Open Math›Year 11›Mathematics Advanced›Working with functions›Variation and modelling

Variation and modelling

Identify direct and inverse variation from equations and tables, determine the constant of variation from data, and construct and evaluate variation models from two data points.

Worked examples

Identifying the type of variation

Straightforward

Problem

State whether each relationship represents direct variation, inverse variation, or neither, and give the general form in each case.

(a)

y = 4x^2

(b)

y = \dfrac{9}{x^3}

(c)

y = 2x + 5

Check (a): compare

y = 4x^2

with the direct variation form

y = kx^n

y = 4x^2 \text{ matches } y = kx^n \text{ with } k = 4,\ n = 2.

\text{This is direct variation ($y$ varies directly as $x^2$).}

Check (b): compare

y = \dfrac{9}{x^3}

with the inverse variation form

y = \dfrac{k}{x^n}

y = \frac{9}{x^3} \text{ matches } y = \frac{k}{x^n} \text{ with } k = 9,\ n = 3.

\text{This is inverse variation ($y$ varies inversely as $x^3$).}

Check (c):

y = 2x + 5

has a non-zero constant term added. Compare with both forms.

y = 2x + 5 \text{ is neither } y = kx^n \text{ nor } y = \frac{k}{x^n}.

\text{The constant term }+5\text{ means it is neither direct nor inverse variation.}

Answer

(a) Direct variation,

y = 4x^2

. (b) Inverse variation,

y = 9/x^3

. (c) Neither.

Finding the constant of variation and using the model

Moderate

Problem

y

varies directly as

x^2

. When

x = 4

y = 80

. Find the constant of variation

k

and use the model to find

y

when

x = 7

Write the general direct variation equation for

y

varying as

x^2

y = kx^2

Substitute the known values

x = 4

and

y = 80

to find

k

80 = k(4)^2 = 16k \implies k = \frac{80}{16} = 5

Write the specific model and substitute

x = 7

y = 5x^2 \implies y = 5(7)^2 = 5 \times 49 = 245

Answer

k = 5

; when

x = 7

y = 245

Constructing a variation model from two data points

Challenging

Problem

It is known that

y

varies inversely as

x

. Two measurements give

y = 20

when

x = 3

, and

y = 15

when

x = 4

. Verify that the model is consistent with both data points, then predict

y

when

x = 12

and comment on the reliability of the prediction.

Write the inverse variation model and find

k

from the first data point.

y = \frac{k}{x} \implies k = xy = 3 \times 20 = 60

Verify

k

using the second data point.

k = xy = 4 \times 15 = 60 \checkmark

\text{Both data points give } k = 60,\text{ so the model is consistent.}

Write the specific model and substitute

x = 12

y = \frac{60}{x} \implies y = \frac{60}{12} = 5

Comment on the reliability of the prediction.

\text{The model is internally consistent (both data points agree).}

\text{However, } x = 12 \text{ is well beyond the observed range of 3 to 4,}

\text{so this is extrapolation. The prediction is reliable only if the}

\text{inverse variation relationship continues to hold for larger values of } x.

Answer

k = 60

; the model is

y = 60/x

; when

x = 12

y = 5

. The prediction involves extrapolation beyond the observed data range and should be verified with additional measurements.

Practise

Q1·Straightforward

Which type of variation does the equation

y = 5x^3

represent?

Direct variationInverse variationNeither direct nor inverse variationLinear variation

Explanation

The equation

y = 5x^3

has the form

y = kx^n

with

k = 5

and

n = 3

. This is direct variation —

y

varies directly as the cube of

x

Q2·Straightforward

Which type of variation does the equation

y = \dfrac{8}{x^2}

represent?

Direct variationInverse variationNeither direct nor inverse variationExponential variation

Explanation

The equation

y = \dfrac{8}{x^2}

has the form

y = \dfrac{k}{x^n}

with

k = 8

and

n = 2

. This is inverse variation — as

x

increases,

y

decreases.

Q3·Straightforward

The table below shows values of

x

and

y

.

|

x

| 1 | 2 | 3 | 4 |
|---|---|---|---|---|
|

y

| 5 | 10 | 15 | 20 |

Which type of variation does the table represent?

