Bird - Higher Engineering Mathematics - 5e - Instructor Manual.pdf

HIGHER ENGINEERING

MATHEMATICS

5TH EDITION

INSTRUCTORS MANUAL

JOHN BIRD

2006 John Bird. All rights reserved. Published by Elsevier.

2

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 Instructors

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.


3

CONTENTS

Page

ASSIGNMENT 1 (chapters 1 to 5) 1




















4

ASSIGNMENT 1 (PAGE 50)

This assignment covers the material contained in chapters 1 to 5.

Problem 1. Factorise 3 2x 4x x 6+ + using the factor theorem. Hence solve the equation

3 2x 4x x 6+ + = 0 Marks

Let f (x) = 3 2x 4x 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 3 2x 4x x 6+ + = (x - 1)(x + 2)(x + 3) 3 If 3 2x 4x x 6+ + = 0 then (x - 1)(x + 2)(x + 3) = 0 from which, x = 1, -2 or -3 2

Total: 5

Problem 2. Use the remainder theorem to find the remainder when 3 22x x 7x 6+ is divided by (a) (x - 2) (b) (x + 1). Hence factorise the cubic expression.

Marks

(a) When 3 22x x 7x 6+ is divided by (x - 2), the remainder is given by: 3 2ap bp cp d+ + + , where a = 2, b = 1, c = -7, d = -6 and p = 2,

i.e. the remainder is: ( ) ( )3 22 2 1 2 7(2) 6+ = 16 + 4 - 14 - 6 = 0 2 hence (x - 2) is a factor of 3 22x x 7x 6+ (b) When 3 22x x 7x 6+ is divided by (x + 1), the remainder is given

by: ( ) ( ) ( )3 22 1 1 1 7 1 6 + = - 2 + 1 + 7 - 6 = 0 2 hence (x + 1) is a factor of 3 22x x 7x 6+


5

Either by dividing 3 22x x 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 3 22x x 7x 6+ = (x - 2)(x + 1)(2x + 3) 3 Total: 7

Problem 3. Simplify 26x 7x 52x 1+ by dividing out

Marks

3x + 5 2x - 1 26x 7x 5+ 26x 3x 10x 5 10x 5 . .

Hence 26x 7x 52x 1+ = 3x + 5 4

Total: 4

Problem 4. Solve the following inequalities:

(a) 2 5x 9 + 2x (b) 3 2t+ 6 (c) x 13x 5

+ > 0

(d) ( )23t 2+ > 16 (e) 22x x 3 < 0 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 3

(a) Since x 13x 5

+ > 0 then

x 13x 5

+ must be positive


6

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 ( )23t 2+ > 16 then 3t + 2 > 16 or 3t + 2 < - 16 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 22x 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) 2x 11

x x 2 (b) ( )( )2

3 xx 3 x 3

+ + (c)

3

2

x 6x 9x x 2 ++

Marks

(a) Let 2x 11

x x 2

x 11 A B A(x 1) B(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 2x 11 4 3

x x 2 (x 1) (x 2) = + 4


7

(b) Let ( )( )23 x

x 3 x 3

+ + ( )( )

( )( )2

2 2

(Ax B)(x 3) C x 3Ax B C(x 3)x 3 x 3 x 3

+ + + ++ + =++ + + 3

Hence 3 - x = (Ax + B)(x + 3) + C 2(x 3)+

Let x = -3: 6 = 0 + 12C hence C = 12

2x coefficients: 0 = A + C hence A = 12

x coefficients: -1 = 3A + B hence -1 = 32

+ B and B = 12

Hence ( )( )23 x

x 3 x 3

+ + = ( )21 1 1x2 2 2

(x 3)x 3

++ ++ or ( )2

1 x 12(x 3)2 x 3

+ ++ 5

(c) Dividing out gives: x - 1

2 33 2

x x 2 x 6x 9x x 2x

+ ++

2x 4x 9 + 2x x 2 + - 3x + 7

Hence 3

2

x 6x 9x x 2 ++ x - 1 + 2

3x 7x x 2 ++ 4

Let 23x 7

x x 2 ++

A B A(x 1) B(x 2)x 2 x 1 (x 2)(x 1)

+ ++ + + 2

from which, -3x + 7 = A(x - 1) + B(x + 2)

Let x = -2: 13 = -3A hence A = 133

Let x = 1: 4 = 3B hence B = 43

Hence 23x 7

x x 2 ++ =

13 43 3

(x 2) (x 1)

++ 3

and 3

2x 6x 9 13 4x 1x x 2 3(x 2) 3(x 1) + = ++ + 1

Total: 24


8

Problem 6. Evaluate, correct to 3 decimal places, 0.9825e

3ln 0.0173

Marks

0.9825e 1.872806134

3ln 0.0173 12.17114633

= = - 0.154 2

Total: 2

Problem 7. Solve the following equations, each correct to 4 significant figures:

(a) ln x = 2.40 (b) x 1 x 23 5 = (c) x25 8 1 e

=

Marks

(a) Since ln x = 2.40 then x = 2.40e = 11.02 2

(b) Since x 1 x 23 5 = then x 1 x 2lg 3 lg 5 = 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)

from which, x = 2 lg5 lg3 0.92081875...lg 5 lg3 0.221848749...

=

i.e. x = 4.151 4

(c) Since x25 8 1 e

= then

x25 1 e

8=

and x2 5 3e 1

8 8 = =

Thus x2 8e

3=

and x 8ln2 3=

i.e. x = 82ln3

= 1.962 4

Total: 10


9

Problem 8. The pressure p at height h above ground level is given by: kh0p p e

= where 0p is the

pressure at ground level and k is a constant. When 0p 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 0p is 101 kilopascals

Marks

Since kh0p p e= then 3 3 1500k100 10 101 10 e =

and 1500k100 e101

= or 1500k101 e100

=

Hence ln 1.01 = 1500 k and k = 1 ln1.011500

= 66.6336 10 4 A table of values is drawn up as shown

h (m) 0 2000 4000 6000 8000 10000 12000

kh0p p e= (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


10

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 = 2.47 2.47e e

2

= 5.869 (or by calculator) 1

(b) tanh 0.6439 = 0.6439 0.6439

0.6439 0.6439

e e 1.378651608e e 2.429131585

=+ = 0.5675 (or by calculator) 1

(c) sech 1.385 = 1.385 1.3851 2

cosh1.385 e e= + = 0.4711 2

(d) cosech 0.874 = 0.874 0.8741 2

sinh 0.874 e e= = 1.011 (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: t sinh t sin t2 cosh t cos t + =

Calculate , correct to 5 significant figures, when = 1.08 and t = 1. Marks

= t sinh t sin t (1.08)(1) sinh1.08 sin1.082 cosh t cos t 2 cosh1.08 cos1.08 + + =

= ( ) 1.302542013 0.8819578060.541.642137538 0.471328364

+ 3

= ( ) 2.1844998190.541.170809174

= 1.0075 2

Total: 5

Problem 11. If A ch x - B sh x x x4e 3e determine the values of A and B.

