Example of Motion of a Particle in Space
Let r = (x, y, z) be a position vector from the origin to a particle located by the
position vector, r , given by
r = t i + sin t j + cos t k where i , j, k are unit vectors (base vectors)
along the x, y, and z axes.
where t is any arbitrary time
Find the velocity of the particle, v, the unit tangential vector, T , the unit normal
vector, N, the curvature of the particle’s path, κ , and the acceleration, a, of the
The derivative of the position vector, r , gives the velocity of the particle.
So v = dr / dt = i + cos t j - sin t k (result)
Since the velocity is always tangent to its path, the unit tangential vector can be
found by dividing the velocity vector by its magnitude.
T = v / | v | = [ i + cos t j - sin t k ] / √ (12 + cos2t + sin2t )
Therefore | v | = v = √2
So T = [ i + cos t j - sin t k ] / √2
Recall that the rate of change of the unit tangential vector along its path, dT/ds
is given by the product of the curvature of the path, κ , with its unit normal vector, N.
So dT/ds = κ N
and | dT/ds | = κ since | N | = 1
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