# Vectors that follow the same path? (should be easy)

1. Sep 17, 2008

### imsoconfused

here's the problem:
which vectors follow the same path as R=ti + t^2j? The speed along the path may be different.
(a) 2ti + 2t^2j (b) 2ti + 4t^2j (c) -ti + t^2j (d) t^3i + t^6j

I think the answer is (a) and (b), and then (c) is in the opposite direction. I believe this because the velocities of (a) and (b) are (a): v=2 + 4t and (b): v=2 + 8t which is similar to the original: v=1+2t. (c): v=-1 + 2t is negative so it is in the opposite direction.

am I correct or at least on the right train of thought?
thanks!

2. Sep 17, 2008

### Dick

This would probably be clearer if the first path was labeled si+s^2j. To show two paths are the same you have to find a change of variables s->t such that the equations become the same. (b) is a good choice because it corresponds to s=2t. Which other one is good. I don't think it's (a).

3. Sep 17, 2008

### imsoconfused

I agree it would be clearer, I just put what the book said. the text is so confusing!!

I don't understand how (b) is a good choice but (a) isn't. why does (b) correspond to s=2t? the way my novice eyes see it is that (a) is just twice the length of s. why is that not so?

4. Sep 17, 2008

### Dick

(b) corresponds to s=2t because (2t)i+(2t)^2j=2ti+4t^2j. If (a) were correct then I should be able to solve s=2t and s^2=2^t^2 for s and t simultaneously for all s and t. And I can't.

5. Sep 17, 2008

### imsoconfused

does that mean differentiating bears no relevance? I think I almost understand. =)

6. Sep 17, 2008

### Dick

The problem said "the speed along the path may be different", so, no, the velocity at any one point doesn't matter. What matters is that for any s along the original path there is a t in the second path such that the two points are equal. Now which other path in the second group is the same as the first.

7. Sep 18, 2008

### imsoconfused

OH!!!!! it's (c) because if I set s=-t, I get -ti + t^2j just like with (b) I get 2ti + 4t^2j when I let s=2t.
yay I understand!!!

8. Sep 18, 2008

### Dick

Good job! You caught one I missed. But there's another one.

9. Sep 18, 2008

### imsoconfused

I guess it has to be d. =) that would happen when I let s=t^3, correct? I think I just assumed (d) wasn't even pertinent since it was so different, but it makes sense that it does now.

10. Sep 18, 2008

### Dick

Yes, yes, yes.