Vector valued function: What part of the train is always moving backward? So I found out that on a train some part of the wheel will always be moving backward. Thinking about it in terms of a space curve, its the section of the path drawn out that drops below the x axis that corresponds to the part moving backward. Is that correct? Can this be shown using an equation for a vector valued function?

This does not require a very difficult computation. Every point of the wheel is subject to two motions: the global motion of the train (which is the same for all points) and the rotational relative motion due to the turning of the wheel. For the latter, the horizontal component of the velocity is simply proportional the the vertical component of the position relative to the axis. This can be seen from differentiating the vector-valued function $(x(t),y(t))=(r\cos(\omega t),r\sin(\omega t))$ with respect to $t$; one finds $x'(t)=\omega\,y(t)$.
Now the two components of the velocity are exactly opposite for the point of the wheel that is in contact with the rail, given that the wheel is not slipping. Given the mentioned dependence of the horizontal component of the rotational velocity on the position (and the obvious fact that for some points of the wheel above the rail, for instance those at the top of the wheel, the sum of the horizontal components is positive), one sees that the net horizontal velocity is negative precisely for the points of the wheel below the point of contact (of course one must know that train wheels, contrary to most other types of wheels, do have such points): a point of the flange starts and ends moving backwards precisely when it passes the level of contact. In fact for all points of the wheel, that horizontal velocity component is proportional to the current vertical position, relative to the point of contact (if the speed of the train is constant).
