Curvature and Torsion for a Space Curve

Question

Calculate the curvature and torsion of

x= $\theta - \sin \theta, y = 1 - \cos \theta, z = 4 \sin (\theta/2)$

$\vec r = (θ - \sin θ)i+(1-\cos θ)j + 4 \sin (θ/2) \ k$

$\vec dr/dt = (1- \cos θ)i + \sin θ \ j+2 \cos (θ/2) \ k$

$d^2\vec r/dt^2 = \sin θ \ i + \cos θ \ j - sin (θ/2) \ k$

Since $k = |\vec dr/dt \times d^2\vec r/dt^2| \ / \ |\vec dr/dt|^3$

I calculated $\vec dr/dt \times d^2\vec r/dt^2$, which upon simplification gave,

$\vec dr/dt \times d^2\vec r/dt^2 = (\sin θ \sin (θ/2) - 2 \cosθ \cos(θ/2)) \ i + (3 \cos θ + 1) \sin (θ/2) \ j + (\cos θ-1) \ k$

In calculating |$\vec dr/dt$ x $d^2$$\vec r/d$$t^2$| is where I'm starting to face complications. I'm not sure if I am doing this correctly or not. But I understand the application of the formulas of torsion and curvature. Kindly guide me for the same.

Answer 1

$\vec r (\theta) = (\theta - \sin \theta, 1 - \cos \theta, 4 \sin \frac{\theta}{2})$

There is another formula for curvature that is easier here.

$r'(\theta) = (1 - \cos \theta, \sin\theta, 2 \cos \frac{\theta}{2})$

$||r'(\theta)|| = \sqrt{(1-\cos\theta)^2 + \sin^2\theta + 4 \cos^2 \frac{\theta}{2}}$

$ = \sqrt{(1 + \cos^2\theta - 2 \cos\theta + \sin^2\theta + 4 \cos^2 \frac{\theta}{2}}$

$ = \sqrt{2 - 2 \cos\theta + 2 + 2 \cos\theta} = 2$

So unit tangent vector, $ \ T(\theta) = \frac{1}{2} (1 - \cos \theta, \sin\theta, 2 \cos \frac{\theta}{2})$

Derivative of unit tangent is $\ T'(\theta) = \frac{1}{2} (\sin \theta, \cos\theta, - \sin \frac{\theta}{2})$

$||T'(\theta)|| = \dfrac{\sqrt{3 - \cos\theta}}{2 \sqrt2}$

Curvature $ \ k = \dfrac{||T'(\theta)||}{||r'(\theta)||} = \dfrac{\sqrt{3 - \cos\theta}}{4 \sqrt2}$

Answer 2

We have the curve $\gamma(t)=(\theta-\sin\theta,1-\cos\theta,4\sin(\theta/2))$.
First we compute its derivatives (observe tha the curve is not parametrized by arc lenght): $$\dot\gamma=(1-\cos\theta,\sin\theta,2\cos(\theta/2))$$ $$\ddot\gamma=(\sin\theta,\cos\theta,-\sin(\theta/2))$$ $$\dddot\gamma=(\cos\theta,-\sin\theta,-\dfrac{1}{2}\cos(\theta/2))$$ Now we are ready to find $\tau$ and $k$: $$\dot\gamma\wedge\ddot\gamma=\det\bigg(\begin{pmatrix}\bf i&&\bf j&&\bf k\\1-\cos\theta&&\sin\theta&&2\cos(\theta/2)\\\sin\theta&&\cos\theta&&-\sin(\theta/2) \end{pmatrix}\bigg)=\textbf i(-\sin\theta\sin(\theta/2)-2\cos\theta\cos(\theta/2))-\textbf j((\cos\theta-1)\sin(\theta/2)-2\sin\theta\cos(\theta/2))+\textbf k(\cos\theta-\cos^2\theta-\sin^2\theta)=$$ $$=(-\sin\theta\sin(\theta/2)-2\cos\theta\cos(\theta/2),(\cos\theta-1)\sin(\theta/2)-2\sin\theta\cos(\theta/2),\cos\theta-1)$$ $$k=\dfrac{\Vert\dot\gamma\wedge\ddot\gamma \Vert}{\Vert\dot\gamma\Vert^3},\tau=-\dfrac{\langle\dot\gamma\wedge\ddot\gamma,\dddot\gamma \rangle}{\Vert \dot\gamma\wedge\ddot\gamma\Vert^2}.$$ $\Vert\dot\gamma \Vert=\sqrt{1+\cos^2\theta-2\cos\theta+\sin^2\theta+4\cos^2(\theta/2)}=\sqrt{2(1-\cos\theta+2\cos^2(\theta/2))}$.