Exponential Growth and Decay
Explore the differential equation dy/dt = ky, learn about half-life and doubling time, and model real-world phenomena involving exponential change.
The Exponential Growth/Decay Model
Many real-world quantities change at a rate proportional to their current value. This is modelled by the differential equation:
dN/dt = kN
The solution is: N(t) = N0ekt
- N0 = initial quantity (when t = 0)
- k > 0 = exponential growth
- k < 0 = exponential decay
Visual: Growth vs Decay
Growth (k > 0)
Curve rises steeply
Decay (k < 0)
Curve falls toward zero
Half-Life and Doubling Time
Two important time measures arise from the exponential model:
Half-Life (Decay)
The time taken for a quantity to reduce to half its initial value.
t1/2 = ln 2/|k|
Doubling Time (Growth)
The time taken for a quantity to double its initial value.
td = ln 2/k
Note: Both formulas use ln 2 (approximately 0.693). These are derived by setting N(t) = N0/2 or N(t) = 2N0 in the equation N(t) = N0ekt and solving for t.
Real-World Applications
The exponential model appears in many contexts:
Population Growth
Bacteria doubling at a constant rate; unrestricted population models.
Radioactive Decay
Unstable atoms decaying with a constant half-life (e.g., Carbon-14 dating).
Cooling/Heating
Newton's Law of Cooling: temperature difference decays exponentially.
Compound Interest
Continuous compounding follows A = A0ert.
Key Vocabulary
Growth Constant (k)
The proportionality constant in dN/dt = kN. Positive for growth, negative for decay.
Half-Life
The time for a decaying quantity to reduce to half its current value: t = ln 2 / |k|.
Doubling Time
The time for a growing quantity to double: t = ln 2 / k.
Initial Value (N0)
The value of the quantity at time t = 0. It determines the vertical scale of the solution curve.
Worked Examples
A population of 500 bacteria doubles every 3 hours. Find the population after 9 hours.
Step 1: N0 = 500, doubling time = 3 h. Find k: k = ln 2/3 = 0.2310
Step 2: Use N(t) = 500e0.2310t
Step 3: N(9) = 500e0.2310 x 9 = 500e2.079 = 500 x 8 = 4000
Answer: After 9 hours there are 4000 bacteria. (This makes sense: 3 doubling periods means 500 x 2^3 = 4000.)
A radioactive substance has a half-life of 10 years. If the initial mass is 80 g, find the mass after 30 years.
Step 1: Half-life = 10, so |k| = ln 2/10 = 0.0693, thus k = -0.0693
Step 2: N(t) = 80e-0.0693t
Step 3: N(30) = 80e-0.0693 x 30 = 80e-2.079 = 80 x 0.125 = 10
Answer: After 30 years, 10 g remains. (3 half-lives: 80 -> 40 -> 20 -> 10.)
The mass of a substance satisfies dM/dt = -0.05M. If M(0) = 200 g, find the half-life.
Step 1: Here k = -0.05, so |k| = 0.05
Step 2: Half-life = ln 2 / |k| = 0.693 / 0.05 = 13.86
Answer: The half-life is approximately 13.86 time units.
Knowledge Check
Select the correct answer for each question. Click "Check Answer" to see if you are right.
Question 1
The solution to dN/dt = kN with N(0) = N0 is:
Question 2
A quantity has a half-life of 5 years. What is the decay constant k (to 4 d.p.)?
Question 3
A colony of 1000 bacteria grows with k = 0.2 per hour. How many bacteria are there after 5 hours (nearest whole number)?
Question 4
If a population doubles every 4 years, what is the doubling time?
Question 5
In the equation N = N0ekt, if k < 0, what happens to N as t increases?
Key Concepts Summary
- ● The equation dN/dt = kN models quantities that change at a rate proportional to their current value.
- ● Its solution is N = N0ekt where k > 0 gives growth and k < 0 gives decay.
- ● Half-life = ln 2 / |k| and doubling time = ln 2 / k.
- ● Applications include population growth, radioactive decay, cooling, and compound interest.
- ● In decay, N never reaches zero but gets arbitrarily close (asymptotic to the t-axis).