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Finding the equation of a parametric curve

  1. Dec 8, 2012 #1
    1. if y(t)= (a/t, b/t, c/t)



    2. Prove that this curve is a straight line. Find the equation of the line



    3. i found the first part without a problem, i just am not sure how to find the equation f the line.
     
  2. jcsd
  3. Dec 8, 2012 #2

    LCKurtz

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    Apparently ##y(t)## is a vector instead of the second component of the right side? How did you show it is a straight line without finding its equation? And what does "the equation f " mean? What forms do you know for straight line equations in 3D?
     
  4. Dec 8, 2012 #3
    y(t) is the parametric curve, and i proved its a straight line by proving the curvature of the line. the "f" is supposed to be of
     
  5. Dec 8, 2012 #4

    Dick

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    Proving the curvature is 0 is probably the hard way. Like LCKurtz said, "What forms do you know for straight line equations in 3D?"
     
  6. Dec 8, 2012 #5
    x-xo/a=y-yo/b=z-zo/c
     
  7. Dec 8, 2012 #6

    Dick

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    That's a good one. It would be even better with pararentheses. Your parametric form is x=a/t, y=b/t, z=c/t. So?
     
  8. Dec 8, 2012 #7
    so do i sub in a point for t in the domain? and that will give <xo,yo, zo> and then how do i find the direction <a,b,c>.
     
  9. Dec 8, 2012 #8

    Dick

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    Try subbing <xo,yo,zo>=<0,0,0>.
     
  10. Dec 8, 2012 #9
    and then the direction (a,b,c) would be point on the line or would it just be (a,b,c)?
     
  11. Dec 8, 2012 #10

    Dick

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    Well, x/a=1/t, y/b=1/t, z/c=1/t. So x/a=y/b=z/c. Direction vector? Point on the line? Or BOTH?
     
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