Section 16.6 : Conservative Vector Fields
4. Find the potential function for the following vector field.
\[\vec F = \left( {6{x^2} - 2x{y^2} + \frac{y}{{2\sqrt x }}} \right)\vec i - \left( {2{x^2}y - 4 - \sqrt x } \right)\vec j\]Show All Steps Hide All Steps
Start SolutionNow, by assumption from how the problem was asked, we could assume that the vector field is conservative but let’s check it anyway just to make sure.
So,
\[\begin{align*}P & = 6{x^2} - 2x{y^2} + \frac{y}{{2\sqrt x }} & \hspace{0.5in}{P_y} & = - 4xy + \frac{1}{{2\sqrt x }}\\ Q & = - \left( {2{x^2}y - 4 - \sqrt x } \right) & \hspace{0.5in}{Q_x} & = - 4xy + \frac{1}{{2\sqrt x }}\end{align*}\]Okay, we can see that \({P_y} = {Q_x}\) and so the vector field is conservative as the problem statement suggested it would be.
Be careful with these problems and watch the signs on the vector components. One of the biggest mistakes that students make with these problems is to miss the minus sign that is in front of the second component of the vector field. There won’t always be a minus sign of course, but on occasion there will be one and if we miss it the rest of the problem will be very difficult to do. In fact, if we miss it we won’t be able to find a potential function for the vector field!
Show Step 2Okay, to find the potential function for this vector field we know that we need to first either integrate \(P\) with respect to \(x\) or integrate \(Q\) with respect to \(y\). It doesn’t matter which one we use chose to use and, in this case, it looks like neither will be any harder than the other.
So, let’s go with the following integration for this problem.
\[\begin{align*}f\left( {x,y} \right) & = \int{{Q\,dy}}\\ & = \int{{ - 2{x^2}y + 4 + \sqrt x \,dy}}\\ & = - {x^2}{y^2} + 4y + y\sqrt x + g\left( x \right)\end{align*}\]Don’t forget that, in this case, because we were integrating with respect to \(y\) the “constant of integration” will be a function of \(x\)!
Show Step 3Next, differentiate the function from the previous step with respect to \(x\) and set equal to \(P\) since we know the derivative and \(P\) should be the same function.
\[{f_x} = - 2x{y^2} + \frac{y}{{2\sqrt x }} + g'\left( x \right) = 6{x^2} - 2x{y^2} + \frac{y}{{2\sqrt x }} = P\hspace{0.25in} \Rightarrow \hspace{0.25in}g'\left( x \right) = 6{x^2}\]Now, recall that because we integrated with respect to \(y\) in Step 2 \(g\left( x \right)\), and hence \(g'\left( x \right)\), should only be a function of \(x\)’s (as it is in this case). If there had been any \(y\)’s in \(g'\left( x \right)\) we’d know there was something wrong at this point. Either we’d made a mistake somewhere or the vector field was not conservative. Of course, we verified that it was conservative in Step 1 and so this would in fact mean we’d made a mistake somewhere!
Show Step 4We can now integrate both sides of the formula for \(g'\left( x \right)\) above to get,
\[g\left( x \right) = 2{x^3} + c\]Don’t forget the “\(+c\)” on this!
Show Step 5Finally, putting everything together we get the following potential function for the vector field.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{f\left( {x,y} \right) = - {x^2}{y^2} + 4y + y\sqrt x + 2{x^3} + c}}\]