Parametic y= sqrt(t +1) and y = sqrt(t-1)

1. Nov 5, 2013

Jbreezy

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

I have to eliminate the parameter and write in Cartesian form.

2. Relevant equations

y= sqrt(t +1) and y = sqrt(t-1)
3. The attempt at a solution

If you were to just go for the gusto and square it out you will end up with 0 = 2. Clearly I'm missing something here. Just stumped. I don't think the right way to go is squaring and setting them equal.

2. Nov 5, 2013

cepheid

Staff Emeritus
Can you post the *exact* problem statement please?

3. Nov 5, 2013

Jbreezy

Yep,

Eliminate the parameter to find a Cartesian equation of the curve.

y = sqrt(t +1) , y = sqrt(t -1)

4. Nov 5, 2013

cepheid

Staff Emeritus
I think perhaps one of the y's is supposed to be an x.

5. Nov 5, 2013

Jbreezy

Ha It is possible. There was a misprint on one of the graphs where it was supposed to say seconds and it was like hours. So I suppose. I just didn't know if I was missing something. Seems like you would just do as I said and square it and yatta but you get 0 = 2.

6. Nov 5, 2013

cepheid

Staff Emeritus
Unless I am missing something obvious, I don't think it has a solution as given.

7. Nov 5, 2013

Jbreezy

Not sure maybe someone else has ideas too?

8. Nov 6, 2013

Dick

Yeah, cepheid is right. There is clearly a typo in the problem. It's just expressing y in terms of two contradictory equations in t. There should be an x in there somewhere.

9. Nov 6, 2013

HallsofIvy

Staff Emeritus
These are NOT "parametric equations" and CANNOT be put in "Cartesian" form without an "x". As others have suggested, it must be $x= \sqrt{t+ 1}$ and $y= \sqrt{t- 1}$. Yes, square both, then solve for t and set the two equations for t equal.

10. Nov 6, 2013

Jbreezy

OK then there is a typo in my book. Thanks for the help I will bring it up in lecture.