MATH U101 · Quiz 1 (08 Sep 2026) · post-mortem

Quiz 1, question by question, from zero

The paper is in the repo (four variants, instructor's key). This page rebuilds each question from first principles, names the exact step where each variant is won or lost, gives the answers for all four variants, and ends every question with a twin to try cold. Read it for the mid-semester exam on 5 Oct, which reuses these three patterns with more room to work.

Start here · the marks arithmetic, before anything else

Quiz 1 is one of three quizzes worth 20% together, best two of three. A bad Quiz 1 is dropped if Quiz 2 (29 Sep) and Quiz 3 (17 Nov) go reasonably. Nothing about this paper is permanent. The exam that decides the course is the mid-sem (25%, open book, 5 Oct) and then the compre (45%).

Honest difficulty read: Q2 and Q3 were standard and gettable (14 of 20 marks) — the lesson and the doubt clinic teach both patterns. Q1 had one twist nobody's tutorial sheet showed: the line touches the curve at one point besides crossing it. That twist is now in the problem clinic. The format — final answer only, overwriting scores zero — is what made a fair paper feel brutal; the plan at the bottom is about that.

The format Q1 · line cuts a limaçon Q2 · T and N Q3 · domain and partials All four variants Next time: the 40-minute plan

The format, and what it rewards

Duration	Marks	Type	Answer format	Partial credit
40 min	20 (6 + 6 + 8)	Closed book	Fill in the blank: limits, vectors, a set, two numbers	None — "any overwriting/cutting … will carry ZERO marks"

Three questions, four variants in the room (the curve's sign, the line's side, the value of t, the constant in front of the log). Every variant has the same structure, so a method learned on one is the method for all.

Q1 · Area inside a limaçon on one side of a line §11.5 · 6 marks

Variant 1, as set The area on the right side of the line x = −1 and inside the curve r = 3 + 2 cos θ is

∫_▢^▢ (3 + 2 cos θ)² dθ + ∫_▢^▢ sec²θ dθ.

Fill in the four limits. (Key: 0, 2π/3 and 2π/3, π — "Q1 has other correct answers".)

Where this lives Problem clinic §2 · the limits routine, example E (a line as r = csc θ) and new example G (this question's shape) · Module 1 notes · Thomas §11.5

What is really being asked

The integrals are given, the ½ is already absorbed, so the whole 6 marks are for four angles. Those angles come from one routine: convert the line to polar, solve for where line and curve meet, then decide on which θ-range the region is bounded by the curve and on which by the line. The twist is in the "meet" step.

Step 1Turn the line into a polar curve

x = −1 means r cos θ = −1, i.e. r = −sec θ. That is only a positive radius where cos θ < 0, so the line lives on π/2 < θ < 3π/2 — the left half-plane, which is where a line x = −1 should be. (Same move for the other variants: x = 1 → r = sec θ; y = −1 → r = −cosec θ; y = 1 → r = cosec θ.) The paper's sec²θ integrand is this line squared — the sign disappears, which is why the paper could print it without telling you the sign.

Step 2Where do line and curve meet — including the touch

Set 3 + 2 cos θ = −sec θ and multiply through by cos θ (write c = cos θ):

3c + 2c² = −1 ⇒ 2c² + 3c + 1 = 0 ⇒ (2c + 1)(c + 1) = 0

Unpack this step: the factorisation

Look for two numbers with product 2 · 1 = 2 and sum 3: they are 1 and 2. Split the middle term, 2c² + c + 2c + 1 = c(2c + 1) + (2c + 1), and factor. Check by expanding: (2c + 1)(c + 1) = 2c² + 3c + 1 ✓.

Two roots, and they mean different things:

cos θ = −½: θ = 2π/3 and 4π/3. There r = 3 − 1 = 2, the points (−1, ±√3). The line genuinely crosses the curve.
cos θ = −1: θ = π, r = 1, the point (−1, 0). The curve touches the line here and comes back. This is the twist. Between 2π/3 and 4π/3 the limaçon dips slightly left of x = −1 (its leftmost x is −9/8, at cos θ = −¾), returns to the line exactly at θ = π, and dips left again.

The touch point does not add a boundary switch — the region is bounded by the line all the way from 2π/3 to 4π/3 — but it is exactly the kind of root that makes a student doubt the algebra ("why three angles?") and lose time. Three angles is correct; one of them is a tangency.

Shaded: inside r = 3 + 2 cos θ and right of x = −1. The line crosses at θ = 2π/3, 4π/3 and touches at θ = π.

