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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 $1$?

Joonas Ilmavirta
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Shuzheng
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1 Answers1

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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 $r(t)=\pm1$.
Joonas Ilmavirta
  • 25,809
  • Thank you, Joonas !

    Can you guide me how to solve this differential equation ? The textbook I'm studying don't require the prerequisite of Differential Equations, so it must be easy to solve ? (Maybe you could describe the steps or show some of them ?)

     – Shuzheng Jan 29 '15 at 17:25
  • And why is $\cos(t+C)$ not a solution ? – Shuzheng Jan 29 '15 at 17:26
  • @user111854, $\cos(t+C)$ is a solution since $\cos(t+C)=\sin(t+\pi/2+C)$. I didn't list it separately since it's included in the answer I gave. If you are unfamiliar with differential equations, it might not be as easy. A short sketch: $(dr/dt)^2+r^2=1\implies dr/dt=\sqrt{1-r^2}\implies$ $ dr/\sqrt{1-r^2}=dt\implies\int dr/\sqrt{1-r^2}=\int dt\implies$ $\arcsin(r)=t+C\implies r=\sin(t+C)$. Alternatively you could observe that $\sin(t+C)$ is a solution by calculation and argue that the solution is unique. If you want more details, ask a new question (and maybe add a link to this question). – Joonas Ilmavirta Jan 29 '15 at 18:34
  • Thank you very much @JoonasIlmavirta – Shuzheng Jan 30 '15 at 08:54