Interesting Graphs: Catenary

If a flexible string is suspended under gravity by its two ends, the shape resembles a chain (Latin catena). That's why the curve is called a catenary. The equation of a catenary in Cartesian coordinates: $y=a\cosh \frac{x}{a}$, where $a$ is a constant; $a$ depends on the mass per unit length and tension of the string. The derivation of equation for the curve can be found here.

Catenaries occur naturally, since they minimize the gravitational potential energy of a string or rope whose location is fixed at two ends, which is equivalent to minimizing the area under the string. But they are also optimal for architects when a flexible cable (or its equivalent) is subject to a uniform force (e.g., gravity or the weight of a bridge, etc.).

Catenoids
The catenoid is a surface of revolution of a catenary curve rotated around its directrix.



Claim: Catenoids are minimal surfaces that minimizes area for a given boundary.

Definition: A minimal surface is a surface $M$ with mean curvature $H=0$ at all points $p \in M$.

Mean curvature $\large H=\frac{Eg+Ge-2Ff}{2(EG-F^2)}$ [Proof: later]

Proof of the claim:

A catenoid can be parametrized by $$x(u,v)=(a\cosh v \cos u,a\cosh v \sin u,av).$$ We then evaluate the partial derivatives of $x$: $$x_u = (-a\cosh v \sin u,a\cosh v \cos u,0)\\
x_v = (-a\sinh v \cos u,a\sinh v \sin u,a)$$ We know that $$n=\frac{x_u \times x_v}{|x_u \times x_v|},$$ and so we have the coefficients of the first fundamental form: $$E=x_u \cdot x_u=a^2\cosh^2 u\\
F=x_u \cdot x_v=0\\
G=x_v \cdot x_v=a^2\cosh^2 u,$$ and the coefficients of the second fundamental form: $$e=n \cdot x_{uu}=-a\\
f=n \cdot x_{uv}=0\\
g=n \cdot x_{vv}=a.$$ Substituting the values to $H$, we have $$H=\frac{Eg+Ge-2Ff}{2(EG-F^2)}=0$$ Thus, catenoid is a minimal surface.

2nd method:
http://www.princeton.edu/~rvdb/WebGL/catenoid_explanation.html

Similarity of catenary and parabola
In the previous post, we have proved that $\cosh x=\frac{e^x+e^{-x}}{2}$, then we have $\cosh x=1+\frac{x^2}{2!}+\frac{x^4}{4!}+...$
For $x\approx 0$, $\cosh x \approx 1+\frac{x^2}{2}$.
RHS represents a parabola, so we conclude that for small $x$, a catenary can be approximated by a parabola.

A common misconception is that a parabola can be used to construct an arch. If we look at the function for a parabola, we see that the slope at any point is given by $2x$ and is changing linearly. On the other hand if we look at the function $\cosh x$, we see that the slope at any point is given by $\frac{e^x-e^{-x}}{2}$, which means the slope of a catenary curve is changing exponentially. In other words, the legs of an inverted catenary curve will be straighter at the base of the arch compared to an inverted parabola, giving the structure more horizontal support.

Applications
Real life examples of catenaries include the cables of a suspension bridge, a rope hanging between two posts, each strand of a spider web, and the Gateway Arch in St. Louis.

If you use physics to model the differential equation describing the effect of a uniform force on a flexible cable, its solution will be of the form $A\cosh ax$.

Reference:
http://www.ias.ac.in/resonance/Volumes/11/08/0081-0085.pdf
http://aleph0.clarku.edu/~djoyce/ma131/gallery.pdf
http://www.princeton.edu/~rvdb/WebGL/catenoid_explanation.html