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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.

Curious
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    You are differentiating $|\alpha(t)|^2$, first of all. If $C=0$, then what do you conclude about $\alpha'(t)$? – Ted Shifrin Aug 29 '20 at 16:59

2 Answers2

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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

$\vert \alpha(t) \vert = C > 0. \tag 9$

Robert Lewis
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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.

Square
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