# How to show curve is closed on non-closed interval

I want to show that $$\boldsymbol{r}(t)=\left(\frac{5}{13}\cos(t),\frac{8}{13}-\sin(t),-\frac{12}{13}\cos(t)\right)$$ is a closed curve, where $$0 \leq t < 2\pi$$. The definition of closed curves I have is a curve defined either on $$\mathbb{R}$$ or on a closed interval. But intuitively this curve should be simple and closed, though I'm not sure how to rigorously justify that.

• If you want to show that the curve is closed, showing periodicity would be a step to take. – Ninad Munshi Apr 19 '20 at 4:30
• @NinadMunshi If we pretend that the curve is defined on all of $\mathbb{R}$, and show that it is $2\pi$-periodic, can we conclude that it is closed on the interval specified above? I have no definition for what it means for a curve to be closed on such an interval; is there a definition that you would use in this case? – A.B Apr 19 '20 at 4:38
• There are two equivalent definitions of "closed" for parameterized curves $\gamma$ with domain $[a,b]$. Both depend on what you mean by "curve" (how smooth? injective or not?). One is that there exists a periodic curve of the right kind of period $b-a$ defined on $\mathbb{R}$ which agrees with $\gamma$ on $[a,b]$. Another is that $\gamma$ is of the right kind and has $\gamma(a)=\gamma(b)$ and sufficiently many derivatives match i.e. $\gamma'(a)=\gamma'(b)$, $\gamma''(a)=\gamma''(b)$ etc. – Max Apr 19 '20 at 6:47
• @Max My confusion is with how to characterize closed curves for the interval $0 \leq t < 2\pi$ which is not a closed interval. Regardless of which dentition of closed curve I use, the domain is either all of $\mathbb{R}$ or $[0,2\pi]$. So I need to know how you can ensure that a curve is closed when you don't actually reach the starting point again. It intuitively makes sense in this case, but I am unable to rigorously justify it without a proper definition that takes the domain into account. – A.B Apr 19 '20 at 14:24
• Well, it seems the most reasonable definition would require that the map extends continuously to the endpoint (in your case $t=2\pi$), and then the resulting curve is closed. (Usually one wants to show things for a purpose, so whatever that purpose is this definition will either be appropriate for it or not; it seems for most purposes this should be the appropriate definition.) – Max Apr 19 '20 at 17:04

$$(x,y,z)=\left(\frac{5}{13}\cos t,\frac{8}{13}-\sin t,-\frac{12}{13}\cos t\right)$$
Rotate the coordinates according to $$\sin\theta = -\frac{12}{13}$$ and $$\cos\theta =\frac{5}{13}$$,
\begin{align} & x’ (t)= x\cos\theta + z\sin\theta = \cos t\\ & z’ (t)= -x\sin\theta + z\cos\theta =0\\ & y’(t)=y -\frac 8{13}= -\sin t\\ \end{align} which leads to
$$(x’)^2+(y’)^2 = 1, \>\>\>\>\>z’=0$$
Moreover, $$x’(0)= x’(2\pi)$$ and $$z’(0)= z’(2\pi)$$. Thus, in the new coordinations, the curve equation explicitly shows that it is a unit circle in the $$x’y’$$ plane, with the same starting and ending point, i.e. a closed curve.