I have found parametrizations for the level curve $y^2-x^2=1$, however, I have a question regarding one of them.
From the Pythagorean trigonometric identity $\cos^2 x + \sin^2 x =1$ we obtain $$\sec^2 x – \tan^2 x =1, x \neq \pi(k+1/2), k \in \mathbb{Z},$$ $$\csc^2 x – \cot^2 x =1, x \neq k\pi, k \in \mathbb{Z}$$ by dividing by $\cos^2 x$ and $\sin^2 x$, respectively.
A parametrization for the given level curve is therefore $$\gamma(t) = (\tan t, \sec t), \ \ -\pi/2 < t < \pi/2, \quad \pi/2 < t < 3\pi/2.$$
A plot shows that this is indeed the case. However, for the other parametrization $$\gamma(t)=(\cot t, \csc t), \ 0 < t < \pi, \quad \pi < t < 2\pi$$ it doesn't seem to work (plotting in Maple). Am I way off here on the second parametrization?
Other trigonometric functions satisfying this relationship is obviously $\cosh$ and $\sinh$ as in $\cosh^2 x – \sinh^2 x = 1, x \in \mathbb{R}$, but this only takes care of the upper branch of the level curve (by parametrizing $\gamma_1(t) = (\sinh t, \cosh t), t \in \mathbb{R}$). Would a complete parametrization then be a union of $\gamma_1(t)$ and $\gamma_2(t) = (\sinh t, -\cosh t), t \in \mathbb{R}$?
Best Answer
I don't have Maple, but Mathematica does just fine with the first and second parametrizations:
and
give the same correct result. The third parametrization using hyperbolic functions does only give the upper half for $\gamma_1$. Unfortunately, trying to take the union of both halves no longer makes it a well-defined parametrization, since for each $t \in \mathbb R$, you now have two distinct ordered pairs $(x_1, y_1) \ne (x_2, y_2)$ on the curve.
As for why you are having problems in Maple, I can only state that your math is correct. Whether your difficulty has to do with user error or a defect/shortcoming in Maple, I cannot say.