Let's solve the problem step by step. --- ### **Part (a):** **Which of the following \( \tilde{f} \) is \( C = \{\tilde{f}(x, y) = k\} \) for some real constant \( k \)? Explain, and find the value of \( k \) where applicable.** Recall: \( C = \{f(x, y) = 1\} \). Let's check each option: #### **(A) \( \tilde{f}(x, y) = e^{f(x, y)} \)** Set \( \tilde{f}(x, y) = k \), so \( e^{f(x, y)} = k \implies f(x, y) = \ln k \). But for points on \( C \), \( f(x, y) = 1 \implies k = e^{1} \). **So, \( C = \{\tilde{f}(x, y) = e\} \). This works.** --- #### **(B) \( \tilde{f}(x, y) = \frac{1}{\ln(f(x, y))} \)** Set \( \tilde{f}(x, y) = k \), so \( \frac{1}{\ln(f(x, y))} = k \implies \ln(f(x, y)) = \frac{1}{k} \implies f(x, y) = e^{1/k} \). But for points on \( C \), \( f(x, y) = 1 \implies k = \frac{1}{\ln(1)} = \frac{1}{0} \) (undefined!). **So, this does NOT work.** --- #### **(C) \( \tilde{f}(x, y) = f(x, y)^2 - 1 \)** Set \( \tilde{f}(x, y) = k \), so \( f(x, y)^2 - 1 = k \implies f(x, y)^2 = k + 1 \implies f(x, y) = \pm\sqrt{k + 1} \). But for points on \( C \), \( f(x, y) = 1 \implies 1^2 - 1 = 0 \). So, \( k = 0 \). **So, \( C = \{\tilde{f}(x, y) = 0\} \). This works.** --- ### **Final Answer for (a):** - **(A) and (C) work:** - For (A): \( C = \{\tilde{f}(x, y) = e \} \) - For (C): \( C = \{\tilde{f}(x, y) = 0 \} \) --- ### **Part (b):** **Draw a replacement for the green portion of \( C \) such that if your new resultant curve \( C' \) can be described as \( \{g(x, y) = 0\} \) (\( \nabla g \neq 0 \)), then there are infinitely many points on \( C' \) where \( \nabla g \) is parallel to \( \langle 1, 1 \rangle \).** #### **Explanation** - \( \nabla g \) parallel to \( \langle 1, 1 \rangle \) means at those points, the level curve is tangent to the direction \( \langle -1, 1 \rangle \) (since the gradient is normal to the curve). - To get infinitely many such points, the curve must have a segment where the tangent direction is always \( \langle -1, 1 \rangle \). #### **Drawing** - Replace the green part with a straight line segment with slope 1 (i.e., direction \( \langle 1, 1 \rangle \)), so the normal vector at every point is parallel to \( \langle 1, 1 \rangle \). - **On the figure:** Draw a straight line segment (with slope 1) connecting the endpoints of the green portion. --- ### **Part (c):** **At how many points of the form \( p = (x, y, 0) \) on \( S \) is \( \nabla F(p) = \langle 0, m, 0 \rangle \) for some value \( m \neq 0 \)? Indicate these points on Figure 3.** #### **Explanation** - \( \nabla F(x, y, z) = (f_x(x, y), f_y(x, y), 0) \) - \( \nabla F = \langle 0, m, 0 \rangle \) with \( m \neq 0 \) ⇒ \( f_x(x, y) = 0 \), \( f_y(x, y) \neq 0 \). - This means at those points, the gradient is vertical in the \( y \)-direction. So, on the curve \( C \), find points where the tangent is horizontal (so the normal/gradient is vertical). #### **On the Figure** - **Number of points:** From the illustration, there are **two** such points: the points on \( C \) where the curve is "flat" (horizontal tangent), i.e., at the topmost and bottommost points of the curve. - **Indicate:** These are the top and bottom of the curve. --- ### **Part (d):** **Find a continuous function \( \tilde{F}(x, y, z) \) such that \( C \) is part of the intersection of the surface \( \tilde{S} = \{\tilde{F}(x, y, z) = 1\} \) with \( \{z = 0\} \), but \( \tilde{S} \neq S \).** #### **Explanation** - At \( z = 0 \), want \( \tilde{F}(x, y, 0) = 1 \) if and only if \( (x, y) \in C \). - But \( \tilde{S} \neq S \) (so \( \tilde{F}(x, y, z) \neq f(x, y) \) globally). **One way:** Let \( \tilde{F}(x, y, z) = f(x, y) + z^2 \). - At \( z = 0 \), \( \tilde{F}(x, y, 0) = f(x, y) \), so the intersection with \( \{\tilde{F} = 1\} \) at \( z = 0 \) is exactly \( C \). - But for \( z \neq 0 \), \( \tilde{F}(x, y, z) \neq f(x, y) \) (since \( z^2 \) is added). --- ### **Final Answers:** #### **(a)** - **(A):** \( \tilde{f}(x, y) = e^{f(x, y)} \), \( k = e \) - **(C):** \( \tilde{f}(x, y) = f(x, y)^2 - 1 \), \( k = 0 \) #### **(b)** - Replace the green part with a straight line segment of slope 1. #### **(c)** - **2 points** on \( C \) where the tangent is horizontal (topmost and bottommost points). #### **(d)** - \( \tilde{F}(x, y, z) = f(x, y) + z^2 \) (or any continuous function that equals \( f(x, y) \) at \( z = 0 \) but differs elsewhere).

