Engineering Maths Assignment on Statistics and Calculus

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Added on 2022/08/18

AI Summary

This Engineering Maths assignment, completed by a student, addresses several key concepts in engineering mathematics. Task 1 involves collecting and presenting data on employee distribution within a company using pie and bar charts, followed by the creation of frequency distributions, histograms, and cumulative frequency curves from provided data. Task 2 focuses on calculating mean, median, and mode for given sales data and grouped frequency data. Task 3 covers differentiation and integration of various functions. Task 4 involves solving quadratic equations using both factorization and the quadratic formula. Task 5 delves into an RC circuit analysis, calculating the time constant, plotting charging voltage, and determining voltage gradients. Finally, Task 6 analyzes power and displacement functions using calculus, including graphing, calculating energy, and determining displacement over a specified time interval. The assignment demonstrates the application of statistical methods and calculus in solving various engineering problems.

Running head: ENGINEERING MATHS
ENGINEERING MATHS
Name of the Student
Name of the University
Author Note

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1ENGINEERING MATHS
Task 1:
a) The chosen company is the popular MNC Google, the pioneer of search engine. Now, in
the Google headquarter, United States the total number of employees are 2365 by the year
2018. The breakdown of employees by different departments are given below.
Department name Members
Engineering 764
Research and
development
387
Infrastructure
development
71
Storage analytics group 1143
Total 2365
Pie chart:
32%
16%
3%
48%
Pie chart
Engineering
Research and development
Infrastructure developmemt
Storage analytics group
Bar chart:

2ENGINEERING MATHS
Engineering Research and
development Infrastructure
developmemt Storage analytics
group
0
200
400
600
800
1000
1200
1400
Bar chart
Department name
Members
b)
1.
Frequency distribution of data:
Classes frequency
Upto 66 5
more than 66 to 92 15.00
more than 92 to
118
11.00
more than 119 to
144
8.00
more than 145 to
170
1.00
Total 40

3ENGINEERING MATHS
2.
Histogram:
Upto 66 more than 66 to 92 more than 92 to
118 more than 119 to
144 more than 145 to
170
0
2
4
6
8
10
12
14
16
Histogram
Classes of Money earned(to nearest £)
frequency
3.
Cumulative frequency distribution less than type:
Classes frequency Cumulative
frequency
Upto 66 5 5
more than 66 to
92
15.00 20.00
more than 92 to
118
11.00 31.00

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4ENGINEERING MATHS
more than 119 to
144
8.00 39.00
more than 145 to
170
1.00 40.00
4.
66 86 106 126 146 166
0
5
10
15
20
25
30
35
40
45
Ogive
Classes
Cumulative frequency
c)
The frequency distribution of the 30 masses of ingots in 5 classes are given below.
1.
Class Frequency
less than 7.4 2
more than 7.4 to
7.8
8

5ENGINEERING MATHS
more than 7.8 to
8.2
8
more than 8.2 to
8.6
8
more than 8.6 4
Total 30
2.
Histogram:
less than 7.4 more than 7.4 to
7.8 more than 7.8 to
8.2 more than 8.2 to
8.6 more than 8.6
0
1
2
3
4
5
6
7
8
9
Frequency
Class
Frequency
3.
Cumulative frequency distribution:
Class Frequency Cumulative
frequency
less than 7.4 2 2

6ENGINEERING MATHS
more than 7.4 to
7.8
8 10
more than 7.8 to
8.2
8 18
more than 8.2 to
8.6
8 26
more than 8.6 4 30
Total 30
4.
Ogive:
less than 7.4 more than 7.4 to
7.8
more than 7.8 to
8.2
more than 8.2 to
8.6
more than 8.6
0
5
10
15
20
25
30
35
Less than Ogive
Class
Cumulative frequency
Task 2:

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7ENGINEERING MATHS
a)
Given sales data is
Month Sales
Jan 400
Feb 500
Mar 500
Apr 600
May 700
Jun 800
mean 583.3333
median 550
mode 500
Mean = (400+500 + 500 + 600 + 700 + 800)/6 = 583.3333
Median is the middle most value when assigned in order. In this case median is the mean of
3rd and 4th as number of values are even. Hence median = (500 + 600)/2 = 550
Mode is the value with highest number of occurrence which 500 which occurs 2 times in the
data.
b)
Gr
ou
ps:
Frequen
cy(fi)
Mid
val(xi)
Cum
ulati
ve
freq
xi*fi

8ENGINEERING MATHS
uen
cy
20.
5–
20.
9
6 20.7 6 6*20.7 = 124.2
21.
0–
21.
4
8 21.2 14 8*21.2=169.6
21.
5–
21.
9
12 21.7 26 12*21.7=260.4
22.
0–
22.
4
8 22.2 34 8*22.2=177.6
22.
5–
22.
9
4 22.7 38 4*22.7=90.8
Tot
al
6+8+12+
8+4=38
124.2+169.6+260.
4+177.6+90.8=82
2.6
M
ea
n
sum(xi*
fi)/
sum(fi)
822.6/3
8=21.64
737
M
ed
ia
n
Mean
of 19th
and
20th
value
21.7
M
od
e
Value
with
highest
frequen
21.7

