demo disc teach a level maths vol. 2: a2 core modules © christine crisp
TRANSCRIPT
Demo DiscDemo Disc
““Teach A Level Maths”Teach A Level Maths”Vol. 2: A2 Core ModulesVol. 2: A2 Core Modules
© Christine Crisp
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 xey
43: Partial Fractions
51: The Vector Equation of a Line
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.
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.
xy sin13 xy
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
One-to-one and many-to-one functions
is an example of a one-to-one function
13 xy is an example of a many-to-one function
xy sin
xy sin13 xy
Consider the following graphs
and
Inverse Functions
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.
e.g.On the same axes, sketch the graph of
and its inverse.
2,)2( 2 xxy
N.B!
)0,2(
)1,3(
xy
)4,4(x
Solution:
)2,0(
)3,1(
Graphs of Inverse Functions
e.g.On the same axes, sketch the graph of
and its inverse.
2,)2( 2 xxy
N.B!xy
2)2( xy
Solution:
N.B. Using the translation of we can see the inverse function is .
x2)(1 xxf
2 xy
Graphs of Inverse Functions
2)2( xy
2 xy
A bit more on domain and range
The domain of is . 2x
)(xf
Since is found by swapping x and y,
)(1 xf
the values of the domain of give the values of the range of .
)(xf)(1 xf
2)2()( xxf 2xDomain
2y2)(1 xxf Range
2,)2()( 2 xxxfThe previous example used .
Graphs of Inverse Functions
2)2( xy
A bit more on domain and range
2,)2()( 2 xxxfThe previous example used .
The domain of is . 2x
)(xf
Similarly, the values of the range of )(xf
)(1 xf give the values of the domain of
2 xySince is found by swapping x and y,
)(1 xf
Graphs of Inverse Functions
the values of the domain of give the values of the range of .
)(xf)(1 xf
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 xey
More Indices and Logs
yxey ex log
xey 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 yxy x
10log10 ( since an index is a log )
We write as ( n for natural ) so,
elog ln
yxey x ln
Logs with a base e are called natural logs
The Function
xey
xey
xy
xey
The Inverse of xey
We can sketch the inverse by reflecting in y = x.
is a one-to-one function so it has an inverse function.
xexf )(
Finding the equation of the inverse function is easy!
yxey x lnSo, xxf ln)(1
0xN.B. The domain is .
We’ve already done the 1st step of
rearranging:
Now swap letters:
xy ln
The Function
xey
xy ln
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.
xy sin
What domain would you use?We need to make sure that we have all the y-values without any being repeated.
9090 x( or, in degrees, as ).
The domain is defined as 22
x
Inverse Trig Ratios
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.
Combined Transformationse.g. 1 Describe the transformations of
that give the function . Hence sketch the function.
xey 12 xey
Solution:• x has been replaced by
2x:so we have a stretch of s.f.
• 1 has then been added:
xx ee 2
122 xx ee
so we have a translation of
parallel to the
x-axis
21
1
0
More Transformations
The point on the y-axis . . .
We do the sketch in 2 stages:xey xey 2
xey xey
xey 2
doesn’t move with a stretch parallel to the x-axis
21
More Transformations
12 xey
xey 2
12 xey1
1
1
We do the sketch in 2 stages:xey xey 2
xey xey
xey 2
21
More Transformations
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.
2
2 xy
x
y
1y
e.g. 2 Solve the inequality . 12 x
2x is above 1 in two regions.
Method 1: Sketch the graphs and .
2 xy 1y
The Modulus Function
2
2 xy
x
y
1y
The right hand branch of the modulus graph is the same as y = x – 2 . . .
12x 3x
e.g. 2 Solve the inequality . 12 x
2x is above 1 in two regions.The sketch doesn’t show the xcoordinates of A and B, so we must find them.
A B
2xy
so B is given by
Method 1: Sketch the graphs and .
2 xy 1y
x x
The gradient of the right hand branch is +1 and the left hand is 1, so the graph is symmetrical.So
, 3112 xxx or
The Modulus Function
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.
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 we’ll look at an equation which has 2 roots and the iteration produces a surprising result.
02 xe xThe equation is .We’ll try the simplest iterative formula first :
21 nx
n ex
xe x 2
Iteration Diagrams and Convergence
2 xey
xy
21 nx
n exLet’s try with close to the positive root.
0x
Let 510 x
Iteration Diagrams and Convergence
2 xey
xy
0x
Using the iterative formula,
,4821 x 9692 x
1x
The staircase moves away from the root.
21 nx
n exLet’s try with close to the positive root.
0x
The sequence diverges rapidly.
