Question

Prove the property for vector fields \(\mathbf{F}\) and \(\mathbf{G}\) and scalar function \(f .\) (Assume that the required partial derivatives are continuous.) $$ \nabla \times(f \mathbf{F})=f(\nabla \times \mathbf{F})+(\nabla f) \times \mathbf{F} $$

Short answer

The property \(\nabla \times(f \mathbf{F})=f(\nabla \times \mathbf{F})+(\nabla f) \times \mathbf{F}\) is proven by expanding the determinants on both sides, computing the curl of \(f \mathbf{F}\), the curl of \(\mathbf{F}\) times \(f\), and ultimately computing the cross product of the gradient of \(f\) with \(\mathbf{F}\). The equivalence of the components on both sides of the equation then proves the identity.

Step-by-step solution

Write out the full expressions for the vector fields and scalar function

Let's denote the vector fields with \(\mathbf{F} = F_i \mathbf{i} + F_j \mathbf{j} + F_k \mathbf{k}\) and the scalar function with \(f = f(x, y, z)\). The goal is to prove that \(\nabla \times(f \mathbf{F})=f(\nabla \times \mathbf{F})+(\nabla f) \times \mathbf{F}\).

Compute the curl of \(f \mathbf{F}\)

Using the definition of the curl as a determinant and applying the scalar function, we get: \[ \nabla \times(f \mathbf{F}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ fF_i & fF_j & fF_k \end{vmatrix} \] We expand this determinant to calculate the components.

Calculate the first term on RHS

The first term is obtained by computing the curl of the vector \(\mathbf{F}\) and then multiplying by scalar function \(f\): \[f(\nabla \times \mathbf{F}) = f \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ F_i & F_j & F_k \end{vmatrix} \] We expand this determinant to find the respective components.

Compute the second term on the RHS

The second term is obtained by computing the gradient of the scalar function and the cross product with the vector \(\mathbf{F}\): \[ (\nabla f) \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \ F_i & F_j & F_k \end{vmatrix} \] We expand this determinant to compute the respective components.

Compare results

Compare the components of \(\nabla \times(f \mathbf{F})\) from the left side of the equation with those that come from the computations of \(f(\nabla \times \mathbf{F})\) and \((\nabla f) \times \mathbf{F}\). If all components on both sides of the equation match, then the property is proven.