parallel vector field geodesic parallel coordinates

Let $X$ be a parallel vector field along the curve $\gamma$. $X(s)=cos(\phi(s))x_1(y^1(s),y^2(s))+\frac{sin(\phi(s))}{h}x_2(y^1(s),y^2(s))$
$x:U\rightarrow M$ is a surface and the metric tensor is given by $g_{ij}=\begin{pmatrix} 1 & 0\\ 0 & h^2 \end{pmatrix} \;\;$ ($h\gt0$)
Prove that $\dot\phi(s)=-h_1(y^1(s),y^2(s))\dot y^2(s)$

Note: $\gamma(s)=x(y^1(s),y^2(s))$ is curve in $M$.

Let $R$ be a polygon bounded by $\gamma$

I did find the solution to the first part but my follow up questions would be, what is the connection between the geodesic curvature of $\gamma$ and $\phi(s)$ ?

• Have you calculated the Christoffel symbols and written down the differential equations that the components of a vector field must satisfy if it is parallel along a curve? – Ted Shifrin Jan 9 '17 at 22:03
• Can you please clarify what each of these quantities are? Presumably $(y^1, y^2) \in U$ are linear coordinates, $(x_1, x_2)$ are local coordinates in $M$, but what is the relationship between $\gamma, \phi$, and $h$? What is the dependence of $h$ on the location in the manifold? – Mnifldz Jan 9 '17 at 22:44
• @Mnifldz: No, $x_i$ is shorthand for $\partial x/\partial y^i$. $h$ is a smooth function on the surface, $\gamma$ and $\phi$ are functions of $s$. I agree that the OP should have made things a bit clearer, but ... Similarly, $h_1 = \partial h/\partial y^1$. – Ted Shifrin Jan 9 '17 at 22:55
• @TedShifrin: I did find the solution but I got a follow up question (What is the connnection of the geodesic curver of $\gamma$ and $\phi(s)$) – XPenguen Jan 10 '17 at 18:08

You've arrived at a vector field $X$ that is parallel along $\gamma$. You need to look at the angle $\psi$ between $X$ and $\dot\gamma$ and see at what rate it is changing. That is, $\dot\psi$ will be the geodesic curvature of $\gamma$, assuming that $\gamma$ is arclength-parametrized.

• Could you explain why the geodesic curvature is the rate of change of the angle between $\dot \gamma$ and $X$ ? – XPenguen Jan 10 '17 at 19:29
• What is your definition of geodesic curvature? Why not start with that and do the computation? – Ted Shifrin Jan 10 '17 at 19:32