# Derivative of Vector Cross Product of Vector-Valued Functions

## Theorem

Let:

$\mathbf r: t \mapsto \begin {bmatrix} x \\ y \\ z \end{bmatrix}$
$\mathbf q: t \mapsto \begin {bmatrix} \chi \\ \gamma \\ \zeta \end{bmatrix}$

be differentiable vector-valued functions, where:

$x, y, z, \chi, \gamma, \zeta$

are (images of) differentiable real functions.

The derivative of the vector cross product of $\mathbf r$ and $\mathbf q$ is given by:

$D_t \left({\mathbf r \left({t}\right) \times \mathbf q \left({t}\right)}\right) = \mathbf r' \left({x}\right) \times \mathbf q \left({x}\right) + \mathbf r \left({x}\right) \times \mathbf q'\left({x}\right)$

## Proof

 $\displaystyle \mathbf r \left({t}\right) \times \mathbf q \left({t}\right)$ $=$ $\displaystyle \begin{bmatrix} x \\ y \\ z \end{bmatrix} \times \begin{bmatrix} \chi \\ \gamma \\ \zeta \end{bmatrix}$ $\displaystyle$ $=$ $\displaystyle \begin{bmatrix} y \zeta - z \gamma \\ z \chi - x \zeta \\ x\gamma -y \chi \end{bmatrix}$ Definition of Vector Cross Product $\displaystyle \leadsto \ \$ $\displaystyle D_t \left({\mathbf r \left({t}\right) \times \mathbf q \left({t}\right)}\right)$ $=$ $\displaystyle \begin {bmatrix} D_t\left({y \zeta - z \gamma}\right) \\ D_t\left({z \chi - x \zeta}\right) \\ D_t\left({x\gamma -y \chi}\right) \end{bmatrix}$ Differentiation of Vector-Valued Function Componentwise $\displaystyle$ $=$ $\displaystyle \begin {bmatrix} \dfrac{\d y} {\d t}\zeta + y \dfrac {\d \zeta} {\d t} - \dfrac {\d z} {\d t} \gamma - z \dfrac {\d \gamma} {\d t} \\ \dfrac {\d z} {\d t} \chi + z \dfrac {\d \chi} {\d t} - \dfrac {\d x} {\d t} \zeta - x \dfrac {\d \zeta} {\d t} \\ \dfrac {\d x} {\d t} \gamma + x \dfrac {\d \gamma} {\d t} - \dfrac {\d y} {\d t} \chi - y \dfrac {\d \chi} {\d t} \end{bmatrix}$ Product Rule for Derivatives, Linear Combination of Derivatives $\displaystyle \mathbf r' \left({x}\right) \times \mathbf q \left({x}\right) + \mathbf r \left({x}\right) \times \mathbf q'\left({x}\right)$ $=$ $\displaystyle \begin {bmatrix} \dfrac {\d x} {\d t} \\ \dfrac {\d y} {\d t} \\ \dfrac {\d z} {\d t} \end {bmatrix} \times \begin {bmatrix} \chi \\ \gamma \\ \zeta \end {bmatrix} + \begin {bmatrix} x \\ y \\ z \end {bmatrix} \times \begin {bmatrix} \dfrac {\d \chi} {\d t} \\ \dfrac {\d \gamma} {\d t} \\ \dfrac {\d \zeta} {\d t} \end{bmatrix}$ $\displaystyle$ $=$ $\displaystyle \begin {bmatrix} \dfrac {\d y} {\d t} \zeta - \dfrac {\d z} {\d t} \gamma \\ \dfrac {\d z} {\d t} \chi - \dfrac {\d x} {\d t} \zeta \\ \dfrac {\d x} {\d t} \gamma - \dfrac {\d y} {\d t} \chi \end {bmatrix} + \begin {bmatrix} y \dfrac {\d \zeta} {\d t} - z \dfrac {\d \gamma} {\d t} \\ z \dfrac {\d \chi} {\d t} - x \dfrac {\d \zeta} {\d t} \\ x \dfrac {\d \gamma} {\d t} - y \dfrac {\d \chi} {\d t} \end {bmatrix}$ Definition of Vector Cross Product $\displaystyle$ $=$ $\displaystyle \begin {bmatrix} \dfrac {\d y} {\d t} \zeta - \dfrac {\d z} {\d t} \gamma + y \dfrac {\d \zeta} {\d t} - z \dfrac {\d \gamma} {\d t} \\ \dfrac {\d z} {\d t} \chi - \dfrac {\d x} {\d t} \zeta + z \dfrac {\d \chi} {\d t} - x \dfrac {\d \zeta} {\d t} \\ \dfrac {\d x} {\d t} \gamma - \dfrac {\d y} {\d t} \chi + x \dfrac {\d \gamma} {\d t} - y \dfrac {\d \chi} {\d t} \end {bmatrix}$ $\displaystyle$ $=$ $\displaystyle \begin {bmatrix} \dfrac {\d y} {\d t} \zeta + y \dfrac {\d \zeta} {\d t} - \dfrac {\d z} {\d t} \gamma - z \dfrac {\d \gamma} {\d t} \\ \dfrac {\d z} {\d t} \chi + z \dfrac {\d \chi} {\d t} - \dfrac {\d x} {\d t} \zeta - x \dfrac {\d \zeta} {\d t} \\ \dfrac {\d x} {\d t} \gamma + x \dfrac {\d \gamma} {\d t} - \dfrac {\d y} {\d t} \chi - y \dfrac {\d \chi} {\d t} \end {bmatrix}$ $\displaystyle$ $=$ $\displaystyle D_t \left({\mathbf r \left({t}\right) \times \mathbf q \left({t}\right)}\right)$

$\blacksquare$