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Grade 7 General Iterative Processes Work with general iterative processes. If you have any questions regarding these resources or come across any errors, please contact [email protected] 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