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Math Help - Shortest distance

  1. #1
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    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.


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  2. #2
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    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
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  3. #3
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    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.
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  4. #4
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    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!
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