Finding the equation of a straight line in 3 dimensions.

[seeingstars63]

Homework Statement

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'.

Homework Equations

Stated them above:]

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

 

[clamtrox]

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

What's the equation for a straight line? Can you find for example dx/dy from these equations?

 

[seeingstars63]

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.

 

[Steely Dan]

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.

Yes, but perhaps a more useful way to write that equation is y = \frac{dy}{dx}x + y(0).

 

[clamtrox]

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.

Yes sorry, that was bad notation. I of course mean partial derivatives: ∂x/∂y = x'(u)/y'(u)