Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/29

 if $T$  and $V$ have the same sign. On the other hand, one has the greatest convex curvature, the other the greatest concave curvature, if $T$  and $V$ have opposite signs. These conclusions contain almost all that the illustrious Euler was the first to prove on the curvature of curved surfaces.

V. The measure of curvature at the point $A$ on the curved surface takes the very simple form

$$k = TV,$$ whence we have the

The measure of curvature at any point whatever of the surface is equal to a fraction whose numerator is unity, and whose denominator is the product of the two extreme radii of curvature of the sections by normal planes.

At the same time it is clear that the measure of curvature is positive for concavo-concave or convexo-convex surfaces (which distinction is not essential), but negative for concavo-convex surfaces. If the surface consists of parts of each kind, then on the lines separating the two kinds the measure of curvature ought to vanish. Later we shall make a detailed study of the nature of curved surfaces for which the measure of curvature everywhere vanishes.

9.

The general formula for the measure of curvature given at the end of Art. 7 is the most simple of all, since it involves only five elements. We shall arrive at a more complicated formula, indeed, one involving nine elements, if we wish to use the first method of representing a curved surface. Keeping the notation of Art. 4, let us set also

$$\begin{aligned} \frac{\partial^{2} W}{\partial x^{2}} &= P', & \frac{\partial^{2} W}{\partial y^{2}} &= Q', & \frac{\partial^{2} W}{\partial z^{2}} &= R' \\ \frac{\partial^{2} W}{\partial y. \partial z} &= P'', & \frac{\partial^{2} W}{\partial x. \partial z} &= Q'', & \frac{\partial^{2} W}{\partial x. \partial y} &= R'' \end{aligned}$$

so that

$$\begin{alignedat}{3}

dP &= P'\, &&dx + R\, &&dy + Q\, &&dz \\ dQ &= R\, &&dx + Q' \, &&dy + P\, &&dz \\ dR &= Q\, &&dx + P\, &&dy + R' \, &&dz \end{alignedat}$$

Now since $t = -\frac{P}{R},$ we find through differentiation

$$R^{2}\, dt = -R\, dP + P\, dR = (PQ - RP')\, dx + (PP - RR)\, dy + (PR' - RQ)\, dz$$