Transcript Document

“Teach A Level Maths”
Vol. 2: A2 Core Modules
Demo Disc
© Christine Crisp
The slides that follow are samples from the 55
presentations that make up the work for the A2 core
modules C3 and C4.
2: Inverse Functions
3: Graphs of Inverse
Functions
4: The Function
y  ex
13: Inverse Trig Ratios
15: More Transformations
16: The Modulus Function
18: Iteration Diagrams and
Convergence
22: Integrating the
Simple Functions
28: Integration giving
Logs
31: Double Angle
Formulae
43: Partial Fractions
47: Solving Differential
Equations
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.
Inverse Functions
One-to-one and many-to-one functions
Consider the following graphs
y  sin x 
y  x 1
3
and
Each value of x maps to
only one value of y . . .
and each y is mapped
from only one x.
Each value of x maps to
only one value of y . . .
BUT many other x values
map to that y.
Inverse Functions
One-to-one and many-to-one functions
Consider the following graphs
y  sin x 
y  x 1
3
and
y  x  1 is an example
of a one-to-one function
3
y  sin x
is an example
of a many-to-one
function

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.
Graphs of Inverse Functions
e.g. On the same axes, sketch the graph of
y  ( x  2) ,
x2
2
and its inverse.
Solution:
N.B!
yx
(4, 4)
x
(1, 3)
(0, 2)
( 3, 1)
( 2, 0)
Graphs of Inverse Functions
e.g. On the same axes, sketch the graph of
y  ( x  2) ,
2
x2
and its inverse.
Solution:
N.B!
yx
y x 2
y  ( x  2) 2
N.B. Using the translation of x
inverse function is f 1 ( x ) 
we can see the
x 2 .
Graphs of Inverse Functions
A bit more on domain and range
The previous example used
y  x 2
y  ( x  2) 2
f ( x )  ( x  2) 2 , x  2 .
The domain of f ( x ) is
x2.
1
Since f ( x ) is found
by swapping x and y,
the values of the domain
of f ( x ) give the values of
1
the range of f ( x ).
f ( x )  ( x  2) 2 Domain x  2
1
f ( x )  x  2 Range y  2
Graphs of Inverse Functions
A bit more on domain and range
The previous example used
y  x 2
y  ( x  2) 2
f ( x )  ( x  2) 2 , x  2 .
The domain of f ( x ) is
x2.
1
Since f ( x ) is found
by swapping x and y,
the values of the domain
of f ( x ) give the values of
1
the range of f ( x ).
Similarly, the values of the range of
f ( x)
give the values of the domain of f 1 ( x )
4: The Function
ye
x
Demo version note:
The exponential function has been
defined and here we build on earlier work
to find the inverse and its graph.
The Function
ye
x
More Indices and Logs
We know that
y  10 x 
x  log 10 y
( since an index is a log )
The function y  e x contains the index x, so x is a
log.
BUT the base of the log is e not 10, so
Logs with a base e
x
y  e  x  log e y are called natural logs
We write log e as ln ( n for natural ) so,
y  e x  x  ln y
The Function
The Inverse of
ye
x
y  ex
f ( x )  e x is a one-to-one function so it has an
inverse function.
ye
x
yx
y  ln x
We can sketch the inverse
by reflecting in y = x.
Finding the equation of
the inverse function is
easy!
We’ve already done the 1st step of rearranging:
y  e x  x  ln y
Now swap letters:
y  ln x
N.B. The domain is
x
So, f
 0.
1
( x )  ln x
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.
Inverse Trig Ratios
y  sin x
What domain would you use?
We need to make sure that we have all the y-values
without any being repeated.


