Paul's Online Notes
Paul's Online Notes
Home / Calculus II / Parametric Equations and Polar Coordinates / Polar Coordinates
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 9.6 : Polar Coordinates

6. Convert the following equation into an equation in terms of polar coordinates.

\[{x^2} = \frac{{4x}}{y} - 3{y^2} + 2\] Show Solution

Basically, 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 (which won’t do us a lot of good in this problem) it will save a lot of work!

First let’s substitute in the equations as needed.

\[{\left( {r\cos \theta } \right)^2} = \frac{{4\left( {r\cos \theta } \right)}}{{r\sin \theta }} - 3{\left( {r\sin \theta } \right)^2} + 2\]

Finally, as we need to do is take care of little simplification to get,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{{r^2}{{\cos }^2}\theta = 4\cot \theta - 3{r^2}{{\sin }^2}\theta + 2}}\]