4 Math Questions

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Q1	Perform the two basic operations of multiplication and division to a complex number in both rectangular and polar form, to demonstrate the different techniques.	· Dividing complex numbers in rectangular and polar forms. · Converting complex numbers between polar and rectangular forms and vice versa. Save Time On Research and Writing Hire a Pro to Write You a 100% Plagiarism-Free Paper. Get My Paper
Q2	Calculate the mean, standard deviation and variance for a set of ungrouped data	· Completing a tabular approach to processing ungrouped data.
Q3	Calculate the mean, standard deviation and variance for a set of grouped data	· Completing a tabular approach to processing grouped data having selected an appropriate group size.
Q4	Sketch the graph of a sinusoidal trig function and use it to explain and describe amplitude, period and frequency.	· Calculate various features and coordinates of a waveform and sketch a plot accordingly. · Explain basic elements of a waveform.
Q5	Use two of the compound angle formulae and verify their results.	· Simplify trigonometric terms and calculate complete values using compound formulae.
Q6	Find the differential coefficient for three different functions to demonstrate the use of function of a function and the product and quotient rules	· Use the chain, product and quotient rule to solve given differentiation tasks.
Q7	Use integral calculus to solve two simple engineering problems involving the definite and indefinite integral.	· Complete 3 tasks; one to practise integration with no definite integrals, the second to use definite integrals, the third to plot a graph and identify the area that relates to the definite integrals with a calculated answer for the area within such.
Q8	Use the laws of logarithms to reduce an engineering law of the type y = axn to a straight line form, then using logarithmic graph paper, plot the graph and obtain the values for the constants a and n.					· See Task.
Q9	Use complex numbers to solve a parallel arrangement of impedances giving the answer in both Cartesian and polar form
Q10	Use differential calculus to find the maximum/minimum for an engineering problem.
Q11	Using a graphical technique determine the single wave resulting from a combination of two waves of the same frequency and then verify the result using trig formulae.
Q12	Use numerical integration and integral calculus to analyse the results of a complex engineering problem

Level of Detail in Solutions: Need to show work leading to final answer

Need

Question 1

(a) Find:

(4 + i2)

(1 + i3)

Use the rules for multiplication and division of complex numbers in rectangular form.

(b) Convert the answer in rectangular form to polar form

(c) Repeat Q1a by first converting the complex numbers to polar form and then using the rules for multiplication and division of complex numbers in polar form.

(d) Convert the answer in polar form to rectangular form.

Question 2

The following data within the working area consists of measurements of resistor values from a production line.

For the sample as a set of ungrouped data, calculate (using at least 2 decimal places):

1. arithmetic mean

1. standard deviation

1. variance

The following data consists of measurements of resistor values from a production line:

51.4

54.1

53.7

55.4

53.1

53.5

54.0

56.0

53.0

55.3

55.0

52.8

55.9

52.8

50.5

54.2

56.2

55.6

52.7

56.1

52.1

54.2

50.2

54.7

56.2

55.6

52.7

52.1

56.1

54.2

50.2

54.7

55.1

54.8

56.5

55.8

55.3

54.5

57.0

56.0

53.9

57.3

55.3

54.4

49.6

54.1

51.6

53.2

54.6

56.4

53.9

50.9

54.0

51.8

56.1

53.2

54.6

56.4

53.9

50.9

54.0

51.8

56.1

‘n’

No.

Value (X)

Mean

(X – Mean)

(X-Mean)2

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

55

56

57

58

59

60

61

62

63

= Total X

Mean = Total X / ‘n’(max) =

‘n’ max – 1 (Y) =

Total (X-Mean)2 ……. let’s call this (Z)

Variance

Variance = (Total(X-Mean)2) / (‘n’ max -1)

= Z / Y =

Standard Deviation

Standard Deviation = square-root of Variance

=

Question 3

Using the data in Q2, we now need to arrange the data into groups so that we can “tally” the data accordingly.

No.

