Example of Motion of a Particle in Space

 1 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. 2 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 3 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   Click here to continue with this example.