MATH U101 · Lesson · before the 5 Oct mid-sem (open book)

How to start a calculus question

What to write in the first two minutes, when you've read a mid-sem question and nothing comes. Five moves that need no idea, only recognition and translation. Then eleven question families, each shown on the real past-paper questions, stopping exactly where the routine hands over to algebra you already know.

Start here — three facts

1. Following a solution and starting one are two different skills. Reading worked solutions trains the first: checking that each line follows from the last. Starting needs the second: producing line one from a blank page. If you've mostly practised the first, "I understand it once I see it, but I can't begin" is exactly what that practice predicts. It says nothing about ability, and it is fixable by drilling openings, which is what this page and the start drill are for.

2. The opening lines are marks. The 2025 paper (open book, 90 minutes, the format closest to yours) says in its instructions: "Please show all the work. Answers without proper reasoning will not be awarded marks." Naming the formula, computing the derivatives at the point and setting up the integral are "proper reasoning". A question you only open still earns something; a blank earns nothing.

3. Your paper is open book, so the formulas are on the desk. What the exam tests is knowing which formula a question needs and what to feed it. That is the recognition step below, and it's trainable. Portions are Modules 1–6. Every past-paper question in those portions falls into one of eleven families, and two past-paper parts are off this year's syllabus (see the skip list).

The routine
Trigger words
F1 Polar area
F2 Polar tangents
F3 Curvature
F4 Osculating circle
F5 Continuity
F6 Derivative by definition
F7 Gradient
F8 Tangent plane
F9 Local extrema
F10 Absolute extrema
F11 Lagrange
Skip list
Exam room

1 · The routine: five moves before any thinking

When you freeze on a question, the freeze is usually a search for "the trick". The routine replaces that search with five small writing jobs. Each one is always possible, and after the fourth, the question has turned into a calculation.

Name the familyRead the question for its trigger words ("area", "osculating", "continuous at", "directional derivative", "absolute maximum" …) and write the family name at the top of your answer. The table below maps words to families.
Write the starting formulaEach family starts from one formula or definition. Write it down in general form, with the book open at its page (the tab list says where). For example, A = ½∫(r_out² − r_in²) dθ or D_uf = ∇f · u.
Translate the dataTurn what the question gives into that formula's ingredients: the derivatives at the point, the gradient at the point, the intersection angles, the shifted variables. This is mechanical differentiation or solving, nothing new.
Set up before computingWrite the whole thing with blanks filled but not yet evaluated: the integral with its limits, the system of equations, the list of candidate points. Most lost marks come from computing something that was set up wrongly.
Plan the checkOne line on how you'll know the answer is sane: an area must be positive; a direction vector must be a unit vector; the centre of an osculating circle must be on the concave side; a candidate point must lie in the region.

Moves 1–4 are what the rest of this page drills, family by family. Move 5 takes ten seconds and catches the classic slips (the unit-vector slip below cost marks on a real 2023 script).

2 · Trigger words → family

If the question says…	Family	Starts from
"area", "inside … outside …", two polar equations, "shared / common region"	F1 Polar area	½∫(r_out² − r_in²) dθ
"tangent … horizontal / vertical / parallel to the x-axis" on a polar curve	F2 Polar tangents	dy/dθ = 0 or dx/dθ = 0
"curvature", "radius of curvature", dT/ds	F3 Curvature	κ = \|y″\|/(1 + y′²)^3/2 or \|v × a\|/\|v\|³
"osculating circle", "circle of curvature"	F4 Osculating circle	centre = P + ρN
a piecewise function (one formula off a point, a value at it); "continuous at"	F5 Continuity	two paths / polar squeeze
the same piecewise function, and "partial derivatives exist?" or "directional derivative at that point"	F6 Derivative by definition	lim [f(P + hu) − f(P)]/h
"directional derivative … in the direction of", "increases most rapidly" (a smooth f)	F7 Gradient	∇f · u, u a unit vector
"tangent plane", "normal line", "normal to the surface", "plane parallel to the xy-plane"	F8 Tangent plane	∇F is the normal
"critical points", "classify", "local max / min / saddle"	F9 Local extrema	∇f = 0, D = f_xxf_yy − f_xy²
"absolute maximum and minimum" on a region (a triangle, a disc, \|x\| + \|y\| ≤ 1 …)	F10 Absolute extrema	interior + edges + corners
"subject to", "on the curve / sphere", "Lagrange multipliers"	F11 Lagrange	∇f = λ∇g, g = c
"torsion", "binormal", dB/ds, any Σ series, projectile, "estimate the error"	Off syllabus	skip (not in this year's portions)

Check yourself: which family? "Let f(x,y) = x²y/(x⁴ + y²) off the origin and f(0,0) = 0. Find the directional derivative of f at (0,0) in the direction of u = (3/5, 4/5)."

