Question

Consider the following statements: 
I. If \(f\) is a differentiable scalar field, then \[ \operatorname{curl}(\operatorname{grad} f)=\vec{0}. \] 
II. If \(\vec{f}\) is a differentiable vector field, then \[ \operatorname{div}(\operatorname{curl}\,\vec{f})=0. \] 
Which one of the following is correct?

• A. I is true, but II is false
• B. I is false, but II is true
• C. Both I and II are true
• D. Neither I nor II is true

Hint:

Remember two fundamental rules of vector calculus:
1. The curl of any gradient field is always zero ($\nabla \times \nabla f = \vec{0}$).
2. The divergence of any curl field is always zero ($\nabla \cdot (\nabla \times \vec{f}) = 0$).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks us to verify two important vector calculus identities involving gradient, curl, and divergence operators. 

Step 2: Key Formula or Approach: 
We use the definitions of vector differential operators: 

Gradient of a scalar function: \[ \text{grad } f = \nabla f \] Curl of a vector field: \[ \text{curl } \vec{A} = \nabla \times \vec{A} \] Divergence of a vector field: \[ \text{div } \vec{A} = \nabla \cdot \vec{A} \] 

Step 3: Detailed Explanation: 

Statement I Analysis: 
We need to evaluate: \[ \text{curl}(\text{grad } f) \] or, \[ \nabla \times (\nabla f) \] For a scalar field \(f(x,y,z)\), the gradient is: \[ \nabla f = \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j} + \frac{\partial f}{\partial z}\hat{k} \] 
The curl of this gradient field is: \[ \nabla \times (\nabla f) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z} \end{vmatrix} \] 
Expanding the determinant: \[ \nabla \times (\nabla f) = \left( \frac{\partial^2 f}{\partial y \partial z} - \frac{\partial^2 f}{\partial z \partial y} \right)\hat{i} \] \[ - \left( \frac{\partial^2 f}{\partial x \partial z} - \frac{\partial^2 f}{\partial z \partial x} \right)\hat{j} \] \[ + \left( \frac{\partial^2 f}{\partial x \partial y} - \frac{\partial^2 f}{\partial y \partial x} \right)\hat{k} \] 
By Clairaut's theorem, the mixed partial derivatives are equal: \[ \frac{\partial^2 f}{\partial y \partial z} = \frac{\partial^2 f}{\partial z \partial y} \] and similarly for other terms. Therefore: \[ \nabla \times (\nabla f) = \vec{0} \] Hence, Statement I is true. 

Statement II Analysis: 
We need to evaluate: \[ \text{div}(\text{curl } \vec{f}) \] or, \[ \nabla \cdot (\nabla \times \vec{f}) \] 
A standard vector calculus identity states: \[ \nabla \cdot (\nabla \times \vec{f}) = 0 \] This means the divergence of the curl of any vector field is always zero. Therefore, Statement II is also true. 

Step 4: Final Answer: 
Both statements I and II are true.