Parametrizing a circle in a counterclockwise direction

Question

How do I parametrize a circle in a clockwise direction? For instance, if the circle is in a counterclockwise direction, the parametrization would be $$c(t) = (r \cos t,r \sin t).$$

I've seen a lot of different answers when it comes to parametrizing a circle in a clockwise direction. Some wrote parametrized it as $$c(t) = (-r\cos t,r\sin t)$$ and some did $$c(t) = (r\cos t,-r\sin t).$$ Please clarify.

Answer 1

A parametrization is not unique. Suppose you parametrize the circle as $(r\cos{t},r\sin{t})$ with $t\in[0,2\pi)$. Then, when $t=0$ you begin at the point $(r,0)$. Then $t$ increases until $t=\pi/2$ and you arrive to the point $(0,r)$, so you are going counterclockwise. If you want to go clockwise you can use $(r\sin{t},r\cos{t})$, and similar things. To check if you are doing it okay, you can do what I did at the begining of the answer. For example, if you use $(-r\cos{t},r\sin{t})$, you begin at $(-r,0)$ and when $t$ increases until $t=\pi/2$ you arrive to $(0,r)$ so you are going clockwise.

You can also use tangent vectors. Suppose we wanted to know if $c(t)=(-r\cos{t},r\sin{t})$ is clockwise. First, we check at what point we are when $t=0$, in this case $(-r,0)$. Now we have that $c^\prime(t)=(r\sin{t},r\cos{t})$, so $c^\prime(0)=(0,r)$. So from $(-r,0)$ you go "up" because the vector $(0,r)$ is a vertical vector that goes "up", and you are clockwise.

Answer 2

Replace $t$ by $-t$ in given $c(t)$

Answer 3

Whether or not the parametrization traces a circle in clockwise direction or anti-clockwise direction depents on the convention of handed-ness you are using for your Cartesian coordinate system.

Let $e_x$ and $e_y$ be unit vectors of such a system where $x$ labels the 1st coordinate and $y$ the second.

If your convention is that a rotation of $90^\circ$ that turns $e_x$ into $e_y$ is clockwise, then $c(t) = (\cos t, \sin t)$ is a parametrization of the unit circle in clockwise direction. The reason is that $c(0^\circ) = (1,0) = e_x$ and $c(90^\circ) = (0,1) = e_y$ is then a clockwise turn by definition.

Most 2D coordinate systems draw $e_x$ to the right and $e_y$ upwards, and use the convention that turning $e_x$ onto $e_y$ by $90\circ$ rotation is counter-clockwise (i.e. you are looking at the clock from the front. This means $c(t)$ from above runs counter-clockwise.

If in such a setup you define a turn of $90^\circ$ that maps $e_x$ to $e_y$ is clockwise (you are looking at the clock from the "rear side"), then $c(t)$ is clockwise, too.