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Homework Help: Shortest curve enclosing given area

  1. Mar 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that the area enclosed by a closed curve {x(t); y (t)} is given by
    [itex]A=\frac{1}{2} \int_{t_1}^{t_2} (x\dot{y}-y\dot{x})dt[/itex]

    Show that the expression for the shortest curve which encloses a given area, A, may be found by minimising the expression

    [itex]s=\int_{t_1}^{t_2} \sqrt{\dot{x}^2 + \dot{y}^2 }dt +\lambda(\frac{1}{2} \int_{t_1}^{t_2} (x\dot{y}-y\dot{x})dt - A)[/itex]

    Hence show that [itex]x \propto \ddot{y}[/itex] and [itex]y \propto \ddot{x}[/itex].

    2. Relevant equations
    Euler-Lagrange equations

    3. The attempt at a solution
    I don't know how to get the equation for the area. A hint says to consider a triangle formed by (0, 0), (x, y ) and (x + dx, y + dy ) - but I'm not sure how to use it.

    But assuming the expression is correct, the expression to be minimised follows simply from applying the Lagrange multiplier technique to the length mimimisation problem. The Euler-Lagrange equation applied to s for x gives
    (See posts below)

    A similar equation comes for y, with x and y's interchanged in the above one. For λ,

    I'm unsure on how to use these three to show the proportionality relations.
    Last edited: Mar 25, 2013
  2. jcsd
  3. Mar 25, 2013 #2


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    Staff Emeritus
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    The equation for the area can be derived from Green's Theorem in the plane. What is the area of the triangle with the coordinates specified in the hint.
  4. Mar 25, 2013 #3
    I do not think the equation you obtained is correct. Unless I am mistaken: $$ \frac {d} {dt} \frac {\partial L} {\partial \dot{x}} = \frac {d} {dt} (\frac {\dot{x}} {\sqrt{\dot{x}^2 + \dot{y}^2}} - \frac 1 2 \lambda y)
    = \frac {d} {dt} (\frac {\dot{x}} {\sqrt{\dot{x}^2 + \dot{y}^2}}) - \frac 1 2 \lambda \dot{y}


    \frac {\partial L} {\partial x} = \frac 1 2 \lambda \dot{y}


    \frac {d} {dt} \frac {\partial L} {\partial \dot{x}} - \frac {\partial L} {\partial x}
    = \frac {d} {dt} (\frac {\dot{x}} {\sqrt{\dot{x}^2 + \dot{y}^2}}) - \lambda \dot{y}
    = \frac {d} {dt} (\frac {\dot{x}} {\sqrt{\dot{x}^2 + \dot{y}^2}} - \lambda y + C) = 0
  5. Mar 25, 2013 #4
    SteamKing, I'm not sure how to get the area of the triangle - what can I choose as the base and height? Or is there some other way?

    Voko, you are right, in my calculation I forgot that y is also a function of t. I will try to get the proportionality result by using the fact that the expression you put in brackets is a constant. Is this approach fine, or is there some simpler way to approach it?
  6. Mar 25, 2013 #5
    To be honest, I am not sure what approach would be most direct here. Curiously, the Beltrami identity seems to yield a different equation.

    For the triangle, describe the minimal rectangle about the triangle. Observe that the area of the rectangle will the sum of four triangles, one of them the original one, another three right rectangles, so you can easily find the area of the original triangle in terms of its coordinates.
  7. Mar 25, 2013 #6
    For the triangle, I am finding that the area is

    [itex]\frac{1}{2} (xdy-ydx)[/itex]

    I think I need to integrate this over some limits to get the enclosed area - but I'm not sure what the limits are.
  8. Mar 25, 2013 #7
    x and y are parametrized by t, so the limits will be some values of t between which the curve makes a full cycle. Because the parametrization is arbitrary, the limits can be any two different values.
  9. Mar 25, 2013 #8
    Right, that makes sense. But one thing I have a doubt over is why the integrating will yield the area enclosed. I can't see how this triangle integrated over the area will cover areas which are not touching the origin, for example.
  10. Mar 25, 2013 #9
    Imagine a circle completely contained in the first quadrant. Then any straight line from the origin will cross the curve twice (except in two points where it will be just tangent). That means any such line can be used to build two triangles. One with the "farthest" curve, which will include the area enclosed by the curve, another with the "nearest" curve, which will only include the area not enclosed by the curve. If these triangles are very narrow, then one could reasonably state that we need to compute their difference to find the the area enclosed by the curve. Now all you need to do is show that the sign of the integrand on the "nearest" curve will be opposite to that on the "farthest" curve.
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