How to solve $\dot{\mathbf{r}}(\theta)^2 + \mathbf{r}(\theta)^2 = 1$?

Question

I'm trying to determine all smooth functions $\mathbf{r}(\theta)$ such that the curve $\gamma(\theta) = (\mathbf{r}(\theta)\cos(\theta), \mathbf{r}(\theta)\sin(\theta))$ is unit-speed curve, i.e. that $\|\dot{\gamma}(\theta)\| = 1 \implies $$\dot{\mathbf{r}}(\theta)^2 + \mathbf{r}(\theta)^2 = 1$. According to my source, which omits intermediate steps, the solution ends up being $\pm\sin(\theta + \alpha)$ for some constant $\alpha$. But how can I show this myself?

Answer 1

Note that since $\alpha$ is constant, the solution can be written as

$$ \sin(\theta +\alpha)=C_1\cos(\theta)+C_2\sin(\theta)$$

for constants $A$ and $B$ using the angle addition formula.

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Differentiating the ODE we obtain

$$2r'(\theta)(r''(\theta)+r(\theta))=0$$

So we have $r'(\theta)=0\implies r(\theta)=C$, and substituting into the equation we obtain $C=\pm 1$. Or we have

$$r''(\theta)+r(\theta)=0$$ and solving gives the solution $$r(\theta)=C_1\cos(\theta)+C_2\sin(\theta)$$

substituting into the ODE gives $C_1^2+C_2^2=1$, thus

$$r(\theta)=\pm \sin(\theta\pm\alpha)$$

for some constant $\alpha$.