Data Range

Gap

Freq. (F)

Mid-Point (X)

(F x X)

Gap x F

(Bar Area)

(X – Mean)

(X-Mean)2

(X-Mean)2 x F

1

2

3

4

5

6

7

8

9

10

11

Total F…Let’s call this (Y)

Total (F x X)

Mean = (Total (F x X)) / (Y)

Total (X-Mean)2 x F ……. let’s call this (Z)

Variance

Variance = (Total(X-Mean)2 x f) / (Total F)

= Z / Y

Standard Deviation

Standard Deviation = square-root of Variance

=

Question 4

1. For a sinusoidal trigonometric function, explain what is meant by:

· amplitude

· periodic time

· frequency

Use diagrams or paragraphs if you like.

(b) An alternating current voltage is given by:

V = 310 Sin (285t + 0.65)

1. sketch the waveform, marking on all main values

1. state whether the waveform is leading or lagging

1. state the phase angle in degrees

1. state the amplitude

1. calculate the periodic time

1. calculate the frequency

Note – for an accurate plot of the waveform you will need to carry out the following steps:

Using the values in the equation work out the Periodic Time. Divide this into 4 quarters:

0 x t

0.25 x t

0.5 x t

0.75 x t

1 x t

Use these values to work out the amplitude for each value of t. Then sketch the plot.

Question 5

Using compound angle formulae, simplify:

(a) Sin (θ – 90o)

(b) Cos (θ + 270o)

In each case, verify your answer by substituting θ = 30o

Question 6

Differentiate:

1. y = (3×2 – 2x)7

1. y = 6×3 .sin4x

1. y = 5 e6x

x – 8

Question 7

(a) Integrate the following:

(i) (4Cos 3θ+ Sin 6θ) dθ

(ii) (2 + Cos 0.83θ) dθ

(b) Evaluate:

(i)

(ii)

(c)
(i) Plot the curve y = 3×2 + 6 between x = 1 and 4
(ii) Find the area under the curve between x = 1 and 4 using integral calculus

Question 8
Create a document titled Laws of Logarithms
Use the laws of logarithms to reduce an engineering law of the type y = axn to a straight line form, then using logarithmic graph paper, plot the graph and obtain the values for the constants a and n.
The following set of results was obtained during an experiment:
X: 2 2.5 3 3.5 4
Y: 8 6.4 5.3 4.6 4
The relationship between the two quantities is of the form y = axb
Complete the law:
(i) using the laws of logarithms to reduce the law to a straight line form and determine the gradient and intercept.
(ii) graphically, using logarithmic graph paper

Question 9
Use complex numbers to solve a parallel arrangement of impedances giving the answer in both Cartesian and polar form
Two impedances Z1 and Z2 are given by the complex numbers:
Z1 = 2 + j10
Z2 = j14
Find the equivalent impedance Z if:
(i) Z = Z1 + Z2 when Z1 and Z2 are in series
(ii) 1 = 1 + 1 when Z1 and Z2 are in parallel
Z Z1 Z2

Question 10
Use complex numbers to solve a parallel arrangement of impedances giving the answer in both Cartesian and polar form Use differential calculus to find the maximum/minimum for an engineering problem.
A sheet of metal is 340 mm x 225 mm and has four equal squares cut out at the corners so that the sides and edges can be turned up to form an open topped box shape.
Calculate:
(i) The lengths of the sides of the cut out squares for the volume of the box to be as big as possible.
(ii) The maximum volume of the box. Z Z1 Z2

Question 11
Using a graphical technique determine the single wave resulting from a combination of two waves of the same frequency and then verify the result using trig formulae.
A sheet of metal is 340 mm x 225 mm and has four equal squares cut out at the corners so that the sides and edges can be turned up to form an open topped box shape.
Calculate:
1. Using a vector or phasor diagram add the following waveforms together and present your answer in the form Vr = V sin(ωt+α).
Show all working including phasor diagrams.
Add the following together:
V1=10sin(ωt)

V2=20sin(ωt+)
1. Verify your result by using trigonometric formulae.

