show that $\frac{d}{dt}\langle \gamma(t),\eta(t)\rangle =\langle\gamma '(t),\eta (t)\rangle +\langle \gamma (t),\eta ' (t)\rangle$

Question

Let $\gamma,\eta:[a,b]\to \mathbb R^n$ be continuous, differentiable, curves.

show that $$\frac{d}{dt}\langle \gamma(t),\eta(t)\rangle =\langle\gamma '(t),\eta (t)\rangle +\langle \gamma (t),\eta ' (t)\rangle$$

My answer:

I derived the left hand side using the definition of the derivative

$$\frac{d}{dt}\langle \gamma(t),\eta(t)\rangle = \lim_{\delta \to 0} \frac{\langle \gamma(t+\delta),\eta(t+\delta)\rangle - \langle \gamma(t),\eta(t)\rangle}{\delta}$$

And here I am stuck. I don't see any manipulation I can do with these inner products since they have nothing in common.

Answer 1

$$ \begin{align*} \frac{1}{h} & \left[ \langle \gamma(t+h),\, \eta(t+h)\rangle - \langle \gamma(t),\, \eta(t) \rangle \right] \\ & = \frac{1}{h} \left[ \langle \gamma(t+h),\, \eta(t+h)\rangle - \langle \gamma(t),\, \eta(t+h)\rangle \right] + \frac{1}{h} \left[ \langle \gamma(t),\, \eta(t+h)\rangle - \langle \gamma(t),\, \eta(t)\rangle \right] \\ &= \left\langle \frac{1}{h} \left[ \gamma(t+h) - \gamma(t) \right],\, \eta(t+h) \right\rangle + \left\langle \gamma(t),\, \frac{1}{h} \left[ \eta(t+h) - \eta(t) \right] \right\rangle. \end{align*} $$ As $h\to 0$ the first expression goes to $$\frac{d}{dt} \langle \gamma(t), \eta(t) \rangle$$ The last expression converges to $$\langle \gamma^{\prime}(t), \eta(t) \rangle + \langle \gamma(t), \eta^{\prime}(t) \rangle $$ This is done by definition of the derivative, by continuity of $\eta$ and by continuity of the scalar product. The desired equality follows.

Alternatively, in terms of components, $\langle\gamma(t),\eta(t)\rangle=\sum^n_{i=1}\gamma^i(t)\eta^i(t)$. Use the product rule: $$\frac{d}{dt}\langle\gamma(t),\eta(t)\rangle=\sum^n_{i=1}\frac{d}{dt}(\gamma^i(t))\eta^i(t)+\sum^n_{i=1}\gamma^i(t)\frac{d}{dt}(\eta^i(t))=\langle\gamma^{'}(t),\eta(t)\rangle+\langle\gamma(t),\eta^{'}(t)\rangle$$ Q.E.D.