Mathematics for Data Science II · End Term · Sep 2025 · FN

**1. Analyze the conditions for $\text{rank}(A) = 1$ vs. $\text{rank}(A) = 2$:** Given the $2 \times 3$ matrix: $$A = \begin{bmatrix} 1 & -2 & 3 \\ 4 & k_1 & k_2 \end{bmatrix}$$ Since the first row $\begin{bmatrix} 1 & -2 & 3 \end{bmatrix}$ is non-zero, $\text{rank}(A)$ is either $1$ or $2$. - $\text{rank}(A) = 1$ if and only if the second row is a scalar multiple of the first row. Since the first entry of row 2 is $4 = 4 \times 1$, row 2 must equal $4 \times (\text{row 1})$: $$k_1 = 4(-2) = -8 \quad \text{and} \quad k_2 = 4(3) = 12$$ **2. Evaluate each option:** - If $\text{rank}(A) = 1$, then $k_1 = -8$ and $k_2 = 12$, so $k_1 + k_2 = 4$ (Option A says $20$) and $k_1 - k_2 = -20$ (Option B says $4$). - If $k_1 = k_2$, then we cannot have both $k_1 = -8$ and $k_2 = 12$ simultaneously (since $-8 \ne 12$). Therefore, the two rows are linearly independent for every $k_1 = k_2$, which means **$\text{rank}(A) = 2$**. **Answer: Option D** A: **Incorrect:** If $\text{rank}(A) = 1$, then $k_1 = -8$ and $k_2 = 12$, so $k_1 + k_2 = 4$, not $20$. B: **Incorrect:** If $\text{rank}(A) = 1$, then $k_1 - k_2 = -8 - 12 = -20$, not $4$. C: **Incorrect:** $\text{rank}(A) = 1$ requires $k_1 = -8$ and $k_2 = 12$, which is impossible when $k_1 = k_2$. D: **Correct:** When $k_1 = k_2$, the second row can never be $4$ times the first row (since $-8 \ne 12$), so $\text{rank}(A) = 2$.

**1. Find the matrix representation $A$ of $T(x, y) = (2x + y, 3x + y)$:** - Domain ordered basis: $\mathcal{B}_1 = \{(1, 0), (0, 1)\}$ - Co-domain ordered basis: $\mathcal{B}_2 = \{(1, 1), (-1, 1)\}$ - Evaluate $T$ on the domain basis vectors and express them in basis $\mathcal{B}_2$: - $T(1, 0) = (2, 3) = \frac{5}{2}(1, 1) + \frac{1}{2}(-1, 1)$ - $T(0, 1) = (1, 1) = 1(1, 1) + 0(-1, 1)$ - Therefore, the matrix $A$ is: $$A = \begin{bmatrix} 5/2 & 1 \\ 1/2 & 0 \end{bmatrix}$$ **2. Check statements S1 and S2 against $B = \begin{bmatrix} 5 & 2 \\ 1 & 0 \end{bmatrix}$:** - **Statement S1 ($A$ is similar to $B$):** Similar matrices must have the same trace and determinant. Here, $\text{tr}(A) = \frac{5}{2}$ while $\text{tr}(B) = 5$ (and $\det(A) = -\frac{1}{2}$ while $\det(B) = -2$). Thus, $A$ is **not** similar to $B$, so **S1 is false**. - **Statement S2 ($A$ is equivalent to $B$):** Two matrices of the same size ($2 \times 2$) are equivalent if and only if they have the same rank. Since $\det(A) = -\frac{1}{2} \ne 0$ and $\det(B) = -2 \ne 0$, both have rank $2$ (in fact, $B = 2I_2 \cdot A$). Thus, **S2 is true**. **Answer: Option C** A: **Incorrect:** S1 is false because $\text{tr}(A) = 5/2 \ne \text{tr}(B) = 5$. B: **Incorrect:** S1 is false and S2 is true. C: **Correct:** $A = \frac{1}{2}\begin{bmatrix} 5 & 2 \\ 1 & 0 \end{bmatrix}$ has the same rank (2) as $B$ so it is equivalent to $B$ (S2 is true), but it has a different trace and determinant so it is not similar to $B$ (S1 is false). D: **Incorrect:** S2 is true because any two invertible $2 \times 2$ matrices are equivalent.