Question 12
In calculating the capacity of absorption towers in Chemical Engineering it is necessary to evaluate certain definite integrals by approximate methods. Evaluate the following definite integrals by use of Simpson’s rule and the trapezoid rule using the given data (xi is an empirical function of x, yi is an empirical function of y):

1
)
x
–
x
(
=

s
Deviation

dard
tan
S
1
–
n
)
x
–
x
(
s

Variance
2
2
2
–
å
å
=
n
1
)
x
–
x
(
=

s
Deviation

dard
tan
S
1
–
n
)
x
–
x
(
s

Variance
2
2
2
–
å
å
=
n
Variance s
x
–
x
)
n
S
dard Devia
tion s
=
x
–
x
)
2
2
2
=
å
å
(
tan
(
n
(
)
ò
+
+
4
1
2
7
6
2
dx
x
x
(
)
dx
e
x
ò
6
2
125
.
0
4
2
p

Que

s

tio

n

1

.

a

)

(

i)

(

4

+i

2

).(1+i

3

)

4(1+i3)+i2(1+i3)

4+i

12

+i2

–

6

=

-2+i

14

(ii)

(4+i2)/(1+i3)

numerator

4(1-i3)+i2(1-i3)

4-i12+i2-(i)

26

4-i

1

0

+6

10-i10

Denominator

1(1-i3)+i3(1-i3)

1-i3+i3-

9

(i)2

1+9=10

Quotient=

=1-i

b)

r=/z/=√(-2)2+1

42

=√

20

0

Argument of z

Tanθ=b/a

=14/-2

=-

7

Θ=

tan

-1(-7)

=-

8

1.87ᵒ×πᶜ/

18

0ᵒ

=πᶜ….4th quadrant

ϴ1=- πᶜ+πᶜ

ϴ2= πᶜ….2nd quadrant(principal arg)

Therefore,arg z= πᶜ+2πᶜn where n=intergers

Z==√200[cos ( πᶜ)+isin( πᶜ)]

c)

(4+i2) to polar form

R=/z/=√42+

22

=

√20

Arg z1

Tanθ1=2/4

Θ=tan-

11

/2

=26.6ᵒ×πᶜ/180ᵒ

=πᶜ

….1st quadrant

Arg z1= πᶜ+2πᶜn…..where n =interger

Z1=√20[cos(πᶜ)+isin(πᶜ)]

(1+i3) to polar form

R=/z/=√12+

32

=√10

Argz2

Tanθ2=3/1

Θ=tan-

13

=71.

5

6

ᵒ×πᶜ/180ᵒ

=πᶜ

Therefore z2=πᶜ+2πᶜn….n=integers

Z2=√10[cos(πᶜ)+isin(πᶜ)]

(i)

Product in polar representation

Z1.z2=√20[cos(πᶜ)+isin(πᶜ)].√10[cos(πᶜ)+isin(πᶜ)]

=√20.√10[cos(πᶜ+πᶜ)+isin(πᶜ+πᶜ)]

=

10√2[cos(πᶜ)+isin(πᶜ)]

(ii)

Quotient in polar representation

Z1/z2=[cos(πᶜ – πᶜ)+isin(πᶜ – πᶜ)]

= [cos(πᶜ )+isin((πᶜ)]

=

√2[cos(πᶜ )-isin((πᶜ)]

d(i)

10√2[cos(πᶜ)+isin(πᶜ)]

=10√2[-0.1

39

+i

0.9

9]

=-1.968+i14

(ii)

√2[cos(πᶜ )-isin((πᶜ)]

=√2[0.803-i0.7

19

]

=1.1

35

-i1.0

17

Qn 9

Z1=2+j10

Z2=j14

(i)

z=z1+z2 in series

=2+j10+j14

=(2+0)+j(10+14)

=2+j

24

(ii)

Z=z1+z2 in parallel

zE=

z1.z2=(2+j10).(0+ j14)

=2(0+j14)+j10(0+j14)

=0+j

28

+0+1

40

(j)2

=j28-140

=-140+j28

zE=×

numerator

=-140(2-j24)+j28(2-j24)

=-280+j

33

60

+j

56

-672(j)2

=-280+j

34

16

+672

=392+j3

41

6

Denominator

zE=2(2-j24)+j24(2-j24)

=4-j

48

+j48-

57

6j2

=4+576

=

58

0

zE=

=

=

0.6

76+j5.890

Question 2

The following data within the working area consists of measurements of resistor values from a production line.

