If a curve is defined like this:
$$
\rho = 1 - \cos(x)
$$
find natural parametrization of it.
So this is how the curve looks like when when I use polar coordinates $x = \rho \cos(t)$, $y = \rho\sin(t)$: $$ \alpha(t) = \bigl((1 - \cos(t)) \cos(t), (1 - \cos(t)) \sin(t)\bigr). $$
Now I found natural parameter to be $s = 4 \arcsin(\sqrt{s/8})$.
Can anyone tell me if this is good?
How does my curve look alike when I plug natural parameter in it?
Am I doing something wrong?
Sorry for the bad post. :/
The polar graph $\rho = f(\theta)$ parametrized by $$ (x, y) = \bigl(f(t)\cos t, f(t) \sin t\bigr) \tag{1} $$ has speed $$ \frac{ds}{dt} = \sqrt{f(t)^{2} + f'(t)^{2}}. $$ For $f(t) = 1 - \cos t$, the speed is $$ \sqrt{(1 - \cos t)^{2} + (\sin t)^{2}} = \sqrt{2(1 - \cos t)} = 2|\sin \tfrac{t}{2}|. $$ If $0 \leq t \leq 2\pi$, then $\sin \frac{t}{2}$ is non-negative, so the arclength function is $$ s = \int_{0}^{t} 2\sin \tfrac{u}{2}\, du = 4(1 - \cos \tfrac{t}{2}), $$ or $t = 2\arccos(1 - \frac{s}{4})$.
The double-angle formulas for the circular functions read \begin{align*} \cos(2\theta) &= \cos^{2}\theta - \sin^{2}\theta, \\ \sin(2\theta) &= 2\sin\theta \cos\theta, \end{align*} and $\cos(\arccos u) = u$ while $\sin(\arccos u) = \sqrt{1 - u^{2}}$.
Substituting $\theta = \arccos(1 - \frac{s}{4})$ gives $$ \left. \begin{aligned} \cos t &= (1 - \tfrac{s}{4})^{2} - 1 + (1 - \tfrac{s}{4})^{2} = 2(1 - \tfrac{s}{4})^{2} - 1, \\ \sin t &= 2(1 - \tfrac{s}{4}) \sqrt{1 - (1 - \tfrac{s}{4})^{2}} = 2(1 - \tfrac{s}{4})\sqrt{\tfrac{s}{4}(2 - \tfrac{s}{4})}. \end{aligned} \right\} \tag{2} $$
The arc length parametrization is found by substituting (2) into (1). As a numerical check, the diagram shows the trig parametrization (red) overlaid by the putative arc length parametrization (blue).