Direct variationInverse variationNeither direct nor inverse variationQuadratic variation

Explanation

Dividing each

y

-value by the corresponding

x

-value gives

5/1 = 5

10/2 = 5

15/3 = 5

20/4 = 5

. The ratio

y/x = 5

is constant, so this is direct variation with

k = 5

, that is

y = 5x

Q4·Straightforward

The table below shows values of

x

and

y

.

|

x

| 2 | 4 | 5 | 10 |
|---|---|---|---|---|
|

y

| 15 | 7.5 | 6 | 3 |

Which type of variation does the table represent?

Direct variationInverse variationNeither direct nor inverse variationLinear variation

Explanation

Multiplying each pair:

2 \times 15 = 30

4 \times 7.5 = 30

5 \times 6 = 30

10 \times 3 = 30

. The product

xy = 30

is constant, so this is inverse variation with

k = 30

, that is

y = \dfrac{30}{x}

Q5·Moderate

y

varies directly as

x

. When

x = 6

y = 24

. Find the constant of variation

k

Explanation

Since

y

varies directly as

x

, write

y = kx

. Substituting

x = 6

and

y = 24

24 = 6k

, so

k = 4

Q6·Moderate

y

varies directly as

x^2

. When

x = 3

y = 45

. Find the constant of variation

k

Explanation

Since

y

varies directly as

x^2

, write

y = kx^2

. Substituting

x = 3

and

y = 45

45 = k(3)^2 = 9k

, so

k = 5

Q7·Moderate

y

varies inversely as

x

. When

x = 4

y = 9

. Find the value of

y

when

x = 6

Explanation

Since

y

varies inversely as

x

, write

y = \dfrac{k}{x}

. From

x = 4

y = 9

k = 4 \times 9 = 36

. The model is

y = \dfrac{36}{x}

. When

x = 6

y = \dfrac{36}{6} = 6

Q8·Moderate

y

varies directly as

x^2

with constant of variation

k = 3

. Find the value of

y

when

x = 5

Explanation

The model is

y = 3x^2

. When

x = 5

y = 3 \times 5^2 = 3 \times 25 = 75

Q9·Challenging

It is known that

y = kx^2

. Two measurements give

y = 12

when

x = 2

, and

y = 27

when

x = 3

. Verify that both data points give the same value of

k

, then find

y

when

x = 5

Explanation

From

(2, 12)

k = \dfrac{12}{2^2} = \dfrac{12}{4} = 3

. From

(3, 27)

k = \dfrac{27}{3^2} = \dfrac{27}{9} = 3

. Both give

k = 3

, confirming the model

y = 3x^2

is consistent. When

x = 5

y = 3 \times 5^2 = 3 \times 25 = 75

Q10·Challenging

It is known that

y = \dfrac{k}{x}

. Two measurements give

y = 6

when

x = 4

, and

y = 4

when

x = 6

. Verify that both data points give the same value of

k

, then find

y

when

x = 12

Explanation

From

(4, 6)

k = 4 \times 6 = 24

. From

(6, 4)

k = 6 \times 4 = 24

. Both give

k = 24

, so the model

y = \dfrac{24}{x}

is consistent. When

x = 12

y = \dfrac{24}{12} = 2

Q11·Challenging

The time

T

hours to complete a task varies inversely with the number of workers

n

. When

n = 4

T = 6

hours, and when

n = 6

T = 4

hours. Verify that both data points give the same value of

k

, then find

T

when

n = 8

Explanation

Write

T = \dfrac{k}{n}

. From

(n = 4, T = 6)

k = 4 \times 6 = 24

. From

(n = 6, T = 4)

k = 6 \times 4 = 24

. Both give

k = 24

, so the model

T = \dfrac{24}{n}

is consistent with both data points. When

n = 8

T = \dfrac{24}{8} = 3

hours. The model's prediction is reliable here since the relationship holds across both observed data points.

Q12·Challenging

The braking distance

d

metres of a car varies directly as the square of its speed

v

km/h. When

v = 20

d = 8

, and when

v = 30

d = 18

. Verify that both data points give the same value of

k

, then find

d

when

v = 50

Explanation

Write

d = kv^2

. From

(v = 20, d = 8)

k = \dfrac{8}{20^2} = \dfrac{8}{400} = 0.02

. From

(v = 30, d = 18)

k = \dfrac{18}{30^2} = \dfrac{18}{900} = 0.02

. Both give

k = 0.02

, confirming the model

d = 0.02v^2

. When

v = 50

d = 0.02 \times 50^2 = 0.02 \times 2500 = 50

metres. Because both data points are consistent, the model is reliable for interpolation, though extrapolation to higher speeds should be treated with caution.