Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/105

 we erect a normal whose direction is expressed by the point $\mathfrak{L}$ of the auxiliary sphere. By moving along the curved line, $\lambda$  and $L$ will therefore change their positions, while $\mathfrak{L}$  remains constant, and $\lambda L$  and $\lambda\mathfrak{L}$  are always equal to $90^{\circ}.$  Therefore $\lambda$  describes the great circle one of whose poles is $\mathfrak{L}.$  The element of this great circle will be equal to $\frac{ds}{r},$  if $r$  denotes the radius of curvature of the curve. And again, if we denote the direction of this element upon the sphere by $\lambda',$ then $\lambda'$  will evidently lie in the same great circle and be $90^{\circ}$  from $\lambda$  as well as from $\mathfrak{L}.$  If we now set

$$\cos \lambda'(1) = \xi',\qquad \cos \lambda'(2) = \eta',\qquad \cos \lambda'(3) = \zeta',$$

then we shall have

$$d\xi = \xi'\, \frac{ds}{r},\qquad d\eta = \eta'\, \frac{ds}{r},\qquad d\zeta = \zeta'\, \frac{ds}{r},$$

since, in fact, $\xi,$  $\eta,$  $\zeta$ are merely the coordinates of the point $\lambda$  referred to the centre of the sphere.

Since by the solution of the equation $f(x, y, z) = 0$ the coordinate $z$  may be expressed in the form of a function of $x,$  $y,$  we shall, for greater simplicity, assume that this has been done and that we have found

$$z = F(x, y).$$

We can then write as the equation of the surface

$$z - F(x, y) = 0,$$

or

$$f(x, y, z) = z - F(x, y).$$

From this follows, if we set

$$\begin{array}{c} dF(x, y) = t\, dx + u\, dy, \\ P = -t,\qquad Q = -u,\qquad R = 1, \end{array}$$

where $t,$  $u$ are merely functions of $x$  and $y.$  We set also

$$dt = T\, dx + U\, dy,\qquad du = U\, dx + V\, dy.$$

Therefore upon the whole surface we have

$$dz = t\, dx + u\, dy$$

and therefore, on the curve,

$$\zeta = t\xi + u\eta.$$

Hence differentiation gives, on substituting the above values for $d\xi,$  $d\eta,$  $d\zeta,$

$$\begin{aligned} (\zeta' - t\xi' - u\eta') \frac{ds}{r} &= \xi\, dt + \eta\, du \\ &= (\xi^{2}T + 2\xi\eta U + \eta^{2}V)\, ds, \end{aligned}$$