The domain is defined as 
 x 
2
2
( or, in degrees, as  90   x  90  ).
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.
More Transformations
Combined Transformations
e.g. 1 Describe the transformations of y  e that
give the function y  e 2 x  1. Hence sketch
the function.
Solution:
x
•
x has been replaced by 2x:
e x  e2x
so we have a stretch of s.f. 1
2
parallel to the x-axis
•
1 has then been added:
e
2x
 e
0
so we have a translation of  
1 
2x
1
More Transformations
We do the sketch in 2 stages:
y  ex
y  e2x
y  e2x
ye
x

1
2
y  ex
The point on the y-axis . . . doesn’t move
with a stretch parallel to the x-axis
More Transformations
We do the sketch in 2 stages:
y  ex
y  e2x
y  e2x
ye
x

1
2
y  ex
y  e2x
1
y  e2x 1
y  e2x 1
1
1
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.
The Modulus Function
e.g. 2 Solve the inequality
x2  1.
Method 1: Sketch the graphs y  x  2 and y  1 .
y
x  2 is above 1 in
two regions.
y  x2
y 1
x
2
The Modulus Function
e.g. 2 Solve the inequality
x2  1.
Method 1: Sketch the graphs y  x  2 and y  1 .
y
x  2 is above 1 in
y 1
y  x2
y  x2
A
B
x
x
x
1
2
3
two regions.
The sketch doesn’t
show the xcoordinates
of A and B, so we must
find them.
The right hand branch of the modulus graph is the
same as y = x – 2 . . . so B is given by
x21  x 3
The gradient of the right hand branch is +1 and the
left hand is 1, so the graph is symmetrical.
So,
x  2  1  x  1 or x  3
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.
Iteration Diagrams and Convergence
In the next example we’ll look at an equation which
has 2 roots and the iteration produces a surprising
result.
The equation is
e  x20.
x
We’ll try the simplest iterative formula first :

ex  2  x
x n1  e xn  2
If you have Autograph or another graph plotter you
may like to try to find both roots before you see
my solution.
Iteration Diagrams and Convergence
Let’s try x n1
positive root.
e
x
n
 2 with x 0 close to the
y  ex  2
yx

Let
x0  1  5

Iteration Diagrams and Convergence
Let’s try x n1
positive root.
e
x
n
 2 with x 0 close to the
The staircase moves
away from the root.

y  ex  2
yx

x0
x1
The sequence diverges rapidly.
Using the iterative formula, x1  2  48 , x 2
 9  96
Iteration Diagrams and Convergence
Suppose we try a value for
x0
on the left of the root.
y  ex  2
yx


x0
Iteration Diagrams and Convergence
Suppose we try a value for
x0
on the left of the root.
y  ex  2
yx


x1
Iteration Diagrams and Convergence
Suppose we try a value for
x0
on the left of the root.
y  ex  2
yx

x2

Iteration Diagrams and Convergence
Suppose we try a value for
x0
on the left of the root.
y  ex  2
yx


x3
Iteration Diagrams and Convergence
Suppose we try a value for
x0
on the left of the root.
y  ex  2
yx


The sequence now converges . . . but to the other
root !
Integrating the Simple Functions
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.
(b)


Integrating the Simple Functions
cos x dx
0
  sin x 0

  sin 
0

Radians!

 sin 0 
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 positive integral
y  cos x
This part gives a negative integral
How would you find
the area?
Ans: Find the
integral from 0 to

and double it.
2
28: Integration giving Logs
Demo version note:
Inspection is used to find integrals of the
form
f / ( x)
dx
f ( x)

In this example, the method is applied to a
question where the integrand needs
adjusting.
Integration giving Logs
e.g. 5

3 sin x
dx
2  cos x
What do we want in the numerator?
Ans:
 sin x
We now need to get rid

sin
x
...
dx of the minus and replace
2  cos x
the 3.

Integration giving Logs
e.g. 5

3 sin x
dx
2  cos x
What do we want in the numerator?
Ans:
 3

 sin x
We now need to get rid
 sin x
dx of the minus and replace
2  cos x
the 3.
Always check by multiplying the numerator by the
constant outside the integral:  3 (  sin x )  3 sin x
Integration giving Logs
e.g. 5

