# A particle moves in a circle in such a way that the x- and y-coordinates of its motion are given in meters as functions of time t in seconds by:

x=5cos(3t)
y=5sin(3t)
What is the period of revolution of the particle?

Here's what my train of thought was... dy/dt over dx/dt would equal dy/dx which equals velocity. So velocity would be -cot(3t). But then how do I find r? Plus none of the answers here have a trig function in them... ???????

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1. Hmmmm. You are not asked for r, you are asked for period, in time.

3t= 2PI*n n=0, 1, ...

t= 2/3 PI n try n=1 for the first rev.

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bobpursley
2. I was looking for r because period = [2(pi)(r)]/v......

And how did you get to 3t=2Pi*n???? What equation is that?????

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3. solve t by the method I gave, for get r.

3t=0, 2PI, 4PI....is not the period of a sin function =2PI or a multiple thereof?

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bobpursley
4. Well, just think about it logically: if the equation were x=5cos(t) and y=5sin(t), then the period of revolution (time for it to complete one cycle) would be 2 pi, because that's the period of the sin/cos function. However, since it's 3t instead of t, the cycle of rotation would be much faster: set 3t=2*pi, so t, or the new period of revolution, is 2*pi/3.

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5. Well, I am 14 years late to answer this, but just hear me out!
We can see that the x component of the co-ordinate is x=2 cos(3t)
Again, recall that in a circular motion the co-ordinates can be written as:
A cos wt (for x co-ordinate) {Here w is omega, the angular velocity}

See the relation? A cos wt <=> 2 cos 3t?
omega=3

Also, omega = 2*pi*frequency
Or omega=(2*pi)/Time period

3=2pi/T
T=2/3 pi or 2*pi/3

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6. Sorry, I meant 5 cos wt and not 2 cos wt