Let's solve **Problem 1(a)** step by step: --- ## 1(a) Find and sketch the domain of \( F(x, y) = 1 + \sqrt{4 - y^2} \). ### **Step 1: Identify restrictions on the function** - The function \( F(x, y) \) involves a square root: \( \sqrt{4 - y^2} \). - The expression inside the square root must be **non-negative** (since we can't take the square root of a negative number in real numbers): \[ 4 - y^2 \geq 0 \] **Explanation:** We require \( 4 - y^2 \geq 0 \) for real values of the function. --- ### **Step 2: Solve the inequality and describe the domain** - Rearranging the inequality: \[ 4 \geq y^2 \] \[ -2 \leq y \leq 2 \] - There is **no restriction on \( x \)** (since \( x \) does not appear in the square root or anywhere else in the function). **Explanation:** The domain is all \( (x, y) \) such that \( y \) is between \(-2\) and \(2\) (inclusive), and \( x \) can be any real number. --- ### **Final Answer:** The domain of \( F(x, y) \) is: \[ \boxed{\{(x, y) \in \mathbb{R}^2 \mid -2 \leq y \leq 2\}} \] or in interval notation: \[ x \in (-\infty, \infty), \quad y \in [-2, 2] \] This region is a horizontal strip in the \(xy\)-plane, with no restriction on \(x\) and \(y\) between \(-2\) and \(2\). --- If you would like the answer for part (b) or another problem, let me know!

## 1(b) Find the range of \( F(x, y) = 1 + \sqrt{4 - y^2} \). ### **Step 1: Determine the maximum and minimum values of \( F \)** - From the domain established in 1(a), \( y \) ranges from \(-2\) to \(2\). - The expression \( \sqrt{4 - y^2} \) achieves its maximum when \( y^2 \) is minimized (i.e., when \( y = 0 \)): \[ \sqrt{4 - 0^2} = \sqrt{4} = 2 \] - Therefore, the minimum occurs when \( y^2 \) is maximized (i.e., when \( y = \pm 2 \)): \[ \sqrt{4 - 2^2} = \sqrt{0} = 0 \] **Explanation:** Thus, the minimum value of \( F \) is \( 1 + 0 = 1 \) and the maximum value is \( 1 + 2 = 3 \). --- ### **Step 2: Describe the range of \( F \)** - Since \( F(x, y) \) is continuous and takes all values between its minimum and maximum: \[ 1 \leq F(x, y) \leq 3 \] **Explanation:** Thus, the range of \( F \) is all values between \(1\) and \(3\), inclusive. --- ### **Final Answer:** The range of \( F(x, y) \) is: \[ \boxed{[1, 3]} \] This indicates that the function takes on all values from \(1\) to \(3\) as \(y\) varies between \(-2\) and \(2\).

## Problem 2 Let \( G(x, y) = \frac{1}{x^2 + y^2} \). ### **2(a) Find and sketch the domain of the function** ### **Step 1: Identify restrictions on the function** - The function \( G(x, y) \) has a denominator of \( x^2 + y^2 \). - This expression cannot be zero, as division by zero is undefined: \[ x^2 + y^2 \neq 0 \] **Explanation:** The only point where \( x^2 + y^2 = 0 \) is at the origin \( (0, 0) \). --- ### **Step 2: Describe the domain of \( G \)** - The domain includes all points \( (x, y) \) in the plane except for the origin: \[ \text{Domain} = \mathbb{R}^2 \setminus \{(0, 0)\} \] **Explanation:** This means \( G(x, y) \) is defined for all real values of \( x \) and \( y \), except at the point where both are zero. --- ### **Final Answer for 2(a):** The domain of \( G(x, y) \) is: \[ \boxed{\mathbb{R}^2 \setminus \{(0, 0)\}} \] This indicates that the function is defined everywhere in the plane except at the origin. --- ### **2(b) Sketch the domain** - The domain can be visualized as the entire \( xy \)-plane with the origin removed, represented by a punctured plane. - You can sketch this by drawing the entire Cartesian plane and marking the origin with a small circle or dot to indicate it is excluded. **Explanation:** The domain is visually a full plane minus the single point at the origin. --- If you have further questions or need assistance with another part, let me know!

