Solving Parametric Equations: Cartesian Equations from Parametric Equations

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To convert the given parametric equations into Cartesian form, the key step is to eliminate the parameter 't'. This can be achieved by solving one of the equations for 't' and substituting it into the other equation. The discussion highlights the importance of recognizing that the Cartesian form may involve relationships beyond a simple y=f(x) format, potentially requiring a more complex function. Additionally, the use of logarithmic transformations and hyperbolic functions is suggested for handling the equations effectively. Ultimately, the goal is to express the relationships between x and y without the parameter 't'.
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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...
 
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Solve for 't' in the first equation, and plug it in the second. Basically you have to eliminate 't'
 
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))?
 
Graph it on a graphing calculator. Nice Graph !

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

The results for x&y should include cosh(u) & sinh(u) respectively.
 
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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)
 
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Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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