

    Next, using \large{ m_1 m_2 = -1} , the gradient of the normal is \large{ \frac{-1}{(\frac{1}{\sqrt{3}})} = -\sqrt{3}}

    Finally, using \large{y - y_1 = m(x - x_1) } gives

    \large{y - \frac{\sqrt{3}}{2} = -\sqrt{3}(x - \frac{1}{2})} which rearranges to \large{y = -\sqrt{3}x + \sqrt{3}}

    EDIT: beaten to it
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    (Original post by fatuous_philomath)

    EDIT: beaten to it
    lol, welcome to the maths subforum during exam season :rolleyes:
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    Indeed.

    Although procrastinating by typing posts requiring extensive use of LaTeX is far more enjoyable than revising for the A level exams I have next week, whether or not my post is useful, and at least I get to learn LaTeX a little better.
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    Can you please explain how you differentiate x and y in this question. I can't seem to get the answer you are getting
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    x = 1-cos2t
    dx/dt = 0 + - (-1)(2)(sin2t)

    y = sin2t
    dy/dt = (2)cos2t
 
 
 
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