Marks

A ch x - B sh x x x4e 3e

i.e. x x x xe e e eA B

2 2

+ x x4e 3e

i.e. x x x xA A B Be e e e2 2 2 2

+ + x x4e 3e 2 2006 John Bird. All rights reserved. Published by Elsevier.

11

Hence A B 42 2 = i.e. A - B = 8 (1)

A B 32 2+ = 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. x x x xe e e e3.52 8.42 5.32

2 2

+ + =

i.e. x x x x1.76e 1.76e 4.21e 4.21e 5.32 + + = x x5.97e 2.45e 5.32 0 =

Hence ( )2x x5.97 e 2.45 5.32e 0 = ( )2x x5.97 e 5.32e 2.45 0 = 3 and

( ) ( )( )( )

2x ( 5.32) 5.32 4 5.97 2.45 5.32 9.3171e

2 5.97 11.94 = =

= 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


12



Problem 1. Determine the 20th term of the series 15.6, 15, 14.4, 13.8,

Marks

The 20th term is given by: a + (n 1)d

i.e. 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

[ ]n nS 2a (n 1)d2= + i.e. [ ]13286 2a (13 1)32

= + i.e. [ ]13286 2a 362

= + 1

Thus, 286 2 2a 3613 = = i.e. 44 36 = 2a

from which, first term, a = 44 362 = 4 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 nth term, n = 9

Then salary in 9th year = a + (n 1)d = 21000 + (9 1)(600)

= 21000 + 8(600) = 25800 2

Total earnings in first 11 years, [ ] [ ]11 n 11S 2a (n 1)d 2(21000) (11 1)6002 2= + = + [ ]11 42000 6000

2= + = 264,000 3

Total: 5


13

Problem 4. Determine the 11th term of the series 1.5, 3, 6, 12, .

Marks

The 11th term is given by: n 1a r where a = 1.5, and the common ratio, r = 2,

i.e. 11th term = 11 1(1.5)(2) = 1536 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, ( ) ( )n 8

8

a r 1 1 2.5 1 1524.879..S(r 1) (2.5 1) 1.5

= = =

= 1016.6 4

Total: 4

Problem 6. Determine the sum to infinity of the series 5, 1, 15

, .

Marks

aS1 r

= where a = 5 and r = 15

,

thus, the sum to infinity, 5 5 25S 1 4 415 5

= = = = 16

4 or 6.25 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.


14

Marks

The G.P. of n terms is given by: 2 n 1a, ar, ar ,...ar

The first term, a = 25 rev/min.

The seventh term is given by: 7 1a r which is 500 rev/min,

i.e. 6a r 500= from which, 6 500 500r 20a 25

= = =

thus, the common ratio, r = 6 20 = 1.64755 2

The first term is 25 rev/min

The second term, ar = (25)(1.64755) = 41.19

The third term, 2 2a r (25)(1.64755)= = 67.86

The fourth term, 3 3a r (25)(1.64755)= = 111.80

The fifth term, 4 4a r (25)(1.64755)= = 184.20

The sixth term, 5 5a r (25)(1.64755)= = 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 ( )62a 3b . Marks

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )6 6 5 4 2 3 3(6)(5) (6)(5)(4)2a 3b 2a 6 2a 3a 2a 3a 2a 3a2! 3!

= + + +

( ) ( ) ( )( ) ( )2 4 5 6(6)(5)(4)(3) (6)(5)(4)(3)(2)2a 3b 2a 3b 3b4! 5!

+ + + 4

= 6 5 4 2 3 3 2 4 5 664a 576a b 2160a b 4320a b 4860a b 2916ab 729b + + + 3 Total: 7

Problem 9. Determine the middle term of 18

13x3y

.


15

Marks

The middle term is the 10th 1

i.e. ( )9

9(18)(17)(16)(15)(14)(13)(12)(11)(10) 13x9! 3y

= 9

9

x48620y

or 9

x48620y

5

Total: 6

Problem 10. Expand the following in ascending powers of t as far as the term in 3t :

(a) 11 t+ (b)

11 3t

For each case, state the limits for which the expansion is valid.

Marks

(a) ( ) 1 2 31 ( 1)( 2) ( 1)( 2)( 3)1 t 1 ( 1)t t t ...1 t 2! 3!

= + = + + ++

= 2 31 t t t ... + + 4 The expansion is valid when t < 1 or -1 < t < 1 2

(b) 11 3t = ( ) ( ) ( ) ( )

1/ 2 2 3

1 3 1 3 51 2 2 2 2 21 3t 1 3t 3t 3t ...2 2! 3!

= + + +

= 2 33 27 1351 t t t ...2 8 16

+ + + + 4

The expansion is valid when 3t < 1 i.e. t < 13

or 13

< t < 13

2

Total: 12

Problem 11. When x is very small show that: ( )21 31 x

21 x 1 x +

Marks

( ) ( ) ( ) [ ]122

21 11 x 1 x 1 ( 2)x ... 1 x ...

21 x 1 x = + = + + + +

1 11 x ( 2)x ... 1 x 2x ...2 2

+ + + + = 31 x2

= 5

when powers of 2 and above are neglected.

Total: 5


16

Problem 12. The modulus of rigidity G is given by 4RGL= where R is the radius, the

angle of twist and L the length. Find the approximate percentage error in G when R is measured

1.5% too large, is measured 3% too small and L is measured 1% too small. Marks

The new values of R, and L are: (1 + 0.015)R, (1 0.03) and (1 0.01)L

New modulus of rigidity = [ ] [ ][ ]4(1 0.015)R (1 0.03)

(1 0.01)L+

3

= [ ] [ ][ ]4 1(1 0.015)R (1 0.03) (1 0.01)L + = 4 4 1 1(1 0.015) R (1 0.03) (1 0.01) L + = 4 1 4 1(1 0.015) (1 0.03)(1 0.01) R L +

= [1 + 4(0.015)][1 - 0.03][1 - (-1)(0.01)]4R

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 Maclaurins series to determine a power series for 2xe cos3x as far as the term in 2x .

