01 · Write a decay equation
An amount decreases at a rate proportional to itself, with positive constant λ. Write the differential equation.
Hint
Decrease requires a negative rate.
Worked solution
dN/dt = −λN, with λ > 0.
Understand · explore · practise
Differential equation models: growth and decay
Build and solve exponential growth and decay models. Determine rate constants from data, calculate doubling time and half-life, convert time units and explain model limitations.
Before you startSeparable differential equations, exponential functions and logarithms.
01 / Translate “rate proportional to amount”
The rate changes as the amount changes.
dN/dt = kN → N(t) = N₀eᵏᵗ
N₀ is the amount at t = 0. Positive k gives growth; negative k gives decay.
Proportionality means the ratio of instantaneous rate to current amount stays constant. A larger amount then has a larger rate magnitude. If N is an amount and t is measured in hours, k has units per hour.
The rate depends on the current amountExplore
At t = 0 hours, N = 100.
dN/dt = 20 amount units per hour.
Each hour multiplies N by 1.221403.
Doubling time = 3.465736 hours.
The constant k is a proportional instantaneous rate. It is not the same as the fractional change over one whole hour. This model assumes k stays constant and the amount remains meaningful over the period considered.
02 / Separate and apply the initial amount
The exponential family follows from a logarithmic integral.
Solve dN/dt = kN with N(0) = N₀ > 0.Worked example
dN/N = k dt
Work with the positive amount.
ln N = kt + C
Integrate.
N = Aeᵏᵗ
A is positive for this physical branch.
N(0) = A = N₀
Apply the initial amount.
N = N₀eᵏᵗ
If N₀ = 0 is allowed, check the separate equilibrium N = 0.
Watch: equal times give equal multipliers
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Find k from two measurements
Use a ratio to eliminate the unknown initial scale.
A culture model has N(0) = 1200 and N(3) = 1800. Time is in hours. Find N(t).Worked example
1800 = 1200e³ᵏ
Substitute the second measurement.
e³ᵏ = 3/2
Divide by 1200.
k = ln(3/2)/3
Take natural logarithms.
N(t) = 1200e^[t ln(3/2)/3]
Equivalently, 1200(3/2)^(t/3). The model predicts N(6) = 2700.
02 · A second growth data set
N(0) = 80 and N(4) = 160. Find k.
Hint
Use e⁴ᵏ = 2.
Worked solution
k = ln 2 / 4 per time unit.
03 · Measurements away from zero
If N(2) = 150 and N(5) = 300 in the same model, find k.
Hint
Divide N(5) by N(2).
Worked solution
e³ᵏ = 2, so k = ln 2 / 3.
04 / Find doubling time from a factor of two
The initial amount cancels.
Find the doubling time for k = ln(3/2)/3.Worked example
2N = Neᵏᴰ
Compare amounts D hours apart.
eᵏᴰ = 2 → D = ln 2 / k
This requires k > 0.
D = 3ln 2 / ln(3/2)
About 5.13 hours, not 6 hours merely because the initial measurement grew by half in three hours.
04 · Tripling time
If k = ln 2 / 4, find the tripling time.
Hint
Solve eᵏᵀ = 3.
Worked solution
T = 4ln 3 / ln 2.
05 · Independence from initial amount
Do two positive initial amounts have different doubling times when k is the same?
Hint
The amount cancels from the ratio.
Worked solution
No. Both have doubling time ln 2 / k.
05 / Use half-life for exponential decay
Choose a positive decay constant or a negative k consistently.
A quantity starts at 48 units and has half-life 6 hours.Worked example
N = 48e⁻λᵗ with λ > 0
Use the decay convention.
24 = 48e⁻⁶λ
At six hours the amount has halved.
λ = ln 2 / 6
Hence k = −ln 2 / 6.
N(t) = 48 × 2^(−t/6)
At t = 18 hours the prediction is 6 units.
06 · Quarter-life
How long does this model take to reach one quarter of its initial amount?
Hint
Two halvings give one quarter.
Worked solution
12 hours.
07 · Find a decay constant
An amount falls to 80% of its initial value in 5 days. Find k in dN/dt = kN.
Hint
e⁵ᵏ = 0.8.
Worked solution
k = ln(0.8)/5 per day, which is negative.
06 / Solve target amounts using logarithms
Check the requested time lies in the physical domain.
