Let's proceed step by step as requested. --- ## Step 1: Geometric Nature of \( M \) (Part (a) and (b)) ### (a) Geometrically, the set \( M \) is: We are given two equations in three variables \((x, y, z)\): \[ \begin{cases} 2xz + y^4 + y^3 + y^2 = -4, \\ e^y + 3z = -2 \end{cases} \] Each equation defines a surface in \(\mathbb{R}^3\). The intersection of two surfaces in three dimensions is typically a **curve** (1-dimensional object), provided the surfaces are not tangent everywhere they intersect. **Answer:** \(\boxed{\text{a smooth curve}}\) **Explanation:** Two independent equations in \(\mathbb{R}^3\) generally define a 1-dimensional locus (a curve). --- ## Step 2: Justification and Tangent Space at the Given Point (Part (b) and (c)) ### (b) Justification Let \( F(x, y, z) = \begin{pmatrix} 2xz + y^4 + y^3 + y^2 + 4 \\ e^y + 3z + 2 \end{pmatrix} \), so \( M = F^{-1}() \). \( F: \mathbb{R}^3 \to \mathbb{R}^2 \). If the Jacobian matrix of \( F \) has rank 2 everywhere on \( M \), the **Implicit Function Theorem** says that \( M \) is a smooth manifold of dimension \( 3-2 = 1 \), i.e., a smooth curve. **Explanation:** Locally, the solution set to two independent equations in three variables is a smooth 1-dimensional manifold (curve) if the equations are independent (their gradients are linearly independent). --- ### (c) Tangent Space at the Point \(\begin{pmatrix} 2 \\ \\ -1 \end{pmatrix}\) #### 1. Compute the Jacobian of \( F \): Let \( p = (x_, y_, z_) = (2, , -1) \). - \( F_1(x, y, z) = 2xz + y^4 + y^3 + y^2 + 4 \) - \( F_2(x, y, z) = e^y + 3z + 2 \) Jacobian: \[ J = \begin{bmatrix} \frac{\partial F_1}{\partial x} & \frac{\partial F_1}{\partial y} & \frac{\partial F_1}{\partial z} \\ \frac{\partial F_2}{\partial x} & \frac{\partial F_2}{\partial y} & \frac{\partial F_2}{\partial z} \end{bmatrix} \] Calculate partial derivatives at \( (2,,-1) \): - \( \frac{\partial F_1}{\partial x} = 2z \) → \( 2 \times (-1) = -2 \) - \( \frac{\partial F_1}{\partial y} = 4y^3 + 3y^2 + 2y \) → - \( \frac{\partial F_1}{\partial z} = 2x \) → \( 2 \times 2 = 4 \) - \( \frac{\partial F_2}{\partial x} = \) - \( \frac{\partial F_2}{\partial y} = e^y \) → \( e^ = 1 \) - \( \frac{\partial F_2}{\partial z} = 3 \) So at \( (2,,-1) \): \[ J = \begin{bmatrix} -2 & & 4 \\ & 1 & 3 \end{bmatrix} \] #### 2. The tangent space is the null space of \( J \): Let \( v = (a, b, c)^T \) be a tangent vector: \[ J v = \implies \begin{cases} -2a + b + 4c = \\ a + 1b + 3c = \end{cases} \] Solve: - \( -2a + 4c = \implies a = 2c \) - \( b + 3c = \implies b = -3c \) So, tangent vectors are \[ v = (2c, -3c, c) = c \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} \] This is a 1-dimensional subspace (a line). **Explanation:** The tangent space to \( M \) at the point is the set of all vectors of the form \( \lambda \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} \), for \( \lambda \in \mathbb{R} \). --- ## **Final Answers** - **(a):** \( M \) is a **smooth curve**. - **(b):** Because the solution set of two independent smooth equations in \(\mathbb{R}^3\) is generally a 1-dimensional smooth manifold (curve), by the Implicit Function Theorem. - **(c):** The affine subspace tangent to \( M \) at \(\begin{pmatrix} 2 \\ \\ -1 \end{pmatrix}\) is \[ \left\{ \lambda \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} \,:\, \lambda \in \mathbb{R} \right\} \] or, equivalently, the line through \(\begin{pmatrix} 2 \\ \\ -1 \end{pmatrix}\) in the direction \(\begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}\).