$\DeclareMathOperator{\sech}{sech}$I've been trying to understand a question in Barrett O'Neill's Elementary Differential Geometry that goes as follows:
"Show that the geometric surface in Exercise 2 is isometric to a helicoid. (Hint: use $$ x(u,v) = \bigl(f(u)\cos v, f(u)\sin v, v\bigr) $$ for a suitable function $f(u)$.)"
Exercise 2 references what the book calls a conformal structure (an abstract surface with an inner product that goes as follows: $\langle v,w\rangle = (v \cdot w)/h^2$) on the entire $\mathbf{R}^2$ plane, such that $h$ here is $\sech(uv)$.
What I was hoping was that I could prove this by showing that the First Fundamental Forms of each surface (both the abstract one and the helicoid) are equal.
From my calculations on the abstract surface: $$ E = G = \frac{1}{\sech(uv)^2},\ \text{and}\quad F = 0. $$ And from my calculations on the helicoid: $$ E = f'(u)^2,\quad F = 0,\quad G = f(u)^2 + 1. $$ Setting the $E$s, $F$s and $G$s equal, I arrived at $f(u) = \sinh(u)$. This almost seems correct but I dont think it quite works.
My questions are, is it valid for me to assume that these two surfaces are isometric if their First Fundamental Forms are equal?
If not, is there a way to come up with a mapping that goes from the abstract surface to the helicoid?
If this is the correct approach, did I make a mistake somewhere?
Maybe the helicoid should also be using inner product defined in exercise 2?
Any insight or suggestions would be helpful!