CURVATURE OF EQUIPOTENTIAL SURFACES 85 
dk 
— = (¢ — r) sin 2A + 25 Cos 2X. 
A Ca) 
(15) 
If this derivative is set equal to zero, and solved for X, we find: 
2s 
(16) tan 2\ = -< 
Tone, 

There are two values of \ lying between o° and 180° which satisfy 
equation (16), and their difference is 90°.' That is to say, if A; is one 
solution of (16) lying in value between 0° and 180°, then \;+90° is 
Lg 

FIG. 3 
the other, that one being chosen which lies in value in the same range. 
This result may be stated in the form of a theorem: At every non- 
singular point (not umbilical) of an analytic surface there are two normal 
sections of the surface, at right angles to each other, for which the curva- 
tures attain extreme values. 
Combining this theorem with the preceding one, in which we have 
seen that the sum of the curvatures for two normal sections at right 
angles to each other is a constant (cf. (14)), we must conclude that 
.at one of these sections the curvature is a maximum and at the other 
it is a minimum. 
* To show that d?k/d\?o when dk/d\=0, we note that, since 
oe 2 cos 2\[(¢ — r) — 2s tan 2a], 
its value when dk/d\=o is obtained by substituting the value of \ of equation (16) in 
this last equation. After simplification, this reduces to 
eran/ (rit)? 1 As) 
which can vanish when and only when r=? and s=o. A point at which this occurs is 
an umbilical point; and, referring back to equation (11), we see that the curvatures of 
the normal sections to a surface at a point of this type are all equal. Such points will be 
excluded from this discussion. 
1 Tf y= and so, the values of \ are 45° and 135°. 
415 