Zoom on the strip near x = −1 (horizontal axis stretched 4×): the two slivers cut off by the line, total area ≈ 0.26, pinched at the touch point.

Step 3Walk θ round once and name the boundary on each stretch

Sweep from θ = −2π/3 to 4π/3 (one full turn, started at a crossing):

−2π/3 ≤ θ ≤ 2π/3 (the right-hand two-thirds of the turn): the ray from the pole exits the region through the curve. Boundary r = 3 + 2 cos θ.
2π/3 ≤ θ ≤ 4π/3: the ray hits the line before the curve (the curve is at or left of x = −1 there). Boundary r = −sec θ.

A = ½ ∫_−2π/3^2π/3 (3 + 2 cos θ)² dθ + ½ ∫_2π/3^4π/3 sec²θ dθ

The paper's form has no ½. Both pieces are symmetric about the x-axis, so halve each θ-range and drop the ½:

A = ∫₀^2π/3 (3 + 2 cos θ)² dθ + ∫_2π/3^π sec²θ dθ

That is the key's answer. The "other correct answers" are the mirror halves — −2π/3 to 0 and π to 4π/3, or 4π/3 to 2π for the first — any pair that covers exactly half of each region once.

Not asked, but worth one minute: the value. (3 + 2 cos θ)² = 11 + 12 cos θ + 2 cos 2θ, antiderivative 11θ + 12 sin θ + sin 2θ; from 0 to 2π/3 that is 22π/3 + 11√3/2. And ∫_2π/3^π sec² = [tan θ] = 0 − (−√3) = √3. Total 22π/3 + 13√3/2 ≈ 34.30, against the whole limaçon's 11π ≈ 34.56 — so the line shaves off only ≈ 0.26, the two thin slivers in the zoom. Sanity check passed.

Classic traps · this exact question

1) Doubting the third root. A quadratic in cos θ can give a tangency root (here cos θ = −1). Keep it, mark it as a touch, and let the crossings set the limits. 2) Using the ½ twice. The paper's form has already dropped it by halving the ranges; writing −2π/3 to 2π/3 in that form doubles the answer. 3) Sign of the secant. x = −1 is r = −sec θ. It squares away in the integrand, but if you solve 3 + 2 cos θ = +sec θ you get 2c² + 3c − 1 = 0 and ugly angles — that ugliness is your alarm. 4) Reading "right of the line" as a θ-range on its own. The θ-range is found from the crossings, never guessed from "right".

Try the twin (variant 4): the area below the line y = 1 and inside r = 3 − 2 sin θ. Fill the four limits of ∫(3 − 2 sin θ)² dθ + ∫ cosec²θ dθ.

Line: y = 1 ⇒ r = cosec θ (sin θ > 0, upper half-plane). Meet: (3 − 2s)s = 1 ⇒ 2s² − 3s + 1 = 0 ⇒ (2s − 1)(s − 1) = 0: crossings at sin θ = ½, θ = π/6 and 5π/6; touch at sin θ = 1, θ = π/2. Region below the line: bounded by the line for π/6 ≤ θ ≤ 5π/6, by the curve for the rest of the turn, 5π/6 ≤ θ ≤ 13π/6. Symmetric about the y-axis (the ray θ = π/2 … 3π/2), so take the left half: A = ∫_5π/6^3π/2 (3 − 2 sin θ)² dθ + ∫_π/2^5π/6 cosec²θ dθ — the key's answer. Its value is the same 22π/3 + 13√3/2, because every variant is this region rotated or reflected.

Q2 · Unit tangent and unit normal of ⟨t, eᵗ⟩ §13.4 · 2 + 4 marks

Variant 1, as set Consider the curve r(t) = t i + e^t j. (a) The unit tangent vector T at t = 1 is ▢. (b) The unit normal vector N at t = 1 is ▢.

Where this lives Notes B4 · T, N · Lesson · rotating T to get N · Doubt clinic D1 (same recipe) · Thomas §13.4

Step 1T: differentiate, then divide by the length

v = r′(t) = ⟨1, e^t⟩, |v| = √(1 + e^2t). At t = 1:

T(1) = 1√(1 + e²) i + e√(1 + e²) j

Leave it exact — the blank wants the expression, and any decimal here invites a rounding argument. (2 marks: this is a two-line question.)

