What happens when $r \to \infty$? Will it be a line? (partial circle)

Question

Let $a$ be a arc of particle circles, which is constant. What happens when $r \to \infty$? Will it be a line?

Radius of partial circle : $r$,

Arc of partial circle : $a$ and constant,

For $r=r_0$

For $r\simeq 2.r_0$

For $r>>>r_0$, like a line,

For ,$\quad\lim\limits_{r\to \infty}\quad$ ,$\quad a$ ,can be partial line?

Answer 1

Let me modify the parametrization in William's answer and give another sense in which the answer is yes. In his answer, he parametrized the arc for a given $r$ by the function $$f_r(s)=\left(r \cos\left(\frac{s}{r}\right), r \sin\left(\frac{s}{r}\right)\right)$$ where $s\in[0,a]$. These functions do not converge pointwise as $r\to\infty$, essentially because the starting points of the arcs are all different and are going out to infinity ($f_r(0)=(r,0)$). However, if we translate them so that they all start at the same point, then they do converge pointwise, and they converge to a line. More precisely, define $$g_r(s)=\left(r \cos\left(\frac{s}{r}\right)-r, r \sin\left(\frac{s}{r}\right)\right).$$

Then for any fixed $s$, $$\lim\limits_{r\to \infty}g_r(s)=(0,s).$$ (This can be deduced from the limits $\lim\limits_{x\to 0}\frac{\cos x-1}{x}=0$ and $\lim\limits_{x\to 0}\frac{\sin x}{x}=1$ by taking $x=s/r$. With a little more work, the convergence can actually be shown to be uniform in $s$.)

That is, the functions $g_r(s)$ parametrizing these arcs converge to the function $g(s)=(0,s)$ parametrizing a line segment.

Answer 2

The arc will flatten out as $r\to \infty$.

Specifically, it flattens out closer and closer into a line because its curvature approaches zero. By this we mean that we parametrize the arc in $\mathbb{R}^2$ by

$$f(s)=\left(r \cos\left(\frac{s}{r}\right), r \sin\left(\frac{s}{r}\right)\right)$$

for $s\in [0,a]$ for some $a>0$. Notice then that this is an arc-length parametrization. Computing the tangent vector

$$T(s)=f'(s) = \left(-\sin\left(\frac{s}{r}\right),\cos\left(\frac{s}{r}\right)\right),$$

we notice that the curvature is given by

$$\kappa = \|T'(s)\|=\frac{1}{r}.$$

Taking $r\to\infty$ we obtain $\kappa\to 0$. Curves of zero curvature are lines and points, so it does approach a (in some weird sense) line or point.

However, the length of this curve is given by $L(f) = ar$. Hence, the length is never non-zero so this would be a line rather than a point.

In one of the other comments you can see that you get a line by translating the above.

Answer 3

Alright, make $r$ as big as you want. Let's say, for example, $r$ = $10^{10}a$.

It turns out that the arc length $a$ actually remains an arc if you didn't change your viewport's size (i.e. the distance from an hypothetical orthographic camera to the paper in which you are drawing).

So to supply you with an answer, let's say your (orthographic) camera is $d=10^{10}$ units over your drawing paper. From this perspective, the two radius will actually be parallel to one another (and one on top of the other), and your arc $a$ will be projected as a point on your new viewport canvas.