Discussion
Let \(\alpha : \left(\dfrac{1}{2}, 2\right) \to \mathbf{R}^{3}\) be defined by \(\alpha(t) = (t, -t, t^{3})\). Then the torsion of \(\alpha\) at \(t = 1\) is :
Answer: C
The torsion is 0.
For \(\alpha(t) = (t, -t, t^{3})\): \(\alpha' = (1, -1, 3t^{2})\), \(\alpha'' = (0, 0, 6t)\), \(\alpha''' = (0, 0, 6)\). Then \(\alpha' \times \alpha'' = (-6t, -6t, 0)\), and
\[ \begin{aligned} \tau &= \frac{(\alpha' \times \alpha'') \cdot \alpha'''}{|\alpha' \times \alpha''|^{2}} \\ &= \frac{0}{72t^{2}} = 0 \end{aligned} \]
The geometric reason: \(y = -x\) at every point, so the whole curve lies in the plane \(x + y = 0\), and a plane curve has zero torsion everywhere, not only at \(t = 1\). The values \(\pm 1\) and 2 cannot occur.
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