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Parametic y= sqrt(t +1) and y = sqrt(t-1)

  1. Nov 5, 2013 #1
    1. The problem statement, all variables and given/known data

    I have to eliminate the parameter and write in Cartesian form.

    2. Relevant equations


    y= sqrt(t +1) and y = sqrt(t-1)
    3. The attempt at a solution

    If you were to just go for the gusto and square it out you will end up with 0 = 2. Clearly I'm missing something here. Just stumped. I don't think the right way to go is squaring and setting them equal.
     
  2. jcsd
  3. Nov 5, 2013 #2

    cepheid

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    Can you post the *exact* problem statement please?
     
  4. Nov 5, 2013 #3
    Yep,

    Eliminate the parameter to find a Cartesian equation of the curve.

    y = sqrt(t +1) , y = sqrt(t -1)
     
  5. Nov 5, 2013 #4

    cepheid

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    I think perhaps one of the y's is supposed to be an x.
     
  6. Nov 5, 2013 #5
    Ha It is possible. There was a misprint on one of the graphs where it was supposed to say seconds and it was like hours. So I suppose. I just didn't know if I was missing something. Seems like you would just do as I said and square it and yatta but you get 0 = 2.
     
  7. Nov 5, 2013 #6

    cepheid

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    Unless I am missing something obvious, I don't think it has a solution as given.
     
  8. Nov 5, 2013 #7
    Not sure maybe someone else has ideas too?
     
  9. Nov 6, 2013 #8

    Dick

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    Yeah, cepheid is right. There is clearly a typo in the problem. It's just expressing y in terms of two contradictory equations in t. There should be an x in there somewhere.
     
  10. Nov 6, 2013 #9

    HallsofIvy

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    These are NOT "parametric equations" and CANNOT be put in "Cartesian" form without an "x". As others have suggested, it must be [itex]x= \sqrt{t+ 1}[/itex] and [itex]y= \sqrt{t- 1}[/itex]. Yes, square both, then solve for t and set the two equations for t equal.
     
  11. Nov 6, 2013 #10
    OK then there is a typo in my book. Thanks for the help I will bring it up in lecture.
     
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