Curvature function and rate of change of angle

Question

Let $\gamma:(a,b)\rightarrow \mathbb{R}^2$ be a smooth curve with $\| \dot{\gamma}(s)\|=1$ for all $s\in (a,b)$. Fix $s_0\in (a,b)$ and let the unit vector $\dot{\gamma}(s_0)$ be represented by $(\cos \phi_0,\sin\phi_0)$. Then there is smooth function $\phi$ with $\phi(s_0)=\phi_0$ such that $$\dot{\gamma}(s)=(\cos\phi(s),\sin\phi(s))$$ for all $s\in (a,b)$.

The proof goes as follows: let $$\dot{\gamma}(s)=(f(s),g(s))$$ so that $f(s)^2+g(s)^2=1$ for all $s$. Define $$\phi(s)=\phi_0 + \int_{s_0}^s (f\dot{g}-g\dot{f})du$$ It is then shown that this is required $\phi$ in the theorem.

Q. I didn't get intuition for choice (definition) of $\phi$. How do we justify the choice of $\phi$ above?

Reference: Elementary differential geometry by Pressley, Proposition 2.2.1 (New edition)

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Using the explicitly defined angular function $\phi$, the curvature function is given by $$\kappa_s =\frac{d\phi}{ds}.$$

Answer 1

Note that

$$\gamma'(s)=(f(s),g(s))$$ and $$\gamma'(s)=(\cos\phi(s),\sin\phi(s))$$ gives

$$\phi(s)=\arctan \dfrac{g(s)}{f(s)}$$ (if $f(s)\ne 0.$)

Taking derivatives we get

$$\phi'(s)=\dfrac{1}{1+\dfrac{g(s)^2}{f(s)^2}}\dfrac{f(s)g'(s)-g(s)f'(s)}{f(s)^2}=f(s)g'(s)-g(s)f'(s).$$

Answer 2

Intuition is that you want to get $\phi$ by integration of $\kappa=\frac{d\phi}{ds}$. You get $$\phi(s)=\phi_0 + \int_{s_0}^s \kappa(u) du$$ by fundamental theorem of calculus. From serret formulae you can derive $$\kappa=\langle \ddot \gamma,J \dot \gamma \rangle$$ where $\dot\gamma$ interpreted as tangential vector field $T$ and $J$ a rotation anticlockwise by $1/2 \pi$. This will yield you $$\phi(s)=\phi_0 + \int_{s_0}^s \langle \dot T(u),JT(u) \rangle du=\phi_0 + \int_{s_0}^s (f\dot{g}-g\dot{f}) du$$ as desired.