Pseudosphere covered by upper half-plane

Consider the (half) pseudosphere (with radius 1), which is the surface of revolution in $\mathbb{R}^3$ generated by the tractrix parametrized by (for $t \geqslant 0$) $$t \mapsto (t -\operatorname{tanh} t, \operatorname{sech} t)~.$$

According to Wikipedia (see pseudosphere), this surface is covered by the region $\{y \geqslant 1\}$ of the upper half-plane, with covering map given explicitely by: $$(x,y)\mapsto \big(v(\operatorname{arcosh} y)\cos x, v(\operatorname{arcosh} y) \sin x, u(\operatorname{arcosh} y)\big)$$ where $t \mapsto (u(t),v(t))$ is the arclength parametrization of the tractrix above.

Ok. So in particular, we should have $$dX^2 + dY^2 + dZ^2 = \frac{dx^2 + dy^2}{y^2}~,$$ right? Here I have denoted of course $$X = v(\operatorname{arcosh} y)\cos x \qquad Y = v(\operatorname{arcosh} y) \sin x \qquad Z = u(\operatorname{arcosh} y)~.$$

But without even bothering to compute the arclength parametrization, isn't it the case that: $$dX^2 + dY^2 + dZ^2 = \big(v(\operatorname{arcosh} y)\big)^2 \, dx^2 \, + \, \frac{\big(u'(\operatorname{arcosh} y))^2 + \big(v'(\operatorname{arcosh} y)\big)^2}{y^2-1}dy^2$$ so in particular already the coefficient of $dy^2$, namely $\dfrac{1}{y^2-1}$, seems to be wrong.

Am I going wrong somewhere, or is Wikipedia?

$\DeclareMathOperator{\sech}{sech}\DeclareMathOperator{\arcosh}{arcosh}$The term arclength parametrization in the Wikipedia link is out of place: If $$(u(t), v(t)) = (t - \tanh t, \sech t),$$ then $$(u'(t), v'(t) = (1 - \sech^{2} t, \sech t \tanh t). \tag{1}$$ Since $\sech(\arcosh y) = 1/y$ and $\tanh(\arcosh y) = \sqrt{y^{2} - 1}/y$, (1) gives $$(u'(\arcosh y), v'(\arcosh y)) = \left(\frac{y^{2} - 1}{y^{2}}, \frac{\sqrt{y^{2} - 1}}{y^{2}}\right), \tag{2}$$ and therefore $$u'(\arcosh y)^{2} + v'(\arcosh y)^{2} = \frac{(y^{2} - 1)^{2}}{y^{4}} + \frac{y^{2} - 1}{y^{4}} = \frac{y^{2} - 1}{y^{2}}. \tag{3}$$
The induced metric is therefore $$dX^{2} + dY^{2} + dZ^{2} = \frac{dx^{2}}{y^{2}} + \frac{u'(\arcosh y)^{2} + v'(\arcosh y)^{2}}{y^{2} - 1}\, dy^{2} = \frac{dx^{2} + dy^{2}}{y^{2}}.$$