Compound growth, depreciation and half-life | Year 11

01 / LEARN

Understand the idea.

Convert the rate and count the periods

Compounding applies interest to the current balance each period. For a nominal annual decimal rate $r$ compounded $n$ times yearly, divide the rate by $n$ and count $n t$ periods in $t$ years. Match the rate period and exponent before calculating.

The model assumes the rate is fixed and no fees, payments or withdrawals alter the balance. Its effective annual rate is ${(1 + \frac{r}{n})}^{n} - 1$ , which includes the effect of interest earned on earlier interest.

V = P {(1 + \frac{r}{n})}^{n t}

Half-life and depreciation are multiplier models

A half-life counts periods differently: after every $h$ time units, the amount is multiplied by one half. The exponent $\frac{t}{h}$ counts half-lives and may be fractional when the model treats change continuously.

Percentage depreciation also multiplies the remaining amount, rather than subtracting a fixed sum. For threshold questions, distinguish a continuous crossing time from the first completed month or year at which the required condition holds.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Quarterly compounding

8000 dollars is invested at 6% per annum, compounded quarterly. Find its value after 2 years, ignoring fees.

Worked solution

1
Convert the annual rate to a quarterly rate.
$r_{q} = \frac{0.06}{4} = 0.015$
A nominal annual rate compounded quarterly is divided across four equal periods.
2
Count the compounding periods.
$n = 4 (2) = 8$
Two years contain eight quarters.
3
Apply the repeated multiplier.
$V = 8000 {(1 + 0.015)}^{8} = 8000 {(1.015)}^{8}$
Each quarter keeps the balance and adds 1.5 percent of that current balance.
4
Round the final balance to cents.
$V \approx 9011.94 dollars$
Currency is reported to two decimal places after evaluating the complete power.

Answer

9011.94 dollars.

EXAMPLE 02

A fractional number of half-lives

A 240 mg sample has an 8-hour half-life. Find its mass after 20 hours.

Worked solution

1
Write the half-life model.
$M (t) = 240 {(\frac{1}{2})}^{t / 8}$
The initial mass is 240 mg, and each eight-hour period halves it.
2
Count the half-lives elapsed.
$\frac{20}{8} = 2.5$
Twenty hours is two complete half-lives plus another half of one.
3
Evaluate the remaining mass.
$M (20) = 240 {(\frac{1}{2})}^{2.5} \approx 42.43 mg$
The continuous exponential model permits a fractional number of half-lives.

Answer

Approximately 42.43 mg.

03 / YOUR TURN

Try it yourself.

Check your understanding

What fraction remains after three half-lives?

Show answer & reasoning

$1 / 8$

Each half-life contributes a factor of $\frac{1}{2}$ . After three half-lives the remaining fraction is ${(\frac{1}{2})}^{3} = \frac{1}{8}$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Find 5000 dollars compounded annually at 4% for 3 years.
2 marks
A 20,000 dollar asset loses 15% yearly. Find its value after 3 years.
2 marks
Write a model for a 96 g sample with a 5-day half-life.
2 marks
For the previous sample, find the mass after 15 days.
2 marks
An account earns 1% each month. Find its effective annual percentage rate.
3 marks
A saver adds 100 dollars at the end of every month. Explain why $P b^{number of months}$ does not describe the whole account balance.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find 5000 dollars compounded annually at 4% for 3 years.
2 marks
$5000 {(1.04)}^{3} = 5624.32$ .
5624.32 dollars.
A 20,000 dollar asset loses 15% yearly. Find its value after 3 years.
2 marks
$20000 {(0.85)}^{3} = 12282.50$ .
12,282.50 dollars.
Write a model for a 96 g sample with a 5-day half-life.
2 marks
The exponent is the number of five-day half-life periods.
$M (t) = 96 {(\frac{1}{2})}^{\frac{t}{5}}$ , $t$ in days.
For the previous sample, find the mass after 15 days.
2 marks
$96 {(\frac{1}{2})}^{3} = 12$ .
12 g.
An account earns 1% each month. Find its effective annual percentage rate.
3 marks
One year multiplies by ${1.01}^{12}$ .
The effective rate is $100 ({1.01}^{12} - 1) \approx 12.68 %$ .
Approximately 12.68%.
A saver adds 100 dollars at the end of every month. Explain why $P b^{number of months}$ does not describe the whole account balance.
3 marks
Each deposit earns interest for a different number of periods. The balance requires extra terms or a recurrence, not one initial lump-sum exponential.
It omits the later deposits.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: exponential and logarithmic functions, growth models and numerical/algebraic prerequisites.

Official curriculum source ↗

In this lesson

Understand the idea.

Convert the rate and count the periods

Half-life and depreciation are multiplier models

See the method in action.

Quarterly compounding

Convert the annual rate to a quarterly rate.

Count the compounding periods.

Apply the repeated multiplier.

Round the final balance to cents.

A fractional number of half-lives

Write the half-life model.

Count the half-lives elapsed.

Evaluate the remaining mass.

Try it yourself.

Check your understanding

Keep practising