Let $\mathbb{R}^2_+:=\{(x,y)\in\mathbb{R}^2\mid y>0\}$ and the metric $g=\frac{1}{y}(dx^2+dy^2)$. Prove $(\mathbb{R}^2_+,g)$ is not complete.
I guess the proper way is to find a divergent curve $\alpha:[0,\infty)\to \mathbb{R}^2_+$ such that $\int_0^\infty|\alpha'(t)|dt<\infty$.
My first attempts were to try $\alpha_1(t)=(f(t),y_0)$ and $\alpha_2(t)=(x_0,f(t))$ for constants $x_0,y_0$. For $\alpha_1,\alpha_2$ to be divergent, we must assume $\lim_{t\to\infty}f(t)=\infty$. These don't work, because: \begin{align*} \int_0^\infty|\alpha_1'(t)|dt=\int_0^\infty\frac{f'(t)}{\sqrt{y_0}}dt=\left.\frac{f(t)}{\sqrt{y_0}}\right|_0^\infty=\infty\\ \int_0^\infty|\alpha_2'(t)|dt=\int_0^\infty\frac{f'(t)}{\sqrt{f(t)}}dt=\left.2\sqrt{f(t)}\right|_0^\infty=\infty \end{align*} For a generic $\alpha(t)=(f(t),g(t))$, the integral becomes $\int_0^\infty\sqrt{\frac{f'(t)^2+g'(t)^2}{g(t)}}dt$, but it doesn't really help me.
How should I approach this?
The integral $\int_0^\infty|\alpha'(t)|\>dt$ just computes the euclidean length of the path. But you want the $g$-length of this path. The $g$-length of a horizontal path is its euclidean length, up to a factor. For vertical paths things are different. Consider a vertical path starting at $(0,1)$ going downwards. A parametrization of this path would be $$\gamma:\quad t\mapsto\alpha(t):=(0,1-t)\qquad(0\leq t<1)$$ with $|\alpha'(t)|\equiv1$. Therefore it its $g$-length computes to $$L(\gamma)=\lim_{b\to1-}\int_0^b{1\over\sqrt{1-t}}\>dt=2\ .$$