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Finding parametric equations for the tangent line

  1. Oct 2, 2005 #1
    Hello everyone, i'm having troubles seeing how this works. The directions are:
    Find parametric equations for the tagent line to the curve with the given parametric equations at the specified point.
    Here is my work and problem:
    http://show.imagehosting.us/show/750696/0/nouser_750/T0_-1_750696.jpg
    The answer they have in the back is:
    x = 1-t, y = t, z = 1-t
    Thanks.
     
  2. jcsd
  3. Oct 2, 2005 #2

    TD

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    The point (1,0,1) corresponds with the parameter t = 0.
    Now, in your derivative, let t = 0 to get (-1,1,-1), this gives the direction.

    Combining it with the point where it has to go through will give the line:
    (1,0,1) + t (-1,1,-1) = (1-t,t,1-t)
     
  4. Oct 2, 2005 #3
    Ohhh!! thanks again TD! But i'm alittle confused, from the point they gave you: (1,0,1). How did you know that corresponds to t = 0?
     
  5. Oct 2, 2005 #4

    TD

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    Because the z-co├Ârdinate (e^(-t)) can only by 1 for t = 0. You can check for x & y too :smile:
     
  6. Oct 2, 2005 #5
    ohh i c now, so really u just got guess a t, that corresponds to the point they say? right? so if they had like (0,0,0) u would have to find a t that satisfied all them right?
     
  7. Oct 2, 2005 #6

    TD

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    Correct, but since t = 0 was the only value that was correct for the z-value, it HAD to be correct for x and y too (if not, the point wouldn't have been on the curve).
     
  8. Oct 2, 2005 #7
    Awesome, thanks for the explantion! it helped greatly!
     
  9. Oct 2, 2005 #8

    TD

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    No problem :smile:
     
  10. Oct 6, 2005 #9
    im a bit lost now. i get the general idea but what about this one:

    x = cos t, y = 3e^(2t), z = 3e^(-2t) and the point is (1, 3, 3)

    in this case t is supposed to be 0, but z does not equal 1 when t is 0

    so how do they come up with that?? And also x would be 1, y would be 3.
     
  11. Oct 6, 2005 #10

    TD

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    But z wouldn't have to be 1, it has to be 3 as well, no?

    [tex]\left( {\cos t,3e^{2t} ,3e^{ - 2t} } \right)\mathop \to \limits^{t = 0} \left( {\cos 0,3e^0 ,3e^0 } \right) = \left( {1,3,3} \right)[/tex]
     
  12. Oct 6, 2005 #11
    so what we really have to do is pick a t that gives us the point given? so in this case the point ( 1, 3, 3) is given or found by plugging in t = 0 into the parametric equations?

    i think that makes since now

    thanks very much
     
  13. Oct 6, 2005 #12

    TD

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    Well yes, and if it would happen that you cannot find a t so that the parametric equation gives you a certain point, then that point just isn't part of the curve.
     
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