Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/108

 where we may assume that $E$  has the same sign as $\frac{A - C}{2},$ then we have

$$\frac{1}{r} = \tfrac{1}{2}(A + C) + E\cos 2(\phi - \theta).$$

It is evident that $\phi$  denotes the angle between the cutting plane and another plane through this normal and that tangent which corresponds to the direction $M.$ Evidently, therefore, $\frac{1}{r}$  takes its greatest (absolute) value, or $r$  its smallest, when $\phi = \theta;$  and $\frac{1}{r}$  its smallest absolute value, when $\phi = \theta + 90^{\circ}.$  Therefore the greatest and the least curvatures occur in two planes perpendicular to each other. Hence these extreme values for $\frac{1}{r}$ are

$$\tfrac{1}{2}(A + C) \pm \sqrt{\left(\frac{A - C}{2}\right)^{2} + B^{2}}.$$

Their sum is $A + C$ and their product $AC - B^{2},$  or the product of the two extreme radii of curvature is

$$= \frac{1}{AC - B^{2}}.$$

This product, which is of great importance, merits a more rigorous development. In fact, from formulæ above we find

$$AC - B^{2} = (\alpha\beta' -\beta\alpha')^{2}(TV - U^{2})Z^{2}.$$

But from the third formula in [Theorem] 6, Art. 7, we easily infer that

$$\alpha\beta' - \beta\alpha' = \pm Z,$$

therefore

$$AC - B^{2} = Z^{4}(TV - U^{2}).$$

Besides, from Art. 8 ,

$$\begin{aligned} Z &= \pm\frac{R}{\sqrt{P^{2} + Q^{2} + R^{2}}} \\ &= \pm\frac{1}{\sqrt{1 + t^{2} + u^{2}}}, \end{aligned}$$

therefore

$$AC - B^{2} = \frac{TV - U^{2}}{(1 + t^{2} + u^{2})^{2}}.$$

Just as to each point on the curved surface corresponds a particular point $L$ on the auxiliary sphere, by means of the normal erected at this point and the radius of