Respect physical boundaries and singular times in changing-quantity models
Use separation on the permitted interval, stop a model at an empty vessel or pole and check whether an algebraic continuation still satisfies the original law.
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
- Integrate powers and logarithmic composites after separation.
- Check non-negative quantities, original denominators and elapsed-time bounds.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
If √V = 3 − t with V ≥ 0, what restriction follows?
What you will learn
- Solve changing-quantity equations with square-root or variable-denominator terms.
- Identify a physical boundary or singular time and interpret a solution only on its valid interval.
Squaring can hide the end of a physical branch
Suppose a supplied vessel model is V′ = −2√V, V(0) = 9 L, with t in minutes and V ≥ 0. For V > 0, separation gives 2√V = −2t + C, so √V = 3 − t. The square-root relation requires t ≤ 3. Thus V = (3 − t)² on 0 ≤ t ≤ 3, reaching an empty vessel at 3 min. The coefficient has units √L per minute so the right-hand side is a volume rate.
The same squared expression at t > 3 increases and no longer satisfies V′ = −2√V: its derivative is positive, while the original right-hand side is negative for a positive volume. If the model has no inflow, staying at V = 0 after emptying satisfies the original rate law. Check this boundary directly, since dividing by √V excluded zero. A convenient polynomial formula is not permission to interpret a vessel as refilling.
A pole and an undefined physical denominator are different boundaries
For y′ = y² and y(0) = 1, integration gives y = 1/(1 − t). The interval through the initial point is t < 1, and forward time uses 0 ≤ t < 1. The solution grows without bound as t approaches 1 from below. The negative formula value at t = 2 lies beyond an undefined pole; it is not a finite continuation of the same initial-value solution. The singularity also warns against indefinite interpretation of a supplied physical model.
In a well-mixed vessel losing 2 L/min with no inflow, an initial 20 L gives V = 20 − 2t. If the amount of dissolved material is S grams, the supplied uniform concentration is S/V g/L, so the loss law is S′ = −2S/(20 − 2t). It is defined only while liquid remains, 0 ≤ t < 10. Starting at 10 g gives S = 10 − t, and the concentration is 1/2 g/L before emptying. At t = 10 the ratio is 0/0, so the concentration formula is undefined even though its limit is 1/2.
Pause and explain
For y = 1/(1 − t), y(0) = 1, is the value at t = 2 a continuation through t = 1?
Worked example
Solve V′ = −2√V, V(0) = 9 L, and explain the valid draining interval and what happens at its end.
Show the worked solution
- For positive V, integrate dV/√V = −2dt to obtain √V = 3 − t.
- Non-negativity of the square root gives V = (3 − t)² for 0 ≤ t ≤ 3; the vessel empties at 3 min.
- Beyond that time the same squared formula violates the original rate sign; with no inflow, V = 0 is a valid continuation.
Answer V = (3 − t)² for 0 ≤ t ≤ 3; empty at 3 min, then stays zero under the no-inflow law.
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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.
V′ = −3√V, V(0) = 16 L, with time in minutes. Find the draining solution, V(2) and the emptying time.
Give me a hint
Integrate for V > 0 and keep the non-negative square-root relation.
Compare my reasoning
- Separation gives 2√V = −3t + 8, so √V = 4 − 3t/2.
- Thus V = (4 − 3t/2)² on 0 ≤ t ≤ 8/3; at t = 2 the volume is 1 L.
- The vessel empties at 8/3 min; continuing the squared polynomial past that boundary would reverse the rate sign.
V = (4 − 3t/2)², 0 ≤ t ≤ 8/3; V(2) = 1 L; empty at 8/3 min.
Look for these in your work
- Check the coefficient in the square-root primitive.
- Keep the physical interval when evaluating and interpreting the result.
Solve y′ = y² with y(0) = 1. State its forward interval and behaviour near the singular time, and explain why evaluating the same formula at 2 does not continue this solution.
Give me a hint
Integrate −1/y = t + C before applying the condition.
Compare my reasoning
- The initial condition gives y = 1/(1 − t).
- The forward solution has 0 ≤ t < 1 and grows without bound as t approaches 1 from below.
- At t = 1 the formula is undefined; its value −1 at t = 2 belongs to a separate branch beyond the pole.
y = 1/(1 − t), 0 ≤ t < 1; diverges as t approaches 1 from below and cannot continue finitely through it.
Look for these in your work
- Use the initial condition with the correct reciprocal sign.
- State the interval and distinguish a separate algebraic branch from continuation.
A vessel starts with 20 L containing 10 g of dissolved material, uniformly mixed. No liquid enters; 2 L/min leaves. Concentration is S/V g/L. Form and solve the amount equation, find S(8) and state what is undefined at emptying.
Give me a hint
Outflow of material equals liquid outflow times its uniform concentration.
Compare my reasoning
- V = 20 − 2t, so S′ = −2S/(20 − 2t) for 0 ≤ t < 10.
- Integration gives ln S = ln(20 − 2t) + C; the initial condition gives S = (20 − 2t)/2 = 10 − t.
- At 8 min, S = 2 g and V = 4 L, so concentration is 1/2 g/L. At emptying the amount tends to zero, but S/V is 0/0 and undefined.
S′ = −2S/(20 − 2t); S = 10 − t for 0 ≤ t < 10; S(8) = 2 g; concentration is undefined at the empty vessel.
Look for these in your work
- Use the amount-to-volume ratio in the stated loss law.
- Retain the pre-emptying interval and distinguish a ratio limit from its endpoint value.
Common mistakes
- Preserve a square-root sign condition after squaring.
- An initial-value solution cannot pass through an undefined pole.
- A concentration ratio is undefined when both the amount and liquid volume are zero.
What can you explain now?
Why is (3 − t)² invalid as the same draining solution for t > 3?
Compare with the explanation
Its derivative becomes positive while −2√V is negative for positive V.
Check the original equation and the non-negative square-root relation, not the squared expression alone.
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Make it stick
Solve a changing-quantity model, identify the first physical or mathematical boundary and check the original rate law before continuing an algebraic expression past it.
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.