Question

Determine whether the vector field is conservative. If it is, find a potential function for the vector field. $$ \mathbf{F}(x, y)=x e^{x^{2} y}(2 y \mathbf{i}+x \mathbf{j}) $$

Short answer

The vector field is conservative, and its potential function is \(f(x, y) = 2xye^{x^2y}\)

Step-by-step solution

Formulate Vector Field Components

Firstly, separate the components of the vector field function. In this case we have the component in \(i\) direction \(M = 2yxe^{x^2y}\) and in \(j\) direction \(N = x^2e^{x^2y}\)

Calculate Partial Derivatives

Next, the partial derivatives must be computed. The task is to find \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\). Using the product and chain rule of derivatives we get \(\frac{\partial M}{\partial y} = 2xe^{x^2y} + 4x^3y^2e^{x^2y}\) and \(\frac{\partial N}{\partial x} = 2xe^{x^2y} + 4x^3y^2e^{x^2y}\)

Compare Partial Derivatives

As established earlier, if the vector field is conservative, the two previously computed partial derivatives should be equal, which they are in this case. Hence, the vector field is conservative.

Find a potential function

Nextly, a potential function, \(f\), must be found for this conservative vector field, by integrating the vector field components, \(M\) and \(N\), with respect to their associated variables. The partial integral of \(N\) with respect to \(y\) yields a function of \(x\) and \(y\) alike. Adding to this the integral of \(M\) with respect to \(x\), not including the terms that already exist in the former integral, gives us the potential function: \(f(x, y) = 2xye^{x^2y}\)