**1. Analyze the relationship between continuous partial derivatives (S1) and continuity (S2):** - **S1 $\implies$ S2:** By a fundamental theorem of multivariable calculus, if the partial derivatives $f_x$ and $f_y$ of $f: \mathbb{R}^2 \to \mathbb{R}$ exist and are continuous at all points (i.e., $f$ is of class $C^1$), then $f$ is differentiable everywhere, and differentiability implies that $f$ is continuous everywhere. Thus, **S1 implies S2**. - **S2 $\not\implies$ S1:** A function can be continuous everywhere without having partial derivatives at all points. For example, $f(x, y) = |x| + |y|$ (or $\sqrt{x^2 + y^2}$) is continuous at all points in $\mathbb{R}^2$, but its partial derivatives do not exist at $(0, 0)$. Thus, **S2 does not imply S1**. **Answer: Option C** A: **Incorrect:** Continuous partial derivatives (S1) guarantee differentiability and hence continuity (S2). B: **Incorrect:** Continuity (S2) does not guarantee that partial derivatives exist (e.g., $f(x,y) = |x|$ at $x = 0$). C: **Correct:** Having continuous partial derivatives everywhere implies differentiability and continuity (S1 $\implies$ S2), but a continuous function need not have partial derivatives everywhere (S2 $\not\implies$ S1). D: **Incorrect:** This reverses the direction of implication.

**1. Evaluate each statement for the $m \times n$ linear system $Ax = b$:** - **Option A:** If $A$ is $n \times n$ and $Ax = b$ has a unique solution, then $A$ is invertible ($\det(A) \ne 0$). Since $A \cdot \text{adj}(A) = \det(A)I$, $\text{adj}(A) = \det(A)A^{-1}$ is also invertible, so $\text{adj}(A)x = b$ has a unique solution. (**Correct**) - **Option B:** The product $Ax$ is a linear combination of the columns of $A$. Therefore, $Ax = b$ has a solution if and only if $b$ belongs to the column space of $A$. (**Correct**) - **Option C:** If $\text{rank}(A) = \min(m, n)$, $Ax = b$ does not necessarily have a unique solution: if $m < n$, there are free variables (infinitely many solutions), and if $m > n$, the system may be inconsistent for some $b$. (**Incorrect**) - **Option D:** If $\text{rank}(A) = m$, the column space of $A$ has dimension $m$ and therefore equals all of $\mathbb{R}^m$. Hence, for every $b \in \mathbb{R}^m$, there exists $x \in \mathbb{R}^n$ such that $Ax = b$. (**Correct**) - **Option E:** If the $n$ columns of $A$ are linearly independent in $\mathbb{R}^m$, then $\text{rank}(A) = n \le m$, so $n \le m$, not $n > m$. (**Incorrect**) **Answer: Options A, B, D** A: **Correct:** If $Ax = b$ has a unique solution, $\det(A) \ne 0$, which implies $\text{adj}(A)$ is invertible and $\text{adj}(A)x = b$ has a solution. B: **Correct:** By definition, $Ax = b$ is consistent if and only if $b \in \text{Column space}(A)$. C: **Incorrect:** Full rank $\min(m, n)$ does not guarantee uniqueness when $m < n$ or consistency when $m > n$. D: **Correct:** When $\text{rank}(A) = m$, the columns of $A$ span $\mathbb{R}^m$, so $Ax = b$ has a solution for every $b$. E: **Incorrect:** Linearly independent columns in $\mathbb{R}^m$ require $n \le m$, not $n > m$.

