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Generalized helix curve

MathReer1

New member
Feb 28, 2018
2
Problem:
Let $\gamma: (\alpha,\beta)\to\mathbb{R}^3$ be a regular curve with torsion and curvature that are never $0.$
Show that $\gamma$ is a generalized helix curve if and only if the binormal $b(s)$ makes a fixed angle with a constant vector $a \in \mathbb{R}^3, a\not=0.$

Definition: A regular curve $\gamma$ in $\mathbb{R}^3$ with positive curvature is called a generalized helix if its tangent vector makes a fixed angle $\theta$ with a fixed unit vector $a$.

I know that:
the curvature $\kappa > 0$
the torsion $\tau >0 $
$n = b \times t$
$t = b \times n$
$b = t \times n$
$t' = \kappa n$
$b' = - \tau n$

I've tried this so for:

⟹:
$a$ is constant so $a' =0$. The curvature $k > 0$.

$(a\cdot t)'= a'\cdot t + a\cdot t' = a\cdot t' = a\cdot (\kappa n).$

We know that $\kappa > 0,$ so: $a\cdot n$ must be $0,$ and $n = b \times t$ so $a\cdot(b \times t)$ must be zero - and I'm stuck here.

⟸:
$(a\cdot b)'=a'\cdot b+a\cdot b'=a\cdot b'=-a\cdot (\tau n).$ Since $\tau >0$ then $-a\cdot n$ must be zero (once again i'm stuck).
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,177
Rochester, MN
I think you're closer than you think. Take a step back and see what it is you're trying to prove. For the forward direction, you assume $\gamma$ is a generalized helix curve, $t\cdot a=\text{const},$ and you want to show that $b\cdot a=\text{const}$. You have arrived at
$$a\cdot n=0.$$
In the other direction, you showed that $(a\cdot b)'=-\tau(a\cdot n)$. But that derivation didn't depend on the assumption $a\cdot b=\text{const},$ did it? How about this:
$$(a\cdot t)'=a'\cdot t+a\cdot t'=a\cdot t'=\kappa(a\cdot n)=-\frac{\kappa}{\tau}\left[-\tau(a\cdot n)\right]=-\frac{\kappa}{\tau}(a\cdot b')=-\frac{\kappa}{\tau}\left[a'\cdot b+a\cdot b'\right]=-\frac{\kappa}{\tau}(a\cdot b)'.$$
Can you finish?