Differential Geometry and Its Application - Exercise 3.2.18

Problem:

Solution:

This is a very long question, and we will tackle this part-by-part. This is for part (a)

Let's start with the formula:

$ x_u = (\beta'(u) + v\delta'(u)) $
$ x_v = (\delta(u)) $
$ x_{uu} = (\beta''(u) + v\delta''(u)) $
$ x_{uv} = (\delta'(u)) $
$ x_{vv} = (0) $

The observation is that $ n = U \cdot  x_{vv} = U \cdot 0 = 0 $. So we have got the first equality.

Now we need to compute $ x_u \times x_v $

$ x_u \times x_v = (\beta'(u) + v\delta'(u)) \times \delta(u) = \beta'(u) \times \delta(u) + v\delta'(u) \times \delta(u) $

So that explain the denominator, as per the hint. Finally we compute $ m $

$ U \cdot x_{uu}  = (\beta'(u) \times \delta(u) + v\delta'(u) \times \delta(u)) \cdot \delta'(u) = \beta'(u) \times \delta(u) \cdot \delta'(u) $. So we finally also explain the numerator!

Notice the numerator is not exactly the same form I had, but it is the same because it is a circular shift of the scalar triple product.

Now we moved on to part (b), we can parametrize $ (x, y, xy) = (u, 0, 0) + v(0, 1, u) $ to make it a ruled surface.

The code for computing the Gaussian curvature is as follow:

syms u
syms v
beta = [u; 0; 0];
delta = [0; 1; u];

beta_u = diff(beta, u);
delta_u = diff(delta, u);

n = -(beta_u.' * cross(delta, delta_u))^2
D = cross(beta_u, delta) + v * cross(delta_u, delta);
d = (D.' * D)^2;

K = simplify(n/d)

We get the answer as $ -\frac{1}{(u^2 + v^2 + 1)^2} $.

For part (c) and (d), the Gaussian curvatures for both cone and cylinder are 0, this because either $ \beta' = 0 $ or $ \delta' = 0 $.

For part (e), the helicoid has the parametrization as $ (v \cos u, v \sin u, bu) = (0, 0, u) + v(\cos u ,\sin u , 0) $, so we use essentially the same code above except

syms b;
beta = [0;0;b*u];
delta= [cos(u);sin(u);0];

So we get the answer as $ -\frac{b^2}{(b^2 + v^2)^2} $

For part (f), we will use the ruling patch we found in Exercise 2.1.22, so we simply put in yet another $ \beta $ and $ \delta $ into the program and get the answer:

$ -\frac{a^2b^2c^2}{(a^2b^2v^2 + a^2c^2(v\cos u - \sin u)^2 + b^2c^2(v\sin u + \cos u)^2)^2} $

For part (g), in some sense, we have already done with it, for the saddle $ z = xy $ is a hyperbolic paraboloid.

For a more general hyperbolic paraboloid, we consider

$ \frac{z}{c} = \frac{y^2}{b^2} - \frac{x^2}{a^2} $

Now let $ u = \frac{y}{b} + \frac{x}{a} $ and $ v = \frac{y}{b} - \frac{x}{a} $

$ (\frac{a}{2}(u - v), \frac{b}{2}(u + v), cuv) $

Now we obtain the ruled patch $ (\frac{a}{2}u, \frac{b}{2}u, 0) + v(-\frac{a}{2}, \frac{b}{2}, cu) $

Because we wanted the expression to show values in terms of $ x $ and $ y $ this time, so we modified the program a bit as follow:

% defining the ruled patch
syms a
syms b
syms c
beta =  [ 0.5 * a * u; 0.5 * b * u; 0  ];
delta = [-0.5 * a    ; 0.5 * b    ; u/c];

% back substitute the x, y values

syms x
syms y
simplify(subs(subs(K, u, y/b + x/a), v, y/b - x/a))

So the answer $ -\frac{4a^6b^6c^2}{(a^4b^4c^2 + 4a^4y^2 + 4b^4x^2)^2} $

Phew, finally!