The problem statement of this HW problem is quoted below:
Problem $7 .$ Let $\mathbb{H}=\{(u, v): v>0\}$ be the hyperbolic plane and use $\mathbf{x}(u, v)=(u, v)$ as the trivial coordinates. Consider the hyperbolic metric $$ \left\langle\mathbf{x}_{u}, \mathbf{x}_{u}\right\rangle=\frac{1}{v^{2}}, \quad\left\langle\mathbf{x}_{u}, \mathbf{x}_{v}\right\rangle= 0, \quad\left\langle\mathbf{x}_{v}, \mathbf{x}_{v}\right\rangle=\frac{1}{v^{2}} $$ (1) Compute the length of the curve $\alpha:[0, a] \rightarrow$ H given by $\alpha(t)=(0,1-t) .$ What happens with its length when $t \rightarrow 1^{-} ?$
The problem was that I could not understand how to use the metric as defined above. From my understanding $\mathbf{x}_u$ is simply $(1, 0)$. Therefore $\left\langle\mathbf{x}_{u}, \mathbf{x}_{u}\right\rangle = \left\langle(1, 0), (1, 0)\right\rangle$ but this is $\frac{1}{v^2}$ (assuming $v$ is the common $v$ coordinate the two tangent vectors share) which is not defined.
I guess what I want is $\Vert\alpha'(t)\Vert$ (for the line length), and this expands to $\langle (-\sin (t),\cos (t)), (-\sin (t),\cos (t))\rangle$ and I do not see how to expand this using the definition above.
Edit: I just realized what I wrote above for $\alpha'(t)$ makes no sense whatsoever. I have no idea how I got that :(.
Although $\mathbf{x}_{u}=(1,0)$ and $\mathbf{x}_{v}=(0,1)$ are the "constant" vectors, we think of both $\mathbf{x}_{u}$ and $\mathbf{x}_{v}$ as vector fields on the upper half plane. So for each $(u, v)$ on the upper half plane, there are the tangent vectors $\mathbf{x}_{u}$ and $\mathbf{x}_{v}$ at $(u, v)$.
In particular, you can define the length of $\mathbf{x}_{u}$, $\mathbf{x}_{v}$ depending on the point $(u, v)$. The equation
$$ \langle \mathbf{x}_{u}, \mathbf{x}_{u}\rangle = \frac{1}{v^2}$$
means the following: at the point $(u, v)$, we define the length (square) of $\mathbf{x}_{u}$ at that point $(u, v)$ to be $1/v^2$. For example, at the point $(2, 3)$, the length (square) of $\mathbf{x}_{u}$ is
$$ \langle \mathbf{x}_{u}, \mathbf{x}_{u}\rangle = \frac{1}{3^2} = \frac 19.$$
Back to your question, $\alpha (t) = (0,1-t)$. So $\alpha'(t) = (0,-1) = -\mathbf{x}_{v}$ (again we do not think of $\alpha'(t)$ as a constant vector, but a vector fields along $\alpha$). Thus for each $t$, we have $$ \|\alpha'(t)\|^2 = \| -\mathbf{x}_{v}\|^2_{\alpha(t)} = \frac{1}{(1-t)^2}. $$