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?
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 $$...$$