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Points of intersection of Parametric Lines

  1. Apr 25, 2013 #1
    1. The problem statement, all variables and given/known data
    I'm told to find the 2 points the two curves P and Q will intersect on and the parametric equations are:
    [tex] P (x=t, y=2t-1)[/tex]
    [tex]Q (x=3t-t^2, y=t+1)[/tex]



    3. The attempt at a solution
    I know i'm supposed to set x-equations and y-equations equal to each and solve so that

    [tex]t=3t-t^2[/tex] for x
    [tex]2t-1=t+1[/tex] for y

    and when I solve them I get [tex]t=0[/tex] and [tex]t=2[/tex] for x
    and
    [tex]t=2[/tex] for y
    the problem is I can't seem to find another t-value for y

    Also I'm not completely sure if I can use t interchangeably between the two equations when solving for them or if I should consider the t to be two separate and unique variables like [tex]t_1[/tex] and [tex]t_2[/tex]
     
  2. jcsd
  3. Apr 25, 2013 #2

    LCKurtz

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    That is the problem. You can't assume the curves cross for the same value of the parameters. So call one parameter ##t## and the other ##s## and try setting the ##x## values and ##y## values equal.
     
  4. Apr 25, 2013 #3
    I did just that quite a while ago when my instructor had hinted at that idea and this is what I came up with

    [tex]t=3s-s^2[/tex]
    [tex]t=s(3-s)[/tex]
    so that for x t=s and t=3-s
    so that any same two values fit in the first equality(?) and only 1.5 solves the equality in the second

    for y it is
    [tex]2t-1=s+1[/tex]
    [tex]2t=s+2[/tex]
    so that for y t=2 and s=2

    I'm still not sure how to find a second point of intersection. Since the first point is (1.5,2)?
     
  5. Apr 25, 2013 #4

    LCKurtz

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    Ignoring the other bad logic, you have two equations in two unknowns. Solve them correctly.
     
  6. Apr 25, 2013 #5
    I'm no good when trying to solve for 2 unknowns algebraically like this, because first thought is to substitute [tex]t=3s-s^2[/tex] into [tex]2t-1=s+1[/tex] but that won't work. And by graph for the first equation t=s=0 or 2 and the second t=s=2
     
  7. Apr 25, 2013 #6

    LCKurtz

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    Yes it will. Try it.
     
  8. Apr 25, 2013 #7
    ok I did that so that [tex]2(3s-s^2)-1=s+1[/tex]
    setting it to 0 gives me [tex]-2s^2+5s-2=0[/tex]
    and solving for that I get s=1/2 and s=2

    then I plug those answers into [tex]t=3s-s^2[/tex]
    so that t=1.25 and t=2

    so then my points of intersection are at
    P (1.25,1.5) and (2,3)
    Q (1.25,1.5) and (2,3)
     
    Last edited: Apr 25, 2013
  9. Apr 25, 2013 #8

    LCKurtz

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    So check your work. Do your s and t values work in their equations? If so, do the two curves go through those two points for the corresponding values of s and t?
     
  10. Apr 25, 2013 #9
    Yep I checked it and the numbers work.
    Thanks for the nudges in the right direction.
     
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