Engineering Math 4th ed [Instructors Manual/Solutions]

Engineering Mathematics

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ENGINEERING MATHEMATICS 4 TH EDITION INSTRUCTOR’S MANUAL WORKED SOLUTIONS TO THE ASSIGNMENTS JOHN BIRD
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INTRODUCTION In ‘ENGINEERING MATHEMATICS 4 TH EDITION’ there are 61 chapters; each chapter contains numerous fully worked problems and further problems with answers. In addition, there are 16 Assignments at regular intervals within the text. These Assignments do not have answers given since it is envisaged that lecturers could set the Assignments for students to attempt as part of there course structure. The worked solutions to the Assignments are contained in this instructor’s manual and with each is a full suggested marking scheme. As a photocopiable resource the main formulae are also included.
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CONTENTS Page ASSIGNMENT 1 (chapters 1 to 4) 1 ASSIGNMENT 2 (chapters 5 to 8) 8 ASSIGNMENT 3 (chapters 9 to 12) 14 ASSIGNMENT 4 (chapters 13 to 16) 20 ASSIGNMENT 5 (chapters 17 to 20) 26 ASSIGNMENT 6 (chapters 21 to 23) 35 ASSIGNMENT 7 (chapters 24 to 26) 42 ASSIGNMENT 8 (chapters 27 to 31) 47 ASSIGNMENT 9 (chapters 32 to 35) 56 ASSIGNMENT 10 (chapters 36 to 39) 60 ASSIGNMENT 11 (chapters 40 to 43) 68 ASSIGNMENT 12 (chapters 44 to 46) 75 ASSIGNMENT 13 (chapters 47 to 49) 80 ASSIGNMENT 14 (chapters 50 to 53) 84 ASSIGNMENT 15 (chapters 54 to 58) 90 ASSIGNMENT 16 (chapters 59 to 61) 97 LIST OF FORMULAE 103 ASSIGNMENT 1 (Page 33)
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This assignment covers the material contained in Chapters 1 to 4. Problem 1. Simplify (a) 2 2 3 ÷ 3 1 3 (b) 1 4 7 2 1 4 × ÷ 1 3 1 5 + + 2 7 24 Marks (a) 2 2 3 ÷ 3 1 3 = 8 3 10 3 ÷ = 8 3 3 10 × = 8 10 = 4 5 4 (b) 1 4 7 2 1 4 × ÷ 1 3 1 5 + + 2 7 24 = 1 4 7 9 4 5 3 15 2 7 24 × ÷ + ⎟ + 2 = 1 9 7 8 15 2 7 24 ÷ + = 7 9 15 8 2 7 24 × + 1 = 35 24 2 7 24 + = 1 11 24 2 7 24 + = 3 18 24 = 3 3 4 2 total : 9 Problem 2. A piece of steel, 1.69 m long, is cut into three pieces in the ratio 2 to 5 to 6. Determine, in centimetres, the lengths of the three pieces. Marks Number of parts = 2 + 5 + 6 = 13 Length of one part = 169 13 . m = 169 13 cm = 13 cm 1 Hence 2 parts 2 × 13 = 26 5 parts 5 × 13 = 65 6 parts 6 × 13 = 78 i.e. 2 : 5 : 6 :: 26 cm : 65 cm : 78 cm 3 total : 4 1 Problem 3. Evaluate 57629 193 . . (a) correct to 4 significant figures (b) correct to 1 decimal place
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Marks 57629 193 . . = 29.859585... by calculator Hence (a) 57629 193 . . = 29.86 , correct to 4 significant figures 1 (b) 57629 193 . . = 29.9 , correct to 1 decimal place 1 total : 2 Problem 4. Determine, correct to 1 decimal place, 57% of 17.64 g. Marks 57% of 17.64 g = 57 100 1764 × . g = 10.1 g, correct to 1 decimal place 2 total : 2 Problem 5. Express 54.7 mm as a percentage of 1.15 m, correct to 3 significant figures. Marks 54.7 mm as a percentage of 1.15 m is: 547 1150 100% . × = 4.76% , correct to 3 significant figures 3 total : 3 Problem 6. Evaluate the following: (a) 2 2 2 2 3 2 4 × × (b) ( ) ( ) 2 16 8 2 3 2 3 × × (c) 1 4 2 1 (d) (27) 1 3 (e) 3 2 2 9 2 3 2 2 2 Marks (a) 2 2 2 2 3 2 4 × × = 2 3 1 = 2 = 4 2 2 4 + + 2
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(b) ( ) ( ) 2 16 8 2 3 2 3 × × = ( ) ( ) 2 2 2 2 3 4 2 3 3 × × = ( ) ( ) 2 2 7 2 4 3 = 2 2 14 12 = 2 14 = 2 = 4 3 12 2 (c) 1 4 2 1 = (4 ) = 4 2 = 16 3 2 + 1 (d) (27) 1 3 = 1 27 1 3 = 1 27 3 = 1 3 3 (e) 3 2 2 9 2 3 2 2
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