Directional Derivatives, the Gradient and Tangent Planes, taught from zero

1 · A slope in any direction §14.5

Module 4 gave the slope of a surface z = f(x, y) in exactly two directions: walking east (f_x) and walking north (f_y). But a hiker can walk north-east, or in any compass direction at all, and in each one the ground rises or falls at its own rate. That rate is the directional derivative.

A direction is described by a unit vector u = u₁i + u₂j, meaning u₁² + u₂² = 1. Starting from the point (a, b), walk a distance h in direction u: you arrive at (a + hu₁, b + hu₂). Rise over run, then let the step shrink:

Why must u be a unit vector? Because then h is exactly the distance walked, and "rise ÷ distance" is a genuine slope. With a longer vector, say length 5, a step of h covers a distance of 5h. The quotient comes out 5 times too big and measures nothing meaningful.

Two special cases show this is not a new idea, just a wider one. With u = i = (1, 0) the formula is exactly Module 4's definition of f_x, and with u = j it is f_y. Partial derivatives are directional derivatives along the axes.

2 · The shortcut: one dot product §14.5

Computing that limit every time would be slow. For the functions you normally meet there is a shortcut. Package the two partials into one vector, the gradient:

Then, whenever f is differentiable at the point (Module 4, §4: for example, when its partials are continuous there),

Watch it once. Take f(x, y) = x² + 2y² at (1, 1), walking towards the direction of the vector 3i + 4j.

1. Gradient: f_x = 2x and f_y = 4y, so ∇f(1, 1) = 2i + 4j.
2. Normalise the direction first. |3i + 4j| = √(9 + 16) = 5, so u = (3/5, 4/5).
3. Dot: D_uf = 2·(3/5) + 4·(4/5) = 6/5 + 16/5 = 22/5 = 4.4.

3 · Which way is steepest? §14.5

A dot product has a second form: a · b = |a||b| cos φ, where φ is the angle between the vectors. Since |u| = 1,

Everything about direction follows from cos φ running between −1 and 1:

• Steepest ascent: φ = 0. Walk along ∇f itself; the rate is +|∇f|, the largest possible.
• Steepest descent: φ = π. Walk along −∇f; the rate is −|∇f|.
• No change: φ = π/2. Walk perpendicular to ∇f (there are two such directions); the rate is 0.

So the gradient is a compass and a speedometer in one: it points uphill, steepest way, and its length is how steep. Try it below on f = x² + 2y², whose level curves are ellipses. Turn the direction u and watch the rate follow cos φ.

Things to notice as you turn u: the rate peaks exactly when u lines up with the blue arrow, and it passes through 0 exactly when u runs along the green level curve. That second fact is the subject of section 5. At (1, 1) the numbers are |∇f| = |(2, 4)| = √20 = 2√5 ≈ 4.47, and the steepest direction is (1, 2)/√5.

4 · Working backwards: find the direction from the rate

A favourite exam twist runs the formula in reverse: "the directional derivative at this point is k; find u." The 2023 mid-sem did exactly this (Q2a). The unknowns are u₁ and u₂, and you have two facts about them:

• the rate: f_xu₁ + f_yu₂ = k (a straight line in the (u₁, u₂) plane);
• unit length: u₁² + u₂² = 1 (the unit circle).

A line meets a circle in 2, 1 or 0 points, so expect two answers in general. There is one answer when |k| = |∇f| (the steepest direction itself), and none when |k| > |∇f|, as in the check above.

Twin of the exam question. f = x²y at (1, 1) with D_uf = 1.

1. ∇f = 2xy i + x² j, so ∇f(1, 1) = (2, 1). Since |∇f| = √5 > 1, expect two answers.
2. Rate equation: 2u₁ + u₂ = 1, so u₂ = 1 − 2u₁.
3. Substitute into the circle: u₁² + (1 − 2u₁)² = 1.
4. Solve: 5u₁² − 4u₁ = 0, so u₁ = 0 or u₁ = 4/5.
5. Back-substitute: u = (0, 1) or u = (4/5, −3/5). Check both: 2·0 + 1 = 1 ✓, 8/5 − 3/5 = 1 ✓, and both have length 1 ✓.

5 · The gradient is perpendicular to the level curves §14.5

Walk along a level curve and, by definition, f doesn't change: the rate in the direction of the curve is 0. Section 3 says the rate is 0 only for directions perpendicular to ∇f. Put the two together: at every point, ∇f is perpendicular to the level curve through that point, pointing towards higher values.

The same argument works one dimension up. For a function F(x, y, z), the gradient ∇F = (F_x, F_y, F_z) is perpendicular to the level surface F = c. That single fact builds every tangent plane in section 7.

6 · At a bad point, go back to the definition

This is the past papers' favourite question, and the one most students lose marks on. The setting: a function defined by a formula everywhere except one point, where it is given a separate value, like

At the origin the shortcut ∇f · u is not allowed. It needs differentiability, which is exactly what is in doubt. So use the definition from section 1 with (a, b) = (0, 0):

1. Substitute the walk. f(hu₁, hu₂) = h³u₁²u₂h⁴u₁⁴ + h²u₂² = hu₁²u₂h²u₁⁴ + u₂² (cancel h² top and bottom).
2. Form the quotient. Subtract f(0, 0) = 0 and divide by h: u₁²u₂h²u₁⁴ + u₂².
3. Let h → 0, splitting cases. If u₂ ≠ 0, the h² term vanishes: D_uf(0, 0) = u₁²u₂/u₂² = u₁²/u₂. If u₂ = 0, the quotient is 0 for every h, so the limit is 0.

