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## Catenary. Amazing properties

By definition, the catenary is the shape taken by a uniform and flexible wire which hangs from two fixed points by the action of gravity only, its weight being uniformly distributed along its length.
1) Find the equation of the catenary.
2) Prove that the catenary is the curve with fixed endpoints that generates the minimal surface of revolution around a given axis.
3) Prove that the catenary has the lowest centroid among the curves with fixed arclength and endpoints.
4) Prove that the solution to this problem is a series of reversed catenaries glued at endpoints. Prove the same thing if the shape of the wheel is replaced by any regular $n$-gon, with $n \geq 3$.

Solutions: 1) At each point there are three forces acting, and their result is zero. The forces are:
i) The tension of the wire
ii) The weight of the wire $w$.
iii) The horizontal pull at the origin, whose magnitude we denote by $H$.
The slope of the tangential force is $w/h$ because the three forces form a right triangle and the slope is the hypotenuse of this right triangle. The weight $w$ can be expressed as $\mu s$ where $\mu$ is the wire density and $s$ is the arclength from the origin to our point. Therefore, we get the equation $y'(x)=\displaystyle \frac{\mu s}{h}=\frac{\mu}{h}\int_0^x \sqrt{1+(f'(t))^2}dt$.

Differentiating, we get the differential equation $y''=\alpha \sqrt{1+(y')^2}$ where $\alpha=\mu/h >0$.

For solving this differential equation we have at least two methods
a) Integrate a function like $\displaystyle \int\frac{u}{\sqrt{1+u^2}}$ making the substitution $u=\sinh v$.
b) Square the relation obtained, differentiate and get $2y''y'''=2\alpha^2 y'y''$. From our relation we see that $y''>0$ everywhere, so we can divide by it and get $y'''=\alpha^2 y'$. Denote $y'$ by $z$ and reach the differential equation $z''=\alpha^2 z$. This implies $y(x)=\displaystyle \frac{e^{\alpha x}+e^{-\alpha x}}{2\alpha}+c=\cosh (\alpha x)/ \alpha +c$.

2) For this section, there are also multiple choices of solve: Euler-Lagrange equations and mean curvature. I choose the first one, which provides quickly the result.
To find the critical points of a functional like $\displaystyle J(f)=\int_a^b F(x,f(x),f'(x))dx$ it is equivalent to solving the Euler-Lagrange equation, namely: $\displaystyle \frac{ \partial F}{\partial f}=\frac{d}{dx}\frac{\partial F}{\partial f'}$. In our case, if the rotated function is $f:[a,b] \to \Bbb{R}_+$, its area equals $\displaystyle \int_a^b 2\pi f(x)\sqrt{1+(f'(x))^2} d x$, and therefore $F(x,f(x),f'(x))=2\pi f(x)\sqrt{1+(f'(x))^2}$. After some calculations we reach the differential equation $f(x)f''(x)=1+f'(x)^2$, which is the equation of the catenary.

3) To prove this it is enough to take a look at Pappus’s Theorem which tells us that the area of a surface of revolution generated by a plane curve $C$ around an axis external to $C$ and in the same plane as $C$ is equal to the product of the arclength of $C$ and the distance traveled by its geometric centroid. This means the area generated of a curve with given arclength is minimum when the geometric centroid is as low as possible. Since the catenary reaches the minimum, it follows that its geometric centroid is the lowest among given curves with fixed endpoints and fixed length.
Another argument would be that by definition, the catenary hangs by the action of gravity in equilibrium, and if we imagine it as a flexible wire fixed at its endpoints, gravity acts on its geometric centroid pulling it down as far as it goes, but catenary is as low as it gets, therefore, its centroid should be the lowest.
4)
See here a successful experiment of a square wheel tricycle which rides smoothly on a bumpy road made out of catenaries. Suppose we want a smooth ride with a polygonal wheel such that the center of the wheel is at height $\delta$. The result follows if we start from a “wheel” which is a segment, such that the segment is tangent to the curve at $(t,y(t))$ and the distance from the tangent line at $(t,y(t))$ to the point $(t,\delta)$ is always $a$.
Denote $L_t: Y-y=y'(X-t)$ the tangent line to the curve at $(t,y(t))$, and $(X_0,Y_0)$ be the closest point to $(t,\delta)$. Then we have $a^2=(X_0-t)^2+(Y_0-\delta)^2$. A straight forward computation gives us $X_0-t=\displaystyle \frac{y'}{1+(y')^2}(\delta-y)$ and $Y_0-\delta=\displaystyle \frac{-1}{1+(y')^2}(\delta - y)$. Therefore, we find the differential equation $a^2=\displaystyle \frac{(y')^2}{(1+(y')^2)^2}(\delta-y)^2+\frac{1}{(1+(y')^2)^2}(\delta -y)^2$, which can be simplified to $\displaystyle y'=-\frac{1}{c}\sqrt{1+(y')^2}$. This is the differential equation of an inverted catenary.
Therefore, for a regular polygonal wheel, every section of the road must be an inverted catenary having arclength equal to the side of the polygon.