Expert: Steve Holleran - 5/25/2008

Question
Hi Steve,

A particle moves along a circular path that is centered at the origin in the x-y plane such that at time t>=0 the    position is given by the vector function s(t)=<(1-t^2)/1+t^2, 2t/(1+t^2)>.

  1. Find the radius of the particles path.
  2. Find the velocity vector of the particle.
  3. Is the particle ever at rest?
  4. If t was allowed to increase without bound, at what point would the particle be?

Answer
Hi Vince,

I haven't done one of these in years, but I think this is the procedure:

1. Since s(t) = (1-t^2 / 1+t^2)i + (2t / 1+t^2)j

think of x(t) = 1-t^2 / 1+t^2   and y(t) = 2t / 1+t^2

then r = sqrt(x(t)^2 + y(t)^2)

      = sqrt( [1-t^2/1+t^2]^2 + [2t/1+t^2]^2)

      = sqrt( 1 - 2t^2 + t^4 / (1+t^2)^2 + 4t^2/(1+t^2)^2)

     = sqrt( [1 + 2t^2 +t^4]/[1+t^2]^2)

    =  sqrt( [1+t^2]^2 / [1+t^2]^2) = sqrt(1) = 1

2.  v(t) = s'(t) = x'(t)i + y'(t)j

and so x'(t) = [(1+t^2)(-2t)-((1-t^2)(2t)] / (1+t^2)^2

            = [-4t] / (1+t^2)^2

and y'(t) = [(1+t^2)(2) -(2t)(2t)] / (1+t^2)^2

         = [2-2t^2] / [1+t^2]^2


3.  Particle at rest if v = 0 , so I think this means the x and y components of the velocity vector have to both = 0, and since x'(t) = 0 at x = 0, but y'(t) = 0 at t = 1, I would think that the velocity is not zero (but I'm not positive about this).

4.  As t --> inf, x(t) -> -1 and y(t)-> 0, so I think the point would
   be (-1, 0)

I'm pretty rusty on this stuff and the improper integrals.

This is the best I can come up with.
Steve