Relevance of Mathematical Methods in Engineering Examples

Table of Contents
LO1: Identify the relevance of mathematical methods to a variety of conceptualized
engineering examples...........................................................................................................................2
LO2: Investigate applications of statistical techniques to interpret, organise and present
data by using appropriate computer software packages.............................................................5
LO3: Use analytical and computational methods for solving problems by relating
sinusoidal wave and vector functions to their respective engineering applications............8
LO4: Examine how differential and integral calculus can be used to solve engineering
problems.................................................................................................................................................11

LO1: Identify the relevance of mathematical methods to a variety of conceptualized
engineering examples.
a. Equating the dimensions of both sides of the equation, we obtain,
Comparing the two sides of the equations, we get
From the third equation, x=0.5, substituting in the second equation, we get y=-0.5 and making
the two substitutions in equation 1, we get z=0
From the values
b. Let
F α ra
F α vb
F α nc
Alternatively
F α ra vb nc
F =k ra rb rc
Where k is a proportionality constant1
[M1L1T-2]=[M0L1T0]a[M0L1T-2]c[M1L-1T-1]c
[M1L1T-2]=[McLa+b-cT-b-c]
Equating the powers of L,M,T on both sides,
C=1
a+b-c=1
-b-c=-2
Solving the three equations, we get
a=1, b=1, c=1
1 Bender, C. M., & Orszag, S. A. (2013). Advanced Mathematical Methods for Scientists and Engineers
Tang, K. (2016). Mathematical Methods for Engineers and Scientists
Occhiogrosso, M. (2010). Sequences and Series: Precalculus

Substituting in the equation,
F=krv
c.
a=3
The sum of nth terms in a GP= n
2 (2 a+ ( n−1 ) d ) where a is the first term and d is the common
difference.
The first 5 terms
S5=2.5(6+4d)
The firs 8 terms
S8=4(6+7d)
But from the statement,
S8=2s5
Hence 2.5*2(6+4d)=4(6+7d)
30+20d=24+28d
Collecting the like terms
6=8d
Hence d=0.75
The series =8,-4, 2, -1, +---
The common ratio r=-4/8=-0.5
Since r<1
Sum of nth terms,
Sn= a ( 1−r n )
1−rn
Where r≠ 1
Sn= 8 ( 1−−0.55 )
1−−0.55 =8
d. From the time speed relation, the distance d is computed as shown below
d=600*1/600=10 miles
The tangents of the given angles are then expressed as shown below
Tan 20=h/(d+x) -----1
Tan 60=h/60---------2
Substituting d and eliminating x from the second equation
X=h/tan 60
Hence
Tan 20=h/(10+h/tan60)
From which h=4.6 miles

e.
1.
0 2 4 6 8 10 12
0
5
10
15
20
25
f(x) = 40 exp( − 0.693147180559945 x )
R² = 1
Radioactivity (counts per second)
2. N= 40e-0.693t
3. t=3
N= 40e-0.693*3
=5.00counts per second
f.
Using time series index from 2000 with an increment of 1 every year, the plot below shows the obtained
graph