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AVprozaik [17]
3 years ago
15

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

Mathematics
1 answer:
Kitty [74]3 years ago
6 0

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}.

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