MATH U101 · Multivariable Calculus · Module 5

Directional Derivatives, the Gradient & Tangent Planes

Thomas §14.5–14.6*, lectures L10–L11. Module 4's partial derivatives, put to work: slopes in every direction, the steepest direction, and the plane that touches a surface. Error estimation (§14.6) is excluded by the handout.

🎧 Narrated walkthroughs for this module

Gradient & directional derivatives (8.3 min, T6 Q8–Q12) · Tangent planes, normal lines & intersections (6.9 min, T7 Q1–Q3). Each one is sound-on and pauses for your next move.

Start here

This block is on every recent mid-sem. 2023: Q2, 15 of 60 marks (find u from a given rate; tangent plane parallel to the xy-plane). 2024: Q3, 20 of 60 (directional derivative from the definition at a bad point; unit normal to a level surface). 2025: Q1, 18 of 50 (continuity, partials and directional derivatives of a piecewise function). This year's mid-sem (5 Oct, open book) covers Modules 1–6.

The computing is Module 4's partial derivatives plus one dot product. The new ideas, the gradient and the tangent plane, are new to everyone in the hall. First time? Start with the Module 5 lesson (slow build, direction widget); this page is the compressed map for revision.

1 · The directional derivative and the gradient §14.5

Intuition: the slope of the surface when you walk from (a, b) in the direction of a unit vector u = (u₁, u₂). Definition:

D_uf(a, b) = lim_h→0 f(a + hu₁, b + hu₂) − f(a, b)h

Shortcut, valid when f is differentiable at the point (e.g. continuous partials; Module 4 §4):

∇f = (f_x, f_y), D_uf = ∇f · u = f_xu₁ + f_yu₂

Normalise first: "in the direction of v" means u = v/|v|. Dotting with an unnormalised v multiplies the answer by |v|.
Three variables work the same way: ∇f = (f_x, f_y, f_z) and u has three components.
Axis directions give back the partials: D_if = f_x, D_jf = f_y.

2 · Steepest, flattest, and working backwards §14.5

Since |u| = 1, D_uf = |∇f| cos φ, where φ is the angle from ∇f to u.

Want	Direction	Rate
Fastest increase	u = ∇f/\|∇f\|	+\|∇f\|
Fastest decrease	u = −∇f/\|∇f\|	−\|∇f\|
No change	the two unit vectors ⟂ ∇f: for ∇f = (p, q), u = ±(−q, p)/\|∇f\|	0
A given rate k	solve f_xu₁ + f_yu₂ = k with u₁² + u₂² = 1	2 answers if \|k\| < \|∇f\|, 1 if equal, none if larger

Worked example 12023 mid-sem, Q2(a) [8 marks]: find u from the rate

As set: "If the directional derivative of the function f(x, y) = 2xy − 3x² at the point (5, 5) along the vector u = u₁i + u₂j is −4, then find u."

Gradient. f_x = 2y − 6x and f_y = 2x, so ∇f(5, 5) = (10 − 30, 10) = (−20, 10). Since |∇f| = √500 ≈ 22.4 > 4, expect two answers.
Two equations. Rate: −20u₁ + 10u₂ = −4, so u₂ = 2u₁ − 2/5. Unit length: u₁² + u₂² = 1.
Substitute and clear fractions. u₁² + (2u₁ − 2/5)² = 1 expands to 5u₁² − (8/5)u₁ + 4/25 = 1. Multiply by 25: 125u₁² − 40u₁ − 21 = 0.
Quadratic formula. u₁ = (40 ± √(1600 + 10500))/250 = (40 ± 110)/250, so u₁ = 3/5 or u₁ = −7/25.
Back-substitute: u₂ = 6/5 − 2/5 = 4/5, or u₂ = −14/25 − 10/25 = −24/25. Answer: u = (3/5)i + (4/5)j or u = −(7/25)i − (24/25)j.
Check (the open-book minute that saves marks): −20(3/5) + 10(4/5) = −12 + 8 = −4 ✓ and −20(−7/25) + 10(−24/25) = (140 − 240)/25 = −4 ✓. Both have length 1 ✓.

