Explanation 114 : We can represent a curve on the surface x(u1,u2) by the equation φ(u1,u2) = λ ; for example (u cos(v), u sin(v), u3) and u2 + v = 4 give the curve with parameter representation (u cos(4-u2), u sin(4-u2), u3).
When a family of curves is such that through every point of the surface there goes exactly one of these curves, and it is given by the equations φ(u1,u2) = λ, λ ∈ I (where φ is a fixed function and I an interval on the real axis), we call it a system of curves on the surface.
Now we can ask, for instance, for another system of curves on the surface such that through every point of the surface there goes exactly one curve of both systems, whilst these two curves intersect under a fixed angle.
Then we call the pair of systems an isogonal net. In the special case that this fixed angle is 90 degrees, we call it an orthogonal net and the two systems each other's orthogonal trajectories.

Explanation 115 : Let the curve x(u1(t),u2(t)) on the surface also be given by an equation φ(u1,u2) = λ.
Then we have for all t : φ(u1(t),u2(t)) = λ, and hence φu1du1/dt + φu2du2/dt = 0.
So the tangent vector xu1du1/dt + xu2du2/dt has the same or opposite direction as -φu2xu1 + φu1xu2.

Explanation 116 : We find the isogonal trajectories of a system φ(u1,u2) = λ on the surface x(u1,u2) as follows:
let x(u1(t),u2(t)) be such a trajectory, and α the fixed angle; then we must have (see 107 ii))

cos(α) = (-φu2du1 a1 1u2du2 a1 2u1du1 a1 2u1du2 a2 2)/(√(a1 1φu22 - 2a1 2φu1φu2 + a2 2φu12).√(a1 1du12 + 2 a1 2du1 du2 + a2 2du22)).

When we solve this differential equation, we get the isogonal trajectories with angle α in the form ψ(u1,u2) = μ.
In the special case that α is ninety degrees, the differential equation becomes

du1/du2 = (a2 2φu1 - a1 2φu2)/(a1 1φu2 - a1 2φu1).

Problem 117 : Determine the orthogonal trajectories of the v-lines u=λ of (u+v,u-v,uv).

Problem 118 : Determine the curves on the sphere that make a fixed angle α with the parallels of latitude (we call them the loxodromes of the sphere).

Explanation 119 : A quadratic equation in du1 and du2, say c1 1du12 + 2 c1 2du1 du2 + c2 2du22 = 0, determines in each point (u1,u2) two directions (if the roots of the equation are real).
Each of both solutions du1/du2 = ψui(u1,u2) (i=1,2) determines a system of curves on the surface, so the quadratic form determines a net . For example, du1 du2 = 0 determines the net of the parameter lines.

Proposition 120 : The net c1 1du12 + 2 c1 2du1 du2 + c2 2du22 = 0 on a surface with first fundamental form a1 1du12 + 2 a1 2du1 du2 + a2 2du22 is orthogonal if and only if a1 1c2 2 - 2 a1 2c1 2 + a2 2c1 1 = 0.

Proof : In 'each' point the quadratic equation c1 1 + 2 c1 2du2/du1 + c2 2(du2/du1)2 = 0 has two solutions du2/du1 = ψ1 and du2/du1 = ψ2, where ψ12 = -2c1 2/c2 2 and ψ1ψ2 = c1 1/c2 2.
The requirement of orthogonality becomes (xu1 + xu2ψ1).(xu1 + xu2ψ2) = 0, so a1 1 - a1 212) + a2 2ψ1ψ2 = 0, so a1 1c2 2 - 2 a1 2c1 2 + a2 2c1 1 = 0.