- #1

stephenklein

- 6

- 3

- Homework Statement
- Imagine we have a string of length ##l## in the ##x y## plane with endpoints ##x=-a## and ##x=a##. The endpoints can move to maximize this area, but the total length of the string is fixed.

a) Show that ##dx = \sqrt {1-y'^2}## where ##ds## is a tiny length of the curve such that ##ds = \sqrt {dx^2+dy^2}## as in class.

b) The area of the shaded rectangle above is ##ydx##, so the sum of all of those areas between ##x=-a## and ##x=a## gives us the total area. However, it’s easier to incorporate the fact that the length is fixed as ##l## by integrating with respect to s, the path length of moving along the string instead. Using the expression in part (a) to convert the ##ydx## integral to an integral with respect to ##ds##, express the area under the string.

c) Using the Euler-Lagrange equations prove that the optimal shape of the string is a semicircle.

A couple hints:

1) The Euler-Lagrange equation should yield ##\frac {dy} {ds} = \sqrt {1- \frac {y^2} {C^2}}##, which you can solve using separation of variables (and probably looking up the integral).

2) A semicircle of radius R has the equation ##x^2+y^2=R##

- Relevant Equations
- Euler-Lagrange equation: $$\frac {\partial f} {\partial y} - \frac {d} {ds} \frac {\partial f} {\partial y'} = 0$$

I was able to work through parts (a) and (b). For part (c), I got $$\frac {\partial f} {\partial y} = \sqrt {1-y'^2}$$ and $$\frac {\partial f} {\partial y'} = \frac {-y y'} {\sqrt {1-y'^2}}$$ Taking ##\frac {d} {ds}## of the latter, I used the product rule for all three terms ##y, y', (\sqrt{1-y'^2})^{-1/2}## and my result was $$\sqrt {1-y'^2} + \frac {y y''+y'^2} {\sqrt {1-y'^2}} + \frac {y y'^2 y''} {(1-y'^2)^{3/2}} = 0$$ I'm unsure (even more, skeptical) that this result simplifies to the one given in the question. I'm confident in everything up until the last derivative, which has a lot of moving parts. Any thoughts?

EDIT: I simplified the above result to ##y'=\sqrt {1+y y''}##, which agrees with a Wolfram widget I found that simplifies Lagrangian equations. I feel like I'm very close, but my result has a sum instead of a difference, and that pesky ##y''## term isn't in the given answer.

EDIT: I simplified the above result to ##y'=\sqrt {1+y y''}##, which agrees with a Wolfram widget I found that simplifies Lagrangian equations. I feel like I'm very close, but my result has a sum instead of a difference, and that pesky ##y''## term isn't in the given answer.

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