# 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.
$\displaystyle x = \tan{t}$

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

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.

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

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.
Code:
                    |
|*
|
| *
|  *
|    *
|       *
|             *
- - - - - - - - + - - - - - - - -
*             |
*       |
*    |
*  |
* |
|
*|
|

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.