<s>[INST] Curvature and Torsion for a Space Curve 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.
 [/INST] $\vec r (\theta) = (\theta - \sin \theta, 1 - \cos \theta, 4 \sin \frac{\theta}{2})$
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}$
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