Why is the metric on a hyperboloid different than the metric on a sphere? I am talking about the unit sphere and unit hyperboloid in $\mathbb{R}^3$. To get a metric you might use the riemann metric and the length of a curve. To calculate the length of a curve $\gamma$ from $a$ to $b$ we have $$l_{g}(\gamma) = \int_{a}^{b} \sqrt{g_{\gamma(t)}(\gamma^{'}(t),\gamma^{'}(t))} dt$$ with a dot product $g$.
Somehow in the spherical case this dot product is induced by the standard dot product from $\mathbb{R}^3 : \langle x,y\rangle = x_1y_1+x_2y_2+x_3y_3$ while in the hyperbolic case we use the Minkowski dot product from the Minkwoski space $\langle x,y\rangle_{\rm Minkowski} = x_1y_1+x_2y_2-x_3y_3$
Although the euclidean, spherical und hyperbolic surface do not share any curves, the same curve $\gamma$ would have the same length in euclidean and spherical geometry and a different length in hyperbolic geometry. 
This makes us especially bad in guessing hyperbolic distances while we are okay in the spherical case.
But why is this the case? I cannot think of a difference of the hyperbolic surface and spherical surface that would justify the different dot products. They are both quadrics. One has positive curvature the other has negative. Does this make a difference?
 A: Here you need to think of the pair (surface, metric) as the geometric object, not only of the surface. The upshot here is that the pairs (hyperboloid, euclidean metric) and (hyperboloid, Minkowski metric) are distinct geometric objects. The way to measure lengths and areas is not the same in these two "worlds". The first one has non-constant positive curvature, and the second one has constant negative curvature $-1$.
The reason is that the hyperboloid plays the role of a sphere with respect to the Minkowski metric, since it can be written as the solution set of $\langle p,p\rangle_L=-1$. Compare that with the sphere equation $\langle p,p \rangle_E=1$. If by now you're wondering whether the one-sheeted hyperboloid defined by $\langle p,p\rangle_L=1$ has something special about it, I'll tell you it does: it has constant curvature $1$ when equipped with the Minkowski metric (the one who correctly expresses it as a sphere).
In fact, here's an exercise for you to understand what happens: define $$g((x_1,y_1,z_1),(x_2,y_2,z_2))= 3x_1x_2+5y_1y_2+2z_1z_2$$and show that $M=\{(x,y,z)\in \Bbb R^3\mid 3x^2+5y^2+2z^2=1\}$ has constant curvature when equipped with $g$.

Solution for suggested exercise. Note that $M = \{ p \in \Bbb R^3\mid g(p,p)=1\}$. If $F(p) = g(p,p)$, then $M = F^{-1}(1)$, meaning that the $g$-gradient of $F$ is always $g$-normal to $M$. Since $${\rm d}F_p(v) = 2g(p.v) = g(2p,v),$$we get ${\rm grad}_gF(p)=2p$, and so $N(p)=p$ is an $g$-unit $g$-normal vector along $M$, just like for spheres in Euclidean space. Indeed, $M$ is a $g$-sphere. Note that the curvature of the Levi-Civita connection of $g$ in $\Bbb R^3$ is zero (indeed, all the Christoffel symbols w.r.t. the usual rectangular coordinates are zero -- this actually says that the Levi-Civita connection is the usual one). So, Gauss' Formula says that $$K(v,w) = \frac{g(I\hspace{-.1cm}I(v,v))g(I\hspace{-.1cm}I(w,w)) - g(I\hspace{-.1cm}I(v,w),I\hspace{-.1cm}I(v,w))}{g(v,v)g(v,w)-g(v,w)^2}.$$Just in the same way we do it for the sphere, you can check (or see e.g. page 101 in Chapter 4 in O'Neill's Semi-Riemannian Geometry with Applications to Relativity) that $I\hspace{-.1cm}I_p(v,w) = -g(v,w)N(p)$. Taking $(v,w)$ to be an orthonormal basis for the tangent plane $T_pM$, plugging in above gives $K(p)=1$ as expected.
