MATH U101 · Multivariable Calculus · Mid-sem prep
Mid-sem past papers, from the first move
Three past mid-sems (Oct 2023, Oct 2024, Oct 2025), every question quoted as set. Each one has its opening moves first, then a full worked solution, the answer, and the traps. Every answer was recomputed independently.
Your exam: 5 Oct, 2:00–3:30 pm, open book, 25%, portions Modules 1–6 (§11.3 to §14.8).
The papers below: 2023 and 2024 were MATH F111 (the old course): closed book, 60 marks, 90 minutes. 2025 was MATH F101 Multivariable Calculus: open book, 50 marks, 90 minutes. That is the closest match to your paper. Its header says: "Answers without proper reasoning will not be awarded marks." So the opening lines you write (which tool, which formula, the set-up) are exactly what gets marked. That is why every question here starts with them. The routine behind them is on How to start a calculus question.
The pattern is strong. A polar area question appeared in all three papers, and a curvature/osculating-circle question in all three. Directional derivatives, tangent planes and absolute extrema (Modules 5–6) took 30 of 60 marks in 2023 and 26 of 60 in 2024. Two questions are not on your syllabus this year; they are greyed out below, with the reason.
Where the marks went
| Topic | Module | 2023 | 2024 | 2025 |
|---|---|---|---|---|
| Polar curves: area of a region, tangents | 1 §11.3–11.5 | Q4 · 15 | Q2(b) · 10 | Q3 · 16 |
| Curvature, osculating circle | 2 §13.3–13.4 | Q1 · 15 | Q1(b) · 8 | Q2(a) · 10 |
| Continuity, partials, directional derivative at a "bad" point | 3–5 §14.2–14.5 | — | Q3(a) · 14 | Q1 · 18 |
| Gradient, directional derivative, tangent plane | 5 §14.5–14.6 | Q2 · 15 | Q3(b) · 6 | — |
| Absolute max/min on a closed region | 6 §14.7 | Q3 · 15 | Q2(a) · 12 | — |
| Binormal and torsion: skip | §13.5 not listed | — | Q1(a) · 10 | — |
| Infinite series: skip | 11 §10, after the mid-sem | — | — | Q2(b) · 6 |
Why the two "skip" rows are safe to skip: your handout lists Module 2 as §13.1–13.4, and binormal/torsion live in §13.5. Series are Module 11, lectures 35–40, well after 5 Oct. Everything else above is squarely in Modules 1–6.
Cover everything below a question's Opening moves box and try to write those moves yourself first: the tool, the formula, the set-up. That is the skill the exam tests, and it is the part that feels impossible until you have rehearsed it a few times. Then read the solution. If you can't start at all, the how-to-start routine gives you a fixed first question to ask for every topic on this page.
Paper 1 · 06 Oct 2025 · MATH F101 · open book · 50 marks
*Please show all the work. Answers without proper reasoning will not be awarded marks. Answer all parts of each question together in the answer booklet.
(The paper also opens with a self-declaration: "I declare that I am not carrying with me any mobile phone, communication, or data storage devices.")
Q1 · A function with a hole at (1, 1) · 6 + 4 + 8 marks §14.2–14.5
Where this lives M3 §5 · two-path test · M4 §1 · partials from the definition · M5 · directional derivative from the definition
- Spot the shape. A formula with a separately defined value at one point means every part is done from the definition (a limit), not from derivative rules.
- Simplify first. The point is (1, 1), so shift it to the origin: write X = x − 1, Y = y − 1 and complete the squares.
- Write the three definitions you'll use before computing anything: the limit along two paths for (a); fx(1,1) = limh→0 [f(1+h, 1) − f(1,1)]/h for (b); Duf(1,1) = limh→0 [f(1+hu1, 1+hu2) − f(1,1)]/h for (c).
