Loav · VCE Specialist Mathematics · Units 3 & 4
S34-E02 · Recommended sequence: Unit 4
1. Translate a rate statement before solving it
A differential-equation model begins with a defined quantity, its units, an independent variable and a rule describing change. “The rate is proportional to the amount present” leads to Q′ = kQ. “The amount approaches a fixed level L at a rate proportional to the shortfall” leads to Q′ = k(L − Q) for k > 0. The equations differ even though both can produce exponential terms in their solutions.
Choose a sign convention and state assumptions. A decay model uses Q′ = −kQ with k > 0; a growth model uses Q′ = kQ. If a quantity is measured in grams and time in hours, k has units h⁻¹. A parameter's numerical value has no complete meaning until the time unit is specified. Changing hours to minutes changes the rate constant.
Before integrating, identify equilibria by setting the right-hand side to zero. Then inspect its sign above and below those levels. This predicts whether a solution rises or falls and provides a check on the final formula. A valid mathematical model can still be a poor real-world description, so keep its assumptions and intended interval explicit.
2. Exponential and limiting-level models
The solution of Q′ = −kQ with Q(0) = Q₀ is Q = Q₀e−kt. Equal time intervals multiply the amount by the same factor. The absolute amount lost per hour decreases as the quantity shrinks, even though the proportional rate remains constant. A constant percentage loss is therefore different from a constant number of units lost each hour.
The half-life solves e−kT = 1/2, giving T = ln 2/k. This interval does not depend on the starting amount within the model. For growth, the doubling time is the same expression with a positive growth constant. Use logarithms to solve threshold questions; a graph can estimate the time but does not replace an exact expression when one is requested.
For Q′ = k(L − Q), separation gives Q = L + (Q₀ − L)e−kt. The difference from the limiting level decays exponentially. If Q₀ < L, the quantity rises toward L; if Q₀ > L, it falls toward L; if Q₀ = L, it remains constant. A finite positive time does not make the exponential exactly zero, so the level is approached rather than first attained.
Worked example 1: Infer a decay parameter from observations
A stored quantity is modelled by Q′ = −kQ, with Q in grams and t in days. Initially Q = 80, and after 5 days Q = 60. The solution is Q = 80e−kt. Substitution gives 60/80 = e−5k, so k = ln(4/3)/5 day⁻¹.
To find when Q reaches 30 g, solve 30/80 = e−kt. Thus t = ln(8/3)/k = 5ln(8/3)/ln(4/3), approximately 17.05 days. This is the time measured from the original observation, not an additional interval after day 5.
The model predicts a positive quantity at every finite nonnegative time. That is a property of the idealised continuous model, not a claim that every physical stored quantity can be measured indefinitely. The parameter was estimated from the observations under an assumed constant proportional decay law.
3. Logistic change combines growth with a limiting factor
A standard logistic equation is N′ = rN(1 − N/K), where r > 0 and K > 0. The rate is proportional to N when N is small, but the factor 1 − N/K reduces the rate near K. The equilibria are N = 0 and N = K. The symbols are convenient labels; understanding their roles matters more than memorising parameter names.
For 0 < N < K, both factors are positive and the solution rises. For N > K, the second factor is negative and the solution falls toward K. The equation itself also permits negative mathematical values, but a context describing a nonnegative amount normally excludes them. Do not use a formula outside its contextual domain simply because the calculator produces a number.
To solve away from the equilibria, separate and use partial fractions:
ln|N| − ln|K − N| = rt + C
N(t) = K/[1 + Ae−rt], with A = (K − N₀)/N₀ for N₀ ≠ 0.
The minus sign before the second logarithm comes from integrating 1/(K − N). The initial condition determines A, which is positive for 0 < N₀ < K and negative for N₀ > K. The equilibrium N = 0 must be retained separately; it is not obtained by inserting N₀ = 0 into a formula that divides by N₀. The equilibrium N = K corresponds to A = 0.
4. Explain the characteristic curve
Differentiate the logistic rate along a solution: N″ = rN′(1 − 2N/K). For a growing solution between zero and K, N′ > 0. It follows that N″ > 0 below K/2 and N″ < 0 above K/2. If the solution passes through K/2 in the relevant time domain, it has an inflection there and its growth rate is greatest at that instant.
The greatest rate is r(K/2)(1/2) = rK/4. This is a rate, with units of quantity per time; K/2 is an amount, not a time. The inflection time must be found by solving N(t) = K/2. For 0 < N₀ < K, it is t = ln A/r. If N₀ > K/2, that time is negative, so a model beginning at t = 0 has no future inflection even though the full mathematical curve has one.
A solution starting above K decreases toward K and is concave up for nonnegative time: both N′ and 1 − 2N/K are negative, so their product is positive. It does not follow the familiar increasing S-shaped segment. A sketch must reflect the actual initial value rather than a memorised generic logistic picture.
Worked example 2: Locate the logistic inflection and maximum rate
A model uses N′ = 0.3N(1 − N/600) and N(0) = 60, with time in weeks and N measured in model units. Here r = 0.3 week⁻¹, K = 600 and A = (600 − 60)/60 = 9. Therefore N = 600/(1 + 9e−0.3t).
The solution increases toward 600. Its inflection occurs at N = 300. Solving 300 = 600/(1 + 9e−0.3t) gives 9e−0.3t = 1, hence t = ln 9/0.3, approximately 7.324 weeks. The greatest rate is 0.3 × 600/4 = 45 model units per week.
