📐 Concept diagram

06-06 — Directional Derivatives and the Gradient

Phase: 6 — Calculus III: Multivariable Calculus Subject: 06-06 Prerequisites: 06-03 — Partial Derivatives, dot products, unit vectors Next subject: 06-07 — Optimization (Multivariable)

Learning Objectives

By the end of this subject, you will be able to:

1. Define and compute the directional derivative D_u f in the direction of any unit vector u
2. Define the gradient vector ∇f and express the directional derivative as D_u f = ∇f · u
3. Prove that the gradient points in the direction of steepest ascent and has magnitude equal to the maximum rate of change
4. Show that the gradient is perpendicular (normal) to level curves and level surfaces
5. Apply gradients to find tangent lines to level curves and tangent planes to level surfaces

Core Content

1. Directional Derivative — Definition

⚠️ CRITICAL FOUNDATION: D_u f = ∇f · u is the single most important formula in multivariable calculus. It connects the gradient to directional rates of change and leads directly to steepest ascent, level curves, and optimization.

Partial derivatives f_x and f_y measure rates of change in the x and y directions only. The directional derivative generalizes this to measure the rate of change in any direction.

Definition: Let u = ⟨a, b⟩ be a unit vector (so a² + b² = 1). The directional derivative of f at (x₀, y₀) in the direction of u is:

$                    f(x₀ + ha, y₀ + hb) − f(x₀, y₀)
D_u f(x₀, y₀) = lim ——————————————————————————————————
               h→0                h
$

Geometric interpretation: Take the vertical plane that contains the point (x₀, y₀, f(x₀, y₀)) and is parallel to the vector u. This plane intersects the surface in a curve. D_u f is the slope of the tangent line to that curve.

Key case: If u = ⟨1, 0⟩ (the unit vector in the x-direction), D_u f = f_x. If u = ⟨0, 1⟩, D_u f = f_y. So partial derivatives are special cases of directional derivatives.

2. Computing D_u f — The Gradient Formula

Theorem: If f is differentiable at (x₀, y₀), then for any unit vector u = ⟨a, b⟩:

$D_u f(x₀, y₀) = f_x(x₀, y₀)·a + f_y(x₀, y₀)·b
$

Or more compactly, using the gradient vector:

∇f = ⟨f_x, f_y⟩      (also written as grad f)

D_u f = ∇f · u       (dot product: f_x·a + f_y·b)

This is the most important formula in multivariable calculus.

For three variables: ∇f = ⟨f_x, f_y, f_z⟩, and D_u f = ∇f · u for any unit vector u ∈ R³.

Example 1: Find the directional derivative of f(x, y) = x²y³ − 4y at (2, −1) in the direction of v = ⟨2, 5⟩.

Step 1: Compute the gradient.

$f_x = 2xy³        →    f_x(2, −1) = 2(2)(−1)³ = −4
f_y = 3x²y² − 4   →    f_y(2, −1) = 3(4)(1) − 4 = 12 − 4 = 8

∇f(2, −1) = ⟨−4, 8⟩
$

Step 2: Make v into a unit vector.

$‖v‖ = √(2² + 5²) = √29
u = v/‖v‖ = ⟨2/√29, 5/√29⟩
$

Step 3: Compute the dot product.

$D_u f = ⟨−4, 8⟩ · ⟨2/√29, 5/√29⟩ = (−8 + 40)/√29 = 32/√29 ≈ 5.94
$

3. Steepest Ascent and the Gradient

Recall the dot product identity: ∇f · u = ‖∇f‖ ‖u‖ cos θ = ‖∇f‖ cos θ, where θ is the angle between ∇f and u.

Since cos θ has maximum value 1 (when θ = 0), the directional derivative is maximized when u points in the same direction as ∇f. The maximum value is ‖∇f‖.

Theorem — Properties of the Gradient:

1. Steepest ascent: ∇f points in the direction in which f increases most rapidly. The maximum rate of increase is ‖∇f‖.
2. Steepest descent: −∇f points in the direction in which f decreases most rapidly. The maximum rate of decrease is −‖∇f‖.
3. Zero change: If u ⊥ ∇f (perpendicular), then D_u f = 0. Along level curves, the change is zero.

Example 2: For f(x, y) = 4 − x² − 2y² at (1, 1):

$∇f = ⟨−2x, −4y⟩ → ∇f(1, 1) = ⟨−2, −4⟩
‖∇f‖ = √(4 + 16) = √20 = 2√5 ≈ 4.47
$

The direction of steepest ascent from (1, 1) is ⟨−2, −4⟩ (or equivalently, the normalized vector ⟨−1/√5, −2/√5⟩). The maximum rate of increase is 2√5 ≈ 4.47.

In which direction should you walk to go downhill fastest? Opposite the gradient: the direction ⟨2, 4⟩ (normalized: ⟨1/√5, 2/√5⟩).

