Showing a Map to the Circle is Surjective and Periodic if it Satisfies a Set of Differential Equations Suppose $\gamma \in C^1(\mathbb{R} , \mathbb{R}^2)$ satisfies
$$ ||\gamma|| = 1$$
and
$$ ||\gamma ' || = 1$$
where $|| \cdot ||$ is the usual Euclidean norm on $\mathbb{R}^2$, and $\gamma'$ is the derivative of $\gamma$. 
Is there a way to show that $\gamma(\mathbb{R}) = S^1$ and that $\gamma$ is periodic directly from the above equations?
 A: I think the following works:
Your conditions imply the slightly more convenient conditions $||\gamma(t)||^2 = 1$ and $||\gamma'(t)||^2 = 1$ for all $t$.
$||\gamma(t)||^2 = \gamma(t) \cdot \gamma(t) = 1$
By the product rule, $2 \gamma'(t) \cdot \gamma(t) = 0 \implies \gamma'(t) \cdot \gamma(t) = 0 \implies x(t)x'(t)+y(t)y'(t) = 0$
Explicitly the system of equations is 
$\begin{cases}
x(t)x'(t)+y(t)y'(t) = 0 \\
x'(t) x'(t) + y'(t)y'(t) = 1
\end{cases}$
$\begin{pmatrix}
x(t) & y(t) \\
x'(t) & y'(t)
\end{pmatrix} 
\begin{pmatrix}
x'(t) \\ y'(t) 
\end{pmatrix} = 
\begin{pmatrix}
0 \\ 1 
\end{pmatrix}$
So
$\begin{pmatrix}
x'(t) \\ y'(t) 
\end{pmatrix} =
\frac{1}{x(t)y'(t) - y(t)x'(t)}
\begin{pmatrix}
y'(t) & -y(t) \\
-x'(t) & x(t)
\end{pmatrix} 
\begin{pmatrix}
0 \\ 1 
\end{pmatrix}$
i.e. $\begin{pmatrix}
x'(t) \\ y'(t) 
\end{pmatrix} =
\frac{1}{x(t)y'(t) - y(t)x'(t)}
\begin{pmatrix}
-y(t) \\
x(t)
\end{pmatrix}$
Note that by the identity
$( \gamma \times \gamma' )^2 = ||\gamma||^2||\gamma'||^2 - (\gamma \cdot \gamma')^2$
(which is readily checked --- here the cross product is the two-dimensional version), it follows that
 $\gamma \times \gamma'  = x(t)y'(t) - y(t)x'(t) = \pm 1$
The system of equations 
$\begin{pmatrix}
x'(t) \\ y'(t) 
\end{pmatrix} =
\begin{pmatrix}
-y(t) \\
x(t)
\end{pmatrix}$
has the family of solutions $x(t) = \cos(t+\phi), y(t)=\sin(t+\phi)$ (counterclockwise).
Similarly, the system of equations
$\begin{pmatrix}
x'(t) \\ y'(t) 
\end{pmatrix} =
-
\begin{pmatrix}
-y(t) \\
x(t)
\end{pmatrix}$
has the family of solutions $x(t) = \cos(-t+\phi), y(t)=\sin(-t+\phi)$ (clockwise).
So it seems the conditions are enough to imply $2\pi$-periodicity even.
