Selected Problems from Homework #4

Chapter 2: Exercise 1.3

Consider the Monge "patch" given by

x[u_, v_]  :=  {u, v, Sqrt[u^2 + v^2]}

The basis for the tangent space are given by the 
      following :

D[x[u, v], u]

D[x[u, v], v]

However,  
    neither is defined at the origin so x is not a coordinate 
        patch at (u, v)  =  (0, 0) .

Exercise 1.8

x[u_, v_]  := {a v Cos[u],  a v Sin[u],  b u}

Cross[D[x[u, v], u], D[x[u, v], v]]

Simplify[%]

The norm of the above is :

Sqrt[%  .  %]

PowerExpand[Simplify[%]]

This never vanishes .

Exercise 1.9

x[u_, v_]  := 
       {u  -  u^3/3  +  u v^2,  v  -  v^3/3  +  v u^2,  
        u^2  -  v^2}

Cross[D[x[u, v], u], D[x[u, v], v]]

Sqrt[% . %]

Simplify[%]

PowerExpand[%]

This never vanishes .  Hence,  
    x[u, v] is a regular map .  We now need to show that it is 
          one to one is u^2  +  v^2  <  3.

Now,  plug in u  =  r Cos[t];  v  =  r Sin[t] :

x[r Cos[t],  r Sin[t]]

Let' s denote this mess by M :

M  =  %

The first component of Mess is :

M[[1]]

The second is :

M[[2]]

The third is :

M[[3]]

Furthermore,  notice that

M[[1]]^2  +  M[[2]]^2  +  4/3 M[[3]]^2

Simplify[%]

as was foretold in our book . Since this increases with r,  
    images of points on two different circles about the origin in 
      the (u, v) plane can' t be mapped to the same 
      point .

On the other hand,  
    consider two points on a circle of radius r <  
      3 in the (u, v) plane,

Let' s consider :

x[r Cos[t],  r Sin[t]]

Simplify[%]

Please look at the handwritten solutions for the rest of this
       problems .

Exercise 1.13

beta[u_]  :=  {a Cos[u],  b Sin[u], 0}

delta[u_]  :=  beta'[u]  +  {0, 0, c}

delta[u]

x[u_, v_]  :=  beta[u]  +  v delta[u]

x[u, v]

Simplify[%]

Call this M :

M  =  %

Let' s verify that this satisfies the defining equation 
      (Recall that M[[1]] means the first component of the 
            vector M,  etc .. ) :

M[[1]]^2 / a^2  +  M[[2]]^2/b^2  -  M[[3]]^2/c^2

Simplify[%]

...  as advertised .

Another ruling can be obtained by

delta[u_]  :=  beta'[u]  +  {0, 0, -c}

x[u, v]

Simplify[%]

M  =  %

M[[1]]^2 /a^2  +  M[[2]]^2/b^2  -  M[[3]]^2/c^2

Simplify[%]

...  so this is another ruling for the 
      hyperboloid .

Exercise 2.6:

x[u_, v_]  := 
       {(R  +  r Cos[u]) Cos[v],  (R  +  r Cos[u]) Sin[v], 
         r Sin[u]}

\(\(Simplify[Cross[D[x[u, v], u], D[x[u, v], v]]]
\)\)

The unit normal vector is given by :

Simplify[% / Sqrt[%  .  %]]

PowerExpand[%]

U  =  %

S[(Subscript[x, u])] is then given by

-D[U, u]

All we need to show is that the following vanishes :

Simplify[%  -  1/r D[x[u, v], u]]

Similarly,

Simplify[
      - D[U, v]  -  Cos[u]/(R  +  r Cos[u]) D[x[u, v], v]]

NOTE : 
      Both disagree with the expressions in the book by a 
        sign .  (Typo!)

Exercise 2.7

x[u_, v_]  :=  {u, v, u v}

Cross[D[x[u, v], u] ,  D[x[u, v], v]]

The unit normal is then

% / Sqrt[%  .  %]

U  = %

Simplify[-D[U, u]]

The previous is the same as :

Simplify[
      - u v/(1  +  u^2  +  v^2)^(3/2) D[x[u, v], u]  +  
        (1  +  v^2)/(1  +  u^2  +  v^2)^(3/2) 
          D[x[u, v], v]]

Similarly,

Simplify[-D[U, v]]

is the same as :

Simplify[
      (1 + u^2)/(1  +  u^2  +  v^2)^(3/2) D[x[u, v], u]  - 
         (u v)/(1  +  u^2  +  v^2)^(3/2) D[x[u, v], v]]

Exercise 3.3 (b):

Consider the 2 x 2 symmetric matrix given by

{{a, b}, {b, c}}

Its eigenvalues can be computed by :

Eigenvalues[%]

FullSimplify[%]

Both of these eigenvalues are clearly real numbers .