Example 3 (temperature): If T(x, y) = 60/(1 + x² + y²) represents temperature on a metal plate, find the direction of greatest temperature increase at (2, 1).

∇T = ⟨60·(−2x)/(1+x²+y²)², 60·(−2y)/(1+x²+y²)²⟩ = ⟨−120x/(1+x²+y²)², −120y/(1+x²+y²)²⟩

At (2, 1): 1 + x² + y² = 1 + 4 + 1 = 6
∇T(2, 1) = ⟨−240/36, −120/36⟩ = ⟨−20/3, −10/3⟩

Direction of greatest increase: toward the origin (negative components point radially inward).
Rate: ‖∇T‖ = √(400/9 + 100/9) = √(500/9) = 10√5/3 ≈ 7.45 °/unit

The temperature increases fastest toward the origin (where T = 60 is maximum). Makes sense: T(x, y) has a peak at the origin.

4. Gradient Perpendicular to Level Curves

Theorem: At any point (x₀, y₀), ∇f(x₀, y₀) is perpendicular to the level curve f(x, y) = k passing through that point.

Proof sketch: Parameterize the level curve as r(t) = ⟨x(t), y(t)⟩ with f(x(t), y(t)) = k. Differentiate:

$∂f/∂x · dx/dt + ∂f/∂y · dy/dt = 0
∇f · r′(t) = 0
$

∇f is perpendicular to the tangent vector r′(t), so ∇f is normal to the level curve.

Application — Tangent line to a level curve: The tangent line to f(x, y) = k at (x₀, y₀) is:

$f_x(x₀, y₀)(x − x₀) + f_y(x₀, y₀)(y − y₀) = 0
$

Example 4: Find the tangent line to the ellipse x²/4 + y² = 1 at (√2, 1/√2).

$f(x, y) = x²/4 + y² = 1

∇f = ⟨x/2, 2y⟩ → ∇f(√2, 1/√2) = ⟨√2/2, 2/√2⟩ = ⟨1/√2, √2⟩

Tangent line: (1/√2)(x − √2) + √2(y − 1/√2) = 0
→ (x − √2)/√2 + √2 y − 1 = 0
→ x/√2 + √2 y = 2
$

5. Gradient for Three Variables — Level Surfaces

For f(x, y, z), ∇f = ⟨f_x, f_y, f_z⟩ is perpendicular to the level surface f(x, y, z) = k.

Tangent plane to a level surface: At (x₀, y₀, z₀) on f(x, y, z) = k:

$f_x(x₀, y₀, z₀)(x − x₀) + f_y(x₀, y₀, z₀)(y − y₀) + f_z(x₀, y₀, z₀)(z − z₀) = 0
$

Or compactly: ∇f(x₀, y₀, z₀) · ⟨x − x₀, y − y₀, z − z₀⟩ = 0.

Example 5: Find the tangent plane to the sphere x² + y² + z² = 14 at (1, 2, 3).

$f(x, y, z) = x² + y² + z²
∇f = ⟨2x, 2y, 2z⟩ → ∇f(1, 2, 3) = ⟨2, 4, 6⟩

Tangent plane: 2(x − 1) + 4(y − 2) + 6(z − 3) = 0
→ 2x − 2 + 4y − 8 + 6z − 18 = 0
→ 2x + 4y + 6z = 28
→ x + 2y + 3z = 14
$

6. Directional Derivative via the Chain Rule

Directional derivatives are connected to the chain rule. If we parameterize motion along the direction u from (x₀, y₀) as:

$x(t) = x₀ + t a,    y(t) = y₀ + t b
$

Then z(t) = f(x(t), y(t)) and:

$dz/dt|_{t=0} = f_x·a + f_y·b = ∇f · u = D_u f
$

This is why the formula works — it's the chain rule evaluated at t = 0.

Key Terms

• 06 06 Directional Derivatives And Gradient
• **Chain Rule

Directional derivatives - Computing D_u f — The Gradient Formula - Correct: A) 3x + 4y = 25 - Correct: A) ⟨2, 2⟩ - Correct: B) 5 - Correct: B) In the direction of ∇f - Correct: B) ⟨6, 8⟩ - Correct: D) (3√2)/2 - Directional Derivative via the Chain Rule - Directional Derivative — Definition - End-of-Subject Quiz**

Worked Examples

Worked Example 1: Directional Derivative Computation

Problem: Find the directional derivative of f(x, y) = e^y sin x at (π/6, 0) in the direction from (π/6, 0) to (π/3, 1).

Solution: First, find the direction vector: v = ⟨π/3 − π/6, 1 − 0⟩ = ⟨π/6, 1⟩.

Unit vector: ‖v‖ = √(π²/36 + 1) = √(π² + 36)/6. u = ⟨π/6, 1⟩ · 6/√(π² + 36) = ⟨π, 6⟩/√(π² + 36).

