Imperial College Engineering Mathematics Solutions

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Engineering Maths
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Table of Contents
Introduction:............................................................................................................................................3
LO1.............................................................................................................................................................4
LO2...........................................................................................................................................................12
LO3............................................................................................................................................................18
LO4...........................................................................................................................................................26
Conclusion.................................................................................................................................................32
References.................................................................................................................................................33
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Introduction:
Engineering mathematics is the way to apply mathematics to solve complex problems in the real
world. It is basically a branch of applied mathematics which consists of mathematical techniques
and methods which are used in engineering and industry. Engineering mathematics consists of
differential equations, real and complex analysis, approximation theory, linear algebra,
probability and many more. There are various organisations from which we can learn about
engineering mathematics. Among them, the most suitable organisation is Imperial College
situated in London. This organisation helps in learning the concepts of engineering mathematics.
In this brief, we will come to know about the various mathematical models, applications of
statistical techniques, analytical and computational methods and differential and integral
calculus.
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LO1
a) Solution- Equate the dimensions on L.H.S and R. H.S
Compare the exponents of [L] [M], [T]
1 equals minus x minus 3y plus 3z
0 equals x plus 3y
1 equals minus 2x
After solving
x=1/2
y=-1/2
z=0
Hence
4

b).Solution: Let F ∝ ra
F ∝ vb
F ∝ η c
So F ∝ ra vb η c
F = K ra vb η c ………….(1)
K refers to proportionality constant
Substitute dimension formula of each and every quantity in equation (1)
[M1L1T-2] = [M0L1T0]a [M0L1T-1]b [M1L-1T-1]c
= [M1L1T-2] = [M0La+b-cT-b-c]
Compare powers of M, L, T on LHS and RHS
We get
c = 1
a + b - c = 1
On solving these equations we get
a equals1
b equals 1
c equals 1
Substitute a, b, c in equation (i)
F= k r v η , which is the viscous force.
c). Solution- S8 = 2(S5)
a=3
5

Sn = n/2 (2a plus (n minus 1) d)
8/2 (2(3) + (8-1) d) = 2 (5/2 (2(3) + (5d-d))
8/2(6+7d) =5(6+4d)
28d = 20d + 30-24
24-30= 20d-28d
d equals 3/4
Sn equals a1 (1-Rn) divides (1-r)
= 8 *(1- (-1/2) 5) / 3/2
= 5.5
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d).
Sol: Calculate the distance d by using speed and time
I min equals 1/60 hr
= 10 miles = 1/60 multiply 600
tan (20o) equals h/ (d + x) …..(1)
tan (60o) equals h/x …..(2)
On solving eq (1) and eq (2)
We get
h equals d (-tan (60o) - tan (20o)) / tan (60o)* tan (20o)
equals 4.6 miles
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e). Solution: 1-
Time (seconds) Radioactivity level
0 20
1 10
2 5
3 2.5
4 1.25
Graph:
Source-http://gladwin.glk12.org/mod/book/view.php?id=5912&chapterid=5667
2- Formula= xt =x0 *(1/2) time/halftime
Where xt = remaining radioactivity at time t
x 0 = initial radioactivity amount
3- As we know x0=20
t= 3 *7 =21 days
x t = x0*(1/2)21/7
= 20*(1/2)3
8

= 20/8
= 2.5 counts/sec
9

f).
Year 2000 2002 2004 2006 2008
Number of
cell-phone
Owners
(thousands)
225 375 623 1037 1725
Solution: 1- Exponential equation of the form y=c.ax
Let y= no. of years since 2000
x= no. of people who own cell phone
a= (375/225)1/2
= 1.67(0.5)
=1.29y= c. (1.29) x
c= take (8, 1725), sub x=8, y=1725 into y= c. (1.29) x
1725= c. (1.29)8
c=225
Now put the value of c and a in y=c.ax
y= 225(1.29) x
2). Put x=25 in y= 225(1.29) x
y= 225(1.29)25
y= 130895.6685
3). Put x=6 and x=10 in y= 225(1.29) x
y= 225(1.29)6 =1036.86
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y= 225(1.29)10 =2971.307
Avg. rate of change = y2-y1/x2-x1
=2971.307-1036.86/10-6
=1934.447/4
=483.61
4). Year = 2006 x=6
Instantaneous rate of change is determined as:
Select two closest points x=5.999, x=6.001
Sub x=5.999 and x= 6.001 in y= 225(1.29) x
y= 225(1.29)5.999
=1036.597579
y= 225(1.29)6.001
=1037.125636
Instantaneous rate of change is: y2-y1/x2-x1
= 1037.125636-1036.597579/6.001-5.999
= 264.0285
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LO2
a- 44, 50, 38, 96, 42, 47, 40, 39, 46, 50
To calculate mean = (38 plus 44 plus 50 plus 42 plus 96 plus 47 plus 46 plus 40 plus 39 plus
50) /10
= 492/10
=49.2
= 44-(49.2)2+50-(49.2)2+38-(49.2)2+96-(49.2)2+42-(49.2)2+47-(49.2)2+40-(49.2)2+39-(49.2)2
=27.04+0.64+125.44+2190.24+51.84+4.84+84.64+104.04+10.24+0.64
=2600.4
2600.4/ 9
=288.7
Variance = 289
Standard deviation = 17.
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