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:

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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.