TITLE: Vector field on $S^2$ whose integral curve is $\left(\frac{t}{\sqrt{1+t^2}},\frac{\cos(t)}{\sqrt{1+t^2}},\frac{\sin(t)}{\sqrt{1+t^2}} \right)$
QUESTION [1 upvotes]: Explicitly find a vector field $V$ on $S^2$ such that the curve $$\gamma(t) = \left(\frac{t}{\sqrt{1+t^2}},\frac{\cos(t)}{\sqrt{1+t^2}},\frac{\sin(t)}{\sqrt{1+t^2}} \right)$$ is an integral curve of $V$.

How can we solve this in an easy way? The brute-force method of assuming that $$V = f_1 \frac{\partial}{\partial x} + f_2\frac{\partial}{\partial y} + f_3 \frac{\partial}{\partial z} $$ and forcing the "integral curve condition" leads to the system of differential equations $$\left\{ \begin{array}{ll}f_1(\gamma(t)) = \frac{d}{dt} \left(\frac{t}{\sqrt{1+t^2}} \right) \\ \\
f_2(\gamma(t)) = \frac{d}{dt} \left(\frac{\cos(t)}{\sqrt{1+t^2}} \right) \\ \\
f_3(\gamma(t)) = \frac{d}{dt} \left(\frac{\sin(t)}{\sqrt{1+t^2}} \right)  
 \end{array}\right., $$ which seems just too difficult to solve.
Edit: based on a deleted comment, the curve is actually injective (because the first component function is injective). However, I don't know how to determine its inverse.

REPLY [1 votes]: The following is a result of a lengthy succession of trial and errors, loosely based on the idea that will follow.
Our $\gamma$ is the projection of the curve $\gamma^* = (t,\cos t, \sin t)$ onto $S^2$. Here projection is
$$ x \to \frac{x}{|x|} \ .$$
We have $\gamma^{*'}(t)=(1, - \sin t, \cos t)$. It easy (well, not so easy!.Initially I had $(1,-z,y)$ which just wouldn't work!) to guess that $\gamma^*$ is tangent to (integral curve to) $W(x,y,z)=(\sqrt{y^2+z^2},-z,y)$. 
Focus on $W$ along $S^2$. We must project this $W$ onto $S^2$. For this, we must remove from $W$ its projection onto $(x,y,z)$-direction.
$$ W - proj_{x,y,z} W  = (\sqrt{y^2+z^2},-z,y) - ((\sqrt{y^2+z^2},-z,y)\cdot (x,y,z))(x,y,z) \ .$$
You can double-check to see that this is in fact perpendicular to $(x,y,z)$ vector.
Guess: $V= (\sqrt{y^2+z^2},-z,y) -x \sqrt{y^2+z^2}(x,y,z)$ on $S^2$ is the desired vector field.
Note that, as expected, this $V$ is tangent to $S^2$, because its dot with $(x,y,z)$ vector is zero (by construction.) 
It remains to show that for any $t$, 
$$ V(\gamma(t)) = \gamma'(t) \ ?$$
And it is!

The original idea was to find a vector field that worked for $\gamma^*$ and the project that field via (the derivative of) the map $x \to \frac{x}{|x|}$ onto $S^2$. It turned out that some choices of $V^*$ working for $\gamma^*$ worked while others didn't. For instance $V=(1,-z,y)$ did not. I still wonder why this is so. It could be that in projecting out to $R^3$ and back to $S^2$ some key data is lost. But where and how? How could we have taken the guess work out of this solution?!
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