Eliminating the parameter

opticaltempest

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Nov 19, 2005
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I am having some trouble seeing how to elimate the parameter from these equations.

\(\displaystyle \L x = t - \sin t\) and \(\displaystyle \L y = 1 - \cos t\)


\(\displaystyle \L\sin t = t - x\) and \(\displaystyle \L \cos t = 1 - y\)

How can I fully elimate the parameter from \(\displaystyle \L\sin t = t - x\)?

Thanks
 
Hello, opticaltempest!

This is not an easy one . . .


Elimate the parameter: x=tsinty=1cost\displaystyle \,\begin{array}{cc}x \:= \:t\,-\,\sin t \\ y \:= \:1\,-\, \cos t\end{array}

The second equation gives us: cost=1y        t=arccos(1y)\displaystyle \,\cos t \:=\:1\,-\,y\;\;\Rightarrow\;\;t \:=\:\arccos(1\,-\,y)

Substitute into the first equation: x  =  arccos(1y)sin[arccos(1y)]\displaystyle \,x\;=\;\arccos(1\,-\,y)\,-\,\sin[\arccos(1\,-\,y)]

    \displaystyle \;\;The second expression can be simplified: sin[arccos(1y)]=2yy2\displaystyle \,\sin[\arccos(1\,-\,y)] \:=\:\sqrt{2y\,-\,y^2} *

Therefore: x  =  arccos(1y)2yy2\displaystyle \,x\;=\;\arccos(1\,-\,y)\,-\,\sqrt{2y\,-\,y^2}

. . . and that's the best we can do.

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*

We have: t=arccos(1y)        cost=1y=1y1\displaystyle \,t\:=\:\arccos(1\,-\,y)\;\;\Rightarrow\;\;\cos t \:=\:1\,-\,y \:=\:\frac{1\,-\,y}{1}

So t\displaystyle t is an angle in a right triangle with: adj=1y,  hyp=1\displaystyle \,adj \:= \:1\,-\,y,\;hyp\:=\:1

Then: opp2=12(1y)2=2yy2        opp=2yy2\displaystyle \,opp^2\:=\:1^2\,-\,(1\,-\,y)^2 \:=\:2y\,-\,y^2\;\;\Rightarrow\;\;opp \:=\:\sqrt{2y\,-\,y^2}

Hence: sint=opphyp=2yy21=2yy2\displaystyle \,\sin t \:=\:\frac{opp}{hyp}\:=\:\frac{\sqrt{2y\,-\,y^2}}{1} \:=\:\sqrt{2y\,-\,y^2}

 
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