Directional Derivatives and the Gradient

Learning Objectives

After studying this chapter, you should be able to:

• Calculate the directional derivative of a function $f(x, y)$ or $f(x, y, z)$ in any given direction.
• Compute the gradient vector $\nabla f$ for scalar functions.
• Interpret the geometric significance of the gradient (direction of steepest ascent).
• Apply the gradient to find tangent lines to level curves and tangent planes to level surfaces.
• Determine the maximum and minimum rates of change at a specific point.

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Key Terms & Glossary

• Gradient ($\nabla f$): A vector field whose components are the partial derivatives of $f$. It points in the direction of the greatest rate of increase.
• Directional Derivative ($D_{\mathbf{u}}f$): The rate at which a function changes at a point in the direction of a unit vector $\mathbf{u}$.
• Unit Vector: A vector with a magnitude (length) of 1. In directional derivatives, the direction must be normalized to a unit vector.
• Level Curve: A curve defined by $f(x, y) = c$. The gradient $\nabla f$ at a point $(x_0, y_0)$ is always perpendicular (orthogonal) to the level curve passing through that point.
• Normal Vector: A vector perpendicular to a surface or curve; the gradient acts as the normal vector to level sets.

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The "Big Idea"

In single-variable calculus, the derivative $f'(x)$ gives the slope in one dimension. In multivariable calculus, partial derivatives $f_x$ and $f_y$ only give the slopes along the axes. The Gradient is the "master key" that unlocks the slope in any direction. It represents the local "uphill" direction on a topographic map. If you are standing on a mountain, the gradient tells you which way is straight up, and the directional derivative tells you how steep the path is if you choose to walk at a specific angle.

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Formula / Concept Box

[!IMPORTANT] Always verify that your direction vector is a unit vector. If you are given a vector $\mathbf{v}$, you must use $\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|}$ before calculating the dot product.

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Hierarchical Outline

1. The Gradient Vector
   • Definition: $\nabla f = \langle f_x, f_y, \dots \rangle$.
   • Computation: Requires finding all first-order partial derivatives.
2. The Directional Derivative
   • Limit definition vs. the Dot Product Theorem.
   • Relationship to partials: $f_x$ is $D_{\mathbf{i}}f$ and $f_y$ is $D_{\mathbf{j}}f$.
3. Properties of the Gradient
   • Steepest Ascent: Points in the direction of maximum increase.
   • Steepest Descent: Points in the direction of maximum decrease ($-\nabla f$).
   • Zero Change: Directions orthogonal to $\nabla f$ result in zero change (moving along a level curve).
4. Applications
   • Tangent lines to level curves $f(x, y) = c$.
   • Normal lines to surfaces $f(x, y, z) = c$.

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Visual Anchors

The Gradient Relationship

Geometric Interpretation in 2D

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Definition-Example Pairs

1. Directional Cosines

• Definition: The components of a unit vector in 3D, expressed as $\cos \alpha, \cos \beta, \cos \gamma$, representing the angles made with the $x, y,$ and $z$ axes.
• Example: A unit vector pointing equally between the axes would have $\cos \alpha = \cos \beta = \cos \gamma = 1/\sqrt{3}$.

2. Steepest Descent

• Definition: The direction exactly opposite the gradient vector where the function's value decreases most rapidly.
• Example: If $\nabla f = \langle 3, 4 \rangle$ at a point, the direction of steepest descent is $\langle -3, -4 \rangle$ with a rate of $-5$.

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Worked Examples

Example 1: Finding the Directional Derivative

Problem: Find $D_{\mathbf{u}}f(1, 2)$ for $f(x, y) = x^2y$ in the direction of $\mathbf{v} = \langle 3, -4 \rangle$.

Solution:

1. Find the Gradient: $f_x = 2xy \implies f_x(1, 2) = 2(1)(2) = 4$ $f_y = x^2 \implies f_y(1, 2) = (1)^2 = 1$ So, $\nabla f(1, 2) = \langle 4, 1 \rangle$.
2. Normalize the Direction Vector: $\|\mathbf{v}\| = \sqrt{3^2 + (-4)^2} = 5$ $\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} = \langle \frac{3}{5}, -\frac{4}{5} \rangle$.
3. Compute the Dot Product: $D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u} = \langle 4, 1 \rangle \cdot \langle 0.6, -0.8 \rangle$ $D_{\mathbf{u}}f = (4)(0.6) + (1)(-0.8) = 2.4 - 0.8 = 1.6$.

Example 2: Direction of Maximum Increase

Problem: At the point $(0, 0, 0)$, in what direction does $f(x, y, z) = e^x + yz$ increase most rapidly?

Solution:

1. Calculate the Gradient: $\nabla f = \langle e^x, z, y \rangle$
2. Evaluate at $(0, 0, 0)$: $\nabla f(0, 0, 0) = \langle e^0, 0, 0 \rangle = \langle 1, 0, 0 \rangle$.
3. Result: The function increases most rapidly in the direction of the positive $x$-axis ($\mathbf{i}$). The maximum rate of change is $\|\langle 1, 0, 0 \rangle\| = 1$.

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Checkpoint Questions

1. If $\nabla f(P) = \langle 0, 0 \rangle$, what is the directional derivative in any direction $\mathbf{u}$?
2. True or False: The directional derivative can be larger than the magnitude of the gradient.
3. Find the gradient of $f(x, y) = \ln(x^2 + y^2)$ at the point $(1, 1)$.
4. If you are walking along a level curve of a mountain, what is the value of your directional derivative in the direction of your path?

▶Click to see Answers

1. 0. Because $D_{\mathbf{u}}f = \langle 0, 0 \rangle \cdot \mathbf{u} = 0$.
2. False. The maximum value is $\|\nabla f\|$.
3. $\nabla f = \langle \frac{2x}{x^2+y^2}, \frac{2y}{x^2+y^2} \rangle$. At $(1, 1)$, $\nabla f = \langle 1, 1 \rangle$.
4. 0. Level curves represent constant height, so the rate of change is zero.