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Eliminate the Parameter

  1. Jan 8, 2012 #1
    1. The problem statement, all variables and given/known data

    a) Eleminate the parameter to find a Cartesian equation of the curve.

    b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases.

    2. Relevant equations

    x = 2cosh(t) y = 5sinh(t)
    cosh^2(x) - sinh^2(x) = 1

    3. The attempt at a solution

    cosh(t) = x/2 sinh(t) = y/5

    (x^(2)/4) - (y^(2)/25) = 1

    y = +/- sqrt(25/4(x^2) - 25)

    This is the best guess I have here, and the graphs look similar... I just really don't know what to do from here (or even if I'm going about it the right way). Any help would be appreciated.
     
  2. jcsd
  3. Jan 8, 2012 #2

    Dick

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    I think you are doing fine so far. What part of this are you having a problem with?
     
  4. Jan 8, 2012 #3
    I graphed the Cartesian equation(s) I got, and the graphs didn't exactly match up.
     
  5. Jan 8, 2012 #4

    Dick

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    In what way didn't they match up? I hope in y = +/- sqrt(25/4(x^2) - 25) you meant y = +/- sqrt((25x^2)/4 - 25). You didn't use enough parentheses in the 25/4(x^2) to make it clear what you meant.
     
  6. Jan 8, 2012 #5
    I'm sorry; that is what I was meaning to type. Well, the parametric graph only takes up the first quadrant; however, the Cartesian graph takes up all four.

    Cartesian graph:

    http://www3.wolframalpha.com/Calculate/MSP/MSP419219ih3hd5eb510179000016g700d6idee9f8f?MSPStoreType=image/gif&s=4&w=200&h=204&cdf=RangeControl [Broken]
     
    Last edited by a moderator: May 5, 2017
  7. Jan 8, 2012 #6

    Dick

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    That sort of thing would be normal. It happens when you square things. You wind up getting more solutions than you started with. If you take 0<t<infinity you are only going to get the first quadant of this graph. If you take -infinity<t<infinity then you should get both quadrants with x>0. The parametric form won't give you any of the points where x<0 since cosh(t)>0. If you really want the cartesian form to reflect that then you could solve for x in terms of y and then just take the positive square root.
     
    Last edited by a moderator: May 5, 2017
  8. Jan 8, 2012 #7
    Oh, I see. Thank you very much for your help.
     
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