F6, derivative by definition, not F7. The trigger is "at (0,0)" for a piecewise function defined separately at that point: you can't use ∇f · u there, because f isn't differentiable at the origin. The limit definition gives u₁²/u₂ = (9/25)/(4/5) = 9/20, while ∇f(0, 0) · u = 0, which is wrong.

3 · F1 — Polar area between curves

🎧 Watch it done first: narrated walkthrough (about 5 min, sound on). It acts out these moves on 2023 Q4 and stops three times to ask for your next move.

The family

Trigger words: "area of the region inside … and outside …", two polar equations, "shared region".

Intersections: set r₁ = r₂ and solve for θ. Also ask whether both curves pass through the origin (a second meeting point that the equation doesn't show).
Which is outer on each interval: test one angle inside each interval. The outer curve is the one farther from the origin along that ray.
Split the integral wherever the outer or inner curve changes (including where a curve stops existing, e.g. r = 6 cos θ for θ > π/2), then ½∫(r_out² − r_in²) dθ on each piece.

Common wrong first move: writing ½∫(r_out − r_in)² dθ (square of the difference instead of difference of squares), or using one integral across a point where the inner curve changes.

Opening on 2023 Q4

As set · 2023 mid-sem, Q4 · 5 + 10 marks

4. For the given curve C₁ : r = 6 cos θ, and C₂ : r = 2 + 2 cos θ, R is the region above the x−axis that lies outside C₁ and inside C₂.

(a) Sketch and shade the region R. [5]
(b) Find the area of the region R. [10]

2023 Q4, drawn from the equations. The shaded region lies inside the cardioid and outside the circle; above the axis it starts at θ = π/3 and runs to θ = π.

Family"Area … outside C₁ and inside C₂", polar equations: F1.
Start formulaA = ½∫(r_out² − r_in²) dθ (Thomas §11.5).
TranslateIntersection: 6 cos θ = 2 + 2 cos θ ⇒ cos θ = ½ ⇒ θ = π/3 above the axis. Test angles: at θ = 0 the circle (r = 6) is outside the cardioid (r = 4), so R has nothing on [0, π/3]. On [π/3, π/2] the cardioid is outer and the circle inner. On [π/2, π] the circle is gone (6 cos θ ≤ 0: those rays meet it only at the origin), so the inner radius is 0.
Set upA = ½∫_π/3^π/2 [(2 + 2cos θ)² − 36 cos²θ] dθ + ½∫_π/2^π (2 + 2cos θ)² dθ.
Check planBoth pieces must come out positive; the total must be smaller than the cardioid's upper half, 3π.

Hand-over: from here it is cos²θ = (1 + cos 2θ)/2 and routine integration. The area is π/2. Full working: past papers, 2023 Q4.

Opening on 2024 Q2(b)

As set · 2024 mid-sem, Q2(b) · 10 marks

(b) Sketch the shared region x² + y² + 2y = 0 and x² + y² + 2x = 0, then express the area of the shared region in integral form (no need to evaluate the integral but write integral with limits, do not use double or triple integrals). [10]

Family"Shared region", "area … integral with limits", no double integrals: F1 (convert to polar first).
Start formula½∫(r_out² − r_in²) dθ; for a lens through the origin the inner radius is 0 and the outer radius is the nearer of the two curves.
Translatex² + y² = r², y = r sin θ: the circles are r = −2 sin θ (centre (0, −1)) and r = −2 cos θ (centre (−1, 0)). They meet at the origin and where sin θ = cos θ, i.e. θ = 5π/4 (the point (−1, −1)). The lens lies in the third quadrant, π ≤ θ ≤ 3π/2.
Set upA = ½∫_π^5π/4 4 sin²θ dθ + ½∫_5π/4^3π/2 4 cos²θ dθ — near θ = π the sine circle is nearer the origin, near 3π/2 the cosine one is.