For the sample as a set of ungrouped data, calculate (using at least 2 decimal places):

1. arithmetic mean

1. stan

dard

deviation

1. variance

The following data consists of measurements of resistor values from a production line:

y

>

52.8

54.2

56.2

55.6

52.7

52.1

56.1

54.2

50.2

54.7

55.3

56.0

55.3

54.1

53.9

54.0

56.1

53.2

54.6

56.4

53.9

50.9

54.0

51.8

56.1

	51 .4				54 .1		53 .7		55 .4		53.1		5 3.5			54.0
	56.0	53.0						5 5.3	55.0				52 .8		55.9
	50 .5						54.2				56.2				55.6				52.7							56.1				52.1
			5 0.2				54.7
	55.1		54.8		56.5
	55.8		54.5	57.0						53.9		57.3
	54.4		49 .6		51.6				53.2				5 4.6
			5 6.4				50.9				51.8

51.4

58.0413

50.5

58.0413

54.2

58.0413

56.1

58.0413

55.8

58.0413

55.3

58.0413

56.4

58.0413

54.6

58.0413

0.115873016

0.013426556

54.1

58.0413

58.0413

0.115873016

0.013426556

54.2

58.0413

1.2

15873016

1.478347191

50.2

58.0413

54.2

58.0413

55.3

58.0413

2.315873016

5.3

63

267826

54.4

58.0413

53.9

58.0413

0.115873016

0.013426556

56.4

58.0413

2.015873016

4.063744016

53.7

58.0413

1.715873016

2.944220207

‘n’	No.	Value (X)	Mean		(X – Mean)		(X-Mean)2
										1																																																															58.0413				0.1 15 87 30 16				0.013426556
					2				56				2.015873 016				4.06 37 44 016
					3		1.715873016		2.944220207
				4			1. 21 5873016			1. 47 8347191
						5		2. 31 5873016		5. 36 3267826
		6		0.515873016		0.26 61 24969
	7		0.015873016		0.000 25 1953
		8	-1.084126984	1.175331318
			9
	10		-3.884126984		15.086442 43
			11			53
12
13	0.315873016	0.0997757 62
		14			-0.184126984			0.03390 27 46
		15
16	-0. 38 4126984	0.14755354
17
18
19

55.3

58.0413

1.215873016

1.478347191

56.2

58.0413

54.7

58.0413

50.2

58.0413

-3.884126984

15.08644243

54.5

58.0413

49.6

58.0413

50.9

58.0413

53.9

58.0413

-0.184126984

0.033902746

55.4

58.0413

58.0413

55.6

58.0413

56.2

58.0413

2.115873016

4.476918619

54.7

58.0413

0.615873016

0.379299572

58.0413

54.1

58.0413

0.015873016

0.000251953

58.0413

50.9

58.0413

-3.184126984

10.13866465

53.1

58.0413

52.8

58.0413

52.7

58.0413

55.6

58.0413

1.515873016

2.297871

20
21		2.115873016		4.476918619
22		0.615873016		0.379 29 9572
23
24	0.415873016	0.172950365
25	-4.484126984	20.10739481
26		-3.184126984		10.13866465
27
28	1.315873016	1.731521794
29			55	0.915873016	0.838823381
30		1.515873016		2.297871
31
32
33			57	2.915873016	8.5023154 45
34
35					54			-0.084126984			0.007077349
36
37	-0.984126984	0.96850 59 21
38		-1.284126984		1.648982111
39		-1.384126984		1.915807508
40

