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

Presentation on theme: "Rates of Change Rates are an important area of Maths and can be found in everyday life, business and science."— Presentation transcript:

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

Common rates RateUnits Pay rate\$ per hour (\$/h) Heart rateBeats per minute (beat/min) Interest ratePercentage per annum (%/yr) Speedkm per hour (km/h) Water flow rateLitres/min (L/min) Fuel economyLitres per 100km (L/100km)

Method for working out constant rates from worded problems Write out the unit Write out the relevant numbers as a division Work out your answer Example 1 If Colin earns \$300 in 6 hours, what is his rate of pay?

Method for working out constant rates from worded problems Write out the unit Write out the relevant numbers as a division Work out your answer Example 2 If a heater uses 20000 Joules of energy every 10 seconds what is its rate of energy use? Which can be written as 2 kJ/s

Method for working out constant rates from worded problems Write out the unit Write out the relevant numbers as a division Work out your answer Example 3 What is the rate of expansion when a temperature change of 10 o C produces an 0.5mm of expansion in a steel beam? The first part of a rate unit always comes from the name of the rate. Unit for expansion = mm so rate unit = mm/ o C

Method for working out constant rates from worded problems Write out the unit Write out the relevant numbers as a division Work out your answer Example 4 If 6 litres of water can dissolve up to 30 grams of a powder what is the maximum solution rate in grams/litre?

What We Will Be Looking At In This Unit The Maths in the previous examples is 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 1 Length (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 CDE (a) In what section A, B, C, D or E, does the digger work at its highest rate? Section E

Introductory Rates Question 1 Length (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 CDE (b) What aspect of the graph did you use to arrive at your answer to part a) ? Gradient

Introductory Rates Question 1 Length (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 CDE (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 1 Length (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 CDE (d) When was the digger not working? Sections B & D

Introductory Rates Question 1 Length (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 CDE (e) What is the digging rate in section A? 30 2

Introductory Rates Question 1 Length (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 CDE (e) What is the digging rate in section B?

Introductory Rates Question 1 Length (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 CDE (e) What is the digging rate in section C? 10 4

Introductory Rates Question 1 Length (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 CDE (e) What is the digging rate in section D?

Introductory Rates Question 1 Length (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 CDE (e) What is the digging rate in section E? 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? 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

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

Introductory Question 2 Length (m) Time (hours) (c) What is the digging rate at 9 hours? 3 3

Introductory Question 2 Length (m) Time (hours) (d) Estimate the digging rate at 6 hours? 26 9

Introductory Question 2 Length (m) Time (hours) (e) What is the average digging over the 10 hours? 39 10

Introductory Question 3 (a) What is the average digging rate over the first 5 hours? 20 5

Introductory Question 3 (b) Estimate the digging rate at 4.5 seconds? 30 5

Introductory Question 3 (c) What is the final digging rate? 18 9

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

Introductory Question 4 (e) What is the average rate of change in the first 12 hours?

Introductory Question 4 - 10 9 (f) Estimate the instantaneous rate of change at 10 hours?

Introductory Question 5 Write out the unit Write out the relevant numbers as a division Work out your answer a) What is the average growth rate (per month) over the total time interval ? 1 st of MonthJanFebMarAprMayJune Plant Height 0mm12mm26mm40mm44mm54mm

Introductory Question 5 Write out the unit Write out the relevant numbers as a division Work out your answer b) What is the average growth rate (per month) over the last two months ? 1 st of MonthJanFebMarAprMayJune Plant Height 0mm12mm26mm40mm44mm54mm

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

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 t 1 = 0, h(t 1 ) = 40 0 – 5 0 2 = 0 at t 2 = 2, h(t 2 ) = 40 2 – 5 2 2 = 60

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 t 1 = 2, h(t 1 ) = 40 2 – 5 2 2 = 60 at t 2 = 2.01, h(t 2 ) = 40 2.01 – 5 2.01 2 = 60.1995 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 1/100 th 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 x 1 = 0, h(x 1 ) = 0.02 0 2 – 0.2 0 + 9 = 9 at x 2 = 3, h(x 2 ) = 0.02 3 2 – 0.2 3 + 9 = 8.58

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 (b) What is the slope (change in height rate) of the bridge at the point 3 horizontal metres across the bridge? at x 1 = 3, h(x 1 ) = 0.02 3 2 – 0.2 3 + 9 = 8.58 at x 2 = 3.01, h(x 2 ) = 0.02 3.01 2 – 0.2 3.01 + 9 = 8.5792

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

Similar presentations