Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/27

Setting, as above,

$${\frac {\partial z}{\partial x}}=t,\quad {\frac {\partial z}{\partial y}}=u$$

and also

$${\frac {\partial ^{2}z}{\partial x^{2}}}=T,\quad {\frac {\partial ^{2}z}{\partial x.\partial y}}=U,\quad {\frac {\partial ^{2}z}{\partial y^{2}}}=V$$

or

$$dt=T\,dx+U\,dy,\quad du=U\,dx+V\,dy$$

we have from the formulæ given above

$$X=-tZ,\quad Y=-uZ,\quad (1-t^{2}-u^{2})Z^{2}=1$$

and hence

$${\begin{array}{c}{\begin{alignedat}{2}dX&=-Z\,dt&&-t\,dZ\\dY&=-Z\,du&&-u\,dZ\end{alignedat}}\\(1+t^{2}+u^{2})\,dZ+Z(t\,dt+u\,du)=0\end{array}}$$

or

$${\begin{aligned}dZ&=-Z^{3}(t\,dt+u\,du)\\dX&=-Z^{3}(1+u^{2})\,dt+Z^{3}tu\,du\\dY&=+Z^{3}tu\,dt-Z^{3}(1+t^{2})\,du\end{aligned}}$$

and so

$${\begin{aligned}{\frac {\partial X}{\partial x}}&=Z^{3}{\bigl (}-(1+u^{2})T+tuU{\bigr )}\\{\frac {\partial X}{\partial y}}&=Z^{3}{\bigl (}-(1+u^{2})U+tuV{\bigr )}\\{\frac {\partial Y}{\partial x}}&=Z^{3}{\bigl (}tuT-(1+t^{2})U{\bigr )}\\{\frac {\partial Y}{\partial y}}&=Z^{3}{\bigl (}tuU-(1+t^{2})V{\bigr )}\end{aligned}}$$

Substituting these values in the above expression, it becomes

$${\begin{aligned}k&=Z^{6}(TV-U^{2})(1+t^{2}+u^{2})=Z^{4}(TV-U^{2})\\&={\frac {TV-U^{2}}{(1+t^{2}+u^{2})^{2}}}\end{aligned}}$$

8.

By a suitable choice of origin and axes of coordinates, we can easily make the values of the quantities ${\textstyle t,}$ ${\textstyle u,}$ ${\textstyle U}$ vanish for a definite point ${\textstyle A.}$ Indeed, the first two