Directional Derivatives Study Notes

Directional Derivatives Study Notes Institution: California State University, Northridge（Northridge, CA） Course: Math 250（Vector Calculus） Instructor: David Klein Instructor Time: Last Tuesday I. Course Introduction: The Core Significance of Directional Derivatives In single-variable calculus, the derivative f(x) describes the instantaneous rate of change of the function along the x axis. However, for multivariable functions z = f (x, y) or w = f (x, y, z), only partial derivatives ∂f/∂x and ∂f/∂y can reflect the rate of change along the coordinate axes, which cannot meet the need for the rate of change in any direction in practical problems. Professor David Klein emphasized in class that directional derivatives (DirectionalDerivatives) are a natural extension of partial derivatives. They can precisely describe the trend of change of multivariable functions in any specified direction in space, serving as the foundation for subsequent learning of gradients, optimization problems, and vector field-related theorems, with widespread applications in fields such as physics, engineering, and economics. For example, in physics, they can be used to calculate the rate of change of velocity in the direction of force, and in engineering, they can analyze the slope change of a surface in a specific direction. II. Strict Definition of Directional Derivatives 2.1 Definition of Directional Derivatives for Binary Functions Let the function z = f (x, y) be defined in a neighborhood of the point P₀(a, b) and let the vector u = (cosθ, sinθ) be a unit vector in the plane (where θ is the angle between the vector u and the positive direction of the x axis). If the limit: Dᵤf (a, b) = lim (h→0) [f (a + h・cosθ, b + h・sinθ) - f (a, b)] /h exists, then this limit is called the directional derivative of the function f (x, y) at the point P₀(a, b) in the direction of the vector u . 2.2 Directional derivative of a ternary function Extending to a ternary function w = f (x, y, z), let the vector u = (cosα, cosβ, cosγ) be a unit vector in space (where α, β, γ are the angles between the vector u and the positive directions of the x, y, z axes, known as direction cosines). Then the directional derivative of the function at the point P₀(x₀, y₀, z₀) in the direction of u is: Dᵤf (x₀, y₀, z₀) = lim (h→0) [f (x₀ + h・cosα, y₀ + h・cosβ, z₀ + h・cosγ) - f (x₀, y₀, z₀)] /h 2.3 Geometric Meaning of the Definition Professor David Klein intuitively explains through three-dimensional surface graphs: For a binary function z = f (x, y), its graph is a surface in space. Through the point P₀(a, b, f (a, b)) draw a plane parallel to the direction u and perpendicular to the xy plane. The intersection line of this plane with the surface is a space curve. The directional derivative Dᵤf (a, b) is the slope of the tangent line to this curve at the point P₀ . 2.4 Key Considerations in the Definition Direction vector u must be a unit vector: If the given direction vector is not a unit vector, it must first be normalized (divided by the magnitude of the vector). Existence of Directional Derivatives: The existence of a directional derivative of a function at a point in a certain direction does not imply that the function is differentiable at that point, nor does it imply the existence of partial derivatives; however, conversely, if the function is differentiable at that point, then the directional derivative in any direction must exist (a subsequent theorem will prove this). Partial derivatives are a special case of directional derivatives: When u = (1, 0) (positive x-axis direction), Dᵤf = ∂f/∂x； When u = (0, 1) (positive y-axis direction), Dᵤf = ∂f/∂y。 