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Math Help - graphing sets of parametric curves

  1. #1
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    graphing sets of parametric curves

    another test prep problem I don't completely understand.

    Graph each of the following sets of parametric equations by eliminating the parameter to get a cartesian equation. Explain the difference between how the two parametric curves are traced out.

    f(x) { x = tan(t) & y = cot(t) on the interval 0 < t < pi

    I understand how to get the cartesian equation:
    y = cot(tan inv(x))

    and i can graph x = tan(t) and y = cot(t) on an x and t axis, y and t axis respectively, but i don't understand why the parametric graph (on my calculator) and the cartesian graph look different, aren't they just different ways to write the same equation?
    a good explanation from beginning to end would be so greatly appreciated, i really want to understand this.
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  2. #2
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    Quote Originally Posted by 1sttime View Post
    another test prep problem I don't completely understand.

    Graph each of the following sets of parametric equations by eliminating the parameter to get a cartesian equation. Explain the difference between how the two parametric curves are traced out.

    f(x) { x = tan(t) & y = cot(t) on the interval 0 < t < pi

    I understand how to get the cartesian equation:
    y = cot(tan inv(x))

    and i can graph x = tan(t) and y = cot(t) on an x and t axis, y and t axis respectively, but i don't understand why the parametric graph (on my calculator) and the cartesian graph look different, aren't they just different ways to write the same equation?
    a good explanation from beginning to end would be so greatly appreciated, i really want to understand this.
    x = \tan{t}

    y = \cot{t} = \frac{1}{\tan{t}} = \frac{1}{x}
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  3. #3
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    Hello, 1sttime!

    Graph each of the following sets of parametric equations
    by eliminating the parameter to get a cartesian equation.
    Explain the difference between how the two parametric curves are traced out.

    . . \begin{Bmatrix}x \:= \:\tan\theta \\ y \:=\: \cot\theta \end{Bmatrix}\quad\text{on }0 < t < \pi

    We have: . \begin{Bmatrix}x \:=\:\tan\theta \\ \\[-4mm] y \:=\:\dfrac{1}{\tan\theta} \end{Bmatrix}

    Hence: . y \:=\:\frac{1}{x}

    This is a hyperbola.
    Code:
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        - - - - - - - - + - - - - - - - -
          *             |
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    With the cartesian equation, we trace the curve as x goes from -\infty to +\infty.

    The curve goes from (-\infty,0) to (0,-\infty), forming the left branch.
    The function is undefined at x = 0.
    Then the curve goes from (0,+\infty) to (+\infty,0), forming the right branch.


    With the parametric equations, we trace the curve as \theta goes from 0 to \pi.

    When \theta=0, we have: . x = 0,\:y = +\infty . . . the point (0,+\infty)
    When \theta = \tfrac{\pi}{2}, we have: . x = +\infty,\:y = 0 . . . the point (+\infty,0)
    . . Hence, as \theta goes from 0 to \tfrac{\pi}{2}, the right branch is traced out.

    And we find that, as \theta goes from \tfrac{\pi}{2} to \pi, the left branch is traced out.


    And that is the difference in the tracing of the two equivalent functions.

    Last edited by Soroban; June 14th 2009 at 05:59 PM.
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