Area between Smooth Curve and Line with Fixed Endpoints is Maximized by Arc of Circle

From ProofWiki
Jump to navigation Jump to search


This article needs to be linked to other articles.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{MissingLinks}} from the code.


Theorem

Let $y$ be a smooth curve, embedded in $2$-dimensional Euclidean space.

Let $y$ have a total length of $l$.

Let it be contained in the upper halfplane with an exception of endpoints, which are on the x-axis and are given.

Suppose, $y$, together with a line segment connecting $y$'s endpoints, maximizes the enclosed area.


Then $y$ is an arc of a circle.


Proof

Without loss of generality, we choose our point of reference such that $y$ intersect x-axis at points $\tuple {-a, 0}$ and $\tuple {a, 0}$ for some $a > 0$.

The area below the curve $y$ is a functional of the following form:

$\ds A \sqbrk y = \int_{-a}^a y \rd x$

Furthermore, $y$ has to satisfy the following conditions:

$\map y {-a} = \map y a = 0$
$\ds L \sqbrk y = \int_{-a}^a \sqrt {1 + y'^2} \rd x = l$

By Simplest Variational Problem with Subsidiary Conditions, there exists a constant $\lambda$ such that the functional:

$\ds A \sqbrk y + \lambda L \sqbrk y = \int_{-a}^a \paren {y + \lambda \sqrt {1 + y'^2} } \rd x$

is extremized by the mapping $y$.

The corresponding Euler's Equation reads:

$1 + \lambda \dfrac \d {\d x} \dfrac {y'} {\sqrt {1 + y'^2} } = 0$

Integrating with respect to $x$ once yields:

$x + \lambda \dfrac {y'} {\sqrt {1 + y'^2} } = C_1$

Solve this for $y'$:

$\ds y' = \pm \frac {c_1 - x} {\sqrt {\lambda^2 - c_1^2 + 2 c_1 x - x^2} }$

Integration yields:

$\paren {x - C_1}^2 + \paren {y - C_2}^2 = \lambda^2$

This is an equation for a circle with radius $\lambda$ and center $\tuple {C_1, C_2}$.

To find $C_1, C_2, \lambda$, apply boundary conditions and the length constraint.

From the boundary conditions we have that:

\(\ds \paren {-a - C_1}^2\) \(=\) \(\ds \lambda^2\)
\(\ds \paren {a - C_1}^2\) \(=\) \(\ds \lambda^2\)

Take the difference of these two equations:

$4 a C_1 = 0 \implies C_1 = 0$

because $a > 0$.

Apply one of the boundary conditions again, that is, at $\tuple {a, 0}$:

$a^2 + C_2^2 = \lambda^2$

Then:

$C_2 = \pm \sqrt {\lambda^2 - a^2}$.

which can be used to get rid of $C_2$.

The last parameter to find is $\lambda$.

We have two cases:

the curve is an arc of the upper semicirle;
the curve is a union of upper semicirle with two arcs of lower semicirle.

In the first case the length constraint is:

$l = 2 \lambda \map \arctan {\dfrac a {\sqrt {\lambda^2 - a^2} } }$

For real $\lambda$, if $\lambda \ge a$, then $l \in \R$.

To find extremal values, consider the derivate $\dfrac {\d l} {\d \lambda}$:

\(\ds \dfrac {\d l} {\d \lambda}\) \(=\) \(\ds 2 \paren {\map \arctan {\frac 1 {\sqrt {\lambda^2 - 1} } } - \frac 1 {\sqrt {\lambda^2 - 1} } }\)
\(\ds \) \(<\) \(\ds 0\) Tangent Inequality

Hence the domain of $l$ is determined by boundary values.

At the boundary of $\lambda = a$ we have:

\(\ds \lim_{\lambda \mathop \to a^+} l\) \(=\) \(\ds \lim_{\lambda \mathop \to a^+} 2 \lambda \map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } }\)
\(\ds \) \(=\) \(\ds 2 a \lim_{\lambda \mathop \to a^+} \map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } }\) Product Rule for Limits of Real Functions

To calculate the limit of such composite function denote:

\(\ds \map f y\) \(=\) \(\ds \map \arctan y\)
\(\ds \map g {\lambda}\) \(=\) \(\ds \frac a {\sqrt {\lambda^2 - a^2} }\)

It follows that:

$\ds \lim_{\lambda \mathop \to a^+} \map g \lambda = + \infty$
$\ds \lim_{y \mathop \to \infty} \map f y = \frac \pi 2$

Arctangent is continuous for all $x \in \R$.

Then, by Limit of Composite Function:

$\ds 2 a \lim_{\lambda \mathop \to a^+} \map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } } = \pi a$

At the boundary of $\lambda = + \infty$ we have:

\(\ds \lim_{\lambda \mathop \to \infty} l\) \(=\) \(\ds \lim_{\lambda \mathop \to \infty} 2 \lambda \map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } }\)
\(\ds \) \(=\) \(\ds \lim_{\lambda \mathop \to \infty} 2 \frac {\map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } } } {\frac 1 \lambda}\) Indeterminate limit $\infty \cdot 0$
\(\ds \) \(=\) \(\ds \lim_{\lambda \mathop \to \infty} 2 \frac {-\frac a {\lambda \sqrt {\lambda^2 - a^2} } } {-\frac 1 {\lambda^2} }\) L'Hôpital's Rule
\(\ds \) \(=\) \(\ds \lim_{\lambda \mathop \to \infty} 2 a \frac \lambda {\sqrt {\lambda^2 - a^2} }\)
\(\ds \) \(=\) \(\ds 2 a\)

In the second case the length constraint is:

$l = 2 \lambda \paren {\pi - \arctan \dfrac a {\sqrt {\lambda^2 - a^2} } }$

Similarly to the previous case:

\(\ds \dfrac {\d l} {\d \lambda}\) \(=\) \(\ds 2 \pi - 2 \map \arctan {\frac 1 {\sqrt {\lambda^2 - 1} } } + \frac 1 {\sqrt {\lambda^2 - 1} }\)
\(\ds \) \(>\) \(\ds 0\) Tangent Inequality

Hence the domain of $l$ is determined by boundary values.

At the boundary of $\lambda = a$ we have:

\(\ds \lim_{\lambda \mathop \to a^+} l\) \(=\) \(\ds 2 \lambda \paren {\pi - \arctan \dfrac a {\sqrt {\lambda^2 - a^2} } }\)
\(\ds \) \(=\) \(\ds 2 \pi a - \pi a\)
\(\ds \) \(=\) \(\ds \pi a\)

As $\lambda$ approaches the infinity we have:

\(\ds \lim_{\lambda \mathop \to \infty} l\) \(=\) \(\ds \lim_{\lambda \mathop \to \infty} 2 \lambda \paren {\pi - \arctan \dfrac a {\sqrt {\lambda^2 - a^2} } }\)
\(\ds \) \(=\) \(\ds \lim_{\lambda \mathop \to \infty} 2 \lambda \pi - 2 a\)
\(\ds \) \(=\) \(\ds \infty\)

Therefore:

$\forall l \ge 2 a: \exists \lambda \ge a$

Hence, within these constraints the real solution maximizing the area with fixed endpoints is an arc of a circle.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Area_between_Smooth_Curve_and_Line_with_Fixed_Endpoints_is_Maximized_by_Arc_of_Circle&oldid=542972"