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Demo Disc Teach A Level Maths Vol. 2: A2 Core Modules © Christine Crisp.

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Presentation on theme: "Demo Disc Teach A Level Maths Vol. 2: A2 Core Modules © Christine Crisp."— Presentation transcript:

1 Demo Disc Teach A Level Maths Vol. 2: A2 Core Modules © Christine Crisp

2 47:Solving Differential Equations 28:Integration giving Logs 31:Double Angle Formulae 18:Iteration Diagrams and Convergence 15:More Transformations 13:Inverse Trig Ratios 3:Graphs of Inverse Functions 2:Inverse Functions The slides that follow are samples from the 55 presentations that make up the work for the A2 core modules C3 and C4. 16:The Modulus Function 22:Integrating the Simple Functions 4:The Function 43:Partial Fractions 51:The Vector Equation of a Line

3 Explanation of Clip-art images An important result, example or summary that students might want to note. It would be a good idea for students to check they can use their calculators correctly to get the result shown. An exercise for students to do without help.

4 2: Inverse Functions Demo version note: The students have already met the formal definition of a function and the ideas of domain and range. In the following slides we prepare to introduce the condition for an inverse function.

5 One-to-one and many-to-one functions Each value of x maps to only one value of y... Consider the following graphs Each value of x maps to only one value of y... BUT many other x values map to that y. and each y is mapped from only one x. and Inverse Functions

6 One-to-one and many-to-one functions is an example of a one-to-one function is an example of a many-to-one function Consider the following graphs and Inverse Functions

7 3: Graphs of Inverse Functions Demo version note: By this time the students know how to find an inverse function. The graphical link between a function and its inverse has also been established and this example follows.

8 e.g.On the same axes, sketch the graph of and its inverse. N.B! x Solution: Graphs of Inverse Functions

9 e.g.On the same axes, sketch the graph of and its inverse. N.B! Solution: N.B.Using the translation of we can see the inverse function is. Graphs of Inverse Functions

10 A bit more on domain and range The domain of is. Since is found by swapping x and y, the values of the domain of give the values of the range of. Domain Range The previous example used. Graphs of Inverse Functions

11 A bit more on domain and range The previous example used. The domain of is. Similarly, the values of the range of give the values of the domain of Since is found by swapping x and y, Graphs of Inverse Functions the values of the domain of give the values of the range of.

12 Demo version note: The exponential function has been defined and here we build on earlier work to find the inverse and its graph. 4: The Function

13 More Indices and Logs The function contains the index x, so x is a log. BUT the base of the log is e not 10, so We know that ( since an index is a log ) We write as ( n for natural ) so, Logs with a base e are called natural logs The Function

14 The Inverse of We can sketch the inverse by reflecting in y = x. is a one-to-one function so it has an inverse function. Finding the equation of the inverse function is easy! So, N.B. The domain is. Weve already done the 1 st step of rearranging: Now swap letters: The Function

15 13: Inverse Trig Ratios Demo version note: This presentation on inverse trig ratios revises the idea of a domain. It also reminds students that a one-to-one relationship is needed in order to find the inverse function.

16 What domain would you use? We need to make sure that we have all the y -values without any being repeated. ( or, in degrees, as ). The domain is defined as Inverse Trig Ratios

17 15: More Transformations Demo version note: As new topics are introduced, the presentations revise earlier work. Here, the exponential function is used to show the result of combining transformations.

18 Combined Transformations e.g. 1 Describe the transformations of that give the function. Hence sketch the function. Solution: x has been replaced by 2 x : so we have a stretch of s.f. 1 has then been added: so we have a translation of parallel to the x -axis More Transformations

19 The point on the y -axis... We do the sketch in 2 stages: doesnt move with a stretch parallel to the x -axis More Transformations

20 We do the sketch in 2 stages: More Transformations

21 16: The Modulus Function Demo version note: Students are encouraged to sketch functions whenever possible rather than always using graphical calculators. A translation has been used to obtain the function shown and the inequality is found using the sketch.

22 x y e.g. 2 Solve the inequality. is above 1 in two regions. Method 1: Sketch the graphs and. The Modulus Function

23 x y The right hand branch of the modulus graph is the same as y = x – 2... e.g. 2 Solve the inequality. is above 1 in two regions. The sketch doesnt show the x coordinates of A and B, so we must find them. AB so B is given by Method 1: Sketch the graphs and. xx The gradient of the right hand branch is +1 and the left hand is 1, so the graph is symmetrical. So, The Modulus Function

24 18: Iteration Diagrams and Convergence Demo version note: By this stage the students understand how to rearrange equations in a variety of ways in order to find an iterative formula. They have met cobweb and staircase diagrams in simple cases of convergence and divergence.

