Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/122

[10]

${\textstyle \displaystyle \left\{{\begin{alignedat}{2}\xi '&=\eta Z&&-\zeta Y,\\\eta '&=\zeta X&&-\xi Z,\\\zeta '&=\xi Y&&-\eta X;\end{alignedat}}\right.}$

[11]

${\textstyle \displaystyle \left\{{\begin{alignedat}{2}\xi &=Y\zeta '&&-Z\eta ',\\\eta &=Z\xi '&&-X\zeta ',\\\zeta &=X\eta '&&-Y\xi '.\end{alignedat}}\right.}$

Likewise, ${\textstyle {\frac {\partial \xi }{\partial s}},}$ ${\textstyle {\frac {\partial \eta }{\partial s}},}$ ${\textstyle {\frac {\partial \zeta }{\partial s}}}$ are proportional to ${\textstyle X,}$ ${\textstyle Y,}$ ${\textstyle Z,}$ and if we set

$${\frac {\partial \xi }{\partial s}}=pX,\qquad {\frac {\partial \eta }{\partial s}}=pY,\qquad {\frac {\partial \zeta }{\partial s}}=pZ,$$

where ${\textstyle {\frac {1}{p}}}$ denotes the radius of curvature of the line ${\textstyle s,}$ then

$$p=X\,{\frac {\partial \xi }{\partial s}}+Y\,{\frac {\partial \eta }{\partial s}}+Z\,{\frac {\partial \zeta }{\partial s}}.$$

By differentiating (7) with respect to ${\textstyle s,}$ we obtain

$$-p=\xi \,{\frac {\partial X}{\partial s}}+\eta \,{\frac {\partial Y}{\partial s}}+\zeta \,{\frac {\partial Z}{\partial s}}.$$

We can easily show that ${\textstyle {\frac {\partial \xi '}{\partial s}},}$ ${\textstyle {\frac {\partial \eta '}{\partial s}},}$ ${\textstyle {\frac {\partial \zeta '}{\partial s}}}$ also are proportional to ${\textstyle X,}$ ${\textstyle Y,}$ ${\textstyle Z.}$ In fact, [from 10] the values of these quantities are also [equal to]

$$\eta \,{\frac {\partial Z}{\partial s}}-\zeta \,{\frac {\partial Y}{\partial s}},\qquad \zeta \,{\frac {\partial X}{\partial s}}-\xi \,{\frac {\partial Z}{\partial s}},\qquad \xi \,{\frac {\partial Y}{\partial s}}-\eta \,{\frac {\partial X}{\partial s}},$$

therefore

$${\begin{aligned}Y\,{\frac {\partial \xi '}{\partial s}}-X\,{\frac {\partial \eta '}{\partial s}}&=-\zeta \left({\frac {Y\,\partial Y}{\partial s}}+{\frac {X\,\partial X}{\partial s}}\right)+{\frac {\partial Z}{\partial s}}(Y\eta +X\xi )\\&=-\zeta \left({\frac {X\,\partial X+Y\,\partial Y+Z\,\partial Z}{\partial s}}\right)+{\frac {\partial Z}{\partial s}}(X\xi +Y\eta +Z\zeta )\\&=0,\end{aligned}}$$

and likewise the others. We set, therefore,

$${\frac {\partial \xi '}{\partial s}}=p'X,\qquad {\frac {\partial \eta '}{\partial s}}=p'Y,\qquad {\frac {\partial \zeta '}{\partial s}}=p'Z,$$

whence

$$p'=\pm {\sqrt {\left({\frac {\partial \xi '}{\partial s}}\right)^{2}+\left({\frac {\partial \eta '}{\partial s}}\right)^{2}+\left({\frac {\partial \zeta '}{\partial s}}\right)^{2}}}$$