Hand-over: the question stops here (no evaluation needed). It equals π/2 − 1 if you want to check. Past papers, 2024 Q2.

4 · F2 — Polar tangents

The family

Trigger words: "tangent line is horizontal / vertical", "a line parallel to the x-axis tangent to the curve".

Parametrise by θ: x = r(θ) cos θ, y = r(θ) sin θ. Multiply out before differentiating.
Horizontal: dy/dθ = 0 with dx/dθ ≠ 0. Vertical: the other way round. Factor the derivative and solve each factor on [0, 2π).
Drop cusps (both derivatives 0 at once, e.g. where r = 0 on a cardioid), and for "a line tangent at two points", pick the solutions that share the same y (or x). A horizontal line y = c is r = c csc θ in polar form.

Common wrong first move: setting dr/dθ = 0. That finds where the curve is farthest from the origin, not where its tangent is horizontal.

Opening on 2025 Q3

As set · 2025 mid-sem (open book), Q3 · 6 + 10 marks

(3) Consider the polar curve r = 4(1 − sin θ), 0 ≤ θ ≤ 2π. Draw a straight line L parallel to the X-axis which is tangent to the curve at two distinct points P and Q. Then

(a) find the polar coordinates of P and Q. [6M]
(b) Sketch the region ~~below the~~ enclosed by straight line L and outside the given polar curve. Find the area of this region. [10M] (correction handwritten on the paper)

2025 Q3, drawn from the equation. The line y = 1 touches both upper lobes; the shaded sliver between it and the cusp is part (b)'s region. The bottom point (8, 3π/2) also has a horizontal tangent, but it is only one point.

Family"Parallel to the X-axis … tangent": a horizontal tangent, F2. Part (b) then becomes F1, with the line as the outer curve.
Start formulay = r sin θ; horizontal where dy/dθ = 0, dx/dθ ≠ 0.
Translatey = 4(sin θ − sin²θ), so dy/dθ = 4 cos θ(1 − 2 sin θ).
Set upRoots on [0, 2π): cos θ = 0 gives π/2 (r = 0, the cusp: drop it) and 3π/2 (r = 8, a single bottom point, not two). sin θ = ½ gives π/6 and 5π/6, both with r = 2 and y = 2 · ½ = 1: two points at the same height. So P = (2, π/6), Q = (2, 5π/6) and L: y = 1, i.e. r = csc θ. For (b): A = ½∫_π/6^5π/6 [csc²θ − 16(1 − sin θ)²] dθ.
Check planAt π/6 both radii are 2 (the curves meet there); the area must be small and positive (it's the sliver in the figure).

Hand-over: ∫csc²θ dθ = −cot θ plus the expanded square. Area 15√3 − 8π ≈ 0.848. Past papers, 2025 Q3; the method is in Module 1 notes §4.

Check yourself: on r = 2(1 + cos θ), where is the tangent vertical? Just the opening: which derivative, and its factors.

x = 2(cos θ + cos²θ), so dx/dθ = −2 sin θ(1 + 2 cos θ). Roots: θ = 0 (r = 4), θ = π (r = 0, the cusp, drop), and cos θ = −½: 2π/3, 4π/3 with r = 1.

5 · F3 — Curvature and radius of curvature

The family

Trigger words: "curvature", "radius of curvature", "dT/ds = CN".

Pick the formula by how the curve is given: a graph y = f(x) → κ = |y″|/(1 + y′²)^3/2; a vector function or a velocity → κ = |v × a|/|v|³. Radius ρ = 1/κ.
Compute the ingredients at the point (or in general if the question asks "where").
If the question gives ρ and asks for the point, set the formula equal to it and solve for x.

Common wrong first move: not recognising dT/ds = κN as the definition of N, and trying to differentiate T with respect to arc length directly.

Opening on 2023 Q1(a)

As set · 2023 mid-sem, Q1 · 5 + 10 marks

1. For the given curve C : y = x², find

(a) the point(s) at which the radius of curvature is √2. [5]
(b) the equation of the osculating circle at the point (√2, 2). [10]

Family"Radius of curvature", a graph: F3.
Start formulaρ = (1 + y′²)^3/2/|y″|.
Translatey′ = 2x, y″ = 2, so ρ = (1 + 4x²)^3/2/2.
Set up(1 + 4x²)^3/2/2 = √2, i.e. (1 + 4x²)^3/2 = 2√2.

