Good day, so the question is from the MIT open courseware page and comes from here which is the "problem set 3". The problem that I don't understand is as follows:
A circular disk of radius 2 has a dot marked at a point halfway between the center and the circumference. Denote this point by $P$. Suppose that the disk is tangent to the $x$-axis with the center initially at $(0, 2)$ and P initially at $(0, 1)$, and that it starts to roll to the right on the x-axis at unit speed. Let C be the curve traced out by the point P.
b) Use vectors to find the parametric equations for $\vec{OP}$ as a function of time $t$.
The provided solution claims that there is a graph, however, I don't see one. From the mentioned answer, I read the following: (b) To get to the point $P$, start at the origin, add the vector $(0, 2)$ to go up to the center of the disk at $t = 0$; then add the vector $(t, 0)$ to get to the center of the disk at time t; and finally (do what it takes to) shift over by one unit at an angle $θ = t/2$, measured clockwise from the vertical (see below). How to do that last move? Just add the vector $−(\sin(t/2), \cos(t/2))$, as can be seen from a sketch. [note. there is no sketch]. Combining, we get $\vec{OP}(t) = (0, 2) + (t, 0) +( − \sin(t/2), − \cos(t/2))$ or $r = r(t) = (t − \sin(t/2), 2 − \cos(t/2))$.
Everything else being clear I just do not see how subtracting $(−\sin(t/2),−\cos(t/2))$ would do the job.
Here's your sketch:
Initially the center of the circle is at $A(0,2)$ and the dot is at $P(0,1),$ as shown in the circle at the left. Note that the coordinates of these points imply the coordinates of the vector $\vec{AP} = \langle 0,-1 \rangle.$
Now we roll the circle along the $x$ axis until its center is at $A'.$ If the circle had slid along the $x$ axis instead of rolling, the dot would now be at $Q,$ directly below $A'.$ But because of the rolling, the circle rotated clockwise and now the dot is at $P'$ with $\angle QA'P' = t/2$ (measured in radians).
Note that $A'R = \cos(t/2)$ (the difference in $y$ coordinates between $A'$ and $P'$) and $P'R = \sin(t/2)$ (the difference in $x$ coordinates). In the figure as shown both $\cos(t/2)$ and $\sin(t/2)$ are positive. But because we go downward from $A'$ to $R$ and leftward from $R$ to $P',$ we need to add negative values to the $x$ and $y$ coordinates of $A'$ in order to arrive at the $x$ and $y$ coordinates of $P'.$ Specifically, we need the coordinates of the vector between these points to be $\vec{A'P'} = \langle -\sin(t/2), -\cos(t/2) \rangle.$
In short, when the angle $t/2$ is acute (and therefore both its sine and cosine are positive), the vector $\vec{A'P'}$ points downward and to the left, so it requires both coordinates to be negative. To get negative numbers from positive numbers you need to apply a $-$ sign.
It is not clear why you assume that $P$ should have coordinates $(\sin(1/t),\cos(1/t))$. (Surely, to start with, you meant $t/2$ rather than $1/t$.)
The points on a unit circle with center at $(0,0)$ (not $(0,2)$) are often described as having coordinates $(\cos\theta, \sin\theta)$ (note that the cosine comes first!), but that's assuming that we start with the point $(1,0)$ (directly to the right of the circle's center) when $\theta = 0$ and proceed counterclockwise around the circle. In this problem the motion around the center of the circle is quite different: we start below the center and move in a clockwise direction.
Since the starting point (when $\theta = 0$) and direction of rotation (as $\theta$ increases) are different from the usual unit-circle arrangement, we have to work out the coordinates from first principles.
By the way, if the solution had said, "as can be seen in the sketch," I would expect to see a sketch. But instead it says, "as can be seen in a sketch," which is something people may write when they are inviting the readers to draw their own sketches. Drawing a sketch like the one above and inserting it in an on-line document (like the problem set, or like this answer) is a lot more effort than doodling one on a piece of scrap paper, so perhaps we can forgive the authors of open courseware from putting the burden on the students to make those doodles. It's true, however, that a picture is worth a lot of words of explanation, so I made the effort to upload one.