Reeds Marine Engineering and Technology Volume 1: Mathematics for Marine Engineers 9781472987518, 9781408176122
This indispensible guide to ship stability covers topics such as flotation and buoyancy, small angle, large angle and lo
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PREFACE This book provides the mathematical knowledge that is needed to pursue a career as a Merchant Navy Engineering Officer. The UK system of certification utilises Higher Education programmes to meet the academic knowledge required for the operational and management level Certificates of Competency. This book covers the fundamental topics of mathematics (algebra, trigonometry and calculus) in detail and also gives an introduction to higher-level topics (Laplace Transforms and Fourier Series). It will be an invaluable aid to all students of technicalbased courses in which mathematics and analytical methods play an important part. Basic principles are introduced from an elementary stage to facilitate self-study. Each chapter includes fully worked examples interwoven into the text, as well as test examples set at the end of each chapter. This will ensure that readers can monitor the progress they are making in this study of mathematics. Each chapter concludes with typical examination questions. Non-programmable calculators may be used and a comprehensive set of formulae are provided. The author has gone beyond the normal practice of merely supplying bare answers to the test examples and examination questions by providing fully worked step-by-step solutions leading to the final answers. It is this sequential learning method that will enable readers to understand the topics and apply them to other analytical situations in all disciplines of engineering, including marine engineering. This new edition is a major update in the subject and learning methods, and brings the material for study completely up to date.
1
INDICES AND LOGARITHMS Indices A prime number can only be divided by two factors, 1 and itself. Any number can be written as a multiplication of its prime factors. Example 200 = 2 × 2 × 2 × 5 × 5 These factors can be written as 23 × 52 , where 2 and 5 are called bases and the numbers 3 and 2 are called indices. Whole number indices are called powers. When an integer is written without an index, the power is 1. Example: 6 = 61 . The reciprocal of a number means 1 divided by the number and has the power −1. 1 So the reciprocal of 5 is 5−1 and has a value of , or 0.2. 5 If the base number is a fraction its reciprocal is simply the fraction turned upside down (or inverted). Example
−1 −1 3 5 5 3 = and, naturally, = 3 5 5 3
A quotient means the division of two quantities so, as in the above example, the 3 quotient of 3 and 5 is = 0. 6. 5 A product means the multiplication of two (or more) quantities.
2
•
Mathematics
Behaviour of Indices Important: The following rules apply only to terms having the same base. Consider quantities with the base ‘a’. Example (a) a3 × a2 = (a × a × a) × (a × a) = a5 a×a×a×a×a×a = a4 (b) a6 ÷ a2 = a×a 3 (c) a2 = a2 × a2 × a2 = a(2+2+2) = a6
So a3 × a2 = a(3+2) = a5 So a6 ÷ a2 = a(6−2) = a4 3 So a2 = a(2×3) = a6
These three statements are examples of the three laws of indices: •
am × an = a(m+n)
•
am ÷ an = a(m−n)
•
(am )n = amn
These laws can be used to find other identities. Example (d) ab ÷ ab = a(b−b) = a0 but ab ÷ ab = 1, as any value divided by the same value must be equal to 1 So a0 = 1 So: any value (numerical or algebraic) to the power zero equals 1. − (e) ab × a−b = a(b+( b)) = a0 = 1
Divide both sides by ab gives a−b =
1 ab
So: a negative power means 1 divided by the positive power (and vice versa) (f ) A root of a value (square, cube, . . . ) can also be written as a power. √ Let a = ap . √ 2 Since a = a1 it follows that (ap )2 = a1 . The third law means that (ap )2 = a2p . 1 So a2p = a1 so 2p = 1 ⇒ p = 2 √ Therefore a = a1/2
Indices and Logarithms
•
3
√ Similarly 3 a = a1/3 etc. If the power is a fraction it means the following – the denominator is the type of root and the numerator is the power. b √ b c a c = c ab or a (g) The square root of a number has an index of 12 , so the square root of 16 can be √ written as 16 or 161/2 . √ √ As 4 × 4 = 16 then 16 = 4, but (−4) × (−4) = 16 so 16 = −4. Therefore every square root has two possible values, one positive, the other √ negative, so 16 = ±4. This is the case for any even root. Example (h) Find the value of 53 × 55 ÷ 54 53 × 55 ÷ 54 = 53 × 55 ÷ 54 = 5(3+5) ÷ 54 = 58 ÷ 54 = 5(8−4) = 54 = 625 or 53 × 55 ÷ 54 = 5(3+5−4) = 54 = 625 2 (i) Find the value of 53 ÷ 54 3 2 5 ÷ 54 = 5(3×2) ÷ 54 = 56 ÷ 54 = 5(6−4) = 52 = 25 2 or 53 ÷ 54 = 5(3×2−4) = 52 = 25 3 3 5 55 × 56 3 3 5 5(3×3) 59 1 1 = = = 5(9−11) = 5−2 = 2 = 55 × 56 511 5 25 5(5+6)
(j) Find the value of
(k) Evaluate 642/3 64 /3 = 2
√ 2 √ √ 3 3 3 2 64 = 42 = 16 or 64 /3 = 642 = 4096 = 16
Obviously the first way is much easier! These rules can be combined to give some useful tips. 2 6 Consider the following: 2 2 2 62 36 6 6 =9 Now = 32 = 9, but = 2 = 2 2 2 4
4
•
Mathematics
This means that ‘the power of a quotient gives the same result as the quotient of the powers’. Since a power can also represent the root of a number the same rule applies. 36 Consider the following: 4 √ 6 36 √ 36 36 = 9 = ±3, but = √ = ± = ±3 Now 4 4 2 4 A root of a number that does not have an exact value is called a surd. For example, √ √ 2, 5. Applying powers to the product of terms gives the same type of result. √ √ √ √ √ √ So 12 = 4 × 3 = 4 × 3 = ± 2 × 3 = ±2 3 8 Consider the following ratio: √ 12 Since
√ √ 8 4 8 √ = ±√ 12 = ±2 3, it follows that √ = ±2 3 3 12
Although this is numerically correct, it is not considered to be written correctly as roots are never left as terms on the denominator. The problem is removed by multiplying numerator and denominator by an appropriate root. √ √ √ 4 4 3 4 4 3 3 √ √ √ Therefore ± =± × = ± √ 2 = ± 3 3 3 3 3 Similarly
√ √ 3 3 2 3 2 √ =√ ×√ = 2 2 2 2
Questions Attempt the following questions, giving the answer in index form: 1.
44 × 42
2.
23 × 22 × 2−4
3.
66 ÷ 62
4.
5 2 3
5.
23 × 24 22
6.
9.
7. 10.
33 × 9
3
(27 × 3)
4
24 × 2−3 2
8. 11.
(16 × 32)3 (8 × 16)
5
52 × 53 × 5−4 53 × 50 × 5−2
12.
23 21 × 2−3
37 × 33 −2 2 2 −3 2 × 2 5 −1 4 3−4
Indices and Logarithms
•
5
Answers 1.
46
2.
21
3.
64
4.
310
5.
25
6.
25
7.
3−1
8.
2−8
9.
38
10.
20
11.
50
12.
2−20
Logarithms If N is a number such that N = ax then x is the logarithm of N to the base a. This is written as x = loga N Example: 1000 = 103 . Therefore log10 1000 = 3.
Questions In each of the following find the value of x:
1.
logx 9 = 2
2.
logx 81 = 4
3.
log2 16 = x
4.
log5 125 = x
5.
log3 x = 2
6.
log4 x = 3
7.
log10 x = 2
8.
log7 x = 0
9.
logx 8 = 3
10.
logx 27 = 3
Answers 1.
3
2.
3
3.
4
4.
3
5.
9
6.
64
7.
100
8.
1
9.
2
10.
3
The usual base of a logarithm is 10, where the abbreviation used is log. This is called a common logarithm. Another base is the number 2. 718 · · · (given the symbol ‘e’): this is called the natural (Naperian) logarithm and is abbreviated as ln.
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•
Mathematics
Rules for the use of logarithms 1. The logarithm of two numbers multiplied together is given by the sum of the individual logarithms: log(xy) = log(x) + log(y) 2. The logarithm of the division of two numbers is given by subtracting the individual logarithms: log(x ÷ y) = log(x) − log(y) 3. The logarithm of a number raised to a power is given by the logarithm of the number multiplied by the power: log(x y ) = y · log(x)
Relation between logarithms of different bases logb (N) =
loga (N) loga (b)
Example Find the logarithm of 8 to base 2 using common logs. log2 (8) =
0. 9030 log (8) = =3 log (2) 0. 3010
This is obviously correct as 23 = 8.
Questions Without using a calculator, find the value of: 1.
log 8 + log 12. 5
2.
log 30 − log 3
3.
log2 28 − log2 7
4.
log5 12. 5 + log5 10
5.
2 log 2 + log 75 − log 3
6.
log 25 ÷ log 5
7.
log 8 ÷ log 2
8.
log 27 ÷ log 3
9.
3 log 2 + log 200 − log 16
10.
log 30 − log 3
Indices and Logarithms
11.
log2 16 − log2 8
12.
log3 2. 7 + log3 10
13.
log 4 + 2 log 5
14.
log 32 ÷ log 2
15.
log 7 + 2 log 5 − log 1. 75
16.
log 18 + log 5 − log 9
17.
log6 42 − log6 7
18.
log 3 + log 15 − log 4. 5
19.
log5 40 − log5 1. 6
20.
log8 72 − log8 1. 125
•
7
Answers 1.
2
2.
1
3.
2
4.
3
5.
2
6.
2
7.
3
8.
3
9.
2
10.
1
11.
1
12.
3
13.
2
14.
5
15.
2
16.
1
17.
1
18.
1
19.
2
20.
2
Test Examples 1 Questions 1 to 5 to be answered without the use of a calculator: 103/2 × 104 × 103/4 103 × 101/4 × 102 2 42 × 4 /3 × 4−2 Evaluate 41/2 × 41/6 8−2 Find the value of −5 8 2/5 5/3 √ 3 × 3×3 Evaluate 3−4/3 × 33 2 √ 2 4 + 200 Find the value of 1 + 3 81
1. Find the value of 2. 3. 4. 5.
For help with the next questions refer to Appendix 1. 6. The ratio of the volumes of two spheres is equal to the ratio of the cubes of their diameters; the volume of one sphere is 24.25 cm3 , find the volume of another sphere whose diameter is twice as much. 7. A pump can empty a tank in 12 h, another pump can empty the same tank in 4 h and a third can empty this tank in 9 h. If all three pumps are set to work together on emptying the tank, how long would it take to empty it?
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•
Mathematics
8. The strength of a beam varies directly as its breadth, directly as the square of its depth and inversely as its length. A beam is 5 m long, 40 mm broad and 100 mm deep, find the breadth of another beam of similar material, 3 m long and 80 mm deep, to have equal strength. 9. A piece of mild steel 50 mm long between gauge points and 80 mm2 crosssectional area, was tested in a tensile testing machine and broke when the gauge length was 62.5 mm and cross-sectional area 48 mm2 . Find the percentage elongation in length and the percentage reduction in cross-sectional area. 10. In a certain three cylinder engine, the power developed in No. 1 cylinder is 15% more than in No. 3, and 5% less power is developed in No. 2 than in No. 3. What percentage of the total engine power is developed in each cylinder? Give the answers correct to one decimal place. 11. A brass casting is composed of 71% copper, 1% tin, 3% lead and the remainder zinc. Find the mass required of each constituent to make 500 kg of this alloy. 12. The heights of an indicator diagram measured at regular intervals along its length are as follows: 27, 39, 47, 51, 48, 32, 20, 11, 8 and 5 mm respectively. Find the mean height of the diagram in mm. 13. 200 tonnes of oil were bought in one port $600/tonne and 600 tonnes of oil at another port at $700/tonne. What is the average cost of the oil per tonne?
2
ALGEBRA
Algebra is a convenient system of notation in which letters or symbols are used to represent quantities, instead of words. Example Suppose a box contained 25 bolts, 34 nuts and 47 washers. To the contents 15 bolts and 18 washers were added but 14 nuts were removed. This could be set down as follows: 25b + 34n + 47w add 15b − 14n + 18w result 40b + 20n + 65w Example The area of a rectangle is found by its length multiplied by its width. If the area is A, the length L and the width W, then A = L × W. If the length and width are subtracted then the answer is L − W, if they are added L + W L and if they are divided L ÷ W or . W Note: the ‘×’ symbol is often ignored so L × W becomes LW (or L · W) When a quantity, y, is multiplied by a number, for example 3, the answer is written 3y. If a quantity appears by itself, for example k, it means one times k. If a combination of letters and numbers are multiplied together the order does not matter so d4e is the same as ed4 and so forth. However, it is usual to write the number first and the letters alphabetical afterwards, i.e. 4de.
10
•
Mathematics
An expression which consists of two terms is called a binomial expression, and one consisting of three terms is called a trinomial expression.
Simplification of Algebraic Expressions Note: Different symbols can be combined by multiplication or division, but cannot be combined by addition or subtraction.
Addition of terms Example x + 3x = 4x but c + d stays as c + d. Example Add 2a + 7b + 3c and 6c + 5a − 2b 2a + 7b + 3c Add 5a − 2b + 6c 7a + 5b + 9c Example Add 3x − 5y + 6z and 4x + 2y − 8z 3x − 5y + 6z Add 4x + 2y − 8z 7x − 3y − 2z
Subtraction of terms Example From 8x subtract 5x 8x − 5x = 3x
Algebra
•
11
From 8x subtract −5x 8x − (−5x) = 8x + 5x = 13x From −8x subtract 5x −8x − 5x = −13x From −8x subtract −5x −8x − (−5x) = −8x + 5x = −3x Example Subtract 6c − 3b + 4a − 7d from 5b + 3d + 2a − 9c 4a − 3b + 6c − 7d
subtract
2a + 5b − 9c + 3d 2a − 8b + 15c − 10d However, 2abc + 5bac = 7abc as each term has the same set of letters although the order differs. m2 n and mn2 are different as the first term is m × m × n and the second is m × n × n.
Questions Simplify, where possible:
1.
p+q
2.
a + 3a
3.
d + 7d − 3d
4.
s + 5s − 3t
5.
6y − (−5y)
6.
c×d
7.
2x × 5x
8.
6s × 2t
9.
q × 7q3
10.
6xy ÷ 3
11.
9x 2 y ÷ 3x
12.
j − (−3k)
13.
16q − 8qt
14.
2a2 b − 6a2
15.
nm + n + 2m + 2
16.
6 + 2d − cd − 3c
17.
m2 n − mn2
12
•
Mathematics
Answers 1.
p+q
2.
4a
3.
5d
4.
6s − 3t
5.
11y
6.
cd
7.
10x 2
8.
12st
9.
7q4
10.
2xy
11.
3xy
12.
j + 3k
13.
8q(2 − t)
14.
2a2 (b − 3)
15.
(n + 2)(m + 1)
16.
(3 + d)(2 − c)
17.
mn(m − n)
An algebraic expression surrounded by a bracket with a number in front means all the terms inside the bracket are to be multiplied by the number outside the bracket. Example 3 (4s + 5t) = 3 × 4s + 3 × 5t = 12s + 15t One very important rule is necessary when removing brackets. If there is a negative sign in front of the bracket, ALL the signs inside the bracket change when the bracket is removed. Example 24x − 3(6y + 4x) = 24x − 18y − 12x = 12x − 18y
SIGN CHANGE
Just as brackets can be removed from an expression, they can equally well be inserted.
Method •
Find out the common terms that can be divided out from the expression. Set this outside the brackets and place inside the brackets whatever is left when the factor is removed.
•
It is often useful to put 1 in front of a term that appears by itself. Example, x = 1x, a2 b = 1a2 b . . .
Example 18xy + 12y = 6y(3x + 2) as 6y divides exactly into 18xy and 12y leaving 3x and 2 respectively.
Algebra
•
13
Questions Multiply out the brackets and collect like terms:
1.
4x(a + b + 2c)
4.
x x2 + x + 2
7.
u2 2u3 + 5u2 − 4u − 3
10.
a (a − m) − m (a + m)
2.
3m(x − 3y + 2z)
3.
−2a (p − q − r + s)
5.
−3m m2 − 2m + 4
6.
−2a 3a2 + 4a − 2
8.
3d2 d2 + 4d + 1
9.
−4m2 m2 − 2m − 3
12.
b (2b − 3a) + x (4b − x)
15.
2a a2 + 3a + 3a2 (a − 4)
11.
m (m + 2u) − u (2m − u)
13.
2c (c + 2d) − 3d (c − 2d)
14.
x2 (3x − 1)− 2x x2 − x + 2
16.
5mn (m − 3n)−2mn (m + 4n)
17.
3u u2 − v2 + 2uv (u − 2v)
18.
3a2 (2a − b) − ab(a + 2b) + b2 (2a − 3b)
19.
4x2 (x − 3y) + 2xy (6x + 3y) − 3y2 (2x − 3y)
20.
4x2 (x − 3y) − 2xy (6x + 3y) + 3y2 (2x − 3y)
Answers 1.
4ax + 4bx + 8cx
2.
3mx − 9my + 6mz
3.
−2ap + 2aq + 2ar − 2as
4.
x3 + x2 + 2x
5.
−3m3 + 6m2 − 12m
6.
−6a3 − 8a2 + 4a
7.
2u5 + 5u4 − 4u3 − 3u2
8.
3d4 + 12d3 + 3d2
9.
−4m4 + 8m3 + 12m2
10.
a2 − 2am − m2
11.
m 2 + u2
12.
2b2 − 3ab + 4bx − x2
13.
2c2 + cd + 6d2
14.
x3 + x2 − 4x
15.
5a3 − 6a2
16.
3m2 n − 23mn2
17.
3u3 + 2u2 v − 7uv 2
18.
6a3 − 4a2 b − 3b3
19.
4x3 + 9y3
20.
4x3 − 24x2 y − 9y3
Questions Factorise the following: 1.
ab + ac
2.
5x + 25
3.
b2 − bc
4.
4b2 − 2bc
5.
6ab − 12a
6.
8a2 b − 32ab
7.
5a3 b2 c − 10a2 b2 c − 15abc2
8.
6ab2 c + 9abc + 15a2 bc3
14
•
Mathematics
Answers 1.
a(b + c)
2.
5(x + 5)
3.
b(b − c)
4.
2b(2b − c) 5abc a2 b − 2ab − 3c
5.
6a(b − 2) 3abc 2b + 3 + 5ac2
6.
8ab(a − 4)
7.
8.
When terms are divided, division can only take place when the term on the denominator (bottom line) is part of every term on the numerator (top line). Example Divide 9a3 b2 + 12a2 b3 − 18a5 b2
by
3a2 b2
Solution Each term has to be divided by 3a2 b2 so: 9a3 b2 + 12a2 b3 − 18a5 b2 9a3 b2 12a2 b3 18a5 b2 = 2 2+ − = 3a + 4b − 6a3 3a2 b2 3a b 3a2 b2 3a2 b2
Questions Divide
1.
4a7 + 2a5 − 4a3 by −2a2
2.
3b3 m − 12b2 m + 6bm by 3bm
3.
4u3 v − 8u2 v2 + 16uv3 by 4uv
4.
6m4 n − 2m3 n2 − 8m2 n3 + 4mn4 by −2mn
5.
−3x6 − 9x5 + 6x4 − 3x3 by −3x2
6.
−8a4 u2 + 4a3 u3 − 12a2 u4 by 4a2 u
7.
6p4 q3 − 9p3 q2 − 3p2 q3 + 6p3 q4 by −3pq2
8.
−8h3 k3 − 6h2 k2 + 2h2 k4 − 4h4 k2 by −2h2 k2
9.
3x5 y2 − 9x4 y3 + 3x3 y4 − 6x2 y5 −3x2 y2
10.
4a3 m2 u2 + 6am3 u2 − 2a3 m2 u − 4am2 u3 2am2 u
Answers 1.
−2a5 − a3 + 2a
2.
b2 − 4b + 2
3.
u2 − 2uv + 4v 2
4.
−3m3 + m2 n + 4mn2 − 2n3
5.
x 4 + 3x 3 − 2x 2 + x
6.
−2a2 u + au2 − 3u
Algebra
7.
−2p3 q + 3p2 + pq − 2p2q2
9.
−x 3 + 3x 2 y − xy2 + 2y3
8. 10.
•
15
4hk + 3 − k2 + 2h2 2a2 u + 3mu − a2 − 2u2
Highest Common Factor (HCF) and Lowest Common Multiple (LCM) Example Find the HCF of 8ab2 c3 , 6a2 b2 and 10ab2 c2 8ab2 c3 = 2 × 2 × 2 × a × b × b × c × c × c 6a2 b2 = 2 × 3 × a × a × b × b 10ab2 c2 = 2 × 5 × a × b × b × c × c × c The biggest collection of factors that is contained in all three expressions is 2×a×b ×b. So the HCF is 2ab2 .
Questions Find the HCF of: 1.
amxy and bmdx
2.
9jhk and 6ghk
3.
mn2 and m2 n
4.
8d2 e and 6def
5.
10km2 n3 and 15k3 m2 n
6.
24a2 b4 and 16a3 b3
7.
uvw, vwx and wxy
8.
8ab3 , 2a2 b2 and 6a3 b
9.
3c4 d3 , 9c2 d5 and 7c3 d4
10.
12ax 3 y2 , 4a3 y4 and 8a2 xy3
11.
15abm2 n, 6cmp2 b and 12b2 dmp
12.
4f 3 gh3 , 8g2 h4 k3 and fg3 h2 k
13.
18a2 bm2 n3 , 24a3 b2 m2 and 12a2 mn2
14.
16uv 2m3 n2 , 32u3 mn3 and 24v 3 m2 n2
15.
16a4 b3 x 3 , 24b2 m3 x 4 y and 20a2 b3 nx 3
16
•
Mathematics
Answers 5km2 n
1.
mx
2.
3hk
3.
mn
4.
2de
5.
6.
8a2 b3
7.
w
8.
2ab
9.
c2 d 3
10.
4ay2
11.
3bm
12.
14.
8mn2
15.
4b2 x 3
gh2
13.
6a2 m
Example Find the LCM of 4a2 bc, 6ac and 8bc2 4a2 bc = 2 × 2 × a × a × b × c 6ac = 2 × 3 × a × c 8bc2 = 2 × 2 × 2 × b × c × c The smallest collection of factors that contains all the factors of the three expressions is: 2×2×2×3×a×a×b×c×c So the LCM is 24a2 bc2 .
Questions Find the LCM of: 1.
mnu and muv
2.
3abm and 2am2
3.
ab, bc and cd
4.
4x 2 y and 6xy3
5.
9u2 mn and 6vm2 n
6.
3a2 b, abc and 6b2 c
7.
m2 n2 , 5m and 10n
8.
3a2 b5 , 6a3 c2 and 12b3 c
9.
3f 2 g, 6g2 h and 4h2 k
10.
8pqr3 , 6p2 qs and 12q3 rs2
11.
a2 bm3 n, ab2 m and abn2
12.
2cdu2 , 3d2 uv and 6c2 uv 2
13.
9c2 d3 , 3bcd2 e and 6bde
14.
6am2 x, 4anx 2 and 8am2 nx 2
15.
3a2 uv 4 , a4 mu2 v 3 and 4a3 m2 v 3
Algebra
•
17
Answers 1.
mnuv
2.
6abm2
3.
abcd
4.
12x2 y3
6.
6a2 b2 c
7.
10m2 n2
8.
12a3 b5 c2
9.
13.
18bc2 d3 e
14.
11.
a2 b2 m3 n2
12.
6c2 d2 u2 v2
5.
18m2 nu2 v
12f 2 g2 h2 k
10.
24p2 q3 r3 s2
24am2 nx2
15.
12a4 m2 u2 v4
Algebraic Fractions Example What is the value of ? if
bd ? = 2 2 2ac 4a bc
4a2 bc2 is obtained by multiplying 2ac by some quantity, call it Q So the numerator bd must also be multiplied by Q to give ? It follows that Q =
4a2 bc2 = 2abc. 2ac
So ? = bd × 2abc = 2ab2 cd. To add or subtract algebraic fractions, bring them all to a common denominator by finding the LCM of the denominators of the given fractions, then add or subtract, like terms being combined. Example Simplify
3x + 1 x + 2 2x − 3 − − 2x 3x 6x
The LCM of the denominators is 6x. For the first fraction, the denominator 2x has to be multiplied by 3 to make the LCM of 6x, and so the numerator also has to be multiplied by 3 to give 3(3x + 1). For the second fraction, the denominator 3x has to be multiplied by 2 to make the LCM of 6x, and so the numerator also has to be multiplied by 2 to give 2(x + 2). For the third fraction, the denominator is already 6x, so the numerator stays as (2x − 3). So
3x + 1 x + 2 2x − 3 3 (3x + 1) 2 (x + 2) (2x − 3) − − = − − 2x 3x 6x 6x 6x 6x 3 (3x + 1) − 2 (x + 2) − (2x − 3) 9x + 3 − 2x − 4 − 2x + 3 5x + 2 = = = 6x 6x 6x
18
•
Mathematics
3x + 1 x + 2 2x − 3 − − contains no brackets. The reason for 2x 3x 6x this is that the division lines showing the fraction act as a double purpose. First they indicate that the numerator is divided by the denominator but second, they show that everything above (or below) the line is one complete term.
Note: the given expression
So
3x + 1 (3x + 1) could be written as 2x 2x
Example Simplify
1 3 + 3x 5y
The LCM of 3x and 5y is 15xy: So
3 1 × 5y + 3 × 3x 5y + 9x 1 + = = 3x 5y 15xy 15xy
Example Simplify
y x + yz xz2
The LCM of yz and xz2 is xyz2 So
y x x × xz + y × y x 2 z + y2 + 2= = yz xz xyz2 xyz2
Example Simplify
2 +5 n 2 2 n 2 5n 2 +5= +5×1= +5× = + =n n n n n n n
Example Simplify
p +3 qr p p qr p 3qr p + 3qr p +3= +3×1= +3× = + = qr qr qr qr qr qr qr
Algebra
•
Example Simplify
a+b b+c + ab bc
a + b b + c (a + b) c + (b + c) a (ac + bc + ba + ca) 2ac + ab + bc + = = = ab bc abc abc abc
Questions Find the expression for the ? in the following:
? 5a = 3b 12bc
1. 4.
? a4 b2 c
=
3d a3 b2
2.
5ab a = bc ?
3.
a 2ab = ? 7bc
5.
? 2a + b = a 3ab
6.
? a + 2b = 5ab 15a2 b
Simplify as far as possible:
7.
2 3 + 4a 3a
8.
3 2 − ab bc
9.
ab +3 c
10.
2 3 − ab bc
11.
3 1 2 + − a b ab
12.
3+a 1 − 2 a
13.
a + b a + 2b 2a + b + + 2 3 4
14.
2 1 + +3 ab bc
15.
a 1 c − − ab bc abc
Answers 1.
20ac
2.
5b2 c
3.
14b2 c
4.
3acd
5.
3b(2a + b)
6.
3a(a + 2b)
7.
17 12a
8.
2c − 3a abc
9.
ab + 3c c
10.
3c − 2a abc
11.
b + 2a − 3 ab
12.
a2 + 3x − 2 2a
13.
16a + 17b 12
14.
c + 2a + 3abc abc
15.
c2 − a2 − 1 abc
19
20
•
Mathematics
Expansions and Factors Expansion of (x + 3y)(x + 2y) means that each term in the first bracket has to be multiplied by each term in the second. So far only the removal of brackets has taken place when only one term exists outside the bracket, for example: 3(5x + 2). However, it is very commonplace to have two (or more) terms outside the bracket, for example: (x + 3y)(x + 2y). This is called a binomial expression because there are two terms in each bracket. It is best to follow the same procedure each time two sets of brackets have to be expanded.
Method First times first, first times last, last times first, last times last. Example Expand (x + 3y)(x + 2y) 3 2
(x + 3y)(x + 2y) = x 2 + 2xy + 3yx + 6y2 1
1
4
2
3
4
or x 2 + 5xy + 6y2
Questions Expand the following and collect like terms: 1.
(x + 1)(x + 3)
2.
(x + 2)(x + 3)
3.
(2x + 1)(x − 1)
4.
(2x + 2)(2x − 1)
5.
(x − 4)(x − 1)
6.
(3x − 7)(4x − 3)
7.
(x + 6)(5x − 6)
8.
(x − 3)(4x + 2)
9.
(4x − 1)(x + 3)
10.
(3x − 6)(3x + 5)
11.
(4x + y)(x − 2y)
12.
(2x + 5y)(2x − 5y)
Algebra
•
21
Answers 1.
x 2 + 4x + 3
2.
x 2 + 5x + 6
3.
2x 2 − x − 1
4.
4x 2 + 2x − 2
5.
x 2 − 5x + 4
6.
12x 2 − 37x + 21
7.
5x 2 + 24x − 36
8.
4x 2 − 10x − 6
9.
4x 2 + 11x − 3
10.
9x 2 − 3x − 30
11.
4x 2 − 7xy − 2y2
12.
4x 2 − 25y2
With practice it will become unnecessary to write down all the working, and the answer can be written down at once. Example, in the expansion (x − 2y) (x − 3y) = x2 − 3xy − 2xy + 6y2 the terms. −3xy and −2xy may be combined mentally and the answer x 2 − 5xy + 6y2 can be written down at once.
Polynomial Division A polynomial is an expression of the form f (x) = a + bx + cx 2 + dx 3 + · · · . The method of dividing polynomials is best shown by looking at worked examples. Example Divide 3x 2 + x − 4 by x − 1 3x 2 + x − 4 is called the dividend and x − 1 the divisor. The usual layout is shown below with both the dividend and divisor shown in descending order of powers.
3x + 4
x − 1 3x 2 + x − 4 3x 2 − 3x 4x − 4 4x − 4 0
22
So
•
Mathematics
3x 2 + x − 4 = 3x + 4 x−1
Reasoning → Dividing the first term of the dividend by the first term of the divisor. 3x 2 gives 3x, which is put above the dividend as shown. → So x → The divisor is then multiplied by 3x, i.e., 3x(x − 1) = 3x 2 − 3x, which is then placed under the dividend as shown. → Subtracting gives 4x and the −4 is ‘dropped down’ to give 4x − 4. → The process is then repeated, i.e., the first term of the divisor, x, is divided into +4x, giving +4, which is placed above the dividend as shown. → Then +4(x − 1) = +4x − 34 which is placed under the 4x − 4. → The remainder on subtraction is 0, which completes the division. → Therefore 3x 2 + x − 4 ÷ (x − 1) = (3x + 4) A check can be made by multiplying x − 1 by 3x + 4.
Example Divide x 2 + 5x + 4 by x + 4
x+1
x + 4 x 2 + 5x + 4 x 2 + 4x x+4 x+4 0
So
x 2 + 5x + 4 =x+1 x+4
Algebra
•
23
Example Divide 6x 2 − 11x − 12 by 3x + 2 2x − 5
3x + 2 6x 2 − 11x − 12 6x 2 + 4x −15x − 12 −15x − 10 −2
So
2 6x 2 − 11x − 12 = 2x − 5 − 3x + 2 3x + 2
Questions Divide 1.
a2 + 6a + 8
by
a+2
2.
b2 − 5b + 6
by
b−3
3.
c2 + 2c − 8
by
c−2
4.
d2 − 2d − 15
by
d+3
5.
e2 + 2e − 9
by
e+4
6.
3f 2 + 5f − 2
by
f +2
7.
2g2 − 13g + 17
by
g−5
8.
3h2 + 14h + 12
by
h+3
9.
j2 + 3jk + 2k2
by
j+k
10.
3m2 + 7mn − 6n2
by
3m − 2n
10p2 − 2pq − 10q2
by
5p − 6q
12.
9r2 − 27rs + 24s2
by
3r − 5s
11.
Answers 1.
a+4
2.
b−2
4.
d−5
5.
e−2−
7.
2g − 3 +
8.
3h + 5 −
11.
2p + 2q +
10.
m + 3n
2 g−5
2 e+4 3 h+3 2q2 5p − 6q
3.
c+4
6.
3f − 6 +
9.
j + 2k
12.
10 f +2
3r − 4s +
4s2 3r − 5s
24
•
Mathematics
Factorisation Since 3 × 7 = 21, 3 and 7 are said to be the factors of 21. Similarly, (x − 3) (x + 4) = x 2 + x − 12, so (x − 3) and (x + 4) are said to be the factors of x 2 + x − 12. An algebraic expression will not always have factors, for example x 2 + x + 1 does not factorise. Example Factorise x 2 + 5x + 6 The idea is to write x 2 + 5x + 6 = (
)(
)
The first term in the expression is x 2 which can only be x multiplied by x, so each bracket begins with x, i.e., (x )(x ) The next term to consider is +6. As this is a positive value it means that both the numbers are positive or both the numbers are negative. The sign in front of the ‘x’ term determines which is the correct sign. In this case, both numbers are positive. So (x+ )(x+ ) The two value are those that multiply to give 6 but add to give 5, i.e., 2 and 3. The answer is therefore (x + 2) (x + 3) Example Factorise x 2 − 11x + 10 Again the first term in each bracket is x. The last number is +10 so the sign of the ‘x’ term means that the factorisation is of the form (x− )(x− ). The required numbers multiply to give 10 and add to give 11, i.e., 1 and 10. So x 2 − 11x + 10 = (x − 1) (x − 10) Example Factorise x 2 − 4x − 12 The first term in each bracket is x.
Algebra
•
25
As the last number is negative, it can only be the multiple of one positive and one negative number, giving (x− )(x+ ). The required numbers multiply to give −12 but add to give −4, i.e., +2 and −6 So x 2 − 4x − 12 = (x − 6) (x + 2) Example Multiply out the brackets: (x + y) (x − y) (x + y) (x − y) = x 2 − xy + xy − y 2 = x 2 − y2 This means that x 2 − y2 factorises as (x + y) (x − y) This is known as the difference of two squares That is, the factors are (the square root of the first term plus the square root of the second) and (the square root of the first term minus the square root of the second) Example Factorise 4a2 − 9b2 √ √ 4a2 = 2a and 9b2 = 3b, so 4a2 − 9b2 = (2a + 3b) (2a − 3b)
Notice that •
When the number term is positive, the two numbers in the brackets are of the same sign, and their total is the x coefficient;
•
When the sign of the x term is negative, the two numbers are of opposite signs, so that the coefficient of x is their difference.
Questions Factorise the following where possible. If there are no factors say so.
1.
x 2 + 4x + 3
2.
x 2 + 3x + 2
3.
x 2 + 13x + 12
4.
x 2 − 3x + 2
5.
x 2 − 7x + 12
6.
x2 − x − 6
7.
x 2 + 3x − 10
8.
x 2 + 2x − 3
9.
x2 + x + 1
10.
x 2 + 5x − 14
11.
x 2 + 3x − 2
12.
x 2 + 7x + 12
26
•
Mathematics 13.
x 2 + x − 12
14.
x 2 − 10x + 21
15.
x 2 + 5x − 6
16.
x 2 − 9x + 20
17.
x 2 + 6x + 9
18.
x 2 − 8x + 12
19.
x 2 − 25
20.
x 2 + 25
Answers 1.
(x + 1)(x + 3)
2.
(x + 2)(x + 1)
3.
(x + 12)(x + 1)
4.
(x − 2)(x − 1)
5.
(x − 3)(x − 4)
6.
(x − 3)(x + 2)
7.
(x + 5)(x − 2)
8.
(x − 2)(x − 1)
9.
Cannot factorise
10.
(x − 2)(x + 7)
11.
Cannot factorise
12.
(x + 3)(x + 4)
13.
(x + 4)(x − 3)
14.
(x − 3) (x − 7)
15.
(x + 6)(x − 1)
16.
(x − 5)(x − 4)
17.
(x + 3)2
18.
(x − 2) (x − 6)
19.
(x + 5)(x − 5)
20.
Cannot factorise
Example Expand x 2 − 3x − 4 (2x − 5) 2 x − 3x − 4 (2x − 5) = 2x 3 − 6x 2 − 8x − 5x 2 + 15x + 20 = 2x 3 − 11x 2 + 7x + 20
Example Simplify 2 (2x − y)2 − (3x + 2y) (x − 2y) − 4y 2 2 (2x − y)2 − (3x + 2y) (x − 2y) − 4y 2 = 2 (2x − y) (2x − y) − (3x + 2y) (x − 2y) − 4y 2 = 2 4x 2 − 4xy + y2 − 3x 2 − 4xy − 4y2 − 4y2 = 8x 2 − 8xy + 2y2 − 3x 2 + 4xy + 4y2 − 4y2 = 5x 2 − 4xy + 2y2
Algebra
•
27
Questions Simplify
1.
(x + 3) (x − 4) + (x − 1) (x + 5)
2.
(x + 6)(x − 4) − (x − 2)(x + 1)
3.
(2x + 3) (x + 1) − 2x (x − 1)
4.
3x (2x − 3) − (3x − 1) (2x + 1)
5.
(3x − 1)2 + (2x − 1)2
6.
(2x + y)2 − (x − 2y)2
7.
(3x + 1) (x + 3) − 3 (x + 1) (x − 4)
8.
5xy + (3x − 2y) (3x + 2y) − (2x + 3y) (3x − 2y)
9.
(x − 1) (x + 1) − (x + 2) (x − 2) + (x + 5) (x − 5)
10.
2 (2x + y) (x − 2y) − (x + y) (x − y) − (2x − y)2
Answers 1.
2x 2 + 3x − 17
2.
3x − 22
3.
7x + 3
4.
1 − 10x
5.
13x 2 − 10x + 2
6.
3x 2 + 8xy − 3y2
7.
19x+ 15
8.
3x 2 + 2y2
9.
x 2 − 22
10.
−x 2 − 2xy − 4y2
Evaluation Evaluation is the process of substituting the numerical values of the algebraic symbols and working out the value of the whole expression. The usual rules of arithmetic apply (BIDMAS is one way of remembering the order of operations). Quantities in Brackets should be solved first, followed by Indices (powers and roots). Next Division and Multiplication must be calculated before Addition and Subtraction. Example Evaluate 3xy + x 2 − 4y when x = 2 and y = 3 3xy + x 2 − 4y = 3 × 2 × 3 + 22 − 4 × 3 = 18 + 4 − 12 = 10
28
•
Mathematics
Example Evaluate (a + b)2 − (c + d) + x 3 − y if a = 3, b = 4, c = 5, d = 6, x = −2, y = −3 (a + b)2 − (c + d) + x 3 − y = (3 + 4)2 − (5 + 6) − (−2)3 − (−3) = 72 − 11 + (−8) − (−3) = 49 − 11 − 8 + 3 = 33
This is much more relevant when the equations used are those that may occur in other units.
Questions 1.
E=
w × v2 2g
Find E if w = 120, v = 33 and g = 32.
2.
p=
2Li t
Find p if L = 230, i = 362 and t = 12. 6.
3.
A=
πd2 4
Find A if d = 71.
4.
S = 4πR2
Find S if R = 18.
4 V = πR3 3
Find V if π = 3. 142 and R = 19.
T = 2π
Find T if π = 3. 142, l = 21. 7 and g = 9. 81.
5.
6.
7.
R=
3
8.
N=
9.
10.
d=
3
l g
3V π
Find R if π = 3. 142 and V = 1046.
36VR 100TP
Find N if V = 85, R = 19, T = 14. 6 and P = 11. 53.
16T πf
Find d if π = 3. 142, T = 136 and f = 52. 8.
A = 2πr + πr2
Find A if π = 3. 142 and r = 5. 3.
Algebra
•
29
Answers 1.
4083.75
2.
13215.9
3.
3595.2
4.
4071.50
5.
28730.9
6.
9.35
7.
10.0
8.
1.86
9.
2.36
10.
121.56
Questions Force × Distance Time Evaluate the power when a force of 3760N raises an object a distance of 4.73 m in 35 s.
1. Power =
2. The potential difference, V volts, available at battery terminals is given by V = E − i. R. Evaluate V when E = 5. 62, i = 0. 70 and R = 4. 30. m 2 3. Given force F = v − u2 , find F when m = 18. 3, v = 12. 7 and u = 8. 24. 2 nE . 4. The current, i amperes, flowing in a number of cells is given by i = R + nr Evaluate the current when n = 56, E = 2. 20, R = 2. 80 and r = 0. 50. L . 5. The time, t seconds, of oscillation for a simple pendulum is given by t = 2π g Determine the time when L = 54. 32 and g = 9. 81. L. i2 . 6. Energy, E joules, is given by the formula E = 2 Evaluate the energy when L = 5. 5 and i = 1. 2. V 7. The current, I amperes, in an a.c. circuit is given by i = . 2 R + X2 Evaluate the current when V = 250, R = 11. 0 and X = 16. 2. 8. The area, A, of any triangle is given by A = s (s − a) (s − b) (s − c) where a+b+c . s= 2 Evaluate the area when a = 3. 60 cm, b = 4. 00 cm and c = 5. 20 cm. 9. Given that a = 0. 290, b = 14. 86, c = 0. 042, d = 31. 8 and e = 0. 650, evaluate v
a. b d − . given that v = c e
b n 10. Calculate the value of X if X = T 1 + when T = b = n = 3/4. 100
30
•
Mathematics
2 1 11. Calculate Z if Z = R2 + ωL − ωC when R = 24, L = 0. 30, ω = 352 and c = 6. 5 × 10−5 . T 12. Calculate a where a = 2 ku − Lv 2 and T = 17. 42, k = 0. 073, u = 5. 46, L = 0. 0890 and v = 3. 21.
Answers 1. Power = 508.1 W 2. V = 2. 61 v 3. F = 854. 5 4. I = 4 A 5. t = 14. 89 s 6. E = 3. 96 J 7. I = 12. 77 A 8. A = 7. 184 cm2 9. v = 7. 327 10. X = 0. 7542 11. Z = 66. 4 12. a = ±15. 52
Test Examples 2 1.
(i) Add 3x + 4y − 5z and −2x − 5y + 4z (ii) Add 2a2 b − ab + 3ab2 and 5ab2 − a2 b + ab (iii) Subtract 2x + 5y − 3z from 5x − 4z + 3y (iv) Subtract −4c − 5b − a from 3a + 6c − 2b
2. Simplify the following expressions by collecting like terms: (i) 5x − 3z − 4x − 2y + 4y + 2z − y (ii) 2. 5a + c − 1. 2a + 2. 5b − 3c + b (iii) b2 − 3ab2 + 2a2 b − 4a2 − 2b2 + 5a2 − 2ab2
Algebra 3. Simplify the following: a × a3 × a5 (i) a2 × a4 5 (ii) b × b−3 × b−2 √ (iii) c × c1/2 × c−1/3 d 4 e2 (iv) 3 √ 3 (v) f 2 × f 1/2 × f −5 4.
(i) Multiply (x + 2y) by (2x + y) (ii) Multiply (2x + y) by (3x − 2y) (iii) Multiply (3x − 4y) by (2x − 3y)
5. Work out the following: (i) (a + b)2 (ii) (a − b)2 (iii) (a + b)3 (iv) (a − b)3 6.
(i) Divide 8a2 − 8ab − 6b2 by (2a − 3b) (ii) Divide 9x 3 − 9x 2 y − 10xy2 + 8y3 by (3x − 4y) (iii) Divide x 3 − y3 by (x − y)
7. Simplify the following expressions by removing brackets and collecting terms: (i) (a + b) + (c − d) − (a − b) − (c + d) + (a − b) (ii) 2 {a − 3 (a + 2) + 4 (2a − 1) + 5} (iii) 2x − [2x − {2x − (2x − 2) − 2} − 2] − 2 8. Factorise the following expressions: (i) 3b2 − 6b + 9 (ii) pv + pvx (iii) ax 3 − 2bx 2 + 3cx (iv) 12a3 b3 c3 − 8a2 b2 c2 + 4abc 9. Factorise the following expressions: (i) D2 − d2 (ii) 1 − a2
•
31
32
•
Mathematics
(iii) 4x 2 y2 − 9z2 (iv) T14 − T24 10. Factorise the following expressions: (i) a2 + 8a + 16 (ii) d2 − 10d + 25 (iii) 9v 2 + 12v + 4 (iv) 4x 2 − 12xy + 9y2 11. Simplify the following expressions: x x 2x + − (i) 2 4 3 2a − 1 2a + 3 3a − 8 − − (ii) b 2b 3b 18 3 4 + − (iii) x + 2 x 2 − 2x − 8 x − 4 12. Find the value of: 2x 3 − x 2 y + xy2 − 3y3 when x = 2 and y = −2.
3
SIMPLE EQUATIONS
An equation is an expression consisting of two ‘sides’, one being equal in value to the other. A simple equation contains one hidden value of the first order (e.g. x, and not x 2 , or x 3 , etc.) which is usually referred to as the unknown and to solve the equation means to find the value of the unknown. Since one side of the equation is always equal to the other side, it follows that if, in the process of solving the equation, a change is made to one side of the equation the same change must be made to the other side in order to maintain equality. The most important thing to remember is that the equality of the equation must be maintained. This means if the values on the left-hand side of the equals sign are doubled then the values on the right-hand side must also be doubled. Any mathematical operation can be used, so long as the ‘balance’ of the equation is maintained. So equality will be maintained if (i) the same quantity is added to both sides, (ii) the same quantity is subtracted from both sides, (iii) every term on both sides is multiplied by the same value, (iv) every term on both sides is divided by the same value, (v) both sides are raised to the same power, (vi) the same root is taken of both sides. Methods of approach (i) eliminate fractions by multiplying all the terms by the LCM of the denominators, (ii) remove brackets, following the rules of Algebra,
34
•
Mathematics
(iii) place all the terms that involve the unknown on one side of the equation, and all other terms on the other side of the equation, (iv) collect and summarise terms on each side, (v) find the value of the unknown.
This is a rough guide only. Example Solve the equation 6x = 24 Dividing both sides of the equation by 6 gives
6x 24 = 6 6
Cancelling gives x = 4. Every equation should be checked by substituting the answer into the original equation. In this case, 6 × 4 = 24, so the answer is correct. Example Solve the equation
2x = 12 3
Multiplying both sides by 3 gives
2x × 3 = 12 × 3 3
Cancelling gives 2x = 36 Dividing both sides by 2 gives
2x 36 = 2 2
Cancelling gives x = 18. Check:
2 × 18 36 = = 12. Therefore answer is correct. 3 3
Example Solve the equation x − 8 = 3 Adding 8 to both sides gives x − 8 + 8 = 3 + 8 So, x = 11 Check: 11 − 8 = 3, so correct. Example Solve the equation x + 5 = 9
Simple Equations
•
35
Subtracting 5 from both sides gives x + 5 − 5 = 9 − 5 So, x = 4 Check: 4 + 5 = 9, so correct. Example Solve the equation 4x + 10 = 18 Subtracting 10 from both sides gives 4x = 8 Dividing both sides by 4 gives x = 2 Check: 4 × 2 + 10 = 8 + 10 = 18, so correct. Example Solve the equation 5x + 6 = 3x + 12 In questions like this, all the terms involving the unknown quantity (in this case, x) must be grouped on one side of the equals sign with everything else on the other side. When this is done, changing from one side of an equals sign to the other must be accompanied by a change of sign. 5x + 6 = 3x + 12 5x − 3x = 12 − 6 2x = 6 so x =
6 =3 2
Check: (5 × 3) + 6 = 21, (3 × 3) + 12 = 21, so correct. Example Solve the equation 6 − 3x = 4x − 20 In order to keep the x terms positive, move the x’s to the right-hand side of the equals sign, everything else to the left-hand side. 6 − 3x = 4x − 22 6 + 22 = 4x + 3x 28 = 7x x=
28 =4 7
Check: 6 − 3 × 4 = 6 − 12 = −6, 4 × 4 − 22 = 16 − 22 = −6, so correct.
