Circle rotating within a circle (roulette) This was something in a course of mine I'm a bit too thick to see. If one takes a circle of radius $3$ and a circle of radius $1$, and rolls the smaller circle smoothly inside the larger one until the point $P$ which starts out in contact with the larger circle is in the same place, how can one prove that the smaller circle goes through exactly two rotations?
The curve generated by the path of the vector $\mathbf{P}$ (the vector from the center of the larger circle to $P$) is called a roulette - more specifically, a deltoid.

If you trust the picture, then you can see that the smaller circle does go through two rotations: letting $\mathbf{C}$ be the center of the smaller circle, the little vector pointing from $\mathbf{C}$ to $\mathbf{P}$ rotates around to pointing due right exactly twice, so the circle rotates twice also. But I don't know how to prove this.
This question came up because we wanted to find the parametric equation of $\mathbf{P}$. It'll be the sum of the vectors $\mathbf{C}$ and $\mathbf{P- C}$; the first one has equation $(2\cos(t), 2\sin(t))$, and if the smaller circle rotates around (clockwise) exactly two times, it'll have the equation $(\cos(2t), -\sin(2t))$. This would give the parametrization of the curve immediately. But how to prove that?
(More generally, it seems like a circle of radius $1$ rotating in a circle of radius $n$ will go around $n-1$ times.)
 A: From the following figure it can immediately been read off that when the point  of contact $Q$ moves counterclockwise by the angle $\phi$ then the inner circle (as a rigid structure) rotates by the amount $2\phi$ clockwise. Therefore after a full turn of $Q$ the point $P$ has made $2$ turns.

A: Three answers here; the last one is probably the best. 
Answer 1
The length of the outer circle is $6\pi$; the length of the inner circle is $2\pi$. That sounds as if it should make for THREE revolutions. But if you consider the case where the inner and outer circles are of identical size, you see that "one revolution comes for free." 
To say that with more detail: start with the two circles tangent at the top point of each. Draw a horizontal tangent to the outer circle. Now roll the inner circle for a distance $t$ along that tangent line to the right, and then "wrap" the tangent line onto the inner circle. In the first step (rolling a distance $t$), the inner circle rotates counterclockwise by an angle $t$ (definition of angle measure as "length of a unit circle's arc that's subtended by a pair of rays emanating from its center", or something like that). When the tangent is wrapped onto the large circle, the small circle, being attached to it, rotates CLOCKWISE a certain amount. How much? The angle subtended by an arc of length $t$ at the large-circle center, i.e., $t/3$. Net rotation? $t - t/3 = \frac{2}{3}t$. When the total distance travelled is $t = 6\pi$, the circumference of the large circle, the amount rotated by the small circle is $\frac{2}{3} 6 \pi = 4\pi$. 
Answer 2
Let me try with vectors and stuff. I'm going to say that the large circle's centered at the origin, $P = (0, 0)$ and that at time $t = 0$, the small circle is tangent to the large one at $(3, 0)$. And the small circle's center is at location $R(t)$, which moves counter clockwise. The rightmost point of the small circle at the start I'll call $S$, with $S(0) = (3, 0)$, and $S(t)$ denoting its position at time $t$; I'll let $r(t) = S(t) - P(t)$. So far so good? 
OK. I'm going to suppose that the point of contact, $C(t)$, moves around the large circle, CCW, so that at time $t$ its distance from the starting point $(3, 0)$ is $3t$ units of arclength. That means that 
$$
C(t) = (3 \cos t, 3 \sin t).
$$ 
(Check that: the length of $C'(t)$ is 3, so we're good.)
According to my claim above, the position $r(t)$ of the marked point, relative to the center, should be rotated by angle $2t$ instead of $3t$, and clockwise, so 
$$
r(t) = (\cos (-2t,) \sin (-2t))
$$
And since $R(t)$ is on the line from the origin to the point of contact $C(t)$, and is 2 units from the origin, we have
$$
R(t) = (2 \cos t, 2\sin t)
$$
so that 
$$
S(t) = (2 \cos(t) + \cos(-2t), 2\sin(t) + \sin(-2t)).
$$
The only question now is "Is the small circle rolling against the large circle without slipping?" 
