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Help with parametric curves

  1. Mar 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider the parametric curve
    x=ln(t) and
    y= 1+t^2

    i need to eliminate the parameter to find the cartesian equation

    2. Relevant equations


    3. The attempt at a solution

    if i solve for t using x I get that t=e^x
    but if i solve for t using y i get t=(y-1)^(1/2)

    when i plug in x into y i get that y=1+e^2x

    but plugging y into x i get that x= ln(y-1)^(1/2)

    both of which have different domains of parameterization and i need to find the domain of parameterization

    how can i know which to use? which is the correct one?

    then part b says that: "since dx/dt=______ it is _____ along the domain of parameterization, the curve goes through ________________" I HAVE NO IDEA WHAT TO PUT THERE

    because i thought i had to just put like dx/dt= 1/t and then i thought i had to put concave up or down but if i use y=1+e^x^2's domain i cant do that
     
    Last edited: Mar 22, 2016
  2. jcsd
  3. Mar 23, 2016 #2

    andrewkirk

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    I don't think they do.

    Check your work. If, having done that, you still think that the two approaches give different domains, post your reasons why.
     
  4. Mar 23, 2016 #3

    HallsofIvy

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    Don't solve the second equation for t- just replace the "t" in the second equation by e^x.

    You mean y= 1+ e^(2x) but, yes, that is correct.

    You get the "domains of parameterization from the original parametric equations, not from the Cartesian equations.
    x= ln(t) so t must be positive.

    [/quote]then part b says that: "since dx/dt=______ it is _____ along the domain of parameterization, the curve goes through ________________" I HAVE NO IDEA WHAT TO PUT THERE

    because i thought i had to just put like dx/dt= 1/t and then i thought i had to put concave up or down but if i use y=1+e^x^2's domain i cant do that[/QUOTE]
    Does "it" refer to the graph of the Cartesian equation? You had previously determined that that function was y= 1+ e^(2x), NOT y= 1+ e^(x^2).
     
    Last edited by a moderator: Mar 23, 2016
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