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?
 A: Points of the wheel of a train touching the track are following a cycloid,



while points near the center follow a curtate cycloid



and points that reach below the track follow a prolate cycloid.



You can see that points below the track will infact travel backwards on a prolate cycloid.
A: This is a question about cycloids. When a wheel of radius $a>0$ rolls on the $x$-axis to the right its center moves according
 to
$$ \left.\eqalign{x(t)&:=a\> t \cr y(t)&:=a \cr}\right\}\qquad (-\infty<t<\infty)\ ,\tag{1}$$
where we have assumed that the angular velocity is $1$. Consider now a point $P$ which is rigidly attached to the wheel at distance $\lambda a>0$ from the axis. The movement of $P$ is a superposition of the translational movement $(1)$ and a rotational movement
$$t\mapsto(-\lambda a\sin t, -\lambda a\cos t)\ .$$
Here we have assumed that at time $t=0$ the point $P$ is at the bottom. (Note that the rotation is clockwise.) In all we obtain the following parametric representation of the curve $\gamma$ described by $P$:
$$\gamma:\qquad  \left.\eqalign{x(t)&:=a\> (t-\lambda \sin t) \cr y(t)&:=a\>(1-\lambda\cos t) \cr}\right\}\qquad (-\infty<t<\infty)\ .$$
Curves of this sort are called cycloids.
Now we look at 
$$\dot x(t)=a(1-\lambda\cos t)\ .$$
When $\lambda>1$ (i.e., $\gamma$ is an elongated cycloid) then there are periodically time intervals in which $\dot x(t)<0$, so that during such intervals the point $P$ moves actually backwards.
Can we find such points $P$ on a train? Yes we do: on the rims of the trains wheels. The rims have a larger radius than the circle rolling on the track.
A: If you have a curve $\gamma$ representing the position of a point (let's say a point on the wheel of a train) and if $v$ is the vector representing the velocity of the train, then the points moving backwards are the ones which satisfty:
$$
 \gamma' \cdot v < 0.
$$
A: Backwards relative to what ? someone on the train or a "stationary" observer ? (Never forget Mr. Einstein). Obviously so for someone on the train: the top goes forward and the bottom goes backward because it is rotating. So I assume you mean relative to a stationary observer, and I think in that case it is not correct.
Let the velocity of the train be magnitude v in a horizontal direction left to right.
Let r be the radius of the wheel and $\theta$ be the clockwise angle from the bottom of the wheel (point of contact) to a point on the wheel's circumference. The rotational velocity of the wheel is $r\theta'$ in a tangential direction. 
The components of the rotational velocity are $r\theta' sin(\theta) $ vertically and $-r\theta' cos(\theta) $ horizontally left to right.
The horizontal component of velocity of the point of contact is 0 = v - $r\theta' cos(0) $, so that $r\theta'$ = v
The horizontal component of velocity at a point at angle $\theta$ on the wheel is v - $r\theta' cos(\theta) $ = $v(1-cos(\theta)$ and this is always >=0.
But I did overlook the fact that this is a train wheel not a car wheel and it does have a rim of larger radius than the rolling surface and as in other posts there are points on this rim moving backwards.
A: Yes. We can model the velocity of any point on the wheel as $$\mathbf{v}=\omega[( a- r\cos\theta)\mathbf i +r\sin\theta]\mathbf j,$$ where $\omega a$ is the forward velocity of the train, $a$ is the rolling radius of the wheel (the distance from the centre of the axle to the rail, so that $\omega$ is the angular velocity of the wheel), $r$ and $\theta$ are respectively the radial distance and the angle forward from the downward vertical from the centre of the axle to the point on the wheel, and $\mathbf i$ and $\mathbf j$ are respectively unit vectors in the forward and upward directions. From this formula, we can see that the $\mathbf i$ component is negative whenever $$r\cos\theta>a.$$ For example, in SI units, taking $a=0.5$, $r=0.55$, $\theta=0$, and $\omega a=30$, we get that a point on the wheel below the contact point with the rail is moving backwards with velocity $3$ metres per second.  
A: 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).