41

55.1

58.0413

1.015873016

1.031997984

42

56

58.0413

1.915873016

3.670569413

43

51.6

58.0413

-2.484126984

6.170886873

44

51.8

58.0413

-2.284126984

5.21723608

45

54

58.0413

-0.084126984

0.007077349
46

53.5

58.0413

-0.584126984

0.341204334

47

55.9

58.0413

1.815873016

3.29739481

48

56.1

58.0413

2.015873016

4.063744

016
49

52.7

58.0413

-1.384126984

1.915807508
50

54.8

58.0413

0.715873016

0.512474175

51

53.9

58.0413

-0.184126984

0.033902746
52

53.2

58.0413

-0.884126984

0.781680524

53

56.1

58.0413

2.015873016

4.063744016
54

51.8

58.0413

-2.284126984

5.21723608
55

54

58.0413

-0.084126984

0.007077349
56

52.8

58.0413

-1.284126984

1.648982111
57

52.1

58.0413

-1.984126984

3.936759889

58

52.1

58.0413

-1.984126984

3.936759889
59

56.5

58.0413

2.415873016

5.836442429

60

57.3

58.0413

3.215873016

10.34183925

61

54.6

58.0413

0.515873016

0.266124969

62

53.2

58.0413

-0.88413

0.781681

63

56.1

58.0413

2.015873 4.063744

3407.3

= Total X

58.0413

54.95645

215.4841

Mean = Total X / ‘n’(ma

x

) =

‘n’ max – 1 (Y) =

Total (X-Mean)2 ……. let’s call this (Z)

Variance

Variance = (Total(X-Mean)2) / (‘n’ max -1)

= Z / Y =

3.920998

S

tan

dard Devia

tion

Standard

Deviation

= square-root of Variance

=

1.980151

Question 3

Using the data in Q2, we now need to arrange the data into groups so that we can “tally” the data accordingly.

No.

(X – Mean)

(X-Mean)2

1

1

5

50

5

2

1

3

51

3

3

1

5

52

5

4

1

9

53

9

5

1

14

54

14

6

1

11

55

11

7

1

15

56

15

8

1

2

57

2

9

10

11

63

Data Range	Gap	Freq. (F)	Mid-Point (X)	(F x X)	Gap x F (Bar Area)	(X-Mean)2 x F
49.5 – 50.5	250	-4.84127	23.4379	117.1895
50.5 – 51.5	153	-3.84127	14.75536	44.26607
51.5 – 5 2.5	260	-2.84127	8.072815	40.36408
52.5 – 53.5	477	-1.84127	3.390275	30.51248
53.5 – 54.5	756	-0.84127	0.707735	9.908293
54.5 – 55.5	605	0.15873	0.025195	0.277147
55.5 – 56.5	840	1.15873	1.342655	20.13983
56.5 – 57.5	114	2.15873	4.660115	9.32023
Total F…Let’s call this (Y)	3455	54.84127	271.97759
Total (F x X)
Mean = (Total (F x X)) / (Y)
Total (X-Mean)2 x F ……. let’s call this (Z)

	Variance	Variance = (Total(X-Mean)2 x f) / (Total F) = Z / Y 4.95936
	Standard Deviation	Standard Deviation = square-root of Variance =2.2269

Question 4

i)

Amplitude is the height of the wave

Periodic time is the time taken to complete one complete oscillation

Frequency refers to the number of oscillations made per second of time

In the above, for time T, the wave oscillates once, hence has a frequency of one.

