How does the length of the minor semi-axis of the moon's disc's inner terminator vary through the cycle? The curve that separates the bright and dark parts of the moon's disc as viewed from the earth is called the inner lunar terminator, which is semi-elliptical in shape. We can model the disc as a unit circle centred on the origin, in which the inner terminator runs from (0,1) to (0,-1) and its major semi-axis always has length 1.

How does its minor semi-axis vary during the lunar cycle? Please assume that orbits are circular.

In other words, if the point halfway along the inner terminator is X, what is the equation for the motion of X back and forth along the line-segment from (-1,0) to (1,0)?

Note that the shape of the visible disc is not a lune, except in the two trivial cases.
The moon is rotating wrt to the sun. As the moon rotates, the longitude of the terminator increases linearly. Treating the moon as being at "great distance, so the moon appears as an orthographic projection, the position of the middle of the terminator varies sinusoidally, it moves fastest when the moon is at first or third quarter. 

In the orthographic projection, the longitude and latitude $(\lambda,\varphi)$ are mapped to (x,y) by
$$
\begin{align}
x &amp;= R\,\cos\varphi \sin\left(\lambda - \lambda_0\right) \\
y &amp;= R\big(\cos\varphi_0 \sin\varphi - \sin\varphi_0 \cos\varphi \cos\left(\lambda - \lambda_0\right)\big)
\end{align}$$

For points on the equator, and treating the central longitude $\lambda_0$ as the lunar prime meridian, and a scale of R=1, that simplifies to
$x = \sin(\lambda)$. As the terminator moves, its longitude increases linearly, so x varies sinusoidally. 

Correcting for elliptical orbits, librations of the moon and a general perspective view would only marginally alter this.