When length equals area

The graph of hyperbolic cosine is called a catenary. A catenary has the following curious property: the length of a catenary between two points equals the area under the catenary between those two points.

The proof is surprisingly simple. Start with the following:

\sqrt{1 +\left(\frac{d}{dx} \cosh(x) \right)^2} = \sqrt{1 + \sinh^2(x)} = \cosh(x)

Now integrate the first and last expressions between two points a and b. Note that the former integral gives the arc length of cosh between a and b, and the later integral gives the area under the graph of cosh between a and b.

By the way, the most famous catenary may be the Gateway Arch in St. Louis, Missouri.

Gateway Arch at night

7 thoughts on “When length equals area

  1. Hey John – found this interesting detail about the St. Louis arch.

    “The Gateway Arch in St. Louis, Missouri, United States is sometimes said to be an (inverted) catenary, but this is incorrect.[19] It is close to a more general curve called a flattened catenary, with equation y = A cosh(Bx), which is a catenary if AB = 1. While a catenary is the ideal shape for a freestanding arch of constant thickness, the Gateway Arch is narrower near the top. According to the U.S. National Historic Landmark nomination for the arch, it is a “weighted catenary” instead. Its shape corresponds to the shape that a weighted chain, having lighter links in the middle, would form.”

  2. I never knew that about catenaries. It vaguely feels like some kind of dual to Kepler’s Law that says planets sweep out equal area per time. Any insight on that?

  3. Maybe somewhat related, what I found extremely interesting in school was the Neile-Parabola, in which the arc length is the function itself composed with a polynomial.

  4. I wondered about the profile curve of the higher dimensional analogue, i.e “when volume equals area”, and mentioned to a friend. He shared the paper “Two Generalizations of a Property of the Catenary”, which you can download here:


    Turns out the [(n+1)-volume/n-volume] always results in the catenary, and the [(n+1)-volume/arc length] results in an n-catenoid (minimal hypersurfaces of revolution in higher dimensions).

  5. So…

    How do I understand the connection between lengths and areas? It seems like a change in units would mess that up. Measure the St Louis arch in inches, then in meters. Is it true that the number of square inches and the number of square meters come out the same as the lengths?

  6. I am building a kiln using a catenary arch. The rear wall and front wall/door will be vertical and fill in the space under the arch, which has the dimensions of 41″W x 39.5″H. I need the area within this arch in order to calculate how many bricks I need to construct the two walls. Can you help me calculate that area? My math skills never advanced beyond algebra.

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