36
•
Mathematics
Example Solve the equation 5 (x − 2) = 25 Removing the bracket gives 5x − 10 = 25 Rearranging gives 5x = 25 + 10 5x = 35 x=
35 =7 5
Check: 5 (7 − 2) = 5 × 5 = 25, so correct. Alternatively, divide both sides by 5, then add 2 to both sides. Example Solve the equation 3 (2x − 6) + 2 (x + 3) = 4 (x − 5) Removing the brackets gives 6x − 18 + 2x + 6 = 4x − 20 Collecting like terms gives 6x + 2x − 4x = −20 + 18 − 6 Simplifying gives 4x = −8 So x =
−8 = −2 4
Check: 3 (−4 − 6) + 2 (−2 + 3) = 3 × −10 + 2 × 1 = −30 + 2 = −28, 4 (−2 − 5) = 4 × −7 = −28, so correct. Example Solve the equation
3x + 2 5 (x − 1) 5x 5 − x − = − 2 6 4 2
LCM of denominators is 12, so multiply throughout by 12 to give 12
3x + 2 2
− 12
5 (x − 1) 6
= 12
5x 4
− 12
6 (3x + 2) − 2 (5 (x − 1)) = 3 (5x) − 6 (5 − x) (18x + 12) − (10x − 10) = 15x − 6 (5 − x) 18x + 12 − 10x + 10 = 15x − 30 + 6x 8x + 22 = 21x − 30
5−x 2
Simple Equations
•
37
22 + 30 = 21x − 8x 52 = 13x x=
52 13
=4 Example Solve the equation 8 −
9x 6 − = 2 (4 − 3x) 2 x
LCM of denominators is 2x, so multiplying every term by 2x gives
9x 2x (8) − 2x 2
6 − 2x = 2x [2 (4 − 3x)] x
16x − 9x 2 − 12 = 4x (4 − 3x) 16x − 9x 2 − 12 = 16x − 12x 2 12x 2 − 9x 2 = 12 3x 2 = 12 x2 = 4 √ x= 4 x = ±2 Note: the answer can be +2 or −2 as (+2)2 = 4 and (−2)2 = 4. So the square root of a number always has two values, one positive, the other negative; this is written using the symbol ±. There is never just one way to solve an equation. Consider the following: 3x 2 + 10x + 3 x 2 − 4x + 3 = x+3 x−1 The numerators look as though they might factorise and, if this is the case, the denominators may cancel with one of the factors. This is worth trying as the result is a much simpler equation to solve. 3x 2 + 10x + 3 x 2 − 4x + 3 = x+3 x−1
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•
Mathematics (x + 3) (3x + 1) (x − 1) (x − 3) = x+3 x−1 3x + 1 = x − 3 3x − x = −3 − 1 2x = −4 x = −2
Questions Solve the following equations:
1.
2x + 5 = 7
2.
8 − 3x = 2
3.
2 x−1= 3 3
4.
2x − 1 = 5x + 11
5.
7 − 4x = 2x − 3
6.
2. 6x − 1. 3 = 0. 9x + 0. 4
7.
2x + 6 − 5x = 0
8.
3x − 2 − 5x = 2x − 4
9.
20x − 3 + 3x = 11x + 5 − 8
10.
2 (x − 1) = 4
11.
16 = 4 (x + 2)
12.
5 (x − 2) − 3 (2x + 5) + 15 = 0
13.
2x = 4 (x − 3)
14.
6 (2 − 3x) − 42 = −2 (x − 1)
15.
2 (3x − 5) − 5 = 0
16.
4 (3x + 1) = 7 (x + 4) − 2 (x + 5)
17.
10 + 3 (x − 7) = 16 − (x + 2)
18.
8 + 4 (x − 1) − 5 (x − 3) = 2 (5 − 2x)
Answers
1. x = 1
2. x = 2
3. x = 6
7. x = 2
8. x =
9. x = 0
1 2
13. x = 6 14. x = −2 15. x = 2 12
4. x = −4
5. x = 1 23
6. x = 1
10. x = 3
11. x = 2
12. x = −10
16. x = 2
17. x = 6 14
18. x = −3
Simple Equations
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39
Further Examples on Simple Equations Example Solve the equation
3 4 = x 5
The lowest common multiple (LCM) of the denominators is 5x, so both sides of the equation are multiplied by 5x. 4 3 × 5x = × 5x x 5 15 = 4x x= Check:
3 15 =3 4 4
4 12 4 3 =3× = = , so correct. 15/4 15 15 5
Example Solve the equation
2x 3 1 3x + +5= − 5 4 20 2
The LCM of the denominators is 20 so multiplying every term by 20 gives 3 1 3x 2x × 20 + × 20 + 5 × 20 = × 20 − × 20 5 4 20 2 8x + 15 + 100 = 1 − 30x 8x + 30x = 1 − 15 − 100 38x = −114 −114 = −3 38 −6 3 11 1 3 × −3 1 −9 1 9 11 2 × −3 3 + +5 = + +5 = 4 , − = − = + =4 , Check: 5 4 5 4 20 20 2 20 2 20 2 20 so correct. x=
Example Solve the equation
3 4 = x − 2 3x + 4
Cross multiplying gives 3 (3x + 4) = 4 (x − 2) Removing the brackets 9x + 12 = 4x − 8
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Mathematics
So 9x − 4x = −8 − 12 Giving 5x = −20 So x =
−20 = −4 5
Check:
3 3 1 4 4 4 1 = =− , = = = − , so correct. −4 − 2 −6 2 3 × −4 + 4 −12 + 4 −8 2
Example √ Solve the equation 2 x = 8
√ 2 x 8 = Dividing both sides by 2 gives 2 2 √ So x = 4 √ 2 Squaring both sides gives x = 42 So x = 16 √ Check: 2 16 = 2 × 4 = 8, so correct. Example Solve the equation
√ x+3 √ =2 x
To remove the fraction both sides are multiplied by √ √ x+3 √ × x =2× x So √ x √ √ Giving x + 3 = 2 x √ √ √ Rearranging gives 3 = 2 x − x = x Squaring both sides gives x = 9 √ 9+3 3+3 6 Check: √ = = 2, so correct. = 3 3 9 Example Solve the equation
16 2 = 9 3x 2
Multiply both sides by the LCM which is 9x 2 So
2 16 × 9x 2 = 2 × 9x 2 9 3x
√ x
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41
So 2x 2 = 16 × 3 = 48 Dividing by 2 gives x 2 = 24 √ So x = 24 = ±4. 889 to 3 decimal places Check:
16 2 16 16 = = , so correct. = 2 3 × 24 48 9 3 × (4. 889)
Summary of the Important Points 1. The equality of an equation must be maintained at every step. 2. Fractions are best removed. This is done by multiplying both sides by the lowest common multiple of the denominators. 3. Brackets should be removed at the earliest possible opportunity. 4. The unknown terms should be grouped on one side of the equals sign and everything else on the other side. 5. When the square root of the unknown is involved it is best to isolate that term on one side of the equals sign, and then to square both sides. 6. When the square of the unknown is involved it is best to isolate that term on one side of the equals sign, and then to take the square root of both sides. •
When taking the square root there are always two answers, one positive and the other negative.
7. All solutions should be checked. 8. Do not try to take more than one step at a time.
Questions Solve the following equations:
1.
1 x+3=4 5
2.
2 5 3 2+ x =1+ x+ 4 3 6
3.
1 1 (2x − 1) + 3 = 4 2
4.
1 2 1 (2x − 3) + (x − 4) + =0 5 6 15
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Mathematics
5.
1 1 (3x − 6) − (5x + 4) 5 4 1 + (2x − 9) = −3 5 x x x =3+ − 3 3 6
6.
x x − =2 3 5
8.
2 3 = x 8
7.
1−
9.
1 7 1 + = 3x 4x 24
10.
x+3 x−3 = +2 4 5
11.
3x 6 − x 2x 3 = + − 20 12 15 2
12.
7 5−y x + = 5 20 4
13.
3 x−2 = 2x − 3 2x + 1
14.
3 2 = x − 3 2x + 1
15.
x x+6 x+3 − = 4 5 2
16.
√ 3 x=9
17.
√ 2 x=5
19.
√ 3 x √ = −6 1− x
20.
18.
3 +3 4= x 10 = 5
21.
16 =
x2 9
22.
23.
6 2x = x 3
24.
x
−1
2
x+2 x−2
=
1 2
8 11 =5+ 2 2 x
Answers 1. x = 5
2. x = −2
3. x = −4 12
7. x = −4
8. x = 5 13
9. x = 2
4. x = 2
5. x = 12
6. x = 15
10. x = 13
11. x = −10 12. x = 2
13. x = 3
14. x = −11 15. x = −6
16. x = 9
17. x = 6 14
18. x = 3
19. x = 4
20. x = 10
22. x = −3 13
23. x = ±3
24. x = ±4
21. x = ±12
Simple Equations
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Practical Problems Involving Simple Equations Example A copper wire has a length, L, of 1.5 km, a resistance, R, of 5Ω and a resistivity, ρ, of 17. 2 × 10−6 Ω mm Find the cross-sectional area, A, of the wire, given that R =
ρL . A
Solution Since R =
ρL then A
17. 2 × 10−6 Ωmm × 1. 5 × 103 m 17. 2 × 10−6 Ωmm × 1. 5 × 106 mm 5Ω = = A A From these units A is measured in mm2 . So 5A = 17. 2 × 10−6 × 1. 5 × 106 giving 17. 2 × 10−6 × 1. 5 × 106 25. 8 × 100 25. 8 = = = 5. 16 mm2 . A= 5 5 5 Remember 100 = 1 The cross-sectional area A is 5.16 mm2 . Example The temperature coefficient of resistance, α, may be calculated from the formula Rt = R0 (1 + αt) where t is the time in seconds, R0 , is the initial resistance and Rt is the resistance after t seconds. Find α given that Rt = 0. 928, Ro = 0. 8 and t = 40. Solution Since Rt = R0 (1 + αt) then 0. 928 = 0. 8 (1 + α (40)) So
0. 928 = 0. 8 + (0. 8) α (40)
0. 928 − 0. 8 = 32α
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•
Mathematics 0. 128 = 32α α=
0. 128 = 0. 004 32
The temperature coefficient of resistance is 0.004. Example 1 The distance s metres travelled in t seconds is given by the formula s = ut + at2 , where 2 u is the initial velocity in m/s and a is the acceleration in m/s2 . Find the acceleration of the body if it travels 168 m in 6 s, with an initial velocity of 10 m/s. Solution 1 s = ut + at2 where s = 168, u = 10 and t = 6. 2 So
36 1 168 = (10) (6) + a (6)2 → 168 = 60 + a → 168 = 60 + 18a 2 2 108 18a = 168 − 60 → 18a = 108 → a = =6 18
So the acceleration is 6 m/s2 . Example When three springs are joined in series the total stiffness, KT , is given by 1 1 1 1 = + + where K1 is the stiffness of the first spring etc. KT K1 K2 K3 Find the total stiffness if K1 = 2 kN/m, K2 = 3 kN/m and K3 = 5 kN/m. Solution 1 1 1 1 1 1 1 1 = + + so = + + KT K1 K2 K3 KT 2 3 5 The LCM of all the denominators is (KT ) (2) (3) (5) = 30KT so multiplying by the LCM gives KT = 1. 43 kN/m (to 3 significant figures).
Questions ρL . A Given that R = 1. 25, L = 2500 and A = 2 × 10−4 find the values of ρ.
1. A formula used for calculating resistance of a cable is R =
2. Force F newtons is given by F = ma, where m is the mass in kg and a is the acceleration in m/s2 . Find the acceleration when a force of 4 kN is applied to a mass of 500 kg.
Simple Equations
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45
3. PV = Mrt is the characteristic gas equation. Find the value of m when P = 100 × 103 , V = 3. 00, R = 288 and T = 300. 4. When three resistors are connected in parallel the total resistance RT is given by 1 1 1 1 = + + where R1 is the resistance of the first resistor etc. RT R1 R2 R3 (a) Find the total resistance when R1 = 3Ω, R2 = 6Ω and R3 = 18Ω. (b) Find the value of R3 when RT = 3Ω, R1 = 5Ω and R2 = 10Ω. V 5. Ohm’s law may be represented by I = , where I is the current in amperes, V is the R voltage and R is the resistance in ohms. A soldering iron takes a current of 0.30 A from a 240 V supply. Find the resistance of the soldering iron.
Answers 1.
10−7
2.
8 m/s2
3.
3.472
4.
(a) 1.8 Ω,
(b) 30 Ω
5.
800 Ω
Further Problems Example The extension, x metres, of an aluminium tie bar of length L metres and cross-sectional area A m2 when carrying a load of F newtons is given by the modulus of elasticity FL . E= Ax Find the extension of the tie bar (in mm) if E = 70 × 109 N/m2 , F = 20 × 106 N, A = 0. 1 m2 and L = 1. 4 m. Solution E=
20 × 106 N × (1. 4m) FL , , so 70 × 109N/m2 = (0. 1) (x) Ax
(the unit of x is therefore the metre) 70 × 109 × 0. 1 × (x) = 20 × 106 × (1. 4) 20 × 106 × (1. 4) 28 × 106 = = 4 × 10−3 x= 70 × 109 × 0. 1 7 × 109 The extension of the tie bar is 4 × 10−3 m = 4 mm.
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Mathematics
Example Power in a d.c. circuit is given by P = V 2 /R where V is the supply voltage and R is the circuit resistance. Find the supply voltage if the circuit resistance is 1.25 Ω and the power measured is 320 W. Solution Since P =
V2 V2 then 320 = R 1. 25
(320) (1. 25) = V 2 i. e. V 2 = 400 √ V = 400 = ±20 volts Supply voltage = ±20 volts. Example A formula relating initial and final pressures P1 and P2 , volumes V1 and V2 and temP1 · V1 P2 · V2 peratures T1 and T2 of an ideal gas is = . Find the value of P3 given T1 T2 3 P1 = 100 × 10 , V1 = 1. 0, V2 = 0. 266, T1 = 423 and T2 = 293. Solution
100 × 103 · (1. 0) P2 · (0. 266) P1 · V1 P2 · V2 Since = = then T1 T2 423 293 Rearranging gives (100 × 103) · (1. 0) × (293) = P2 · (0. 266) × (423) 100 × 103 · (1. 0) × (293) P2 = (0. 266) × (423) P2 =
29300 × 103 = 260. 4 × 103 = 2. 604 × 105 112. . 518
To 2 significant figures P2 = 2. 6 × 105 or 260000. Example D The stress f in a material of a thick cylinder can be obtained from = d Calculate the stress given that D = 21. 5, d = 10. 75 and p = 1800.
f +p . f −p
Simple Equations
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47
Solution
21. 5 D f +p f + 1800 then = Since = d f −p 10. 75 f − 1800 f + 1800 . i.e. 2 = f − 1800 Squaring both sides gives: 4=
f + 1800 f − 1800
4 (f − 1800) = f + 1800 4f − 7200 = f + 1800 4f − f = 1800 + 7200 3f = 9000 f = 3000 So stress f = 3000.
Questions 1. A rectangle has a length of 20 cm and a width b cm. When its width is reduced by 4 cm its area becomes 160 cm2 . Find the original width and area of the rectangle. 2. Given R2 = R1 (1 + αt), find α given R1 = 5. 0, R2 = 6. 03 and t = 51. 5. 3. If v 2 = u2 + 2as, find u given that v = 24, a = −40 and s = 4. 05. 4. The relationship between the temperature on a Fahrenheit scale and a Celsius scale is F = 95 C + 32. Express 113◦F in degrees Celsius. w , find the value of S given that w = 1. 219, g = 9. 81 and t = 0. 3132. 5. If T = 2π Sg 6. An alloy contains 60% by weight of copper, the remainder being zinc. How much copper must be mixed with 50 kg of alloy to give an alloy containing 75% copper? 7. A rectangular lab has a length equal to one and a half times its width and a perimeter of 40 m. Find its length and width.
Answers 1. 12 cm, 240 cm2 2. 0.004 3. 30 4. 45◦ C 5. 50 6. 30 kg 7. 12 m, 8 m
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Mathematics
Problems Involving Simple Equations From the facts given in the problem and what is required to be found, the first task is to make up a straightforward equation. Make a sketch of the problem if possible and show all the given data on the sketch, indicating the unknown quantity to be found. If a sketch is not practical, it may help to write down the given quantities. It helps to give the unknown quantity a symbol that is representative of the quantity: for example, ‘let t be the time taken, in seconds’, ‘let m be the mass in kg’. Example A port (left) tank and a starboard (right) tank, each of 200 tonne capacity, are each half full with oil. Find what mass of oil must be pumped out of the port tank and into the starboard tank so that there will be four times the mass of oil in one tank as in the other. Port
Starboard M tonne pumped in
Capacity of each tank 200 tonnes Half full 100 tonnes
M tonne taken out 100 – M tonne left
Initial mass in each tank = 100 tonne Let M be the mass of oil pumped over, then Final mass of oil in port tank = 100 − M tonne Final mass in starboard tank = 100 + M tonne
100 + M tonne finally in tank
Simple Equations
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49
Mass in starboard tank is four times that of mass in port tank, so 100 + M = 4 (100 − M) 100 + M = 400 − 4M M + 4M = 400 − 100 5M = 300 M = 60 Therefore, mass to be transferred = 60 tonne. This will leave 40 tonne in the port tank and have the starboard tank holding 160 tonne (four times as much). Example An electric train leaves station A bound for station B at the same instant that a diesel train leaves B for A on parallel lines along the same route. The speed of the electric train is 80 km/h, that of the diesel train 64 km/h and the distance between stations is 72 km. Find the distance from A where the trains will pass each other. Electric train 80 km/h
Passing point C
Diesel train 64 km/h
A
B x km
72 – x km 72 km
Let x = distance of passing place from A, then (72 − x) = distance of passing place from B. As the trains leave at the same instant, the time taken to reach the passing point must be the same for each train. ⎫ ⎧ ⎬ ⎨ Time for electric train Time for diesel train = ⎭ ⎩ to travel x km to travel (72 − x) km Substituting, Time =
Distance for each train Speed
distance travelled by electric train distance travelled by diesel train = speed of electric train speed of diesel train 72 − x x = 80 64 LCM = 320, multiplying both sides by 320
50
•
Mathematics 4x = 5 (72 − x) 4x = 360 − 5x 4x + 5x = 360 9x = 360 x = 40
Therefore the trains pass each other 40 km from A. Example A ship travels upriver against the current for a distance of 25 nautical miles and then downriver with the current over the same distance, taking 3.5 h for the double journey. If the normal speed of the ship in still water is 15 knots, find the speed of the current. Let x = speed of current (knot) speed of ship against current = (15 − x) knot then, speed of ship running with cuirrent = (15 + x) knot Now time to go upriver + time to go downriver = total time So Therefore
distance upriver distance downriver = = 3. 5 h speed upriver speed downriver 25 25 + = 3. 5 (15 − x) (15 + x)
Multiply both sides by the LCM of the denominators, that is, by (15 − x)(15 + x). 25 (15 − x) (15 + x) (15 − x) 25 (15 − x) (15 + x) = 3. 5 [(15 − x) (15 + x)] + (15 + x) ←− 25 (15 + x) + 25 (15 − x) = 3. 5 152 − x 2 ←− {difference of two squares} 375 + 25x + 375 − 25x = 3. 5 152 − x 2 750 = 787. 5 − 3. 5x 2 3. 5x 2 = 787. 5 − 750 3. 5x 2 = 37. 5 37. 5 37. 5 2 and so x = = ±3. 273 This gives x = 3. 5 3. 5 Therefore speed of current is 3.273 knots.
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51
Transposition of Formulae A formula is a group of algebraic symbols which expresses a rule for finding the value of one of the symbols when values are substituted for the others. To transpose means to change the order of the symbols to produce a re-arranged equation giving the expression for one of the other symbols in terms of the rest. The rules for the transposition of formulae are the same as those used for the solution of simple equations. Basically, the equality of the formula must be maintained. In order to transpose a formula it is necessary to use inverse operations. An inverse operation is one where, when it is applied immediately after some mathematical operation, the expression remains unchanged. Example if the mathematical operation is ‘multiply by 6’ the inverse operation is ‘divide by 6’ 6X because (X × 6) ÷ 6 = = X. 6 So the inverse of ‘multiply’ is ‘divide’ and naturally, the inverse of ‘divide’ is ‘multiply’. The following table gives some of the operations and their inverses:
Operation
Inverse
Addition
Subtraction (and vice versa)
Multiplication
Division (and vice versa)
Square
Square root (and vice versa)
Cube
Cube root etc. (and vice versa)
Invert (turn upside down) Invert Make negative
Consider the formula S = ut +
Make negative
at2 and what is more, the values of u, t and a are known. 2
It is required that this formula is transposed so ‘a’ is the new subject.
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•
Mathematics
Imagine the steps taken on a calculator to work out the value of s, starting with the value of a (the letter that is to be the new subject). Once the steps have been established all that is needed is the opposite operations to be used working in reverse. The diagram should explain. Start with a
Multiply by t 2
Divide by 2
Result a
Divide by t 2
Multiply by 2
Add ut
Result S
Subtract ut
Start with S
All that is needed is for the instructions to be carried out •
Start with S
S
•
Subtract ut
S − ut
•
Multiply by 2 2(S − ut) 2 (S − ut) Divide by t2 t2 2 (S − ut) Result a= t2
• •
It must be pointed out there is a slight problem with algebraic expressions: namely, there is often more than one way of writing the same formula. It looks different but would give the same answer when numbers were substituted for the symbols. Consider the answer found above: a = Now
2 (S − ut) 2S − 2ut = t2 t2 2S − 2ut 2S 2ut = 2 − 2 t2 t t 2S 2u − t2 t
So either a =
2 (S − ut) t2
(multiply out the bracket) (split into two fractions with a common denominator) (cancel top and bottom of second fraction by ‘t’)
2 (S − ut) 2S 2u or a = 2 − , and BOTH are equally correct. 2 t t t
Here is the problem: suppose this question was set in one of the exercises to be done 2S 2u and, furthermore, you had arrived at the answer a = 2 − . However, on checking t t 2 (S − ut) the answers, the solution was given as a = . Unless you can show the two t2 expressions mean the same, the natural conclusion is that you have made a mistake. If in any doubt, ask for clarification. Alternatively, the following sections give examples of the different types of formula that may occur and how they can be transposed.
Simple Equations
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53
Type 1: Symbols connected by plus or minus signs Example 1 Transpose c = d + e + f to make e the subject of the formula. The aim is to end up with e by itself on the left-hand side of the equals sign. Changing the equation around so that e is on the LHS gives d+e+f =c Subtracting d from both sides gives d+e+f −d = c−d
so
e+f =c−d
so
e= c−d−f
Subtracting f from both sides gives e+f −f = c−d−f
As with simple equations, if a term changes from one side of the equals sign to the other either by addition or subtraction, there is an appropriate sign change. So directly d+e+f =c
gives
e = c−d−f
Example 2 If 2c + 3d = f − g + 4h, make g the subject. In this case the subject is negative, 2c + 3d = f − g + 4h A negative term is always going to cause confusion so it is best to make the term positive, and the easiest way is to move the term to the other side of the equals sign. Therefore 2c + 3d + g = f + 4h It is now a simple task to remove 2c and 3d from the left-hand side by subtraction, giving g = f + 4h − 2c − 3d
Type 2: Formulae involving multiplications Example 3 Transpose a = ω · r to make ω the subject.
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Mathematics
Dividing both sides by r gives
a ω·r = r r
so ω =
a r
Example 4 Suppose (a + b)2 = 10cd Express this formula with d as the subject. In reality this is no different from the previous example: two terms d and 10c are multiplied together to make a third term (a + b)2 , so dividing both sideds by 10c gives (a + b)2 = d. 10c Therefore d =
(a + b)2 . 10c
Type 3: Formulae involving fractions (or divisions) Example 5 If Z =
V , rearrange to make V the subject. R
Multiplying both sides by R gives V = Z · R Example 6 If Z =
V , rearrange to make R the subject. R
Multiply both sides by R, as in Example 5, so V = Z · R then divide both sides by Z. Therefore R =
V Z
Example 7 Transpose the formula Z =
αV to make (a) R the subject; (b) V the subject R
The only difference between this and the previous examples is the introduction of a multiplying constant, α. (a) Multiplying both sides by R gives RZ = αV αV Dividing by Z gives R = . Z (b) From part (a): αV = RZ
Simple Equations Dividing by α gives V =
RZ . α
Example 8 If s = ut +
at2 transpose the equation to make a the subject 2 ut +
at2 =s 2
at2 = s − ut 2 2 at Multiplying by 2 gives: 2 = 2 (s − ut) 2 at2 2 (s − ut) = And dividing by t2 gives: t2 t2 Moving ut gives:
therefore at2 = 2 (s − ut) therefore a =
2 (s − ut) t2
Questions Make the symbol shown in brackets the subject of the formula shown. Express each answer in its simplest form.
1.
a+b =c−d
(d)
2.
x + 3y = t
(y)
3.
c = 2πr
(r)
4.
y = mx + c
(x)
5.
i = PRT
(T)
6.
7.
S=
(r)
8.
1.
d = c−a−b
2.
3.
c 2π
4.
5.
T=
a 1−r
E (R) R 9 F = C + 32 (C) 5 i=
Answers
7.
i PR a S−a or R = 1 − R= S S
6. 8.
t−x 3 y−c x= m E R= i 5 C = (F − 32) 9 y=
•
55
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Mathematics
Type 4: Formulae with the new subject in a bracket Example 9 Transpose the equation R = R0 (1 + αt) to make α the subject. It is probably best to isolate the bracket on the LHS. R0 (1 + αt) = R so Therefore (1 + αt) =
R0 (1 + αt) R = R0 R0
R R → 1 + αt = R0 R0 R R0 R αt = −1 R0 αt R − R0 = t R0 t R−S So α = St
1 + αt =
→ →
R − R0 R0 R − R0 α= R0 t
αt =
Example 10 (u + v) · t to make v the subject 2 (u + v) ·t = s 2 (u + v) . t = 2s → (u + v) . t = 2s Multiplying by 2 gives: 2 2 (u + v) 2s 2s .t = → u+v = Dividing by t gives: t t t 2s 2s Subtracting u gives: u+v−u = −u → v = −u t t Transpose s =
So v =
2s −u t
Type 5: Formulae with powers and roots Example 11 Transpose k =
mv 2 to make v the subject. 2
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57
mv 2 = 2k → mv 2 = 2k 2 mv 2 2k 2k mv 2 = 2k → = → v2 = m m m 2k 2k 2k → → v= v2 = v2 = m m m 2k So v = m mv 2 =k 2
→
2
Example 12 Pythagoras Theorem states: ‘In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.’ This can be expressed algebraically as A c b
B
C
a
AB2 = AC2 + BC2
or c2 = a2 + b2
Transpose Pythagoras Theorem to make b the subject a2 + b2 = c2 b2 = c2 − a2
→ a2 + b2 − a2 = c2 − a2 b2 = c2 − a2 →
→ b2 = c2 − a2 → b= c2 − a2 So b = c2 − a2
Example 13 If a simple pendulum consisting of a mass, m kg, attached to a mass-less string of length, L metres, is allowed to swing freely the periodic time, T seconds, is given by the formula L T = 2π , where g is the acceleration due to gravity. Transpose the formula to make L g the subject: L 2π T T L L g 2π =T → = → = g 2π 2π g 2π
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Mathematics
2 2 T L L T = → = g 2π g 2π 2 2 L L T T = → g× =g× g 2π g 2π 2 gt2 T So L = g or L = 2π 4π 2
T 2 2π 2 T → L=g 2π →
L = g
Questions Make the symbol shown in brackets the subject of the formula shown. Express each answer in its simplest form. b (c + d) d
1.
a=
2.
y3 =
5L2 PN
(P)
3.
1 1 1 = + R R1 R2
(R2 )
4.
V=
5.
a2 b2 + =1 x 2 y2
(x)
6.
A=
r2 . θ 2
(θ )
7.
A=
8.
π 3 (D − d3 ) 6
πR2 θ 360 Z = R2 + (2πfL)2
(c)
(d)
(R) (f )
Answers 1.
c=
ad d (a − b) or c = d − b b
2.
P=
5L2 y3 N
3.
R2 =
R · R1 R1 − R
Simple Equations 4.
d=
6.
θ=
7.
3
2A r2
R=
8.
f=
D3 −
6V π
360A πθ
Z 2 − R2 2πL
Type 6: Formulae where factorisation is needed Example 14
2b3 x + 3b3 y = Ft
2b3 x + 3b3 y to make b the subject t 3 2b x + 3b3 y → · t = Ft → 2b3 x + 3b3 y = Ft t → b3 (2x + 3y) = Ft
b3 (2x + 3y) = Ft
→
Transpose the formula F = 2b3 x + 3b3 y =F t
Ft → (2x + 3y) Ft So b = 3 (2x + 3y) b3 =
b3 (2x + 3y) = Ft (2x + 3y) √ Ft 3 3 b = 3 (2x + 3y)
Ft → b3 = (2x + 3y) Ft → b= 3 (2x + 3y)
Example 15 If t = √
a−1 , make m the subject of the formula mx + 2m
√ a−1 a−1 √ =t → √ × mx + 2m mx + 2m mx + 2m √ √ =t× mx + 2m → a − 1 = t mx + 2m √ √ √ t mx + 2m a − 1 a−1 = → mx + 2m = t mx + 2m = a − 1 → t t y
•
59
60
•
Mathematics
√ 2 a − 1 2 (a − 1)2 a−1 2 mx + 2m = → mx + 2m = → m(x + 2) = t t t2 So m =
(a − 1)2 t2 (x + 2)
Type 7: Formulae where the new subject appears in more than one term Example 16 If x =
y , make y the subject 1+y
The only way to transpose for a subject that appears more than once in the formula is to rearrange the terms so that all the terms that contain the repetition are on one side of the equals sign and that all the terms that do not contain the repeated subject are on the other side of the equals sign. The repetition is then factorised out, so it appears to be written only once. Then proceed as usual. y =x 1+y y = x + xy So y =
→ y = x(1 + y) → y = x + xy → y − xy = x
→ 1y − xy = x
→
y (1 − x) = x
x (1 − x)
Example 17 If
1 4K − 2b = , make K the subject x 3c + 5K
1 4K − 2b = → 1(3c + 5K) = (4K − 2b)x → 3c + 5K = 4Kx − 2b x 3c + 5K → 3c + 2b = 4Kx − 5K → 3c + 2b = K(4x − 5) So K =
3c + 2b (4x − 5)
Example 18 Transpose the following expression to give the value of i in terms of P and t. Find the value of i when P = 50 and t = 4 P=
100 (1 + i)t
Simple Equations
•
61
Therefore P (1 + i)t = 100 100 (1 + i)t = P t 100 so 1+i= P 1 100 /t t 100 i= − 1 or i = −1 P P √ 100 − 1 = 4 2 − 1 = 0. 1892 Inserting the numerical values gives i = 4 50 Example 19 An adiabatic process is one where heat does not enter or leave a system and is typified by equations of the type p1 · V1n = p2 · V2n Transpose the expression to give V1 in terms of the other quantities. Find the value of V1 when p1 = 1. 035, p2 = 4. 14, V2 = 0. 5 and n = 1. 3 p1 · V1n = p2 · V2n p2 · V2n V1n = p 1 p2 · V2n so V1 = n p1 p2 V1 = n × n V2n p1 p2 V1 = V2 × n p1 1 p2 n or V1 = V2 × p1 Substituting the values gives V1 = 0. 5 ×
1 4. 14 1. 3 = 0. 5 × 40.769 = 1. 452 1. 035
correct to 3 decimal places.
Questions Transpose each equation so that the symbol indicated is the subject, and express the answers in their simplest form.
62
•
Mathematics
1.
A = π · r2
2.
V=
3.
S = 4πR
4.
f=
5.
d = 2 h (2r − h)
6.
V=
7.
v=
8.
4π · r3 3
(r)
(r) r2 + R2 2
(r)
uv u+v
(u)
(the subject appears more than once)
(r)
πh 3R2 + h2 6
(R)
iR E − iR 2gh q = A 2 A −1 α
(i)
(A)
Answers 1.
r=
A π
2.
3.
3V 4π S 2 r= 2 − R2 4πR r=
4.
u=
5.
r=
6.
fv f −v
d2 + 4h2 8h 6V − πh3 R= 3πh
⇒
r=
2S2 − R2 16π 2 R2
⇒
r=
S2 − R2 8π 2 R2
Simple Equations
•
63
vE R (1 + v)
7.
i=
8.
qα A= 2 q − 2ghα 2
More practise Questions Make the symbol(s) indicated the subject of the formulae, and express the answers in their simplest form.
1.
R=
N−M D
(D)
2.
T = a + bN2
4.
v2 = u2 − 2as
(s,u)
5.
E=
7.
T −W =
W v2 gx
(W)
8.
(Q)
11.
(p)
14.
10.
d=
3
13.
16.
19.
22.
P Q−P
m2 − 2p2 2m2 + p2 nsh s2 C= R2 − 6 4 D f +p = d f −p 1 V = i r2 + (ωC)2 n=
(R)
17.
(p)
20.
(C)
23.
m 2 v − u2 2g
2 V = π r2 h + r 3 h2 + k2 T = 2π 2gh
(N)
3.
P + 3Q x = Q − 3P y
(u)
6.
e=
(h)
9.
v=
(k)
12.
e=
(a)
15.
C = 2 2hr − h2
P−p PT − pt
2 (S − an) n(n − L) √ Mn2 4L2 + b2 x= K
(p)
gd 1 +
d=
(Q)
ir V 1− ZpΦ V R2 1 − ω= LC 4L2 n=
18.
P=
(V)
21.
Z=
(R)
24.
t−k k (1 + kt)
(b)
3h d
ω 2 M2 −
(h)
(t)
(r)
r2 2π L
r2 + (2π fL)2
2V r i= L R+ n
()
(f)
(r)
Answers 1.
M = N − RD
4.
s=
√ u2 − v 2 , u = ± v2 + 2as 2a
T −a b
2.
N=±
5.
u = ± v2 −
2Eg m
3.
Q=
P(y + 3x) x − 3y
6.
p=
P (1 − eT) 1 − et
64
•
Mathematics
W=
7.
Tgx gx + v2
8.
h=
V 2 − r π r2 3
P Q = 3 +P d 1 − 2n2 p = ±m 2 + n2 s2 36c2 R= + 4 n2 s2 h2 f D2 − d 2 p= D2 + d 2
10.
13.
16.
19.
C=
22.
i ω
1 V 2 − i2 r 2
11.
k=±
14.
a=
17.
20.
s d (n − L) − h 2 K 2 x2 b= − 4L2 M2 n4 V = ir + nZpΦ
23.
T 2 gh − h2 2π 2
R=
4L − 4ω2 L2 C 2 C
9.
h=
12.
t=
15.
r=
18.
21.
24.
As it is an essential skill, more practice questions are provided.
Questions
4G K+ 3 1. v = ρ Make G the subject of the formula.
2. P = f (b − nd) t Make f the subject of the formula. 6M bd2 Make d the subject of the formula.
3. f =
WL Ax Make x the subject of the formula.
4. E =
5.
θ − θO Rt = R θ − θd Make θ the subject of the formula. a (V − b) = RT v2 Make p the subject of the formula.
6.
p+
v2 − gd 3g k 1 + e2 1 − e2 k2 c2 + 4h2 8h
1√ 2 2 2 2 4π L P + r M Z 2 − r2 f= 4π 2 L2 ω=
r=
2Vn nR − iL L
Simple Equations
7.
1 1 m = + Y μ YG Make YG the subject of the equation. πPr 4 t 8vL Make v the subject of the formula.
8. η =
9.
μ CZ 4 λ5/2 N = ρ A Make λ the subject of the formula.
10. T = 4π
(M + 3m) L 3 (M + 2m) g
Make m the subject. 11. p =
y+k k y + p1 p2
Make y the subject. 12. Make D2 the subject of the equation L =
πh 2 D2 − D21 . 2 4d
Answers 1.
3.
3 2 v ρ −K 4 6M d= bf G=
Rt · θd − R · θO Rt − R
5.
θ=
7.
YG =
9.
11.
μY μ − mY 2 Aμ 5 λ= ρCZ 4 N y=
kp1 (p2 − p1 ) p2 (p − p1 )
2.
f=
P (b − nd) t
4.
x=
WL EA
6.
p=
a RT − 2 V −b v
8.
v=
10.
12.
π Pr4 t 8ηL M 16π 2 L − 3gT 2 m= 6 gT 2 − 8π 2 L 4Ld2 − D21 D2 = πh
•
65
66
•
Mathematics
A technique where algebraic fractions, for example,
3x − 1 , are ‘transposed’ is (x + 2) x 2 + 4
that of Partial Fractions. To demonstrate the concept consider the question: 5 3 + ? (x + 2) (x − 4) Now
What is the result of
3 5 3 (x − 4) + 5 (x + 2) 3x − 12 + 5x + 10 8x − 2 + = = 2 = 2 (x + 2) (x − 4) (x + 2) (x − 4) x − 4x + 2x − 8 x − 2x − 8
Therefore
5 8x − 2 3 + = 2 (x + 2) (x − 4) x − 2x − 8
and obviously
8x − 2 x 2 − 2x − 8
=
5 3 + (x + 2) (x − 4)
The procedure of ‘breaking up’ partial fractions.
8x − 2 x 2 − 2x − 8
into
5 3 + is called resolving into (x + 2) (x − 4)
Partial Fractions Certain conditions must be in place before an algebraic expression can be resolved into partial fractions. •
The denominator must factorise, as in the above example x 2 − 2x − 8 = (x − 2) (x − 4). It makes perfect sense that, if the denominator does not factorise, then it cannot be resolved into partial fractions.
•
The degree of the numerator must be at least one less than the degree of the numerator. This means, the highest power of the variable in the numerator has to be at least one less than the highest power in the denominator. In the above example, the numerator is of degree 1 (the ‘8x’ term) and the degree of the denominator is 2 ( the ‘x 2 ’ term). ◦ If this is not the case then the numerator must first be divided by the denominator using the technique of polynomial division. The remainder of this division can then be resolved into partial fractions.
Partial fractions can be described as being one of three main types.
Simple Equations
•
67
1. Linear denominators Similar to the example above, where the denominator can be described as being the product of several terms of degree 1. In general, the denominator is of the form (x − a) (x − b) · · · (x − k), and the numerator is a function of lesser degree. For example,
2x − 1 (x − 1) (x + 2) (x + 4)
This would be resolved as follows: A B C 2x − 1 = + + (x − 1) (x + 2) (x + 4) (x − 1) (x + 2) (x + 4) When resolving, the numerator of each individual fraction has to be one degree lower than the numerator. In this case, all the denominators are of degree 1, so the numerators have to be of degree 0 (a constant). So A B C 2x − 1 = + + (x − 1) (x + 2) (x + 4) (x − 1) (x + 2) (x + 4) =
A (x + 2) (x + 4) + B (x − 1) (x + 4) + C (x − 1) (x + 2) (x − 1) (x + 2) (x + 4)
Therefore 2x − 1 = A (x + 2) (x + 4) + B (x − 1) (x + 4) + C (x − 1) (x + 2) To obtain the basic equation it is quicker to use the following procedure: •
The common denominator is denominator of the left-hand side ◦ In this case (x − 1) (x + 2) (x + 4)
•
All the individual fractions on the right-hand side must have this denominator in common ◦ First term – has (x − 1), missing (x + 2)(x + 4). Hence A. (x + 2)(x + 4)
•
◦
Second term – has (x + 2), missing (x − 1)(x + 4). Hence B. (x − 1)(x + 4)
◦
Third term – has (x + 4), missing (x − 1)(x + 2). Hence C. (x − 1)(x + 2)
Therefore 2x − 1 = A (x + 2) (x + 4) + B (x − 1) (x + 4) + C (x − 1) (x + 2).
This equation is true for all values of x, so a suitable choice of this value to substitute into the equation will be crucial.
68
•
Mathematics
Method: Choose a value of x that removes a bracket from the system. 2x − 1 = A (x + 2) (x + 4) + B (x − 1) (x + 4) + C (x − 1) (x + 2) 2 (1) − 1 = A (1 + 2) (1 + 4)
Let x = 1:
⇒
1 = 15A
⇒
Let x = −1: 2 (−2) − 1 = B (−2 − 1) (−2 + 4)
⇒
−5 = −6B ⇒
Let x = −4: 2 (−4) − 1 = C (−4 − 1) (−4 + 2)
⇒
−9 = 10C
⇒
1 15 5 B= 6
A=
C=−
9 10
Therefore 1 15 56 −9 10 2x − 1 = + + (x − 1) (x + 2) (x + 4) (x − 1) (x + 2) (x + 4) 5 9 1 + − = 15 (x − 1) 6 (x + 2) 10 (x + 4)
2. Quadratic denominator One (or more) of the terms in the denominator will be of the form ax 2 + bx + c , where the value of b could be 0. The numerator, being one degree less, would therefore be of the form (Ax + B). For example,
x 2 + 2x − 2 (x + 1) x 2 + 1
Bx + C A x 2 + 2x − 2 = + 2 2 (x + 1) x +1 (x + 1) x + 1 This gives x 2 + 2x − 2 = A x 2 + 1 + (Bx + C)(x + 1)
So
3 let x = −1: (−1)2 + 2 (−1) − 2 = A (−1)2 + 1 ⇒ −3 = A (2) ⇒ A = − 2 The next choice of x is 0, since this will remove B from the equation let x = 0:
1 3 −2 = A (1) + C (1) ⇒ −2 = A + C ⇒ −2 = − + C ⇒ C = − 2 2
There are now no other values of x that remove any of the unknowns from the equation. Therefore, either any value of x can be taken, which will result in an equation involving A, B and C. A and C can then be substituted, giving the value of B. Or, the coefficients of powers of x can be equated.
Simple Equations Coefficient of x 2 :
•
69
5 3 1 = A+B ⇒ 1 = − +B ⇒ B = 2 2
Therefore
5 2x − 1 2 −3 2 x 2 + 2x − 2 = + (x + 1) (x 2 + 1) (x + 1) x 2 + 1 =−
5x − 1 3 + 2 (x + 1) 2 x + 1
3. Repeated linear factors The denominator will contain a linear term raised to a power, (x + a)n For example,
x2 + 2 (x − 1)3
This is resolved into partial fractions whose denominators are the linear term raised to the power 1 up to the linear term raised to the power n. In the case of the example, n = 3. So
x2 + 2 (x − 1)3
=
B A C + + (x − 1) (x − 1)2 (x − 1)3
This gives x 2 + 2 = A (x − 1)2 + B (x − 1) + C Let x = 1:
(1)2 + 2 = C ⇒ C = 3
Coefficient of x 2 : 1 = A ⇒ A = 1 Choose another value of x: Let x = 0 : 2 = A (−1)2 + B (−1) + C ⇒ 2 = A − B + C ⇒ B = A + C − 2 = 1 + 3 − 2 So B = 2 Therefore
2 1 3 x2 + 2 + = + 3 2 (x − 1) (x − 1) (x − 1) (x − 1)3
Examples Find partial fractions for the following: (i)
x+3 (x − 3)(x − 4)
(ii)
4x 2 − 7x − 5 x 2 − 2x − 5 (x − 2)
(iii)
3x − 4 (x + 1) (x − 2)2
70
•
Mathematics
Solutions (i)
A B x+3 = + ⇒ x + 3 = A (x − 4) + B (x − 3) (x − 3) (x − 4) (x − 3) (x − 4) Let x = 4 ⇒ 7 = 1B ⇒ B = 7 Let x = 3 ⇒ 6 = −1A ⇒ A = −6 Therefore −6 7 x+3 = + (x − 3) (x − 4) (x − 3) (x − 4)
(ii)
(Ax + B) 4x 2 − 7x − 5 C + = 2 2 (x − 2) x − 2x − 5 (x − 2) x − 2x − 5 2 2 ⇒ 4x − 7x − 5 = (Ax + B) (x − 2) + C x − 2x − 5
3 5 Let x = 0 ⇒ −5 = −2B − 5C ⇒ −5 = −2B − 3 ⇒ B = 1 17 Coefficient of x 2 ⇒ 4 = A + C ⇒ A = 5 Therefore 17 3 x + 1 (17x + 5) 3 4x 2 − 7x − 5 5 + 5 = + = 2 (x − 2) 5 x 2 − 2x − 5 5 (x − 2) x 2 − 2x − 5 (x − 2) x − 2x − 5
Let x = 2 ⇒ 16 − 14 − 5 = −5C ⇒ −5C = −3 ⇒ C =
(iii)
B C A 3x − 4 + + = 2 (x + 1) (x − 2) (x − 2)2 (x + 1) (x − 2) ⇒ 3x − 4 = A (x − 2)2 + B (x + 1) (x − 2) + C (x + 1) 2 Let x = 2 ⇒ 2 = 3C ⇒ C = 3 −7 Let x = −1 ⇒ −7 = 9A ⇒ A = 9 7 Coefficient of x 2 ⇒ 0 = A + B ⇒ B = 9 Therefore 7 2 −7 3x − 4 + + = 2 (x (x 9 + 1) 9 − 2) (x + 1) (x − 2) 3 (x − 2)2
Questions Find partial fractions for the following: (i)
x+7 (x − 2)(x − 5)
(ii)
8x 2 − 19x − 24 x 2 − 2x − 5 (x − 2)
(iii)
5x 2 − 13x + 5 (x − 2)3
Simple Equations
•
71
Answers (i)
A B x+7 = + ⇒ x + 7 = A (x − 5) + B (x − 2) (x − 2) (x − 5) (x − 2) (x − 5) x = 5 ⇒ 12 = 3B ⇒ B = 4 x = 2 ⇒ 9 = −3A ⇒ A = −3 Therefore −3 4 x+7 = + (x − 2) (x − 5) (x − 2) (x − 5)
(ii)
(Ax + B) C 8x 2 − 19x − 24 + = 2 2 (x − 2) x − 2x − 5 (x − 2) x − 2x − 5 2 ⇒ 8x − 19x − 24 = (Ax + B) (x − 2) + C x 2 − 2x − 5
x = 2 ⇒ 32 − 38 − 24 = −5C ⇒ C = 6 x = 0 ⇒ −24 = −2B − 5C ⇒ −24 = −2B − 30 ⇒ B = −3 Coefficient of x 2 ⇒ 8 = A + C ⇒ A = 2 Therefore (2x − 3) 6 8x 2 − 19x − 24 + = 2 2 (x − 2) x − 2x − 5 (x − 2) x − 2x − 5 (iii)
B A C 5x 2 − 13x + 5 + = + 3 2 (x − 2) (x − 2) (x − 2) (x − 2)3 2 ⇒ 5x 2 − 13x + 5 = A (x − 2) + B (x − 2) + C x = 2 ⇒ 20 − 26 + 5 = C ⇒ C = −1 Coefficient of x 2 ⇒ 5 = A ⇒ A = 5 x = 0 ⇒ 5 = 4A − 2B + C ⇒ 5 = 20 − 2B − 1 ⇒ B = 7 Therefore 5x 2 − 13x + 5 (x − 2)
3
=
7 7 1 + − 2 (x − 2) (x − 2) (x − 2)3
Logarithmic or Indicial Equations Some equations result in the unknown value being a power, or index. Example Given that p2 = 1.16.
T1 = T2
p1 p2
n find the value of n when T1 = 570, T2 = 250, p1 = 19 and
72
•
Mathematics
The method in cases like this is to take logarithms of both sides. n p1 T1 = log So log T2 p2 The third law of logarithms states: The logarithm of a number raised to a power is given by the logarithm of the number multiplied by the power: log(x y ) = y · log(x). p1 T1 = n × log Therefore log T2 p2 T1 log log (T1 ) − log (T2 ) T 2 = , using the second law of logarithms and so n = p1 log (P1 ) − log (P2 ) log p2 log (570) − log (250) = 0. 2948 correct to 4 decimal places log (19) − log (1. 16)
Therefore n =
Note: either base 10 or natural logarithms may be used. Example T1 = Given that T2
p1 p2
n−1 n
find the value of n when T1 = 710, T2 = 260, p1 = 30 and p2 = 0. 95
710 = 260
710 log 260
but
30 0. 95
n−1 n
n−1 n
⇒ log
710 260
⎡ ⎤ n−1 n 30 ⎦ = log ⎣ 0. 95
30 = × log So 0. 95 710 log n−1 260 = 0. 29097 = Therefore 30 n log 0. 95
n−1 = 0. 29097 ⇒ n − 1 = 0. 29097n ⇒ n − 0. 29097n = 1 n
Therefore 0. 70903n = 1 ⇒ n =
1 = 1. 410 correct to 3 decimal places. 0. 70903
Simple Equations
•
73
Test Examples 3 1. Find the value of x which satisfies the following equation: 8 + 5x − 7 = 3x + 9 2. Find the value of a in the equation: 2(a + 3) + 3(2a − 4) = 4(11 − 3a) 3. Solve the equation: 3[3 − {x + 2(1 − x)} − 4x] = 2[x − 3(2 + x) − 4] 4. Find x for the following equation: 2x x 5 4x 1 + − = + 3 4 6 5 3 5. Find a in the following:
6. If
1 − 2a2 1 a 1 5 − + = − 6a 4 3 6 2a
3 4 2 + = 2 , find x. x−3 x+2 x −x−6
7. A ship travelling at 17.5 knot leaves one port bound for another 4.5 h after a another ship, whose speed is 16 knots, leaves the same port set on the same course. After how many hours and at what distance from port will the fast ship overtake the slower? 8. A rectangular plate is to be cut so that its length is four times the breadth and having an area of 1 m2 . Find the length and breadth. R2 (1 + αθ2 ) to find an expression for θ2 . = 9. Transpose R1 (1 + αθ1 ) Calculate the value of θ2 when: R1 = 200, R2 = 240, θ1 = 15 and α = 0. 0042. 10. The diameter in mm of coupling bolts should not be less than that given by the formula: D3 d= 3. 5 × n × R where D = diameter of shafting in mm; n = number of bolts per coupling; and R = pitch circle radius in mm. Express R in terms of the other quantities and find the pitch circle radius when d = 82. 5 mm, D = 381 mm and n = 8 bolts per coupling.
