Mathematics Advanced • Year 11 • Module 3 • Lesson 15
Module 3 Review
Apply Module 3 calculus tools to fencing optimisation, projectile motion, container design, traffic-flow totals, and tabulated approximations.
Problem 1 — Maximum-area fenced paddock
A farmer has 80 m of fencing and wants to enclose a rectangular paddock against a long, straight stone wall (so only three sides need fencing).
Set up: What are we solving for?
(i) Let the side perpendicular to the wall have length w. Express the side parallel to the wall in terms of w, and hence write the area A(w) as a quadratic in w. 2 marks
(ii) Use calculus to find the value of w that maximises A. Confirm using the second-derivative test. 3 marks
(iii) State the maximum enclosed area in m². 1 mark
Stuck? Revisit lesson § Worked Example 2 — optimisation requires expressing one variable in terms of another, then differentiating.Problem 2 — Vertical projectile motion
A ball is thrown vertically upward from the top of a 20 m cliff with initial velocity 15 m/s. Its height above the cliff base at time t seconds is
h(t) = 20 + 15t − 5t² (m), for 0 ≤ t ≤ T (when ball lands).
Set up: What are we solving for?
(i) Find the velocity function v(t) = h′(t) and use it to find the time at which the ball reaches its maximum height. 2 marks
(ii) Find the maximum height reached above the cliff base. 1 mark
(iii) Find T (the time at which the ball lands, i.e. h(T) = 0) and the impact speed |v(T)| in m/s. 3 marks
Problem 3 — Engineered water bottle (volume of revolution)
A water bottle is designed by rotating the curve y = (1/2)x + 1 on [0, 6] about the x-axis (all in cm). The x-axis is the axis of symmetry.
Set up: What are we solving for?
(i) Set up V = π ∫ [f(x)]² dx and expand the integrand. 2 marks
(ii) Evaluate V in exact form (in terms of π) and to the nearest cm³. 3 marks
(iii) State the bottle's capacity in mL (1 mL = 1 cm³), and comment on whether it could hold a standard 600 mL drink. 2 marks
Stuck? Revisit § Worked Example 2 (Module 3 Review) — expand the squared linear before integrating.Problem 4 — Total cars from rate samples
A loop sensor records the rate r(t) (cars per minute) crossing a bridge at 10-minute intervals through a busy 60-minute period:
| t (min) | 0 | 10 | 20 | 30 | 40 | 50 | 60 |
|---|---|---|---|---|---|---|---|
| r(t) | 22 | 34 | 50 | 58 | 46 | 30 | 18 |
Set up: What are we solving for?
(i) Use the trapezoidal rule with all 6 strips to estimate the total cars over the hour: ∫₀⁶⁰ r(t) dt. 2 marks
(ii) Identify the time at which the flow is greatest (using the table), and use a difference-quotient estimate over a 10-minute window to estimate how quickly the rate is changing at t = 20 min. 2 marks
(iii) The traffic operator wants the total rounded to the nearest hundred cars for a public report. State the figure and one limitation of using the trapezoidal rule on 10-minute samples. 2 marks
Problem 5 — Curved deck for a coffee shop
A new deck has a curved edge described by y = 4 − x²/4 (in metres) for −4 ≤ x ≤ 4, joined to a straight back wall along the x-axis (so the deck is the region between the curve and the x-axis).
Set up: What are we solving for?
(i) Use integration to find the deck area exactly, in m². 3 marks
(ii) Decking timber costs $185 per square metre installed. Calculate the total decking cost. 2 marks
(iii) A railing follows the curved edge of the deck. Estimate the curve's length using a "5 trapezoid endpoints" method: take points on the curve at x = −4, −2, 0, 2, 4, then sum the straight-line distances between consecutive points. 3 marks
Stuck on (iii)? For each segment, length = √((Δx)² + (Δy)²). Compute y at each x first.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — 80 m fencing against a wall
Set up. 3 sides of fencing: 2w + l = 80 ⇒ l = 80 − 2w. Maximise A.
