- C(1/2,0) = ( 5/4, 0, 1/2 )
[ 2 s cos(t) -(1+s^2) sin(t) ]
DC(s,t)= [ 2 s sin(t) (1+s^2) cos(t) ]
[ 1 0 ]
[ 1 0 ]
DC(1/2,0)=[ 0 5/4 ]
[ 1 0 ]
[ 1 ] [ 1 ]
T1 = DC(1/2,0) [ ] = [ 0 ]
[ 0 ] [ 1 ]
[ 0 ] [ 0 ]
T2 = DC(1/2,0) [ ] = [ 5/4]
[ 1 ] [ 0 ]
- The span of {T1, T2} is a plane since they are linearly independent.
If we base the vectors at C(1/2,0), they span the tangent space
to the image of C at that point.
- They are linearly independent because I cannot write
either one as a linear combination of the other. Multiples of
T1 will never contain a non-zero second component;
scalar multiples of T2 will never contain a non-zero first component.
[ 1 0 ]
DN(0,0) = [ 0 0 ]
[ 1 0 ]
The images under DN(0,0)
of the unit vectors in the coordinate directions are
the first and second columns of DN(0,0).
These images are not linearly independent because
the second vector is zero times the first vector.
The span of these two vectors is just a line in the direction
of the first image vector. Geometrically, this implies that
the image surface does not have a well-defined tangent plane
at N(0,0).