Begin2.DVI

From this result the center of curvature is found to have the position vector

C(x) = r (x) + αN(x) = xˆe 1 +f(x)ˆe 2 +1 + [f

′(x)] 2 f′′(x)

[−f′(x)ˆe 1 +ˆe 2 ]. (6 .73)

Note the position vector can also be expressed in the form

C(x) = r (x) + ρˆen, where ρ=^1

κ, (6 .74)

and consequently the coordinates of the center of curvature can be determined.

These coordinates are given by

h=x−

f′(x) f′′(x)(1 + [f

′(x)] (^2) ) and k=f(x) +^1
f′′(x)(1 + [f
′(x)] (^2) ), (6 .75)

provided that f′′(x)= 0.For f′′(x) = 0,there is a point of inflection, and the circle of

curvature degenerates into a straight line which is the tangent line to the point of

inflection of the curve. Consider the set of all circles which have their centers on

the normal line to the curve and which pass through the point where the normal

line intersects the curve (i.e., circles are tangent to the tangent vectors). Of all the

circles, there is only one which has a contact of the second order and this circle has

its center at the center of curvature (h, k ).A contact of second order means that not

only does the circle and curve have a common point of intersection and a common

first derivative but also that they have a common second derivative. A proof of

these statements is now offered. Let the equation of the circle be denoted by

(ξ−h)^2 + (η−k)^2 =ρ^2 , (6 .76)

where the (ξ, η )axes coincide with the (x, y )axes and h, k, ρ are the functions of x

derived above. If one considers xas being held constant and treats η as a function

of ξ, then by differentiating the equation of the circle (6.76) twice one produces the

derivatives

(ξ−h) + (η−k)dη dξ

= 0 and 1 +

( dη dξ

) 2 + (η−k)d

(^2) η
dξ^2
= 0 (6 .77)

At the common point of intersection where (ξ, η ) = (x, f (x)) one finds

ξ−h=