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Thread: 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.
    $\displaystyle x = \tan{t}$

    $\displaystyle y = \cot{t} = \frac{1}{\tan{t}} = \frac{1}{x}$
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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.

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

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

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

    This is a hyperbola.
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    With the cartesian equation, we trace the curve as $\displaystyle x$ goes from $\displaystyle -\infty$ to $\displaystyle +\infty.$

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


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

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

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


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

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