Step 1Complete the squares: the function is simpler than it looks
Numerator: x2 − 2x = (x − 1)2 − 1 and y2 − 2y = (y − 1)2 − 1, so x2 − 2x − (y2 − 2y) = (x − 1)2 − (y − 1)2 (the two −1s cancel). Denominator: (x − 1)2 − 1 + (y − 1)2 − 1 + 2 = (x − 1)2 + (y − 1)2. With X = x − 1, Y = y − 1:
Step 2(a) Two paths, two different limits
Along the horizontal line y = 1 (Y = 0): f = X2/X2 = 1, so the limit is 1. Along the vertical line x = 1 (X = 0): f = −Y2/Y2 = −1. Two paths give different limits, so lim(x,y)→(1,1) f does not exist, and f is not continuous at (1, 1).
Step 3(b) The partial-derivative quotients blow up
(f(1 + h, 1) = 1 for every h ≠ 0, by Step 2.) 1/h has no finite limit, so fx(1, 1) does not exist. The same way, f(1, 1 + h) = −1 gives the quotient −1/h, so fy(1, 1) does not exist either.
Step 4(c) Directional derivative from the definition
Take a unit vector u = (u1, u2), so u12 + u22 = 1. Moving a distance h along u means X = hu1, Y = hu2:
The numerator is a fixed number. If it is not zero, the quotient blows up as h → 0 and Duf does not exist. If u12 = u22, the quotient is 0 for every h, so the limit is 0. With u12 + u22 = 1 that means u12 = u22 = ½: the four diagonal unit vectors.
1) Writing Duf = ∇f · u and solving ∇f · u = 0. That formula needs f to be differentiable at the point, and here the partials don't even exist, so there is no ∇f. At a "bad" point, always use the limit definition. 2) In (a), testing only lines through the point and stopping when two of them agree. Agreement proves nothing; one disagreement is enough. 3) Forgetting that u must be a unit vector: list all four, (±1/√2, ±1/√2).
Q2 · Osculating circle of r(t) = t i + (1/t) j · 10 marks (+ a series part to skip) §13.4
Where this lives M2 notes Part B · curvature · Curvature doubt clinic · osculating circles
- Name the three ingredients of any osculating circle: the point P, the radius ρ = 1/κ, and the direction N toward the centre. Centre = P + ρN.
- Write the curvature formula you'll use. For a plane curve, κ = |v × a| / |v|3 = |x′y″ − y′x″| / (x′2 + y′2)3/2.
- Get v and a at t = 1 and the point: P = (1, 1).
Step 1Velocity, acceleration, curvature
v = (1, −1/t2), a = (0, 2/t3); at t = 1: v = (1, −1), a = (0, 2), |v| = √2.
Step 2Which way is the centre?
T = (1, −1)/√2. N is perpendicular to T, so it is (1, 1)/√2 or (−1, −1)/√2. It points to the side the curve bends toward. y = 1/x is concave up for x > 0 (the acceleration (0, 2) points up), so N = (1, 1)/√2. As a check, the part of a perpendicular to v is (0, 2) − (0,2)·(1,−1)2(1, −1) = (0, 2) + (1, −1) = (1, 1), which points the same way.
Step 3Centre and equation
Check: (1, 1) is on it, since 1 + 1 = 2 ✓.
1) Putting the centre on the wrong side, at (0, 0). It must sit on the concave side; the figure is the instant check. 2) Writing the radius in the equation as √2 instead of its square: the right-hand side is ρ2 = 2.
Q3 · Cardioid r = 4(1 − sin θ), a double tangent and a notch · 6 + 10 marks §11.3–11.5
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
Where this lives M1 notes §6 · horizontal tangents, lines in polar · M1 notes §5 · area in polar · Polar problem clinic
- Translate the words. "Parallel to the X-axis and tangent" means a horizontal tangent: dy/dθ = 0 (with dx/dθ ≠ 0).
- Write y in terms of θ: y = r sin θ = 4 sin θ − 4 sin2 θ, then differentiate.
- For the area, write the line in polar form (y = c becomes r = c csc θ) and set up ½∫(outer2 − inner2) dθ between the rays through P and Q.
