Ask your own question, for FREE!
OCW Scholar - Multivariable Calculus 20 Online
OpenStudy (anonymous):

Can someone please explain why the cycloid curve has a vertical movement on the cusps? Thank you!

OpenStudy (anonymous):

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.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!