V =

310

Sin (285t + 0.65)

X

y

0

0.2		0
0.6	-310
0.9
1.2	310

ii)The wave form is leading

iii) The phase angle in degrees =0.65

iv)The amplitude is 310

v)Periodic time T is

wt=2π0

t=2850/(360*πc)

=1.58πr0

T =2αc /1.581 αc

=1.262 seconds

vi)Frequency

f=1/t

=1/1.263

=0.792hz

With t=1.263

Divide by 4 threfore we have

t=0,A=310

t=0.3, A=0

t=0.6, A=-310

t=0.9,A=0

t=1.2,A=310

hence the graph

Question 6

Differentiate:

1. y = (3×2 – 2x)7 ans:7(3×2-2x) 6(6x-2)

1. y = 6×3 .sin4x ans:18x2sin4x+24x3cos4x

(C)
y=5e^6x/x-8

Let u=5e^6x and t=6x

dt/dx=6

du/dt=5e^t

.: du/dx=du/dt*dt/dx

=5e^t.6

Let v=x-8

Dv/dx=1

By quotient rule

Dy/dx=(Vdu/dx-Udv/dx)/V^2

={(x-8).30e^6x – 5e^6x.1 }/(x-8)(x-8)

/ (x-8)^2

Question 7

(a)

(i) ∫ (4cos3ϴ+sin6 ϴ)dϴ

=4∫cos3ϴdϴ+∫sin6ϴdϴ

For 4∫cos3ϴdϴ

Let u-3ϴ

1/3du=dϴ

Substituting for dϴ

4∫cos u.1/3du

=4/3(sin u)+c….eqn(i)

For ∫sin6ϴdϴ

Let u-6ϴ

1/6du=dϴ

Substituting or dϴ

∫sin u.1/6du

=1/6∫(sin u)du

=1/6(-cos u)+c

=-1/6cos 6ϴ +c ….eqn(ii)

Adding eqn (i) and (ii)

=

(ii) ∫(2+cos0.83)dϴ

=2∫dϴ+∫cos0.83ϴ dϴ

For 2∫dϴ

=2ϴ+c…eqn(i)

For ∫cos0.83ϴ dϴ

=1/0.83sin0.83+c…eqn(ii)

Adding eqn (i) and (ii)

C(I)
y=3x^2+6

Values of x and y

When x=0 ; y=3(0)^2+6 =6

x=1 ; y=3(1)^2+6 =9

x=2; y=3(2)^2+6 =18

x=3 ; y=3(3)^2+6 =33

x=4 ; y=3(4)^2+6 =54

NB Using these values draw the required graph pliz

(ii)

A =∫

=

=3[1/3 +c

=3[64/3-1/3]

=63….(i)

6

=6[x+c]14

=6[4-1]

=18…eqn(ii)

Adding (i) and (ii)

=63+18= 81 sq units

Question 8

y=axn

Introducing logs

log y= log(axn )

loga+logxn

loga+nlogx this is the straight line form

Taking a tablex:a,0,1,2,…

Since y intercept=0,the equation becomes:

logy=nlogx+0

log1=nlog1

-0.3=0.3n

n=-1

0.9=loga+0.3(-1)

loga=1.2

gradient= n=-1

intersept= loga=1.2

sketching the graph:

1

2

8

2

6

3

3

4

5

4

4

n	x	y	n logx	antilog
0.301	1.9999
2.5	6.4	0.7959
5.3	1.4314	27.00023
3.5	4.6	2.1763	150.072
3.0103	1024.0001

Qn 9

Z1=2+j10
Z2=j14
(i)
z=z1+z2 in series
=2+j10+j14
=(2+0)+j(10+14)
=2+j24

(ii)
Z=z1+z2 in parallel
zE=
z1.z2=(2+j10).(0+ j14)
=2(0+j14)+j10(0+j14)
=0+j28+0+140(j)2
=j28-140
=-140+j28
zE=×
numerator
=-140(2-j24)+j28(2-j24)
=-280+j3360+j56-672(j)2
=-280+j3416+672
=392+j3416
Denominator
zE=2(2-j24)+j24(2-j24)
=4-j48+j48-576j2
=4+576
=580
zE=
=
=0.676+j5.890
0.2 0.60000000000000064 0.9 1.2 0 -310 0 310 0 0.30000000000000032 0.60000000000000064 0.9 1.2 1.5 310 0 -310 0 310 0 Y-Values 0.70000000000000062 1.8 2.6 2.7 3.2 0.8 1
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