The osculating circle is the circle that kisses the path. It lies in the osculating plane, tangent to the path, and it turns at the same rate as the path. It just fits. The radius of the osculating circle is the radius of curvature. This isn't a given; it's something we need to prove. Hang on.
As before, we assume p′′ is continuous. Three points determine a circle, just as they determine a plane. As the three points approach t, the circle and the plane both approach the osculating plane. Thus the osculating circle, the circle that just fits, is contained in the osculating plane. We only need establish its radius.
Again, I will present the concept, without the formal deltas and epsilons. Recall that the vector n(t) indicates the change in direction, how fast the particle is turning, as a function of time. With continuous accelleration, the rate of turn, n(t), does not jump wildly. Near a given point on the path, you can almost call it constant. The same holds for direction, velocity, and speed.
Let c(t) be n(t) divided by speed. Think of this as a curvature vector. If we took the norm of c(t), we'd have curvature, as defined in the last section. We already showed that c(t) is independent of speed, so reparameterize the path by arc length. Now the particle moves at unit speed. Time and arc length have become synonymous, and c(t) gives the change in direction with respect to arc length, as well as the change in direction with respect to time. As our three points approach their limit, the circles defined by these points have curvature approaching c(t). The osculating circle has curvature c(t).
What kind of circle is that? Let's consider the general circle r×cos(θ),r×sin(θ), where r is the radius. Reparameterize θ = t/r. In 2πr seconds we traverse the circumference, hence the speed is always 1. The curvature is the norm of the accelleration, which is 1/r. This is the reciprocal of r. therefore the radius of the osculating circle is 1 over the curvature.
We are, at this point, overdue for an example. Return to the parabola y = x2. We computed its arc length earlier. Recall that speed was sqrt(4x2+1). Call this s(x). The velocity vector is [1,2x], hence the direction d(x] is [1,2x)/s(s). Differentiate this to get:
n(x) = d′(x) = (-4x,2) over s3
Take the norm and divide by s, giving sqrt(16x2+4)/s4, or 2/s3. This gives the curvature for each value of x. Take the reciprocal to get the radius of curvature. Picture a circle, tangent to the parabola, with this radius. The circle that fits into the origin has radius 1/2. The circle at x = ±1 has radius 5×sqrt(5)/2 = 5.59. As x increases, the curvature approaches 0, and the radius of curvature approaches infinity.