# Finding a parametric equation for y = f(x)

1. Feb 15, 2004

### AngelofMusic

I've searched the web for information on Parametric Equations, and most of them only give me information on how to find y = f(x) when given y = y(t) and x = x(t).

Is there any sort of method for doing the reverse? I'm told that there are theoretically an infinite number of parametric equations for a given curve, but how do you go about finding any of them?

One of the questions in our textbook says:

Parametrize the curve y = f(x), x between [a,b].

No further information was given. I think I need to know this in order to do the questions that come afterwards. For example, $$y^2=x^3$$, from (4,8) to (1,1). The answer ends up being y = (2-t)^3 and x = (2-t)^2. The solution manual gives me no explanations as to how they arrived at that answer.

Help?

2. Feb 15, 2004

### lethe

there are an infinite number of parametrizations. one easy choice is the following:

given a curve f(x)=y

let y=f(t) and x=t.

this is a sort of trivial parametrization. there are lots more. what exactly are you looking for?

3. Feb 15, 2004

### AngelofMusic

I'm basically looking for the thought process behind choosing a certain type of parametrization based on a restricted domain for t. The example I gave in my question:

$$y^2=x^3$$

The question restricted the domain of t to be between 0 and 1. And one of the answers ended up being $$y(t) = (2-t)^3$$ and $$x(t) = (2-t)^2$$.

How did they come up with that? Is it just sort of guessing and checking?

4. Feb 16, 2004

### NateTG

Well I would be incliend to say:
$$y^2=x^3=t$$
So
$$y=t^{\frac{1}{2}}$$
$$x=t^{\frac{1}{3}}$$
but that can be a bit ugly, so the people replaced by $$t'=t^6$$ so that the exponents are integers.

With the new $$t$$ you get
$$y=t^3$$
$$x=t^2$$

The $$2-$$ is added for the final values, but that could also be done by chaning the range for $$t$$.

5. Feb 16, 2004

### HallsofIvy

Staff Emeritus
NateTG is, as usual, exactly right.

There exist many "parametrizations" for any given curve. Given that y2= x3, the simplest to find is to set
y2= x3= t and solve for x and y separately.

If you don't like fractions (and who does?) you might think instead
( )2= ( )3 and not that those will be "obviously" equal if you put t3 in the first parenthesis and t2 in the second. That is: x= t2 and y= t3 just as NateTG gave.

That's a perfectly good parametrization- in particular, when t= 1, it gives the point (1, 1) and when t= 2, it gives the point (4,8).

Once you have that, you could replace the "t" by any formula (in particular "2- t") and still have a parametrization for the same curve. With x= (2-t)2 and y= (2-t)3, when t= 1, we get (1,1) and when t= 0, we get (4,8), just the reverse of what we had before.

6. Feb 16, 2004

### AngelofMusic

Oooh, thank you both very much NateTG and HallsofIvy! That made perfect sense! :-)