74
•
Mathematics
11. Given T1 = T2
p1 p2
n−1 n
find the value of n when T1 = 797, T2 = 301, p1 = 34. 4 and p2 = 1. 05 12. Find the values of x in each of the following: (a) 40.59x = 56. 36 (b) x 1.95 = 12. 4x 0.53 (c) e5x = 46. 382.6 √ (d) 3 x = e2.5
4
SIMULTANEOUS EQUATIONS It has been seen that only one equation is needed to find the value of a single unknown quantity. However, if one equation has two unknown quantities, for example, 3x + y = 12, there is no longer a unique answer. In this example, x could be 4 with y equalling 0 or x could equal 0 with y equalling 12.
If two different equations involving the same two unknowns are available then a single solution can be found. Similarly, if there are 3 unknowns there must be three different equations etc. When such sets of equations have to be solved for the unknowns, they are known as simultaneous equations. There are several methods available but the method shown below is that of elimination.
Two Unknowns Where the Coefficients Are Whole Numbers Example Solve the following equations for x and y : 3x + 4y = 5 and 12 + 2x − 5y = 0: •
The two equations should be written down with similar unknown terms under each other. It is usual to put the unknown terms on the LHS of the equation with the
76
•
Mathematics
constant term on the RHS. So 3x + 4y = 5 2x − 5y = −12 •
The two equations are numbered. This makes reference to the equations much easier. So 3x + 4y = 5
(1)
2x − 5y = −12
(2)
•
The equations have to be multiplied in such a way that the coefficients of one of the unknowns become equal. The choice of either x or y is arbitrary.
•
If, in this example, the x coefficients are to be made equal then equation (1) has to be multiplied by the x coefficient in equation (2) (i.e. 2) and equation (2) has to be multiplied by the x coefficient in equation (1) (i.e. 3). This gives 6x + 8y = 10
(3)
6x − 15y = −36
(4)
The new equations are renumbered. •
By adding or subtracting these new equations one unknown can be eliminated. If the sign of the common coefficients is the same (i.e. both are positive or both are negative) then one equation should be subtracted from the other. If the signs of the coefficients are different then the equations should be added.
•
As the x coefficients are both positive the equations should be subtracted. So equation (4) is subtracted from equation (3) giving: 6x + 8y = 10 6x − 15y = −36 0 + 23y = 46
•
This new equation has only one unknown and can therefore be solved. 23y = 46 gives y = 2
•
This value should be substituted into one of the original equations So, in (1) 3x + 8 = 5, so 3x = −3 giving x = −1
•
The solution to the simultaneous equation is x = −1, y = 2.
Simultaneous Equations
•
77
Example Solve 8x − 3y = 39
(1)
7x + 5y = −4
(2)
To eliminate the y’s, equation (1) is multiplied by 5 and equation (2) by 3. This gives 40x − 15y = 195
(3)
21x + 15y = −12
(4)
When equation (3) is added to equation (4) (added because the y coefficients have different signs) the result is 61x = 183 So x =
183 =3 61
Substituting this value into equation (1) gives 24 − 3y = 39 −3y = 15 y=
15 −3
= −5 Check: In equation (2), 7 × 3 + 5 × (−5) = 21 − 25 = −4. Therefore correct. Example Solve 4x − 18 = 3y 1 + x + 2y = 0 Rearranging gives 4x − 3y = 18
(1)
x + 2y = −1
(2)
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Mathematics
The simplest way to make the coefficients the same is to multiply equation (2) by 4. So 4x − 3y = 18
(3)
4x + 8y = −4
(4)
Subtracting (3) from (4) gives 8y − (−3y) = −4 − 18 11y = −22 y = −2 Substituting into equation (3) gives 4x − 3 × (−2) = 18 4x + 6 = 18 So 4x = 18 − 6 = 12 x=
12 =3 4
Check: In equation (2), 3 + 2 × (−2) = 3 − 4 = −1. Therefore correct.
Questions 1.
a+b = 7
2.
a−b = 3 3.
3s + 2t = 12
x + 3y = 4 4.
4s − t = 5 5.
5m − 3n = 11 5x = 2y 3x + 7y = 41
3x − 2y = 13 2x + 5y = −4
6.
3m + n = 8 7.
2x + 5y = 7
8a − 3b = 51 3a + 4b = 14
8.
1 − 3d = 5c 2d + c + 4 = 0
Simultaneous Equations
•
79
Answers 1.
a = 5, b = 2
2.
x = 1, y = 1
3.
s = 2, t = 3
4.
x = 4, y = −2
5.
1 1 m=2 , n= 2 2
6.
a = 6, b = −1
7.
x = 2, y = 5
8.
c = 2, d = −3
Two Unknowns Where the Coefficients Are Fractions Example Solve a b − =0 6 8 a b + =6 3 4
(1) (2)
It is best to get rid of the fractions before the solution of the simultaneous equations is started. Multiply each equation by the LCM of its denominators. So, multiply (1) by 24 and (2) by 12, giving 4a − 3b = 0
(3)
4a + 3b = 72
(4)
Adding equations (3) and (4) will eliminate b, giving 8a = 72, so a = 9. Substituting into equation (3) gives (4 × 9) − 3b = 0 3b = 36 so b = 12
Checking by substituting into equation (2) gives So correct.
9 12 3 3 − = − = 0. 6 8 2 2
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Mathematics
Two Unknowns Where the Coefficients Are Decimals Example Solve 1. 2x − 1. 8y = −21
(1)
2. 5x + 0. 6y = 65
(2)
It is best to get rid of the decimals before the solution of the simultaneous equations is started. Multiply each equation by the appropriate power of 10, that is, 1 decimal place multiply by 10, 2 decimal places multiply by 100 etc. So multiplying equations (1) and (2) by 10 gives 12x − 18y = −210
(3)
25x + 6y = 650
(4)
To eliminate y, multiply equation (4) by 3 giving 12x − 18y = −210
(5)
75x + 18y = 1950
(6)
Adding (5) and (6) gives 87x = 1740 x=
1740 87
x = 20 Substituting in (4) gives (25 × 20) + 6y = 650 500 + 6y = 650
Simultaneous Equations
•
81
So 6y = 650 − 500 6y = 150 y=
150 6
y = 25 Checking by substituting into equation (1) gives (1. 2 × 20) − (1. 8 × 25) = 24 − 45 = −21. So correct. It is important to realise that numbers must not be rounded at any stage. The full result of each multiplication and division must be used if accuracy errors are to be avoided.
Two Unknowns Involving Reciprocals Example Solve 3 6 + =4 a b 9 12 − = −3 a b The following substitutions should be made: let
1 1 = X and = Y. a b
This gives 3X + 6Y = 4
(1)
9X − 12Y = −3
(2)
These can now be solved as normal. Equation (1) multiplied by 2 gives 6X + 12Y = 8
(3)
9X − 12Y = −3
(4)
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•
Mathematics
Adding (3) and (4) gives 15X = 5 5 15 1 X= 3 X=
Substituting into (4) gives 1 − 12Y = −3 9× 3 3 − 12Y = −3 −12Y = −6 So
But
Y=
−6 1 = −12 2
1 1 1 1 1 1 = X and = Y so = a and = b giving a = = 3 and b = = 2. a b X Y 1/3 1/2
Checking by substituting into the original first equation gives
3 6 + = 1 + 3 = 4. 3 2
So correct. Example Solve x−1 y+2 2 + = 3 5 15 1−x 5+y 5 + = 6 2 6
(1) (2)
Before the equations can be solved simultaneously the fractions have to be removed and the equations rearranged. Equation (1) multiplied by 15 and equation (2) multiplied by 6 gives 5 (x − 1)) + 3 (y + 2) = 2
(3)
(1 − x) + 3 (5 + y) = 5
(4)
Simultaneous Equations
•
83
Expanding the brackets gives 5x − 5 + 3y + 6 = 2
(5)
1 − x + 15 + 3y = 5
(6)
Rearranging gives 5x + 3y = 1
(7)
−x + 3y = −11
(8)
Equation (7) subtract equation (8) gives 6x = 12 so x = 2 Substituting into equation (7) gives (5 × 2) + 3y = 1 10 + 3y = 1 3y = −9 y = −3 Substituting in equation (8) gives (−2) + (3 × −3) = −2 + (−9) = −11, so correct.
Questions In Questions 1–7 solve the following pairs of simultaneous equations and check the results:
1.
3 2 + = 14 x y
2.
5 3 − = −2 x y 3.
3 1 + =5 2p 5q 1 35 5 − = p 2q 2
4 3 − = 18 a b 2 5 + = −4 a b
4.
5 3 + = 1. 1 x y 3 7 − = −1. 1 x y
84
•
Mathematics c+1 d+2 − +1=0 4 3
5.
3r + 2 2s − 1 11 − = 5 4 5
6.
1 − c 3 − d 13 + + =0 5 4 20 7.
20 5 = x + y 27
8.
If 5x −
3 + 2r 5 − s 15 + = 4 3 4
16 4 = 2x − y 33
4 5 xy + 1 3 = 1 and x + = find the value of y y 2 y
Answers 1.
1 1 x= , y= 2 4
2.
1 1 a= , b=− 3 2
5.
c = 3, d = 4
6.
r = 3, s =
1 2
3.
1 1 p= , q= 4 5
4.
x = 10, y = 5
7.
x = 5, y = 1
3 4
8.
1
Practical Problems: Simultaneous Equations in Two Unknowns Example The law connecting friction F and load L for an experiment is of the form F = aL + b, where a and b are constants. When F = 5. 6, L = 8. 0 and when F = 4. 4, L = 2. 0. Find the values of a and b and the value of F when L = 6. 5. Substituting the values into the law gives 5. 6 = 8. 0a + b
(1)
4. 4 = 2. 0a + b
(2)
Subtracting equation (2) from equation (1) gives 1. 2 = 6a so a =
1. 2 = 0. 2 6
Simultaneous Equations
•
85
Substituting this value into equation (1) gives 5. 6 = (8. 0 × 0. 2) + b 5. 6 = 1. 6 + b b=4 Checking these values in equation (2) gives, (2. 0 × 0. 2) + 4 = 0. 4 + 4 = 4. 4 so correct. When L = 6. 5, F = (0. 2 × 6. 5) + 4 = 1. 3 + 4 = 5. 3. Example When Kirchhoff’s laws are applied to the electrical circuit shown, the currents i1 and i2 are connected by the equations: 27 = 1. 5i1 + 8 (i1 − i2 )
(1)
−26 = 2i2 − 8 (i1 − i2 ) i1
i2 (i1 – i2)
27
1.5 Ω
(2)
8Ω
26
2Ω
Solve the equations for the currents i1 and i2 . Removing the brackets from equation (1) gives 27 = 1. 5i1 + 8i1 − 8i2 Rearranging gives 9. 5i1 − 8i2 = 27
(3)
Removing the brackets from equation (2) gives −26 = 2i2 − 8i1 + 8i2 Rearranging gives −8i1 + 10i2 = −26
(4)
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•
Mathematics
Multiplying equation (3) by 10 gives 95i1 − 80i2 = 270
(5)
−64i1 + 80i2 = −208
(6)
Multiplying equation (4) by 8 gives
Adding equations (5) and (6) gives 31i1 = 62 i1 =
62 =2 31
Substituting in equation (5) gives 190 − 80i2 = 270 −80i2 = 80 So
i2 = −1
Checking in equation (4) gives (−8 × 2) + (10 × −1) = (−16) + (−10) = −26 So correct. Example The distance, s metres from a fixed point, of a vehicle travelling in a straight line with constant acceleration, a m/s2 , is given by s = ut + 12 at2 , where u is the initial velocity in m/s and t is the time in seconds. Determine the initial velocity and acceleration given that s = 42 m when t = 2 s and s = 144 m when t = 4 s. Find also the distance travelled after 3 s. 1 1 Substituting s = 42 and t = 2 into s = ut + at2 gives 42 = 2u + a (2)2 2 2 So 42 = 2u + 2a
(1)
1 1 Substituting s = 144 and t = 4 into s = ut + at2 gives 144 = 4u + a (4)2 2 2 So
144 = 4u + 8a
(2)
Multiplying equation (1) by 2 gives 84 = 4u + 4a
(3)
Simultaneous Equations
•
87
Subtracting equation (3) from equation (2) gives 60 = 4a a=
60 4
a = 15 Substituting into equation (1) gives 42 = 2u + (2 × 15) 42 = 2u + 30 So 2u = 12 u=
12 =6 2
Checking in equation (2) gives (4 × 6) + (8 × 15) = 24 + 120 = 144 So correct. The distance travelled after 3 seconds is found by substituting t = 3, u = 6 and a = 15 1 into s = ut + at2 . 2 So s = (6 × 3) +
1 15 × 9 (15) (3)2 = 18 + = 18 + 67. 5 = 85. 5 m. 2 2
Questions 1. In a system of pulleys, the effort P required to raise a load W is given by P = aW + b, where a and b are constants. If W = 40 when P = 12 and W = 90 when P = 22, find the values of a and b. 2. Applying Kirchhoff’s laws to an electrical circuit produces the following equations: 5 = 0. 2i1 + 2 (i1 − i2 ) 12 = 3i2 + 0. 4i2 − 2 (i1 − i2 ) Determine the values of i1 and i2 · i1 and i2 . 3. Velocity v is given by the formula v = u + at. If v = 20 when t = 2 and v = 40 when t = 7 find the values of u and a. Find also the value of v when t = 3. 5.
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4. The molar heat capacity of a solid compound is given by the equation c = a + bt. When c = 52, T = 100 and when c = 172, T = 400. Find the values of a and b. b 5. In an engineering process two variables p and q are related by q = ap + , where a p and b are constants. Evaluate a and b if q = 13 when p = 2 and q = 22 when p = 5. 6. Equations connecting parallel resistances in a circuit are 4 6 9 + + =6 R1 R2 R3 2 1 15 11 + + =8 R1 R2 R3 12 If R2 = R3 , find the values of R1 , R2 and R3 . 7. x grams of cartridge brass (70% copper, 30% zinc, by mass) and y grams of naval brass (62% copper, 37% zinc, 1% tin, by mass) when fused together give a new alloy containing 342.6 grams of copper and 165.6 grams of zinc. Find x and y.
Answers 1.
a=
1 , b=4 5
2.
i1 = 6. 47, i2 = 4. 62
3.
u = 12, a = 4, v = 26
5.
a = 4, b = 10
6.
R1 = 4, R2 = R3 = 3
7.
X = 330, Y = 180
4.
a = 12, b = 0. 40
A More Complicated Situation Given the simultaneous equations 1. 5x × 2y = 18 4x × 1. 5y = 54
find the values of x and y. As seen earlier in the text, a method to deal with unknown values that are powers is to use logarithms.
Simultaneous Equations
•
89
Taking log10 of both sides of the equations gives: log (1. 5x × 2y ) = log (18) log (4x × 1. 5y ) = log (54)
which gives
log (1. 5x ) + log (2y ) = log (18) log (4x ) + log (1. 5y ) = log (54) x × log (1. 5) + y × log (2) = log (18)
therefore and
x × log (4) + y × log (1. 5) = log (54) Inserting the log values gives 0. 1761x + 0. 301y = 1. 2253
(i)
0. 6021x + 0. 1761y = 1. 7324
(ii)
Dividing (i) by 0.1761, and (ii) by 0.6021 gives x + 1. 709y = 7. 127
(iii)
x + 0. 2925y = 2. 877
(iv)
Subtracting (iv) from (iii) gives 1. 4165y = 4. 25 y=3 Substituting y = 3 into (iii) gives x + 1. 709 × 3 = 7. 127 x = 7. 127 − 5. 127 x=2 So x = 2 and y = 3. There are other methods of approach to the solution of simultaneous linear equations. A second method is described below: •
Transpose one of the equations to write one unknown in terms of the other quantities.
•
Substitute this expression in place of the variable in the second equation, producing a single equation with one unknown value.
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•
Mathematics
•
Solve this equation.
•
Calculate the value of the second unknown.
Example 2x + 5y = 34 4x − 3y = 16 Transpose the first equation to write x in terms of the other quantities 2x + 5y = 34 2x = 34 − 5y x = 17 − 2. 5y
(i)
Substitute this expression into the second equation 4x − 3y = 16 4 (17 − 2. 5y) − 3y = 16 68 − 10y − 3y = 16 68 − 13y = 16 68 − 16 = 13y 52 = 13y y=4 Substitute y = 4 into (i) gives x = 17 − 2. 5 × 4 x = 17 − 10 x=7 The values are x = 7 and y = 4. A third method is: •
Transpose both equations to find expressions for the same unknown.
•
Equate these expressions.
•
Solve for the unknown.
•
Substitute this value into one of the original equations andcalculate the second unknown.
Simultaneous Equations
•
91
Using the same example: 2x + 5y = 34
(i)
4x − 3y = 16 means that x = 17 − 2. 5y x = 4 + 0. 75y and so 17 − 2. 5y = 4 + 0. 75y 17 − 4 = 0. 75y + 2. 5y 13 = 3. 25y y=
13 3. 25
y=4 Substituting into (i) gives 2x + 5 × 4 = 34 2x = 34 − 20 2x = 14 x=7 No single method is better or worse than another. Use the method that is most ‘understandable’. If you find algebra to be really confusing there is always Cramer’s Rule which works for two unknowns and is totally arithmetical. Using the same example 2x + 5y = 34 4x − 3y = 16 write down just the numbers.
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Mathematics
So the equations give 2
5
4 −3
34 16
Using the first two columns: calculate (top left times bottom right) minus (top right times bottom left). So
(2 × −3) − (5 × 4) = −6 − 20 = −26.
Replace the first column with the third column and perform the same calculation: 34
5
16 −3
gives (34 × −3) − (5 × 16) = −102 − 80 = −182
Divide this answer by the first answer, that is, variable, x.
−182 = 7. This is the value of the first −26
Replace the second column of the original block by the third column and repeat the process to find the value of the second variable, y. 2
34
4
16
gives (2 × 16) − (34 × 4) = 32 − 136 = −104.
−104 =4=y −26
So the solution is x = 7, y = 4.
Three unknowns Two unknowns need two equations so three unknowns need three equations. Example Find the values for x, y and z which satisfy the equations: 3x + 2y − z = 4
(i)
2x + y + z = 7
(ii)
x−y+z =2
(iii)
Eliminate one variable at a time, in this case, z seems to be the easiest.
Simultaneous Equations
•
93
Add (i) and (ii) 3x + 2y − z = 4
(i)
2x + y + z = 7
(ii)
So
5x + 3y = 11
(iv)
Subtract (iii) from (ii) 2x + y + z = 7
(ii)
x−y+z = 2
(iii)
x + 2y = 5
(v)
Multiply (v) by 5 and subtract (iv) 5x + 10y = 25 5x + 3y = 11
(iv)
7y = 14 y=2 Substitute y = 2 into (v) x+4=5 x=1 Substitute x = 1 and y = 2 into (iii) 1−2+z= 2 z =2+2−1 z=3 Therefore x = 1, y = 2 and z = 3.
Test Examples 4 1. When 21 2 times one number are added to 31 2 times another the result is 19; and when 31 2 times the first number is subtracted from 21 2 times the second, the result is 3. Find the numbers.
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•
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2. Find the value of x and y in the simultaneous equations: 2x 3y 3 − = 3 5 4
and
x y 13 − = 2 4 16
3. Find the values of a and b which satisfy the equations: a (1 + 2b) = 3
and
a (1 − 3b) = 0. 5
4. A man and his wife are 72 and 68 years old respectively, and have one grandson and one grand-daughter. The man’s age is equal to the sum of four times the grandson’s age and three times the grand-daughter’s. The woman age is equal to the sum of three times the grandson’s age and four times the grand-daughter’s. Find the ages of the grandchildren. 5. The difference between two numbers is 2 and the difference between their squares is 6. Find the numbers. 6. The linear law of a simple lifting machine is given by F = a + bm where m = mass lifted, F = effort applied, a and b being constants. In a certain machine it was found that when m = 30 kg, F = 35 N and when m = 70 kg, F = 55 N. Find the constants a and b, write the law for the lifting machine and find the effort required to lift a mass of 60 kg. 7. Two ships, A and B, leave one port bound for another on the same course. B leaves one hour later than A and overtakes in 8 h. If the speeds of each ship had been 4 knots slower, B would have overtaken A 2 h earlier. Find the original speeds of the ships. 8. Given the simultaneous equations: x 6 + =4 2−y x
and
2x 9 − =1 2−y x
find the values of x 1 , , x and y. 2−y x 9. Find the values of x and y in the simultaneous equations: 2x = 4y
and
4x−1 = 2y+1
10. Given the simultaneous equations 1. 259x+1 × 1. 175y−1 = 2. 323 3. 162x × 1. 778y = 25. 12 find the values of x and y.
Simultaneous Equations
•
95
11. Given the following relationship: d2 x 2 y2 = 2= 2 2 x y D express x in terms of d and D, and find the values of x and y when D = 75 and d = 25. 12. Find the values for a, b and c in the simultaneous equations: 3a + 6b − 2c = 7. 25 2a + 3b + 4c = 26 4a − 2b + c = 10. 25
5
QUADRATIC EQUATIONS A quadratic equation is one where the highest power of the unknown is 2. For example, x 2 − 3x − 6 = 0 is a quadratic equation. There are 3 main methods of solving of this type of equation: 1. by Factorisation (not always possible), 2. by ‘Completing the square’, 3. by Using the ‘quadratic formula’. The solutions of the quadratic equation are called the roots of the equation.
Solution by Factorisation When (x + 3) and (x − 2) are multiplied together the result is x 2 + x − 6. The reverse process, called factorisation, means that x 2 + x − 6 = (x + 3) (x − 2). For example, the solution of x 2 + x − 6 = 0 is the same as the solution of (x + 3) (x − 2) = 0 This means that either (x + 3) = 0, in which case x = −3, or (x − 2) = 0, in which case x = 2. Example Solve x 2 − 4x − 12 = 0 by factorisation.
Quadratic Equations
•
97
The first term in each bracket is x. −12 is the product of, either, −12 and +1, +12 and −1, −6 and +2, +6 and −2, −4 and +3 or +4 and −3. The only pair that gives −4 as the x coefficient is −6 and +2. So x 2 − 4x − 12 = (x − 6) (x + 2) Either
x−6= 0
i.e. x = 6,
x+2=0
i.e. x = −2
or
So the roots of x 2 − 4x − 12 = 0 are x = 6 and x = −2. Example Solve 3x 2 − 11x − 4 = 0 by factorisation. The factors of 3x 2 are 3x and x, so the brackets are (3x
) and (x
).
The factors of −4 are −4 and +1, +4 and −1 and +2 and −2. The x coefficient is given by the outside product added to the inside product which, in this case, is −11. As a diagram, this means that (3x
?) and (x=
#
?) when added together give −11x.
The only possible option is (3x + 1) (x − 4) giving either 3x = −1 or x = 4. 1 The roots of 3x 2 − 11x − 4 = 0 are therefore x = − and x = 4. 3 Example By factorising, find the roots of x 2 − 8x + 16 = 0. x 2 − 8x + 16 = 0 → (x − 4) (x − 4) = 0
i. e. (x − 4)2 = 0
The LHS is called a ‘perfect square’ and so x = 4 is the only root of x 2 − 8x + 16 = 0. Example By factorising, find the roots of 9x 2 − 49 = 0. The LHS is called ‘the difference of two squares’ as 9x 2 − 49 = (3x)2 − 72 . 7 7 9x 2 − 49 factorises to give (3x + 7) (3x − 7) = 0 and so the roots are − and + . 3 3
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Mathematics
Given the roots If the two roots are α and β then (x − α) (x − β) = 0. Example The roots of a quadratic equation are If α =
1 and −2. Determine the equation. 3
1 1 (x − (−2)) = 0 and β = −2 then x − 3 3
So 1 (x + 2) = 0 x− 3 2 1 x 2 + 2x − x − = 0 3 3 2 5 x2 + x − = 0 3 3 Multiplying by 3 eliminates the fractions so 3x 2 + 5x − 2 = 0.
Questions Solve the following questions by factorisation (if possible):
1.
x 2 + 6x + 8 = 0
2.
x 2 + 10x + 16 = 0
3.
x 2 + 12x + 27 = 0
4.
x 2 + 8x + 15 = 0
5.
x 2 + 7x + 10 = 0
6.
x 2 + 3x − 18 = 0
7.
x 2 + 3x − 10 = 0
8.
x 2 − 11x + 28 = 0
9.
x 2 + 4x + 3 = 0
10.
x 2 − 6x + 5 = 0
11.
x 2 + 11x + 24 = 0
12.
x 2 + 15x + 36 = 0
13.
t2 + 19t + 34 = 0
14.
t2 + 18t + 65 = 0
15.
t2 − 7t − 44 = 0
16.
s2 + s − 30 = 0
17.
m2 + 5m − 36 = 0
18.
q2 − 5q − 66 = 0
19.
c2 − 11c + 18 = 0
20.
x2 − x − 1 = 0
Quadratic Equations
•
99
Answers 1.
(x + 2)(x + 4)
2.
(x + 2)(x + 8)
3.
(x + 3)(x + 9)
4.
(x + 3)(x + 5)
5.
(x + 2)(x + 5)
6.
(x + 6)(x − 3)
7.
(x + 5)(x − 2)
8.
(x − 4)(x − 7)
9.
(x + 3)(x + 1)
10.
(x − 5)(x − 1)
11.
(x + 3)(x + 8)
12.
(x + 12)(x + 3)
13.
(t + 17)(t + 2)
14.
(t + 13)(t + 5)
15.
(t − 11)(t + 4)
16.
(s + 6)(s − 5)
17.
(m + 9)(m − 4)
18.
(q − 11)(q + 6)
19.
(c − 9)(c − 2)
20.
Does not factorise
Solution by ‘Completing the Square’ As shown earlier, expressions like (x − 4)2 , x 2 and (x + 1)2 are perfect squares. √ If x 2 = 5, then x = ± 5. √ √ If (x − 4)2 = 5, then x − 4 = ± 5 and therefore x = +4 ± 5. √ √ If (x + 1)2 = 5, then x + 1 = ± 5 and therefore x = −1 ± 5. So if the quadratic equation can be rearranged so the LHS of the equation is a perfect square and the RHS is a number, then the roots can be found, as in the examples above. The process of rearranging the equation into a perfect square is called ‘completing the square’. The method is shown by using the following example. Example Solve 2x 2 + 5x = 3 by ‘completing the square. The process is as follows: 1. Rearrange the equation so all the terms are on the same side of the equals sign and that the x 2 term is positive Hence 2x 2 + 5x − 3 = 0 2. Make the coefficient of the x 2 term 1. In this example, this is done by dividing everything by 2. Hence 3 2 2 5 x + x− =0 2 2 2 3 5 i. e. x 2 + x − = 0 2 2
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•
Mathematics
3. Rearrange the equation so the x 2 and x terms are on the LHS of the equals sign and the constant is on the RHS. So 3 5 x2 + x = 2 2 4. Take the coefficient of x, half it and square, the answer. Add this value to both sides 5 5 of the equation. In this case the coefficient is , halving it gives and squaring it 2 4 2 5 gives . So 4 5 x2 + x + 2
2 2 3 5 5 = + 4 2 4
5 The LHS is now a perfect square, that is, x 2 + x + 2 5. Calculate the value of the RHS. So
2 5 2 5 = x+ 4 4
5 2 3 25 24 + 25 49 = = x+ = + 4 2 16 16 16 5 7 5 49 = →x+ =± 6. Taking the square root of both sides gives x + 4 16 4 4 5 7 7. Solving this simple equation gives x = − ± 4 4 i. e. and
So the roots of 2x 2 + 5x = 3 are
5 7 2 1 x=− + = = 4 4 4 2 5 7 12 x = − − = − = −3 4 4 4 1 and −3. 2
Example Solve 2x 2 + 9x + 8 = 0 by completing the square. 2x 2 + 9x + 8 = 0 9 x2 + x + 4 = 0 2 9 x 2 + x = −4 2
Quadratic Equations
•
2 2 9 9 9 x2 + x + = −4 + 2 4 4 2 9 81 x+ = −4 + 4 16 2 9 −64 + 81 x+ = 4 16 2 9 17 x+ = 4 16 9 17 x+ =± 4 16 = ±1. 031 x = −2. 25 ± 1. 031 x = −1. 22 or − 3. 28 Answers given are correct to 3 significant figures. Example Solve 4. 6x 2 + 3. 5x − 1. 75 = 0 by completing the square. Answers to be given correct to 3 decimal places. 4. 6x 2 + 3. 5x − 1. 75 = 0 4. 6x 2 + 3. 5x = 1. 75 3. 5 x= 4. 6 3. 5 x2 + x= 4. 6 3. 5 3. 5 2 x+ = x2 + 4. 6 9. 2 x2 +
1. 75 4. 6 1. 75 4. 6 1. 75 + 4. 6
3. 5 9. 2
2
3. 5 2 The LHS is now a perfect square so x + = 0. 5251654 9. 2 x+ That is,
3. 5 √ 3. 5 = 0. 5251654 = ±0. 7246830 → x = − ± 0. 7246830 9. 2 9. 2
x = 0. 344 or − 1. 105
101
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Mathematics
Questions Solve the following equations using the method of ‘completing the square’. Give the answers correct to 3 decimal places. 1.
x 2 + 4x + 1 = 0
2.
2x 2 + 5x − 4 = 0
3.
3x 2 − x − 5 = 0
4.
5x 2 − 8x + 2 = 0
5.
4x 2 − 11x + 3 = 0
6.
2x 2 + 5x = 2
Answers 1.
−3. 732, −0. 268
2.
−3. 137, 0. 637
3.
1. 468, −1. 135
4.
1. 290, 0. 310
5.
2. 443, 0. 307
6.
−2. 851, 0. 351
Solution of Quadratic Equations by Formula The general form of a quadratic equation is ax 2 + bx + c = 0, where a, b and c are constants. If the same procedure is applied to this equation as that applied to Examples 5, 6 and 7 the result is known as the quadratic formula and is given by: √ −b ± b2 − 4ac x= 2a Example Using the quadratic formula, solve x 2 + 2x − 8 = 0. Comparing x 2 + 2x − 8 = 0 with ax2 + bx + c = 0 gives a = 1, b = 2 and c = −8. √ 2 −2 ± (2)2 − 4 × 1 × (−8) −b ± b − 4ac gives x = Substituting these values into x = 2a 2×1 √ √ −2 + 6 −2 − 6 −2 ± 4 + 32 −2 ± 36 −2 ± 6 = = so x = or = 2 or −4 So x = 2 2 2 2 2 (The equation could have been solved by factorisation as x 2 + 2x − 8 = (x + 4) (x − 2) So (x + 4) (x − 2) = 0 gives x = −4 or x = 2.)
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Example Using the quadratic formula, solve 3x 2 − 11x − 4 = 0. This gives a = 3, b = −11 and c = −4. √ √ − (−11) ± (−11)2 − 4 × 3 × (−4) 11 ± 121 + 48 11 ± 169 11 ± 13 = = = So x = 2×3 6 6 6 This gives x =
2 1 24 or − , i.e. x = 4 or − 6 6 3
Example Using the quadratic formula, solve 4x 2 + 7x + 2 = 0, giving the answers correct to 2 decimal places. This gives a = 4, b = 7 and c = 2. √ √ √ −7 ± 72 − 4 × 4 × 2 −7 ± 49 − 32 −7 ± 17 −7 ± 4. 123 = = = So x = 2×4 8 8 8 −7 − 4. 123 −7 + 4. 123 = −0. 36 or x = = −1. 39 x= 8 8 Example Using the quadratic formula, solve significant figures.
3 x+2 + = 7, giving the answers correct to 4 4 x−1
Multiplying throughout by 4(x − 1) gives 4 (x − 1) × 4 (x − 1) × 7
x+2 3 + 4 (x − 1) × = 4 x−1
So (x − 1) (x + 2) + 12 = 28 (x − 1) x 2 + 2x − x − 2 + 12 = 28x − 28 So
x 2 − 27x + 38 = 0
Putting a= 1, b = −27 and c = 38 into the formula gives x = − (−27) ± (−27)2 − 4 × 1 × 38 2×1 √ √ 27 ± 729 − 152 27 ± 577 27 ± 24. 0208 = = So x = 2 2 2 This gives x =
51. 0208 2. 9792 = 25. 51 or = 1. 490. 2 2
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Mathematics
Questions Solve the following equations using the quadratic formula. Give the answers correct to 3 decimal places.
1.
2x 2 + 5x − 4 = 0
4.
4x + 5 =
3 x
2.
5. 76x 2 + 2. 86x − 1. 35 = 0
3.
2x 2 − 7x + 4 = 0
5.
(2x + 1) =
5 x−3
6.
x+1 = x−3 x−1
Answers 1.
0.637, −3. 137
2.
0.296, −0. 792
3.
2.781, 0.719
4.
0.443, −1. 693
5.
3.608, −1. 108
6.
4.562, 0.438
Practical questions involving quadratic equations Example The area of a rectangle is 23.6 cm2 and its width is 3.10 cm shorter than its length. Determine the dimensions of the rectangle, correct to 3 significant figures. Let the length of the rectangle be L cm. Then the width is (L − 3. 10) cm. Area = length × width = L × (L − 3. 10) = 23. 6 So L2 − 3. 10L − 23. 6 = 0 Using the formula: √ − (−3. 10) ± (−3. 10)2 − 4 (1) (−23. 6) 3. 10 ± 9. 61 + 94. 4 3. 10 ± 10. 20 = = L= 2×1 2 2 So L =
7. 10 13. 30 = 6. 65 cm or − = −3. 55 cm. 2 2
Length cannot be negative so L = 6. 65 cm and width = 6. 65 − 3. 10 = 3. 55 cm. Check: 6. 65 × 3. 55 = 23. 6075 = 23. 6 correct to 3 significant figures.
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105
Example Calculate the diameter of a solid cylinder which has a height of 82.0 cm and a total surface area of 2.0 m2 . Total surface area = curved surface area + area of both circular ends. So A = 2πrh + 2πr2 where r = radius and h = height. Since the area is in m2 the height should be in metres. Therefore h = 0. 82 m. So 2. 0 = 2πr (0. 82) + 2πr2 → 2πr2 + 2πr (0. 82) − 2. 0 = 0 Dividing by 2 throughout gives πr2 + πr (0. 82) − 1. 0 = 0 Using the quadratic formula with a = π, b = 0. 82π and c = −1 gives − (0. 82π) ± (0. 82π)2 − 4π (−1) − (0. 82π) ± (0. 82π)2 + 4π = r= 2π 2π √ −2. 5761 ± 19. 2027 = 6. 2832 This gives either r = 0. 2874 m or r = −1. 1074 m. So the radius of the cylinder is 0.2874 m, making the diameter 0.575 m, correct to 3 significant figures. Example The height, s metres, of a mass projected vertically upwards at time t seconds is 1 s = ut − gt2 , where u is the initial velocity and g is the acceleration due to gravity. 2 Take u as 30 m/s g as 9.81 m/s2 . Determine how long the mass will take to reach a height of 16 m (a) on the ascent, (b) on the descent When s = 16 m, 16 = 30t −
1 (9. 81) t2 2
16 = 30t − 4. 905t2 4. 905t2 − 30t + 16 = 0
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Using the formula − (−30) ± (−30)2 − 4 (4. 905) (16) 2 (4. 905)
So either t =
√ √ 30 ± 900 − 313. 92 30 ± 586. 08 = 9. 81 9. 81 30 ± 24. 21 = 9. 81
=
54. 21 = 5. 53 s or t = 0. 59 s, correct to 2 decimal places. 9. 81
The answers are therefore (a) 0.59 s and (b) 5.53 s. Example Two ships sail from one port to another, a distance of 825 nautical miles, on the same course, the speed of one ship being 4 knots faster than the other. The fast ship leaves port 2 h after the slow ship and arrives at their destination 14 h sooner. Find the speeds of the two ships. Let x knots = speed of slow ship then (x + 4) knots = speed of fast ship. Difference in times for the two ships to cover the journey = 2 + 14 = 16 h An equation can be formed from the times taken on the journey: Time taken by slow ship − Time taken by fast ship = 16 h Time =
Distance Speed
therefore, So
Distance travelled by slow ship Distance travelled by fast ship − = 16 Speed of slow ship Speed of fast ship
825 825 = 16 − (x + 4) x
Multiplying every term by x(x + 4) and simplifying, 825 825 · x · (x + 4) = 16 · x · (x + 4) · x · (x + 4) − (x + 4) x 825 (x + 4) − 825x = 16x 2 + 64x 825x + 3300 − 825x = 16x 2 + 64x 3300 = 16x 2 + 64x 6x 2 + 64x − 3300 = 0 Dividing throughout by 16, x 2 + 4x − 206. 25 = 0
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Solving this quadratic by formula: 42 − 4 × 1 × (−206. 25) x= 2×1 √ √ −4 ± 16 + 825 −4 ± 841 = = 2 2 33 −4 ± 29 25 = or − = 12. 5 or − 16. 5 So x = 2 2 2 −4 ±
The minus quantity is not realistic, the practical value of x is therefore 12.5. Speed of slow ship = 12.5 knots, speed of fast ship = 16.5 knots.
Questions Practical problems involving quadratic equations: 1 1. The angle a rotating shaft turns through in t seconds is given by θ = ωt + αt2 . 2 Determine the time taken to complete 4 radians if ω = 3. 0 rad/s and α = 0. 60 rad/s2 . 2. The power P developed in an electrical circuit is given by P = 10i − 8i2 , where I is the current in amperes. Determine the current necessary to produce a power of 2.5 W in the circuit. 3. The sag, x metres, in a cable stretched between two supports distance L metres apart 12 is given by L = + x. x Determine the sag when the distance between the supports is 20 m. x2 , determine x given that v = 10. 4. The constant K is 1. 8 × 10−5 . Given that K = v (1 − x) 3x (20 − x) 5. The bending moment M at a point in a beam is given by M = , where x 2 metres is the distance from the point of support. Determine the distance x when the bending moment is 50 Nm. 6. If two resistances, x and y, are connected in series their total resistance is x + y. 1 1 1 When connected in parallel their total resistance is such that = + . R x y Two resistors, when connected in series, have a total resistance of 40 ohms. When connected in parallel their total resistance is 8.4 ohms. If one of these resistors is Rx ohms: show that R2x − 40Rx + 336 = 0 and calculate the resistance of each.
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Mathematics
Answers 1.
1.191 s
2.
0.345 A or 0.905 A
3.
0.619 m or 19.38 m
4.
0.0133
5.
1.835 m or 18.165 m
6.
12 ohms, 28 ohms
Simultaneous Equations – One Linear and One Quadratic Two simultaneous equations, one linear and the other quadratic, usually have two pairs of solutions. Graphically, the equations give rise to a straight line and a curve. One possibility is shown in Figure 5.1. The solutions to the equations are the coordinates of the points of intersection. If, however, the graphs resembled Figure 5.2, there would only be one pair of answers. If the graph resembled Figure 5.3 there would be no solutions. It must be noted that graphical means, unless performed by a computer, are generally inaccurate and time consuming, so an algebraic method is used. The form of the equations defines the method to be used. Example (a) y = 2x 2 + 3x − 5 and y = 4x − 2 (1) y = 2x 2 + 3x − 5 and y = 4x − 2, so 2x 2 + 3x − 5 = 4x − 2 (2) Rearranging this equation gives: 2x 2 + 3x − 5 − 4x + 2 = 0
Figure 5.1
Quadratic Equations
Figure 5.2
Figure 5.3
(3) Resulting in the quadratic equation: 2x 2 − x − 3 = 0 (4) This factorises to give: (2x − 3) (x + 1) = 0 3 (5) Therefore x = or x = −1 2 3 3 (6) y = 4x − 2, so when x = , y = 4 × − 2 = 4 2 2 (7) x = −1, y = 4 × (−1) − 2 = −6 3 , 4 and (−1, −6) (8) The two pairs of answers are 2 −b ± b2 − 4ac (9) Using the formula x = 2a [a] In this case, a = 2, b = −1 and c = −3, giving
(−1)2 − 4 × 2 × (−3) − (−1) ± x= 2×2
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109
110
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Mathematics
[b] So x = −4 4
+1 ±
(+1 + 24) 1 ± (25) 1 ± 5 1 + 5 1−5 6 = = = and = and 4 4 4 4 4 4
[c] This gives x = above
3 and x = −1, the question then continues as shown at (6) 2
(b) 2x 2 + y2 = 3 and x + y = 2 (1) The substitution method works best for this type of question. (2) x + y = 2 means that y = 2 − x (3) Substituting this into 2x 2 + y2 = 3 gives 2x 2 + (2 − x)2 = 3 (4) Expanding the brackets and collecting terms gives (5) 2x 2 + (2 − x)2 = 3 ⇒ 2x 2 + 4 − 4x + x 2 = 3 ⇒ 3x 2 − 4x + 1 = 0 1 (6) This factorises to give (x − 1) (3x − 1) = 0 so x = 1 and x = (or use formula) 3 1 1 5 (7) y = 2 − x so when x = 1, y = 2 − 1 = 1, and when x = , y = 2 − = 3 3 3 1 5 , (8) The two pairs of answers are (1, 1) and 3 3 Before the section on the solution of cubic equations is discussed, it will be useful to go over the work on ‘Polynomial division’.
Recap: Polynomial Division A polynomial is an expression of the form f (x) = a + bx + cx 2 + dx 3 + K. Polynomial division is sometimes required when solving equations with high powers. Example Divide 2x 2 + x − 3 by x − 1 2x 2 + x − 3 is called the dividend and x − 1 the divisor. The usual layout is shown below with both the dividend and divisor shown in descending order of powers.
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111
2x + 3 x − 1 2x 2 + x − 3 2x 2 − 2x 3x − 3 3x − 3 0 Reasoning (1) Dividing the first term of the dividend by the first term of the divisor. 2x 2 (2) That is, gives 2x, which is put above the first term of the dividend as shown. x (3) The divisor is then multiplied by 2x, i.e., 2x(x − 1) = 2x 2 − 2x, which is then placed under the dividend as shown. (4) Subtracting gives 3x and the −3 is ‘dropped down’ to give 3x − 3. (5) The process is then repeated, that is, the first term of the divisor, x, is divided into 3x, giving +3, which is placed above the dividend as shown. (6) Then 3(x − 1) = 3x − 3 which is placed under the 3x − 3. (7) The remainder on subtraction is 0, which completes the division. (8) Therefore 2x 2 + x − 3 ÷ (x − 1) = (2x + 3) A check can be made by multiplying x − 1 by 2x + 3. For further examples see Chapter 2.
Function Notation In most cases seen thus far an equation has usually been given as y = · · · , where the expression following the equals sign usually is an equation in terms of x. In simple terms, this is saying that y is a function of x, that is, the value of y is related to the value of x by the equation. This can be written as y = f (x) where f (x) = · · · . For example, saying y = x 2 is the same as saying f (x) = x 2 . If the value of y, when x = 3, has to be found it is much simpler to write ‘What is f (3)?’ than ‘Find the value of y when x = 3’.
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Mathematics
Questions For f (x) = x 2 − 3x + 4, find: 1.
f (1)
2.
f (2)
3.
f (−1)
4.
f (0)
9.
f (1) ÷ f (0)
5.
f (−3)
For f (x) = x 3 + 2x 2 + 3x + 1, find: 6.
f (−1)
7.
f (−2)
8.
f (3) ÷ f (2)
10.
f (3) × f (−3)
Cubic Equations A cubic equation is one which contains the cube of the unknown, so x 3 = 8 is a cubic equation of the simplest form and is solved simply by taking the cube root of both sides. √ Therefore x = 3 8 = 2. However, the form usually associated with the title cubic equation contains either or both of the first and second powers of the unknown as well as its cube. Example x 3 + 2x 2 − x = 2. One method of solution is to bring all terms of the equation to one side of the equals sign, leaving 0 on the other side. This expression can, hopefully, be resolved into its three factors and each of these factors are then, in turn, equated to 0 which produces a value of the unknown that will satisfy the given equation. Example x 3 + 2x 2 − x − 2 = 0 The three factors of this expression are: (x − 1), (x + 1) and (x + 2) By equating each factor to 0, the three roots are obtained. x−1= 0
∴x=1
x+1=0
∴ x = −1
x+2=0
∴ x = −2
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113
The roots of this equation are +1, −1 and −2, and any one of these values of x will satisfy the given equation. If the equation is not a very easy one like the above, the factors will not be readily seen. The following theorem proves to be very useful.
The Factor Theorem A statement of the theorem is ‘if x = α is a root of the equation f (x) = 0, then (x − α) is a factor of f (x)’ Example Factorise x 3 − 7x − 6 and use the answers to solve the cubic equation x 3 − 7x − 6 = 0. Solution Let f (x) = x 3 − 7x − 6 If
x = 1, then f (1) = 13 − 7(1) − 6 = −12
If
x = 2, then f (2) = 23 − 7(2) − 6 = −12
If
x = 3, then f (3) = 33 − 7(3) − 6 = 0
Since f (3) = 0, (x − 3) is a factor of f (x). At this point there is a choice of methods to follow. Either f (x) can be divided by (x − 3) to give x 2 + 3x + 2, and this can be factorised on sight to give the other two factors (x + 2) and (x + 1). Alternatively, other values of x can be tried. As x gets larger, the cubed term predominates and f (x) gets larger. It therefore makes sense to try negative values of x. If x = −1, f (−1) = (−1)3 − 7 (−1) − 6 = −1 + 7 − 6 = 0 So (x = −1) = (x + 1) is a factor of f (x). If x = −2, f (−2) = (−2)3 − 7 (−2) − 6 = −8 + 14 − 6 = 0 So (x = −2) = (x + 2) is a factor of f (x).
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Therefore x 3 − 7x − 6 = (x − 3) (x + 1) (x + 2) which gives the solutions to f (x) = 0 as x = 3, −1 and −2.
Questions Use the factor theorem to factorise: 1.
x 2 + 2x − 3
2.
x 3 + x 2 − 4x − 4
3.
2x 3 + 5x 2 − 4x − 7
4.
2x 3 − x 2 − 16x + 15
5.
x 3 + 4x 2 − x − 6
6.
x 3 − 2x 2 − x + 2
Answers 1.
f (1) = 0; f (−3) = 0
so
(x − 1) (x + 3)
2.
f (−1) = 0; f (−2) = 0; f (2) = 0
so
3.
f (−1) = 0; no other integers
so
(x + 1) (x + 2) (x − 2) (x + 1) 2x 2 + 3x − 7
4.
f (1) = 0; f (−3) = 0; f (2. 5) = 0
so
(x − 1) (x + 3) (2x − 5)
5.
No integer values so does not factorise
6.
f (−1) = 0; f (2) = 0; f (1) = 0
so
(x + 1) (x − 2) (x − 1)
In such cases, attempt to get one solution by trial and error and from this get the first factor. Dividing the cubic equation by this factor will produce a quadratic equation which can be solved by factors or by quadratic formula. The following examples show how this is done. Example Find the roots of the equation: x 3 − x 2 = 5. 75x − 7. 5. Re-arrange with all terms on one side, in descending powers of x x 3 − x 2 − 5. 75x + 7. 5 = 0 Find the first root by trial Try x = 1, 13 − 12 − 5. 75 × 1 + 7. 5 = 1 − 1 − 5. 75 + 7. 5 = 1. 75 Try x = 2, 23 − 22 − 5. 75 × 2 + 7. 5 = 8 − 4 − 11. 5 + 7. 5 = 0 so this is one solution. x = 2 satisfies the equation, therefore x − 2 must be a factor.
