Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/40

 Furthermore, the area of the surface element in the form of a parallelogram between the two lines of the first system, to which correspond $q,$  $q + dq,$ and the two lines of the second system, to which correspond $p,$  $p + dp,$  will be

$$\sqrt{EG - F^{2}}\, dp. dq.$$

Any line whatever on the curved surface belonging to neither of the two systems is determined when $p$  and $q$ are supposed to be functions of a new variable, or one of them is supposed to be a function of the other. Let $s$  be the length of such a curve, measured from an arbitrary initial point, and in either direction chosen as positive. Let $\theta$  denote the angle which the element

$$ds = \sqrt{E\, dp^{2} + 2F\, dp. dq + G\, dq^{2}}$$

makes with the line of the first system drawn through the initial point of the element, and, in order that no ambiguity may arise, let us suppose that this angle is measured from that branch of the first line on which the values of $p$ increase, and is taken as positive toward that side toward which the values of $q$  increase. These conventions being made, it is easily seen that

$$\begin{aligned} \cos \theta. ds &= \sqrt{E}. dp + \sqrt{G}. \cos \omega. dq = \frac{E\, dp + F\, dq}{\sqrt{E}} \\ \sin \theta. ds &= \sqrt{G}. \sin \omega. dq = \frac{\sqrt{(EG - F^{2})}. dq}{\sqrt{E}} \end{aligned}$$

18.

We shall now investigate the condition that this line be a shortest line. Since its length $s$ is expressed by the integral

$$s = \int \sqrt{E\, dp^{2} + 2F\, dp. dq + G\, dq^{2}}$$

the condition for a minimum requires that the variation of this integral arising from an infinitely small change in the position become equal to zero. The calculation, for our purpose, is more simply made in this case, if we regard $p$ as a function of $q.$  When this is done, if the variation is denoted by the characteristic $\delta,$  we have

$$\begin{aligned} \delta s &= \int \frac{\left(   \frac{\partial E}{\partial p} . dp^{2}  + 2\frac{\partial F}{\partial p} . dp . dq  + \frac{\partial G}{\partial p} . dq^{2}  \right) \delta p  + (2E\, dp + 2F\, dq)\, d\, \delta p}{2\, ds} \\ &= \frac{E\, dp + F\, dq}{ds}. \delta p\,+\end{aligned}$$