Attachment 26774

please read the question above.

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.

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- Jan 31st 2013, 12:05 AMhappymatthematicsParametrization of the curve
Attachment 26774

please read the question above.

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. - Jan 31st 2013, 01:24 AMchiroRe: Parametrization of the curve
Hey happymatthematics.

Does the curve meet both constraints? (If so, hint: get rid of the z term first). - Jan 31st 2013, 03:30 PMhappymatthematicsRe: Parametrization of the curve
is

{z=t

{y=s

{x=(1-t)/s

correct? - Jan 31st 2013, 04:07 PMchiroRe: Parametrization of the curve
Can you outline how you got the answer (the method is always more important than the final answer in isolation)?

- Feb 1st 2013, 03:00 AMhappymatthematicsRe: Parametrization of the curve
Hello chiro.

I've tried in two ways.

Please read the picture below.

Attachment 26797

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! - Feb 1st 2013, 08:48 AMHallsofIvyRe: Parametrization of the curve
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.

Quote:

oh...

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