Mathematical Methods in Engineering: A Practical Approach

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Table of Contents
LO1: Identify the relevance of mathematical methods to a variety of conceptualised
engineering examples.................................................................................................................2
LO2: Investigate applications of statistical techniques to interpret, organise and present data
by using appropriate computer software packages....................................................................9
LO3: Use analytical and computational methods for solving problems by relating sinusoidal
wave and vector functions to their respective engineering applications..................................12
LO4: Examine how differential and integral calculus can be used to solve engineering
problems...................................................................................................................................18

LO1: Identify the relevance of mathematical methods to a variety of conceptualised
engineering examples.
a) In the given equation:
v=C px δ y V z
v = speed of sound
p = pressure
δ = density
C = dimensionless constant
V = Volume of gas
The dimension equation of the formula: -
[ L ] / [ T ] = ( [ M ] / [ T ]2 [ L ])X * ( [ M ] / [ L ]3 )Y * [ L3]Z
On comparing the exponents on both the sides of the equation we will get the
expression : -
1 = -x – 3y + 3z
0 = x + y
-1 = -2x
Solving the above equations we get the value of x, y and z as:-
x = ½ , y = -1/2 , z = 0
Therefore,
p = ( [ M ] / [ T ]2 [ L ])X
δ = ( [ M ] / [ L ]3 )Y
V = [ L3]Z
The answer obtained is : -
V = C √ p
δ
b) Let us first consider,

F is proportional to ra,
F is proportional to vb,
F is proportional to nc,
Combining all we get,
F is proportional to ra vb nc
F = kra vb nc ……………( i )
k = constant
Determining the dimension equation for the above equation: -
[ M1 L1 T-2] = [ M0 L1 T0 ]a [ M0 L1 T-1 ]b [ M1 L-1 T-1 ]c
Which can be written as: -
[ M1 L1 T-2] = [ Mc La+b-c T-b-c ]
Equating the exponents on both the side of the equations: -
c = 1
1 = b – c
-b -c = -2
Solving the above we get,
a = b = c = 1
by putting the value of a,b and c in equation (i) we get,
= > F = k r v n
c) The formula used to find the addition of the series is given by: -
Sn = n/2 (2a + (n -1) d
Where,

Sn = sum of the nth term
a = first term in the series
d = common difference between two consecutive terms
n = number of terms
according to the question
S8 = Twice of S5
a = 3
S8 = (8 / 2) ( ( 2 * 3 ) + ( ( 8 – 1 ) * d ) )
S8 = 4 ( 6 + 7 d )
S8 = 24 + 28 d
S5 = (5 / 2) ( ( 2 * 3 ) + ( ( 5 – 1 ) * d ) )
S5 = 5 / 2 ( 6 + 4 d )
S5 = 15 + 10 d
S8 = 2 * S5 …………..(i)
Substituting value of S8 and S5 in equation ( i )
( 24 + 28 d ) = 2 * ( 15 + 10 d )
24 + 28 d = 30 + 20 d
28 d – 20 d = 30 – 24
8 d = 6
d = ¾
Sum is calculated by : -
Sn = a1 ( 1 – r n ) / ( 1 – r )
= 8 ( 1 – ( 3 / 4 )5 ) / ( 1 – ( 3 / 4 ) )
= 8 ( 1 – ( 243 / 1024 ) / ( 4 – 3 ) / 4 )
= 24.40

d) For calculating h, we first have to calculate d by using speed and time while
considering ( 1 min = 1/60 hour )
d is equal to = 600 * ( 1 / 60 ) = 10
Now, to calculate the tangent of the angle given due elevation: -
Tan ( 20 0 ) = h / ( d + x )
Tan ( 60 0 ) = h / x
After eliminating x from the equation above we get,
h = d / [ ( 1 / tan ( 20 0 ) ) – ( 1 / tan ( 60 0 )) ]
The answer we get from the above equation is
= 4.6 Miles
e) (i) the graph below shows the radioactivity per week : -

ii) The formula for radioactivity in term of time is: -
Xt = X0 * ( 1 / 2 ) time / halflife
Where,
Xt represent the radioactivity amount remaining at ( t )
X0 represent initial radioactivity
iii) As we know: -
X0 = 20
t = 3 * 7 = 21 days
half-life = 7days
Xt = X0 * ( 1 / 2 ) 21/7 = 20 multiply ( 1 / 2 ) 3 is 20 / 8 = 2.5 counts / sec
f) i) The equation that is required is => y = c multiplied by ax
now assuming,
x be the people on the phones
a = (375 which is divided by 225 ) 1 / 2
= 1.29
y = c multiplied by ( 1.29 )8
c multiplied by ( 1.29 ) 8 = 1725
y = c * ( 1.29 ) x
1725 = c * ( 1.29 ) 8
1725 when divided by 7.669c
c = 225
now, putting the value of c = 225 and a = 1.29 in the equation.
Y = 225 multiplied by ( 1.29 )x

Now the equation we got is: =
y = 225 multiplied by ( 1.29 )25
ii) year 2000 that is 0.25 year
putting value of x = 25
in equation y = 225 multiplied by ( 1.29 )25
the value of y comes,
y = 130895.6685
hence, (130895.6685) thousands of people in Canada owns a phone during 2025.
iii) Now put the value of x = 6 , x = 10 in equation y = 225 * ( 1.29 )x
y = 225 multiplied by ( 1.29 ) 6 = 1036.86  61036.86
y = 225 multiplied by ( 1.29 ) 10 = 2971.307  102971.307
A .R .O .C. = ( y2 -y1 ) / ( x2 -x1 )
= 2971.307 minus 1036.86 / ( 10 minus 6 ) = 483.61
Therefore, the rate of change b/w 2006 and 2010 is equal to ( 483.61 ) thousands of
people in a year.
iv) considering two nearest points
say x = 5.999 , x = 6.001
in y = 6.001 in the eq. y.
y = 225 multiplied by ( 1.29 ) to the power 5.999  1036.597579 ( 5.999,
1036.597579)

y = 225 multiplied by ( 1.29 ) to the power 6.001  1037.125636 ( 6.001,
1037.125636)
A .R .O .C . = (y2 minus y1 ) / ( x2 minus x1 )
= ( 1037.125636 minus 1036.597579 ) divided by ( 6.001 minus 5.999)
= 264.0285
Hence the change rate in 2006 is ( 264.0285 ) thousand people in a year.

