Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/46

 and also

$$N - M' = \psi.$$ Then the equation just found can be thrown into the following form:

$$\begin{alignedat}{1} \sqrt{E}. dp. \sin(M' &- \omega + \psi) + \sqrt{G}. dq. \sin(M' + \psi) \\ &= \sqrt{E'}. dp'. \sin M' + \sqrt{G'}. dq'. \sin(M' + \omega') \end{alignedat}$$

or

$$\begin{alignedat}{1} \sqrt{E}. dp. \sin(N' &- \omega - \omega' + \psi) - \sqrt{G}. dq. \sin(N' - \omega' + \psi) \\ &= \sqrt{E'}. dp'. \sin(N' - \omega') + \sqrt{G'}. dq'. \sin N' \end{alignedat}$$

And since the equation evidently must be independent of the initial direction, this direction can be chosen arbitrarily. Then, setting in the second formula $N' = 0,$ or in the first $M' = 0,$  we obtain the following equations:

$$\begin{aligned} \sqrt{E'}. \sin \omega'. dp' &= \sqrt{E}. \sin(\omega + \omega' - \psi). dp  + \sqrt{G}. \sin(\omega' - \psi). dq \\ \sqrt{G'}. \sin \omega'. dq' &= \sqrt{E}. \sin(\psi - \omega). dp + \sqrt{G}. \sin\psi. dq \end{aligned}$$

and these equations, since they must be identical with

$$\begin{alignedat}{2} dp' &= \alpha\, dp &&+ \beta\, dq \\ dq' &= \gamma\, dp &&+ \delta\, dq \end{alignedat}$$

determine the coefficients $\alpha,$  $\beta,$  $\gamma,$  $\delta.$ We shall have

$$\begin{aligned} \alpha &= \sqrt{\frac{E}{E'}}. \frac{\sin(\omega + \omega' - \psi)}{\sin\omega'}, & \beta &= \sqrt{\frac{G}{E'}}. \frac{\sin(\omega' - \psi)}{\sin\omega'} \\ \gamma &= \sqrt{\frac{E}{G'}}. \frac{\sin(\psi - \omega)}{\sin\omega'}, & \delta &= \sqrt{\frac{G}{G'}}. \frac{\sin\psi}{\sin\omega'} \end{aligned}$$

These four equations, taken in connection with the equations

$$\begin{aligned} \cos\omega &= \frac{F}{\sqrt{EG}}, & \cos\omega' &= \frac{F'}{\sqrt{E'G'}}, \\ \sin\omega &= \sqrt{\frac{EG - F^{2}}{EG}}, & \sin\omega' &= \sqrt{\frac{E'G' - F'^{2}}{E'G'}}, \end{aligned}$$

may be written

$$\begin{aligned} \alpha\sqrt{E'G' - F'^{2}} &= \sqrt{EG'}. \sin(\omega + \omega' - \psi) \\ \beta \sqrt{E'G' - F'^{2}} &= \sqrt{GG'}. \sin(\omega' - \psi) \\ \gamma\sqrt{E'G' - F'^{2}} &= \sqrt{EE'}. \sin(\psi - \omega) \\ \delta\sqrt{E'G' - F'^{2}} &= \sqrt{GE'}. \sin \psi \end{aligned}$$

Since by the substitutions

$$\begin{alignedat}{2} dp' &= \alpha\, dp &&+ \beta\, dq, \\ dq' &= \gamma\, dp &&+ \delta\, dq \end{alignedat}$$