MathematicalRigor

Theorem: Provided |T 0 | = 6 0 and |r00 | = 6 0, then |r00 × T 0 | = 0 if and only if d 0 dt |r | = 0 By use of the division rule one can find: T0 = d 0 r0 dt |r | d r0 r00 = − 0 0 0 dt |r | |r | |r |2 Cross product is distributive over addition, thus r00 × T 0 = d 0 1 00 00 dt |r | · (r00 × r0 ) · (r × r ) − |r0 | |r0 |2 Simplifying further we reach d 0 |r | r00 × T 0 = − dt 0 2 · (r00 × r0 ) |r | d 0 00 0 00 0 dt |r | = 0, then |r × T | = 0. Conversely if |r × T | = 0, then d 0 |r | d 0 as |T 0 × r0 | = 6 0, one can conclude − dt|r0 |2 = 0 and dt |r | = 0. If 1 |r00 × r0 | = 6 0