How Does the Euler-Lagrange Equation Apply to Parametric Solutions?

[Another]

Homework Statement

I am wondering about definition of a function

Relevant Equations

$F = \frac{(1+(y_x)^2)^{\frac{1}{2}}}{(y_1-y)^{\frac{1}{2}}}$

My question : I am wondering about definition of a function. when $y_x = (\frac{b+y}{a-y})^2$

Why in this book is defined solution $y = y(x)$ in from $y = y(θ(x))$ . And have a relationship in the form

$y = \frac{1}{2} (a-b) - \frac{1}{2} (a+b) cosθ$ ?

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In this book is defined $F = \frac{(1+(y_x)^2)^{\frac{1}{2}}}{(y_1-y)^{\frac{1}{2}}}$

The E-L equation in case F = $F(y,y_x)$ ===> $F - y_x \frac{∂F}{∂y_x} = c$ when c is constant.

 

[TSny]

My question : I am wondering about definition of a function. when $y_x = (\frac{b+y}{a-y})^2$

The exponent should be 1/2 rather than 2.

Why in this book is defined solution $y = y(x)$ in from $y = y(θ(x))$ . And have a relationship in the form

$y = \frac{1}{2} (a-b) - \frac{1}{2} (a+b) cosθ$ ?

This is just a very clever substitution that allows you to fairly easily find a solution of the differential equation, $y_x = (\frac{b+y}{a-y})^{1/2}$. It yields a solution in parametric form: $y(\theta)$ ; $x(\theta)$, where $\theta$ is a parameter.

 

[Another]

The exponent should be 1/2 rather than 2.This is just a very clever substitution that allows you to fairly easily find a solution of the differential equation, $y_x = (\frac{b+y}{a-y})^{1/2}$. It yields a solution in parametric form: $y(\theta)$ ; $x(\theta)$, where $\theta$ is a parameter.

thank you very much