Level 2: Circle with 2 Points
Again, determine the proportionality constraints of each point, as functions of t. This time, they're on a circle instead of a line. You'll find that the solutions are similar to level 1, with one key distinction: proportionality values can logically be negative on a circle. Because the proportionality on a circle is cyclic, negative values will simply move backwards.
| t | ||||
| 0 | 0 | 6.2831853 | ||