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Distance Functions to Plane CurvesFor each point C , let
Demonstration 5: Distance Functions to Plane Curves
For which points
C
does the function have a horizontal inflection point?
We may consider the family of circles centered at the point
C
of increasing radii
r
. How does the circle of radius
r
with center
C
intersect the curve, and how is this related to the intersection of the
graph of
fC
with the line
y=r2
? ...and this is related to curvature because
We define a distance function
fc(t)=(X(t)-C)(X(t)-C)
from a point C to a
curve X in the plane.
We want C to represent the center of the circle with the greatest
possible order of contact with X(to). We know that the circle will have to be tangent to the curve at
X(to). This implies that to will be a
critical point of fc. If the circle is too small, this point will be a local minimum. If
the circle is too large, the function will have a local maximum. When
C is the center of the osculating circle, the distance function will have an inflection point
at to, that is, its first two derivatives will be zero. For the condition on the first derivative,
we get
2X'(t)(X(t)-C)=0.
Thus we know that
X(to)-C
is a multiple of the normal
to the curve at to and therefore we can
write it as
X(to)-C=mU(to)
. For the condition on the second derivative, we get
2X''(t)
X(to)-C+2X'(to)X'(to)=0. Substituting for
X(to)-C, we have
fc''(t)=2X'(to)X'(to)+X''(to)mU(to)=0. But
X''(to)U(to)=s' |