1. (60 points.) Consider a rope of uniform mass density ) = dm/ds hanging from two points, (x1, Y1) and (x2, y2), as shown in Figure 1. The gravitational potential

energy of infinitely tiny element of this rope at point (x, y) is given byan d U=d m g y=\lambda g d s y where d s^{2}=d x^{2}+d y^{2} A catenary is the curve that the rope assumes, that minimizes the total potential energyof the rope. (a) Show that the total potential energy U of the rope hanging between points x1 andX2 is given by U[x]=\lambda g \int_{\left(x_{1}, y_{1}\right)}^{\left(x_{2}, y_{2}\right)} y d s=\lambda g \int_{y_{1}}^{y_{2}} d y y \sqrt{1+\left(\frac{d x}{d y}\right)^{2}} (b) Since the curve passes through the points (x1, Yı) and (x2, Y2), we have no variationsat these (end) points. Thus, show that (c) Using the extremum principle show that the differential equation for the catenary is \frac{d x}{d y}=\frac{a}{\sqrt{y^{2}-a^{2}}} where a is an integration contant. (d) Show that integration of the differential equation yields the equation of the catenary y=a \cosh \frac{x-x_{0}}{a} where xo is another integration constant. (e) For the case y1 = Y2 we have \frac{y_{1}}{a}=\cosh \frac{x_{1}-x_{0}}{a} \frac{y_{2}}{a}=\cosh \frac{{x}_{2}-{x}_{0}}{a} which leads to the solution, assuming x1 7 x2, {x}_{0}=\frac{\boldsymbol{x}_{\mathbf{1}}+\boldsymbol{x}_{\mathbf{2}}}{\mathbf{2}} Identify xo in Figure 1. (f) Next, derive \frac{y_{1}}{a}=\frac{y_{2}}{a}=\cosh \frac{x_{2}-x_{1}}{2 a} which, in principle, determines a. However, this is a transcendental equation in aand does not allow exact evaluation of a, and one depends on numerical solutions.Observe that, if x = xo in Eq. (6), then y = a. Identify a in Figure 1. \frac{\delta U[x]}{\delta x(y)}=-\lambda g \frac{d}{d y}\left[y \frac{\frac{d x}{d y}}{\sqrt{1+\left(\frac{d x}{d y}\right)^{2}}}\right]