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A regular curve $\textbf{$\gamma$}$ in $\mathbb{R}^3$ with curvature $> 0$ is called a generalized helix if its tangent vector makes a fixed angle $\theta$ with a fixed unit vector $\textbf{a}$. Show that the torsion $\tau$ and curvature $\kappa$ of $\textbf{$\gamma$}$ are related by $\tau = ±\kappa \cot \theta$. Show conversely that, if the torsion and curvature of a regular curve are related by $\tau = \lambda \kappa$ where $\lambda$ is a constant, then the curve is a generalized helix.

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I have done the following:

$\textbf{a}$ is a fixed unit vector: $||\textbf{a}||=1$ and since it is fixed we have that $\textbf{a}'=0$.

$$\textbf{a} \cdot \textbf{t}=||\textbf{a}|| ||\textbf{t}|| \cos \theta = ||\textbf{t}|| \cos \theta \\ \Rightarrow (\textbf{a} \cdot \textbf{t})'=0 \Rightarrow \textbf{a}' \cdot \textbf{t}+\textbf{a} \cdot \textbf{t}'=0 \Rightarrow \textbf{a} \cdot \kappa \textbf{n}=0 \overset{ \kappa >0 }{ \Rightarrow } \textbf{a} \cdot \textbf{n}=0 \Rightarrow \textbf{a} \bot \textbf{n}$$

$\textbf{t}$, $\textbf{n}$ and $\textbf{b}$ consists an orthonormal basis of $\mathbb{R}^3$.

So, $$\textbf{a}=A \textbf{t}+ B \textbf{n}+ C\textbf{b}$$

We have that $\textbf{a}$ is on the plane spanned by $\textbf{t}$ and $\textbf{b}$.

So, $B=0$.

Therefore, $$\textbf{a}=A \textbf{t}+C\textbf{b}$$

How could we continue?

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1 Answer 1

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You know $\mathbf{a \cdot t}=\cos{\theta}$, so you also have $A=\cos{\theta}$. Differentiating gives $$ 0 = \cos{\theta} \, \mathbf{t}' + C' \mathbf{b}+ C \mathbf{b}'. $$ We have $\mathbf{t}'=\kappa \mathbf{n}$, and $\mathbf{b}'=-\tau \mathbf{n}$, so one equation is $$ 0 = (\kappa\cos{\theta}-\tau C) \mathbf{n} + C' \mathbf{b}, $$ and so $$ C'=0, \quad \kappa \cos{\theta} = \tau C. $$ Now we have to get to $C$. But this is easy: $\mathbf{a}$ is a unit vector, so $(\mathbf{a} \cdot \mathbf{b})^2+(\mathbf{a} \cdot \mathbf{t})^2 = 1$, and therefore $1=\cos^2{\theta}+C^2$, or $C=\pm \sin{\theta}$, which gives you the first part.


For the second part, let's try and cook up a constant vector $\mathbf{a}$ so that $$\mathbf{a} \cdot \mathbf{t}=\text{const.}$$

First, differentiating this equation gives $ \kappa \mathbf{a} \cdot \mathbf{n} = 0 $. Assume $\kappa \neq 0$, so the equations are nontrivial, and we again have your equation $$ \mathbf{a} = A \mathbf{t}+C\mathbf{b}, $$ and we then have, differentiating, $$ 0 = (\kappa A-\tau C) \mathbf{n} + A' \mathbf{t} + C' \mathbf{b} = 0, $$ which immediately implies that $A'=C'=0$, and $\kappa A -\tau C=0 $. The relationship $\tau=\lambda \kappa$ then implies $A=\lambda C$, and hence $$ \mathbf{a} = C(\lambda \mathbf{t}+\mathbf{b}). $$ Imposing that $\mathbf{a}$ is a unit vector gives $C^2(1+\lambda^2)=1$. This $\mathbf{a}$ clearly exists and is unique up to sign, and was constructed so that $ (\mathbf{a} \cdot \mathbf{t})'=0 $, so we have that the curve is a generalised helix.

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    $\begingroup$ Yes, $\mathbf{t}$ is defined to be the unit tangent vector. $\endgroup$
    – Chappers
    Commented Oct 13, 2015 at 18:18
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    $\begingroup$ 1) Yes. 2) You have to show that $C$ is constant: it's not necessarily obvious from the previous working. The point is that $\mathbf{b} \perp \mathbf{b}'$ and $\mathbf{t}'$, so you can (say) dot the equation with $\mathbf{b}$ and conclude that $C'=0$. $\endgroup$
    – Chappers
    Commented Oct 13, 2015 at 18:43
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    $\begingroup$ Because $\mathbf{t}'=\kappa \mathbf{n}$ and $\mathbf{n} \perp \mathbf{b}$. $\endgroup$
    – Chappers
    Commented Oct 14, 2015 at 16:20
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    $\begingroup$ $$ 0 = (\kappa A-\tau C) \mathbf{n} + A' \mathbf{t} + C' \mathbf{b} = 0. $$ Dot this with $\mathbf{n}$ and you get $\kappa A-\tau C=0$. You can solve this to find $\tau C = \kappa A$, and the relation $\tau=\lambda \kappa$ allows you to substitute and find $\kappa \lambda C= \kappa A$, so $A=\lambda C$. Inserting this in the equation for $\mathbf{a}$ gives $\mathbf{a}=C(\lambda \mathbf{t}+\mathbf{b})$, and then you can solve for $C$ by choosing $\mathbf{a} \cdot \mathbf{a}=1$. (Note there was an error in my answer, which hopefully I have now fixed.) $\endgroup$
    – Chappers
    Commented Oct 14, 2015 at 17:50
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    $\begingroup$ We are trying to prove the existence of a constant unit vector $\mathbf{a}$ such that $\mathbf{a} \cdot \mathbf{t}=$ const., so first I derive the consequences of $\mathbf{a}$ being constant (i.e. the properties it has to have), and then show that a vector satisfying these exists. $\endgroup$
    – Chappers
    Commented Oct 15, 2015 at 8:04

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