Let's solve each question step-by-step: --- ### **Question 7** Given: - \( \phi = x^2 y z^3 \) - \( \vec{A} = xz \hat{i} - y^2 \hat{j} + 2x^2y \hat{k} \) You are asked to find the divergence of **A** at \( (1, 1, 2) \). #### **Step 1: Find the divergence** The divergence of \( \vec{A} = (A_x, A_y, A_z) \) is: \[ \text{div } \vec{A} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z} \] Given: - \( A_x = xz \) - \( A_y = -y^2 \) - \( A_z = 2x^2y \) Calculate each partial derivative: - \( \frac{\partial}{\partial x}(xz) = z \) - \( \frac{\partial}{\partial y}(-y^2) = -2y \) - \( \frac{\partial}{\partial z}(2x^2y) = 0 \) (since no \( z \) in this term) So: \[ \text{div } \vec{A} = z - 2y + 0 = z - 2y \] #### **Step 2: Evaluate at \( (1, 1, 2) \)** Plug in \( x = 1, y = 1, z = 2 \): \[ \text{div } \vec{A} = 2 - 2 \times 1 = 2 - 2 = 0 \] **Final Answer for Q7:** \[ \boxed{0} \] --- ### **Question 8** Given: - \( \vec{A} = [3xz^2,\ -yz,\ x+2z] \) - Need to find \( |\text{curl}(\text{curl}\ \vec{A})| \) at \( (1,-1,2) \) to 2 decimal places. #### **Step 1: Find curl(A)** Let \( \vec{A} = (A_x, A_y, A_z) \): \[ \text{curl}\ \vec{A} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \end{array} \right| \] So: - \( A_x = 3xz^2 \) - \( A_y = -yz \) - \( A_z = x + 2z \) Let's compute each component: **i-component:** \[ \left(\frac{\partial}{\partial y}(x + 2z) - \frac{\partial}{\partial z}(-yz)\right) = (0 - (-y)) = y \] **j-component:** \[ - \left(\frac{\partial}{\partial x}(x + 2z) - \frac{\partial}{\partial z}(3xz^2)\right) = - (1 - 6xz) \] \[ = -1 + 6xz \] **k-component:** \[ \frac{\partial}{\partial x}(-yz) - \frac{\partial}{\partial y}(3xz^2) = (0 - 0) = 0 \] So, \[ \text{curl}~\vec{A} = y \hat{i} + (6xz - 1)\hat{j} + 0\hat{k} \] #### **Step 2: Find curl(curl(A))** Let \( \vec{B} = \text{curl}~\vec{A} = (y, 6xz - 1, 0) \) Compute \( \text{curl}~\vec{B} \): **i-component:** \[ \frac{\partial}{\partial y}(0) - \frac{\partial}{\partial z}(6xz - 1) = 0 - 6x = -6x \] **j-component:** \[ - \left( \frac{\partial}{\partial x}(0) - \frac{\partial}{\partial z}(y) \right ) = - (0 - 0 ) = 0 \] **k-component:** \[ \frac{\partial}{\partial x}(6xz - 1) - \frac{\partial}{\partial y}(y) = 6z - 1 \] So, \[ \text{curl}(\text{curl}\ \vec{A}) = (-6x, 0, 6z - 1) \] #### **Step 3: Magnitude at \( (1, -1, 2) \)** - \( x = 1 \) - \( z = 2 \) So, \[ (-6x, 0, 6z-1) = (-6, 0, (6 \times 2) - 1) = (-6, 0, 11) \] Magnitude: \[ |\text{curl}(\text{curl}\ \vec{A})| = \sqrt{(-6)^2 + 0^2 + 11^2} = \sqrt{36 + 121} = \sqrt{157} \] \[ \sqrt{157} \approx 12.53 \] **Final Answer for Q8:** \[ \boxed{12.53} \] --- ### **Summary of answers** **Q7:** \( 0 \) **Q8:** \( 12.53 \)