Unpack this step: freeing the bracket from the 3/2 power

2√2 = 2¹ · 2^1/2 = 2^3/2. So (1 + 4x²)^3/2 = 2^3/2, and raising both sides to the power 2/3 gives 1 + 4x² = 2.

Hand-over: x = ±½, so the points are (±½, ¼). Part (b) is F4 (next section).

Opening on 2024 Q1(b) (and why Q1(a) is skipped)

As set · 2024 mid-sem, Q1 · 10 + 8 marks

1. A particle moves in space with velocity vector given by v(t) = 3t²i + (1 + t)j + (1 − t)k. Determine the constants C₁ and C₂ at t = 1 in the following identities:

(a) dB/ds = C₁N. [10]
(b) dT/ds = C₂N. [8]

Family(a) has B, the binormal: that's torsion, Thomas §13.5, off this year's syllabus, so skip it. (b) dT/ds = κN is the definition of N, so C₂ = κ: F3.
Start formulaκ = |v × a|/|v|³.
Translatev(1) = (3, 2, 0); a = v′ = (6t, 1, −1), so a(1) = (6, 1, −1).
Set upv × a as a 3×3 determinant with rows i j k / 3 2 0 / 6 1 −1; then the two lengths.

Hand-over: v × a = (−2, 3, −9), |v × a| = √94, |v| = √13, so C₂ = √94/(13√13) ≈ 0.207. Past papers, 2024 Q1.

6 · F4 — Osculating circle

The family

Trigger words: "osculating circle", "circle of curvature" at a point.

Get a graph if you can: r(t) = ti + (1/t)j is just y = 1/x. Compute y′ and y″ at the point.
Sign of y″ tells you the concave side: positive means the centre is above the tangent line.
Centre and radius: (x₀ − y′(1 + y′²)/y″, y₀ + (1 + y′²)/y″) and ρ² = (1 + y′²)³/y″². This is "point + ρ times the unit normal" written out. Answer as (x − h)² + (y − k)² = ρ².

Common wrong first move: putting the centre on the wrong side of the curve (the convex side). The sign of y″ decides it; the formula above handles it automatically.

Opening on 2023 Q1(b)

Family"Osculating circle at the point (√2, 2)": F4.
Start formulaCentre (x₀ − y′(1 + y′²)/y″, y₀ + (1 + y′²)/y″), ρ = (1 + y′²)^3/2/|y″|.
TranslateAt x = √2: y′ = 2√2, y″ = 2 (positive: centre above the tangent), 1 + y′² = 9.
Set upρ = 27/2; centre (√2 − 2√2 · 9/2, 2 + 9/2).

Hand-over: centre (−8√2, 13/2), so (x + 8√2)² + (y − 13/2)² = (27/2)². Past papers, 2023 Q1.

Opening on 2025 Q2(a)

As set · 2025 mid-sem (open book), Q2 · 10 + 6 marks

(2) (a) Find the equation of the osculating circle of the curve r(t) = ti + (1/t)j at t = 1. [10M]

(b) Suppose aₙ > 0 and Σ aₙ converges. Determine the convergence/divergence of the series Σ ln(1 + aₙ) and Σ ln(2 + aₙ). [6M]

Family(a) "osculating circle": F4. (b) is series, off this year's syllabus.
Start formulaThe same centre and ρ² formulas.
Translatex = t, y = 1/t ⇒ y = 1/x; at (1, 1): y′ = −1/x² = −1, y″ = 2/x³ = 2 (concave up), 1 + y′² = 2.
Set upCentre (1 − (−1)(2)/2, 1 + 2/2), ρ² = 2³/2².

Hand-over: (x − 2)² + (y − 2)² = 2. More on this family: curvature doubt page.

7 · F5 — Limit and continuity at a point

The family

Trigger words: a function given by one formula away from a point and a separate value at it; "is f continuous at …".

Shift to the bad point first if it isn't the origin: X = x − a, Y = y − b. Complete the squares; the formula usually collapses.
Straight lines Y = mX and the axes. Different values ⇒ no limit ⇒ not continuous.
If every line gives the same value, try a curve y = x^k that makes the numerator and denominator the same size; or, if you expect the limit to exist, go polar and squeeze: |f| ≤ (power of r) → 0.

