HIGHER ENGINEERING
MATHEMATICS
5 EDITION
TH
INSTRUCTOR’S MANUAL
JOHN BIRD
© 2006 John Bird. All rights reserved. Published by Elsevier.
INTRODUCTION
In ‘Higher Engineering Mathematics 5th Edition’ there are 74 chapters; each chapter contains
numerous fully worked problems and further problems with answers, the latter being contained
within some 250 Exercises arranged evenly throughout the text. In addition, there are 19
Assignments at regular intervals within the text. These Assignments do not have answers given
since it is envisaged that lecturers could set the Assignments for students to attempt as part of their
course structure. The worked solutions to the Assignments are contained in this Instructor’s
Manual and with each is a full suggested marking scheme.
Some remedial algebra is made available on the web, together with a Remedial Algebra
Assignment, the solutions of which are included in this manual.
© 2006 John Bird. All rights reserved. Published by Elsevier.
2
CONTENTS
Page
ASSIGNMENT 1 (chapters 1 to 5)
1
ASSIGNMENT 2 (chapters 6 to 8)
12
ASSIGNMENT 3 (chapters 9 to 11)
19
ASSIGNMENT 4 (chapters 12 to 14)
27
ASSIGNMENT 5 (chapters 15 to 18)
35
ASSIGNMENT 6 (chapters 19 to 22)
42
ASSIGNMENT 7 (chapters 23 to 26)
52
ASSIGNMENT 8 (chapters 27 to 31)
58
ASSIGNMENT 9 (chapters 32 to 36)
67
ASSIGNMENT 10 (chapters 37 to 39)
75
ASSIGNMENT 11 (chapters 40 to 42)
82
ASSIGNMENT 12 (chapters 43 to 45)
87
ASSIGNMENT 13 (chapters 46 to 49)
92
ASSIGNMENT 14 (chapters 50 to 53)
101
ASSIGNMENT 15 (chapters 54 to 56)
113
ASSIGNMENT 16 (chapters 57 to 60)
121
ASSIGNMENT 17 (chapters 61 to 63)
128
ASSIGNMENT 18 (chapters 64 to 68)
138
ASSIGNMENT 19 (chapters 69 to 74)
145
© 2006 John Bird. All rights reserved. Published by Elsevier.
3
ASSIGNMENT 1 (PAGE 50)
This assignment covers the material contained in chapters 1 to 5.
Problem 1. Factorise x 3 + 4x 2 + x − 6 using the factor theorem. Hence solve the equation
x 3 + 4x 2 + x − 6 = 0
Marks
Let f (x) = x 3 + 4x 2 + x − 6
then f (1) = 1 + 4 + 1 - 6 = 0, hence (x - 1) is a factor
f (2) = 8 + 16 + 2 - 6 = 20
f (-1) = - 1 + 4 - 1 - 6 = - 4
f (-2) = - 8 + 16 - 2 - 6 = 0, hence (x + 2) is a factor
f (-3) = - 27 + 36 - 3 - 6 = 0, hence (x + 3) is a factor
Thus x 3 + 4x 2 + x − 6 = (x - 1)(x + 2)(x + 3)
3
If x 3 + 4x 2 + x − 6 = 0 then (x - 1)(x + 2)(x + 3) = 0
2
from which, x = 1, -2 or -3
Total: 5
Problem 2. Use the remainder theorem to find the remainder when 2x 3 + x 2 − 7x − 6 is divided by
(a) (x - 2) (b) (x + 1). Hence factorise the cubic expression.
Marks
(a) When 2x 3 + x 2 − 7x − 6 is divided by (x - 2), the remainder is given
by: ap3 + bp 2 + cp + d , where a = 2, b = 1, c = -7, d = -6 and p = 2,
i.e. the remainder is: 2 ( 2 ) + 1( 2 ) − 7(2) − 6 = 16 + 4 - 14 - 6 = 0
3
2
2
hence (x - 2) is a factor of 2x 3 + x 2 − 7x − 6
(b) When 2x 3 + x 2 − 7x − 6 is divided by (x + 1), the remainder is given
by:
2 ( −1) + 1( −1) − 7 ( −1) − 6 = - 2 + 1 + 7 - 6 = 0
3
2
2
hence (x + 1) is a factor of 2x 3 + x 2 − 7x − 6
© 2006 John Bird. All rights reserved. Published by Elsevier.
4
Either by dividing 2x 3 + x 2 − 7x − 6 by (x - 2)(x + 1) or by using the factor or remainder
theorems the third factor is found to be (2x + 3)
Hence 2x 3 + x 2 − 7x − 6 = (x - 2)(x + 1)(2x + 3)
3
Total: 7
Problem 3. Simplify
6x 2 + 7x − 5
by dividing out
2x − 1
Marks
3x + 5
2x - 1 6x 2 + 7x − 5
6x 2 − 3x
10x − 5
10x − 5
.
.
Hence
6x 2 + 7x − 5
= 3x + 5
2x − 1
4
Total: 4
Problem 4. Solve the following inequalities:
(a) 2 – 5x ≤ 9 + 2x
(b) 3 + 2t ≤ 6
(d) ( 3t + 2 ) > 16
(e) 2x 2 − x − 3 < 0
2
(c)
x −1
>0
3x + 5
Marks
(a) Since 2 – 5x ≤ 9 + 2x then 2 – 9 ≤ 2x + 5x
i.e.
–7 ≤ 7x from which, -1 ≤ x or x ≥ -1
2
(b) Since 3 + 2t ≤ 6 then -6 ≤ 3 + 2t ≤ 6
-6 ≤ 3 + 2t becomes –9 ≤ 2t i.e. –4.5 ≤ t
and 3 + 2t ≤ 6 becomes 2t ≤ 3 i.e. t ≤ 1.5
Hence, –4.5 ≤ t ≤ 1.5
(a) Since
3
x −1
x −1
> 0 then
must be positive
3x + 5
3x + 5
© 2006 John Bird. All rights reserved. Published by Elsevier.
5
hence, either (i) x – 1 > 0 and 3x + 5 > 0
or (ii) x – 1 < 0 and 3x + 5 < 0
(i) x > 1 and x > -5/3 thus, x > 1
(ii) x < 1 and x < -5/3 thus, x < -5/3
Thus, x > 1 or x < -5/3
3
(d) Since ( 3t + 2 ) > 16 then 3t + 2 > 16 or 3t + 2 < - 16
2
i.e. 3t + 2 > 4 or 3t + 2 < -4
i.e. 3t > 2 or 3t < -6 hence, t > 2/3 or t < -2
3
(e) Since 2x 2 − x − 3 < 0 then (2x – 3)(x + 1) < 0
hence, either(i) (2x – 3) > 0 and (x + 1) < 0
or (ii) (2x – 3) < 0 and (x + 1) > 0
(i) x > 3/2 and x < -1 and it is not possible to satisfy both
(ii) x < 3/2 and x > -1 i.e. -1 < x < 1.5
3
Total : 14
Problem 5. Resolve the following into partial fractions:
(a)
x − 11
2
x −x−2
(b)
3− x
2
( x + 3) ( x + 3)
(c)
x 3 − 6x + 9
x2 + x − 2
Marks
(a) Let
x − 11
x − 11
A
B
A(x + 1) + B(x − 2)
=
+
=
≡
(x − 2)(x + 1)
x − x − 2 (x − 2)(x + 1) (x − 2) (x + 1)
2
Hence
x - 11 = A(x + 1) + B(x - 2)
Let x = 2:
Let x = -1:
Hence
2
- 9 = 3A
hence A = -3
- 12 = - 3B hence
x − 11
4
3
=
−
x − x − 2 (x + 1) (x − 2)
2
B=4
4
© 2006 John Bird. All rights reserved. Published by Elsevier.
6
(b) Let
(Ax + B)(x + 3) + C ( x 2 + 3)
3− x
Ax + B
C
≡
+
=
( x 2 + 3) ( x + 3) ( x 2 + 3) (x + 3)
( x 2 + 3) ( x + 3)
3
3 - x = (Ax + B)(x + 3) + C (x 2 + 3)
Hence
hence C =
1
2
x 2 coefficients: 0 = A + C
hence A = −
1
2
x coefficients: -1 = 3A + B
hence -1 = −
Let x = -3:
6 = 0 + 12C
3
1
+ B and B =
2
2
1
1
1
− x+
3−x
1− x
1
or
Hence 2
= 22 2 + 2
+
2
2 ( x + 3 ) 2(x + 3)
( x + 3 ) ( x + 3 ) ( x + 3) (x + 3)
5
(c) Dividing out gives:
x - 1
x 2 + x − 2 x3
− 6x + 9
x 3 + x 2 − 2x
− x 2 − 4x + 9
−x 2 − x + 2
- 3x + 7
Hence
Let
x 3 − 6x + 9
−3x + 7
≡ x-1+ 2
2
x +x−2
x +x−2
4
−3x + 7
A
B
A(x − 1) + B(x + 2)
+
≡
≡
2
(x + 2)(x − 1)
x + x − 2 x + 2 x −1
from which,
2
-3x + 7 = A(x - 1) + B(x + 2)
Let x = -2:
13 = -3A hence A = −
Let x = 1:
4 = 3B hence B =
13
3
4
3
13
4
−3x + 7
3 + 3
Hence 2
=
x + x − 2 (x + 2) (x − 1)
−
and
x 3 − 6x + 9
13
4
= x −1−
+
2
x +x−2
3(x + 2) 3(x − 1)
3
1
Total: 24
© 2006 John Bird. All rights reserved. Published by Elsevier.
7
Problem 6. Evaluate, correct to 3 decimal places,
5e −0.982
3ln 0.0173
Marks
5e −0.982
1.872806134
=
= - 0.154
3ln 0.0173 −12.17114633
2
Total: 2
Problem 7. Solve the following equations, each correct to 4 significant figures:
(a) ln x = 2.40
x −1
(b) 3
=5
x
− ⎞
⎛
(c) 5 = 8 ⎜ 1 − e 2 ⎟
⎝
⎠
x −2
Marks
(a) Since ln x = 2.40 then x = e 2.40 = 11.02
2
(b) Since 3x −1 = 5x − 2 then lg 3x −1 = lg 5x − 2
and
(x - 1)lg 3 = (x - 2)lg 5
i.e.
x lg 3 - lg 3 = x lg 5 - 2 lg 5
Hence
2 lg 5 - lg 3 = x lg 5 - x lg 3
and
2 lg 5 - lg 3 = x(lg 5 - lg 3)
2 lg 5 − lg 3 0.92081875...
=
lg 5 − lg 3 0.221848749...
from which,
x=
i.e.
x = 4.151
4
x
x
−
− ⎞
⎛
5
(c) Since 5 = 8 ⎜1 − e 2 ⎟ then = 1 − e 2
8
⎝
⎠
and
e
−
x
2
x
Thus
and
i.e.
5 3
= 1− =
8 8
e2 =
8
3
x
8
= ln
2
3
x = 2 ln
8
= 1.962
3
4
Total: 10
© 2006 John Bird. All rights reserved. Published by Elsevier.
8
Problem 8. The pressure p at height h above ground level is given by: p = p0 e − kh where p0 is the
pressure at ground level and k is a constant. When p0 is 101 kilopascals and the pressure at a height
of 1500 m is 100 kilopascals, determine the value of k. Sketch a graph of p against h (p the vertical
axis and h the horizontal axis) for values of height from zero to 12000 m when p0 is 101 kilopascals
Marks
Since p = p0 e − kh then 100 × 103 = 101× 103 e −1500k
and
100
101 1500k
=e
= e −1500k or
101
100
Hence
ln 1.01 = 1500 k and k =
1
ln1.01 = 6.6336 × 10−6
1500
4
A table of values is drawn up as shown
h (m)
0
2000 4000 6000 8000 10000 12000
p = p0 e − kh (kPa) 101 99.7 98.4
97.1 95.8
94.5
93.3
2
A graph of p/h is shown in Fig. 1
4
Figure 1
Total: 10
© 2006 John Bird. All rights reserved. Published by Elsevier.
9
Problem 9. Evaluate correct to 4 significant figures:
(a) sinh 2.47
(b) tanh 0.6439
(c) sech 1.385
(d) cosech 0.874
Marks
(a) sinh 2.47 =
e 2.47 − e −2.47
= 5.869
2
(b) tanh 0.6439 =
(c) sech 1.385 =
(or by calculator)
e0.6439 − e −0.6439 1.378651608
=
= 0.5675
e0.6439 + e −0.6439 2.429131585
1
(or by calculator)
1
1
2
= 1.385 −1.385 = 0.4711
cosh1.385 e
+e
(d) cosech 0.874 =
1
2
= 0.874 −0.874 = 1.011
sinh 0.874 e
−e
2
(or use calculator)
2
Total: 6
Problem 10. The increase in resistance of strip conductors due to eddy currents at power
frequencies is given by:
λ=
α t ⎡ sinh αt + sin αt ⎤
2 ⎣⎢ cosh αt − cos αt ⎦⎥
Calculate λ, correct to 5 significant figures, when α = 1.08 and t = 1.
Marks
λ=
α t ⎡ sinh αt + sin αt ⎤ (1.08)(1) ⎡ sinh1.08 + sin1.08 ⎤
=
⎢⎣ cosh1.08 − cos1.08 ⎥⎦
2 ⎢⎣ cosh αt − cos αt ⎥⎦
2
⎡1.302542013 + 0.881957806 ⎤
= ( 0.54 ) ⎢
⎣1.642137538 − 0.471328364 ⎥⎦
3
⎛ 2.184499819 ⎞
= ( 0.54 ) ⎜
⎟ = 1.0075
⎝ 1.170809174 ⎠
2
Total: 5
Problem 11. If A ch x - B sh x ≡ 4e x − 3e − x determine the values of A and B.
Marks
A ch x - B sh x ≡ 4e x − 3e − x
i.e.
⎛ e x + e− x
A⎜
⎝ 2
⎞
⎛ e x − e− x ⎞
x
−x
B
−
⎟
⎜
⎟ ≡ 4e − 3e
⎠
⎝ 2 ⎠
i.e.
A x A −x B x B −x
e + e − e + e ≡ 4e x − 3e − x
2
2
2
2
2
© 2006 John Bird. All rights reserved. Published by Elsevier.
10
Hence
A B
− =4
2 2
A B
+ = −3
2 2
i.e.
A-B=8
(1)
i.e. A + B = -6
(2)
2
Adding equations (1) and (2) gives: 2A = 2 from which, A = 1
1
Substituting in equation (1) or (2) gives: B = -7
1
Total: 6
Problem 12. Solve the following equation: 3.52 ch x + 8.42 sh x = 5.32 correct to 4 decimal places.
Marks
3.52 ch x + 8.42 sh x = 5.32
i.e.
i.e.
⎛ e x + e− x
3.52 ⎜
⎝ 2
⎞
⎛ e x − e− x
8.42
+
⎟
⎜
⎠
⎝ 2
⎞
⎟ = 5.32
⎠
1.76 e x + 1.76 e − x + 4.21e x − 4.21e− x = 5.32
5.97 e x − 2.45e − x − 5.32 = 0
5.97 ( e x ) − 2.45 − 5.32 e x = 0
2
Hence
5.97 ( e x ) − 5.32 e x − 2.45 = 0
2
and
e =
x
−(−5.32) ±
3
( −5.32 ) − 4 ( 5.97 )( −2.45) 5.32 ± 9.3171
=
2 ( 5.97 )
11.94
2
= 1.22589 or -0.33477
2
Hence x = ln 1.22589 = 0.2036671.. and x = ln(-0.33477) which has no solution,
i.e. x = 0.2037, correct to 4 significant figures.
2
Total: 7
TOTAL ASSIGNMENT MARKS: 100
© 2006 John Bird. All rights reserved. Published by Elsevier.
11
ASSIGNMENT 2 (PAGE 75)
This assignment covers the material contained in chapters 6 to 8.
Problem 1. Determine the 20th term of the series 15.6, 15, 14.4, 13.8, …
Marks
The 20th term is given by:
i.e.
a + (n – 1)d
15.6 + (20 – 1)(-0.6) = 15.6 – 19(0.6) = 15.6 – 11.4 = 4.2
3
Total: 3
Problem 2. The sum of 13 terms of an arithmetic progression is 286 and the common difference is
3. Determine the first term of the series.
Marks
Sn =
n
[ 2a + (n − 1)d ]
2
Thus,
i.e. 286 =
286 × 2
= 2a = 36
13
from which, first term, a =
13
[ 2a + (13 − 1)3]
2
i.e. 286 =
13
[ 2a + 36]
2
1
i.e. 44 – 36 = 2a
44 − 36
=4
2
3
Total: 4
Problem 3. An engineer earns £21000 per annum and receives annual increments of £600.
Determine the salary in the 9th year and calculate the total earnings in the first 11 years.
Marks
If first term, a = £21000 and the n’th term, n = 9
Then salary in 9th year = a + (n – 1)d = 21000 + (9 – 1)(600)
2
= 21000 + 8(600) = £25800
Total earnings in first 11 years, S11 =
=
n
11
[ 2a + (n − 1)d ] = [ 2(21000) + (11 − 1)600]
2
2
11
[ 42000 + 6000] = £264,000
2
3
Total: 5
© 2006 John Bird. All rights reserved. Published by Elsevier.
12
Problem 4. Determine the 11th term of the series 1.5, 3, 6, 12, ….
Marks
The 11th term is given by:
a r n −1 where a = 1.5, and the common ratio, r = 2,
11th term = (1.5)(2)11−1 = 1536
i.e.
2
Total: 2
Problem 5. Find the sum of the first eight terms of the series 1, 2.5, 6.25, …. , correct to 1 decimal
place.
Marks
In the series 1, 2.5, 6.25, … , the common ratio, r = 2.5
and the sum of the first eight terms, S8 =
a ( r n − 1)
(r − 1)
=
1( 2.58 − 1)
(2.5 − 1)
=
1524.879..
1.5
4
= 1016.6
Total: 4
Problem 6. Determine the sum to infinity of the series 5, 1,
1
, ….
5
Marks
S∞ =
a
1
where a = 5 and r = ,
1− r
5
thus, the sum to infinity, S∞ =
5
1−
1
5
=
5 25
1
=
= 6 or 6.25
4 4
4
5
3
Total: 3
Problem 7. A machine is to have seven speeds ranging from 25 rev/min to 500 rev/min. If the
speeds form a geometric progression, determine their value, each correct to the nearest whole
number.
© 2006 John Bird. All rights reserved. Published by Elsevier.
13
Marks
The G.P. of n terms is given by: a, ar, ar 2 ,...ar n −1
The first term, a = 25 rev/min.
The seventh term is given by: a r 7 −1 which is 500 rev/min,
i.e.
a r 6 = 500 from which,
thus, the common ratio, r =
6
r6 =
500 500
=
= 20
a
25
20 = 1.64755
2
The first term is 25 rev/min
The second term, ar = (25)(1.64755) = 41.19
The third term, a r 2 = (25)(1.64755) 2 = 67.86
The fourth term, a r 3 = (25)(1.64755)3 = 111.80
The fifth term, a r 4 = (25)(1.64755) 4 = 184.20
The sixth term, a r 5 = (25)(1.64755)5 = 303.48
Hence, correct to the nearest whole number, the speeds of the machine are:
25, 41, 68, 112, 184, 303 and 500 rev/min.
6
Total: 8
Problem 8. Use the binomial series to expand ( 2a − 3b ) .
6
Marks
( 2a − 3b )
6
= ( 2a ) + 6 ( 2a ) ( −3a ) +
6
5
(6)(5)
(6)(5)(4)
4
2
3
3
( 2a ) ( −3a ) +
( 2a ) ( −3a )
2!
3!
(6)(5)(4)(3)
(6)(5)(4)(3)(2)
2
4
5
6
( 2a ) ( −3b ) +
( 2a )( −3b ) + ( −3b )
4!
5!
4
= 64a6 − 576a 5b + 2160a4b 2 − 4320a 3b 3 + 4860a 2b 4 − 2916ab 5 + 729b 6
3
+
Total: 7
18
⎛
1 ⎞
Problem 9. Determine the middle term of ⎜ 3x − ⎟ .
3y ⎠
⎝
© 2006 John Bird. All rights reserved. Published by Elsevier.
14
Marks
The middle term is the 10th
1
9
⎛x⎞
x9
(18)(17)(16)(15)(14)(13)(12)(11)(10)
1 ⎞
9⎛
i.e.
( 3x ) ⎜ − ⎟ = −48620 9 or −48620 ⎜ ⎟
y
9!
⎝ 3y ⎠
⎝y⎠
9
5
Total: 6
Problem 10. Expand the following in ascending powers of t as far as the term in t 3 :
(a)
1
1+ t
1
1 − 3t
(b)
For each case, state the limits for which the expansion is valid.
Marks
(a)
1
(−1)(−2) 2 (−1)(−2)(−3) 3
−1
= (1 + t ) = 1 + (−1)t +
t +
t ...
1+ t
2!
3!
= 1 − t + t 2 − t 3 + ...
4
The expansion is valid when t < 1 or -1 < t < 1
(b)
1
−1/ 2
= (1 − 3t )
1 − 3t
2
⎛ 1 ⎞⎛ 3 ⎞
⎛ 1 ⎞⎛ 3 ⎞⎛ 5 ⎞
− ⎟⎜ − ⎟
⎜
⎜ − ⎟⎜ − ⎟⎜ − ⎟
2
3
2 ⎠⎝ 2 ⎠
⎛ 1⎞
⎝
= 1 + ⎜ − ⎟ ( −3t ) +
( −3t ) + ⎝ 2 ⎠⎝ 2 ⎠⎝ 2 ⎠ ( −3t ) ...
2!
3!
⎝ 2⎠
3
27
135 3
t + ...
= 1 + t + t2 +
2
8
16
The expansion is valid when 3t < 1 i.e.
t <
4
1
3
or −
1
1
<t<
3
3
2
Total: 12
Problem 11. When x is very small show that:
1
(1 + x )
2
3
≈ 1− x
2
1− x
Marks
1
(1 + x )
2
1− x
= (1 + x )
−2
(1 − x )
−
1
2
⎛ ⎛ 1⎞
⎞
= [1 + (−2)x + ...] ⎜1 − ⎜ − ⎟ x + ... ⎟
⎝ ⎝ 2⎠
⎠
1
3
⎛ 1⎞
≈ 1 − ⎜ − ⎟ x + (−2)x + ... ≈ 1 + x − 2x + ... = = 1 − x
2
2
⎝ 2⎠
5
when powers of 2 and above are neglected.
Total: 5
© 2006 John Bird. All rights reserved. Published by Elsevier.
15
Problem 12. The modulus of rigidity G is given by G =
R4 θ
where R is the radius, θ the
L
angle of twist and L the length. Find the approximate percentage error in G when R is measured
1.5% too large, θ is measured 3% too small and L is measured 1% too small.
Marks
The new values of R, θ and L are: (1 + 0.015)R, (1 – 0.03)θ and (1 – 0.01)L
[ (1 + 0.015)R ] [(1 − 0.03)θ]
[(1 − 0.01)L]
4
New modulus of rigidity =
3
= [ (1 + 0.015)R ] [ (1 − 0.03)θ][ (1 − 0.01)L ]
−1
4
= (1 + 0.015) 4 R 4 (1 − 0.03)θ(1 − 0.01) −1 L−1
= (1 + 0.015) 4 (1 − 0.03)(1 − 0.01) −1 R 4 θ L−1
= [1 + 4(0.015)][1 - 0.03][1 - (-1)(0.01)]
R 4θ
L
neglecting products of small terms
= [1 + 0.06 – 0.03 + 0.01] G = (1 + 0.04) G
i.e. G is increased by 4%
4
Total: 7
Problem 13. Use Maclaurin’s series to determine a power series for e2x cos 3x as far as the term in
x2 .
Marks
Let f(x) = e 2x cos 3x
f(0) = e0 cos 0 = 1
1
f '(0) = −3e0 sin 0 + 2 e0 cos 0 = 2
3
f '(x) = ( e 2x ) ( −3sin 3x ) + ( cos 3x ) ( 2 e 2x )
= −3e 2x sin 3x + 2 e 2x cos 3x
f ''(x) = ( −3e 2x ) ( 3cos 3x ) + ( sin 3x ) ( −6 e 2x ) + ( 2 e 2x ) ( −3sin 3x ) + ( cos 3x ) ( 4 e 2x )
f ''(0) = −9 e0 cos 0 − 6 e0 sin 0 − 6 e0 sin 0 + 4 e0 cos 0 = -9 – 0 – 0 + 4 = -5
3
x2
f ''(0) + ...
The Maclaurin’s series is: f(x) = f(0) + x f '(0) +
2!
i.e.
5
e2x cos 3x = 1 + 2x − x 2 + ...
2
3
Total: 10
© 2006 John Bird. All rights reserved. Published by Elsevier.
16
Problem 14. Show, using Maclaurin’s series that the first four terms of the power series for cosh 2x
2
4
is given by: cosh 2x = 1 + 2x 2 + x 4 + x 6
3
45
Marks
f(x) = cosh 2x
f(0) = cosh 0 = 1
f ′(x) = 2 sinh 2x
f ′(0) = 2 sinh 0 = 0
f ′′(x) = 4 cosh 2x
f ′′(0) = 4
f ′′′(x) = 8 sinh 2x
f ′′′(0) = 0
f iv (x) = 16 cosh 2x
f iv (0) = 16
f v (x) = 32 sinh 2x
f v (0) = 0
f vi (x) = 64 cosh 2x
f vi (0) = 64
Since f (x) = f (0) + x f '(0) +
x2
x3
f ''(0) + f '''(0) + ...
2!
3!
f(x) = cosh 2x = 1 + x(0) +
then
5
x2
x3
x4
x5
x6
(4) + (0) + (16) + (0) + (64) + ...
2!
3!
4!
5!
6!
2
4
cosh 2x = 1 + 2x 2 + x 4 + x6
3
45
i.e.
6
Total: 11
Problem 15. Expand the function x 2 ln (1 + sin x ) using Maclaurin’s series and hence evaluate:
∫
1/ 2
0
x 2 ln (1 + sin x ) dx correct to 2 significant figures.
Marks
f(0) = ln(1 + sin 0) = 0
Let f(x) = ln(1 + sin x)
f ′(x) =
f ′′(x) =
=
cos x
1 + sin x
f ′(0) =
(1 + sin x)(− sin x) − (cos x)(cos x)
(1 + sin x )
2
=
cos 0
=1
1 + sin 0
1
1
− sin x − sin 2 x − cos 2 x
− sin x − 1
(1 + sin x)
−1
=−
=
2
2
(1 + sin x)
(1 + sin x) 1 + sin x
(1 + sin x )
f ′′(0) =
2
−1
= -1
1 + sin 0
3
© 2006 John Bird. All rights reserved. Published by Elsevier.
17
f ′′′(x) =
(1 + sin x)(0) − (−1)(cos x)
cos x
=
2
(1 + sin x)
(1 + sin x) 2
Hence, ln(1 + sin x) = f (0) + x f '(0) +
= 0 + x(1) +
f ′′′(0) =
cos 0
=1
(1 + sin 0) 2
2
x2
x3
f ''(0) + f '''(0) + ...
2!
3!
x2
x3
x2 x3
(−1) + (1) + ... = x − + − ...
2!
3!
2
6
2
⎛
⎞
x2 x3
x4 x5
Thus, x 2 ln ( 2 + sin x ) = x 2 ⎜ x − + − ... ⎟ = x 3 − + − ...
2
6
2
6
⎝
⎠
and
∫
1/ 2
0
x 2 ln(1 + sin x) dx = ∫
1/ 2
0
⎛ 3 x 4 x5
⎞
⎜ x − + − ... ⎟ dx
2
6
⎝
⎠
1/ 2
⎡ x4 x5 x6 ⎤
= ⎢ − + ⎥
⎣ 4 10 36 ⎦ 0
⎛ ⎛ 1 ⎞ 4 ⎛ 1 ⎞5 ⎛ 1 ⎞ 6 ⎞
⎜⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎟
2
2
2
= ⎜ ⎝ ⎠ − ⎝ ⎠ + ⎝ ⎠ ⎟ − (0)
⎜ 4
10
36 ⎟
⎟⎟
⎜⎜
⎝
⎠
2
= 0.015625 – 0.003125 + 0.000434
= 0.013, correct to 2 significant figures.
2
Total: 13
TOTAL ASSIGNMENT MARKS: 100
© 2006 John Bird. All rights reserved. Published by Elsevier.
18
ASSIGNMENT 3 (PAGE 114)
This assignment covers the material contained in chapters 9 to 11.
Problem 1. Use the method of bisection to evaluate the root of the equation x 3 + 5x = 11 in the
range x = 1 to x = 2, correct to 3 significant figures.
Marks
Let f (x) = x 3 + 5x − 11
f (1) = 13 + 5 − 11 = −5
and
f (2) = 23 + 5(2) − 11 = 7
Hence a root lies between x = 1 and x = 2
2
Using a tabular approach:
x1
x2
x3 =
x1 − x 2
2
f ( x3 )
1
-5
1
+7
1
2
1.5
-0.125
1.5
2
1.75
+3.109
1.5
1.75
1.625
+1.416
1.5
1.625
1.5625
+0.627
1.5
1.5625
1.53125
+0.2466
1.5
1.53125
1.515625
+0.0597
1.5
1.515625
1.5078125
-0.0329
1.51171875
+0.0133
1.5078125 1.515625
Hence, x = 1.51, correct to 3 significant figures
10
Total: 12
Problem 2. Solve x 3 + 5x = 11 using an algebraic method of successive approximations. Correct
to 3 significant figures.
Marks
f (x) = x 3 + 5x − 11
f (1) = 13 + 5 − 11 = −5
and
f (2) = 23 + 5(2) − 11 = 7
2
Hence a root lies between x = 1 and x = 2
© 2006 John Bird. All rights reserved. Published by Elsevier.
19
First approximation
1
Let the first approximation be 1.4
Second approximation
Let the true value of the root, x2 , be (x1 + δ1)
Let f(x1 + δ1) = 0, then since x1 = 1.4,
(1.4 + δ1)3 + 5(1.4 + δ1) - 11 = 0
Neglecting terms containing products of δ1 and using the binomial series gives:
[(1.4)3 + 3(1.4)2 δ1] + 7 + 5δ1 - 11 ≈ 0
2.744 + 5.88δ1 + 7 + 5 δ1 - 11 ≈ 0
10.88δ1 ≈ 11 – 2.744 – 7
δ1 ≈
1.256
≈ 0.11544
10.88
Thus x2 ≈ 1.4 + 0.11544 = 1.51544
4
Third approximation
Let the true value of the root, x3 , be (x2 + δ2)
Let f(x2 + δ2) = 0, then since x2 = 1.51544,
(1.51544 + δ2)3 + 5(1.51544 + δ2) - 11 = 0
Neglecting terms containing products of δ2 gives:
(1.51544 )
3
+ 3 (1.51544 ) δ2 + 7.5772 + 5δ 2 − 11 ≈ 0
2
3.480296 + 6.889675δ2 + 7.6772 + 5δ2 - 11 ≈ 0
11.889675δ2 ≈ 11 – 3.480296 – 7.5772
δ2 ≈
−0.057496
≈ −0.004836
11.889675
Thus x3 ≈ (x2 + δ2) = 1.51544 – 0.004836 ≈ 1.5106
4
Fourth approximation
Let the true value of the root, x4 , be (x3 + δ3)
© 2006 John Bird. All rights reserved. Published by Elsevier.
20
Let f(x3 + δ3) = 0, then since x3 = 1.5106,
(1.5106 + δ3)3 + 5(1.5106 + δ3) - 11 = 0
Neglecting terms containing products of δ3 gives:
(1.5106 )
3
+ 3 (1.5106 ) δ3 + 7.553 + 5δ3 − 11 ≈ 0
2
3.447057 + 6.845737δ3 + 7.553 + 5δ3 - 11 ≈ 0
11.845737δ2 ≈ 11 – 3.447057 – 7.553
δ2 ≈
−0.000057
≈ −0.0000048
11.845737
Thus x4 ≈ (x3 + δ3) = 1.5106 – 0.0000048 ≈ 1.5106
4
Since x3 and x4 are the same when expressed to the required degree of accuracy, then
the required root is 1.51, correct to 3 decimal places.
