Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/21

which is infinitely small; and finally, let ${\textstyle \lambda }$ be the point on the sphere representing the direction of the element ${\textstyle AA'.}$ Then we shall have

$$dx=ds.\cos(1)\lambda ,\quad dy=ds.\cos(2)\lambda ,\quad dz=ds.\cos(3)\lambda$$

and, since ${\textstyle \lambda L}$ must be equal to ${\textstyle 90^{\circ },}$

$$X\cos(1)\lambda +Y\cos(2)\lambda +Z\cos(3)\lambda =0$$

By combining these equations we obtain

$$X\,dx+Y\,dy+Z\,dz=0.$$

There are two general methods for defining the nature of a curved surface. The first uses the equation between the coordinates ${\textstyle x,}$ ${\textstyle y,}$ ${\textstyle z,}$ which we may suppose reduced to the form ${\textstyle W=0,}$ where ${\textstyle W}$ will be a function of the indeterminates ${\textstyle x,}$ ${\textstyle y,}$ ${\textstyle z.}$ Let the complete differential of the function ${\textstyle W}$ be

$$dW=P\,dx+Q\,dy+R\,dz$$

and on the curved surface we shall have

$$P\,dx+Q\,dy+R\,dz=0$$

and consequently,

$$P\cos(1)\lambda +Q\cos(2)\lambda +R\cos(3)\lambda =0$$

Since this equation, as well as the one we have established above, must be true for the directions of all elements ${\textstyle ds}$ on the curved surface, we easily see that ${\textstyle X,}$ ${\textstyle Y,}$ ${\textstyle Z}$ must be proportional to ${\textstyle P,}$ ${\textstyle Q,}$ ${\textstyle R}$ respectively, and consequently, since

$$X^{2}+Y^{2}+Z^{2}=1,$$

we shall have either

$${\begin{aligned}X&={\frac {P}{\sqrt {P^{2}+Q^{2}+R^{2}}}},&Y&={\frac {Q}{\sqrt {P^{2}+Q^{2}+R^{2}}}},&Z&={\frac {R}{\sqrt {P^{2}+Q^{2}+R^{2}}}}\end{aligned}}$$

or

$${\begin{aligned}X&={\frac {-P}{\sqrt {P^{2}+Q^{2}+R^{2}}}},&Y&={\frac {-Q}{\sqrt {P^{2}+Q^{2}+R^{2}}}},&Z&={\frac {-R}{\sqrt {P^{2}+Q^{2}+R^{2}}}}\end{aligned}}$$

The second method expresses the coordinates in the form of functions of two variables, ${\textstyle p,}$ ${\textstyle q.}$ Suppose that differentiation of these functions gives

$${\begin{alignedat}{2}dx&=a\,dp&&+a'\,dq\\dy&=b\,dp&&+b'\,dq\\dz&=c\,dp&&+c'\,dq\end{alignedat}}$$