This is from Sakai's Riemannian Geometry:
Let $(r, \theta)$ be polar coordinates of the plane. We define a Riemannian metric $g$ on the plane by $g(\frac{\partial}{\partial r}, \frac{\partial}{\partial r}) = 1$, $g(\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta}) = 0$, and $g(\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta}) = f^{2}(r, \theta)$ where $f(r, \theta)$ is of class $C^2$ with $f(r\neq 0, \theta) > 0$, $f(0, \theta)=0$, and $\frac{\partial}{\partial r} f(0, \theta) = 1$. Let $\gamma_{\theta}$ are geodesics (so $\theta$ is constant) and let $V(r)$ is a parallel vector field along $\gamma_{\theta}$ perpendicular to $\gamma_{\theta}$. Show that $Y(r) = f(r, \theta) V(r)$ is a Jacobi field along $\gamma_{\theta}$.
I think the notation is causing a fair bit of confusion for me. I want to show that $Y$ satisfies the Jacobi field equation. So since $V(r)$ is parallel, $\nabla_{\partial/\partial r} V \equiv 0$ and also $\nabla_{\dot{\gamma}} \dot{\gamma} = 0$, so the equation reduces to
$\begin{align*} \nabla_{\partial/\partial r} \nabla_{\partial/\partial r} Y + R(Y, \dot{\gamma})\dot{\gamma} &= \nabla_{\partial/\partial r} \nabla_{\partial/\partial r} Y + \nabla_{Y} \nabla_{\dot{\gamma}} \dot{\gamma} - \nabla_{\dot{\gamma}} \nabla_{Y} \dot{\gamma} - \nabla_{[Y, \dot{\gamma}]} \dot{\gamma} \\ &= \nabla_{\partial/\partial r} \left(\frac{\partial f}{\partial r} V \right) + \nabla_{Y} \nabla_{\dot{\gamma}} \dot{\gamma} - \nabla_{\dot{\gamma}} \nabla_{Y} \dot{\gamma} - \nabla_{[Y, \dot{\gamma}]} \dot{\gamma} \\ &= \frac{\partial^2 f}{\partial r^2} V - \nabla_{\dot{\gamma}} \nabla_{Y} \dot{\gamma} - \nabla_{[Y, \dot{\gamma}]} \dot{\gamma}. \end{align*}$
But I'm unsure where to go from there to ensure that I get zero. I tried playing around with some properties of Levi-Civita connections, but I couldn't come up with anything. Can anyone help me finish up the proof? Many thanks.