LO2: Investigate applications of statistical techniques to interpret, organise and present
data by using appropriate computer software packages.
a) Mean can be calculated using the formula: -
Addition of total observation divided by the total number of observations =
(sum of observation/number of observations)
 (40, 50, 38, 96, 42, 47, 40, 39, 46, 50) divided by 10
 488 divided by 10 = 48.8
The standard deviation is given below: -
44 48.8 – 44 4.8
50 50 – 48.8 1.2
38 48.8 – 38 10.8
96 96 – 48.8 47.2
42 48.8 – 42 6.8
47 48.8 - 47 1.8
40 48.8 – 40 8.8
39 48.8 – 39 9.8
46 48.8 – 46 2.8
50 50 – 48.8 1.2
( ( minus 4.8)2 + ( 1.2 ) + ( minus 10.8 ) 2 + ( 47.2 ) 2 + ( 6.8 ) 2 + ( minus 1.8 )2 +
( minus 8.8 )2 + ( minus 9.8 )2 + ( minus 2.8 )2 + ( 1.2) 2 )/ 10 = 260.12
b) Given that,
n = 20
p = 0.25
μ = n * p = 20 * 0.25
= 5 * σ 2 = n * p = 20 * 0.25

= 20 * 0.25 * 0.75
= 3.75 multiplied by σ = root of ( 3.75 ) = 1.94
n * p = 15 greater than equal to 5
n * ( 1 minus p ) = 15 greater than equal to 5
As per the corollary concept, 1 B ( 20, 0.25) approximately equals to N ( 5, 1.94 )
c) Replacing 3 % of the defected motors the manufacturer will take around years long
guarantee of 6.24 which should be offered by the manufacturer.
μ = average life of motor which is up to 10m years
standard deviation ( σ ) is equal to 2
formula required to calculate:-
z = ( x minus μ ) / σ
P ( Z < z ) = 3 / 100
By calculating we get,
Z = ( minus 188) / 100,
Putting the value of σ, μ and z in above equation
We get ,
 (minus 188) / 100 = ( x minus 10) / 2 = ( minus 376 divided by 100) = X minus
10
 X is equal to 10 minus 3.76 = 6.24 years.
d) Let rejected pistons = x
Success is rejections according to the questions.
Given,
n = 10, p = 0.12 & q = 0.88

i) When x = zero there is no rejections
p of ( x ) = Cnx px qn-x = C100 ( 0.12 )0 ( 0.88 )10 = 0.2785
on one rejection,
p of ( x ) = C101 ( 0.12 )1 ( 0.88 )9 = 0.37977
on two rejections,
p of ( x ) = C102 ( 0.12 )2 ( 0.88 )8 = 0.23304
the probabilities of not getting more than 2 rejection
is equals to = P ( x less than equal to 2)
= 0.2785 + 0.37977 + 0.23304 = 0.89131
ii) Least piston getting rejected = 2
P ( two rejects ) = 1 minus P (x less than equal to 1 )
= 1 minus ( p of x0 adding to x1) = 1 minus ( 0.2785 + 0.37977) = 0.34173
iii) The results of probabilities above are out of ten. The non-rejection
probabilities are 0.89. the probability of two pistons getting rejected with
manufacturer making100 percent profit is equal to 0.341 and manufacturer
will be providing benefit.
e) Using the formula : =
Z = ( x minus μ ) divided by σ
Converting 1250 to ( 1250 minus 1000 ) divided by 125 = 2
Converting 980 to ( 980 minus 1000 ) divided by 125 = minus 0.16
Converting 1150 to ( 1150 minus 1000 ) divided by 125 = 1.2

LO3: Use analytical and computational methods for solving problems by relating
sinusoidal wave and vector functions to their respective engineering applications.
a) i) let,
P ( t ) = a * cos ( b * ( t – d ) + c
Pmax = 20
Pmin = 4
c = ( Pmax plus Pmin ) divided by 2 = ( 20 plus 4 ) divided by 2 is equals to 12
|a| = ( ( min of P) minus ( max of P ) ) divided by 2 is equals to = ( 20 minus 4 )
divided by 2 equals 8.
Days after 1st January where the P ( t ) is maxed is given by: -
Number of days from January to May + 21 days from June
Adding the t = 31, 28, 31, 30, 31, 21, is equals to 172
Using period to find b at a greater than 0.
A period is 365 which equal to 2 * π divided by b
Therefore, b = 2 multiplied by π divided by 365
Cos function. Has max when ( t = 0 ) and P ( t ) is max at t =172
P ( t ) = 8 multiplied by cos ( ( 2 π divided by 365 ) multiplied by ( t – 172 ) ) + 12
Now, P ( t ) = max ( at t =172 )
Putting t =172 in equation.
 P ( 172 ) = 8 * cos ( ( 2 multiplied by π / 365 ) multiplied by ( 172 minus 172 ) )
plus 12 equals to 8 .
 Cos ( 0 ) + 12 = 20