Engineering Mathematics: Practical Applications and Solutions

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ENGINEERING MATHEMATICS

Contents
LO1............................................................................................................................................3
LO2............................................................................................................................................8
LO3..........................................................................................................................................11
LO4..........................................................................................................................................16
References................................................................................................................................20

LO1
(A) According to both side dimension equitation result is
(L)/(T) = {(M)/(T)2 (L)}X { [(M)/(L)]3}Y {(L)3}Z
Done the compression on both equitation and the result are:
1= -X-3Y+3Z
X+Y = 0
-2X = -1
Result values are
X = 1/2, Y= -1/2 and Z= 0
According to equation value
p = {(M)/(T)2(L)}X
OR
(p) = {(M)/(L)3}Y {(L)3}Z
Result:
V=C√p/(p)
(b) According to conceder value
ra ∝ F
vb∝ F
nc ∝F so

ravbnc ∝ F
In other terms
F = K ravbnc………………………… (a)
K= Constant
Put the value in (a) equation
(M1 L1 T-2) = {(M0 L1 T0)}a{(M0 L1 T-1)}b {(M1 L-1 T-1)}c
In other way
(M1 L1 T-2) = {Mc La+b-c T-b-c}
Both side powers is equating
C equal to = 1
1 is = +b-c
Minus b –c is equal -2
Valve of a, b, c is = 1
Substitutes the equation (a)
Result F=K r v n
(c) According to the equation Sn=n/2{2a+(n-a)d}, by given equation s8
=2{S5} in which a =3 let us
8 dived by 2 * {2*(3)} + (8-1) *d} is give 2*{5/2}*{(2*(3)+(5-1)*d}
4{6+7* d} put 5 *(6+4*d)
{24+28}*d = {30+20}*d

-8d =-6
d= and equation sum is¾
Sn = a1{1-rn}/{1-r}
Sn = 8{1-(3/4)5}/{1-(3/4)}
= 8{1-(243/1024)/ (4-3)/4}
= 24.400
(d) Find the value of h, identify the d, use the time and speed function
(Where 1 Min. =1/60 Hr)
D= 600 multiple {1/60} =10
Equation Tangent is
Tan 20˚ =h/{d+x} or Tan 60˚ = h/x
Eliminate the value of x from the equations
h =d/{1 / Tan 20˚ -1/ Tan 60˚}
h =4.60 Miles
(e) (1) According to the graph

(2) According to Equation formula
X{sub t} = X(sub 0)* {1/2}TIME/Half Life In which
X{sub t} are remaining amount of substance in time t and
factor X {sub 0} is the last amount of substance.
(3) In the equation formula
X {sub 0} Equals to 20
T (time) =3* days 7 is equal to 21 day.
Half time life is 7 day.
X {sub t} = to X {sub 0}*{1/2}21/7put 20 *(1 dived 2)3 /20 in which 8
provide 2.50 count/sec.
(f)
(1) According to the equation need c=y multiple of ax
Now assume

X= People who have phone (in 1000)
A= {375/225}1/2
= 1.29
Y = Multiply of c {1.29}8
C multiply {1.29}8= 1725, and y = c{1.29}x
1725 = C *{1.29}8
1725 is divided by 7.669c= 7.669c divided 7.669c
Where c is equal to 225
Substitute by c= 225 or a= 1.29 from the equation
Y= 225 multiple (1.29)x
In this equation y = 225 *{1.29}x
(2) In the year = 2000 which is consist as .25 year
Substitute by x= 25 and y =225*{1.29}25
In which y = 130895.65
So 130895.65 thousand is the cell phone in 2025 (Canada)
(3) Substitute by x= 6and x= 10 in y = 225 *{1.29}x
Y= 225x{1.29}6 = 1036.86 where(6,1036.86)
Y=225x{1.29}10= 2971.30 where (10, 2971.305)
According to equation = {y2-y1}/{x2-x1}
{2971.30 subtract by 1036.86}/{10-6}
Average change in 2010 and 2006 is 483.62 thousand per year.

