The Catenary curve Question: what is the shape of the St. Louis Gateway Arch? Question: what is the shape of the St. Louis Gateway Arch? Not a parabola! New Question: find the shape of a hanging chain. 1. A mass hanging from two chains: −mg T1 T1 sin(θ1) T2 θ 1 θ2 m T1 cos(θ1) Balance forces! I x-direction: T1 cos(θ1) = T2 cos(θ2). I y-direction: T1 sin(θ1) + T2 sin(θ2) = mg. 2. Hanging a chain: the set-up y f (x) x Coordinate system: f (0) = 0; f 0(0) = 0. I ρ = density of chain I A = cross-sectional areas I Z x q s(x) = arclength of f from 0 to x = 1 + (f 0(t)) dt 0 Play the balancing forces game for the hanging chain! Let ∆x > 0. Balance the forces for a mass at position x with horizontal length ∆x: T(x + ∆x) y f (x) T(x) θ(x) x x x-direction: balance horizonal forces T(x) cos(θ(x)) = T(x + ∆x) cos(θ(x + ∆x)) for all x! =) T(x) cos(θ(x)) = B constant Play the balancing forces game for the hanging chain! Let ∆x > 0. Balance the forces for a mass at position x with horizontal length ∆x: T(x + ∆x) y f (x) θ(x + ∆x) −θ(x) x x x + ∆x T(x) x-direction: balance horizonal forces T(x) cos(θ(x)) = T(x + ∆x) cos(θ(x + ∆x)) for all x! =) T(x) cos(θ(x)) = B constant y-direction: balance vertical forces T(x + ∆x) sin(θ(x + ∆x)) − T(x) sin(θ(x)) = mg = ρAg(arclength x to x + ∆x) = ρAg (s(x + ∆x) − s(x)) Set equal and divide by ∆x: T(x + ∆x) sin(θ(x + ∆x)) − T(x) sin(θ(x)) ρAg (s(x + ∆x) − s(x)) = ∆x ∆x d d ∆x ! 0 =) T(x) sin(θ(x)) = ρAg s(x) dx dx d d T(x) sin(θ(x)) = ρAg s(x) dx dx B Recall: T(x) cos(θ(x)) = B =) T(x) = . Substitute: cos(θ(x)) d sin(θ(x)) d B = ρAg s(x) dx cos(θ(x)) dx d ρAg d tan(θ(x)) = s(x) dx B dx d ρAg d f 0(x) = s(x) dx B dx d Z x q f 00(x) = K 1 + (f 0(t))2 dt dx 0 q f 00(x) = K 1 + (f 0(x))2 Solve the second order, nonlinear, separable differential equation to get the curve that models a hanging chain: q f 00(x) = K 1 + (f 0(x))2 Step 1. Let u(x) = f 0(x), so u0(x) = f 00(x). Exercise. Replace f 0 with u and solve to find a relationship between u and x. q u0(x) = K 1 + (u(x))2 Z du Z p = K dx 1 + u2 Z 2 sec (φ) dφ 2 p = Kx + C u = tan(φ); du = sec (φ) 1 + tan2(θ) Z sec(φ) dφ = Kx + C ln j sec(φ) + tan(φ)j = Kx + C p ln j 1 + u2 + uj = Kx + C right triangle p 1 + u2 + u = CeKx C > 0 How to write u as a function of x? p 1 + u2 u φ 1 u opp u = = tan(φ) = 1 adj p hyp 1 + u2 sec(φ) = = adj 1 Write u as a function of x: p u + 1 + u2 = CeKx p p p CeKx(u − 1 + u2) = (u + 1 + u2)(u − 1 + u2) p CeKx(u − 1 + u2) = u2 − (1 + u2) = −1 p =) u − 1 + u2 = −Ce−Kx Putting these together: p p u + 1 + u2 + u − 1 + u2 = CeKx − Ce−Kx 2u = CeKx − Ce−Kx C(eKx − e−Kx) u = 2 What is C? We have an initial condition! u(0) = f 0(0) = 0 p =) 0 + 1 + 0 = Ce0 =) C = 1 Step 2. Now, eKx − e−Kx u = f 0(x) = 2 is another separable differential equation! Solve by integrating: Z 1 Z f (x) = f 0(x)dx = eKx − e−Kx dx 2 1 = (eKx + e−Kx) + C 2K 1 Solve for C1: 1 0 = f (0) = (1 + 1) + C 2K 1 1 0 = + C K 1 1 C = − 1 K ρAg Recall: K = . The final solution is: B ρ ρ Ag x − Ag x B e B + e B B f (x) = − 2ρAg ρAg The curve is a hyperbolic cosine function! x −x I 2 cosh(x) = e + e x −x I 2 sinh(x) = e − e d2 d cosh(x) = sinh(x) = cosh(x) dx2 dx How to build the St. Louis arch? A hanging chain is perfectly balanced with respect to tension and gravity: flip it over! 630 ft 630 ft x f (x) = −630 cosh + 1260 239:2 defined on the domain jxj ≤ 315. This is a catenary curve!.