6 Variational Methods

Now recall that infinitesimal arc length is given by√∆s ≈ √ (∆x) 2 +(∆y) 2 =1+( ) 2 ∆y∆x.∆xThus we obtain∫ x2S =(2π) y(x) √ 1+(y ′ (x)) 2 dx.x 1Thus we haveF (x, y, y ′ )=2πy √ 1+(y ′ ) 2and the Euler equation (6.1.3) becomes[]d yy ′− (1+(y ′ ) 2 ) 1/2 =0.dx (1+(y ′ ) 2 ) 1/2If we use (6.1.5) then we getTo solve this equation we letyy ′′ − (y ′ ) 2 − 1=0.p = y ′ so that y ′′ = dpdx = dp dydy dx = pdp dyand this givespy dpdy = p2 +1.This equation is separable, and is integrated as followspdpp 2 +1 = dyy⇒1 2 ln(p2 + 1) = ln(y)+C.Exponentiating both sides we getSolving for dy/dx we get(y = α 1+( ) ) 2 1/2dy.dxdydx = ( y2α − 1 ) 1/2,6