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Divide this factor into the cubic expression, this will produce a quadratic whose roots can be found. 2
x + x − 3. 75 3 x − 2 x − x 2 − 5. 75x + 7. 5
x 3 − 2x 2 +1x 2 − 5. 75x + 7. 5 +1x 2 − 2x −3. 75x + 7. 5 −3. 75x + 7. 5 0
Equate the resulting quadratic to 0 and, if the factors cannot be seen, solve by formula, x 2 + x − 3. 75 = 0. So a = 1, b = 1 and c = −3. 75
x=
−1 ±
(1)2 − 4 × 1 × (−3. 75)
2×1 √ √ −1 ± 1 + 15 −1 ± 16 = = 2 2
So x =
5 3 and − 2 2 x 2 + x − 3. 75 = 0
The roots of the given equation are 2,
5 3 and − . 2 2
The result could be checked by substituting each of these roots into the given cubic equation. Example Solve the following equation: 2x 2 = 17 +
3 x
Multiply throughout by the LCM, which is x, to eliminate the fraction: 2x 3 = 17x + 3
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Arrange all terms on left-hand side and equate to 0, 2x 3 − 17x − 3 = 0 The roots must multiply together to give −3, so choose the factors of 3 as first choices. Find the first root by trial, try x = 1, 2 × 13 − 17 × 1 − 3 = 2 − 17 − 3 = −18 Try x = 3, 2 × 33 − 17 × 3 − 3 = 54 − 51 − 3 = 0 Hence x = 3 is a root and x − 3 must be a factor. Divide the cubic equation by this factor to obtain a quadratic which can be solved, 2
2x + 6x + 1 x − 3 2x 3 − 0x 2 − 17x − 3
2x 3 − 6x 2 + 6x 2 − 17x − 3 + 6x 2 − 18x +x − 3 +x − 3 0 The other two roots can be found by equating the resulting quadratic to zero and b2 − 4ac −b ± solving by using the formula x = 2a −6 ± 62 − 4 × 2 × 1 In this case, a = 2, b = 6 and c = 1, giving x = 2×2 −6 ± (36 − 8) −6 ± (28) −6 ± 5. 2915 = = = −0. 1771 and −2. 8229 So x = 4 4 4 Therefore, the roots of the given cubic equation are 3, −0. 1771 and −2. 8229. Note the last two examples. In the former the last term of the cubic equation is 7.5 and easy factors of this number which come readily to mind are 1 and 7.5, 2 and 3.75, 3 and 2.5, 5 and 1.5; all being
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either positive or negative, and the first root of this equation, by trial, should be one of these. Trial figures for the first root should therefore be chosen with this in mind. If the first root is not a number which can readily be found by trial as in the examples, the process can be quite labour intensive and the equation would probably be more easily solved by plotting an accurate graphical.
Test Examples 5 1. Find the value of x in each of the following equations: (i) (2x + 8)(3x − 5) = 0 (ii) (0. 5x − 10)(0. 25x + 5) = 0 (iii) (5x + 0. 5)(4x + 0. 8) = 0 2. Solve the following equations by the method of factorising: (i) 3x 2 + 2x − 33 = 0 (ii) 4x 2 − 17x + 4 = 0 (iii) 12x 2 + 10x − 12 = 0 3. Solve the following equations by the method of completing the square: 15 (i) x 2 − x − =0 4 (ii) 3x 2 + 2x − 1 = 0 (iii) 4x 2 − 9x + 2 = 0 4. Solve the following equations by the quadratic formula: (i) 3x 2 − 2x + 0. 25 = 0 (ii) 5x 2 + 4x − 5. 52 = 0 (iii) 10x 2 − x − 0. 2 = 0 5. Find the value of x when: log (0. 5x) = 2 × log (x − 6) 6. Find the value of b in the equation: 6b4 − 2. 46b2 + 0. 24 = 0 7. Find the value of V in the equation: V 2.8 − 5. 1V 1.4 + 5. 6 = 0
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8. Find the values of x and y in the following simultaneous equations: x 2 − xy + 2y2 = 1 x + 2y = 8 9. In 5 h less time than it takes a certain ship to travel 330 nm, another ship which is 31/2 knots faster can travel 4 nm further. What are the speeds of the two ships? 10. The area of a rectangle is 76 cm2 and the perimeter is 350 mm. Find the length and breadth. 11. Find the roots of the following cubic equation: 2x 3 + 3x 2 − 17x = 30. 3 16 2− = 3. 12. Find the values of x to satisfy the following equation: 2x − x x
6
GRAPHS A graph is a diagram which shows the relationship between two quantities. Graphs are usually plotted on squared paper (metric). Taking the related quantities to be x and y, Figure 6.1 shows the elements of plotting graphs.
Both quantities vary in value throughout the graph and the value of one depends upon the value of the other. In practice there are often cases where the values of one of the quantities are entirely dependent upon how the other is varied. Take for example the stretching of a spring, if a load is suspended on the spring a certain amount of stretch takes place, if a heavier load is suspended a greater stretch occurs, so the amount of stretch depends on the magnitude of the load. This is called Hooke’s Law. In this example, the load would be referred to as the independent variable and the stretch as the dependent variable. In the drawing of a graph of a function of x such as x 2 − 2x + 5, let the value of this be denoted by y and so y = x 2 − 2x + 5. For a series of chosen values of x the values of y are calculated to obtain a number of plotting points to enable the graph to be drawn. Choose the values of x and the values of y depend upon these, hence x is the independent variable and y the dependent variable. The larger the scale to which the graph is drawn the more accurate the results obtained when reading points from the graph, therefore the scale chosen should be as large as possible depending upon the highest and lowest values to be represented. The various values of x and y (or other similar pairs of variable quantities) are plotted thus, referring to Figure 6.1. The point of intersection of the horizontal and vertical base lines, marked 0, is the common 0 point for both quantities and therefore represents 0 value for both x and y.
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Mathematics
Positive values of y
120
4
y
3 2 1
–4 –2 Negative values of x
Negative values of y
x –6
–1
2 4 Positive values of x
6
–2 –3 –4
Figure 6.1
Positive values of x are measured horizontally to the right from 0 and negative values are measured horizontally to the left. The horizontal base line is referred to as the xx axis. Horizontal measurements are called abscissae (singular: abscissa). Positive values of y are measured vertically above 0 and negative values are measured vertically downwards from 0. The vertical base line is the yy axis. Vertical measurements are called ordinates.
Plotting of Straight-Line Graphs Consider an equation of the form y = a + bx where a and b are any constant quantities and may be positive or negative. Example Let a = 2 and b = 1. 5, so y = 2 + 1. 5x For various values of x between the limits of, say, x = −2 and x = +4, the y values are calculated as: When x = −2, y = 2 − 3 = −1 When x = −1, y = 2 − 1. 5 = +0. 5 When x = 0,
y = 2 + 0 = +2
Graphs 10
•
121
y
8
6
y = 2 + 1.5x
4 2 x –2
–1
1
2
3
4
–2
Figure 6.2
When x = +1, y = 2 + 1. 5 = +3. 5 When x = +2, y = 2 + 3 = +5 When x = +3, y = 2 + 4. 5 = +6. 5 When x = +4, y = 2 + 6 = +8 These co-ordinates can be plotted and a graph drawn through the plotted points as in Figure 6.2. Note (i) The graph is a straight line, and with this knowledge the graph of any equation of the form y = a + bx may be plotted with two pairs of values only to give two points through which to draw the straight line. (Although it is best to use three in case a mistake has been made in the calculations.) (ii) From left to right the graph slopes upwards and the value of b determines the gradient of the slope. In this case, for every increase of 1 in x value there is an increase of 1.5 in the y value; if b had been greater than 1.5 the slope would have been greater and vice-versa. The graph slopes upwards because b is positive, if b had been negative the graph would have sloped downwards. (iii) The value of y when x is 0 is 2, and this is the value of a. It is the intercept on the y axis.
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•
16
Mathematics y
y=4+x
14 12 10
y = 3 + 0.5x
8 6 4 2
y = 2 – 0.25x 1
2
3
4
5
6
–2
x 7
8
9
10
11
12
y = –0.5x
–4 –6
y = –2 – 0.5x
–8
Figure 6.3
Now consider the equations, •
y =4+x
•
y = 3 + 0. 5x
•
y = 2 − 0. 25x
•
y = −0. 5x
•
y = −2 − 0. 5x
All of these are of the general form y = a + bx. Graphs of these equations are plotted in Figure 6.3 between the limits x = 0 and x = 12. Examine these graphs carefully.
Determining the Equation to a Straight-Line Graph Straight-line graphs are examples of direct or indirect proportion. Another way the equation of a straight line can be written is y = mx + c where y is the ‘vertical’ variable and x is the ‘horizontal’ variable. The value of m gives the slope of the graph and the value of c gives the intercept on the y axis that is, the line where x = 0.
Graphs
•
123
If the graph slopes from bottom left to top right, the slope is positive; if it slopes from top left to bottom right the slope is negative. To calculate the slope once the graph has been drawn: •
draw a right-angled triangle using the straight line as the hypotenuse;
•
calculate the lengths of the vertical and horizontal sides (not their physical lengths but what their lengths represent on the scales on the axes);
•
find the value of m by dividing the vertical figure by the horizontal.
If the graph does not have the vertical axis at x = 0, the value of c cannot be read off the scale but has to be calculated. •
Re-arrange the equation to give c = y − mx.
•
Choose any point on the straight line and find its x and y co-ordinates.
•
Substitute these values along with the value of the slope, m, into the formula to find the value of c.
Example The following table shows the cost of printing circuit boards (Figure 6.4).
175
Number printed
5
10
15
25
30
Cost (pence)
35
60
85
135
160
Cost, y Cost of printing circuit boards
150
x B
125 100 75 50
x A
25
–5
Figure 6.4
5
Number of printed, x 10
15
20
25
30
124
•
Mathematics
Represent the information graphically using appropriate scales. (i) Where does the line cross the x and y axes? (ii) What is the equation of the line? (i) By accurate plotting, the values where the line crosses the axes are (−2, 0) and (0, 10) for the x and y axis respectively. (ii) To find the slope, m, choose any two points, A and B say. The co-ordinates of A are (5, 35) and B (25, 135). The vertical difference is 135 − 35 = 100. The horizontal difference is 25 − 5 = 20. Vertical difference 100 The slope is m = = =5 Horizontal difference 20 The intercept has already been found, i.e., c = 10 The equation of the line is therefore y = 5x + 10 (Check the equation by choosing a point, say (15, 85). 85 = 5 × 15 + 10)
Questions Calculate the equation of the straight line that best represents the data given. 1. The table gives a set of readings obtained when a mass of x (g) is hung from a spring, causing it to extend to a length y (cm):
x (g)
10
40
80
100
120
150
y (cm)
116
123
144
152
160
173
Answer: y = 0. 44x + 107 2. The electrical resistance of various lengths of wire are given in the following table:
Length x (mm)
110
120
150
160
190
210
Resistance y (ohms)
3.3
3.6
4.5
4.8
5.7
6.3
Answer: y = 0. 03x
Graphs
•
125
3. The following is a comparison table between temperature measured in degrees Fahrenheit and degrees Celsius:
◦ C(x)
10
25
50
60
75
100
◦ F(y)
50
77
122
140
167
212
Answer: y = 1. 6x + 32 4. The table shows the length of a metal rod at different temperatures. It is assumed that the equation is L = m · t+ c t (◦ C)
0
10
20
30
40
50
60
L (mm)
100
100.02
100.04
100.06
100.08
100.10
100.12
Answer: L = 0. 02t + 10
Example Find the equation of a straight line graph which passes through the points (2, 4) and (10, 7). The slope is m =
Vertical difference 7−4 3 = = = 0. 375. Horizontal difference 10 − 2 8
The intercept c = y − mx where (x, y) is one of the points, say (2, 4). So c = 4 − 2 × 0. 375 = 4 − 0. 75 = 3. 25. The equation is therefore y = 0. 375x + 3. 25. There are many practical applications of determining the equation to a given graph such as when performing a series of experiments on a machine, an engine, or a piece of material under a strength test, to ascertain how one quantity varies with another. For instance, various loads can be hung from the lifting hook of a lifting machine and the effort required to lift each load is found experimentally and tabulated. A graph is then drawn showing how the effort varies with the load. The equation to the graph is determined and this is the law of the particular machine.
126
•
Mathematics
Example In an experiment on a certain machine the following data were observed:
Load, m
20
40
60
80
100
120
Effort, F
4.5
6.1
7.8
8.9
10.4
12.1
If the law of this machine expressing the relationship between effort applied (F) and load lifted (m) can be expressed as the linear equation F = a + bm, plotting a graph representing the above experimental values with a horizontal scale of load will determine the linear law of this machine. After plotting the experimental values a straight line is drawn as near as possible through these points as shown in Figure 6.5. Those points not exactly on this line are probably due to irregularities of the machine or errors of observation during the experiment. The value of a reads 3 The value of b is the increase of effort per unit increase of load. Choosing two points on the line, P1 and P2 , Q to P2 measures 7.5, P1 to Q measures 100, therefore b = 7. 5 ÷ 100 = 0. 075. Hence the linear law of this machine is F = 3 + 0. 075m. 15 F
Load (m) against effort (F ) P2
F = 0.075m + 3
10
7.5 5
Q P1
100
m 20
Figure 6.5
40
60
80
100
120
140
Graphs
•
127
Graphical Solution of Simultaneous Linear Equations The method of solving simultaneous equations by graphical means can best be demonstrated by an example: To find the values of x and y which satisfy the equations, 2x + 5y = 34 and
4x − 3y = 16
Find the value of y in the equation 2x + 5y = 34 5y = 34 − 2x y = 6. 8 − 0. 4x
(i)
Find the value of y in the second equation 4x − 3y = 16 −3y = 16 − 4x 4 16 y= x− 3 3 Two points are required for each line so for y = 6. 8 − 0. 4x when x = 0,
y = 6. 8 − 0 = 6. 8
when x = 10, y = 6. 8 − 4 = 2. 8 which gives the points (0, 6.8) and (10, 2.8) 16 4 For y = x − 3 3 16 when x = 0, y = − 3 16 36 16 20 4 = − = when x = 9, y = × 9 − 3 3 3 3 3 20 16 and 9, which gives the points 0, − 3 3
(ii)
128
•
Mathematics
8 y 6
y = 6.8 – 0.4x
4 2 x 2 –2
4
6
8
10
12
y = 1.33x – 5.33
–4 –6
Figure 6.6
The points x = 0 and x = 10 for the first equation, and x = 0 and x = 9 for the second equation were chosen simply because they appeared to be easy figures for substituting and to produce reasonably sized graphs. x = 0 is obviously the first choice for a plotting point of any graph, the other point could be any value of x but it should be chosen with a view to produce simple figures. The graph is now plotted as shown in Figure 6.6. At the point of intersection of the lines the values are x = 7 and y = 4 and these are the only values which are true for both equations. Hence, x = 7 and y = 4.
Graphical Solution of Quadratic Equations Graphs of equations containing x to the first power are all straight lines. Graphs of equations containing x to other powers, such as x 2 (quadratic equations) and x 3 (cubic equations) are curves, and obviously more than two plotting points are
Graphs
•
129
necessary as a guide to the drawing of a curve; the more plotting points the more accurate the curve can be drawn. As in previous cases the equation to be solved is first simplified, all terms are brought to one side in order of descending powers of the unknown, say x, leaving 0 on the side. Replace the nought by y, that is, y = the given expression and plot a graph for a series of values of x. Where this graph intersects the x axis (the line y = 0) are the values of x which will satisfy the equation. The values of y in proximity to these points therefore change from positive to negative or from negative to positive, the trial values of x for calculating the plotting points for the curve should therefore be chosen with this is in mind. The following worked examples will clarify this explanation.
Example Find, graphically, the values of x which solve the equation, x 2 + 2x = 5. 5x − 1. 96 Simplify and bring all terms to one side, so Let y = x 2 − 3. 5x + 1. 96 Calculate values of y for selected values of x When x = 0, y = 0 − 0 + 1. 96 = +1. 96 When x = 1, y = 1 − 3. 5 + 1. 96 = −0. 54 When x = 2, y = 4 − 7 + 1. 96 = −1. 04 When x = 3, y = 9 − 10. 5 + 1. 96 = +0. 46 When x = 4, y = 16 − 14 + 1. 96 = +3. 96 Note that the value of y changes sign between x = 0 and x = 1, and again between x = 2 and x = 3, therefore the two values of x will be obtained from the graph at these two intersections of the x axis and hence there is no need to plot the graph beyond these limits. The larger the scale the more accurate will be the reading of the values of x. Therefore the graph should be plotted to the largest scale possible on the paper.
130
•
Mathematics
3 y
2
1 x 1
2
3
–1
–2
Figure 6.7
Results of greater accuracy could be obtained by calculating values of y for a few points between x = 0 and x = +1, and also between x = +2 and x = +3 and drawing to a larger scale only those two parts of the curve which cross the x axis. By drawing the graph as shown in Figure 6.7, when y = 0, x = 0. 7 and 2.8, the graph is concave upwards (i.e. +x 2 ). Example Solve x 2 − 0. 8x − 3. 84 = 0 Let y = x 2 − 0. 8x − 3. 84 When x = 0, y = 0 − 0 − 3. 84 = −3. 84 When x = 1, y = 1 − 0. 8 − 3. 84 = −3. 64 When x = 2, y = 4 − 1. 6 − 3. 84 = −1. 44 When x = 3, y = 9 − 2. 4 − 3. 84 = +2. 76 Note that the value of y changes sign from negative to positive between x = 2 and x = 3, therefore there is no need to proceed further in this direction. The other value where change of sign takes place, that is, from positive to negative, must be when x is a negative quantity.
Graphs
•
131
Proceeding then in this direction, When x = −1, y = 1 + 0. 8 − 3. 84 = −2. 04 When x = −2, y = 4 + 1. 6 − 3. 84 = +1. 76 Knowing that y = 0 between x = −1 and −2, there is sufficient data to plot the graph. It may be of assistance to clear up the above by tabulating these results before attempting to plot them. Thus −2
x
y +1. 76
−1
0
+1
+2
+3
−2. 04
−3. 84
−3. 64
−1. 44
+2. 76
Plotting the graph as in Figure 6.8 and reading the values of x when y = 0, x = 2. 4 or −1. 6 Another method of solving quadratic equations, and one which can often be applied to solve cubic and more complicated equations is demonstrated by the following example. Example To solve x 2 − 4x + This
7 =0 4
can be written as x 2 −
7 =0 4x − 4
3
y
2 1 x = – 1.6 –2
x = 2.4
–1
1 –1 –2 –3 –4 –5
Figure 6.8
2
x 3
132
•
Mathematics
7 So let y1 and let y2 = 4x − 4 more importantly, y1 = y2 . = x2
then
x2 −
7 =0 4x − 4
becomes y1 − y2 = 0 or,
In other words the solutions are the x ordinates of the intersection of the graphs of y1 and y2 Graph plotting points are found for each part: y1 = x 2
(i)
When x = 0, y1 = 0 When x = 1, y1 = 1 When x = 2, y1 = 4 When x = 3, y1 = 9 When x = 4, y1 = 16 y2 = 4x −
7 4
(ii)
This is a straight line equation and two plotting points only are required plus one for checking purposes. When x = 0, y2 = −
7 4
7 25 = 4 4 7 57 When x = 4, y2 = 16 − = 4 4
When x = 2, y2 = 8 −
The curve of y1 = x 2 and the straight line representing y2 = 4x − shown in Figure 6.9.
7 are now plotted as 4
From the graph it is seen that the curves intersect when x = 0. 5 and x = 3. 5. This means x 2 − 4x +
7 = 0 when x = 0. 5 and x = 3. 5. 4
Always check results by substituting them into the original equation to see if they are correct.
Graphs
•
133
18 y 16
y2 = 4 x −
14
7 4
12 10 8 6
y1 = x 2
4 2 1
2
3
4
5
–2 –4
x = 3.5
x = 0.5
Figure 6.9
Graphical Solution of Simultaneous Quadratic and Linear Equations These are dealt with in a similar manner as the last example. Two quadratic equations could be involved, or one quadratic and one linear equation. Example Find the values of x and y which satisfy the equations y = 12 + 3x − 0. 5x 2
and
y = 14 − 1. 25x The extreme values of x to be taken for calculating the plotting points are chosen by making estimates to where the two graphs cross.
134
•
Mathematics
Taking the linear equation first, being a straight line graph only three points are needed: y = 14 − 1. 25x when x = 0,
y = 14 − 0 = 14
when x = 5,
y = 14 − 6. 25 = 7. 75
when x = 10, y = 14 − 12. 5 = 1. 5 Taking the quadratic: y = 12 + 3x − 0. 5x 2 when x = 0, y = 12 + 0 − 0 = 12 when x = 2, y = 12 + 6 − 2 = 16 when x = 4, y = 12 + 12 − 8 = 16 when x = 6, y = 12 + 18 − 18 = 12 when x = 8, y = 12 + 24 − 32 = 4 when x = 9, y = 12 + 27 − 40. 5 = −1. 5 This appears to be sufficient. The graph is convex upwards because the x 2 term is negative.
18 y 16 14
P1
12 10 8 6 4 P2
2
x
1
–2 –4
Figure 6.10
2
3
4
5
6
7
8
9
10
11
12
Graphs
•
135
Drawing the graph as in Figure 6.10, the points of intersection P1 and P2 produce the following values of x and y: x = 0. 5 x=8
and and
y = 13. 375 or y=4
Determination of laws One case has already been shown, that of determining the law connecting effort and load in a lifting machine. This relation was expressed by a straight line equation and the law found quite simply. In experimental work when two variables are thought to be connected, a set of measurements can be taken, plotted as a graph and examined to see if such a link exists. If a link does exist, the equation of the link can be found and used to determine further values. If the points, when plotted, lie on a straight line then the relationship is, y = a + bx, where b is the slope and a the intercept on the y axis. However, in the majority of cases the points will not lie on a straight line but on a curve. In such cases the nonlinear equation can be modified to produce a linear one, enabling the constants in the equation to be found. Some of the more common types are given below. If it suspected that the link is: 1. y = ax 2 + b Plotting y against x 2 should result in a straight line where the slope is a and the intercept is b. a 2. y = + b x 1 Plotting y against should result in a straight line where the slope is a and the x intercept is b. 3. y = a · x n Taking logarithms (usually base 10, but not essentially) gives: log(y) = log a · x n so log(y) = log(a) + log x n so log(y) = log(a) + n. log (x)
136
•
Mathematics
Plotting log(y) against log(x) should result in a straight line where the slope is n and the intercept is log(a). 4. y = a · bx Taking logarithms (usually base 10, but not essentially) gives: log(y) = log a · bx so log(y) = log(a) + log bx so log(y) = log(a) + x. log(b) Plotting log(y) against x should result in a straight line where the slope is log(b) and the intercept is log(a). 5. y = A · ebx Taking natural logarithms gives: ln(y) = ln A · ebx so ln(y) = ln(A) + bx. ln(e) so ln(y) = ln(A) + bx Plotting ln(y) against x should result in a straight line with slope b and intercept ln(A). One typical example is that of a mass of a gas being expanded or compressed in a cylinder. The law connecting the variation of pressure as the volume of the gas is increased or decreased by the movement of the piston is: p · V n = constant, C. Writing this equation in log form (as in 3. above): log P + n log V = log C and transposing to express log p in terms of the other quantities: log P = log C − n log V It is now reduced to a straight line equation similar to y = a + bx the variable log p taking the place of the variable y the variable log V taking the place of the variable x the constant log C taking the place of the constant a the constant n taking the place of the constant b
Graphs
•
137
A graph of log p against log V is now plotted and the values of n and C are determined from the slope and intercept. Example The following ordinates and abscissae were measured from part of the expansion curve of an indicator diagram off an I.C. engine, where p is the pressure and V the volume of the gases in the cylinder. If the law of expansion can be expressed by p · V n = C estimate the value of n. p
28
24
16.6
13
9.7
6.8
4.7
V
0.6
0.7
0.9
1.1
1.4
1.8
2.4
Tabulating the values of p and V with their respective logarithms:
p
log(p)
V
log(V)
28.0
1.4472
0.6
−0.2218
24.0
1.3802
0.7
−0.1549
16.6
1.2201
0.9
−0.0458
13.0
1.1139
1.1
0.0414
9.7
0.9868
1.4
0.1461
6.8
0.8325
1.8
0.2553
4.7
0.6721
2.4
0.3802
The graph is now plotted as shown in Figure 6.11. Note that the lowest value of log p is 0.6721, the graph can be drawn to a larger scale by starting with a value of log p just a little lower than this, say 0.6, instead of commencing with 0 origin. Choosing two points on the graph, n=
decrease in log (p) 0. 65 = = 1. 3 increase in log (V) 0. 5
Therefore n = 1. 3 Thus, log p = log C − 1. 3 log V In the original equation p · V 1.3 = C
138
•
Mathematics 1.5
log(p)
1.4 1.3 1.2 1.1
0.65
1 0.9 0.8
0.5 0.7 log(V ) 0.6
0
Figure 6.11
Curve Sketching It is often useful to sketch the graph by inspection of the algebraic equation. From previous work, the general form y = a + bx: this is a straight line graph, +b slopes upwards, −b slopes downwards, the greater the value of b the steeper the slope, a is the intercept on the y axis (when x = 0). For the quadratic equation, general form is y = ax 2 + bx + c: this is a parabolic graph, +a convex upwards, the greater the a value the steeper the slope (or the ‘narrower’ the curve): The line of symmetry of the parabola is the straight (vertical) line x = −
b : 2a
If the parabola intersects the x axis, the roots of the equation (i.e. when y = 0) are at the √ b2 − 4ac values to the left and right of the line of symmetry. 2a √ −b ± b2 − 4ac In other words, at the values x = (see section on solution of quadratic 2a equations).
Graphs
•
139
Test Examples 6 1. On the same set of axes, plot the following four graphs represented by the equations: (i) y = 2 + x (ii) y = 12 − 1. 5x (iii) y = −1 − 0. 5x (iv) y = −4 + 1. 25x all between the limits of x = 0 and x = 12. 2. A straight line passes through the pair of points (−2, 14.5) and (8, −3). From it derive the equation to the graph: 3. Plot a graph using the following values and find the equation of the graph. x
−2
−1
0
1
2
3
y
10
7
4
1
−2
−5
4. The following data were taken during an experiment on a small turbine where P represents the power developed and m the rate of consumption of steam. Assuming that the relationship between P and m can be represented by the straight line equation m = a + bP, draw a straight line as near as possible through the plotted points of the experimental results, estimate the values of a and b and hence the law connecting p and m. P
20
25
30
35
40
45
50
m
220
265
315
365
410
455
505
5. Find, graphically, the values of x and y which satisfy the simultaneous equations, 3x + 5y = 23 and,
5x − 2y = 12. 5
6. Find, graphically, the values of p and q in the simultaneous equations, 5p − 2q = 5. 6 and 2p − 3q = −4. 8 7. Find, graphically, the value of x in the equation, x 2 − 5x +
21 = 0. 4
140
•
Mathematics
8. Draw the graph of y = 0. 5x 2 − 2x − 6 between the values of x = −4 and x = +8. From the graph read the values of x in the following equations: (i) 0. 5x 2 − 2x − 6 = 0 (ii) 0. 5x 2 − 2x − 4 = 0 (iii) 0. 5x 2 − 2x − 1 = 0 (iv) 0. 5x 2 − 2x = 0 (v) 0. 5x 2 − 2x + 1 = 0 9. On the same set of axes draw graphs of y1 = x 2 and y2 = 3. 5 + 2. 5x between the values of x = −2 and x = +4; find the values of x in the equation x 2 −2. 5x −3. 5 = 0. 10. Find, by graphical means, the values of x and y which satisfy the simultaneous equations, y = 0. 4x 2 − 3x + 2 and y = 1. 4x − 2 11. Using graphical means only and taking values of x between 0 and +6, solve the following pair of simultaneous equations for x and y: y 2 = 16x
and
y = 5 + 8x − 2x 2 √ Note: y2 = 16x can be expressed as y = ±4 x Suggested scales: x axis 2 cm = 1 unit, y axis 2 cm = 4 units. 12. By drawing a graph of y = e−x and y = x 2 solve the equation: x 2 · ex = 1 Note: plot values between x = 0 and x = 1.
7
TRIGONOMETRY AND GEOMETRY
An angle is the corner of two joining lines and the magnitude of an angle is measured in either degrees or radians.
Measurement of Angles The most common unit of angle measurement that is known is probably the degree. There are 360◦ in a circle and so a degree is one three-hundred-and-sixtieth part of a circle. The symbol for a degree is ◦ and so there are 360◦ in a circle. Figure 7.1 shows a quarter of a circle which is 90◦ and is termed a right-angle, an angle which is less than 90◦ (Figure 7.2) is called an acute angle, greater than 90◦ but less than 180◦ (Figure 7.3) is an obtuse angle, and greater 180◦ (Figure 7.4) is a reflex angle.
Right angle 90°
Figure 7.1
142
•
Mathematics
Acute angle 180°
Figure 7.4
One sixtieth part of a degree is termed one minute and the sixtieth part of a minute is one second. Symbols are used to represent degrees, minutes and seconds. 60 s = 60 = 1 min; 60 min = 60 = 1◦ ; 360◦ = 1 circle. An angle of 35 degrees 23 minutes and 15 seconds is written as 35◦ 23 15 . Notation depends on the circumstance: some situations require angles to be given in decimal form accurate to, say, 3 decimal places; others require answers correct to the nearest tenth of one minute, for example 35◦ 25.7 ; and others may require an accuracy correct to the nearest second, for example 35◦ 23 15 .
Circular measure A radian is the angle subtended at the centre of a circle by an arc equal in length to the radius (Figure 7.5).
Trigonometry and Geometry Therefore 2π radians = 360◦ 180◦ ∼ 1 radian = = 57. 3◦ π
or
57◦17 45
1◦ =
•
143
π ∼ = 0. 01745 radians 180
s
A
r
θ
Figure 7.5
θ is in radians Length of arc, s = r · θ Area of sector, A =
1 2 ·r ·θ 2
Similarly, if a wheel of 0.3 m radius turns through 4 rad in 1 s, then a point on the rim moves at a rate of 4 × 0. 3 = 1. 2 m/s. In symbols: If ω = angular velocity, v = linear velocity and r = radius then v = ωr The angular velocity of a rotating part of a mechanism is often more conveniently expressed in rad/s rather than the more practical rev/min (rpm).
To convert from one to the other Suppose that N = velocity in rev/min and ω = velocity in rad/s So N revolutions per minute = N × 2π = 2πN rad/min = Therefore ω =
2πN . 60
2πN rad/s 60
144
•
Mathematics
Example Express in radians and degrees the angles subtended at the centre of a circle of 50 mm radius, by arc length of, (i) 50, and (ii) 140 mm respectively. (i) Arc length of 50 mm: Since S = rθ ,
θ=
180 s 50 = = 1 rad, 1 rad = ≈ 57. 3◦ r 50 π
(ii) Arc length of 140 mm: θ=
180 s 140 = = 2. 8 rad, 2. 8 rad = 2. 8 × ≈ 160. 4◦ r 50 π
Example A flywheel of 1 m diameter is rotating at 120 rev/min. (i) Express this in rad/s, (ii) Find the linear velocity, in m/s, of a point on the rim. 2πN 2π × 120 240π (iii) ω = = = = 4π rad/s. 60 60 ‘60 It is acceptable to leave the answer in terms of π but as a number ω ≈ 12. 57 rad/s. (iv) v = ωr = 4π × 0. 5 ≈ 6. 28 m/s (Remember the diameter is 1 m)
Questions 1. Express as radians, to 4 decimal places: (a) 36◦
(b) 42◦ 24
(c)
86◦ 45
(d) 228◦ 16
2. Express the following angles, stated in radians, in degrees and minutes: (a) 1.8
(b) 0.7942 (c) 3.2106 (d) 1.7763
3. Express in degrees: 2π 5π 9π π (b) (c) (d) (a) 4 5 9 8 4. Express in radians, as fractions of π: (a) 80◦
(b) 54◦
(c)
135◦
(d) 165◦
5. The pendulum of a longcase clock is 50 cm long and swings through an arc of 8 cm. Through what angle does the pendulum swing? 6. Find the area of the sector of a circle of diameter 12 cm bounded by two radii and an arc length of 10 cm. 7. The mass of a circular brass disc is 280 g. The mass of a sector sheared out of it is 63 g. Find the angle of the sector. 8. An electric motor runs at 1500 rpm. What is the angular speed in radians per second?
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9. A 20 cm diameter flywheel is rotating at 5400 rpm. What is its angular speed and what is the linear speed of a point on its rim? 10. A car with a 50 cm diameter wheel is travelling at 72 km/h. What is the angular speed of rotation of the wheels and how many rpm does this represent? 11. The cross-section of a piece of corrugated plastic used for roofing consists of a series of circular arcs, each 8 cm across and 2 cm deep. Each sheet is 80 cm wide. If the sheet could be flattened, how wide would it be then? 8 cm 2 cm
12. A tramline AXYB consists of two straight portions AX and YB each 500 m long, the angle between them being 132◦ , connected by a circular arc of radius 80 m. What is the total length of the line from A to B? X
A
Y
B
Answers 1.
(a) 0.6284
(b) 0.7399
2.
(a) 103◦ 6
(b) 45◦ 31
3.
(a) 45◦
4.
(a)
5.
9◦ 10
6.
30 cm2
7.
81◦
8.
157 rad/s
9.
566 rad/s, 56.6 m/s
10.
80 rad/s, 764 rpm
11.
92.7 cm
12.
1067 m
4π 9
(b) 72◦ (b)
3π 10
(c) 1.514
(c) 183◦ 57
(c) 100◦ (c)
3π 4
(d) 3.983 (d) 101◦ 46
(d) 202.5◦ (d)
11π 12
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Trigonometric Ratios A right-angled triangle is one which contains an angle of 90◦ , the longest side (opposite the right angle) is termed the hypotenuse, the other two sides are termed the opposite and adjacent depending upon which of the other two angles are under consideration. Considering angle θ (Figure 7.6), side AC is opposite the angle and therefore referred to as such, the other side BC is the adjacent.
Hy
po
Opposite
ten us
e
A
B
θ
C
Adjacent
Figure 7.6
A
B
Adjacent
Hy
po ten us e
α
C
Opposite
Figure 7.7
If angle α is being considered (Figure 7.7), BC is the opposite and AC is the adjacent.
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The ratios of the lengths of the sides of a right-angled triangle are expressed by sine (abbreviated sin), cosine (abbreviated cos) and tangent (abbreviated tan), as follows: sine of angle =
opposite hypotenuse
cosine of angle =
adjacent hypotenuse
tangent of angle =
opposite adjacent
Referring to Figure 7.6, AC AB BC cos θ = AB AC tanθ = BC sin θ =
These are the ratios most often used. Another three ratios are the reciprocals of the three above: namely the cosecant (cosec), the secant (sec) and the cotangent (cot) where 1 AB = sin θ AC AB 1 = sec θ = cos θ BC BC 1 = cotθ = tanθ AC
cosec θ =
Every angle has its own value of sine, cosine and tangent. The sum of the angles in any triangle is 180◦, therefore if one angle of a right-angled triangle is 30◦ the other must be 60◦ , and if one angle is 45◦ the other must be 45◦ . A calculator will give the values of these ratios for any given angle. Example sin 30◦ = 0. 5 This means that
opposite = 0. 5 (from Figure 7.8) hypotenuse
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If the hypotenuse is taken as, say, 200 cm, then
opposite = 0. 5 200
and so opposite= 0. 5 × 200 = 100 cm
200 100 30°
Figure 7.8
Similarly cos 30◦ = 0. 8660 which means that
adjacent = 0. 866, 200
giving adjacent = 173. 2 cm
60° 200 100 30° 173.2
Figure 7.9
As the base angle is 30◦ , the angle at the top must be 60◦ . From Figure 7.9 it can be seen that sin 60◦ = Similarly cos 60◦ =
100 = 0. 5 = sin 30◦ 200
173. 2 = 0. 866 = cos 30◦ 200
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Alternative Notation for the Right-Angled Triangle, ABC Using A as the reference angle:
B
( e, c nus
te ypo
H
A
Adjacent, b
Opposite, a
e)
t sid
es long
C
Hypotenuse: the side AB opposite the right angle, denoted c Opposite:
the side BC opposite to the angle A, denoted a
Adjacent:
the side AC adjacent to the angle A, denoted b
Note: That is, capital letters for angles, small case letters for sides opposite the angle.
Other terms used when describing angles Complementary angles If the sum of any two angles is 90◦ they are said to be complementary angles. For example, 60◦ is the complementary angle to 30◦ , 50◦ is complementary to 40◦ , 70◦ is complementary to 20◦ , and so on. In all such cases the sine of an angle is equal to the cosine of its complement. The notes referring to Figure 7.11 show that the sine of 30◦ is the same as the cosine of 60◦ , and cos 30◦ is the same as sin 60◦ .
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α°
θ°
Figure 7.10
In general, referring to Figure 7.10, sin θ = cos α = cos (90 − θ) cos θ = sin α = sin (90 − θ) tanθ = cot α = cot(90 − θ)
Supplementary angles If the sum of two angles is 180◦ they are said to be supplementary to each other. Thus 80◦ is the supplementary angle of 127◦ , and so on. In order that angles can be more easily understood it is useful to know the following basic facts about straight lines and angles.
Properties of Angles and Straight Lines (i) The total angle on a straight line is 180◦ . Angles A and B are called adjacent angles (being next to each other). They are also supplementary.
A
B
(ii) When two straight lines cross, the opposite angles are equal. χ α
β δ
α=β χ=δ
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Parallel lines When two parallel lines are cut by a transversal the following facts are true: •
the corresponding angles are equal, a=l:b=m:c=p:d=q: a d
b c
l m q
•
the alternate angles are equal, d=m:c=l:
•
the interior angles are supplementary, d + l = 180◦ : c + m = 180◦
p
(a) An acute-angled triangle has all its angles less than 90◦ . (b) A right-angled triangle has one of its angles equal to 90◦ . (c) An obtuse-angled triangle has one angle greater than 90◦ . (d) A scalene triangle has all three sides of different length and hence all three angles are different. (e) An isosceles triangle has two sides and two angles equal. (f ) The three angles add up to 180◦ . (g) The longest side is opposite the largest angle and the shortest side opposite the smallest angle. (h) An equilateral triangle has all three sides the same length and all three angles equal. The angles are therefore 60◦ .
Theorem of Pythagoras This theorem refers only to right-angled triangles. Pythagoras’ Theorem states:
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•
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‘In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.’ A
c b
B
C
a
Figure 7.11
This can be expressed algebraically as AB2 = AC2 + BC2
or c2 = a2 + b2
Pythagoras’ Theorem is generally used to find the length of one side of a right-angled triangle when the other two sides are known (Figure 7.11). Examples (i)
(ii)
x 5
12
x 2 = 52 + 122 = 25 + 144 = 169 √ x = 169 = 13
10
7
x
102 = x 2 + 72 x 2 = 102 − 72 = 100 − 49 = 51 √ x = 51 = 7. 14
It follows that if three given sides of a triangle obey Pythagoras’ Theorem the triangle must be a right-angled triangle.
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This may be shown by a small square of side c contained in a larger square of side (a + b) arranged as in Figure 7.12. The area of a right-angled triangle is
Base × Perpendicular height 2
In the triangles of sides a, b and c, the area of one triangle = So the area of four triangles = 4 ×
ab 2
ab = 2ab 2
The area of large square = (a + b)2 = a2 + 2ab + b2 and the area of small square = c2 .
b c c a
b
a
Figure 7.12
Area of small square = Area of large square − Area of 4 triangles c2 = (a + b)2 − 2ab = a2 + 2ab + b2 − 2ab = a2 + b2 Example Find the side marked x.
x
5
3
Let the height of the triangles be h.
10
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•
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Then
so
h2 = 52 − 32 = 25 − 9 = 16
x 2 = 102 + h2 = 100 + 16 = 116 √ Therefore x = 116 = 10. 77 to 2 decimal place accuracy. Also
Questions In questions 1 to 8, find the value of the length, x.
1.
2. x
x
5
6
7
8
3.
4. 9.4
x
12.5 x
3.4
8.4
5.
6. 15
x
7
x
3
x
7.
10 16
8. 33 8
8 6
x 56
x
9.
Calculate the length of the diagonal of a square with sides of 10 cm.
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Answers 1.
10
2.
8.60
3.
8.76
4.
9.26
5.
10.61
6.
11.83
7.
10.58
8.
63.00
9.
14.14
Examples Referring to Figure 7.13. C
A
B
Figure 7.13
1. If angle A = 42◦ and side AC = 12 cm, what is the length of side BC? Since BC is the side opposite to angle A and AC is the hypotenuse, the sine ratio is used. BC = sin 42◦ AC So
BC = AC × sin 42◦ = 12 × 0. 6691 cm = 8. 03 cm
2. If angle C = 17. 8◦ and A = 5 cm, find the length of side AC. AB is the side opposite ∠C and AC is the hypotenuse, so the sine ratio is used. AB = sin C AC AB 5 5 So AC = = = 16. 4 cm = ◦ sin C sin 17. 8 0. 3057 The two previous examples have allowed a side to be found given a side and an angle. It is also possible to find an angle given two sides. To do this the inverse trigonometric ratios must be used.
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The inverse ratios are written sin−1 (. . .), cos−1 (. . .) and tan−1 (. . .) In some texts they are called arcsin(. . .), arcos(. . .) and arctan(. . .). 3. Side AC = 11 cm and side BC = 8 cm. What are angles A and C? Since BC is the side opposite angle A and AC is the hypotenuse sin A = So
BC 8 = = 0. 7273 to 4 decimal places. AC 11
A = sin−1 0. 7273 = 46. 7◦
Since the three angles add up to 180◦, angle C = 180◦ − (A + 90◦) = 43. 3◦ .
Questions 1. Find the length of the sides marked x in the following diagrams: (ii)
(i)
20
18
x
35°
42°
x
2. Find the size of the angles marked y in the following diagrams: (i)
(ii) θ° 12
5
θ°
3
8
3. In the triangle ABC, ∠B = 90◦ , ∠C = 26. 35◦ and b = 13. 4 cm. Calculate the length of side c of the triangle. 4. In the triangle ABC, ∠C = 90◦ , ∠A = 69. 3◦ and a = 3. 4 cm. Calculate the length of side c of the triangle. 5. An equilateral triangle has a vertical height of 20 cm. Calculate the length of the equal sides. 6. Calculate the length of the equal sides of an isosceles triangle whose altitude (vertical height) is 15 cm and whose equal angles are 48.6◦ .
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7. Find the values of the letters in the following table (AB is the hypotenuse): Angle A
Angle B
Angle C
Side a
Side b
(i)
A
90◦
15 cm
25 cm
(ii)
A
90◦
38 cm
42 cm
(iii)
15◦
90◦
a
40 cm
90◦
a
15 cm
63◦
(iv) (v)
90◦
A 19◦
(vi) (vii)
26◦
90◦
12 cm a
90◦
(viii)
B
90◦
Side c
23 cm 32.5 cm
39 cm 955 mm
c 1000 mm
Answers 1.
(i) 16.38
(ii) 12.04
2.
(i) 36.9◦
(ii) 33.7◦
3.
c = 5. 95
4.
c = 3. 63
5.
23.09 cm
6.
20 cm
7.
(i) A = 31. 0◦ (v) 58.6◦
(ii) A = 42. 1◦
(vi) a = 30. 73 cm
(iii) a = 10. 72 cm (vii) 43.4◦
(iv) 7.64 cm
(viii) 5.7◦
Examples Where Trigonometry can be Used (i) From a point 52 m away from the foot of a vertical building, the angle of elevation of the top of the building is 20◦ . What is the height of the building?
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The angle of elevation of an object is the angle through which the eye is raised above the horizontal to see the object. The angle of depression is the angle below the horizontal when looking down on an object. D
E
A diagram of the situation is crucial.
h
20° 52 m
∴ tan 20 =
h 52
∴ h = 52 × tan20 = 52 × 0. 3640 So
h = 18. 93 m to 2 decimal places
Bearings Bearings refer to the direction of one point from another point. Sometimes they are given in the form north-east (NE), west (W) etc. but usually they are given as three-figure numbers. The bearing represents the angle turned from North in a clockwise direction.
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Some examples are shown below: (a)
(b)
N
(c)
N
N
270°
045°
200° (Due west)
To solve some questions involving bearings it is necessary to have a suitable diagram showing angles and distances clearly, calculating some angles using basic geometry, before even considering Pythagoras’ Theorem or basic trigonometry. Example A ship sails for a distance of 15 km on a bearing of 250◦. How far south from the original position is the ship? N
S
250°
20° d 15 km F
S is the start position and F the final position. Let d represent the distance south of S. sin 20 =
d ⇒ d = 15 × sin20 = 15 × 0. 3420 = 5. 13 km 15
Questions 1. A ship is 3 km from the foot of a 150 m high cliff. What is the angle of depression of the ship from the cliff top?
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2. A square has sides of length 7 cm. Find the distance from the mid-point of one of the sides to either end of the opposite side. 3. From the top of a 90 m high cliff the angle of depression of a boat at sea is 15.3◦ . How far is the boat (i) from the cliff? (ii) from the observer? 4. What is the angle of elevation of the top of a tree 8 m high from a point on the ground 9.2 m away from the foot of the tree? 5. Find the sides of the right-angled triangle ABC if the sides of the square are 8 cm.
20°
6. A man walks 3 km from A to B on a bearing of 020◦ and then 2 km from B to C on a bearing of 110◦ . What is the distance and bearing of C from A?
Answers The angle of depression = d. This is the
d
1. 150 m Cliff
x Ship
3000 m
same as angle x. 150 = 0. 05 ⇒ x = d = 2. 9◦ tanx = 3000
7
2.
3.5 d
3.
d=
x = 15. 3◦
15.3° 90 m Cliff
√ (7)2 + (3. 5)2 = 61. 25 = 7. 83 cm
90 90 ⇒b= = 329 m b tan15. 3 90 90 sin 15. 3 = ⇒d= = 341 m d sin 15. 3
tan15. 3 =
d x b
Boat
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4.
8m
tanx =
8 = 0. 8696 ⇒ x = 41◦ 9. 2
x° 9.2 m
5.
h tan20 = ⇒ h = 2. 912 ⇒ height = 8 10. 912 8 tan20 = ⇒ d = 21. 98 ⇒ base = 29. 98 d By Pythagoras’ Theorem, hypotenuse = 31.9 cm
h 20° 8 8
8 20°
8 d
6.
B
110°
2 km
20°
C
3 km a
The shaded angle is 90◦ . 2 tan(a) = ⇒ a = 33. 7◦ ⇒ Bearing = 3 033. 7 √ √ d = 22 + 32 = 13 = 3. 61 km
d
A
Coordinate Systems The usual system of coordinates (x, y) is called Cartesian or rectangular coordinates. The term ‘rectangular’ is used because the x value and the y value represent the base length (x) and the height (y) of a rectangle (Figure 7.14). Another way of identifying the point is to use Polar coordinates where the coordinates represent a distance and angle.
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(x, y) y
x
Figure 7.14
∠r, θ r θ
Figure 7.15
By comparing Figures 7.14 and 7.15 it can be seen that y • r= x 2 + y2 and θ = tan−1 x • x = r · cos θ and y = r · sin θ These formulae allow conversion between one system and the other. Example In rectangular coordinates a point is described as being at (3, 4). What are its Polar coordinates (angle in degrees)? √ √ r = 32 + 42 = 9 + 16 = 25 = 5 4 = tan−1 (1. 333) = 53. 13◦ θ = tan−1 3 Therefore
∠5, 53. 13◦
Example In Polar coordinates a point is described as being ∠ 8, 30◦
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What are its rectangular coordinates? x = 8 cos 30◦ = 8 × 0. 866 = 6. 93 y = 8 sin 30◦ = 8 × 0. 5 = 4 (6. 93, 4).
Therefore
Calculators have these conversion functions as standard. It is a very useful tool so check how your particular calculator operates. Sines, cosines and tangents should be quite familiar but how exactly do these functions change for different angles? Graphs of each of the three functions will help to explain. Imagine a rod, length 1 m, fixed at one end, and free to rotate in a circle about that fixed end. Initially the rod is held horizontally pointing to the right. As the rod rotates anticlockwise the angle the rod is to the horizontal increases, as does the height of the end of the rod above the initial position. If the height of the end of the rod is plotted against the angle the rod makes with the horizontal, a smooth curve is formed. This is the sine wave. (Heights below the initial position are negative.) Once the rod has made one complete rotation the curve is repeated. The wave is said to be periodic. 90°
α
180°
h
h 0°
0°
α
90°
180°
270°
270°
If instead of the vertical height above the horizontal being measured and plotted, the horizontal distance from the vertical is plotted, the resulting curve is that of the cosine wave. 1
0
–1
90°
180°
270°
360°
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It can be shown that tan θ =
sin θ . cos θ
This has an interesting effect on the graph because when θ = 90◦ , cos θ = 0, and so tan θ does not exist, as it is not possible to divide by 0. The full graph looks like the following: 10 8 6 4 2 0°
90°
180°
270°
360°
–2 –4 –6 –8 –10
It is obvious that this graph is periodic but repeats after 180◦ whereas the sine and cosine curves repeat after 360◦. If the rod is rotating at a fixed angular velocity, ω rad/s, the above graphs can be plotted where time is the horizontal variable, not angle.