9ENGINEERING MATHS
cy
When assigned in order then the values gives like the following.
20.7
20.7
20.7
20.7
20.7
20.7
21.2
21.2
21.2
21.2
21.2
21.2
21.2
21.2
21.7
21.7
21.7
21.7
21.7
21.7
21.7
21.7
21.7
21.7
21.7
21.7
22.2
22.2
22.2
22.2
22.2
22.2
22.2
22.2
22.7
22.7
22.7
22.7

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10ENGINEERING MATHS
Hence, from the list median = Mean of 19th and 20th value = 21.7
Mode = 21.7 as this value has the highest frequency of occurrence of 12 times.
Task 3:
i) y=2 x4 +3 x−2+9
dy
dx =8 x3−6 x−3
ii) y=5 e3 x – 8 e−2 x
dy/dx = 5∗3 e3 x – 8∗(−2)e−2 x = 15 e3 x +16 e−2 x
iii) y = 6sin(3x) – cos(2x)
dy/dx = 18cos(3x) + 2sin(2x)
b)
i) y = ∫ ( 4 x2+ x−2+4 ) dx=( 4
3 ) x3− 1
x +4 x +c (where c is the constant of integration)
ii) y = ∫ ( 5 sin ( 3 x ) +7 cos 4 x ) dx = (−5
3 )cos ( 3 x ) + ( 7
4 )sin ( 4 x ) +c (where c is the constant of
integration)
Task 4:
a)
x^2 + 5x -14 = 0
 x^2 +7x – 2x – 14=0
 x(x+7) – 2(x+7) = 0

11ENGINEERING MATHS
 (x+7)(x-2) = 0
 x = -7 or 2
b) 2x^2 +2x – 15 = 0
x = −2 ± √4−4∗2∗(−15 )
2∗2 = −2 ± √120
4 = (-2± 10.954)/4 = -3.2385 or 2.2385.
Task 5:
C = 80 nF = 80*10-9 F
R = 40 kΩ = 40*10^3 Ω
V = 5 V
i) time constant T = RC = 40*10^3 *80*10-9
= 0.0032 Ohm-Farad or seconds.
ii)
Charging voltage plot:

12ENGINEERING MATHS
0 0.005 0.01 0.015 0.02 0.025
0
1
2
3
4
5
6
voltage(in volts)
time in secs
Voltage in volts
iii)
The rate of change of voltage at t = 6msec = 0.006 sec is
(dV/dt) | (t=0.006) = d
dt (V (1−e
−t
T )) = V ( e
−t
T
T ) = 5 ( e
−0.006
0.0032
0.0032 ) = 239.617 volts/sec = 0.2396
volts/msec
iv) Now, from the plot when t = 6 msec or 0.006 sec then V = 4.233 volts
In the next instant at t = 7 msec or 0.007 sec V = 4.59 volts
Hence, gradient at t = 6 msec is
(4.59-4.233)/(7-6) = 0.20579 volts/msec.
v) It can be seen that the gradient at t = 0.006 sec or 6 msec from the curve is sufficiently
close to the actual gradient and the error approaches zero as the interval size decreases.
vi) Given, v=V ∗e
−t
RC

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13ENGINEERING MATHS
dv/dt = −V
RC e
−t
RC
dv
dt ∨t =T = −V
RC e
−T
RC = −5/0.0032 e(−1)=−574.8116 volts / sec
Task 6:
Power P = 120t^(1.8) + 8t
i)
Graph against time:
0 2 4 6 8 10 12 14 16
0
2000
4000
6000
8000
10000
12000
14000
16000
18000
P=120t^1.8 + 8t(in W)
time in secs
Power (in W)
ii)
Area representing energy between 5 and 10 secs:

14ENGINEERING MATHS
iii) The energy between 5 and 10 seconds is
∫
5
10
P ( t ) dt =∫
5
10
(120 t1.8 +8 t )dt= [ ( 120
2.8 )∗t2.8 +4 t2
]5
10
= ( 120
2.8 )∗102.8 +4 ¿ 102 -
( ( 120
2.8 )∗52.8 +4 ¿ 52 ) = 23458.28 Joules.
iv) Given velocity of the machine v = 10t^2 + 0.6t mm/sec.
The displacement over first 5 secs of motion is given by,
∫
0
5
(10 t2+0.6 t )dt = [ ( 10
3 )t3 +0.3 t2
]0
5
=( 10
3 )53 +0.3∗52−( ( 10
3 )03 +0.3∗02 ) = 424.167 mm.
Hence, the displacement of the machine over first 5 seconds of motion is 424.167 mm.

15ENGINEERING MATHS

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