基础工程数学：第五版 - 约翰·伯德

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"Basic Engineering Mathematics, Fifth Edition" 是一本由 John Bird 编写的工程数学教材，适用于10至12年级的学生。这本书以其独特的教学方法和丰富的练习题著称，旨在帮助不同能力层次的工程学学生，特别是对理论数学感到困难的学生。第五版在满足GCSE工程学、文凭课程和BTEC First规格的基础上，还涵盖了BTEC国家和A Level工程课程的必要数学概念。本书的特点在于其渐进式教学法，通过大量详尽的示例和练习题（包括750个已解问题和1550个额外问题）来逐步引导学生掌握工程数学的基础知识。书中的14个复习测试则提供了自我评估和巩固学习的机会。所有复习测试的完整答案都可以在配套的教师网站上找到。作者John Bird是一位在数学和工程领域有着深厚背景的专业人士，他的著作考虑到了非数学专业学生的需要，起始于低于GCSE水平的起点，使学生能够逐步适应并理解工程数学的复杂概念。这使得这本书成为各级别工程学学生的重要参考资料，无论他们是处于Level 2还是Level 3的学习阶段。此书的内容涵盖了广泛的工程数学主题，可能包括代数、微积分、几何、概率与统计等，这些都是工程学科中不可或缺的部分。每个主题都以实践为导向，强调数学在实际工程问题中的应用，帮助学生建立将理论应用于实际情境的能力。 “Basic Engineering Mathematics, Fifth Edition”是一部全面而实用的教材，不仅提供了丰富的学习材料，而且注重培养学生的数学思维和问题解决技能，是工程学入门学生和需要强化数学基础的高级学生不可或缺的工具书。

Basic arithmetic 3

50 38

120

110

Figure 1.1

1.3 Revision of multiplication and

division

You can probably already multiply two numbers

together and divide one number by another. However, if

you need a revision then the followingworked probl ems

should be helpful.

Pr oblem 7. Determine 86 ×7

HTU

× 7

602

(i) 7 ×6 = 42. Place the 2 in the unit s (U) column

and ‘carry’ the 4 int o the tens (T) column.

(ii) 7 ×8 = 56;56 +4 (carried) = 60. Place the 0 in

the tens column and the 6 in the hundreds (H)

column.

Hence, 86 ×7 =602

A good grasp of multiplication tables is needed when

multiplying such numbers; a reminder of the multipli-

cation table up to 12 ×12 is shown below. Conﬁdence

with handling numbers will be greatly improved if this

table is memorized.

Pr oblem 8. Determine 764 ×38

764

× 38

6 112

2 2 920

2 9 032

Multiplication table

× 2 3 4 5 6 7 8 9 10 11 12

2 4 6 8 10 12 14 16 18 20 22 24

3 6 9 12 15 18 21 24 27 30 33 36

4 8 12 16 20 24 28 32 36 40 44 48

5 10 15 20 25 30 35 40 45 50 55 60

6 12 18 24 30 36 42 48 54 60 66 72

7 14 21 28 35 42 49 56 63 70 77 84

8 16 24 32 40 48 56 64 72 80 88 96

9 18 27 36 45 54 63 72 81 90 99 108

10 20 30 40 50 60 70 80 90 100 110 120

11 22 33 44 55 66 77 88 99 110 121 132

12 24 36 48 60 72 84 96 108 120 132 144

4 Basic Engineering Mathematics

(i) 8 ×4 =32. Place the 2 in the units column and

carry 3 int o the t ens column.

(ii) 8 ×6 =48;48 +3 (carried) = 51. Place the 1 in

the tens column and carry the 5 into the hundreds

column.

(iii) 8 ×7 =56;56 +5 (carried) = 61. Place 1 in the

hundreds column and 6 in the t housands column.

(iv) Place 0 i n the unit s column under t he 2.

(v) 3 ×4 = 12. Place the 2 in the tens column and

carry 1 into the hundreds column.

