Solving Parametric Equations: Cartesian Equations from Parametric Equations

[Michael_Light]

Homework Statement

I have [PLAIN]http://img543.imageshack.us/img543/1608/msp481619fbgebbd2f2fg34.gif and [PLAIN]http://img153.imageshack.us/img153/121/msp69719fbh6if8c7b729c0.gif as my parametric equations with ''t'' as parameter. How to find its Cartesian equation?

Homework Equations

The Attempt at a Solution

I know i have to eliminate the ''t'', but i have no ideas how to eliminate it. Can anyone help me? Thanks...

 

[praharmitra]

Solve for 't' in the first equation, and plug it in the second. Basically you have to eliminate 't'

 

[Disconnected]

That fact that they say Cartesian makes me think that this is not just a y=f(x) question but a f(x,y) parametrized as f(t). Michael, is there more to the question that you posted?

Or is it just a case of finding t in terms of some f(x) and then replacing all of the t's in the y=g(t) to get y=f(g(x))?

 

[SammyS]

Graph it on a graphing calculator. Nice Graph !

Let u = ln(t) for t > 0 → t = e^u. (Take care of t < 0 later.)

The results for x&y should include cosh(u) & sinh(u) respectively.

 

[SammyS]

Multiply the numerators & denominators of both expressions by 1/t.

$$x(t)=\frac{2at(1/t)}{3(t^2+1)(1/t)}=\frac{2a}{3(t+\frac{1}{t})}$$

$$y(t)=\frac{-2bt(1/t)}{3(t^2-1)(1/t)}=\frac{-2b}{3(t-\frac{1}{t})}$$

Therefore, x(1/t) = x(t) and y(1/t) = -y(t)