Model an approach to a limiting value from above or below
Solve a rate proportional to an excess or deficit, use data to calibrate the approach and interpret the limit without claiming it is reached.
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
- Separate variables using a shifted dependent quantity.
- Solve exponential equations with exact logarithms.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
At 50°C in a 20°C environment, what is the excess temperature?
What you will learn
- Solve y′ = −λ(y − a) and calibrate it from excess or deficit data.
- Interpret the direction, threshold time and limiting value of the solution.
Solve for the excess or deficit
With a fixed limiting value a and λ > 0, dy/dt = −λ(y − a) separates to dy/(y − a) = −λ dt away from equilibrium. Integration gives y = a + Ae^(−λt), and y(0) = y₀ sets A = y₀ − a. If A > 0 the solution decreases from above a; if A < 0 it increases from below a. A = 0 gives the constant equilibrium that division initially excluded.
For an environment at 20°C, T(0) = 80°C and T(10) = 50°C, the excess falls from 60 to 30 in ten minutes. Hence λ = ln 2/10 per minute and T = 20 + 60·2^(−t/10). Halving applies to the excess above 20, not to the whole temperature. At t = 20 min the temperature is 35°C, not 20°C or half of its original temperature.
The equilibrium is approached from the starting side
For the same cooling model, reaching 25°C means the excess is 5. Thus 60·2^(−t/10) = 5, giving t = 10ln 12/ln 2, approximately 35.8 min. The exact environment temperature is only approached as t tends to infinity when A ≠ 0. Because an exponential never changes sign, the solution does not cross the fixed limiting value under this law.
A balance law can have this same form. V′ = 6 − V/10 equals −(V − 60)/10. Starting at 20 L gives V = 60 − 40e^(−t/10), increasing towards 60 L from below. This predicts a limiting volume under the stated inflow and outflow assumptions, not the size of an unstated tank. Identify the relevant excess or deficit before calibrating and state the assumptions for long-term predictions.
Pause and explain
Which quantity halves regularly in the model T′ = −λ(T − 20)?
Worked example
A supplied cooling law has environment 20°C, T(0) = 80°C and T(10) = 50°C, with time in minutes. Find T(t) and the time to 25°C.
Show the worked solution
- Use T = 20 + 60e^(−λt); the ten-minute observation gives e^(−10λ) = 1/2.
- Thus T = 20 + 60·2^(−t/10).
- At 25°C the excess is 5, so t = 10ln 12/ln 2 ≈ 35.8 min.
Answer T(t) = 20 + 60·2^(−t/10); reaches 25°C at 10ln 12/ln 2 ≈ 35.8 min.
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.
An environment is 18°C. A supplied cooling model has T(0) = 90°C and T(6) = 54°C. Find T(t), T(12) and the first time to 27°C.
Give me a hint
The excess halves from 72 to 36 in six minutes.
Compare my reasoning
- The model is T = 18 + 72·2^(−t/6).
- At 12 min the excess is 18, giving T = 36°C.
- At 27°C the excess is 9 = 72/8, so three excess halvings take 18 min.
T = 18 + 72·2^(−t/6); T(12) = 36°C; reaches 27°C at 18 min.
Look for these in your work
- Use the environment offset in both calibration observations.
- Solve the requested excess threshold and return to the full temperature.
A supplied warming law is T′ = −λ(T − 24), λ > 0. T(0) = 8°C and T(4) = 16°C. Find T(t), T(8) and its limiting behaviour, with time in minutes.
Give me a hint
The signed initial deviation is −16, and its magnitude halves in four minutes.
Compare my reasoning
- Use T = 24 − 16e^(−λt); at t = 4 the deficit is 8, so λ = ln 2/4.
- Thus T = 24 − 16·2^(−t/4), giving T(8) = 20°C.
- The derivative is positive below 24, and the model approaches 24°C from below without crossing or reaching it at finite time.
T = 24 − 16·2^(−t/4); T(8) = 20°C; limit 24°C from below.
Look for these in your work
- Keep the negative initial deviation.
- Interpret the rate sign and the limiting value on the forward interval.
A supplied volume balance is V′ = 6 − V/10, V(0) = 20 L, with t in minutes. Find V(t), the first time to 50 L and the limiting volume.
Give me a hint
Rearrange the law as a rate proportional to the deficit from 60.
Compare my reasoning
- Integration and the initial value give V = 60 − 40e^(−t/10).
- At V = 50 the deficit is 10, so e^(−t/10) = 1/4 and t = 10ln 4 ≈ 13.9 min.
- The volume increases towards 60 L from below, assuming the inflow and outflow law remain valid.
V = 60 − 40e^(−t/10); reaches 50 L at 10ln 4 ≈ 13.9 min; limit 60 L.
Look for these in your work
- Check the starting value and the balance equation.
- Distinguish a conditional limiting volume from an unstated tank capacity.
Common mistakes
- Calibrate using excess or deficit rather than the full quantity.
- A solution starting below the limiting value can increase under the same negative-excess law.
- A limiting value is not automatically reached at finite time or a physical capacity.
What can you explain now?
For y = a + Ae^(−λt), λ > 0 and A < 0, how does y approach a?
Compare with the explanation
It increases towards a from below.
The negative deviation becomes smaller in magnitude while retaining its sign.
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
Solve a shifted proportional-rate equation, calibrate its deviation and explain whether the solution approaches the limit from above or below and when a threshold is reached.
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.