Step 1(a) Horizontal tangents
So cos θ = 0 (θ = π/2 or 3π/2) or sin θ = ½ (θ = π/6 or 5π/6).
- θ = 3π/2: r = 8, the single bottom point (0, −8). Only one point on that line.
- θ = π/2: r = 0, the cusp at the origin, where dx/dθ is also 0. Not a tangent point of this kind.
- θ = π/6, 5π/6: r = 4(1 − ½) = 2. Cartesian points (2 cos π/6, 2 sin π/6) = (√3, 1) and (−√3, 1): two distinct points at the same height, so L: y = 1. Check dx/dθ = −4 sin θ − 4 cos 2θ = −2 − 2 = −4 ≠ 0 at π/6 ✓ (and at 5π/6, by symmetry).
Step 2(b) The region, and the line in polar form
The region is the small notch between the two top lobes: above the curve (outside it) and below y = 1, for rays θ from π/6 to 5π/6. On each ray it runs from the curve r = 4(1 − sin θ) (inner) out to the line. In polar, y = 1 is r sin θ = 1, that is r = csc θ (outer).
The curve really is inside the line on these rays: 4(1 − s) ≤ 1/s ⇔ 4s − 4s2 ≤ 1 ⇔ (2s − 1)2 ≥ 0 for s = sin θ > 0. It is always true, with equality exactly at P and Q.
Step 3The two pieces
Line part: ½[−cot θ]π/65π/6 = ½[√3 − (−√3)] = √3. (Sanity check: that's the area of triangle OPQ, ½ × base 2√3 × height 1.)
Curve part: expand and use sin2 θ = (1 − cos 2θ)/2:
Unpack the evaluation
3θ/2 changes by (3/2)(2π/3) = π. 2 cos θ goes from √3 to −√3: change −2√3. −sin 2θ/4 goes from −(√3/2)/4 to −(−√3/2)/4: change +√3/4. Then 8 × (π − 7√3/4) = 8π − 14√3.
1) Taking the bottom point θ = 3π/2 as "the" horizontal tangent. It is a single point, and the question asks for two. 2) Integrating only the curve, ½∫16(1 − sin θ)2, and forgetting that the line is the outer boundary. 3) Writing the line as r = 1. That is a circle; a horizontal line is r = csc θ.
Paper 2 · 03 Oct 2024 · MATH F111 · closed book · 60 marks
• This question paper contains 3 questions with 2 sub-parts for each of the question.
• Answer all the questions in the given order with all sub-parts together in appropriate places.
• All notations carry their usual meaning as given in the Text book.
Q1 · dT/ds and dB/ds for a moving particle · 8 marks on your syllabus (+10 to skip) §13.4
Where this lives M2 notes Part B · T, N, κ · M2 vector-functions lesson
- Recognise the definition. N is defined as (dT/ds) / |dT/ds| and κ = |dT/ds|, so dT/ds = κN: C2 is the curvature.
- Choose the formula that needs no arc length: κ = |v × a| / |v|3. You are handed v directly, so a = v′.
Step 1v, a and their cross product at t = 1
v(1) = (3, 2, 0), a = v′ = (6t, 1, −1), so a(1) = (6, 1, −1).
Step 2The curvature
Using |v|2 in the denominator instead of |v|3. Check the units: |v × a| carries speed², and curvature must be 1/length, so you need speed³ below.
Q2 · Extrema on a diamond; the area shared by two circles · 12 + 10 marks §14.7, §11.5
🎧 Part (a) has a narrated walkthrough (about 5 min): the candidate routine animated, with three pause-and-commit questions.
Where this lives M6 · absolute extrema on a closed region · M1 notes §5 · area in polar
(a) Absolute extrema on |x| + |y| ≤ 1
- Closed and bounded region + continuous f means absolute max and min exist, and they occur at an interior critical point or on the boundary. Say this sentence in the answer.
- Draw the region: |x| + |y| = 1 is a diamond with corners (±1, 0), (0, ±1), so the boundary is four straight edges.