Marks

Let f(x) = 2xe cos3x f(0) = 0e cos 0 = 1 1

( )( ) ( )( )2x 2xf '(x) e 3sin 3x cos3x 2e= + = 2x 2x3e sin 3x 2e cos3x + f '(0) = 0 03e sin 0 2e cos 0 + = 2 3 ( )( ) ( )( ) ( )( ) ( )( )2x 2x 2x 2xf ''(x) 3e 3cos3x sin 3x 6e 2e 3sin 3x cos3x 4e= + + + f ''(0) = 0 0 0 09e cos 0 6e sin 0 6e sin 0 4e cos 0 + = -9 0 0 + 4 = -5 3

The Maclaurins series is: f(x) = f(0) + x f '(0) + 2x f ''(0) ...

2!+

i.e. 2x 25e cos 3x 1 2x x ...2

= + + 3 Total: 10


17

Problem 14. Show, using Maclaurins series that the first four terms of the power series for cosh 2x

is given by: cosh 2x = 2 4 62 41 2x x x3 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

ivf (x) = 16 cosh 2x ivf (0) = 16 vf (x) = 32 sinh 2x vf (0) = 0 vif (x) = 64 cosh 2x vif (0) = 64 5

Since 2 3x xf (x) f (0) x f '(0) f ''(0) f '''(0) ...

2! 3!= + + + +

then f(x) = cosh 2x = 1 + x(0)2 3 4 5 6x x x x x(4) (0) (16) (0) (64) ...

2! 3! 4! 5! 6!+ + + + + +

i.e. cosh 2x = 2 4 62 41 2x x x3 45

+ + + 6 Total: 11

Problem 15. Expand the function ( )2x ln 1 sin x+ using Maclaurins series and hence evaluate:

( )1/ 2 20

x ln 1 sin x dx+ correct to 2 significant figures. Marks

Let f(x) = ln(1 + sin x) f(0) = ln(1 + sin 0) = 0 1

f (x) = cos x1 sin x+ f (0) =

cos 01 sin 0+ = 1 1

f (x) = ( ) ( )2 2

2 2(1 sin x)( sin x) (cos x)(cos x) sin x sin x cos x

1 sin x 1 sin x+ =+ +

= 2 2sin x 1 (1 sin x) 1

(1 sin x) (1 sin x) 1 sin x + = =+ + + f (0) =

11 sin 0

+ = -1 3


18

f (x) = 2 2(1 sin x)(0) ( 1)(cos x) cos x(1 sin x) (1 sin x)+ =+ + f (0) = 2

cos 0(1 sin 0)+ = 1 2

Hence, ln(1 + sin x) = 2 3x xf (0) x f '(0) f ''(0) f '''(0) ...

2! 3!+ + + +

= 2 3x x0 x(1) ( 1) (1) ...

2! 3!+ + + + =

2 3x xx ...2 6

+ 2

Thus, ( )2x ln 2 sin x+ = 2 3 4 52 3x x x xx x ... x ...2 6 2 6

+ = +

and 4 51/ 2 1/ 22 3

0 0

x xx ln(1 sin x)dx x ... dx2 6

+ = +

=

4 5 6

1/ 24 5 6

0

1 1 1x x x 2 2 2 (0)4 10 36 4 10 36

+ = + 2

= 0.015625 0.003125 + 0.000434

= 0.013, correct to 2 significant figures. 2

Total: 13



19



Problem 1. Use the method of bisection to evaluate the root of the equation 3x 5x 11+ = in the range x = 1 to x = 2, correct to 3 significant figures.

Marks

Let f (x) = 3x 5x 11+ f (1) = 31 5 11 5+ = and f (2) = 32 5(2) 11 7+ = Hence a root lies between x = 1 and x = 2 2

Using a tabular approach:

1x 2x 1 23x xx

2= ( )3f x

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

1.5078125 1.515625 1.51171875 +0.0133

Hence, x = 1.51, correct to 3 significant figures 10

Total: 12

Problem 2. Solve 3x 5x 11+ = using an algebraic method of successive approximations. Correct to 3 significant figures.

Marks

f (x) = 3x 5x 11+ f (1) = 31 5 11 5+ = and f (2) = 32 5(2) 11 7+ = 2 Hence a root lies between x = 1 and x = 2


20

First approximation Let the first approximation be 1.4 1 Second approximation Let the true value of the root, x2 , be (x1 + 1) Let f(x1 + 1) = 0, then since x1 = 1.4, (1.4 + 1)3 + 5(1.4 + 1) - 11 = 0 Neglecting terms containing products of 1 and using the binomial series gives: [(1.4)3 + 3(1.4)2 1] + 7 + 51 - 11 0 2.744 + 5.881 + 7 + 5 1 - 11 0 10.881 11 2.744 7

1 1.256 0.1154410.88 Thus x2 1.4 + 0.11544 = 1.51544 4 Third approximation Let the true value of the root, x3 , be (x2 + 2) Let f(x2 + 2) = 0, then since x2 = 1.51544, (1.51544 + 2)3 + 5(1.51544 + 2) - 11 = 0 Neglecting terms containing products of 2 gives: ( ) ( )3 2 2 21.51544 3 1.51544 7.5772 5 11 0+ + + 3.480296 + 6.8896752 + 7.6772 + 52 - 11 0 11.8896752 11 3.480296 7.5772

2 0.057496 0.00483611.889675

Thus x3 (x2 + 2) = 1.51544 0.004836 1.5106 4 Fourth approximation Let the true value of the root, x4 , be (x3 + 3)


21

Let f(x3 + 3) = 0, then since x3 = 1.5106, (1.5106 + 3)3 + 5(1.5106 + 3) - 11 = 0 Neglecting terms containing products of 3 gives: ( ) ( )3 2 3 31.5106 3 1.5106 7.553 5 11 0+ + + 3.447057 + 6.8457373 + 7.553 + 53 - 11 0 11.8457372 11 3.447057 7.553

2 0.000057 0.000004811.845737

Thus x4 (x3 + 3) = 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:

( )x xy 2.5 e e x 25= + Use Newtons method to determine the value of x, correct to 2 decimal places, for which the value

of y is zero

Marks

If y = 0, x x2.5e 2.5e x 25 0 + = Let f (x) = x x2.5e 2.5e 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 1r = 2.2 4


