Given the scalar function $\phi(x, y, z) = e^{xy} - x^2yz^2$. The gradient of a scalar function $\phi(x, y, z)$ is given by: $$\nabla \phi = \frac{\partial \phi}{\partial x} \mathbf{i} + \frac{\partial \phi}{\partial y} \mathbf{j} + \frac{\partial \phi}{\partial z} \mathbf{k}$$ a) Evaluate $\nabla \phi (x, y, z)$ Step 1: Calculate the partial derivative with respect to $x$. $$\frac{\partial \phi}{\partial x} = \frac{\partial}{\partial x} (e^{xy} - x^2yz^2) = y e^{xy} - 2xyz^2$$ Step 2: Calculate the partial derivative with respect to $y$. $$\frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y} (e^{xy} - x^2yz^2) = x e^{xy} - x^2z^2$$ Step 3: Calculate the partial derivative with respect to $z$. $$\frac{\partial \phi}{\partial z} = \frac{\partial}{\partial z} (e^{xy} - x^2yz^2) = -x^2y(2z) = -2x^2yz$$ Step 4: Combine the partial derivatives to form the gradient vector. $$\nabla \phi (x, y, z) = (y e^{xy} - 2xyz^2) \mathbf{i} + (x e^{xy} - x^2z^2) \mathbf{j} + (-2x^2yz) \mathbf{k}$$ The gradient is: $$\boxed{\nabla \phi (x, y, z) = (y e^{xy} - 2xyz^2) \mathbf{i} + (x e^{xy} - x^2z^2) \mathbf{j} - 2x^2yz \mathbf{k}}$$ b) Evaluate $\nabla \phi (2, -1, 1)$ Substitute $x=2$, $y=-1$, and $z=1$ into the expression for $\nabla \phi (x, y, z)$ from part (a). Step 1: Calculate the $\mathbf{i}$ component. $$y e^{xy} - 2xyz^2 = (-1) e^{(2)(-1)} - 2(2)(-1)(1)^2 = -e^{-2} + 4$$ Step 2: Calculate the $\mathbf{j}$ component. $$x e^{xy} - x^2z^2 = (2) e^{(2)(-1)} - (2)^2(1)^2 = 2e^{-2} - 4$$ Step 3: Calculate the $\mathbf{k}$ component. $$-2x^2yz = -2(2)^2(-1)(1) = -2(4)(-1)(1) = 8$$ Step 4: Combine the components. $$\nabla \phi (2, -1, 1) = (-e^{-2} + 4) \mathbf{i} + (2e^{-2} - 4) \mathbf{j} + 8 \mathbf{k}$$ The gradient at $(2, -1, 1)$ is: $$\boxed{\nabla \phi (2, -1, 1) = (4 - e^{-2}) \mathbf{i} + (2e^{-2} - 4) \mathbf{j} + 8 \mathbf{k}}$$ c) Evaluate $|\nabla \phi (1, 1, -1)|$ First, evaluate $\nabla \phi (1, 1, -1)$ by substituting $x=1$, $y=1$, and $z=-1$ into the expression for $\nabla \phi (x, y, z)$. Step 1: Calculate the $\mathbf{i}$ component. $$y e^{xy} - 2xyz^2 = (1) e^{(1)(1)} - 2(1)(1)(-1)^2 = e - 2(1)(1) = e - 2$$ Step 2: Calculate the $\mathbf{j}$ component. $$x e^{xy} - x^2z^2 = (1) e^{(1)(1)} - (1)^2(-1)^2 = e - 1(1) = e - 1$$ Step 3: Calculate the $\mathbf{k}$ component. $$-2x^2yz = -2(1)^2(1)(-1) = -2(1)(-1) = 2$$ Step 4: Form the gradient vector at $(1, 1, -1)$. $$\nabla \phi (1, 1, -1) = (e - 2) \mathbf{i} + (e - 1) \mathbf{j} + 2 \mathbf{k}$$ Step 5: Calculate the magnitude of the vector. The magnitude of a vector $A\mathbf{i} + B\mathbf{j} + C\mathbf{k}$ is $\sqrt{A^2 + B^2 + C^2}$. $$|\nabla \phi (1, 1, -1)| = \sqrt{(e-2)^2 + (e-1)^2 + (2)^2}$$ $$|\nabla \phi (1, 1, -1)| = \sqrt{(e^2 - 4e + 4) + (e^2 - 2e + 1) + 4}$$ $$|\nabla \phi (1, 1, -1)| = \sqrt{e^2 - 4e + 4 + e^2 - 2e + 1 + 4}$$ $$|\nabla \phi (1, 1, -1)| = \sqrt{2e^2 - 6e + 9}$$ The magnitude of the gradient at $(1, 1, -1)$ is: $$\boxed{|\nabla \phi (1, 1, -1)| = \sqrt{2e^2 - 6e + 9}}$$