6.6 Exponential growth and decay in modelling — Learn

Prerequisites: the model $y = Ae^{kt}$ from Section 6.2; solving an exponential equation from Section 6.4; substituting into a formula.

By the end of this section you can: build an exponential model from a written description; use it to predict a value; find a doubling time or a half-life; find the time for a quantity to reach a given level; and state the limitations of an exponential model.

1. The growth and decay model

A quantity that changes at a rate proportional to its current size follows

y = Ae^{kt},

where $A = y(0)$ is the starting value and $k$ is the rate constant. The context decides the meaning of each symbol:

Context	$A$	$k$	Behaviour
Money, continuous interest	initial amount	interest rate per year	grows
Radioactive decay	initial mass	$-\,$ decay constant	decays
Drug concentration	initial dose	$-\,$ elimination rate	decays
Population growth	initial population	growth rate per year	grows

A positive $k$ describes growth, and a negative $k$ describes decay. The same mathematics covers every row of the table; only the story changes.

Example 1. £ $2000$ is invested with continuous compounding at $3\%$ per year, so $A = 2000e^{0.03t}$ . Find the amount after $5$ years.

Substitute $t = 5$ :

A = 2000e^{0.15} = 2323.67\ldots,

so the investment is worth £ $2323.67$ to the nearest penny.

Example 2. A radioactive isotope has a half-life of $20$ years. Find the decay constant $k$ in $m = m_0e^{-kt}$ .

Half-life means $m = \dfrac{m_0}{2}$ when $t = 20$ . Divide by $m_0$ and substitute:

e^{-20k} = \frac{1}{2}.

Take logarithms: $-20k = -\ln 2$ , so

k = \frac{\ln 2}{20} \approx 0.0347.

Example 3. A drug is modelled by $A = 80e^{-0.4t}$ mg, where $t$ is in hours. Find the time for the amount to fall to $10$ mg.

Divide by $80$ :

e^{-0.4t} = \frac{10}{80} = 0.125.

Take logarithms: $-0.4t = \ln 0.125$ , so

t = \frac{\ln 0.125}{-0.4} = 5.198\ldots \approx 5.2\text{ hours}.

Example 4. A population grows as $P = 500e^{0.15t}$ . Find its doubling time.

Doubling means $P = 2 \times 500$ , so $e^{0.15t} = 2$ . Take logarithms:

t = \frac{\ln 2}{0.15} \approx 4.62.

The population doubles about every $4.6$ years. The starting value $500$ cancels, so the doubling time does not depend on it.

2. Limitations and refinements

A model is a simplification, and an exponential model has limits.

Growth cannot continue forever. A population with unlimited food would follow $P = Ae^{kt}$ , but in reality food, space and disease slow the growth. A model can be refined by adding a limit, or by using it only over the range where the data were collected.
Decay approaches zero but never reaches it. Radioactive mass and drug concentration get closer and closer to zero. The model predicts a very small positive amount at any time, not an exact zero.
The constants must come from data. The value of $k$ is estimated from measurements, and a different sample gives a different estimate. The estimate is only as reliable as the data.
The model is only valid over the observed range. Predicting far beyond the data, called extrapolation, is risky. A model fitted over ten years may not describe the next fifty.
Units and time must match. If $k$ is per year, then $t$ must be in years. Mixing units changes the prediction.

Progress check

Return to the mould culture in the chapter problem. The fitted model was $N = 200\left(1.5\right)^t$ , and it predicted $N = 2000$ at about $5.7$ hours.

Write the model in the form $N = 200e^{kt}$ , and find $k$ .
Explain why the number $1.5$ becomes $k = \ln 1.5$ .
Give one reason why the exponential model may fail after a long time.
Suggest one refinement that would make the prediction more realistic.

Watch out. These mistakes are common.

Using a percentage as a decimal in the wrong place. A growth rate of $3\%$ per year is $k = 0.03$ , not $k = 3$ and not $k = 0.3$ . A rate of $3$ would multiply the amount by $e^{3} \approx 20$ each year.

Forgetting that a half-life gives a negative $k$ . Decay means the exponent is negative, so write $e^{-kt}$ with $k > 0$ . If you use $e^{kt}$ , the value of $k$ must itself be negative.

Doubling $A$ instead of setting $y = 2A$ . For a doubling time, substitute $2A$ for $y$ and let $A$ cancel. Do not change the value of $A$ itself.

Treating the model as exact. The model predicts, it does not guarantee. A value from the model is an estimate, and it should be reported with sensible rounding.

Re-check: substitute your value of $k$ back into the model and confirm one known data point. For the mould, $N(1) = 200e^{\ln 1.5} = 200 \times 1.5 = 300$ , which matches the table.

You should now be able to

build a model $y = Ae^{kt}$ from a written description;
decide the sign of $k$ from the context;
find a value by substitution;
find a half-life or a doubling time by solving an exponential equation;
find the time for a quantity to reach a stated level;
state the limitations of an exponential model and suggest a refinement.