(i) A(w) = w · (80 − 2w) = 80w − 2w².
(ii) A′(w) = 80 − 4w = 0 ⇒ w = 20. A″(w) = −4 < 0, so this is a maximum. So w = 20 m, l = 80 − 40 = 40 m.
(iii) Max area = 20 × 40 = 800 m².
Problem 2 — Projectile h(t) = 20 + 15t − 5t²
Set up. Velocity = h′(t); max height at v = 0; landing when h = 0.
(i) v(t) = h′(t) = 15 − 10t. v = 0 at t = 1.5 s.
(ii) h(1.5) = 20 + 22.5 − 11.25 = 31.25 m.
(iii) Solve 20 + 15t − 5t² = 0, i.e. 5t² − 15t − 20 = 0 ⇒ t² − 3t − 4 = 0 ⇒ (t − 4)(t + 1) = 0. Take t = 4 (the positive root), so T = 4 s. Impact speed: |v(4)| = |15 − 40| = 25 m/s.
Problem 3 — Bottle from y = x/2 + 1 on [0, 6]
Set up. Disc method about x-axis.
(i) [f(x)]² = (x/2 + 1)² = x²/4 + x + 1. V = π ∫₀⁶ (x²/4 + x + 1) dx.
(ii) V = π [x³/12 + x²/2 + x]₀⁶ = π (216/12 + 36/2 + 6) = π(18 + 18 + 6) = 42π cm³ ≈ 132 cm³.
(iii) 132 cm³ = 132 mL — well short of 600 mL. The bottle could not hold a standard 600 mL drink as designed; you would need a wider or longer profile (or a larger constant term).
Problem 4 — Traffic over 60 min
Set up. Total cars = ∫ r dt; use trapezoidal rule.
(i) h = 10; interior sum = 34 + 50 + 58 + 46 + 30 = 218. Total ≈ (10/2)·[22 + 2(218) + 18] = 5·[22 + 436 + 18] = 5 · 476 = 2380 cars.
(ii) Greatest flow at t = 30 min (58 cars/min). Rate of change of r at t = 20 estimated over [20, 30]: ≈ (58 − 50) / 10 = 0.8 cars/min per min (i.e. flow is still rising at about 0.8 cars/min²). (Or over [10, 20]: (50 − 34)/10 = 1.6; or symmetric over [10, 30]: (58 − 34)/20 = 1.2 cars/min². Any reasonable estimate is acceptable.)
(iii) Public-report figure: ≈ 2400 cars (to the nearest hundred). Limitation: 10-minute samples cannot capture short-term spikes between samples (e.g. a 90-second surge from a school zone or traffic signal cycle), so the smooth trapezoidal approximation can miss extreme but brief rates.
Problem 5 — Deck y = 4 − x²/4 on [−4, 4]
Set up. Area = ∫ y dx; for railing, sum segment lengths.
(i) Area = ∫₋₄⁴ (4 − x²/4) dx = [4x − x³/12]₋₄⁴ = (16 − 64/12) − (−16 + 64/12) = 16 − 16/3 − (−16 + 16/3) = 16 − 16/3 + 16 − 16/3 = 32 − 32/3 = (96 − 32)/3 = 64/3 m² ≈ 21.33 m².
(ii) Cost = 64/3 × $185 = $11 840/3 ≈ $3946.67.
(iii) y-values at x = −4, −2, 0, 2, 4: y = 0, 3, 4, 3, 0. Segments (each Δx = 2):
(−4,0) to (−2,3): √(4 + 9) = √13 ≈ 3.606
(−2,3) to (0,4): √(4 + 1) = √5 ≈ 2.236
(0,4) to (2,3): √(4 + 1) = √5 ≈ 2.236
(2,3) to (4,0): √(4 + 9) = √13 ≈ 3.606
Total ≈ 2(√13 + √5) ≈ 2(3.606 + 2.236) = 2(5.842) ≈ 11.68 m of railing. (True arc length is somewhat larger because the curve bows outward between sample points.)