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Calculus of Variations; Maximum enclosed area problem.

  1. Oct 26, 2016 #1
    The problem reads: "You are given a string of fixed length l with one end fastened at the origin O, and you are to place the string in the (x, y) plane with its other end on the x axis in such a way as to maximise the area between the string and the x axis. Show that the required shape is a semicircle."
    Hint: Show that the area can be written as
    $$J(y) = \frac{1}{c}\int_{0}^{l} y(s)\sqrt{1 + (y'(s))^2}ds$$
    Where s is the arc-length along the string, and then use the following first integral of the Euler-Lagrange equation
    $$F - y'\frac{\partial F}{\partial y'} = C$$
    Where ##y' = dy/ds## and ##F=y\sqrt{1 + (y')^2}##

    My attempt at a solution:
    I know that the rules of this forum state that saying I really don't have a clue what to do is not allowed but I genuinely have been stuck on this for all of today and for hours last night as well. I know that if y were a function of x, then the following would work
    $$\frac{\triangle{s^2}}{\triangle{x^2}} = 1 + \frac{\triangle{y^2}}{\triangle{x^2}}$$
    and then taking the limit as x tends to 0 would give
    $$ds = \sqrt{1 + (y')^2}dx$$

    But due to the question stating that y is a function of s, the arc-length. I have absolutely no clue where to go, I simply can not see how it is shown. Any help would be much appreciated, thank you!
     
  2. jcsd
  3. Oct 26, 2016 #2

    stevendaryl

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    Your expression for [itex]ds[/itex] is confusing you. The relationship between [itex]ds[/itex] and [itex]dx[/itex] is:

    [itex]ds = \sqrt{1 + (\frac{dy}{dx})^2} dx[/itex]

    What you wrote is incorrect, because [itex]y' \equiv \frac{dy}{ds}[/itex], not [itex]\frac{dy}{dx}[/itex]. And what you want is [itex]dx[/itex] in terms of [itex]ds[/itex], not [itex]ds[/itex] in terms of [itex]dx[/itex]. So how do you find [itex]dx[/itex] in terms of [itex]ds[/itex] and [itex]dy[/itex]?

    Now, you plug in your answer for [itex]dx[/itex] in the expression for the area under the curve, which is just [itex]\int y dx[/itex].
     
    Last edited: Oct 26, 2016
  4. Oct 26, 2016 #3

    Ray Vickson

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    I don't understand the formulation. If I were doing it I would regard it as a maximization problem with a constraint. For a string of length ##c## we have the problem
    $$\begin{array}{rl}
    \max &J = \int_0^{x_1} y \, dx \\
    \text{subject to} & \int_0^{x_1} \sqrt{1 + y'^2} \, dx = c\\
    \text{and}& y(0) = y(x_1) = 0
    \end{array}
    $$
    We could convert this to an unconstrained problem using a Lagrange multiplier method;
    $$\max K = \int_0^{x_1} L(y,y') \, dx \equiv \int_0^{x_1} y \, dx + u \int_0^{x_1} \sqrt{1+y'^2} \, dx $$
    with boundary conditions ##y(0) = y(x_1) = 0##.
    Here, the real constant ##u## is the Lagrange multiplier and ##L(y,y') = y + u \sqrt{1 + y'^2}.##
     
  5. Oct 26, 2016 #4

    stevendaryl

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    But you can convert [itex]\int y dx = \int y \frac{dx}{ds} ds = \int y \sqrt{1-(\frac{dy}{ds})^2} ds[/itex]. Presumably, that is equivalent to doing the Lagrange multiplier thing.
     
  6. Oct 29, 2016 #5

    Ray Vickson

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    Yes, but that is not the formulation originally given in the question, which is to maximize ##c^{-1} \int_0^{l} y \sqrt{1+y'(s)^2} \, ds##. When I said I did not understand the formulation, that is the one I was referring to.
     
  7. Oct 29, 2016 #6

    stevendaryl

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    Oh, I didn't notice the sign. I think the sign of [itex]y'[/itex]. I think the sign is wrong in the original post.
     
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