Chapter 3: Problem 25
*The method of Lagrange multipliers (see Appendix A, Problem 11) can be extended to the calculus of variations. To find the maxima and minima of an integral \(I\), subject to the condition that another integral \(J=0\), we have to find the stationary points of the integral \(I-\lambda J\) under variations of the function \(y(x)\) and of the parameter \(\lambda\). Apply this method to find the catenary, the shape in which a uniform heavy chain hangs between two fixed supports. [The required shape is the one that minimizes the total potential energy, subject to the condition that the total length is fixed. Show that this leads to a variational problem with $$ f\left(y, y^{\prime}\right)=(y-\lambda) \sqrt{1+y^{\prime 2}} $$ and hence to the equation $$ (y-\lambda) y^{\prime \prime}=1+y^{\prime 2} $$ Solve this equation by introducing the new variable \(u=y^{\prime}\) and solving for \(u(y) .]\)
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