TITLE: Parameterizing same curve in different directions
QUESTION [6 upvotes]: $$\textbf{r}_1(t)= \sin t \textbf{i}+ \cos t \textbf{j}$$ 
and
$$\textbf{r}_2(t)= \cos t \textbf{i}+ \sin t \textbf{j}$$
represent the same curve but traversed in different directions. $\textbf{r}_1$ clockwise and $\textbf{r}_2$ anticlockwise. It is not obvious to me just why this is case? Is there a way to determine  direction other than trial and error?

REPLY [5 votes]: $\newcommand{\Vec}[1]{\mathbf{#1}}$Two paths
$$
\Vec{r}_{1}(t) = f(t)\, \Vec{i} + g(t)\, \Vec{j},\quad
\Vec{r}_{2}(t) = g(t)\, \Vec{i} + f(t)\, \Vec{j}
$$
differ by reflection across the line $y = x$, which acts on Cartesian coordinates of points by $(x_{0}, y_{0}) \mapsto (y_{0}, x_{0})$.
Because your $\Vec{r}_{2}$ is the standard (counterclockwise, unit-speed) parametrization of unit circle in the plane, $\Vec{r}_{1}$ parametrizes the image of the unit circle after reflection across the diagonal, namely the unit circle itself. It's geometrically clear the direction of travel changes.

Incidentally, an arbitrary path can be reparametrized "in the opposite direction" by making time run backward. i.e., replacing $t$ by $-t$ (or more generally by $c - t$ for some constant $c$).
For example, the path
$$
\Vec{r}_{3}(t) = \Vec{r}_{2}(-t)
= \cos(-t)\, \Vec{i} + \sin(-t)\, \Vec{j}
= \cos t\, \Vec{i} - \sin t\, \Vec{j}
$$
is a clockwise, unit-speed parametrization of the unit circle.
The same is true of $\Vec{r}_{2}(\frac{\pi}{2} - t)$, which you can easily check is $\Vec{r}_{1}(t)$.
Generally, if you have a path $\Vec{r}$ with domain $[a, b]$, the customary idiom for reversing the direction is
$$
\Vec{r}^{-}(t) = \Vec{r}(a + b - t),\quad a \leq t \leq b.
$$
(Can you see why? :)