extra credit

exam-2/exam2.md

@@ -15,8 +15,6 @@ author: |
 \newcommand{\now}[1]{\textcolor{blue}{#1}}
 \newcommand{\todo}[0]{\textcolor{red}{\textbf{TODO}}}
 
-[ 3 8 9 ]
-
 ## Reflection and Refraction
 
 1. \c{Consider a sphere $S$ made of solid glass ($\eta$ = 1.5) that has radius
@@ -186,7 +184,7 @@ author: |
    $z$-axis, the rotation matrix would look like:
 
    $$
-   M =
+   M_1 =
    \begin{bmatrix}
      \cos(23.5^\circ) & \sin(23.5^\circ) & 0 & 0 \\
      -\sin(23.5^\circ) & \cos(23.5^\circ) & 0 & 0 \\
@@ -205,15 +203,15 @@ author: |
    $y$-axis, so the matrix looks like:
 
    $$
-   M' =
+   M_2 =
-   M^{-1}
+   M_1^{-1}
    \begin{bmatrix}
      \cos(\theta(t)) & 0 & \sin(\theta(t)) & 0 \\
      0 & 1 & 0 & 0 \\
      -\sin(\theta(t)) & 0 & \cos(\theta(t)) & 0 \\
      0 & 0 & 0 & 1 \\
    \end{bmatrix}
-   M
+   M_1
    $$
 
    c. \c{[5 points extra credit] What series of rotation matrices could you use
@@ -222,7 +220,39 @@ author: |
 
    In the image, the globe itself does not rotate, but I'm going to assume it
    revolves around the sun at a different angle $\phi(t)$. The solution here
-   would be to
+   would be to rotate the globe _after_ the translation to whatever its position
+   is.
+
+   - First, rotate the globe to the $23.5^\circ$ tilt, using $M_2$ as described
+     above.
+
+   - Then, translate the globe to its position, which I'm going to assume is
+     $(r, 0, 0)$.
+
+     $$
+     M_3 =
+     \begin{bmatrix}
+       1 & 0 & 0 & r \\
+       0 & 1 & 0 & 0 \\
+       0 & 0 & 1 & 0 \\
+       0 & 0 & 0 & 1 \\
+     \end{bmatrix}
+     $$
+
+   - Finally, rotate by $\phi(t)$ around the $y$-axis _after_ the translation,
+     using:
+
+     $$
+     M_4 =
+     \begin{bmatrix}
+       \cos(\phi(t)) & 0 & \sin(\phi(t)) & r \\
+       0 & 1 & 0 & 0 \\
+       -\sin(\phi(t)) & 0 & \cos(\phi(t)) & 0 \\
+       0 & 0 & 0 & 1 \\
+     \end{bmatrix}
+     $$
+
+   The resulting transformation matrix is $M = \boxed{M_4 M_3 M_2}$.
 
 ## The Camera/Viewing Transformation