1
Total: 16
Problem 3. The solution to a differential equation associated with the path taken by a projectile for
which the resistance to motion is proportional to the velocity is given by:
y = 2.5 ( e x − e − x ) + x − 25
Use Newton’s method to determine the value of x, correct to 2 decimal places, for which the value
of y is zero
Marks
If y = 0,
2.5e x − 2.5e − x + x − 25 = 0
Let f (x) = 2.5e x − 2.5e − x + x − 25
f (0) = 2.5 – 2.5 + 0 – 25 = -25
f (1) = -18.12
f (2) = -4.866
f (3) = 28.09
Hence a root lies between x = 2 and x = 3. Let r1 = 2.2
4
© 2006 John Bird. All rights reserved. Published by Elsevier.
21
f ′(x) = 2.5e x + 2.5e − x + 1
r2 = r1 −
f (r1 )
2.5e 2.2 − 2.5e −2.2 + 2.2 − 25
= 2.2 −
f '(r1 )
2.5e2.2 + 2.5e −2.2 + 1
= 2.2 −
r3 = 2.222 −
−0.514474147
= 2.222
23.83954164
f (2.222)
0.01542962
= 2.222 −
= 2.221
f '(2.222)
24.33539016
Hence, x = 2.22, correct to 2 decimal places.
7
Total: 11
Problem 4. Convert the following binary numbers to decimal form:
(a) 1101
(b) 101101.0101
Marks
(a) 11012 = 1× 23 + 1× 22 + 0 × 21 + 1× 20 = 8 + 4 + 0 + 1 = 1310
2
(b) 101101.01012 = 1× 25 + 0 × 24 + 1× 23 + 1× 22 + 0 × 21 + 1× 20 + 0 × 2−1 + 1× 2−2
+ 0 × 2−3 + 1× 2−4
= 32 + 0 + 8 + 4 + 0 + 1 + 0 +
1
1
+0+
= 45.312510
4
16
3
Total: 5
Problem 5. Convert the following decimal numbers to binary form: (a) 27
(b) 44.1875
Marks
(a)
Hence
2710 = 110112
3
© 2006 John Bird. All rights reserved. Published by Elsevier.
22
(b)
Hence 4410 = 1011002
3
Hence 44.187510 = 101100.00112
3
Total: 9
Problem 6. Convert the following denary numbers to binary, via octal:
(a) 479
(b) 185.2890625
Marks
(a)
From Table 10.1, page 88, 7378 = 111 011 1112
Hence
47910 = 111 011 1112
3
© 2006 John Bird. All rights reserved. Published by Elsevier.
23
(b)
From Table 10.1, page 88, 2718 = 010 111 0012
3
0.2890625 × 8 = 2.3125
0.3125 × 8
= 2.5
0.5 × 8
= 4.0
i.e. 0.289062510 = .2248 = .010 010 1002 from Table 10.1, page 88.
Hence 185.289062510 = 10 111 001.010 010 12
3
Total: 9
Problem 7. Convert (a) 5F16 into its decimal equivalent
(b) 13210 into its hexadecimal equivalent
(c) 110 101 0112 into its hexadecimal equivalent
Marks
(a) 5F16 = 5 ×161 + F × 160 = 5 ×161 + 15 ×160 = 80 + 15 = 9510
2
(b)
i.e.
13210 = 8 416
3
(c) Grouping bits in 4’s from the right gives: 1101010112 = 0001 1010 1011
and assigning hexadecimal symbols to each group gives: 1
Hence,
A
B
110 101 0112 = 1AB16
3
Total: 8
Problem 8. Use the laws and rules of Boolean algebra to simplify the following expressions:
(a) B. ( A + B ) + A.B
(b) A.B.C + A.B.C + A.B.C + A.B.C
© 2006 John Bird. All rights reserved. Published by Elsevier.
24
(a) B. ( A + B ) + A.B = A.B + B.B + A.B = A.B + 0 + A.B = A.B + A.B
= A. ( B + B ) = A. (1) = A
Marks
4
(b) A.B.C + A.B.C = A.B. ( C + C ) = A.B. (1) = A.B
and A.B.C + A.B.C = A.B. ( C + C ) = A.B. (1) = A.B
Thus, A.B.C + A.B.C + A.B.C + A.B.C = A.B + A.B = A. ( B + B ) = A. (1) = A
5
Total: 9
Problem 9. Simplify the Boolean expression A.B + A.B.C using de Morgan’s laws.
)(
(
A.B + A.B.C = A.B.A.B.C = A + B . A.B + C
(
Marks
)
1
)
= ( A + B ) . A.B + C = ( A + B ) . ( A + B + C )
1
= A.A + A.B + A.C + A.B + B.B + B.C
1
= A + A.B + A.C + A.B + 0 + B.C = A + A ( B + B ) + A.C + B.C
1
= A + A (1) + A.C + B.C = A + A.C + B.C = A (1 + C ) + B.C = A + B.C
1
Total: 5
Problem 10. Use a Karnaugh map to simplify the Boolean expression:
A.B.C + A.B.C + A.B.C + A.B.C
Marks
4
The horizontal couple gives: A.B.C + A.B.C = A.C
The vertical couple gives: A.B.C + A.B.C = A.B
The bottom right-hand corner square cannot be coupled and is A.B.C
Hence A.B.C + A.B.C + A.B.C + A.B.C = A.C + A.B + A.B.C
4
Total: 8
© 2006 John Bird. All rights reserved. Published by Elsevier.
25
Problem 11. A clean room has two entrances, each having two doors, as shown in Figure A3.1. A
warning bell must sound if both doors A and B or doors C and D are open at the same time. Write
down the Boolean expression depicting this occurrence, and devise a logic network to operate the
bell using NAND-gates only.
Figure A3.1
Marks
The Boolean expression which will ring the warning bell is: A.B + C.D
2
A circuit using NAND-gates only is shown in Figure 2.
Figure 2
6
Total: 8
TOTAL ASSIGNMENT MARKS: 100
© 2006 John Bird. All rights reserved. Published by Elsevier.
26
ASSIGNMENT 4 (PAGE 146)
This assignment covers the material contained in chapters 12 to 14.
Problem 1. A 2.0 m long ladder is placed against a perpendicular pylon with its foot 52 cm
from the pylon. (a) Find how far up the pylon (correct to the nearest mm) the ladder reaches.
(b) If the foot of the ladder is moved 10 cm towards the pylon how far does the top of the
ladder rise?
Marks
(a) From Figure 3,
h=
22 − 0.522 = 1.931 m
by Pythagoras’ theorem.
3
Figure 3
(b) New height, h′ =
22 − 0.422 = 1.955 m
h′ - h = 1.955 – 1.931 = 0.024 m = 24 mm = how far the top of the ladder rises.
4
Total: 7
Problem 2. Evaluate correct to 4 significant figures: (a) cos 124°13′
(b) cot 72.68°
Marks
O
13 ⎞
⎛
(a) cos 124°13′ = cos ⎜124 ⎟ = -0.5623
60 ⎠
⎝
(b) cot 72.68° =
2
1
= 0.3118
tan 72.68°
2
Total: 4
Problem 3. From a point on horizontal ground a surveyor measures the angle of elevation of
a church spire as 15°. He moves 30 m nearer to the church and measures the angle of
elevation as 20°. Calculate the height of the spire.
© 2006 John Bird. All rights reserved. Published by Elsevier.
27
Marks
From the sketch in Figure 4, tan 20° =
tan15° =
and
h
x
from which, h = x tan 20°
h
30 + x
2
from which, h = (30 + x) tan 15°
2
Figure 4
Hence,
x tan 20° = (30 + x) tan 15°
i.e.
0.36397 x = 0.26795(30 + x) = 8.0385 + 0.26795 x
0.36397 x – 0.26795 x = 8.0385
i.e.
0.09602 x = 8.0385
and
x=
8.0385
= 83.717
0.09602
Hence, the height of the spire, h = 83.717 tan 20° = 30.47 m
5
Total: 9
Problem 4. If secant θ = 2.4613 determine the acute angle θ.
Marks
1
= 0.406289
2.4613
2
θ = cos −1 (0.406289) = 66.03°
2
Since sec θ = 2.4613 then cos θ =
and
Total: 4
Problem 5. Evaluate, correct to 3 significant figures:
3.5 cos ec 31°17 '− cot(−12°)
3 sec 79°41'
Marks
3.5 cos ec 31°17 '− cot(−12°) 3.5 (1.925778428 ) − ( −4.70463011)
= 0.683
=
3 sec 79°41'
3 ( 5.583834323)
5
Total: 5
© 2006 John Bird. All rights reserved. Published by Elsevier.
28
Problem 6. A man leaves a point walking at 6.5 km/h in a direction E 20° N (i.e. a bearing of
70°). A cyclist leaves the same point at the same time in a direction E 40° S (i.e. a bearing of
130°) travelling at a constant speed. Find the average speed of the cyclist if the walker and
cyclist are 80 km apart after 5 hours.
Marks
After 5 hours the walker has travelled 5 × 6.5 = 32.5 km (shown as AB in Figure 5).
If AC is the distance the cyclist travels in 5 hours then BC = 80 km.
Figure 5
Using the sine rule:
80
32.5
32.5sin 60°
= 0.351823
=
from which, sin C =
80
sin 60° sin C
Hence, C = sin −1 (0.351823) = 20.60° (or 159.40° which is impossible in this case),
Thus, B = 180° - 60° - 20.60° = 99.40°
Using the sine rule again:
3
80
b
=
sin 60° sin 99.40°
b=
from which,
80sin 99.40°
= 91.14 km
sin 60°
3
Since the cyclist travels 91.14 km in 5 hours then:
average speed =
dis tan ce 91.14
=
= 18.23 km/h
time
5
2
Total: 8
© 2006 John Bird. All rights reserved. Published by Elsevier.
29
Problem 7. A crank mechanism shown in Figure A4.1 comprises arm OP, of length 0.90 m,
which rotates anti-clockwise about a fixed point O, and connecting rod PQ of length 4.20 m.
End Q moves horizontally in a straight line OR. (a) If ∠POQ is initially zero, how far does
end Q travel in
1
revolution. (b) If ∠POQ is initially 40° find the angle between the
4
connecting rod and the horizontal and the length OQ. (c) Find the distance Q moves (correct
to the nearest cm) when ∠POR changes from 40° to 140°.
Figure A4.1
Marks
(a) When θ = 0, OQ = 4.20 + 0.90 = 5.10 m
When θ = 90°, OQ =
4.22 − 0.902 = 4.10 m by Pythagoras’s theorem.
Hence, Q moves 5.10 – 4.10 = 1.0 m
(b) Using the sine rule:
and
4
4.2
0.90
0.90sin 40°
=
from which, sin Q =
= 0.13774
sin 40° sin Q
4.20
Q = sin −1 (0.13774) = 7.92° or 7°55′′
3
Hence, angle P = 180° - 40° - 7.92° = 132.08°
Using the sine rule again gives:
from which,
(c) Using the sine rule:
OQ =
OQ
4.20
=
sin132.08° sin 40°
4.20sin132.08°
= 4.85 m
sin 40°
3
4.20
0.90
=
sin140° sin Q
© 2006 John Bird. All rights reserved. Published by Elsevier.
30
from which,
and
sin Q =
0.90sin140°
= 0.13774
4.20
Q = sin −1 (0.13774) = 7.92°
Hence, angle P = 180° - 140° - 7.92° = 32.08°
Using the sine rule again gives:
from which,
OQ =
2
OQ
4.20
=
sin 32.08° sin140°
4.20sin 32.08°
= 3.47 m
sin140°
2
Hence, Q moves from 4.85 m to 3.47 m, i.e. 1.38 m
2
Total: 16
Problem 8. Change the following Cartesian co-ordinates into polar co-ordinates, correct to 2
decimal places, in both degrees and in radians: (a) (-2.3, 5.4)
(b) (7.6, -9.2)
Marks
(a) From Figure 6, r =
⎛ 5.4 ⎞
2.32 + 5.42 = 5.87 and α = tan −1 ⎜
⎟ = 66.93°
⎝ 2.3 ⎠
Hence, θ = 180° - 66.93° = 113.07°
Thus, (-2.3, 5.4) = (5.87, 113.07°)
Figure 6
(b) From figure 7, r =
or (5.87, 1.97 rad)
5
Figure 7
⎛ 9.2 ⎞
7.62 + 9.22 = 11.93 and α = tan −1 ⎜
⎟ = 50.44°
⎝ 7.6 ⎠
Hence, θ = 360° - 50.44° = 309.56°
Thus, (7.6, -9.2) = (11.93, 309.56°)
or (11.93, 5.40 rad)
5
Total: 10
Problem 9. Change the following polar co-ordinates into Cartesian co-ordinates, correct to 3
decimal places: (a) (6.5, 132°)
(b) (3, 3 rad)
© 2006 John Bird. All rights reserved. Published by Elsevier.
31
Marks
(a) (6.5, 132°) = (6.5 cos 132°, 6.5 sin 132°) = (-4.349, 4.830)
3
(b) (3, 3 rad) = (3 cos 3, 3 sin 3) = (-2.970, 0.423)
3
Total: 6
Problem 10. (a) Convert 2.154 radians into degrees and minutes.
(b) Change 71°17′ into radians
Marks
180° ⎞
⎛
(a) 2.154 rad = ⎜ 2.154 ×
⎟ = 123.415° = 123°25′
π ⎠
⎝
(b) 71°17′ = 71
.
π
17
= 71.283 ×
rad = 1.244 rad
60
180
2
2
Total: 4
Problem 11. 140 mm of a belt drive is in contact with a pulley of diameter 180 mm, which is
turning at 300 revolutions per minute. Determine (a) the angle of lap, (b) the angular velocity
of the pulley, and (c) the linear velocity of the belt assuming no slipping occurs.
Marks
(a) With reference to Figure 8, arc length, s = r θ
from which,
angle of lap, θ =
.
s
140
=
= 1.5 5 or 89.13° or 89°8′
r 180 / 2
3
Figure 8
© 2006 John Bird. All rights reserved. Published by Elsevier.
32
⎛ 300 ⎞
(b) Angular velocity, ω = 2π ⎜
⎟ = 10π rad/s or 31.42 rad/s
⎝ 60 ⎠
3
rad ⎞
⎛
−3
(c) Linear velocity, v = ωr = ⎜10π
⎟ ( 90 ×10 m ) = 0.9π m/s or 2.83 m/s
s
⎝
⎠
3
Total: 9
Problem 12. Figure A4.2 shows a cross-section through a circular water container where the
shaded area represents the water in the container. Determine: (a) the depth, h, (b) the area of
the shaded portion, and (c) the area of the unshaded area.
Figure A4.2
Marks
(a) Since the triangular part is an equilateral triangle, from Figure 9,
OX = 122 − 62 = 10.39 cm from Pythagoras’ theorem.
Hence, the depth of water, h = 12 + 10.39 = 22.39 cm.
3
Figure 9
(b) Area of shaded portion = area of triangle + area of major sector
=
1
1
60 × π ⎞
⎛
(12)(10.39) + (12) 2 ⎜ 2π −
⎟
2
2
180 ⎠
⎝
= 62.34 + 376.99 = 439.33 cm 2
3
2
© 2006 John Bird. All rights reserved. Published by Elsevier.
33
(c) Area of unshaded portion = π (12 ) - 439.33 = 452.39 – 439.33 = 13.06 cm 2
2
3
Total: 11
Problem 13. Determine (a) the co-ordinates of the centre of the circle, and (b) the radius,
given the equation x 2 + y 2 − 2x + 6y + 6 = 0
Marks
x 2 + y 2 − 2x + 6y + 6 = 0 is of the form x 2 + y 2 + 2ex + 2fy + c = 0
(a) If (a, b) are the co-ordinates of the centre of the circle,
then
a=−
−2
=1
2
and
b= −
6
= -3
2
Hence, (1, -3) are the co-ordinates of the centre of the circle.
(b) Radius, r =
a 2 + b 2 − c = (1) 2 + (−3) 2 − 6 = 4 = 2
4
3
Total: 7
TOTAL ASSIGNMENT MARKS: 100
© 2006 John Bird. All rights reserved. Published by Elsevier.
34
ASSIGNMENT 5 (PAGE 189)
This assignment covers the material contained in chapters 15 to 18.
Problem 1. Solve the following equations in the range 0° to 360°
(a) sin −1 (-0.4161) = x
(b) cot −1 (2.4198) = θ
Marks
(a) If sin −1 (-0.4161) = x then from Figure 10, α = 24.59°
and x = 204.59° and 335.41°
Figure 10
4
Figure 11
⎛ 1 ⎞
(b) If cot −1 (2.4198) = θ, then tan −1 ⎜
⎟ = θ , and from Figure 11,
⎝ 2.4198 ⎠
α = 22.45°
and θ = 22.45° and 202.45°
4
Total: 8
Problem 2. Sketch the following curves labelling relevant points:
(a) y = 4 cos(θ + 45°)
(b) y = 5 sin(2t - 60°)
Marks
(a) y = 4 cos(θ + 45°) is sketched in Figure 12.
(b) y = 5 sin(2t - 60°) is sketched in Figure 13.
© 2006 John Bird. All rights reserved. Published by Elsevier.
35
Figure 12
4
Figure 13
4
Total: 8
Problem 3. The current in an alternating current circuit at any time t seconds is given by:
i = 120 (100πt + 0.274 ) amperes
Determine: (a) the amplitude, periodic time, frequency and phase angle (with reference to
120 sin 100πt), (b) the value of current when t = 0, (c) the value of current when t = 6 ms,
(d) the time when the current first reaches 80 A. Sketch one cycle of the oscillation.
Marks
1
(a) Amplitude = 120 A
ω = 100π, hence, periodic time, T =
2π
1
= s = 20 ms
100π 50
1
© 2006 John Bird. All rights reserved. Published by Elsevier.
36
Frequency, f = 50 Hz
1
Phase angle = 0.274 rad = 15.70° leading
1
(b) When t = 0, i = 120 sin 0.274 = 32.47 A
3
(c) When t = 6 ms, i = 120 sin (100π× 6 × 10−3 + 0.274 ) = 120 sin 2.1589556 = 99.84 A
3
(d) When i = 80 A, 80 = 120 sin (100πt + 0.274 )
from which,
80
= sin (100πt + 0.274 )
120
⎛ 80 ⎞
and arcsin ⎜
⎟ = 100πt + 0.274
⎝ 120 ⎠
i.e.
0.72972766 = (100πt + 0.274 )
Hence, time, t =
0.72972766 − 0.274
= 1.451 ms
100π
5
One cycle of the current waveform is shown in Figure 14.
4
Figure 14
Total: 19
Problem 4. A complex voltage waveform v is comprised of a 141.1 V r.m.s. fundamental
voltage at a frequency of 100 Hz, a 35% third harmonic component leading the fundamental
voltage at zero time by π/3 rad, and a 20% fifth harmonic component lagging the fundamental
at zero time by π/4 rad. (a) Write down an expression to represent voltage v.
(b) Draw the complex voltage waveform using harmonic synthesis over one cycle of the
fundamental waveform using scales of 12 cm for the time for one cycle horizontally and
1 cm = 20 V vertically.
© 2006 John Bird. All rights reserved. Published by Elsevier.
37
Marks
(a) Voltage v =
2 (141.1) sin ( 2π100t ) + 0.35 2 (141.1) sin ( 2 π300t + π / 3 ) + 0.2 2 (141.1) sin ( 2π500t − π / 4 )
i.e. v = 200 sin 200πt + 70 sin(600πt + π/3) + 40 sin (1000πt - π/4)
5
(b) The complex waveform is shown in Figure 15.
Time for 1 cycle, T = 1/f = 1/100 = 10 ms.
10
Figure 15
Total: 15
© 2006 John Bird. All rights reserved. Published by Elsevier.
38
Problem 5. Prove the following identities:
(a)
⎡1 − cos 2 θ ⎤
⎢ cos 2 θ ⎥ = tan θ
⎣
⎦
⎛ 3π
⎞
(b) cos ⎜ + φ ⎟ = sin φ
⎝ 2
⎠
(c)
sin 2 x
1
= tan 2 x
1 + cos 2x 2
Marks
(a) L.H.S. =
⎡1 − cos 2 θ ⎤
⎡ sin 2 θ ⎤
=
⎢ cos 2 θ ⎥
⎢ cos 2 θ ⎥
⎣
⎦
⎣
⎦
=
since cos 2 θ + sin 2 θ = 1
sin θ
= tan θ = R.H.S.
cos θ
3π
3π
⎛ 3π
⎞
(b) L.H.S. = cos ⎜ + φ ⎟ = cos cos φ − sin sin φ
2
2
⎝ 2
⎠
3
from compound angles
= 0 – (-1) sin φ = sin φ = R.H.S.
(c) L.H.S. =
sin 2 x
sin 2 x
sin 2 x
1 ⎛ sin x ⎞
=
=
= ⎜
⎟
2
2
1 + cos 2x 1 + ( 2 cos x − 1) 2 cos x 2 ⎝ cos x ⎠
=
3
2
1
1
2
( tan x ) = tan 2 x = R.H.S.
2
2
3
Total: 9
Problem 6. Solve the following trigonometric equations in the range 0° ≤ x ≤ 360°:
(a) 4 cos x + 1 = 0
(b) 3.25 cosec x = 5.25
(c) 5sin 2 x + 3sin x = 4
(d) 2sec 2 θ + 5 tan θ = 3
Marks
(a) Since 4 cos x + 1 = 0 then cos x = −
1
4
and x = cos −1 (-0.25)
i.e. x = 104.48° (or 104°29′) and 255.52° (or 255°31′)
(b) Since 3.25 cosec x = 5.25
then cosec x =
5.25
3.25
and sin x =
3.25
5.25
⎛ 3.25 ⎞
i.e. x = sin −1 ⎜
⎟ = 38.25° (or 38°15′) and 141.75° (or 141°45′)
⎝ 5.25 ⎠
(c) Since 5sin 2 x + 3sin x = 4 then
and sin x =
4
4
5sin 2 x + 3sin x − 4 = 0
−3 ± 32 − 4(5)(−4) −3 ± 89
=
= 0.6434 or -1.2434
2(5)
10
Ignoring the latter, sin x = 0.6434 and x = sin −1 0.6434 = 40.05° or 139.95°
5
© 2006 John Bird. All rights reserved. Published by Elsevier.
39
(d) Since 2sec 2 θ + 5 tan θ = 3
Hence,
then
2 tan 2 θ + 5 tan θ − 1 = 0
2 (1 + tan 2 θ ) + 5 tan θ = 3
from which,
−5 ± 52 − 4(2)(−1) −5 ± 33
=
tan θ =
= 0.18614066 or -2.68614066
2(2)
4
and θ = 10.54°, 190.54°, 110.42° and 290.42°
5
Total: 18
Problem 7. Solve the equation: 5sin ( θ − π / 6 ) = 8cos θ for values of 0 ≤ θ ≤ 2π.
Marks
5sin ( θ − π / 6 ) = 8cos θ
i.e. 5[sin θ cos π/6 – cos θ sin π/6] = 8 cos θ
2
4.33 sin θ - 2.5 cos θ = 8 cos θ
1
Thus,
4.33 sin θ = 10.5 cos θ
and
Hence,
sin θ 10.5
=
= 2.42494226
cos θ 4.33
2
i.e.
tan θ = 2.42494226
1
θ = tan −1 (2.42494226) = 67.59° and 247.59°
and
2
Total: 8
Problem 8. Express 5.3 cos t – 7.2 sin t in the form R sin(t + α). Hence solve the equation:
5.3 cos t – 7.2 sin t = 4.5 in the range 0 ≤ t ≤ 2π.
Marks
Let 5.3 cos t – 7.2 sin t = R sin(t + α) = R[sin t cos α + cos t sin α]
= (R cos α) sin t + (R sin α) cos t
Hence 5.3 = R sin α
and
-7.2 = R cos α
i.e. sin α =
5.3
R
i.e. cos α = −
7.2
R
© 2006 John Bird. All rights reserved. Published by Elsevier.
40
Figure 16
There is only one quadrant where sine is positive and cosine is negative, i.e. the
second, as shown in Figure 16.
R = 5.32 + 7.22 = 8.94
hence,
and
⎛ 5.3 ⎞
φ = tan −1 ⎜
⎟ = 0.6346 rad
⎝ 7.2 ⎠
α = π - 0.6346 = 2.507 rad.
6
Thus, 5.3 cos t – 7.2 sin t = 8.94 sin(t + 2.507)
If
and
5.3 cos t – 7.2 sin t = 4.5
sin(t + 2.507) =
then
8.94 sin(t + 2.507) = 4.5
4.5
= 0.50336
8.94
t + 2.507 = sin −1 (0.50336) = 0.5275 rad or 2.6141 rad
and
t = 0.5275 – 2.507 = -1.9795 ≡ -1.9795 + 2π = 4.304 s
or
t = 2.6141 – 2.507 = 0.107 s
6
Total: 12
Problem 9. Determine
∫ 2 cos 3t sin t dt
Marks
⎧1
⎫
2 cos 3t sin t = 2 ⎨ [sin(3t + t) − sin(3t − t)]⎬ = sin 4t − sin 2t
⎩2
⎭
Hence
∫ 2 cos 3t sin t dt = ∫ sin 4t − sin 2t dt
1
1
= − cos 4t + cos 2t + c
4
2
Total: 3
TOTAL ASSIGNMENT MARKS: 100
© 2006 John Bird. All rights reserved. Published by Elsevier.
41
ASSIGNMENT 6 (PAGE 247)
This assignment covers the material contained in chapters 19 to 22.
Problem 1. Sketch the following graphs, showing the relevant points:
(a) y = ( x − 2 )
2
(d) 9x 2 − 4y 2 = 36
(c) x 2 + y 2 − 2x + 4y − 4 = 0
(b) y = 3 – cos 2x
⎧
⎪ −1
⎪
⎪
(e) f (x) = ⎨ x
⎪
⎪
⎪ 1
⎩
π
2
π
π
− ≤x≤
2
2
π
≤x≤π
2
−π≤ x≤−
Marks
(a) A graph of y = ( x − 2 ) is shown in Figure 17.
2
Figure 17
3
(b) A graph of y = 3 – cos 2x is shown in Figure 18.
Figure 18
42
3
© 2006 John Bird. All rights reserved. Published by Elsevier.
(c) x 2 + y 2 − 2x + 4y − 4 = 0 is a circle. If the co-ordinates of the circle are (a, b) then
a=-
−2
4
= 1 and b = - = -2
2
2
and radius, r =
a 2 + b 2 − (−4) = 12 + ( −2 ) − ( −4 ) = 3
2
The circle is shown in Figure 19.
3
Figure 19
(d) 9x 2 − 4y 2 = 36 is equivalent to
9x 2 4y 2
x 2 y2
−
= 1 i.e. 2 − 2 = 1 which is a
36
36
2
3
hyperbola as shown in Figure 20.
3
Figure 20
⎧
⎪ −1
⎪
⎪
(e) f (x) = ⎨ x
⎪
⎪
⎪ 1
⎩
π
2
π
π
− ≤x≤
is shown in Figure 21.
2
2
π
≤x≤π
2
−π≤ x≤−
© 2006 John Bird. All rights reserved. Published by Elsevier.
43
3
Figure 21
Total: 15
Problem 2. Determine the inverse of f (x) = 3x + 1
y −1
Let f (x) = y then y = 3x + 1 and transposing for x gives: x =
3
and interchanging x and y gives: y =
Marks
x −1
3
Hence, the inverse of f (x) = 3x + 1 is: f −1 (x) =
x −1
3
3
Total: 3
Problem 3. Evaluate, correct to 3 decimal places:
2 tan −1 1.64 + sec −1 2.43 – 3 cos ec −1 3.85
Marks
2 tan −1 1.64 + sec −1 2.43 – 3 cos ec −1 3.85
⎛ 1 ⎞
−1 ⎛ 1 ⎞
= 2 tan −1 1.64 + cos −1 ⎜
⎟ − 3sin ⎜
⎟
⎝ 2.43 ⎠
⎝ 3.85 ⎠
= 2(1.02323409 …) + 1.14667223… - 3(0.26275322…) = 2.405
3
Total: 3
Problem 4. Determine the asymptotes for the following function and hence sketch the curve:
y=
44
( x − 1)( x + 4 )
( x − 2 )( x − 5)
© 2006 John Bird. All rights reserved. Published by Elsevier.
Rearranging y =
( x − 1)( x + 4 )
( x − 2 )( x − 5)
Marks
gives:
y ( x 2 − 7x + 10 ) = x 2 + 3x − 4
x 2 ( y − 1) + x ( −7y − 3) + 10y + 4 = 0
and
For asymptotes parallel to the x-axis:
For asymptotes parallel to the y-axis:
y – 1 = 0 i.e. y = 1
x 2 − 7x + 10 = 0
(x – 2)(x – 5) = 0 from which, x = 2 and x = 5
i.e.
2
2
To check for other asymptotes, let y = mx + c
(mx + c)( x 2 − 7x + 10 ) = x 2 + 3x − 4
then
i.e.
mx 3 − 7mx 2 + 10mx + cx 2 − 7cx + 10c − x 2 − 3x + 4 = 0
and
mx 3 + (c − 1 − 7m)x 2 − 7mx + 10mx − 7cx + 10c − 3x + 4 = 0
Hence
m = 0 and c – 1 – 7m = 0 from which, c = 1
Thus, y = mx + c, i.e. y = 0x + 1 i.e. y = 1 is an asymptote, as determined above.
A sketch of y =
Figure 22
( x − 1)( x + 4 )
( x − 2 )( x − 5)
is shown in Figure 22.
4
Total: 8
© 2006 John Bird. All rights reserved. Published by Elsevier.
45
Problem 5. Plot a graph of y = 3x 2 + 5 from x = 1 to x = 4. Estimate, correct to 2 decimal places,
using 6 intervals, the area enclosed by the curve, the ordinates x = 1 and x = 4, and the x-axis by (a)
the trapezoidal rule, (b) the mid-ordinate rule, and (c) Simpson’s rule.
Marks
A table of values is shown below and a graph plotted as shown in Figure 23.
X
1.0
1.5
2.0
2.5
3.0
3.5
4.0
y = 3x 2 + 5
8.0 11.75 17.0 23.75 32.0 41.75 53.0
Figure 23
3
(a) Since 6 intervals are used, ordinates lie at 1, 1.5, 2, 2.5, …
By the trapezoidal rule,
⎧1
⎫
shaded area = ( 0.5 ) ⎨ ( 8.0 + 53.0 ) + 11.75 + 17.0 + 23.75 + 32.0 + 41.75⎬
⎩2
⎭
= 78.38 square units
3
(b) With the mid-ordinate rule, ordinates occur at 1.25, 1.75, 2.25, 2.75, 3.25 and 3.75
x
1.25
1.75
2.25
2.75
3.25
3.75
y = 3x 2 + 5 9.6875 14.1875 20.1875 27.6875 36.6875 47.1875
By the mid-ordinate rule,
shaded area = ( 0.5 ){9.6875 + 14.1875 + 20.1875 + 27.6875 + 36.6875 + 47.1875}
= 77.81 square units
46
3
© 2006 John Bird. All rights reserved. Published by Elsevier.
(c) By Simpson’s rule,
shaded area ≈
=
1
( 0.5) {(8.0 + 53.0 ) + 4 (11.75 + 23.75 + 41.75) + 2 (17.0 + 32.0 )}
3
1
( 0.5){61 + 309 + 98} = 78 square units
3
3
Total: 12
Problem 6. A circular cooling tower is 20 m high. The inside diameter of the tower at different
heights is given in the following table:
Height (m)
0
Diameter (m) 16.0
5.0
10.0
15.0
20.0
13.3
10.7
8.6
8.0
Determine the area corresponding to each diameter and hence estimate the capacity of the tower in
cubic metres.
