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

  1. Sep 17, 2008 #1
    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. jcsd
  3. Sep 17, 2008 #2

    Dick

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    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).
     
  4. Sep 17, 2008 #3
    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?
     
  5. Sep 17, 2008 #4

    Dick

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    (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.
     
  6. Sep 17, 2008 #5
    does that mean differentiating bears no relevance? I think I almost understand. =)
     
  7. Sep 17, 2008 #6

    Dick

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    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.
     
  8. Sep 18, 2008 #7
    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!!!
     
  9. Sep 18, 2008 #8

    Dick

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    Good job! You caught one I missed. But there's another one.
     
  10. Sep 18, 2008 #9
    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.
     
  11. Sep 18, 2008 #10

    Dick

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    Yes, yes, yes.
     
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