
Shortest distance
Consider the integral I with integrand F(x,y,y'), limits a,b with y(b) undetermined.
(1) Derive both the EulerLagrange equation and the endpoint condition that 'partial' df/dy' vanishes at b.
(2) Use this result to show that the shortest distance from the origin to the line x = 1 lies
along the line y = 0.
I have done question (1). I included it for context. For (2) I am not sure what the functional is.

Re: Shortest distance
It's the arc length integral. See here for details:
Arc length  Wikipedia, the free encyclopedia
Assuming that $\displaystyle f$ is differentiable, the length of an arc over an interval $\displaystyle [a;b]$ described by a function $\displaystyle f$ is given by
$\displaystyle \int_a^b \sqrt{1+f'(x)} dx$

Re: Shortest distance
Thanks for that. So in this case are my limits (endpoints) x=0 and x=1 with boundary condition y(0)=0? Then I need to use natural boundary condition that df/dy'= 0 when x=1? I got the right answer just need to know it was through correct working.

Re: Shortest distance
Frankly, I can't tell. Your approach looks reasonable. But I don't know about the "natural boundary condition", nor do I understand why this condition should necessarily hold in your case. I'm no expert in variational calculus. Sorry!