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I wonder if someone can help me with the following question:

Determine wether the image of the curve $\alpha(t)=(2\cos^2(t)-3,\sin(t)-8,3\sin^2(t)+4), t \in (-\frac{\pi}{2},\frac{\pi}{2}) $ is contained in:

a) a plane or not,

b) a straight line or not.

For a) I need to find the torsion and for b) the curvature, but then I need to reparametrize $\alpha$ by arc length and I'm not sure how to do this.

I tried calculating $\dot{\alpha}(t)=(4\cos(t)\sin(t),\cos(t),6\sin(t)\cos(t))$ to get the arc length function

$s(t)= \int_{-\frac{\pi}{2}}^t \lVert \dot{\alpha}(u) \rVert du= \int_{-\frac{\pi}{2}}^t \sqrt{ 13\sin^2(2u) +\cos^2(u) } du $, and then plugg $s(t)$ into $\alpha$, but the expression gets super messy . Am I doing something wrong? Is there any easier way to go about it?

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    $\begingroup$ There's no need to reparametrize to unit-speed just to compute curvature and torsion. Go back to your formulae, and recall the chain rule. $\endgroup$ – J. M. is a poor mathematician Oct 23 '17 at 10:59
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Have a look here

You have

$\alpha(t) =\left(2 \cos ^2(t)-3,\sin (t)-8,3 \sin ^2(t)+4\right)$

$\alpha'(t)=\left(-2 \sin (2 t),\cos (t),3 \sin (2 t)\right)$

$\alpha''(t)=\left(-4 \cos (2 t),-\sin (t),6 \cos (2 t)\right)$

Curvature is defined as

$$\kappa(t)=\frac{\alpha'(t)\times\alpha''(t)}{||\alpha'(t)||^3}$$ where $\times$ is the cross product

and I got

$$\kappa (t)=\left(\frac{6 \cos ^3(t)}{\left(13 \sin ^2(2 t)+\cos ^2(t)\right)^{3/2}},0,\frac{4 \cos ^3(t)}{\left(13 \sin ^2(2 t)+\cos ^2(t)\right)^{3/2}}\right)$$

In a similar way you can find torsion

By the way a graph of the curve is below

Hope it is useful

Edit

The torsion is zero because $$\left| \begin{array}{rrr} -2 \sin (2 t) & \cos (t) & 3 \sin (2 t) \\ -4 \cos (2 t) & -\sin (t) & 6 \cos (2 t) \\ 8 \sin (2 t) & -\cos (t) & -12 \sin (2 t) \\ \end{array} \right|=0$$ therefore $\alpha$ is a plane curve $$...$$ enter image description here

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  • $\begingroup$ the curvature is a scalar $\endgroup$ – janmarqz Nov 8 '17 at 14:38
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There is a shortcut:

By inspection, $3x+2z=5$, and the curve is contained in that plane.

And, ignoring the unessential constants, $3y^2=z$, which cannot represent a straight line in the plane (it is a parabola, the intersection of a parabolic cylinder with the plane).

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  • $\begingroup$ Thanks, this was interesting! I didn't really get how you got to the equation $3x+2z=5$, do you think you could elaborate a little? $\endgroup$ – user202542 Oct 23 '17 at 13:51
  • $\begingroup$ @user202542: compute $3x+2z$. $\endgroup$ – Yves Daoust Oct 23 '17 at 14:01
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On arc parametrization, we can neatly differentiate thrice to evaluate $\tau$.

By definition of torsion the scalar triple product

$$( r^{'} \cdot r^{''} \times r^{'''} ) $$

should vanish for a planar curve as checked by Raffaele. The determinant ( in which third column is $-1\frac12$ times the first ) vanishes so curve has zero torsion.

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