Relevance of Mathematical Methods in Engineering Examples

LO1: Identify the relevance of mathematical methods to a variety of conceptualized
engineering examples.
a. Speed dependence on pressure, density and volume is provided in the
equation below. From the [L], [M] and [T] systems, the two sides of the
equations are equated to yield equation 2 shown below.
.............................1
..................2
In order to solve for x, y and z, the corresponding polynomials of [L], [M] and
[T] in the left and right hand side of the equations are compared to obtain
equations 3-5 shown below.
..................3
...............................4
.................................5
From equation 5,
x=−1
−2 =0.5
Substituting x=0.5 in eqn 4,
0= y +0.5
y=−0.5
From the values of x and y, substituting in eqn 1,
1=−0.5+1.5+3 z
z=0
Substituting the values of x, y and z in equation 1, we get
b. From the problem statement,
let
F α ra ... ... ... ... ... ... 1
F α vb ... ... ... ... ... ... 2
F α nc ... ... ... ... ... ...3
The three equations can be rewritten as equation shown in equation 4

F α ra vb nc ... ... ... ... ... ... 4
Using a proportionality constant k, equation 4 can be rewritten as equation 5
shown below
F=k ra vb nc ... ... ... ... ... ... 5
Introducing dimensional analysis on both sides of equation 4, we obtain
[ M1 L1 T 2 ]=[M 0 L1 T 0 ]a [ M 0 L1 T−2]b [M 1 L−1 T−1 ]c ... ... ... ... ... ... 6
Simplifying equation 5, we obtain,
[ M 1 L1 T2 ] = [ M c La+b−c T−b−c ]
1
... ... ... ... ... ..7
Equating the powers of L,M,T on both sides, we obtain
c=1...............8
a+b-c=1..............9
-b-c=-2................10
From 8, c=1, substituting c in 10, b=-(-2+1)=1
Substituting c=1 and b=1 in equation 9, a=1
Hence
a=1, b=1, c=1
Substituting in the equation,
F=krv
c. The first term=a, d=common difference and n= number of terms
a=3
In a geometric progression, sum of nth terms Sn= n
2 (2 a+ ( n−1 ) d ) where a is
the first term and d is the common difference.
For the first 5 terms
S5=2.5(6+4d)...............1
The firs 8 terms
S8=4(6+7d)....................2
But from the statement,
S8=2S5.............................3
Hence 2.5×2(6+4d)=4(6+7d)
30+20d=24+28d..............4
From equation 4, collecting the unknown and the known on one side, we
obtain
30-24=28d-20d
Hence d=0.75
The given series =8,-4, 2, -1, +---

The common ratio r=-−4
8 =-0.5
Because r<1
Sum of nth terms,
Sn= a ( 1−r n )
1−r n
Where r≠ 1
Sn= 8 ( 1−−0.55 )
1−−0.55 =8
d.
Consider the diagram shown below
The distance d is easily computed from the time distance relation
d=600×1/600=10 miles........................1
From the diagrams, tan 20 and tan 60 can be expressed as shown in equation
2 and 3
Tan 20=h/(d+x) -----2
Tan 60=h/60---------3
Substituting d and eliminating x from the third equation
X=h/tan 60..........4
Hence
Tan 20=h/(10+h/tan60)
From which h=4.60 miles
e.

a. A graph of amount of radioactivity against time in weeks
e 0 2 4 6 8 10 12
0
5
10
15
20
25
f(x) = 40 exp( − 0.69 x )
R² = 1
Radioactivity (counts per second)
b. N= 40e-0.693t
c. t=3
N= 40e-0.693*3
=5.00counts per second
f.
Assigning the year 2000 a time series index of 1 and an increment of 1 in the
subsequent years, the graph below shows the growth of telphones with time. The
graph is obtained from Microsoft Excel 2013.