Marks
A table showing the areas corresponding to height is shown below:
Height (m)
0
5.0 10.0 15.0 20.0
2
⎛ πd 2 ⎞
Area ⎜ =
m ⎟ 201.06 138.93 89.92 58.09 50.27
⎝ 4
⎠
3
Using Simpson’s rule,
capacity of tower ≈
=
1
( 5.0 ) {( 201.06 + 50.27 ) + 4 (138.93 + 58.09 ) + 2 (89.92 )}
3
1
( 5.0 ){251.33 + 788.08 + 179.84} = 2032 m 3
3
3
Total: 6
Problem 7. A vehicle starts from rest and its velocity is measured every second for 6 seconds, with
the following results: Time t (s)
0
1
2
3
4
5
6
Velocity v (m/s) 0 1.2 2.4 3.7 5.2 6.0 9.2
Using Simpson’s rule, calculate (a) the distance travelled in 6 s (i.e. the area under the v/t graph)
and (b) the average speed over this period.
© 2006 John Bird. All rights reserved. Published by Elsevier.
47
Marks
(a) Distance travelled =
=
(b) Average speed =
1
(1) {( 0 + 9.2 ) + 4 (1.2 + 3.7 + 6.0 ) + 2 ( 2.4 + 5.2 )}
3
1
{9.2 + 43.6 + 15.2} = 22.67 m
3
22.67 m
= 3.78 m/s
6s
4
2
Total: 6
Problem 8. Four coplanar forces act at a point as shown in Figure A6.1. Determine the value and
direction of the resultant force by (a) drawing (b) by calculation.
Figure A6.1
Marks
(a) From Figure 24, by drawing, resultant, R = 8.7 N and θ = 230°.
Figure 24
5
(b) By calculation:
Total horizontal component,
H = 4 cos 90° + 5 cos 180° + 8 cos 225° + 7 cos 315° = -5.7071
Total vertical component,
V = 4 sin 90° + 5 sin 180° + 8 sin 225° + 7 sin 315° = -6.6066
Hence, resultant, R =
48
( −5.7071) + ( −6.6066 )
2
2
= 8.73 N
© 2006 John Bird. All rights reserved. Published by Elsevier.
θ = tan −1
and
V
⎛ −6.6066 ⎞
= tan −1 ⎜
⎟ = 229.18°
H
⎝ −5.7071 ⎠
(or use complex numbers)
5
Total: 10
Problem 9. The instantaneous values of two alternating voltages are given by:
π⎞
⎛
v1 = 150sin ⎜ ωt + ⎟ volts
3⎠
⎝
and
π⎞
⎛
v 2 = 90sin ⎜ ωt − ⎟ volts
6⎠
⎝
Plot the two voltages on the same axes to scales of 1 cm = 50 volts and 1 cm =
π
rad. Obtain a
6
sinusoidal expression for the resultant v1 + v 2 in the form R sin(ωt + α): (a) by adding ordinates at
intervals and (b) by calculation.
Marks
(a) From Figure 25, by adding ordinates at intervals, the waveform of v1 + v 2 is seen
to have a maximum value of 175 V and is leading by 30° or
Hence,
v1 + v 2 = 175sin ( ωt + 0.52 ) volts
π
rad i.e. 0.52 rad.
6
7
Figure 25
(b) By calculation:
At time t = 0, the phasors v1 and v 2 are shown in Figure 26.
© 2006 John Bird. All rights reserved. Published by Elsevier.
49
Total horizontal component, H = 150 cos 60° + 90 cos-30° = 152.942
Total vertical component, V = 150 sin 60° + 90 sin-30° = 84.904
Resultant, v1 + v 2 = 152.9422 + 84.9042 = 174.93 volts
97
Figure 26
Direction of v1 + v 2 = tan −1
84.904
= 29.04° or 0.507 rad.
152.942
Hence, v1 + v 2 = 174.93sin ( ωt + 0.507 ) volts
6
Total: 13
Problem 10. If a = 2i + 4j – 5k and b = 3i – 2j + 6k determine:
(ii) a + b
(i) ab
(iii) a × b
(iv) the angle between a and b
Marks
(i)
ab = (2)(3) + (4)(-2) + (-5)(6) = 6 – 8 – 30 = -32
3
(ii)
a + b = 5i + 2 j + k =
3
i
j
52 + 22 + 12 =
30 or 5.477
k
(iii) a × b = 2 4 −5 = i (24 – 10) – j (12 + 15) + k (-4 –12) = 14i – 27j – 16k
3 −2 6
(iv) cos θ =
4
a b
−32
−32
=
=
= -0.68147
2
2
2
2
2
2
a b
45 49
2 + 4 + (−5) 3 + (−2) + 6
Hence, θ = arccos(-0.68147) = 132.96°
4
Total: 14
50
© 2006 John Bird. All rights reserved. Published by Elsevier.
Problem 11. Determine the work done by a force of F newtons acting at a point A on a body, when
A is displaced to point B, the co-ordinates of A and B being (2, 5, -3) and (1, -3, 0) metres
respectively, and when F = 2i – 5j + 4k newtons.
Marks
Work done = F d , where d = (i – 3j) – (2i + 5j – 3k) = -i – 8j + 3k
Hence, work done = (2)(-1) + (-5)(-8) + (4)(3) = -2 + 40 + 12 = 50 Nm or 50 J
4
Total: 4
Problem 12. A force F = 3i – 4j + k newtons act on a line passing through a point P. Determine
moment M and its magnitude of the force F about a point Q when P has co-ordinates (4, -1, 5)
metres and Q has co-ordinates (4, 0, -3) metres.
Marks
Moment, M = r × F where r = (4i – j + 5k) – (4i – 3k) = -j + 8k
Hence, moment M = (-j + 8k) × (3i – 4j + k)
i
j
k
= 0 −1 8 = i(-1 +32) – j(-24) + k(3) = 31i + 24j + 3k Nm
3 −4 1
Magnitude of M, M = r × F =
( r r )( FF ) − ( r F )
3
2
where r r = (0)(0) + (-1)(-1) + (8)(8) = 65
F F = (3)(3) + (-4)(-4) + (1)(1) = 26
r F = (0)(3) + (-1)(-4) + (8)(1) = 12
Hence, M =
( 65)( 26 ) − (12 )
2
= 1546 = 39.32 Nm
3
Total: 6
TOTAL ASSIGNMENT MARKS: 100
© 2006 John Bird. All rights reserved. Published by Elsevier.
51
ASSIGNMENT 7 (PAGE 286)
This assignment covers the material contained in chapters 23 to 26.
Problem 1. Solve the quadratic equation x 2 − 2x + 5 = 0 and show the roots on an Argand diagram.
Marks
2
− − 2 ± ⎡( −2 ) − 4 (1)( 5 ) ⎤ 2 ± −16 2 ± j4
⎣
⎦
=
=
x=
= 1 ± j2
2 (1)
2
2
5
The two roots are shown on the Argand diagram in Figure 27.
4
Figure 27
Total: 9
Problem 2. If Z1 = 2 + j5 , Z2 = 1 − j3 and Z3 = 4 − j determine, in both Cartesian and polar forms,
the value of
Z1 Z2
+ Z3 , correct to 2 decimal places.
Z1 + Z2
Marks
( 2 + j5 )(1 − j3) = 2 − j6 + j5 − j215 = (17 − j) = (17 − j) × (3 − j2)
Z1 Z2
=
Z1 + Z2 ( 2 + j5 ) + (1 − j3)
(3 + j2)
(3 + j2) (3 + j2) (3 − j2)
=
Hence,
51 − j34 − j3 + j2 2 49 − j37
=
= 3.77 − j2.85
32 + 22
13
Z1 Z2
+ Z3 = (3.77 – j2.85) + (4 – j) = 7.77 – j3.85
Z1 + Z2
or 8.67∠ − 26.36°
5
2
2
Total: 9
52
© 2006 John Bird. All rights reserved. Published by Elsevier.
Problem 3. Three vectors are represented by A, 4.2∠45° , B, 5.5∠ − 32° and C, 2.8∠75° .
Determine in polar form the resultant D, where D = B + C – A.
Marks
D = B + C – A = 5.5∠ − 32° + 2.8∠75° - 4.2∠45°
= (4.664 – j2.915) + (0.725 + j2.705) – (2.970 + j2.970)
5
= (2.419 – j3.180) = 4.0∠-52.74°
3
Total: 8
Problem 4. Two impedances, Z1 = ( 2 + j7 ) ohms and Z2 = ( 3 − j4 ) ohms are connected in series to a
supply voltage V of 150∠0° V. Determine the magnitude of the current I and its phase angle
relative to the voltage.
Marks
Current, I =
V
150∠0°
150∠0°
150∠0°
= 25.72∠-30.96°
=
=
=
Z1 + Z2 (2 + j7) + (3 − j4)
5 + j3
34∠30.96°
Hence, the current magnitude is 25.72 A and its phase angle is –30.96°
6
Total: 6
Problem 5. Determine in both polar and rectangular forms:
(a) [ 2.37∠35°]
4
(b) [3.2 − j4.8]
5
(c)
[ −1 − j3]
Marks
(a) [ 2.37∠35°] = 2.37 4 ∠ ( 4 × 35° ) = 31.55∠140° = -24.17 + j20.28
4
4
(b) [3.2 − j4.8] = [5.769∠ − 56.31°] = 5.7695 ∠ ( 5 × −56.31° )
5
5
= 6390∠-281.55° = 6390∠78.45° = 1279 + j6261
6
1
(c)
1
1
2
⎡
⎤
[ −1 − j3] = ⎣ 10∠ − 108.435°⎦ = 10 2 ∠ ⎛⎜ × −108.435° ⎞⎟
⎝2
⎠
= 1.778∠-54.22° and 1.778∠125.78° = ±(1.04 – j1.44)
5
Total: 15
© 2006 John Bird. All rights reserved. Published by Elsevier.
53
⎛ −5 2 ⎞ ⎛ 1 6 ⎞
Problem 6. Determine ⎜
⎟×⎜
⎟
⎝ 7 −8 ⎠ ⎝ −3 −4 ⎠
Marks
⎛ −5 2 ⎞ ⎛ 1 6 ⎞ ⎛ −11 −38 ⎞
⎜
⎟×⎜
⎟ =⎜
⎟
⎝ 7 −8 ⎠ ⎝ −3 −4 ⎠ ⎝ 31 74 ⎠
4
Total: 4
j3
(1 + j2) ⎞
⎛
Problem 7. Calculate the determinant of ⎜
⎟
− j2 ⎠
⎝ (−1 − j4)
Marks
j3
(1 + j2)
= − j2 6 − (1 + j2 )( −1 − j4 ) = 6 – [-1 – j4 – j2 - j2 8 ]
− j2
(−1 − j4)
= 6 – [ 7 – j6 ] = -1 + j6
4
Total: 4
⎛ −5 2 ⎞
Problem 8. Determine the inverse of ⎜
⎟
⎝ 7 −8 ⎠
Marks
⎛ −5 2 ⎞
1 ⎛
−1
If B = ⎜
⎟ then B =
⎜
40 − 14 ⎝
⎝ 7 −8 ⎠
⎛ 4
⎜ − 13
−8 −2 ⎞
1 ⎛ − 8 −2 ⎞
⎟ =
⎜
⎟ or ⎜
− 7 −5 ⎠ 26 ⎝ −7 −5 ⎠
⎜− 7
⎜
⎝ 26
1 ⎞
13 ⎟
⎟
5 ⎟
− ⎟
26 ⎠
−
4
Total: 4
⎛ −1 3 0 ⎞ ⎛ 2 −1 3 ⎞
Problem 9. Determine ⎜⎜ 4 −9 2 ⎟⎟ × ⎜⎜ −5 1 0 ⎟⎟
⎜ −5 7 1 ⎟ ⎜ 4 −6 2 ⎟
⎝
⎠ ⎝
⎠
Marks
4
−3 ⎞
⎛ −1 3 0 ⎞ ⎛ 2 −1 3 ⎞ ⎛ −17
⎟
⎜
⎟ ⎜
⎟ ⎜
⎜ 4 −9 2 ⎟ × ⎜ −5 1 0 ⎟ = ⎜ 61 −25 16 ⎟
⎜ −5 7 1 ⎟ ⎜ 4 −6 2 ⎟ ⎜ −41
6 −13 ⎟⎠
⎝
⎠ ⎝
⎠ ⎝
9
Total: 9
54
© 2006 John Bird. All rights reserved. Published by Elsevier.
⎛ 2 −1 3 ⎞
⎜
⎟
Problem 10. Calculate the determinant of ⎜ −5 1 0 ⎟
⎜ 4 −6 2 ⎟
⎝
⎠
Marks
2 −1 3
−5
1
4
−6
0 =3
2
−5
1
4
−6
+2
2
−1
−5
1
using the third column
= 3(30 – 4) + 2(2 – 5) = 3(26) + 2(-3) = 78 – 6 = 72
3
3
Total: 6
Problem 11. Solve the following simultaneous equations:
4x – 3y = 17
x+y+1=0
using matrices.
Marks
Since
4x – 3y = 17
x + y = -1
then
⎛ 4 −3 ⎞⎛ x ⎞ ⎛ 17 ⎞
⎜
⎟⎜ ⎟ = ⎜ ⎟
⎝ 1 1 ⎠⎝ y ⎠ ⎝ −1⎠
1
⎛ 4 −3 ⎞
1 ⎛ 1 3 ⎞ 1⎛ 1 3 ⎞
The inverse of ⎜
⎜
⎟= ⎜
⎟
⎟ is
4 − −3 ⎝ −1 4 ⎠ 7 ⎝ −1 4 ⎠
⎝1 1⎠
Hence
2
1 ⎛ 1 3 ⎞⎛ 4 −3 ⎞⎛ x ⎞ 1 ⎛ 1 3 ⎞⎛ 17 ⎞
⎜
⎟⎜
⎟⎜ ⎟ = ⎜
⎟⎜ ⎟
7 ⎝ −1 4 ⎠⎝ 1 1 ⎠⎝ y ⎠ 7 ⎝ −1 4 ⎠⎝ −1⎠
⎛ 1 0 ⎞⎛ x ⎞ 1 ⎛ 14 ⎞
⎜
⎟⎜ ⎟ = ⎜
⎟
⎝ 0 1 ⎠⎝ y ⎠ 7 ⎝ −21⎠
⎛ y⎞ ⎛ 2 ⎞
⎜ ⎟ = ⎜ ⎟ i.e. x = 2 and y = -3
⎝ y ⎠ ⎝ −3 ⎠
and
3
Total: 6
Problem 12. Use determinants to solve the following simultaneous equations:
4x + 9y + 2z = 21
-8x + 6y – 3z = 41
3x + y – 5z = -73
© 2006 John Bird. All rights reserved. Published by Elsevier.
55
Marks
4x + 9y + 2z – 21 = 0
-8x + 6y – 3z – 41 = 0
3x + y – 5z + 73 = 0
Hence
−y
x
z
=
=
=
9
2 −21
4 2 −21
4 9 −21
4
6 −3 −41
−8 −3 −41
−8 6 −41
−8
1 −5 73
3 −5 73
3 1 73
3
−1
9 2
6 −3
1 −5
4
x
−y
z
=
=
9(−424) − 2(479) − 21(−27) 4(−424) − 2(−461) − 21(49) 4(479) − 9(−461) − 21(−26)
=
x
−y
z
−1
=
=
=
−4207 −1803 6611 −601
i.e.
Hence,
−1
4(−27) − 9(49) + 2(−26)
x=
−4207
= -7
601
y=
3
1803
6611
= 3 and z =
= 11
601
601
3
(or use Cramer’s rule or Gaussian elimination).
Total: 10
Problem 13. The simultaneous equations representing the currents flowing in an unbalanced,
three-phase, star-connected, electrical network are as follows:
2.4 I1 + 3.6 I 2 + 4.8 I3 = 1.2
−3.9 I1 + 1.3I 2 − 6.5 I3 = 2.6
1.7 I1 + 11.9 I 2 + 8.5 I3 = 0
Using matrices, solve the equations for I1 , I 2 and I3 .
Marks
⎛ 2.4 3.6
⎜
⎜ −3.9 1.3
⎜ 1.7 11.9
⎝
4.8 ⎞ ⎛ I1 ⎞ ⎛ 1.2 ⎞
⎟⎜ ⎟ ⎜
⎟
−6.5 ⎟ ⎜ I 2 ⎟ = ⎜ 2.6 ⎟
8.5 ⎟⎠ ⎜⎝ I3 ⎟⎠ ⎜⎝ 0 ⎟⎠
The inverse of the 3 by 3 matrix is:
26.52 −29.64 ⎞
⎛ 88.4
1
⎜
⎟
22.1
12.24
−3.12 ⎟
2.4(88.4) − 3.6(−22.1) + 4.8(−48.62) ⎜⎜
⎟
⎝ −48.62 −22.44 17.16 ⎠
56
4
© 2006 John Bird. All rights reserved. Published by Elsevier.
Hence
26.52 −29.64 ⎞ ⎛ 1.2 ⎞ ⎛ 3 ⎞
⎛ 1 0 0 ⎞ ⎛ I1 ⎞
⎛ 88.4
1 ⎜
⎜
⎟⎜ ⎟
⎟⎜ ⎟ ⎜ ⎟
12.24
−3.12 ⎟ ⎜ 2.6 ⎟ = ⎜ 1 ⎟
⎜ 0 1 0 ⎟ ⎜ I 2 ⎟ = 58.344 ⎜ 22.1
⎜
⎟⎜ ⎟
⎜
⎟⎜ ⎟ ⎜ ⎟
⎝ 0 0 1 ⎠ ⎝ I3 ⎠
⎝ −48.62 −22.44 17.16 ⎠ ⎝ 0 ⎠ ⎝ −2 ⎠
i.e.
I1 = 3, I 2 = 1 and I 3 = −2
6
Total: 10
TOTAL ASSIGNMENT MARKS: 100
© 2006 John Bird. All rights reserved. Published by Elsevier.
57
ASSIGNMENT 8 (PAGE 329)
This assignment covers the material contained in chapters 27 to 31.
Problem 1. Differentiate the following with respect to the variable:
(a) y = 5 + 2 x 3 −
1
x2
(b) s = 4 e 2 θ sin 3θ
(c) y =
3ln 5t
cos 2t
(d) x =
2
(t
2
− 3t + 5 )
Marks
(a) y = 5 + 2 x 3 −
1
= 5 + 2x 3/ 2 − x −2
2
x
dy
2
⎛3
⎞
= 0 + ( 2 ) ⎜ x1/ 2 ⎟ − ( −2x −3 ) = 3 x + 3
dx
x
⎝2
⎠
(b) s = 4 e 2 θ sin 3θ
3
i.e. a product
ds
= ( 4 e 2θ ) ( 3cos 3θ ) + ( sin 3θ ) ( 8e 2θ ) = 4e 2 θ ( 3cos 3θ + 2sin 3θ )
dθ
(c) y =
3ln 5t
cos 2t
dy
=
dt
(d) x =
3
i.e. a quotient
( cos 2t ) ⎛⎜
3⎞
3
cos 2t + 6 ln 5t sin 2t
⎟ − ( 3ln 5t )( −2sin 2t )
⎝t⎠
t
=
2
cos 2 2t
( cos 2t )
2
(t
2
− 3t + 5 )
Let u = t 2 − 3t + 5 then
Hence, x =
dx
dx du ⎛ 1 ⎞
×
= ⎜−
=
⎟ ( 2t − 3) =
du dt ⎝
dt
u3 ⎠
3
du
= 2t – 3
dt
2
dx
1
= 2u −1/ 2 and
= −u −3/ 2 = −
du
u
u3
3 − 2t
(t
2
− 3t + 5
)
3
4
Total: 13
Problem 2. If f(x) = 2.5x 2 − 6x + 2 find the co-ordinates at the point at which the gradient is –1.
Marks
f(x) = 2.5x 2 − 6x + 2
Gradient = f ′(x) = 5x – 6 = -1 from which, 5x = 5 and x = 1
3
When x = 1, f (x) = f (1) = 2.5(1) 2 − 6(1) + 2 = -1.5
© 2006 John Bird. All rights reserved. Published by Elsevier.
58
Hence, the gradient is –1 at the point (1, -1.5)
2
Total: 5
Problem 3. The displacement s cm of the end of a stiff spring at time t seconds is given by:
s = a e − kt sin 2πft . Determine the velocity and acceleration of the end of the spring after 2 seconds if
a = 3, k = 0.75 and f = 20.
Marks
s = a e − kt sin 2πft
Velocity =
i.e. a product
ds
= ( a e − kt ) ( 2πf cos 2πft ) + ( sin 2πft ) ( − k a e − kt )
dt
2
When t = 2, a = 3, k = 0.75 and f = 20,
velocity = ( 3e − (0.75)(2) ) ( 2π× 20 cos 2π(20)(2) ) − ( sin 2π(20)(2) ) ( (0.75) 3e− (0.75)(2) )
= 120π e
Acceleration =
−( 0.75) (2)
− 0 = 84.12 cm/s
2
d 2s
2
= ( a e− kt ) ⎡ − ( 2πf ) sin 2πft ⎤ + ( 2πf cos 2πft ) ( − k a e − kt )
2
⎣
⎦
dt
+ ( sin 2πft ) ( k 2 a e − kt ) + ( −k a e− kt ) ( 2πf cos 2πft )
3
When t = 2, a = 3, k = 0.75 and f = 20,
acceleration = 0 + ( 40π ) ( −2.25e−1.5 ) + 0 + ( −2.25e−1.5 ) ( 40π ) = −180π e−1.5
= -126.2 cm / s 2
3
Total: 10
Problem 4. Find the co-ordinates of the turning points on the curve y = 3x 3 + 6x 2 + 3x − 1 and
distinguish between them.
Marks
dy
= 9x 2 + 12x + 3 = 0 for a turning point
dx
Since y = 3x 3 + 6x 2 + 3x − 1 then
= (3x + 3)(3x + 1)
from which,
x = -1 or x = −
1
3
3
When x = -1, y = 3(−1)3 + 6(−1)2 + 3(−1) − 1 = -3 + 6 – 3 – 1 = -1
© 2006 John Bird. All rights reserved. Published by Elsevier.
59
3
2
1
1 2
4
⎛ 1⎞
⎛ 1⎞
⎛ 1⎞
When x = − , y = 3 ⎜ − ⎟ + 6 ⎜ − ⎟ + 3 ⎜ − ⎟ − 1 = − + − 1 − 1 = −1
3
9 3
9
⎝ 3⎠
⎝ 3⎠
⎝ 3⎠
4⎞
⎛ 1
Hence, turning points occur at (-1, -1) and ⎜ − , −1 ⎟
9⎠
⎝ 3
2
d2 y
= 18x + 12
dx 2
When x = -1,
d2 y
is negative, hence, (-1, -1) is a maximum point.
dx 2
1 d2 y
When x = − ,
is positive, hence,
3 dx 2
1
4⎞
⎛ 1
⎜ − 3 , −1 9 ⎟ is a minimum point.
⎝
⎠
1
Total: 7
Problem 5. The heat capacity C of a gas varies with absolute temperature θ as shown:
C = 26.50 + 7.20 × 10−3 θ − 1.20 × 10−6 θ2
Determine the maximum value of C and the temperature at which it occurs.
Marks
dC
= 7.20 × 10−3 − 2.40 × 10−6 θ = 0 for a maximum or minimum value,
dθ
7.20 × 10−3
= 3000
from which, θ =
2.40 ×10−6
2
d 2C
= −2.40 ×10−6 , which is negative, and hence θ = 3000 gives a maximum value.
2
dθ
Cmax = 26.50 + ( 7.20 ×10−3 ) ( 3000 ) − (1.20 × 10−6 ) ( 3000 )
2
= 26.50 + 21.6 – 10.8 = 37.3
3
Hence, the maximum value of C is 37.3, which occurs at a temperature of 3000.
Total: 5
Problem 6. Determine for the curve y = 2x 2 − 3x at the point (2, 2): (a) the equation of the
tangent, (b) the equation of the normal.
Marks
(a) Gradient, m =
dy
= 4x - 3
dx
At the point (2, 2), x = 2 and m = 4(2) – 3 = 5
2
© 2006 John Bird. All rights reserved. Published by Elsevier.
60
Hence, equation of tangent is: y − y1 = m ( x − x1 )
i.e.
y – 2 = 5(x – 2)
i.e.
y – 2 = 5x - 10
y = 5x – 8
or
(b) Equation of normal is:
y − y1 = −
1
( x − x1 )
m
1
( x − 2)
5
i.e.
y−2= −
i.e.
1
2
y−2= − x+
5
5
or
2
5y – 10 = -x + 2
5y + x = 12
or
2
Total: 6
Problem 7. A rectangular block of metal with a square cross-section has a total surface area of 250
cm 2 . Find the maximum volume of the block of metal.
Marks
The rectangular block is shown in Figure 28 having dimensions x by x by y.
Surface area, A = 2x 2 + 4xy = 250
(1)
Volume, V = x 2 y
Figure 28
From equation (1), 4xy = 250 − 2x 2 and y =
250 − 2x 2
4x
⎛ 250 − 2x 2 ⎞
1 3
Hence, V = x 2 ⎜
⎟ = 62.5x − x
4x
2
⎝
⎠
and
3
dV
3
= 62.5 − x 2 = 0 for a maximum or minimum value,
dx
2
© 2006 John Bird. All rights reserved. Published by Elsevier.
61
i.e. 62.5 =
3 2
x from which, x =
2
2(62.5)
= ±6.455cm
3
d2V
d2V
=
−
3x
and
when
x
=
+6.455,
is negative, indicating a maximum value.
dx 2
dx 2
Hence, maximum volume = 62.5x -
1 3
1
3
x = 62.5(6.455) - ( 6.455 ) = 269cm 3
2
2
4
Total: 7
Problem 8. A cycloid has parametric equations given by: x = 5(θ - sin θ) and y = 5(1 – cos θ).
Evaluate (a)
dy
dx
(b)
d2 y
when θ = 1.5 radians. Give answers correct to 3 decimal places.
dx 2
Marks
(a) x = 5(θ - sin θ)
hence,
y = 5(1 – cos θ) hence,
Thus,
dx
= 5 − 5cos θ = 5 (1 − cos θ )
dθ
1
dy
= 5sin θ
dθ
1
dy
dy dθ
5sin θ
sin θ
=
=
=
dx dx 5 (1 − cos θ ) (1 − cos θ )
dθ
When θ = 1.5 radians,
dy
sin1.5
0.99749499
= 1.073, correct to 3
=
=
dx (1 − cos1.5 ) 1 − 0.07073720
decimal places
d ⎛ dy ⎞ d ⎛ sin θ ⎞ (1 − cos θ ) cos θ − ( sin θ )( sin θ )
2
1 − cos θ )
(
d 2 y dθ ⎝⎜ dx ⎠⎟ dθ ⎝⎜ 1 − cos θ ⎠⎟
(b)
=
=
=
dx
dx 2
5 (1 − cos θ )
5 (1 − cos θ )
dθ
cos θ − cos 2 θ − sin 2 θ
(1 − cos θ )
5 (1 − cos θ )
2
=
=
=
cos θ − ( cos 2 θ + sin 2 θ )
5 (1 − cos θ )
− (1 − cos θ )
5 (1 − cos θ )
3
=
3
2
2
=
cos θ − 1
5 (1 − cos θ )
−1
5 (1 − cos θ )
2
3
1
© 2006 John Bird. All rights reserved. Published by Elsevier.
62
When θ = 1.5 radians,
d2 y
−1
−1
=
=
= 0.232, correct to 3
2
2
dx
5 ( 0.863529348 )
5 (1 − cos1.5 )
1
decimal places
Total: 8
Problem 9. Determine the equation of (a) the tangent, and (b) the normal, drawn to an ellipse
x = 4 cos θ, y = sin θ at θ =
π
3
Marks
(a) Equation of tangent is: y − y1 =
dy1
( x − x1 )
dx1
At point θ, x1 = 4 cos θ hence,
y1 = sin θ
hence,
dx1
= −4sin θ
dθ
1
dy1
= cos θ
dθ
1
dy1
dy1 dθ
cos θ
1
=
=
= − cot θ
dx
dx1
−4sin θ
4
1
dθ
1
1
Hence, the equation of the tangent is: y − sin θ = − cot θ ( x − 4 cos θ )
4
At θ =
π
,
3
y − sin
π
π⎛
π⎞
1
= − cot ⎜ x − 4 cos ⎟
3
4
3⎝
3⎠
y – 0.86603 = -0.14434(x – 2)
i.e.
y – 0.86603 = -0.14434x + 0.28868
i.e.
(b) Equation of the normal is: y − y1 = −
1
1
x − x1 ) = y − sin θ = −
(
( x − 4 cos θ )
dy1
⎛ 1
⎞
⎜ − cot θ ⎟
dx1
⎝ 4
⎠
y − sin θ = 4 tan θ ( x − 4 cos θ )
i.e.
At θ =
2
y = 1.093 – 0.144x
π
,
3
y − sin
1
π
π⎛
π⎞
= 4 tan ⎜ x − 4 cos ⎟
3
3⎝
3⎠
y – 0.88603 = 6.92820(x – 2)
© 2006 John Bird. All rights reserved. Published by Elsevier.
63
i.e.
y – 0.86603 = 6.92820x – 13.8560
i.e.
2
y = 6.928x - 12.990
Total: 8
Problem 10. Determine expressions for
(a) z = 5y 2 cos x
dz
for each of the following functions:
dy
(b) z = x 2 + 4xy − y 2
Marks
(a) If z = 5y 2 cos x then
dz
dx
= ( 5y 2 ) ( − sin x ) + ( cos x )(10y )
dy
dy
= − ( 5y 2 sin x )
or
(b) If z = x 2 + 4xy − y 2 then
dx
+ 10y cos x
dy
⎛
dx ⎞
5y ⎜ 2cos x − y sin x ⎟
dy ⎠
⎝
2
⎛ dx ⎞ ⎤
dz
dx ⎡
= 2x
+ ⎢ 4x(1) + y ⎜ 4 ⎟ ⎥ − 2y
dy
dy ⎣
⎝ dy ⎠ ⎦
= 2x
dx
dx
+ 4x + 4y
− 2y
dy
dy
or
( 2x + 4y )
dx
+ 4x − 2y
dy
3
Total: 5
Problem 11. If x 2 + y 2 + 6x + 8y + 1 = 0 , find
dy
in terms of x and y.
dx
Marks
If x 2 + y 2 + 6x + 8y + 1 = 0 then 2x + 2y
dy
dy
+6+8 +0 = 0
dx
dx
i.e.
( 2y + 8)
i.e.
dy −2x − 6
=
dx 2y + 8
dy
= −2x − 6
dx
or
dy − x − 3
=
dx y + 4
3
Total: 3
Problem 12. Determine the gradient of the tangents drawn to the hyperbola x 2 − y 2 = 8 at x = 3.
© 2006 John Bird. All rights reserved. Published by Elsevier.
64
Marks
Since x 2 − y 2 = 8
When x = 3,
Hence,
then
dy
=0
dx
2x − 2y
i.e.
dy x
=
dx y
9 − y 2 = 8 from which, y = ±1.
dy
x 3
= =
= ±3
y ±1
dx
3
Total: 3
( x + 1) ( x − 2 )
y=
4
( 2x − 1) 3 ( x − 3)
2
Problem 13. Use logarithmic differentiation to differentiate:
with respect to
x.
Marks
⎧ ( x + 1)2
⎪
ln y = ln ⎨
⎪⎩ ( 2x − 1)
1
4
( x − 2 ) ⎫⎪
2
= ln ( x + 1) + ln ( x − 2 ) 2 − ln ( 2x − 1) − ln ( x − 3) 3
⎬
4
3
( x − 3) ⎪⎭
1
4
= 2 ln ( x + 1) + ln ( x − 2 ) − ln ( 2x − 1) − ln ( x − 3)
2
3
4
1 dy
2
1
2
4
=
+
−
−
y dx ( x + 1) 2 ( x − 2 ) ( 2x − 1) 3 ( x − 3)
( x + 1)
dy
=
dx ( 2x − 1)
2
Hence,
( x − 2 ) ⎪⎧ 2
1
2
4
⎪⎫
+
−
−
⎨
⎬
4
3
( x − 3 ) ⎩⎪ ( x + 1) 2 ( x − 2 ) ( 2x − 1) 3 ( x − 3 ) ⎭⎪
2
Total: 6
Problem 14. Differentiate y =
when θ =
3eθ sin 2θ
θ
5
and hence evaluate
dy
, correct to 2 decimal places,
dθ
π
.