**1. Determine the rank of the $3 \times 3$ matrix $A$:** - Since $A$ is a non-zero matrix, $\text{rank}(A) \ge 1$. - Since $A^2 = 0$, the column space (image) of $A$ is contained in the null space (kernel) of $A$, which gives: $$\text{rank}(A) \le \text{nullity}(A)$$ - By the Rank-Nullity Theorem for a $3 \times 3$ matrix, $\text{rank}(A) + \text{nullity}(A) = 3$, so: $$2\,\text{rank}(A) \le 3 \implies \text{rank}(A) \le 1.5$$ - Combining $1 \le \text{rank}(A) \le 1.5$ for an integer rank gives $\text{rank}(A) = 1$. **2. Use the invariance of rank under similarity and equivalence:** - Since $A$ is similar to $B$, $\text{rank}(B) = \text{rank}(A) = 1$. - Since $B$ is equivalent to $C$, $\text{rank}(C) = \text{rank}(B) = 1$. **Answer: 1**

**1. Check the basis vectors of the subspace $W$:** Let $u_1 = (1, 0, 1)$ and $u_2 = (1, 4, -1)$. Compute their dot product: $$u_1 \cdot u_2 = 1(1) + 0(4) + 1(-1) = 0$$ Thus, $\{u_1, u_2\}$ is already an orthogonal basis for the plane $W$! **2. Find the orthogonal projection of $P = (0, 3, 0)$ onto $W$ (or onto $W^\perp$):** A normal vector to the plane $W$ is given by the cross product: $$n = u_1 \times u_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ 1 & 4 & -1 \end{vmatrix} = (-4, 2, 4) = 2(-2, 1, 2)$$ Let $w_\perp = (-2, 1, 2)$, which has length $\|w_\perp\| = \sqrt{(-2)^2 + 1^2 + 2^2} = \sqrt{9} = 3$. The minimum distance from $P = (0, 3, 0)$ to the subspace $W$ (which passes through the origin) is the length of the projection of $P$ onto the unit normal vector: $$d(P, W) = \frac{|P \cdot w_\perp|}{\|w_\perp\|} = \frac{|0(-2) + 3(1) + 0(2)|}{3} = \frac{3}{3} = 1$$ **Answer: 1**

**1. Use the continuity of the outer function $g(t) = t^2$:** We are given that: $$\lim_{(x,y) \to (0,0)} f(x, y) = 3$$ Since $g(t) = t^2$ is a polynomial, it is continuous at $t = 3$ (and everywhere on $\mathbb{R}$). **2. Evaluate the limit of the composite function:** By the limit law for composite functions: $$\lim_{(x,y) \to (0,0)} (g \circ f)(x, y) = g\left(\lim_{(x,y) \to (0,0)} f(x, y)\right) = g(3) = 3^2 = 9$$ **Answer: 9**

**1. Apply the orthogonality conditions $A^T A = I$:** For the matrix $A = \begin{bmatrix} \frac{1}{\sqrt{2}} & b \\ c & -\frac{1}{\sqrt{2}} \end{bmatrix}$ to be orthogonal, its columns must be orthonormal: 1. Unit length of column 1: $$\left(\frac{1}{\sqrt{2}}\right)^2 + c^2 = 1 \implies \frac{1}{2} + c^2 = 1 \implies c^2 = \frac{1}{2} \implies c = \pm \frac{1}{\sqrt{2}}$$ 2. Unit length of column 2: $$b^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 = 1 \implies b^2 + \frac{1}{2} = 1 \implies b^2 = \frac{1}{2} \implies b = \pm \frac{1}{\sqrt{2}}$$ 3. Orthogonality of columns 1 and 2: $$\frac{1}{\sqrt{2}}b + c\left(-\frac{1}{\sqrt{2}}\right) = 0 \implies \frac{1}{\sqrt{2}}(b - c) = 0 \implies b = c$$ **2. Count the valid pairs $(b, c)$:** Since $b = c$ and $b \in \left\{\frac{1}{\sqrt{2}},\; -\frac{1}{\sqrt{2}}\right\}$, there are exactly **$2$** such orthogonal matrices in $S$: $$(b, c) = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \quad \text{and} \quad (b, c) = \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$$ **Answer: 2**

