Applications of Mathematical Methods in Engineering: Unit 2 Homework

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Unit 2
Engineering Mathematics
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Contents
Introduction......................................................................................................................................3
LO1: Identify the relevance of mathematical methods to a variety of conceptualized engineering
examples..........................................................................................................................................4
LO2: Investigate applications of statistical techniques to interpret, organise and present data by
using appropriate computer software packages.............................................................................13
LO3: Use analytical and computational methods for solving problems by relating sinusoidal
wave and vector functions to their respective engineering applications.......................................17
LO4: Examine how differential and integral calculus can be used to solve engineering problems.
.......................................................................................................................................................24
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Introduction
Engineering mathematics can be defined as application of mathematical methods to solve the
real world problems. In the course of this document the dos and don’ts of engineering
mathematics would be discussed. The topics that are to be covered here are finding out how
mathematical methods are used to different concepts of engineering example, to gather
information about application of the statistical methods to predict, organize and present data by
using the correct packages of computer software and usage of methods problems by using vector
functions. In the course of this document, the learner would be able to know about the real life
implementation of engineering mathematics and also its application in scientific research.
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LO1: Identify the relevance of mathematical methods to a variety of
conceptualized engineering examples.
Answer a) Given:
Speed of Sound in the gas= v
Gaseous Pressure= p
Gaseous Density= P
Volume of the gas= V
To find= Variable x, y, z in equation:
v= CpxPyVz____________________________ (Equation 1)
Solution: Using dimensional analysis (Lira, 2013).
Where C= Dimensionless Constant
We know that,
Dimensions of speed= L/T
Dimensions of Pressure= M/ (T2L)
Dimensions of Density= M/ L3
Dimensions of Volume= L3
Where L= Liter
M= Meter
T= Time
Hence, by rule of Units and Dimension
L/ T= (M/T2L)x(M/L3)y(L3)z__________________________ (Equation 2)
Comparing the powers of L, M and T in equation 2
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-x -3*y + 3*z= 1
x + y = 0
-2x= -1
By solving
x= 1/2
y= -1/2
z= 0_______________ (Answer)
Putting the value of x, y and z in equation 1 we get
v= C (p/P)½
Answer b)
Given= Force of Viscosity= F
Velocity of sphere moving in liquid= v
Radius of sphere= r
Coefficient of Viscosity= n
F is dependent upon v, r and n
To find: Expression for F using dimensional analysis.
Solution: We know that,
F α ra
F α nb
F α vc
Hence
F α ravbnc
Let F = kravbnc___________________________ (Equation 1)
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Where k is the constant of proportionality.
We know that,
Using Dimensional Analysis (Lira, 2013).
Dimension of F= M1L1T-2
Dimension of r= L1
Dimension of v= L1/T-1
Dimension of n= M1L-1T-1
Equating the dimensions in equation 1
[M1L1T-2]= [L1]a [L-1T-1]b [M1L-1T-1]c__________________________________ (Equation 2)
Equating the values of M, L and T in equation 2 we get
c = 1
a + b – c = 1
-b – b = 2
a = 1, b = 1, c = 1
Putting the values of a, b and c in equation 1 we get
F = krvn ________ (Answer)
Answer c) Part 1
Given-
First term: a1= 3
Let the addition of 8 terms from the beginning= s8
Let the addition of 5 terms from the beginning = s5
S8 = 2s5___________________________ (Equation 1)
To find: Common difference (d)
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Additon of n terms in AP= sn = n/2 (2a + (n-1) d)_________ (Rabago, 2015).
For s8, s8 = 8/2 (2(3) + (8-1)d)
For s5 = 5/2 (2(3) + (5-1)d)
From equation 1
8/2(2(3) + (8-1)*d) = 2*{5/2*((3) + (5-1)*d)}
4 * ( 6 + 7 * d ) = 5 * ( 6 + 4 * d )
24+28 * d = 30 + 20 * d
28d – 20 * d = 30-24
8d = 6
d = ¾ _____________________________ (Answer)
Part 2-
Given: Series 8-4+2-1…..
