Rates of Change Rates are an important area of Maths and can be found in everyday life, business and science.

Rates of Change

Rates are an important area of Maths and can be found in everyday life, business and science.

Common rates

Rate Units

Pay rate $ per hour ($/h)

Heart rate Beats per minute (beat/min)

Interest rate Percentage per annum (%/yr)

Speed km per hour (km/h)

Water flow rate Litres/min (L/min)

Fuel economy Litres per 100km (L/100km)

Method for working out constant rates from worded problems

Write out the unitWrite out the relevant numbers as a divisionWork out your answer

Example 1If Colin earns $300 in 6 hours, what is his rate of pay?

hour

hour

hourratepay

/50$

/$50

/$6

300

Method for working out constant rates from worded problems

Write out the unitWrite out the relevant numbers as a divisionWork out your answer

Example 2If a heater uses 20000 Joules of energy every 10 seconds what is its rate of energy use?

sec/2000

sec/10

20000

J

Jratepay

Which can be written as 2 kJ/s

Method for working out constant rates from worded problems

Write out the unitWrite out the relevant numbers as a divisionWork out your answer

Example 3What is the rate of expansion when a temperature change of 10oC produces an 0.5mm of expansion in a steel beam?

Cmm

Cmmrateexapnsion

/05.0

/10

5.0

The first part of a rate unit always comes from the name

of the rate. Unit for expansion = mm so rate unit = mm/oC

Method for working out constant rates from worded problems

Write out the unitWrite out the relevant numbers as a divisionWork out your answer

Example 4If 6 litres of water can dissolve up to 30 grams of a powder what is the maximum solution rate in grams/litre?

Lg

Lgratesolution

/5

/6

30

What We Will Be Looking At In This Unit

The Maths in the previous examplesis fairly straight forward. In this unit we are going to look at:- finding rates from graphs and tables. - visualising graphs for a variety of rate

situations

Introductory Rates Question 1Length

(m)

Time (hour)s)

10 8.0 6.0 4.0

100

90

80

60

70

50

40

30

10

20

0 12 2.0

110

A B C D E

(a) In what section A, B, C, D or E, does the digger work at its highest rate?

Section E

Introductory Rates Question 1Length

(m)

Time (hour)s)

10 8.0 6.0 4.0

100

90

80

60

70

50

40

30

10

20

0 12 2.0

110

A B C D E

(b) What aspect of the graph did you use to arrive at your answer to part a) ?

Gradient

Introductory Rates Question 1Length

(m)

Time (hour)s)

10 8.0 6.0 4.0

100

90

80

60

70

50

40

30

10

20

0 12 2.0

110

A B C D E

(c) What are the units for the digging rate?

Unit for digging = m so digging rate unit

is m/hour Or metres per hour.

This unit can also be deduced from gradient

rule of rise/run

Introductory Rates Question 1Length

(m)

Time (hour)s)

10 8.0 6.0 4.0

100

90

80

60

70

50

40

30

10

20

0 12 2.0

110

A B C D E

(d) When was the digger not working?

Sections

B & D

Introductory Rates Question 1Length

(m)

Time (hour)s)

10 8.0 6.0 4.0

100

90

80

60

70

50

40

30

10

20

0 12 2.0

110

A B C D E

(e) What is the digging rate in section A?

hourmetres

run

risemAinratedigging

/152

30

30

2

Introductory Rates Question 1Length

(m)

Time (hour)s)

10 8.0 6.0 4.0

100

90

80

60

70

50

40

30

10

20

0 12 2.0

110

A B C D E

(e) What is the digging rate in section B?

0mBinratedigging

Introductory Rates Question 1Length

(m)

Time (hour)s)

10 8.0 6.0 4.0

100

90

80

60

70

50

40

30

10

20

0 12 2.0

110

A B C D E

(e) What is the digging rate in section C?

hourmetres

run

risemCinratedigging

/5.24

10

10

4

Introductory Rates Question 1Length

(m)

Time (hour)s)

10 8.0 6.0 4.0

100

90

80

60

70

50

40

30

10

20

0 12 2.0

110

A B C D E

(e) What is the digging rate in section D?

0mDinratedigging

Introductory Rates Question 1Length

(m)

Time (hour)s)

10 8.0 6.0 4.0

100

90

80

60

70

50

40

30

10

20

0 12 2.0

110

A B C D E

(e) What is the digging rate in section E?

hourmetres

run

risemEinratedigging

/67.163

50

50

3

Instantaneous and Average Rates of Change

When you are asked for an average rate of change you are being asked for a summary of the rate of change a time interval (or other interval).

Average rate = gradient between two points

When you are asked the rate of change at a point you are being asked for the instantaneous rates of change at a point.

Instantaneous rate = gradient of tangent at that point

Notes

Example 1

What is the average toy production rate in the first 12 minutes?

min/5

min12

60

toys

toysrun

risemrateproductionave

toys

Time (min)

60 toys

12 minutes

Example 2

Estimate the production rate at 8 minutes?

toys

Time (min)

52 toys

8 minutes

8 minutes point

min/5.6

min8

52

min8

toys

toysrun

risematrateproduct

Introductory Question 2

Length(m)

Time(hours)

(a) When is the digger digging at a constant rate?

Constant rate = constant gradient

Constant rate in 0 to 5 hours and 7 to 10 hours

Introductory Question 2

Length(m)

Time(hours)

(b) What is the digging rate at 2 hours?

