# Shortest distance

• Dec 29th 2011, 07:37 AM
Duke
Shortest distance
Consider the integral I with integrand F(x,y,y'), limits a,b with y(b) undetermined.

(1) Derive both the Euler-Lagrange 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.

• Dec 30th 2011, 02:56 AM
jens
Re: Shortest distance
It's the arc length integral. See here for details:

Arc length - Wikipedia, the free encyclopedia

Assuming that $f$ is differentiable, the length of an arc over an interval $[a;b]$ described by a function $f$ is given by

$\int_a^b \sqrt{1+f'(x)} dx$
• Dec 31st 2011, 02:17 AM
Duke
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.
• Jan 2nd 2012, 01:48 PM
jens
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!