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 viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 16.6 : Conservative Vector Fields
2. Determine if the following vector field is conservative.
\[\vec F = \left( {2x\sin \left( {2y} \right) - 3{y^2}} \right)\vec i + \left( {2 - 6xy + 2{x^2}\cos \left( {2y} \right)} \right)\vec j\] Show SolutionThere really isn’t all that much to do with this problem. All we need to do is identify \(P\) and \(Q\) then run through the test.
So,
\[\begin{align*}P & = 2x\sin \left( {2y} \right) - 3{y^2} & \hspace{0.5in}{P_y} & = 4x\cos \left( {2y} \right) - 6y\\ Q & = 2 - 6xy + 2{x^2}\cos \left( {2y} \right)& \hspace{0.5in}{Q_x} & = - 6y + 4x\cos \left( {2y} \right)\end{align*}\]Okay, we can clearly see that \({P_y} = {Q_x}\) and so the vector field is conservative.