01 · Find the growth constant
A quantity Q grows at 15 units/day when Q = 250 units, with growth proportional to Q. Form its equation.
Hint
15 = k·250.
Worked solution
dQ/dt = 0.06Q, with t in days and k = 0.06 day⁻¹.
Understand · explore · practise
Forming differential equations
Turn descriptions of growth, decay, cooling, inflow and geometric change into differential equations. Interpret signs, constants, units and equilibrium using an interactive rate-law model and original practice.
Before you startRates of change; proportionality; chain rule
01 / Translate a rate statement into an equation
The derivative describes how a quantity changes.
“The rate of increase of y is proportional to y” → dy/dt = ky, k > 0
A differential equation contains a derivative of an unknown function. When forming a model, name the changing quantity, choose its time unit and translate the statement about its rate.
Forming an equation is different from solving it. You can often interpret signs and equilibrium states before finding a formula for the whole motion.
Read a rate from the current stateExplore
V = 10 L; dV/dt = 5 L/min. The volume is increasing.
Equilibrium: V = 20 L. Below it the volume increases; above it the volume decreases.
The blue line plots rate against volume, not volume against time. The arrow shows the direction of change, not an exact change after one minute. Assume the tank has enough capacity for the states shown.
02 / Model proportional growth
“Proportional to the amount” describes its rate, not its value.
A population P grows at a rate proportional to its current size.Worked example
dP/dt = kP, with k > 0
P depends on time t.
If P = 300 and dP/dt = 12 per day, then k = 12/300 = 0.04 day⁻¹
Use simultaneous state-and-rate information.
dP/dt = 0.04P
The rate changes as P changes.
When P = 500, the model predicts 20 per day
A constant proportionality factor does not mean a constant absolute rate.
02 · Avoid a wrong translation
Why does P = kt not express “the growth rate of P is proportional to P”?
Hint
Differentiate P = kt.
Worked solution
It gives dP/dt = k, a constant absolute rate. Proportional growth requires dP/dt = kP instead.
03 / Represent proportional loss with the correct sign
Use a positive constant and show the minus sign explicitly.
A material loses mass at a rate proportional to its mass M.Worked example
dM/dt = −kM, with k > 0
For M > 0 this gives a negative mass rate.
At M = 80 g, the loss is 6 g/hour
The signed rate is −6 g/hour.
−6 = −80k, so k = 0.075 hour⁻¹
Both negatives cancel.
dM/dt = −0.075M
The magnitude of the loss decreases as the remaining mass decreases.
03 · Another decay model
A mass loses 7 g/hour when 140 g remains, with loss proportional to mass. Form its equation.
Hint
The signed mass rate is −7.
Worked solution
dM/dt = −0.05M, with t measured in hours.
04 · Interpret the sign
For dM/dt = −0.05M, find the rate when M = 60 g and state the loss per hour.
Hint
Substitute M = 60.
Worked solution
The signed rate is −3 g/hour. The mass is being lost at 3 g/hour.
04 / Translate inverse proportionality carefully
Put the stated quantity in the denominator.
The speed of increase of x is inversely proportional to x², for x > 0.Worked example
dx/dt = k/x², with k > 0
It is the rate that is inversely proportional to the square.
If x = 3 m and dx/dt = 2 m/s, then 2 = k/9
Use the stated instant.
k = 18 m³/s
The units make k/x² a length rate.
dx/dt = 18/x²
At x = 6, the rate is 1/2 m/s.
05 · Inverse length
The rate of increase of x is inversely proportional to x. It is 3 cm/s when x = 4 cm. Form the equation for x > 0.
Hint
dx/dt = k/x.
Worked solution
k = 12 cm²/s, so dx/dt = 12/x.
06 · Compare input sizes
Under dx/dt = 18/x², what happens to the rate when x is tripled?
Hint
Compare x² with (3x)².
Worked solution
The rate becomes one ninth of its previous value, while x remains positive.
05 / Use the difference from the surroundings
The same model also describes warming below ambient temperature.
An object’s temperature T approaches a constant room temperature of 18°C.Worked example
dT/dt = −k(T − 18), with k > 0
The rate opposes the temperature difference.
At T = 58°C, it cools at 4°C/min
The signed rate is −4.
−4 = −k·40, so k = 0.1 min⁻¹
Use the difference 58 − 18, not T alone.
dT/dt = −0.1(T − 18)
Above 18 it cools; below 18 it warms; at 18 the rate is zero.
07 · Form a cooling equation
A room is at 22°C. An object at 62°C cools at 2°C/min, with the rate proportional to the temperature difference. Form the equation.
Hint
−2 = −k(62 − 22).
Worked solution
dT/dt = −0.05(T − 22), with time in minutes.
08 · Below the room temperature
For question 07, find the temperature rate when T = 12°C.
Hint
12 − 22 is negative.
Worked solution
dT/dt = −0.05(−10) = 0.5°C/min. The object warms in this model.
06 / Combine inflow and loss
Net change equals input minus output.
Water enters a tank at 10 L/min. Outflow is proportional to its current volume V, with factor 0.5 min⁻¹.Worked example
dV/dt = 10 − 0.5V
Both terms have units L/min.
At V = 10 L, dV/dt = 5 L/min
Inflow exceeds outflow.
At V = 30 L, dV/dt = −5 L/min
Outflow exceeds inflow.
At V = 20 L, dV/dt = 0
The equilibrium volume balances the two flows.
Compare rates below, at and above equilibrium
Pause, replay or seek freely. The notes explain the same idea and stay in view.
