Dot Product of Constant Magnitude Vector-Valued Function with its Derivative is Zero/Proof 1
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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.
Let $\map {\mathbf f} x$ be such that its magnitude is constant:
- $\size {\map {\mathbf f} x} = c$
for some $c \in \R$.
Then the dot product of $\mathbf f$ with its derivative is zero:
- $\map {\mathbf f} x \cdot \dfrac {\d \map {\mathbf f} x} {\d x} = 0$
Proof
\(\ds \map {\mathbf f} x \cdot \dfrac {\d \map {\mathbf f} x} {\d x}\) | \(=\) | \(\ds \size {\map {\mathbf f} x} \dfrac {\d \left\lvert{\map {\mathbf f} x}\right\rvert} {\d x}\) | Dot Product of Vector-Valued Function with its Derivative | |||||||||||
\(\ds \) | \(=\) | \(\ds \size {\map {\mathbf f} x} \dfrac {\d c} {\d x}\) | Magnitude of $\mathbf f$ is constant | |||||||||||
\(\ds \) | \(=\) | \(\ds \size {\map {\mathbf f} x} \times 0\) | Derivative of Constant |
Hence the result.
$\blacksquare$