Answer on Question #72278 – Math – Differential Geometry | Topology Question
Compute the torsion of the following curves
(i) γ ( t ) = ( 4 5 cos t , 1 − sin t , − 3 5 cos t ) \gamma(t) = \left(\frac{4}{5} \cos t, 1 - \sin t, -\frac{3}{5} \cos t\right) γ ( t ) = ( 5 4 cos t , 1 − sin t , − 5 3 cos t )
(ii) γ ( t ) = ( t , cosh t ) \gamma(t) = (t, \cosh t) γ ( t ) = ( t , cosh t )
(iii) γ ( t ) = 4 5 ( cos 3 t , sin 3 t ) \gamma(t) = \frac{4}{5} (\cos^3 t, \sin^3 t) γ ( t ) = 5 4 ( cos 3 t , sin 3 t )
Solution
Torsion: τ ( t ) = ( γ ′ ( t ) → × γ ′ ′ ( t ) → ) ⋅ γ ′ ′ ′ ( t ) → ∥ γ ′ ( t ) → × γ ′ ′ ( t ) → ∥ 2 \tau(t) = \frac{(\overrightarrow{\gamma'(t)} \times \overrightarrow{\gamma''(t)}) \cdot \overrightarrow{\gamma'''(t)}}{\|\overrightarrow{\gamma'(t)} \times \overrightarrow{\gamma''(t)}\|^2} τ ( t ) = ∥ γ ′ ( t ) × γ ′′ ( t ) ∥ 2 ( γ ′ ( t ) × γ ′′ ( t ) ) ⋅ γ ′′′ ( t )
(i) γ ( t ) = ( 4 5 cos t , 1 − sin t , − 3 5 cos t ) \gamma(t) = \left(\frac{4}{5} \cos t, 1 - \sin t, -\frac{3}{5} \cos t\right) γ ( t ) = ( 5 4 cos t , 1 − sin t , − 5 3 cos t )
We need to differentiate γ ⃗ ( t ) \vec{\gamma}(t) γ ( t )
γ ′ ( t ) → = ( − 4 5 sin t , − cos t , 3 5 sin t ) γ ′ ′ ( t ) → = ( − 4 5 cos t , sin t , 3 5 cos t ) γ ′ ′ ′ ( t ) → = ( 4 5 sin t , cos t , − 3 5 sin t ) \begin{array}{l}
\overrightarrow{\gamma'(t)} = \left(-\frac{4}{5} \sin t, -\cos t, \frac{3}{5} \sin t\right) \\
\overrightarrow{\gamma''(t)} = \left(-\frac{4}{5} \cos t, \sin t, \frac{3}{5} \cos t\right) \\
\overrightarrow{\gamma'''(t)} = \left(\frac{4}{5} \sin t, \cos t, -\frac{3}{5} \sin t\right) \\
\end{array} γ ′ ( t ) = ( − 5 4 sin t , − cos t , 5 3 sin t ) γ ′′ ( t ) = ( − 5 4 cos t , sin t , 5 3 cos t ) γ ′′′ ( t ) = ( 5 4 sin t , cos t , − 5 3 sin t ) γ ′ ( t ) → × γ ′ ′ ( t ) → = ∣ i ⃗ j ⃗ k ⃗ − 4 5 sin t − cos t 3 5 sin t − 4 5 cos t sin t 3 5 cos t ∣ = i ⃗ ( − 3 5 cos 2 t − 3 5 sin 2 t ) + + j ⃗ ( − 12 25 sin t cos t + 12 25 sin t cos t ) + k ⃗ ( − 4 5 sin 2 t − 4 5 cos 2 t ) = = i ⃗ ( − 3 5 ) + j ⃗ ( 0 ) + k ⃗ ( − 4 5 ) \begin{array}{l}
