Q:

Reparametrization of a curve in opposite direction and find intersection of two particles travelling in opossite directions on the same curve.

Hi I need help with this problem:

Given $$r(t)=(e^t,e^t\cos t, e^t \sin t),\quad t\in[0,2\pi]$$
  It represents the trajectory of a particle $P$.

Draw the curve.
Reparametrize $r$ such that it represents a particle $Q$ moving  in the opposite direction.
If $P$ starts at one end of the curve and $Q$ starts at the other end (at the same time), find the point of the trajectory when they both crash.

1.- I first tried to draw the curve, but I don't know how to exactly do it with those $e^t$ in there. I tried by factoring them out like $e^t(1, \cos t, \sin t)$, so I think that the that it would be like an spiral, because of $e^t$, but I really don't know how to draw it.
2.- Then I tried to reparametrize r, so I put $-t$ instead of $t$ on $r$, but I don't know if I'm correct.
3.- I don't know how to find the point when both particles collide, I was thinking of setting both parametrizations as equal and then finding the time $t$ that satisfies that, but I'm not sure.

A:

The particle has to be at the end point $(e^{2\pi},e^{2\pi}\cos(2\pi),e^{2\pi}\sin(2\pi))$ for $t=0$ and at the start point at $t=2\pi$. We need a function of $t$, substituting itself, taking the values of $2\pi$ at $t=0$ and $0$ at $t=2\pi$. $f(t)=2\pi-t$ will do the job:
$$r(t)=(e^{2\pi-t},e^{2\pi-t}\cos(2\pi-t), e^{2\pi-t} \sin(2\pi-t)),\quad t\in[0,2\pi]$$
The reparametrization is not unique, of course, but with this one, the time when the particles collide must satisfy $t=2\pi-t$ or $t=\pi$
For the drawing, maybe it's worth noting that the projection on the $y-z$ plane is a logarithmic spiral 