Common wrong first move: trying only y = mx, getting 0 every time, and declaring the limit 0. Lines agreeing proves nothing.

Opening on 2025 Q1(a)

As set · 2025 mid-sem (open book), Q1 · 6 + 4 + 8 marks

(1) Consider the function f(x, y) = (x² − y² − 2x + 2y)/(x² + y² − 2x − 2y + 2) if (x, y) ≠ (1, 1), and f(x, y) = 0 if (x, y) = (1, 1).

(a) Determine whether f is continuous at the point (1, 1). [6M]
(b) Determine whether the partial derivatives f_x(1, 1) and f_y(1, 1) exist. [4M]
(c) Find all unit vectors u for which the directional derivative D_uf(1, 1) = 0. [8M]

FamilyPiecewise, "continuous at (1, 1)": F5. Parts (b), (c) are F6.
Start formulaContinuous ⟺ lim_{(x,y)→(1,1)} f = f(1, 1) = 0; two paths with different limits kill it.
TranslateX = x − 1, Y = y − 1: numerator (X² − 1) − (Y² − 1) = X² − Y², denominator X² + Y².
Set upAlong Y = 0: value 1. Along X = 0: value −1.

Unpack this step: completing the square in the numerator

x² − 2x = (x − 1)² − 1 = X² − 1, and −y² + 2y = −[(y − 1)² − 1] = −Y² + 1. The ±1s cancel. The denominator works the same way, with +2 absorbing both −1s.

Hand-over: the limit does not exist, so f is not continuous at (1, 1). Module 3 notes §5 has the two-path test.

Opening on 2024 Q3(a), continuity part

As set · 2024 mid-sem, Q3 · (8 + 6) + (4 + 2) marks

3. (a) Let u = u₁i + u₂j be a unit vector with u₂ ≠ 0 and f : ℝ² → ℝ be a function defined by f(x, y) = x²y/(x⁶ + y²) if (x, y) ≠ (0, 0), and f(x, y) = 0 if (x, y) = (0, 0). Calculate the directional derivative of f at (0, 0) in the direction of u. Is the function f continuous at (0, 0)? Justify your answer mathematically. [8+6]

(b) Find a unit vector u which is normal to the level surface f(x, y, z) = 0 at the point (−1, −1, 2), where f(x, y, z) = xyz² − 4. Find the directional derivative of f at (−1, −1, 2) in the direction of u. [4+2]

Family"Continuous at (0, 0)?" for a piecewise function: F5.
Start formulaShow a path whose limit is not f(0, 0) = 0.
TranslateLines y = mx: mx³/(x⁶ + m²x²) = mx/(x⁴ + m²) → 0. Every line agrees, so move on to curves.
Set upNumerator x²y, denominator has y²: try y = x², giving x⁴/(x⁶ + x⁴) = 1/(x² + 1) → 1.

Hand-over: 1 ≠ 0, so not continuous. The directional-derivative half is F6 below.

8 · F6 — Partial or directional derivative from the definition

The family

Trigger words: the same piecewise function, and "do f_x, f_y exist at …" or "directional derivative at the special point".

Write the limit definition: D_uf(P) = lim_h→0 [f(P + hu) − f(P)]/h. For f_x, use u = i.
Substitute: f(P + hu) from the "≠" branch, f(P) from the "=" branch.
Pull out the lowest power of h, cancel, then let h → 0.

Common wrong first move: computing ∇f · u. That shortcut needs f to be differentiable at the point, and these functions aren't even continuous there.

Opening on 2024 Q3(a), derivative part

Family"Directional derivative of f at (0, 0)" for a piecewise function: F6.
Start formulaD_uf(0, 0) = lim [f(hu₁, hu₂) − 0]/h.
Translatef(hu₁, hu₂) = h³u₁²u₂/(h⁶u₁⁶ + h²u₂²).
Set upDivide by h and take h² out of the denominator: u₁²u₂/(h⁴u₁⁶ + u₂²), then h → 0 (fine because u₂ ≠ 0).

Hand-over: D_uf(0, 0) = u₁²/u₂. Every directional derivative exists, yet f is discontinuous: that contrast is the point of the question.