\overrightarrow{\gamma'(t)} \times \overrightarrow{\gamma''(t)} = \left| \begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
-\frac{4}{5} \sin t & -\cos t & \frac{3}{5} \sin t \\
-\frac{4}{5} \cos t & \sin t & \frac{3}{5} \cos t
\end{array} \right| = \vec{i} \left(-\frac{3}{5} \cos^2 t - \frac{3}{5} \sin^2 t\right) + \\
+ \vec{j} \left(-\frac{12}{25} \sin t \cos t + \frac{12}{25} \sin t \cos t\right) + \vec{k} \left(-\frac{4}{5} \sin^2 t - \frac{4}{5} \cos^2 t\right) = \\
= \vec{i} \left(-\frac{3}{5}\right) + \vec{j}(0) + \vec{k} \left(-\frac{4}{5}\right) \\
\end{array} γ ′ ( t ) × γ ′′ ( t ) = ∣ ∣ i − 5 4 sin t − 5 4 cos t j − cos t sin t k 5 3 sin t 5 3 cos t ∣ ∣ = i ( − 5 3 cos 2 t − 5 3 sin 2 t ) + + j ( − 25 12 sin t cos t + 25 12 sin t cos t ) + k ( − 5 4 sin 2 t − 5 4 cos 2 t ) = = i ( − 5 3 ) + j ( 0 ) + k ( − 5 4 ) ( γ ′ ( t ) → × γ ′ ′ ( t ) → ) ⋅ γ ′ ′ ′ ( t ) → = ( − 3 5 ) ( 4 5 sin t ) + ( 0 ) ( cos t ) + ( − 4 5 ) ( − 3 5 sin t ) = 0 ∥ γ ′ ( t ) → × γ ′ ′ ( t ) → ∥ 2 = ( ( − 3 5 ) 2 + ( 0 ) 2 + ( − 4 5 ) 2 ) 2 = 1 τ ( t ) = 0 \begin{array}{l}
\left(\overrightarrow{\gamma'(t)} \times \overrightarrow{\gamma''(t)}\right) \cdot \overrightarrow{\gamma'''(t)} = \left(-\frac{3}{5}\right) \left(\frac{4}{5} \sin t\right) + (0)(\cos t) + \left(-\frac{4}{5}\right) \left(-\frac{3}{5} \sin t\right) = 0 \\
\left\| \overrightarrow{\gamma'(t)} \times \overrightarrow{\gamma''(t)} \right\|^2 = \left(\sqrt{\left(-\frac{3}{5}\right)^2 + (0)^2 + \left(-\frac{4}{5}\right)^2}\right)^2 = 1 \\
\tau(t) = 0 \\
\end{array} ( γ ′ ( t ) × γ ′′ ( t ) ) ⋅ γ ′′′ ( t ) = ( − 5 3 ) ( 5 4 sin t ) + ( 0 ) ( cos t ) + ( − 5 4 ) ( − 5 3 sin t ) = 0 ∥ ∥ γ ′ ( t ) × γ ′′ ( t ) ∥ ∥ 2 = ( ( − 5 3 ) 2 + ( 0 ) 2 + ( − 5 4 ) 2 ) 2 = 1 τ ( t ) = 0
If the torsion of a regular curve with non-vanishing curvature is identically zero, then this curve belongs to a fixed plane.
(ii) γ ( t ) = ( t , cosh t ) \gamma(t) = (t, \cosh t) γ ( t ) = ( t , cosh t )
A plane curve with non-vanishing curvature has zero torsion at all points.
