Natural parameterization of the following curves:

Question

I am having trouble finding the natural parameterization of these curves:

$$\alpha(t)=\left(\sin^2\left(\frac{t}{\sqrt{2}}\right),\frac{1}{2}\sin \left(t\sqrt{2}\right), \left(\frac{t}{\sqrt{2}}\right)\right)$$

The thing is when finding $$\|\alpha'(t)\|=\sqrt{\frac{3}{2}\sin^2\left({\sqrt{2}t}\right)+1}$$ I do not know how to integrate this. The second one I have is

$$\beta(t)=\left(\frac{4}{5}\cos t,1-\sin{t},-\frac{3}{5}\cos t\right)$$

I get $s=4\left(1-\cos\left(\frac{t}{2}\right)\right)$ or $t=2\arccos({4-s})$

I am to find the Tangent, normal, binormal, tangent and curvature of the curves, but I am at a block, because when I try to naturally parameterize then I come to problems:

1. For the first one, I cannot figure out the integral of $\|\alpha'(t)\|$
2. For the second one I think I have made a mistake because finding the derivative when putting $t$ in dependence of $s$ in $\beta(t)$ would be very messy business to find the derivative for example.

Answer 1

$\alpha(t)=\left(\sin^2\left(\frac{t}{\sqrt{2}}\right),\frac{1}{2}\sin \left(t\sqrt{2}\right), \left(\frac{t}{\sqrt{2}}\right)\right) $

Since $\sin^2(x) =\frac12(1-\cos(2x)) $, $\sin^2\left(\frac{t}{\sqrt{2}}\right) =\frac12(1-\cos(2\frac{t}{\sqrt{2}})) =\frac12(1-\cos(t\sqrt{2})) $, so $\alpha(t) =\left(\frac12(1-\cos(t\sqrt{2})) ,\frac{1}{2}\sin \left(t\sqrt{2}\right) , \left(\frac{t}{\sqrt{2}}\right)\right) $.

From this, $\alpha'(t) =\left((\sqrt{2}/2)\sin(t\sqrt{2}) ,(\sqrt{2}/2)\cos \left(t\sqrt{2}\right) , \left(\sqrt{2}/2\right)\right) =(\sqrt{2}/2)\left(\sin(t\sqrt{2}) ,\cos \left(t\sqrt{2}\right) ,1\right) $ so that $||\alpha'(t)||^2 =(1/2)(1+1) =1 $.

Answer 2

$\beta(t)=\left(\frac{4}{5}\cos t,1-\sin{t},-\dfrac{3}{5}\cos t\right)$

$\beta'(t)=\left(-\dfrac45\sin t,-\cos t,\dfrac 35\sin t\right)$ Then,

$\Vert\beta'(t)\Vert^2=\left(\dfrac{4^2}{5^2}\sin^2t+\cos^2t+\dfrac{3^2}{5^2}\sin^2t\right)=\left(\dfrac{4^2+3^2}{5^2}\sin^2t+\cos^2t\right)=1$

So, $s=\int_{0}^t1\mathrm d\tau=t$ The curve was already parametrized by arclength.