(vi) 3 ×6 = 18;18 +1 (carried) = 19. Place the 9 in

the hundreds column and carry t he 1 into the

thousands column.

(vii) 3 ×7 = 21;21 +1 (carried) = 22. Place 2 in the

thousands column and 2 in the ten thousands

column.

(viii) 6112 +22920 = 29032

Hence, 764 ×38 = 29032

Again, knowing multiplication tables is rather important

when multiplying such numbers.

It is appreciated, of course, that such a multiplication

can,and probably will, beperformed usinga calculator.

However, there are times when a calculator may not be

available and it is then useful to be able to calculate the

‘long way’.

Pr oblem 9. Multiply 178 by −46

When the numbers have different signs, the result will

be negative. (With this i n mind, the problem can now

be solved by multiplying 178 by 46). Following the

procedure of Problem 8 gives

178

× 46

1068

7120

8188

Thus, 178 ×46 =8188 and 178 ×(−46) =−8188

Problem 10. Determine 1834 ÷7

262



1834

(i) 7 into 18 goes 2, remainder 4. Place the 2 above

the 8 of 1834 and carry the 4 remainder to the

next digit on the right, making i t 43.

(ii) 7 into 43 goes 6, remainder 1. Place the 6 above

the 3 of 1834 and carry the 1 remainder to the

next digit on the right, making i t 14.

(iii) 7 int o 14 goes 2, remainder 0. Place 2 above the

4 of 1834.

Hence, 1834 ÷7 = 1834/7 =

1834

= 262.

The method shown is called short division.

Problem 11. Determine 5796 ÷12

483



5796

(i) 12 into 5 won’t go. 12 int o 57 goes 4; place 4

above the 7 of 5796.

(ii) 4 ×12 = 48; place the 48 below t he 57 of 5796.

(iii) 57 −48 = 9.

(iv) Bring down the 9 of 5796 to give 99.

(v) 12 into 99 goes 8; place 8 above the 9 of 5796.

(vi) 8 ×12 = 96; place 96 below the 99.

(vii) 99 −96 = 3.

(viii) Bring down the 6 of 5796 to give 36.

(ix) 12 int o 36 goes 3 exactly.

(x) Place the 3 above the ﬁnal 6.

(xi) Place the 36 below the 36.

(xii) 36 −36 = 0.

Hence, 5796 ÷ 12 = 5796/12 =

5796

= 483.

The method shown is called long division.

Basic arithmetic 5

Now try the following Practice Exercise

Prac tice Exercise 2 Further problems on

multiplication and division (answers on

page 340)

Determine the values of the expressions given in

problems 1 to 9, wit hout using a calculator.

1. (a) 78 × 6 (b) 124 × 7

2. (a) £261 × 7 (b) £462 × 9

3. (a) 783 kg × 11 (b) 73 kg × 8

4. (a) 27 mm × 13 (b) 77 mm × 12

5. (a) 448 × 23 (b) 143 ×(−31)

6. (a) 288 m ÷ 6 (b) 979 m ÷ 11

(a)

1813

(b)

896

(a)

21424

(b) 15900 ÷ 15

(a)

88737

(b) 46858 ÷ 14

10. A screw has a mass of 15 grams. Calculate,

in kilograms, the mass of 1200 such screws

(1 kg = 1000 g).

1.4 Highest common factors and

lowest common multiples

When two or more numbers are multiplied together, the

individualnumbers are called factors. Thus, a factor is a

number which divides into another number exactly. The

highest common factor (HCF) is the largest number

which divides into two or more numbers exactly.

For example, consider the numbers 12 and 15.

The factors of 12 are 1, 2, 3, 4, 6 and 12 (i.e. all the

numbers that divide into 12).

The factors of 15 are 1, 3, 5 and 15 (i.e. all the numbers

that divide into 15).

1and3aretheonlycommon factors; i.e., numbers

which are factors of both 12 and 15.

Hence, the HCF of 12 and 15 is 3 since 3 is the highest

number which divides into both 12 and 15.