3
Marks
⎧ 3eθ sin 2θ ⎫
θ
5/ 2
ln y = ln ⎨
⎬ = ln 3 + ln e + ln ( sin 2θ ) − ln θ
5
θ
⎩
⎭
5
= ln 3 + θ + ln ( sin 2θ ) − ln θ
2
Hence,
3
1 dy
2 cos 2θ 5
= 0 +1+
−
y dx
sin 2θ 2θ
© 2006 John Bird. All rights reserved. Published by Elsevier.
65
dy 3eθ sin 2θ ⎧
2
5⎫
=
− ⎬
1+
⎨
dx
θ5 ⎩ tan 2θ 2θ ⎭
and
When θ =
3
π
3
⎫
⎛π⎞⎧
3e sin 2 ⎜ ⎟ ⎪
⎪
dy
⎝ 3 ⎠ ⎪1 + 2 − 5 ⎪
=
⎨
⎬
5
dx
⎪ tan 2θ 2 ⎛⎜ π ⎞⎟ ⎪
⎛π⎞
⎜ ⎟
⎪⎩
⎝ 3 ⎠ ⎪⎭
⎝3⎠
π
,
3
=
7.40362
[1 − 1.1547005 − 2.387324146]
1.1222033
= -16.77, correct to 2 decimal places.
3
Total: 9
Problem 15. Evaluate
d ⎡t
( 2t + 1) ⎤⎦ when t = 2, correct to 4 significant figures.
dt ⎣
Marks
If y =
Hence,
1
t
( 2t + 1) = ( 2t + 1) t
1 dy
=
y dt
and
When t = 2,
then
1
1
ln y = ln ( 2t + 1) t = ln ( 2t + 1)
t
( t ) ⎛⎜
2 ⎞
⎟ − ln ( 2t + 1)(1)
⎝ 2t + 1 ⎠
by the quotient rule
t2
⎡ 2t
⎤
− ln ( 2t + 1) ⎥
⎢
( 2t + 1)
dy
⎥=
= y⎢
dt
t2
⎢
⎥
⎢
⎥
⎣
⎦
dy
=
dt
t
3
⎡ 2t
⎤
⎢ ( 2t + 1) − ln ( 2t + 1) ⎥
⎥
( 2t + 1) ⎢
t2
⎢
⎥
⎢
⎥
⎣
⎦
⎡4
⎤
− ln 5 ⎥
⎢
2
5⎢5
⎥ = -0.4525, correct to 4 significant figures.
⎢ 4 ⎥
⎣
⎦
2
Total: 5
TOTAL ASSIGNMENT MARKS: 100
© 2006 John Bird. All rights reserved. Published by Elsevier.
66
ASSIGNMENT 9 (PAGE 365)
This assignment covers the material contained in chapters 32 to 36.
Problem 1. Differentiate the following functions with respect to x:
(b) 3ch 3 2x
(a) 5 ln(sh x)
(c) e 2x sec h 2x
Marks
(a)
⎡⎛ 1 ⎞
⎤
d
{5ln(sh x)} = 5 ⎢⎜
⎟ ch x ⎥ = 5 coth x
dx
⎣⎝ sh x ⎠
⎦
2
(b)
d
{3ch 3 2x} = 3 ( 3ch 2 2x ) ( 2sh 2x ) = 18ch2 2x sh 2x
dx
2
(c)
d 2x
e sec h 2x} = ( e2x ) ( −2 sec h 2x tanh 2x ) + ( sec h 2x ) ( 2e 2x )
{
dx
= 2e 2x sec h 2x [1 − tanh 2x ]
3
Total: 7
Problem 2. Differentiate the following functions with respect to the variable:
1
x
(a) y = cos −1
5
2
(d) y = 3sinh −1
(b) y = 3e
( 2x
2
2sec −1 5x
(c) y =
x
sin −1 t
− 1)
Marks
1
x
(a) y = cos −1
5
2
hence
or
(b) y = 3e
sin −1 t
hence
⎡
⎢
1
−
dy 1 ⎢
2
= ⎢
dx 5 ⎢ ⎡
2
⎤
⎢ ⎢1 − ⎛⎜ x ⎞⎟ ⎥
⎢
2
⎣ ⎣⎢ ⎝ ⎠ ⎦⎥
−1
⎛ 4 − x2 ⎞
10 ⎜
⎟
⎝ 4 ⎠
or −
⎛
dy
sin −1 t ⎜
= 3e
⎜⎜
dx
⎝
⎤
⎥
⎥
−1
⎥ =
⎥
⎛
x2 ⎞
⎥ 10 ⎜ 1 − ⎟
4 ⎠
⎝
⎥
⎦
1
4 − x2
10
2
⎞
⎟
2 ⎟
( 1 − t ) ⎟⎠
1
or −
1
5 ⎡⎣ 4 − x 2 ⎤⎦
3
3
© 2006 John Bird. All rights reserved. Published by Elsevier.
67
2sec −1 5x
(c) y =
x
hence
⎛
( x ) ⎜⎜
dy
⎝ 5x
=
dx
( 2 ) 5 ⎞⎟ − 2sec−1 5x 1
(
)( )
2
( 5x ) − 1 ⎟⎠
x2
2
− 2sec −1 5x
25x − 1
x2
2
=
(d) y = 3sinh −1
⎡
⎢
dy
= 3⎢
⎢
dx
⎢
⎢⎣
( 2x
2
− 1)
4
hence
⎤
1
6x
−
1
2
2 4x ⎥
−
2x
1
( )
(
)
2x 2 − 1 =
2
⎥=
2
2x 2 − 1 + 1
⎡
⎤⎥
2
⎢⎣ ( 2x − 1) + 1⎥⎦ ⎥⎥
⎦
=
6x
2x 2 − 1 =
2x 2
6x
x ⎡⎣ 2 ( 2x 2 − 1) ⎤⎦
=
6x
⎡ 2x 2 ( 2x 2 − 1)⎤
⎣
⎦
6
(
4
)
⎡ 2 2x 2 − 1 ⎤
⎣
⎦
Total: 14
Problem 3. Evaluate the following, each correct to 3 decimal places:
(a) sinh −1 3
(b) cosh −1 2.5
(c) tanh −1 0.8
Marks
⎧⎪ 3 + 12 + 32
(a) sinh 3 = ln ⎨
1
⎩⎪
−1
⎫⎪
⎬ = ln 3 + 10 = 1.818
⎭⎪
(
)
2
⎪⎫
⎪⎧ 5 + 21 ⎪⎫
⎬ = ln ⎨
⎬ = 1.567
⎪⎭
⎩⎪ 2 ⎭⎪
2
(b) cosh −1 2.5 = cosh −1
5
⎪⎧ 5 + 52 − 22
= ln ⎨
2
2
⎪⎩
(c) tanh −1 0.8 = tanh −1
4 1 ⎛ 5+ 4⎞ 1
= ln ⎜
⎟ = ln 9 = 1.099
5 2 ⎝ 5−4 ⎠ 2
(or by calculator)
2
Total: 6
∂z ∂z ∂ 2 z ∂ 2 z ∂ 2 z
∂2z
Problem 4. If z = f (x, y) and z = x cos(x + y) determine
,
,
,
,
and
∂x ∂y ∂x 2 ∂y 2 ∂x∂y
∂y∂x
Marks
z = x cos(x + y)
∂z
= ( x ) [ − sin(x + y) ] + cos(x + y) = -x sin(x + y) + cos(x + y)
∂x
68
2
© 2006 John Bird. All rights reserved. Published by Elsevier.
∂z
= -x sin(x + y)
∂y
2
∂2z ∂
= [ − x sin(x + y) + cos(x + y) ] = ( − x ) cos(x + y) + sin(x + y) ( −1) − sin(x + y)
∂x 2 ∂x
= -x cos(x + y) – 2 sin(x + y)
2
∂2z ∂
= [ − x sin(x + y) ] = -x cos(x + y)
∂y 2 ∂y
2
∂2z
∂
= [ − x sin(x + y) ] = ( − x ) cos(x + y) + sin(x + y) ( −1)
∂x∂y ∂x
= -x cos(x + y) – sin(x + y)
2
∂2z
∂
= [ − x sin(x + y) + cos(x + y) ] = -x cos(x + y) – sin(x + y)
∂y∂x ∂y
2
Total: 12
Problem 5. The magnetic field vector H due to a steady current I flowing around a circular wire of
radius r and at a distance x from its centre is given by: H = ±
⎞
I ∂ ⎛
x
⎜ 2
⎟
2 ∂x ⎝ r + x 2 ⎠
r 2I
Show that H = ±
2
(r
2
+ x2 )
3
Marks
1
⎡ 2
⎤
1 2
2
2 −2
+
−
+
r
x
1
x
r
x
(
)
(
)
( 2x ) ⎥
(
)
⎞
I ∂ ⎛
x
I⎢
2
H=±
⎥
⎜ 2
⎟=± ⎢
2
2
2 ∂x ⎝ r + x ⎠
2⎢
⎡ r2 + x2 ⎤
⎥
⎣
⎦
⎢⎣
⎥⎦
⎡ 2
x2
2
+
−
r
x
I ⎢⎢
r2 + x2
= ±
2⎢
r2 + x2 )
(
⎢
⎣
⎡ ( r2 + x2 ) − x2
⎤
⎢
⎥
I
r2 + x2
⎥=± ⎢
2 ⎢ ( r2 + x2 )
⎥
⎢
⎥
⎦
⎢⎣
3
⎤
⎡
⎤
r2
⎥
⎢
⎥
⎥ = ± I ⎢ r2 + x2 ⎥
⎥
2 ⎢ ( r2 + x2 ) ⎥
⎥
⎢
⎥
⎥⎦
⎣
⎦
I⎡
r2
=± ⎢
2 ⎢ (r2 + x2 ) r2 + x2
⎣
⎤
⎥= ±
⎥
2
⎦
r 2I
(r
2
+x
2
)
4
3
Total: 7
© 2006 John Bird. All rights reserved. Published by Elsevier.
69
⎛ dx dy ⎞
Problem 6. If xyz = c, where c is constant, show that: dz = −z ⎜ + ⎟
y ⎠
⎝ x
Marks
If xyz = c then z =
c
= c x −1 y −1
xy
Total differential, dz =
=
∂z
∂z
dx + dy = ( −c x −2 y −1 ) dx + ( −c x −1 y −2 )
∂x
∂y
3
⎛ dx dy ⎞
c
1 ⎤
−c
−c ⎡ 1
dx −
dy =
dx + dy ⎥ = − z ⎜
+
⎟
⎢
2
2
x y
xy
x y ⎣x
y ⎦
y ⎠
⎝ x
3
Total: 6
y
Problem 7. An engineering function z = f (x, y) and z = e 2 ln ( 2x + 3y ) . Determine the rate of
increase of z, correct to 4 significant figures, when x = 2 cm, y = 3 cm, x is increasing at 5 cm/s and
y is increasing at 4 cm/s.
Marks
y
z = e 2 ln ( 2x + 3y )
Rate of increase of z,
dz ∂z dx ∂z dy
=
+
dt ∂x dt ∂y dt
y
y
⎛
⎞
⎡
⎤
2
2
⎛ 1 2y ⎞ ⎥ dy
2
e
dx
2
e
⎜
⎟
⎢
+
+ ln(2x + 3y) ⎜ e ⎟
=
⎜ 2x + 3y ⎟ dt ⎢ 2x + 3y
⎜
⎟
⎝ 2 ⎠ ⎥⎥ dt
⎝
⎠
⎣⎢
⎦
When x = 2 cm, y = 3 cm,
4
dx
dy
= 5cm / s and
= 4 cm / s,
dt
dt
3
3
⎛
⎞
⎡
⎤
2
⎛ 1 32 ⎞ ⎥
dz ⎜
2e
2 e2
⎟
⎢
=
+ ln[2(2) + 3(3)] ⎜ e ⎟ (4)
(5) +
⎢ 2(2) + 3(3)
dt ⎜⎜ 2(2) + 3(3) ⎟⎟
⎝ 2 ⎠ ⎥⎥
⎢
⎝
⎠
⎣
⎦
= 3.44745 + (0.68949 + 5.747653)(4) = 29.20 cm/s
4
Total: 8
70
© 2006 John Bird. All rights reserved. Published by Elsevier.
Problem 8. The volume V of a liquid of viscosity coefficient η delivered after time t when passed
through a tube of length L and diameter d by a pressure p is given by V =
p d4t
.
128 η L
If the errors in V, p and L are 1%, 2% and 3% respectively, determine the error in η.
Marks
If V =
p d4t
p d4t
then η =
128 η L
128 L V
and error in η,
δη =
∂η
∂η
∂η
δV + δp +
δL
∂V
∂p
∂L
⎛ −p d 4 t ⎞
⎛ d4t ⎞
⎛ −p d 4 t ⎞
=⎜
δ
+
δ
+
V
p
⎜
⎟
⎜
⎟ δL
2 ⎟
2
⎝ 128 L V ⎠
⎝ 128 LV ⎠
⎝ 128 L V ⎠
=
4
p d4t ⎡ 1
1
1 ⎤
− δV + δp − δL ⎥
⎢
128 L V ⎣ V
p
L ⎦
⎡ 1
⎤
1
1
= η ⎢ − ( 0.01V ) + ( 0.02 p ) − ( 0.03L ) ⎥
p
L
⎣ V
⎦
= η [-0.01 + 0.02 – 0.03] = -0.02 η
Hence, the error in η is –2%
4
Total: 8
Problem 9. Determine and distinguish between the stationary values of the function
f (x, y) = x 3 − 6x 2 − 8y 2 and sketch an approximate contour map to represent the surface
f (x, y).
Marks
Let z = x 3 − 6x 2 − 8y 2 then
For stationary points,
∂z
∂z
= −16y
= 3x 2 − 12x and
∂y
∂x
∂z
= 3x 2 − 12x = 0
∂x
from which, 3x(x – 4) = 0
and
and
∂z
= −16y = 0
∂y
x = 0 or x = 4
from which, y = 0
Hence, the stationary points are at (0, 0) and (4, 0)
5
© 2006 John Bird. All rights reserved. Published by Elsevier.
71
∂2z
∂2z
= −16
=
−
,
6x
12
∂y 2
∂x 2
∂2z
=0
∂x∂y
and
For the point (0, 0),
∂ 2z
∂2z
∂2z
16
=
−
and
=0
=
−
,
12
∂x 2
∂y 2
∂x∂y
For the point (4, 0),
∂ 2z
∂2z
∂2z
16
=
−
and
=0
12
=
,
∂y 2
∂x∂y
∂x 2
3
2
⎛ ∂2z ⎞
For the stationary points, ⎜
⎟ =0
⎝ ∂x∂y ⎠
2
∆ (0,0)
⎛ ∂ 2z ⎞ ⎛ ∂ 2z ⎞ ⎛ ∂ 2z ⎞
2
=⎜
⎟ − ⎜ 2 ⎟ ⎜ 2 ⎟ = ( 0 ) − (−12)(−16) = −192
⎝ ∂x∂y ⎠ ⎝ ∂x ⎠ ⎝ ∂y ⎠
∆ (4,0) = ( 0 ) − (12)(−16) = +192
2
⎛ ∂2z ⎞
⎜ 2 ⎟ < 0, the point (0, 0) is a maximum
⎝ ∂x ⎠(0,0)
Since
∆ (0,0) < 0
Since
∆ (4,0) > 0, the point (4, 0) is a saddle point
and
4
The maximum value of z is 0 and the value of z at the saddle point (4, 0) is:
( 4)
3
− 6 ( 4 ) − 8 ( 0 ) = −32
2
2
An approximate contour map representing the surface f (x, y) is shown in Figure 29.
Figure 29
8
Total: 20
Problem 10. An open, rectangular fish tank is to have a volume of 13.5 m3 . Determine the least
surface area of glass required.
72
© 2006 John Bird. All rights reserved. Published by Elsevier.
Marks
Let the dimensions of the fish tank be as shown in Figure 30.
Figure 30
Volume, V = xyz = 13.5 m3
(1)
Surface area, S = 2xy + xz + 2yz
(2)
From equation (1), z =
13.5
xy
⎛ 13.5 ⎞
⎛ 13.5 ⎞
Substituting in equation (2) gives: S = 2xy + x ⎜
⎟ + 2y ⎜
⎟
⎝ xy ⎠
⎝ xy ⎠
i.e.
S = 2xy +
13.5 27
+
y
x
2
∂S
27
= 2y − 2 = 0 for a stationary point. Hence, x 2 y = 13.5
∂x
x
(3)
13.5
∂S
13.5
= 2x − 2 = 0 for a stationary point. Hence, xy 2 =
∂y
y
2
(4)
x 2 y 13.5
Equation (3) divided by equation (4) gives:
=
xy 2 13.5
2
( 2y )
Substituting x = 2y in equation (3) gives:
4y3 = 13.5
i.e.
and
y=
3
2
i.e.
x
=2
y
i.e. x = 2y
y = 13.5
13.5
= 1.5 m
4
Since x = 2y, then x = 3.0 m
From equation (1), (3.0)(1.5) z = 13.5, from which, z = 3.0 m
∂ 2S 54
∂ 2S 27
=
,
=
,
∂y 2 y3
∂x 2 x 3
5
∂ 2S
=2
∂x∂y
∂ 2S
= 2,
When x = 3.0 and y = 1.5 then
∂x 2
∂ 2S
∂ 2S
27
=2
=
= 8 and
∂x∂y
∂y 2 (1.5 )3
2
⎛ ∂2 y ⎞ ⎛ ∂2z ⎞ ⎛ ∂2z ⎞
2
∆=⎜
⎟ − ⎜ 2 ⎟ ⎜ 2 ⎟ = ( 2 ) − ( 2 )( 8 ) = −12
⎝ ∂x∂y ⎠ ⎝ ∂x ⎠ ⎝ ∂y ⎠
© 2006 John Bird. All rights reserved. Published by Elsevier.
73
Since ∆ < 0 and
∂ 2S
> 0 then the surface area S is a minimum.
∂x 2
Hence, the minimum dimensions are 3.0 m by 3.0 m by 1.5 m.
3
Therefore, the minimum surface area = 2(3.0)(1.5) + (3.0)(3.0) + 2(1.5)(3.0)
= 9.0 + 9.0 + 9.0 = 27 m 2
2
Total: 12
TOTAL ASSIGNMENT MARKS: 100
74
© 2006 John Bird. All rights reserved. Published by Elsevier.
ASSIGNMENT 10 (PAGE 396)
This assignment covers the material contained in chapters 37 to 39.
Problem 1. Determine (a)
∫3
t 5 dt
(b)
∫
2
3
x
2
dx
(c)
∫ ( 2 + θ)
2
dθ
Marks
5
2
(a) ∫ 3 t 5 dt = 3∫ t dt = (3)
(b)
(c)
2
∫
3
x
2
dx = 2 ∫ x
∫ ( 2 + θ)
2
−
2
3
7
2
t
6 7
+c=
t +c
7
7
2
dt = (2)
3
1
3
x
+c = 63 x +c
1
3
dθ = ∫ ( 4 + 4θ + θ2 ) dθ = 4θ +
3
4θ2 θ3
1
+ + c = 4θ + 2θ 2 + θ 3 + c
2
3
3
3
Total: 9
Problem 2. Evaluate the following integrals, each correct to 4 significant figures:
(a)
∫
π
3
0
3sin 2t dt
(b)
∫
2
1
⎛ 2 1 3⎞
⎜ 2 + + ⎟ dx
x 4⎠
⎝x
(c)
∫
3
dt
0 e 2t
1
Marks
(a)
∫
π
3
0
3sin 2t dt = −
3
3
2π
3
π/3
[cos 2t ] 0 = − ⎡⎢cos − cos 0⎤⎥ = − [ −0.5 − 1] = 2.250
2
2⎣
3
2
⎦
5
2
(b)
∫
2
1
2⎛
⎡ 2x −1
1 3⎞
3 ⎤
⎛ 2 1 3⎞
−2
+
+
=
+
+
=
+ ln x + x ⎥
dx
2x
dx
⎜ 2
⎟
⎜
⎟
⎢
∫
1
x 4⎠
x 4⎠
4 ⎦1
⎝x
⎝
⎣ −1
6⎞ ⎛ 2
3⎞
⎛ 2
= ⎜ − + ln 2 + ⎟ − ⎜ − + ln1 + ⎟ = 2.443
4⎠ ⎝ 1
4⎠
⎝ 2
(c)
∫
1
1
3
3
3
dt = ∫ 3e−2t dt = ⎡⎣e−2t ⎤⎦ = − ⎡⎣e −2 − e0 ⎤⎦ = 1.297
2t
0
0 e
0
2
−2
1
5
5
Total: 15
Problem 3. Calculate the area between the curve y = x 3 − x 2 − 6x and the x-axis.
© 2006 John Bird. All rights reserved. Published by Elsevier.
75
Marks
y = x 3 − x 2 − 6x = x ( x 2 − x − 6 ) = x ( x − 3)( x + 2 )
When y = 0, x = 0 or x = 3 or x = -2
When x = 1, y = 1 – 1 – 6 i.e. negative, hence the curve is as shown in Figure 31.
4
Figure 31
Shaded area =
∫ (x
0
−2
3
− x 2 − 6x ) dx − ∫ ( x 3 − x 2 − 6x ) dx
3
0
0
3
⎡ x 4 x 3 6x 2 ⎤
⎡ x 4 x 3 6x 2 ⎤
= ⎢ − −
−
− −
2 ⎥⎦ − 2 ⎢⎣ 4 3
2 ⎥⎦ 0
⎣4 3
⎡
⎤ ⎡ 1⎤ ⎡
8
3⎤
⎛
⎞ ⎤ ⎡⎛ 81
⎞
= ⎢( 0 ) − ⎜ 4 + − 12 ⎟ ⎥ − ⎢⎜ − 9 − 27 ⎟ − ( 0 ) ⎥ = ⎢5 ⎥ − ⎢ −15 ⎥
3
4⎦
⎝
⎠ ⎦ ⎣⎝ 4
⎠
⎣
⎦ ⎣ 3⎦ ⎣
= 21
1
or 21.08 square units
12
6
Total: 10
Problem 4. A voltage v = 25 sin 50πt volts is applied across an electrical circuit. Determine, using
integration, its mean and r.m.s. values over the range t = 0 to t = 20 ms, each correct to 4 significant
figures.
Marks
1
Mean value =
20 ×10−3
=−
∫
20×10−3
0
( 50 )( 25)
50π
20×10−3
⎡ cos 50πt ⎤
25sin 50πt dt = ( 50 )( 25 ) ⎢ −
50π ⎦⎥ 0
⎣
20×10−3
[ cos 50πt ] 0
=−
25
50
[ −1 − 1] = = 15.92 volts
π
π
5
© 2006 John Bird. All rights reserved. Published by Elsevier.
76
1
⎧
⎨
−3
⎩ 20 × 10
R.m.s. value =
∫
20×10−3
0
( 25 )
2
⎫
sin 2 50πt dt ⎬
⎭
=
−3
⎧
⎫
2 20×10 1 − cos 2 ( 50πt )
50
25
dt ⎬
(
)
⎨
∫
0
2
⎩
⎭
=
⎧⎪ 50 ( 25 )2
⎨
⎪⎩ 2
⎡ sin100πt ⎤
⎢⎣ t − 100π ⎥⎦
0
=
⎧ 50 ( 25 )2
⎪
⎨
⎪⎩ 2
⎡⎛
⎤ ⎫⎪
sin100π ( 20 × 10−3 ) ⎞
−3
⎢⎜ 20 ×10 −
⎟ − ( 0)⎥ ⎬
⎟
100π
⎢⎜⎝
⎥⎪
⎠
⎣
⎦⎭
=
⎧⎪ 50 ( 25 )2
⎫⎪ 25
20 ×10−3 ) ⎬ =
= 17.68 volts
(
⎨
2
2
⎩⎪
⎭⎪
20×10−3
⎫⎪
⎬
⎪⎭
7
Total: 12
Problem 5. Sketch on the same axes the curves x 2 = 2y and y 2 = 16x and determine the coordinates of the points of intersection. Determine (a) the area enclosed by the curves, and
(b) the volume of the solid produced if the area is rotated one revolution about the x-axis.
Marks
The curves are equal at the points of intersection. Thus, equating the two y values
gives:
i.e.
x2
=4 x
2
or
x ( x 3 − 64 ) = 0
x4
= 16x
4
x 4 = 64x and x 4 − 64x = 0
from which,
from which, x = 0 or x = 4
Hence, (0, 0) and (4, 8) are the co-ordinates of the points of intersection.
5
The curves are shown in Figure 32.
4
(a) Shaded area =
∫
4
0
⎡ 3
⎤
⎢ 4x 2 x 3 ⎥
⎛
⎛ 8 3 43 ⎞
x2 ⎞
4
x
−
dx
=
−
=
4 − ⎟ − (0)
⎜
⎟
⎢ 3
⎥
⎜
2 ⎠
6
3
6⎠
⎝
⎝
⎢
⎥
⎣ 2
⎦0
= 10.67 square units
4
4
(b) Volume =
∫
4
0
πy dx = π ∫
2
4
0
⎛
⎡16x 2 x 5 ⎤
x4 ⎞
− ⎥
⎜ 16x − ⎟ dx = π ⎢
4 ⎠
20 ⎦ 0
⎝
⎣ 2
© 2006 John Bird. All rights reserved. Published by Elsevier.
77
⎡⎛
⎤
45 ⎞
2
= π ⎢⎜ 8 ( 4 ) − ⎟ − ( 0 ) ⎥ = π [128 − 51.2] = 76.8π cubic units
20 ⎠
⎣⎝
⎦
or 241.3 cubic units
4
Figure 32
Total: 13
Problem 6. Calculate the position of the centroid of the sheet of metal formed by the x-axis and the
part of the curve y = 5x - x 2 which lies above the x-axis.
Marks
y = 5x - x 2 = x(5 – x) and when y = 0, i.e. the x-axis, x = 0 or x = 5
2
_
A sketch of y = 5x - x 2 is shown in Figure 33, where x = 2.5 by symmetry.
2
Figure 33
2
1 5 2
1 5
1 5
y dx
5x − x 2 ) dx
25x 2 − 10x 3 + x 4 ) dx
(
(
∫
∫
∫
0
0
0
=2 5
=2
y= 2 5
5
2
2
∫ y dx
∫ ( 5x − x ) dx
∫ ( 5x − x ) dx
_
0
0
0
© 2006 John Bird. All rights reserved. Published by Elsevier.
78
5
=
1 ⎡ 25x 3 10x 4 x 5 ⎤
−
+ ⎥
2 ⎢⎣ 3
4
5 ⎦0
5
⎡ 5x 2 x 3 ⎤
⎢ 2 − 3⎥
⎣
⎦0
3
4
5
1 ⎡ 25 ( 5 ) 10 ( 5 ) ( 5 ) ⎤
−
+
⎢
⎥
2 ⎢⎣ 3
4
5 ⎥⎦ 52.08333
=
=
= 2.5
20.8333
⎡ 5 ( 25 ) 125 ⎤
−
⎢
⎥
2
3 ⎦
⎣
Hence, the co-ordinates of the centroid are at (2.5, 2.5)
4
1
Total: 9
Problem 7. A cylindrical pillar of diameter 400 mm has a groove cut around its circumference as
shown in Figure A10.1. The section of the groove is a semicircle of diameter 50 mm. Given that the
centroid of a semicircle from its base is
4r
, use the theorem of Pappus to determine the volume of
3π
material removed, in cm3 , correct to 3 significant figures.
Figure A10.1
Marks
Distance of the centroid of the semicircle =
4r 4 ( 25 ) 100
mm
=
=
3π
3π
3π
100 ⎞
⎛
Distance of the centroid from the centre of the pillar = ⎜ 200 −
⎟ mm
3π ⎠
⎝
100 ⎞
⎛
Distance moved by the centroid in one revolution = 2π ⎜ 200 −
⎟ mm
3π ⎠
⎝
200 ⎞
⎛
= ⎜ 400π −
⎟ mm
3 ⎠
⎝
4
From Pappus, volume = area × distance moved by centroid
i.e.
200 ⎞
⎛1
⎞⎛
3
volume = ⎜ π 252 ⎟ ⎜ 400π −
⎟ = 1168251 mm
2
3
⎝
⎠⎝
⎠
© 2006 John Bird. All rights reserved. Published by Elsevier.
79
i.e. volume of material removed = 1168 cm 3
4
Total: 8
Problem 8. A circular door is hinged so that it turns about a tangent. If its diameter is 1.0 m find its
second moment of area and radius of gyration about the hinge.
Marks
From Table 38.1, page 385, second moment of area about a tangent
=
5π 4 5π
4
r =
( 0.5 ) = 0.245m 3
4
4
5
5
r=
( 0.50 ) = 0.559 m
2
2
Radius of gyration, k =
3
2
Total: 5
Problem 9. Determine the following integrals:
(a)
∫ 5 ( 6t + 5)
7
(b)
dt
∫
3ln x
dx
x
(c)
2
∫ ( 2θ − 1) dθ
Marks
(a)
∫ 5 ( 6t + 5)
Hence,
(b)
∫
dt
Let u = 6t + 5 then
5 ( 6t + 5 ) dt = ∫ 5u 7
7
3ln x
dx
x
Hence,
(c)
∫
7
∫
5
du 5 u 8
8
=
+c =
( 6t + 5 ) + c
48
6 6 8
Let u = ln x then
3ln x
dx =
x
2
∫ ( 2θ − 1) dθ
∫
du
du
= 6 and dt =
dt
6
du 1
=
dx x
3
and dx = x du
3u
3u 2
3
2
+ c = ( ln x ) + c
x du = ∫ 3u du =
2
x
2
Let u = 2θ - 1 then
3
du
du
= 2 and dθ =
dθ
2
1
Hence,
2
∫ ( 2θ − 1) dθ = ∫
1
−
2 du
u2
= ∫ u 2 du =
+c = 2 u +c
1
u 2
2
= 2 ( 2θ − 1 ) + c
3
Total: 9
© 2006 John Bird. All rights reserved. Published by Elsevier.
80
Problem 10. Evaluate the following definite integrals:
(a)
∫
π/2
0
π⎞
⎛
2sin ⎜ 2t + ⎟ dt
3⎠
⎝
(b)
∫
1
0
3 x e 4x
2
−3
dx
Marks
(a)
∫
π/2
0
π⎞
⎛
2sin ⎜ 2t + ⎟ dt
3⎠
⎝
Let u = 2t +
π⎞
⎛
∫ 2sin ⎜⎝ 2t + 3 ⎟⎠ dt = ∫ 2sin u
Hence,
π
du
du
then
= 2 and dt =
3
dt
2
du
= sin u du = − cos u + c
2 ∫
π⎞
⎛
= − cos ⎜ 2t + ⎟ + c
3⎠
⎝
Thus,
∫
π/2
0
π/ 2
π⎞
⎡
π ⎞⎤
⎡ ⎛
π⎞
π⎤
⎛
⎛
2sin ⎜ 2t + ⎟ dt = ⎢ − cos ⎜ 2t + ⎟ ⎥ = − ⎢ cos ⎜ π + ⎟ − cos ⎥
3⎠
3 ⎠⎦ 0
3⎠
3⎦
⎝
⎝
⎣
⎣ ⎝
= -[-0.5 – 0.5] = 1
(b)
∫
1
0
3 x e 4x
−3
∫
1
0
Let u = 4x 2 − 3
dx
∫ 3x e
Hence,
Thus,
2
4x 2 −3
3 x e4x
2
−3
dx = ∫ 3x eu
then
du
= 8x
dx
and dx =
5
du
8x
du 3 u
3
3 2
= ∫ e du = e u + c = e4x −3 + c
8x 8
8
8
1
2
3
3
dx = ⎡e 4x −3 ⎤ = ⎡⎣e1 − e −3 ⎤⎦ = 1.001
⎣
⎦
0
8
8
5
Total: 10
TOTAL ASSIGNMENT MARKS: 100
© 2006 John Bird. All rights reserved. Published by Elsevier.
81
ASSIGNMENT 11 (PAGE 417)
This assignment covers the material contained in chapters 40 to 42.
