# Do we have to show it for both cases?

A plane curve is given by $\gamma (\theta)=(r \cos \theta , r \sin \theta )$, where r is a smooth function of $\theta$ (so that $(r, \theta )$ are the polar coordinates of $\gamma (\theta )$). Under what conditions is $\gamma$ regular? Find all functions $r(\theta )$ for which $\gamma$ is unit-speed. Show that, if $\gamma$ is unit-speed, the image of $\gamma$ is a circle; what is its radius?

I have done the following:

$\gamma$ is regular if the following condition doesn't hold $$r' (\theta )=r(\theta )=0$$

$\gamma$ is unit-speed when $r(\theta )=\pm \sin (\theta+c)$, where $c$ a constant.

If $\gamma$ is unit-speed, then $r(\theta)=\pm \sin (\theta +c)$, so $$\gamma (\theta)=(\pm \sin (\theta+c) \cos \theta, \pm \sin (\theta+c)\sin \theta)$$

Does this mean that we have two $\gamma$ ? So do we have to show for both of them, one for $+$ and one for $-$, that its image is a circle? One for only one of them?

You can choose $C=c+\pi$, and then $\sin{(\theta+C)}=-\sin{(\theta+c)}$, so having the $\pm$ there doesn't create any more solutions if you allow any $c \in [0,2\pi)$.
• So the curve $$\gamma (\theta)=(\pm \sin (\theta+c) \cos \theta, \pm \sin (\theta+c)\sin \theta)$$ becomes $$\gamma (\theta)=( \sin (\theta+C) \cos \theta, \sin (\theta+C)\sin \theta), \text{ where } C=c+\pi \text{ and } c \in [0, 2\pi )$$ Is this correct? Oct 1, 2015 at 17:03
• Your first equation is two curves, $\gamma_c(t)=(\sin{(\theta+c)})\cos{\theta},\sin{(\theta+c)}\sin{\theta})$ and $-\gamma_c(t)$. But $-\gamma_c(t) = \gamma_{c+\pi}(t)$, so it's another curve in the same family as $\gamma_c$ (and since $\gamma_{c+2\pi}=\gamma_{c}$, if $c$ can be any number between $0$ and $2\pi$, it covers all of them anyway. Oct 1, 2015 at 17:11