Question

Verify that the vector field is conservative. \(\mathbf{F}(x, y)=12 x y \mathbf{i}+6\left(x^{2}+y\right) \mathbf{j}\)

Short answer

Yes, the given vector field is conservative as the curl of F is zero.

Step-by-step solution

Identifying the Vector Components

First, it's important to assign each part of the vector field to its corresponding component in two dimensions, \(x\) and \(y\). So, \(F_{x} = 12xy\) and \(F_{y} = 6x^2 + 6y\).

Verifying the Conservative Condition

Next, a vector field is conservative if and only if the curl of F is zero, which means the partial derivative of \(F_y\) with respect to \(x\) equals the partial derivative of \(F_x\) with respect to \(y\). Let's calculate these partial derivatives.

Calculate the Partial Derivatives

The partial derivative of \(F_y\) with respect to \(x\) is \(\frac{\partial F_y}{\partial x} = \frac{\partial}{\partial x} (6x^2 + 6y) = 12x\). The partial derivative of \(F_x\) with respect to \(y\) is \(\frac{\partial F_x}{\partial y} = \frac{\partial}{\partial y} (12xy) = 12x\). It can be observed that \(\frac{\partial F_y}{\partial x} = \frac{\partial F_x}{\partial y}\).

Final Analysis

Since the curl of F is zero, as \(\frac{\partial F_y}{\partial x} = \frac{\partial F_x}{\partial y}\), we can conclude that the given vector field F is conservative.