Let's solve both problems step by step. --- ### **(1) Directional Derivative** Given: - \( f(x, y) = \frac{1}{2}x^2 + \sqrt{3}y^2 \) - Point: \( (2, 1) \) - The directional derivative achieves **half** the maximum possible rate of change in the direction \( \vec{u} \). #### **Step 1: Compute the Gradient** \[ \nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \] \[ \frac{\partial f}{\partial x} = x, \quad \frac{\partial f}{\partial y} = 2\sqrt{3}y \] At \( (2,1) \): \[ \nabla f(2, 1) = (2, 2\sqrt{3}) \] #### **Step 2: Maximum Rate of Change** The maximum rate of change at \( (2, 1) \) is the magnitude of the gradient: \[ |\nabla f(2, 1)| = \sqrt{2^2 + (2\sqrt{3})^2} = \sqrt{4 + 4 \cdot 3} = \sqrt{16} = 4 \] #### **Step 3: Half the Maximum Rate** So, half the maximum rate is \( 2 \). The directional derivative in direction \( \vec{u} \) is: \[ D_{\vec{u}} f = \nabla f \cdot \vec{u} = 2u_1 + 2\sqrt{3}u_2 \] We want: \[ |2u_1 + 2\sqrt{3}u_2| = 2 \] \[ |u_1 + \sqrt{3}u_2| = 1 \] #### **Step 4: \( \vec{u} \) is a Unit Vector** \[ u_1^2 + u_2^2 = 1 \] Let's check the options: --- **Option (A):** \( \vec{u} = \frac{1}{\sqrt{5}}\langle -1, 2 \rangle \) - \( u_1 = -\frac{1}{\sqrt{5}},\quad u_2 = \frac{2}{\sqrt{5}} \) - \( u_1^2 + u_2^2 = \frac{1}{5} + \frac{4}{5} = 1 \) ✔️ - \( u_1 + \sqrt{3}u_2 = -\frac{1}{\sqrt{5}} + \sqrt{3} \cdot \frac{2}{\sqrt{5}} = \frac{-1 + 2\sqrt{3}}{\sqrt{5}} \) \[ |u_1 + \sqrt{3}u_2| = \left| \frac{-1 + 2\sqrt{3}}{\sqrt{5}} \right| \approx \frac{-1 + 3.464}{2.236} = \frac{2.464}{2.236} \approx 1.1 \] Not equal to 1. **No.** --- **Option (B):** \( \vec{u} = \langle -\sqrt{3}, 1 \rangle \) - Not a unit vector. **No.** --- **Option (C):** \( \vec{u} = \langle 1, 0 \rangle \) - \( u_1 = 1, u_2 = 0 \) - \( u_1^2 + u_2^2 = 1 \) - \( u_1 + \sqrt{3}u_2 = 1 \) - \( |u_1 + \sqrt{3}u_2| = 1 \) ✔️ So **(C)** is correct. --- **Option (D):** \( \vec{u} = \langle 0, 1 \rangle \) - \( u_1 = 0, u_2 = 1 \) - \( u_1 + \sqrt{3}u_2 = \sqrt{3} \approx 1.732 \neq 1 \) **No.** --- **Final Answer for (1):** \[ \boxed{(C)\ \langle 1, 0 \rangle} \] --- ### **(2) Tangent Planes to yz-plane** Given: The surface \( S = \{ ef(y, z) = g(x) \} \) in \( \mathbb{R}^3 \). A tangent plane is parallel to the yz-plane if its normal is parallel to the x-axis (i.e., no \( \hat{i} \) component). #### **Step 1: Implicit Surface Equation** \[ ef(y, z) - g(x) = 0 \] Let \( F(x, y, z) = ef(y, z) - g(x) \). The gradient: \[ \nabla F = \left( -g'(x), e \frac{\partial f}{\partial y}, e \frac{\partial f}{\partial z} \right) \] For the tangent plane to be parallel to yz-plane, normal should be parallel to x-axis, so the y and z components of the gradient must be zero: \[ e \frac{\partial f}{\partial y} = 0, \quad e \frac{\partial f}{\partial z} = 0 \] So, \[ \frac{\partial f}{\partial y} = 0, \quad \frac{\partial f}{\partial z} = 0 \] So at some point \( (y, z) \), both partial derivatives must be zero. #### **Step 2: Analyze Options** **(A) \( f(y, z) = y + z \)** \[ \frac{\partial f}{\partial y} = 1, \quad \frac{\partial f}{\partial z} = 1 \] Never zero. **Impossible for tangent plane to be parallel to yz-plane.