I want to find the arclength some interval bounded by points $P$ and $Q$ on a parametric object. This parametric object is defined by the 2D stereographic function: $f(x, y) = (tx, 1 + t(y-1))$ . If you plug in all of the points of a centered-at-origin unit circle and for each point, transform its $t$ value from $1 \le t \le \frac{1}{1 - y}$ in a linear fashion, you'll get a nice transformation from the circle to the $y = 0$ line or the x-axis. If you look at this parametric object at the "half way" position, perhaps at $t = \frac{1}{2(1 - y)}$ for each point, you'll then understand what I tried to do with this drawing:enter image description here
I think it would be useful to convert our function from multivariable to singlevariable. Therefore $f(\theta) = (tcos(\theta), 1 + t(sin(\theta) - 1))$
I think it would then be useful to get $t$ in terms of $\theta$. The best I can do is get $t$ in terms of $\theta$ and some other constant, call it $a$. The constant $a$ represents how far into the transformation you are. When $a = 0$, the circle is itself and when $a = 1$, the circle is fully transformed into the y=0 line. Therefore after following the linear $t = t_i + a(t_f - t_i)$ format and knowing that we want $1 \le t \le \frac{1}{1 - sin(\theta)}$ this gives us $t_i = 1$ and $t_f = \frac{1}{1 - sin(\theta)}$ Therefore $t = 1 + a(\frac{1}{1 - sin(\theta)} - 1)$
Therefore our original function where $t$ is now in terms of $\theta$ turns into: $f(\theta) = ((1 + a(\frac{1}{1 - sin(\theta)} - 1))cos(\theta), 1 + (1 + a(\frac{1}{1 - sin(\theta)} - 1))(sin(\theta) - 1))$
This then means that $f'(\theta)_x = \frac{a}{1 - sin(\theta)} + (a - 1)sin(\theta)$. and $f'(\theta)_y = (1 - a)cos(\theta)$
Following the format of $arclength = \int_{\theta i}^{\theta f} \sqrt{(f'(\theta)_x)^2 + (f'(\theta)_y)^2}\, dθ $ is how I arrived at my answer so far:
$$arclength = \int_{\theta i}^{\theta f} \sqrt{(\frac{a}{1 - sin(\theta)} + (a - 1)sin(\theta))^2 + ((1 - a)cos(\theta))^2}\, dθ $$
When a = 0, you're finding the arc length of a centered-at-origin unit circle. When a = 1, you're finding the "arclength" of the y = 0 line (which is just a linear distance). When a = anything in between, you're finding the arclength of a curvy line interval such as the one in the image attached to this post. The problem with this answer is that it is only integrable at $a = 0$, so circle segments are the only arc lengths that I can actually find
From here on out, I will attempt to interpret @SV-97's answer to see if it will help me get an integrable function at all values of a.
"Okay, if you express the way t changes as a function of θ you can consider this as a "curve on a manifold""
Alright so I do have t in terms of $\theta$ I got $t = 1 + a(\frac{1}{1 - sin(\theta)} - 1)$ and I explain where it came from above
I'm having a hard time understanding why you bumped things up to a higher dimensional manifold. I'll do more reading on geodesics of riemann manifolds