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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.

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  • $\begingroup$ If you want to show that the curve is closed, showing periodicity would be a step to take. $\endgroup$ – Ninad Munshi Apr 19 '20 at 4:30
  • $\begingroup$ @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? $\endgroup$ – A.B Apr 19 '20 at 4:38
  • $\begingroup$ 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. $\endgroup$ – Max Apr 19 '20 at 6:47
  • $\begingroup$ @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. $\endgroup$ – A.B Apr 19 '20 at 14:24
  • $\begingroup$ 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.) $\endgroup$ – Max Apr 19 '20 at 17:04
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$$(x,y,z)=\left(\frac{5}{13}\cos t,\frac{8}{13}-\sin t,-\frac{12}{13}\cos t\right)$$

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.

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