2.5: Distance Functions to Plane CurvesFor each point C, letThe Curve window shows a curve X(t), which you can define. There is also a hotspot for the point Q which can be moved around in the plane. The second window shows the graph of distance from the point C to the curve, plotted against time t.
1.For which points C does the function have a horizontal inflection point?
2. 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 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' |