Question

A fluid flow field is given by the velocity vector \( \vec{V} = e^{xyz}(x\hat{i} + z\hat{k}) \). The curl of velocity at (1, 2, 3) is

• A. \( e^6 (9\hat{i} - 16\hat{j} - 3\hat{k}) \)
• B. \( e^6 (9\hat{i} - 3\hat{k}) \)
• C. \( e^6 (9\hat{i} + 16\hat{j} - 3\hat{k}) \)
• D. \( e^6 (-16\hat{i} + 9\hat{j} - 3\hat{k}) \)

Hint:

When calculating the curl, write out the vector components \(V_x, V_y, V_z\) clearly first. Be very careful with the signs in the determinant expansion, especially the minus sign on the \(\hat{j}\) component.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The curl of a vector field \( \vec{V} \) is a vector operator that describes the infinitesimal rotation of the field. It is calculated as the cross product of the del operator (\(\nabla\)) and the vector field \( \vec{V} \).
Step 2: Key Formula or Approach:
The curl is calculated using the determinant of a matrix: \[ \nabla \times \vec{V} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}
V_x & V_y & V_z \end{vmatrix} \] First, we identify the components of the velocity vector \( \vec{V} \): \( \vec{V} = e^{xyz}x\hat{i} + 0\hat{j} + e^{xyz}z\hat{k} \) So, \( V_x = xe^{xyz} \), \( V_y = 0 \), and \( V_z = ze^{xyz} \).
Step 3: Detailed Calculation:
We expand the determinant: \[ \nabla \times \vec{V} = \hat{i} \left(\frac{\partial V_z}{\partial y} - \frac{\partial V_y}{\partial z}\right) - \hat{j} \left(\frac{\partial V_z}{\partial x} - \frac{\partial V_x}{\partial z}\right) + \hat{k} \left(\frac{\partial V_y}{\partial x} - \frac{\partial V_x}{\partial y}\right) \] Now we compute the partial derivatives (using the product rule where necessary):
- \( \frac{\partial V_z}{\partial y} = \frac{\partial}{\partial y}(ze^{xyz}) = z . (xze^{xyz}) = xz^2e^{xyz} \)
- \( \frac{\partial V_y}{\partial z} = \frac{\partial}{\partial z}(0) = 0 \)
- \( \frac{\partial V_z}{\partial x} = \frac{\partial}{\partial x}(ze^{xyz}) = z . (yze^{xyz}) = yz^2e^{xyz} \)
- \( \frac{\partial V_x}{\partial z} = \frac{\partial}{\partial z}(xe^{xyz}) = x . (xye^{xyz}) = x^2ye^{xyz} \)
- \( \frac{\partial V_y}{\partial x} = \frac{\partial}{\partial x}(0) = 0 \)
- \( \frac{\partial V_x}{\partial y} = \frac{\partial}{\partial y}(xe^{xyz}) = x . (xze^{xyz}) = x^2ze^{xyz} \)
Substitute these back into the curl expression: \[ \nabla \times \vec{V} = \hat{i}(xz^2e^{xyz} - 0) - \hat{j}(yz^2e^{xyz} - x^2ye^{xyz}) + \hat{k}(0 - x^2ze^{xyz}) \] \[ \nabla \times \vec{V} = e^{xyz} [ (xz^2)\hat{i} - (yz^2 - x^2y)\hat{j} - (x^2z)\hat{k} ] \] Now, evaluate the curl at the point (1, 2, 3), so \(x=1, y=2, z=3\).
The term \(e^{xyz} = e^{1 . 2 . 3} = e^6\).
- \(\hat{i}\) component: \(xz^2 = (1)(3^2) = 9\)
- \(\hat{j}\) component: \( -(yz^2 - x^2y) = -((2)(3^2) - (1^2)(2)) = -(18 - 2) = -16 \)
- \(\hat{k}\) component: \( -x^2z = -(1^2)(3) = -3 \)
Combining the components: \[ \nabla \times \vec{V} |_{(1,2,3)} = e^6 (9\hat{i} - 16\hat{j} - 3\hat{k}) \] Step 4: Why This is Correct:
The calculation correctly applies the formula for the curl of a vector field. The partial derivatives were computed correctly using the product and chain rules, and the resulting expression was evaluated accurately at the specified point. The final result matches option (A).