**1. Find the dimension of $W$ by row-reducing the matrix of spanning vectors:** Form the $4 \times 4$ matrix whose rows are the four spanning vectors of $W$: $$M = \begin{bmatrix} 2 & 0 & 2 & 0 \\ 0 & -3 & -3 & 0 \\ 1 & 1 & 2 & 1 \\ 4 & 0 & 4 & -3 \end{bmatrix}$$ Notice that for every row $(x_1, x_2, x_3, x_4)$, the third coordinate equals the sum of the first two coordinates ($x_3 = x_1 + x_2$): - Row 1: $2 + 0 = 2$ - Row 2: $0 + (-3) = -3$ - Row 3: $1 + 1 = 2$ - Row 4: $4 + 0 = 4$ Thus, the columns are linearly dependent ( specifically $C_3 = C_1 + C_2$), so $\text{rank}(M) \le 3$. Furthermore, the first three rows are clearly linearly independent (or looking at columns 1, 2, 4 of rows 1, 2, 4: $\det\begin{bmatrix} 2 & 0 & 0 \\ 0 & -3 & 0 \\ 4 & 0 & -3 \end{bmatrix} = 18 \ne 0$), so $\dim(W) = \text{rank}(M) = 3$. **2. Compute $\dim(W^\perp)$:** In $\mathbb{R}^4$, the dimension of the orthogonal complement $W^\perp$ is: $$\dim(W^\perp) = 4 - \dim(W) = 4 - 3 = 1$$ (Indeed, $W^\perp = \text{span}\{(1, 1, -1, 0)\}$.) **Answer: 1**

**1. Check the level set $L_{z_0}(f) = \{(x, y) \in D \mid f(x, y) = z_0\}$ for each option:** - **Option A:** $f(x, y) = e^{x+y},\; z_0 = -1$. Since $e^{x+y} > 0$ for all $(x, y) \in \mathbb{R}^2$, the equation $e^{x+y} = -1$ has no solutions ($L_{-1}(f) = \emptyset$), which is not a line. - **Option B:** $f(x, y) = y - 2x,\; z_0 = -3$. The level set is $\{(x, y) \in \mathbb{R}^2 \mid y - 2x = -3\}$, i.e., $y = 2x - 3$, which is a **straight line** in $\mathbb{R}^2$. (**Correct**) - **Option C:** $f(x, y) = \frac{1}{x - y},\; z_0 = -1$ (defined on $D = \{(x, y) \in \mathbb{R}^2 \mid x \ne y\}$). Setting $\frac{1}{x - y} = -1$ gives $x - y = -1$, or $y = x + 1$. Every point on the line $y = x + 1$ satisfies $x \ne y$, so the entire **straight line** $y = x + 1$ is the level set. (**Correct**) - **Option D:** $f(x, y) = \sin(2x - 3y),\; z_0 = 3$. Since $|\sin(\theta)| \le 1$ for all real $\theta$, $\sin(2x - 3y) = 3$ has no solutions ($L_3(f) = \emptyset$). **Answer: Options B, C** A: **Incorrect:** $e^{x+y}$ is strictly positive, so $L_{-1}(f) = \emptyset$ (empty set). B: **Correct:** $y - 2x = -3$ defines the straight line $y = 2x - 3$ in $\mathbb{R}^2$. C: **Correct:** $\frac{1}{x-y} = -1 \iff x - y = -1$, which defines the straight line $y = x + 1$ in $\mathbb{R}^2$. D: **Incorrect:** $\sin(2x - 3y)$ lies in $[-1, 1]$, so $L_3(f) = \emptyset$.

**1. Analyze the definition of injectivity and level sets:** - By definition of the image $\text{Image}(f)$, for every $z_0 \in \text{Image}(f)$, the level set $L_{z_0}(f) = \{(x, y) \in D \mid f(x, y) = z_0\}$ is non-empty (it contains at least one element). - **S1 $\implies$ S2:** If $f$ is injective (one-to-one), no two distinct points in $D$ can map to the same $z_0$. Thus, for every $z_0 \in \text{Image}(f)$, $L_{z_0}(f)$ contains **at most** one element, and since $z_0 \in \text{Image}(f)$, it contains **exactly one** element. - **S2 $\implies$ S1:** Conversely, if every $z_0 \in \text{Image}(f)$ has a unique pre-image in $D$, then $f(p_1) = f(p_2) = z_0$ implies $p_1 = p_2$, so $f$ is injective. Therefore, **S1 holds if and only if S2 holds**. **Answer: Option B** A: **Incorrect:** S1 and S2 are logically equivalent definitions of a one-to-one function. B: **Correct:** A function is injective if and only if the pre-image (level set) of every element in its image is a singleton set. C: **Incorrect:** S2 also implies S1. D: **Incorrect:** S1 also implies S2.