(4) Nearest point of x is equal 5.99 and x is equal to 6.001
Substitute by x = 5.99,x= 6.001 in the equation y = 6.001 from equation
y.
Y =225 multiply by (1.29)5.99 = 1036.57 (5.9, 1036.59)
Y = 225 multiply by (1.29)6.01 =1037.1256 (6.001, 1037.12)
According to the equation {y2-y1}/{x2-x1}
{1037.12 subtract by 1036.59}/{6.001 subtract 5.99}
264.02 (thousand)rates is change per year.
LO2
(A) According to the given data value are
(40,50,38,96,42,47,40,39,46,50) divided by the 10 = 488/10 is equal
=48.8
According to deviation
{(-4.8)2}+{1.2}+{(-10.8)2}+{47.2}2+{6.8}2+{-1.8}2+{-8.8}2+{-9.8}2+{-
2.8}2+{1.2}2/10= 260.123
(b) According to equation parameter

μ = np = 20 multiply .255
= 5σ2 =np{1-p}
=20 mutiply .255 * 0.750
=3.75 σ=√3.75-1.94
So gain the np= 15>=5 and nx (1 subtract by p) = 15>=5
Concept define B{20,0.250} –N {5,1.94}

(C) According to the replacement change rate is 3% for the motor it
should be fail in the 6.24 year, guaranty is provided by the manufacture
According to the equation
μ = 10 year = Average life of motor
then σ= 2
and z, equal to the X subtract/ σ
P (Z<z) equal to 3/100
With the use of calculation Z = negative 188 / by 188 by 100 enter the
value from the equation μ, σ, Z
-188 divided to 100 =X subtract 10/2
= (- 376/by 100) =x is subtracted by 10 from the x equal 10 mins 3.76 in
the year of 6.24

(d) According to the position of X equation considered
In which n is = to 10, p is = to 0.88
(1) When the put value x is = to 0
P (x) = C10{.12}0 {.88} 16 = 0.27856
From the rejection value
P{x} =C10{.012}1{.88}9 = .37977
Both rejections is equal to x
P{x} = C210{.12}2{.88}8 =.23305
Probability of the rejection is
Probability = P{x<=2}
=.27+.38+.235 =.895
(2) When the two pistons is reject
Probability is conceder as P {x<=1}
= 1 minus {p of(x0) in 1 min. (.28+.375)} = .3415
(3) Probabilities result is 10 in which not get the rejection from 0.89
than piston manufacture according to the 100% benefit and get the
probability structure from two piston is rejected from 0.341 than also
conceded the benefits.
(4) According to the formula
Z equal {x- μ/ σ} in question
1250 converted into {1250-1000} divided 125
980 converted to the {980-1000} divided 125 give {-.16}

1150 change to the {1150-1000} divided bt 125 give 1.2
LO3
(a) (1) According to P{t} equal to a, multiple cos{bx(t-d)}+c provide
equation is
Pmini = 4 and value of Pmax equal to 20.
C { Pmini + Pmax }/2 equal to {20+4}/2 equal 12 and value of a ={Mini value
of P} –{Max. value of P}/2 = {20-4}/2 equal 8
Total days from 1 jan. in which value of P(t) is maximum and counting
according to day January to may month and add the 21 day of month of
june
Total value of (t) provide = 31,28,31,30,31,21 equal to 172
According to the finding value b{b>0}
Time period 365 equal to 2*Π/b in which b is equal to 2*x divided by 365
Maximum value of the Cos function is equal to the 0 and the value of P(t)
get the max in point t is equal to the 172
Cos{x} is transfer to the 172 from the right value
P{t}= 8 multiplied to cos{2x/365) multiplied {t-172}+12
Get the maximum value of P {t} in point t =172 by putting the value of t
P {172} = -8 cos {2 * Π /by the 365}x{172 minus the value 172} plus the
12 give 8.
Cos {0} +12 give the value 20.
(2) According to the graph value