By Murray Gerstenhaber, James D. Stasheff

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The x, y are not independent functions of s, however, because x˙ 2 + y˙ 2 = 1 at every point on the curve. Here a dot denotes a derivative with respect to s. (a) Introduce infinitely many Lagrange multipliers λ(s) to enforce the x˙ 2 + y˙ 2 constraint, one for each point s on the curve. From the resulting functional derive two coupled equations describing the catenary, one for x(s) and one for y(s). By thinking about the forces acting on a small section of the cable, and perhaps by introducing the angle ψ where x˙ = cos ψ and y˙ = sin ψ, so that s and ψ are intrinsic coordinates for the curve, interpret these equations and show that λ(s) is proportional to the position-dependent tension T (s) in the chain.

14 A rod used as: (a) a column, (b) a cantilever. By considering small deformations of the form ∞ y(z) = an sin n=1 nπ z L show that the column is unstable to buckling and collapse if Mg ≥ π 2 YI /L2 . (b) Leonardo da Vinci’s problem: The light cantilever. 14b). The rod is used as a beam or cantilever and is fixed into a wall so that y(0) = 0 = y (0). A weight Mg is hung from the end z = L and the beam sags in the (−y)-direction. We wish to find y(z) for 0 < z < L. We will ignore the weight of the beam itself.

Then, looking back at the derivation of the time-independence of the first integral, we see that if L does depend on time, we instead have dE ∂L =− . 116) so that − d dE J˙ · A d 3 x = = (Field Energy) − dt dt ˙ + J˙ · A d 3 x. 3 Lagrangian mechanics 25 ˙ we find Thus, cancelling the duplicated term and using E = −A, d (Field Energy) = − dt J · E d 3 x. 118) Now J · (−E) d 3 x is the rate at which the power source driving the current is doing work against the field. The result is therefore physically sensible.