22

f (x) = x x2.5e 2.5e 1+ + 2.2 2.2

12 1 2.2 2.2

1

f (r ) 2.5e 2.5e 2.2 25r r 2.2f '(r ) 2.5e 2.5e 1

+ = = + +

= 0.5144741472.223.83954164 = 2.222

3f (2.222) 0.01542962r 2.222 2.222f '(2.222) 24.33539016

= = = 2.221

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) 3 2 1 021101 1 2 1 2 0 2 1 2= + + + = 8 + 4 + 0 + 1 = 1013 2 (b) 5 4 3 2 1 0 1 22101101.0101 1 2 0 2 1 2 1 2 0 2 1 2 0 2 1 2

= + + + + + + + 3 40 2 1 2 + +

= 32 + 0 + 8 + 4 + 0 + 1 + 0 + 14

+ 0 + 116

= 1045.3125 3

Total: 5

Problem 5. Convert the following decimal numbers to binary form: (a) 27 (b) 44.1875

Marks (a)

Hence 10 227 11011= 3


23

(b)

Hence 10 244 101100= 3

Hence 10 244.1875 101100.0011= 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, 8 2737 111 011 111=

Hence 10 2479 111 011 111= 3


24

(b)

From Table 10.1, page 88, 8 2271 010 111 001= 3

0.2890625 8 = 2.3125 0.3125 8 = 2.5 0.5 8 = 4.0 i.e. 10 8 20.2890625 .224 .010 010 100= = from Table 10.1, page 88. Hence 10 2185.2890625 10 111 001.010 010 1= 3 Total: 9

Problem 7. Convert (a) 165F into its decimal equivalent

(b) 10132 into its hexadecimal equivalent

(c) 2110 101 011 into its hexadecimal equivalent

Marks

(a) 1 0 1 0165F 5 16 F 16 5 16 15 16= + = + = 80 + 15 = 1095 2 (b)

i.e. 10 16132 84= 3 (c) Grouping bits in 4s from the right gives: 2110101011 0001 1010 1011= and assigning hexadecimal symbols to each group gives: 1 A B

Hence, 2 16110 101 011 1AB= 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+ + +


25

Marks (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 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 Morgans laws. Marks ( ) ( )A.B A.B.C A.B.A.B.C A B . A.B C+ = = + + 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


26

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



27



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 = 2 22 0.52 = 1.931 m by Pythagoras theorem. 3

Figure 3

(b) New height, h = 2 22 0.42 = 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 12413 (b) cot 72.68 Marks

(a) cos 12413 = cos O13124

60 = -0.5623 2

(b) cot 72.68 = 1tan 72.68 = 0.3118 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.


28

Marks

From the sketch in Figure 4, htan 20x

= from which, h = x tan 20 2

and htan1530 x

= + 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 8.0385x0.09602

= = 83.717

Hence, the height of the spire, h = 83.717 tan 20 = 30.47 m 5 Total: 9

Problem 4. If secant = 2.4613 determine the acute angle . Marks

Since sec = 2.4613 then cos = 1 0.4062892.4613

= 2

and = 1cos (0.406289) = 66.03 2 Total: 4

Problem 5. Evaluate, correct to 3 significant figures: 3.5 cos ec 31 17 ' cot( 12 )3 sec79 41'

Marks

( ) ( )( )

3.5 1.925778428 4.704630113.5 cos ec 31 17 ' cot( 12 )3 sec79 41' 3 5.583834323

= = 0.683 5

Total: 5


29

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 32.5sin 60 sin C

= from which, 32.5sin 60sin C 0.351823

80= =

Hence, C = 1sin (0.351823) = 20.60 (or 159.40 which is impossible in this case), Thus, B = 180 - 60 - 20.60 = 99.40 3

Using the sine rule again: 80 bsin 60 sin 99.40

=

from which, 80sin 99.40b 91.14 kmsin 60

= = 3

Since the cyclist travels 91.14 km in 5 hours then:

average speed = dis tan ce 91.14time 5

= = 18.23 km/h 2

Total: 8


30

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 14

revolution. (b) If POQ is initially 40 find the angle between the

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 = 0, OQ = 4.20 + 0.90 = 5.10 m

When = 90, OQ = 2 24.2 0.90 = 4.10 m by Pythagorass theorem. Hence, Q moves 5.10 4.10 = 1.0 m 4

(b) Using the sine rule: 4.2 0.90sin 40 sin Q

= from which, 0.90sin 40sin Q 0.13774

4.20= =

and Q = 1sin (0.13774) = 7.92 or 755 3 Hence, angle P = 180 - 40 - 7.92 = 132.08

Using the sine rule again gives: OQ 4.20sin132.08 sin 40

=

from which, OQ = 4.20sin132.08sin 40

= 4.85 m 3

(c) Using the sine rule: 4.20 0.90sin140 sin Q

=


31

from which, 0.90sin140sin Q 0.137744.20

= =

and Q = 1sin (0.13774) = 7.92 Hence, angle P = 180 - 140 - 7.92 = 32.08 2 Using the sine rule again gives: OQ 4.20

sin 32.08 sin140=

from which, OQ = 4.20sin 32.08 3.47 msin140

= 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 = 2 22.3 5.4+ = 5.87 and 1 5.4tan 66.932.3

= =

Hence, = 180 - 66.93 = 113.07 Thus, (-2.3, 5.4) = (5.87, 113.07) or (5.87, 1.97 rad) 5

Figure 6 Figure 7

(b) From figure 7, r = 2 27.6 9.2 11.93+ = and 1 9.2tan 50.447.6

= =

Hence, = 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)


32

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 7117 into radians Marks

(a) 2.154 rad = 1802.154 123.415 = = 12325 2

(b) 7117 = .1771 71.28360 180

= rad = 1.244 rad 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 from which, angle of lap, = s 140

r 180 / 2= = .1.55 or 89.13 or 898 3

Figure 8


33

(b) Angular velocity, = 300260

= 10 rad/s or 31.42 rad/s 3

(c) Linear velocity, v = r = ( )3rad10 90 10 ms = 0.9 m/s or 2.83 m/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,

2 2OX 12 6= = 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

= 21 1 60(12)(10.39) (12) 22 2 180

+ 3

= 62.34 + 376.99 = 439.33 2cm 2


34

(c) Area of unshaded portion = ( )212 - 439.33 = 452.39 439.33 = 13.06 2cm 3 Total: 11

