# 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 $$P$$.

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

• 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. Mar 15 '19 at 11:51
• 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. Mar 15 '19 at 13:08
• 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. Mar 15 '19 at 13:14
• 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 $2\pi -t$? Mar 15 '19 at 13:21
• Yes, I did it in GeoGebra, but how can I sketch it easily in case I need to it by hand? Mar 15 '19 at 13:22