The Equation of a Sine Wave The general formula for a sine wave is: R · sin(ωt + α). where R is the amplitude, ω is the angular velocity and α is the phase shift relative to sin(ωt). From these values the frequency, periodic time and phase angle may be found since ω • the frequency f = Hz, 2π 1 2π • the periodic time T = or s and ω f 180◦ • the phase angle is α × π
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Example The current in a circuit at any time t seconds is given by i = 75. 0 sin(100 πt + 0. 320) amperes. Find the amplitude, periodic time, frequency and phase angle, relative to sin(100 πt). Solution (i) Amplitude = 75.0 amperes 2π 2π • Periodic time = = = 0. 02 s ω 100π 100π ω • Frequency = = = 50 Hz 2π 2π • Phase angle = 0. 32 radians = 18. 3◦ (leading). •
Questions 1. An alternating voltage is given by v = 63. 0 sin(250πt + 0. 240) V. Find the amplitude, periodic time, frequency and phase angle, relative to sin(250 πt). 2. The instantaneous value of current in an alternating-current circuit at any time t is given by i = 200. 0 sin(50πt − 0. 683) amperes. Find the amplitude, periodic time, frequency and phase angle, relative to sin(50 πt). 3. The current in an a.c. circuit is given by i = 40 sin(100πt − 0. 32) amperes. Find the amplitude, the periodic time, the frequency and the phase angle, relative to sin(100 πt).
Answers 1.
63, 125 Hz, 8 ms, 13.8◦
2.
200, 40 ms, 25 Hz, −39. 1◦
3.
40, 20 ms, 50 Hz, −18. 3◦
Sinusoidal Waveforms Here are some examples of sine waves and their equations: 1. y = sin(x) Amplitude: 1
Period: 360◦
Frequency:
360◦ =1 360◦
Phase angle: 0◦
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•
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–90
90
180
270
360
–1 –2 –3 –4
Figure 7.16
2. y = 3sin(x) Amplitude: 3
Period: 360◦
Frequency:
360◦ =1 360◦
Phase angle: 0◦
4 y 3 2 1 x –90
90
180
270
–1 –2 –3 –4
Figure 7.17
3. y = 3sin(2x) Amplitude: 3
Period: 180◦
Frequency:
360◦ =2 180◦
Phase angle: 0◦
360
Trigonometry and Geometry
•
4 y 3 2 1 x –90
90
180
270
360
–1 –2 –3 –4
Figure 7.18
4. y = 3sin(x + 45◦ ) Amplitude: 3
Period: 360◦
4
Frequency:
360◦ =1 360◦
Phase angle: +45◦
y
3 2 + 45° 1 x –90
90
180
270
–1 –2 –3 –4
Figure 7.19
5. y = 3sin(x − 45◦ ) Amplitude: 3
Period: 360◦
Frequency:
360◦ =1 360◦
Phase angle: −45◦
360
167
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•
Mathematics 4 y 3 2 – 45° 1 x
–90
90
180
270
360
–1 –2 –3 –4
Figure 7.20
6. y = 3sin(2(x − 30◦)) or y = 3sin(2x − 60◦ ) Amplitude: 3
Period: 180◦
Frequency:
360◦ =2 180◦
Phase angle: −60◦
4 y 3 2 1
– 30° x
–90
90
180
270
–1 –2 –3 –4
Figure 7.21
7. y = 3sin(2(x + 60◦)) or y = 3sin(2x + 120◦ ) 360◦ Amplitude: 3 Period: 180◦ Frequency: =2 180◦
Phase angle: +120◦
360
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4 y 3 2 1
+ 60° x
–90
90
180
270
360
–1 –2 –3 –4
Figure 7.22
It is very important to remember that equations whose variable is t (time) must be calculated taking the angles as radians and not degrees. Example The current in a circuit at any time t seconds is given by i = 75. 0 sin(100πt + 0. 320) amperes. Find: (i) the amplitude, periodic time, frequency and phase angle, relative to sin(100 πt). (ii) the value of the current when t = 0; (iii) the value of the current when t = 6 ms, that is, 0.006 s; (iv) the time when the current first reaches 50.0 amperes; (v) the time when the current is a maximum. Sketch one cycle of the waveform. Solution (i) Amplitude = 75. 0 amperes Periodic time = Frequency =
2π 2π = = 0. 02 s ω 100π
100π ω = = 50 Hz 2π 2π
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•
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Phase angle = 0. 32 rad = 18. 3◦ (leading). 0. 32 α = 0. 00102 s = 1. 02 ms . This will be needed later! In addition = ω 100π (ii) When t = 0, i = 75. 0 sin(0 + 0. 320) = 75 sin(0. 320) = 75 × 0. 314567 = 23. 6 amperes (iii) When t = 0. 006 s, i = 75. 0 sin(100π × 0. 006 + 0. 320) = 75 sin(2. 205) = 75 × 0. 80557 = 60. 4 amperes (iv) When i = 50 amperes, 50 = 75 sin(100πt + 0. 320) 0. 66667 = sin(100πt + 0. 320) 100πt + 0. 320 = sin−1 (0. 66667) 100πt + 0. 320 = 0. 7297 t=
0. 7297 − 0. 320 100π
= 0. 0013 s = 1. 3 ms (v) At a maximum i = 75 sin(100πt + 0. 320) = 1 100πt + 0. 320 = 1. 5708 t=
1. 5708 − 0. 320 100π
= 0. 00398 s = 3. 98 ms
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80 i, current (amps) 60 This is the same time as the calculation for the time to reach a maximum
40
18.98 ms
20 M 5 Quarter wave
–20 –40
t, time (ms)
B
A 10
15 Half wave
20
3.98 ms
–60
8.98 ms
–80
1.02 ms
Figure 7.23
There is a way of checking whether everything has been drawn correctly. α For the graph in Figure 7.23, = 1. 02 ms is the time shift for one cycle of the waveform. ω Therefore the time value for point A is 20 − 1. 02 = 18. 98 ms. Between points A and B there is a half wave. Since the periodic time is 20 ms, a half wave must take 10 ms and so B is at time 8.98 ms. Similarly there is a quarter wave between B and M (the time for a maximum current). A quarter wave takes 5 ms and so M is at time 8. 98 − 5 = 3. 98 ms. The answer to (v), the time to a maximum works out at 3.98 ms, totally independent of the graphical answer.
Questions 1. An alternating voltage is given by v = 63. 0 sin(250πt + 0. 240) V. Find: (i) the amplitude, periodic time, frequency and phase angle, relative to sin(250 πt).
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(ii) the value of the voltage when t = 0; (iii) the value of the voltage when t = 2 ms; (iv) the time when the voltage first reaches 30.0 V; (v) the time when the voltage is first a maximum. Sketch one cycle of the waveform. 2. The instantaneous value of current in an a.c. circuit at any time t is given by i = 200. 0 sin(50πt − 0. 683) amperes. Find: (i) the amplitude, periodic time, frequency and phase angle, relative to sin(50 πt). (ii) the value of the current at t = 0; (iii) the value of the current at t = 10 ms; (iv) the time that the current first reaches 100.0 amperes; (v) the time that the current first reaches −58.0 amperes; (vi) the time in the first cycle where the current is a maximum. Sketch the curve showing all the relevant details. 3. The current in an a.c. circuit is given by i = 40 sin(100πt − 0. 32) amperes. Find: the amplitude, the periodic time and frequency, the phase angle, relative to sin(100 πt) and sketch one cycle of the waveform.
Solutions 1.
(i) 63, 125 Hz, 8 ms, 13.8◦ ,
2.
(i) 200, 40 ms, 25 Hz, −39. 13◦ ,
3.
(i) 40
α = 0. 3 ms ω
α = −4. 3 ms ω
(ii) 14.98
(ii) −126.2
(ii) 20 ms, 50 Hz
(iii) 61.2
(iii) 155.1
(iii) −18. 3◦
(iv) 0.33 ms
(iv) 7.7 ms
i0 = −12. 6
(v) 1.7 ms
(v) 2.5 ms
tmax = 6. 01
(vi) 14.3 ms
α = −1. 01 ms ω
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173
200 q2 150 100 50 t, time (s) 0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
–50 q3
–100
q1
–150 –200
Figure 7.24
Addition of Sine Waves of Equal Frequency If the two waves y = 12 sin(50πt) and y = 18 cos(50πt) are added the result is a third sine wave (see Figure 7.25) 25 y 12sin(50πt ) + 18cos(50πt )
20 15 10
12sin(50πt )
5
t (ms) 5
10
–5 –10 –15 –20 –25
Figure 7.25
18cos(50πt )
15
20
25
30
35
40
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•
Mathematics
The resultant wave 12 sin(50πt) + 18 cos(50πt) is obviously just another sine wave. Its equation is therefore y = R · sin(ωt + α) It is necessary to find the amplitude, R, and the phase shift, α, of the resultant wave compared with the sine wave.
Theory If the two waves a · sin(ωt) and b · cos(ωt) are added the resulting wave can be written √ b −1 . in the form R · sin(ωt + α) where R = a2 + b2 and α = tan a Example Taking the waves 12 sin(50πt) and 18 cos(50πt): they are of the same frequency since both have angular velocity 50π rad/s √ The resultant amplitude R = 122 + 182 √ √ So R = 144 + 324 = 468 = 21. 633 to 3 decimal place accuracy. 18 = tan−1 (1. 5) In radians, since the variable is time, α = tan−1 12 Therefore α = 0. 983 rad The resultant equation is therefore y = 21. 633 sin(50πt + 0. 983) In practice, if a and b, in that order, are put through the calculator’s rectangular to polar conversion facility then the answers produced are R and α (remember that the angles are in radians). Therefore 12 sin(50πt) + 18 cos(50πt) = 21. 63 sin(50πt + 0. 9828) [Keep R to 2 decimal places and α to 3 decimal places.]
Questions Find the single wave when the following pairs of waves are added.
Trigonometry and Geometry Answers 1.
3 sin(t)
4 cos(t)
5 sin(t + 0. 9273)
2.
4 sin(2t)
6 cos(2t)
7. 21 sin(2t + 0. 9828)
3.
12 sin(5t)
5 cos(5t)
13 sin(5t + 0. 3948)
4.
34 sin(50πt)
67 cos(50πt)
75. 13 sin(50πt + 1. 1012)
5.
100 sin(5πt)
45 cos(5πt)
109. 66 sin(5πt + 0. 4229)
Relationships Between Sine, Cosine and Tangents of Angles Referring to Figure 7.26 A
B
θ
C
Figure 7.26
BC AC and cos θ = AB AB AC BC AC AB AC sin θ = ÷ = × = cos θ AB AB AB BC BC sin θ =
So
but
AC = tan θ BC
•
175
176
•
Mathematics tanθ =
Therefore
sin θ cos θ
(i)
By Pythagoras’ Theorem (AB)2 = (AC)2 + (BC)2 Dividing every term by (AB)2 gives Therefore 1 = but sin θ =
AC AB
2
+
BC AB
(AB)2 (AB)2
=
(AC)2 (AB)2
+
(BC)2 (AB)2
2
BC AC and cos θ = AB AB
Therefore 1 = sin2 θ + cos2 θ which is usually written as sin2 θ + cos2 θ = 1
(ii)
In the same manner, the links between the reciprocal functions can be found. (AB)2 = (AC)2 + (BC)2 and dividing every term by (AC)2 gives
(AB)2 (AC)2
=
(AC)2 (AC)2
+
(BC)2 (AC)2
cosec2 θ = 1 + cot2 θ
Therefore
(AB)2 = (AC)2 + (BC)2 and dividing every term by (BC)2 gives
(iii) (AB)2 (BC)2
=
(AC)2 (BC)2
+
(BC)2 (BC)2
sec2 θ = tan2 θ + 1
(iv)
Example Without using calculator or tables, find the cosine and tangent of an angle whose sine is 0.8. sin2 θ + cos2 θ = 1 so cos2 θ = 1 − sin2 θ Therefore
cos θ =
Also
tanθ =
√ √ 1 − sin2 θ = 1 − (0. 8)2 = 1 − 0. 64 = 0. 36 = 0. 6 0. 8 sin θ = = 1. 333 correct to 3 decimal places. cos θ 0. 6
Be careful in how you write the powers of trigonometric functions. Example sin θ × sin θ is written as sin2 θ , not sin θ 2
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177
sin θ 2 means the sine of θ 2 Similarly, cos2 θ , tan2 θ etc.
Identities An identity is an equation that is true for all values of the quantities involved in the equation. Example x 2 − y2 = (x − y)(x + y) (x + y)2 = x 2 + 2xy + y2 are examples of identities because they are true for any values of x and y. A trigonometric identity is an equation that is true for any angle, thus the four expressions derived in the previous section: tan θ =
sin θ cos θ
sin2 θ + cos2 θ = 1 cosec2 θ = 1 + cot2 θ sec2 θ = tan2 θ + 1 are all trigonometric identities because they are true for any value of angle θ and many more can be proved from knowledge of these fundamental identities and the trigonometric ratios. A few simple examples are shown below. Examples (i) Prove that
1 + cos θ sin θ Simplifying the left-hand side of the equation: cosec θ + cot θ =
cos θ 1 + sin θ sin θ 1 + cosθ = sin θ
cosec θ + cot θ =
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(ii) Prove that sec A − cos A = tanA sin A Simplifying the left-hand side of equation: 1 − cos A sec A − cos A cos A = sin A sin A cos2 A 1 − cos A = cos A sin A 1 − cos2 A = cos A sin A 1 − cos2 A = cos A sin A sin2 A cos A sin A sin A = cos A =
= tanA (iii) Prove that tanθ sin θ = 1 + tan2 θ Simplifying the right-hand side of equation: tanθ tanθ =√ sec2 θ 1 + tan2 θ tan θ = sec θ 1 sin θ ÷ = cos θ cos θ sin θ cos θ × = cos θ 1 = sin θ Note: See also Compound and Double Angles in Chapter 8.
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Introduction to Hyperbolic Functions There are three main functions, corresponding to the sine, cosine and tangent of trigonometry. ex − e−x 2 ex + e−x • hyperbolic cosine of x, cosh(x) = 2 ex − e−x • hyperbolic tangent of x, tanh(x) = ex + e−x •
hyperbolic sine of x, sinh(x) =
Most calculators allow these functions to be evaluated by pressing the ‘hyp’ key before ‘sin’, ‘cos’ or ‘tan’.
Questions Evaluate, correct to 4 significant figures. 1. (i) sinh(0. 64)
(ii) sinh(2. 182)
2. (i) cosh(0. 72)
(ii) cosh(2. 4625)
3. (i) tanh(0. 65)
(ii) tanh(1. 81)
x . 50 Evaluate, to 4 significant figures, the value of y when x = 25.
4. A wire hangs so that its shape is described by y = 50 cosh
5. The length l of a heavy cable hanging under gravity is given by l = 2c · sinh
L . 2c
Find the value of l when c= 40 and L = 30. 6. 3d 2 is a formula for velocity of waves over the bottom of 6. V = 0. 55L · tanh L shallow water, where d is the depth and L is the wavelength. If d = 8. 0 and L = 96, calculate V.
Answers 1.
(i) 0.6846,
(ii) 4.376
2.
(i) 1.271,
(ii) 5.910
3.
(i) 0.5715,
(ii) 0.9478
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•
Mathematics 4.
5638
5.
30.71
6.
5.042
Hyperbolic Identities There are several ‘links’ between hyperbolic functions. Some of them are: •
cosh(x) + sinh(x) = ex
•
cosh(x) − sinh(x) = e−x
•
cosh2(x) – sinh2(x) = 1.
Note that cos2(x) + sin2(x) = 1
Be careful with the signs.
Solving Equations Using Hyperbolic Functions a · sinh(x) + b · cosh(x) = c where a, b and c are constants can be solved by following these steps: x x e + e−x e − e−x and cosh(x) to . 1. Change sinh(x) to 2 2 2. Arrange the equation into the form p · ex + q · e−x + r = 0, where p, q and r are constants. 3. Multiply each term by ex , which gives p · (ex )2 + r · ex + q = 0 since (ex ) · e−x = 1. 4. Solve the resulting quadratic equation by factorisation or formula. 5. This gives values for ex , so take natural logarithms (ln) to find the values of x. Example Solve the equation 2. 6 cosh(x) + 5. 1 sinh(x) = 8. 73.
Trigonometry and Geometry Solution
ex + e−x 2. 6 2
ex − e−x + 5. 1 2
•
181
= 8. 73
1. 3ex + 1. 3e−x + 2. 55ex − 2. 55e−x = 8. 73 3. 85ex − 1. 25e−x − 8. 73 = 0 2 3. 85 ex − 8. 73ex − 1. 25 = 0 This quadratic equation has a = 3. 85, b = −8. 73 and c = −1. 25 which, on substitution into the quadratic formula, gives ex = 2. 4027 or −0. 1351 Therefore x = ln(2. 4027) or ln(−0. 1351) Therefore x = 0. 8766 to 4 significant figures as ln(−0. 1351) has no solution. Example x . Determine, correct to 4 significant 40 figures, (a) y when x is 25, and (b) x when y = 54. 30 A chain hangs in the form given by y = 40 cosh
Solution
x 25 (a) Since y = 40 cosh , y = 40 cosh = 40 cosh(0. 625) = 40 × 1. 20175 = 40 40 48. 07 x 54. 30 25 (b) When y = 54. 30, 54. 30 = 40 cosh ⇒ cosh = = 1. 3575 40 40 40 Following the procedure 1–4 given previously: x/40 + e−x/40 e = 1. 3575 ⇒ ex/40 + e−x/40 = 2. 715 ⇒ ex/40 + e−x/40 − 2. 715 = 0 2 Multiplying by ex/40 gives (ex/40 )2 − 2. 715ex/40 + 1 = 0 Using a = 1, b = −2. 715 and c = 1 in the quadratic formula gives ex/40 = 2. 2756 or x = ln(2. 2756) or ln(0. 43945). 0.43945. So 40 Therefore x = 40 ln(2. 2756) = 32. 89 or 40 ln(0. 43945) = −32. 89
Questions In questions 1–5 solve the given equations correct to 4 decimal places. 1. sinh(x) = 1
[0.8814]
2. 2 cosh(x) = 3
[±0. 9624]
3. 3. 5 sinh(x) + 2. 5 cosh(x) = 0
[−0. 8959]
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Mathematics
4. 2 sinh(x) + 3 cosh(x) = 5
[0. 6389, −2. 2484]
5. 4 tanh(x) − 1 = 0
[0.2554]
6. A flexible cable suspended between two horizontal points hangs in the form of a catenary,
the xequation of the curve being given by y = c cosh − 1 , where y is the sag of the cable, x the horizontal distance from c the midpoint to one end and c is a constant. Determine the sag when c = 20 and 2x = 16 [1.622m]
Answers 1.
0.8814
2.
±0. 9624
3.
−0.8959
4.
0.6389, −2.2484
5.
0.2554
6.
1.622 m
Addition of sinh(ax) and cosh(ax) Example Find the number of sinh and the number of cosh functions that add together to give 3e2x + 5e−2x . Solution Suppose there are A sinh functions and B cosh functions.
2x e2x − e−2x e + e−2x +B· 2 2 B A 2x A −2x B =e + +e − 2 2 2 2
So A · sinh(2x) + B · cosh(2x) = A ·
Equating this to 3e2x + 5e−2x gives A B + =3 2 2 Addition of these two equations gives B = 8; subtraction gives A = −2 B A − =5 2 2 Therefore 3e2x + 5e−2x = −2 sinh(2x) + 8 cosh(2x)
Trigonometry and Geometry
•
Basic Hyperbolic Trigonometric Identities sinh(A ± B) = sinh A · cosh B ± coshA · sinh B cosh(A ± B) = coshA · cosh B ± sinh A · sinh B If B = A, sinh 2A = 2 sinh A · cosh A and cosh 2A = cosh2 A + sinh2 A
Questions Solve the following equations for x:
1.
sinh(x) =
4 3
2.
tanh(x) =
1 2
3.
sinh(x) + 4 = 4 cosh(x)
4.
7 + 2 cosh(x) = 6 sinh(x)
5.
2 sinh(x) + 6 cosh(x) = 9
6.
5 cosh(x) + sinh(x) = 7
7.
2 cosh(2x) + 3 cosh(x) − 8 = 0
8.
4 cosh(x) − e−x = 3
9.
20 cosh(2x) − 21 sinh(x) = 200
Answers 1.
1.0986
2.
0.5493
3.
0, 0.511
4.
1.386
5.
−1.386, 0.693
6.
−1.099, 0.693
7.
±0.693
8.
−0.693, 0
9.
−1.386, 1.609
Latitude and Longitude Figure 7.27 represents the earth, centre O: N and S are the north and south poles. •
All circles on the earth’s surface centred at O are great circles.
•
All great circles passing through the poles are meridians of longitude.
183
184
•
Mathematics N
C Q
P
W
θ
O
E
X
Y
S
Figure 7.27
•
All circles whose plane is at right angles to the line NOS are parallels of latitude: of these only the equator is a great circle.
•
The longitude of any point on the earth’s surface is measured by its angular displacement east or west of the meridian which passes through Greenwich.
•
The latitude of any point is given by its angular displacement, θ , north or south of the plane of the equator.
In Figure 7.28, if NPXS represents the meridian of Greenwich and NQYS another meridian, the circle centre C through P and Q a parallel of latitude, and the circle WXYE the (or XOY), and the latitude of Q is the equator, then the longitude of Q is the angle PCQ (or POX). angle θ , QOY = 16◦ and QOY = 54◦ , then the longitude of Q is 16◦ E and its If, for example, PCQ ◦ latitude is 54 N. The metre was originally defined as one ten-millionth part of the quadrant of the meridian which passes through Paris. If R is the earth’s radius, then 1 of 2πR = 107 m 4
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•
185
∴ 2πR = 4 × 107 m ∴ 2πR = 4 × 104 km The circumference of the Earth is therefore 40,000 km. If P is in latitude θ and r is the radius of the circle of latitude, then r = R · cos θ . This relationship, together with the value of R, will help in the calculation of distances on the earth’s surface. N r
P θ
C
R O X
S
Figure 7.28
Example Find the distance, measured along the parallel of latitude, between two places in latitude 42◦ N, whose longitudes are 23◦ W and 17◦ E. Solution = XOY = 23◦ + 17◦ = 40◦ The differences in longitude of the two places (A and B) = ACB The radius of the small circle, the parallel of latitude, r = R · cos 42◦ Therefore, arc AB of circle of longitude = AB =
1 40 × 2πr = × 2πR cos 42◦ 360 9
1 × 4 × 104 × cos 42◦ km ≈ 3300 km 9
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•
Mathematics N
N C
r C
A B
R 42
O X
Y
A
42
Y
r O, C
O
B 40° A
S
X
S
Figure 7.29
Longitude and Time The earth rotates on its axis once every 24 h spinning from west to east. This means that a difference of one degree of longitude causes a difference of four minutes of time, and that places to the west reach noon later than places to the east.
The nautical mile One nautical mile (nm) is defined as the length of an arc on a great circle which subtends an angle of 1 at the earth’s centre. Hence 360 × 60 nm = 4 × 104 km. Therefore 1 nm = 1.852 km A speed of 1 nm/h is called one knot: this unit of speed is almost universally used by seamen and airmen. Example P and Q are two points on the earth’s surface in latitude 56◦ N: their longitudes are 25◦E and 95◦ E respectively. Take 2πR = 4 × 104 km. Find the distance PQ. (a) in km along the parallel of latitude, (b) in km along a great circle, (c) in nautical miles along a great circle.
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•
187
Solution N
N (i)
(ii) C P
r 56
C
Q
R 56
O X
C
P
O
Y
S
O
S
(iii)
(iv) θ
35° 35°
θ R
r
Q
P
P
Q
7 70 7 × 2π · r = × 2π · R cos(56◦ ) = × 4 × 104 × cos(56◦) 360 36 36 So PQ ≈4350 km
(a) In (iii), arc PQ =
(b) From (iii) PQ = 2 · r · sin(35) = 2 · R cos(56) · sin(35) From (iv) PQ = 2R · sin(θ ) ∴ sin (θ ) = cos (56) · sin (35)
⇒
θ = sin−1 (0. 3207) = 18◦ 42
Therefore ∠POQ = 2θ = 37◦24 = 37. 4◦ 37. 4 37. 4 × 2π · R = × 4 × 104 ≈ 4160 km So in (iv), arc PQ = 360 360 (c) ∠POQ = 37◦ 24 = 2244 ⇒ arc PQ = 2244 nm As a check, 2244 × 1. 852 ≈ 4160 km.
Questions Take 2πR = 4 × 104 km. Answers should be to 3 significant figures. 1. On a globe of radius 15 cm find the length of the parallel of latitude 62◦ S. 2. Two places on the same meridian have latitudes 23◦ S and 41◦ S. What is their distance apart, measured along the meridian?
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•
Mathematics
3. As in question 2, but the latitudes are 23◦ S and 41◦ N. 4. Two places on the equator have longitudes 63◦ E and 132◦E. How far apart are they, measured along the equator? 5. As in question 4, but the places have longitudes 132◦E and 126◦ W. 6. Two places on the same meridian are 3940 km apart. If one of them is in latitude 18◦ N and the other is in the southern hemisphere what is the latitude of the other? 7. What is the length of the Arctic Circle (latitude 66◦ 30 N)? How far from the North Pole is any point on this circle? 8. Two places are both in latitude 37◦ S and their longitudes are 34◦W and 29◦E. What is their distance apart, measured along the parallel of latitude?
Answers 1.
44.24 cm
2.
2000 km
3.
7111.1 km
5.
11333.3 km
6.
17.5◦S
7.
15950 km
2611 km
4.
7666.7 km
8.
5590.4 km
Test Examples 7 1. (i) Express the following angles in rad: 114◦ 36 and 286◦ 30 . (ii) Express in (a) radian and degree, the angles subtended by arc lengths of 10 m, and 30.4 m respectively, on a circle of 10 m diameter. 2. A disc flywheel is rotating at a speed of 10.52 rad/s. Calculate the linear velocity, in m/s, of points on the wheel at radii of 100 and 500 mm respectively. Find the linear velocity in m/s of the rim at 2 m diameter when rotating at 125 rev/min. 3. In a right-angled triangle ABC, length AC is 36 mm, length BC is 27 mm, and the angle at C is 90◦ . Calculate the length of the hypotenuse AB and the sine, cosine and tangent of the angle at B. 4. Find, without the use of calculator or tables, the sine and tangent of an angle whose cosine is 0.4924. 5. Write down the sine, cosine and tangent of the following angles: 10◦ 33 46◦ 55 150◦ 47 201◦ 21 287◦ 14 Sketch sine and cosine curves for angles between 0◦ and 360◦. 6. The following refer to angles between 0◦ and 360◦: (a) find the angles whose sines are: 0.3783, −0.7005,
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•
189
(b) find the angles whose cosines are: 0.9687, −0.8769, (c) find the angles whose tangents are: 0.2010, −3.2006, 7. If θ = 80◦ , find the values of: sin θ
cos θ
sin 2θ
cos 2θ
sin2 θ
cos2 θ
8. Find the values of θ between 0◦ and 360◦ which satisfy the equation: cos2 θ − sin2 θ = 1 9. If a = 2 sin θ , and b = 5 cos θ , find the values of θ for angles between 0◦ and 180◦ when: a2 2b2 = −4 2 5 10. The velocity of the piston in a reciprocating engine is given by: sin 2θ m/s v = ωr sin θ + 2n and the acceleration is given by: cos 2θ m/s2 a = ω r cos θ + n 2
where
ω = angular velocity of crank in rad/s, r = throw of crank (1/2 stroke) in m, θ = angle of crank past top centre, in deg, n = ratio of connecting rod length to crank length.
Calculate (i) the velocity, and (ii) the acceleration of the piston of an engine of 1 m stroke, connecting rod length 2 m, at the instant the crank is 80◦ past top centre and running at 150 rev/min. 11. The Bernoulli equation may be written as P2 v22 P1 v12 + + h1 = + + h2 ρ 2g ρ 2g
If (h1 − h2 ) = 2, v12 − v22 = 8. 4, P1 = 350, ρ = 10, transpose the formula in a suitable way to find the value of the pressure P2 · (g = 9. 81) 12. The increase in resistance of strip conductors due to eddy currents at power frequencies is given by: α · t sinh(αt) + sin(αt) λ= 2 cosh(αt) − cos(αt)
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•
Mathematics
Calculate λ, correct to 5 significant figures, when α = 1. 5 and t = 1. 2 13. The signal to noise ratio is given by the formula: A X = 10 × log10 B Y where X is signal power and Y is noise power in watts. A Determine Y given = 15 and X = 2. B 14. The signal to noise ratio is given by the formula: Psignal S = A · ln N Pnoise where P is signal and noise power in watts. S Determine Pnoise correct to 3 significant figures, given = 4. 0, A = 2. 0 and N Psignal = 2. 3 15. If 6ex − 3e−x = A · cosh(x) + B · sinh(x) find the values of A and B. 16. A polytropic process is one in which the pressure and volume of a gas are related by the formula p · v n = c, where c is a constant. The temperatures (T) and specific volumes (v) at the beginning and end of such a process are related by the formula T2 = T1
v2 v1
1−n
Find an expression for the index, n, in terms of T1 , T2 , v1 and v2 . Calculate the value of n for such a process in which the specific volume of a vapour is 0.0292 m3 kg−1 at 211 K and 0.00399 m3 kg−1 at 300 K. 17. The increase in resistance of strip conductors due to eddy currents at power frequencies is given by: α. t sinh(αt) + sin(αt) λ= 2 cosh(αt) − cos(αt) Calculate λ, correct to 5 significant figures, when α = 1. 8 and t = 1. 25
R U2 find the value of U2 (to 2 decimal places) given that 18. If θt − θi = · ln J U1 θt = 2. 5, θi = 1. 5, R = 0. 415, J = 0. 3, U1 = 50
Circle Theorems Useful to know
Trigonometry and Geometry
•
191
Figure 7.30 (1) The angle subtended at the centre of a circle is always twice the angle subtended at the circumference. (2) If AB forms a diameter then the angle θ is always 90◦ .
θ
2θ
Figure 7.30
Figure 7.31 (3) The angles made in the same segment are equal.
θ θ
Figure 7.31
192
•
Mathematics
Definition Figure 7.32 A cyclic quadrilateral is a four-sided figure whose vertices (corners) all lie on the circumference of the same circle.
Supplementary (Add to 180°)
Figure 7.32
Figure 7.33 (4) Chords of equal length are equidistant from the centre of the circle.
L
d
O d
L
Figure 7.33
Trigonometry and Geometry
•
193
Figure 7.34 (5) If two chords intersect, the product of the parts of one chord is equal to the product of the parts of the other.
a d c b
Figure 7.34
Figure 7.35 (6) The last result is true even if the chords intersect outside the circle, that is, a · b = c · d
a
b
c
Figure 7.35
d
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•
Mathematics
Figure 7.36 (7) Extended chords are called secants. If a tangent and secant intersect as shown then AO · OB = OC 2
A
B
O
C
Figure 7.36
Figure 7.37 (8) If a radius is drawn to meet a tangent at its point of tangency the two lines meet at right angles.
O Radius
Figure 7.37
Tangent
8
SOLUTION OF TRIANGLES As seen in the previous chapter, the ratios of the lengths of the sides of a right-angled triangle are expressed by sine, cosine and tangent, as follows: sine of angle =
opposite hypotenuse
cosine of angle =
adjacent hypotenuse
tangent of angle =
opposite adjacent
Example A light on a cliff is viewed from a position at sea, the angle of elevation being 32◦ 15 . At a point 100 m further away from the cliff and in direct line with the first observation point, the light is viewed again and the angle of elevation is now 20◦ 44 . Find (a) the height of the light above sea level, and (b) the horizontal distance from the second observation point to the light. Referring to Figure 8.1, Therefore Similarly Therefore
CD = tan 32◦ 15 x CD = x × 0. 63095
(i)
CD = tan 20◦ 44 100 + x CD = (100 + x) × 0. 37853 CD = 37. 853 + 0. 37853x
(ii)
196
•
Mathematics D
Light
32°15´ C
x
20°44´ B
100 m
A
Figure 8.1
Equating (i) and (ii) gives 0. 63095x = 37. 853 + 0. 37853x 0. 25242x = 37. 853 x=
37. 853 0. 25242
x = 149. 96 m Therefore AC = 100 + 149. 96 Horizontal distance = 249.96 m From (i): CD = x × 0. 63095 = 149. 96 × 0. 63095 = 94. 62 m
Inverse Trigonometric Functions The calculator will only give the principal value, α. Consider sin(x) = 0. 5 So x = sin−1 (0. 5) = 30◦ However, it can be seen that there is another angle that also has a sine of 0.5. An accurate graph will give this angle as 150◦, but how can it be found without an accurate plot?
•
Solution of Triangles 1.2
197
y
1 0.8 0.6 0.4 0.2 x –120
–90
–60
–30
–0.2
30
60
90
120
150
180
210
240
270
300
330
360
–0.4 –0.6 –0.8 –1 –1.2
Figure 8.2
Answer: because the graph is symmetrical about the x = 90◦ line, the second angle is simply 180◦ minus the angle given by the calculator. So for sin−1 (x). The second value for 0◦ ≤ x ≤ 360◦ is given by 180◦ − α, where a is the principal value. To find the second value for the angle in the range 0◦ ≤ x ≤ 360◦ a different ‘rule’ is needed for cosine and tangent. cos−1 (x) The second value for 0◦ ≤ x ≤ 360◦ is given by 360◦ − α. tan−1 (x) The second value for 0◦ ≤ x ≤ 360◦ is given by α + 180◦. If values of x>360◦ are required they are found by adding 360◦ to the values already obtained. Example sin(x) = 0. 6 From the calculator x = 36. 9◦ , and the alternative value is 180 − 36. 9, that is, 143.1◦. sin(x) = −0. 3 From the calculator x = −17. 5◦, and the alternative is 180 − (−17. 5)) = 197. 5.
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The next values are −17. 5 + 360 and 197. 5 + 360. That is, 342.5◦ and 557.5◦ . Therefore the required solutions are 197.5◦ and 342.5◦. cos(x) = 0. 34. The calculator gives 70.1◦ and the alternative solution is 360◦ − 70. 1◦ , that is, 289.9◦.
Cosine Rule and the Sine Rule for Non-Right-Angled Triangles The Cosine rule applies if you are given either (i) two sides and the included angle, for example, b, c and A; or (ii) three sides.
The rule(s) (i) a2 = b2 + c2 − 2 × b × c × cos(A) b2 + c2 − a2 (ii) cos(A) = 2×b×c The Sine rule applies when you are given a ‘matching pair’ of values and one other dimension.
The rule sin(A) sin(B) sin(C) = = , if you are finding an angle, a b c a b c (ii) = = , if you are finding a side. sin(A) sin(B) sin(C) (i)
Remember that there is the possibility of two values for the angle when the sine function is used. It is preferable to use the Cosine rule (ii) if at all possible.
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Plane Trigonometry – Sine and Cosine Rules 1. The triangle XYZ has y = 11 cm, z = 15 cm and X = 55◦. X
55° 11 cm
15 cm
Z Y
Using the cosine rule x 2 = y2 + z2 − 2 · y · z · cos(X) Therefore x = 112 + 152 − 2 × 11 × 15 × cos(55) = 12. 5 cm. (The exact value of x should be either written down or, more preferably, stored in the calculator’s memory.) Re-arranging the cosine rule gives cos(Z) = Therefore cos (Z) =
x 2 + y 2 − z2 2·x·y
12. 52 + 112 − 152 = 0. 19142 2 × 12. 5 × 11
Therefore Z = cos−1 (0. 19142) = 78. 96◦ Similarly cos(Y) =
x 2 + z2 − y2 12. 52 + 152 − 112 = = 0. 69421 2·x·z 2 × 12. 5 × 15
Therefore Y = cos−1 (0. 69421) = 46. 0◦ . Example The vertical post of a jib crane is 10 m long. The angle between jib and post is 40◦, and between jib and tie the angle is 45◦ . Find the length of the tie and the length of the jib.
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TIE
45°
Post = 10 m
B
JIB 40°
A
Figure 8.3
Referring to Figure 8.3, Therefore a =
c c · sin (A) a = so a = sin(A) sin(C) sin(C)
10 × 0. 6428 = 9. 090 m 0. 7071
Thus B = 180◦ − (40◦ + 45◦) = 95◦ Also
c c · sin (B) b = so b = sin(B) sin(C) sin(C)
Therefore b =
10 × 0. 9962 = 14. 088 m 0. 7071
Length of TIE = 9. 09 m Length of JIB = 14. 09 m Example Two ships leave the same port at the same time on courses which diverge at 29◦ . When one ship had travelled 40 nm the two ships were then 21 nm apart, find how far the other ship had travelled from port. Referring to Figure 8.4, Therefore sin (C) =
a c c · sin (A) = so sin (C) = sin(A) sin(C) a
40 × 0. 48481 = −0. 92345 21
So C = 67◦ 26 or 180◦ − 67◦ 26 = 112◦ 34 Therefore there are two possible answers to this problem, as illustrated in Figure 8.4.
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201
C
21 nm
C
a 21 nm a
b
A
29°
c 40 nm
Port
B
Figure 8.4
If C = 67◦ 26 , B = 180◦ − (29◦ + 67◦26 ) = 83◦ 34 But
b a · sin (B) a = so b = sin(A) sin(B) sin(A)
Therefore
b=
21 × 0. 99370 = 43. 04 0. 48481
(i)
21 × 0. 62160 = 26. 93 0. 48481
(ii)
If C = 112◦34 , B = 180◦ − (29◦ + 112◦34 ) = 38◦26 Therefore
b=
For the two ships to be 21 nm apart and one having travelled 40 nm, the other ship has travelled either 43.04 or 26.93 nm. Example A ship going due North at 16 knots runs into a 5 knots current moving North-East. Find the resultant speed and the direction of the ship. Reference Figure 8.5, Angle A = 90◦ + 45◦ = 135◦ a2 = b2 + c2 − 2 × b × c × cos(A) = 162 + 52 − 2 × 16 × 5 × cos(135) = 256 + 25 − 160 × (−0. 7071) = 349. 14
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c=5 A
b = 16
135°
a
C
Figure 8.5
Therefore a = Now
√ 349. 14 = 19. 85 knots
c c · sin(A) a = so sin (C) = sin(A) sin(C) a
Therefore sin (C) =
5 × 0. 70711 = 0. 17811 19. 85
So C = 10◦ 16 Resultant speed = 19.85 knots on a bearing of 010◦16 (10◦16 East of North).
Application of Trigonometry Questions 1. Two voltage phasors are shown in Figure 8.6. If z = 60 V and w = 90 V, calculate the value of their resultant (i.e. the length AB) and the angle the resultant makes with z. 2. Two powerboats, A and B, leave a port at the same time. A sails at a steady speed of 52 km/h on a bearing of 212◦ and B at 38 km/h on a bearing of 156◦. Find their distance apart after 21/2h. 3. A room 9 m wide has a span roof which slopes at 32◦ on one side and 41◦ on the other. Find the length of the roof slopes to the nearest centimetre.
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203
B
w = 90 V
30° A
z = 60 V
Figure 8.6
Y
75° X
Z
Figure 8.7
4. A jib crane consists of a vertical stanchion AB, 5.2 m in length, the inclined jib BC, 12.8 m in length, and a tie AC. The angle BAC is 122◦. Calculate (a) the length of the tie and (b) the inclination of the jib to the vertical. 5. A reciprocating engine mechanism is shown in Figure 8.7, where XY represents the rotating crank, YZ is the connecting rod and Z is the piston which moves horizontally along the line XZ. If the rotating crank is 1.24 cm in length and the connecting rod is 6.48 cm, calculate for the position shown (a) the inclination of the connecting rod to the vertical and (b) the distance XZ. 6. Figure 8.8 shows the jib of a crane, consisting of three members plus the pulley wire supporting the load. Find the length of the strut AC and determine its angle from the vertical. 7. A crank mechanism of a petrol engine is shown in Figure 8.9. Arm OA is 12.0 cm long and rotates clockwise about O. The connecting rod AB is 35.0 cm long and end B is constrained to move horizontally. (i) For the position shown above, determine the angle between AB and the horizontal. (ii) Calculate the length of OB.
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•
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110° b
Load
θ A
Figure 8.8
A 35.0 cm B
12.0 cm 57° O
Figure 8.9
8. PQ and QR are the phasors representing the alternating currents in two branches of a circuit. Phasor PQ is 20.0 A and is horizontal. Phasor QR (which is joined to the end of PQ to form a triangle PQR) is 29.0 A and is at an angle of 32◦ to the horizontal. Determine the resultant phasor PR and the angle it makes with PQ.
Answers 1.
145 V, 18.1◦
2.
110 km
3.
6.17 m, 4.99 m
4.
(a) 9.26 m
5.
(a) 10.7◦
(b) 6.689 cm
6.
10.24, 55.7◦
7.
(a) 16.7◦
(b) 40.1 cm
8.
47.2, 19.0◦
(b) 37.9◦
Example The three sides of a triangle measure 8, 6 and 4 cm respectively. Find the angles (Figure 8.10).
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C
6 cm
b 4 cm
a c 8 cm
B
A
Figure 8.10
b2 + c2 − a2 2bc 2 4 + 82 − 62 = 2×4×8
cos A =
= 0. 6875 Therefore angle A = 46◦ 34 (The alternative angle for A is 180◦ − 46◦34 = 133◦26 . This cannot be the case as angle C is larger than angle A because side c is longer than side a. This would make angle C larger than 133◦ 26 making the angles in the triangle greater than 180◦.) b a = sin(A) sin(B) So sin(B) =
so
sin (B) =
b · sin (A) a
4 × 0. 7262 = 0. 4841 6
Therefore angle B = 28◦ 57 This gives angle C = 180◦ − (46◦34 + 28◦57 ) = 180◦ − 75◦31 Therefore angle C = 104◦29 Summarising the formulae given so far for solving triangles, the sine, cosine and tangent ratios of sides are used for right-angled triangles, and either the sine rule or cosine rule for other triangles. The sine rule is easier to calculate than the cosine rule but there is the problem with each inverse sine calculation resulting in two possible values for the angle. This problem never occurs with the cosine function and so the cosine rule may be preferable.
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The cosine rule must be used in the situations where •
two sides and the angle between them (the included angle) are given, or
•
three sides are given.
Areas of Triangles Area of triangle =
1 (base × perpendicular height) 2
Referring to Figure 8.11, Perpendicular height = a × sin (C) 1 So Area = (b × a × sin C) 2 Therefore Area =
ab · sin C 2
In general terms this means that ‘the area is half the product of two sides and the sine of the angle between these sides’. An easy formula to apply and a very useful one. Another important formula for finding the area of a triangle is one which calculates the area directly from the three sides of the triangle: Area = s (s − a) (s − b) (s − c) where a, b and c are the sides of the triangle and (a + b + c) s= 2
B
a
C
Figure 8.11
Perp. c height b Base
A
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Example The three sides of a triangle measure 8, 10 and 14 cm respectively, find the area enclosed. s=
(a + b + c) (8 + 10 + 14) = = 16 2 2
s − a = 16 − 8 = 8 s − b = 16 − 10 = 6 s − c = 16 − 14 = 2 √ √ Area = s (s − a) (s − b) (s − c) = 16 × 8 × 6 × 2 = 1536 = 39. 2 cm2 , correct to 1 decimal place.
Equilateral triangle An equilateral triangle is one which has three equal sides and three equal angles (Figure 8.12). Each angle is therefore 60◦ . Perpendicular height = Side × sin 60◦
√ 3 = Side × 2
1 Area of triangle = (Base × Perpendicular height) 2
60°
Sid
de
e
Si
h
60°
60° Side
Figure 8.12
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Therefore Area =
√ 3 Side × Side × 2 2
√ 3 = × (Side)2 ≈ 0. 433 × (Side)2 4
Isosceles triangle An isosceles triangle is one which has two equal sides and two equal angles. In Figure 8.13 angles A and C are equal, therefore the lengths of the sides a and c are equal.
c
a
It is thus an easy matter to calculate the perpendicular height or other data required to find the area by any of the general formulae.
h
A
C
Figure 8.13
Compound Angles In some problems it is often convenient to express the relationship between the trigonometrical ratios of a compound angle and those of the two single component angles. Referring to Figure 8.14, Area of CED = area of CFD + area of CFE e · [d · sin (A + B)] g · [e sin (A)] · g · [d sin (B)] · = + 2 2 2
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209
D e g
A B
C
a F
c
b d
E
Figure 8.14
Dividing throughout by
ed , 2 sin (A + B) =
g g · sin A + · sin B d e
= cos B · sin A + cos A · sin B Usually written as, sin (A + B) = sin A · cos B + cos A · sin B
(i)
By a similar process it can be shown that, sin (A − B) = sin A · cos B − cos A · sin B
(ii)
Again referring to Figure 8.14 and applying the cosine rule, cos C = cos (A + B) =
e2 + d 2 − c 2 2ed e2 + d2 − (a + b)2 e2 + d2 − a2 − 2ab − b2 = 2ed 2ed
but, by Pythagoras, e2 − a2 = g2 and d2 − b2 = g2 Therefore
Therefore
cos (A + B) =
2g2 − 2ab g2 ab g g a b = − = × − × 2ed ed ed e d e d
cos (A + B) = cos A · cos B − sin A · sin B
(iii)
By a similar kind of process it can also be shown that, cos (A − B) = cos A · cos B + sin A · sin B
(iv)
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The tangent of angle being the sine divided by its cosine, the relationship for the tangent of a compound angle can be obtained by dividing sin(A + B) by cos(A + B), tan(A + B) =
sin (A + B) sin A · cos B + cos A · sin B = cos(A + B) cos A · cos B − sinA · sin B
Dividing all terms, top and bottom, by cos A cos B, sin A · cos B cos A · sin B + cos A cos B tan(A + B) = cos A cos B cos A · cos B sin A · sin B − cos A cos B cos A cos B Therefore tan (A + B) =
tan A + tanB 1 − tanA · tanB
(v)
tan (A − B) =
tan A − tanB 1 + tanA · tanB
(vi)
and by a similar process,
In summary sin (A ± B) = sin A · cos B ± cos A · sin B cos (A ± B) = cos A · cos B ± sin A · sin B tan(A ± B) =
tan A ± tanB 1 ± tanA · tan B
Double Angles Formulae expressing the relationship between the ratios of angles and double angles are useful and can be obtained from those above by letting A = B; sin (A + B) = sin A · cos B + cos A · sin B
gives
sin 2A = 2 sin A cos A
cos (A + B) = cos A · cos B − sin A · sin B
gives
cos 2A = cos2 A − sin2 A (#)
gives
tan 2A =
tan (A + B) =
tan A + tanB 1 − tanA · tanB
Also, since cos2 A + sin2 A = 1 then cos2 A = 1 − sin2 A and sin2 A = 1 − cos2 A
2 tanA 1 − tan2 A
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Substituting for cos2 A and sin2 A in turn into (#) gives: cos 2A = cos2 A − sin2 A = 1 − sin2 A − sin2 A Therefore cos 2A = 1 − 2 sin2 A cos 2A = cos2 A − sin2 A = cos2 A − 1 − cos2 A Therefore cos 2A = 2 cos2 A − 1 Note: Also see Identities in Chapter 7.
Test Examples 8 1. The top of a vertical mast is viewed from a position at 15 m from its base on level ground and the angle of elevation measured to be 45◦ 34 . Find the height of the mast. 2. A boat is sighted from a point on a cliff 95 m above the sea, the angle of depression of the line of view being 14◦ 25 . Find (i) the horizontal distance from cliff to boat, and (ii) the distance from observation point to boat. 3. A right-angled triangular plate has one angle of 28◦ 37 and the length of the hypotenuse is 120 mm. Find the lengths of the other two sides and the area of the triangle. 4. The top of a flagstaff is viewed from a point on a level ground at some distance from the foot of the staff, and the angle of elevation is 48◦ 30 . At another point 10 m further away from the foot, the staff top is viewed again and this time the angle of elevation is 37◦ 38 . Find the distance from the foot of staff to the first observation point and also the height of the flagstaff. 5. The length of the sides of a cube is 60 mm. Find the length of the diagonal across the face and the length of cross diagonal from one corner to the opposite corner passing through centre of the cube. 6. The length of the vertical post of a jib crane is 15 m. The angle between jib and post is 35◦ 30 and between tie and post the angle is 105◦ 30 . Calculate the length of the jib and tie. 7. In a reciprocating engine of 800 mm stroke, the connecting rod is 1600 mm long. Find how far the crosshead has moved from the top of its stroke when the crank is 35◦ past its top dead.
