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?

Re: Surface, Meridian & Geodesics

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.

Re: Surface, Meridian & Geodesics

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 $\displaystyle k_{g}=0$

However, I dont really understand your second method?

Re: Surface, Meridian & Geodesics

The point (x(u), 0, u), rotated around the z-axis, moves around the circle $\displaystyle y^2+ x^2= x(u)^2$, z= u and so can be written in parametric eqations $\displaystyle x= x(u)^2cos(\theta)$, $\displaystyle y= x(u)^2 sin(\theta)$, $\displaystyle z= u$, with the parameters u and $\displaystyle \theta$

Re: Surface, Meridian & Geodesics