Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/122

 : [10] $\displaystyle \left \{ \begin{alignedat}{2} \xi'  &= \eta  Z &&- \zeta Y, \\ \eta' &= \zeta X &&- \xi   Z, \\ \zeta' &= \xi  Y &&- \eta  X; \end{alignedat}\right.$
 * [11] $\displaystyle

\left\{ \begin{alignedat}{2} \xi  &= Y\zeta' &&- Z\eta', \\ \eta &= Z\xi'   &&- X\zeta', \\ \zeta &= X\eta' &&- Y\xi'. \end{alignedat}\right. $

Likewise, $\frac{\partial \xi}{\partial s},$ $\frac{\partial \eta}{\partial s},$  $\frac{\partial \zeta}{\partial s}$  are proportional to $X,$  $Y,$  $Z,$  and if we set

$$\frac{\partial \xi}{\partial s}  = pX,\qquad \frac{\partial \eta}{\partial s} = pY,\qquad \frac{\partial \zeta}{\partial s} = pZ,$$

where $\frac{1}{p}$ denotes the radius of curvature of the line $s,$  then

$$p = X\, \frac{\partial \xi}{\partial s} + Y\, \frac{\partial \eta}{\partial s}  + Z\, \frac{\partial \zeta}{\partial s}.$$

By differentiating (7) with respect to $s,$ we obtain

$$-p = \xi\, \frac{\partial X}{\partial s}  + \eta\, \frac{\partial Y}{\partial s}   + \zeta\, \frac{\partial Z}{\partial s}.$$

We can easily show that $\frac{\partial \xi'}{\partial s},$ $\frac{\partial \eta'}{\partial s},$  $\frac{\partial \zeta'}{\partial s}$  also are proportional to $X,$  $Y,$  $Z.$  In fact, [from 10] the values of these quantities are also [equal to]

$$\eta\, \frac{\partial Z}{\partial s} - \zeta\, \frac{\partial Y}{\partial s},\qquad \zeta\, \frac{\partial X}{\partial s} - \xi\, \frac{\partial Z}{\partial s},\qquad \xi\, \frac{\partial Y}{\partial s} - \eta\, \frac{\partial X}{\partial s},$$

therefore

$$\begin{aligned} Y\, \frac{\partial \xi'}{\partial s} - X\, \frac{\partial \eta'}{\partial s} &= - \zeta\left(\frac{Y\, \partial Y}{\partial s} + \frac{X\, \partial X}{\partial s}\right) + \frac{\partial Z}{\partial s}(Y\eta + X\xi) \\ &= - \zeta\left(\frac{X\, \partial X + Y\, \partial Y + Z\, \partial Z}{\partial s}\right) + \frac{\partial Z}{\partial s}(X\xi + Y\eta + Z\zeta) \\ &= 0, \end{aligned}$$

and likewise the others. We set, therefore,

$$\frac{\partial \xi'}{\partial s} = p'X,\qquad \frac{\partial \eta'}{\partial s} = p'Y,\qquad \frac{\partial \zeta'}{\partial s} = p'Z,$$

whence

$$p' = \pm \sqrt{\left(\frac{\partial \xi'}{\partial s}\right)^{2} + \left(\frac{\partial \eta'}{\partial s}\right)^{2} + \left(\frac{\partial \zeta'}{\partial s}\right)^{2}}$$