Mathematics Advanced • Year 12 • Module 6 • Lesson 15
Module Synthesis — Further Calculus
Apply the full Module 6 toolbox to architecture, pharmacology, finance, ecology and physics word problems.
Problem 1 — Stadium roof (volume of revolution)
A small architectural feature on a stadium roof has cross-section y = √x for 0 ≤ x ≤ 4 (units: metres). It is rotated about the x-axis to form a solid.
Set up: What are we solving for?
(i) Set up the disk-method volume integral V = π ∫₀⁴ [f(x)]² dx and identify [f(x)]². 1 mark
(ii) Evaluate V exactly, then to 2 d.p. in cubic metres. 2 marks
(iii) If the rendering material costs $180 per cubic metre, find the cost (to nearest dollar). 2 marks
Stuck? Revisit lesson § Formula Reference — Volume (disk).Problem 2 — Drug elimination + separable DE
A drug obeys dC/dt = −k C with half-life t1/2 = 4 hours. A patient takes a 200 mg dose at t = 0.
Set up: What are we solving for?
(i) Use separation of variables to derive C(t) = 200 e−kt and find k (exact). 2 marks
(ii) Find C(10) to the nearest mg. 2 marks
(iii) The total drug exposure over the first 12 hours is ∫₀¹² C(t) dt (units mg·hour, often called "AUC"). Evaluate it exactly and to 1 d.p. 3 marks
Problem 3 — Saving rates (FTC + exponential)
An investor's savings account balance satisfies dA/dt = 0.04 A + 1000 (continuous compounding at 4% per year plus a deposit rate of $1000/year), with A(0) = 10000. (You may take the particular solution as known.)
A(t) = 35000 e0.04 t − 25000
Set up: What are we solving for?
(i) Verify A(0) = 10000 and verify A(t) satisfies the DE. 3 marks
(ii) Find A(20) to the nearest dollar. 2 marks
(iii) The average balance over the 20 years is (1/20) ∫₀²⁰ A(t) dt. Evaluate to the nearest dollar. 3 marks
Stuck? Use ∫ e0.04 t dt = (1/0.04) e0.04 t = 25 e0.04 t.Problem 4 — Particle motion with sign change
A particle moves along a straight line with velocity v(t) = 3t² − 12t + 9 m/s for t ≥ 0.
Set up: What are we solving for?
(i) Factor v(t) and find the times at which the particle is instantaneously at rest. 2 marks
(ii) Find the displacement from t = 0 to t = 4. 2 marks
(iii) Find the total distance travelled from t = 0 to t = 4, and explain in one sentence why it differs from the displacement. 3 marks
Problem 5 — Choose the right technique (mixed grid)
For each integral, name the best technique (power rule / substitution / by parts / standard / partial fractions / FTC + sketch), then evaluate.
Set up: What are we solving for?
| # | Integral | Technique | Value |
|---|---|---|---|
| A | ∫₀¹ x² dx | ||
| B | ∫ x ex² dx | ||
| C | ∫ x ln(x) dx | ||
| D | ∫₀π sin(x) dx |
Marks: 1 mark per row for technique + 1 mark per row for the value. Total 8 marks.
(i) Complete the grid above. 8 marks
(ii) Which row required the LIATE rule? Explain in one sentence. 1 mark
How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Stadium roof volume
Set up. Disk method volume of y = √x rotated about x-axis.
(i) [f(x)]² = (√x)² = x. So V = π ∫₀⁴ x dx.
(ii) V = π [x²/2]₀⁴ = π · 8 = 8π m³ ≈ 25.13 m³.
(iii) Cost = 8π × 180 = 1440π ≈ $4524.
Problem 2 — Drug elimination + AUC
Set up. Solve the separable DE, evaluate at t = 10, and compute the area under C(t) on [0, 12].
(i) dC/C = −k dt ⇒ ln|C| = −kt + C₁ ⇒ C(t) = 200 e−kt (using C(0) = 200). Half-life: (ln 2)/k = 4 ⇒ k = (ln 2)/4 ≈ 0.1733 / hour.
(ii) C(10) = 200 e−10k = 200 · 2−10/4 = 200 · 2−2.5 ≈ 200 × 0.1768 ≈ 35 mg.
(iii) ∫₀¹² 200 e−kt dt = (200/(−k))[e−kt]₀¹² = (200/k)(1 − e−12k) = (200/k)(1 − 2−3) = (200/k)(7/8). With k = (ln 2)/4: = (200 × 4 / ln 2)(7/8) = (800/ln 2)(7/8) = 700/ln 2 ≈ 1009.9 mg·hr (exact form: 700/ln 2).
Problem 3 — Savings (DE verification + averages)
Set up. Verify a given particular solution, then compute the time-averaged value.
(i) A(0) = 35000 e0 − 25000 = 35000 − 25000 = 10000 ✓. dA/dt = 35000 · 0.04 e0.04 t = 1400 e0.04 t. RHS = 0.04 A + 1000 = 0.04(35000 e0.04 t − 25000) + 1000 = 1400 e0.04 t − 1000 + 1000 = 1400 e0.04 t = LHS ✓.
(ii) A(20) = 35000 e0.8 − 25000 ≈ 35000 × 2.2255 − 25000 ≈ 77,894 − 25,000 ≈ $52,894.
(iii) ∫₀²⁰ A(t) dt = ∫₀²⁰ (35000 e0.04 t − 25000) dt = 35000 · 25 · [e0.04 t]₀²⁰ − 25000 · 20 = 875000 (e0.8 − 1) − 500000 ≈ 875000 × 1.2255 − 500000 ≈ 1,072,344 − 500,000 ≈ 572,344. Average = 572,344 / 20 ≈ $28,617.
Problem 4 — Particle with sign change
Set up. Sign analysis of v(t), displacement integral, distance integral (split at sign changes).
(i) v(t) = 3(t² − 4t + 3) = 3(t − 1)(t − 3). Rest at t = 1 and t = 3.
(ii) Displacement = ∫₀⁴ (3t² − 12t + 9) dt = [t³ − 6t² + 9t]₀⁴ = 64 − 96 + 36 = 4 m.
(iii) Sign: + on [0, 1], − on [1, 3], + on [3, 4]. Using x(t) = t³ − 6t² + 9t: x(0) = 0, x(1) = 4, x(3) = 0, x(4) = 4. Distance = |x(1) − x(0)| + |x(3) − x(1)| + |x(4) − x(3)| = 4 + 4 + 4 = 12 m. Distance > displacement (12 > 4) because the particle moved backward from t = 1 to t = 3; the backward motion is added to total distance but cancels in net displacement.
Problem 5 — Technique grid
Set up. Match technique to integrand structure, then evaluate.
(i) A: power rule / FTC; ∫₀¹ x² dx = [x³/3]₀¹ = 1/3. B: substitution u = x²; ∫ x ex² dx = (1/2) ex² + C. C: integration by parts (LIATE: Log first, so u = ln x); ∫ x ln(x) dx = (x²/2) ln(x) − ∫ (x²/2)(1/x) dx = (x²/2) ln(x) − x²/4 + C. D: standard integral + FTC; ∫₀π sin(x) dx = [−cos(x)]₀π = −(−1) − (−1) = 2.
(ii) Row C used LIATE: with both a log and an algebraic factor present, LIATE prioritises Log, so u = ln(x) — which simplifies after one differentiation, while x² is easy to integrate.