The reciprocal of the curvature κis called the radius of curvature ρ. Note that straight
lines have a constant angle θbetween a unit tangent vector and the x-axis and hence
the curvature of straight lines is zero since the curvature is a measure of how fast
the tangent vector is changing with respect to arc length.
To understand the meaning of the radius of curvature, consider the vectors N(x)
and N(x+ ∆ x)which are normal to the curve y =f(x) at the points (x, f (x)) and
(x+ ∆ x, f (x+ ∆ x)). These vectors are illustrated in figure 6-15. For appropriate
scalars αand β, the vector equations
C(x) = r (x) + αN(x) and C(x) = r (x+ ∆ x) + βN(x+ ∆ x)
depict the common point of intersection (h, k ) of these normal lines to the plane
curve, provided these normal lines are not parallel. The scalars αand βare related
by the vector equation
r (x) + αN(x) = r (x+ ∆ x) + βN(x+ ∆ x). (6 .69)If in the limit as ∆x→ 0 , the point of intersection (h, k )approaches a specific value,
this limit point is called the center of curvature. To find the center of curvature
(h, k ),the scalar α(or β) must be determined. This is accomplished by expanding
the above equations relating αand β. When the vector equation (6.69) is expanded,
one finds have
xˆe 1 +f(x)ˆe 2 −αf ′(x)ˆe 1 +αeˆ 2 = (x+ ∆ x)ˆe 1 +f(x+ ∆ x)ˆe 2 −βf ′(x+ ∆ x)ˆe 1 +βˆe 2.Equate like components the two scalar equations and show
x−αf ′(x)−(x+ ∆ x) + βf ′(x+ ∆ x) = 0 and
f(x) + α−f(x+ ∆ x)−β= 0.(6 .70)By eliminating β from these two equations one finds
α[
f′(x+ ∆ x)−f′(x)
∆x]
= 1 +[
f(x+ ∆ x)−f(x)
∆x]
f′(x+ ∆ x). (6 .71)In this equation let ∆x→ 0 and solve for αand find
α=1 + [f′(x)]^2
f′′(x) , f′′(x)= 0. (6 .72)