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Section 9.6 : Polar Coordinates

7. Convert the following equation into an equation in terms of Cartesian coordinates.

\[6{r^3}\sin \theta = 4 - cos\theta \] Show Solution

There is a variety of ways to work this problem. One way is to first multiply everything by \(r\) and then doing a little rearranging as follows,

\[6{r^4}\sin \theta = 4r - rcos\theta \hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}6{r^3}\left( {r\sin \theta } \right) = 4r - rcos\theta \hspace{0.25in}\]

We can now use the following formulas to finish this problem.

\[x = r\cos \theta \hspace{0.25in}\hspace{0.25in}y = r\sin \theta \hspace{0.25in}\hspace{0.25in}r = \sqrt {{x^2} + {y^2}} \]

Here is the answer for this problem,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{6y{{\left[ {\sqrt {{x^2} + {y^2}} \,} \right]}^3} = 4\sqrt {{x^2} + {y^2}} - x}}\]