**1. Find the matrix of $T\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{bmatrix} a + b \\ c + d \end{bmatrix}$ with respect to $\mathcal{B}_1$ and the standard basis of $\mathbb{R}^2$:** Evaluate $T$ on each matrix in $\mathcal{B}_1$: - $T\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ - $T\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ - $T\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$ - $T\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$ Putting these as columns gives $\begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}$. (**Option A is correct**, Option C is incorrect.) **2. Find the matrix of $T$ with respect to $\mathcal{B}_2$ and the standard basis of $\mathbb{R}^2$:** Evaluate $T$ on each matrix in $\mathcal{B}_2$: - $T\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ - $T\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$ - $T\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ - $T\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$ Putting these as columns gives $\begin{bmatrix} 1 & -1 & 1 & -1 \\ 1 & 1 & 1 & 1 \end{bmatrix}$. (**Option B is correct**, Option D is incorrect.) **Answer: Options A, B** A: **Correct:** The images of the four standard basis matrices in $\mathcal{B}_1$ are $(1,0), (1,0), (0,1), (0,1)$. B: **Correct:** The images of the four matrices in $\mathcal{B}_2$ are $(1,1), (-1,1), (1,1), (-1,1)$. C: **Incorrect:** Swaps the first and second rows of the matrix representation. D: **Incorrect:** The signs in the first row are $1, -1, 1, -1$, not $-1, 1, -1, 1$.

**1. Analyze $S: \mathbb{R}^2 \to M_{2\times 2}$ and $T: M_{2\times 2} \to \mathbb{R}^2$:** Given $S(a, b) = \begin{bmatrix} a & a \\ b & b \end{bmatrix}$ and $T\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{bmatrix} a + b \\ c + d \end{bmatrix}$: - **Check Statement 1 ($S \circ T$ is an isomorphism):** $S \circ T: M_{2\times 2} \to M_{2\times 2}$ maps a 4-dimensional space through a 2-dimensional space $\mathbb{R}^2$, so $\text{rank}(S \circ T) \le 2 < 4$. Thus $S \circ T$ is not an isomorphism. (**Statement 1 is false**) - **Check Statement 2 ($T \circ S$ is an isomorphism):** For any $(a, b) \in \mathbb{R}^2$: $$(T \circ S)(a, b) = T\begin{pmatrix} a & a \\ b & b \end{pmatrix} = \begin{bmatrix} a + a \\ b + b \end{bmatrix} = \begin{bmatrix} 2a \\ 2b \end{bmatrix} = 2I_{\mathbb{R}^2}\begin{bmatrix} a \\ b \end{bmatrix}$$ Since $2I_{\mathbb{R}^2}$ is invertible, $T \circ S: \mathbb{R}^2 \to \mathbb{R}^2$ is a bijective linear transformation, i.e., an isomorphism! (**Statement 2 is true**) - **Check Statement 3 ($S \circ T$ is injective):** Since $\begin{pmatrix} 1 & -1 \\ 0 & 0 \end{pmatrix} \ne 0$ is in $\ker(T)$, it is also in $\ker(S \circ T)$, so $S \circ T$ is not injective. (**Statement 3 is false**) **2. Conclude:** Only **1** statement (Statement 2) is correct, so $n = 1$. **Answer: 1**