Problem 13. Determine (a) the co-ordinates of the centre of the circle, and (b) the radius,

given the equation 2 2x y 2x 6y 6 0+ + + =

Marks 2 2x y 2x 6y 6 0+ + + = is of the form 2 2x y 2ex 2fy c 0+ + + + =

(a) If (a, b) are the co-ordinates of the centre of the circle,

then 2a2= = 1 and b = 6

2 = -3

Hence, (1, -3) are the co-ordinates of the centre of the circle. 4

(b) Radius, r = 2 2 2 2a b c (1) ( 3) 6 4+ = + = = 2 3

Total: 7



35



Problem 1. Solve the following equations in the range 0 to 360

(a) 1sin (-0.4161) = x (b) 1cot (2.4198) =

Marks

(a) If 1sin (-0.4161) = x then from Figure 10, = 24.59 and x = 204.59 and 335.41 4

Figure 10 Figure 11

(b) If 1cot (2.4198) = , then 1 1tan2.4198

= , and from Figure 11,

= 22.45 and = 22.45 and 202.45 4 Total: 8

Problem 2. Sketch the following curves labelling relevant points:

(a) y = 4 cos( + 45) (b) y = 5 sin(2t - 60) Marks

(a) y = 4 cos( + 45) is sketched in Figure 12. (b) y = 5 sin(2t - 60) is sketched in Figure 13.


36

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 120 100 t 0.274= + amperes Determine: (a) the amplitude, periodic time, frequency and phase angle (with reference to

120 sin 100t), (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

(a) Amplitude = 120 A 1

= 100, hence, periodic time, T = 2 1 s100 50

= = 20 ms 1


37

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 ( )3100 6 10 0.274 + = 120 sin 2.1589556 = 99.84 A 3 (d) When i = 80 A, 80 = 120 sin ( )100 t 0.274 + from which, 80

120 = sin ( )100 t 0.274 + and 80arcsin 100 t 0.274

120 = +

i.e. 0.72972766 = ( )100 t 0.274 + Hence, time, t = 0.72972766 0.274

100

= 1.451 ms 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 /3 rad, and a 20% fifth harmonic component lagging the fundamental at zero time by /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.


38

Marks

(a) Voltage v =

( ) ( ) ( ) ( ) ( )2 (141.1)sin 2 100t 0.35 2 141.1 sin 2 300t / 3 0.2 2 141.1 sin 2 500t / 4 + + + i.e. v = 200 sin 200t + 70 sin(600t + /3) + 40 sin (1000t - /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


39

Problem 5. Prove the following identities:

(a) 2

2

1 cos tancos

= (b) 3cos sin

2 + = (c)

22sin x 1 tan x

1 cos 2x 2=+

Marks

(a) L.H.S. = 2 2

2 2

1 cos sincos cos

= since 2 2cos sin 1+ =

= sincos

= tan = R.H.S. 3

(b) L.H.S. = 3 3 3cos cos cos sin sin2 2 2 + = from compound angles

= 0 (-1) sin = sin = R.H.S. 3

(c) L.H.S. = ( )22 2 2

22

sin x sin x sin x 1 sin x1 cos 2x 2cos x 2 cos x1 2cos x 1

= = = + +

= ( )2 21 1tan x tan x2 2

= = R.H.S. 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) 25sin x 3sin x 4+ = (d) 22sec 5 tan 3+ = Marks

(a) Since 4 cos x + 1 = 0 then cos x = 14

and x = 1cos (-0.25)

i.e. x = 104.48 (or 10429) and 255.52 (or 25531) 4 (b) Since 3.25 cosec x = 5.25 then cosec x = 5.25

3.25 and sin x = 3.25

5.25

i.e. x = 1sin 3.255.25

= 38.25 (or 3815) and 141.75 (or 14145) 4

(c) Since 25sin x 3sin x 4+ = then 25sin x 3sin x 4 0+ =

and 23 3 4(5)( 4) 3 89sin x2(5) 10

= = = 0.6434 or -1.2434

Ignoring the latter, sin x = 0.6434 and x = 1sin 0.6434 = 40.05 or 139.95 5


40

(d) Since 22sec 5 tan 3+ = then ( )22 1 tan 5 tan 3+ + = Hence, 22 tan 5 tan 1 0+ = from which,

25 5 4(2)( 1) 5 33tan2(2) 4

= = = 0.18614066 or -2.68614066

and = 10.54, 190.54, 110.42 and 290.42 5 Total: 18

Problem 7. Solve the equation: ( )5sin / 6 8cos = for values of 0 2. Marks

( )5sin / 6 8cos = i.e. 5[sin cos /6 cos sin /6] = 8 cos 2 Thus, 4.33 sin - 2.5 cos = 8 cos 1 and 4.33 sin = 10.5 cos Hence, sin 10.5

cos 4.33 = = 2.42494226 2

i.e. tan = 2.42494226 1

and = 1tan (2.42494226) = 67.59 and 247.59 2 Total: 8

Problem 8. Express 5.3 cos t 7.2 sin t in the form R sin(t + ). Hence solve the equation: 5.3 cos t 7.2 sin t = 4.5 in the range 0 t 2. Marks

Let 5.3 cos t 7.2 sin t = R sin(t + ) = R[sin t cos + cos t sin ] = (R cos ) sin t + (R sin ) cos t Hence 5.3 = R sin i.e. sin = 5.3

R

and -7.2 = R cos i.e. cos = 7.2R


41

Figure 16

There is only one quadrant where sine is positive and cosine is negative, i.e. the

second, as shown in Figure 16.

2 2R 5.3 7.2 8.94= + = and 1 5.3tan7.2

= = 0.6346 rad

hence, = - 0.6346 = 2.507 rad. Thus, 5.3 cos t 7.2 sin t = 8.94 sin(t + 2.507) 6

If 5.3 cos t 7.2 sin t = 4.5 then 8.94 sin(t + 2.507) = 4.5

and sin(t + 2.507) = 4.58.94

= 0.50336

t + 2.507 = 1sin (0.50336) = 0.5275 rad or 2.6141 rad

and t = 0.5275 2.507 = -1.9795 -1.9795 + 2 = 4.304 s or t = 2.6141 2.507 = 0.107 s 6

Total: 12

Problem 9. Determine 2cos3t sin t dt Marks

2 cos 3t sin t = [ ]12 sin(3t t) sin(3t t) sin 4t sin 2t2

+ =

Hence 2cos3t sin t dt = sin 4t sin 2t dt = 1 1cos 4t cos 2t c4 2 + + Total: 3



42



Problem 1. Sketch the following graphs, showing the relevant points:

(a) ( )2y x 2= (b) y = 3 cos 2x (c) 2 2x y 2x 4y 4 0+ + =

(d) 2 29x 4y 36 = (e) 1 x

2

f (x) x x2 2

1 x2

=

Marks

(a) A graph of ( )2y x 2= is shown in Figure 17.

