Is it correct to state that $\langle x(t),x(t)\rangle' = 2\langle x'(t),x(t)\rangle$ for an arbitrary inner product?

Question

Let $x:\textbf{R}\to\textbf{R}^{3}$ be a differentiable function, and let $r:\textbf{R}\to\textbf{R}$ be the function $r(t) = \|x(t)\|$, where $\|x\|$ denotes the length of $x$ as measured in the usual $l^{2}$ metric. Let $t_{0}$ be a real number. Show that if $r(t_{0})\neq 0$, then $r$ is differentiable at $t_{0}$, and \begin{align*} r'(t_{0}) = \frac{\langle x'(t_{0}),x(t_{0})\rangle}{r(t_{0})} \end{align*}

MY ATTEMPT

Since $r^{2}(t) = \|x(t)\|^{2} = \langle x(t),x(t)\rangle = x^{2}_{1}(t) + x^{2}_{2}(t) + x^{2}_{3}(t)$, we conclude that \begin{align*} r(t)r'(t) = x_{1}(t)x'_{1}(t) + x_{2}(t)x'_{2}(t) + x_{3}(t)x'_{3}(t) = \langle x'(t),x(t)\rangle \end{align*}

Since $r(t_{0}) \neq 0$, the result follows:

\begin{align*} r'(t_{0}) = \frac{\langle x'(t_{0}),x(t_{0})\rangle}{r(t_{0})} \end{align*}

Based on this exercise, I would like to know if it is correct to state that

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}t}\langle x(t),x(t)\rangle = 2\langle x'(t),x(t)\rangle \end{align*}

for an arbitrary inner product space.

Answer 1

Yes, that's true in general. Even more generally, if $\omega:E \times E \to F$ is a bounded bilinear map between normed vector spaces $E$ and $F$, and if $x: \Bbb{R} \to E$ is a differentiable map, then \begin{align} \dfrac{d}{dt} \bigg|_t \omega(x(t), x(t)) = \omega(x(t), x'(t)) + \omega(x'(t), x(t)). \end{align} If you further assume $\omega$ is symmetric then this reduces to: \begin{align} \dfrac{d}{dt} \bigg|_t \omega(x(t), x(t)) = 2\omega(x'(t), x(t)). \end{align} What you proved is the special case where $E = \Bbb{R}^3$ and $F = \Bbb{R}$ and $\omega(\cdot, \cdot) = \langle \cdot, \cdot \rangle$.

Answer 2

Yes. Notice that \begin{align*} \frac{d}{dt} \langle x(t), x(t) \rangle &= \lim_{h\to 0}\frac{\langle x(t+h),x(t+h) \rangle - \langle x(t), x(t) \rangle}{h}\\ &=\lim_{h\to 0} \frac{\langle x(t+h),x(t+h) \rangle - \langle x(t+h),x(t) \rangle + \langle x(t+h),x(t) \rangle - \langle x(t),x(t) \rangle}{h}\\ &= \lim_{h\to 0} \frac{\langle x(t+h), x(t+h) - x(t)\rangle + \langle x(t+h)-x(t),x(t)\rangle}{h}\\ &= \lim_{h\to 0} \left[ \left \langle x(t+h), \frac{x(t+h)-x(t)}{h}\right \rangle + \left\langle \frac{x(t+h)-x(t)}{h}, x(t) \right\rangle\right]. \end{align*} Now you just need to prove that if $y_h \to y$ and $z_h \to z$ as $h\to 0$, then $$\lim_{h\to 0} \langle y_h,z_h\rangle = \langle y,z\rangle,$$ and you will be done.