Engineering Mathematics: Dimensional Analysis and Progressions

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Engineering Maths
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Contents
Introduction......................................................................................................................................3
LO1..................................................................................................................................................3
P1. Apply dimensional analysis techniques to solve complex problems....................................3
M1. Use dimensional Analysis to derive expression...................................................................5
P2. Generate Ans.s from contextualized arithmetic and geometric progressions.......................7
P3. Determine Sol.s of equations using exponential, trigonometric and hyperbolic functions...9
D1. Present statistical data in a method that can be understood by non- technical audience....11
LO2................................................................................................................................................15
P4 Summarize data by calculating mean and standard deviation and simplify data into
graphical form............................................................................................................................15
P5...............................................................................................................................................16
M2..............................................................................................................................................17
LO3................................................................................................................................................19
P6...............................................................................................................................................19
P7...............................................................................................................................................22
M3..............................................................................................................................................25
LO4................................................................................................................................................28
P8...............................................................................................................................................28
P9...............................................................................................................................................29
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D3..............................................................................................................................................31
References......................................................................................................................................32
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LO1
P1. Apply dimensional analysis techniques to solve complex problems.
Ans. a) Stated:
Velocity of Sound in gaseous environment = v
Pressure applied by the gas= p
Density of the gas= P
Gaseous volumetric value= V
To evaluate = x, y, z :
v= C p x Py Vz____________________________ (Equation 1)
Sol.: By the help of dimensional analysis (Lira, 2013).
Here C= Constant with no dimension
It is already known that,
Velocity Dimensions= L/T
Pressure Dimensions= M/ (T2L)
Density Dimensions= M/ L3
Volumetric Dimensions= L3
Here L= Liters
M= Meter
T= Times
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Therefore due to the help of unit and dimension
L/ T= ( M /T2L)x( M/L3)y( L3)z__________________________ (Equation 2)
Exponential Comparison
−x−3∗y +3∗z=1
x + y=0
−2 x=−1
By determining
x=1/2
y=-1/2
z=0_______________ (Ans.)
Placing the magnitude of the variables x, y, z in the Stated equation 1
v=C(p/P)½
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M1. Use dimensional Analysis to derive expression.
Ans. b)
Provided= The value of the viscous force= F
Speed of moving spherical body in the fluid= v
Radius = r
Viscous Co- efficient = n
The value of F is depending upon the values of n, v and r
To evaluate: Equation for F.
Sol.: It is already known that,
F α ra
F α nb
F α vc
Therefore
Fα ra vb nc
Hence
F=k ra vb nc
__________________________ (Eqn. 1)
k= constant.
It is already known that,
By the use of dimensional Analysis (Lira, 2013).
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Dimensions of Force (F)= M1 L1 T-2
Dimensions of radius (r) =L1
Dimension of velocity (v) =L1/T-1
Dimension of n=M1L-1T-1
Putting the above dimension in eqn. 1
[M1L1T-2]=[L1]a [L-1T-1]b [M1L-1T-1]c__________________________________ (Equation 2)
Putting the value of M,L as well as T in the equation 2 it is found that
c=1
a+b–c=1
-b–b=2
a=1,b=1,c=1
Putting the value of a,b as well as c in eqn. 1 it can be said that
F=krvn________ (Ans.)
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P2. Generate Ans. from contextualized arithmetic and geometric progressions.
Ans. c)
Stated-
Initial value: a1=3
Assuming the sum of initial 8 values= s8
Assuming the sum of initial 5 values= s5
S8=2s5___________________________ (Equation 1)
To find: Value of the common difference (d)
Sum 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 _____________________________ (Ans.)
(ii)-
Stated: Series8-4+2-1…..
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To find: Sum of this series up to 5 terms.
Sol.: 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_____________________(Ans.)
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P3. Determine Sol. of equations using exponential, trigonometric and hyperbolic functions.
Ans. d) (Yin et. al., 2014)
Stated: 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
Sol.: Speed of plane=600 miles/hour
Hence distance d=600*(1/60)______________________________{1minute=1/60hours}
Distance d=10miles
Here tan(20o)=h/(d+x)_____________________________(Equation 1)
And tan(60o)=h/x
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x=h/tan(60o)
Putting value of x in equation 1 we get
tan(20o)=h/(d+(h/tan(60o)))
To find h,
h=d/[1/tan(20o)-1/tan(60o)]
h=4.6miles__________________________ (Ans.)
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D1. Present statistical data in a method that can be understood by non- technical audience.
Ans. e) Equation for radio activity.
It is already known that
Units of radio activities is Bq=Changes in number of non- decayed nucleous/Variation in time
1Bq=∆N/∆t(Silverman, 2015)
Applying Decay constant
∆N/∆T=-λN
Activity(A):A=λA
Equation of radioactive decay is:N=Noe-λt_________________ (Equation 1)
A=λN
From Equation 1
A=λNoe-λt
Hence the equation for evaluating the value of radioactivity:
A=Aoe-λt
Here Ao=λNo
here N= Non- decayed nucleus at‘t’
No= Non- decayed nucleus at t=0
λ=Constants of Decaying(s-1)
Stated half-life of the object=1week
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