25 If you have Autograph or another graph plotter you may like to try to find both roots before you see my solution. In the next example well look at an equation which has 2 roots and the iteration produces a surprising result. The equation is. Well try the simplest iterative formula first : Iteration Diagrams and Convergence

26 Lets try with close to the positive root. Let Iteration Diagrams and Convergence

27 Using the iterative formula, The staircase moves away from the root. Lets try with close to the positive root. The sequence diverges rapidly. Iteration Diagrams and Convergence

28 Suppose we try a value for on the left of the root. Iteration Diagrams and Convergence

29 Suppose we try a value for on the left of the root.

30 Iteration Diagrams and Convergence Suppose we try a value for on the left of the root.

31 Iteration Diagrams and Convergence Suppose we try a value for on the left of the root.

32 The sequence now converges... but to the other root ! Iteration Diagrams and Convergence Suppose we try a value for on the left of the root.

33 22: Integrating the Simple Functions Demo version note: The opportunity is taken here to remind the students about using integration to find areas, whilst applying the work to a trig integration. Integrating the Simple Functions

34 (b) Radians! The definite integral can give an area, so this result may seem surprising. However, the graph shows us why it is correct. This part gives a negative integral This part gives a positive integral How would you find the area? Ans: Find the integral from 0 to and double it. Integrating the Simple Functions

35 28: Integration giving Logs Demo version note: Inspection is used to find integrals of the form In this example, the method is applied to a question where the integrand needs adjusting.

36 e.g. 5 What do we want in the numerator? Ans: We now need to get rid of the minus and replace the Integration giving Logs

37 What do we want in the numerator? Ans: We now need to get rid of the minus and replace the 3. Always check by multiplying the numerator by the constant outside the integral: e.g. 5 Integration giving Logs

38 What do we want in the numerator? Ans: We now need to get rid of the minus and replace the 3. e.g. 5 Integration giving Logs

39 31: Double Angle Formulae Demo version note: Summaries are given at appropriate points in the presentations. The notebook icon suggests that students might want to copy the slide. Double Angle Formulae

40 SUMMARY The double angle formulae are: Double Angle Formulae

41 N.B. The formulae link any angle with double the angle. For example, they can be used for and We use them to solve equations to prove other identities to integrate some functions and Double Angle Formulae

42 43: Partial Fractions Demo version note: Introductory exercises follow most sections of theory and examples. The next slide shows the first exercise on finding Partial Fractions. The full solution is given and the cover-up method used to check the result. Students would use their textbooks for further practice.

43 Express each of the following in partial fractions. 1. Exercises Partial Fractions

44 Solutions: 1. Multiply by : So, Check: Partial Fractions

45 Solutions: 1. Multiply by : So, Check: Partial Fractions

46 Solutions: 1. Multiply by : So, Check: ( You dont need to write out the check in full ) Partial Fractions

47 47: Solving Differential Equations Demo version note: The slides here show the introduction to the method of separating the variables to solve some differential equations. Solutions are applied later to applications of growth and decay functions.

48 Before we see how to solve the equation, its useful to get some idea of the solution. e.g. (2) The equation tells us that the graph of y has a gradient that always equals y. We can sketch the graph by drawing a gradient diagram. For example, at every point where y = 2, the gradient equals 2. We can draw a set of small lines showing this gradient. 2 1 We can cover the page with similar lines. Solving Differential Equations

49 We can now draw a curve through any point following the gradients. Solving Differential Equations

50 However, we havent got just one curve. Solving Differential Equations

51 The solution is a family of curves. Can you guess what sort of equation these curves represent ? ANS: They are exponential curves. Solving Differential Equations

52 51: The Vector Equation of a Line Demo version note: By this time, the students have practised using vectors and are familiar with the notation. The equation of a straight line in vector form is developed.

53 Finding the Equation of a Line In coordinate geometry, the equation of a line is e.g. The equation gives the value (coordinate) of y for any point which lies on the line. The vector equation of a line must give us the position vector of any point on the line. We start with fixing a line in space. We can do this by fixing 2 points, A and B. There is only one line passing through these points. The Vector Equation of a Line

54 x x x We consider several more points on the line. x x x A and B are fixed points. We need an equation for r, the position vector of any point R on the line. r1r1 a Starting with R 1 : The Vector Equation of a Line

55 x x x We consider several more points on the line. x x x x a r2r2 A and B are fixed points. Starting with R 1 : We need an equation for r, the position vector of any point R on the line. The Vector Equation of a Line

56 x x x We consider several more points on the line. x x x x a r3r3 A and B are fixed points. Starting with R 1 : We need an equation for r, the position vector of any point R on the line. The Vector Equation of a Line

57 x x x x x x x a So for R 1, R 2 and R 3 For any position of R, we have t is called a parameter and can have any real value. It is a scalar not a vector. The Vector Equation of a Line

58 Full version available from:- Chartwell-Yorke Ltd. 114 High Street, Belmont Village, Bolton, Lancashire, BL7 8AL England, tel (+44) (0) , fax (+44) (0)


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