Let's carefully analyze each part of the problem with detailed reasoning, calculations, and explanations. --- ## Given data and setup: - \( C \) is a curve in the \( yz \)-plane, defined as \( C = \{ (0, y, z) \mid (y, z) \text{ satisfy the curve } C \} \). - The surface \( S \subset \mathbb{R}^3 \) is formed by translating \( C \) along the \( x \)-direction, i.e., \( S = \{ (x, y, z) \mid (y,z) \in C, x \in \mathbb{R} \} \). - \( f(z, y) \) is a differentiable function on \( \mathbb{R}^2 \) with \( \nabla f \neq 0 \) everywhere. - The curve \( C \) is given as \( \{ (0, y, z) \mid f(z, y) = 1 \} \). - The function \( F(z, y, x) = f(z, y) \) extends \( f \) into \( \mathbb{R}^3 \). - The surface \( S \) can be written as \( S = \{ (x, y, z) \mid F(z, y, x) = 1 \} \). --- ## Part (a): Which of the given functions \( \tilde{f}(x, y) \) correspond to level sets \( C = \{ \tilde{f}(x, y) = k \} \)? ### Goal: Determine for each \( \tilde{f} \), whether the set \( C \) corresponds to some constant level set \( \{ \tilde{f} = k \} \). ### **Given options:** 1. \( \tilde{f}_1(x, y) = e^{f(x, y)} \) 2. \( \tilde{f}_2(x, y) = \frac{1}{\ln(f(x, y))} \) 3. \( \tilde{f}_3(x, y) = f(x, y)^2 - 1 \) --- ### **Step-by-step reasoning:** **Recall:** \( C = \{ (0, y, z) \mid f(z, y) = 1 \} \). --- ### **Option 1: \( \tilde{f}_1(x, y) = e^{f(x, y)} \)** - To represent \( C \), we want \( \tilde{f}_1(0, y) = k \), i.e., \[ e^{f(0, y)} = k. \] - But on \( C \), \( f(0, y) = 1 \). So, \[ e^{1} = e, \] hence, \[ \boxed{\tilde{f}_1(0, y) = e} \quad \text{for all points of } C. \] - **Conclusion:** \( C = \{ (x, y) \mid \tilde{f}_1(x, y) = e \} \). - **Answer:** Yes, \( C \) is the level set \( \{\tilde{f}_1 = e\} \). --- ### **Option 2: \( \tilde{f}_2(x, y) = \frac{1}{\ln(f(x, y))} \)** - For \( C \), \( f(0, y) = 1 \). - Check what \( \tilde{f}_2 \) equals there: \[ \tilde{f}_2(0, y) = \frac{1}{\ln(1)}. \] - Since \( \ln(1) = 0 \), the expression is undefined (infinite). So, **this does not produce a finite constant \( k \)** at the points of \( C \). - **Conclusion:** \( \tilde{f}_2 \) is not well-defined on \( C \), so it cannot be a level set \( \{ \tilde{f}_2 = k \} \). --- ### **Option 3: \( \tilde{f}_3(x, y) = f(x, y)^2 - 1 \)** - For \( C \), \( f(0, y) = 1 \). Substituting: \[ \tilde{f}_3(0, y) = 1^2 - 1 = 0. \] - Therefore, on \( C \), \( \tilde{f}_3 = 0 \). - **Answer:** \( C = \{ (x, y) \mid \tilde{f}_3(x, y) = 0 \} \). --- ### **Final conclusion for (a):** - **Functions \( \tilde{f}_1 \) and \( \tilde{f}_3 \) define level sets containing \( C \):** - \( \boxed{ \text{(A) } \tilde{f}(x, y) = e^{f(x, y)}, \quad \text{level } e } \) - \( \boxed{ \text{(C) } \tilde{f}(x, y) = f(x, y)^2 - 1, \quad \text{level } 0 } \) --- ## Part (b): Replace a portion of \( C \) with a curve \( C' \) such that the new curve can be described as \( g(z, y) = 0 \), and there are infinitely many points where \( \nabla g \) is parallel to \( \langle 1, 1 \rangle \). ### Goal: Construct \( g(z, y) \) such that: - The level set \( g(z, y) = 0 \) contains the replaced portion, - The gradient \( \nabla g \) at infinitely many points is parallel to \( \langle 1, 1 \rangle \). ### **Understanding the problem:** - \( \nabla g = \left( \frac{\partial g}{\partial z}, \frac{\partial g}{\partial y} \right) \). - \( \nabla g \parallel \langle 1, 1 \rangle \) means: \[ \nabla g = \lambda \langle 1, 1 \rangle \] at those points, for some scalar \( \lambda \). - **Condition for \( \nabla g \parallel \langle 1, 1 \rangle \):** \[ \frac{\partial g}{\partial z} = \frac{\partial g}{\partial y} \neq 0 \] at infinitely many points. ### **Constructing \( g(z, y) \):** - To ensure \( \nabla g \) is parallel to \( \langle 1, 1 \rangle \) on a curve, \( g \) must have: \[ \frac{\partial g}{\partial z} = \frac{\partial g}{\partial y} = h(z, y). \] - This suggests \( g(z, y) \) is of the form: \[ g(z, y) = z - y + \text{(constant)}. \] - For simplicity, choose: \[ g(z, y) = z - y. \] - Then, \[ \nabla g = (1, -1), \] which is **not** parallel to \( \langle 1, 1 \rangle \). To get the gradient parallel to \( \langle 1, 1 \rangle \), the partial derivatives must be equal: \[ \frac{\partial g}{\partial z} = \frac{\partial g}{\partial y} \neq 0. \] - Therefore, choose: \[ g(z, y) = z + y, \] then \[ \nabla g = (1, 1), \] which **is exactly** parallel to \( \langle 1, 1 \rangle \). ### **Answer:** - **Replace the green portion with the line:** \[ g(z, y) = z + y = 0, \] which is a straight line with slope \( -1 \), and **every point** on this line has \( \nabla g = (1, 1) \), always parallel to \( \langle 1, 1 \rangle \). - Since we want **infinitely many points** with this property, choosing a line \( z + y = 0 \) ensures this. --- ## Part (c): Points \( p = (x, y, 0) \in S \) where \( \nabla F(p) = (0, m, 0) \), \( m \neq 0 \). ### **Understanding:** - \( F(z, y, x) = f(z, y) \) (independent of \( x \)). - Gradient: \[ \nabla F = \left( \frac{\partial F}{\partial z}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial x} \right) = (f_z(z, y), f_y(z, y), 0). \] - Condition: \[ \nabla F(p) = (0, m, 0), \quad m \neq 0, \] at some point \( p = (x, y, 0) \). - Since \( \nabla F \) has third component zero, and the partial derivatives are independent of \( x \), the points \( p \) satisfy: \[ f_z(z, y) = 0, \] and \[ f_y(z, y) = m \neq 0. \] - **Note:** The \( x \)-coordinate is arbitrary here; the gradient depends only on \( y \) and \( z \). ### **How many such points?** - These points correspond to points on \( C \) where: \[ f_z(z, y) = 0, \] and \( f_y(z, y) \neq 0 \). - On the curve \( C \), these are points where the tangent is **horizontal in the \( yz \)-plane**, i.e., the tangent vector is in the \( y \)-direction. - The number of such points depends on the shape of \( C \). ### **In the figure:** - Usually, the topmost and bottommost points are candidates, where the tangent is horizontal (assuming \( C \) is a smooth closed curve). - Therefore, **there are two points** where \( \nabla F \) is of the form \( (0, m, 0) \), with \( m \neq 0 \): the top and bottom points of \( C \). - **Indicate these points** on the figure at the highest and lowest points of \( C \). --- ## Part (d): Find a continuous function \( \tilde{F}(x, y, z) \) such that: - \( C \subseteq \tilde{S} = \{\tilde{F} = 1\} \), - \( \tilde{S} \neq S \), - and at \( z=0 \), the intersection \( \{\tilde{F} = 1\} \cap \{z=0\} \) contains \( C \). ### **Approach:** - The key is to construct \( \tilde{F} \) so that: \[ \tilde{F}(x, y, 0) = f(0, y). \] - Since \( C \) is defined by \( f(z, y) = 1 \), at \( z=0 \): \[ f(0, y) = 1, \] for all \( y \) in the domain