# Midterm Review

1. Sep 30, 2014

### Shackleford

1. The problem statement, all variables and given/known data

Consider the surface M parametrized by σ(u, v) = (sin u cos v, sin u sin v, 3 cos u), 0 < u < π, 0 < v < 2π.

(ii) Show that the surface is regular.

(iii) Let P = ($\sqrt{2}/2, \sqrt{2}/2, 0)$ in M. Find the matrix Sp with respect to the basis {σu(P), σv(P)} in TP(M).

2. Relevant equations

Surface is regular if σu x σv does not equal zero at any point in the domain.

3. The attempt at a solution

σu = (cos v cos u, sin v cos u, -3 sin u)

σv = (-sin u sin v, sin u cos v, 0)

σu x σv = (3 cos v sin2u, 3 sin2u sin v, sin u cos u)

||σu x σv|| = sqrt{9 sin4u + sin2u cos2u}

Of course, it looks like the norm of the vector product should simplify further.

I'm also unsure about the given basis for the matrix of the shape operator. By direction calculation, the matrix contains expressions of the first and second fundamental forms. Do I evaluate the partials at the point P and use those in the inner product calculations?

Last edited: Sep 30, 2014
2. Sep 30, 2014

### HallsofIvy

Staff Emeritus
If u= 0 then $$\sigma_u\times \sigma_v= 0$$.

3. Sep 30, 2014

### Shackleford

Yes. I'm a bit confused. It says to show that the surface is regular.

4. Oct 1, 2014

### Ray Vickson

The original description excluded the boundaries $u =0$, $u = \pi$, $v = 0$, $v = 2 \pi$.

5. Oct 1, 2014

### Shackleford

Okay. It doesn't look like any combination of u and v in the domain yields (0, 0, 0), so I can state that it is regular. However, it seems that the norm argument should simplify a bit more.

6. Oct 1, 2014

### Ray Vickson

Who cares? You just want to know if it = 0 or not.

7. Oct 1, 2014

### Shackleford

(iii) Let P = $(\sqrt{2}/2, \sqrt{2}/2,0)$ in M. Find the matrix Sp with respect to the basis {σu(P), σv(P)} in TP(M).