For which functions $r$ is the curve $r(t)(\cos t,\sin t)$ regular or unit speed?

Question

Decide conditions on smooth function $r$ such that $\nu$ is a regular curve and decide functions $r$ for which $\nu$ has speed $1$.

Let $r: \mathbb R \rightarrow \mathbb R$ be a smooth function and define $\nu: \mathbb R \rightarrow \mathbb R^2$ by $\nu(t) = (r(t)\cos(t), r(t)\sin(t))$.

I want to determine the conditions on $r$ that make $\nu$ into a regular curve ($\nu'(t) \neq 0$).

I see that if I suppose that $\nu(t) = 0$ then $$(r'(t)\cos(t) - r(t)\sin(t)) = (r'(t)\sin(t) + r(t)\cos(t)) =0$$ which implies $(r(t) - r'(t))(\cos(t)+\sin(t)) = 0$ which is true for $t = -\pi/4+ k\pi$ or $r'(t) = r(t)$.

Is the condition $r'(t) \neq r(t)$ enough to guarantee that $\nu$ is regular?

How do I determine all functions $r$ such that $\nu$ has speed $1$?

Answer 1

The derivative of $\nu$ is $$ \nu'(t) = (r'(t)\cos(t)-r(t)\sin(t),r'(t)\sin(t)+r(t)\cos(t)) $$ so $$ \|\nu'(t)\|^2 =\dots=r'(t)^2+r(t)^2. $$

Therefore:

• $\nu'(t)=0\iff r(t)=r'(t)=0$.
• If $r'(t)\neq r(t)$ for all $t$, then $\nu'(t)\neq0$ for all $t$. (This condition is sufficient but not necessary.)
• The curve $\nu$ has speed one if and only if $r$ satisfies the differential equation $r'(t)^2+r(t)^2=1$. This differential equation is not too hard to solve; you get $r(t)=\sin(t+C)$ or $r(t)=\pm1$.