If I get (6-a, a) as the parameter, is a the parameter, or is (6-a) the parameter?
When the textbook asks me to 'describe these solutions using a parameter' for two lines with infinite solutions, do I need to still include the two variables of the domain and codomain (x and y) in my answer?
I.e. x + y = 6
create a parameter:
let y = a
(6-a, a)
...
x = (6-a, a) where a is a real number. ?
Many thanks,
Corey
a is the parameter. It's like you said before - a parameter is basically a variable, so it makes sense from that, that a would be the parameter, not 6-a which is an expression involving the parameter - just like 6-x is an expression involving the variable, and the variable is x. And in the same ways, parameter equations are just like normal equations - so you wouldn't say, x=(6-a,a). You'd instead say, x=6-a, and y=a, listing the two equations separately as a solution.
This is going to be a massive tangent, but eh, YOLO. Parameterization is confusing when brought up this way, because you may honestly be thinking - "what's the point? Why not just say the solutions are on the line y=6-x?" And the answer for that is while you can do this for a straight line, you can't do this for other shapes.
For example, consider a particle moving around in a circle. It follows the path x^2+y^2=1. Cool. Where does the particle start? At x=0? Well, if it's x=0, then is it at y=1 or y=-1? What about y=0? Then is it x=1, or x=-1? Maybe it's when x=y - but is that at 1/sqrt(2), or -1/sqrt(2)? Here, x and y are positions, so we have no clue where the particle starts. Functions, like y=6-x, don't have this ambiguity - we know that everything starts at the lowest part of the domain, and finishes at the highest part. Well, what if I parameterise my equations like this:
x=cos(t)
y=sin(t)
Where t is time. Suddenly, we have two functions which tell you the x and y position separately based on what time it is. So, at t=0, the particle is at the position (cos(0),sin(0))=(1,0). At t=pi/2, the position is now (cos(pi/2),sin(pi/2))=(0,1). In fact, here's desmos graph for you to play around with, where you can change the value of t and look at what the particle's position is:
https://www.desmos.com/calculator/icx6fwpe1fNotice what shape it makes as t increases? It's still a circle - but now, there's no ambiguity about where the particle starts, and where it might be at some other time. In fact, now we know something that x^2+y^2=1 could NEVER tell us - the particle is moving anti-clockwise around the circle as time goes forward. And using parameterisation like this, we can make even more, weirder, shapes, like this one:
https://www.desmos.com/calculator/cw1dafgcgvThis is why parameterising is useful, and why it makes sense to do it the way we are. The reason the textbook has introduced it this way to you is to teach it to you in a familiar context - as a way of drawing lines (and fun fact: in higher dimensions, the ONLY way to draw a line is to parameterise equations in the same way you are are learning). But, as you can see from these examples, the way of defining a parameterisation is exactly the same way you would do any other equation - by putting y=something, but also adding in x=something, and vice-versa.