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4 Nonlinear relationships
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Level 10

4.02 Visualising compound interest

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Introduction

When calculating compound interest using  repeated applications of simple interest,  we noticed that the total amount of an investment that uses compound interest increases at an increasing rate. While we can see this in how the interest amount increases year by year, it becomes a lot more obvious once we plot the investment's growth on a graph.

Compound and simple interest curves

When plotting compound interest, we want to see how the investment total changes over time. This means that our independent variable for the x-axis will be the number of periods that have passed and the dependent variable for the y-axis will be the investment total.

If we calculate what the investment total will be after some number of periods, we have both the x and y-values for a point that can be plotted.

Once we have plotted a few points, we can start to see how the curve joining the points for compound interest growth is non-linear. As expected, the curve will be increasing at an increasing rate.

If we compare the curve for a compound interest investment to a simple interest investment we can clearly see how compound interest results in a non-linear curve.

Examples

Example 1

Roxanne used repeated applications of simple interest to calculate how much an investment of \$200 would grow over 3 years if it earned compound interest at a rate of 22 \% p.a., compounding annually.

\text{No. of years}0123
\text{Investment value (\$)}200244297.68363.17

Which of the following graphs correctly shows the relationship between the number of years passed and the value of the investment?

A
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\text{No. of years}
100
200
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400
\text{Value} (\$)
B
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\text{No. of years}
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400
\text{Value}(\$)
C
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\text{No. of years}
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\text{Value} (\$)
D
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\text{No. of years}
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\text{Value}(\$)
Worked Solution
Create a strategy

Choose the graph that has the points from the table plotted.

Apply the idea

Looking at the points on each graph, the one that matches with the table is Option C. So Option C is the correct answer.

Example 2

The graph below shows both an investment with simple interest, and one with compound interest, labelled Investment A and Investment B respectively.

2
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\text{Time (years)}
2
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\text{Investment } \left(\$1000\right)
a

Which investment has a higher principal amount?

Worked Solution
Create a strategy

Choose the graph with the higher y-intercept.

Apply the idea

The graph for Investment A has a higher y-intercept than the graph for Investment B. So Investment A has a higher principal amount.

b

Which investment has a higher final amount after 10 years?

Worked Solution
Create a strategy

Find the point on each graph that lines up with x=10.

Apply the idea
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\text{Time (years)}
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\text{Investment } \left(\$1000\right)

From the coordinate plane we can see that when x=10 the value of Investment B is higher that the value of a Investment A.

So Investment B has a higher final amount after 10 years.

c

After how many years are the investments equal in value?

Worked Solution
Create a strategy

Find the intersection point of the graphs.

Apply the idea
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\text{Time (years)}
2
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\text{Investment } \left(\$1000\right)

From the the coordinate plane we can see that the two graphs intersect at a point where x=9.

So this means that after 9 years the investments are equal in value.

Idea summary
2
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\text{Time (years)}
2
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\text{Investment } \left(\$1000\right)

Compound interest curves are non-linear because they increase at an increasing rate.

Simple interest graphs are linear because they increase at a constant rate.

Aspects of compound interest curves

When plotting a curve that represents compound interest growth, the different aspects of the compound interest scenario will be reflected in the different aspects of the graph.

A compound interest scenario has two key factors: the principal amount and the interest rate.

Since the principal amount is the amount invested at the start, it will always be the investment total after zero periods. In other words, the y-value of the y-intercept will always be the principal amount on a compound interest curve.

The interest rate will determine how steeply the curve increases. The greater the interest rate, the faster the investment will increase with respect to time. We can check whether an interest rate matches a graph by comparing two points that are one period apart. The increase from one y-value to the other should match the interest rate.

Examples

Example 3

Luke invests \$50 into an account which accumulates interest at a rate of 8 \, \% p.a., compounding annually.

a

Which of the following graphs correctly represents the relationship between the number of years passed and Luke's account balance?

A
5
10
15
20
\text{No. of years}
50
100
150
200
250
300
350
\text{Value (\$)}
B
5
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\text{No. of years}
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\text{Value (\$)}
C
5
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\text{No. of years}
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\text{Value (\$)}
D
5
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\text{No. of years}
50
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\text{Value (\$)}
Worked Solution
Create a strategy

Check the y-intercept and another point on the graph.

Apply the idea

Since it is compound interest, the graph should be non-linear. This means Option D cannot be correct, since it is linear.

The principal is \$50 so the y-intercept of the graph should be 50, which is true for all the graphs.

We can find the value of the investment after 5 years and see which option matches it. So we need to increase \$50 by 8\% five times, which means we need to multiply by 1.08 five times.

\displaystyle \text{Amount of investment}\displaystyle =\displaystyle \$50 \times 1.08 \times 1.08 \times 1.08 \times 1.08 \times 1.08 Increase by 8\% \, \, 5 times
\displaystyle =\displaystyle \$73.47Evaluate

So for 5 on the horizontal axis we want the curve to be at \$73.47 which is approximately half way between 50 and 100.

So the correct answer is Option B.

b

How many years will it take for the account to reach a total value of \$200? Give your answer to the nearest whole number.

Worked Solution
Create a strategy

Use the graph from part (a).

Apply the idea

Using the correct graph from part (a), we can see that we reach a value of \$200 at 18 years.

5
10
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\text{No. of years}
50
100
150
200
250
300
350
\text{Value (\$)}

So the amount invested will be \$200 after 18 years.

Idea summary

The y-value of the y-intercept will always be the principal amount on a compound interest curve.

The greater the interest rate, the faster the curve will increase with respect to time.

Outcomes

VCMNA328

Connect the compound interest formula to repeated applications of simple interest using appropriate digital technologies

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