\documentclass{article}
\usepackage{amsmath}

\title{Notes}
\author{Don Yang}
\begin{document}

\section{de Casteljau's algorithm}

Interpolating between control points for a single component ($a$, $b$,
$c$, and $d$ are control point values):
\begin{align}
ab &= a + t(b - a) \nonumber\\
bc &= b + t(c - b) \nonumber\\
cd &= c + t(d - c) \nonumber\\
abc &= ab + t(bc - ab) \nonumber\\
bcd &= bc + t(cd - bc) \nonumber\\
P(t) &= abc + t(bcd - abc)
\end{align}

\section{First derivative of cubic b\'ezier curves}

Expand $P(t)$:
\begin{align}
ab &= (1 - t)a + t \cdot b \nonumber\\
bc &= (1 - t)b + t \cdot c \nonumber\\
cd &= (1 - t)c + t \cdot d \nonumber\\
abc &= (1 - t)ab + t \cdot bc \nonumber\\
    &= (1 - t)((1 - t)a + t \cdot b) + t((1 - t)b + t \cdot c) \nonumber\\
    &= (1 - t)^2 \cdot a + t(1 - t) \cdot b
      + t(1 - t) \cdot b + t^2 c \nonumber\\
    &= (1 - t)^2 \cdot a + 2t(1 - t) \cdot b + t^2 c \nonumber\\
bcd &= (1 - t)^2 \cdot b + 2t(1 - t) \cdot c + t^2 d \nonumber\\
P(t) &= (1 - t)abc + t \cdot bcd \nonumber\\
     &= (1 - t)((1 - t)^2 a + 2t(1 - t) b + t^2 c)
      + t((1 - t)^2 b + 2t(1 - t) c + t^2 d) \nonumber\\
     &= (1 - t)^3 a + 2t(1 - t)^2 b + t^2(1 - t) c
      + t(1 - t)^2 b + 2t^2(1 - t) c + t^3 d \nonumber\\
     &= (1 - t)^3 a + 3t(1 - t)^2 b + 3t^2(1 - t)c + t^3 d
\end{align}

Derivatives of each part:
\begin{align}
\frac{d}{dt} (1 - t)^3 &= 3(1 - t)^2 (-1) \nonumber\\
                       &= -3(1 - t)^2 \nonumber\\
\frac{d}{dt} 3t(1 - t)^2 &= 3(1 - t)^2 + 3t \cdot 2(1 - t)(-1) \nonumber\\
                         &= 3(1 - t)^2 - 6t(1 - t) \nonumber\\
\frac{d}{dt} 3t^2(1 - t) &= 6t(1 - t) + 3t^2(-1) \nonumber\\
                         &= 6t(1 - t) - 3t^2 \nonumber\\
\frac{d}{dt} t^3 &= 3t^2 \nonumber
\end{align}

First derivative:
\begin{align}
P'(t) &= -3(1 - t)^2 a + (3(1 - t)^2 - 6t(1 - t)) b
       + (6t(1 - t) - 3t^2) c + 3t^2 d \nonumber\\
      &= 3(-(1 - t)^2 a + ((1 - t)^2 - 2t(1 - t)) b
       + (2t(1 - t) - t^2) c + t^2 d)
\end{align}

\section{Cross product of two vectors}

\begin{align}
\begin{vmatrix}
   i & j & k \\
   a_x & a_y & a_z \\
   b_x & b_y & b_z
\end{vmatrix}
&= i \begin{vmatrix} a_y & a_z \\ b_y & b_z \end{vmatrix} -
   j \begin{vmatrix} a_x & a_z \\ b_x & b_z \end{vmatrix} +
   k \begin{vmatrix} a_x & a_y \\ b_x & b_y \end{vmatrix} \nonumber\\
&= i(a_y b_z - b_y a_z) - j(a_x b_z - b_x a_z) + k(a_x b_y - b_x a_y)
\end{align}

\section{Rotation around unit vector}

Let $\boldsymbol{u} = (x, y, z)^T$ be unit vector of rotation axis,
and $\theta$ be the rotation angle, the rotation matrix is:
\begin{align}
\begin{bmatrix}
   0 & -z & y \\
   z & 0 & -x \\
   -y & x & 0
\end{bmatrix}
(sin \theta) + uu^T + (I - uu^T)(cos \theta)
\end{align}
Let $c\theta = cos \theta$ and $s\theta = sin \theta$:
\begin{align}
&
   \begin{bmatrix}
      0 & -z & y \\
      z & 0 & -x \\
      -y & x & 0
   \end{bmatrix}
   s\theta + uu^T + (I - uu^T)c\theta \nonumber\\
&=
   \begin{bmatrix}
      0 & -z s\theta & y s\theta \\
      z s\theta & 0 & -x s\theta \\
      -y s\theta & x s\theta & 0
   \end{bmatrix} +
   \begin{bmatrix}
      x^2 & xy & xz \\
      xy & y^2 & yz \\
      xz & yz & z^2 
   \end{bmatrix} +
   \begin{bmatrix}
      1-x^2 & -xy & -xz \\
      -xy & 1-y^2 & -yz \\
      -xz & -yz & 1-z^2
   \end{bmatrix} c\theta \nonumber\\
&=
   \begin{bmatrix}
      0 & -z s\theta & y s\theta \\
      z s\theta & 0 & -x s\theta \\
      -y s\theta & x s\theta & 0
   \end{bmatrix} +
   \begin{bmatrix}
      x^2 & xy & xz \\
      xy & y^2 & yz \\
      xz & yz & z^2 
   \end{bmatrix} +
   \begin{bmatrix}
      (1-x^2)c\theta & -xy c\theta & -xz c\theta \\
      -xy c\theta & (1-y^2)c\theta & -yz c\theta \\
      -xz c\theta & -yz c\theta & (1-z^2) c\theta
   \end{bmatrix} \nonumber\\
&=
   \begin{bmatrix}
      x^2 + (1 - x^2)c\theta &
      -z s\theta + xy - xy c\theta &
      y s\theta + xz - xz c\theta \\
      z s\theta + xy - xy c\theta &
      y^2 + (1 - y^2)c\theta &
      -x s\theta + yz - yz c\theta \\
      -y s\theta + xz - xz c\theta &
      x s\theta + yz - yz c\theta &
      z^2 + (1 - z^2) c\theta
   \end{bmatrix} \nonumber\\
&=
   \begin{bmatrix}
      x^2 + (1 - x^2)c\theta &
      -z s\theta + (1 - c\theta)xy &
      y s\theta + (1 - c\theta)xz \\
      z s\theta + (1 - c\theta)xy &
      y^2 + (1 - y^2)c\theta &
      -x s\theta + (1 - c\theta)yz \\
      -y s\theta + (1 - c\theta)xz &
      x s\theta + (1 - c\theta)yz &
      z^2 + (1 - z^2)c\theta
   \end{bmatrix} \nonumber\\
&=
   \begin{bmatrix}
      c\theta + (1 - c\theta)x^2 &
      -z s\theta + (1 - c\theta)xy &
      y s\theta + (1 - c\theta)xz \\
      z s\theta + (1 - c\theta)xy &
      c\theta + (1 - c\theta)y^2 &
      -x s\theta + (1 - c\theta)yz \\
      -y s\theta + (1 - c\theta)xz &
      x s\theta + (1 - c\theta)yz &
      c\theta + (1 - c\theta)z^2
   \end{bmatrix}
\end{align}

\end{document}