** --- **(B) \( f(y, z) = z^3 \)** \[ \frac{\partial f}{\partial y} = 0, \quad \frac{\partial f}{\partial z} = 3z^2 \] Both zero only if \( z = 0 \), so at \( z = 0 \): \[ \frac{\partial f}{\partial y} = 0, \quad \frac{\partial f}{\partial z} = 0 \] Thus, possible at \( z = 0 \). **So, it is possible for tangent plane to be parallel to yz-plane.** --- **(C) \( f(y, z) = y^2 + z^2 \)** \[ \frac{\partial f}{\partial y} = 2y, \quad \frac{\partial f}{\partial z} = 2z \] Both zero at \( y = 0, z = 0 \). So, again, possible for tangent plane to be parallel to yz-plane. --- #### **Conclusion** Only (A) makes it **impossible** for the tangent plane to be parallel to the yz-plane. --- **Final Answer for (2):** \[ \boxed{(A)\ f(y, z) = y + z} \] --- ### **Summary** 1. **(C) \( \langle 1, 0 \rangle \)** 2. **(A) \( f(y, z) = y + z \)** --- **If you need further explanation of any steps, let me know!**

Let's analyze both problems step-by-step, including the reasoning for each answer. --- ## **(1) Directional Derivative** **Given:** - Function: \(f(x, y) = \frac{1}{2}x^2 + \sqrt{3} y^2\) - Point: \((2,1)\) - The directional derivative achieves **half** the maximum rate of change in some direction \(\vec{u}\). ### **Step 1: Find the gradient at \((2,1)\)** \[ \nabla f(x,y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \] Calculating partial derivatives: \[ \frac{\partial f}{\partial x} = x \] \[ \frac{\partial f}{\partial y} = 2 \sqrt{3} y \] At \((2,1)\): \[ \nabla f(2,1) = (2, 2 \sqrt{3}) \] ### **Step 2: Magnitude of the gradient** \[ |\nabla f| = \sqrt{2^2 + (2 \sqrt{3})^2} = \sqrt{4 + 4 \times 3} = \sqrt{4 + 12} = \sqrt{16} = 4 \] **Maximum rate of change** at the point is the magnitude of the gradient, which is 4. ### **Step 3: Find the directional derivative in direction \(\vec{u}\)** The directional derivative in the direction of a unit vector \(\vec{u} = (u_1, u_2)\): \[ D_{\vec{u}} f = \nabla f \cdot \vec{u} = 2 u_1 + 2 \sqrt{3} u_2 \] Since \(\vec{u}\) is a unit vector: \[ u_1^2 + u_2^2 = 1 \] **Given:** The directional derivative achieves **half** the maximum change, so: \[ |D_{\vec{u}} f| = \frac{1}{2} |\nabla f| = 2 \] Thus: \[ |2 u_1 + 2 \sqrt{3} u_2| = 2 \] Divide both sides by 2: \[ |u_1 + \sqrt{3} u_2| = 1 \] --- ## **Step 4: Check given options for \(\vec{u}\)** - **Option (A):** \(\vec{u} = \frac{1}{\sqrt{5}}(-1, 2)\) Check if this is a unit vector: \[ \left( \frac{-1}{\sqrt{5}} \right)^2 + \left( \frac{2}{\sqrt{5}} \right)^2 = \frac{1}{5} + \frac{4}{5} = 1 \] So it **is** a unit vector. Compute: \[ u_1 + \sqrt{3} u_2 = \frac{-1}{\sqrt{5}} + \sqrt{3} \times \frac{2}{\sqrt{5}} = \frac{-1 + 2\sqrt{3}}{\sqrt{5}} \] Calculate numerator: \[ -1 + 2 \times 1.732 \approx -1 + 3.464 = 2.464 \] Magnitude: \[ \left| \frac{2.464}{\sqrt{5}} \right| = \frac{2.464}{2.236} \approx 1.10 \] **Not exactly 1**, so **option (A)** does *not* satisfy the condition. - **Option (B):** \(\vec{u} = (-\sqrt{3}, 1)\) Is this a unit vector? \[ (\sqrt{3})^2 + 1^2 = 3 + 1 = 4 \] Not a unit vector. So, discard. - **Option (C):** \(\vec{u} = (1, 0)\) Check: \[ 1^2 + 0^2 = 1 \] Yes, unit vector. Compute: \[ u_1 + \sqrt{3} u_2 = 1 + \sqrt{3} \times 0 = 1 \] Absolute value is 1, exactly as required. - **Option (D):** \(\vec{u} = (0, 1)\) Check: \[ 0^2 + 1^2 = 1 \] Yes, unit vector. Compute: \[ 0 + \sqrt{3} \times 1 = \sqrt{3} \approx 1.732 \] Not equal to 1, so **not** satisfying the condition. --- ## **Final conclusion:** **The only direction among options that yields a directional derivative of half the maximum is \(\boxed{(1,0)}\).** **Answer: \(\boxed{(C)}\)** --- ## **(2) About the surface \(S = \{ e^{f(y,z)} = g(x) \}\)** **Given:** - \(F(x, y, z) = e^{f(y,z)} - g(x) = 0\) - \(f, g\) are \(C^1\), and \(g'(x) \neq 0\). --- ## **Question:** *The surface \(S\) cannot have a tangent plane parallel to the yz-plane if:* - (A) \(f(y,z) = y + 2\) - (B) \(f(y,z) = z^3\) - (C) \(f(y,z) = y^2 + z^2\) --- ## **Step 1: When is the tangent plane parallel to yz-plane?** - The normal vector to the surface at a point \((x,y,z)\): \[ \nabla F = \left( -g'(x), e^{f(y,z)} \frac{\partial f}{\partial y}, e^{f(y,z)} \frac{\partial f}{\partial z} \right) \] - For the tangent plane to be parallel to yz-plane, its normal vector must be **parallel to the x-axis**, i.e., \[ \text{y-component} = 0, \quad \text{z-component} = 0 \] which implies: \[ e^{f(y,z)} \frac{\partial f}{\partial y} = 0 \] \[ e^{f(y,z)} \frac{\partial f}{\partial z} = 0 \] Since \(e^{f(y,z)} \neq 0\), these reduce to: \[ \frac{\partial f}{\partial y} = 0, \quad \frac{\partial f}{\partial z} = 0 \] Meaning, at the point, both partial derivatives must be zero. --- ## **Step 2: For each \(f(y,z)\):** - **(A) \(f(y,z) = y + 2\)** \[ \frac{\partial f}{\partial y} = 1 \neq 0 \] Thus, the gradient's y-component is never zero; the tangent plane **cannot** be parallel to yz-plane at any point. - **(B) \(f(y,z) = z^3\)** \[ \frac{\partial f}{\partial y} = 0 \] \[ \frac{\partial f}{\partial z} = 3z^2 \] The derivatives are both zero only when: \[ z = 0 \] At \(z=0\), both derivatives are zero, so the tangent plane **can** be parallel to yz-plane at points where \(z=0\). - **(C) \(f(y,z) = y^2 + z^2\)** \[ \frac{\partial f}{\partial y} = 2y \] \[ \frac{\partial f}{\partial z} = 2z \] Both zero at \(y=0, z=0\), so the tangent plane **can** be parallel to yz-plane at that point. --- ## **Conclusion:** - The surface **cannot** have a tangent plane parallel to yz-plane **if** the derivatives are **never zero** simultaneously. - Only **(A)** has derivatives that are **never zero** simultaneously, so the tangent plane can **never** be parallel to yz-plane. --- ## **Final answer:** \(\boxed{\text{(A)}\ f(y,z) = y + 2}\) --- # **Summary:** - **Problem (1):** The direction \(\langle 1, 0 \rangle\) (Option C) yields half the maximum rate of change. - **Problem (2):** The function \(f(y,z) = y + 2\) (Option A) makes it impossible for the tangent plane to be parallel to the yz-plane. --- **If further clarification is needed, feel free to ask!**