The handwritten answer on this paper wrote "(3, 4)" and "(−7, −24)": right directions, but not unit vectors. A directional derivative is only defined along a unit vector, so give (3/5, 4/5) and (−7/25, −24/25).

3 · The gradient is normal to level curves and surfaces §14.5

Along a level curve f doesn't change, so the rate along it is 0. That means the curve's direction is ⟂ ∇f. The same holds in 3D: ∇F is perpendicular to the level surface F(x, y, z) = c, and it points towards larger F.

Level curves of f = x² + 2y² with ∇f (scaled down) at points on them: always perpendicular, always pointing to larger f.

4 · At a special point of a piecewise function: use the definition

For f defined by a formula except at one point P (given a separate value there), differentiability at P is exactly what's in doubt, so ∇f · u is off-limits. The routine:

Shift if needed. If the special point isn't the origin, set X = x − a, Y = y − b and rewrite (complete the squares).
Substitute the walk (hu₁, hu₂) and simplify, cancelling powers of h. Use u₁² + u₂² = 1.
Subtract the special value, divide by h, let h → 0. Split cases (e.g. u₂ = 0 vs u₂ ≠ 0).
State for which u the limit exists, and its value.

Two outcomes the papers use:

Exists in every direction, yet f is discontinuous. 2024 Q3(a): x²y/(x⁶ + y²) gives D_uf(0, 0) = u₁²/u₂ (for u₂ ≠ 0). But along y = x², f = x⁴/(x⁶ + x⁴) = 1/(x² + 1) → 1 ≠ 0. Directional derivatives only test straight lines; the two-path test with a curve catches the discontinuity.
Exists only in some directions. 2025 Q1(c): after the shift, (X² − Y²)/(X² + Y²). The quotient is (u₁² − u₂²)/h, which has a limit only when u₁² = u₂² (worked example 2).
Diagnostic: if the definition's answer differs from ∇f · u (computed from the partials at P), then f is not differentiable at P.

Worked example 22025 mid-sem, Q1 [6 + 4 + 8 marks]: continuity, partials and directional derivatives at a special point

As set: "Consider the function f(x, y) = (x² − y² − 2x + 2y)/(x² + y² − 2x − 2y + 2) if (x, y) ≠ (1, 1), and 0 if (x, y) = (1, 1). (a) Determine whether f is continuous at the point (1, 1). (b) Determine whether the partial derivatives f_x(1, 1) and f_y(1, 1) exist. (c) Find all unit vectors u for which the directional derivative D_uf(1, 1) = 0."

Shift to the special point. Complete the squares: x² − 2x = (x − 1)² − 1 and y² − 2y = (y − 1)² − 1. With X = x − 1, Y = y − 1, the numerator is (X² − 1) − (Y² − 1) = X² − Y², and the denominator is (X² − 1) + (Y² − 1) + 2 = X² + Y². So f = (X² − Y²)/(X² + Y²), with value 0 at X = Y = 0.
(a) Two paths. Along Y = 0: f = X²/X² = 1. Along X = 0: f = −1. Different limits, so the limit at (1, 1) does not exist and f is not continuous at (1, 1).
(b) Partials by definition. f(1 + h, 1) = h²/h² = 1, so the quotient is (1 − 0)/h = 1/h, which is unbounded: f_x(1, 1) does not exist. Likewise f(1, 1 + h) = −1 gives −1/h: f_y(1, 1) does not exist.
(c) The walk. f(1 + hu₁, 1 + hu₂) = h²(u₁² − u₂²)/h²(u₁² + u₂²) = u₁² − u₂², using u₁² + u₂² = 1. The quotient is (u₁² − u₂²)/h.
When does the limit exist? If u₁² ≠ u₂² it blows up, so D_uf doesn't exist. If u₁² = u₂², the quotient is 0 for every h, so D_uf(1, 1) = 0.
Unit vectors. u₁² = u₂² and u₁² + u₂² = 1 give u₁² = u₂² = ½. Answer: u = (±1/√2, ±1/√2), all four sign choices. In every other direction the directional derivative doesn't exist, so it can't equal 0.

