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.


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.