Page:General Investigations of Curved Surfaces, by Carl Friedrich Gauss, translated into English by Adam Miller Hiltebeitel and James Caddall Morehead.djvu/100

of curvature becomes infinite or the curvature vanishes. Then, generally speaking, since here

$$-p\sin \phi +q\cos \phi$$

will change its sign, we have here a point of inflexion.

5.

The case where the nature of the curve is expressed by setting ${\textstyle y}$ equal to a given function of ${\textstyle x,}$ namely, ${\textstyle y=X,}$ is included in the foregoing, if we set

$$V=X-y.$$

If we put

$$dX=X'\,dx,\qquad dX'=X''\,dx,$$

then we have

$${\begin{array}{c}p=X',\qquad q=-1,\\P=X'',\qquad Q=0,\qquad R=0,\end{array}}$$

therefore

$$\pm {\frac {1}{r}}={\frac {X''}{(1+X'^{2})^{3/2}}}.$$

Since ${\textstyle q}$ is negative here, the upper sign holds for increasing values of ${\textstyle x.}$ We can therefore say, briefly, that for a positive ${\textstyle X''}$ the curve is concave toward the same side toward which the ${\textstyle y}$-axis lies with reference to the ${\textstyle x}$-axis; while for a negative ${\textstyle X''}$ the curve is convex toward this side.

6.

If we regard ${\textstyle x,}$ ${\textstyle y}$ as functions of ${\textstyle s,}$ these formulæ become still more elegant. Let us set

$${\begin{alignedat}{2}{\frac {dx}{ds}}&=x',\qquad &{\frac {dx'}{ds}}&=x'',\\{\frac {dy}{ds}}&=y',\qquad &{\frac {dy'}{ds}}&=y''.\end{alignedat}}$$

Then we shall have

$${\begin{alignedat}{2}x'&=\cos \phi ,\qquad &y'&=\sin \phi ,\\x''&=-{\frac {\sin \phi }{r}},\qquad &y''&={\frac {\cos \phi }{r}};\end{alignedat}}$$

or

$${\begin{alignedat}{2}y'&=-rx'',\qquad &x'&=ry'',\end{alignedat}}$$