Let's break down the solution step by step. ### **Step 1: Understanding the Directional Derivative** The **directional derivative** of \( f \) at point \( P \) in the direction of a unit vector \( \mathbf{v} \) is: \[ D_{\mathbf{v}} f(P) = \nabla f(P) \cdot \mathbf{v} \] It represents the rate of change of the function \( f \) at \( P \) in the direction of \( \mathbf{v} \). --- ### **Step 2: Estimating the Gradient from the Contour Map** - The gradient \( \nabla f \) points in the direction of steepest increase of \( f \). - Its magnitude is greatest where contours are closest together. - The gradient is **perpendicular** to the contour lines. --- ## **Part (a): \( D_{\mathbf{v}} f(\mathbf{B}) \) where \( \mathbf{v} = \mathbf{i} + \mathbf{j} \)** #### **1. Find the location and values:** - Point \( B \) is at (1, 2). - Nearby contours: At \( (1, 1) \) (point \( A \)): \( f = 0 \) - At \( (2, 3) \): \( f = -1 \) #### **2. Find the Direction:** - The vector \( \mathbf{v} = \mathbf{i} + \mathbf{j} \). - **Unit vector:** \( \mathbf{u} = \frac{1}{\sqrt{2}}\mathbf{i} + \frac{1}{\sqrt{2}}\mathbf{j} \) #### **3. Estimate rate of change:** Move from \( B = (1,2) \) to \( (2,3) \): - Change in \( f \): \( -1 - 0 = -1 \) - Distance: \( \sqrt{(2-1)^2 + (3-2)^2} = \sqrt{2} \) So, the **rate of change** is: \[ \text{Rate} = \frac{-1}{\sqrt{2}} \] Thus, \[ D_{\mathbf{v}} f(\mathbf{B}) \approx -\frac{1}{\sqrt{2}} \] --- ## **Part (b): \( D_{\mathbf{v}} f(\mathbf{D}) \) where \( \mathbf{v} = -\mathbf{i} \)** #### **1. Find the location and values:** - Point \( D \) is at (3, 5). - Move left (negative x-direction): from \( (4,5) \) to \( (3,5) \). - Contours at \( x = 4 \): \( f = -2 \) - Contours at \( x = 3 \): \( f = -1 \) #### **2. Direction:** - Direction is \( -\mathbf{i} \) (unit vector). #### **3. Estimate rate of change:** Move from \( (4,5) \) to \( (3,5) \): - Change in \( f \): \( -1 - (-2) = 1 \) - Distance: \( 1 \) So, \[ D_{\mathbf{v}} f(\mathbf{D}) = \frac{1}{1} = 1 \] --- ## **Final Answers** ### **(a) \( D_{\mathbf{v}} f(\mathbf{B}) \) in direction \( \mathbf{i} + \mathbf{j} \):** \[ \boxed{ -\frac{1}{\sqrt{2}} } \] ### **(b) \( D_{\mathbf{v}} f(\mathbf{D}) \) in direction \( -\mathbf{i} \):** \[ \boxed{ 1 } \] --- **Explanation:** - The directional derivative is estimated by the change in \( f \) over the distance between two points in the given direction (following the contours). - The negative sign in (a) means \( f \) decreases in that direction. - The positive sign in (b) means \( f \) increases as you move in the negative \( x \)-direction at \( D \).