γ ′ ‾ ( t ) = ( 1 , sinh t , 0 ) γ ′ ′ ‾ ( t ) = ( 0 , cosh t , 0 ) γ ′ ′ ′ ‾ ( t ) = ( 0 , sinh t , 0 ) \begin{array}{l}
\overline{\gamma'} (t) = (1, \sinh t, 0) \\
\overline{\gamma''} (t) = (0, \cosh t, 0) \\
\overline{\gamma'''} (t) = (0, \sinh t, 0) \\
\end{array} γ ′ ( t ) = ( 1 , sinh t , 0 ) γ ′′ ( t ) = ( 0 , cosh t , 0 ) γ ′′′ ( t ) = ( 0 , sinh t , 0 ) γ ′ ‾ ( t ) × γ ′ ′ ‾ ( t ) = ∣ i ⃗ j ⃗ k ⃗ 1 sinh t 0 0 cosh t 0 ∣ = k ⃗ ( cosh t ) \overline{\gamma'} (t) \times \overline{\gamma''} (t) = \left| \begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
1 & \sinh t & 0 \\
0 & \cosh t & 0
\end{array} \right| = \vec{k} (\cosh t) γ ′ ( t ) × γ ′′ ( t ) = ∣ ∣ i 1 0 j sinh t cosh t k 0 0 ∣ ∣ = k ( cosh t ) ( γ ′ ‾ ( t ) × γ ′ ′ ‾ ( t ) ) ⋅ γ ′ ′ ′ ‾ ( t ) = ( 0 ) ( 0 ) + ( 0 ) ( sinh t ) + ( cosh t ) ( 0 ) = 0 \left(\overline{\gamma'} (t) \times \overline{\gamma''} (t)\right) \cdot \overline{\gamma'''} (t) = (0)(0) + (0)(\sinh t) + (\cosh t)(0) = 0 ( γ ′ ( t ) × γ ′′ ( t ) ) ⋅ γ ′′′ ( t ) = ( 0 ) ( 0 ) + ( 0 ) ( sinh t ) + ( cosh t ) ( 0 ) = 0 ∥ γ ′ ‾ ( t ) × γ ′ ′ ‾ ( t ) ∥ 2 = ( ( 0 ) 2 + ( 0 ) 2 + ( cosh t ) 2 ) 2 = ( cosh t ) 2 > 0 \left\| \overline{\gamma'} (t) \times \overline{\gamma''} (t) \right\|^2 = \left(\sqrt{(0)^2 + (0)^2 + (\cosh t)^2}\right)^2 = (\cosh t)^2 > 0 ∥ ∥ γ ′ ( t ) × γ ′′ ( t ) ∥ ∥ 2 = ( ( 0 ) 2 + ( 0 ) 2 + ( cosh t ) 2 ) 2 = ( cosh t ) 2 > 0 τ ( t ) = 0 \tau(t) = 0 τ ( t ) = 0 ( i i i ) γ ( t ) = 4 5 ( cos 3 t , sin 3 t ) (iii) \ \gamma(t) = \frac{4}{5} (\cos^3 t, \sin^3 t) ( iii ) γ ( t ) = 5 4 ( cos 3 t , sin 3 t )
A plane curve with non-vanishing curvature has zero torsion at all points.
γ ′ ( t ) → = ( − 12 5 sin t cos 2 t , 12 5 sin 2 t cos t , 0 ) \overrightarrow{\gamma'(t)} = \left(- \frac{12}{5} \sin t \cos^2 t, \frac{12}{5} \sin^2 t \cos t, 0\right) γ ′ ( t ) = ( − 5 12 sin t cos 2 t , 5 12 sin 2 t cos t , 0 ) γ ′ ′ ( t ) → = ( − 12 5 cos 3 t + 24 5 sin 2 t cos t , − 12 5 sin 3 t + 24 5 sin t cos 2 t , 0 ) \overrightarrow{\gamma''(t)} = \left(- \frac{12}{5} \cos^3 t + \frac{24}{5} \sin^2 t \cos t, -\frac{12}{5} \sin^3 t + \frac{24}{5} \sin t \cos^2 t, 0\right) γ ′′ ( t ) = ( − 5 12 cos 3 t + 5 24 sin 2 t cos t , − 5 12 sin 3 t + 5 24 sin t cos 2 t , 0 ) γ ′ ′ ′ ( t ) → = ( 84 5 sin t cos 2 t − 24 5 sin 3 t , − 84 5 sin 2 t cos t + 24 5 cos 3 t , 0 ) \overrightarrow{\gamma'''(t)} = \left( \frac{84}{5} \sin t \cos^2 t - \frac{24}{5} \sin^3 t, -\frac{84}{5} \sin^2 t \cos t + \frac{24}{5} \cos^3 t, 0 \right) γ ′′′ ( t ) = ( 5 84 sin t cos 2 t − 5 24 sin 3 t , − 5 84 sin 2 t cos t + 5 24 cos 3 t , 0 ) γ ′ ( t ) → × γ ′ ′ ( t ) → = = ∣ i ⃗ j ⃗ k ⃗ − 12 5 sin t cos 2 t 12 5 sin 2 t cos t 0 − 12 5 cos 3 t + 24 5 sin 2 t cos t − 12 5 sin 3 t + 24 5 sin t cos 2 t 0 ∣ = = − 144 25 k ⃗ ( sin 4 t cos 2 t + sin 2 t cos 4 t ) = − 144 25 ( sin 2 t cos 2 t ) k ⃗ \begin{array}{l}
\overrightarrow{\gamma'(t)} \times \overrightarrow{\gamma''(t)} = \\
= \left| \begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
-\frac{12}{5} \sin t \cos^2 t & \frac{12}{5} \sin^2 t \cos t & 0 \\
-\frac{12}{5} \cos^3 t + \frac{24}{5} \sin^2 t \cos t & -\frac{12}{5} \sin^3 t + \frac{24}{5} \sin t \cos^2 t & 0
\end{array} \right| = \\
= -\frac{144}{25} \vec{k} (\sin^4 t \cos^2 t + \sin^2 t \cos^4 t) = -\frac{144}{25} (\sin^2 t \cos^2 t) \vec{k}
\end{array} γ ′ ( t ) × γ ′′ ( t ) = = ∣ ∣ i − 5 12 sin t cos 2 t − 5 12 cos 3 t + 5 24 sin 2 t cos t j 5 12 sin 2 t cos t − 5 12 sin 3 t + 5 24 sin t cos 2 t k 0 0 ∣ ∣ = = − 25 144 k ( sin 4 t cos 2 t + sin 2 t cos 4 t ) = − 25 144 ( sin 2 t cos 2 t ) k ( γ ′ ( t ) → × γ ′ ′ ( t ) → ) ⋅ γ ′ ′ ′ ( t ) → = = ( 0 ) ( 84 5 sin t cos 2 t − 24 5 sin 3 t ) + ( 0 ) ( − 84 5 sin 2 t cos t + 24 5 cos 3 t ) + + ( − 144 25 ( sin 2 t cos 2 t ) ) ( 0 ) = 0 ∥ γ ′ ( t ) → × γ ′ ′ ( t ) → ∥ 2 = ( ( 0 ) 2 + ( 0 ) 2 + ( − 144 25 ( sin 2 t cos 2 t ) ) 2 ) 2 = = ( 144 25 ) 2 sin 4 t cos 4 t \begin{array}{l}
\left( \overrightarrow{\gamma'(t)} \times \overrightarrow{\gamma''(t)} \right) \cdot \overrightarrow{\gamma'''(t)} = \\
= (0) \left( \frac{84}{5} \sin t \cos^2 t - \frac{24}{5} \sin^3 t \right) + (0) \left( -\frac{84}{5} \sin^2 t \cos t + \frac{24}{5} \cos^3 t \right) + \\
+ \left( -\frac{144}{25} (\sin^2 t \cos^2 t) \right) (0) = 0 \\
\left\| \overrightarrow{\gamma'(t)} \times \overrightarrow{\gamma''(t)} \right\|^2 = \left( \sqrt{ (0)^2 + (0)^2 + \left( -\frac{144}{25} (\sin^2 t \cos^2 t) \right)^2 } \right)^2 = \\
= \left( \frac{144}{25} \right)^2 \sin^4 t \cos^4 t \\
\end{array} ( γ ′ ( t ) × γ ′′ ( t ) ) ⋅ γ ′′′ ( t ) = = ( 0 ) ( 5 84 sin t cos 2 t − 5 24 sin 3 t ) + ( 0 ) ( − 5 84 sin 2 t cos t + 5 24 cos 3 t ) + + ( − 25 144 ( sin 2 t cos 2 t ) ) ( 0 ) = 0 ∥ ∥ γ ′ ( t ) × γ ′′ ( t ) ∥ ∥ 2 = ( ( 0 ) 2 + ( 0 ) 2 + ( − 25 144 ( sin 2 t cos 2 t ) ) 2 ) 2 = = ( 25 144 ) 2 sin 4 t cos 4 t τ ( t ) = 0 \tau(t) = 0 τ ( t ) = 0
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#340153 on Dec 2023