Derivative of Vector Cross Product of Vector-Valued Functions/Proof 1

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Theorem

Let $\mathbf a: \R \to \R^3$ and $\mathbf b: \R \to \R^3$ be differentiable vector-valued functions in Cartesian $3$-space.


The derivative of their vector cross product is given by:

$\map {\dfrac \d {\d x} } {\mathbf a \times \mathbf b} = \dfrac {\d \mathbf a} {\d x} \times \mathbf b + \mathbf a \times \dfrac {\d \mathbf b} {\d x}$


Proof

Let:

$\mathbf a: x \mapsto \begin {bmatrix} a_1 \\ a_2 \\ a_3 \end {bmatrix}$
$\mathbf b: x \mapsto \begin {bmatrix} b_1 \\ b_2 \\ b_3 \end {bmatrix}$


Then:

\(\ds \mathbf a \times \mathbf b\) \(=\) \(\ds \begin {bmatrix} a_1 \\ a_2 \\ a_3 \end {bmatrix} \times \begin {bmatrix} b_1 \\ b_2 \\ b_3 \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end {bmatrix}\) Definition of Vector Cross Product
\(\ds \leadsto \ \ \) \(\ds \map {\dfrac \d {\d x} } {\mathbf a \times \mathbf b}\) \(=\) \(\ds \begin {bmatrix} \map {\dfrac \d {\d x} } {a_2 b_3 - a_3 b_2} \\ \map {\dfrac \d {\d x} } {a_3 b_1 - a_1 b_3} \\ \map {\dfrac \d {\d x} } {a_1 b_2 - a_2 b_1} \end {bmatrix}\) Differentiation of Vector-Valued Function Componentwise
\(\ds \) \(=\) \(\ds \begin {bmatrix} \dfrac {\d a_2} {\d x} b_3 + a_2 \dfrac {\d b_3} {\d x} - \dfrac {\d a_3} {\d x} b_2 - a_3 \dfrac {\d b_2} {\d x} \\ \dfrac {\d a_3} {\d x} b_1 + a_3 \dfrac {\d b_1} {\d x} - \dfrac {\d a_1} {\d x} b_3 - a_1 \dfrac {\d b_3} {\d x} \\ \dfrac {\d a_1} {\d x} b_2 + a_1 \dfrac {\d b_2} {\d x} - \dfrac {\d a_2} {\d x} b_1 - a_2 \dfrac {\d b_1} {\d x} \end {bmatrix}\) Product Rule for Derivatives, Linear Combination of Derivatives
\(\ds \dfrac {\d \mathbf a} {\d x} \times \mathbf b + \mathbf a \times \dfrac {\d \mathbf b} {\d x}\) \(=\) \(\ds \begin {bmatrix} \dfrac {\d a_1} {\d x} \\ \dfrac {\d a_2} {\d x} \\ \dfrac {\d a_3} {\d x} \end {bmatrix} \times \begin {bmatrix} b_1 \\ b_2 \\ b_3 \end {bmatrix} + \begin {bmatrix} a_1 \\ a_2 \\ a_3 \end {bmatrix} \times \begin {bmatrix} \dfrac {\d b_1} {\d x} \\ \dfrac {\d b_2} {\d x} \\ \dfrac {\d b_3} {\d x} \end{bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} \dfrac {\d a_2} {\d x} b_3 - \dfrac {\d a_3} {\d x} b_2 \\ \dfrac {\d a_3} {\d x} b_1 - \dfrac {\d a_1} {\d x} b_3 \\ \dfrac {\d a_1} {\d x} b_2 - \dfrac {\d a_2} {\d x} b_1 \end {bmatrix} + \begin {bmatrix} a_2 \dfrac {\d b_3} {\d x} - a_3 \dfrac {\d b_2} {\d x} \\ a_3 \dfrac {\d b_1} {\d x} - a_1 \dfrac {\d b_3} {\d x} \\ a_1 \dfrac {\d b_2} {\d x} - a_2 \dfrac {\d b_1} {\d x} \end {bmatrix}\) Definition of Vector Cross Product
\(\ds \) \(=\) \(\ds \begin {bmatrix} \dfrac {\d a_2} {\d x} b_3 - \dfrac {\d a_3} {\d x} b_2 + a_2 \dfrac {\d b_3} {\d x} - a_3 \dfrac {\d b_2} {\d x} \\ \dfrac {\d a_3} {\d x} b_1 - \dfrac {\d a_1} {\d x} b_3 + a_3 \dfrac {\d b_1} {\d x} - a_1 \dfrac {\d b_3} {\d x} \\ \dfrac {\d a_1} {\d x} b_2 - \dfrac {\d a_2} {\d x} b_1 + a_1 \dfrac {\d b_2} {\d x} - a_2 \dfrac {\d b_1} {\d x} \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} \dfrac {\d a_2} {\d x} b_3 + a_2 \dfrac {\d b_3} {\d x} - \dfrac {\d a_3} {\d x} b_2 - a_3 \dfrac {\d b_2} {\d x} \\ \dfrac {\d a_3} {\d x} b_1 + a_3 \dfrac {\d b_1} {\d x} - \dfrac {\d a_1} {\d x} b_3 - a_1 \dfrac {\d b_3} {\d x} \\ \dfrac {\d a_1} {\d x} b_2 + a_1 \dfrac {\d b_2} {\d x} - \dfrac {\d a_2} {\d x} b_1 - a_2 \dfrac {\d b_1} {\d x} \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \map {\dfrac \d {\d x} } {\mathbf a \times \mathbf b}\)

$\blacksquare$


Sources

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