Dot Product of Vector-Valued Function with its Derivative

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Theorem

Let:

$\map {\mathbf f} x = \ds \sum_{k \mathop = 1}^n \map {f_k} x \mathbf e_k$

be a differentiable vector-valued function.


The dot product of $\mathbf f$ with its derivative is given by:

$\map {\mathbf f} x \cdot \dfrac {\map {\d \mathbf f} x} {\d x} = \size {\map {\mathbf f} x} \dfrac {\d \size {\map {\mathbf f} x} } {\d x}$

where $\size {\map {\mathbf f} x} \ne 0$.


Proof

\(\ds \map {\mathbf f} x \cdot \dfrac {\map {\d \mathbf f} x} {\d x}\) \(=\) \(\ds \map {\mathbf f} x \cdot \sum_{k \mathop = 0}^n \dfrac {\map {\d f_k} x} {\d x} \mathbf e_k\) Definition of Derivative of Vector-Valued Function
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n \map {f_k} x \dfrac {\map {\d f_k} x} {\d x}\) Definition of Dot Product
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^n \map {f_k} x \dfrac {\map {\d f_k} x} {\d x}\) Definition of Dot Product


Then:

\(\ds \dfrac {\d \size {\map {\mathbf f} x} } {\d x}\) \(=\) \(\ds \dfrac \d {\d x} \sqrt {\sum_{k \mathop = 0}^n \paren {\map {f_k} x}^2}\) Definition of Vector Length
\(\ds \) \(=\) \(\ds \frac 1 {2 \sqrt {\ds \sum_{k \mathop = 0}^n \paren {\map {f_k} x}^2} } \dfrac \d {\d x} \sum_{k \mathop = 0}^n \paren {\map {f_k} x}^2\) Chain Rule for Derivatives, Derivative of Power
\(\ds \) \(=\) \(\ds \frac 1 {\size {\map {\mathbf f} x} } \frac 1 2 \sum_{k \mathop = 0}^n \dfrac \d {\d x} \paren {\paren {\map {f_k} x}^2}\) Definition of Vector Length, Sum Rule for Derivatives: General Result
\(\ds \) \(=\) \(\ds \frac 1 {\size {\map {\mathbf f} x} } \frac 1 2 \sum_{k \mathop = 0}^n 2 \map {f_k} x \dfrac {\map {\d f_k} x} {\d x}\) Chain Rule for Derivatives, Derivative of Power
\(\ds \) \(=\) \(\ds \frac 1 {\size {\map {\mathbf f} x} } \map {\mathbf f} x \cdot \dfrac {\map {\d \mathbf f} x} {\d x}\) simplification, and from $(1)$

Hence the result.

$\blacksquare$


Sources

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