$$\textbf{r}_1(t)= \sin t \textbf{i}+ \cos t \textbf{j}$$ 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?

  • $\begingroup$ Not obvious to you why it's the same curve, or not obvious that the orientation is changed? $\endgroup$
    – GFauxPas
    Jul 5, 2017 at 11:23
  • $\begingroup$ Why the orientation is changed? $\endgroup$
    – Eiraus
    Jul 5, 2017 at 11:25

3 Answers 3


$\newcommand{\Vec}[1]{\mathbf{#1}}$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? :)


Maybe representing the two curves slightly differently will help. \begin{eqnarray} r_1(t)=\begin{pmatrix}\sin(t)\\\cos (t)\end{pmatrix} \quad r_2(t)=\begin{pmatrix}\cos (t)\\\sin(t)\end{pmatrix} \end{eqnarray} in your $\textbf{i},\textbf{j}$ basis. Also the derivatives are
\begin{eqnarray} \dot{r}_1(t)=\begin{pmatrix}\cos(t)\\-\sin (t)\end{pmatrix} \quad \dot{r}_2(t)=\begin{pmatrix}-\sin(t)\\\cos(t)\end{pmatrix} \end{eqnarray} Let's also find a common point for which the curves are the same,for example $r_1(\pi/2)=r_2(0)$. Now the direction of the derivatives at this point is opposite. \begin{eqnarray} \dot{r}_1((\pi/2)=\begin{pmatrix}\cos((\pi/2)\\-\sin ((\pi/2)\end{pmatrix}=\begin{pmatrix}0\\-1\end{pmatrix} \quad \dot{r}_2(0)=\begin{pmatrix}0\\1\end{pmatrix} \end{eqnarray}


To determine direction compute $r_i'(t)$. Recall that this is the displacement vector at $r_i(t)$ i.e it tells you the direction. For instance,

$$r_2'(0.01) \approx (-.01,1)$$

Hence at $r_2(0.01)$ we are moving in the direction of $(-.01,1)$ i.e if we are to stay on the circular path, we must be going in the counter-clockwise direction. I hope this makes sense. To detect the change in orientation, you can again look at the velocity vectors. Compare $r_1'(s_0)$ to $r_2'(t_0)$ where $r_1(s_0) = r_2(t_0)$.


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