## Problem 2(b) Find and sketch the domain of the function \( G(x,y) = \frac{1}{x^2 + y^2} \) ### **Step 1: Identify restrictions on the function** - The function \( G(x, y) \) has a denominator of \( x^2 + y^2 \). - This expression cannot be zero, leading to the restriction: \[ x^2 + y^2 \neq 0 \] **Explanation:** The only point where \( x^2 + y^2 = 0 \) occurs is at the origin \( (0, 0) \). --- ### **Step 2: Describe the domain of \( G \)** - The function is defined for all points \( (x, y) \) in the plane except at the origin. - Therefore, the domain can be expressed as: \[ \text{Domain} = \mathbb{R}^2 \setminus \{(0, 0)\} \] **Explanation:** This means \( G(x, y) \) is defined for all real values of \( x \) and \( y \), except where both are zero. --- ### **Final Answer for 2(b):** The domain of \( G(x, y) \) is: \[ \boxed{\mathbb{R}^2 \setminus \{(0, 0)\}} \] ### **Sketch of the Domain:** - The domain can be visualized as the entire \( xy \)-plane with the origin removed. - To sketch: - Draw the Cartesian plane. - Mark the origin \( (0, 0) \) with an open circle (or a small dot) to indicate it is excluded. **Explanation:** This visual representation shows that the function is defined everywhere in the plane except at the origin. --- If you have more questions or need further assistance with other problems, feel free to ask!

## Problem 3: Textbook 9.1 #27 ### **Step 1: Understand the problem** - Without specifics from the textbook, we need to identify what the problem is asking. - Typically, problems in this section may involve evaluating a function, finding derivatives, or analyzing a surface. **Explanation:** Review the problem statement thoroughly to clarify what is being asked. If it involves a specific function or graph, note the relevant equations or properties. --- ### **Step 2: Solve the problem** 1. **Identify the function or expression involved.** 2. **Determine what needs to be calculated** (e.g., limits, derivatives, critical points, etc.). 3. **Apply appropriate mathematical techniques** based on the problem's requirements. **Explanation:** Follow standard procedures for the type of problem presented. For example, if the problem involves finding a derivative: - Use differentiation rules (product, quotient, chain rules as applicable). - Evaluate any limits if required. --- ### **Final Answer:** - As I don’t have the specific details of the question from the textbook, ensure to follow the problem's requirements and apply the right mathematical methods to arrive at the solution. - If you provide the specifics of the problem, I can assist further in detailing the solution process and final answer. --- If you have more details or need assistance with another problem, let me know!

