Let us construct a parametric equation of a straight line
through a
point
on the algebraic curve
(5.93)
where
and
are constants and
is a parameter
[437,95,107]. We find the intersections
between
and the
algebraic curve
by determining the roots of
. Since
, Taylor expansion of the
left hand side gives
(5.94)
where partial derivatives of
are evaluated at
.
When
and
are not both zero (
)
at
, (5.94) has a simple root
and every line through
has a single intersection
with the algebraic curve at
except for one case where
for certain values of
and
. In such cases (5.94) has a double root
,
provided at least one of the second order partial derivatives is not
zero (
), and
is tangent to
the curve at
.
When
is a singular point (
), and at least one of
,
,
is not zero (
), then
is a
double root and has at least two intersections at
except
for the values of
and
which satisfy
(5.95)
In such cases,
is a triple root, provided at least one of the
third order partial derivatives is not zero (
). We can solve the quadratic equation
(5.95) for
or
which leads to the following three
possibilities:
(1) Two real distinct roots: These values correspond to two
distinct tangent directions at the singular point, which implies the
algebraic curve has a self-intersection.
The Folium of Descartes, which is shown in Fig. 1.1,
has such singularity at the origin.
(2) One real double root: This value corresponds to one tangent
direction at the singular point, which implies a cusp. An illustrative
example, which is a semi-cubical parabola, is given in Fig.
2.3.
(3) Two complex roots: No real tangents at the singular point
imply an isolated point. An example of an isolated point is
given in Example 5.8.1 (see Fig. 5.21).
Example 5.8.1.
Let the algebraic curve be
[437], then
The
-turning points can be found by finding the roots of
and
. We immediately deduce
. Upon substitution to
we obtain
. Since
, (0,0) is not a
-turning
point. Therefore (-1,0) is the only
-turning point. On the other
hand
-turning points, which satisfy
and
, have
no real solutions. It is apparent from the above discussion that
is the only singular point.
Tangents at
can be obtained from
, which gives
, and hence no real
solution. Therefore,
is an isolated point. If the domain of
interest is
, border points are
. The above algebraic curve is depicted in Fig.
5.21.
Figure 5.21:
Algebraic curve with an isolated point at (0,0)
Example 5.8.2.
We have studied the semi-cubical parabola
in Example
2.1.1. The curve has a singular point at
. Since
we have
. At (0,0) we have a double
root
. Thus, at the singular point (0,0) we have a cusp whose
tangent direction is along the
axis as shown in
Fig. 2.3.
Example 5.8.3.
Let us consider the equation
within the domain
, taken from Geisow [124]. This
is a degree 6 algebraic curve illustrated in Fig. 5.22. On
every border line segment, there are three border points. The curve
has no singular points, but involves two (internal) loops and six
border-to-border branches. The algebraic curve
in this
example has degrees
in
and
. Consequently, using the
previous formulae the number of
turning points,
turning points
and singular points (in the entire complex plane) is bounded by
,
, and
. However, as we can see in
Fig. 5.22, these numbers overestimate the
actual number of such points in the real square
.
Figure 5.22:
A degree six algebraic curve (adapted from [124])
Next: 5.8.1.5 Computing starting points
Up: 5.8.1 Rational polynomial parametric/implicit
Previous: 5.8.1.3 Characteristic points
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December 2009