Model exponential growth decay and threshold times
Calibrate an exponential model from two observations, find a threshold time and distinguish positive decay from reaching zero.
Section 3.8 · Differential equations
Exam scope and assessment notes
Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027, section 3.8. Paper 1 calculus and Paper 3 Integration are assumed. This pathway uses first-order separable equations, including Paper 3 integration methods, with all context assumptions supplied. Second-order equations and integrating-factor methods are beyond this mapped requirement. Check factors before dividing, retain equilibrium solutions, and state the interval and physical meaning of an initial-value solution. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Solve y′ = ky and use initial conditions.
- Use logarithms to solve exponential equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
If e^(2k) = 3/2, which value is k?
What you will learn
- Solve and calibrate a proportional growth or decay equation.
- Interpret its constant, doubling or halving time and finite-time predictions.
Use observations to calibrate the constant
The law dP/dt = kP has family P = Ae^(kt). A starting value P(0) = P₀ gives A = P₀. If P(a) = P₁ with positive amounts, then k = ln(P₁/P₀)/a. The ratio must use matching units and the elapsed interval a. Growth gives k > 0 and decay gives k < 0. In a supplied population model with P₀ = 400 and P(2) = 600, k = (1/2)ln(3/2) per hour.
This model is P(t) = 400(3/2)^(t/2), and its doubling time is ln 2/k = 2ln 2/ln(3/2), approximately 3.42 hours. The doubling time is a duration, whereas k is a reciprocal-time constant. Changing from hours to minutes changes the numerical value of k. Different time units do not produce different physical predictions when converted consistently.
A model has assumptions as well as a formula
For M′ = −λM with λ > 0, M = M₀e^(−λt). Its half-life is ln 2/λ. Starting from a positive M₀, this exponential stays positive at every finite t and approaches zero as t grows without bound. A practical detection threshold can be reached in finite time; exact zero cannot under this law. Solve the particular threshold rather than replacing it with a half-life.
Calibration to two observations does not establish that the rate law will remain valid indefinitely. State that predictions depend on a constant proportional rate and the supplied conditions continuing to hold. A continuously modelled population need not give an integer prediction; use the rounding requested by the problem and do not turn an approximate model value into a guaranteed count. Retain exact logarithms until the final numerical result.
Pause and explain
Under positive exponential decay, is exact zero reached at a finite time?
Worked example
A supplied population model has P′ = kP, P(0) = 400 and P(2) = 600, with time in hours. Find P(t) and its doubling time.
Show the worked solution
- The starting value gives P = 400e^(kt). Use e^(2k) = 3/2 to obtain k = ln(3/2)/2.
- Thus P(t) = 400(3/2)^(t/2).
- Doubling requires e^(kt) = 2, so t = 2ln 2/ln(3/2) ≈ 3.42 hours.
Answer P(t) = 400(3/2)^(t/2); doubling time 2ln 2/ln(3/2) ≈ 3.42 h.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A supplied decay model starts at 80 g and has 40 g after 3 hours. Find M(t), its positive decay constant λ and M(9).
Give me a hint
The amount halves in three hours.
Compare my reasoning
- Use M = 80e^(−λt) and e^(−3λ) = 1/2.
- Thus λ = ln 2/3 per hour and M(t) = 80·2^(−t/3).
- After 9 hours there have been three halvings, so M(9) = 10 g.
λ = ln 2/3 h⁻¹; M(t) = 80·2^(−t/3); M(9) = 10 g.
Look for these in your work
- Use a positive λ with the negative sign in the rate law.
- Check the initial amount and repeated-halving prediction.
P′ = kP, P(0) = 150 and P(4) = 240, with time in hours. Find k and the time when P first reaches 300.
Give me a hint
Use the ratio 240/150 before solving the doubling condition.
Compare my reasoning
- The ratio gives e^(4k) = 8/5, hence k = ln(8/5)/4.
- Reaching 300 requires e^(kt) = 2.
- Thus t = 4ln 2/ln(8/5), approximately 5.90 hours, under the unchanged law.
k = ln(8/5)/4 h⁻¹; threshold time 4ln 2/ln(8/5) ≈ 5.90 h.
Look for these in your work
- Calibrate using the actual elapsed interval.
- Retain exact logarithms until final rounding.
A measured signal follows a supplied exponential-decay law. It starts at 90 units and is 30 after 2 hours. Find the first time it reaches 1 unit, and explain whether the model then predicts zero.
Give me a hint
Calibrate the decay from a factor of one third over two hours.
Compare my reasoning
- The solution is q(t) = 90·3^(−t/2), with positive decay constant ln 3/2.
- Set q = 1 to obtain t = 2ln 90/ln 3, approximately 8.19 hours.
- The amount remains positive afterwards at every finite time; reaching the threshold is not reaching zero, and the prediction assumes the supplied law continues.
q(t) = 90·3^(−t/2); reaches 1 at 2ln 90/ln 3 ≈ 8.19 h; exact zero is not reached at finite time.
Look for these in your work
- Distinguish the requested threshold from a halving time.
- State the conditional interpretation and the limiting rather than finite zero.
Common mistakes
- The logarithm uses the later-to-earlier quantity ratio.
- A half-life is a duration and is not the decay constant itself.
- Positive exponential decay approaches zero without reaching it at finite time.
What can you explain now?
For M′ = −λM with λ > 0, what is the half-life?
Compare with the explanation
ln 2/λ.
The ratio M(t)/M(0) is e^(−λt); set it to 1/2 and solve for the duration.
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Make it stick
Calibrate a proportional-rate model, solve an exact threshold equation and explain the difference between a constant, a doubling or halving duration, and a limiting value.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.