Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/99

 curve toward the right, at right angles to the tangent, that is, in the direction $\phi + 90^{\circ};$ and let the length of this normal be $\delta\rho.$  Then, evidently, we have

$$\begin{aligned} \delta x &= \delta\rho. \cos(\phi + 90^{\circ}), \\ \delta y &= \delta\rho. \sin(\phi + 90^{\circ}), \end{aligned}$$

or

$$\begin{aligned} \delta x &= -\delta\rho. \sin\phi, \\ \delta y &= +\delta\rho. \cos\phi. \end{aligned}$$

Since now, when $\delta\rho$  is infinitely small,

$$\begin{aligned} \delta V &= p\, \delta x + q\, \delta y \\ &= (-p\sin\phi + q\cos\phi)\, \delta\rho \\ &= \mp\delta\rho\sqrt{p^{2} + q^{2}} \end{aligned}$$

and since on the curve itself $V$  vanishes, the upper signs will hold if $V,$ on passing through the curve from left to right, changes from positive to negative, and the contrary. If we combine this with what is said at the end of Art. 2, it follows that the curve is always convex toward that side on which $V$  receives the same sign as

$$Pq^{2} - 2Qpq + Rp^{2}.$$

For example, if the curve is a circle, and if we set

$$V = x^{2} + y^{2} - a^{2}$$

then we have

$$\begin{array}{c} p = 2x,\qquad q = 2y, \\ P = 2,\qquad Q = 0,\qquad R = 2, \\ Pq^{2} - 2Qpq + Rp^{2} = 8y^{2} + 8x^{2} = 8a^{2}, \\ (p^{2} + q^{2})^{3/2} = 8a^{3}, \\ r = \pm a \end{array}$$

and the curve will be convex toward that side for which

$$x^{2} + y^{2} > a^{2},$$

as it should be.

The side toward which the curve is convex, or, what is the same thing, the signs in the above formulæ, will remain unchanged by moving along the curve, so long as

$$\frac{\delta V}{\delta\rho}$$

does not change its sign. Since $V$  is a continuous function, such a change can take place only when this ratio passes through the value zero. But this necessarily presupposes that $p$  and $q$ become zero at the same time. At such a point the radius