# Thread: [SOLVED] Calculas of Variations problem

1. ## [SOLVED] Calculas of Variations problem

Hey all

I do Calculas of Variations, and have an exam in around nine hours, so would appreciate help before then. After is good too thhough.

I got a solution, but it has a bit I don't understand, was hoping you guys could help.

My problem is:
Find extremal of J(y)=$\displaystyle \int_0^1(y(x))dx$,subject to $\displaystyle \int_0^1\sqrt(1+y'(x)^2)dx = L$.
$\displaystyle y(0) = y(1) =0$

Solution that I have gotten is:
$\displaystyle F'= F + \lambda*G$
$\displaystyle F= y(x), G =\sqrt(1+(y')^2).$

$\displaystyle \frac{dF*}{dy}-\frac{d}{dx}(\frac{dF*}{dy'})=0$
$\displaystyle 1-\frac{d}{dx}(\lambda*y'/\sqrt(1+(y')^2)$
Integrate:
$\displaystyle \rightarrow \lambda*y'/\sqrt(1+(y')^2)=x+C$ C arbitary constant
so
$\displaystyle (y')^2/(1+(y')^2)=(x+C)^2/\lambda^2$
$\displaystyle y'^2 = ((x+c)^2/\lambda^2)/(1-(x+C)^2/\lambda^2);$
$\displaystyle y'=(x+C/\lambda)/(\sqrt(1+(x+C)^2)/\lambda^2).$
$\displaystyle u=1-(x+C)^2/\lambda^2$

Put u in, integrate, change the result around a small bit and you get
$\displaystyle (x+C)^2/\lambda^2+(y+D)^2/\lambda^2=1$ D arbitary constant
inset bound conditions, you get $\displaystyle C = -1/2$
$\displaystyle D=\sqrt(\lambda^2-1/4) =\sqrt(\lambda^2-C^2)$

So, $\displaystyle (x-1/2)^2+(y+\sqrt(\lambda^2-1/4)^2=\lambda^2$
Diffretiate with respect to x;
$\displaystyle (x+1/2)+(y+\sqrt(\lambda^2-1/4)*y'=0 => y'=-(x-1/2)/(y+\sqrt(\lambda^2-1/4).$

Then (And herin lies the problem), I get
$\displaystyle 1+y'^2=1+(x-1/2)^2/((y+\sqrt(\lambda^2-1/4))^2) = \lambda^2/(\lambda^2-(x-1/2)^2.$

How does he change $\displaystyle 1+(x-1/2)^2/((y+\sqrt(\lambda^2-1/4))^2)$ into $\displaystyle \lambda^2/(\lambda^2-(x-1/2)^2)?$