I am again having problems solving some exercise for my classical differential geometry class:
Let $\alpha : I \longrightarrow \mathbb{R}^2$ be a curve (parameterised by arc length) and let $\dot{\alpha}(s) = \big( f(s), g(s) \big)$. Show that there exists a smooth function $\varphi : I \longrightarrow \mathbb{R}$ such that $\dot{\alpha}(s) = \big( \cos \varphi(s), \sin \varphi(s) \big)$. Approach this problem as follows:
a) $\quad$ Set
$$\varphi(s) = \int_{s_0}^s \big( f \dot{g} – \dot{f} g \big)$$
and show that
$$F := f \cos \varphi + g \sin \varphi$$
as well as
$$G := f \sin \varphi – g \cos \varphi$$ are constants.
b) $\quad$ Calculate the values of $F$ and $G$ and show that this implies the assumption.
What I have so far is, that because of $\; || \, \dot{\alpha}(s) \,|| = 1 \;$, the equations
$$ f^2 + g^2 = 1 \quad , \qquad f \dot{f} + g \dot{g} = 0 \qquad \qquad (\ast)$$
both hold. Now, when calculating the derivatives of $F$ and $G$ with respect to $s$, one finds that by using equations $(\ast)$ the derivatives vanish and therefore $F$ and $G$ are constant.
Moreover, I assume that $F = 1$ has to hold. Why is this? Because if $\dot{\alpha}$ can be expressed as $\big( \cos \varphi(s), \sin \varphi(s) \big)$, then $F$ is nothing but $|| \, \dot{\alpha}(s) \, ||^2$.
Furthermore, I assume that $G = 0$ has to hold. Why is this? Because if $\dot{\alpha}$ can be expressed as $\big( \cos \varphi(s), \sin \varphi(s) \big)$, then $G$ is nothing but a scalar product of $\dot{\alpha}$ with a perpendicular vector of same length.
However, I have difficulties showing that $F = 1$ and $G = 0$. I have already done some lengthy calculations — all of them lead nowhere. I have the feeling that I am missing some obvious and easy point/approach here. Does anyone have a hint for me?
Best Answer
Assuming $s_0$ is a real number such that $(\cos s_0, \sin s_0) = (1,0)$, calculate $F(s)$ and $G(s)$ at $s=s_0$. They are the constants you are looking for.