Question

Compute the torsion of the circular helix γ(t)=(acos t, asin t, bt)

Answer 1

Question #72105, Math / Geometry

Compute the torsion of the circular helix $y(t) = (\text{acos } t, \text{asin } t, \text{bt})$

Answer.

Arc length parametrization: $y = \frac{1}{c}\left(acos\frac{s}{c}, asin\frac{s}{c}, b\frac{s}{c}\right)$, where $c = \sqrt{a^2 + b^2}$.

$$y'(t) = \left(- \frac {a}{c} \sin \frac {s}{c}, \frac {a}{c} \cos \frac {s}{c}, \frac {b}{c}\right).$$ $$y''(t) = \left(- \frac {a}{c^2} \cos \frac {s}{c}, - \frac {a}{c^2} \sin \frac {s}{c}, 0\right).$$ $$y'''(t) = \left(\frac {a}{c^3} \sin \frac {s}{c}, - \frac {a}{c^3} \cos \frac {s}{c}, 0\right).$$

Torsion $\tau = \frac{\left(u' u'' u'''\right)}{u'' \cdot u''}$.

$$\left(y' y'' y'''\right) = \left| \begin{array}{ccc} - \frac {a}{c} \sin \frac {s}{c} & \frac {a}{c} \cos \frac {s}{c} & \frac {b}{c} \\ - \frac {a}{c^2} \cos \frac {s}{c} & - \frac {a}{c^2} \sin \frac {s}{c} & 0 \\ \frac {a}{c^3} \sin \frac {s}{c} & - \frac {a}{c^3} \cos \frac {s}{c} & 0 \end{array} \right| = \frac {b}{c} \left| \begin{array}{ccc} - \frac {a}{c^2} \cos \frac {s}{c} & - \frac {a}{c^2} \sin \frac {s}{c} \\ \frac {a}{c^3} \cos \frac {s}{c} & - \frac {a}{c^3} \sin \frac {s}{c} \end{array} \right| = \frac {a^2 b}{c^6}.$$ $$y'' \cdot y'' = \left(- \frac {a}{c^2} \cos \frac {s}{c}, - \frac {a}{c^2} \sin \frac {s}{c}, 0\right) \cdot \left(- \frac {a}{c^2} \cos \frac {s}{c}, - \frac {a}{c^2} \sin \frac {s}{c}, 0\right) = \frac {a^2}{c^4}.$$ $$\text{So } \tau = \frac {a^2 b}{c^6} * \frac {c^4}{a^2} = \frac {b}{c^2} = \frac {b}{a^2 + b^2}.$$

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