Formulate a differential equation from a stated rate of change
Identify the changing quantity, translate proportionality with the correct sign and determine a rate constant from supplied data.
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
- Interpret a derivative as an instantaneous rate.
- Rearrange a proportional relationship and use matching units.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
An amount is decreasing by 3 units per hour. What is its signed derivative?
What you will learn
- Translate a rate statement into a first-order differential equation.
- Calculate a proportionality constant and check its sign and units.
Name the dependent quantity and its rate
If M(t) is an amount in grams and t is elapsed hours, dM/dt is its instantaneous change in grams per hour. A rate proportional to the amount takes the form dM/dt = kM. The proportionality constant is fixed within the supplied model; it is not generally the measured rate itself. A growing amount has k > 0, while a decreasing positive amount has k < 0. Alternatively write a decay model as dM/dt = −λM with λ > 0, and keep that convention throughout.
When M = 20 g and the amount is decreasing at 3 g per hour, dM/dt = −3, so k = −3/20 = −0.15 per hour. Thus dM/dt = −0.15M. The rate changes as M changes. Substituting the calibration observation back into the equation is a simple sign and arithmetic check. An average change over an interval is not automatically an instantaneous rate observation.
Translate the quantity to which the rate is proportional
If a temperature T is approaching a fixed environment temperature 20, the relevant difference is T − 20. A supplied cooling law is dT/dt = −λ(T − 20), with λ > 0. It predicts cooling above 20, warming below 20 and zero change at 20. A law proportional to T itself would describe a different situation. Define the environment and time unit before finding λ from a given rate.
For a vessel with constant inflow 6 L/min and outflow proportional to its current volume V, conservation gives dV/dt = 6 − kV. If the net increase is 2 L/min at V = 40 L, then k = (6 − 2)/40 = 0.1 per minute. Zero net rate occurs at V = 60 L, provided the stated law and vessel assumptions continue to apply. This is a model prediction, not a claim about an unstated vessel capacity.
Pause and explain
Cooling is proportional to the excess above a fixed temperature a. Which law has the right sign above a, for λ > 0?
Worked example
An amount M in grams decreases at a rate proportional to M. At M = 20 g, the decrease is 3 g/h. Form its differential equation.
Show the worked solution
- Use dM/dt = kM and record a decrease as dM/dt = −3.
- Substitute −3 = 20k to obtain k = −3/20 per hour.
- The equation dM/dt = −(3/20)M returns −3 g/h at M = 20 g.
Answer dM/dt = −(3/20)M, with t in hours.
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.
A population N grows at a rate proportional to N. When N = 100, dN/dt = 8 per hour. Form the equation and give the constant's unit.
Give me a hint
Use k = rate divided by current quantity.
Compare my reasoning
- Write dN/dt = kN.
- Use 8 = 100k, so k = 0.08 per hour.
- The model dN/dt = 0.08N gives a positive growth rate and recovers the observation.
dN/dt = 0.08N; k = 0.08 h⁻¹.
Look for these in your work
- Use the rate observation with the matching quantity.
- Check the sign and reciprocal-time unit.
A temperature T approaches an environment at 20°C by a law proportional to T − 20. At T = 60°C it is cooling at 8°C/min. Form its equation using a positive decay constant.
Give me a hint
A cooling rate is negative; the current excess is 40°C.
Compare my reasoning
- Write dT/dt = −λ(T − 20), with λ > 0.
- Substitute −8 = −λ·40 to find λ = 1/5 per minute.
- At T = 60 the resulting equation returns −8; at T = 20 it returns zero.
dT/dt = −(T − 20)/5; λ = 1/5 min⁻¹.
Look for these in your work
- Use excess temperature rather than absolute temperature.
- Verify both the measured rate and equilibrium state.
A vessel receives 6 L/min and loses liquid at kV L/min, where V is its volume in litres. Its net increase at V = 40 L is 2 L/min. Form dV/dt and find the zero-net-rate volume under this law.
Give me a hint
Net change equals inflow minus outflow.
Compare my reasoning
- Use dV/dt = 6 − kV and 2 = 6 − 40k.
- Hence k = 1/10 per minute and dV/dt = 6 − V/10.
- Setting the net rate to zero gives V = 60 L, conditional on the stated law remaining valid.
dV/dt = 6 − V/10; zero net rate at V = 60 L.
Look for these in your work
- Form the balance before calibrating the constant.
- Interpret the equilibrium as conditional on the supplied assumptions.
Common mistakes
- A decrease requires a negative derivative for a positive amount.
- Use the stated excess or deficit, not a different quantity.
- A proportionality constant is not the rate measured at one particular state.
What can you explain now?
What is the rate variable when y depends on elapsed time t?
Compare with the explanation
dy/dt.
It measures the instantaneous change of y per unit time, rather than y itself.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Name a dependent quantity and its derivative, translate a proportional-rate statement, and check the constant's sign, units and calibration observation.
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.