Local polar coordinates for a curve with nonvanishing position in the plane: a differentiable angle function with prescribed derivative

By [L1], $r$ is differentiable with $r^{\mathrm{\prime }}=(\rho {\rho }^{\mathrm{\prime }}+\sigma {\sigma }^{\mathrm{\prime }})\mathrm{/}r$. Since $\rho ,\sigma$ are twice differentiable and $r>0$, this quotient is differentiable by the product and quotient rules; hence $r$ is twice differentiable.

The vector $(\rho (t_0)\mathrm{/}r(t_0),\sigma (t_0)\mathrm{/}r(t_0))$ satisfies $(\rho \mathrm{/}r{)}^2+(\sigma \mathrm{/}r{)}^2=1$ by the definition of $r$, so by [L3] there is a real ${\theta }_0$ with $\cos {\theta }_0=\rho (t_0)\mathrm{/}r(t_0)$ and $\sin {\theta }_0=\sigma (t_0)\mathrm{/}r(t_0)$.

Define $g:I\to \mathbb{R}$ by $g:=(\rho {\sigma }^{\mathrm{\prime }}-\sigma {\rho }^{\mathrm{\prime }})\mathrm{/}{r}^2$. The numerator is continuous and $r^2>0$ is continuous, so $g$ is continuous on $I$ by [L4]; hence by [L4] the function $\theta (t):={\theta }_0+{\int }_{t_0}^{t}g(s)ds$ is differentiable on $I$ with ${\theta }^{\mathrm{\prime }}=g$. Moreover $g$ is differentiable by the product and quotient rules, since $\rho ,\sigma$ are twice differentiable and $r^2>0$; therefore $\theta$ is twice differentiable with ${\theta }^{\mathrm{\prime }}=g$.

Define $A:=\rho \cos \theta +\sigma \sin \theta$ and $B:=\sigma \cos \theta -\rho \sin \theta$ on $I$. Evaluating at $t_0$ with the choice of ${\theta }_0$ and [L3] gives $A(t_0)=\rho (t_0{)}^2\mathrm{/}r(t_0)+\sigma (t_0{)}^2\mathrm{/}r(t_0)=r(t_0)$ and $B(t_0)=\sigma (t_0)\rho (t_0)\mathrm{/}r(t_0)-\rho (t_0)\sigma (t_0)\mathrm{/}r(t_0)=0$.

Put $c:=r^{\mathrm{\prime }}\mathrm{/}r$ on $I$. Differentiating with [L2] and substituting ${\theta }^{\mathrm{\prime }}=g$ gives the two identities ${\rho }^{\mathrm{\prime }}+\sigma {\theta }^{\mathrm{\prime }}=c\rho$ and ${\sigma }^{\mathrm{\prime }}-\rho {\theta }^{\mathrm{\prime }}=c\sigma$: indeed ${\rho }^{\mathrm{\prime }}+\sigma {\theta }^{\mathrm{\prime }}={\rho }^{\mathrm{\prime }}+\sigma (\rho {\sigma }^{\mathrm{\prime }}-\sigma {\rho }^{\mathrm{\prime }})\mathrm{/}{r}^2=({\rho }^{\mathrm{\prime }}{\rho }^2+\rho \sigma {\sigma }^{\mathrm{\prime }})\mathrm{/}{r}^2=\rho (\rho {\rho }^{\mathrm{\prime }}+\sigma {\sigma }^{\mathrm{\prime }})\mathrm{/}{r}^2=c\rho$, and ${\sigma }^{\mathrm{\prime }}-\rho {\theta }^{\mathrm{\prime }}={\sigma }^{\mathrm{\prime }}-\rho (\rho {\sigma }^{\mathrm{\prime }}-\sigma {\rho }^{\mathrm{\prime }})\mathrm{/}{r}^2=({\sigma }^{\mathrm{\prime }}{\sigma }^2+\rho \sigma {\rho }^{\mathrm{\prime }})\mathrm{/}{r}^2=\sigma (\rho {\rho }^{\mathrm{\prime }}+\sigma {\sigma }^{\mathrm{\prime }})\mathrm{/}{r}^2=c\sigma$, using [L1]. Hence $A^{\mathrm{\prime }}=({\rho }^{\mathrm{\prime }}+\sigma {\theta }^{\mathrm{\prime }})\cos \theta +({\sigma }^{\mathrm{\prime }}-\rho {\theta }^{\mathrm{\prime }})\sin \theta =cA,$ $B^{\mathrm{\prime }}=({\sigma }^{\mathrm{\prime }}-\rho {\theta }^{\mathrm{\prime }})\cos \theta -({\rho }^{\mathrm{\prime }}+\sigma {\theta }^{\mathrm{\prime }})\sin \theta =cB\mathrm{.}$

By the quotient rule of [L2] and $r^{\mathrm{\prime }}=cr$ from [L1], $(A\mathrm{/}r{)}^{\mathrm{\prime }}=\frac{A^{\mathrm{\prime }}r-A{r}^{\mathrm{\prime }}}{r^2}=\frac{cAr-Acr}{r^2}=0,(B\mathrm{/}r{)}^{\mathrm{\prime }}=\frac{cBr-Bcr}{r^2}=0\text{on }I\mathrm{.}$ The functions $A\mathrm{/}r$ and $B\mathrm{/}r$ are continuous on $I$; by [L5] and step 2.2 they are the constants $1$ and $0$: $A=r$ and $B=0$ on $I$.

Solving the two equations $A=\rho \cos \theta +\sigma \sin \theta =r$ and $B=\sigma \cos \theta -\rho \sin \theta =0$ for $(\rho ,\sigma )$ gives $\rho =A\cos \theta -B\sin \theta =r\cos \theta$ and $\sigma =A\sin \theta +B\cos \theta =r\sin \theta$ on $I$: the displayed linear system has the invertible coefficient matrix $\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)$ of determinant ${\cos }^2\theta +{\sin }^2\theta =1$.

Let $\vartheta :I\to \mathbb{R}$ be twice differentiable with $\rho =r\cos \vartheta$, $\sigma =r\sin \vartheta$. Differentiating the two identities and eliminating $r^{\mathrm{\prime }}$ gives $\rho {\sigma }^{\mathrm{\prime }}-\sigma {\rho }^{\mathrm{\prime }}=r\cos \vartheta (r^{\mathrm{\prime }}\sin \vartheta +r{\vartheta }^{\mathrm{\prime }}\cos \vartheta )-r\sin \vartheta (r^{\mathrm{\prime }}\cos \vartheta -r{\vartheta }^{\mathrm{\prime }}\sin \vartheta )=r^2{\vartheta }^{\mathrm{\prime }},$ so ${\vartheta }^{\mathrm{\prime }}=(\rho {\sigma }^{\mathrm{\prime }}-\sigma {\rho }^{\mathrm{\prime }})\mathrm{/}{r}^2$ on $I$.