37. The Tangent Vector1
Dated: 05-07-2025
\[
\vec P^\prime(u)
=
\begin{bmatrix}
\frac d {du} x(u) \hat i &
\frac d {du} y(u) \hat j &
\frac d {du} z(u) \hat k
\end{bmatrix}
\]
\[
\vec P^\prime(u)
=
\begin{bmatrix}
x^\prime & y^\prime & z^\prime
\end{bmatrix}
\]
The set2 of vectors1 \(p_0, p_1, p_0^\prime\) and \(p_1^\prime\) are called boundary conditions.
This method itself is called the cubic Hermite interpolation, after C. Hermite (1822-1901) the French mathematician who made significant contributions to our understanding of cubic and quadratic polynomials.
\[x(u) = a_x u^3 + b_x u^2 + c_x u + d_x\]
\[\frac d {du} x(u) = \frac d {du} \left(a_x u^3 + b_x u^2 + c_x u + d_x\right)\]
\[x^\prime = 3a_x u^2 + 2b_x u + c_x\]
\[x(0) = x_0 = d_x\]
\[x(1) = a_x + b_x + c_x + d_x\]
\[x^\prime(0) = x^\prime_0 = c_x\]
\[x^\prime(1) = x^\prime_1 = 3a_x + 2b_x + c_x\]
Solving for \(a_x, b_x, c_x\) and \(d_x\), we get
\[a_x = 2(x_0 - x_1) + x_0^\prime + x_1^\prime\]
\[b_x = 3(-x_0 + x_1) - 2x_0^\prime - x_1^\prime\]
\[c_x = x_0^\prime\]
\[d_x = x_0\]
Substituting in \(\vec x(u)\), we get
\[
\vec x(u) = (2x_0 - 2x_1 + x_0^\prime + x_1^\prime) u^3 + (-3x_0 + 3x_1 - 2x_0^\prime - x_1^\prime x_0^\prime) u^2 + x_0^\prime + x_0
\]
\[
\vec x(u) = (2u^3 - 3u^2 + 1)x_0 + (-2u^3 + 3u^2)x_1 + (u^3 - 2u^2 + u)x_0^\prime + (u^3 - u^2)x_1^\prime
\]
Because \(\vec y(u)\) and \(\vec z(u)\) have equivalent forms
\[
\vec P(u) = (2u^3 - 3u^2 + 1)p_0 + (-2u^3 + 3u^2)p_1 + (u^3 - 2u^2 + u)p_0^\prime + (u^3 - u^2)p_1^\prime
\]
Let
\[
\begin{aligned}
& F_1 = 2u^3 - 3u^2 + 1 \\
& F_2 = -2u^3 + 3u^2 \\
& F_3 = u^3 - 2u^2 + u \\
& F_4 = u^3 - u^2
\end{aligned}
\]
\[
F =
\begin{bmatrix}
F_1 & F_2 & F_3 & F_4
\end{bmatrix}
\]
\[
B =
\begin{bmatrix}
p_0 & p_1 & p_0^\prime & p_1^\prime
\end{bmatrix}^T
\]
\[
\vec P(u) = F_1 p_0 + F_2 p_1 + F_3 p_0^\prime + F_4 p_1^\prime
\]
\[\vec P(u) = FB\]
\(F\) can be further decomposed into
\[F = UM\]
where
\[
U = \begin{bmatrix} u^3 & u^2 & u & 1 \end{bmatrix}
\]
\[
M =
\begin{bmatrix}
2 & -2 & 1 & 1 \\
-3 & 3 & -2 & -1 \\
0 & 0 & 1 & 0 \\
1 & 0 & 0 & 0
\end{bmatrix}
\]
\[\vec P(u) = UMB\]
\[\vec p_0^\prime = \hat {p_0^\prime} \cdot |p_0^\prime|\]
\[\vec p_1^\prime = \hat {p_1^\prime} \cdot |p_1^\prime|\]
\[
\therefore \vec P(u) = (2u^3 - 3u^2 + 1)p_0 + (-2u^3 + 3u^2)p_1 + (u^3 - 2u^2 + u)\hat {p_0^\prime} \cdot |p_0^\prime| + (u^3 - u^2)\hat {p_1^\prime} \cdot |p_1^\prime|
\]
\[
\implies
B =
\begin{bmatrix}
p_0 \\
p_1 \\
\hat {p_0^\prime} \cdot |p_0^\prime|\\
\hat {p_1^\prime} \cdot |p_1^\prime|\\
\end{bmatrix}
\]