Figure 17 3

(b) A graph of y = 3 cos 2x is shown in Figure 18.

Figure 18 3


43

(c) 2 2x y 2x 4y 4 0+ + = is a circle. If the co-ordinates of the circle are (a, b) then

a = - 22 = 1 and b = - 4

2 = -2

and radius, r = ( ) ( )22 2 2a b ( 4) 1 2 4+ = + = 3 The circle is shown in Figure 19.

Figure 19 3

(d) 2 29x 4y 36 = is equivalent to 2 29x 4y 1

36 36 = i.e.

2 2

2 2

x y 12 3

= which is a

hyperbola as shown in Figure 20.

Figure 20 3

(e)

1 x2

f (x) x x2 2

1 x2

=

is shown in Figure 21.


44

Figure 21 3

Total: 15

Problem 2. Determine the inverse of f (x) = 3x + 1

Marks

Let f (x) = y then y = 3x + 1 and transposing for x gives: x = y 13

and interchanging x and y gives: y = x 13

Hence, the inverse of f (x) = 3x + 1 is: 1 x 1f (x)3

= 3 Total: 3

Problem 3. Evaluate, correct to 3 decimal places:

2 1tan 1.64 + 1sec 2.43 3 1cos ec 3.85

Marks

2 1tan 1.64 + 1sec 2.43 3 1cos ec 3.85

= 1 1 11 12 tan 1.64 cos 3sin2.43 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:

( )( )( )( )x 1 x 4

yx 2 x 5 +=


45

Marks

Rearranging ( )( )( )( )x 1 x 4

yx 2 x 5 += gives: ( )2 2y x 7x 10 x 3x 4 + = +

and ( ) ( )2x y 1 x 7y 3 10y 4 0 + + + = For asymptotes parallel to the x-axis: y 1 = 0 i.e. y = 1 2

For asymptotes parallel to the y-axis: 2x 7x 10 0 + = i.e. (x 2)(x 5) = 0 from which, x = 2 and x = 5 2

To check for other asymptotes, let y = mx + c

then (mx + c)( 2x 7x 10 + ) = 2x 3x 4+ i.e. 3 2 2 2mx 7mx 10mx cx 7cx 10c x 3x 4 0 + + + + = and 3 2mx (c 1 7m)x 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 ( )( )( )( )x 1 x 4

yx 2 x 5 += is shown in Figure 22.

Figure 22 4

Total: 8


46

Problem 5. Plot a graph of 2y 3x 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) Simpsons 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

2y 3x 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,

shaded area = ( ) ( )10.5 8.0 53.0 11.75 17.0 23.75 32.0 41.752

+ + + + + +

= 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

2y 3x 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 3


47

(c) By Simpsons rule,

shaded area ( ) ( ) ( ) ( ){ }1 0.5 8.0 53.0 4 11.75 23.75 41.75 2 17.0 32.03

+ + + + + +

= ( ){ }1 0.5 61 309 983

+ + = 78 square units 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 5.0 10.0 15.0 20.0

Diameter (m) 16.0 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

Area 2

2d m4

= 201.06 138.93 89.92 58.09 50.27 3

Using Simpsons rule,

capacity of tower ( ) ( ) ( ) ( ){ }1 5.0 201.06 50.27 4 138.93 58.09 2 89.923

+ + + +

= ( ){ }1 5.0 251.33 788.08 179.843

+ + = 2032 3m 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 Simpsons 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.


48

Marks

(a) Distance travelled = ( ) ( ) ( ) ( ){ }1 1 0 9.2 4 1.2 3.7 6.0 2 2.4 5.23

+ + + + + +

= { }1 9.2 43.6 15.23

+ + = 22.67 m 4

(b) Average speed = 22.67 m6s

= 3.78 m/s 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 = 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 = ( ) ( )2 25.7071 6.6066 + = 8.73 N


49

and = 1 1V 6.6066tan tanH 5.7071

= = 229.18 5 (or use complex numbers)

Total: 10

Problem 9. The instantaneous values of two alternating voltages are given by:

1v 150sin t 3 = + volts and 2v 90sin t 6

= volts

Plot the two voltages on the same axes to scales of 1 cm = 50 volts and 1 cm = 6 rad. Obtain a

sinusoidal expression for the resultant 1 2v v+ in the form R sin(t + ): (a) by adding ordinates at intervals and (b) by calculation.

Marks

(a) From Figure 25, by adding ordinates at intervals, the waveform of 1 2v v+ is seen

to have a maximum value of 175 V and is leading by 30 or 6 rad i.e. 0.52 rad.

Hence, ( )1 2v v 175sin t 0.52+ = + volts 7

Figure 25

(b) By calculation:

At time t = 0, the phasors 1 2v and v are shown in Figure 26.


50

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, 1 2v v+ = 2 2152.942 84.904+ = 174.93 volts 97

Figure 26

Direction of 1 2v v+ = 1 84.904tan 29.04 or 0.507 rad.152.942 =

Hence, ( )1 2v v 174.93sin t 0.507 volts+ = + 6 Total: 13

Problem 10. If a = 2i + 4j 5k and b = 3i 2j + 6k determine:

(i) a b (ii) a b+ (iii) a b (iv) the angle between a and b

Marks

(i) a b = (2)(3) + (4)(-2) + (-5)(6) = 6 8 30 = -32 3

(ii) a b+ = 5i 2 j k+ + = 2 2 25 2 1+ + = 30 or 5.477 3

(iii) a b = i j k2 4 53 2 6

= i (24 10) j (12 + 15) + k (-4 12) = 14i 27j 16k 4

(iv) 2 2 2 2 2 2

a b 32 32cosa b 45 492 4 ( 5) 3 ( 2) 6

= = =+ + + + = -0.68147

Hence, = arccos(-0.68147) = 132.96 4 Total: 14


51

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 k0 1 83 4 1

= i(-1 +32) j(-24) + k(3) = 31i + 24j + 3k Nm 3

Magnitude of M, M r F= = ( )( ) ( )2r r F F r F 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 = ( )( ) ( )265 26 12 1546 = = 39.32 Nm 3 Total: 6



52



Problem 1. Solve the quadratic equation 2x 2x 5 0 + = and show the roots on an Argand diagram. Marks

x = ( ) ( )( )

( )22 2 4 1 5 2 16 2 j4

2 1 2 2

= = = 1 j2 5

The two roots are shown on the Argand diagram in Figure 27.

