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Paramatrize a peicewise function?

  1. Dec 11, 2004 #1
    Can someone hepl me find a "formual" way to paramatrize a peicewise function? Up till now i've been doing it with guess and check and am now stuck on the last part.

    The line goes from (1,3) to (0,0)
    So far I have

    [itex]
    7 \leq t \leq 10
    [/itex]

    [itex]
    y(t) = 10 - t
    [/itex]

    and am stuck on find a function for [itex]x(t)[/itex]


    Thanks.
     
  2. jcsd
  3. Dec 11, 2004 #2

    Hurkyl

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    One way is to find an ordinary expression for the line, something of the form y = mx+b, then use your known expression for y to get an expression for x.


    However, a more common way to parametrize lines is to identify each point in terms of the "proportion" from the first to the last point.

    That is, you put your variable in the range [0, 1], then the corresponding point on the line segment AB is given by:

    P = (1 - t) A + t B
     
  4. Dec 11, 2004 #3
    Brilliant!

    I don't know why that eluded me....

    -Burg
     
  5. Dec 11, 2004 #4
    So now in 3-Space,

    The line goes from (0,1,0) to (0,0,1) on the function [itex] y^2+z^2 = 1[/itex]

    i have [itex]x(t) = 0 [/itex] which is obvious, but how do i find the other two, [itex] y(t) [/itex] and [itex] z(t)[/itex]

    i can't sovle a system like i did with the last one, and since its not linear, im not sure if that second method works either. Unfourtunetlly my book is lacking in 3-space examples...

    -Thanks!
     
  6. Dec 11, 2004 #5

    Hurkyl

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    Can you reduce it to a 2-space problem?
     
  7. Dec 11, 2004 #6
    I could "pretend". But since both y and z are changing i'm have trouble seeing what one would be to sub in for the other. Any hints on where to start?

    -Burg
     
  8. Dec 11, 2004 #7

    Hurkyl

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    You should already be familiar with a parametrization of an equation of the form [itex]p^2 + q^2 = 1[/itex]... do you recognize the geometric object it defines?
     
  9. Dec 11, 2004 #8
    I just got it, after i posted that last one. It's a circile (duh, brain fart) so i use y(t)= cos t and z(t) = sin t. But there was a piece right before it, so [itex] 3 \leq t \leq [/itex]? So does that mean

    [itex] y(t) = cos ((t-3)\frac{\pi}{2} )
    [/itex]
    [itex]z(t) = sin ((t-3)\frac{\pi}{2} )[/itex]
    [itex]3 \leq t \leq 4
    [/itex]?
     
    Last edited: Dec 11, 2004
  10. Dec 11, 2004 #9

    Hurkyl

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    If that parametrization lies on the circular arc, and begins and ends in the right place, then it's good!


    I'm not sure what you mean by "a piece right before it" -- I guess you're talking about another part of the problem you've already solved.
     
  11. Dec 11, 2004 #10
    Ya, there was another piece of the piecewise goes from (3,0,0) so [itex] 0 \leq t \leq 3 [/itex] then the next piece (this one) starts at 3. I guess i could just start at 0 for each one, but this is how our prof tought it, and probablly how he wants to see it done.

    Thanks for the help! Final's on monday!!
     
  12. Dec 11, 2004 #11

    Hurkyl

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    The way you did it is right -- you wouldn't have a very well defined parametrization if it takes t=0 to two different points!
     
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