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 $t,$  $u,$  $U$ vanish for a definite point $A.$  Indeed, the first two