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A curve that intersects itself at some point w/o trig functions

  1. Dec 30, 2013 #1
    In order for an equation to be a function, it has to pass the vertical line test.

    A circle is not a function because it does not pass the vertical line test.

    A curve containing a loop does not pass the vertical line test and to me that means it is not a function.

    However, if I am given two parametric equations, x(t) and y(t) that both seem to be functions (passing the vertical line test), and am provided a picture of the graph of the parametric equations, if the curve of the parametric equations contains a loop, I would assume right away that the combined equation, y(x), can not be a function because the graph has a loop.

    So if I have x(t) = t^{5} + 4t^{3} and y(t) = t^{2}

    And because x(t) is not solvable, I can't sub the t(x) into y(t), but I can sub t(y) into x(t):

    [itex]x(y) = ((+/-)y^{5/2}- 4(+/-)y^{3/2}[/itex]

    [itex]x= y^{5/2}- 4y^{3/2}[/itex] or [itex]x= -y^{5/2}+ 4y^{3/2}[/itex]

    My question is here:

    Is there anything specific about this function or a function of similar form that enables me to see that A) it is not a function B) It creates a loop when graphed (implying there is at least some point at which the curve intersects itself)?
  2. jcsd
  3. Dec 30, 2013 #2


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    The vertical line test you mentioned simply means finding out whether there is a single x that gives two different ys or not.You can figure that out by some calculation too.
    Consider the equation of a circle: [itex] x^2+y^2=R^2 [/itex]. I can write [itex] y=\pm \sqrt{R^2-x^2} [/itex]. It is obvious that there exits some x that give two different ys. In fact all x gives two different ys except x=R. So from the equation of a circle,you can get a function!
  4. Dec 30, 2013 #3

    So when given the parametric equations:

    x(t) = [itex]t^{5} + 4t^{3}[/itex] and [itex]t^{2}[/itex]

    At what point can I determine the equation will or will not pass the vertical line test (be or not be a function)?

    Question 1) Can I tell by looking at the set of parametric equations given, or can I only tell by looking at the combined x(y) function?

    Question 2) If x(t) can not be solved in terms of t(x), then how can x(y) be solved in terms of y(x)?

    That is, are you saying that it is possible to isolate y to one side in this(these) equation(s)?

    [itex]x(y) = ((+/-)y^{5/2}- 4(+/-)y^{3/2}[/itex]

    [itex]x= y^{5/2}- 4y^{3/2}[/itex] or [itex]x= -y^{5/2}+ 4y^{3/2}[/itex]

    I believe I just learned that a function of this particular form can not be solved for the independent variable, so it doesn't seem to follow the example with the circle you provided, due to being a 5th degree polynomial, unless I'm neglecting some other facts that I'm not yet aware of.
  5. Dec 30, 2013 #4


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    This you may consider pedantic but you wait till you see the guys who come next.

    At least in my time a function is not what you think - what you think used to be called a "single-valued function". A function in my day was just a rule, give x calculate something, not necessarily only one thing, called f(x), by the rule. And yes, the most usual, useful ones tended to be single valued, and you tended to think they are all like that. But even when you defined them to be like that, their inverses often aren't.

    Functions have tended to be replaced by "mappings" which I think is almost the same thing (:uhh: we'll soon hear :biggrin:) and which may have been more evocative if I hadn't got used to functions, that maybe emphasizes more calculation.
    Last edited: Dec 30, 2013
  6. Dec 30, 2013 #5

    I think I kind of understand what you mean. I don't consider an equation a function if one x produces more than one y's.

    Suppose I have a curve that creates a loop. The equation that describes that curve is not considered a function because of the rule that one x must produce one y.

    However, can a curve of this type be described by some sort of equation? That is, an equation which does not include trig functions?
  7. Dec 30, 2013 #6


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    Did you read post #2 in this thread?
  8. Dec 30, 2013 #7

    But I thought [itex] y=\pm \sqrt{R^2-x^2} [/itex] is not a function exactly because one x value gives two different y values. I thought it can't be a function unless you write it expressly as two different equations, at which point there are now two separate functions.
  9. Dec 30, 2013 #8


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    I think Lebombo has problem with my post because of the following line:
    It was just a miss-typing.In fact I meant:
    Going with the definitions,I can say there can exist a relation between the members of two sets(which can be the same). If the relation gets one member of the input set and gives one member of the output set,it is called a binary relation. If this binary relation, relates each member of the input set to only one member of the output set, it is called a function. If this function relates each member of the output set to only one member of the input set,it is called a one to one function. If for every member of the output set,there exists a member of the input set which is related to it by our function,our function is called an onto or Surjective function.
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