Dot Product of Constant Magnitude Vector-Valued Function with its Derivative is Zero/Proof 2
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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
\(\ds \map {\mathbf f} x \cdot \map {\mathbf f} x\) | \(=\) | \(\ds c^2\) | Given assumption | |||||||||||
\(\ds \dfrac {\d f} {\d x} \paren {\map {\mathbf f} x \cdot \map {\mathbf f} x}\) | \(=\) | \(\ds 0\) | Derivative of Constant | |||||||||||
\(\ds \dfrac {\d \map {\mathbf f} x} {\d x} \cdot \map {\mathbf f} x + \map {\mathbf f} x \cdot \dfrac {\d \map {\mathbf f} x} {\d x}\) | \(=\) | \(\ds 0\) | Derivative of Dot Product of Vector-Valued Functions | |||||||||||
\(\ds 2 \map {\mathbf f} x \cdot \dfrac {\d \map {\mathbf f} x} {\d x}\) | \(=\) | \(\ds 0\) | Dot Product Operator is Commutative | |||||||||||
\(\ds \map {\mathbf f} x \cdot \dfrac {\d \map {\mathbf f} x} {\d x}\) | \(=\) | \(\ds 0\) |
$\blacksquare$