This is a very basic question, and its funny that I'm able to solve more advanced problems like this, but I was presented with a basic one and got stumped. I have the equation $$y=-\frac{3}{4}x+6.$$ In $\mathbb{R}^2$, this is a line. I want to find a parametrization of this line. I guess the problem here is I never really understood the concept of parametrization. I'm just a robot-I follow the steps the book tells me, but I really dont understand the intuition behind it. What is the point of parametrization (in laymen's terms, which is hard to find anywhere), and how would I do it for this equation?

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    $\begingroup$ Please don't use the slash for fractions. As written, $3/4x$ could be interpreted as $3/(4x)$ (well, if you ignore the fact that you say you are talking about a line), though you clearly mean $\frac{3}{4}x$. $\endgroup$ – Arturo Magidin Feb 15 '11 at 5:43

Think of a parametrization as describing the "trace" of the curve, with $t$ representing time. You want to write equations \begin{align*} x &= f(t),\\ y &= g(t) \end{align*} that describe someone tracing the line as $t$ varies.

How would you trace the graph of a function $y=g(x)$? Well, the points on the graph are all of the form $(x,g(x))$, so the simplest is to use a parametrization like \begin{align*} x &= t,\\ y &= g(t). \end{align*} If you think about what the points $(x(t),y(t))$ look like as $t$ ranges from $a$ to $b$, you'll see that you are giving the graph of $y=g(x)$ from $x=a$ to $x=b$.

That gives you one way to give the line.

Another is to pick a point on the line, and just say "go along this direction, backwards or forwards". A direction is given by a vector $(u,v)$. So if your line goes through the point $(a,b)$, and has direction $(u,v)$, then one possible parametrization just says: "start at $x=a$ and $y=b$, and then move $x$ and $y$ in the direction $(u,v)$: if you move in the $x$ direction by $ku$, then you need to move in the $y$ direction by $kv$." That is, \begin{align*} x &= a + tu\\ y &= b + tv. \end{align*} You can do that with the graph you have above by determining a point and a direction. For the direction: the slope of the line with direction $(u,v)$ is $\frac{v}{u}$, because $v$ is the rise and $u$ is the run.

  • $\begingroup$ But so whats the difference between using x and t in the equation? $\endgroup$ – Snowman Feb 15 '11 at 5:55
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    $\begingroup$ @mohabitar: For parametrizing the graph of a function, not much, except the point of view; we think of the graph of $y=g(x)$ as something that is whole and complete, we tend to think of parametrizations as describing a dynamic situation where a point is "moving over time." And parametrizations can be used to describe more general curves that cannot be given by simple functions, curve that cross themselves multiple times, etc. $\endgroup$ – Arturo Magidin Feb 15 '11 at 6:02

I may be misunderstanding your question, but you have a parametrization given by your equation:

$$ x\mapsto (x,-\frac 34 x+6) $$

or if you want something with integer coefficients:

$$ t\mapsto (4t, 6-3t).$$

The point of paramterization is that on one hand you reduce the number of variables you;re working with (in this case from two: $x,y$ to one $t$), but more importantly you make an implicit situation, that is, one defined by equations into an explicit one, that is, a way to generate the solutions.

In this case the point is perhaps not obvious because the equation is so simple, but imagine that if you have a very difficult equation then how much easier it is to work with the solutions when you are given a way to generate them (that is, by a parametrization) than when they are just given as the solutions of the equation.

Perhaps a better example to work with is the solutions to the equation $$ x^2+y^2=1.$$ A possible parametrization is $$t\mapsto (\cos t, \sin t).$$

The sad truth about parametrizations is that they actually rarely exist. I guess you could interpret this as saying that if you can find a parametrization, then you should be happy.

Addendum To answer the question in the comments: Imagine that you have to solve a system of equations in two variables. In other words you have two curves given by their equation on the plane and you have to find their intersection points. If you can parametrize one of them, then you can plug the parametrization into the other and solve an equation of a single variable. Here is an example:

Say you need to find the solution to the equation system $$ x^2+y^2=0 \qquad 2x^3y-2xy^3=1.$$ (this is a rather random choice)

If you just go head-to-the-wall approach, you might solve the first as a quadratic equation and plug that into the second an end up with a degree six equation. Good luck with that. (Actually it might be solvable in this case, I have no idea, but in general a degree six equation cannot be solved with a formula).

On the other hand, if you use the parametrization of the first equation given above, you end up with the equation $$ 2\cos^3t\sin t-2\cos t\sin^3 t=1.$$ This can easily be reduced to $$\sin 4t =\frac 12$$ which can be easily solved.

The point is that parametrization helped you solve a system of equations that would normally be very difficult or impossible to solve explicitly.

  • $\begingroup$ How has my equation been reduced to one variable with parametrization? $\endgroup$ – Snowman Feb 15 '11 at 5:59
  • $\begingroup$ You can write both $x$ and $y$ in terms of $t$. For example, relating to your question, you have $t\mapsto (t,-\frac{3}{4}t+6 )$ $\endgroup$ – Justin Feb 15 '11 at 6:02

There are an infinite amount of ways to parametrize this line. A good way here would be to let say
$$\displaystyle x=t$$
replacing x with t in the first equation, you know that
$$\displaystyle y=-\frac{3}{4}t+6$$
And there you are!
When you want to parametrize an equation, start by letting one of your variables equal to t, and then solve for the other variable(s), replacing t in the equation.
Note the great explanation from Arturo Magidin, you want to think of the position of a particle at some time $t$.


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