Calculus of Variations: Shortest distance between two points in 3D space

[Esran]

Homework Statement

Show that the shortest distance between two points in three dimensional space is a straight line.

Homework Equations

Principally, the Euler Lagrange equation.

The Attempt at a Solution

I understand how to do this for a plane, but when we move into three dimensions, our distance element is then ds=(dx²+dy²+dz²)^(1/2), which throws me off with the added variable. How do I handle calculus of variation problems like this in more than two variables? What will be my functional?

 

[diazona]

If I remember correctly, the procedure should be essentially identical whether you're working with two or three (or more) coordinates. What problem specifically did you run into that is caused by the third variable?

 

[Esran]

Ah, never mind. I got it.

 

[hhhmortal]

I came across this question myself. I used the constrained Euler equation and solved for two dependent variables i.e. y and z.

The functional I need to solve is (1 + y'² + z'²)^(1/2)

I can't seem to get the equation for the straight line, what is its form?

 

[Dickfore]

Minimize the functional:

<br /> L[x(t), y(t), z(t)] = \int_{t_{0}}^{t_{1}}{\sqrt{\dot{x}^{2} + \dot{y}^{2} + \dot{z}^{2}} \, dt}<br />

What are the Euler equations for each of the functions x(t), y(t), z(t)?

 

[hhhmortal]

Aha, I tried this and I got the Euler equation to be:

d/dt [ x' / (x' + y'² + z'²)^(1/2) ] for the function x(t) I also got it for y(t) and z(t) the same way.


I tried differentiating w.r.t 't' for all x, y and z, but I don't think I got it right.

 

[Dickfore]

Aha, I tried this and I got the Euler equation to be:

d/dt [ x' / (x' + y'² + z'²)^(1/2) ]

Since there is no equality sign, this is not an equation.