IB Maths AA HL Topic 5 — Calculus Paper 1 & 2 HL ~12 min read

Modelling with Differential Equations

Real-world rate problems translate naturally into DEs. The phrase “rate of change of P” is dP/dt; “proportional to” is “= k ×”; “decreasing” means a negative sign on k. Once you’ve translated to a DE, pick the right technique from earlier notes (separation, integrating factor) to solve it, then interpret the answer back in context.

📘 What you need to know

“Rate of change of X with respect to time” means dX/dt. Time is the usual independent variable in modelling.
“Proportional to Y” means “= k · Y” for some constant k. The constant of proportionality is found from data (e.g., a half-life or a measurement at a known time).
Sign of k: assume k > 0 always; use −k for decreasing quantities (cooling, decay) so the negative is explicit.
Standard models you should recognize: exponential growth/decay dN/dt = ±kN; Newton’s cooling dT/dt = −k(T − T_a); logistic dN/dt = kN(a − N) (next note).
“Rate in − rate out”: for tanks and mixing, dS/dt = (rate flowing in) − (rate flowing out). Both rates need their UNITS to match (e.g., kg/min).
Initial conditions come from context: “starts with 100 g” gives M(0) = 100; “at t = 5 min, T = 70°C” gives T(5) = 70.
Workflow: translate words → set up DE → solve using earlier techniques → check answer makes sense in context (units, limits, growth direction).
Long-term behaviour: take t → ∞ to find the equilibrium / terminal / steady-state value of the modelled quantity.

Translating words into a DE

Translation dictionary “rate of change of X” → dXdt “proportional to Y” → = kY “decreasing” → use −k

The four canonical modelling templates. Most IB exam problems fit one of these patterns. Once you spot the pattern, the DE practically writes itself.

Solution techniques recap

Separation of variables

dy/dx = g(x)·h(y)

Use for growth dN/dt = kN, decay dM/dt = −kM, Newton’s cooling dT/dt = −k(T − T_a). All separable.

Integrating factor

dy/dx + P(x)y = Q(x)

Use for mixing problems dS/dt + (rate/V)S = constant, falling with air resistance dv/dt + (k/m)v = g.

🧭 Recipe — modelling problems

Identify variables and rates: name the quantity (P, T, S, …), the independent variable (usually t), and the rate d…/dt.
Translate the verbal relationship into a DE. Use the dictionary: “proportional to” → “= k ×”; “decreasing” → use −k; “difference” → (A − B).
Solve using the right technique: separation of variables for most growth/decay/cooling, integrating factor for in–out flow problems.
Apply initial/boundary conditions to determine the constants (both k and c). Usually one IC gives c; a second data point gives k.
Interpret the answer in context: check units, take t → ∞ for long-term behaviour, answer the specific question asked.

Worked examples

WE 1

Setup only — bacteria growth

A bacteria population N grows so that its rate of change at any time t is proportional to the current population. Write down a differential equation modelling this.

Step 1 — identify the rate rate of change of N → dN/dt Step 2 — translate “proportional to N” proportional to N → = k · N (for some constant k) Step 3 — check sign population is GROWING → k > 0, positive sign dNdt = kN (k > 0) this is the classical exponential growth model. Solving (by separation) gives N = N₀ e^(kt) — unlimited growth, only valid for small populations relative to environmental limits.

WE 2

Setup — Newton’s Law of Cooling

An object at temperature T cools so that its rate of cooling is proportional to the difference between T and the ambient temperature T_a. Assuming the object is hotter than its surroundings, write down a differential equation modelling this.

Step 1 — identify the rate rate of cooling = dT/dt Step 2 — translate “proportional to (T – T_a)” dT/dt = ± k(T – T_a) Step 3 — fix the sign object hotter than surroundings: T – T_a > 0 object is cooling: dT/dt < 0 ⟹ sign on the right must be NEGATIVE: dT/dt = -k(T – T_a) dTdt = −k(T − T_a) (k > 0) the minus sign is essential: dT/dt must be negative when T > T_a (the object cools toward ambient). Same form also models a cold object warming up — the equation handles both cases automatically.

WE 3

Radioactive decay — solve and predict

A radioactive isotope decays so that its rate of decay is proportional to the mass remaining. The initial mass is 100 g and the half-life is 8 days. Find the mass after 24 days.

Step 1 — set up DE dM/dt = -k M, M(0) = 100 Step 2 — solve by separation of variables (1/M) dM = -k dt ln M = -kt + c (M > 0, drop modulus) M = A e^(-kt) Step 3 — apply IC M(0) = 100 100 = A · 1 ⟹ A = 100 M(t) = 100 e^(-kt) Step 4 — use half-life to find k M(8) = 50 ⟹ 50 = 100 e^(-8k) e^(-8k) = 1/2 ⟹ -8k = -ln 2 ⟹ k = (ln 2)/8 Step 5 — evaluate at t = 24 M(24) = 100 e^(-24 · ln(2)/8) = 100 e^(-3 ln 2) = 100 · 2^(-3) = 100/8 M(24) = 12.5 g three half-lives have passed (24/8 = 3), so mass is divided by 2³ = 8. Going through the exponential form is the rigorous method; the “halving” shortcut works when t is a whole-number multiple of the half-life.

WE 4

Newton’s cooling — solve fully

A cup of coffee at 90°C is placed in a room at 20°C. After 5 minutes the coffee has cooled to 70°C. Find an expression for the coffee’s temperature T at time t minutes.

