Is the arc length always irrational between two rational points?

Question

Recently I was wondering: Why does pi have an irrational value as it is simply the ratio of diameter to circumference of a circle? As the value of diameter is rational then the irrationality must come from the circumference.

Then I used calculus to calculate the arc length of various functions with curved graphs (between two rational points) and found the arc length two be irrational again.

Do all curved paths have irrational lengths?

My logic is that while calculating the arc length (calculus) we assume that the arc is composed of infinitely small line segments and we are never close the real value and unlike the area under a curve, there do not exist an upper and lower limit which converges to the same value.

If yes, are these the reasons irrational values exist in the first place?

Answer 1

Obviously, a straight line between two rational points can have rational length $-$ just take $(0,0)$ and $(1,0)$ as your rational points.

But a curved line can also have rational length. Consider parabolas of the form $y=\lambda x(1-x)$, which all pass through the rational points $(0,0)$ and $(1,0)$. If $\lambda=0$, then we get a straight line, with arc length $1$. And if $\lambda=4$, then the curve passes through $(\frac12,1)$, so the arc length is greater than $2$.

Now let $\lambda$ vary smoothly from $0$ to $4$. The arc length also varies smoothly, from $1$ to some value greater than $2$; so for some value of $\lambda$, the arc length must be $2$, which is a rational number.

Answer 2

An example of a curve with rational arc lengths between at least some pairs of rational points is a cardioid.

Down to scaling and rotation, a cardioid may be rendered in polar coordinates by the equation

$$r=1-\cos\theta$$

with arc length differential

$$ds=\left(\sqrt{r^2+(dr/d\theta)^2}\right)d\theta=\sqrt{2-2\cos\theta}~d\theta=2\sin(\theta/2)d\theta$$

Integrating this from $\theta=0$ to an arbitrary value of $\theta$ gives the arc length function

$$s=4(1-\cos(\theta/2))$$

Thus the arc length from the origin to $(-2,0)$ ($\theta=\pi$) is $4$. Moreover, suppose we select $\theta=2\cos^{-1}(a/c)$ where $a^2+b^2=c^2$ is a Pythagorean triple. Then we have

$$\cos\theta=2(a^2/c^2)-1$$

$$\sin\theta=2(b/c)(a/c)=2ab/c^2$$

Clearly giving rational values for the Cartesian coordinates $x=(1-\cos\theta)\cos\theta$ and $y=(1-\cos\theta)\sin\theta$. The arc length from the origin is then the rational quantity

$$s=4(1-\cos(\theta/2))=4(1-a/c)$$

Answer 3

So, my question is that do all curved path have irrational lengths?

Of course not. A circle with radius $\frac{1}{2\pi}$ is a curved path and has length $1$ which is a rational number. If you put the center of the circle to $(-\frac1{2\pi}, 0)$, then $(0,0)$, a "rational" point, is on the circle, and the circle can be seen as a path from $(0,0)$ to $(0,0)$.

Answer 4

Consider the two points $(-\frac12,0)$ and $(\frac12,0)$. For any real value of $y_0$, we can draw a circular arc between these two points which is centered at $(0,y_0)$ and which lies entirely in the upper half-plane. As $y \to - \infty$, the length of this arc approaches 1 (since the arc approaches a straight line); as $y \to +\infty$, the arc length approaches $\infty$. Since the arc length varies continuously with $y_0$, it must be the case that the arc length can be any real number greater than 1, including all rational lengths greater than 1.

Answer 5

No.

Take any smooth curve between two rational points and deform it to change its length by a finite amount. During deformation, you will cross infinitely many rational lengths.

A simple example is a polynomial having two rational roots, times a variable factor.

---

Now consider the curve of parametric equations

$$\begin{cases}x=\dfrac{t^3}3-t,\\y=t^2\end{cases}$$

(a modified Tschirnhausen cubic).

We have

$$s=\int_a^b\sqrt{(t^2-1)^2+4t^2}\,dt=\int_a^b(t^2+1)\,dt=\frac{b^3-a^3}3+b-a,$$

so that the length between two rational $t$ (giving rational endpoints) is always rational.

Answer 6

Contributing another simple counterexample, let $f(x)=-\cos x$ with $x\in[0,\pi]$. Then, the length of $f$ between $A(0,-1)$ and $B(\pi,1)$ is given by: $$\ell(f)=\int_0^\pi|f'(t)|dt=\int_0^\pi\sin tdt=[-\cos t]_0^\pi=2.$$ Note, also, that the ratio between the curve and the "diameter" of it, $AB$ is: $$\frac{\ell(f)}{(AB)}=\frac{2}{\sqrt{\pi^2+2}},$$ which is, again, irrational.