# Geodesics in the Hyperbolic Plane

In my the class the definition given for a curve $$c(t)$$ to be a geodesic is that $$c''(t)$$ is orthogonal to the surface at the point $$c(t)$$. The hyperbolic plane is the upper half plane with the metric: $$$$ds^2 = \frac{1}{y^2}(dx^2 + dy^2)$$$$ I have a fundamental misunderstanding somewhere, because I cannot think why the following curve is not a geodesic in $$\mathbb{H}$$: $$c(t) = (t, t)$$ for $$t > 0$$. This curve has second derivative as zero, but we have learnt the geodesics of $$\mathbb{H}$$ and this is not one of them. Where am I reasoning this incorrectly?

"A curve $$c$$ is geodesic when $$c''$$ is orthogonal to the surface" is for the case when your surface is isometrically embedded in $$\Bbb R^n$$ (usually $$\Bbb R^3$$) and $$\Bbb R^n$$ has the usual metric. Hyperbolic plane is not embedded in $$\Bbb R^n$$.