Interpret a saturating model and check it against further data
Calibrate a separable saturation model, retain both equilibria, interpret its limiting value and use an additional observation to judge the fit.
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
- Use partial fractions for y(1 − y) and apply initial conditions.
- Read exponential limits and compare a model prediction with observed data.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which constant amounts solve P′ = rP(1 − P/L), r > 0, L > 0?
What you will learn
- Solve and calibrate a supplied first-order saturation law.
- Interpret growth or decline towards equilibrium and compare a prediction with a further observation.
Separate the supplied saturation law
For P′ = rP(1 − P/L), with r > 0 and L > 0, first check P = 0 and P = L as constant equilibria. Non-equilibrium branches separate using 1/[P(1 − P/L)] = 1/P + 1/(L − P). Integration gives ln|P/(L − P)| = rt + C and P = L/(1 + Ae^(−rt)), where A = L/P₀ − 1 for a non-zero starting amount. Retain a non-vanishing denominator on the chosen interval. The quantities and assumptions are supplied; no specialist context is needed.
For L = 200, P(0) = 20 and P(5) = 50, A = 9. The later observation gives 50 = 200/(1 + 9e^(−5r)), so e^(−5r) = 1/3 and r = ln 3/5 per hour. Hence P = 200/(1 + 9·3^(−t/5)). It predicts P(10) = 100 and P(15) = 150, with limit 200 from below. The limiting value is conditional on the supplied law, rather than a guarantee established by the two observations.
Interpret direction and test predictions
When 0 < P < L the rate is positive; when P > L it is negative. A solution starting above L can approach it from above rather than making the formula automatically invalid. Within 0 < P < L, the rate rP(1 − P/L) is a quadratic in P with maximum rL/4 at P = L/2. The rate then slows as P approaches L. For the plotted model the maximum rate occurs at P = 100, which is reached at t = 10 h.
A fitted model should be checked against further data that were not used to find its constants. With supplied L = 80, P(0) = 10 and P(2) = 20, the fitted rate constant is r = (1/2)ln(7/3) and the prediction at t = 4 is 35. If the observed value then is 42, the model underpredicts by 7. That discrepancy is evidence to review the rate-law assumptions or data quality; fitting the first two observations does not prove universal accuracy.
Pause and explain
A model fits its two calibration observations. A third observation differs from its prediction. What is justified?
Worked example
A supplied saturation law is P′ = rP(1 − P/200), with P(0) = 20 and P(5) = 50. Find its solution, P(10), P(15) and limit.
Show the worked solution
- Separation gives P = 200/(1 + Ae^(−rt)); the starting value gives A = 9.
- The five-hour value gives e^(−5r) = 1/3, so r = ln 3/5.
- Thus P = 200/(1 + 9·3^(−t/5)), giving 100 at 10 h, 150 at 15 h and limit 200 from below.
Answer P = 200/(1 + 9·3^(−t/5)); P(10) = 100; P(15) = 150; limit 200 from below.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Solve y′ = y(1 − y), y(0) = 1/4. Find the first non-negative input where y = 1/2 and name the other constant solutions.
Give me a hint
The initial ratio y/(1 − y) is 1/3.
Compare my reasoning
- Partial fractions and the initial condition give y = e^x/(3 + e^x).
- Setting y = 1/2 gives e^x = 3, so x = ln 3.
- The original equation also has equilibria y = 0 and y = 1; this initial-value solution lies strictly between them.
y = e^x/(3 + e^x); reaches 1/2 at x = ln 3; equilibria 0 and 1.
Look for these in your work
- Solve the threshold on the branch selected by the initial value.
- Retain both equilibrium solutions of the original equation.
A supplied model is P′ = 0.2P(1 − P/100), P(0) = 150, with t in hours. Find P(t), P(5ln 2), the initial rate and its forward limiting behaviour.
Give me a hint
The initial amount is above the limiting value, making the denominator constant A negative.
Compare my reasoning
- A = 100/150 − 1 = −1/3, so P = 100/[1 − (1/3)e^(−t/5)], defined for all t ≥ 0.
- At t = 5ln 2 the exponential is 1/2, giving P = 120; the initial rate is 0.2·150·(−1/2) = −15 per hour.
- The solution decreases towards 100 from above without crossing it; its forward denominator stays positive.
P = 100/[1 − (1/3)e^(−t/5)]; P(5ln 2) = 120; initial rate −15 h⁻¹ in amount units; limit 100 from above.
Look for these in your work
- Keep the signed constant and check the forward denominator.
- Interpret decline rather than assuming every saturation model must start below its limit.
A supplied law is P′ = rP(1 − P/80), with P(0) = 10 and P(2) = 20. Find r and predict P(4). A further observation is 42 at t = 4. State the discrepancy and what this check establishes.
Give me a hint
Use the first two observations to fit the model, then reserve the third for comparison.
Compare my reasoning
- The starting value gives A = 7. At t = 2, 1 + 7e^(−2r) = 4, so r = (1/2)ln(7/3).
- At t = 4, e^(−4r) = (3/7)², giving P(4) = 80/(1 + 9/7) = 35.
- The observed 42 exceeds the prediction by 7; the model underpredicts and its assumptions or the measurements need review.
r = (1/2)ln(7/3); prediction P(4) = 35; underprediction by 7 against the further observation.
Look for these in your work
- Calibrate only with the two stated fitting observations.
- Compare the independent observation and avoid claiming that a fitted law is guaranteed accurate.
Common mistakes
- Dividing by P(L − P) loses two equilibrium solutions.
- A limiting value does not mean the solution reaches it at a finite time.
- Calibration data cannot also serve as an independent check of the model.
What can you explain now?
For r > 0 and L > 0, what is the sign of rP(1 − P/L) when P > L?
Compare with the explanation
Negative.
The positive P factor multiplies a negative saturation factor, predicting decline towards the limit.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Separate a supplied saturation law, retain its equilibria and interpret direction, maximum rate and limit. Use a further observation to test a calibrated prediction.
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.