# The curve has constant torsion.

Question:

Show that when the curve $c_1=c_1(t)$ has constant torsion $\tau$, the curve $$c_2=c_2(t)=-\frac{1}{\tau}N+\int_{t_0}^{t}B(u)du$$ has constant curvature $-\tau$ or $+\tau$.

What I know:

$$\dot T=\kappa N$$ $$\dot{N}=-\kappa T +\tau B$$ $$\dot B=-\tau N$$

In fact, I posted what i did. But, I am not sure. There may be some mistakes.

• To B11b: Interesting question, but I suggest you edit it a bit; for example, don't you mean $\dot N = -\kappa T + \tau B$ etc.? And shouldn't you have $c_2(t) = \frac{1}{\tau}N + \int_{t_0}^t B(u)du$? I'll edit it if you like. I'm concerned if you don't do these edits, the question is in danger of closure, which I don't think it deserves! By the way, +1! Commented Oct 20, 2013 at 18:57
• Dear @RobertLewis I Will be happy if you will correct :) Commented Oct 20, 2013 at 18:58
• To B11b: OK, I'm on it . . . let me know if I get it right, OK? Commented Oct 20, 2013 at 18:59
• Okay okay @RobertLewis Commented Oct 20, 2013 at 19:00
• To B11b: Glad to help out! Cheers! Commented Oct 20, 2013 at 19:09

You are on the right track. Remember unit tangent $T=c_2^\prime$, so when you get $$c_2^\prime=\frac{\kappa}{\tau}T$$ simply do the substitution and get $$T=\frac{\kappa}{\tau}T$$ then square both sides: $$1=\left(\frac{\kappa}{\tau}\right)^2$$

• I cannot understand what u said. Please can you explain more. $c'_2=\frac{\kappa t}{\tau}$ that's this equal to $T$. Commented Oct 20, 2013 at 21:46
• I dont know how to conclude my solution. Commented Oct 20, 2013 at 21:48
• @B11b: is it better now? Commented Oct 20, 2013 at 21:48
• Then I need to write $\kappa=-\kappa^2 n-\ddot {n}$ right? Commented Oct 20, 2013 at 21:50
• Sorry sorry I understand. Yeup:)) Commented Oct 20, 2013 at 21:52