## Problem 4: Contours and Contour Maps ### **4(a) Function: \( f(x, y) = 2x - y \)** #### **Step 1: Find the equations of contours** To find contours for a specific value \( z_0 \), set: \[ 2x - y = z_0 \] For three values of \( z_0 \), we can use \( z_0 = 0, 1, 2 \). 1. **For \( z_0 = 0 \):** \[ 2x - y = 0 \quad \Rightarrow \quad y = 2x \] 2. **For \( z_0 = 1 \):** \[ 2x - y = 1 \quad \Rightarrow \quad y = 2x - 1 \] 3. **For \( z_0 = 2 \):** \[ 2x - y = 2 \quad \Rightarrow \quad y = 2x - 2 \] **Explanation:** These equations represent straight lines in the \( xy \)-plane, with varying intercepts. #### **Step 2: Describe the contours** - The contours are parallel lines with a slope of 2. - As \( z \) increases, the lines shift downwards (the y-intercept decreases). - This indicates a linear relationship; as \( x \) increases, \( y \) changes linearly. --- ### **4(b) Function: \( g(x, y) = y - x^2 \)** #### **Step 1: Find the equations of contours** Set: \[ y - x^2 = z_0 \] For three values of \( z_0 \), let's choose \( z_0 = 0, 1, 2 \). 1. **For \( z_0 = 0 \):** \[ y - x^2 = 0 \quad \Rightarrow \quad y = x^2 \] 2. **For \( z_0 = 1 \):** \[ y - x^2 = 1 \quad \Rightarrow \quad y = x^2 + 1 \] 3. **For \( z_0 = 2 \):** \[ y - x^2 = 2 \quad \Rightarrow \quad y = x^2 + 2 \] **Explanation:** These equations represent parabolas that open upwards. As \( z \) increases, the entire parabola shifts upward. #### **Step 2: Describe the contours** - The contours are parabolas with vertex points at \( (0, 0), (0, 1), (0, 2) \). - This shows that as \( z \) increases, the height of the parabolas increases, indicating a concave upward shape. --- ### **4(c) Function: \( h(x, y) = xy \)** #### **Step 1: Find the equations of contours** Set: \[ xy = z_0 \] For three values of \( z_0 \), choose \( z_0 = 1, 2, -1 \). 1. **For \( z_0 = 1 \):** \[ xy = 1 \quad \Rightarrow \quad y = \frac{1}{x} \] 2. **For \( z_0 = 2 \):** \[ xy = 2 \quad \Rightarrow \quad y = \frac{2}{x} \] 3. **For \( z_0 = -1 \):** \[ xy = -1 \quad \Rightarrow \quad y = \frac{-1}{x} \] **Explanation:** These equations represent hyperbolas. For positive \( z_0 \), the hyperbolas are in the first and third quadrants. For negative \( z_0 \), they are in the second and fourth quadrants. #### **Step 2: Describe the contours** - The contours for positive \( z_0 \) are hyperbolas that approach the axes but never touch them. - The contours for negative \( z_0 \) show a similar shape but in opposite quadrants. - This indicates that the product \( xy \) is constant, suggesting a relationship where increasing \( x \) results in decreasing \( y \) in the hyperbolic regions. --- ### **Final Notes:** - For contour maps, plot the obtained curves on the \( xy \)-plane corresponding to each \( z_0 \). - The visual representation helps in understanding how the function behaves in three dimensions based on the contours derived. If you need further assistance or specific visualizations, let me know!

## Final Answers for Problem 4 ### **4(a) Function: \( f(x, y) = 2x - y \)** - **Contours:** - For \( z_0 = 0 \): \( y = 2x \) - For \( z_0 = 1 \): \( y = 2x - 1 \) - For \( z_0 = 2 \): \( y = 2x - 2 \) - **Description:** The contours are straight lines with a slope of 2, indicating a linear relationship. As \( z \) increases, the lines shift downward. --- ### **4(b) Function: \( g(x, y) = y - x^2 \)** - **Contours:** - For \( z_0 = 0 \): \( y = x^2 \) - For \( z_0 = 1 \): \( y = x^2 + 1 \) - For \( z_0 = 2 \): \( y = x^2 + 2 \) - **Description:** The contours are parabolas that open upwards. As \( z \) increases, the parabolas shift upward, indicating a concave upward shape. --- ### **4(c) Function: \( h(x, y) = xy \)** - **Contours:** - For \( z_0 = 1 \): \( y = \frac{1}{x} \) - For \( z_0 = 2 \): \( y = \frac{2}{x} \) - For \( z_0 = -1 \): \( y = \frac{-1}{x} \) - **Description:** The contours are hyperbolas. For positive \( z_0 \), they lie in the first and third quadrants, while for negative \( z_0 \), they are in the second and fourth quadrants. This indicates the inverse relationship between \( x \) and \( y \). --- ### **Visual Representation:** - **Contour Map Sketches**: Create a visual representation for each function based on the derived equations, plotting them on the \( xy \)-plane. Use appropriate tools or graphing software to visualize the contours effectively. If you require any more information or further assistance on this topic, feel free to ask!

