Calculus of Variations; Maximum enclosed area problem.

[Aaron Curran]

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!

 

[stevendaryl]

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!

Your expression for ds is confusing you. The relationship between ds and dx is:

ds = \sqrt{1 + (\frac{dy}{dx})^2} dx

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

Now, you plug in your answer for dx in the expression for the area under the curve, which is just \int y dx.

 

[Ray Vickson]

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!

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 $L(y,y') = y + u \sqrt{1 + y'^2}.$

 

[stevendaryl]

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 $L(y,y') = y + u \sqrt{1 + y'^2}.$

But you can convert \int y dx = \int y \frac{dx}{ds} ds = \int y \sqrt{1-(\frac{dy}{ds})^2} ds. Presumably, that is equivalent to doing the Lagrange multiplier thing.

 

[Ray Vickson]

But you can convert \int y dx = \int y \frac{dx}{ds} ds = \int y \sqrt{1-(\frac{dy}{ds})^2} ds. Presumably, that is equivalent to doing the Lagrange multiplier thing.

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.

 

[stevendaryl]

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.

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