III. Methods for Calculating Directional Derivatives 3.1 Direct Calculation Using the Definition (Basic Method) Example 1: Let the function f (x, y) = x² + xy, find the f (x, y) directional derivative at the point (1, 2) in the direction u = (cos (π/3), sin (π/3)) = (1/2, √3/2) direction. Solution: According to the definition of the directional derivative: Dᵤf (1, 2) = lim (h→0) [f (1 + h/2, 2 + (√3 h)/2) - f (1, 2)] /h Step 1: Calculate f (1 + h/2, 2 + (√3 h)/2)： f (1 + h/2, 2 + (√3 h)/2) = (1 + h/2)² + (1 + h/2)(2 + (√3 h)/2) = 1 + h + h²/4 + 2 + (√3 h)/2 + h + (√3 h²)/4 = 3 + (2 + √3/2) h + [(1 + √3)/4] h² Step 2: Calculate f (1 + h/2, 2 + (√3 h)/2) - f (1, 2)： f (1, 2) = 1² + 1×2 = 3 Difference = 3 + (2 + √3/2) h + [(1 + √3)/4] h² - 3 = (2 + √3/2) h + [(1 + √3)/4] h² Step 3: Substitute into the limit calculation: Dᵤf (1, 2) = lim (h→0) [(2 + √3/2) h + [(1 + √3)/4] h²] /h = lim (h→0) [2 + √3/2 + [(1 + √3)/4] h] = 2 + √3/2 Conclusion: The directional derivative of the function at the point (1, 2) in the specified direction is 2 + √3/2 ≈ 2.866. 3.2 Using the dot product formula for gradients (efficient method) Professor David Klein explained the formula for calculating the directional derivative of differentiable functions, which is more concise than the definition method and is one of the core focus points in the Math 250 course. 3.2.1 Core Theorem If the function z = f (x, y) is differentiable at the point (x, y) then the directional derivative of the function in that point along any unit vector u = (u₁, u₂) exists and is given by: Dᵤf (x, y) = ∇ ∇f (x, y)・u where f (x, y) = (∂f/∂x, ∂f/∂y) is called the gradient (Gradient) of the function f (x, y) and“・” denotes the dot product of vectors. 3.2.2 Generalization of the Formula (for a function of three variables) For a differentiable function w = f (x, y, z), the directional derivative along the unit vector u = (u₁, u₂, u₃) is: Dᵤf (x, y, z) = ∇f (x, y, z)・u = (∂f/∂x) u₁ + (∂f/∂y) u₂ + (∂f/∂z) u₃ 3.2.3 Formula Proof (Binary Function) Since the function f (x, y) is differentiable at the point (x, y) according to the definition of differentiability： f (x + Δx, y + Δy) - f (x, y) = (∂f/∂x)Δx + (∂f/∂y)Δy + o (√(Δx² + Δy²)) Let Δx = h・u₁，Δy = h・u₂（since u is a unit vector, √(Δx² + Δy²) = h），then： [f (x + h・u₁, y + h・u₂) - f (x, y)] /h = (∂f/∂x) u₁ + (∂f/∂y) u₂ + o (h)/h As h→0 o (h)/h → 0, thus: Dᵤf (x, y) = (∂f/∂x) u₁ + (∂f/∂y) u₂ = ∇f・u 3.2.4 Steps for Applying the Formula Calculate the partial derivatives ∂f/∂x, ∂f/∂y (or the three partial derivatives of a ternary function); Construct the gradient vector ∇f； Verify that the direction vector v is a unit vector; if not, normalize it to u = v / v（vdenotes the magnitude of vector v ）； Calculate the dot product of the gradient and the unit direction vector; the result is the directional derivative. Example 2: Recompute the directional derivative of Example 1 using the gradient formula. Solution: Step 1: Calculate the partial derivatives: ∂f/∂x = 2x + y，∂f/∂y = x Step 2: Calculate the gradient at the point (1, 2) ： ∇f (1, 2) = (2×1 + 2, 1) = (4, 1) Step 3: The direction vector u = (1/2, √3/2) is already a unit vector, calculate the dot product： Dᵤf (1, 2) = (4, 1)・(1/2, √3/2) = 4×(1/2) + 1×(√3/2) = 2 + √3/2 The result is consistent with the definition method, verifying the validity of the formula. Example 3: Let the function f (x, y, z) = xyz + x², find the directional derivative of f (x, y, z) at the point (1, -1, 2) in the direction of the vector v = (2, -1, 1) direction. Solution: Step 1: Determine if the direction vector is a unit vector, calculatev： v = √(2² + (-1)² + 1²) = √(4 + 1 + 1) = √6 ≠ 1 Normalize the direction vector: u = v / ||v|| = (2/√6, -1/√6, 1/√6) Step 2: