Let $\gamma:I\subset\mathbb R\to\mathbb R^n$ differentiable. Show that $\|\gamma(t)\|$ is constant iff $\gamma(t)\cdot\gamma'(t)=0$ for all $t\in I$.
This is my proof: $\implies]$ Let $\gamma:I\subset\mathbb R\to\mathbb R^n$ differentiable and suppose that there is $k\in\mathbb R^+$ such that $\|\gamma(t)\|=k$ for all $t\in I$. Then, $$\gamma(t)\cdot\gamma(t)=\|\gamma(t)\|=k^2$$ $$\implies\gamma'(t)\cdot\gamma(t)+\gamma(t)\cdot\gamma'(t)=0$$ $$\implies2\gamma'(t)\cdot\gamma(t)=0$$ $$\implies \gamma'(t)\cdot\gamma(t)=0,$$ for all $t\in I$.
$\impliedby]$ ?
I don't know if these steps are invertible to prove the other implication. I want to be a little formal, but I need help to see why I can integrate. Thank you. :)
For the other direction, suppose that $\gamma(t) \cdot \gamma'(t) = 0$ for all $t \in I$. In order to show that $\lVert \gamma(t) \rVert$ is constant, show that its square is constant, which you can do by showing that its derivative vanishes: \begin{align*} \tfrac{d}{dt} \lVert \gamma(t) \rVert^2 &= \tfrac{d}{dt} \bigl[ \gamma(t) \cdot \gamma(t) \bigr] \\ &= 2 \, \gamma(t) \cdot \gamma'(t) \\ &= 0, \end{align*} where the last expression is zero by hypothesis.
Thus, $\lVert \gamma(t) \rVert^2 = c \geq 0$. Let $k = c^{1/2}$ so that $k^2 = c$. Then, $$ \lVert \gamma(t) \rVert = k. $$