# Prove that $X$ parametrizes a regular surface $M$ in $\mathbb{R}^3$ and determine for which values $p$ the curve $y$ is geodesic on $M$.

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 $<X_t , y''>$ and $<X_p, y''>$ to equal zero, and then solve for angle $p$. I think this is supposed to work because the vectors in the tangent plane should be orthogonal to the second derivative of a geodesic. This process involves a lot of expanding and multiplying, creating lots of opportunities for making mistakes, and I am still not sure how to simplify further and also solve for $p$. Can anyone tell me the $p$ values.. so I can check my work? Is there a different way I am supposed to find all $p$ values that doesn't involve so many calculations?

$y = [e^{-t} \cos(t \cos p), e^{-t} \sin(t \cos p), g(t) ]$

$y' = [-e^{-t} \cos(t \cos p) - e^{-t} \sin(t \cos p) \cos p$,

$-e^{-t} \sin(t \cos p) + e^{-t} \cos(t \cos p) \cos p$,

$\sqrt{1-e^{-2t}} ]$

$y'' = [e^{-t} \cos(t \cos p) \sin^2 p + 2e^{-t} \sin(t \cos p) \cos p$,

$e^{-t} \sin(t \cos p) \sin^2 p - 2e^{-t} \cos(t \cos p) \cos p, \frac{(-e^{-2t})}{ \sqrt{1-e^{-2t}}} ]$

Set $0 = <X_t, y''>$ $= [\cos(t \cos p) + \sin(t \cos p) \cos p] * [e^{-2t} \cos(t \cos p) \sin^2 p - 2e^{-2t} \sin(t \cos p) \cos p ] +$ $[-e^{-2t} \sin^2 p - 2e^{-2t} \cos^2 p] + [\sin(t \cos p) \cos(t \cos p)(2e^{-2t} +e^{-2t} \sin^2 p)] + [-e^{-2t}]$ $...$

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You should not set $<X_t,y \prime\prime>$ and $<X_p,y \prime\prime>$ equal to $0$. A geodesic is defined by $<S,y \prime\prime>=0$ 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 $<S,y \prime\prime>=0$ – Tyler Brabham Nov 2 '12 at 6:14