# Thread: graphing sets of parametric curves

1. ## 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.

2. Originally Posted by 1sttime
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.
$x = \tan{t}$

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

3. 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.
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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.