Step 2N: rotate T a quarter turn, toward the bend

For a plane curve you do not need to differentiate T. Rotate ⟨p, q⟩ by 90° to get ⟨−q, p⟩ or ⟨q, −p⟩ — swap and flip one sign — then keep the one pointing to the concave side. y = e^x is concave up everywhere (y″ = e^x > 0), so N must have a positive j-component:

N(1) = −e√(1 + e²) i + 1√(1 + e²) j

Check: T · N = (−e + e)/(1 + e²) = 0 ✓, length 1 ✓, j-component positive ✓.

Unpack this step: the same N from the definition N = (dT/dt)/|dT/dt|

T = ⟨1, e^t⟩(1 + e^2t)^−1/2. Product rule and factor out (1 + e^2t)^−3/2: dT/dt = [⟨0, e^t⟩(1 + e^2t) − e^2t⟨1, e^t⟩](1 + e^2t)^−3/2 = e^t⟨−e^t, 1⟩(1 + e^2t)^−3/2. Its direction is ⟨−e^t, 1⟩ — the rotation answer, with the sign chosen for you. Slower, but it is the proof that the rotate-toward-the-bend rule is right.

y = e^x at (1, e): T along the curve, N a quarter-turn anticlockwise, into the bowl.

Classic traps

1) The wrong perpendicular. ⟨e, −1⟩/√(1 + e²) is perpendicular and unit — and scores zero, because N points into the bend. Always check the sign against concavity. 2) Forgetting to normalise. ⟨1, e⟩ is a tangent vector; the unit tangent divides by √(1 + e²). 3) Substituting t = 1 before differentiating (if you take the definition route): T(1) is a constant and its derivative is 0.

Try the twin: r(t) = ⟨t, ln t⟩. Find T and N at t = 1.

v = ⟨1, 1/t⟩, at t = 1: ⟨1, 1⟩, length √2, so T = ⟨1, 1⟩/√2. Rotations: ⟨−1, 1⟩/√2 or ⟨1, −1⟩/√2. y = ln x is concave down (y″ = −1/x² < 0), so N has a negative j-component: N = ⟨1, −1⟩/√2. This is the variant that punishes "always take the one with positive y".

Q3 · Domain and partial derivatives of a log of a ratio §14.1, 14.3 · 3 + 3 + 2 marks

Variant 1, as set f(x, y) = ln √(x² + y²) − x√(x² + y²) + x. (a) Domain of f is ▢. (b) f_x(0, 1) = ▢. (c) f_y(0, 1) = ▢. (Other variants multiply f by 3, 5 or 7.)

Where this lives Module 3 notes · domains · Module 4 notes · partials · Thomas §14.1, §14.3 · kit: ln(a/b) = ln a − ln b, chain rule on √ and ln

Step 1Domain: where is the thing inside the log positive?

Write R = √(x² + y²), the distance from the origin. The one fact that decides everything: R ≥ |x|, with equality only when y = 0 (then R = |x|).

Off the x-axis (y ≠ 0): R > |x|, so R − x > 0 and R + x > 0. The ratio is positive; the log exists.
On the x-axis: for x > 0, R − x = x − x = 0 (log of 0); for x < 0, R + x = 0 (division by 0); at the origin, 0/0. Every point of the axis fails.

Domain = {(x, y) ∈ ℝ² : y ≠ 0} = ℝ² minus the x-axis

All four key answers on the paper say this in different notation — ℝ² ∖ {y = 0}, ℝ² ∖ {(x, 0)}, "{y ≠ 0}", "ℝ² ∖ x-axis". Any of them is full marks; a half-answer like "x ≠ 0" or "(0, 0) excluded" is zero.

The domain is two open half-planes. The dashed x-axis is removed: the numerator dies on its right half, the denominator on its left half.

Step 2Split the log before differentiating

f = ln(R − x) − ln(R + x), with ∂R∂x = xR, ∂R∂y = yR

Unpack this step: ∂R/∂x

R = (x² + y²)^1/2; treat y as a constant; chain rule: ½(x² + y²)^−1/2 · 2x = x/R. Same for y.

Splitting the ratio into a difference of logs turns one quotient-rule mess into two chain-rule lines. Do this every time a log of a ratio appears.

Step 3f_x — and a simplification that makes the whole question one line

f_x = x/R − 1R − x − x/R + 1R + x

Now notice x/R − 1 = (x − R)/R = −(R − x)/R, so the first fraction is just −1/R; likewise x/R + 1 = (R + x)/R makes the second 1/R. Hence, everywhere on the domain,

f_x = −2R = −2√(x² + y²) ⇒ f_x(0, 1) = −2

If you don't spot the simplification, substitute (0, 1) directly: R = 1, first fraction (0 − 1)/(1 − 0) = −1, second (0 + 1)/(1 + 0) = 1, difference −2. Same answer, thirty seconds longer. With the constant k in front (variants 2–4): −2k, i.e. −6, −10, −14.

