MathHelp

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

particle.

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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