185 
because limit of - = 0, since Ox is a tangent to the pro- 
h 
jection of OP' on the plane of xy; 
.•. R' = R cos 9. 
Cor. Hence it follows that the osculating circles of all 
sections of a curve surface that have a common tangent 
line, are situated on the surface of a sphere the radius of 
which is the radius of curvature of the normal section pass 
ing through the same tangent line; for OA' = OA cos AO A' 
(and consequently A A O is a right angle) if OA be the 
diameter of the circle of curvature of the normal, and OA' 
that of the circle of curvature of the oblique section ; the 
latter circle is therefore situated on a sphere whose diameter 
is AO. 
216. To find the radius of curvature of any normal 
section of a surface at a given point, in terms of the co 
ordinates of that point. 
Let x, y, z be the co-ordinates of the given point P 
(fig. 63) of the surface, p, q, r, &c. the partial differential 
coefficients of z expressed in terms of those co-ordinates; 
PT the tangent line through which the normal section is 
to pass, which, since it lies in the tangent plane at P, 
will be determined by the equation to its projection on the 
plane of xy, viz. y — y = m {x — x). Then if we determine 
a sphere having a contact of the first order with the surface 
at P, and whose section by the vertical plane PQT has a 
contact of the second order with the section of the surface 
by the same plane, all planes passing through PT will cut 
the sphere in circles, which are the circles of curvature to 
the corresponding sections of the surface, and therefore the 
radius of the sphere will be the radius of curvature of the 
normal section ; and this we are enabled to do, because the 
four disposable constants in the equation to a sphere will 
enable us to satisfy the four requisite conditions, which are,