Thread: Surface, Meridian & Geodesics

Hi,

I am trying to solve the following question:

Show that for a surface given by revolving the curve (x(u),0,u) around the z-axis, the meridians are geodesics.


My attempt:
Firstly, I am trying to get a visual picture for the function (x(u),0,u). So I am seeing the function as x=z, which gives a line through the origin, and if I rotate it about the z-axis then it should be a cone? Is this correct?

After that I should paramatise the surface? Am I heading in the right direction?

x=x(u), z=u, how did you see x=z?
To prove that meridians of a revolving surface are geodesics, notice that the surface is symmetric about the generatrix( the meridian). To be more precise, a reflection about the plane spanned by the z-axis and a generatrix, keeps the surface unchanged. Pick two nearby points p, q at the generatrix, the generatrix must be the shortest path between these two points, otherwise suppose there is another curve c connecting p, q, the reflection c' has the same length of c, so c' is shortest too. This contradicts the assertion that geodesics are locally unique.

Firstly, thank you for replying.

Secondly, what do you mean my generatrix?

I am still trying to visulise the graph. So if i let u=0,1,2,3..., then i get z=0,1,2,3..., but x=x(0),x(1),x(2),x(3)...??? What is the x-axis?

Now, regardless of the values of x, if I rotate around the z-axis, this will give me circles...right? So I need to parametrise this circle and show that

However, I dont really understand your second method?

The point (x(u), 0, u), rotated around the z-axis, moves around the circle , z= u and so can be written in parametric eqations , , , with the parameters u and

Thank you