Mathematics Advanced • Year 12 • Module 6 • Lesson 3
Integrating Exponentials and Logarithms
Apply exponential and reciprocal antiderivatives to medical, biological, financial and electrical contexts.
Problem 1 — Radioactive tracer (medical imaging)
A radioactive tracer in a patient's bloodstream decays at rate
dA/dt = −2 e−0.5 t mg/h,
with A(0) = 4 mg initial dose.
Set up: What are we solving for?
(i) Find A(t), the amount of tracer in the body at time t. 2 marks
(ii) How much tracer remains after 4 hours? Give the answer to 2 decimal places. 1 mark
(iii) Describe what happens to A(t) as t → ∞ and explain in one sentence what this means medically. 2 marks
Stuck? Revisit lesson § Real-World Anchor — Tumour Growth and Medical Imaging.Problem 2 — Population growth rate (biology)
A small bacterial population grows so that its rate of increase (in thousands of cells per hour) is
dP/dt = 3 e0.2 t, t in hours, t ≥ 0,
and initially P(0) = 5 (thousand cells).
Set up: What are we solving for?
(i) Find P(t). 2 marks
(ii) Predict the population (in thousands) after 5 hours, to 2 decimal places. 1 mark
(iii) Explain in one sentence why P(t) grows without bound under this model and identify one biological reason the model would eventually break down. 2 marks
Problem 3 — Continuously-compounded interest (finance)
A savings account is credited interest continuously at rate
dB/dt = 240 e0.04 t dollars per year,
with B(0) = $6000 (initial balance equal to the starting principal at this rate).
Set up: What are we solving for?
(i) Find B(t), the balance at time t years. 2 marks
(ii) Find B(10), to the nearest dollar. 2 marks
(iii) Verify in one line that dB/dt = 0.04 · B(t), which is the standard continuous-compounding ODE. 2 marks
Stuck on (iii)? Compute 0.04 · B(t) using your formula from (i) and compare with the given dB/dt.Problem 4 — Capacitor discharge (electrical)
The current flowing out of a discharging capacitor is
I(t) = 0.1 e−2 t amperes, t in seconds.
The total charge Q(t) that has flowed out by time t is found by integrating I(t), with Q(0) = 0.
Set up: What are we solving for?
(i) Find Q(t). 2 marks
(ii) Find the total charge delivered as t → ∞. 2 marks
(iii) Explain in one sentence why even though current flows for all t ≥ 0, the total charge is finite. 1 mark
Problem 5 — Why ln|x| appears (anti-derivative diagnostic)
Three students try to integrate ∫ 5/x dx. Audit each answer.
Set up: What are we solving for?
(i) For each student, decide ✓ or ✗ and give a one-line reason. 3 marks
| Student | Claim | ✓ / ✗ and reason |
|---|---|---|
| Asha | ∫ 5/x dx = 5 · x⁰/0 + C, "undefined" | |
| Ben | ∫ 5/x dx = 5 ln(x) + C | |
| Cleo | ∫ 5/x dx = 5 ln|x| + C |
(ii) Write a one-sentence rule for marking ∫ k/x dx in any test. 2 marks
Stuck? The "+ C" and the "|x|" are both required.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Radioactive tracer
Set up. Antidifferentiate dA/dt to get A(t), fix C from A(0) = 4, then evaluate at t = 4 and analyse the limit as t → ∞.
(i) A(t) = ∫ −2 e−0.5t dt = (−2) · (1/−0.5) e−0.5t + C = 4 e−0.5t + C. A(0) = 4 ⇒ 4 + C = 4 ⇒ C = 0. A(t) = 4 e−0.5t.
(ii) A(4) = 4 e−2 = 4/e² ≈ 0.54 mg.
(iii) As t → ∞, e−0.5t → 0, so A(t) → 0. Medically: the tracer is essentially completely eliminated from the body in the long run.
Problem 2 — Bacterial growth
Set up. Antidifferentiate dP/dt to obtain P(t), fix C using P(0) = 5, evaluate at t = 5, and interpret long-term behaviour.
(i) P(t) = ∫ 3 e0.2t dt = 3 · (1/0.2) e0.2t + C = 15 e0.2t + C. P(0) = 5 ⇒ 15 + C = 5 ⇒ C = −10. P(t) = 15 e0.2t − 10.
(ii) P(5) = 15 e1 − 10 = 15e − 10 ≈ 40.77 − 10 = ≈ 30.77 thousand cells.
(iii) Since 15 e0.2t → ∞ as t → ∞, P(t) grows without bound. In reality the model breaks down because of finite resources (food, space, oxygen), waste accumulation, or immune response — none of which are accounted for in the simple exponential model.
Problem 3 — Continuous interest
Set up. Antidifferentiate dB/dt to obtain B(t), use B(0) = 6000 to fix C, evaluate at t = 10, then verify the differential relationship.
(i) B(t) = ∫ 240 e0.04t dt = 240 · (1/0.04) e0.04t + C = 6000 e0.04t + C. B(0) = 6000 ⇒ 6000 + C = 6000 ⇒ C = 0. B(t) = 6000 e0.04t.
(ii) B(10) = 6000 e0.4 ≈ 6000 × 1.4918 ≈ $8951.
(iii) 0.04 · B(t) = 0.04 · 6000 e0.04t = 240 e0.04t = dB/dt ✓ — confirms the standard continuous-compounding ODE dB/dt = rB with r = 0.04.
Problem 4 — Capacitor discharge
Set up. Antidifferentiate I(t) to obtain Q(t) with Q(0) = 0, then take the limit as t → ∞.
(i) Q(t) = ∫ 0.1 e−2t dt = 0.1 · (1/−2) e−2t + C = −0.05 e−2t + C. Q(0) = 0 ⇒ −0.05 + C = 0 ⇒ C = 0.05. Q(t) = 0.05 (1 − e−2t).
(ii) As t → ∞, e−2t → 0, so Q(t) → 0.05 C (= 50 mC).
(iii) The current decays so quickly (exponentially) that the cumulative charge delivered between t = 0 and ∞ is bounded — the infinite sum of exponentially decreasing pieces converges to a finite value.
Problem 5 — Auditing ∫ 5/x dx
Set up. For each claim, identify the rule used and check both the sign/coefficient and whether the absolute value and constant of integration are present.
(i)
Asha: ✗ — applied the power rule, which fails at n = −1; the case x⁰/0 is the warning sign that a different rule is needed.
Ben: ✗ — the antiderivative is correct only for x > 0; without the absolute value the formula does not cover x < 0, where 1/x is also defined.
Cleo: ✓ — both ln|x| (covers x > 0 and x < 0) and + C are present.
(ii) "For any nonzero constant k, ∫ k/x dx = k ln|x| + C — always include the absolute value and the constant of integration."