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Finding the equation of a straight line in 3 dimensions.

  1. May 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that the shortest path between two points in three dimensions is a straight line. Write the path in the parametric form:

    x=x(u) y=y(u) z=z(u)

    and then use the 3 Euler-Lagrange equations corresponding to ∂f/∂x=(d/du)∂f/∂y'.


    2. Relevant equations
    Stated them above:]


    3. The attempt at a solution
    I found all of the answers in relation to the Euler-Lagrange equations, but I am not sure where to go from there. For each coordinate, ∂f/∂x,∂f/∂y,∂f/∂z, they all equal 0 so that means that d/du(∂f/∂x,y,z) are all also zero. As a result, I get constants for each and hence don't know how to implement these constants into a straight line equation.

    The constants are :
    ∂L/∂x'=x'(u)/(x'(u)^2 +y'(u)^2 +z'(u)^2)^(1/2)=C_1
    ∂L/∂y'=y'(u)/(x'(u)^2 +y'(u)^2 +z'(u)^2)^(1/2)=C_2
    ∂L/∂z'=z'(u)/(x'(u)^2 +y'(u)^2 +z'(u)^2)^(1/2)=C_3
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 11, 2012 #2
    What's the equation for a straight line? Can you find for example dx/dy from these equations?
     
  4. May 11, 2012 #3
    Thanks for the reply, clamtrox. I get what the equation of a straight line is: y=mx+b, but I'm not sure what you mean for finding dx/dy from those equations. There is also a dz.
     
  5. May 11, 2012 #4
    Yes, but perhaps a more useful way to write that equation is [itex]y = \frac{dy}{dx}x + y(0)[/itex].
     
  6. May 14, 2012 #5
    Yes sorry, that was bad notation. I of course mean partial derivatives: ∂x/∂y = x'(u)/y'(u)
     
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