Parameterizing a surface

Question

The question I was asked goes like this:

The part of the hyperboloid $5x^2 − 5y^2 − z^2 = 5$ that lies in front of the yz-plane. Let x, y, and z be in terms of u and/or v. Find a parametric representation for the surface.

So, by fixing $x$ as a constant $u$, I find that the slice is an ellipse which can be parameterized as $y = \sqrt{\frac{5u^2-5}{5}}\cos{v}$ and $z = \sqrt{5u^2-5}\sin{v}$, where $u$ ranges from $1$ to infinity, and $v$ ranges from $0$ to $2\pi$.

However, I was told that this is not the correct parameterization. The correct answer is

$$ \begin{align*} x&=\sqrt{1+u^2+\frac{1}{5}v^2} \\ y&= u \\ z&= v \end{align*} $$

I believe that this is just another way of parameterizing the same surface, and I don't see why my answer is wrong. Any idea?

Answer 1

I interpret being 'in front of the $yz$-plane' as the first coordinate being non-negative, thus the surface $S$ in question $\{(x,y,z)\in \mathbb R^3\colon x\ge 0\land 5x^2 − 5y^2 − z^2 = 5\}$.

It can easily be proved that this set equals $\left\{\left(\sqrt{1+y^2+\dfrac{z^2}5}, y,z\right)\colon y,z\in \mathbb R\right\}$, therefore $S=\vec r\left[\mathbb R^2\right]$, where and $\vec r\colon \mathbb R^2\to \mathbb R^3, (u,v)\mapsto \left(\sqrt{1+u^2+\dfrac{v^2}5}, u,v\right)$. So this $\vec r$ is certainly a suitable candidate for parametrization (you need to check that it satisfies the other properties for being one). In my mind this is the most natural option.

You suggest taking $\vec s\colon D\to \mathbb R^3, (\rho,\theta)\mapsto \left(\rho, \sqrt{\rho ^2-1}\cos(\theta), \sqrt 5\sqrt{\rho^2-1}\sin(\theta)\right)$, where $D=[1,+\infty[\times [0,2\pi[$.

I don't see how to make any variation of this parametrization become injective (I think it's not possible) and that might be why yours was marked wrong.

Anyway, I'll prove that $S=\vec s[D]$. The inclusion $\supseteq$ is the easy one. I'll help out with the other one.

Let $(x,y,z)\in S$. Due to $x^2=1+y^2+\dfrac{z^2}5$ it's easy to conclude that $x\ge 1$.

One wishes to find $(\rho, \theta)\in [1,+\infty[\times[0,2\pi[$ such that $(x,y,z)= \left(\rho, \sqrt{\rho ^2-1}\cos(\theta), \sqrt 5\sqrt{\rho^2-1}\sin(\theta)\right)$.

If $x=1$, then $y=0=z$ and $(x,y,z)=\vec s(1,\theta)$, for any $\theta$.

Suppose $x>1$. If such $(\rho, \theta)$ exists, necessarily $\rho =x$ and $y=\sqrt{\rho ^2-1}\cos(\theta)$. This ensues the equality $\cos(\theta)=\dfrac{y}{\sqrt{x^2-1}}$. So it's necessary to check that $-1\leq \dfrac{y}{x^2-1} \leq 1$ or equivalently that$\dfrac{y^2}{x^2-1}\leq 1$, and this in fact holds because $x^2=1+y^2+\dfrac{z^2}5$.

So take $\theta=\arccos\left(\dfrac y{\sqrt{x^2-1}}\right)\lor \theta=2\pi-\arccos\left(\dfrac y{\sqrt{x^2-1}}\right)$, depending on whether you're in $[0,\pi]$ or $[\pi, 2\pi]$.

All that is left is to check that this choice of $(\rho, \theta)$ works. I'll check the first one.

In fact $$\require{cancel} \begin{align} \vec s(\rho, \theta)&=\left(\rho, \sqrt{\rho ^2-1}\cos(\theta), \sqrt 5\sqrt{\rho^2-1}\sin(\theta)\right)\\ &=\left(x, \sqrt{x ^2-1}\cos\left(\arccos\left(\dfrac y{\sqrt{x^2-1}}\right)\right), \sqrt 5\sqrt{x^2-1}\sin\left(\arccos\left(\dfrac y{\sqrt{x^2-1}}\right)\right)\right)\\ &=\left(x,y,\sqrt 5\sqrt{x^2-1}\sqrt{1-\left(\dfrac y{\sqrt{x^2-1}}\right)^2}\right)\\ &=\left(x,y,\sqrt 5\cancel{\sqrt{x^2-1}}\sqrt{\dfrac{x^2-1}{\cancel{x^2-1}}-\dfrac{y^2}{\cancel{x^2-1}}}\right)\\ &=\left(x,y,\sqrt 5\sqrt{y^2+\frac{z^2}5-y^2}\right)\\ &=(x,y,z). \end{align} $$

To ensure that you get an appropriate $z$ just choose one of the two possible values of $\theta$.

Answer 2

Usually, conic sections are parametrized best with some kind of cylindrical or spherical coordinate transformation. Given $$5x^2 - 5y^2 - z^2 = 5, \quad x > 0,$$ this suggests a cylindrical parametrization with $$(y,z) = (u \cos v, \sqrt{5} u \sin v),$$ which easily gives $$5x^2 = 5 + 5y^2 + z^2 = 5 + 5u^2 (\cos^2 v + \sin^2 v) = 5(u^2+1),$$ or $$x = \sqrt{u^2 + 1},$$ where we take the positive square root since we were asked to provide the portion of the surface "in front of" the $yz$-plane (which I assume to mean $x > 0$). Thus one parametrization is $$\begin{align*} x &= \sqrt{u^2 + 1}, \\ y &= u \cos v, \\ z &= \sqrt{5}u \sin v, \end{align*}$$ for $u \in [0,\infty)$ and $v \in [0, 2\pi)$. Now, the above parametrization works just fine, but we can make it a little nicer and consider $$\begin{align*} x &= \sec u, \\ y &= \tan u \cos v, \\ z &= \sqrt{5} \tan u \sin v, \end{align*}$$ for $u \in [0,\pi/2)$, $v \in [0,2\pi)$, which gives the same result: this was motivated by the familiar trigonometric identity $$\tan^2 \theta + 1 = \sec^2 \theta.$$ Clearly, we can see that parametrizations are not unique, and that in general, infinitely many legitimate parametrizations of the same surface are possible.