**1. Compute the partial derivatives of $f(x, y) = \frac{1}{1 + x^2 + y^2}$:** $$f_x(x, y) = \frac{-2x}{(1 + x^2 + y^2)^2}, \qquad f_y(x, y) = \frac{-2y}{(1 + x^2 + y^2)^2}$$ The tangent line $L_{(x_0, y_0)}(u_1, u_2)$ to the graph of $f$ at $(x_0, y_0, f(x_0, y_0))$ in direction $(u_1, u_2)$ is parameterized by: $$\mathbf{r}(t) = \big(x_0 + t u_1,\; y_0 + t u_2,\; f(x_0, y_0) + t(u_1 f_x(x_0, y_0) + u_2 f_y(x_0, y_0))\big), \quad t \in \mathbb{R}$$ **2. Check Options A and B for $L_{(1, 1)}(1, 0)$:** - At $(x_0, y_0) = (1, 1)$: $f(1, 1) = \frac{1}{1 + 1 + 1} = \frac{1}{3}$ and $f_x(1, 1) = \frac{-2}{3^2} = -\frac{2}{9}$. - Along direction $(u_1, u_2) = (1, 0)$: $$\mathbf{r}(t) = \left(1 + t,\; 1,\; \frac{1}{3} - \frac{2}{9}t\right)$$ - Setting $1 + t = 4 \implies t = 3$ gives $z = \frac{1}{3} - \frac{2}{9}(3) = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}$. Thus, **$(4, 1, -\frac{1}{3})$ lies on $L_{(1, 1)}(1, 0)$** (**Option B is correct**, Option A is incorrect). **3. Check Options C and D for $L_{(1, -1)}(0, 1)$:** - At $(x_0, y_0) = (1, -1)$: $f(1, -1) = \frac{1}{3}$ and $f_y(1, -1) = \frac{-2(-1)}{3^2} = \frac{2}{9}$. - Along direction $(u_1, u_2) = (0, 1)$: $$\mathbf{r}(t) = \left(1,\; -1 + t,\; \frac{1}{3} + \frac{2}{9}t\right)$$ - For $t = 3$: $\mathbf{r}(3) = \left(1,\; 2,\; \frac{1}{3} + \frac{6}{9}\right) = (1, 2, 1)$ (**Option C is correct**). - For $y = -2 \implies t = -1$: $z = \frac{1}{3} - \frac{2}{9} = \frac{1}{9} \ne 1$ (**Option D is incorrect**). **4. Check Option E:** The direction vector of the line $L_{(x, y)}(1, 0)$ is $(1, 0, f_x(x, y))$. Therefore, the difference $v_1 - v_2$ of any two points on this line is a scalar multiple of $(1, 0, f_x(x, y))$, i.e., lies in $\text{span}\{(1, 0, f_x(x, y))\}$ (**Option E is correct**). **Answer: Options B, C, E** A: **Incorrect:** At $x = 4$ ($t = 3$), the $z$-coordinate on $L_{(1,1)}(1,0)$ is $1/3 - 2/3 = -1/3$, not $1/3$. B: **Correct:** Setting $t = 3$ in $(1 + t, 1, 1/3 - 2t/9)$ gives the point $(4, 1, -1/3)$. C: **Correct:** Setting $t = 3$ in $(1, -1 + t, 1/3 + 2t/9)$ gives the point $(1, 2, 1)$. D: **Incorrect:** Setting $y = -2$ ($t = -1$) gives $z = 1/3 - 2/9 = 1/9 \ne 1$. E: **Correct:** The tangent line $L_{(x,y)}(1,0)$ has direction vector $(1, 0, f_x(x,y))$, so $v_1 - v_2 \in \text{span}\{(1, 0, f_x(x,y))\}$.

**1. Find the normal vector $v_n$ to the tangent plane at $A_n = \left(\frac{1}{n}, \frac{1}{n}\right)$:** The tangent plane to $z = f(x, y)$ at $(x_n, y_n)$ is given by $f_x(x_n, y_n)(x - x_n) + f_y(x_n, y_n)(y - y_n) - (z - z_n) = 0$, so the unique normal vector with third component $-1$ is: $$v_n = \big(f_x(A_n),\; f_y(A_n),\; -1\big)$$ **2. Take the limit as $n \to \infty$:** As $n \to \infty$, $A_n = \left(\frac{1}{n}, \frac{1}{n}\right) \to (0, 0)$. Since $f_x(x, y) = \frac{-2x}{(1 + x^2 + y^2)^2}$ and $f_y(x, y) = \frac{-2y}{(1 + x^2 + y^2)^2}$ are continuous, we have: $$\lim_{n \to \infty} f_x(A_n) = f_x(0, 0) = 0, \qquad \lim_{n \to \infty} f_y(A_n) = f_y(0, 0) = 0$$ Therefore: $$\lim_{n \to \infty} v_n = (0, 0, -1)$$ Since $v_n$ converges to $(0, 0, -1)$, the angle $\theta_n$ (in degrees) between $v_n$ and $(0, 0, -1)$ satisfies: $$\lim_{n \to \infty} \theta_n = 0$$ **Answer: 0**

