How to find if particle is moving in clockwise direction or not? I've been given a vector equation in the form of r(t), how do i find if the particle is going in clockwise direction or in anticlockwise direction?
 A: I assume your question is about a motion on the plane. You can just look at $r'(t)\times r''(t)$. This vector is perpendicular to the plane. Assuming you choose a right framework in 3D, it is parallel to your third unit vector $\mathbf{k}$. If its third coordinate is positive, the motion is counterclockwise, otherwise it is clockwise. This is because $r''(t)$ always points to the side where the center of curvature is located.
A: It is accepted convention that anticlockwise rotation is positive.
In the oval/loop below with above accepted convention (around pole/origin at O) the rotation
$ A\to B$ is positive and  $ B\to A$ is negative.
and they are reversed if convention is made positive for clockwise rotation.

So an arbitrary sense of rotation is inconclusive regarding its sign.
A: I assume you are dealing with two dimensional motion.
You're given a vector equation in the form $\displaystyle \vec{r(t)} = x(t)\mathbb i + y(t)\mathbb j$
Letting $\displaystyle |\vec{r(t)}| = r$ and $\displaystyle \theta$ be the argument of the particle's position vector measured in a counterclockwise sense from the origin, you can determine the polar equation as follows:
$\displaystyle x(t) = x = r\cos\theta$
and $\displaystyle y(t) = y = r\sin\theta$
Dividing one by the other, you get $\displaystyle \tan\theta = \frac yx$
Differentiating implicitly with respect to time $\displaystyle t$,
$\displaystyle \sec^2\theta \ \dot\theta = \frac{x\dot y - y\dot x}{x^2}$
Noting that $\displaystyle \sec^2\theta = 1 + \tan^2 \theta$,
$\displaystyle (1 + \frac {y^2}{x^2}) \dot \theta = \frac{x\dot y - y\dot x}{x^2}$
$\displaystyle \dot \theta = \frac{x\dot y - y\dot x}{x^2} \cdot \frac{x^2}{x^2+y^2} = \frac{x\dot y - y\dot x}{r^2}$
As the denominator is always non-negative, you only need to compute the sign of the numerator. If it's positive, the particle is moving counterclockwise, and vice versa.