of \( C \). - To ensure \( C \subseteq \{\tilde{F} = 1\} \), define: \[ \boxed{ \tilde{F}(x, y, z) = f(z, y) + \phi(z), } \] where \( \phi(z) \) is continuous, with: \[ \phi(0) = 0, \] but \( \phi(z) \neq 0 \) for some \( z \neq 0 \), so that: \[ \tilde{F}(x, y, 0) = f(0, y) + \phi(0) = 1 + 0 = 1, \] which matches \( C \). - For \( z \neq 0 \): \[ \tilde{F}(x, y, z) = f(z, y) + \phi(z), \] which differs from \( S \) (since \( \phi(z) \neq 0 \) for some \( z \neq 0 \)), ensuring \( \tilde{S} \neq S \). ### **Example:** Choose: \[ \boxed{ \tilde{F}(x, y, z) = f(z, y) + \sin(z). } \] - At \( z=0 \): \[ \tilde{F} = f(0, y

Let's continue with the detailed explanation for part (d): --- ### **Part (d):** Construct a continuous function \( \tilde{F}(x, y, z) \) such that: - \( C \subseteq \tilde{S} = \{\tilde{F} = 1\} \), - \( \tilde{S} \neq S \), - and \( \tilde{F} \) matches \( f(z, y) \) at \( z=0 \) on \( C \). --- ### **Construction:** Recall that: - \( C \) is given by \( f(z, y) = 1 \), - \( S = \{ (x, y, z) \mid f(z, y) = 1 \} \), - At \( z=0 \), on \( C \), we have: \[ f(0, y) = 1. \] To **embed \( C \) in the intersection** of the surface \( \tilde{S} \) with \( z=0 \), we require: \[ \tilde{F}(x, y, 0) = 1, \] for all points \( (x, y, 0) \) on \( C \). --- ### **Step 1: Define \( \tilde{F} \) to match \( f(z, y) \) at \( z=0 \).** A natural choice: \[ \boxed{ \tilde{F}(x, y, z) = f(z, y) + \psi(z), } \] where \( \psi(z) \) is a continuous function with: \[ \psi(0) = 0, \] but \( \psi(z) \neq 0 \) for some \( z \neq 0 \). - **At \( z=0 \):** \[ \tilde{F}(x, y, 0) = f(0, y) + \psi(0) = 1 + 0 = 1, \] ensuring that \( C \subseteq \{\tilde{F}=1\} \). - **For \( z \neq 0 \):** \[ \tilde{F}(x, y, z) = f(z, y) + \psi(z), \] which differs from \( f(z, y) \) unless \( \psi(z) = 0 \). --- ### **Step 2: Picking \( \psi(z) \) to ensure \( \tilde{S} \neq S \).** - Choose \( \psi(z) \neq 0 \) for some \( z \neq 0 \); for example: \[ \boxed{ \psi(z) = \sin(z), } \] which is continuous, satisfies \( \psi(0)=0 \), and is non-zero for many \( z \). - Then, \[ \tilde{F}(x, y, z) = f(z, y) + \sin(z), \] - The surface: \[ \tilde{S} = \{ (x, y, z) \mid f(z, y) + \sin(z) = 1 \}. \] - Since \( \sin(z) \neq 0 \) for some \( z \neq 0 \), \( \tilde{S} \neq S \). --- ### **Final answer:** \[ \boxed{ \tilde{F}(x, y, z) = f(z, y) + \sin(z). } \] **This function:** - Satisfies \( \tilde{F}(x, y, 0) = f(0, y) + 0 = 1 \), - Contains \( C \) in \( \{\tilde{F} = 1 \} \) at \( z=0 \), - Is continuous, - And ensures \( \tilde{S} = \{\tilde{F} = 1\} \) is not the same as \( S \). --- ## **Summary of final answers:** ### **(a)** - \( C = \{\tilde{f}_1 = e\} \) with \( \tilde{f}_1 = e^{f(x, y)} \), - \( C = \{\tilde{f}_3=0\} \) with \( \tilde{f}_3 = f(x, y)^2 - 1 \). ### **(b)** - Replace a portion of \( C \) with the line \( g(z,y) = z + y = 0 \), - which has gradient \( (1, 1) \), parallel to \( \langle 1, 1 \rangle \). ### **(c)** - There are **two points** on \( C \) where \( \nabla F = (0, m, 0) \) with \( m \neq 0 \), corresponding to the highest and lowest points of \( C \). ### **(d)** - Define \[ \boxed{ \tilde{F}(x, y, z) = f(z, y) + \sin(z), } \] - ensuring the intersection with \( z=0 \) contains \( C \), but \( \tilde{S} \neq S \). --- If you have further questions or need clarification, feel free to ask!