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8. The fuel valve of a diesel engine closes when the piston has moved one-tenth of its stroke, find the angular position of the crank from top dead centre when the valve closes. 9. A ship going due South at 17 knots runs into a 4 knots current running 50◦ East of North. Find the resultant speed and direction of the ship. 10. Two ships approach a port, their courses converging at an angle of 23◦ . At a certain time one ship is twice as far from port as the other and their distance apart is 32 nm; find how far each ship is from port. 11. Two ships are 50 nm apart and going towards the same port at the same speed of 18 knots, on courses which converge at 73◦ 39 . If one ship will arrive in port half-an-hour before the other, find their distances from port. 12. Three sides of a triangle measure 16.4, 10.2 and 9.8 m respectively. Find the angles. 13. Two sides of a triangle measure 6.5 and 7.5 m respectively and the angle included between these two sides is 46◦ 51 . Find the area of the triangle. 14. Find the area in cm2 of a triangular plate of sides 71, 42 and 53 mm long respectively. 15. The area of an equilateral triangle is 57.27 cm2 . Find the length of its sides.
9
MENSURATION OF AREAS Units Areas are measured in (unit)2 , e.g. m2 , cm2 . It is often necessary to change units. The conversion factors are given in the table below: Linear
Area
1 cm = 10 mm
1 cm2 = 102 = 100 mm2
1 m = 100 cm
1 m2 = 1002 = 10000 cm2
1 m = 1000 mm
1 m2 = 10002 = 1000000 mm2
1 km = 1000 m
1 km2 = 10002 = 1000000 m2
In general, if the linear units are in the ratio 1 : r, the areas are in the ratio 1 : r2 . It follows that, for example, as 1 cm2 = 100 mm2
then 1 mm2 =
1 cm2 = 0. 01 cm2 = 10−2 cm2 100
A rectangle is a four-sided figure whose opposite sides are parallel and equal in length to each other (Figure 9.1). Each of the angles is 90◦ .
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Height
Base
Figure 9.1
Area of rectangle = base × height. A parallelogram is a four-sided figure whose opposite sides are parallel and equal in length to each other. It therefore follows that opposite angles are equal, one pair of opposite angles being obtuse and the other pair acute and supplementary to the obtuse angles. It may be considered as a rectangular framework leaning over to one side as in Figure 9.2 where it can be seen that the outer triangular area (shown dotted) at one end is equal to the inner triangular area at the other and therefore the area of the parallelogram is equal to that of a rectangle of the same base and same perpendicular height.
Perpendicular height
Base
Figure 9.2
Area of parallelogram = Base × Perpendicular height. Also, if a diagonal is drawn from one corner to the opposite corner, it will bisect the parallelogram into two equal parts, each part being a triangle, the area of each being half that of the parallelogram (Figure 9.3).
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215
Perpendicular height
Base
Figure 9.3
Area of triangle =
1 × Base × Perpendicular height 2
ab · sin C 2 (a + b + c) Area of triangle = s (s − a) (s − b) (s − c) where s = 2
Area of triangle =
A rhombus is a special kind of parallelogram. It is a diamond-shaped, four-sided figure with all sides of equal length and opposite sides parallel to each other (see Figure 9.4). The diagonals of a rhombus are perpendicular to each other and their intersection forms right angles; the diagonals bisect each other and each bisects its corner angles.
Figure 9.4
The area of a rhombus is half of the product of its diagonals. A trapezium is a four-sided figure, of which only two sides are parallel. Referring to Figure 9.5, if a and b are the respective lengths of the two parallel sides and h is the perpendicular height, then,
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Mathematics
Area of trapezium = Average length × Perp. height =
(a + b) h 2
a a+b
h
2
b
Figure 9.5
The equivalent rectangle of the trapezium is shown in dotted lines.
Polygons A polygon is a figure bounded by more than four straight sides. A regular polygon has all its sides equal in length and all its angles equal; any regular polygon is therefore made up of as many equal triangles as there are sides and the method of finding the area of a polygon is to first find the area of one triangle and then multiply by the number of triangles which make up the polygon (Figure 9.6). Special names are given to some regular polygons as follows: Number of sides
Name
5
Pentagon
6
Hexagon
7
Heptagon
8
Octagon
9
Nonagon
10
Decagon
A polygon with n sides is made up from n triangles. Central angle of each triangle, α =
360◦ n
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•
217
α r
r
Figure 9.6
If the radius of the circle that surrounds the polygon is r, then r is also the slant height of each triangle. Note that all the triangles are isosceles. Area of each triangle
ab · sin C r × r × sin α r2 · sin α = = 2 2 2
As there are n triangles the area of polygon
n · r2 · sin α 2
It is, however, also convenient to express the area of the more common polygons in terms of the length of the ‘sides’ or ‘flats’. If the central angle is α, then the base angle β = base angles are equal).
h β Side
Figure 9.7
From Figure 9.7
h = tanβ Side/2
180 − α (as the triangle is isosceles the 2
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•
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So perpendicular height h =
(Side) tan β 2
The area of the triangle is therefore and the area of the polygon is If there are n sides then α =
(Side) tan β (Side)2 · tan β 1 × (Side) × = 2 2 4
n · (Side)2 · tan β 4
360 for the most common polygons. n
Number of sides
α
β
5
72◦
54◦
6
60◦
60◦
8
45◦
671/2◦
10
36◦
72◦
Therefore the areas of these polygons are: Number of sides
Area
5
1. 7205 × (side)2
6
2. 5981 × (side)2
8
4. 8284 × (side)2
10
7. 6942 × (side)2
The circle The circumference is the outer rim of the circle, an arc is part of the Circumference, other common terms are illustrated in Figure 9.8. Circumference = π × diameter = πd = π × 2 × radius = 2πr where π = 3. 142 to 3 decimal places.
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219
Sector Chord Arc Diameter Segment Radius
Tangent
Figure 9.8
The definition of π is the result when the circumference of a circle is divided by its diameter. The area of a circle is given by the formula Area = π · r2 . Many formulae are just stated, but in some cases, they can be proved. In other words a reason can be given for the way they look. Take the circle as an example. A circle can be considered to be a regular polygon with an infinitely large number of sides. Therefore the circle is made up from a great number of small triangles with vertical height r. Each triangle has an area equal to area of all the triangles.
r × Base and so the area of the circle is equal to the 2
Therefore, area of circle = sum of areas of the triangle = sum of
r × Base
2 r × Base r {Base} = but 2 2 So Area
where
r × Base 2
=
{Base} = Circumference
r × 2π · r = π · r2 2
2 d π · d2 d Also, since r = , area = π · = 2 2 4 It is often more practical to use the formulae containing d, rather than r, because it is easier in reality to measure the diameter of a circle than its radius.
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Annulus or circular ring Area of annulus = Area of outer circle − Area of inner circle
R
r
Figure 9.9
Area = π · R2 − π · r2 = π R2 − r2
(see Figure 9.9)
or
2
2 d D −π · Area = π · 2 2 π 2 D − d2 = 4 By factorisation R2 − r2 = (R + r)(R − r) and D2 − d2 = (D + d)(D − d) Use of the factors makes calculations of annular areas much quicker. Example Find the effective under-face area of a reciprocating pump piston 41.5 mm diameter if the piston rod diameter is 8.5 mm. π π 2 Area of annulus = D − d2 = (D + d) (D − d) 4 4 π Area = (41. 5 + 8. 5) (41. 5 − 8. 5) 4 π = × 50 × 33 4 = 1296 mm2 to the nearest whole number
Mensuration of Areas
Sector of a Circle A sector of a circle is shown in Figure 9.10. Arc
Area θ
r
Figure 9.10
If θ is the angle at the centre, in degrees, area of circle of 360◦ = π · r2 area of sector of 1◦ =
π · r2 360
area of sector of θ ◦ =
θ · π · r2 360
Similarly arc length =
θ πd θ πr = 360 180
Further, if the angle θ at the centre be measured in radians, area of circle of 2π radians = π · r2 area of sector of 1 radian =
π · r2 r2 = 2π 2
area of sector of θ radians = Similarly arc length = rθ
θ · r2 2
•
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•
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Segment of a Circle Referring to Figure 9.10,
Area of segment
θ
Figure 9.10
Area of segment = Area of sector − Area of triangle From previous section, area of sector =
θ · r2 2
where θ is the centre angle in radians. Area of triangle =
ab · sin C 2
where, in this case,
a = r, b = r and C is the centre angle θ in radians. Therefore area of triangle =
r2 · sin θ 2
Area of segment = area of sector − area of triangle r2 · θ r2 · sin θ − 2 2 2 r = · [θ − sin θ] , θ in radians 2 =
Example Calculate the area of segment which subtends an angle of 150◦ at the centre of a circle of 150 mm diameter.
Mensuration of Areas Converting θ to radians, θ = 150 ×
•
223
5π π = 180 6
Also sin 150◦ = 0. 5 r2 · [θ − sin θ] 2 752 5π − 0. 5 = 2 6
Area of segment =
= 5957mm2
Ellipse Circumference = π× average diameter (see Figure 9.11)
Minor axis d
Major axis D
Figure 9.11
Therefore C = π Area =
D+d 2
π ·D·d 4
Surface of a Cylinder Imagine the curved surface of a cylinder being unrolled as in Figure 9.12: it can be seen that the curved surface area of a cylinder A = πd · h or A = 2π · r · h.
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h
Circumference = π·d
Figure 9.12
Surface of a Sphere The curved surface area of a sphere is equal to the curved surface area of its circumscribing cylinder, that is, a cylinder of equal diameter and height. Referring to Figure 9.13,
d
d
Figure 9.13
Curved surface area = π × diameter × height Therefore area = π · d2 or area = 4π · r2 The curved surface area of a segment of the sphere or any such sliced portion is equal to the curved surface area of the corresponding slice of the circumscribing cylinder. Referring to Figure 9.14, Curved surface area of segment = π · d · h = πd × h
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•
225
h
d
Figure 9.14
Surface area of a cone If a cone has a base of radius r and the length of a side (i.e. the slant height) is L then the curved surface area = π · r · L.
L
r
Theorem of Pappus, or Guldinus This theorem is one of the most useful to employ in finding areas and volumes of objects of circular shapes, or if the area or volume is known it can be used to determine the position of the centroid, or centre of gravity (mass), of many sections.
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Mathematics
With regard to areas the theorem states: If a line, lying totally on one side of a fixed axis, is rotated about that axis in its own plane, it will sweep out a surface area equal to the length of the line multiplied by the distance moved by its centre of gravity. Axis
L
L r
Axis
Area swept out
Figure 9.15
Consider a ‘line’ such as a straight piece of wire of length L (Figure 9.15), positioned at a distance r from an axis parallel to the wire. If the wire is moved around through a complete circle about the axis, a surface area like a thin cylinder will be swept out. The centre of gravity of the piece of wire is at its midpoint. The distance moved by the centre of gravity in one revolution is the circumference of the cylinder. Therefore, distance moved = 2π · r So the area swept out = Length of line × Distance moved by centre of gravity = L×2π ·r Thus area swept out = 2π · r · L
Surface area of a frustum of a cone (or pyramid) The frustum of a cone (or pyramid) is the shape remaining when the top section of the cone (or pyramid) is removed by a cut parallel to the base (Figure 9.16). The only measurements available are the radius (or diameter) of the top surface, the radius (or diameter) of the base surface, the height of the frustum and the length of the sloping side.
Mensuration of Areas
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227
r
L
h
R
Figure 9.16
Curved surface area of frustum = L × 2π ×
(R + 2) = π · L · (R + r) 2
Surface area of a torus Consider a circle of radius r whose centre is R units from a fixed vertical axis (Figure 9.17). When this circle is swept through a complete revolution about the axis, the shape created is called a torus (basically, a doughnut). The surface area of this torus, that is the area exposed to the air is given by the formula: curved surface area = 4π 2 · R · r.
r
R
Figure 9.17
If the circle was replaced by an ellipse of major axis D and minor axis d, curved surface area π 2 · R · (D + d).
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Similar figures Areas of similar figures vary as the square of their corresponding linear dimensions. Similar figures mean that they are of the same shape and proportions, although their sizes are different.
d
2d
Figure 9.18
Consider two circles, one of diameter d, the other twice as big in diameter = 2d, as in Figure 9.18. Area of small circle =
π · d2 4
Area of large circle =
π · (2d)2 π · 4d2 = =4· 4 4
π · d2 4
Therefore the ratio of their areas is 1 : 4, that is 1 : 22 Therefore multiplying a linear dimension, for examples, diameter, by 2 increases the area by a factor of 4. So if lengths are increased by a factor of 3, areas are increased by a factor of 9. In general, if the linear units are in the ratio 1 : r, the areas are in the ratio 1 : r2 . Example A hexagonal plate is cut out of a thin sheet of metal. Due to an error in marking off, the sides were all made 10% longer than intended. What is the percentage of error in area? Ratio correct dimensions to wrong = 100 : 110 = 1 : 1. 1
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Therefore lengths are a factor of 1.1 too large. So area will be a factor of 1.12 too large, that is, a factor of 1.21 Therefore the area is 21% too large.
Irregular figures There are occasions when an area cannot be calculated precisely because it is not a regular shape. In other words it is not made up of triangles, rectangles, etc. To find the area of an irregular shape an approximate method has to be used. One such method is Simpson’s First Rule. In essence the area is divided up into an even number of equally spaced strips and a formula applied to estimate the area. (The more strips, the more accurate the estimate: doubling the number of strips will increase the accuracy by a factor of 16.)
Simpson’s First Rule Simpson’s First Rule is based on the fact that a parabola y = ax 2 + bx + c can be found so that it passes through three points (Figure 9.19). So, in the following diagram, the curve AD can be represented as a parabola which then allows the area of ABCD to be approximated as: h Area of ABCD ≈ (1 · y0 + 2 · y1 + 1 · y2 ) 3 A D y0
y1
y2
B
C h
h
Figure 9.19
This rule can be applied repeatedly to adjacent pairs of areas to approximate the area of larger shapes such as WXYZ (Figure 9.20).
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W Z 1 y0
y1
2 y2
y3
3 y4
y5
X
y6 Y
h
Figure 9.20
1 =≈ Area
h (1 · y0 + 4 · y1 + 1 · y2 ) 3
h (1 · y2 + 4 · y3 + 1 · y4 ) 3 h 3 ≈ (1 · y4 + 4 · y5 + 1 · y6 ) Area 3
2 ≈ Area
1 + Area 2 + Area 3 Area WXYZ = Area
h h (1 · y0 + 4 · y1 + 1 · y2 ) + (1 · y2 + 4 · y3 + 1 · y4 ) 3 3 h + (1 · y4 + 4 · y5 + 1 · y6 ) 3 h ≈ (1 · y0 + 4 · y1 + 2 · y2 + 4 · y3 + 2 · y4 + 4 · y5 + 1 · y6 ) 3 ≈
This method can be extended to any even number of strips, each of width h. It is easiest to set the calculations in a table, the values 1, 4, 2, 4, . . . being called the ‘Simpson multipliers’.
Waterplane area When calculating the area of a waterplane it is usual to divide the length of the ship into about 10 equal sections. It is convenient to make the measurements from the centreline to the ship side, giving half ordinates. The answer from Simpson’s rule can then be doubled to give the waterplane area.
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Example The equally spaced half ordinates of a watertight flat 27 m long are 1.1, 2.7, 4.0, 5.1, 6.1, 6.9 and 7.7 m respectively. Calculate the area of the flat.
Solution Half ordinate
Multiplier Product for area
1.1
1
1.1
2.7
4
10.8
4.0
2
8.0
5.1
4
20.4
6.1
2
12.2
6.9
4
27.6
7.7
1
7.7
Total
87.8
As there are 7 ordinates there must be 6 equal intervals. Therefore h = Area =
27 = 4. 5 m 6
h 4. 5 × Total = × 87. 8 = 132. 2 m2 3 3
The waterplane is therefore 132. 2 × 2 = 264. 4 m2 . Example The ordinates measured from the centreline to the hull across a ship at the load water line are: 0.2, 9, 15.5, 20, 21.5, 20.5, 18.5, 12.5 and 1.3 m respectively, and the length is 180 m. Find the water plane area.
Solution Half ordinate
Multiplier Product for area
0.2
1
0.2
9.0
4
36.0
15.5
2
31.0
20.0
4
80.0
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Multiplier Product for area
21.5
2
43.0
20.5
4
82.0
18.5
2
37.0
12.5
4
50.0
1.3
1
1.3
Total
360.5
As there are 9 ordinates there must be 8 equal intervals. Therefore h = Area =
180 = 22. 5 m 8
h 22. 5 × Total = × 360. 5 = 2703. 75 m2 3 3
The waterplane is therefore 2703. 75 × 2 = 5407. 5 m2 .
Test Examples 9 1. In a parallelogram ABCD, the opposite parallel sides AD and BC are each 100 mm long, the other sides are each 60 mm long, and the diagonal AC is 140 mm. Calculate the angles, the short diagonal, the perpendicular height and the area. 2. The sides of a rhombus are each 32 mm long and the length of the long diagonal is 48 mm. Calculate the angles, the length of the short diagonal and the area. 3. In a trapezium ABCD, the two parallel sides are AB and CD, and their lengths are 100 and 60 mm respectively. Side BC is perpendicular to the parallel sides and its length is 50 mm. Find the area of the trapezium (cm2 ). 4. The lengths of the sides of a four-sided figure ABCD are, in m, AB = 1, BC = 2, CD = 1. 5, DA = 3. 5, and the angle BCD is 117◦ 17 . Find the area of the figure. 5. The length of the sides of a regular hexagonal plate is 80 mm. The plate is cut parallel to one of its sides and this reduces the area by 10%, calculate the thickness of the piece cut off. 6. An octagonal plate, the sides of which are each 30 mm long, has a circular hole 50 mm diameter cut out of it. Find the net area of the plate in mm2 .
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7. Find the length of the sides and the area of the largest equilateral triangular plate that can be cut out of a circular plate 120 mm diameter. 8. The outer and inner diameters of the collar of a single-collar thrust shaft are 755 and 415 mm respectively, and the effective area of contact with the thrust pads is 0.7 of the face of the collar. Calculate (i) the effective area of contact, in m2 , and (ii) the total force on the collar, in kN, when the thrust pressure is 2000 kN/m2 . 9. Find the area, in cm2 , of the smaller segment of a circle of 200 mm diameter if the length of the chord is 180 mm. 10. Find the diameter of a solid hemisphere whose total surface area (including the flat circular base) is 58.9 cm2 . 11. The ball of a Brinell hardness testing machine is 10 mm diameter. Calculate the depth and curved surface area of an indentation in a material under test when the surface diameter of the indentation is 5 mm. 12. It is required to make a hollow cone out of thin flat sheet steel, the base diameter of the cone to be 150 mm and the perpendicular height 125 mm. Find the dimensions of the sector to be cut out of the sheet to make this cone. 13. A lampshade has the form of a frustum of a cone, the diameters at the base and top are 320 and 180 mm respectively and the perpendicular height is 170 mm. Calculate the curved surface area. 14. A circular ring made of round bar is 640 mm outside diameter and 440 mm inside diameter. Calculate the surface area to be painted. 15. An equilateral triangular plate has sides 125 mm long, and another similarly shaped plate has sides 175 mm long. By what percentage is the larger plate greater in area than the smaller plate? 16. Regularly spaced semi-ordinates measured transversely across a ship at the load water line are as follows: 0.1, 3, 5.85, 7.2, 8.1, 8.4, 8.4, 8.25, 8.1, 7.5, 6.3, 3.75 and 0.5 m respectively, and the length is 150 m. Find the area of the waterplane by Simpson’s rule.
10
VOLUME – MASS, CENTRE OF GRAVITY, MOMENT Volume (V) is the result of the product of three dimensions measured in similar units. Units m3 , mm3 and, often, cm3 . Linear
Volume
1 cm = 10 mm
1 cm3 = 103 = 1, 000 mm3
1 m = 100 cm
1 m3 = 1003 = 106 = 1, 000, 000 cm3
1 m = 1000 mm
1 m3 = 10003 = 109 = 1, 000, 000, 000 mm3
1 km = 1000 m
1 km3 = 10003 = 109 = 1, 000, 000, 000 m3
1000 cm3 = 1litre so 1 cm3 = 1millilitre (ml) In general, if the linear units are in the ratio 1 : r, the volumes are in the ratio 1 : r3 It follows that, for example, as 1 cm3 = 1, 000 mm3 then mm3 =
1 cm3 = 0. 001 cm3 = 10−3 cm3 1, 000
Mass (m) is the quantity of matter possessed by a body and is proportional to the volume and the density of the body.
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It is a constant quantity, that is, the mass can only be changed by adding more matter or taking matter away. The S.I. unit for mass is the kilogram (kg). (1 kg = 1, 000 grams (g); 1, 000 kg = 1 tonne) Density (ρ) is a measure of the mass per unit volume. The unit of density is kg/m3 . Other units are: g/cm3 for solid materials, g/ml for liquids, g/l for gases. In some cases t/m 3 (tonne per m 3 ) and kg/1 (kg per litre) may be used. The formula connecting these three quantities is: Mass = Volume × Density m =V ·ρ The density of pure water may be taken as 1000 kg/m3 (This is equal to 1 tonne/m3 , 1 kg/l and 1 g/ml.) Units must be consistent throughout, such as, Mass [kg] = Volume [m3 ] × Density [kg/m3 ] or Mass [kg] = Volume [cm3 ] × Density [g/cm3 ] Relative density (specific gravity) of a substance is the ratio of the mass of a volume of the substance to the mass of an equal volume of pure water. In other words, it is the ratio of the density of the substance to the density of pure water.
Prisms A regular prism is a bar of regular cross-section, some examples are given in Figure 10.1.
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Figure 10.1
In all these cases, Volume = Area of cross-section × Length Hence, to find the volume of a prism, calculate the area of the end and multiply this by length (or height) of the prism. Example A brass bar 250 mm long has a constant hexagonal cross-section measuring 90 mm across the face from one corner to the opposite corner. Find (i) the volume of the bar, (ii) the mass in kg if the density of brass is 8.4 g/cm3 . 90 mm
45 mm
45 mm
60° 45 mm
Figure 10.2
Side of each equilateral triangle = 45 mm (see Figure 10.2) Perpendicular height = 45 × sin 60◦ Area of each equilateral triangle =
1 × Base × Perpendicular height 2
Volume – Mass, Centre of Gravity, Moment =
•
237
1 × 45 × 45 sin60◦ 2
= 876. 85 mm2 Therefore area of end face = 6 × 876.85 = 5261.1 mm2 Volume of prism = Area × Length Therefore V = 5261. 1 × 250 = 1315275 mm3 = 1315 cm3 (nearest whole number) Mass (m) = Volume (V) × Density (p) Therefore m = 1315 × 8. 4 = 11046 g So mass = 11.05 kg to 2 decimal places.
Pyramids A pyramid is a body standing on a triangular, square or polygonal base, its sides tapering to a point at the apex, some examples are illustrated in Figure 10.3. The cone may be considered as a pyramid with a circular base.
Figure 10.3
The volume of a pyramid is one-third of the volume of its circumscribing prism.
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Therefore, the volume of a cone is one-third of the volume of a solid cylinder having the same height and base diameter as the cone. Therefore, volume of a cone =
π · r2 · h . 3
Similarly, the volume of a square pyramid is one-third the volume of a bar of square base with height equal to that of the pyramid. In all cases: 1 Volume of pyramid = (Area of base × Perpendicular height) 3
Oblique Prisms and Pyramids If the prism or pyramid be imagined as being made up of a number of discs or laminations and pushed over to one side, it can be seen by reference to Figure 10.4 that the same rule for finding the volumes of regular prisms or regular pyramids can be applied provided the perpendicular height is used.
Figure 10.4
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239
Frustums A frustum of a cone or pyramid is the bottom piece left, after a portion has been sliced off the top (Figure 10.5). The volume can be found by subtracting the volume of the sliced-off top part from the volume of the complete cone.
r
h
L
R
Figure 10.5
Example Find the volume of a frustum of a cone if the radius of the base is 5 cm, the radius of the top surface is 3 cm and the frustum has a height of 10 cm.
3 cm
10 cm
10 cm
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Mathematics
Looking at the frustum in cross-section gives: 3 10 5
It is necessary to project the sides to form the original cone and then find the height of the projected section, h.
h
3 10 5
By similar triangles So
h h + 10 = 3 5
5h = 3 (h + 10) ⇒ 3h = 5h + 30 ⇒ 2h = 30
Therefore h = 15 cm The volume of the whole cone is VW =
π · r2 · h π × 52 × (15 + 10) = = 654. 50 cm3 3 3
The volume of the top cone is VT =
π · r2 · h π × 32 × 15 = = 141. 37 cm3 3 3
So the volume of the frustum is 654. 50 − 141. 37 = 513. 13 cm3
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241
Sphere Volume of sphere =
π · d3 4π · r3 = 3 6
Volume of hollow sphere = Volume of outer sphere − Volume of inner sphere Therefore V =
π · D3 π · d 3 π 3 − = D − d3 6 6 6
where D = outer diameter and d = inner diameter. Example A solid lead cone, 40 mm diameter at the base and 120 mm perpendicular height is to be melted down and cast into a hollow sphere of 10 mm uniform thickness. Find the inside and outside diameters of the sphere.
Solution Working in centimetre: Let d = inside diameter, then (d + 2) cm = outside diameter. π 3 Volume of hollow sphere = D − d3 6 π (d + 2)3 − d3 = 6 π 3 = d + 6d2 + 12d + 8 − d3 6 π 2 6d + 12d + 8 = 6 π · r2 · h Volume of cone = 3 π × 22 × 12 = 3 = 16π cm3 Therefore π 2 6d + 12d + 8 = 16π 6 6d2 + 12d + 8 = 96 6d2 + 12d − 88 = 0 3d2 + 6d − 44 = 0
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Solving this quadratic: 62 − 4 × 3 × (−44) d= 2×3 √ −6 ± 564 = 6 −6 ± 23. 749 = 6 −6 ±
= 2. 958 and − 4. 958 Obviously d has to be positive, so the inner diameter is 2.96 cm and the outer diameter is 4.96 cm, both to 2 decimal places.
Theorem of Pappus applied to volumes With regard to volumes the theorem states: If an area, lying totally on one side of a fixed axis, is rotated about that axis in its own plane, it will sweep out a volume equal to its area multiplied by the distance moved by its centre of gravity, or centroid.
Axis
r R
Figure 10.6
In the majority of cases areas are being swept around one complete revolution and the resultant volumes are referred to as ’solids of revolution’. For example, consider a flat circular disc of radius r, its centre being at R from the axis as in Figure 10.6. If this area is swept around the axis through one complete revolution, it
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will sweep out a solid ring of circular section (a torus), the mean radius of the ring being R, and the radius of the cross-section of the material being r. A practical example of this is the volume of a ring, therefore: Volume swept out = Area × Distance moved by the centroid Volume of ring = π · r2 × 2πR = 2π 2 · R · r2 Now consider an elliptical area of major diameter D, minor diameter d, swept around through one revolution, R being the radius from the axis to the centre of the ellipse (Figure 10.7).
Axis
R d
D
Figure 10.7
The volume swept out is the shape of a circular lifebuoy of elliptical cross-section. Volume swept out = Area × Distance moved by the centroid Volume of lifebuoy =
π2 π D·d×π ·R = D·d·R 4 4
It is best to work from first principles making use of this theorem rather than applying formulae. Example The inside diameter of a solid, circular, cork lifebuoy is 500 mm and the section is elliptical 160 mm major diameter by 120 mm minor diameter. Find (i) its volume in m3 , (ii) its mass if the density of cork is 240 kg/m3 .
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Referring to Figure 10.7, working in metres: D = 160 mm = 0. 16 m d = 120 mm = 0. 12 m R=
(Internal diameter) (Major axis) + = 250 + 80 = 330 mm = 0. 33 m 2 2 Volume swept out = Area × Distance moved by the centroid Volume =
π × 0. 16 × 0. 12 × 2π × 0. 33 4
= 0. 031267 m3 Mass = Volume × Density = 0. 031267 × 240 = 7. 504 kg
Force, weight and centre of gravity Force is that which produces or tends to produce motion in a body. The unit of force is the newton (N). It may be defined as ‘the force required to give unit acceleration (a gain of velocity of 1 m/s every second the force is applied) to unit mass (one kg)’. The weight (W) of a body is the gravitational force on the mass of that body, that is, the force of attraction exerted on the body by the earth. If a body is allowed to fall freely, it will fall with an acceleration of 9.81 m/s2 , this is termed gravitational acceleration and is represented by g. Since 1 N of force will give 1 kg of mass an acceleration of 1 m/s2 , then the force (N) to give m kg of mass an acceleration of 9.81 m/s2 is m × 9. 81. Hence, at the earth’s surface, the gravitational force on a mass of m kg is mg newtons, therefore: Weight (N) = Mass (kg) × g m/s2 W = m·g The centre of gravity of a body is that point through which the whole weight of the mass can be considered as acting (Centre of Mass).
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For instance, imagine a body to be compressed in volume into one tiny particle without losing any mass, the position of this small heavy particle would be at the centre of gravity of the body to have the same effect. If the body was suspended from this point, or supported on it, it would balance perfectly without tilting. When dealing with an area instead of a solid, an area theoretically has no mass, therefore it is not strictly correct to use the term centre of gravity, in such cases the term centroid can be used.
Parallelogram A parallelogram would balance if laid on a knife edge along either one of its two diagonals, therefore its centre of gravity (for a plate) or centroid (for a plane area) is at the intersection of its diagonals as shown in Figure 10.8.
Figure 10.8
Trapezium The centroid of a trapezium is at the intersection of EF and HG as shown in Figure 10.9, found graphically as follows: Join EF (the midpoints) Produce AD to G, length DG being equal to BC Produce BC to H, length BH being equal to AD Join HG
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Mathematics
The intersection of EF and HG is the centroid. B
H
A
E
C
F
D
G
Figure 10.9
Triangle The centroid of a triangle lies on a line joining a corner of the triangle with the midpoint of its opposite side, at a point at one-third of the height from that side (see Figure 10.10).
h
h 3
Figure 10.10
Pyramid The centre of gravity of a pyramid or cone is at one-quarter of the height above the base (see Figure 10.11).
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247
h h 4
Figure 10.11
Hemisphere The centre of gravity of a hemisphere is at three-eighths of the radius above the diameter.
Volume of the frustum of a cone Let r = radius at top, R = radius at bottom and h = perpendicular height. See Figure 10.12. r
h
R
Figure 10.12
π ·h 2 R + R · r + r2 3 d π ·h 2 D D + D · d + d2 As R = and r = , Volume = 2 2 12
Total volume =
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Similar solids Areas of similar solids vary according to the cube of their corresponding linear dimensions. Similar solids mean that they are of the same shape and proportions, although their sizes are different.
2d
d
Figure 10.13
Consider two spheres, one of diameter d, the other twice as big in diameter = 2d, as in Figure 10.13.
Volume of small sphere = Volume of large sphere =
π · d3 6
π · (2d)3 π · 8d3 π · d3 = =8· 6 6 6
Therefore the ratio of their areas is 1 : 8, that is 1 : 23 . Therefore multiplying a linear dimension, for example, diameter, by 2 increases the volume by a factor of 8. So if lengths are increased by a factor of 3, volumes are increased by a factor of 27. In general, if the linear units are in the ratio 1 : r, the areas are in the ratio 1 : r3 It follows that, if the solids were made of the same material, their masses would also vary in the same ratios.
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Simpson’s first rule applied to volumes The procedure of finding the volume of an irregular object by Simpson’s rule is the same as for the area of an irregular figure, it is merely a matter of substituting cross-sectional areas for ordinates. Example The immersed cross-sectional areas through a ship 180 m long, at equal intervals, are 5, 118, 233, 291, 303, 304, 304, 302, 283, 171 and 0 m2 respectively. Calculate the displacement of the ship in sea water of 1.025 tonne/m3 .
Solution Cross-sectional areas Multiplier
Product for volume
5
1
5
118
4
472
232
2
466
291
4
1164
303
2
606
304
4
1216
304
2
608
302
4
1208
283
2
566
171
4
684
0
1
0
Total
6995
180 = 18 m 10 h 18 Volume of displacement = × Total = × 6995 = 41970 m3 3 3 Common interval h =
Displacement = Volume of displacement × Density = 41970 × 1. 025 = 43019 tonne
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Mathematics
Example A casting of light alloy, 750 mm long, has a variable cross-sectional area throughout its length. At regular distances of 125 mm apart, starting from one end, the sectional areas are, 12.2, 17.5, 23.2, 27.9, 21.0, 11.2 and 0 cm2 respectively. Find the volume and its mass if the density of the material is 3.2 g/cm3 .
Solution Cross-sectional areas Multiplier
Product for volume
12.2
1
12.2
17.5
4
70.0
23.2
2
46.4
27.9
4
111.6
21.0
2
42.0
11.2
4
44.8
0
1
0
Total
327.0
Common interval h = 12. 5 cm Volume =
12. 5 h × Total = × 327. 0 = 1362. 5 cm3 3 3
Mass = Volume × Density = 1362. 5 × 3. 2 = 4360 g = 4. 36 kg Example Plot the graph xy = 6 between the limits x = 2 and x = 6. If the area under this graph is rotated about its x axis through one complete revolution, calculate the volume swept out by Simpson’s rule.
Solution x·y = 6 ⇒ y =
6 x
Plotting points for values of x between the limits x = 2 and x = 6,
Volume – Mass, Centre of Gravity, Moment
x
2 3
y
3 2
4
5
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251
6
1.5 1.2
1
4 y
3
2
1
0
1
2
3
4
5
x 6
Figure 10.14
The graph is shown in Figure 10.14. When the area under this graph is rotated about its x axis through one revolution, the volume swept out appears as shown in Figure 10.15. y
x
Figure 10.15
The y ordinates of the graph become the radii of the solid at regular intervals along its length. Putting the cross-sectional areas at these regular intervals through Simpson’s rule as in the previous example:
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Mathematics Radii r
Cross-sectional areas = π · r2
Multiplier
Product for volume
3
9π
1
9π
2
4π
4
16π
1.5
2. 25π
2
4. 5π
1.2
1. 44π
4
5. 76π
1
1π
1
1π
Total
36. 26π
Common interval between ordinates = 1 Volume =
1 × 36. 26π = 37. 97 units3 3
Flow of liquid through pipes Volume flow (V) is the volume of a fluid flowing past a given point in unit time (m3 /s). The velocity or speed of flow is the ’length’ of liquid which passes in a given time. For example, if the velocity of the liquid is 2 m/s it means that a column of the liquid 2 m long passes every second. So Volume flow m3 /s = Cross-sectional area of the liquid m2 × Velocity (m/s) .
Therefore V = A · v Mass flow (m) is the mass of fluid flowing past a given point in unit time. Since density is the mass per unit volume, then the mass flow is the product of the volume flow and the density. So Mass flow (kg/s) = Volume flow m3 /s × Density kg/m3 Therefore m = V · ρ
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Example Oil of density 0.85 g/ml flows full bore through a pipe 50 mm diameter at a velocity of 1.5 m/s. Find the quantity flowing (i) in m3 /h, (ii) kg/h, (iii) tonne/h.
Solution Diameter of pipe = 0.05 m Velocity = 1.5 m/s = 1. 5 × 3600 = 5400 m/h (i) Volume flow m3 /h = Area m2 × Velocity (m/h) V=
π · (0. 05)2 × 5400 4
= 10. 603 m3 (ii) Density = 0. 85g/ml = 0. 85 × 106 g/m3 0. 85 × 106 kg/m3 = 1000 = 850 kg/m3 Mass flow (kg/h) = Volume flow m3 /h × Density kg/m3 Therefore m = 10. 603 × 850 = 9012. 4 kg (iii) m = 9.0124 tonne/h
Flow through valves When a liquid flows out of the open end of a pipe, the maximum quantity of liquid escaping depends upon the area of the bore of the pipe end. The flow can be restricted by a cover over the pipe end so that the area of escape is less than the area of the pipe bore. Referring to Figure 10.16, the area of escape is the surface area of a cylinder whose diameter is the same as that of the pipe and whose height is the ‘Lift’. In other words, the area of escape is that of a rectangle whose length is the circumference of the pipe (π · d) and width is the lift.
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Mathematics
Lift
Figure 10.16
The maximum effective lift will be when this area of escape is equal to the area of the bore. (In other words, when the amount of water flowing in the pipe has the same area through which to escape.) Therefore, Area of escape = Area of bore, π · d2 4 d lift = 4
(π × d) × lift =
So the maximum effective lift is one-quarter of the valve diameter. See following example for practical use. If the lift is more than this, no more liquid can flow through than that which is allowed by the area of the bore of the seat, but if the lift is less than one-quarter of the diameter, the area of escape is less than the area of the bore, and the quantity of liquid escaping depends upon the area p × d×lift. Example Calculate the quantity of water flowing, in litre/minute through a valve 100 mm diameter when the lift is 15 mm and the velocity of the water is 3 m/s, assuming that the wings of the valve take up one-sixth of the circumference.
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One-sixth of the area is obstructed by the wings, this leaves five-sixths of the area for the water to flow through. Working in metres: 5 × (π × 0. 1) × 0. 015 m2 6 5 Volume of water = × π × 0. 1 × 0. 015 × 3 m3 /s 6
Area of escape between valve and pipe =
As 1 m3 = 1000 litres, the volume of water =
5 × π × 0. 1 × 0. 015 × 3 × 1000 × 60 6
= 706. 86 litre/min.
Centres of gravity by first moments When an object is subjected to a force which acts in line with its centre of mass there is no turning effect produced. However, if the force acts along any other line, the force will have a turning effect. The greater the perpendicular distance between the line of action of the force and the centre of mass or pivot, the larger the turning effect (see Figure 10.17). The turning effect is called the turning moment of a force and can be found by: Moment = Force × Perpendicular distance from force to pivot. The unit of measurement of a moment is the newton metre (Nm)
Perpendicular distance between force and pivot
d
F Pivot
Figure 10.17
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The moment of a force F about the pivot is F × d. Turning moments act in either a clockwise or anticlockwise direction. If an object is balanced then the sum of all the clockwise moments equals the sum of all the anticlockwise moments. Example Find the missing force X which is needed to hold the following balanced beam in equilibrium (Figure 10.18). Find also the total force on the pivot, R.
4 kN
8 kN
2m
X
2m
R
3m
Space diagram
Figure 10.18
Solution For equilibrium: Clockwise turning moments about R = Anticlockwise turning moments about R (X × 3) = (4 × 4) + (8 × 2) 3X = 8 + 16 3X = 24 So X = 8 kN For equilibrium: Total upward force on beam = Total downward force on beam So
R =4+8+X R = 4 + 8 + 8 = 20 kN
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•
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Centre of gravity of compound shapes Consider a piece of plate cut to shape shown in Figure 10.19. Imagine this plate supported horizontally on one single support: call this support the fulcrum. The fulcrum (F) must be positioned exactly at the centre of gravity of the plate if the plate is to be perfectly balanced because ’the centre of gravity is that position through which the whole weight can be considered as acting’. Now take moments about the end O: in other words, imagine the plate is temporarily hinged at this end. For perfect equilibrium the moments of all the forces tending to turn the plate clockwise around the hinge must be equal to the moments of the forces tending to turn the plate anticlockwise about the hinge. Let W1 , W2 and W3 represent the weights of the top, centre and bottom parts. Let X1 , X2 and X3 represent the distances of the centres of gravity of these parts from 00. Let X represent the position of the fulcrum (F) from 00, and as previously stated this is the centre of gravity of the whole plate. Moments about 00: Clockwise moments = Anticlockwise moments (W1 × X1 ) + (W2 × X2 ) + (W3 × X3 ) = F × X Since the total upward force Fmust be equal to the total downward force, then F = W1 + W2 + W3 . Therefore (W1 × X1 ) + (W2 × X2 ) + (W3 × X3 ) = (W1 + W2 + W3 ) × X Therefore X =
(W1 × X1 ) + (W2 × X2 ) + (W3 × X3 ) (W1 + W2 + W3 )
In other words :
X=
Moments of weights Weights
As previously explained, weight = mass × g, and since g is constant it can be cancelled from numerator and denominator to give: Moments of masses X= Masses
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Similarly, if the same material is used throughout the shape, then Mass = Volume × Density. Therefore the density can be cancelled out giving: Moments of volumes X= Volumes Furthermore, if the plate is of uniform thickness, Volume = Area × Thickness, and therefore the thickness can be cancelled giving: Moments of areas X= Areas So for convenience take moments of masses, moments of volumes or moments of areas. The above expressions provide methods of finding the position of the centre of gravity of objects and figures made up from regular shapes. Moments as used above are usually referred to as ’first moments’ to distinguish them from ‘second moments’ which are used in applied mechanics. 120 20
80
CG
150
100
Y 30
160
Figure 10.19
Example: To find the position of the centre of gravity of the plate illustrated in Figure 10.19 which is symmetrical and of uniform thickness throughout, the dimensions shown being all in mm. Working in centimetre: Area of top flange = 12 × 2 = 24 cm2
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Area of centre web = 10 × 8 = 80 cm2 Area of bottom flange = 16 × 3 = 48 cm2 Distance of centre of gravity of top flange from base = 14 cm Distance of centre of gravity of centre web from base = 8 cm Distance of centre of gravity of bottom flange from base = 1. 5 cm Taking moments about base,
Y=
Moments of areas Areas
(24 × 14) + (80 × 8) + (48 × 1. 5) 24 + 80 + 48 336 + 640 + 72 = 152 =
Therefore Y =
1048 = 6. 89 cm above the base. 152
Example A hole 30 mm diameter is bored through a solid disc 90 mm diameter, the centre of the hole being 25 mm from the centre of the disc (Figure 10.20). Find the position of the centre of gravity of the disc after the hole has been cut out.
90 mm 25 mm 30 mm 20 mm
Figure 10.20
Moments about base, working in centimetre: Moments of areas y= Areas
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π · × 32 π · × 92 × 4. 5 − ×2 4 4 =
π · × 32 π · × 92 − 4 4 3 2 9 × 4. 5 − 3 × 2 = 92 − 32 346. 5 = 72
cancel by
π 4
= 4. 8125 cm Therefore centre of mass is 48.125 mm from the base, or 3.125 mm above the disc centre. (Alternatively, moments can be taken about centre of disc.) Note: In this example area is lost by boring the hole, therefore the summation of areas is the net area obtained by subtracting the area of the hole from the area of the disc; also, the summation of moments of areas is the difference between the moments of areas of the disc and hole. In each of the above two cases, it is obvious that the centre of gravity lies on the vertical centre line because the figures are symmetrical, therefore it is sufficient to calculate the position of the centre of gravity in one direction only. For figures that are not symmetrical, it is necessary to express the position of the centre of gravity in two directions at right angles to each other, say from the base and from one side; this is done by taking moments about these two datum lines separately.
Irregular figures Simpson’s rule can be employed to find the moment of an irregular area in a similar manner to which it is applied in finding the area.
Moment of area using Simpson’s Rule This is found in the longitudinal direction by multiplying each of the ordinates by its distance from the axis and then using Simpson’s rule.
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Example The distances from the centre-line to the hull for a 240 m long ship were taken at intervals of 30 m, the measurements starting from the aft of the vessel and numbered 0 to 8 inclusive. The values were:
Position number
0
1
2
3
4
5
6
7
8
Half-width
0
12.2
15.8
16.0
16.0
15.9
13.9
9.8
0
Find the position of the centre of gravity of the vessel. (Moments are taken from the amidships position.)
Solution Position
Half-width, w
Multiplier
A=w× multiplier
0 (aft)
0
1
0
−4
1
12.2
4
48.8
−3
−146.4
2
15.8
2
31.6
−2
−63.2
3
16.0
4
64.0
−1
−64.0
4 (amid)
16.0
2
32.0
0
5
15.9
4
63.6
1
63.6
6
13.9
2
27.8
2
55.6
7
9.8
4
39.2
3
117.6
0
1
0
4
0
8 (fore) Total
Common interval h = Area = A ×
Position from midships, p M = A × p
309.0
240 = 30 m 8
h 30 × 2 (for both sides) = 309. 0 × × 2 = 6180 m2 3 3
Excess moment = −36. 8 (aft)
0
−36.8
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Mathematics
h ×M×h×2 M×h = Centre of gravity from amidships = 3 h A ×A×2 3 Therefore centre of gravity =
−36. 8 × 30 = −3. 57 m (aft of amidships) 309. 0
Example Take a right-angled triangle of 80 mm base and 48 mm height as illustrated in Figure 10.21, to find the area and position of centroid from base by Simpson’s rule. Taking a regular shape such as this allows a comparison to be made with results calculated from formulae.
20 40
48 mm
60 80 mm
Figure 10.21
Ordinate
Multiplier
0
1
Area = Ordinate × Multiplier
Distance of ordinate from base
M =A×p
4.8
0
3.6
28.8
2.4
19.2
1.2
28.8
0
0
0 2
4 8
4
2 8
6
4
8
1
24 8
Total Common interval h =
48 4. 8 = 1. 2 cm 4
76.8
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•
263
48 × 1. 2 = 19. 2 cm2 3
Moment of area about base =
76. 8 × 1. 2 = 30. 72 cm3 3
Distance of centroid from base =
Moment of area 30. 72 = = 1. 6 cm Area 19. 2
From Figure 10.10 the position of the centroid is a distance of one-third of the height from the base. As the height of the triangle is 4.8 cm the centroid will be 1.6 cm from the base, a value which matches that obtained by Simpson’s rule. Further, if this triangle is swept through one complete revolution about its base it will sweep out a cone with dimensions 4.8 cm radius of base and 8 cm perpendicular height. Therefore the volume of this cone =
π × 4. 82 × 8 = 193. 02 cm3 3
By Theorem of Pappus: Volume swept out = Area × Distance moved by the centroid Therefore Volume = 19. 2 × 2π × 1. 6 = 193. 02 cm3
Shift of centre of gravity due to shift of loads Consider a system composed of loads which weigh w1 , w2 and w3 as shown in Figure 10.22, the centre of gravity of each being h1 , h2 and h3 respectively from the base.
w3
w2
h2 h1
w3
w1
Figure 10.22
d
w2
h3
w1
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Let y1 be the distance of the centre of gravity of the whole system from the base, then: Moments of weights y1 = Weights =
w1 h1 + w2 h2 + w3 h3 w1 + w2 + w3
If w3 is lifted into the position shown, through a height of d, the new centre of gravity of the whole system from the base, represented by y2 , is: w1 h1 + w2 h2 + w3 (h3 + d) w1 + w2 + w3 w1 h1 + w2 h2 + w3 h3 + w3 d = w1 + w2 + w3
y2 =
The shift of the centre of gravity of the whole system in the direction measured from the base is y2 − y1 which is: w1 h1 + w2 h2 + w3 h3 + w3 d w1 h1 + w2 h2 + w3 h3 − w1 + w2 + w3 w1 + w2 + w3 w3 d = w1 + w2 + w3
y2 − y1 =
In words this is: Shift in centre of gravity =
Weight shifted × Distance moved Total weight
As shown previously, weight can be represented by mass, so Shift in centre of gravity =
Mass shifted × distance moved Total mass
Test Examples 10 1. An I-section steel girder of 150 mm overall depth has unequal flanges, the top flange is 100 mm wide by 12 mm thick and the bottom flange is 140 mm wide by 14 mm thick. The centre web is 10 mm thick. Considering the flanges as rectangular in section by neglecting radii and fillets, calculate the mass in kg/m run if the density of the material is 7.86 g/cm3 . 2. A hollow steel shaft, 400 mm outside diameter and 200 mm inside diameter, has a coupling 75 mm thick and 760 mm diameter at each end, and the overall length
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•
265
is 6 m. Neglecting fillets and coupling bolt holes, find the mass of the shaft in tons taking the density of steel as 7. 86 × 103 kg/m3 . 3. A cylinder and sphere and base of a cone are all the same diameter, and the heights of the cylinder and cone are each equal to the diameter of the sphere. Find the ratio of the volumes of the cylinder and sphere relative to the volume of the cone. 4. A piece of flat steel plate, having a mass of 6.5 kg/m2 , is cut to the shape of a sector of a circle of radius 180 mm and subtended angle at the centre 240◦ , and the sector is rolled into a cone. Find (i) the mass of material used, (ii) the diameter of the base of the cone, (iii) the perpendicular height of the cone, (iv) the capacity of the cone in litres. 5. An object is constructed by brazing the base of a solid cone to the flat surface of a solid hemisphere, the diameter of the base of the cone and the diameter of the hemisphere both being 60 mm, and the perpendicular height of the cone 50 mm. Find the mass of the object if the density of the materials is 8.4 g/cm3 . 6. A hollow lead sphere has a uniform thickness of 10 mm and its mass is 3 kg. Taking the specific gravity of lead as 11.4, find its outside diameter. 7. A hole 24 mm diameter is bored centrally through a sphere 51 mm diameter. Calculate the volume of the drilled sphere in cm3 and its mass if the density of the material is 7. 86 × 103 kg/m3 . 8. A tapered hole is bored through a right circular cone, concentric with the axis of the cone. The base diameter of the cone is 64 mm and the perpendicular height is 60 mm. The diameter of the hole at the base of the cone is 28 mm and the diameter where it breaks through the surface of the cone is 16 mm. Calculate the volume and mass of the remaining hollow frustum, taking the density of the material as 8.4 g/cm3 . 9. The lengths of the sides of the base of a regular hexagonal pyramid is 25 mm and the perpendicular height is 60 mm. Find the volume in cm3 . If this pyramid is cut through a plane parallel to its base at half the height, find the volume of the remaining frustum. 10. A vessel in the form of a hollow cone with vertex downwards is partially filled with water. The volume of the water is 200 cm3 and the depth of the water is 50 mm. Find the volume of water which must be added to increase the depth to 70 mm. 11. The diameter of the base of a hollow cone is 300 mm and its perpendicular height is 500 mm. It is partly filled with water so that when resting on its base the depth of the water is 250 mm. If the cone is inverted and balanced on its apex, what will then be the depth of the water?
