Welcome back — missed you this week. Given the scalar function $f(x,y,z) = x^2 + y$. Step 1: Find the gradient of $f$. The gradient of a scalar function $f$ is given by: $$ \nabla f = \frac{\partial f}{\partial x} \mathbf{i} + \frac{\partial f}{\partial y} \mathbf{j} + \frac{\partial f}{\partial z} \mathbf{k} $$ First, calculate the partial derivatives: $$ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2 + y) = 2x $$ $$ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2 + y) = 1 $$ $$ \frac{\partial f}{\partial z} = \frac{\partial}{\partial z}(x^2 + y) = 0 $$ Substitute these into the gradient formula: $$ \nabla f = 2x \mathbf{i} + 1 \mathbf{j} + 0 \mathbf{k} $$ $$ \nabla f = 2x \mathbf{i} + \mathbf{j} $$ The gradient of $f$ is $\boxed{\mathbf{2x \mathbf{i} + \mathbf{j}}}$. Step 2: Find the divergence. Divergence is defined for a vector field. Assuming the question asks for the divergence of the gradient of $f$, which is $\nabla \cdot (\nabla f)$. Let $\mathbf{F} = \nabla f = 2x \mathbf{i} + \mathbf{j} + 0 \mathbf{k}$. The divergence of a vector field $\mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k}$ is given by: $$ \nabla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} $$ Here, $P = 2x$, $Q = 1$, and $R = 0$. Calculate the partial derivatives: $$ \frac{\partial P}{\partial x} = \frac{\partial}{\partial x}(2x) = 2 $$ $$ \frac{\partial Q}{\partial y} = \frac{\partial}{\partial y}(1) = 0 $$ $$ \frac{\partial R}{\partial z} = \frac{\partial}{\partial z}(0) = 0 $$ Substitute these into the divergence formula: $$ \nabla \cdot (\nabla f) = 2 + 0 + 0 = 2 $$ The divergence of the gradient of $f$ is $\boxed{\mathbf{2}}$. Step 3: Find the curl. Curl is defined for a vector field. Assuming the question asks for the curl of the gradient of $f$, which is $\nabla \times (\nabla f)$. Let $\mathbf{F} = \nabla f = 2x \mathbf{i} + \mathbf{j} + 0 \mathbf{k}$. The curl of a vector field $\mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k}$ is given by: $$ \nabla \times \mathbf{F} = \left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right) \mathbf{i} + \left(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\right) \mathbf{j} + \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) \mathbf{k} $$ Here, $P = 2x$, $Q = 1$, and $R = 0$. Calculate the partial derivatives: $$ \frac{\partial R}{\partial y} = \frac{\partial}{\partial y}(0) = 0 $$ $$ \frac{\partial Q}{\partial z} = \frac{\partial}{\partial z}(1) = 0 $$ $$ \frac{\partial P}{\partial z} = \frac{\partial}{\partial z}(2x) = 0 $$ $$ \frac{\partial R}{\partial x} = \frac{\partial}{\partial x}(0) = 0 $$ $$ \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(1) = 0 $$ $$ \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(2x) = 0 $$ Substitute these into the curl formula: $$ \nabla \times (\nabla f) = (0 - 0) \mathbf{i} + (0 - 0) \mathbf{j} + (0 - 0) \mathbf{k} = 0 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k} $$ $$ \nabla \times (\nabla f) = \mathbf{0} $$ The curl of the gradient of $f$ is $\boxed{\mathbf{0}}$. Send me the next one 📸