ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: mano91 on October 29, 2009, 11:40:44 am

Title: find the cartesian eqn of...
Post by: mano91 on October 29, 2009, 11:40:44 am: x= t/(t^2 +1)

y=1/(t^2+1)

this is from mav 08. a little help please
Title: Re: find the cartesian eqn of...
Post by: mano91 on October 29, 2009, 11:46:57 am: never mind. i figured it out :)
Title: Re: find the cartesian eqn of...
Post by: Hooligan on October 29, 2009, 01:11:47 pm: Is the answer $y = \frac{x}{t}$ ?
Title: Re: find the cartesian eqn of...
Post by: humph on October 29, 2009, 01:13:38 pm: You can't have the parametric variable $t$ in the cartesian equation, you have to express it in terms of $x$ and/or $y$ .
Title: Re: find the cartesian eqn of...
Post by: Mxbn0 on October 29, 2009, 01:24:20 pm: good question eh?

for those who don't know, you have to let x= t*(1/t^2+1) and thus x= t*(y). then u can go from there :)

Answer: x^2 + ( y - 1/2 )^2 = ( 1/2 )^2
Title: Re: find the cartesian eqn of...
Post by: NE2000 on October 29, 2009, 03:06:43 pm: Yeah I think I just said:
$t = x(t^2+1)$
$t = \frac {x}{y}$

And then subbed that into the y-expression and just rearranged. Same result as Mxbn0
Title: Re: find the cartesian eqn of...
Post by: Hooligan on October 29, 2009, 03:13:58 pm: Quote from: humph on October 29, 2009, 01:13:38 pm
You can't have the parametric variable $t$ in the cartesian equation, you have to express it in terms of $x$ and/or $y$ .

LOL. *headdesk* that was stupid of me. :buck2:

anyways... I can't get it now. I got the $t = \frac{x}{y}$ but then what do I do from there? sub it back into y and then re-arrange? I tried that ... but don't get 1/2 anywhere... ???
Title: Re: find the cartesian eqn of...
Post by: sachinmachin on October 29, 2009, 03:15:36 pm: yeh this question had me stumped. didnt realise that t=x/y..
would have made things so much easier if id realised..
Title: Re: find the cartesian eqn of...
Post by: GerrySly on October 29, 2009, 03:26:58 pm: Quote from: Hooligan on October 29, 2009, 03:13:58 pm
anyways... I can't get it now. I got the $t = \frac{x}{y}$ but then what do I do from there? sub it back into y and then re-arrange? I tried that ... but don't get 1/2 anywhere... ???
$\begin{align*} x&=\frac{t}{t^2+1}\\ t&=\frac{x}{y}\\ y&=\frac{1}{t^2+1}\\ &=\frac{1}{(\frac{x}{y})^2 + 1}\\ &=\frac{y^2}{x^2+y^2}\\ 0&=y^3-y^2+yx^2\\ x^2&=-\frac{y^3-y^2}{y}\\ &=-(y^2-y)\\ &=-((y-\frac{1}{2})^2-\frac{1}{4})\\ x^2+(y-\frac{1}{2})^2&=\frac{1}{4} \end{align*}$
Title: Re: find the cartesian eqn of...
Post by: Hooligan on October 29, 2009, 03:33:17 pm: Quote from: GerrySly on October 29, 2009, 03:26:58 pm
Quote from: Hooligan on October 29, 2009, 03:13:58 pm
anyways... I can't get it now. I got the $t = \frac{x}{y}$ but then what do I do from there? sub it back into y and then re-arrange? I tried that ... but don't get 1/2 anywhere... ???
$\begin{align*} x&=\frac{t}{t^2+1}\\ t&=\frac{x}{y}\\ y&=\frac{1}{t^2+1}\\ &=\frac{1}{(\frac{x}{y})^2 + 1}\\ &=\frac{y^2}{x^2+y^2}\\ 0&=y^3-y^2+yx^2\\ x^2&=-\frac{y^3-y^2}{y}\\ &=-(y^2-y)\\ &=-((y-\frac{1}{2})^2-\frac{1}{4})\\ x^2+(y-\frac{1}{2})^2&=\frac{1}{4} \end{align*}$

whoa... that's crazy!! Would never thought of that. I got up the 6th line, and then I just got stuck. Thanks a bunch. :D