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Math Help - Parametric tangent lines

  1. #1
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    Parametric tangent lines

     x = cos(t) +2cos(2t)
     y = sin(t) +2sin(2t)

     dx = -4sin(2t)-sin(t)
     dy = 4cos(2t)-cos(t)

    I have found the crossing point to be at (-2,0) by looking at my graphing calc, is there a more precise way?

    How do i find the tangent lines to this point now?

    Thanks
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  2. #2
    MHF Contributor Calculus26's Avatar
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    You're question doesn't really makes sense to me as the 2 equations

    are one curve -- that's the idea of a parametrically defined curve.

    Could you send the original question?

    What is actually crossing Here?

    A general answer to slope for a parametrically defined curve is:

    dy/dx = (dy/dt)/(dx/dt)
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  3. #3
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    Quote Originally Posted by Calculus26 View Post
    You're question doesn't really makes sense to me as the 2 equations

    are one curve -- that's the idea of a parametrically defined curve.

    Could you send the original question?

    What is actually crossing Here?

    A general answer to slope for a parametrically defined curve is:

    dy/dx = (dy/dt)/(dx/dt)

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  4. #4
    MHF Contributor Calculus26's Avatar
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    Ok now it makes sense the curve crosses itself.

    to get the tangent lines

    you need to find the times when it crosses itself

    t = arccos(-1/4) which is the same as pi -arcos(1/4) and

    also the 3d quadrant angle pi +arcos(1/4)


    use dy/dx = dy/dt/dx/dt as mentioned before

    To solve this I got the pt(-2,0) as you did graphically

    Set cos(t) + 2cos(2*t) =-2

    cos(t) +4cos^2(t) -2 = -2

    cos(t)(1+4cos(t)) = 0

    cos(t) = 0 yields pi/2,3pi/2

    the second term gives

    cos(t) = 1/4 t = arccos(-1/4) whiwhich yields the same results I mentioned above

    plug in the values in y to see which ones yield 0 0r even easier

    Now I'll let you do the same thing with y(t) = sin(t) +2sin(2t) = 0
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  5. #5
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    Hello, silencecloak!

    This is a long a tedious problem . . .


     x \:= \cos t +2\cos2t
     y \:=\: \sin t +2\sin2t

    \frac{dx}{dt} \:=\: \text{-}\sin t -4\sin2t
    \frac{dy}{dt} \:=\: \cos t + 4\cos2t

    I have found the crossing point to be at (-2,0) by looking at my graphing calc.
    Is there a more precise way? . . . . I don't know

    How do i find the tangent lines to this point now?
    Assuming that (-2,0) is indeed the intersection, we can work backwards.


    Since x = -2, we have: . \cos t + 2\cos2t \:=\:-2 \quad\Rightarrow\quad \cos t + 2(2\cos^2t-1) \:=\:-2

    . . \cos t + 4\cos^2\!t - 2 \:=\:-2 \quad\Rightarrow\quad \cos t + 4\cos^2\!t \:=\:0 \quad\Rightarrow\quad \cos t(1 + 4\cos t) \:=\:0

    And we have: . \begin{array}{cccc}\cos t \:=\:0 & \Rightarrow & t \:=\:\pm\frac{\pi}{2} \\<br />
1+4\cos t \:=\:0 & \Rightarrow & t \:=\:\arccos\left(\text{-}\frac{1}{4}\right) \end{array}


    Since y = 0, we have: . \sin t + 2\sin2t \:=\:0 \quad\Rightarrow\quad\sin t + 2(2\sin t\cos t) \:=\:0

    . . \sin t + 4\sin t\cos t \:=\:0 \quad\Rightarrow\quad \sin t(1 + 4\cos t) \:=\:0

    And we have: . \begin{array}{cccc}\sin t \:=\:0 & \Rightarrow & t \:=\:0,\pi \\<br />
1 + 4\cos t \:=\:0 & \Rightarrow & t \:=\:\arccos\left(\text{-}\frac{1}{4}\right) \end{array}

    The only root common to both sets is: . t \:=\:\arccos\left(\text{-}\tfrac{1}{4}\right)


    We can work out the trig values for t and 2t.

    . . We have: . t \:=\:\arccos\left(\text{-}\tfrac{1}{4}\right) \quad\Rightarrow\quad \boxed{\cos t \:=\:-\frac{1}{4}}

    . . \sin t \:=\:\pm\sqrt{1-\cos^2\!t} \:=\:\pm\sqrt{1-\left(\text{-}\tfrac{1}{4}\right)^2} \:=\:\pm\sqrt{\tfrac{15}{16}} \quad\Rightarrow\quad\boxed{ \sin t\:=\:\pm\frac{\sqrt{15}}{4}}

    . . \sin2t \:=\:2\sin t\cos t \:=\:2\left(\pm\tfrac{\sqrt{15}}{4}\right)\left(\t  ext{-}\tfrac{1}{4}\right) \quad\Rightarrow\quad \boxed{\sin2t \:=\:\mp\frac{\sqrt{15}}{8}}

    . . \cos2t \:=\:2\cos^2\!t - 1\:=\:2\left(\text{-}\tfrac{1}{4}\right)^2-1 \quad\Rightarrow\quad \boxed{\cos2t \:=\:-\frac{7}{8}}


    \frac{dy}{dx} \:=\:\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \;=\;\frac{\cos t + 4\cos2t} {\text{-}\sin t - 4\sin 2t}

    Substitute the values we found above and get the two slopes of the tangents.

    Then you can write the equations of the two tangents through (-2, 0).

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