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Find a piecewise smoother parameterization of C

  1. Dec 4, 2012 #1
    I am having a real tough time trying to build a parametrization of the path c. its a really simple one but my professor just briefly went over it, and I cant find the thought process to put it together. I know the equation of a plane curve r(t)=x(t)i+y(t)j and everything. But its just not making sense to me. I would really like to understand how the book got this answer, I know there are many answers but if I can figure this one out i might be able to see other ways. If you know a good vid for this stuff please link it.[PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/IMAG0094.jpg[/PLAIN]
     
  2. jcsd
  3. Dec 4, 2012 #2

    Simon Bridge

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    Well,

    c1: (x,y) = (t,0) : 0 < t ≤ 3
    c2: (x,y) = (3,t-3) : 3 < t ≤ 6
    c3: ... you should be able to figure the rest.

    Then you can integrate from t=0 to t=12.

    This is pretty much what the book says - but they use unit vectors.
    To understand it, put a dot on the graph at the t=0 position, then watch where it goes as t gets bigger.
     
  4. Dec 4, 2012 #3

    Mark44

    Staff: Mentor

    Since you already have the answer in the book, maybe it would make more sense to see why their parameterization works.

    For the first leg, r(t) = ti, 0 ≤ t ≤ 3
    Each value of t will get you to a point on the x-axis between 0 and 3. When t = 3, you're at the point (3, 0).

    For the second leg, all the x values are 3, which explains the 3i component.
    For the vertical component we want to map the t values from 3 to 6 to the values 0 to 3, so we subtract 3. This gives us r(t) = 3i + (t - 3)j, 3 ≤ t ≤ 6.
    When t = 3, we're at the point (3, 0). When t = 4.5, we're at the point (3, 1.5). When t = 6, we're at (3, 3).

    For the third leg, the y-value stays fixed at 3, but the x-value has to go from 3 down to 0, while the t values range from 6 to 9. If we were mapping [6, 9] to [0, 3] we could use t - 6, but we want to end up with [3, 0], so t - 6 won't work. Switching the order from t - 6 to 6 - t maps the t values in [6, 9] to x values of 0 to -3, so they're off. If we adjust by adding 3 to 6 - t, we get 9 - t, and this correctly maps t values between 6 and 9 to x values from 3 down to 0. So r(t) = (12 - t)i + 0j, 6 ≤ t ≤ 9.

    The last leg keeps x fixed at 0, but maps t values from 9 to 12 onto y values from 3 down to 0. I leave it for you to figure out a formula for this part.
     
  5. Dec 5, 2012 #4
    ok I think I kind of got it now, I tried this one. And found a piecewise, does this one work?
    [PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/IMAG0095.jpg[/PLAIN]
     
  6. Dec 5, 2012 #5

    Mark44

    Staff: Mentor

    Test it and see. You have c1(t) = 1 - t for 0 <= t <= 2. You need to include i in your formula.

    Does c1(0) give you the point (1, 0)?
    Does c1(2) give you the point (-1, 0)?

    Same thing for the other two legs.
     
  7. Dec 5, 2012 #6
    I think you made a mistake on the second one. C1 ends at -i but C2 starts you at -i + j
     
  8. Dec 5, 2012 #7
    k found my mistake on C2 and believe it to be (t-2)j

    so I now have c1 r(t)=(1-t)i 0≤t≤2
    C2 r(t)=(t-3)i +(t-2)j 2≤t≤3
    c3 r(t)=(t-3)i +(4-t)j 3≤t≤4

    I think this is correct.
     
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