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 ${\textstyle E}$ has the same sign as ${\textstyle {\frac {A-C}{2}},}$ then we have

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

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

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

Their sum is ${\textstyle A+C}$ and their product ${\textstyle AC-B^{2},}$ 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 ${\textstyle L}$ on the auxiliary sphere, by means of the normal erected at this point and the radius of