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Math Help - Parametric question...confused on trig =/

  1. #1
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    Parametric question...confused on trig =/

    Question:
    A curve has parametric equations
    x = 3 cos2 t, y = sin 2t, 0 ≤ t < π.
    (i) Show that
    = \frac{dy}{dx} = -\frac{2}{3}cot 2t
    (ii) Find the coordinates of the points where the tangent to the curve is parallel to the x-axis

    Ok, the first bit i did no problem, but its the second part because the gradient must be 0 correct?
    meaning cot 2t = 0
    However you cannot do the inverse of this because when cot t = 0...its an asymptote...

    Thanks for any help
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  2. #2
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    Quote Originally Posted by AshleyT View Post
    Question:
    A curve has parametric equations
    x = 3 cos2 t, y = sin 2t, 0 ≤ t < π.
    (i) Show that
    = \frac{dy}{dx} = -\frac{2}{3}cot 2t
    (ii) Find the coordinates of the points where the tangent to the curve is parallel to the x-axis

    Ok, the first bit i did no problem, but its the second part because the gradient must be 0 correct?
    meaning cot 2t = 0
    However you cannot do the inverse of this because when cot t = 0...its an asymptote...

    Thanks for any help
    For (i), I find \frac{dy}{dx} = -\frac{1}{3}\cot 2t.

    As for (ii), you must be confusing with something else: you indeed need to find points where 0=\frac{dy}{dx}, which means 0=\cot 2t=\frac{\cos 2t}{\sin 2t}. This is 0 when \cos 2t=0. We have 0\leq 2t<2\pi, hence the solutions are 2t=\frac{\pi}{2} and 2t=\frac{3\pi}{2}. It remains to find (x,y) for these t.
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  3. #3
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    Quote Originally Posted by Laurent View Post
    For (i), I find \frac{dy}{dx} = -\frac{1}{3}\cot 2t.

    As for (ii), you must be confusing with something else: you indeed need to find points where 0=\frac{dy}{dx}, which means 0=\cot 2t=\frac{\cos 2t}{\sin 2t}. This is 0 when \cos 2t=0. We have 0\leq 2t<2\pi, hence the solutions are 2t=\frac{\pi}{2} and 2t=\frac{3\pi}{2}. It remains to find (x,y) for these t.
    Thanks! I think i just managed to confuse myself.

    For part one though: How?
    Cuz you divide
    2cos 2t by -3sin 2t
    and get -2/3 cot 2t
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  4. #4
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    Quote Originally Posted by AshleyT View Post
    Thanks! I think i just managed to confuse myself.

    For part one though: How?
    Cuz you divide
    2cos 2t by -3sin 2t
    and get -2/3 cot 2t
    You divide 2\cos 2t by -{\color{red}6}\sin 2t (there is the same factor 2 when differentiating \cos 2t or \sin 2t).
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