Problem 1. Determine the following integrals:
(a)
∫ cos
3
x sin 2 x dx
∫
(b)
2
( 9 − 4x )
2
(c)
dx
2
∫
( 4x
2
− 9)
dx
Marks
(a)
(b)
∫ cos
3
x − cos x sin 4 x ) dx =
∫ ( cos x sin
( 9 − 4x )
2
2
∫
2
=
dx = ∫
=∫
(c)
x sin 2 x dx = ∫ cos x (1 − sin 2 x ) sin 2 x dx
∫ cos x cos
2
∫
2
x sin 2 x dx =
( 4x
2
− 9)
∫
dx =
=∫
2
dx = ∫
⎡ ⎛9
2 ⎞⎤
⎢4 ⎜ 4 − x ⎟⎥
⎠⎦
⎣ ⎝
1
⎡⎛ 3 ⎞ 2
⎤
2
⎢⎜ ⎟ − x ⎥
⎢⎣⎝ 2 ⎠
⎥⎦
⎡ ⎛ 2 9 ⎞⎤
⎢4 ⎜ x − 4 ⎟⎥
⎠⎦
⎣ ⎝
1
⎡ 2 ⎛ 3 ⎞2 ⎤
⎢x − ⎜ ⎟ ⎥
⎝ 2 ⎠ ⎥⎦
⎢⎣
2
dx = ∫
x
2x
+ c = sin −1
+c
3
⎛3⎞
⎜ ⎟
⎝2⎠
2
⎡ 2 ⎛ 3 ⎞2 ⎤
2 ⎢x − ⎜ ⎟ ⎥
⎝ 2 ⎠ ⎦⎥
⎣⎢
dx = cosh −1
4
dx
⎡⎛ 3 ⎞ 2
⎤
2 ⎢⎜ ⎟ − x 2 ⎥
⎣⎢⎝ 2 ⎠
⎦⎥
dx = sin −1
2
1 3
1
sin x − sin 5 x + c
3
5
5
dx
x
2x
+ c = cosh −1
+c
3
⎛3⎞
⎜ ⎟
⎝2⎠
5
Total: 14
Problem 2. Evaluate the following definite integrals, correct to 4 significant figures:
(a)
∫
π/ 2
0
3sin 2 t dt
(b)
∫
π/3
0
3cos 5θ sin 3θ dθ
∫
(c)
2
0
5
dx
4 + x2
Marks
(a)
∫
π/ 2
0
2
3sin t dt =
∫
π/2
0
π/ 2
3 ⎡ sin 2t ⎤
⎡1
⎤
3 ⎢ (1 − cos 2t ) ⎥ dt = ⎢ t −
2⎣
2 ⎥⎦ 0
⎣2
⎦
© 2006 John Bird. All rights reserved. Published by Elsevier.
82
⎡⎛
⎤
⎛π⎞⎞
sin 2 ⎜ ⎟ ⎟
⎢
⎥
⎜
3 π
⎝ 2 ⎠ ⎟ − ⎛ 0 − sin 0 ⎞ ⎥
= ⎢⎜ −
⎜
⎟
2 ⎢⎜ 2
2
2 ⎠⎥
⎟ ⎝
⎟
⎢⎜⎝
⎥
⎠
⎣
⎦
=
(b) ∫
π/3
0
3π
4
or 2.356 correct to 4 significant figures.
3cos 5θ sin 3θ dθ = ∫
π/3
0
5
3
⎡sin ( 5θ + 3θ ) − sin ( 5θ − 3θ ) ⎤⎦ dθ
2⎣
π/3
3 π/3
3 ⎡ cos8θ cos 2θ ⎤
= ∫ ( sin 8θ − sin 2θ ) dθ = ⎢ −
+
0
2
2⎣
8
2 ⎥⎦ 0
⎡⎛
⎤
8π
2π ⎞
cos
cos
⎥
⎟
3 ⎢⎜
cos
0
cos
0
⎛
⎞
3 +
3 − −
+
= ⎢⎜ −
⎟ ⎜
⎟⎥
2 ⎢⎜
8
2 ⎟ ⎝
8
2 ⎠⎥
⎜
⎟
⎠
⎣⎢⎝
⎦⎥
=
3
⎡( 0.0625 − 0.25 ) − ( −0.125 + 0.5 ) ⎤⎦
2⎣
= -0.8438 correct to 4 significant figures.
(c)
∫
2
0
6
2
2
5
1
x⎤
5
⎡5
dx = 5∫ 2
dx = ⎢ tan −1 ⎥ = ⎡⎣ tan −1 1 − tan −1 0 ⎤⎦
2
2
0 2 +x
4+ x
2
2⎦0
⎣2
= 1.963 correct to 4 significant figures.
4
Total: 15
Problem 3. Determine (a)
∫x
x − 11
dx
−x−2
2
(b)
∫ (x
2
3− x
dx
+ 3) ( x + 3)
Marks
(a) Let
A ( x + 1) + B ( x − 2 )
x − 11
x − 11
A
B
≡
=
+
=
x − x − 2 ( x − 2 )( x + 1) ( x − 2 ) ( x + 1)
( x − 2 )( x + 1)
2
Hence,
x – 11 = A(x + 1) + B(x – 2)
Let x = 2:
-9 = 3A hence, A = -3
Let x = -1:
-12 = -3B hence, B = 4
Hence,
x − 11
4
3
=
−
x − x − 2 ( x + 1) ( x − 2 )
2
5
© 2006 John Bird. All rights reserved. Published by Elsevier.
83
∫x
x − 11
4
3
dx = ∫
dx = 4 ln(x + 1) – 3 ln(x – 2) + c
−
−x−2
( x + 1) ( x − 2 )
2
⎧⎪ ( x + 1)4 ⎫⎪
+c
ln ⎨
3⎬
−
x
2
(
)
⎩⎪
⎭⎪
or
4
( Ax + B )( x + 3) + C ( x 2 + 3)
3− x
Ax + B
C
(b) Let 2
≡
+
=
( x + 3) ( x + 3) ( x 2 + 3) ( x + 3)
( x 2 + 3) ( x + 3)
3 – x = (Ax + B)(x + 3) + C ( x 2 + 3)
Hence,
Let x = -3:
6 = 0 + 12C
hence,
C=
x 2 coefficients: 0 = A + C hence, A = −
Hence,
∫
1
2
hence, -1 = −
x coefficients: -1 = 3A + B
3− x
dx =
2
( x + 3) ( x + 3) ∫
=
∫
1
2
3
+B
2
B=
and
1
2
1
1
1
− x+
2
2 + 2 dx
2
( x + 3) ( x + 3)
6
1
1
1
− x
2
+ 22
+ 2 dx
2
( x + 3) ( x + 3) ( x + 3)
1
1⎛ 1
x ⎞ 1
= − ln ( x 2 + 3) + ⎜
tan1
⎟ + ln ( x + 3) + c
4
2⎝ 3
3⎠ 2
1
1
x 1
tan −1
+ ln ( x + 3 ) + c
= − ln ( x 2 + 3 ) +
4
2 3
3 2
6
Total: 21
Problem 4. Evaluate
∫
2
1
3
dx correct to 4 significant figures.
x ( x + 2)
2
Marks
A x ( x + 2 ) + B ( x + 2 ) + Cx 2
3
A B
C
Let 2
≡ + +
=
x ( x + 2) x x 2 ( x + 2)
x 2 ( x + 2)
Hence,
3 = Ax(x + 2) + B(x + 2) + Cx 2
Let x = 0:
3 = 0 + 2B + 0
hence,
B=
3
2
Let x = -2:
3 = 0 + 0 + 4C
hence,
C=
3
4
© 2006 John Bird. All rights reserved. Published by Elsevier.
84
x 2 coefficients:
Hence,
∫
2
1
0=A+C
3
dx =
2
x ( x + 2)
∫
hence, A = −
3 3
3
4 + 2 + 4 dx =
x x 2 ( x + 2)
−
3
4
6
2
3 3
⎡ 3
⎤
−
−
+ ln ( x + 2 ) ⎥
ln
x
⎢⎣ 4
2x 4
⎦1
⎛ 3
⎞ ⎛ 3
3
3
3 3
⎞
+ ln ( 4 ) ⎟⎟ − ⎜ − ln1 − + ln 3 ⎟
= ⎜⎜ − ln 2 −
2 ( 2) 4
2 4
⎠
⎝ 4
⎠ ⎝ 4
= (-0.23014) – (-0.67604) = 0.4459 correct to 4 significant
figures.
6
Total: 12
Problem 5. Determine:
dx
∫ 2sin x + cos x
Marks
If tan
x
2t
1− t2
2 dt
then sin x =
,
cos
x
=
and dx =
2
2
2
1+ t
1+ t
1+ t2
dx
=
Thus ∫
2sin x + cos x
=
∫t
2
2 dt
2 dt
2
2
2 dt
∫ ⎛ 2t 1 ⎞+ t ⎛ 1 − t 2 ⎞ = ∫ 4t1++1t− t 2 = ∫ 1 + 4t − t 2
+
2⎜
2 ⎟ ⎜
2 ⎟
1+ t2
⎝ 1+ t ⎠ ⎝ 1+ t ⎠
−2 dt
−2 dt
=∫
=
− 4t − 1
(t − 2) 2 − 5 ∫
( 5)
3
2 dt
2
− (t − 2) 2
⎡ 1
⎧⎪ 5 + (t − 2) ⎫⎪⎤
ln ⎨
= 2⎢
⎬⎥ + c
⎢⎣ 2 5 ⎪⎩ 5 − (t − 2) ⎪⎭⎥⎦
x⎫
⎧
⎪ 5 − 2 + tan 2 ⎪
1
dx
i.e. ∫
=
ln ⎨
⎬+c
2sin x + cos x
5 ⎪ 5 + 2 − tan x ⎪
2⎭
⎩
5
Total : 8
Problem 6. Evaluate:
∫
π/2
π/3
dx
correct to 3 decimal places.
3 − 2sin x
Marks
If tan
x
2t
then sin x =
2
1+ t2
and dx =
2 dt
1+ t2
© 2006 John Bird. All rights reserved. Published by Elsevier.
85
Then
∫
dx
=
3 − 2sin x ∫
=
2 dt
2 dt
2
2 dt
1+ t
1+ t2
=∫
=
∫
3 + 3t 2 − 4t
⎛ 2t ⎞
3 (1 + t 2 ) − 4t
3 − 2⎜
2 ⎟
⎝ 1+ t ⎠
(1 + t 2 )
∫ 3t
2
2 dt
2 dt
2
dt
2
dt
=∫
= ∫
= ∫
2
2
4
− 4t + 3
⎛
⎞ 3 ⎛ 2⎞
4 3 ⎛ 2⎞ 5
3 ⎜ t 2 − t + 1⎟
⎜ t − ⎟ +1−
⎜t − ⎟ +
3
⎝
⎠
9
⎝ 3⎠
⎝ 3⎠ 9
⎡
⎧ ⎛ 2 ⎞⎫
⎛
2 ⎞⎤
t − ⎟⎥
⎢
⎜
⎪ 3 ⎜ t − 3 ⎟ ⎪⎪
2
dt
2 1
2
⎠
−1
−1 ⎪ ⎝
3
⎟⎥ =
tan ⎜
tan ⎨
= ⎢
= ∫
⎬
2
2
3 ⎛ 2⎞ ⎛ 5 ⎞
3⎢ 5
5
5 ⎪
⎜ 5 ⎟⎥
⎪
⎜
⎟
⎟
⎜t − ⎟ +⎜
⎪⎩
⎝ 3 ⎠ ⎦⎥
⎭⎪
⎣⎢ 3
⎝ 3⎠ ⎝ 3 ⎠
5
Hence,
π/2
⎡
⎡
⎧ ⎛ 2 ⎞ ⎫⎤
⎧ ⎛ π 2 ⎞⎫
⎧ ⎛ π 2 ⎞ ⎫⎤
3 ⎜ t − ⎟ ⎪⎥
3⎜ − ⎟ ⎪
⎢
⎢
⎪
⎪
⎪ 3 ⎜ − 3 ⎟ ⎪⎪⎥
π/2
dx
2
2
⎪
⎪
3
3
⎠
⎠
⎠
−1 ⎪ ⎝
−1 ⎪ ⎝ 2
−1 ⎪ ⎝ 3
∫ π / 3 3 − 2sin x = 5 ⎢⎢ tan ⎨ 5 ⎬⎥⎥ = 5 ⎢⎢ tan ⎨ 5 ⎬ − tan ⎨ 5 ⎬⎥⎥
⎪
⎪
⎪
⎪
⎪
⎪
⎢
⎢
⎪⎩
⎪⎭⎥⎦ π / 3
⎪⎩
⎪⎭
⎪⎩
⎪⎭⎥⎦
⎣
⎣
=
2
[0.38424 − 0.16856] = 0.193. correct to 3 decimal places.
5
5
Total : 10
TOTAL ASSIGNMENT MARKS: 80
© 2006 John Bird. All rights reserved. Published by Elsevier.
86
ASSIGNMENT 12 (PAGE 441)
This assignment covers the material contained in chapters 43 to 45.
Problem 1. Determine the following integrals: (a)
∫ 5x e
2x
dx
(b)
∫t
2
sin 2t dt
Marks
(a)
∫ 5x e
2x
Hence,
dx
∫ 5x e
2x
Let u = 5x then
du
=5
dx
and dv = e 2x dx
1
then v = ∫ e 2x dx = e 2x
2
⎛1
⎞
dx = ( 5x ) ⎜ e 2x ⎟ − ∫
⎝2
⎠
=
(b)
∫t
2
sin 2t dt
Hence,
∫t
2
5 2x
e ( 2x − 1) + c
4
or
du
= 2t
dt
and dv = sin 2t dt
du = 5 dx
5 2x 5 2x dx
⎛ 1 2x ⎞
by parts
⎜ e ⎟ ( 5dx ) = x e − ∫ e
2
2
⎝2
⎠
5 2x 5 2x
xe − e + c
2
4
Let u = t 2 then
from which,
from which, du = 2t dt
then v =
⎛ 1
⎞
sin 2t dt = ( t 2 ) ⎜ − cos 2t ⎟ − ∫
⎝ 2
⎠
1
∫ sin 2t dt = − 2 cos 2x
⎛ 1
⎞
⎜ − cos 2t ⎟ ( 2t dt )
⎝ 2
⎠
1
= − t 2 cos 2t + ⎡ ∫ t cos 2t dt ⎤
⎣
⎦
2
∫ t cos 2t dt
Hence,
Let u = t then
du
=1
dt
and dv = cos 2t
then
5
(1)
4
from which, du = dt
v=
1
∫ cos 2t dt = 2 sin 2t
⎛1
⎞
⎛1
⎞
∫ t cos 2t dt = ( t ) ⎜⎝ 2 sin 2t ⎟⎠ − ∫ ⎜⎝ 2 sin 2t ⎟⎠ dt =
1
1
t sin 2t + cos 2t
2
4
Substituting in equation (1) gives:
∫t
2
1
1
1
sin 2t dt = − t 2 cos 2t + t sin 2t + cos 2t + c
2
2
4
4
Total: 13
© 2006 John Bird. All rights reserved. Published by Elsevier.
87
Problem 2. Evaluate correct to 3 decimal places:
∫
4
1
x ln x dx
Marks
∫
4
1
x ln x dx
Let u = ln x
du 1
=
dx x
then
from which,
du =
dx
x
3
and dv =
then
⎛ 3
2 3
2 ⎜ x2
=
x ln x − ⎜
3
3⎜ 3
⎝ 2
Thus,
∫
v=
∫
x 2 2 23
x dx =
= x
3 3
2
1
⎛2 3⎞
⎛ 2 3 ⎞ du 2 3
2
x ln x − ∫ x 2 dx
x ln x dx = ( ln x ) ⎜ x 2 ⎟ − ∫ ⎜ x 2 ⎟ =
3
3
⎝3 ⎠
⎝3 ⎠ x
∫
Hence,
x dx
1
2
⎞
⎟
2 3
4 3
x ln x −
x +c
⎟+c =
3
9
⎟
⎠
5
4
4 3⎤
⎡2 3
x ln x −
x ⎥
x ln x dx = ⎢
9
⎣3
⎦1
4
1
4 3⎞ ⎛2 3
4 3⎞
⎛2 3
4 ln 4 −
4 ⎟ − ⎜ 1 ln1 −
1 ⎟
=⎜
9
9
⎝3
⎠ ⎝3
⎠
32 ⎞ ⎛ 2
4⎞
⎛ 16
= ⎜ ln 4 − ⎟ − ⎜ ln1 − ⎟
9 ⎠ ⎝3
9⎠
⎝ 3
= (3.83801)-(-0.44444) = 4.282 correct to 3 decimal places.
5
Total: 10
Problem 3. Use reduction formulae to determine: (a)
∫x
3
e3x dx
(b)
∫t
4
sin t dt
Marks
(a) For
∫x
n
e3x dx let u = x n then
and let dv = e3x dx
Then
∫x
Hence,
88
n
⎛1 ⎞
e3x dx = ( x n ) ⎜ e3x ⎟ − ∫
⎝3 ⎠
∫x
n
du
= n x n −1 and du = n x n −1dx
dx
1
from which, v = ∫ e3x dx = e3x
3
1 n 3x n 3x x −1
⎛ 1 3x ⎞
n −1
⎜ e ⎟ ( n x dx ) = x e − ∫ e x dx
3
3
⎝3 ⎠
1
n
e3x dx = I n = x n e3x − I n −1
3
3
3
© 2006 John Bird. All rights reserved. Published by Elsevier.
1
3
When n = 3, I3 = x 3e3x − I 2
3
3
1
2
I 2 = x 2 e3x − I1
3
3
1
1
I1 = x e3x − I0
3
3
1
I0 = ∫ e0 e3x dx = e3x
3
and
∫x
Hence,
3
e3x dx =
1 3 3x 3 ⎡ 1 2 3x 2 ⎧ 1 3x 1 ⎛ 1 3x ⎞ ⎫ ⎤
x e − ⎢ x e − ⎨ x e − ⎜ e ⎟ ⎬⎥
3
3 ⎣3
3 ⎩3
3 ⎝ 3 ⎠ ⎭⎦
=
1 3 3x ⎡ 1 2 3x 2 3x 2 3x ⎤
x e −⎢ x e − xe + e ⎥
3
9
27 ⎦
⎣3
=
1 3 3x 1 2 3x 2 3x 2 3x
x e − x e + xe − e + c
3
3
9
27
or
1
2
2⎤
⎡1
e3x ⎢ x 3 − x 2 + x − ⎥ + c
3
9
27 ⎦
⎣3
4
(b) From equation (3), page 426,
∫x
n
∫t
Hence,
4
sin x dx = I n = − x n cos x + n x n −1 sin x − n(n − 1)I n − 2
sin t dt = I 4 = − t 4 cos t + 4t 3 sin t − 4 ( 3) I 2
I 2 = − t 2 cos t + 2t1 sin t − 2(1)I0
I0 =
and
∫t
Hence,
4
∫ sin t dt = − cos t
sin t dt = − t 4 cos t + 4t 3 sin t − 12 ⎡⎣ − t 2 cos t + 2t sin t − 2 ( − cos t ) ⎤⎦
= −t 4 cos t + 4t 3 sin t + 12t 2 cos t − 24t sin t − 24cos t + c
6
Total: 13
Problem 4. Evaluate
∫
π/ 2
0
cos 6 x dx using a reduction formula.
Marks
From equation (6), page 430,
Hence,
∫
π/ 2
0
cos6 x dx = I6 =
∫
π/2
0
cos n x dx = I n =
n −1
In −2
n
6 −1
5
I4 = I4
6
6
© 2006 John Bird. All rights reserved. Published by Elsevier.
89
and
I4 =
Thus,
∫
π/2
0
3
I2 ,
4
I2 =
1
I0 and
2
∫
π/2
0
1dx = [ x ] 0 =
π/ 2
π
2
5 ⎡ 3 ⎧ 1 ⎛ π ⎞ ⎫⎤ 15
π
⎢ ⎨ ⎜ ⎟ ⎬⎥ =
6 ⎣ 4 ⎩ 2 ⎝ 2 ⎠ ⎭⎦ 96
cos 6 x dx =
6
Total: 6
Problem 5. Evaluate
∫
3
1
5
dx using (a) integration (b) the trapezoidal rule (c) the mid-ordinate
x2
rule (d) Simpson’s rule. In each of the approximate methods use 8 intervals and give the answers
correct to 3 decimal places.
Marks
(a)
∫
3
1
3
⎡ 5x −1 ⎤
⎡1⎤
⎡1 ⎤
5
x
dx
=
⎢ −1 ⎥ = −5 ⎢ x ⎥ = −5 ⎢ 3 − 1⎥
∫1
⎣ ⎦1
⎣
⎦
⎣
⎦1
5
dx =
x2
3
3
−2
= 3.333, correct to 3 decimal places.
(b) With the trapezoidal rule, width of interval =
4
3 −1
= 0.25, hence the ordinates
8
occur at 1, 1.25, 1.5, 1.75, …
x
1
1.25 1.5
5
x2
5
3.2
Hence
1.75
2.0
2.25
2.5
2.75
3.0
.
∫
3
1
.
2.2 1.6327 1.25 0.9877 0.8 0.6612 0.5
5
dx
x2
.
.
⎧1 ⎛
⎫
⎞
= 0.25 ⎨ ⎜ 5 + 0.5 ⎟ + 3.2 + 2.2+ 1.6327 + 1.25 + 0.9877 + 0.8 + 0.6612 ⎬
⎠
⎩2 ⎝
⎭
= 0.25(13.5316) = 3.383
5
(c) With the mid-ordinate rule, the mid-ordinates occur at 1.125, 1.375, 1.625, …
x
5
x2
90
1.125
1.375
1.625
1.875 2.125
2.375
2.625
2.875
3.9506 2.6446 1.8935 1.4222 1.1073 0.8864 0.7256 0.6049
© 2006 John Bird. All rights reserved. Published by Elsevier.
Hence
∫
3
1
5
dx = 0.25{3.9506 + 2.6446 + 1.8935 + 1.4222 + 1.1073 + 0.8864
x2
+ 0.7256 + 0.6049}
= 0.25(13.2351) = 3.309
5
(d) Using Simpson’s rule, using the table of values from part (b) above,
∫
3
1
5
dx
x2
=
.
.
1
⎧
⎫
( 0.25 ) ⎨⎛⎜ 5 + 0.5 ⎞⎟ + 4 ( 3.2 + 1.6327 + 0.9877 + 0.6612 ) + 2 ⎛⎜ 2.2+ 1.25 + 0.8 ⎞⎟ ⎬
3
⎠
⎝
⎠⎭
⎩⎝
=
.
1
( 0.25) 5.5+ 25.9264 + 8.5444 = 3.336
3
{
}
5
Total: 19
Problem 6. An alternating current i has the following values at equal intervals of 5 ms:
Time t (ms)
0
5 10
15
20 25 30
Current i (A) 0 4.8 9.1 12.7 8.8 3.5 0
Charge q, in coulombs, is given by q = ∫
30×10−3
0
i dt
Use Simpson’s rule to determine the approximate charge in the 30 ms period.
Marks
q=∫
30×10−3
0
i dt ≈
=
1
5 ×10−3 ) {( 0 ) + 4 ( 4.8 + 12.7 + 3.5 ) + 2 ( 9.1 + 8.8 )}
(
3
1
( 5 ×10−3 ){0 + 84 + 35.8} = 199.7 ×10−3 = 0.1997 C
3
4
Total: 4
TOTAL ASSIGNMENT MARKS: 65
© 2006 John Bird. All rights reserved. Published by Elsevier.
91
ASSIGNMENT 13 (PAGE 474)
This assignment covers the material contained in chapters 46 to 49.
Problem 1. Solve the differential equation: x
dy
+ x 2 = 5 given that y = 2.5 when x = 1.
dx
Marks
Since x
dy
+ x2 = 5
dx
dy 5
= −x
dx x
then
and y =
⎛5
y = 5ln x −
i.e.
⎞
∫ ⎜⎝ x − x ⎟⎠ dx
x2
+c
2
2
y = 2.5 when x = 1, hence 2.5 = 5 ln 1 – 0.5 + c from which, c = 3
y = 5 ln x −
Hence, the particular solution is:
x2
+3
2
2
Total: 4
Problem 2. Determine the equation of the curve which satisfies the differential equation
2xy
dy
= x 2 + 1 and which passes through the point (1, 2).
dx
Marks
Since 2xy
dy
= x2 +1
dx
2y
then
dy
1
=x+
dx
x
⎛
1⎞
and
∫ 2y dy = ∫ ⎜⎝ x + x ⎟⎠ dx
i.e.
x2
y =
+ ln x + c
2
2
4=
When x = 1, y = 2 hence
Hence, the particular solution is:
or
y2 =
1
+ ln1 + c
2
3
from which, c = 3
1
2
x2
1
+ ln x + 3
2
2
2y 2 = x 2 + 2 ln x + 7
2
Total: 5
92
© 2006 John Bird. All rights reserved. Published by Elsevier.
Problem 3. A capacitor C is charged by applying a steady voltage E through a resistance R. The
p.d. between the plates, V, is given by the differential equation: CR
dV
+V=E
dt
(a) Solve the equation for E given that when time t = 0, V = 0.
(b) Evaluate voltage V when E = 50 V, C = 10 µF, R = 200 kΩ and t = 1.2 s.
Marks
(a) CR
dV
+V=E
dt
hence
dV E − V
=
dt
CR
and
dV
1
∫ E − V = ∫ CR dt
t
+k
CR
3
When time t = 0, V = 0 hence -ln E = k
1
i.e.
Therefore
from which,
i.e.
− ln ( E − V ) =
− ln ( E − V ) =
t
− ln E
CR
ln E − ln ( E − V ) =
t
CR
t
⎛ E ⎞
ln ⎜
⎟=
⎝ E − V ⎠ CR
3
hence
t
E
= e CR
E−V
and
t
−
E−V
= e CR
E
E −V = Ee
and
−
t
CR
t
−
⎛
⎞
V = E ⎜ 1 − e CR ⎟
⎝
⎠
4
(b) When E = 50 V, C = 10µF, R = 200 kΩ and t = 1.2 s,
1.2
−
⎛
10×10−6 × 200×103
V = 50 ⎜1 − e
⎜
⎝
⎞
⎟⎟ = 22.56 V
⎠
2
Total: 14
© 2006 John Bird. All rights reserved. Published by Elsevier.
93
Problem 4. Show that the solution to the differential equation: 4x
(
3y 2 = x 1 − x 3
)
dy x 2 + y 2
is of the form
=
dx
y
given that y = 0 when x = 1
Marks
Rearranging 4x
Let y = vx, then
dy x 2 + y 2
=
dx
y
gives:
dy x 2 + y 2
=
which is homogeneous in x and y.
dx
4xy
dy
dv
= v+x
dx
dx
2
2
dv x 2 + v 2 x 2 x (1 + v ) 1 + v 2
dy
=
=
=
gives: v + x
Substituting for y and
dx
4x(vx)
x 2 (4v)
4v
dx
i.e.
x
(1 + v2 ) − 4v2 = 1 − 3v2
dv 1 + v 2
=
−v=
dx
4v
4v
4v
4v
∫ 1 − 3v
Separating the variables gives:
2
1
dv = ∫ dx
x
2
− ln (1 − 3v 2 ) = ln x + C
3
Integrating both sides gives:
y
Replacing v by gives:
x
2 ⎛ 3y 2 ⎞
− ln ⎜1 − 2 ⎟ = ln x + C
3 ⎝
x ⎠
When y = 0, x = 1, hence,
2
− ln (1) = ln1 + C
3
Hence the solution of 4x
dy x 2 + y 2
=
dx
y
−
2
3
is:
−
2
3
i.e.
⎛ x2 ⎞3
=x
⎜ 2
2 ⎟
⎝ x − 3y ⎠
or
⎛ x2 ⎞
= x 3/ 2
⎜ 2
2 ⎟
⎝ x − 3y ⎠
and
⎛ x2 ⎞
2
2
⎜ 3/ 2 ⎟ = x − 3y
x
⎝
⎠
i.e.
x1/ 2 = x 2 − 3y 2
i.e.
3y 2 = x 2 − x1/ 2
or
2
2
2 ⎛ 3y 2 ⎞
− ln ⎜1 − 2 ⎟ = ln x
3 ⎝
x ⎠
or
= ln x
2
from which, C = 0
⎛ x 2 − 3y 2 ⎞
⎜
⎟
2
⎝ x
⎠
i.e.
⎛ x 2 − 3y 2 ⎞
ln ⎜
⎟
2
⎝ x
⎠
4
(
=x
3y 2 = x 1 − x 3
)
4
Total: 12
94
© 2006 John Bird. All rights reserved. Published by Elsevier.
Problem 5. Show that the solution to the differential equation
x cos x
is given by:
dy
+ ( x sin x + cos x ) y = 1
dx
xy = sin x + k cos x where k is a constant.
Marks
Dividing x cos x
dy
+ ( x sin x + cos x ) y = 1
dx
by x cox x
gives:
dy ⎛ x sin x + cos x ⎞
1
+⎜
⎟y =
dx ⎝
x cos x
x cos x
⎠
i.e.
dy ⎛
1⎞
sec x
+ ⎜ tan x + ⎟ y =
dx ⎝
x⎠
x
which is of the form
dy
1⎞
sec x
⎛
+ Py = Q where P = ⎜ tan x + ⎟ and Q =
dx
x⎠
x
⎝
⎛
4
1⎞
∫ P dx = ∫ ⎜⎝ tan x + x ⎟⎠ dx = ln ( sec x ) + ln x = ln ( x sec x )
2
P dx
Integrating factor = e ∫ = eln(x sec x ) = x sec x
Hence,
⎛ sec x ⎞
y ( x sec x ) = ∫ ( x sec x ) ⎜
⎟ dx
⎝ x ⎠
xy sec x = ∫ sec 2 x dx = tan x + k
i.e.
xy =
i.e.
tan x
k
+
sec x sec x
5
i.e. the general solution is: xy = sin x + k cos x
Total: 11
Problem 6. (a) Use Euler’s method to obtain a numerical solution of the differential equation:
dy y
= + x 2 − 2 given the initial conditions that x = 1 when y = 3, for the range x =1.0(0.1)1.5
dx x
(b) Apply the Euler-Cauchy method to the differential equation given in part (a) over the same
range.
(c) Apply the integrating factor method to solve the differential equation in part (a) analytically.
(d) Determine the percentage error, correct to 3 significant figures, in each of the two numerical
methods when x = 1.2
© 2006 John Bird. All rights reserved. Published by Elsevier.
95
Marks
(a) Using Euler’s method, let
dy
y
= y ' = + x2 − 2
dx
x
3
2
Initially, x = 1 when y = 3, hence ( y ')0 = + (1) − 2 = 2
1
Hence, line 1 of Table 13.1 is completed.
Table 13.1
From line 2, where x 0 = 1.1 and h = 0.1, y1 = y 0 + h ( y ' ) = 3 + 0.1(2) = 3.2
and
( y ' )0 =
y1
3.2
2
2
+ ( x0 ) − 2 =
+ (1.1) − 2 = 2.119090909
x0
1.1
Table 13.1 is completed by similar calculations.
(b) Using the Euler-Cauchy method, let y ' =
8
y
+ x2 − 2
x
3
2
Initially, x = 1 when y = 3 hence ( y ')0 = + (1) − 2 = 2
1
y P1 = y 0 + h ( y ' )0 = 3 + 0.1(2) = 3.2
(
)
1
y C 1 = y0 + h ⎡( y ')0 + f x1 , y P1 ⎤
⎦
2 ⎣
yP
⎤
1 ⎡
1
2
⎡ 3.2
⎤
2
y C 1 = y0 + h ⎢( y ' )0 + 1 + x1 − 2 ⎥ = 3 + ( 0.1) ⎢ 2 +
+ (1.1) − 2 ⎥ = 3.2059545
2 ⎣
x1
2
⎣ 1.1
⎦
⎦
( y ') 1 =
y P1
x1
+ x1 − 2 =
2
3.2059545
2
+ (1.1) − 2 = 2.124504091
1.1
The first two lines of Table 13.2 are thus completed; the remaining calculations
are achieved in a similar way to that shown above.
