Engineering Math Unit 2: Solved Problems and Concepts

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Unit 2: Engineering Math

Contents
LO1..................................................................................................................................................4
A).................................................................................................................................................4
B).................................................................................................................................................4
C).................................................................................................................................................5
D).................................................................................................................................................6
E)..................................................................................................................................................7
1)..............................................................................................................................................7
2)..............................................................................................................................................8
3)..............................................................................................................................................8
F)..................................................................................................................................................8
LO2................................................................................................................................................11
A)...............................................................................................................................................11
B)...............................................................................................................................................11
C)...............................................................................................................................................12
D)...............................................................................................................................................13
E)................................................................................................................................................14
LO3................................................................................................................................................15
A)...............................................................................................................................................15
B)...............................................................................................................................................19
C)...............................................................................................................................................20
D)...............................................................................................................................................21
E)................................................................................................................................................22
F)................................................................................................................................................23
G)...............................................................................................................................................23
LO4................................................................................................................................................25
A)...............................................................................................................................................25
B)...............................................................................................................................................25
C)...............................................................................................................................................26
D)...............................................................................................................................................27
E)................................................................................................................................................29

References......................................................................................................................................30

LO1
A)
Answer: Given, v = C px py Vz
And, C = dimensional constant
Now, equating all the dimensions of both side of the equation, we will get;
Then, comparing all the exponents of [L], [M] and [T] from both side of the of the equation, we
will get;
1 = -x-3y+3z
0 = x+y
-1 = -2x
After calculating above three equation, we get;
x = 1/2
y = -1/2
z = 0
Now, putting the value of x, y and z on the first equation, we get;
v=C ❑
√ p
q
B)
Answer: Let’s assume;
F ∝ va, F ∝ rb and F ∝ ηc …... (i)
So, F = Kvarb ηc

Where, ‘K’ is a dimensional constant.
Now,
Dimensional formula for F = [M1L1T-2]
Dimensional formula for v = [L1T-1]
Dimensional formula for r = [L1]
Dimensional formula for η = [M1L-1T-1]
Putting above dimensional formulae in the equation (1), we get;
[M1L1T-2] = [L1T-1]a [L1]b [M1L-1T-1]c
[M1L1T-2] = [Mc La+b+c T-a-c] ……... (ii)
Using principle of homogeneity on the equation (ii), we get;
c = 1 …... (iii)
a+b-c = 1 …... (iv)
-a – c = -2 …... (v)
Now, Putting c = 1 in (v) and a = 1 & c = 1 in (iv), we get;
a = 1
b = 1
Hence, putting value of a, b and c in equation (i), we get;
F = kηrv
C)
Given, S8 = 2(S5)
A = 3
We know that,

Sn ¿ n
2 (2 a+ ( n−1 ) d )
Now, putting the given value of S8 and A in the above equation, we get;
8
2 ( 2 ( 3 ) + ( 8−1 ) d )=2 ¿
4(6 + 7d) = 5(6 + 4d)
24 + 28d = 30 + 20d
-6 = - 8d
d = 3
4
The given series in the question is: 8 − 4 + 2 − 1 + . . .
Total number of terms in series = 5
So, using above equation, we get;
S5 = 5
2 (2(8)+ ( 5−1 ) 3
4 )
= 5
2 (19)
= 95
2
= 47.5
D)
Answer: Given,

First, we will calculate the value of d;
d = 600 * ( 1
60) miles
= 10 miles
Now, we will find the tangent of given elevation angle;
tan (200) = h
(d + x )
And, tan (600) = h
x
So, using above two equations to eliminate x, we get;
h = d / [tan (200) – 1 / tan (600)]
= 4.6 miles (which is rounded up to 2 decimal places)
E)
Answer:
1) The graph for the amount of radioactivity against times in weeks has been shown below
(Lumenlearning, 2019).

0 1 2 3 4 5 6 7 8 9 10
0
5
10
15
20
25
Amount of radioactivity
Amount of radioactivity
Time (Half life a week)
(Source: Created by the author)
2)
Answer:
The formula to find out the total number of radioactivity in terms of time has been given below
(Lumenlearning, 2019).
A (sub t) = A (sub 0) x (1 / 2) ̂ (time / half-life)
Where, A(t) = amount of radioactivity remaining at t time
A (sub 0) = Initial amount of radioactivity at time 0 (Zero)
3)
Answer:
Given,
A (sub 0) = 20 counts per second
Time = 3 weeks = 27 days
Putting the value in above equation, we get;
A (sub 3 weeks) = A (sub 0) x (1 / 2) ̂ ( 27 / 7)

= (20 counts per second) x (1 / 2) ̂ ( 3)
F)
Answer a): We know that equation for exponential calculation is:
Y = (C)*(Ax)
where, y = No. of years
x = No. of person using cell phone which is in thousands
Now, taking the given value, we get;
A = (375/225)(1/2)
= 1.29
Putting the value of A in the above equation, we get;
Y = (C)*(1.29x)
C = 1725/ (1.298)
= 225
Hence, putting value of A and C in the above equation, we get;
Y = (225)*(1.29x)
It is required equation.
Answer b): First, we calculated total number of year = 2025-2000 years
= 25 years (which is x)
Putting the value of x as 25 years in the above equation, we get;
Y= (225)*(1.2925)
= 130895.669
Answer c): Given,

Year to be calculated = 2006 and 2010
So, x will have two value 6 and 10, using both value in the above equation;
Y1 = (225)*(1.296) = 1036.86
Y2 = (225)*(1.2910) = 2971. 307
Now, the average rate of change in thousands person per year is;
(Y1 - Y2) / (x2-x1)
= (2971.307-1036.86) / (10-6)
= 483.61 (thousands person per year)
Answer d): Given year for calculation = 2016
So, x = 6
Assuming two close points near the value of x which is 6 are: 5.999 and 6.001
Using above two values, we get;
Y1 = (225)*1.295.999) = 1036.597
Y2 = (225)*(1.296.001) = 1037.125
Hence, instantaneous rate of change = (Y1 - Y2) / (x2-x1)
= (1037.125-1036.597) / (6.001-5.999)
= 264.02

LO2
A)
Answer: Given;
Total number of sample = 10
Data = 44, 50, 38, 98, 42, 47, 40, 39, 46, 50
We know that, mean is given by;
x= 1
n ∑
i=1
n
x
Where, x = Mean
x = mid value
n = total number of data
Hence, putting the given value in above equation, we get;
x= 1
10 ∑
i=1
n
x
= 49.2
Now, the standard deviation is given by;
S= √ 1
( n−1) ∑
i=1
n
( x−x)2
Putting the given value of data and n = 10 in the above equation, we get;
S = 17
B)
Answer: Normal approximation for the binomial distribution
Normal distribution consist of two real number which define mean and standard deviation for the
given data resulting in a complex distribution for the analysis of the data. As a result, correct

normal distribution needs to be defined to simplify the complex distribution of data. It is done
using mathematical formula and calculations to determine the normal approximation value of the
binomial distribution for the given data (Taylor, 2018).
Given; n= 20 and p = 0.25
We know that,
μ = np
= (20)*(0.25)
= 5
Also, σ2 = np (1-p)
= 3.75
σ = 1.94
C)
Answer: Given;
The average life of motor (μ) = 10 years
Standard deviation (σ) = 2
We know that;
Z = X−μ
σ
So, P (Z≤z) = 0.03
After calculation, we get; Z = -1.88
Now, putting the value of Z, μ and σ in the above equation, we get;
-1.88 = X−10
2
X - 10 = (-1.88*2)