Question

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

Answer 1

Answer:

The torsion of the helix is $\tau=\frac{1}{2}$.

Step-by-step explanation:

To complete this exercise we need to recall the formula for the torsion of a curve. Given a parametrization $r(t) = (x(t), y(t), z(t))$ the torsion of the curve is given by

$\tau = \frac{(r'\times r'')\cdot r'''}{\|r'\times r''\|^2}$.

So, the first step is to find the derivatives of the vector function $r$.

Thus,

$r(t)=(\cos t,\sin t, t)$,

$r'(t) = (-\sin t, \cos t, 1)$,

$r''(t) = (-\cos t, -\sin t, 0)$,

$r'''(t)=(\sin t, -\cos t,0)$.

Now, we must calculate the cross product of the vector functions $r'$ and $r''$.

$r'(t)\times r''(t)=\begin{vmatrix}i& j & k\\ -\sin t & \cos t & 1\\ -\cos t & -\sin t & 0\end{vmatrix} = i\begin{vmatrix} \cos t & 1\\ -\sin t & 0\end{vmatrix} -j\begin{vmatrix} -\sin t & 1\\ -\cos t & 0\end{vmatrix}+k\begin{vmatrix} -\sin t & \sin t\\ -\cos t & -\sin t\end{vmatrix}$

$r'(t)\times r''(t) = i\sin t -j\cos t +k(\sin^2t+\cos^2t) = i\sin t -j\cos t +k$.

Now we calculate $\|r'\times r''\|^2$:

$\|r'\times r''\|^2 = \sin^2t+\cos^2t+1=2$

Recall that the norm of a vector in the space $\mathbb{R}^3$ is

$\|(x,y,z)\|^2 = x^2+y^2+z^2$.

At this point we have

$\tau = \frac{1}{2}\left((i\sin t -j\cos t +k)\cdot (i\sin t -j\cos t+0k)\right) = \frac{1}{2}\left(\sin^2t+\cos^2t\right) =\frac{1}{2}$.