Iteration Diagrams and Convergence
2 xey
xy
Suppose we try a value for on the left of the root. 0x
0x
Iteration Diagrams and Convergence
2 xey
xy
1x
Iteration Diagrams and Convergence
Suppose we try a value for on the left of the root. 0x
2 xey
xy
2x
Iteration Diagrams and Convergence
Suppose we try a value for on the left of the root. 0x
2 xey
xy
3x
Iteration Diagrams and Convergence
Suppose we try a value for on the left of the root. 0x
2 xey
xy
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. 0x
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
xy cos
dxx
0cos(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.2
xsin 0
0sinsin 0
Integrating the Simple Functions
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.
dxxf
xf )(
)(/
e.g. 5 dxx
x cos2
sin3
What do we want in the numerator?
Ans: xsin
We now need to get rid of the minus and replace the 3.
dxx
x
cos2
sin. . .
Integration giving Logs
What do we want in the numerator?
Ans: xsin
We now need to get rid of the minus and replace the 3.
dxx
x
cos2
sin3
Always check by multiplying the numerator by the constant outside the integral: xx sin3)sin(3
e.g. 5 dxx
x cos2
sin3
Integration giving Logs
What do we want in the numerator?
Ans: xsin
We now need to get rid of the minus and replace the 3.
3
Cx )cos2ln(3
e.g. 5 dxx
x cos2
sin3
dxx
x
cos2
sin
Integration giving Logs
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
SUMMARY
The double angle formulae are:
AAA cossin22sin )1(
AAA 22 sincos2cos )2(
1cos2 2 A )2( a
A2sin21 )2( b
A
AA
2tan1
tan22tan
)3(
Double Angle Formulae
N.B. The formulae link any angle with double the angle.For example, they can be used for x2 xan
d x
2
xand
y32
3 yand
We use them • to solve equations
• to prove other identities• to integrate some functions
4 and
2
Double Angle Formulae
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.
Express each of the following in partial fractions.
1.
Exercises
)2)(3(
55
xx
x 2.)3)(12(
5
xx
3.
1
22 x
4.)1(
37
xx
x
Partial Fractions
Solutions:1.
23)2)(3(
55
x
B
x
A
xx
x
)2)(3( xxMultiply by :)3()2(55 xBxAx
:2x B55 1 B:3x A520 4 A
2
1
3
4
)2)(3(
55
xxxx
xSo,
5
20
Check:
gives3x
Partial Fractions
Solutions:1.
23)2)(3(
55
x
B
x
A
xx
x
)2)(3( xxMultiply by :)3()2(55 xBxAx
:2x B55 1 B:3x A520 4 A
2
1
3
4
)2)(3(
55
xxxx
xSo,
gives2x,4Check:
gives3x5
20
5
5
Partial Fractions
Solutions:1.
23)2)(3(
55
x
B
x
A
xx
x
)2)(3( xxMultiply by :)3()2(55 xBxAx
:2x B55 1 B:3x A520 4 A
2
1
3
4
)2)(3(
55
xxxx
xSo,
5
5gives2x,4 1Check
:gives3x
( You don’t need to write out the check in full )
5
20
Partial Fractions
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.
Before we see how to solve the equation, it’s useful to get some idea of the solution.
ydx
dye.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
ydx
dy
We can now draw a curve through any point following the gradients.
Solving Differential Equations
ydx
dy
However, we haven’t got just one curve.
Solving Differential Equations
ydx
dy
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
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.
Finding the Equation of a LineIn coordinate geometry, the equation of a line is cmxy e.g. 32 xy
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
x
B
A
x
Ox
We consider several more points on the line.
2Rx
3Rx
1Rx
A and B are fixed points.
We need an equation for r, the position vector of any point R on the line.
r1a
Starting with R1: 1r aAB2
1
The Vector Equation of a Line
x
B
A
x
Ox
B
AWe consider several more points on the line.
x
1R
2R
3R
x
x
xa
r2
A and B are fixed points.
Starting with R1:
2r
1r aAB2
1
a AB2
We need an equation for r, the position vector of any point R on the line.
The Vector Equation of a Line
x
B
A
x
Ox
B
AWe consider several more points on the line.
Ox
1R
2R
3R
x
x
xa
r3
A and B are fixed points.
Starting with R1:
2r
1r aAB2
1
a AB2
3r a AB)( 41
We need an equation for r, the position vector of any point R on the line.
The Vector Equation of a Line
x
B
A
x
Ox
B
A
x
1R
2R
3R
x
x
xa
O
So for R1, R2 and R3
2r
1r a
21
a 2
3r a )( 41
AB
AB
AB
For any position of R, we have
r a
t ABt is called a parameter and can have any real value.It is a scalar not a vector.
The Vector Equation of a Line
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