Download General Iterative Processes

Grade 7
General Iterative
Processes
Work with general iterative processes.
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Lesson Plan
Lesson Overview
Objective(s)
Work with general iterative processes
Grade
Prior Knowledge
Rearranging equations
Substitution
Duration
Provided prior knowledge (in particular rearranging) is secure content can be taught with practice time within 70 minutes.
Resources
Print slides: 4, 6, 9, 13, 16, 19
Equipment
7
Calculator
Progression of Learning
What are the students learning?
How are the students learning? (Activities & Differentiation)
Rules of substitution
Give students slide 4 printed. This is a recap of substitution. Make sure students are confident with the rules
and are able to complete without a calculator. Show solutions on slide 5 – students can self mark their work.
10
How the iterative process can be used to solve an equation – where
the solution exists
Give students slide 6 printed. Explain how the solution can be seen graphically. Once you have some limits
can substitute into the equation to find more accurately to a specified number of decimal places.
10
Work with general iterative processes
Give students slide 9. Allow students to start independently. Applying knowledge from lesson so far to
attempt answering the questions. Ensure that students are confident using the associated vocabulary and
can rearrange efficiently and accurately. Use slides 10 -12 to go through the answers with students.
20
Work with general iterative processes in contextualised problems
Give students slide 13. Allow students to attempt question on their own for 2 minutes. Review question
together and model answer.
10
Calculating arc lengths and areas of sectors in exam questions (from
specimen papers)
Give students slide 16 and 19. This includes 5 exam questions related to objective. Students need to use
notes from lesson to answer the questions. Ensure that all steps are shown. Relate to mark scheme to show
how the marks are allocated.
20
Next Steps
Assessment
PLC/Reformed Specification/Target 7/Ratio, Proportion & Rates of Change/General Iterative
Processes
Key Vocabulary
Bisection
Iterative
Processes
Y axis
X axis
Solution
Equation
Function
f=8
Work out 2f + 7
g = -2
h=4
Work out 3g + 5h
a=4
b = -5
Work out the value 2a + 3b
Substitution
This formula is used for working out the cost, £C, of
repairing a car.
n is the number of hours worked
C = nL + 1.2P L is the labour rate (£)
P is the cost of parts (£)
(a) Work out the cost of repairing a car when
n=3
L = 18
P = 110
(b) Complete this table for another repair
x=3
Work out 4x2
Student Sheet 1
C
£235
n
L
P
£22
£150
Substitution Solutions
Iterative Process to solve an equation.
The solutions of the equation formed when the function is equal to 0.
1) x2 + 6x + 2 = 0
The solution is
where the graph
crosses the x
axis and y = 0.
Here we can see it crosses the x axis between
0 and -1.
Therefore the solution to the equation must
also lie between 0 and -1.
To find the solution to 1 decimal place:
X value
Student Sheet 2
Y value
Solution =
Substitute into y = x2 + 6x + 2, x values between the limits 0 and -1.
How to use the Iterative Process to solve an
equation.
The solutions of the equation formed when the function
is equal to 0.
The solution is
where the graph
1) y = x2 + 6x + 2
crosses the x
axis and y = 0.
Here we can see it crosses
the x axis between
0 and -1.
Therefore the solution to
the equation must also lie
between 0 and -1.
How to use the Iterative Process to solve an
equation.
What is the solution to the equation x2 + 6x + 2 = 0 to 1 decimal
place?
Substitute into y = x2 + 6x + 2, x values between the limits 0 and -1.
X value
Y value
-0.1
1.41
-0.2
0.84
-0.3
0.29
-0.4
-0.24
The solution is -0.4
The sign has changed from positive
to negative so we know that the
solution lies between -0.3 and -0.4
Iterative Form
Show that x3 + 7x - 100 = 0 can be written as:
𝑥𝑛+1 =
3
100 − 7𝑥𝑛
Use the iteration 𝑥𝑛+1 =
3
100 − 7𝑥𝑛
to find the solution to x3 + 7x - 100 = 0 correct to 2dp.
Use a starting value of x0= 4
Xn
Xn+1
4
4.1601
Solution:
Re-arrange these equations to make xn+1 the subject
x3 + 5x – 10 only has one solution, show that this solution lies
between 1<x<1.5
x3 + 3x - 1 = 0
2x3 + 4x = 0
x3 - x + 3 = 0
3x3 + 9x - 27 = 0
Student Sheet 3
x3 - 2x – 8 only has one solution, show that this solution lies
between 2<x<2.5
How to re-arrange an equation so that it can
be written in an iterative form.
Show that x3 + 7x - 100 = 0 can be written as:
This means the
next term.
𝑥𝑛+1 =
3
This is the
starting term.
