How to show curve is closed on non-closed interval

Question

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.

Answer 1

$$(x,y,z)=\left(\frac{5}{13}\cos t,\frac{8}{13}-\sin t,-\frac{12}{13}\cos t\right)$$

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.