Step 4f_y

f_y = y/RR − x − y/RR + x

At (0, 1): R = 1, so 1/1 − 1/1 = 0. (In general the bracket is (y/R) · 2x/(R² − x²) = 2x/(Ry), zero along the whole y-axis — which is why every variant asks at x = 0.) The constant k changes nothing: k · 0 = 0.

Classic traps

1) Domain too small. "x ≠ 0" or "origin excluded" — the whole x-axis goes, both halves, for two different reasons. 2) Differentiating the ratio as one lump. Legal, but the quotient rule on √-expressions under a clock is where the arithmetic dies. Split the log first. 3) Dropping ∂R/∂x. Treating √(x² + y²) as a constant when differentiating in x gives f_x = −1/(R−x) − 1/(R+x) = −2 at (0,1) — the right number by accident, and the wrong number at every other point. Do not learn from the lucky case. 4) The constant. 7 ln(…) at (0, 1) gives −14, not −2 — read the front of the function.

Try the twin: same f (k = 1), find f_x(1, 1) and f_y(1, 1).

R = √2. From the general forms above: f_x = −2/R = −2/√2 = −√2 ≈ −1.41421; f_y = 2x/(Ry) = 2/(√2 · 1) = √2 ≈ 1.41421. Without the general forms, substitute into the two-fraction expressions: f_y = (1/√2)/(√2 − 1) − (1/√2)/(√2 + 1) = (1/√2)·[(√2 + 1) − (√2 − 1)]/(2 − 1) = (1/√2)·2 = √2 ✓ (conjugate multiplication in the middle step).

All four variants, one table

Variant	Q1 curve · line	Q1 limits (key)	Q2 t	Q2 T, N	Q3 constant	Q3 (a) · (b) · (c)
1	r = 3 + 2 cos θ · right of x = −1	0 → 2π/3 ; 2π/3 → π	1	⟨1, e⟩/√(1 + e²) ; ⟨−e, 1⟩/√(1 + e²)	1	{y ≠ 0} · −2 · 0
2	r = 3 − 2 cos θ · left of x = 1	π/3 → π ; 0 → π/3	2	⟨1, e²⟩/√(1 + e⁴) ; ⟨−e², 1⟩/√(1 + e⁴)	3	{y ≠ 0} · −6 · 0
3	r = 3 + 2 sin θ · above y = −1	π/2 → 7π/6 ; 7π/6 → 3π/2	3	⟨1, e³⟩/√(1 + e⁶) ; ⟨−e³, 1⟩/√(1 + e⁶)	5	{y ≠ 0} · −10 · 0
4	r = 3 − 2 sin θ · below y = 1	5π/6 → 3π/2 ; π/2 → 5π/6	4	⟨1, e⁴⟩/√(1 + e⁸) ; ⟨−e⁴, 1⟩/√(1 + e⁸)	7	{y ≠ 0} · −14 · 0

Two designed-in checks, usable in the room: every Q1 region is the same shape rotated or reflected (area 22π/3 + 13√3/2 in all four), and every Q3 (c) is 0 because the point sits on the y-axis. If your variant breaks either pattern, re-check before writing.

Next time: the 40-minute plan for an answer-only paper

Order by marks-per-minute, not by number. Q2 (6 marks, 5 minutes) first, then Q3 (8 marks, 12 minutes), then Q1 (6 marks, the rest). The hardest question goes last so it cannot eat the easy marks.
Rough work on the question paper margin, answer in pencil-light first, ink once. Overwriting is zero; a wrong first ink is worse than a blank.
One check per blank before inking. T·N = 0 and the sign of N; the domain covers both halves of the axis; Q1 limits swept once, no ½ doubled; an ugly quadratic means a sign slip.
Leave a blank rather than guess wildly only when the check fails and time is gone — a blank and a wrong answer both score zero, but a half-finished overwrite can cost the neighbouring answer's legibility.

Ask an AI well · about this paper

"Thomas 15e §11.5 only: give me the area inside r = 3 + 2 sin θ and below the line y = 1, as ∫(…)² dθ + ∫cosec²θ dθ with limits, and check mine: I say …" · "Thomas §14.3: for f = ln((√(x²+y²) − x)/(√(x²+y²) + x)), check that f_x simplifies to −2/√(x²+y²) and show me the two lines I might be skipping." Keep to §11.5, §13.4, §14.1–14.3 — the quiz's own scope.