96
© 2006 John Bird. All rights reserved. Published by Elsevier.
8
Table 13.2
(c) Rearranging
∫
dy y
dy ⎛ 1 ⎞
− ⎜ ⎟ y = x2 − 2
= + x 2 − 2 gives:
dx x
dx ⎝ x ⎠
− ln x dx
ln x −1 dx
1
1
− dx = − ln x hence, integrating factor = e ∫
= e∫
= x −1 or
x
x
⎛1⎞
Thus y ⎜ ⎟ = ∫
⎝x⎠
2
y
2⎞
x2
⎛
⎛1⎞ 2
x
dx
i.e.
=
−
=
− 2 ln x + c
−
x
2
dx
(
)
⎜
⎟
⎜ ⎟
x ∫⎝
x⎠
2
⎝x⎠
x = 1 when y = 3, hence
Hence,
2
y x2
5
=
− 2 ln x +
x 2
2
(d) When x = 1.2, y =
3 1
5
= − 2 ln1 + c from which, c =
1 2
2
or
y=
1 3
5
x − 2x ln x + x
2
2
5
1
5
3
(1.2 ) − 2 (1.2 ) ln1.2 + (1.2 ) = 3.426428264
2
2
From Table 13.1, with the Euler method, when x = 1.2, y = 3.41190909
1
Hence, percentage error in Euler method
⎛ 3.41190909 − 3.426428264 ⎞
=⎜
⎟ × 100% = -0.426%
3.41190909
⎝
⎠
2
From Table 13.2, with the Euler-Cauchy method, when x = 1.2, y = 3.426613242
Hence, percentage error in the Euler-Cauchy method
⎛ 3.426613242 − 3.426428264 ⎞
=⎜
⎟ × 100% = 0.00540%
3.426613242
⎝
⎠
2
Total: 30
© 2006 John Bird. All rights reserved. Published by Elsevier.
97
Problem 7. Use the Runge-Kutta method to solve the differential equation:
dy
y
= + x2 − 2
dx
x
in the range 1.0(0.1)1.5, given the initial conditions that at x = 1, y = 3. Work to an accuracy of 6
decimal places
Marks
Using the Runge-Kutta procedure:
1. x 0 = 1, y 0 = 3 and since h = 0.1, and the range is from x = 1.0 to x = 1.5, then
x1 = 1.1, x 2 = 1.2, x 3 = 1.3, x 4 = 1.4, and x 5 = 1.5
2
Let n = 0 to determine y 1 :
2. k1 = f ( x 0 , y 0 ) = f (1, 3); since
dy
y
3
= + x 2 − 2 , f (1, 3) = + 12 − 2 = 2.0
x
1
dx
h
h ⎞
0.1
⎛
⎛ 0.1
⎞
, 3+
(2.0) ⎟ = f (1.05, 3.1)
3. k 2 = f ⎜ x 0 + , y 0 + k1 ⎟ = f ⎜ 1 +
2
2 ⎠
2
2
⎝
⎝
⎠
=
3.1
+ 1.052 − 2 = 2.054881
1.05
h
h ⎞
0.1
⎛
⎛ 0.1
⎞
, 3+
(2.054881) ⎟ = f (1.05, 3.102744 )
4. k 3 = f ⎜ x 0 + , y0 + k 2 ⎟ = f ⎜1 +
2
2 ⎠
2
2
⎝
⎝
⎠
=
3.102744
+ 1.052 − 2 = 2.057494
1.05
5. k 4 = f ( x 0 + h, y 0 + hk 3 ) = f (1 + 0.1, 3 + 0.1(2.057494) ) = f (1.1, 3.205749)
=
6. y n +1 = y n +
y1 = y0 +
3.205749
+ 1.12 − 2 = 2.124317
1.1
h
{k1 + 2k 2 + 2k 3 + k 4 } and when n = 0:
6
h
{k1 + 2k 2 + 2k 3 + k 4 }
6
=3+
0.1
{ 2.0 + 2(2.054881) + 2(2.057494) + 2.124317 }
6
=3+
0.1
{ 12.349067 } = 3.205818
6
7
A table of values may be constructed as shown in Table 13.3. The working has
been shown for the first two rows.
98
© 2006 John Bird. All rights reserved. Published by Elsevier.
n
xn
k1
k2
k4
0
1
1
1.1
2.0
2.054881
2.057494
2.124317
3.205818
2
1.2
2.124380 2.202532
2.205930
2.295343
3.426429
3
1.3
2.295358 2.395458
2.399462
2.510289
3.666354
3
4
1.4
2.510272 2.631291
2.635773
2.767094
3.929879
3
5
1.5
2.767056 2.908177
2.913043
3.064122
4.221106
3
k3
yn
3
Table 13.3
Let n = 1 to determine y2 :
2. k1 = f ( x1 , y1 ) = f (1.1, 3.205818); since
dy
y
= + x2 − 2 ,
dx
x
f (1.1, 3.205818) =
3.205818
+ 1.12 − 2 = 2.124380
1.1
h
h ⎞
0.1
0.1
⎛
⎛
⎞
,3.205818 +
(2.124380) ⎟
3. k 2 = f ⎜ x1 + , y1 + k1 ⎟ = f ⎜1.1 +
2
2 ⎠
2
2
⎝
⎝
⎠
= f (1.15, 3.312037) =
3.312037
+ 1.152 − 2 = 2.202532
1.15
h
h ⎞
0.1
0.1
⎛
⎛
⎞
, 3.205818 +
(2.202532) ⎟
4. k 3 = f ⎜ x1 + , y1 + k 2 ⎟ = f ⎜1.1 +
2
2 ⎠
2
2
⎝
⎝
⎠
= f (1.15, 3.315945 ) =
3.315945
+ 1.152 − 2 = 2.205930
1.15
5. k 4 = f ( x1 + h, y1 + hk 3 ) = f (1.1 + 0.1, 3.205818 + 0.1(2.205930) )
= f (1.2, 3.426411) =
6. y n +1 = y n +
y 2 = y1 +
3.426411
+ 1.22 − 2 = 2.295343
1.2
h
{k1 + 2k 2 + 2k 3 + k 4 } and when n = 1:
6
h
{k1 + 2k 2 + 2k 3 + k 4 }
6
= 3.205818 +
0.1
{ 2.124380 + 2(2.202532) + 2(2.205930) + 2.295343 }
6
= 3.205818 +
0.1
{ 13.236647 } = 3.426429
6
6
This completes the third row of Table 13.3. In a similar manner y3 , y 4 and y5 can be
© 2006 John Bird. All rights reserved. Published by Elsevier.
99
calculated and the results are as shown in Table 13.3.
Total: 24
TOTAL ASSIGNMENT MARKS: 100
100
© 2006 John Bird. All rights reserved. Published by Elsevier.
ASSIGNMENT 14 (PAGE 525)
This assignment covers the material contained in chapters 50 to 53.
Problem 1. Find the particular solution of the following differential equations:
(a) 12
(b)
d2 y
dy 1
− 3y = 0 given that when t = 0, y = 3 and
=
2
dt
dt 2
dy
d2 y
dy
=1
+ 2 + 2y = 10 e x given that when x = 0, y = 0 and
2
dx
dx
dx
Marks
(a) The auxiliary equation is: 12 m 2 − 3 = 0
from which, m 2 =
3
1
and m = ±
12
2
1
t
2
Hence, the general solution is:
y = A e + Be
When t = 0, y = 3 thus
3=A+B
1
− t
2
3
(1)
1
t
dy 1
1 − 12 t
2
= A e − Be
dt 2
2
When t = 0,
dy 1
=
dt 2
thus
2 × equation (2) gives:
1 1
1
= A− B
2 2
2
(2)
1=A–B
(3)
Equation (1) – equation (3) gives: 2 = 2 B and B = 1
From either equation (1) or (2),
A=2
1
Hence, the particular solution is:
t
y = 2e 2 + e
1
− t
2
5
(b) The auxiliary equation is: m 2 + 2m + 2 = 0
m=
and
−2 ± 22 − 4(1)(2)
= −1 ± j1
2
Hence, the complimentary function, u = e − x ( A cos x + Bsin x )
3
Let the particular integral, v = k e x
then
i.e.
(D
2
+ 2D + 2 ) ⎡⎣ k e x ⎤⎦ = 10 e x
k e x + 2k e x + 2k e x = 10 e x
© 2006 John Bird. All rights reserved. Published by Elsevier.
101
5k e x = 10 e x from which, k = 2
i.e.
and the particular integral, v = 2 e x
y = u + v = e − x ( A cos x + B sin x ) + 2e x
Thus,
When x = 0, y = 0
thus
0=A+2
5
from which, A = -2
dy
= e − x ( −A sin x + Bcos x ) + ( A cos x + Bsin x ) ( −e − x ) + 2 e x
dx
When x = 0,
dy
= 1 thus 1 = B – A + 2
dx
i.e.
1=B+2+2
and
B = -3
Hence, the particular solution is: y = e − x ( −2cos x − 3sin x ) + 2e x
or
y = −e − x ( 2cos x + 3sin x ) + 2e x
4
Total: 20
Problem 2. In a galvanometer the deflection θ satisfies the differential equation:
d 2θ
dθ
+2 +θ= 4
2
dt
dt
Solve the equation for θ given that when t = 0, θ = 0 and
dθ
= 0.
dt
Marks
The auxiliary equation is: m 2 + 2m + 1 = 0
from which,
i.e.
(m + 1)(m + 1) = 0
m = -1 twice
Thus, the complimentary function is: u = ( At + B ) e − t
Let the particular integral, v = k
from which,
then
(D
2
+ 2D + 1) k = 4
k = 4 and thus, v = 4
2
Hence, the particular solution is:
θ = u + v = ( At + B ) e − t + 4
When t = 0, θ = 0,
0 = B + 4 from which, B = -4
thus
3
dθ
= ( At + B ) ( −e − t ) + ( e − t ) (A)
dt
© 2006 John Bird. All rights reserved. Published by Elsevier.
102
When t = 0,
dθ
=0
dt
hence
0 = -B + A
i.e.
0 = -(-4) + A
Hence, the particular solution is:
θ = ( −4t − 4 ) e − t + 4
or
from which,
A = -4
θ = 4 − 4 ( t + 1) e − t
7
Total: 12
Problem 3. Determine y (n ) when y = 2x 3e 4x
Marks
With y = 2x 3e4x , let v = 2x 3 , since its fourth derivative is zero,
u = e 4x since the n’th derivative is 4n eax (from equation (1), page 555)
and
2
Using Leinbiz’s theorem,
y (n ) = u (n ) v + nu (n −1) v(1) +
n(n − 1) (n − 2) (2) n(n − 1)(n − 2) (n −3) (3)
u
v +
u
v + ....
2!
3!
where in this case v = 2x 3 , v (1) = 6x 2 , v(2) = 12x , v (3) = 12 and v( ) = 0
4
Hence,
n(n − 1) n − 2 4x
( 4 e ) (12x)
2!
n(n − 1)(n − 2) n −3 4x
n(n − 1)(n − 2)(n − 3) n − 4 4x
+
4 e ) (12 ) +
(
( 4 e ) ( 0)
3!
4!
y (n ) = ( 4n e 4x )( 2x 3 ) + n ( 4n −1 e4x )( 6x 2 ) +
n(n − 1) 2
n(n − 1)(n − 2)
⎛
4 ) (12x ) +
= 4n − 4 e 4x ⎜ 44 2x 3 + n(43 )(6x 2 ) +
( 4 )(12 ) + 0 ⎞⎟
(
2
6
⎝
⎠
i.e. y (n) = e4x 4n − 4 ( 512x 3 + 384n x 2 + 96 x n(n − 1) + 8n(n − 1)(n − 2) )
8
Total: 10
Problem 4. Determine the power series solution of the differential equation:
d2 y
dy
+ 2x
+y=0
2
dx
dx
using Leibniz-Maclaurin’s method, given the boundary conditions that at x = 0, y = 2 and
dy
= 1.
dx
Marks
The differential equation may be rewritten as: y′′ + 2xy′ + y = 0 and from the Leibniz
© 2006 John Bird. All rights reserved. Published by Elsevier.
103
theorem, each term is differentiated n times, which gives:
y (n + 2) + { y(n +1) (2x) + n y(n ) (2) + 0} + y(n ) = 0
y (n + 2) + 2x y(n +1) + (2n + 1) y(n ) = 0
i.e.
(1)
2
y (n + 2) + (2n + 1) y(n ) = 0
At x = 0, equation (1) becomes:
y (n + 2) = −(2n + 1) y (n )
from which, the recurrence relation is:
2
Substituting n = 0, 1, 2, 3, … gives:
For n = 0,
( y '')0 = − ( y )0
n = 1,
( y ''')0 = −3 ( y ')0
n = 2,
(y )
n = 3,
(y )
n = 4,
(y )
0
= −9 ( y(4) ) = −9 {1× 5 ( y )0 } = −1× 5 × 9 ( y )0
n = 5,
(y )
0
= −11( y (5) ) = −11{3 × 7 ( y ')0 } = −3 × 7 × 11( y ')0
n = 6,
(y )
0
= −13 ( y (6) ) = −13{−1× 5 × 9 ( y )0 } = 1× 5 × 9 × 13 ( y )0
{
}
0
= −5 ( y '' )0 = −5 − ( y )0 = 1× 5 ( y )0
0
= −7 ( y ''' )0 = −7 {−3 ( y ')0 } = 3 × 7 ( y ')0
(4)
(5)
(6)
(7)
(8)
0
0
0
Maclaurin’s theorem states: y = ( y )0 + x ( y ')0 +
5
x2
x3
x4
( y '')0 + ( y ''')0 + ( y(4) )0 + ....
2!
3!
4!
Substituting the above values into Maclaurin’s theorem gives:
x2
x3
x4
x5
y = ( y )0 + x ( y ' )0 + {− ( y )0 } + {−3 ( y ')0 } + {1× 5 ( y )0 } + {3 × 7 ( y ' )0 }
2!
3!
4!
5!
x6
x7
x8
+ {−1× 5 × 9 ( y )0 } + {−3 × 7 × 11( y ')0 } + {1× 5 × 9 × 13 ( y )0 }
6!
7!
8!
5
Collecting similar terms together gives:
⎧ 1x 2 1× 5 x 4 1× 5 × 9 x 6 1× 5 × 9 × 13 x 8
⎫
y = ( y )0 ⎨1 −
+
−
+
− ...⎬
2!
4!
6!
8!
⎩
⎭
⎧
⎫
3 x 3 3 × 7 x 5 3 × 7 × 11x 7
+ ( y ')0 ⎨ x −
+
−
+ ...⎬
3!
5!
7!
⎩
⎭
The boundary conditions are: at x = 0, y = 2 and
dy
= 1, i.e. ( y )0 = 2 and ( y ')0 = 1
dx
© 2006 John Bird. All rights reserved. Published by Elsevier.
104
Hence, the power series solution of the differential equation:
d2 y
dy
+ 2x
+ y = 0 is:
2
dx
dx
1 × 5 4 1 × 5 × 9 6 1 × 5 × 9 × 13 8
⎧ 1
⎫
x −
x +
x − ...⎬
y = 2 ⎨1 − x 2 +
4!
6!
8!
⎩ 2!
⎭
3
3 × 7 5 3 × 7 × 11 7
⎧
⎫
x −
x + ...⎬
+ ⎨x − x3 +
3!
5!
7!
⎩
⎭
6
Total: 20
Problem 5. Use the Frobenius method to determine the general power series solution of the
differential equation:
d2 y
+ 4y = 0
dx 2
Marks
The differential equation may be rewritten as: y′′ + 4y = 0
Let a trial solution be of the form: y = xc{a0 + a1x + a2x2 + a3x3 + .. + arxr+ ..} (1)
where a0 ≠ 0,
i.e.
y = a0 xc + a1xc+1 + a2xc+2 + a3xc+3 + … + arxc+r +…
(2)
Differentiating equation (2) gives:
y′ = a0cxc-1 + a1(c + 1)xc + a2(c + 2)xc+1 + …. + ar(c + r)xc+r-1 + …
and y′′ = a0c(c – 1)xc-2 + a1c(c + 1)xc-1 + a2(c + 1)(c + 2)xc + ….
+ ar(c + r - 1)(c + r)xc+r-2 + …
Replacing r by (r + 2) in ar(c + r - 1)(c + r)xc+r-2 gives: ar+2(c + r + 1)(c + r+ 2)xc+r
Substituting y and y′′ into each term of the given equation y′′ + 4y = 0 gives:
y′′ + 4y = a0c(c – 1)xc-2 + a1c(c + 1)xc-1 + [a2(c + 1)(c + 2) + 4a0] xc + ….
+ [ar+2(c + r + 1)(c + r + 2) + 4ar] xc+r + … = 0
6
The indicial equation is obtained by equating the coefficient of the lowest power of
x to zero.
Hence,
a0c(c – 1) = 0
from which,
c = 0 or c = 1
since a0 ≠ 0
2
For the term in xc-1, i.e. a1c(c + 1) = 0
With c = 1, a1 = 0; however, when c = 0, a1 is indeterminate, since any value of a1
© 2006 John Bird. All rights reserved. Published by Elsevier.
105
combined with the zero value of c would make the product zero.
For the term in xc, a2(c + 1)(c + 2) + 4a0 = 0 from which, a 2 =
For the term in xc+r,
−4a 0
(c + 1)(c + 2)
(3)
2
ar+2(c + r + 1)(c + r + 2) + 4ar = 0
a r+2 =
from which,
−4 a r
(c + r + 1)(c + r + 2)
2
When c = 0: a1 is indeterminate, and from equation (3)
a2 =
In general, a r + 2 =
−4a 0 −4a 0
=
(1× 2)
2!
−4 a r
−4a1
−4a1
−4a1
=
=
and when r = 1, a 3 =
(r + 1)(r + 2)
(2 × 3) (1× 2 × 3)
3!
when r = 2, a 4 =
−4 a 2
−4 ⎛ −4 a 0 ⎞ 16a 0
=
⎜
⎟=
3 × 4 3 × 4 ⎝ 2! ⎠
4!
4a
16a 0 4
4a
⎧
⎫
Hence, y = x 0 ⎨a 0 + a1x − 0 x 2 − 1 x 3 +
x + ...⎬
2!
3!
4!
⎩
⎭
from equation (1)
5
⎧ 4x 2 16x 4
⎫
⎧
⎫
4x 3 16x 5
= a 0 ⎨1 −
+
+ ...⎬ + a1 ⎨ x −
+
+ ...⎬
2!
4!
3!
5!
⎩
⎭
⎩
⎭
Since a 0 and a1 are arbitrary constants depending on boundary conditions, let
a 0 = A and a1 = B, then:
⎧
⎫
⎧
⎫
4
42
4
42
y = A ⎨1 − x 2 + x 4 + ...⎬ + B ⎨ x − x 3 + x 5 + ...⎬
4!
3!
5!
⎩ 2!
⎭
⎩
⎭
(4)
4
[When two solutions of the indicial equation differ by an integer, as in this problem
where c = 0 and 1, and if one coefficient is indeterminate, as with when c = 0,
then the complete solution is always given by using this value of c. Using the second
value of c, i.e. c = 1 will merely give a series which is one of the series in the first
solution, i.e. in equation (4).]
Total: 21
Problem 6. Determine the general power series solution of Bessel’s equation:
x2
d2 y
dy
+x
+ ( x 2 − v 2 ) y = 0 and hence state the series up to and including the term
2
dx
dx
in x 6 when v = +3.
Bessel’s equation may be written as: x y ''+ xy '+ ( x − v ) y = 0
2
2
Marks
2
© 2006 John Bird. All rights reserved. Published by Elsevier.
106
Let a trial solution be of the form y = xc{a0 + a1x + a2x2 + a3x3 + … + arxr+…}
where a0 ≠ 0,
i.e.
y = a0 xc + a1xc+1 + a2xc+2 + a3xc+3 + … + arxc+r +…
(1)
Differentiating equation (1) gives:
y′ = a0cxc-1 + a1(c + 1)xc + a2(c + 2)xc+1 + …. + ar(c + r)xc+r-1 + …
and y′′ = a0c(c – 1)xc-2 + a1c(c + 1)xc-1 + a2(c + 1)(c + 2)xc + ….
+ ar(c + r - 1)(c + r)xc+r-2 + …
Substituting y, y′ and y′′ into each term of x 2 y ''+ xy '+ ( x 2 − v 2 ) y = 0 gives:
a0c(c – 1)xc + a1c(c + 1)xc+1 + a2(c + 1)(c + 2)xc+2 + …. + ar(c + r - 1)(c + r)xc+r + …
+ a0cxc + a1(c + 1)xc+1 + a2(c + 2)xc+2 + …. + ar(c + r)xc+r + …
+ a0 xc+2 + a1xc+3 + a2xc+4 +...+ arxc+r+2 +…- a0 v 2 xc - a1 v 2 xc+1 -...- ar v 2 xc+r +…= 0
6
The indicial equation is obtained by equating the coefficient of the lowest power of
x to zero, i.e. a0c(c – 1) + a0c - a0 v 2 = 0
from which,
a 0 ⎡⎣c 2 − c + c − v 2 ⎤⎦ = 0
i.e.
a 0 ⎡⎣ c 2 − v 2 ⎤⎦ = 0 from which, c = +v or c = -v since a0 ≠ 0
2
For the term in xc+r, ar(c + r - 1)(c + r) + ar (c + r) + ar-2 - ar v 2 = 0
ar [(c + r – 1)(c + r) + (c + r) - v 2 ] = - ar-2
i.e.
ar [ (c + r)(c + r − 1 + 1) − v 2 ] = - ar-2
i.e.
ar [( (c + r) 2 − v 2 ] = - ar-2
i.e. the recurrence relation is:
For the term in xc+1,
ar =
ar − 2
v − (c + r)2
2
for r ≥ 2
(2)
3
a1 [c(c + 1) + (c + 1) - v 2 ] = 0
i.e.
a1 [ (c + 1) 2 - v 2 ] = 0
but if c = v
a1 [ (v + 1) 2 - v 2 ] = 0
i.e.
a1 [ 2v + 1 ] = 0
Similarly, if c = -v
a1 [ 1 − 2v ] = 0
© 2006 John Bird. All rights reserved. Published by Elsevier.
107
The terms (2v + 1) and (1 –2v) cannot both be zero since v is a real constant, hence
a1 = 0. Since a1 = 0, then from equation (2)
and
a2 =
a0
v − (c + 2) 2
a6 =
a0
and so on.
⎡⎣ v − (c + 2) ⎤⎦ ⎡⎣ v − (c + 4) 2 ⎤⎦ ⎡⎣ v 2 − (c + 6) 2 ⎤⎦
a4 =
2
2
2
When c = +v, a 2 =
a4 =
a 3 = a 5 = a 7 = ..... = 0
a0
⎡⎣ v − (c + 2) ⎤⎦ ⎡⎣ v 2 − (c + 4) 2 ⎤⎦
2
2
=
3
2
a0
a0
−a 0
−a
= 2
=
= 2 0
2
2
v − (v + 2)
v − v − 4v − 4 4 + 4v 2 ( v + 1)
2
a0
a0
a0
=
= 5
2
3
2
2
⎡⎣ v − (v + 2) ⎤⎦ ⎡⎣ v − (v + 4) ⎤⎦ [−2 (v + 1)][−2 (v + 2)] 2 (v + 1)(v + 2)
2
2
=
a6 =
2
a0
2 × 2(v + 1)(v + 2)
4
a0
a0
= 4
2
2
2
⎡⎣ v − (v + 2) ⎤⎦ ⎡⎣ v − (v + 4) ⎤⎦ ⎡⎣ v − (v + 6) ⎤⎦ ⎡⎣ 2 × 2(v + 1)(v + 2) ⎤⎦ [ −12(v + 3)]
2
2
2
−a 0
−a 0
= 6
and so on.
2
2 × 2(v + 1)(v + 2) × 2 × 3(v + 3) 2 × 3!(v + 1)(v + 2)(v + 3)
4
The resulting solution for c = +v is given by:
⎧
⎫
x2
x4
x6
y = A x v ⎨1 − 2
+ 4
− 6
+ ...⎬
⎩ 2 (v + 1) 2 × 2!(v + 1)(v + 2) 2 × 3!(v + 1)(v + 2)(v + 3)
⎭
6
which is valid provided v is not a negative integer and where A is an arbitrary constant.
When v = +3,
⎧
⎫
x2
x4
x6
y = A x ⎨1 − 2
+ 4
− 6
+ ...⎬
⎩ 2 (3 + 1) 2 × 2!(3 + 1)(3 + 2) 2 × 3!(3 + 1)(3 + 2)(3 + 3)
⎭
3
⎧ x2
⎫
⎫
x4
x6
x2 x4
x6
3 ⎧
i.e. y = A x ⎨1 − 4 + 7
− 6
+ ...⎬ or y = A x ⎨1 − +
−
+ ...⎬
⎩ 16 640 46 080
⎭
⎩ 2 2 × 5 2 × 6!
⎭
3
4
Total: 26
Problem 7. Determine the general solution of
∂u
= 5xy
∂x
© 2006 John Bird. All rights reserved. Published by Elsevier.
108
Marks
Since
∂u
= 5xy then integrating partially with respect to x gives:
∂x
u=
∫
x2
5
5xy dx = ( 5y ) + f (y) = x 2 y + f (y)
2
2
2
Total: 2
Problem 8. Solve the differential equation
= 0,
∂2u
= x 2 ( y − 3) given the boundary conditions that at x
2
∂x
∂u
= sin y and u = cos y.
∂x
Marks
Since
∂2u
= x 2 ( y − 3) then integrating partially with respect to x gives:
2
∂x
∂u
x3
= ∫ x 2 ( y − 3) dx = ( y − 3) ∫ x 2 dx = ( y − 3) + f (y) where f (y) is an
3
∂x
arbitrary function.
From the boundary conditions, when x = 0,
Hence,
Now
sin y = ( y − 3)
(0)
3
1
∂u
= sin y
∂x
3
+ f (y) from which, f (y) = sin y
∂u
x3
= ( y − 3) + sin y
3
∂x
2
Integrating partially with respect to x gives:
⎡
⎤
x4
x3
u = ∫ ⎢( y − 3) + sin y ⎥ dx =
( y − 3) + x(sin y) + F(y)
12
3
⎣
⎦
1
From the boundary conditions, when x = 0, u = cos y, hence
cos y =
(0) 4
( y − 3) + (0) sin y + F(y)
12
from which, F(y) = cos y
1
∂2u
= x 2 ( y − 3) for the given boundary conditions is:
Hence, the solution of
2
∂x
x4
u=
( y − 3 ) + x sin y + cos y
12
1
Total: 6
© 2006 John Bird. All rights reserved. Published by Elsevier.
109
Problem 9. Figure A14.1 shows a stretched string of length 40 cm which is set oscillating by
displacing its mid-point a distance of 1 cm from its rest position and releasing it with zero velocity.
∂2u 1 ∂2u
= 2 2 where c2 = 1, to determine the resulting
Solve the wave equation:
2
∂x
c ∂t
motion u(x, t).
Figure A14.1
Marks
Following the procedure for the solution of the wave equation:
1. The boundary and initial conditions given are:
u(0, t) = 0 ⎫
⎬ i.e.fixed end po int s
u(40, t) = 0 ⎭
u(x, 0) = f (x) =
1
x
20
=−
0 ≤ x ≤ 20
1
40 − x
x+2=
20
20
1
20 ≤ x ≤ 40
2
⎡ ∂u ⎤
⎢⎣ ∂t ⎥⎦ = 0 i.e. zero initial velocity
t =0
2. Assuming a solution u = XT, where X is a function of x only, and T is a function
∂u
∂2u
= X 'T and
= X ''T and
of t only, then
∂x
∂x 2
Substituting into the partial differential equation,
gives:
X ''T =
1
XT ''
c2
∂2u 1 ∂2u
=
∂x 2 c2 ∂t 2
i.e. X ''T = XT '' since c 2 = 1
X '' T ''
X ''
=
then µ =
X
T
X
2
X '' T ''
=
X
T
3. Separating the variables gives:
Let constant, µ =
∂u
∂2u
= XT ' and
= XT ''
∂y
∂y 2
and
µ=
T ''
T
© 2006 John Bird. All rights reserved. Published by Elsevier.
110
X′′ - µX = 0 and T′′ - µ T = 0
from which,
2
4. Letting µ = - p 2 to give an oscillatory solution gives
X′′ + p 2 X = 0
and
T′′ + p 2 T = 0
The auxiliary equation for each is: m 2 + p 2 = 0 from which, m = − p 2 = ± jp
2
5. Solving each equation gives:
X = A cos px + B sin px
and
T = C cos pt + D sin pt
Thus, u(x, t) = {A cos px + B sin px}{C cos pt + D sin pt}
1
6. Applying the boundary conditions to determine constants A and B gives:
(i) u(0, t) = 0, hence 0 = A{C cos pt + D sin pt} from which, A = 0
Therefore, u(x, t) = B sin px {C cos pt + D sin pt}
(a)
1
(ii) u(40, t) = 0, hence 0 = B sin 40p{C cos pt + D sin pt}
B ≠ 0 hence sin 40p = 0 from which, 40p = nπ and p =
7. Substituting in equation (a) gives: u(x, t) = B sin
∞
u n (x, t) = ∑ sin
or, more generally,
n =1
nπ
40
2
nπt
nπt ⎫
nπx ⎧
+ D sin
⎨C cos
⎬
40
40 ⎭
40 ⎩
nπx ⎧
nπt
nπt ⎫
+ Bn sin
⎨A n cos
⎬
40 ⎩
40
40 ⎭
(b)
1
where A n = BC and Bn = BD
8. From equation (8), page 517,
An =
=
2 L
nπx
f (x) sin
dx
∫
L 0
L
40 ⎛ 40 − x ⎞
2 ⎡ 20 ⎛ 1 ⎞
nπx
nπx ⎤
+
x
sin
dx
dx ⎥
⎜
⎟
⎜
⎟ sin
⎢
∫
∫
0
20
40 ⎣ ⎝ 20 ⎠
40
40
⎝ 20 ⎠
⎦
Each integral is determined using integration by parts with the result:
An =
8
nπ
sin
2
n π
2
4
2
From equation (9), page 518, Bn =
2 L
nπx
g(x) sin
dx
∫
0
cnπ
L
© 2006 John Bird. All rights reserved. Published by Elsevier.
111
⎡ ∂u ⎤
⎢⎣ ∂t ⎥⎦ = 0 = g(x) thus, Bn = 0
t =0
1
Substituting into equation (b) gives:
∞
u n (x, t) = ∑ sin
n =1
∞
=
∑ sin
n =1
Hence,
u(x, t) =
8
π2
∞
nπx ⎧
nπt
nπt ⎫
+ Bn sin
⎨A n cos
⎬
40 ⎩
40
40 ⎭
nπx ⎧ 8
nπ
nπt
nπt ⎫
+ (0) sin
⎨ 2 2 sin cos
⎬
40 ⎩ n π
2
40
40 ⎭
1
∑n
n =1
2
sin
nπx
nπ
nπt
sin
cos
40
2
40
4
Total: 23
TOTAL ASSIGNMENT MARKS: 140
© 2006 John Bird. All rights reserved. Published by Elsevier.
112
ASSIGNMENT 15 (PAGE 551)
This assignment covers the material contained in chapters 54 to 56.
Problem 1. A company produces five products in the following proportions:
Product A 24
Product B 16
Product C 15
Product D 11
Product E 6
Present these data visually by drawing (a) a vertical bar chart (b) a percentage component bar chart
(c) a pie diagram.
Marks
(a) A vertical bar chart is shown in Figure 34.
3
Figure 34
(b) For the percentage bar chart, 24 + 16 + 15 + 11 + 6 = 72, hence,
A≡
24
16
15
× 100% = 33.3% , B ≡ × 100% = 22.2% , C ≡ × 100% = 20.8% ,
72
72
72
D≡
11
6
×100% = 15.3% and E ≡ × 100% = 8.3% .
72
72
3
A percentage component bar chart is shown in Figure 35.
2
Figure 35
(c) Total number of products = 24 + 16 + 15 + 11 + 6 = 72
© 2006 John Bird. All rights reserved. Published by Elsevier.
