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

Since by the solution of the equation ${\textstyle f(x,y,z)=0}$ the coordinate ${\textstyle z}$ may be expressed in the form of a function of ${\textstyle x,}$ ${\textstyle y,}$ 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 ${\textstyle t,}$ ${\textstyle u}$ are merely functions of ${\textstyle x}$ and ${\textstyle y.}$ 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 ${\textstyle d\xi ,}$ ${\textstyle d\eta ,}$ ${\textstyle d\zeta ,}$

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