Step 1 — set up DE dT/dt = -k(T – 20), T(0) = 90, T(5) = 70 Step 2 — separate and integrate 1/(T – 20) dT = -k dt ln|T – 20| = -kt + c (T > 20, drop modulus) T – 20 = A e^(-kt) T(t) = 20 + A e^(-kt) Step 3 — apply IC T(0) = 90 90 = 20 + A ⟹ A = 70 T(t) = 20 + 70 e^(-kt) Step 4 — use T(5) = 70 to find k 70 = 20 + 70 e^(-5k) 50 = 70 e^(-5k) ⟹ e^(-5k) = 5/7 -5k = ln(5/7) ⟹ k = (1/5) ln(7/5) T(t) = 20 + 70 (5/7)^t/5 °C long-term: as t → ∞, T → 20°C (coffee reaches room temperature). The (5/7)^(t/5) form is equivalent to e^(-kt) but cleaner because the base 5/7 is the cooling ratio over a 5-minute period.

WE 5

Mixing tank — uses integrating factor

A tank contains 1000 L of pure water. Brine with salt concentration 0.05 kg/L flows in at 10 L/min, and the well-mixed solution flows out at 10 L/min. Find the amount of salt S(t) in the tank at time t minutes.

Step 1 — set up DE (rate in – rate out) Rate in: 10 L/min × 0.05 kg/L = 0.5 kg/min Concentration in tank: S/1000 kg/L Rate out: 10 L/min × S/1000 = S/100 kg/min dS/dt = 0.5 – S/100, S(0) = 0 Step 2 — rewrite in standard form for IF dS/dt + S/100 = 0.5; P = 1/100, Q = 0.5 Step 3 — find IF μ = e^(t/100) Step 4 — multiply and integrate d/dt (S e^(t/100)) = 0.5 e^(t/100) S e^(t/100) = 0.5 · 100 e^(t/100) + C = 50 e^(t/100) + C S(t) = 50 + C e^(-t/100) Step 5 — apply IC S(0) = 0 0 = 50 + C ⟹ C = -50 S(t) = 50(1 − e⁻^t/100) kg long-term: as t → ∞, S → 50 kg, matching the concentration of incoming brine (0.05 kg/L × 1000 L = 50 kg). The tank asymptotically reaches the inflow concentration — makes physical sense.

WE 6

Falling with air resistance — terminal velocity

A skydiver of mass 80 kg falls from rest. Air resistance is proportional to velocity, with constant of proportionality 16 kg/s. Taking g = 9.8 m/s², so that Newton’s second law gives 80 dv/dt = 80g − 16v, find v(t) and the terminal velocity.

Step 1 — simplify and standard form 80 dv/dt = 80(9.8) – 16v Divide by 80: dv/dt = 9.8 – 0.2v Rearrange: dv/dt + 0.2v = 9.8; P = 0.2, Q = 9.8 Step 2 — IF μ = e^(0.2t) Step 3 — multiply and integrate d/dt (v e^(0.2t)) = 9.8 e^(0.2t) v e^(0.2t) = 9.8/0.2 e^(0.2t) + C = 49 e^(0.2t) + C v(t) = 49 + C e^(-0.2t) Step 4 — apply IC v(0) = 0 0 = 49 + C ⟹ C = -49 v(t) = 49(1 – e^(-0.2t)) Step 5 — terminal velocity lim t→∞ v(t) = 49(1 – 0) = 49 m/s v(t) = 49(1 − e⁻^0.2t) m/s; terminal velocity = 49 m/s at terminal velocity, dv/dt = 0, so 80g – 16v = 0 → v = 80(9.8)/16 = 49 m/s. The skydiver approaches but never quite reaches 49 m/s — typical of “approach to equilibrium” models.

💡 Top tips

Always check the sign: is the quantity growing or shrinking? Match the sign of the RHS so dN/dt has the right sign.
Assume k > 0 and use −k for decreasing. Cleaner than letting k be signed and trying to remember which way it goes.
Two unknowns need two data points: c (from IC) and k (from a second measurement, like a half-life or temperature at t = 5).
“Approach to equilibrium” form y = A + Be^−kt: very common in cooling, mixing, and resisted-motion problems. The constant A is the long-term value (t → ∞).
Check units: rates must have consistent units (kg/min, L/min, °C/min, m/s²). Mismatched units = wrong DE.

⚠ Common mistakes

Wrong sign on k: writing dT/dt = k(T − T_a) for a cooling object would predict T increasing (since T > T_a). Must be dT/dt = −k(T − T_a).
Confusing concentration with amount: in a tank problem, S/V is the concentration; S is the total mass of dissolved substance. The outflow rate is (flow rate) × (concentration) = (flow rate) × S/V.
Forgetting to find both constants: “c” from the IC and “k” from a second data point. Many students stop after finding c.
Not interpreting the answer: the question often asks for terminal velocity, time to reach a certain value, or long-term equilibrium — not just the formula. Answer what’s asked.
Mixing time units: if k is in “per minute” but you evaluate at t in seconds, the answer is wrong. Keep units consistent throughout.

Up next: The Logistic Equation. Pure exponential growth dN/dt = kN predicts unlimited growth — unrealistic. Real populations have a carrying capacity a, the maximum size the environment can sustain. The logistic DE dN/dt = kN(a − N) models this: rapid growth when N is small, slowing as N approaches a. Solving uses partial fractions.

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