I'm studying for a qualifying exam, and I'm working on the following exercise which was on the most recent qualifying exam:
Let $\mathbb{R}^3$ have coordinates $(x,y,z)$ and the standard Euclidean structure. Let $S\subset\mathbb{R}^3$ be the surface parametrized locally by $x=t+s, y=t^2+2ts, z=t^3+3st^2$, where $s,t>0$. Using any method you please, determine the Gauss curvature function $K(s,t)$.
Now, I've attempted this problem a couple of different ways (Namely through the Maurer-Cartan form), and I finally got that $K\equiv 0$ after pulling out my old undergraduate differential text (do Carmo) and getting my hands dirty with computations involving E,F,G,e,f,g.
After taking several hours finding a correct method and running through the computations only to get a simple answer, and seeing that this exercise was intended for a qualifying exam (a timed exam), I'm led to believe that there might be some obvious sign that this surface might have Gaussian curvature identically zero everywhere (Assuming my calculations are indeed correct).
So my question is just that: upon receiving a parametrization for a surface, is there some quick and easy test or observation that may lead one to quickly conclude that $K\equiv 0$?
Thanks in advance.
Note that
$$X(t, s) = f(t) + sf'(t),$$
where $f(t) = (t, t^2, t^3)$ (The precise forms does not matter). Then
$$X_t = f' + sf'' , \ \ X_s = f'.$$
This implies that $f''$ lies in the tangent plane of the surface. Since $X_{ss} = 0$, and $X_{st} = f''$ lie in the tangent plane, we have $$K = \frac{eg - f^2}{EG-F^2} = \frac{e\times 0 - 0^2}{EG-F^2} = 0.$$