To find Sum of this series up to 5 terms.
Solution: Common difference in this series: r = -0.5
Sum of a geometric progression = a1 (1 - rn)/(1 - r) _______________ (Rabago, 2015)
Here a1= 8, n = 5 and r = -0.5
Sum of GP up to 5 terms (S5) is
S5= 8[ (1 – (-0.5)5)/(1 – (-0.5))]
S5 = 8[ (1-(-.03125))/1+0.5))]
S5 = 8[ ( 1+ 0.03125 ) / 1.5 ]
S5 = 8[(1. 03125) / 1.5 ]
S5 = 8* 0. 6875
S5 = 5.5 _____________________(Answer)
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Answer d) (Yin et. al., 2014)
Given: Angle of Elevation from A at 10:00= 20o
Angle of elevation from point A at 10:01= 60o
Current altitude= h
Speed of Plane= 600 miles/hour
Distance covered in 1 minute= d
Distance from point A to C = x
To find: Value of altitude h
Solution: Speed of plane= 600 miles/hour
Hence distance d = 600*(1/60)______________________________{1 minute = 1/60 hours}
Distance d= 10 miles
Here tan(20o) = h/(d+x)_____________________________(Equation 1)
And tan(60o) = h/x
x= h/tan(60o)
Putting value of x in equation 1 we get
tan(20o)= h/(d+(h/tan(60o)))
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To find h,
h = d/[1/tan(20o)-1/tan(60o)]
h= 4.6 miles__________________________ (Answer)
Answer e) Deriving formulae for radioactivity.
We know that
Unit of radioactivity is Bq = Change in number of undecayed nuclei/Change in time
1Bq= ∆N/∆t (Silverman, 2015)
Applying Decay constant
∆N/∆T= -λN
Activity (A): A= λA
Equation of radioactive decay is: N=No e-λt_________________ (Equation 1)
A=λN
From Equation 1
A=λ No e-λt
Hence the formulae for radioactivity is
A=Ao e-λt
Where Ao=λNo
Where N= the number of Undecayed nuclei at time‘t’
No= No. of undecayed nuclei at t=0
λ= decay constant(s-1)
Given half-life of the substance= 1 week
Initial amount of radioactivity= 20 counts per second.
To find radioactivity of the substance after 3 weeks:
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Since half-life of a radioactive substance is 1 week, then after one week the amount of
radioactivity of the substance would be= 20/2= 10
After 2 weeks= 10/2= 5
And After 3 weeks= 5/2= 2.5.
Graph for the Radioactive decay of the substance would be.
Answer f)
Exponential equation is found in the form of y=c*ax. (Ricker and Rosen, 2018)
Let y= the number of years since 2000
Let x= No. of people who own the phone/1000
Here ‘a’ will be
a= (375/225)(1/2)
a= (1.67)(1/2)
a≈ 1.29
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Now for c
Taking year 2008, hence
x= 8 and y= 1725
1725= c*1.298
c≈ 225
Hence exponential equation= y = 225*(1.29)x
Part2: In case of increased sales,
Year from 2000 where 2000 is 0-25 years
x= 25
Hence y would be,
y=225(1.29)25
y=130895.6685
Part3: For average rate of change, from 2006 to 2010
x1= 6, x2=10
Average Rate of change= (y2-y1)/(x2-x1)
Avg= [{225(1.29)10}-{225(1.29)6}]/(10-6)
Avg= 1934.447/4
For average rate of change= 483.61__________________ (Answer)
Part4:
Instantaneous Rate of Change in 2006
Let x1= 5.999 and x2=6.001
From the exponential equation
Inst= (y2-y1)/(x2-x1)
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Inst= [{225(1.29)5.999}-{225(1.296.001)}]/(6.001-5.999)
Instantaneous Rate of change= 264.0285__________________________ (Answer)
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