A multiple i s a number which contains another number

an exact number of times. The smallest number which

is exactly divisible by each of two or more numbers is

called the lowest common multiple (LCM).

For example, the multiples of 12 are 12, 24, 36, 48,

60, 72,... and the multiples of 15 are 15, 30, 45,

60, 75,...

60 is a common multiple (i.e. a multiple of both 12 and

15) and there are no lower common multiples.

Hence, the LCM of 12 and 15 i s 60 since 60 is the

lowest number t hat both 12 and 15 divide into.

Here are some further problems involving t he determi-

nation of HCFs and LCMs.

Problem 12. Determine the HCF of t he numbers

12, 30 and 42

Probably the simplest way of determining an HCF is to

express each number in terms of its lowest factors. This

is achieved by repeatedly dividing by the prime numbers

2, 3, 5, 7, 11, 13, … (where possible) in turn. Thus,

12 = 2 × 2 × 3

30 = 2 × 3 × 5

42 = 2 × 3 × 7

The factors which are common to each of the numbers

are 2 in column 1 and 3 in column 3, shown by the

broken lines. Hence, the HCF is 2 × 3; i.e., 6.Thatis,

6 i s the largest number which will divide into 12, 30

and 42.

Problem 13. Determine the HCF of t he numbers

30, 105, 210 and 1155

Using t he method shown in Problem 12:

30 = 2 × 3 × 5

105 = 3 × 5 × 7

210 = 2 × 3 × 5 × 7

1155 = 3 × 5 × 7 × 11

The factors which are common to each of the numbers

are 3 in column 2 and 5 in column 3. Hence, the HCF

is 3 × 5=15.

Problem 14. Determine the LCM of the numbers

12, 42 and 90

6 Basic Engineering Mathematics

The LCM is obtained by ﬁnding the lowest factors of

each of the numbers, as shown in Problems 12 and 13

above, and then selecting the largest group of any of the

factors present. Thus,

12 = 2 ×2 × 3

42 = 2 × 3 × 7

90 = 2 × 3 ×3 × 5

The largest group of any of t he factors present is shown

by the broken lines and are 2 ×2 in 12, 3 ×3 in 90, 5 in

90 and 7 in 42.

Hence, the LCM is 2 ×2 ×3 ×3 ×5 ×7 = 1260 and

is t he smallest number which 12, 42 and 90 will all

divide into exactly.

Problem 15. Determine the LCM of t he numbers

150, 210, 735 and 1365

Using t he method shown in Problem 14 above:

150 = 2 × 3 × 5 ×5

210 = 2 × 3 × 5 × 7

735 = 3 × 5 × 7 ×7

1365 = 3 × 5 × 7 × 13

Hence, the LCM is 2 × 3 × 5 × 5 × 7 × 7 × 13 =

95550.

Now try the following Practice Exercise

Prac tice Exercise 3 Further problems on

highest common factors and lowest common

multiples (answers on page 340)

Find (a) t he HCF and (b) the LCM of the following

groups of numbers.

1. 8, 12 2. 60, 72

3. 50, 70 4. 270, 900

5. 6, 10, 14 6. 12, 30, 45

7. 10, 15, 70, 105 8. 90, 105, 300

9. 196, 210, 462, 910 10. 196, 350, 770

1.5 Order of precedence and brackets

1.5.1 Order of precedence

Sometimes addition, subtraction, multiplication, divi-

sion, powers and brackets may all be inv olved in a

calculation. For example,

5 −3 ×4 +24 ÷(3 +5 ) −3

This is an extreme example but will demonstrate the

order that is necessary when ev aluating.

When we read, we read from left to right. However,

with mathematics there is a deﬁnite order of precedence

which we need to adhere to . The order is as fol lows:

Brackets

Order (or pOwer)

Division

Multiplication

Addition

Subtraction

Notice that the ﬁrst letters of each word spell BOD-

MAS, a handy aide-m ´emoire. Order means pOwer. For

example, 4

= 4 ×4 = 16.

