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Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(x)

  1. May 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Solve the following parametric equations for y = f(x)

    x(t) = e[tex]^{-t}[/tex] + t
    y(t) = e[tex]^{t}[/tex] - t


    3. The attempt at a solution

    I'm completely stuck on this one. I've noticed x + y = e[tex]^{-t}[/tex] + e[tex]^{t}[/tex] and also

    y = [tex]\frac{1}{x-t}[/tex] [tex]- t [/tex]

    but thats about as far as I've gotten.
     
  2. jcsd
  3. May 20, 2010 #2
    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    Plug -t into the y equation. Like this:

    y(-t) instead of y(t).

    I hope that's what you're looking for, I assume, by reading the problem statement, that you want y as a function of x?
     
  4. May 20, 2010 #3
    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    Well I've never seen anything like that, every example I've seen involved elimintating t from the system of equations using algebraic manipulations. But yes I am trying to find y as a function of only x
     
    Last edited: May 20, 2010
  5. May 20, 2010 #4

    gabbagabbahey

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    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    I don't see how that works. Sure, you can say [itex]y(-t)=x(t)[/itex], but that equation still makes explicit reference to [itex]t[/itex].
     
  6. May 20, 2010 #5

    gabbagabbahey

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    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    Take a look at [itex]x(t)y(t)[/itex] and [itex]x(t)-y(t)[/itex]:wink:
     
  7. May 20, 2010 #6

    Mark44

    Staff: Mentor

    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    And if I understand what gabbagabbahey is driving at, sinh(t) = (1/2)(e^t - e^(-t)) and cosh(t) = (1/2)(e^t + e^(-t)) might be helpful.
     
  8. May 20, 2010 #7
    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    I'm sorry but I'm just not seeing it.

    x(t) - y(t) = e[tex]^{-t}[/tex] -e[tex]^{t}[/tex] + 2t

    x(t)y(t) = 1 + t(-e[tex]^{-t}[/tex] + e[tex]^{t}[/tex] - t) = 1 - t(e[tex]^{-t}[/tex] - e[tex]^{t}[/tex] + t) = 1 - t(x - y - t)

    I've tried combining both equations in a couple of different ways but I always end up with this equality. Any more help?
     
  9. May 21, 2010 #8

    gabbagabbahey

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    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    Oops! When I quickly calculated it out on a scrap of paper, I had incorrectly written down x-y=2t....which would have made things nice and easy.

    The only way that I can think of is to find [itex]y'(x)[/itex] and [itex]y''(x)[/itex] using the chain rule, in order to get an ODE that doesn't depend on [itex]t[/itex], and then solve that....but it gets messy pretty quick.
     
  10. May 21, 2010 #9
    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    I talked to my professor today and he said that we are supposed to use the inverse cosh function.

    x + y = e[tex]^{-t}[/tex] + e[tex]^{t}[/tex] = 2cosh(t). Therefore t = arccosh((x + y)/2)

    then subsitute t into y = [tex]\frac{1}{x-t}[/tex]-t

    I never would have thought that inverse cosh would be an acceptable answer, in my experience if you have to rely on inverse functions you've got the wrong answer lol.
     
    Last edited: May 21, 2010
  11. May 21, 2010 #10

    Char. Limit

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    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    I suppose Mark44 was on the right track then.
     
  12. May 21, 2010 #11

    Mark44

    Staff: Mentor

    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    Actually, I don't think I was. You can get to x + y = 2cosh(t), and then use cosh^(-1) to solve for t, but I don't see any simplification that ends with y as a function of x.
     
    Last edited: May 21, 2010
  13. May 21, 2010 #12
    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    The answer he is expecting is y = [tex]\frac{1}{x - arccosh((x+y)/2)}[/tex] - arccosh((x+y)/2). I know that its not an explict function of x but y is implicitly defined by the previous equation.
     
    Last edited: May 21, 2010
  14. May 21, 2010 #13

    Mark44

    Staff: Mentor

    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    Well, then maybe I was on the right track.

    x + y = et + e-t = 2cosh(t)
    ==> (x + y)/2 = cosh(t)
    ==> cosh-1((x + y)/2) = t
    or t = arccosh((x + y)/2)

    Substituting for t in y = et - t, we have
    [tex]y = e^{arccosh((x + y)/2)} - arccosh((x + y)/2)[/tex]

    This seems to be going in the right direction, but I don't know how to deal with the exponential term. If this involved triangle trig inverses, I would draw a right triangle and see if I could rewrite the first term above, but I'm not as familiar with the geometry in the hyperbolic trig functions.
     
  15. May 21, 2010 #14
    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    Once I'd found t I did some rearranging before I substituted it back into the equations.

    x = e[tex]^{-t}[/tex] + t ==> x-t = e[tex]^{-t}[/tex]

    y = e[tex]^{t}[/tex] - t ==> y = [tex]\frac{1}{x-t}[/tex]-t

    then I substituted for t. It may not be any prettier but at least it doesn't have an exponential in it
     
  16. May 21, 2010 #15
    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    I am taking a bit of a different approach:

    x + y = exp(-t) + exp(t)

    Multiply both sides by exp(t) to get:

    (x + y) ext(t) = 1 + exp(t)^2

    So 0 = exp(t)^2 - (x+y)exp(t) + 1

    You could use the quadratic formula to solve for exp(t) = a big huge two-valued mess. Then take the ln of THAT and have t?

    I also tried this:

    (x - t)(y + t) = exp(-t) exp(t)
    xy - ty + tx - t^2 = 1

    Rearrange a little to get
    0 = t^2 - t(x - y) + 1 - xy

    And maybe you could solve this quadratic?
     
  17. May 21, 2010 #16
    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    It's impossible to solve

    [tex]
    x = e^{-t} + t
    [/tex]

    for t in terms of elementary functions.
     
  18. May 21, 2010 #17
    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    [tex] \cosh^{-1} x = \ln (x - \sqrt{x^2-1}) [/tex] (for real-valued functions) according to Wikipedia, which gets rid of the exponential term but doesn't obviously simplify to the term in the answer given.

    http://en.wikipedia.org/wiki/Inverse_hyperbolic_function
     
  19. May 21, 2010 #18
    Re: Solving Parametric Equations: x(t) = exp(-t) + t and y(t) = exp(t) - t for y = f(

    Not much help, but I think it's cool that we can go around in a circle: you can simplify the arcosh expression to be equivalent to the "minus" root of the radical.

    [tex] \cosh^{-1} \frac{x+y}{2} = \ln \left( \frac{x+y}{2} - \sqrt{ \frac{(x+y)^2}{4} - 1} \right) [/tex]

    And as Mark44 said,

    [tex] \cosh^{-1} \frac{x+y}{2} = t [/tex]

    [tex] t = \ln \left( \frac{x+y}{2} - \sqrt{ \frac{(x+y)^2}{4} - 1} \right) [/tex]

    [tex] e^t = \frac{x+y}{2} - \frac{\sqrt{(x+y)^2 - 4}}{4} [/tex]

    [tex] e^t = \frac{x+y - \sqrt{ (x+y)^2 - 4}}{2} [/tex], which is a solution to 0 = exp(t)^2 - (x + y) exp(t) + 1.
     
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