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The space curve $$\mathbf x (t) = \begin{pmatrix} \cosh t \\ \sinh t \\ t \end{pmatrix}$$ is an example of a generalized helix, meaning that its tangent vector makes a constant angle $\theta$ with a fixed unit vector $\mathbf A$, so that $\mathbf T \cdot \mathbf A = \cos \theta$. Find the unit vector $\mathbf A$ in this case and the angle $\theta$.

Attempt:

I could write the unit vector $\mathbf A = \cos \theta \mathbf T + \sin \theta \mathbf B$ since then the magnitude is one and the condition $\mathbf T \cdot \mathbf A = \cos \theta$ is easily recovered. Then sub in expressions for $\mathbf T$ and $\mathbf B$, thereby expressing $\mathbf A$ in terms of the basis vectors in Euclidean space, but this doesn't help with finding the angle. Thanks for any pointers.

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  • $\begingroup$ You have that $\mathbf{T}(t)$ and $\mathbf{A}$, so just substitute this back into your defining equation. $\endgroup$ Commented Sep 27, 2014 at 12:38
  • $\begingroup$ Hi Travis, I have $\mathbf T (t)$ in terms of the Euclidean basis, but $\mathbf A$ I have in terms of the frenet-serret basis. So, I subbed in $\mathbf T$ and $\mathbf B$ into my equation for $\mathbf A$ and then took the dot product but that doesn't seem to help. Or did I misunderstand what you meant? $\endgroup$
    – CAF
    Commented Sep 27, 2014 at 12:48
  • $\begingroup$ That's even better, since $\bf{T}$ is orthogonal to $\bf{N}$ and $\bf{B}$, so $\cos \theta$ is just the coefficient of $\bf A$ in the Frenet-Serret basis (at any point). $\endgroup$ Commented Sep 27, 2014 at 12:53
  • $\begingroup$ What coefficient do you mean here? The $\mathbf A$ I am working with is $\mathbf A = \cos \theta \mathbf T + \sin \theta \mathbf B$. Thanks $\endgroup$
    – CAF
    Commented Sep 27, 2014 at 12:56
  • $\begingroup$ Ah, I understand your explanation now, you haven't actually found $\bf A$. $\endgroup$ Commented Sep 27, 2014 at 12:59

1 Answer 1

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Computing gives that (in the Euclidean basis) $${\bf T} = \frac{{\bf x}'}{|{\bf x}'|} = \frac{1}{\sqrt{2}}(\tanh t, 1, \text{sech} t).$$

Also in the Euclidean basis, decompose $${\bf A} = (a, b, c).$$ Now, we know that $\cos \theta = {\bf T} \cdot {\bf A}$ is constant, so $$({\bf T} \cdot {\bf A})' = 0.$$ If we expand this quantity in the Euclidean basis using the above expressions, we get constraints on $a, b, c$, and these together with the fact that $|{\bf A}| = 1$ determines $\bf A$ up to sign.

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  • $\begingroup$ So, $(\mathbf T \cdot \mathbf A)' = \mathbf T ' \cdot \mathbf A = 0 \Rightarrow a/c = \sinh t$ together with the fact that $|\mathbf A| =1 \Rightarrow \sqrt{c^2 \cosh^2 t + b^2} = 1$ $\endgroup$
    – CAF
    Commented Sep 27, 2014 at 13:21
  • $\begingroup$ Since $\bf A$ is constant, its components must also be constant. $\endgroup$ Commented Sep 27, 2014 at 13:40
  • $\begingroup$ If $\mathbf A$ is constant, $(\mathbf T \cdot \mathbf A)' = \mathbf T' \cdot \mathbf A = 0$. Computing $\mathbf T'$ and taking the dot product with $\mathbf A$ gives the constraint $asech^2t = c $sech$ t \tanh t \Rightarrow a/c = \sinh t $. Then I rearranged this for $a$ and subbed into the unity modulus condition and that gave me $c^2 \cosh^2 t + b^2 = 1$. $\endgroup$
    – CAF
    Commented Sep 27, 2014 at 13:48
  • $\begingroup$ It's true that $\frac{a}{c} = \sinh t$ if $c \neq 0$. Whether $c = 0$ or not, we can still say that $a = c \sinh t$. Now, $a$ is constant, so the RHS is constant---what does this tell us about $a$ and $c$? $\endgroup$ Commented Sep 27, 2014 at 13:57
  • $\begingroup$ That would mean $a$ and $c$ are both zero, since we could vary the RHS without affecting the LHS which would not make sense given their established equality. Is that right? $\endgroup$
    – CAF
    Commented Sep 27, 2014 at 14:06

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