Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/107

plane, therefore, is passed through the normal of the curved surface. Hence we have for the radius of curvature the simple formula

$${\frac {1}{r}}=Z(\xi ^{2}T+2\xi \eta U+\eta ^{2}V).$$

11.

Since an infinite number of planes may be passed through this normal, it follows that there may be infinitely many different values of the radius of curvature. In this case ${\textstyle T,}$ ${\textstyle U,}$ ${\textstyle V,}$ ${\textstyle Z}$ are regarded as constant, ${\textstyle \xi ,}$ ${\textstyle \eta ,}$ ${\textstyle \zeta }$ as variable. In order to make the latter depend upon a single variable, we take two fixed points ${\textstyle M,}$ ${\textstyle M'}$ ${\textstyle 90^{\circ }}$ apart on the great circle whose pole is ${\textstyle L.}$ Let their coordinates referred to the centre of the sphere be ${\textstyle \alpha ,}$ ${\textstyle \beta ,}$ ${\textstyle \gamma ;}$ ${\textstyle \alpha ',}$ ${\textstyle \beta ',}$ ${\textstyle \gamma '.}$ We have then

$$\cos \lambda (1)=\cos \lambda M.\cos M(1)+\cos \lambda M'.\cos M'(1)+\cos \lambda L.\cos L(1).$$

If we set

$$\lambda M=\phi ,$$

then we have

$$\cos \lambda M'=\sin \phi ,$$

and the formula becomes

$${\begin{aligned}\xi &=\alpha \cos \phi +\alpha '\sin \phi ,\end{aligned}}$$

and likewise

$${\begin{aligned}\eta &=\beta \cos \phi +\beta '\sin \phi ,\\\zeta &=\gamma \cos \phi +\gamma '\sin \phi .\end{aligned}}$$

Therefore, if we set

$${\begin{aligned}A&=(\alpha ^{2}T+2\alpha \beta U+\beta ^{2}V)Z,\\B&=(\alpha \alpha 'T+(\alpha '\beta +\alpha \beta ')U+\beta \beta 'V)Z,\\C&=(\alpha '^{2}T+2\alpha '\beta 'U+\beta '^{2}V)Z,\end{aligned}}$$

we shall have

$${\begin{aligned}{\frac {1}{r}}&=A\cos ^{2}\phi +2B\cos \phi \sin \phi +C\sin ^{2}\phi \\&={\frac {A+C}{2}}+{\frac {A-C}{2}}\cos 2\phi +B\sin 2\phi .\end{aligned}}$$

If we put

$${\begin{aligned}{\frac {A-C}{2}}&=E\cos 2\theta ,\\B&=E\sin 2\theta ,\end{aligned}}$$