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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 $P$.

  1. Draw the curve.
  2. Reparametrize $r$ such that it represents a particle $Q$ moving in the opposite direction.

  3. 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.

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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

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  • $\begingroup$ So, can I always use this method to parametrize a function but with opposite direction? When can I just write $-t$? I Still don't understand how to draw it. $\endgroup$ Mar 15 '19 at 11:51
  • $\begingroup$ In fact, as you can check, I used $-t$! because, as you think correctly, we need this minus sign to go backwards, simply we need too to consider the starting point: it is at a different place! Without the $2\pi$ offset the starting point would be the same. $\endgroup$ Mar 15 '19 at 13:08
  • $\begingroup$ It's a complicated curve to draw. Use any online program to draw surfaces and curves (e.g. GeoGebra 3D) and you can check for yourself. $\endgroup$ Mar 15 '19 at 13:14
  • $\begingroup$ So, in case I only need to find a parametrization that goes in opposite direction, I can just change $t$ to $-t$? And in case I need to make both of them to start at the same time, I do what you did by writing $2\pi -t$? $\endgroup$ Mar 15 '19 at 13:21
  • $\begingroup$ Yes, I did it in GeoGebra, but how can I sketch it easily in case I need to it by hand? $\endgroup$ Mar 15 '19 at 13:22

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