Question

Vector field \(\vec{V}\) is defined as
\[ \vec{V} = 3x^2yz\,\hat{i} - 5xy\,\hat{j} + 6yz^2\,\hat{k} \] The curl of \(\vec{V}\) at point \((2,-1,1)\) is

• A. \(6\hat{i} - 12\hat{j} - 7\hat{k}\)
• B. \(-12\hat{i} - 10\hat{j} - 12\hat{k}\)
• C. \(-34\)
• D. \(\begin{bmatrix} -12 & 12 & -12 \\ 5 & -10 & 0 \\ 0 & 6 & -12 \end{bmatrix}\)

Hint:

Use \(\text{curl}\,\vec{V}=\nabla\times\vec{V}\) as a \(3\times3\) determinant, find the six needed partial derivatives, then plug in \((2,-1,1)\).

Answer 1

The Correct Option is A

Step 1: Write down the curl formula.
For a vector field \(\vec{V} = V_x\hat{i}+V_y\hat{j}+V_z\hat{k}\), the curl is defined as the determinant
\[ \text{curl}\,\vec{V} = \nabla \times \vec{V} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ V_x & V_y & V_z \end{vmatrix} \] Expanding this determinant gives three components:
\[ \text{curl}\,\vec{V} = \left(\frac{\partial V_z}{\partial y}-\frac{\partial V_y}{\partial z}\right)\hat{i} - \left(\frac{\partial V_z}{\partial x}-\frac{\partial V_x}{\partial z}\right)\hat{j} + \left(\frac{\partial V_y}{\partial x}-\frac{\partial V_x}{\partial y}\right)\hat{k} \] Here \(V_x=3x^2yz\), \(V_y=-5xy\), \(V_z=6yz^2\).

Step 2: Find each partial derivative needed.
\(\dfrac{\partial V_z}{\partial y} = 6z^2\), and \(\dfrac{\partial V_y}{\partial z} = 0\) (since \(V_y\) does not contain \(z\)).
\(\dfrac{\partial V_z}{\partial x} = 0\) (since \(V_z\) does not contain \(x\)), and \(\dfrac{\partial V_x}{\partial z} = 3x^2y\).
\(\dfrac{\partial V_y}{\partial x} = -5y\), and \(\dfrac{\partial V_x}{\partial y} = 3x^2z\).

Step 3: Assemble the curl as a function of \(x,y,z\).
\[ \text{curl}\,\vec{V} = (6z^2 - 0)\hat{i} - (0 - 3x^2y)\hat{j} + (-5y - 3x^2z)\hat{k} = 6z^2\hat{i} + 3x^2y\hat{j} + (-5y-3x^2z)\hat{k} \]
Step 4: Substitute the point \((x,y,z)=(2,-1,1)\).
i-component: \(6(1)^2 = 6\).
j-component: \(3(2)^2(-1) = 3(4)(-1) = -12\).
k-component: \(-5(-1) - 3(2)^2(1) = 5 - 12 = -7\).
So \(\text{curl}\,\vec{V} = 6\hat{i} - 12\hat{j} - 7\hat{k}\) at this point.

Step 5: Check why the other options are wrong.
Option (B) has the wrong numbers for each component, coming from mixing up which partial derivative belongs to which term. Option (C), \(-34\), is not even a vector, so it cannot be a curl (curl of a vector field is always a vector, never a scalar). Option (D) is the full \(3\times 3\) matrix of all nine partial derivatives (the Jacobian of \(\vec{V}\)), not the antisymmetric combination that defines curl, so it is a different mathematical object entirely.

Final Answer:
\[ \boxed{\text{curl}\,\vec{V}\Big|_{(2,-1,1)} = 6\hat{i} - 12\hat{j} - 7\hat{k}} \]