# Parametric equations

1. Dec 1, 2011

### wuffle

Hello forum! I have yet another question concerning calculus and the topic we're doing right now is extremely confusing for me.

1. The problem statement, all variables and given/known data

Eliminate the parameter from the parametric equations x=3t/(1+t3), y=3t2/(1+t3), and hence find an Cartesian equation in x and y for this
curve.
2. Relevant equations

3. The attempt at a solution

ummm, i have no idea how to approach this question, i assume with parametric equation you need to express t in terms of y and plug in it another equation where x is defined.... however, these two equations are pretty complicated and you cant really express t in terms of y and plug it in another equation, what do you do?

I notice the x and y equations are very similar except that y=t^2 in numerator and im certain you should use that somehow....but I have no idea how

Last edited: Dec 1, 2011
2. Dec 1, 2011

### dextercioby

So it's y(x) = x t(x), but t(x) can't be expressed in a simple manner, since it should come from a cubic.

3. Dec 1, 2011

### wuffle

I guess...

turns out my x=y^2 was wrong, now im completely lost.

actually yeah, i just fond out y(x)=x * t(x), i just plugged in some numbers and it turned out y/x=t, what do i do with that?

Last edited: Dec 1, 2011
4. Dec 2, 2011

### wuffle

Still can't get it.

I guess writing it as y(x)=xt(x) works but shouldnt we get rid of t in our equation?

5. Dec 2, 2011

### dextercioby

You should get rid of t, of course, but it ain't easy. To get somewhere, you should first have x and y defined as functions of t on a certain interval, i.e. look for the inverse function t =t(x) only when this exists, i.e. on the set of x's on which the mapping is single valued and well-bahaved.

But then even this can't guarantee you that t=t(x) can be found without solving the cubic with Cardano's formulae.