Real valued functions $f,g:\mathbb{R}_+ \rightarrow \mathbb{R}$ are $f(u) = e^{-u}$ and $g(u) = \int \sqrt{(1-e^{-2t}} dt$.
Given $X:\mathbb{R}_+ \rightarrow \mathbb{R}^2$ with $X(u,v) = [f(u) \cos v, f(u) \sin v, g(u)]$, prove that $X$ parametrizes a regular surface $M$ in $\mathbb{R}^3$. Second, determine for which values $p$ the curve $y$ is geodesic on $M$. The curve is $y(t)=X(t,t \cos p)$.
For the first part, I calculated $X_u$ and $X_v$, and their non-zero cross product.
$X_u \times X_v = [-e^{-u} \cos v\sqrt{1-e^{-2u}}, -e^{-u} \sin v \sqrt{1-e^{-2u}}, -e^{-2u} ]$. Since the last component of the cross product is zero, then the whole vector can never be zero. Must anything else be shown about the functions, etc to prove $X$ parametrizes a regular surface?
For the second part, I aimed to set the inner product values $
$y = [e^{-t} \cos(t \cos p), e^{-t} \sin(t \cos p), g(t) ]$
$y' = [-e^{-t} \cos(t \cos p) - e^{-t} \sin(t \cos p) \cos p$,
$-e^{-t} \sin(t \cos p) + e^{-t} \cos(t \cos p) \cos p$,
$\sqrt{1-e^{-2t}} ]$
$y'' = [e^{-t} \cos(t \cos p) \sin^2 p + 2e^{-t} \sin(t \cos p) \cos p$,
$e^{-t} \sin(t \cos p) \sin^2 p - 2e^{-t} \cos(t \cos p) \cos p, \frac{(-e^{-2t})}{ \sqrt{1-e^{-2t}}} ]$
Set $0 =
=0$ where $S$ is the intrinsic normal vector. In other words, if $T$ is the tangent for the curve $y$, and $n$ is the normal vector to the surface, then $S = n\times T$. You should probably figure out $T$. You already found $n$. Then just calculate $S$ and solve for $p$ in $=0$