Worth writing in the answer: along the diagonals the function is constantly 0 (since X² = Y² there), which is why those four directions are flat. The paper says "Answers without proper reasoning will not be awarded marks", so state each case.

5 · Tangent planes and normal lines §14.6

For a level surface F(x, y, z) = c at P = (x₀, y₀, z₀) on the surface (check this first), the gradient ∇F(P) is the normal:

plane: F_x(P)(x − x₀) + F_y(P)(y − y₀) + F_z(P)(z − z₀) = 0

normal line: (x₀, y₀, z₀) + t·∇F(P)

Graph z = f(x, y): use F = f − z, which gives z − z₀ = f_x(x − x₀) + f_y(y − y₀).
Plane parallel to the xy-plane: normal ∥ k, so F_x = F_y = 0 (and F_z ≠ 0). Parallel to the xz-plane: F_x = F_z = 0. Parallel to the yz-plane: F_y = F_z = 0. Then solve those equations together with the surface equation.
Unit normal: ±∇F/|∇F|, both signs. 2024 Q3(b): F = xyz² − 4 at (−1, −1, 2) has ∇F = (yz², xz², 2xyz) = (−4, −4, 4). So u = ±(−1, −1, 1)/√3, and the directional derivative along it is ±|∇F| = ±4√3 (along the gradient's own direction, the rate is the maximum).

Worked example 32023 mid-sem, Q2(b) [7 marks]: tangent plane parallel to the xy-plane

As set: "Find the point(s) on the surface (x − y)² + y² + (y + z)² = 1, where the tangent plane is parallel to the xy-plane."

Set up. F = (x − y)² + y² + (y + z)². Parallel to the xy-plane means normal ∥ k: we need F_x = 0 and F_y = 0.
Partials (chain rule on each square): F_x = 2(x − y), F_y = −2(x − y) + 2y + 2(y + z), F_z = 2(y + z).
Solve. F_x = 0 gives x = y. Then F_y = 0 + 2y + 2(y + z) = 4y + 2z = 0, so z = −2y.
Back into the surface. 0 + y² + (y − 2y)² = 2y² = 1, so y = ±1/√2.
Points: (1/√2, 1/√2, −√2) and (−1/√2, −1/√2, √2). Check F_z ≠ 0: F_z = 2(y + z) = −2y = ∓√2 ≠ 0 ✓, so the normal is a genuine vertical vector at both points.

6 · The tangent line to a curve of intersection §14.6

Two surfaces F(x, y, z) = c₁ and G(x, y, z) = c₂ usually meet in a curve. At a point P on both, the curve lies inside each surface, so its tangent vector lies in both tangent planes. That makes it perpendicular to both normals, ∇F(P) and ∇G(P). The vector perpendicular to two given vectors is their cross product:

direction v = ∇F(P) × ∇G(P), tangent line: P + t v

Written out, with the determinant layout that keeps the signs straight (note the minus on the middle term):

	i	j	k
∇F	F_x	F_y	F_z
∇G	G_x	G_y	G_z

∇F × ∇G = (F_yG_z − F_zG_y) i − (F_xG_z − F_zG_x) j + (F_xG_y − F_yG_x) k

1 · Check the point is on both surfaces: substitute it into each equation.
2 · Both gradients at the point: write each surface as F = c; a plane such as y = 1 is G = y, with ∇G = (0, 1, 0).
3 · The cross product, component by component, using the layout above.
4 · The parametric line x = x₀ + v₁t, y = y₀ + v₂t, z = z₀ + v₃t.
5 · Simplify the direction by scaling: any nonzero multiple of v gives the same line. Check v · ∇F = v · ∇G = 0.

Worked example 4Sphere meets plane: the tangent line at (1, 2, −1)

Find parametric equations for the line tangent to the curve of intersection of x² + y² + z² = 6 and x + y + z = 2 at (1, 2, −1).

