Show that $|α(t)|$ is a nonzero constant if and only if $α(t)$ is orthogonal to $α'(t)$ for all $t ∈ I$ . Let $\alpha: I → \mathbb{R^3}$ be a parametrized curve, with $α'(t) \neq 0$ for all $t \in I$ . Show
that $|α(t)|$ is a nonzero constant if and only if $α(t)$ is orthogonal to $α'(t)$
for all $t ∈ I$ .
my attempt:
For the second implication: Suppose that $ \alpha (t) \cdot \alpha'(t) = 0 $. I should show that $ | \alpha (t) | = C \neq 0 $, for all $ t \in I $, where $ C $ is a constant. Now $( | \alpha (t) |^2)' = 2 \alpha (t) \alpha'(t) = 0 $. This implies that $ | \alpha (t)| = C$ for some constant $C$. However I could not show that that $ C \neq 0 $.
I need suggestions for the other implication please.
 A: If $\vert \alpha(t) \vert$ is a non-zero constant,
$\vert \alpha(t) \vert = C > 0, \tag 1$
then
$\langle \alpha(t), \alpha(t) \rangle = \vert \alpha(t) \vert^2 = C^2 \tag 2$
is also a constant, whence
$\langle \alpha'(t), \alpha(t) \rangle = \dfrac{1}{2} \dfrac{d}{dt}(\langle \alpha(t), \alpha(t) \rangle) = \dfrac{1}{2} \dfrac{d(C^2)}{dt} = 0, \tag 3$
that is, $\alpha(t)$ is orthogonal to $\alpha'(t)$.
Conversely,
$\langle \alpha'(t), \alpha(t) \rangle = 0 \tag 4$
implies, via
$\langle \alpha'(t), \alpha(t) \rangle = \dfrac{1}{2} \dfrac{d}{dt}(\langle \alpha(t), \alpha(t) \rangle) \tag{4.5}$
and
$\langle \alpha(t), \alpha(t) \rangle \ge 0 \tag 5$
that there exists some non-negative real constant $C$ with
$\langle \alpha(t), \alpha(t) \rangle = C^2; \tag{5.5}$
we may assume $C \ne 0$ for otherwise
$\alpha(t) = 0, \tag 6$
which implies
$\alpha'(t) = 0, \tag 7$
contrary to hypothesis.  It now follows that
$ \vert \alpha(t) \vert^2 = \langle \alpha(t), \alpha(t) \rangle = C^2 > 0, \tag 8$
whence
$\vert \alpha(t) \vert = C > 0. \tag 9$
A: I think you are very close. You know that differentiating $\alpha \cdot \alpha = |\alpha|^2$ gives us this equation.
$$2 \alpha ' \cdot \alpha = \frac{d}{dt}|\alpha|^2$$
If $\alpha'$ and $\alpha$ are always orthogonal, the left side of the equation will be $0$, meaning the derivative of the magnitude is $0$, so the magnitude is constant. If the magnitude is constant, its' derivative is $0$, so the right side will be $0$, meaning $\alpha'$ and $\alpha$ are always orthogonal. At least to me, it seems like we don't need two directions; the proof just falls out of this equation.