For a threshold of 480, the denominator must equal 600/480 = 5/4. Hence 9e−0.3t = 1/4 and t = ln 36/0.3, approximately 11.945 weeks. Distinguish the threshold value from its corresponding time, and retain exact logarithms until the final rounding step.
5. Mixing models come from a balance equation
For a well-mixed tank, the amount of dissolved material changes according to “rate in minus rate out”. If inflow is qin litres per minute with concentration cin grams per litre, its material inflow rate is qincin grams per minute. If the tank contains Q grams in V litres, the outflow concentration is Q/V under the perfect-mixing assumption.
V(t) = V₀ + (qin − qout)t.
Equal liquid inflow and outflow make volume constant, but do not make the dissolved amount constant. Unequal flows make V depend on time. A denominator V₀ used after the volume has changed gives the wrong model. Restrict the time interval so that the volume remains positive and any tank-capacity limit is respected. No solution should describe negative water volume.
Worked example 3: Solve a constant-volume mixing model
A tank contains 120 L of solution with 6 kg of dissolved material. Solution of concentration 0.2 kg/L enters at 3 L/min, while well-mixed solution leaves at 3 L/min. The volume remains 120 L. The incoming material rate is 0.6 kg/min and the outgoing rate is 3Q/120 = Q/40 kg/min.
Thus Q′ = 0.6 − Q/40 = (24 − Q)/40. With Q(0) = 6, separation gives Q = 24 − 18e−t/40. The limiting amount is 24 kg, corresponding to the incoming concentration 0.2 kg/L. This is a useful physical check on the algebra.
When Q = 18, we have e−t/40 = 1/3, so t = 40ln 3 minutes, approximately 43.944 minutes. At that moment Q′ = (24 − 18)/40 = 0.15 kg/min. A positive net rate is consistent with the tank concentration still being below the incoming concentration.
6. Assess what the model can support
A solution is only as reliable as its assumed rate law and parameter estimates. Check its starting value, signs, equilibrium levels, asymptotes, units and physically allowed values. Do not interpret an extrapolation far beyond observed conditions as a guaranteed prediction. In particular, a logistic curve alone does not establish a biological mechanism or justify an ecological or medical conclusion.
For a technology check, compare the analytic solution's derivative with the original rate expression and plot both the quantity and its rate. In a logistic model, the peak of the rate should line up with the quantity's inflection when that point lies in the chosen interval. In a mixing model, compare the long-term concentration with the inflow concentration. These checks test interpretation as well as calculation.
A useful non-routine question is which parameter can change without shifting the equilibrium. In N′ = rN(1 − N/K), changing r alters the time scale but preserves the equilibrium levels. Changing K changes the upper equilibrium and the halfway inflection level. Identify what the question holds fixed before comparing curves; the same initial value may represent different fractions of K.
Practise, then check
- A quantity satisfies Q′ = −0.08Q, Q(0) = 50. Find its formula and half-life.
- Solve H′ = 0.4(70 − H), H(0) = 20, and state the limit.
- Find equilibria and the sign of N′ for N′ = 0.2N(1 − N/500) with nonnegative N.
- For that logistic equation with N(0) = 100, find the solution and inflection time.
- For the same equation with N(0) = 400, is there a future inflection for t ≥ 0?
- A 100 L tank receives 2 L/min at 3 g/L and loses 2 L/min. Write the amount equation and its equilibrium amount.
- A tank initially holds 100 L, receives 3 L/min at 2 g/L and loses 1 L/min. Write V(t) and Q′, and explain why the volume matters.
Fully explained answers
- Separation gives Q = 50e−0.08t. Half of the initial amount requires e−0.08T = 1/2, hence T = ln 2/0.08, approximately 8.664 time units. The equation's time unit must be supplied by the context.
- The difference H − 70 decays exponentially, giving H = 70 − 50e−0.4t. It starts at 20, has positive derivative and approaches 70 from below as t increases. It does not reach 70 at a finite time.
- The equilibria are N = 0 and 500. The rate is positive for 0 < N < 500 and negative for N > 500. The sign of 1 − N/500 determines whether the nonnegative quantity rises or falls.
- Here A = (500 − 100)/100 = 4, so N = 500/(1 + 4e−0.2t). Its inflection occurs at N = 250, giving t = ln 4/0.2, approximately 6.931 time units. The positive time is consistent with starting below the halfway level.
- No. The initial value 400 already exceeds 250, and the solution increases toward 500, so it remains concave down for t ≥ 0. Algebraically A = 1/4 gives inflection time ln(1/4)/0.2 < 0, outside the stated time domain.
- The incoming material rate is 2 × 3 = 6 g/min. Outflow removes 2Q/100 = Q/50 g/min, so Q′ = 6 − Q/50. The equilibrium is Q = 300 g, giving concentration 300/100 = 3 g/L, equal to the incoming concentration.
- The volume is V = 100 + 2t L. Hence Q′ = 6 − Q/(100 + 2t) in g/min. The outgoing concentration uses the current volume, not the initial 100 L. The model applies before any capacity limit and assumes complete mixing throughout.
Curriculum and sources
VCE Specialist Mathematics Units 3 and 4, Areas of Study 4 as applicable, with Outcomes 1–3. Scope follows the VCAA Mathematics Study Design 2023–2027 and the current Specialist Mathematics FAQ. Curriculum checked 15 October 2026. Teaching reference: Cambridge Senior Mathematics VCE Specialist Mathematics 3&4, second edition (2023), sections 11D–11G (pp493–515). All teaching prose, examples, practice questions and diagrams are original. Independent teaching material; not a VCAA publication.
