Parameterisation of a circle?

Question

If I'm parametersing a unit circle about the origin for a line integral in a clockwise direction should it be $(x,y)=(\sin(t),\cos(t))$ or $(x,y)=(\cos(t),-\sin(t))$?

Does it depend on whether I start on the $x$ axis or the $y$ axis? Does it also matter if t is between $0$ and $2\pi$ or just a section of the circle?

I'm just trying to find a general case for all these things.

Answer 1

So, to answer the clockwise/anticlockwise thing, find the velocity of your parametrisation! (differentiate with respect to $t$). With a little abuse of notation,

$$\frac{d(\sin t, \cos t)}{dt} = (\cos t, -\sin t)$$

So, see that when $t = 0$ in particular, your position is $(0, 1)$ and your velocity is $(1, 0)$. So, this is starting at the top of the circle, moving clockwise (since its velocity is pushing it "forward" from the top)

You can repeat this for different parametrisations.

In general, you can consider the parametrisation to be

$$ s(t) = (\cos (t + \phi), \pm \sin (t + \phi)) $$ where $\phi$ is some global "phase factor" that you can use to rotate the particle around.

Your choice of $+\sin t$ or $-\sin t$ will determine the velocity (that is, the direction of rotation)

For a detailed analysis, I'd recommend getting used to the general form of the sine wave

Answer 2

If the counterclockwise parametrisation of the unit circle, begining at $t= 0+2k\pi$ is $\mathbf x(t)= (\cos t, \sin t)$, then $\mathbf x (-t) $ would do it clockwise. That is $\mathbf x (t) = (\cos (-t), \sin (-t))=(\cos t , -\sin t)$.

If you want to begin at another angle $\theta$, just do $\mathbf x(t+\theta)$, with $t\in[0, 2\pi)$.

Answer 3

Either $(x,y)=(\sin(t),\cos(t))$ or $(\cos(t),-\sin(t))$ will work just fine. In fact, the difference isn't likely to effect your calculations much (because sines and cosines are so closely related).

As for your second question, the domain you choose for $t$ is really up in the air. Any $2\pi$-length interval will do, so $0 \leq t \leq 2\pi$ or $-\pi \leq t \leq \pi$ or $\sqrt{2}-\pi/2 \leq t \leq \sqrt{2}+3\pi/2$ or ... all work. The interval you choose merely determines where your circle "starts" and "ends". Since the choice of start/end point has no effect on the line integral, this doesn't matter.

All of this said, my own approach would be to use the standard parameterization: $(x,y)=(\cos(t),\sin(t))$ with $0 \leq t \leq \pi$. Then adjust for orientation. [For a line integral through a vector field, this flips the sign. For a line integral with respect to arc length, this has no effect.]