How can we find the angle at which two parametrised differentiable curves \(\alpha: I \rightarrow S\) and \(\beta : I \rightarrow S\) make on the surface \(S\) of a torus? We can take a parametrisation of a torus, \[ \phi(u,v) = ( \sin u \cos v, \sin u \sin v, \cos u) \].

and let \[cos \theta = \frac{\left< \phi_u, \phi_v\right>}{\lvert \phi_u \rvert \lvert \phi_v \rvert} = \frac{F}{\sqrt{EG}} = \]. The coefficients of the first fundamental form can be calculated as follows: The differential of the map \(\phi \) with respect to \(u\) is \( \frac{\partial \phi}{ \partial u} \), written as \( \phi_u \) for convenience, \[ \phi_u = ( \cos u \cos v, \cos u \sin v, -\sin u). \] Similarly, \[ \phi_v = ( -\sin u \sin v, \sin u \cos v , 0). \] Hence \(E= \left< \phi_u, \phi_u \right> = \cos^2(u) + \sin^2(u) = 1\), \(F= \left< \phi_u, \phi_v \right> = 0\), and \(G= \left< \phi_v, \phi_v \right> = \sin^2(u). \) Hence the angle at which the two coordinate curves of the torus meet are: \[ \frac{F}{\sqrt{EG} = 0.} \] Now the coordinate curves of a parametrisation are orthogonal if and only if \(F(u,v) = 0\) for all \( (u,v)\). And such a parametrisation is called an orthogonal projection. Therefore, we see that the above parametrisation \(\phi\) is an orthogonal projection.

The angle at which the two curves meet at \(t= t_0 \) can be expressed as \[ cos \theta_1 = \frac{ \left< \alpha'(t_0), \beta' (t_0) \right>}{\lvert \alpha'(t_0) \rvert \lvert \alpha'(t_0) \rvert} \] where \( \alpha ' (t) = \phi_u a + \phi_v b \) and \(\beta '(t) = \phi_u c + \phi_v d \). Hence, we have

\[ \frac {E (ac) + F (bc) + F (ad) + G ( bd)}{\sqrt{a^2 + b^2\sin^2(u)} \sqrt{c^2 + d^2\sin^2(u) }} =\frac { ac+ \sin^2(u)( bd)}{\sqrt{(a^2 + b^2\sin^2(u))(c^2 + d^2\sin^2(u)) }} .\]

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