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Section 9.6 : Polar Coordinates
5. Convert the following equation into an equation in terms of polar coordinates.
\[\frac{{4x}}{{3{x^2} + 3{y^2}}} = 6 - xy\] Show SolutionBasically, what we need to do here is to convert all the \(x\)’s and \(y\)’s into \(r\)’s and \(\theta \)’s using the following formulas.
\[x = r\cos \theta \hspace{0.25in}\hspace{0.25in}y = r\sin \theta \hspace{0.25in}\hspace{0.25in}{r^2} = {x^2} + {y^2}\]Don’t forget about the last one! If it is possible to use this formula (and you can see where we’ll use it in the case can’t you?) it will save a lot of work!
First let’s substitute in the equations as needed.
\[\frac{{4\left( {r\cos \theta } \right)}}{{3{r^2}}} = 6 - \left( {r\cos \theta } \right)\left( {r\sin \theta } \right)\]Finally, as we need to do is take care of little simplification to get,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{4\cos \theta }}{{3r}} = 6 - {r^2}\cos \theta \sin \theta }}\]