I am working on a question and I don't have the slightest idea where to begin. Any nudge in the right direction would be very helpful.
Here is the question:
A bicycle wheel has radius R. Let P be a point on the spoke of a wheel at a distance d from the center of the wheel. The wheel begins to roll to the right along the x-axis. The curve traced out by P is given by the following parametric equations:
$x=22\theta-15sin(\theta)$
$y=22-15cos(\theta)$
What must we have for R and d?
On
We measure the angle $\theta$ as the angle that a line from the center of the wheel to the point $P$ makes with the negative $y$-axis. Without Loss of Generality, the point $P$ is at $(0,R-d)$ when $\theta =0$.
As the center of the wheel moves to the right by a distance $x_C$, the wheel has traveled a distance $R \theta$. Therefore, the parametric equation for the center of the wheel is simply $x_C=R\theta$ and $y_C=R$.
If we observe the motion of $P$ from the frame of reference of the center of the wheel, we observe simple circular motion. Thus, from the center of the wheel the parametric description of the motion of $P$ is $x=-d\sin \theta$ and $y=-d\cos \theta$.
Putting the two motions together we have
$$\begin{align} x&=R\theta -d\sin \theta\\\\ y&=R-\cos \theta \end{align}$$
Thus, $R=22$ and $d=15$.
Lets find $d$ first. Our point is moving with simple harmonic motion in the vertical direction, and it has a maximum of $22+15 = 37$ and a minimum of $22-15 = 7$. So, its distance from the centre of the wheel is $d = (37 - 7)/2 = 15$.
After the wheel has made one full revolution, the point $P$'s $x$-coordinate has moved from $0$ to $22\cdot 2\pi$. This must be the same as the circumference of the wheel. So, $R=22$ must be the radius of the wheel.