On both? 1 + 4 + 1 = 6 ✓ and 1 + 2 − 1 = 2 ✓.
Gradients. F = x² + y² + z²: ∇F = (2x, 2y, 2z) = (2, 4, −2). G = x + y + z: ∇G = (1, 1, 1).
Cross product. i: 4·1 − (−2)·1 = 6. j: −(2·1 − (−2)·1) = −4. k: 2·1 − 4·1 = −2. So v = (6, −4, −2).
Scale. Divide by 2: direction (3, −2, −1). Line: x = 1 + 3t, y = 2 − 2t, z = −1 − t.
Check. (3, −2, −1)·(2, 4, −2) = 6 − 8 + 2 = 0 ✓ and (3, −2, −1)·(1, 1, 1) = 0 ✓: perpendicular to both normals.

The sphere and the plane meet in a circle. At P, ∇F (sphere's normal) and ∇G (plane's normal) point different ways; the tangent line runs along the circle, perpendicular to both. Computed orthographic view.

Tutorial 7 Q2–Q3 are this exact routine, solved on the tutorial companion: Q2 (a cylinder and a paraboloid meet in a horizontal circle, so the tangent comes out horizontal) and Q3 (the plane y = 1, whose gradient (0, 1, 0) has two zero components; still just cross it).

Classic traps · curves of intersection

1) Using one gradient as the direction. ∇F alone gives the normal line of one surface, not the curve's tangent. 2) The middle sign. The j component is −(F_xG_z − F_zG_x); check with the two dot products. 3) Zero components. A gradient like (0, 1, 0) is fine; the cross product still works. 4) A zero cross product means the gradients are parallel: the surfaces touch there and this method gives no line. 5) Skipping the on-both check.

7 · Linearization and total differential §14.6*

L(x, y) = f(a, b) + f_x(a, b)(x − a) + f_y(a, b)(y − b) is the tangent plane used to approximate values. The total differential is df = f_x dx + f_y dy. Example: f = x² + 2y² at (1, 1) gives L(1.1, 0.9) = 3 + 0.2 − 0.4 = 2.8 (true value 2.83). Error estimation is excluded by the handout.

Classic traps · the module's mark-losers

1) Not normalising the direction vector, or reporting a direction like (3, 4) when a unit vector is asked for. 2) Using ∇f · u at a special point of a piecewise function. That shortcut needs differentiability; use the limit definition. 3) Stopping at one answer when solving for u: a line meets the unit circle twice. 4) "Every directional derivative exists, so f is continuous" is false (2024 Q3a); test a curved path like y = x². 5) Tangent plane at a point not on the surface: substitute the point into the equation first. 6) Graph surfaces: forgetting the −1 from F = f − z, which flips the plane's z-term. 7) Unit normal with one sign only: say ±.

Your minimal prerequisite kit for this module

Unit vectors: v/|v|, and |(a, b)| = √(a² + b²).
Dot product, both forms: a₁b₁ + a₂b₂ = |a||b| cos φ; perpendicular exactly when it is 0.
Solving a line with a circle: substitute, expand a square, quadratic formula (worked example 1).
Plane through a point with a given normal (A, B, C): A(x − x₀) + B(y − y₀) + C(z − z₀) = 0.
Module 4 partials (fast and accurate) and Module 3's path test for limits.
Completing the square, to shift a special point to the origin (worked example 2).
Cross product of two 3-vectors by the i/j/k determinant, with the minus on the middle term (section 6).

What to practise

Exercise ranges are approximate for the 15th edition, so check against your copy. Prefer odd numbers (answers in the back).

Skill	Where	How many
Gradients; directional derivative towards a given vector (normalise!)	Thomas §14.5 exercises ~1–30	6–8 odd
Directions of fastest increase/decrease and zero change	§14.5 exercises ~31–38	3–4 odd
Tangent planes and normal lines to level surfaces and graphs	§14.6 exercises ~1–18	5–6 odd
Tangent line to a curve of intersection (∇F × ∇G)	§14.6 exercises on curves of intersection (check your copy) · Tutorial 7 Q2, Q3	3–4
Linearization and total differential	§14.6 linearization exercises (skip error estimates)	2–3
Past mid-sem questions on this block	2023 Q2 · 2024 Q3 · 2025 Q1	all, timed