Opening on 2025 Q1(b)(c)

Family"Do the partial derivatives exist at (1, 1)", "unit vectors with D_uf(1, 1) = 0": F6.
Start formulaf_x(1, 1) = lim [f(1 + h, 1) − f(1, 1)]/h, and the general D_u limit.
TranslateIn the shifted form (X² − Y²)/(X² + Y²): f(1 + h, 1) = h²/h² = 1, f(1, 1 + h) = −1, and f(1 + hu₁, 1 + hu₂) = u₁² − u₂² (using u₁² + u₂² = 1).
Set upQuotients 1/h, −1/h, and (u₁² − u₂²)/h.

Hand-over: neither partial exists; D_uf(1, 1) exists only when u₁² = u₂², and then it is 0: u = (±1/√2, ±1/√2), all four. Past papers, 2025 Q1.

9 · F7 — Directional derivative through the gradient

The family

Trigger words: "directional derivative … in the direction of", "increases most rapidly", a polynomial-type f (differentiable everywhere).

Gradient in general, then at the point.
Make the direction a unit vector: u = v/|v|. If u = (u₁, u₂) is unknown, u₁² + u₂² = 1 is your second equation.
Dot product, or for "most rapidly": direction ∇f/|∇f|, rate |∇f|.

Common wrong first move: dotting with a vector that isn't unit length, or giving a direction like "(3, 4)" when the question wants u itself.

Opening on 2023 Q2(a)

As set · 2023 mid-sem, Q2 · 8 + 7 marks

2. (a) 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. [8]

(b) 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. [7]

Family(a) directional derivative of a polynomial: F7. (b) tangent plane: F8.
Start formulaD_uf = ∇f · u, |u| = 1.
Translate∇f = (2y − 6x, 2x), so ∇f(5, 5) = (−20, 10).
Set up−20u₁ + 10u₂ = −4 and u₁² + u₂² = 1. From the first, u₂ = 2u₁ − 0.4; substitute.
Check planBoth answers must have length 1.

Hand-over: 5u₁² − 1.6u₁ − 0.84 = 0, so u = (3/5, 4/5) or (−7/25, −24/25). The answers written on the photographed 2023 script say "(3, 4)" and "(−7, −24)". Those are directions, not unit vectors, and the check in move 5 would have caught it.

Check yourself: f(x, y) = x² + 3xy at (1, 2), direction v = (3, 4). Opening only.

∇f = (2x + 3y, 3x) = (8, 3) at (1, 2). u = (3, 4)/5. D_uf = (24 + 12)/5 = 36/5.

10 · F8 — Tangent plane, normal line, normal to a level surface

The family

Trigger words: "tangent plane", "normal line", "normal to the (level) surface", "tangent plane parallel to the xy-plane".

Write the surface as F(x, y, z) = c and compute ∇F: it is the normal.
At a given point: plane ∇F(P) · (r − r₀) = 0; line r₀ + t∇F(P); unit normal ∇F/|∇F|.
"Parallel to the xy-plane": the normal must be vertical, so F_x = 0 and F_y = 0 (with F_z ≠ 0). Solve these together with the surface equation.

Common wrong first move: setting F_z = 0 for "parallel to the xy-plane". That makes the plane vertical.

Opening on 2023 Q2(b)

Family"Tangent plane parallel to the xy-plane": F8.
Start formulaNormal ∇F vertical ⟺ F_x = F_y = 0.
Translate∇F = (2(x − y), −2(x − y) + 2y + 2(y + z), 2(y + z)).
Set upF_x = 0 ⇒ x = y; then F_y = 4y + 2z = 0 ⇒ z = −2y; substitute into the surface: y² + y² = 1.

Hand-over: (1/√2, 1/√2, −√2) and (−1/√2, −1/√2, √2); F_z = −2y ≠ 0 at both. Past papers, 2023 Q2.

Opening on 2024 Q3(b)

Family"Unit vector normal to the level surface … directional derivative in that direction": F8, then F7.
Start formulau = ∇f/|∇f|; D_uf = ∇f · u.
Translate∇f = (yz², xz², 2xyz) = (−4, −4, 4) at (−1, −1, 2).
Set up|∇f| = 4√3, u = (−1, −1, 1)/√3 (or its negative).

Hand-over: D_uf = |∇f| = 4√3 (or −4√3 for the opposite normal). Both normals are correct answers; say which one you took.

11 · F9 — Local extrema and the second-derivative test

The family

Trigger words: "find and classify the critical points", "local maximum / minimum / saddle point".

