1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Asymptotes as the lines y=x an y=-x

  1. Nov 2, 2006 #1
    Hi there.

    I'm a maths teacher and today was having a discussion with my Head of Department about asymptotes (as you do!)

    She was me if I could think on an equation of a graph(s) which has an asymptotes at the line y=x and another at y=-x.

    Thinking about it, neither of us could come up with anything. It's really bugging me now. I ain't too sure if there is such a function...though it seems silly there not being.

    The only way I managed to get a graph like what I wanted was to rotate the graph of y=1/x by various angles.

    So can such a graph exist? Is it possible to transform equations of graphs via rotations (something I can't ever remember doing)?

    Hope I can come up with an answer by tomorrow :)
     
  2. jcsd
  3. Nov 2, 2006 #2
    How about

    [tex]y = x \tanh{x}[/tex] ?

    as

    [tex] x \tanh{x} = x \frac{e^x - e^{-x}}{e^x + e^{-x}}[/tex]

    therefore, for large positive x,

    [tex]x \tanh{x} \approx x \frac{e^x}{e^x} = x[/tex]

    and for large negative x

    [tex] x \tanh{x} \approx x (\frac{-e^{-x}}{e^{-x}}) = -x[/tex]
     
  4. Nov 2, 2006 #3

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    You certainly can rotate curves!

    For example, the curve described by the equation:
    [tex]x^{2}-y^{2}=1[/tex]
    has asymptotes y=x and y=-x.

    Note, however, that in this case, there exists no function of x, by which the y-coordinates of the curve could be computed, and the curve in question cannot be regarded as the graph of some function, i.e, the set of points describable as (x,f(x)), where f is some function.
     
  5. Nov 3, 2006 #4

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Doesn't one branch of that hyperbola still have those lines as asymptotes?
    In a physics context, the worldline of a uniformly accelerated observer [which can be regarded as function (t,x(t)), where x(t)=sqrt(1+t^2)] is asymptotic to a light cone.
     
  6. Nov 3, 2006 #5
    Thanks for the help guys :D

    My head of department was so sure you could each of the 4 sections between the said asymptotes filled with a curve.

    Using what the first two posts said, I drew the graphs of

    y = sqrt(1+x^2)
    y = -sqrt(1+x^2)
    x = sqrt(1+y^2)
    x = -sqrt(1+y^2)

    to get what I needed (at least I think that is what I i if I remember correctly back to last night....)

    But I only went and forgot all about this today until just now...so I never i speak to my HofD about it...never mind :D
     
  7. Nov 3, 2006 #6

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You can combine the four curves to get one that fills the whole thing. :smile:

    We already know that

    x² - y² = 1

    fills in two sections, and

    y² - x² = 1

    fills in the other two.

    We can combine them into one equation:

    (x² - y² - 1) (y² - x² - 1) = 0

    which will fill in all four quadrants.
     
  8. Nov 3, 2006 #7

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    ...which can be simplified to read (x² - y²)² = 1.
    Of course, the easy way to see this is to note that
    y² - x² = 1 is the same as x² - y² = -1.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Asymptotes as the lines y=x an y=-x
  1. F:x->y ? (Replies: 2)

  2. Solve sin(y/2) = y/4 (Replies: 2)

  3. Solving for y (Replies: 3)

Loading...