When does the 48-unit half-life model first fall below 5 units?Worked example
48 × 2^(−t/6) < 5
Use t ≥ 0.
−(t/6)ln 2 < ln(5/48)
Both logarithm arguments are positive.
t > 6ln(48/5)/ln 2
Reverse the inequality when dividing by the negative coefficient.
Equality gives the crossing time; below requires strictly later
Do not silently replace a strict threshold by equality.
08 · A growth threshold
For N(t) = 80 × 2^(t/4), when is N = 640?
Hint
The multiplier is 8.
Worked solution
t = 12 time units.
09 · Can decay reach zero?
Does N₀e⁻λᵗ with N₀ > 0 and λ > 0 reach zero at a finite time?
Hint
The exponential is positive for every finite input.
Worked solution
No. It approaches zero as t tends to infinity. Real measurements may have a detection threshold that the ideal model does not represent.
07 / Convert k when the time unit changes
The exponent must be dimensionless.
A model uses k = 0.12 per day. Rewrite it with time h in hours.Worked example
t in days = h/24
Convert the independent variable.
N = N₀e^(0.12h/24)
Substitute into the exponent.
k = 0.005 per hour
The same physical model uses a different numerical rate constant.
10 · Hours to minutes
Convert k = −0.3 per hour to a rate per minute.
Hint
One hour contains 60 minutes.
Worked solution
k = −0.005 per minute.
08 / Distinguish instantaneous rate from interval change
A unit-time multiplier is eᵏ, not 1 + k.
If k = 0.2 per hour, what is the one-hour percentage increase?Worked example
N(t + 1)/N(t) = e^0.2
Use the ratio of exponential values.
Fractional increase = e^0.2 − 1
Subtract the original amount fraction.
Percentage increase ≈ 22.14%
It is not exactly 20%; k describes the instantaneous proportional rate.
11 · Calibrate from a percentage change
A model increases by 10% over each hour. Find its continuous k.
Hint
Set eᵏ = 1.1.
Worked solution
k = ln 1.1 per hour.
12 · Equal intervals
Why are percentage changes over equal time intervals constant in this model?
Hint
Form N(t + Δt)/N(t).
Worked solution
The ratio is e^(kΔt), independent of the starting time t.
09 / Calculate an instantaneous rate from the current amount
Use kN at the requested time.
In the culture model, find the growth rate at t = 3 hours.Worked example
N(3) = 1800 and k = ln(3/2)/3
Use the calibrated values.
dN/dt = kN = 600ln(3/2)
This is about 243.28 amount units per hour.
The average increase over the first three hours was 200 per hour
An instantaneous rate differs from an interval average.
13 · A decay rate
For N′ = −0.1N, what is the instantaneous rate when N = 70?
Hint
Substitute the current amount.
Worked solution
−7 amount units per time unit.
10 / State the assumptions behind the exponential
A fitted model can fail outside the observed range.
Constant k assumes the same proportional mechanism persists. Growth may slow because of limited nutrients, space or other constraints; decay may involve several mechanisms. Data can be noisy, and a smooth real-valued model may only approximate a count of individual objects. Explain these limitations in the context of the question.
14 · A limiting resource
Why can unlimited exponential population growth become unrealistic?
Hint
Consider the environment sustaining the growth.
Worked solution
Resources and space may become limiting, so the proportional growth constant need not remain fixed.
15 · Extrapolation
Does matching two measurements prove the model will remain accurate much later?
Hint
A fit does not verify future assumptions.
Worked solution
No. Later conditions may change; predictions need contextual validation.
11 / Translate the rate, calibrate, then interpret
Keep signs, units and assumptions visible.
- Use N′ = kN for proportional growth or decay.
- Apply the initial value in N = N₀eᵏᵗ.
- Find k from ratios and logarithms.
- Use ln 2/k for growth doubling, or ln 2/λ for decay half-life.
- Convert the time unit and k together.
- Distinguish instantaneous rate from finite percentage change.
- Check physical time domains and model limitations.
16 · A final calibration
A quantity starts at 200 and falls to 50 after 8 hours. Give its half-life and model.
Hint
The amount has undergone two halvings.
Worked solution
Half-life 4 hours; N(t) = 200 × 2^(−t/4), for t ≥ 0 within the model’s valid period.
Section 1 of 11 · Translate “rate proportional to amount”