5 − 3 × 4 + 24 ÷ (3 + 5) − 3

is evaluated as

follows:

5 −3 ×4 +24 ÷(3 +5) −3

= 5 −3 ×4 +24 ÷8 −3

(Bracket is removed and

3 +5 replaced with 8)

= 5 −3 ×4 +24 ÷8 −9(Order means pOwer; in

this case, 3

=3 ×3 = 9)

= 5 −3 ×4 +3 −9(Division: 24 ÷8 = 3)

= 5 −12 +3 −9(Multiplication: −3 ×4 =−12)

= 8 −12 −9(Addition: 5 +3 = 8)

=−13 (Subtraction: 8 −12 −9 =−13)

In practice, it does not matter if multiplication is per-

formed before division or if subtraction is performed

bef ore addition. What is i mportant is t hat the pro-

cess of multiplication and division must be completed

bef ore addition and subtraction.

1.5.2 Brack ets and oper ator s

The basic laws governing the use of brackets and

operators are shown by the following examples.

Basic arithmetic 7

(a) 2 +3 = 3 +2; i .e., the order of numbers when

adding does not matter.

(b) 2 ×3 = 3 ×2; i.e., the order of numbers when

multiplying does not matter.

when adding does not affect the result.

(d) 2 ×(3 ×4) = (2 ×3) ×4; i.e., the use of brackets

when multiplying does not affect the result.

(e) 2 ×(3 +4) = 2(3 +4) = 2 ×3 +2 ×4; i .e., a

number placed outside of a bracket indicates

that the whole contents of the bracket must be

multiplied by that number.

(f ) (2 +3 )(4 +5) = (5)(9) = 5 ×9 =45; i .e., adja-

cent brackets indicate multiplication.

(g) 2[3 +(4 ×5

)] = 2[3 +20] = 2 ×23 = 46; i.e.,

when an expression contains inner and outer

brackets, the inner brackets are removed

ﬁrst.

Here are some further problems in which BODMAS

needs to be used.

Problem 16. Find the value of 6 +4 ÷(5 −3)

The order of precedence of operations is remembered

by the word BODMAS. Thus,

6 +4 ÷(5 −3) = 6 +4 ÷2(Brackets)

= 6 +2(Division)

= 8 (Addition)

Problem 17. Determine the value of

13 −2 ×3 +14 ÷(2 +5)

13 −2 ×3 +14 ÷(2 +5) = 13 −2 ×3 +14 ÷7(B)

= 13 −2 ×3 +2(D)

= 13 −6 +2(M)

= 15 −6(A)

= 9 (S)

Problem 18. Evaluate

16 ÷(2 +6) +18[3 +(4 ×6) −21]

=16 ÷(2 +6) +18[3 +24 −21] (B: inner bracket

is determined ﬁrst)

=16 ÷8 +18 ×6(B)

=2 +18 ×6(D)

=2 +108 (M)

=110 (A)

Note that a number outside of a bracket multiplies all

that is inside the brackets. In this case,

18[3 +24 −21] = 18[6], which means 18 ×6 =108

Problem 19. Find the value of

23 −4(2 ×7) +

(144 ÷4)

(14 −8)

23 −4(2 ×7) +

(144 ÷4)

(14 −8)

=23 −4 ×14 +

(B)

=23 −4 ×14 +6(D)

=23 −56 +6(M)

=29 −56 (A)

=−27 (S)

Problem 20. Evaluate

3 +





−3



1 +(4 ×6) ÷(3 ×4)

15 ÷3 +2 ×7 −1

3 ×

√

4 +8 −3

3 +





−3



1 +(4 ×6) ÷(3 ×4)

15 ÷3 +2 ×7 −1

3 ×

√

4 +8 −3

3 +4 +8

1 +24 ÷12

15 ÷3 +2 ×7 −1

3 ×2 +8 −9 +1

3 +4 +8

1 +2

5 +2 ×7 −1

3 ×2 +8 −9 +1

5 +14 −1

6 +8 −9 +1

= 5 +

= 5 +3 = 8

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