Example of Motion of a Particle in Space  (continued)

 4 Next find the unit normal vector,  N  ;  recall   N  =   [dT/ds] / κ   But  κ  =  |  dT/ds  |   Therefore   N  =  dT/ds  /  | dT/ds |   =  [dT/dt] / | dT/dt |     and        dT/dt    =  [- sin t j  -  cos t k ]  / √2        | dT/dt |   =    1 / √2   So    [dT/dt] / | dT/dt |   =  [- sin t j  -  cos t k ]   =   N                     (result)     Check:  The unit tangential and normal vectors are perpendicular.  Therefore their dot product should be zero.              Is        T  ∙  N   =  0 ?        { [ i   +  cos t j  -  sin t k ]  / √2 }  ∙  {  [- sin t j  -  cos t k ] }                =  - cos t  sin t  +  sint t  cos t  =  0    (check) 5 Next find the curvature,  κ .             κ    =    | dT/ds |     =    | dT/dt |  dt/ds   using the chain rule            Now   dt/ds  =  1/v            κ    =    | dT/ds |     =    | dT/dt |  ( 1 / v )   =  (1 / √2) (1 / √2)  =  ½      (result 6 Finally, the acceleration of the particle is the rate of change of the velocity of the particle,   dv/dt.   Recall   v   =     i   +  cos t j  -   sin t k   .   So       a  =  - sin t j  -   cos t k     (result in Cartesian components)   Click here to continue with calculation of the tangential  and normal components of acceleration.