Problem 2: A piece of string is length L. One end is taped to the origin and the other end is taped somewhere on the x axis. Find the path of the string that produces the maximum area. Note that we are not trying to find a minimum (like most of the examples in lecture), but the maximum. The Euler-Lagrange works in this case because it finds extremum, which can produce either max or min. a)The area is usually found by integrating A=] ydx, but in this case, you are not given y(x). 50 rather than using x as the independent variable, use the pathlength s as the independent variable. So start by writing the area A as integral over f(y',y)ds. Hint: ds'=dx2+dy?, so dx2=ds2- dy, so dx = ds 1- dy ds. = ds /1 -y'2 b) Find y as a function of s, y(s). Since f only depends on y'(s) and y(s), but not s explicitly, you can use the simplification of Euler-Lagrange that I put in lecture notes, f - y' df dy - = Constant = C c)Next, you know that y=0 when s=L because the string is taped somewhere on the x axis. Use your answer to part b to show that L/C=1. d) Finally, we would really like y(x) instead of y($). The idea is to get x(s) and eliminate s from x(s) and y(s). How do you get x(s)? Start by integrating x=[ dx = [ ds\\/1 - y'2 and use y(s) from part b to get y'(s). You should be able to do the integral to get x($) e) Once you eliminate s, show that (x-C) +y?=C?. Of course, this describes the path of a circle of radius C, with center of circle at C, and from part c, C=L. The path is a semicircle in this case