- Make the candidate list in order: interior (fx = fy = 0), then each edge as a one-variable function, then the four corners. Compare values at the end.
Step 1Interior critical point
fx = 4x − 3y = 0, fy = 1 − 3x = 0 ⟹ x = 1/3, y = 4/9. It is inside, since 1/3 + 4/9 = 7/9 < 1. f(1/3, 4/9) = 2/9 + 4/9 − 4/9 = 2/9.
Step 2The four edges (substitute, then 1-variable calculus)
| Edge | Substitute | f on the edge | Critical point on it |
|---|---|---|---|
| x + y = 1, 0 ≤ x ≤ 1 | y = 1 − x | 5x2 − 4x + 1 | 10x − 4 = 0 ⟹ (2/5, 3/5), f = 1/5 |
| −x + y = 1, −1 ≤ x ≤ 0 | y = 1 + x | −x2 − 2x + 1 | −2x − 2 = 0 ⟹ x = −1, a corner |
| x + y = −1, −1 ≤ x ≤ 0 | y = −1 − x | 5x2 + 2x − 1 | 10x + 2 = 0 ⟹ (−1/5, −4/5), f = −6/5 |
| x − y = 1, 0 ≤ x ≤ 1 | y = x − 1 | −x2 + 4x − 1 | −2x + 4 = 0 ⟹ x = 2, off the edge |
For example, on the first edge: 2x2 + (1 − x) − 3x(1 − x) = 2x2 + 1 − x − 3x + 3x2 = 5x2 − 4x + 1. The other three are the same kind of expansion.
Step 3Corners, then compare everything
Corners: f(1, 0) = 2, f(0, 1) = 1, f(−1, 0) = 2, f(0, −1) = −1. All candidates: 2/9, 1/5, −6/5, 2, 1, 2, −1.
1) Checking only the corners. The minimum here is in the middle of an edge. 2) Classifying the interior point with the second-derivative test and stopping there. For absolute extrema you just evaluate f at every candidate and compare (here fxxfyy − fxy2 = −9 < 0 anyway, a saddle). 3) Solving an edge's derivative and keeping a point outside the edge's range (x = 2 on the last edge).
(b) The lens shared by two circles
- Complete the squares to see the circles: x2 + (y + 1)2 = 1 (centre (0, −1)) and (x + 1)2 + y2 = 1 (centre (−1, 0)), both radius 1.
- Go polar (single integral, as demanded): substitute x2 + y2 = r2, y = r sin θ, which gives r = −2 sin θ and r = −2 cos θ.
- Find where they cross and which is inner on each ray. The area is ½∫rinner2 dθ, split at the crossing ray.
Step 1Intersections
Subtract the equations: 2y − 2x = 0, so y = x. Then 2x2 + 2x = 0, so x = 0 or −1: the points (0, 0) and (−1, −1). The lens lies in the third quadrant, π ≤ θ ≤ 3π/2, and the crossing ray is θ = 5π/4.
Step 2Which curve bounds each ray
A ray from the origin leaves the lens through whichever circle it meets first, the one with the smaller r. For π ≤ θ ≤ 5π/4 (close to the negative x-axis), |sin θ| ≤ |cos θ|, so r = −2 sin θ is smaller. For 5π/4 ≤ θ ≤ 3π/2, r = −2 cos θ is smaller.
Not asked, but as a check: the two halves are equal by symmetry about y = x, and the total is π/2 − 1 ≈ 0.571 (confirmed numerically).
1) Using limits 0 to 2π, or giving the circles positive r in the first quadrant. They live in the third quadrant, where −2 sin θ and −2 cos θ are positive. 2) One integral with the wrong curve for half the lens. Always split at the intersection ray.
Q3 · Directional derivatives, discontinuity, and a normal vector · 8 + 6 + 4 + 2 marks §14.2, §14.5
Where this lives M5 · directional derivatives, gradient, normals · M3 §5 · two-path test
(a) Directional derivative at the origin, and continuity
- Piecewise function at the special point, so use the definition: Duf(0,0) = limh→0 [f(hu1, hu2) − f(0,0)]/h.
