Minimal surfaces and gaussian and normal curvaturess If $M$ is the surface $$x(u^1,u^2) = (u^2\cos(u^1),u^2\sin(u^1), p\,u^1)$$ then I am trying to show that $M$ is minimal. $M$ is referred to as a helicoid.
Also I am confused on how $p$ affects the problem
 A: Note that $$x_{1}=\frac{\partial x}{\partial u^1}=(-u^2\sin(u^1),u^2\cos(u^1),p)$$ and 
$$x_{2}=\frac{\partial x}{\partial u^2}=(\cos(u^1),\sin(u^1),0).$$
This implies that the unit normal is given by
$$x_1\times x_2=(-p\sin(u^1),p\cos(u^1),-u^2)$$
which implies that
$$n=\frac{x_1\times x_2}{\|x_1\times x_2\|}=\frac{1}{\sqrt{p^2+(u^2)^2}}(-p\sin(u^1),p\cos(u^1),-u^2).$$
Also, 
$$E=x_1\cdot x_1=(u^2)^2+p^2, F=x_1\cdot x_2=0, G=x_2\cdot x_2=1$$
On the other hand, we have
$$x_{11}=(-u^2\cos(u^1),-u^2\sin(u^1),0),$$
$$x_{12}=x_{21}=(-\sin(u^1),\cos(u^1),0).$$
$$x_{22}=(0,0,0).$$
Therefore, we have 
$$e=x_{11}\cdot n=0, f=x_{12}\cdot n=p, g=x_{22}\cdot n=0.$$
Therefore, we can calculate the mean curvature using the formula (see equation (6) here): 
$$H=\frac{eG+gE-2fF}{2(EG-F^2)}=0$$
which shows that $M$ is minimal. 
A: There is a good reason that the value of $p$ does not matter, as long as $p \neq 0.$ 
If you begin with a sphere of radius $R$ and blow it up to a sphere of radius $SR,$ the result is to multiply the mean curvature by $\frac{1}{S}.$ This is a general phenomenon. A map, which is also linear, given by moving every point $(x,y,z)$ to $(\lambda x, \lambda y, \lambda z)$ for a positive constant $\lambda,$ is called a homothety. A homothety takes any surface and divides the mean curvature (at matching points, of course)) by $\frac{1}{\lambda}.$ This can be done in any $\mathbb R^n,$ I guess we are sticking with $\mathbb R^3.$
So, what I need to do is show you that your helicoid with parameter $p,$ expanded or shrunk by a homothety, is the helicoid with a different parameter, call it $q.$ I'm going to use $u = u^1, v = u^2.$ And that is just
$$ \frac{q}{p} x(u, \frac{pv}{q}) = \frac{q}{p} \left(\frac{pv}{q} \cos u , \frac{pv}{q} \sin u, p u   \right) = (v \cos u , v \sin u, q u).  $$ 
Well, the mean curvature of the original helicoid is $0$ everywhere. So the new helicoid is still minimal. 
There is a bit of work showing that a homothety changes the mean curvature in the way I described, no easier than your original problem. True, though.
