I'm set to calculate the arc length of a circle after a full revolution by using line integrals. This should be easy but im trying not to leave behind checking the hypothesis before i start. I made this parameterization of the curve:
$$\alpha (t)=\left ( 2\cdot cos(t),2\cdot sin(t),2 \right )$$
As i need $\alpha (t)$ to be a regular parameterization of the curve, one of the checkmarks is for it to be injective within the interval $\left ( a,b \right )$ in this case $\left ( 0,2\pi \right )$.
It's clear to see that a circle is a simple curve, as it doesn't cross itself besides on it's borders.
Nevertheless, i'm trying to solve it analytically:
So, being $t_1$ and $t_2$ real numbers which belong to $\left ( 0,2\pi \right )$,$\alpha (t)$ is injective if:
$$\alpha (t_1)=\alpha (t_2)\Rightarrow t_1 =t_2$$
This is where i'm stuck:
$$cos(t_1) = cos(t_2)$$
$$sin(t_1) = sin(t_2)$$
Taking the inverse of either cosine or sine in the respective equation would solve it, but i think that the yielded solution would only be true for an interval where the function from which you want to use this said inverse is injective (Because it needs to be -along being surjective- in order to apply it's inverse). This interval would be for example, $(0,\pi/2)$ but then the problem here is that i can't prove injectivity for the whole curve for $t\in \left ( 0,2\pi \right )$, just for this portion. Correct me if i'm wrong in any of these assumptions :)
I could take this interval for $\alpha (t)$, calculate the length and then mulitply it by four because its symmetric.
The thing is, i want to be able to solve this system of equations and prove $\alpha (t)$ is injective for $t\in \left ( 0,2\pi \right )$, while not taking a shortcut like benefit from its symmetric properties.
I remember i saw in class a way to solve this algebraically but i cant recall how.