Path of Light- Calculus of Variation

[Newtons Balls]

Homework Statement

Let y(x) represent the path of light through a variable transparent medium. The speed of light at some point (x,y) in the medium is a function of x alone and is written c(x). Write down an expression for the time T taken for the light to travel along some arbitrary path y(x) from the point (a,yI) to the point (b,yF) in the form:
T=[tex]\int L(y,y')dx[/tex]
(between a and b, can't seem to add in limits on the integral sign)

where L is a function you should determine.

Homework Equations

S = [tex]\int \sqrt{1+y'^{2}}dx[/tex]

I derived this though, don't think its meant to be a known equation.

The Attempt at a Solution

Well using calculus of variation I know you can formulate a path length as:
S = [tex]\int \sqrt{1+y'^{2}}dx[/tex]

So the time for the light to travel along this path should just be:

T= [tex]\int \frac{ \sqrt{1+y'^{2}}}{c(x)}dx [/tex]

However, this has a dependence on x where the question states it must be only in terms of y and y'. This is because in the first part of the question I had to derive the Euler-Legrange equation and as light takes the path which minimises time, you must plug this function L into the equation.

So, how do I get rid of the x dependence? If there's a trick or a technique involved could someone just inform me of the name so I can look it up for myself? I've read the classical mechanics textbook which is meant to accompany this course and can find nothing which helps with this.

Thank.

 

[granpa]

can x be written as a function of y?

 

[Newtons Balls]

Well that's what has been running through my head the whole time, however everything that I have been given in the question I've posted. The only thing I can come up with is c(x) = c(y[tex]^{-1}[/tex](y)) as in the inverse function of y...of y. I don't think that helps...

Sorry if I'm being dense and there's something obvious going on here.