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I have solved a simple exercise, but in solving it I came to a contradiction. I want to know if it's correct.

In the exercise I have a curve parameterized by the arc length such that:

$\alpha (0)=\frac{1}{\sqrt{3}}(1, 1, 0)$

The binormal vector is $B(t)=\frac{1}{\sqrt{6}}(\sin t-\cos t,\sin t+\cos t,2)$

And the torsion is always negative. ($\tau(t)<0$)

I have to find the curvature of the curve.

I derived the binormal vector and then with the third equation of Frenet and knowing that the torsion is negative I determined the torsion (I took the norm from both sides). Knowing the torsion and using Frenet's third equation again, I could find the normal vector N. Then, I used $ N\wedge B = T $ to find the tangent vector. Finally I used the first Frenet equation to find the curvature. The problem is that when doing the whole process I got to the curvature is negative which is impossible. I want to know if something is wrong in my reasoning.

Pentachito
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  • When a three-dimensional curve is parameterized by the arc length the curvature should be the norm of the second derivative. – Pentachito Oct 20 '19 at 14:29
  • It wouldn’t be the first time that a textbook or other course material had an error in it. What’s the source of this exercise? – amd Oct 20 '19 at 17:41
  • It's from an exam that I had not long ago. – Pentachito Oct 23 '19 at 02:32

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