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Let $\alpha: I\to \Bbb R^3$ be a regular para curve. I want to prove that:

$|\alpha(t)|=c\ne 0 \iff \langle \alpha(t),\alpha'(t)\rangle = 0,\forall t\in I$

Now $|\alpha(t)|=c\ne 0$ means that this traces some subset or an entire path on the surface of a sphere radius $c$, in some direction. I can see that then the tangent to any point on the sphere is definitely an orthogonal plane(it's flat on the surface). What I can't do is prove either direction mathematically.

$(\implies)$ $|\alpha(t)|=c\ne0$ gives us $x(t)^2+y(t)^2+z(t)^2=c^2$ which is our surface transversal as we know. No idea how to get the orthogonality.

Actually the fact that $xx'+yy'+zz'=0$ looks like it implies that whatever was nonzero in $\alpha$ is zero in $\alpha'$, so then $\alpha'=0$

But the definition of orthogonality comes from nonzero vectors, so I don't know.

Differential Geometry
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    Hint: what's the derivative of $|\alpha(t)|^2$? – Anthony Carapetis Aug 11 '15 at 12:20
  • You've got an answer, but note that "$xx' + yy' + zz' = 0$" says precisely that the dot product of the position $(x, y, z)$ and the velocity $(x', y', z')$ is zero, i.e., these vectors are orthogonal. (It looks as if you were interpreting the left-hand side term by term, which is not productive.:) – Andrew D. Hwang Aug 11 '15 at 13:25

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Hint: $|\alpha(t)| = \sqrt{\langle \alpha(t), \alpha(t) \rangle} = c$, so $$\langle \alpha(t), \alpha(t) \rangle = c^2.$$

What happens when you derive both sides with respect to $t$?

Huy
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  • $\langle\alpha(t),\alpha'(t)\rangle+\langle\alpha'(t),\alpha(t)\rangle=0\implies$ they negate eachother, or both are zero, but they are equal right, so then they must both be zero, and that solves going to the right, is that right? – Differential Geometry Aug 11 '15 at 12:26
  • Yes, both summands on the LHS of your equation are equal because of symmetry of the inner product, so they must both be zero. You can basically go along the exact same chain of thought - but backwards - to get the other direction of the statement you want to prove. – Huy Aug 11 '15 at 12:29
  • Thanks very muchly, helped a bunch. – Differential Geometry Aug 11 '15 at 12:30