Let's break the problem into the required two-step answers for both (a) and (b), with explanations at each step and a final answer at the end. --- ## (a) \( f(x_1, x_2, x_3) = x_1^2 + x_2^2 + x_3^2 - 1 \) ### **Step 1: Compute \( Df(P) \) and show \( Df(P) \neq \) for all \( P \in S \)** - \( S \) is the solution set to \( x_1^2 + x_2^2 + x_3^2 = 1 \), i.e., the **unit sphere** in \( \mathbb{R}^3 \). - Compute the gradient (Jacobian): \[ Df(x_1, x_2, x_3) = \left( \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \frac{\partial f}{\partial x_3} \right) = (2x_1, 2x_2, 2x_3) \] - For any point \( P = (x_1, x_2, x_3) \in S \), we have \( x_1^2 + x_2^2 + x_3^2 = 1 \), so at least one coordinate is nonzero. Thus, the gradient is never the zero vector on \( S \). **Explanation:** The Jacobian (gradient) is nonzero everywhere on the sphere because at least one coordinate is always nonzero. This satisfies the condition for the Implicit Function Theorem. --- ### **Step 2: Conclude \( S \) is a smooth 2D surface** - By the Implicit Function Theorem, since \( Df(P) \neq \) for all \( P \in S \), the solution set \( S \) is a smooth manifold of dimension \( 3-1 = 2 \) (a surface) in \( \mathbb{R}^3 \). **Explanation:** The unit sphere is a smooth 2D surface (manifold) in \( \mathbb{R}^3 \) because the conditions of the Implicit Function Theorem are met everywhere on \( S \). --- ### **Final answer for (a):** > **The unit sphere is a smooth 2-dimensional manifold in \( \mathbb{R}^3 \) because the gradient of \( f \) is nonzero everywhere on the sphere.** --- ## (b) \( f(x_1, x_2, x_3) = x_1^4 + x_2^4 + x_3^4 + 4x_1^2x_2^2 + 3 \), \( S \) is the solution set to \( f = \) ### **Step 1: Compute \( Df(P) \) and check if \( Df(P) \neq \) on \( S \)** - \( f(x_1, x_2, x_3) = x_1^4 + x_2^4 + x_3^4 + 4x_1^2x_2^2 + 3 \) - Compute the gradient: \[ \frac{\partial f}{\partial x_1} = 4x_1^3 + 8x_1x_2^2 \] \[ \frac{\partial f}{\partial x_2} = 4x_2^3 + 8x_2x_1^2 \] \[ \frac{\partial f}{\partial x_3} = 4x_3^3 \] So, \[ Df(x_1, x_2, x_3) = (4x_1^3 + 8x_1x_2^2, \; 4x_2^3 + 8x_2x_1^2, \; 4x_3^3) \] - For points on \( S \), \( f(x_1, x_2, x_3) = \implies x_1^4 + x_2^4 + x_3^4 + 4x_1^2x_2^2 = -3 \). - All terms on the left except the constant are **non-negative**, so the only way this can happen is if all terms are zero and \( 3 = -3 \), which is impossible **unless** negative values are allowed via complex numbers. - Let's check for real solutions: - All terms \( x_1^4, x_2^4, x_3^4, 4x_1^2x_2^2 \geq \) for real numbers. - Their sum \( \geq \), so the left side is always \( \geq 3 \), never \( \). - Therefore, **there are no real solutions** for \( S \) in \( \mathbb{R}^3 \). **Explanation:** The equation \( x_1^4 + x_2^4 + x_3^4 + 4x_1^2x_2^2 + 3 = \) has **no real solutions** because the sum of non-negative terms plus 3 cannot be zero. --- ### **Step 2: Attempt to identify the surface and convert to cylindrical coordinates** - Set \( x_1 = r\cos\theta \), \( x_2 = r\sin\theta \), \( x_3 = z \). - \( x_1^2 = r^2\cos^2\theta \), \( x_2^2 = r^2\sin^2\theta \). - \( x_1^4 + x_2^4 + 4x_1^2x_2^2 = (x_1^2 + x_2^2)^2 = r^4 \). - \( x_3^4 = z^4 \). - So \( f = r^4 + z^4 + 3 \). - The equation \( r^4 + z^4 + 3 = \) again cannot have real solutions, as all terms are non-negative. **Explanation:** Even in cylindrical coordinates, the equation has no real solutions. Thus, \( S = \emptyset \) in \( \mathbb{R}^3 \). --- ### **Final answer for (b):** > **The solution set \( S \) is empty in \( \mathbb{R}^3 \), so there is no smooth surface in real space defined by this equation.** --- Let me know if you'd like further clarification or if you want to consider complex solutions!