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Is it possible to convert the set of Cartesian equations:

$$x(t) = (20-30)*\cos(2t)+45*\cos(2t*(20-30)/20))$$ $$y(t) = (20-30)*\sin(2t)+45*\sin(2t*(20-30)/20))$$

where

$$t \in [0,2\pi)$$

Into a single Polar-Coordinates equation, so that $f(\theta) = r$?

If so, how?

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We know that

$$r=\sqrt[]{x^2+y^2}.$$

Since you have equations for $x$ and $y$ in terms of the angle $t$, you can use the above formula to find a function of $t$ that gives $r$.

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Your expressions simplify to $$x(t)=45 \cos (t)-10 \cos (2 t)$$ $$y(t)=-45 \sin (t)-10 \sin (2 t)$$ So, since $$r^2(t) = x^2(t) + y^2(t)$$ after some minor expansions and simplifications, you should arrive to $$r^2(t)=2125-900 \cos (3 t)$$ and then ...

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  • $\begingroup$ I don't understand that last step, can't the r($theta$) equation be found by simply taking the square root of both sides of your second-last equation? $\endgroup$ – Kent Jul 6 '14 at 17:24
  • $\begingroup$ Sorry ! I thought that you wanted $t$. I fix it ! Cheers. $\endgroup$ – Claude Leibovici Jul 6 '14 at 17:47
  • $\begingroup$ Somehow, it's not right. Graphing the two versions (Cartesian and Polar) gives completely different graphs. $\endgroup$ – Kent Jul 7 '14 at 22:52

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