![Question 1.
a) (i)
(4+i2).(1+i3)
4(1+i3)+i2(1+i3)
4+i12+i2-6
=-2+i14
(ii)
(4+i2)/(1+i3)
numerator
4(1-i3)+i2(1-i3)
4-i12+i2-(i)26
4-i10+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+142
=√200
Argument of z
Tanθ=b/a
=14/-2
=-7
Θ=tan-1(-7)
=-81.87ᵒ×πᶜ/180ᵒ
=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)]](https://image.slidesharecdn.com/question1-221031120244-8458e989/85/question-1-a-i4i21i341i3i21i34i12i26docx-1-320.jpg)
![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
=71.56ᵒ×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[-0.139+i0.99]
=-1.968+i14](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Recommended
Recommended
More Related Content
Similar to Question 1. a) (i)(4+i2).(1+i3)4(1+i3)+i2(1+i3)4+i12+i2-6.docx
Similar to Question 1. a) (i)(4+i2).(1+i3)4(1+i3)+i2(1+i3)4+i12+i2-6.docx (8)
More from IRESH3
More from IRESH3 (20)
Recently uploaded
Recently uploaded (20)
Question 1. a) (i)(4+i2).(1+i3)4(1+i3)+i2(1+i3)4+i12+i2-6.docx
- 1. Question 1. a) (i) (4+i2).(1+i3) 4(1+i3)+i2(1+i3) 4+i12+i2-6 =-2+i14 (ii) (4+i2)/(1+i3) numerator 4(1-i3)+i2(1-i3) 4-i12+i2-(i)26 4-i10+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+142 =√200 Argument of z Tanθ=b/a =14/-2 =-7 Θ=tan-1(-7) =-81.87ᵒ×πᶜ/180ᵒ =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)]
- 2. 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 =71.56ᵒ×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[-0.139+i0.99] =-1.968+i14
- 3. (ii) √2[cos(πᶜ )-isin((πᶜ)] =√2[0.803-i0.719] =1.135-i1.017 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 Question 2
- 4. 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
- 6. ‘n’ No. Value (X) Mean (X – Mean) (X-Mean)2 1 51.4 58.0413 0.115873016 0.013426556 2 56 58.0413 2.015873016 4.063744016 3 50.5 58.0413 1.715873016 2.944220207 4 54.2 58.0413
- 19. 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 = 3.920998 Standard Deviation 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.
- 20. 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 49.5 – 50.5 1 5 50 250 5 -4.84127 23.4379 117.1895 2 50.5 – 51.5 1 3 51 153 3 -3.84127 14.75536 44.26607 3 51.5 – 52.5 1 5 52 260
- 21. 5 -2.84127 8.072815 40.36408 4 52.5 – 53.5 1 9 53 477 9 -1.84127 3.390275 30.51248 5 53.5 – 54.5 1 14 54 756 14 -0.84127 0.707735 9.908293 6 54.5 – 55.5 1 11 55 605 11 0.15873 0.025195 0.277147 7 55.5 – 56.5
- 22. 1 15 56 840 15 1.15873 1.342655 20.13983 8 56.5 – 57.5 1 2 57 114 2 2.15873 4.660115 9.32023 9 10
- 23. 11 Total F…Let’s call this (Y) 63 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
- 24. 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 0.2 0 0.6 -310 0.9 0 1.2
- 25. 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 = (3x2 – 2x)7ans:7(3x2-2x) 6(6x-2) 1. y = 6x3 .sin4xans:18x2sin4x+24x3cos4x (C) y=5e^6x/x-8 Let u=5e^6x and t=6x dt/dx=6
- 26. 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)
- 27. 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)
- 28. loga=1.2 gradient= n=-1 intersept= loga=1.2 sketching the graph: n x y n logx antilog 1 2 8 0.301 1.9999 2 2.5 6.4 0.7959 6 3 3 5.3 1.4314 27.00023 4 3.5 4.6 2.1763 150.072 5 4 4 3.0103 1024.0001
- 29. 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 0.20.600000000000000640.91.20- 310031000.300000000000000320.600000000000000640.91.21.5
- 32. n S dard Devia tion s = x - x ) 2 2 2 = å å ( tan ( n 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. 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.
- 33. 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 · See Task. Q10 Use differential calculus to find the maximum/minimum for an
- 34. engineering problem. · See Task. 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. · See Task. Q12 Use numerical integration and integral calculus to analyse the results of a complex engineering problem · See Task. Level of Detail in Solution s: 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
- 35. 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
- 38. 50.9 54.0 51.8 56.1 ‘n’ No. Value (X) Mean (X – Mean) (X-Mean)2 1 2
- 39. 3 4 5
- 40. 6 7 8 9
- 41. 10 11 12
- 42. 13 14 15 16
- 43. 17 18 19
- 44. 20 21 22
- 45. 23 24 25 26
- 46. 27 28 29
- 47. 30 31 32
- 48. 33 34 35 36
- 49. 37 38 39
- 50. 40 41 42
- 51. 43 44 45 46
- 52. 47 48 49
- 53. 50 51 52 53
- 54. 54 55 56
- 55. 57 58 59
- 56. 60 61 62 63
- 57. = Total X Mean = Total X / ‘n’(max) = ‘n’ max – 1 (Y) = Total (X-Mean)2 ……. let’s call this (Z) Variance
- 58. 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)
- 60. 4 5
- 61. 6 7
- 62. 8 9 10
- 63. 11 Total F…Let’s call this (Y) Total (F x X)
- 64. 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
- 65. = 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:
- 66. 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 = (3x2 – 2x)7
- 67. 1. y = 6x3 .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 = 3x2 + 6 between x = 1 and 4 (ii) Find the area under the curve between x = 1 and 4
- 68. 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
- 69. 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
- 70. 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.
- 71. 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 -
- 74. Variance s x - x ) n S dard Devia tion s = x - x ) 2 2 2 = å å ( tan ( n (
- 76. 2 p