At time $t$ the point of contact is $C(t)$; its tangent vector is 
$$C'(t) = (-3 \sin t, 3 \cos t)$$.
At time $t$, the speed of any point on the small circle (in a frame of reference based at $R(t)$) is the same as the speed of $r(t)$ in that frame, which is $2$. And the direction is tangent to the small circle, clockwise. For the point at the point of contant, that clockwise vector of length $2$ is 
$$
(-\sin(t), \cos(t))  
$$
But in the frame of the circle-center, we must add to this the velocity vector of $R(t)$, namely
$$
R'(t) = (-2 \sin(t), 2 \cos t).
$$
So the velocity of that point of the small circle, at the mooment of tangency, is the sum of these, i.e.
$$
(-3\sin t, 3 \cos t)
$$
which is the same velocity with which the poitn fo tangency is moving along the large circle, so they're rolling without slipping. 
(I've probably lost a sign or two here, but perhaps not...)
Answer 3
Using the notation of the previous answer, under the motion I described for the small circle, the point of the small circle that's farthest to the right at time 0, which I've called "S", is at location 
$$
S(t) = P + R(t) + r(t) = (0,0) + (2 \cos t, 2 \sin t)  + (\cos(-2t),  \sin(-2t))
$$
at time $t$. 
The point of tangency with the big circle is at location $C(t) = (3 \cos t, 3 \sin t)$.
Since there's nothing special about the point $S$ -- any other point of the small circle describes a path just like S's path, but rotated a bit inside the big circle --- I just need to show that at the moment when $S$ is at the point of tangency, the velocity vector of $S(t)$ is zero; that's what it means to roll without slipping. Well
\begin{align}
S'(t) &= (-2 \sin t, 2 \cos t)  + (2\sin(-2t),  -2\cos(-2t))\\
S'(0) &= (0, 2)  + (0,  -2)\\
&= (0, 0)
\end{align}
This also, of course, matches the "deltoid" picture: at the moment of tangency, the marked point has stopped moving out towards the large circle and is about to move in towards its center; it has no circumferential velocity at all. 
So: since my moving point (a) is on the little circle, and (b) has velocity zero at the moment of tangency, my description of the motion of the little circle must be correct. 
At what time $t$ has the point of tangency $C(t)$ moved all the way around the big circle CCW? Ans: when $t = 2\pi$. And how much rotation (relative to its moving center) has the small circle undergone? Well, since $r(t) =  (\cos(-2t),  \sin(-2t))$, it's made two full clockwise turns. 
A: Notice the center of the inner radius is at distance $R-r$ from the center of the outer radius, where $R$ is the radius of the outer circle, and $r$ the radius of the inner circle.
So, when it does one turn, it runs the distance $d=2\pi (R-r)$.
Now, regarding only the rotation of a fixed point $P$ on the circle, it's the same, that it rolls round a circle, or on a line. But on a line it's simpler: the center advances by distance $d$, so a point on the circumference advances by the same amount, or $d=\theta r$. So it turns by $\theta=2\pi\frac{R-r}{r}$, and here, $4\pi$, or two turns.
A: The smaller circle is actually making 3 rotations in the larger Circle: when the Red Dot on the smaller Circle points to the right for the first time, it's actually half way through the second rotation. 
The Red Dot pointing  due right is not actually significant of a full rotation. A full rotation of the smaller Circle is actually signified by the Red Dot touching the Outer Circle again. 
It's more easily seen that there's three rotations of the smaller Circle if you start the smaller Circle in the bottom of the larger Circle, marking the two where they touch together and rotating the larger circle around until the two points come back together. You'll see that the smaller Circle actually next three rotations,
