Question

Determine whether or not the vector field is conservative. $$ \mathbf{F}(x, y)=15 x^{2} y^{2} \mathbf{i}+10 x^{3} y \mathbf{j} $$

Short answer

Yes, the given vector field is conservative as the curl of the vector field is zero, indicated by the equality of partial derivatives.

Step-by-step solution

Compute partial derivatives

The first step is to calculate the partial derivative of \( F_{i} \) with respect to y and \( F_{j} \) with respect to x. Here \( F_{i} = 15x^{2}y^{2} = P \) and \( F_{j} = 10x^{3}y = Q \). So, \( \frac{\partial P}{\partial y} = 30x^{2}y \) and \( \frac{\partial Q}{\partial x} = 30x^{2}y \).

Check if the vector field is conservative

The crucial step in determining if a vector field is conservative is to check if the curl of the vector field is zero. In other words, if \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \). As both our computed partial derivatives are equal ( \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} = 30x^{2}y \)), this verifies that the given vector field is indeed conservative.