30

5

hourm

run

risemhoursatratedigging

/65

30

2

Introductory Question 2

Length(m)

Time(hours)

(c) What is the digging rate at 9 hours?

33

hourm

run

risemhoursatratedigging

/13

3

9

Introductory Question 2

Length(m)

Time(hours)

(d) Estimate the digging rate at 6 hours? 26

9

hourm

run

risemhoursatratedigging

/89.29

26

6

Introductory Question 2

Length(m)

Time(hours)

(e) What is the average digging over the 10 hours?

39

10 hourm

run

risemratedig

/9.310

39

Introductory Question 3

(a) What is the average digging rate over the first 5 hours?

20

5hourm

run

risemratedig

/45

20

Introductory Question 3

(b) Estimate the digging rate at 4.5 seconds?

30

5

hourm

run

risemratedig

/65

30

Introductory Question 3

(c) What is the final digging rate? 18

9

hourm

run

risemratedig

/29

18

Introductory Question 4

(a) At what instant is the rate of change of depth equal to zero

Answer

At 6 seconds and at 18 seconds

Introductory Question 4

(b) At what instant is the rate of change a maximum.

Answer

At 0 seconds and at 24 seconds

Introductory Question 4

(c) The magnitude of the rate of change at 0 and 12 seconds is exactly the same but one is positive and the other is negative. What is the significance of the sign?

AnswerA positive rate means that the depth is increasing with time and a negative rate means that it is decreasing with time.

m = 1.3 m/h m = –1.3 m/h

Introductory Question 4

4.3

8

hourm

run

risemratedig

/54.08

3.4

(d) What is the average rate of change in the first 8 hours?

Introductory Question 4

hourmrun

risemratedig

/0

(e) What is the average rate of change in the first 12 hours?

Introductory Question 4

- 10

9

hourm

run

risemratedig

/11.19

10

(f) Estimate the instantaneous rate of change at 10 hours?

Introductory Question 5Write out the unitWrite out the relevant numbers as a divisionWork out your answer

a) What is the average growth rate (per month) over the total time interval?

monthmm

monthmmrategrowth

/8.10

/5

54

1st of Month Jan Feb Mar Apr May June

Plant Height 0mm 12mm 26mm 40mm 44mm 54mm

Introductory Question 5Write out the unitWrite out the relevant numbers as a divisionWork out your answer

b) What is the average growth rate (per month) over the last two months?

monthmm

monthmmrategrowth

/7

/2

14

1st of Month Jan Feb Mar Apr May June

Plant Height 0mm 12mm 26mm 40mm 44mm 54mm

Rates from Graphs

Textbook Exercises

Average rates of change

Ex18C p515 Q2, 6 Instantaneous rates of change using tangents

Ex 18D p519 Q1, 2

Working Out Rates From Formulae

If you have a rule for a relationship the simplest way to find a rate is to use the gradient formula

With the variables above replaced by the appropriate variables from the problem

12

12

12

12 )()(

xx

xfxf

xx

yymrate

Rates from formulae

Problem 1

When a ball is fired upwards its height can be calculated with the formula

h(t) = 40t – 5t 2

(a) What is the average speed of the ball in the first 2 seconds?

at t1 = 0, h(t1) = 40 0 – 5 02 = 0

at t2 = 2, h(t2) = 40 2 – 5 22 = 60

sm

tt

ththmrateave

/3002

060

)()(

12

12

Rates from formulae

Problem 1

When a ball is fired upwards its height can be calculated with the formula

h(t) = 40t – 5t 2

(b) What is the approximate speed of the ball at 2 seconds?

at t1 = 2, h(t1) = 40 2 – 5 22 = 60

at t2 = 2.01, h(t2) = 40 2.01 – 5 2.012 = 60.1995

sm

tt

ththmrateave

/95.190.201.2

601995.60

)()(

12

12

Average rates are an indication of the round about rate over an interval. If the interval is very small then the average

will be close to the actual instantaneous rate for all the points in the interval. In general moving on about 1/100th of the magnitude of the initial point will give a

good approximation to an instantaneous rate at the initial point.

Rates from formulae

Problem 2

The height of a bridge above a river is closely modelled by the equation

h(x) = 0.02x 2 – 0.2x +9

(a) What is the average slope (change in height rate) of the bridge in the first 3 metres?

at x1 = 0, h(x1) = 0.02 02 – 0.2 0 + 9 = 9

at x2 = 3, h(x2) = 0.02 32 – 0.2 3 + 9 = 8.58

14.0

/14.003

958.8

)()(

12

12

mm

xx

xhxhmrateave

Rates from formulae

Problem 2

The height of a bridge above a river is closely modelled by the equation

h(x) = 0.02x2 – 0.2x +9

(b) What is the slope (change in height rate) of the bridge at the point 3 horizontal metres across the bridge?

at x1 = 3, h(x1) = 0.02 32 – 0.2 3 + 9 = 8.58

at x2 = 3.01, h(x2) = 0.02 3.012 – 0.2 3.01 + 9 = 8.5792

08.0

/08.0301.3

58.85792.8

)()(

12

12

mm

tt

xhxhmrateave

Rates from formulae

Textbook Problems

Ex18C p513

Q1, 3

Ex18D p521

Q9, 10, 11,15

Estimating Rates from formulae

Examples & Questions

Example 1 p503

Exercise 8A p505

Q6, 7

Rates Revision and Test

Use the Rates revision sheet to study for the Rates test. No solutions will be supplied for the revision sheet since the test is very similar to the revision sheet