Solving Geodesic Equations on Surfaces of Revolution

[mooshasta]

Homework Statement

I'm given the surface of revolution parametrized by \psi (t, \theta ) = (x(t), y(t)cos \theta, y(t)sin \theta ) where the curve \alpha (t) = (x(t),y(t)) has unit speed. Also given is that \gamma (s) = \psi (t(s), \theta (s)) is a geodesic which implies the following equations hold:

<br /> \ddot{\theta} = -2 \frac{y&#039;}{y} \dot{\theta} \dot{t} and \ddot{t} = y y&#039; \dot{\theta}^2<br />

where

<br /> y&#039; = \frac{dy}{dt}, \dot{\theta} = \frac{d \theta}{ds}, \ddot{\theta} = \frac{d^2 \theta}{ds^2}, \dot{t} = \frac{dt}{ds}, \ddot{t} = \frac{d^2 t}{ds^2}<br />

I have to show that the following quantities are independent of s:

\dot{t}^2 + y^2 \dot{\theta}^2 = E
y^2 \dot{\theta} = A

Homework Equations

All that I can think may be of relevance that isn't already listed is that for a unit speed curve, y&#039;^2 + x&#039;^2 = 1. Not sure that this matters here, though.

The Attempt at a Solution

I've tried rearranging the equations to try to resemble the desired equations, but it's been pretty unfruitful. I was thinking about maybe differentiating one of the equations w.r.t. s, but I'm not sure how one would deal with the third derivative of t or \theta.

Any help (or hints) are is appreciated! Thanks!

 

[gabbagabbahey]

In this case, I think the easiest way to tackle this is to work backwards from the two equations you are trying to prove:

What is \frac{d}{ds}(y^2 \dot{\theta})?

 

[mooshasta]

Thank you! Your hint was just what I needed. For some reason I was forgetting that \frac{dy}{ds} = y&#039; \dot{t}, but now it makes perfect sense.

Thanks again :)