This figure shows another view of the tangent surface of the twisted cubic.
Please recall that the surface is given parametrically by:
t ---> (x,y,z) = (t, t2, t3),
so that the surface is given parametrically by:
(t,u) --> (t+u, t2 + 2tu, t3 + 3t2u).
In the portion that is shown here, we have -1.5 < t
< 1.5, while the range of u-values varies with
t in such a way that:
- The portion of tangent line that is shown is between the planes
x = -1 and x = 1.
{Thus, the point of tangency is not included
for any t-value with t > 1
or t < -1.}
- Each segment has one endpoint on the plane
x = -1 and/or one endpoint on the plane
x = 1.
{Some segments have been "trimmed", to make
their length < 2.}
Portions of the following curves are traced in the figure:
- The twisted cubic, shown in dark blue.
- The intersections of the tangent surface with the planes
x = 1 and x = -1,
shown in blue-green and
dark green respectively.
Each of these curves has a cusp
at the point where it meets the twisted
cubic.
The implicit equation of the tangent surface.
By substituting from the parametrization, we obtain:
y - x² = - u² and
z - 3xy + 2x³ = 2u³.
This leads to the following version of the implicit equation:
4(y - x²)³ +
(z - 3xy + 2x³)² = 0.
We can use this version of the equation to prove that
the twisted cubic, i.e.
y = x² and z = z³,
is the singular locus of the tangent surface. On the other hand,
the terms of degrees 5 and 6 are the same on both sides of the equation, so that the
implicit equation can be simplified to the following degree 4
version:
z² - 6xyz + 4x³z + 4y³
- 3x²y² = 0.
The osculating plane at a point of the twisted cubic,
and the tangent cone (of the tangent surface) at the same point.
For simplicity, let's consider the situation at the origin.
The lowest degree homogeneous component of the defining equation of the
surface is z², so the tangent cone at the origin is
given by the equation z² = 0. On the other hand,
the parametrization of the curve is given by the vector valued function
f(t) = (t,t²,t³),
so that f'(0) = (1,0,0)
is a tangent vector at the origin. Furthermore, the osculating plane to the curve
at the origin corresponds to the vector space spanned by
f'(0) and
f"(0) = (0,2,0).
{In general, the principal normal to the curve at the origin
is perpendicular to f'(0)
and lies in the subspace of R³ spanned by
f'(0) and
f"(0) = (0,2,0).
The fact that f"(0) happens to
be perpendicular to f'(0)
is specific to the parameter value t = 0.}
Anyway, the osculating plane at the origin is the plane
z = 0. This means that the tangent cone to the surface
at the origin coincides with the osculating plane to the curve at the
origin -- or more precisely, the tangent cone is equal to the
osculating plane, counted with multiplicity = 2.
In the case of the twisted cubic, one can check directly
that at every point of the curve, the tangent cone to the tangent surface
is equal to the osculating plane of the curve at the same point --
counted with multiplicity 2 of course. The verification is somewhat intricate
and is not presented here. I suspect that this is a particular
instance of a theorem about space curves, but I don't know a reference for such
a theorem.
Return to the initial view of
the tangent surface of the twisted cubic.
Return to the view in which
all tangent line segments have length = 2.
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