Stationary path for integral with sqrt(x(1-y'^2)) integrand

[Kidphysics]

Homework Statement

Path between O=(0,0) and P=(0,4) for which the integral

int. O to P (x (1-y'^2))^(1/2) dx

becomes stationary.

Homework Equations

http://en.wikipedia.org/wiki/Calculus_of_variations#Example

The Attempt at a Solution

Okay guys,

so mimicking the example from wiki but with L= (x)^1/2 (1-y'^2)^1/2

dL/dy=0 and dL/dy'= -(x)^1/2 y' / (1-y'^2)^1/2 and then d/dx dL/df' = - y'/[(2x^1/2)(1-y'^2)^1/2]

and since d/dx dL/df' =0 this implies y'=0 and we integrate to get y=c, adhering to boundary conditions y must equal 0... was that the correct way sorry for not writing in latex if I must I will redo the post

 

[haruspex]

dL/dy'= -(x)^1/2 y' / (1-y'^2)^1/2 and then d/dx dL/df' = - y'/[(2x^1/2)(1-y'^2)^1/2]

That last expression looks too simple. There should be a y" in there, and two different powers of x. Remember it's not a partial derivative this time.

 

[Kidphysics]

That last expression looks too simple. There should be a y" in there, and two different powers of x. Remember it's not a partial derivative this time.

ahhhh you are correct! y is a function of x..

 

[Kidphysics]

well.. I used the product rule and the chain rule and now I am looking at an unbelievable

2 x^1/2 y'^2 y'' - 1/2 x^-1/2 y' - x^1/2 y'' + 1/2 x^-1/2 y'^3 =0 .

I believe I took the derivative correctly this time but now I am lost as what to do with this..

 

[Kidphysics]

That last expression looks too simple. There should be a y" in there, and two different powers of x. Remember it's not a partial derivative this time.

well.. I used the product rule and the chain rule and now I am looking at an unbelievable

2 x^1/2 y'^2 y'' - 1/2 x^-1/2 y' - x^1/2 y'' + 1/2 x^-1/2 y'^3 =0 .

I believe I took the derivative correctly this time but now I am lost as what to do with this.

 

[haruspex]

Allow me to put that in LaTex:

$$2 x^{1/2} y'^2 y'' - (1/2)x^{-1/2} y' - x^{1/2} y'' + (1/2) x^{-1/2} y'^3 =0$$
Simplifying:
$$4 x y'^2 y'' - y' - 2x y'' + y'^3 =0$$
But that's not quite what I get. Writing K = (1-y'²)^1/2:
L = x^1/2K
∂K/∂y' = -y'/K
∂L/∂y' = -x^1/2y'/K
$$0 = -\frac{d(∂L/∂y')}{dx} = \frac{x^{-1/2}y'}{2K} + \frac{x^{1/2}y''}{K} - \frac{x^{1/2}y'\frac{∂K}{∂y'}y''}{K^2}$$
$$0 = y'+ 2xy'' - 2\frac{xy'\frac{∂K}{∂y'}y''}{K}$$
$$0 = y'+ 2xy'' + 2\frac{xy'^2y''}{K^2}$$
$$0 = (y'+ 2xy'')(1-y'^2) + 2xy'^2y''$$
$$0 = y'+ 2xy'' - y'^3$$
Is that right or did I get sign wrong somewhere?

 

[Kidphysics]

Allow me to put that in LaTex:

$$2 x^{1/2} y'^2 y'' - (1/2)x^{-1/2} y' - x^{1/2} y'' + (1/2) x^{-1/2} y'^3 =0$$
Simplifying:
$$4 x y'^2 y'' - y' - 2x y'' + y'^3 =0$$
But that's not quite what I get. Writing K = (1-y'²)^1/2:
L = x^1/2K
∂K/∂y' = -y'/K
∂L/∂y' = -x^1/2y'/K
$$0 = -\frac{d(∂L/∂y')}{dx} = \frac{x^{-1/2}y'}{2K} + \frac{x^{1/2}y''}{K} - \frac{x^{1/2}y'\frac{∂K}{∂y'}y''}{K^2}$$
$$0 = y'+ 2xy'' - 2\frac{xy'\frac{∂K}{∂y'}y''}{K}$$
$$0 = y'+ 2xy'' + 2\frac{xy'^2y''}{K^2}$$
$$0 = (y'+ 2xy'')(1-y'^2) + 2xy'^2y''$$
$$0 = y'+ 2xy'' - y'^3$$
Is that right or did I get sign wrong somewhere?

hey thanks, jumped out of bed to respond.. I'm sure you have the math correct but I can't see how this helps us get a solution the the original problem. At this step we have a nonlinear ode so I'm a little frightened we are on the wrong track? Thank you for helping thus far

 

[haruspex]

$$0 = y'+ 2xy'' - y'^3$$

It's not hard from there. There are no y terms, so it's first order, and you can get it into the form f(x) = g(t)t' where t = y', so then it's just an integration to get an equation relating x to y'.