113
A≡
24
16
15
× 360° = 120° , B ≡ × 360° = 80° , C ≡ × 360° = 75° ,
72
72
72
D≡
11
6
× 360° = 55° and E ≡ × 360° = 30° .
72
72
3
A pie diagram is shown in Figure 36.
2
Figure 36
Total: 13
Problem 2. The following lists the diameters of 40 components produced by a machine, each
measured correct to the nearest hundredth of a centimetre:
1.39 1.36 1.38 1.31 1.33 1.40 1.28 1.40 1.24 1.28 1.42 1.34 1.43 1.35
1.36 1.36 1.35 1.45 1.29 1.39 1.38 1.38 1.35 1.42 1.30 1.26 1.37 1.33
1.37 1.34 1.34 1.32 1.33 1.30 1.38 1.41 1.35 1.38 1.27 1.37
(a) Using 8 classes form a frequency distribution and a cumulative frequency distribution.
(b) For the above data draw a histogram, a frequency polygon and an ogive.
Marks
(a) Range = 1.24 to 1.47 i.e. 0.23 Hence, let classes be 1.24-1.26, 1.27-1.29,
1.30-1.32, … A frequency distribution and cumulative frequency distribution is
shown in the following table:
8
© 2006 John Bird. All rights reserved. Published by Elsevier.
114
(b) A histogram and frequency polygon are shown in Figure 37.
8
Figure 37
An ogive is shown in Figure 38.
5
Figure 38
Total: 21
© 2006 John Bird. All rights reserved. Published by Elsevier.
115
Problem 3. Determine for the 10 measurements of lengths shown below: (a) the arithmetic mean,
(b) the median, (c) the mode, and (d) the standard deviation.
28 m, 20 m, 32 m, 44 m, 28 m, 30 m, 30 m, 26 m, 28 m, and 34 m.
Marks
(a) Arithmetic mean =
28 + 20 + 32 + 44 + 28 + 30 + 30 + 26 + 28 + 34
= 30 m
10
2
(b) Ranking gives: 20 26 28 28 28 30 30 32 34 34
Median =
28 + 30
= 29 m
2
2
1
(c) The mode is 28 m
(d) Standard deviation, σ =
⎡ ( 28 − 30 )2 + ( 20 − 30 )2 + ... + ( 34 − 30 )2 ⎤
344
⎢
⎥=
10
10
⎢⎣
⎥⎦
= 5.865
5
Total: 10
Problem 4. The heights of 100 people are measured correct to the nearest centimetre with the
following results:
150-157 cm 5
158-165 cm 18
166-173 cm 42
174-181 cm 27
182-189cm 8
Determine for the data (a) the mean height and (b) the standard deviation.
Marks
(a) Mean height =
=
153.5 × 5 + 161.5 ×18 + 169.5 × 42 + 177.5 × 27 + 185.5 × 8
100
17070
= 170.7
100
4
(b) Standard deviation,
⎡ 5 (153.5 − 170.7 ) + 18 (161.5 − 170.7 ) + 42 (169.5 − 170.7 )
_
⎡
⎛
⎞⎤
⎢
f
x
x
−
2
2
⎟⎥
⎢∑ ⎜
⎢
+ 27 (177.5 − 170.7 ) + 8 (185.5 − 170.7 )
⎝
⎠
⎢
⎥ = ⎢
100
⎢
∑ f ⎥ ⎢⎢
⎢⎣
⎥⎦
⎣⎢
2
σ=
=
2
2
⎤
⎥
⎥
⎥
⎥
⎥
⎦⎥
4
⎡1479.2 + 1523.52 + 60.48 + 1248.48 + 1752.32 ⎤
⎢⎣
⎥⎦
100
© 2006 John Bird. All rights reserved. Published by Elsevier.
116
=
⎛ 6064 ⎞
⎜
⎟ = 60.64 = 7.787 cm
⎝ 100 ⎠
4
Total: 12
Problem 5. Draw an ogive for the data of component measurements given below, and hence
determine the median and the first and third quartile values for the distribution.
Class intervals (mm)
Frequency
Cumulative frequency
1.24-1.26
2
2
1.27-1.29
4
6
1.30-1.32
4
10
1.33-1.35
10
20
1.36-1.38
11
31
1.39-1.41
5
36
1.42-1.44
3
39
1.45-1.47
1
40
Marks
The ogive (i.e. cumulative frequency/upper class boundary values) is shown in
Figure 39.
The first quartile value Q1 occurs at a cumulative frequency value of
40
i.e. 10,
4
hence, Q1 = 1.325mm .
1
The median value Q 2 occurs at a cumulative frequency value of
40
i.e. 20,
2
hence, Q 2 = 1.355mm .
1
The third quartile value Q3 occurs at a cumulative frequency value of
hence, Q 3 = 1.382mm
3
× 40 i.e. 30
4
1
© 2006 John Bird. All rights reserved. Published by Elsevier.
117
7
Figure 39
Total: 10
Problem 6. Determine the probabilities of:
(a) drawing a white ball from a bag containing 6 black and 14 white balls
(b) winning a prize in a raffle by buying 6 tickets when a total of 480 tickets are sold
(c) selecting at random a female from a group of 12 boys and 28 girls
(d) winning a prize in a raffle by buying 8 tickets when there are 5 prizes and a total of 800
tickets are sold.
Marks
(a) p =
14
14
=
20
6 + 14
or 0.70
2
(b) p =
6
1
or 0.0125
=
480
80
2
(c) p =
28
28
7
or 0.70
=
=
12 + 28 40 10
2
5
5
or 0.05
=
800 100
2
(d) p = 8 ×
Total: 8
© 2006 John Bird. All rights reserved. Published by Elsevier.
118
Problem 7. The probabilities of an engine failing are given by: p1 , failure due to overheating;
p 2 , failure due to ignition problems;
p3 , failure due to fuel blockage.
1
1
2
When p1 = , p 2 = and p3 = , determine the probabilities of:
8
5
7
(a) all three failing (b) the first and second but not the third failure occurring (c) only the second
failure occurring (d) the first or the second failure occurring but not the third.
Marks
Since p1 =
__
by: p1 =
1
1
2
, p 2 = and p3 = , the probabilities of an engine not failing are given
5
7
8
__
7 __ 4
5
, p2 =
and p3 = .
8
5
7
(a) The probability of all three failures occurring
1 1 2
1
or 0.00714
= p1 × p 2 × p3 = × × =
8 5 7
140
3
(b) The probability of the first and second but not the third failure occurring
__
1 1 5
1
or 0.0179
= p1 × p 2 × p3 = × × =
8 5 7
56
3
(c) The probability of only the second failure occurring
__
__
7 1 5
1
or 0.125
= p1× p 2 × p3 = × × =
8 5 7
8
3
(d) The probability of the first or the second failure occurring but not the third
__
⎛ 1 1 ⎞ 5 13 5 13
or 0.232
= ( p1 + p 2 ) × p3 = ⎜ + ⎟ × = × =
56
⎝ 8 5 ⎠ 7 40 7
3
Total: 12
Problem 8. In a box containing 120 similar transistors 70 are satisfactory, 37 give too high a gain
under normal operating conditions and the remainder give too low a gain.
Calculate the probability that when drawing two transistors in turn, at random, with replacement,
of having (a) two satisfactory, (b) none with low gain, (c) one with high gain and one satisfactory,
(d) one with low gain and none satisfactory.
Determine the probabilities in (a), (b) and (c) above if the transistors are drawn without
replacement.
© 2006 John Bird. All rights reserved. Published by Elsevier.
119
Marks
With replacement
70 70
49
×
=
or 0.3403
120 120 144
2
70 + 37 70 + 37 ⎛ 107 ⎞
×
=⎜
(b) p =
⎟ = 0.7951
120
120
⎝ 120 ⎠
2
(a) p =
2
(c) p =
37 70 70 37
×
+
×
= 0.3597
120 120 120 120
2
(d) p =
13 50 50 13
×
+
×
= 0.0903
120 120 120 120
2
Without replacement
(a) p =
70 69
×
= 0.3382
120 119
2
(b) p =
107 106
×
= 0.7943
120 119
2
(c) p =
37 70 70 37
×
+
×
= 0.3627
120 119 120 119
2
Total: 14
TOTAL ASSIGNMENT MARKS: 100
© 2006 John Bird. All rights reserved. Published by Elsevier.
120
ASSIGNMENT 16 (PAGE 576)
This assignment covers the material contained in chapters 57 to 60.
Problem 1. A machine produces 15% defective components. In a sample of 5, drawn at random,
calculate, using the binomial distribution, the probability that:
(a) there will be 4 defective items
(b) there will be not more than 3 defective items
(c) all the items will be non-defective.
Draw a histogram showing the probabilities of 0, 1, 2, … , 5 defective items.
Marks
Let p = 0.15, q = 0.85 and n = 5.
(q + p)
5
= q 5 + 5q 4 p +
5 × 4 3 2 5 × 4 × 3 2 3 5× 4 × 3× 2 4
qp +
qp +
q p + p5
2!
3!
4!
(a) The probability of 4 defective items =
5× 4 × 3× 2 4
qp
4!
= 5 ( 0.85 )( 0.15 )
4
= 0.00215
4
(b) Not more than 3 defective items means the sum of the first 4 terms
= ( 0.85 ) + 5 ( 0.85 ) ( 0.15 ) + 10 ( 0.85 ) ( 0.15 ) + 10 ( 0.85 ) ( 0.15 )
5
4
3
2
2
3
= 0.4437 + 0.3915 + 0.1382 + 0.0244
8
= 0.9978
(c) The probability that all items will be non-defective is 0.4437
2
A histogram showing the probabilities of defective items is shown in Figure 40.
© 2006 John Bird. All rights reserved. Published by Elsevier.
121
6
Figure 40
Total: 20
Problem 2. 2% of the light bulbs produced by a company are defective. Determine, using the
Poisson distribution, the probability that in a sample of 80 bulbs: (a) 3 bulbs will be defective, (b)
not more than 3 bulbs will be defective, (c) at least 2 bulbs will be defective.
Marks
λ = 2% of 80 = 1.6
The probability of 0, 1, 2, .. defective items are given by:
(a) The probability of 3 defective bulbs =
e−λ , λ e−λ ,
λ 2 e −λ
, ...
2!
λ 3e −λ 1.63 e −1.6
=
= 0.1378
3!
6
3
(b) The probability of not more than 3 defective bulbs is given by:
e−1.6 + 1.6 e −1.6 +
1.62 e−1.6 1.63 e −1.6
+
2!
3!
= 0.2019 + 0.3230 + 0.2584 + 0.1378 = 0.9211
6
(c) The probability that at least two bulbs will be defective is given by:
1 − ( e −λ + λ e−λ ) = 1 – (0.2019 + 0.3230) = 0.4751
4
Total: 13
© 2006 John Bird. All rights reserved. Published by Elsevier.
122
Problem 3. Some engineering components have a mean length of 20 mm and a standard deviation
of 0.25 mm. Assume that the data on the lengths of the components is normally distributed. In a
batch of 500 components, determine the number of components likely to:
(a) have a length of less than 19.95 mm (b) be between 19.95 mm and 20.15 mm (c) be longer
than 20.54 mm.
Marks
__
(a)
x − x 19.95 − 20
=
= -0.2 standard deviations.
z=
σ
0.25
From Table 58.1, page 561, when z = -0.2 the partial area under the standardised
curve is 0.0793 (i.e. the shaded area of Figure 41 is 0.0793 of the total area).
The area to the left of the shaded area = 0.5 – 0.0793 = 0.4207
Thus, for 500 components, 0.4207 × 500 are likely to have a length less than
19.95 mm, i.e. 210.
5
Figure 41
(b) When length is 20.15 mm, z =
20.15 − 20
= 0.6 and from Table 31.1, the area
0.25
under the standardisation curve is 0.2257
Hence, the total partial area between z = -0.2 and z = 0.6 is:
0.0793 + 0.2257 = 0.3050 as shown shaded in Figure 42.
It is likely that 0.3050 × 500 components will lie between 19.95 mm and
20.15 mm, i.e. 153.
5
© 2006 John Bird. All rights reserved. Published by Elsevier.
123
Figure 42
(c) When the length is 20.54 mm, z =
20.54 − 20
= 2.16 and the partial area
0.25
corresponding to this z-value is 0.4846
The area to the right of the shaded area shown in Figure 43 is 0.5 – 0.4846,
i.e. 0.0154
Hence, 0.0154 × 500 components are likely to be greater than 20.54 mm, i.e. 8.
5
Figure 43
Total: 15
Problem 4. In a factory, cans are packed with an average of 1.0 kg of a compound and the masses
are normally distributed about the average value. The standard deviation of a sample of the contents
of the cans is 12 g. Determine the percentage of cans containing (a) less than 985 g (b) more than
1030 g (c) between 985 g and 1030 g.
Marks
(a) The z-value for 985 g is
985 − 1000
= -1.25
12
From Table 58.1, page 561, the corresponding area under the standardised normal
curve is 0.3944. Hence, the area to the left of 1.25 standard deviations is:
0.5 – 0.3944 = 0.1056, i.e. 10.56% of the cans contain less than 985 g.
4
© 2006 John Bird. All rights reserved. Published by Elsevier.
124
(b) The z-value for 1030 g is
1030 − 1000
= 2.5
12
From Table 58.1, the area under the normal curve is 0.4938. Hence, the area to
the right of 2.5 standard deviations is 0.5 – 0.4938 = 0.0062, i.e. 0.62% of the
cans contain more than 1030 g.
4
(c) The area under the normal curve corresponding to –1.25 to 2.5 standard
deviations is 0.3944 + 0.4938 = 0.8882, i.e. 88.82% of the cans contain
between 985 g and 1030 g.
2
Total: 10
Problem 5. The data given below gives the experimental values obtained for the torque output, X,
from an electric motor and the current, Y, taken from the supply.
Torque X
0 1 2 3 4 5
6
7
8
9
Current Y 3 5 6 6 9 11 12 12 14 13
Determine the linear coefficient of correlation for this data.
Marks
Using a tabular approach:
14
© 2006 John Bird. All rights reserved. Published by Elsevier.
125
Coefficient of correlation, r =
101.5
∑ xy
=
( ∑ x )( ∑ y ) (82.5)(132.9 )
2
= 0.969
4
2
There is therefore good correlation between X and Y.
Total: 18
Problem 6. Some results obtained from a tensile test on a steel specimen are shown below:
Tensile force (kN) 4.8 9.3 12.8 17.7 21.6 26.0
Extension (mm)
3.5 8.2 10.1 15.6 18.4 20.8
Assuming a linear relationship:
(a) determine the equation of the regression line of extension on force
(b) determine the equation of the regression line of force on extension
(c) estimate (i) the value of extension when the force is 16 kN, and (ii) the value of force when the
extension is 17 mm.
Marks
10
∑Y = a N +a ∑X
(a)
0
and
76.6 = 6 a 0 + 92.2 a1
Hence,
and
1
1418.70 = 92.2 a 0 + 1729.22 a1
∑ XY = a ∑ X + a ∑ X
0
2
1
(1)
(2)
92.2 × (1) gives:
92.2 ( 76.6 ) = 92.2 ( 6 ) a 0 + 92.2 ( 92.2 ) a1
(3)
6 × (2) gives:
6 (1418.70 ) = 6 ( 92.2 ) a 0 + 6 (1729.22 ) a1
(4)
© 2006 John Bird. All rights reserved. Published by Elsevier.
126
−1449.68 = 0 − 1874.48a1
(3) – (4) gives:
a1 =
from which,
1449.68
= 0.773
1874.48
76.6 = 6 a 0 + 92.2 ( 0.773)
Substituting in equation (1) gives:
a0 =
from which,
76.6 − 92.2 ( 0.773)
= 0.888
6
Hence, the regression line of extension on tensile force is:
Y = a 0 + a 1X
i.e.
∑X = b N + b ∑Y
(b)
0
∑ XY = b ∑ Y + b ∑ Y
0
(1)
1418.70 = 76.6 b0 + 1196.06 b1
(2)
Thus, 76.6 ( 92.2 ) = 76.6 ( 6 ) b0 + 76.6 ( 76.6 ) b1
(3)
6 (1418.70 ) = 6 ( 76.6 ) b0 + 6 (1196.06 ) b1
(3) – (4) gives:
2
1
92.2 = 6 b0 + 76.6 a1
Hence,
and
and
1
6
Y = 0.888 + 0.773 X
(4)
−1449.68 = −1308.80 b1
1449.68
= 1.108
1308.80
b1 =
from which,
Substituting in (1) gives: 92.2 = 6 b 0 + 76.6 (1.108 )
b0 =
from which,
92.2 − 76.6 (1.108 )
= 1.221
6
Hence, the regression line of force on extension is:
X = b0 + b1Y
i.e.
X = 1.221 + 1.108 Y
6
(c)(i) Extension when the force is 16 kN is: Y = 0.888 + 0.773(16) = 13.26 mm
1
(ii) Force when the extension is 17 mm is: X = 1.221 + 1.108(17) = 20.06 kN
1
Total: 24
TOTAL ASSIGNMENT MARKS: 100
© 2006 John Bird. All rights reserved. Published by Elsevier.
127
ASSIGNMENT 17 (PAGE 625)
This assignment covers the material contained in chapters 61 to 63.
Problem 1. 1200 metal bolts have a mean mass of 7.2 g and a standard deviation of 0.3 g.
Determine the standard error of the means. Calculate also the probability that a sample of 60 bolts
chosen at random, without replacement, will have a mass of (a) between 7.1 g and 7.25 g, and
(b) more than 7.3 g.
Marks
For the population: number of bolts, N p = 1200
standard deviation, σ = 0.3 g; mean, µ = 7.2 g
For the sample:
number of sample, N = 60
Mean of sampling distributions of means, µ _ = µ = 7.2 g
x
Standard error of the means, µ _ =
x
σ
N
⎛ N p − N ⎞ 0.3 ⎛ 1200 − 60 ⎞
⎜⎜
⎟⎟ =
⎜
⎟
−
N
1
60 ⎝ 1200 − 1 ⎠
p
⎝
⎠
= 0.03776 g
3
__
x−x
7.1 − 7.2
(a) z =
When x = 7.1 g, z =
= -2.65 standard deviations
σ_
0.03776
1
x
When x = 7.25 g, z =
7.25 − 7.2
= 1.32 standard deviations
0.03776
1
From Table 58.1, page 561, the area corresponding to these z-values are 0.4960
and 0.4066. Hence, the probability of the mean mass lying between 7.1 g and
7.25 g is 0.4960 + 0.4066 = 0.9026
(b) When x = 7.3 g, z =
3
7.3 − 7.2
= 2.65
0.03776
1
From Table 58.1, the area corresponding to this z-value is 0.4960 the area lying to
the right of this is 0.5 – 0.4960 = 0.0040 hence, the probability that a sample
will have a mass of more than 7.3 g is 0.0040
3
Total: 12
© 2006 John Bird. All rights reserved. Published by Elsevier.
128
Problem 2. A sample of 10 measurements of the length of a component are made and the mean of
the sample is 3.650 cm. The standard deviation of the samples is 0.030 cm. Determine (a) the 99%
confidence limits, and (b) the 90% confidence limits for an estimate of the actual length of the
component.
Marks
_
For the sample: sample size, N = 10, mean, x = 3.650 cm
standard deviation, s = 0.030 cm
(a) The percentile value corresponding to a confidence coefficient value of t 0.99 and
a degree of freedom value of ν = 10 – 1 = 9, is 2.82 from Table 61.2, page 587.
Estimated value of the mean of the population
_
=x±
tc s
( N − 1)
= 3.650 ±
( 2.82 )( 0.0030 ) = 3.650 ± 0.0282
(10 − 1)
Thus, the 99% confidence limits are 3.622 cm to 3.678 cm.
5
(b) For t 0.90 , ν = 9, t c = 1.38 from Table 61.2, page 587.
Estimated value of the 90% confidence limits
_
= x±
tc s
( N − 1)
= = 3.650 ±
(1.38)( 0.0030 ) = 3.650 ± 0.0138
(10 − 1)
Thus, the 90% confidence limits are 3.636 cm to 3.664 cm.
5
Total: 10
Problem 3. An automated machine produces metal screws and over a period of time it is found that
8% are defective. Random samples of 75 screws are drawn periodically.
(a) If a decision is made that production continues until a sample contains more than 8
defective screws, determine the type I error based on the decision for a defect rate of 8%.
(b) Determine the magnitude of the type II error when the defect rate has risen to 12%.
The above sample size is now reduced to 55 screws. The decision now is to stop the machine
for adjustment if a sample contains 4 or more defective screws.
(c) Determine the type I error if the defect rate remains at 8%.
© 2006 John Bird. All rights reserved. Published by Elsevier.
129
(d) Determine the type II error when the defect rate rises to 9%.
Marks
N = 75, p = 0.08, q = 0.92
Since Np and Nq are both > 5, a normal approximation to the binomial distribution is
used.
(a) Mean of the normal distribution, Np = (75)(0.08) = 6
Standard deviation of the normal distribution =
( Npq ) =
(75)(0.08)(0.92)
= 2.35
A type I error is the probability of stopping the machine, i.e. the probability of
getting greater than 8 defective screws in a sample, even though the defect rate
is still 8%.
z-value corresponding to 8 defective screws =
8−6
= 0.85
2.35
The area between the mean and a z-value of 0.85 is 0.3023 from Table 58.1, page
561. Thus the probability of more than 8 defective screws is the area to the right
of the z-ordinate at 0.85, i.e. 0.5 – 0.3023 = 0.1977
5
Hence, the type I error = 19.8%
(b) Type II error is the probability of a sample containing less than 8 defective screws
even though the defect rate has risen to 12%.
Now N = 75, p = 0.12, q = 0.88
Np and Nq are both > 5, hence a normal approximation to a binomial distribution
is used, where mean = Np = (75)(0.12) = 9
and standard deviation =
z-value is given by
( Npq ) =
(75)(0.12)(0.88) = 2.81
8−9
= -0.36
2.81
From Table 58.1, page 561, the area between the mean and z = -0.36 is 0.1406
Hence, the probability of getting less than 8 defective screws is
0.5 – 0.1406 = 0.3594, i.e. the type II error = 35.9%.
5
© 2006 John Bird. All rights reserved. Published by Elsevier.
130
(c) N = 55, p = 0.08
When N ≥ 50 and Np < 5, the Poisson approximation to a binomial distribution is
used. λ = Np = (55)(0.08) = 4.4
Probability of a sample containing 0 defective screws = e −λ
= 0.0123
Probability of a sample containing 1 defective screws = λ e −λ
= 0.0540
Probability of a sample containing 2 defective screws =
λ 2 e −λ
= 0.1188
2!
Probability of a sample containing 3 defective screws =
λ 3e −λ
3!
= 0.1743
Probability of a sample containing 0, 1, 2 or 3 defective screws = 0.3594
Hence, the probability of a sample containing 4 or more defective screws
= 1 – 0.3594 = 0.6406
i.e. the type I error = 64.1% of stopping the machine for adjustment when it
6
should continue running.
(d) When the defect rate has risen to 9%, p = 0.09 and Np = λ = 4.95
Since N ≥ 50 and Np < 5, the Poisson approximation to a binomial distribution is
used.
Probability of a sample containing 0 defective screws = e−λ
= 0.0071
Probability of a sample containing 1 defective screws = λ e −λ
= 0.0351
Probability of a sample containing 2 defective screws =
λ 2 e −λ
= 0.0868
2!
Probability of a sample containing 3 defective screws =
λ 3e −λ
= 0.1432
3!
Probability of a sample containing 0, 1, 2 or 3 defective screws = 0.2722
Hence, the type II error of leaving the machine running when it should be
stopped = 27.2%
6
Total: 22
© 2006 John Bird. All rights reserved. Published by Elsevier.
131
Problem 4. In a random sample of 40 similar light bulbs drawn from a batch of 400 the mean
lifetime is found to be 252 hours. The standard deviation of the lifetime of the sample is 25 hours.
The batch is classed as inferior if the mean lifetime of the batch is less than the population mean of
260 hours. As a result of the sample data, determine whether the batch is considered to be inferior at
a level of significance of (a) 0.05, and (b) 0.01
Marks
Population size, N p = 400, population mean, µ = 260 hour,
_
mean of sample, x = 252 h, standard deviation of sample, s = 25 h,
size of sample, N = 40
_
_
H1 : x < µ
H0 : x = µ
_
z=
x− µ
s ⎛ Np − N ⎞
⎜
⎟
N ⎝⎜ N p − 1 ⎠⎟
=
252 − 260
25 ⎛ 400 − 40 ⎞
⎜
⎟
40 ⎝ 400 − 1 ⎠
=
−8
= -2.13
( 3.9528 )( 0.9499 )
3
(a) For a level of significance of 0.05 and a one-tailed test (see Table 62.1, page 594),
all values to the left of the z-ordinate at –1.645 indicate that the results are ‘not
significant’, i.e. they differ significantly from the null hypothesis. Since the
z-value is –2.13 is less than –1.645, the batch is considered to be inferior at a
3
level of significance of 0.05
(b) The z-value for a level of significance of 0.01 for a one-tailed test is –2.33 (see
Table 62.1, page 594). Since the z-value of –2.13 lies to the right of this ordinate,
it does not differ significantly from the null hypothesis and the batch is not
considered to be inferior at a level of significance of 0.01
3
Total: 9
Problem 5. The lengths of two products are being compared.
Product 1: sample size = 50, mean value of sample = 6.5 cm,
standard deviation of whole batch = 0.40 cm.
© 2006 John Bird. All rights reserved. Published by Elsevier.
132
Product 2: sample size = 60, mean value of sample = 6.65 cm,
standard deviation of whole batch = 0.35 cm.
Determine if there is any significant difference between the two products at a level of significance
of (a) 0.05, and (b) 0.01
Marks
H 0 : µ1 − µ 2 = 0
H1 : µ1 − µ 2 ≠ 0
The z-value for the difference of the sample mean is given by:
__
z=
__
x1 − x 2
⎛ σ12 σ 2 2 ⎞
+
⎜
⎟
⎝ N1 N 2 ⎠
=
6.5 − 6.65
⎛ 0.402 0.352 ⎞
+
⎜
⎟
60 ⎠
⎝ 50
=
−0.15
= -2.07
0.0724
3
(a) For a two-tailed test at a 0.05 level of significance, z = ± 1.96 from Table 62.1,
page 594. Hence, product 1 is significantly different from product 2 at a level
2
of significance of 0.05
(b) For a two-tailed at a 0.01 level of significance, z = ± 2.58
Hence, there is no significant difference between product 1 and product 2 at a
2
level of significance of 0.01
Total: 7
Problem 6. The resistance of a sample of 400 resistors produced by an automatic process have the
following resistance distribution.
Resistance (Ω) 50.11 50.15 50.19 50.23 50.27 50.31 50.35
Frequency
9
35
61
102
89
83
21
Calculate for the sample: (a) the mean, and (b) the standard deviation. (c) Test the null hypothesis
that the resistance of the resistors are normally distributed at a level of significance of 0.05, and
determine if the distribution gives a ‘too good’ fit at a level of confidence of 90%
Marks
_
(a) Sample mean, x =
(9 × 50.11) + (35 × 50.15) + ... + (21× 50.35)
= 50.246
400
3
© 2006 John Bird. All rights reserved. Published by Elsevier.
133
(b) Sample standard deviation,
s=
=
⎧ 9(50.11 − 50.246) 2 + 35(50.15 − 50.246) 2 + ... + 21(50.35 − 50.246) 2 ⎫
⎨
⎬
400
⎩
⎭
⎛ 1.3248 ⎞
⎜
⎟ = 0.05755
⎝ 400 ⎠
4
(c) The probability for each class and the expected frequency is calculated as in the
table below.
__
x − x x − 50.246
For the third column, z =
=
s
0.05755
The area in column 4 is obtained from Table 58.1, page 561.
The expected frequency in the last column is given by the value of the area in the
fifth column × 400
7
© 2006 John Bird. All rights reserved. Published by Elsevier.
134
To determine the χ 2 -value, see the table below.
7
To test the significance of χ 2 -value:
ν=N–1–M=7–1–2=4
From Table 63.1, page 609, the χ p -value corresponds to χ 0.95 , ν 4 is 9.49.
2
2
Hence, since 11.7375 > 9.49, the null hypothesis that the resistances are normally
distributed is rejected.
2
For χ 0.10 , ν 4 , χ p = 1.06. hence, since 11.7375 > 1.06, the fit is not ‘too good’.
2
2
2
Total: 25
Problem 7. A fishing line is manufactured by two processes, A and B. To determine if there is any
difference in the mean breaking strengths of the lines, 8 lines by each process are selected and
tested for breaking strength. The results are as follows:
Process A
8.6 7.1 6.9 6.5 7.9 6.3 7.8 8.1
Process B
6.8 7.6 8.2 6.2 7.5 8.9 8.0 8.7
Determine if there is a difference between the mean breaking strengths of the line manufactured by
the two processes, at a significance level of 0.10, using (a) the sign test,
(b) the Wilcoxon sign-rank test, (c) the Mann-Whitney test.
© 2006 John Bird. All rights reserved. Published by Elsevier.
135
Marks
(a) Sign test
(i)
H1 : B.S.A ≠ B.S.B
H 0 : B.S.A = B.S.B
(ii) α 2 = 0.10 = 10% since it is a two-tailed test
(iii) (A - B) = +1.8 -0.5 -1.3 +0.3 +0.4 -2.6 -0.2 -0.6
(iv) There are 3 + signs and 5 – signs; taking the smaller number, S = 3
(v) From Table 63.3, page 614, with n = 8 and α 2 = 10%, S ≤ 1
Since from (iv), S is not equal or less than 1, the null hypothesis cannot be
rejected, i.e. there is no difference between the breaking strengths of the
5
lines.
(b) Wilcoxon signed-rank test
(i)
H 0 : B.S.A = B.S.B
H1 : B.S.A ≠ B.S.B
(ii) α 2 = 0.10 = 10% since it is a two-tailed test
(iii) (A - B) = +1.8 -0.5 -1.3 +0.3 +0.4 -2.6 -0.2 -0.6
(iv) Ranking gives: -0.2 +0.3 +0.4 -0.5 -0.6 -1.3 +1.8 -2.6
1
2
3
4
5
6
7
8
(v) There are 3 + signs with rankings 2, 3 and 7. Hence, T = 2 + 3 + 7 = 12
(vi) From Table 63.4, page 617, with n = 8 and α 2 = 10%, T ≤ 5
From (v), T is not equal or less than 5, hence, the null hypothesis cannot be
rejected, i.e. there is no difference between the breaking strengths of the
5
lines.
(c ) The Mann-Whitney test
(i)
H 0 : B.S.A = B.S.B
H1 : B.S.A ≠ B.S.B
(ii) α 2 = 0.10 = 10% since it is a two-tailed test
(iii) Arranging the 16 pieces of data in order gives:
© 2006 John Bird. All rights reserved. Published by Elsevier.
136
6.2 6.3 6.5 6.8 6.9 7.1 7.5 7.6 7.8 7.9 8.0 8.1 8.2 8.6 8.7 8.9
B A A
B A A B B A A B
A B A B B
(iv) The number of B’s preceding the A’s in the sequence is as follows:
B A A B A A B B
1
1
2
2
A
A
4
4
B
A
5
B
A
B
B
6
(v) Adding the numbers from (iv) gives:
U = 1 + 1 + 2 + 2 + 4 + 4 + 5 + 6 = 25
(vi) From Table 63.5, pages 621/2, with n1 = 8, n 2 = 8, and α 2 = 10%, U ≤ 15
From (v), U is not equal or less than 15, hence, the null hypothesis cannot
be rejected, i.e. there is no difference between the breaking strengths of
the lines.
5
Total: 15
TOTAL ASSIGNMENT MARKS: 100
© 2006 John Bird. All rights reserved. Published by Elsevier.
137
ASSIGNMENT 18 (PAGE 655)
This assignment covers the material contained in chapters 64 to 68.