So the directional derivative exists in every direction. Now the surprise: this function is not even continuous at the origin. Along any straight line y = mx the values tend to 0, but along the parabola y = x² they are x⁴/(x⁴ + x⁴) = ½ exactly, all the way into the origin. That is Module 3's two-path test, with a curve as the second path.

Three lessons, all examinable:

• Directional derivatives in every direction do not imply continuity. Directional derivatives only test straight lines, and a function can misbehave along curves.
• The shortcut gives the wrong answer here. Along the axes f is 0, so f_x(0, 0) = f_y(0, 0) = 0 and ∇f · u = 0. But the true value is u₁²/u₂. When the definition and the shortcut disagree, the function is not differentiable at that point.
• At a special point, always use the definition. The 2024 mid-sem (Q3a) asked exactly this with x⁶ in place of x⁴. The same three steps give the same u₁²/u₂.

The other way it can go: the limit exists in only some directions. Take f = xy/(x² + y²) with f(0, 0) = 0. Then f(hu₁, hu₂) = u₁u₂ (the h² cancels, and u₁² + u₂² = 1), so the quotient is u₁u₂/h. It has a finite limit only when u₁u₂ = 0, that is, only along the four axis directions, where it is 0. The 2025 mid-sem (Q1c) used this pattern, disguised by centring the function at (1, 1). A shift X = x − 1, Y = y − 1 strips the disguise.

7 · Tangent planes and normal lines §14.6

Zoom in on a smooth surface and it looks flat. That flat sheet is the tangent plane, and section 5 hands us its orientation for free. Write the surface as a level surface F(x, y, z) = c. Then ∇F at the point P = (x₀, y₀, z₀) is perpendicular to the surface, so it is the plane's normal vector. A plane through P with normal (A, B, C) is A(x − x₀) + B(y − y₀) + C(z − z₀) = 0. So:

Example. The sphere x² + y² + z² = 14 at P = (1, 2, 3) (check: 1 + 4 + 9 = 14 ✓, so P is on it). With F = x² + y² + z², ∇F = (2x, 2y, 2z) = (2, 4, 6) at P. The tangent plane is 2(x − 1) + 4(y − 2) + 6(z − 3) = 0, which simplifies to x + 2y + 3z = 14. The normal line is (1 + 2t, 2 + 4t, 3 + 6t); at t = −½ it passes through the centre (0, 0, 0), as a sphere's normals should.

When the surface is a graph z = f(x, y), move everything to one side: F = f(x, y) − z = 0. Then ∇F = (f_x, f_y, −1), and the plane rearranges to

For z = x² + 2y² at (1, 1, 3): z = 3 + 2(x − 1) + 4(y − 1).

"Where is the tangent plane parallel to the xy-plane?" (2023 mid-sem, Q2b.) The xy-plane has normal k = (0, 0, 1). Parallel planes have parallel normals, so ∇F must point straight up or down: F_x = 0 and F_y = 0, with F_z ≠ 0. Solve those together with the surface equation. Twin: on (x − 1)² + (y − 2)² + z² = 9, F_x = 2(x − 1) = 0 and F_y = 2(y − 2) = 0 give x = 1, y = 2. The surface then forces z² = 9, so the points are (1, 2, ±3): the top and bottom of the sphere, as you'd expect.

A unit normal (2024 mid-sem, Q3b) is ∇F divided by |∇F|, and its negative is equally valid. Give both, or say "±".

Where two surfaces meet

Two surfaces usually cross along a curve: a cylinder cut by a plane gives a circle or an ellipse, and a sphere cut by a plane gives a circle. What is the tangent line to that curve at a point P on both?

Zoom in at P. Each surface looks like its tangent plane, so near P the curve looks like the place where the two tangent planes cross, and two planes cross in a straight line. That line is the tangent we want. It lies in the first plane, so it is perpendicular to the first normal ∇F; it lies in the second, so it is perpendicular to ∇G too. The one direction perpendicular to two given vectors is their cross product, so the direction is ∇F × ∇G. The recipe, a worked example and a 3D figure are in the notes, section 6; Tutorial 7 Q2–Q3 use exactly this.

8 · Linearization and the total differential §14.6*

The tangent plane is also the best flat approximation to the surface near the point. Its formula, used for estimating values, is called the linearization of f at (a, b):

For f = x² + 2y² at (1, 1): L(1.1, 0.9) = 3 + 2(0.1) + 4(−0.1) = 2.8, against the true value 1.21 + 1.62 = 2.83. The total differential packages the same idea as a change: df = f_x dx + f_y dy (here 2(0.1) + 4(−0.1) = −0.2). The handout excludes §14.6's error-estimation part, so the formulas and one numeric use like this are all you need.

9 · You're ready: what to do next

The whole module in five sentences. The directional derivative is the slope along a unit vector. For differentiable functions it is ∇f · u = |∇f| cos φ, so the gradient points steepest-uphill with length equal to the steepest slope. The gradient is perpendicular to level curves and surfaces, which makes it the normal of every tangent plane. At a special point of a piecewise function, the shortcut is off-limits: go back to the limit definition. Normalise every direction before you dot.

• Rehearse the exam patterns on the notes page: three past mid-sem questions worked in full.
• Then the real papers: the mid-sem past papers page has this block's questions from 2023, 2024 and 2025.
• Mechanics: the notes page's exercise table maps §14.5–14.6 (odd numbers, answers in the back).