**1. Determine the maximum and minimum values of $f(x, y) = \cos(x^2 + y^2)$ on $D = \mathbb{R}^2$:** - As $(x, y)$ ranges over all of $\mathbb{R}^2$, the quantity $r^2 = x^2 + y^2$ takes all non-negative real values $[0, \infty)$. - Since $\cos(t)$ on $[0, \infty)$ attains its maximum value of $1$ (e.g., at $t = 0$, i.e., $(x, y) = (0, 0)$) and its minimum value of $-1$ (e.g., at $t = \pi$, i.e., $x^2 + y^2 = \pi$), we have: $$M_b = 1, \qquad m_b = -1$$ **2. Compute $M_b - m_b$:** $$M_b - m_b = 1 - (-1) = 2$$ **Answer: 2**

**1. Find all critical points of $f(x, y) = \cos(x^2 + y^2)$:** Compute the gradient $\nabla f(x, y)$: $$f_x(x, y) = -2x\sin(x^2 + y^2) = 0$$ $$f_y(x, y) = -2y\sin(x^2 + y^2) = 0$$ A point $(x, y)$ is a critical point if and only if either: 1. $(x, y) = (0, 0)$, or 2. $\sin(x^2 + y^2) = 0 \iff x^2 + y^2 = k\pi$ for some positive integer $k = 1, 2, 3, \dots$. **2. Determine the maximum value of $R > 0$ for a unique critical point in $D = \{(x, y) \mid x^2 + y^2 < R\}$:** - The origin $(0, 0)$ (where $x^2 + y^2 = 0 < R$) is always in $D$ for any $R > 0$. - The next closest critical points lie on the circle $x^2 + y^2 = \pi$. - For $D = \{(x, y) \mid x^2 + y^2 < R\}$ to contain **only** $(0, 0)$ and none of the points on $x^2 + y^2 = \pi$, we need $R \le \pi$. - Thus, the maximum value of $R$ is $\pi \approx 3.14$. **Answer: 3.14**

**1. Compute the second partial derivatives of $f(x, y) = \cos(x^2 + y^2)$:** - $f_{xx}(x, y) = -2\sin(x^2 + y^2) - 4x^2\cos(x^2 + y^2)$ - $f_{yy}(x, y) = -2\sin(x^2 + y^2) - 4y^2\cos(x^2 + y^2)$ - $f_{xy}(x, y) = -4xy\cos(x^2 + y^2)$ **2. Evaluate $\det(H_f(x, y))$ at every critical point of $f$:** - **At $(0, 0)$:** $\sin(0) = 0$ and $x = y = 0$, so $f_{xx}(0, 0) = f_{yy}(0, 0) = f_{xy}(0, 0) = 0$, which gives: $$\det(H_f(0, 0)) = 0$$ - **At any critical point with $x^2 + y^2 = k\pi$ ($k \in \mathbb{Z}^+$):** $\sin(k\pi) = 0$, so: $$f_{xx} = -4x^2\cos(k\pi), \quad f_{yy} = -4y^2\cos(k\pi), \quad f_{xy} = -4xy\cos(k\pi)$$ Therefore: $$\det(H_f(x, y)) = f_{xx}f_{yy} - (f_{xy})^2 = 16x^2y^2\cos^2(k\pi) - 16x^2y^2\cos^2(k\pi) = 0$$ Since $\det(H_f(x, y)) = 0$ at every critical point, the set of all Hessian determinants at the critical points is $\{0\}$, and its minimum value is **$0$**. **Answer: 0**