## Problem 5: Function \( z = \sqrt{x^2 + y^2} \) ### **5(a) Find the contours for \( z = 1, 2, \) and \( 3 \)** #### **Step 1: Set up the contour equations** To find the contours for specific values of \( z \), we set up the equations: 1. **For \( z = 1 \)**: \[ \sqrt{x^2 + y^2} = 1 \quad \Rightarrow \quad x^2 + y^2 = 1 \] - This represents a circle of radius 1 centered at the origin. 2. **For \( z = 2 \)**: \[ \sqrt{x^2 + y^2} = 2 \quad \Rightarrow \quad x^2 + y^2 = 4 \] - This represents a circle of radius 2 centered at the origin. 3. **For \( z = 3 \)**: \[ \sqrt{x^2 + y^2} = 3 \quad \Rightarrow \quad x^2 + y^2 = 9 \] - This represents a circle of radius 3 centered at the origin. **Explanation:** Each contour at a specific \( z \) value forms a circle in the \( xy \)-plane, with the radius equal to the value of \( z \). --- #### **Step 2: Contour Map** - The contour map consists of three concentric circles, centered at the origin: - Circle of radius 1 for \( z = 1 \) - Circle of radius 2 for \( z = 2 \) - Circle of radius 3 for \( z = 3 \) **Sketch:** Draw three circles with radii 1, 2, and 3, all centered at the origin. --- ### **5(b) Visualize the plot of the full surface** #### **Step 1: Understanding the Surface** - The equation \( z = \sqrt{x^2 + y^2} \) represents a cone opening upwards. - It is symmetric around the z-axis, extending infinitely as \( z \) increases. #### **Step 2: Using Desmos 3-D** 1. **Plot the surface \( z = \sqrt{x^2 + y^2} \)** in Desmos 3-D or similar graphing software. 2. **Trace the levels for \( z = 1, 2, \) and \( 3 \)**: - Locate the circles on the \( xy \)-plane corresponding to each \( z \) value. - The traces should appear as circular cross-sections of the cone at heights \( z = 1, 2, \) and \( 3 \). **Sketch:** 1. **Draw the cone**: Visualize it as a 3D shape extending from the origin and pointing upwards. 2. **Indicate the traces**: Mark where the circles at each \( z \) level intersect the cone. --- ### **Final Answer:** - The contours for the function \( z = \sqrt{x^2 + y^2} \) at levels \( z = 1, 2, \) and \( 3 \) are circles of radii 1, 2, and 3, respectively. - The surface represented by the function is a cone opening upwards from the origin. If you need any further clarification or assistance, feel free to ask!

## Final Answers for Problem 5 ### **5(a) Contours for \( z = \sqrt{x^2 + y^2} \)** - **Contours:** 1. **For \( z = 1 \)**: - Equation: \( x^2 + y^2 = 1 \) - Shape: Circle of radius 1 centered at the origin. 2. **For \( z = 2 \)**: - Equation: \( x^2 + y^2 = 4 \) - Shape: Circle of radius 2 centered at the origin. 3. **For \( z = 3 \)**: - Equation: \( x^2 + y^2 = 9 \) - Shape: Circle of radius 3 centered at the origin. ### **Contour Map Description:** - The contour map consists of three concentric circles: - **Circle 1**: Radius 1 (for \( z = 1 \)) - **Circle 2**: Radius 2 (for \( z = 2 \)) - **Circle 3**: Radius 3 (for \( z = 3 \)) ### **5(b) Visualization of the Surface** - **Surface Description**: - The function \( z = \sqrt{x^2 + y^2} \) represents a **circular cone** that opens upwards from the origin. - **Using Desmos 3-D**: 1. **Plot the surface** in Desmos or similar graphing software. 2. **Sketch the traces** for \( z = 1, 2, \) and \( 3 \): - At each height \( z \), the intersection with the cone is a circle of the respective radius. ### **3D Visualization Sketch:** - **Cone**: A 3D shape extending upwards from the origin. - **Traces**: Indicate the circular cross-sections at \( z = 1, 2, \) and \( 3 \). --- This concludes the answers for Problem 5. If you have further questions or need additional assistance, feel free to ask!