- For continuity, look at the degrees. The x6 against y2 suggests a path where y ~ x3 or y = x2, not just straight lines.
Step 1The directional derivative
(Cancel h2 from the fraction's top and bottom, then the remaining h against the 1/h. As h → 0 the h4 term vanishes, and u2 ≠ 0 keeps the denominator safe.) So Duf(0, 0) = u12/u2.
Step 2Continuity: straight lines lie, a parabola tells the truth
Along any line y = mx: f = mx3/(x6 + m2x2) = mx/(x4 + m2) → 0, and along x = 0, f = 0. Every line agrees on 0. But along the parabola y = x2:
So the limit at (0, 0) is not 0 (it doesn't exist), and f is not continuous at (0, 0). Worth saying in the answer: this is the standard example that directional derivatives can exist in every direction while f is discontinuous.
1) "All lines give 0, so f is continuous." Straight lines are not all paths. 2) "The directional derivatives exist, so f is differentiable (and continuous)." Existence of every Duf proves neither. 3) Answering ∇f(0,0) · u = 0: the partials here are 0, but u12/u2 is not ∇f · u, because f is not differentiable at the origin.
(b) Unit normal and the directional derivative along it
- "Normal to a level surface" means the gradient. ∇f at the point is perpendicular to the level surface through it.
- Check the point is on the surface ((−1)(−1)(4) − 4 = 0 ✓), then compute ∇f and divide by its length.
Step 1Gradient and unit normal
∇f = (yz2, xz2, 2xyz) = (−4, −4, 4) at (−1, −1, 2), and |∇f| = √48 = 4√3. So u = (−1, −1, 1)/√3 (its negative is also a unit normal).
Step 2Directional derivative along it
Duf = ∇f · u = (4 + 4 + 4)/√3 = 12/√3 = 4√3, which is |∇f|, as it must be along the gradient's own direction. For the opposite normal it is −4√3.
Paper 3 · 09 Oct 2023 · MATH F111 · closed book · 60 marks
1. This question paper contains 4 questions. Answer all the questions in the given order with all sub-parts together in appropriate places.
2. All notations carry their usual meaning as per the Text book.
3. Calculators are not allowed.
(Our copy is a photo with someone's handwritten answers on it. They are all right except the one flagged in Q2(a).)
Q1 · Curvature of the parabola y = x² · 5 + 10 marks §13.4
Where this lives M2 notes Part B · curvature · Curvature doubt clinic (T4 Q3 is this exact family)
- Parametrise the curve: r(t) = ti + t2j, so v = (1, 2t), a = (0, 2).
- Write κ as a function of t using |v × a| / |v|3, then radius ρ = 1/κ.
- For (b), list the three ingredients: point, ρ, N. Then centre = point + ρN.
Step 1(a) Radius of curvature as a function of t
Set ρ = √2: (1 + 4t2)3/2 = 2√2 = 23/2, so 1 + 4t2 = 2, so t = ±½. Points (±½, ¼).
Unpack: from the 3/2 power to 1 + 4t² = 2
2√2 = 2 · 21/2 = 23/2. Both sides are positive numbers raised to the same power 3/2, so the bases are equal (raise both sides to the power 2/3).
Step 2(b) The three ingredients at (√2, 2), where t = √2
v = (1, 2√2), |v| = √(1 + 8) = 3. ρ = 93/2/2 = 27/2. T = (1, 2√2)/3. The parabola opens upward, so N points up and in toward its axis: N = (−2√2, 1)/3 (perpendicular to T, unit length, positive y-part). Check: the part of a = (0, 2) perpendicular to v is (0, 2) − (4√2/9)(1, 2√2) = (−4√2/9, 2/9), which is a positive multiple of (−2√2, 1) ✓.