Gradient:

$f_x = e^y cos x    →    f_x(π/6, 0) = 1·cos(π/6) = √3/2
f_y = e^y sin x    →    f_y(π/6, 0) = 1·sin(π/6) = 1/2

∇f = ⟨√3/2, 1/2⟩
$

Directional derivative:

$D_u f = ⟨√3/2, 1/2⟩ · ⟨π, 6⟩/√(π² + 36)
      = (π√3/2 + 3)/√(π² + 36)
      = (π√3 + 6)/(2√(π² + 36))
$

Worked Example 2: Maximum Rate of Change

Problem: The height of a mountain is given by h(x, y) = 2000 − 0.001x² − 0.004y² (in meters, x and y in meters from the peak). You are at (500, 300). In what direction should you go to climb most steeply, and what is that rate?

Solution:

$∇h = ⟨−0.002x, −0.008y⟩
∇h(500, 300) = ⟨−1, −2.4⟩
‖∇h‖ = √(1 + 5.76) = √6.76 = 2.6 meters per meter

Direction of steepest ascent: ⟨−1, −2.4⟩ (toward the origin/peak).
Unit direction: ⟨−1/2.6, −2.4/2.6⟩ ≈ ⟨−0.385, −0.923⟩.
$

Worked Example 3: Tangent Line to Level Curve

Problem: Find the tangent line to the curve x²y + xy² = 6 at (1, 2).

Solution:

$f(x, y) = x²y + xy²
∇f = ⟨2xy + y², x² + 2xy⟩
∇f(1, 2) = ⟨2(1)(2) + 4, 1 + 2(1)(2)⟩ = ⟨4 + 4, 1 + 4⟩ = ⟨8, 5⟩

Tangent line: 8(x − 1) + 5(y − 2) = 0
→ 8x − 8 + 5y − 10 = 0
→ 8x + 5y = 18
$

Worked Example 4: Tangent Plane to Level Surface

Problem: Find the tangent plane to the ellipsoid x² + 2y² + 3z² = 21 at (1, 2, 2).

Solution:

$f(x, y, z) = x² + 2y² + 3z²
∇f = ⟨2x, 4y, 6z⟩
∇f(1, 2, 2) = ⟨2, 8, 12⟩

Tangent plane: 2(x − 1) + 8(y − 2) + 12(z − 2) = 0
→ 2x − 2 + 8y − 16 + 12z − 24 = 0
→ 2x + 8y + 12z = 42
→ x + 4y + 6z = 21
$

Quiz

Q1: The directional derivative D_u f requires u to be:

A) Any vector B) A unit vector (‖u‖ = 1) C) A vector parallel to ∇f D) A vector perpendicular to ∇f

Correct: B)

• If you chose B: D_u f = ∇f · u is only valid when u is a unit vector. Using a non-unit vector scales the result by ‖v‖. Correct!
• If you chose A: Using any vector without normalizing gives the wrong rate of change.
• If you chose C: That maximizes D_u f but isn't required.
• If you chose D: That makes D_u f = 0, which is a valid direction but not a requirement.

Q2: For f(x, y) = x² + y², the gradient ∇f(3, 4) is:

A) ⟨3, 4⟩ B) ⟨6, 8⟩ C) ⟨9, 16⟩ D) ⟨5, 5⟩

Correct: B)

• If you chose B: f_x = 2x → 6, f_y = 2y → 8. ∇f = ⟨6, 8⟩. Correct!
• If you chose A: This is the position vector at (3,4), not the gradient.
• If you chose C: This is f evaluated componentwise as 3²=9, 4²=16 — didn't differentiate.
• If you chose D: Random values.

Q3: The gradient ∇f at a point is always:

A) Tangent to the level curve through that point B) Perpendicular (normal) to the level curve through that point C) Zero D) Parallel to the x-axis

Correct: B)

• If you chose B: ∇f is perpendicular to level curves. The tangent line equation is ∇f · ⟨x−x₀, y−y₀⟩ = 0. Correct!
• If you chose A: The tangent vector is perpendicular to the gradient.
• If you chose C: Only at critical points.
• If you chose D: No reason for ∇f to align with the x-axis.

Q4: The direction of steepest ascent is:

A) Any direction perpendicular to ∇f B) The direction of ∇f C) The direction of −∇f D) The direction of the x-axis

Correct: B)

• If you chose B: D_u f = ‖∇f‖ cos θ is maximized when cos θ = 1, i.e., u is parallel to ∇f. Correct!
• If you chose A: Perpendicular to ∇f gives D_u f = 0 (direction of no change).
• If you chose C: −∇f is steepest descent.
• If you chose D: Only if ∇f happens to point along the x-axis.

