1. **State the problem:** Given the scalar function $$Q(x,y,z) = xy^2 + y^2z$$, we need to find: 1. The gradient of $$Q$$ at the point $$(1,1,-1)$$. 2. The divergence of the gradient of $$Q$$ at the same point. 3. The curl of the gradient of $$Q$$ at the same point. 2. **Recall formulas and rules:** - The gradient of a scalar function $$Q$$ is $$\nabla Q = \left(\frac{\partial Q}{\partial x}, \frac{\partial Q}{\partial y}, \frac{\partial Q}{\partial z}\right)$$. - The divergence of a vector field $$\mathbf{F} = (F_x, F_y, F_z)$$ is $$\nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}$$. - The curl of a vector field $$\mathbf{F}$$ is $$\nabla \times \mathbf{F} = \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right)$$. - Important rule: The curl of a gradient of any scalar function is always the zero vector. 3. **Calculate the gradient $$\nabla Q$$:** - $$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(xy^2 + y^2z) = y^2 + 0 = y^2$$ - $$\frac{\partial Q}{\partial y} = \frac{\partial}{\partial y}(xy^2 + y^2z) = 2xy + 2yz$$ - $$\frac{\partial Q}{\partial z} = \frac{\partial}{\partial z}(xy^2 + y^2z) = 0 + y^2 = y^2$$ So, $$\nabla Q = (y^2, 2xy + 2yz, y^2)$$. 4. **Evaluate $$\nabla Q$$ at $$(1,1,-1)$$:** - $$y^2 = 1^2 = 1$$ - $$2xy + 2yz = 2 \times 1 \times 1 + 2 \times 1 \times (-1) = 2 - 2 = 0$$ - $$y^2 = 1$$ Thus, $$\nabla Q(1,1,-1) = (1, 0, 1)$$. 5. **Calculate the divergence of $$\nabla Q$$:** - Let $$\mathbf{F} = \nabla Q = (F_x, F_y, F_z) = (y^2, 2xy + 2yz, y^2)$$. - Compute partial derivatives: - $$\frac{\partial F_x}{\partial x} = \frac{\partial}{\partial x}(y^2) = 0$$ (since $$y^2$$ does not depend on $$x$$) - $$\frac{\partial F_y}{\partial y} = \frac{\partial}{\partial y}(2xy + 2yz) = 2x + 2z$$ - $$\frac{\partial F_z}{\partial z} = \frac{\partial}{\partial z}(y^2) = 0$$ (since $$y^2$$ does not depend on $$z$$) - So, $$\nabla \cdot \nabla Q = 0 + (2x + 2z) + 0 = 2x + 2z$$. 6. **Evaluate divergence at $$(1,1,-1)$$:** - $$2 \times 1 + 2 \times (-1) = 2 - 2 = 0$$ 7. **Calculate the curl of $$\nabla Q$$:** - The curl of a gradient is always zero: $$\nabla \times (\nabla Q) = \mathbf{0}$$ **Final answers:** - Gradient at $$(1,1,-1)$$: $$\boxed{(1, 0, 1)}$$ - Divergence of gradient at $$(1,1,-1)$$: $$\boxed{0}$$ - Curl of gradient at $$(1,1,-1)$$: $$\boxed{(0, 0, 0)}$$