Step 3Centre and equation
Check with the point (√2, 2): (9√2)2 + (−9/2)2 = 162 + 81/4 = 729/4 ✓.
1) Giving only x = ½: the parabola is symmetric, so there are two points. 2) Centre on the convex side, at (√2, 2) − ρN. 3) Stopping at ρ and the centre without writing the circle's equation, which is what's asked.
Q2 · A directional derivative backwards; a horizontal tangent plane · 8 + 7 marks §14.5–14.6
Where this lives M5 · gradient, directional derivative, tangent planes
(a) Find u from Duf = −4
- Formula: Duf = ∇f · u with u a unit vector (f is a polynomial, so it is differentiable and the dot product is allowed).
- Two unknowns need two equations: ∇f(5,5) · u = −4 and u12 + u22 = 1.
Step 1Gradient and the linear equation
∇f = (2y − 6x, 2x) = (−20, 10) at (5, 5). So −20u1 + 10u2 = −4, that is u2 = 2u1 − 2/5.
Step 2Substitute into the unit condition
u12 + (2u1 − 2/5)2 = 1 ⟹ 5u12 − (8/5)u1 + 4/25 − 1 = 0 ⟹ (× 25) 125u12 − 40u1 − 21 = 0.
Unpack the discriminant
b2 − 4ac = (−40)2 − 4(125)(−21) = 1600 + 10 500 = 12 100 = 1102, a perfect square. Exam numbers are usually designed to come out like this, so a non-square here would be a hint to recheck ∇f.
Then u2 = 2(3/5) − 2/5 = 4/5 or u2 = −14/25 − 10/25 = −24/25.
Step 3Check both
−20(3/5) + 10(4/5) = −12 + 8 = −4 ✓ and −20(−7/25) + 10(−24/25) = (140 − 240)/25 = −4 ✓. Both have length 1 (3-4-5 and 7-24-25 triangles).
It gives "(3, 4)" and "(−7, −24)". Those point the right way, but they are not unit vectors, and with them ∇f · u = −20 and −100, not −4. The directional derivative is defined for unit vectors, so the answer must be (3/5, 4/5) and (−7/25, −24/25). Writing the unnormalised pair loses marks.
(b) Where the tangent plane is horizontal
- Surface as a level surface: F(x, y, z) = (x − y)2 + y2 + (y + z)2 − 1 = 0; the tangent plane's normal is ∇F.
- Translate "parallel to the xy-plane": normal parallel to k, i.e. Fx = 0 and Fy = 0 (with Fz ≠ 0). Three equations: these two plus the surface.
Step 1The partials
Fx = 2(x − y), Fy = −2(x − y) + 2y + 2(y + z), Fz = 2(y + z).
Step 2Solve
Fx = 0 ⟹ x = y. Then Fy = 2y + 2(y + z) = 4y + 2z = 0 ⟹ z = −2y. On the surface: 0 + y2 + (y − 2y)2 = 2y2 = 1 ⟹ y = ±1/√2.
Points: (1/√2, 1/√2, −√2) and (−1/√2, −1/√2, √2). There Fz = 2(y − 2y) = −2y = ∓√2 ≠ 0 ✓, so the normal really is vertical.
Setting Fz = 0. That gives tangent planes containing the vertical direction (vertical planes), the opposite of what is asked.
Q3 · Absolute extrema on a slice of a disc · 15 marks §14.7
Where this lives M6 · absolute extrema on a closed region
- Draw the region. x2 + (y − 1)2 ⩽ 1 is the disc of radius 1 centred at (0, 1). 0 ⩽ x ⩽ y keeps the wedge between the y-axis and the line y = x. Corners: (0, 0), (1, 1), (0, 2).
- Say the existence sentence (closed, bounded, f continuous), then list candidates: interior critical points, then the three boundary pieces (two segments and an arc).
- Optional shortcut: complete the square, f = x2 + (y + 1)2 + 1: squared distance from (0, −1), plus 1.
Step 1Interior
fx = 2x = 0, fy = 2y + 2 = 0 ⟹ (0, −1), which is not in the region. So both extremes are on the boundary.