100 − 7𝑥𝑛
x3 + 7x - 100 = 0
x3 = 100 - 7x
3
x = 100 − 7𝑥
Cube root
both sides
-7x + 100
To both sides
𝑥𝑛+1 =
3
100 − 7𝑥𝑛
One hundred minus seven multiplied by
the starting term, all cube rooted, is equal
to the next term.
Using iterations to find a solution
Use the iteration 𝑥𝑛+1 =
correct to 2dp.
3
100 − 7𝑥𝑛 to find the solution to x3 + 7x - 100 = 0
Use a starting value of x0= 4
Xn
Xn+1
4
4.1602
4.1601
4.1385
4.1385
4.141415
Start with one value and
keep using the outcomes as
your next starting value.
Therefore 4.14 2dp is the solution to x3 + 7x - 100
These round to
the same
answer to 2dp.
(4.14). Hence
this is the
solution.
Using iterations to find a solution –
Now you try …
1)
Re-arrange these equations to make xn+1 the subject
a) x3 + 3x - 1 = 0
𝒙𝒏+𝟏 =
𝟑
𝟏 − 𝟑𝒙𝒏
b) 2x3 + 4x = 0
𝒙𝒏+𝟏 =
𝟑
−𝟐𝒙𝒏
x3 - x + 3 = 0
𝒙𝒏+𝟏 =
𝟑
𝒙𝒏 − 𝟑
d) 3x3 + 9x - 27 = 0 𝒙𝒏+𝟏 =
𝟑
𝟗 − 𝟑𝒙𝒏
c)
2) x3 + 5x – 10 only has one solution, show that this solution lies between 1<x<1.5
Substitute x = 1 to get, -4
Substitute x = 1.5 to get, 0.875
As this is a negative and positive value we know that the solution lies between 1 and 1.5
3) x3 - 2x – 8 only has one solution, show that this solution lies between 2<x<2.5
Substitute x = 2 to get, -4
Substitute x = 2.5 to get, 2.625
As this is a negative and positive value we know that the solution lies between 2 and 2.5
Problem Solving and Reasoning
(a) Show that x3 + 5x – 50 = 0 can be written as: 𝑥𝑛+1 =
(b) Use the iteration 𝑥𝑛+1 =
3
3
50 − 5𝑥𝑛
50 − 5𝑥𝑛 to find the solution to x3 + 5x – 50 = 0 correct to 2dp. Use a starting value of with x0 = 3.
(c) How would x2 + x –2 = 0 be re- arranged into its iteration formula?
(d) How would x3 + 2x2 + 5x + 10 = 0 be re- arranged into its iteration formula?
(e) What do xn+1 and xn mean?
(f) Why do we want to look out for a sign switch (positive to negative) during the iterative process?
Student Sheet 4
Problem Solving and Reasoning
1)
a) Show that x3 + 5x – 50 = 0 can be written as: 𝑥𝑛+1 = 3 50 − 5𝑥𝑛
x3 = 50 - 5x
𝟑
𝒙 = 𝟓𝟎 − 𝟓𝒙
Therefore: 𝒙𝒏+𝟏 = 𝟑 𝟓𝟎 − 𝟓𝒙𝒏
b) Use the iteration 𝑥𝑛+1 = 3 50 − 5𝑥𝑛 to find the solution to x3 + 5x – 50 = 0
correct to 2dp.
Use a starting value of with x0 = 3.
Xn
Xn+1
3
3.271
3.271
3.228
3.228
3.235
3.235
3.234
3.234
3.234
Substitute in our xn
values in the rearranged formula.
The solution is 3.23 (2dp)
Reason and explain
 How would x2 + x –2 = 0 be re- arranged into its iteration
formula?
𝒙
= 𝟐−𝒙
𝒏+𝟏
𝟐
𝒏
 How would x3 + 2x2 + 5x + 10 = 0 be re- arranged into its
iteration formula?
𝒙𝒏+𝟏 =
𝟑
−𝟏𝟎 − 𝟓𝒙𝒏 − 𝟐𝒙𝒏 𝟐
 What do xn+1 and xn mean?
𝑵𝒆𝒙𝒕 𝒕𝒆𝒓𝒎
Current 𝒕𝒆𝒓𝒎
 Why do we want to look out for a sign switch (positive to
negative) during the iterative process?
𝑻𝒉𝒊𝒔 𝒊𝒔 𝒘𝒉𝒆𝒓𝒆 𝒕𝒉𝒆 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒍𝒊𝒆𝒔
Exam Questions – Specimen Papers - 1
Student Sheet 5
Exam Questions – Specimen Papers
Exam Questions – Specimen Papers
Exam Questions – Specimen Papers - 2
Student Sheet 6
Exam Questions – Specimen Papers
Exam Questions – Specimen Papers
Exam Questions – Specimen Papers