# Thread: Parametrization of the curve

1. ## Parametrization of the curve

I don't know what I'm doing.
x^2+y+1-xy=2
x^2-1+y-xy=0
(x-1)(x+1)-y(x-1)=0
(x-1)(x+1-y)=0
.
.
oh...
can't find any clue to mfind the parametrimization of the curves.

2. ## Re: Parametrization of the curve

Hey happymatthematics.

Does the curve meet both constraints? (If so, hint: get rid of the z term first).

is
{z=t
{y=s
{x=(1-t)/s
correct?

4. ## Re: Parametrization of the curve

Can you outline how you got the answer (the method is always more important than the final answer in isolation)?

5. ## Re: Parametrization of the curve

Hello chiro.
I've tried in two ways.

Does any one of my two attempts correct?
I don't know the reason why we need to do parametrization.
sometimes, when I encounter some simple system of equations, I can just let x=t, y=s and then write z in terms of t and s.
but most of the time, it is not the case.
how do I know or how can I check if my parametrization is correct?
Thank you.
by the way, since I'm afraid if I wrote a long question then no one would answer me, I raise a question as short as possible.
It's happy to heard from you that outlining my scratch work!

6. ## Re: Parametrization of the curve

Originally Posted by happymatthematics

I don't know what I'm doing.
x^2+y+1-xy=2
x^2-1+y-xy=0
(x-1)(x+1)-y(x-1)=0
(x-1)(x+1-y)=0
x- 1= 0 only when x= 1 so as long as x is not 1, you have x+1- y= 0 or y= x+1.
From the original equations, you already have z= 1- xy= 1- x(x+ 1)= 1- x- x^2.
Now take x itself as parameter: x= t, y= t+ 1, z= 1- t- t^2.

You need to remember that there are many different "parameterizations" for the same curve so at some point you need to make a "choice" as I did in setting x= t.

oh...
can't find any clue to mfind the parametrimization of the curves.