Library / Foundations: Maths / FMA-38
Foundations: Maths
Exponential growth and decay
A quantity changes at a rate proportional to its size.
- Ref
- FMA-38
- Rev
- v1
- Field
- Foundations: Maths
- Result
- N
N₀ is the starting amount, k the rate constant (positive for growth, negative for decay) and t the time, with k and t in matching units. The rate of change at any instant is kN, proportional to the current size. The doubling time is ln 2 / k and the half-life ln 2 / |k|.
A quantity changes at a rate proportional to its size. Positive k gives growth, negative k gives decay.
Calculator
- Doubling time (k > 0) or half-life (k < 0)13.86 s
- Change since the start64.87 %
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A constant percentage change per unit time always gives this curve: slow at first, then faster for growth, and steep at first then flat for decay.
- The doubling time ln 2 / k does not depend on the starting amount, so growth from 100 or from a million doubles in the same time.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing () | Local step | Change in result |
|---|---|---|---|---|---|---|
| k Rate constant | −1 to 1 1/s | 0.00454 | 2,203,000 | 2,203,000 | +0.04 1/s | +49.2 % |
| N0 Starting amount | 1 to 10,000 | 1.649 | 16,490 | 16,490 | +10% | +10 % |
| t Time | 0 to 100 s | 100 | 14,840 | 14,740 | +10% | +5.13 % |
Sweep of Starting amount
| Starting amount | Amount at time t |
|---|---|
| 1 | 1.649 |
| 3.162 | 5.214 |
| 10 | 16.49 |
| 31.62 | 52.14 |
| 100 | 164.9 |
| 316.2 | 521.4 |
| 1,000 | 1,649 |
| 3,162 | 5,214 |
| 10,000 | 16,490 |
Sweep one variable
Variables and how each one contributes
Amount at time t
What the equation gives you. Everything else feeds it.
Starting amount
The amount at t = 0. Directly proportional: doubling the start doubles the amount at every later time.
Rate constant
The rate constant per unit time (negative for decay). Acts through the exponent: with kt = 0.5 a 10 % rise in k raises N by about 5 %, and the effect grows as kt grows.
Time
The elapsed time. Also acts through the exponent: each extra second multiplies N by e^k, here about 5 % per second.
Worked example with the values above
- Exponent: kt = 0.05 × 10 = 0.5.
- Growth factor: e0.5 = 1.6487.
- N = 100 × 1.6487 = 164.9.
Assumptions and limits
- The rate is proportional to the current amount, with k constant.
- No limit on growth (no saturation) and no outside inputs.
Common mistakes
- Putting the rate as a percentage into k: 5 % per second is k = 0.05, not 5.
- Mixing the units of k and t, such as k per hour with t in minutes.
Statistics and uncertainty
Because k multiplies t in the exponent, a given relative error in k becomes an error in N of about kt × (relative error in k); fitting k from the logarithm of the data is the usual way to reduce the effect of scatter.
Where it is used
- Radioactive decay, cooling and discharge toward a limit, and charging curves.
- Early-stage population or cost growth.
Maths you need
- Exponential function: e^x and its inverse ln
- Rates of change: rate proportional to size