Figure 27 4

Total: 9

Problem 2. If 1Z 2 j5= + , 2Z 1 j3= and 3Z 4 j= determine, in both Cartesian and polar forms,

the value of 1 2 31 2

Z Z ZZ Z

++ , correct to 2 decimal places.

Marks

( )( )( ) ( )

21 2

1 2

2 j5 1 j3Z Z 2 j6 j5 j 15 (17 j) (17 j) (3 j2)Z Z 2 j5 1 j3 (3 j2) (3 j2) (3 j2) (3 j2)

+ + = = = = + + + + + +

= 2

2 2

51 j34 j3 j 2 49 j37 3.77 j2.853 2 13

+ = = + 5

Hence, 1 2 31 2

Z Z ZZ Z

++ = (3.77 j2.85) + (4 j) = 7.77 j3.85 2

or 8.67 26.36 2 Total: 9


53

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, ( )1Z 2 j7= + ohms and ( )2Z 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 = 1 2

V 150 0 150 0 150 0Z Z (2 j7) (3 j4) 5 j3 34 30.96

= = =+ + + + = 25.72-30.96

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) [ ] 42.37 35 (b) [ ]53.2 j4.8 (c) [ ]1 j3 Marks

(a) [ ] 42.37 35 = ( )42.37 4 35 = 31.55140 = -24.17 + j20.28 4 (b) [ ]53.2 j4.8 = [ ] ( )5 55.769 56.31 5.769 5 56.31 = = 6390-281.55 = 639078.45 = 1279 + j6261 6

(c) [ ]1 j3 = 1 12 2 110 108.435 10 108.4352

=

= 1.778-54.22 and 1.778125.78 = (1.04 j1.44) 5 Total: 15


54

Problem 6. Determine 5 2 1 67 8 3 4

Marks

5 2 1 67 8 3 4 =

11 3831 74

4

Total: 4

Problem 7. Calculate the determinant of j3 (1 j2)

( 1 j4) j2+

Marks

( )( )2j3 (1 j2) j 6 1 j2 1 j4( 1 j4) j2

+ = + = 6 [-1 j4 j2 -2j 8 ]

= 6 [ 7 j6 ] = -1 + j6 4

Total: 4

Problem 8. Determine the inverse of 5 27 8

Marks

If B = 5 27 8 then

1 8 21B7 540 14

= = 4 1

8 21 13 13or26 7 5 7 5

26 26

4

Total: 4

Problem 9. Determine 1 3 0 2 1 34 9 2 5 1 05 7 1 4 6 2

Marks

1 3 0 2 1 34 9 2 5 1 05 7 1 4 6 2

=

17 4 361 25 1641 6 13

9

Total: 9


55

Problem 10. Calculate the determinant of 2 1 35 1 04 6 2

Marks

2 1 35 1 2 1

5 1 0 3 24 6 5 1

4 6 2

= + using the third column 3

= 3(30 4) + 2(2 5) = 3(26) + 2(-3) = 78 6 = 72 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 171 1 y 1

= 1

The inverse of 4 31 1

is 1 3 1 31 11 4 1 44 3 7

= 2

Hence 1 3 4 3 x 1 3 171 11 4 1 1 y 1 4 17 7

=

1 0 x 1410 1 y 217

=

and y 2y 3

= i.e. x = 2 and y = -3 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


56

Marks

4x + 9y + 2z 21 = 0

-8x + 6y 3z 41 = 0

3x + y 5z + 73 = 0

Hence x y z 19 2 21 4 2 21 4 9 21 4 9 26 3 41 8 3 41 8 6 41 8 6 31 5 73 3 5 73 3 1 73 3 1 5

= = =

4

x y z9( 424) 2(479) 21( 27) 4( 424) 2( 461) 21(49) 4(479) 9( 461) 21( 26)

= =

14( 27) 9(49) 2( 26)

= +

i.e. x y z 14207 1803 6611 601

= = = 3

Hence, x = 4207601

= -7 y = 1803601

= 3 and z = 6611601

= 11 3

(or use Cramers 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:

1 2 3

1 2 3

1 2 3

2.4 I 3.6 I 4.8I 1.23.9 I 1.3I 6.5I 2.6

1.7 I 11.9 I 8.5I 0

+ + = + =

+ + =

Using matrices, solve the equations for 1I , 2I and 3I .

Marks

1

2

3

2.4 3.6 4.8 I 1.23.9 1.3 6.5 I 2.61.7 11.9 8.5 I 0

=

The inverse of the 3 by 3 matrix is:

88.4 26.52 29.64

1 22.1 12.24 3.122.4(88.4) 3.6( 22.1) 4.8( 48.62)

48.62 22.44 17.16

+ 4


57

Hence 1

2

3

1 0 0 I 88.4 26.52 29.64 1.2 310 1 0 I 22.1 12.24 3.12 2.6 1

58.3440 0 1 I 48.62 22.44 17.16 0 2

= =

i.e. 1 2 3I 3, I 1 and I 2= = = 6 Total: 10



58



Problem 1. Differentiate the following with respect to the variable:

(a) 3 21y 5 2 xx

= + (b) 2s 4e sin 3= (c) 3ln 5tycos 2t

= (d) ( )22x

t 3t 5=

+

Marks

(a) 3 3/ 2 221y 5 2 x 5 2x xx

= + = +

dydx

= ( ) ( )1/ 2 330 2 x 2x2 + = 323 x x+ 3 (b) 2s 4e sin 3= i.e. a product

dsd = ( )( ) ( )( )2 24e 3cos3 sin 3 8e + = ( )24e 3cos 3 2sin 3 + 3

(c) 3ln 5tycos 2t

= i.e. a quotient

dydt

= ( ) ( )( )

( )23cos 2t 3ln 5t 2sin 2tt

cos 2t

= 23 cos 2t 6ln 5t sin 2tt

cos 2t

+ 3

(d) ( )22x

t 3t 5=

+ Let u = 2t 3t 5 + then du

dt = 2t 3

Hence, x = 1/ 22 2uu

= and 3/ 23

dx 1udu u

= =

dxdt

= ( )3

dx du 1 2t 3du dt u

= = ( )32

3 2t

t 3t 5

+

4

Total: 13

Problem 2. If f(x) = 22.5x 6x 2 + find the co-ordinates at the point at which the gradient is 1. Marks

f(x) = 22.5x 6x 2 + Gradient = f (x) = 5x 6 = -1 from which, 5x = 5 and x = 1 3

When x = 1, f (x) = f (1) = 22.5(1) 6(1) 2 + = -1.5 2006 John Bird. All rights reserved. Published by Elsevier.