09 · A new balance model
A tank receives 8 L/min and loses 0.2V L/min when its volume is V litres. Form the equation and find the equilibrium volume.
Hint
Set the net rate equal to zero.
Worked solution
dV/dt = 8 − 0.2V. Equilibrium occurs at V = 40 L, provided this fits within the tank’s operating range.
10 · Interpret both sides
For question 09, find the rates at V = 25 L and V = 50 L.
Hint
Evaluate the net flow at each state.
Worked solution
The rates are 3 L/min and −2 L/min respectively. Below 40 L the volume increases; above 40 L it decreases.
07 / Convert the model to the variable requested
A volume equation is not yet a radius equation.
A sphere gains volume at a rate proportional to its surface area, with coefficient a > 0.Worked example
dV/dt = a(4πr²)
The given proportionality concerns volume rate.
V = (4/3)πr³, so dV/dt = 4πr²(dr/dt)
Use the geometric chain rule.
4πr²(dr/dt) = 4aπr²
Equate the expressions.
dr/dt = a for r > 0
The resulting radius rate is constant; a has units length/time.
11 · Convert proportional volume growth
A sphere satisfies dV/dt = kV for k > 0. Form the equation for its radius r > 0.
Hint
Substitute V = (4/3)πr³ and dV/dt = 4πr²(dr/dt).
Worked solution
4πr²(dr/dt) = (4/3)kπr³, so dr/dt = kr/3.
12 · Convert a cone model
Water in a cone has V = πh³/12 and volume rate dV/dt = 6π. Form the equation for h > 0.
Hint
dV/dh = πh²/4.
Worked solution
(πh²/4)(dh/dt) = 6π, so dh/dt = 24/h². Use consistent centimetre and second units.
08 / Check the constant’s units
The two sides of the equation must describe the same kind of rate.
If dy/dt = kyⁿ, then [k] = [y]^(1−n) / [time].
For dM/dt = −kM with M in grams and t in hours, k has units hour⁻¹. For dx/dt = k/x² with x in metres and t in seconds, k has units m³/s.
State the time unit: a coefficient per hour cannot be used unchanged when t is measured in minutes.
13 · Change the time unit
A model is dM/dt = −0.12M with t in hours. Write the corresponding equation when time τ is measured in minutes.
Hint
t = τ/60, so dt/dτ = 1/60.
Worked solution
dM/dτ = −0.002M. The coefficient is 0.002 min⁻¹.
09 / Separate the rate law from an initial condition
The initial value identifies a particular modelled history.
dP/dt = 0.04P and P(0) = 300Worked example
The differential equation supplies the rate law
It describes how the rate depends on the current population.
P(0) = 300 supplies a starting value
It does not say that the starting rate is 300.
The initial rate is 0.04·300 = 12 per day
Evaluate the equation at the starting state.
A full solution would describe P at every time in its valid range
Solving differential equations is developed with integration later.
14 · Read the initial condition
For dV/dt = 10 − 0.5V with V(0) = 6 L, find the initial rate.
Hint
Substitute V = 6 into the rate law.
Worked solution
The initial rate is 7 L/min. The initial volume is 6 L; these are different quantities.
10 / Check a proposed solution by differentiation
A function must satisfy both the equation and any initial condition.
Check P(t) = 300e^(0.04t) for dP/dt = 0.04P, P(0) = 300Worked example
P′(t) = 12e^(0.04t)
Differentiate the proposed function.
0.04P(t) = 12e^(0.04t)
The right-hand side matches.
P(0) = 300
The starting condition is also satisfied.
This verifies the proposal; it does not derive it from the equation
Keep verification and solution methods distinct.
15 · Test a proposed tank solution
Check V(t) = 20 − 14e^(−t/2) against dV/dt = 10 − 0.5V and V(0) = 6.
Hint
Differentiate the exponential and substitute into the right-hand side.
Worked solution
V′ = 7e^(−t/2), and 10 − 0.5V = 7e^(−t/2). Also V(0) = 20 − 14 = 6, so both conditions hold.
11 / State assumptions and physical restrictions
A model can be mathematically consistent and physically limited.
Proportional growth may stop fitting when space or resources become limited. A cooling model assumes a fixed ambient temperature and an approximately constant coefficient. A tank model must allow for its capacity and the conditions under which its outflow law is reasonable.
For equations such as dx/dt = k/x², the expression is undefined at x = 0. A derivation that cancels r² assumes r ≠ 0. Record those restrictions instead of silently extending the formula.
An equilibrium is a constant state at which the rate is zero. For the balance model, nearby states move towards the equilibrium; the rate-against-volume graph makes that direction visible.
16 · Find a modelling error
A cooling equation is written dT/dt = −kT in a room at 22°C, with T measured in °C. What essential term is missing?
Hint
The rate should be zero at room temperature.
Worked solution
The model should use the difference T − 22: dT/dt = −k(T − 22), with k > 0. The proposed equation incorrectly predicts a nonzero cooling rate at 22°C.
12 / Form the equation before trying to solve it
Name the variables, translate the rate, check the model.
- Write a derivative for the changing rate.
- Translate direct and inverse proportionality exactly.
- Use the correct sign for growth or loss.
- For cooling, use the difference from ambient temperature.
- For flows, subtract output from input.
- Use geometry and the chain rule to change the dependent variable.
- Determine constants with state-and-rate data, and check their units.
- Keep initial values separate from initial rates.
- Check domains, equilibria and the physical range.
Section 1 of 12 · Translate a rate statement into an equation