Both first partials, set to 0 together; solve the system (one equation often gives one variable in terms of the other).
Second partials f_xx, f_yy, f_xy in general, then D = f_xxf_yy − f_xy².
A table: point · D · f_xx · verdict (D > 0 and f_xx > 0: min; D > 0, f_xx < 0: max; D < 0: saddle; D = 0: no conclusion).

Common wrong first move: solving f_x = 0 alone and dividing by a variable that might be 0, which silently loses a critical point (e.g. at the origin).

No past paper has asked this as a separate question yet. It appears inside F10: 2024 Q2(a)'s interior critical point (1/3, 4/9) is where f_x = 4x − 3y and f_y = 1 − 3x vanish. For an absolute extremum you don't need to classify it, only evaluate it. It is in your portions (§14.7), so it can appear on its own.

Check yourself: f(x, y) = x³ − 3x + y². Opening only.

f_x = 3x² − 3 = 0 ⇒ x = ±1; f_y = 2y = 0 ⇒ y = 0. D = 6x · 2 − 0 = 12x: at (1, 0) D = 12 > 0, f_xx = 6 > 0: local min; at (−1, 0) D = −12: saddle.

12 · F10 — Absolute extrema on a closed region

🎧 Watch it done first: the narrated walkthrough (about 5 min, sound on) runs the candidate routine on 2024 Q2(a). A point walks each edge while f is plotted, and the candidates fill a table.

The family

Trigger words: "absolute maximum and absolute minimum values" on a named region (a triangle, a square, a disc piece, |x| + |y| ≤ 1).

Sketch the region and name every boundary piece and every corner.
Interior: solve ∇f = 0; keep a solution only if it lies inside.
Each boundary piece: substitute its equation to get a one-variable function, and find where its derivative is 0 within that piece's range. On a circle, substituting x² + y² often kills the squares.
One candidate table: every point from 2 and 3, plus the corners, with its f value. Largest entry = absolute max, smallest = absolute min.

Common wrong first move: running the second-derivative test (F9). It classifies interior points but says nothing about the boundary, where the absolute extremes often are.

Opening on 2023 Q3

As set · 2023 mid-sem, Q3 · 15 marks

3. Find the absolute maximum and absolute minimum values of f(x, y) = x² + y² + 2y + 2 on the region {(x, y) ∈ ℝ² : 0 ⩽ x ⩽ y and x² + (y − 1)² ⩽ 1}. [15]

2023 Q3's region, drawn to scale: the piece of the disc x² + (y − 1)² ≤ 1 between the y-axis and the line y = x. Three boundary pieces, three corners.

Family"Absolute maximum and minimum … on the region": F10.
Start formulaCandidates: interior critical points, critical points on each boundary piece, corners.
TranslateInterior: ∇f = (2x, 2y + 2) = 0 ⇒ (0, −1), outside the region, so no interior candidate. Boundary pieces: ① x = 0, 0 ≤ y ≤ 2; ② y = x, 0 ≤ x ≤ 1 (it meets the circle where 2x² − 2x = 0); ③ the arc from (1, 1) to (0, 2). Corners (0, 0), (1, 1), (0, 2).
Set up① f = y² + 2y + 2; ② f = 2x² + 2x + 2; ③ on the circle x² + y² = 2y, so f = 4y + 2. All three are increasing on their ranges, so only corners survive: table of f(0, 0), f(1, 1), f(0, 2).

Hand-over: f(0, 0) = 2, f(1, 1) = 6, f(0, 2) = 10: absolute max 10 at (0, 2), min 2 at (0, 0). Past papers, 2023 Q3.

Opening on 2024 Q2(a)

As set · 2024 mid-sem, Q2 · 12 + 10 marks

2. (a) Find the absolute maximum and minimum values of the function z = f(x, y), where f(x, y) = 2x² + y − 3xy in the plane region R bounded by the lines |x| + |y| = 1. [12]

Family"Absolute maximum and minimum … in the plane region": F10.
Start formulaThe candidate list.
TranslateThe region is a diamond with corners (±1, 0), (0, ±1) and four edges, one per quadrant: x + y = 1, −x + y = 1, −x − y = 1, x − y = 1. Interior: f_x = 4x − 3y = 0, f_y = 1 − 3x = 0 ⇒ (1/3, 4/9), inside (1/3 + 4/9 < 1).
Set upOn each edge put y in terms of x (e.g. y = 1 − x for 0 ≤ x ≤ 1 gives 5x² − 4x + 1), find the vertex if it's inside the edge's range, and fill one table: interior point, edge vertices, four corners.