59

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:

kts a e sin 2 ft= . Determine the velocity and acceleration of the end of the spring after 2 seconds if a = 3, k = 0.75 and f = 20.

Marks kts a e sin 2 ft= i.e. a product

Velocity = ( )( ) ( )( )kt ktds a e 2 f cos 2 ft sin 2 ft k a edt = + 2 When t = 2, a = 3, k = 0.75 and f = 20,

velocity = ( )( ) ( )( )(0.75)(2) (0.75)(2)3e 2 20cos 2 (20)(2) sin 2 (20)(2) (0.75)3e = ( )0.75 (2)120 e 0 = 84.12 cm/s 2

Acceleration = ( ) ( ) ( )( )2 2kt kt2d s a e 2 f sin 2 ft 2 f cos 2 ft k a edt = + ( )( ) ( )( )2 kt ktsin 2 ft k a e k a e 2 f cos 2 ft + + 3 When t = 2, a = 3, k = 0.75 and f = 20,

acceleration = ( )( ) ( )( )1.5 1.50 40 2.25e 0 2.25e 40 + + + = 1.5180 e = -126.2 2cm / s 3

Total: 10

Problem 4. Find the co-ordinates of the turning points on the curve 3 2y 3x 6x 3x 1= + + and distinguish between them.

Marks

Since 3 2y 3x 6x 3x 1= + + then 2dy 9x 12x 3 0dx

= + + = for a turning point

= (3x + 3)(3x + 1)

from which, x = -1 or x = 13

3

When x = -1, 3 2y 3( 1) 6( 1) 3( 1) 1= + + = -3 + 6 3 1 = -1


60

When x = 13

, 3 21 1 1y 3 6 3 1

3 3 3 = + + =

1 2 41 1 19 3 9

+ =

Hence, turning points occur at (-1, -1) and 1 4, 13 9

2 2

2

d ydx

= 18x + 12

When x = -1, 2

2

d ydx

is negative, hence, (-1, -1) is a maximum point. 1

When x = 13

, 2

2

d ydx

is positive, hence, 1 4, 13 9

is a minimum point. 1

Total: 7

Problem 5. The heat capacity C of a gas varies with absolute temperature as shown:

3 6 2C 26.50 7.20 10 1.20 10 = + Determine the maximum value of C and the temperature at which it occurs.

Marks

3 6dC 7.20 10 2.40 10 0d

= = for a maximum or minimum value,

from which, 3

6

7.20 102.40 10

= = 3000 2 2

62

d C 2.40 10d

= , which is negative, and hence = 3000 gives a maximum value.

maxC ( )( ) ( )( )23 626.50 7.20 10 3000 1.20 10 3000 = + = 26.50 + 21.6 10.8 = 37.3

Hence, the maximum value of C is 37.3, which occurs at a temperature of 3000. 3

Total: 5

Problem 6. Determine for the curve 2y 2x 3x= at the point (2, 2): (a) the equation of the tangent, (b) the equation of the normal.

Marks

(a) Gradient, m = dydx

= 4x - 3

At the point (2, 2), x = 2 and m = 4(2) 3 = 5 2


61

Hence, equation of tangent is: ( )1 1y y m x x = i.e. y 2 = 5(x 2)

i.e. y 2 = 5x - 10

or y = 5x 8 2

(b) Equation of normal is: ( )1 11y y x xm = i.e. ( )1y 2 x 2

5 =

i.e. 1 2y 2 x5 5

= + or 5y 10 = -x + 2

or 5y + x = 12 2

Total: 6

Problem 7. A rectangular block of metal with a square cross-section has a total surface area of 250

2cm . 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 = 22x 4xy 250+ = (1) Volume, V = 2x y

Figure 28

From equation (1), 24xy 250 2x= and y = 2250 2x

4x

Hence, 2

2 3250 2x 1V x 62.5x x4x 2

= = 3

and 2dV 362.5 x 0dx 2

= = for a maximum or minimum value,


62

i.e. 62.5 = 23 x2

from which, x = 2(62.5) 6.455cm3

= 2

2

d V 3xdx

= and when x = +6.455, 2

2

d Vdx

is negative, indicating a maximum value.

Hence, maximum volume = 62.5x - 31 x2

= 62.5(6.455) - ( )31 6.4552

= 3269cm 4

Total: 7

Problem 8. A cycloid has parametric equations given by: x = 5( - sin ) and y = 5(1 cos ).

Evaluate (a) dydx

(b) 2

2

d ydx

when = 1.5 radians. Give answers correct to 3 decimal places.

Marks

(a) x = 5( - sin ) hence, ( )dx 5 5cos 5 1 cosd

= = 1

y = 5(1 cos ) hence, dy 5sind

= 1

Thus, ( ) ( )dy

dy 5sin sinddxdx 5 1 cos 1 cosd

= = =

When = 1.5 radians, ( )dy sin1.5 0.99749499dx 1 cos1.5 1 0.07073720

= = = 1.073, correct to 3 2 decimal places

(b) ( )( ) ( )( )

( )( )

22

2

1 cos cos sin sind dy d sin1 cosd y d dx d 1 cos

dxdx 5 1 cos 5 1 cosd

= = =

2

= ( )( )( )( ) ( )

2 2

2 2 2

3 3

cos cos sincos cos sin1 cos cos 1

5 1 cos 5 1 cos 5 1 cos

+ = =

= ( )( ) ( )3 21 cos 1

5 1 cos 5 1 cos = 1


63

When = 1.5 radians, ( ) ( )2

22

d y 1 1dx 5 0.8635293485 1 cos1.5

= = = 0.232, correct to 3 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 , y = sin at = 3

Marks

(a) Equation of tangent is: ( )11 11

dyy y x xdx

=

At point , 1x 4cos= hence, 1dx 4sind = 1

1y sin= hence, 1dy cosd = 1

1

1

11

dydy cos 1d cotdxdx 4sin 4

d

= = =

1

Hence, the equation of the tangent is: ( )1y sin cot x 4cos4

=

At = 3 , 1y sin cot x 4cos

3 4 3 3 =

y 0.86603 = -0.14434(x 2) i.e. y 0.86603 = -0.14434x + 0.28868 i.e. y = 1.093 0.144x

Bird - Higher Engineering Mathematics - 5e - Instructor Manual.pdf

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