Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/38

 Let $S$  denote the first member of this equation, which will also be a function of $r,$  $\phi.$ Differentiation of $S$  with respect to $r$  gives

$$\begin{aligned} \frac{\partial S}{\partial r} &= \frac{\partial^{2} x}{\partial r^{2}}. \frac{\partial x}{\partial \phi} + \frac{\partial^{2} y}{\partial r^{2}}. \frac{\partial y}{\partial \phi} + \frac{\partial^{2} z}{\partial r^{2}}. \frac{\partial z}{\partial \phi} + \tfrac{1}{2}. \frac{\partial \left(      \left(\tfrac{\partial x}{\partial r}\right)^{2}     + \left(\tfrac{\partial y}{\partial r}\right)^{2}     + \left(\tfrac{\partial z}{\partial r}\right)^{2}   \right)}{\partial \phi} \\ &= \frac{\partial \xi}{\partial r}. \frac{\partial x}{\partial \phi} + \frac{\partial \eta}{\partial r}. \frac{\partial y}{\partial \phi} + \frac{\partial \zeta}{\partial r}. \frac{\partial z}{\partial \phi} + \tfrac{1}{2}. \frac{\partial(\xi^{2} + \eta^{2} + \zeta^{2})}{\partial \phi} \end{aligned}$$

But

$$\xi^{2} + \eta^{2} + \zeta^{2} = 1,$$

and therefore its differential is equal to zero; and by the preceding article we have, if $\rho$  denotes the radius of curvature of the line $r,$

$$\frac{\partial \xi}{\partial r} = \frac{X}{\rho},\quad \frac{\partial \eta}{\partial r} = \frac{Y}{\rho},\quad \frac{\partial \zeta}{\partial r} = \frac{Z}{\rho}$$

Thus we have

$$\frac{\partial S}{\partial r} = \frac{1}{\rho}. (X\xi' + Y\eta' + Z\zeta'). \frac{\partial v}{\partial \phi} = \frac{1}{\rho}. \cos L\lambda'. \frac{\partial v}{\partial \phi} = 0$$

since $\lambda'$  evidently lies on the great circle whose pole is $L.$ From this we see that $S$  is independent of $r,$  and is, therefore, a function of $\phi$  alone. But for $r = 0$ we evidently have $v = 0,$  consequently $\frac{\partial v}{\partial \phi} = 0,$  and $S = 0$  independently of $\phi.$  Thus, in general, we have necessarily $S = 0,$  and so $\cos\lambda\lambda' = 0,$  ''i. e. '', $\lambda\lambda' = 90^{\circ}.$ From this follows the

If on a curved surface an infinite number of shortest lines of equal length be drawn from the same initial point, the lines joining their extremities will be normal to each of the lines.

We have thought it worth while to deduce this theorem from the fundamental property of shortest lines; but the truth of the theorem can be made apparent without any calculation by means of the following reasoning. Let $AB,$  $AB'$ be two shortest lines of the same length including at $A$  an infinitely small angle, and let us suppose that one of the angles made by the element $BB'$  with the lines $BA,$  $B'A$  differs from a right angle by a finite quantity. Then, by the law of continuity, one will be greater and the other less than a right angle. Suppose the angle at $B$ is equal to $90^{\circ} - \omega,$  and take on the line $AB$  a point $C,$  such that

$$BC = BB'. \operatorname{cosec} \omega.$$

Then, since the infinitely small triangle $BB'C$ may be regarded as plane, we shall have

$$CB' = BC. \cos \omega,$$