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Eliminate Parameter with Sin and Cos

  1. May 1, 2015 #1
    x = 2 − π cos t
    y = 2t − π sin t
    −π ≤ t < π
    I understand how to eliminate parameter using sin^2 + cos^2 = 1.
    I cant figure out how to deal with the "2t" in the y equation, if you solve for sin(t) and square, you get
    ((2t-y)/π )^2 which leaves the parameter. Is there a way to get it into the argument of sin?
    I need this to then figure out intersection points and tangents later. Thanks in advance for any insight!
     
  2. jcsd
  3. May 1, 2015 #2

    fzero

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    You will not be able to completely eliminate ##t##, but you can use these 2 equations to write a single equation that depends on ##x,y## and ##t## (but not ##\sin t,\cos t##). There is an interesting geometric interpretation of the equation that we can discuss after you try to find it.
     
  4. May 1, 2015 #3
    Thank you for the help! I can't figure what that equation would be ( I can see one involving arccos and substitution but I think that would be a dead end.)
    I thought maybe finding the derivative dy/dx for the parametric eqn and realizing that at the point of intersection the 2 slopes would be dy/dx and -dy/dx might lead me closer?
     
  5. May 1, 2015 #4

    fzero

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    Using the derivatives might be of help later, but for now do something simpler. Rewrite the equations in the form

    $$\begin{split}
    x -2 & = -\pi \cos t, \\
    y -2t & = -\pi \sin t\end{split}$$

    and then square both and add them (you were close to this in your original attempt). Can you interpret the resulting quadratic equation?
     
  6. May 1, 2015 #5
    (x-2)^2 = (pi^2)(cos^2(t))
    (y-2t)^2 = (pi^2)(sin^2(t))
    added =
    (x-2)^2 + (y-2t)^2 = (pi^2)(cos^2(t)) + (pi^2)(sin^2(t))
    (x-2)^2 + (y-2t)^2 = (pi^2)[(cos^2(t)) + (sin^2(t))]
    (x-2)^2 + (y-2t)^2 = (pi^2)
    [(x-2)^2]/(pi^2) + [(y-2t)^2]/(pi^2) = 1

    It seems to look like an ellipse adjusted upwards by the parameter? I really appreciate the help!
     
  7. May 1, 2015 #6

    fzero

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    Correct. To be precise, it's actually a circle with radius ##\pi##.
     
  8. May 2, 2015 #7
    So it seems at this point to find where the graph would intersect itself, we would ask what value for t would give us 2 sets of identical x,y coordinates?
     
  9. May 2, 2015 #8

    fzero

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    Actually you want to find values ##t_1## and ##t_2## for which ##( x(t_1),y(t_1))=( x(t_2),y(t_2))##. In this case there is only one pair of such ##t_1,t_2##, but in general there can be more.

    Also, I should warn that the graph itself is not a circle, despite the appearance of that equation above.
     
  10. May 2, 2015 #9

    Svein

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