Let's solve the two parts step by step: --- ## **Part (a):** ### **Given Vector Field in Cylindrical Coordinates:** \[ A = \rho z \sin\phi\, \mathbf{a}_\rho + 3\rho z^2 \cos\phi\, \mathbf{a}_\phi \] Evaluate at \((\rho, \phi, z) = (5, \frac{\pi}{2}, 1)\). ### **Divergence in Cylindrical Coordinates** \[ \nabla \cdot \mathbf{A} = \frac{1}{\rho} \frac{\partial}{\partial \rho}(\rho A_\rho) + \frac{1}{\rho} \frac{\partial A_\phi}{\partial \phi} + \frac{\partial A_z}{\partial z} \] Here, \(A_z = 0\). ### **Step 1: Identify Components** \[ A_\rho = \rho z \sin\phi \\ A_\phi = 3\rho z^2 \cos\phi \\ A_z = 0 \] --- ### **Step 2: Compute Each Term** #### **First Term:** \[ \frac{1}{\rho} \frac{\partial}{\partial \rho}(\rho A_\rho) = \frac{1}{\rho} \frac{\partial}{\partial \rho}(\rho \cdot \rho z \sin\phi) = \frac{1}{\rho} \frac{\partial}{\partial \rho}(\rho^2 z \sin\phi) \] \[ \frac{\partial}{\partial \rho} (\rho^2 z \sin\phi) = 2\rho z \sin\phi \] \[ \Rightarrow \frac{1}{\rho}(2\rho z \sin\phi) = 2z \sin\phi \] #### **Second Term:** \[ \frac{1}{\rho} \frac{\partial A_\phi}{\partial \phi} = \frac{1}{\rho} \frac{\partial}{\partial \phi}(3\rho z^2 \cos\phi) \] \[ = \frac{1}{\rho} (3\rho z^2 (-\sin\phi)) = -3 z^2 \sin\phi \] #### **Third Term:** \[ \frac{\partial A_z}{\partial z} = 0 \] --- ### **Step 3: Combine Terms** \[ \nabla \cdot \mathbf{A} = 2z \sin\phi - 3z^2 \sin\phi \] --- ### **Step 4: Evaluate at \((\rho, \phi, z) = (5, \frac{\pi}{2}, 1)\):** \[ z = 1, \quad \phi = \frac{\pi}{2} \implies \sin\phi = 1 \] \[ \nabla \cdot \mathbf{A} = 2(1)(1) - 3(1)^2(1) = 2 - 3 = -1 \] --- ## **Final Answer (a):** \[ \boxed{-1} \] --- ## **Part (b):** ### **Given Vector Field in Spherical Coordinates:** \[ B = 2r \sin\theta \cos\phi\, \mathbf{a}_r + r \cos\theta \cos\phi\, \mathbf{a}_\theta - r \sin\phi\, \mathbf{a}_\phi \] Evaluate at \( (r = 1.5, \theta = 30^\circ, \phi = 50^\circ) \). ### **Divergence in Spherical Coordinates:** \[ \nabla \cdot \mathbf{B} = \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 B_r) + \frac{1}{r \sin\theta} \frac{\partial}{\partial\theta}(\sin\theta B_\theta) + \frac{1}{r \sin\theta} \frac{\partial B_\phi}{\partial\phi} \] --- ### **Step 1: Identify Components** \[ B_r = 2r \sin\theta \cos\phi \\ B_\theta = r \cos\theta \cos\phi \\ B_\phi = -r \sin\phi \] --- ### **Step 2: Compute Each Term** #### **First Term:** \[ \frac{1}{r^2} \frac{\partial}{\partial r}(r^2 B_r) = \frac{1}{r^2} \frac{\partial}{\partial r}[r^2 \cdot 2r \sin\theta \cos\phi] = \frac{1}{r^2} \frac{\partial}{\partial r}[2r^3 \sin\theta \cos\phi] \] \[ = \frac{1}{r^2} [6r^2 \sin\theta \cos\phi] = 6 \sin\theta \cos\phi \] #### **Second Term:** \[ \frac{1}{r \sin\theta} \frac{\partial}{\partial\theta}(\sin\theta B_\theta) \] \[ = \frac{1}{r \sin\theta} \frac{\partial}{\partial\theta}[\sin\theta \cdot r \cos\theta \cos\phi] = \frac{1}{r \sin\theta} \frac{\partial}{\partial\theta}[r \sin\theta \cos\theta \cos\phi] \] \[ = \frac{1}{r \sin\theta} [r \cos\phi \frac{\partial}{\partial\theta}(\sin\theta \cos\theta)] \] Recall: \( \frac{d}{d\theta}(\sin\theta \cos\theta) = \cos^2\theta - \sin^2\theta = \cos 2\theta \) \[ = \frac{1}{r \sin\theta} [r \cos\phi \cos 2\theta] = \frac{\cos\phi \cos 2\theta}{\sin\theta} \] #### **Third Term:** \[ \frac{1}{r \sin\theta} \frac{\partial B_\phi}{\partial\phi} \] \[ = \frac{1}{r \sin\theta} \frac{\partial}{\partial\phi}[-r \sin\phi] = \frac{1}{r \sin\theta} [-r \cos\phi] = -\frac{\cos\phi}{\sin\theta} \] --- ### **Step 3: Combine Terms** \[ \nabla \cdot \mathbf{B} = 6 \sin\theta \cos\phi + \frac{\cos\phi \cos 2\theta}{\sin\theta} - \frac{\cos\phi}{\sin\theta} \] \[ = 6 \sin\theta \cos\phi + \frac{\cos\phi (\cos 2\theta - 1)}{\sin\theta} \] Recall: \( \cos 2\theta - 1 = -2\sin^2\theta \) \[ = 6 \sin\theta \cos\phi - \frac{2\cos\phi \sin^2\theta}{\sin\theta} = 6 \sin\theta \cos\phi - 2\cos\phi \sin\theta \] \[ = 4 \sin\theta \cos\phi \] --- ### **Step 4: Evaluate at \( r = 1.5, \theta = 30^\circ, \phi = 50^\circ \)** - \( \sin 30^\circ = 0.5 \) - \( \cos 50^\circ \approx 0.6428 \) \[ \nabla \cdot \mathbf{B} = 4 \cdot 0.5 \cdot 0.6428 = 2 \cdot 0.6428 = 1.2856 \] --- ## **Final Answer (b):** \[ \boxed{1.29} \] (Rounded to two decimal places)

(a) **–1**; (b) **1.29**

(a) –1; (b) 1.29

The divergence of the vector field in part (a) evaluated at the given point is **–1**. The divergence of the vector field in part (b) evaluated at the specified coordinates is approximately **1.29**.