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Vector Funtions

by mit_hacker
Tags: funtions, vector
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mit_hacker
#1
Oct21-07, 10:15 AM
P: 93
1. The problem statement, all variables and given/known data
Each of the following paths describes the motion of a particle having the same path, namely the unit circle x^2 + y^2 =1. Although the path for each particle is the same, the behavior of each particle is different. For each particle, answer the following questions:

i. .....
ii .......
iii Does the particle move counterclockwise or clockwise?
iiii .........

2. Relevant equations



3. The attempt at a solution

The logic I used to answer part 3 for each particle was to check the position at t=0 and then again at t=pi/2. The change in position will tell me whether it has moved clockwise or counterclockwise. What I need help is n confirming whether or not my method is correct or whether there is some other neater way to do this question.

Thank-you all for the help!!
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HallsofIvy
#2
Oct21-07, 12:45 PM
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P: 39,534
?? You haven't given the formula for the motion so it is not clear whether your method works or not. It is possible that the particle is moving clockwise at t= 0, then before t= pi/2, turns and moves back counter clockwise. If the particle is moving either always counterclockwise or always clockwise, as is implied by the fact that the question does not ask about a particular time, yes, that method works. But what if it had asked which way the particle was moving at t= 0?
mit_hacker
#3
Oct21-07, 08:34 PM
P: 93
It is always moving in the same direction. For instance, one of the formulas was
(Cos t)i - (Sin t)j.

You raised an interesting question. What if the question asked which way is the particle moving at t=0. What method would you apply in that case? Please tell me if I am right:

You find the tangent to the curve at that point and find the direction of that vector with respect to the xy-plane.

Am I right?

HallsofIvy
#4
Oct22-07, 05:53 AM
Math
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Vector Funtions

Quote Quote by mit_hacker View Post
It is always moving in the same direction. For instance, one of the formulas was
(Cos t)i - (Sin t)j.
It would have helped if you told us that to begin with!

You raised an interesting question. What if the question asked which way is the particle moving at t=0. What method would you apply in that case? Please tell me if I am right:

You find the tangent to the curve at that point and find the direction of that vector with respect to the xy-plane.

Am I right?[/QUOTE]
Yes, that would work. Another way would be to just look at one component: y= -sin(t) so y'= cos(t). At t= 0, the point is (1, 0) and y is increasing: counter-clockwise.


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