# Partial Solutions to Brainfood #8

These are partial answers so that you can check to see if you are on target. We expect your actual answers to be "fleshed out" a bit more.
1. C(1/2,0) = ( 5/4, 0, 1/2 )
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  ]

```
3. ```                      [  1 ]   [  1 ]
T1 = DC(1/2,0) [    ] = [  0 ]
[  0 ]   [  1 ]

[  0 ]   [  0 ]
T2 = DC(1/2,0) [    ] = [ 5/4]
[  1 ]   [  0 ]

```
4. 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.
5. 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.
6. ```
[  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).