Problem 1. Find the Laplace transforms of the following functions: (a) 2t 3 − 4t + 5
(b) 3e −2t − 4sin 2t
(c) 3 cosh 2t
(d) 2t 4 e −3t
(e) 5e 2t cos 3t
(f) 2 e3t sinh 4t
Marks
⎛ 3! ⎞ 4 5 12 4 5
(a) Λ {2t 3 − 4t + 5} = 2 ⎜ 3+1 ⎟ − 2 + = 4 − 2 +
s
s
s
⎝s ⎠ s s
3
3
8
⎛ 2 ⎞
=
− 4⎜ 2
− 2
2 ⎟
s+2
⎝s +2 ⎠ s+2 s +4
(b) Λ {3e−2t − 4sin 2t} =
or
(c) Λ{3 cosh 2t} =
3
3 (s2 + 4 ) − 8 (s + 2 )
( s + 2) (s2 + 4 )
3s 2 − 8s − 4
( s + 2 ) s2 + 4
=
(
3
)
3s
3s
= 2
2
s −4
s −2
2
2
⎛
4! ⎞
48
(d) Λ {2t 4 e −3t } = 2 ⎜
⎟ =
4 +1
⎜ ( s + 3) ⎟ ( s + 3 )5
⎝
⎠
3
⎛
⎞
5 (s − 2)
5 (s − 2)
s−2
= 2
(e) Λ {5e 2t cos 3t} = 5 ⎜
⎟= 2
2
2
⎜ ( s − 2 ) + 3 ⎟ s − 4s + 4 + 9 s − 4s + 13
⎝
⎠
3
⎛
⎞
4
8
=
(f) Λ {2 e3t sinh 4t} = 2 ⎜
⎟
2
2
⎜ ( s − 3) − 42 ⎟ s − 6s − 7
⎝
⎠
2
Total: 16
Problem 2. Find the inverse Laplace transforms of the following functions: (a)
(b)
12
s5
(c)
4s
s +9
2
(d)
5
s −9
2
(e)
3
(s + 2)
4
(f)
s−4
s − 8s − 20
2
(g)
5
2s + 1
8
s − 4s + 3
2
Marks
© 2006 John Bird. All rights reserved. Published by Elsevier.
138
⎧
⎫
⎧
⎫
⎪
⎪⎪ 5
⎪
⎪ 5 − 12 t
5
5
1
⎪
⎧
⎫
−1
−1
−1
(a) Λ ⎨
⎬= Λ ⎨ 1⎬ = e
⎬=Λ ⎨
⎩ 2s + 1 ⎭
⎪s + ⎪ 2
⎪ 2 ⎛⎜ s + 1 ⎞⎟ ⎪ 2
⎩ 2⎭
⎪⎩ ⎝ 2 ⎠ ⎪⎭
3
⎧ 4!⎫
⎧12 ⎫ 1
(b) Λ −1 ⎨ 5 ⎬ = t 4 hence Λ −1 ⎨ 5 ⎬ = t 4
⎩s ⎭ 2
⎩s ⎭
2
s ⎫
⎧ 4s ⎫
−1 ⎧
(c) Λ −1 ⎨ 2
⎬ = 4 Λ ⎨ 2 2 ⎬ = 4 cos 3t
⎩s + 3 ⎭
⎩s + 9 ⎭
2
⎧ 5 ⎫ 5 −1 ⎧ 3 ⎫ 5
(d) Λ −1 ⎨ 2
⎬ = Λ ⎨ 2 2 ⎬ = sinh 3t
⎩s − 9 ⎭ 3
⎩s − 3 ⎭ 3
2
⎧⎪ 3! ⎫⎪ −2t 3
⎧⎪ 3 ⎫⎪ 1 −2t 3
= e t hence Λ −1 ⎨
(e) Λ −1 ⎨
= e t
3+1 ⎬
4⎬
⎩⎪ ( s + 2 ) ⎭⎪
⎩⎪ ( s + 2 ) ⎭⎪ 2
2
⎧
⎫⎪
s−4
⎧ s−4 ⎫
−1 ⎪
4t
(f) Λ −1 ⎨ 2
⎬ =Λ ⎨
⎬ = e cos 6t
2
2
⎩ s − 8s − 20 ⎭
⎪⎩ ( s − 4 ) − 6 ⎪⎭
3
⎧
⎫⎪
⎧
⎫⎪
8
8
1
⎧
⎫
−1 ⎪
−1 ⎪
2t
(g) Λ −1 ⎨ 2
=
=
Λ
Λ
8
⎬
⎨
⎬
⎨
⎬ = 8e sinh t
2
2
2
2
⎩ s − 4s + 3 ⎭
⎪⎩ ( s − 2 ) − 1 ⎪⎭
⎪⎩ ( s − 2 ) − 1 ⎭⎪
3
Total: 17
Problem 3. Use partial fractions to determine the following:
⎧ 5s − 1 ⎫
(a) Λ −1 ⎨ 2
⎬
⎩s − s − 2 ⎭
⎧ 2s 2 + 11s − 9 ⎫
(b) Λ −1 ⎨
⎬
⎩ s (s − 1)(s + 3) ⎭
⎧⎪
⎫⎪
13 − s 2
(c) Λ −1 ⎨ 2
⎬
⎩⎪ s ( s + 4s + 13) ⎭⎪
Marks
A ( s + 1) + B ( s − 2 )
5s − 1
5s − 1
A
B
≡
=
+
=
s 2 − s − 2 ( s − 2 )( s + 1) ( s − 2 ) ( s + 1)
( s − 2 )( s + 1)
(a)
Hence
5s – 1 = A(s + 1) + B(s –2)
2
Let s = 2 then
9 = 3A and A = 3
1
Let s – 1 then
-6 = -3B and B = 2
1
⎧ 3
2 ⎪⎫
⎧ 5s − 1 ⎫
−1 ⎪
2t
−t
Thus, Λ −1 ⎨ 2
+
⎬=Λ ⎨
⎬ = 3e + 2e
s
2
s
1
−
−
s
s
2
−
+
) ( ) ⎭⎪
⎩
⎭
⎩⎪ (
(b)
2
2s 2 + 11s − 9 A
B
C
A(s − 1)(s + 3) + Bs (s + 3) + Cs (s − 1)
= +
+
=
s ( s − 1)( s + 3) s ( s − 1) ( s + 3)
s ( s − 1)( s + 3)
Hence,
2s 2 + 11s − 9 = A(s − 1)(s + 3) + Bs (s + 3) + Cs (s − 1)
3
© 2006 John Bird. All rights reserved. Published by Elsevier.
139
Let s = 0 then
-9 = -3A and A = 3
Let s = 1 then
4 = 4B and
Let s = -3 then
1
B=1
1
-24 = 12C and C = -2
1
⎧⎪ 2s 2 + 11s − 9 ⎫⎪
⎧
1
2 ⎫⎪
−1 ⎪ 3
t
−3t
Hence, Λ −1 ⎨
−
⎬=Λ ⎨ +
⎬ = 3 + e − 2e
⎪⎩ s ( s − 1)( s + 3) ⎪⎭
⎪⎩ s ( s − 1) ( s + 3) ⎪⎭
3
A ( s 2 + 4s + 13) + ( Bs + C ) s
13 − s 2
A
Bs + C
≡
+
=
2
s ( s 2 + 4s + 13) s s + 4s + 13
s ( s 2 + 4s + 13)
(c)
13 − s 2 = A ( s 2 + 4s + 13) + ( Bs + C ) s
Hence
3
Let s = 0
13 = 13A and A = 1
1
s 2 coefficients:
-1 = A + B hence B = -2
1
s coefficients:
0 = 4A + C hence C = -4
1
⎧⎪
⎫⎪
⎧
2s + 4 ⎫
2s + 4 ⎫⎪
13 − s 2
−1 ⎧ 1
−1 ⎪ 1
Thus, Λ −1 ⎨ 2
⎬=Λ ⎨ −
⎬
⎬=Λ ⎨ − 2
2
2
⎩ s s + 4s + 13 ⎭
⎩⎪ s ( s + 2 ) + 3 ⎭⎪
⎩⎪ s ( s + 4s + 13) ⎭⎪
⎧⎪ 1
2 ( s + 2 ) ⎫⎪
−2t
= Λ −1 ⎨ −
⎬ = 1 − 2e cos 3t
2
2
s
⎪⎩ ( s + 2 ) + 3 ⎪⎭
3
Total: 24
Problem 4. In a galvanometer the deflection θ satisfies the differential equation:
d 2θ
dθ
+2 +θ= 4
2
dt
dt
Use Laplace transforms to solve the equation for θ given that when t = 0, θ = 0 and
dθ
= 0.
dt
Marks
Taking the Laplace transforms of each term gives:
⎧ d 2θ ⎫
⎧ dθ ⎫
Λ ⎨ 2 ⎬ + 2 Λ ⎨ ⎬ + Λ{θ} = Λ{4}
⎩ dt ⎭
⎩ dt ⎭
i.e.
4
⎡⎣s 2 Λ{θ} – s y(0) - y '(0) ] + 2[sΛ{θ} – y(0)] + Λ{θ} =
s
2
y(0) = 0 and y′(0) = 0
Hence,
s 2 Λ{θ} + 2sΛ{θ} + Λ{θ} =
4
s
© 2006 John Bird. All rights reserved. Published by Elsevier.
140
(s
i.e.
2
+ 2s + 1) Λ{θ} =
Λ{θ} =
i.e.
4
s
4
4
=
2
s ( s + 2s + 1) s ( s + 1)
2
⎧⎪ 4 ⎫⎪
θ = Λ −1 ⎨
2⎬
⎩⎪ s ( s + 1) ⎭⎪
and
3
A ( s + 1) + Bs ( s + 1) + Cs
A
B
C
≡ +
+
=
2
2
s ( s + 1) ( s + 1)
s ( s + 1)
2
4
s ( s + 1)
2
Hence
4 = A ( s + 1) + Bs ( s + 1) + Cs
2
Let s = 0 then
4=A
1
Let s = -1 then
4 = -C and C = -4
1
s 2 coefficients:
0 = A + B from which B = -4
1
Therefore
⎧⎪ 4
4
4 ⎫⎪
−
= 4 − 4 e− t − 4 e− t t
θ = Λ −1 ⎨ −
2⎬
⎩⎪ s ( s + 1) ( s + 1) ⎭⎪
i.e.
θ = 4 − 4 (1 + t ) e− t
2
3
Total: 13
Problem 5. Solve the following pair of simultaneous differential equations:
dx
= 3x + 2y
dt
dy
2 + 3x = 6y
dt
3
given that when t = 0, x = 1 and y = 3.
Marks
Rearranging gives:
3
dx
− 3x − 2y = 0
dt
(1)
2
dy
+ 3x − 6y = 0
dt
(2)
Taking Laplace transforms gives:
3 [s Λ{x} – x(0) ] − 3 Λ{x} - 2Λ{y} = Λ{0}
(1′)
1
2 [s Λ{y} – y(0) ] + 3 Λ{x} - 6Λ{y} = Λ{0}
(2′)
1
© 2006 John Bird. All rights reserved. Published by Elsevier.
141
x(0) = 1 and y(0) = 3
Hence,
3sΛ{y} – 3 - 3Λ{x} - 2Λ{y} = 0
(1′′)
and
2sΛ{y} – 6 + 3Λ{x} - 6Λ{y} = 0
(2′′)
i.e.
(3s – 3)Λ{x} - 2Λ{y} = 3
(3)
3Λ{x} + (2s - 6) Λ{y} = 6
(4)
(s – 1) × equation (4) gives:
3(s – 1)Λ{x} + (2s - 6) (s – 1)Λ{y} = 6(s – 1)
3(s – 1)Λ{x} - 2Λ{y} = 3
(5)
(3)
Equation (5) – equation (3) gives:
( 2s
2
− 8s + 6 + 2 ) Λ{y} = 6s – 6 – 3
( 2s
i.e.
2
− 8s + 8 ) Λ{y} = 6s – 9
Λ{y} =
and
6s − 9
3s − 4.5
= 2
2s − 8s + 8 s − 4s + 4
3
2
A (s − 2) + B
3s − 4.5
3s − 4.5
A
B
=
=
+
=
2
2
2
s − 4s + 4 ( s − 2 )
(s − 2) (s − 2)
(s − 2)
2
Hence
3s – 4.5 = A(s – 2) + B
1.5 = B
1
3=A
1
⎧⎪ 3
1.5 ⎫⎪
+
= 3e 2t + 1.5e 2t t
y = Λ −1 ⎨
2⎬
⎩⎪ ( s − 2 ) ( s − 2 ) ⎭⎪
2
Let s = 2 then
Equating s coefficients gives:
Thus,
2
(s – 3) × equation (3) gives:
3(s – 1)(s – 3)Λ{x} – 2(s – 3)Λ{y} = 3(s – 3)
and
3Λ{x} + 2(s - 3)Λ{y} = 6
(6)
(4)
Equations (4) + (6) gives:
( 3s
and
2
− 12s + 9 + 3) Λ{x} = 3s – 9 + 6
Λ{x} =
3s − 3
s −1
= 2
3s − 12s + 12 s − 4s + 4
2
3
A (s − 2) + B
s −1
s −1
A
B
=
=
+
=
2
2
2
s 2 − 4s + 4 ( s − 2 )
(s − 2) (s − 2)
(s − 2)
Thus,
Let s = 2 then
s – 1 = A(s – 2) + B
1=B
2
1
© 2006 John Bird. All rights reserved. Published by Elsevier.
142
s coefficients:
1=A
1
⎧ 1
1 ⎫⎪
⎧ s −1 ⎫
−1 ⎪
= e2t + e2t t
+
x = Λ −1 ⎨ 2
⎬ =Λ ⎨
2⎬
⎩ s − 4s + 4 ⎭
⎪⎩ ( s − 2 ) ( s − 2 ) ⎭⎪
Hence,
2
Total: 20
Problem 6. Determine the poles and zeros for the transfer function: F(s) =
(s + 2)(s − 3)
and
(s + 3)(s 2 + 2s + 5)
plot them on a pole-zero diagram.
Marks
For the numerator to be zero, (s + 2) = 0 and (s - 3) = 0
zeros occur at s = -2 and at s = +3
Hence,
2
Poles occur when the denominator is zero, i.e. when (s + 3) = 0 i.e. s = -3
1
−2 ± 22 − 4(1)(5) −2 ± −16 −2 ± j4
and when s + 2s + 5 = 0, i.e. s =
=
=
2
2
2
2
2
= (-1 + j2) or (-1 – j2)
2
The poles and zeros are shown on the pole-zero map of F(s) in Figure 44.
Figure 44
3
Total: 10
TOTAL ASSIGNMENT MARKS: 100
© 2006 John Bird. All rights reserved. Published by Elsevier.
143
© 2006 John Bird. All rights reserved. Published by Elsevier.
144
ASSIGNMENT 19 (PAGE 704)
This assignment covers the material contained in chapters 69 to 74.
Problem 1. Obtain a Fourier series for the periodic function f (x) defined as follows:
⎧−1 when − π ≤ x ≤ 0
f (x) = ⎨
⎩ 1 when 0 ≤ x ≤ π
The function is periodic outside of this range with period 2π.
Marks
The square wave function is shown in Figure 45.
Figure 45
∞
The Fourier series is given by: f (x) = a 0 + ∑ ( a n cos nx + b n sin nx )
n =1
Mean value, a0 = 0 from Figure 45.
2
an = 0 , i.e. the series contains no cosine terms since Figure 45 is an odd function.
2
bn =
1 π
1
f (x) sin nx dx =
∫
π −π
π
{∫
0
−π
}
π
(−1) sin nx dx + ∫ (1) sin nx dx
0
0
π
1 ⎧⎪ ⎡ cos nx ⎤
⎡ − cos nx ⎤ ⎫⎪ 1 ⎧ ⎡ cos 0 cos− πn ⎤ ⎡ − cos πn − cos 0 ⎤ ⎫
= ⎨⎢
+
−
+
−
⎬= ⎨
⎬
π ⎩⎪ ⎣ n ⎥⎦ −π ⎢⎣ n ⎥⎦ 0 ⎭⎪ π ⎩ ⎢⎣ n
n ⎥⎦ ⎢⎣ n
n ⎥⎦ ⎭
1 ⎧⎡ 1 ⎛ 1 ⎞⎤ ⎡ ⎛ 1 ⎞ ⎛ 1 ⎞⎤ ⎫ 1 ⎛ 2 2 ⎞ 4
⎨ − ⎜− ⎟ + −⎜− ⎟ −⎜− ⎟ ⎬ = ⎜ + ⎟ =
π ⎩ ⎣⎢ n ⎝ n ⎠ ⎦⎥ ⎣⎢ ⎝ n ⎠ ⎝ n ⎠ ⎦⎥ ⎭ π ⎝ n n ⎠ nπ
When n is odd: b n =
When b1 =
4
,
π
b3 =
4
,
3π
When n is even: b n =
Hence, f (x) =
b5 =
3
2
4
,…
5π
1 ⎧⎛ 1 1 ⎞ ⎛ 1
1 ⎞⎫
⎨⎜ − ⎟ + ⎜ − − − ⎟ ⎬ = 0
n ⎠⎭
π ⎩⎝ n n ⎠ ⎝ n
2
4
4
4
sin x + sin 3x + sin 5x + ...
π
3π
5π
© 2006 John Bird. All rights reserved. Published by Elsevier.
145
f (x) =
or
4⎛
1
1
⎞
sin x + sin 3x + sin 5x + ... ⎟
⎜
3
5
π⎝
⎠
2
Total: 13
Problem 2. Obtain a Fourier series to represent f (t) = t in the range -π to +π.
Marks
A sketch of the waveform is shown in Figure 46.
Figure 46
∞
The Fourier series is given by: f (x) = a 0 + ∑ ( a n cos nx + b n sin nx )
n =1
Since the mean value of the waveform is zero, then a0 = 0
2
and since the waveform represents an odd function, i.e. an = 0
2
1 ⎡ − t cos nt
1 π
1 π
b n = ∫ f (t) sin nt dt = ∫ t sin nt dt = ⎢
−∫
π⎣
π −π
π −π
n
π
⎛ − cos nt ⎞ ⎤
⎜
⎟ dt ⎥ by parts
⎝ n ⎠ ⎦ −π
π
1 ⎡ − t cos nt sin nt ⎤
= ⎢
+ 2 ⎥
π⎣
n
n ⎦ −π
1 ⎡⎛ −π cos nπ sin nπ ⎞ ⎛ −(−π) cos n ( −π ) sin n(−π) ⎞ ⎤
= ⎢⎜
−
+
⎟⎥
⎟−⎜
π ⎢⎣⎝
n
n2 ⎠ ⎝
n
n2
⎠ ⎥⎦
=
1 ⎡ −π cos nπ π cos− nπ ⎤
2
−
= − cos nπ
⎢
⎥
π⎣
n
n
n
⎦
5
since cos(-πn) = cos nπ
When n is odd, b n =
2
n
When n is even, b n = −
thus b1 =
2
n
2
2
2
, b3 = , b5 = , …
1
3
5
1
1
1
thus b 2 = − , b 4 = − , b6 = − , …
1
2
3
2
1
2
1
Thus f (t) = t = 2sin t − sin 2t + sin 3t − sin 4t + sin 5t − sin 6t + ...
3
2
5
3
© 2006 John Bird. All rights reserved. Published by Elsevier.
146
1
1
1
1
1
⎛
⎞
i.e. t = 2 ⎜ sin t − sin 2t + sin 3t − sin 4t + sin 5t − sin 6t + ... ⎟
2
3
4
5
6
⎝
⎠
for values of t between -π and +π
4
Total: 13
Problem 3. Expand the function f (θ) = θ in the range 0 ≤ θ ≤ π into (a) a half range cosine series,
and (b) a half range sine series.
Marks
(a) The function f (θ) = θ is shown in Figure 47 as an even function, and for a half
∞
range cosine series f ( θ ) = a 0 + ∑ a n cos nθ
n =1
Figure 47
π
1 π
1 π
1 ⎡ θ2 ⎤
π
a 0 = ∫ f ( θ ) dθ = ∫ θ dθ = ⎢ ⎥ =
π 0
π 0
π ⎣ 2 ⎦0 2
2
π
an =
=
2 π
2 π
2 ⎡ θ sin nθ cos nθ ⎤
f ( θ ) cos nθ dθ = ∫ θ cos nθ dθ = ⎢
+
by parts
∫
π 0
π 0
π⎣ n
n 2 ⎥⎦ 0
2 ⎡⎛ π sin nπ cos nπ ⎞ ⎛
cos 0 ⎞ ⎤ 2 ⎛
cos nπ cos 0 ⎞
+
− 2 ⎟
⎜
⎟ − ⎜ 0 + 2 ⎟⎥ = ⎜ 0 +
⎢
2
n ⎠ ⎝
n ⎠⎦ π ⎝
n2
n ⎠
π ⎣⎝ n
=
When n is odd:
an =
When n is even: a n =
2
( cos nπ − 1)
π n2
4
2
−4
−4
−4
hence, a1 =
−1 − 1) =
, a 3 = 2 ,…
2 (
2
πn
πn
π
π3
2
(1 − 1) = 0
π n2
Hence, the Fourier half range cosine series is:
f (θ) = θ =
π 4⎛
1
1
⎞
− ⎜ cos θ + 2 cos 3θ + 2 cos 5θ + ... ⎟
2 π⎝
3
5
⎠
4
(b) The function f (θ) = θ is shown in figure 48 as an odd function, and for a half
© 2006 John Bird. All rights reserved. Published by Elsevier.
147
∞
range sine series f ( θ ) = ∑ b n sin nθ
n =1
Figure 48
π
bn =
2 π
2 π
2 ⎡ −θ cos nθ sin nθ ⎤
f ( θ ) sin nθ dθ = ∫ θ sin nθ dθ = ⎢
by parts
+
∫
n
n 2 ⎥⎦ 0
π 0
π 0
π⎣
2 ⎡⎛ −π cos nπ sin nπ ⎞
⎤
2
+
⎜
⎟ − ( 0 + 0 ) ⎥ = − cos nπ
⎢
2
n
n ⎠
n
π ⎣⎝
⎦
=
When n is odd, b n =
2
n
When n is even, b n = −
thus b1 =
2
n
4
2
2
2
, b3 = , b5 = , …
1
3
5
2
2
2
thus b 2 = − , b 4 = − , b 6 = − , …
2
4
6
Hence, the Fourier half range sine series is:
1
1
1
1
⎛
⎞
f ( θ ) = θ = 2 ⎜ sin θ − sin 2θ + sin 3θ − sin 4θ + sin 5θ − ... ⎟
2
3
4
5
⎝
⎠
4
Total: 18
⎧ 0 when − 4 ≤ x ≤ −2
⎪
Problem 4. (a) Sketch a waveform defined by: f (x) = ⎨ 3 when − 2 ≤ x ≤ 2 and is periodic
⎪ 0 when 2 ≤ x ≤ 4
⎩
outside of this range of period 8.
(b) State whether the waveform in (a) is odd, even or neither odd nor even.
(c) Deduce the Fourier series for the function defined in (a).
Marks
(a) The waveform is shown in Figure 49.
(b) The waveform represents an even function.
2
(c) Since the waveform is an even function, i.e. bn = 0
2
© 2006 John Bird. All rights reserved. Published by Elsevier.
148
2
Figure 49
a0 =
2
4
1 L/2
1 4
1 −2
f (x) dx = ∫ f (x) dx = ⎡ ∫ 0 dx + ∫ 3dx + ∫ 0 dx ⎤
∫
−2
2
⎦⎥
L −L / 2
8 −4
8 ⎣⎢ − 4
=
an =
1
1
3
2
[3x ] −2 = [6 − −6] =
8
8
2
2
2 L/2
2 4
2 2
⎛ 2πnx ⎞
⎛ 2πnx ⎞
⎛ πnx ⎞
f
(x)
cos
dx
=
f
(x)
cos
dx
3cos ⎜
=
⎜
⎟
⎜
⎟
⎟ dx
∫
∫
∫
2
−
−
−
L
/
2
4
L
8
8
⎝ L ⎠
⎝ 8 ⎠
⎝ 4 ⎠
2
⎡ ⎛ πnx ⎞ ⎤
sin
3 ⎢ ⎜⎝ 4 ⎟⎠ ⎥
3 ⎡ ⎛ 2πn ⎞
−πn ⎤
⎛ −2πn ⎞ ⎤ 3 ⎡ πn
⎢
⎥ =
=
− sin ⎜
=
− sin
sin ⎜
sin
⎟
⎟
⎢
⎥
⎢
πn ⎣ ⎝ 4 ⎠
4 ⎢ ⎛ πn ⎞ ⎥
2
2 ⎦⎥
⎝ 4 ⎠ ⎦ πn ⎣
⎜
⎟
⎢⎣ ⎝ 4 ⎠ ⎥⎦
−2
3
When n is even, a n = 0
a1 =
3⎛
π
−π ⎞ 3
6
⎜ sin − sin
⎟ = (1 − −1) =
π⎝
π
2
2 ⎠ π
a3 =
3 ⎛ 3π
−3π ⎞ 3
6
⎜ sin − sin
⎟ = ( −1 − 1) = −
3π ⎝
2
2 ⎠ 3π
3π
a5 =
−5π ⎞ 3
3 ⎛ 5π
6
⎜ sin − sin
⎟ = (1 − −1) =
5π ⎝
2
2 ⎠ 5π
5π
Hence, f (x) =
i.e. f (x) =
3 6
⎛ πx ⎞ 6
⎛ 3πx ⎞ 6
⎛ 5πx ⎞
+ cos ⎜ ⎟ − cos ⎜
⎟ + cos ⎜
⎟ − ...
2 π
⎝ 4 ⎠ 3π
⎝ 4 ⎠ 5π
⎝ 4 ⎠
3 6⎧
πx 1
3 πx 1
5πx
⎫
+ ⎨cos
− cos
+ cos
− ...⎬
2 π⎩
4 3
4
5
4
⎭
4
Total: 15
Problem 5. Displacement y on a point on a pulley when turned through an angle θ degrees is given
by: θ
30
60
90
120 150 180 210 240 270 300 330 360
y 3.99 4.01 3.60 2.84 1.84 0.88 0.27 0.13 0.45 1.25 2.37 3.41
Sketch the waveform and construct a Fourier series for the first three harmonics.
© 2006 John Bird. All rights reserved. Published by Elsevier.
149
Marks
The waveform is shown sketched in Figure 50.
Figure 50
2
The data is tabulated as shown in Table 19.1.
Table 19.1
11
f ( θ ) = a 0 + a1 cos θ + a 2 cos 2θ + a 3 cos 3θ + ... + b1 sin θ + b 2 sin 2θ + b3 sin 3θ + ...
© 2006 John Bird. All rights reserved. Published by Elsevier.
150
1 p
1
a 0 ≈ ∑ y k = ( 25.04 ) = 2.09
p k =1
12
an ≈
2 p
∑ yk cos nθk
p k =1
hence, a1 =
and
2 p
b n ≈ ∑ y k sin nθk
p k =1
a3 =
hence, b1 =
and
b3 =
1
2
2
( 7.37 ) = 1.23 , a 2 = ( 0.36 ) = 0.06
12
12
2
2
( 0.24 ) = 0.04
12
1
2
2
( 9.48) = 1.58 , b2 = ( 0.09 ) = 0.015
12
12
2
2
( 0.04 ) = 0.007
12
1
Hence, y = 2.09 + 1.23cos θ + 0.06cos 2θ + 0.04cos 3θ + ... + 1.58sin θ
+0.015sin 2θ + 0.007 sin 3θ + ...
3
Total: 23
Problem 6. A rectangular waveform is shown in Figure A19.1.
Figure A19.1
(a) State whether the waveform is an odd or even function. (b) Obtain the Fourier series for the
waveform in complex form. (c) Show that the complex Fourier series in (b) is equivalent to:
f(x) =
20 ⎛
1
1
1
⎞
⎜ sin x + sin 3x + sin 5x + sin 7x + ... ⎟
π⎝
3
5
7
⎠
Marks
(a) The square wave shown is an odd function since it is symmetrical about the
origin.
1
(b) The period of the waveform, L = 2π.
From equa (16), page 695: c n = − j
2 L2
⎛ 2πnx ⎞
f (x) sin ⎜
⎟ dx
∫
L 0
⎝ L ⎠
= −j
2 π
5 π
⎛ 2πnx ⎞
5 sin ⎜
dx = − j ∫ sin nx dx
⎟
∫
π 0
2π 0
⎝ 2π ⎠
© 2006 John Bird. All rights reserved. Published by Elsevier.
151
π
5 ⎡ − cos nx ⎤
5
= − j ⎡⎣( − cos πn ) − ( cos 0 ) ⎤⎦
= −j ⎢
⎥
π ⎣ n ⎦0
πn
cn = − j
i.e.
5
[1 − cos πn ]
πn
(1)
4
From equation (11), page 691, the complex Fourier series is given by:
∞
f(x) =
∑c
n =−∞
n e
j
2 πnx
L
∞
=
5
∑ − j nπ (1 − cos nπ ) e
jnx
(2)
1
n =−∞
(c) From equation (1) above,
when n = 1, c1 = − j
5
5
j10
(1 − cos π ) = − j (1 − −1) = −
(1)π
π
π
when n = 2, c2 = − j
5
(1 − cos 2π ) = 0 ; in fact, all even values of cn will be zero
2π
when n = 3, c3 = − j
5
5
j10
(1 − cos 3π ) = − j (1 − −1) = −
3π
3π
3π
By similar reasoning, c5 = −
j10
j10
, c7 = −
, and so on.
5π
7π
2
When n = -1, c−1 = − j
5
5
j10
1 − cos ( −π ) ) = + j (1 − −1) = +
(
(−1)π
π
π
When n = -3, c−3 = − j
5
5
j10
(1 − cos(−3π) ) = + j (1 − −1) = +
(−3)π
3π
3π
By similar reasoning, c−5 = +
j10
j10
, c−7 = +
, and so on.
5π
7π
2
Since the waveform is odd, c0 = a 0 = 0
∞
From equation (2) above, f(x) =
1
5
∑ − j nπ (1 − cos nπ ) e
jnx
hence,
n =−∞
f(x) = −
j10 jx j10 j3x j10 j5x j10 j7 x
e −
e −
e −
e − ...
π
3π
5π
7π
+
j10 − jx j10 − j3x j10 − j5x j10 − j7x
e +
e
e
e
+
+
+ ...
3π
5π
7π
π
⎛ j10 jx j10 − jx ⎞ ⎛ j10 3x j10 −3x ⎞ ⎛ j10 5x j10 −5x ⎞
= ⎜−
e +
e ⎟+⎜−
e +
e ⎟+⎜−
e +
e ⎟ + ....
3π
5π
π
⎝ π
⎠ ⎝ 3π
⎠ ⎝ 5π
⎠
= −
=
j10 jx
j10 3x
j10 5x
e − e − jx ) −
e − e−3x ) −
e − e−5x ) + ....
(
(
(
3π
5π
π
10 jx
10 3x
10 5x
e − e− jx ) +
e − e −3x ) +
e − e−5x ) + ....
(
(
(
jπ
j3π
j5π
© 2006 John Bird. All rights reserved. Published by Elsevier.
152
by multiplying top and bottom by j
=
20 ⎛ e jx − e − jx
⎜
π⎝
2j
=
20
20
20
sin x + sin 3x + sin 5x + .... from equation (4), page 690
π
3π
3x
i.e. f(x) =
⎞ 20 ⎛ e j3x − e − j3 ⎞ 20 ⎛ e j5x − e − j5x
⎟+ ⎜
⎟+ ⎜
2j
2j
⎠ 3π ⎝
⎠ 5π ⎝
⎞
⎟ + .... by rearranging
⎠
5
20 ⎛
1
1
1
⎞
sin x + sin 3x + sin 5x + sin 7x + ... ⎟
⎜
π⎝
3
5
7
⎠
∞
Hence, f(x) =
5
∑ − j nπ (1 − cos nπ ) e
n =−∞
jnx
≡
20 ⎛
1
1
1
⎞
sin x + sin 3x + sin 5x + sin 7x + ... ⎟
⎜
π⎝
3
5
7
⎠
2
Total: 18
TOTAL ASSIGNMENT MARKS: 100
© 2006 John Bird. All rights reserved. Published by Elsevier.
153