Exponential Growth and Decay Revision Notes for VCE SSCE Mathematical Methods

Exponential Growth and Decay

Understanding exponential change

When the rate at which a quantity increases or decreases is proportional to its current value, the quantity follows the law of exponential change. This is one of the most important mathematical models in science and finance.

The fundamental relationship can be expressed as:

A = A_0e^{kt}

where:

$A$ is the quantity at time $t$
$A_0$ is the initial quantity (at time $t = 0$ )
$k$ is the rate constant
$t$ is time

infoNote

The exponential function $e^{kt}$ describes how quantities change over time when the rate of change depends on the current amount. This simple yet powerful relationship appears throughout nature, from bacterial growth to radioactive decay.

Growth versus decay

The sign of the rate constant $k$ determines whether we have growth or decay:

When $k > 0$ , the model shows exponential growth:

Growth of cells and bacteria
Population growth
Continuously compounded interest

When $k < 0$ , the model shows exponential decay:

Radioactive decay
Cooling of materials

Alternative form of the model

We can also write the exponential model in the form:

A = A_0b^t

where $b = e^k$

In this alternative form:

Growth occurs when $b > 1$
Decay occurs when $b < 1$

infoNote

This form can sometimes be more intuitive, especially when dealing with percentage changes. For example, a 5% annual increase corresponds to $b = 1.05$ .

Cell growth and generation time

Bacterial cells often reproduce by dividing into two new cells at regular intervals. This type of growth can be modelled using a specific exponential formula.

When a particular type of bacteria cell divides into two new cells every $T_D$ minutes, the number of cells after $t$ minutes is:

N = N_0 2^{\frac{t}{T_D}}

where:

$N_0$ is the initial number of cells
$N$ is the number of cells after time $t$
$T_D$ is called the generation time (the time it takes for the population to double)

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Worked Example: Finding Generation Time

Question: What is the generation time of a bacterial population that increases from 5000 cells to 100,000 cells in four hours of growth?

Solution:

We know that $N_0 = 5000$ and $N = 100,000$ when $t = 240$ minutes (four hours).

Using the formula $N = N_0 2^{\frac{t}{T_D}}$ :

100,000 = 5000 \times 2^{\frac{240}{T_D}}

Dividing both sides by 5000:

20 = 2^{\frac{240}{T_D}}

To solve for $T_D$ , we take logarithms of both sides. Using base-2 logarithms:

\log_2 20 = \frac{240}{T_D}

Therefore:

T_D = \frac{240}{\log_2 20} \approx 55.53 \text{ minutes}

The generation time is approximately 55.53 minutes.

Radioactive decay and half-life

Radioactive materials decay in a predictable exponential manner. The amount of radioactive material $A$ present at time $t$ (measured in years) follows:

A = A_0e^{-kt}

where:

$A_0$ is the initial amount of material
$k$ is a positive constant (note: the negative sign in the exponent makes this a decay model)
$k$ depends on the type of radioactive material

chatImportant

Notice that $k$ itself is positive, but the negative sign in the exponent $-kt$ makes this a decay model. This is why we write $A = A_0e^{-kt}$ rather than $A = A_0e^{kt}$ with a negative $k$ value.

Understanding half-life

A radioactive substance is commonly described by its half-life, which is the time required for half the material to decay. The half-life is a characteristic property of each radioactive element.

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Worked Example: Calculating Half-Life

Question: After 1000 years, a sample of radium-226 has decayed to 64.7% of its original mass. Find the half-life of radium-226.

Solution:

We use the formula $A = A_0e^{-kt}$ .

When $t = 1000$ , we have $A = 0.647A_0$ . Therefore:

0.647A_0 = A_0e^{-1000k}

Dividing both sides by $A_0$ :

0.647 = e^{-1000k}

Taking the natural logarithm of both sides:

-1000k = \log_e 0.647

k = \frac{-\log_e 0.647}{1000} \approx 0.000435

Now to find the half-life, we need to find when $A = \frac{1}{2}A_0$ :

A_0e^{-kt} = \frac{1}{2}A_0

e^{-kt} = \frac{1}{2}

-kt = \log_e\left(\frac{1}{2}\right)

t = -\frac{\log_e\left(\frac{1}{2}\right)}{k} \approx 1591.95

The half-life of radium-226 is approximately 1592 years.