3 sin x
dx
2  cos x
What do we want in the numerator?
Ans:
 3

 sin x
We now need to get rid
 sin x
dx of the minus and replace
2  cos x
the 3.
  3 ln( 2  cos x )  C
Double Angle Formulae
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:
sin 2 A  2 sin A cos A
      (1)
cos 2 A  cos 2 A  sin 2 A       (2)
 2 cos 2 A  1
      ( 2a )
 1  2 sin 2 A
      ( 2b)
tan 2 A 
2 tan A
1  tan 2 A
      ( 3)
Double Angle Formulae
N.B. The formulae link any angle with double the
angle.
For example, they can be used for
2x
x
and
x
and
x
2
4
and
2
and
3y
2
3y
We use them
•
•
•
to solve equations
to prove other identities
to integrate some functions
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.
Partial Fractions
Exercises
Express each of the following in partial fractions.
1.
3.
5x  5
( x  3)( x  2)
2
x2  1
2.
5
( 2 x  1)( x  3)
4.
7x  3
x( x  1)
Solutions:
1.
Partial Fractions
5x  5
A
B


( x  3)( x  2) x  3 x  2
Multiply by ( x  3)( x  2) :
5 x  5  A( x  2)  B( x  3)
x  2:
5  5B
 B 1
x  3 :
 20  5 A  A  4
5x  5
4
1
So,


( x  3)( x  2) x  3 x  2
Check:
 20
x  3 gives
5
Solutions:
1.
Partial Fractions
5x  5
A
B


( x  3)( x  2) x  3 x  2
Multiply by ( x  3)( x  2) :
5 x  5  A( x  2)  B( x  3)
x  2:
5  5B
 B 1
x  3 :
 20  5 A  A  4
5x  5
4
1
So,


( x  3)( x  2) x  3 x  2
Check:
 20
5
x  3 gives
 4 , x  2 gives
5
5
Solutions:
1.
Partial Fractions
5x  5
A
B


( x  3)( x  2) x  3 x  2
Multiply by ( x  3)( x  2) :
5 x  5  A( x  2)  B( x  3)
x  2:
5  5B
 B 1
x  3 :
 20  5 A  A  4
5x  5
4
1
So,


( x  3)( x  2) x  3 x  2
Check:
 20
5
x  3 gives
 4 , x  2 gives  1
5
5
( You don’t need to write out the check in full )
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.
Solving Differential Equations
e.g. (2)
dy
 y
dx
Before we see how to solve the equation, it’s useful
to get some idea of the solution.
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
2
gradient equals 2. We can
1
draw a set of small lines
showing this gradient.
We can cover the page with similar lines.
Solving Differential Equations
dy
 y
dx
We can now draw a curve through any point following
the gradients.
Solving Differential Equations
dy
 y
dx
However, we haven’t got just one curve.
Solving Differential Equations
dy
 y
dx
The solution is a family of curves.
Can you guess what sort of equation these curves
represent ?
ANS: They are exponential curves.
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.
The Vector Equation of a Line
Finding the Equation of a Line
In coordinate geometry, the equation of a line is
y  mx  c
e.g.
y  2 x  3
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
A and B are fixed points.
We consider several
more points on the line.
We need an equation
for r, the position
vector of any point R on
x
the line.
O
Starting with R1:
R3
x
xA
x R1
a
x
r1
B
x
r1  a 
1
2

AB
R2
The Vector Equation of a Line
A and B are fixed points.
We consider several
more points on the line.
We need an equation
for r, the position
vector of any point R on
x
the line.
O
Starting with R1:
R3
x
xA
x R1
a
r1  a 
x
1
2

AB

r 2  a  2 AB
B
r2
x
R2
The Vector Equation of a Line
R3
A and B are fixed points.
We consider several
more points on the line.
x
r3
We need an equation
for r, the position
vector of any point R on
x
the line.
O
Starting with R1:
xA
x R1
a
x
B
x
r1  a 
1
2

AB

r 2  a  2 AB
r3  a 

( 14 ) AB
R2
The Vector Equation of a Line
So for R1, R2 and R3
r1  a 
1
2
R3

x
AB

x R1
r 2  a  2 AB
r 3  a  (
1
)
4
xA
a

AB
x
x
B
x
O
For any position of R, we have

r  a  t AB
t is called a parameter and can have any real value.
It is a scalar not a vector.
R2
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