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."