Register to reply 
A curve that intersects itself at some point w/o trig functions 
Share this thread: 
#1
Dec3013, 10:21 AM

P: 136

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
Dec3013, 11:20 AM

P: 831

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^2x^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! 


#3
Dec3013, 12:18 PM

P: 136

Shyan,
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. 


#4
Dec3013, 01:13 PM

HW Helper
P: 1,985

A curve that intersects itself at some point w/o trig functions
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 "singlevalued 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 ( we'll soon hear ) and which may have been more evocative if I hadn't got used to functions, that maybe emphasizes more calculation. 


#5
Dec3013, 02:02 PM

P: 136

epenguin,
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? 


#6
Dec3013, 02:05 PM

P: 949




#7
Dec3013, 06:07 PM

P: 136

But I thought [itex] y=\pm \sqrt{R^2x^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. 


#8
Dec3013, 11:27 PM

P: 831

I think Lebombo has problem with my post because of the following line:



Register to reply 
Related Discussions  
Proving that a curve intersects a surface at a right angle  Calculus & Beyond Homework  8  
Point at which a line intersects a plane  General Math  1  
Points that a curve's normal line intersects  Calculus & Beyond Homework  2  
Tangent line and points it intersects a curve  Calculus & Beyond Homework  4  
Point on a parabola where the tangent intersects a point  Calculus & Beyond Homework  0 