Hand-over: absolute max 2 at (±1, 0), absolute min −6/5 at (−1/5, −4/5). The interior point gives only 2/9. Past papers, 2024 Q2.

13 · F11 — Lagrange multipliers

The family

Trigger words: "subject to", "on the circle / sphere / plane …", "use Lagrange multipliers", "nearest point".

Name f (what to optimise) and g = c (the constraint).
Write the system: each component of ∇f = λ∇g, plus g = c. That's one equation per unknown, λ included.
Eliminate λ by dividing or equating two equations; get a relation like y = 2x; put it into the constraint. Then evaluate f at every point found.

Common wrong first move: dividing by a variable that could be 0, and losing solutions. Check the zero case separately.

No past paper in these three years used it, but it is §14.8 and inside your Module 6 portions. Also, on a curved boundary piece in F10 (like 2023 Q3's arc), Lagrange is an alternative to substitution.

Check yourself: extreme values of f = x + 2y on x² + y² = 5. Opening only.

1 = 2λx, 2 = 2λy, x² + y² = 5. Dividing: y = 2x, so 5x² = 5, x = ±1: max 5 at (1, 2), min −5 at (−1, −2).

14 · Recognise and skip: what's off this year's portions

Your mid-sem covers Modules 1–6, sections §11.3–§14.8, with the handout's exclusions. Two parts of the past papers fall outside that. Recognising them quickly is itself a skill: in practice, don't spend time on them; on your paper they shouldn't appear at all.

You see	It's	Why it's out
"torsion", "binormal B", dB/ds — 2024 Q1(a)	Thomas §13.5	The handout's curve sections stop at §13.4.
any Σ series, "converges / diverges" — 2025 Q2(b)	§10.1–10.8	Module 11, taught after the mid-sem this year.
projectile motion	§13.2	Explicitly excluded in the handout.
"estimate the error in the linear approximation"	§14.6	Error estimation is excluded; linearisation itself is in.

15 · In the exam room

Tab the book by family, before the day. Open book only helps if you can find the page in ten seconds. Check the exact rule on the course notice (textbook only, or notes too), then tab:

Family	Thomas 15e	Our pages
F1 polar area · F2 polar tangents	§11.5 area formula · slope of a polar curve (§11.4)	Module 1 notes §5, §4
F3 curvature · F4 osculating circle	§13.4	Module 2 notes B4, curvature doubts
F5 continuity	§14.2 (two-path test)	Module 3 notes §5–§7
F6 derivative by definition	§14.3 definition of f_x, §14.5 definition of D_uf	Module 4 notes §1
F7 gradient · F8 tangent plane	§14.5, §14.6	Module 5 notes
F9 local · F10 absolute extrema · F11 Lagrange	§14.7, §14.8	Module 6 notes
Trig and integration identities (cos²θ = (1 + cos 2θ)/2 …)	endpapers	formula flashcards

Budget the clock by marks. In 2025 it was 90 minutes for 50 marks: about 1.8 minutes a mark, so a 10-mark part gets about 18 minutes.
Open every question before solving any. Write moves 1–4 for each part first (ten minutes for the whole paper), then compute. A blank page is the only answer guaranteed zero.
Stuck mid-calculation? Leave the set-up visible, write one line on what you'd do next, and move on. The set-up already carries marks.
Move 5 at the end: unit vectors have length 1, areas are positive, candidate points are inside the region, the circle's centre is on the concave side.

Stuck on a family? Ask an AI well

Pin it to your syllabus and ask for openings, not solutions:

"I'm revising Thomas' Calculus 15e §14.7 for an open-book exam. Give me four 'absolute max/min on a closed region' questions with different regions (triangle, square, disc, |x| + |y| ≤ 1). For each, I'll write only the candidate list set-up: interior equations, each boundary piece with its range, corners. Check my set-up before showing any answer."

"Quiz me on recognising question families from Thomas §11.5, §13.4 and §14.2–14.8. Give me one-line question stems; I'll name the family and the formula it starts from. Tell me when I'm wrong and why."

Cross-check any number an AI gives you against the book or our pages; AI arithmetic can be confidently wrong.