(1) Find the torsion of the helix a(t) = (cos(t), sin(t), t)

Answer:The torsion of the helix is [tex]\tau=\frac{1}{2}[/tex].Step-by-step explanation:To complete this exercise we need to recall the formula for the torsion of a curve. Given a parametrization [tex] r(t) = (x(t), y(t), z(t))[/tex] the torsion of the curve is given by[tex] \tau = \frac{(r'\times r'')\cdot r'''}{\|r'\times r''\|^2}[/tex].So, the first step is to find the derivatives of the vector function [tex]r[/tex].Thus,[tex]r(t)=(\cos t,\sin t, t)[/tex],[tex] r'(t) = (-\sin t, \cos t, 1)[/tex],[tex]r''(t) = (-\cos t, -\sin t, 0)[/tex],[tex]r'''(t)=(\sin t, -\cos t,0)[/tex].Now, we must calculate the cross product of the vector functions [tex]r'[/tex] and [tex]r''[/tex].[tex]r'(t)\times r''(t)=\begin{vmatrix}i& j & k\\ -\sin t & \cos t & 1\\ -\cos t & -\sin t & 0\end{vmatrix} = i\begin{vmatrix} \cos t & 1\\ -\sin t & 0\end{vmatrix} -j\begin{vmatrix} -\sin t & 1\\ -\cos t & 0\end{vmatrix}+k\begin{vmatrix} -\sin t & \sin t\\ -\cos t & -\sin t\end{vmatrix}[/tex][tex]r'(t)\times r''(t) = i\sin t -j\cos t +k(\sin^2t+\cos^2t) = i\sin t -j\cos t +k[/tex].Now we calculate [tex]\|r'\times r''\|^2[/tex]:[tex]\|r'\times r''\|^2 = \sin^2t+\cos^2t+1=2[/tex]Recall that the norm of a vector in the space [tex]\mathbb{R}^3[/tex] is[tex]\|(x,y,z)\|^2 = x^2+y^2+z^2[/tex].At this point we have[tex]\tau = \frac{1}{2}\left((i\sin t -j\cos t +k)\cdot (i\sin t -j\cos t+0k)\right) = \frac{1}{2}\left(\sin^2t+\cos^2t\right) =\frac{1}{2}[/tex].