Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/120

 $$\xi = \frac{\partial x}{\partial s},\qquad \eta = \frac{\partial y}{\partial s},\qquad \zeta = \frac{\partial z}{\partial s}.$$

The extremities of all shortest lines of equal lengths $s$ correspond to a curved line whose length we may call $t.$  We can evidently consider $t$  as a function of $s$  and $\theta,$  and if the direction of the element of $t$  corresponds upon the sphere to the point $\lambda'$  whose coordinates are $\xi',$  $\eta',$  $\zeta',$  we shall have

$$\xi'. \frac{\partial t}{\partial \theta} = \frac{\partial x}{\partial \theta},\qquad \eta'. \frac{\partial t}{\partial \theta} = \frac{\partial y}{\partial \theta},\qquad \zeta'. \frac{\partial t}{\partial \theta} = \frac{\partial z}{\partial \theta}.$$

Consequently

$$(\xi\xi' + \eta\eta' + \zeta\zeta')\, \frac{\partial t}{\partial \theta} = \frac{\partial x}{\partial s}. \frac{\partial x}{\partial \theta} + \frac{\partial y}{\partial s}. \frac{\partial y}{\partial \theta} + \frac{\partial z}{\partial s}. \frac{\partial z}{\partial \theta}.$$ This magnitude we shall denote by $u,$ which itself, therefore, will be a function of $\theta$  and $s.$

We find, then, if we differentiate with respect to $s,$

$$\begin{aligned} \frac{\partial u}{\partial s} &= \frac{\partial^{2} x}{\partial s^{2}}. \frac{\partial x}{\partial \theta} + \frac{\partial^{2} y}{\partial s^{2}}. \frac{\partial y}{\partial \theta} + \frac{\partial^{2} z}{\partial s^{2}}. \frac{\partial z}{\partial \theta} + \tfrac{1}{2}\, \frac{\partial \left\{ (\tfrac{\partial x}{\partial s})^{2} + (\tfrac{\partial y}{\partial s})^{2} + (\tfrac{\partial z}{\partial s})^{2}\right\}}{\partial \theta} \\ &= \frac{\partial^{2} x}{\partial s^{2}}. \frac{\partial x}{\partial \theta} + \frac{\partial^{2} y}{\partial s^{2}}. \frac{\partial y}{\partial \theta} + \frac{\partial^{2} z}{\partial s^{2}}. \frac{\partial z}{\partial \theta}, \end{aligned}$$

because

$$\left(\frac{\partial x}{\partial s}\right)^{2} + \left(\frac{\partial y}{\partial s}\right)^{2} + \left(\frac{\partial z}{\partial s}\right)^{2} = 1,$$

and therefore its differential is equal to zero.

But since all points [belonging] to one constant value of $\theta$ lie on a shortest line, if we denote by $L$  the zenith of the point to which $s,$  $\theta$  correspond and by $X,$  $Y,$  $Z$  the coordinates of $L,$  [from the last formulæ of Art. 13 ],

$$\frac{\partial^{2} x}{\partial s^{2}} = \frac{X}{p},\qquad \frac{\partial^{2} y}{\partial s^{2}} = \frac{Y}{p},\qquad \frac{\partial^{2} z}{\partial s^{2}} = \frac{Z}{p},$$

if $p$  is the radius of curvature. We have, therefore,

$$p. \frac{\partial u}{\partial s} = X\, \frac{\partial x}{\partial \theta} + Y\, \frac{\partial y}{\partial \theta} + Z\, \frac{\partial z}{\partial \theta} = \frac{\partial t}{\partial \theta}(X\xi' + Y\eta' + Z\zeta').$$

But

$$X\xi' + Y\eta' + Z\zeta' = \cos L\lambda' = 0,$$

because, evidently, $\lambda'$  lies on the great circle whose pole is $L.$ Therefore we have

$$\frac{\partial u}{\partial s} = 0,$$