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Math Help - Two particles travel along the space curves...

  1. #1
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    Two particles travel along the space curves...

    The Question:

    Two particles travel along the space curves:
    r1(t) = < t , t^2, t^3 >
    r2(t) = < 1 + 2t, 1 + 6t, 1 + 14t >

    (a) Do the particles collide?
    (b) Do their paths intersect?

    My Attempt:

    (a)
    For the two particles to collide, there must be a value of t such that r1(t) = r2(t). So, I set the components of the positions functions equal to each other:

    t = 1 + 2t => t = -1
    t^2 = 1 + 6t => t = 3 + or - sqrt(10)
    I can see already that because t is equal to different quantities in the first two components, the particles do not collide.

    (b)
    I'm not sure how I might find if the space curves ever intersect...Other than using the same method I used above, which just can't apply for both parts.

    Thanks for any responses!
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  2. #2
    Junior Member
    Joined
    Sep 2009
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    You've got the right idea so far as i can tell, but what you need to understand for the second one is that for their paths to intersect, both points must have passed through the intersection point at some point in time. So one might have passed through when t=3 while the other passed through when t=17. What you can do here is to leave r1(t) in terms of the parameter t but express r2(t) in terms of parameter s. NOW equate your components like you did in part 1, except now you'll have 3 equations which link s and t. Try to solve these - if you get a contradiction then they don't intersect (i.e. there is no t value you can put in for r1 nor s value for r2 which result in the same coordinates), if you don't get a contradiction (so there do exist some t and corresponding s values which result in the same coords) then the paths DO intersect (and you've got the parameter value at this point).

    Not certain if this is the most efficient method but it usually works for me.
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  3. #3
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    Oct 2009
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    Perfect! Thanks for the response. Hopefully there will be a question just like this on the exam tomorrow, because I understand it perfectly now.
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