Clockwise or counterclockwise motion?

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The particle's path is defined by the equation r = <cos(t), sin(t)>, indicating it moves in a circular motion. As t increases, the particle starts at <1,0> and travels counterclockwise, confirmed by graphing the function. To determine the direction of motion for any parametric curve, the acceleration r''(t) can be analyzed using the cross product of r' and r''. A positive result indicates a left turn (counterclockwise), while a negative result indicates a right turn (clockwise). This method effectively assesses the motion direction in the xy plane.
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Homework Statement



Th path of the particle P is given by the following equation:

r = &lt;cos(t),sin(t)&gt; defined for all t>0

Is the particle traveling clockwise or counterclockwise as t increases?

Homework Equations


The Attempt at a Solution



I graphed the function, and it is a circle, initially the particle is at the point <1,0> and it moves counterclockwise according to the graph.

But how could I find whether it's traveling clockwise or counterclockwise in general, for any parametric curve?

Might it have anything to do with cross products? This is all assuming the particle's motion lies strictly in the xy plane.

BiP
 
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Bipolarity said:

Homework Statement



Th path of the particle P is given by the following equation:

r = &lt;cos(t),sin(t)&gt; defined for all t>0

Is the particle traveling clockwise or counterclockwise as t increases?

Homework Equations





The Attempt at a Solution



I graphed the function, and it is a circle, initially the particle is at the point <1,0> and it moves counterclockwise according to the graph.

But how could I find whether it's traveling clockwise or counterclockwise in general, for any parametric curve?

Might it have anything to do with cross products? This is all assuming the particle's motion lies strictly in the xy plane.

BiP
Look at
$$ \frac{d \bf{r}}{dt}$$

It will be pointing in the direction of increasing t.
 
Mark44 said:
Look at
$$ \frac{d \bf{r}}{dt}$$

It will be pointing in the direction of increasing t.

Hey Mark, thanks. I know that v is parallel to the direction of the particle for increasing t. I can calculate it by differentiating each component of r with respect to t.

However, once I know r and v (as functions of t), how can I use them to decide whether the particle travels clockwise, or how can I determine for what values of t the particle travels clockwise?

My friend suggested polar coordinates but I don't know if that works. My idea was cross product of r and v, but is that correct?

BiP
 
For the example you gave, you just need to know the basic properties of sine and cosine. But in general, if you want to know whether your curve is turning left or right, you need to calculate the acceleration r''(t). If r' cross r'' is positive then that means you are turning left. If r' cross r'' is negative that means it is turning to the right. The cross product comes out to be: x' y'' - y' x''. So you just check whether that is positive (counterclockwise=left turn) or negative (clockwise=right turn)
 
Bipolarity said:
Hey Mark, thanks. I know that v is parallel to the direction of the particle for increasing t. I can calculate it by differentiating each component of r with respect to t.

However, once I know r and v (as functions of t), how can I use them to decide whether the particle travels clockwise, or how can I determine for what values of t the particle travels clockwise?

My friend suggested polar coordinates but I don't know if that works. My idea was cross product of r and v, but is that correct?

BiP

If you are dealing with a plane curve, then sure. The sign of the k component of rxv will tell you which direction you are traveling around the origin.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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