Given $x^{2/3}+y^{2/3}=1$, we can solve for $y$ and find two functions. In terms of nomenclature, do we have one or two parametrizations of the curve?

Question

Consider the equation $$x^{2/3}+y^{2/3}=1\tag{1}$$

If we assume that $y$ is defined implicitly in terms of $x$, we can differentiate implicitly

$$\frac{2}{3}x^{-1/3}(x)+\frac{2}{3}y^{-1/3}(x)y'(x)=0$$

$$y'(x)=-\frac{y^{1/3}(x)}{x^{1/3}}\tag{2}$$

At this point we don't really know yet where this derivative is defined. We have simply assumed it is defined and computed what it should be.

From $(2)$, we not that $x=0$ isn't part of the domain of $y'$, but since $y$ is a function of $x$, there could be other constraints on the domain of $y'$.

$(1)$ is simple enough that we can solve for $y$ explicitly

$$y=\pm \sqrt{(1-x^{2/3})^3}$$.

The points that satisfy $(1)$ form a set. This set does not necessarily represent a single function, and we see this because we obtained two functions. The two functions have the same domain $[-1,1]$.

Here we can see what the curve, and the two specific functions obtained that generate the curve, are

We can also try to represent the curve represented by $(1)$ as a parametric equation, ie as a vector-valued function $c(t)$.

The most obvious and trivial parametric representation is something like

$$\vec{c_1}(t)=(t, \sqrt{(1-x^{2/3})^2})$$

$$\vec{c_2}(t)=(t, -\sqrt{(1-x^{2/3})^2})$$

One question I have is about nomenclature. Is it correct to say that $\vec{c_1}$ and $\vec{c_2}$ represent a parametrization of $(1)$? It seems slightly odd to have two functions with the same domain parametrizing a single curve.

Note that a more clear-cut case is if we use an alternative parametrization

$$\vec{d}(t)=(x,y)=(u(t),v(t))=(\cos^3{t}, \sin^3{t})$$

Let's visualize what this means

If we start at the purple dot, the parametrization means that as $t$ increases from $-\frac{\pi}{2}$ we move counter-clockwise around the curve.

This is the sense in which I asked about nomenclature in the case of the first parametrization. In that case, we can't really think about moving counter-clockwise for $x$ in some interval. We can only think about it like that for each individual portion of the curve.

A few more details

Note that $u'(t)=-3\cos^2{t}\sin{t}$.

Consider intervals $(k\frac{\pi}{2},(k+1)\frac{\pi}{2})$ for $t$, with $k \in \mathbb{Z}$. $u'\neq 0$ in these intervals.

It can be shown that the derivative of $y$ relative to $x$ is

$$y'(x)=\frac{v'(t)}{u'(t)}$$

Thus,

$$y'(x)=-\frac{\sin{t}}{\cos{t}}\tag{3}$$

So what does $(3)$ mean?

It's giving us the derivative of $y$ in relation to $x$, or $u$ in relation to $v$. For example, suppose we are in the first quadrant. As $t$ increases, we are moving northwest. At these points, $\cos{t}$ and $\sin{t}$ are both positive and so $y'(x)$ is negative, which makes sense.

Answer 1

In my opinion it is not correct to say that $\vec c_1$ and $\vec c_2$ both parametrize the curve. A parametrization of a curve is some continuous map $\vec \phi$ whose image is the entire curve. It does not refer to a collection of maps the union of whose images is the entire curve.

Having the same domain is not an issue at all. The fact that there are two functions instead of one is an issue. When you see mathematicians referring to a parametrization, they are generally speaking of a single map. So results about parametrizations will not directly apply to your pair of maps. In most cases this will not be a problem. But there may be some theorem that requires a single map for its proof. So you would need to examine the results individually to figure out which is which. By using the same terminology for a different case, you may confuse your readers, who may not realize the discrepancy.

There is nothing wrong with using different parametrizations for different parts of the curve. Differential Geometry does something very similar in the definition of a manifold, except there the parametrizations are required to overlap, and play nicely with each other on that overlap. But we use a different terminology. Each parametrization is called a "chart", and the collection of all of them is called an "atlas".

If you want have a single parametrization for the entire curve, you need to shift one of them to an adjacent domain, and piece them together: $$\vec c(t) = \begin{cases}(t, \sqrt{(1-t^{2/3})^2})&t\in[-1,1]\\ (2-t, -\sqrt{(1-(2-t)^{2/3})^2})&t\in (1,3]\end{cases}$$

Of course, I began this with "in my opinion" because that is all it is. There is no arbitrating body of mathematical terminology. If you just called the pair a parametrization, you likely would only confuse neophytes and merely cause a very mild cognitive dissonance in everybody else.

Answer 2

I think it's good to take a look at the notion of a manifold. It's a space locally homeomorphic to some Euclidean space. The word 'locally' shows that we generally need more than one parametrization of a manifold.

• The set $\{(x,y)\,|\,y=x^2\}$ admits only one parametrization.
• The set $S=\{(x,y)\,|\,x^2+y^2=1\}$ (and your curve) doesn't admit only one parametrization, it needs at least two, since it's not homeomorphic to any subset of $\Bbb R$. For example we can consider parametrizations $c_1\colon (-\pi,\pi)\to S$ and $c_2\colon (\pi/2,3\pi/2)\to S$ given by $c_i(t)=(\cos t,\sin t)$. They are both homeomorphisms between domain and some open subsets of $S$ and these subsets cover $S$. We can also consider four parametrizations: $t\mapsto (t,\pm\sqrt{1-t^2})$ and $t\mapsto (\pm\sqrt{1-t^2},t)$, $|t|<1$.

If we consider the parametrizations only for integration, we don't care about the assumption that parametrizations are to be 'local homeomorphisms'. Particularly the parametrizations can be defined on closed subsets of $\Bbb R^n$ and don't need to be injections, provided we control the 'duplicate points'. For example for integration purpose we can consider only one parametrization of a circle: $c\colon [-\pi,\pi]\to S$ given by $c(t)=(\cos t,\sin t)$. We see that $c(-\pi)=c(\pi)$ but this doesn't change the integral. One can also think of a parametrization $c\colon [-\pi,\pi)\to S$. Here it is bijective and continuous, but it's inverse isn't continuous. For some application this isn't a problem but for some other is.