Population growth modelling

In many situations, population growth can be modelled using exponential functions. This assumes that the growth rate remains constant over the period considered.

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Worked Example: Town Population Growth

Question: The population of a town was 8000 at the beginning of 2007 and 15,000 at the end of 2014. Assume that the growth is exponential.

a) Find the population at the end of 2016.

b) In what year will the population be double that of 2014?

Solution:

Let $P$ be the population at time $t$ years, measured from 1 January 2007. Then:

P = 8000e^{kt}

At the end of 2014, $t = 8$ and $P = 15,000$ . Therefore:

15,000 = 8000e^{8k}

\frac{15}{8} = e^{8k}

k = \frac{1}{8}\log_e\left(\frac{15}{8}\right) \approx 0.079

The rate of increase is 7.9% per annum.

a) When $t = 10$ :

P = 8000e^{10k} \approx 17,552.6049 \approx 17,550

The population is approximately 17,550.

b) When does $P = 30,000$ ? We solve:

30,000 = 8000e^{kt}

\frac{30,000}{8000} = e^{kt}

\frac{15}{4} = e^{kt}

t = \frac{1}{k}\log_e\left(\frac{15}{4}\right) \approx 16.82

The population reaches 30,000 approximately 16.82 years after the beginning of 2007, which means during the year 2023.

chatImportant

When continuing calculations, it's important to store the exact value of $k$ in your calculator rather than using the rounded approximation 0.079. This ensures accuracy in subsequent calculations.

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Worked Example: Changing Animal Populations

Question: There are approximately ten times as many red kangaroos as grey kangaroos in a certain area. If the population of grey kangaroos increases at a rate of 11% per annum while that of the red kangaroos decreases at 5% per annum, find how many years must elapse before the proportions are reversed, assuming the same rates continue to apply.

Solution:

Let $G_0$ be the initial population of grey kangaroos.

Then the number of grey kangaroos after $n$ years is $G = G_0(1.11)^n$ (11% increase means multiply by 1.11 each year).

The number of red kangaroos after $n$ years is $R = 10G_0(0.95)^n$ (5% decrease means multiply by 0.95 each year).

When the proportions are reversed, the grey kangaroo population will be ten times the red kangaroo population:

G = 10R

G_0(1.11)^n = 10 \times 10G_0(0.95)^n

(1.11)^n = 100(0.95)^n

Taking natural logarithms of both sides:

\log_e((1.11)^n) = \log_e(100(0.95)^n)

n\log_e 1.11 = \log_e 100 + n\log_e 0.95

Rearranging to isolate $n$ :

n\log_e 1.11 - n\log_e 0.95 = \log_e 100

n = \frac{\log_e 100}{\log_e 1.11 - \log_e 0.95} \approx 29.6

The proportions of the kangaroo populations will be reversed after 30 years.

bookmarkSummary

Key Points to Remember:

The law of exponential change is $A = A_0e^{kt}$ , where $k > 0$ indicates growth and $k < 0$ indicates decay.
For cell growth, the generation time $T_D$ represents the time taken for the population to double, using the formula $N = N_0 2^{\frac{t}{T_D}}$ .
Half-life is the time required for half of a radioactive substance to decay. To find half-life, solve $A_0e^{-kt} = \frac{1}{2}A_0$ .
When solving exponential problems, always store exact values in your calculator rather than using rounded approximations for intermediate calculations.
Population models can use either base $e$ (as in $P = P_0e^{kt}$ ) or percentage multipliers (as in $P = P_0(1.11)^t$ for 11% growth).

Exponential Growth and Decay (VCE SSCE Mathematical Methods): Revision Notes