As far as I understand it. . . the instantaneous direction of motion is vertical because the tangent to the graph of the motion at the cusp is vertical. Direction: The vertical tangent is explained in both the lecture (using approximations) and the Related Reading (using calculus). The slope (dy/dx) of the tangent to the curve signifies the direction of the curve, which in this case is the direction of motion. So the instantaneous direction of motion is vertical. Velocity: However, the Related Reading explains that the velocity of the point at the cusp is 0. So, the point has no instantaneous movement at the cusp (v=change in position/change in time=0). (We have position$OP=<x,y>=<a \theta-asin \theta, a-acos \theta>$. So $velocity =d(OP)/d \theta=<a-acos \theta,-asin \theta>$ which at the cusp is$<a-acos0, -asin0>=<0,0>.)$ So at the cusp, the direction of motion is vertical, while the instantaneous velocity is 0.