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12. The surface area of a solid sphere is 1.5 times the surface area of a smaller sphere, and the difference in their volumes is 10 cm3 . Find the volume and diameter of the smaller sphere. 13. The diameters of a barrel are 395 mm at each end, 477 mm at quarter and three-quarter lengths, 500 mm at mid-length. The total length is 581 mm. Using Simpson’s rule calculate the capacity of the barrel in litres. 14. A water trough has a regular isosceles triangular section, the angle at the bottom being 80◦ . Calculate the volume flow of water along the trough, in m3 /h, when the depth of the water in the trough is 180 mm and it is flowing at a velocity of 0.5 m/s. 15. Find the height of the centre of gravity of a frustum of a cone which is 80 mm diameter at the base, 60 mm diameter at the top and 40 mm perpendicular height.
11
DIFFERENTIAL CALCULUS (DIFFERENTIATION)
The development and application of the differential calculus may be explained by the relation between equations and their graphs. The meaning of the gradient at any point on such a graph is of particular importance in this context and it is therefore dealt with first. The symbol Δ represents a change in a quantity. So, for example, Temperature change
ΔT = T2 − T1
Change in time
Δt = t2 − t1
Change in angle
Δθ = θ2 − θ1
Change in distance
Δx = x2 − x1
Change in velocity
Δv = v2 − v1
The symbol δ means a small but measurable change in a quantity such as δT, δt, δθ etc.
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Mathematics
Gradient of a Straight Line It will be recalled, from previous work on graphs, that the gradient (or slope) of a line is change in y the between any two points on the line. change in x B
y
δy
Δy
δx
Δx
x
Figure 11.1
For the line shown in Figure 11.1, Gradient =
Change in y Δy = between A and B. Change in x Δx
This would be the same for a small change. The ratio
δy Δy = δx Δx
δy Δy is the same as the ratio and the ratio is the gradient of the straight line. Δx δx
Infinitesimally Small Changes ‘d’ The symbol d is used to denote a change that is infinitesimally small. In other words, the points A and B are essentially the same point. On the graph the ratios are all the same. The value is the same at any point on a straight-line graph. dy δy Δy = = dx δx Δx This ratio holds true even when the changes approach zero.
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269
The following examples are given by way of revision, to show how the gradient of a line is calculated. Example Plot the graph of the equation y = 3x between the limits x = 0 and x = 4 and find the gradient of the graph. 15 y
Change in y = 12
10
5
Change in x = 4 1
2
x 3
4
Figure 11.2
Referring to Figure 11.2, Change in y Change in x 12 = 4
Gradient =
=3 Note: The change in y is three times the change in x. Thus the gradient indicates the rate of change of one quantity, with respect to another related quantity. Example Plot the graph of the equation y = 2 − 0. 25x between the limits x = 0 and x = 8. Find the gradient of this graph. The graph is shown in Figure 11.3.
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•
Mathematics 2.5 y 2
1 0.5
Change in y = 2
1.5
Change in x = 8 1
2
3
x
4
5
6
7
8
Figure 11.3
Change in y Change in x −2 = 8
Gradient =
= −0. 25 (Note that the value of y changes from 2 to zero, that is, the change in y is −2.) Two important points are noted from the previous examples: (i) If y increases as x increases the gradient is positive (ii) If y decreases as x increases the gradient is negative.
Zero Gradient The gradient of a horizontal straight line is zero, since the change in y is zero for any given change in the value of x (Figure 11.4). y c
y=c
x
Figure 11.4
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•
271
Gradient of a Curve The gradient of a curve is not constant, but is changing from point to point along the length of the curve. The gradient of a curve at any given point is the gradient of the tangent to the curve at that point P.
b
tP
P
o
tt
r cu
ve
a
n ge
n Ta a
c
Figure 11.5
Referring to Figure 11.5, Gradient of the curve at P = Gradient of the tangent at P The gradient of a curve may be obtained by drawing the tangent and measuring the gradient, or mathematically, by the differential calculus.
Differentiation from First Principles From Figure 11.5 it was seen that the gradient of the tangent at P was required, in order to determine the gradient of the curve at P. Chord PA is now constructed, as shown (Figure 11.6). δ is used, for convenience, to express mathematically the change in the value of y and x between points P and A. Thus, δx, pronounced ‘delta ex’, means a small change in the value of x.
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•
Mathematics A
δy P
e rv
t
en
to
δx
P at
cu
ng Ta
Figure 11.6
Similarly, δy means the corresponding change in the value of y. From Figure 11.6, δy δx The gradient of this chord is obviously quite different to the gradient of the tangent to the curve. Gradient of chord PA =
However, if point A is now allowed to move along the curve towards point P, it is seen (Figure 11.6) that the slope of the chord more and more resembles the slope of the tangent as A gets closer to P (Figure 11.7).
A δy
P
e rv
nt
to
at
P
δx
cu
e ng
Ta
Figure 11.7
In fact, when point A gets very close to point P, the slope of the chord and the tangent are virtually the same. As A gets very close to P, so length δx is becoming very small, until it finally approaches zero value. That is, δx approaches zero value (denoted by δx → 0).
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•
273
From the above reasoning, it follows that, when δx has become infinitely small, the ratio δy/δx has reached a special value, or limit, where it represents the gradient of the curve at point P. δy it is identified by changing the δx δy dy notation from to . This relationship is expressed mathematically as: δx dx δy dy = lim it δx→0 δx dx In order to indicate that this is a special value of
dy is called the differential coefficient of y with respect to x and the process of dx dy finding is called differentiation. dx
The term
Example From first principles, derive an expression for the gradient of the curve y = x 2 at any point on the curve. The sketch of the graph of y = x 2 is shown in Figure 11.8. Point P(x, y) represents any point on the curve. y
A
(x + δx, y + δy)
P (x, y)
x
Figure 11.8
Chord PA is drawn as shown. The co-ordinates for point P are (x, y). Since A is very close to P, the co-ordinates for point A are (x + δx, y + δy) Now, the equation of this curve is
y = x2
(i)
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•
Mathematics
Therefore, at point P, y = x 2 and, at point A, y + δy = (x + δx)2 y + δy = x 2 + 2x · δx + (δx)2
So
(ii)
Subtracting equation (i) from equation (ii) gives (y + δy) − y = x 2 + 2x · δx + (δx)2 − x 2 δy = 2x · δx + (δx)2
Therefore
Therefore the gradient But
(iii)
δy 2x · δx + (δx)2 δy is therefore = δx δx δx
2x · δx + (δx)2 2x · δx (δx)2 = + = 2x + δx δx δx δx
If point A now moves very close to point P, so length δx will approach zero value. δy = 2x δx→0 δx
So lim it
Therefore
dy = 2x dx
Hence, at any point on the curve y = x 2 the gradient of the curve is 2x. Example Calculate the gradient of the curve y = x 2 at the points x = 3 and x = −2 From the previous example, any point on the curve has gradient
dy = 2x dx
When x = 3, gradient = 2 × 3 = 6 When x = −2, gradient = 2 × −2 = −4 These results can be confirmed by drawing tangents to the curve y = x 2 at x = 3 and x = −2 The gradient of each of these tangents, when measured, should be reasonably close to the calculated values. Example From first principles, find the differential coefficient of y, with respect to x, for the equation y = 3x 3 .
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•
275
(The differential coefficient or gradient may be calculated without actually drawing a graph, although a sketch of the graph is often useful when solving problems.) y = 3x 3 , therefore (y + δy) = 3 (x + δx)3 = 3 x 3 + 3x 2 (δx) + 3x · (δx)2 + (δx)3 So (y + δy) = 3x 3 + 9x 2 (δx) + 9x · (δx)2 + 3 (δx)3 It follows that (y + δy) − y = 3x 3 + 9x 2 (δx) + 9x · (δx)2 + 3 (δx)3 − 3x 3 So δy = 9x 2 (δx) + 9x · (δx)2 + 3 (δx)3 and
δy 9x 2 (δx) + 9x · (δx)2 + 3 (δx)3 = = 9x 2 + 9x · (δx) + 3 (δx)2 δx δx
Therefore lim it
δx→0
δy = 9x 2 δx
So for the curve y = 3x 3 the differential coefficient is 9x 2
General Rule for Differentiating a Power From the previous section, the following results were obtained: For the curve y = x 2 ,
dy dy = 2x and for the curve y = 3x 3 , = 3x 2 dx dx
From the form of these results (which is confirmed by solving further examples), the following general rule is noted: •
when y = a · x n the differential coefficient
So, if y = 9x 4 (i.e. a = 9 and n = 4),
dy − n · a · x n−1 dx
dy = 4 × 9x 4−1 = 36x 3 dx
This is true for any value of n. Example Differentiate the following equations with respect to x: (a)
y = x8
(d)
y=x
(b) y = 2x −3 1 (e) y = x
(c) y = −7x −4 x3 (f ) y = 12
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•
Mathematics
Answers (a)
y = x8
dy = 8x 7 dx
(b)
y = 2x −3
6 dy = −3 × 2x −3−1 = −6x −4 = − 4 dx x
(c)
y = −7x −4
28 dy = −4 × −7x −4−1 = 28x −5 = 5 dx x
(d)
y = x or y = x 1
dy = 1 · x 1−1 = 1 · x 0 = 1 dx
(e)
y=
1 = x −1 x
1 dy = −1 × x −1−1 = −1x −2 = − 2 dx x
(f )
y=
x3 12
x 3−1 3x 2 x 2 dy =3· = = dx 12 12 4
Differential Coefficient of a Constant As shown previously, when y is a constant, the resulting graph has zero gradient (see Figure 11.4). That is, if y = a (where a is any constant),
dy = 0. dx
Differential Coefficient of a Sum of Terms The differential coefficient of a sum of terms is obtained by differentiating each term separately. Example If y = x 3 − 2x 2 + 4x − 9, then
dy = 3x 2 − 4x + 4 dx
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•
277
Example Differentiate the equation y = Rearrange the equation: y = Therefore
3 x3 − 2 + 3x with respect to x. 3 x
3 x3 x3 − 2 + 3x = − 3x −2 + 3x 3 x 3
6 dy = x 2 3x 2 = − (−2) × 3x −2−1 + 3 = x 2 + 6x −3 + 3 = x 2 + 3 + 3 dx 3 x
Example Differentiate the equation s = 3t2 − 20t + 40 with respect to t. (Note: Symbols s and t represent two related quantities, as do x and y.) s = 3t2 − 20t + 40 so
ds = 6t − 20 dt
Second Differential Coefficient In some cases, having differentiated an expression once, it is necessary to differentiate a second time. The notation
d2 y is used to denote the second differentiation of an expression. dx 2
Example Consider the equation y = x 4 Differentiating once: Differentiating twice:
dy = 4x 3 dx d2 y = 12x 2 dx 2
Example Determine the second differential coefficient of the following equation: y = x 3 − 4x 2 + 3x − 7 dy = 3x 2 − 8x + 3 dx d2 y = 6x − 8 dx 2
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•
Mathematics
Distance, Velocity and Acceleration For a moving body: Average velocity =
Distance travelled Time taken
Using symbol v for velocity, s for distance and t for time, this equation may be expressed δs in calculus notation: average velocity = . δt If the velocity at a given instant (the instantaneous velocity) is required, the time interval δt must be very small, that is, δt must approach zero value. δs ds = δt→0 δt dt
Thus: lim it
Hence, at any given time t, the instantaneous velocity of a moving body may be ds expressed as: v = dt Similarly: Average acceleration =
Change in velocity Time taken
By the same reasoning as that above, the instantaneous acceleration a is given by the dv expression: a = . dt Example The distance s moved by a body in time t is given by the formula s = t3 − 3t2 . Find expressions for the velocity and acceleration of the body at any instant. s = t3 − 3t2 ds = 3t2 − 6t dt dv = 6t a= dt Note: The expression for the acceleration of the body is seen to be the second differential equation of the original equation relating distance s to time t. v=
s = t3 − 3t2
ds = 3t2 − 6t dt
d2 s = 6t dt2
Example A body moves s metres in t seconds according to the relationship s = t3 − 7t2 − 3.
Differential Calculus (Differentiation)
•
279
(a) Derive expressions for the velocity and acceleration of the body at any instant. (b) Use these expressions to find the velocity and acceleration of the body after 5 s.
Answer (a) s = t3 − 7t2 − 3 so v = 3t2 − 14t and a = 6t − 14 (b) When t = 5, v = 3 (5)2 − 14 × 5 = 5 m/s and a = 6 × 5 − 14 = 16 m/s2 Example The distance (displacement) s moved by a body in time t is given by the expression: s = 40t − 5t2 Calculate (a) the velocity after 2 s; (b) the time taken for the body to come to rest. (a) s = 40t − 5t2 so v = 40 − 10t When t = 2 s, v = 40 − 20 = 20 m/s (b) When the body comes to rest, v = 0 40 Therefore 40 − 0t = 0 so t = =4s 10
∴t=
40 10
Maxima and Minima An important application of differential calculus involves functions which have maximum or minimum values. The graph shown below (Figure 11.9) represents a function which has a local maximum value of y at point A and a local minimum value of y at B. Such points are called turning points and have zero gradient, that is, at a turning point dy = 0. dx At a maximum value the gradient changes from positive to negative, so the gradient is decreasing. Therefore the rate of change of the gradient is negative. dy d d2 y d (gradient) dx = = 2. The rate of change of the gradient is given by dx dx dx
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g ra d
ient
dy =0 dx Ne
tiv
gr
ad
Posit
ive g
radie
nt
e
Po s itive
ga
ie
nt
dy =0 dx
Figure 11.9
So if
d2 y < 0, the turning point is a maximum. dx 2
It follows that if
d2 y > 0, the turning point is a minimum. dx 2
The case where
d2 y = 0 will be looked at later in the text. dx 2
Example The curve y = 3x − x 2 + 1 has a turning point. (a) Find the value of x at which this occurs. (b) Determine whether the turning point is a maximum or minimum value. (c) Find the value of y at the turning point.
Answer dy = 3 − 2x dx dy At a turning point =0 dx Therefore 3 − 2x = 0 giving x = 1. 5
(a) y = 3x − x 2 + 1 so
(b)
d2 y = −2 and so the turning point is a maximum. dx 2
(c) To find the maximum value of y, substitute x = 1. 5 in the original equation:
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d2 y d2 y : if = 0 the curve has a point of inflexion. The gradient dx 2 dx 2 does not change from negative to positive, or vice-versa, so this point is stationary.
Returning to the value of
A point of inflexion resembles the following diagram (Figure 11.10).
Figure 11.10
Example (a) Determine the values of x which give maximum or minimum values of y in the equation y = x 3 − 6x 2 + 9x (b) Calculate these maximum or minimum values of y.
Answer dy d2 y = 3x 2 − 12x + 9 and = 6x − 12 dx dx 2 dy =0 For maximum or minimum values dx Therefore 3x 2 − 12x + 9 = 0 or x 2 − 4x + 3 = 0 This factorises to give (x − 3)(x − 1) = 0 Therefore the turning points occur when x = 1 and x = 3. d2 y (b) When x = 1: = (6 × 1) − 12 = −6 : dx 2 When x = 1: y = (1)3 − 6 (1)2 + 9 × 1 = 4: therefore ‘maximum’ at (1, 4) (a)
d2 y = (6 × 3) − 12 = 6: dx 2 When x = 3: y = (3)3 − 6 (3)2 + 9 × 3 = 0: therefore ‘minimum’ at (3, 0)
When x = 3:
Example A rectangular sheet of metal measures 16 cm by 10 cm.
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From each corner a square of side x cm is removed, and the flaps formed are bent up to make an open box. Find its maximum volume. 16 cm
x cm x cm 10 cm
x cm
(16 – 2x) cm (10 – 2x) cm
Solution The volume of the ‘box’ is length × width × height Therefore volume = (16 − 2x) (10 − 2x) x = 160x − 52x 2 + 4x 3 cm3 So
dV = 160 − 104x + 12x 2 = 0 at a turning point dx
Cancelling by 4 gives 3x 2 − 26x + 40 = 0 Solving, using the quadratic formula or by factorising, gives
20 and 2. 3
Differential Calculus (Differentiation) The first value can be rejected since 10 − 2 × cannot be a negative value.
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283
10 20 =− cm and, obviously a length 3 3
This gives the solution as x = 2. The dimensions of the box are therefore: Length = 16 − 2 × 2 = 12 cm Width = 10 − 2 × 2 = 6 cm Height = 2 cm giving a volume of 12 × 6 × 2 = 144 cm3 d2 V = −104 + 24x = −104 + 24 × 2 = −56 when x = 2, making the turning point dx 2 a maximum.
Now
Questions 1. 80 metres of fencing is used to enclose a rectangular area, using a straight wall for one side. Find the maximum area that can be enclosed. 2. Panels of length 2 m are used to enclose a rectangular area 512 m2 . If one side of the pen is a straight wall and needs no fencing, find the minimum number of panels needed. 3. Sheet metal is used for making an oil-tank, open at the top, on a square base. If the volume of the tank is to be 4 m3 , find the least area of sheet metal required.
Solutions 1. A = (80 − 2x) · x = 80x − 2x 2 dA = 80 − 4x = 0 at turning point dx d2 A Therefore 4x = 80 and so x = 20 2 = −4 and since this is obviously negative, the dx turning point is a maximum.
x
x
80 – 2x
The maximum area is A = (80 − 2 × 20) 20 = 40 × 20 = 800 m2
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2. Let the total length of panels be L Then L = 2x + y 512 But x · y = 512 and so y = x
x
512 m2
x
y Therefore L = 2x + 512x −1 dL = 2 − 512x −2 = 0 at a turning point dx 512 = 2 so 512 = 2x 2 giving x 2 = 256 x2 So x = 16 m d2 L = 1024x −3 which is positive when x = 16 so this is a minimum value. dx 2 512 = 32 making L = 64 m. If x = 16, y = 16 As each panel is 2 m wide the requirement is for 32 panels. 4 3. Volume gives x 2 · h = 4 so h = 2 x 4 16 + x2 Surface area A = 4xh + x 2 = 4x · 2 + x 2 = x x dA = −16x −2 + 2x So A = 16x −1 + x 2 giving dx 16 16 So 2x − 2 = 0 giving 2x = 2 so 2x 3 = 16 x x Therefore x 3 = 8 giving x = 2 4 When x = 2, h = 2 = 1 2 32 d2 A = 32x −3 + 2 = 3 + 2 > 0 giving a minimum value for A. 2 dx 2 Therefore A = 4 × 2 × 1 + 22 = 8 + 4 = 12 m2 .
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Differentiation of sin(x) and cos(x) For any angle x and any value n: If y = sin (x) , then
dy = cos (x) dx
If y = sin (nx) , then If y = cos (x) , then
dy = n · cos (nx) dx
dy = − sin (x) dx
If y = cos (nx) , then
dy = −n · sin (nx) dx
Note: In calculus, angles are always measured in radians. These differential coefficients can be demonstrated by reference to the sketch of the graphs of sin(x) and cos(x) (Figure 11.11). Gradient = 0 y = cos(x) y = sin(x) Gradient = –1
π
Figure 11.11
For example, π when x = rad, the gradient of the sine curve is zero: 2 π this gives the point , 0 on the curve of the differential. 2
2π
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when x = π rad, the gradient of the sine curve is −1: this gives the point (π, −1) on the curve of the differential. If this procedure is continued for different values of x on the sine curve, the points dy plotted form the cosine curve, showing that if y = sin (x), then = cos (x). dx Similarly, the gradient at any point on the cosine curve is numerically equal to the negative of the corresponding value of the sine curve. Example Differentiate the following expressions with respect to x: (a) y = sin(3x) (b) y = 2 sin(5x) (c) y = 4 sin(x) − 3 sin(2x) (d) y = 4 cos(2x) (e) y = 6 sin(4x) + 5 cos(3x)
Answers (a) (b) (c) (d) (e)
dy dx dy dx dy dx dy dx dy dx
= 3 cos (3x) = 10 cos (5x) = 4 cos (x) − 6 cos (2x) = −8 sin (2x) = 24 cos (4x) − 15 sin(3x)
Differentiation of ln(x) and ex if y = ex , then
dy = ex dx
if y = enx , then
dy = n · enx dx
Differential Calculus (Differentiation) if y = ln (x) , then
•
287
dy 1 = dx x
if y = ln (nx) , then
dy 1 = dx x
Note carefully the last statement: the value of n does not appear in it does occur in all other cases.
dy for y = ln(nx) but dx
This can be explained as follows. Using the laws of logarithms: ln (nx) = ln (n) + ln (x) Therefore:
d d d [ln (nx)] = [ln (n)] + [ln (n)] dx dx dx
but ln(n) is a constant and so Therefore
d [ln (n)] = 0 dx
d d 1 1 [ln (n)] + [ln (n)] = 0 + = dx dx x x
Naperian logarithms (In) have special applications, for example, when finding the work done by an expanding gas. Example Obtain the second differential coefficient of the equations y = ln (x) and y = ex .
Answer If y = ln (x), then
dy 1 = = x −1 dx x
d2 y 1 = (−1) × x −1−1 = −x −2 = − 2 dx 2 x dy = ex if y = ex , then dx d2 y so 2 = ex dx So
Function Notation In an equation such as y = 2x or y = 5x 3 + 3x, the value of y obviously depends upon the value chosen for x. Hence, y is said to be a function of x and the general expression
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for such a relationship is: y = f (x) Functional notation may also be used to indicate the differentiation process, using the symbols shown below: y = f (x) dy = f (x) dx d2 y = f (x) dx 2 Example If y = 5x 3 + 3x then f (x) = 5x 3 + 3x f (x) = 15x 2 + 3 f (x) = 30x
Text Examples 11 1. Differentiate the following equation with respect to x: (a) y = x 3 + 3x 2 − 9x + 4 7 2x 3 − 2 +x (b) y = 3 x √ 5 (c) y = x 3 + 1 (d) y = 3 cos (x) + 2 sin(x) 2. Calculate the gradient of the curve y = x 2 + 3x − 7 when x = 3 and x = −2. 3. The displacement s metres of a body from a fixed point is given by the equation s = 20t − 5t2 + 4 where t is the time in seconds. Find: (a) The velocity after 2 s. (b) The displacement when the velocity is zero. (c) The acceleration. 4. Determine the gradient of a tangent to any point in the curve y = 4x +
1 x
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Show that there are two points where the gradient is zero. 5. Find the co-ordinates of the point on the graph of y = 3x 2 − x + 2 where the gradient is equal to −7. 6. The angle θ radian through which a shaft has turned after time t seconds is given by θ = 2 + 16t −
t2 2
Find the angular velocity after2 s and the time for the shaft to come to rest. dθ Note: angular velocity ω = dt 7. (a) Determine the second differential coefficient of the expression y = x 3 + 3x + ln (x) with respect to x. (b) Determine the second differential coefficient of the expression y = 3 cos (θ ) − 7 sin (θ ) + θ with respect to θ 8. Determine the maximum value of y in the equation y = 12x + 3x 2 − 2x 3 . 9. For a beam of length l, the bending moment M at any distance x from one end is given by: M=
ωLx ωx 2 − 2 2
where ω is the uniform load per unit length. Show that bending moment is a maximum at the centre of the beam. 10. Differentiate the equations: following
4 1 (a) t = 2. 1 × 3 2 − √ 5 θ θ a · un − 1 c (c) y = 3x · x 2 − 4
(b) z =
(d) y = 2 cos (θ ) + 5 x3 − 2x 2 + 3x + 1. 3 12. Power (P) and Voltage (V) of a lamp are related by P = a · V b where a and b are constants. Find an expression for:
11. Determine the minimum value of f (x) for the equation f (x) =
(a) the rate of change of power with voltage and (b) power per volt at 100 volts, when a = 0. 5 × 10−10 and b = 6. 13. A line of length L is to be cut up into four parts and put together as a rectangle. Show that the area of the rectangle will be a maximum if each of its sides is equal to one quarter of L (i.e. a square).
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14. Determine the second differential coefficient of: (a) f (θ ) = cos (θ ) − ln(θ ) (b) f (t) = a · t2 + 2 · ln(t) (c) f (x) = 5ex 15. A body moves so that its displacement (distance) x metres, which it travels from a certain point O, is given by: x = 0. 2t2 + 10. 4 where t is the time in seconds. Find the velocity and acceleration (a) 5 s after the body begins to move and (b) when the displacement is 100 m. 16. Angular displacement (θ rad) from rest of a revolving wheel is given by: θ = 2. 1 − 3. 3t + 4. 8t2 where t is the time in seconds. Find the angular velocity and angular acceleration after 1.5 s. 17. Verify that the equation f (x) = x 5 − 5x has a maximum and a minimum value and determine the value of x and f (x) at these points. dy = 4x + 2. 18. The gradient function of y = a · x 2 + bx + c is dx The function has a minimum value of 1. Find the values of a, b and c.
12
INTEGRAL CALCULUS (INTEGRATION) Integration may be considered as reversing the process of differentiation.
That is, given the differential coefficient of a function, we are required to find the original function. The symbol is used to denote the integration process. This is the old-fashioned letter ’S’ and the reason for its use becomes evident when integration is used to find the ’sum’ of a number of quantities.
Constant of Integration Consider the three equations below: y = x 2 , y = x 2 + 3, y = x 2 + 7 For each of these equations
dy = 2x, which equation is the integral of 2x? dx
Obviously, when reversing the differentiation process (i.e. integrating), provision must be made for the possibility of a constant in the original equation. This is achieved by adding a constant, C, called the constant of integration.
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Therefore
2x dx = x 2 + C
which may be interpreted as: (the integral of) 2x (with respect to x) = x 2 + C. Every integral must be concluded with ‘d#’ where # is the letter corresponding to the variable in the equation.
General Rule for Integration of Powers of x Consider the following differential equations: x2 , 2 x3 If y = , 3 x4 If y = , 4
If y =
dy =x dx dy = x2 dx dy = x3 dx
From these equations, the following integrals are obtained: x2 +C x dx = 2 x3 x 2 dx = +C 3 x4 x 3 dx = +C 4 From these cases it can be seen that the general rule is ‘add 1 to the power, then divide by this new power’. Expressing this mathematically, a general rule for integration is obtained, which includes a constant of integration: x n+1 x n dx = +C n+1 Note: there is an important exception to this rule. 1 cannot be integrated using this rule as the new power would x be 0, and division by 0 is impossible.
The integral of x −1 or
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Definition
1 dx = ln (x) + C x
Example Evaluate the following integrals: −4 x (a) x 7 dx (b) dx 3x 5 dx (c) 3
(d)
8 dx
Solutions
x8 +C 8 x6 3x 6 (b) 3x 5 dx = +C = +C 6 2 −4 x −3 x −3 1 x dx = +C =− +C =− 3 +C (c) (−3) 3 3× 9 9x (d) 8 dx = 8x + C (a)
x 7 dx =
This last result may be confirmed by differentiating the answer, or by applying the general rule, as shown: x1 8 dx = (8 × 1) dx = 8 × x 0 dx as x 0 = 1 = 8 × + C = 8x + C 1
Integration of a Sum of Terms The integral of a sum of terms is equal to the sum of their separate integrals. Example
2 3x + 7x − 10 dx =
3x 2 dx +
= x3 +
7x dx −
10dx
7x 2 − 10x + C 2
Note: The constants which should be added to each separate term are combined as a single constant. Example Evaluate the integral:
x 2 − 3x 3 + 2 dx
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2 x 3 3x 4 − + 2x + C x − 3x 3 + 2 dx = 3 4
Example Integrate the expression 3t2 + t + 1 with respect to t. 2 t2 3t + t + 1 dt = t3 + + t + C 2 Example 1 with respect to x. 2 1 x 2 3x − 6x + dx = x 3 − 3x 2 + + C 2 2
Integrate the expression 3x 2 − 6x +
Questions Work out the following:
x 5 dx
1.
x
4.
−3
2.
dx
5.
10.
1 √ dx x 1 + y3 dy y2
x 3 − 2x dx
3x − 5x + 6 dx 2
7.
8. 11.
1+x dx x4
y4 − 2y3 − 9 dy
3.
x 3 + 6 dx
(x + 1) (x + 2) dx
6.
x + x4 dx x3 1 dy y 1+ y
9. 12.
Answers 1.
x6 +c 6
4.
−
7.
√ x 1/2 +c =2 x+c 1/2
9.
−x −1 +
11.
x 5 y4 − − 9y + c 5 2
x −2 +c 2
2.
x4 − x2 + c 4
5.
x3 −
8.
−
x2 1 x2 +c = − + c 10. 2 2 x 12.
3.
5x 2 + 6x + c 6. 2
x4 + 6x + c 4 x 3 3x 2 + + 2x + c 3 2
1 x −3 x −2 1 − +c=− 3 − 2 +c 3 2 3x 2x
y2 1 − +c 2 y y2 +y+c 2
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Evaluating the Constant of Integration The value of the constant of integration, for a given function, can be calculated, provided a corresponding pair of values of x and y are known. Example The gradient of a curve is 4x + 5. If the curve passes through the point (x = 0, y = −4), find the equation of the curve. Gradient of curve Therefore
dy = 4x + 5 dx dy y= dx = (4x + 5) dx =2x 2 + 5x + C dx
(i)
To find the constant of integration C, substitute x = 0, y = −4, in equation (i): −4 = 2 × (0)2 + 5 × 0 + C Therefore C = −4 The equation of the curve is y = 2x 2 + 5x − 4 Example The curve of a graph has a gradient of 10x − x 2 If the curve passes through the point (3, 52), find the equation of the graph. Gradient of curve Therefore
dy = 10x − x 2 dx x3 dy dx = 10x − x 2 dx = 5x 2 − + C y= dx 3
To find the constant of integration C, substitute x = 3, y = 52, in equation (ii): 52 = 5 × (3)2 − Therefore 52 = 45 − 9 + C
so C = 16
The equation of the curve is y = 5x 2 −
x3 + 16 3
(3)3 +C 3
(ii)
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Integration of sin(x), cos(x), ln(x) and ex The following differential coefficients are known: if y = sin (x) , then
dy = cos (x) dx
if y = cos (x) , then
dy = − sin (x) dx
if y = ln (x) , then if y = ex , then
dy 1 = dx x
dy = ex dx
Reversing the process, the following integrals are obtained: cos(x)dx = sin(x) + C sin (x) dx = − cos (x) + C 1 dx = ln (x) + C x ex dx = ex + C When x is scaled by a value, for example, y = sin(3x) or in general, y = sin(nx) the following integrals are obtained: sin(nx) cos(nx)dx = +C n cos (nx) +C sin (nx) dx = − n ln (x) 1 dx = +C nx n enx enx dx = +C n Example Determine the integral [2 cos (x) − sin (x)] dx [2 cos (x) − sin (x)] dx = 2 sin (x) − cos (x) + C = 2 sin (x) + cos (x) + C
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Example Determine the integral [cos (2θ) + 5 sin(3θ)] dθ with respect to θ . sin (2θ) 5 cos (3θ ) [cos (2θ) + 5 sin (3θ)] dθ = − +C 2 3 Example
5 dx x 5 1 dx = 5 × dx = 5 ln (x) + C x x
Determine the integral
Definite Integration: Area Under a Curve Theory In Figure 12.1, the curve represents the sketch of a function f (x). The area abc may be divided into a large number of strips, or elements, one of which is shown. Let A = area abc δA = area of one strip If δx is made very small, the effect of the curve at the top of the strip becomes negligible and the strip can then be considered to be a rectangle of height y and width δx. Therefore the area of the strip, δA = y · δx Dividing both sides by δx gives Letting δx reduce to 0 gives
δA =y δx
dA δA = lim it =y dx δx→0 δx
Integrating both sides of this expression with respect to x: dA dx = (y) dx dx Therefore A = y dx between the limits x = a and x = b.
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•
Mathematics y c
y a
b
δx
x
Figure 12.1
This is written using the notation: b A=
y dx a
To evaluate the integral: •
Integrate f (x) (the constant, C, can be omitted).
•
The integral is inserted in square brackets, with upper and lower limits of x placed after the second bracket with the lower limit at the bottom of the bracket and the upper limit at the top.
•
Evaluate this expression using the top value, b, and call it B.
•
Evaluate this expression using the bottom value, a, and call it A.
•
Work out B − A.
•
C can be ignored because it would automatically cancel when B − A is calculated.
Example Find the area between the curve y = x 2 , the x axis and the ordinates x = 2 and x = 4 (Figure 12.2).
4 x 2 dx = 2
x3 3
4
=
2
43 3
−
23 3
=
64 8 56 − = = 18. 67 3 3 3
When an integral has limits, it is called a definite integral.
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y c y = x2
2
4
x
Figure 12.2
60 50 40 30 20 10 1
2
3
4
5
6
Figure 12.3
Example Find the area enclosed by the curve y = x 2 + 2x + 1, the x axis, and the ordinates x = 2 and x = 5 (Figure 12.3). 5 Area =
x 2 + 2x + 1 dx
2
=
x3 + x2 + x 3
5 2
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•
Mathematics
3 53 2 + 52 + 5 − + 22 + 2 3 3 8 125 + 25 + 5 − +4+2 = 3 3 8 125 + 30 − − 6 = 3 3 117 = + 24 3
Area =
= 63 units2 One important result from definite integration is that areas of a graph that appear below the x axis have a negative value.
Questions 1. Find the area under the curve y = x + 3x 2 between x = 1 and x = 2. 1 2. Find the area under the curve y = 2 between x = 1 and x = 2. x 3. Find the area under the curve y = 2x − x 2 between x = 0 and x = 2. √ 4. Find the area under the curve y = x between x = 1 and x = 4.
Answers 2
x + 3x
1. A =
2
1
2 2. A =
x2 + x3 dx = 2
2
=
1
3 1 1 4 +8 − + 1 = 10 − = 8 2 2 2 2
2
1 1 − (−1) = x −2 dx = −x −1 1 = − 2 2
1
2 3. A = 0
4 4. A = 1
2 8 4 1 x3 23 2 2 −0= 4− = = 1 2x − x dx = x − = 2 − 3 0 3 3 3 3
2
3 √ 2 14 2 3 4 2 42 −1 = =4 x dx = x 2 = 3 3 3 3 1
The following questions involve the calculation of areas but also require an application of algebra to their solution.
Questions 1. The diagram below shows the graph of y = x 2 − 4
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•
301
(a) Find the co-ordinates of P, Q and R. (b) Find the area of the region enclosed by the curve and the x axis. y
x P
Q
R
Answers (a) At the points P and Q, y = 0 Therefore x 2 − 4 = 0 so x 2 = 4 Therefore x = ±2, giving P at (−2, 0) and Q at (2, 0) At R, x = 0, so y = −2 giving R at (0, −2)
1 3 1 2 (−1)3 (1)3 x (b) Area = − 4x − 4 (1) − − 4 (−1) x − 4 dx = = 3 3 3 −1 −1 1 22 22 1 − 4 − − + 4 = − , so actual area = Therefore area = 3 3 3 3 2. (a) Sketch the curve y = x(3 − x). (b) Find the area of the region enclosed by the curve and the x axis.
Answers (a)
3
y
2 1 x –1
1
2
3
–1 –2 –3
3
3 x (3 − x) dx =
(b) Area = 0
3x − x
0
2
3x 2 x 3 − dx = 2 3
3
=
0
27 27 − 2 3
−0 =
9 2
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3. The curve y = 9 − x 2 and the line y = 2x + 6 intersect at P and Q. (a) Find the co-ordinates of P and Q. (b) Find the area of the region enclosed by the curve and the x axis. y Q
x P
Answers (a) At P and Q 9 − x 2 = 2x + 6 so 0 = x 2 + 2x − 3 = (x + 3) (x − 1) Therefore at P, x = −3, giving y = 0. So P(−3, 0) At Q, x = 1, giving y = 8. So Q (1, 8) 1 1 1 2 2 (b) Area = 9 − x dx − (2x + 6) dx = −x − 2x + 3 dx −3
−3 −3
3 1 (−3)3 x 1 2 2 − (−3) + 3 (−3) Area = − − x + 3x = − −1+3 − − 3 3 3 −3 22 5 − (9 − 9 − 9) = Area = 3 3 1 4. The diagram shows the curve y = √ and the line y = x. x The line and the curve intersect at M. (a) Find the coordinates of M.
(b) Find the area of the shaded region. y
M
0
4
x
Integral Calculus (Integration) Answers √ 1 (a) At M, √ = x so x · x = 1 giving x = 1 x When x = 1, y = 1 so M (1, 1) 2 1 3/2 4 1 4 √ x 2x x dx = + (b) Area = x dx + 2 0 3 1 0
1
Area =
1 14 31 2×8 2×1 1 − (0) + − = + = 2 3 3 2 3 6
5. The diagram shows the curve y = 2x + There is a minimum point at M (a) Find the coordinates of M.
1 x2
(b) Find the area of the shaded region. y
M
x 0
2
Answers (a) For a minimum value
dy =0 dx y = 2x +
Therefore
1 → y = 2x + x −2 x2
dy = 2 − 2x −3 = 0 at a turning point dx 2−
2 =0 x3 2 2= 3 x
2x 3 = 2
•
303
•
304
Mathematics x3 = 1 √ 3 x= 1 x=1
1 =3 12 Therefore M is the point (1, 3) 2
2 1 2 −2 −1 2 2 2x + x dx = x − x 1 = x − (b) Area = x 1 1 1 7 1 1 − 12 − = 4− − (0) = = 3. 5 Area = 22 − 2 1 2 2 When x = 1, y = (2 × 1) +
Definite integration of other functions is performed in exactly the same manner. There is one critical point that must be remembered – any operations involving Calculus and trigonometric functions must be performed using radian measure. Example π Evaluate the definite integral sin (x) dx 0
π
sin (x) dx = [− cos (x)]π0 = (− cos (π)) − (− cos (0)) = (− (−1)) − (− (1)) = 1 + 1 = 2
0
Questions Calculate the values of:
1.
2 x3 + 4 dx
2.
1
5.
1
6.
2 cos(x) dx 3
3.
0
π/3 2t dt 2 sin 3
π/2 sin(2t)dt
π cos
7. π/2
π
1
2
0
2 1 + 2e0.3x dx 1
ex dx
11.
0
3 14. 2
t dt 2
e−2t dt
0.5
3 0
(1 + 0. 6 cos(0. 2t))dt
8. −0.2
1 12.
1
4 ex + e−x dx 15.
π/2 3 sin(4φ)dφ 0
π/2
(sin(x) − sin(3x))dx 10.
4.
0
π/6
9.
13.
2 2 x + x3 dx x
1 eθ/3 − θ/3 e
0.5
2e−0.4t dt
1 16. −1
4 dx 5e1.4x
dθ
Integral Calculus (Integration)
•
305
Answers 1.
7.75
2.
4.67
3.
0.561
4.
0
5.
0.521
6.
0
7.
0.586
8.
1.12
9.
1.33
10.
1.72
11.
0.0585
12.
0.253
13.
4.15
14.
2.18
15.
0.811
16.
0.732
Integration as a Summation In the previous section, it was shown that the area under a curve could be determined by dividing the area into very narrow strips. y c
y a
b
x
δx
Figure 12.4
One such strip is shown in Figure 12.4. The width of the strip may be represented by symbol dx if its width approaches 0 value. Area of one strip = y × dx The total area abc could be obtained by adding together all these small areas, such as the one shown. That is, area abc = the summation of all the areas such as y · dx between the limits x = a and x = b.
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•
Mathematics b
Therefore Area =
y dx a
Thus, the symbol
may be interpreted as meaning the summation of the quantities to
which it is applied.
Volume of a solid of revolution If a curve is rotated about the x axis as shown (Figures 12.5 and 12.6), the shape generated is called a solid of revolution.
y dx
Figure 12.5
y
dx
Figure 12.6
Integral Calculus (Integration)
•
307
The area under the curve is divided into a large number of elemental strips, such as the one shown (Figure 12.5). By rotating this element about the x axis, a solid disc is generated (Figure 12.6). Volume of one disc = Cross-sectional area × Thickness = π · y2 · dx The total volume is the summation of the volumes of all such discs between the limits x = a and x = b. b Therefore Volume =
πy2 dx a
Since the constant factor π is not affected by the integration process, the equation may be written as: b Volume of a solid of revolution = π y2 dx a
Example The curve y = x 2 is rotated about the x axis between the limits x = 1 and x = 4. Determine the volume of the solid of revolution produced (Figure 12.7).
Figure 12.7
Solution b V=
4 π · y2 dx =
a
=π
5 4
x 5
1
1
2 π · x 2 dx = π
4 1
4 x dx
308
•
Mathematics V =π
45 5
−
15 5
= 204. 8π = 642. 8 cubic units (1 decimal place)
Example Calculate the volume generated by rotating the curve y = x 3 about the x axis between x = 0 and x = 2.
Solution b
2 2
V=
π · y dx = a
2 π · x 3 dx = π
0
2 0
7 2 7 6 128π x 2 = units3 x dx = π =π 7 0 7 7
So the volume of revolution is 57.4 units3 to 1 decimal place. Example Use integral calculus to prove that the volume of a cone is of the base and h is the vertical height (Figure 12.8).
π · r2 · h where r is the radius 3
r h
Figure 12.8
The gradient of the sloping line is
Volume of revolution = π
h 0
Therefore volume =
r r and so it has equation y = x h h
r 2 x dx = π h
h 0
π · r2 r2 2 x dx = h2 h2
h x 2 dx 0
h 3 π · r2 h3 π · r2 h π · r2 x 3 π · r2 h (0) − = 2 · = = · 2 2 h 3 0 h 3 h 3 3
Integral Calculus (Integration)
•
309
Distance and velocity by integration In the previous chapter, it was shown that the relation between distance s, time t and velocity v could be expressed as: ds =v dt Integrating both sides of this equation with respect to t, gives
ds dt = vdt dt s=
Therefore
vdt
(i)
Similarly the relationship between acceleration, velocity and time was defined as dv =a dt v=
so it follows that
a dt
(ii)
Example The velocity v of a body after time t is given by the expression v = 10 − 2t. If displacement (distance) s = 24 when t = 2, obtain an expression for distance s in terms of t.
Solution
s=
v dt =
(10 − 2t) dt = 10t − t2 + C
but when t = 2, s = 24; therefore 24 = (10 × 2) − 22 + C This gives 24 = 20 − 4 + C and so C = 8 Therefore s = 10t − t2 + 8 Example The velocity v m/s after t seconds for a body is given by v = 4 + 7t.
•
310
Mathematics
Find the distance travelled by the body in the interval from t = 0 to t = 5 s. 5 s= 0
5 175 7t2 7 × 52 (4 + 7t) dt = 4t + − (0) = 20 + = 107. 5 m = (4 × 5) + 2 0 2 2
Example A body moves such that its acceleration, a m/s2 after time t seconds is given by a = 18 − 2t. (a) Derive an expression for the velocity v m/s of the body, given that v = 20 m/s when t = 0. (b) Use the expression to find the velocity of the body after 3 s.
Solution
(a) v =
adt =
(18 − 2t) dt = 18t − t2 + C
but when t = 0 v = 20; therefore 20 = 0 − 0 + C, so C = 20 and v = 18t − t2 + 20 (b) When t = 3, v = (18 × 3) − 32 + 20 = 54 − 9 + 20 = 65 m/s.
Test Examples 12 1. Evaluate the following integrals: 4 (a) x − x 2 − 8x + 5 dx (b) 3 (c)
1 4 dx + x2 x
2x − 3x 2 − 1 dx
1
dy = x 2 + x − 2. dx If the curve passes through the point x = 2, y = 5, find its equation.
2. A curve has a gradient of
3. Determine the integrals: (3 cos (x) − 2 sin(x) + 4) dx (a) (4 cos (x) − sin (x) + x) dx (b)
Integral Calculus (Integration)
•
311
π/2 (c) cos (x) dx 0
4. Find the area between the curve y = x 3 −4x 2 + 3x and the x axis between the limits x = 1 and x = 3. 5. The curve shown below (Figure 12.9) was plotted during the isothermal expansion of a gas, following the law pV = C, between volume V1 and V2 . V2 Show that the area under the graph = pV · ln V1
p pV = C
V V1
V2
Figure 12.9
6. The acceleration a m/s2 of a body is given by the equation a = 6t. (a) Obtain an expression for velocity v after t seconds given that v = 0 when t = 0. (b) Calculate the average velocity during the period t = 2 to t = 3. (c) Calculate the instantaneous velocity at t = 2. 5. 7. The velocity v m/s of a body after time t seconds is given by the equation v = 3t2 + 8t + 12. Find the displacement of s metres for the body after 10 s, given that s = 10 when t = 0. x 8. A cone is generated by rotating the line y = about the x axis from x = 0 to x = 6. 2 Calculate the volume of the cone. 9. Find the volume generated by rotating about the x axis that part of the curve y = x 2 − x which lies between its intersections with the x axis.
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•
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10. Determine the following integrals: (x + 1) (x + 2) dx (a) 1 1 − 2 dx (b) x x 2 (a + 2b) (x + 1) dx (c) 1
3ex dx
(d)
11. Evaluate the area enclosed by the curve f (x) = 0. 06x 2 + 10, the x axis and the ordinates x = 6 and x = 8. 12. A gas is compressed from a volume of V1 to a volume of V2 according to the law p · V n = C, where p is the pressure and C is a constant. Show that the area under the compression curve on a pressure–volume diagram, between the stated volume limits, is given by: C Area = − n−1
1 V2n−1
−
1
V1n−1
13. Determine the following integrals: 1 1 sin (θ ) − dθ (a) 3 2 π sin (x) dx (b) 0 π/2
(5 cos (x) + 3 sin (x) − x) dx
(c) 0
14. Find the volume of the ellipsoid formed by rotating the ellipse: major axis 2a and minor axis 2b, about major axis.
y2 x2 + = 1, of a2 b2
15. A particle is projected witha horizontal velocity (u) into a resisting medium so that dv equals a constant (k) multiplied by its velocity (v). at time (t) its acceleration dt v Show that: ln = −kt u 16. Sketch the curve and find an expression for the area between limits x = 0 and x = 1, b the curve y = and the x axis. Evaluate this area when a = 1 and b = 2. (x + a)
Integral Calculus (Integration)
•
313
17. Evaluate the following integrals: 3 dx (a) x 3 x (b) dx a 3 (c) 4z + 3z2 + 2z + 1 dz (2 cos (θ ) − 5 sin (θ )) dθ (d) 18. A particle P starts from rest at O with a velocity of 5 m/s and moves along a straight line OX with an acceleration of −2t2 at time t seconds after leaving O. Describe the motion after 3 s.
13
ADVANCED CALCULUS: METHODS OF DIFFERENTIATION Differentiation – The Product Rule When two separate functions of a variable are multiplied they can only be differentiated if the product rule is applied.
The Product Rule Let y = u · v where u and v are functions of the same variable, x, for instance. Then to differentiate y the following rule applies: dv du dy =u· +v· dx dx dx or sometimes dy = u · dv + v · du dx
Advanced Calculus: Methods of Differentiation
•
315
Example (set all questions this way to aid memory) Let y = x 3 · sin (x)
Solution v = sin(x)
u = x3 du dx
dv
= 3x2
dx
= cos(x)
so
dy = x 3 · cos(x) + 3x 2 · sin(x) dx
If possible the answer should be simplified as far as possible. So
dy = x 2 (x · cos(x) + 3 sin(x)) dx
Questions Differentiate the following: 1. 4. 7. 10.
y = 3x 2 · cos(x) y = 3x 2 + 2 · x 2 − 2 y = x 3 · ln(2x) y = e2x 4x 2 − 4x + 1
√ x · cos(x)
2.
y = sin(x) · cos(x)
3.
y=
5.
y = 5e2x · sin(x) √ y = 3 x · e4x √ y = 3 x 7 · ln(5x)
6.
y = e7x · cos(x)
9.
y = ex · ln(x)
8. 11.