Calculate the partial derivatives: ∂f/∂x = yz + 2x，∂f/∂y = xz，∂f/∂z = xy Step 3: Calculate the gradient at the point (1, -1, 2) ： ∂f/∂x (1, -1, 2) = (-1)×2 + 2×1 = 0 ∂f/∂y (1, -1, 2) = 1×2 = 2 ∂f/∂z (1, -1, 2) = 1×(-1) = -1 ∇f (1, -1, 2) = (0, 2, -1) Step 4: Calculate the dot product: Dᵤf (1, -1, 2) = (0, 2, -1)・(2/√6, -1/√6, 1/√6) = 0×(2/√6) + 2×(-1/√6) + (-1)×(1/√6) = -3/√6 = -√6/2 Conclusion: The directional derivative of this ternary function at the specified point in the specified direction is -√6/2 ≈ -1.225, and the negative sign indicates that the function is monotonically decreasing in that direction. 3.3 Special case simplification calculations Along the positive direction of the coordinate axis: u = (1, 0) when, Dᵤf = ∂f/∂x；u = (0, 1) when, Dᵤf = ∂f/∂y。 Along the negative direction of the coordinate axis: u = (-1, 0) when, Dᵤf = -∂f/∂x；u = (0, -1) when, Dᵤf = -∂f/∂y。 Along the gradient direction: It will be shown later that the directional derivative achieves its maximum value along the gradient direction, i.e., f (the magnitude of the gradient)。 ∇ Four, the relationship between directional derivatives and gradients 4.1 The Geometric Meaning of the Gradient ∇ Professor David Klein emphasized in class that the gradient f is a vector whose direction is the direction of the maximum rate of change of the function f (x, y) at that point, and the magnitude of the gradient f is the value of this maximum rate of change; while the opposite ∇ ∇ direction of the gradient (− f) is the direction of the minimum rate of ∇ change of the function, with the minimum rate of change being - f. 4.2 Range of Directional Derivatives From the properties of the dot product a・b = abcosφ（where φ is the angle between vectors a and b ），for the directional derivative formula： Dᵤf = ∇f・u = ∇fucosφ = ∇fcosφ (Since u is a unit vector, u = 1). Thus, the range of the directional derivative is: ∇ - f ≤ Dᵤf ≤ ∇f When φ = 0 (u is in the same direction as value); ∇f)时，Dᵤf = ∇f (maximum When φ = π (u is in the opposite direction to (minimum value); When φ = π/2 (u is perpendicular to change in this direction). ∇f)时，Dᵤf = -∇f ∇f), Du f = 0 (the function has no 4.3 The Relationship Between the Gradient and Level Curves For a binary function z = f (x, y), the equation of the level curve is f (x, y) = C (C is a constant). Take the total differential of both sides of the level curve equation: (∂f/∂x) dx + (∂f/∂y) dy = 0 ∇ That is f・(dx, dy) = 0. Here, (dx, dy) is the tangent vector of the contour line at that point, so the gradient f is perpendicular to the tangent vector of the contour line, meaning the gradient direction is the ∇ normal vector direction of the contour line. Professor David Klein uses a topographic map analogy: contour lines represent points of the same altitude, the gradient direction is the“steepest uphill” direction, and the tangent direction of the contour line is the“horizontal direction” (no altitude change), which is consistent with the conclusion that the directional derivative is 0 when φ = π/2 . Example 4: Let the function f (x, y) = x² - y², find: （1）The gradient at the point (2, 1) ; (2) The directional derivative of this point in the direction of the gradient; (3) The directional derivative of this point in the direction of the tangent to the contour line. Solution: (1) Calculate the gradient: ∂f/∂x = 2x, ∂f/∂y = -2y ∇f (2, 1) = (2×2, -2×1) = (4, -2) (2) The unit vector in the direction of the gradient u = -2) / √(4² + (-2)²) = (4, -2)/√20 = (2/√5, -1/√5) ∇f / ||∇f|| = (4, Directional derivative: Dᵤf (2, 1) = ∇f (2, 1) = √(4² + (-2)²) = √20 = 2√5 (3) The direction of the contour line is perpendicular to the gradient, so the directional derivative is 0. 0.