Eliminating the parameter

[opticaltempest]

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

 

[soroban]

Hello, opticaltempest!

This is not an easy one . . .

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

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

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

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

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

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

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*

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

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

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

Hence: \(\displaystyle \,\sin t \:=\:\frac{opp}{hyp}\:=\:\frac{\sqrt{2y\,-\,y^2}}{1} \:=\:\sqrt{2y\,-\,y^2}\)