Q5: Find D_u f at (1, 1) for f(x, y) = x²y in the direction ⟨1, 1⟩.

A) 3 B) √2 C) 3√2 D) (3√2)/2

Correct: D)

• If you chose D: ∇f = ⟨2xy, x²⟩ → ∇f(1,1) = ⟨2, 1⟩. u = ⟨1,1⟩/√2. D_u f = 2/√2 + 1/√2 = 3/√2 = 3√2/2. Correct!
• If you chose A: Forgot to normalize u — used v = ⟨1,1⟩ directly: 2·1 + 1·1 = 3.
• If you chose B: That's ‖∇f‖ but ∇f(1,1) norm is √(4+1) = √5, not √2.
• If you chose C: Multiplied 3 by √2 instead of dividing.

Q6: Given ∇f(2, −1) = ⟨−4, 8⟩, what is the maximum rate of increase of f at that point?

A) 4 B) √80 = 4√5 C) 12 D) 8

Correct: B)

• If you chose B: Maximum rate = ‖∇f‖ = √(16 + 64) = √80 = 4√5 ≈ 8.94. Correct!
• If you chose A: That's just |f_x|.
• If you chose C: That's |f_x| + |f_y| = 4 + 8 = 12, not the magnitude.
• If you chose D: That's just |f_y|.

Practice Problems

(Answers are below. Try each problem before checking.)

Problem 1: Find the directional derivative of f(x, y) = x³ − 3xy + 4y² at (1, 2) in the direction of the vector ⟨3, 4⟩.

Problem 2: Find the direction in which f(x, y) = x² + xy + y² increases most rapidly at (−1, 1). What is the maximum rate of increase?

Problem 3: Find the tangent line to the level curve x³ + y³ = 9 at (2, 1).

Problem 4: Find the tangent plane to the surface z = 9 − x² − y² at (1, 2, 4). (Hint: write as F(x, y, z) = x² + y² + z − 9 = 0, a level surface of F.)

Problem 5: If T(x, y) = 100 − x² − 2y² describes temperature, in which direction from (3, 2) should an insect move to cool down fastest?

Problem 6: Compute D_u f(0, 0) for f(x, y) = x²y/(x² + y²) if (x, y) ≠ (0, 0), f(0, 0) = 0, in the direction u = ⟨1/√2, 1/√2⟩. Use the limit definition.

Problem 7: Find the points on the ellipsoid x² + 2y² + 3z² = 1 where the tangent plane is parallel to the plane 3x − y + 3z = 1.

Summary

1. The directional derivative D_u f = ∇f · u gives the rate of change of f in the direction of the unit vector u — it generalizes partial derivatives to arbitrary directions
2. The gradient ∇f = ⟨f_x, f_y⟩ (or ⟨f_x, f_y, f_z⟩ in R³) is the central vector in multivariable calculus: D_u f = ∇f · u = ‖∇f‖ cos θ
3. ∇f points in the direction of steepest ascent with magnitude equal to the maximum rate of change; −∇f points in the direction of steepest descent
4. The gradient is always perpendicular to level curves (R²) and level surfaces (R³), providing an effortless way to find tangent lines and tangent planes to implicitly defined curves and surfaces
5. The directional derivative exists even for some non-differentiable functions, highlighting that it probes only one path while differentiability requires good behavior in all directions

Pitfalls

• Forgetting to normalize the direction vector. The directional derivative formula D_u f = ∇f · u requires u to be a UNIT vector. Using the raw direction vector v = ⟨a, b⟩ without normalizing by ‖v‖ gives a result that is off by a factor of ‖v‖. Always compute u = v/‖v‖ first.
• Confusing steepest ascent with steepest descent. ∇f points in the direction of maximum INCREASE. The direction of maximum DECREASE is −∇f. Students often lose points by giving the gradient as the answer when asked "which way is downhill?".
• Thinking the gradient points along level curves. The gradient is PERPENDICULAR (normal) to level curves and level surfaces. If you think ∇f lies tangent to the level curve, you'll write the wrong tangent line equation — the tangent line uses ∇f as the normal vector, not the direction vector.
• Misapplying the dot product interpretation. D_u f = ‖∇f‖ cos θ only equals ‖∇f‖ when u is parallel to ∇f. Students sometimes set D_u f = ‖∇f‖ without checking whether u is actually in the gradient direction, or neglect to normalize u and get a mismatch between formula and computation.
• Assuming D_u f exists implies differentiability. The directional derivative can exist in every direction at a point even when the function is not differentiable (or even continuous) there. If all directional derivatives exist but the function is not differentiable, the gradient formula D_u f = ∇f · u may fail.

Next Steps

Move on to 06-07 — Optimization (Multivariable) to learn how to find local maximums and minimums of multivariable functions using critical points, the second derivative test (Hessian determinant), and the Extreme Value Theorem on closed bounded regions.