1. **Problem (i):** Find a unit normal vector to the surface defined by $f(x,y) = x^3$ at the point $(2,-1,8)$. 2. Compute the partial derivatives: $$f_x = \frac{\partial}{\partial x} (x^3) = 3x^2$$ $$f_y = \frac{\partial}{\partial y} (x^3) = 0$$ 3. Evaluate at the point $(2,-1)$: $$f_x(2,-1) = 3(2)^2 = 12$$ $$f_y(2,-1) = 0$$ 4. The surface can be written as $z = f(x,y)$, so the gradient vector of the surface is: $$\nabla f = (f_x, f_y, -1) = (12, 0, -1)$$ 5. The normal vector to the surface is perpendicular to the tangent plane, so it is: $$\mathbf{n} = (12, 0, -1)$$ 6. Find the magnitude of $\mathbf{n}$: $$|\mathbf{n}| = \sqrt{12^2 + 0^2 + (-1)^2} = \sqrt{144 + 1} = \sqrt{145}$$ 7. The unit normal vector is: $$\hat{\mathbf{n}} = \frac{1}{\sqrt{145}} (12, 0, -1) = \left(\frac{12}{\sqrt{145}}, 0, \frac{-1}{\sqrt{145}}\right)$$ --- 8. **Problem (ii):** Find a unit normal vector to the surface defined by $f(x,y,z) = \ln(xy - z)$ at the point where $x = y = 1$. 9. Rewrite the function as $z = f(x,y)$: Given $f(x,y,z) = \ln(xy - z)$, we consider the implicit surface: $$g(x,y,z) = \ln(xy - z) = \text{constant}$$ To find the normal vector, compute the gradient of $g$: $$\nabla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z} \right)$$ 10. Compute partial derivatives: $$\frac{\partial g}{\partial x} = \frac{1}{xy - z} \cdot y = \frac{y}{xy - z}$$ $$\frac{\partial g}{\partial y} = \frac{1}{xy - z} \cdot x = \frac{x}{xy - z}$$ $$\frac{\partial g}{\partial z} = \frac{1}{xy - z} \cdot (-1) = \frac{-1}{xy - z}$$ 11. Evaluate at $x = y = 1$ and $z = f(1,1)$: Since $z = xy - e^{g} = xy - e^{\ln(xy - z)}$, but here we just use the point where $x = y = 1$ and $z$ is such that $xy - z > 0$. Assuming $z = 1$ (since $z = 1 \cdot 1 - z$ implies $z$ is consistent with the point), then: $$xy - z = 1 \cdot 1 - z = 1 - z$$ Given the problem states $x = y = 1$, but $z$ is not explicitly given, assume $z = 0$ for evaluation (or the problem is incomplete). For the sake of the problem, assume $z = 0$. Then: $$xy - z = 1 - 0 = 1$$ 12. Evaluate partial derivatives at $(1,1,0)$: $$f_x(1,1) = \frac{1}{1} = 1$$ $$f_y(1,1) = \frac{1}{1} = 1$$ $$f_z(1,1) = \frac{-1}{1} = -1$$ 13. The gradient vector (normal vector) is: $$\mathbf{n} = (1, 1, -1)$$ 14. Find the magnitude: $$|\mathbf{n}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$$ 15. The unit normal vector is: $$\hat{\mathbf{n}} = \frac{1}{\sqrt{3}} (1, 1, -1) = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right)$$