Step 2The three boundary pieces
- Segment on x = 0, 0 ≤ y ≤ 2: f = y2 + 2y + 2, increasing for y > −1, so from f(0, 0) = 2 up to f(0, 2) = 10.
- Segment on y = x, 0 ≤ x ≤ 1: f = 2x2 + 2x + 2, increasing, from 2 at (0, 0) to 6 at (1, 1).
- Arc of the circle from (1, 1) to (0, 2): on the circle x2 = 1 − (y − 1)2 = 2y − y2, so f = 2y − y2 + y2 + 2y + 2 = 4y + 2, for 1 ≤ y ≤ 2: from 6 to 10.
Step 3Compare
Largest value 10 at (0, 2); smallest 2 at (0, 0). The distance picture agrees: (0, 0) is the region's nearest point to (0, −1) (distance 1, f = 1 + 1), and (0, 2) the farthest (distance 3, f = 9 + 1).
1) Keeping the interior critical point (0, −1) even though it is outside the region. 2) Using the whole circle, or the lower arc from (0, 0) to (1, 1). That arc has y < x, so it is outside the wedge. 3) Forgetting the corner (1, 1) where two pieces meet.
Q4 · Inside a cardioid, outside a circle · 5 + 10 marks §11.4–11.5
🎧 This question has a narrated walkthrough (about 5 min): the four opening moves animated, with three pause-and-commit questions.
Where this lives M1 notes §5 · area between polar curves · Polar problem clinic · Polar pipeline drill
- Identify both curves: r = 6 cos θ is a circle of diameter 6 through the origin (centre (3, 0)); r = 2 + 2 cos θ is a cardioid (cusp at the origin, reaching (4, 0)).
- Intersect: 6 cos θ = 2 + 2 cos θ ⟹ cos θ = ½ ⟹ θ = π/3 (above the axis).
- On each range of rays, decide outer and inner, then ½∫(outer2 − inner2) dθ.
Step 1Which rays carry the region
- 0 ≤ θ < π/3: 6 cos θ > 2 + 2 cos θ (because cos θ > ½), so the cardioid is inside the circle. Nothing is outside C1 and inside C2.
- π/3 ≤ θ ≤ π/2: the region runs from the circle (inner) out to the cardioid (outer).
- π/2 ≤ θ ≤ π: the circle has r = 6 cos θ ≤ 0; it lies entirely in x ≥ 0. So the region is the whole cardioid there, from 0 out.
Step 2Set up
Step 3Evaluate
With cos2 θ = (1 + cos 2θ)/2: ½(2 + 2 cos θ)2 = 3 + 4 cos θ + cos 2θ, so
Unpack the square and the identity
(2 + 2 cos θ)2 = 4 + 8 cos θ + 4 cos2 θ = 4 + 8 cos θ + 2(1 + cos 2θ) = 6 + 8 cos θ + 2 cos 2θ; halve it. The same half-angle identity is on the formula flashcards.
The √3 terms cancel exactly, a sign the set-up was right. (Numerical integration agrees: 1.5708.)
1) Subtracting the circle all the way to θ = π. Past π/2, 6 cos θ is negative and those points are below the axis, not part of the circle above it. 2) Starting at 0 instead of π/3. 3) Sketch marks: label the intersection and shade only above the x-axis.
After this page
- Re-do the openings cold. Tomorrow, read only the verbatim cards and write the opening moves for each question on paper, then compare. That is what How to start a calculus question drills.
- Open book means tab your pages. Tab the curvature formula and the osculating-circle recipe (curvature clinic), polar area and horizontal tangents (M1 §6), the directional-derivative definition (M5) and the extrema routine (M6). Looking things up costs minutes you won't have, so decide in advance where each formula lives.
- Time: 90 minutes for 50 marks is about 1.8 minutes per mark. A 10-mark polar area is worth about 18 minutes: sketch first (it earns marks and prevents wrong limits).