Integrate $(6\mathrm{.}2),$ simplify the result and integrate again to get $(6\mathrm{.}3)$ where $c$ and $c^{\text{'}}$ are constants of integration. If $x_1=-\sqrt[\phantom{2}]{3},x_2=\sqrt[\phantom{2}]{3},$ and $l=4\frac{}{3},$ show that the center and radius o

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Calculus of Variations

Example $1.$ Given two points $x_1$ and $x_2$ on the $x$ axis, and an arc length $l>x_2-x_{1,f}$ the shape of the curve of length $l$ joining the given porins which, with the $x_{\text{d}\text{x}\text{i}\text{s}}$ incloses the largest area.

We want to maximize $I={\displaystyle \underset{x_1}{\overset{x_2}{}}}ydx$ subject to the condition $J={\displaystyle \underset{x_1}{\overset{x_2}{}}}{d}_8=1.$ Here $F=y$ and $G=\sqrt[\phantom{2}]{1+y^{\text{'}2}}$ so

$F+G=y+\sqrt[\phantom{2}]{1+y^{\text{'}2}}$

We want the Euler equation for $F+G.$ Since

$\frac{\text{d}\text{e}\text{l}}{del{y}^{\text{'}}}(F+G)=\frac{y^{\text{'}}}{\sqrt[\phantom{2}]{1+y^{\text{'}2}}}$ and $\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{y}}(F+G)=1$

the Euler equation is

$\frac{d}{dx}(\frac{y^{\text{'}}}{\sqrt[\phantom{2}]{1+y^{\text{'}2}}})-1=0$

The solution of $(6\mathrm{.}2)$ is $($Problem $7)$:

$(x+c{)}^2+(y+c^{\text{'}}{)}^2={}^2$

We see that the answer to our problem is an arc of a circle passing through the tro given points, and the Lagrange multiplier $$ is the radius of the circle. The center and radius of the circle are determined by the given points $x_1$ and $x_2,$ and the given arc length $l($Problem $7).$f the circle are $(0,-1)$ and $=$ radius $=2.$