What is a smooth curve in $\mathbb{R}^2$ intuitively?

Question

While studying for my exam, I've run into some problems understanding what a smooth curve in $\mathbb{R}^2$ is.

I first thought that, intuitively, I could think of a piece of string on a piece of paper, where the string is the curve and my paper is the $\mathbb{R}^2$ space.

But in my book, it says the following:

A smooth curve in $\mathbb{R}^2$ is every subset $\Gamma$ of $\mathbb{R}^2$ which can be written as $\Gamma = \phi[a,b]$, where $\phi:[a,b] \rightarrow \mathbb{R}^2$ $(a<b)$ has the following properties:

1. $\phi$ is a bijection of $[a,b]$ on $\Gamma$
2. $\phi \in C^1[a,b]$
3. $\phi'(t) \neq 0$ $\forall a<t<b$

Now my problem lies in these properties:

I googled examples of smooth curves and was returned images of ellipsoids and circles (among other shapes).
As far as I know, to represent a circle or ellipsoid, every value in $[a,b]$ would need have 2 values in $\mathbb{R}^2$ and because of this it's not a bijection.
Another problem arises in the fact that the derivative cannot be 0, as this is the case in 2 points in aforementioned shapes.

Answer 1

You've got a couple of mis-interpretations here; let me clear them up.

1. You are thinking of graphs $\{(x,f(x))\mid x\in[a,b]\}$, where $f$ is a function $[a,b]\to\mathbb{R}$, instead of more general subsets of the plain. Here, $\phi$ is a parameterization of your curve: it is a function $\phi(t)=(x(t),y(t))$. The condition that this be a bijection has nothing at all to do with whether or not each $x$-coordinate has at most one corresponding $y$-coordinate; rather, it is simply whether each point $(x,y)$ has at most one $t$ such that $\phi(t)=(x,y)$.

2. The derivative of $\phi$ is not the slope of the tangent line; rather, it is a direction vector for the tangent line. If we write $\phi(t)=(x(t),y(t))$ as above, then $\phi'(t)=(x'(t),y'(t))$. The condition that $\phi'$ be non-zero is actually the condition that the derivative is never the zero vector. At no point on the unit circle, with its usual parameterization $\phi(t)=(\cos t,\sin t)$ for $0\leq t\leq 2\pi$, is the derivative the $0$ vector.

I will note, however, that the definition you're given DOES preclude considering circles as smooth curves; not for the reasons you suggest, however. The problem with circles is that there is no way to come up with a parameterization from a closed interval to a circle without at least two $t$-values giving the same point; in the classic parameterization on $[0,2\pi]$, these are $t=0$ and $t=2\pi$.

You can think of the definition you've been given as being for a non-closed smooth curve. You can give a (barely!) more general definition which wouldn't exclude circles/other closed curves, as follows: simply relax the bijection condition to say that we need the restriction of $\phi$ to $[a,b)$ to be a bijection, rather than the entire map to be a bijection. This allows us to start and end at the same point.

Answer 2

After doing some more research, I've come to the conclusion that the images google returned were images of PIECEWISE smooth curves, which is why I got confused. I'm going to leave this question though since it could help out someone who made the same mistake as me

Answer 3

Just to add a word about condition 3 (this is usually stated by saying that your curve is regular). This is not absolutely necessary and one could, in principle, develop a theory of curves having also non-regular points. What is interesting is to consider which properties depend on regularity. The most important one is the fact that a differentiable path admits arclength parametrization if and only if it is regular at every point. The second one is that at non regular points one does not have a Frénet frame.

The usual example is the cuspidal curve $F(t)=(t^3,t^2)$ which is not regular at $t=0$ (remark that one may still define a tangent line at that point by using a concept of weak tangent, i.e. the common limit, if it exists of left and right tangent lines).