12.
y = 6e5x · sin(x)
Answers 1.
dy = 6x cos(x) − 3x 2 sin(x) = 3x (2 cos(x) − x sin(x)) dx
2.
dy = cos2 (x) − sin2 (x) dx
3.
√ cos(x) − 2x sin(x) dy 1 − 1 √ = x 2 cos(x) − x sin(x) = dx 2 2 x
4.
dy = (6x) x 2 − 2 + 3x 2 + 2 (2x) = 6x 3 − 12x + 6x 3 + 4x = 12x 3 − 8x dx
5.
dy = 10e2x sin(x) + 5e2x cos(x) = 5e2x (2 sin(x) + cos(x)) dx
6.
dy = 7e7x cos(x) − e7x sin(x) = e7x (7 cos(x) − sin(x)) dx
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•
Mathematics
7. 8. 9. 10.
1 dy = 3x 2 ln(2x) + x 3 · = x 2 (3 ln(2x) + 1) dx x √ 3 1 3e4x (8x + 1) dy √ = 3 x4e4x + x − 2 · e4x = dx 2 2 x 1 1 dy = ex ln(x) + ex · = ex ln(x) + dx x x dy = 2e2x 4x 2 − 4x + 1 + e2x (8x − 4) = 2e2x 4x 2 − 1 dx
The Quotient Rule When two functions of the same variable are divided the quotient rule of differentiation has to be applied. The quotient rule is:
u If y = then v
dv du dy v · dx − u · dx = dx v2
or sometimes
Example Let y =
3x 3 sin(x)
Solution u = 3x 3
v = sin(x)
so du = 9x 2 dx
Therefore
dv = cos(x) dx
dy sin(x) · 9x 2 − 3x 3 · cos(x) = dx sin2 (x)
Or, if possible,
dy 3x 2 · (3 sin(x) − x · cos(x)) = dx sin2 (x)
dy v · du − u · dv = dx v2
Advanced Calculus: Methods of Differentiation
•
317
Questions Differentiate the following: 4x
1. y =
x2 − 1
4 sin(x) 5x 2 + 2x ln(6x) 9. y = 6 sin(x) 5. y =
2x − 1 3x 2 + 5x 6 cos(x) 6. y = 3 x +4 √ 3 x+x 10. y = 7 2 ln(4x) 2. y =
3. 7. 11.
3e2x 4x 2 − 3 √ x3 y= cos(x) x2 y= 3 sin(x) y=
3x 4 + 2x 2 − 1 4e5x 4e6x 8. y = sin(x) 3 ln 52 x 12. y = 2 x + 2x 4. y =
Answers 1.
2 4 x2 + 1 x − 1 4 − 4x(2x) dy −4x2 − 4 = = 2 2 = − 2 dx x2 − 1 x2 − 1 x2 − 1
2.
2 3x + 5x 2 − (2x − 1) (6x + 5) 6x2 + 10x − 12x2 − 4x + 5 dy = = 2 2 dx 3x2 + 5x 3x2 + 5x So
dy −6x2 + 6x + 5 = 2 dx 3x2 + 5x
3.
2 4x − 3 6e2x − 3e2x (8x) 6e2x 4x2 − 4x − 3 dy = = 2 2 dx 4x2 − 3 4x2 − 3
4.
4e5x 12x3 + 4x − 20e5x 3x4 + 2x2 + 1 dy = 2 dx 4e5x 3 12x + 4x − 5 3x4 + 2x2 + 1 dy −15x4 + 12x3 − 10x2 + 4x − 5 = = 5x dx 4e 4e5x
5.
2 5x + 2x 4 cos(x) − 4(10x + 2) sin(x) dy = 2 dx 5x2 + 2x
6.
3 x + 4 (−6 sin(x)) − 6 cos(x) 3x2 dy = 2 dx x3 + 4
7.
dy = dx
9.
dy = dx
11.
√ 3 x 2 cos(x) + x. sin(x) cos2 x 1 x
sin(x) − ln(6x) cos(x) 6 sin2 x
dy 3x(2 sin(x) − x cos(x)) = dx 9 sin2 (x)
So
8.
10.
12.
−6 x3 + 4 sin(x) + 3x2 cos(x) dy = 2 dx x3 + 4 dy 4e6x (6 sin(x) − cos(x)) = dx sin2 x 3 √3 + 1 √ + 1 ln(4x) − dy 2 x x = 2 7 dx 2 (ln(4x) 3(x + 2) − 6(x + 1) ln dy = 2 dx x2 + 2x
5x 2
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•
Mathematics
The function of a function rule (chain rule) The type of function defined as ‘a function of a function’ is probably best looked at with an example. The function y = (4x − 3)7 can be differentiated by first multiplying (4x − 3) by itself seven times and then differentiating each term in turn. This is a long process and is, more often than not, subject to error. In this type of function a substitution is made. Example Let u = 4x − 3, then instead of y = (4x − 3)7 the equation is y = u7 .
Solution The rule for differentiating by substitution is
dy dy du = × dx du dx
In other words, differentiate the ‘substituted’ function, differentiate the substitution, multiply these two answers and, finally, replace the substitution by the original expression. From the above substitution, Therefore,
dy du = 7u6 and = 4. du dx
dy = 7u6 × 4 = 28u6 = 28 (4x − 3)6 dx
Questions Differentiate
1.
y = sin (6x + 1)
2.
5 y = 3t 4 − 2t
3.
y=
4.
y = (x + 1)4
5.
y = (1 − 2x)7
6.
y = sin2 (x)
7.
y=
8.
y = 6 cos3 (x)
9.
y=
10.
y=
1 (2x − 1)6
13.
1 2x2 − 3x + 1 3 y = 2x2 + 3x + 1 2
16.
y = sin 3x2
17.
y = cos (sin (2x))
19.
y = (sin (3x) + cos (2x))4
20.
y=
4x2 + x − 3
4x3 + 2x2 − 5x
12.
3 y= 2 3x − 1
sin (3x)
15.
y = (cos (6x)) 3
18.
y = sin
11.
y = cos7 (x)
14.
y=
3
5
x3 − 5x2 − 6x + 4
√ x
Advanced Calculus: Methods of Differentiation
•
319
Answers 1.
dy = 6 cos(6x + 1) dx
2.
4 dy = 5 · 12t3 − 2 3t4 − 2t dt
3.
8x + 1 dy = √ dx 2 4x 2 + x − 3
4.
dy = 4(x + 1)3 dx
5.
dy = −14(1 − 2x)6 dx
6.
dy = 2 sin(x) cos(x) dx
7.
−12 dy = dx (2x − 1)7
8.
dy = −18 cos2 (x) · sin(x) dx
9.
12x 2 + 4x − 5 dy = √ dx 2 4x 3 + 2x 2 − 5x
10.
4x − 3 dy = − 2 2 dx 2x − 3x + 1
11.
dy = −7 cos6 (x) · sin(x) dx
12.
18x dy = − 2 dx 3x 2 − 1
13.
1/2 dy 3 = (4x + 3) 2x 2 + 3x + 1 dx 2
14.
3 cos (3x) dy = dx 2 sin (3x)
15.
dy = −10 (cos (6x))2/3 · sin (6x) dx
16.
17.
dy = −2 sin (sin (2x)) · cos (2x) dx
18.
19.
dy = 4 [sin (3x) + cos (2x)]3 · [3 cos (3x) − 2 sin (2x)] dx
20.
3x 2 − 10x − 6 dy =
2 dx 3 x 3 − 5x 2 − 6x + 4 2
dy = 6x · cos 3x 2 dx √ dy cos x √ = dx 2 x
Differentiating inverse trigonometric functions There is more than one way of approach, but this is possibly ‘easier’ to follow. Example Differentiate y = sin−1 (x)
320
•
Mathematics
Solution y = sin−1 (x) means sin(y) = x or that x = sin (y) · · · · · · (#) So
cos (y) dy 1 dx = cos(y) = , inverting both sides gives = dy 1 dx cos(y)
However, this gives the differential as a function of y, not x. One of the standard trigonometric identities is cos2 (y) + sin2 (y) = 1
Re-arranging gives cos2 (y) = 1 − sin2 (y) ⇒ cos(y) = 1 − sin2 (y) Substituting for sin(y) [see (#) above] gives cos(y) = Therefore, if y = sin−1 (x),
√ 1 − x2
1 dy =√ dx 1 − x2
Example Differentiate y = sin−1 (5x)
Solution y = sin−1 (5x) means dx cos(y) = dy 5
sin (y) = 5x
⇒
x=
sin (y) 5
dy 5 = dx cos(y)
As in the above example cos(y) = 1 − sin2 (y)
√ Substituting for sin(y) gives cos(y) = 1 − (5x)2 = 1 − 25x 2
So
and so
Therefore, if y = sin−1 (5x),
5 dy =√ dx 1 − 25x 2
Example Differentiate y = cos−1 (3x)
Solution y = cos−1 (3x) means cos (y) = 3x So
dx − sin(y) = dy 3
and so
⇒
x=
cos (y) 3
dy 3 =− dx sin(y)
Re-arranging cos2 (y) + sin2 (y) =1 gives sin2 (y) =1 − cos2 (y) ⇒ sin(y) =
Substituting for cos(y) gives sin(y) = 1 − (3x)2
1 − cos2 (y)
Advanced Calculus: Methods of Differentiation Therefore, if y = cos−1 (3x),
•
321
3 dy = −√ dx 1 − 9x 2
Example Differentiate y = cosh−1 (x)
Solution y = cosh−1 (x) means x = cosh(y) So
dy 1 dx = sinh(y), therefore = dy dx sinh(y)
One of the standard hyperbolic identities is cosh2 (y) − sinh2 (y) = 1
Re-arranging gives sinh2 (y) = cosh2 (y) − 1 ⇒ sinh(y) = cosh2 (y) − 1 Substituting for cosh(y) gives sinh(y) = Therefore, if y = cosh−1 (x),
√ x2 − 1
1 dy =√ 2 dx x −1
Questions 1.
y = cos−1 (x) 2.
y = sinh−1 (x)
Find the second derivatives of 3.
y = cos−1 (x) 4.
y = cosh−1 (x)
Find the first derivative of
Answers 1.
1 dy = −√ dx 1 − x2
2.
1 dy =√ dx x2 + 1
3.
x dy = − 3 dx 1 − x2
4.
x dy = − 3 dx x2 + 1
Methods of Integration Integration by substitution Functions which require integrating are not usually in the form where they can be looked up in a set of tables. However, there is a way around this problem if a suitable substitution is made.
322
•
Mathematics
An important rule to follow is that once the substitution has been made the original variable is replaced entirely. The following examples show the method. Example Find cos(5x + 2) dx
Solution du du = 5, that is, du = 5dx ⇒ dx = dx 5 du Substituting into the original integral gives cos(u) 5 1 1 1 cos(u)du = sin(u) + c = sin(5x + 2) + c So 5 5 5 Let u = 5x + 2; then
Example Find 3x(x 2 + 2)6 dx
Solution du du = 2x ⇒ dx = dx 2x du 3 = u6 du as the x’s cancel. Substituting gives 3x(u)6 2x 2
Let u = x 2 + 2; then
Therefore the integral is
3 u7 3 · +c= · (x 2 + 2)7 + c 2 7 14
Example et dt Find 3 + et
Solution du du = et ⇒ dt = t dt e t e du du · = ln(u) + c = ln(3 + et ) + c Substituting gives = u et u Let u = 3 + et ; then
Advanced Calculus: Methods of Differentiation
•
323
Questions Integrate by substitution:
1.
sin(3x + 2)
2.
2 cos(4t + 1)
3.
4 sin(6x − 3)
4.
1 (9x + 5)8 12
5.
4e7t−1
6.
5t(t2 − 1)7
7.
x(3x 2 + 4)8
8.
(4x 2 − 1)
9.
4x 3 − 3x
6x + 2
5 3x 2 + 2x − 1
Answers
1 cos(3x + 2) + c 3
1.
u = 3x + 2 ⇒ I =
2.
1 u = 4t + 1 ⇒ I = − sin(4t + 1) + c 2
3.
u = 6x − 3 ⇒ I =
2 cos(6x − 3) + c 3
4.
u = 9x + 5 ⇒ I =
1 (9x + 5)9 + c 972
5.
4 u = 7t − 1 ⇒ I = e7t−1 + c 7
6.
u = t2 − 1 ⇒ I =
7.
u = 3x 2 + 4 ⇒ I =
8. 9. 10.
8 5 2 t −1 +c 16
9 1 2 3x + 4 + c 54 √ 3 2 2 3x + 2 2 + c u = 3x 2 + 2 ⇒ I = u du = 3 3 √ du 2 3 = 4x − 3x 2 + c u = 4x 3 − 3x ⇒ I = u 3 9 1 u = 3x 2 + 2x − 1 ⇒ I = u−5 du = − 4 + c 2 4 3x + 2x − 1
If a substitution is unavailable, and for the most case these are quite obvious, the next approach is ‘integration by parts’.
324
•
Mathematics
Integration by parts The formula for integration by parts is:
u dv = uv−
v du
It is best demonstrated by an example. Example Find 3x · cos (x) dx
Solution As far as possible, always let the variable ‘u’ equal the term that includes the ‘x’. In this case u = 3x so the term ‘dv’ is whatever follows the ‘x’ expression: so for this example dv = cos(x) Now it is necessary to determine ‘du’ and ‘v’ •
to determine ‘du’, differentiate ‘u’,
•
to determine ‘v’ integrate ‘dv’.
So du = 3 and v = sin(x)
These terms are now inserted into the formula
u dv = uv −
v du giving
3x · sin(x) −
sin(x) · 3dx
It is now necessary to determine the final integral.
sin(x) · 3 dx =
3 sin(x) dx = −3 cos(x)
Substituting this expression into the formula gives 3x · sin(x) − (−3 cos(x)) So 3x · cos (x) dx = 3x · sin(x) + 3 cos(x) + C
Advanced Calculus: Methods of Differentiation Recap
•
325
To determine
3x · cos (x) dx
∫3x.cos(x)dx u
dv
u = 3x
dv = cos(x)
du = 3 v = sin(x)
3x · cos (x) dx = 3x · sin(x) −
sin(x) · 3 dx = 3x · sin(x) + 3 cos(x) + C
Example Find x · e2x dx
Solution
If x · e2x dx is compared with the left-hand side of the rule it is possible to put u = x and dv = e2x 1 So du = 1 and v = e2x 2 These expressions are substituted into the rule giving 1 1 2x 1 1 e · 1dx = x · e2x − e2x + C x · e2x dx = x· e2x − 2 2 2 4 The main problem lies in deciding which part of the expression is ‘u’ and which is ‘dv’. •
Always choose the ‘u’ function so that it will become simpler on differentiation.
•
Always choose the ‘dv’ function so that it can be integrated easily.
Questions Using integration by parts, find the following indefinite integrals and evaluate the definite integrals: 1.
x · e dx x
2.
3x · ln(5x) dx
3.
2x · e3x dx
326
•
Mathematics
π
2
π x · cos(x) dx 5.
4. 0
1 x·e
7.
−x
dx
x · sin(x) dx
ln(x) dx
6.
π 2
1
π
2 (π − x) · cos(x) dx 9.
8.
0
2
0
Hint: ln(x) = 1 × ln(x)
√ x · ln(x) dx
1
Answers 1.
x · ex − ex + c
2.
3 2 x (2 ln(5x) − 1) + c 4
3.
2 2 x · e3x − e3x + c 3 9
4.
0.571
5.
2.142
6.
0.386
7.
0.265
8.
2
9.
0.494
A third procedure is to use partial fractions. This involves splitting an algebraic fraction into its component parts and integrating each in turn. The forms of partial fractions are summarised as: Denominators containing
Expression example
Form of partial fraction
Linear factors
f (x) (x + a) (x − b) (x + c)
B C A + + x+a x−b x+c
Repeated linear factors
f (x) (x + a)3
B C A + + (x + a) (x + a)2 (x + a)3
Quadratic factors
f (x)
ax 2 + bx + c
(x + d)
Ax + B
ax 2 + bx + c
+
D (x + d)
Integration using partial fractions Once the partial fractions have been obtained the following integrals have to be used to obtain the final solution: 1 1 • dx = · ln (ax + b) ax + b a 1 1 −1 x • · tan dx = x 2 + a2 a a
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f (x) dx = ln [f (x)] f (x)
An example may help to explain the third integral: cos (x) dx = ln [sin (x)] If f (x) = sin(x) then f (x) = cos(x) and so sin (x) Example
Determine
Solution
11 − 3x dx (linear factors) (x − 1) (x + 3) A B A(x + 3) + B(x − 1) 11 − 3x ≡ + ≡ (x − 1) (x + 3) (x − 1) (x + 3) (x − 1)(x + 3)
Hence 11 − 3x ≡ A(x + 3) + B(x − 1) Let x = 1, then 8 = 4A, from which A = 2. Let x = −3, then 20 = −4B, from which B = −5 5 11 − 3x 2 dx = − dx = 2 ln (x −1) − 5 ln (x + 3) + c Hence (x − 1) (x + 3) (x − 1) (x + 3) (x − 1)2 + c, by the laws of logarithms. or ln (x + 3)5 Example Determine
2x + 3 dx (repeated linear factor) (x − 2)2
Solution
2x + 3 (x − 2)
2
≡
B A A(x − 2) + B + ≡ 2 (x − 2) (x − 2) (x − 2)2
Therefore 2x + 3 ≡ A(x − 2) + B Let x = 2, then 7 = B Choose any other x value Let x = 4, then 11 = 2A + B, then 4 = 2A, then A = 2. 2x + 3 7 7 2 So + +c dx = 2 ln(x − 2) − dx = (x − 2) (x − 2)2 (x − 2) (x − 2)2
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7 (x − 2)
Example
Determine
dx is determined using the substitution u = (x − 2) 2
2 dx (x − 1)(x 2 + 1)
Solution (Bx + C) A 2 + 2 = (x − 1)(x 2 + 1) (x − 1) x +1 So
2 = A x 2 + 1 + (Bx + C) (x − 1)
Let x = 1, then 2 = A(1 + 1), then 2 = 2A, then A = 1 Let x = 0, then 2 = A(1) + (C)(−1), then 2 = A − C, then 2 = 1 − C, then C = −1 Comparing the coefficients of x2 : 0 = A + B, then B = −1 (−x − 1) x 1 1 1 2 − dx + − 2 dx = dx = So 2 2 2 (x − 1) (x − 1) (x − 1)(x + 1) x +1 x +1 x +1 1 2 dx = ln (x − 1) − ln x 2 + 1 − tan−1 (x) + C Therefore (x − 1)(x 2 + 1) 2
Questions By the use of partial fractions find the following integrals and evaluate the definite integrals. 1. 3. 5.
5x − 3 dx (x − 3)(x + 3) 9x 2 + 34x + 29 dx (x + 1)(x + 2)(x + 3) 6 − 15x + 7x 2 dx 2x(1 − x)(2 − 3x)
4 2. 3
1.5 4. 1.1
5 6. 2
x dx (x − 1)(x − 2) 3x 2 − 7x + 2 dx (2x − 1)(x + 1)(x − 1) 2x + 4 dx (1 + x)(x − 1)(2x + 1)
Advanced Calculus: Methods of Differentiation Answers 1.
A B 5x − 3 ≡ + ⇒ 5x − 3 = A(x + 3) + B(x − 3) (x − 3)(x + 3) x − 3 x + 3 Choose values of x to eliminate a bracket Let x = 3
so 12 = 6A
A=2
Let x = −3 so −18 = −6B B = 3 3 2 Therefore I = + dx = 2 ln(x − 3) + 3 ln(x + 3) + c x−3 x+3 2.
x A B ≡ + ⇒ x = A(x − 2) + B(x − 1) (x − 1)(x − 2) x − 1 x − 2 Let x = 2 so
2=B
Let x = 1 so
1 = −A A = −1 4 2 −1 + dx = [− ln(x − 1) + 2 ln(x − 2)]43 Therefore I = x−1 x−2 3
I = (− ln(3) + 2 ln(2)) − (− ln(2) + 2 ln(1)) = 3 ln(2) − ln(3) − 2 ln(1) = 0. 981 3.
9x 2 + 34x + 29 A B C ≡ + + (x + 1)(x + 2)(x + 3) x + 1 x + 2 x + 3 So 9x 2 + 34x + 29 = A(x + 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x + 2) Let x = −1 so 4 = A(1)(2)
2A = 4
A=2
Let x = −2 so −3 = B(−1)(1)
−B = −3 B = 3
Let x = −3 so 8 = C(−2)(−1) 2C = 8 C=4 3 4 2 + + dx Therefore I = x+1 x+2 x+3 Integral = 2 ln(x + 1) + 3 ln(x + 2) + 4 ln(x + 3) + c 4.
3x 2 − 7x + 2 A B C ≡ + + (2x − 1)(x + 1)(x − 1) 2x − 1 x + 1 x − 1 So 3x 2 − 7x + 2 = A(x + 1)(x − 1) + B(2x − 1)(x − 1) + C(2x − 1)(x + 1) 3 1 1 3 so − = A − A=1 Let x = 2 4 2 2 Let x = −1 so 12 = B(−3)(−2)
6B = 12
B=2
Let x = 1 so −2 = C(1)(2) 1.5 2 1 1 So I = + − dx 2x − 1 x + 1 x − 1
2C = −2
C = −1
1.1
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•
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Mathematics 1.5 1 ln(2x − 1) + 2 ln(x + 1) − ln(x − 1) = 2. 8723 − 3. 8776 = −1. 005 2 1.1
A B C 6 − 15x + 7x 2 ≡ + + 2x(1 − x)(2 − 3x) 2x 1 − x 2 − 3x So 6 − 15x + 7x 2 = A(1 − x)(2 − 3x) + B(2x)(2 − 3x) + C(2x)(1 − x) Let x = 0 6 = A(1)(2)
2A = 6
A=3
Let x = 1 −2 = B(2)(−1)
−2B = −2 B = 1
Let x = 2 4 = A(−1)(−4) + B(4)(−4) + C(4)(−1) 4 = 4A − 16B − 4C I= 6.
4 = 12 − 16 − 4C so 4C = 12 − 16 − 4 = −8 C = −2 1 2 3 2 3 + − dx = ln(x) − ln(1 − x) + ln(2 − 3x) + c 2x 1 − x 2 − 3x 2 3
A B C 2x + 4 ≡ + + (1 + x)(x − 1)(2x + 1) 1 + x x − 1 2x + 1 So 2x + 4 = A(x − 1)(2x + 1) + B(1 + x)(2x + 1) + C(1 + x)(x − 1) Let x = −1
2 = A(−2)(−1)
2A = 2
A=1
Let x = 1
6 = B(2)(3) 6B = 6 B=1 1 3 3 1 Let x = − 3=C − 3=C − C = −4 2 2 2 4 5‘ 5 1 4 1 I= + − dx = ln(1 + x) + ln(x − 1) − 2 ln(2x + 1) 2 1 + x x − 1 2x + 1 2
I = (ln(6) + ln(4) − 2 ln(11)) − (ln(3) + ln(1) − 2 ln(5)) = 0. 503
Integration of trigonometric products The four basic identities are: 1 [sin(A + B) + sin(A − B)] 2 1 cos(A) sin(B) = [sin(A + B) − sin(A − B)] 2 1 cos(A) cos(B) = [cos(A + B) + cos(A − B)] 2 1 sin(A) sin(B) = − [cos(A + B) − cos(A − B)] 2 sin(A) cos(B) =
(1) (2) (3) (4)
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The method is to use the above identities to rewrite trigonometric products into trigonometric additions/subtractions. Example Find 8 sin(5t) cos(2t) dt
Solution
Using (1)
1 [sin (7t) + sin(3t)] dt 2 1 So 8 × [sin (7t) + sin (3t)] dt = 4 [sin (7t) + sin (3t)]dt 2 cos (7t) cos (3t) − +C =4 − 7 3
8 sin(5t) cos(2t) dt =
8×
Questions 1.
Find
2.
Find
5 cos(6t) · cos(2t) dt
3.
Evaluate
5 sin(7t) cos(3t) dt π/4 cos(4x) cos(2x) dx 0
4.
5.
sin(8t) cos(t) dt
Find
Evaluate
π/3 cos(5x) cos(2x) dx 0
6.
sin(2t) cos(t) dt
Find π
7.
cos(5x) cos(2x) dx
Evaluate 0
8.
9.
sin(3t) cos(t) dt
Find
Evaluate
π/2 cos(2x) cos(3x) dx 0
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Answers 1.
5 sin(8t) sin(4t) + +C 2 8 4
2.
3.
0.167
4.
5.
0.062
6.
7.
0
8.
9.
0.6
cos(10t) cos(4t) 5 − − +C 2 10 4 cos(9t) cos(7t) 1 − − +C 2 9 7 cos(3t) 1 − − cos(t) + C 2 3 cos(4t) cos(2t) 1 − − +C 2 4 2
Applications of Integration Basic formulae If y is a function of x and the range is a ≤ x ≤ b then the following formulae apply: Areas and volumes of revolution Area : A =
b
y dx a
y y = f(x)
Area A
x 0
Figure 13.1
x=a
x=b
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With reference to Figure 13.1, the volume of revolution, V, obtained by rotating Area A through one revolution about the x axis is given by: b V=
π · y2 dx a
360◦
If a curve x = f (y) is rotated about the y axis between limits y = c and y = d then the volume generated is given by: d V = π · x 2 dy c
Example The curve y = x 2 + 4 is rotated about the x axis between the limits x = 1 and x = 4. Determine the volume of the solid of revolution produced (Figure 13.2).
Figure 13.2
Solution b
V=
4
π · y dx = 2
a
=π V =π
2 π · x + 4 dx = π
1
x5
5
+
8x 3 3
4
4
2
x 4 + 8x 2 + 16 dx
1
+ 16x 1
1 5 8 13 45 8 · 43 + + 16 · (4) − + + 16 (1) = 420. 6π cubic units. 5 3 5 3
Questions 1. The curve xy = 3 is rotated one revolution about the x axis between the limits x = 2 and x = 3. Determine the volume of the solid produced.
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2. The curve y = 2x 2 + 3 is rotated about (a) the x axis between the limits x = 0 and x = 3, and (b) the y axis, between the same limits. Determine the volumes generated in each case.
Answers 1.
1.5π cubic units
2.
(a) 329.4π cubic units (b) 81π cubic units
Centroids of plane areas The centroid of an area is the point at which the total area is considered to be situated for calculation purposes. (If the area was a thin lamina the centroid would indicate the position of the centre of mass.) If x and y represent the coordinates of the centroid of area A, then: b
1 2
x · y dx x=
a
and y =
b
b y2 dx a
b
y dx
y dx
a
a
Example Find the position of the centroid of the area bounded by the curve y = 3x 2 , the x axis and the ordinates x = 0 and x = 2.
Solution
2
2 x · 3x 2 dx
x=
0
=
2
0
=
2
3x 2 dx 0
3x 3 dx
3x 2 dx 0
3x 4 4
2
[x 3 ]20
0
=
12 = 1. 5 8
Advanced Calculus: Methods of Differentiation
1 2 y=
2
3x
2 2
dx
0
9x 4 dx 0
=
2
2
1 2
2
3x 2 dx
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2 9 x5 9 32 · 18 2 5 0 = 2 5 = = 3. 6 = 8 8 5
3x 2 dx
0
0
Therefore the centroid lies at (1.5, 3.6) Example Find the position of the centroid of the area enclosed by the curve y = 5x − x 2 , the x axis and the ordinates x = 0 and x = 5
Solution 5
5
x · 5x − x 2 dx
5x 2 − x 3 dx
5x 3 x 4 − 3 4
5
625 0 = 5 = x= 5 = 12 = 2. 5 125 2 3 5 x 5x 2 2 − 6 5x − x dx 5x − x dx 2 3 0 0
0
0
0
and 1 2 y=
5
5x − x
2 2
dx
0
5
5x − x
0
2
1 2 =
5
25x 2 − 10x 3 + x 4 dx
0
5
dx
5x − x 2 dx
5 1 25x 3 10x 4 x 5 − + 2 3 4 5 0 = 125 6
0
1 25 · (125) 10(625) 3125 · − + 2 3 4 5 y= = 2. 5 125 6 Therefore the centroid lies at (2.5, 2.5).
Questions For questions 1 and 2, find the position of the centroids of the areas bounded by the given curves, the x axis and the given ordinates. 1. y = 3x + 2, x = 0 and x = 4. 2. y = 5x 2 , x = 1 and x = 4. 3. Find the position of the centroid of a sheet of metal formed by the curve y = 4x − x 2 , the x axis and the lines x = 0 and x = 4.
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4. Find the co-ordinates of the centroid of the area which lies between the curve y = x 2 − 2x, the x axis and the lines x = 0 and x = 2.
Answers 1.
(2.5, 4.75)
2.
(3.036, 24.36)
3.
(2, 1.6)
4.
(1, −0.4)
Second moment of area It is the measure of resistance to the bending of a structural member. The further the majority of the mass of a body is from its centroid the more resistive it is to bending. If Ix and Iy represent second moments of area about the x-axis and y-axis respectively, then: 1 Ix = 3
b
b 3
y dx
Iy =
and
a
x 2 · ydx a
Example Calculate the second moment of area about the y axis of the function 1 y= for values of x between 3 and 4. (x + 2) (x + 1) (x − 1)
Solution
4
Moment = 3
x2 dx = (x + 2) (x + 1) (x − 1)
4 3
1/2 1/6 4/3 − + dx (x + 2) (x + 1) (x − 1)
(The evaluation of the partial fractions is not shown here.) Therefore
4 1 1 4 Moment = · ln (x + 2) − · ln (x + 1) + · ln (x − 1) 3 2 6 3 4 · ln(5) ln(4) ln(2) 4 · ln(6) ln(5) ln(3) − + − − + = 3 2 6 3 2 6 = 0. 199 to 3 decimal places
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Second moment of mass (moment of inertia) If the mass per unit volume of the volume of revolution generated by the rotation of area A about the x axis is m, then the moment of inertia of the solid about the x axis is: 1 Ix = 2
b y4 dx a
Theorem of Pappus If x and y represent the coordinates of the centroid of area A, then: Volume of revolution obtained by rotating area A about the x axis = 2πyA Volume of revolution obtained by rotating area A about the y axis = 2πxA Example From an earlier example, the position of the centroid of the area bounded by the curve y = 3x 2 , the x axis and the ordinates x = 0 and x = 2 was found to be x = 1. 5, y = 3. 6 and the area enclosed was 8 unit2 . Therefore Volume of revolution obtained by rotation about the x axis = 2π × 3. 6 × 8 = 180. 96 unit3 Volume of revolution obtained by rotation about the y axis = 2π × 1. 5 × 8 = 75. 40 unit3
The area between curves The area enclosed between the curves y = f1 (x) and y = f2 (x) (the shaded area in the diagram, Figure 13.3) is given by: b b b shaded area = f2 (x)dx − f1 (x)dx = [f2 (x) − f1 (x)] dx a
a
a
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1
y = f1(x)
2
4 x
3 x=b
x=a
Figure 13.3
Example Determine the area enclosed between the curves y = x 2 + 1 and y = 7 − x
Solution It is first of all necessary to find the points of intersection of the two curves (see Figure 13.4). 12
y
x –4
Figure 13.4
–3
–2
–1
1
2
3
Advanced Calculus: Methods of Differentiation At these points x 2 + 1 = 7 − x
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339
y =7−x
Therefore x 2 + x − 6 = 0 ⇒ (x + 3) (x − 2) = 0 ⇒ x = −3 and x = 2 2 2 2 (7 − x) − x + 1 dx = Shaded area = 6 − x − x 2 dx −3
−3
3 2
9 −27 4 8 x2 x − −18 − − = 12 − − Therefore area = 6x − − 2 3 −3 2 3 2 3 27 125 22 − − = = 3 2 6
Mean and root mean square values With reference to Figure 13.5, 1 mean value: y = b−a
b ydx a
⎧ ⎫ b ⎨ ⎬ 1 y2 dx r.m.s. value: ⎭ ⎩b − a a
y y = f(x)
y
x 0
x=a
x=b
Figure 13.5
Example A sinusoidal voltage has equation y = 50 sin(t). Find the mean and r.m.s. values over half a cycle.
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Solution Half a cycle means that a = 0 and b = π. 1 Mean value y = π −0 Therefore y =
π 50 sin(t)dt = 0
1 [−50 cos(t)]π0 π
50 100 50 [(− cos(π) − (− cos(0))] = [− (−1) − (−1)] = π π π
The mean value is 31.83 to 2 decimal places. ⎧ ⎫ ⎫ ⎧ ⎨ 1 π ⎬ ⎨ 2500 π ⎬ (50 sin(t))2 dt = The r.m.s. value is sin2 (t)dt ⎭ ⎭ ⎩π −0 ⎩ π 0
0
Now sin2 (t)dt cannot be found directly, but the table of integrals gives sin(2t) 1 sin2 (t)dt = t − 2 2 π
sin (2π) sin (0) π 1 1 π− − 0− = sin (t) dt = 2 2 2 2 2 2
So 0
$
Therefore the r.m.s. value =
2500 π √ · = 1250 = 35. 36 v. π 2
Note: For a sine wave, mean value =
1 2 × peak value; r.m.s. = √ × peak value π 2
Example Calculate the r.m.s. value of y = 250 sin (25πt) for the first 2 ms.
Solution The r.m.s. value is ⎧ ⎫ ⎫ ⎧ 0.002 0.002 ⎨ ⎨ ⎬ ⎬ 1 (250 sin(25πt))2 dt = 31250000 sin2 (25πt)dt ⎭ ⎭ ⎩ 0. 002 − 0 ⎩ 0
The trigonometric identity gives sin2 (25πt) = 2
0
1 (1 − cos (50πt)) 2
sin (50πt) 0.002 sin (0. 1π) 1 1 t− 0. 002 − − (0) sin (25πt) dt= = 2 50π 2 50π 0
0.002
So
0
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0.002
sin2 (25πt) dt = 0. 000016368
So 0
Therefore the r.m.s. value =
√ √ 31250000 × 0. 000016368 = 511. 5 = 22. 62
Questions 1. The velocity, v ms−1 , of a particle is given by the equation v = 4t + 2, where t is the time in seconds. Find the average value of v between t = 2 and t = 4 s. 2. Find the r.m.s. value of the function y = x (3 − x) for values of x from 0 to 3. 3. A current i = 5 sin(t) A flows in an electrical circuit. Determine the mean and r.m.s. values, over the range t = 0 to t = π s.
Answers 1 1. y = b−a
b
1 ydt = 4−2
a
mean value =
42 + 4
4 (4t + 2) dt = 2
4 4 4 1 4t2 1 · + 2t = · 2t2 + 2t 2 = t2 + t 2 2 2 2 2
− 22 + 2 = 20 − 6 = 14 ms−1
3
3 x 2 (3 − x)2 dx [x (3 − x)] dx = 2
2. The squared value = 0
3 =
0
x 2 9 − 6x + x 2 dx
0
3 9x 3 6x 4 x 5 − + 9x − 6x + x dx = 3 4 5 0 0 9 × 27 6 × 81 243 − + − (0) = 3 4 5 8. 1 So the squared value = 8.1, giving the mean value as = 2. 7 3−0 √ The r.m.s. value is therefore 2. 7 = 1. 643 to 4 significant figures. 3
2
1 3. y = π −0
3
4
π 5 sin(t)dt = 0
1 [−5 cos(t)]π0 π
5 10 5 So y = [(− cos(π) − (− cos(0))] = [− (−1) − (−1)] = π π π
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Therefore the mean value is 3.18 to 2 decimal places. ⎧ ⎫ ⎫ ⎧ ⎨ 1 π ⎬ ⎨ 25 π ⎬ (5 sin(t))2 dt = The r.m.s. = sin2 (t)dt ⎭ ⎭ ⎩π − 0 ⎩π
0
0
sin(2t) 1 t− 2 2 π sin (2π) sin (0) π 1 1 π− − 0− = So sin2 (t) dt = 2 2 2 2 2 0 $ 25 π √ · = 12. 5 = 3. 54 V, to 2 decimal places. Therefore the r.m.s. value = π 2 sin2 (t)dt =
But
Questions Determine the root mean square values of the following functions between the specified limits: 1.
y = x 2 from x = 1 to 3.
2.
y = x from x = 0 to 2.
3.
y = sin(x) + 2 from x = 0 to 2π
4.
y = sin(2x) from x = 0 to π
5.
y = ex from x = −1 to +1
Answers 1.
4.92
2.
1.15
3.
5.317
4.
0.707
5.
1.35
Application of Differentiation Rates of change Example If the volume of a spherical balloon increases by 4 cm3 each second, what is the rate of growth of the radius?
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Solution 4 If the radius of the balloon is r, then the volume V = π · r3 3 The rate of change of volume with respect to time, that is, The Chain rule gives:
dV is 4 cm3 dt
dV dV dV dr = · where = 4π · r2 dt dr dt dr
Therefore 4 = 4π · r2 ×
dr 4 dr 1 and so = = cm/s dt dt 4π · r2 π · r2
Example A container in the shape of a right circular cone of height 20 cm, and base radius 2 cm is catching the drips from a tap leaking at the rate of 0.2 cm3 /s. Find the rate at which the surface area of water is increasing when the water is halfway up the cone.
Solution Suppose the height at any time is h cm, and that the radius of the surface of the water at that time is r cm. 2
r
20
h
Figure 13.6
By similar triangles (Figure 13.6),
h h r = , and so, r = 2 20 10
The surface area of the water A = π · r2 = h = 10 cm.
dA π · h2 and it is necessary to find when 100 dt
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By the Chain rule:
dA dA dh = · dt dh dt dA 2πh dh dA 2πh = and so = · dh 100 dh 100 dt
But
(1)
1 π · h3 The volume of water, V = π · r2 · h = 3 300 Therefore
dV dV dh 3π · h2 dh dV 3πh2 = and so = · = · dh 300 dt dh dt 300 dt
Therefore
dV 3π · h2 dh 3π · h2 dh dh 20 = · ⇒ 0. 2 = · ⇒ = dt 300 dt 300 dt dt πh2
From equation 1 and equation 2 and when h = 10 cm,
(2)
2 dA 2π · h 20 = · = dt 100 π · h2 5h
2 dA = = 0. 04 cm2 /s. dt 50
Questions 1. The area of the surface of a sphere is 4π · r2 , r being the radius. Find the rate of change of the area in cm2 /s when r = 2 cm, given that the radius increases at the rate of 1 cm/s. 2. The volume of a cube is increasing at the rate of 2 cm3 /s. Find the rate of change of the length of the side of the base when its length is 3 cm. 3. The area of a circle is increasing at a rate of 3 cm2 /s. Find the rate of change of the circumference when the radius is 2 cm. 4. At a given instant the radii of two concentric circles are 8 and 12 cm. The radius of the outer circle increases at the rate of 1 cm/s and that of the inner at 2 cm/s. Find the rate of change of the area enclosed between the two circles. 5. A hollow right circular cone is held vertex downwards beneath a tap leaking at the rate of 2 cm3 /s. Find the rate of rise of water level when the depth is 6 cm given that the height of the cone is 18 cm and its radius is 12 cm. 1 6. An ink blot on a piece of paper spreads at the rate of cm2 /s. 2 Find the rate of increase of the radius of the circular blot when the radius is 1/2 cm. 7. A hemispherical bowl is being filled with water at auniform rate. When the height 1 of the water is h cm the volume is π r · h2 − h3 cm3 , r cm being the radius of 3 the hemisphere. Find the rate at which the water level is rising when it is halfway to the top, given that r = 6 cm and that the bowl fills in 1 min.
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8. An inverted right circular cone with a vertical angle of 120◦ is collecting water from a tap at a steady rate of 18π cm3 /min. Find (i) the depth of the water after 12 min, (ii) the rate of increase of the depth at this instant. 9. From the formula v = (60s + 25) the velocity, v, of a particle can be calculated when its distance, s, from the origin is known. Find the acceleration when v = 10. dy 1 2 dx when x = 2 given =1 10. If y = x − , find x dt dt
Answers 1.
16π cm2 /s
2.
2 cm/s 27
3.
3 cm/s 2
4.
Decreasing, 8π cm2 /s
5.
1 cm/s 8π
6.
1 cm/s 2π
7.
4 cm/s 45
8.
6 cm,
9.
30
10.
4 15
Test Examples 13 Determine the following integrals: 1. x · ex dx 2. π x · sin(x)dx 4. 3. π
2 5. 7.
9x 2 + 34x + 29 dx (x + 1)(x + 2)(x + 3) 1
1 − x 2 dx x·
6. 4
7x − 2x 2 (2 − x)2 (1 + x) 3
3
sin (θ ) · cos2 (θ ) dθ
5x − 3 dx (x − 3)(x + 3)
10.
x
8.
0
0
2x · e3x dx
π 9.
x2 − 7
dx
dx
5 sin(7t) cos(3t)dt
1 cm/min 6
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Mathematics
π/4 11. cos(4x) cos(2x)dx 0
12. (i) Determine the value of y for the area enclosed between the curve y = x 2 + 1, the ordinates x = 1 and x = 3 and the x axis. (ii) Calculate the volume of revolution when this shape is rotated by 360◦ about the x axis (see Figure 13.7). 12 y 10 8 6 4 2 x 1
2
3
4
Figure 13.7
13. (i) Determine the area enclosed by the two curves y = x 2 and y = Figure 13.8).
(8x) (see
(ii) If this area is rotated through 360◦ about the x axis determine the volume of the solid of revolution produced.
4 3
y = x2
2
y = √(8x)
1 1 –1
Figure 13.8
2
Advanced Calculus: Methods of Differentiation
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347
14. A particle moves in such a way that the displacement, s, from a certain point is given by the formula s = 2t − 1. 4 ln(0. 5 + 4t) (i) Determine expressions for the velocity and acceleration of the particle. (ii) Calculate the displacement and acceleration of the particle when the velocity is 0, giving the answer correct to 3 decimal places.
14
STATISTICS In general, the purpose of statistics is to take the evidence contained in a numerical record of certain events and to present particular features of the data which can then be used in a management context.
Definitions •
A population is the group from which the data is collected.
•
A sample is a part of the population.
•
An individual is a member of the population (this does not necessarily mean a person).
•
A variable is a feature characteristic of any individual that differs from individual to individual.
•
A variable that differs in quantity is called a quantitative variable, for example, height.
•
A variable that differs in description is called a qualitative variable for exampe, colour.
•
A discrete variable is one that can only take certain values within its range, for example, the number of men making a ship’s crew.
•
A continuous variable is one that can take any value within its range, for example, the length of a ship may be measured to any degree of accuracy.
Tabular Presentation of Data The advantages are: •
the required figures can be located more readily,
•
comparisons can be made more easily,
•
patterns may be revealed,
•
Statistics •
tables are easier for further calculations,
•
tables take up less space.
349
It should always be borne in mind that when presenting a report some explanation should accompany it, pointing out the most significant features. Points worth remembering are as follows: •
Have the purpose of the table clearly in mind from the start. Typical aims may be: ◦
to show the original figures in an orderly manner,
◦
to show a distinct pattern in the data,
◦
to summarise the information,
◦
to publish statistics that others may wish to use, for example, Lloyd’s statistics.
•
Aim at simplicity so the table is easy to understand.
•
Always give the table a full explanatory heading.
•
State the units of measurement.
•
State the source of the data.
•
Make sure that headings are unambiguous.
An example illustrating some of these points is shown in Table 14.1. Table 14.1 The number and gross tonnage of trading vessels of over 100 gross registered tons in ‘flag of convenience’ fleets, 1954–1976 1954 Country of registration
1976
Gross tonnage Gross tonnage Number in million tons Number in million tons
Liberia
245
2.4
2, 600
73.5
Panama
595
4.1
2, 680
15.6
Honduras
130
0.4
57
0.1
Cyprus
—
–.–
765
3.1
Singapore
—
–.–
722
5.5
Lebanon
—
–.–
136
0.2
Somalia
—
–.–
255
1.8
Total
970
6.9
7, 215
99.8
32, 358
97.4
65, 887
372.0
3
7
11
27
World total Percentage of the world total Source: Lloyd’s Register of Shipping.
350
•
Mathematics
The original purpose of Table 14.1 was to show the growth in the number and tonnage of vessels in the main countries considered as offering the facility of flag of convenience over the period 1954–1976.
Frequency Distributions The most popular form of tabular presentation of statistical data is the use of a frequency distribution. This is a table which shows, for each separate value or group of values of the variable, the number of times which it occurs. Table 14.2 Frequency distribution of the number of days in a month, for Liverpool, with 0.3 mm or more of rainfall Number of days in a month with 0.3 mm or more of rain
Number of months
13
5
14
0
15
3
16
0
17
2
18
1
19
1
Total
12
Table 14.3 Frequency distribution of the number of ships passing the Felixstowe Ledge buoy per hour in a survey from 1100 to 1800 Time
Number of ships
1100 but before 1200
4
1200 but before 1300
9
1300 but before 1400
5
1400 but before 1500
7
1500 but before 1600
3
1600 but before 1700
6
1700 but before 1800
9
Total
43
Statistics
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351
Table 14.4 Number of stars with given sidereal hour angles Sidereal hour angles Number of stars of magnitude less than 3 0◦ –29◦ 3 ◦ ◦ 30 –59 3 60◦ –89◦ 4 ◦ ◦ 90 –119 6 120◦ –149◦ 7 ◦ ◦ 150 –179 7 180◦ –209◦ 3 ◦ ◦ 210 –239 4 240◦ –269◦ 6 ◦ ◦ 270 –299 7 300◦ –329◦ 4 ◦ ◦ 330 –359 6 Total
60
Examples Obviously the choice of classes depends on the particular circumstances but there are a few general hints. •
Make the classes of equal size where possible.
•
Make the class intervals multiples of numbers in which people are used to thinking, such as, 5, 10, 100 etc. and avoid such numbers as 13, 17 etc.
•
The number of classes should usually be between 5 and 20.
•
If there are a few extreme values at either end of a distribution it is better to put them together in an open-ended class rather than having a series of classes with very few observations in them.
Remember that these are only a guide and do not necessarily have to be adhered to.
Pictorial Representation of Data There are a few general rules. •
Unless you have the deliberate intention of misleading it is important to give the correct visual impression and to avoid optical illusions. So, if a graph is drawn, it is important to state the scale used on each axis.
352
•
Mathematics
•
The graph should have a clear comprehensive title.
•
The variable being measured should be on the horizontal axis with the results of the measurement, or frequencies, on the vertical axis.
•
The axes should be labelled.
•
The source of the graph should be stated.
•
It is important to keep any form of visual representation simple, so that overcrowding is avoided.
Histograms These are used to represent frequency distributions, and consist of a series of touching rectangles with bases on a horizontal axis. Each rectangle represents a value or class in the distribution and its height is drawn so that the area of the rectangle is proportional to the frequency it represents. Thus the total area of the histogram represents the total frequency in the distribution. Figure 14.1 shows the histogram of the sidereal hour angle data given in Table 14.4. The labelling of the classes is important since this is a discrete frequency distribution and yet the histogram demands a series of touching rectangles drawn on a continuous scale. It is as though each rectangle is extended by 1/2◦ in each direction, thus the second
8
Frequency
6 4
2 Sidereal hour angle 30
Figure 14.1
60
90
120
150
180
210
240
270
300
330
360
Statistics
10
•
353
Number of ships passing Felixstowe Ledge buoy per hour
9 8 7 6 5 4 3 2 1
Time of day
0 1100
1200
1300
1400
1500
1600
1700
1800
Figure 14.2
one is drawn from 291/2◦ to 391/2◦ . In a continuous distribution there is no need for this extension and it is most usual to label the end of the classes. Hence Figure 14.2 shows the histogram for the ships’ arrival patterns shown in Table 14.3.
Frequency Polygons Instead of a histogram, which is rather tedious to draw, a frequency polygon may be used, as shown in Figure 14.3. This is just a straight line graph joining the data points together. The points are plotted at the midpoint of the class, for example, the point for the data representing 13.00 to 14.00 is plotted at 13.30.
Cumulative Frequency Curves (Ogives) This plots the running total of the frequencies against bits variable. The data representing the sidereal hour angles (Table 14.4) becomes Table 14.5.
354
•
Mathematics
10
Number of ships passing Felixstowe Ledge buoy per hour
9 8 7 6 5 4 3 2 1
Time of day
0 1100
1200
1300
1400
1500
1600
1700
1800
Figure 14.3
60
Frequency
50 40 30 20 10 Sidereal hour angle 30
60
90
120
150
180
210
240
270
300
330
360
Figure 14.4
Table 14.5 Hour angles Cum. freq