# Homework Help: Parametric Equations and direction

1. Oct 1, 2008

### UMich1344

The problem statement, all variables and given/known data

Consider the parameterization of the unit circle given by $$x=cos(3t^{2}-t)$$, $$y=sin(3t^{2}-t)$$ for t in $$(-\infty,\infty)$$.

In which intervals of t is the parameterization tracing the circle out in a clockwise direction?

In which intervals of t is the parameterization tracing the circle out in a counter-clockwise direction?

The attempt at a solution

I know this can't be too difficult. I'm just really struggling to understand the intervals in terms of the direction of motion. So far I've been able to conclude that the entire unit circle is in fact traced out by this parameterization and that when t=0, the point being traced out on the circle is at (1,0). I can see that initially for t>0, the motion is counter-clockwise but can't determine when the motion changes direction again. The same applies for the other direction. I see that initially for t<0, the motion is clockwise but am struggling to see when motion in that direction changes and begins going the other direction.

2. Oct 1, 2008

### HallsofIvy

You know (I hope!) that sine and cosine have period $2\pi$. The parametric equations x= cos(u), y= sin(u) will describe a circle (counter-clockwise) as u goes from 0 to $2\pi$. Now, what if u= 3t2- t? that is equal to t(3t- 1) which has zeros at t= 0 and t= 1/3. It's not too hard to see that its vertex is at t= 1/6 and, at that point, u= 3(1/6)2- 1/6= 1/12- 2/12= -1/12. Of course x= cos(u), y= sin(u) trace the circle in counter- clockwise direction as long as u is increasing, clockwise as long as u is decreasing. Can you get the answer now?

3. Oct 2, 2008

### UMich1344

Would these intervals be correct?

Clockwise for t: $$(-\infty, \frac{1}{6})$$

Counter-clockwise for t: $$(\frac{1}{6}, \infty)$$

$$\frac{1}{6}$$ is not included in the interval because the derivative of u (see below) is 0 at that point, so it is neither increasing nor decreasing. Right?

Work:

$$u=3t^{2}-t$$

$$du=6t-1$$

For clockwise, u must be decreasing, so du must be less than 0.

$$du<0$$

$$6t-1<0$$

$$t<\frac{1}{6}$$

For counter-clockwise, u must be increasing, so du must be greater than 0.

$$du>0$$

$$6t-1>0$$

$$t>\frac{1}{6}$$