My syllabus defines smooth plane curves as follows
A smooth curve in $\mathbb{R}^2$ is every subset $\Gamma$ of $\mathbb{R}^2$ that can be written as $\Gamma = \mathbf{r}[a,b]$, with $\mathbf{r}:[a,b] \to \mathbb{R}^2 \quad(a<b)$ a vector valued function with following characteristics:
- $\mathbf{r}$ is a bijection from $[a,b]$ to $\Gamma$
- $\mathbf{r} \in C^1[a,b]$
- $\mathbf{r}'(t) \ne \mathbf{0}$ for all $a \le t \le b$.
I have a few question regarding this definition (it's OK if you can only answer one of them, please do so down below):
- $\mathbf{r} \in C^1[a,b]$ is a bijection $[a,b] \to \Gamma$. Does this imply that $\mathbf{r}^{-1} \in C^1(\Gamma)$ is a bijection that (locally) maps the curve to a straight line (in $[a,b] \subset \mathbb{R}$)?
- Characteristic 3 in the definition means that the derivative of $\mathbf{r}$ with respect to $t$ cannot be the zero-$\underline{vector}$. Suppose $\mathbf{r}(t) = (x(t),y(t))$. Characteristic 3 actually states that $ \frac{dx}{dt}\mathbf{e}_x+\frac{dy}{dt}\mathbf{e}_y \ne \mathbf{0}$. Does this imply that one of the derivatives $x'(t), y'(t)$ can be zero, but not both of them?
- I don't see why $\mathbf{r}'(a)$ and $\mathbf{r}'(b)$ cannot be equal to the zero-vector. Is it because $a$ and $b$ are boundary points? I believe we can take a restriction $\mathbf{r}_{\varepsilon}:[a+\varepsilon,b-\varepsilon] \to \mathbb{R}^2$, which (after $\varepsilon \to 0+$) will give us the same properties of the original curve (and $\mathbf{r}$). Isn't this the same as allowing $\mathbf{r}'(t)$ to be $\mathbf{0}$ in $a$ and $b$?
Thanks for answering and helping me out.
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