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Math Help - Finding Equation of Tangent Line (Parametric) - # 2

  1. #1
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    Finding Equation of Tangent Line (Parametric) - # 2

    x = \cos t + \cos 2t

    y = \sin t + \sin 2t

    (-1,1)

    (y - 1) = m(x + 1)

    \dfrac{dy/dt}{dx/dt} = \dfrac{2\sin^{2}t + \cos t}{\sin t - 2\cos^{2} + \sin t} = \dfrac{2\sin^{2}t + \cos t}{2\sin t - 2\cos^{2}t} = \sin t - \dfrac{1}{2}cos^{-1} t ???

    Find t

    -1 = \cos t + \cos 2t

    -1 = \cos 3t

    \arccos(-1) = 3t

    180 = 3t

    t = 30
    Last edited by Jason76; July 26th 2014 at 12:41 AM.
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  2. #2
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    Re: Finding Equation of Tangent Line (Parametric) - # 2

    Quote Originally Posted by Jason76 View Post
    x = \cos t + \cos 2t

    y = \sin t + \sin 2t

    (-1,1)

    (y - 1) = m(x + 1)

    \dfrac{dy/dt}{dx/dt} = \dfrac{2\sin^{2}t + \cos t}{\sin t - 2\cos^{2} + \sin t} = \dfrac{2\sin^{2}t + \cos t}{2\sin t - 2\cos^{2}t} = \sin t - \dfrac{1}{2}cos^{-1} t ???
    I presume the "???" is because you just can't believe that is correct! You are right \frac{a+ b}{c+ d} is NOT equal to \frac{a}{c}+ \frac{b}{d}


    Find t

    -1 = \cos t + \cos 2t

    -1 = \cos 3t
    Oh! Ouch! Surely you know that cos(t)+ cos(2t) is NOT equal to cos(3t)?
    [/quote]

    \arccos(-1) = 3t

    180 = 3t

    t = 30[/QUOTE]
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  3. #3
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    Re: Finding Equation of Tangent Line (Parametric) - # 2

    How can we solve for t, since we know that this cos stuff cannot be added?

    Quote Originally Posted by HallsofIvy View Post
    I presume the "???" is because you just can't believe that is correct! You are right \frac{a+ b}{c+ d} is NOT equal to \frac{a}{c}+ \frac{b}{d}



    Oh! Ouch! Surely you know that cos(t)+ cos(2t) is NOT equal to cos(3t)?
    \arccos(-1) = 3t

    180 = 3t

    t = 30[/QUOTE][/QUOTE]
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    Re: Finding Equation of Tangent Line (Parametric) - # 2

    Quote Originally Posted by Jason76 View Post
    x = \cos t + \cos 2t

    y = \sin t + \sin 2t

    (-1,1)

    (y - 1) = m(x + 1)

    \dfrac{dy/dt}{dx/dt} = \dfrac{2\sin^{2}t + \cos t}{\sin t - 2\cos^{2} + \sin t} = \dfrac{2\sin^{2}t + \cos t}{2\sin t - 2\cos^{2}t} = \sin t - \dfrac{1}{2}cos^{-1} t ???

    Find t

    -1 = \cos t + \cos 2t <snip>
    $-1=\cos(t)+\cos^2(t)-\sin^2(t)=\cos(t)+2\cos^2(t)-1$

    $0=\cos(t)+2\cos^2(t)$

    Let $u=\cos(t)$

    $0=2u^2+u =u(2u+1)$

    $u=0 \vee u=-\dfrac 1 2$

    $\cos(t)=0 \Rightarrow t=\dfrac {\pi}{2} + \pi k, ~k \in \mathbb{Z}$

    $\cos(t)=-\dfrac 1 2 \Rightarrow t = \dfrac {2\pi}{3}+2\pi k \vee t=\dfrac{4\pi}{3} + 2\pi k, ~k\in \mathbb{Z}$
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