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  "body": "\n## Problem\n\nA uniform, perfectly flexible and inextensible chain of length $\\ell$ hangs under constant gravity from two fixed points $(a, A)$ and $(b, B)$.\nHere $a \u003c b$, and $\\ell$ is larger than the distance between the points.\nWhat shape does the chain take?\n\nAs a variational problem: among $y \\in C^1([a, b])$ with $y(a) = A$, $y(b) = B$ and [[arc length|0TH78S7GBZ]]\n$$\\int_a^b \\sqrt{1 + y'(x)^2} \\, dx = \\ell,$$\nfind one that minimises the potential energy, which is proportional to\n$$U(y) = \\int_a^b y(x) \\sqrt{1 + y'(x)^2} \\, dx.$$\n\n## Remarks\n\nThis is an [[isoperimetric problem|6FY0KHMK5A]].\nWith a multiplier $\\lambda$, the Lagrangian is $(y - \\lambda) \\sqrt{1 + y'^2}$, which does not depend on $x$.\nThe [[Beltrami identity|F7EKTEESQ5]] gives $(y - \\lambda) / \\sqrt{1 + y'^2} = c$, and the solutions with $c \u003e 0$ are the catenaries\n$$y(x) = \\lambda + c \\cosh\\Big(\\frac{x - d}{c}\\Big),$$\nwhere the constants $c$, $d$, $\\lambda$ are fixed by the two end conditions and the length.\nHere $\\cosh$ is the [[hyperbolic cosine|036696EMG2]].\nJacob Bernoulli posed the problem in Acta Eruditorum, and solutions by Leibniz, Huygens and Johann Bernoulli were published in 1691.\n\n## Explanations\n\n### L1 human\n\nHold a chain or a necklace by its two ends and let it hang.\nThe curve it forms looks like a parabola but is a different curve, the catenary, given by the hyperbolic cosine.\nThe chain settles in the shape that puts its centre of mass as low as possible, given its length.\n",
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