Question

COMPUTE THE TORSION OF γ(t) = (cos^3 t,sin^3 t)

Answer 1

Answer on Question #72485, Math / Geometry

Q. Compute the torsion of

$$\gamma(t) = (\cos^3 t, \sin^3 t)$$

Solution

Torsion: $\tau(t) = \frac{(\overrightarrow{\gamma'(t)} \times \overrightarrow{\gamma''(t)}) \cdot \overrightarrow{\gamma'''(t)}}{\|\overrightarrow{\gamma'(t)} \times \overrightarrow{\gamma''(t)}\|^2}$

$\gamma(t) = (\cos^3 t, \sin^3 t)$

A plane curve with non-vanishing curvature has zero torsion at all points.

$\overrightarrow{\gamma'(t)} = (-3\sin t\cos^2 t, 3\sin^2 t\cos t, 0)$

$\overrightarrow{\gamma''(t)} = (-3\cos^3 t + 6\sin^2 t\cos t, -3\sin^3 t + 6\sin t\cos^2 t, 0)$

$\overrightarrow{\gamma'''(t)} = (21\sin t\cos^2 t - 6\sin^3 t, -21\sin^2 t\cos t + 6\cos^3 t, 0)$

$$\begin{array}{l} \overrightarrow{\gamma'(t)} \times \overrightarrow{\gamma''(t)} = \\ = \left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ -3\sin t\cos^2 t & 3\sin^2 t\cos t & 0 \\ -3\cos^3 t + 6\sin^2 t\cos t & -3\sin^3 t + 6\sin t\cos^2 t & 0 \end{array} \right| = \\ = -9\vec{k}(\sin^4 t\cos^2 t + \sin^2 t\cos^4 t) = -9(\sin^2 t\cos^2 t)\vec{k} \end{array}$$ $$\begin{array}{l} \left(\overrightarrow{\gamma'(t)} \times \overrightarrow{\gamma''(t)}\right) \cdot \overrightarrow{\gamma'''(t)} = \\ = (0)(21\sin t\cos^2 t - 6\sin^3 t) + (0)(-21\sin^2 t\cos t + 6\cos^3 t) + \\ + (-9(\sin^2 t\cos^2 t))(0) = 0 \\ \left\| \overrightarrow{\gamma'(t)} \times \overrightarrow{\gamma''(t)} \right\|^2 = \left(\sqrt{(0)^2 + (0)^2 + (-9(\sin^2 t\cos^2 t))^2}\right)^2 = \\ = 81\sin^4 t\cos^4 t \end{array}$$ $$\tau(t) = 0$$